I " r- . : LIBRARY UNIVERSITY OF CALIFORNIA. Class Strength of Materials BY H. E. MURDOCK, M. E., C. E. MEMBER OF THE SOCIETY FOR THE PROMOTION OF ENGINEERING EDUCATION AND OF THE DEPARTMENT OF THEORETICAL AND APPLIED MECHANICS IN THE UNIVERSITY OF ILLINOIS FIRST EDITION FIRST THOUSAND NEW YORK JOHN WILEY & SONS LONDON: CHAPMAN & HALL, LIMITED 1911 COPYRIGHT, 1911, BY H. E. MURDOCK Stanbope ipress F. H. GILSON COMPANY BOSTON. U.S.A. PREFACE IN preparing this book the author has had in mind primarily the needs of his own students in strength of materials. He hopes, however, that it will meet a real want in other colleges and technical schools also. This book has been written with the aim of making intelligible the fundamental principles of the strength of materials without the formal use of the calculus. The works which do not use the ordinary calculus treatment usually omit some important parts such as the deflec- tion of beams, strength of columns, horizontal shear, combined stresses, impact loads, etc. This book is designed to give a fairly complete course in the subject for students who have not had the calculus, or when graphical presentations are preferred. However, a sepa- rate chapter giving the derivation of the elastic curve of beams by the calculus method has been included for those who desire such treatment. Effort has been made to present the derivation of the formulas in a clear and concise manner, in such a way as to enable the student to obtain an adequate compre- hension of the principles involved. While the aim is to emphasize the elementary principles and to develop independent reasoning in the student, the ground covered is that usually given in a college course for engineering students. Many illustrative examples and problems are given for the purpose of making clear the application of the theory. Answers to some of the problems are given in order that the student may occasionally check his iii 226183 IV PREFACE numerical work. The order of the arrangement is one that has given good satisfaction. In the deduction of the shear formula it is brought out at the first that the shearing stress is not uniformly dis- tributed over the sectional area of the beam and that the maximum stress is greater than that obtained by dividing the vertical shear by that area. A chapter on graphic integration is included and the graphical method of determining the deflection of beams is utilized. The graphical method appeals to the eye as well as to the reason, and thus supplies an additional avenue of con- ception. It also shows to advantage the meaning of the constants of integration. The graphical method is also much more readily applicable to beams carrying non- uniform distributed loads, and to beams for which the moment of inertia of the cross section is not constant. When one set of curves is drawn for a given beam carry- ing a given system of loading, those curves may be used for all similar beams with similar loading. In the chapter on the calculus method an attempt is made to give the physical conception of the constants of integration rather than to treat them simply as mathematical symbols. As the nature of the behavior of columns under load is very uncertain, the treatment given to columns is largely empirical. Emphasis is laid on the straight line formula, although the Euler and the Rankine formulas are also given. The author wishes to acknowledge his indebtedness to the following professors and instructors in the College of Engineering of the University of Illinois: to Dr. N. Clifford Ricker for the interest shown in the preparation of this work, and the use of tables and data prepared by him; to Professor A. N. Talbot and Professor H. F. Moore for many suggestions as to the form, arrange- PREFACE v ment, and subject matter, and much assistance in the preparation of the book; to Mr. G. P. Boomsliter for his criticizing and checking the examples and problems; to Mr. C. R. Clarke and Mr. C. E. Noerenberg for their criticisms and help in preparing the manuscript for the publishers. Although the work has been carefully checked, errors may exist, and for any intimation of these I shall be obliged. H. E. MURDOCK. TABLE OF CONTENTS CHAPTER I MATERIALS OF CONSTRUCTION ART. PAGE 1. Introduction i 2. Mechanical and Physical Properties 2 3. Masonry 3 (a) Stone Masonry 3 (fe) Brick Masonry 3 (c) Concrete, Plain and Reinforced 4 4. Timber 4 5. Cast Iron 4 6. Wrought Iron 5 7. Steel 5 8. Other Materials 6 Example 7 Problems . . 8 CHAPTER II DIRECT STRESSES 9. Definitions 9 10. Tension 10 11. Compression 12 12. Shear 13 13. Oblique Shear 13 14. Stress-Deformation Diagrams 15 15. Elastic Limit and Yield Point 16 16. The Modulus of Elasticity 17 17. Resilience 18 18. The Shearing Modulus of Elasticity 18 19. Poisson's Ratio 19 20. Reduction of Area 19 vii viii TABLE OF CONTENTS ART. PAGE 21. Uses of the Modulus of Elasticity 20 22. Stresses Used in Design 20 Problems 22 CHAPTER III DIRECT STRESSES APPLICATIONS 23. Simple Cases of Direct Stresses 25 24. Stresses in Thin Cylinders 25 25. Stresses in a Hoop 26 26. Stresses Due to Change in Temperature 27 27. Stresses in Thin Spheres 28 28. Thick Cylinders under Interior Pressure 28 29. Cylinders under Exterior Pressure 29 Examples 30 Problems 31 CHAPTER IV RIVETED JOINTS 30. Riveted Joints 32 31. Kinds of Riveted Joints 32 32. Methods of Failure of Riveted Joints 33 33. Computation of Unit Stresses Developed in Riveted Joints. . . 34 34. Single-Riveted Lap Joint 35 35. Double-Riveted Lap Joint 36 36. Lap Joint with More than Two Rows of Rivets 37 37. Butt Joint 38 38. Compression Loads for Riveted Joints 38 39. Efficiency of Riveted Joints 38 40. Design of Riveted Joints 39 Examples 42 Problems 44 CHAPTER V BEAMS EXTERNAL FLEXURAL FORCES 41. Definitions 47 42. Methods of Loading Beams / . . . . 48 43. Forces Acting on a Beam as a Whole 49 TABLE OF CONTENTS IX ART. PAGE 44. Forces Acting on a Portion of a Beam, Internal Stresses 51 45. Vertical Shear 52 46. Sign and Unit of Vertical Shear 53 47. Value of the Vertical Shear for Cantilever and Simple Beams . . 53 48. Load and Shear Diagrams 54 49. Relation between the Load and the Shear 56 50. The Rate of Change of the Vertical Shear 57 51. Relation between the Load and Shear Diagrams 57 52. Bending Moment 58 53. Sign and Unit of Bending Moment 59 54. The Values of the Bending Moment at the Section MN, Dis- tant x from the Left Support or Origin for Cantilever and Simple Beams 60 55. Bending Moment Diagrams 61 56. Relation between the Vertical Shear and the Bending Moment 62 57. The Rate of Change of the Bending Moment 63 58. The Maximum Vertical Shear and Bending Moment 64 59. Load, Shear, and Moment Diagrams for Cantilever and Simple Beams. Maximum Shear and Moment 65 60. Relative Strength of Cantilever and Simple Beams 73 61. Moving Concentrated Loads on a Beam 74 CHAPTER VI BEAMS INTERNAL FLEXURAL STRESSES 62. Forces and Stresses 83 63. Resisting Shear. The Shear Formula 84 64. The Value of k in the Shear Formula 85 65. Resting Moment 86 66. Assumptions for the Resisting Moment 86 67. Distribution of the Fiber Stresses 87 68. Position of the Neutral Surface and the Neutral Axis 88 69. The Moment Formula 90 70. Units 91 71. Total Horizontal Compressive and Tensile Stresses 91 72. The Three Problems 93 73. Modulus of Rupture 97 74. Maximum Stress Diagrams 97 75. Beams of Uniform Strength 99 Examples 99 Problems 101 TABLE OF CONTENTS CHAPTER VII STRESSES IN SUCH STRUCTURES AS CHIMNEYS, DAMS, WALLS, AND PIERS ART. PAGE 76. Kinds of Stresses 106 77. Eccentric Loads on Short Prisms 107 78. Eccentricity of a Load that Will Produce Zero Stress in the Outside Fiber 108 79. The Kern no 80. Case of Eccentric Loads Caused by a Combination of the Weight of the Material and Lateral Pressure no 81. Effect when the Line of Action of the Resultant Falls Outside of the Kern in 82. The Maximum Stress when the Line of Action of the Resultant Falls Outside the Middle Third for Rectangular Prisms which take no Tension 112 Examples 113 Problems 114 CHAPTER VIII GRAPHIC INTEGRATION 83. Definitions 116 84. The First Method of Obtaining the Second Integrated Curve . . 117 85. The Second Method of Obtaining the Second Integrated Curve 1 20 86. Constant of Integration 124 87. Units 126 Examples 127 Problems 128 CHAPTER IX DEFLECTION OF BEAMS ELASTIC CURVE 88. Bending 130 89. The Radius of Curvature of Beams 130 90. The Slope of the Neutral Surface 132 91. The Slope Curve 134 92. The Rate of Change of the Slope 135 93. The Deflection of Beams. The Elastic Curve 135 TABLE OF CONTENTS xi ART. PAGE 94. The Rate of Increase of the Deflection 137 95. Relations between the Five Curves 137 96. The Units for the Five Curves 137 Example 138 Problems 139 CHAPTER X ELASTIC CURVE CANTILEVER AND SIMPLE BEAMS AND BEAMS FIXED AT BOTH ENDS 97. Cantilever Beam, Concentrated Load at the End 140 98. Cantilever Beam, Concentrated Load away from the Free End 144 99. Cantilever Beam, Uniform Load 144 ico. Cantilever Beam, Various Loading 147 101. Simple Beam, Concentrated Load at the Center 148 102. Simple Beam, Uniform Load 150 103. Beam Fixed at Both Ends, Concentrated Load at Center. ... 154 104. Points of Inflection 157 105. Beam Fixed at Both Ends, Uniform Load 158 106. Relative Strength and Stiffness of Beams 160 107. Maximum Stress and Deflection 163 108. Relation between the Maximum Stress and the Maximum Deflection 164 Example 167 Problems 168 CHAPTER XI ELASTIC CURVE OVERHANGING, FIXED AND SUPPORTED, CONTINUOUS BEAMS 109. Overhanging Beam, Concentrated Loads 170 no. Overhanging Beam, Uniform Load 172 in. Beam Fixed and Supported, Concentrated Load at Center 172 112. Beam, Both Ends Fixed, Concentrated Load at any Point. . 176 113. Continuous Beams 178 114. The Theorem of Three Moments 181 115. Hinging Points for Continuous Beams 185 Problems 187 xii TABLE OF CONTENTS CHAPTER XII ELASTIC CURVE OF BEAMS DETERMINED BY THE ALGEBRAIC METHOD ART. PAGE 116. The Algebraic Relations between the Five Curves 189 1 1 7. -The Choice of Coordinate Axes 192 118. The Constants of Integration 192 119. Determination of the Constants of Integration 193 Section of Zero Vertical Shear 195 Section of Zero Bending Moment 196 Section of Zero Slope 196 Section of Zero Deflection 196 120. Essential Quantities to be Known about Beams 196 Examples 197 Problems 206 CHAPTER XIII SECONDARY STRESSES 121. Horizontal Shear in Beams 208 122. The Magnitude of the Horizontal and Vertical Shearing Unit- Stress at a Point 209 123. Plate Girders, First Method 212 124. Plate Girders, Second Method 212 125. Combined Flexure and Tension or Compression 214 126. Combined Shearing Stresses and Tensile or Compressive Stresses 216 Examples 217 Problems 219 CHAPTER XIV COLUMNS AND STRUTS 127. Discussion 221 128. Stiffness of Columns 221 129. The Strength of Columns 222 130. The Straight-line Formula 225 131. Eccentric Loads on Columns 228 132. The Method of Transmitting Loads to Columns 229 TABLE OF CONTENTS Xlii OTHER COLUMN FORMULAS ART. PAGE 133. Comparative Strength and Stiffness of Long, Ideal Columns. Condition of the Ends 230 134. Rankine's Formula. Columns of Intermediate Length 231 135. Euler's Formula. Long Columns 233 136. The Three Problems 234 137. Eccentric Loads on Columns 234 138. Behavior of Columns under Load 237 Examples 239 Problems 243 CHAPTER XV TORSION 139. Stress and Deformation. Round Shafts 246 140. The Torsion Formula. Round Shafts 247 141. Stiffness of Shafts 248 142. Other Shapes of Cross Section of Shafts 249 143. Power Transmitted by Shafts 250 144. Combined Twisting and Bending 251 Examples 251 . Problems 252 CHAPTER XVI REPEATED STRESSES, RESILIENCE, HYSTERESIS IMPACT 145. Repeated Stresses 255 146. Resilience . 256 147. Resilience of a Bar under Direct Stress 258 148. Resilience of a Beam 258 149. Mechanical Hysteresis 259 150. Lag 260 151. The Effect of Rest 260 152. Suddenly Applied Loads 261 153. Impact Loads 262 154. Drop Loads 264 Examples 264 Problems 265 XIV TABLE OF CONTENTS APPENDIX A CENTROIDS AND MOMENTS OF INERTIA OF AREAS ART. PAGE Ai 267 A 2 . Centroids of Areas 267 AS. Axis of Symmetry 268 AI. Centroid of a Triangle 268 Ag. Centroid of a Sector of a Circular Area 269 Ae. Centroids of Composite Areas 271 AT. Moment of Inertia 272 Ag. Radius of Gyration 273 Ag. Polar Moment of Inertia. The Relation between the Polar Moment of Inertia and I x and I y 273 AIQ. Relation between Moments of Inertia about Parallel Axes in the Plane of the Area 274 An. The Moment of Inertia of a Parallelogram about a Centroidai Axis in the Plane of the Area 275 Ai2. The Moment of Inertia of a Triangle about the Centroidai Axis 279 AIS. The Moment of Inertia of a Circular Area 279 AH. Moment of Inertia of Composite Areas 281 Example 282 Problems 283 Tables 287 Index 299 STRENGTH OF MATERIALS CHAPTER I MATERIALS OF CONSTRUCTION i. INTRODUCTION. Strength of Materials treats of the action of the parts or members of structures or ma- chines in resisting loads and other forces which come upon them. By the use of the principles of mechanics and the properties of materials, it determines the internal forces or stresses which are developed in the simpler forms of construction, as beams and columns, when they are subjected to loads. The properties of the engineer- ing materials are obtained through experimental tests. Many of the formulas derived in strength of materials are based on both theoretical analysis and experimental data, and the subject, therefore, is of a semi-empirical nature. In architectural and engineering construction, sta- bility, strength, durability, and economy are essential elements. The proper proportioning, spacing, and con- nection of the parts are important. Too little material in a member would make the structure unsafe, and too much would mean a waste. In general, one member should not be designed in such a way that it will be weaker than others in the structure. Proper design, then, takes into account the properties and qualities of materials and the mechanics of their action in a struc- ture in such a way as to insure safety and economy. i 2 MATERIALS OF CONSTRUCTION [CHAP. I 2. MECHANICAL AND PHYSICAL PROPERTIES. The materials of construction possess characteristic proper- ties known as mechanical and physical properties. These properties measure the fitness and ability of the material to sustain external loads or forces under given con- ditions. Different materials possess these properties in different degrees, and, of course, different grades of the same material differ in their properties. Some of these characteristic properties can be expressed quantitatively between fairly well defined limits which are determined by test, while others may be specified in terms of ability to withstand certain tests and fulfill certain require- ments. The mechanical properties include strength, elasticity, stiffness, and resilience. Other physical prop- erties frequently referred to are toughness, ductility, malleability, hardness, fusibility, and weldableness. When a load is applied to a piece or member of a structure the material undergoes a change in size and shape. If on the removal of the load the original size and shape are resumed the material is said to be elas- tic. Elasticity, then, is the property of a material by which it will regain its original size and shape on the removal of an applied load. A material which will not recover its original dimensions after deformation is termed plastic. If it will only partially recover its original di- mensions after deformation it is said to be partially elastic and partially plastic. Most constructional materials are nearly or quite perfectly elastic up to a certain limit of deformation, beyond which they are partly elastic and partly plastic. The ability to resist change in shape and size when a load is applied is termed stiffness. In elastic materials the amount of change in size and shape is generally proportional to the amount of the load applied. Materials will differ in their tensile, compressive, ART. 3] MASONRY 3 and shearing strengths. The strength of a material is ordinarily determined under the application of a static load applied in a slowly increasing amount. The effect of permanent loads, of suddenly applied loads, and of impact loads, and of the repetition of a load many times, requires separate consideration. A material possesses the property of ductility if the length can be increased and the cross section decreased considerably before rupture occurs. Toughness is that property by which a material will not rupture until it has deformed considerably under loads at or near its maximum strength. This deformation may be pro- duced by stretching, bending, twisting, etc. A tough material gives warning of failure. It will resist impact and will permit rougher treatment in the manipulations which attend fabrication and use. A brittle material will rupture without developing much deformation and without giving warning. Brittle materials are unfitted to resist shock or sudden application of load. 3. MASONRY. Masonry is mostly used to carry compression loads, such as come on foundations, walls, piers, chimneys, etc. (a) Stone Masonry. The kinds of stone that are best adapted to building and construction purposes are those that can be worked satisfactorily, can be obtained in suitable size, have great compressive strength, and are durable. Sandstone, limestone, marble, granite, trap, and slate are those in most common use. Stone masonry is laid up in mortar, and the quality and character specified will depend upon the purpose and need of the structure. The weight of stone masonry is about 1 60 pounds per cubic foot. (b) Brick Masonry. Many grades of brick are used. This great variety affords the designer opportunity for 4 MATERIALS OF CONSTRUCTION [CHAP. I selecting the kind specially adapted to his purpose. Special kilns are required for burning bricks to fulfill special requirements, such as paving brick, fire brick, pressed brick, etc. The range in the quality of brick is indicated in part by the compressive strength which varies from 400 pounds per square inch to 15,000 pounds per square inch. The strength of brick masonry depends largely upon the kind of mortar used in the joints and upon the workmanship, but it is much smaller than that of the individual brick, ranging from one-sixth to one- third as much. The weight of brick masonry is about 125 pounds per cubic foot. (c) Concrete, Plain and Reinforced. In recent years concrete has come into common use for building and structural purposes. The convenience with which it can be made into the required form, its durability, and its fireproofing qualities make it a desirable material. For foundations and for places where only compression comes on the structure the plain concrete is more gener- ally used, but where tension exists steel is embedded in the concrete to take the tension. 4. TIMBER. Timber has been used extensively for building purposes. There are many varieties and qual- ities on the market, affording good opportunity for the selection of the timber most suitable for the desired purpose. The cost of timber is gradually increasing, and some species have disappeared and others are dis- appearing from the market for structural purposes. The strength depends upon the species, the condition of growth, the seasoning, the defects in the timber, etc. 5. CAST IRON. Cast iron is a brittle metal. Its cheapness, the ease with which it is cast into special forms and machined into exact shapes, and its high ART. 7] STEEL $ compressive strength make it valuable for a great many purposes; but its low tensile strength, compared with that of other metals, and its brittleness make it an undesirable material for resisting shock or tension. Cast iron is made by smelting ore in a blast furnace. In its crude form it is called pig iron. The strength of cast iron and its other properties vary widely and depend upon the amount and condition of the carbon and other ingredients which it contains. 6. WROUGHT IRON. Wrought iron is made from pig iron in a reverberatory furnace by what is called the puddling process. The puddled balls obtained in the process are run through a squeezer and much of the cinder expelled. The material is rolled into muck bars which are cut, piled together, heated, and finally rolled into the shapes desired. The strength and other qual- ities depend upon the quality of the pig iron used, and upon the details of manufacture. Because wrought iron can be easily worked and welded it is adaptable to many uses, but its use has diminished and steel has taken its place until very little is now manufactured. 7. STEEL. The term steel covers a wide range of material, soft steel, mild steel, medium steel, hard steel, tool steel, etc., all being expressions used in con- nection with the various steels. In structural steel the element carbon is the one generally used to control strength and hardness, though other elements like phos- phorus, sulphur, and manganese exercise important in- fluence upon other properties. The best structural steel is made by the open-hearth process. In this process pig iron, together with scrap steel and some iron ore, are melted in an open-hearth furnace, the carbon, silicon, and other elements are 6 MATERIALS OF CONSTRUCTION [CHAP. I burned out, and a recarburizer is added to give the proper carbon content and to remove the iron oxide and increase the manganese, the final product being molded into ingots. Acid open hearth steel is produced in a furnace which has a siliceous lining; no reagent is added to remove the phosphorus, and hence the phosphorus content of the product is the same as that of the charge. Basic open hearth steel is produced in a furnace having a dolomitic lining (giving a basic chemical reaction), and lime is added to remove the phosphorus. In the Bessemer process, melted pig iron is placed in a Bessemer converter and, by the action of air which is blown through the charge, most of the carbon, silicon, and manganese are burned out, and a recarburizer is later added to give the proper carbon content and to remove the iron oxide and to increase the manganese, the final product being molded into ingots. As used in the United States the Bessemer converters have siliceous linings, and no phosphorus is removed. Relatively little Bessemer steel is now used for structural purposes. In the crucible process, crude wrought iron is fused with a carbon flux in a sealed air-tight vessel. The crucible process is in use for making hard steel, like tool steel, spring steel, etc. The carbon content of steel varies from less than one- tenth per cent for the softest steels to more than one and one-half per cent for the hardest carbon steels. Metals like nickel, tungsten, vanadium, etc., are also added to give special amounts of strength or hardness and produce grades of steel which have special adapta- bility for various purposes. 8. OTHER MATERIALS. Many other materials used by the engineer and architect are specially adapted to the purpose for which they are intended. Rope is made ART. 8] OTHER MATERIALS of fibrous materials such as hemp, manilla, cotton, etc., and of wire. Belting is made of leather, canvas and rubber, and of metallic links. Several alloys having copper as a basic element are made, such as phosphor bronze, brass, etc. Several kinds of artificial stone are manufactured, for most of which sand and hydraulic cement are used as the basic constituents. Metals such as lead and aluminum are also used for various purposes. Table I gives average values of the weights of various materials used in constructional work, but variation from the tabulated values is to be expected. TABLE i WEIGHTS OF VARIOUS MATERIALS USED IN CONSTRUCTION Material. Weight, lb. per cu. ft. Material. Weight, lb. per cu. ft. Timber 2S to 4X Sandstone I ZO Cast iron 450 Granite 1 70 Wrought iron . 480 Marble 1 70 Steel . 4OO Slate 1 7 C Brass ci c Terra cotta facing I IO Copper, Bronze Aluminum. . . . Brick Limestone. . . . 550 1 60 100 to 150 165 Terra cotta, fireproof- ing Book tile Concrete 50 60 150 EXAMPLE What is the weight of a solid stone masonry pier with uni- formly sloping sides and rectangular section, 4 feet by 8 feet at the top and 8 feet by 16 feet at the bottom and 20 feet high? This example is most easily worked by using the prismoidal formula to obtain the volume. This is V = - (A + 4 B + C) in 6 which V is the volume, h is the altitude, A and C are the areas of the two bases, and B is the sectional area at the middle point. MATERIALS OF CONSTRUCTION [CHAP. I For the given example, A =4X8 = 32 square feet. 5 = 6X12=72 square feet. C = 8 X 1 6 = 128 square feet. Therefore the weight of the pier is = i6o X - 2 S - (32 + 4 X 72 + 128) = 160 X i493l = 239,000 Ib. PROBLEMS 1. What is the weight of a wrought iron rod of i square inch sectional area i yard long? Ans. 10 Ib. 2. What is the weight of a hollow log 3^ feet in external diam- eter, 2 feet in internal diameter, and 16 feet long? 3. What is the weight per lineal foot of a concrete dam 4 feet high, i foot thick at the top, and 2 feet thick at the base? 4. What is the weight of a solid granite obelisk 40 feet high, i foot square at the top, and 3 feet square at the base? Ans. 29,500 Ib. 5. What is the weight of a square brick chimney 30 feet high, the inside dimensions being 2 feet at the top and 2\ feet at the bottom, and the thickness of the walls uniformly 8 inches? 6. A certain white oak log 12 feet long and 2 feet in diameter weighed 1885 pounds, what was the weight of a cubic foot of that white oak? CHAPTER II DIRECT STRESSES 9. DEFINITIONS. Force is an action of one body upon another which tends to change its shape and to produce a change of motion in the body. In this book the use of the term will generally be restricted to forces which are externally applied to the member. Stress is an internal action which is set up between the adjacent particles of a body when forces or loads are applied to the body. It is developed whenever the body undergoes a change in shape. Stress may be considered an internal force. A unit-stress is obtained by dividing the total stress by the area over which it acts if the stress is uniformly distributed. In the case of this uniform distribution the unit-stress is the amount of stress per unit of area of the sectional area. If the stress is not uniformly distributed, the unit-stress, or the intensity of stress, at a point of the sectional area is equal to the amount of stress that would be developed upon a unit of area if the stress were uniform over the area and if its intensity were the same as that at the point. Deformation is a change in a dimension of a specimen. Shortening is a decrease in the length of a specimen. Elongation is an increase in the length of a speci- men. Detrusion is a lateral deformation in which the par- ticles apparently slip past each other. It is caused by a shearing force. 9 10 DIRECT STRESSES [CHAP. II An axial load is one whose line of action coincides with the axis of the member. The axial load may be the resultant of several loads. An axial stress is one developed by an axial load. If a plane is passed perpendicular to the axis of a bar, its intersection with the bar is called the cross section, or the section, and its area the sectional area. 10. TENSION. When a load tends to pull the par- ticles of a material directly apart in the direction of the load the material is under tension and the load is a tension load. The internal stresses developed are tensile stresses. The resulting deformation is an elongation. As long as rupture does not occur, the forces acting on all, or on a part of the specimen, are in equilibrium. By the principles of theoretical mechanics it is shown that the conditions of equilibrium are that there shall be no resultant force and no resultant moment. These conditions are expressed in three fundamental equations 2F x = o, (i) 2 Fy = O, (2) 2 M = o. (3) These conditions of equilibrium are essential for deter- mining the internal stresses produced by external forces. For a homogeneous specimen in direct tension, under (a) (6) FIG. i. an axial load the stress is uniform over the entire sec- tional area A. Let Fig. 1(0) represent the member carrying the tension load P. Imagine the member cut as indicated. Fig. i(b) shows the left portion of ART. 10] TENSION II the member with the forces and stresses acting upon it indicated.* The total resisting stress is jA, where / is the tensile unit-stress developed. The resisting stress is treated as an external force in the free-body diagram, then by taking as the J^-axis the axis of the piece or member, the summation of the ^-forces gives 2 F x = fA - P = o, If the load becomes great enough to cause rupture the maximum unit-stress developed at any time before rupture is called the ultimate tensile strength. In some materials, such as wrought iron and soft steel, the load will increase to a maximum value, then decrease before rupture occurs. The ultimate strength differs for differ- ent specimens of the same material, and for purposes of design the value should be determined for each material used in the structure. The unit-stress at the point of rupture is called the rupturing strength. The rupturing strength is of no prac- tical value. For brittle materials the rupturing strength and the ultimate strength are equal. When a specimen is broken by a tension load, its final length will be greater than its original length. The ratio of the increase in length to the original length is called the ultimate elongation. For ductile materials the length of the specimen has an influence upon this ratio. So for purposes of uniformity the ultimate deformation is usually obtained for specimens of stand- ard size, either two or eight inches in gauge length. The average values of the ultimate tensile strength and of the ultimate elongation for specimens of eight- inch gauge length are given in Table 2. * Fig. i (&) is called a free-body diagram. 12 DIRECT STRESSES [CHAP. II TABLE 2 ULTIMATE TENSILE STRENGTH AND ULTIMATE ELONGATION OF MATERIALS Material. Ultimate tensile strength, Ib. per sq. in. Ultimate elongation, per cent. Timber 6,000 to 10,000 I (T Cast iron 2O,OOO 2 Wrought iron 50,000 3O.O Structural steel 6o,OOO 25 .0 to 30.0 Steel wire 60 ooo to 250 ooo 10 o to 25 o II. COMPRESSION. When a force acting on a member tends to push the particles closer together in the direc- tion of the force the member is in compression. The stresses arising are compressive stresses. For compres- sion there is a shortening. If the load is axial and is applied in such a manner that the stress developed is uniformly distributed over a section of the member, the p compressive unit-stress developed is / = -j The aver- age values of the ultimate compressive strength are given in Table 3. The table does not include values of the ultimate compressive strengths of malleable materials. Their values, however, should not be considered greater than the ultimate tensile strengths. TABLE 3 ULTIMATE COMPRESSIVE STRENGTH OF MATERIALS Material. Ultimate compressive strength, Ib. per sq. in. Timber 7.OOO Cast iron 9O,OOO Brick. . .... 6,OOO Brick masonry i, t;oo Rich concrete 2,COO Stone 10,000 ART. 13] OBLIQUE SHEAR 12. SHEAR. When external forces tend to cause two adjacent sections of a member to slip past each other the member is in shear. Stresses resisting such forces are shearing stresses. When the two shearing forces are near together the shear is considered as a simple (a) (c) FIG. 2. stress. Fig. 2 shows cases of direct shear. If a force P tends to shear a specimen along an area A the average p shearing unit-stress is s values of the ultimate strength in shear. ~j' Table 4 gives average TABLE 4 ULTIMATE SHEARING STRENGTH OF MATERIALS Material. Ultimate shearing strength, Ib. per sq. in. Timber: Along grain 400 Across grain 3,000 Cast iron 2O,OOO Wrought iron 40,000 Structural steel 50,000 Rivet steel 45,OOO 13. OBLIQUE SHEAR. Shearing stresses are developed in structural members which are subjected to direct tension or compression. Let Fig. 3 represent a speci- men under the compression force P, and imagine it cut along the plane AB. The two plane surfaces made 14 N DIRECT STRESSES [CHAP. II by the cut would slip past each other under the action of the force. This tendency of the sections to slip past each other, which gives rise to shearing stresses, always exists in members under load. To deduce the value of 0v, FIG. 3. FIG. 4. the shearing unit-stress developed along an oblique plane making an angle 6 with the axis, let Fig. 4 repre- sent the free-body diagram of one end of the specimen. The resisting stresses acting on the fibers of the plane may be resolved into their components Q and N parallel and normal to the plane respectively. Taking the X-axis along the plane there results 2 F x = P cos - Q = o, /. Q = Pcos0. The component Q parallel to the plane is the force that keeps this end from slipping past the other one, and therefore Q is the resultant of the shearing stresses, which act parallel to the plane. If A is the area of the cross ^ section of the specimen, -: - is the area of the section sin cut. The shearing unit-stress then is A P s = P cos 6 -T- - - = -r sin 6 cos 6. sm0 A This is the value of the shearing unit-stress along any oblique plane. To find the value of 6 for the maximum shearing stress developed in a specimen the relation sin 2 6 + cos 2 0=1 exists. It is shown by algebra that when the sum of two variables is constant the product of those variables is a maximum when they are equal; * * See "Higher Algebra," by John F. Downey, page 252. ART. 14] STRESS-DEFORMATION DIAGRAMS 15 therefore for the maximum shearing unit-stress sin 6 = cos 6 which is true when 6 = 45 then _ P i i P_ Sm ~ ' ~/=- * T=- ~ . ' A V2 V2 2 A 14. STRESS-DEFORMATION DIAGRAMS. Whenever a load is applied to a specimen of any material there is a corresponding deformation. A graphical representa- tion showing the values of the unit-stress developed in 60000 or Unit Deformation , 0.3 Structural Steel FIG. 5. the specimen along one axis and the corresponding values of its unit deformation along the other axis is called a stress-deformation diagram. Fig. 5 is such a diagram for a specimen of soft steel. The unit-stress p f = -r is plotted along the vertical axis and the unit g deformation e = j is plotted along the horizontal axis. i In these equations P is the total load on the specimen, A is the cross-sectional area, e is the total deformation 16 DIRECT STRESSES [CHAP. II at the load P, and / is the original length of the speci- men. In the diagram shown, the stresses measured upward from are tension and those measured down- ward are compression, and the deformations measured to the right are elongations and those measured to the left shortenings. 15. ELASTIC LIMIT AND YIELD POINT. For stresses between the two points A and A', Fig. 5, the deform- ation is proportional to the stress, while for stresses beyond these points the proportionality does not hold. These points are the elastic limits, A in tension, A' in compression. As long as the stress is between the two values corresponding to A and A', the specimen will return to its original length upon the removal of the load. If the stress becomes greater than these values, however, the length after removing the load will not be the same as before; the difference or the change in length is the permanent set. The elastic limit is the point on the stress-deformation curve where the curve departs from a straight line, or the elastic limit is that unit-stress beyond which permanent set is developed. At the point B, Fig. 5, the deformation increases markedly with but little increase in the stress. That point is the yield point. The yield point then may be defined as the unit-stress at which there is a marked increase in the deformation with but little or no increase in the stress. Table 5 gives average values of the elastic limits of wrought iron and steel as commonly determined in the laboratory. The values for tension and compres- sion are about the same. Values for timber and cast iron have not been included on account of the uncer- tainty in their determination, but when used they may be taken to be about one-third to two-thirds of the ultimate strength. ART. 1 6] MODULUS OF ELASTICITY TABLE 5 ELASTIC LIMIT OF WROUGHT IRON AND STEEL Material. Elastic limit, Ib. per sq. in. Wrought iron Structural steel Hard steel 25,000 35.000 rn OOO i6. THE MODULUS OF ELASTICITY. For values of the stress less than the elastic limit the rate at which the unit deformation increases with the increase in the unit- stress is constant,* i.e., the unit deformation is propor- tional to the unit-stress (Fig. 6). This is commonly called Hooke's Law. Then for stresses below the elastic limit the unit-stress divided by the unit deformation gives a constant. This ratio is the modulus of elasticity or the coefficient of elasticity. Young's modulus is the modulus of elasticity for direct tension or compression. FlGt 6> If is the modulus of elasticity, / the unit-stress below the elastic limit, and e the corresponding unit deform- ation the value of the modulus of elasticity is P Ae * Experiments indicate that the increase of deformation is not abso- lutely proportional to the increase in stress, but for practical purposes they may be taken as varying directly with each other. i8 DIRECT STRESSES [CHAP. II In the formula for E, t is an abstract number; conse- quently the unit for E is the same as that for /, pounds per square inch, tons per square foot, etc. In Table 6 are given average values of the modulus of elasticity in tension and compression for some materials. TABLE 6 MODULUS OF ELASTICITY Material. Modulus of elasticity, Ib. per sq. in. Timber. I,5OO,OOO Cast iron 15,000,000 Wrought iron 25,000,000 Steel 30,000,000 17. RESILIENCE. The stress-deformation diagrams show that a force acts through a distance and thus does work on the speci- men. When the load is released the specimen gives up energy stored in it. This energy a speci- men under stress is ca- pable of giving up in returning to its original dimensions is called re- silience. 18. THE SHEARING MODULUS OF ELASTICITY. Under shearing forces a specimen will undergo a detrusion. The unit de- FIG. 7. trusion is obtained by dividing the total detrusion by the length over which it occurs. The ratio of the ART. 20] REDUCTION OF AREA unit-stress developed in the specimen to the unit de- trusion is called the shearing modulus of elasticity. It is also called the modulus of transverse elasticity, and the modulus of rigidity. When a specimen is sub- ject to shearing stresses as in Fig. 7 the unit-stress can be calculated and the detrusion measured. If E 8 is the shearing modulus of elasticity, e s the unit detrusion, s the shearing unit-stress, In Table 7 are given average values of the shearing modulus of elasticity. TABLE 7 SHEARING MODULUS OF ELASTICITY Material. Shearing modulus of elasticity, Ib. per sq. in. Timber, across grain. . . . Cast iron 400,000 6,OOO,OOO Wrought iron IO,OOO,OOO Steel I2,OOO,OOO 19. POISSON'S RATIO. As the length of a specimen is increased by a tension load the lateral dimensions decrease. For a compression load the lateral dimensions increase. For stresses below the elastic limit the ratio of the lateral unit deformation to the longitudinal unit deformation is called Poisson's ratio. This ratio is considerably less than I and for most metals ranges between J and |. 20. REDUCTION OF AREA. When a specimen is rup- tured by a tension force the final sectional area is less than the original area. The ratio of the amount the 20 DIRECT STRESSES [CHAP. II section at rupture is decreased to the original area is called the reduction of area. Thus, if A i is the original area and A 2 is the final area at rupture, the reduction f . AI A 2 of area is -= --- 21. USES OF THE MODULUS OF ELASTICITY. The Pi maximum unit -stress for which the formula E = - A Ae may be used in calculations is the elastic limit. For stresses below the elastic limit E may be calculated from data observed in the laboratory; and the change in length of a specimen may be calculated by the for- Pl mula e = -j=, . The modulus of elasticity has an impor- tant application also in determining the deflection of beams and the strength of columns. 22. STRESSES USED IN DESIGN. In making a design safety and economy must be considered. Experiments indicate that at stresses slightly beyond the elastic limit there is a marked change in the structure of the material, and therefore the working stresses should not be carried beyond that value. At least there should not be noticeable permanent set. Working stresses are the allowable stresses used for designing; they should always be well below the elastic limit. The method of fixing upon values for allowable working stresses is by making a set of experiments in which the elastic limit and ultimate strength of a number of specimens are determined, and then by taking a cer- tain per cent either of the elastic limit or of the ulti- mate strength as the working stress. A knowledge of the behavior of the material under stress is essential for a proper determination of working stresses. ART. 22] WORKING STRESSES 21 TABLE 8 SAFE WORKING STRESSES IN POUNDS PER SQUARE INCH FOR STEADY LOADS Material. Tension. Shear. Compression. Bending (fiber). Perpendic- ular to grain. Parallel to grain. Timber: Cedar, white 800 600 1,000 I,2OO I,OOO 800 800 I,OOO 800 900 1,000 1,000 100 100 240 IOO 200 80 120 2OO 1 60 1 60 2OO 125 IOO IOO IOO IOO 2,500 9.500 10,000 ( 8,000 ( 10,000 1 80 1 80 300 300 340 1 80 240 800 500 500 600 240 2OO 2OO 2OO 2OO ,IOO ,IOO ,2OO ,600 ,300 ,OOO ,2OO ,800 ,4OO ,2OO ,750 ,40O I,OOO I,2OO I,2OO I,2OO I,OOO I,OOO I,2OO I,2OO I, IOO 800 1,300 1, 800 1,200 I,2OO 1,400 I,2OO I,OOO I, IOO 1,000 1,000 6,000 I 2 ,OOO l6,OOO Cypress . . Elm Fir, Washington .... Gum Hemlock Larch Maple, sugar (hard). Maple (average) Oak, red white Pine, longleaf loblolly shortleaf yellow, (Ark., etc.) 800 800 3,000 12,000 15,000 Spruce Cast iron I 2 ,OOO I 2 ,OOO 1 2 ,OOO 18,000 (Bearing) no 250 350 Wrought iron' Steel, structural rivet Brickwork (in lime) Brickwork (in Portland cement) Concrete (Portland cement) Table 8 is a modified extract from a table of allowable working stresses compiled by Ricker from building or- dinances. A few additions are given. A few changes also have been made to agree with recent building ordinances. The factor of safety is defined as the ratio of the ulti- mate strength to the working stress. This value varies 22 DIRECT STRESSES [CHAP. II for the different materials and for the kind of loading. Variable loads produce higher stresses than steady loads of the same magnitude. Suddenly applied loads and shocks produce higher stresses than variable loads of the same magnitude. Therefore the factor of safety for variable loads is usually taken about one-half greater than that for steady loads, and for sudden loads or shocks it is two or three times that for steady loads. In speci- fications and building ordinances the allowable stresses are usually given and also the tests for the materials specified. For such cases a factor of safety has been considered. PROBLEMS 1. What must be the height of a brick tower if the compressive unit-stress on the lowest brick is one-tenth of its ultimate strength? 2. Determine the shearing unit-stress tending to shear off the head of a i|-inch wrought iron bolt under a tension of 10,000 pounds, if the head is f inch deep. 3. A wrought iron plate \ inch thick requires a force of 80,000 pounds to punch a round hole | inch in diameter through it. Find the ultimate shearing strength of the plate. 4. What force is required to punch a i-inch hole in a -inch structural steel plate ? 5. In a tension test of a 0.19 per cent carbon steel specimen the diameter was 0.5 inch, and the gauge length was 1.25 inches. Each scale division on the extensometer represented inch. 125,000 P is the load in pounds and e is the reading on the exten- someter in scale divisions. The following readings were made; P 2000, e = 60; P = 6000, e = 180; P = 8000, e = 240; PROB.] DIRECT STRESSES 23 P = 8000 -|- , e = 290; P = 8100, e = 6 10. The maximum load was 15,600 pounds, the corresponding length between punch marks was 1.59 inches. The load at rupture was 12,000 pounds, and the corresponding length was 1.76 inches. The diameter at the fracture was 0.313 inch. (a) Calculate the unit-stress for each load. Ans. 30,600 Ib. per sq. in. 40,800 Ib. per sq. in. (b) Calculate the unit elongations for each load. Ans. 0.00115, 0.00154. (c) Plot the stress-deformation diagram. (d) What is the elastic limit ? Ans. 40,800 Ib. per sq. in. (e) What is the reduction of area? Ans. 60.7%. (f) What is the modulus of elasticity ? Ans. 26,600,000 Ib. per sq. in. 6. A wrought iron rod 2 inches square and 10 feet long length- ened 0.02 inch by suspending a load from its lower end. Determine the load. - 7. How much will a roo-ft. steel tape, | inch wide and $V inch thick, stretch under a pull of 40 pounds ? Ans. 0.16 in. 8. A vertical wooden bar 50 feet long and 6 inches square carries a load of 18,000 pounds at its lower end. Find the unit- stress at the upper end and the elongation of the bar due to the combined weight of bar and load. 9. Determine the elongation of a i-inch wrought iron rod 10 feet long, under a tensile load of 20,000 pounds. 10. How many ^-inch square rods of strong steel would be required for the suspension of a platform loaded with 15 tons, if the stretching of the rods is limited to one-half their elon- gation at the elastic limit? Each rod carries equal shares of the load. 11. What shearing load will a rivet inch in diameter safely carry ? A rivet f inch in diameter ? 12. What should be the depth of the head of a bolt f inch in diameter to carry safely the shear ? 13. What must be the bearing area to carry safely a load of 20,000 pounds on a Washington fir beam ? 14. What should be the sectional area of a steel member (a x) DIRECT STRESSES [CHAP. II of the truss shown in Fig. 8 ? If (a y) is a short compression longleaf pine member, what should be its section ? 15. Member (a y), Fig. 9, is a steel rod i inches in diameter stressed to its safe working stress. What should be the corre- sponding sectional area of the short white cedar compression members (a x) ? FIG. 8 FIG. 9. 16. Design a cast iron washer for the member shown in Fig. 10 17. What must be the cross section of specimens of the follow- ing materials in order that the unit-stress may be about one-third the elastic limit under a load of 30,000 pounds in tension? (a) Wrought iron; (b) Steel. 1 8. In a compression test of a 6-inch concrete cube the load was 107,000 pounds at rupture. What was the shearing unit- stress along a plane inclined 30 to the axis? What was the maximum shearing stress developed ? Ans. 1290 Ib. per sq. in. 1490 Ib. per sq. in. 19. What will be the elongation at the elastic limit and at rupture of an 8-inch specimen of the following materials ? (a) Wrought iron; (b) Structural steel; (c) Hard steel. 20. What should be the sectional area of a steel rod if it is to take a tension load of 70,000 pounds ? 21. If a cast iron specimen 1X2 inches in sectional area breaks under a tensile load of 42,000 pounds, what load will probably break a cast iron rod 2 inches in diameter? CHAPTER III DIRECT STRESSES APPLICATIONS 23. SIMPLE CASES OF DIRECT STRESSES. The sim- plest cases of direct stresses are such as exist in eyebars, belts, ropes, cables, tension members in trusses, etc. For such members the unit-stress developed is obtained by dividing the load the member carries by the sectional area of the member. There are other cases for which the stresses developed are practically direct stresses although the line of action of the load may not be along the axis of the member that carries the load. Such cases will be considered in this chapter. 24. STRESSES IN THIN CYLINDERS. When a thin cylinder is under interior pressure, as a steam boiler, water pipe, etc., the forces tending to burst the cylinder act normally to the inside surface, Fig. n. These forces develop internal tensile stresses in the metal of the cylinder. In order to determine the magnitude of these stresses, imagine the plane AB, Fig. n, passed perpendicular to the page and containing the axis of the cylinder. The portion ABD is in equilibrium under the forces and stresses acting upon it, and if that half of the cylinder were filled with some solid substance the interior forces acting upon it would be normal to the plane AB\ the resisting stresses also would be normal to that plane. The internal stresses actually developed in the cylinder are the same as would be developed under the imaginary condition. Fig. 12 is a free- body diagram of the part ABD. Let r be the radius, t the thickness, 25 26 DIRECT STRESSES APPLICATIONS [CHAP. Ill and / the length of cylinder, Q the interior pressure per unit of area, and / the resisting unit-stress, which is approximately uniform over the resisting area. Then the force tending to rupture the cylinder is 2 Qlr, and the FIG. ii. resisting stress on one side is /// and on both sides is 2 ftl. Then for equilibrium 2 Qlr = 2 ftl, .' Qr=fl. This formula is true for thin cylinders. 25. STRESSES IN A HOOP. If hoops are heated and then shrunk onto cylinders, the unit-stress can be ob- tained by the application of the formula for the modulus PI fl of elasticity, E = -r- = - if the difference between the Ac 6 normal diameter of the hoop when cool and the one to which it is shrunk can be obtained. The effect is the same as if the hoop is stretched from its normal diam- eter di to the diameter d of the hoop when shrunk on the cylinder. The difference in the length of the hoop will ART. 26] TEMPERATURE STRESSES be TT (d di) and the unit elongation will be TT (d di) d d\ d d\ . _ . -T- irdi = -j or -j approximately, ror steel tires CL i CL a common rule is to make -, - about Actually d 1500 the final diameter d of the hoop and cylinder will be slightly less than the original diameter of the cylinder, as the metal of the cylinder will deform under the pres- sure due to the hoop. The amount of this deformation will depend upon the ratio of the moduli of elasticity of the materials of the cylinder and the hoop, and upon the thickness of each. 26. STRESSES DUE TO CHANGE IN TEMPERATURE. When a material is heated it will expand if free, and when cooled it will contract. If / is the change in temperature and c is the coefficient of expansion, or change per unit of length for one degree rise or fall, the change per unit of length will be e = ct. If the TABLE 9 COEFFICIENTS OF EXPANSION PER DEGREE FAHR. Material. Coefficient of expansion. Masonry Cast iron Wrought iron Steel . 0000050 .0000062 .0000067 .0000065 member is brought back to its original length by an external force the unit-stress developed will be/ = eE = ctE. If, instead of being allowed to change in length and then being brought back to its original length when a change in temperature takes place, the specimen is rigidly held in its original position, a unit-stress of / = ctE will be developed. The effect is the same as if 28 DIRECT STRESSES APPLICATIONS [CHAP. Ill the specimen were allowed to change in length and then were brought back to its original length by an external force. Table 9 gives the values of the coefficient of expansion for each degree of change in temperature Fahr. 27. STRESSES IN THIN SPHERES. Internal pressure in domes or other thin spheres tends to cause rupture around a circumference, (Fig. 13). By using the same nomenclature as for thin cylinders, the force tending to push off the dome is Qirr 2 and the stress resisting this force is/27iT/, .'. Qr = 2ft. This formula also applies when an interior pressure acts upon a cylinder head. FIG. 13. FIG. 14. 28. THICK CYLINDERS UNDER INTERIOR PRESSURE. If the metal is thick in comparison with the diameter of the pipe the stresses developed are not direct stresses and are not uniform throughout the thickness of the metal of the wall of the cylinder, and the formulas of the previous paragraphs cannot be used for such cases, (Fig. 14). Various expressions for the value of the ART. 29] CYLINDERS EXTERIOR PRESSURE 29 maximum stress developed have been deduced. For the case when there are longitudinal and transverse stresses due to the interior pressure Lame* gives the formula for the maximum tensile stress developed, which comes on the inner surface of the pipe, where r\ is the internal radius and r 2 is the external radius, f = Q W + n 2 ) ' Calvarino* gives tor the same case, f= Birnie* gives the formula for thick guns when no longi- tudinal stress is developed, Any of these formulas may be used to investigate or design thick pipes or guns. 29. CYLINDERS UNDER EXTERIOR PRESSURE. Recent experiments on the collapse of tubes under exterior pressure indicate that for a length of tube greater than six times its diameter the rupturing pressure is indepen- dent of the length. The formulas deduced by Carman and Carr in the University of Illinois Engineering Exper- iment Station Bulletin No. 5 are as follows: for thin tubes where -. is less than 0.025, And for thick tubes or ^ greater than 0.03, Q-K'-C. (2) * See "Strength of Materials," by A. Morley. 30 DIRECT STRESSES APPLICATIONS [CHAP. Ill In these formulas / is the thickness of the walls, d is the outside diameter, Q is the external pressure in pounds per square inch causing collapse, and K, K', and C are experimental constants. Table 10 gives the values of these constants. TABLE 10 VALUES OP K, K'.AND C FOR PIPES UNDER EXTERIOR PRESSURE Material. K. K'. C. Cold-drawn seamless steel Brass 50,200,000 25,150,000 95.5 2 Q? ?6< 2,090 2 4.74. Lap- welded steel 83,270 I O2 tj EXAMPLES 1. What internal pressure will probably rupture a cast iroh pipe 8 inches in diameter and inch thick ? @4 = 20,000 X J, Q = 1250 Ib. per sq. in. 2. If a steel rail of sectional area 8.8 square inches is subjected to a drop in temperature of 50 Fahr. and is prevented from shortening, what is the force exerted upon it if the initial force was zero? The unit-stress developed is .0000065 X 50 X 30.000,000 = 9750 Ib. per sq. in. in tension. The total force is 8.8 X 9750 = 85,800 Ib. PROB.] DIRECT STRESSES APPLICATIONS 31 PROBLEMS 1. What is the maximum tensile unit-stress in a pipe 24 inches in diameter, the plate being f inch thick, and the internal pressure 80 pounds per square inch? Ans. 2560 Ib. per sq. in. 2. What internal pressure will rupture a 1 2-inch steel pipe | inch thick? Ans. 3750 Ib. per sq. in. 3. What should be the thickness of the lower plates of a steel stand-pipe 20 feet in diameter carrying a water pressure of 80 pounds per square inch? Use a unit-stress of 16,000 pounds per square inch and the efficiency of the joint 75 per cent. Ans. About 0.8 in. 4. What stress is developed in a spherical steam dome 10 inches in diameter, \ inch thick, under a steam pressure of 120 pounds per square inch ? 5. What external pressure will cause a 2-inch cold-drawn steel tube | inch thick to collapse ? 6. What internal pressure will burst a wrought iron cylinder of 48 inches inside diameter and f inch thickness? 7. Determine the thickness of a wrought iron steam pipe 18 inches inside diameter to resist a pressure of 200 pounds per square inch with an allowable stress of 6000 pounds per square inch. 8. What should be the minimum thickness of a cast iron sphere 12 inches inside diameter to withstand safely a steady internal pressure of 200 pounds per square inch ? 9. What internal pressure will burst a cast iron sphere 24 inches inside diameter and ^ inch thick? 10. A short wrought iron bar i inches in sectional area has its ends fixed immovably between two walls with no stress when the temperature is 50 Fahr. What pressure will be exerted on the walls when the temperature is 1 20 Fahr. ? 11. Steel railroad rails, each 30 feet long, are laid at a temper- ature of 40 Fahr. What space must be left between them in order that their ends shall just meet when the temperature is 100 Fahr. ? If the rails had been laid with their ends in contact, what would be the unit-stress in them at 100 Fahr. ? 12. A wrought iron tie rod 20 feet in length and 2 inches in diameter is screwed up to a tension of 10,000 pounds in order to tie together two walls of a building. Find the stress in the rod when the temperature falls 30 Fahr, Also when it rises 20 Fahr. CHAPTER IV RIVETED JOINTS 30. RIVETED JOINTS. In pipes, tanks, and boilers made of rolled plates, the plates are usually connected by rivets, and stress is transmitted from one plate to the other through the rivets. Such joints may be called boiler, tank, or pipe joints. Connections of bridge members, and connections between the members of roof trusses, columns, beams, etc., are also made by means of rivets. Such joints may be called structural joints. Wherever pieces of metal are connected by rivets the design should give the most efficient and economical connection consistent with the given con- ditions. A joint will fail at its weakest part, and the most efficient design will have all parts of the joint of equal strength. Although the actual stresses developed in a riveted joint may be complex, the usual method is to simplify the calculation by assuming conditions giving direct tensile, compressive, and shearing stresses. As such the stresses are easily computed. The treatment of boiler joints and of structural joints is essentially the same. 31. KINDS OF RIVETED JOINTS. Riveted joints may also be classified according to the method of connecting the plates and the number of rows of rivets used. In Fig. 15 are shown two styles of lap joints; the main plates overlap each other and are connected by the 32 ART. 32] FAILURE OF RIVETED JOINTS 33 rivets. Fig. 15 (a) and (b) represent a single riveted lap joint, and (c) and (d) represent a lap joint with two rows of rivets, with the rivets staggered. Fig. 18 shows two styles of butt joint; the edges of the main plate almost or wholly butt against each other, and the connection is made through cover plates. F.ig. 18 (b) shows a butt joint with a single cover plate, and Fig. 1 8 A H \iii_- J \ G H \ J B (a) .p pH (C) (6) FIG. 15. (c) shows one with a double cover plate. There are other styles of riveted joints, but the general method of treat- ment is the same for all kinds. 32. METHODS OF FAILURE OF RIVETED JOINTS. The three principal ways in which riveted joints may fail are (i) (for tension loads) by tension in the plates along AB, Fig. 15; (2) by crushing of the rivets along CD or EF\ (3) by shear of the rivets along ED. Besides these, failure may occur, (4), by shearing of the plate along GH and //; (5) by bending of the rivets; (6) by bending of the plates, thus allowing the rivet heads to shear off or the rivets to fail in tension ; (7) by failure of the plate 34 RIVETED JOINTS [CHAP. IV in tension along KL and MN for staggered rivets. Failure by shear in the plates, number (4), is avoided by making the lap GH large enough to insure safety. A rule sometimes followed is to make the lap one and one-half times the diameter of the rivet. Bending of the rivets, and of the plates, numbers (5) and (6), may occur in a single- rive ted lap joint. Failure number (7) (6) - ) ( ( - - - - - (d) FIG. 16. is avoided by making the distance between the rows of rivets large enough so that KL + MN is somewhat greater than QR. Fig. 1 6 indicates the various ways in which a riveted joint may fail. (a) shows a failure due to tensile stress in the plate. shows a failure due to bearing stress in the rivet. shows a failure due to shearing stress in the rivet. shows a failure due to the shearing stress in the plate between the edge of the plate and the rivet hole. (e) shows a failure by tension along a staggered line. (b) (c) (d) 33. COMPUTATION OF UNIT-STRESSES DEVELOPED IN RIVETED JOINTS. The calculation of the unit- stresses developed in a riveted joint is made by assuming that the stress is uniformly distributed over the particular area which is in tension, compression, or shear, and ART. 34] SINGLE-RIVETED LAP JOINT 35 p hence that the unit- stress is / = -r , where P is the total A load coming on the area A which resists tension, compres- sion, or shear. In order to determine the tensile stress developed, the area of the section subjected to tension should be obtained. To determine the compressive or bearing stress, the area subjected to bearing should be found. For bearing it is assumed that the stress on one rivet is uniformly distributed over an area equal to that obtained by multiplying the thickness of the plate by the diameter of the rivet. This area is the projection of the rivet on the thickness of the plate. If the entire stress transmitted by a rivet is taken by a section of the rivet at one face of the plate the rivet is said to be in single shear, and the resisting area is equal to the cross-sectional area of the rivet. If the stress transmitted is taken by sections at two faces of a plate the rivet is said to be in double shear, and the resisting area is equal to twice the cross-sectional area of the rivet. The method of calculating the stresses will be given for a few cases. 34. SINGLE-RIVETED LAP JOINT. Let Fig. 17 repre- sent a single- rive ted lap joint in which the load P is to be transmitted from one plate to the other by n rivets in one row. The load must be transmitted by tension through the plate past the row of rivets. The greatest tensile unit-stress developed will come along the section AB. Let ft be the tensile unit-stress de- veloped, t the thickness of the plate, d r the diameter of the rivet hole, and b the width of the plate. Then the smallest area in tension taking the load is t (b nd'), p and the unit-stress is f t = . ,-, JTV / (b nd) and P=f t t(b-nd'). (i) RIVETED JOINTS [CHAP. IV The load brings compression on the side of the rivet as shown in Fig. 17 (b). If d is the diameter of the rivet, the total area upon which the load is assumed to be distributed in compression or bearing is ntd, and the p bearing unit-stress developed is /. = -? and P = f c ntd. (2) The load tends to shear the rivets along the plane CD between the two plates. If f a is the shearing unit- D (a) (6) (c* FIG. 17. stress, and A the sectional area of each rivet, the total resisting area in shear is nA, and the shearing unit- p stress is f a = j and P=f s nA. (3) By use of the three formulas just developed the unit- stresses existing in a single- rive ted lap joint under a given load can be calculated, or the load a given joint will carry can be determined, or a joint can be designed to carry a given load. If the plates connected are of different thickness the smaller value of t should be used in the formulas. 35. DOUBLE-RIVETED LAP JOINT. The equations representing the relation between the load transmitted ART. 36] LAP JOINT 37 and the unit-stress developed for a double- rive ted lap joint are similar to those given for the lap joint with a single row of rivets. For this case let n be the number of rivets in one row, and HI the total number of rivets, then the formulas become P=f t t(b-nd') 9 (i') P=f e nitd, (2') P = f 8 n 1 A. (3') (a) (b) (c) FIG. 18. 36. LAP JOINT WITH MORE THAN Two Rows OF RIVETS. If there are more than two rows of rivets the assumption is generally made that the load is distributed evenly among the rivets, and the load each rivet carries is obtained by dividing the total load by the number of rivets carrying the load. With the same nomen- clature as in the last article the same formulas hold for 38 RIVETED JOINTS [CHAP. IV this case (n would usually be the number of rivets in the outside row). 37. BUTT JOINT. Fig. 18 (b) shows a butt joint with a single cover plate, for which kind of joint the formulas of Art. 34 may be applied, if n refers to the number of rivets on one side of the seam. The similarity between this joint and a lap joint is readily seen by considering the cover plate with one side of the main plate. For a butt joint with two cover plates, using the same nomenclature as in Art. 35, the formulas become P = f t t(b-nd'), (i") P = f c n l d, (2"} P = f s n l2 A. (3") Two sections of each rivet are brought into shear. 38. COMPRESSION LOADS FOR RIVETED JOINTS. For the foregoing analyses the loads have been considered in tension. If the member is a compression member the formulas for shear and for compression in the rivets will remain unchanged; the formula for compression in the rivet need not ordinarily be considered. 39. EFFICIENCY OF RIVETED JOINTS. The efficiency of a riveted joint is the ratio of the strength of the joint to the strength of the unpunched plate. In figuring the strength working stresses are usually used in the formulas to determine the load P. When a joint is de- signed, its strength under tension in the plate calculated by formulas (i) Art. 34, 35, and 37, the bearing strength of the rivets calculated by formulas (2), and the shearing strength of the rivets calculated by formulas (3) should be obtained and the smallest value of the load P taken as the strength of the joint; for if a greater load is put on the joint one of the safe stresses would be exceeded. ART. 40] DESIGN OF RIVETED JOINTS 39 This value divided by the working strength of the un- punched plate is the efficiency. In boiler lap joints, for the pitch p, which is the distance from center to center of adjacent rivets in one row, the strength of the unpunched plate is f t tp. Therefore, the efficiency for tension is p-d') = p-d' fttp _ P _ The efficiency for compression in the rivets is f c ntd = f c nd UP ftp ' And the efficiency for shear in the rivets is = ' _f.nA *' ~ UP The actual efficiency of the joint is the smallest one of the above values. Similar expressions for the various other types of riveted joints may be deduced. Table II gives the range of values for efficiency for the types of boiler joints listed for ordinary design. TABLE ii EFFICIENCY OF JOINTS Kind of joint. Efficiency, per cent. Single-riveted lap joint 50-65 Double-riveted lap joint 65-75 Single-riveted butt joint 65-75 Double-riveted butt joint 70-80 Triple- rive ted joints are frequently used and high effi- ciencies are obtained. 40. DESIGN OF RIVETED JOINTS. For use with ordi- nary thickness of plates in structural work f-inch and f-inch rivets are the prevailing sizes. For light work 40 RIVETED JOINTS [CHAP. IV ^-inch and f-inch rivets are used. In specially heavy sections larger rivets are used, ij-inch rivets being occasionally used. The pitch of the rivets to be used in a design depends upon the kind of joint used and the purpose of the joint. For tanks and boilers the joint must be tight as well as strong; therefore the spacing should be small. For structural members strength is the main object to be accomplished. The diameter of the rivet hole is somewhat larger than the diameter of the rivet. In boiler joints the diameter of the rivet hole is usually assumed equal to or T V inch larger than that of the rivet, and in structural joints it is assumed f inch larger. The ideal joint for strength and economy would be the one that would be of equal strength in tension in the plate, bearing in the rivets, and shear in the rivets. For this to be the case the three values of the allowable resisting stresses as calculated by the three formulas, Art. 34, 35, and 37, would be the same; therefore, for single- riveted lap joints, f t t(b-nd') = f e ntd,' and f c ntd =f s nA, and f t t(b-nd') = f s nA. Similar equations can be written for the other types of riveted joints. * In practice it is not usually necessary or practicable to make such ideal joints, and the resulting efficiency will be somewhat less than that of the ideal joint. Limitations of the size of rivet, conditions for tight- ness of joints, convenience for shop work, and many other items may prevent making joints of equal strength in tension, bearing, and shear, and small variations from ART. 40] DESIGN OF RIVETED JOINTS 41 the ideal conditions will not materially decrease the efficiency. For the design of boiler, tank, or pipe joints the follow- ing procedure is a convenient one and is recommended. (a) Decide upon the working stresses for tension, shear, and bearing, and calculate f, :f t , a.ndf c :f t . (b) Select the type of joint to suit the conditions. (c) In the first calculations assume the efficiency, calculate the necessary thickness, and then select a commercial thickness of plate. (d) Determine the limiting size of rivet for shear or bearing. The general limiting size for any style of joint may well be expressed in terms of the thick- ness of the plate. For example, for f a 'ft = 2:3 and fc'-ft = 3'- 2 in the design of riveted lap joints any diameter of rivet less than 2.87 / will not involve a question of bearing strength. Also in ordinary butt joints having double cover plates, the bearing strength of the rivet will not need consideration if the diameter of the rivet does not exceed 1 .43 t. (e) Select a working size of rivet within this limit. (f) With / and d determined calculate the pitch by equating the strength in tension and the strength in shear or bearing, using shear or bearing according to which controls the strength for the type of joint used. Only in special or unusual types of joints will tension and shear govern. (g) Calculate the efficiency of the joint and the stresses in the rivets and plate to see that the working stresses are not exceeded. Practice is not uniform in regard to the values of the allowable unit-stresses to be used in design. The following values may be used in solving problems in this course: /, equals 8000 pounds per square inch, ft equals 12,000 pounds per square inch, and f c equals 42 RIVETED JOINTS [CHAP. IV 18,000 pounds per square inch. The resulting ratios are ~ = ~ = - The assumption made for the bearing Jt Jc 3 area of the rivets is only approximate and experiments show high values for the ultimate crushing strength figured on that basis ; therefore a high value for the crush- ing unit-stress can be assumed. EXAMPLES i. Select two channels for the lower chord of a truss in which the maximum stress is 49,300 pounds in tension. Also determine the number of f-inch rivets, if |-inch gusset plates are used, for the connection. Use ft = 15,000 pounds per square inch,/ c = 18,000 pounds per square inch, and / = 8000 pounds per square inch. The net sectional area required is A t = 49,300-:- 15,000 = 3. 29 square inches. By use of a hand book, the section is found first. Try two 2-inch by 2-inch by ^-inch angles. The gross area is A g = 2 X 1.75 = 3.5 square inches. By counting the diameter of the hole -inch larger than that of the rivet, the effective tension area will be reduced by the amount Ai= 2 X f X \ = .875 square inch. The effective area then would be A e = A g Ai = 3.500 - .875 = 2.625 square inches. This is too small. Try two 3^-inch by 3-inch by A-inch angles. A g = 2 X 1.94 = 3.88 square inches. Ai = 2XjiXr5 = .547 square inch. The effective area then is A e = A g - Ai = 3.88 - .55 = 3.33 square inches. This is a little in excess of the required area; therefore use two 3 1-inch by 3-inch by A-inch angles. To determine the number of rivets necessary for bearing, the EXAM.] RIVETED JOINTS 43 required bearing area is A c = = 2.74 square inches. The 18,000 greatest bearing stress will be developed between the gusset plate and the rivet, since the thickness of the two angles is greater than that of the gusset plate. The bearing area for one rivet is dt = f X f = .281 square inch. Therefore, the number of rivets required for bearing is 2.74 n c = ~ = 10 rivets. .281 To determine the number of rivets necessary for shear, the required area for shear is A 8 - = 6.16 square inches. 8000 Each rivet is in double shear and the total area of each rivet in shear is 2 A = 2 X .442 = .884 square inch. Therefore, the number of rivets required for shear is 6.16 n 8 = - = 7 rivets. .004 Since bearing requires 10 rivets that number must be used. The shearing stress then is below the allowable. 2. Calculate the unit-stresses developed in a triple-riveted lap joint of a boiler 4 feet in diameter carrying no pounds per square inch pressure, if the pitch is 3 inches, the thickness of the plate % inch, and the diameter of the rivets f inch. What is the effi- ciency of the joint ? The load transmitted through the three rivets in the pitch length of 3 inches is D QDp no X48 X 3 P = - - = 7920 pounds. The tension area carrying the load is found by assuming the diam- eter of the rivet hole the same as that of the rivet. *./<= = - = 7040 pounds per square inch. At 1.125 44 RIVETED JOINTS [CHAP. IV The bearing area is Ac = 3/d = 3X^Xf = 1.125 square inches. .*. f c = -zi- = 7040 pounds per square inch. The shearing area is A 8 = 3 X .442 = 1.326 square inches. .*. fs = =5970 pounds per square inch. 1.326 By using for the allowable unit-stresses /. = 8000 pounds per square inch, ft = 12,000 pounds per square inch, and/ c = 18,000 pounds per square inch, the efficiency of the joint can be calculated. The load the unpunched plate would carry in the pitch length is P = 12,000 X I X 3 = 18,000 pounds. The load the punched plate will safely carry in the pitch length is Pt 12,000 X 1.125 = I 3.5oo pounds. The load the rivets will carry in compression is Pb = 18,000 X 1.125 = 20,250 pounds. The load the rivets will carry in shear is P a = 8000 X 1.326 = 10,600 pounds. The allowable load will be the least of these three values, which is 10,600 pounds, and the efficiency is 10,600 = e a = - - = 59 per cent. 10,000 The efficiencies for tension and bearing are higher. A larger rivet would give a higher efficiency. PROBLEMS i. A column bracket consists of a 6-inch X 6-inch X f-inch angle, and is riveted to the column, which is a 1 2-inch, 30-pound channel, whose thickness is 0.513 inch. It carries a load of 20,000 pounds and is riveted to the column with 5 rivets I inch PROB.] RIVETED JOINTS 45 in diameter. Determine the unit-stresses developed in bearing and shear. 2. If two plates 4 inches wide and f inch thick are connected by four f-inch rivets in two rows, what load will the joint safely carry ? 3. Determine the required number of rivets for a joint to carry 20,000 pounds, using -inch plates. What is the efficiency of the joint ? 4. Determine the required number of rivets for a joint to carry 25,000 pounds, using f-inch plates. What is the efficiency of the joint ? 5. Design an angle bracket to be riveted to a column which consists of two 1 2-inch, 3o-pound channels latticed together with the channel flanges extending outwards. The bracket is to support one end of a simple beam which carries a total uniform load of 60,000 pounds. Use rivets | inch in diameter. Neglect bending in outstanding leg of the bracket angle. 6. Design a splice to connect two plates 10 inches wide and f inch thick which are subjected to a tension of 78,000 pounds. Use two splice plates and rivets which are | inch in diameter. 7. Select two angles to carry a tension load of 24,500 pounds and determine the number of rivets necessary if f-inch gusset plates are used. Use the allowable stresses given in example i. Ans. 2 2 in. X 2 in. X A in. A n = 5. 8. Two f-inch plates are connected by three f-inch rivets. What tension load will the joint safely carry ? If two of the rivets are in one row what should be the width of the plates? Ans. 10,600 Ib. 4.1 in. 9. In a butt joint with a double cover plate the main plates are | inch thick and the cover plates are each ^ inch thick. Design the joint to take a lension load of 20 tons. If the load is in com- pression how will the design be changed ? 10. In a boiler 60 inches in diameter carrying steam pressure at 120 pounds per square inch the plate is inch thick, the rivets are inch in diameter and the pitch is 2\ inches. The joint is a double-riveted lap joint. What are the tensile, shearing, and the compressive unit-stresses coming in the joint? What is the efficiency of the joint ? 46 RIVETED JOINTS [CHAP. IV 11. Determine the efficiency of a double-riveted lap joint where / = ^ inch, p = 3^ inches, and the diameter of the rivet is ^| inch. 12. Determine the efficiency of a single-riveted, two-strap butt joint, if t = f inch, p = 2 inches, and the diameter of the rivet is $ inch. 13. Determine the pitch for a double- riveted, two-strap butt joint in which t = \ inch, and the diameter of the rivets is \\ inch, so that the strength of the joint against tearing the plates between the rivet holes shall equal the compressive strength of the rivets. What is the efficiency of this joint? 14. Find the thickness of plates for a boiler shell 8 feet in diameter to carry a pressure of 160 pounds per square inch, if the efficiency of the joint is 80 per cent, and the stress in the plates is 5 tons per square inch. 15. Determine the efficiency of a single- riveted lap joint if t = \ inch, d = | inch, and p = 2 & inches. 16. Calculate the efficiency of a double-riveted lap joint if / = ^ inch, d = i inch, and p = 3 f inches. 17. Determine the pitch for a single-riveted, two-strap butt joint in which / = f inch and d = i ^ inches, so that the strength of the joints against tearing the plates between the rivet holes shall equal the compressive strength of the rivets. Determine also the efficiency of the joint. 18. Design a double-riveted, two-strap butt joint for ^-inch plates and find its efficiency. CHAPTER V BEAMS EXTERNAL FLEXURAL FORCES 41. DEFINITIONS. Flexure occurs in a member when the load has a component normal to the axis of the member which causes the member to bend. A beam is a bar subjected to flexure. Usually the applied forces are normal to the axis of the beam, as when a horizontal bar resting on supports at its ends sustains vertical loads along its length. However, the term is also applicable when the direction of the applied forces is not at right angles to the axis. The loads on a beam cause it to bend and thus produce internal stresses which resist the bending. These stresses are called flexural stresses. The curve assumed by the axis of the beam under load is the elastic curve. The following treatment considers the beam to be horizontal and the loads vertical, but with slight modifications it may be adapted to beams in any position and with loads in any direction. The -XT-axis will be taken to coincide with the axis of the beam before bending. The F-axis will be taken at right angles to the X-axis through the left support or left end. A cantilever beam is one which has one end free and the other end fixed in such a manner that the tan- gent to the elastic curve at the fixed end remains hori- zontal. The elastic curve and the beam itself may be spoken of as being horizontal at the fixed end. A simple beam is one which rests upon two end supports. 47 48 BEAMS EXTERNAL FLEXURAL FORCES [CHAP. V An overhanging beam has one or both of its supports, away from the ends of the beam. A continuous beam is one that rests, on more than two- supports. The end of a beam is said to be fixed if it is restrained in such a manner that the elastic curve remains hori- zontal when the load is applied. It follows from the FIG. 19. Concentrated load. definitions above that a cantilever beam has one sup- port, a simple beam has two supports, and a continuous beam has more than two supports. 42. METHODS OF LOADING BEAMS. According to the distribution of the loads, a beam may carry concentrated, uniform, and nonuniform or varying loads. When the load is transmitted to the beam through a comparatively small area it is said to be concentrated. Fig. 19 shows a concentrated load at the center of the span. If the load is distributed evenly over the beam it is a uniform ART. 43] FORCES ACTING ON BEAMS load. Fig, 20 shows a load that Is practically uniform. If the load is distributed over the beam and is not of the same intensity throughout, it is said to be non- uniform or varying. According to the method of application, loads are said to be dead or live loads. A dead load is one that the member always supports, such as its weight or FIG. 20. Uniform load. (Loaned by the Leonard Construction Company, Chicago.) loads due to the weight of other portions of the struc- ture of which the member is a part. Live loads are those that come upon the member temporarily, such as a train passing over a bridge, a crowd of people assembled in an auditorium, or machinery or a stock of goods. The loads shown in Fig. 19 and 20 are live loads. 43. FORCES ACTING ON A BEAM AS A WHOLE. The external forces acting on a beam are in equilibrium. The loads supported by the beam are usually known. 50 BEAMS EXTERNAL FLEXURAL FORCES [CHAP. V The forces supporting the beam the reactions may not be known at the start, but when possible should be determined before making other calculations. The loads and the reactions of the supports form a system of parallel forces. From theoretical mechanics the con- ditions of equilibrium for such a system are that there shall be no resultant force and no resultant moment. These conditions are expressed in two equations: 2^ = 0, (i) S M = o. (2) These formulas are used in determining the reactions. ILLUSTRATIVE EXAMPLE Let it be required to determine the reactions on a beam 8 feet long, carrying a uniform load of 4000 pounds and a concentrated load of 6000 pounds 3 feet from the left support. (See Fig. 21.) 6000* -3- 4000* -8'- FlG. 21. Let Ri be the reaction of the left support and R* the reaction of the right support, then S F = Ri + R* 4000 6000 = o, Ri + Rz = 10,000 pounds, 2 M a = Rz X 8 - 4000 X 4 - 6000 X 3 = o, R 2 = 34000 = 4250 pounds, and Ri = 10,000 4250 = 5750 pounds. As a check take moments about the right reaction, 2 M b = 6000 X 5 + 4000 X 4 - Ri X 8 = o, Ki = 5750 pounds. The line of action of the distributed load is taken through its center of gravity. ART. 44] EXTERNAL FORCES INTERNAL STRESSES $1 44. FORCES ACTING ON A PORTION OF A BEAM. INTERNAL STRESSES. In Fig. 22 is shown a beam with its loads. An effect of loading the beam is to produce internal stresses in the beam at all sections. To study the nature of these stresses imagine the beam to be cut along the vertical section AB. Then in order that equilibrium in the left portion of the beam be main- tained with the external loads and reactions acting upon it, forces such as Vi, Hi, and H z must be supplied. ? w 2 T r T R l T I' : (a) HI >H 2 RI E F ; (b) IG. 22. It is evident that before the section was cut internal stresses must have existed at this section, which acted upon the left portion of the beam. For present purposes these stresses may be considered to be replaced by Vi, HI, and Hz which will later be found to be the resultants of the internal stresses. All the forces shown in Fig. 22 (b) are external with respect to the left por- tion, but when the whole beam is considered, FI, Hi, and H 2 are replaced by internal stresses. The reaction and the loads on the left portion of the beam tend to 52 BEAMS EXTERNAL FLEXURAL FORCES [CHAP. V cause motion upward (or downward) and rotation in the clockwise direction about an axis in the section. Both of these tendencies are counteracted by the internal stresses at this section, or by their resultants, FI, HI, and H 2 . The right portion of the beam could be treated in a similar manner, but the direction of FI, HI, and H 2 would be opposite to that shown for the left portion, and the magnitude would be the same. The system of forces acting to one side of the sec- tion of the beam is coplanar, nonparallel, nonconcurrent. The equations of equilibrium are, 2 F y = o, (i) 2/^ = 0, (2) 2 M = o. (3) To satisfy equation (i) the vertical resisting force FI in Fig. 22 (b) must act downward (or upward). To satisfy equation (2) HI must equal H 2 . And to satisfy equation (3) the resisting forces must produce an anti- clockwise moment, hence there must be compression in the top fibers of the simple beam shown. In other cases, as in a cantilever beam, tension may exist in the top and compression in the bottom of the beam. 45. VERTICAL SHEAR. From Art. 44 it is seen that the external forces to the left of the section tend to cause the left portion of the beam to slip upward (or downward) past the portion on the right. Whenever this is the case vertical shear is said to exist at the section. Ver- tical shear is the force that tends to move the left por- tion of a beam past the right portion or to cut the beam along a vertical plane. The magnitude of this force is measured by the summation of the vertical forces acting on the beam to the left of the section. (The vertical forces acting on the beam to the right of the section could be used as well.) ART. 47] VERTICAL SHEAR 53 (i) -1- If V represents the vertical shear at a section the distance x from the left support, in which 2 RL represents the sum of all the reactions to the left of the section, and 2 WL is the sum of all the loads to the left of the section. The magnitude of V is equal to that of the resultant of all the external forces acting on the left portion of the beam. 46. SIGN AND UNIT OF VERTICAL SHEAR. The sign of the vertical shear depends upon the relative values of the reactions and the loads to the left of the section. If 2 RL is the greater, the IA 1 sign is plus, in which case the left portion tends to slip in the upward or positive di- rection past the right portion. The unit of vertical shear is the same as that used for force, and is usually the pound. 47. THE VALUES OF THE VERTICAL SHEAR AT THE SECTION AB, DISTANT & FROM THE LEFT END OR ORIGIN FOR CANTILEVER AND SIMPLE BEAMS, (i) For a cantilever beam with a concentrated load W at the end (Fig. 23 (i)), (2) I 1 1 R,=-j (3) R,-*1 ^ *k R<-{ B ^ Ri= M FIG. 23. 54 BEAMS EXTERNAL FLEXURAL FORCES [CHAP. V (2) For a cantilever beam with a uniform load of w pounds per inch of length (Fig. 23 (2)), V = wx. (3) For a simple beam with a concentrated load W in the center of the span (Fig. 23 (3)), W V = to the left of the center 2 or V = to the right of the center. (4) For a simple beam with a uniform load of w pounds per inch of length (Fig. 23 (4)), T7 l V = wx. 2 ILLUSTRATIVE EXAMPLES 1. If a cantilever beam of 8-ft. span carries a load of 500 pounds at the free end, the value of the vertical shear at any section is 500 pounds. The left reaction is zero and the load to the left is 500 pounds; therefore V = o 500 = 500 pounds. 2. For a simple beam of lo-ft. span carrying a uniform load of 6000 pounds, the left reaction is 3000 pounds. If x is ex- pressed in inches w = 6000 -f- 120 = 50 pounds per inch and the expression for the vertical shear is V = 3000 50 x. By substituting values of x in this equation we obtain values of the vertical shear for the sections considered; thus at 25 inches, V = 3000 1250 = 1750 pounds. At the distance 6 feet or 72 inches from the left support the vertical shear is V ="3000 3600 = 600 pounds. 48. LOAD AND SHEAR DIAGRAMS. It is convenient and useful to indicate the value of the load and the shear at all points along the beam by means of vertical dis- tances measured from a horizontal axis. These verti- ART. 48] LOAD AND SHEAR DIAGRAMS 55 cal distances are called ordinates. With the length of the beam and the loading known, the axis OX, Fig. 24, can be drawn to scale to represent the length of the beam. Thus, if the beam is 12 feet or 144 inches long, and, if it is convenient to make the axis OX 3 inches long, the scale of the length would be I inch equals 48 inches. From the axis in Fig. 24 at any point A a FIG. 24. length AB can be erected perpendicular to OX to repre- sent the intensity of load at that point. If the load at that point is 240 pounds per foot of length or 20 pounds per inch of length, the ordinate AB can be made I inch, in which case I inch would represent a load of 100 pounds per inch. In the same way the value of the intensity of load at all other points along the beam can be repre- sented by ordinates. The continuous curve connecting the ends of these ordinates is called the load curve, and the whole figure the load diagram, because the inten- sities of the load are shown at every point along the beam. The locations of the concentrated loads are indicated by arrows. Positive values are measured to the right of and up from the origin 0, negative to the left of and down from the origin O. Loads act down and consequently are negative. Knowing the loading, the reactions can be calculated and the values of the vertical shear can be obtained for all sections along the beam by the formula V = 56 BEAMS EXTERNAL FLEXURAL FORCES [CHAP. V By a procedure similar to that used in making the load diagram the values of the vertical shear at all points along the length of the beam can be indicated by ordi- nates from a reference axis OX. Thus, Fig. 25 is the shear diagram for the loading shown in Fig. 24. In the shear dia- gram the distance from the axis OX to the curve at any point A' shows the value A'B' of the shearing force for that section of the FIG. 25. beam. (a) 49. RELATION BETWEEN THE LOAD AND THE SHEAR. For distributed loads there is a definite relation between the load and the shear. To deduce this relation let Fig. 26 (a) be the load diagram and Fig. 26 (b) the shear diagram for a given case. Let u be a small length (called an element of length) measured along the OX axis, and let w be the average load per unit of length over this portion. Let x be the distance from the left support or origin, and V the verti- FlG - 26 - cal shear at the left end of the element of length. The load over this length is wu, and is equal to the small shaded area in the load diagram. Then the difference ART. 51] THE LOAD AND SHEAR DIAGRAMS 57 in the vertical shear at the two sides of the small length is v = wu. 50. THE RATE OF CHANGE OF THE VERTICAL SHEAR. The rate at which the vertical shear changes is equal to the amount of change divided by the length in which the change is made, and is wu = w. u may be taken so small that A C in Fig. 26 (b) approaches a straight line. When u is made indefinitely small AC coincides with the tangent to the shear curve. It is also seen that - equals tan ACB, which represents the slope of the shear curve at the given section when u is indefinitely small, w is the intensity of the load at that section. Therefore, the rate of change of the vertical shear at any sec- tion of a beam equals the R intensity of the load at that section. It is also represented by the slope of the shear curve at that section. 51. RELATION BE- TWEEN THE LOAD AND SHEAR DIAGRAMS. The relation given in Art. 49 affords a convenient graphical method of de- termining the change in the vertical shear between any two sections AB and CD of a beam, Fig. 27. Divide the length AC into several FIG. 27. 58 BEAMS EXTERNAL FLEXURAL FORCES [CHAP. V small lengths. Then v\ equals area ABn, v z equals area 1 122, va equals area 2233, etc. The total change v r be- tween the two sections is the summation of all the w's between them. This summation is equal to the area under the load curve between the sections, which is the total load between the two sections. Therefore, the change in the vertical shear between any two sections of a beam equals the area under the load curve between the two sections. (Distributed loads.) 52. BENDING MOMENT. Moment of a force is the product of the force and the perpendicular distance from the line of action of the force to the origin of moments. The moment measures the tendency of the force to cause the ob- ject acted upon by the force to turn about an axis FIG. 28. ,, 1^1 r through the origin of mo- ments. Fig. 28 indicates a force F acting upon an ob- ject which it tends to turn about the point 0. The moment in this case is Fa where a is the perpendicular distance from the origin to the line of action of the force. For a section of a beam the bending moment is the sum of the moments of all external forces acting on the beam to one side of the section about an axis in the section. A result of a bending moment is a tendency to cause the portion of the beam considered to rotate about an axis in the section. In determining the bending moment, the portion of the beam to the right or the one to the left of the section may be used with the same results. It is the common convention to use the left portion of the beam and this convention will be followed in this book. The bending moment at any section then is equal ART. 53] SIGN AND UNIT OF BENDING MOMENT 59 to the summation of the moments of all the forces acting upon the beam to the left of the section about an axis in the section. Since the external forces are T 1 Y 2 ^ Y 3 Y 4 made up of the reac- tions and loads, the bending moment M - equals the moments R i of the reactions minus the moments of the loads to the left of the section, or expressed in a formula (see Fig. 29), M = 2RLX- 2W(x-p), in which %RLX is the summation of the moments of the reactions, and 2W (x p) is the summation of the moments of the loads to the left of the section. 53. SIGN AND UNIT OF BENDING MOMENT. When there is compression in the top fibers of the beam at a section the bending moment is positive at the section; when there is tension in the top fibers the sign of the bending moment is negative. The radius of curvature of the elastic curve is positive for a positive bending moment and negative for a negative bending moment. The sign 'of the bending moment due to a force is posi- tive when the force itself would produce compression in the top fibers. Hence in all cases the bending moment of a reaction is positive, and that of a load is negative. The unit in which the bending moment is measured will depend upon the units of force and length employed. The pound-inch* is in most common use for beams and will be employed here. * The term " inch-pound," which is also used for the unit of moment, does not make a distinction between the unit of work and the unit of moment. 60 BEAMS EXTERNAL FLEXURAL FORCES [CHAP. V A r,_W R rir IB (i) -I- 54. THE VALUES OF THE BENDING MOMENT AT THE SECTION AB, DISTANT a? FROM THE LEFT END OR ORIGIN FOR CANTILEVER AND SIMPLE BEAMS. (1) Cantilever beam, concentrated load W at the end M = -Wx. Left reaction is zero (Fig. 30 (i)). (2) Cantilever beam, uniformly distributed load of w pounds per inch ir X ^ M = wx- = w 2 2 The left reaction is zero. The load to the left of the section is wx and its (2) W arm is - (Fig. 30 (2)). (3) Simple beam, con- centrated load Wat center. W The left reaction is 2 x for the left half, 2 (3) FIG. 30. M = and = Wl Wx 2 2 for the right half (Fig. 30(3)). ART. 55] BENDING-MOMENT DIAGRAMS 61 (4) Simple beam, uni- form load of w pounds per inch. The left reaction is 2 ,, wl x M = x wx - 2 2 _ wlx wx 2 2 2 The load to the left of the section is wx and its arm is (Fig. 30 (4)). M FIG. 30. ILLUSTRATIVE EXAMPLES 1. For a cantilever beam of g-it. span carrying a load of 2000 pounds at the free end the bending moment at the section distant x from the free end is M = 2000 x. If x is 50 inches, M = 2000 X 50 = 100,000 Ib.-in. 2. For a simple beam of 2o-ft. span with a concentrated load of 10,000 pounds at the center the left reaction is 5000 pounds and the bending moment at 8 feet or 96 inches is M ge = 5000 X 96 = 480,000 Ib.-in. At] 1 6 feet the bending moment is Mm = 5o X 192 10,000 X 72 = 240,000 Ib.-in. 3. For a simple beam of i2-ft. span carrying a uniform load of 100 pounds per inch Rt = 7200 pounds and the bending moment at the center is IPO X 72 X 72 M 72 = 7200 X 72 - - - = 259,200 Ib.-in. 55. BENDING-MOMENT DIAGRAMS. The bending- moment diagram shows the value of the bending mo- ment at all points along the beam. Fig. 31 (d) is such a diagram. OX represents to scale the length of the beam, and the ordinate M represents the bending 62 BEAMS EXTERNAL FLEXURAL FORCES [CHAP. V moment at the section AB. The values of the bending moment may be calculated by the formula A C B 6 (d) FIG. 31. or by the method of the following article. 56. RELATION BE- TWEEN THE VERTICAL SHEAR AND THE BEND- ING MOMENT. At a section distant x from the left support the bending moment is M, and at a section distant x + u it is M + m, where u is an element of length along the OX- axis and m is the difference in the bending moment at the two sections . I n the free- body diagram of the element of length of the beam between the two sections shown in Fig. 31 (a) M is the bending moment of the external forces to the left of the section AB. This mo- ment is transferred to the element of the beam, and the bending moment trans- ferred from this element to the right portion is M + m. The increase m is due to the external forces acting on the small portion of the beam. The external forces with respect to this portion acting upon it are V the vertical shear at the left, wu the weight, where w is the weight ART. 57] CHANGE OF THE BENDING MOMENT 63 per unit of length, and V + v the vertical shear at the right, as shown in the diagram. Taking moments about an axis in the section CD, there results u u z M + m = M -\-Vu wu- = M + Vu w ; m = Vu w 2 u may be taken so small that the load over this elemen- tary length may be considered uniform. From Fig. 31 (c) it is seen that Vu is the area of the rectangle EFHI u 2 and w is that of the triangle EGI. Therefore, m = EFHI EGI, which is equal to the area between the axis and the shear curve. If the sections are taken far apart the distance between the two sec- tions should be divided into a large number of elementary lengths, and the total change in the bending moment will equal the summation of all the elementary changes. This leads to the conclusion that the change in the bend- ing moment between two sections equals the area under the shear curve between the two sections. Having the vertical shear diagram drawn and knowing the value of the bending moment for any section of the beam, that for any other section may be obtained by getting the area under the shear curve between the two sections and adding it to the known moment. Areas above the axis are positive and those below are negative. 57. THE RATE OF CHANGE OF THE BENDING MOMENT. From the equation for the change of the bending moment u? between two sections, m = Vu w the expression for the rate at which the bending moment changes at any section may be deduced by allowing u to become 64 BEAMS EXTERNAL FLEXURAL FORCES [CHAP. V so small that the two sections AB and CD will be con- secutive ; then w is so small compared to Vu that it may be considered equal to zero. Therefore the rate of change is m _ Vu _ w u That is, the rate of change of the bending moment at any section equals the vertical shear at that section. 58. THE MAXIMUM VERTICAL SHEAR AND BENDING MOMENT. The greatest shearing stress will be at the section for which the vertical shear is the greatest and the greatest tensile and compressive moment stresses will be where the bending moment is the greatest. In any kind of beam the greatest shear occurs just to one side of a support. Because beams usually fail at the sec- tion of maximum bending moment, that section is called the danger section. From the relation existing between the shear area and the bending moment there is found a simple method of locating the danger sec- tion. By Art. 56 the area under the shear curve be- tween two sections repre- sents the change in the bending moment between those two sections. As we go along the beam the (d) bending moment increases FlG as long as the shear area is positive, and when the area becomes negative the moment grows less (Fig. 32). Shear area above the axis is positive and that below is ART. 59] LOAD, SHEAR, AND MOMENT negative, the sign of the shear changing where the shear curve crosses the axis. Therefore, the maximum bending moment in a beam occurs where the shear curve crosses the axis, i.e., where the vertical shear is zero. This point may be obtained by plotting the shear curve. It may also be obtained from the equation of the vertical shear for the portion of the beam in which the shear passes through zero, by equating] V to zero and solving for x. Note that the equation representing V must be for the portion of the beam in which the shear actually does pass through zero. Whenever the shear passes through zero at a concentrated load, the maximum bend- ing moment will be under that load. Since the bending moment is zero at both supports of a simple beam the shear area above the axis is equal to that below the axis. 59. LOAD, SHEAR, AND MOMENT DIAGRAMS FOR CANTILEVER AND SIMPLE BEAMS. MAXIMUM SHEAR AND MOMENT. For all cases let x be the dis- tance from the origin to the section AB con- sidered, W the total load on the beam, / the span, V m the maximum shear, and M m the max- imum moment. (i) For a cantilever beam with a load at ol the end (Fig. 33), M = -Wx. The area in shear dia- gram is negative and 66 BEAMS EXTERNAL FLEXURAL FORCES [CHAP. V equals Wx. The maximum moment occurs at the wall and equals the entire area in the shear diagram which is M m = - Wl. (2) For a cantilever beam with a uniform load (Fig. 34) . The load per unit of length is W F = -y*, V m =-W, which comes at the wall. t _ r2 __Wtf 2 s ' T 2" This is equal to the shear area to the left of the section. ART. 59] ' LOAD, SHEAR, AND MOMENT 6 7 The maximum moment comes at the wall and equals the entire area under the shear curve and is 2 2 (3) For a simple beam with a concentrated load at the center of the span (Fig. 35) , w V=to the left of the load, W V = to the right of the load, 68 BEAMS EXTERNAL FLEXURAL FORCES [CHAP. V V m = w w M = x to the left of the load W w/ l \ x- W(x - -j = ^ - x) to the right of the load. The shear passes through zero under the load; therefore the maximum moment occurs at the center of the beam. The area under the shear curve to the left of the center is W l = Wl 22 4 ' ,, Wl M m =' 4 (4) For a simple beam with a uniform load (Fig. x 36), the load per unit of W length is w = =- = wl = W 2 2 wl W W V = wx = j-Xj 2 2 I v - m 2 The average ordinate in the shear area to the sec- tion at the distance x from the left end is wl wl _ wx _ ART. 59] LOAD, SHEAR, AND MOMENT 69 therefore the area under the curve to the section equals wlx wx 2 M 2 2 wlx wx 2 Wx 2 2 2 2 / The shear passes through zero at the center of the span ; therefore the maximum moment occurs at that section and equals the shear area to the left of the center. . m ~222~8 8 (5) For a simple beam with a concentrated load at JM FIG. 37. any point distant kl from the left support, in which k is less than i (Fig. 37), Ri= W(i -k), V = W(i - k) to the left of load, V = - Wk to the right of load, M = W(i - k) x to the left of load, M= W(i - k)x -W(x-kl) = Wk(l- x) to the right of load. 70 BEAMS EXTERNAL FLEXURAL FORCES [CHAP. V The shear passes through zero under the load. The area to the load is (6) A uniform and a concentrated load on a simple beam (Fig. 38). For cases of this kind the shear will be a maximum at one of the supports and will pass through zero under the concentrated load or between that load and the center of the beam. The point at which it passes through zero de- x pends on the ratio of the loads and the posi- tion of the concentrated load. The shear curve should be plotted to obtain that point, then the area under that curve above the axis calculated for the maxi- mum bending moment. (7) For several concentrated loads the maximum shear occurs at one of the reactions and it passes through zero at one of the loads (Fig. 39). The point of zero shear may be obtained by plotting the shear curve. Then the maximum moment may be obtained by cal- culating the area in the shear diagram above the axis. (8) For the case of several concentrated loads and a uniform load the shear may pass through zero at one of the concentrated loads or between any two of them, the position of zero shear depending upon the rela- tive values of the loads and their positions (Fig. 40). ART. 59] LOAD, SHEAR, AND MOMENT LJ 1 (a) (b) (c) FIG. 39. r r W vf/in. (a) (6) (c) FIG. 40. 72 BEAMS EXTERNAL FLEXURAL FORCES [CHAP. V The shear diagram should be plotted to locate the section of zero shear. To find the exact location when the sec- tion of zero shear falls between two loads the shear formula for a section between these two loads may be equated _to zero and the corresponding value of x deter- mined. Then the moment equation applied for that point will give the value of the maximum moment, or the area in the shear diagram above the axis may be obtained for the maximum moment. (9) For the case of a beam overhanging one or both supports the general principles hold (Fig. 41). The w, (a) (6) |W 3 IR, maximum shear generally is not equal to a reaction, but it may be obtained by the application of the general equation for shear. It will come just to one side of a support. The shear passes through zero at the supports, and also between them, consequently the maximum ART. 60] CANTILEVER AND SIMPLE BEAMS 73 bending moment may occur at any one of the three points. The moment must be obtained for all three and the greatest one taken for the maximum. 60. RELATIVE STRENGTH OF CANTILEVER AND SIMPLE BEAMS. The shearing unit-stresses developed in a beam are directly proportional to the vertical shear, and the unit-stresses resisting the bending moment are directly proportional to the bending moment; therefore the shearing strength of a certain type of beam is in- versely proportional to the maximum vertical shear developed in that type, and the strength to resist bend- ing is inversely proportional to the maximum bending moment developed. Table 12 gives the values of the maximum vertical shear and the maximum bending mo- ment, and the relative strengths in shear and bending for simple and cantilever beams. TABLE 12 RELATIVE STRENGTHS IN SHEAR AND BENDING Kind of beam. Maximum vertical shear. Maximum bending moment Relative strength in shear. Relative strength in bending. 1 I TT/ TT// there are three vari- c / ables: V, s, and A in the former and M, /, and- in the G latter. Any one of the three variables in each equation may be determined if the other two are known. This gives rise to three problems that may be investigated by the use of the shear and moment formulas. PROBLEM I. Investigation of Beams. Given a beam with its load, to calculate the maximum unit-stresses. 94 BEAMS INTERNAL FLEXURAL STRESSES [CHAR VI By the principles developed in Chapter V the values of the maximum vertical shear and the bending moment may be calculated. A and - may be determined from c the dimensions of the cross section of the beam, by the methods of Appendix A. To obtain the maximum y shearing stress, s = -7 T- may be used, and to obtain the ,,, , Me fiber stress,/ = -y- may be used. ILLUSTRATIVE EXAMPLE Calculate the maximum shearing stress and the maximum fiber stress developed in a longleaf pine beam of lo-ft. span, breadth 4 inches, depth 8 inches, when carrying a concentrated load of 12,000 pounds 4 feet from the left support. Ri = 720 pounds, V m = 720 pounds, S m = \ (720 -T- 32) = 33. 7 pounds per square inch, M m = 720 X 48 = 34, 560 pound-inches, ^4X8X8X8^512. 12 3 c = 4 inches, ... f = 34,56o X 4 X 3 = glo pounds per square inch< PROBLEM II. Safe Loads for Beams. By the use of the shear and moment formulas, the load which a given beam will safely carry may be obtained. M = gives c the value of the maximum allowable bending moment, from which values the load may be selected. After de- termining the load by use of the moment formula, the beam should be investigated for the maximum shearing stress developed by that load by use of the shear formula V= ksA. The allowable shearing stress or bending mo- ment as calculated should not be exceeded. ART. 72] THE THREE PROBLEMS 95 ILLUSTRATIVE EXAMPLE What uniform load will a lo-inch, 25-pound I-beam carry when used as a simple beam of i6-ft. span with an allowable fiber stress of 16,000 pounds per square inch? From Table 20 which gives values of section factors of I-sections, - = 24.4 inches 3 . c The maximum allowable resisting moment is 16,000 X 24.4 = 390,400 pound-inches. The maximum bending moment for the uniformly Wl distributed load on a simple beam occurs at the center and is o W X 192 - = 390,400, o W = 1 6, 2 70 pounds. To get the approximate shearing stress, the area of the web is 10 X 0.31 = 3.1 square inches. /. s = (16,270 -5-2)^- 3.1 = 2630 pounds per square inch, which is safe. PROBLEM III. The Design of Beams. The loading of a beam and the maximum allowable stress may be specified, to select or design a beam to carry the load safely. The design of beams is the problem most generally met with by the engineer and architect. It admits of many solutions, and the designer must use his best judgment in choosing the form and size of the cross section to be used. The material most remote from the neutral axis is all that is stressed to the maximum, while that at the neutral axis has no fiber stress in it. The material is most efficiently used when the largest pro- portion of it is stressed nearly to the maximum stress, and obviously this condition exists in a section having a large part of the material well away from the neutral axis. -Necessarily there must also be such a distribu- tion of the material as will insure safety against shear, buckling, and twisting. In steel I-beams there is a large portion of the material near the outside fiber, and yet the web is large enough to resist the shear. 96 BEAMS INTERNAL FLEXURAL STRESSES [CHAP. VI Generally, rupture is due to bending rather than to shear, and occurs at the danger section. In the deter- mination of the safe loads for beams and the design of beams, the moment formula M = is the governing c y formula, the shear formula 5 = T-J being used after- ward to investigate the beam for the shearing unit- stress. If the shearing unit-stress developed is too large another design must be made, but this is seldom neces- sary, except for short deep beams. ILLUSTRATIVE EXAMPLE Design a square loblolly pine cantilever beam for a span of 8 feet with a concentrated load of 500 pounds at the free end. From the table of allowable stresses (No. 8) / = 1000 pounds per square inch. M m = 500 X 96 = - 48,000 pound-inches. (The nega- tive sign may be neglected, as that simply means that the stress in the top fibers is tension.) If b is the breadth of the section, I - = -, c 6' 73 .*. 48,000 = 1000 X > 6 b 3 = 288, b = 6. 6 1 inches. The maximum shearing unit-stress for this load is s = 3jf = 3 x 5oo = lb . 2 A 2 43.7 which is safe. While the above beam satisfies the condition of the problem it is not a standard section and probably would be replaced by a 6-inch by 8-inch beam, for which case the fiber stress is = 48,000X4X12 = 6X8X8X8 The maximum shearing stress is s = - X ^ = 15.6 lb. per sq. in. 2 48 ART. 74] MAXIMUM STRESS DIAGRAMS 97 73. MODULUS OF RUPTURE. The moment formula M = is not applicable to beams of material for which the stress is not proportional to the deformation, or for non-homogeneous beams, or for beams under stresses greater than the elastic limit of the material. However, it is frequently used to calculate a nominal unit-stress developed in a beam when the bending moment is great enough to cause failure. The unit- stress thus calculated is called the modulus of rupture. This usually lies between the ultimate compressive strength and the ulti- mate tensile strength of the material. Table 13 gives average values for the modulus of rupture. TABLE 13 MODULUS OF RUPTURE Material. Modulus of rupture Ib. per sq. in. Timber 7000 to oooo Cast iron 35.000 74. MAXIMUM STRESS DIAGRAMS. The value of the maximum shearing unit-stress for a section can be obtained by dividing the vertical shear for the section by kA, the product of the sectional area and a constant k depending upon the shape of the section. The value of the maximum fiber stress for the section can be ob- tained by dividing the bending moment at that section by -, the section modulus. If the shear and moment c diagrams are drawn, the values of V and M may be taken directly from the diagrams; thus, in Fig. 53 (b), CD divided by kA (assumed constant) gives the shear- ing unit-stress. The stress is always proportional to 98 BEAMS INTERNAL FLEXURAL STRESSES [CHAP. VI the ordinate CD. in Fig. 53 (c) is the bending moment at the section which divided by - gives the maximum fiber stress, which comes in the extreme fiber of the beam at that section. The fiber stress is pro- (a) portional to the ordinate of the moment curve. Thus for a beam of uniform section the ordinates to the vertical shear curve represent the maximum shearing stresses, and the ordinates to the bending moment curve represent the maximum fiber stresses existing in the beam. Consequently the shear and moment diagrams may be considered stress diagrams. The above reason- ing is for beams of constant section, but with modifica- tions similar reasoning may be applied to beams in which the section is not uniform. EXAM.] BEAMS INTERNAL FLEXURAL STRESSES 99 75. BEAMS OF UNIFORM STRENGTH. If a beam is of uniform section the maximum fiber stress occurs only in the outside fiber at the section of the greatest bending moment. The stress varies with the bending moment along the length of the beam. In order to have the most efficient beam all the material in it should be stressed to the allowable stress, and to approach this state, besides keeping the material near the outside surface, the cross section is sometimes made to vary with the bending moment. This is done in plate girders where extra cover plates are added toward the center of the span. EXAMPLES i. If a 4-inch by 6-inch by o. 4-inch channel is used as a simple beam of 8-ft. span with a concentrated load of 2000 pounds three feet from the left support, (a) what is the maximum fiber stress developed? (b) What is the stress developed on a fiber 2 inches from the top and 2 feet from the right support? RI = 1 2 50 pounds. (a) The maximum moment is under the load and is M m = 1250X3X12= 45,000 pound-inches. The centroid is 1.29 inches from the back (see Fig. A 6 , Appen- dix A). Therefore c = 4 1.29 = 2.71 inches, A =5.28 square inches, / = 8.41 inches 4 , , Me 45,000 X 2.71 f =: __ = ___ = I4?soo lb per sq m> (b) Mn = 1250 X -8 X 12 - 2000 x 3 X 12 = 18,000 Ib.-in. y = 2.71 2 = .71 inch. f 18.000 X 0.71 fv = = 1520 lb. per sq. in. 100 BEAMS INTERNAL FLEXURAL STRESSES [CHAP. VI 2. What uniform load will a 4-inch by 6-inch yellow pine timber safely carry when used as a simple beam of lo-ft. span? w; / 12 bd? 1X6x6 - C = T = T = - - 2 The allowable stress is 1000 pounds per square inch. (Table 8), = looo X 24 = 24,000 pound-inches, c which is the allowable resisting moment. The maximum bending moment is Wl W X 10 X 12 = - - = 24,000, o o W = 1600 pounds. The load per lineal foot is ^ a = 160 pounds per foot. Inves- tigating for the shearing stress, s = f X -V? = 5 pounds per square inch. 3. Using the results of Example No. 2, what is the total com- pressive stress at a section 30 inches from the left end, and where is the line of action of the resultant ? Ri = 800 pounds. MM = 800 X 30 - -W X 30 X 15 = 18,000 Ib.-in. The area in compression is 12 square inches. The stress on the centroid of this area is , 18,000 X i.<5 fv = - = 375 ib. per sq. in. The total compressive stress is 12X375 = 4500 pounds. The line of action of the resultant is I the depth from the top or i inch from the top or 2 inches above the neutral surface. 4. Design a simple cypress timber beam to carry a uniform load of 8000 pounds on a span of 1 2 feet. ^.w/. 8000x144. o o PROB.] BEAMS INTERNAL FLEXUfcAE- &TR2SSE& \ ^M The beam must be large enough to take this moment without exceeding the allowable stress which is 1000 pounds per square inch. (Table 8), M 144,000 . , , 7 _ = _ _ = IAA inches 3 = - / 1000 C An indefinite number of cross sections will satisfy this, but the one chosen should be economical and suited for the purpose. It must be wide enough to prevent lateral bending. If 6 is assumed equal to - then 7 bd? d* - = = = 144, -Q O & +* . o o 8 . O Q OTJ ** 0^3 d Its y ~ +J OJ i2^ S Ill ^J 3 d 1 Simple Beams 29 JQ 10 12 2,OOO ? o o o o o o ? 10" 25# 21 14 O 28,000 7 o 23 14 o ? 7 o 9" 2I# 33 16 1,000 10,000 8 o o 24. 16 ? 20,000 8 o o IS" 9Q# ?c 16 2,000 ? 8 o o 15" 8s# 36 18 o I 2 ,OOO 4 7,000 9 27 18 o 9,000 3 9 7" 20f 38 39 40 20 20 24 1,000 1,000 ? 20,000 ? 12,000 6 8 o o o o o 12" 3 i# 15" 8o# 41 2O 1,000 5,000 6 6,000 IO ? 42 2O ? 4,000 8 5,000 12 12" ioo# Cantilever Beams 43 . . 8 4,OOO o o o o ? 44 IO ? o o o o 10" 25# 45 12 O 10,000 12 O o 46 12 2,OOO 10,000 12 O o ? CHAPTER VII w STRESSES IN SUCH STRUCTURES AS CHIMNEYS, DAMS, WALLS, AND PIERS 76. KINDS OF STRESSES. For structures that sus- tain a side thrust and a direct weight, as chimneys, dams, etc., there is a combination of direct and flexural stresses. The treatment given in this chapter is based upon the as- sumption that the direct and flexural stresses act independ- ently of each other, and that the side thrust does not cause ap- preciable deflection. It is also assumed that the stress is pro- portional to the deformation and that the material of the struc- ture is elastic. For bearing on soil this assumption may be only approximately true. When there is no side thrust and the vertical section is sym- metrical, the total weight above a horizontal section is resisted by direct compressive stresses on the sections and the unit- W stress is/ = -j (Fig. 54 (a)). A normal force P produces flexural 106 /A Ti mmm (a) w ART. 77] ECCENTRIC LOADS ON SHORT PRISMS 107 stresses. A shearing stress p of s = T-T acts along the sec- Ks\. tion. The bending moment due to the load increases the com- pression on the opposite side from the force and decreases it on the nearer side (Fig. 54 (6)). This lateral pressure P may be due to the wind, water, embank- ments, etc. It may not be horizontal, in which case the horizontal and vertical compon- ents of the resultant of the weights and lateral forces should be taken as producing the flex- ure! stresses and the direct stresses respectively. 77. ECCENTRIC LOADS ON SHORT PRISMS. Let the load W have the eccentricity e (Fig. 55 (a)). This load may be re- placed by its components, W, Wi, and Wz shown in Fig. 55 (6), Wi and Wz being taken as acting along the axis of the prism and equal in magnitude to W. The two equal and opposite forces W and W 2 form a couple, the moment of which is We. The ef- fect of this couple is to produce a bending or moment stress. In Fig. 55 (c) W and W 2 are replaced by the equivalent moment C. 108 STRESSES IN CHIMNEYS, DAMS, WALLS [CHAP. VII The stresses developed by the eccentric load W will be the same as those produced by its components which are the axial load W and the couple We. The stress due to the axial load is the same at all points of the sec- W tional area and is -j- . The stress developed by the moment of the couple may be considered as a fiber j Me Wee , . , ,. stress and is equal to -y- = j , where c is the distance of the most remote fiber from the centroidal axis of the section, and / is the moment of inertia of the section about that axis. If r is the radius of gyration, / = Ar 2 . Fig. 55 (d) is a free-body diagram showing the stresses developed by the eccentric load. The maximum stress developed is in the fiber most remote from the centroidal axis on the side nearer the load while the minimum stress is developed in the most remote fiber on the opposite side. The maximum stress equals the sum of the direct and moment stresses, and the minimum stress equals the difference. Therefore, W . Wee W f *=A+ = A W Wee W Then, the stress developed in the outside fiber is 78. ECCENTRICITY OF A LOAD THAT WILL PRODUCE ZERO STRESS IN THE OUTSIDE FIBER. If the eccentric- ity is increased /2 becomes greater and/i becomes smaller. After a certain point is passed /i reverses if the material will take tension. If W is a compression load, just before the tensile stresses act, /i becomes zero. To ART. 78] ECCENTRICITY 109 obtain the eccentricity e\ that will make j\ zero, equate /i to zero. Then W where e\ is the greatest eccentricity the load may have before tension is produced on the side away from the compression load. For a rectangle the eccentricity to give zero stress in the outside fiber will now be found, / bd* d 2 d r ~7 ~n = an d c = - , A 12 bd 12 2 # 12 _ d ** Cl = 7 " 6 ' Therefore, as long as a compression load is kept on the middle third of a solid rectangular prism the stress over the entire area will be compression. For a circle, 2 6 4 & A d r 2 = -j; = and c = - > Trfir lo 2 4 * 16 _ d ei ~~d~s' 2 Therefore, as long as the load is kept on the middle quarter of a solid circular prism all the stress will be compression. 110 STRESSES IN CHIMNEYS, DAMS, WALLS [CHAP. VII 79- THE KERN. If the line of action of the load falls within a certain part of the cross section of a prism, all stresses in the section will be of one kind; and if the line of action of the load falls outside of that area the stresses will be partly tension and partly com- pression. This area is called the kern or kernel. In a solid circular prism the kern is a circular area whose diameter is one- fourth the diameter of the prism. And for a rectangular prism the kern is a diamond-shaped fig- ure whose diagonals are one-third the lateral di- f FIG. 56. mensions of the prism. (See Fig. 56.) 80. CASE OF ECCENTRIC LOADS CAUSED BY A COM- BINATION OF THE WEIGHT OF THE MATERIAL AND LATERAL PRESSURE. Call the weight of the material above the section AB being considered, W, and call the lateral pressure on the prism above the section P, Fig. 57- The magnitude and direction of the resultant, R, of these two forces depend upon the forces. Its line of action passes through the intersection of W and P and intersects the section AB in some point C usually not at the centroid, thus producing an eccentric load on the section AB. The eccentricity e is DC and can be calculated by taking moments about the centroidal axis at D. Resolve R into its vertical and horizontal com- ponents, Y and H, at C where it intersects the cross sec- tion for which the stresses are to be found. (Fig. 57 ART. 81] RESULTANT OUTSIDE OF THE KERN III H produces a shearing unit-stress along the plane of TT magnitude 5 = T-T , and F is an eccentric load producing the stress f-ifr.*59 in which the plus sign is used to obtain the maximum compressive stress, /2, and the minus sign is used to obtain the minimum stress, /i. 81. EFFECT WHEN THE LINE OF ACTION OF THE RE- SULTANT FALLS OUTSIDE OF THE KERN. If the result- ant of all loads above a section of a prism under compression falls outside the kern, the minimum stress 112 STRESSES IN CHIMNEYS, DAMS, WALLS [CHAP. VII 1 /i = f I A becomes negative or tension if the material will take tension, Fig. 58. For materials that will not take tension, such as masonry, the joints on the side opposite the eccentric load will tend to open, but failure will not necessarily follow. If the structure is subject to water pressure the water may get into the cracks and produce an upward pres- sure which tends to help overturn the structure. In chimneys and walls where there is no upward pressure due to water getting into the cracks if they do form, the tendency for them to form is not FIG. 58. so objectionable. The safe limit for the compressive stress should not be ex- ceeded by the maximum stress developed. 82. THE MAXIMUM STRESS WHEN THE LINE OF ACTION OF THE RE- SULTANT FALLS OUTSIDE THE MIDDLE THIRD FOR RECTANGULAR PRISMS WHICH TAKE NO TENSION. The stresses will be distributed as shown in Fig. 59 forming FIG. 59. EXAM.] STRESSES IN CHIMNEYS, DAMS, WALLS 113 a wedge-shaped prismatic stress volume BCDEFG, there being no stress on the area ABGH. The vertical com- ponent, F, of the resultant equals the summation of the vertical resisting stresses. The line of action of F passes through the centroid of the prism which is ^ (CB) from CF. If b is the breadth CF, the area over which the resisting stress is distributed is If d is the depth AC, and e the eccentricity, F is at the distance e from the edge CF. The length BC is Id \ 3! e]. If / is the maximum compressive stress, the average on the stressed area is - and the total resisting stress is f-KH-t-">' (d-2e), 4Y / = EXAMPLES i. Find the maximum stress at the foot of a stone wall 20 feet high and 4 feet thick when there is a wind pressure of 35 pounds per square foot; also when there is a wind pressure of 45 pounds per square foot if the masonry weighs 150 pounds per cubic foot. Consider a portion of the wall / feet long, Y=W= isox 20X4X/ = i2,oooX/ pounds, H=P = 35 X 2oX/ = 7ooX/ pounds, 3 X 7 X = Ie g pounds per square inch. 2X4X 144 XJ 114 STRESSES IN CHIMNEYS, DAMS, WALLS [CHAP. VII The eccentricity is found by taking moments about a centroidal axis of the base. e = 70QOX/XIO = _7 foot< Thig . g ^ ^ middle third 1 2,000 Xl 12 , _ F Fee _ 12,000 X / , i2,ooo/7X i2X 24X 12 ~A I 48X12x2 12 X 12 /x 48X48X48 = 21 + 1 8 = 3 9 pounds per square inch. For second part, F=i2,ooox/; H=gooxl; e == QO Xl = -g- 12,000 12 This is outside the middle third. 3 (--) = 3 (24 ~ 9) = 45 inches. .'. A' = 45 X 1 2 X / = 540 X / square inches. , 2 F 2X 12,000 X / . , / = j = - = 44 pounds per square inch. A 540 X / 2. What should be the thickness of a rectangular wall 15 feet high to resist a wind pressure of 40 pounds per square foot with- out any tension in the windward side, if the material weighs 140 pounds per cubic foot? Let d be the thickness. The weight of each lineal foot isPF=i5XiX 140 d = 2100 d pounds. The wind pressure for each lineal foot is# = i5XiX4o = 600 pounds, 600 X 7-5 _ 15 For zero stress e\ = - . 6 2iood 7 d d is Jo oo 1 = j d = = I2 - 86 6 id 7 d = 3. 59 feet. PROB.] STRESSES IN CHIMNEYS, DAMS, WALLS 115 PROBLEMS 1. Find the maximum and minimum unit-stress in a rod 2 inches in diameter under a tension load of 16,000 pounds if it is applied at a point | inch from the center of the cross section. 2. What are the dimensions of the kern in a rectangle 3 inches by 8 inches? In a hollow circular chimney of inner diameter 8 feet and outer diameter 10 feet? 3. In a brick wall 20 feet high and 4 feet thick, weighing 115 pounds per cubic foot, (a) What horizontal wind pressure will cause zero stress on the windward side of the base ? (b) With that wind pressure what will be the maximum stress? (c) If the wind pressure is 40 pounds per square foot what will be the max- imum and minimum stresses? (d) With a wind pressure of 40 pounds per square foot what will be the stress on the windward side 10 feet above the ground? 4. A square compression piece 9X9 inches carries an eccen- tric load of 16,200 pounds so applied that the stress on one edge equals o. Determine the application point of the resultant load. 5. What must be the thickness of a wall 25 feet high, weighing 120 pounds per cubic foot, if a maximum unit-stress of 47.2 pounds per square inch is developed when the wind pressure is 40 pounds per square foot? Ans. 4.5 feet. 6. Would a brick wall 30 feet high, weighing 120 pounds per cubic foot, 3 feet thick at the top and 4 feet thick at the base, with one side vertical, be safe if subject to a wind pressure of 40 pounds per square foot? CHAPTER VIII GRAPHIC INTEGRATION* 83. DEFINITIONS. In Chapter V the relations be- tween the load, shear, and moment diagrams are given as follows : the difference between the ordinates at any two sections in the moment diagram repre- sents the area in the shear diagram between the two sections, and the difference between the two ordinates of the shear diagram repre- sents the area in the load diagram between the or- dinates at the same sec- x tions; thus, in Fig. 60, LN in the moment dia- gram represents the area EHIG of the shear diagram, and EF in the shear dia- gram represents the area ABCD of the load diagram. The first integrated curve is defined as one in which the ordinates represent the area under a given curve. Thus, the moment curve is the first integrated curve of the shear curve, and the shear curve is the first integrated curve of the load curve. Since the moment curve is the * For students who have had integral calculus and who do not intend to follow the graphical method of determining deflections of beams, this chapter may be omitted. 116 ART. 84] OBTAINING SECOND INTEGRATED CURVE 117 first integrated curve of the first integrated curve of the load curve, it is called the second integrated curve of the load curve. The second integrated curve is one in which the ordinates represent areas under the first integrated curve. The integrated curve of the second FIG. 61. FIRST METHOD OF GRAPHIC INTEGRATION integrated curve is the third integrated curve. Simi- larly, the nth integrated curve is one in which ordinates represent the area in the (n i)th integrated curve. The graphical method of deriving the integrated curves from given ones will be deduced before applying them to the theory of beams. 84. THE FIRST METHOD OF OBTAINING THE SEC- OND INTEGRATED CURVE. In the following methods, if the given areas are not bounded by straight lines, the n8 GRAPHIC INTEGRATION [CHAP. VIII greater the number of points secured on the resulting curve, the more nearly accurate that curve will be. The equations representing the curves will be deduced by making the number of points secured infinite. Let Fig. 6 1 (a) be the given curve of which it is desired to obtain the first and second integrated curves. The A FIG. 61. FIRST METHOD or GRAPHIC INTEGRATION. curve is taken below the axis making the area between the curve and the axis negative, which corresponds to the load curves already described. To obtain the first integrated curve, divide the area into a number of parts as indicated, and measure each area by any method. Then from an arbitrarily chosen axis O'X' in (b) lay off AB to a selected scale to represent the area AI, then lay off CD to represent the area A^. Continue this process for the entire area under curve (a), then connect by a ART. 84] OBTAINING SECOND INTEGRATED CURVE 119 continuous curve O'G'D' all the points thus obtained. This is the first integrated curve. Any ordinate to this curve as G\G' represents the total area accumulated from the left end to the section GH. In physical problems there is always a constant as 00' to be added to the area under the curve. This constant, which is called the constant of integration is the value of the ordinate to the integrated curve at the origin. In the usual case it can be determined by the conditions of the problem. It frequently is zero. In the shear curve, the constant is the vertical shear at the left sup- port as 00' in (b). In the moment curve it is the moment at the left support. After determining the value of this constant, draw the axis OX which is the true axis of reference for the integrated curve (b) ; the true value of the function represented in curve (b) is then H'G' at the section GH. The area to be considered in the integrated curve is that between the curve and the axis OX. To obtain the second integrated curve, divide the area between the curve O'G'D' and the axis OX into small parts as indicated in Fig. 61 (b), and from some chosen axis OX in (c) erect, to some scale, the ordinate A'B' equal to the area A\ in (6), C'D' in (c) equal to the area AZ in (b), and so on until the entire area in (b) is covered, then the ordinate Gi'Hi represents the accumulated area in (b) from the origin to the section G'H'. The con- stant of integration will depend upon the conditions of the problem. In the illustration it is assumed to be zero. Curve (b) is the first integrated curve of (a), and (c) is the first integrated curve of (b) and a second integrated curve of (a). As long as the constant can be determined, a higher integrated curve can be obtained by the fore- going method, the nth process giving the nth integrated curve. 120 GRAPHIC INTEGRATION [CHAP. VIII 85. THE SECOND METHOD OF OBTAINING THE SEC- OND INTEGRATED CURVE. The method given in the previous article cannot be employed unless the constants can be determined independently. For cases when the FIG. 62. SECOND METHOD OF GRAPHIC INTEGRATION. constants cannot be determined a second method must be combined with the first. Draw the first integrated curve Fig. 62 (b) in the same manner as in the first method, using the arbitrary axis O'X'. Project the ART. 85] OBTAINING SECOND INTEGRATED CURVE 12 1 areas AI, A Zj A s , etc., in (b) to the vertical axis X'L as indicated. If it is desired to obtain the integrated curve for (b) referred to any axis as 0"X" , thereby assuming a constant of integration 0"0', take a pole P f on that axis a distance H from the axis X'L. H, which is called the pole distance, is measured in the same units and to the same scale as distances along the axis OX. Con- nect P' to the ends of the lengths representing the areas on the line X'L, as P'X', P'B, P'C, etc. These lines are called rays, and the polygon P'X'L is called the ray or vector polygon. A i is to be replaced by its compon- ents X'P f and P'B, A 2 by its components BP' and P'C, etc. Draw through the mean ordinate of the area AI in (a) a vertical line of indefinite length, as A\A' . Also draw through the mean ordinate of the area A z in (a) the vertical line A 2 B', and draw through the mean ordinate of A 3 the line A 3 C'. Continue this process for all the elementary areas in curve (a). From the origin 0' in (c 1 ) draw O'A' parallel to P'X' in (b), and from the point A ' in (c'} where the line O'A ' intersects the vertical line through the mean ordinate of A\ in (a) draw A'B' parallel to P'B in (b), and from the intersection, B' in (c'} of A'B' with the vertical through the mean ordinate of A 2 , draw B'C' parallel to P'C in (b). Continue this process until the entire polygon 'A'B' E' X' is drawn. This polygon is called the string or funicular polygon and the lines O'A', A'B', B'C', etc., are called strings. The ordinate measured from the horizontal axis O'X" in curve (c r ) represents the integrated area of diagram (b) between the curve and the axis O ff X ff , i.e., the ordinate of a second integrated curve of (a), The constant 0"0' of (b) for the axis 0"X" call AS. Take any section MN, then y in (c r ) represents the accu- mulated or integrated area in (b) from the origin to the section. 122 GRAPHIC INTEGRATION [CHAP. VIII Proof: From the similar triangles P'X'X" and O'Aid Ai in the ray and funicular polygons respectively, = ri (f^). from triangles P'X'B and A'AiBi, ^ = ^L?2 ; H pi from triangles P'BC and B'BiCi, jj = ( ^ lCl) ; from triangles P'CD and C'CiDi,-=j = ; from triangles ri P 3 P'DE and D'DiEi, ~ = * By clearing the above equations of fractions, the following are obtained: Ai'x = H (Aid), (i) (2) (3) (4) Aip* = H(DiEi). (5) By subtracting the left members of the last four equations from the left member of the first equation, and the right members of the last four equations from the right member of the first equation there results the following equation : Ai'x Aip i Azpz A s pz A 4 p4 = PI (Aid AiBi - BiCi - CiDi - DiEi), and Ai'x- 2Ap = Hy, since A\d AiBi Bid C\D\ DiEi = y. It can be seen from curve (b) that AI'X 2 Ap equals the algebraic sum of the area under the curve repre- sented by the shaded portion; therefore, Hy equals the area between the axis and the curve from the origin to the section, and by use of a proper scale y represents the area. Hence, (c r ) is a first integrated curve of (b) and is a second integrated curve of (a). ART. 85] OBTAINING SECOND INTEGRATED CURVE 123 The greater the number of parts into which the area under the curve is divided the more nearly the true curve will the funicular polygon be. When the number of parts becomes infinite the funicular polygon becomes FIG. 62. SECOND METHOD or GRAPHIC INTEGRATION. a smooth curve inscribed within the broken funicular polygon shown. This smooth curve is the true inte- grated curve and can be drawn inscribed in the broken funicular polygon. In Fig. 62 (b), by choosing the 124 GRAPHIC INTEGRATION [CHAP. VIII pole P' on the axis 0"X" it assumes the constant of integration equal to 0"0' ', and if this is not the true value, curve (c f ) is not the true curve, and the process so far is only tentative. 86. THE CONSTANT OF INTEGRATION. The constant of integration in every case depends on the conditions of the given problem, and unless it has been determined from the given conditions of the problem, the assumed one as 0"0', Fig. 62 (b), generally is not the true value, but the true value can be determined by use of the curve (c f ). Whatever quantity is represented by the area under curve (b) or by its equivalent, the ordinate of the curve (c f ), Fig. 63, there will be two points along the X-axis at which the values of the ordinates are known or can be determined. For beams these points will usually be at the ends, supports, or center. When the constant AI in Fig. 63 (b) cannot be determined from the curves (a) and (b) curve (c') must be employed. For the arbitrary constant A\ chosen in Fig. 63 (b) curve (c f ) shows a value of zero at the left end and a negative value X"X' at the right end. This shows that the negative area exceeded the positive area by the amount X"X' '. Suppose that the values at each end should have been zero, the curve then should end at X n ', as would the moment curve for a simple beam. For this to be the case the negative area in curve (b) must be decreased and the positive area increased. To ac- complish this the reference axis must be lowered, thus making the constant larger than the value assumed, A\ . The method of obtaining the value of the constant that will make the positive area equal to the negative area in (b) is to draw O'X 1 the closing line of the funic- ular polygon in (c f ), then draw P'X in the vector polygon parallel to O'X 1 in the funicular polygon. Then through ART. 86] THE CONSTANT OF INTEGRATION 125 X draw the horizontal axis OX, giving XX' or 00' in (b) the true value of the constant. FIG. 63. To DETERMINE THE CONSTANT OF INTEGRATION. Proof: By lowering the reference axis in (b) to OX the positive area is increased the amount O"X"XO which equals / multiplied by XX" . This must equal H 126 GRAPHIC INTEGRATION [CHAP. VIII multiplied by X"X' in (<;') to have the positive and negative areas in (b) equal. From similar triangles P'X"X in (b) and 0'X"X' in (c r ) : TT I vv" ~ X'X" The difference between the two axes O'X" and O'X' in (c') should be added to the ordinates in the original curve to obtain the true curve. Proof: From similar triangles P'X"X in (b) and O'SR in (c') : r H(RS) is the value represented by the ordinate (RS) in (c r ) and x(XX") is the shaded area in (b) which is the increase over the former value. Each ordinate now may be increased to its true value from a horizontal axis, or the true curve may be obtained by taking a new pole on the axis OX in (b) . (The pole distance may be changed if desired.) Then by using the new pole P and proceeding by the same method used in drawing (c'), the true integrated curye (c) is obtained. The student should supply the proof that the ordinates in (c) represent the area between the curve and the axis OX in (6). 87. UNITS. The units for the ordinates of the inte- grated curves will depend on the units used for x and y, and the units to be used for x and y will depend on the problem to be solved. The unit for x is the same for all curves. The unit for the ordinates of curve (b) is the product of the x unit and the y unit or the unit formed by the product xy; and that for curve (c) is the product of the unit of the 'ordinate for (b) and the x unit, i.e., the unit formed by the product x 2 y. In problems for beams x will represent a length. EXAM.] GRAPHIC INTEGRATION 127 EXAMPLES i. By the method of graphic integration draw the shear and the moment curves for a simple beam of i2-ft. span carry- ing a concentrated load of 2000 pounds 9 feet from the left support. The shear diagram is drawn in the usual way, Fig. 64. Select- ing the pole P with the pole distance equal to 72 inches the moment 1000* 1167* 2000* ^_ -12- (a) (b) Co) FIG. 64. curve (c] is drawn, (p a) in (c) is parallel to (P A} in (6), (p - b) in (c) is parallel to (P - B} in (6), and ( - c) in (c) is parallel to (P C) in (6). To get the moment at any section as at MN measure MN, using the same ^cale as is used in draw- ing the shearing forces, and multiply by the pole distance, in this case 72 inches. This gives the bending moment represented by MN equal to 66,000 pound-inches. 2. Determine the vertical shear at the left support (constant of integration) by the graphical method for a simple beam of i4-ft. span carrying a uniform load of 500 pounds per foot and a concentrated load of 3500 pounds 5 feet from the left support. 128 GRAPHIC INTEGRATION [CHAP. VIII By using the O'X'-axis in Fig. 65 (b) lay off distances as AB to represent the area in the load diagram to the left of the section. Lay off the concentrated load BC to the same scale in pounds. Select any pole P f with a pole distance say 60 inches' and draw (c r ) as described in the text. Draw P'X in (b) parallel to the closing line O'X' in (c r ). Then draw OX horizontal giving 00' equal to 5750 pounds which is the vertical shear at the left support. PROBLEMS i. By the graphical methods draw the shear and moment diagrams for the following cases: (a) Cantilever beam of Q-ft. span, concentrated load of 5000 pounds at the free end and one of 6000 pounds at the center. (b) Simple beam of i6-ft. span with a total uniform load of 18,000 pounds. PROB.] GRAPHIC INTEGRATION 129 (c) Simple beam of i6-ft. span, uniform load of 1000 pounds per foot and a concentrated load of 10,000 pounds 8 feet from the left support. 2. By the graphical methods determine the vertical shear at the left end for each of the following systems of loading: (a) Simple beam of lo-ft. span with a uniform load of 400 pounds per foot and a concentrated load of 1000 pounds 3 feet from the left support. (b) Simple beam of 2o-ft. span, concentrated loads of 5000 pounds 7 feet, 2500 pounds 10 feet, and 10,000 pounds 15 feet from left support. (c) A beam 16 feet long overhanging the right support 4 feet with a uniform load of 1 500 pounds per foot. (d) A cantilever beam of lo-ft. span with a uniform load of 300 pounds per foot and a concentrated load of 800 pounds at the free end. (NOTE. The value of the moment at the free end is zero, and that at some other point should be calculated and laid off to scale to draw the closing line. The vertical scale for the moment curve equals the product of the vertical scale of the shear curve multiplied by the pole distance.) CHAPTER IX DEFLECTION OF BEAMS ELASTIC CURVE 88. BENDING. The elastic curve is the curve assumed by the neutral surface of a beam under load. The deflec- tion of beams can be obtained only from the elastic curve. For certain kinds of beams the reactions, the maximum shear and the maximum moment can be obtained only by the use of the elastic curve, while for cantilever, simple, and overhanging o' v beams the reactions, the shear, and the moment may be obtained without its use, also for beams fixed at both ends and loaded sym- metrically the reactions and shear may be obtained without its use. 89. THE RADIUS OF CURVATURE OF BEAMS. In Fig. 66 let l\ be an ele- ment of length of a beam under load. The loads FIG. 66. cause a bending moment M at the section CE. This bending moment may be considered constant over the element of length /i. The deformation due to the shear- ing stresses will not be considered in this discussion. 130 ART. 89] RADIUS OF CURVATURE OF BEAMS 13! The neutral surface AB, which was straight before the load was applied, is bent to a curve under the influence of the bending moment M. Assume that a normal sec- tion of the beam before bending remains a normal section after bending, that the moduli of elasticity of the material in compression and in tension are equal, and that the stresses developed are below the elastic limit. Let CE and DF be two sections normal to the beam and parallel to each other before bending. After bending, their planes will intersect at some point 0, the center of curvature of AB. Let the radius of curvature BO be r, and the dis- tance of the most remote fiber from the neutral surface BF be c. Draw GH parallel to CE, then DG is the short- ening of the top fiber in the length l\ and HF is the elongation of the bottom fiber in that length. For a symmetrical section these deformations are equal. Let this deformation be e f '. Then the unit deformation is e' e f r and the unit-stress developed is/ = T E. From the /i k f 1 * Me moment formula / = -y- e' Me e' Me Therefore, T& = ~T' or T~~^J' l\ 1 l\ rLl Since h is very small the triangles OAB and BHF may be considered to be similar. HF BF e' c Hence ' = > r = ' Mc _ ~ ( = In this equation M is the bending moment for the element of length /i, r is the radius of curvature of AB, E is the modulus of elasticity of the material, and / is the moment of inertia of the cross section of the beam 132 DEFLECTION OF BEAMS [CHAP. IX about the neutral axis. It is seen from the equation that the radius of curvature of a portion of a beam varies inversely with the bending moment. 90. THE SLOPE OF THE NEUTRAL SURFACE. The slope of a curve at a given point is the measure of the tangent of the angle the curve makes with the hor- izontal axis. Thus, in Fig. 67, tan a is the slope of the FIG. 67. curve AB at the point P. In Fig. 68, which is greatly exaggerated, let a be the angle the neutral surface of a beam at A makes with the horizontal, a' is the increase in the angle over the element of length AB. Since the angles are very small for ordinary beams, the tangents or slopes of the angles may be considered equal to the angles themselves without appreciable error; therefore, the increase in the slope is equal to the increase in the angle measured in radians, and the slope and the angle may be interchanged. From the figure it 7 / is seen that a' = ; hence, the increase in the slope over r\ the element of length // is , l^ M'h' . El a = = -prr- since r = T7 ' TI El M Similarly the increase in the slope a" over the element of M"l " length BK is * . The increase in the slope over ILL ART. 90] SLOPE OF THE NEUTRAL SURFACE 133 any element of length has a similar form. The total increase of the slope between any two sections of the FIG. 68. beam is obtained by adding all the increases between the sections, which is KA I ~^- + . El El El If the slope at any section is known, the slope at another section may be found by adding the increase, between the two sections. 134 DEFLECTION OF BEAMS [CHAP. IX 91. THE SLOPE CURVE. The relation in the last article affords the means of deriving the graphical method of determining the slope curve for a beam. The slope curve (a curve) is one in which the ordinates show the values of the slope at every point along the beam. In which 14? lj I is the bending moment for the section divided by the product of the modulus of elasticity of the material and the moment of inertia of the cross section about the neutral axis. 93. THE DEFLECTION OF BEAMS. THE ELASTIC CURVE. In Fig. 70 (e) let APB represent the position assumed by a portion of the neutral surface of a beam under load. Divide the length of the curve into the ele- ments //, /i", /i'", etc. If at any point P the value of the deflection y is known, that for any other point Q may be determined by calculating the increase in the deflection be- tween the two points. Let the angles made with the hori- zontal by the lengths //, //', /i'", etc., be a', a", a"', etc. 136 DEFLECTION OF BEAMS [CHAP. IX Then the increase in y over the length // is y' = u' tan a! where u' is the horizontal projection of //, y" = u" tan a", y'" = u'" tan a", etc. The tangent of the angle the elastic curve makes with the horizontal axis is the O r FIG. 70 slope of the curve at that point; therefore, the increase in y between the two points is = M'tana'+ u" tan a" =ua u"a" + From Fig. 70 (d) it is seen that ua equals a small shaded area under the slope curve, and the summation of all the areas, 2 ua, equals the total area under the slope curve between the two points. Therefore, the increase in the deflection of the elastic curve of a beam between any two points is equal to the area under the slope curve between those two points. The deflection at the sup- ports is usually known. ART. 96] UNITS FOR THE FIVE CURVES 137 94. THE RATE OF INCREASE OF THE DEFLECTION. The rate of increase of the deflection is = a. The u rate of increase in the deflection at a section is equal to the slope of the elastic curve at the section. 95. RELATIONS BETWEEN THE FIVE CURVES. Com- bining the relations between the load, shear, and moment curves deduced in Chapter V, with those between the moment, slope, and elastic curves, important results are obtained. Since these five curves are the principal ones for beam problems, and since they form a con- tinuous chain between the load and the deflection of a beam, they will be referred to as the five curves. The relation existing between them may be stated as follows : Between any two sections of the beam : (1) The increase in the vertical shear equals the area under the load curve between the sections. (2) The increase in the bending moment equals the area under the shear curve between the sections. (3) The increase in the slope equals the area under the moment curve between the sections divided by El. (4) The increase in the deflection equals the area under the slope curve between the sections. Thus it follows that the question of determining the elastic curve is one of determining constants of integration and of obtaining areas under curves. These principles will be applied to various kinds of beams, and the constants determined and the areas obtained. 96. THE UNITS FOR THE FIVE CURVES. In the fol- lowing discussions (a) will refer to the load curves, (b) the vertical shear curves, (c) the bending moment curves, (d) the slope curves, and (e) the deflection or elastic curves. The modulus of elasticity E will be 138 DEFLECTION OF BEAMS [CHAP. IX taken in pounds per square inch, / in (inches) 4 . For all the five curves and for the pole distances one inch along the X-axis will represent m inches of length measured parallel to the beam. The scale of ordinates of the curves will be: Curve (a) i inch = w' pounds per inch run. I square inch area = w'm pounds. Curve (b) i inch = n square inches from (a) = nw'm pounds. I square inch area = nw'm 2 pound-inches. Curve (c) i inch = p square inches from (b) = pnw'm 2 pound-inches. I square inch area = pnw'm 3 pound- (inches) 2 . r- /j\ 1. Q. square inches from (c) qpnw'm* Curve (d) I inch = * H pr -" = ^ ^ T Ll L1 ,. , . ,. . , pqnw'm* . , which is a ratio. I square inch area = ^ r inches. Ll Curve (e) I inch = r square inches from (d) = iLl inches. For an illustration of the method of determining the scales of the curve see Example I at the end of the next chapter. EXAMPLE i. What will be the increase in the slope from the left end to the middle of a Q-inch, 2i-pound I-beam of i2-ft. span with the concentrated load at the center that will produce a maximum fiber stress of 16,000 pounds per square inch? The maximum moment is developed at the center and is M = J = 16,000 X 18.9 = 302, 400 pound-inches. Since the moment increases directly from zero at the end to the maximum, the moment curve is as drawn in Fig. 71. The area PROB.] DEFLECTION OF BEAMS 139 under the moment curve to the left of the center is | X 72 X 302,400 = 10,886,400 pound-inches 2 . Therefore the change in the slope over this length is 10,886,400 30,000,000 X 84.9 = -0043. 302 FlG. 71. PROBLEMS 1. What is the radius of curvature at the ends and at the center of a 3-inch X 4-inch stick 8 feet long used as a simple beam with a load of 350 pounds concentrated at the center? Ans. 2860 inches. 2. What is the change in the slope from the left end to the center in the beam in Problem No. i ? What is the change in the slope over the first two feet? 3. What is the total change in the slope over the entire length of a cantilever beam of ip-ft. span carrying a concentrated load of 3000 pounds at the end, if the beam- is of a standard I-section and the stress does not exceed 12,000 pounds per square inch? 4. What is the rate of increase of slope at every two feet of length along the beams in Problems No. 2 and 3 ? CHAPTER X CANTILEVER AND SIMPLE BEAMS AND BEAMS FIXED AT BOTH ENDS 97. CANTILEVER BEAM, CONCENTRATED LOAD AT THE END. By use of the method analyzed in Art. 85 and 86 the curves in Fig. 72 are drawn for a cantilever beam with a load W at the end. From these curves the shear, moment, slope, and deflection at any section may be scaled off directly. Algebraic expressions for these quantities will now be deduced. From the definition of vertical shear, V = -W. Since the bending moment at the left end is zero, that at the section AB is equal to the area under the shear curve between the origin and the section, which is Wx. M = -Wx. The increase in the slope from the left end is equal to the area in the moment diagram from the origin to the Wx 2 section divided by El, and equals =r T - The free end 2 JtltjL of the beam deflects. The beam remains horizontal at the wall, thereby making the slope zero at the wall. Wl 2 The total area under the moment curve is This 2 divided by El gives the total change in the slope from one end to the other. If the slope at the left end of the beam is a\, it is 140 ART. 97] CANTILEVER BEAM, CONCENTRATED LOAD 14! changed from this value to zero at the right end. There- fore, FIG. 72. CANTILEVER BEAM, CONCENTRATED LOAD To get the slope at the section AB add to ai the change from the left end to the section which gives W 2EI 142 CANTILEVER AND SIMPLE BEAMS [CHAP. X The change in the deflection of the elastic curve is equal to the area under the slope curve. The change in the deflection is shown in (d) by the area OO'HI -.x area which equals area OO'GI area O'GH or Wl 2 2 O'GH since OO' = pj' Fig. 73 represents the area O'GH drawn to a large scale. Divide the length x into FIG. 73- a great number of parts n parallel to GH and let A BCD be any division such as the pth from the apex 0'. /v* The distance of this strip from 0' is p-t the breadth is x W /px\ 2 n - and the depth is =r T { ) . The area of the strip n 2 El \n / W p 2 x* then is A BCD = .. ~"|" In this p represents any and all numbers to n. The area O'GH, then, is the sum of all such areas as A BCD. W By algebra it can be shown that 2 ^ + 3^ + n _. * See "Higher Algebra," by John F. Downey, page 373. ART. 97] CANTILEVER BEAM 143 W _ Area GH= 2 El W x*/2n* + i>n* + n\ -;( - ^ - n*\ 6 / Since the side O'H is a continuous curve, as n is increased the assumed broken line will approach the actual curve and they will coincide when n equals infinity. By assuming this to be the case, the actual area O'GH is obtained, and and 7 -become zero; therefore Wx 3 Area O'GH and Area 00' HI 6 El Wl 2 x Wx 3 2 El 6 El Since the area under the slope curve represents the change in the deflection the expression for the area OO'HI is the increase in the deflection from the left end to the section AB. In Fig. 72 (e) if the J^'-axis were used as reference, the deflection at the left end would be zero, and the above equation would give the value of the deflec- tion at any section AB. The axis of reference is usually taken in the position of the neutral surface before any load is on the beam, for which case the deflection at the left end is 00' equal to XX', Fig. 72 (e), which equals the total change in the deflection over the entire length of the beam. This change is obtained by letting x equal / in the expression for the area under the slope curve, WPx Wy? WP oo> = - wp = y 3 El /2 -y \A/ / Y'3 lA/I v *v rvJt W v 2 El 6EI 144 CANTILEVER AND SIMPLE BEAMS [CHAP. X which is the equation of the elastic curve of a cantilever beam with a load W at the free end. The maximum deflection occurs at the free end and is WP A = 3 El 98. .' CANTILEVER BEAM, CONCENTRATED LOAD AWAY FROM THE FREE END. The solution of the problem for a concentrated load at the end of a cantilever beam can be extended to cover the problem when the load is away from the end. The dotted lines in Fig. 72 indicate the extension of the solution for the previous case to cover this case. The load, shear, and moment curves would be similar to those given. The slope has the con- stant value 00' from the free end to the load. The addi- tional deflection of the free end equals the area KJO'O. The student may deduce the equations for this case. 99. CANTILEVER BEAM, UNIFORM LOAD. In Fig. 74 are ' drawn the curves for a cantilever beam with a uniform load. The expressions for the values repre- sented by the different curves at the section AB the distance x from the free end will be deduced. The load per unit of length of the beam equals w. Vertical shear V= wx. Bending moment M= The slope at the right end is zero as the elastic curve there is horizontal. The area under the moment curve to the section AB is (see Art. 97). The total area o *2)J^ *7P) J^ is , hence the total change in the slope is -r^- r mak- D O ILL >7fjJ3 ing the slope at the left end a\ equal to ^=r^ ; therefore, O JLLJ. Q1 wl 3 blope a ~ ART. 99] CANTILEVER BEAM, UNIFORM LOAD 145 FIG. 74. CANTILEVER BEAM, UNIFORM LOAD. 146 CANTILEVER AND SIMPLE BEAMS [CHAP. X The deflection at the right end is zero. In order to find the deflection at any section AB it is necessary to obtain the area under the slope curve since the change in the deflection is equal to that area which is O'HIO. In Fig. 74 W, area O'HIO = area O'GIO - area O'GR ivfix 6 El - area O'GH. FIG. 75- Let Fig. 75 represent the area O'GH drawn to a larger scale. Divide it into a great number of strips n parallel to GH as A BCD. The width of each strip is - Let n this be any strip as the pth from 0', then its distance X IV /'tflx^\ from 0' is p and its depth is 'T~pT\~r)' Therefore the area A BCD equals l! r 4 p represents all numbers from i to n. The total area O'GH equals the sum of all such small areas. Area O'GH = Area u uw From algebra S (n) s w 4 + 2 w 3 + w 2 * See " Higher Algebra," by John F. Downey, page 373. ART. 100] CANTILEVER BEAM, VARIOUS LOADING 147 Area O'GH - * (" 4 + 2 ** 3 + O '6 EM 4 WX* /I . __ I ~ ~~ since and 5 reduce to zero when n equals infinity. 2 n 4 n 2 Therefore area O'GIO = ~ - -^ This equals the O Ll 24 & change in the deflection from the free end to the section AB. The total change over the entire length is obtained by letting x equal / in the expression for the change. This reduces to wl* wl* wl* wl* WP 8 El ~ 8 El which is the maximum deflection and occurs at the free end. W is the total load. w/ 4 wl s x wx* ~8^E7 + 6^E7~247' which is the equation of the elastic curve for a cantilever beam with a uniform load. 100. CANTILEVER BEAM, VARIOUS LOADING. If the end projects beyond the uniform load, the load, shear, and moment diagrams will be similar' to those of Fig. 74. The slope will be constant from the free end to the load as indicated by the dotted lines OJKO', Fig. 74 (d). The additional deflection will be equal to the area O'OJK. If the beam has a concentrated load at the end and a uniform load the equations for the two cases may be combined. Any other combination of uniform and concentrated loads may be made, and corresponding equations derived similarly to the foregoing deductions. 148 CANTILEVER AND SIMPLE BEAMS [CHAP. X loi. SIMPLE BEAM, CONCENTRATED LOAD AT THE CENTER. The curves in Fig. 76 are drawn for a simple beam with a load W at the center. The vertical shear to the left of the load is W V = -- 2 The bending moment to the left of the load is M M = -- 2 The slope curve must pass through the axis at the center of the beam, since the elastic curve is horizontal there, making the slope at that point equal to zero. The change in the slope from the end to the middle equals the area to the left of the center in the moment diagram Wl 2 divided by El, which is , ^ r ; therefore, the slope at Wl 2 the left end is -- T~^T an d the slope at the section AB lo Li to the left of the center is Wx 2 Wl 2 4 El i6EI The deflection at the end is zero. The change in the deflection from the end to the section AB to the left of the center can be obtained by calculating the area under the slope curve (d). Area OABO' = area OACO' - area O'BC /Wl 2 x wx* \ WVx . wx* \_ WPx I ,t T? r i \i6EI 12 El I 16EI ' 12 El (see Art. 97). Since the area is below the axis it is negative. WPx Wx* y ~i6EI^i2El' ART. 101] SIMPLE BEAM, LOAD AT CENTER 149 o k FIG. 76. SIMPLE BEAM, CONCENTRATED LOAD AT THE CENTER. 150 CANTILEVER AND SIMPLE BEAMS [CHAP. X The maximum deflection occurs at the center and equals the area below the axis in the slope diagram, which may be found by letting x equal - in the expres- sion for the deflection; therefore the maximum deflection is WP A =- 48 El The foregoing equations are for a section to the left of the center. Equations representing the curves to the right of the center may be deduced similarly to the above. The student may derive those equations. 102. SIMPLE BEAM, UNIFORM LOAD. The curves in Fig. 77 are drawn for the case of a simple beam with a uniform load. For the section AB the vertical shear is V = -- wx. 2 The bending moment is wlx wx 2 M= ----- 22 x The slope at the center is zero; the change in the slope from the end to the center equals the area under the moment curve to the left of the center divided by El, which is ^.-.^ wl 2 ^ I ~vn P W P W P Area ODG = X - - area OFD = - = 82 ID 48 24 (see Art. 97). The slope is changed from <*i at the left end to zero at the center; therefore, wl 3 1 24 El 24 El The change in the slope to the section AB equals the ART. 102] SIMPLE BEAM, UNIFORM LOAD 151 FIG. 77. SIMPLE BEAM, UNIFORM LOAD, CANTILEVER AND SIMPLE BEAMS [CHAP. X area OB C divided by El. By referring to Fig. 78 we see that Area OBC = area ODG - area CBDG wl* 24 -area CBDG. * Area CBDG = area. CEDG - area BED = -- - x - area BED = ~ area BED. The ordinate BE represents the triangular area MNQ in Fig. 77 (6). A D777- Area BED = 2X3 w/l V = [ X] 6\2 J 6 \ 4 Area OJ5C = area ODG - area CEDG + area BED, wl 3 wl 3 ~ 24~T6 _ wlx 2 _ wx 3 ~4~ ~& wl 2 x wl 3 _ wl 2 x wlx 2 _ wx* ~8~ h 48 " "8~" "~4~ '"6" The slope at the section AB then is wl 3 a = 24E7 ' 4 7 6 El Since the deflection at the end is zero, that at the section AB equals the area under the slope curve between ART. 102] SIMPLE BEAM, UNIFORM LOAD 153 the origin and that section. In Fig. 77 (d) the deflec- tion is shown by Area 00'BC= -(area OO'AC-area O'AB) area 'AB\ ,24 El Divide the area O'AB into a great number of strips n. OC The width of each strip is - If the strip shown is px the pth strip its distance from 0' is > and its depth . wl (pW\ w tp*x*\ _ u is ^7^-1 ^^7 M-T-j- The small area then is 4 El \ n 2 / 6EI \n z I wl /^\ w /p 3 x*\ . Z^\ w 3 / ~ 6~EI I*?/ 1 m P represents all num- bers to n, and to obtain the total area O'BA all such areas must be added ; therefore 4 EM 6 EM wlx 3 (2 n 3 + 3 n 2 + n\ rv ~6~ ) + 2n 3 + ri _ 2 ^ 4 V 12 El 24 El When n is infinitely large 7 5 and - reduce to 2 n o n A.n 2 zero. 154 CANTILEVER AND SIMPLE BEAMS [CHAP. X \24EI 12 El 24 El wl 3 x wly? ivx* 24 El ' 12 7 24 El The maximum deflection occurs at the center or when x = - in the expression for y. Substituting this value for x the maximum deflection is found to be ~^/ = ~^El' 103. BEAM FIXED AT BOTH ENDS, CONCENTRATED LOAD W AT THE CENTER. A beam with fixed ends has restraining moments at the walls which keep the beam horizontal at those points. These moments must be determined in the solution of the problem. From the symmetry of the beam the reactions at the walls are equal, and the restraining moments are equal. In Fig. 79 the shear and moment curves are drawn in the usual manner; then by using the pole P' in Fig. 79 (c) a slope curve (d') is drawn. Since the beam is hori- zontal at the ends and at the center the true slope curve must pass through the axis at those three points. There- fore connect the ends by O'X'. Now in (c) draw P'X parallel to O'X' in (d f ), and through X draw the hori- zontal axis OX which is the true moment axis giving the moment OO' at the left support, which equals XX' ', the moment at the right support. Then with a new pole P on the true axis in the moment diagram the true slope curve (d) is drawn. To prove that the slope in (d) will be reduced to zero at the right end, draw the closing line OX in (d) and draw OZ parallel to O'X' in (d'), then the angle ZOX must be equal to a = a' in (d') because Z"X' and O'X" are both parallel to P'X' in (c) ; i.e., hori- ART. 103] BEAM FIXED AT BOTH ENDS FIG. 79. BEAM FIXED AT BOTH ENDS, CONCENTRATED LOAD AT THE CENTER, zontal. Y'XO and VOX in (d) are equal because Y'X and YO are drawn parallel to PX in (c). X'UXm(c) = YOZin(d), YOX'm (d) = X'PX'm (c), UX'P' + UP'X' in (c) = X'PX + ^ r P r ^ in (c) = 70^ in (d) + a in (rf'), 156 CANTILEVER AND SIMPLE BEAMS [CHAP. X /. YOZ in (d) = VOX in (d) + a = YOX in (<*) + XOZ in (d), .: ZOX in (d) =a'm (d f ) = X'P'X in (c). To obtain the bending moment at the walls and at the center of the beam, it is known that the slope is zero at the ends and at the center. Hence the total change in the slope from the end to the center is zero; there- fore, the positive and negative areas to the left of the center are equal. As the moment varies directly along the length of the beam that at the end is equal to minus the moment at the center. The total change in the moment equals the area under the shear curve Wl to the left of the center which is -- Consequently, 4 11 Wl , , the moment at the wall is and that at the center o Wl is -5-- The deflection at the ends is zero. Since the o constants have been determined the equations of the curves for a section AB to the left of the center can be written : W V= 2 ' Wl . Wx Wlx Wx* ~ Wlx z _ Wx* I6EI* I2EI' In order to obtain the deflection y the area under the slope curve is determined by the same method as that by which the area under the moment curve for a simple beam uniformly loaded was determined (Art. 102). The maximum deflection is at the center and equals ART. 104] POINTS OF INFLECTION 157 the area under the slope curve to the left of the center, which is f that area : which is found by letting x = - in the expression for WP Wl* WP 64 El ' 96.E7 192 El 104. POINTS OF INFLECTION. In Fig. 79 (c) are shown two points marked / where the moment is zero. At these points the moment changes from negative to positive in going toward the center of the beam. The stresses also change from tension to compression in the top fibers and from compression to tension in the bottom fibers. These points where the fiber stresses change are called the points of inflection or points of contraflex- ure. Outside these points the beam curves downward, and inside them it curves upward. Since there are no flexural stresses at the points of inflection, the beam could be hinged at those points without affecting the stresses at the other sections of the beam.* For a beam fixed at both ends with a concentrated load at the center the inflection points occur at the two outside quarter points, hence it may be considered as a simple beam of length - with the load W at the center and two / W cantilevers each of length - with the load of at the ends. The simple beam may be considered as resting on the two cantilevers. Wherever the tensile stresses in a beam are to be taken by steel, as in reinforced con- crete beams, part of the steel is bent down somewhere near the inflection points. The inflection points are located where the greatest positive and negative slopes occur. * On account of secondary stresses and horizontal shear which have not yet been considered, the behavior of the beam may be somewhat different if hinged at the points of inflection. 158 CANTILEVER AND SIMPLE BEAMS [CHAP. X 105. BEAM FIXED AT BOTH ENDS, UNIFORM LOAD. The reactions equal one-half the total load for a beam fixed at both ends and carrying a uniform load. The restraining moments at the ends keep the beam hori- zontal at these points. It is also horizontal at the center. In Fig. 80 curves (a), (b), and (c) are drawn in the usual manner. The curve (d') is then drawn by using the pole P' in (c). Connect O'X' in (d'), then, in (c), P'X is drawn parallel to O'X' in (d f ) , and OX, which is the true moment axis, is drawn, giving the bending moments 00' and XX' at the walls. By selecting a new pole P on the moment axis the true slope curve (d) is drawn, from which in turn the elastic curve (e) is drawn. In order to determine the bending moment at the wall and at the center it is known that the slope is zero at both sections, and, therefore, that the positive moment area ABC equals the negative area OA 0' . The total change in the moment from the end to the center equals the area under the shear curve between those two sec- tuft tions, which is By methods similar to those already o given, A ^ A r\t Area OAO = 4 3 and Area ABC = areaABCD - areaACD, (BC) = (AD) = W (area HIF in (b)). Area ABC = w (DC) = - - * Area 0.40'= area^^C, W_Xi_ _ WXi_ m I ^1 _|__J?ll _ 2X1* \. 4 3 6[8 42 J ART. 105] BEAM FIXED AT BOTH ENDS 159 Collecting and reducing, #1= - Vi2, which is the distance from the end to the inflection point. FIG. 80. BEAM FIXED AT BOTH ENDS, UNIFORM LOAD. (AD) in (c) represents the area FHI in (6), and (O'E) in (c) represents the area FGO in (b) ; 160 CANTILEVER AND SIMPLE BEAMS [CHAP. X therefore II V (AD) (DC) 2 \z Xl ) (Q'E) ~ (EC}' (AD) (O'E) ~ // //y Hence, the bending moment (BC)-at the center is J of = ' and that at the ends is --- The deflec- tion at the ends is zero. Since all the constants are determined, by methods similar to those already given the following equations for the curves at the section AB are deduced. The load per unit of length of the beam equals w, ir w l V = -- wx, ,, wl 2 . wlx wx 2 M = --- - 12 2 2 wl z x wlx 2 wx* ~ + wl 2 x 2 wlx 3 ~ The deflection is the greatest at the center and is ob- tained by letting x = - in the value for y, or = w/ 4 Wl 3 384 / 384^ 106. RELATIVE STRENGTH AND STIFFNESS OF BEAMS. The strength of a beam is proportional to the load it ART. 106] RELATIVE STRENGTH AND STIFFNESS 161 will carry with an assigned value of the maximum stress. For a beam of given section the allowable resisting shear and resisting moment are fixed by the allowable stresses. If the shearing stress or the deflection is not the con- trolling factor in the design of a beam, the strength depends upon the allowable resisting moment The allowable bending moment is equal to this resisting moment. The strength of a certain type of beam is inversely proportional to the maximum bending moment produced in the beam by a given load. For a beam of length /, the load W to develop the fiber stress / may be obtained by use of the moment formula M = in which M is the maximum bending moment, here to be expressed in terms of W and /, and - is the section modulus, which c is fixed for a given section. For example, in a cantilever beam with a concentrated load at the end M = Wl (the sign being neglected) ; hence, Wl = and W ' c cl Column two in Table 14 contains the expressions for the maximum bending moment developed in the various types of beams given, and column two in Table 15 gives the value of the load to produce the fiber stress /. If beams of the same material for the various types given are of equal length and section, their relative strengths will be proportional to the coefficients of 7 given in cl Table 15, as , will be the same for all the beams. cl The stiffness of a beam is proportional to the load necessary to produce a given maximum deflection. The load W to cause a maximum deflection may be obtained by solving for W in terms of the maximum deflection A. 162 CANTILEVER AND SIMPLE BEAMS [CHAP. X For example, the maximum deflection of a cantilever beam with a concentrated load at the end is A = Wl 3 El (the sign being neglected) ; hence, W = 3 -- A. TABLE 14 MAXIMUM MOMENTS AND MAXIMUM DEFLECTIONS Kind of beam. Maximum moment M. Maximum deflection A. I 1 f ' 1 i Wl Wl 3 3 El W/^in. t Wl Wl 3 L i k ^ ^ 2 SEI S ' Wl Wl 3 4 48 El vy/J^in. m s Wl 3 r : ^ 8 384 / f~ \ ^ 8 192 / y/, W 7 / W 7 / W 7 / 3 / / y. Q 77 7 f" * 3 Column three in Table 14 gives the expression for the maximum deflection of each type of beam there shown, and column three of Table 15 gives the value of the load to produce the deflection A. If beams of the same material for the types given are of equal length and section their relative stiffnesses will be proportional 7? T T?T to the coefficients of -^- A, since -^- A is the same for all types, assuming equal deflections. ART. 107] MAXIMUM STRESS AND DEFLECTION 163 TABLE 15 LOAD TO CAUSE A GIVEN MAXIMUM STRESS AND A GIVEN MAXIMUM DEFLECTION Kind of beam. Load TF to cause stress /. Load tF to cause deflection A 1 ! I < 7 > ^ // ' hi 3- -/T A '/> //-, \N/l */ in. | , fl '* IW < ^i '/ y d / 8 \ 1 4- - 48. f A vy/^in. 8 -f 76i f A 1 r 1 8 {I / . 1 t F 1 ' d {T IQ2. p A Pr g J W/^/in. | -i 384. f A f Y/ General type. f, -^ 107. MAXIMUM STRESS ANIJ DEFLECTION. From Art. 1 06 it is seen that if a given beam used as a canti- lever will safely carry a given load at the end, it would carry twice that load uniformly distributed on the cantilever, four times that load if used as a simple beam with the load concentrated at the center, eight times that load if used as a simple beam with a uniform load, eight times that load if both ends are fixed and the load is concentrated at the center, and twelve times that load if both ends are fixed and the load is uniformly distributed. It is also seen that if a given load at the end of a canti- lever beam will cause a given maximum deflection, to 164 CANTILEVER AND SIMPLE BEAMS [CHAP. X cause the same maximum deflection it will take two and two-thirds times that load uniformly distributed over the cantilever, sixteen times that load if the beam is used as a simple. beam with the load concentrated at the center, twenty-five and one-fifth times that load if uni- formly distributed over a simple beam, sixty-four times that load if concentrated at the center of a fixed-ended beam, and one hundred and twenty-six times that load if uniformly distributed over a beam with both ends fixed. In all the above cases the stresses are supposed not to exceed the elastic limit. 108. RELATION BETWEEN THE MAXIMUM STRESS AND THE MAXIMUM DEFLECTION. In column 2 of Table 15 appears the maximum stress / developed under the load W, and in column 3 the maximum deflection. By equating these two expressions for the load the rela- tion of the maximum deflection to the maximum stress for a given load W is obtained. Let a represent the co- fT 77 T efficients of^y > and let/3 represent the coefficients of -=- A. cl l Then by equating the two expressions for W there results The last equation gives the maximum deflection in terms of the maximum stress. A=0.63 in. (e) Scale Inches 1 FIG. 81. xj* X 2" SIMPLE OAK BEAM, SPAN 6', LOAD 150! 2' FROM LEFT SUPPORT. 166 EXAM.] CANTILEVER AND SIMPLE BEAMS 167 EXAMPLE Draw the elastic curve for a if -inch by 2-inch oak beam of 6-ft. span carrying a load of 150 pounds 2 feet from the left support. Solution: / - ^ - '* X ** 2 X 2 = x (inch)', E = 1,500,000 pounds per square inch. In Fig. 8 1 the horizontal scale is i inch equals 12 inches. (On the diagram the length representing i inch is indicated at the bottom.) (a) represents the beam, (b) is the shear diagram in which the vertical scale is i inch equals 100 pounds. To draw the moment curve the pole distance is taken equal to 24 inches; the vertical scale of the moment curve then is i inch equals 24 X 100 = 2400 pound-inches. In drawing the slope curve the pole distance was taken equal to 20 inches; the vertical scale of the slope curve then is i inch equals 20 X 2400 -5- El = 20 X 2400 -j- 1,500,000 X i = 0.032. As yet the slope is not known for any point along the beam; consequently an arbitrary axis O'X' is assumed in curve (d) and the pole P' taken with a pole distance equal to 3 1 \ inches giving values of the deflection scale to be i inch equal to 0.032 X 31! = i.o inch. With the pole P t the curve (e'} is drawn. This gives a deflection at the right support equal to 0.53 inch. It should be zero. The closing line O'X' in (e f ) is drawn, and parallel to this line the ray P'X in (d) is drawn, then the true axis OX in (d) is drawn, and with the pole P on this axis, with a pole distance of 32! inches, draw the true elastic curve (e}. The deflections can be measured directly from this curve. The maximum deflection occurs at the point of zero slope which is 31 inches from the left support. The deflection at that point is A = 0.63 inch. If it is desired to find the maximum deflection for a 5o-pound load divide the value for 150 pounds by 3; this gives 0.21 inch. For any beam with any concentrated load at the one-third point this set of curves can be used simply by changing the scale to agree with the data of the given beam. i68 CANTILEVER AND SIMPLE BEAMS [CHAP. X PROBLEMS 1. Draw the load, shear, moment, slope, and deflection curves and determine the maximum deflection and the maximum fiber stress for the following beams: (a) A 4-inch by 8-inch timber beam used as a cantilever of 8-ft. span with a concentrated load of 500 pounds at the end. (b) A i5-inch, 42-pound, cantilever I-beam of zo-ft. span carrying a uniform load of 15,700 pounds. (c) A simple timber beam of i4-ft. span, 8 inches wide and 14 inches deep carrying a load of 1550 pounds concentrated at the center. (d) Same as (c) with an additional uniform load of 200 pounds per foot. (e) A i2-inch, 31. 5-pound I-beam of i6-ft. span when fixed at both ends and carrying a concentrated load of 2400 pounds at the center. (f) An 1 8-inch, 55-pound I-beam used as a simple beam of 2o-ft. span carrying a uniform load of 60,000 pounds. (g) A simple timber beam of lo-ft. span, 10 inches wide and 12 inches deep carrying a uniform load of 8000 pounds and a concentrated load of 2000 pounds at the center. 2. In a test of a i^-inch by 2-inch yellow pine beam of 6-ft. span the following maximum deflections for the corresponding loads at the center were observed: Load, pounds. Deflection, inches. 50 .218 100 150 2OO 374 .562 .718 250 905 What is the modulus of elasticity of the yellow pine ? Ans. 2,040,000 Ib. per sq. in. 3. What is the bending moment at the walls and at the center of a beam fixed at both ends of i6-ft. span, and carrying a con- centrated load of 8300 pounds at the center ? What is the maxi- mum fiber stress developed if a zo-inch, 25-pound I-beam is PROB.] CANTILEVER AND SIMPLE BEAMS 169 used? What is the fiber stress 4 feet from the walls? What is the shearing stress at the section 4 feet from the walls ? 4. Design a longleaf pine beam with both ends fixed to carry a uniform load of 6000 pounds on a span of 12 feet. What will be the maximum fiber stress developed at the center of the span ? Locate the inflection points. 5. What steel I-beam with fixed ends is required for a span of 20 feet to support a uniform load of 20,000 pounds, with a maximum unit-stress of 15,000 pounds per square inch ? Find also the maxi- mum deflection. 6. For a simple beam with a load concentrated at the distance kl from the left support, k being a fraction, show that the equation of the elastic curve to the left of the load is 6/ v 6/ v 7. Designabeamof 2o-ft.spanto carry 18,000 pounds, fixed ends. 8. Calculate the maximum deflection of a steel bar, supported at its ends, i in. sq., 6 ft. long, with a load of 100 pounds at its center. 9. A floor is to support a uniform load of 100 pounds per square foot. The lo-inch, 2 5 -pound I-beams have a span of 20 feet and are spaced 6 feet apart between centers. Does the maxi- mum deflection of the beams exceed alo of the span? 10. Deduce the equation of the elastic curve and the expression for the maximum deflection for a beam on which the load varies uniformly from zero at the ends to w pounds per lineal unit at the center. Given 2 (n 4 = [n (n + i) (6 n 3 + 9 n 2 + n i)] -f- 30. Ans. To the left of the center, wlx? wx* $wPx . WP y ~ ' 2 4 E7 60 Ell 192 El 60 El In the following problems write the special equations of the elastic curve and obtain the maximum deflections. 11. Ai2-in.,3if-poundl-beamusedasacantileverbeamof 20-ft. span and carrying a concentrated load of 1000 pounds at free end. 12. A lo-inch, 25-pound I-beam used as a cantilever beam of i5-ft. span and carrying a uniform load of 500 pounds per foot. 13. A 20-inch, 65-pound I-beam used as a simple beam of 24-ft. span carrying a concentrated load of 20,000 pounds at the center. 14. An i8-inch, 55-pound I-beam used as a simple beam of i5-ft. span carrying a uniform load of 4000 pounds per foot. CHAPTER XI OVERHANGING, FIXED AND SUPPORTED, AND CONTINUOUS BEAMS 109. OVERHANGING BEAM, CONCENTRATED LOADS. In Fig. 82 are drawn the shear, moment, slope, and deflec- tion diagrams for two concentrated loads on an overhang- ing beam, W\ at the left end which overhangs the support, and W 2 between the supports. After drawing curve (b), curve (c) is obtained by use of the pole P in (b). The bending moment is zero at the ends and also at the point of inflection 7. By use of the pole P in (c) the slope curve (d) is drawn. Since the value of the slope is not known at any point, the curve (e r ) is drawn by using the pole P' in (d), thus assuming the slope at the left end to be O'A. The supports A' and B' should be on a hori- zontal line. Therefore, to obtain the true elastic curve and the correct value of the slope at the left end, connect A'B' in (e r ), then draw P'X in (d) parallel to A'B'\ through X draw the horizontal axis OX, which is the true axis of reference for the slope curve. This gives the slope at the left end to be OA . Then by use of any pole P on the axis OX in (d) the true elastic curve (e) is drawn. This method is general and may be employed for any system of loading for cases in which the beam rests on two supports. If desired, the equations for the different parts of the elastic curve can be obtained by methods similar to those in Chapter X and the ex- pressions for the maximum moment, the maximum de- flection, and the location of the inflection point may be obtained. 170 FIG. 82. OVERHANGING BEAM, CONCENTRATED LOADS. 171 172 BEAMS, OVERHANGING, FIXED, ETC. [CHAP. XI no. OVERHANGING BEAM, UNIFORM LOAD. In Fig. 83 are drawn the curves for a beam overhanging both sup- ports and carrying a uniform load . The bending moment is zero at both ends and at two points between the supports. These points, marked I in curve (c), are the points of inflection. The value of the slope is not known for any point, so the curve (e f ) is drawn by using the pole P' in (d). The supports should be on the same horizontal line. By connecting A' and B', the points of support, it is seen that A 'B r is horizontal and thus the true axis in the slope curve was assumed correctly. The equa- tions of the elastic curve, and the expression for the maxi- mum moment, the maximum deflection, and the position of the points of inflection may be determined. in. BEAM FIXED AND SUPPORTED, CONCENTRATED LOAD AT THE CENTER. For beams of this kind the values of the reactions cannot be obtained without resort to the elastic curve. Referring to Fig. 84 the curves marked by letters with the subscript I are drawn as if the beam were a simple beam. To make the beam hori- zontal at the right end the restraining moment at that end must be great enough to make the deflection /i for the first element of length in Fig. 84 (ei) equal to zero, in which case the slope <*i at the right end is reduced to zero. The resisting moment at the wall decreases the left reaction and changes the shear, moment, slope, and elastic curves; from the definition of bending moment, the fixing moment at the wall is due to a force at the left reaction equal to the amount that reaction is decreased by the fixing moment. In order to determine the amount of this force, a force W\ is assumed to be acting at the left end of the beam. The shear, moment, slope, and elastic curves marked by the letters with the subscript 2 are drawn for the load W\. Since the two ends remain on 174 BEAMS, OVERHANGING, FIXED, ETC. [CHAP. XI a horizontal line the beam would curve upward. For this case the deflection would be / 2 for the first element of length shown in the curve (e 2 ), and the slope at the right end would be a 2 . If the assumed force W\ is of FIG. 84. BEAM FIXED AND SUPPORTED, CONCENTRATED LOAD AT THE CENTER. the proper magnitude so that when the effect of this force is combined with the effect of W on the simply supported beam, / 2 of curve (e%) would be equal to /i of (ei) in order that the deflection of the first element of length be zero. Or, expressed otherwise, 0:2 would be equal in magnitude to ai in order that the slope at the ART. in] BEAM FIXED AND SUPPORTED 175 right end be zero under the combination. The true force W necessary to make the moment at the wall great enough to bring the beam horizontal at that point is to the force W\ as the ratio of the slopes OL\ to a 2 ; therefore the true force is W' = Wi. By use of the definitions and equations in the previous chapters, the shear, moment, slope, and deflection of a given beam are directly proportional to the load causing them. Consequently, the reaction at the left end of the beam is lessened by the amount W. The fixing moment at the wall is W'l, or it is equal to the moment M 2 multiplied by the ratio Therefore, the fixing moment 0:2 is M'= M 2 =W l l= 2 0=2 For the true reaction at the left end of the beam with one end fixed and the other supported, the left reaction of the simple beam is reduced by the amount W , after which the true curves (&), (c), (d), and (e) may be drawn. In the foregoing solution the pole distances were taken equal, for which case the actual lengths for i and 2 may be taken for the reduction ratio. If all the pole distances were not taken equal, the actual values repre- sented by ai and 0.1 must be used in the ratio. To obtain the values of the reactions, the moment under the load, and the restraining moment at the wall, it is known that the positive area and negative area in Fig. 84 (J 2 ) are equal, since the total change in deflection over the entire length of the beam is zero. In order that these areas be equal; 176 BEAMS, OVERHANGING, FIXED, ETC. [CHAP. XI 2 = 20:3 (Art. 105). 2 -f 0:3 = jjj X - = ,(area in (c 2 ) divided by El). Wl 2 ( area -4-B-X" in (ci) divided by 7). 0:2 16 16 ' M f =-^-rWl(at the wall). ID M = Wl (under the load). The inflection point occurs at the point of zero moment which is f\ I from the fixed end. 112. BEAM, BOTH ENDS FIXED, CONCENTRATED LOAD AT ANY POINT. The method of the last article may be followed for a beam fixed at one end and supported at the other with any system of loading. When both ends are fixed and the loading is not symmetrical a method quite similar to that of the last article may be followed, but instead of finding the fixing moment at one end only, it is necessary to determine' the fixing moment at both ends. In Fig. 85 draw the curves (61), (ci), ( and - , respectively. The reaction equals 10 2 10 the algebraic difference of the shear to the right and to the left of the support. Therefore, K _ _ 4 wl j\i j\i > 10 ART. 114] THE THEOREM OF THREE MOMENTS 183 The maximum positive bending moment occurs between the supports where the vertical shear is zero. To obtain the position of zero shear let V equal zero. T/ V = 10 o. 1, 10 4 ^v 4/ ^/4A 2 _ A 10 10 2 \IO/ 200 Since the moment is zero at the left support it will be zero again where the shear areas above and below the axis are equal, o which is at the distance 2x1 or / from the left support. This 10 gives the inflection point. In the middle span the shear is FIG. 88. zero at the center. The area under the shear curve to the center ; then the bending moment at the center is is -- X - 22 --- 1 -- = - . The inflection points occur where the bend- 10 4 20 ing moment passes through zero. Therefore, the distance from 184 BEAMS, OVERHANGING, FIXED, ETC. [HAP. XI the second support to the inflection point is found from: W? , Wl WXi The slope and the deflection may be obtained by graphical methods or by the calculation of areas as is done in the previous chapters. In Fig. 88 (6) is drawn the shear diagram and in Fig. 88 (c) is drawn the moment diagram for this beam. By a method similar to that employed in the fore- going example the coefficients for wl for the vertical shears at each side of the supports for continuous beams TABLE 16 COEFFICIENTS OF wl FOR THE VERTICAL SHEAR AT THE SUPPORTS OF CONTINUOUS BEAMS. 1 -fif _^_ 8 Spa t_ 1 it' 3"" n ft -; foilo" TojTo" n 1 -, *$< 1 A ~1 ^ " ft r i 17; 15 28! 28 _13|M 28!28 - i ? l .JfiUI 28128 28 z 1 str 1 $< 3 4 * 23|20 __J 38| 38 : J9|18_ 38|38 __20|23_ 38[38 38~" t j ~ 7 - r ~T~ tr- "* J p * 4L 63 1 55 104 ~104!lOl >|< 1 49! 51 104; 104 n < i 53 ! 53 104J104 *P J1IJL _ 1011104 J5!J3_ 1041104 _ JL r t 1 it* 1 >1V 1 4r 4< 1 i >T carrying uniform loads over the entire length were ob- tained as given in Table 16. The negative coefficients of wl 2 for the bending moments at the supports were also obtained as given in Table 17. Tables 16 and 17 may be extended in the following manner: By following down to the right or to the left a line of similar supports for the different spans, to obtain ART. 115] HINGING POINTS FOR CONTINUOUS BEAMS 185 the coefficients for a beam having an odd number of spans, as five, for the second support, the moment coeffi- cient is ^V The 4 is obtained by adding the 3 of ^ to the i of T V- The 38 is obtained by adding 28 of ft- to the 10 of T V. This method may be employed for any TABLE 17 COEFFICIENTS OF -rf FOR THE BENDING MOMENT AT THE SUPPORTS OF CONTINUOUS BEAMS. No. Spans K -1 H< 1 H Ho Ho o a j o ^28 ^8 ^28 o 2 2 J 4 ^ 3 /^8 Hj Moi M(M 1 Mo4 Q G ^-.^^-^^-t-^^-^-^ support of any beam with an odd number of spans. For a beam with an even number of spans, as four, the coefficient is gV The 3 is obtained by multiplying the i of T V by 2 and adding the i of \. The 28 is obtained by multiplying the 10 of T T o by 2 and adding the 8 of \. This method can be followed for extending either Table 16 or 17. 115. HINGING POINTS FOR CONTINUOUS BEAMS. If a continuous beam is to be made of several parts, it is necessary to know at what points the various parts should be hinged, in order that the " continuous " effect may be secured, as a continuous beam is stronger than several simple beams over the various spans. Any given continuous beam may be hinged at the inflection points, and the bending moment would be unchanged along the beam. i86 BEAMS, OVERHANGING, FIXED, ETC. [CHAP, xi An economical method is to hinge the beam at such points as to make the maximum negative bending moments at the supports and the maximum positive bending moments in the spans equal in magnitude. The portion of the beam between the hinges in a span acts as a simple beam and the portions from the support to the hinges act as cantilever beams. For the case of Hinge-. Wf/m. FIG. 89. uniform loads and equal spans, Fig. 89, each hinge carries one-half the load on the intermediate length. If / is the length of one span, /i the distance between the hinges, 1 2 the distance from the support to a hinge, and w the load per unit of length, the maximum bending 7 2 moment in the center is -~- and the maximum bending o moment at "the support is ( - -\ -- 1- For equal maximum bending moments II --- T= = .14644 /. 2 -v/8 - -J= = r= .70712 . 4 + v 8 v 2 PROB.] BEAMS, OVERHANGING, FIXED, ETC. 187 From this relation the maximum bending moment is found to be ,, _ wlS _ Wl ~8~ = 76" Thus it is seen that if the beams are hinged at the proper points the efficiency is increased from thirty to sixty per cent. To use beams hinged in this way they should be fixed at the end supports. With uniform load the beams would remain horizontal at the supports, but if the load is not uniform at any time, the beam should be fixed at all the supports. Two lengths of beams could be used, one length about three- tenths the length of one span, to be used over the supports, and the other length about seven-tenths the length of one span, to be used between the hinges. If the loads are concentrated at the middle of the spans the lengths should be made equal. PROBLEMS 1. Draw the shear, moment, slope, and elastic curves for a 9-inch, 2i-pound I-beam of length 20 feet, overhanging each support 4 feet, carrying concentrated loads of 10,000 pounds at the left end, 12,000 pounds 8 feet from the left support, and 15,000 pounds at the right end. From the curves determine the deflection at each load and the maximum deflection. 2. What are the maximum shearing and fiber stresses developed in the beam of Problem No. i ? 3. Design a rectangular Washington fir beam 18 feet long, overhanging one support 4 feet, to carry a total uniform load of 9000 pounds. The shearing unit-stress is not to exceed 100 pounds l83 BEAMS, OVERHANGING, FIXED, ETC. [CHAP. XI per square inch, the maximum fiber stress is not to exceed 1200 pounds per square inch, and the maximum deflection is not to exceed sl& of the span between the supports. 4. Draw the shear, moment, slope, and elastic curves of a beam fixed at one end and supported at the other, of length /, carrying a uniform load of w pounds per lineal inch, and determine the value of the reactions, the restraining moment at the wall, the maximum positive moment, and the elastic curve. WX* wlx ~r 24 El i6EI 48 El 5. Draw the shear, moment, slope, and elastic curves for an 8-inch by lo-inch beam of 12 -ft. span fixed at one end and sup- ported at the other, carrying a concentrated load of 8000 pounds 7 feet from the restrained end. What are the maximum shearing and fiber stresses developed in the beam? 6. Solve Problem No. 5 if both ends are fixed. 7. A continuous beam of two spans carries a load of 100 pounds per foot over one span of 1 2 feet and 200 pounds per foot over the other span of 8 feet. Determine the moment at the middle support and the reactions. Ans. Mz 20,640 lb.-in., Ri =457 lb., R% = 1758 lb., R 3 = 585 lb. 8. Determine the bending moment at the middle support and the maximum positive bending moments in each span of a beam 24 feet long, one span being 10 feet and the uniform load for that span 24,000 pounds, the other span being 14 feet and the uniform load for that span 28,000 pounds. Select the proper I-beam for this loading. 9. Select the proper continuous I-beam to carry a uniform load of 144,000 pounds uniformly distributed over six spans of 12 feet each. 10. If the beam of Problem No. 9 were fixed at the supports and hinged so as to make the bending moment at the supports equal to that at the middle of the span, what I-beam would be required ? What would be the length of each section ? CHAPTER XII ELASTIC CURVE OF BEAMS DETERMINED BY THE ALGEBRAIC METHOD* 116. THE ALGEBRAIC RELATIONS BETWEEN THE FIVE CURVES. As deduced in Art. 89 the expression for the radius of curvature of a beam is El where E is the modulus of elasticity, / is the moment of inertia of the cross section about the neutral axis, and M is the bending moment. The algebraic expression for the radius of curvature for a curve as deduced in the calculus is r = (2) where x and y are the coordinates of the point of the given curve, for which r is the radius of curvature and f is the slope of the tangent to the curve at the given point. For beams the X-axis is horizontal and the F-axis is vertical, and since the slope of the elastic curve is small at all points of the beam the value of f -^- * This chapter introduces the calculus method for the only time and is intended only for students who have had courses in differential and integral calculus. 180 100 ELASTIC CURVE ALGEBRAIC METHOD [CHAP. XH generally is very small, and in comparison with i may be neglected, with which approximation dr* By substituting this value of r in equation (i) there results /g = Jf. (4) By combining this relation with those given in Chapters V and IX the following values for the section distant x from the origin result: The ordinate to the elastic curve is *=/(*)- (a) The slope of the elastic curve is <=. o.) The bending moment is (C) The vertical shear is The load per unit of length is = dx ~ dr* dx* dx* If the value of any one of the variables is known for the above equations, the values of those lower in the scale may be determined by differentiation as indicated. but usually it is necessary to start with the lower equa- tions and derive the higher ones by integration. In problems concerning beams the operation of integration between definite limits is not generally applied, con- each operation introduces a constant of Aft. ittSl ALGEBRAIC RELATIONS OF FIVE CURVES 191 integration which must be determined born the known conditions governing: the case- For deriving higher curves the equations may be written in the following form: The load per unit of length is = (i) The vertical shear is The bending moment is M = fvd* + M l =ffwd**+fvjx + Mi. (3) The slope is _, The deflection is 5 The method of evaluating these expressions will be given later. The latter set of equations is the one to be employed in determining the elastic deflections. Any one of the equations may be used to start with, if the variables can be expressed in terms of -jr. The load, shear, and moment equations can usually be written by applying the definitions. If the moment equation is used to start with, one integration and the determination of IQ2 ELASTIC CURVE ALGEBRAIC METHOD [CHAP. XII constant of integration are avoided, but since these operations are of the simplest in calculus there is no advantage in starting with any other than the load or shear equation. The constants should be determined as they appear if convenient. In the case of concen- trated loads the equation of the load curve is zero, and the shear curve probably would be the best with which to start. 117. THE CHOICE OF COORDINATE AXES. In the deduction of the formula for the radius of curvature Tf-r r = -jj the X-axis was taken parallel to the axis of the beam before bending, and the F-axis at right angles to the X-axis. The origin may be chosen arbitrarily, and for some particular cases it is more convenient to take the origin at the center of the beam, but in this book the X-axis will be taken to coincide with the axis of the beam before the beam is bent, and the F-axis will be taken at right angles to the X-axis at the left end of the span under consideration. In the solutions the proper algebraic signs should be observed. 118. THE CONSTANTS OF INTEGRATION. In all cases an approximate diagram of the deflected beam will be of value in determining the constants of integration. For problems in the determination of the deflection of beams, the constant of integration for any curve is the value of the variable at the origin, as here treated. Thus, Fi, introduced in equation (2), Art. 116, is the value of the vertical shear at the origin. See Fig. 90. Mi introduced in equation (3) is the value of the bending moment at the origin; ai introduced in equation (4) is the value of the slope at the origin; y\ introduced in equation (5) is the value of the deflection at the origin. ART. 119] CONSTANTS OF INTEGRATION 193 119. DETERMINATION OF THE CONSTANTS OF INTE- GRATION. If the values of the constants can be deter- mined, they may be inserted into the equations at once; thus, Vi and MI can be determined in many cases at first. For other cases it may be known where the shear is zero, and then the value of zero, for V and the corresponding value of x may be substituted in equation (2) to give the value of V\. If the position of zero bending moment is known, the value of M and the corresponding value of x substituted in equation (3) will give MI. Likewise to determine a\ it may be known where the slope is zero, i.e. where the beam is horizontal, and that value of a and the corresponding value of x substituted in equation (4) will give the 194 ELASTIC CURVE ALGEBRAIC METHOD [CHAP. XII value of a\. For determining y\ it is known where the deflection is zero, which is at the supports for the usual cases. ILLUSTRATIVE EXAMPLE Deduce the equation of the elastic curve and the value of the maximum deflection for a cantilever beam with a concentrated load at the end. See Fig. 91. FIG. 91. The load per unit of length = o. V = V l = -W, M = _ J Wdx -f M l = - Wx + Mi, M = o when x = o, .*. MI = o. (Zero moment), = - 1=7 f Wxdx + <*!=- --Wx* + ai . Ll J 2 rLl The slope a equals zero when x equals /, as the beam is horizontal at the wall, therefore Wl 2 2 El Wl 2 2 El (Zero slope) , Wfx y = o, for x = /, WP WP 6EI + 2EI Wx* . Wl 2 x o = w/ 3 - -^~- (Zero deflection), 3 El wi* 6 El 2 El The maximum deflection occurs where x = o, and is WP A = ART. 1 19] CONSTANTS Of INTEGRATION When there are concentrated loads on the beam the shear equation changes at every concentrated load, consequently the equations for the moment, slope, and elastic curves have different expressions on each side of the load, and for each of the curves there is one more equation than there are concentrated loads on the beam. Care should be taken in substituting values of x in these equations to see that the equation is true for the particular value of x used. With concentrated loads the two sections of the beam on each side of the load have a common slope at the load, and also a com- mon deflection. For continuous beams and overhanging beams the two sections on each side of a support have a common slope and a common deflection at the support. FIG. 92. Thus, in Fig. 92 the portions of the elastic curve AB and BC have a common tangent (i.e. a common slope) and a common deflection at the point B. Also the portions BC and CD have a common tangent and a common deflection at the point C. The beam is fixed at the point D, hence the slope of the portion CD is zero at the wall. The following principles, then, may be used in the determination of the constants of integration: (a) The section of zero vertical shear can be obtained by drawing the shear diagram, and if it occurs at a 196 ELASTIC CURVE ALGEBRAIC METHOD [CHAP. XII point where there is no concentrated load or reaction the corresponding value of x may be used in the shear equation, with V equal to zero and the value of the constant Vi determined. However, this substitution will seldom be necessary, as the value of V\ will usually be determined by other methods. (b) The section of zero bending moment will be at the free ends of beams, as at A, Fig. 92, and at the ends supported without restraint; also at the points of inflec- tion for overhanging, continuous, and restrained beams, as at /, Fig. 92, but the inflection points in such beams cannot be obtained by inspection. (c) The section of zero slope is at the horizontal por- tion of the beam, as at D, Fig. 92. For symmetrical beams carrying symmetrical loads the beam is hori- zontal at the axis of symmetry (at the center). By definition beams with fixed ends are horizontal at the fixed ends. (d) For the axes chosen the section of zero deflection is at the supports. For overhanging and continuous beams there may be one or two positions in a span where the deflection is zero, but these points cannot be determined by inspection. 120. ESSENTIAL QUANTITIES TO BE KNOWN ABOUT BEAMS. In all kinds of beams the important things to be obtained are the position and magnitude of the maximum stresses and the maximum deflection. For overhanging, continuous, and fixed beams the inflection points need to be found. When the maximum vertical shear is determined, the maximum shearing stress is y then obtained by use of the shear formula J = rr ' When the maximum bending moment is found the maximum fiber stress developed may be obtained by use of the EXAM.] ELASTIC CURVE ALGEBRAIC METHOD 197 moment formula/ = =- These formulas may be used to determine the safe load for a given beam, and also to design beams. Building specifications usually state that the maximum deflection shall not exceed a given amount, therefore it is necessary to be able to determine the maximum deflection for beams. EXAMPLES i. Deduce the equation of the elastic curve and the maximum deflection for a cantilever beam with a uniform load of w pounds per lineal unit. See Fig. 93. The load per lineal unit is w. FIG. 93. When x = l, a = o, /. o = When x o, V = o, /. M When x = o, 1,1 = o, . F x = o. (Zero shear), =- Cwxdx + M l = - * + J 2 x = o. (Zero moment), gj + i and i=^gj-- (Zero slope), '*' a 6EI + 6EI' 198 ELASTIC CURVE ALGEBRAIC METHOD [CHAP. XII /, C wxfdx . C AH-* J -67+J When x = /, y = o, 6EI wl* = and y l = - - (Zero deflection), 8 hii SEI (Elastic curve). The maximum deflection is at the end, where x = o, and is w^__ WP_ &EI SEI' 2. Deduce the equation of the elastic curve and the maximum deflection for a simple beam with a uniform load. See Fig. 94. The load per lineal unit is w. o '* FIG. 94. When s = o the vertical shear is equal to the left reaction, which . wl \ wxdx + \ - 22 When oj = o, M = o, .'. Jkfj = o. (Zero moment) . EXAM.] ELASTIC CURVE ALGEBRAIC METHOD 199 The beam is horizontal at the center, hence w? when x = -, a = o. 2 ^ f wx?dx ~ m _ wx* wlx* wPx ~ 24EI i2EI~ 24EI yi ' When x = o the deflection is zero, or y = o, /. y { = o. (Zero deflection), wx* wlx? The maximum deflection occurs at the center and is obtained by letting x = - in the equation for y, and it is found to be _^ ^ = J. WL 384 El 384 El ' 3. Deduce the equation of the elastic curve, and the value of the maximum deflection for a simple beam with a concentrated FIG. 95. load at the distance kl from the left support in which k is a fraction. See Fig. 95. Ri = W(i- k}. The load curve is at zero. To the left of the load, V = W(i-k}. U = f W(i - k} dx + M, = W (i - k) x + Mi, when x = o, M = o, /. M : = o, ... M = W(i-k)x. (b) (c) 200 ELASTIC CURVE ALGEBRAIC METHOD [CHAP. XII The beam is horizontal between the load and the center but for what value of x it is not known, hence we must let on remain in the equations till enough conditions are obtained to determine its value. y = -gj (i - k) ^ dx + J" aidx + yi, When x = o, y = o, .: y^ = o, W (d) This is all that is known about the equations to the left of the load, hence we must use those to the right of the load. To the right of the load, V=-Wk. (i) M = - f Wkdx + M / = - Wkx + MS. When x = l, M = o, /. MS = Wkl, and M = Wk (I - x) (2) Wkr n / = J (l-xJdx + aS, The value of a. for this curve is not known for any value of x. Hence / must be kept in the equation till its value can be determined. TTTl 7 / I* &V It When x = /, y = o, .: y t = -- ai I, 3 El , Wkl 3 EXAM.] ELASTIC CURVE ALGEBRAIC METHOD 201 To determine i and / it is known that the slope of both portions of the elastic curve at the load are the same. Therefore (c) and (3) are equal when x equals kl. W, .v* 1 / 1 This gives one relation between ' S^> 7 7 "~~ ' A Ib bd 3 b 2 4 2bd 2 A 12 This is one-half greater than the average vertical shear- ing stress. For circular sections the maximum shearing stress is V 64 This is one-third greater than the average vertical shearing stress. For built-up and I -sections the maximum shearing stress is approximately equal to that obtained by divid- ing the vertical shear by the area of the web AI. Rectangle I-Section The variation in the intensity of the shearing unit- stress for various sections is shown by the diagrams of Fig. 1 01. 212 SECONDARY STRESSES [CHAP. XIII Because of their small strength in shear parallel to the grain, timber beams frequently fail by shearing along the neutral surface. Beams should always be investi- gated for the maximum shearing stress developed. 123. PLATE GIRDERS, FIRST METHOD. One method of the design of plate girders is to consider all plates and angles acting as a homogeneous beam. The girder may then be designed by the use of the moment and shear formulas, and the pitch of the rivets can be determined by the use of the formula for horizontal y shear. In Sh = jrA'y', s h is the stress developed upon a unit of area of the horizontal plane. Multiplying this by b gives the total force which would be transmitted from the upper section to the lower one in one unit of length of the girder. For built-up sections the stress must be transmitted from the upper plates to those next below, through the rivets connecting the plates. If p is the pitch of the rivets, and there are n rivets in the dis- tance p, the force that each rivet must carry will be . n In The greatest number of rivets will be required where the product VA' is greatest. 124. PLATE GIRDERS, SECOND METHOD. Another method of design for plate girders is to assume that all the tensile and compressive stresses are taken by the flanges, and that the stress is uniform over the section of the flanges, and that the shear is taken by the web. The stresses calculated in this way are probably a little in excess of those actually developed, but the error is on the side of safety. ART. 124] PLATE GIRDERS, SECOND METHOD 213 FIG. 102. Let Fig. 1 02 be the cross section of a girder, A the effective area of one flange, d the distance between the centroids of the flanges, and / the unit-stress developed in the flanges, then the compressive and tensile forces equal Af and the resultants act at the centroids of the flanges. The moment of these internal stresses resists the bending mo- ment due to the external forces. For equilibrium, then, M = Afd, where M is the bending moment and Afd is the resisting moment. The bending moment increases toward the center of the span, and to increase the resisting moment with the same unit-stress the area A is increased by add- ing cover plates in the center. The pitch p of the rivets connecting the flanges to the web may now be found. (See Fig. 103.) The change in the stress FIG. 103. in the flanges between any two sections must be trans- mitted through the rivets to the web. This difference in a unit of length is - Mi V d d' since Mt MI is the rate of change of the bending moment, as the distance between the sections was taken as unity. If the rate of change of the bending moment, 214 SECONDARY STRESSES [CHAP. XIII or F, is constant, the change for the distance p is p -j- If there are n rivets in the pitch p, and R is the allowable force one rivet will transmit, _nRd P " 125. COMBINED FLEXURE AND TENSION OR COM- PRESSION. When a beam is subject to axial loads in connection with the flexural loads the maximum stress developed may be considered as made up of two parts that due to bending and J Load L that due to the axial load. The axial load increases R i R!S the tensile oj compressive B & F stress due to the bending. "/ Fig. 104 (b) is a free-body diagram of a portion of the c D E. beam under a compression (b) load P. If the deflection FIG. 104. of the beam is small the moment due to P may be neglected. If M is the bending moment due to the flexural loads, A the sectional area, and - the section modulus, Me the maximum flexural stress developed is /i = -j- (indicated by AB in compression). The compressive p unit-stress due to the axial load is /2 = -r (indicated by A FA). It is seen that the maximum compressive stress is developed in the most remote fiber in compression ART. 125] COMBINED FLEXURE AND TENSION, 215 and equals the sum of the flexural and direct stres"s and is Me P If P is a tension load the maximum tensile stress is of the same form as that given for compression. In longer beams, where the deflection is appreciable, there is an additional moment M' due to the load P. The moment is decreased if P /-"-^ P is tension and increased if P is compression. (See Fig. 105.) If Ai is the max- imum deflection due to both p IGi I05 the transverse and longi- tudinal loads the moment of P is PAi, and the total moment is M M' = M PAi. , MIC (M PAi)c h= ~T ~T The plus sign is for a compression load and the minus is for a tension load. In order to find the stress AB,or /i due to the moment of both loads, Ai must be expressed in terms of /i, the maximum stress for the deflection AI. From Art. 107, Ai = ^j~ - pJlLC Me from which fi - p To this add the direct stress -j due to the axial load. 2l6 SECONDARY STRESSES [CHAP. XIII Then the maximum stress developed is found to be f f .P Me i i \ . p J-Ji + A-~T 7 ! \ - s The minus sign is used for P in compression and the plus for P in tension. 126. COMBINED SHEARING STRESSES AND TENSILE OR COMPRESSIVE STRESSES. Let Fig. 106 represent a small portion of a beam where the c ;< x -*~~ >[ B known unit-stresses are s in shear ~ anc ^ / m tension or compression, 5 fv z \\ ^^'^ r uz an d / being at right angles to each A TJfc u *^ other. Along all diagonal planes, FIG. 106. as AB, there are normal and tan- gential components of the stresses. s p is the shearing unit-stress along the plane and / is the tensile or compressive unit-stress normal to the plane. In more advanced texts it is shown that the value of to give the maximum s p is such that tan 2 <$> = , and 2 S the corresponding maximum shearing stress is S-o The value of < to give the maximum f n is such that cot 20= , and the maximum tensile stress is In the latter equation the maximum stress will be obtained with the plus sign, and if f n is a tensile stress, the maximum / n will be a tensile stress. If / is a com- pressive stress the maximum / will be a compressive EXAM.] SECONDARY STRESSES' 217 stress. If the minus sign is used before the radical the resulting f n will be negative, which indicates that it is of opposite sign from/, i.e., / is compression for/ tension, and/ n is tension for/ compression. EXAMPLES. i. Determine the maximum horizontal shearing unit-stress in a timber beam 8 inches by 14 inches under a load of 24,000 pounds applied at the third points. Ri = jR 2 = 1 2, ooo pounds. 2. Compute the pitch of |-inch rivets for a plate girder of 72-ft. span and 7 feet 2 inches deep, the cover plates being 14 inches wide with a total thickness of | inch at the center, being connected to angles | inch thick. (NOTE: This girder was designed to carry a live train load together with the weight of the tracks and the girder.) The maximum vertical shear at the ends was found to be 137,000 pounds; at 5 feet from the ends, 121,000 pounds; at 10 feet from the ends, 102,000 pounds, etc. Each rivet will carry .601 X 8000 =4810 pounds in shear, and f X | X 18,000 = 5900 pounds in bearing. The shear governs in this case. Taking 2 rivets in the pitch p, 2 X 4810 X 86 p = - - = 6 inches at the ends. 137,000 p = 6.8 inches at 5 feet, p = 8.05 inches at 10 feet, etc. For concentration of the loads on the girder the maximum allowable pitch would be about 6 inches. 2l8 SECONDARY STRESSES [CHAP. XIII 3. A timber 8 inches by 10 inches is used as a simple beam of i2-ft. span to carry a uniform load of 4000 pounds and end com- pression loads of 40,000 pounds. What is the maximum stress developed? By assuming the deflection negligible, ,. P Me _ 40,000 4000 X 144 X 5 X 12 -~ A ~T 80 8X8X loXioX 10' = 500 + 540 = 1040 pounds per square inch. By use of the formula assuming the deflection not negligible, = 500 + 540 8 X 40,000 X I44 2 X 3 I 76.8 X 1,500,000 X 2000' = 500+ 54o( ^-37- ) = 5+ -^~z V i -.08647 9 I 36 = 500 + 590 = 1090 pounds per square inch. Thus it is seen that the moment of the axial load about the central section increases the stress about 5 per cent. 4. A bolt i inch in diameter is subjected to a tension of 3000 pounds and at the same time to a cross shear of 5000 pounds. Determine the maximum tensile and shearing unit-stresses. 5 = 5000 4- .7854 = 6370 pounds per square inch. / = 3000 -5- .7854 = 3820 pounds per square inch. By substitution in the formulas for the maximum tensile and shearing stresses, i/ V f n = + i6370 2 + - = 8560 Ib. per sq. in., tension. ,-v/ 6370* -f- = 6650 Ib. per sq. in., shear. PROB.] SECONDARY STRESSES 219 PROBLEMS 1. A simple rectangular timber beam 8 inches by 12 inches and of lo-ft. span carries a uniform load of 2000 pounds per foot. Determine the horizontal shearing unit-stress at the following points: (a) At the neutral surface over a support, (b) 3 inches from the neutral surface over a support, (c) 4 inches from the neutral surface at a quarter point. 2. What is the maximum shearing unit-stress developed in an I-beam of the largest standard section which carries a uniform load over a span of 20 feet if the maximum fiber stress does not exceed 16,000 pounds per square inch? 3. A girder of 55^-ft. span is built up of -inch by 4 feet lo-inch web plate, four 5-inch by 6-inch by f-inch angles, with the 5-inch leg riveted to the web, and four cover plates at the quarter points 14 inches by \ inch. The rivets are | inch in diameter and spaced 3 inches apart, and there are two rows in each 5-inch leg. Determine the maximum shearing and bearing unit-stresses that would probably come on the rivets. (Section is similar to that shown in Fig. 102 with cover plates added. Girder is to carry a train load.) 4. Determine the maximum stress developed in a 6-inch, 15-pound I-beam of 6-ft. span with both ends fixed, carrying a uniform load of 8 tons and tension loads at the ends of 6 tons. 5. What will be the maximum fiber stress developed in a simple timber beam 6 inches by 8 inches of 8-ft. span, with a concen- trated load of 1500 pounds at the center and end compression loads of 10,000 pounds? 6. A i2-inch, 4O-pound I-beam of 6-ft. span carries a uniform load of 1 200 pounds per foot, and is subjected to an axial compres- sion of 60,000 pounds. Find the maximum stress developed. 7. Find the size of a square maple simple beam for a simple span of 12 feet to carry a load of 500 pounds at the middle, when it is also subjected to an axial compression of 2000 pounds. 8. A bar of iron is under a direct tensile stress of 4000 pounds per square inch and a shearing stress of 3500 pounds per square inch. Find the maximum tensile and shearing unit-stresses. 9. Design a white oak beam with both ends fixed, for a span of 220 SECONDARY STRESSES [CHAP. XIII 12 feet, which is to carry a concentrated load of 4 tons at the center and a tension load of 5 tons. 10. What I-beam would be required for the loading given in Problem No. 6 if the unit-stress is not to exceed 16,000 pounds per square inch? n. What will be the maximum shearing and tensile unit- stresses developed in a |-inch bolt if it is subjected to a tension load of 5000 pounds and a cross shearing load of 5000 pounds? CHAPTER XIV COLUMNS AND STRUTS 127. DISCUSSION. The terms columns and struts are usually applied to prismatic members designed to carry compression loads when the length has an effect on the strength of the member. For short compression mem- bers, lateral deflection is inappreciable under load, while for longer ones (i.e. columns) it may be of consequence. Long columns will not carry so great loads as shorter ones of the same mate- rial and section, since the lateral bend- ing of the column causes the stress to be distributed unevenly over the cross section of the column and makes it greater on the concave side than the value obtained by dividing the load by the sectional area (see Fig. 107). The formulas used in designing columns, and in calculating the stress developed in them, are to a large extent empirical. ' I07 ' A large number of formulas have been developed by different investigators, and those in most common use will be given. 128. STIFFNESS OF COLUMNS. If a flat board is used as a column, bending will occur about an axis parallel to the longer side of a section. In all columns free to 221 222 COLUMNS AND STRUTS [CHAP. XIV bend in any direction bending will occur in the direction in which the column is least stiff. In other words, the bending will occur about the axis for which the moment of inertia and the radius of gyration are the least. The most economical column section, therefore, would be one for which the tendency to bend would be the same for all axes. The slenderness ratio is the ratio of the length of a column to the least radius of gyration of the cross section, and equals - where / is the length of the column and r is the least radius of gyration as determined by the principles of Appendix A. / and r should be in the same units, and the inch is the unit most commonly employed. 129. THE STRENGTH OF COLUMNS. The yield point of the material, which is somewhat higher than the elastic limit, is practically the ultimate strength for col- umns built of structural steel or other similar material. When a column is sensibly bent, the bending moment at the section of greatest deflection increases rapidly with a small increase of load. The moment of the load at the danger section will cause the column to fail under a load somewhat greater than that load which will develop a stress equal to the elastic limit of the material. Fig. 108 shows characteristic failures for compression specimens of timber. The short one shows oblique shear failure, the intermediate ones show failure in com- pression, and the longest one shows failure due to bend- ing of the column. The condition of the ends also has an effect on the strength of a column. Fig. 109 shows the position assumed by long homogeneous columns under load, with different end conditions: (a) with both ends round ART. 129] THE STRENGTH OF COLUMNS 223 FIG. 108. COLUMNS AND STRUTS [CHAP. XIV or pivoted, (b) with one end round and the other end fixed, and (c) with both ends fixed. Fixing the ends increases the strength of a column. In Fig. 109 it may be seen that about two-thirds of the column in () is in a condition similar to that in (a), and that half of the column in (c) is in a condition similar to that in (a). (a) (b) (c) FIG. 109. It is commonly assumed that fixing one end is equivalent to decreasing the length to f and leaving the ends round, and fixing both ends is equivalent to decreasing the length to f and leaving both ends round. In very long columns the column may fail by sidewise deflection without any portion of the material being injured. This action occurs at a lower slenderness ratio in a material like, timber than in a material like steel. The phenomenon of sidewise failure can be illustrated by the blade of a tee-square. ART. 130] THE STRAIGHT-LINE FORMULA 225 130. THE STRAIGHT-LINE FORMULA. By examining the data of tests of columns it is found by plotting points p to represent the average stress -7 at rupture, for various values of the slenderness ratio - that a straight line can be drawn which will fairly represent average ulti- p mate values of -j for the different slenderness ratios. A p Thus, in Fig. no values of -j are given along the vertical axis and values of - are given along the horizontal axis. An equation representing this straight line is of the P C* I form r =/i In this formula PI is the load at A f rupture. A similar formula may be used to determine the safe load for a given column or to determine the proper sectional area to carry a given load. Of course, the safe load P will be considerably - lower than the rup- turing load PI. The straight-line formula for the average stress ~ over the section of the FlG - IIO< P Cl column is -r = / where P is the safe load and the A T values of / and C are to be specified. In the straight- line formula / C- is considered as the allowable safe unit-stress. This formula shows that the strength of a column becomes less as the length increases and as the radius of gyration decreases. It is purely empirical, as 226 COLUMNS AND STRUTS [CHAP. XIV it is based entirely upon experimental data, but it is con- sidered to be as reliable as any and is quite generally employed. The values of /i and C\ for various materials can be derived from experimental data, and they may be used as a guide to determine values of / and C to be used in design. The greatest value of the slenderness ratio to be used in the application of the straight-line formula is usually given as I oo to 125. In any case it should not be greater than 150. P I Cooper gives the formula -r = 17,000 90- as safe yl T for soft steel columns of a through railroad bridge. p i Ketchum gives the formula -j = 16,000 70- for steel A. T columns in building frames. The limit of - is 125. The Chicago building ordinance as revised in 1910 may be taken to be illustrative of architectural practice. Its requirements are as follows : P I For steel columns -r = 16,000 70 - J\. T p I For wrought iron columns . . . -r = 12,000 6o- ^T. T p I For cast iron columns -r = 10,000 40 - A T The maximum allowable compressive stress shall not exceed the values given in Table 18. - shall not exceed 120. For timber columns the following is a modification of the formula used by Ricker and of that given in the Chicago building ordinance : P I For timber columns. . . . -r =/ .00367 - ART. 130] THE STRAIGHT-LINE FORMULA 227 in which / is the value of the allowable compressive stress parallel to the grain given in Table 8. The ordinance provides that the slenderness ratio - shall not exceed 120. The original form of this formula was for columns of rectangular section. TABLE 1 8 MAXIMUM ALLOWABLE COMPRESSIVE STRESS IN POUNDS PER SQUARE INCH, CHICAGO BUILDING ORDINANCE, 1910. Material. Compressive stress, pounds per square inch. Steel 14,000 Wrought iron Cast iron IO.OOO IO,OOO Two problems can be solved by the use of the straight- line formula, (i) the safe load a given column will carry, and (2) the design of columns. For these problems the above formulas may be used unless the specifications state otherwise. Of these problems the design of col- umns is the One most commonly met by the engineer and the architect, and it admits of many solutions. ILLUSTRATIVE EXAMPLES i. What load would a 1 5-inch, 42-pound I-beam 9 feet long safely carry if used as a column in a bridge ? From Table No. 21 giving properties of I-sections, A = 12.48 square inches, the least r = 1.08 inches. / _ QX 12 _ 108 r ~ i. 08 ~ i. 08 " By the use of Cooper's formula P = A ^17,000 - 90 A P = 12.48 X (17,000 9000) = 12.48 X 8000 = 99,840 pounds. 228 COLUMNS AND STRUTS [CHAP. XIV 2. Design a square shortleaf pine column n feet long to carry a load of 10,000 pounds. Let d be the dimension of a side and r the radius of gyration. A = d*. The least value of r may be such that / 12 X ii 132 - = 120, least r = = -*- = i.i inch. r 120 120 = v7 +A V -r n fll 12 Least d = r Vi2 = i.i X 3.46 = 3.81 inches. P I For the formula = / .00367 - , A T / = 1200 from Table 8. = 1200 - .0036 X 1200 X /IX X I2> 10,000 = i2ood z igjod, d 2 - i.64-, then c Af I = Ar 2 , and - is the slenderness ratio. The factor is a fraction TABLE 19 VALUES OF USED IN RANKINE'S FORMULA Conditions of the ends. Timber. Cast iron. Wrought iron. Steel. 4 4 4 4 3000 I. 7 8 5000 1.78 36,000 I. 7 8 25,000 I. 7 8 3000 I 5000 I 36,000 I 25,000 I 3000 5000 36,000 25,000 ART. 135] EULER'S FORMULA 233 which is determined partly by experiment and partly from the theory of Art. 133. Having experimental data available and assuming the relative strengths as given in Art. 133 the values of were found to be as given in Table 19. It may be noted that trie numerator of the fraction indicates the condition of the ends and the denominator is the characteristic for the material. / should be the allowable working stress for problems in design. 135. EULER'S FORMULA. Long Columns. Euler deduced a formula for long, ideal, homogeneous columns. For such columns it has been found that when the load reaches a certain limit, if a lateral deflection occurs the load will hold the column in equi- librium in that position. If the load is decreased the column will come back to a straight position, and if the load is increased the deflection increases until failure finally follows. From analogy with beams, the deflection of a column from the straight position varies inversely with the modulus of elasticity and the moment of inertia of the section, and directly as the square of the length (Art. 108); consequently the load a given long column will carry is directly proportional to the modulus of elasticity and to the moment of inertia and. is inversely proportional to the square of the length. Therefore the formula based on these principles has the general form P = - For n Euler deduced the theoretical value ir z for columns with both ends round, 2|7r 2 for columns with one end fixed and the other end round, 4?r 2 for columns with both ends fixed. Therefore, the values of the load P which will cause failure as determined by Euler's formula, are: Both ends round, ir*EA \ r i One end fixed, the other end round, 234 COLUMNS' AND STRUTS [CHAP. XIV Both ends fixed, \ r ) One end fixed, the other round and free to move, In these formulas E is the modulus of elasticity, 7 = Ar 2 is the least moment of inertia of the cross section, r is the least radius of gyration of the cross section, and - is the slenderness ratio. Euler's formula as given is for the critical load, and it should be modified before being used for design, and it should be used only for columns of which the slenderness ratio is not less than about 200. For design the formula would be modified by intro- ducing a " factor of safety." (See Example 4.) 136. THE THREE PROBLEMS. Three typical problems may be investigated by the use of the column formulas: (i) The in- vestigation of columns, which consists of determining the maxi- mum unit-stress developed in a given column under a given load. (This can be done only with Rankine's formula, for which case the stress is only nominal.) (2) The load which a given column will carry safely. (3) The design of a column to carry a given load. 137. ECCENTRIC LOADS ON COLUMNS. Rankine's and Euler's formulas as given above are to be used only when the load is axial. In Rankine's formula, p for axial loads, the part of the stress is due to direct compres- jrL P /A 2 sion and the part X {-) is due to the bending moment in the A \rj column. If the load has the eccentricity e, the increase in the stress due to this eccentricity by Art. 77 is ; consequently A Y EXAM.] COLUMNS AND STRUTS 235 the stress developed in a column under an eccentric load will be the sum of the three stresses and is P , P /A 2 , P ec P which may be used for columns to which Rankine's formula would ordinarily be applied for an axial load. For columns of large slenderness ratio a nearer approximation may be made as follows: If P is the load with the eccentricity e and the maximum deflection A, the total eccentricity of p the load at the point of maximum deflection is ei = e + A (Fig. 113), and by considering the stress on the cross sec- tion at that point as the result of the eccentric load P, the maximum unit-stress from Art. 77 is in which ei must be calculated. Let P be the load ob- tained by the use of Euler's formula for the column, and imagine it placed concentric with the column when the deflection is A, then the column will be in equilibrium under that load. As the column is in equilibrium under either the eccentric load P or the concentric load P , the moments at the danger section for both loads may be equated, hence P A=P(*+A) or, A--** and and / = - { i + _ gP *L . ,- II3 . This formula for eccentric loads on columns may be used for long columns for which Euler's formula could be applied for axial loads. EXAMPLES i. If two 8-inch, 1 8-pound I-beams, latticed together so that the distance between their centroids is 6^ inches, are used as a 236 COLUMNS AND STRUTS [CHAP. XIV column 20 feet long with both ends round, what is the unit-stress developed by an axial load of 40 tons ? By Rankine's formula /-5(.+.m A\ VJ I P = 80,000 pounds, A = 10.66 square inches, = , / = 20 X 12 = 240 inches, 25,000 / _ ^ tl+^-fw} 1 } _ 7500 x 1.86 10.66 V 25,000 \3.27/ / = 14,000 pounds per square inch. 2. If the load in Example No. i is applied 2 inches from the cen- troid of the section what would be the maximum stress developed ? For this case the formula is c = (x -\- width of flange) -5- 2 = (6.32 + 4) -5- 2 = 5.18 inches. = So.ooo/ + 4 /^4Q\ 2 + 2 X 5 .i8\ , 10.66 V 25,ooo\3.27/ (3-27)* / / = 7500 (i -f- .86 + .96) = 7500 X 2.82 = 21,200 pounds per square inch. These results show that when a load is axial the stress may be within the safe limit, while a slight shifting of the load may cause dangerously high stresses. 3. Design a square timber column 10 feet long with one end fixed and the other end round to carry a load of 5000 pounds safely. Let d be one dimension of the section, then A = d 2 I "d* d r = i/ = = 7 inches, / = 10 X 12 = 120 inches. V 12 -r- d 2 via For Rankine's formula = ^J and / = 800 pounds per square inch. 3000 5000 800 = 800 d 2 d* " 1.78 X 120 X 120 X 12 d 2 + 102.5 3000 Xd 2 8d 4 50 d 2 5125 = o. ART. 138] BEHAVIOR OF COLUMNS UNDER LOAD 237 Solving this as a quadratic equation, d z = 29.4, d = 5. 4 inches. For Euler's formula, by using a factor of safety of 10, 5000 X io = 2 ' 25 X 9 ' 87 X I - 5 ' 000 X = 192 d', 120 X 120 X 12 d* = 260, d = 4+ inches. As the slenderness ratio of the column is in the neighborhood of 100, the result obtained by Euler's formula is not so reliable as the other. 4. What will be the maximum stress developed in a cast iron column 4 inches in diameter and 1 2 feet 6 inches long, with both ends round, carrying a load of 70x30 pounds placed i inches from the center of the end? By the use of Rankine's modified formula for eccentric loading, we find the stress. i inch, / = 150 inches, A = 12.57 square inches, , 7000 / T 4 /i5oA , 1.5 X 2\ 7000 / T T Q . .\ Jf = i + 1 ) H 1 = - (i + io + 3) 12. 57V 5000 V i / i / 12.57 = 12,300 pounds per square inch. 138. BEHAVIOR OF COLUMNS UNDER LOAD. In columns of ordinary length used in construction the stresses set up by eccentricity of loading due to non- straightness, unevenness of bearing at ends, and other causes due to shop and erection processes, often are so great that the effect of the length of the column is almost negligible. This is especially true of columns built up 238 COLUMNS AND STRUTS [CHAP. XIV of several parts (e.g., a column built up of two channels connected by lattice work). Due to bends in the com- ponent parts of such built-up columns, slip of rivets and other causes, the extreme fiber stress, even in short columns, may be as much as 50 per cent greater than the average stress.* Furthermore, in designing columns great care should be taken that they are not built up of so thin metal that there is danger of failure by "wrinkling" of plates under load. So much uncer- tainty exists as to the action of built-up columns that low stresses should be used in designing them, and care should be taken to see that any new column is not built up of parts relatively thinner and more liable to "wrinkling" failure than are the parts of existing successful columns. A formula depending upon experimental data for its constants should be used only in designing columns similar to those from which the data were derived. For example, if a series of experiments is made upon columns of one shape of cross section, the data should not be re- lied upon in designing columns of a different shape of cross section, although the material and slenderness ratio may be the same. Whether the results of tests of small columns can be used for determining the allowable stresses in similar large columns is a disputed question among engineers. Such a procedure is sometimes necessary, and in such a case working stresses in the large columns should be low. * See Bulletin No. 44 of the Engineering Experiment Station of the University of Illinois. EXAM.] COLUMNS AND STRUTS 239 EXAMPLES 1. What should be the distance from center to center of two 8-inch, 1 8-pound I-beams latticed together for a column section to make the radii of gyration equal for the two principal axes? (Fig. 114.) The moments of inertia about the X and Y axes must be made equal. From the table of properties of I-sections, the mo- ment of inertia of one of the sections about the X-axis is 56.9 (in.) 4 , and the moment of inertia about the F'-axisis 3.78 (in.) 4 , and the area of one section is 5.33 pj G square inches. The moment of inertia of one section about the F-axis is I+Ad and equals 3.78 + 5.33 f-.l . (fx\*\ 3.78 + 5.33 f - J j. For equal moments of inertia of the built-up section about the X and F axes, 2 (56.9) = 2 (3-78 + 5-33 (-J, .'. x = 6.32 inches. 2. A column built up of two io-in., i5-lb. channels laced together with a distance of 6.33 inches between the backs, is 1 8 feet long. Determine the load the column will carry with eccen- tricities of o inch, i inch, 2 inches, 4 inches, 6 inches, 8 inches, 10 inches, 12 inches, and 14 inches, respectively, the point of application of the load being on the centroidal axis which is per- pendicular to the web of the channel, and plot the curve showing the relation between the load and the eccentricity. The moments of inertia about both principal axes are equal for the given spacing and r = 3.87 inches, A = 8.92 square inches, / = 12 X 18 = 216 inches, l =^ = 55.8, c = 6 ' 33 + 2 X 2 ' 6 r 3.87 2 = 5.77 inches, - &S'. .385. By the use of Ketchum's formula, ( i + J = 16,000 - 70 - = 16,000 - 70 X 55.8 = 12,100 pounds per square inch, 240 COLUMNS AND STRUTS [CHAP. XIV P (i + .385 e) = 8.92 X 12,100 = 107 ,800 pounds. .'. P = 107,800 pounds, P! = 107,800 -T- 1.385 = 78,000 pounds, P 2 = 107,800 -T- 1.77 = 60,900 pounds, P 4 = 107,800 -j- 2.54 = 42,500 pounds, P 6 = 107,800 -7-3.31 = 32,600 pounds, P 8 = 107,800 -T- 4.08 = 26,400 pounds, P 10 = 107,800 -7- 4.85 = 22,200 pounds, P 12 = 107,800 -T- 5.62 = 19,200 pounds, P 14 = 107,800 -7- 6.39 = 16,900 pounds. These results are plotted in Fig. 115. 100000 80000 1 360000 o PW a 1340000 o i-l 20000 Load - eccentricity Curve for a column 18 ft. long of 2 10 in. 15 Ib. channels latticed together. K 1 *!^ 16000 -^ \ \ t | \ \ s X ^v ^s ^ ) ^-t >^^ "-^<- u. 2 i 6 8 10 12 1 Eccentricity hi Inches. FIG. 115. 3. Design the upper chord of a roof truss hi which the maxi- mum stress is 65,100 pounds compression, and the length between supported points is 5 feet. Use two angles connected by f-in. gusset plates and f-in. rivets. EXAM.] COLUMNS AND STRUTS 241 Since the member is in compression the rivet holes need not be deducted. The least allowable r is 5 X I2 = - inch. I2O 2 By use of a handbook we find the properties of the angles. For direct compression alone the required area is 65,100 * 16,000 = 4.07 square inches. For the column section the area must be somewhat greater. Try two 5-inch by 3-inch by A-inch angles placed with the short legs outstanding. The least r is 1.22 inches, A = 4.82 square inches. The allowable stress is = 16,000 70 X - - = 1 2, 560 pounds per square inch. The /I 1.22 actual stress is = 5>I = 13,500 pounds per square inch. A. 4- 20 This is not safe. Try two 5-inch by 3^-inch by ^-inch angles. The least r is 1.47 inches, A = 5.12 square inches. The allowable stress is - = 16,000 - 70 X = 13,140 pounds per square inch. A 1.47 The actual stress is = 5> IO = 12,750 pounds per square inch. A. 5- 12 This is safe. The actual average unit-stress is nearly equal to the allowable, so use two 5-inch by 3^-inch by ^-inch angles with short legs outstanding. 4. Draw the diagrams representing the relation between the load and the length of columns of hemlock for the common rec- tangular sections. P I By use of the formula = / .00367- the values are obtained. A T From Table 8, / = 1000 pounds per square inch. For a 2-inch by 2-inch column, A = 4 square inches, d = 3.46 r. The maxi- mum length for which this section may be used is / = 1 20 r = When - = o, P = 4000 pounds. When / = 5 feet = 60 inches, 242 COLUMNS AND STRUTS [CHAP. XIV Safe Loads for Hemlock Columns by =1000-3.64 10 15 Length of Column in Feet. Fig. 116. ( I000 _ > I000 x / = 6o X 346 = IQ3 = 2500 pounds. The other values given in the diagram, Fig. 116, were obtained in the same manner as those for a 2 -inch by 2 -inch column. PROB.] COLUMNS AND STRUTS 243 PROBLEMS 1. Determine the distance between the backs of two 9-in., i3.25-lb. channels latticed back to back, for equal radii of gyration. Ans. 5.62 in. 2. What will be the radii of gyration with respect to the two principal axes of a column section built up of two io-in., 25-lb. I-beams and two f-in. cover plates 12 inches wide? 3. In a compression test of specimens of different lengths of the same piece of red oak of cross section i| inches by 2 inches the following values were obtained : Length, inches. Maximum load, pounds. 6 22,000 12 18,000 24 14,200 36 8,000 Plot a curve showing the relation between the average unit- stress and the slenderness ratio, and determine the value of /i and Ci at rupture, in the straight-line formula. Ans.fi = 8000 Ib. per sq. in. Ci = 63 Ib. per sq. in. 4. Design a square longleaf pine column 14 feet long to carry a load of 6 tons. 5. Design a latticed column 18 feet long built up of two steel channels to carry a load of 20 tons. 6. What safe load will a hollow cast iron column 10 feet long carry if the outside dimensions are 6 inches by 7 inches and the inside dimensions are 4 inches by 5 inches ? 7. Design a steel column 14 feet long to carry an eccen- tric load of 20 tons applied 2 inches from the outside of the column. 8. What should be the spacing of 2 -inch by 4-inch timber posts 6 feet long to carry a platform on which the maximum load is to be 200 pounds per square foot ? 244 COLUMNS AND STRUTS [CHAP. XIV 9. Find the load by Rankine's formula that would probably rupture a cast iron column with fixed ends, 18 feet long and 6 inches in diameter. 10. By the use of Rankine's modified formula for eccentric loads on columns calculate the load that would develop a unit- stress of 1000 pounds per square inch in a 6-inch by 6-inch column 10 feet long with round ends for the following eccentricities: (a), o; (b), i inch; (c), 2 inches; (d), 4 inches; and (e), 6 inches. Plot a curve showing the relation between the load and the eccentricity. 11. By use of the straight-line formula solve Problem No. 10 if the column is of Washington fir. 12. If a i2-in., 4o-lb. I-beam 18 feet long is used as a column with round ends, what is the slenderness ratio? According to Euler's formula, what load would cause rupture ? 13. What safe load will a column 27 feet long built up of two g-in., 13.25-^. channels latticed together and placed 6 inches back to back, safely carry if used in a bridge? Ans. 67,300 Ib. 14. What should be the greatest length for which timber columns of the following sections may be used? 2 inches by 2 inches, 4 inches by 4 inches, 4 inches by 6 inches, 6 inches by 6 inches, 6 inches by 10 inches, 6 inches by 12 inches, 8 inches by 8 inches, 10 inches by 12 inches, 12 inches by 12 inches. Ans. I = 34.7 where d is the least lateral dimension. 4 in. X 4 in., n ft., 7 in. 54 in. X 6 in., n ft., 7 in.; 10 in. X 12 in., 28 ft, 11 in. 15. Determine the safe load for 4-inch by 4-inch red oak columns, which are 3 feet, 7 feet, and n feet, 7 inches long respectively. Plot a curve showing the relation between the load and the length of the column. Do the same for various other sections of oak columns carrying the curves to the maximum allowable length of column. (This set of curves may be made to include all commercial sizes of sections and put on one diagram. Then the red oak column necessary for any load and any length can be selected directly from the diagram.) 16. Determine the safe load for various lengths and various sections of columns of the different kinds of timber given in Table 8, and plot the curves as in Problem No. 15. PROB.] COLUMNS AND STRUTS 245 17. Design a strut 12 feet, 9 inches long in a roof truss to carry a compression load of 12,000 pounds. Use two angles with a f-in. gusset plate between them, and f-in. rivets. Ans. 4 in. X 3 in. X TS in. [s_. 1 8. Design a strut 5 feet, 9 inches long in a roof truss to carry a load of 20,800 pounds. Ans. Two 2 \ in. X 2 in. X T\i n - ls_, short legs outstanding, f-in. gusset plate. 19. What four angles with the long legs outstanding would be required to be riveted to a iVin. plate for a column 18 feet long to carry a load of 27,370 pounds? Ans. 4 in. X 3 in. X T'S in. [s_, width of plate 8 in. 20. A wooden stick 3 -inch by 4-inch in cross section and 10 feet long is used as a column with fixed ends. Find by Rankine's formula the unit-stress developed under a load of | ton. 21. Find the safe load for a hollow cast iron column of outside dimensions 8 inches by 6 inches, inside dimensions 6 inches by 4 inches and 12 feet long. 22. A hollow yellow pine column of square section, 5 inches outside dimensions, and 4 inches inside dimensions, has a length of 1 6 feet. What load could the column safely carry? 23. A cylindrical steel column with round ends is 36 feet long and 6 inches in diameter. Calculate by Euler's formula the axial load that would probably produce rupture. 24. Determine the safe load for a hollow round cast iron column of external diameter 1 2 inches, thickness i inch, and length 12 feet. 25. A square white oak column 12 feet long is to support a load of 16 tons. What must be the size of the column? 26. Determine the size of a rectangular loblolly column 20 feet long to carry safely a load of 24 tons. Ans. 8 in. by 10 in. 27. A round solid cast iron strut 15 feet long carries a load of 10 tons. What should be its diameter? CHAPTER XV TORSION 139. STRESS AND DEFORMATION. ROUND SHAFTS. When a couple, as indicated by Pa in Fig. 117, in a FIG. 117. plane perpendicular to the axis of a shaft acts upon the shaft, it is twisted, and one cross section tends to slip by the section next to it. This tendency is resisted by the torsional stresses set up in the shaft. The stresses developed are shearing stresses. If AB in Fig. 115 is the original position and AB' the final position of an element of the surface of the shaft, the end of the shaft has twisted through the angle < or BOB' , which is pro- portional to the couple acting on the shaft and to the length of the shaft, when the stresses developed are within the elastic limit. The element will have twisted through the angle 6 or BAB', which is proportional to the couple but independent of the length of the shaft. 24 6 ART. 140] THE TORSION FORMULA 247 140. THE TORSION FORMULA. ROUND SHAFTS. In Fig. 118 let the forces producing the couple be P and the arm between them be p, then the couple C equals Pp. Under the influence of this couple the radius OA will have swept through the angle AOA' to the position OA ' while it still remains a straight line. The deformation FIG. 1 1 8. of a fiber of the section is proportional to the distance from the center of the shaft to the fiber. Therefore the unit-stress developed on the fibers when the greatest stress is below the elastic limit is proportional to the dis- tance of the fiber from the center. If s is the maximum unit-stress developed upon the outer fiber of the shaft, and r is the radius of the shaft, the unit-stress on the A/ fiber a distance y from the axis is s y = s - The total stress on the elementary area a is - ya, and the moment of this stress about the axis of the shaft is - yay = - ay 2 . r r The moment of the stresses acting on the entire cross section is the sum of all such expressions, and for equi- librium, sJ Cr C - , or 5 = -j , 248 TORSION [CHAP. XV where C = Pp and is the twisting moment, and 2 ay z = J and is the polar moment of inertia of the section about the axis. For solid circular shafts, Fig. 119, r ird* Trf 4 , A . . A x J = = (Appendix A), c 2r 2 16 i6C FIG. 119. FIG. 1 20. For hollow circular shafts of outer radius r and inner radius ri, Fig. 120, STT - ri 57T - s 2r 16 d i6Cd 7T - Three typical problems may be investigated by the use of the torsion formula: (i) The investigation, (2) determining the allowable couple, and (3) the design of a shaft to transmit a given couple. 141. STIFFNESS OF SHAFTS. The relation between the angle of twist and the shearing modulus of elasticity ART. 142] CROSS SECTION OF SHAFTS 249 may now be deduced. Since BB ' in Fig. 121 is small, BB' = r$ = Id, and 6 being in radians. The detru- sion of a fiber on the surface in the length I is BB' ', the FIG. 121. D T3t i unit detrusion is j = -y- The shearing modulus of elasticity is Cr Unit-stress s J Cl E. Unit detrusion e r$ an ut In the formulas, E a is in pounds per square inch, C is in pound-inches, / is in inches, /is in (inches) 4 , and is in radians. In tests, if the angle of twist is measured in degrees the value must be reduced to radians by the relation One radian = 57.3 degrees. 142. OTHER SHAPES OF CROSS SECTION OF SHAFTS. For any other than circular sections the foregoing formulas cannot be applied. Experiment has shown that if the section has two axes of symmetry the fibers 250 TORSION [CHAP. XV at the ends of the shorter axis have the greatest dis- tortion, and consequently the greatest unit-stress will occur at those points. Along the cor- ners of rectangular shafts there is no relative distortion of the fibers, and those fibers have no stress developed in them. Saint Venant has investigated the subject and devised formulas which reduce to the following forms for the maximum stress, where C is the twisting moment and s is the maximum stress: (a) Square shaft,* C 0.208 d 3 s. s is at the middle of the side. (a) C = 1- atfs. 16 (b) Elliptical shaft,* at the end of the shorter axis. (c) Rectangular shaf t , * C = ( s s i.Sb s is at the middle of the longer side. Merrimanf gives for rectangular cross sections the formula C = f ab 2 s. 143. POWER TRANSMITTED BY SHAFTS. The pri- mary purpose of shafting is the transmission of power. The pulleys are frequently fastened to the shaft by keys and keyways, in which case the formula for the relation between the maximum stress and the twisting moment is complex. However, the power a circular shaft with- out a keyway will transmit can easily be obtained if the allowable stress is known. If C is the couple acting on the shaft the work done by turning the shaft through an angle B is CO. Proof: Let P be the force of the couple * See "History of Elasticity," Vol. II, parti, by Todhunter and Pearson. t See Merriman's "Mechanics of Materials." EXAM.] TORSION 251 and p the arm. The distance through which P will move in turning through an angle is pd, and the work done is PpB, or C6, as Pp = C. If the shaft makes N revolutions per minute the work done in one minute will be C2wN. C2irN sJ2irN = sJN ~ 33,000 X 12 ~ 39 6 >oo r ~ 63,030 r ' 144. COMBINED TWISTING AND BENDING. If a bend- ing moment is developed in the shaft as well as a twist- ing moment, there is a combination of stresses. The maximum fiber stress developed by the bending moment may be obtained by the use of the moment formula / = , and the maximum shearing stress may be ob- tained from the torsion formula s = -=- . These stresses may be combined to obtain the maximum shearing stress and the maximum tensile or compressive stress by the formulas Sr, = EXAMPLES i. A solid circular steel shaft 10 feet long and 2 inches in diameter has a couple of 126,000 pound-inches acting upon it. (a) What is the maximum unit-stress developed in the shaft? (b) What is the unit-stress f-inch from the axis of the shaft? (c) At 300 R.P.M. what is the horse power developed ? (d) What is the angle through which one end would twist past the other? (e) Through what angle would a line on the surface twist? 252 TORSION [CHAP. XV (a) , = = l6 X > 6 = 8000 Ib. per sq. in. (b) s v = X = 6000 Ib. per sq. in. i 4 (c) H.P. = ",6QXarX3oo . go. H.P. 33,000 X 12 (e) = i X( 4 35') -120 = 2^'. 2. What should be the diameter of a solid shaft to transmit 500 horse power at 80 revolutions per minute if the maximum torsional stress is not to exceed 9000 pounds per square inch? sJN H.P. = 500 = 63,030 r 9000 Trtf X 80 63,030 X 1 6 d = 6 inches. PROBLEMS 1. What maximum unit-stress will be developed in a hollow shaft of 3 inches outside and 2 inches inside diameter when twisted by a force of 3000 pounds at a distance of i foot from the axis? What is the minimum stress developed ? Ans. 8460 Ib. per sq. in. 2. What horse power will be transmitted by the shaft in Problem No. i when making 90 revolutions per minute? 3. What must be the diameter of a solid steel shaft to transmit 1 20 horse power at 80 revolutions per minute if the allowable unit- stress is 10,000 pounds per square inch. Ans. 3.6 in. PROS.] TORSION 253 4. If the shaft of Problem No. 2 is 20 feet long between the pulleys, what will be the angle of twist when transmitting the re- quired power? 5. A wrought iron shaft 7 feet, 6 inches long and 2 inches in diameter twists through an angle of 10 30' under the influence of a couple produced by a force of 2500 pounds at a distance of i foot from the axis. Compute the shearing modulus of elasticity. 6. What are the maximum shearing and tensile stresses de- veloped in a shaft 2\ inches in diameter under a twisting moment of 12,000 pound-inches and at the same time under a bending moment of 800 pound-feet ? 7. What will be the maximum stress developed in a rectangular shaft of dimensions i inch by i \ inches if the twisting moment is 400 pound-feet? 8. Determine the maximum stress developed in a shaft i inch square if the twisting moment is produced by a force of 75 pounds at a distance of 14 inches from the axis. 9. What stress will be developed in an elliptical shaft of dimen- sions i inch by i \ inches if the twisting couple is 400 pound-feet? 10. What should be the diameter of a steel shaft to transmit safely 500 horse power at 150 revolutions per minute? 11. Calculate the horse power that a round, wrought iron shaft 8 inches in diameter and making 150 revolutions per minute will safely transmit. 12. A hollow steel shaft of outside diameter 6 inches safely transmits 450 horse power at 100 revolutions per minute. Find the inside diameter. Ans. di = 3.82 in. 13. Find the shearing modulus of elasticity of a cast iron bar 10 inches long and 0.82 inch in diameter if twisted through an angle of 1.3 by a twisting moment of 50 pound-feet. 14. A structural steel shaft 120 feet long and 16 inches in diam- eter transmits 8000 horse power at 20 revolutions per minute. Find the angle of twist and the stress developed. 15. A solid shaft 6 inches in diameter is coupled by bolts i inch in diameter on a flange coupling. The centers of the bolts are 5 inches from the axis. Find the required number of bolts. 16. A wrought iron shaft is subjected simultaneously to a 254 TORSION [CHAP. XV bending moment of 10,000 pound-inches and a twisting moment of 12,000 pound-inches. Determine the least diameter of the shaft if the maximum tensile stress is not to exceed 10,000 pounds per square inch and the shearing stress is not to exceed 8000 pounds per square inch. 17. Find the horse power that can be transmitted safely by a. cast iron shaft 3 inches in diameter and making 60 revolutions per minute. 18. A steel wire 0.18 inch in diameter and 10 inches long is twisted through an angle of 9.2 by a moment of 20 pound-inches. Determine the shearing modulus of elasticity of the wire. CHAPTER XVI REPEATED STRESSES, RESILIENCE, HYSTERESIS, IMPACT 145. REPEATED STRESSES. The behavior of mate- rials under repeated stresses and impact is somewhat different from that for static or slowly applied stresses. The experiments of Wohler, Bauschinger, and others for repeated stresses show that a material will fail TABLE 20 TESTS ON WROUGHT IRON [Wohler.] Number of applications. Unit-stress producing rupture. 800 107,000 52,800 48,400 450,000 39,000 10,140,000 35.000 under stresses lower than the ultimate strength of the material. For an enormous number of applications of a stress about equal to the elastic limit, the material ruptured. When the stress was reversed and carried to about one-half to two-thirds the elastic limit for each reversal, an enormous number of applications of the stress caused rupture. These experiments were carried on in such a manner that the time between each appli- cation or reversal of stress was so short that the specimen had no time to rest. It is interesting to note in Table 20 the variation in the maximum applied stress with the 255 256 REPEATED STRESSES, ETC. [CHAP. XVI number of applications for wrought iron. Fig. 123 shows graphically the number of applications of a given stress necessary to produce rupture in wrought iron. 100000 200000 300000 400000 500000 Number of Applications of Stress. FIG. 123. Instances in which repeated stress and reversed stresses would influence the design of the members would be shafting, car axles, piston rods, all rolling or vibrating members, etc. 146. RESILIENCE. When a load is applied to a member it will deform. On removing the load the mem- ber will resume its former size and shape for stresses below the elastic limit. And when the elastic limit has been exceeded the material will partly recover its original size and shape. The load applied to the material does work on it, and in turn when the load is being released the material gives out energy. Resilience is the amount of potential energy stored in a material when it is under stress. Elastic resilience is the amount of potential energy stored in a member when the stress is within the elastic limit. The modulus of resilience is the amount of energy stored in a unit of volume of a member when the stress is at the elastic limit. Resilience can be recovered to do work. ART. 146] RESILIENCE 257 When the stress is carried beyond the elastic limit permanent set is developed. In such cases a larger amount of work has been done upon the specimen than it will give out upon releasing the load. The work that cannot be recovered is used in permanently distorting the material, and is converted into heat. Fig. 124 shows G H F FIG. 124. a typical soft steel stress-deformation diagram. In this diagram the ordinates represent the unit-stress and the abscissas represent the unit deformation . The work done on the material is the average force times the distance through which the force acts. Since the stress-deforma- tion diagram shows the unit-stress developed in a speci- men and the corresponding unit deformation under that stress, the area between the curve and the horizontal axis represents the work done on a unit of volume of the material. When the stress is not carried beyond the elastic limit all the work done can be recovered. The triangular area ACB represents the modulus of resilience. When the point D is reached the work done on a unit of volume of the material is represented by the area ACDH, and the work that can be recovered (the resilience) for that point is represented by the area GJDH. When the point of rupture E is reached, the 258 REPEATED STRESSES, ETC. [CHAP. XVI total work done on a unit of volume of the material is represented by the area A CDEF. 147. RESILIENCE OF A BAR UNDER DIRECT STRESS. For tension or compression, let the load on the bar be P, the sectional area A, the length /, the deformation e. For stresses below the elastic limit the work done is Pe _fAfl_P_ A1 2 " 2E 2E This work done on the specimen equals the resilience stored in the specimen. Therefore the resilience is f 2 The resilience per unit of volume is ^~- If / is equal to the elastic limit the resilience per unit of volume is the modulus of resilience. 148. RESILIENCE OF A BEAM. An expression for the resilience of a beam may be deduced similarly to the following method. Take the case of a cantilever beam of length / with a concentrated load W at the end, W Fig. 125. The average force will be and the deflec- FIG. 125. tion will be A. The work done on the beam which is equal to the resilience is W W wv ART. 149] MECHANICAL HYSTERESIS 259 This can be expressed in terms of the maximum stress on the outer fiber from the formula , Me Wlc , J7 fl /-- or fF-jg. fz -~ is the same expression as obtained in Art. 147. In the case of a uniform load each elementary load does an amount of work equal to one-half the load times the distance it deflects, and is - y where w is the load per unit of length, u is an element of length, and y is the deflection at the point (see Fig. 126). As the load is uniform the work done by each element is pro- 0| portional to its deflection y. From Fig. 126 it is seen that uy is a small FlG I26 area between the X-axis and the elastic curve of the bent beam. The total work done is the summation of all such expressions as y and equals ^ wu w^ w ^ , . 7, y 2^uy = X (area between the X-axis ^^ 22 2 and the elastic curve) since 2 uy is the area between the X-axis and the elastic curve. This area may be determined by the same method as is used in finding the deflection curves. 149. MECHANICAL HYSTERESIS. In Fig. 127 is shown the stress-deformation curve for the case where the elastic limit has been exceeded. After the point A had been reached the load was removed. The curve is convex downward, as ADC indicates. On reapplying the load the 260 REPEATED STRESSES, ETC. [CHAP. XVI curve will be convex upward, as CEA. The resilience or work obtained from the material is A B CD, and that put into it is ABCE. The energy represented by the loop ADCE is lost as heat and is called mechanical hysteresis. 150. LAG. For some ~" materials at stresses be- yond the elastic limit when c FIG. 127. the load is stopped the specimen will continue to deform for some time. The metal yields while the load is not increased. This phe- nomenon is known as lag, and Fig. 128 is a stress- deformation diagram in which lag is shown. 151. THE EFFECT OF REST. By allowing a FIG. 128. specimen to rest after being stressed beyond the elastic limit, it will partly re- cover its elastic prop- erties. Fig. 129 shows the stress-def- ormation curves for steel before and after resting. In the one marked "before rest- ing" the stress had been carried beyond the elastic limit, and reversed several times, the speci- men being heated by the work done on it. FIG. 129. ART. 152] SUDDENLY APPLIED LOADS 261 152. SUDDENLY APPLIED LOADS. In the foregoing portion of the book the load was considered to be grad- ually applied to the specimen or member. If the load is suddenly applied the stresses are much higher than when the load is gradually applied. In order to get the relation between the stress produced by a gradually applied and a suddenly applied load let the deforma- tion under the load gradually applied be e, and when suddenly applied be e\. Having the deformation, the corresponding unit-stress developed can be determined, since the stress below the elastic limit is proportional to the deformation. The work done on the member when the load is gradually applied is equal to the product of We the average force and the deformation and is - . The force varies from o to W. The area OA B in Fig. 130 w represents the work done. When the load is suddenly applied the total load acts through the entire deforma- tion, as indicated by the line AB in Fig. 131, but the internal resisting stresses vary from zero to the value of FB along the line OB. When the point B is reached the external work done is We, while the work stored in We the member or the resilience is - . According to the principle of the conservation of energy the load will not stop until the resilience equals the work done, conse- quently the deformation and the stress in the member 262 REPEATED STRESSES, ETC. [CHAP. XVT will still increase. When the deformation is e the work done in excess to the resilience stored in the material is represented by the area OAB which equals^. There- fore the excess of resilience over the work done beyond w FIG. 131. the point B which is represented by the area BCD, must We be equal to . The external work done is represented by the area A CEO and the resilience stored in the mate- rial is represented by the area ODE. As the two triangles BOA and BDC are similar and equal, the similar sides must be equal, therefore, BC = AB = e, and consequently f\ = 2 f where /i is the stress due to the suddenly applied load and / is the stress developed when the load is gradually applied. This shows that the deformation and the stress developed by a load when suddenly applied are twice what they would be if the load is gradually applied. 153. IMPACT LOADS. A load W moving horizontally Wv 2 with a velocity v possesses the kinetic energy - , which ART. 153] IMPACT LOADS 263 is equal to Wh, where g is the acceleration due to gravity and h is the vertical distance the weight would fall to acquire the velocity v. This energy must be overcome by the resilience stored in the member. This energy may be equated to the resilience of the member for any given case to obtain the stress developed. For example, if the member is in direct tension or compression , 4 1 2 WhE f== \-^T J fl l E' A method more generally applied is to obtain the relation between the deformation the load would pro- duce when gradually applied and the deformation pro- duced under impact. The stress is proportional to the deformation. Let Q be the maximum total resisting force under the impact load and e\ the deformation produced by the impact load. The work done by the resisting force is L since the resisting force varies directly from zero to Q. This work is the resilience and equals the external work, 2 If the deformation under the static load W is e the following proportion results: e e\ Solving these two equations for Q and e\, e\= A/2 he, (3 = 1 264 REPEATED STRESSES, ETC. [CHAP. XVI From these equations it is seen that the deformation, and the resisting force, and the unit-stress developed, increase directly with the velocity of the load, or with the square root of the height h. 154. DROP LOADS. If the impact load falls verti- cally onto a member through the height h before im- pinging upon it, the load also does work through the deformation of the member. Then, using the same nomenclature as given in the previous article we have, Q W and - = d e Solving these equations for e\ and Q, e\ = e + \/2 he + e 2 , n TJ/ , W A/2 he + e 2 and Q = W H -- - c It is seen from these two equations that a drop of a short distance develops a high stress compared with that developed under the static load W. EXAMPLE. i. Find the amount of work necessary to stress a bar of wrought iron 5 feet long and i inch in diameter, from zero to the elastic limit 100 times. P = 25,000 X .7854 = 19,635 pounds. 25,OOO,OOO Work = - PeN = 1^35 x >005 x I00 = 4009 ft> . lb> 2 2 PROB.] REPEATED STRESSES, ETC. 265 2. If a force of 50 pounds is suddenly applied at the center of a 2-inch by 2 -inch simple timber beam of 6-ft. span what will be the deflection and what will be the maximum stress developed ? bd 3 _ 2 X8 4 L - = j C = I. 12 12 3 The deflection and stress developed are the same as those developed by twice the static load, or 100 pounds. 3 *>"* f Me 100 X 72 X i X 3 /i.- - - ' ^ = 1350 lb. per sq. in. i 4X4 3. If the weight in Example No. 2 falls i inch before impinging on the beam what stress will be developed and what will be the maximum deflection? Ai = .194 + A2 X.IQ4 X i +.04 = .194 + .65 = .844 inch. The stress developed is proportional to the deflection and is 844 /i = 1350 X ^ = 2920 pounds per square inch. Or the stress is the same as that developed by a static load of 194 218 X 72 X 3 -- - - - = 2920 pounds per square inch. 4X4 PROBLEMS i. What is the resilience stored in a cubic inch of the follow- ing materials when the stress is at the elastic limit (modulus of resilience)? (a) Wrought iron, (b) Structural steel. Ans. 12. $ in.-lb.; 20.4 in.-lb. 266 REPEATED STRESSES, ETC. [CHAP. XVI 2. What horse power is required to stress a structural steel rod 2 inches in diameter and 6 feet long from zero to the elastic limit 120 times per minute? 3. Solve Problem No. 2 if the stress is carried from one-half the elastic limit to the elastic limit each time. 4. If a load of 2000 pounds is suddenly applied to the end of a steel rod 3 feet long and 1.5 inches in diameter, what will be the deformation and the unit-stress developed? 5. If the load in Problem No. 4 is moving horizontally with a velocity of 5 feet per second at the instant of impinging on the rod, what deformation and unit-stress will be developed ? 6. If the load in Problem No. 4 is falling with a velocity of 5 feet per second at the instant of impinging on the rod, what will be the deformation and the unit-stress developed ? 7. What is the work required to deflect a 2 -inch by 4-inch timber beam of 8-ft. span by a central load that will produce a maximum stress equal to 1 200 pounds per square inch ? Solve this problem for both cases, when the beam is on the edge and when it is lying flat. 8. If a load of 2 tons falls through a distance of ^ foot, and strikes at the center of a io-in., 25-lb. I-beam of i6-ft. span, what deflection and stress will be developed ? 9. A structural steel rod is required to support a suddenly applied load of 10,000 pounds. What is the minimum diameter of the rod if permanent set is avoided? APPENDIX A CENTROIDS AND MOMENTS OF INERTIA OF AREAS AI. Such expressions as S ay and S ay 2 will occur in finding the stresses developed in beams under load, where a is an element of area and y is the distance of that element from a reference line or axis. It is neces- sary to be able to evaluate these expressions for the various shapes of cross sections found in beams. A 2 . CENTROIDS OF AREAS. The centroid of an area is the point at which a very thin homogeneous plate FIG. of the shape of the area would balance: it is the point at which, if the area were concentrated, its moment about any axis would be equal to the moment of the area as originally distributed. Calling y the distance from the ^T-axis to the centroid of the area A, a an element 267 268 APPENDIX [App. A of the area, and y the distance of that element from the X-axis : Ay = 2 ay, And calling x the distance of the centroid of the area from the F-axis, and x the distance of an element of the area from that axis, ~ S ax Ji . A The axes may be chosen arbitrarily (Fig. AI). For solids the term "centroid" is synonymous with "center of gravity," and the latter term is also frequently used with areas. A 3 . AXIS OF SYMMETRY. If a straight line can be drawn through an area dividing it into two exactly similar halves, that line is called an axis of symmetry, and an area that can be divided in this manner is called a symmetrical area. The areas shown in Fig. A 2 are FIG. A 2 . symmetrical areas and the axes shown are axes of sym- metry. If there is an axis of symmetry in an area, the centroid is located on that axis. This fact simplifies the solution for locating the centroids of a large number of areas. A 4 . CENTROID OF A TRIANGLE. Imagine the triangle to be made up of a large number of strips of very small ART. AS] CENTROID OF A SECTOR 269 width parallel to the base. Each strip may be considered an element of the area. The centroid of any strip CD, Fig. AS, is at the middle point of its length. The cen- troids of all the other strips parallel to the base come at the middle of their lengths. The line joining the centers of all these strips is a straight line and is called a median. FIG. A 3 . The centroid of the entire triangle falls on the median AB, since the centroids of all its elements fall on that line. If the triangle is considered as being made up of strips parallel to another side it is shown by the same reasoning that the centroid of the triangle lies on another median. Therefore, the centroid of a triangle is at the intersection of the medians, which is at a distance of one- third the altitude from the base. A 5 . CENTROID OF A SECTOR OF A CIRCULAR AREA. The centroid of a circular sector may be located in the following manner: Let the angle at the center subtended by the radii be 2 a, and r be the radius (Fig. A 4 ). Take the Jf-axis as the axis of symmetry. Then ~y = o. Con- sider the sector as being made up of a great number of triangular elements, as OAB. The distance of the cen- troid of the triangle from is f r, and the distance from the F-axis or x is f r cos 6. Draw the arc CED with radius equal to f r. The centroids of all elements of the sector fall on this arc. The total area of the sector may 270 APPENDIX [App. A be represented by the length of the arc CED, and the area OAB may be represented in the same way by the arc GF. The centroid of the sector would be the same as if the en- tire area were concen- trated on the arc CED. Draw GH perpendicu- lar to the X-axis and HF parallel to that axis. Since FG is very small it may be taken as a straight line. The angle FGH equals 6. The -moment of the area of the triangle OAB about the F- axis is x (GF) = f rcos0(G70,andthe moment of the total sector equals the sum- mation of all such expressions, and is 2 r cos 6 (GF) . This sum divided by the total area gives x. The area of the sector equals CED, which equals f r 2 a = f ra, - = 2%rcos6(GF) = 2 (GF) cos _ g_(gg) m %ra 2 a 2 a (GH) , and the summation of all such lengths FIG. A 4 . (GF) cos 6 equals (CD) = 2-f rsina r sn a, _ _ 2 (GF) cos _ | sin a _ 2 sin a = - r 2 a 2a 3 a The angle a must be expressed in radians in applying this formula. 7T * 4. 7" When a = - the sector is a semicircle and x 2 3^ ART. A 6 ] CENTROIDS OF COMPOSITE AREAS 271 A 6 . CENTROIDS OF COMPOSITE AREAS. For such sections as often occur in practice, where they are built up of several different parts, or when the area may be divided into simpler areas, the centroid of the area can be ob- tained by applying the fundamental formula y = -p- to the section. The method is most readily understood from an example. Let it be required to locate the cen- troid of the channel section shown in Fig. A 5 . Divide A" tfL. FIG. A 6 . the section into three rectangles A, B, and C. The following tabulated values are then found: Part. Area = a. y ay A 4X-4= 1-6 2.O 3.200 B 4X.4 = 1-6 2.O 3.200 C 5.2 X-4 = 2.08 0.2 0.416 A = 5.28 square inches, 6.816 2ay= 6.816, y = 5-28 = 1.29 inch. Since the F-axis is an axis of symmetry x = o. Another method easily applied for certain sections results from subtracting moments. The solution of the above example by this method is to consider the whole 272 APPENDIX [App. A rectangle 4 inches by 6 inches with the rectangle 3.6 inches by 5.2 inches cut away from the top as indicated FIG. A 6 . in Fig. A 6 . The following table is then obtained : Part. Area = a. y ay A 6X4 =24.00 2 48.000 B 5.2 X 3-6 = 18.72 2 . 2 41 . 184 I A = 5.28, S ay = 6.816, y = -^-r = 1.29 inches. A 7 . MOMENT OF INERTIA. The moment of inertia of an area is the summation of the products obtained by FIG. A 7 . multiplying each elementary part of the area by the square of its distance from an axis. The axis taken is ART. A 9 ] POLAR MOMENT OF INERTIA 273 called the inertia axis. Thus the moment of inertia of an area shown in Fig. A? with respect to the X-axis is: I x = 2 ay 2 , and the moment of inertia with respect to the F-axis is I y = 2 ax 2 . These expressions are for the moment of inertia of the area about axes in the plane of the area. Since the moment of inertia is the product of an area and a length squared, the units in which it is expressed are L 2 X L 2 = L 4 , a length to the fourth power. A 8 . THE RADIUS OF GYRATION. The radius of gyra- tion with respect to an axis is denned as the square root of the quotient obtained by dividing the moment of inertia of the area with respect to the same axis by the area. Thus, if / is the moment of inertia and A is the area, the radius of gyration is v/i It is seen that the radius of gyration gives the position for which a concentration of the area would give the same moment of inertia as is found for the distributed area. The value of r should not be confused with the distance to the centroid of the area. A 9 . POLAR MOMENT OF INERTIA. THE RELATION BETWEEN THE POLAR MOMENT OF INERTIA AND I x AND I y . The moment of inertia of an area about an axis perpendicular to the plane of the area is the polar mo- ment of inertia and is obtained by taking the sum of the products formed by multiplying each element of the area by the square of its perpendicular distance to the axis. If the axis is perpendicular to the plane at in Fig. AS, the distance to an element is p = Vy 2 + x 2 . 274 APPENDIX [App. A The polar moment of inertia of the area equals 2 ap 2 . If / is the polar moment of inertia of the area about that axis, j = 2 ap 2 = 2 a (y 2 + x 2 ) = S ay 2 + S ax*, since S a/ = 7* and 2 ax 2 = I v . A" A'" = r ,= 2 + r 2 . FIG. A 8 . The polar moment of inertia of a plane area about an axis perpendicular to the plane of the area equals the sum of the moments of inertia of the area about two rectangular axes in the plane of the area intersecting the given axis. This is the relation between the moments of inertia about three mutually perpendicular axes, two of which lie in the plane of the area. Aw. RELATION BETWEEN MOMENTS OF INERTIA ABOUT PARALLEL AXES IN THE PLANE OF THE AREA. In Fig. A 9 let be the centroid of the area, / the moment of inertia of the area about the J^f-axis, and I' the moment of inertia of the area about the Jf'-axis at a distance d from the centroidal axis. Then /' = S ay 2 + 2 ad 2 + 2 d S ay, 2 ay 2 = I, S aJ 2 = yl^ 2 , and 2 d 2 03; = 2 <14y = o, since y = o, ft T /I j2 .-./' = 7+ A'/. ' - 1 ART. An] I OF A PARALLELOGRAM 275 The moment of inertia of an area about an axis parallel to a centroidal axis in the plane of the area is equal to the FIG. A 9 . moment of inertia about the centroidal axis plus the area times the square of the distance between the two axes. If the moment of inertia of an area about any axis is given, that for any other parallel axis can be obtained. First, the moment of inertia about the centroidal axis must be obtained by the formula 7 = /' Ad . Second, the moment of inertia about the parallel axis can be obtained by the use of the formula I" = /+ Ad . An. THE MOMENT OF INERTIA OF A PARALLELO- GRAM ABOUT A CENTROIDAL Axis IN THE PLANE OF THE AREA. The inertia axis is taken parallel to opposite sides, b is the breadth of the parallelogram and d is the depth perpendicular to the chosen axis, Fig. AID. Let the area be divided into a large number n, of equal strips par- allel to the axis, each strip being taken so small in width that it is an element of the given area. The width of each strip is - , and the area of each strip is d, a = -b. n Let the strip shown be the pth one from the axis in which p is any number up to - , then the distance from the axis 276 APPENDIX to the element of the area shown is [App. A The moment of inertia of the parallelogram about the centroidal axis is FIG. Aw. To obtain the moment of inertia for the area of the paral- lelogram above the axis, p must represent all numbers up to - . The same is true for the area below the axis, therefore, ART. An] I OF A PARALLELOGRAM, 277 From algebra, _ n 3 + 3 n 2 -f 2 n ~^~ + 3 n* _ = 12 > The greater w is made the more nearly is the true value for / obtained, and when n becomes infinitely great the exact value of the moment of inertia is obtained. For ^ 2 this condition - and 2 become zero, - _ bd* " 12* The radius of gyration with respect to the centroidal axis is II /bd* ,. /d 2 = \/-r = I/ '- bd = I/ = V A V 12 V 12 2 The moment of inertia of a parallelogram about one of its sides is often needed, and by the application of the formula I r =I + Ad\ The corresponding radius of gyration is d r = ^' The rectangle is the usual form of parallelogram for which the moment of inertia is needed. 2 7 8 APPENDIX [App. A FIG. An. x' FIG, A 12 , ART. Ai 3 ] I OF A CIRCULAR AREA 279 Ai 2 . THE MOMENT OF INERTIA OF A TRIANGLE ABOUT ITS CENTROIDAL AXIS. The moment of inertia bd? of the parallelogram about the J^'-axis, Fig. A n , is The rectangle may be considered as being made up of two triangles ABC and ABD, both equal and similar. Consequently the moments of inertia of the two tri- angles about the ^'-axis are equal. Therefore, the moment of inertia of one of the triangles about that axisis r , U* U* I x ' = -- -T- 2 = 12 24 This axis is at the distance - from the centroidal axis o of the triangle (Fig. Ai 2 ), /. J x = I x r - Ad\ j _ bd 3 bd d* _ bd 3 J. x X ~^ ~^ * 24 2 36 36 The corresponding value of the radius of gyration is d ' = 3 V2 ; b is the base of the triangle and d is the altitude. Ais. THE MOMENT OF INERTIA OF A CIRCULAR AREA. Let d be the diameter of the circle. Let the area be divided into a great number, n, of elementary annular strips concentric with the entire area, Fig. Ai 3 . The width of each strip will be --- Let the strip shown in 2 n Fig. A 13 be the pih strip from the center, then the radius of this strip is 2 = M. 2 n The area of the element is *d d -K& n 2n 280 APPENDIX [APP. A The polar moment of inertia of the entire area about the axis perpendicular to the area at O is in which p represents all numbers up to n. = S (I 3 + 2 3 + - - ' + p* + FIG. AM. From algebra, 2 (l 3 + 2 3 + ' + P* + + W 2 ) n 2 (n + i) 2 w 4 + 2w 3 + w 2 = - = > 4 4 ... / = ^ 4 /rc 4 + 2rc 3 + rc 2 \ = 7rJ 4 /I [ T [ r Since w should be made infinitely great to obtain the true - and 2 n 4 n moment of inertia, - - and r reduce to zero, and 2 32 ART. I OF COMPOSITE AREAS 28l From Art. A 9 J = I x -(- I y and from the symmetry of the figure I x = I y , - - J *# r x TV = - I x is the moment of inertia of the circle about a diameter. 9.4" Ai 4 . MOMENT OF INERTIA OF COMPOSITE AREAS. In order to obtain the moment of inertia of a built-up section (composite area) for a given axis the area should be divided into its simpler parts, and the moment of inertia of each part with respect to the given axis ob- tained. The moment of inertia of the entire area with respect to the axis equals the sum of the moments of inertia of its component parts with respect to the axis. The ap- plication can be understood by an example. Let it be required to determine the moment of inertia and the radius of gyration of a T-section 8 inches by 9.4 inches by 0.4 inch with respect to a centroidal axis parallel to the flange of the T, Fig. A 14 . By taking moments about the X-axis, _ 8 X .4 X 9.2 + 9 X .4 X 4.5 y '- 8 X 4 + 9X4 = 6 ' 71 mches - For the part A , bd* : X FIG. A] 12 8 X .064 12 X 2.492 .04 + 19.84 = 19.88 inches. 282 For the part B, APPENDIX [App. A + 3-6 X 2.2I 2 = 24.3 + 17.58 = 41.88 inches 4 . Therefore, the moment of inertia of the section about the centroidal axis is 7 = 19.88 -f 41.88 = 61.76 inches 4 . 3.01 inches. 6.8 EXAMPLE i. Determine the moment of inertia and the radius of gyration with respect to the axis through the base and the centroidal axis, of a channel section, 4 inches by 6 inches by 0.4 inch, Fig. Ai 6 . The moment of inertia can be ob- tained for either axis and then trans- ferred to the other, or it can be obtained for each one independently. That for the axis through the base will be obtained, then the transfer to the centroidal axis made. 0.4" 1.29" A B c 4- 6- -> FIG. Ai 6 . Part. Area. J d A$ bd* 12 I A B C 1.6 1.6 2.08 2.O 2.O O. 2 2.O 2.O 0. 2 6.4 6.4 .083 2-133 2-133 .027 8-533 8-533 . II 5-28 12.883 4-2Q3 17.176 I x = 17.176 inches 4 , say 17.18, and r* = \ n = 1.8 inches. > 5.28 d = y = 1.29 inches. (Ex. p. 271.) 1 = 1- Ad* = 17.18 - 5.28 X i.2Q 2 = 17.18 - 8.77 = 8.41 inches 4 and r = \/ ^ = 1.26 inches. V 5 >2 PROB.] APPENDIX 283 PROBLEMS 1. Find the distance of the centroid of a trapezoid with one base a, the other b, and the altitude h, from the base whose length is a. An,. y=^\h. 3<* + 3& 2. Determine the moment of inertia and radius of gyration with respect to the Jf-axis arid with respect to the F-axis passing through the centroid of the area shown in Fig. A 16 . Ans. I x = $9.3 in. 4 , I y = 29. 8 in. 4 , r y = 1.68 in. :0.4' 5"9.75*Is- 0.4: JW LLL. FIG. An. 3. Locate the centroid and determine the moment of inertia and the radius of gyration with respect to the X and Y axes through the centroid for the T-section shown in Fig. An. Ans. y = 4.01 in.,/ x = 17. 4 in. 4 4. Calculate the moment of inertia and radius of gyration of a circular area of diameter 4 inches with respect to the diameter and with respect to a tangent. Also find the polar moment of inertia with respect to the center. 5. A section is built up of two 15 -in.. 33-lb. channels placed back to back. What should be the distance between them to have the moments of inertia of the section equal with respect to the two rectangular axes passing through the centroid of the section? Ans. 9.5 in. from back to back. 284 APPENDIX [APP. A 6. Find the moment of inertia and radius of gyration with respect to the centroidal X and Y axes of the I-section shown in Fig. Ai 8 . 7. A girder is built up of four 6-inch by 6-inch by i-inch FIG. A is. FIG. A 19. FIG. A 20 . angles and a 30-inch by i-inch plate; determine the moment of inertia of the section with respect to the centroidal X and Y axes. See Fig. A 19 . 8. Locate the centroid of the shaded area shown in Fig. A 20 , and find the moment of inertia with respect to the axis parallel to a side and passing through the centroid. Ans. x=y = .776 r in., I X =I V = .1368 r* in. 4 , I x =J y = .0075 r 4 in. 4 Prove that the moment of inertia of each of the following areas about the centroidal axis and the corresponding radius of gyration are as given : 9. /* = bd 3 - Mi 3 12 bd 3 - Mi 3 12 (bd -MO PROB.] APPENDIX I x = 64 285 -Si :: (b 13. L _ J bd* -k*(b- t) Tx V I2 [bd-h(b-)] 14. + hf 2 sb 3 + /i/ 3 286 APPENDIX [APP. A TABLE 21 PROPERTIES OF STANDARD LIGHT STEEL I-BEAM SECTIONS Axis perpendicular to web. Axis parallel to web. Weight Section Depth, I__l_ __ per foot, area, ncnes. Pounds. Sq. in. I I/c r / r Inches*. Inches 3 . Inches. Inches 4 . Inches. 24 80 23.32 2088.0 174.0 9.46 42.86 .36 20 80 23-73 1467.0 146.7 7.86 45-81 39 20 65 19.08 1170.0 117.0 7.83 27.86 .21 18 55 15-93 795-6 88.4 7.07 21.19 -IS 15 80 23-57 789.1 105.2 5-79 41.31 -32 IS 60 17.67 609.0 81.2 5-87 25.96 .21 IS 42 12.48 441.8 58.9 5-95 14.62 .08 12 40 11.84 268.9 44-8 4-77 13.81 08 12 31* 9.26 215.8 36.0 4-83 9-50 .OI IO 25 7-37 122. 1 24.4 4.07 6.89 0.97 9 21 6.31 84.9 18.9 3-67 S.I6 0.90 8 18 5-33 56-7 14.2 3-27 3-78 0.84 7 IS 4-42 36.2 10.4 2.86 2.6 7 0.78 6 I2j 3-6i 21.8 7-3 2.46 1.85 0.72 5 9f 2-37 12. I 4-8 2.05 1.23 0.65 4 1\ 2.21 6.0 3-o i .64 0.77 0-59 3 sl I-6 3 2.5 i-7 1.23 0.46 0-53 TABLE 22 PROPERTIES OP STANDARD LIGHT STEEL CHANNEL SECTIONS Axis perpendicular to web. Axis parallel to web. Depth, Inches. Weight per foot, Pounds. Section area, Sq. in. 7 r I r y Inches 4 . Inches. Inches 4 . Inches. Inches. IS 33 9.90 312.6 5-62 8.2 3 0.91 0.79 12 20* 6.03 128.1 4.61 3-91 0.81 0.70 IO IS 4-46 66.9 3-87 2.30 0.72 0.64 9 i3i 3-89 47-3 3-49 1.77 0.67 0.61 8 "i 3-35 32.3 3.10 1-33 0.63 0.58 7 9i 2-85 21 .1 2.72 0.98 o-59 o-SS 6 8 2.38 13.0 2-34 0.70 0-54 0.52 5 6| i-95 7-4 i-95 0.48 0.50 0.49 4 si i-SS 3-6 i-S6 0.32 o-4S 0.46 3 4 1.19 1.6 1.17 O.2O 0.41 0.44 TABLES 287 TABLE i WEIGHTS OF VARIOUS MATERIALS USED IN CONSTRUCTION Material. Weight, lb. per cu. ft. Material. Weight, lb. per cu. ft. Timber Cast iron Wrought iron . Steel 25 to 45 450 480 400 Sandstone Granite Marble Slate ISO 170 170 I7C Brass Copper, Bronze Aluminum .... 515 550 1 60 Terra cotta, facing. . . . Terra cotta, fireproof- ing . . no 5 Brick 100 to 150 Book tile 60 Limestone. . . . 165 Concrete 150 TABLE 2 ULTIMATE TENSILE STRENGTH AND ULTIMATE ELONGATION OF MATERIALS Material. Ultimate tensile strength, lb. per sq. in. Ultimate elongation, per cent. Timber 6,000 to 10,000 I ^ Cast iron 2O OOO Wrought iron 5O,OOO 2Q O Structural steel 60 ooo 2C n tO ?O O Steel wire 60 ooo to 250 ooo 10 o to 25 o 288 STRENGTH OF MATERIALS TABLE 3 ULTIMATE COMPRESSIVE STRENGTH OF MATERIALS Material. Ultimate compressive strength, Ib. per sq. in. Timber 7 OOO Cast iron OO OOO Brick . 6,OOO Brick masonry Rich concrete Stone 1,500 2,500 IO,OOO TABLE 4 ULTIMATE SHEARING STRENGTH OF MATERIALS Material. Ultimate shearing strength, Ib. per sq. in. Timber: Along grain 4.OO Across grain 3.OOO Cast iron 2O,OOO Wrought iron Structural steel Rivet steel 40,000 50,000 45,000 TABLE 5 ELASTIC LIMIT OF WROUGHT IRON AND STEEL Material. Elastic limit, Ib. per sq. in. Wrought iron Structural steel Hard steel 25,000 35.000 *\O OOO TABLES 289 TABLE 6 MODULUS OF ELASTICITY Material. Modulus of elasticity, Ib. per sq. in. Timber I,5OO,OOO Cast iron 15,000,000 Wrought iron 25,000,000 Steel 30,000,000 TABLE 7 SHEARING MODULUS OF ELASTICITY Material. Shearing modulus of elasticity, Ib. per sq. in. Timber, across grain. . . . Cast iron 4OO,OOO 6,OOO,OOO Wrought iron IO,OOO,OOO Steel I2,OOO,OOO 2 go STRENGTH OF MATERIALS TABLE 8 SAFE WORKING STRESSES IN POUNDS PER SQUARE INCH FOR STEADY LOADS Material. Tension. Shear. Compression. Bending (fiber). Perpendic- ular to grain. Parallel to grain. Timber: Cedar, white 800 600 I,OOO I,2OO I,OOO 800 800 I,OOO 800 900 1,000 1,000 IOO IOO 240 IOO 200 80 120 200 160 1 60 200 I2S IOO IOO IOO IOO 2,500 9.500 10,000 ( 8,000 ( 10,000 1 80 1 80 300 300 340 1 80 240 800 500 500 600 240 20O 20O 200 200 1,100 I, IOO I,2OO 1, 6oO 1,300 I,OOO I,2OO 1, 800 I,40O I,20O 1,750 ,400 ,000 ,200 ,200 ,200 I,OOO 1,000 I,2OO I,2OO I, IOO 800 1,300 1, 800 I,2OO 1,200 I,4OO I,2OO I,OOO I, IOO I,OOO I,OOO 6,000 I 2 ,OOO 16,000 Cypress Elm Fir, Washington .... Gum . . ... Hemlock Larch . ... Maple, sugar (hard). Maple (average) Oak, red white Pine, longleaf loblolly shortleaf yellow, (Ark., etc.) Spruce 800 800 3,000 12,000 15,000 Cast iron 12,000 I 2 ,000 12,000 18,000 (Bearing) no 250 350 Wrought iron .... Steel, structural rivet Brickwork (in lime) . . . Brickwork (in Portland cement) Concrete (Portland cement) TABLES 291 TABLE 9 COEFFICIENTS OF EXPANSION PER DEGREE FAHR. Material. Coefficient of expansion. Masonry . 0000050 Cast iron .0000062 Wrought iron Steel .0000067 .0000065 TABLE 10 VALUES OF K, K,' AND C FOR PIPES UNDER EXTERIOR PRESSURE (For use in the Carman and Carr Formula.) Material. K. K'. C. Cold-drawn seamless steel Brass Lap- welded steel 50,200,000 25,150,000 95.520 93.365 83,270 2,090 2,474 I,O2< TABLE ii EFFICIENCY OF JOINTS Kind of joint. Efficiency, per cent. Single-riveted lap joint ^0-65 Double-riveted lap joint Single-riveted butt joint 65-75 6^-75 Double-riveted butt joint 70-80 2Q2 STRENGTH OF MATERIALS TABLE 12 RELATIVE STRENGTHS IN SHEAR AND BENDING Kind of beam. Maximum vertical shear. Maximum bending moment Relative strength in shear. Relative strength in bending. I I TT77 ; I >P W I I it Wl j w/(*/in. | W 2 I 2- rfcf W Wl r f 2 4 4 w/mn. | W Wl 2 8 f f TABLE 13 MODULUS OF RUPTURE Material. Modulus of rupture Ib. per sq. in. Timber 7000 to oooo Cast iron. . 3<\ OOO TABLES 2 93 TABLE 14 MAXIMUM MOMENTS AND MAXIMUM DEFLECTIONS Kind of beam. Maximum moment Maximum deflection A. i PFJ IF/ 3 ^ Y' 3 El W/^/in. fy Wl Wl 3 1 IW t ^=3 // 2 BEI r ' Wl Wl 3 4 48 El Y// 1 (/in. Wl SWP r 1 8 .JW 384 / Tf/3 i 8 192^7 W/i $/in W 7 / 3 / / 12 ^ 2A 78/1 F7 304 /ii 294 STRENGTH OF MATERIALS TABLE 15 LOAD TO CAUSE A GIVEN MAXIMUM STRESS AND A GIVEN MAXIMUM DEFLECTION Kind of beam. Load tF to cause stress /. Load W to cause deflection A w ^ f i 8 , e. 3 - 7 A / P d 3 /3 ^ V!/l y in. ^ , A 8 / A J k V zi" " 5/28 2 /28 - >)< 1 s As f r~ f T ^ ~3 Hs 4 /3R i ^ _ 8 /<04 %4 = 296 STRENGTH OF MATERIALS TABLE 18 MAXIMUM ALLOWABLE COMPRESSIVE STRESS IN POUNDS PER SQUARE INCH, CHICAGO BUILDING ORDINANCE, 1910. Material. Compressive stress, pounds per square inch. Steel. . I4,OOO Wrought iron IO,OOO Cast iron IO,OOO TABLE 19 VALUES OF < USED IN RANKINE'S FORMULA Conditions of the ends. Timber. Cast iron. Wrought iron. Steel. 4 4 4 4 3000 I. 7 8 5000 I. 7 8 36,000 I. 7 8 25,000 1.78 Tlnth pnrlc; fiTPrl 3000 I 5000 I 36,000 I 25,000 I 3000 5000 36,000 25,000 TABLE 20 TESTS ON WROUGHT IRON UNDER REPEATED STRESSES [Wohler.} Number of applications. Unit-stress producing rupture. 800 107,000 450,000 10,140,000 52,800 48,400 39,000 35.000 TABLES 297 TABLE 21 PROPERTIES OF STANDARD LIGHT STEEL I-BEAM SECTIONS Axis perpendicular to web. Axis parallel to web. Depth, Inches. Weight per foot, Pounds. Section area, Sq. in. I I/c r 7 r Inches*. Inches 3 . Inches. Inches 4 . Inches. 24 80 23-3 2 2088.0 174.0 9.46 42.86 -36 2O 80 23-73 1467.0 146.7 7.86 45-81 39 2O 65 19.08 1170.0 II7.O 7.83 27.86 .21 18 55 15-93 795-6 88.4 7.07 21.19 -IS 15 80 23-57 789.1 105.2 5-79 41.31 32 15 60 17.67 609.0 8l.2 5-87 25.96 .21 15 42 12.48 441.8 58.9 5-95 14.62 .08 12 40 11.84 268.9 44-8 4-77 13.81 I 08 12 3*i 9.26 215.8 36.0 4-83 9-50 1. 01 IO 25 7-37 122. 1 24.4 4.07 6.89 0.97 9 21 6.31 84.9 18.9 3-67 5-16 0.90 8 18 5-33 56-7 14.2 3-27 3-78 0.84 7 15 4.42 36.2 10.4 2.86 2.67 0.78 6 12;- 3-6i 21.8 7-3 2.46 1-85 0.72 5 9i 2-37 12. I 4.8 2.05 1.23 0.65 4 7i 2.21 6.0 3-o 1.64 0.77 o-59 3 51 I-6 3 2-5 i-7 1-23 0.46 o-53 TABLE 22 PROPERTIES OF STANDARD LIGHT STEEL CHANNEL SECTIONS Axis perpendicular to web. Axis parallel to web. Weight Section Depth, Inches. per foot, Pounds. area, Ski. in. 7 Inches 4 . r Inches. 7 Inches 4 . r Inches. Inches. IS 33 9.90 312.6 5.62 8.23 0.91 0-79 12 20i 6.03 I28.I 4.61 3-91 0.81 0.70 10 15 4.46 66.9 3-87 2.30 0.72 0.64 9 I3l 3-89 47-3 3-49 1.77 0.67 0.61 8 IIj 3-35 32.3 3-io i-33 0.63 0.58 7 9f 2-85 21. I 2.72 0.98 -59 o-55 6 8 2.38 13.0 2-34 0.70 0-54 0.52 5 6* !-95 7-4 i-95 0.48 0.50 0-49 4 5i i-55 3-6 1-56 0.32 0-45 0.46 3 4 1.19 1.6 1.17 o. 20 0.41 0.44 INDEX PAGE Area, reduction of 19 sectional 10 symmetrical 268 Areas, centroids of 267 centroids of composite 271 moment of inertia of circular 279 moment of inertia of composite 281 Axes, choice of coordinate 192 Axis, inertia 273 neutral 87, 88 of symmetry 268 Bar, resilience of a, under direct stress 258 Beam 47 cantilever 47, 65, 73, 140, 144, 147 continuous 48, 1 78 fixed 48 fixed and supported 1 70 fixed at both ends 154, 158, 176 forces acting on a, as a whole 49 forces acting on a portion of a 51 overhanging 48, 168, 170 resilience of a 258 simple 47, 65, 73, 148, 150 Beams, continuous 1 78 deflection of 135 design of 95 essential quantities to be known about 196 hinging points for continuous 185 horizontal shear in 208 investigation of 93 methods of loading 48 of uniform strength 99 radius of curvature of 130 299 300 INDEX PAGE Beams, relative strength and stiffness of 160 safe loads for 94 the three problems 93 Bending 130 combined twisting and 251 Bending moment 58 diagrams 61 diagrams for cantilever, and simple beams 65 relation between the vertical shear and the 62 sign and unit of 59 the maximum 64, 65, 74 the rate of change of 63 the values of 60 values of the maximum 64, 74, 164 Brittleness 3 Butt joint 33, 38 Cantilever beam 47, 65, 73, 140, 144, 147 Cast iron 4 Centroid, of a triangle 268 of a sector of a circular area 269 Centroids, of areas 267 of composite areas 271 Chimneys 106 Circular area, centroid of sector of a 269 moment of inertia of a 279 Closing line 124 Coefficients of expansion 27 Columns and struts 221 behavior of, under load 237 comparative strength of; condition of the ends 230 eccentric loads on 228, 234 long, Euler's formula 233 Rankine's formula 231 stiffness of 221 the methods of transmitting loads to 229 the straight-line formula 225 the strength of 222 the three problems 234 Combined stresses 216, 251 Comparative strength of columns 230 Composite areas, centroids of 271 INDEX 301 PAGE Composite areas, moments of inertia of 281 Compression 12 combined flexure and tension or 214 Concentrated load 48 Concrete, plain and reinforced 4 Constant of integration 119, 124, 192 determination of 193 Continuous beam 48 Contraflexure, points of 157 Coordinate axes, choice of 192 Curve, elastic 47, 130, 135 load 55 Cylinders, thick, under interior pressure 28 under exterior pressure 29 Dams 106 Danger section 64 Dead loads 49 Deflection of beams 135 maximum 162 rate of increase of " 137 relation between the maximum stress and maximum deflection. . 164 Deformation 9 Design of riveted joints 39 Detrusion 9 Diagrams, bending moment 61 free-body 1 1 load and shear 54 load, shear, and moment, for cantilever and simple beams 65 maximum stress 97 relation between the load and shear 57 stress-deformation 15 Direct stresses 9 simple cases of 25 Drop loads 264 Ductility 3 Eccentric loads, case of, caused by a combination of the weights of the material and lateral pressure no Eccentric loads on columns 228 Eccentric loads on short prisms 107 302 INDEX PAGE Eccentricity of a load that will produce zero stress in the outside fiber 108 Efficiency of riveted joints 38 Elastic curve 47, 130, 135 Elastic limit 16 Elastic resilience 256 Elasticity 2 coefficient of 17 modulus of 17 Elongation 9 ultimate 1 1 Euler's formula, long columns 233 Expansion, coefficients of 27 Factor of safety 21 Fiber stresses, distribution of 87 First integrated curve 116 Five curves, relation between the 137 Flexure, combined, and tension or compression 214 Force 9 Free-body diagram , 1 1 Funicular polygon 121 Girders, plate 212 Gyration, radius of 273 Hinging points for continuous beams 185 Hooke's law 17 Hoop, stresses in a 26 Horizontal shear in beams 208 Hysteresis, mechanical 259 Impact loads 262 Inertia axis 273 Inflection points 15? Integrated curves 116, 117 Integration, constant of 119, 124, 192 Internal stresses 5 1 Joints, riveted 3 2 boiler 3 2 butt.. 33. 3 8 INDEX 303 PAGE Joints, compression loads for 38 design of riveted 39 efficiency of riveted 38 kinds of riveted 32 lap, single-riveted 35 lap, double-riveted 36 lap, with more than two rows of rivets 37 methods of failure of riveted 33 pipe 32 structural 32 tank 32 Kern, the no effect when the resultant falls outside of the in the maximum stress when the line of action of the resultant falls outside the middle third for rectangular prisms which take no tension 112 Lag 260 Lap joint 32 Live loads 49 Load, axial 10 case of eccentric, caused by a combination of the weight of the material and lateral pressure no diagrams 54 diagrams for cantilever and simple beams 65 Loads, drop 264 eccentric, on columns 228 eccentric, on short prisms 107 moving, on beams 74 impact 262 suddenly applied 261 Masonry 3 brick 3 concrete 4 stone 3 Mechanical hysteresis 259 Mechanical properties 2 Modulus of elasticity 17 uses of 20 Modulus, shearing 18 304 INDEX PAGE Modulus of resilience 256 Modulus of rigidity ig Modulus of rupture 97 Modulus of transverse elasticity 19 Modulus, section go Young's 17 Moment, assumptions for the resisting 86 bending 58 maximum 64, 74, 162 resisting 86 resisting, assumptions for the 86 Moment formula, the . go Moment of inertia 272 of a circular area 279 of composite areas 281 of a parallelogram about a centroidal axis 275 of a triangle about a centroidal axis 27g polar 273 relation between, about parallel axes 274 Moments, the theorem of three 181, 205 Moving concentrated loads on a beam 74 Neutral axis 87 position of the 88 Neutral surface 87, 132 position of the 88 slope of the 132 Non-uniform loads 4g Overhanging beam 48, 168, 1 70 Parallelogram, moment of inertia of 275 Permanent set 16 Physical properties 2 Piers 106 Pipe, stresses in a thin 25 Plasticity 2 Plate girders 212 Points, of inflection 157 hinging, for continuous beams 185 Poisson's ratio ig Polar moment of inertia 273 INDEX 305 PAGE Pole 121 Pole distance 121 Prisms, eccentric loads on short 107 Properties, mechanical and physical 2 Radius of curvature of beams 130 Radius of gyration 273 Rankine's formula 231 Ray polygon 121 Rays 121 Reduction of area 19 Relations between the five curves 137 algebraic 189 Relative strength and stiffness of beams 160 Repeated stresses 255 Resilience 18, 256 elastic 256 modulus of 256 of a bar under direct stress 258 of a beam 258 Resisting moment 86 Resisting shear 84 Rest, the effect of 260 Riveted joints 32 computation of unit-stresses developed in 34 efficiency of 38 kinds of 32 methods of failure of 33 Rupture, modulus of 97 Second integrated curve, the first method of obtaining the 117 the second method of obtaining the 119 Section, cross 10 Section modulus 90 Shafts, round, stress and deformation 246 other shapes of cross section of 249 power transmitted by . . 250 stiffness of 248 Shear formula, the . 84 values of k in the 85 Shear 13 double 35 306 INDEX PAGE Shear, horizontal, in beams 208 oblique 13 resisting 84 single 35 vertical 52 vertical, sign and unit of 53 vertical, values of 53 Shearing modulus of elasticity 18 Shearing stress, the maximum 209 combined, and tensile or compressive stress 214 Shell, stresses in a thin 25 Shortening 9 Simple beam 47, 65, 73, 148, 150 Slenderness ratio 222 Slope curve 134 Slope of the neutral surface 132 Slope, the rate of change of 135 Spheres, stresses in thin 28 Steel 5 Stiffness 2 Stiffness of beams, relative 160 Stiffness of shafts 248 Straight-line formula, the 225 Strength of beams, relative 73 Strength, rupturing 1 1 ultimate compressive 12 ultimate shearing 13 ultimate tensile n Stress 9 axial 10 combined shearing and tensile, or compressive 216 fiber 90 maximum, diagrams 97 maximum horizontal and vertical shearing unit, at a point 209 relation between the maximum stress and the maximum deflection 164 Stress-deformation diagrams 15 Stresses, compressive 12 due to change in temperature 27 fiber, distribution of : 87 flexural 47 in a hoop 26 internal 5 1 INDEX 307 PAGE Stresses, in thin cylinders 25 in thin spheres 28 maximum 209 repeated 255 shearing 13 tensile 10 total horizontal compressive and tensile 91 used in design 20 working 20 String polygon 121 Strings 121 Struts, columns and 221 Suddenly applied loads 261 Surface, neutral 87, 88 Symmetry, axis of 268 Temperature, stresses due to change in 27 Tension 10 combined flexure and, or compression 214 Theorem of three moments 181. 205 Timber ' 4 Torsion 246 the, formula 247 Toughness 3 Triangle, centroid of a 268 moment of inertia of a 279 Twisting, combined, and bending 251 Uniform load 49 Unit-stress 9 Units for the five curves 137 Varying load 49 Vector polygon 121 Vertical shear 52 diagrams 54 diagrams for cantilever and simple beams 65 maximum, the 64, 65 rate of change of 57 relation between, and the bending moment 62 sign and unit of 53 values of 53 values of the maximum 64, 65 308 INDEX PAGE Walls 106 Wrought iron 5 Yield point 16 Young's modulus " 17 Zero bending moment, the section of 196 Zero deflection, the section of 196 Zero slope, the section of 196 Zero vertical shear, the section of 195 UNIVERSITY OF CALIFORNIA LIBRARY THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW is 5 1920 226.183