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WORKS of Professor MANSFIELD MERMAN 
 
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A TEXT-BOOK 
 
 ON THE 
 
 MECHANICS OF MATERIALS 
 
 AND OF 
 
 BEAMS, COLUMNS, AND SHAFTS. 
 
 v >;.** ;.' -..' ;\ ;'>; ; 
 MANSFIELD MERRIMAN, 
 
 PROFESSOR OF CIVIL ENGINEERING IN LEHIGH UNIVERSITY. 
 
 NINTH EDITION, REVISED. 
 FOURTH THOUSAND. 
 
 NEW YORK: 
 
 JOHN WILEY & SONS. 
 
 LONDON: CHAPMAN & HALL, LIMITED. 
 
 1903. 
 
COPYRIGHT, 1885, 1890, 1895, 
 'MANSFIELD ' M'ERRIMAN. 
 
 G'<f ft. Vc r * (>*c r r r e 
 
 I TT ^ T 
 
 v <^ 
 
 \ c\ 
 
 ROBERT DRUMHONO, KLECTROTYPBR AND PRINTER, NBW YORK 
 
PREFACE. 
 
 The following pages contain an elementary course of study 
 in the resistance of materials and the mechanics of beams, 
 columns and shafts, designed for the use of classes in technical 
 schools and colleges. It should be preceded by a good train- 
 ing in mathematics and theoretical mechanics, and be followed 
 by a special study of the properties of different qualities of 
 materials, and by detailed exercises in construction and design. 
 
 As the plan of the book is to deal mainly with the mechanics 
 of the subject, extended tables of the results of tests on different 
 kinds and qualities of materials are not given. The attempt, 
 however, has been made to state average values of the quanti- 
 ties which express the strength and elasticity of what may be 
 called the six principal materials. On account of the great 
 variation of these values in different grades of the same material 
 the wisdom of this attempt may perhaps be questioned, but 
 the experience of the author in teaching the subject during the 
 past eleven years has indicated that the best results are attained 
 by forming at first a definite nucleus in the mind of the student, 
 around which may be later grouped the multitude of facts 
 necessary in his own particular department of study and work. 
 
 As the aim of all education should be to develop the powers 
 of the mind rather than impart mere information, the author 
 has endeavored not only to logically set forth the principles 
 and theory of the subject, but to so arrange the matter that 
 students will be encouraged and required to think for them- 
 selves. The problems which follow each article will be found 
 
 iii 
 
 M23925 
 
IV PREFACE. 
 
 useful for this purpose. Without the solution of many numer- 
 ical problems it is indeed scarcely possible for the student to 
 become well grounded in the theory. The attempt has been 
 made to give examples, exercises, and problems of a practical 
 nature, and also of such a character as to clearly illustrate the 
 principles of the theory and the methods of investigation. 
 
 MANSFIELD MERRIMAN. 
 
 SOUTH BETHLEHEM* PA., December, 1889. 
 
 NOTE TO THE NINTH EDITION. 
 
 The sixth edition contained double the matter of previous 
 ones, Chapters VIII to XI being new. In the seventh edi- 
 tion Arts. 6 1, 62, 117, 1 1 8, 149, 150, and 151 were rewritten. 
 In the eighth edition medium steel was substituted for wrought 
 iron in the discussion of all rolled I beams. In this edition 
 all known errors have been corrected and some former state- 
 ments and problems replaced by better ones. New matter 
 on eccentric loads on columns will be found in Art. 62, on 
 formulas for repeated stresses in Art. 92, on spherical and 
 cylindrical rollers in Arts. 106 and 107, on the oscillations of 
 bars and beams under impact in Arts. 103 and m, on com- 
 bined stresses in Art. 137, and on the International Associa- 
 tion for Testing Materials in Art. 152, several of these articles 
 being entirely rewritten. These changes and other minor 
 ones have been made not only for the purpose of keeping the 
 book abreast with modern progress, but also to better adapt 
 it to the use of students and engineers. 
 
 M. M. 
 
CONTENTS. 
 
 CHAPTER I. 
 
 PAGES 
 
 RESISTANCE AND ELASTICITY OF MATERIALS.... 1-21 
 
 Art. i. Average Weights. 2. Stresses and Deformations. 
 3. Experimental Laws. 4. Elastic Limit and Coefficient of 
 Elasticity. 5. Tension. 6. Compression. 7. Shear. 8. Fac- 
 tors of Safety and Working Stresses. 
 
 CHAPTER II. 
 
 PIPES, CYLINDERS, AND RIVETED JOINTS 22-35 
 
 Art. 9. Water and Steam Pipes. 10. Thin Cylinders and 
 Spheres, u. Thick Cylinders. 12. Investigation of Riveted 
 Joints. 13. Design of Riveted Joints. 14. Miscellaneous 
 Exercises. 
 
 CHAPTER III. 
 
 CANTILEVER BEAMS AND SIMPLE BEAMS 36-84 
 
 Art. 15. Definitions. 16. Reactions of the Supports. 17. 
 The Vertical Shear. 18. The Bending Moment. 19. Inter- 
 nal Stresses and External Forces. 20. Experimental and 
 Theoretical Laws. 21. The Two Fundamental Formulas. 
 22. Center of Gravity of Cross-sections. 23. Moment of Iner- 
 tia of Cross-sections. 24. The Maximum Bending Moment. 
 25. The Investigation of Beams. 26. Safe Loads for Beams. 
 27. Designing of Beams. 28. The Modulus of Rupture. 29. 
 Comparative Strengths. 30. Iron and Steel I Beams. 31. 
 Iron and Steel Deck Beams. 32. Cast-iron Beams. 33. Gen- 
 eral Equation of the Elastic Curve. 34. Deflection of Canti- 
 lever Beams. 35. Deflection of Simple Beams. 36. Compar- 
 ative Deflection and Stiffness'. 37. Relation between Deflec- 
 tion and Stress. 38. Cantilever Beams of Uniform Strength. 
 39. Simple Beams of Uniform Strength. 
 
Vi CONTENTS. 
 
 CHAPTER IV. 
 
 PACKS 
 
 RESTRAINED BEAMS AND CONTINUOUS BEAMS. 85-110 
 
 Art. 40. Beams Overhanging One Support. 41. Beams 
 Fixed at One End and Supported at the Other. 42. Beams 
 Overhanging Both Supports. 43. Beams Fixed at Both 
 Ends. 44. Comparison of Restrained and Simple Beams. 
 45. General Principles of Continuity. 46. Properties of 
 Continuous Beams. 47. The Theorem of Three Moments. 
 48. Continuous Beams with Equal Spans. 49. Continuous 
 Beams with Unequal Spans. 50. Remarks on the Theory 
 of Flexure. 
 
 CHAPTER V. 
 
 COLUMNS OR STRUTS 111-134 
 
 Art. 51. Cross-sections of Columns. 52. General Princi- 
 ples. 53. EULER'S Formula. 54. HODGKINSON'S Formu- 
 las. 55. RANKINE'S Formula. 56. Radius of Gyration of 
 Columns. 57. Investigation of Columns. 58. Safe Loads 
 for Columns. 59. Designing of Columns. 60. The Straight- 
 line Formula. 61. RITTER'S Rational Formula. 62. Ec- 
 centric Loads. 
 
 CHAPTER VI. 
 
 TORSION AND SHAFTS I35-I43 
 
 Art. 63. The Phenomena of Torsion 64. The Funda- 
 mental Formula for Torsion. 65. Polar Moments of Iner- 
 tia. 66. The Constants of Torsion. 67. Shafts for Trans- 
 mission of Power. 68. Round Shafts. 69. Hollow Shafts. 
 70. Miscellaneous Exercises. 
 
 CHAPTER VII. 
 
 COMBINED STRESSES 144-161 
 
 Art. 71. Combined Tension and Compression. 72. 
 Stresses due to Temperature. 73. Combined Tension and 
 Flexure. 74. Combined Compression and Flexure. 75. 
 Shear Combined with Tension or Compression. 76. Com- 
 
CONTENTS. Vll 
 
 PAGES 
 
 bined Flexure and Torsion. 77. Combined Compression 
 and Torsion. 78. Horizontal Shear in Beams. 79. Maxi- 
 mum Internal Stresses in Beams. 
 
 CHAPTER VIII. 
 
 THE STRENGTH OF MATERIALS.. 162-196 
 
 Art. 80. Mean Constants. 81. Historical. 82. Testing 
 Machines. 83. Timber. 84. Brick. 85. Cement and Mor- 
 tar. 86. Stone. 87. Cast Iron. 88. Wrought Iron. 89. 
 Steel. 90. Other Materials. 91. The Fatigue of Materials. 
 92. Repeated Stresses. 
 
 CHAPTER IX. 
 
 THE RESILIENCE OF MATERIALS 197-221 
 
 Art. 93. Sudden Loads and Impact. 94. The Modulus of 
 Resilience. 95. External Work and Resilience. 96. Elastic 
 Resilience of Beams. 97. Ultimate Resilience. 98. Early 
 History of Resilience. 99. Modern Experiments. 
 
 CHAPTER X. 
 
 TENSION AND COMPRESSION 222-240 
 
 Art. ico. Elongation under own Weight. 101. Bar of 
 Uniform Strength. 102. Longitudinal Impact. 103. Oscil- 
 lations after Impact. 104. Centrifugal Stress. 105. Shrink- 
 age of Hoops. 1 06. Spherical Rollers. 107. Cylindrical 
 Rollers. 108. Eccentric Loads. 
 
 CHAPTER XL 
 
 FLEXURE OF BEAMS 241-271 
 
 Art. 109. The Work of Flexure, no. Static and Sudden 
 Deflections, in. Deflection under Impact. m|. Oscilla- 
 tions after Impact. 112. Pressure due to Impact. 113. Cen- 
 trifugal Stress. 114. Live-load Velocity. 115. Work of the 
 Vertical Shear. 116. Deflection due to Shearing. 117. Flex- 
 ure and Compression. 118. Flexure and Tension. 
 
viii CONTENTS. 
 
 CHAPTER XII. 
 
 PAGES 
 
 SHEAR AND TORSION ,272-287 
 
 Art. 119. Stresses due to Shear. 120. Resilience under 
 Shear. 121. The Coefficient of Elasticity. 122. Resilience 
 under Torsion. 123. Hollow Shafts. 124. Shaft Couplings. 
 125. A Crank Pin and Shaft. 126. A Triple-crank Pin. 
 
 CHAPTER XIII. 
 
 APPARENT STRESSES AND TRUE STRESSES... 288-309 
 Art. 127. Mathematical Theory of Elasticity. 128. Lateral 
 Deformation. 129. True Tensile and Compressive Stresses. 
 130. Normal and Tangential Stresses. 131. Resultant 
 Stresses. 132. The Ellipsoid of Stress. 133. The Three 
 Principal Stresses. 134. A Numerical Case. 135. Maxi- 
 mum Shearing Stresses. 136. The Ellipse of Stress. 137. 
 Formulas for True Combined Stresses. 
 
 CHAPTER XIV. 
 
 STRESSES IN GUNS. 310-327 
 
 Art. 138. LAMP'S Formulas. 139. A Solid Gun. 140. A 
 Compound Cylinder. 141. CLAVARINO'S Formulas. 142. 
 BIRNIE'S Formulas. 143. Hoop Shrinkage. 144. Hooped 
 Guns. 
 
 CHAPTER XV. 
 
 PLATES, SPHERES, AND COLUMNS.. . 328-343 
 
 Art. 145. Circular Plates. 146. Elliptical Plates. 147. 
 Rectangular Plates. 148. Hollow Spheres. 149. Experi- 
 ments on Columns. 150. EULER'S Modified Formula. 151. 
 The Deflection of Columns. 
 
 APPENDIX 344-362 
 
 International Association for Testing Materials. The Ve- 
 locity of Stress. Advanced Problems. Answers to Prob- 
 lems. Description of Tables. Tables of Logarithms, Squares, 
 and Circles. Weights of Wrought-iron Bars. 
 
 INDEX , 363-368 
 
CONTENTS. IX 
 
 TABLES. 
 
 PAGES 
 
 Average Weights and Specific Gravities i 
 
 Constants for Tension 9 
 
 Tensile Test of a Wrought-iron Bar 1 1 
 
 Constants for Compression 14 
 
 Constants for Shearing 15 
 
 Factors of Safety 1 8 
 
 Centers of Gravity of Cross-sections 52 
 
 Moments of Inertia of Cross-sections 53 
 
 Properties of I Beams 65 
 
 Properties of Deck Beams 67 
 
 Cantilever Beams and Simple Beams 79 
 
 Cantilever, Simple, and Restrained Beams 95 
 
 Continuous Beams with Equal Spans 104, 105 
 
 Average Constants for Columns 122 
 
 Radii of Gyration of Cross-sections 1 23 
 
 The Straight-line Column Formula f . . . . 128 
 
 Experiments on Rupture of Columns 130 
 
 Weights and Coefficients of Expansion 163 
 
 Elastic Limits and Coefficients of Elasticity 163 
 
 Ultimate Tensile and Compressive Strengths 164 
 
 Ultimate Shearing Strength and Modulus of Rupture 164 
 
 Strength of Timber 171 
 
 Strength of Cement and Mortar 175 
 
 Strength of Stone 177 
 
 Strength of Wrought Iron 183 
 
 Strength of Steel 187 
 
 Ultimate Resilience of a Steel Bar 206 
 
 Velocities of Stress 347 
 
 Logarithms of Numbers 356 
 
 Squares of Numbers 358 
 
 Areas of Circles 360 
 
 Weights of Wrought-iron Bars 362 
 
Evolvi varia problemata. In scientiis enim ediscendis 
 prosunt exempla magis quam praecepta. Qua de causa in his 
 fusius expatiatus sum. NEWTON. 
 
 Nous avons pour but, non dc donner un traitecomplet, mais 
 de montrer, par dcs exemples simples et varies, I'utilit6 et 1'im- 
 portance de la theorie mathematique de 1'elasticite. LAM* 
 
MECHANICS OF MATERIALS. 
 
 CHAPTER I. 
 THE RESISTANCE AND ELASTICITY OF MATERIALS. 
 
 ARTICLE i. AVERAGE WEIGHTS. 
 
 The principal materials used in engineering constructions 
 are timber, brick, stone, cast iron, wrought iron, and steeJ. 
 The following table gives their average unit-weights and aver- 
 age specific gravities. 
 
 
 Average Weight. 
 
 
 Material. 
 
 
 Average 
 Specific 
 Gravity. 
 
 Pounds per 
 Cubic Foot. 
 
 Kilos per 
 Cubic Meter. 
 
 Timber 
 
 4 
 
 600 
 
 0.6 | 
 
 Brick 
 
 125 
 
 2000 
 
 2.0 
 
 Stone 
 
 TOO 
 
 2 560 
 
 2.6 
 
 Cast Iron 
 
 450 
 
 7 200 
 
 7.2 
 
 Wrought Iron 
 
 480 
 
 7700 
 
 7-7 
 
 Steel 
 
 49 
 
 7 800 
 
 7.8 
 
 These weights, being mean or average values, should be care- 
 fully memorized by the student as a basis for more precise 
 knowledge, but it must be noted that they are subject to more 
 or less variation according to the quality of the material. 
 Brick, for instance, may weigh as low as 100, or as high as 150 
 pounds per cubic foot, according as it is soft or hard pressed* 
 
RESISTANCE AND ELASTICITY OF MATERIALS. CH. I. 
 
 Unless otherwise stated the above average values will be used 
 in the examples and problems of this book. In all engineering 
 reference books are given tables showing the unit-weights far 
 different qualities of the above six principal materials, and also 
 for copper, lead, glass, cements, and other materials used in 
 construction. 
 
 \ For\cpiji^*4t4ng the weights of bars, beams, and pieces of uni- 
 t t e form < cross-section,. J:he following approximate simple rules will 
 A I ii' bft^n, be; fauhd^CenVenient. 
 
 A wrought iron bar one square inch in section and one 
 
 yard long weighs ten pounds. 
 
 Steel is about two per cent heavier than wrought iron. 
 Cast iron is about six per cent lighter than wrought iron. 
 Stone is about one-third the weight of wrought iron. 
 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 i X 3 inches and 
 
 12 feet long; its cross-section is 4.5 square inches, hence its 
 weight is 45 X 4 = 180 pounds. A steel bar of the same 
 dimensions will weigh 180 + 0.02 X 180 = about 184 pounds, 
 and a cast iron bar will weigh 180 0.06 X 180 about 169 
 pounds. 
 
 By reversing the above rules the cross-sections of bars are 
 readily computed from their weights per yard. Thus, if a stick 
 of timber 15 feet long weigh 120 pounds, its weight per yard is 
 24 pounds, and its cross-section is 12 X 2.4 = about 28.8 square 
 inches. 
 
 Problem I. How many square inches in the cross-section of a 
 wrought iron railroad rail weighing 24 pounds per linear foot ? 
 In a steel rail? In a wooden beam? 
 
 Prob. 2. Find the weights of a wooden beam 6x8 inches in 
 section and 13 feet long, of a steel bar one inch in diameter and 
 
 13 feet long, and of a common brick 2X4 inches and 8 inches 
 long. 
 
ART. 2. STRESSES AND DEFORMATIONS. 3 
 
 ART. 2. STRESSES AND DEFORMATIONS. 
 
 A 'stress ' is a force which acts in the interior of a body and 
 resists the external forces which tend to change its shape. If 
 a weight of 400 pounds be suspended by a rope, the stress in 
 the rope is 400 pounds. This stress is accompanied by an 
 elongation of the rope, which increases until the internal molec- 
 ular stresses or resistances are in equilibrium with the exterior 
 weight. Stresses are measured in pounds, tons, or kilograms. 
 A 'unit-stress' is the amount of stress on a unit of area; this is 
 expressed either in pounds per square inch, or in kilograms per 
 square centimeter. Thus, if a rope of two square inches cross- 
 section sustains a stress of 400 pounds, the unit-stress is 200 
 pounds per square inch, for the total stress must be regarded 
 as distributed over the two square inches of cross-section. 
 
 A ' deformation ' is the amount of change of shape of a 
 body caused by the external forces. If a load be put on a 
 column its length is shortened, and the amount of shortening 
 is a deformation. So in the case of the rope, the amount of 
 elongation is a deformation. Deformations are generally meas- 
 ured in inches, or centimeters. 
 
 The word ' strain ' is often used in technical literature as 
 synonymous with stress, and sometimes it is also used to desig- 
 nate the deformation, or change of shape. On account of this 
 ambiguity the word will not be employed in this book. 
 
 Three kinds of simple stress are produced by forces which 
 tend to change the shape of a body. They are, 
 
 Tensile, tending to pull apart, as in a rope. 
 Compressive, tending to push together, as in a column. 
 Shearing, tending to cut across, as in punching a plate. 
 
 The nouns corresponding to these three adjectives are Tension, 
 Compression, and Shear. The stresses which occur in beams, 
 
4 RESISTANCE AND ELASTICITY OF MATERIALS. CH. I. 
 
 columns, and shafts are of a complex character, but they may 
 always be resolved into the three kinds of simple stress. The 
 first effect of an applied force is to cause a deformation. 
 This deformation receives a special name according to the kind 
 of stress which accompanies it. Thus, 
 
 Tension produces an elongation. 
 Compression produces a shortening. 
 Shear produces a detrusion. 
 
 This change of shape is resisted by the stresses between the 
 molecules of the body, and as soon as these internal resistances 
 balance the exterior forces the change of shape ceases and the 
 body is in equilibrium. But if the external forces be increased 
 far enough the molecular resistances are finally overcome and 
 the body breaks or ruptures. 
 
 In any case of simple stress in a body in equilibrium the 
 total internal stresses or resistances must equal the external 
 applied force. Thus, in the above instance of a rope from 
 which a weight of 400 pounds is suspended, let it be imagined 
 to be cut at any section ; then equilibrium can only be main- 
 tained by applying at that section an upward force of 400 
 pounds ; hence the stresses in that section must also equal 400 
 pounds. In general, if a steady force P produce either ten- 
 sion, compression, or shear, the total stress produced is also P+ 
 for if not equilibrium does not obtain. In such cases, then, the 
 word * stress ' may be used to designate the external force as 
 well as the internal resistances. 
 
 Tension and Compression are similar in character but differ 
 in regard to direction. A tensile stress in a bar occurs when 
 two forces of equal intensity act upon its ends, each in a direc- 
 tion away from the other. In compression the direction of the 
 forces is reversed and each acts toward the bar. Evidently a 
 simple tensile or compressive stress in a bar is to be regarded 
 as evenly distributed over the area of its cross-section, so that 
 
ART. 2. STRESSES AND DEFORMATIONS. 5 
 
 if P be the total stress in pounds and A the area of the cross- 
 
 P 
 
 section in inches, the unit-stress is -j in pounds per square inch. 
 
 A 
 
 Shear requires the action of two forces exerted in parallel 
 planes and very near together, like the forces in a pair of 
 shears, from which analogy the name is derived. Here also 
 the total shearing stress P is to be regarded as distributed 
 
 P 
 
 uniformly over the area A, so that the unit-stress is -r- And 
 
 A 
 
 conversely if ^S represent the uniform unit-stress the total stress 
 Pis AS. 
 
 In any case of simple stress acting on a body let P be the 
 total stress, A the area over which it is uniformly distributed, 
 and 5 the unit-stress. Then, 
 
 (i) P=AS. 
 
 Also let A be the total linear deformation produced by the 
 stress, / the length of the bar, and s the deformation per unit 
 of length. Then this deformation is to be regarded as uni- 
 formly distributed over the distance /, so that also, 
 
 (i)' \ = Is. 
 
 The laws implied in the statement of these two formulas are 
 confirmed by experiment, if the stress be not too great. 
 
 Unit-stress in general will be denoted by 5, whether it be 
 tension, compression, or shear. S, will denote tensile unit- 
 stress,. S c compressive unit-stress, and S t shearing unit-stress, 
 when it is necessary to distinguish between them. 
 
 Prob. 3. A wrought iron rod ij inches in diameter breaks 
 under a tension of 67610 pounds. Find the breaking unit- 
 stress. 
 
 Prob. 4. If a cast-iron bar i X 2 inches in size breaks 
 under a tension of 60 ooo pounds, what tension will break a 
 bar if X if inches in size? 
 
RESISTANCE AND ELASTICITY OF MATERIALS. CH. L 
 
 ART. 3. EXPERIMENTAL LAWS. 
 
 Numerous tests or experiments have been made to ascertain 
 the strength of materials and the laws that govern stresses and 
 deformations. The resistance of a rope, for instance, may be 
 investigated by suspending it from one end and applying 
 weights to the other. As the weights are added the rope will 
 be seen to stretch or elongate, and the amount of this deforma- 
 tion may be measured. When the load is made great enough 
 the rope will break, and thus its ultimate tensile stress is- 
 known. For stone, iron, or steel, special machines, known as 
 testing machines, have been constructed by which the effect of 
 different stresses on different qualities and forms of materials 
 may be accurately measured. 
 
 All experiments, and all experience, agree in establishing the 
 five following laws for cases of simple tension and compression, 
 which may be regarded as the fundamental principles of the 
 science of the strength of materials. 
 
 (A) When a small stress is caused in a body a small de- 
 formation is produced, and on the removal of the stress 
 the body springs back to its original form. For small 
 stresses, then, materials may be regarded as perfectly 
 elastic. 
 
 (B] Under small stresses the deformations are approxi- 
 mately proportional to the forces, or stresses, which pro- 
 duce them, and also approximately proportional to the 
 length of the bar or body. 
 
 (C) When the stress is great enough a deformation is 
 produced which is partly permanent, that is, the body 
 does not spring back entirely to its original form on re- 
 moval of the stress. This permanent part is termed a 
 set. In such cases the deformations are not proportional 
 to the stresses. 
 
ART. 4. ELASTIC LIMIT AND COEFFICIENT OF ELASTICITY. / 
 
 (D] When the stress is greater still the deformation 
 rapidly increases and the body finally ruptures. 
 
 (E) A sudden stress, or shock, is more injurious than a 
 steady stress or than a stress gradually applied. 
 
 The words small and great, used in stating these laws, have, as 
 will be seen later, very different values and limits for different 
 kinds of materials and stresses. 
 
 The 'ultimate strength ' of a material under tension, com- 
 pression, or shear, is the greatest unit-stress to which it can be 
 subjected. This occurs at or shortly before rupture, and its 
 value is very different for different materials. Thus if a bar 
 whose cross-section is A breaks under a tensile stress P, the 
 ultimate tensile strength of the material is P -r- A. 
 
 Prob. 5. If the ultimate strength of wrought iron is 55 ooo 
 pounds per square inch, what tension will rupture a bar 6 feet 
 long which weighs 60 pounds? 
 
 Prob. 6. If a bar I inch in diameter and 8 feet long elon- 
 gates 0.05 inch under a stress of 15000 pounds, how much, 
 according to law (B), will a bar of the same size and material 
 elongate whose length is 12 feet and stress 30000 pounds? 
 
 ART. 4. ELASTIC LIMIT AND COEFFICIENT OF ELASTICITY. 
 
 The ' elastic limit ' is that unit-stress at which the permanent 
 set is first visible and within which the stress is directly propor- 
 tional to the deformation. For stresses less than the elastic 
 limit bodies are perfectly elastic, resuming their original form 
 on removal of the stress. Beyond the elastic limit a permanent 
 alteration of shape occurs, or, in other words, the elasticity of 
 the material has been impaired. It is a fundamental rule in all 
 engineering constructions that materials can not safely be 
 strained beyond their elastic limit. 
 
 The ' coefficient of elasticity ' of a bar for tension, compres- 
 sion, or shearing, is the ratio of the unit-stress to the unit- 
 
8 RESISTANCE AND ELASTICITY OF MATERIALS. CH. I. 
 
 deformation, provided the elastic limit of the material be not 
 exceeded. Let 5 be the unit-stress, s the unit-deformation, and 
 E the coefficient of elasticity. Then by the definition, 
 
 (2) E = - and 5 = Es. 
 
 By law (B) the quantity E is a constant for each material, until 
 S reaches the elastic limit. Beyond this limit s increases more 
 rapidly than S and the ratio is no longer constant. Equation 
 (2) is a fundamental one in the science of the strength of 
 materials. Since E varies inversely with s, the coefficient of 
 elasticity may be regarded as a measure of the stiffness of the 
 material. The stiffer the material the less is the change in 
 length under a given stress and the greater is E. The values 
 of E for materials have been determined by experiments with 
 testing machines and their average values will be given in the 
 following articles. E is necessarily expressed in the same unit 
 as the unit-stress S. Some authors give the name ' modulus of 
 elasticity* to the quantity E. 
 
 Another definition of the coefficient of elasticity for the case 
 of tension is that it is the unit-stress which would elongate a 
 bar to double its original length, provided that this could be 
 done without exceeding the elastic limit. That this defini- 
 tion is in agreement with (2) may be shown by regarding a bar 
 of length / which elongates the amount A under the unit- 
 
 P \ 
 
 stress -j. Here the unit-elongation is 7 and (2) becomes, 
 
 P A PI 
 (20 -_=-+-=-, 
 
 p 
 
 and if \ be equal to /, E is the same as the unit-stress -r. 
 
 A 
 
 Prob. 7. Find the coefficient of elasticity of a bar of wrought 
 iron \\ inches in diameter and 16 feet long which elongates \ 
 inch under a tensile stress of 21 ooo pounds. 
 
ART. 5. 
 
 TENSION. 
 
 Prob. 8. If the coefficient of elasticity of cast iron is 1 5 ooo ooo 
 pounds per square inch, how much will a bar 2X3 inches and 
 6 feet long stretch under a tension of 5 ooo pounds ? 
 
 ART. 5. TENSION. 
 
 The phenomena of tension observed when a gradually in- 
 creasing stress is applied to a bar, are briefly as follows : When 
 the unit-stress .V is less than the elastic limit 5,, the unit-elon- 
 gation s is small and proportional to 5. Within this limit the 
 ratio of 5 to s is the coefficient of elasticity of the material. 
 After passing the elastic limit the bar rapidly elongates and 
 this is accompanied by a reduction in area of its cross-section. 
 Finally when 5 reaches the ultimate tensile strength S ( , the 
 bar tears apart. Usually S t is the maximum unit-stress on the 
 bar, but in some cases the unit-stress reaches a maximum 
 shortly before rupture occurs. 
 
 The constants of tension for timber, cast iron, wrought iron 
 and steel are given in the following table. The values are 
 average ones and are liable to great variations for different 
 grades and qualities of materials. Brick and stone are not 
 here mentioned, as they are rarely or never used in tension. 
 
 Material. 
 
 Coefficient of 
 Elasticity, E. 
 
 Elastic 
 Limit, S e . 
 
 Ultimate 
 Tensile 
 Strength, St . 
 
 Ultimate 
 Elongation. 
 
 
 Pounds per square 
 inch. 
 
 Pounds per 
 square inch. 
 
 Pounds per 
 square inch. 
 
 Per cent. 
 
 Timber, 
 
 I 500000 
 
 3 000 
 
 IO OOO 
 
 1-5 
 
 Cast Iron, 
 
 15 ooo ooo 
 
 6 000 
 
 20 000 
 
 o-5 
 
 Wrought Iron, 
 
 25 oooooo 
 
 25 ooo 
 
 55 ooo 
 
 20 
 
 Steel, 
 
 30 ooo ooo 
 
 50 ooo 
 
 100 000 
 
 IO 
 
 The values of the coefficients of elasticity, elastic limits, and 
 breaking or ultimate strengths are given in pounds per square 
 inch of the original cross-section of the bar. The ultimate elon- 
 gations are given in percentages of the original length ; these 
 
10 
 
 RESISTANCE AND ELASTICITY OF MATERIALS. CH. I. 
 
 express elongations per linear unit; they should be regarded 
 as very rough averages, which are subject to great variations 
 depending on the shape, size, and quality of the specimen. 
 
 The ultimate elongation, together with the reduction in area 
 of the cross-section, furnishes the means of judging of the duc- 
 tility of the material. The reduction of area in cast iron and in 
 hard steel is very small, while in ductile materials like wrought 
 iron and soft steel it may be as high as 50 or 60 per cent of 
 the original section. 
 
 A graphical illustration of the principal phenomena of tension, 
 is given in Fig. I. The unit-stresses are taken as ordinates and 
 
 100,0001 
 
 0390,000 
 g 80,000 
 
 a 
 
 70,000 
 
 "60,000 
 
 I 
 
 JS 50,000 
 
 p, 40,000 
 
 a 
 
 S 30,000 
 
 | 20,000 
 
 10,000 
 
 
 
 Elongations per unit of length= s 
 
 :ht Iron 
 
 0.02 
 
 0.04 
 
 0.06 0.08 
 
 Fig. i. 
 
 0.10 
 
 0.12 
 
 0.14 
 
 0.16 
 
 the unit-elongations as abscissas. For each unit-stress the cor- 
 responding unit-elongation as found by experiment is laid off, 
 and curves drawn through the points thus determined. The 
 curve for each of the materials is a straight line from the origin 
 until the elastic limit is leached, as should be the case accord- 
 ing to the law (S). The tangent of the angle which this line 
 makes with the axis of abscissas is equal to S -=- s, which is the 
 same in value as the coefficient of elasticity of the material. 
 At the elastic limit a sudden change in the curve is noticed and 
 the elongation rapidly increases. The termination of the curve 
 
ART. 5. 
 
 TENSION. 
 
 II 
 
 indicates the point of rupture. These curves show more 
 plainly to the eye than the values in the table can do the 
 differences in the properties of the materials. It will be seen 
 that the elastic limit is not a well defined point, but that its 
 value is more or less uncertain, particularly for cast iron and 
 timber. It should be also clearly understood that individual 
 curves for special cases would often show marked variations 
 from their mean forms as represented in the diagram. 
 
 As a particular example a 
 tensile test of a wrought iron 
 bar f inches in diameter and 
 12 inches long made at the 
 Pencoyd Iron Works will be 
 considered. In the first col- 
 umn of the following table 
 are given the total stresses 
 which were successively ap- 
 plied, in the second the 
 stresses per square inch, in 
 the third the total elonga- 
 tions, and in the fourth the 
 elongations or sets after re- 
 moval of the stress. The 
 unit-elongations are found by 
 dividing those in the table 
 by 12 inches, the length of 
 the specimen. Then the 
 coefficient of elasticity can 
 be computed for different 
 values of .$ and s. Thus for the fourth and seventh cases, 
 
 Total 
 Stress 
 in 
 Pounds. 
 
 Stress 
 per 
 Square 
 Inch. 
 
 Elongation. 
 
 Load 
 on. 
 
 Load 
 off. 
 
 2 245 
 
 5 000 
 
 .001 
 
 .000 
 
 4490 
 
 10 000 
 
 .004 
 
 .000 
 
 6 735 
 
 15 ooo 
 
 .005 
 
 .000 
 
 8 980 
 
 2O OOO 
 
 .008 
 
 .000 
 
 9878 
 
 22 OOO 
 
 .009 
 
 .000 
 
 10 776 
 
 24 ooo 
 
 .OIO 
 
 .000 
 
 ii 674 
 
 26 ooo 
 
 .0105 
 
 .000 
 
 12 572 
 
 28 ooo 
 
 .Oil 
 
 .000 
 
 13470 
 
 30000 
 
 .013 
 
 .000 
 
 14368 
 
 32 000 
 
 .014 
 
 .000 
 
 15 266 
 
 34 ooo 
 
 .015 
 
 .002 
 
 16 164 
 
 36 ooo 
 
 .022 
 
 .007 
 
 17 062 
 
 38 ooo 
 
 .416 
 
 3995 
 
 17 960 
 
 40 ooo 
 
 5445 
 
 523 
 
 25 450 
 
 50 ooo 
 
 1.740 
 
 1.707 
 
 21 17^ 
 
 CT fOO 
 
 2 j68 
 
 
 ~J */D *; - ~~~ 
 
 Specimen broke with 
 
 51 600 pounds 
 
 per square inch. 
 
 Stretch in 12 inches, 2 468 inches. 
 
 Stretch in 8 inches, 1.812 inches. 
 
 Stretch in 8 inches, 22.65 P er cent. 
 
 Fractured area, o 297 square inches. 
 
 for S = 20 ooo, s 
 
 0.008 
 
 12 
 
 0.0105 
 
 for S = 26 ooo, s = - 
 
 12 
 
 and = 30000000; 
 and E = 29 700 ooo. 
 
12 RESISTANCE AND ELASTICITY OF MATERIALS. LH. 1. 
 
 The elastic limit was reached at about 33 ooo pounds per 
 square inch, indicated by the beginning of the set and the rapid 
 increase of the elongations. The ultimate tensile strength of 
 the specimen was 5 1 600 pounds per square inch. The ultimate 
 unit-elongation in 8 inches of the length was 0.226 inches per 
 linear inch. It hence appears that this bar of wrought iron was 
 higher than the average as regards stiffness, elastic limit and 
 ductility, and lower than the average in ultimate strength. 
 
 The ' working stress ' for a material is that unit-stress to 
 which it is, or is to be, subjected. This should not be greater 
 than the elastic limit of the material, since if that limit be ex- 
 ceeded there is a permanent set which impairs the elasticity. 
 In order to secure an ample margin of safety it is customary to 
 take the working stresses at from one-third to two-thirds the 
 "elastic limit S e . The reasons which govern the selection of 
 proper values of the working stresses will be set forth in the 
 following articles. 
 
 To investigate the security of a piece subjected to a tension 
 P t it is necessary first to divide P by the area of the cross-sec- 
 tion and thus determine the working stress. Then a com- 
 parison of this value with the value of S e for the given material 
 will indicate whether the applied stress is too great or whether 
 the piece has a margin of safety. For example, if a tensile 
 stress of 4 500 pounds be applied to a wrought iron bar of f 
 inches diameter the working unit-stress is, 
 
 P 4500 
 
 o -j= - = 10 ooo pounds per square inch, nearly. 
 A 0.442 
 
 As this is less than one-half the elastic limit of wrought iron the 
 bar has a good margin of security. 
 
 To design a piece to carry a given tension P it is necessary 
 to assume the kind of material to be used and its allowable 
 
 p 
 ^working stress S. Then -^ is the area of the cross-section 
 
ART. 6. COMPRESSION. 13 
 
 of the piece, which may be made of such shape as the circum- 
 stances of the case require. For example, if it be required to 
 design a wooden bar to carry a tensile load of 4 500 pounds, 
 the working stress may be assumed at I ooo pounds per 
 square inch and the required area is 4.5 square inches, so that 
 the bar may be made 2 X 2j inches in section. 
 
 The elongation of a bar within the elastic limit may be com- 
 puted by the help of formula (2). For instance, let it be re- 
 quired to find the elongation of a wooden bar 3X3 inches and 
 12 feet long under a tensile stress of 9000 pounds. From the 
 formulas (2) and (i), 
 
 SP X Pl 
 
 Substituting in this the values E = i 500000, A = g, t= 144, 
 and P = 9 ooo, the probable value of the elongation X is found 
 to be 0.096 inches. 
 
 Prob. 9. Find the size of a square wooden bar to safely 
 carry a tensile stress of 20000 pounds. 
 
 Prob. 10. Compute the elongation of a wrought-iron bar, 
 ii inches in diameter and 24 feet long, under a tension of 
 10000 pounds. Also that of a cast-iron bar. 
 
 ART. 6. COMPRESSION. 
 
 The phenomena of compression are similar to those of ten- 
 sion, provided that the length of the specimen does not exceed 
 about five times its least diameter. The piece at first shortens 
 proportionally to the applied stress, but after the elastic limit is 
 passed the shortening increases more rapidly, and is accom- 
 panied by a slight enlargement of the cross-section. When the 
 stress reaches the ultimate strength of the material the specimen 
 cracks and ruptures. If the length of the piece exceeds about 
 ten times its least diameter, a sidewise bending or flexure of 
 the specimen occurs, so that it fails under different circum- 
 
14 RESISTANCE AND ELASTICITY OF MATERIALS. CH. I. 
 
 stances than those of direct compression. All the values given 
 in this article refer to specimens whose lengths do not exceed 
 about five times their least diameter. Longer pieces will be 
 discussed in Chapter V under the head of 'columns.' Owing 
 to the difficulty of making experiments on short specimens, the 
 phenomena of compression are not usually so regular as those 
 of tension. 
 
 The constants of compression for short specimens are given 
 in the following table, the values, like those for tension, being 
 rough average values liable to much variation in particular 
 cases. 
 
 Material. 
 
 Coefficient of 
 Elasticity, E. 
 
 Elastic 
 Limit, S e . 
 
 Ultimate 
 Compressive 
 Strength, S c . 
 
 Timber, 
 
 Lbs. per sq. in. 
 
 i 500 ooo 
 
 Lbs. per sq. in. 
 3 ooo 
 
 Lbs. per sq. in. 
 8 000 
 
 Brick, 
 
 
 
 2 500 
 
 Stone, 
 
 6 OOO OOO 
 
 
 6 ooo 
 
 Cast Iron, 
 
 15 ooo ooo 
 
 20 OOO 
 
 90 ooo 
 
 Wrought Iron, 
 
 25 ooo ooo 
 
 25 ooo 
 
 55 ooo 
 
 Steel, 
 
 30 ooo ooo 
 
 50 ooo 
 
 150 ooo 
 
 The values of the coefficient of elasticity and the elastic limit 
 for timber, wrought iron, and steel here stated are the same as 
 those for tension, but the same reliance cannot be placed upon 
 them, owing to the irregularity of experiments thus far made. 
 There is reason to believe that both the elastic limit and the 
 coefficient of elasticity for compression are somewhat greater 
 than for tension. 
 
 The investigation of a piece subjected to compression, or the 
 design of a short piece to be subjected to compression, is 
 effected by exactly the same methods as for tension. Indeed 
 it is customary to employ these methods for cases where the 
 length of the piece is as great as ten times its least diameter. 
 
ART. 7. 
 
 SHEAR. 
 
 Prob. ii. Find the height of a brick tower which crushes 
 under its own weight. Also the height of a stone tower. 
 
 Prob. 12. Compute the amount of shortening in a wrought 
 iron specimen I inch in diameter and 5 inches long under a load 
 of 6000 pounds. 
 
 ART. 7. SHEAR. 
 
 Shearing stresses are developed whenever two forces, act- 
 ing like a pair of shears, tend to cut a body between them. 
 When a plate is punched the ultimate shearing strength of the 
 material must be overcome over the surface punched. When 
 a bolt is in tension the applied stress tends to shear off the 
 head and also to strip or shear the threads in the nut and screw. 
 When a rivet connects two plates which transmit tension the 
 plates tend to shear the rivet across. 
 
 The ultimate shearing strength of materials is easily deter- 
 mined by causing rupture under a stress P, and then dividing 
 P by the area A of the shorn surface. The value of this for 
 timber is found to be very much smaller along the grain than 
 across the grain ; for the first direction it is sometimes called 
 longitudinal shearing strength and for the second transverse 
 shearing strength. The same distinction is sometimes made 
 in rolled wrought iron plates and bars where the process of 
 manufacture induces a more or less fibrous structure. The 
 elastic limit and the amount of detrusion for shearing are dif- 
 
 Material. 
 
 Coefficient of 
 Elasticity, f 
 
 Ultimate 
 Shearing 
 Strength, S t . 
 
 Timber, Longitudinal, 
 
 40OOOO 
 
 600 
 
 Timber, Transverse, 
 
 
 3000 
 
 Cast Iron, 
 
 6000000 
 
 20 OOO 
 
 Wrought Iron, 
 
 10 000 OOO 
 
 50000 
 
 Steel, 
 
 II 000000 
 
 700OO 
 
16 RESISTANCE AND ELASTICITY OF MATERIALS. CH. L 
 
 ficult to determine experimentally. The coefficient of elas- 
 ticity, however, has been deduced by means of certain calcula- 
 tions and experiments on the twisting of shafts, explained in 
 Chapter VI under the head of torsion. 
 
 The investigation and design of a piece to withstand shear- 
 ing stress is made by means of the equation P = AS, in 
 the same manner as for tension and compression. As an 
 
 instance of investigation, 
 consider the cylindrical 
 wooden specimen shown 
 in Fig. 2, which has the 
 
 following dimensions : length ab = 6 inches, diameter of ends 
 = 4 inches, diameter of central part = 2 inches. Let this 
 specimen be subjected to a tensile stress in the direction of its 
 length. This not only tends to tear it apart by tension, but 
 also to shear off the ends on a surface whose length is ab and 
 whose diameter is that of the central cylinder. The force P 
 required to cause this longitudinal shearing is, 
 
 P = AS, = 3.14 X 2 X 6 X 600 = 22 600 pounds, 
 
 while the force required to rupture the specimen by tension is, 
 
 P AS t = 3.14 X i 2 X 10 ooo = 31 400 pounds. 
 
 As the former resistance is only about two-thirds that of the 
 latter the specimen will evidently fail by the shearing off of 
 the ends. 
 
 When a bar is subject either to tension or to compression 
 a shear occurs in any section except those perpendicular and 
 
 parallel to the axis of the bar. 
 Let Fig. 3 represent a bar of 
 cross-section A subject to the 
 
 tensile stress P which produces 
 7 
 
 F ig. 3 . in every section perpendicular 
 
 to the bar the unit-stress . Let mn be a plane making an* 
 
 A 
 
ART. 8. FACTORS OF SAFETY AND WORKING STRESSES. 17 
 
 angle 6 with the axis, and cutting from the bar a section whose 
 area is A l . On the left of the plane the stress P may be resolved 
 'nto the components P l and P t , respectively parallel and nor- 
 mal to the plane, and the same may be done on the right. 
 Thus it is seen that the effect of the tensile stress P on the 
 plane mn is to produce a tension P^ normal to it, and a shear 
 P, along it, for the two forces P l and P l act in parallel planes 
 and in opposite directions. The shearing stress P t has the 
 value P cos 6, which is distributed over the area A l whose 
 value is A -+- sin 0. Hence the shearing unit-stress in the 
 given section is, 
 
 P P 
 
 S l =- = - f s 
 /i , A 
 
 When 6 = o, or 6 = 90, the value of 5 X is zero. The maxi- 
 
 p 
 
 mum value of 5, occurs when = 45, and then 5, = J , or 
 
 A 
 
 a tensile unit-stress 5 on a bar produces a shearing unit-stress 
 of $S along every section inclined 45 degrees to the axis of the 
 bar. The above investigation applies also to compression if 
 the direction of P be reversed, and it is sometimes observed in 
 experiments on the compression of short specimens that rup- 
 ture occurs by shearing along oblique sections. 
 
 Prob. 13. A hole } inches in diameter is punched in a wrought 
 iron plate J inches thick by a pressure on the punch of 78 ooo 
 pounds. What is the ultimate shearing strength of the iron? 
 
 Prob 14. A wrought iron bolt i inches in diameter has a 
 head inches long. Find the unit-stress tending to shear off 
 ihe head when a tension of 3 ooo pounds is applied to the bolt. 
 
 ART. 8. FACTORS OF SAFETY AND WORKING STRESSES. 
 
 The factor of safety for a body under stress is the ratio of its 
 ultimate strength to the actual existing unit-stress. The factor 
 of safety for a piece to be designed is the ratio of the ultimate 
 
i8 
 
 RESISTANCE AND ELASTICITY OF MATERIALS. CH. I. 
 
 strength to the proper allowable working stress. Thus if S t 
 be the ultimate, 5 the working stress, and / the factor of 
 safety, then 
 
 / = , and S t =fS. 
 
 The factor of safety is hence always an abstract number, which 
 indicates the number of times the working stress may be mul- 
 tiplied before the rupture of the body. 
 
 The law (E) in Art. 3 indicates that working stresses should 
 be lower for shocks and sudden stresses than for steady loads 
 and slowly varying stresses. In a building the stresses on the 
 walls are steady, so that the working strength may be taken 
 high and hence the factor of safety low. In a bridge the 
 stresses in the several members are more or less varying in 
 character which requires a lower working strength and hence 
 a higher factor of safety. In a machine subject to shocks the 
 working strength should be lower still and the factor of safety 
 very high. The law (E) from which these conclusions are de- 
 rived is not merely the result of experience, but can be con- 
 firmed by theoretical discussion (Art. 103). 
 
 The following are average values of the allowable factors of 
 safety commonly employed in American practice. These values 
 
 Material. 
 
 For Steady 
 Stress. 
 (Buildings.) 
 
 For Varying 
 Stress. 
 (Bridges.) 
 
 For Shocks. 
 (Machines.) 
 
 Timber, 
 
 8 
 
 IO 
 
 15 
 
 Brick and Stone, 
 
 15 
 
 25 
 
 30 
 
 Cast Iron, 
 
 6 
 
 15 
 
 2O 
 
 Wrought Iron, 
 
 4 
 
 6 
 
 10 
 
 Steel, 
 
 5 
 
 7 
 
 15 
 
 are subject to considerable variation in particular instances, not 
 only on account of the different qualities and grades of the 
 
ART. 8. FACTORS OF SAFETY AND WORKING STRESSES. 19 
 
 material, but also on account of the varying judgment of 
 designers. They will also vary with the range of varying stress, 
 so that different parts of a bridge may have very different 
 factors of safety. 
 
 The proper allowable working stress for any material in 
 tension, compression, or shearing, may be at once found by 
 dividing the ultimate strength by the proper factor of safety. 
 Regard should also be paid to the elastic limit in selecting the 
 working stresses, particularly for materials whose elastic limit 
 is well defined. For wrought iron and steel the working 
 strength should be well within the elastic limit, as already in- 
 dicated in previous articles. For cast iron, stone, brick, and 
 timber it is often difficult to determine the elastic limit, and 
 experience alone can guide the proper selection of the working 
 strength. The above factors of safety indicate indeed the con- 
 clusions of experiment and experience extending over the past 
 hundred years. 
 
 The student should clearly understand that the exact values 
 given in this and the preceding articles would not be arbitrarily 
 used in any particular case of design. For instance, if a given 
 lot of wrought iron is to be used in an engineering structure, 
 specimens of it should be tested to determine its coefficient of 
 elasticity, elastic limit, ultimate strength, and percentage of 
 elongation. Then the engineer will decide upon the proper 
 working stresses, being governed by its qualities as shown by 
 the tests, the character of the stresses that come upon it, and 
 the cost of workmanship. 
 
 The two fundamental principles of engineering design are 
 stability and economy, or in other words : 
 
 First, the structure must safely withstand all the stresses 
 
 which are to be applied to it. 
 Second, the structure must be built and maintained at the 
 
 lowest possible cost. 
 
20 RESISTANCE AND ELASTICITY OF MATERIALS. CH. 1. 
 
 The second of these fundamental principles requires that all 
 parts of the structure should be of equal strength, like the 
 celebrated 'one-hoss shay' of the poet. For, if one part is 
 stronger than another, it has an excess of material which might 
 have been spared. Of course this rule is to be violated if the 
 cost of the labor required to save the material be 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, as a rule, be less than the 
 values given above. .For the design of important structures 
 specifications are prepared which state the lowest allowable 
 unit-stresses that can be used. 
 
 The factors of safety stated above are supposed to be so 
 arranged that, if different materials be united, the stability 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 wrought iron with a factor of 4 or stone with a factor of 15, 
 provided the stresses are due to steady loads. 
 
 The assignment of working stresses with regard to the 
 elastic limits of materials is more rational than that by means 
 of the factors of safety, and in time it may become the more 
 important and valuable method. But at present the ultimate 
 strengths are so much better known and so much more definitely 
 determinable than the elastic limits that the empirical method 
 of factors of safety seems the more important for the use of 
 students, due regard being paid to considerations of stiffness, 
 elastic limit, and ductility. 
 
 As an example, let it be required to find the proper size of a 
 wrought iron rod to carry a steady tensile stress of 90 ooo 
 pounds. In the absence of knowledge regarding the quality 
 of the wrought iron, the ultimate strength S t is to be taken as 
 
ART. 8. FACTORS OF SAFETY AND WORKING STRESSES. 21 
 
 the average value, 55 ooo pounds per square inch. Then, for a 
 factor of safety of 4, the working stress is, 
 
 S = - =13 750 pounds per square inch. 
 
 The area of cross-section required is hence, 
 
 90000 
 A = 6.6 square inches, 
 
 which may be supplied by a rod of 2|$ inches diameter. 
 
 Prob. 15. Determine the size of a short steel piston rod when 
 the piston is 20 inches in diameter and the steam pressure upon 
 it is 67.5 pounds per square inch. 
 
 Prob. 16. A wooden frame ABC forming an equilateral tri- 
 angle consists of short pieces 2X2 inches jointed at A, B, and 
 C. It is placed in a vertical plane and supported at B and C 
 so that BC is horizontal. Find the unit-stress and factor of 
 safety in each of the three pieces when a load of 5 890 pounds 
 is applied at A. 
 
22 PIPES, CYLINDERS, AND RIVETED JOINTS. CH. LL 
 
 CHAPTER II. 
 PIPES, CYLINDERS, AND RIVETED JOINTS. 
 
 ART. 9. WATER AND STEAM PIPES. 
 
 The pressure of water or steam in a pipe is exerted in every 
 direction, and tends to tear the pipe apart longitudinally. This 
 is resisted by the internal tensile stresses of the material. If 
 / be the pressure per square inch of the water or steam, d the 
 diameter of the pipe and / its length, the force P which tends 
 to cause longitudinal rupture is/ . Id. This is evident from the 
 fundamental principle of hydrostatics that the pressure of water 
 in any direction is equal to the pressure on a plane perpen- 
 dicular to that direction, or may be seen by imagining the pipe 
 to be filled with a solid substance on one side of the diameter, 
 which would receive the pressure / on each square inch of the 
 area Id and transmit it into the pipe. If / be the thickness of 
 the pipe and 5 the tensile stress which is uniformly distributed 
 over it, as will be the case when / is not large compared with 
 d, the resistance on each side is //. 5. As the resistance must 
 equal the pressure, 
 
 pld 2tlS, or pd = 2/5, 
 which is the formula for discussing pipes under internal pressure. 
 
 The unit-pressure/ for water may be computed from a given 
 head h by finding the weight of a column of water one inch 
 square and h inches high. Or if h be given in feet, the pressure 
 in pounds per square inch may be computed from / = 0.434^. 
 
 Water pipes maybe made of cast or wrought iron, the former 
 being more common, while for steam the latter is preferable. 
 
ART. 9. WATER AND STEAM PIPES. 23 
 
 Wrought iron pipes are sometimes made of plates riveted to- 
 gether, but the discussion of these is reserved for another 
 article. A water pipe subjected to the shock of water ram 
 needs a high factor of safety, and in a steam pipe the factors 
 should also be high, owing to shocks liable to occur from con- 
 densation and expansion of the steam. The formula above 
 deduced shows that the thickness of a pipe must increase 
 directly as its diameter, the internal pressure being constant. 
 
 For example, let it be required to find the factor of safety 
 for a cast iron water pipe of 12 inches diameter and f inches 
 thickness under a head of 300 feet. Here /, the pressure per 
 square inch, equals 130.2 pounds. Then from the formula the 
 unit-stress is, 
 
 pd 130.2 X 12 
 5 = * = - 5 = i 250 pounds per square inch, 
 
 21 2 X 
 
 and hence the factor of safety is, 
 
 20000 
 
 which indicates ample security under ordinary conditions. 
 
 Again let it be required to find the proper thickness for a 
 wrought iron steam pipe of 18 inches diameter to resist a pres- 
 sure of 1 20 pounds per square inch. With a factor of safety 
 of 10 the working strength 5 is about 5 500 pounds per square 
 inch. Then from the formula, 
 
 pd 120 X 18 
 
 / = = - = 0.2 inches. 
 25 2 X 5 500 
 
 In order to safely resist the stresses and shocks liable to occur 
 in handling the pipes, the thickness is often made somewhat 
 greater than the formula requires. 
 
 Prob. 17. What should be the thickness of a cast iron pipe 
 of 1 8 inches diameter under a head of 300 feet? 
 
 Prob. 1 8. A wrought iron pipe is 3 inches in internal diame- 
 
24 PIPES, CYLINDERS, AND RIVETED JOINTS. CH. II. 
 
 ter and weighs 24 pounds per linear yard. What steam pres- 
 sure can it carry with a factor of safety of 8 ? 
 
 ART. 10. THIN CYLINDERS AND SPHERES. 
 
 A cylinder subject to the interior pressure of water or steam 
 tends to fail longitudinally exactly like a pipe. The head of 
 the cylinder however undergoes a pressure which tends to 
 separate it from the walls. If d be the diameter of the cylin- 
 der and / the' internal pressure per square unit, the total pres- 
 sure on the head is ^nd* . p. If 5 be the working unit-stress 
 and t the thickness of the cylinder, the resistance to the pres- 
 sure is approximately ndtS, if t be so small that 5 is uniformly 
 distributed. Since the resistance must equal the pressure, 
 
 \nd* .p = 7tdt.S, or pd tfS. 
 
 By comparing this with the formula of the last article it is seen 
 that the resistance of a pipe to transverse rupture is double the 
 resistance to longitudinal rupture. 
 
 A thin sphere subject to interior pressure tends to rupture 
 around a great circle, and it is easy to see that the conditions 
 are exactly the same as for the transverse rupture of a cylin- 
 der, or that pd 4tS. For thick spheres and cylinders the 
 formulas of this and the last article are only approximate. 
 
 A cylinder under exterior pressure is theoretically in a simi- 
 lar condition to one under interior pressure as long as it re- 
 mains a true circle in cross-section. A uniform interior pres- 
 sure tends to preserve and maintain the circular form of the 
 cylindrical annulus, but an exterior pressure tends at once to 
 increase the slightest variation from the circle and render it 
 elliptical. The distortion when once begun rapidly increases, 
 and failure occurs by the collapsing of the tube rather than by 
 the crushing of the material. The flues of a steam boiler are 
 the most common instance of cylinders subjected to exterior 
 
ART. 10. 
 
 THIN CYLINDERS AND SPHERES. 
 
 pressure. In the absence of a rational method of investigating 
 such cases recourse has been had to experiment. Tubes of 
 various diameters, lengths, and thicknesses have been subjected 
 to exterior pressure until they collapse and the results have 
 been compared and discussed. The following for instance are 
 the results of three experiments by FAIRBAIRN on wrought 
 iron tubes. 
 
 Length 
 in 
 Inches. 
 
 Diameter, 
 in 
 Inches. 
 
 Thickness 
 in 
 Inches. 
 
 Pressure 
 per 
 Sq. Inch. 
 
 37 
 
 9 
 
 0.14 
 
 378 
 
 60 
 
 14* 
 
 0.125 
 
 125 
 
 61 
 
 iSf 
 
 0.25 
 
 420 , 
 
 From these and other similar experiments it has been concluded 
 that the collapsing pressure varies directly as some power of the 
 thickness, and inversely as the length and diameter of the tube. 
 For wrought iron tubes WOOD gives the empirical formula for 
 the collapsing pressure per square inch, 
 
 /2.l8 
 
 / == 9600000--. 
 
 The values of / computed from this formula for the above 
 three experiments are 397, 120, and 409, which agree well with 
 the observed values. 
 
 The proper thickness of a wrought iron tube to resist ex- 
 terior pressure may be readily found from this formula after 
 assuming a suitable factor of safety. For example, let it be 
 required to find / when p = 120 pounds per square inch, / = 72 
 inches, d = 4 inches and the factor of safety = 10. Then 
 
 ,,.., = 10X.20X 7 2>M = g 
 
 9600000 
 
 from which with the help of logarithms the value of / is found 
 to be 0.22 inches. 
 
26 PIPES, CYLINDERS, AND RIVETED JOINTS. CH. IL 
 
 Prob. 19. What interior pressure per square inch will burst 
 a cast iron sphere of 24 inches diameter and | inches thickness. 
 
 Prob. 20. What exterior pressure per square inch will col- 
 lapse a wrought iron tube 72 inches long, 4 inches diameter and 
 0.25 inches thickness? 
 
 ART. ii. THICK CYLINDERS. 
 
 When the walls of a cylinder are thick compared with its 
 interior diameter it cannot be supposed, as in the preceding 
 articles, that the stress is uniformly distributed over the thick- 
 ness /. Let Fig. 4 represent one-half of a section of a thick 
 
 cylinder subject to interior pressure 
 over the length /, tending to produce 
 longitudinal rupture. Let r and r l 
 be the interior and exterior radii, 
 then T-, r = t the thickness. Let 
 5 and S, be the tensile unit-stresses 
 at the inner and outer edges of the 
 annulus. Before the application of 
 the pressure the volume of the annulus is n(r* r 2 )/, after the 
 pressure is applied the radius r l is increased to r l -\- y l and r to 
 r + y, so that its volume is n(r l + y$l n(r + yJL The an- 
 nulus is however really changed only in form, so that the two 
 expressions for the volume are equal, and equating them gives,, 
 
 or, since y and 7, are small compared with r and r, their 
 squares may be neglected, and hence 
 
 or - = ~ 
 
 Now if the material is not stressed beyond the elastic limit the 
 unit-stresses 6" and ^ are proportional to the corresponding 
 unit-elongations. The elongation of the inner circumference 
 is 27ty and that of the outer circumference is 27ty lt and divid- 
 
ART. II. THICK CYLINDERS. 2/ 
 
 ing these by 2nr and 2nr l respectively the unit-elongations are 
 found; then, 
 
 = 2 -*.** = ?* . L. 
 
 S, r ' r, r ' y^ 
 Substituting in this the value of the ratio as above found, 
 
 that is, the unit-stresses in the walls of the cylinder .vary in- 
 versely as the squares of their distances from the center. 
 
 The total stress acting over the area 2t . / is now to be found 
 by summing up the unit-stresses. Let S x be any unit-stress at 
 a distance x from the center, and 5, as before, be that at the 
 inner circumference, which is the greatest of all the unit-stresses* 
 Then by the law of variation, 
 
 The stress acting over the area /. dx ?*s then 
 
 5, Idx = SSl?jr, 
 and the total stress over the area 2t . I is 
 
 2 5rV r~, = 
 J r ** 
 
 - = 2SlJ* 
 
 This is the value of the internal resisting stress in the walls of 
 the pipe ; if / be neglected in comparison with r it reduces to 
 2Slt which is the same as previously found for thin cylinders ; 
 if t = r it becomes Sit or only one-half the resistance of a thin 
 cylinder. 
 
 The total interior pressure which tends to rupture the cylin- 
 der longitudinally is 2rl . /, if p be the unit-pressure (Art. 9). 
 
28 PIPES, CYLINDERS, AND RIVETED JOINTS. CH. II. 
 
 Equating this to the total internal resisting stress gives 
 
 from which one of the quantities S, p, r, or / can be computed 
 when the other three are given. 
 
 The above formula was deduced by BARLOW. Although 
 more accurate for thick cylinders than the formula of Art. 10, 
 it is not in practice considered so reliable as the formula of 
 LAME which is deduced in Chapter XIV. If ' r = 6 inches, 
 / i inch, and/ = 400 pounds per square inch, BARLOW'S 
 formula gives 5=2800, while LAME'S formula (page 314) 
 gives 5 = 2620 pounds per square inch. 
 
 Prob. 21. Prove when the thickness of a pipe equals its in- 
 terior radius that the exterior circumference elongates one-half 
 as much as the interior circumference. 
 
 Prob. 22. If a gun of 3 inches bore is subject to an interior 
 pressure of I Soc pounds per square inch, what should be its 
 thickness so that the greatest stress on the material may not 
 exceed 3 ooo pounds per square inch ? 
 
 ART. 12. INVESTIGATION OF RIVETED JOINTS. 
 
 When two plates which are under tension are joined togethei 
 by rivets, these must transfer that tension from one plate to 
 another. A shearing stress is thus brought upon each rivet 
 which tends to cut it off. A compressive stress is also brought 
 sidewise upon each rivet which tends to crush it ; this particu- 
 lar kind of compression is often called "bearing stress." The 
 exact manner in which it acts upon the cylindrical surface of 
 the rivet is not known, but it is usually supposed to be equiva- 
 lent to a stress uniformly distributed over the projection of the 
 
 surface on a plane through the axis of the rivet. 
 
 ft 
 Case I. Lap Joint with single riveting. Let P be the tensile 
 
 stress which is transmitted from one plate to the other by 
 
ART. 12. INVESTIGATION OF RIVETED JOINTS. 
 
 2 9 
 
 means of a single rivet, t the thickness of the plates, d the 
 diameter of a rivet, and a the pitch of the rivets. Let S t , S ft 
 and S c be the unit-stresses in 
 tension, shear, and compression ^N 
 
 produced by P upon the plates " s^ 
 
 and rivets. Then for the ten- 
 sion on the plate, -* 
 
 P = t(a- d)S t , 
 for the shear on the rivet, 
 
 and for compression on the rivet, 
 
 Fig. 5- 
 
 From these equations the unit-stresses may be computed, when 
 the other quantities are known, and by comparing them with 
 the proper working unit-stresses the degree of security of the 
 joint is estimated. 
 
 Case II. Lap Joint with double riveting. In this arrange- 
 ment the plates have a wider lap, 
 and there are two rows of rivets. 
 Let a be the pitch of the rivets in 
 one row, then the tensile stress P is 
 distributed over two rivets, and the 
 three formulas are, 
 
 P=t(a-d}S ty 
 
 \ 
 
 from which the unit-stresses may be 
 
 computed and the strength of the 
 
 joint be investigated. The loss of Fig . 6 
 
 strength is here generally less than in the previous case since a 
 
 can be made larger with respect to d. 
 
30 PIPES, CYLINDERS, AND RIVETED JOINTS. CH. II. 
 
 Case III. Butt Joint with single riveting. For this arrange- 
 ^ ^ / x. ment the shear on each of 
 
 the rivets comes on two 
 
 X^X X^x cross-sections, which is said 
 
 Fig. 7. to be a case of double shear, 
 
 and the formulas are, 
 
 P= t (a- d)S ty = 2 . i-7r</ 2 S s = tdS e . 
 
 Accordingly, a lap joint with double riveting has the same ten- 
 sile and shearing strength as a butt joint with single riveting, 
 if the values of a, d, and t be equal in the two cases ; the bear- 
 ing resistance, however, is only one half as large. 
 
 For example, let it be required to investigate a single riveted 
 butt joint consisting of plates 0.75 inches thick with covers 
 0.375 inches thick, and rivets of I inch diameter and 4 inches 
 pitch, when a tension of 8 ooo pounds is transmitted through 
 one rivet. First, the working tensile unit-stress on the plate is 
 found to have the value, 
 
 8000 
 
 ~~^ ^l = 3 56 P oun ds per square inch. 
 
 Next the shearing unit-stress on the rivets is, 
 
 8000 
 o s = -Q = 5 100 pounds per square inch. 
 
 2 X. 0.785 
 
 Lastly, the bearing compressive unit-stress on the rivets is, 
 
 8000 
 S e = = 10 700 pounds per square inch. 
 
 It thus appears that the joint has the greatest factor of safety 
 against tension and the least against compression of the rivets. 
 It should be said, however, that for wrought iron plates and 
 rivets the highest allowable working stresses for tension, shear, 
 and bearing are generally considered to be about 9 ooo, 7 500, 
 and 1 2 ooo pounds per square inch respectively; hence the 
 
ART. 12. INVESTIGATION OF RIVETED JOINTS. 31 
 
 joint has proper security under the given conditions although 
 the degree of security is quite different for the different 
 stresses. 
 
 The 'efficiency' of a joint is defined to be the ratio of its 
 highest allowable stress to the highest allowable stress of the 
 unriveted plate. The highest allowable stresses in tension, 
 shear, and compression are the three expressions for P, using, 
 for wrought iron, the values above mentioned ; and the highest 
 allowable stress of the unriveted plate isatSt. Thus result the 
 following values of the efficiency, 
 
 a-d 
 For tension, e = - - 
 
 For shear, 
 
 
 . M . dS c 
 For compression, e = ^ 
 
 in which m denotes the number of rivets in the width a which 
 transmit the tension P and n denotes the number of rivet-sec- 
 tions in the same space over which the shear is distributed. 
 The smallest of these values of e is to be taken as the efficiency 
 of the joint. Thus for the above numerical example the three 
 values are 0.75, 0.44, and 0.33 ; the working strength of the 
 joint is, hence, only 33 per cent of that of the unriveted plate. 
 
 If in the above formulas, S t , S t , and S c be taken as the ulti- 
 mate strengths, the resulting values of e will be the efficiencies 
 at the moment of rupture of the joint. For the same numerical 
 example the three ultimate efficiencies are 0.75, 0.48, and 0.25. 
 
 Prob. 23. A boiler is to be formed of wrought iron plates f 
 inches thick, united by single lap joints with rivets f inches 
 diameter and if inches pitch. Find the efficiency of the joint. 
 Find the factor of safety of the boiler if it is 30 inches in di- 
 ameter and carries a steam pressure of 100 pounds. 
 
32 PIPES, CYLINDERS, AND RIVETED JOINTS. CH. IL 
 
 ART. 13. DESIGN OF RIVETED JOINTS. 
 
 A theoretically perfect joint with regard to strength is one 
 so arranged that all parts, (like the one-hoss shay) have the 
 same degree of security. Thus the resistance of the plate to 
 tension must equal the resistance of the rivets to shearing, and 
 each of these must equal the resistance of the rivets to com- 
 pression. The three expressions for P of the last Article should 
 hence be equal, or, what amounts to the same thing, the three 
 efficiencies should be equal. Equating then the second to the 
 third and solving for </, gives 
 
 from which d can be computed when t is assumed. Again, 
 equating the first and third and eliminating d gives, 
 
 , s; 
 
 from which the pitch of the rivets can be obtained. Inserting 
 these values of d and a in either of the expressions for e fur- 
 nishes the formula, 
 
 e = - 
 
 r mS e 
 
 from which the efficiency can be ascertained. The best 'joint 
 will be that which has the least loss of strength due to the 
 riveting, or that which has a value of e as near unity as possible. 
 
 Using for wrought iron plates and rivets the working unit- 
 stresses S t = 9000, 5 S = 7 500, and S c 12000 pounds per 
 square inch, the above formulas for a lap joint with single row 
 of rivets where m = I and n = I, reduce to, 
 
 d = 2.04*, a = 4.75*, e = 0.57, 
 
 so that, if the thickness of the plate be given and the diameter 
 and pitch of the rivet be made according to these rules, the 
 
ART. 13. DESIGN OF RIVETED JOINTS. 33 
 
 joint has about 57 per cent of the strength of the unholed 
 plate. For a lap joint with double riveting where m = 2 and 
 n = 2, they become 
 
 d 2.04/, a = 7.48*, e = 0.73. 
 
 This example shows clearly the advantage of double over single 
 riveting, and by adding a third row the efficiency will be raised 
 to about 80 per cent. 
 
 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 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 so that the formulas above 
 deduced are to be regarded only as a rough guide. The old 
 rules which are still often used for determining the pitch in butt 
 joints, are 
 
 the first being for single and the second for doubje riveting. 
 These are deduced by making the strength of the joint equal 
 in tension and shear, and taking S s = S t . 
 
 It may be required to arrange a joint so as to secure eithei 
 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 
 lap of the plates, and the distance between the rows of rivets, 
 is determined by practical considerations rather than by 
 theoretic formulas. 
 
34 PIPES, CYLINDERS, AND RIVETED JOINTS. CH. II. 
 
 Prob. 24. A lap joint with double riveting is to be formed 
 of plates % inches thick with rivets -J inches diameter. Find 
 the pitch so that the strength of the plate shall equal the 
 shearing strength of the rivets, and compute the efficiency of 
 the joint. 
 
 ART. 14. MISCELLANEOUS EXERCISES. 
 
 It will be profitable to the student to thoroughly perform the 
 following exercises and problems and to write upon each a 
 detailed report, which should contain all the sketches and 
 computations necessary to clearly explain the data, the reason- 
 ing, the computations, and the conclusions. Problem 26 is 
 intended for students proficient in the use of calculus. 
 
 Exercise I. Visit an establishment where tensile tests are 
 made. Ascertain the kind of machine employed, its capacity, 
 the method of applying the stresses, the method of measuring 
 the stresses, the method of measuring the elongations. Ascer- 
 tain the kind of material tested, the reason for testing it, and 
 the conclusions derived from the tests. Give full data for the 
 tests on four different specimens, compute the values of co- 
 efficient of elasticity, ultimate strength and ultimate elongation 
 for them, and state your conclusions. 
 
 Exercise 2. Procure a wrought iron bolt and nut. Measure 
 the diameter of bolt, length of head, and length of nut. State 
 the equation of condition that the head of the bolt may shear 
 off at the same time the bolt ruptures under tension. Com- 
 pute the length of head for a given diameter. Explain why 
 the length of the head is made greater than theory apparently 
 requires. Compile a table giving dimensions of bolts and nuts 
 of different diameters. 
 
 Exercise 3. Go to a boiler shop and witness operations upon a 
 boiler in process of construction. Ascertain length and diameter 
 of boiler, thickness, pitch and diameter of rivets, method of 
 forming holes, method of doing the riveting. Compute the loss 
 of strength caused by the riveting. Compute the steam pressure 
 
ART. 14. MISCELLANEOUS EXERCISES. 35 
 
 which would cause longitudinal rupture of the plate along a 
 line of rivets. Ascertain whether the joint is proportioned in 
 accordance with theory 
 
 Prob. 25. A wrought iron pipe f inches thick and 20 inches 
 in diameter is to be subjected to a head of water of 345 feet. 
 Compute the probable increase in diameter due to the interior 
 pressure, regarding the pipe as thin. 
 
 Prob. 26. Let a pier whose top width is b and length / support 
 a uniformly distributed load P. Let the width of the pier at a 
 distance y below the top be x, its constant length being / at all 
 horizontal sections. Let w be the weight of the masonry per 
 cubic foot. Prove that, in order to make the compressive 
 unit-stress, the same for all horizontal sections, the profile of 
 
 the pier must be such as to satisfy the equation y = -- log, 
 
 w b 
 
 in which .S = -. 
 ol 
 
36 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 CHAPTER III. 
 CANTILEVER BEAMS AND SIMPLE BEAMS. 
 
 ART. 15. DEFINITIONS. 
 
 Transverse stress, or flexure, occurs when a bar is in a 
 horizontal position upon one or more supports. The weight 
 of the bar and the loads upon it cause it to bend and induce 
 in it stresses and strains of a complex nature which, as will be 
 seen later, may be resolved into those of tension, compression, 
 and shear. Such a bar is called a beam. 
 
 A simple beam is a bar resting upon supports at its ends. 
 A cantilever beam is a bar on one support in its middle, or the 
 portion of any beam projecting out of a wall or beyond a sup* 
 port may be called a cantilever beam. A continuous beam is 
 a bar resting upon more than two supports. In this book the 
 word beam, when used without qualification, includes all kinds, 
 whatever be the number of the supports, or whether the ends 
 be free, supported, or fixed. 
 
 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 unsupported at the other is a cantilever beam. 
 
 The loads on beams are either uniform or concentrated. A 
 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 
 
ART l6. REACTIONS OF THE SUPPORTS. 37 
 
 pounds per linear foot. The uniform load per linear unit is 
 designated by w, then wx will represent the load over any 
 distance x. If / be the length of the beam, the total uniform 
 load is wl which may be represented by W. A concentrat- 
 ed load is a weight applied at a definite ppint and is desig- 
 nated by P. 
 
 In this chapter cantilever and simple beams will be princi- 
 pally discussed, although all the fundamental principles and 
 methods hold good for restrained and continuous beams as 
 well. Unless otherwise stated the beams will be regarded as 
 of uniform cross-section throughout, and in computing their 
 weights the rules of Art. i will be found of service. 
 
 Prob. 27. Find the diameter of a round steel bar which 
 weighs 48 pounds, its length being 4 feet. 
 
 ART. 16. REACTIONS OF THE SUPPORTS. 
 
 The points upon which a beam is supported react upward 
 against the beam an amount equal to the pressure of the beam 
 upon them. The beam, being at rest, is a body in equilibrium 
 under the action of a system of forces which consist 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 
 fundamental conditions of static equilibrium, which for a 
 system of vertical forces, are, 
 
 2 of all vertical forces = o, 
 
 2 of moments of all forces = o. 
 
 The first of these conditions says that the sum of all the 
 loads on the beam is equal to the sum of the reactions. Hence 
 if there be but one support, this condition gives at once the 
 reaction. 
 
 For two supports the second condition must be used in con- 
 nection with the first. The center of moments may be taken 
 
38 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 anywhere 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 
 y moments, with the center at the 
 **' left support, is 1 3/? a 4/> = o, 
 
 ^ 
 
 from which R^ ^P. Again the 
 
 equation of moments, with the center at the right support, is 
 13^ gP=o, from which R l = ^P. As a check it may be 
 observed that R, + R, = 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. 
 
 The reactions due to both uniform and concentrated loads 
 on a simple beam may be obtained by adding together the 
 reactions 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. 9 represent a simple beam 12 feet 
 in length between the supports and weighing 35 pounds per 
 . linear foot, its total weight 
 
 I 300 60l 1501 t IT 
 
 , , t ... > J being 420 pounds. Let 
 
 H 3 -4^-2 4<- 3- -*[; 4 
 
 A A' there be three concentrated 
 
 \E! -K 2 | loads of 300, 60, and 150 
 
 Fig - 9 - pounds placed at 3, 5, and 
 
 8 feet respectively from the left support. To find the right 
 
 reaction R^ the center of moments is taken at the left support, 
 
 and the weight of the beam regarded as concentrated at its 
 
 middle ; then the equation of moments is, 
 
 R t X 12 = 420 X 6+ 300 X 3+6o X 5 + 150 X 8 
 from which ^ = 410 pounds. In like manner to find R } the 
 center of moments is taken at the right support, and 
 
 R l x 12 = 420 x6 
 
ART. 17. THE VERTICAL SHEAR. 39 
 
 from which R t = 520 pounds. As a check the sum of R l and 
 R 9 is seen 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 
 continuous beams may be deduced, as will be explained in 
 Chapter IV. 
 
 Prob. 28. A simple beam 12 feet long weighs 20 pounds per 
 linear foot and carries a load of 500 pounds. Where should 
 this load be put so that one reaction may be double the other ? 
 
 Prob. 29. A simple beam weighing 30 pounds per linear foot 
 is 1 8 feet long. A weight of 700 pounds is placed 5 feet from 
 the left end and one of 500 pounds at 8 feet from the right 
 end. Find the reactions due to the total load. 
 
 ART. 17. THE VERTICAL SHEAR. 
 
 When a beam is short it sometimes fails by shearing in a 
 vertical section as shown in Fig. 10. The external force which 
 produces this shearing on any section i 
 
 is the resultant of all the vertical forces | ^J J^ 
 
 on one side of that section. Thus, in 
 the second diagram the resultant of all 
 these external forces is the loads and 
 the weight of the part of the beam on 
 the left of the section ; in the third dia- 
 gram the resultant is the loads and the 
 weight on the right, or it is reaction at pi g . i . 
 
 the wall minus the weight of the beam between the wall and 
 the section. 
 
 'Vertical Shear ' is the name given to the algebraic sum of 
 all external forces on the left of the section considered. Let 
 
 , . 
 
 T2~ 
 
40 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 it be denoted by V, then for any section of a simple or can 
 tilever, beam, 
 
 V= Left reaction minus all loads on left of section. 
 Here upward forces are regarded as positive and downward 
 forces as negative. V is hence positive or negative according 
 as the left reaction exceeds or is less than the loads on the left 
 of the section. To illustrate, consider a simple beam loaded 
 
 in any manner and cut at any sec- 
 
 P\ I IP tion by a vertical plane mn. Let 
 
 ^ rh * * 1 R 1 be the left and R, the right re- 
 action. Let 2P, denote the sum 
 of all the loads on the left of the 
 
 Fig - " section and 2P 9 the sum of those 
 
 on the right. Then, from the definition, 
 
 r=R l --sp l . 
 
 Since R, + R, = 2P, + 2P> it is clear if R, 2P, = + V 
 that R t 2P 9 = V, or that the resultant of all the external 
 forces on one side of the section is equal and opposite to the 
 resultant of those on the other side. They form, in short, a 
 pair of shears acting on opposite sides of the section and tend- 
 ing to cause a sliding or detrusion along the section. The 
 value of the vertical shear for any section of a simple beam or 
 cantilever is readily found by the above equation. When R 1 
 exceeds 2P^ , the vertical shear V is positive, and the left part 
 of the beam tends to slide upward relative to the right part. 
 When R 1 is less than 2P t , the vertical shear V is negative, 
 and the left part tends to slide downward relative to the other. 
 In the upper diagram of Fig. 10 the shear in the left hand sec- 
 tion is positive and that in the right hand section is negative. 
 
 The vertical shear varies greatly in value at different sections 
 of a beam. Consider first a simple beam / feet long and weigh- 
 ing w pounds per linear foot. Each reaction is then \wL 
 Pass a plane at any distance x from the left support, then from 
 
ART. 17. THE VERTICAL SHEAR. 41 
 
 the definition the vertical shear for that section is F \wl wx. 
 Here it is seen that V has its greatest value \wl when x = o, 
 that V decreases as x increases, 
 and that V becomes o when x \L I I 
 When x is greater than /, V is 
 
 negative and becomes \wl when 
 
 x = /. The equation F= ^ee'/ wx 
 
 is indeed the equation of a straight Fig. ia. 
 
 line, the origin being at the left support, and may be plotted 
 
 so that the ordinate at any point will represent the vertical 
 
 shear for the corresponding section of the beam, as shown 
 
 in Fig. 12. 
 
 Consider again a simple beam as in Fig. 13 whose span is 12 
 feet and having three loads of 240, 90, and 120 pounds, situated 
 3, 4, and 8 feet respectively from 
 the left support. By Art. 16 the 
 left reaction is found to be 280 
 and the right reaction 170 pounds. 
 Then for any section between the 
 left support and the first load 
 the vertical shear is F=-|-28o 
 pounds, for a section between the 
 first and second loads it is V = 280 240 = + 40 pounds, for 
 a section between the second and third loads V= 280 240 
 - 90 = 50 pounds, and for a section between the third load 
 and the right support V = 280 240 90 120 = 170 
 pounds, which has the same numerical value as the right reac- 
 tion. By laying off ordinates upon a horizontal line a graphical 
 representation of the distribution of vertical shears throughout 
 the beam is obtained. 
 
 For any section of a simple beam distant x from the left 
 support, let R t denote the left reaction, w the weight of the 
 uniform load per linear unit, and 2P t the sum of all the con- 
 
42 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. IIL 
 
 centrated loads between the section and that support. Then 
 the definition gives, 
 
 as a general expression for the vertical shear at that section. 
 
 A cantilever beam can be so drawn that there is no reaction 
 at the left end, and for any section V = wx 2P 1 . 
 Thus, in Fig. 14, the vertical 
 
 shear for a section in the 
 space a is V = wx, and for 
 a section in the space b it is 
 V = wx P, and the graphi- 
 cal representation is as shown 
 Fig. i 4 . below the beam. 
 
 The vertical shear for any section of a beam is a measure of 
 the tendency to shearing along that section. The above ex- 
 amples show that this is greatest near the supports. It is rare 
 that beams actually fail in this manner, but it is often necessary 
 to investigate the tendency to such failure. 
 
 Prob^. 30. A simple beam 12 feet long and weighing 20 
 pounds per linear foot has loads of 600 and 300 pounds at 2 
 and 4 feet respectively from the left end. Find the vertical 
 shears at several sections throughout the beam, and draw a 
 diagram to show their distribution. 
 
 ART. 1 8. THE BENDING MOMENT. 
 
 The usual method of failure of beams is by cross-breaking or 
 transverse rupture. This is caused by the external forces 
 producing rotation around some point in the section of failure, 
 Thus, in Fig. 14, let a be the distance between the end and the 
 load P, and b be the distance between P and the wall. Then 
 the tendency of Pto cause rotation around a point in the section 
 at the wall is measured by its moment Pb ; if, however, the 
 
ART. 1 8. THE BENDING MOMENT. 43 
 
 load were at the end its tendency to produce rotation around 
 the same point would be measured by the moment f\a "+ b). 
 
 1 Bending moment ' is the name given to the algebraic sum 
 of the moments of the external forces on the left of the sec- 
 tion with reference to a point in that section. Let it be de- 
 noted by M. Then, for a cantilever or simple beam, 
 
 M =. moment of reaction minus sum of moments of loads. 
 
 Here the moment of -upward forces is taken as positive and 
 that of downward forces as negative. M may hence be positive 
 or negative according as the first or second term is the greater. 
 
 For a simple beam of length /, uniformly loaded, each reaction 
 is \wl. For any section distant x from the left support the 
 
 moment is M = \wl . x wx . \x, , 
 
 j^- X" 
 
 x being the lever arm of the re- r-f 
 
 action Jze'/, and \x the lever arm 
 
 of the load wx. Here M = o 
 
 when x = o and also when x = /, 
 
 and M is a maximum when x = \l. Fi s- s- 
 
 The equation, in short, is that of a parabola whose maximum 
 
 ordinate is w/ f and whose graphical representation is as given 
 
 in Fig. 1 5, each ordinate showing the value of M for the cor- 
 
 responding value of the abscissa x. 
 
 Consider next a simple beam loaded with only three weights 
 P, , P^ , and P 3 . Here for any section between the left support and 
 the first load M = Rx, and for any izj |P 2 ,;> 
 
 section between the first and second _t! 11 \ 
 
 loads M = kxP, (xd). Each of ; 
 these expressions is the equation of 
 a straight line, x being the abscissa 
 and J/the ordinate, and the graphical 
 
 representation of bending moments is as shown in Fig. 16. It 
 is seen that for a simple beam all the bending moments are 
 positive. 
 
44 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 For a cantilever there is no reaction at the left end and all 
 the bending moments are negative, the tendency to rotation 
 thus being opposite in direction to that in a simple beam. For 
 instance, for a cantilever beam uniformly loaded and having a 
 load at the end the bending moment is M = Px^wx 1 . 
 Here the variation of moments may be represented by a parab- 
 ola, M being o at the free end and a maximum at the wall. 
 
 For any given case the bending moment at any section may 
 be found by using the definition given above. The external 
 forces on the left of the . section are taken merely for conven- 
 ience, for those upon the right have also 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 the beam around a point 
 in that section. 
 
 The bending moment is a compound quantity resulting from 
 the multiplication of a force by a distance. Usually the forces 
 are expressed in pounds and the distances in feet or inches ; 
 then the bending moments are pound-feet or pound-inches. 
 Thus if a load of 500 pounds be at the middle of a simple 
 beam of 8 feet span, the bending moment under the load is, 
 
 M = 250 X 4 i ooo pound-feet = 12 ooo pound-inches. 
 Again let a simple beam of 8 feet span be uniformly loaded 
 with 500 pounds and have a weight of 200 pounds at the mid- 
 dle. Then the bending moment at the middle is, 
 
 M = 350 X 4250 X 2 = 900 pound-feet. 
 Hence the tendency to rupture is less in the second case than 
 in the first. 
 
 Prob. 31. A beam 6 feet long and weighing 20 pounds per 
 foot is placed upon a single support at its middle. Compute the 
 bending moments for sections distant I, 2, 3, 4, and 5 feet from 
 the left end, and draw a curve to show the distribution of mo- 
 ments throughout the beam. 
 
ART. 19. INTERNAL STRESSES AND EXTERNAL FORCES. 45 
 
 Prob. 32. A simple beam of 6 feet span weighs 20 pounds 
 per linear foot and has a load of 270 pounds at 2 feet from the 
 left end. Find the vertical shears for sections one foot apart 
 throughout the beam, and draw the diagram of shears. Find 
 the bending moments for the same sections and draw the dia- 
 gram to represent them. 
 
 ART. 19. 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 ex- 
 ternal forces and the internal stresses ; or, since the effect of 
 the external forces upon any section is represented by the ver- 
 tical shear (Art. 17) and by the bending moment (Art. 18), the 
 problem is to find the relation between these quantities and the 
 internal stresses in that section. 
 
 Consider a beam of any kind which is loaded in any manner. 
 Imagine a vertical plane mn cutting the beam at any cross-sec- 
 tion. In that section there are act- i | I I 
 ing unknown stresses of various in- 
 tensities and directions. Let the 
 beam be imagined to be separated 
 into two parts by the cutting plane 
 and let forces X, Y, Z, etc., equiv- 
 
 Lu; i 
 
 x 
 
 T 
 
 alent to the internal stresses, be 
 
 /\ j 
 
 applied to the section as shown in , i 
 
 Fig. 17. Then the equilibrium of x *j m *- *- 
 
 each part of the beam will be undis- r^x^ 1 
 
 turbed, for each part will be acted ~f^ 
 
 upon by a system of forces in equi- 
 librium. Hence the following fun- Pig . I7 . 
 damental principle is established. 
 
 The internal stresses in any cross-section of a beam hold irk 
 equilibrium the external forces on each side of that section. 
 
 I 
 
46 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 This is the most important principle in the theory of flexure. 
 It applies to all beams, whether the cross-section be uniform or 
 variable and whatever be the number of the spans or the na- 
 ture of the loading. 
 
 Thus in the above figure the internal stresses X, Y, Z, etc., 
 hold in equilibrium the loads and reactions on the left of the 
 section, and also those on the right.' Considering one part only 
 a system of forces in equilibrium is seen, to which the three 
 necessary and sufficient conditions of statics apply, namely, 
 2 of all horizontal components o, 
 2 of all vertical components = o, 
 2 of moments of all forces = o. 
 
 From these conditions can be deduced three laws concerning 
 the unknown stresses in any section. Whatever be the inten- 
 sity and direction of these stresses, let each be resolved into 
 its horizontal and vertical components. The horizontal com- 
 ponents will be applied at different points in the cross-section, 
 v II some acting in one direc- 
 
 1 1 tion and some in the other, 
 
 or in other words, some of 
 the horizontal stresses are 
 tensile and some compres- 
 Fjg. 18. sive ; by the first condition 
 
 the algebraic sum of these is zero. The vertical components 
 will add together and form a resultant vertical force F which, 
 by the second condition, equals the algebraic sum of the exter- 
 nal forces on the left of the section. As this internal force 
 Facts in contrary directions upon the two parts into which the 
 beam is supposed to be separated, it is of the nature of a shear. 
 Hence for any section of any beam the following laws concern- 
 ing the internal stresses may be stated. 
 
 1st. The algebraic sum of the horizontal stresses is zero ; or 
 the sum of the horizontal tensile stresses is equal to the 
 sum of the horizontal compressive stresses. 
 
 
 I 
 
ART. IQ. INTERNAL STRESSES AND EXTERNAL FORCES. 47 
 
 2nd. The algebraic sum of the vertical stresses forms a re- 
 sultant shear which is equal to the algebraic sum of the 
 external vertical forces on either side of the section. 
 
 3rd. The algebraic sum of the moments of the internal 
 stresses is equal to the algebraic sum of the moments of 
 the external forces on either side of the section. 
 
 These three theoretical laws are the foundation of the theory 
 of the flexure of beams. Their expression may be abbreviated 
 by introducing the following definitions. 
 
 4 Resisting shear* is the name given to the algebraic sum of 
 the internal vertical stresses in any section, and ' vertical shear' 
 is the name for the algebraic sum of the external vertical forces 
 on the left of the section. 'Resisting moment' is the name 
 given to the algebraic sum of the moments of the internal hori- 
 zontal stresses with reference to a point in the section, and 
 * bending moment ' is the name for the algebraic sum of the 
 moments of the external forces on either side of the section 
 with reference to the same point. Then the three laws may 
 be thus expressed for any section of any beam, 
 
 Sum of tensile stresses = Sum of compressive stresses. 
 Resisting shear = Vertical shear. 
 
 Resisting moment == Bending moment. 
 
 The second and third of these equations furnish the funda- 
 mental laws for investigating beams. They state the relations 
 between the internal stresses in any section and the external 
 forces on either side of that section. For the sake of uniform- 
 ity the external forces on the left hand side of the section will 
 generally be used, as was done in Arts. 17 and 18. 
 
 Prob. 33. A beam of weight W which is 6 feet long is sus- 
 tained at one end by a force of 280 pounds acting at an angle 
 of 60 degrees with the vertical, and at the other end by a ver- 
 tical force Y and a horizontal force X. Find the values of X 
 and Y, and the weight of the beam. 
 
48 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 ART. 20. EXPERIMENTAL AND THEORETICAL LAWS. 
 
 From the three necessary conditions of static equilibrium, as 
 stated in Art. 19, three important theoretical laws regarding 
 internal stresses were deduced. These alone, however, are not 
 sufficient for the full investigation of the subject, but recourse 
 must be had to experience and experiment. Experience teaches 
 that when a beam deflects one side becomes concave and the 
 other convex, and it is reasonable to suppose that the hori- 
 zontal tensile stresses are on the convex side and the compres- 
 sive stresses on the concave. By experiments on beams this is 
 confirmed and the following laws deduced. 
 
 (F) The horizontal fibers on the convex side are elongated 
 and those on the concave are shortened, while near the 
 center is a ' neutral surface ' which is unchanged in length. 
 
 (G) The amount of elongation or compression of any fiber 
 is directly proportional to its distance from the neutral 
 surface. Hence by law (B) the horizontal stresses are 
 also directly proportional to their distances from the 
 neutral surface, provided the elastic limit of the material 
 be not exceeded. (See Art. 50.) 
 
 From these laws there will now be deduced the following im- 
 portant theorem regarding the position of the neutral surface : 
 
 The neutral surface passes through the centers of gravity 
 of the cross-sections. 
 
 To prove this let a be the area of any elementary fiber and z its 
 distance from the neutral surface. Let 5 be the unit-stress 
 on the fiber most remote from the neutral surface at the 
 distance c. Then by law 
 
 = unit-stress at the distance unity, 
 c 
 
 z = unit-stress at the distance 2, 
 
ART. 21. THE TWO FUNDAMENTAL FORMULAS. 49 
 
 therefore - az = the total stress on any fiber of area a, 
 
 Sciz 
 and 2 = algebraic sum of all horizontal stresses. 
 
 But by the first law of Art. 19 this algebraic sum is zero, and 
 since 5 and rare constants the quantity 2azmust be zero. This, 
 however, is the condition which makes the line of reference 
 pass through the center of gravity as is plain from the defini- 
 tion of the term 'center of gravity.' Therefore, the neutral 
 surface of beams passes through 
 the centers of gravity of the 
 cross-sections. 
 
 -!-_,_.,_ _Neutr.lSurf. 
 
 The * neutral axis ' of a cross- 
 
 section is the line in which the 
 
 neutral surface intersects the 
 
 plane of the cross-section. On the left of Fig. 19 is shown the 
 
 neutral axis of a cross-section and on the right a trace of the 
 
 neutral surface. 
 
 Prob. 34. A beam 3 inches wide and 6 inches deep is loaded 
 so that the unit-stress at the remotest fiber of a certain cross- 
 section is 600 pounds per square inch. Find the sum of all the 
 tensile stresses on the cross-section. 
 
 Prob. 35. A wooden beam 6 x 12 inches and five feet long is 
 supported at one end and kept level by two horizontal forces 
 X and Z acting at the other end in the median line of the cross- 
 section, the former at 2 inches from the top and the latter at 2 
 inches from the base. Find the values of X and Z. 
 
 ART. 21. THE Two FUNDAMENTAL FORMULAS. 
 
 Consider again any beam loaded in any manner and cut at 
 any section by a vertical plane. The internal stresses in that 
 section hold in equilibrium the external forces on the left of 
 
5O CANTILEVER BEAMS AND SIMPLE BEAMS. CK. HI, 
 
 the section, and as shown in Art. 19, the followirg fundamental 
 laws obtain, 
 
 Resisting shear = Vertical shear, 
 Resisting moment = Bending moment. 
 
 The principles established in the preceding pages can now be 
 applied to the algebraic expression of these four quantities. 
 
 The resisting shear is the algebraic sum of all the vertical 
 components of the internal stresses at any section of the beam. 
 If A be the area of that section and S s the shearing unit-stress, 
 regarded as uniform over the area, then from formula (i), 
 
 Resisting shear = AS S . 
 
 The vertical shear for the same section of the beam being V 
 (Art. 17), the first of the above fundamental laws becomes, 
 
 (3) AS. = F, 
 
 which is the first fundamental formula for the discussion of 
 beams. 
 
 The resisting moment is the algebraic sum of the moments 
 of the internal horizontal stresses at any section with reference 
 to a point in that section. To find an expression for its value 
 let 5 be the horizontal unit-stress, tensile or compressive as the 
 case may be, upon the fiber most remote from the neutral axis 
 and let c be the shortest distance from that fiber to said axis. 
 Also let z be the distance from the neutral axis to any fiber 
 having the elementary area a. Then by law (G) and Fig. 19, 
 
 
 
 - = unit-stress at a distance unity, 
 
 
 
 z = unit-stress at distance z, 
 
 hence -- = total stress on any fiber of area a, 
 
ART. 21. THE TWO FUNDAMENTAL FORMULAS. $1 
 
 and - = moment of this stress about neutral axis. 
 
 c 
 
 Hence 2 -- = resisting moment of horizontal stresses. 
 
 Since 5 and c are constants this expression may be written 
 
 5 
 
 -2az\ But 2aif, being the sum of the products formed by 
 
 multiplying each elementary area by the square of its distance 
 from the neutral axis, is the moment of inertia of the cross- 
 section with reference to that axis and may be denoted by /. 
 
 Therefore, 
 
 57 
 
 Resisting moment = -- 
 
 The bending moment for the same section of the beam being 
 M (Art. 1 8), the second of the above fundamental laws becomes, 
 
 (4) ~ = M, 
 
 which is the second fundamental formula for the discussion of 
 beams. 
 
 Experience and experiment teach that simple beams of uni- 
 form section break near the middle by the tearing or crushing 
 of the fibers and very rarely at the supports by shearing. 
 Hence it is formula (4) that is mainly needed in the practical 
 investigation of beams. The following example and problem 
 relate to formula (3) only, while formula (4) will receive detailed 
 discussion in the subsequent articles. 
 
 As an example, consider a wrought iron I beam 15 feet long 
 and weighing 200 pounds per yard, over which roll two locomo- 
 tive wheels 6 feet apart and each bearing 12 ooo pounds. The 
 maximum vertical shear at the left support will evidently occur 
 when one wheel is at the support (Art. 16). The reaction will 
 then be 500 -f- 12 coo + T V X 12 ooo = 19 700 pounds. Ac- 
 cordingly the greatest value of V in the beam is 19 700 
 
52 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 pounds. As the area of the cross-section is 20 square inches 
 the average shearing unit-stress by formula (3) is 985 pounds, 
 so that the factor of safety is about 50. 
 
 Prob. 36. A wooden beam 6x9 inches and 12 feet in span 
 carries a uniform load of 20 pounds per foot besides its own 
 weight and also two wheels 6 feet apart, one weighing 4 ooo 
 pounds and the other 3 ooo pounds. Find the factor of safety 
 against shearing. 
 
 ART. 22. CENTER OF GRAVITY OF CROSS-SECTIONS. 
 
 The fundamental formula (4) contains c, the shortest dis- 
 tance from the remotest part of the cross-section to a hori- 
 zontal axis passing through the center of gravity of that cross- 
 section. The methods of rinding c are explained in books on 
 theoretical mechanics and will not here be repeated. Its val- 
 ues for some of the simplest cases are however recorded for 
 reference. 
 
 For a rectangle whose height is d, c \d. 
 
 For a circle whose diameter is d, c = %d. 
 
 For a triangle whose altitude is d, c = \d 
 
 For a square with side d having one diagonal 
 
 vertical, c = d yf" 
 
 For a I whose depth is d, c = %d. 
 
 For a JL whose depth is d, thickness of flange , 
 width of flange d, and thickness of web t', 
 
 t 'd + t(b - t'} 
 For a trapezoid whose depth is d, upper base , 
 
 b + 2b' a 
 
 and lower base b , c = -j. 77 
 
 b + b 3 
 
 The student should be prepared to readily apply the principle 
 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 areas, 
 
ART. 23. MOMENT OF INERTIA OF CROSS-SECTIONS. 53 
 
 whose centers of gravity are known, so that the statement of 
 the equation for finding c is very simple. 
 
 Prob. 37. Find the value of c for a rail headed beam whose 
 section is made up of a rectangular flange } X 4 inches, a rect- 
 angular web \ X 5 inches, and an elliptical head J inches deep 
 and i inches wide. 
 
 ART. 23. MOMENT OF INERTIA OF CROSS-SECTIONS. 
 
 The fundamental formula (4) contains /, the moment of in- 
 ertia of the cross-section of the beam with reference to a hori- 
 zontal axis passing through the center of gravity of that cross- 
 section. Methods of determining / are explained in works on 
 elementary mechanics and will not here be repeated, but the 
 values of some of the most important cases are recorded for 
 reference. 
 
 r / 
 
 For a rectangle of base b and depth d, I = -- 
 
 nd % 
 
 For a circle of diameter d y I =. = 
 
 64 
 
 For an ellipse with axes a and b, the latter 
 
 nab* 
 
 vertical, I? - 
 
 64 
 
 bd* 
 For a triangle of base b and depth d, /= 
 
 For a square with side d, having one diag- 
 
 onal vertical, /= 
 
 For a I with base b, depth d, thickness of 
 flanges t and thickness of web /' ', 
 
 bd* (b f)(d 2t) % 
 
 12 
 
 For a JL with base b, depth d, thickness 
 of flange /, thickness of web /' and 
 
 area A, /= - * - k*^, 4 Ac* 
 
54 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 The value of / for any given section may always be computed 
 by dividing the figure into parts whose moments of inertia are 
 known and transferring these to the neutral axis by means of 
 the familiar rule /! = / + Ah*, where 7 is the primitive value 
 for an axis through the center of gravity, I I the value for any 
 parallel axis, A the area of the figure and h the distance be- 
 tween the two axes. 
 
 Prob. 38. Compute the least moment of inertia of a trape- 
 zoid whose altitude is 3 inches, upper base 2 inches, and lower 
 base 5 inches. 
 
 Prob. 39. Find the moment of inertia of a triangle with ref- 
 erence to its base, and also with reference to a parallel axis 
 passing through its vertex. 
 
 ART. 24. THE MAXIMUM BENDING MOMENT. 
 
 The fundamental equation (4), namely = M, is true for 
 
 any section of any beam, / being the moment of inertia of that 
 section about its neutral axis, c the vertical distance from that 
 axis to the remotest fiber, 5 the tensile or compressive unit- 
 stress on that fiber, and M the bending moment of all the ex- 
 ternal forces on one side of the section. For a beam of con- 
 stant cross-section S varies directly as M, and the greatest 5 
 will be found where M is a maximum. The place where M 
 has its maximum value may hence be called the * dangerous 
 section,' it being the section where the horizontal fibers are 
 most highly strained. 
 
 For a simple beam uniformly loaded with w pounds per 
 linear unit, the dangerous section is evidently at the middle, 
 
 wl* 
 and, as shown in Art. 18, the maximum Mis -$- 
 
 o 
 
 For a simple beam loaded with a single weight P at the dis- 
 
 l p 
 tance/ from the left support, the left reaction is R P , 
 
ART. 24. THE MAXIMUM BENDING MOMENT. 55 
 
 p\p 
 
 and the maximum moment is - ~ . If Pbe movable the 
 distance/ will be variable, and when the load is at the middle 
 the maximum M is 
 
 For a beam loaded with given weights, either uniform or 
 concentrated, it may be shown that the dangerous section is at 
 the point where the vertical shear passes through zero. To 
 prove this let P^ be any concentrated load on the left of the 
 section and/ its distance from the left support, and w the uni- 
 form load per linear unit. Then, for any section distant x 
 from the left support, 
 
 M=R l x-wx.-- 2P,(x -/). 
 
 To find the value of x which renders this a maximum, the first 
 derivative must be put equal to zero ; thus, 
 
 ~ = R.-wx 2P, = o. 
 dx 
 
 But RI wx 2P { is the vertical shear V for the section x 
 (see Art. 17). Therefore the maximum moment occurs at the 
 section where the vertical shear passes through zero. 
 
 To find the dangerous section for any given case the reac- 
 tions are first to be computed by Art. 16, and then the verti- 
 cal shears are to be investigated by Art. 17. For a cantilever, 
 however it be loaded, it is seen that the dangerous section 
 is at the wall. For a simple beam with concentrated loads 
 the point where the vertical shear passes through zero must 
 usually be ascertained by trial. Thus, referring to Fig. 9 and 
 the example in Art. 16, the vertical shear just at the left of 
 the first load is V = 520 3 X 35 = + 4 r 5 pounds, and just 
 at the right of the first load it is F= 520 3X35 300 
 + 115 pounds. Again for the second load the vertical shear 
 just at the left is V= 520 5 X 35 30x3 = -f 45 pounds, and 
 just at the right it is V= 520 5 X 35 360 = 15 pounds. 
 Hence in this case the vertical shear changes sign, or passes 
 
56 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 through zero, under the second load, and accordingly this is 
 the position of the dangerous section. 
 
 When the dangerous section has been found the bending 
 moment for that section is to be computed by the definition 
 of Art. 1 8, and this will be the maximum bending moment for 
 the beam. Thus, for the numerical example of the last para- 
 graph, the maximum bending moment is, 
 M= 520 X 5 175 X 2j 300 X 2 = + 1562.5 pound-feet. 
 Again, let a cantilever beam 8 feet long be loaded with 40 pounds 
 per linear foot and carry a weight of 1 50 pounds at the free end ; 
 then the maximum bending moment is, 
 
 M = 320 X 4 150x8= 2 480 pound-feet. 
 The bending moment for simple beams is seen to be always 
 positive and for cantilever beams always negative. That is to 
 say, in the former case the exterior forces on the left of the 
 section cause compression in the upper and tension in the lower 
 fibers of the beam, while in the latter case this is reversed ; or 
 the upper side of a deflected simple beam is concave and the 
 upper side of a deflected cantilever beam is convex. 
 
 Prob. 40. A simple beam 12 feet long carries a load of 150 
 pounds at 5 feet from the left end and a load of 150 pounds at 
 5 feet from the right end. Find the dangerous section, and the 
 maximum bending moment. 
 
 Prob. 41. A simple beam 12 feet long weighs 20 pounds per 
 foot and carries a load of 100 pounds at 4 feet from the left end 
 and a load of 50 pounds at 7 feet from tlie left end. Find the 
 dangerous section, and the maximum bending moment. 
 
 ART. 25. THE INVESTIGATION OF BEAMS. 
 
 The investigation of a beam consists in deducing the greatest 
 horizontal unit-stress 5 in the beam from the fundamental for- 
 mula (4). This may be written, 
 
 <?- 
 
 T' 
 
ART. 25. THE INVESTIGATION OF BEAMS. 57 
 
 First, from the given dimensions find, by Art. 22, the value of c 
 and by Art. 23 the value of 7. Then by Art. 24 determine the 
 value of maximum M. From (4) the value of S is now known. 
 Usually c and 7 are taken in inches, and M in pound-inches ; 
 then the value of 5 will be in pounds per square inch. 
 
 The value of vS will be tension or compression according as 
 the remotest fiber lies on the convex or concave side of the beam. 
 If S' be the unit-stress on the opposite side of the beam and 
 c' the distance from the neutral axis, then from law (G), 
 
 C C' f t 
 
 - = ^r and S' = S-. 
 
 c c c 
 
 If S be tension, S' will be compression, and vice versa. Some- 
 times it is necessary to compute S' as well as S in order to 
 thoroughly investigate the stability of the beam. By comparing 
 the values of S and S f with the proper working unit-stresses 
 for the given materials (Art. 8), the degree of security of the 
 beam may be inferred. 
 
 As an example consider a wrought iron I' beam whose depth 
 is 12 inches, width of flange 4.5 inches, thickness of flange I inch 
 and thickness of web 0.78 inches. It is supported at its ends 
 forming a span of 12 feet, and carries two loads each weighing 
 10 ooo pounds, one being at the middle and the other at one 
 foot from the right end. 
 
 By Art. i, w = 56 pounds per linear foot. 
 
 By Art. 16, R = 6169 pounds. 
 
 By Art. 22, c = 6 inches. 
 
 By Art. 23, 1= 338 inches 4 . 
 
 By Art. 24, x = 6 feet for dangerous section. 
 
 By Art. 24, max. M 36006 X 12 pound-inches. 
 
 Then from formula (4) the unit-stress at the dangerous section is, 
 
 ~ 36000 X 12 x 6 
 
 o = - - - - = 7 
 
 700 pounds per square inch. 
 
58 CANTILEVER BEAMS AND SIMPLE BEAMS- CH. III. 
 
 This is the compressive unit-stress on the upper fiber and also 
 the tensile unit-stress on the lower fiber, and being only about 
 one-third of the elastic limit for wrought iron and about one- 
 seventh of the ultimate strength it appears that the beam is 
 entirely safe for steady loads (Art. 8). It will usually be best 
 in solving problems to insert all the numerical values at first in 
 the formula and thus obtain the benefit of cancellation. 
 
 A short beam heavily loaded should also be investigated for 
 the shearing stress at the supports in the manner mentioned in 
 Art. 21, but in ordinary cases there is little danger from this 
 cause. Thus for the above example the maximum vertical 
 shear occurs at the right end and is 14500 pounds; as the 
 area of the cross-section is 16.8 square inches, the mean shear- 
 ing unit-stress at the right end is from (3), 
 
 S e = ^-jr = 863 pounds per square inch, 
 ID. 8 
 
 so that the factor of safety against shearing is nearly 60. 
 
 Prob. 42. A piece of scantling 2 inches square and 10 feet 
 long is hung horizontally by a rope at each end and three 
 painters stand upon it. Is it safe? 
 
 Prob. 43. A wrought iron bar one inch in diameter and two 
 feet long is supported at its middle and a load of 500 pounds 
 hung upon each end of it. Find its factor of safety. 
 
 ART. 26. SAFE LOADS FOR BEAMS. 
 
 
 
 The proper load for a beam should not make the value of 5 
 at the dangerous section greater than the allowable unit-stress. 
 This allowable unit-stress or working strength is to be assumed 
 according to the circumstances of the case by first selecting a 
 suitable factor of safety from Art. 8 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, by the general formula (4), 
 
ART. 27. DESIGNING OF BEAMS. 59 
 
 the bending moment M may be expressed in terms of the un- 
 known loads on the beam, and thus those loads be found. The 
 sign of the bending moment should not be used in (4), since 
 that sign merely denotes whether the upper fiber of the beam 
 is in tension or compression, or indicates the direction in which 
 the external forces tend to bend it. 
 
 As an example, consider a 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, /= .g, and M = 36 x 6zv. Then from 
 formula (4), * 
 
 800 X 54 
 2i6ze;= j , whence w = 1 1 pounds. 
 
 Since a wooden beam 2X3 inches weighs about 2 pounds per 
 linear foot, the safe load in this case will be about 9 pounds 
 per foot. 
 
 Prob. 44. A wooden beam 8x9 inches and of 14 feet span 
 carries a load, including its own weight, of w pounds per linear 
 foot. Find the value of w for a factor of safety of 10. 
 
 Prob. 45. A steel railroad rail of 2 feet span carries a load 
 P at the middle. If its weight per yard is 56 pounds, /= 12.9 
 inches 4 and c = 2.16 inches, find P so that the greatest horizon- 
 tal unit-stress at the dangerous section shall be 6 ooo pounds 
 per square inch. 
 
 ART. 27. 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. Thus the maximum 
 bending moment may be found. The allowable working strength 
 5 is assumed in accordance with engineering practice. Then 
 formula (4) may be written, 
 
60 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 and the numerical value of the second member be found. The 
 dimensions to be chosen for the beam must give a value of- 
 
 equal to this numerical value, and these in general are deter- 
 mined tentatively, 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 wrought iron beam of 4 feet span is required 
 to carry a rolling load of 500 pounds. Here, by Art. 24, the 
 value of maximum M due to the load of 500 pounds is 6 ooo 
 pound-inches. From Art. 8 the value of 5 for a variable load 
 is about 10 ooo pounds per square inch. Then, 
 
 / 6000 
 
 c 10 ooo 
 
 = 0.6 inches 3 . 
 
 An infinite number of cross-sections may be selected with this 
 value of . If the beam is to be round and of diameter </, it 
 
 is known that c = \d and /= -7 . Hence, 
 
 itd* 
 , =0.6, whence d= 1.83 inches. 
 
 If the cross-section is to be rectangular, the dimensions 1x2 
 inches would give the value of as f which would be a little 
 
 too large, but it would be well to use it because the weight of 
 the beam itself has not been considered in the discussion. If 
 thought necessary these dimensions may now be investigated 
 by Art. 25 in order to determine how closely the actual unit- 
 stress agrees with the value assumed. Thus the rectangular 
 section I x 2 inches weighs 6f pounds per foot ; the maximum 
 
ART. 28. THE MODULUS OF RUPTURE. 6l 
 
 bending moment is then 6 1 60 pound-inches, and the unit-stress 
 is found to be 9 240 pounds per square inch. 
 
 Prob. 46. Design a hollow circular wrought iron beam for a 
 span of 12 feet to carry a load of 320 pounds per linear foot. 
 
 Prob. 47. A rectangular wooden beam of 14 feet span carries 
 a load of I ooo pounds at its middle. If its width is 4 inches 
 find its depth for a factor of safety of 10. 
 
 ART. 28. THE MODULUS OF RUPTURE. 
 
 The fundamental formula (4) is only true for stresses within 
 the elastic limit, since beyond that limit the law (G) does not 
 hold, and .the horizontal unit-stresses 
 are no longer proportional to their ) 
 distances from the neutral axis, but 
 increase in a less rapid ratio. The 
 sketch shows a case where the fiber 
 
 stresses above m and below n have Flg> **' 
 
 surpassed the elastic limit. It is however very customary in 
 practical computations to apply (4) to the rupture of beams. 
 
 The ' modulus of rupture ' is the value of 5 deduced from 
 formula (4) when the beam is loaded up to the breaking point. 
 It is always found by experiment that the modulus of rupture 
 does not agree with either the ultimate tensile or compressive, 
 strength of the material but is intermediate between them. If 
 formula (4) were valid beyond the elastic limit, the value of 5 
 for rupture would agree with the least ultimate strength, with 
 tension in the case of cast iron and with compression in the case 
 of timber. Ductile materials like wrought iron and soft steel 
 have no modulus of rupture because they continue to bend 
 indefinitely under increasing transverse loads and failure does 
 not occur by breaking. The following table gives values of 
 the modulus of rupture, all in pounds per square inch, as 
 found by tests on rectangular beams. 
 
62 
 
 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 Material. 
 
 Tensile 
 
 Strength, S t . 
 
 Modulus of 
 Rupture, S r . 
 
 Compressive 
 
 Strength, S c . 
 
 Timber 
 
 IOOOO 
 
 9000 
 
 8000 
 
 Brick 
 
 
 800 
 
 2 500 
 
 Stone 
 
 
 2000 
 
 6 ooo 
 
 Cast Iron 
 
 20000 
 
 35000 
 
 90000 
 
 Wrought Iron 
 
 55000 
 
 . 
 
 55000 
 
 Steel 
 
 IOOOOO 
 
 
 150000 
 
 By the use of the experimental values off the modulus of rup- 
 ture it is easy with the help of formula (4) 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. 
 
 Prob. 48. What must be the size of a square wooden beam of 
 8 feet span in order to break under its own weight? 
 
 Prob. 49. A cast iron cantilever beam 2 inches square and 6 
 feet long carries a load P at the end. Find the value of P to 
 cause rupture. 
 
 ART. 29. COMPARATIVE STRENGTHS. 
 
 The strength of a beam is measured by the load that it can 
 carry. Let it be required to determine the relative strength of 
 the four following cases, 
 
 1st, A cantilever loaded at the end with W, 
 2nd, A cantilever uniformly loaded with W, 
 3rd, A simple beam loaded at trie middle with W, 
 4th, A simple beam loaded uniformly with W. 
 
 Let / be the length in each case. Then, from Art. 24 and for- 
 mula (4), 
 
 For ist, M=Wt and hence W = -. 
 
ART. 29. COMPARATIVE STRENGTHS. 63 
 
 Wl SI 
 
 For 2nd, M= - and hence W = 2-,-. 
 
 2 1C 
 
 Wl SI 
 
 For 3rd, M = - and hence W = 4-7-. 
 
 4 lc 
 
 Wl SI 
 
 For 4th, M = -5- and hence W= 8-7-. 
 o lc 
 
 Therefore the comparative strengths of the four cases are as 
 the numbers I, 2, 4, 8. That is, if four such beams be of equal 
 size and length and of the same material, the 2nd is twice as 
 strong as the 1st, the 3rd four times as strong, and the 4th eight 
 times as strong. From these equations also result the follow- 
 ing important laws. 
 
 The strength of a beam varies directly as 5, directly as 
 7, inversely as c, and inversely as the length /. 
 
 A load uniformly distributed produces only one-half- as 
 much stress as the same load when concentrated. 
 
 These apply to all cantilever and simple beams whatever be 
 the shape of the cross-section. 
 
 When the "cross-section is rectangular, let b be. the breadth 
 and dfthe depth, then (Art. 23) the above equations become, 
 
 where n is either i, 2, 4, or 8, as the case may be. Therefore, 
 The strength of a rectangular beam varies directly as the 
 breadth and directly as the square of the depth. 
 
 The reason why rectangular beams are put with the greatest 
 dimensions vertical is now apparent. 
 
 To find the strongest rectangular beam that can be cut from 
 a circular log of given diameter D, it is necessary to make bd* 
 a maximum. Or the value of b is to be found which makes 
 b (D* F) a maximum. By placing the first derivative equal 
 to zero this value of b is readily found. Thus, 
 
 b = DV and dDV- 
 
6 4 
 
 CANTILEVER BEAMS AND SIMPLE BEAMS. ClI. III. 
 
 Fig. 21. 
 
 Hence very nearly, b\ d\ : 5 : 7. From this it is evident 
 that the way to lay off the strongest beam on 
 \ the end of a circular log is to divide the diameter 
 j into three equal parts, from the points of divi- 
 J 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 beam thus cut out is, of course, not as strong as the log, 
 and the ratio of the strength of the beam to that of the log is 
 
 that of their values of -, which will be found to be about 0.65. 
 
 Prob. 50. Compare the strength of a rectangular beam 2 
 inches wide and 4 inches deep with that of a circular beam 3 
 inches in diameter. 
 
 Prob. 51. Compare the strength of a wooden beam 4 X 6 
 inches and 10 feet span with that of a wrought iron beam I X 2 
 inches and 7 feet span. 
 
 ' 
 
 ART. 30. IRON AND STEEL I BEAMS. 
 
 Medium-steel I beams are rolled at present in about thirteen 
 different depths or sizes ; of each there is a light and a heavy 
 weight, and weights intermediate in value 
 may also be obtained. They are extensively 
 used in engineering and architecture. The 
 following table gives mean sizes, weights, 
 and moments of inertia of those steel beams 
 most commonly found in the market. The 
 sizes of different manufacturers agree as to 
 depth, but vary slightly with regard to pro- 
 portions of cross-section, weights per foot, 
 Fig. aa. an( j mO ments of inertia. Fig. 22 shows the 
 
 proportions of the light and heavy 6-inch beams. Wrought-iron 
 I beams were extensively used prior to 1890, but are now rarely 
 
ART. 30. 
 
 IRON AND STEEL I BEAMS. 
 
 rolled ; the table for wrought-iron sections given in previous 
 editions of this book is hence here replaced by one for steel. 
 
 The moments of inertia in the fourth column of the table are 
 taken about an axis perpendicular to the web at the center, 
 this being the neutral axis of the cross-section when used as a 
 beam. The values of /' are with reference to an axis coincid- 
 ing with the center line of the web and are for use in Chapter V. 
 
 Depth. 
 
 Weight per 
 Foot. 
 
 Section 
 Area. 
 
 Moment of 
 Inertia. 
 
 Section 
 Modulus. 
 
 Moment of 
 Inertia. 
 
 Inches. 
 
 Pounds. 
 
 A 
 Sq. Inches. 
 
 Inches*. 
 
 c 
 Inches 1 . 
 
 I* 
 Inches*. 
 
 24 
 
 IOO 
 
 29.4 
 
 2380 
 
 198 
 
 48.6 
 
 24 
 
 80 
 
 23.5 
 
 2088 
 
 174 
 
 42.9 
 
 20 
 
 75 
 
 22.1 
 
 I 269 
 
 127 
 
 302 
 
 20 
 
 65 
 
 *I9.I 
 
 I 170 
 
 iif 
 
 27-9 
 
 18 
 
 70 
 
 20.6 
 
 921 
 
 102 
 
 24.6 
 
 18 
 
 55 
 
 15.9 
 
 796 
 
 88.4 
 
 21.2 
 
 15 
 
 55 
 
 15.9 
 
 511 
 
 68.1 
 
 I7.I 
 
 15 
 
 42 
 
 12.5 
 
 442 
 
 58.9 
 
 14.6 
 
 12 
 
 35 
 
 10.3 
 
 228 
 
 38.0 
 
 10. 1 
 
 12 
 
 31* 
 
 9-3 
 
 216 
 
 36.0 
 
 9.50 
 
 10 
 
 40 
 
 u. 8 
 
 159 
 
 31-8 
 
 9.50 
 
 10 
 
 25 
 
 7-4 
 
 122 
 
 24.4 
 
 6.89 
 
 9 
 
 35 
 
 10.3 
 
 112 
 
 24.8 
 
 7-31 
 
 9 
 
 21 
 
 6.3 
 
 85.0 
 
 18.9 
 
 5-16 
 
 8 
 
 25* 
 
 7-50 
 
 68.4 
 
 17.1 
 
 4-75 
 
 8 
 
 18 
 
 5-33 
 
 56.9 
 
 14.2 
 
 3.78 
 
 7 
 
 20 
 
 5-88 
 
 42.2 
 
 12. 1 
 
 3-24 
 
 7 
 
 15 
 
 4.42 
 
 36.2 
 
 10.4 
 
 2.67 
 
 6 
 
 17^ 
 
 5-07 
 
 26.2 
 
 8-73 
 
 2.36 
 
 6 
 
 I2 i 
 
 3-61 
 
 21.8 
 
 7.27 
 
 1.8 5 
 
 5 
 
 I4f 
 
 4-34 
 
 15.2 
 
 6.08 
 
 1.70 
 
 5 
 
 ti 
 
 2.8 7 
 
 12. 1 
 
 4.84 
 
 1.23 
 
 4 
 
 xoi 
 
 3.09 
 
 7-1 
 
 3-55 
 
 1. 01 
 
 4 
 
 7* 
 
 2.21 
 
 6.0 
 
 3.00 
 
 0.77 
 
 3 
 
 74 
 
 2.21 
 
 2.9 
 
 1-93 
 
 0.60 
 
 3 
 
 5i 
 
 1.6 3 
 
 2.5 
 
 1.71 
 
 0.46 
 
66 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 In investigating the strength of a given I beam the value of - 
 
 is taken from the table and 5 is computed from formula (4). 
 In designing an I beam for a given span and loads the value of 
 
 - is found by (4) from the data and then from the table that I 
 
 is selected which has the nearest or next highest corresponding 
 value. Intermediate weights between those given in the table 
 
 can also usually be obtained ; thus, if the computed value of - 
 
 should be 37.0 a 12-inch beam weighing about 33 pounds per 
 foot might be chosen. 
 
 For example, let it" be required to determine which I should 
 be selected for a floor loaded with 1 50 pounds per square foot, 
 the beams to be of 20 feet span and spaced 12 feet apart be- 
 tween centers, and the maximum unit-stress 5 to be 16000 
 pounds per square inch. Here the uniform load on the beam 
 is 12 X 20 X 150 = 36 ooo pounds = W. From formula (4), 
 
 - HL 36 000 X 20 X 12 _ ~ 
 
 c ~~~- S " 8X16000 7 ' 5 ' 
 
 and hence the heavy 1 5-inch I should be selected. 
 
 In the following examples and problems the ultimate tensile 
 strength of medium steel is taken as 65 ooo pounds per square 
 inch, and factors of safety are based on this constant. 
 
 Prob. 52. A heavy 15 inch 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 feet span under the 
 same load. 
 
 Prob. 53. A floor, which is to sustain a uniform load of 175 
 pounds per square foot, is to be supported by heavy 10 inch I 
 beams of 15 feet span. Find their proper distance apart from 
 center to center so that the maximum fiber stress may be 
 1 2 ooo pounds per square inch. 
 
ART. 31. 
 
 IRON AND STEEL DECK BEAMS. 
 
 Q 
 
 ART. 31. IRON AND STEEL DECK BEAMS. 
 Deck beams are used in the construction of buildings, and 
 are of a section such as shown in Fig. 23. The heads are 
 formed with arcs of circles but may be 
 taken as elliptical in computing the values 
 of c and /. The following table gives di- 
 mensions of a few of the steel sections 
 found in the market. 
 
 By means of formula (4) a given deck 
 beam may be investigated or safe loads be 
 determined for it, or one may be selected 
 for a given load and span. Sometimes T 
 irons are used instead of deck beams ; the 
 values of c and / for these are given in the 
 handbooks issued by the manufacturers, or they may be com- 
 puted with an accuracy usually sufficient by regarding the 
 web and flange as rectangular (Arts. 22 and 23). 
 
 Fig. 23. 
 
 Size. Depth. 
 
 Weight 
 per Foot. 
 
 Section 
 Area. 
 
 Moment 
 of Inertia. 
 
 From Top 
 to Axis. 
 
 ~i 
 Section 
 
 Modulus. 
 / 
 
 
 
 A 
 
 / 
 
 e 
 
 
 
 
 
 
 
 c 
 
 Inches. 
 
 Pounds. 
 
 Sq. Inches. 
 
 Inches*. 
 
 Inches. 
 
 Inches*. 
 
 Heavy 9 
 
 30 
 
 8.8 
 
 93-2 
 
 4-75 
 
 19.6 
 
 Light 9 
 
 26 
 
 7 .6 
 
 8 5 .2 
 
 4.81 
 
 17-7 
 
 H 8 
 
 24.5 
 
 7-2 
 
 62.8 
 
 4-45 
 
 I4.I 
 
 L 8 
 
 2O. I 
 
 5-9 
 
 55-6 
 
 4-56 
 
 12.2 
 
 H 7 
 
 23-5 
 
 6.9 
 
 45-5 
 
 3-89 
 
 II.7 
 
 L 7 
 
 18.1 
 
 5-3 
 
 38.8 
 
 4.00 
 
 9-7 
 
 Prob. 54. A heavy 7 inch deck beam is loaded uniformly 
 with 50 ooo pounds. Find its factor of safety for a span of 
 22 feet. Also for a span of 1 1 feet. 
 
 Prob. 55. What uniform load should be placed upon a heavy 
 7 inch deck beam of 22 feet span so that the greatest unit-stress 
 at the dangerous section may be 12 ooo pounds per square inch? 
 
68 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 ART. 32. CAST IRON BEAMS. 
 
 Wrought iron beams are usually made with equal flanges 
 since the resistance of wrought iron is about the same for both 
 tension and compression. For cast iron, however, the flange 
 under tension should be larger than that under compression, 
 since the tensile resistance of the material is much less than its 
 compressive resistance. Let S' be the unit-stress on the re- 
 motest fiber on the tensile side and S that on the compressive 
 side, at the distances c' and c respectively from the neutral axis. 
 Then, from law 
 
 c' ~ S' ' 
 
 Now if the working values of 5 and S' can be selected the 
 ratio of c to c' is known and a cross-section can be designed, 
 but it is difficult to assign these proper values on account of 
 our lack of knowledge regarding the elastic limits of cast iron. 
 
 According to HODGKINSON'S investigations the following 
 are dimensions For a cast iron beam of equal ultimate strength. 
 
 Thickness of web = /, 
 
 Depth of beam = 13.5^, 
 
 Width of tensile flange = \2t, 
 
 Thickness of tensile flange = 2/, 
 
 Width of compressive flange = 5/, 
 Thickness of compressive flange = i%t y 
 
 Value of c = gt, 
 
 Value of / = 923^. 
 
 Here the unit-stress in the tensile flange is one-half that in the 
 compressive flange. Although these proportions may be such 
 as to allow the simultaneous rupture of the flanges, yet it does 
 not necessarily follow that they are the best proportions for 
 ordinary working stresses, since the factors of safety in the 
 flanges as computed by the use of formula (4) would be quite 
 different. The proper relative proportions of the flanges of 
 
ART. 32. CAST IRON BEAMS. 69 
 
 cast iron beams for safe working stresses have never been 
 definitely established, and on account of the extensive use of 
 wrought iron the question is not now so important as formerly. 
 
 As an illustration of the application of formula (4) let it be 
 required to determine the total uniform load W for a cast iron 
 JL beam of 14 feet span, so that the factor of safety may be 6, 
 the depth of the beam being 18 inches, the width of the flange 
 12 inches, the thickness of the stem I inch, and the thickness 
 of the flange i inches. First, from Art. 22 the value of c is 
 found to be 12.63 inches, and that of c' to be 5.37 inches, 
 From Art. 23 the value of / is computed to be I 031 inches 4 , 
 From Art. 24 the maximum bending moment is, 
 
 wl* 
 
 M = - = 2 1 W pound-inches. 
 8 
 
 Now with a factor of safety of 6 the working strength S on the 
 remotest fiber of the stem of the dangerous section is to be 
 
 ^7: pounds per square inch. Hence from formula (4), 
 
 2lW = 9QQQQX 103^ whence w _ 53 300 pounds. 
 
 Again with a factor of safety of 6 the working strength S' on 
 the remotest fiber of the flange at the dangerous section is to be 
 
 - pounds per square inch. Hence from the formula, 
 2 1 w = 2Q ^ OX I031 , whence W = 30 400 pounds. 
 
 The total uniform load on the beam should hence not exceed 
 30400 pounds. Under this load the factor of safety on the 
 tensile side is 6, while on the compressive side it is nearly 12. 
 
 Prob. 56. A cast iron beam in the form of a channel, or 
 hollow half rectangle, is often used in buildings. Suppose the 
 thickness to be uniformly one inch, the base 8 inches, the height 
 
70 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 6 inches and the span 12 feet. Find the values of 5 and ' at 
 the dangerous section under a uniform load of 5 ooo pounds. 
 
 ART. 33. GENERAL EQUATION OF THE ELASTIC CURVE. 
 
 When a beam bends under the action of exterior forces the 
 curve assumed by its neutral surface is called the elastic curve. 
 It is required to deduce a general expression for its equation. 
 
 Let // in the figure be any normal section in any beam. 
 Let mn be any short length dl, measured on the neutral surface, 
 
 and let qmq be drawn parallel to 
 the normal .section through n. 
 Previous to the bending the sec- 
 tions pp and tt were parallel ; 
 now they intorsect at o the cen- 
 ter of curvature. Previous to 
 the bending pt was equal to dl, 
 now it has elongated or shortened 
 the amount pq. The distance 
 Fig - 34> pq will be called A and the dis- 
 
 tance mp is the quantity c (Art. 22). The elongation A is pro- 
 duced by the unit-stress 5, and from (2) its value is, 
 
 A = 
 
 where E is the coefficient of elasticity of the material of the 
 beam. From the similar figures omn and mpq, 
 
 om _ mp R __ c 
 
 where R is the radius of curvature om. Inserting in this the 
 above value of A, it becomes, 
 
 S__E L 
 
 f ~ R ' 
 
ART. 33. GENERAL EQUATION OF THE ELASTIC CURVE. /I 
 
 But the fundamental formula (4) may be written in the form 
 
 S_M 
 
 7~ = 7* 
 
 and hence, by comparison, 
 
 M EI 
 M=. 
 
 This is the formula which gives the relation between the bend- 
 ing moment of the exterior forces and the radius of curvature 
 at any section. Where M is o the radius R is oo ; where M is 
 a maximum R has its least value. 
 
 Now, in works on the differential calculus, the following 
 value is deduced for the radius of curvature of any plane curve 
 whose abscissa is x, ordinate y, and length /, namely, 
 
 ~ 
 
 dx . d*y 
 Hence the most general equation of the elastic curve is, 
 
 dr = I 
 
 dx . d*y ~ M ' 
 
 which applies to the flexure of all bodies governed by the laws 
 of Arts. 3 and 20. 
 
 In discussing a beam the axis of x is taken as horizontal and 
 that of y as vertical. Experience teaches us that the length of 
 a small part of a bent beam does not materially differ from 
 that of its horizontal projection. Hence dl may be placed 
 equal to dx for all beams, and the above equation reduces to 
 the form, 
 
 (5) ^ = - 
 
 dx* EI' 
 
 This is the general equation of the elastic curve, applicable to 
 all beams whatever be their shapes, loads or number of spans. 
 M is the bending moment of the external forces for any sec- 
 
72 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 tion whose abscissa is x, and whose moment of inertia with 
 respect to the neutral axis is /. Unless otherwise stated /will 
 be regarded as constant, that is, the cross-section of the beam 
 is constant throughout its length. 
 
 To obtain the particular equation of the elastic curve for any 
 special case, it is first necessary to express M as a function of 
 x and then integrate the general equation twice. The ordinate 
 y will then be known for any value of x. It should, however, 
 be borne in mind that formula (5), like formula (4), is only true 
 when the unit-stress 5 is less than the elastic limit of the 
 material. 
 
 Prob. 57. A wooden beam inch wide, f inch deep, and 3 
 feet span carries a load of 14 pounds at the middle. Find the 
 radius of curvature for the middle, quarter points, and ends. 
 
 ART. 34. DEFLECTION OF CANTILEVER BEAMS. 
 Case I. A load at the free end. Take the origin of co-ordi- 
 
 ^ I Jp ... [ t _ nates at the free end, and as 
 
 p 
 
 =7 I I/ in Fig. 25, let m be any point 
 
 ^^^_^_ __$ of the elastic curve whose 
 
 Fig- 25- abscissa is x and ordinate jr. 
 
 For this point the bending moment M is Px and the general 
 formula (5) becomes, 
 
 . 
 
 dx* 
 By integration the first derivative is found to be 
 
 E g-=-**+C. 
 
 dx 2 
 
 d * 
 
 But ~- is the tangent of the angle which the tangent to the 
 dx 
 
 elastic curve at m makes with the axis of x, and as the beam is 
 
ART. 34. DEFLECTION OF CANTILEVER BEAMS. 73 
 
 dy 
 
 fixed at the wall the value of -y- is o when x equals /. Hence 
 C = J/Y 2 , and the first differential equation is, 
 
 'dx~ 2 2 
 
 The second integration now gives, 
 
 But y = o, when x = o. Hence C' = o, and 
 6EIy = P($rx - **), 
 
 which is the equation of the elastic curve for a cantilever of 
 length / with a load P at the free end. If x = I the value of 
 y will be the maximum deflection, which may be represented 
 by A. Then, 
 
 and for any point of the beam the deflection is A y. 
 
 Case II. A uniform load. Let the origin be taken at the 
 free end as before, and x and y 
 be the co-ordinates of any point 
 of the elastic curve. Let the load 
 per linear unit be w. Then for Fi - 
 
 any section M = \wx* and formula (5) becomes, 
 
 wx* 
 
 Integrate this, determine the constant of integration by the 
 
 dy 
 consideration tjiat -r- = o when x = /, and then, 
 
 ~-=wr -wx\ 
 dx 
 
74 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 Integrate again, and after determining the constant, the equa- 
 tion of the elastic curve is, 
 
 which is a biquadratic parabola. For x = /, y = A the maxi 
 mum deflection, whose value is, 
 
 where W is the total uniform load on the cantilever. 
 
 Case III. A load at the free end and also a uniform load. 
 Here it is easy to show that the maximum deflection is 
 
 A _ spr + 3 wr 
 
 2 4 7 
 
 which is the sum of the deflections due to the two loads. 
 Hence it appears that, as in cases of stress, each load produces 
 its effect independently of the other. 
 
 In order that the formulas for deflection may be true, the 
 maximum unit-stress 5 produced by all the loads must not 
 exceed the elastic limit of the material. 
 
 Prob. 58. Compute the deflection of a cast iron cantilever 
 beam, 2X2 inches and 6 feet span, caused by a load of 100 
 pounds at the end. 
 
 Prob. 59. In order to find the coefficient of elasticity of a 
 cast iron bar 2 inches wide, 4 inches deep, and 6 feet long, it 
 was balanced upon a support and a weight of 4 ooo pounds 
 hung at each end, causing a deflection of 0.401 inches. Com- 
 pute the value of E. 
 
 ART. 35. DEFLECTION OF SIMPLE BEAMS. 
 
 The deflection of a simple beam due to a load at the middle, 
 or to a uniform load, is readily obtained from the expressions 
 just deduced for cantilever beams. Thus, for a simple beam 
 of span / with a load P at the middle, if Fig. 27 be inverted it 
 
ART. 35. DEFLECTION OF SIMPLE BEAMS. 75 
 
 will be seen to be equivalent to two cantilever beams of length 
 \l with a load \P at each end. The formula for the maximum 
 deflection of a cantilever beam hence applies to this figure, if 
 
 pr 
 
 I be replaced by / and P by \P, which gives J = ,, - for the 
 
 deflection at the middle of the simple beam. It will be well, 
 however, to use the general formula (5) and treat each case in- 
 dependently. 
 
 Case I. A single load P at the middle. Let the origin be 
 taken at the left support. For 
 any section between the left 
 support and the middle the 
 bending moment M is \Px. Fi e- 
 
 Then the general formula (5) becomes, 
 
 dy 
 Integrate this and find the constant by the fact that -7- = o 
 
 when x = J/. Then integrate again and find the constant by 
 the fact that 7 = when x = o. Thus, 
 
 4&EIy = P(AfX* $l*x), 
 
 is the equation of elastic curve between the left hand support 
 and the load. For the greatest deflection make x = J/, then, 
 
 z>/ 8 
 A = 
 
 Case II. A uniform load. Let w be the load per linear unit, 
 then the formula (5) becomes, 
 
 d^y _ wlx wx* 
 
 hl d? = T ~^~- 
 
 Integrate this twice, find the constants as in the preceding para- 
 graph, and the equation of the elastic curve is, 
 
76 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 from which the maximum deflection is found to be, 
 
 = 384^7 ==: 384^7* 
 
 Case III. A load P at any point. Here it is necessary first 
 to consider that there are two elastic curves, one on each side 
 
 of the load, which have distinct 
 equations, but which have a 
 common tangent and ordinate 
 under the load. As in Fig. 28, 
 
 let the load be placed at a distance kl from the left support, 
 k being a number less than unity. Then the left reaction is 
 R = P(\ k). From the general formula (5), with the origin 
 at the left support, the equations are, 
 On the left of the load, 
 
 (a) EI = 
 
 (c) 
 On the right of the load, 
 
 (a)' EI d - = Rx~P(x- kl\ 
 
 (b) r EI = \Rx* - \Px* + Pklx + C 3 , 
 
 (c)' Ely = %Rx* - \Px* + \Pklx* + C,x + C t . 
 
 To determine the constants consider in (c) that y = o when 
 x o, and hence that 7, = o. Also in (c)', y = o when x l\ 
 again since the curves have a common tangent under the load, 
 (b) = (b) f when x = /, and since they have a common ordinate 
 at that point (c) = (c) f when x kl. Or, 
 
 o = 
 
ART. 36. COMPARATIVE DEFLECTION AND STIFFNESS. // 
 
 From these three equations the values of C lt C it and C^ are 
 found. Then the equation of the elastic curve on the left of 
 the load is, 
 
 6EIy = P(i - k)x* - P(2k - 3/P 
 
 To find the maximum deflection, the value of x which renders 
 y a maximum is to be obtained by equating the first derivative 
 to zero. If k be greater than , this value of x inserted in the 
 above equation gives the maximum deflection ; if k be less than 
 , the maximum deflection is on the other side of the load. 
 For instance, if k = , the equation of the elastic curve on the 
 left of the load is, 
 
 = i6/V - 
 
 This is a maximum when x = 0.567, which is the point of great- 
 est deflection. 
 
 Prob. 60. Prove, when k is greater than \ in Fig. 28, that the 
 
 pr 
 
 maximum deflection is A = =r- r (i k)($k 
 
 Prob. 61. In order to find the coefficient of elasticity of 
 Quercus alba a bar 4 centimeters square and one meter long 
 was supported at the ends and loaded in the middle with weights 
 of 50 and 100 kilograms when the deflections were found to be 
 6.6 and 13.0 millimeters. Show that the mean value- of E was 
 74 500 kilos per square centimeter. 
 
 ART. 36. COMPARATIVE DEFLECTION AND STIFFNESS. 
 
 From the two preceding articles the following values of the 
 maximum deflections may now be written and their compari- 
 son will show the relative stiffness of the different cases. 
 
 i wr 
 
 For a cantilever loaded at the end with W, A = - . . 
 
78 CANTILEVER BEAMS AND SIMPLE BEAMS. Cll. III. 
 
 i wr 
 
 For a cantilever uniformly loaded with W, A = - . -=rjr. 
 
 8 El 
 
 i wr 
 
 For a simple beam loaded at middle with W, A = . ~ . 
 
 48 El 
 
 t J^/7 3 
 
 For a simple beam uniformly loaded with W y A = 1~ . _ . 
 
 384 hi 
 
 The relative deflections of these four cases are hence as the 
 numbers I, f , T V, and 
 
 These equations also show that the deflections vary directly 
 as the load, directly as the cube of the length, and inversely as 
 
 E and /. For a rectangular beam I = , and hence the de- 
 
 flection of a rectangular beam is inversely as its breadth and 
 inversely as the cube of its depth. 
 
 The stiffness of a beam is indicated by the load that it can 
 carry with a given deflection. From the above it is seen that 
 the value of the load is, 
 
 where m has the value 3, 8, 48, or ^-f^ as the case may be. 
 Therefore, the stiffness of a beam varies directly as E, directly 
 as 7, and inversely as the cube of its length, and the relative 
 stiffness of the above four cases is as the numbers i, 2f, 16, 
 and 25f . From this it appears that the laws of stiffness are 
 very different from those of strength. (Art. 29.) 
 
 Prob. 62. Compare the strength and stiffness of a joist 3X8 
 inches when laid with flat side vertical and when laid with nar- 
 row side vertical. 
 
 Prob. 63. Find the thickness of a white pine plank of 8 feet 
 span required not to bend more than ^f^th of its length under 
 a head of water of 20 feet. 
 
ART. 37. RELATION BETWEEN DEFLECTION AND STRESS. 79 
 
 ART. 37. RELATION BETWEEN DEFLECTION AND STRESS. 
 
 Let the four cases discussed in Arts. 29 and 36 be again con- 
 sidered. For the strength, 
 
 W = n-, where n = I, 2, 4, or 8. 
 
 For the stiffness, 
 
 EIA 
 
 W=m , where m = 3, 8, 48, or 76^. 
 
 By equating these values of W the relation between A and 5 
 is obtained, thus, 
 
 5 = 
 
 mEcA 
 
 or 
 
 A = 
 
 nl*S 
 
 mcE ' 
 
 These equations, like the general formula (4) and (5), are only 
 valid when 5 is less than the elastic limit of the materials. 
 
 This also shows that the maximum deflection A varies as 
 
 - for beams of the same material under the same unit-stress S. 
 c 
 
 From the preceding articles the following table may also be 
 compiled which exhibits the most important results relating to 
 both absolute and relative strength and stiffness. 
 
 Case. 
 
 Max. 
 Vertical 
 Shear. 
 
 Max. 
 Bending 
 Moment. 
 
 Max. 
 Stress 
 S. 
 
 Max. 
 Deflec- 
 tion. 
 
 Relative 
 Strength. 
 
 Relative 
 Stiffness. 
 
 
 
 
 Wlc 
 
 I Wl* 
 
 
 
 Cantilever loaded at end, 
 
 W 
 
 Wl 
 
 
 
 I 
 
 I 
 
 
 
 
 I 
 
 3 El 
 
 
 
 Cantilever loaded uniformly, 
 
 W 
 
 km 
 
 Wlc 
 
 7.1 
 
 I Wl* 
 
 2 
 
 n 
 
 Simple beam loaded at middle, 
 
 itr 
 
 1 M/7 
 
 Wlc 
 
 I Wl* 
 
 4 
 
 16 
 
 Simple beam loaded uniformly, 
 
 ** 
 
 1 M/7 
 
 Wlc 
 
 5 Wl* 
 
 8 
 
 25* 
 
80 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 Here the signs of the maximum shears and moments are 
 omitted as only their absolute values are needed in computa- 
 tions. Evidently the moments are negative for the first and 
 second cases, and positive for the third and fourth, the direc- 
 tion of the curvature being different. 
 
 Prob. 64. Find the deflection of a medium-steel heavy 10- 
 inch I beam of 9 feet span when stressed by a uniform load 
 up to 30000 pounds per square inch. 
 
 Prob. 65. A wooden beam of breadth &, depth d, and span x 
 is loaded with P at the middle. Find the value of x so that 
 rupture may occur under the load. Find also the value of x 
 so that rupture may occur by shearing at the supports. 
 
 ART. 38. CANTILEVER BEAMS OF UNIFORM STRENGTH. 
 
 All cases thus far discussed have been of constant cross- 
 section throughout their entire length. But in the general 
 formula (4) the unit-stress 5 is proportional to the bending mo- 
 ment M y and hence varies throughout the beam in the same 
 way as the moments vary. Hence some parts of the beam are 
 but slightly strained in comparison with the dangerous section. 
 
 A beam of uniform strength is one so shaped that the unit- 
 stress S is the same in all fibers at the upper and lower surfaces. 
 Hence to ascertain the form of such a beam the unit-stress S 
 
 in (4) must be taken as constant and be made to vary with 
 
 M. The discussion will be given only for the most important 
 practical cases, namely, those where the sections are rectangular. 
 
 / 1 72 
 
 For these equals ?- , and formula (4) becomes, 
 
 ' 
 
 In this bd* must vary with M for forms of uniform strength 
 
ART. 38. CANTILEVER BEAMS OF UNIFORM STRENGTH. 8 1 
 
 Elevation 
 
 For a cantilever beam with a load P at the end, M = Px 
 and the equation becomes \Sbd* = Px, in which P and 5 are 
 constant. If the breadth be taken as con- 
 stant, d* varies with x and the profile is 
 that of a parabola whose vertex is at the 
 load, as shown in Fig. 29. The equation of 
 
 the parabola is d* = x from which d may 
 
 be found for given values of x. The walk- Fig> 29> 
 
 ing beam of an engine is often made approximately of this 
 shape. If the depth of the cantilever beam 
 be constant then b varies directly as x and 
 hence the plan should be a triangle as shown 
 in Fig. 30. The value of b for given values 
 of P, 5, d, and x may be found from the 
 6Px 
 
 expression b = 
 
 Sd'' 
 
 Fig. 30. 
 
 For a cantilever beam uniformly loaded with w per linear 
 unit M = %wx*, and the equation becomes \Sbd* = \wx*, 
 in which w and .S are known. If the 
 breadth be taken as constant then d varies 
 as x and the elevation is a triangle, as 
 in Fig. 31, whose depth at any point is 
 
 If however the depth be 
 
 Fig. 
 
 ifi 
 
 taken constant, then b = -| x* which is the equation of a 
 
 parabola whose vertex is at the free end of the cantilever and 
 whose axis is perpendicular to it. .Or the equation may be 
 satisfied by two parabolas drawn upon opposite sides of the 
 center line as shown in Fig. 32. 
 
 The vertical shear modifies in practice the shape of these 
 forms near their ends. For instance, a cantilever beam loaded 
 
82 
 
 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 at the end with P requires a cross-section at the end equal to 
 
 p 
 
 -^r where S c is the working shearing strength. This cross- 
 
 ^c 
 
 section must be preserved until a value of 
 x is reached, where the same value of the 
 cross-section is found from the moment. 
 
 The deflection of a cantilever beam of 
 uniform strength is evidently greater than 
 that of one of constant cross-section, since 
 the unit-stress 5 is greater throughout. In 
 any case it may be determined from the general formula 
 (5) by substituting for -M and / their values in terms of x, in- 
 tegrating twice, determining the constants, and then making 
 x equal to / for the maximum value of y. 
 
 For a cantilever beam loaded at the end and of constant 
 breadth, as in Fig. 29, formula (5) becomes, 
 
 d*y \2Px _2 
 dx*~ Ebd* ~ E 
 
 Integrating this twice and determining the constants, as in 
 Art. 34, the equation of the elastic curve is found to be, 
 
 In this make x = /, and substitute for 5 its value -- , where 
 
 bd* 
 
 d^ is the depth at the wall. Then, 
 
 spr 
 
 ~^bd~^ 
 which is double that of a cantilever beam of uniform depth d. 
 
 For a cantilever beam loaded at the end and of constant 
 depth, formula (5) becomes, 
 
 d*y _ \2Px _ 2S 
 ~-~Ebd*~~~~Ed* 
 
ART. 39. SIMPLE BEAMS OF UNIFORM STRENGTH. 83 
 
 By integrating this twice and determining the constants as 
 before, the equation of the elastic curve is found, from which 
 the deflection is, 
 
 which is fifty per cent greater than for one of uniform section. 
 
 Prob. 66. 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 hori- 
 zontal unit-stress, and 4000 pounds per square inch for the 
 shearing unit-stress. 
 
 ART. 39. SIMPLE BEAMS OF UNIFORM STRENGTH. 
 
 In the same manner it is easy to deduce the forms of uni- 
 form strength for simple beams of rectangular cross-section. 
 
 For a load at the middle and breadth constant, M = \Px> 
 
 <>p 
 and hence, Sbd* = \Px. Hence d* ^ x, from which values 
 
 00 
 
 of d may be found for assumed values of x. Here the profile 
 o.f the beam will be parabolic, the vertex being at the support, 
 and the maximum depth under the load. 
 
 For a load at the middle and depth constant, M \Px as 
 
 \P 
 
 before. Hence b = -FT^ an d the plan must be triangular or 
 
 Ow 
 
 lozenge shaped, the width uniformly increasing from the sup- 
 port to the load. 
 
 For a uniform load and constant breadth, M = %wlx 
 and hence, ^/ 8 = ^- (Ix x*\ and the profile of the beam must 
 be elliptical, of preferably a half-ellipse. 
 
84 CANTILEVER BEAMS AND SIMPLE BEAMS. CH. III. 
 
 *J*70) 
 
 For a uniform load and constant depth, b = -^ (Ix x*) and 
 
 ow 
 
 hence the plan should be formed of two parabolas having their 
 vertices at the middle of the span. 
 
 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 if 
 the breadth is constant, and is one and one-half times as much 
 if the depth is constant. 
 
 Prob. 67. Draw the profile for a cast iron beam of uniform 
 strength, the span being 8 feet, breadth 3 inches, load at the 
 middle 30,000 pounds, using the same working unit-stresses as 
 in Prob. 66. 
 
 Prob. 68. Find the deflection of a steel spring of constant;, 
 depth and uniform strength which is 6 inches wide at the mid- 
 dle, 52 inches long, and loaded at the middle with 600 pounds, 
 the depth being such that the maximum fiber stress is 20 OOQ 
 pounds per square inch. 
 
ART. 40. BEAMS OVERHANGING ONE SUPPORT. 
 
 __________ ? ________ x-- m 
 
 CHAPTER IV. 
 RESTRAINED BEAMS AND CONTINUOUS BEAMS. 
 
 ART. 40. BEAMS OVERHANGING ONE SUPPORT. 
 
 A cantilever beam has its upper fibers in tension and the 
 lower in compression, while a simple beam has its upper fibers 
 in compression and the lower in tension. Evidently a beam 
 overhanging one support, 
 as in Fig. 33, has its over- / 
 hanging part in the con- & 
 dition of a cantilever, 
 and the part near the 
 other end in the condi- 
 tion of a simple beam. 
 Hence there must be a 
 point i where the stresses 
 change from tension to 
 compression, and where 
 the curvature changes Fg. 33- 
 
 from positive to negative. This point i is called the inflection 
 point ; it is the point where the bending moment is zero. An 
 overhanging beam is said to be subject to a restraint 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 
 found by using the principle of moments as in Art. 16. Thus, 
 if the distance between the supports be /, the length of the 
 
86 RESTRAINED AND CONTINUOUS BEAMS. CH. IV. 
 
 overhanging part be m y and the uniform load per linear foot be 
 w, the two reactions are, 
 
 wl wm* n wl . u'm* 
 
 *> = T- ^' R, = - + wm^--, 
 
 whose sum is equal to the total load wl -\- wm. Here, as in 
 all cases of uniform load, the lever arms are taken to the cen- 
 ters of gravity of the portions considered. 
 
 When the reactions have been found, the vertical shear at 
 any section can be computed by Art. 17, and the bending mo- 
 ment by Art. 1 8, bearing in mind that for a section beyond the 
 right support the reaction R^ must be considered as a force 
 acting upward. Thus, for any section distant x from the left 
 support, 
 
 When x is less than /, When x is greater than /, 
 
 V= R, - wx, JS=R 1 + R,- wx, 
 
 The curves corresponding to these equations are shown on 
 Fig- 33- The shear curve consists of two straight lines ; F R, 
 
 r> 
 
 when x = o, and V = o when x = - ; at the right support 
 
 V =R l wl from the first equation, and V= R l + R t wl 
 from the second ; V = o when x = l-\- m. The moment curve 
 consists of two parts of parabolas ; M = o when x =. o, M is a 
 
 r> 
 
 maximum when x = l -, M = o at the inflection point where 
 
 w 
 
 x = - l , M has its negative maximum when x = /, and M = o 
 
 w 
 
 when x = /+ m. The diagrams show clearly the distribution 
 of shears and moments throughout the beam. 
 
 For example, if I = 20 feet, m = 10 feet, and w = 40 pounds 
 per linear foot, the reactions are R t = 300 and R, = 900 pounds. 
 
ART. 40. BEAMS OVERHANGING ONE SUPPORT. 8/ 
 
 Then the point of zero shear or maximum moment is at 
 x 7.5 feet, the inflection point at x = 15 feet, the maximum 
 shears are -(- 300, 500, and -f- 400 pounds, and the maximum 
 bending moments are -\- I I2 5 and 2 ooo pound-feet. Here 
 the negative bending moment at the right support is numeri- 
 cally greater than the maximum positive moment. The rela- 
 tive values of the two maximum moments depend on the ratio 
 of m to /; if m = o there is no overhanging part and the beam 
 is a simple one ; if m = \l the case is that just discussed ; if 
 m = I the reaction R^ is zero, and each part is a cantilever beam. 
 
 After having thus found the maximum values of Fand M 
 the beam may be investigated by the application of formulas 
 (3) and (4) in the same manner as a cantilever or simple beam. 
 By the use of formula (5) the equation of the elastic curve be- 
 tween the two supports is found to be, 
 
 From this the maximum deflection for any particular case may 
 be determined by putting -~- equal to zero, solving for x, and 
 then finding the corresponding 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. 
 
 Prob. 69. 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 diagrams of shears and moments. 
 
 Prob. 70. 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 180 pounds, and at 6 feet from the right end 
 is a load of 125 pounds. Find the reactions, the inflection 
 point, and draw the shear and moment diagrams. 
 
88 RESTRAINED AND CONTINUOUS BEAMS. CH. IV. 
 
 ART. 41. BEAMS FIXED AT ONE END 'AND SUPPORTED AT 
 
 THE OTHER. 
 
 A beam is said to be fixed at the end when it is so restrained 
 in a wall that the tangent to the elastic curve at the wall is 
 horizontal. Thus, in Fig. 33, if the part m is of such a length 
 
 that the tangent over the right sup- 
 
 --*-- >! i i i port is horizontal, the part / is in the 
 
 same condition as a beam fixed at 
 one end and supported at the other. 
 Fig. 34 shows the practical arrange- 
 ment of such a beam, the left sup- 
 Flg * 34 ' port being upon the same level as 
 
 the lower side of the beam at the wall. The reactions of such 
 a beam cannot be determined by the principles of statics alone, 
 but the assistance of the equation of the elastic curve must be 
 invoked. 
 
 Case I. For a uniform load over the whole beam, as in Fig. 
 34, let R be the reaction at the left end. Then for any section 
 the bending moment is Rx \wx*. Hence the differential 
 equation of the elastic curve is, 
 
 Integrate this once and determine the constant from the neces- 
 
 dv 
 sary condition that ~ = o when x = I. Integrate again and 
 
 find the constant from the fact that y = o when x = o. Then, 
 
 Here also y = o when x = /, and therefore R = 
 
 The moment at any point now is M = \wlx %wx*, and by 
 placing this equal to zero it is seen that the point of inflection 
 is at x = |/. By the method of Art. 24 it is found that the 
 
ART. 41. BEAMS FIXED AT ONE END. 89 
 
 maximum moments are+yl-g-w/ 8 and \wl*, and that the 
 distribution of moments is as represented in Fig. 34. 
 
 dv 
 The point of maximum deflection is found by placing -*- 
 
 equal to zero and solving for x. Thus 8* 8 glx* + / 3 = o, 
 one root of which is x = +0.42157, and this inserted in the 
 value of y gives, 
 
 J = 0.0054 j, 
 
 for the value of the maximum deflection. 
 
 Case II. For a load at the middle it is first necessary to con- 
 sider that there are two elastic curves having a common ordi- 
 nate and a common tangent under the load, since the expres- 
 sions 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, 
 
 (.) 
 
 (*) / 
 
 On the right of the load the similar equations are, 
 (a)' EI^^Rx-P^-V), 
 
 (6)' E/& 
 
 (c)' 
 
 To determine the constants consider in (*:) that y = o when 
 x = o and hence that C, = o. In (&)' the tangent = o when 
 x = / and hence C^= \Rr. Since the curves have a common 
 
90 RESTRAINED AND CONTINUOUS BEAMS. CH. IV. 
 
 tangent under the load (b) = (&)' for x =%!, and thus the value 
 of C v is found. Since the curves have a common ordinate 
 under the load (c) = (c)' when x = /, and thus C t is found. 
 
 Then, 
 
 _ RJ_ Prx Rl*x 
 (c) Ely- g-+ g 2 , 
 
 Rx* Px* Plx* Rl*x , Pr 
 (c)' Ely-- - + ~+' 
 
 From the second of these the value of the reaction is R = 
 
 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 
 maximum negative moment occurs at the wall, and its value .is 
 -f^Pl. The inflection point is at x = -f^l. The deflection un- 
 der the load is readily found from (c) by making x /. The 
 maximum deflection occurs at a less value of x, which may be 
 found by equating the first derivative to zero. 
 
 Case III. For a load at any point whose distance from the 
 left support is kl, the following results may be deduced by a 
 method exactly similar to that of the last case. 
 
 Reaction at supported end = %P(2 $k -f- k?). 
 
 Reaction at fixed end = \P(^k k*\ 
 
 Maximum positive moment = \Plk(2 ^k-^-k*}. 
 
 Maximum negative moment = %Pl(k 3 ). 
 
 The absolute maximum deflection occurs under the load when 
 
 Prob. 71. Draw the diagrams of shears and moments for 
 a load at the middle, taking P =. 600 pounds and /= 12 feet. 
 
 Prob. 72. Find the position of load Pwhich gives the maxi- 
 mum positive moment. Find also the position which gives the 
 maximum negative moment. 
 
ART. 42. BEAMS OVERHANGING BOTH SUPPORTS. QI 
 
 ART. 42. BEAMS OVERHANGING BOTH SUPPORTS. 
 
 When a beam overhangs both supports the bending moments 
 for sections beyond the supports are negative, and in general 
 between the supports there will be two inflection points. If 
 the lengths m and n be 
 equal the reactions will be ^ m -^~ 
 
 equal under uniform load, 
 
 each being one half of the 
 
 total load. In any case, 
 
 whatever be the nature of 
 
 the load, the reactions may 
 
 be found by the principle of moments (Art. 16), and then the 
 
 vertical shears and bending moments may be deduced for all 
 
 sections, after which the formulas (3) and (4) can be used for 
 
 any special problem. 
 
 Under a uniformly distributed load, and m = n, which is the 
 most important practical case, each reaction is wm -\- %wl, the 
 maximum shears at the supports are wm and \wl, the maxi- 
 mum moment at the middle is + w(^r %m*), the maximum 
 moment at each support is $wm*, and the inflection points 
 
 are distant J V T 4m 3 from the middle of the beam. Fig. 35 
 shows the distribution of moments for this case. If m = o, the 
 beam is a simple one ; if / = o, it consists of two cantilever 
 beams. 
 
 Prob. 73. If m = n in Fig. 35, find the ratio of / to m in 
 order that the maximum positive moment may numerically 
 equal the maximum negative moment. 
 
 Prob. 74. 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 shear and moment diagrams. 
 
92 RESTRAINED AND CONTINUOUS BEAMS. CH. IV. 
 
 ART. 43. BEAMS FIXED AT BOTH ENDS. 
 
 If, in Fig. 35, the distances m and n 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 lengths m and n 
 which will cause the curve to be horizontal at the support can 
 be determined by the help of the elastic curve. For uniform 
 load n = m and the bending moment at any section in the 
 span / distant x from the left support is, 
 
 M = (wm + %wl)x %w(m -f- 
 which reduces to the simpler form, 
 M = M f 
 
 in which M' represents the unknown bending moment \wn? 
 at the left support. 
 
 Again, for a single load P at the middle of /in Fig. 35 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, 
 
 which reduces to, 
 
 in which M ' denotes the unknown moment Qm at the left 
 support. The problem of finding the bending moment at any 
 section hence reduces to that of determining M' the moment 
 at the support. 
 
 Case I. For a uniform load the general equation of the elastic 
 curve now is, 
 
ART. 43. BEAMS FIXED AT BOTH ENDS. 93 
 
 Integrating this twice, making -jj- = o when x = o and also 
 
 when x = /, the value of M' 
 
 \ 
 
 ^ --- ^i 
 
 is found to be - , and 
 
 the equation of the elastic ^tll^ 
 
 curve becomes, m? 
 
 2EIy = w(-l*x* + 2lx* - S). Fi *- 36- 
 
 From this the maximum deflection is found to be, 
 
 ~ 
 
 The inflection points are located by making M = o, which 
 gives x 4/ /A / The maximum positive moment is at 
 
 wr 
 
 the middle and its value is - - ; accordingly the horizontal 
 
 24 
 
 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 %w/, at 
 the middle o, and at the right end 
 
 Case II. For a load at the middle the general equation of 
 the elastic curve between the left end and the load is, 
 
 and in a similar manner to that 
 
 of the last case it is easy to find 
 
 that the maximum negative mo- Fi &- 37. 
 
 ments are -J/V, that the maximum positive moment is -J/Y, that 
 
 the inflection points are half-way between the supports and the 
 
 pr 
 
 load, and that the maximum deflection is - -=-=: - 
 
 192^7 
 
 : Case III. For load P at a distance kl from the left end let 
 
94 RESTRAINED AND CONTINUOUS BEAMS. CH. IV. 
 
 M' and V denote the unknown bending moment and vertical 
 shear at that end. Then on the left of the load, 
 
 M=M'+ V'x, 
 and on the right of the load 
 
 M= M' + V'x P(x - kl). 
 
 By inserting these in the general formula (5), integrating each 
 twice and establishing sufficient conditions to determine the 
 unknown M' and V and also the constants of integration, the 
 following results may be deduced, 
 
 Shear at left end = P(\ $tf + 2>P), 
 
 Shear at right end = Pk\^ 2k). 
 
 Moment at left end = Plk(\ 2k + JP), 
 
 Moment at right end = Plk\\ k\ 
 
 Moment under load = + Plk\2 4& -\- 2&). 
 
 If k =. \ the load is at the middle and these results reduce to 
 the values found in Case II. 
 
 Prob. 75. Show from the .results above given for Case III 
 
 kl 
 
 that the inflection points are at the distances - - and 
 
 i + 2k 
 
 th left end 
 
 Prob. 76. What medium-steel I beam is required for a span 
 of 24 feet to support a uniform load of 25 ooo pounds, the ends 
 being merely supported? What one is needed when the ends 
 are fixed? 
 
 ART. 44. COMPARISON OF RESTRAINED AND SIMPLE BEAMS. 
 
 As the maximum moments for restrained beams are less 
 than for simple beams their strength is relatively greater. This 
 was to be expected, since the restraint produces a negative 
 bending moment and lessens the deflection which would other- 
 
ART. 44. COMPARISON OF RESTRAINED AND SIMPLE BEAMS. 95 
 
 wise occur. The comparative strength and stiffness of canti- 
 levers and simple beams is given in Art. 37. To these may 
 now be added four cases from Arts. 41 and 43, and the follow- 
 ing table be formed, in which W represents the total load, 
 whether uniform or concentrated. 
 
 Beams of Uniform Cross-section. 
 
 Maximum 
 Moment. 
 
 Maximum 
 Deflection. 
 
 Relative 
 Strength. 
 
 Relative 
 Stiffness. 
 
 
 
 I Wl* 
 
 
 
 Cantilever, load at end 
 
 Wl 
 
 
 I 
 
 I 
 
 
 
 3 ~I 
 
 
 
 
 
 I Wl* 
 
 
 
 Cantilever, uniform load 
 
 l yyi 
 
 
 2 
 
 2| 
 
 
 
 8 ~f 
 
 
 8 
 
 Simple beam, load at middle 
 
 i f^/y 
 
 48 ~EI 
 
 4 
 
 16 
 
 Simple beam uniformly loaded 
 
 \Wl 
 
 384 El 
 
 8 
 
 25f 
 
 Beam fixed at one end, supported at 
 
 
 Wl* 
 
 
 
 other, load near middle 
 
 0.192^7 
 
 o.o, 8a _ 
 
 5-2 
 
 18.3 
 
 Beam fixed at one end, supported at 
 other, uniform load 
 
 \Wl 
 
 W7 
 0.0054-^ 
 
 8 
 
 62 
 
 Beam fixed at both ends, load at 
 
 
 i Wl* 
 
 
 
 middle 
 
 i Wl 
 
 
 8 
 
 64 
 
 
 
 192 El 
 
 
 
 Beam fixed at both ends, uniform 
 
 
 i Wl* 
 
 
 
 load 
 
 T^f W/ 
 
 
 12 
 
 128 
 
 
 
 p4^7 
 
 
 
 This table shows that a beam fixed at both ends and uni- 
 formly loaded is one and one-half times as strong and five times 
 as stiff as a simple beam under the same load. The advantage 
 of fixing the ends is hence very great. 
 
 Prob. 77. Prove, for a uniformly loaded beam with equal 
 overhanging ends, that the deflection at the middle is given by 
 
 the formula ' ($/' 24^'). 
 
 354/i/ 
 
 'Prob. 78. Find the deflection of a g-inch 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 
 1 6 800 pounds per square inch. 
 
96 RESTRAINED AND CONTINUOUS BEAMS. CH. IV. 
 
 ART. 45. GENERAL PRINCIPLES OF CONTINUITY. 
 
 A continuous beam is one supported upon several points in 
 the same horizontal plane. A simple beam may be regarded 
 as a particular case of a continuous beam where the number of 
 supports is two. The ends of a continuous beam are said to 
 be free when they overhang, supported when they merely rest 
 on abutments, and restrained when they are horizontally fixed 
 in walls. 
 
 The general principles of the preceding chapter hold good 
 for all kinds of beams. If a plane be imagined to cut any 
 beam at any point the laws of Arts. 19 and 20 apply to the 
 stresses in that section. The resisting shear and the resisting 
 moment for that section have the values deduced in Art. 21 
 and the two fundamental formulas for investigation are, 
 
 (3) S,A = V, 
 
 (4) ^=M 
 
 Here 5 S is the vertical shearing unit-stress in the section, and 
 S is the horizontal tensile or compressive unit-stress on the 
 fiber most remote from the neutral axis ; c is the shortest dis- 
 tance from that fiber to that axis ; / the moment of inertia, 
 and A the area of the cross-section. V is the vertical shear of 
 the external forces on the left of the section, and M is the 
 bending moment of those forces with reference to a point in 
 the section. For any given beam evidently 5 S and S may be 
 found for any section as soon as V and M are known. 
 
 The general equation of the elastic line, deduced in Art. 33, 
 is also valid for all kinds of beams. It is, 
 
 (5) 
 UJ 
 
ART. 45. GENERAL PRINCIPLES OF CONTINUITY. 97 
 
 where x is the abscissa and y the ordinate of any point of the 
 elastic curve, M being the bending moment for that section, 
 and E the coefficient of elasticity of the material. 
 
 The vertical shear V is the algebraic sum of the external 
 forces on the left of the section, or, as in Art. 17, 
 
 V = Reactions on left of section minus loads on left of section. 
 
 For simple beams and cantilevers the determination of V 
 for any special case was easy, as the left reaction could be 
 readily found for any given loads. For continuous beams, 
 however, it is not, in general, easy to find the reactions, and 
 hence a different method of determining Fis necessary. Let 
 Fig. 38 represent one span of a continuous beam. Let V be 
 the vertical shear for any L j 
 
 * h^- .* -X- __ ^_ ^4 
 
 section at the distance x __ _J p i 
 
 from the left support, and ) \ 
 
 V the vertical shear at a ^r' 
 
 section infinitely near to Fig. 38. 
 
 the left support. Also let 2P t denote the sum of all the con- 
 
 centrated loads on the distance x, and wx the uniform load. 
 
 Then because V is the algebraic sum of all the vertical forces 
 
 on its left, the definition of vertical shear gives, 
 
 (6) V= V -wx-^P^ 
 
 Hence V can be determined as soon as V is known. 
 
 The bending moment M is the algebraic sum of the mo- 
 ments of the external forces on the left of the section with ref- 
 erence to a point in that section, or, as in Art. 18, 
 
 M -=. moments of reactions minus moments of loads. 
 
 For the reason just mentioned it is in general necessary to de- 
 termine M for continuous and restrained beams by a different 
 method. Let M' denote the bending moment at the left sup- 
 port of any span as in Fig. 38, and M" that at the right sup- 
 
98 RESTRAINED AND CONTINUOUS BEAMS. CH. IV. 
 
 port, while M is the bending moment for any section distant x 
 from the left support. Let P l be any concentrated load upon 
 the space x at a distance / from the left support, k being a 
 fraction less than unity, and let w be the uniform load per lin- 
 ear unit. Let V be the resultant of all the vertical forces on 
 the left of a section in the given span infinitely near to the 
 left support, and let m be the distance of the point of applica- 
 tion of that resultant from that support. Then the definition 
 of bending moment gives, 
 
 But Vm is the unknown bending moment M' at the left 
 support. Hence 
 
 (7) M=M f +V'x-%wx* 2*P l (x-kl), 
 
 from which M may be found for any section as soon as M' and 
 V have been determined. 
 
 The vertical shear V at the support may be easily found if 
 the bending moments M' and M" be known. Thus in equa- 
 tion (7) make x = /, then M becomes M", and hence, 
 
 The whole problem of the discussion of restrained and con- 
 tinuous beams hence consists in the determination of the bend- 
 ing moments at the supports. When these are known the 
 values of M and V may be determined for every section, and 
 the general formulas (3), (4), and (5) be applied as in Chapter 
 III, to the investigation of questions of strength and deflec- 
 tion. The formulas (6), (7), and (8) apply to cantilever and 
 simple beams also. For a simple beam M' M" = o, and 
 V = R. For a cantilever beam M' = o for the free end, and 
 M" is the moment at the wall. 
 
 The relation between the bending moment and the vertical 
 
ART. 46. PROPERTIES OF CONTINUOUS BEAMS. 99 
 
 shear at any section is interesting and important. At the sec- 
 tion x the moment is M and the shear is V. At the next con- 
 secutive section x -)- dx the moment is M -\- dM, which may 
 also be expressed by M-\- Vdx. Hence, 
 
 dM 
 
 ~ dx' 
 
 This may be proved otherwise by differentiating (7) and com- 
 paring with (6). From this it is seen that the maximum mo- 
 ments occur at the sections where the shear passes through 
 zero. 
 
 Prob. 79. A bar of length 2/ and weighing w per linear 
 unit is supported at the middle. Apply formulas (6) and (7) 
 to the statement of general expressions for the moment and 
 shear at any section on the left of the support, and also at any 
 section on the right of the support. 
 
 ART. 46. PROPERTIES OF CONTINUOUS BEAMS. 
 
 The theory of continuous beams presented in the following 
 pages includes only those with constant cross-section having 
 the supports on the same level, as only such are used in engin- 
 eering constructions. Unless otherwise stated, the ends will 
 be supposed to simply rest upon their supports, so that there 
 can be no moments at those points. Then the end spans are 
 somewhat in the _ h _ la _ h _ i* 
 condition of A ? A 3 4 A 
 
 a beam with one 
 
 overhanging end, Ulflflllft^ 
 and the other 
 
 spans somewhat Fig. 39- 
 
 in the condition of a beam with two overhanging ends. At 
 each intermediate support there is a negative moment, and the 
 distribution of moments throughout the beam will be as repre- 
 sented in Fig. 39. 
 
IOO RESTRAINED AND CONTINUOUS BEAMS. CH. IV. 
 
 As shown in Art. 45, the investigation of a continuous beam 
 depends upon the determination of the bending moments at 
 the supports. In the case of Fig. 39 these moments being 
 those at the supports 2, 3, and 4, may be designated M t , M z , 
 and M 4 . Let V^ V^, V 9 , and V^ denote the vertical shear at the 
 right of each support. The first step is to find the moments 
 MI , M 3 , and M 4 . Then from formula (8) the values of V l , F 2 , 
 F 3 , and F< are found, and thus by formula (7) an expression for 
 the bending moment in each span may be written, from which 
 the maximum positive moments may be determined. Lastly, 
 by formulas (3) and (4) the strength of the beam may be in- 
 vestigated, and by (5) its deflection at any point be deduced. 
 
 For example, let the beam in Fig. 39 be regarded as of four 
 equal spans and uniformly loaded with w pounds per linear 
 unit. By a method to be explained in the following articles it 
 may be shown that the bending moments at the supports are, 
 
 From formula (8) the vertical shears at the right of the several 
 supports are, 
 
 etc. 
 
 And from (6) those on the left of the supports 2, 3, 4, etc., are 
 found to be, %%wl, \\wl, ^wl, etc. From formula (7) 
 the general expressions for the bending moments now are, 
 
 For first span, M= 
 
 For second span, M= 
 
 For third span, M= 
 
 ' For fourth span, M = 
 
 From each of these equations the inflection points may be 
 found by putting M=o, and the point of maximum positive 
 
 moment by putting -- = 0. The maximum positive mo- 
 
ART. 46. PROPERTIES OF CONTINUOUS BEAMS. IOI 
 
 ments are found to have the following values, 
 
 T V 2 *W. Ttfcr*"' 1 . iHiw' 1 . and T^W/'. 
 
 For any particular case the beam may now be investigated by 
 formulas (3) and (4). 
 
 The reactions at the supports are not usually needed in the 
 discussion of continuous beams, but if required they may 
 easily be found from the adjacent shears. Thus for the above 
 case, 
 
 = ftwl, etc., 
 and the sum of these is equal to the total load 
 
 The equation of the elastic curve in any span is deduced by 
 inserting in (5) for M its value and integrating twice. When 
 
 dv 
 x = o, the tangent -~ is the tangent of the inclination at the 
 
 left support, and when x = I it is the tangent of the inclination 
 at the right support. When x = o, and also when x = /, the 
 ordinate y = o, and from these conditions the two unknown 
 tangents may be found. In general the maximum deflection 
 in any span of a continuous beam will be found intermediate 
 in value between those of a simple beam and a restrained 
 beam. 
 
 In the following pages continuous beams will only be inves- 
 tigated for the case of uniform load. The lengths of the spans 
 however may be equal or unequal, and the load per linear foot 
 may vary in the different spans. 
 
 Prob. 80. In a continuous beam of three equal spans the 
 negative bending moments at the supports are -fawl*. Find 
 the inflection points, the maximum positive moments and the 
 reactions of the supports. 
 
IO2 RESTRAINED AND CONTINUOUS BEAMS. CH. IV. 
 
 ART. 47. THE THEOREM OF THREE MOMENTS. 
 
 Let the figure represent any two adjacent spans of a continu- 
 ous beam whose lengths are I' and I" and whose uniform loads 
 per linear foot are w' and w" respectively. Let M' , M", and 
 
 M'" represent the 
 three unknown mo- 
 
 af tVi^ cnrrrrfc 
 clL LJnC SUL)L)Ol Lz> 
 
 &* v > ^jv" Let V and V" be the 
 
 vertical shears at the 
 
 Fig - 4 " right of the first and 
 
 second supports. Then, for any section distant x from the 
 left support in the first span, the moment is, 
 
 M=M' + V'x-\wx\ 
 
 If this be inserted in the general formula (5) and integrated 
 twice and the constants determined by the condition that 
 y = o when x o and also when x /, the value of the tan- 
 gent of the angle which the tangent to the elastic curve at any 
 section in the first span makes with the horizontal is found 
 to be, 
 
 dy _ \2M\2x - 
 dx 
 
 Similarly if the origin be taken at the next support the value 
 of the tangent of inclination at any point in the second span is> 
 
 dy_ _ 
 
 dx 24^7 
 
 Evidently the two curves must have a common tangent at the 
 support. Hence make x = I' in the first of these and x = o 
 in the second and equate the results, giving, 
 
 "-*e//"= - \2M"l" - V"l' n w"l"\ 
 
ART. 48. CONTINUOUS BEAMS WITH EQUAL SPANS. 103 
 
 Let the values of V and V" be expressed by (8) in terms of 
 M f , M" , and M"' , and the equation reduces to, 
 
 7f//" 7J/'I"* 
 
 (g) M'l' + 2M"(r + t") + M'"i" = - -, 
 
 4 4 
 
 which is the theorem of three moments for continuous beams 
 uniformly loaded. 
 
 If the spans are all equal and the load uniform throughout, 
 this reduces to the simpler form, 
 
 M' + ^M" + M"' = . 
 
 In any continuous beam of s spans there are s -f- I supports 
 and s I unknown bending moments at the supports. For 
 each of these supports an equation of the form of (g) may be 
 written containing three unknown moments. Thus there will 
 be stated s I equations whose solution will furnish the values 
 of the s I unknown quantities. 
 
 Prob. 8 1. A simple wooden beam one inch square and 15 
 inches long is uniformly loaded with 100 pounds. Find the 
 angle of inclination of the elastic curve at the supports. 
 
 ART. 48. CONTINUOUS BEAMS WITH EQUAL SPANS. 
 
 Consider a continuous beam of five equal spans uniformly 
 loaded. Let the supports beginning on the left be numbered 
 I, 2, 3, 4, 5, and 6. From the theorem of three moments an 
 equation may be written for each of the supports 2, 3, 4, and 
 5 ; thus, 
 
104 RESTRAINED AND CONTINUOUS BEAMS. ClI. IV. 
 
 Since the ends of the beam rest on abutments without restraint 
 M l = M 6 o. Hence the four equations furnish the means 
 of finding the four moments M^, M 3 , M t , M 6 . The solution 
 may be abridged by the fact that M.. = M 6 , and M 3 = M^ 
 which is evident from the symmetry of the beam. Hence, 
 
 From formula (8) the shears at the right of the supports are, 
 
 etc. 
 
 From (7) the bending moment at any point in any span may 
 now be found as in Art. 46, and by (3), (4), and (5) the complete 
 investigation of any special case may be effected. 
 
 In this way the bending moments at the supports for any 
 number of equal spans can be deduced. The following tri- 
 angular table shows their values for spans as high as seven in 
 number. In each horizontal line the supports are represented 
 by squares in which are placed the coefficients of wl*. For 
 example, in a beam of 3 spans there are four supports and the 
 bending moments at those supports are o, -fowl*, 
 and o. 
 
 Equal Spans. 
 
 Moments. 
 
 Fig. 41. 
 
 The vertical shears at the supports are also shown in the 
 following table for any number of spans up to 5. The space 
 representing a support shows in its left-hand division the shear 
 on the left of that support and in its right-hand division 
 
ART. 48. CONTINUOUS BEAMS WITH EQUAL SPANS. 
 
 105 
 
 the shear on the right. The sum of the two shears for any 
 support is, of course, the reaction of that support. For ex- 
 ample, in a beam of five equal spans the reaction at the second 
 support is 
 
 Fig. 42. 
 
 It will be seen on examination that the numbers in any 
 oblique column of these tables follow a certain law of increase 
 by which it is possible to extend them, if desired, to a greater 
 number of spans than are here given. 
 
 As an example, let it be required to select a I beam to span 
 four openings of 8 feet each, the load per span being 14000 
 pounds and the greatest horizontal stress in any fiber to be 
 12000 pounds per square inch. The required beam must 
 satisfy formula (4), or, 
 
 7 = M 
 
 C~~ 12 000 
 
 where M is the maximum moment. From the table it is seen 
 that the greatest negative moment is that at the second sup- 
 port, or ~w/*. The maximum positive moments are, 
 
 2o 
 
 V* 
 
 max M = = 
 
 For fust span, 
 
 For second span, max M = 
 
 77 
 
 r , + = 
 
 2W 
 
106 RESTRAINED AND CONTINUOUS BEAMS. CH. IV. 
 
 The greatest value of M is hence at the second support. Then, 
 
 / = 3 X 14 OOP X 8 X 12 = I2 
 c 28 X 12 ooo 
 
 and from the table in Art. 30 it is seen that a light 7-inch beam 
 will be required. 
 
 Prob. 82. Draw the diagram of shears and the diagram of 
 moments for the case of three equal spans uniformly loaded. 
 
 Prob. 83. Find what I beam is required to span three open- 
 ings of 12 feet each, the load on each span being 6 ooo pounds, 
 and the greatest value of 5 to be 12000 pounds per square 
 inch. 
 
 ART. 49. CONTINUOUS BEAMS WITH UNEQUAL SPANS. 
 
 As the first example, consider two spans whose lengths are 
 /,,/,, and whose loads per linear unit are w l and w^ . The 
 theorem of three moments in (9) then reduces to, 
 
 and hence the bending moment at the middle support is, 
 
 M 
 
 From this the reaction at the left support may be found by (8) 
 and the bending moment at any point by (7). 
 
 Next consider three spans whose lengths are / x , /, , and / 3 , 
 loaded uniformly with w l , w^ , w 3 . The bending moments at 
 the second and third supports are M 9 and M 3 . Then from (9), 
 
 and the solution of these gives the values of M^ and M a . A 
 very common case is that for which ^ = /, /, = / 8 = /, and 
 w l =w 9 = w,= w. For this case the solution gives, 
 
ART. 50. REMARKS ON THE THEORY OF FLEXURE. IO/ 
 
 Here if n = I, these two moments become -fowl*, as also 
 shown in the last article. 
 
 Whatever be the lengths of the spans or the intensity of the 
 loads, the theorem of three moments furnishes the means of 
 finding the bending moments at the supports. Then from (8), 
 (7), and (6) the vertical shears and bending moments at every 
 section may be computed. Finally, if the material be not 
 strained beyond its elastic limit, formula (5) may be used to 
 determine the deflection, while (4) investigates the strength of 
 the beam. 
 
 Prob. 84. A continuous beam of three equal spans is loaded 
 only in the middle span. Find the reactions of the end sup- 
 ports due to this load. 
 
 Prob. 85. A heavy 1 2-inch 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 12000 pounds per square 
 inch. 
 
 ART. 50. REMARKS ON THE THEORY OF FLEXURE. 
 
 The theory of flexure presented in this and the preceding 
 chapter is called the common theory, and is the one universally 
 adopted for the practical investigation of beams. It should not 
 be forgotten, however, that the axioms and laws upon which 
 it is founded are only approximate and not of an exact nature 
 like those of mathematics. Laws (A) and (B) for instance are 
 true as approximate laws of experiment, but probably not as 
 exact laws of science. Law (G) has been established by the 
 observed fact that a vertical line, drawn upon the side of the 
 beam before flexure, remains a straight line after flexure, even 
 when the elastic limit of the material is exceeded. 
 
IO8 RESTRAINED AND CONTINUOUS BEAMS. Ql. IV. 
 
 When experiments on beams are carried to the point of rup- 
 ture and the longitudinal unit-stress 6" computed from formula 
 (4) a disagreement of that value with those found by direct 
 experiments on tension or compression is observed. This is 
 often regarded as an objection to the common theory of flexure, 
 but it is in reality no objection, since law (G) and formula (4) 
 are only true provided the elastic limit of the material be not 
 exceeded. Experiments on the deflection of beams furnish on 
 the other hand the most satisfactory confirmation of the theory. 
 When E is known by tensile or compressive tests the formulas 
 for deflection are found to give values closely agreeing with 
 those observed. Indeed so reliable are these formulas that it 
 is not uncommon to use them for the purpose of computing E 
 from experiments on beams. If however the elastic limit of 
 the material be exceeded, the computed and observed deflec- 
 tions fail to agree. 
 
 On the whole it may be concluded that the common theory 
 of flexure is entirely satisfactory and sufficient for the investi- 
 gation of all practical questions relating to the strength and 
 stiffness of beams. The actual distribution of the internal 
 stresses is however a matter of very much interest and this will 
 be discussed at some length in Chapter VIII. 
 
 The theory of flexure is here applied to continuous beams 
 only for the case of uniform loads. It should be said however 
 that there is no difficulty in extending it to the case of concen- 
 trated loads. By a course of reasoning similar to that of Art. 
 48 it may be shown that the theorem of three moments for 
 single loads is, 
 
 M'l' + 2M"(l' + /") + M'"l" = - P'l'\k - &) 
 
 Here as in Fig. 37 the moments at three consecutive supports 
 are designated by M', M" , and M"' and the lengths of the two 
 spans by /' and I". P' is any load on the first span at a dis- 
 
ART. 50. REMARKS ON THE THEORY OF FLEXURE. IOQ 
 
 tance kl' from the left support and P" any load on the second 
 span at a distance kl" from the left support, k being any frac- 
 tion less than unity and not necessarily the same in the two 
 cases. From this theorem the negative bending moments at 
 the supports for any concentrated loads may be found, and the 
 beam be then investigated by formulas (6) and (4). For ex- 
 ample, if a beam of three equal spans be loaded with P at the 
 middle of each span, the negative moments at the supports 
 
 are each $-Pl. 
 20 
 
 The Journal of the Franklin Institute for March and April, 
 1875, contains an article by the author in which the law of in- 
 crease of the quantities in the tables of Art. 48 is explained 
 and demonstrated. A general abbreviated method of deduc- 
 ing the moments at the supports for both uniform and concen- 
 trated loads on restrained and continuous beams is given in the 
 Philosophical Magazine for September, 1875. See Text-book 
 on Roofs and Bridges, Part IV. 
 
 Exercise 4. Consult BARLOW'S Strength of Materials (Lon- 
 don, 1837), an d write an essay concerning his experiments to 
 determine the laws of the strength and stiffness of beams. Con- 
 sult also BALL'S Experimental Mechanics. 
 
 Exercise 5. Consult Engineering News, Vol. XVIII, pp.309, 
 352, 404, 443; Vol. XIX, pp. 11, 28, 48, 84; and Vol. XXII, 
 p. 121. Write an essay concerning certain erroneous views re- 
 garding the theory of flexure which are there discussed. Con- 
 sult also TODHUNTER'S History of the Elasticity and Strength 
 of Materials. 
 
 Exercise 6. Procure six sticks of ash each -J X f inches and 
 of lengths about 8, 12, and 16 inches. Devise and conduct 
 experiments to test the following laws : First, the strength of a 
 beam varies directly as its breadth and directly as the square 
 of its depth. Second, the stiffness of a beam is directly as its 
 breadth and directly as the cube of its depth. Third, a beam 
 fixed at the ends is twice as strong and four times as stiff as a 
 
110 RESTRAINED AND CONTINUOUS BEAMS. CH. IV. 
 
 simple beam when loaded at the middle. Write a report 
 describing and discussing the experiments. 
 
 Exercise 7. In order to test the theory of continuous beams 
 discuss the following experiments by FRANCIS and ascertain 
 whether or not the ratio of the two observed deflections agrees 
 with theory. "A frame was erected, giving 4 bearings in the 
 same horizontal plane, 4 feet apart, making 3 equal spans, each 
 bearing being furnished with a knife edge on which the beam 
 was supported. Immediately over the bearings and secured to 
 the same frame was fixed a straight edge, from which the de- 
 flections were measured. A bar of common English refined 
 iron, 12 feet 2j inches long, mean width 1.535 inches, mean 
 depth 0.367 inches, was laid on the 4 bearings, and loaded at 
 the center of each span so as to make the deflections the same, 
 the weight at the middle span being 82.84 pounds and at each 
 of the end spans 52.00 pounds. The deflections with these 
 weights were, 
 
 At the center of the middle span 0.281 inches. 
 
 At center of end spans, 0.275 and 0.284 inches, 
 
 mean 0.280 inches. 
 
 A piece 3 feet uf inches long was then cut from each end of 
 the bar, leaving a bar 4 feet 4f inches long, which was replaced 
 in its former position and loaded with the same weight (82.84 
 pounds) as before, when its deflection was found to be 1.059 
 inches." 
 
 Prob. 86. A beam of three spans, the center one being / and 
 the side ones nl, is loaded with P at the middle of each span. 
 Find the value of n so that the reactions at the end may be 
 one-fourth of the other reactions. 
 
 Prob. 87. Let a beam whose cross-section is an isosceles tri- 
 angle have the base b and the depth d. Prove that if 0.13^ be 
 cut off from the vertex the remaining trapezoidal beam will 
 be about 9 per cent stronger than the triangular one. 
 
ART. 51. CROSS-SECTIONS OF COLUMNS. Ill 
 
 CHAPTER V. 
 COLUMNS OR STRUTS. 
 
 ART. 51. CROSS-SECTIONS OF COLUMNS. 
 
 A column is a prism, greater in length than about ten times 
 its least diameter, which is subject to compression. If the 
 prism be only about four or six times as long as its least diam- 
 eter the case is one of simple compression, the constants for 
 which are given in Art. 6. In a case of simple compression 
 failure occurs by the crushing and splintering of the material, 
 or by shearing in directions oblique to the length. In the case 
 of a column, however, failure is apt to occur by a sidewise 
 bending which causes flexural stresses. The word * strut ' is 
 frequently used as synonymous with column, and sometimes 
 also the word ' pillar.' 
 
 Wooden columns are usually square or round and they may 
 be built hollow. Cast iron columns are usually round and they 
 are often cast hollow. Wrought iron columns are made of a 
 great variety of forms. I beams may be used, but most columns 
 are usually made of three or more different shape-irons riveted 
 together. The Phcenix column is made by riveting together 
 flanged circular segments so as to form a closed cylinder. It is 
 clear that a square or round section is preferable to an unsym- 
 metrical one, since then the liability to bending is the same in 
 all directions. For a rectangular section the plane of flexure 
 will evidently be perpendicular to the longer side of the cross- 
 section, and in general the plane of flexure will be perpendicular 
 to that axis of the cross-section for which the moment of in- 
 ertia is the least. In designing a column it is hence advisable 
 
112 
 
 COLUMNS OR STRUTS. 
 
 CH. V. 
 
 that the cross-section should be so arranged that the moments 
 of inertia about the two principal rectangular axes may be 
 approximately equal. 
 
 For instance, let it be required to construct a column with 
 two I shapes and two plates as shown in Fig. 43. The I beams 
 are to be light lo-inch ones weighing 30 
 pounds per linear foot, and having the 
 flanges 4.32 inches wide. The plates are 
 to be inch thick, and it is required to 
 find their length x so that the liability to 
 bending about the two axes shown in the 
 figure may be the same. From the manu- 
 facturer's table it is found that the moment 
 
 43- 
 
 of inertia / of the beam about an axis through its center of 
 gravity and perpendicular to the web is 150, while the moment 
 of inertia I' about an axis through the same point and parallel 
 to the web is nearly 8. Hence, for the axes showji in the 
 figure, the moments of inertia are, 
 For axis perpendicular to plates, 
 
 / * 3 /^ I _ . v > O I , . v > 
 
 12 
 
 For axis parallel to plates, 
 * X 0.5" 
 
 12 
 
 + 2 X 0.5* x 5-25' + 2 X 150. 
 
 Placing these two expressions equal, the value of x is found to 
 be between 14 and 14^ inches. 
 
 Prob. 88. A column is to be formed of two light 1 2-inch eye- 
 beams connected by a lattice bracing. Find the proper distance 
 between their centers, disregarding the moment of inertia of 
 the latticing. 
 
 Prob. 89. Two joists each 2X4 inches are to be placed 6 
 inghes apart between their centers, and connected by two others 
 each 8 inches wide and x inches thick so as to form a closed 
 hollow rectangular column. Find the proper value of x. 
 
ART. 52. 
 
 GENERAL PRINCIPLES. 
 
 ART. 52. GENERAL PRINCIPLES. 
 
 If a short prism of cross-section A be loaded with the weight 
 P, the internal stress is to be regarded as uniformly distributed 
 
 over the cross-section, and hence the compressive unit-stress S e 
 
 p 
 is --. But for a long prism, or column, this is not the case; 
 
 A 
 
 P 
 while the average unit-stress is r , the stress in certain parts of 
 
 the cross-section may be greater and upon others less than this 
 value on account of the transverse stresses due to the sidewise 
 flexure. Hence in designing a column the load P must be 
 taken as smaller for a long one than for a short one, since evi- 
 dently the liability to bending increases with the length. 
 
 Numerous tests on the rupture of long columns have 
 shown that the load causing the rupture is approximately in- 
 versely proportional to the square of the length of the column. 
 That is to" say, if there be two columns of the same material 
 and cross-section and one twice as long as the other, the long 
 one will rupture under about one-quarter the load of the 
 short one. 
 
 The condition of the ends of columns exerts a great influence 
 upon their strength. Class (a) includes those with ' round ends/ 
 or those in such condition that they are free to turn at the ends. 
 Class (c) includes those whose ends 
 are ' fixed ' or in such condition that 
 the tangent to the curve at the ends 
 always remains vertical. Class (b) 
 includes those with one end fixed 
 and the other round. In architecture 
 it is rare that any other than class (c) 
 is used. In bridge construction and 
 in machines, however, columns of 
 classes (b) and (a) are very common. It is evident that class (c) 
 
114 COLUMNS OR STRUTS. CH. V. 
 
 is stronger than (), and that (b) is stronger than (a), and this is 
 confirmed by all experiments. Fig. 44 is intended as a sym- 
 bolical representation of the three classes of columns, and not 
 as showing how the ends are rendered 'round' and 'fixed' in 
 practical constructions. 
 
 The theory of the resistance of columns has not yet been 
 perfected like that of beams, and accordingly the formulas for 
 practical use are largely of an empirical character. The form 
 of the formulas however is generally determined from certain 
 theoretical considerations, and these will be presented in the 
 following articles as a basis for deducing the practical rules. 
 
 Prob. 90. A column of \\rouHit-iron shapes weighing 186 
 pounds per yard is 1 1 inches square and 3 feet long. Whc*c 
 load will it carry with a factor of safety of 5 ? 
 
 ART. 53. EULER'S FORMULA FOR LONG COLUMNS. 
 
 Consider a column of cross-section A loaded with a weight P 
 |p under whose action a certain small sidewise bending 
 4 occurs. Let the column be round, or free to turn at 
 both ends as in Fig. 45. Take the origin at the upper 
 end, and let x be the vertical and y the horizontal co- 
 ordinate of any point of the elastic curve. The gen- 
 eral equation (5) deduced in Art. 33, applies to all 
 bodies subject to flexure provided the bending be 
 slight and the elastic limit of the material be not ex- 
 ceeded. For the column the bending moment is 
 Fig. 45. Py, the negative sign being used because the curve 
 is concave to the axis of x ; hence, 
 
 . 
 
 The first integration of this gives, ( ^ 
 
 
ART. 53. 
 
 EULER'S FORMULA. 
 
 dy 
 But when y = the maximum deflection ^, the tangent ~- = o. 
 
 Hence C = P4*, and by inversion, 
 
 dx = (^^=, 
 
 The second integration now gives, 
 
 =(?)' 
 
 arc sin ~ 4- T'. 
 
 ' 
 
 Here C"' is o because / = o when x = o. Hence finally the 
 equation of the elastic curve of the column is, 
 
 This equation is that of a sinusoid. But also y = o when 
 
 / P\* 
 x = I. Hence if n be an integer, IVJ^T] must equal nn, or, 
 
 p =&-. 
 
 which is EULER'S formula for the resistance of columns. This 
 reduces the equation of the sinusoid to, 
 
 y = A sin nn-j. 
 
 The three curves for n = I, n = 2, and n 3 are shown in 
 Fig. 46. In the first case the 
 curve is entirely on one side of 
 the axis of x, in the second case 
 it crosses that axis at the mid- 
 dle, and in the third case it 
 crosses at \l and \ /, the points 
 of crossing being also inflection 
 points where the bending mo- 
 ment is zero. Evidently the 
 
 Fig. 46. 
 
 greatest deflection will occur for the case where n = i, and 
 
. 
 
 1 16 COLUMNS OR STRUTS. CH. V. 
 
 this is the most dangerous case. Hence, 
 
 n*EI 
 (a) p=j r - ) 
 
 is EULER'S formula for long columns with round ends. 
 
 A column with one end fixed and the other round is closely 
 represented by the portion b'b" of the second case, b' being 
 the fixed end where the tangent to the curve is vertical. Here 
 n = 2, and the length b'b" is three-fourths of the entire length, 
 hence, 
 
 is RULER'S formula for long columns with one end fixed and 
 the other round. 
 
 A column with fixed ends is represented by the portion c'c" 
 of the case c. Here n = 3, and the length c'c" is two-thirds of 
 the entire length, hence, 
 
 ... 
 
 is EULER'S formula for long columns with fixed ends. 
 
 From this investigation it appears that the relative resist- 
 ances of long columns of the classes (a), (), and (c) are as the 
 numbers i, 2j, and 4, when the lengths are the same, and this 
 conclusion is approximately verified by experiments. It also 
 appears that, if the resistance of three columns of the classes 
 (a), (b), and (V) are to be equal, their lengths must be as the 
 numbers I, if, and 2. 
 
 The moment of inertia 7 in the above formulas is taken about 
 a neutral axis of the cross-section perpendicular to the plane of 
 the flexure, and in general is the least moment of inertia of 
 that cross-section, since the column will bend in the direction 
 which offers the least resistance. If A be the area of the 
 
ART. 54. HODGKINSON'S FORMULAS. 117 
 
 cross-section and r its least radius of gyration, the value of / 
 is Ar* t and EULER's formula may be written 
 
 p y 
 
 where m = i, 2j, or 4, for the three end conditions. 
 
 The value of P in EULER'S formula gives the load which 
 holds the column in equilibrium when it is laterally deflected. 
 If the load be less than this value of P the column, if 
 deflected, will return to its original straight position. If the 
 load be 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 bending. The maximum deflection A is 
 indeterminate, since it cannot be found from the equation of 
 the elastic curve. 
 
 EULER'S formula is little used in practice, except in Ger- 
 many. When so used it takes the form 
 
 P _ rnrfE S 
 
 A~ ~7~'T" 
 
 in which f is an assumed factor of security and P is the load 
 on the column. 
 
 Prob. 91. Show that the value of the constant m for a long 
 column which is fixed at one end and entirely free at the other 
 is m = J. 
 
 ART. 54. HODGKINSON'S FORMULAS. 
 
 EULER'S formula gives valuable information regard- 
 ing the laws of flexure of long columns. For cylindrical 
 columns it shows, since /= Ttd'/G^., that the load P which 
 causes lateral failure varies directly as the fourth power of 
 the diameter and inversely as the square of the length. 
 HODGKINSON in his experiments observed that this was 
 approximately but not exactly the case. He therefore 
 
Il8 COLUMNS OR STRUTS. Cli. V. 
 
 wrote for each kind of columns the analogous expression, 
 
 and determined the constants a, /?, and 6 from the results of 
 his experiments, thus producing empirical formulas. 
 
 Let P be the crushing load in gross tons, d the diameter of 
 the column in inches, and / its length in feet. Then HODG- 
 KINSON'S empirical formulas are, 
 
 For solid cast iron cylindrical columns, 
 
 P = 14.9- for round ends, 
 
 for flat ends, 
 
 For solid wrought iron cylindrical columns, 
 
 ^3-76 
 P = 42-7^ for round ends; 
 
 ^3.76 
 P = 134- 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. 
 
 By the help of logarithms it is easy to apply these formulas 
 to the discussion of given cases. Usually P will be given and 
 d required, or d be given and P required. By using assumed 
 factors of safety the proper size of cylindrical columns to carry 
 given loads may also be determined. These formulas, it should 
 be remembered, do not apply to columns shorter than about 
 
ART. 55. RANKINE'S FORMULA. 119 
 
 thirty times their least diameters. The word flat used in this 
 Article is to be regarded as equivalent to fixed. 
 
 Prob. 92. A cast-iron cylindrical column with flat ends is 3 
 inches diameter and 8 feet long. What load will cause it 
 to fail? 
 
 Prob. 93. 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. Find the proper diameter. 
 
 ART. 55. RANKINE'S FORMULA. 
 
 The columns which are generally employed in engineering 
 practice are intermediate in length between short prisms and 
 the long columns to which EULER'S formula ap- 
 plies. They fail under the stresses caused by 
 combined flexure and compression. Fig. 47 
 shows the flexure very much exaggerated. The 
 
 P 
 
 load P produces an average unit-stress on any 
 
 A 
 
 horizontal cross-section whose area is A, but in 
 consequence of the bending this is increased on 
 the concave side and diminished on the convex 
 side by an amount S^ The value of S l depends 
 upon the bending moment PA, where A is the 
 lateral deflection at the middle of the column. 
 
 The total unit-stress on the concave side is then 
 
 p Fig. 47. 
 
 + S lt and it isnatural to suppose that failure 
 
 A. 
 
 will occur when this is equal to the ultimate strength of the 
 material. Many formulas have been proposed for the investi- 
 gation of such columns, but all of them are more or less em- 
 pirical, as certain constants are derived from the results of ex- 
 
120 COLUMNS OR STRUTS. CH. V. 
 
 periments upon the rupture of columns, or assumptions are 
 made regarding the form of the formula. 
 
 The formulas of EULER are unsuitable for practical cases of 
 investigation because they contain no constant indicating the 
 working or ultimate compressive strength of the material, and 
 because they apply only to long columns. HODGKINSON's 
 formulas are unsatisfactory for similar reasons, and because 
 they do not well agree with later experiments. 
 
 The formula which appears to have the best theoretical 
 foundation will now be presented. It is sometimes called 
 GORDON'S formula, and occasionally it is referred to as " GOR- 
 DON'S formula modified by RANKINE," but the best usage 
 gives to it the name of RANKINE'S formula. (The formula 
 deduced by GORDON differs from (10) in that it applies only to 
 rectangular or circular cross-sections, r being replaced by d, the 
 least side or diameter, and q having different values from those 
 given in the table.) 
 
 Let P be the load on the column, / its length, A the area of 
 its cross-section, / the moment of inertia, and r the radius of 
 gyration of that cross-section with reference to a neutral axis 
 perpendicular to the plane of flexure, and c the shortest dis- 
 tance from that axis to the remotest fiber on the concave side. 
 The average compressive unit-stress on 
 
 p 
 
 any cross-section is , but in consequence 
 A 
 
 of the flexure this js increased on the 
 concave side, and decreased on the con- 
 vex side. Thus in Fig. 48 the average 
 
 p 
 
 unit-stress - is represented by cd, but on 
 
 the concave side this is increased to aq, and on the convex 
 side decreased to bq. The triangles pdq and qdp represent 
 
ART. 55- RANKINE'S FORMULA. 121 
 
 the effect of the flexure exactly as in the case of beams, pq 
 indicating the greatest compressive and qp the greatest ten- 
 sile unit-stress due to the bending. Let the total maximum 
 unit-stress aq be denoted by 5 and the part due to the flexure 
 be denoted by S 19 Then, 
 
 New, from the fundamental formula (4) the flexural stress is 
 
 Me 
 
 , where M is the external bending moment, which for a col- 
 
 umn has its greatest value when M = PA, A being the maxi- 
 mum deflection. / = Ar* is the well-known relation between / 
 and r. Hence the value of S l is, 
 
 PAc PAc 
 
 By analogy with the theory of beams, as in Art. 37, the value 
 
 ^ 
 of A may be regarded as varying directly as . Hence if q be 
 
 a quantity depending upon the kind of material and the con- 
 dition of the ends, the total unit-stress is, 
 
 P 
 
 This may now be written in the usual form, 
 
 P _ S 
 (10) A = 
 
 which is RANKINE'S formula for the investigation of columns. 
 
 The above reasoning has been without reference to the ar- 
 rangement of the ends of the column. By Art. 53 it is known 
 that a column with round ends must be one half the length of 
 one with fixed ends in order to be of equal strength, and that 
 a column with one end fixed and the other round must be 
 
122 
 
 COLUMNS OR STRUTS. 
 
 CH. V. 
 
 three fourths the length of one with fixed ends in order to be 
 of equal strength. Therefore if q be the constant for fixed 
 ends, (fq will be the constant for one end fixed and the other 
 round, and 2*q will be the constant for both ends round. 
 
 The values of q to be taken for use in formula (10) for the 
 examples and problems of this chapter may be the following 
 rough values, unless otherwise stated, while the values of 
 the ultimate compressive unit-stress 6" will be taken from 
 the table in Art. 6. 
 
 Material. 
 
 Both Ends 
 Fixed. 
 
 Fixed and 
 Round. 
 
 Both Ends 
 Round. 
 
 Timber, 
 Cast Iron, 
 Wrought Iron, 
 Steel, 
 
 I 
 
 I. 7 8 
 
 4 
 
 3 ooo 
 
 i 
 
 3 ooo 
 1.78 
 
 3 ooo 
 
 4 
 
 5 ooo 
 
 i 
 
 5 ooo 
 i.78 
 36 ooo 
 
 1.78 
 
 5 ooo 
 4 
 
 36 ooo 
 I 
 
 36000 
 4 
 
 25 ooo 
 
 25 ooo 
 
 25 ooo 
 
 Very wide variation in the values of q is found from experi- 
 ments on the rupture of different types of columns. Formula 
 (10) when used with such experimental values is an empirical 
 one only. In Art. 61 it is shown how a theoretic value of q 
 may be derived which renders (10) a rational formula. 
 
 Prob. 94. Plot the curve represented by formula (10) for 
 cases of wrought-iron columns with fixed and with round ends, 
 
 P I 
 
 taking the values of as ordinates, and the values of as ab- 
 
 jrL T 
 
 scissas. 
 
 ART. 56. RADIUS OF GYRATION OF CROSS-SECTIONS. 
 
 The radius of gyration of a surface with reference to an axis 
 is equal to the square root of the ratio of the moment of iner- 
 
ART. 57. INVESTIGATION OF COLUMNS. 123 
 
 tia of the surface referred to the same axis to the area of the 
 figure. Or if r be radius of gyration, / the moment of inertia, 
 and A the area of the surface, then I =Ar*. 
 
 In the investigation of columns by formula (10) the value of 
 r a is required, r being the least radius of gyration. These 
 values are readily derived from the expressions for the mo- 
 ment of inertia given in Art. 23, the most common cases being 
 the following, 
 
 For a rectangle whose least side is d, r* = . 
 
 For a circle of diameter d, r* = -^. 
 
 16 
 
 For a triangle whose least altitude is d, r* = $. 
 
 Io 
 
 For a hollow square section, r 1 = 
 
 For a hollow circular section, r* = 
 
 12 
 
 d'+d" 
 
 16 
 
 For I beams and other shapes, r 9 is found by dividing the least 
 moment of inertia of the cross-section by the area of that cross- 
 section. For instance, by the help of the table in Art. 30, the 
 least value of r a for a light 12-inch I beam is found to be 
 g= 1.02 inches'. 
 
 Prob. 95. Compute the least radius of gyration for a T iron 
 whose width is 4 inches, depth 4 inches, thickness of flange J 
 inches, and thickness of stem f inches. 
 
 ART. 57. INVESTIGATION OF COLUMNS. 
 The investigation of a column consists in determining the 
 maximum compressive unit-stress 5" from formula (10). The 
 values of P, A, /, and r will be known from the data of the given 
 case, and q is known from the results of previous experiments. 
 Then, p 
 
124 COLUMNS OR STRUTS. CH. V. 
 
 and, by comparing the computed value of 5 with the ultimate 
 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 5400 pounds. Here P = 
 5 400, A S square inches, ! = 216 inches. From the table 
 q = 3 -^. From Art. 56, 
 
 - - **** = - 
 
 Then the substitution of these values gives, 
 
 5400/ 2r6X2i6\ 
 
 5 = g (i + 3000X22 J = 5430 pounds per square inch. 
 
 Here the average unit-stress is 675 pounds per square inch, 
 but the flexure has increased that stress on the concave side to 
 5 430 pounds per square inch, so that the factor of safety is 
 only about i. 
 
 Prob. 96. A cylindrical wrought iron column with fixed ends 
 is 12 feet long, 6.36 inches in exterior diameter, 6.02 inches in 
 interior diameter, and carries a load of 98 ooo pounds. Find 
 its factor of safety. 
 
 Prob. 97. A pine 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 be only one foot, what is 
 the factor of safety? 
 
 ART. 58. SAFE LOADS FOR COLUMNS. 
 
 To determine the safe load for a given column it is necessary 
 to first assume the allowable working unit-stress 5. Then from 
 formula (10) the safe load is, 
 
 r 
 
ART. 59. DESIGNING OF COLUMNS. 12$ 
 
 Here A, /, and r are known from the data of the given problem 
 and q is taken from the table in Art. 55. 
 
 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 may be 
 800 pounds per square inch. From the formula, 
 
 800 X 12 
 
 P = - = = about i 300 pounds. 
 
 120' 
 
 3 ooo X | 
 
 A short prism 3X4 inches should safely carry seven times 
 this load. 
 
 Prob. 98. Find the safe load for a heavy steel I beam of 
 15 inches depth and 10 feet length when used as a column with 
 fixed ends, the factor of safety being 4. 
 
 Prob. 99. 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. 
 
 ART. 59. DESIGNING OF COLUMNS. 
 
 When a column is to- be selected or designed the load to be 
 borne will be known, as also its length and the condition of the 
 ends. A proper allowable unit-stress 5 is assumed, suitable for 
 the given material under the conditions in which it is used. 
 Then from formula (i) the cross-section of a short column or 
 
 P 
 
 prism is -=-, and it is certain that a greater value of the cross- 
 section than this will be required. Next assume a form and 
 area A, find r 2 , and from the formula (10) compute 5". If the 
 computed value agrees with the assumed value the correct size 
 has been selected. If not, assume a new area and compute .S 
 again, and continue the process until a proper agreement is 
 attained. 
 
126 COLUMNS OR STRUTS. CH. V. 
 
 For example, a hollow cast iron rectangular column of 18 feet 
 length is to carry a load of 60 ooo pounds. Let the working 
 strength .S be 15 OOO pounds per square inch. Then for a short 
 length the area required would be four square inches. Assume 
 then that about 6 square inches will be needed. Let the sec- 
 tion be square, the exterior dimensions 6x6 inches, and the in- 
 terior dimensions 5^ X 5i inches. Then A = 5.75, 1= 18 X 12, 
 P 60000, q ^5, r 2 5.52, and from (10), 
 
 5 = 
 
 6oooo/ i8 2 X 12' \ 
 
 -- 1 1 + -- j = about 30 ooo, 
 5.75 \ [ 5000 X 5-52' 
 
 which shows that the dimensions are much too small. Again 
 assume the exterior side as 6 inches and the interior as 5 inches. 
 Then A = n, r 2 = 5.08, and 
 
 6oooo/ i8 2 X 12" 
 
 As this is very near the required working stress, it appears that 
 these dimensions very nearly satisfy the imposed conditions. 
 
 In many instances it is possible to assume all the dimensions 
 of the column except one, and then after expressing^ and r in 
 terms of this unknown quantity, to introduce them into (10) 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 sus- 
 tain a load of 100000 pounds with a factor of safety of 10. 
 
 Here let x be the unknown side ; then A = x* and r* = . 
 
 From (10), 
 
 100 ooo/ 24" X i: 
 
 800 = 5 i H 
 
 x v 3 ooo x 
 
 By reduction this becomes, 
 
 Sx 4 i ooo*' = 331 776, 
 the solution of which gives 16.6 inches for the side of the column. 
 
ART. 60. THE STRAIGHT-LINE FORMULA. I2/ 
 
 Prob. ioo. Find the size of a square wooden column with fixed 
 ends and 12 feet in length to sustain a load of 100000 pounds 
 with a factor of safety of 10. Find also its size for round ends. 
 
 ART. 60. TrtE STRAIGHT-LINE FORMULA. 
 
 In 1886 a straight-line formula for columns, as a substitute 
 for RANKINE'S, was proposed by THOMAS H. JOHNSON, which 
 has since been extensively used on account of its simplicity 
 when expressed in numerical form. The notation being the 
 same as in Art. 55, this formula is, 
 
 p -s k l 
 
 ~A~ r' 
 
 in which k is a constant whose value is, 
 
 t S 
 ' 3 
 
 where m is I, 2j, or 4, depending on the condition of the ends, 
 as in EULER'S formula (Art. 53). 
 
 This formula is not a rational one, the equation of the 
 straight line being assumed merely as a good representation 
 of the results of experiments on the rupture of columns. The 
 value of k is deduced by making the straight line tangent to the 
 curve which represents EULER'S formula. Thus, let the values of 
 
 P I 
 
 and - be regarded as the ordinates and abscissas of a curve, 
 
 /i T 
 
 and be designated by u and v respectively. Then the equa- 
 tions of EULER'S curve and of the assumed straight line are, 
 
 ~ 
 u - 5 and u = S kv. 
 
 By placing equal the values of u in these two equations and 
 also the values of the first derivatives, the ordinate and abscissa 
 of the point of tangency are found to be, 
 
 i o 
 * = S and z/ 
 
128 
 
 COLUMNS OR STRUTS. 
 
 CH. V. 
 
 and then the value of k, as above ghnen. results. The value of 
 i\ is the limiting value of -, within which the straight-line 
 
 formula is to be used. 
 
 The values of 5 to be used for cases of rupture are such as 
 make the straight-line agree best with experimental results. 
 The values derived by JOHNSON in his discussion are given in 
 the- following table, together with the corresponding values of 
 
 k and limiting values of - , which are computed by taking m as 
 2j, if, and I for the given cases. 
 
 Kind of Column. 
 
 5 
 
 k 
 
 Limit 
 r 
 
 Wrought iron: 
 
 
 
 
 Flat ends, 
 
 42 ooo 
 
 128 
 
 218 
 
 Hinged ends, 
 
 42 ooo 
 
 157 
 
 178 
 
 Round ends, 
 
 42 ooo 
 
 203 
 
 I 3 8 
 
 Mild steel: 
 
 
 
 
 Flat ends, 
 
 52 500 
 
 J 79. 
 
 195 
 
 Hinged ends, 
 
 52 500 
 
 220 
 
 159 
 
 Round ends, 
 
 52 500 
 
 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, 
 
 5400 
 
 28 
 
 128 
 
 It will be noticed that the values of S in the above table are 
 less than the average values of ultimate strength given in Art. 
 6. For ductile materials, like wrought iron and mild steel, this 
 should be the case in columns, since when the elastic limit is 
 passed a flow of metal begins which causes the lateral deflec- 
 tion to increase, and failure then rapidly follows. 
 
 Reference is made to JOHNSON'S paper in Transactions of 
 
ART. 6l. HITTER'S RATIONAL FORMULA. 129 
 
 American Society of Civil Engineers for July 1886, for a 
 fuller discussion of the straight-line formula. Although much 
 used for computing values of P/A, it is inconvenient for find- 
 ing S, since this quantity is included in k and a cubic equa- 
 tion in 5 results. 
 
 Prob. 101. Solve Problems 99 and 100 by the straight-line 
 formula, using the values given in the table. 
 
 ART. 61. RITTER'S RATIONAL FORMULA. 
 
 Several attempts have been made to establish a formula for 
 columns, which shall be theoretically correct, like formula (4) 
 for beams, when the material is not stressed beyond the 
 elastic limit. The most successful attempt is that of RlTTER, 
 who in 1873 proposed the formula 
 P S 
 
 in which Pis the load on the column, /its length, A the area 
 and r the least radius of gyration of the cross-section, E the 
 coefficient of elasticity, S e the unit-stress at the elastic limit, 
 and 5 the greatest compressive unit-stress on the concave 
 side, while m = 4 when both ends are fixed, m = 2\ when 
 one end is round and the other fixed, m = i when both ends 
 are round, and m = i when one end is fixed and the other 
 free. 
 
 The form of this formula is the same as that of RANKINE'S 
 formula, (10) in Art. 55, but it deserves a special name 
 because it completes the deduction of the latter formula by 
 finding for q a value which is closely correct when the stress 
 5 does not exceed the elastic limit S t . 
 
 To justify RITTER'S formula let it be noted that EULER'S 
 formula (Art. 53) gives the value of P which causes the failure 
 of a long column by lateral bending. In the actual long 
 column, however, the load must be less than given by 
 
130 COLUMNS OR STRUTS. CH. V 
 
 EULER'S formula, since it is required that stable equilibrium 
 shall prevail. Now, if there be written 
 P _ S r* 
 
 A~~s e ' mn '7* 
 
 for the long column, stable equilibrium prevails when S is less 
 than S e , while failure occurs when 5 equals S e , because then 
 the column will not spring back if laterally deflected by a 
 slight force. S e /S is hence the factor of security /for a long 
 stable column, as noted at the end of Art. 53. Now RlTTER' I 
 formula reduces to the above form for long columns when l/r 
 is so great that unity in the denominator may be neglected in 
 comparison with the following term. It also reduces to the 
 form P/A = 5 for short blocks, when / = o. The formula 
 hence satisfies the two limiting conditions, and its general form 
 is justified by the reasoning of Art. 55. 
 
 Another demonstration may be given as follows: Let P/A 
 be denoted by y and l/r by x. Then the formulas of RAN- 
 KINE and of EULER for stable equilibrium are 
 
 5 mifE 
 
 y = , .1. ^ and y = -yy-- 
 
 Now let the curves which these equations represent be tan- 
 gent to each other. For the point of tangency the two 
 ordinates are equal, as also the two derivatives of y with 
 respect to x. Solving these two equations there results 
 x^ = oo for the point of tangency, while 
 
 Sf S< 
 
 mn*E ~ 
 is the constant in RANKINE'S formula. Thus, 
 
 P 5 
 
 A - S e /" 
 
 which is the rational formula of RITTER. 
 
ART. 61. 
 
 HITTER'S RATIONAL FORMULA. 
 
 To find the mean theoretical values of q the mean values 
 of S e and E, as stated in Art. 6, may be used, whence results 
 the following table: 
 
 RATIONAL VALUES OF q FOR FORMULA (to). 
 
 Material. 
 
 Both Ends 
 Fixed. 
 
 Fixed and 
 Round. 
 
 Both Ends 
 Round. 
 
 Timber 
 
 I 
 
 1.78 
 
 4 
 
 
 20000 
 
 20 ooo 
 
 20 ooo 
 
 
 I 
 
 1.78 
 
 4 
 
 
 30000 
 
 30 ooo 
 
 30000 
 
 Wrought Iron 
 
 I 
 40000 
 
 1.78 
 
 40 ooo 
 
 4 
 40000 
 
 C-,,1 
 
 I 
 
 1.78 
 
 4 
 
 
 24 ooo 
 
 24 ooo 
 
 24 ooo 
 
 These theoretical values of q are smaller than the empirical 
 values given in Art. 55, except that for steel. Values of/ 1 
 computed from RITTER'S formula are hence generally larger 
 than those found from the empirical formulas in common use. 
 In specifications, however, the values of q and 5 to be used 
 are generally stated, so that the designer is free from the 
 responsibility of selecting them. 
 
 It should be noted that this rational formula, having been 
 deduced from the laws which govern the behavior of materials 
 within the elastic limit, is not necessarily true for cases of 
 rupture. In RANKINE'S formula (10) the unit-stress 5 may 
 be the ultimate strength or any smaller value, but in RlTTER'S 
 formula 5 cannot exceed the elastic limit S e . This rational 
 formula cannot hence be justified or disproved because it 
 agrees or fails to agree with the results of experiments on the 
 rupture of columns. 
 
 Prob. 1 02. Compute by RITTER'S formula the safe steady 
 load for a hollow cast-iron column with fixed ends, the out- 
 side dimensions being 4X5 inches, the inside dimensions 
 3X4 inches, and the length 1 8 feet. 
 
132 COLUMNS OR STRUTS. CH. V. 
 
 Prob. 103. Find by RlTTER's formula the size of a square 
 wooden strut with fixed ends, and 12 feet long, to carry a 
 load of 100 ooo pounds, taking .S as 800 pounds per square 
 inch. 
 
 ART. 62. ECCENTRIC LOADS. 
 
 In all that precedes, the load on the column has been sup- 
 posed to be applied at the center of gravity of the cross- 
 section. Let now the case be considered where the load P 
 is applied at the horizontal distance / from that center of 
 gravity and in the same vertical plane with the least radius of 
 gyration r. The reacting load at the other end is similarly 
 situated with respect to the cross-section at that end. Let 
 an origin be taken at the upper end of the column and let x 
 be the vertical and y the horizontal co-ordinate of any point 
 in the elastic curve. The column having round ends, the 
 bending moment for the end is PP, and for any other point 
 *\p-\-y)- Then from (5) the differential equation of the 
 elastic curve is 
 
 where the negative sign is used because the curve is concave 
 to the axis of x. This may be written, 
 
 , g =-/?'(/ + .,), where P = ^ ~, 
 
 and by two integrations there is found, 
 p I * 
 
 as may be proved by differentiating the last equation twice. 
 
 The deflection at the middle of the column is found by 
 making x = / and y = A, whence 
 
 "- i). 
 
ART. 62. ECCENTRIC LOADS. 133 
 
 From this equation the load which holds the deflected column 
 in equilibrium with the deflection A is found to be 
 
 which reduces to EULER'S formula for round-ended columns 
 when/= o, since the value of cos^o is \n. 
 
 It thus appears that the deflection of a column is perfectly 
 determinate when the load is applied eccentrically. The 
 above formula for A is, however, a very inconvenient one for 
 practical computations. An approximate formula giving 
 closely the same numerical results may be deduced by assum- 
 ing that the curve of the deflected column is a parabola; the 
 equation of this curve with respect to an origin at the end is 
 y = (lx x*)4.A/r. Inserting this value of y in the first ex- 
 pression of this article, integrating twice, determining the 
 constants, and then making x = %l and y = A, there will be 
 found 
 
 A = 
 
 9.6EIPI" 
 
 as a practical formula for the deflection of a round-ended 
 column under a load P having the eccentricity/. 
 
 Resuming now the reasoning of Art. 55, the maximum 
 compressive unit-stress S on the concave side of the middle 
 of the column is 
 
 P Me P, pc . Ac 
 
 and from this S may be computed for any value of A. By 
 inserting in this expression the above value of A, there 
 results 
 
134 COLUMNS OR STRUTS. CH. V. 
 
 pr 
 
 PI , pc i + v \ 
 = A( I + S'---&)> 
 
 where V '- = 
 
 from which 5 may be found without computing A. 
 
 For example, let a steel column with round ends be loaded 
 with 168000 pounds. Let A = 16.8 square inches, r = 3.01 
 inches, = 4.45 inches, /== 192 inches, and /= I inch. 
 Then P/A = 10000 pounds per square inch, E = 30 ooo ooo 
 pounds per square inch, /= 151.4 inches 4 . Then the second 
 formula for the deflection gives A = 0.197 inches, and the 
 first formula for 5 gives 16900 pounds per square inch. In 
 this case the eccentric load increases the mean unit-stress 
 about 69 per cent. By the use of the exact formula for A the 
 same numerical values- are also obtained. 
 
 The value of P/A cannot be obtained from the above 
 formula for S, since it is contained in v. By transformation, 
 however, a quadratic equation results from which P/A may 
 be computed when 5 is given. A more convenient way will 
 be to compute 5 for an assumed value of P/A, then insert 
 this assumed value and compute again, and so continue until 
 the computed and assigned values of 5 agree, which they will 
 do when the correct value of P/A has been found. 
 
 Prob. 104. Show that (/+ A)c equals r* for a column so 
 deflected that there is no stress on the concave side. 
 
 Prob. 105. A wooden strut with round ends is 18 feet 
 long, 4 inches square, and has a load of 5000 pounds applied 
 half-way between the center and corner of the cross-section. 
 What is the direction and amount of deflection? 
 
ART. 63. 
 
 THE PHENOMENA OF TORSION. 
 
 135 
 
 Fig. 49- 
 
 CHAPTER VI. 
 TORSION, AND SHAFTS FOR TRANSMITTING POWER. 
 
 ART. 63. THE PHENOMENA OF TORSION. 
 
 Torsion occurs when applied forces tend to cause a twisting 
 of a body around an axis. Let one end of a horizontal shaft 
 be rigidly fixed and 
 let the free end have 
 a lever / attached at 
 right angles to its 
 axis. A weight P 
 hung at the end of 
 this lever will twist 
 the shaft so that fibers 
 such as ab, which were 
 
 originally horizontal, assume a spiral form ad like the strands 
 of a rope. Radial lines such as cb will also have moved through 
 a certain angle &cd. 
 
 Experiments have proved, that if P be not so large as to 
 strain the material beyond its elastic limit, the angles bed and 
 bad are proportional to P and that on the removal of the stress 
 the lines cd and ad return to their original positions cb and ab. 
 The angle bed is evidently proportional to the length of the 
 shaft, while bad is independent of the length. If the elastic 
 limit be exceeded this proportionality does not hold, and if the 
 twisting be great enough the shaft will be ruptured. These 
 laws are but a particular case of the general axioms stated 
 in Art. 3. 
 
 The product Pp is the moment of the force P with respect to 
 the axis of the shaft, / being the perpendicular distance from 
 
136 TORSION AND SHAFTS. CH. VI. 
 
 that axis to the line of direction of P, and is called the twisting 
 moment. Whatever be the number of forces acting at the end 
 of the shaft, their resulting twisting moment may always be 
 represented by a single product Pp. 
 
 A graphical representation of the phenomena of torsion may 
 be made as in Fig. I, 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 ultimate elongation, serves 
 to compare the relative ductility of specimens. 
 
 Prob. 1 06. If a force of 80 pounds at 18 inches from the axis 
 twists a shaft 60, what force will produce the same result when 
 acting at 4 feet from the axis ? 
 
 Prob. 107. 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 4 feet 
 long be twisted by a force of 500 pounds acting at a distance 
 of 1 8 inches from the axis? 
 
 ART. 64. THE FUNDAMENTAL FORMULA FOR TORSION. 
 
 The stresses which occur between any two cross-sections of 
 a bar under torsion are similar to those of shearing, each sec- 
 tion tending to shear off from the one ad- 
 jacent to it. When equilibrium obtains the 
 external twisting moment is exactly bal- 
 anced by the sum of the moments of these 
 resisting internal stresses, or, 
 
 Resisting moment = twisting moment. 
 
 The law governing the distribution of these 
 5. internal stresses is to be taken the same as 
 
 in beams, namely, that they vary directly as the distance from 
 
ART. 64. THE FUNDAMENTAL FORMULA FOR TORSION. 137 
 
 the axis, provided that the elastic limit of the material be not 
 exceeded. 
 
 If Pbe the force acting at a distance/ from the axis about 
 which the twisting takes place, the value of the twisting moment 
 is Pp. To find the resisting moment, let c be the distance from 
 the axis to the remotest part of the cross-section where the 
 unit-shear is 5, . Then since the stresses vary as their distances 
 from the axis, 
 
 - =. unit-stress at a unit's distance from axis, 
 
 S z 
 - = unit-stress at a distance z from axis, 
 
 / 
 
 = total stress on an elementary area a y ' 
 
 = moment of this stress with respect to axis, 
 a ^ = internal resisting moment. 
 
 This may be written 2az\ But 2 at? is the polar moment 
 
 of inertia of the cross-section with respect to the axis, and may 
 be denoted by /. Therefore, 
 
 (ii) 
 
 which is the fundamental formula for torsion. 
 
 The analogy of formula (11) with formula (4) for the flexure 
 of beams will be noted. Pp, the twisting moment, is often the 
 resultant of several forces, and might have been expressed by 
 a single letter like the M in (4). By means of (u) a shaft 
 subjected to a given moment may be investigated, or the 
 proper size be determined for a shaft to resist given forces. 
 
 Prob. 108. Three forces of 120, 90, and 70 pounds act at 
 distances of 6, n, and 8 inches from the axis and at different 
 
138 TORSION AND SHAFTS. CH. VI. 
 
 distances 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. 109. 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 center. What 
 unit-stress will be produced in the same shaft by two forces of 
 40 pounds, one acting at 21 and the other at 36 inches from 
 the center? 
 
 ART. 65. POLAR MOMENTS OF INERTIA. 
 
 The polar moment of inertia for simple figures is readily 
 found by the help of the calculus, as explained in works on 
 elementary mechanics. It is also a fundamental principle 
 that, 
 
 /=/, + /,, 
 
 where / is the polar moment of inertia, 7, the least and 7 2 the 
 greatest rectangular moment of inertia about two axes passing 
 through the center. The following are values of J for some 
 of the most common cases. 
 
 For a circle with a diameter d, J = , 
 
 For a square whose side is d, J =. -^-, 
 
 For a rectangle with sides b and d, / == -J . 
 
 12 12 
 
 The value of c in all cases is the distance from the axis 
 about which the twisting occurs, usually the center of figure 
 of the cross-section, to the remotest part of the cross-section. 
 Thus, 
 
 For a circle with diameter d, c -= %d, 
 
 For a square whose side is d, c = d y^, 
 
 For a rectangle with sides b and d, c = % -\/& -(- d*. 
 
ART. 66. THE CONSTANTS OF TORSION. 139 
 
 It is rare in practice that formulas for torsion are needed for 
 any cross-sections except squares and circles. 
 
 Prob. no. Find the values of J and c for an equilateral tri- 
 angle whose side is d. 
 
 Prob. in. Find, from the data in Art. 30, the values of J 
 and c for a light 6 inch / section. 
 
 ART. 66. THE CONSTANTS OF TORSION. 
 
 The constant 5 S computed from experiments on the rupture 
 of shafts by means of formula (u) may be called the modulus 
 of torsion, in analogy with the modulus of rupture as com 
 puted from (4). The values thus found agree closely with the 
 ultimate shearing unit-stress given in Art. 7, viz., 
 
 For timber, S, = 2 ooo pounds per square inch, 
 
 For cast iron, S t = 25 ooo pounds per square inch, 
 For wrought iron, 5, = 50 ooo pounds per square inch, 
 For steel, S, = 75 ooo pounds per square inch. 
 
 By the use of these average values it is hence easy to compute 
 from (11) the load P acting at the distance / which will cause 
 the rupture of a given shaft. 
 
 The coefficient of elasticity for shearing may be computed 
 from experiments on torsion in the following manner. Let a 
 circular shaft whose length is / and diameter d be twisted 
 through an arc 6 by the twisting moment Pp. Here a point 
 on the circumference of one end is twisted relative to a corre- 
 sponding point on the other end through the arc or through 
 the distance \Qd, so that the detrusion per unit of length is 
 
 _ 
 
 From the fundamental definition of the coefficient of elasticity 
 E t as given in (2), 
 
 _S,_2S,l 
 
 * t? 7 ~8P 
 
140 TORSION AND SHAFTS. CH. VI. 
 
 and inserting for S s its value from (u), there results, 
 
 E __ &Ppl 
 
 ' ntid* ' 
 
 from which E t can be computed when all the quantities in the 
 second member have been determined by experiment, pro- 
 vided that the elastic limit of the material be not exceeded. 
 The numerical value of must here be expressed in terms of 
 the same unit as n. 
 
 Prob. 112. What force P acting at the end of a lever 24 
 inches long will twist asunder a steel shaft 1.4 inches in 
 diameter? 
 
 Prob. 113. An iron shaft 5 feet long and 2 inches in diam- 
 eter is twisted through an angle of 7 degrees by a force of 
 5 ooo pounds acting at 6 inches from the center, and on the re- 
 moval of the force springs back to its original position. Find 
 the value of E for shearing. 
 
 ART. 67. SHAFTS FOR THE TRANSMISSION OF POWER. 
 
 Work is the product of a resistance by the distance through 
 which it acts, and is usually measured in foot-pounds. A horse- 
 power is 33000 foot-pounds of work done in one minute. It 
 is required to determine the relation between the horse-power 
 H transmitted by a shaft and the greatest internal shearing 
 unit-stress 5, produced in it. 
 
 Let a shaft making n revolutions per minute transmit H 
 horse-power .The work may be applied by a belt from the motor 
 to a pulley on the shaft, then, by virtue of the elasticity and re- 
 sistance of the material of the shaft, it is carried through other 
 pulleys and belts to the working machines. In doing this the 
 shaft is strained and twisted, and evidently S, increases with H. 
 Let P be the resistance acting at the circumference of the pulley 
 and/ the radius of the pulley. In making one revolution the 
 
ART. 68. ROUND SHAFTS. 141 
 
 force Pacts through the distance 2np and performs the work 
 2npP, and in n revolutions it performs the work 27tpPn. Then 
 if P be in pounds and / in inches, the imparted horse-power is, 
 
 H _ 2npPn 
 33 ooo X 12 
 
 The twisting moment Pp in this expression may be expressed, 
 as in formula (11), by the resisting moment -^-. Hence the 
 equation becomes, 
 
 (12) H = 
 
 198000^ 
 
 This is the formula for the discussion of shafts for the trans- 
 mission of power, and in it / and c must be taken in inches and 
 5 S in pounds per square inch, while n is the number of revolu- 
 tions per minute. 
 
 Prob. 114. A wooden shaft 6 inches square breaks when 
 making 40 revolutions per minute. Find the horse-power then 
 probably transmitted. 
 
 ART. 68. ROUND SHAFTS. 
 
 For round shafts of diameter d, the values of J and c are to 
 be taken from Art. 65 and inserted in the last equation, giving, 
 
 TT 
 
 5, = 321 ooo--, or 
 na 
 
 The first of these may be used for investigating the strengt- 
 of a given shaft when transmitting a certain number of horse- 
 power with a known velocity. The computed values of S s , 
 compared with the ultimate values in Art. 67, will indicate the 
 degree of security of the shaft. Here d must be taken in 
 inches and S, will be in pounds per square inch. 
 
 The second equation may be used for determining the di- 
 ameter of a shaft to transmit a given horse-power with a given 
 
142 TORSION AND SHAFTS. CH. VI. 
 
 number of revolutions per minute. Here a safe allowable 
 value must be assumed for .S s in pounds per square inch, and 
 then d will be found in inches. This equation shows that the 
 diameter of a shaft varies directly as the cube root of the 
 transmitted horse-power and inversely as the cube root of its 
 velocity. 
 
 Prob. 115. Find the factor of safety for a wrought iron shaft 
 2^ inches in diameter when transmitting 25 horse-power while 
 making 100 revolutions per minute. 
 
 Prob. 116. Find the diameter of a wrought iron shaft to 
 transmit 90 horse-power with a factor of safety of 8 when mak- 
 ing 250 revolutions per minute, and also when making 100 
 revolutions per minute. 
 
 ART. 69. HOLLOW SHAFTS. 
 
 Hollow forged steel shafts are now coming into use for ocean 
 steamers, their strength being greater than solid shafts of the 
 same sectional area. If D be the exterior and d the interior 
 diameter, and A the area of the cross-section, the polar moment 
 of inertia is, 
 
 and the discussion of any case can be made by formula (12), c 
 being replaced by %D. 
 
 For example, let it be required to determine the interior 
 diameter of a nickel-steel shaft, when D = 17 inches, to trans- 
 mit 16 ooo horse power at 50 revolutions per minute, with a 
 stress of 25 ooo pounds per square inch on the exterior circum- 
 ference. Here everything is given except d, and by solution 
 its value is found to be n inches nearly. 
 
 Shafts are subject to flexural stresses due to their own weight 
 and to applied loads, as well as to torsional stresses. The effect 
 of these will be discussed in Art. 76. 
 
ART. 70. MISCELLANEOUS EXERCISES. H3 
 
 Prob. 117. Find the diameter of a solid shaft for the con- 
 ditions of the above example, and compare its weight with that 
 of the hollow one. 
 
 Prob. 1 1 8. Find the horse-power transmitted by a hollow 
 shaft, when D 15! inches, d = o> inches, and S s = 12 500 
 pounds per square inch, the number of revolutions per minute 
 being 50. 
 
 ART. 70. MISCELLANEOUS EXERCISES. 
 
 Exercise 8. Make experiments to verify the phenomena of 
 torsion stated in Art. 63. Show by your experiments that the 
 strength of a round shaft varies directly as the cube of its 
 diameter, and is independent of its length. 
 
 Exercise 9. Make a theoretical investigation to ascertain if 
 the strength of a square shaft can be increased by cutting off 
 material from the corners. If such is found to be the case 
 write an essay explaining the reasoning, the computations and 
 the conclusion. 
 
 Exercise 10. Go to a testing room and inspect THURSTON's 
 testing machine for torsion. Ascertain the dimensions and 
 kind of specimens tested thereon. Explain with sketches the 
 construction of the machine, the method of its use, and the 
 torsion diagrams. State how the quality of the specimens is 
 inferred from the torsion diagrams. 
 
 Prob. 119. Compare the strength of a square shaft with that 
 of a circular shaft of equal area. 
 
 Prob. 1 20. 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 25 ooo 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. 
 
144 COMBINED STRESSES. CH. VII. 
 
 CHAPTER VII. 
 COMBINED STRESSES. 
 
 ART. 71. COMBINED TENSION AND COMPRESSION. 
 
 Tensile and compressive forces acting upon a bar in the 
 direction of its length, produce a resultant stress equal to their 
 numerical difference, which may be either tensile or compres- 
 sive. This case is of frequent occurrence in the members of 
 bridge trusses. 
 
 A tensile force acting upon a bar produces a tensile unit- 
 stress 5 and unit-elongation s. It is found by experiment that 
 the lateral unit-contraction of the bar, when 5 is within the 
 elastic limit, is about -J^, and hence the internal compressive 
 unit-stress normal to the length of the bar is about ^S. Thus 
 internal stresses may exist in a body in directions which do not 
 correspond with any of the applied exterior forces. In general, 
 if s be any unit-deformation, the corresponding internal unit- 
 stress is sE (Art. 4). 
 
 If three tensile forces P l , P^ , and P 3 act normally upon the 
 sides of a rectangular prism whose areas are A l , A^ , and A 3 , the 
 unit-stresses apparently produced upon those sides are S l = 
 P 1 -^A 1) S, = P^^-A,, and 5 3 = P 3 -^ A, ; but the real effective 
 internal unit-stresses are much smaller, their values, by the 
 principle of the last paragraph, being T l = S l J5 5 -J5 3 , 
 T, = S, - to - iS, , and T t = S, - fS, - fS t . These for- 
 mulas apply when some or all of the external forces are com- 
 pressive as well as tensile, if the apparent unit-stresses be taken 
 positive when tension and negative when compression. For 
 example, let a cube whose edge is unity be subject to a 
 
ART. 72. STRESSES DUE TO TEMPERATURE. 145 
 
 compression of 60 pounds upon two opposite sides and to 45 
 pounds tension upon two other opposite sides, the third pair 
 of sides having no forces applied to them. Then the effective 
 internal unit-stress normal to the first pair of sides is 75 pounds 
 compression, that for the second pair is 65 pounds tension, and 
 that for the third pair is 5 pounds tension. 
 
 Prob. 121. A common brick 2X4X8 inches is subject to 
 compression of 3 200 pounds upon its top and bottom faces, 500 
 pounds upon its sides, and 60 pounds upon its ends. Find the 
 effective internal unit-stresses in the three directions. 
 
 ART. 72. STRESSES DUE TO TEMPERATURE. 
 
 If a bar be unstrained it expands when the temperature rises 
 and contracts when the temperature falls. But if the bar be 
 under stress, so that the change of length cannot occur, an ad- 
 ditional unit-stress must be produced which will be equivalent 
 to the unit- stress that would cause the same change of length 
 in the unstrained bar. Thus if a rise of temperature elongates 
 a bar of length unity the amount s when free from stress, it 
 will cause the unit-stress S = sE (see Art. 4) when the bar is 
 prevented from expanding by external forces. 
 
 Let / be the length of the bar, a its coefficient of linear ex- 
 pansion for a change of one degree, and X the change of length 
 due to the rise or fall of / degrees. Then, 
 
 ;t = atl. 
 and the unit-deformation s is, 
 
 A 
 s = -, = at. 
 
 The unit-stress produced by the change in temperature hence is, 
 
 5 = atE 
 
 which is seen to be independent of the length of the bar. The 
 total stress on the bar is then AS. 
 
146 COMBINED STRESSES. CH. VII. 
 
 The following are average values of the coefficients of linear 
 expansion for a change in temperature of one degree Fahren- 
 heit. 
 
 For brick and stone, a = o.ooo oo 50, 
 
 For cast iron, a = o.ooo oo 62, 
 
 For wrought iron, a = o.ooo oo 67, 
 
 For steel, a = o.ooo oo 65. 
 
 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 
 building. Let it be required to find the stress in the rod when 
 the temperature falls 10 F. Here, 
 
 5 = o.ooo oo 67 X 10 X 25 ooo ooo = i 675 pounds. 
 The total tension in the rod now is, 
 
 9 ooo + 3.14 X i 675 = 14 ooo pounds. 
 Should the temperature rise 10 the tension in the rod would 
 
 become, 
 
 9 ooo 3.14 X i 675 = 4000 pounds. 
 
 In all cases the stresses caused by temperature are added or 
 subtracted to the tensile or compressive stresses already 
 existing. 
 
 Prob. 122. A cast iron bar is confined between two immovable 
 walls. What unit-stress will be produced by a rise of 40 in 
 temperature ? 
 
 ART. 73. COMBINED TENSION AND FLEXURE. 
 
 Consider a beam in which the flexure produces a unit-stress 
 S l at the fiber on the tensile side most remote from the neutral 
 axis. Let a tensile stress P be then applied to the ends of the 
 
 bar uniformly distributed over the cross-section A. The ten- 
 
 p 
 sile unit-stress at the neutral surface is then - and all the 
 
 longitudinal stresses due to the flexure are increased by this 
 
ART. 73. COMBINED TENSION AND FLEXURE. 147 
 
 P 
 
 amount. Then 5 = + 5, is an approximate value of the 
 
 A 
 
 maximum tensile unit-stress. 
 
 In designing a beam under combined tension and flexure the 
 
 p 
 
 dimensions must be so chosen that 7 -j- 5, shall not exceed 
 
 si 
 
 the proper allowable working unit-stress. For instance, let it 
 be required to find the size of a square wooden beam of 12 
 feet span to hold a load of 300 pounds at the middle while 
 under a longitudinal stress of 2000 pounds, so that the maxi- 
 mum tensile unit-stress may be about I ooo pounds per square 
 inch. Let d be the side of the square. From formula (4), 
 
 c _ 6J/_ 6 X 150 X 72 
 '" ~- 
 
 Then from the conditions of the problem, 
 2 ooo , 64 800 _ 
 
 ~d^ ~^~ ^ 
 
 from which results the cubic equation, 
 
 d* - 2^=64.8, 
 whose solution gives for d the value 4.25 inches. 
 
 In investigating a beam under combined tension and flexure 
 the maximum combined unit-stress 5 is computed, and the 
 factor of safety found by comparing it with the ultimate 
 tensile strength of the material. The method here given is 
 approximate; more accurate methods are in Art. 118. 
 
 Prob. 123. A heavy 12-inch I beam of 6 feet span carries a 
 uniform load of 200 pounds per linear foot, besides its own 
 weight, and is subjected to a longitudinal tension of 80000 
 pounds. Find the factor of safety of the beam. 
 
 Prob. 124. What I beam of 12 feet span is required to carry 
 a uniform load of 200 pounds per linear foot when subjected 
 to a tension of 50000 pounds, the maximum tensile stress at 
 the dangerous section to be loooo pounds per square inch ? 
 
148 COMBINED STRESSES. CH. VII. 
 
 ART. 74. COMBINED COMPRESSION AND FLEXURE. 
 
 Consider a beam in which the flexure produces a unit-stress 
 .S, in the fiber on the compressive side most remote from the 
 neutral axis. Let a compressive stress P be applied in the direc- 
 tion of its length uniformly over the cross-section A. Then at 
 
 p 
 the neutral surface the unit-stress is and at the remotest 
 
 A 
 p 
 
 fiber it is - + 5 1 ,. The discussion of this case is hence exactly 
 A 
 
 similar to that of the last article. If the beam is short the 
 total working unit-stress is to be taken as for a short prism ; 
 if long it should be derived from RANKINE'S formula for 
 columns. 
 
 The method of investigation explained in this and the pre- 
 ceding article is the one ordinarily used in practice on account 
 of the complexity of the formulas which result from the strict 
 mathematical determination of the moments of the applied 
 forces. Although not exact the method closely approximates 
 to the truth, giving values of the stresses a little too large for 
 the case of tension and a little too small for the case of 
 compression. (See Arts. 117 and 118.) 
 
 A rafter of a roof is a case of combined compression and 
 
 flexure. Let b be its width, 
 d its depth, / the length, 
 w the load per linear unit, 
 and the angle of inclina- 
 tion. To find the horizon- 
 tal reaction H the center 
 of moments is to be taken 
 Flg " 5I< at the lower end, and 
 
 TT , . , j I COS 0, rr Wl 
 
 H , I sin = wl . , whence ff= cot 0. 
 
 2 - 2 
 
ART. 74. COMBINED COMPRESSION AND FLEXURE. 149 
 
 For any section whose distance from the upper end is x, the 
 flexural unit-stress now is from (4), 
 
 c _ 6M _ 6(Hx sin \wx* cos 0) 
 
 ^~W ~~bd^~ 
 
 and the uniform compressive unit-stress is, 
 c _ H cos -f- wx sin 
 
 6> - ~w"~ 
 
 The total compressive unit-stress on the upper fiber hence is, 
 c o o 3W/COS0,, ^ , w/cot0cos0 
 = 5 ' + 5 '~ -~^~ (lX ~- 
 
 bd 
 This can be shown to be a maximum when 
 
 * = i/+f</ tan 0, 
 and substituting this, the maximum unit-stress is, 
 
 _ 3 wl* cos w/cosec wsin tan 
 a " ~"~ 
 
 which formula may be used to investigate or to design rafters 
 subject to uniform loads. 
 
 In any inclined rafter let P denote all the load above a sec- 
 tion distant x from the upper end. Then reasoning as before 
 the greatest unit-stress for that section is found to be, 
 c _ Me .PsmQ.H cos 
 : 7" ~A~ ~A~> 
 from which S x may be computed for any given case. 
 
 Prob. 125. A roof with two equal rafters is 40 feet in span 
 and 1 5 feet in height. The wooden rafters are 4 inches wide 
 and each carries a load of 450 pounds at the center. Find 
 the depth of the rafter so that 5 may be 700 pounds per 
 square inch. 
 
 Prob. 126. A wooden beam 10 inches wide and 8 feet long 
 carries a uniform load of 500 pounds per linear foot and is sub- 
 jected to a longitudinal compression of 40 ooo pounds. Find 
 the depth of the beam so that the maximum working unit-stress 
 may be about 800 pounds per square inch. 
 
ISO 
 
 COMBINED STRESSES. 
 
 CH. VIL 
 
 ART. 75 SHEAR COMBINED WITH TENSION OR COMPRESSION. 
 
 Let a bar whose cross-section is A be subjected to the longi- 
 tudinal tension or compression P and at the same time to a 
 shear Fat right angles to its length. The longitudinal unit- 
 
 p 
 stress is - which may be denoted by/, and the shearing unit- 
 
 A 
 
 stress is - which may be denoted by v. It is required to find 
 
 the maximum unit-stresses produced by the combination of p 
 and v. In the following demonstration P will be regarded as 
 a tensile force, 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 tensile force p and p., 
 and by the vertical shear v and v, as 
 shown in Fig. 52. These forces are not in 
 P y equilibrium unless a horizontal couple be 
 
 > applied as in the figure, each of whose 
 
 forces is equal to v. Therefore at every 
 Fig. 52- point of a body under vertical shear there 
 
 exists a horizontal shear, and the horizontal shearing unit- 
 stress is equal to the vertical shearing unit-stress. 
 
 Let a parallelopipedal element have the length dm, the 
 height dn, and a width of unity. The tensile iorc.Qp.dn tends 
 to pull it apart longitudinally. 
 The vertical shear vdn tends to 
 cause rotation and this is resisted, 
 as shown above, by the horizon- 
 tal shear vdm. These forces 
 maybe resolved into rectangular 
 components parallel and perpen- 
 dicular to the diagonal dz, as shown in Fig. 53. The compo- 
 nents parallel to the diagonal form a shearing force sdz, and 
 
 \ 
 
 v.dn 
 
 dm 
 
 v.dn 
 
 dn 
 
 \ 
 
 t.dz 
 
 Fig. 53. 
 
ART. 75. SHEAR COMBINED WITH TENSION. I$I 
 
 those perpendicular to it a tensile force tdz, s being the shear- 
 
 ing and t the tensile unit-stresses. Let be the angle between 
 
 dz and dm. The problem is first to state expressions for sdz 
 and tdz in terms of 0, and then to determine the value of 0, 
 
 or the ratio of dm to dn, which gives the maximum values 
 of s and t. 
 
 By simple resolution of forces, 
 
 sdz = pdn cos -[~ vdm cos vdn sin 0, 
 /*/ = ^/# sin -|- w/w sin -\- vdn cos 0. 
 
 Divide each of these by dz, for -=- put its value sin and for 
 
 -T- its value cos 0. Then the equations take the form, 
 
 s = p sin cos -|- z/(cos a sin a 0), 
 t =. p sin 2 -|- 2z sin cos 0. 
 
 1'hese may be written, 
 
 s = %p sin 20 -f- z; cos 20, 
 
 / = /(i cos 20) + ?' sin 20. 
 
 By placing the first derivative of each of these equal to zero it 
 is found that, 
 
 P 
 
 s is a maximum when tan 20 = , 
 
 2it 
 
 2V 
 
 t is a maximum when tan 20 = -- . 
 
 / 
 
 Expressing sin 20 and cos 20 in terms of tan 20 and inserting 
 them in the above the following values result : 
 
 jm.,==^+ 
 
 / max. t = irf + */* 
 
 These formulas apply to the discussion of the internal stresses 
 in beams, as well as to combined longitudinal stress and vertical 
 shear directly applied by external forces. If p is tension / is 
 
152 COMBINED STRESSES. CH. VII. 
 
 tension, if/ is compression / is also compression. If when/ is 
 tension the negative sign be used before the radical, the re- 
 sultant value of t is the maximum compressive unit-stress. 
 
 Prob. 127. A bolt f-inch in diameter is subjected to a tension 
 of 2 ooo pounds and at the same time to a cross shear of 3 ooo 
 pounds. Find the maximum tensile and shearing unit-stresses 
 and the directions they make with the axis of the bolt. 
 
 ART. 76. COMBINED FLEXURE AND TORSION. 
 
 This case occurs when a shaft for the transmission of power 
 is loaded with weights. Let 5 be the greatest flexural unit- 
 stress computed from (4) and S s the torsional shearing unit- 
 stress computed from (12) or by the special equations of Arts. 
 67 and 68. Then, according to the last article, the resultant 
 maximum unit-stresses are, 
 
 max. ten. or comp. / -J5+ y 5/ 
 max. shear s = 
 
 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 factor of safety of 
 a wrought iron 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 flexural unit-stress is, 
 
 5= -js = 5 40 pounds per square inch. 
 Ttd 
 
 From Art. 68 the torsional unit-stress is, 
 
 IT 
 
 S s = 321 ooo --j- 3 = 4 ooo pounds per square inch. 
 
ART. /6. COMBINED FLEXURE AND TORSION. 153 
 
 The maximum tensile and compressive unit-stress now is, 
 
 / = 2 700 4- ,y/4 ooo 2 +~2~70o a = 7 600 pounds per square in. 
 and the factor of safety is hence over 7. 
 
 As a second example, let it be required to find the size of 
 a square wooden shaft for a water-wheel weighing 3 ooo pounds 
 which transmits 8 horse-power while making 20 revolutions per 
 minute. The length of the shaft is 16 feet, and one-third of 
 the weight is concentrated at the center and the remainder is 
 equally divided between two points, each 6 feet from the center. 
 Here the greatest flexural unit-stress is, 
 
 _ 6(1 500 X 96 i ooo X 72) _ 432000 
 ~~d*~ d* ' 
 
 and from Art. 67 the torsional unit-stress is, 
 _ 267 500 X 8 _ 107000 
 
 J, 7". 7T . 
 
 2od d 
 
 From the formula of the last Article the combined tensile or 
 compressive unit-stress is, 
 
 470400 
 d 3 
 
 Now if the working value of / be taken at 600 pounds per 
 square inch the value of d will be about 9 inches. From for- 
 mula (13) also 
 
 s= 254400 
 
 and if the working value of s be taken at 1 50, the value of d, is 
 found to be about 12 inches. The latter value should hence 
 be chosen for the size of the shaft. 
 
 ,. By similar reasoning it may be proved that the formula for 
 finding the diameter of a round iron shaft is, 
 
 \6M 16 I M* 402 500 ooo//' 8 
 
154 COMBINED STRESSES. CH. VIL 
 
 where M is the maximum bending moment of the transverse 
 forces in pound-inches, H the number of transmitted horse- 
 power, n the number of revolutions per minute, and / the 
 safe allowable tensile or compressive working strength of the 
 material. 
 
 Prob. 128. Find the factor of safety for the data of Prob. 
 115 when the shaft is in bearings 12 feet apart and carries a 
 load of 200 pounds at the middle. 
 
 ART. 77. COMBINED COMPRESSION AND TORSION. 
 
 In the case of a vertical shaft the torsional unit-stress 5, com- 
 bines with the direct compressive stress due to the weights 
 upon the shaft, and produces a resultant compression t and 
 shear s. From formulas (13) the combined unit-stresses are, 
 
 s ~ v S* + i-S; 1 . 
 
 The use of these is the same as those of the last Article, S 9 
 being found from the formulas of Chapter VI, while S c is com- 
 puted from formula (i) if the length of the shaft be less than 
 ten times its diameter and from (10) for greater lengths. 
 
 In order to prevent vibration and flexure it is usual to place 
 bearings at frequent intervals on a vertical shaft so that prob- 
 ably the use of formula (10) will rarely be required, particularly 
 if / be taken at a low value. For a round shaft the expression 
 for t becomes, 
 
 / = -' 7 ~ ~^' 
 
 in which P is the load. From this the diameter d may be found 
 when / and the other data are given. 
 
 Prob. 129. A vertical shaft, weighing with its loads 6000 
 
ART. 78. HORIZONTAL SHEAR IN BEAMS. 1 55 
 
 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. 130. 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 unit-stress against compression as 10000 and 
 against shearing as 7 ooo pounds per square inch. 
 
 ART. 78. HORIZONTAL SHEAR IN BEAMS. 
 
 The common theory of flexure as presented in Chapters III 
 and IV considers that the internal stresses at any section are 
 resolved into their horizontal and vertical components, the 
 former producing longitudinal tension and compression and 
 the latter a transverse shear, and that these act independently 
 of each other. Formula (3) 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 deflects more than a solid timber of the 
 same depth, and this is largely due to the lack of horizontal 
 resistance between the layers. The common theory of flexure 
 in neglecting the horizontal shear generally errs on the side of 
 safety. In a few 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 simulta- 
 neously with the vertical shear is evident from the considera- 
 tions in Art. 75. 
 
 Let Fig. 54 represent a portion of a bent beam of uniform 
 
156 
 
 COMBINED STRESSES. 
 
 CH. VII. 
 
 -r " v y 
 
 *% ^' 
 
 t 
 \ 
 
 m 
 
 m' 
 
 -*. ^ 
 
 1 
 
 i 
 
 fr- 
 
 
 ~- 
 
 
 
 
 section. Let a rectangular notch nmpq be imagined to be cut 
 into it, and let forces be applied to it to preserve the equilib- 
 rium. Let H be the sum of all the horizontal components of 
 
 these forces act- 
 ing on mn and H' 
 the sum of those 
 acting on qp. 
 Now H' is greater 
 or less than H, 
 Fig ' 54> hence the differ- 
 
 ence H' H must act along mq as a horizontal shear. Let the 
 distance mq be dx, the thickness mm! be , and the area mqmm' 
 be at a distance c' above the neutral surface. Let c be the dis- 
 tance from that neutral surface to the remotest fiber where the 
 unit-stress is 5. Let a be the cross-section of any fiber. Let 
 M be the bending moment at the section mn and M' that at 
 the section qp. Now from the fundamental laws of flexure, 
 
 - = unit-stress at a unit's distance from neutral surface, 
 c 
 
 c- 
 
 - y unit-stress at distance y from neutral surface, 
 
 aSy _ 
 c 
 
 = total stress on fiber a at distance y, 
 
 = sum f horizontal stresses between m and . 
 
 M 
 
 The value of H hence is, since - -- , 
 
 and likewise for the other section, 
 
ART. 78. HORIZONTAL SHEAR IN BEAMS. 157 
 
 The horizontal shear therefore is expressed by 
 
 M r -M 
 H' -H=- - ^ay. 
 
 Now since the distance mq is dx, the value of M* M is dM. 
 Also if S k be the horizontal shearing unit-stress upon the area 
 bdx the value of H 1 H is SJbdx. Hence, 
 
 dM 
 
 Again from Art. 45 it is plain that - - is the vertical shear V 
 at the section under consideration. Therefore, 
 
 (14) S h = --^a yi 
 
 lo 
 
 is the formula for the horizontal shearing unit-stress at any 
 point of any section of the beam. 
 
 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 is the statical 
 moment of the area mm'nn' with reference to the neutral axis ; 
 it is zero when cf = c, and a maximum when c' = 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. The 
 formula for the maximum horizontal shearing unit-stress at any 
 section therefore is, 
 
 Here / is the moment of inertia of the whole cross-section with 
 reference to the neutral axis (Art. 23), b is the width of the 
 beam along the neutral surface, and ^ C ay\s the statical mo- 
 ment of the area of the part of the cross-section on one side of 
 the neutral axis. Let A l be the area of the cross-section on 
 
158 COMBINED STRESSES. CH. VII. 
 
 one side of the neutral axis and c l the distance of its center of 
 gravity from that axis; then 2 c ay = A^, and the formula 
 
 becomes, 
 
 VA c A , - 
 (I 4 y S.= -j', 
 
 k i u*rvWH\ A^Mq O 
 
 which gives the maximum shearing unit-stress, both horizontal 
 and vertical, at the neutral surface. The mean unit-stress given 
 by (3) is always less than this maximum. 
 
 For a rectangular beam of breadth b and depth d, the valu* 
 
 . bd* bd d bd* _, 
 
 of / is , and A.c. = . = . Then, 
 
 12 248 
 
 By inserting in this the value of V for any section the co*re- 
 sponding value of 5, at the neutral surface is found. In this 
 particular instance it is seen that the approximate formula (3) 
 gives values of 5 S which are 33 per cent lower than the true 
 maximum value. 
 
 Prob. 131. In the Journal of the Franklin Institute for Feb- 
 ruary, 1883, is detailed an experiment on a spruce joist 3j X 12 
 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 ten- 
 sile and shearing unit-stresses. 
 
 
 ART. 79. MAXIMUM INTERNAL STRESSES IN BEAMS. 
 
 From the last Article it is evident that at every point of a 
 beam there exists a horizontal unit-shear of the intensity S h and 
 also a vertical unit-shear of the same intensity, whose value is 
 given by (14). At every point there also exists a longitudinal 
 tension or compression which may be computed from (4) with 
 the aid of the principle that these stresses vary directly as their 
 
ART. 79. MAXIMUM INTERNAL STRESSES IN BEAMS. 159 
 
 distances from the neutral axis. Let v denote the unit-shear 
 thus determined and / the tensile or compressive unit-stress. 
 Then from Art. 75 the maximum unit-shear at that point is, 
 
 and it makes an angle with the neutral surface such that, 
 
 tan 2<t>=^_. 
 
 2V 
 
 Also the maximum tensile or compressive unit-stress at that 
 point is, 
 
 and it makes an angle with the neutral surface such that, 
 
 tan 20=-^. 
 P 
 
 From these formulas the lines of direction of the maximum 
 stresses may be traced throughout the beam. 
 
 For the maximum shear v is greatest and p is zero at the 
 neutral surface, while v is zero and p is greatest at the upper 
 and lower surfaces. Hence for the neutral surface is o, it 
 increases with p, and becomes 45 at the upper and lower 
 
 surfaces. 
 
 For the maximum tension t is greatest and equal to p on 
 the convex side where v = o and 6 = o. As the neutral sur- 
 face is approached v increases, / decreases, and increases. 
 At the neutral surface v is greatest, / is zero, and = 45. 
 Here the maximum tension and compression are each equal 
 to v. 
 
 For the maximum compression in like manner 6 is o at the 
 concave surface and 45 at the neutral surface. The lines of 
 maximum tension if produced beyond the neutral surface would 
 evidently cut those of maximum compression at right angles 
 and be vertical at the concave surface. 
 
l6o COMBINED STRESSES. CH. VII. 
 
 The following figure is an attempt to represent the lines 
 which indicate the directions of the maximum unit-stress in a 
 
 beam. The full lines 
 above the neutral 
 surface are those of 
 maximum compres- 
 sion, while those be- 
 low are maximum 
 tension. The broken 
 lines are those of 
 Flg ' 55 ' maximum shear. On 
 
 any line the intensity of stress varies with the inclination, being 
 greatest where the line is horizontal and least where its inclina- 
 tion is 45. The lines of maximum shear cut those of maxi- 
 mum tension and compression at angles of 45. The lines of 
 maximum tension above the neutral surface and those of maxi- 
 mum compression below it are not shown ; if drawn they would 
 cut the others at right angles and become vertical at the upper 
 and lower edges of the beam. 
 
 It appears from the investigation that the common theory of 
 flexure gives the horizontal unit-stress correctly at the dan- 
 gerous section of a simple beam where the vertical shear is 
 zero. At other sections the stress S as computed from (4) is 
 correct for the remotest fiber, but for other fibers the unit-stress 
 t is greater. It is hence seen that the main practical value of 
 the theory of internal stress is in showing that the intensity of 
 the shear varies throughout the cross-section of the beam. For 
 a restrained beam, where the vertical shear suddenly changes 
 sign at the dangerous section, the common theory gives the 
 horizontal stress 5 correctly for the remotest fiber only, and 
 it might be possible in some forms of cross-sections for the 
 maximum stress / to be slightly greater than S for a fiber 
 nearer to the neutral surface. All that has here been deduced 
 
ART. 79. MAXIMUM INTERNAL STRESSES IN BEAMS. l6l 
 
 justifies the validity of the common theory of flexure as a 
 correct guide in the practical design and investigation of 
 beams. 
 
 Prob. 132. A joist fixed at both ends is 3 X 12 inches and 
 12 feet long, and is strained by a load at the middle, so that 
 the value of 5 as computed from (4) is 4000 pounds per square 
 inch. Find the value of t for points over the support distant 
 3, 4, and 5 inches from the neutral surface. 
 
 Prob. 133. Show, fora point between the neutral surface and 
 the convex side, that there exists a maximum compression as 
 well as a maximum tension. Deduce an expression for the 
 value of this maximum compression and its direction. Draw 
 a figure showing the curves over the entire beam for both these 
 stresses. 
 
1 62 THE STRENGTH OF MATERIALS. CH. VIII, 
 
 CHAPTER VIII. 
 THE STRENGTH OF MATERIALS. 
 
 ART. 80. MEAN CONSTANTS. 
 
 The term ' strength of materials ' is generally understood 
 to refer to stresses caused by slowly applied loads, where the 
 conditions are those of statics only. The ultimate tensile 
 strength of timber, for example, is obtained by placing a speci- 
 men, like that shown in Fig. 2 of Art. 8, in a testing machine 
 and pulling it by a force which gradually increases until rup- 
 ture occurs. When forces are applied suddenly or with shock, 
 the condition is one of work or resilience (see Chapter IX). 
 
 The following tables recapitulate the mean values of the con- 
 stants of the strength of materials which have been given in 
 the preceding pages. It is here again repeated that these 
 values are subject to wide variations dependent on the kind 
 and quality of the material, and for many other reasons. Tim- 
 ber, for instance, varies in strength according to the climate 
 where grown, the soil, the age of the tree, the season of the 
 year when cut, the method and duration of the process of 
 seasoning, the part of the tree used, the knots and wind shakes, 
 the form and size of the test specimen, and the direction of its 
 fibers, so that it is a difficult matter to state definite numerical 
 values concerning its elasticity and strength. The quality of 
 the material causes a yet wider variation, so wide in fact that 
 in some cases testing machines alone could scarcely distinguish 
 between wrought iron and steel ; for while the higher grades of 
 steel have much greater strength than the tables give, the mild 
 structural and merchant steels may have values almost as low 
 
ART. 80. 
 
 MEAN CONSTANTS. 
 
 as the average constants for wrought iron. In general, there- 
 fore, the following values should not be used in actual cases of 
 investigation and design except for approximate computations. 
 
 Detailed tables giving the results of experiments upon 
 numerous kinds and qualities of materials may be found in the 
 various engineers' pocket books, in THURSTON'S Materials of 
 Engineering (New York, 1884), in BURR'S Elasticity and 
 Strength of Materials (New York, 1888), in JOHNSON'S 
 Materials of Construction (New York, 1897), and in the works 
 noted at the end of the next article. It is, however, impos- 
 sible to ascertain the exact strength of any particular lot of 
 material by reference to books, but tests must be made. 
 
 TABLE I. 
 
 
 Mean Weight. 
 
 Coefficient of Linear Expansion. 
 
 Material. 
 
 Pounds per 
 cubic foot. 
 
 Kilograms per 
 cubic meter. 
 
 For i Fah. 
 
 For i Cent. 
 
 Timber, 
 
 40 
 
 600 
 
 0.0000020 
 
 0.0000036 
 
 Brick, 
 
 125 
 
 2 000 
 
 0.0000050 
 
 0.0000090 
 
 Stone, 
 
 160 
 
 2 560 
 
 0.0000050 
 
 0.0000090 
 
 Cast Iron, 
 
 450 
 
 7 200 
 
 0.0000062 
 
 O.OOOOII2 
 
 Wrought Iron, 
 
 480 
 
 7 700 
 
 0.0000067 
 
 O.OOOOI2I 
 
 Steel, 
 
 490 
 
 7 800 
 
 0.0000065 
 
 O.OOOOII7 
 
 TABLE II. 
 
 Material. 
 
 Elastic Limit. 
 
 Coefficient of Elasticity. 
 
 Pounds per 
 square inch. 
 
 Kilograms 
 per square 
 centimeter. 
 
 Pounds per 
 square inch. 
 
 Kilograms per 
 square centimeter. 
 
 Timber, 
 
 3 000 
 
 210 
 
 I 500000 
 
 105 ooo 
 
 Cast Iron, 
 
 6 000 
 
 420 
 
 15 000000 
 
 i 050 ooo 
 
 Wrought Iron, 
 
 25 ooo 
 
 1750 
 
 25 oooooo 
 
 I 750000 
 
 Steel, 
 
 50000 
 
 3500 
 
 30000000 
 
 2 100 000 
 
104. 
 
 THE STRENGTH OF MATERIALS 
 
 TABLE III. 
 
 CH. VIII 
 
 Material. 
 
 Ultimate Tensile Strength. 
 
 Ultimate Compressive Strength. 
 
 Pounds per 
 square inch. 
 
 Kilograms 
 per square 
 centimeter. 
 
 Pounds per 
 square inch. 
 
 Kilograms 
 per square 
 centimeter. 
 
 Timber, 
 
 10 000 
 
 700 
 
 8 000 
 
 560 
 
 Brick, 
 
 200 
 
 14 
 
 2 500 
 
 175 
 
 Stone, 
 
 
 
 6 000 
 
 420 
 
 Cast Iron, 
 
 2O OOO 
 
 I 400 
 
 go ooo 
 
 6 300 
 
 Wrought Iron, 
 
 55 ooo 
 
 3850 
 
 55 ooo 
 
 3850 
 
 Steel, 
 
 100 000 
 
 7 ooo 
 
 150 ooo 
 
 10 500 
 
 TABLE IV. 
 
 
 Ultimate Shearing Strength. 
 
 Modulus of Rupture. 
 
 Material. 
 
 Pounds per 
 square inch. 
 
 Kilograms 
 per square 
 centimeter. 
 
 Pounds per 
 square inch. 
 
 Kilograms 
 per square 
 centimeter. 
 
 Timber, 
 
 ( 600 ) 
 ( 3 ooo f 
 
 \ 42 I 
 
 ( 210 ) 
 
 9 ooo 
 
 630 
 
 Stone, 
 
 
 
 2 000 
 
 140 
 
 Cast Iron, 
 
 20 000 
 
 I 4OO 
 
 35 ooo 
 
 2 450 
 
 Wrought Iron, 
 
 50 ooo 
 
 3 500 
 
 
 
 Steel, 
 
 70000 
 
 4900 
 
 
 
 ART. 81. HISTORICAL. 
 
 Some topics in the mechanics of materials were discussed 
 and partially developed in advance of precise knowledge re- 
 garding elastic resistance and ultimate strength. The first 
 investigations were those by GALILEI in 1638 on the flexure 
 of beams and on forms of uniform strength. In 1678 HOOKE 
 announced the law " ut tensio sic vis," namely, that the elon- 
 gation of a spring is proportional to the force which causes it ; 
 three years earlier he had performed experiments in the pres- 
 ence of the king of England illustrating the law, which indeed 
 
ART. 81. HISTORICAL. 165 
 
 he had published in 1660 concealed in the anagram 
 " ceiilnosssttu v." 
 
 During the eighteenth century but slight advances were 
 made in practical knowledge, although the theory of flexure 
 was improved by BERNOULLI, LEIBNITZ, COULOMB, EULER 
 and others. The few experimental results during this century 
 were on the rupture of timber by flexure and by tension, ques- 
 tions of ultimate strength being only investigated while that of 
 elastic limit was scarcely recognized. 
 
 Early in the nineteenth century appeared the * Lectures on 
 Natural Philosophy ' by YOUNG in which is found a clear pres- 
 entation of many of the laws of flexure both under static 
 forces and under shock. It also introduces for the first time 
 the coefficient or modulus of elasticity, , but fails to note that 
 it can only be deduced or applied when the elastic limit of the 
 material is not surpassed. A little later BARLOW, TREDGOLD, 
 and HODGKINSON experimented on timber and cast iron, both 
 in the form of beams and of columns ; their methods and re- 
 sults, although now seeming rude and defective, are deserving 
 of praise as the first of real practical value. 
 
 The complete theory of the flexure of beams and the equa- 
 tions of the elastic curve are due to NAVIER, who from 1820 
 to 1833 established the modern mathematical theory of elas- 
 ticity on a solid foundation, which by his followers LAME, ST. 
 VENANT and BOUSSINESQ has been applied to very complex 
 problems, a full account of which is given in TODHUNTER and 
 PIERSON'S History of the Elasticity and Strength of Materials 
 (3 volumes, 1886-1892). 
 
 In 1849 was published the ' Report of the Commissioners on 
 the Application of Iron to Railway Structures/ which may be 
 regarded as the landmark of the beginning of modern methods 
 of testing. It contains the records of valuable tests by WILLIS, 
 JAMES, HODGKINSON and GALTON on the strength of cast 
 
l66 THE STRENGTH OF MATERIALS. CH. VIII. 
 
 and wrought iron as well as upon the resistance to impact, 
 investigations of the increase in stress caused by a rolling load 
 on a beam, experiments on the fatigue of metals, and the evi- 
 dence given by leading British engineers as to their opinions 
 on proper factors of safety under different conditions. The 
 immediate result of this report was the decision by the English 
 board of trade that the factor of safety for cast, iron should 
 be twice as great for rolling loads as for steady ones, while 
 throughout both Europe and the United States it excited a 
 marked interest and impetus in the subject of testing materials. 
 
 A volume would be required to outline the progress and the 
 results of the experimental work done since 1850. The main 
 conclusions will be noted in subsequent articles, and many 
 others will be found in the books noted in Art. 80. The fol- 
 lowing additional references to works of the principal experi- 
 menters will enable students to consult original authorities as 
 well. It should be noted, however, that very important contri- 
 butions have appeared in technical periodicals and in the trans- 
 actions of engineering societies ; for the titles of many of these 
 see * Descriptive Index of Engineering Literature,' published 
 in 1892, 1896, and 1901. 
 
 WADE, W., and RODMAN, T. J.: Reports of Experiments 
 on the Strength and other Properties of Metals for Cannon. 
 Two volumes, quarto: Philadelphia, 1856, pp. 482; Boston, 
 1 86 1, pp. 308. 
 
 KIRKALDY, D.: Experiments on Wrought Iron and Steel. 
 Second edition, London, 1861, octavo, pp. 227, plates xvi. 
 
 FAIRBAIRN, W.: An Experimental Inquiry into the 
 Strength, Elasticity, Ductility and other Properties of Steel. 
 London, 1869, octavo, pp. 51. 
 
 STYFFE, K.: The Elasticity, Extensibility and Tensile 
 Strength of Iron and Steel. London, 1869, octavo, pp. 171, 
 plates ix. 
 
 BAUSCHINGER, J.: Mittheilungen aus dem mechanisch- 
 
ART. 82. TESTING MACHINES. l/ 
 
 technischen Laboratorium der polytechnischen Schule in 
 Munchen. Munich, 1873-1890, folio; usually published annu- 
 ally, each part dealing with a special subject. 
 
 GILMORE, Q. A.: The Compressive Resistance of Free- 
 stone, Brick Piers, Hydraulic Cements, Mortars and Concretes. 
 New York, 1888, octavo, pp. 198, plates viii. 
 
 Reports of the U. S. Board appointed to Test Iron, Steel, 
 and other Metals. Washington, two volumes, 1878 and 1881, 
 octavo, pp. 592 and 684, with many plates. 
 
 Bulletins of the Forestry Division of the U. S. Department 
 of Agriculture since 1892. Also, Report of Tenth Census on 
 Forest Trees of North America. 
 
 U. S. Ordnance Bureau : Tests of Metals. Annual records 
 of tests at Watertown Arsenal since 1883. 
 
 Baumaterialienkunde. A semi-monthly journal published 
 at Stuttgart since 1896; the official organ of the International 
 Association for Testing Materials. 
 
 Prob. 134. Consult Engineering News, Aug. 17, 1899, and 
 ascertain the history and objects of the International Associa- 
 tion for Testing Materials. 
 
 ART. 82. TESTING MACHINES. 
 
 Tests on the flexure and rupture of beams were made in the 
 seventeenth century, and machines for this purpose are com- 
 paratively simple. The load is usually applied at the middle 
 of a beam supported at its ends and gradually increased until 
 rupture occurs, the deflection being measured also as a test of 
 stiffness. The apparatus for determining the deflection should 
 be attached to the supports so that the compression of these 
 may not affect the observed values. In the simplest case 
 weights are hung upon a ring or stirrup placed upon the mid- 
 dle of the beam, and are added in regular increments of 100 
 pounds, more or less, depending upon the size of the beam. 
 When the elastic limit is not exceeded the coefficient of elas- 
 ticity E may be computed from the observed deflection by the 
 
l68 THE STRENGTH OF MATERIALS. CH. VIII. 
 
 formula in Case I of Art. 35. From the breaking load P the 
 modulus of rupture S r may be computed by the formula (4) 
 of Art. 21 ; the value of this lies between the ultimate tensile 
 and compressive strengths of the material. As a method of 
 comparison of different qualities of materials this test is an 
 excellent one. 
 
 Machines for tensile tests are usually arranged so as to 
 operate on a specimen of the shape shown in Fig. 2 of Art. 8. 
 The heads are either griped in jaws or are provided with 
 threads so that they may be screwed into nuts to which the 
 tension is applied. The power may be furnished by a lever in 
 machines of small capacity, in others by a screw or by hydrau- 
 lic pressure, the total tension being weighed off on a compound 
 lever. In commercial tests the ultimate elongation is alone 
 measured ; this is done by marking inch spaces on the speci- 
 men and measuring them after rupture. In scientific tests an 
 extensimeter is attached to the specimen so that the elonga- 
 tion can be read at each increment of weight. The elongation 
 is usually expressed as a percentage of the original length 
 between two marks whose distance apart is more than eight 
 times the diameter of the specimen. In the case of ductile 
 metals a marked diminution in diameter occurs at the point of 
 rupture, and the two parts of the specimen are seen to have a 
 taper on each side of the fracture. On this account the per- 
 centage of ultimate elongation will depend upon the distance 
 between the two marks. As no standard proportions have yet 
 been adopted, it is desirable that the actual distance on which 
 the elongation is measured should be always stated. 
 
 In ductile materials, like wrought iron and mild steel, the 
 final strength is less than the maximum strength, owing to the 
 rapid flow of metal which is the cause of the taper and con- 
 traction. The maximum strength is usually alone recorded, 
 and in this volume the term ' ultimate strength ' is to be un- 
 
ART. 82. TESTING MACHINES. 169 
 
 derstood, not as that at the instant of rupture, but as the maxi- 
 mum unit-stress observed shortly before rupture. 
 
 The contraction of area, introduced by KlRKALDY as an 
 element to be noted in tests of ductile metals, is now regarded 
 as of equal importance with the ultimate elongation, since it is 
 subject to less variation with the length of the specimen. This 
 is expressed as a percentage of the original area of the cross- 
 section. According to KlRKALDY, the tensile strength and 
 ultimate contraction of area, when considered jointly, furnish 
 the best index for judging the quality of wrought iron and 
 steel. 
 
 In all these tensile tests the load is applied gradually, and 
 not suddenly or with impact. A test, however, may be made 
 slowly or quickly, and it is found that the degree of rapid- 
 ity has a marked influence upon the results. In general, a 
 quick test gives a higher elastic limit and a higher strength 
 than a slow one, while the ultimate elongation is less. Attempts 
 have hence been made to specify the degree of rapidity with 
 which the test should be conducted, and undoubtedly some 
 uniformity in this respect will in time be required in standard 
 specifications. (Art. 88.) 
 
 Compressive tests are mainly confined to brick and stone, 
 and are but little used for metals on account of the difficulty 
 of securing a uniform distribution of stress over the surfaces. 
 Rupture in these cases rarely occurs by true crushing, but by 
 a diagonal shearing or splitting, and it is not easy to obtain 
 precise measures of the amount of shortening. Cement, which 
 is always used in compression; is indeed usually tested by ten- 
 sion, and this is found to be the cheaper and more satisfactory 
 method. 
 
 The first testing machines in the United States were those 
 built by WADE and RODMAN between 1850 and 1860 for test- 
 ing gun-metal for the government. About this time the 
 
1/0 THE STRENGTH OF MATERIALS. CH. VIIL 
 
 rapid introduction of iron bridges led to experiments by 
 PLYMPTON and by ROEBLING. About 1865 machines were 
 made by FAIRBANKS, which have since been developed into 
 forms applicable to all classes of work, as also have those made 
 by OLSEN and by RIEHLE. A machine devised by THURSTON 
 soon after 1870 has done excellent work on stresses of torsion. 
 The machine built by EMERY for the United States Board of 
 1878-81, and now in the Watertown arsenal, is the most pre- 
 cise one ever devised, and its work has been of great value in 
 advancing our knowledge of materials. Large machines for 
 testing eye-bars have been built by bridge companies since 
 1880, and several testing laboratories now exist provided with 
 machines for every kind of work. 
 
 The capacity of a testing machine is the tension or pressure 
 which it can exert. A small machine for testing cement need 
 not have a capacity greater than 1000 pounds. Machines of 
 50000, looooo, and 150000 pounds for testing metals are com- 
 mon. The Watertown machine has a capacity of I ooo ooo 
 pounds and can test a heavy bar 30 feet long and a small hair 
 with equal precision. The eye-bar machine at Athens, Pa., 
 has a capacity of I 244 ooo pounds, and that at Phcenixville, 
 Pa., a capacity of 2 160000 pounds. 
 
 For descriptions of the principal testing machines the stu- 
 dent should consult ABBOTT'S papers in Van Nostrand's 
 Magazine, Vol. XXX, reprinted as Van Nostrand's Science 
 Series, No. 74. KENT'S Strength of Materials, No. 41 in that 
 Series, contains also valuable information and discussions of 
 methods of testing. JOHNSON'S Materials of Construction, 
 1897, and MARTENS' Handbook of Testing Materials, .1899, 
 treat fully of testing machines and of the methods of making 
 tests so that the results shall be reliable. 
 
 Prob. 135. A steel eye-bar, 10 X 2f inches, tested at Phce- 
 nixville in 1 893, broke under a tension of I 626 322 pounds, with 
 
ART. 83. 
 
 TIMBER. 
 
 171 
 
 20.5 per cent elongation in 47 feet, and a reduction of area of 
 50.4 per cent. Compute the ultimate tensile strength per 
 square inch of original area, and also per square inch of frac- 
 tured area. 
 
 ART. 83. TIMBER. 
 
 Good timber is of uniform color and texture, free from 
 knots, sapwood, 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 is al- 
 ways greatest in the direction of the grain, the sidewise resist- 
 ance to tension or compression being scarcely one fourth of 
 the longitudinal. 
 
 The following table gives average values of the ultimate 
 strength in pounds per square inch of a few of the common 
 
 Kind. 
 
 Pounds 
 per 
 Cubic Foot. 
 
 Tensile 
 Strength. 
 
 Modulus 
 of 
 Rupture. 
 
 Compress- 
 ive 
 Strength. 
 
 Hemlock 
 
 2C 
 
 8 ooo 
 
 6000 
 
 5 ooo 
 
 White Pine 
 
 27 
 
 AQ 
 
 8000 
 
 12 OOO 
 
 60OO 
 7 OOO 
 
 5500 
 
 e OOO 
 
 Red Oak 
 
 42 
 
 Q OOO 
 
 7 OOO 
 
 6 OOO 
 
 Yellow Pine... 
 White Oak 
 
 45 
 
 48 
 
 I50OO 
 12 OOO 
 
 IIOOO 
 10 OOO 
 
 9OOO 
 8000 
 
 kinds of timber as determined from tests of small specimen 
 carefully selected and dried. Large pieces of timber such ai 
 are actually used in engineering structures will probably have 
 an ultimate strength of from fifty to eighty per cent of these 
 values. Moreover the figures are liable to a range of 25 per 
 cent on account of variations in quality and condition arising 
 from place of growth, time when cut, and method and duration 
 
172 THE STRENGTH OF MATERIALS. Ctl. VIII. 
 
 of seasoning. To cover these variations the factor of safety of 
 10 is not too high. 
 
 The shearing strength of timber is still more variable than 
 the tensile or compressive resistance. White pine across the 
 grain may be put at 2500 pounds per square inch, and along 
 the grain at 500. Chestnut has 1500 and 600 respectively, 
 yellow pine and oak perhaps 4 ooo and 600 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 coefficient 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 computa- 
 tions. The ultimate elongation is small, usually being between 
 I and 2 per cent. 
 
 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 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 compressive strength of 200 pounds per 
 square inch. 
 
 ART. 84. BRICK. 
 
 Brick is made of clay which consists mainly of silicate of 
 alumina with compounds of lime, magnesia, and iron. The clay 
 is prepared by cleaning it carefully from pebbles and sand, 
 mixing it with about one half its volume of water, a'nd temper- 
 ing it by hand stirring or in a pug-mill. It is then moulded in 
 
ART. 84. BRICK. 173 
 
 rectangular boxes by hand or by special machines, and the 
 green bricks are placed under open sheds to dry. These are 
 piled in a kiln and heated for nearly two weeks 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 en- 
 sure toughness and hardness. 
 
 The common size is 2 X 4 X 8J 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 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 per cent. 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 crushing strength of brick is variable ; while a mean 
 value may be 2500 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 15000 pounds per 
 square inch, or even more. Crushing tests are difficult and ex- 
 pensive to make on account of the labor of preparing the speci- 
 men cubes so as to secure surfaces truly parallel. 
 
1/4 THE STRENGTH OF MATERIALS. CH. VIII. 
 
 A flexural test for brick is often used. The brick is sup- 
 ported at the middle of the flat side upon a fixed steel support 
 about one half inch in width, and around its ends are placed 
 stirrups to which is hung a vessel for holding weights. The 
 loads being applied until rupture occurs, the modulus of rup- 
 ture can be computed from the formula 
 
 in which / is the length of the brick between stirrups, b its 
 breadth, d its thickness, and W the breaking load. For ex- 
 ample, if / = 8 inches, b 4 inches, d 2 inches, and W = 
 1000 pounds, then S r = 750 pounds per square inch. This 
 modulus of rupture is intermediate between the ultimate ten- 
 sile and compressive strengths, and although not a physical 
 constant it serves the means of making excellent comparisons 
 of different qualities of brick. 
 
 Tensile and shearing tests of bricks are rarely made and but 
 little is known of their behavior under such stresses. The ulti- 
 mate tensile strength may perhaps range from 50 to 500 pounds 
 per square inch. As for elastic limit and coefficient of elas- 
 ticity, so little is known that nothing can be said. 
 
 ART. 85. CEMENT AND MORTAR. 
 
 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 to 300 pounds per square inch. Its strength slowly in- 
 creases with age, and it may be considerably increased by using 
 a smaller proportion of sand. 
 
 Hydraulic mortar is composed of hydraulic cement and 
 sand in varying proportions. The less the proportion of sand 
 the greater is its strength. If 5 be the strength of neat cement, 
 that is, cement with no sand, and S l be the strength of mortar 
 
ART. 85. 
 
 CEMENT AND MORTAR. 
 
 175 
 
 having/ parts of sand to i part of cement, the formula 
 
 5 - 5 
 
 gives a rough approximation to the strength of the mortar. 
 A common proportion is 3 parts of sand to I of cement, the 
 strength of this being about one-fourth that of the cement it- 
 self. The strength of hydraulic mortar also increases with 
 its age. 
 
 There are two classes of hydraulic cements, the Rosendale 
 or natural cements, and the Portland or artificial 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 square inch of mortars of both 
 classes of different ages and different proportions of sand. 
 
 Propor- 
 tion of 
 Sand. 
 
 ROSENDALE. 
 
 PORTLAND. 
 
 One 
 Week. 
 
 One 
 
 Month. 
 
 Six 
 
 Months. 
 
 One 
 Year. 
 
 One 
 Week. 
 
 One 
 Month. 
 
 Six 
 
 Months. 
 
 One 
 Year. 
 
 O 
 
 1 2O 
 
 200 
 
 300 
 
 350 
 
 300 
 
 400 
 
 450 
 
 500 
 
 I tO I 
 
 70 
 
 125 
 
 200 
 
 250 
 
 125 
 
 200 
 
 300 
 
 350 
 
 2 tO I 
 
 30 
 
 60 
 
 120 
 
 180 
 
 100 
 
 ISO 
 
 250 
 
 300 
 
 3 to i 
 
 20 
 
 30 
 
 70 
 
 130 
 
 80 
 
 5 
 
 200 
 
 250 
 
 4 to i 
 
 10 
 
 2O 
 
 60 
 
 100 
 
 50 
 
 70 
 
 120 
 
 150 
 
 These figures are obtained from tests of briquettes carefully 
 made in molds, and are probably higher than mortar made 
 under usual circumstances would give. The briquette is re- 
 moved from the mold when set, allowed to remain in air for 
 one day, and then put under water for the remainder of the 
 time. 
 
 As cements and mortars are only used in compression, the 
 tensile test may seem an inappropriate one. Compressive tests, 
 however, are expensive and unsatisfactory to make under or- 
 
176 THE STRENGTH OF MATERIALS. CH. VIII. 
 
 dinary conditions. 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 test is frequently employed, 
 the briquette being put under water as soon as it has set. For 
 Rosendale 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, how- 
 ever, the material should be thoroughly rammed into the 
 moulds, 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 much higher than the tensile strength, and it increases more 
 rapidly with age. A common statement is that the compressive 
 strength is from eight to ten times the tensile strength. 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. Rosendale cement mortar when three 
 or four years old has 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 tensile strength usually only 
 about one-fourth. 
 
 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. Tensile tests of briquettes, more than 
 all others, are valuable on account of the ease with which they 
 can be made, and the reliable conclusions that can be drawn 
 from the results. 
 
ART. 86. 
 
 STONE. 
 
 177 
 
 ART. 86. 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 varie- 
 ties form most valuable building material. In general it is 
 easy to cut and dress, but the variety known as Potsdam sand- 
 stone is very hard in some localities. 
 
 Limestone is formed by consolidated marine shells, and is 
 of diverse quality. Marble is limestone which has been re- 
 worked in the laboratory of nature so as to expel the impuri- 
 ties, 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 of aqueous origin metamorphosed under 
 heat and pressure ; 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, usually strong and dur- 
 able, 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, as large blocks cannot be readily obtained 
 and cut to size. It has, however, a high strength, and is re- 
 markable for durability. 
 
 The average weights and ultimate compressive strengths of 
 these four classes are as follows : 
 
 Kind. 
 
 Weight 
 per Cubic 
 Foot. 
 
 Compress- 
 ive 
 Strength. 
 
 Modulus 
 of 
 Rupture. 
 
 Sandstone 
 
 Pounds. 
 150 
 
 Lbs. per 
 sq. in. 
 5000 
 
 Lbs. per 
 sq. in. 
 I 500 
 
 Limestone 
 
 1 60 
 
 7000 
 
 I 5OO 
 
 Granite 
 
 I6 5 
 
 12000 
 
 2000 
 
 Trap 
 
 175 
 
 I6OOO 
 
 
1^8 THE STRENGTH OF MATERIALS. ClI. VIII. 
 
 These figures, however, refer to small specimens such as 
 can be used 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, in planes making an 
 angle of about 45 degrees with the direction of the pressure 
 (see Fig. 3, Art. 8). In order to insure a perfect distribution 
 of pressure over the surfaces, cushions of wood, leather, and 
 lead are often placed upon them, but the advantage of using 
 them is doubtful. 
 
 The coefficient of elasticity of stone has been found to 
 range from 5000000 to 10000000 pounds per square inch. 
 The elastic limit is difficult to observe, if indeed any exists. 
 Little is known of .the tensile and shearing strengths of stone, 
 except that they are smaller than the compressive strength in 
 some varieties. Tests to determine the modulus of rupture by 
 loading a stone beam to destruction are easily made, and will 
 probably serve to compare the quality of different specimens 
 better than other tests of strength. 
 
 Slate is an argillaceous stone consolidated under very heavy 
 pressure so as to form a marked cleavage at right angles to the 
 direction of that pressure. 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 10000 pounds per 
 square inch, and its modulus of rupture about 7 ooo pounds 
 per square inch. Unfortunately it is liable to corrode under 
 the action of the atmosphere, and its marked cleavage and 
 grain render its strength variable in different directions. See 
 Transactions American Society of Civil Engineers, Sept. 1892 
 and Dec. 1894. 
 
 The quality of a building stone cannot be safely inferred 
 from tests of strength, as its durability depends largely upon 
 
ART. 87. CAST IRON. 179 
 
 its capacity to resist the action of the weather. Hence cor- 
 rosion and freezing tests, impact tests, and observations of the 
 behavior of stone under conditions of actual use are more im- 
 portant than the determination of crushing strength in a com- 
 pression machine. See BAKER'S Masonry Construction for 
 full information regarding these subjects. 
 
 ART. 87. 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 
 moulds, 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 mak- 
 ing wrought iron. 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 per cent of carbon ; forge pig has a specific gravity of from 
 7.1 to 7.4, and it contains from 4 to 2 per cent 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 per cent 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 
 
ISO THE STRENGTH OF MATERIALS. CH. VIII. 
 
 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 factor 
 which governs its strength, particularly that percentage which 
 is chemically combined with the iron. For example, the fol- 
 lowing are the results of tests by WADE of three classes of 
 cast iron for guns, the tensile strength being expressed in 
 pounds per square inch : 
 
 VT c & r* Percentage of Carbon. Ultimate 
 
 No. Specific Gravity. Graphite< 8 Combined. Tensile Strength. 
 
 1 7.204 2.06 1.78 28800 
 
 2 7.154 2.30 1.46 24800 
 
 3 7.087 2.83 0.82 20 ioo 
 
 Here it is seen that the total carbon is about the same in the 
 three kinds, but the smaller the percentage of combined carbon 
 the less is the specific gravity and the ultimate strength. 
 
 As average values for the ultimate strength of cast iron, 
 20000 and 90000 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 great variation in quality. 
 For first-class gun iron WADE found a tensile strength of over 
 30000 and a compressive strength of over 150000 pounds per 
 square inch. On the other hand medium-quality castings often 
 have a tensile strength less than 16000 pounds per square inch. 
 
 In tensile tests the elongations increase faster than in the 
 simple ratio of the applied stresses, so that the elastic limit is 
 poorly defined, and the coefficient of elasticity may range from 
 10 ooo ooo to 20 ooo ooo pounds per square inch. HODGKIN- 
 SON deduced formulas for finding the elongation for different 
 stresses, but these are of little practical importance. 
 
 The flexural test is a good one for comparing the strength 
 
ART. 88. WROUGHT IRON. 181 
 
 of different bars of cast iron. A bar 2 X I inch in cross-section 
 and 27 inches long, laid flatwise on two supports 24 inches 
 apart, should carry a load of 2000 pounds, or more, applied at 
 the middle ; that is, the modulus of rupture should be 36 ooo 
 pounds per square inch, or more. 
 
 The high compressive strength and cheapness of cast iron 
 render it a valuable material for many purposes, but its brittle- 
 ness, low tensile strength, and ductility forbid its use in struc- 
 tures subject to variations of load or to shocks. Its ultimate 
 elongation being scarcely one per cent, the work required to 
 cause rupture in tension is small compared to that for wrought 
 iron and steel, and hence as a structural material the use ot 
 cast iron is very limited. POLE'S Iron as a Material of Con- 
 struction (London, 1872) is an elementary work which the 
 student may consult with advantage. 
 
 ART. 88. WROUGHT IRON. 
 
 The ancient peoples of Europe and Asia were acquainted 
 with wrought iron and steel to a limited extent. It is men- 
 tioned in Genesis, iv., 22, and in one of the oldest pyramids of 
 Egypt a piece of iron has been found. It was produced, prob- 
 ably, 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 draught 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 
 mainly by the use of forge pig (Art. 87), the one most exten- 
 sively used being the puddling process. Here the forge pig 
 is subjected to the oxidizing flame of a blast in a reverbera- 
 tory 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 
 
1 82 THE STRENGTH OF MATERIALS. CH. VI I L 
 
 bars are cut, laid in piles, heated, and rolled, forming what is 
 called merchant bar. If this is cut, piled, and rolled again a 
 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 i to I inch thick, and of 
 varying 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 rail form (Art. 31). 
 
 Wrought iron is metallic iron containing less than 0.25 
 per cent of carbon, and which has been manufactured without 
 fusion. Its tensile and compressive strengths are closely equal 
 on the average from 50000 to 55 ooo pounds per square inch. 
 The elastic limit is well defined at about 25 ooo pounds per 
 square inch, and within that limit the law of proportionality of 
 stress to deformation is strictly observed. It is tough and 
 ductile, having an ultimate elongation of from 20 to 30 per 
 cent. It is stiffer than cast iron, the coefficient of elasticity 
 being 25 ooo ooo pounds per square inch. It is malleable, can 
 be forged and welded, and has a high capacity to withstand 
 the action of shocks. It cannot, however, be tempered nor 
 can it be melted at any ordinary temperature. 
 
 The cold-bend test for wrought iron is an important one 
 for judging of general quality. A bar perhaps | X f inches 
 and 15 inches long is bent when cold either by pressure or by 
 blows of a hammer. Bridge iron should bend through an 
 angle of 90 degrees to a curve whose radius is twice the thick- 
 ness of the bar, without cracking. Rivet iron should bend 
 through 1 80 degrees until the sides of the bar are in contact, 
 without showing signs of fracture. Wrought iron that breaks 
 under this test is lacking in both strength and ductility. 
 
 The tensile test on a small specimen not less than 12 inches 
 
ART. 88. 
 
 WROUGHT IRON. 
 
 183 
 
 in length and 0.5 square inches in cross-section is mainly em- 
 ployed. The elongation should be measured on a length not 
 less than 8 inches. The following requirements for structural 
 iron in tension are those recommended by a committee of the 
 American Society of Civil Engineers in 1895 : 
 
 Kind of Wrought Iron. 
 
 Yield Point. 
 Lbs. per 
 sq. in. 
 
 Maximum 
 Strength. 
 Lbs. per 
 sq. in. 
 
 Ultimate 
 Elongation. 
 P.er cent. 
 
 Reduction 
 of Area. 
 Per cent. 
 
 Bars 
 
 26000 
 
 50000 
 
 20 
 
 30 
 
 Tension Plates and Shapes 
 
 26000 
 
 48000 
 
 15 
 
 20 
 
 Compression Plates and Shapes 
 
 26000 
 
 48 ooo 
 
 12 
 
 16 
 
 Web Plates 
 
 26OOO 
 
 46000 
 
 8 
 
 12 
 
 The term 'yield point ' is here used instead of elastic limit, as 
 the latter is strictly the point at which the elongation ceases 
 to be proportional to the applied load, and it is not easy to de- 
 termine its exact value. On a delicate machine, however, the 
 sudden increase of elongation soon after the elastic limit is 
 passed cannot fail to be noticed by the drop of the scale 
 beam, and this is called the yield point. It is further recom- 
 mended that the stress be applied to the specimen at the uni- 
 form rate of from 4 ooo to 5 ooo pounds per square inch per 
 minute below the yield point and 7 ooo to 8 ooo pounds per 
 square inch per minute above the yield point. 
 
 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 per cent 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 in ductility or ultimate elongation. Annealing lowers 
 the ultimate strength, but increases the elongation. Boiler- 
 plate iron will generally have an ultimate strength greater than 
 55 ooo pounds per square inch. Iron wire, owing to the pro- 
 
184 THE STRENGTH OF MATERIALS. CH. VIII. 
 
 cess 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 slightly exceeding the elastic limit is to 
 cause a small permanent set, but the elastic limit will be found 
 to be higher than before. This is decidedly injurious to the 
 quality of the material on account of the accompanying change 
 in structure, and hence it is a fundamental principle that the 
 working unit-stresses should not exceed the elastic limit. For 
 proper security indeed the allowable unit-stress should seldom 
 be greater than one half the elastic limit. 
 
 In a rough general way the quality of wrought iron may be 
 estimated by the product of its tensile strength and ultimate 
 elongation, this product being an approximate measure of the 
 work required to produce rupture. A more precise measure 
 of this work K is, however, given by the formula 
 
 (15) K=WS. + 2$) 
 
 in which s is the ultimate unit-elongation, S e is the elastic limit, 
 and S t the tensile strength. For example, take two wrought- 
 iron specimens, the first having S e = 27 ooo, 5 = 66000, s = 13 
 per cent = 0.13, and the second having S e = 24000, S t = 51 OOO, 
 s = 24 per cent = 0.24. Then the values of K are 
 
 K, 6 890, K., = 10 080, 
 
 which shows that the work required to rupture the second 
 specimen will be much greater than for the first. Thus high 
 tensile strength is not a good quality when accompanied by 
 low elongation. The value of K is the number of inch-pounds 
 of work required for the rupture of a specimen one square inch 
 in section- and one inch long. 
 
ART. 89. STEEL. 185 
 
 Wrought iron has no proper modulus of rupture since, ow- 
 ing to the change of molecular condition after the elastic limit 
 is exceeded, bars can be bent to almost any extent without 
 fracture. The same is true of soft and medium steel. 
 
 ART. 89. 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: 
 
 Per cent of Carbon. Spec. Grav. Properties. 
 
 Cast iron, 5 to 2 7.2 Fusible, not malleable. 
 
 Steel, i.SOtoo.io 7.8 Fusible and malleable. 
 
 Wrought iron, 0.30 to 0.05 7.7 Malleable, not fusible. 
 
 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 
 subject to considerable variation. 
 
 The three principal methods of manufacture are the cruci- 
 ble process, the open-hearth process, and the Bessemer pro- 
 cess. 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 in a 
 Siemens furnace, wrought-iron scrap being added until the 
 proper degree of carbonization is secured. In the Bessemer 
 process pig iron is completely decarbonized in a converter by 
 
186 THE STRENGTH OF MATERIALS. CH. VIII. 
 
 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 guns, armor plates, and for 
 some structural purposes ; the Bessemer process produces 
 steel for railroad rails and also for structural shapes. 
 
 The physical properties of steel depend both upon the 
 method of manufacture and upon the chemical composition, 
 the carbon having the controlling influence upon strength. 
 Manganese promotes malleability and silicon increases the 
 hardness, while phosphorus and sulphur tend to cause brittle- 
 ness. The higher the percentage of carbon within reasonable 
 limits the greater is the ultimate strength and the less the 
 elongation. THURSTON proposes the formula 
 
 (16) S t = 60 ooo -f- 70 oooC 
 
 for the ultimate tensile strength in pounds per square inch, C 
 being the per cent of carbon. Thus, with 0.40 per cent of car- 
 bon the value of S t is 88 ooo pounds per square inch. The 
 ultimate elongation is approximately inversely proportional to 
 the tensile strength, a formula frequently given being : 
 
 150 1500000 
 elongation in per cent = 2- r -^ = F ; 
 
 thus when C = 0.70, the tensile strength is about 109 ooo 
 pounds per square inch and the elongation about 14 per cent. 
 These approximate formulas refer only to unannealed steel. 
 
 A classification of steel according to the percentage of 
 carbon and its physical properties of tempering and welding is 
 as follows : 
 
 Extra hard, i.oo to o.6o# C., takes high temper, but not weldable. 
 
 Hard, 0.70 to 0.40* C., temperable, but welded with difficulty. 
 
 Medium, 0.50 to 0.20* C., poor temper, but weldable. 
 
 Mild, 0.40 to 0.05^ C., not temperable, but easily welded. 
 
ART. 89. 
 
 STEEL. 
 
 187 
 
 It is seen that these classes overlap so that there are no distinct 
 lines of demarcation. The extra-hard steels are used for tools, 
 the hard steels for piston rods and other parts of machines, 
 the medium steels for rails and beams, and the mild or soft 
 steels for plates, rivets, and other purposes. 
 
 The influence of annealing, or keeping the metal in contact 
 with a light fire for some days, is marked in reducing the ulti- 
 mate strength and increasing the elongation. As an example 
 the following table gives some of the results of a large series 
 of tests exhibited by the Bethlehem Iron Company at the 
 World's Columbian Exposition of 1893, all being flat bars of 
 
 Per cent 
 of 
 Carbon. 
 
 Tensile Strength. 
 Pounds per square in. 
 
 Ultimate Elongation. 
 Per cent. 
 
 Unan- 
 nealed. 
 
 Annealed. 
 
 Unan- 
 nealed . 
 
 Annealed. 
 
 0.08 
 
 58 ooo 
 
 56000 
 
 27 
 
 31 
 
 0.25 
 
 &40OO 
 
 75000 
 
 21 
 
 25 
 
 0.50 
 
 125 ooo 
 
 9QOOO 
 
 II 
 
 19 
 
 0.67 
 
 136000 
 
 112 OOO 
 
 6 
 
 16 
 
 1.04 
 
 153000 
 
 I2800O 
 
 3i 
 
 II 
 
 Bessemer steel. The process of annealing is thus seen to 
 greatly improve the capacity of the high-carbon steels to resist 
 work, since the product of tensile strength and elongation is 
 materially increased (Art. 88). 
 
 For bridges and buildings the following requirements are 
 recommended by the committee report of the American So- 
 ciety of Civil Engineers in 1895, for ultimate tensile strength : 
 
 For low steel, 60 ooo 4 ooo, 
 
 For medium steel, 65 ooo 4 ooo, 
 
 For high steel, 70000 4000, 
 
 all being in pounds per square inch. The yield point is re- 
 quired to be 55 per cent of the ultimate strength, the per cent 
 of elongation to be I 500000 divided by the ultimate strength, 
 
188 THE STRENGTH OF MATERIALS. CH. VIII. 
 
 and the per cent of reduction of area to be 2 800 ooo divided 
 by the ultimate strength. Specimens of medium steel cut 
 from bars or plates must stand bending through 180 degrees 
 to an inner radius of one and one half times the thickness of 
 the specimen without sign of fracture, while for the high steel 
 the same must be the case to a radius of twice the thickness of 
 the specimen. 
 
 The strength of steel may be greatly increased by com- 
 pressing it while fluid, by the use of nickel as an alloy, and by 
 the processes of forging. Nickel steel has been made with an 
 elastic limit over 100000 and with an ultimate tensile strength 
 of 277 ooo pounds per square inch. By tempering both 
 strength and hardness are increased, and by annealing its re- 
 sistance to shock is improved. See paper by R. W. DAVEN- 
 PORT in Engineering News for Nov. 23, 1893. 
 
 The compressive strength of steel is always higher than the 
 tensile strength. The maximum value recorded for hardened 
 steel is 392000 pounds per square inch. The expense of com- 
 mercial tests of compression is, however, so great that they 
 are seldom made. The shearing strength is about three-fourths 
 of the tensile strength. Soft and structural steels have no 
 modulus of rupture, since bars can be bent through 180 degrees 
 by transverse pressure. 
 
 The elastic limit is usually well defined and closely coinci- 
 dent with the yield point. In tension, as a rough rule, it may 
 be taken as one-half of the ultimate strength, in compression 
 at somewhat less than one-half, perhaps one-third, for the hard 
 steels. The coefficient of elasticity is subject to but little 
 variation with the percentage of carbon, and the mean value of 
 30000000 pounds per square inch may be used in ordinary 
 computations both for tensile and compressive stresses that do 
 not exceed the elastic limit. In shearing the coefficient of elas- 
 ticity may be taken as a mean at two-fifths of that for tension. 
 
ART. 90. OTHER MATERIALS. 189 
 
 Steel castings are extensively used for axle boxes, cross- 
 heads, and joints in structural work. They contain from 0.25 
 to 0.50 per cent of carbon, ranging in tensile strength from 
 60000 to looooo pounds per square inch. 
 
 Steel has entirely supplanted wrought iron for railroad rails, 
 and largely so for structural purposes. Its price being the 
 same, its strength greater, its structure more homogeneous, the 
 low and medium varieties are coming more and more into use 
 as a satisfactory and reliable material for large classes of engi- 
 neering constructions. 
 
 ART. 90. OTHER MATERIALS. 
 
 Concrete, composed of hydraulic mortar and broken stone, 
 is an ancient material, having been extensively used by the 
 Romans. It is mainly employed for foundations and mono- 
 lithic structures, but in some cases large blocks have been 
 made which are laid together like masonry. Like mortar, its 
 strength increases with age. When six months old its mean 
 compressive strength ranges from 700 to I 500 pounds per 
 square inch, and when one year old it is probably about fifty 
 per cent greater. 
 
 Beton is an artificial stone made of hydraulic cement and 
 sand which has been subject to prolonged trituration. Its 
 strength is about double that of ordinary concrete. 
 
 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 ultimate strengths of about 36000 and 50000 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 increase as the squares of their diameters. 
 
IQO THE STRENGTH OF MATERIALS. CH. VIII. 
 
 Phosphor bronze is an alloy of copper and tin containing 
 from 2 to 6 per cent of phosphorus. It is remarkable for its 
 complete fluidity so that most perfect castings caii 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 40000 to 100000 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 is light, weighing only 168 pounds per cubic 
 foot. Its ultimate tensile strength is about 25 ooo pounds 
 per square inch. It has a low coefficient of elasticity, and its 
 ultimate elongation is also low. Alloys of aluminum and 
 copper have been made with a tensile 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 was ascertained by 
 THURSTON to be that composed of 55 parts of copper, 43 of 
 zinc, and 2 of tin, its ultimate tensile strength of 68 900 pounds 
 per square inch, with an elongation of 48 per cent and a reduc- 
 tion of area of 70 per cent. See THURSTON'S Materials of 
 Engineering, Vol. III. 
 
 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 strength of lead is only about one-tenth of that of 
 brass, and it attains a permanent set under a small tensile 
 
 stress. 
 
ART. 91. THE FATIGUE OF MATERIALS 19! 
 
 Glass has a tensile strength of about 5 ooo and a compres- 
 sive strength of about 8 ooo pounds per square inch. 
 
 ART. 91. THE FATIGUE OF MATERIALS. 
 
 The ultimate strength S u is usually understood to be that 
 steady unit-stress which causes rupture at one application. 
 Experience and experiments, however, teach that if a unit- 
 stress somewhat less than S u be applied a sufficient number of 
 times to a bar rupture will be caused. The experiments of 
 WOHLER have been of great value in establishing the laws 
 which govern the rupture of metals under repeated applica- 
 tions 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 stress decreases as the number of appli- 
 cations increase. In other experiments where the initial stress 
 was not o, but a permanent value 5, 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. From a discussion of these numer- 
 ous experiments the following laws may be stated. 
 
 1. By repeated applications of stress rupture may be caused 
 by a unit-stress less in value 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 
 
I9 2 THE STRENGTH OF MATERIALS. CH. VIII. 
 
 to the elastic limit the number of applications required 
 to rupture it is enormous. 
 
 4. A range of stress from tension into compression, or vice 
 versa, 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 ' used in stating these laws 
 means about 40 millions, that being roughly the number used 
 by WOHLER to cause rupture under the conditions stated. 
 For all practical cases of repeated stress, except in fast moving 
 machinery, this great number would seldom be exceeded during 
 the natural life of the piece. 
 
 In Art. 8 it was recognized that the working stress should 
 be less for pieces subject to varying stresses than for those car- 
 rying steady loads only. For many years indeed it has been 
 the practice of designers to grade the working stress accord- 
 ing to the range of stresses to which it might be liable to be 
 subjected. WOHLER's laws and experiments afford however a 
 means of grading these values in a more satisfactory manner 
 than mere judgment can do, and formulas for that purpose will 
 be deduced in the next Article. After the working stress 
 S w is determined the cross-section of the piece is found in the 
 usual way, if in tension by formula (i), and if in compression 
 by formula (i) or (10) as the case may require. 
 
 Prob. 136. How many years will probably be required for a 
 tie bar in a bridge truss to receive 40 million repetitions of 
 stress ? 
 
 ART. 92. REPEATED STRESSES. 
 
 Consider a bar in which the unit-stress varies from S' to S, 
 the latter being the greater numerically. Both S' and 5 may 
 
ART. 92. 
 
 REPEATED STRESSES. 
 
 193 
 
 be tension or both may be compression, or one may be ten- 
 sion and the other compression ; in the last case the sign of 
 6" is to be taken as minus. Consider the stress to be re- 
 peated an enormous number of times from S' to S and rupture 
 to then occur under the greater unit-stress S. By the sec- 
 ond law above stated 5 is some function of 5 S' ; this is 
 equivalent to saying that 5 is a function of 5(i S'/S), or 
 more simply a function of S'/S. Now if P' and P be the 
 total stresses on the bar the ratio S'/S equals P'/P, and 
 hence the unit-stress 5 which causes rupture after an enor- 
 mous number of repetitions is a function of P'/P. 
 
 LAUNHARDT'S formula for 5 applies to the case where the 
 limiting stresses P r and P are both tension or both compres- 
 sion, so that P'/P is always positive. Let the values of this 
 ratio be taken as abscissas ranging from o to I, and those of 
 S as ordinates. Let the function of P'/P be supposed to rep- 
 resent a straight line whose equation is 
 
 5 = m -f- n-p- , 
 
 in which m and n are constants to be determined. Let u be 
 
 the ultimate strength of the material under one application of 
 
 stress, and e the unit-stress 
 
 at the elastic limit. Now 
 
 if P'/P is unity, then 5 is 
 
 u and hence u = m -f- n. 
 
 Also, from the third law of 
 
 the last article, if P'/P is 
 
 zero, then 5 is e and hence 
 
 e = m. Accordingly the 
 
 value of m is e, that of n is FlG - 56. 
 
 u e> and the equation of the straight line becomes 
 
 1.0 
 
 -0.5 
 
 -H).5 
 
 +1.0 
 
 which gives the unit-stress 5 that ruptures the bar after an 
 
194 THE STRENGTH OF MATERIALS. Cli VIII. 
 
 enormous number of repetitions of stress ranging from P' to P. 
 For mean values of u and ^, this becomes for wrought iron 
 
 S= 25000(1 +ff 
 
 For example, let a bar range in tension from 80000 to 
 160000 pounds; then the value of P'/P is % and, from the 
 formula, S = 40 ooo pounds per square inch is the rupturing 
 unit-stress. 
 
 WEYRAUCH'S formula for 5 applies to the case where the 
 bar ranges in stress from P' to P, one being tension and the 
 other compression, and P being the greater numerically. 
 Here P'/P is always negative, and the law connecting it with 
 S is taken to be the same as before. Let e be the unit-stress 
 at the elastic limit and /the unit-stress which, under the fifth 
 law of the last article, causes rupture when P' and P are nu- 
 merically equal. By the third law, if P'/P is zero, then 5 is 
 e and hence e = m. By the fifth law, if P'/P is I, then 5 
 is /and hence /= m n. 'Accordingly the value of m is e, 
 that of n is e /, and the equation of the straight line is 
 
 For example, taking / = %e, this becomes for wrought iron 
 
 in which the value of P'/P is to be inserted as negative. 
 Thus if P 1 = 80 ooo pounds compression and P= 160000 
 pounds tension, then P'/P= , and 5=18750 pounds 
 per square inch is the rupturing unit-stress after an enormous 
 number of repetitions. 
 
 In Fig. 56 the ordinates u, e, and / represent the values of 
 5 for the values + I, o, and I of the ratio P'/P. The 
 straight line joining the tops of the ordinates u and e repre- 
 sents LAUNHARDT'S formula, while that joining the tops of 
 the ordinates e and f represents WEYRAUCH'S formula. 
 
ART. 92. REPEATED STRESSES. IQ5 
 
 Another formula may be established by assuming the law 
 of variation of 5 to be that represented by a curve joining the 
 tops of the three ordinates u, e, and f. The simplest curve 
 is a parabola whose equation is 
 
 P' P'\* 
 
 To determine ;//, n, and/, consider first that if P'/P= -f- i, 
 then S=u and hence um-\-n-^-p\ second that if 
 P'/P=o, then S=e and hence e = m\ third that if 
 P'/P i, then 5=/and hence/= m n+p. From 
 these three conditions the values of m, n, and p are found 
 and accordingly 
 
 2e 
 
 is a formula for the rupturing unit-stress in a bar whose total 
 stress ranges between the limits P' and P. If P' and P are 
 both tension or both compression the ratio P'/P is positive, 
 if one is tension and the other compression the ratio P'/P is 
 negative. It is seen that (17) always gives values of 5 a little 
 smaller than those found from the straight-line formulas. 
 
 For structural steel where u = 64000, e = 32 ooo, and/ = 
 16000 pounds per square inch, the formula (17) becomes 
 
 For a bar of such steel whose stress ranges from 1 80000 
 pounds tension under dead load to 540000 pounds tension 
 under live load the value of P'/P is + , and the formula 
 gives 5 = 40900 pounds per square inch. If the stress ranges 
 from 180000 pounds compression to 540000 pounds tension, 
 then the value of P'/P is , and the formula gives 5 = 
 24900 pounds per square inch. 
 
 In using these formulas in cases of design a factor of safety 
 
196 THE STRENGTH OF MATERIALS. CH. VIII. 
 
 is applied. Thus with a factor of 4 the straight-line formulas 
 for structural steel take the form 
 
 = 8000(1+ -, 
 
 the first being used when P'/P is positive and the second 
 when it is negative, while the parabolic formula reduces to 
 
 These formulas give the allowable unit-stresses to be used in 
 designing bridge members of structural steel. As an example 
 let it be required to find by help of the last formula the proper 
 sectional area for a bar of structural steel subject to repeated 
 tension ranging between 29000 and 145000 pounds; here 
 P' /P =. -f- O.2, 5 = 9280 pounds per square inch, and 
 
 P 145000 
 A= S'~ 9280 = IS sc l uare mches. 
 
 In like manner if a short steel bar ranges in stress from 
 29000 pounds compression to 145000 pounds tension the 
 value of P'/P is 0.2 and S = 6 880 pounds per square inch ; 
 then the cross-section required is 21.1 square inches. 
 
 Prob. 137. A short bar of wrought iron is subject to re- 
 peated stresses ranging from 16000 pounds compression to 
 80 ooo pounds tension. What should be the area of its cross- 
 section for a factor of safety of 5 ? 
 
 Prob. 138. A wrought-iron bar has a tension of 3000 
 pounds per square inch as its least unit-stress. For a factor 
 of safety of 4 what is the greatest tensile unit-stress 5 to 
 which it should be subjected when the unit-stress is often 
 repeated from 3 ooo up to 5 ? 
 
ART. 93. SUDDEN LOADS AND IMPACT. 197 
 
 CHAPTER IX. 
 THE RESILIENCE OF MATERIALS. 
 
 ART. 93. SUDDEN LOADS AND IMPACT. 
 When a tensile load is slowly and uniformly applied to a bar it 
 increases slowly from o up to the final value P, and the stress 
 in the bar at any instant is equal to the tensile force existing 
 at that instant; the elongation of the bar increases propor- 
 tionally to the stress from o up to the final limit A, if the 
 elastic limit is not exceeded. The work done upon the bar by 
 the external force is then equal to its mean intensity %P multi- 
 plied by the distance A, or \P\ ; the work of the molecular 
 forces is also equal to this same quantity PA. 
 
 A load P is said to be suddenly applied when its intensity is 
 the same from the beginning to the end of the elongation. 
 The stress in the bar, however, increases from o up to a limit Q. 
 Let y be the elongation produced by the sudden load P\ then 
 the work of this external force is Py. If the stresses are 
 within the elastic limit so that they increase proportionally to 
 the elongation, the mean stress is \Q and the work of the re- 
 sisting forces is \Qy. Hence, as these two works must be equal, 
 
 \Qy = Py or Q = 2P. 
 
 Now let A be the elongation due to the load P when gradually 
 applied, then by law (B), 
 
 i=s or A 
 
 Therefore is established the following important theoretical law, 
 
 A suddenly applied load produces double the stress and 
 double the deformation caused by the same load when 
 applied slowly with uniform increments. 
 
198 THE RESILIENCE OF MATERIALS. CH. IX. 
 
 This law is only true when all the stresses are within the elastic 
 limit of the material. The sudden load P thus causes the end 
 of the bar to move from o to 2A, when the stress becomes 2P 
 the resultant force tending to move the end is P 2P or P 
 and hence the end moves backward, until after a series of 
 oscillations it comes to rest with the elongation A due to the 
 static stress P. The time of this oscillation, as also the velocity 
 of the end of the bar at any instant, can be computed by the 
 principles of dynamics. 
 
 Impact is said to be produced upon the end of a bar when 
 a load P falls from a height h upon it. Here the stress in the 
 bar will increase from o up to a certain limit Q and the defor- 
 mation from o up to a certain limit y. If the elastic limit of 
 the material be not exceeded, the stress at any instant will be 
 proportional to the deformation, so that the work of the in- 
 ternal stresses will be %Qy. The work done by the exterior force 
 P in the same time is P(h -\- y). Hence 
 
 But if A be the deformation due to a static load P, the law of 
 proportionality gives 
 
 Q y 
 
 p ~A* 
 
 Combining these two equations there is found, 
 
 If h = o these formulas reduce to Q = 2P and y = 2A, which 
 is the case of a suddenly applied load ; if h = 4A, they become 
 Q = 4P and y = 4/1 ; if h = I2A they give Q = 6P and y = 6A. 
 Since A is a small quantity for any metallic bar, it follows that 
 a load P dropping from a moderate height may produce great 
 
ART. 94. THE MODULUS OF RESILIENCE. 
 
 stresses and deformations. Experiments made upon springs 
 show that the theory here presented is correct, provided the 
 elastic limit of the material is not surpassed by the stress Q. 
 
 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 and the rapid 
 variations in the stresses cause a change in molecular structure 
 which impairs the elasticity of the material. Generally it will 
 be 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. Shocks which produce 
 stresses above the elastic limit cause the material to become 
 stiff and brittle, and hence it is that the working unit-stresses 
 based upon static loads should be taken very low (Art. 8). 
 
 Prob. 139. In an experiment upon a spring a weight 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. 94. THE MODULUS OF RESILIENCE. 
 
 When an applied stress causes a deformation work is done. 
 Thus if a tensile stress P be applied by increments to a bar, 
 so that the stress gradually increases from o to the value P, the 
 work done is the product of the average stress by the total 
 elongation A. This product is termed the resilience of the bar. 
 If the stress does not exceed the elastic limit of the material 
 the average stress is \P y and the work or resilience is \P\. If 
 the cross-section of the bar be A and its length /, the unit- 
 stress is P - A = S and the unit-elongation is X -- / = j, S o 
 that the work of the internal resisting stresses performed 
 
200 THE RESILIENCE OF MATERIALS. CH. IX. 
 
 on each unit of length of the bar per unit of cross-section is 
 From formula (2) the value of s is -, and accordingly this 
 
 work may be written, 
 
 (.8) K= 1 -^. 
 
 If 5 be the unit-stress at the elastic limit, the quantity K is 
 called the modulus of resilience of the material. 
 
 Resilience is often regarded as a measure of the capacity of 
 a material to withstand impact, for if a shock or sudden blow 
 be produced by a falling body, its intensity depends upon the 
 weight and the height through which it has fallen, that is, upon 
 its kinetic energy or work. Hence the higher the resilience 
 of a material the greater is its capacity to endure work that 
 may be performed upon it. The modulus of resilience is a 
 measure of this capacity within the elastic limit only. 
 
 The following are values of the modulus of resilience as 
 computed from (18) by the use of the average constants given 
 in Art. 5. 
 
 For timber, K= 3.0 inch-pounds, 
 
 For cast iron, K= 1.2 inch-pounds, 
 
 For wrought iron, K = 12.5 inch-pounds, 
 
 For steel, K = 41.7 inch-pounds. 
 
 The ultimate resilience of materials cannot be expressed by a 
 
 rational formula, because the law of increase of elongation be- 
 
 yond the elastic limit is unknown. In Fig. I the ultimate 
 
 resilience is indicated by the area between any curve and the 
 
 axis of abscissas, since that area has the same value as the total 
 
 work performed in producing rupture. For timber and cast 
 
 iron the ratio of these areas is about the same as that of the 
 
 values of K, but for wrought iron and steel the areas are nearly 
 
 equal. 
 
 Prob. 140. What horse-power engine is required to strain 250 
 
ART. 95. EXTERNAL WORK AND RESILIENCE. 2OI 
 
 times per minute a bar of wrought iron 18 feet long and 2 
 inches in diameter from o up to 12 500 pounds per square inch. 
 
 ART. 95. EXTERNAL WORK AND RESILIENCE. 
 
 When a body is deformed by applied forces the work done 
 by these forces is called the external work. For example, if 
 a bar is subjected to a tensile force which is slowly applied 
 until it reaches the intensity P, an elongation A is produced, 
 and the external work is %PX. Again, if a beam be subject to 
 a concentrated load P gradually applied, a deflection A occurs 
 under the load and the external work is $PA. If the load is 
 applied suddenly so that its full intensity is P during the entire 
 time of application, then the external works in the two cases 
 are P\ and PA respectively. If P falls from a height // above 
 the top of the bar or beam the external works are P(Ji + X) 
 and P(h + ^) respectively. 
 
 If a beam be uniformly loaded with w per linear unit the 
 load on any short length dx is wdx y and if y be the deflection 
 at the point whose abscissa is x, the elementary external work 
 for a gradually applied load is \wy . dx. The integration of 
 this over the entire length of the beam will give the total ex- 
 ternal work of the uniform load. 
 
 As external force is resisted by internal stress so external 
 work is resisted by internal work. Each elementary stress 
 multiplied by its displacement gives a corresponding elemen- 
 tary work, and the sum of all these products is the total inter- 
 nal work. By the law of conservation of energy, 
 
 Internal Work = External Work, 
 
 provided that no work is lost in heat by the application of the 
 external forces. Now the word * resilience ' is used to denote 
 the internal work of the stresses, and hence 
 
 Resilience = External Work, 
 
202 THE RESILIENCE OF MATERIALS. CH. IX. 
 
 or the resilience of a body is its capacity to resist the work of 
 external forces. 
 
 Elastic resilience is internal work when the body is not 
 stressed beyond the elastic limit. Non-elastic resilience is in- 
 ternal work when the stresses range from the elastic limit to 
 the point of rupture. 
 
 In Art. 94 an expression for resilience within the elastic 
 limit was deduced for a bar of unit length and unit cross-sec- 
 tion. If a bar of cross-section A and length / be subject to a 
 tensile or compressive force P, the deformation A is produced. 
 If the load be gradually applied the external work is ^P\. 
 Let S be the unit-stress produced, and E be the coefficient of 
 elasticity of the material. Then, from Arts. 2 and 4, 
 
 P=AS, K=^; 
 
 and accordingly the internal work or elastic resilience is 
 
 (18)' K=~.Al; 
 
 that is, the elastic resiliences of bars of the same material un- 
 der the same unit-stress are proportional to their volumes. If 
 6" be the stress at the elastic limit the quantity S 1 /2E is the 
 modulus of resilience, and accordingly the maximum elastic 
 resilience of a bar is the product of its modulus of resilience by 
 its volume. 
 
 A theoretic expression for non-elastic resilience cannot be 
 deduced, but in Art. 97 it will be shown how this can be esti- 
 mated when sufficient experimental data are given. Non- 
 elastic resiliences, however, are generally closely proportional 
 to the volumes of bodies, the material and the maximum stress 
 being constant. 
 
 Prob. 141. How many foot-pounds of work are required to 
 strain a wrought-iron bar, 4 inches in diameter and 54 inches 
 long, from 6000 pounds per square inch up to 12000 pounds 
 per square inch ? 
 
ART. 96. ELASTIC RESILIENCE OF BEAMS. 2O3 
 
 ART. 96. ELASTIC RESILIENCE OF BEAMS. 
 
 When a beam deflects under the action of a load the fibers 
 on one side of the neutral surface are elongated while those 
 on the other side are shortened. If the elastic limit is not ex- 
 ceeded the stress in any fiber is proportional to its distance 
 from the neutral surface (Art. 20). The internal work or 
 elastic resilience of the beam is the half-sum of the products 
 formed by multiplying the stress upon each elementary area 
 by its corresponding change of length. The half-sum instead 
 of the sum is taken, because the stress uniformly increases 
 from o up to its maximum value as the load is applied. Thus, 
 if T be the unit-stress under the elongation e, the unit-stress 
 
 Tx 
 
 for an elongation x is , and the work in the distance dx is 
 
 dx; integrating this between the limits o and e gives \Te as 
 
 the internal work. 
 
 Using the same notation as in Chapter III, the horizontal 
 unit-stress upon the remotest fiber at the dangerous section of 
 the beam is called S, and the distance of that fiber from the 
 neutral surface is called c. Let a single concentrated load W 
 be gradually applied to the beam, and let A be the deflection 
 beneath it. The external work of the load is then | WA % and 
 this equals the elastic resilience if the unit-stress 5 does not 
 surpass the elastic limit. If / be the length of the beam, and 
 7 the moment of inertia of the cross-section, the value of W 
 is, from Art. 29, 
 
 *-% 
 
 where n is I for a cantilever loaded at the end and 4 for a 
 simple beam loaded at the middle. Also from Art. 37, 
 
 nsr 
 
 A = pr, 
 
 mcE 
 
204 THE RESILIENCE OF MATERIALS. CH. IX. 
 
 where m is 3 for the cantilever and 48 for the simple beam. 
 The internal work of the beam hence is : 
 
 . 
 
 2 E me* 
 
 or, putting for / its value Ar* where A is the area and r the 
 least radius of gyration of the cross-section, 
 
 o* *-=?" 
 
 which is a general expression for the elastic resilience of a 
 beam under a single concentrated load. 
 
 1 5* 
 
 If the beam be strained to the elastic limit the factor - -7, 
 
 2 E 
 
 is the modulus of resilience of the material (Art. 94). For 
 
 a 
 
 either the cantilever or the simple beam the value of is J. 
 
 m 
 
 r* . 
 For a rectangular beam a is J. Thus for a rectangular beam 
 
 the internal work is 
 
 (19)' K = l 2 ^ Al= L*. Al , 
 
 that is, the work required to deflect a rectangular beam by a 
 concentrated load is proportional to its volume Al y and the 
 work required to cause the stress 5 to reach the elastic limit is 
 the product of the modulus of resilience and one-ninth of its 
 volume. 
 
 For a cantilever loaded uniformly with w pounds per linear 
 foot the load on any short length dx is wdx, and if y be the 
 deflection at that point the elementary external work is \wydx. 
 Inserting for/ its value from Art. 34, there is found 
 
 2 4 o7 8afi/ 
 for the total external work of the uniform load. This must be 
 
ART. 96. ELASTIC RESILIENCE OF BEAMS. 2O5 
 
 equal to the internal work. Substituting for W its value in 
 terms of 5 from Art. 29, the elastic resilience of a rectangular 
 cantilever is then 
 
 which is nine-tenths of that found for the concentrated load, 
 and also proportional to the volume and modulus of resilience. 
 A similar result is easily deduced for a simple beam uniformly 
 loaded. 
 
 Formula (19) shows that the internal work or resilience de- 
 veloped within the elastic limit is proportional to the product 
 of the volume of the beam and the ratio r'/c*. As, however, 
 this ratio always has a numerical value which is the same for 
 similar sections, it may be stated as a general law, that the 
 elastic resiliences of beams of similar cross-section are propor- 
 tional to their volumes. 
 
 As a numerical example let it be required to determine the 
 horse-power necessary to deflect 50 times per second a rectan- 
 gular wrought-iron beam 6 feet long, 2 inches wide, and 3 
 inches deep, so that at each deflection the unit-stress 5 may 
 range from 5000 to 10000 pounds per square inch, the beam 
 being a cantilever with the load applied at the end. From 
 formula (19)' the work in fifty deflections is 
 
 K = S<X2X (ioooo - 5 ooo) = 3 600 inch-pounds, 
 
 which is 300 foot-pounds, and hence the power required is 
 300/550 = 0.52 horse-powers. 
 
 The strength of a rectangular beam increases with the square 
 of its depth and its stiffness with the cube of the depth (Arts. 
 29 and 36). The elastic resilience, however, increases only 
 with the area of the cross-section ; hence for a given unit-stress 
 S it is immaterial whether the short or the long side of a beam 
 
206 
 
 THE RESILIENCE OF MATERIALS. 
 
 CH. IX. 
 
 be placed vertical when its office is the resistance of external 
 work only. 
 
 Prob. 142. Deduce an expression for the elastic resilience 
 K of a beam fixed at both ends and loaded in the middle ; also 
 for a beam fixed at both ends and uniformly loaded. 
 
 ART. 97. ULTIMATE RESILIENCE. 
 
 The ultimate resilience of a body under stress is equal to 
 the total external work required to produce rupture. The 
 elastic resilience is that part of the ultimate resilience in which 
 the stresses do not surpass the elastic limit of the material. 
 In wrought iron and steel the ultimate resilience greatly sur- 
 passes the elastic resilience, being sometimes five hundred 
 times as large. 
 
 In order to show this fact the particular case of a steel speci- 
 men 12 inches long and 0.505 inches in diameter will be taken, 
 which was tested in a tension machine and the elongations 
 observed at certain intervals. In the following table the first 
 
 Load. 
 Pounds. 
 
 Stress. 
 Pounds per 
 square in. 
 
 Elongation. 
 Per cent. 
 
 Partial 
 Work. 
 Inch-Lbs. 
 
 Total Work 
 Inch-Lbs. 
 
 2OO 
 
 IOOO 
 
 0.00 
 
 
 O 
 
 IOOO 
 
 5000 
 
 O.OI 
 
 0.3 
 
 
 
 3OOO 
 
 I5OOO 
 
 O.04 
 
 3-0 
 
 3 
 
 
 
 
 6.0 
 
 
 $000 
 
 2500O 
 
 O.O7 
 
 
 9 
 
 7000 
 
 35000 
 
 O.IO 
 
 9.0 
 
 18 
 
 
 
 
 16.0 
 
 
 9000 
 
 45000 
 
 O.I4 
 
 
 34 
 
 9600 
 
 48000 
 
 0.16 
 
 9-3 
 
 44 
 
 10 000 
 
 5OOOO 
 
 0.7O 
 
 264.6 
 
 308 
 
 12 000 
 
 60000 
 
 I. QO 
 
 660.0 
 
 968 
 
 I4OOO 
 
 7OOOO 
 
 3.62 
 
 II20.O 
 
 2088 
 
 I6OOO 
 
 80000 
 
 8.50 
 
 3660.0 
 
 5748 
 
 16800 
 
 83600 
 
 15-20 
 
 5480.6 
 
 11229 
 
 I5OOO 
 
 75000 
 
 24.50 
 
 7374 9 
 
 18604 
 
ART. 97. ULTIMATE RESILIENCE. 2O/ 
 
 column gives the total applied load, the second the correspond- 
 ing stress per square inch, and the third the elongation ex- 
 pressed in per cent. The elastic limit was observed at 48 ooo 
 pounds per square inch with o.io per cent elongation. The 
 elongation then rapidly increased with the unit-stress (as seen 
 in the diagram Fig. I of Art. 5). At 83 600 pounds per square 
 inch the maximum tensile strength was reached and the mate- 
 rial was elongating very rapidly. The load was then slowly 
 removed, but the elongation continued to increase until rup- 
 ture occurred at 75 ooo pounds per square inch with a total 
 elongation of 24.5 per cent. 
 
 The external work per cubic inch of material may be approxi- 
 mately computed for any interval by multiplying the average 
 stress during that interval by the elongation which occurs. 
 The given elongations divided by 100 give the elongations per 
 linear inch. Thus while the load ranged from i ooo to 3 ooo 
 pounds the elongation per inch increased from o.oooi to 0.0004. 
 Hence the work done upon one cubic inch of the specimen in 
 that interval was, 
 
 (15000 + 5 ooo) (0.0004 0.0001) = 3.0 men-pounds. 
 Similarly the external work per cubic inch performed in the 
 last interval is 
 
 $(83 600 + 75 ooo) (0.245 0.152) = 7374-9 inch pounds, 
 and thus the quantities in the fourth column of the table arc 
 separately computed. 
 
 The summation of the fourth column gives the total exter- 
 nal work per cubic inch required to stress the bar from o up to 
 the point of rupture. In the last column sums are given for 
 each value of the unit-stress, and these are closely equal to the 
 internal works. Thus it is seen in this particular case that for 
 one cubic inch of material 
 
 Elastic Resilience = 44 inch-pounds, 
 Ultimate Resilience = 18604 inch-pounds; 
 
208 THE RESILIENCE OF MATERIALS. CH. IX. 
 
 and therefore that the ultimate resilience is more than 400 
 times as great as the elastic resilience. 
 
 The total external work required to rupture the specimen is 
 found by multiplying the above ultimate resilience by the vol- 
 ume of the specimen, which is 2.4 cubic inches. This work is 
 found to be 3721 foot-pounds. 
 
 The example here given is a case of static resilience, where 
 the internal work is slowly developed under the action of an 
 external force gradually applied. The more common cases of 
 resilience, however, are those developed by the impact of a 
 falling body, and a general discussion of these will be presented 
 in the following articles. 
 
 Prob. 143. Compute the elastic and ultimate static resilience 
 for the specimen of wrought iron whose test is given in Art. 5, 
 page ii. 
 
 ART. 98. EARLY HISTORY OF RESILIENCE. 
 
 The matter in the remainder of this chapter is taken from 
 an address on * The Resistance of Materials under Impact/ 
 delivered by the author in August, 1894, before the Section of 
 Mechanical Science and Engineering of the American Associ- 
 ation for the Advancement of Science. 
 
 In 1807 THOMAS YOUNG announced the fundamental ideas of 
 the resistance of materials under impact. " The action which 
 resists pressure," he said, " is called strength, and that which 
 resists impulse may properly be called resilience." He stated 
 that the resilience of a body is proportional to its strength and 
 extension jointly, and that it is measured by the height through 
 which a given weight must fall to cause rupture. The resili- 
 ence of beams of the same kind he made proportional to their 
 volumes, as also the resilience of shafts, whether solid or hol- 
 low. He further suggested that a very high velocity of the 
 
ART. 98. EARLY HISTORY OF RESILIENCE. 2OO, 
 
 moving weight may rupture a body by impact before its full 
 resilience can be developed. 
 
 Resilience is thus the capacity of a body to resist applied 
 work, or it is the internal work which can be developed by the 
 energy of a moving body. External force is resisted by inter- 
 nal stress, and the resulting deformation is a secondary conse- 
 quence ; but impact is resisted by resilience, where the deforma- 
 tion is of equal importance with the stress, for internal work 
 is the product of these two factors. 
 
 In YOUNG'S time the elastic limit of materials was but 
 vaguely recognized, and HOOKE'S law of proportionality of 
 stress to elongation was often applied to all the phenomena 
 preceding rupture. YOUNG'S statements are valid in a general 
 way, but we now know that it is necessary to distinguish be- 
 tween the two cases of elastic resistance and non-elastic resist- 
 ance. Hence there are two divisions of the subject of impact : 
 first, that of elastic resilience, where the molecular forces^ do 
 not surpass the elastic limit of the material ; and second, that 
 of ultimate resilience, where the elastic limit is exceeded and 
 rupture finally occurs. 
 
 The problems of elastic resilience are largely theoretical and 
 mathematical. Their discussion begins with YOUNG, and has 
 been continued by a long line of investigators to the present 
 time ; it forms, indeed, the most prominent part of the theory 
 of elasticity, which has been so thoroughly set forth in the 
 history of TODHUNTER and PlERSON. Starting with HOOKE'S 
 law of proportionality of stress to deformation, and applying 
 to this the mechanical laws of force, velocity, and work, the 
 stresses, displacements, and resilience of elastic bodies subject 
 to impact have been deduced. The results thus theoretically 
 found have been confirmed by many experiments, as must 
 necessarily be the case, since all the laws of the discussion are 
 those of experiment and experience. 
 
210 THE RESILIENCE OF MATERIALS. CH. IX. 
 
 The first problems of elastic resilience were those of bars or 
 rods subject to the longitudinal impact of a moving weight. 
 The vibrations of stretched wires had been discussed by 
 EULER, LAGRANGE, and others ; but NAVIER, in 1823, appears 
 to have been the first to investigate the oscillations and maxi- 
 mum stresses in a horizontal bar due to a weight impinging on 
 its end. PONCELET, in 1829, treated the same problem for a 
 vertical bar subject to longitudinal impact by a falling weight. 
 These discussions showed that for the case of horizontal im- 
 pact the maximum elongation is a mean proportional between 
 twice the height of fall and the extension due to the same 
 weight when applied gradually, while for vertical impact it is 
 the sum of the static extension and the hypothenuse of a 
 right-angled triangle, whose sides are this static extension and 
 the dynamic extension stated for the horizontal bar. It is a 
 corollary from this that under the sudden application of a load, 
 which is the particular case of impact when the height of fall 
 is zero, the maximum elongation, and hence the maximum in- 
 ternal stress, is twice as great as that caused by the same load 
 when applied gradually. (Art. 93.) 
 
 While speaking of PONCELET, it is well to pause to remark, 
 that his work on industrial mechanics which contains these in- 
 vestigations deserves especial mention as the book that marks 
 the beginning of technical education ; his conceptions and ex- 
 planations of these problems of external work and internal 
 resilience may indeed still be regarded as models of clear and 
 accurate reasoning. 
 
 The solutions of NAVIER and PONCELET for longitudinal 
 impact on a bar were approximate in the sense that its weight 
 was regarded as small compared to that of the striking, body. 
 Later investigations by SAINT VENANT, BOUISSINESQ, and 
 others, have completely resolved the problem when the rela- 
 tive masses are taken into account, and have also disclosed all 
 
ART. 98. EARLY HISTORY OF RESILIENCE. 211 
 
 the attendant circumstances of duration and intensity of the 
 forces at the surface of impact. 
 
 Another important problem is that of transverse impact on 
 a beam giving rise to flexural resilience. The first investiga- 
 tion of this was by HODGKINSON, 1833, in connection with the 
 discussion of experiments, while the full elastic theory for dif- 
 ferent ratios of weights of the beam and falling body was de- 
 veloped by Cox in 1848, and later most fully extended by the 
 French elasticians in a similar manner to that of longitudinal 
 impact. The law of proportionality of resilience to volume 
 was shown to be true only when the latter is increased an 
 amount proportional to one-half the ratio of the weight of 
 the beam to that of the striking body, and similar modifications 
 are necessary regarding deflections and internal stresses. In 
 many problems the velocity of transmission of stress, which 
 is the same as the velocity of sound in the given material, 
 forms an important element, as indeed YOUNG had suggested 
 in 1807, at the very beginning of the consideration of impact 
 resistances. 
 
 While the conclusions of all these investigations of elastic 
 resilience are true and valuable, it should never be forgotten 
 that they are only true when the conditions are observed under 
 which they are deduced, namely, that the elastic limit is not 
 exceeded by the maximum internal stress. Now, while the 
 elastic limit for some materials is as high as one-half the ulti- 
 mate strength, the elongation up to the elastic limit is small 
 compared with the ultimate elongation ; and hence the elastic 
 resilience may be very small less perhaps than one part in a 
 thousand, compared with the total ultimate resilience. The 
 phenomena of elastic resilience form, indeed, such a small por- 
 tion of those occurring in practice, that it is difficult to observe 
 them with precision, while those of non-elastic resilience are 
 ever present. Many a misconception has arisen and many a 
 
212 THE RESILIENCE OF MATERIALS. CH. IX 
 
 paradox has been founded on the assumption that the theory 
 of the smaller part is applicable also to the larger part of the 
 phenomena (Art. loo). 
 
 The investigation of ultimate resilience is necessarily experi- 
 mental, since beyond the elastic limit no theoretical relation 
 between stress and deformation is known. In 1818 TREDGOLD 
 made experiments on wooden beams subject to the impact of 
 a falling ball, and concluded that YOUNG'S law of proportion, 
 ality of resilience to volume was not justified. HODGKI:;SON, 
 in 1835, experimented on cast-iron beams under lateral impact 
 from a pendulum, and found that the deflections were propor- 
 tional to the velocities of impact, that the same work was 
 required to break the beam whether struck at the middle or at 
 the quarter point, and that the weight of the beam increased 
 its ultimate resilience all of which is more nearly in accord- 
 ance with the elastic theory than perhaps might be expected. 
 In 1849 was published the ' Report of the Commissioners 
 appointed to inquire into the application of iron to Railway 
 Structures," which contains some of the most comprehensive 
 and valuable experiments ever undertaken. In view of later 
 practice it was perhaps unfortunate that these were mostly 
 made upon cast iron, but the methods of investigation and the 
 conclusions deduced render this report an epoch-making one 
 in the history of the resistance of materials. Here can only 
 be noted those points relating to impact on which experiments 
 were made by HODGKINSON, WILLIS, and GALTON. Trans- 
 verse impact on beams by oft-repeated blows of a ball swung 
 as a pendulum, and also by pressure applied by a revolving 
 cam, indicated for the first time the* laws of fatigue under re- 
 peated stresses ; transverse impact with a single blow confirmed 
 YOUNG'S theorem of resilience and volume, as the same amount 
 of work was required to rupture a rectangular beam whether 
 struck on its narrow or broad side ; and the inertia of the beajn 
 was shown to be an important factor in increasing its resilience. 
 
ART. 99. MODERN EXPERIMENTS. 213 
 
 Weights suddenly applied without impact gave deflections 
 nearly double the static deflections, and the same ratio was 
 observed in the breaking loads. Experiments on the effects 
 of a load moving with different velocities over a beam showed 
 that the deflection could amount to more than double the 
 static deflection, and the elaborate theoretical discussions of 
 this case by STOKES, together with the evidence given by ex- 
 perienced engineers, form the basis of many practical rules 
 since used in bridge design. The immediate result indeed was 
 the revision of the factors of safety by the British Board of 
 Trade and the establishment of a rule that for cast iron the 
 factor for live load should be double that for dead load. 
 
 ART. 99. MODERN EXPERIMENTS. 
 
 The thirty years following the middle of the century may be 
 designated as the period of development of the modern methods 
 of static tests upon tensile specimens. In the United States 
 this development, begun by WADE, RODMAN, and PLYMPTON, 
 later produced the numerous testing machines of FAIRBANKS, 
 OLSEN, and RIEHLE, and at last has culminated in the precise 
 apparatus of EMERY and in the powerful machines at Athens 
 and Phcenixville. These static tensile tests seem to have 
 diverted attention from the question of impact and to have 
 caused the neglect of the theory of resilience, for the objects 
 of the method have been mainly to determine the elastic limit, 
 maximum strength, and ultimate elongation, considerations of 
 work having been largely overlooked. Yet the diagram formed 
 by laying off the unit-stresses as ordinates and the correspond- 
 ing elongation as abscissas shows by its area the internal work 
 which resists rupture, and thus is approximately a measure of 
 the ultimate resilience per cubic unit of the material. The 
 phenomena in a slow test, however, are not in all respects the 
 same as in a rapid one, the elastic limit and maximum strength 
 being generally greater in the latter, while the elongation is 
 
214 THE RESILIENCE OF MATERIALS. CH. IX. 
 
 less. How the diagram would look under the very rapid de- 
 velopment of stress is yet to be determined, and not until this 
 is done by an autographic device can a full knowledge of the 
 total field of resilience be obtained. High values of the ulti- 
 mate elongation and contraction of area have been regarded as 
 of great importance in judging of the quality of iron and steel, 
 and this has given rise to what may be called an unconscious 
 manipulation of the machine, the power being applied at such 
 a rate as to secure these results consistent with other condi- 
 tions. It is safe to say that stresses are not applied and that 
 rupture does not occur in this manner in practice ; and here it 
 is that static tests fail to give full satisfaction. 
 
 It would be easy to take a great deal of space in criticism of 
 the imperfections of the static method of testing, to show the 
 uncertainties of percentages of elongation, to dwell on the dis- 
 crepant results found from specimens of different shape, and 
 to state the various views regarding contraction of area and 
 the appearance of the fracture. But this is not the place. It 
 may be said again, however, that the work per cubic unit de- 
 veloped before rupture is an important element, by the study 
 of which much can be learned ; for the total resilience depends 
 not merely on maximum strength and ultimate elongation, but 
 upon the rate of variation of these throughout the test. Each 
 material, or specimen, by virtue of its physical and chemical 
 properties and its method of manufacture, has the capacity for 
 resisting a certain amount of applied work, of which the quanti- 
 ties measured in the static test are not complete factors. The 
 full resilience can indeed only be found from an autographic 
 diagram like that used in THURSTON's torsion machine, which 
 has done good service in this direction. 
 
 During all this development of static testing one impact test 
 has survived and everywhere held its own. This is the cold- 
 bend test for wrought iron and steel. In the rolling-mill it is 
 
ART. 99. MODERN EXPERIMENTS. 21$ 
 
 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 reliable method of estimating 
 toughness, ductility, strength, and resilience. It is true that 
 numerical values of these qualities are not obtained, but the 
 indications are so valuable, that if all tests except one were to 
 be abandoned, the simple cold-bend test would probably be 
 the one which the majority of engineers would desire to retain. 
 
 Returning now to impact experiments proper, the experi- 
 ments of KlRKALDY, in 1862, deserve notice. He subjected 
 wrought-iron specimens to sudden tensile stress by loads which 
 were applied without velocity, and found that they broke with 
 82 per cent less load than under static conditions. This has 
 since been quoted as a contradiction of the law that the stress 
 produced by a gradual load is one-half that caused by the same 
 load when suddenly applied ; but really no contradiction exists, 
 since the law is only true when the elastic limit is not exceeded. 
 For ultimate resilience the ratio of gradual to sudden load pro- 
 ducing the same stress is always greater than one-half, and 
 approaches nearer to unity the more ductile the material. 
 KlRKALDY also introduced the ultimate contraction of area as 
 an element of equal importance with the ultimate elongation ; 
 and undoubtedly this is the case, since it is subject to less 
 variation with the length of the specimen. However, little 
 satisfactory evidence exists that contraction of area is a meas- 
 ure of resistance to impact, as some have supposed. 
 
 The impact hammer or ram delivering repeated blows, as 
 used by SANDBERG and STYFFE in 1868 on railroad rails and 
 by the United States Board for Testing Metal in 1878 in ex- 
 periments on chain iron, is perhaps the most common method 
 of conducting tests for resilience. The weight of the ram 
 multiplied by the height of fall measures the work done in one 
 
2l6 THE RESILIENCE OF MATERIALS. CH. IX. 
 
 blow, and the number of foot-pounds required to produce rup- 
 ture furnishes an index of the ultimate resilience of the bar or 
 specimen. Although valuable for comparative purposes, like 
 the cold-bend test, this method fails to give proper numerical 
 results, since the effect of a second blow is not really added to 
 that of the first on account of the dissipation of internal work 
 into heat which follows the release of the stress when this is 
 greater than the elastic limit. That the work of several blows 
 affords little indication of ultimate resilience was well shown 
 by UCHATIUS, in 1874, who experimented on rods of small 
 size under the impact of a small ram, and found that rupture 
 could be caused by one blow of 28 inches fall giving 6 foot- 
 pounds, by four blows of 21 inches fall giving 18 foot-pounds, by 
 37 blows of loj inches fall giving 83 foot-pounds, or finally by 
 2 050 blows of 3J inches fall giving I 536 foot-pounds of work. 
 Indeed, the fact that one strong blow is more efficient than 
 many weak ones is well known to every artisan. The compar- 
 ative weights of the ram and bar are also of importance in 
 such tests, as will be later noticed. 
 
 The rise of the elastic limit after the release of a stress 
 which surpasses it, first shown by THURSTON in 1873 > an ^ the 
 consequent stiffening of the material under several such appli- 
 cations of stress, is also observed in cases of repeated impact. 
 Thus MARTENS, in 1879, tested a railroad rail by a ram falling 
 alternately on the head and base ; the second blow produced 
 only 95 per cent of the deflection caused by the first, and after 
 four or five blows only 90 per cent was obtained. Here, of 
 course, a large part of the work of each blow was dissipated in 
 heat due to the change of molecular structure, so that the total 
 work expended gives no valid numerical measure of the true 
 ultimate resilience of the material. Our knowledge as to what 
 occurs under such repeated impacts is of the most uncertain 
 kind. Most of the stresses are no doubt injurious, as the final 
 result proves, but a few may perhaps be beneficial in improv- 
 
ART. 99. MODERN EXPERIMENTS. 2I/ 
 
 ing the quality of the material like the process of cold-hammer- 
 ing and cold-rolling in manufacture. BAUSCHINGER'S discovery 
 in 1885, that the raising of the tensile elastic limit is accom- 
 panied by a lowering of the compressive elastic limit, is a most 
 important one, which is likely in connection with future inves- 
 tigations to lead to a clearer knowledge of the laws of stress in 
 materials under repeated impact. 
 
 The investigations of WoHLER between 1860 and 1870, and 
 the later ones of SPANGENBERG, on the resistance of materials 
 under repeated stresses applied with little impact, have been 
 of the greatest value in influencing the rational design of 
 bridges and other structures subject to variations of load. 
 They establish the facts that repeated stresses below the elas- 
 tic limit do not injure the material, that repeated stresses be- 
 yond this limit cause injury in proportion to the range between 
 the maximum and minimum limits, and that the greater the 
 range the less the number of repetitions required to produce 
 rupture. (Art. 91.) The prompt adoption of rules for design- 
 ing based on these conclusions, in place of the arbitrary methods 
 of adding 20 or 30 per cent to the static stresses, indicated that 
 the engineering profession appreciated the importance of pro- 
 viding for the resistance to impact in a rational way. Yet prob- 
 ably greater impacts occur on bridges than these repeated 
 stresses, for failures are not infrequent, and the life of a bridge 
 under heavy traffic scarcely exceeds a dozen years. The fur- 
 ther study of the effect of repeated impact is thus imperative, 
 for what we now know is but little compared to what is to be 
 learned. 
 
 MAITLAND, in 1887, showed by many experiments on tensile 
 specimens subject to many blows of a falling ram that the ulti- 
 mate elongation was greater than in static tests, it being in 
 some cases nearly doubled. This might, perhaps, be expected 
 from what has been said regarding the beneficial influence of 
 
2l8 THE RESILIENCE OF MATERIALS. CH. IX. 
 
 some of the stresses. He also exploded powder and gun-cot- 
 ton between two cylinder heads joined together by rods, and 
 found their ultimate elongation to be more than double the 
 corresponding ones under static stress. This, on the other 
 hand, might not have been expected from the general conclu- 
 sions regarding the effect of time on stress and elongation, and 
 it seems not to be the case for the impact of a single blow in 
 later experiments. Thus it appears in problems of ultimate 
 resilience, where no theory has yet been developed, that gen- 
 eral principles derived from static tests should be applied with 
 caution. 
 
 The tests made by E. D. ESTRADA, in 1893, have recently 
 led to much interesting discussion on the subject of impact. 
 Specimens of wrought iron and steel were tested by the usual 
 tension machine, and also under the blow of a ram weighing 
 100 pounds falling through vertical heights ranging from 5 to 
 25 feet. Over 40 specimens were broken under repeated 
 blows, they being so arranged that they were brought into ten- 
 sion by the pressure caused by the impact. The number of 
 blows ranged from 2 to 14, the height of fall varying in differ- 
 ent cases. These valuable experiments show clearly that the 
 ultimate elongation under repeated impact is greater than for 
 static stresses, the average of all being about one-third greater, 
 while the contraction of area shows only a small increase ; but 
 no conclusions regarding the elastic limit or the maximum 
 strength under impact can be derived from them. The impact 
 was not directly applied to the specimen, but through a num- 
 ber of plates and bolts through which it was transmitted, and 
 thus much of the work was spent in acting against their inertia 
 and resilience. This, however, in no way invalidates the im- 
 portant deductions regarding the influence of repeated impact 
 on ultimate elongation and contraction of area, and the con- 
 clusion of KlRKALDY regarding the value of the latter as an 
 index of toughness and ductility is thoroughly confirmed. 
 
ART. 99. MODERN EXPERIMENTS. 2 19 
 
 The elongation before rupture appears to vary approximately 
 as the amount of work expended by the ram, but the work re- 
 quired to produce rupture does not give a satisfactory com- 
 parison of the ultimate resilience of different specimens, except 
 in those cases where the height of fall was the same. 
 
 The experimental work thus briefly reviewed has been done 
 by special improvised apparatus, and with little or no uniform- 
 ity of method ; but the time may come when machines for 
 impact tests will be put on the market and be in common use. 
 At present almost the only one that can be mentioned is that 
 devised by W. J. KEEP, about 1889, for testing jthe resilience 
 of cast iron. The blows are delivered by a hammer weighing 
 25 pounds, falling like a pendulum through heights less than 6 
 inches, and produce horizontal flexure at the middle of a small 
 bar i foot in length. Beginning with a fall of inch, successive 
 blows are applied, each with a fall -J inch greater than the pre- 
 ceding, until rupture occurs. The deflections and sets of the 
 bars are graphically recorded by the machine itself, and thus 
 excellent comparisons of the ultimate resilience of different 
 grades of metal may be obtained. 
 
 Impact tests are most important in the case of railroad rails, 
 car wheels, tires and axles, and other forms liable to shock. 
 Such rail tests have been carried on for many years in Europe, 
 and since 1890 in the United States by P. H. DUDLEY. 
 Already at least three prominent railroads require drop tests of 
 steel rails to be made at the mill. One rail butt 4^ feet long is 
 to be taken from each heat, placed on solid supports with either 
 head or base upwards, and a weight of 2000 pounds is dropped 
 upon it, the height of fall being 16 feet for rails lighter than 
 70 pounds per yard and 20 feet for heavier ones, while the dis- 
 tance between the supports is 3 feet for the former and 4 feet 
 for the latter. Under this test 90 per cent of the rails must 
 not break, and the elongation of the base or head under the 
 
220 THE RESILIENCE OF MATERIALS. CH. IX. 
 
 greatest tension must be more than 5 per cent. This require- 
 ment of testing by a single blow is in every respect more satis- 
 factory than by several repeated ones, as the deflection and de- 
 formation are produced by a known quantity of work and the 
 complex phenomena of stiffening and the loss of work in heat 
 are largely avoided. 
 
 Aside from the physical qualities of the metal, impact tests 
 on rails and wheels promise to give important conclusions re- 
 garding the influence of temperature, of chemical composition, 
 and of methods of manufacture, and thus to lead to a better, 
 more uniform, and cheaper product. The discovery by GOSS, 
 in 1 892, that the driving wheels of a locomotive lift up from the 
 rail during a part of each revolution when moving at high 
 speed shows that impacts are of more common occurrence than 
 generally supposed, and to satisfactorily resist these an in- 
 creased resilience in wheels, rails, and bridges is required. 
 
 At the closing session of the Engineering Congress held in 
 1893 in Chicago, a resolution was offered by DEBRAY that uni- 
 form methods of testing are desirable for purposes of compari- 
 son, and it was adopted unanimously. For common static 
 tests of tension and compression the time has certainly arrived 
 when rules to secure uniformity should be framed and followed. 
 The conferences held at Munich in 1884 and at Dresden in 
 1886 made a good beginning, and the later work done in 1891 
 by the Committee of the American Society of Mechanical 
 Engineers will undoubtedly bear good fruit. With respect to 
 impact, however, our knowledge is not yet sufficient to frame 
 uniform rules. The recommendation that the weight of the 
 anvil, or supporting blocks which hold the specimen, should be 
 at least ten times that of the ram is an excellent one, but 
 others fully as important will doubtless be developed by fur- 
 ther rational investigation and experiment. 
 
 The influence of the inertia of the resisting body in modify- 
 
ART. 99. MODERN EXPERIMENTS. 221 
 
 ing the effect of the impact should not be forgotten. If the 
 ram be light, local damage in the body struck will be the re- 
 sult rather than the development of its resilience. COTTERILL 
 has suggested for the case of transverse impact that if the total 
 work of the ram be taken as proportional to its weight plus 
 one-half the weight of the beam, then the work spent in de- 
 veloping resilience may be taken as proportional to the weight 
 of the ram and that spent in local damage to one-half the 
 weight of the beam. This is a good rule to keep in mind, for 
 longitudinal impact producing pure tension, one-third being 
 the fraction to be used instead of one-half. The fall of the 
 ram should not be too great, for a high velocity of impact is 
 also apt to cause the work to be spent locally, since time is re- 
 quired for the transmission of internal stress. The cutting of 
 hard steel plates by the impact of particles of sand at high 
 velocities is an example familiar to engineers, and that drops 
 of water will wear away stone is known to all ; hence a low 
 fall combined with a heavy ram seems best adapted for con- 
 ducting impact tests where the most satisfactory numerical 
 results are desired. 
 
 It may further be suggested that the blow should be de- 
 livered directly to the beam or specimen without the interven- 
 tion of intermediate blocks or parts which may absorb work. 
 At the same time the surface of contact should be sufficiently 
 large so that the compression of the ram may not, if possible, 
 exceed the elastic limit, and thus loss of work in heat be 
 avoided. In elastic resilience the applied work is not trans- 
 formed into heat, in resilience accompanied by permanent de- 
 formation it is ; and hence the tests should be so conducted 
 that the specimen and not the ram may be heated. Lastly, it 
 may be noted that the rebound of the ram should be subtracted 
 from the total fall to give an exact measure of the work 
 actually performed by it upon the body which is tested. 
 
222 TENSION AND COMPRESSION. CH. X. 
 
 CHAPTER X. 
 TENSION AND COMPRESSION. 
 
 ART. 100. ELONGATION UNDER OWN WEIGHT. 
 
 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 
 stress equal to the weight of the bar. In many practical cases 
 this stress is so small that it can be neglected in comparison 
 with that caused by the applied tension. 
 
 The elongation of a vertical bar under its own weight is one- 
 half that caused by the same load applied at the end. To 
 show this, let / be the length of the bar, A the area of the 
 cross-section, and W the weight ; let x be any distance from 
 
 x 
 the lower end, then Wj is the weight of this portion. The 
 
 elementary elongation caused by it on the length dx is, from 
 Art. 5, 
 
 Wxdx 
 
 and the integral of this between the limits o and / gives 
 
 Wl 
 
 A ' == ~2~AE> 
 which is one-half of that caused by a load W at the end. 
 
 The work performed by gravity in elongating the bar is one- 
 third of that done by the same load applied at the end. For, 
 the elementary work done upon the element dx by the lower 
 portion of the bar is 
 
 \Wx Wx^dx 
 
 ~2~T d ^ - 2AEI* ' 
 
ART. 101. BAR OF UNIFORM STRENGTH. 223 
 
 the integral of which, between the limits o and /, gives 
 
 while that done by Wat the end of the bar is %W\ or 
 
 The total elongation caused by the weight W and a load P 
 at the end hence is 
 
 and the total work of elongation is 
 
 These formulas, let it be remembered, are only valid when the 
 unit-stress (P + W)/A is less than the elastic limit of the 
 material. They apply also to the case of compression within 
 the elastic limit, A. being the amount of shortening. 
 
 Prob. 144. Find the length of a wrought-iron bar, and its 
 elongation, when suspended at the upper end so that the unit- 
 stress at that end may be equal to the elastic limit. 
 
 ART. 10 1. BAR OF UNIFORM STRENGTH. 
 
 A suspension bar of constant cross-section is stressed at the 
 lower end by the load P, and at the upper end by P plus the 
 weight of the rod. When the bar is very long and heavy its 
 section should be less at the lower than at the upper end in 
 order to economize material. The bar in such cases is some- 
 times made in parts, the cross-section of any part being less 
 than that of the one above it. 
 
 The theoretic form for a bar of uniform strength may be 
 determined as follows : Let P be the load applied to the 
 lower end, and let 5 be the allowable working unit-stress. 
 Then the area of the lower end is 
 
224 TENSION AND COMPRESSION. CH. X. 
 
 Let A be the area of any section at a distance y from the 
 lower end, then A + dA will be the area at the distance 
 y + dy, and the area dA must provide for the 
 
 weight in the distance dy. Let w be the weight 
 
 per cubic unit of the bar, then 
 
 dA = ^, 
 
 from which by integration, and observing that 
 A = A, when y = o, there is found 
 
 S* 
 
 iv 
 
 Replacing for S its value P/A 9 , this may be 
 
 FIG. 57. 
 
 written 
 
 wA 
 = log, -4. + --y, 
 
 where the logarithms are in the Naperian system. Passing to 
 common logarithms, it becomes 
 
 log A = log A. + 0.43429 -ply, 
 which is the form for practical computation. 
 
 For example, let P = 10000 pounds, S = $ ooo pounds per 
 square inch, and w = 0.25 pounds per cubic inch. Then A 9 
 = 2 square inches, and the formula becomes 
 
 log A = 0.30103 + O.OOOO2I7/. 
 
 Now for y = 100 inches, log A = 0.30320, and A = 2.01 square 
 inches; for y = 1000 inches, log A = 0.32274 and A =2.10 
 square inches ; while when/ = lOOOO inches, A 3.30 square 
 inches. Thus it is only in the case of very long bars that an 
 appreciable increase in cross-section is found. 
 
 Prob. 145. Let a pier whose top section is a rectangle of 
 length / and breadth b support the load P, as in Prob. 26. 
 Deduce the value of log x in terms of b, /, and P. 
 
ART. 102. LONGITUDINAL IMPACT.' 22$ 
 
 ART. 102. LONGITUDINAL IMPACT. 
 
 By longitudinal impact is understood the impinging of a 
 moving body upon the end of a bar so as to produce in it 
 either tensile or compressive stress. In Art. 93 this subject 
 was discussed and formulas were deduced for the elongation 
 and stress under impact. These formulas, however, take no 
 account of the resisting influence of the inertia of the bar, 
 which may modify the results when the weight of the bar is 
 large compared to that of the moving body. 
 
 Let a body of weight P be free to move along a horizontal 
 bar, and let v be its velocity when it reaches the end of the 
 
 FIG. 58. 
 
 tf 
 bar. The work then stored in it is P } or Ph, if h be the 
 
 height due to the velocity v. This work is expended in over- 
 coming the inertia of the particles of the bar and in elongating 
 it through a distance y. A series of oscillations then results 
 until finally the end of the bar comes to rest at its original 
 position, provided that the elastic limit be not exceeded by the 
 maximum stress produced. 
 
 The load P has the velocity v before it strikes the end of 
 the bar. When in complete contact both P and the end of 
 the bar are moving with the velocity V which is less than v. 
 At this instant any element of the bar dW\s moving with a 
 velocity u. Thus the work stored up by P and the bar at this 
 instant of complete contact is 
 
 F a C l u 9 
 K = P + I dW.. 
 
 ** Jo 2 S 
 
226 TENSION AND COMPRESSION. CH. X. 
 
 Now u o for the fixed end, and u = V for the free end of 
 the bar, and in general u is proportional to the distance x from 
 the fixed end. Thus 
 
 and 
 and introducing these values, the integral in the above expres- 
 
 sion is found to be \W , or one third of that which would 
 obtain if the entire bar were in motion. Then 
 
 is the work stored up by load and bar when the end of the bar 
 and the load are moving with the common velocity V. 
 
 Now at this instant the impulse of the two bodies is equal to 
 the impulse of P before striking, whence 
 
 Pv 
 
 or F =/> + ^ 
 and hence the work K is 
 
 P* v* Pv* Ph 
 
 ~ P + ^W 2g " 2g(i + P) == i + \V 
 
 in which k denotes the ratio of W to P. This work is ex- 
 pended in elongating the bar through the distance y, the inter- 
 nal stress increasing from o up to Q y so that the entire internal 
 work is \Qy. Hence 
 
 ifi^rpp 
 
 is the equation between external and internal work. 
 
 If the elastic limit be not exceeded the forces Q and P are 
 proportional to the elongations they can produce. Let \ be 
 the static elongation due to P. Then Q/P = y/\ and the 
 equation gives 
 
 ~2hK~ I 2h 
 
 y = 
 
ART. 102, LONGITUDINAL IMPACT. 
 
 from which the elongation y and the stress Q are deter- 
 mined. 
 
 The static elongation \ due to a load P and a cross-section 
 A is, from Art. 5, 
 
 and this may be inserted in the above formulas either in literal 
 or numerical form. The formulas apply equally well to the 
 case of compression where the load P impinges upon the end 
 of the bar, provided that / be not so long that lateral flexure 
 occurs. 
 
 If the bar be placed in a vertical position the additional 
 work Py is performed while the load descends through the dis- 
 tance /, and the equation of work is 
 
 Ph 
 
 which leads to the formulas 
 
 (20) 
 
 and these are the same as those of Art. 93 if the ratio k be 
 made equal to o. 
 
 The influence of the weight of the bar is hence to diminish 
 the elongation and stress under impact. For instance, in the 
 case of the horizontal bar let 5" be the unit-stress produced if 
 
 
 the weight of the bar be very small ; then / = is the unit- 
 
 VI + & 
 
 stress when the weight of the bar is k times that of the load. 
 Thus if W = P the unit-stress is 0.8/5, and if W = gP the 
 unit-stress is 0.56"; but these high values of k are unusual. 
 
228 TENSION AND COMPRESSION. CH. X. 
 
 As a numerical example, let it be required to find the stress 
 produced in a vertical wrought-iron bar, one square inch in 
 section and 18 feet long by the impact of a body weighing 30 
 pounds falling through a height of one foot. Here W = 60 
 pounds and k = 2. The static elongation is 
 
 PI 
 
 A = ~r- = o.ooo 2592 inches. 
 
 Then, as h = 12 inches, formulas (20) give 
 
 y = o.ooo 2592 X 236.7 = 0.0614 inches, 
 Q = 30 X 236.7 = 7100 pounds, 
 
 which shows that very high stresses may be produced by the 
 impact of light bodies falling through moderate heights. 
 
 Prob. 146. A weight of 60 pounds moving horizontally im- 
 pinges upon the end of a bar of wrought iron 2 inches in diam- 
 eter and 12 feet long. Find the velocity v which will stress 
 the bar up to the elastic limit. 
 
 ART. 103. OSCILLATIONS AFTER IMPACT. 
 
 Referring to the case of longitudinal impact shown in Fig. 
 58', it is clear that the velocity of the weight P decreases after 
 it strikes the end of the bar, and that the Velocity at any 
 instant is a function of the corresponding elongation x. When 
 x equals the final elongation y, the velocity is zero ; the end 
 of the bar now springs back so that the velocity becomes 
 negative, and this increases numerically until x = o, and then 
 decreases until it becomes zero at x = y ; the oscillation is 
 next performed in the opposite direction. These oscillations 
 would continue indefinitely were it not for resistances of fric- 
 tion, but owing to these they become less and less in ampli- 
 tude until finally the bar comes to rest. 
 
 Let v x be the velocity of the end of the bar when the elon- 
 gation x is attained, and let Q x be the corresponding stress in 
 the bar. The kinetic energy.of the moving bar and weight 
 
ART. 103. OSCILLATIONS AFTER IMPACT. 22Q 
 
 then equals the internal work still to be performed in increas- 
 ing x to y, or 
 
 Replacing Q by P.y/^ and Q x by P. x/\, this becomes 
 
 which gives the velocity of the end of the bar for any value 
 of x. When x = -\- y or x = y this velocity is zero. 
 
 To find the time in which the oscillation from -\- y to y 
 is performed, let / be the time counted from the instant when 
 x = o. Putting v x = dx/dt, the last equation becomes 
 
 dt = 
 
 g vy - **' 
 
 the integration of which gives 
 
 -V 
 
 
 When ^r = y the arc is \n, and when x = y the arc is f TT. 
 Hence the number of seconds in one oscillation is 
 
 =V- 
 
 which is the same as the time of oscillation of a pendulum 
 whose length is A(i + \K). 
 
 As a numerical example let the data of the last problem be 
 considered. Here P = 60 pounds, W= 125.7 pounds, k = 
 2.09, ^ = 0.00011 inches, g = 32.2 X 12 inches per second 
 per second. Then the time of one oscillation is t = 0.022 
 
230 TENSION AND COMPRESSION. CH. X. 
 
 seconds, and about 460 oscillations would be performed in one 
 second were it not for frictional resistances. 
 
 For the case of the vertical bar the first equation of this 
 article will be modified by subtracting P(y x) from the 
 second member, this expressing the potential energy to be 
 expended by the weight in falling through the distance^ x. 
 The expression for the time is found to be 
 
 t = \ / - ' versin 
 
 g y - A' 
 
 which shows that the middle of the oscillation occurs at x = A 
 and that the bar finally comes to rest with the elongation A. 
 The time of one oscillation is the same as before 
 
 If a weight W^ be fixed at the end of the bar and the weight 
 P impinge upon it, then the formulas of this and the last 
 article apply if I + t^ De replaced by I + k, . + \k, in which 
 k, is the ratio WJP. 
 
 ART. 104. CENTRIFUGAL STRESS. 
 
 When a body of weight P revolves around an axis with the 
 uniform velocity v y and r is its distance from the axis, a cen- 
 trifugal force Q is generated whose value, as deduced in theo- 
 retical mechanics, is 
 
 and which acts as a stress in the cord or bar that connects the 
 body with the axis. The case shown in Fig. 59 is that of a 
 
 FIG. 59. 
 bar of uniform cross-section and length / having a weight P 
 
ART. 104. CENTRIFUGAL STRESS. 231 
 
 attached to one end while it revolves around an axis A at the 
 other end. It is required to find the centrifugal stress in the 
 bar at A when the speed of n revolutions per second is main- 
 tained. 
 
 Let x be any distance from the axis the velocity at this 
 distance is 2nxn, or, if GO be the angular velocity, the velocity at 
 
 the distance x is XGO, and 
 
 oo 
 GO = 27tn, or n = . 
 
 Now let Wbe the weight of the bar, and dW 'an element at 
 the distance x. Then the centrifugal stress at A is 
 
 Q = P oo 9 + \ dW 
 
 & I/O 
 
 g* 
 
 But dW = Wdx/l\ inserting this and integrating, 
 
 o 
 
 which gives the centrifugal stress at the axis. 
 
 As an example let a bar of wrought iron 2X2 inches and 6 
 feet long have a weight of 400 pounds attached at 6J feet from 
 the axis of revolution. It is required to find the number of 
 revolutions per second in order to produce rupture. Solving 
 the last equation for a?, there results, 
 
 . Qz 
 - Pr + twr 
 
 in which Q = 2 X 2 X 55 ooo = 220000, P = 400 pounds, 
 W = 80 pounds, g = 32.16 feet per second per second, / = 6 
 feet, r = 6 J feet; then GO = 50.8, and 
 
 50.8 
 n = -- =8.1 revolutions per second, 
 
 27T 
 
 which is the speed required. 
 
 Another case is that of a thin circular rim or hoop of mean 
 radius r and thickness / which revolves uniformly around its 
 
232 TENSION AND COMPRESSION. CH. X. 
 
 center as an axis. Let W be its weight, which is closely 
 equal to 2nrtw y if w be the weight of the material per cubic 
 
 unit and the length perpen- 
 dicular to the plane of the 
 drawing be unity. The total 
 radial centrifugal force due 
 to the angular velocity GO is 
 
 FIG. 60. Q = 
 
 o 
 
 and the centrifugal force per square unit is 
 
 Q 
 
 which acts upon the hoop in the same manner as the internal 
 pressure of fluid in a pipe (Art. 9). Let 6" be the tangential 
 tensile unit-stress caused in the hoop ; then for equilibrium 
 
 2rp = 2/5, 
 and hence the value of 5 is 
 
 Tb w w 
 
 S=f = -^=-(2nrn)\ 
 
 which shows that the tensile unit-stress in a thin hoop is r times 
 the centrifugal force of a cubic unit of the revolving material. 
 
 For example, let it be required to find the unit-stress in a 
 cast-iron hoop 2 inches thick, 4 inches wide, and 62 inches 
 outer diameter when making 300 revolutions per minute. 
 Here w = 450/1728 pounds per cubic inch, g 32.16 X 12 
 inches per second per second, r = 30 inches, n = 5 revolutions 
 per second ; then 5 is found to be 630 pounds per square inch, 
 which is a safe value for cast iron in tension under such con- 
 ditions. 
 
 Prob. 147. A cast-iron bar is 3 X 2 inches in section, and 
 9 feet long. Through the middle and normal to the flat side 
 is a hole f inches in diameter. If the bar be revolved around 
 an axis through this hole, how many revolutions per second 
 will produce rupture? 
 
ART. IO5. SHRINKAGE OF HOOPS. 233 
 
 ART. 105. SHRINKAGE OF HOOPS. * 
 
 Hoops and tires are frequently turned with the interior 
 diameter slightly less than that of the wheels or cylinders upon 
 which they are to be placed. They are then expanded by- 
 heat and placed in position, and upon cooling are held firmly 
 in position by the radial stress thus produced. The effect of 
 this radial stress is to cause tension in the hoop, and compres- 
 sion throughout the mass that it encircles. 
 
 When the hoop is thin compared to the diameter of the 
 cylinder upon which it is to be shrunk, the entire stress due to 
 the shrinkage may be practically regarded as confined to the 
 hoop. The tangential unit-stress in the hoop will then be due 
 only to the increase in length of the circumference, and this 
 will be proportional to the increase in its diameter. 
 
 Let D be the diameter of the cylinder upon which the hoop 
 is to be shrunk, and d be the interior diameter to which the 
 hoop is turned. Supposing that D is unchanged by the shrink- 
 age, d will be increased to D y and the relative change in length 
 or the unit-elongation of the hoop will be 
 
 D-d 
 ~~d~' 
 
 and hence the unit-stress produced will be 
 
 where E is the coefficient of elasticity of the material. 
 
 A very common rule for the case of steel hoop shrinkage is 
 
 to make D d equal to yVfrrA that is, the hoop is turned so 
 that the interior diameter is T^ryth less than the diameter 
 of the cylinder. Then 
 
 D-d i 
 
 or s may be taken also as y^. Thus the tangential unit- 
 
234 TENSION AND COMPRESSION. CH. X. 
 
 stress S will be 20 ooo pounds per square inch. For wrought 
 iron s may be taken as 2 fa Q and S will be 12 500 pounds per 
 square inch. 
 
 These values of 5 are too high, because a part of the effect 
 of shrinkage is expended in producing compression in the 
 cylinder. In Art. 142 the subject will be further discussed, 
 and the final decrease in D as well as the increase in d will be 
 determined. 
 
 Prob. 148. 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. Com- 
 pute the tensile unit-stress in the hoop. 
 
 ART. 1 06. SPHERICAL ROLLERS. 
 
 Let a sphere of radius r be placed between two plates of 
 equal thickness r, and be subject to compression by a load W. 
 
 The vertical diameter DD is 
 thus shortened to BB, while 
 any vertical line dd is short- 
 ened to bb. Let the total 
 shortenings CD be called A, 
 and the shortenings cd at 
 any point be called y. Let 
 the unit-stresses at B and b 
 be denoted by 5" and S y . 
 Then, if the elastic limit is 
 FIG. 61. not exceeded, 
 
 5 X S 
 
 s,=y or ^ = i* 
 
 The unit-stress 5 is evidently the maximum, and it is required 
 to express it in terms of W, r, and r r This will now be done 
 approximately, assuming that each vertical element acts inde- 
 pendently of the others and that no lateral bulging of the 
 sphere occurs. 
 
 
ART. I06. SPHERICAL ROLLERS. 235 
 
 The value of A may be expressed by noting that BD is the 
 shortening of the radius r, and CB that of the thickness r l ; 
 thus 
 
 in which E and E l are the coefficients of elasticity of the 
 sphere and plates respectively. Also the sum of all the verti- 
 cal stresses in the spherical segment PDF must equal the total 
 load, or 
 
 = ydA, 
 
 where dA denotes an elementary area of the circle whose 
 radius is CF. Thus two equations have been found connect- 
 ing the two unknown quantities 5 and A. 
 
 To solve these equations consider that y is any ordinate cd 
 of the spherical segment corresponding to an abscissa Cc or x. 
 
 Thus fydA is the volume of the segment, which is very nearly 
 
 equal to one-half the cylinder having CF as the radius of its 
 base and CD as its altitude, since the arc of contact FBF is 
 very small. Now CD = A and CF = i/2r\ A' = 
 nearly, and hence 
 
 W = T - 7T2r\ = 
 
 A. 2 
 
 Inserting for \ its value, this becomes 
 
 which is an approximate formula for the investigation of 
 spherical rollers when the upper and lower plates are of equal 
 thickness. 
 
 Compression tests made upon spheres between plane sur- 
 faces show that usually the plates are not indented as shown 
 by FBFC in Fig. 61, but that the sphere alone suffers mate- 
 
236 TENSION AND COMPRESSION, CH. X. 
 
 rial deformation. In this case BC is zero, and thus the last 
 formula applies independently of the thickness of the plates 
 if r l /E l be made zero. The practical formula for spherical 
 rollers of radius r then is 
 
 W = nS*r*/E, 
 
 or, the strength of a sphere varies with the square of its 
 radius. Strictly this formula only applies when the elastic 
 limit is not exceeded, but for cases of rupture W= Cr* is an 
 empirical formula, C being a constant found by experiment 
 for each kind of material. 
 
 Prob. 149. How many steel spherical rollers are required 
 for a load of 6 ooo pounds and a working stress of 1 5 ooo 
 pounds per square inch if r = 2 inches, and also if r = 6 
 inches? 
 
 ART. 107. CYLINDRICAL ROLLERS. 
 
 The reasoning of the last article applies also to a cylindrical 
 roller included between two plates of equal thickness. Let 
 Fig. 6 1 represent a cross-section perpendicular to the axis of 
 the roller whose length is /. Then, as before, 
 
 are the two equations for determining 5. The area A is that 
 of a plane with width FF and length /, so that jydA is the 
 
 volume of the cylindrical segment whose cross-section is FDF. 
 Since the area of FDF is very small, it is closely two-thirds of 
 the rectangle whose base is FF and altitude is CD. Now CD 
 = A and FF = 2 ^2r\ very nearly, and hence 
 
 . 2 
 
ART. ID/. CYLINDRICAL ROLLERS. 237 
 
 and this becomes, after inserting the value of \ , 
 
 which is an approximate formula for the investigation of 
 cylindrical rollers. 
 
 In making tests on rollers it is found that the plates are usu- 
 ally but little indented, most of the deformation being in the 
 roller. The above investigation may be adapted to this case by 
 making r l / 1 equal to zero, and the last formula then becomes 
 
 =M/ 
 
 V E 
 
 which shows that the strength of a cylindrical roller varies 
 directly as its radius. If w be the load per unit of length, then 
 
 'Is 5 
 
 Taking 5= 15000 and E = 30000000 pounds per square 
 inch for steel or cold rolled iron, this reduces to w = 6$or or 
 w = 315^, where d is the diameter of the roller. 
 
 It should be noted that a common rule used in proportion- 
 ing cylindrical rollers for bridge seats is to take the safe load 
 in pounds per linear inch as 500 V d, thus making the strength 
 vary with the square root of the diameter. This practice 
 seems to be based upon the authority of GRASHOF, who was 
 the fisst to deduce the formula at the top of this page, but 
 who appears to have placed a doubtful interpretation upon it. 
 In general, the true rule is that the load should be nearly 
 proportional to the diameter of the roller. 
 
 Prob. 150. A load of 192 ooo pounds is carried on wrought- 
 iron rollers 16 inches long and 3 inches in diameter. How 
 many rollers are required if 5 is to be 12 ooo pounds per 
 
 square inch ? 
 
2 3 8 
 
 TENSION AND COMPRESSION. 
 
 CH. X. 
 
 ART. 108. ECCENTRIC LOADS. 
 
 Let a load P be applied to the end of a short bar at a hori- 
 zontal distance / from the center of gravity of its cross-section, 
 as shown in Fig. 62. If two forces equal to P 
 and acting in opposite directions be supposed 
 to be applied to the end of the bar, the equi- 
 librium is undisturbed, and it is seen that the 
 downward force produces pure tension, while 
 the couple formed by P and the other force 
 produces flexure. Let A be the cross-section 
 of the bar and >S the maximum unit-stress 
 caused by the flexure ; then the resultant unit- 
 stresses are 
 
 P _P 
 
 1 ~~ A ' ' * ~~ A '" ' 
 the former being on the side of the bar nearest 
 FIG. 62. to p anc j tne latter on the other side. 
 
 Let / be the moment of inertia of the cross-section with 
 respect to an axis through the center of gravity and normal to 
 the direction of the arm /. Let r be the radius of gyration of 
 the cross-section so that./ = Ar*. Let c l and c t be the dis- 
 tances from the axis to the sides of the section nearest to and 
 farthest from P. Then from the formula (4) of Art. 21 the 
 two values of S are found, and the above expressions reduce 
 to the practical formulas 
 
 which give the greatest and least tensile stresses under the 
 eccentric application of the load. 
 
 These formulas apply also to compression under an eccen- 
 tric load, provided that the bar be short so that no lateral 
 flexure can occur. As an example let a short block have a 
 
ART. 1 08. 
 
 ECCENTRIC LOADS. 
 
 239 
 
 rectangular section, the dimension parallel to the arm p being 
 d, while the one normal to it is b. Then the formulas reduce 
 
 and in Fig. 63 are shown the distribution of stresses for several 
 values of /. In the first sketch P is applied at the center of 
 the section so that the unit-stresses are uniform and both S 
 
 and S, are equal to -j. In the second sketch/ is taken as 
 
 
 p 
 c 
 
 p 
 c 
 
 p 
 
 c 
 
 c /flll 
 
 \ 
 
 
 
 I 
 
 
 1 
 
 m 
 
 JIP 1 * 
 
 
 
 ' 
 
 p 
 
 JP* 
 
 Si 
 
 ( 
 
 f 
 
 f 
 
 
 
 
 
 
 FIG. 63. 
 
 3 P l P 
 
 which gives ,= --- and S, = - . In the third sketch / is 
 
 2 si 2 A 
 
 p 
 
 taken as \d, so that S t = 2 and S 9 ^= o. As the load moves 
 
 A 
 
 further away from the center C, the stress 5", increases, while 
 S t becomes negative, showing that it is tension. Thus in the 
 last sketch where p is equal to \d the compressive unit-stress 
 
 P P 
 
 5, is 3 -r, while the tensile unit-stress 5, is 2 -j. In all these 
 
 ./i A 
 
 P 
 
 cases the unit-stress at the center C is the mean value -r. 
 
 It is thus seen that the eccentric application of a load may 
 materially increase the direct stress of tension or compression. 
 In the case of a rectangular masonry pier the greatest devia- 
 tion of P from the center should never be greater than \d, in 
 
240 TENSION AND COMPRESSION. CH. X. 
 
 order that no tension may be brought upon the joint. In 
 other words the point of application of P should be kept 
 within the * middle third ' of the base. 
 
 In both the cases above discussed the body under tension or 
 compression has been supposed to be very short, so that no 
 material lateral movement can result. If in Fig. 62 the bar be 
 long the arm / will be different for different parts of the 
 length, and the greatest bending moment will occur at the 
 lower end, so that the formulas given apply only to that end, 
 while for other sections they give results too large. For 
 compression, however, the reverse is the case, and application 
 can only be safely made to blocks whose length does not 
 exceed ten times the thickness (see Art. 62). 
 
 Prob. 151. A bar of circular cross-section, 2 inches in diam- 
 eter, is under tension by a load of 12000 pounds which is 
 applied at 0.5 inches from the center. Compute the maximum 
 tensile unit-stress. 
 
ART. 109. THE WORK OF FLEXURE. 241 
 
 CHAPTER XL 
 FLEXURE OF BEAMS. 
 
 ART. 109. THE WORK OF FLEXURE. 
 
 This subject was treated in Art. 96, but it will now be dis- 
 cussed again in order to deduce a more general relation be- 
 tween the work of the external forces and that of the internal 
 stresses. 
 
 Let P be a load upon a beam and A the deflection under it. 
 The load being gradually applied the work done by P is %Pd 9 
 and this must be equal to the internal work or resilience of 
 the molecular stresses. 
 
 For a beam loaded uniformly with w per linear unit let y 
 be the deflection at any point whose abscissa is x. Then the 
 load wdx on the short distance dx deflects the amount j, and 
 the elementary external work is \wydx. Thus the integral of 
 this, if y is known in terms of x, will give the total work of 
 the uniform load, if the integration^ be extended over the 
 entire length of the beam. This will be equal to the internal 
 work, or resilience. 
 
 A general expression for the resilience, or internal work of 
 the horizontal fibers, in terms of the bending moment M of 
 the applied forces will now be deduced ; it is applicable to all 
 cases in which the elastic limit of the material is not surpassed 
 by the maximum fiber stress S. 
 
 When a beam deflects under the action of a load the hori- 
 zontal fibers upon one side of the neutral surface are elongated 
 and upon the other side are compressed. The internal work 
 will be found by taking the sum of the products formed by 
 
242 FLEXURE OF BEAMS. CH. XI. 
 
 multiplying the stress upon any elementary area by its elonga- 
 tion or compression. 
 
 Using the same notation as in Chapter III., the horizontal 
 unit-stress at any distance z from the neutral axis is represented 
 
 by . In the distance dx the elongation or compression due 
 
 C* ^7 
 
 to this unit-stress, is by (2) found to be . The elementary 
 
 work of a fiber of the area a under this gradually applied unit- 
 stress hence is, 
 
 i Saz Szdx 
 ~2'~7~ cE * 
 
 The work done in the distance dx by all the fibers in the cross- 
 
 section now is, 
 
 , S'Zaz* , 
 dK = - irr-dx. 
 
 2?E 
 
 S* . M* 
 Here ^a = /and from formula (4), the value of 5- is - 
 
 TM f 77^- * 
 
 Therefore dK = 
 
 . 
 
 2EI 
 
 This is the formula for the work done in the distance dx. By 
 expressing M as a function of x, and integrating, the total in- 
 ternal work K between assigned limits can be found. 
 
 For example, consider a cantilever beam loaded at the end 
 with a weight P. Here M = Px. Inserting this and in- 
 tegrating between the limits o and /, gives, 
 
 for the total internal work in the beam due to a load which is 
 gradually applied. 
 
ART. 109. THE WORK OF FLEXURE. 243 
 
 The preceding furnishes a new method of deducing the de- 
 flection of a beam loaded with a single weight P. Let A be 
 the deflection under the weight. Then \PA is the external 
 work done by the load P upon the beam, and this must equal 
 the internal work K. Hence the formula, 
 
 PA- C M * dx 
 
 from which A may be found for particular cases. 
 
 For example, consider a cantilever beam loaded at the end 
 with P. Then the internal work is, as shown above, 
 Hence the deflection A is, 
 
 A =-. Pl * 
 
 which is the same as otherwise found in Art. 34. 
 
 Px 
 
 For a simple beam loaded at the middle the value of Mis 
 
 2 
 
 and then 
 
 from which the deflection is, 
 
 which is the same as found in Art. 35 by the use of the elastic 
 curve. 
 
 Prob. 152. Prove that the internal work caused by a uni- 
 formly distributed load on a cantilever beam is ^ths of that 
 caused by the same load applied at the end. 
 
 Prob. 153. Deduce by the method of Art. 35, and also by 
 the use of the principle of internal work, the deflection under 
 a load P which is placed upon a simple beam at a distance / 
 from one end. 
 
244 FLEXURE OF BEAMS. CH. XI. 
 
 ART. 1 10. STATIC AND SUDDEN DEFLECTIONS. 
 
 A static deflection is one produced by the gradual applica- 
 tion of the load, so that at each instant the beam is in a condi- 
 tion of static equilibrium and hence no oscillations occur. The 
 formulas for the deflection of beams deduced in Chapters III 
 and IV, as well as the discussion of the last Article, are all static 
 deflections where the elastic limit of the material is not sur- 
 passed. In such case the load increases from o up to its maxi- 
 mum value P, and simultaneously the deflection increases 
 from o up to its maximum value A. If Q be the load at any 
 instant and 8 the corresponding deflection, then the deflections 
 are proportional to the loads or 6/A = Q/P. 
 
 A load is said to be suddenly applied when it acts with uni- 
 form intensity during the full period. For instance, let a load 
 be attached to a ring which is placed around the middle of a 
 beam, and let the load be supported so .that the ring just 
 touches the upper surface of the beam ; then if the support of 
 the load be suddenly withdrawn the force of gravity acts upon 
 the load with uniform intensity during the entire period of de- 
 flection. In this case the maximum deflection is greater than 
 for a static load, but as soon as it is reached a series of oscilla- 
 tions occurs until finally the beam comes to rest with a deflec- 
 tion due to the static load. The maximum stress in the beam 
 evidently occurs at the instant of greatest deflection. If S be 
 the unit-stress due to the static load P under the deflection A, 
 and if T be the unit-stress due to the sudden load P under the 
 deflection tf, both loads being applied at the same point on the 
 same beam, then from Art. 37 the stresses are proportional to 
 the deflections, or 6/A = 775. 
 
 The deflection under a sudden load is double that under the 
 same static load, and the stress under a sudden load is double 
 that under the same static load. In order to prove this refer 
 
ART. 1 10. STATIC AND SUDDEN DEFLECTIONS. 245 
 
 to Art. 96 and note that the total internal work or resilience 
 due to the maximum unit-stress S in the beam is 
 
 K=\? E .C.Al, 
 
 where Al is the volume of the beam, and C is a constant de- 
 pending upon the shape of its cross-section and the arrange- 
 ment of its ends. Thus for a given beam the internal work 
 varies as S 2 . Now for the gradually applied load P the deflec- 
 tion is A and the external work is \PA. Hence as internal 
 work equals external work, 
 
 which gives the relation between A and 5. Again for the 
 sudden load the deflection is d and the maximum unit-stress is 
 T\ the external work, however, is P3, since the load acts with 
 full intensity during the entire period of deflection, while the 
 stress increases gradually from o up to T\ hence 
 
 i r f 
 PS = - -jCAl, 
 
 which gives the relation between d and T. By comparing 
 these two equations there is found 
 
 and remembering that T/S = <?/^, as shown above, this gives 
 
 6 = 24, and T 25, 
 which proves the proposition as stated. 
 
 Another method of establishing the same truth is to com- 
 pare the sudden load P with the static load W which will pro- 
 duce the same deflection A and hence the same unit-stress 5. 
 The internal work is thus the same in the two cases, but the 
 external work of the sudden load is PA, while that of the static 
 load is %WA. Therefore P = %W\ that is, the sudden load 
 that produces a given stress is one-half of the static load that 
 produces the same stress. 
 
246 FLEXURE OF BEAMS. CH. XL 
 
 These propositions are true only when the maximum stresses 
 caused by the deflections are within the elastic limit of the 
 material, since all the reasoning by which they are established 
 supposes the law of proportionality of stress to deformation to 
 be observed. It is hence not to be expected that they can be 
 verified by the rupture of beams. It has, however, been found 
 to be approximately true for cast iron, but for wrought iron 
 and steel the sudden load that produces rupture is greater 
 than one-half of the static load. 
 
 The following experiments of HODGKINSON on cast-iron 
 beams illustrate very well the agreement of theory with prac- 
 tice near the elastic limit and the disagreement when the 
 elastic limit is exceeded. Each beam was laid on supports 9. 
 feet apart, and a lever at the middle prevented deflection when 
 the load was applied. This was hung to the beam by two 
 rings, one on each side of the lever and placed as close to it as 
 possible without touching. The lever being instantaneously 
 removed the load was brought suddenly into action, and the 
 deflection was registered upon a vertical sheet of paper by 
 means of a pencil screwed to the side of the beam. Before 
 applying the loads in this sudden manner the beams were tested 
 in the usual way by gradually applied loads. The results here 
 given are the mean of two or three tests upon different beams. 
 It will be found that for each beam the first load gives a unit- 
 stress less than the elastic limit, while the second gives a 
 
 Size of Beam. 
 Inches. 
 
 Load. 
 Pounds. 
 
 Deflection 
 Gradual. 
 
 in Inches. Ratio of Deflections, 
 Sudden. Sudden to Gradual. 
 
 I X 2 
 
 112 
 
 0.253 
 
 0.515 
 
 2.04 
 
 I X 2 
 
 224 
 
 O.58O 
 
 0-933 
 
 1.61 
 
 IX 3 
 
 224 
 
 0.163 
 
 0.303 
 
 1.86 
 
 I X 3 
 
 560 
 
 0.4IO 
 
 0.720 
 
 1.76 
 
 4X it 
 
 448 
 
 0.770 
 
 1.510 
 
 1.96 
 
 4 X it 
 
 784 
 
 1.275 
 
 2.225 
 
 i-73 
 
 greater value. Thus for the first beam under 112 pounds at 
 
ART. III. DEFLECTION UNDER IMPACT. 247 
 
 the middle S is 4 540, and for 224 pounds 5 is 9070 pounds 
 per square inch. 
 
 The breaking loads of these beams were also observed under 
 both gradual and sudden loads with the following results : 
 
 Size of Beam. Breaking Load in Pounds. Ratio of Loads, 
 
 Inches. Gradual. Sudden. Gradual to Sudden, 
 
 1X2 IOOO 569 1.76 
 
 i X 3 2008 1219 1.65 
 
 4X4 1911 1082 1.77 
 
 Here the ratio is seen to be about the same as for the heavier 
 
 loads in the cases of deflection. 
 
 KIRKALDY'S experiments on wrought-iron beams show 
 that the ratio of the gradual to the sudden breaking load 
 ranges between 1.2 and 1.3, so that the elastic law does not 
 even approximately apply. There is a very good reason why 
 it should not apply, and for this the student may consult the 
 Proceedings of the Engineers' Society of Western Pennsyl- 
 vania for May, 1894. 
 
 Prob. 154. Compute the static deflection and the unit-stress 
 5 for the third experiment on the cast-iron beams given above. 
 
 ART. in. DEFLECTION UNDER IMPACT. 
 
 As the deflection under a sudden load is greater than for a 
 static load, so the deflection under impact is greater still. A 
 load falling through a small height upon a beam may easily 
 produce a deflection and a stress that may prove dangerous to 
 its stability. In Arts. 93 and 103 rules were deduced for bars 
 subject to impact, and a similar investigation leading to similar 
 results will now be made for beams. 
 
 Let the beam be placed in a horizontal position and let a 
 weight P move horizontally with the velocity v so as to strike 
 it at the middle and cause a lateral deflection. The work 
 
248 FLEXURE OF BEAMS. CH. XI. 
 
 V* 
 
 stored up by the moving weight is P 9 or Ph if h be the 
 
 o 
 
 height due to the velocity. Let W be the weight of the beam, 
 6 the maximum deflection produced by the impact, and T the 
 corresponding maximum unit-stress. Let A be the deflection 
 due to a static load P t and S be the corresponding maximum 
 unit-stress. Then d/A = T/S, or the deflections are propor- 
 tional to the stresses. Also, let Q be a static load which will 
 produce the deflection d, then likewise Q/P = <$/A, or the 
 static loads are proportional to the deflections. The internal 
 work or resilience of the beam now is %Qd, and this must equal 
 the external work Ph, or 
 
 Q6 = .2f.d = Ph. 
 
 Also substituting for tf its value in terms of T, S, and A, an 
 equation between 5 and T results. Thus, 
 
 (22) * 
 
 are the formulas for finding the maximum deflection and cor- 
 responding unit-stress in a simple beam under impact at the 
 middle, A and 5 being given by the expressions of Chapter III, 
 
 ~ mEP ~ n' 
 
 where / is the length of the beam, m and n numbers depending 
 upon the arrangement of its ends, / the moment of inertia of 
 the cross-section, and c the distance from the neutral axis to 
 the fiber whose stress is 5 or T. 
 
 The above results will be somewhat modified by the resist- 
 ance of inertia of the beam, as in the case of longitudinal im- 
 pact discussed in Art. 103. An instant before the contact the 
 weight P has the velocity v and the beam is at rest ; when the 
 contact is complete both P and the middle of the beam are 
 moving with a common velocity V. Plainly V is less than v, 
 
ART. III. DEFLECTION UNDER IMPACT. 249 
 
 V* 
 
 as a portion of the energy P has been communicated to the 
 
 beam whose weight is W. When the velocity V is acquired 
 any element dW is moving with the velocity u, and the com- 
 bined energy of weight and beam then is 
 
 V* f* u* 
 
 K = P-- dW = 
 
 V* f* 
 
 = P-- + / 
 
 *g r J 
 
 and this is the energy which is effective in producing the de- 
 flection. But also, as soon as the velocity V is acquired the 
 condition of impact of must obtain, namely, 
 
 Pv = (P+qW)V. 
 
 Eliminating V from these equations gives 
 _ />V Ph 
 
 ~~~-~- i + qk> 
 
 in which k is the ratio W/P and q is a number whose value is 
 to be determined. 
 
 The above value of K being placed equal to J<2#, as before, 
 there are found 
 
 (22 y * = X /^g, T=S^^ } , , 
 
 as the modified formulas for deflection and stress due to hori- 
 zontal impact. 
 
 It now remains to determine the value of ^, and this will de- 
 pend upon the arrangement of the ends of the beam. The 
 general value of q is seen to be 
 
 and the integration is to be extended over the entire length of 
 the beam. Now dW/W = dx/l\ also if y be the deflection at 
 any point and y^ the deflection at the point whose velocity is 
 F, then u/ V = y/y r Thus the integral becomes 
 
 />v f 
 = J y ;i> 
 
FLEXURE OF BEAMS. CH. XL 
 
 which may be applied to any particular case where y/y^ can be 
 expressed as a function of x. 
 
 For a beam supported at the ends and loaded in the middle 
 the ordinate y of the elastic curve in terms of the maximum 
 deflection^ and the abscissa x is, from Art. 35, 
 
 and for this case the integral becomes 
 
 Therefore for a beam supported at the ends and impinged 
 upon in the middle there obtains 
 
 which is the value to use in formula (22)'. This result was 
 first established theoretically by Cox in 1848, but HODGKIN- 
 SON had previously found by experiments on cast-iron beams 
 that the value of q was about \. 
 
 As a numerical example let a cast-iron beam on supports 9 
 feet apart be I inch wide and 2 inches deep, and have a static 
 load of 50 pounds at the middle. The deflection and stress at 
 the middle due to this load are 
 
 A 0.131 inches, 5 = 2025 pounds per square inch. 
 Now suppose that this load of 50 pounds moves horizontally 
 with a velocity due to a fall of inch. Then from (22) 
 
 d = 0.362 inches, T= 5590 pounds per square inch, 
 which are the results, supposing that the beam has no weight. 
 Taking this into account, the weight W is 56.4 pounds, whence 
 ' = 1.128, and by (22)' 
 
 d = 0.290 inches, T = 4470 pounds per square inch, 
 which are more exact results. 
 
 It is seen by this investigation that very small velocities of 
 
ART. III. DEFLECTION UNDER IMPACT. 2$ I 
 
 impact may produce very high stresses in a beam. Thus in 
 (22) the static deflection A is always small, and if h = 2-J, Tis 
 28. It is also seen that the influence of the resisting inertia of 
 the beam increases with k, that is, with the ratio of the weight 
 of the beam to the impinging weight. 
 
 When the weight falls vertically upon the horizontal beam, 
 h being the height of fall to the top of the beam, the formulas 
 (22) and (22)' are to be modified slightly, since the external 
 work is P(h -f- 6). Thus are found 
 
 which are seen to be the same in form as those deduced for 
 longitudinal impact in Art. 103. For instance, if the load in 
 the above example drops % inch, then $ = 0.295 inches instead 
 of 0.290 inches. 
 
 The interesting experiments made by KEEP in 1899 enable 
 a comparison of theory and practice to be made. A hori- 
 zontal bar of cast iron I X i X 24 inches was loaded with 
 weights of 25, 50, 75, and 100 pounds, and the corresponding 
 static deflections were found to be 0.0448, 0.0896, 0.1344, 
 and o. 1792 inches. They were then struck laterally by ham- 
 mers of the same weights which 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 the theo- 
 retic deflections under impact have been computed from 
 formula (22)', and the following is a comparison of observed 
 and theoretic values : 
 
 Swinging P = 25 50 75 100 pounds 
 
 Observed 3 = 0.122 0.150 0.175 0.200 inches 
 Theoretic d = o.ioo 0.145 0.180 0.209 inches 
 
FLEXURE OF BEAMS. CH. XI. 
 
 Experiments were also made by allowing the same weights to 
 fall vertically on the bar through heights of 2 inches. For- 
 mula (22)" applies to this case, and the following is a com- 
 parison of the dynamic deflections as observed and computed : 
 
 Falling P = 25 50 75 100 pounds 
 
 Observed 8 = 0.130 0.159 0.181 0.209 inches 
 Theoretic S = 0.103 0.152 0.188 0.220 inches 
 
 It is seen that the comparison is very satisfactory ; it would 
 be expected, however, that the observed values should be 
 always slightly less than the theoretic ones, because some of 
 the work of impact is probably expended in producing heat. 
 Prob. 155. Check some of the above values of theoretic <?. 
 
 ART. i n. OSCILLATIONS AFTER IMPACT. 
 
 The formulas of the last article apply to the momentary 
 maximum deflection under the impact of the weight. A 
 series of oscillations then ensues, and finally, if the weight 
 remains on the beam, it comes to rest with a deflection due 
 to the static load. The following figure, illustrating these 
 phenomena in an interesting manner, is from an automatic 
 record taken by P. H. DUDLEY in 1895. A railroad rail 30 
 feet long and weighing 80 pounds per yard was placed at its 
 extreme ends upon rigid supports and had a scale-pan of 210 
 pounds weight suspended from the middle. Secured to the 
 center of the rail was an attachment carrying a pencil which 
 recorded upon paper moving horizontally the deflections and 
 oscillations of the rail. A load of 100 pounds being suddenly 
 applied, the maximum deflection was 0.240 inches, but when 
 the beam came to rest the static deflection was found to be 
 o. 120 inches. The weight of 100 pounds was next dropped 
 from the height of 12 inches, and the maximum deflection was 
 0.910 inches; about 240 oscillations then ensued, as shown in 
 the figure, and in about 42 seconds the rail came to rest. 
 
ART. 112. 
 
 PRESSURE DUE TO IMPACT. 
 
 253 
 
 By an investigation exactly similar to that of Art. 103 
 it may be shown that the time of one oscillation of a simple 
 beam, when struck either horizontally 
 or vertically, is given by 
 
 g 
 
 in which k is the ratio of the weight 
 of the beam to that of the falling 
 load, or k = W/P. If P falls in a 
 scale-pan whose weight is W lt then 
 I _|_ \^k is to be replaced by I + , + 
 \\k, in which k, is the ratio WJP. 
 For the above case P= 100, W l =2\o, 
 and W 800 pounds ; also A =. o. 120 
 inches; and for these data the formula 
 gives / = 0.146 seconds for the time 
 of one oscillation, and accordingly the 
 theoretic time for 240 oscillations is 
 about 35 seconds; owing to air resis- 
 tance, however, this was prolonged to 
 about 42 seconds. 
 
 Prob. 156. Prove that || is to be 
 replaced by ^| for the case of impact 
 against a beam fixed at both ends. 
 
 ART. 112. PRESSURE DUE TO 
 IMPACT. 
 
 When a weight P falls from a 
 height h upon a beam, a pressure is 
 produced at the point of contact. This 
 pressure is variable during the period 
 of deflection, being zero at the first 
 instant of contact, rapidly increasing to a maximum, and then 
 decreasing, until at the moment of greatest deflection there 
 
254 FLEXURE OF BEAMS. CH. XI. 
 
 exists a pressure sufficient to cause the weight to rebound a 
 short distance. The laws governing the variation of the press- 
 ure are not understood, but the mean or average pressure 
 existing during the impact can be satisfactorily ascertained. 
 
 Let h be the height of fall above the top of the rail, and d 
 the deflection produced by P falling through this height. Let 
 R be the mean or average pressure existing between the rail 
 and weight during the impact. The work of the falling ram is 
 P(h -f- ) and the work of the mean pressure R is R3. Equat- 
 ing these values, there is found 
 
 from which R can be computed when d has been measured. 
 
 This formula is correct both for elastic and non-elastic de- 
 flections. If the elastic limit be not exceeded, d can be com- 
 puted by (22)" of the last article. If the elastic limit is 
 exceeded, 8 can only be found by actual measurement. 
 
 To illustrate, let a ram weighing 2 ooo pounds fall from a 
 height of 20 feet upon a railroad rail, this being one method of 
 testing rails at the mill. The rail will be deflected an amount 
 #, depending upon its size, shape, length, and the quality of 
 the material. For 8 = i inch R will be 482 ooo pounds ; for 
 d = 2 inches R will be 242 ooo pounds ; for tf = 4 inches R 
 will be 1 22 ooo pounds. Thus the mean pressure decreases 
 approximately inversely as the deflection. 
 
 The maximum pressure in such cases cannot be theoreti- 
 cally ascertained, but it probably is not as great as double the 
 mean pressure. The maximum pressure per square inch will 
 depend, of course, upon the area of contact between the ram 
 and the rail, and upon the manner in which the total pressure 
 is distributed over that area. 
 
 Prob. 157. Compute the mean pressure on the rail for the 
 data given at the foot of page 252. 
 
ART. 113. CENTRIFUGAL STRESS. 255 
 
 ART. 113. CENTRIFUGAL STRESS. 
 
 The rod that connects the cross-head of an engine with the 
 crank pin is subject to a centrifugal stress owing to the fact 
 that one end revolves in a circle. The horizontal rod, or 
 parallel bar joining two driving wheels of a locomotive is an- 
 other instance of centrifugal stress ; this is simpler than the 
 connecting rod, because all points are revolving with the same 
 velocity, and hence it will be discussed first. 
 
 Let V be the velocity of a locomotive in feet per second, 
 and v the velocity of revolution of the end of the parallel rod 
 around the axle of the driver to which it is attached. Let R 
 be the radius of the driver, and r the radius of the circle of 
 revolution of the end of the parallel rod. Then since the 
 
 B B 
 
 FIG. 64. 
 
 velocity of revolution of the circumference of the driver is the 
 same as that of the speed of the train, the value of v is 
 
 Now, not only the end but every point in the parallel rod is 
 revolving with the velocity v in a circle whose radius is r. 
 Thus a centrifugal force is generated which produces stresses. 
 When the rod is at its lowest position BB, this centrifugal 
 force acts as a downward uniform load producing flexure ; at 
 the highest position AA it acts as an upward uniform load 
 producing flexure ; at the position CC, on the same level as 
 the axles, it produces a compressive stress in the direction of 
 the length of the rod. 
 
256 FLEXURE OF BEAMS. CH. IX. 
 
 Let w be the weight of the parallel rod per linear unit ; then 
 from mechanics the centrifugal force of this weight is 
 
 which may be called the centrifugal load per unit of length, 
 The rod being a beam supported at its ends, having a length /, 
 a breadth b y and a depth d, the maximum unit-stress due to 
 this uniform load is, from (4), 
 
 __ 
 / ~ 4&/ 8 ' 
 
 which is the flexural stress due to centrifugal force when the 
 bar is at its highest or at its lowest position. 
 
 In this formula * is the acceleration of gravity, or 32.16 feet 
 per second per second. In applying it numerically, however, 
 all quantities should be expressed in terms of the same linear 
 unit, the inch being preferable. For example, let a locomo- 
 tive be running at 60 miles per hour, the radius of the drivers 
 being 3 feet and that of the parallel rod I foot, this being of 
 steel, 4 inches deep, 2 inches thick, and 8 feet long. Here V 
 = 88 feet per second =12 X 88 inches per second, g = 32.16 
 X 12 inches per second per second, R = 3 X 12 inches, r = 
 I X 12 inches, / = 8 X 12 inches, b = 2 inches, d = 4 inches, 
 and w = 2.27 pounds per linear inch. The centrifugal load 
 per inch then is 
 
 . 2.27 X 12 X (88 X 12)' 
 f= 32.16 X 12 X (3X12)- = 6l P unds > 
 and then the maximum fiber stress is 
 
 e 3 X 61 X (8 X I2) a 
 
 5 = - f. - = 13 200 pounds per square inch, 
 
 4 X 2 x 10 
 
 which is not probably sufficiently low when it is considered 
 that the parallel rod is subject to vibrations and shocks. 
 
 The connecting rod moves in a circle of radius r at the crank 
 pin, while the other end moves only in a straight line. Thus 
 
ART. 114. LIVE-LOAD VELOCITY. 257 
 
 at the end A there is no centrifugal load, while at B the cen- 
 
 trifugal load is the same as given by the above expression 
 
 for/". When the rod is in the 
 
 position shown in Fig. 65, it is a 
 
 beam acted upon by a centrif- 
 
 ugal load which varies uni- 
 
 formly from o at A to / at B. 
 
 The total load is hence \fl, the ' ' FlQ 6 
 
 reaction at A is \fl and that at 
 
 B is 4/7. The bending moment for any section distant x from 
 
 A is 
 
 and the maximum value of M occurs for x = // 1/3, which 
 gives max. M = 0.0638/7*. Hence from (4), 
 
 // 
 
 ^ = 0.383^,, 
 
 in which f is given by the same expression as above. 
 
 By comparing this with the value of 6" for the parallel rod it 
 is seen that the former is about twice as great, if the length 
 and cross-section be the same in the two cases. The parallel 
 rod needs the greatest cross-section at the middle, while the 
 connecting rod needs the greatest cross-section at about o.6/ 
 from the cross-head. 
 
 Prob. 158. The connecting rod of an engine is 2 feet long 
 and it is attached to a crank pin at a distance of 6 inches from 
 the axis of a fly-wheel. If the wheel makes 750 revolutions 
 per minute, find a square cross-section for the connecting rod 
 so that the centrifugal unit-stress 5 may be 4200 pounds per 
 square inch. 
 
 ART. 114. LIVE-LOAD VELOCITY. 
 
 It is well known that when a live load moves over a beam 
 or bridge the deflections and stresses are greater than those 
 
258 FLEXURE OF BEAMS. CH. XI. 
 
 due to the same load at rest. In general the greater the veloc- 
 ity the greater also are the deflections and the stresses. The 
 exact theoretical investigation of this question is one of very 
 great difficulty, as differential equations arise which cannot be 
 integrated except by an unsatisfactory tentative process. 
 Approximate formulas have, however, been established, and one 
 of these will here be deduced. 
 
 Let a static load P rest on the middle of a simple beam. 
 From Chapter III the deflection and the maximum unit-stress 
 under the load are 
 
 A C - 
 
 ~' : ' 
 
 where / is the length of the beam, 7 the moment of inertia of 
 the cross-section with respect to the neutral axis, c the dis- 
 tance from that axis to the remotest fiber where the unit-stress 
 is S. Now, let the load P be moving horizontally across the 
 beam with the velocity v, and when it reaches the center let 
 the deflection be d and the maximum unit-stress be 71 It is 
 required to find $ and T in terms of A and 5. 
 
 When the load P runs over the beam, the curve in which it 
 moves is found by putting kl=. x in the first equation on page 
 77, or 
 
 Differentiating this twice, and then making x = /, gives 
 d*\ i PI 
 
 as the reciprocal of the radius of curvature of this curve at 
 the middle of the beam. 
 
 It is assumed that when P reaches the middle of the beam 
 it produces the same deflection A as if it were at rest, and also 
 that it causes a downward pressure F due to the centrifugal 
 force arising from motion in the curve. This pressure F gives 
 
ART. 114. LIVE-LOAD VELOCITY. 259 
 
 rise to an additional deflection, thus increasing A to d, and 5 
 to r- Then, 
 
 Now, to find TS the expression for centrifugal force is known 
 from mechanics, and inserting in it the above value of R, 
 there results 
 
 /V _ P'tv* 2P*lh 
 
 ~~ == 
 
 in which i?/2g has been replaced by h, the height of fall which 
 will produce v. In this last expression P/EI may be replaced 
 by its value in terms of A or by its value in terms of 5 from 
 the first equations given. Making these substitutions in the 
 formulas for tf and S, they become 
 
 (23) 
 
 which give the approximate deflection and maximum unit- 
 stress at the middle of the beam due to a load P moving with 
 the velocity v = ^/2gh. 
 
 As an example let a wrought-iron plate girder have a span 
 of 80 feet, a depth of 7 feet 2 inches, a flange cross-section of 
 38 square inches, and a moment of inertia of about 134000 
 inches. Let it be required to find the deflection and maximum 
 unit-stress when a single load of 60 ooo pounds crosses the 
 girder at a velocity of 80 miles per hour. Here P = 60 OOO 
 pounds, /= 960 inches, 7= 134000 inches, c = 43 inches, 
 and E 25 ooo ooo pounds per square inch. Then when the 
 load is at rest at the middle of the beam, 
 
 ^ = -33 inches, S = 4620 pounds per square inch. 
 
26o FLEXURE OF BEAMS. CH. XL 
 
 Now, a speed of 80 miles per hour corresponds to a velocity of 
 117 feet per second, and the velocity head is 
 
 Then from (23) are found the increased deflection and unit- 
 stress, 
 
 6 = 0.33(1 + 0.029) = 0.34 inches, 
 
 T = 4620(1 + 0.029) = 4750 pounds per square inch, 
 which show the influence of the velocity to be small. 
 
 When a uniform live load is moving over the beam or bridge, 
 a similar investigation may be made, regarding the centrifugal 
 force at each point as a vertical load. Let w be the uniform 
 live load per linear unit ; then when this extends over the 
 whole beam, 
 
 5^/ 4 
 ~ 
 
 87 ' 
 
 are the static deflection and maximum fiber stress at the mid- 
 dle. Let 8 be the deflection and T the unit-stress when the 
 entire uniform load is moving with the velocity v, and let h be 
 the head due to this velocity. Then by a method similar in 
 principle to the above, it may be shown that 
 
 ,1 + 
 
 (23/ 
 
 -X. 
 
 which are the approximate deflection and unit-stress at the 
 middle of the beam due to the moving load wl. 
 
 As an example take the plate girder whose data are given 
 above and let a uniform load of 1800 pounds per linear foot be 
 moving over it. Then 
 
 A = 0.495 inches, 5=5 545 pounds per square inch, 
 are the static deflection and unit-stress at the middle. As 
 before, h = 213 feet, and then from (23)' 
 
ART. 114. LIVE-LOAD VELOCITY. 26l 
 
 6 = 0.506 inches, T = 5 670 pounds per square inch, 
 which are only about 2.2 per cent greater than the static values. 
 It should be remarked that the allowance made for velocity 
 of a live load in practice is much greater than the above for- 
 mulas indicate. It is often customary to add from 10 to 30 
 per cent to the statical stresses in order to cover the effect of 
 velocity, the greater values being used for the shorter bridges. 
 It should also be said that experiments indicate a greater 
 increase in deflection than these formulas give. The most ex- 
 tensive set of experiments in this direction is that made by 
 JAMES, WILLIS, and GALTON, for the British board of 1848, 
 and in some of these the statical deflection was more than 
 doubled under heavy loads. 
 
 It may be further noted that a perfect formula for the effect 
 of velocity of live load would show in the case of very high 
 speeds that there would be no increase in deflection, since 
 there would then not be sufficient time for the load to fall 
 through the distance tf. This was recognized by the board 
 above mentioned, and the fact ascertained in several tests. 
 For instance a wrought-iron beam 9 feet long, I inch wide, and 
 3 inches deep was subjected to a load of 1778 pounds moving 
 at different velocities, with the following results : 
 
 Velocity in feet per second, o 15 29 36 43, 
 Deflection in inches, 0.29 0.38 0.50 0.62 0.46. 
 
 Here it will be seen that the deflection for 43 feet per second 
 is less than that for 36 feet per second. Unfortunately these 
 experiments were made without regard to the elastic limit of 
 the material, and hence the results are of little use as a test 
 of theory; for instance, the static load of 1778 pounds causes 
 a unit-stress of 32 ooo pounds per square inch on the middle of 
 the wrought-iron beam, and thus at all velocities the elastic 
 limit was surpassed. 
 
 The formulas of this article are confessedly imperfect, as 
 
262 FLEXURE OF BEAMS. CH. XL 
 
 they do not take into account the influence of the inertia of 
 the beam which will tend to modify them materially. This 
 very complex question cannot here be investigated, but the 
 student is referred to Appendix B of the Report of the Com- 
 missioners on the Application of Iron to Railway Purposes 
 (London, 1849) f r an excellent general discussion. It may 
 also be noted that the above formula (23) agrees with the first 
 two terms of the series given on page 203 of that Report. 
 
 Prob. 1 59. Deduce the values of $ and T given in formula 
 (23)' for the uniform moving load. 
 
 Prob. 160. What velocity v must the load P have so that, 
 in crossing the above plate-girder, there would not be sufficient 
 time for it to fall through the vertical deflection of 0.33 inches ? 
 
 ART. 115. WORK OF VERTICAL SHEARS. 
 
 In the discussion of Art. 109 the entire external work of 
 the load P was supposed to be expended in the work of elon- 
 gating and compressing the horizontal fibers of the beam. In 
 reality, however, a part of the external work is expended in 
 the slipping or detrusion due to the vertical shears throughout 
 the beam. 
 
 As in Chapter III suppose the vertical shear to be uniformly 
 distributed over the cross-section of the beam. 
 Let Fbe the vertical shear at any distance x from 
 an origin and be the angle of detrusion in the 
 distance dx. Then taking the shear to increase 
 slowly from o up to its value V, the work done by 
 it in the distance dx is, since is very small, 
 
 dK=\V.dx. tan = %V(J)dx, 
 in which in the last value is to be taken in 
 circular measure. 
 
 FIG. 66. Let A be the cross-section of the beam, 5 S the 
 
 shearing unit-stress, and E s the coefficient of elasticity for 
 
 dx 
 
ART. US. WORK OF VERTICAL SHEARS. 263 
 
 shearing. As is the amount of detrusion per unit of dis- 
 placement, its value is found by the same process as the unit- 
 deformation s for tension or compression, namely 
 
 s, v 
 *--.=-- ~AE; 
 
 This being inserted, the expression for dK becomes 
 
 V ' d * 
 
 which is the elementary work of shearing in the distance dx. 
 By expressing V as a function of x and integrating over the 
 entire length of the beam, the total work of shearing is found. 
 
 For instance, a simple beam loaded with P at the middle 
 has the shear V constant throughout and equal to \P. Then 
 
 K > 
 
 = 
 
 is the internal work or resilience due to all the shearing forces 
 over the span /. In Art. 109 the internal work of the horizon- 
 tal stress was found to be 
 
 so that the ratio of the former to the latter is 
 K^ 12EI Ef 
 
 K' 
 
 in which r is the radius of gyration of the cross-section with 
 respect to the neutral axis. 
 
 For example, let a cast-iron beam have a square cross-section 
 of side d\ then r a = T VA and E = %E S approximately (Art. 
 121). The ratio of the internal work of shearing to that of 
 the fiber stresses is then 5*/ a /2/ 3 . If the length is 30 times 
 the depth, or / = 30^, then this ratio is -^ ; if / = 6o*/, the 
 ratio is 4 * 4 . For a very short beam, such as / = 2d, the 
 ratio is , showing that in such cases the work of shearing is 
 greater than that of the horizontal fiber stresses. 
 
2b4 FLEXURE OF BEAMS. CH. XI. 
 
 Prob. 161. Deduce an expression for the work of shearing in 
 a beam supported at the ends and uniformly loaded. 
 
 ART. 116. DEFLECTION DUE TO SHEARING. 
 
 The treatment of the deflection of beams in the previous 
 pages has been solely from the standpoint of the horizontal 
 stresses as expressed by the external bending moment. The 
 last Article shows, however, that the resisting work of shearing 
 may be a material amount for short beams, and it hence 
 appears that the former investigations are more or less incom- 
 plete. 
 
 Let a load P produce the deflection A beneath it. The ex- 
 ternal work, if P has been gradually applied, is %P4, and this 
 must equal the internal work, or resilience, of the molecular 
 stresses. The work of the horizontal stresses of tension and 
 compression is deduced in Art. 109 and that of the vertical 
 stresses of shearing in Art. 115. Then the [external work 
 equated to the sum of these, gives 
 
 C M*dx . CVdx 
 J 
 
 in which M is the bending moment and V the vertical shear at 
 any section distant x from the origin. To apply this to a par- 
 ticular case, M and V are to be expressed in terms of x and 
 the integration extended over the entire length of the beam. 
 
 For example, let a simple beam of span / have a load P at 
 the middle. Then M= \Px and V = \P. Inserting these 
 and bearing in mind that each integral is equal to twice the 
 value between the limits o and /, there is found, 
 
 A- Pr . Pl 
 ~ 4 87 ~ h 4^' 
 
 which is the deflection under the load. The second term here 
 gives the deflection due to the vertical shears. By placing 
 
ART. 1 1 6. DEFLECTION DUE TO SHEARING. 265 
 
 A = 7/r a , where r is the radius of gyration, this becomes 
 
 = 
 
 which is the formula for deflection, taking into account the 
 effect of the vertical shears. For long beams the deflection 
 due to shearing is scarcely appreciable ; for short beams, how- 
 ever, it may be larger than that due to the bending moment. 
 
 Attention was first called to the inaccuracy of the ordinary 
 formula for deflection in the case of short beams in a paper by 
 NORTON read before the American Association for the Ad- 
 vancement of Science in 1870. A number of experiments on 
 white pine beams of different lengths and sizes were made, and 
 it was shown that the deflections were directly proportional to 
 the loads and inversely proportional to the breadth of the 
 beam, as the common formula requires. The deflections were, 
 however, not directly proportional to the cubes of the spans 
 nor inversely proportional 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 deduced the formula 
 
 ~~ 
 
 as applicable to beams of breadth b and depth d, where C was 
 a constant whose value he did not theoretically determine. 
 From one series of experiments he found for the white pine 
 beams 
 
 E = 1 428 ooo, C = 0.0000094, 
 
 and using these values the deflections for other series com- 
 puted from the formula agreed very well with those observed. 
 
 From the theoretic formula for A deduced above, it is seen 
 that NORTON'S value of C is 1/4 E s , or 
 
 E s = ^ = 266 ooo pounds per square inch, 
 4C 
 
266 FLEXURE OF BEAMS. CH. XL 
 
 which should be the coefficient 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 experi- 
 ments on torsion, found E s = 220 ooo pounds per square inch 
 for white pine. The experiments of NORTON, therefore, con- 
 firm the theoretic formula above deduced for the true deflec- 
 tion of a simple beam loaded at the middle. 
 
 When a uniform load extends over the beam and it is de- 
 sired to find the deflection at a particular point, let M and V 
 be the bending moment and vertical shear due to the uniform 
 load, and M' and V the bending moment and vertical shear 
 due to a load P' placed at the particular point. Then the 
 deflection at that point is given by 
 
 CV'Vdx 
 
 I ^JP'E^ 
 
 in which both bending moments and vertical shears are to be 
 expressed in terms of x, and the integration extended over 
 the entire length of the beam. For instance, let a simple 
 beam have the uniform load wl, and let it be required to find 
 the deflection at the middle. Here M' = fP 'x, M=\w(lx 
 x*\ V' = \P' > and V '= \wl wx. Inserting and integrating 
 between the limits o and /, there results 
 
 wl* 
 
 which is the deflection at the middle, the first term being the 
 deflection due to the horizontal stresses of tension and com- 
 pression and the second that due to the vertical stresses of 
 shearing. The ratio of the second term to the first is seen to 
 be slightly greater than in the case of a single load at the 
 middle. 
 
 Prob. 162. Prove formula (24)' by considering that the work 
 due to the imaginary load P ' is equal to the stress caused by 
 
ART. II/. FLEXURE AND COMPRESSION. 267 
 
 that load multiplied by the deformation caused by the stress 
 due to the total uniform load. 
 
 ART. 117. FLEXURE AND COMPRESSION. 
 
 Let a beam be subject to flexure by transverse loads and 
 also to a compression in the direction of its length. If the 
 longitudinal compression be not large the combined maximum 
 stress due to flexure and compression may be computed by 
 the approximate method of Art. 74. It is clear, however, 
 that if the compression be large the deflection A will be 
 increased, and hence the effective bending moment and maxi- 
 mum fiber stresses will be greater than given by that method. 
 A closer approximation will now be established. 
 
 Let P be the longitudinal compressive force and M the 
 bending moment of the flexural forces. Let M l be the actual 
 bending moment for the section where the deflection is A ; 
 this is greater than M, on account of the moment PA of the 
 force P, or M^ = M -{- PA. Now the maximum fiber unit- 
 stress 5, which results from this moment M l is, from (4), 
 _ M,c _ (M+PA)c 
 
 * l = / 7~ "' 
 
 where I is the moment of inertia of the cross-section and c 
 the distance from the neutral axis to the remotest fiber. 
 The value of A may be expressed in terms of 5", regarding A 
 to vary with Sj in the same manner as for a beam subject to 
 no longitudinal compression. Inserting then for A its value 
 from Art. 37, and solving for S 19 gives 
 
 Me 
 ~ = 
 
 _ 
 
 mE 
 
 where n and m are numbers depending upon the arrangement 
 of the ends and the kind of loading. This formula was first 
 deduced by J. B. JOHNSON, who regarded m/n as 10 for all 
 kinds of loading. Art. 37 shows, however, that m/n depends 
 
268 FLEXURE OF BEAMS. CH. XL 
 
 on the arrangement of the ends as well as on the load ; for a 
 simple beam uniformly loaded m/n = 9.6, and for a load at 
 the middle m/n =12. 
 
 The maximum compressive unit-stress on the concave side 
 of the beam is 5 = 5 X -|- P/A. For example, let a wooden 
 beam 8 feet long, 10 inches wide, and 9 inches deep be under 
 a compression of 40 ooo pounds, while at the same time it 
 carries a total uniform load of 4000 pounds. Here M = \WL 
 48 ooo pound-inches, c = 4^ inches, / = -r^bd* 
 inches 4 , / = 96 inches, P = 40 ooo pounds, n = 8, m = 
 and E = I 500 ooo pounds per square inch. Inserting these 
 values in the formula the flexural stress S l is found to be 371 
 pounds per square inch. The compressive unit-stress due 
 directly to P is P/A = 40 000/90 = 444, so that the total 
 stress 5 = 371 -{- 444 =815 pounds per square inch. 
 
 While the above method is better than that of Art. 74, it 
 is not exact, and gives in general values of 5 which are too 
 small. The exact method of dealing with combined flexure 
 and compression is by the help of the elastic curve. The 
 results will now be developed for the most common case. 
 
 Let a simple beam of span / be uniformly loaded with in 
 per linear unit, and at the same time be under the longitudinal 
 compression P. The bending moment for any point whose 
 co-ordinates are x and y is, 
 
 M = \wlx %wx* + />, 
 and the differential equation of the elastic curve is, 
 
 where the negative sign of the bending moment is taken 
 because the curve is concave to the axis of x. By two 
 integrations results 
 
 wlx wx* wEI /cos fi(x J/) \ 
 ~~2P " ~2P + ~7*~\ ~ V f 
 
ART. 117. FLEXURE AND COMPRESSION. 269 
 
 in which /?, as in Art. 62, is an abbreviation for (P/EI}*, or 
 
 is an arc expressed in terms of the radius as unity. In this 
 equation of the elastic curve let x = i/, then^ = J, and 
 
 Inserting this in the expression for S t , there is found 
 
 which is an exact expression for the maximum compressive 
 flexural unit-stress. Lastly, 5, -f- P/A is the total unit-stress 
 5 due to the combined flexure and compression. 
 
 To illustrate this method let the data of the above numeri- 
 cal example be again used. Here w = 4000/96 pounds per 
 linear inch, and the other quantities as before. The arc /?/ 
 is found to be 0.318, and the corresponding angle is 18 13', 
 whence sec \ftt = 1.0545. Then from the formula the 
 flexural unit-stress S l is 416 pounds per square inch. Lastly, 
 the total unit-stress is 416 + 444 = 860 pounds per square 
 inch. A comparison for this numerical example shows that 
 the rough method of Art. 74 gives 799, and the method of 
 JOHNSON 815, while the exact method gives 860 pounds per 
 square inch for the maximum unit-stress. 
 
 If w = o in the above formulas the case reduces to that of 
 a column where sec i/?/ = oo , and hence both A and 5, are 
 indeterminate. If on the other hand P = o, the case is that 
 of a simple beam uniformly loaded, and it may be shown that 
 A and 5, will reduce to the expressions for that case. 
 
 Prob. 163. Show, by the calculus method of evaluating 
 indeterminate quantities, that the statement in the last sen- 
 tence is correct. 
 
270 FLEXURE OF BEAMS. CH. XL 
 
 Prob. 164. Let a simple wooden beam 32 feet long, 9 
 inches wide, and 10 inches deep, carry a total uniform load of 
 2000 pounds while at the same time it is under a longitudinal 
 compression of 9000 pounds. Compute the maximum unit- 
 stress S by the three methods. 
 
 ART. 118. 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 tension is to decrease the deflection, and thus also the 
 tensile flexural stress. If M be the bending moment of the 
 transverse loads, and M l that of the combined flexure and 
 tension, then M^ = M PA. Let S, be the resulting unit- 
 stress on the fiber most remote from the neutral surface; 
 then, JOHNSON'S formula in the last article gives S, , if the 
 minus sign in the denominator be changed to plus. Finally, 
 S, -f- PI -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 
 80000 pounds, E being 29000000 pounds per square inch. 
 The weight of the bar is 490 pounds, and M = j- X 490 X 18 
 X 12 = 13 230 pound-inches. Also c = 4 inches, /= 42.67 
 inches 4 , m/n= 9.6, P =. 80000 pounds, /= 216 inches. 
 Then the maximum flexural tensile stress S l is 943 pounds 
 per square inch. Finally, the total tensile stress is 5 = 943 
 -f- 10 ooo = 10 943 pounds per square inch. 
 
 As in the last article, a more accurate way of dealing with 
 this case is by use of the general equation of the elastic curve. 
 The expression for the bending moment is, 
 M = \wlx \ivx* />, 
 
 which is the same as before, except in the sign of P. Hence 
 by changing the sign of P in the expressions for A and 5, , 
 they apply to the case of combined flexure and tension. In 
 
ART. Il8. FLEXURE AND TENSION. 
 
 doing this \fil becomes ^filV-- i, and this changes the cir- 
 cular secant to the hyperbolic secant; thus, 
 
 is the deflection of the beam, and 
 
 S, = ~(i - sech 
 
 is the unit-stress due to the flexure. Finally S t + P/A is the 
 total unit-stress due to the combined flexure and tension. 
 
 Since many students are unacquainted with hyperbolic 
 functions, it may be here noted that they are closely analogous 
 with the circular functions. Thus for circular functions cos* 
 -f- sin a = i, but for hyperbolic functions cosh' sinh" = I. 
 A table of hyperbolic sines, cosines, and tangents is given in 
 " Higher Mathematics" (Wiley & Sons, 1896). In the 
 absence of a table the hyperbolic cosine and secant can be 
 computed from 
 
 e 6 + e~ e 
 cosh 6 r - , sech = 
 
 where e is the base of the Naperian system of logarithms. 
 
 As an example, for the above eye-bar w = 2.267 pounds 
 per linear inch, and i/?/ = 0.868 = #; then sech 6 = 0.714, 
 and the flexural stress S, = 940 pounds per square inch. 
 Lastly, the total unit-stress 5 = 940 -f- 10000 = 10940 
 pounds per square inch, which differs but little from the result 
 found before. 
 
 Prob. 165. Compute the deflection A for the above eye- 
 bar before and after the tension is applied. 
 
 Prob. 166. Show that the deflection of a cantilever beam 
 loaded at the end with W and under the longitudinal tension P 
 is J = WP-\l - ft' 1 tanh /?/), where ft = (///)*. 
 
272 
 
 SHEAR AND TORSION. 
 
 CH. XII 
 
 CHAPTER XII. 
 SHEAR AND TORSION. 
 
 ART. 119. STRESSES CAUSED BY SHEAR. 
 
 It is shown in Art. 8, and also in Art. 75, that forces of ten- 
 sion or compression acting upon a body produce not only 
 internal tensile or compressive stresses, but also internal shear- 
 ing stresses. Conversely, an external shear acting upon a 
 body produces in it not only internal shearing stresses, but also 
 internal tensile and compressive stresses. 
 
 For example, the rectangle ABCD in the web of a plate 
 girder, shown in Fig. 67, 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 F,, which in this case is caused by 
 the resistance of the flange > rivets. Let 
 the thickness of the material be one unit ; 
 then if 5 and S l be the shearing unit- 
 stresses, 
 
 A f} y i A f~V 
 
 ^1Z> ./I// 
 
 and it is now to be shown that 5 and 5 X are equal. Taking 
 either A or D as a center of moments, the equation of mo- 
 ments is, 
 
 V X AD = V, X AB, 
 
 FIG. 67. 
 
ART. 119. STRESSES CAUSED BY SHEAR. 2/3 
 
 and hence by division 
 
 V V 
 
 AB = AD> r 5=5 
 
 that is, the shearing unit-stresses on adjacent sides of the rec- 
 tangle 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 be very small the conclusion is strictly true that the 
 vertical shearing unit-stress is equal to the horizontal shearing 
 unit-stress (Art. 79). 
 
 The resultant of V and F, acts as a tension on the diagonal 
 BD and as a compression on the diagonal AC, thus tending to 
 deform the rectangle into a rhombus. The maximum value of 
 this resultant will be when F, = F, that is, when the rectangle 
 is a square.. The resultant tension or compression then acts 
 at an angle of 45 degrees with the length of the beam, and its 
 value is VjH. The tensile or compressive unit-stress is ob- 
 tained by dividing F\/2 by the area normal to its direction, 
 or 
 
 that is, the tensile or compressive unit-stress caused by shear 
 is equal to the shearing unit-stress. 
 
 This may also be proved from the discussion in Art. 75. 
 Thus in formula (13) let / = o; then max. / = v y which is 
 the same result. The action of tension or compression on a 
 bar produces a shearing unit-stress equal to one half the tensile 
 or compressive unit-stress, but the action of a shear \ oduces 
 tensile and compressive unit-stresses equal to the sheai ng unit- 
 stress itself. This may be regarded as a most fortunate arrange- 
 ment in view of the fact that the shearing strength f materi- 
 als is usually less than the tensile or compressive strength. 
 
 Prob. 167. A square bar is subject to a tension of 6000 
 
274 SHEAR AND TORSION. CH. XII. 
 
 pounds per square inch in the direction of its length and to a 
 lateral compression of 2000 pounds per square inch on two 
 opposite sides. Show that the maximum shearing unit-stress 
 in the bar is 4000 pounds per square inch. 
 
 ART. 120. RESILIENCE UNDER SHEAR. 
 
 The resilience of a body under the action of shear is gov- 
 erned by similar laws to that of tension and flexure, namely, it 
 is proportional to the square of the maximum unit-stress and to 
 the volume of the body. Thus in Fig. 68 let a shearing unit- 
 stress S g act upon a parallelepiped of length / and cross-section 
 A, deforming it into a rhombus and causing each right angle 
 on the side to be increased or decreased by 
 the amount 0. The external work done 
 by the total shear V, supposing it be 
 gradually applied, is VI tan 0, or since 
 is a very small angle, simply F/0, if 
 be expressed in circular measure. This is 
 equal to the internal work or resilience of 
 FIG. 68. all the shearing stresses. But F= AS,, 
 
 and if E t be the coefficient of elasticity for shearing, then 
 5, = ,0(Art. 4). Therefore 
 
 is an expression for the work of shearing. The first factor 
 .St/2E t may be called the modulus of resilience for shearing, 
 in analogy with that for tension (Art. 94), and the second fac- 
 tor Al is the volume of the body. This expression, however, 
 is only valid when the stress S s is within the elastic limit of the 
 material. 
 
 Practically a shear cannot act upon a body of any consider- 
 able size without causing flexure or torsion, and thus only a 
 part of the external work will be expended in the internal 
 work of shearing. Hence the above rule for resilience under 
 
ART. 121. THE COEFFICIENTS OF ELASTICITY. 
 
 275 
 
 shear is of limited application, unless the effect of the accom- 
 panying flexure be considered. In the case of long beams the 
 work of shearing is indeed but a small part of that due to the 
 flexure (Art. 115). 
 
 The ultimate resilience of shearing is far more difficult to 
 estimate than that of tension or flexure. It can, however, be 
 experimentally determined by the power required to punch a 
 hole through a plate, although even in this case some of the 
 applied work is lost in heat and friction. 
 
 Prob. 168. Estimate the horse-power required to punch 50 
 holes per minute in a wrought-iron plate } inch thick, the 
 diameter of the holes being 2 inches. 
 
 ART. 121. THE COEFFICIENTS OF ELASTICITY. 
 
 The coefficient of elasticity for shearing has a certain rela- 
 tion to the coefficient of elasticity for tension, which will now 
 be deduced. Let abed represent one side of a cube which un- 
 der the tensile unit-stress 5 is elongated into the parallelepiped 
 
 o 
 
 D 
 
 FIG. 69. 
 
 A BCD, the length ab being increased to AB, and the breadth 
 ad being decreased to AD. The ratio of the lateral decrease 
 to the longitudinal increase is designated by e, a mean value of 
 which for iron and steel is , as already mentioned in Art. 71. 
 The distance AB ab is the unit-elongation s, and the distance 
 ad AD is the lateral unit-contraction es. 
 
 The distortion of the square into the parallelogram may be 
 regarded as caused by the shearing stresses acting along the 
 
2/6 SHEAR AND TORSION. CH. XII. 
 
 two diagonals AC and BD. The angle cab, originally JTT, is 
 changed into CAB, which is \n J0 ; the total change of 
 angle between the two diagonals of abed being the distortion 
 due to shearing. Now 
 
 BC i es 
 
 , 
 
 and since is very small, the value of tan (J?r 0) is i 0. 
 very nearly. Hence 
 
 I GS 
 
 i = Tqrj' or > = ( r + e ) s > 
 
 since after reduction s can be neglected in comparison with I. 
 Lastly, replacing for its value S S /E S , and for s its value S/E, 
 and remembering that S s = %S, as shown in Art. 7, there is 
 found the important formula, 
 
 ' 2(+)' 
 
 which gives the coefficient of elasticity for shearing in terms of 
 the coefficient of elasticity for tension. 
 
 The abstract number e is called the factor of lateral contrac- 
 tion. For cast iron its value is found to be about i, and thus 
 E s = f E. For wrought iron and steel e is about , and for 
 these materials E s = %E. For fibrous or non-homogeneous 
 materials, however, formula (25) often fails to apply, for the 
 reason that E is not the same in all directions as it is in a 
 homogeneous body. Using the mean values of E given in 
 Art. 80, the mean values of the coefficients of elasticity for 
 shearing are, 
 
 For cast iron, E s = 6 ooo ooo, 
 For wrought iron, E s = 9 400 ooo, 
 For steel, ,= 11 200 ooo, 
 
 all being in pounds per square inch. By means of experi- 
 ments on the torsion of shafts (Art. 66) these values have been 
 verified. 
 
ART. 122. RESILIENCE UNDER TORSION. 277 
 
 Prob. 169. Prove that tan (\TT 0) is I very nearly, 
 when is small. 
 
 Prob. 170. A cast-iron shaft 60 inches long and 2 inches in 
 diameter is twisted through an angle of 7 degrees by a force of 
 2500 pounds acting at 12 inches from the center, and on the 
 removal of the force springs back to its original position. 
 Compute the factor of lateral contraction e. 
 
 ART. 122. RESILIENCE UNDER TORSION. 
 
 When a shaft is twisted by a force P acting with a lever arm 
 /, as in Fig. 49 of Art. 63, each element of the cross-section is 
 subject to a shearing unit-stress S. The stress being slowly 
 developed, the internal work, or resilience, is equal to \S mul- 
 tiplied by its displacement 0, or if E t denote the coefficient of 
 elasticity for shearing, the work of any elementary area a and 
 length dx is 
 
 dK = 5 .adx= l - adx. 
 
 Now let S t be the shearing unit-stress at the part of the cross- 
 section most remote from the axis, and let c be its distance 
 from that axis ; also let z be the distance of S from the axis. 
 
 Then 5 = S, -, and 
 
 . i S; a* 
 
 dK = - -=- . r . dx 
 2 E t c* 
 
 is an expression for the internal work. To integrate this over 
 the entire volume of the shaft all cross-sections being similar, 
 2 at? is the polar moment of inertia J, and 2dx is the length 
 of the shaft /. Thus 
 
 /^\ r l S * J i I S * ^ AJ 
 
 '"iy-p'-ii-?^* 
 
 where r is the polar radius of gyration of the cross-section de- 
 fined byy = Ar*. Now Al is the volume of the shaft, and it 
 is thus seen that the resiliences of shafts of similar cross-sec- 
 
SHEAR AND TORSION. CH. XIL 
 
 tions are proportional to their volumes. Hence the resilience 
 of a shaft under torsion is governed by laws similar to those of 
 a beam under flexure (Art. 96). 
 
 The formula here established is only valid when the greatest 
 unit-stress 5 5 does not surpass the elastic limit for shearing. 
 
 1 >S a 
 When .S s corresponds to the elastic limit, the quantity - ~ may 
 
 2 &* 
 
 be called the modulus of resilience for torsion or shearing, in 
 analogy to the modulus of resilience for tension or compres- 
 sion (Art. 94). 
 
 As an example, let it be required to find the work necessary 
 to strain a steel shaft 12 inches in diameter and 30 feet long 
 up to its elastic limit, supposed to be 30000 pounds per square 
 inch. Here 5 S = 30000 and E s = n 200000 pounds per 
 square inch (Art. 121); also, c = 6 inches, A 113.1 square 
 inches, / = A 7 ^' \Ad* = 2O 3^ inches 4 , / 360 inches. 
 Inserting all values, K is found to be 818000 inch-pounds or 
 68 200 foot-pounds. Thus to produce this stress in the shaft 
 in one minute more than 2 horse-powers are required. 
 
 Prob. 171. Compare the resilience of a square shaft and a 
 round shaft, the cross-sections and lengths being equal. 
 
 ART. 123. HOLLOW AND SOLID SHAFTS. 
 
 It was mentioned in Art. 69 that a hollow shaft is stronger 
 than a solid shaft of the same sectional area. A general com- 
 parison will now be made with respect to strength, stiffness,, 
 and resilience. Let A be the ara of the cross-section in both 
 cases, let D be the outer and d the inner diameter of the hol- 
 low shaft; then A = \n(D* d*\ and the diameter of the 
 solid shaft is d, = -/>' d\ 
 
 The strength of a shaft under torsion is measured by the 
 twisting moment it can carry under a given unit-stress, and by 
 
ART. 123. HOLLOW AND SOLID SHAFTS. 279 
 
 Art. 64 this is seen to vary as J/c, or as its polar moment of 
 inertia divided by its radius. Hence for the hollow shaft 
 
 and for the solid shaft 
 
 L - ?*L - 
 
 c ~ i6d^ ~ 4 
 
 Therefore, dividing the first of these by the second, and letting 
 k denote the value of D/d, there results 
 hollow _ P+i 
 solid ~ kJ~P~-~\ 
 
 which is the ratio of the strength of a hollow shaft to a solid 
 one of the same sectional area. 
 
 The stiffness of a shaft under torsion is measured by the 
 twisting moment it can carry with a given angle of torsion. 
 As seen in Art. 66, this angle is 
 
 SJ_ Ppl_ 
 
 - -- 
 
 and hence the stiffness varies directly as the polar moment of 
 inertia. For the hollow shaft 
 
 / = A*(/> ' - d ') = \A(D ' + </'), 
 and for the solid shaft 
 
 Therefore, dividing the first by the second, and designating 
 the quantity D/d by k, 
 
 hollow k* + i 
 
 solid ~ ' - i 1 
 
 which is the ratio of the stiffness of a hollow shaft to a solid 
 one of the same sectional area. 
 
 The resilience of a shaft is measured by the work required to 
 produce a given unit-stress. From (26) of the last Article this 
 
2$O SHEAR AND TORSION. CH. XII. 
 
 is seen to vary with J/c*. Thus for the hollow shaft 
 
 and for the solid shaft 
 
 Dividing the first by the second, or putting D/d equal to k, 
 hollow k* + i 
 solid k* ' 
 
 which is the ratio of the resilience of a hollow shaft to a solid 
 one of the same sectional area. 
 
 In practice the outer diameter is often about twice the inner 
 diameter. For this case k = 2, and the above formulas show 
 that a hollow shaft has 1.44 times the strength, 1.67 times the 
 stiffness, and 1.2$ times the resilience of a solid shaft of the 
 same sectional area. These conclusions are valid only when 
 the maximum stress is within the elastic limit of the material ; 
 for higher stresses a theoretic comparison cannot be satisfac- 
 torily made. 
 
 Shafts of the same material and length are of the same 
 strength when their values of J/c are equal, of the same stiff- 
 ness when their values of /are equal, and of the same resili- 
 ence when their values of J/c* are equal. It is thus easy to 
 show that the percentage of weight saved by using a hollow 
 shaft instead of a solid one is 
 
 For equal strength, 100 
 
 For equal stiffness, 
 
 \ 
 
 100 
 For equal resilience, , , , , 
 
 K -f- I 
 
 ci which k is the ratio of the outer to the inner diameter of 
 
ART. 124. SHAFT COUPLINGS. 28 1 
 
 the hollow shaft. See a paper by R. W. DAVENPORT in the 
 Transactions of the Society of Naval Architects and Marine 
 Engineers for 1893 (reprinted in Engineering News, Nov. 23, 
 1893) for a discussion of the practical advantages of hollow 
 shafts made of steel having a high elastic limit. 
 
 Prob. 172. Compare the strength of a solid shaft 13 inches 
 in diameter with that of a hollow shaft with outer diameter 17 
 inches and inner diameter 1 1 inches, the elastic strengths being 
 30000 and 50000 pounds per square inch respectively. 
 
 ART. 124. SHAFT COUPLINGS 
 
 At A and B in Fig. 70 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 another are subject to shearing stress, and they must be 
 
 B CD 
 
 FIG. 70. 
 
 of sufficient strength to safely carry it. This shear differs from 
 that in the main body of the shaft only in intensity, and it is 
 the greatest upon the side of the bolt most remote from the 
 axis. 
 
 Let / be the polar moment of inertia of the cross-section of 
 a solid shaft, c its radius, and S t the shearing unit-stress on the 
 outer surface. Let J l be the polar moment of inertia of the 
 cross-section of the bolts, c l 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 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 JJc v The polar moment of inertia of the cross-section 
 of one bolt with respect to the axis of the shaft is equal to its 
 
282 SHEAR AND TORSION. CH. XII. 
 
 polar moment with respect to its own axis plus the area of the 
 cross-section 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 
 of the shaft, and n the number of bolts ; then 
 
 and equating these values there is found 
 
 D \d + 2h) = nd\d* + Stf), 
 
 which is the necessary relation between the quantities in order 
 that the bolts may be equal in strength to the shaft, provided 
 the material be the same. 
 
 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 ^ = h and /, = ^nd^h a . This amounts to 
 the same thing as regarding d as small compared to h, and 
 the expression then reduces to 
 
 , or d= 
 
 In practice the bolts are often made a little larger in diameter 
 than this formula requires. 
 
 The above supposes the shaft to be solid. If it be hollow 
 with outer diameter D and inner diameter d l , the D 3 in the 
 above expressions is to be replaced by Z>* d*/D, if d l be 
 large enough to have any influence. 
 
 The case shown at CD in Fig. 70 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 
 
ART. 125. A CRANK PIN AND SHAFT. 283 
 
 also depend upon the work transmitted by the shaft. This 
 case will be investigated in Art. 126 in connection with the dis- 
 cussion of the pin of a crank shaft. 
 
 Prob. 173. A solid shaft 6 inches in diameter is coupled by 
 bolts li inches in diameter with their centers 5 inches from the 
 axis. How many bolts are necessary ? 
 
 Prob. 174. A hollow shaft 17 inches in outer and II inches 
 in inner 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. 125. A CRANK PIN AND SHAFT. 
 
 A crank pin, CD in Fig. 71, is subject to a pressure W from 
 the connecting rod which is uniformly distributed over nearly 
 its entire length. This pressure varies at different positions in 
 the stroke of the engine, but for ordinary computations may 
 be taken at from 10 to 20 per cent greater than the total mean 
 pressure on the steam piston ; to this may be added the weight 
 of the connecting and piston rods in case these should be ver- 
 tical in position. 
 
 This maximum pressure W causes a cross-shear in the crank 
 pin at the section C, and it also causes a flexural stress at C 
 due to a uniform load over the cantilever 
 
 CD. These may be computed by the 
 
 methods of Chapter III, and their com- { A 
 bined influence can be determined by 
 formula (13) of Art. 75. The compressive 
 or bearing stress on the surface of the pin 
 is usually also to be considered, this being 
 estimated per square unit of diametral j 
 
 area. 
 
 Owing to the constant alternation of these stresses as the 
 crank arm revolves, the allowable working unit-stresses should 
 
 \w 
 
284 
 
 SHEAR AND TORSION. 
 
 CH. XII. 
 
 be taken low in designing the pin, arm, and journal bearing. 
 The crank arm is under flexure as a cantilever loaded at the 
 end, while the part of the shaft AB which rests in the journal 
 is subject to combined shearing, flexure, and torsion. The 
 methods of Chapters VI and VII give all required to thor- 
 oughly discuss these cases. See UNWIN'S Elements of Machine 
 Design for the special formulas generally used in practice. 
 
 Prob. 175. The crank CD in Fig. 71 is 8 inches long and 4 
 inches in diameter, the maximum pressure W being 60000 
 pounds. Compute the bearing unit-stresses, the shearing unit - 
 stress and the flexural unit-stress. Compute the maximum 
 unit-stress due to combined shear and flexure. 
 
 ART. 126. A TRIFLE-CRANK PIN. 
 
 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 
 1 20 degrees with each other, thus securing a uniform action 
 upon the shaft. Fig. 72 shows one of these cranks, AB and 
 
 
 B 
 C 
 
 k 
 
 # 
 
 
 XT' 
 
 C3 
 
 F { 
 
 c 
 
 
 
 ^ 
 
 
 D 
 
 [ (J 
 
 
 i vL 
 
 FIG. 72. 
 
 EF being portions of the shaft resting in journal bearings, CD 
 the crank pin 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 
 
ART. 126. A TRIPLE-CRANK PIN. 28$ 
 
 abstract of such an investigation will, however, here be given 
 for the crank pin. There are three cranks, and the one to be 
 considered 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 24 
 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 7200 horse-power to the 
 shaft EF, of which 4800 horse-power is transmitted through the 
 shaft AB and through the crank pin CD. The maximum 
 pressure W brought by the connecting rod upon the crank pin 
 is 156000 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 
 
 156000 
 5 ' == YffiS _ 6 ) = 7 6 5 pounds per square inch, 
 
 which is a low and safe value. 
 
 The shearing stress due to W, taken as uniformly distributed 
 over the cross-section of the pin, is 
 
 _ 78000 
 
 5 ' = oTs^cTs'"^ 3 ) = 345 pounds per S( * uare mch ' 
 
 which is low, but will be much increased by the other stresses 
 acting on the end section. 
 
 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 this 
 4800 horse-power is found, from the first equation on page 141 
 of Art. 67, to be 
 
 198000 X 4800 
 Pp = 3.1416 X 80 = 3 782 ooo pound-inches, 
 
286 SHEAR AND TORSION. CH. XII. 
 
 and this is equal to the resisting moment of the crank pin, or 
 to SJJc l , in which J l is the polar moment of inertia of the 
 cross-section with respect to the axis of the shaft and c l is the 
 distance from that axis to the side of the pin where the stress 
 6" is to be found. Now, c = 30 + 9 = 39 inches, and then 
 /, = 0.0982(18* 6') + 0.7854(18' - 6') X 39'. Hence, from 
 formula (i i) of Art. 64, 
 
 Pp,c, 3 782 ooo X 39 
 
 o, = =- = - = 420 pounds per square inch, 
 
 /i 354000 
 
 which is the maximum shearing stress due to the transmitted 
 power. 
 
 The flexure of the pin due to the torsion carried through it 
 falls under a case not heretofore considered, except in the brief 
 mention in Art. 124. The twisting moment Pp is equivalent 
 to a force P acting at a distance of 30 inches from the shaft 
 and normal to the crank arms ; the value of P is 
 
 3 782000 
 P = =126 100 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. Hence there is a bending moment 
 M' at each end, opposite in sign but equal in value, and the 
 moment at any section is M =. M' -f- Px ; but when x = /the 
 value of M is M' and therefore M' = J/V, which is the 
 maximum bending moment. Thus from (4) of Art. 21 
 
 . M'c _ 63000 X 18 X 9 _ .. 
 **- I ^,r(i8 4 -6 4 ) 
 
 which is the flexural stress in pounds per square inch. 
 
 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 with ex- 
 actness, but there is a radial compressive unit-stress 5 6 brought 
 
ART. 126. A TRIPLE-CRANK PIN. 287 
 
 by the web upon the pin of probably 3 ooo pounds per square 
 inch, and this is accompanied by a tangential compressive unit- 
 stress S 6 of about 4000 pounds per square inch (Art. 142). 
 These take effect in the fillet of the pin on the inside of the 
 web at D, where also all of the other stresses concentrate 
 except 5,. In Art. 134 it will be shown how these several 
 values may be combined in order to obtain the maximum 
 tensile, compressive, and shearing stresses. 
 
 Prob. 176. Draw the shear and moment diagram for the 
 lateral flexure of the pin due to the transmitted torsion. 
 
288 APPARENT AND TRUE STRESSES. CH. XIII. 
 
 CHAPTER XIII. 
 APPARENT STRESSES AND TRUE STRESSES. 
 
 ART. 127. THE MATHEMATICAL THEORY OF ELASTICITY. 
 
 In Art. 5 are stated several laws, derived from experiment, 
 which are the foundation of the science of Mechanics of Ma- 
 terials. Of these (A) and (B) relate to elasticity and are the 
 basis of all discussions concerning stresses that do not surpass 
 the elastic strength of the material, the latter being usually re- 
 ferred to as HOOKE'S law (Art. 81). All the theoretic formulas 
 of the preceding pages are derived by the help of this law, and 
 these hence constitute a part of the mathematical theory of 
 elasticity. 
 
 This theory is one of vast extent and far-reaching conse- 
 quences, and its full development would require volumes. It 
 includes not only the complete investigation of the stresses 
 and deformations produced in every part of a body by gradu- 
 ally applied exterior forces, but also those arising under condi- 
 tions of impact. It deals not only with elastic solids, but with 
 fluids, gases, and the ether of space. The discussion of 
 stresses and deformations, both in homogeneous and crystalline 
 bodies, leads to the investigation of wave propagations, the 
 time and velocity of elastic oscillations, and numerous other 
 phenomena of physics. In this Chapter will be presented a 
 few of the fundamental principles with reference to homoge- 
 neous materials only. 
 
 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 
 
ART. 127. MATHEMATICAL THEORY OF ELASTICITY. 289 
 
 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 
 materials, it is allowable to neglect the squares and higher 
 powers of a unit-elongation in comparison with the elongation 
 itself. Thus if / be the length of a bar, which under the action 
 of stress is increased to the length l(\ + s\ the square of this 
 new length may be taken as l\i -\- 2s) and the cube as 
 l\i + 3*). For a substance like india rubber, where this as- 
 sumption does not apply, some of the conclusions of the theory 
 of elasticity are not necessarily valid. 
 
 Another axiom derived from experience is the following: 
 Under tensile forces the volume of a body is increased and 
 under compressive forces it is diminished. This is the case for 
 the common materials, although bodies may exist for which it 
 is not true. Thus if a cube be compressed upon two opposite 
 faces the edges parallel to the forces are decreased while those 
 at right angles to the forces are increased in length ; on the 
 whole, however, the volume of the cube is slightly decreased. 
 
 The student should consult the article on Elasticity by 
 KELVIN in the Encyclopaedia Britannica, as also the History 
 of TODHUNTER and PlERSON. The works of CLEBSCH (Elas- 
 ticitat fester Korper, 1862), WlNKLER (Elasticitat und Festig- 
 keit, 1867), GRASHOF (Theorie der Elasticitat und Festigkeit, 
 1878), and FLAMANT (Resistance des Materiaux, 1886) maybe 
 mentioned as treating the subject both from the theoretical 
 and the engineering point of view. 
 
 Prob. 177. Consult TODHUNTER and PlERSON'S History of 
 the Theory of Elasticity and of the Strength of Materials, 
 and ascertain something about the investigations of SAINT 
 
 VENANT. 
 
290 APPARENT AND TRUE STRESSES. CH. XIII. 
 
 ART. 128. LATERAL DEFORMATION. 
 
 It has already been noted, in Arts. 71 and 121, that a bar 
 under tension not only elongates in the direction of its length, 
 but is subject to a lateral contraction. So if a bar be under 
 compression there occurs a longitudinal contraction and a 
 lateral elongation. If P be the applied force and A the area 
 of the cross-section the unit-stress is P/A =5; if / be the 
 length. and A the longitudinal change of length, then A// = s is 
 the unit-elongation or unit-contraction. In the case of tension 
 each unit-length of the bar is increased by s, but each unit at 
 right angles to the length is decreased by the amount es, where 
 e is an abstract number less than unity, called the factor of 
 lateral contraction. In the case of compression es is the lateral 
 unit-elongation. 
 
 Let a cube in the interior of a bar under tension have each 
 of its sides unity before the application of the stress ; under this 
 tension the cube is deformed so that its sides are i -{-s, l es, 
 and I es. The volume of the deformed body is, if the 
 squares and cubes of s be neglected in comparison with the 
 first power, 
 
 (l+*Xl-eO' = I +(1-26)5. 
 
 Now in order that the volume may be increased, (i 2e)s 
 must be positive, or e must be less than J. When e is o no 
 lateral contraction occurs, when e = the lateral contraction is 
 a maximum. Thus for all bodies whose volume is increased 
 under tension the factor of lateral contraction must lie be 
 tween o and J. 
 
 The same reasoning applies in the case of compression, for 
 which as far as known e has the same value as for tension. By 
 precise measurements of bars under stresses within the elastic 
 limit it has been found that e generally lies between 0.2 and 
 0.4 for wrought iron and steel, a mean value extensively used 
 being -J. For cast iron e is about J or a little less. 
 
ART. 129. TRUE TENSILE AND COMPRESSIVE STRESSES. 291 
 
 Prob. 178. A bar of steel 2X2 inches and 6 feet long is 
 pulled by a force of 50 ooo pounds. Compute the percentage 
 of increase of length, and the percentage of increase of volume. 
 
 ART. 129. TRUE TENSILE AND COMPRESSIVE STRESSES. 
 
 If a parallelepiped be subject to a unit-stress 5 in the direc- 
 tion of its length this is the true stress on all planes normal to 
 its length. In directions at right angles to the length there 
 exists, however, a lateral contraction which implies an internal 
 stress of compression. If s be the unit-elongation due to S, 
 the lateral unit-contraction is es, and this is the same as would 
 be produced by a lateral compressive unit-stress eS. Thus it 
 is clear that the deformation at right angles to 5 is the same as 
 that produced by an actual unit-stress eS. In a similar manner 
 if forces act upon all the sides of the parallelepiped the true 
 internal stresses are different from the apparent ones. 
 
 All the stresses computed thus far in this volume are appar- 
 ent stresses, for the influence of lateral deformation has not 
 been taken into account. In this Chapter the letter 5 will 
 denote the apparent unit-stresses, while the true unit-stresses 
 corresponding to the actual deformations will be designated 
 by T. The true resistance of a body depends upon the actual 
 deformations produced, and these are measured by the true 
 internal stresses. 
 
 Let a homogeneous parallelopiped be subject to normal 
 forces of tension or compression 
 upon its six faces, those upon oppo- 
 site faces being equal. Let the 
 edges of the parallelopiped be de- 
 signated by 01, 02, 03, as in Fig. 73. 
 Let S t be the normal unit-stress 
 upon the two faces perpendicular to 
 the edge 01, and 5, and S t those upon the faces normal to 02 
 
APPARENT AND TRUE STRESSES. CH. XIII. 
 
 and 03 ; thus the directions of S l 9 S a , 5, are parallel to 01, 
 02, 03, respectively. Then, supposing that these stresses are 
 all tensile, and that the factor of lateral contraction e is the 
 same in all directions, the true internal unit-stresses in the 
 three directions are, 
 
 T, = S, - eS, - eS, , 
 (27) r, = S f - eS, - eS l9 
 
 T, = S> eS, e-S, , 
 
 in which e lies between o and , as shown in Art. 128. If any 
 apparent stress 5 be compressive, it is to be taken as negative 
 in the formulas, and then the true stresses 7",, J!,, T 9 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 compression 5 ; then the true internal unit-stress T 
 is S(i 2e), or about 5, and its linear deformation is only 
 about one-third of that due to a compressive stress 5 applied 
 upon two opposite faces. Again, if a bar have a tension S^in 
 the direction of its length, and no apparent stresses upon its 
 sides, then T l = 5, while 7", = T t = eS,. 
 
 The true deformations corresponding to the true internaV 
 stresses will be denoted by ^ , /,, /,. If the coefficient of 
 elasticity in all directions be , then 
 
 T T T 
 
 /_j /_!_ /____ 
 
 ** E' ' ' ' "" E 
 
 \ 
 
 are the unit-elongations or unit-contractions parallel to the 
 three coordinate axes. 
 
 As a simple example, let a steel bar 2 feet long and 3X2 
 inches in cross-section be subject to a tension of 60000 pounds 
 in the direction of its length and to a compression of 432 ooo 
 pounds upon the two opposite flat sides. Here S l = 60 000/6 
 = 10 ooo pounds per square inch, 5 a = 432 000/72 = 6000 
 
ART. 130. NORMAL AND TANGENTIAL STRESSES. 293 
 
 pounds per square inch, and .S, = o. Then from (27), taking 
 as the true internal stresses are 
 
 7\ = +12 ooo, 7; = -9330, r, = 
 and it is thus seen that the true tensile unit-stress is 20 per 
 cent greater than the apparent, while the true compressive 
 unit-stress is more than 50 per cent greater than the apparent. 
 
 If the parallelepiped in Fig. 73 be subject to the action of 
 oblique stresses R t , R t , R t , each may be resolved into a stress 
 normal to the face and into shearing stresses parallel to the 
 edges. In such a case the true unit-stresses T l , 7", , 7", can- 
 not be directly found, but it will be shown in the following 
 articles that three planes can be determined upon which the 
 apparent stresses are wholly normal, and that these are the 
 maximum apparent tensile and compressive stresses due to 
 R t , R tJ R % \ these being found, formulas (27) are directly 
 applicable. 
 
 Prob. 179. In Fig. 73 let a plane be passed through the edge 
 02 and through the edge diagonally opposite to it. Let the 
 edges 01, 02, 03 be equal in length. Show that the apparent 
 shearing unit-stress on this plane is J(5, S,). 
 
 ART. 130. 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 forces in directions not normal. Here each oblique 
 stress may be decomposed into three components parallel re- 
 spectively to three coordinate axes, OX, O Y, OZ. Upon each 
 of the faces perpendicular to OX the normal component of the 
 oblique unit-stress is designated by S x and the two tangential 
 components by S^ and S xx . A similar notation applies to each 
 of the other faces. An 5 having but one subscript denotes a 
 tensile or compressive stress, and its direction is parallel to the 
 
294 
 
 APPARENT AND TRUE STRESSES. CH. XIIL 
 
 axis corresponding to that subscript. An 5 having two sub- 
 scripts denotes a shearing stress, the first subscript designating 
 the axis to which the face is perpendicular and the second 
 designating the axis to which the stress is parallel ; thus S gx is 
 
 FIG. 74. 
 
 on the face perpendicular to OZ and its direction is parallel to 
 OX. In Fig. 74 the six components for three sides of the 
 parallelepiped are shown. Neglecting the weight of the par- 
 allelepiped 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. 74 in order that equilibrium may obtain. 
 Let the parallelepiped be a cube with each edge equal to unity ; 
 then if no tendency to rotation exists with respect to an axis 
 through the center of the cube and parallel to OX it is neces- 
 sary that S yz should equal S gy . A similar condition obtains for 
 each of the other rectangular axes, and hence 
 
 (<>Q\ C _ C C _ C C _ C 
 
 \ ^/ *^xy ~~~ *-^yx t *-^yz ~~~ *^xyi ^xz *^sx t 
 
ART. 131. RESULTANT STRESS. 295 
 
 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 
 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 
 parallelepiped such that the resultant stresses upon it are 
 wholly normal. These resultant normal stresses will be 5, , .5, , 
 S 9 , from which by (27) the true normal stresses T t , 7,. T 9 
 can be found. It will later be shown that these stresses 5, , 5, , 
 5", are the maximum apparent stresses of tension or compres- 
 sion resulting from the given normal and tangential stresses. 
 
 Prob. 1 80. Let a, b, c be the angles which a line makes with 
 the axes OX, OY, OZ, respectively. Show that the sum of 
 the squares of the cosines of these angles is equal to unity. 
 
 ART. 131. RESULTANT STRESS. 
 
 The resultant unit-stress upon any face of the parallelepiped 
 in Fig. 74 is the resultant of the three rectangular unit-stresses 
 acting upon that face. Thus for the face normal to OZ the 
 resultant unit-stress is given by 
 
 R: = S:-\-S'^ + S\,, 
 
 and the total resultant stress upon that face is the product of 
 its area and R 9 . 
 
 The resultant unit-stress R upon any elementary plane hav- 
 ing any position can be determined when the normal and tan- 
 gential stresses in the directions parallel to the coordinate axes 
 are known. Let a plane be passed through the corners I, 2, 3, 
 of the parallelepiped in Fig. 74, and let a, b, c be the angles 
 that its normal makes with the axes OX, O Y, OZ, respectively. 
 Let a, ft, y be the angles which the resultant unit-stress R 
 makes with the same axes. Let A be the area of the triangle 
 
296 APPARENT AND TRUE STRESSES. CH. XIII. 
 
 123; then the total resultant stress upon that area is AR, and 
 its components parallel to the three axes are AR cos a, AR 
 cos ft, AR cos y. The triangle whose area is A, together with 
 the three triangles 012, 023, 031, 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 cos a, 
 A cos b, A cos c, and the stresses upon them are the products of 
 the areas by the several unit-stresses. Now the components of 
 these four stresses with respect to each rectangular axis must 
 vanish as a necessary condition of equilibrium. Hence, can- 
 celling out A, which occurs in all terms, there results 
 
 R cos a = S* cos a -f- S yjf cos b -f- S zx cos c, 
 (29) R cos ft = S^y cos a -\- S y cos b -\- S, y cos c, 
 R cos Y = S xz cos a -f- S y , cos b -\- S s cos c, 
 
 in which the second members are all known quantities. 
 
 From these equations the values of R cos <*, R cos ft, R cos y 
 can be computed ; then the sum of the squares of these is R* 
 since cos* a -f- cos a ft-\- cos 3 y = i. The value of cos a is found 
 by dividing that of R cos a by R, and similarly for cos ft and 
 cos /. Now the angle between the directions of R and the 
 normal to the plane is given by 
 
 cos B == cos a cos a + cos b cos /? -|- cos c cos ^, 
 
 and then the tensile or compressive unit-stress normal to the 
 given plane is R cos 0, while the resultant shearing unit-stress 
 is R sin 8. This shearing stress may be resolved into two com- 
 ponents in any two directions on the plane. 
 
 As a simple numerical example, let a bolt be subject to a 
 tension of 12000 pounds per square inch and also to a cross- 
 shear of 8 ooo pounds per square inch. It is required to find 
 the apparent unit-stresses on a plane making an angle of 60 
 degrees with the axis of the bolt. Take OX parallel to the 
 tensile force and OY parallel to the cross-shear. Then S x = 
 
ART. 132. THE ELLIPSOID OF STRESS. 297 
 
 -f 12000, Sjej, = 8000, Sy* = 8000, and the other stresses are 
 zero ; also a = 30, b = 60, and c = 90. Then from (29) 
 R cos a = + 14 390, R cos /? = -f- 6930, 7? cos y = o, 
 and the resultant stress in the plane is, 
 
 R = -/I4 390' + 693' = 15 970 pounds per square inch. 
 The direction made by R with the axis is, 
 
 cos a = -_ = 0.901, a = 
 
 cos ft = = 0.434, ft = 25t, 
 
 and the angle between the resultant R and the normal to the 
 plane is given by 
 
 cos 6 = 0.866 X 0.901 + 0.5 X 0.434 = 0.997. 
 Lastly, the normal tensile stress on the plane is found to be 
 R cos 0= 15 920 pounds per square inch, while the shearing 
 stress on the plane is R sin 6 = 1200 pounds per square inch. 
 
 Prob. 1 8 1. Find for the above example the position of a 
 plane upon which there is no shearing stress. 
 
 ART. 132. THE ELLIPSOID OF STRESS. 
 
 The resultant unit-stress R upon any plane makes an angle 
 B with the normal to that plane. If at any point this plane 
 be supposed to vary its direction the intensity of R will be 
 represented by the radius vector of an ellipsoid. 
 
 Let R t , RI , R t be the resultant unit-stresses upon the three 
 faces of the parallelepiped in Fig. 74, and let 0, , 0, , 0, be the 
 angles which they make with the coordinate axis OX\ then 
 
 cos 0, = J?, cos 0, = ?, cos 0, = %^, 
 K i K * <K 9 
 
 determine the directions of R, , R 9 , R t . 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 
 
298 
 
 APPARENT AND TRUE STRESSES. 
 
 CH. XIII. 
 
 R x , R y , R z be its components parallel to these new axes. 
 Then R cos a is the component of R parallel to OX, and 
 
 R cos a = R x cos 6, -f- R y cos # a + R s cos #, , 
 or, inserting for the cosines their values, 
 
 Comparing this with the first equation in (30), it is seen that, 
 
 R x . R v R. 
 
 cos a = -75-, cos b = , cos c = ^-. 
 
 But the sum of the squares of these cosines is equal to unity ; 
 therefore, 
 
 &? R? R 
 
 in which the numerators are variable coordinates and the de- 
 nominators are given quantities. This is hence the equation 
 of the surface of an ellipsoid with respect to three coordinate 
 axes having the directions of R l , R t , R 3 . 
 
 The ellipsoid is hence a figure whose radius vector repre- 
 sents the resultant unit-stress upon a plane whose normal 
 makes an angle 6 with the direction of that radius vector. If 
 the forces be entirely confined to one plane the ellipsoid re- 
 duces to an ellipse. 
 
 If there be three planes at right angles to each other which 
 are subject only to normal stresses, as in Fig. 75, the normal 
 unit-stresses S x , S yt S, correspond 
 to R 1 , R t , ^ 8 in the above equa- 
 tion of the ellipsoid. In this case 
 S*, S y , S, are the three axes of 
 the ellipsoid. If now shearing- 
 stresses be applied to the faces 
 the ellipsoid will be deformed, and 
 the three axes will take other 
 
 FIG. 75. 
 
 position corresponding to three planes upon which no shear- 
 
ART. 133. THE THREE PRINCIPAL STRESSES. 299 
 
 ing stresses act. The stresses corresponding to the axes of 
 the ellipsoid are called principal stresses. 
 
 Prob. 182. If S y = S z in Fig. 75, show that the ellipsoid be- 
 comes either a prolate spheroid or an oblate spheroid. 
 
 ART. 133. 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. 131). It is evident, 
 however, that planes may exist upon which only normal 
 stresses act, so that is zero and R is pure tension or com- 
 pression. In order to find these planes and the stresses upon 
 them the angles or, ft, y in the equations (29) are to be made 
 equal to a, b, c, respectively. Also replacing R by S, they 
 
 become, 
 
 (S Sr) cos a = S yx cos b + $** cos c, 
 (S S y ) cos b = S xy cos a -f- S ty cos c, 
 (S S g ) cos c = S xg cos a + S yg cos b t 
 
 in which S, cos a, cos b, cos c are quantities to be determined. 
 
 The three angles, however, are connected by the necessary 
 
 relation, 
 
 cos' a + cos 1 b + cos 1 c = i, 
 
 and hence four equations exist between four unknowns. 
 
 Remembering the relation between the shearing stresses ex- 
 pressed in (28), the solution of the equations leads to a cubic 
 equation for S, which is of the form, 
 
 (30) 5' - AS 9 + S - C = o, 
 
 in which the values of the coefficients are, 
 
 A = S Jf + S,+ S., 
 
 B = 5,5, + S f S. + S.S, - S',, - 5% 
 
 a - 
 
 yt 
 
300 APPARENT AND TRUF STRESSES. CH. XIII. 
 
 and the three roots of this cubic are the three normal stresses 
 of tension or compression, often called the three principal 
 stresses. 
 
 The directions of these principal stresses with respect to the 
 axes OX, O Y, OZ, are given by the values of cos a, cos b, 
 cos c, which are found to be, 
 
 cos a = 
 in which the values of the m's are, 
 
 m 9 = (S, - S)(S* - S) - $*, 
 m,= (S x -S)(S y - S) -5^, 
 m = m t -[- m t -f- m 3 ; 
 
 and it will now be shown that each principal stress is perpen- 
 dicular to the plane of the other two. 
 
 Let S lf S. t , 5, be the three roots of the cubic equation (30). 
 Let #, , , , , be the angles which S t makes with the three co- 
 ordinate axes OX, O Y, OZ, and let # a , a , c, be the angles 
 which 5 a makes with the same axes. The angle between the 
 directions of S l and 5 2 is then given by 
 
 cos = cos ^ cos a t + cos ^i cos ^ ~t~ cos c \ cos c * 
 Now in the first set of formulas of this article let 5 be made 
 5, and a, b, c be changed to a l9 ,, ,; let the first equation be 
 multiplied by cos a^ , the second by cos &, , and the third by 
 cos r, ; and let the three equations be added ; then 
 
 5, (cos a, cos <z a + cos b l cos a + cos c l cos ^ 3 ), 
 
 is one term in this sum. Again, let T be made T 9 , and a, b, c, 
 be changed to a 9 , b t , , ; let the equations be multiplied by 
 cos a l , cos , , cos r, , respectively, and added ; then, 
 
 5,(cos a^ cos , -f- cos b l cos <^, -[" cos ^ cos r s ), 
 is one term in the sum, while all the other terms are the same 
 
ART. 134. A NUMERICAL CASE. 3<DI 
 
 as before. Hence if S l and 5, are unequal, the factor in the 
 parenthesis, which is cos 0, must vanish ; is therefore a right 
 angle or 5, and S t are perpendicular. In the same manner it 
 may be shown that 5, is perpendicular to both 5, and 5,. 
 
 The three principal stresses are hence perpendicular to each 
 other, and as the only diameters of the ellipsoid which have 
 this property are its axes, it follows that the directions of the 
 principal stresses S l , S a , S, are those of the axes of the ellip- 
 soid of stress. These principal stresses thus give the maxi- 
 mum normal stresses of tension or compression. 
 
 An interesting property of the three rectangular stresses S, , 
 S y , S t , is that their sum is constant, whatever maybe the posi- 
 tion of the coordinate axes. For the sum of the three princi- 
 pal stresses 5, , S t , S t is equal to the coefficient A in the cubic 
 of (30), and hence, 
 
 S,+ S, + S, = S, + 5, + S,; 
 
 that is, the sum of the normal unit-stresses in any three rec- 
 tangular directions is constant. 
 
 Prob. 183. When two principal stressses are equal, show that 
 the value of each is (AB gC)/(2A* 6B), where A, B, C 
 are the coefficients in (30). 
 
 ART. 134. A NUMERICAL CASE. 
 
 To apply the preceding principles to a particular case a 
 crank pin similar to that investigated in Art. 126 may betaken. 
 The axis OX is assumed parallel to the axis of the pin, O Y 
 parallel to the crank arm, and OZ perpendicular to both. On 
 one side of the crank pin near its junction with the arm there 
 were found the following apparent stresses: A cross shear 
 from the pressure of the connecting rod giving S^ = 300 
 pounds per square inch, a shear due to the transmitted torsion 
 giving S xz = 900 pounds per square inch, a flexural stress due 
 
302 APPARENT AND TRUE STRESSES. CH. XIII. 
 
 to the connecting rod giving S* = -j- 800 pounds per square 
 inch, a flexural stress due to the transmitted torsion giving 
 S x = + l 600 pounds per square inch, and two compressions 
 due to shrinkage giving S y = 4000 and S z = 2 ooo pounds 
 per square inch. 
 
 The two shears having the same direction add together, as 
 also the two tensions, and the data then are, 
 
 Sjfg = I 2OO, S* = -f 2 4OO, S y = 4 OCX), S z = 2 OOO. 
 
 Inserting these in (30) it becomes, 
 
 5 3 + 3 6oo5 2 7 840 oooS 27 200 ooo ooo = o, 
 
 and the three roots of this are the three principal stresses. 
 To solve this put 5 = x i 200, and it reduces to 
 
 x* 1 2 160 OOO.T 1 1 096 ooo ooo = o. 
 
 As this cubic has three real roots, it is to be solved by the help 
 of a table of cosines; thus let 
 
 3?* a = 12 160000, 2r i cos 30 = ii 096000000, 
 
 from which r = 2 013 and cos 30 = 0.680 1. Then from the 
 table 30 = 47 09' whence 0=15 43'. The roots now are 
 
 x^ = 2r cos = + 3 880, 
 
 x^ = 2r cos (0+ 120) = 2 890, 
 x % = 2r cos (0 + 240) = 990 ; 
 
 and finally the three principal stresses are, 
 
 S, = x^ i 200 = + 2 680, 
 5, = x^ i 200 = 4 090, 
 5, = x^ i 200 = 2 190, 
 
 of which Si is the maximum tension and S t the maximum 
 compression. 
 
 These are the apparent stresses. Taking e = J for steel, 
 the true principal stresses are now found by (27) to be, 
 
 T t = +4770, T, = - 4250, T, = - i 720, 
 
ART. 135. MAXIMUM SHEARING STRESSES. 303 
 
 which shows that the maximum true tensile stress is nearly 
 double the apparent, while the maximum true compressive 
 stress is 6 per cent greater than the apparent. 
 
 In any case the method of procedure is similar. The appar- 
 ent stresses being first computed for the point under consider- 
 ation, their values establish the cubic equation (30) whose 
 solution gives the three principal apparent stresses S, , 5,, S, ; 
 then from (27) the true maximum stresses of tension or com- 
 pression are found. It frequently happens that one of the 
 principal apparent stresses is zero ; in this event the last term 
 of the cubic vanishes and the ellipsoid becomes an ellipse. 
 For instance let a bar be under a tension S* and a single cross- 
 shear S*j,. Then from (30) the principal stresses are 
 
 s, = to + 
 
 the first of which agrees with the result given in a different 
 notation in equation (13) of Art. 75. 
 
 Prob. 184. Find the maximum true unit-stresses of tension 
 or compression for the data of Prob. 121. 
 
 ART. 135. MAXIMUM SHEARING STRESSES. 
 
 As there are certain planes upon which the tensile and com- 
 pressive unit-stresses are a maximum, so there are certain 
 other planes upon which the shearing unit-stresses have their 
 maximum values. In order to determine these it is well to 
 take the axes of the ellipsoid as the coordinate axes, and upon 
 the planes normal to these there are no shearing stresses. The 
 stresses 5,, S 9 , S 9 will give apparent shearing stresses on 
 other planes, while T lt T tt T t will give the true shearing 
 stresses. 
 
 Let i 2 3 in Fig. 76 be any plane whose normal makes the 
 angles a, b, c with the coordinate axes. Let R be the result- 
 ant unit-stress upon this plane, and a, ft, y be the angles which 
 
304 
 
 APPARENT AND TRUE STRESSES. 
 
 CH. XIII. 
 
 it makes with the same axes. The angle between R and the 
 
 normal is expressed by, 
 
 cos = cos a cos a -\- 
 
 cos b cos ft + cos c cos Y* 
 
 and the resultant shearing unit- 
 stress on the plane is 
 
 R sin = 
 
 cos 8 0. 
 
 If R be apparent stress, this is the 
 FIG. 76. apparent shearing stress ; if R be 
 
 true stress, this is the true shearing stress. 
 The value of R, as a true stress, is given by 
 
 R 9 = (T, cos of + (T, cos b)' + (T, cos c)\ 
 
 Now, since both R cos a and T v cos a are components of R in 
 the direction OX, they are equal, and hence 
 
 T. 
 
 cos a = i cos #, 
 
 T 
 
 COS /?=:-' COS 
 .A. 
 
 cos y = * cos 
 
 Substituting these in the value of cos 0, the resulting true 
 shearing unit-stress is expressed by 
 
 T 3 = (T, cos of + (T, cos )' + (r, cos <:)' - 
 
 (T, cos 8 + r, cos 8 3+7; cos 8 <:)', 
 
 by the discussion of which the values of #, b, c, which render T 
 a maximum, are deduced. Bearing in mind that the sum of 
 the squares of the three cosines is unity, the discussion gives 
 
 (SO 
 
 and therefore there are six planes of maximum shearing stress, 
 each of which is parallel to one of the principal stresses and 
 bisects the angle between the other two. On each of these 
 planes the shearing unit-stress is one half the difference of the 
 principal unit-stresses whose direction is bisected. 
 
ART. 136. THE ELLIPSE OF STRESS. 305 
 
 The same investigation applies equally well to the apparent 
 shearing unit-stresses, whose maximum values are 
 
 (3 1)' 5 = 4(5, - S,), S= i(5, - S,), S= 4(5, - 5,), 
 a.nd whose directions are the same as those of the maximum 
 true shearing unit-stresses. The sign indicates that the 
 shears have opposite directions on opposite sides of the plane, 
 but in numerical work it is always convenient to take them as 
 positive, or rather, as signless quantities. 
 
 As an example, let a bar be subject to a tension of 3 ooo 
 pounds per square inch in the direction of its length and to a 
 compression of 6000 pounds per square inch upon two oppo- 
 site sides. Here 5, = + 3 OCXD, 5, = 6 ooo, 5, = o ; then from 
 (31)' the maximum apparent shearing stresses are, 
 
 5=4 500, 5=3 ooo, 5 = i 500. 
 
 But from (27), taking e as J, the true tensile and compressive 
 unit-stresses are 7\ = + 5 ooo, T t = 7 ooo, T % = + i ooo, 
 and then from (31) the maximum true shearing stresses are, 
 
 T = 6 ooo, T 4 ooo, T = 2 ooo, 
 which are 33 per cent greater than the apparent ones. 
 
 Prob. 185. Compute the maximum apparent and true shear- 
 ing unit-stresses for the data given in Prob. 121. 
 
 ART. 136. 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 (30) 
 vanishes. An instance of this is where 5, = o, S y ,= o, 5^., 
 = o, which is that of a body subject to the normal stresses 
 S x , S y , and to a cross-shear S^. The cubic equation then 
 reduces to 
 
 5' - (S, + S,)S+ S x S y -5^ = 0; 
 
 and the two roots of this are the two principal apparent stresses 
 whose directions correspond to the two axes of the ellipse. 
 
306 
 
 APPARENT AND TRUE STRESSES. 
 
 CH. XIII. 
 
 Let 5, and 5, be these roots, and in Fig. 77 let OA and OB 
 be laid off at right angles to represent their values. Let an 
 
 B 
 
 FIG. 77- 
 
 ellipse be described upon the axes AA and BB, and let be 
 the angle AON which any line ON makes with OA. Upon a 
 plane normal to ON at O the shearing unit-stress is 
 
 OS = (5, 5 2 ) sin cos 0, 
 and the normal unit-stress of tension or compression is 
 
 ON = S, cos' + S a sin 8 0, 
 while the resultant unit-stress is 
 
 OR = VS> 9 cos 9 + 5 a 2 sin' 0. 
 
 The diagrams in Fig. 77 give graphic representations of these 
 values as the angle varies from o to 360 degrees. In the 
 first diagram S, and 5 9 are both tension or both compression, 
 in the second diagram one is tension and the other compression. 
 The broken curve shows the locus of the point N, and the 
 dotted curve the locus of 5. For every value of the lines 
 OS and ON are at right angles to each other, and OR is their 
 resultant. 
 
 As a numerical illustration take the case of a bolt subject to 
 a tension of 2 ooo and also to a cross shear of 3 ooo pounds 
 
ART. 130. THE ELLIPSE OF STRESS. 307 
 
 per square inch. Here S x = + 2 ooo, S xy = 3 ooo, and S y = o ; 
 the above quadratic equation then gives 5, = +4 160 and 
 S 9 = 2 160 as the two maximum unit-stresses of tension and 
 compression. The direction made by 5, with the axis of the 
 bolt, as found by the value of cos a in Art. 133, is about 54^ 
 degrees. From (31)' the maximum shear is 3 160 pounds per 
 square inch. These are the apparent maximum stresses. 
 
 To find the true maximum stresses formulas (27) give, tak- 
 ing as the factor of lateral contraction, T t = + 4 880, 
 7", = 3 550, 7", = 670 pounds per square inch as the prin- 
 cipal tensions and compressions; then from (31) the greatest 
 shearing stress is 7^=4220 pounds per square inch. Here 
 the true maximum tension is 17 per cent greater than the 
 apparent, the true compression is 64 per cent greater, and the 
 true shearing stress is 33 per cent greater. The true stresses 
 cannot be represented by an ellipse, but an ellipsoid of internal 
 stress results of which the second diagram in Fig. 77 may be 
 regarded as a typical section. 
 
 Cases can, however, be imagined in which one of the true 
 principal stresses is zero. If 5, , 5, , S t are the apparent 
 stresses in three rectangular directions, it. is seen from (27) 
 that when 5, eS l eS, is zero, the true stress T t is also zero. 
 For instance, let a cube be under compression by three nor- 
 mal stresses of 30, 24, and 18 pounds per square inch and let 
 e = ; then T, = 16, T t = 8, and T t = o. Here the ellipse 
 of true stress has its correct application and there are no true 
 stresses in a plane normal to the plane of T t and 7",. 
 
 Prob. 1 86. A body is subject to a tension of 4000 and to a 
 compression of 6000 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. 
 
308 APPARENT AND TRUE STRESSES. CH. XIII. 
 
 ART. 137. FORMULAS FOR TRUE COMBINED STRESSES. 
 
 In the preceding articles it has been shown how the true 
 stresses for all cases may be found. For the most common 
 case, that of shear combined with tension or compression, it 
 is well, in conclusion, to write the formulas by which the true 
 maximum unit-stresses may be computed, as these will in 
 general give higher values than the expressions deduced in 
 Arts. 76 and 77. 
 
 Let S x be the apparent tensile unit-stress and S xy be the 
 apparent shearing unit-stress, the first being found from (4), 
 and the second from (3) for the case of a beam or from (12) 
 for the case of a shaft. Then in the quadratic equation of 
 the last article S y is zero, and the two roots are 
 
 which are the apparent maximum tensile and compressive 
 unit-stresses due to the combination of S x and S xy . Substi- 
 tuting these in (27) there results 
 
 >*y > 
 
 >xy > 
 
 which are the true maximum unit-stresses, the first being ten- 
 sile and the second compressive. These formulas apply also 
 when S* is a compressive stress, its sign being then taken as 
 negative. The true maximum shearing unit-stress is in both 
 cases given by 4(7", 7!,), that is, by the second term of the 
 formulas. For metals, the factor of lateral contraction e lies 
 usually between $ and J. 
 
 For instance, taking the numerical example on page 152, 
 let it be required to find the factor of safety of a wrought- 
 iron shaft 3 inches in diameter and 12 feet between bearings, 
 
ART. 137. TRUE COMBINED STRESSES. 309 
 
 which transmits 40 horse-powers while making 120 revolu- 
 tions per minute, and upon which a *load of 800 pounds is 
 brought by a belt and pulley at the middle. Here the 
 flexural stress on the outer fiber is 5 400 pounds per square 
 inch and the torsional stress is 4 ooo pounds per square inch, 
 both acting at the circumference of the shaft. Thus %S X is 
 
 2 700, and V^S 2 X + Sly ls 49; tnen 
 
 T } = f X 2700 + f X 4900 = + 8 300 pounds per square inch, 
 
 and accordingly the factor of safety is 6.6, whereas by consid- 
 ering the apparent stresses only it was found to be 7.5. 
 
 Finally, it may be observed that the formula deduced in 
 Art. 78 for the maximum horizontal shearing unit-stress in a 
 beam is not changed by the introduction of the idea of true 
 stresses, since there is no longitudinal unit-stress at the 
 neutral axis. Along this axis there exists the horizontal shear 
 and at right angles to it another of equal intensity, these com- 
 bining to cause tensile and compressive unit-stresses of the 
 same intensity in directions bisecting the lines of shear. 
 
 Prob. 187. Solve Problems 129 and 130 by the application 
 of the methods of this Chapter. 
 
STRESSES IN GUNS. 
 
 CH. X.IV, 
 
 CHAPTER XIV. 
 STRESSES IN GUNS. 
 
 ART. 138. LAMP'S FORMULA. 
 
 Let a thick hollow cylinder, shown in longitudinal and 
 cross-section in Fig. 78, be subject to a pressure p l on each 
 square unit of the inner surface and to a pressure /, on each 
 square unit of the outer surface. The inner pressure may be 
 
 
 f 
 
 ? 
 
 FIG. 78. 
 
 produced by the expansion of a gas and the outer pressure by 
 the atmosphere or by other causes. It is required to deter- 
 mine the internal stresses produced by these pressures at any 
 point in the cylindrical annulus. 
 
 The outer pressure on the end of the closed cylinder is 
 nr \P* an d * ne i nn er pressure on the end is nr*p r If the inner be 
 greater than the outer pressure, as is often the case, the differ- 
 ence of these, or 7t(r^p l r, 8 /,) is the longitudinal tension on 
 the annulus. For any part of the cylinder, not very near the 
 end, this must be uniformly distributed over the cross-section. 
 The longitudinal unit-stress on the annulus is hence a constant 
 for all points, and its value is found by dividing the total 
 stress by the area of the cross-section, or 
 
ART. 138. LAMP'S FORMULA. 311 
 
 which may be either tension or compression according as the 
 numerator is positive or negative. 
 
 This longitudinal stress, together with the radial pressures, 
 causes a longitudinal elongation of the cylinder, which also is 
 to be regarded uniform for all parts of the annulus. 
 
 Let x be the distance from the center to any point within 
 the annulus. Any elementary particle is 
 here held in equilibrium by the longi- 
 tudinal unit-stress q, a tangential unit- 
 stress S, and a radial unit-stress /. The 
 value of / is evidently intermediate be- 
 tween /, and /, ; in Fig. 79 it is, like S, 
 drawn as if a tensile stress. Now from FlG - 7 ^ 
 
 Art. 129 the effective longitudinal unit-elongation of the cylin- 
 der due to these three stresses is, 
 
 in which e is the factor of lateral contraction whose mean 
 value is about -J. But, as above noted, both q and / are con- 
 stant for all parts of the annulus, and it hence follows that 
 S-{-p is also a constant, or 
 
 which is one equation between 5 and p. 
 
 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 px upon the inner surface and (/ + dp)(x -f- dx) upon 
 the outer surface of this elementary annulus. Thus, exactly 
 as in the case of a thin pipe (Art. 9), the equation of equilib- 
 rium is, 
 
 + dx) -px = Sdx, 
 
312 STRESSES IN GUNS. CH. XIV. 
 
 and, neglecting the product dp . dx, this reduces to 
 
 xdp -\-pdx Sdx, 
 which is a second equation between 6" and /. 
 
 The solution of these two equations gives for 5 and / the 
 values 
 
 where 7, is a constant of integration. To find the values of 
 7, and C 9 it may be noted that / becomes p l when x = r t , 
 and that / becomes p^ when x = r a : thus, 
 
 and these being inserted give 
 
 (32) 
 
 . _ A ~ A 
 
 P ' 
 
 which are LAME'S formulas for the stresses in hollow cylinders 
 under inner and outer pressures. In deriving these both 
 S and / have been supposed to be tension. This will be the 
 case if the formulas give positive values ; if, however, either 3 
 or / has a negative value the stress is compression instead of 
 tension. 
 
 The tangential unit-stress 5 is usually greater than the radial 
 stress /, and is the controlling factor in the design of guns. It 
 is seen to increase as x decreases, and hence it is the greatest 
 at the inner surface of the cylinder. It may be either tension 
 or compression, depending upon the relative values of the 
 radii and pressures. The radial pressure / is always compres- 
 sion, its value ranging from /, at the inner to /, at the outer 
 surface. 
 
ART. 139. A SOLID GUN. 313 
 
 As a numerical example let a cylinder be one foot in inner 
 and two feet in outer diameter, the inner pressure being 600 
 and the outer 15 pounds per square inch. Here r l = 6 inches, 
 r, 12 inches,/, = 600, /, = 15, and the formulas become, 
 
 28 080 28 080 
 
 For the inner surface x = 6 inches, whence 5 = -f- 960 and 
 p = 600 pounds per square inch ; at the outer surface x == 12 
 inches, whence 5 = + 375 and/ = 15 pounds per square 
 inch, -f- denoting tension and denoting compression. 
 
 Prob. 1 88. A solid cylinder is subject to a unfform outer 
 pressure of p % pounds per square inch. Prove that the radial 
 compression is uniform throughout. Prove that the tangen- 
 tial stress 5 is compressive and equal to the radial compression 
 at all parts of the cylinder. 
 
 ART. 139. A SOLID GUN. 
 
 A gun tube without hoops is a solid hollow cylinder, sub- 
 ject upon the outer surface to atmospheric pressure and upon 
 the inner surface to the pressure of the gas arising from the 
 explosion of the powder. The outer pressure /, is so small 
 compared to the inner pressure /, that it may be neglected. 
 Then making x equal to r l in (32) they become, 
 
 which give the greatest tensile and compressive unit-stresses 
 caused by the interior pressure. The tangential tension and 
 the radial compression decrease as x increases, and at the outer 
 surface where x = r t their values are 
 
 2r/ 
 
 = . ' A / = <>> 
 
STRESSES IN GUNS. 
 
 CH. XIV. 
 
 which are the least tensile and compressive unit-stresses caused 
 by the inner pressure. 
 
 As a special case let the outer radius be twice the inner 
 radius or r 9 = 2r 1 . Then the tension for the inner surface 
 becomes S l = ^p l , and for the outer surfaces it is 5 t = |/ A . 
 
 FIG. 80. 
 
 Thus the different parts of the annulus are quite unequally 
 stressed, and hence the material is not utilized to the best 
 advantage. The shaded area in Fig. 80 shows the manner in 
 which the tangential tension varies in this case throughout 
 the annulus. 
 
 The maximum stress in a solid gun is hence the tangential 
 tension at the inner surface which is given by the formula, 
 
 and this is the expression frequently used in cases of investi- 
 gation and design. 
 
 As an example of investigation let a cast-steel gun have an 
 inner diameter of 7.5 inches at the powder chamber and the 
 thickness of the tube be 1.75 inches. What is the greatest 
 tension produced when the inner pressure arising from the 
 explosion is 10000 pounds per square inch ? Here r, = 3.75 
 inches, r a = 5.50 inches, /, = 10000 pounds per square inch. 
 Then from the formula S t is found to be 27 300 pounds per 
 
ART. 140. A COMPOUND CYLINDER. 315 
 
 square inch, which is less than the elastic limit of steel, and 
 hence not too large. 
 
 As an example of design let the inner diameter be 3.25 
 inches, the inner pressure caused by the explosion 15000 
 pounds per square inch, and the allowable working strengthen 
 tension be 30000 pounds per square inch. What should be 
 the outer diameter? Here r l = 1.625 inches, /, = 15 ooo and 
 S l = 30000 pounds per square inch. Then solving the last 
 formula for r t there results 
 
 from which the outer radius r t is found to be 2.815 inches; 
 thus the thickness of the tube is 1.19 inches, and its outer 
 diameter is 5.63 inches. 
 
 Prob. 189. A solid gun tube is 6 inches in diameter and 3 
 inches thick. What is the inner pressure that will produce a 
 maximum tangential tension of 30000 pounds per square 
 inch? 
 
 ART. 140. A COMPOUND CYLINDER. 
 
 In a solid gun the maximum tension occurs at the inner sur- 
 face during the explosion, rising suddenly from o up to its 
 greatest value 5",. If now the metal near the bore can be 
 brought into compression, this must be overcome before the 
 tension can take effect, and thus the capacity to resist the 
 inner 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 outer pressure /, over the surface of 
 the tube. This arrangement may be called a hollow com- 
 pound cylinder. 
 
 In its normal state of rest the inner cylinder, or tube, has no 
 pressure on its inner surface and/, on its outer surface. Mak- 
 
3l6 STRESSES IN GUNS. CH. XIV. 
 
 m g A = o in (32), and also x = r l and x = r, in succession, 
 there are found 
 
 which are the tangential unit-stresses at the inner and outer 
 surfaces of the tube due to the external pressure/,,. Both of 
 these are compression, but the former is numerically greater 
 than the latter, since 2r, 2 is greater than r a 2 -f- r*. 
 
 If the hoop is shrunk on so as to produce a compressive 
 uniSstress S c at the inner surface of the tube, the pressure p % 
 upon the outer surface must be, 
 
 A= -57-*' 
 
 and, if S c be assumed, the shrinkage may be so regulated as 
 to produce this pressure / a in the normal state of rest. Then 
 the tangential stresses throughout the tube are all compression, 
 while the radial pressures range from /, on the outer surface 
 to o on the inner surface. 
 
 As an example, let r, = 2 inches, r a = 3 inches, and let it be 
 required to find the outer pressure which will cause a tangen- 
 tial compressive unit-stress of 18000 pounds per square inch 
 at the inner surface of the tube. Here, 
 
 and hence the hoop must be shrunk on so as to produce this 
 outer pressure on the hoop. 
 
 When the gun is fired the explosion of the powder causes 
 an internal tangential tension 5 given by (32), whose greatest 
 value is at the inner surface of the tube. Making x = r l , this 
 tensile unit-stress is, 
 
 (32 y St=B fcL 
 
ART. 141. CLAVARINO'S FORMULAS. 317 
 
 which is LAME'S formula for the investigation of the tube of a 
 compound gun. This tension first overcomes the initial com 
 pression S c due to shrinkage, so that the effective tension at 
 the bore during firing is 5, S c . 
 
 For example, let a tube whose inner and outer diameters are 
 4 inches and 6 inches be hooped so that a tangential com- 
 pression of 18000 pounds per square inch is caused at the 
 bore, while the inner pressure caused by the explosion is 25 ooo 
 pounds per square inch. It is required to find the tangential 
 stress at the bore during the explosion. Here r l = 2 and 
 r, = 3 inches,/, = 25 ooo, and/, = 5 ooo, as seen above. Then 
 S l = 47 ooo pounds per square inch is the tension due to the 
 explosion, but before this can take effect the initial compres- 
 sion of 18000 pounds must be overcome. Hence the result- 
 ant tension at the bore during the firing is 47 ooo 18 ooo = 
 29 ooo pounds per square inch. 
 
 If this tube be without a hoop the tension at the inner sur 
 face, found by the method of the last article, is 57 200 pounds 
 per square inch. The very great advantage of the hoop in 
 diminishing the internal stresses during the firing is hence 
 apparent. 
 
 Prob. 190. A gun-tube 3 inches in diameter and 1.5 inches 
 thick is hooped so that the tangential compression on the 
 inner surface is 30 ooo pounds per square inch. What inner 
 pressure/, will produce a resultant tangential tension on the 
 inner surface of 30 ooo pounds per square inch ? 
 
 ART. 141. CLAVARINO'S FORMULAS. 
 
 The preceding method of investigating the strength of gun 
 tubes is defective in that the two stresses 5 and / of formula 
 (32) are only apparently the internal stresses. It was shown 
 in Art. 71 and also in Art. 129 that the true internal stresses 
 are those corresponding to the deformations produced. These 
 
318 STRESSES IN GUNS. CH. XIV. 
 
 deformations are influenced by the factor of lateral contrac- 
 tion of the material, which for steel and gun metal is usually 
 taken as J (Art. 128). 
 
 At any point in the annulus (Fig. 79) the apparent tangen- 
 tial, radial, and longitudinal unit-stress are S, p, and , respec- 
 tively. Let T be the true tangential unit-stress ; then from (27) 
 of Art. 129 its value is, 
 
 r=s-i/-fe. 
 
 Inserting in this the values of S,/, and q from Art. 138, it re- 
 duces to, 
 
 t,,\ T- r -&-ll& + -JfiSL A -A 
 
 3(n'-O 3(r,' -*-,') * 
 
 which is CLAVARINO'S principal formula for the investigation 
 and design of guns. 
 
 This formula shows, as before, that the tangential stress is 
 greatest at the inner surface of the cylinder. Making x = r^ , 
 this maximum tension is 
 
 T - 
 
 which is the practical formula for the discussion of the most 
 common cases. 7i may be either tension or compression, de- 
 pending upon the relative values of the pressures and radii. 
 
 CLAVARINO'S formulas are now frequently used in the 
 investigation and design of guns, instead of those of LAME. 
 In order to compare them, the particular case where the outer 
 diameter is double the inner diameter may be considered. 
 Here r a = 2r t , and (32)' and (33)' reduce to 
 
 c __ 5A 8/ T - *#' - 2QA 
 
 Of ---- 2 -- . 
 
 3 9 
 
 Now if / 3 = o, the first formula gives a smaller stress than the 
 second ; if /, = p lt the first gives a stress three times as large 
 as the second ; if /, = o, the first gives a little larger stress 
 
ART. 142. BIRNIE'S FORMULAS. 319 
 
 than the second. Thus, since the second formula gives un- 
 doubtedly a better representation of the true stress than the 
 first, it follows that LAMP'S method errs on the side of danger 
 in a solid gun and on the side of safety in a hooped tube. 
 
 The value here used as the coefficient of lateral contrac- 
 tion is that employed in the United States by both the Army 
 and the Navy in ordnance formulas, and also generally in 
 Europe. In France, however, the value e = J is adopted. 
 
 Prob. 191. Solve Problem 189 by the formulas of this ar- 
 ticle, and compare the two results. 
 
 ART. 142. BIRNIE'S FORMULAS. 
 
 The preceding articles present an outline of the methods of 
 investigating stresses in guns by the formulas of LAM and 
 CLAVARINO. The formulas of LAM refer to apparent stresses 
 only ; those of CLAVARINO, although referring to true stresses, 
 contain an error which renders them not strictly correct for 
 hooped guns. This error lies in taking for the longitudinal 
 unit-stress q the value given in first equation of Art. 138. That 
 value of q is correct for a hollow cylinder subject to pressure 
 upon its ends as well as upon its curved surfaces. For a gun, 
 however, q has a different value. When the gun is at rest q is 
 zero, for then no exterior longitudinal forces act upon it. 
 
 For the investigation of a gun at rest the true tangential 
 stress T should be deduced by making q = o. Thus, in Art. 
 141 the correct value of T is 5 \p, or, 
 
 3 W - o r 
 
 which is BIRNIE'S formula for the investigation and design 
 of hooped guns. Making x = r t , this becomes, 
 
 / 7 ,y T - 
 
 (34) r '^~ 3(r.'-0 
 
320 STRESSES IN GUNS. CH. XIV. 
 
 which is the tangential unit-stress at the inner surface of a 
 hoop whose radii are r l and r a . These formulas are used in 
 the Ordnance Bureau of the United States army, not only for 
 hoops, but for jackets and tubes, both at rest and during the 
 firing. 
 
 Strictly speaking, BIRNIE'S formulas apply only to hoops 
 and tubes upon which the exterior longitudinal stress q is zero. 
 For a solid gun, or for a tube attached to the breech block, a 
 more correct formula may be found by considering the actual 
 value of q due to the inner pressure. Here the longitudinal 
 pressure is nr^p l , and this produces longitudinal tension upon 
 the area n(r* r, a ), so that 
 
 q ~ r** l r* 
 
 *9 T \ 
 
 is the apparent longitudinal unit-stress. Then the true tan- 
 gential stress T is .S \p \q, or, 
 
 r?r; *. -A 
 '" 
 
 and making in this x = r l , it becomes 
 
 which is the true tangential unit-stress upon the inner surface 
 of the bore. 
 
 To compare the formulas of CLAVARINO and BIRNIE the 
 particular case where r a = 2r, may be considered. Then (33)' 
 and (34)' reduce to 
 
 __ i/A ~ so/, T _ i8A - 24A 
 "T" "T" 
 
 Now, if A = o> as f r a solid gun during firing, the second for- 
 mula gives a stress 6 per cent larger than the first ; if A = > 
 as for a hooped gun at rest, the second gives a stress 25 
 
ART. 143. HOOP SHRINKAGE. 321 
 
 per cent greater than the first. Thus for this case, at least, 
 CLAVARINO'S formulas give the stresses somewhat too low. 
 
 Prob. 192. Solve Problem 189 by the formulas of this ar- 
 ticle and compare the results with those of Problem 191. 
 
 ART. 143. HOOP SHRINKAGE. 
 
 Let A be the elongation or contraction of any radius x due to 
 inner or outer pressure, then 2n\ is the elongation or contrac- 
 tion of any circumference 2nx. Now 2n\/2nx is the change 
 in the circumference per unit of length due to the unit-stress 
 T\ hence \/x = T/E, or 
 
 is the change in length of any radius to the circle where the 
 tangential unit-stress is 71 If x = r t the deformation A, is 
 that of the radius of the bore due to the unit-stress T v : if 
 x = r 9 the change A, is that of the outer radius of the tube 
 where the unit-stress is T f 
 
 Suppose a compound cylinder to be formed by shrinking a 
 hoop upon a tube. The inner radius of the tube is r l and its 
 outer radius r, ; the inner radius of the hoop is r, and its outer 
 radius r t . In consequence of the shrinkage the radial pressure 
 /> a is produced between the two surfaces ; this causes the tube 
 to be under tangential compression and 
 the hoop to be under tangential tension. 
 It is required to find these stresses when 
 the original inner radius of the hoop is less 
 than that of the outer radius of the tube 
 by the amount A. 
 
 Let A, be the decrease in the outer 
 radius of the tube and A,' the increase in 
 
 the inner radius of the hoop ; then A = A, -f- ^/ ^ n Fig- 8 1, 
 which is much exaggerated, cd represents A, and be repre- 
 
322 STRESSES IN GUNS. CH. XIV. 
 
 sents \J. Also, let T t be the tangential compression at the 
 outer surface of the tube due to the shortening A a , and let 
 77 be the tangential tension at the inner surface of the hoop 
 due to the elongation A/. Then 
 
 or 
 
 T, + T,' = Q, 
 
 't 
 
 which gives one equation between 7!, and 7",'. 
 
 Formula (34) is applied to the tube by making p l = o and 
 x = r t ; thus the tangential compression Js, 
 
 Formula (34) is applied to the hoop by replacing /, by /, , /, 
 by o, r, by r, , and r t by r, ; then making x = r, , there results, 
 
 2 
 ~ 
 
 which is the tangential tension. Dividing the first of these by 
 the second, gives 
 
 L. f = a - or r,=7>, 
 
 which is a second equation between 7", and 7 1 /. 
 The solution of these equations gives the values 
 A a , _E\. b 
 
 ' r, 'a + 6' ' r, ' 
 
 in which a and b are the known quantities, 
 _2 2r t a + r, g ,__^ r, a 
 
 " * 
 
 and thus the tangential compression at the outer surface of the 
 tube and the tangential tension at the inner surface of the hoop 
 may be computed. The tangential compression at the bore is, 
 
ART. 143. HOOP SHRINKAGE. 323 
 
 however, greater than 7" a , and it may be found from (34)' by 
 substituting the value of/,, now known ; thus, 
 
 T 
 
 J 
 
 which is the greatest compression in the tube. 
 
 As a numerical example let a compound cylinder be formed 
 of a steel tube whose inner radius is 3 inches and outer radius 5 
 inches, with a steel hoop whose thickness is 2 inches. It is re- 
 quired to find the stresses produced when the original difference 
 between the outer radius of the tube and the inner radius of 
 the hoop is 0.004 inches. First, the sum of the two tangential 
 stresses at the surface of contact is, 
 
 E\ _ 30 ooo ooo X 0.004 _ 
 
 -- -- 24 ooo. 
 
 r* 5 
 
 Second, since r t = 3, r, = 5, and r t = 7 inches, 
 
 l8 + 2 5 __ 43 h - 2 5 + 9 8 _ 82 
 " 3 25 - 9 "" 24' "349-25 "24 
 
 Third, the tangential compression at the outer surface of the 
 tube is, 
 
 T % = X 24 ooo = 8 260 pounds per square inch. 
 
 43 -r 82 
 
 Fourth, the tangential tension at the inner surface of the 
 
 hoop is, 
 
 ft? 
 
 TV = - - X 24 ooo =15 740 pounds per square inch. 
 43 + 82 
 
 Thus it is seen that the hoop tension is nearly double the com. 
 pression on the outer surface of the tube. At the bore of the 
 tube, however, the tangential compression is found to be 14 400 
 pounds per square inch. 
 
 The decrease in the outer radius of the tube is next com- 
 puted; thus, 
 
 A, = ^ = 0.00138 inches, 
 
324 STRESSES IN GUNS. CH. XIV. 
 
 and the increase in the inner radius of the hoop is, 
 
 T'r 
 
 A,' = ~ 0.00262 inches. 
 h 
 
 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 outer radius of 5.0014 inches, and the hoop to an inner 
 radius of 4.9974 inches. The radius of the bore, however, will 
 then be less than 3 inches by the quantity, 
 
 T 
 
 A, = ^ = 0.00144 inches, 
 , 
 
 and hence if its final radius is to be exactly 3 inches, it must 
 be turned to a radius of 3.0014 inches. 
 
 If this example be solved by using the formulas of CLAVA- 
 RINO instead of those of BlRNlE, the following values will be 
 found: J", = 7 030, 7 1 / = 16970, 7^ 14400 pounds per 
 square inch; A a 0.00117, \' = 0.00283, an d ^, 0.00144 
 inches. The shrinkages thus agree within 0.0002, which is as 
 close as measurements can be relied upon. 
 
 Prob. 193. A solid steel shaft, 6 inches in radius, is to be 
 hooped so that the greatest tensile stress in the hoop and the 
 greatest compressive stress in the shaft shall be 15 ooo pounds 
 per square inch. Find the thickness of the hoop and the 
 radius to which it should be turned. 
 
 ART. 144. 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 sim- 
 ple case of a tube with one hoop can here only be considered. 
 If the radii be given, as also the inner pressure /, due to the 
 explosion, it may be desired to find the shrinkages so that this 
 requirement will be fulfilled. As p l is very large, it is desir- 
 
ART. 144. 
 
 HOOPED GUNS. 
 
 325 
 
 able 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 dia- 
 gram in Fig. 82 represents the distribution of the internal 
 stresses under this supposition. O is the center of the gun, 
 
 b 
 
 FIG. 82. 
 
 OA the inner radius r l , while AB is the thickness of the tube 
 and BC that of the hoop. The shaded areas show the stresses 
 due to the hoop shrinkage, Aa and Bb being the tangential 
 compressions T l and 7", of the last article, while Bb' is the tan- 
 gential tension 7 1 /, and Cc is the tangential tension at the outer 
 surface of the hoop. When the explosion occurs the two 
 cylinders are thrown into tangential tension, Aa l and Bb v 
 being those at the Inner surfaces of the tube and hoop. The 
 above principle indicates that both Aa t and Bb l should be 
 equal to the maximum allowable unit-stress 7*, which for 
 guns is often taken nearly as high as the elastic limit of the 
 material. 
 
 In designing a 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 y 
 Aa l , and Bb l in Fig. 82 are each equal to the elastic limit of 
 the material. Or, given the radius of the bore and the outer 
 
326 STRESSES IN GUNS. CH. XIV. 
 
 radius of the hoop, it may be required to find the intermediate 
 radius under the same conditions. These problems can be 
 solved as well as more complex ones relating to guns with 
 several hoops. 
 
 The stresses in guns are greatest near the powder chamber, 
 since the gas expands and its pressure decreases as the pro- 
 jectile moves outward. Hence the hoops increase in number 
 toward the powder chamber, each being shorter than the one 
 beneath it. Guns with seven or more hoops have been built. 
 Wire has been used instead of hoops, but not with good results. 
 
 The longitudinal stress is mostly borne by the tube if that 
 be attached to the breech block. In this case the longitudinal 
 internal pressure during firing is 7tr*p^ and this divided by 
 n(r* r?) gives the apparent longitudinal unit-stress q. This, 
 however, is decreased by the influence of the tangential and 
 radial pressures. Thus, from Arts. 129 and 138, 
 
 which is the true longitudinal unit-stress on the tube. When 
 the jacket is attached to the breech block, which is more often 
 the case, it is subject to the inner pressure / from the tube, 
 to the outer pressure /, from the hoop, and it carries the 
 entire longitudinal stress 7tr?p l ; thus 
 
 T 
 
 is the longitudinal unit-stress on the jacket. 
 
 Reference is made to the excellent work of MEIGS and 
 INGERSOLL, The Elastic Strength of Guns (Baltimore, 1885), 
 for a detailed presentation of the formulas and methods used 
 in the United States Navy for the design of guns. The for- 
 mulas used by the Ordnance Bureau of the Army will be 
 
ART. 144. HOOPED GUNS. 327 
 
 found set forth in a thorough manner in STORY'S Elements 
 of the Elastic Strength of Guns (Fort Monroe, 1894). 
 
 Prob. 194. Prove that a gun tube with one hoop is most 
 advantageously designed when the common radius of tube and 
 hoop is a mean proportional between the other two radii. (To 
 solve this problem derive an expression for /, in terms of r lf 
 r 9 ,r t , and T e . Then the most advantageous value of r, is 
 that which renders/, a maximum.) 
 
328 PLATES, SPHERES, AND COLUMNS. ClI. XV. 
 
 CHAPTER XV. 
 PLATES, SPHERES, AND COLUMNS. 
 
 ART. 145. CIRCULAR PLATES. 
 
 Let a circular plate of radius r and uniform thickness d be 
 subject on one side to a pressure / on each square unit of 
 area and be supported or fixed around the circumference. The 
 head of a cylinder under the pressure of water or steam is a 
 circular plate in such a condition. The maximum stress 
 caused by the flexure will evidently be at the middle, and this 
 is required to be determined. 
 
 As the simplest case let the plate be merely supported 
 around the circumference. The total load on the plate being 
 7tr*p, the total reaction of the support 
 is also nr*p, or the reaction per linear 
 unit is \rp. Now let a strip having the 
 small width b be imagined to be cut out 
 of the plate, so that its central line co- 
 incides with a diameter. The reaction 
 UUIUUUU at each end of this strip is b.^rp, and the 
 
 j^m T ** : nj jt rp? load on the strip is b.2r.p. The sum of 
 
 the two reactions being only one half the 
 
 load, an upward shearing force equal to 
 
 b.r.p must act along the sides of the strip to maintain the 
 
 equilibrium. At the center of the circle there can be no 
 
 shearing stress and the most probable assumption regarding its 
 
 distribution on the sides of the strip is to take it as varying 
 
 uniformly from the center to the circumference, as shown by 
 
 the dotted lines in Fig. 83. 
 
ART. 145. CIRCULAR PLATES. 329 
 
 The strip whose width is b and length 2r is thus a beam 
 acted upon by two vertical reactions, each equal to tyrp, a 
 downward load 2brp y and two vertical shears on the sides 
 each equal to \brp. The bending moment at the middle of 
 this imaginary beam hence is, 
 
 M \brp . r + \brp . \r brp . \r = \br*p, 
 
 and the maximum unit-stress on the upper or lower fiber at 
 the middle of the strip is, 
 
 _ Me _ 6M _ r^ 
 S >~ f = bd'- 2p <T 
 
 This value of S l is not the real horizontal fiber stress at the 
 center of the circle, but only the apparent stress due to con- 
 sidering the elementary strip. At the center the fiber stresses 
 are acting in all directions. If a second strip be passed in 
 Fig. 83 at right angles to the first, a stress 5, equal in value 
 but normal in direction to the first will be found. The true 
 fiber-stress T will be determined from the principle of Art. 
 129, taking into account the factor of lateral contraction e; 
 thus on the upper side of the plate, 
 
 T=(S,-eS,+ ef) 
 
 and on the lower side of the plate, 
 
 which is the value to be used, since rupture will generally be- 
 gin on the tensile side. 
 
 For iron and steel the mean value of the factor of lateral 
 contraction e is . Hence 
 
 is the practical formula for the discussion of iron and steel 
 plates under a uniform pressure when simply supported at the 
 circumference. The unit-pressure that such a plate can carry 
 
330 PLATES, SPHERES, AND COLUMNS. CH. XV. 
 
 with a given unit-stress hence varies directly as the square of 
 its thickness and inversely as the square of its radius. 
 
 The more common case of a circular plate fixed around its 
 circumference cannot be solved without determining the elastic 
 curve into which a diameter deflects. The investigation is too 
 difficult and lengthy for this elementary book, but it can be 
 said that the general conclusion is that the true effective unit- 
 stress T is about three-fourths of that for the supported plate. 
 Using this result the coefficient in the last formula will be 
 unity, or 
 
 is the practical formula for iron and steel plates under uniform 
 pressure when fixed around the circumference. 
 
 Assuming a safe working stress T, the proper thicknesses 
 of plates under uniform pressure then are 
 
 =r ~ and d = 
 
 the first being for supported and the second for fixed circum- 
 ference. For example, let a fixed cast-iron cylinder head of 3 
 feet diameter be required to sustain a uniform pressure of 250 
 pounds per square inch, so that the maximum tensile unit- 
 stress T may be 3 600 pounds per square inch ; then 
 
 which is the required thickness. 
 
 The formulas derived by GRASHOF from a more extended 
 theoretical analysis, are, 
 
 2p 
 
 and d = 
 
 for the thickness of supported and fixed plates respectively. 
 
ART. 146. ELLIPTICAL PLATES. 331 
 
 The second of these, applied to the above numerical example, 
 gives d = 3$ inches. 
 
 Prob. 195. Deduce a formula for a circular plate sup- 
 ported around the circumference and carrying a single load P 
 at the center. 
 
 ART. 146. ELLIPTICAL PLATES. 
 
 Elliptical plates are commonly used for the covers of man- 
 holes in boilers and stand-pipes. Let/ be the uniform unit- 
 pressure on the plate, a the major axis and b the minor axis of 
 the ellipse. It is required to find the 
 maximum unit-stress 5 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. 84, one along the major axis and the other 
 along the minor axis. Let W l and W t be the loads on these 
 strips, and R t and R t the reactions at their ends. At the center 
 they have a common deflection, which by Art. 36 is 
 
 __ _ W? 
 
 ~'mEl ~ 
 
 and hence Wj? = WJ?\ or, since the reactions are proper- 
 tional to the loads, 
 
 j = or 
 
 that is, the reactions at the ends of the axes are inversely as 
 the cubes of the lengths of the axes. Hence the total weight 
 nabp is not uniformly distributed on the support around the 
 circumference, 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 fiber 
 stresses near the center are the greatest in directions parallel 
 
332 PLATES, SPHERES, AND COLUMNS. ClI. XV. 
 
 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 veri- 
 fied by tests. 
 
 A satisfactory solution of this very difficult problem has 
 not been made. From the discussion given by BACH (Elas- 
 ticitat und Festigkeit, 1894) the following approximate formula 
 results, which may advantageously be used for elliptical plates 
 in the absence of other rules : t 
 
 Ca*b*p 
 " (a* -f tf}d" 
 
 in which 5 is th~ maximum unit-stress at the center, d is the 
 thickness of the plate, and C is a number which is probably 
 about f when the circumference is supported and about J 
 when it is fixed. 
 
 Prob. 196. A common proportion for manhole covers is to 
 make a i.$b. If the thickness be f inches, the major axis 
 16 inches, and the material wrought iron, find the safe pressure 
 per square inch. 
 
 ART. 147. RECTANGULAR PLATES. 
 
 A rectangular plate of length a and breadth , subject to a 
 uniform pressure^ per square unit, distributes that pressure 
 over the supports in a similar manner to the elliptical plate. 
 The reaction per linear unit is less on the ends than on the 
 sides, 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 
 by BACH for such plates is, 
 
 e_ C 
 ~ 
 
 in which 5" is the maximum unit-stress at the middle, and d is 
 the thickness of the plate. The value of C as found by BACH, 
 by experiments on square plates, ranged from f- to f , according 
 
ART. 148. HOLLOW SPHERES. 333 
 
 as the condition of the edges approached that of a mere sup- 
 port or a state of fixedness. 
 
 For a square plate whose side is a this formula gives 
 
 c _ and c _ 
 
 " - 
 
 for free and fixed supports respectively. A theoretic analysis 
 by GRASHOF for square plates fixed at the middle of each side 
 gives the true unit-stress as 
 
 <*>- gd> 
 
 if the factor of lateral contraction e be taken at -J, as usual for 
 iron plates (Art. 128). 
 
 While the numerical coefficients, as deduced by different 
 authors, vary somewhat, it is well established that the unit- 
 stress at the middle of the plate varies directly as its area and 
 the unit-pressure /, and inversely as the thickness of the plate. 
 The strength of a plate as measured by the pressure / that it 
 can carry varies directly as the square of the thickness and in- 
 versely as the area. 
 
 Prob. 197. Prove that the maximum unit-stress caused by a 
 given uniform load Wis independent of the size of the plate. 
 
 ART. 148. HOLLOW SPHERES. 
 
 Hollow spheres are used in certain forms of boilers under 
 inner steam-pressure. The ends of steam and water cylinders 
 are sometimes made hemispherical instead of plane, in order 
 to avoid flexure (Art. 145). If the thickness of the sphere is 
 small compared to its radius, the investigation is simple. Let 
 r be the radius and t the thickness. Let p be the inner press- 
 ure per square unit, and 5 the tensile unit-stress on the an- 
 nulus. Then on any great circle the total pressure is nr'p, and 
 
334 PLATES, SPHERES, AND COLUMNS. CH. XV, 
 
 this is resisted by the tension 2nrtS in the section of the an 
 nulus. Equating these gives 
 
 rp = 2/5, or S = 2, 
 
 which is the formula generally used for thin spheres under 
 inner pressure. But in strictness 5 is the apparent stress, 
 while another equal in intensity acts at right angles to it. 
 Thus from Art. 129 the true stress on the inner surface is 
 
 which gives the maximum true unit-stress, since that on the 
 outer surface is f S. The usual formula thus errs on the side 
 of safety. 
 
 The investigation of a thick hollow sphere under inner and 
 outer pressure will be similar to that of the thick cylinder in 
 Art. 138. Let r l be the inner and r t the outer radius, p l and 
 /, being the corresponding pressures per square unit of surface. 
 Fig. 79 of Art. 138 may represent a partial section of the 
 sphere, x being the radius of any elementary annulus where 
 the radial unit-stress is / and the tangential unit-stress is 5. 
 From the symmetry of the sphere it is seen that another stress 
 5 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 /, S, and S. The sum 
 of these is regarded as constant throughout the annulus, or 
 
 is one equation between 5" and /. 
 
 Now the inner pressure on a great circle whose radius is x 
 is nx?p, and the outer pressure on a great circle whose radius 
 is x-\-dx is n(x -f- dx)\p + dp). The difference of these is 
 equal to the resistance of the elementary annulus, which is 
 Zitxdx.S, or 
 
 (x + dx)\p + dp) - x*p = *Sxdx\ 
 
ART. 148. HOLLOW SPHERES. 335 
 
 and this, omitting quantities of the second order, reduces to 
 
 xdp + 2pdx = 2$dx, 
 which is a second equation between 5 and /. 
 
 Substituting in the second equation the value of 5 from the 
 first, and integrating, the value of / in terms of x is found, 
 and thus 5 also ; the results are 
 
 s=c ' + ^' * = c >- 2 i?< 
 
 in which C 9 is a constant of integration. To find the values of 
 C l and C % > the value of/ becomes /, when x = r lt and /, 
 when x = r % ; thus 
 
 and these are the formulas for thick hollow spheres deduced 
 by LAME. It is seen that they are analogous to those for 
 thick cylinders, the radii being cubed instead of squared. It 
 is also seen that the formula for 5 is the important one and 
 that its greatest value occurs at the inner surface of the 
 sphere. 
 
 For the common case of inner pressure, let/ t =o, and make 
 x = r l9 then the greatest tangential stress is 
 
 .* * <**** 
 
 which is the common formula for hollow spheres. This gives 
 the apparent tensile unit-stress at the inner surface. The true 
 unit-stress at the inner surface is, by Art. 129, 
 
 T = S - IS 4- \i> = 2r * "*" T * - ^ 
 ? l r* r' 3 
 
 which will usually be found to be less than S,. 
 
 As an example, let a cylinder 4 inches in inner and 8 inches 
 in outer radius have a hemispherical end with the same radii, 
 
336 PLATES, SPHERES, AND COLUMNS. CH. XV. 
 
 and be subject to an inner water-pressure of 4000 pounds per 
 square inch. Then the apparent tensile stress on the inner 
 surface of the hemisphere is 
 
 256 -4- 64 
 5, = ^- . 4000 = 2 860 pounds per square inch, 
 
 while the true tensile stress is 
 
 i 024 + 64 4 ooo , 
 
 T.= ~^^ ~ = 2640 pounds per square inch, 
 512 04 3 
 
 which shows that the true value is about 8 per cent less than 
 the apparent. 
 
 For the cylinder the apparent and true tensile unit-stresses 
 at the inner surface are, from Arts. 138 and 140, 
 
 _ r ; 4- r' 4r,* + r,S. 
 
 s '-^ r ^ 7/ " l= '"^^7' 
 
 which L *ive S l = 6 700 and T t = 7 600 pounds per square 
 inch, so that the true stress is 13 per cent greater than the 
 apparent. 
 
 If the end of the cylinder be a flat plate of the same thick- 
 ness as the cylinder, or 4 inches, and be fixed around the cir- 
 cumference, the true stress on the outer side is 
 
 T = -f^ X 4 ooo = 16 ooo pounds per square inch, 
 which is 6 times as great as for the hemisphere, and more 
 than double the greatest stress in the cylinder. The advan- 
 tage of hemispherical ends in reducing the stresses is thus seen 
 to be very great. It may be remarked, in conclusion, that the 
 theory of internal stress in cylinders and spheres is not perfect, 
 for it fails to give the same results for the common surface of 
 junction of a cylinder and hemisphere. 
 
 Prob. 198. A hollow sphere is to be subject to a steam- 
 pressure of 600 pounds per square inch, its inner radius being 
 8 inches. Find its thickness so that the greatest stress may 
 be i ooo pounds per square inch. 
 
ART. 149. EXPERIMENTS ON COLUMNS. 337 
 
 Prob. 199. Investigate the discrepancy between the formu- 
 las for hollow cylinders and hollow spheres for the following 
 numerical case. A hollow cylinder with hollow hemispherical 
 ends, the inner diameters being 8 inches and the outer 
 diameters 12 inches, is subject to an inner water pressure of 
 2400 pounds per square inch. Compute, by Art. 141 and by 
 this article, the true maximum unit-stress T for the common 
 plane of junction of cylinder and hemisphere. 
 
 ART. 149. EXPERIMENTS ON COLUMNS. 
 
 It is impossible to present here even a summary of the many 
 experiments that have been made to determine the laws of re- 
 sistance of columns. The interesting tests made by CHRISTIE 
 in 1883 f r tne Pencoyd Iron Works will however be briefly 
 described on account of their great value and completeness as 
 regards wrought iron struts, embracing angle, tee, beam, and 
 channel sections. See 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 300, 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 
 49600 pounds per square inch, ultimate elongation 18 per cent 
 in 8 inches. The length of the specimens varied from 6 inches 
 up to 1 6 feet, and the ratio of length to least radius of gyra- 
 tion varied from 20 to 480. Each specimen was placed in a 
 Fairbanks' testing machine of 50 ooo pounds capacity and the 
 power applied by hand through a system of gearing to two 
 
338 
 
 PLATES, SPHERES, AND COLUMNS. 
 
 CH. XV. 
 
 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- 
 
 Length divided by 
 Least Radius of 
 Gyration. 
 
 Flat Ends. 
 
 Fixed Ends. 
 
 Hinged Ends. 
 
 Round Ends. 
 
 20 
 
 46000 
 
 46000 
 
 46000 
 
 44000 
 
 40 
 
 4000O 
 
 40000 
 
 40000 
 
 36500 
 
 60 
 
 36000 
 
 36000 
 
 36000 
 
 30500 
 
 80 
 
 32000 
 
 32000 
 
 31500 
 
 25000 
 
 100 
 
 29800 
 
 30000 
 
 28000 
 
 20500 
 
 1 20 
 
 26300 
 
 28000 
 
 24300 
 
 16 500 
 
 ,140 
 
 23500 
 
 25500 
 
 21 000 
 
 12800 
 
 160 
 
 20000 
 
 23000 
 
 16500 
 
 95oo 
 
 180 
 
 16800 
 
 20 OOO 
 
 12800 
 
 7500 
 
 200 
 
 14500 
 
 17500 
 
 I0800 
 
 6000 
 
 220 
 
 12 700 
 
 15000 
 
 8800 
 
 5000 
 
 240 
 
 II 200 
 
 13000 
 
 7500 
 
 4300 
 
 260 
 
 9800 
 
 II OOO 
 
 6500 
 
 3800 
 
 280 
 
 8 500 
 
 10 000 
 
 5700 
 
 3200 
 
 300 
 
 7200 
 
 9000 
 
 5000 
 
 2800 
 
 320 
 
 6000 
 
 8000 
 
 4500 
 
 2 5OO 
 
 540 
 
 5 ioo 
 
 7000 
 
 400O 
 
 2 IOO 
 
 360 
 
 4300 
 
 6500 
 
 3500 
 
 I 9OO 
 
 380 
 
 3500 
 
 5 800 
 
 3000 
 
 I 700 
 
 4 00 
 
 3 ooo 
 
 5 200 
 
 . 2500 
 
 I 500 
 
 420 
 
 2 500 
 
 4800 
 
 2300 
 
 I 300 
 
 440 
 
 2200 
 
 4300 
 
 2 IOO 
 
 
 460 
 
 2000 
 
 3 800 
 
 I 900 
 
 
 480 
 
 I 900 
 
 
 I 800 
 
 
ART. 150. EULER'S MODIFIED FORMULA. 339 
 
 fully conducted experiments. The first column gives the 
 values of l/r and the other columns the values of P/A, the 
 latter being the ultimate load in pounds per square inch. 
 From the results it will be seen that there is little practical 
 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 3^ times that of those with round 
 ends. 
 
 EULER'S formula fairly represents the results of the tests 
 on the long round-ended columns. Taking E = 25 ooo ooo 
 pounds per square inch for wrought iron, and * = 10, Art. 
 
 53 g^es, 
 
 P S 
 
 -^ = 250000000^,, 
 
 from which are computed, 
 
 for l/r 300, 340, 380, 400, 
 P/A = 2800, 2 200, 1700, 1400, 
 while the experiments give the values 
 
 P/A = 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. 
 
 Prob. 200. Plot the above experiments on round-ended 
 columns, taking P/A as abscissa and l/r as ordinate. Also 
 plot on the same sheet EULER'S curve and the straight line 
 given by T. H. JOHNSON'S formula. 
 
 ART. 150. EULER'S MODIFIED FORMULA. 
 
 EULER'S formula for columns expresses the condition of 
 indifferent equilibrium or that state which borders between 
 stable and unstable equilibrium. In all cases of indifferent 
 equilibrium slight causes produce marked effects, and hence 
 it seems important to inquire whether EULER'S formula, as 
 
340 PLATES, SPHERES, AND COLUMNS. CH. XV. 
 
 given in Art. 53, represents the exact relation between P/A 
 and l/r. It will be apparent on reflection that, while the 
 formula contains but one length /, there are really three 
 different lengths that should be taken into consideration. 
 Let / represent the length of the straight column in its un- 
 strained state, /, the length of the straight column after com- 
 pression by the unit-stress P/A, and / 8 the chord of the 
 deflected curve after lateral bending. 
 
 Referring now to formula (5) of Art. 33, it is seen that this 
 does not include the effect of the longitudinal compression 
 P/A. To introduce this let 5 be the total unit-stress pro- 
 duced by both flexure and compression ; then in the demon- 
 stration the first four formulas will be thus modified : 
 
 dl .* S ~ P ' A E S^P/A M 
 di,~\> ~7~ ~ R' c ~ r 
 
 and from these there results, 
 
 dx*~ Eldl, "~ Ell? 
 
 which is the correct differential equation of the elastic curve 
 for a body under combined flexure and compression. 
 
 Passing now to EULER'S deduction in Art. 53, Mis replaced 
 by Py, and the equation of the elastic curve for a round- 
 ended column is 
 
 A ' 
 
 y----A sin 
 
 Here y = o when x = 7 9 ; hence by the same reasoning as 
 before, 
 
 P=n>EI or 5=*'^ 
 
 is the exact condition for the state of indifferent equilibrium. 
 
 To complete the investigation /, and / a are to be expressed 
 
 in terms of /. Now / /, is the deformation due to the 
 
 longitudinal compression P/A, and hence from the fundamen- 
 
ART. 150. EULER'S MODIFIED FORMULA. 341 
 
 tal definition of the coefficient of elasticity (Art. 4), 
 
 which gives the length of the straight column after longitu- 
 dinal compression. 
 
 To find /, it is plain that /, /, is the fall of the end of the 
 column due to the lateral flexure and that P(/ t /,) is the 
 work done in this fall. This external work must equal the 
 internal work of the flexural stresses. From Art. 109, 
 
 JT s_ _ _ 
 
 '' ~~ 2EI ~~ 2EI s 
 
 and integrating this between the limits x = o and x = /, , the 
 internal work K is found to be P*d*lJ^EI. Thus, equating 
 the two works, 
 
 and from this, in connection with the above value of /, , 
 
 i-P/AE , 
 
 ~ i P4'Ar** 
 
 which gives the length of the chord of the deflected curve. 
 
 The quantities P/AE and PA*/^Ar* are small in comparison 
 with unity, and hence their squares and their product may be 
 neglected; also the reciprocal of I P/AE may be taken as 
 i -f- P/AE. Introducing the values of /, and /, into the above 
 expression for P it reduces to 
 
 P -r* P P 
 
 and since n*Er*/l* is a close approximation to the value of 
 
 P/A it may replace the latter in the parenthesis, giving finally 
 
 P 
 
 which is EULER'S modifie'd formula for round-ended columns, 
 By writing 2\n* instead of nr a it applies to a column with one 
 
342 PLATES, SPHERES, AND COLUMNS. CH. XV. 
 
 end round and the other fixed, and by writing 4^ a instead of 
 7r a it applies to a column with both ends fixed. 
 
 A modification of EULER'S formula by use of the / t and / 
 was given by WlNCKLER in 1867, by GRASHOF in 1878, 
 and by PRICHARD in 1897 (see Engineering News, May 6, 
 1897). The formula as given by them can be obtained from 
 the above by making / 2 = /, or by making A = o. Since the 
 quantities TrV 2 // 2 and ifA/T are small compared with unity, 
 EULER'S modified formula gives numerical results but little 
 larger than the usual ones. It shows, however, that P really 
 increases slightly as A increases, and that hence A is in strict- 
 ness not wholly indeterminate, as the common theory teaches. 
 
 Prob. 201. Show that the fall of the load P, due to direct 
 compression of the column, equals the fall due to the lateral 
 flexure when A = 2r. 
 
 ART. 151. THE DEFLECTION OF COLUMNS. 
 
 An ideal column always remains straight under the action 
 of an axial load W, provided that this load is less than the 
 value of P given by EULER'S formula. If a slight lateral 
 force be applied it will deflect, recovering its straight condi- 
 tion, however, as soon as this force is removed. Under the 
 action of such a slight lateral force there is a definite relation 
 between the deflection A and the maximum unit-stress 5 on 
 the concave side. This relation, as given by Art. 55, is 
 
 s 
 
 and it shows that 5 increases with A. 
 
 Now if ^becomes so great that the column does not spring 
 back, but remains in indifferent equilibrium, its value is the P 
 given by EULER'S formula, and 
 
 SI' 
 
 Here 5 is indeterminate, and the column may remain in in- 
 
ART. 151. DEFLECTION OF COLUMNS. 343 
 
 different equilibrium with many different values of A. How- 
 ever, if A be so great that 5 becomes equal to the elastic 
 limit S e , failure at once follows, and hence the greatest possi- 
 ble deflection is found by using S e for 5 in the last equation. 
 When a steadily increasing load W is applied to an actual 
 column it may, on account of being non-homogeneous or not 
 perfectly straight, begin to bend without the application of 
 any lateral force long before W reaches the value P given by 
 EULER'S formula. Suppose, for instance, that the column is 
 slightly crooked so that it has an initial deflection 6. The 
 actual deflection A will now be measured, not from the axis 
 of ordinates, but from the original position of the axis of the 
 column. Then for an ideal column with the deflection A the 
 bending moment is PA, while for the actual column the bend- 
 ing moment is W(d + J), and equating these there is found, 
 
 W 
 * = pW*< 
 
 Thus here the actual deflection increases with W, and no in- 
 determinateness occurs. This case in fact is closely analogous 
 with that of a column when the load is eccentric with respect 
 to the axis, the deflection A being always perfectly determi- 
 nate (Art. 62). Even if d be very small A may become a 
 considerable quantity, and as it increases the unit-stress 5 also 
 increases, failure being practically complete as soon as .S 
 reaches the elastic limit. 
 
 These conclusions, it will be seen, are essentially the same 
 as those already derived in Art. 62. For a detailed discussion 
 of columns based on these considerations see an article by 
 MONTCRIEFF in Proceedings of American Society of Civil 
 Engineers for March, 1900, or in its Transactions, Vol. XLIII, 
 1900. 
 
344 APPENDIX. 
 
 APPENDIX. 
 
 INTERNATIONAL ASSOCIATION FOR TESTING MATERIALS. 
 
 In 1882 a number of German professors of engineering met 
 at Munich and discussed the question as to how uniformity 
 in the methods of testing materials might be promoted. 
 Formal conferences were held in other German cities in 
 1884, 1886, 1888, and 1893, at which engineers frpm other 
 Europ'ean countries were present. The reports of these 
 conferences attracted wide attention and the movement 
 assumed an international character. , 
 
 In 1895 the fifth conference met at Zurich, all European 
 countries, except Turkey, being represented, as also the 
 United States. At this meeting the International Associa- 
 tion for Testing Materials was formally organized, its object 
 being " the development and unification of standard methods 
 of testing for the determination of the properties of materials 
 of construction and of other materials, and also the perfection 
 of apparatus for that purpose." This may be called the first 
 congress of the Association. The second congress met at 
 Stockholm in 1897, there being present 361 members, repre- 
 senting eighteen countries. The third congress was intended 
 to be held at Paris in 1900, but this was omitted on account 
 of the Exposition Congress on the same subject which had 
 been organized by the French government. 
 
 In 1899 the total number of members of the International 
 Association was about I 700, of which 393 are credited to 
 Russia, 384 to Germany, 213 to Austria, 140 to the United 
 States, 87 to England, 82 to Switzerland, 77 to France, 60 
 to Sweden, 42 to Holland, 41 to Norway, and the remainder 
 to twelve other countries. 
 
VELOCITY OF STRESS. 345 
 
 The official organ of the Association is the journal Bau- 
 materialienkunde, published semi-monthly at Stuttgart, Ger- 
 many. The American Section of the Association, organized 
 in 1898, publishes occasional Bulletins containing its papers 
 and proceedings. Bulletin No. 4, published in September, 
 1899, contains an address on the work of the Association by 
 the Chairman of the American Section, delivered at the sec- 
 ond annual meeting of the Section, in which may be found 
 detailed information regarding the work and plans of the 
 Association. Bulletin No. 5 contains a valuable report on the 
 present state of knowledge regarding impact and impact tests. 
 
 The technical work of the Association is done by Interna- 
 tional Committees which study definite problems and make 
 reports on them to the congresses. At the congress to be held 
 in 1901 or 1902 it is expected that many of these committees 
 will make valuable reports. The report of the American 
 branch of the Committee on International Specifications for 
 Testing Iron and Steel may be found in Bulletins Nos. 8-18 
 of the American Section, issued in May, 1900. These are 
 accompanied by tables showing the requirements of American 
 manufacturers and consumers regarding the chemical compo- 
 sition and physical properties of wrought iron and various 
 grades of steel, and they also contain standard specifications 
 proposed by the Committee. 
 
 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 wiL be 
 shown to depend upon the stiffness and density of the mate- 
 rial. In fact, a sudden stress is propagated through a body 
 in the same manner as sound is propagated through the air. 
 
346 APPENDIX. 
 
 Let v be the velocity with which stress is transmitted in a 
 body whose coefficient of elasticity is E, and whose weight per 
 cubic unit is w at a place where the acceleration of gravity is 
 g. It is required to find v in terms of E, w y and g. 
 
 If F be a force which acting continuously for one second 
 produces the velocity #, and if a body of the weight W ac- 
 quires under the action of gravity the velocity^ in one second, 
 then the forces are proportional to the accelerations that they 
 produce, or 
 
 # 
 Fg = Wu, whence F = W, 
 
 o 
 
 which is one of the well-known formulas of mechanics. 
 
 Now let a unit-stress 5 be applied to the end of a bar, pro- 
 ducing 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 
 elongation 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 sv. Thus in one second the 
 center of gravity of the bar is moved the distance %sv, and its 
 velocity u at the end of the second is sv. Now referring 
 to the formula Fg = Wu, the value of F is S, which is equal to 
 Es, the value of W is wv, since the cross-section considered is 
 unity, and hence Esg = wv . sv, whence 
 
 v = 
 
 which is the formula for the velocity of wave propagation in 
 elastic media first deduced by NEWTON. 
 
 Taking for E and w their mean values given in Art. 80, and 
 also g as 32.16 feet per second per second, since the unit-weights 
 w are given for the surface of the earth, the mean values of the 
 velocity of transmission of stress for different materials are 
 found to be as follows : 
 
ADVANCED PROBLEMS. 347 
 
 For timber, v = 13 200 feet per second, 
 
 For stone, v = 13 200 feet per second, 
 
 For cast iron, v = 12 400 feet per second, 
 For wrought iron, v = 15 500 feet per second, 
 For steel, v = 17 200 feet per second. 
 
 In making this computation E must be taken in pounds per 
 square foot, since both w and g are expressed in terms of feet. 
 
 The velocity of sound, light, and all wave propagations in 
 elastic media is given by the above formula for v. The ratio 
 of w to g is a constant for the same material at any point in 
 space, and it expresses the density, while E is an index of stiff- 
 ness. The ether that transmits waves of light must be lighter 
 than air and stiffer than steel, in order that v may be the high 
 value found by observation. 
 
 Prob. 203. What time is required for sound to travel a dis- 
 tance of 5 miles in water, the linear unit-compression for water 
 being 0.00005 ? 
 
 ADVANCED PROBLEMS. 
 
 Many questions relating to flexure and torsion have not 
 been treated in the preceding pages, as their discussion prop- 
 erly belongs to special works on special branches of applied 
 mechanics. A few of these are here noted as advanced exer- 
 cises that may be assigned by teachers as prize problems. 
 
 Prob. 204. Prove that the maximum bending moment caused 
 by two equal loads rolling over a simple beam occurs at the 
 section distant a from the middle, a being the distance be- 
 tween the two loads. 
 
 Prob. 205. Prove that the maximum bending moment at a 
 given section in a simple beam, due to a given system of mov- 
 ing loads, occurs when the sum of those on the left of the sec- 
 tion divided by the distance of the section from the left end 
 equals the total load divided by the length of the span. 
 
348 APPENDIX. 
 
 Prob. 206. Prove that the maximum maximorum bending 
 moment in a simple beam, due to a given system of moving 
 loads, occurs at a section so located that the distance between 
 it and the center of gravity of the loads is bisected by the 
 middle of the span. 
 
 Prob. 207. Let a helical spring consist of round wire, let r 
 be the radius of the coil, d the diameter of the wire, and P the 
 tensile and compressive load upon the spring. Show that 
 
 i6Pr 
 S ~'-~^d* 
 is the shearing or torsional unit-stress in the wire. 
 
 Prob. 208. The data being the same as in the last problem, 
 and n being the number of coils, show that 
 
 is the elongation or compression of the spring. 
 
 Prob. 209. A simple beam of given rectangular cross-section 
 carries a load of w pounds per linear unit in addition to its own 
 weight. If 5 be the allowable working stress show that the 
 greatest possible length of the beam is 
 
 /= - - s * 
 
 + bdu) ' 
 
 where u is the weight of a cubic unit of the material, b is the 
 breadth and d the depth of the beam. 
 
 Prob. 210. If the flanges of an I beam be considered to 
 carry all the bending moment and the web all the vertical 
 shear, show that the most advantageous proportions are such 
 that the cross-section of the flanges equals the cross-section of 
 the web. 
 
 Prob. 211. If y be the elongation of a spring or bar under 
 longitudinal impact, prove that 
 
 is the time of one oscillation. 
 
ADVANCED PROBLEMS. 349 
 
 Prob. 212. Show that the percentage of weight saved by 
 using a hollow instead of a solid shaft is -TTT if tnev are made 
 
 of equal resilience, n being the ratio of the outer to the inner 
 diameter of the hollow section. 
 
 Prob. 213. Show, for a shaft of square cross-section, that 
 the formulas for investigation and design are 
 
 S t =26 7S oo^-,, d-- 
 
 in which the letters have the same meaning as in Art. 68. 
 
 Prob. 214. A load P is supported by three strings of equal 
 size lying in the same vertical plane. The middle string is 
 vertical and each of the others makes an angle with it. If S 
 be the stress on the middle string and 5, the stress on each of 
 the others, show that 
 
 5 - P S - P COS> 
 
 " I + 2 COS' ' l ~ I -f- 2 COS' P 
 
 (Note : To solve this problem the condition must be intro- 
 duced that the internal work of all the stresses is a minimum.) 
 
 Prob. 215. A load P is supported by three strings of equal 
 size lying in the same plane. The middle string is vertical, one 
 string makes with it the angle 6 on one side, and the second 
 string makes with it the angle on the other side. Find the 
 stresses in the strings. 
 
 Prob. 216. A circular ring of radius R is pulled in the direc- 
 tion of a diameter by two tensile forces each equal to P. 
 Show that the maximum bending moment is at the section 
 where P is applied, and that its value is 
 
 M- PR * 
 
 in which r is the radius of gyration of the cross-section of the 
 ring. Deduce also an expression for the maximum negative 
 bending moment. 
 
 Prob. 217. A continuous beam of five equal spans has a load 
 
350 APPENDIX. 
 
 P on the second span at a distance kl from the second support. 
 Show that the reaction of the first support is 
 
 and that the reaction of the second support is 
 J?, = ^(209 - 45^ - 38i 
 
 Prob. 218. Discuss the formula tan20=//2z/ in Art. 75 
 and show that it represents two sets of shear curves, each 
 being at right angles to the other. Draw the two sets of 
 curves for the case of a simple beam of rectangular cross- 
 section. 
 
 Prob. 219. A beam of two equal spans has a joint at the 
 middle of the first span so that the moment there is always 
 zero. Find the reactions due to a load P on the first span ; 
 (a) when k is less than ; (b) when k is greater than \. Draw 
 for each case the diagrams of shears and moments. 
 
 Prob. 220. A cantilever bridge has three spans, the length of 
 each end span being / and that of the middle span m. The 
 middle span has two joints, the distance of each from the 
 nearest support being n. Find the reactions at the supports ; 
 (a) when the load P is on an end span ; (b) when it is on the 
 middle span between a pier and the nearest joint ; (c) when it 
 is on the middle span between the two joints. 
 
 Prob. 221. A beam is fixed at the ends A and C, and loaded 
 at B with a load P ; the end C, however, being free to deflect, 
 while B and C are kept on the same level. Show that the re- 
 actions at A and C are 
 
 4* -3* R __ P fr-sr 
 
 * - p ~^Jt' R '- P -J=W 
 
 in which k represents the ratio of BC to AC. 
 
 Prob. 222. A continuous beam of three spans has each end 
 span of length / and the middle span of length /. Find the 
 reactions due to a load P in an end span. 
 
ANSWERS TO PROBLEMS. 35 l 
 
 ANSWERS TO PROBLEMS. 
 
 Below will be found the answers to about nine-tenths of the 
 problems stated in the preceding pages, the number of the 
 problem being in parenthesis and the answer immediately fol- 
 lowing. It has been thought well that some answers should 
 be omitted in order that the student may struggle with them 
 to ascertain the truth, according to his best knowledge of the 
 subject, rather than to make his numerical results agree with 
 given figures. However satisfactory it may be to the student 
 to know the result of an exercise he is to. solve, let him remem- 
 ber that after commencement day the answers to problems will 
 never be given. 
 
 The unit-stresses to be employed in solutions will be, unless 
 otherwise stated in the problem, uniformly taken from the 
 tables given in the text and in Art. 80. Considering the great 
 variation in these data it has not been thought best to carry 
 the numerical answers to more than three significant figures, 
 but in making the solution four significant figures should be 
 retained through the work in order that the third may be correct 
 in the final result. 
 
 Chapter I. ( A ) 7.2, 7.06, and 86.4 square inches. (2) 173, 
 34.7, and 4.44 pounds. (3) 55 100 pounds per square inch. 
 (4) 44700 pounds. (5) 165000 pounds. (6) 0.15 inches 
 (7) 26250000 pounds per square inch. (8) 0.004 inches. 
 (9) About 4^ inches X 4i inches. (10) 0.065 and 0.108 inches. ', 
 (12) 0.00153 inches. (13) 52900 pounds per square inch. 
 (14) 849 pounds per square inch. (15) About 1} inches in di 
 ameter. (16) 9 for AB and 23 for BC. 
 
 Chapter II. (17) 0.88 inches if/= 15. (18) I 170 pounds 
 per square inch. (19) 2 500 pounds per square inch. (20) I 620 
 pounds per square inch. (22) 2| inches. (23) 57 per cent ; 
 /= 7.7. (24) 3^ inches; about 0.72. (25) 0.0032 inches. 
 
352 APPENDIX. 
 
 Chapter III. (27) 2\ inches. (29) 998 and 742 pounds. 
 (3) + 800, + * 60, and 180 at I, 3, and 5 feet from left end. 
 (31) 10, 40, 90, 40, 10 pound-feet. (33) Y= 140 and 
 X = 242 pounds. (34) 2700 pounds. (35) X Z 375 
 pounds. (36) About 28. (37) 4.20 inches. (38) c 1.714 
 inches, 7= 7.39 inches 4 . (39) -fabd* and %bd\ (41) At 5.37 
 feet from left end ; M= 689 pound-feet. (42) No. (43) The 
 bar will break. (44) 294 pounds per linear foot. (45) About 
 6000 pounds. (47) 8.87 inches. (48) 0.0178 inches. (49) About 
 610 pounds. (51) Ultimate strengths about as 4 to I, while 
 working strengths for a steady load are about as 3.7 to i. 
 (52) 3.0 and 1.5. (53) 6 feet, 5 inches. (54) The beam will 
 break. (56) 5=5610 and S' = 3 170 pounds per square 
 inch. (57) 209 inches, 418 inches, and oo. (58) 0.622 inches. 
 (59) J 4 5 o pounds per square inch. (62) As 8 to 3 ; as 64 
 
 7 72 
 
 to 9. (63) 7^ inches. (64) 0.243 inches. (65) x 6 ooo p 
 
 for the first case ; the shear at supports is independent of x. 
 (68) 0.72 inches. 
 
 Chapter IV. (69) The diagrams should always be drawn on 
 cross-section paper. (70) R i = 290. (72) =0.366, and 
 =0.577. (73) /= 2.828 m. (74) Max. positive M= 120 
 pound-feet. (76) A light 1 5-inch beam ; a heavy 12-inch beam. 
 (78) 0.0269 inches. (80) R^ R. = -&wl\ R, = R 9 = \\wl. 
 
 (81) o 25' 47". (83) - = 7.2 which requires the light 6-inch 
 
 beam. (84) -fowl. (86) n =0.6095. 
 
 Chapter V. (88) 9.42 inches. (89) 2 inches. (90) 205 ooo 
 pounds. (92) 69.7 tons. (93) 5.05 inches. (95) r = 0.86 
 inches. (96) if. (97) 2.35 and 24. (98) 168 ooo pounds. 
 (99) 23 ooo pounds. (100) 13^ and i6f inches square. (104) 
 Draw Fig. 48 so as to make bq = O', then state the equation 
 of moments and reduce it by the relation between the similar 
 triangles. 
 
ANSWERS TO PROBLEMS. 353 
 
 Chapter VI. (106) 30 pounds. (107) 105 degrees. (108) 
 720, 270, and 290 pound-inches. (109) I 876 pounds per square 
 inch. (no) /= 0.0361^* and <:= 0.577^. (in) J = 23.6 
 and =3. 41. (112) I 680 pounds, (i 13) 9 380000 pounds 
 per square inch, (i 14) 65 horse-power. (115)9.7. (116)2.65 
 and 3.58 inches. (118)6500. ( 1 19) As V^n to 3. ( 1 20) As 
 100 to 1 06. 
 
 Chapter VII. (122) 3720 pounds per square inch. v I2 3) 
 8 100 pounds per square inch. (124) The light 9-inch beam. 
 (125) Nearly 8 inches. (126) 9 inches. (127) / = 9 420, 
 - 54 13'; S== 7160, 0-9 13'. (129) 5.4. (130) 2j 
 inches. (131) S = 5 660 and S s = 205 pounds per square inch. 
 (132) At 3 inches from neutral surface S = 2OOO, S k = 250, 
 and / = 2 030 pounds per square inch. 
 
 Chapter IX. (139) Theoretic stress is 3.3 per cent and 
 theoretic elongation is 3.5 per cent greater than the observed 
 values. (140) 1.34 horse-power. (141) 122 foot-pounds. 
 (142) For the second case K= S*Al/\$E if section is rect- 
 angular. 
 
 Chapter X. (144) 7 500 feet and 3.75 feet. (145) log x 
 
 wbl 
 lg b + -4343 p-y- (146) About 101 feet per second. (148) 
 
 Nearly 28000 pounds per square inch. (149) 64 rollers 2 
 inches in diameter, or 7 rollers 6 inches in diameter. (150) 6. 
 (151) 1 1 500 pounds per square inch. 
 
 Chapter XI. (152) For uniform load M = \wx* and 
 K = W*l*/4pEI. (153) Deflection = T \th of that for load at 
 middle. (154)^ = o. 1 74 inches, 5 4030 pounds per square 
 inch. (157) 1420 pounds, (158) About 1.4 inches. 
 (160) Over 600 miles per hour. (161) W t l/2^E s A. (166) See 
 Higher Mathematics,' Chap. IV, Art. 36. 
 
 Chapter XII. (167) (6000) *( 2 ooo) = 4000. (168) 
 About 12 horse-power. (170) , = 9380000, then find e 
 
354 APPENDIX. 
 
 from formula (25). (171) As 2 to 3. (172) As 100 to 307. 
 (173) 8 bolts. (174) 2 T V inches. (175) Bearing unit-stress 
 i 880 pounds per square inch. (176) Shear is uniform through- 
 out, while moment is zero at middle. 
 
 Chapter XIII. (178) 0.042 and 0.014 per cent. (181) S6I9' 
 with axis of bolt. (183) See Theory of Equations in Algebra. 
 (184) Greatest tension = 36.3, compression = 87.1 pounds per 
 square inch. (185) Apparent = 53.7, true = 61.7 pounds per 
 square inch. (186) 54 44' with greater and 3 5 46' with lesser 
 stress. (187) Maximum true compression = 12900 pounds 
 per square inch, or 27 per cent more than the apparent. 
 
 Chapter XIV. (188) Make r l = o. (189) 18000 pounds per 
 square inch. (190) 54000 pounds per square inch. (191) 
 15900 pounds per square inch. (192) 15000 pounds per 
 square inch. (194) Deduce an expression for p v in terms of 
 the given radii and S e ; then find the value of r 9 which renders 
 p l a maximum. 
 
 Chapter XV. (195) See the books mentioned in Art. 127. 
 (196) About 50 pounds per square inch with factor of 10. (198) 
 Thickness = 1.6 inches. (201) See the demonstrations in 
 Arts. 55 and 108. 
 
 Appendix. (202) See Railroad Gazette, June 7, 1895. (205) 
 See Roofs and Bridges, Part I. (217) See London Philosoph- 
 ical Magazine, September, 1875. (218) See Engineering News, 
 August I, 1895. (221) See article by J. L. GREENLEAF in 
 Journal of Franklin Institute for July, 1895. 
 
 ^DESCRIPTION OF TABLES. 
 
 Tables I, II, III, and IV, in Art. 80 (pages 163 and 164) 
 give mean constants of the elasticity and strength of the prin- 
 cipal materials used in engineering. Other tables, not num- 
 bered, are noted in the Table of Contents (page ix). 
 
DESCRIPTION OF TABLES. 355 
 
 Table V, on the next two pages, gives four-place logarithms 
 of numbers which will be found very useful and sufficiently 
 accurate for all computations in the mechanics of materials. 
 
 Table VI gives four-place squares of numbers from i.oo to 
 9.99, the arrangement being the same as that of the logarithmic 
 table. By properly moving the decimal point 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, correct to four significant figures. 
 
 Table VII gives four-place areas of circles for diameters 
 ranging from i.oo to 9.99, arranged in the same manner. By 
 properly moving the decimal point four-place areas for all circles 
 maybe 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 12.3 feet, the 
 area is 116.9 square feet, all correct to four significant figures. 
 
 Table VIII 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 -J to lof 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 per cent ; 
 
 For steel, add 2 per cent. 
 
 For example, a cast-iron bar 6 inches square and 8 feet long 
 weighs 8(157.6 0.06 X 157.6) = I 185 pounds. In like man- 
 ner a steel bar 2 T 3 T inches in diameter and 4 feet 9 inches in 
 length weighs 44(12.53 +0.02 X 12.53) = 57'9 pounds. 
 
356 
 
 APPENDIX 
 
 TABLE V. COMMON LOGARITHMS. 
 
 n 
 
 01234 
 
 56789 
 
 Diff. 
 
 IO 
 
 oooo 0043 0086 0128 0170 
 
 0212 0253 0294 0334 0374 
 
 42 
 
 ii 
 
 0414 0453 0492 0531 0569 
 
 0607 0645 0682 0719 0755 
 
 38 
 
 12 
 
 0792 0828 0864 0899 0934 
 
 0969 1004 IO 38 1072 1106 
 
 35 
 
 14 
 
 1139 1173 1206 1239 1271 
 
 1461 1492 1523 1553 1584 
 
 !33 T 335 !3 6 7 1399 '43 
 1614 1644 1673 J 7Q3 1732 
 
 32 
 3 
 
 15 
 
 1761 1790 .1818 1847 J 875 
 
 1903 1931 1959 1987 2014 
 
 28 
 
 16 
 
 2O4I 2O68 2095 2122 2148 
 
 2175 2201 2227 2253 2279 
 
 27 
 
 17 
 
 2304 2330 2355 2380 2405 
 
 2430 2455 2480 2504 2529 
 
 2 5 
 
 18 
 
 2 553 2577 266l 2625 2648 
 
 2672 2695 27*8 2742 2765 
 
 24 
 
 T 9 
 
 2788 28lO 2833 2856 2878 
 
 2900 2923 2945 2967 2989 
 
 22 
 
 20 
 
 3010 3032 3054 3075 3096 
 
 3 IlS 3139 3 J 6o 3181 3201 
 
 21 
 
 21 
 
 3222 3243 3263 3284 3304 
 
 3324 3345 33 6 5 3385 3404 
 
 2O 
 
 22 
 
 3424 3444 3464 3483 3502 
 
 3522 354i 35 60 3579 3598 
 
 19 
 
 2 3 
 
 3617 3636 3655 3674 3692 
 
 37ii 3729 3747 3766 3784 
 
 18 
 
 24 
 
 3802 3820 3838 3856 3874 
 
 3892 3909 3927 3945 3962 
 
 18 
 
 25 
 
 3979 3997 4014 4031 4048 
 
 4065 4082 4099 4116 4133 
 
 17 
 
 26 
 
 4150 4166 4183 4200 4216 
 
 4232 4249 4265 4281 4298 
 
 17 
 
 27 
 
 4314 4330 4346 4362 4378 
 
 4393 4409 4425 4440 4456 
 
 16 
 
 28 
 
 4472 4487 4502 4518 4533 
 
 4548 4564 4579 4594 4609 
 
 15 
 
 2 9 
 
 4624 4639 4654 4669 4683 
 
 4698 4713 4728 4742 4757 
 
 '5 
 
 30 
 
 4771 .4786 4800 4814 4829 
 
 4843 4857 4871 4886 4900 
 
 14 
 
 
 4914 4928 4942 4955 4969 
 
 4983 4997 5011 5024 5038 
 
 14 
 
 32 
 
 5051 5065 5079 5092 5105 
 
 5119 5132 5145 5159 5172 
 
 13 
 
 33 
 
 5185 5198 5211 5224 5237 
 
 5250 5263 5276 5289 5302 
 
 13 
 
 34 
 
 53*5 5328 5340 5353 5366 
 
 5378 5391 5403 5416 5428 
 
 13 
 
 35 
 
 544i 5453 5465 5478 5490 
 
 5502 5514 5527 5539 555 1 
 
 12 
 
 36 
 
 5563 5575 5587 5599 5^n 
 
 5 6 23 5635 5 6 47 5658 5670 
 
 12 
 
 37 
 
 5682 5694 5705 5717 5729 
 
 5740 575 2 5763 5775 5786 
 
 12 
 
 38 
 
 5798 5809 5821 5832 5843 
 
 5855 5866 5877 5888 5899 
 
 II 
 
 39 
 
 5911 5922 5933 5944 5955 
 
 5966 5977 5988 5999 6010 
 
 II 
 
 40 
 
 6021 6031 6042 6053 6064 
 
 6075 6085 6096 6107 6117 
 
 II 
 
 4 1 
 
 6128 6138 6149 6160 6170 
 
 6180 6191 6201 6212 6222 
 
 II 
 
 42 
 
 6232 6243 6253 6263 6274 
 
 6284 6294 6304 6314 6325 
 
 IO 
 
 43 
 44 
 
 6335 6 345 6355 6365 6375 
 6435 6444 6454 6464 6474 
 
 6385 6395 6405 6415 6425 
 6484 6493 6 53 6 5 r 3 6522 
 
 10 
 IO 
 
 ^!5 
 
 6532 6542 6551 6561 6571 
 
 6580 6590 6599 6609 6618 
 
 10 
 
 46 
 
 6628 6637 6646 6656 6665 
 
 6675 6684 6693 6702 6712 
 
 9 
 
 47 
 
 6721 6730 6739 6749 6758 
 
 6767 6776 6785 6794 6803 
 
 9 
 
 48 
 
 6812 6821 6830 6839 6848 
 
 6857 6866 6875 6884 6893 
 
 9 
 
 49 
 
 6902 6911 6920 6928 6937 
 
 6946 69^5 6964 6972 6981 
 
 9 
 
 5 
 
 6990 6998 7007 7016 7024 
 
 7033 7042 7050 7059 7067 
 
 9 
 
 
 7076 7084 7093 7101 7110 
 
 7118 7126 7135 7143 7152 
 
 8 
 
 5 2 
 
 7160 7168 7177 7185 7193 
 
 7202 7210 7218 7226 7235 
 
 8 
 
 53 
 
 7243 7251 7 2 59 7267 7275 
 
 7284 7292 7300 7308 7316 
 
 8 
 
 54 
 
 73 2 4 7332 7340 7348 735 6 
 
 7364 7372 7380 7388 7396 
 
 8 
 
 n 
 
 01234 
 
 56789 
 
 Diff. 
 
COMMON LOGARITHMS. 
 
 357 
 
 TABLE V. COMMON LOGARITHMS. 
 
 n 
 
 01234 
 
 56789 
 
 Diff. 
 
 55 
 
 7404 7412 7419 7427 7435 
 
 7443 7451 7459 7466 7474 
 
 8 
 
 56 
 
 7482 7490 7497 755 75 '3 
 7559 7566 7574 7 582 75 8 9 
 7634 7642 7649 7657 7664 
 
 7520 7528 7536 7543 7551 
 7597 7604 7612 7619 7627 
 7672 7679 7686 7694 7701 
 
 
 59 
 
 7709 7716 7723 7731 7738 
 
 7745 775 2 776o 7767 7774 
 
 
 60 
 61 
 
 7782 7789 7796 7803 7810 
 7853 7860 7868 7875 7882 
 
 7818 7825 7832 7839 7846 
 7889 7896 7903 7910 7917 
 
 7 
 
 62 
 
 79 2 4 793 1 7938 7945 795 2 
 
 7959 7966 7973 798o 7987 
 
 
 63 
 
 7993 8000 8007 8014 8021 
 
 8028 8035 8041 8048 8055 
 
 
 
 64 
 
 8062 8069 8075 8082 8089 
 
 8096 8102 8109 8116 8122 
 
 
 65 
 
 8129 8136 8142 8149 8156 
 
 8162 8169 8176 8182 8189 
 
 7 
 
 66 
 
 8195 8202 8209 8215 8222 
 
 8228 8235 8241 8248 8254 
 
 
 67 
 
 8261 8267 8274 8280 8287 
 
 8293 8299 8306 8312 8319 
 
 
 68 
 
 832? 8331 8338 8344 8351 
 
 8357 8363 8370 8376 8382 
 
 
 69 
 
 8388 8395 8401 8407 8414 
 
 8420 8426 8432 8439 8445 
 
 
 70 
 
 8451 8457 8463 8470 8476 
 
 8482 8488 8494 8500 8506 
 
 6 
 
 7 r 
 
 8513 8519 8525 8531 8537 
 
 8543 8549 8555 8561 8567 
 
 
 72 
 73 
 
 8573 8579 8585 8591 8597 
 8633 8639 8645 8651 8657 
 
 8603 8609 8615 8621 8627 
 8663 8669 8675 8681 8686 
 
 
 74 
 
 8692 8698 8704 8710 8716 
 
 8722 8727 8733 8739 8745 
 
 
 75 
 
 8751 8756 8762 8768 8774 
 
 8779 8785 8791 8797 8802 
 
 6 
 
 76 
 
 8808 8814 8820 8825 8831 
 
 8837 8842 8848 8854 8859 
 
 
 77 
 
 8865 8871 8876 8882 8887 
 
 8893 8899 8904 8910 8915 
 
 
 78 
 79 
 
 8921 8927 8932 8938 8943 
 8976 8982 8987 8993 8998 
 
 8949 8954 8960 8965 8971 
 9004 9009 9015 9020 9025 
 
 
 80 
 
 9031 9036 9042 9047 9053 
 
 9058 9063 9069 9074 9079 
 
 5 
 
 81 
 82 
 
 9085 9090 9096 9101 9106 
 9i3 9M3 9M9 9 T 54 9'59 
 
 9112 9117 9122 9128 9 r 33 
 9165 9170 9175 9180 9186 
 
 
 83 
 
 9191 9196 9201 9206 9212 
 
 9217 9222 9227 9232 9238 
 
 
 84 
 
 9243 9248 9253 9258 9263 
 
 9269 9274 9279 9284 9289 
 
 
 85 
 
 9294 9299 9304 9309 9315 
 
 9320 9325 9310 9375 9340 
 
 5 
 
 86 
 
 9345 9350 9355 936o 9365 
 
 9370 9375 938o 93 g 5 9390 
 
 
 87 
 
 9395 94oo 9405 94:0 9415 
 
 9420 9425 9430 9435 9440 
 
 
 88 
 
 9445 9450 9455 946 > 9465 
 
 9469 9474 9479 9484 9489 
 
 
 89 
 
 9494 9499 954 959 95 ' 3 
 
 95 l8 9523 9528 9533 953 s 
 
 
 90 
 
 9542 9547 9552 9557 9562 
 
 9566 9571 9576 9581 9586 
 
 5 
 
 91 
 
 9590 9595 96oo 9605 9609 
 
 9614 9619 9624 9628 9633 
 
 
 92 
 
 9638 9643 9647 9652 9657 
 
 9661 9666 9671 9675 9680 
 
 
 93 
 
 9685 9689 9694 9699 9703 
 
 9708 9713 9717 9722 9727 
 
 
 94 
 
 973 i 9736 9741 9745 975 
 
 9754 9759 9763 9768 9773 
 
 
 95 
 
 9777 9782 9786 9791 9795 
 
 9800 9805 9809 9814 9818 
 
 4 
 
 96 
 
 9823 9827 9832 9836 9841 
 
 9845 9850 9854 9859 9863 
 
 
 97 
 
 9868 9872 9877 9881 9886 
 
 9890 9894 9899 9907 9908 
 
 
 98 
 
 9912 9917 9921 9926 9930 
 
 9934 9939 9943 994 995 2 
 
 
 99 
 
 9956 9961 9965 9969 9974 
 
 9978 9983 9987 9991 9996 
 
 
 n 
 
 01234 
 
 5 6 7 8 9 
 
 Diff. 
 
358 
 
 APPENDIX. 
 
 TABLE VI. SQUARES OF NUMBERS. 
 
 7Z. 
 
 01234 
 
 56789 
 
 Diff. 
 
 .O 
 
 i.ooo 1.020 1.040 1.061 1.082 
 
 1.103 I - I2 4 I - I 45 1.166 i.iSS 
 
 22 
 
 .1 
 
 1.210 1.232 1.254 1.277 I -3 
 
 1.323 1.346 1.369 1.392 1.416 
 
 24 
 
 .2 
 
 1.440 1.464 1.488 1.513 1.538 
 
 1.563 1.588 1.613 1.638 1.664 
 
 26 
 
 3 
 
 1.690 1.716 1.742 1.769 1.796 
 
 1.823 1.850 1.877 1.904 1.932 
 
 28 
 
 -4 
 
 1.960 1.988 2.016 2.045 2.074 
 
 2.IO3 2.132 2.l6l 2.I9O 2.22O 
 
 3 
 
 i 
 
 2.250 2.280 2.310 2.341 2.372 
 2.560 2.592 2.624 2.657 2.690 
 
 2.403 2.434 2.465 2.496 2.528 
 2.723 2.756 2.789 2.822 2.856 
 
 32 
 34 
 
 7 
 
 2.890 2.924 2.958 2.993 3- 02 8 
 
 3.063 3.098 3.133 3.168 3.204 
 
 36 
 
 1.8 
 
 3.240 3.276 3.312 3.349 3.386 
 
 3.423 3.460 3.497 3.534 3.572 
 
 
 1.9 
 
 3.610 3.648 3.686 3.725 3.764 
 
 3.803 3.842 3.881 3.920 3.960 
 
 40 
 
 2.O 
 
 4.000 4.040 4.080 4.121 4.162 
 
 4.203 4.244 4.285 4.326 4.368 
 
 42 
 
 2.1 
 
 4.410 4.452 4.494 4.537 4.580 
 
 4.623 4.666 4.709 4.752 4.796 
 
 44 
 
 2.2 
 
 4.840 4.884 4.928 4.973 5.018 
 
 5.063 5.108 5.153 5.198 5.244 
 
 46 
 
 2-3 
 
 2.4 
 
 5-290 5-336 5.382 5.429 5.476 
 5.760 5.808 5.856 5.905 5.954 
 
 5-523 5-570 5.617 5.664 5.712 
 
 6.003 6.052 6. i oi 6.150 6.200 
 
 4 8 
 5 
 
 2 -5 
 
 6.250 6.300 6.350 6.401 6.452 
 
 6.503 6.554 6.605 6.656 6.708 
 
 52 
 
 2.6 
 
 6.760 6.8 1 2 6.864 6.917 6.970 
 
 7.023 7.076 7.129 7.182 7.236 
 
 54 
 
 2.7 
 
 7-290 7-344 7-398 7-453 7-5 8 
 
 7.563 7.618 7.673 7.728 7.784 
 
 56 
 
 2.8 
 
 7.840 7.896 7.952 8.009 8.066 
 
 8.123 8.180 8.237 8.294 8.352 
 
 
 2.9 
 
 8.410 8.468 8.526 8.585 8.644 
 
 8.703 8.762 8.821 8.880 8.940 
 
 60 
 
 3- 
 
 9.000 9.060 9.120 9.181 9.242 
 
 9-303 9-3 6 4 9-425 9-4S6 9.548 
 
 62 
 
 3.1 
 
 9.610 9.672 9.734 9.797 9.860 
 
 9.923 9.986 10.05 I0 - ir IO - T 8 
 
 6 
 
 3-2 
 
 10.24 10.30 10.37 10.43 IO -5 
 
 10.56 10.63 10.69 10.76 10.82 
 
 7 
 
 3-3 
 
 10.89 10.96 11.02 11.09 ii-i6 
 
 11.22 11.29 IT -36 1142 11.49 
 
 7 
 
 34 
 
 11.56 11.63 II -7 JI -7 ^1-83 
 
 11.90 11.97 12.04 I2 .II 12. l8 
 
 7 
 
 3-5 
 
 12.25 12.32 12.39 12.46 12.53 
 
 1 2.60 12.67 I2 74 12.82 12.89 
 
 7 
 
 3-6 
 
 12.96 13-03 13.10 13.18 13.25 
 
 !3-3 2 1340 1347 13-54 13-62 
 
 7 
 
 3-7 
 
 13.69 13.76 13.84 13.91 13.99 
 
 14.06 14.14 14.21 14.29 14.36 
 
 8 
 
 3-8 
 
 14.44 M-S 2 H-59 M-67 14-75 
 
 14.82 14.90 14.98 15.05 15.13 
 
 8 
 
 3-9 
 
 15.21 15.29 15.37 15.44 15.52 
 
 15.60 15.68 15.76 15.84 15.92 
 
 8 
 
 4.0 
 
 16.00 16.08 16.16 16.24 16.32 
 
 16.40 16.48 16.56 16.65 J 6-73 
 
 8 
 
 4.1 
 
 16.81 16.89 J 6-97 17.06 17.14 
 
 17.22 17.31 17.39 17.47 17-56 
 
 8 
 
 4.2 
 
 17.64 17.72 17.81 17.89 17.98 
 
 1 8.06 18.15 18.23 l8 -3 2 18.40 
 
 9 
 
 4-3 
 
 18.49 18.58 18.66 18.75 l8 - 8 4 
 
 18.92 19.01 19.10 19.18 19.27 
 
 9 
 
 44 
 
 19.36 19.45 19.54 19.62 19.71 
 
 19.80 19.89 19.98 20.07 20.16 
 
 9 
 
 4-5 
 
 20.25 20.34 20.43 20.52 20.61 
 
 20.70 20.79 20.88 20.98 21.07 
 
 9 
 
 4.6 
 
 21. 16 21.25 21.34 21.44 21.53 
 
 21.62 21.72 21.81 21.90 22.00 
 
 9 
 
 4-7 
 
 22.09 22.18 22.28 22.37 22.47 
 
 22.56 22.66 22.75 22.85 22.94 
 
 10 
 
 4.8 
 
 23.04 23.14 23.23 23.33 23.43 
 
 23.52 23.62 23.72 23.81 23.91 
 
 IO 
 
 4-9 
 
 24.01 24.11 24.21 24.30 24.40 
 
 24.50 24.60 24.70 24.80 24.90 
 
 IO 
 
 5-o 
 5- 1 
 
 25.00 25.10 25.20 25.30 25.40 
 26.01 26.11 26.21 26.32 26.42 
 
 25.50 25.60 25.70 25.81 25.91 
 26.52 26.63 26.73 26.83 26.94 
 
 IO 
 IO 
 
 5-2 
 
 27.04 27.14 27.25 27.35 27.46 
 
 27.56 27.67 27.77 27.88 27.98 
 
 II 
 
 5-3 
 
 28.09 28.20 28.30 28.41 28.52 
 
 28.62 28.73 28.84 28.94 29.05 
 
 II 
 
 5-4 
 
 29.16 29.27 29.38 29.48 29.59 
 
 29.70 29.81 29.92 30.03 30.14 
 
 II 
 
 n. 
 
 01234 
 
 56789 
 
 Diff. 
 
SQUARES OF NUMBERS. 
 
 359 
 
 TABLE VI. SQUARES OF NUMBERS 
 
 n. 
 
 01234 
 
 56789 
 
 Diff. 
 
 5-5 
 
 30.25 30.36 30.47 30.58 30.69 
 
 30.80 30.91 31.02 31.14 31.25 
 
 I! 
 
 5.6 
 
 31.36 31.47 31.58 31.70 31.81 
 
 31.92 32.04 32.15 32.26 32.38 
 
 II 
 
 5-7 
 
 32.49 32.60 32.72 32.83 32.95 
 
 33.06 33.18 33.29 33.41 33.52 
 
 
 5.8 
 
 33.64 33.76 33.87 33.99 34. rr 
 
 34.22 34.34 34.46 34.57 34.69 
 
 
 5-9 
 
 34-8i 34.93 35.05 35.16 35.28 
 
 35-40 35.52 35.64 35.76 35.88 
 
 
 6.0 
 
 36.00 36.12 36.24 36.36 36.48 
 
 36.60 36.72 36.84 36.97 37.09 
 
 
 6.1 
 
 37-21 37.33 37.45 37.38 37.70 
 
 37.82 37.95 38.07 38.19 38.32 
 
 
 6.2 
 
 38.44 38.56 38.69 38.81 38.94 
 
 39.06 39.19 39.31 39.44 39.56 
 
 13 
 
 6-3 
 
 39-69 39-82 39.94 40.07 40.20 
 
 40.32 40.45 40.38 40.70 40.83 
 
 13 
 
 6.4 
 
 40.96 41.09 41.22 41.34 41.47 
 
 41.60 41.73 41.86 41.99 42.12 
 
 U 
 
 6-5 
 
 6.6 
 
 42.25 42.38 42.51 42.64 42.77 
 43-56 43-69 43-82 43-96 44-09 
 
 42.90 43-03 43- '6 43-3 43-43 
 44.22 44.36 44.49 44.62 44.76 
 
 13 
 U 
 
 6.7 
 
 44.89 45.02 45.16 45.29 45.43 
 
 45.56 45.70 45- 8 3 45-97 46.10 
 
 U 
 
 6.8 
 
 46.24 46.38 46.51 46.65 46.79 
 
 4692 47.06 47.20 47.33 47.47 
 
 14 
 
 6.9 
 
 47.61 47.75 47-89 48.02 48.16 
 
 48.30 48.44 48.58 48.72 48.86 
 
 14 
 
 70 
 
 49.00 49.14 49.28 49.42 49.56 
 
 49.70 49.84 49.98 50.13 5027 
 
 U 
 
 7- 1 
 
 50.41 50.55 50.69 50.84 50.98 
 
 51.12 51.27 51.41 51.55 51.70 
 
 M 
 
 7.2 
 
 51.84 51.98 52.13 52.27 52.42 
 
 52.56 52.71 52.85 53.00 53.14 
 
 15 
 
 7-3 
 
 53-29 53-44 53-58 53.73 53.88 
 
 54.02 54.17 54.32 54.46 54.61 
 
 15 
 
 7-4 
 
 54.76 54-91 55-o6 55- 2 o 55.35 
 
 55-50 55.65 55.80 55.95 56.10 
 
 15 
 
 # 
 ft 
 
 56.25 56.40 56.55 56.70 56.85 
 57-76 57.91 58.06 58.22 58.37 
 59.29 59.44 59.60 59.75 59.91 
 60.84 61.00 61.15 61.31 61.47 
 
 C7.oo 57.15 57.30 57.46 57.61 
 58.52 58.68 58.83 58.98 59.14 
 60.06 60.22 60.37 60.53 60.68 
 61.62 61.78 61.94 62.09 62.25 
 
 IS 
 15 
 
 1 6 
 16 
 
 79 . 
 
 62.41 62.57 62.73 62.88 63.04 
 
 63.20 63.36 63.52 63.68 63.84 
 
 16 
 
 8.0 
 8.1 
 
 64.00 64.16 64.32 64.48 64.64 
 65.61 65.77 65.93 66. 10 66.26 
 
 64.80 64.96 65.12 65.29 65.45 
 66.42 66.59 66.75 66.91 67.08 
 
 16 
 16 
 
 8.2 
 
 67.24 67.40 67.57 6773 67.90 
 
 68.06 68.23 68.39 68.56 68.72 
 
 17 
 
 8-3 
 
 68.89 69.06 69.22 69.39 69.56 
 
 69.72 69.89 70.06 70.22 70.39 
 
 17 
 
 8.4 
 
 70.56 70.73 70.90 71.06 71.23 
 
 71.40 71.57 71.74 71.91 72.08 
 
 17 
 
 & 
 
 72.25 72.42 72.59 72.76 72.93 
 
 73- 10 73-27 73-44 73-62 73-79 
 
 17 
 
 8.6 
 8.7 
 
 73-96 74-13 743 74-48 74.65 
 75.69 75.86 76.04 76.21. 76.39 
 
 74.82 75.00 75.17 75.34 75-52 
 76.56 76.74 76.91 77.09 77.26 
 
 11 
 
 8.8 
 8.9 
 
 77.44 77.62 77.79 77.97 78.15 
 79.21 79.39 79.57 79.74 79.92 
 
 78.32 78.50 78.68 78.85 79.03 
 80. 10 80.28 80.46 80.64 80.82 
 
 18 
 18 
 
 9.0 
 
 8 1. oo 81.18 81.36 81.54 81.72 
 
 81.90 82.08 82.26 82.45 82.63 
 
 18 
 
 9.1 
 
 82.81 82.99 83.17 83.36 83.54 
 
 83.72 83.91 84.09 84.27 84.46 
 
 18 
 
 9.2 
 
 84.64 84.82 85.01 85.19 85.38 
 
 85.56 85.75 85.93 86.12 86.30 
 
 i9 
 
 9-3 
 
 86.49 86.68 86.86 87.05 87.24 
 
 87.42 87.61 87.80 87.98 88.17 
 
 19 
 
 94 
 
 88.36 88.55 88.74 88.92 89.11 
 
 89.30 89.49 89.68 89.87 90.06 
 
 19 
 
 9-5 
 
 90.25 90.44 90.63 90.82 91.01 
 
 91.20 91.39 91.58 91.78 91.97 
 
 19 
 
 9.6 
 
 92.16 92.35 92.54 92.74 92.93 
 
 93- i 2 93.32 93.51 93.70 93.90 
 
 19 
 
 9-7 
 
 94.09 94.28 94.48 94.67 94.87 
 
 95.06 95.26 95.45 95.65 95.84 
 
 20 
 
 9.8 
 
 96.04 96.24 96.43 96.63 96.83 
 
 97.02 97.22 97.42 97.61 97.81 
 
 20 
 
 9-9 
 
 98.01 98.21 98.41 98.60 98.80 
 
 99.00 99.20 99.40 99.60 99.80 
 
 20 
 
 n. 
 
 01234 
 
 56789 
 
 Diff. 
 
APPENDIX. 
 
 TABLE VII. AREAS OF CIRCLES. 
 
 d 
 
 01234 
 
 56789 
 
 Diff. 
 
 .0 
 
 .7854 .8012 .8171 .8332 .8495 
 
 .8659 .8825 .8992 .9161 .9331 
 
 
 .1 
 
 .9503 .9677 .9852 1.003 1. 021 
 
 1.039 L057 1-075 1.094 1. 112 
 
 
 .2 
 
 1.131 1.150 1.169 1.188 1.208 
 
 1.227 1.247 1.267 1.287 1.307 
 
 19 
 
 -3 
 
 1.327 1.348 1.368 1.389 1.410 
 
 1.431 1.453 1.474 1-496 I.5I7 
 
 21 
 
 4 
 
 1.539 I -5 61 I -5 8 4 i. 606 1.629 
 
 1.651 1.674 1-697 1.720 1.744 
 
 22 
 
 5 
 
 1.767 1.791 1.815 1.839 1-863 
 
 1.887 1.911 1.936 1.961 1.986 
 
 24 
 
 .6 
 
 2.0II 2.036 2.061 2.087 2. 112 
 
 2.158 2.164 2.190 2.217 2.243 
 
 26 
 
 7 
 
 2.270 2.297 2.324 2.351 2.378 
 
 2.405 2.433 2.461 2.488 2.516 
 
 27 
 
 .8 
 
 2-545 2.573 2.602 2.630 2.659 
 
 2.638 2.717 2.746 2.776 2.806 
 
 2 9 
 
 9 
 
 2.835 2.865 2.895 2.926 2.956 
 
 2.986 3.017 3.048 3.079 3.110 
 
 30 
 
 2.0 
 
 3.142 3.173 3.205 3.237 3.269 
 
 3-301 3.333 3.365 3.398 3.431 
 
 32 
 
 2.1 
 
 3.464 3.497 3.530 3.563 3.597 
 
 3.631 3.664 3.698 3.733 3.767 
 
 34 
 
 2.2 
 
 3.801 3.836 3.871 3.906 3.941 
 
 3.976 4.012 4.047 4.083 4.119 
 
 35 
 
 2-3 
 
 4.155 4.191 4.227 4.264 4.301 
 
 4-337 4-374 4-412 4.449 4.486. 
 
 36 
 
 2. 4 
 
 4.524 4.562 4.600 4.638 4.676 
 
 4.7M 4-753 4-792 4-831 4-8/0 
 
 38 
 
 2-5 
 
 4.909 4.948 4.988 5.027 5.067 
 
 5.107 5.147 5.187 5 228 5.269 
 
 40 
 
 2.6 
 
 5.309 5.350 5.391 5.433 5.474 
 
 5-5I5 5-557 5-599 5-641 5.683 
 
 4i 
 
 2.7 
 
 5.726 5.768 5.811 5.853 5.896 
 
 5.940 5.983 6.026 6.070 6.114 
 
 43 
 
 2.8 
 
 6.158 6.202 6.246 6.290 6.335 
 
 6.379 6.424 6.469 6.514 6.560 
 
 44 
 
 2.9 
 
 6.605 6.651 6.697 6.743 6.789 
 
 6.835 6.881 6.928 6.975 7.022 
 
 46 
 
 3-0 
 
 7.069 7.116 7.163 7.211 7.258 
 
 7.306 7.354 7.402 7.451 7.499 
 
 48 
 
 3-1 
 
 7.548 7.596 7.645 7.694 7-744 
 
 7.793 7.843 7.892 7.942 7.992 
 
 49 
 
 3-2 
 
 8.042 8.093 8.143 8.194 8.245 
 
 8.296 8.347 8.398 8.450 8.501 
 
 5i 
 
 3-3 
 
 8.553 8.605 8.657 8.709 8.762 
 
 8.814 8.867 8.920 8.973 9.026 
 
 52 
 
 3-4 
 
 9.079 9,133 9.186 9.2.40 9.294 
 
 9.348 9.402 9.457 9.511 9.566 
 
 54 
 
 3-5 
 
 9.621 9.676 9.731 9.787 9.842 
 
 9.898 9-954 IO.OI IO.O7 IO.I2 
 
 56 
 
 3-6 
 
 10.18 10.24 10.29 10.35 10.41 
 
 10.46 10.52 10.58 10.64 10.69 
 
 6 
 
 3-7 
 
 10.75 10-81 10.87 10.93 10.99 
 
 II.O4 II- IO II. l6 11.22 11.28 
 
 6 
 
 3-3 
 
 11.34 H-4O 11.46 11.52 11.58 
 
 11.64 11.70 11.76 11.82 11.88 
 
 6 
 
 3-9 
 
 11.95 12. or 12.07 12.13 12.19 
 
 12.25 12-32 12.38 12.44 12.50 
 
 6 
 
 4.0 
 
 12.57 !2.63 12.69 12.76 12.82 
 
 12.88 12.95 13.01 13.07 13.14 
 
 7 
 
 4.1 
 
 13.20 13.27 13.33 13-40 13-46 
 
 13-53 13-59 13-66 13.72 13-79 
 
 7 
 
 4.2 
 
 13.85 13.92 13.99 14-05 14.12 
 
 14.19 14.25 14.32 14.39 14-45 
 
 7 
 
 4-3 
 
 14.52 14.59 14-66 14.73 14.79 
 
 14.86 14.93 15.00 15.07 15.14 
 
 7 
 
 4-4 
 
 15.21 15.27 15.34 15.41 15.43 
 
 15.55 15.62 15.69 15.76 15.83 
 
 7 
 
 4-5 
 
 15.90 15.98 16.05 16.12 16.19 
 
 16.26 16.33 16.40 16.47 16.55 
 
 7 
 
 4-6 
 
 16.62 16.69 16.76 16.84 16.91 
 
 16.98 17.06 17.13 17.20 17.28 
 
 7 
 
 4-7 
 
 17.35 17.42 17.50 17.57 17.65 
 
 17.72 17.80 17.87 17.95 18.02 
 
 8 
 
 4.8 
 
 18.10 18.17 18.25 18.32 18.40 
 
 18.47 18.55 18.63 18.70 18.78 
 
 8 
 
 4-9 
 
 18.86 18.93 19.01 19.09 19.17 
 
 19.24 19.32 19.40 19.48 19.56 
 
 8 
 
 5-0 
 
 19.63 19.71 19.79 19.87 19.95 
 
 2O.O3 2O.II 2O.I9 2O.27 2O.35 
 
 8 
 
 5-1 
 
 20.43 20.51 20.59 20.67 20.75 
 
 20.83 20.91 20.99 21.07 21. 16 
 
 8 
 
 5-2 
 
 21.24 21.32 21.40 21.48 21.57 
 
 21.65 21-73 2I.8I 2I.9O 21.98 
 
 8 
 
 5-3 
 
 22.06 22.15 22.23 22.31 22.40 
 
 22.48 22.56 22.65 22.73 22.82 
 
 8 
 
 5-4 
 
 22.90 22.99 23.07 23.16 23.24 
 
 23.33 23-41 23.50 23.59 23.67 
 
 9 
 
 d 
 
 01234 
 
 5 6 7 89 
 
 Diff. 
 
AREAS OF CIRCLES. 
 
 3 6i 
 
 TABLE VII. AREAS OF CIRCLES. 
 
 d 
 
 01234 
 
 56789 
 
 Diff. 
 
 5-5 
 
 23.76 23.84 23.93 24.02 24.11 
 
 24.19 24.28 24.37 24.45 24.54 
 
 9 
 
 5-6 
 
 24.63 24.72 24.81 24.89 24.98 
 
 25.07 25.16 25.25 25.34 25.43 
 
 9 
 
 5-7 
 
 25.52 25.61 25.70 25.79 2 5- 88 
 
 25.97 26.06 26.15 26.24 26.33 
 
 9 
 
 5-8 
 
 26.42 26.51 26.60 26.69 26.79 
 
 26.88 26.97 27.06 27.15 27.25 
 
 9 
 
 5.9 
 
 27.34 27.43 27.53 27.62 27.71 
 
 27.81 27.90 27.99 28.09 27.18 
 
 9 
 
 6.0 
 
 28.27 28.37 28.46 28.56 28.65 
 
 28.75 28.84 28.94 29.03 29.13 
 
 9 
 
 6.1 
 
 29.22 29.32 29.42 29.51 29.61 
 
 29.71 29.80 29.90 30.00 30.09 
 
 10 
 
 6.2 
 
 30.19 30.29 30.39 30.48 30.58 
 
 30.68 30.78 30.^8 30.97 31.07 
 
 10 
 
 6-3 
 
 31.17 31.27 31.37 31.47 31.57 
 
 31.67 31.77 31.87 31.97 32.07 
 
 10 
 
 6.4 
 
 32.17 32.27 32.37 32.47 32.57 
 
 32.67 32.78 32.88 32.98 33-08 
 
 10 
 
 6.5 
 
 33.18 33.29 33.39 33.49 33.59 
 
 33.70 33.80 33.90 34 oo 34.ii 
 
 10 
 
 6.6 
 
 34.21 34.32 34.42 34-52 34-63 
 
 34-73 34.84 34-94 35.05 35.15 
 
 10 
 
 6.7 
 
 35.26 35.36 35.47 35.57 35.68 
 
 35.78 35.89 36.00 36.10 36.21 
 
 10 
 
 6.8 
 
 36.32 36.42 36.53 36.64 36.75 
 
 36.85 36.96 37.07 37.18 37-23 
 
 ii 
 
 6.9 
 
 37-39 37.50 37-6r 37.72 37.83 
 
 37.94 38.05 38.16 38.26 38.37 
 
 ii 
 
 7.0 
 
 38.48 3S.59 3?- 70 38.82 38.93 
 
 39.04 39.15 39.26 39.37 39.48 
 
 ii 
 
 7-1 
 
 39-59 39-70 39.82 39.93 40.04 
 
 40.15 40.26 40.38 40.49 40.60 
 
 ii 
 
 7.2 
 
 40.72 40.83 40.94 41.06 41.17 
 
 41.28 41.40 41.51 41.62 41-74 
 
 ii 
 
 7-3 
 
 41.85 41.97 42.08 42.20 42.31 
 
 42.43 42.54 42.66 42.78 42.89 
 
 ii 
 
 7-4 
 
 43.01 43.12 43.24 43.36 43.47 
 
 43-59 43.71 43.83 43-94 44-o6 
 
 12 
 
 7-5 
 
 44.18 44-30 44-41 44-53 44.65 
 
 44.77 44.89 45.01 45.13 45.25 
 
 12 
 
 7-6 
 
 45.36 45-48 45.60 45.72 45.84 
 
 43.96 46.08 46.20 46.32 46.45 
 
 12 
 
 7-7 
 
 46.57 46.69 46.81 46.93 47.05 
 
 47-17 47-29 47.42 47.54 47-66 
 
 12 
 
 7-8 
 
 47.78 47.91 48.03 48.15 48.27 
 
 48.40 48 52 48.65 48.77 48.89 
 
 12 
 
 7-9 
 
 49.02 49.14 49.27 49.39 49.51 
 
 49.64 49.76 49.89 50.01 50.14 
 
 12 
 
 8.0 
 
 50.27 50.39 50.52 50.64 50.77 
 
 50.90 51.02 51.15 51.28 51.40 
 
 13 
 
 8.1 
 
 51.53 51.66 51.78 51.91 52.04 
 
 52.17 52.30 52.42 52.55 52.68 
 
 13 
 
 8.2 
 
 52.81 52.94 53.07 53.20 53.33 
 
 53.46 53.59 53.72 53.85 53.98 
 
 13 
 
 8-3 
 
 54.11 54.24 54.37 54.50 54.63 
 
 54.76 54.89 55.02 55.15 55-29 
 
 13 
 
 8.4 
 
 55.42 55.55 55.68 55.81 55.95 
 
 56.08 56.21 56.35 56.48 56.61 
 
 13 
 
 8.5 
 
 56.75 56.88 57.01 57.15 57.28 
 
 57-41 57.55 57.68 57.82 57.95 
 
 13 
 
 8.6 
 
 58.09 58.22 58.36 58.49 58.63 
 
 58.77 58.90 59.04 59.17 59.31 
 
 14 
 
 8.7 
 
 59-45 59.58 59-72 59-86 59-99 
 
 60.13 60.27 60.41 60.55 60.68 
 
 M 
 
 8.8 
 
 60.82 60.96 61.10 61.24 61.38 
 
 61.51 61.65 61.79 61.93 62.07 
 
 14 
 
 8.9 
 
 62.21 62.35 62.49 62.63 62.77 
 
 62.91 63.05 63.19 63.33 63.48 
 
 14 
 
 9.0 
 
 63.62 63.76 63.90 64.04 64.18 
 
 64.33 64.47 64.61 64.75 64-90 
 
 14 
 
 9.1 
 
 65.04 65.18 65.33 65.47 65.61 
 
 65.76 65.90 66.04 66.19 66.33 
 
 14 
 
 9.2 
 
 66.48 6662 66.77 66.91 67.06 
 
 67.20 67.35 67.49 67.64 67.78 
 
 15 
 
 9-3 
 
 67.93 68.08 68.22 68.37 68.51 
 
 68.66 68.81 68.96 69.10 69.25 
 
 15 
 
 9.4 
 
 69.40 69.55 69.69 69.84 69.99 
 
 70.14 70.29 70.44 70.58 70.73 
 
 15 
 
 9-5 
 
 70.88 71.03 71.18 71.33 71.48 
 
 71.63 71.78 71.93 72.08 72.23 
 
 15 
 
 9.6 
 
 72.38 72.53 72.68 72.84 72.99 
 
 73 14 73.29 73.44 73.59 73.75 
 
 15 
 
 9-7 
 
 73.90 74.05 74.20 74.36 74-51 
 
 74.66 74.82 74.97 75.12 75.28 
 
 15 
 
 9.8 
 
 75-43 75.58 75.74 75.89 76.05 
 
 76.20 76.36 76.51 76.67 76.82 
 
 16 
 
 9-9 
 
 76.98 77.13 77.29 77.44 77.60 
 
 77-76 77.91 78.07 78.23 78.38 
 
 16 
 
 d 
 
 G I 2 3 4 
 
 56789 
 
 Diff. 
 
362 
 
 APPENDIX. 
 
 TABLE VIII. WEIGHT OF WROUGHT-IRON BARS. 
 
 Side or 
 Diam- 
 eter. 
 Inches. 
 
 Pounds per Linear 
 Foot. 
 
 Side or 
 Diam- 
 eter. 
 Inches. 
 
 Pounds per Linear 
 Foot. 
 
 Side or 
 Diam- 
 eter. 
 Inches. 
 
 Pounds per Linear 
 Foot. 
 
 Square 
 Bars. 
 
 Round 
 Bars. 
 
 Square 
 Bars. 
 
 Round 
 Bars. 
 
 Square 
 Bars. 
 
 Round 
 Bars. 
 
 
 
 
 
 2 
 
 13.33 
 
 10.47 
 
 5 
 
 83.33 
 
 65.45 
 
 ft 
 
 0.013 
 
 O.OIO 
 
 ft 
 
 14.18 
 
 11.14 
 
 i 
 
 87.55 
 
 68.76 
 
 i 
 
 0.052 
 
 O.O4I 
 
 1 
 
 15,05 
 
 11.82 
 
 * 
 
 91.88 
 
 72.16 
 
 ft 
 
 O.II7 
 
 0.092 
 
 ft 
 
 15.95 
 
 12.531 
 
 1 
 
 96.30 
 
 75.64 
 
 i 
 
 0.208 
 
 0.164 
 
 i 
 
 16.88 
 
 13.25 
 
 i 
 
 100.8 
 
 79.19 
 
 & 
 
 0.326 
 
 0.256 
 
 ft 
 
 17.83 
 
 14.00 
 
 f 
 
 IQ5.5 
 
 82.83 
 
 1 
 
 0.469 
 
 0.368 
 
 1 
 
 18.80 
 
 14-77 
 
 i 
 
 110. 2 
 
 86.56 
 
 ft 
 
 0.638 
 
 0.501 
 
 TV 
 
 19.80 
 
 15.55 
 
 1 
 
 II5-I 
 
 90.36 
 
 * 
 
 0.833 
 
 0.654 
 
 i 
 
 20.83 
 
 16.36 
 
 6 
 
 120.0 
 
 94-25 
 
 A 
 
 1.055 
 
 0.828 
 
 & 
 
 21.89 
 
 17.19 
 
 i 
 
 I25.I 
 
 98.22 
 
 f 
 
 1.302 
 
 1.023 
 
 f 
 
 22.97 
 
 18.04 
 
 i 
 
 130.2 
 
 102.3 
 
 tt 
 
 1.576 
 
 1.237 
 
 H 
 
 24.08 
 
 18.91 
 
 1 
 
 135.5 
 
 106.4 
 
 f 
 
 1.875 
 
 1-473 
 
 i 
 
 25.21 
 
 19.80 
 
 1 
 
 140.8 
 
 no. 6 
 
 it 
 
 2.2OI 
 
 1.728 
 
 if 
 
 26.37 
 
 20.71 
 
 f 
 
 146.3 
 
 114.9 
 
 1 
 
 2-552 
 
 2.004 
 
 1 
 
 27-55 
 
 21.64 
 
 f 
 
 I5I.9 
 
 II9-3 
 
 if 
 
 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 
 
 TV 
 
 3.763 
 
 2-955 
 
 i 
 
 32.55 
 
 25.57 
 
 i 
 
 175-2 
 
 137.6 
 
 i 
 
 4.219 
 
 3-3I3 
 
 i 
 
 35-21 
 
 27.65 
 
 1 
 
 187.5 
 
 147-3 
 
 ft 
 
 4-701 
 
 3.692 
 
 f 
 
 37-97 
 
 29.82 
 
 f 
 
 200.2 
 
 157-2 
 
 i 
 
 5.208 
 
 4.091 
 
 i 
 
 40.83 
 
 32.07 
 
 8 
 
 213.3 
 
 167.6 
 
 T 5 * 
 
 5.742 
 
 4.510 
 
 f 
 
 43.8o 
 
 34.40 
 
 i 
 
 226.9 
 
 178.2 
 
 1 
 
 6.302 
 
 4.950 
 
 f 
 
 46.88 
 
 36.82 
 
 i 
 
 240.8 
 
 189.2 
 
 T 5 * 
 
 6.888 
 
 5.410 
 
 1 
 
 50.05 
 
 39-31 
 
 f 
 
 255.2 
 
 200.4 
 
 i 
 
 7.500 
 
 5.890 
 
 4 
 
 53-33 
 
 41.89 
 
 9 
 
 270.0 
 
 212. 1 
 
 ft 
 
 8.138 
 
 6.392 
 
 i 
 
 56.72 
 
 44-55 
 
 i 
 
 285.2 
 
 224.0 
 
 1 
 
 8.802 
 
 6.913 
 
 i 
 
 60.21 
 
 47-29 
 
 i 
 
 300.8 
 
 236.3 
 
 
 
 9.492 
 
 7-455 
 
 1 
 
 63.80 
 
 50.11 
 
 f 
 
 316.9 
 
 248.9 
 
 f 
 
 10.21 
 
 8.018 
 
 i 
 
 67.50 
 
 53-01 
 
 10 
 
 333-3 
 
 261.8 
 
 it 
 
 10.95 
 
 8.601 
 
 1 
 
 71.30 
 
 56.00 
 
 i 
 
 350.2 
 
 275-1 
 
 1 
 
 11.72 
 
 9.204 
 
 f 
 
 75-21 
 
 59-07 
 
 i 
 
 367.5 
 
 288.6 
 
 it 
 
 12,51 
 
 9.828 
 
 1 
 
 79.22 
 
 62.22 
 
 f 
 
 385.2 
 
 302.5 
 
INDEX. 
 
 363 
 
 INDEX. 
 
 Advanced problems, 347 
 Aluminum, 190 
 Angle iron, 127 
 Angular velocity, 233 
 Annealing, 183, 187 
 Answers to problems, 351 
 Apparent and true stresses, 288-309 
 
 shears, 305 
 Appendix, 344-362 
 Areas of circles, 355, 360 
 Army, gun formulas, 319, 320 
 Average constants, 163 
 
 BACH'S formulas, 332 
 Bariron. 182, 362 
 BARLOW'S formula, 28 
 Bars of uniform strength, 227 
 resilience of, 202 
 under centrifugal stress, 233 
 under impact, 229, 231 
 weights of, 2, 355, 362 
 Beams, 36-110, 146-161, 243-271 
 
 bending moments, 42, 347 
 
 cantilevers, 36-94 
 
 cast iron, 60 
 
 center of gravity of sections, 52 
 
 combined stresses, 146-149, 267-271 
 
 continuous, 99-110 
 
 deck, 67 
 
 definitions, 36 
 
 deflection, 72-77, 95,245, 249, 265 
 
 deflection and stiffness, 74 
 
 deflection and stress, 79 
 
 designing of, 58 
 
 elastic curve, 70 
 
 elastic resilience, 203 
 
 experimental laws, 48 
 
 fixed, 88-98 
 
 flexure of, 107, 243-271 
 
 flexure and torsion, 152 
 
 fundamental formulas, 49 
 
 GALILEI'S investigations, 162 
 
 horizontal shear, 155 
 
 impact on, 249 
 
 internal stresses, 45, 158 
 
 internal work, 203 
 
 maximum moments, 52, 347 
 
 modulus of rupture, 61 
 
 moments of inertia, 53 
 
 Beams, moving loads, 259, 348 
 
 overhanging, 85, 91 
 
 reactions, 37, 85 
 
 resilience of, 96, 243 
 
 restrained, 85, 95 
 
 safe loads for, 58 
 
 simple, 36-84 
 
 stiffness, 74 
 
 sudden loads, 248 
 
 theoretical laws, 48 
 
 uniform strength, 80-84 
 
 vertical shear, 39, 262 
 
 weights of, 2, 355, 362 
 Bending moment, 42-44 
 
 maximum, 54 
 
 maximum maximorum, 348 
 
 tables of, 79, 95, 104 
 
 triangular load, 257 
 Beton, 189 
 
 BIRNIE'S formulas, 319 
 Boilers, 25, 31, 333 
 
 ends of, 328 
 
 joints in, 28-34 
 
 tubes in, 25 
 Bolts, 17, 34, 281, 307 
 Books of reference, 163, 166, 289, 326 
 Brass, 190 
 Brick, 172-174 
 
 modulus of rupture, 62, 174 
 
 strength of, 14, 173 
 
 weight of, 2, 173, 355 
 Brick tower, 15 
 Bridge iron, 182 
 
 rollers, 239 
 Briquettes, 175 
 Bronze, 190 
 Butt joints, 30, 31 
 
 Cantilever beams, 36-110 
 
 deflection of, 72, 82 
 
 elastic curve, 73 
 
 fundamental formulas, 49 
 
 internal work, 244 
 
 resilience, 204 
 
 tables for, 79, 95 
 
 uniform ftrength, 80 
 Carbon in steel, 180, 183, 186 
 Castings. 179, 189 
 Cast iron, constants for, 163, 179 
 
INDEX. 
 
 Cast iron, factors of safety, 18 
 
 in compression, 14, 180 
 in shear, 15 
 in tension, 9, 180 
 modulus of rupture, 61, 181 
 pipes, 22, 23 
 resilience of, 199, 219 
 weight of, i, 355 
 Cements, 174 
 Center of gravity, 52 
 Centrifugal stress, 232, 255 
 Chestnut, 171 
 
 CHRISTIE'S experiments, 129, 341 
 Circles, areas of, 355, 360 
 Circular plates, 328 
 CLAVARINO'S formulas, 317 
 Coefficient of elasticity 7, 8, 163 
 
 compression, 14 
 
 shear, 15, 275 
 
 tension, 9, 163, 165 
 Coefficient of expansion, 145 
 Cold bend test, 182, 214, 224 
 
 rolling, 183, 217 
 Columns, 111-134, 337~343 
 
 deflection of, 133, 340 
 
 design of, 125, 337 
 
 eccentric loads, 133, 343 
 
 ends of, 113, 120 
 
 EULER'S formula, 114, 340 
 
 experiments on, 337 
 
 GORDON'S formula, 119 
 
 HODGKINSON'S formula, 117 
 
 investigation of, 123, 337 
 
 JOHNSON'S (T. H.) formula, 127 
 
 modified EULER formula, 339 
 
 radius of gyration, 122 
 
 RANKINE'S formula, 119 
 
 RITTER'S formula, 132 
 
 rupture of, 113, 127 
 
 safe loads for, 124 
 
 sections of, in 
 
 theory of, 131 
 Combined stresses, 144-161, 288-309 
 
 compression and flexure, 148, 269 
 
 flexure and torsion, 152 
 
 shear and tension, 150, 296 
 
 tension and compression, 144, 290 
 
 tension and flexure, 146, 267 
 
 torsion and compression, 154 
 Comparison of beams, 62, 77, 94 
 
 shafts, 279 
 
 Compound cylinder, 315, 323 
 Compression, 3, 4, 13, 226 
 
 and flexure, 148, 269 ^ 
 
 and shear, 150 
 
 and tension, 144, 290 
 
 and torsion, 154 
 
 Compression, cast iron, 180 
 
 cement, 176 
 
 eccentric loads, 241, 342 
 
 mortar, 176 
 
 steel, 188 
 
 stone, 177 
 Concrete, 189 
 Connecting rod, 256 
 Constants, tables of, 163 
 Continuity, 96 
 Continuous beams, 36, 76-110 
 
 equal spans, 106 
 
 properties of, 99 
 
 tables of, 104, 105 
 
 three moments, 102 
 
 unequal spans, 106 
 Contraction of area, 169, 215 
 Couplings for shafts, 281 
 Crank arm, 283 
 
 pin, 283, 285, 301 
 Cross-sections, 52, in, 124 
 Cubic equation, 299, 302 
 Curve, elastic, 71 
 
 of stresses, 10 
 Cylinders, 22, 118, 310 
 
 compound, 310, 315 
 
 exterior pressure, 24, 316 
 
 interior pressure, 22, 313 
 
 thick, 26, 310 
 
 thin, 24 
 
 with hoops, 235, 321 
 Cylindrical rollers, 238 
 
 Deflection of beams, 36, 79, 95, 173 
 
 cantilever beams, 72-75 
 
 restrained beams, 85-95 
 
 simple beams, 75-99 
 
 sudden loads, 246 
 
 under impact, 251, 253 
 
 under moving load, 259 
 
 under shearing, 264 
 Deflection of columns, 115, 133, 340 
 Deformations, 3-6, 290-292 
 Description of tables, 355 
 Designing, 12, 195 
 
 beams, 58 
 
 columns, 125, 337 
 
 guns, 325 
 
 shafts, 141, 280 
 Detrusion, 3, 15, 262 
 Ductile materials, 162 
 DUDLEY'S tests, 219, 344 
 
 Eccentric loads, 134, 240, 342 
 Efficiency of a joint, 31 
 Elastic curve, 36, 165 
 cantilever beams, 73 
 
INDEX. 
 
 Elastic curve, columns, 115, 343 
 
 continuous beams, 73 
 general equation, 70 
 restrained beams, 85-95 
 simple beams, 75 
 Elastic limit, 7, 165 
 
 cast iron, 180 
 
 compression, 14 
 
 shear, 15 
 
 steel, 188 
 
 tension, 9 
 
 timber, 172 
 
 wrought iron, 183 
 Elasticity, laws of, 6 
 
 theory of, 288 
 Elastic resilience, 197-225 
 Ellipse of stress, 305 
 Ellipsoid of stress, 297 
 Elliptical plates, 331 
 Elongation, 3s 5, 8, 183 
 
 ultimate, 9, 13, 168 
 
 under impact, 218, 229 
 
 under own weight, 226 
 ESTRADA'S tests, 218, 223 
 EULER'S formula, 114, 340 
 Exercises, 34, 109. 143, 347 
 Experimental laws, 6, 48, 191 
 Experiments, 130, 134, 143, 164, 213 
 External work, 201, 243 
 
 Factor of lateral contraction, 276, 290, 
 
 319 
 
 of safety, 17-21, 166, 213 
 Fatigue of materials, 191 
 Flexure, 36-110, 146-161, 243-271 
 
 and compression, 148, 269 
 
 and tension, 144, 267 
 
 and torsion, 152, 283 
 
 erroneous views, 109 
 
 of brick, 174 
 
 of cast iron, 180 
 
 of crank pin, 286 
 
 under impact, 210. 251 
 
 under live load, 259 
 
 work of, 201, 243, 263 
 Floor beams, 66, 87 
 Forge pig, 179 
 Formulas, principal: 
 
 (1) P=AS, 5 
 
 (2) S = s 8 
 
 (3) AS, = V, 50 
 
 Formulas, principal 
 
 : 
 
 
 (id) P Sc a 
 
 
 . 123 
 
 * ' A 7 ' 
 
 1 
 
 
 (.,)^=// 
 
 C 
 
 
 . 137 
 
 
 (I2) *=-^L 
 
 198 ooof 
 
 , 
 
 . 141 
 
 (13), 151 
 
 (M), 
 
 157 
 
 (15), 
 
 184; 
 
 (16), 186 
 
 (17), 
 
 194 
 
 (18), 
 
 200; 
 
 (19), 204 
 
 (20), 
 
 231 
 
 (21), 
 
 240; 
 
 (22), 250 
 
 (23), 
 
 259 
 
 (24), 
 
 264 ; 
 
 (25), 276 
 
 (26;, 
 
 277 
 
 (27), 
 
 292 ; 
 
 (28), 294 
 
 (29), 
 
 296 
 
 (30), 
 
 299; 
 
 (31), 304 
 
 (32), 
 
 312 
 
 (33), 
 
 318; 
 
 (34), 319 
 
 (35), 
 
 337- 
 
 
 Foundry pig, 179 
 
 
 
 (4) 7 = 
 
 51 
 
 (5) ^.T=^7 71 
 
 (6), (7), (8), (9), 97-103 
 
 Glass, 191 
 
 GORDON'S formula, 119 
 
 Granite, 177 
 
 GRASHOF'S formulas, 239, 330 
 
 Gravity, center of, 52 
 
 specific, i, 180 
 Gun metal, 169 
 Guns, 28, 310-327 
 
 hooped, 315, 324 
 
 solid, 313, 320 
 Gyration, radius of, 124 
 
 Helical springs, 348 
 
 Hemispheres, 333 
 
 Hemlock, 171 
 
 Historical notes, 162, 208 
 
 HODGKINSON'S formula, 117 
 
 Hollow cylinders, 24, 26, 310-327 
 
 shafts, 142, 278 
 
 spheres, 24, 333 
 HOOKE'S law, 6, 164, 288 
 Hoops, centrifugal stress, 234 
 
 for guns, 315, 321, 324 
 
 shrinkage of, 235, 316 
 Horizontal impact, 210, 229 
 
 shear, 155, 273 
 Horse-power, 140, 205 
 Hydraulic cement, 174 
 
 I beams, 51, 87, 106 
 India rubber, 289 
 Inertia of a bar, 220, 229 
 
 of a beam, 250, 262 
 moment of, 53, 138 
 Impact, 162-225 
 on bars, 210, 229 
 on beams, 211, 249 
 pressure due to, 222, 253 
 
366 
 
 INDEX. 
 
 Impact, tests on, 215-220, 248, 344 
 Impulse, 230 
 Inflection point, 85 
 Internal stresses, 4, 45, 158, 288 
 work, 201, 209, 213, 244 
 Investigation, 12, 14 
 
 of beams, 56, 259 
 
 of columns, 125, 337 
 
 of guns, 314, 323 
 
 of joints, 28 
 
 of shafts, 141, 284 
 Iron, 168, 179, 181, 329 
 
 Jacket for guns, 315 
 Joints, riveted, 28-34 
 efficiency of, 31 
 
 KEEP'S impact machine, 219, 251 
 KIRKALDY'S tests, 215, 249 
 
 LAME'S formulas, 310, 335 
 Lateral contraction, 275, 290 
 
 factor of, 276 
 Lateral deflection, 134, 341 
 
 deformation, 290 
 LAUNHARDT'S formula, 194 
 Laws, experimental, 6, 7, 48 
 of fatigue, 191 
 of internal stresses, 46, 288 
 of resilience, 204, 208 
 Lead, 190 
 Limestone, 177 
 Live loads, 258 
 Loads, 36, 240 
 
 safe, for beams, 58 
 
 safe, for columns, 126, 337 
 
 sudden, 245 
 Locomotive, 255 
 X-ogarithms, 356 
 Longitudinal impact, 229 
 
 Manholes, 331 
 
 Materials, constants of, 162 
 
 factors of safety, 18 
 fatigue of, 191 
 resilience of, 197-225 
 strength of, 162-196 
 weights of, i, 355 
 Maximum internal stresses, 158, 299, 
 
 303, 306 
 
 moments, 54, 347 
 strength, 168 
 Modulus of elasticity, 8 
 
 resilience, 165, 199, 278 
 rupture, 61, 174, 181, 185 
 Moment of inertia, 53 
 for beams, 65, 67 
 
 Moment of inertia, for columns, 116 
 
 for shafts, 138 
 Moments, 42-107 
 
 bending, 42 
 
 for continuous beams, 104 
 
 maximum, 54, 347 
 
 resisting, 50, 136 
 
 theorem of three, 102 
 
 twisting, 137 
 Mortar, 174 
 
 Navy, gun formulas, 319, 326 
 Neutral axis, 49, 116 
 
 surface, 48 
 Nickel steel, 188 
 
 Oak, 171 
 
 One-hoss shay, 20 
 
 Ordnance formulas, 310-327 
 
 Ores of iron, 179 
 
 Oscillations of a bar, 199, 210, 229 
 
 of a beam, 246, 252 
 Overhanging beams, 85, 91 
 
 Parabola, 43, 81, 133, 194 
 Parallel rod, 255 
 Paving brick, 173 
 Phenomena of compression, 13 
 shear, 15 
 tension, 9 
 torsion, 135 
 Phosphor bronze, 190 
 Piers, 35, 228 
 Pig iron, 179 
 Pine, 171 
 Pipes, 22-27 
 thick, 22 
 thin, 26 
 Piston rod, 21 
 Plate girder, 259, 272, 308 
 
 iron, 182 
 Plates, 328-333 
 on cylinders, 238 
 on spheres, 237 
 Polar moments, 138 
 Portland cement, 175 
 Power, shafts for, 140 
 Pressure due to impact 222, 253 
 Principal stresses, 299 
 Problems, 2-350 
 answers to, 351 
 Puddling furnace, 181 
 
 Radius of gyration, 122, 277 
 Rafters, 148 
 
 Railroad rails, 2, 59, 219, 252, 254 
 Range of stresses, 191-196 
 
INDEX. 
 
 367 
 
 Reactions, 37, 85, 101, 148 
 Rectangular beams, 63, 205 
 
 plates, 332 
 
 Repeated stresses, 191-196, 217 
 Resilience, 197-225 
 
 of bars, 202 
 
 of beams, 203, 243 
 
 of shearings, 274 
 
 of torsion, 277 
 
 Resistance of materials, 1-22, 162-196 
 Resisting moment, 50, 136 
 Restrained beams, 85-95 
 
 comparison of, 94 
 Resultant stress, 295 
 Riveted joints, 28-35 
 
 design of, 32 
 
 efficiency of, 31 
 Rivet iron, 182 
 Rivets, 28-34, 272 
 Rollers, 236, 238 
 Ropes, 189 
 
 Round shafts, 141, 279 
 Rosendale cement, 175 
 Rupture, 4, 7, 9, 165 
 
 of beams, 51 
 
 of columns, 129 
 
 in repeated stress, 191 
 
 modulus of, 61, 164 
 
 Safe loads, 58, 123, 337 
 
 Safety, factors of, 17-21, 213 
 
 Sandstone, 177 
 
 Set, 6 
 
 Shaft couplings, 281 
 
 Shafts, 135-143, 277-287 
 
 for power, 140 
 
 hollow, 142, 278 
 
 round, 141, 279 
 
 square, 143 
 
 resilience of, 280 
 
 stiffness of, 279 
 
 strength of, 141, 278 
 Shape iron, 66, 182 
 Shear, 3, 5, 15, 16, 172, 272-287 
 
 and tension, 16, 150 
 
 horizontal, 155 
 
 longitudinal, 15 
 
 resilience of, 274 
 
 resisting, 47 
 
 stresses due to, 272 
 
 vertical, 39 
 Shearing stresses, 294, 303 
 
 coefficient of elasticity, 266, 275 
 
 deflection due to, 264 
 
 ultimate strength, 15, 308 
 
 work of, 263 
 Shears for cantilevers, 42 
 
 Shears for continuous beams, 105 
 
 for simple beams, 40 
 Shocks, 7, 1 8, 199 
 Short beams, 265 
 Shrinkage of hoops, 235, 321 
 Simple beams ; see Beams. 
 Slate, 178 
 
 Solid shafts, 141, 278 
 Sound, 347 
 Specific gravities, I 
 Spheres, 24, 236, 333 
 Spherical rollers, 236 
 Springs, 199. 348 
 Square plates, 333 
 shafts, 143 
 
 Squares of numbers, 355, 358 
 Static deflections, 246 
 
 resilience, 208 
 Statics, 288 
 Steam boilers, 24, 31, 333 
 
 pipes, 22, 23 
 Steel, 185-189 
 constants of, 163 
 factors of safety, 18 
 hoop shrinkage, 235 
 resilience, 200, 206 
 Steel beams, 66 
 cranks, 284 
 plates, 329 
 rollers, 239 
 spheres, 237 
 Stiffness of beams, 77 
 of shafts, 279 
 Stone, 163, 177 
 Straight- line formula, 127 
 Strain, 3 
 
 Strength of materials, 1-21, 162-196 
 history of, 162 
 tables, 164 
 ultimate, 7, 168 
 Stress, ellipse of, 305 
 
 ellipsoid of, 297 
 velocity of, 345 
 Stresses, 3-21 
 apparent, 288-309 
 centrifugal, 232, 255 
 combined, 144-161, 290, 308 
 in guns, 310-327 
 repeated, 193 
 sudden, 197, 247 
 temperature, 145 
 true, 288-309 
 working, 17-21, 194 
 Sudden deflections, 246 
 loads, 197 
 
 Tables ; see page ix 
 
368 
 
 INDEX. 
 
 Temperature, 145, 235 
 Tensile tests, n, 168 
 Tension, 3, 4, 9, 226-242 
 and flexure, 146 
 and shear, 150 
 apparent and true, 288-309 
 cast iron, 9, 180 
 cement and mortar, 175 
 eccentric, 240 
 steel, 9, 187 
 tangential, 312 
 timber, 9, 171 
 wrought iron, 9, 183 
 Testing machines, 167-170, 213-220 
 
 capacity of, 170 
 Tests, ash sticks, 109 
 brick, 173, 174 
 cast iron, 180 
 cement, 175 
 cold bend, 182 
 columns, 127 
 compression, 169 
 continuous beams, no 
 fatigue, 191 
 impact, 211-224 
 steel, 187 
 stone, 178 
 tension, 6, n 
 timber, 172 
 torsion, 143, 170 
 uniform methods, 220 
 wrought iron, n, 183 
 Theorem of three moments, 102 
 Thick hollow cylinders, 26, 310 
 
 spheres, 334 
 
 THURSTON'S torsion machine, 143, 214 
 Timber, 9, 14, 171 
 factors of safety, 18 
 modulus of rupture, 62 
 resilience, 200 
 weight, i, 171, 355 
 Torsion, 135-143, 272-287 
 coefficient of elasticity, 139 
 combined, 152, 154, 285 
 modulus of, 139 
 phenomena of, 135 
 resilience of, 277 
 
 Transmission of power, 140, 285 t 
 Trap rock, 177 
 True deformations, 292 
 stresses, 288-309 
 Tubes for boilers, 25 
 for guns, 315 
 
 Ultimate resilience, 200, 206 
 Ultimate strength, 7, 168 
 
 compression, 14, 169 
 
 shear, 15 
 
 tension, 14, 168 
 Uniform strength, 20 
 
 bars, 227 
 
 beams, 35, 80, 83 
 Unit-stress, 3, 5 
 
 repeated, 193 
 
 working, 17 
 
 Velocity of live load, 257 
 
 stress, 345 
 Vertical bar, 226 
 Vertical shear, 39-42, 105 
 deflection due to, 264 
 stresses caused by, 272 
 work of, 262 
 Vibration strength, 196 
 Vibrations of a beam, 252 
 Volume, change of, 289 
 
 resilience of, 205, 212 
 
 Water pipes, 22, 23, 190 
 Wave propagation, 346 
 Weights of bars, 2, 355, 362 
 materials, i, 163 
 WEYRAUCH'S formula, 194 
 Wheel, revolving, 234 
 Wire, 183 
 Wire guns, 326 
 W^HLER'S tests, 191 
 Work of flexure, 243 
 rupture, 184 
 shearing, 274 
 tension, 202 
 torsion, 278 
 vertical bar, 226 
 vertical shear, 262 
 Working stress, 12, 17, 192 
 Wrought iron, 163, 181 
 factors of safety, 18 
 resilience, 200 
 shear, 15 
 tension, 9, 183 
 weight of, i, 362 
 work of, 184 
 Wrought-iron bars, 2, 36* 
 
 Yield point, 183 
 YOUNG'S laws, 208 
 
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