Class / ft <. Rnnk - n<^ Gopyiightl^?. /^/^ COPYRIGHT DEPOSn^ THE ELEMENTS OF MECHANICS OF MATERIALS A TEXT FOR STUDENTS IN ENGINEERING COURSES BY C. E. HOUGHTON, A.B., M.M.E. ASSOCIATE PROFESSOR OF MECHANICAL ENGINEERING NEW YORK UNIVERSITY ILLUSTRATED SECOND EDITION REVISED AND ENLARGED NEW YORK D. VAN NOSTRAND COMPANY 1915 J -rK^?^ Copyright, 1909, By D. van NOSTRAND COMPANY. Copyright, 1915, By D. van NOSTRAND COMPANY. THE SCIENTIFIC PRESS ROBERT DRUMMOND AND COMPANY BROOKLYN. N. Y. QGT-4I9I5 PREFACE This is not a treatise on the Mechanics of Materials. The efforts of Merriman, Burr, Lanza, and others cover the field so thoroughly that there is no present need of such a work. It is designed to be an elementary text-book for students in the engineering courses in colleges and universities, where the time allotted to the subject does not exceed three or four recitations per week, for one half year, and where the course is preceded by college courses in mathe- matics, through integral calculus, mechanics, and physics. The extreme mathematical treatment of the subject has been avoided, but where the use of higher mathematics leads to clearness they have been freely used. As it is intended as a text-book, the general cases are discussed fully, leaving the student to derive the formulas for special cases as part of the regular problem work. At the end of each chapter there are review questions covering the more important parts of the subjects dis- cussed and problems illustrating the s^^me. The solution of one problem of each type has been given to show the application of the general formulas. The appendix contains tables giving the values of the engineering constants of materials and the formulas com- monly used in design, in addition to the tables usually found in books of this character. The notation has been made uniform with that of Mer- riman's works, so that his more complete treatise on the subject may be conveniently used as a reference book. Nbw York, January, 1909. lU TABLE OF CONTENTS CHAPTER I Applied Mechanics AJBTIOLE PAGE 1. Forces in structures 1 2. Axial forces 3 3. A bar . 3 4. Internal forces 3 6. Tensile or compressive stresses 4 6. Unit stress 4 7. Maximum tensile or compressive stresses .... 5 8. Shearing stresses , . 5 9. External and internal forces 6 10. Deformation of elastic bodies 7 11. Unit deformations . . . . ... . . 7 12. Modulus or coefficient of elasticity . . . . .8 13. The elastic limit 8 14. Ultimate strength 9 15. Resilience 10 16. Ductility. 11 17. Elastic resilience .12 18. Use of formulas . . 12 19. Constants of materials 13 20. Units 14 21. Working stresses ; factors of safety 14 22. Accuracy of calculations 15 Examination questions 16 Problems 18 CHAPTER II Applications 23. Bars of uniform strength , .22 24. Thin pipes, cylinders, and spheres 25 VI TABLE OF CONTENTS 25. Thick pipes .... 26. Riveted joints . . • 27. Tension in plates . 28. Shear on rivets 29. Compression on rivets or plates 30. General case of a riveted joint 31. Kinds of riveted joints . 32. Efficiency of a riveted joint . 33. Stresses due to change of temperature Problems PAOK ^7 30 30 32 33 33 35 37 39 40 CHAPTER III Beams 34. Kinds of beams 46 35. Reactions at the supports . . . . . . .47 36. Uniform and concentrated loads 47 37. Vertical shear . . .49 38. Bending moment . .50 39. Resisting shear 51 40. Resisting moment .52 41. Use of formula 55 42. Shear and moment diagrams 58 43. Shear diagrams 59 44. Moment diagrams 60 45. The relation between the vertical shear and bending moment 63 46. Relative strengths of simple and cantilever beams . . 64 47. Overhanging beams 65 48. Beams of uniform strength 66 49. Moving loads . . 67 50. Use of formula 69 51. Examination questions 70 Problems 73 CHAPTER IV Torsion 52. Derivation of formula 53. Modulus of section 54. Square sections 83 86 86 TABLE OF CONTENTS Yll ARTICLE 55. Illustrations . . . . 56. Twist of shafts 57. Relative strengths and stiffness 58. Horse power of shafts . 59. Shaft couplings 60. Modulus of rupture for torsion 61. Helical springs Examination questions . Problems .... PAGE 87 89 90 92 93 94 95 CHAPTER V The Elastic Curve 62. Definition .99 63. The equation of the elastic curve 99 64. Deflection of beams 102 65. Fixed or restrained beams . 105 66. Beams fixed at both ends 107 67. Continuous beams . . . 109 Examination questions 114 Problems . . . ^ 116 CHAPTER VI Long Columns 68. Stresses in long columns 69. Euler's formula for long columns . 70. Columns with round or pin ends . 71. Columns with square, flat, or fixed ends 72. Columns with round and square ends 73. Rankine's formula for long columns 74. Applications Examination questions . Problems 123 124 125 126 128 131 135 187 139 CHAPTER Vn Combined Stresses 75. Stresses due to force 76. Tension or compression combined with bending 143 143 viii TABLE OF CONTENTS ARTICLE PAGE 77. Roof rafters . 147 78. Eccentric axial loads . , , . . , , . 149 79. Shear and axial stress ..,,,,,. 149 80. Maximum stresses • . , . 150 81. Horizontal shear in beams 152 82. Maximum stresses in beams . 156 Examination questions 158 Problems 160 CHAPTER Vm Compound Bars and Beams 83. Definition 163 84. Compound columns, alternate layers 163 85. Compound columns, longitudinal layers .... 163 86. Compound beams . 165 87. Reenforced concrete beams ....... 166 88. Straight line fornmla for reenforced concrete beams . . 166 Examination questions 173 Problems . . 174 Tables, Explanation of 176 Table 1. Notation . . . . • 176 Table 2. Fundamental formulas 178 Table 3. Derived formulas 179 Table 4. Properties of beams 182 Table 5. Average constants of materials 183 Table 6. Properties of sections 184 MECHANICS OF MATERIALS CHAPTER I APPLIED MECHANICS Article 1. Forces in Structures. One of the problems that confronts the engineer called upon to design any machine or structure is to so propor- tion the various parts that they will resist the forces that act on them. To do this, he must apply the laws of mechanics to the forces to be resisted, and study the action of the materials under the same forces. This application of mechanics may be termed Applied Mechanics or the Mechanics of Materials. If we consider any structure or any member of the structure to. be at rest, according to the laws of mechanics the forces that act on the structure or member must be in equilibrium. The various parts of a machine often have relative motion, but by introducing a force equal and op- posite to the force Avhich produces that motion, the forces that act on the member of a machine may also be treated as a system of forces in equilibrium. In the more extended treatment of this subject the forces are taken as acting in different planes. The simpler theory that treats the forces as coplanar is the one that will be used here. 1 2 MECHANICS OF MATERIALS Since each member must be designed separately, if the forces that act on any member are determined, we will find that the forces may be resolved into : Forces tending to lengthen the member, Forces tending to shorten the member, Forces tending to bend the member, Forces tending to shear the member. Imagine any body, acted on by a system of forces in equilibrium, denoting the sum of the components of the forces parallel to some line as the -X" forces, and those per- pendicular to the same line as the ]r forces; the sum of the X forces is zero and the sum of the Y forces is zero. If we take as the line of reference the axis of the body pass- ing through the center of gravity of the body, the result- ant of the X forces will be a couple, unless the lines of action of the forces of the couple coincide with the axis. Each force of this couple may be replaced by a single force of equal magnitude, acting in the line of the axis and a couple whose moment is the moment of the force about a point in the axis. The X forces acting in the line of the axis will tend to either lengthen or shorten the member in the line of the axis, and the couple as well as the Y forces will tend to bend the member. If the member be cut by a plane perpendicular to the axis, the Y forces on either side of the section will be opposite in sign, and in general tend to slide one part of the member, relative to the other, along the plane of the section. Any of the resultants may be zero, and in that case there would be no tendency to deformation in that line. The force acting on any member is always transmitted by a surface of finite area, but by considering that each APPLIED MECHANICS elementary area of the surfaces in contact transmits the same amount of pressure, we may use the resultant of these elementary pressures passing through the center of gravity of the areas in contact, as the force applied. Art. 2. Axial Forces. When the lines of action of the acting forces lie in the axis of the body, passing through the centers of gravity of all sections perpendicular to the axis, the forces are called Axial Forces^ and their effect is to either lengthen or shorten the member. V \r a \^ Fig. 4. a Art. 3. A Bar. The member may have any shape whatever, JC the simplest be- ing a prismatic or cylindrical form, where any section perpendicular to the axis has the same shape and area. This form will be called a Bar. Art. 4. Internal Forces or Stresses. Let Fig. 4 represent a bar under the action of the axial forces P and P. Suppose the bar to be cut by any plane through the axis into the segments a and 6, and consider the segment a. 4 MECHANICS OF MATERIALS This segment is acted on by the external force P, and as a part of the whole bar it is in equilibrium ; hence there must be forces acting in the plane section X-X^ whose resultant acts in the same line, and is equal and opposite to the force P. As the same is true of the segment 5, the internal forces, called stresses, acting between a and 5, hold the segments a and h in equilibrium against the external forces. Therefore, in any plane section of the bar there exists a pair of equal and opposite forces or stresses, each of which are induced by and resist the external forces. In general, the stresses may be resolved into components parallel and perpendicular to the plane section. The components parallel to the plane of the section prevent the sliding of the segments along the plane, and are termed Shearing Stresses, while those perpendicular to the plane are called either Tensile or Compressive Stresses, depending on whether they tend to extend or compress the particles on which they act. Art. 5. Tensile or Compressive Stresses. When the external forces are axial, and the section per- pendicular to the axis, the stresses can have no component parallel to the plane of the section; hence axial forces can produce only tensile or compressive stresses in planes per- pendicular to the axis. The plane dividing the bar into the segments a and b was any plane ; hence the reasoning holds true for all such planes, and there are only tensile or compressive stresses equal to the external force P, in all sections perpendicular to the axis. Art. 6. Unit Stress. Since the force P is the resultant of all the equal unit pressures on the areas in contact, it is reasonable to assume APPLIED MECHANICS 5 that the stresses on each unit of area of the plane section are also equal, and if S equals the sum of the stresses acting on each unit of area of the plane section, and A is the area of that section, then F = AS. (a) S is termed the unit stress, and is the resisting force per square unit of area ; hence S must be expressed in the same units as P and A. Art. 7. Maximum Tensile or Compressive Stresses. When the cutting plane is not perpendicular to the axis, the resultant stress may be resolved into components parallel and perpendicular to the plane of the section, those perpendicular being either tensile or compressive stresses, while those parallel are shearing stresses. As neither component can equal the resultant, it is evi- dent that the maximum tensile or compressive stresses v^ill be found in a section perpendicular to the axis. In such a section there are no shearing stresses, and when the bar has. a uniform section area A, the formula (tx) will determine the maximum tensile or compressive unit stress induced by the axial force P. If the areas of all sections perpendicular to the axis are not equal, the greatest unit stress will be found where the section area is the least, and the value of A to be used in formula (a) is the area of the least section. Art. 8. Shearing Stresses. When the external forces act in adjacent parallel lines, since the stresses can have no component perpendicular to the line of action of the forces, the stress in a section parallel to the line of action of the forces must be a 6 MECHANICS OF MATERIALS Shearing Stress^ as the external forces tend to slide the two sections of the bar along the plane of the section. (Fig. 8.) Assuming that the stress is uniformly distributed over the section, and that Sg is the unit stress in shear, then formula {a) P = AS^^ where A is the area of the section 5 Fig. 8. parallel to the line of the forces and P the forces pro- ducing the shear, will always give the relation between the external forces and the maximum unit shearing stress in the section. Art. 9. External and Internal Forces. The external forces on any member of a machine or structure are the weights or loads that member has to support and the pressure it receives from the adjacent members, while the internal forces are those that transmit the external force from element to element through the member. These latter forces are stresses, and the inter- nal force per unit of area is the Unit Stress^ and will be designated by the letter aS', with a subscript ^, c, or s, as the stress is tension, compression, or shear. The formula P = AS is a general one and applies to all cases where . the stress is uniformly distributed over the area of any plane section A^ and S, the kind of unit stress. APPLIED MECHANICS 7 Art. 10. Deformation of Elastic Bodies. In the study of mechanics, the forces were assumed to act on rigid bodies; that is, on bodies whose shape was not altered by the application of the force. As there are no rigid bodies in nature, every force applied produces some deformation or change of shape. This fact, however, does not prevent the application of the laws of mechanics to elastic bodies under the action of force after the deforma- tion has taken place, since equilibrium must exist at that time. Art. 11. Unit Deformations. Consider a bar, I units in length, and A square units in section, under the action of an axial force P. From equa- P tion (a), S = --, S\^ constant, since P and A are constant A for all sections perpendicular to the axis, or each square unit of every section is acted on by a force S. Suppose the bar to be divided into bars, each one unit in section and I units long, then each of these bars is acted on by a force S. The change in the length ?, that may take place under the force S^ being e, since all particles are under the same force, the change in the length must be equal to the length Z, multiplied by the unit load S^ and some number which depends on the nature of the material. Calling this number -— , the value of e must be e =—. This may be written - = — . Letting r = e equal -tj I IE I the change in the length of a bar one unit in length, or the unit deformation, it follows that e = — or ■p _S_ unit stress ^7 >. € unit deformation 8 MECHANICS OF IVLATERIALS Art. 12. Modulus or Coefficient of Elasticity. If the value of S does not exceed a critical value which varies for different materials, U will be constant for all values of S, and this constant is called tlie Modulus or Co- efficient of Elasticity. This constant is the value of tlie ratio of the unit stress to the unit deformation, as may be seen by the inspection of formula (6). PI P Equation (5) may be written E = — -, since S = ~ and e = -, and as A and I are constants for any bar under axial forces, E will be constant when e varies directly with P. Art. 13. Elastic Limit. If a bar length I is subjected to a small axial force P, it is observed that the length has changed a certain small amount. If Pg is twice P, experiment shows that the change in length is twice that due to P, and if P^ is n times P, the change in the length is n times that due to P, provided that P^ does not exceed a certain limiting value which varies for each material and bar. That is, within that limit the change in length of any bar is pro- portional to the external force applied. If P^ is the limiting value for a given bar, then the corresponding value of S as derived from formula {a) must be the limiting value of the unit stress. This value of S^ being a unit stress, is independent of the dimensions of the bar and depends only on the material of the bar, is called the Elastic Limit of the material or the limit of elasticity. The elastic limit of a material may be defined as the unit stress for which the deformations cease to be propor- tional to the applied force. APPLIED MECHANICS 9 To determine this value of S for any material, axial loads are applied to a bar of the material, the loads being applied in small equal increments, and the change in length due to each increment of load measured. For any total load less than the load producing a stress equal to the elastic limit, the last increment of load should produce the same change in length as any previous increment. Therefore, when an increase in the change in length for any equal increment of load is noted, the elastic limit has been passed. It will be noticed that the exact value of elastic limit depends on the accuracy with which the loads and especially the de- formations are measured. It has also been observed that if the stress in any bar is less than the elastic limit, the bar will return to its origi- nal length when the load is removed, and if the stress is slightly above the elastic limit, there will be some perma- nent change in the length of the bar. The unit stress at which this yielding takes place is called the Yield Point or the Commercial Elastic Limit. The latter term comes from the commercial practice of determining the elastic limit by the drop of the beam in the testing machine. While the yield point or commercial elastic limit is from 3 to 5 % higher than the true elastic limit, the ease with which the latter value may be determined and the fact that the allowable value of aS' for any engineering structure rarely exceeds one half of the elastic limit, com- bine to make it the one in general use. Art. 14. Ultimate Strength. After the elastic limit is reached, the change in length increases more and more rapidly as the loads are increased, and finally a load is reached that causes the bar to rupture. 10 MECHANICS OF MATERIALS If P is the load that causes rupture and A is the original area of the bar, then the value of S obtained from formula (a) is called the Ultimate Strength of the material. Art. 15. Resilience. If a bar of wrought iron, whose section area is A^ and the length I as measured between two punch marks on the bar, is placed in a testing machine, the tensile loads and their corresponding extensions being measured and plotted to scale on section paper, the result would be a diagram similar to that in Fig. 15. § b a /T — ^.^^^^^^'^^ / /I / 1 1 1 1 y f Change in /en gth, or unit- deformation. Fig. 15. Since P = AS^ the ordinates representing loads may, by a change of scale, represent unit stresses, and the ab- scissa representing total elongations may also represent unit elongations. Such a diagram is called a Stress-strain diagram. The use of the word "strain" gives it the meaning of "unit deformation." Authorities do not agree on the use of the word, some giving it the meaning as above, while others use it to mean load. On account of APPLIED MECHANICS 11 this ambiguity, the term "unit deformation" will be used in its place. The unit stress for any load is obtained by dividing the load by the original area, and the unit elon- gations by dividing the elongation for any load by the original length I, The point a on the curve is the elastic limit, and b is the ultimate strength of the material. After a load corresponding to h is reached, the bar begins to reduce at some point very rapidly, finally breaking at a load less than the load at h. This load is called the Breaking Load^ and has but little significance in engineering work. Since one coordinate represents force, and the other space, the area o ah c e, when measured in the proper units, is the work done in breaking a bar of unit volume. If we define the Ultimate Resilience as the work done in breaking a bar of unit volume, the area represents that quantity. Art. 16. Ductility. As the whole curve is rarely ever determined, the Duc- tility^ a term that is defined by its method of calculation and is proportional to the ultimate resilience, is generally used in its place. The ductility of any material is calculated as follows: After the bar has been broken by a tensile load, the pieces are removed from the testing machine and the broken ends placed together. The distance between tlie original punch marks, being ^, has now increased to Z+^; then -t. is called the Ductility^ and is usually stated as a percentage of the original length. 12 MECHANICS OF MATERIALS Art. 17. Elastic Resilience. The area oaf is easily calculated, as it is the area of a triangle, and when measured in work units is called the Elastic Resilience. This work is evidently the unit stress at a x the unit deformation at a 2 Calling this value Ar, it is the elastic resilience of the material, or the work done in raising the unit stress in a bar of unit volume from to the elastic limit. The same reasoning is true for any stress less than the elastic limit, giving a method of calculating the work necessary to produce any given stress less than the elastic limit in a bar of unit volume. If S is any stress less than the elastic limit, and e the unit deformation at that stress, then ^ = J aS'c is the work done per unit of volume, and the work done on any bar whose volume is F' is 1^ times that quantity. Taking the work done on any bar as ^ Se x V, substituting for S its value --, for e, - and for F", Al, the expression for the work done on any bar reduces to K = ^ Pe which is in terms of the total load and deformation, and is true when P is less than the load, causing a stress within the elastic limit. Art. 18. Use of Formulas. Whenever any stress can be assumed to be uniformly distributed over the area of any section of a bar, formula (a), P = AS, gives the relation between the load and the stresses on that area, and any two of these quantities involved being given, the other may be found. The relation between the load and the deformation for APPLIED MECHANICS 13 >S PI such cases is given by U = — = — — , which is true when e Ae aS' = — is less than the elastic limit, and is applicable to .A. all problems involving the deformations of a bar under axial loads. Also k = ^ Se and K = ^ Pe can be used under the same conditions when the data for the problem include a consideration of the work done in deforming a bar. When the load is axial and the stress may be taken as uniformly distributed over the area of any section, these three equa- tions furnish the means for the design and investigations of the strength of all members of a structure or machine. Writing (a) aS' = ^, (&) ^=-, and A; = 1 Se, it will be noticed that the equations are simply the algebraic ex- pressions of the definitions of Unit Stress, Modulus of Elasticity, and the Unit Resilience, making it easier to remember either the formula or the definition. Art. 19. Constants of Material. These three equations make use of the following con- stants of materials: Elastic Limit, Ultimate Strength, Modulus of Elasticity, and Modulus of Elastic Resilience, all of which have to be determined by experimental work on bars of the various materials. As this work cannot be done for each problem, a table containing average values of these constants for the more common materials will be found in the Appendix, and the question of the units in which each is expressed becomes important. 14 MECHANICS OF MATERIALS Art. 20. Units. In American practice, the linear unit is the inch ; the square unit, the square inch, and the unit of weight, the pound. In the tables, the values of E. L., U., and E., are given in pounds / square inch and k in inch pounds. Therefore when the solution of any problem requires the use of any of these constants, all of the quantities involving weights or loads must be in pounds, and those involving linear or square measure, in inches. Art. 21. Working Stresses ; Factors of Safety. No material is entirely free from flaws and imperfections, which tend to diminish the area that is effective in resist- ing the external force, and in no case should the stress in any member be greater than the elastic limit, as such a stress would cause some permanent deformation. Sud- denly applied loads, shocks, and loads producing alternate tension and compression, all produce stresses that are p greater than the value of S = — , which is the value of S "" A for the same loads gradually and steadily applied. When any of these conditions occur, they tend to reduce the allowable or safe value of S to be used in the equation P = AS. Therefore, if U is the ultimate strength of the material, the allowable value of the unit stress can be found by dividing the ultimate strength by some number. Calling this number the Factor of Safety, F, S = —is tlie value of F fS to be used in connection with the formula F = AS, and is termed the Safe or Working Stress. The factor F depends — APPLIED MECHANICS 15 1. On the reliability of the material ; that is, the lia- bility of flaws or imperfections that may reduce the effect- ive area of the section. 2. On the way in which the loads are applied. The first part has been termed a factor of ignorance, while the second may be determined more or less accurately from theoretical considerations. The factors of safety as given in the tables in the Appen- dix are those to be used in the solution of the problems given in this book, and it must be remembered that the values are only approximate. The safety of any structure calling for a large factor, while the consideration of cost always demands the smallest one, the final choice of the factor of safety to be used in any given case must be largely a question of engineering judgment. In some cases, as in buildings, the allowable stress under the various kinds of loadings is a part of the building laws, and the engineer has to conform to the law\ For steady loads and reliable material the smallest fac- tor in general use is about four. On account of the danger of permanent injury to the material, no stress should exceed the elastic limit ; hence, it would seem better engineering to base the factor of safety on the elastic limit rather than on the ultimate strength, but such practice is not general in engineering work. Art. 22. Accuracy of Calculations. As the use of a factor of safety of four will result in an area of section about twice what it would have been had the allowable stress been equal to the elastic limit, and the values of E. L., U., and E., being determined by experi- 16 MECHANICS OF MATERIALS ment, are liable to an error of from 3 to 5 %, there is no necessity for absolute accuracy in the calculations for the design of the different parts of a structure. Students are urged to use a slide rule in the numerical solution of the problems given in this text, not only to save time during the college course, but because they will find the use of a slide rule almost necessary in the prac-^ tice of their profession. The slide rule should always give results that are not in error more than 1, or at the outside 2 %, which is close enough for the greater part of engineering calculations. They are warned that it is the significant figures in a number that is to be used as a factor, and not the decimal point, that is of importance. If in one case the area was given as 12,500 sq. in. and as .00125 sq. in. in another, the figure 5 is of equal importance in each case. The use of the latter area as .0012 sq. in. will result in an error of 4 9^ in the value of the stress. EXAMINATION QUESTIONS 1. There is no relative motion between the different parts of a bridge, therefore each part must be in equi- librium. How is it that the same laws may be applied to machine parts, which we know have relative motion ? 2. When may forces be considered as Axial ? 3. Define the term "Bar" as used in the text. 4. What is a stress ? If there are stresses in every section of a bar, why is it that there is no relative motion between the different parts ? 5. When may stresses be termed Tensile or Compres- sive Stresses ? 6. What is a Unit Stress ? APPLIED MECHANICS 17 7. If a member of any structure varies in size at dif- ferent parts of its length, how can you find the maximum tensile or compressive unit stress due to an axial load ? 8. Give some examples of forces that produce tensile, compressive, and shearing stresses. 9. What are the external forces for any member of a structure ? What are the internal forces ? 10. Show that P = AS, give the units involved, and the limits of use for the formula. 11. The science of mechanics is based on the action of forces on rigid bodies. Why is it that the same laws may be applied to forces acting on elastic bodies ? 12. What is meant by the expression, Unit Deformation ? 13. If the modulus of elasticity for steel is 30,000,000 and for wrought iron is 25,000,000, and one bar of each is the same size and carries the same tensile load, which bar will stretch the most? 14. The value of U may be termed a measure of the rigidity of a material. Why ? 15. Define Elastic Limit and Ultimate Strength. 16. Why does the Commercial Elastic Limit or Yield Point differ. from the true elastic limit? 17. What is a Stress-strain diagram ? 18. If a load corresponding to the ultimate strength of the material is placed on a bar, why is not that load called the Breaking Load ? Article 14 says otherwise. Explain. 19. What is meant by the term Ductility ? 20. Define Elastic Resilience. 21. State the formulas for calculating the elastic resili- ence and modulus of elasticity ; give the units involved and the limits of use for each formula. 18 MECHANICS OF MATERIALS 22. What is a factor of safety ? A working stress ? What is the difference between a working stress and a safe stress ? What do you understand by a safe load ? 23. Show why, if the calculations for the stresses in any member are not in error more than 2%, they are substantially correct. PROBLEMS 1. A square steel bar 2x2 in. in section and 4 ft. long, carries a tensile load of 60,000 lb. Required the unit tensile stress. Solution. The relation between the load, area, and unit stress, when the load is axial, is always given hj P = AS, hence substituting for F and A, from the data given in the problem, 60,000 = 4: X S, or S = 15,000 Ib./sq. in. 2. A round wooden column, 16 in. in diameter and 12 ft. 6 in. long, supports a load of 20 tons. Required the unit stress. 3. What is the value of the maximum tensile load the bar in problem i will carry ? 4. A wrought iron bar is 2 in. in diameter and 5 ft. long. What tensile load may be carried if the unit stress does not exceed 10,000 Ib./sq. in. ? 5. What is the maximum tensile load the bar in prob- lem 4 will support ? 6. A square cast iron column is hollow, 10 x 10 in. on the outside and 8x8 in. on the inside. Required the maximum compressive load that may be carried. 7. In problem 6, keeping the outside dimensions the same, required the inside dimensions if the load is 360,000 lb. and the unit stress is 10,000 Ib./sq. in. 8. A punch is 1 in. in diameter. Required the prob- able pressure necessary to force the punch through a steel plate, ^ in. thick. APPLIED MECHANICS 19 Solution. In order to force the punch through the plate, the unit shearing stress on an area equal to the cylindrical surface of the punched hole must be the ultimate shearing stress of the material ; hence, P = 7r(^«x»S' = 7rxlx^x 50,000 = 80,000 lb. approximately. 9. Using a punch 1 in. in diameter and an available force of 80,000 lb., what is the thickest wrought iron that can be punched ? 10. An iron casting is bolted to the floor by four one-inch bolts, and a force tends to slide the casting along the floor. Neglecting friction, what is the probable magnitude of the force when the unit shearing stress in the bolts is 10,000 lb. / sq. in. ? 11. If the force in problem 10 was 24,000 lb., select four standard bolts, so that the unit shearing stress will not exceed 10,000 lb. / sq. in. 12. If steel costs five cents per pound and wrought iron four cents, which will it be cheaper to use to carry a tensile load if the same factor of safety is used in each case ? Note. Assume the weights per cubic foot are the same for each ; then the weights in each case will be proportional to the areas of the sections. 13. With wrought iron at four cents per pound, and other conditions the same as in 12, how much can you afford to pay for steel ? 14. Required the probable elongation of the bar in problem 1. Solution. The relation between the elongation and an axial load PI is given by ^ — — , and substituting the data as given in the prob- Ae lem, 30,000,000 = ^Q^QQQ x ^^ , ... g := .024 in. 4e 20 MECHANICS OF MATERIALS 15. How much work is done by the force in prob- lem 1 ? 16. How much work is done by the force in prob- lem 4 ? 17. A concrete pier 3 ft. by 4 ft. in area carries a load of 300 tons. Required the unit stress. 18. A brick pier carries the same load with a unit stress of 18 tons / sq. ft. Required the area of the sec- tion. 19. The thickness of the head of a standard bolt is approximately equal to the diameter of the bolt. Com- pare the unit tensile stress in the bolt with the unit shear- ing stress in the head. 20. A standard steel bolt IJ in. in diameter supports a tensile load of 9800 lb. Required the factors of safety for the tensile and shearing stresses. (The least area to resist tension is at the root of the thread. See tables.) 21. If the bar in problem l was 4 ft. 2 in. long and the elongation was .025 in., required the modulus of elasticity. 22. If the modulus of elasticity of wood is 1,500,000, required the shortening of the column in problem 2. 23. A steel bar 1 in. in diameter has two punch marks 8 in. apart marked on it. The bar is placed in a testing machine and it is found that there is a rapid change in the rate of elongation, when the load was 24,000 lb. and after a load of 47,000 lb., no more load could be added, the bar finally breaking between the punch marks, when the load was 42,000 lb. The broken pieces were placed end to end, and the distance between the punch marks was found to be 10.4 in. Required the elastic limit, ultimate strength, and the ductility of the material. APPLIED MECHANICS 21 24. If the bar in problem 1 has the load increased from 60,000 to 120,000 lb., how much more work is done ? 25. How much should the bar in problem 24 stretch while the additional load is being added ? 26. A certain grade of piano wire has an elastic limit of 100 tons / sq. in. If the diameter of a piece of the wire is .05 in. in diameter, required the diameter of a wrought iron wire to carry the same load when the unit stress is equal to the elastic limit in each case. 27. Show that the work done by an axial force on 1 S^ any bar is -Sr= - — x volume, provided the elastic limit is not passed. 28. If the load in problem 1 is suddenly applied, what will be the value of the maximum stress induced in the bar? 29. If a square wrought iron bar is to sustain a sud- denly applied load of 60,000 lb. and the stress is not to exceed 15,000 lb. / sq. in., required the dimensions of the bar. 30. A round steel rod is to carry a tensile load of 37,700 lb. with a factor of safety of five ; required the diameter of the bar. 31. Find the factor of safety in problem 1. 32. A steel rod 2 in. in diameter in a bridge truss has a unit stress due to the weight of the bridge of 4000 lb. / sq. in. A heavily loaded truck, if placed on the bridge, will add 51,000 lb. load to that already on the rod. Is it safe for the truck to cross ? CHAPTER II APPLICATIONS Article 23. Bars of Uniform Strength. In the previous chapter, the bar was one of uniform section and no account was taken of its weight. When the bar was short, the effect of the weight of the bar could be neglected in comparison with the applied loads, and the unit stress found was that due to the loads alone. If the bar under axial forces is very long, the stress due to its own weight becomes too large to be neglected, and the stress in the bar is that due to the loads plus that due to its own weight. Take the case of a wire rope used to hoist a bucket from a deep mine shaft. The weight of the bucket and its contents produces a certain unit stress in the rope that is equal at all sections of the rope. If P is the weight of the bucket and its contents, and P A the area of the section of the rope, this stress is S = —-, The sectional area of the rope at any point has to support the weight of rope below that point, as well as the weight of the bucket and its contents; therefore the section of the rope at the upper end being A^ and IF the weight of the rope, S= — —^ — instead of — —. A A It is readily seen that if W is small, the value of JP + TF 22 APPLICATIONS 23 is not sensibly greater tlian the value of P ; the value of S p will not be materially changed from — -. ■^ In the case where P + TF is much greater than P, and the rope is the same size throughout, it must be large enough to carry a load of P -\-W, When a long vertical bar of uniform section is under an axial load, it follows that every section, except one, is larger than necessary, and if the section area is varied so that the unit stress in each section is the same, the weight of the bar could be reduced. Such a bar is called a Bar of Uniform Strength, This does not mean that the strength of the bar is the same at all sections, but that the change of section area makes the unit stress the same at all sections, and might better be termed a bar of uni- form stress, or a uniformly safe bar. Consider such a vertical bar, length ?, and an axial load P P, The smallest possible area, JLq, is given by ^ = -^» where S is the allowable unit stress, and design the bar so that S shall be constant. In Fig. 23, let A^ be the area at the end where the load is applied, and A be the area at any distance y from that section. Then at a dis- tance y •\- dy the area must be increased to ^ + dA. Let w equal the weight of a cubic unit of the material; then the additional weight to be carried on the area A + dA is Aw dy^ since the term containing dA dy can be neglected in comparison to the term containing only dy. Since 8 is constant, and this weight is to be carried on the area dA^ ■.A Aw dy -J jS dA rt N 24 MECHANICS OF MATERIALS gives the relation between the increase of length and the increase of area. If (1) is integrated, w (2) S and since A = J.^, when «/ = , 0= log A^. Substituting this value for C in (2) and transposing, log,^ = -^+log,^oOrlogio^ = 0.434-y + logio^ (3) is an expression for the relation between the least area and Fig. 23. the area at any distance y from that section. In the application of the formula to a given case, different values might be assigned to ^, and the corresponding values of A found, enough values being calculated to enable the pro- file to be drawn. In this case tlie outline of a vertical section is slightly curved. If the vertical section is made trapezoidal, and the bar is a masonry pier, the top of APPLICATIONS 25 the trapezoid is made proportional to Aq^ and the base to the value of A in equation (3), when ?/ is the height of the pier. Such a pier would require more material than one of uniform strength ; but, although the unit stress would be only approximately equal at all sections, it would repre- sent common practice. Art. 24. Thin Pipes, Cylinders, and Spheres. Take a pipe, internal diameter _D, thickness of the pipe wall t, carrying a water pressure of H lb. / sq. in., to find Fig. 24 a. the unit stress in the walls of the pipe. As each unit of length of the pipe is under the same forces, we may take the length as unity. Suppose a length of pipe equal to unity (Fig. 24 a) to be cut by a diametral plane X-X; then the stresses acting on the pipe walls at the section cut by the diametral plane must resist the pressure of the water tending to force the two halves of the pipe apart, and if ^ is small, the stress may be considered as being uniformly distributed over the sections of the pipe walls cut by the diametral plane. Therefore, if we can calcu- late the value of P, the force tending to separate the two halves of the pipe, the formula P = AS^ where A is the 26 MECHANICS OF MATERIALS area of the section of the pipe walls, will give the required unit stress. A principle of hydraulics states that the pressure of water is the same in all directions and normal to the sur- face. Let Fig. 24 h represent a half section, perpendic- ular to the axis of the pipe. If 6 is any angle, then ?^ :RrdeCo36 Fig. 24 6. for a length of pipe equal to unity, the radius of the pipe being r, an area of the internal surface of the pipe, equal to rd6^ carries a pressure of H lb. / sq. in., or the total radial force on that area is Rrdd. This force may be resolved into components,* parallel and perpendicular to the line X-X^ which is the trace of the diametral cutting plane. The components are Rr cos Odd and Rr sin 6d0. The sum of the Rr cos Odd forces for one half of the pipe is zero, and the sum of the Rr sin Odd forces for the same half is 2 Rr^ or RB^ which is the force per unit of length perpendicular to the cutting plane, resulting from the internal pressure R. Substituting this value in the general formula P = AS, APPLICATIONS 27 and noting that the area of the section of the pipe walls is 2 1, we have BI) = 2 St (1) as a general expression for the unit stress induced in a longitudinal section of a pipe whose walls are thin. If the ends of the pipe are closed, the internal pressure of the water on the ends of the pipe tends to rupture the pipe in a plane perpendicular to the axis. The force acting on the ends of the pipe is evidently , and the area to resist this force is irDt ; hence a 4 substitution of these values in P = AS gives HI) = 4 St, showing that the stress in a plane perpendicular to the axis is only one half that on a plane through the axis. For a sphere with thin walls, the water pressure tends to produce rupture on the line of a great circle. It is readily seen that the pressures and areas are the same as for a plane section perpendicular to the axis of a cylin- der ; hence the same relation holds true. Akt. 25. Thick Pipes. If t is large, the stress in a plane through the axis is no longer uniformly distributed over the area of the sec- tion, but is greater on the internal radius. Many formulas have been proposed for finding the maximum unit stress in this case, the one given here being due to Barlow. The results are in simple form, and the value of the maximum unit stress being greater than that given by the more exact discussions, places the error on the side of safety. Barlow's formula assumes that when the fluid pressure acts on the internal surface of the pipe, while the diameter 28 MECHANICS OF MATERIALS is increased, the volume of the pipe walls for a unit of length remains unchanged. If we let I) be the internal, and D^ be the external, diameters of a pipe whose walls are thick (Fig. 25), and Fig. 25. the thickness of the pipe walls ^, before the pressure is applied the volume of a ring ona unit in length is Let e and e-^ be the extensions of the diameters due to the fluid pressure and the volume becomes TT TT l(i>^ + e,y-'^(D + ey. (2) Expanding (2), neglecting the e^, as « is a very small quantity, and equating (1) and (2), the equation reduces *o D^e^^De. (3) APPLICATIONS 29 The unit elongations are — and -J-, since the change in the circumference of the thin shells of diameters D and D^ are ire and tt^^ while the original circumferences are ttD and irD^ The unit stresses in the thin shells of the diameters D and i)^, being S and S^^ as the unit stresses are proportional to the unit deformations within the elas- tic limit, ^ = ^=:^. (4) A But I>xH — -^^' hence, e, D' and substituting this value of — in (4) gives or the unit stresses are inversely proportional to the squares of the diameters or radii. Let S^ be the unit stress at a radius x. Then, from (5), and the total force exerted over the area dx times 1 is SJx^^Sr^^' (6) The integral of the left hand member of (6) is the sum- mation of all the stresses on one side of the pipe, and is 30 MECHANICS OF MATERIALS equal to one half of the total fluid pressure tending io rupture the pipe ; hence the total force is DE=2Sr^r^='^r^^^^^. (7-5 *^r x^ r + t D-\-2t ^ ^ Equation (7) reduces to ED^ = 2 St instead of EJ) = 2 St for thin pipes. Formula (7) is the one to use in all cases where the value of ED^ is enough larger than ED to cause serious error. The error in Barlow's formula increases as the internal radius decreases, and for thick pipes where the diameter is small, the more exact formulas of Lami and others should be used. (See Merriman's "Treatise on the Mechanics of Materials.") Art. 26. Riveted Joints. In the determination of simple stress, such as tension, P . compression and shear, the formula 'S=—r always gives the relation between the force P and the unit stress S. The area A must always be the area over which the stress is induced, and will, for tensile or compressive stresses, be a section of the bar perpendicular to the line of action of the force P, and parallel to the same line for shearing stresses. If the area of the section of the bar varies for different cutting planes, the plane that gives the least area should always be chosen. When two plates are joined together by means of rivets, the joint is called a Eiveted Joint. Art. 27. Tension in the Plates. Let A and B (Fig. 27 a) be two plates joined together by means of rivets passing through the cover plates a and b. Let P be the tensile force tending to separate the plates APPLICATIONS 31 >< B ( a ( ) ) V A and B, w the width of the plates, t the thickness of the plates A and B^ t^ the thick- ness of the plates a and 5, and (^ the diameters of the rivets. Since P is a tensile force, and the greatest number of rivets in line is two, if we pass a plane perpendicular to the line of action of the force P through the line of the two rivets, cutting either of the plates A or B, or the cover plates a and b, the stress which acts in such a plane to resist separation is the product of the unit stress induced and the area cut. The area" cut by the plane is either the thickness of the two cover plates a and b times the width of the plates less the diameters of the rivets in line, or the thickness of the plates A or B into the same quantity. Hence the relation between the tensile force P and the unit stress induced in the plates is, 2ti(w — 2 d) iSf — P, or t(w ~2 d) S^ = P, depending on whether the failure is in the plates ^ or ^ or the covfer plates a and b. As there is no reason why one of these sections should be stronger than the other, 2 t^ is generally made equal to t. Since any other cutting plane would cut a larger area, the value of S as given in the above formula is the largest Fig. 27 a. 32 MECHANICS OF MATERIALS value possible with the force P. Figure 27 h shows the failure by tension in the plates. B rty=c^ V Fig. 27 6. Art. 28. Shear on the Rivets. If the plates do not yield in tension, in order to pull the plate A away from B^ and the cover plates, the num- ber of rivets passing through A must be sheared off in two sections parallel to the line of action of the force P, and perpendicular to the axis of the rivets. Therefore the area to resist shear on the rivets caused by the force P must be twice the sectional area of each rivet times the number of rivets passing through the plate A. These values substituted in the general formula, P = AS give (See Fig. 28.) for this joint, 2 x —^ — S^ p. APPLICATIONS 33 Art. 29. Compression on the Rivets or Plates. Suppose that the plates resisted the ten- sile stress, and that the rivets, the shearing stress, the force P acting on A^ causes the plates to bear on the cylindrical surface of the rivets. The exact effective area of each rivet, or the plate through which it passes, is not known ; but it is assumed to be the projected area of the rivet; that is, the diameter of the rivet times the thickness of the plate through which it passes. On this assumption, taking 2t-^ = t^ the area to resist compression on each rivet is dt. This area times the number of rivets passing through A^ when substituted in the general formula, P = AS, gives for this joint, 2 dtS, = P„ as the relation between the force P and the unit compressive stress on the rivets or plates. (See Fig. 29.) As there is no other way that the joint can fail, the equation that gives the least value of P determines the way in which the joint is most liable to fail. Art. 30. General Case of Riveted Joint. In general, while the joint may be very long, the rivets are regularly spaced. In this case, the distance between 34 MECHANICS OF MATERIALS the centers of any two rivets in line is called the " pitcla " of the rivets, and P is taken as that proportion of the load on the entire joint that the pitch is of the length; or in other words, P is the load or force on the joint for a distance eqnal to the pitch of the rivets. Tension in the Plates, Taking P as above, and letting p be the pitch of the rivets, the relation between the tensile unit stress in the solid plate and the force P is, tp8,=^P, (1) where, if aS' is the safe unit stress, P is the safe load. For all joints in tension, since there can never be but one rivet in line in the distance p^ tip-d)8, = P,. (2) If, as before, aS'^ is the safe tensile unit stress, P^ is the safe load when failure is considered as taking place by tension in the punched plates. If the values of Sf are the same in (1) and (2), it is easily seen that P^ can never equal P, Shear on the J^ivets. ^ The relation between the load P and the unit shearing stresses will depend on the nature of the joint. In any given case, the product of the number of times each rivet niay shear, the number of rivets in the distance p, and the area of the section of the rivet perpendicular to its axis, will be the area over which the shearing stresses act. Letting e be the number of rivets times the number of sections in the distance p^ then ^ecPS;=P.. (3) APPLICATIONS 35 AVlien safety of the joint against failure by the shearing of the rivets is considered, if Sg is the safe unit shearing stress, then P^ is the safe load. Considering (2) and (3), the values of p and d can be chosen so that when safe values of the unit stresses S^ and Ss are used, P^ = P^, but they are not necessarily equal. Compression on the Rivets or Plates. Taking S^ as the safe unit stress in compression, and td times the number of rivets in the distance p as the area resisting compression, and letting c-^ be the number of '^i^e*''' c,tdS, = P, (4) is the relation between the unit stress in compression and the load P^. As before, P^ may have different values from either P^ or P^, but if they are assumed to be equal, and safe values of aS^, S^^ and S^ are used in equations (2), (3), (4), as there are three equations and three variable quantities, p^ t, and d can always be determined. If the values of p^ t, and d are found in this manner,, the joint will be equally safe against failure in all ways. In general, the equation which gives the least value of P will show the way in which failure is most liable to occur. Art. 31. Kinds of Riveted Joints. Lap Joints. Here the two plates to be joined together lap by each other and the rivets pass through both plates. The rivets tend to shear on but one section, and are said to be in "single shear." Putt Joints with Single Cover Plates. Here the plates are both in the same plane, and the joint between them is covered by a plate of the same thickness as the plates. Any rivet passes through the cover plate and one of the 36 MECHANICS OF MATERIALS Double riveted lap joint. Double riveted butt joint with single cover plate. I 11 1^ Triple riveted lap joint. Double riveted butt joint with double cover plates. FiQ. 31. APPLICATIONS 37 plates that are to be joined together. The conditions for shear and compression are evidently the same as for lap joints. Butt Joints with Double Cover Plates. In this kind of a joint the plates are in the same plane, and the cover plates, each one half the thickness of the plates to be joined, are placed on either side of the joint. Any rivet passes through both cover plates and one of the plates that are to be joined. An inspection of the figure w^ill show that each rivet is liable to be sheared in two sections and is said to be in " double shear," while the conditions for compression are the same as for the one with single cover plates. Either type of a joint may have one or more rows of rivets, and the pitch in all rows is generally the same. The joint is said to be Single., Double., or triple riveted, as there are one, two, or three rows of rivets. The figures show the details of the various joints and styles of riveting. It is evident that if lines are drawn passing through any two adjacent rivets in the same row, and parallel to the line of action of P, the rivets included between these lines will be the number of rivets that are to be considered as resisting the shear and compression. Art. 32. Efficiency of a Riveted Joint. When P^, P^, and P^ are the maximum safe loads a riveted joint will carry, the efficiency of that joint may be defined as the ratio of the least of the above values, to the load the unpunched plate of the same length will carry under the same conditions. From this definition it is evident that the efficiency of any joint is P^, P^, or P^., divided by P, depending on the 38 MECHANICS OF MATERIALS relative values of P^, P^,, and P^. Of all the ways in which a riveted joint may fail, the failure b3^ compression of the rivets or plates is the least understood, and many engineers design the joint for equal strength in tension and shear, and simply check the resulting dimensions for the compressive stress. This practice has resulted in the efficiency of a riveted joint being given as , but this P is only approximately true unless P^= P^=i P^. In general, when the values of t and d are calculated, the nearest commercial sizes have to be chosen, and the values of P^, P^, and P^ are rarely ever equal. In many cases the pitch is fixed by the conditions for the tightness of the joint against leakage, as for boilers, tanks, and pipes, and in such cases only two conditions can be satisfied. In the development of the preceding formula no account has been taken of the friction that must exist between the plates through which the rivets must pass. As there is no good theoretical way of introducing the resistances due to friction in the formulas for strength, riveted joints have been pulled apart in testing machines, and the accuracy of the formulas checked by the breaking load as determined by the test. While the results in many cases seem to show that the theory that has been given here does not hold true, the conditions that are con- current with the rupturing load not being the same as when all the stresses are within the elastic limit, there seems to be no good reason why the formulas as developed will not giye reliable results. In the design of a riveted joint for a pipe or a boiler to carry a pressure of R lb. /sq. in., as one half of the APPLICATIONS 39 internal pressure tending to disrupt the pipe or boiler is carried on one joint of the shell the value of P to be used in the formulas is one half of the total pressure acting over a length jt?, or, — —^ = P. Art. 33. Stresses Due to Change of Temperature. All metals tend to change in length as their temperature changes. If the change is resisted, that resistance must cause a stress in the material. Consider a bar I units in length, A units in area, free to change its length as the temperature changes. If the change in length due to a given change of temperature is 6, and a force is exerted to restore the bar to its original length, the unit stress induced by that force will be given e Therefore if a force prevents the change from taking place, it must induce an equal unit stress, and this unit stress is independent of the area of the bar. Knowing the change' in unit of length for a change of 1° of tem- perature, or the coefficient of linear expansion, the unit stress in any bar corresponding to any change of tem- perature may be found provided the unit stress is within the elastic limit of the material. If the bar is under an initial unit stress before the change of temperature, the change will increase or de- crease that stress, depending on the nature of the initial and temperature stresses. 40 MECHANICS OF MATERIALS PROBLEMS 1. How long will a bar of wood have to be, in order that its own weight will produce a unit stress of 300 lb./ sq. in. ? The bar is hanging vertically. Solution. Taking the weight of a bar of wood, 1 sq. in. in sec- tion and 3 ft. long, as |f lb., the bar will have to be as long as 300 divided by if, equal 360 yd. Ans. 2. What is the length of a vertical steel bar a square inch in area, that carries a tensile load of 4000 lb., at the lower end when the maximum unit stress is 15,000 Ib./sq. in. ? 3. Find the probable elongation in problem 1. Solution. Since the maximum unit stress is 300 lb. / sq. in. and the minimum 0, the average unit stress must be 150 lb. / sq. in., S I and we have given that E = — , e 1,500,000 = 150 X 360 x 36^ .^ ^ ^ ^ 296 in. e 4. Find the total elongation in problem 2. (Total elongation is that due to the load and its own weight.) 5. Find the height of a brick chimney of uniform section, when the maximum compressive unit stress is 18 tons /sq. ft. 6. Suppose that the sectional area of the base of a chimney was twice that at the top, and that the change in area was uniform, how high could the chimney be built if the limiting value of the unit stress was 18 tons /sq.ft.? 7. Find the areas of the top and bottom section of a stone pier, 100 ft. high, to carry a load of 240 tons, the unit stress in all sections to be constant. 8. If the pier in problem 7 had the top and bottom areas as found and the vertical section was trapezoidal, find the unit stress at the bottom of the pier. APPLICATIONS 41 9. The wire rope used for hoisting in a certain mine is 1^ in. in diameter, and weighs 2.5 lb. /ft. If the mine is 800 ft. deep and the safe working load for the rope is 5^ tons, what weight may be raised ? 10. A pipe 6 in. in diameter is to carry water under a pressure of 1000 Ib./sq. in., with a factor of safety of 6; required the thickness of the pipe walls. 11. A standard 2-inch steel pipe is 2.375 in. outside diameter and 2.067 inside. This size is tested under a pressure of 500 lb. /sq. in.; required the unit stress in the pipe walls. 12. A steel pipe 10 in. in diameter is to carry water under 2770 ft. head. The factor of safety is to be 10. Find the thickness of the pipe. (A column of water 1 ft. high and 1 sq. in. in area weighs .434 lb.) 13. Check the results in problem 12 by Barlow's for- mula and find the unit stress. 14. Compare the maximum unit stress in the pipe of problem 11, as determined by the formulas for thick and thin pipes. • 15. Write the formulas for determining the strength of the following riveted joints intension, compression, and shear. The pitch is p, the thickness of the plates t, diam- eter of the rivets c?, and the safe unit stresses in tension, compression, and shear are aS'^, /S'^, and Sg. (a) Single riveted lap joint. (5) Double riveted lap joint. (c) Single riveted butt joint with one cover plate. (d) Double riveted butt joint with one cover plate. (e) Single riveted butt joint with two cover plates. (/) Double riveted butt joint with two cover plates. (^) Triple riveted butt joint with two cover plates. 42 MECPIANICS OF MATERIALS Solution for (a). ^ If Pt is a tensile force acting on the joint for a distance equal to the pitch, then since P = AS, and the area to resist the tensile stress is t(p — d), Pt = t(p — d)St gives the relation between the load and the unit tensile stress. Let Pc be the tensile force that brings compression on the rivets. As there is but one rivet in the distance j9, the area to resist compres- sion is td, and since P = .4^', P^ = tdS^ is the relation between the load and the compressive unit stress, and if Pg is the tensile force that produces shear on the rivets, as there is only one rivet in the distance », and it can shear in but one section, as P = AS, P, = Ss is the relation betw^een the load and the unit stress in shear. 16. Assume that the values of p, t, d, St, Sc, and >Ss for any given joint are taken so that Pt = Pc = Ps show that CiSc the efficiency is given by — - — ^—^ ciSc~]~ St Solution. Since Pt = Ps t{p-d)St = citdSc (ciSc-\- St)d P = ^ Since Pt, Pc and Ps are all equal, efficiency = . P Substituting for p, ciScD efficiencv = ^ ^' - ""'^^ (ciSci-St) ciSc-^St St ^ which is the efficiency for any joint for which the values of Pt, Ps and Pc are all equal. 17. A steam boiler, 60 in. in diameter, carrying 120 Ib./sq. in., is to have the longitudinal seams double riv- eted butt, joints with two cover plates. Take S^ = 12,000 Ib./sq. in., aS'^ = 10,000 Ib./sq. in., and make the joint equally safe against failure by either tension or shear. If APPLICATIONS 43 the efficiency is to be approximately 75%, required the thickness of the plates, the diameter, and pitch of the rivets. Solution. The thickness of the plate is given by RD = .75 • 2 St as the unit stress in the unpunched part of the plates can be only 75% of the allowable unit stress, or, 120 X 60 = .75 X 2 X 12,000 t 120 X 60 _ Q ^ .75 X 2 X 12,000 ' ' and the expression for the efficiency in tension being efficiency P^^ = .75, P p = 4:d. Take y'g as the nearest market size for the required thickness of the plate, for equal strength, t(p-d)St = 2x2 x^Ss, tV (^d- rf) 12,000 = 7rd^ X 10,000, d = I'' approximately ; then t = ^\", d = i", and p = 2". 18. Using the approximate values for p, t, and d as deter- mined in problem 17 and assuming aS = 10,000 Ib./sq. in., find the other allowable unit stresses, considering the joint to be equally safe against failure by tension, compression and shear. 19. A boiler, 30 in. in diameter, has double riveted lap joints, plates ^ in. thick, rivets ^ in. diameter, pitch of the rivets 2.5 in. Taking S^ as 60,000 Ib./sq. in., find the pressure per square inch that may be carried with a factor of safety of 6, considering failure by tension of the plates alone. 20. What are the values of S^ and jS^, and the efficiency of the joint, in problem 19? 21. A triple riveted butt joint with two equal cover plates is to have an efficiency of 80 %. Using S^ = 10,000 44 MECHANICS OF MATERIALS lb. /sq. in. as the safe unit stress in tension, what witi be the values of S^ and aS'^., when the joint is equally safe against failure bj tension, compression, or shear ? 22. If Sf is taken as 10,000 Ib./sq. in., and S^ as 15,000 lb. /sq. in., what is the highest possible efficiency of a double riveted lap joint, designed for equal strength against tension, compression, and shear? 23. A steam boiler with double riveted lap joints is to carry 125 lb. /sq. in. pressure. The allowable tensile unit stress is 10,000 lb. /sq. in. The plates are | in. thick, rivets 1 in. diameter, and the pitch 3J in. What will be the largest possible diameter of the boiler ? 24. Compare the values of S^, S^, and S^ in problem 23. 25. A 30-foot steel railroad rail undergoes a change of temperature of 100° . If the change in length is pre- vented, what unit stress will be set up in the rail ? Solution. The coefficient of linear expansion for steel is .0000065 per degree ; hence the unit elongation for 100° is .00065 in., and since S = Ec, S = 30,000,000 X .00065 == 19,500 lb. / sq. in. 26. For electric railway work, the steel rails are often welded together. Assuming that there is no change of length, what is the maximum range of temperature allow- able if the unit stress is not to be greater than the elastic limit? 27. The walls of a building had bulged out, and to pull them into place, five steel rods each two sq. in. in area were passed through from one wall to the other. The temperature was then raised 100° and the nuts on the rods tightened, so that the load on each bolt was" 1000 lb. When the rods are at the original temperature, what is the maximum pull they could exert on the walls ? APPLICATIONS 45 28. At St. Louis, Mo., a battery of steam boilers was connected together by a pipe in which there was no pro- vision made for expansion. The temperature of the steam was about 360° F. and that of the room, 100". Assuming that there was no change in length, what was the maximum unit stress in the pipe due to the change of temperature ? (a) Material of the pipe steel ? (5) Material of the pipe cast iron ? CHAPTER III BEAMS Art. 34. Kinds of Beams. When a bar is placed in a horizontal position, and acted on by forces perpendicular to the axis of the bar, it is called a Beam. C 3 Cantilever Beam. Fig. 34 a. Cantilever Beam. Fig. 34 &. m wm. I ^flj ■ Hi fl; Simple Beam. Fig. 34 c. Continuous Beam. Fig. 34 d. I If the beam has two supports on which it merely rests, it is called a Simple Beam. A cantilever beam has only one support, which is at the middle, or, what is the same thing, has one end firmly fixed in the wall, leaving the other end free. When a beam has both ends firmly fixed in the walls, or one end fixed and the other merely supported, it is called a Fixed or Restrained Beam. A beam supported at more than two places is called a Continuous Beam, 46 BEAMS 47 Art. 35. Reactions at the Supports. The reactions at the supports are the forces acting between the supporting walls and the beam, and so far as the beam is concerned, they may be treated as vertical forces acting upward. Art. 36. Uniform and Concentrated Loads. The loads on a beam are the weights that the beam carries, and since the attraction of gravity always acts downward, they may be represented as vertical forces. When the load is distributed uniformly over the entire length of the beam so that each element of the length of the beam carries the same load, the load is said to be a Uniform Load. When a load is carried on so small a portion of the length of the beam that the effect of the weight acting as it does over that small portion may be assumed to have the same effect as a single force acting at the center of the load, it is called a Concentrated Load. Since a simple or cantilever beam under any loads may be considered as a body acted on by forces which keep it at rest, the laws relating to the equilibrium of forces must be satisfied. In general, all the forces, loads and reactions, will be vertical, and the magnitude and position of the loads will be known, so that the magnitude of the reactions may be determined by applying the laws relating to the equilibrium of parallel forces. These laws are: The algebraic sum of all the forces equals zero, and the algebraic sum of the moments of all the forces about any point equals zero. 48 MECHANICS OF MATERIALS These two equations are sufficient to determine £he reactions for simple and cantilever beams, as there are not more than two quantities to be determined. For all other beams another condition is derived by the use of the theory of the Elastic Curve. (See Chapter IV.) The length of a simple beam is the distance between the supporting walls and the distance the beam projects beyond the wall for a cantilever beam. While in any case the beam must rest on the supporting wall for a finite distance, the point of application of the single force that is to replace the resultant of the forces acting between the beam and the wall is taken at the edge of the wall beyond which the beam projects. Let Fig. 36 represent ^ J ^ "p-A"-^ ^ simple beam, length Z, I^^ weight TT, carrying two ^m m--i"-^ i ^ concentrated loads P^ and ^^ n£ 1*2 at distances 'p-^ and 'p^, ^i^- ^' from the right reaction, and the values of the reactions i?i and H^ are required. The weight W may be considered as a uniform load, and for equilibrium, i?i + i^2-A-^2-TF=0. (1) Taking moments about a point in the line of action of 7^2, and giving the moment a positive sign when it tends to produce clockwise rotation, Ii,l^B,^-P^p,-P,p,-^^(i. (2) The term containing H^ is zero, therefore It^ may be found, and by substituting for H^ in equation (1), i?2 ^^^.y also be determined. BEAMS 49 R^ may also be found by taking moments about a point in the line of action of i?i, and this value used as a check. If the weight of the beam, TFJ included both the weight of the beam and a uniform load, the equations would have been the same. Art. 37. Vertical Shear. Let Fig. 37 a represent a simple beam, loaded with both uniform and concentrated loads, and Fig. 37 5 a cantilever beam with the same loading. Suppose either beam to be cut by a plane X—X perpen- dicular to the axis of the beam, at any distance x from the ccmrrrr^ K Fig. 37 a. R. r Fig. 37 6. left end of the beam, and consider the end marked E. This end is acted on by known forces, since, when the loads are known, the reactions can be found. Relative to the end marked F^ these forces tend to produce translation either up or down, the magnitude of the resultant force being the algebraic sum of all the forces acting on the part E. Similarly, considering the part marked F^ the resultant of all the forces acting on F tends to produce translation relative to F^ and these two resultants being equivalent to all the forces acting on the beam, must be equal and opposite in sign, since the beam is in equilibrium. These two resultants are a pair of shearing forces, and either one is the force producing shear in the plane X—X^ and if we wish to call one of these forces the Vertical 50 MECHANICS OF MATERIALS Shear at the section X-X, it will be necessary to define vertical shear so that the sign will be determined as well as the magnitude. The vertical shear at any section of a beam is defined as follows : The Vertical Shear for any section of a beam is the algebraic sum of all the forces acting on that portion of the beam, lying to the left of that section. When the resultant force acts upward, the vertical sliear is considered positive, and negative when it acts downward. Art. 38. Bending Moment. If we take the sum of the moments of all the forces that act on the right and also those that act to the left of the section X-X (Figs. 37 a and 37 ¥) about a point in that section, the resulting moments must be equal and opposite in sign, and as either is a measure of the tend- ency of rotation to take place about a point in the plane X-X^ they are the bending moments for that section. In order to fully determine the bending moment both as to sign and magnitude, it is defined as follows : The Bending Moment at any section of a beam is the algebraic sum of the moments of all the forces, acting on that portion of the beam lying to the left of the section, moments being taken about a point in that section.* It is considered positive when the moment tends to produce clockwise motion and negative for counterclock- wise motion. The " section of the beam " as used in the definition of both vertical shear and bending moment *■ A cantilever beam is always considered as being fixed at the right end, leaving the left end free. If the beam projects toward the right, look at it from the other side. BEAMS 51 refers to any plane section perpendicular to the axis of the beam, and the " forces acting to the left of the sec- tion " includes both the loads and reactions whose points of application are on the left of the section. Art. 39. Resisting Shear. Since the part of the beam on the left of the section X-X (Figs. 37 a and 37 5) is acted on by external forces, and as a part of the whole beam it is in equilibrium, there must be internal forces acting in the section X-X^ which taken with the external forces acting to the left of the section, constitute a system of forces in equilibrium. Suppose these unknown forces to be resolved into their horizontal and vertical components. Then, since equi- librium exists, the algebraic sum of the vertical compo- nents of the internal forces must equal the sum of the vertical forces, and since the external forces have no horizontal components, the sum of the horizontal compo- nents of the internal forces must be zero, and also, the algebraic sum of the moments of the external forces must equal the sum of the moments of the internal forces, mo- ments being taken about a point in the section X-X. From the first condition, if we give the name of Resist- ing Shear to the sum of the vertical components of the internal forces, The Vertical Shear = the Resisting Shear., and assuming the shearing forces to be uniformly dis- tributed over the area of the section, then F= AS, (3) where A is the area of the section and S is the unit shear- ing stress, and F"the vertical shear for the section. 52 MECHANICS OF MATERIALS Art. 40. Resisting Moment. From the second condition, since the horizontal com- ponents of the. internal forces must be either tensile or compressive forces, the sum of the tensile Forces = the sum of the compressive Forces. Giving the name of Resisting Moment to the moments of the internal forces about a point in the section, the third condition for equilibrium states that, The Bending Moment = the Resisting Moment. The relation between the bending moment of the exter- nal forces and the unit stresses in the section considered, cannot be found by the laws of mechanics alone, as the distribution of the internal forces is unknown. The in- formation necessary may be derived from the results of experimental observations on beams while under the action of bending forces. When a beam is under the action of bending forces, it is observed that along the concave surface of the beam the fibers * of the beam are shortened, while those on the convex surface are lengthened, and that along a certain plane section of the beam there is no change in length. We know that a compressive force shortens, and that a tensile force lengthens, any bar on which it acts, and that where there is no deformation there can be no force act- ing ; therefore the stress on the concave surface must be compression, and that on the convex surface, a tensile * The word "fiber" as used here may be defined as a bar of elemen- tary sectional area and a length equal to that of the beam, the whole beam being composed of a bundle of such fibers. It is not necessary that the beam should be of a fibrous material in order that this conception should be true. BEAMS 53 stress, while at the certain plane, called the Neutral Sur- face, there is no stress of any kind. When the loads were such that there was no unit stress greater than the elastic limit, it was observed that the deformation of any fiber was proportional to its distance from the neutral surface. If we call the trace of this neutral plane on any plane section of the beam perpendicular to the axis, the Neutral Axis of the section, since there is no unit stress greater than the elastic limit, it is evident that the unit stress at any point in that section, and consequently the forces pro- ducing that unit stress, must vary directly as the distance from the neutral axis. This assumes that JE is constant, since U = — , and when the location of the axis is known, e the unit stress at any point may be found. Let Fig. 40 represent any cross section perpendicular to the axis of the beam, and the line X-X the neutral axis. From the experimental observations we know that the greatest unit stress must be at the greatest distance from the neutral / ^ n \ axis ; and letting e be the distance / i 1 V from the neutral axis to the fiber / \ most distant from that axis, A the area of the section, dA the area of Fig. 40. any fiber, ^ the distance of that fiber from the neutral axis, and S the unit stress at a distance e from the neutral axis, then since the force varies as the distance from this neu- tral axis, the force at any distance ?/, acting on the area of any elementary fiber dA is —yd A, and — ) ydA is the sum- c c^ mation of the horizontal forces acting on the whole section. 54 MECHANICS OF MATERIALS From the conditions of the problem, this sum is equal to zero, and as neither S or c can be zero, I ^dA must be zero. If we assume the density to be constant, this is the con- dition where the axis of moments passes through the cen- ter of gravity of the section; hence the neutral axis passes through the center of gravity of the section. The V force on any fiber being — ?/c? J., the moment of this force c about the neutral axis is — y'^dA, and the sum of the mo- c ments of these forces about a point in the section is — I 'if'dA, which is the Resisting Moment by definition. The Cy'^dA is defined in mechanics as the moment of in- ertia of the section about a gravity axis, and is repre- sented by the symbol I. Therefore, since the Bending Moment = the Resisting Moment, aj M = —. (d) c In this formula ikTis the bending moment of the exter- nal forces that act on the left of any section, Zthe moment of inertia of the section about a gravity axis perpendicu- lai" to the direction of bending, c the greatest distance of any fiber from the neutral axis, while S is the maximum unit stress in that section. The formula expresses the relation between the bend- ing moment and the unit stress in the section, and if the maximum unit stress in a beam is desired, M must be the maximum bending moment for that beam under the given loads. BEAMS 56 As an aid to memory, attention is called to the simi- larity between this expression and the one derived for axial stress. P and M are the external forces, S in each case is the unit stress induced; and as - depends on the shape and c area of the section for its value, it may be considered as replacing A in the formula for axial stress. Art. 41. Use of Formula. In the derivation of the formula for axial stress, there was no consideration taken of the intensity of the stress or of the nature of the material, the formula holding true for all unit stresses and materials. When the formula Mc S = —r was derived, certain conditions were specified. They are : (1) The material was to be elastic, and since we assumed that the forces were proportional to the deformations, the modulus of elasticity must also be constant. (2) In order for I y'^dA to be the sum of the moments of the differential areas about the gravity axis, the material of the beam must have a uniform density. (3) No unit stress to be greater than the elastic limit. If the material and loading of a beam does not satisfy these three conditions, the formula S = — - will not give the true unit stresses. The Modulus of Rupture is the value of S as derived from S = -— , when M is large enough to rupture the beam. Since the formula only holds true for unit stresses within the elastic limit, the value of the modulus of rup- 56 MECHANICS OF MATERIALS ture as an engineering constant is at least doubtful. In testing cast iron bars in bending, the breaking load at the center, which is of course proportional to the modulus of rupture, is taken as a measure of the quality of the material. The results of such tests are useful for comparison only when the tests are made on bars of the same length and size. Mc The formula S — -— , expressing as it does the relation between the bending moment and the unit stress in a beam, is used in all calculations for the strength, safety, and design of beams. When sufficient data are given to fully determine the value of M, the value of either S or — may be found. ^ I The value of - depends on the form and area of the sec- e tion, and is called the Section Modulus. Mc In the application of the formula S = — — to any given case, the question of units becomes one of great importance. It does not make any difference in the effect of M, whether it is expressed in inch pounds, or foot pounds; but as S is the unit stress and is usually given in pounds/ square inch, the value of M must be expressed in inch pounds and c and /in inches, in order to get correct results. The ton could just as well be used as the unit of weight and the foot as a unit of length, but such practice would require a special table of the constants of materials, the one given in the Appendix being based on the pound and inch. As there are many different sections that have the same section modulus, the designer is called on to choose a form of section best suited to the conditions that exist in the case in hand. BEAMS 57 The value of / for a rectangular section, breadth 5, and depth c?, about an axis through the center of gravity parallel to the side 5, is — -^, and ^ is -• Substituting these values in tlie gen- Mc eral formula S = -—- gives the value of iif as 6 I Therefore, when the section is rec- tangular we see that the value of M for any given value of S increases directly as h and as c?^, and any in- crease in the value of d increases the strength more than a proportional in- - crease in the value of 5, while the weight of the beam will be the same for either case. a -2:02— CD Fig. 41 a. A practical limit of the ratio of t is about 6. r Z. 61 WT. 29.3 TO 34.6 LBS. ti. -244^ /if iO H-^^ From the known condition that the unit stress in any point in the section varies as the dis- tance from the neutral axis, it is evident that the form of sec- tion which presents the great- est area where the unit stresses are large and a minimum area where they are small will be the best from an economic Fig. 41 h. standpoint. The common steel I beam, so called from the form of the section, is an example of this distribution of area. There KfC 1_ '^ -3-k- 58 MECHANICS OF MATERIALS _ T. 97 WT. 9.3 LBS-; "W dk is always a maximum vertical shear to be resisted in all beams, and care should be taken that the area of the sec- tion chosen is large enough to resist the shearing forces. In most cases if the beam is safe against the bending forces, it will be safe against the shearing forces, but the unit shearing stress should always be inves- tigated before the final decision is made as to the size of a beam. The hollow box or cored sections in cast iron machine parts subjected to bending forces represent the best practice on account of the large value of I relative to the weight. Plate girders, made by riveting angle irons to a steel plate called the web, making a beam whose section resembles that of the common I beam, are in common use in structural steel construction. > -^ -J Ol Ol - .52'^' 1-88 Fig. 41 d Plate Girder. Fig. 41 e. Art. 42. Shear and Moment Diagrams. If a line which may be either straight, curved, or broken be drawn so that the ordinate to that line from any point of a straight line representing the length of the beam equals the vertical shear or bending moment for that section of the beam, the resulting diagram is called a Shear or Moment Diagram, depending on whether the vertical shears or the bending moments are used as ordinates.* * This line will be called the shear or moment line. BEAMS 59 To draw either diagram, the shears or moments for each unit of length of the beam might be calculated and the results plotted to scale, using as ordinates the values of the moments or shears and as abscissa the distances of the sections from the left end of the beam. A line through the points so located would be the shear or moment line, as the case might be. This is a tedious process and may be shortened by a study of the effect of the different kinds of loads on the form of the shear or moment line. Art. 43. Shear Diagram. For a concentrated load, the difference between the shears for sections taken just to the right and left of the point of application of the load is the magnitude of that load, since in the former case the " negative forces acting to the left of the section " are increased by the magnitude of the load over those for a section taken just to the left of the load. Therefore, the shear line will always contain a line perpendicular to the length of the beam under each concentrated load. When only concentrated loads are considered, the shear line between any two concentrated loads will be a straight line parallel to the length of the beam, since there is no change in the forces acting to the left of any section taken between the two loads. For uniform loads of w pounds per linear unit, the shear line will be a straight line inclined toward the right, as the resultant force on the beam due to the uni- form load is decreased by wx^ where x is the distance of the section from the left end. If there are concentrated loads as well as uniform loads, 60 MECHANICS OF MATERIALS the shear line will be straight and vertical under the loads and inclined between the concentrated loads. To apply these principles, let Fig. 43 represent a simple beam, carrying two concentrated loads, P^ and P^^ and a uniform load of w pounds per linear unit. Remembering the definition of vertical shear, it is easily noted that the shear at the left end is equal to the left reaction R^ which is plotted as AO. The shear between the left end and P^ is R^ — wx^ where x is the distance of the section considered from the left end. For a section distant p-^ from the left end, taken just to the left of P-^, the shear is less than the shear to the left end by wp^.^ and the ordinate to the line AB at any point will be the vertical shear for that section. For a section just to the right of P-^, the equation of the shear has become R^ — wx — P^, as the two sections taken to the right and left of P^ are considered so close together that the uniform load has not increased. The shear line will therefore drop to (7 on a vertical line through the point of application of P^. Between P^ and P^ the shear will decrease at the same rate as between R^ and Pj, since the load increases di- rectly as the distance and CD will be parallel to AB. Then comes the drop due to P2, and EF is parallel to AB and CD. FO must be the value of the right reaction, since the sum of the vertical forces must be equal to zero. It is evident that the consideration of more concen- FiG. 43. BEAMS 61 trated loads would simply extend the diagram in a similar manner, and also, if the magnitude of the uniform load per unit of length should change at any point, the inclination of the shear line would also change at the same point. Art. 44. Moment Diagrams. The general equation of the bending moment for any section of a beam may be written from its definition. For a section of a beam distant x from the left end, the moment of the left reaction R-^ about a point in that sec- tion is R^x^ and the moment of the uniform load on the left of the section about a point in that section is the X . tux arm - times wx. or -— -, hence, 2 2 n.jr^2 fthe sum of the moments of the loads"! M = R^x — - — j that act to the left of the section I "^ [about a pomt in that section. J If there are no concentrated loads, the term ^containing the sum of the moments, etc., is zero, and for a cantilever beam the term containing the reaction R^ is zero, since there is no left reaction. In some cases the supports of a beam are not at the ends, and in that case the moment of the left reaction would be R-^(x—thQ distance of the reaction from the left end). When there are uniform loads on the beam, the above equation shows that M varies with x^^ hence the moment line will be a curve. If there are no concentrated loads, the equation of the curve will be the same at all sections, being a parabola whose equation is n,^ -r. wx^ M— R.x ^ 2 When there are concentrated loads in connection with the uniform load, the form of the equation changes at each 62 MECHANICS OF MATERIALS concentrated load, and the moment line, while still para- bolic in form, has a different equation between each pair of concentrated loads. If only concentrated loads are considered, the equation of the bending moment is {the sum of the moments of the loads acting] to the left of the section about a point in l that section. J If a^j is the distance from P^ of any section of a beam taken between any two concentrated loads P^ and P^, the moment of P^ about a point in that section is P-^Xy It is evident that the equation of the bending moment for any section will be changed by the addition of PiX-^ the instant the section is taken to the right of P^.^ and as at that point x-^ is very small there will be no abrupt change in the value of the bending moment as the section passes under Py As the value of il[f depends on the first power of a?, it is evident that the form of the equation is that of a straight line. Therefore, w^hen only concentrated loads are considered the moment line will consist of a series of straight lines whose inclination changes at each concentrated load. If the loads are all concentrated, the bending moments may be calculated for sections under the loads and plotted to scale. Joining the points so plotted by straight lines will accurately determine the form of the moment line. If uniform loads are to be considered, the moment line be- tween any two loads being a parabola, the bending moments for enough sections between any two loads must be found to enable the curve to be drawn (Fig. 43). The value of Mh\ the general equation for the bending moment is zero when x is zero, hence M\^ zero at the left end. The moments of all the forces about any point being BEAMS 63 zero, M must also be zero at the right end. That this is true for simple beams is so apparent that no proof is needed. In the case of cantilever beams with the right end fixed in the wall, if we remember that such a beam is but one half of a beam supported at the middle and that the right end of such a beam is, strictly speaking, the middle, the truth of the statement is e\ddent. Therefore the moment diagram will be a closed figure in all cases. Art. 45. The Relation between the Vertical Shear and the Maximum bending Moment. Given a beam carrying both uniform and concentrated loads, the value of the bending moment at any section to the right of all the concentrated loads considered is : wx M = Rix — - — Pi{x — pi)—P2{x — p2) . . . -Pn{x-pn) where pi, P2 - - - Vn are the distances of the loads Pi, P2 ' . • Pn from the left end of the beam and x is the distance of the section considered from the same end. The value of x for which M is a maximum is the value that renders —— =0 OT Rix — wx — Pi — P2 . . . Pn = 0. dx The left hand member of the last question is the ex- pression for the vertical shear for any section of a beam, therefore the value of x which makes the vertical shear zero renders the bending moment a maximum. As the equation for i!i^ was a general one and will apply to all kinds of beams and loadings, the results are true for all cases. The section of a beam where the bending moment is maximum is called the Dangerous Section^ and the prob- 64 MECHANICS OF MATERIALS lem of finding this section is simply one of finding where the vertical shear passes through zero. , Drawing the shear diagram, if the shear passes through zero under a concentrated load, the dangerous section is determined at once. When the shear becomes zero between two concentrated loads, the general equation for the vertical shear may be written for that part of the beam and equated to zero. The value of x which satisfies this equation determines the dangerous section. Having found the dangerous section, the bending moment may be calculated for that section, and when this value of Mi^ substituted in the formula i)[f= ~, the value c of aS' will be the maximum unit stress in the beam. Art. 46. Relative Strengths of Simple and Cantilever Beams. Let the uniform load on either kind of a beam be W and a single concentrated load at the middle of a simple beam or at the end of a cantilever beam also be W', then if a be some number whose value depends on the kind of a beam and the way in which it is loaded, the maximum bending moment for the beam may be expressed as This value substituted in the general formula M= — , c gives = — , which may be written W= — — . The a c cl strength of a beam may be defined as the weight it will carry with a given unit stress. From the above equation for TFit is evident that the weight a beam will carry with a given unit stress depends on the value of ot, hence the BEAMS 65 relative strengths of simple and cantilever beams loaded as above are directly proportional to cc. If the expres- sions for the maximum bending moments in simple and cantilever beams loaded with W as above are written, an inspection of the results will show that for a cantilever beam loaded with TFat the end a = 1, a cantilever beam loaded uniformly with W a = 2, a simple beam loaded with TTat the middle a= 4, a simple beam loaded uniformly with W «= 8. Art. 47. Overhanging Beams. Beams that overhang the supports are called Overhang- ing Beams. The fact that the reactions do not have their points of application at the ends of the beam does not prevent the application of the laws of mechanics to the de- termination of the magnitude of two reactions. Consider a beam that overhangs one or both supports and loaded in any way. Taking moments about a point in the line of action of one of the reactions as i^g' ^^^^ moment of H^ — the moments of the loads on the left of M^ + the moment of the loads on the right of M^ = 0, and as the sum of all the forces is zero, M^ -\- M^ = the sum of all the loads. These two equations will suffice to fully determine the reactions It^ and H^ when the magnitude and position of the loads are known. The shear and moment diagrams can be drawn by the same principles that were applied to simple and cantilever beams. In general, the vertical shear will pass through zero at two or more points, giving more than one value of x for which the bending moment is a maximum. The value of the bending moment at each of these points must be 66 MECHANICS OF MATERIALS calculated in order to find the greatest bending moment in the beam. The maximum moments may be either positive or negative, but when using the greatest value of M in the formula M = — , the substitution is to be made witliout c regard to sign. As the sign of the bending moment changes from posi- tive to negative, its value must pass through zero, and the position of the section of a beam for which the bending moment is zero is called an Inflection Point. The position of an inflection point may be approximately located by an inspection of the shear or moment diagrams and the general expression for the bending moment for that part of the beam written. Equating this expression to zero gives the position of the inflection point accurately. Akt. 48. Beams of Uniform Strength. When the maximum unit stress in all sections of a beam is constant, the beam is said to be one of Uniform Strength. The beams so far discussed have all had uniform sections, and the value of - was the same for all sections. c The bending moment iHi" varies for all sections, and if S is to be constant, - must vary with M, since M = For c c any beam loaded in any way the bending moment for a section at any distance from the left end may be expressed in terms of a variable distance x and this expression equated to — c Assigning different values to x, the corresponding values BEAMS 67 of - may be determined and the section of the beam at that "" . I point chosen to satisfy the value of - as found above. Beams of uniform strength may have any form of section, but they are usually made either rectangular in section or an approximation to such a section. I hcP' For beams of rectangular section - equals — ;- and M = ' in which either h or c? may be variable. Ex- 6 pressing ilf^in terms of the variable distance x^ the law of the variation of h or d^ as the case may be, determines the shape of the beam. At any section of the beam where iH/ = 0, there is no moment to be resisted, and so far as bending is concerned the area of that section can be made zero. In addition to the unit stress due to bending, there is at all sections of the beam a shearing unit stress due to the vertical shear at that section. If aS' is the allowable unit stress in shear the area of the section where iHf = is given by ^ = — -, where V is the vertical shear at that section. Art. 49. Moving Loads. In many cases the position of the loads is not fixed, and as the loads may occupy various positions, the value of M for finding the greatest unit stress in a beam must be de- termined from the position of the loads which gives the greatest bending moment. When the loads on a beam may change their positions they are called moving or Live Loads to distinguish them from stationary or Dead Loads. Assume a beam, length Z, and Pj and P^ two unequal loads that pass over the beam. See Fig. 49, 68 MECHANICS OF MATERIALS Let z be the distance of the greater load P^ from the left end of the beam, and p be the distance between the two Q loads. The dangerous sec- -P— $ * tion Avill always occur under one of the loads, and assum- . 5 Sk ■ J ing that it occurs under P-^ ^^^- ^^- Let Q be the magnitude of the resultant of P^ and P^^ and x be the distance of its line of action from P^ ; then P^x =^ P^{p - x'), Pi + A Q Therefore the resultant of P^ and Pg acts at a distance P V z -\ — -^ from the left end of the beam. Taking moments about a point in the line of action of the reaction P^^ and B^z^Qz-^- -^P-z = M. M will be a maximum when dM ===(?_2^^__^ = 0, dz II I P^p Therefore when ill/" is a maximum the middle of the beam is halfway between the resultant of tlie loads and the dangerous section of the beam. BEAMS 69 It is evident from the form of the equations that the same results would have been obtained had there been any number of forces. The results were obtained on the assumption that the dangerous section occurred under the first load, and if this assumption is true, the vertical shear must change sign as the section is taken to the right or left of the load Pj. When there are but two loads, the dangerous section al- ways occurs under the left hand load when the two loads are equal, and under the heavier load when they are not equal. When there are more than two loads, the position ^of the loads that gives the greatest bending moment does not always have the dangerous section under the maxi- mum load, but the general law holds true that When the middle of the beam is halfway between the re- sultant of the loads and the dangerous section, a maximum bending moment occurs. Art. 50. Use of Formula. To find the position of a system of loads that causes the greatest bending moment, assume the loads to be so placed that when the dangerous section is assumed to occur under any load, the above criterion is satisfied and the vertical shear passes through zero, and calculate the bending moment for that position. The result will be the maximum bending moment that can occur and have the dangerous section under the load as assumed. Assuming the dangerous section to occur under any other load, the maximum moment for that position may be found. The position that gives the greatest value of M will be 70 MECHANICS OF MATERIALS the one where the greatest bending moment occurs as the loads move over the beam. When a uniform live load moves over a beam, the greatest bending moment occurs when the load extends over the entire length of the beam. When the load only partially covers the beam, the maximum moment occurs at the dangerous section and the above criterion holds true. The greatest vertical shear caused by any moving load is found at the supports when the resultant of the loads is nearest to that support. Art. 51. examination 1. When is a bar called a beam? A simple beam? a cantilever beam ? a continuous beam ? 2. What is meant by, " the reactions at the supports " ? 3. Define the term Uniform Load; Concentrated Load. 4. Name the laws of mechanics that are used to deter- mine the reactions for a simple beam. 5. Define Vertical Shear and Bending Moment. 6. As, "the sum of the forces," as well as, "the sum of the moments of the forces," acting on each side of the sec- tion are equal, why is it necessary to use the expression, " acting to the left of the section," in giving the above definitions ? 7. From the definitions of the bending moment and vertical shear, write the expression for each in terms of a variable distance from the left end of the beam. 8. Why is it necessary to say, " moments being taken about a point in that section," in defining the bending moment at any section ? 9. Define Resisting Shear; Resisting Moment. BEAMS 71 10. Certain laws are deduced from experimental obser- vations made on beams under the action of bending forces. W hat are they ? 11. What is the Neutral Surface of a beam? the Neutral Axis of a section of a beam ? 12. Show that the neutral axis passes through the center of gravity of the section. 13. Prove that S=^^, 14. State clearly what each symbol in the equation S = means, and the units that should be used in sub- stituting for each. 15. What conditions as to loads and material must any Mc beam satisfy in order that >S'=— r-will be true for that beam ? 16. What is a Modulus of Rupture ? 17. Define the term Strength of a beam. 18. Show that the strength of any beam depends on the value of c£, where a is a number depending on the kind of a beam and the nature of tlie loading. 19. What is a Shear diagram ? a Moment diagram ? 20. Define the Shear and Moment line. 21. The shear line for a beam carrying only concentrated loads consists of horizontal and vertical straight lines. Why? 22. When only uniform loads are considered, the shear line is a straight line from end to end. Why ? 23. Show that for a beam with both concentrated and uniform loads, the moment line will be a series of curved lines. If there are no uniform loads, show that the moment line will consist of a succession of straight lines at different inclinations. 72 MECHANICS OF MATERIALS 24. If any part of the moment line is straight and parallel to the line representing the beam, what can you say of the bending moments for any sections taken in that part of the beam ? 25. Define the expression, 'Hhe Dangerous Section of a beam." 26. Give the relation that exists between the maximum bending moment for any beam, and the vertical shear for that section. 27. Show that the problem of finding the section of a beam where the bending moment is a maximum, is the same as that of finding the section where the vertical shear passes through zero. 28. What are overhanging beams? 29. How do they differ from simple beams? 30. Show why the reactions may be found in the same manner as for simple beams. 31. What is meant by the term Inflection Point ? 32. If the vertical shear is zero at more than one sec- tion of a beam, how can you find the greatest bending moment for that beam ? 33. When is a beam said to be one of Uniform Strength? 34. Show that if the section of any beam is varied so that — varies with M^ the beam will be one of uniform e strength. 35. When are the loads on a beam called Moving Loads? 36. Give the criterion for the position of a system of moving loads that causes a maximum bending moment. , 37. If there is more than the one position of the loads that satisfies the criterion, how can you tell which position causes the largest bending moment? BEAMS 73 PROBLEMS 1. Find the reactions for a simple beam carrying a uniform load of w lb. /in. The length is I in. and the whole weight TF lb. Solution. As the whole load is W, the sum of the two reactions must equal W or B A- Ti — W Taking moments about a point in the line of action of itg, since the resultant of the uniform loads act at the center of the beam, ^ 2 R.= — = — and R^ = — . ^22 ^2 2. Find the reactions for a simple beam, length ?, carry- ing a single concentrated load P at a distance p from the right support. Solution. The sum of the reactions equals the loads ; hence i?i + 7^2 = P. Taking moments about a point in the line of action of jRg, R^l - Pp = 0, Substituting the value of R^, Pp R -^P + R2 = P or R^ = p(l-P\ 3. Find the reactions for a simple beam 10 ft. long carrying a uniform load of 500 lb. /ft. 4. Find the reactions for a simple beam 30 ft. long carrying a load of 10 tons at a distance of 20 ft. from the left end of the beam. 5. Find the reactions for a simple beam 30 ft. long carrying two equal loads of 5 tons each at 10 ft. from either end, and a uniform load of 500 lb. / ft. 74 MECHANICS OF MATERIALS 6. Find the vertical shears at the middle and at the ends of the beam in problem 3. Solution. From the definition of vertical shear, F at a section taken just to the right of R^ is the value of that reaction, or 2500 lb. At the middle of the beam the force acting upward is simply the left reaction, and the forces on the left of the section that act down are the unit loads on that part of the beam, or one half the total uni- form load. Therefore the vertical shear at the middle is 2500 - 2500 = 0. At the right end just to the left of the right reaction 2500 - 5000 = - 2500 lb. or the vertical shear at R2 is — 2500 lb. 7. Find the vertical shears in the beam given in prob- lem 4, at sections taken just to the right and left of the load and at each end of the beam. 8. Find the vertical shears in the beam in problem 5, at each end of the beam and at sections taken just to the right and left of each load. 9. Find the bending moment at the middle of a simple beam, length I in., (a) for a load P at the middle. (5) for a uniform load of w lb. /in. P Solution for (a). The reactions are each—, hence by the defi- nition of the bending moment, for a section - from the left end of P I PI the beam M = — x - = — , as there are no other forces acting between R, and P. 2 2 4 10. Find the bending moments at section just to the right and left of the load on the beam given in problem 2. 11. Find the bending moment at the wall for a canti- lever beam, length I in., when the beam carries, («) a uniform load of w lb. / in. (5) a single load P at the free end. (c) a single load P at ^ in. from the free end. BEAMS 75 Solution for (a). As there is no left reaction, the bending moment is the moment of the forces acting on the beam about a point in a section taken at the wall. The moments of all the uniform forces being equal to the moment of their resultant, let the resultant of the forces wl equal W ; then 12. Find the bending moment at each load and at the middle of the beam given in problem 5. 13. A cast iron bar 1 by 1 in. in section and 36 in. long is broken as a beam. The modulus of rupture is 35,000 Ib./sq. in. Required the maximum bending moment. 14. Draw the shear dia- grams approximately to scale for simple beams 30 ft. in length loaded with (a) a uniform load of 100 lb. /ft. (5) a concentrated load of 3000 lb. at the middle. ((?) a uniform load of 100 lb. /ft. and a load of 3000 lb. at the middle. (c?) two equal concentrated loads at 5 and 10 ft. from the left end. 3000 .0CO00C)QQ000000. tsoo]^ I B 00 Problem 14 a. B Solution for (a), from the left end is The vertical shear at any section distant x V = 1500 - 100 X. This is the equation of a straight line, and as V is 1500 lb. at the left support and — 1500 lb. at the right support, if AB is 30 ft., AC = 1500 lb., and BD = — 1500 lb., then the line through the points C and D will be the shear line, and the diagram ABDC will be the shear diagram. 15. Draw the shear diagram approximately to scale for a cantilever beam 10 ft. long loaded with {a) a uniform load of 500 lb. /ft. (J)) a concentrated load of 500 lb. at the left end. 76 MECHANICS OF MATERIALS so o ^ K s -500 C jfi' ' *500 (c) a uniform load of 500 lb. /ft. and a concentrated load of 5000 lb. at the free end. (c?) two equal loads 2500 lb. each at 5 and 8 ft. from the free end of the beam. Problem 15 6. Solution for (b). The equation of the shear line is F= — 500, since there is no left reaction and the only load is 500 lb. at the end. Therefore, the shear line will be a straight line parallel to AB and at a distance — 500 from that line. A BCD is the shear diagram. 16. Draw the moment diagrams approximately to scale for the beams as given in problems 14 and 15. Solution for the beam in 14 a. The value of the bending moment at any section x from the left end is M = R^x wx^ = 1500 a; - 100 — 2' where x is in feet. Giving any values to x, the corresponding value of M may be found. When a: = 0, X = ^ ft., a; = 10 ft., a: = 15 ft.. M = 6250 ft. lb. M = 10,000 ft. lb. M = 11,250 ft. lb. 3000 nnnnnnnnnnno 1500 As the loads are symmetrical with the middle of the beam, the values of M for x equal 20, 25, and 30 ft. will be the same as for x equal 10, 5, and ft. Let AB = 80 ft. and CD, EF, GH, etc., repre- sent on some scale the values of M corresponding to x equal 5, 10, 15 . ., . 30 ft. ; then a smooth curve passing through ADF, etc., will be the moment line, and the figure AHB will be the moment diagram. Problem 16. BEAMS 77 17, Write the expressions for the value of M and V for any section of the beam, and determine the maximum bending moments and vertical shears, («) for a simple beam loaded uniformly with W lb. (S) for a simple beam loaded at the middle with W lb. (tf) for a cantilever beam loaded uniformly with W lb. (c?) for a cantilever beam loaded with W at the end. The length of all the beams being I ft. Solution fok, (a). The expressions for M and Fmay be obtained from the general formula by making " the concentrated loads to the left of the section " equal zero. Therefore, M = R^x - wx"^ 2 » and F = -Rj — WXy P wl Hence T7- wl V = — — wx. This expression is zero when x = -, and as ikf is a maximum when V is zero, substituting the value of x, which renders F= in the expression for M, gives T^ _ wlx _ wx^ _ wZ^ _ wP _ wl"^ _ Wl Evidently F is a maximum when x is zero. 18. A simple beam is 20 ft. long and carries two con- centrated loads, one 100 lb. at 5 ft. and the other 500 lb. at 8 ft. from the left end, and a uniform load of 100 lb. / ft. extending over the beam for a distance of 12 ft. from the right end of the beam. Draw the shear and moment diagrams and calculate the maximum bending moment and vertical shear. 19. A simple wooden beam 20 ft. long, 8 in. wide, 10 in. deep, carries a uniform load of 80 lb. /ft. Required the maximum unit stress in the beam. 78 MECHANICS OF MATERIALS Solution. For a uniformly loaded simple beam the maximum value of M is : Wl 80 X 20 X 20 X 12 .^ ^^^ . ,, = — = 48,000 m. lb., 8 8 ' ' and d c 2 6 6 6 I M^ bd'^ Sx 102 800 12 e Mc 48,000 X 6 oftm. / 6 = = — = odO lb. / sq. in. / 800 ' ■ 20. A simple wooden beam, rectangular in section, and 20 ft. long, is to be designed to carry a load of 240 lb. at the middle with a maximum unit stress of 300 lb. /sq. in. (d may be assumed to be equal to 6 h.) Solution. The maximum bending moment is : ,^ PI 240 X 20 X 12 , / bd'^ d^ M = — ■ = and - = — = — , 4 4 c 6 36 , M I Mx3Q .^ and - — = -. .-. = d^. S c S Whence d^ = 240 x 20 x 12 x 36 ^ -^^^S, or d = 12, b = 2. 4 X 300 ' 21. A simple wooden beam is 1 foot square and 10 yd. long. What uniform load can it carry if the unit stress is not to exceed 300 lb. / sq. in. ? 22. If the beam in problem 21 also carried a load of 1000 lb. at the middle, what unifoim load may also be C9;rried if the unit stress is not to exceed 400 lb. / sq. in.? 23. Two planks, 12 in. wide and 2 in. thick, are placed one on top of the other and used as a simple beam. They support a uniform load of 1000 lb. What is the maximum unit stress in the material? Assume that the surfaces in contact are frictionless and that I =30 ft. 24. A simple beam of wrought iron is 4 in. wide, 6 in. deep, 12 ft. long, and carries a uniform load of 32,000 lb. Is it safe? BEAMS 79 25. If the beam in problem 24 was a cantilever beam, what uniform load will it carry with a maximum unit stress not greater than the elastic limit ? 26. A simple steel 10 in. I beam weighing 25 lb. / ft. is 18 ft. long and carries a uniform load, including its own weight of 15,000 lb. Required the maximum unit stress. Solution. The table gives the value of - for this beam as 24.4. The maximum value of M is ^ Wl 15,000 X 18 X 12 , e ^^ — =: ■ — and S = . 8 8 / TT o 15,000 X 18 X 12 X 1 -,a «AA lu / Hence S = — ^ = 16,600 lb. / sq. m. 8 X 24.4 ^ ^ 27. Show that when the weight TF of a simple beam is 2 % of the load at the center, the error in the unit stress as found by neglecting the uniform load due to the weight of the beam is about 1 %. 28. A common rule states that when the load at the center of a simple beam is greater than five times the weight of the beam the weight may be neglected when making the calculations for strength. What maximum error will this rule allow ? 29. If W is the w^eight of a cantilever beam and -P the P load at the end, find the ratio of -^ when the error in the unit stress caused by the neglect of W is 5 %. 30. Select a simple I beam of structural steel 24 ft. long to carry a load of 12,500 lb. at the middle with a factor of safety of 4. The maximum value of M is M = ^' = IMOO ^ 24 X 12. s = 6W0 ^ ^, ,b / _ .^ 4 4 4 M^I^ 12,500 X 24 X 12 ^ ^^ S c 4 X 15,000 The table gives the value of / for a 15 in. beam weighing 45 lb. /ft. as 60.8, and as P = 15 W, this beam will satisfy the conditions. 80 MECHANICS OF iAIATERIALS 31. Select a standard steel I beam 10 ft. long to be used as a cantilever beam, to carry a load of 4 tons at the free end. Factor of safety 4. 32. Select a simple steel I beam for a span of 20 ft. to carry two equal loads of 2 tons each at 5 ft. from either end and a uniform load of 300 lb. /ft., the unit stress not to be greater than 16,000 lb. /sq. in. 33. Select a standard steel I-beam to be used as a simple beam 16 ft. long and carry a uniform load of 10,000 lb. Safe unit stress equal 16,000 Ib./sq. in. 34. A simple steel I-beam 10 in. deep, 30 ft. long, weigh- ing 40 lb. /ft. supports a uniform load of 11,250 lb. If the allowable unit stress is 16,000 Ib./sq. in., is the beam safe? 35. A cantilever beam 20 ft. long carries a uniform load of 50001b. Assume the unit stress equal to 16,000 Ib./sq. in. and select a steel I-beam to carry the load. 36. Select a standard steel channel 12 ft. long to be placed with the flanges vertical and used as a simple beam to carry a uniform load of 15,000 lb. Factor of safety 4. 37. A beam 30 ft. long is supported at points 10 and 5 ft. from the right and left ends. There is a uniform load of 500 lb. / ft. between the supports and concentrated loads of 450 lb. at either end of the beam. (a) Draw the shear and moment diagrams. (^) Find the greatest bending moments and vertical shears. (c) Find the inflection points. (c?) Select a steel I beam to carry the loads with a maximum unit stress of 16,000 lb. /sq. in. 38. A beam supported at two points 18. ft. apart over- hangs each support 6 ft. The overhanging ends carry a BEAMS 81 uniform load of 300 lb. / ft. and there is a concentrated load of 12,000 lb. at the middle of the beam. (a) Draw the shear and moment diagrams. (5) Find the greatest bending moment and vertical shear. (c?) Find the inflection points. (d) Select a steel I beam to carry the loads with a maximum unit stress of 16,000 lb. /sq. in. 39. Three men carry a stick of timber 12 x 12 in. x 12 ft. long. One man is at one end and the other two are at such a point that each of the three men carries an equal load. Find that point. 40. A cantilever beam of uniform strength rectangular in section is 12 ft. long and carries a load of 1200 lb. at the free end. The material is cast iron and the factor of safety is 10. Find the largest and smallest sections and make sketch showing the plan and elevation of the beam when, (a) the width is constant at 4 in. (5) the depth is constant at 12 in. 41. A simple beam of uniform strength is rectangular in section and 12 ft. long and carries a uniform load of 9600 lb. The material is cast iron and the factor of safety is 10. Find the smallest and largest sections and make a sketch showing the plan and elevation of the beam when, (a) the width is constant at 4 in. (5) the depth is constant at 12 in. 42. If the beam in problem 38 was a rectangular steel beam of uniform strength and constant depth, find the proper size for the largest section when h = d for that sec- tion and the safe working unit stress is 16,000 lb. / sq. in. If the allowable unit stress in shear is 10,000 lb. / sq. in., find the area of the least section possible. Make a sketch of the plan of the beam. 82 MECHANICS OF MATERIALS - 43. Two equal loads are 6 ft. apart. Find their position as they are moved over a simple beam 20 ft. long that gives the greatest bending moment in the beam. 44. Three loads each 4 ft. apart are moved over a beam 18 ft. long. From left to right the loads are 4000, 2000, and 2000 lb. Find the position of the loads that gives the greatest bending moment in the beam. 45. Two loads 6000 and 4000 lb., 5 ft. apart are moved over a simple beam 32 ft. long. What size steel I-beam will be required? The unit stress is 16,000 Ib./sq.in. and the stress due to the weight of the beam is neglected. 46. As the loads given in problem 45 pass over the beam, asssume that the weight of the beam is 1600 lb. and draw a diagram to scale, using as ordinates the bending moment under the 6000 lb. load and as abscissa the dis- tance of that load from the left end of the beam. If the unit stress is 16,000 Ib./sq. in., determine the si^e of the steel I-beam required. 47. If Q is the resultant of a system of loads moving over a simple beam, x the distance of Q from the left end, and a the distance of Q from the middle of the beam, show that the conditions for maximum moment require that I I — X = X — 2 a when a: > -, 2 I — X = X -}- 2a when x<-^l> and if this condition is satisfied, that the value of the maximum moment is given by wr ^ ( 9 ^2 I ^^® moments of the loads on 1 -^'^max == y V^ ± ^ ^^ "■ 1 the left of the dangerous section.] CHAPTER IV TORSION Article 52. Derivation of Formula. In the previous chapters the forces were assumed to act in a plane passing through the axis and were either parallel or perpendicular to the axis. The forces that produce the stress known as torsion act in planes that are perpendicular to the axis of the bar, and while the lines of action of the forces are perpen- dicular to it, they do not pass through the axis. The effect of such forces must be to twist the bar. Assume a cylindrical bar, one end of which is firmly fixed in the wall, to be acted on by a couple ly- ing in a plane perpendic- ular to the axis of the bar at a distance I from the wall, and whose moment about that axis is Pp. A fiber of the bar that before the application of the force occupied the position of the line ad (Fig. 52 a), after the force has been applied will occupy the position of the helix ah^ and a point d on the surface of the bar will have moved to the position h. It is evident that the angle had^ the angle of the helix, is independent of the length of the bar and depends only 83 Fig. 52 a. 84 MECHANICS OF MATERIALS on the twisting forces and the material of the bar. Since any plane section perpendicular to the axis of the bar between the wall and the couple would contain an arc similar to hd^ and the length of that arc is proportional to the distance from the wall, the angle hod is propor- tional to the length of the bar and the twisting forces. Tills angle is called the angle of twist, and will be denoted by 6, By analogy with tension, the distance a point on the end of the bar moves under the action of the twisting forces being similar to the distance a point on the end of a tension bar moves under the action of the tensile forces, the ai'c hd may be taken as a measure of the deformation of the surface fibers of the bar due to the twisting forces. As the arc hd was proportional to the length, — can be taken as representing the unit deformation. Experiment has proven that when a bar is circular in section and no stress is greater than the elastic limit, the line od. moved to anv new position during the twisting of the bar, re- mains a straight line. This being true, the length of any arc d^-^ (Fig. 52 5), with a center at and a radius «/, is proportional to its radius, and as the arc is propor- tional to the deformation at the radius «/, the force Fig. 52&. producing this deforma- TORSION 85 tion is proportional to the same distance. This reasoning is true for any perpendicular section of the bar between the plane of the couple and the wall. Assume the bar to be cut by a plane perpendicular to the axis between the wall and the twisting couple, and introduce forces in that section to render the free end in equilibrium. These forces must all act in a plane per- pendicular to the axis, since the external couple has no component perpendicular to such a plane, and their re- sultant must be equivalent to a couple whose moment is equal to that of the twisting moment. Considering the two faces of the bar in any section, it is evident that the face on the right end tends to slide against the face on the left end as the former tends to rotate about the axis of the bar, thus producing a shearing stress throughout the section. Since the force, and con- sequently the unit shearing stress, is proportional to the distance from the center of the bar, if we let S^ be the unit shearing stress in the surface fibers, e be the distance of those fibers from the axis of the bar, and ?/ be the dis- tance of any fiber from the same axis, then — ? is the unit c stress at a unit's distance from the axis and _^?/ is the c unit stress at any distance y. Considering the stress uniformly distributed over any small area dA at the distance ?/, — ^dA is the force acting c on that elementary fiber, and — ? ^'^dA is the moment of c this force about the axis of the bar. But the sum of the moments of the forces acting in the section is equal to the twisting moment Pp, hence -^ / y^dA = Pp. I yMA is 86 MECHANICS OF MATERIALS the expression for the polar moment of inertia about an axis through the center of gravity of the section, and writing J for „ ^ ^ I y^dA gives Pp = — , {e) which shows the relation between the maximum unit stress in shear and the twisting forces. Art. 53. Modulus of Section. Comparing formula (e) with M— — , it will be noticed c that they are of the same general form, ilf and Pp repre- sent the effect of the external forces, S in each case is the maximum unit stress in the section, and by analogy with -, - may be called a modulus of the section. Formula c c (e) is only true for bars whose sections are circular, and where the material and loading of the bar satisfies the conditions stated for M = — c Art. 54. Square Sections. For rectangular sections the formula is only approxi- mately true, as a radial line drawn from the corner does not remain straight during the twisting of the bar, as was the case with the circular section. The investigations of St. Venant cover the rectangular sectioli, and his results give for a rectangular shaft sub- jected to torsion, g ■^P^ ^'^ (nearly), in which d is the side of the square. From the form of J d^ the expression the effective value of - must be --— instead ^ G 4.8 TORSION 87 of the calculated value — - — , hence the value of Pp as 6 found from the above formula is less than that obtained from equation (e), when equal values of S^ and d are used. Square sections are apt to be weaker than St. Venant's formula would indicate, since the maximum stress is carried on the edge of the square and any slight defect reduces the effective diameter of the bar. For this reason square sections are rarely used to resist torsion alone. Art. 55. Illustrations. The resistance at the wall may be assumed as another couple, whose moment is equal and opposite to that of the twisting moment, without altering the conditions as assumed when formula (e) was developed. In the case of a line shaft, where the belt from the engine produces a twisting moment at one end of the shaft, the resistance of the belt on the pulley at the other end is equivalent to an equal moment. If there are several pulleys on the same shaft, each producing a moment by its resistance to turning, it is evident that the resisting moment of the shaft will not be constant throughout the length, but will vary with the resistance that it has to overcome. To illustrate, assume a shaft with three pulleys, one at each end and one at the middle of the shaft. The driving moment is applied at the left end and the resisting moment of the other two pulleys, P^Pi and P^p^-, must be equal to the driving moment Pp. The moment to be resisted by the shaft between the driving pulley and the one at the middle is Pp = P^fx + ^ilPv After passing 88 MECHANICS OF MATERIALS the middle pulley the resistance to be transmitted by the shaft is only the twisting force of the third pulley, and consequently the resisting moment between the second and third pulleys is equal to P^ip^,' Art. 56. Twist of Shafts. In Fig. 52 Z», dh was taken as a measure of the defor- mation of the surface fiber, and — the unit deformation of that fiber. From the figure dh = 6c and by the defi- nition of the modulus of elasticity, if S^ is the unit stress in the surface fibers, the shearing modulus of elasticity i^^^st be Q ^n T> 1 ^_S,_SJ. Ppl I The latter expression is obtained by substituting for S^ its value from /S'„ = —^- Equation (/) gives the relation between the modulus of elasticity for shear, the twisting moment or the unit shearing stress, and the angle of twist. When the data given in any problem is sufficient to determine two of the three quantities, the twisting moment, the unit shearing stress, or the section modulus, the cr J formula Pp=— ^— will completely determine the other c one. Or when the given data will determine three of the following quantities, the twisting moment or the unit shearing stress, the dimensions of the bar, the angle of twist, and the shearing modulus of elasticity, formula (/) can be used to solve any problem involving the twisting moment or the unit stress and the angle of twist. TORSION 89 Since all of the tabulated values of the constants of materials are given in pounds and inches, all dimensions of weight and linear or square measure must be reduced to pounds and inches before making the substitutions in the formulas. As the value of 6 used in the development of the formula was in circular measure, 6 must be ex- pressed in radians. Art. 57. Relative Strengths and Stiffness of Shafts. The strength of a shaft may be defined as the twisting, moment it will carry with a given unit stress. Formula (e) shows that the strengths of shafts vary directly as the value of -^— for each shaft, and when the shafts are ^ . J of the same material, as — c Defining the stiffness of a shaft as the angle of twist for a given value of P^, since = -^ or _^, it is evident that the stiffness of two shafts of the same material varies directly as — or -, depending on whether the twisting force is given directly or in terms of the stress. Art. 58. Horse Power of Shafts. A horse power being defined as 33,000 ft. -lb. per minute, if H is the horse power to be delivered by a shaft making N revolutions per minute, the value of Pp in terms of the horse power may be found from the equality of the work done by the twisting force per minute and the work represented by J?^ horse powers. Assuming that J^ is a force acting at a radius jt?, the work done by that force in one revolution must be 2'TrPp in. -lb. and as a horse power is 33,000 ft. -lb. per minute, or 396,000 90 MECHANICS OF MATERIALS in. -lb. per minute, 2 TrPpiV^ = 396,000 H, wliicli re- PpN duces to ■g"= /^ (approximately). Substituting for Pp its value in terms of S^ from formula (e), gives jy= ^t) ^rvr. — These two expressions may be used to de- termine the horse power that a given shaft will transmit when the number of revolutions per minute and the twist- ing moment or the maximum unit stress are known. The values of either Pp or Sg may be found from (/) and the angle of twist for a given horse power determined. Art. 59. Shaft Couplings. When two lengths of shafting are to be joined together, the connection is often made as in Fig. 59. The moment Fig. 59. of the shearing stresses in the bolts must be equal to the twisting moment of the shaft, and the relation between the twisting moment of the shaft and the resisting moment of the bolts may be stated as Pp = SJ — —^ where J' is the polar moment of inertia of the section of a bolt about the axis of the shaft, n the number of bolts, SJ the maximum unit stress in the bolts, and ^ sile or compressive force P and the dimensions of the spring. The length of one coil of the spring is approximately ttD, and if the wire is twisted through a small angle 0, by a moment that increases uniformly from to —^^ the PD work done on the wire of one coil is 6 — ■ — , where 9 is in 1 circular measure. If we let A be the total deflection of the spring, that is the amount of shortening or lengthening, 94 MECHANICS OF MATERIALS and B the deflection for one coil, as the force acting varied uniformly from to P, the work done on one coil is Since these two expressions represent the same quantity of work, they must be equal, or = — — or = — . As 6 is the angle of twist, formula (/) gives as its value FJ Fd^ ' Equating the two values of ^, lb - — — = — or = , Fd^ D Fd^ ' which is the deflection of one coil in terms of the load P. Taking • e=^=~ Fc i>' we have h = — ^— — • as the relation between the unit stress Fd and the deflection for one coil. Since the strength of any bar under the action of twist- ing forces is independent of the length of the bar, the formula for the strength of a spring is also independent of the number of coils. If n is the number of coils, the total deflection of any spring must be n times the deflection for one coil, or ^^^3^ ^^^^2^ A = or -^ ■• Fd"^ Fd EXAMINATION 1. What is a torsional force ? How does it differ from a force that produces a shearing stress that is uniform over the section of a bar ? 2. Can a torsional force produce any tensile or com- pressive stresses ? TORSION 95 3. " It is evident that the angle had is independent of the length of the bar." Prove it. 4. Show that the angle of twist depends on the length of the bar. 5. If a straight radial line drawn on the end of a bar remains straight when the bar is being twisted, show that this fact proves that the unit stress is proportional to the distance from the axis. 6. Show that Ss=-^r-- 7. State under what conditions of load and material the formula Pp = — ^— is true. c 8. Define the terms "strength of a shaft"; "stiffness of a shaft." 9. Does the strength of a shaft depend on its length ? Is the same true of the stiffness ? 10. Two shafts of the same diameter and length are of different materials. What is their relative strength ? What is their relative stiffness ? 11. Show how to find the expression for the modulus of elasticity in shear. 12. Why was it necessary to reduce the value of a horse power, 33,000 ft. -lb., to inch-pounds in order to obtain the expression SJJST ^ 63,000 c' PROBLEMS 1. One end of a circular bar 2 in. in diameter and 10 ft. long is fixed in a wall, and at the other end there is a couple whose moment is 300 ft.-lb. Required the unit stress in the bar ? Solution. The value of J = ^ — and c = — ; hence - = — — . Pp = 300 ft.-lb. = 3600 in.-lb. ^^ . ^ c 16 96 MECHANICS OF MATERIALS S J Substituting these values in Pp = - — , and solving for Ss, gives c o 3600 X 16 nnnrv ii, / Ss = = 2290 Ib./sq. m. TT X 8 2. A circular bar 7 in. in diameter, 10 ft. long, is acted on by a force of 10 tons perpendicular to, and at a distance of 3.14 ft. from tlie axis. Required the unit stress induced. 3. Find the diameter of a circular steel bar to carry a twisting moment of 20 ft. -tons. 4. Find load that can be applied at the end of an arm 6 ft. long so that the maximum unit stress induced in a cir- cular bar 2 in. in diameter will not exceed 12,000 Ib./sq. in. 5. A circular steel bar 2 in. in diameter is twisted by a force at the end of an arm 6 ft. long. Required the force if the unit stress due to torsion is equal to the elastic limit. 6. If the bar in problem l was soft steel, find the angle of twist. Solution. Taking the values as found for 1 and substituting in = ^, Z rr 10 X 12 in. and F = 12,000,000, Fc ^ ^ 2290 X 10 X 12 ^ 22,900 ^ ^229 radian, 12,000,000 X 1 1,000,000 or approximately 1° 18'. 7. A soft steel bar is 6 in. in diameter and 20 ft. long. What force acting tangent to the surface will twist the bar through an angle of 1° ? 8. A steel shaft 2 in. in diameter and 50 ft. long is transmitting a torsional moment that causes a unit stress equal to the elastic limit. Required the angle one end is twisted through relative to the other. 9. Find the diameter of a steel shaft 10 ft. long to carry a twisting moment of 81,700 ft. -lb., if the unit stress is not to be greater than 10,000 Ib./sq. in., and the angle of twist less than 1.15°. TORSION 97 10. Find the diameter of a steel shaft making 100 revs. /inin. and transmitting 200 H.P., the unit stress being 6300 lb. /sq. in. S,JN 6300 X TT X c?3 X 1 00 nnn Solution. ^ = 0^^= 63,000x16 "^^^^ - ^3 ^ 200 X 63,000 X 16 ^ ^ g^ .^^ 6300 X TT X 100 The shaft chosen would be either 4^ or 5 in. in diameter. 11. The Allis Chalmers Co. base their tables for the strength of mild steel shafting on the formula J£= cd^ JV, where d is the diameter of the shaft in inches, and c a number which has the following values: For heavy or main shafts c = .008 For shafts carrying gears c= .010 For light shafts carrying pulleys c = .013 Find the unit working stress allowable in each case. 12. A hollow steel shaft is 12 in. external and 10 in. internal diameter. Find the H.P. that may be trans- mitted at 100 R.P.M. when the unit stress due to torsion does not exceed 6000 Ib./sq. in. 13. A flange coupling is to be used to connect two lengths of 6 in. steel shafting. The maximum allowable shearing unit stress in the shaft and bolts is 6000 lb. /sq. in. Assume the diameter of the bolt circle to be 8 in. and find: (a) By the use of the approximate formula the total bolt area required. (6) Assuming the number of bolts as 6, determine the true unit stress in the bolts. 14. A hollow shaft has the outside diameter twice the inside diameter. Compare its strength with that of a solid shaft of the same material and section area. 15. If the elastic limit of the material in one shaft is 60,000 lb. / sq. in. and costs 10 ^ per pound, what can 98 MECHANICS OF MATERIALS you afford to pay for a shaft to do the same work, if the elastic limit of the material is*30,000 lb. /sq. in. ? 16. A helical spring is made of wire whose diameter is 1 in., the mean diameter of the coil 4 in., and has 30 coils. If the value of F is 12,000,000 and the working unit stress 60,000 lb. /sq. in., required the load it will carry and the deflection under that load. 17. Find the size of the wire and the mean diameter of the coils for a steel helical spring with 30 coils to carry a maximum load of 6000 lb., have a deflection of 10 in. under that load and a limiting unit stress of 60,000 Ib./sq. in. 18. D. K. Clark gives as the deflection for one coil of a helical spring h— ^^ , where d is the diameter of the wire in sixteenths of an inch. CHAPTER V THE ELASTIC CURVE Article 62. Definition. When the beam is bent, the neutral surface assumes a curved form; and the projection of this surface on a vertical plane parallel to the axis of the beam is called the Elastic Curve. If the equation of this curve for any beam is expressed as y^f(x), where y is the deflection of the beam at any point distant x from the left end of the beam, the deflection of the beam at any point can be easily found. Art. 63. Equation of the Elastic Curve. To derive this equation of the elastic curve let Fig. 63 represent a portion of a bent beam. Let ji measured along the axis of the beam be tak- en as representing dl., where I is the length of the beam and the curve dl a differential part of Fig. 63. 99 100 MECHANICS OF MATERIALS the elastic curve, then if o is the center of curvature, oi and oj are radii. Before any bending took place these radii were parallel to each other, and eh may be assumed to have been in the position U parallel to af. Let the greatest distance of the neutral axis to the surface fiber be c and assume that jh = c. The deformation of the outer fiber is kb and the unit deformation is — , since Jcb is the change dl ^ in the length dl; J^ is the modultis of elasticity, and S the unit stress in that fiber, hence kb = ^-—-. From the similar triangles jkb and oij — = ^. ri) ji of But oj = r = the radius of curvature, ji = c?Z, jb = c, and kb — -^7- ; hence, substituting these values in (1), we have Sdl c ^ ^ \^ , CI -^^ Edl r E r I . M E El ,oN hence -_ = — , or r = — -, (2) I r M ^ ^ is the equation of the elastic curve of a beam in terms of the modulus of elasticity, the bending moment at any point of the beam, the moment of inertia, and the radius of curvature of the elastic curve at that point. The value of r expressed in rectangular coordinates is 1+r^ '•^ § ^^^ dx^ Since the degree of curvature of any beam in an en- gineering structure is very small, the value of the tangent THE ELASTIC CURVE 101 of the angle which the tangent at any point on the curve makes with the axis of ^ is a very small quantity and -^ j may be neglected in comparison with unity. Equa- tion (3) then reduces to If dx is assumed to be equal to dl^ an assumption which is approximately true when the degree of curvature is small, and the value of r as just found is inserted in equation (2), we have dx^ which is the differential equation of the elastic curve of any beam. M is the bending moment for any part of the beam for which the equation represents the curve, therefore must be expressed in terms of x any distance from the left end of the beam, and y is the ordinate to the curve or the deflection of the beam. Since the condition that jE^ = — was introduced in the € derivation, the unit stress must be within the elastic limit, Mc and as the formula aS' = -— was also used, all the condi- tions of the materials that are necessary to the correct use of that formula obtain with the one just developed. The assumptions that dl = dx^ and that the maximum value of -^ was so small that ( — ) might be neglected as dx \dxj compared to unity, introduce errors that, while they are small, increase as the degree of curvature increases. The 102 MECHANICS OF MATERIALS use of the formula under ordinary engineering conditions gives results that agree well with those obtained in exper- imental work, the discussion of the limitations to its use being given to show that the formula is not applicable to all cases. Art. 64. Deflection of Beams. Let M = f(x) be the expression for the bending moment for a portion of any beam loaded in any way; then mfi=fix) (1) is the differential equation of the elastic curve for that part of the beam for which f(x) represents the bending moment. Integrating (1), EI^=fix-)+0, (2) where (7 is a constant of integration. The value of C may generally be found by noting that -^ is the tangent of the angle which the tangent at x makes with the axis of X, and from the conditions of the problem, finding a value of X for which -^ is either zero or known. dx Integrating again, EIy=f"(ix')^-Ox+ C,. (3) The value of (7^, the constant of integration, can in most cases be determined by finding a value for x for which 1/ is either zero or known. When O and C^ are determined, equation (3) may be used to find the deflection at any point of the beam for which the bending moment is equal to /(a;). THE ELASTIC CURVE 103 As EI— is the differential coefficient of Ely, the value of dx dy X that makes EI— = will determine the maximum value dx of y, provided that this value of x falls within the limiting values of x, for which f{x) =M, in which case 2/ = maximum deflection. In the determination of the constants of integration, the student is reminded that when the moment diagram is symmetrical about a vertical line at the middle of the beam, — is zero for a value of a; =— , and for the special dx 2 dv case of cantilever beams ~ is zero at the wall, as the re- dx strainment keeps that part of the beam fixed and horizontal. Since there is no deflection at the supports, 1/ will always be zero at the points where the beam is supported. When there are concentrated loads on the beam, the expression for itfwill take a different form for each part of the beam between the loads or between the loads and reactions. If there are n concentrated loads on the beam, there will he n -\- 1 forms that the expression for the bending moment may take, making n -\- 1 equations of the elastic curve, tlie double integration of each bringing into the problem 2 (^ + 1) constants of integration. For simple beams the value of x that will make -^ = dx is not known unless the loads are symmetrical with the middle of the beam, but ^ is always equal to zero at the supports where x is either or I. Any two of the 7^ + 1 equations representing consecu- tive portions of the beam will have the expressions for — ^ dx and i/ equal for a value of x at the load where the equa- tions meet. As there are but 2 n -\- 2 constants of inte- 104 MECHANICS OF MATERIALS gration, and two known conditions are always to be had from the fact that ?/ = when x= or x = I, the 2 n equations resulting from the equating of the values of — ^ and y under each load will suffice to determine the dx ^ other 2 n constants of integration. In using equation (3) for the determination of the maximum deflection of a beam, the student is again reminded that the equation only holds true for tliat portion of the beam represented by /(a;), and where there are several concentrated loads, each portion may have to be investigated, in order to find the greatest value of ^ for the beam. In general, an inspection of the distribution of the loads on the beam will show the portion of the beam where the greatest deflection is likely to occur. If the bending moments for simple and cantilever beams loaded uniformly with TFJ simple beams loaded at the middle with TF, and cantilever beams loaded with TTat the end, are expressed in terms of x^ these moments may be substituted in UI —^ = iHf, forming four differential equa- dx^ tions. Integrating each equation twice gives the value of y, the deflection of the beam for any value of x. The maximum value of the deflection y may be expressed as 1 WP A= — — --, where /3 is a constant depending on the kind of a beam and the nature of the loads. Transposing, W= /6— -— . The load a similar beam will carry was ST given as TF= a — -, and if these two expressions for W ^^ a SP are equated we find that A = — — -, giving the relation THE ELASTIC CURVE 105 between the maximum deflection, the dimensions of the beam, the unit stress, and the modulus of elasticity. Attention is called to the way A varies in the two expressions for the maximum deflection. When the load W is considered, A varies directly as P and inversely as Z, and when the unit stress is given, A varies directly as P and inversely as e. For rectangular sections, this latter variation makes the maximum deflec- tion independent of the breadth of the beam. Aet. 65. Restrained or Fixed Beams. A beam is said to be restrained or fixed when one or both ends are so firmly imbedded in the wall that the tan- gent to the elastic curve at the fixed ends always remains horizontal during the flexure of the beam. A cantilever beam under this definition is a fixed beam when it projects from the wall ; but since there was no reaction at the free end, the bending moments and shears can be determined without reference to the fixed end. When the free end of a cantilever beam has a support placed under it, the con- ditions are. changed. The magnitude of the reactions can not be determined from the conditions for mechanical equilibrium that were used for simple beams. The value of Jf for any section of the beam will contain a term that includes the unknown reaction, and this value of M sub- stituted in the general equation of the elastic curve will give the differential equation of the curve for this beam. When this equation is integrated twice, the value of the unknown reaction may be found by applying the known conditions that the resulting equations must satisfy. Let Fig. 65 represent a beam fixed at one end and supported at the other, loaded in any way. If B^ is the 106 MECHANICS OF MATERIALS reaction at the left end of the beam, the bending moment for a section distant x from that end is M= R^x - "ff- rthe sum of the moments of the loads' to the left of the section with refer- ence to a point in that section and putting this value of M\xv the general equation of the elastic curve gives ( sum of the moments ) \ of the loads, etc. \ ' which is the general equation of the elastic curve for a beam fixed at one end and supported at the other. If there are no concen- ^^^^ trated loads, the last zz^U/ l®^''^^ o^ ^^ right hand member will be zero, ... and if there is no uni- will be form load, IVX" 2 Fig. 65. z^^:^^ zero. In a previous ^^ article it was shown ., that in a sfeneral case Wy'^////^. there were 2 (w + 1) conditions for the deter- mination of an equal number of integration constants, and in this case there is the additional condition that -^ = when x = 1, which ax may be used to determine i^^, since the tangent to the curve is horizontal at the wall. Stated briefly, the con- ditions that ?/ = at the fixed and supported ends, -^ == at the fixed end, and the ^n equations resulting from the THE ELASTIC CURVE 107 equating of the values of -^ and y under each load, will be sufficient to completely determine the values of the 2 (n + 1) constants of integration, and that of the unknown reaction at the left end of the beam. Art. 66. Beams fixed at Both Ends. Let Fig. ^^ represent a beam fixed at both ends and loaded in any way. The forces which act between the beam and the walls supporting the beam, keeping the tan- gent to the elastic curve at the wall horizontal, are un- known. The un- known systems of '^'^^ forces acting in each wall may each A be replaced by a single vertical force acting up- ward at the face of each wall, and a couple whose mo- ment is sufficient to keep the tangent at the wall horizon- tal. The beam, under the action of these forces and the loads, may then be considered as a body in equilibrium. From the mechanics of equilibrium of parallel forces, it is evident that tlie sum of the two vertical forces acting up- ward must be equal to the sum of the loads on the beam, and if the moments of the couples are determined, the values of the two vertical forces may be found. Fig. 66. 108 MECHANICS OF MATERIALS Let R^ (Fig. 66) be the force at the left end and R^ the force at the right end, and the moments of the couples at the left and right ends of the beam be M^ and M^^ respec- tively. Since the moment of a couple about any point is constant, the moment of the couple on the left end about a point in a section distant x from the left end is M^^ and the value of the bending moment for that section is ■n/r n/r , -r> WOiP' (the suiii of the moments) J- ■■■ 2 (of the loads, etc. ) This value of iJf substituted in EI—^ — M'\^ the differ- ential equation of the elastic curve for a beam fixed at both ends. If there are no concentrated loads, the last term of the expression for M will be zero, and when there is no uniform load, the term containing w will disappear. If there are n concentrated loads, there will be ^ + 1 values for M^ and each expression will contain the unknown moment My The double integration of the 71+1 equations will bring into the problem 2(n + 1) con- stants of integration, making 2 ti + 6 unknown constants to be determined, as My, M^, R^ and R^ are all unknown. Noting that R^ + R^ equals the sum of the loads, and that the moments of all the forces must be zero will give two, the values of y and — ^ being zero at each wall will dx ^ give four more, the derived values for y and -^ in the CbX adjacent expressions for M are equal for values of x under any concentrated load will supply 2 n equations, the required 2^ + 6 equations may be written. If the loads are symmetrical with the middle of the beam, the solution will be much simplified, as — ^ is zero at the dx THE ELASTIC CURVE 109 middle of the beam, and R^ = R^^ and M^ = M^. The methods given are general and will completely determine the bending moments or deflections for any restrained beams. Having fully determined the expression for the values of M and V for any restrained beam, the shear and moment diagrams may be drawn, and the maximum mo- ments as well as the inflection points determined, as for simple beams. The value of x which makes the vertical shear equal zero gives the dangerous section for the beam, and the formula aS'= — with the value of ili'a maximum can be used for all investigations of the strength and safety, as well as the design of restrained beams. Art. 67. Continuous Beams. A continuous beam was defined as one having more than two supports. The supports are as- 1 00X^)^0000^^000 sumed to be on the I Nil <-x-^ Uz same level and rigid, I ^_____.^ and the section of the VJ" beam uniform. ' Fig. 67. Let Fig. 67 represent any two intermediate spans of a continuous beam or girder whose lengths are l^ and l^ and loads per linear unit w-^ and w^ respectively. There is an unknown bending moment at each support. Let Ny, iV^, and iVg represent these moments. The moment N-^ is produced by the resultant of all the forces acting to the left of the first support. This result- ant and the reaction at the left-hand support may be re- placed by a couple whose moment is iV^ and a vertical force V^ acting at the left-hand support. Evidently F^ is the sum of the resultant force and the reaction at the left support. 110 MECHANICS OF MATERIAL Similarly F^, and a couple whose moment is iV^g' ^^^^^ ^s and a couple whose moment is iVg, can replace the forces acting at and to the left of the second and third supports. Taking the origin at the left-hand support and x as any distance in the left-hand span from that support, the equa- tion of the elastic curve for that span becomes ^jg=Jf = iVi+Fl^-^'. (1) Integrating twice and determining the constants by the known conditions that ^ = when x equals either or ?p . UI^^=W,x + ^-'^+0, (2) and m^=^^^-'^+0,x + C,. (3) (72=0 and 0^=-^ 2 6~ ' Substituting the value of 0-^ in (2), ■^^^"■^i'^ + ~2 6~+^ 2 6~ ^^^ Taking the origin at the second support and x as any distance in the second span, the differential equation of the elastic curve for that span becomes ^jj=i(f=iv, + r,^-^. (6) Integrating twice and determining constants from the known conditions that ?/ = when x equals either or l^^ v/e can write THE ELASTIC CURVE 111 The values of -j- in (4) and (6) are equal when x=^l^ in (4) and in (6). (4) reduces to — -o-H Q 5—' \'J ax x=L and (6) to ax "^2^2 -^2^2 ^2^2 :r=o 24 2 From (1) M= N^ when x = ly. (8) Similarly^ from (5) M = N"^ when x — l^^ w 1 '^ TV — TV ?/; 7 if=iV3 = iV2+F2?2-^. .•.r2 = ^^iy^2+^^ (10) 2 Equating (7) and (8) and substituting values of V^ and V^ from (9) and (10) and reducing, we have N,l, + 2 N, (I, + ?,) + N,l, = - ^^^'° + "'^^''' , (11) which gives the relation between the unknown bending moments iV^, iVg, and iVg, and the uniform loads w-^ and w^. One equation similar to (11) may be written for any three consecutive supports, and if n is the number of the sup- ports, n — 2 such equations may be written. When the ends of the beam are merely supported, the moments are iV^, iV2 • • • ^n-i ^^^ the values of iV^ and iV„ are each equal to zero, so that the n — 2 equations that may be written will completely determine the values of the 7^ — 2 unknown moments. ' When both ends of the beam are to be fixed, the two additional conditions that -^ = at either end support. 112 MECHANICS OF MATERIALS will furnish two more equations that will be sufficient to determine the unknown bending moments at the end supports. When the values of the bending moments at the sup- ports are determined, the maximum bending moment in any span may be found by writing for the desired span an equation similar to tliat given for M in (1). The forces Fj, 7^, and F^ are equal to the vertical shears just to the right of the first, second, and third supports respectivelyv Let R with subscripts 1, 2, and 3 represent the reactions at the corresponding supports. Then for the first support From the definition of vertical shear Fi - M>i?i + i?2 = Fj or iZ^ = Fj - ( Fj - w^l{). (12) As an equation similar to (12) can be written for any span, the reaction at any support can be determined. When there are concentrated loads on each span, the discussion is complicated by the fact that there are two or more forms for the equation of the elastic curve for any span instead of only one, bringing into the problem two constants of integration for each concentrated load. clii Notinor the additional conditions; that the values of -^ and y^ where two equations meet at the point of application of any concentrated load are equal for the value of x at that point, it is simply a question of algebra to find the relation between the three moments. The expression as given in equation (11) is known as the Theorem of Three Moments and was published in 1857. Knowing the reactions and the loads, the shear and moment diagrams may be drawn as for simple beams. THE ELASTIC CURVE US The maximum bending and shearing stresses may be found by using the maximum values of M and V in the Mc V fundamental formulas S= — — and S=—. I A The maximum bending moments occur where the shear passes through zero; hence, by writing the expression for the vertical shear and equating it to zero, the position of the maximum moment may be found and its magnitude determined from the equation for the bending moment. Equating the expression for the bending moment to zero will give the point of inflection. The caution given before may well be repeated here : " When substituting the value of x which renders the verti- cal shear zero to determine the value of the maximum bending moment^ substitute in the particular expression for the value of M that applies to that portion of the beam^ and when equating the expression for M to zero in order to find the in- flection points^ the expression which applies to that portion of the beam must be used. ^^ As w^s the case with overhanging beams, the inflection points and the points of zero shear may be approximately located by inspection of the moment and shear diagrams. The maximum stress in a continuous beam with a num- ber of equal spans may be less than the maximum stress for simple beams for the same spans ; hence, when using continuous beams care must be taken to insure that all the supports are on the same level and practically rigid, otherwise the maximum stress may be greater than for a number of simple beams. Take the case of two equal spans with a uniform load of w lb. /ft. If the beam is truly continuous, that is the supports on the same level and practically rigid, the maximum unit stress is, — 4 wl'^. 114 MECHANICS OF MATERIALS This is numerically equal to the maximum moment in a simple beam of the same span. Suppose the middle support of the continuous beam to sink so that there is no reaction, then the maximum moment is ^ w 4:P = ^ wl?^ instead of \ wl\ or the unit stress will be four times as large as for the simple beam. As the deflection of beams as used in engineering structures is a very small quantity, the stress may easily approximate this maximum value when the supports do not remain precisely at the same level. EXAMINATION What is meant by the expression, "the elastic curve of a beam " ? Derive the differential equation for the elastic curve of any beam. Name all the conditions that must be satisfied if the use of the equation will give results that are approximately correct. What is the " deflection " of a beam ? Show how the differential equation of the elastic curve may be used to determine the maximum deflection for any beam. Prove that if the loading of any simple beam is symmet- rical with the middle of the beam, -^ = at that point. dx Why is it that when there are n concentrated loads on a beam, the expression for the bending moment may take n-\-\ different forms ? Show that the maximum deflections of simple and canti- lever beams rectangular in section, and uniformly loaded, vary as — and ^3. What is a restrained beam ? THE ELASTIC CURVE 115 Why is it that the laws of mechanical equilibrium can not be used to determine the reactions for restrained beams ? Explain how the differential equation of the elastic curve may be used as an aid in the determination of the reactions for restrained beams. Explain how to find the maximum bending moment for any restrained beam. Will the maximum bending moment and the maximum deflection always be found at the same point in the beam ? State under what conditions they will be found at the, same point. What is a continuous beam ? State the various steps in the method used in deriving the equation known as " the equation of three moments." Write "the three-moment equation" for a continuous beam with equal spans and a uniform load on each span. Show how to find the reactions for a continuous beam after the equation of three moments has been found. Show that in the case of a continuous beam uniformly loaded the maximum unit stress may accidentally be greater than the maximum unit stress for a number of simple beams, one for each span. When the size of a beam for a given span has been found on the assumption that the beam was to be fixed at both ends, it is necessary to take precautions to preserve the restrainment. Why ? Show that in any uniformly loaded beam, simple, restrained, or continuous, if Vis the vertical shear on the right of any left hand support, and iV is the bending moment at that support, the maximum bending moment in that span is J^Max. = ^ + -^ — , in which w is the uniform load per inch. 116 MECHANICS OF MATERIALS PROBLEMS ' 1. Show that if the depth of a rectangular beam of uniform strength is constant, the elastic curve is a circle. Solution. We have the relation _- = £, fro!n which r is constant E r when c is constant, but c = — ; hence, r is constant and the curve a circle. 2. Show that if a simple beam carries two equal loads at equal distances from either end of the beam, the elastic curve between the loads is a circle. 3. A simple beam 80 ft. long carries a uniform load of 160 lb. /ft. The beam is 4 in. wide and 6 in. deep. Modulus of elasticity 1,600,000. Required, the inclina- tion of the beam with the horizontal at the supports. Solution. Find the value of -H from the differential equation dx of the elastic curve ; put x = and solve. 4. Find the maximum deflection for a simple beam, length Z, modulus of elasticity E^ moment of inertia Z, loaded with (a) a uniform load of w lb. /in. (5) a single load P at the middle. Solution for (a). The expression for the bending moment at any section of the beam is M = R^x and Ri = — ; hence EI — - — — • is the differential equation of the elastic curve for dx'' 2 2 ^ this beam. Integrating, £:/ ^ = ^' - ^' + C; but^ = when x = l, therefore C = - 4'; dx 4: Q ' dx 2 24' J u i-i. *.♦ -i- 1 -r-i dy wlx^ wx^ wl^ and substituting^ its value, hi -^ = . dx 4 6 24 Integrating again, ETy='^- — -'^+ 0,; but as v = when x = 0, C. = 0, and ■^ 12 24 24 ^' ^ ' 1 ' the equation of the elastic curve becomes 24 Ely = 2 wlx^ — wx^ —wl^x. THE ELASTIC CURVE 117 Equating the expression for -^ to zero, we find that x = -, — .365 /, cix ^ and 1.365 l. The latter values have no meaning for this problem, and y is a maximum when x —-. Substituting this value for x in the expression will give the value of the maxinmm deflection. 5. Find the maximum deflection of a cantilever beam, length I inches, modulus of elasticity ^, moment of inertia Z, loaded with (a) a uniform load of w lb. /in. (6) a single load at the end. Solve by taking the origin at the wall and measuring x toward the free end, and also at the free end of the beam. (c) a single load P, at a distance M from the wall. Suggestion for (c). For any section on the left of P, M = 0; hence that part of the beam is straight. The value of -^ for x = kl (Ix gives the slope of the straight portion. When the deflection under the load is known, the deflection of the end of the beam may easily be found. Take the origin at the wall. 6. A steel cantilever beam rectangular in section is 18 in. in length, and is to carry a load of 1000 lb. at the free .end, and deflect .36 in. under that load. The unit stress may be taken as 30,000 Ib./sq. in. Find the size of the beam. 7. A beam fixed at one end and supported at the other is I in. long and carries, (^a) a uniform load of w lb. /in. (h) a single load P at the middle ; find the expressions for, (1) the deflection at any point of the beam. (2) the maximum deflection and the point where it occurs, (3) the maximum bending moment and where it occurs, (4) the reaction at the supported end of the beam, and make a sketch showing the form of the shear and moment diagrams. 118 MECHANICS OF iMATERIALS Solution for (a). Taking the origin at the supported end let R^ be the unknown reaction at the supported end. The value of M at any point of the beam is Af= R^x , as there are no con- centrated loads on the beam, and the differential equation of the elastic curve becomes Ei'^.^n..--4. (1) Integrating, ij/ 1 = ^' _ !|! + c. (2) dy The tangent at the fixed end is zero, hence -^ = when x = 1, and substituting x=^l in (2) gives C = -^ ^. Substitute this value of C in (2) and integrate, ^- R,x^ wx^ wPx Rd'^x ^ .„. This equation must be true for a; = when y = 0, therefore C^ = 0. It must also be true for y = and x = l; making y = and x = I, (3) becomes Substituting for Cj and R^ in (3), we have 48 Ely = 3 ivlx^ - 2 wx^ - wPx (3') as the equation of the elastic curve for this beam. The value of M was R,x , and as it, = , ~ 8 T"' ^ ^ also V = R^ — tvx = tvxy (5) 8 and from these two equations the shear and moment diagrams may be drawn. 3 I V = for X — — ; therefore the maximum moment occurs at | Z, 8 and this value of x substituted in (4) gives M - ^ ^^' THE ELASTIC CURVE 119 I'here will be a negative moment at the wall, and its value can be obtained by making x = I in (4), or *-' 8 Making M = and solving (4) for x gives ,^ 3 irlx wx^ r. 3 ? 31 = = 0, or X = — , 8 2' 4 as the inflection point. Substituting for C and R^ in equation (2) gives ■pT^IjL — ^ it'/a:^ ^ijx^ tvl^ 3 wl^ This expression is the differential coefficient of (3) ; therefore, equating the right hand member to zero and solving for x gives the value of X for which (3) is a maximum. This results in a cubic equation, 8 x^ — 9 Ix^ -{- P = 0. The roots of this equation are x = I, .42 Z, and — .298/. The latter value has no meaning for this beam, and X = I is a minimum; therefore the maximum deflection occurs at X = .42 l. Substituting x, .42 I in (3') gives 2/max. = as the maximum deflection. Suggestion for (&). There are two differential equations for the elastic curve and the two curves have a common tangent and common deflection for a value of a; = -. 2 8. A beam I in. long is fixed at both ends and carries, (a) a uniform load of w lb. /in., (6) a single load P at the middle ; find (1) the expression for the deflection at any point of the beam, (2) the expression for the maximum bending moment and its position, (3) the expression for the maximum deflection and where it occurs. (4) Make a sketcli showing the form of the shear and moment diagrams. 120 MECHANICS OF MATERIALS Suggestion for (b). Since the loads are symmetrical with the middle of the beam M^ = M^ and i2j = i?2 = "o' ^"^ TT^ ~ ^> when X = I, X = —, and x = 0. 2 9. A rectangular beam, 20 ft. long fixed at both ends carries a uniform load of 100 lb. /ft. If the modulus of elasticity is 1,500,000, the safe unit stress is 500 Ib./sq. in. and d = 4:b, find the size of the beam. 10. If the beam in 9 carried a load of 2000 lb. at the middle, and the other data the same, find the size of the beam. 11. Find the maximum deflections for the beams in problems 9 and 10. 12. Draw the shear and moment diagrams for the beams as given in problems 9 and 10. 13. If the beams in problems 9 and 10 were to be beams of uniform strength w4th a constant depth, sketch the plan and elevation for each beam. 14. Sketch the shear and moment diagrams for the beams in problem 13, using as ordinates the unit shearing stresses instead of the vertical shears, and the unit bending stress instead of the bending moments. 15. Select a standard steel I beam to be used as a con- tinuous girder for four equal spans of 8 ft. each. The ends are simply supported and the beam is to carry a uni- form load of 7000 lb. / ft. Unit stress not greater than 16,000 Ib./sq. in. Solution. As the ends are supported, N^ = N^ = and from the symmetry of the spans and loads, iVg = N^ and R^ = R^, R^ = R^. The three-moment equation for equal loads and spans reduces to Beginning with the second span, N,+ ^N, + N,= -'f, (2) THE ELASTIC CURVE 121 and with the third span, Wi 72 N,^4.N,-\-N, = --^. (3) Since N2 = N^, (2) becomes 2iY, + ^N, = -~, (4) AT - _f^ and as N^ = 0, eliminating N^ from (3) and (4), we find that IV ['■^ 14 Making iV^ = and substituting for iVg in (1), ^ 28 V For the first span, F = F, — ivx or V = when x = — 1. w The bending moment for any section of the first span [see equa- tion (1), art. 67] is M = N,-\- V,x - ^' and !/„,,,. = ^1 + |^. (5) Y Similarly, for the second span V = V^ — wx, or F= when x =—^, and as the value of the bending moment for any section in the second span is M=N,-\- V2X - ^, the iH,,ax. =N2 + ^. (6) 2 2 w Equations (9) and (10) of Art. 67 give value of Vi and V^- Sub- stituting these values in (5) and (6), we have •^^max in the first span = -^^ tvP = approx. ^^^ ivP, ^ 1568 ^^ 28 -^^max in the second span = ivP = approx. — wl'^, and as the moments in the third and fourth spans are the same as in the second and first, the greatest moment occurs at the second sup- port and is —^ , which may be written ^ , where TF is the total 28 28 load on one span. Substituting this value of i^f in — = - , we find the value of - as S c c 36 in.3. From the tables we find that a 12-in. I beam weighing 31.5 lb. /ft. has a value of - = 36, and that beam will be chosen. c 122 MECHANICS OF MATERIALS 16. I beams are to be chosen to cover two equal spans of 10 ft. each. The uniform load for each span is 5000 lb. /ft. and the maximum allowable unit stress is 15,000 lb. /sq. in. Choose a standard I beam on the assumption that the beam is continuous. What beam must be used if two simple beams are used instead ? 17. Select a standard steel I beam to cover three spans of 8 ft. each and carry a uniform load of 7000 lb. /ft. The unit stress is not to exceed 16,000 Ib./sq. in. What size beam would it be necessary to use if each span was covered by a simple beam ? How much weight of steel will the use of continuous beams save ? 18. In problem 17 find the reactions at the supports for each kind of a beam. 19. Select a continuous steel I beam to cover two equal spans of 12 ft. each, and carry a load of 36,000 lb. at the middle of each span. Unit stress 12,000 Ib./sq. in. Suggestion. There are four forms of the differential equation of the elastic curve, each two having a common tangent and deflection at the load or reaction where they meet. The three reactions are all unknown but R^ = R^, and there is an unknown bending moment at R^., the middle reaction. CHAPTER VI LONG COLUMNS Art. 68. Stresses in Long Columns. A theoretical bar under the action of axial compression should always yield by crushing, since the resultant of the stresses and the axial loads are in the same line. In dealing with the actual materials and loads, the force due to the load is not always axial, no matter how much care is taken to make it so, and no material in common use is absolutely uniform. In nearly every case of bars under axial compression, there is more or less of a couple caused by the bar not being straight and uniform in struc- ture, or the loads not axially applied. The effect of this couple is to produce bending in the bar. As the amount of bending or deflection of a beam under the action of bending forces varies directly with the cube of th^ length, it is easy to see that the danger of bending a bar under the action of compressive forces which are not exactly axial, increases very rapidly with the length, and also that any bending increases the moment of the axial forces. Experiment has proven that when the length of the bar does not exceed ten times its least dimension, it is more liable to fail by crushing, than by bending and crushing combined. 128 124 MECHANICS OF MATERIALS Therefore, when the length of a bar under axial com- pression does not exceed ten times the least dimension, the formula P = AS may be used to determine the rela- tions between the load and unit stresses. Such a bar is called a short column or strut. When the length of the bar exceeds this limit, it is called a long column; and as the stress which is a combination of bend- ing and compressive stresses is not uniformly distributed over the area of the cross section, the formula P =AS no longer applies. The formulas expressing the relations between the loads and the dimensions of a long column that are in more or less general use are: Euler's Formula. This formula is derived from theoret- ical assumptions. The magnitude of the load given by its use depends on the elastic properties of the material and not on the unit stress induced. Rankine's Formula. In the derivation of this formula certain assumptions more or less theoretically true are made. The loads obtained by its use are made to agree with the results of experiment by the introduction of an empirically determined constant. The Parabolic Straight-line Formulas and their modi- fications. These formulas are entirely empirical. They are simply equations for curves whose coordinates satisfy experimental data. Art. 69. Euler's Formula. The derivation of this formula assumes : that the column is perfectly straight, the resistance of the material in any cross section uniform, and that the load acts along the gravity axis of the column. LONG COLUMNS 125 Such a column would never bend under any load, but would fail by crushing. To produce incipient bending he assumed a slight lateral force to be exerted against the column while it was under the axial load, and determined the value of the axial load P that would keep the column bent after the lateral force was removed. The elastic curve of a bent column may take any one of several forms, depending on the condition of the ends. Columns with '' Round " or " Pin " Ends. If the ends of the column are de- signed so that there will be no restraint to the tendency to rotation about a point on the end, the column is said to have '-'- Round " or "Pm " ends. If the ends are rounded as in Fig. 69, it will bend in a single curve, and the elastic curve may be represented as in the figure. Taking the origin at (9, the moment of P about a point in any section dis- tant X from i^ M = Pa — Py, and the equation of the elastic curve for this case is EI^=Pa dx^ Py. (1) Multiplying by 2 -^ and integrating with respect to 2/, El{^^ = 2 Pay - Pf + C^. (2) Evidently -^ is when y = 0, hence C. = 0. dx ^ 126 MECHAiNICS OF MATERIALS Extracting the square root and transposing, dx =Vf dy -P V2 ay — 2/2 integrating, ay-y^ X = V"^ vers"! ^ + C'g. When a^ = 0, then y =^ a and vers"! — = a (3) (4) 2' hence 01, = \/ — • 2 2 ^ P Also when a: = - , y = 0, and vers"! ^ = 0, hence L= -TlJ^^ or P = 2 ^ P which gives the value of the load P that will keep a column with round ends bent, after the lateral force has been removed. As the deflection y has disappeared from the expression for P, its value is independent of the amount of bending. Colixmns with Square, Flat, or Fixed Ends. When the column is so designed that the tangent to the elastic curve at each end will be parallel to the length of the column, the column is said to have square, flat, or fixed ends. A column of this type may have a cap, which on account of the large surface it presents, tends to prevent free bend- FiG. 69a. ing about a point in that end, or it may LONG COLUMNS 127 have the ends firmly imbedded in masonry. The form that the elastic curve takes in this case is similar to that of a beam fixed at both ends when there is a single load at the middle of the span. Let Fig. 69a represent a column with fixed ends as bent by the load P. The moments that keep the ends of the column vertical being denoted by M^ and M^^ the moment due to M^ and the load P, at any section of the column distant x from O, is M= M^+Pa-Py, and the equation of the elastic curve is El'^=M^-\-Pa-Py, (5) Integrating and determining the values of Mi and the constant of integration from the conditions that dy/dx = when y = or a, X = when y = a and x = 1/2 when y = 0, we have : or P = P which is Euler's formula for a long column with both ends fixed. Columns with Round and Square Ends. When one end of a column is restrained so that the tangent to the elastic curve at that end is always vertical, and the other end is left free to rotate about a point in that end, the column is said to be one with round and square ends. The elastic curve will take some such form as shown in Fig. 696. 128 MECHANICS OF MATERIALS Taking the origin at Oi, we have EI^^ = Pa- Py + Hx, which reduces to P = 2EI IT P (nearly) . If the effect of the couple H-H is neglected, the value of P is given by ^ 4: P ' which is the value usually given for Pwhen the column has one end round and the other square. In the solution of problems by the use of these formulas, I is always the moment of inertia about a gravity axis perpendicular to the direction of bending. If the column is free to bend in any direction, the least value of I for that section must be used. E is the modulus of elasticity, and as the tabulated values are in inch pounds, P must be expressed in pounds and I and I in inches. It will be noticed that the formulas for the three types of long columns differ only by a constant. If the strength of a column is defined as the load it will carry, and the strength of a column with square ends is taken as unity, it is easily shown that the strengths are as: 1 for a column with square ends, ^ for a column with round ends, -^Q for a column with round and square ends. LONG COLUMNS 129 Ar^ may be substituted for Tin the general formula for long columns and the formula written A Z2 ' where m has values of 1, 4, |^, according to the condition of the ends of the columns, and r is the least radius of gyration. * If the stress was uniformly distributed over the section p of the column, the value of — would be the unit stress. p A Taking -— as the unit stress, m = l, tt^ = 10, and the , I ratio of - = 100, the value of jS for a wrought iron column is 25,000 lb. /sq. in., which is practically the elastic limit of the material. If - is taken as 50, the value of S becomes greater than r the ultimate strength of the material. As the majority of columns in engineering structures have the ratios of - from 50 to 150, it is plain that Euler's formula will not always give satisfactory results. The results of experimental work on long columns show that the formula can not be depended on unless the ratio of - is nearly 200. The ratio of the length to the least value of the radius of gyration is called the " slenderness ratio " of a columi and when this ratio is greater than 35, the column may be considered as a long column. * P/A is often referred to as the unit load on the column. 130 MECHANICS OF MATERIALS Euler's formula not being satisfactory for the usual range of the values of - has led to the adoption of a r formula, first proposed by Gordan and later modified by Rankine. Rankine's modification of Gordan's formula is in com- mon use among American engineers, while the European engineers prefer Euler's formula or some modification of it. If Euler's formula is used for the design of a long col- umn, the resulting dimensions should be checked by the formula for axial compression to determine the value of the unit stress. Art. 70. Rankine's Formula. This formula is based on tlie assumption that the maxi- mum unit stress in any section of a long column under axial compression is a combination of the compressive unit stress due to the axial compressive force and the maximum unit bending stress due to the probable moment of the saine force. Let Sc be the maximum unit compressive stress in any p section of a long column, aS = — be the unit compressive Mc stress due to the load P, Sd = — — the maximum unit com- pressive stress due to the probable moment of the axial force, and A the area of the section of the column; then P Mc Sc = S + S,ovSc = ^-\-^^, (1) Ji. 1 from which the maximum unit stress Ss may be found P Mc P when — and — r- are known. The value of — may easily A I A LONG COLUMNS 131 Mc be found, but the value of — - is indeterminate, owing to the lack of knowledge of the probable eccentricity of the force P. If we assume that A is the maximum deflection due to the unknown moment M, the value of that moment may be expressed as PA. Writing Ar^ for / and making M = PA, equation (1) reduces to By analogy with beams where the maximum deflection varies as — , A may be taken as proportional to — . If c c is some number depending on the material of the column for its value, and n is a number whose value depends on the conditions at the ends of the column, then A = n4> -. c Substituting this value of A in equation (2) , we have ■& = f(l + «0 5),OrP = ^i^, (3) 1 + n0- which is Rankine's formula for the solution of problems relating to long columns. Since S^ is usually given in pounds per square inch, P must be expressed in pounds, and I and r in inches. The value of r to be used will depend on the direction in which bending takes place. If there is no external restraint to bending, the least value of r for the section considered must be used. The value of (f) has to be determined by experimental work on long columns. The method generally taken to deter- mine the value of (j) is to find by experiment the load 132 MECHANICS OF MATERIALS that will cause the column to fail, and substitute that value for P in Equation (3). S^ is taken as the ultimate compressive strength of the material and, as n^'l^ A, and r are all known, the value of ^ can be found. The relia- bility of the formula depends on the accuracy with which the value of (/> is determined. The following table gives the usual value of (j> as found for columns where the ratio of - varied from 20 to 200, and therefore can be applied ^ . Z . . . to problems where the ratio of - is within these limits. r For hard steel 4> = 1 20000 > for mild steel <^ = 1 . 30000 ' for wrought iron <#> = 1 36000 ' for cast iron 4> = 1 . 6400 ' for timber ^ = 1 3000* From Euler's formulas for columns with ends square, round, round and square and equal moments of inertia it i-s easy to see that the strength of a column with round ends is J that of the one with square ends, while the one with the ends round and square has but ye ^^e strength of the one with square ends. ' If P, h and h are the lengths of three columns with equal moments of inertia, equal strengths, same materials and ends square, round, and round and square respectively, smce F - ^2 - ;,2 - 4 1^2 > it is evident that I = 2li = ^h. AS Substituting these values for lmP = — successively, l-\-n4>~ LONG COLUMNS 133 there results a numerical coefficient for the last term in the denominator of 1, 4, and y-. In Equation (3) n was a number depending for its value on the conditions at the ends of the columns, hence n must equal 1 for columns with square ends, 4 for columns with round ends, and ^^- for columns with round and square ends. In the use of Rankine's formula the value of So may be taken as the safe working unit stress, and then P will be P the load that can be carried with safety. — is always less 12 than aS'c, as 1+^0 — is always greater than unity; hence, no matter how short a column may be, Rankine's formula will give a safe value for P when the safe value of S\ is used. Since nxj) is a small quantity, if the ratio of - is small, l-\-n(f)-- may be practically unity and the value of P very nearly equal to AS^. The value of (/> being derived from experiment, the formula is limited in its use to the range in the values of - covered by the exper- r imental work in its determination. The values of , as given, apply very well to all columns where the ratio of - lies between 20 and 200. Above tliis ratio the load r given by Rankine's formula will still be safe, as the formula gives the value of P too small. Art. 71. RittCi's Formula. The following formula for the value of the unit load on a long column was proposed by Ritter in 1873. P ^ S 'A 1 , j^i!' 134 MECHANICS OF MATERIALS S, the maximum unit stress, is assumed to be equal to or less than the elastic limit of the material, *Se the normal elastic limit, E the modulus of elasticity, and ix is 47r^, fTr^, and TT^ for columns with square, round, and round and square ends respectively. This formula has the same general form as Rankine's, Se the empirical constant being replaced by fxE^' S The values of — ^ for materials having a well-defined }jlE elastic limit, which is approximately one-half the ultimate strength, are practically the same as the experimentally determined values for for the same materials. For materials where the ratio of the ultimate strength to the elastic limit is not well defined and greater than 2, such as S cast iron and timber the values of — ^ differ so widely from jjiE the empirical values for = 4^7r2 may justify the use of the latter expression for when the value of that constant has not been experi- mentally determined for the conditions obtaining in any given problem. The following table gives the comparative values of Seiq - 1) and ^TT^E with the data used in calculating its value. Material Elastic Limit Lbs./Sq. In. Ultimate Strength Lbs./Sq. in. Modulus of Elasticity- million Lb./Sq. In. Se(q-l) Cast Iron Timber 4,000 3,300 25,000 30,000 80,000 10,000 55,000 60,000 12 1.5 28 30 1 6,300 1 8,950 1 1 6,400 1 Wrought Iron. . . Structural steel. . 3,000 1 37,000 1 36,000 1 40,000 30,000 136 MECHANICS OF MATERIALS Art. 72. The Parabolic Formula. I P Writing x for — and y for — , Euler's may be expressed as 2/ = — 7r~ ^^^ the curve representing the equation plotted using the values of x and y as coordinates. P If on the same diagram the values of — for varying values of -, as determined by experiments on long columns, r are plotted it would be noticed that for values of - some- r what greater than 150, Euler's formula satisfies the exper- mental data very closely. Using the same notation, Rankine's formula becomes Sc y = 1 + (^^2 which as the value of (f)X^ is generally less than unity is approximately equal to y = Sc — Sc^x'^' The last equation may be written P P y = a — hx^ or — = a — h —, A r^ which is Professor J. B. Johnson's possible formula for the strength of long columns. Professor Johnson assumed that for very short columns I P where th6 ratio of - approached zero, — should be equal to r A the elastic limit of the material and that the parabola representing his formula should be tangent to the curve LONG COLUMNS 137 given by Euler's formula, and determined value for a and b to satisfy these conditions. These conditions give a equal to the elastic limit of the material and b = - — ^— -. In the following table, taken from Johnson's Modern Framed Structures, a is the observed elastic limit and the value of b is calculated on the assumption that m-K^ =16 and 25 for hinged and square ends respectively. Material Lb./Sq. In. 6 I = 7< Wrought Iron: Hinged ends Square ends • Mild Steel: Hinged ends 34,000 34,000 42,000 42,000 60,000 ' 60,000 a 25,000 3,300 4,000 3,500 0.67 0.43 0.97 0.62 6.25 2.25 h 0.6 0.7 0.8 0.8 170 210 150 Square ends Cast Iron: Round ends 190 70 Square ends Timber (square ends only) : White pine • •• • Yellow pine, short leaf Yellow pine, long leaf White* oak 120 I d 60 60 60 60 p p The parabolic formula for timber is — = a — b — in A d? which d is the least dimension of the column. When the P values of a and b given in the table are used, — is the value of the unit load. To determine the safe unit load, the values of a and b are to be divided by a factor of safety. 138 MECHANICS OF MATERIALS Art. 73. Straight Line Formulas. Mr. T. H. Johnson after a study of the experimental data from tests on long columns found that within the usual limits of - the equation r y = c — ax or — = c — d- A r could be made to satisfy the experimental data very closely. The straight line was assumed to be tangent to the curve representing Euler's formula and the values of c and d determined so as to satisfy this condition and make the p values of — agree with the experiments. The following table taken from Merriman's Mechanics 3f Materials gives the values of c and d for different Material Lb./Sq. In. d I = 7< Wrought Iron: Square ends 42,000 42,000 42,000 52,500 52,500 52,500 80,000 80,000 80,000 5,400 128 157 203 179 220 284 438 537 693 28 218 Hinged ends 178 Round ends 138 Structural Steel Square ends 195 Hinged ends 159 Round ends 123 Cast Iron: Square ends 122 Hinged ends 99 Round ends 77 Oak square ends 128 LONG COLUMNS 139 materials and the limiting value of — for each case. For r value of - greater than the limiting value, Euler's formula may be used to determine the unit load. On account of the simphcity of this formula it has been used to a considerable extent in certain kinds of work. The formula gives the value of the Hmiting unit load for the column. To determine the safe unit load the values for c and d as given in the table should be divided by the factor of safety. Many formulas of the same form are made parts of specifications and building laws. In the majority of such cases the values assigned to c to d are such as will determine the safe allowable unit load. Art. 74. Comparison of Formula. The purely theoretical formula of Euler does not give limiting loads that agree well with practice for columns of ordinary lengths. When the length of the column is such that the strength depends more on the stresses due to bending forces than those due to the axial compressive forces, the conditions approximate the assumptions made by Euler, and his formula will give reliable results. Although Rankine's formula is derived from assumptions that obtain in every long column, the formula must be con- sidered empirical since it contains the experimentally deter- mined constant . When the experimental work from which also depends on the ratio of the ultimate strength to the elastic limit of the material, should give a wider range to the use of Rankine's formula. The parabolic and straight line formulas for the limiting loads for long columns are entirely empirical and the value of the results obtained by their use depends entirely on the accuracy with which the straight line or parabola represent- ing these formulas satisfies the experimental data. In each case the value of the slenderness ratio for which the straight line or parabola becomes tangent to Euler's curve is the limiting value of the ratio for their use. For greater ratios Euler's formula should give reliable results. The formulas for the strength of beams and bars are derived on assumptions that can be closely approximated in practice, but the formulas proposed for the unit load for a long column give the limiting value of that load under conditions as nearly ideal as it is possible to obtain experi- mentally. As it is impossible to make any assumption as to the variation of the actual conditions from the ideal, safety requires the use of a larger factor of safety with column formulas than with the formulas for bars and beams. LONG COLUMNS 141 In the choice of a factor of safety it must be remembered that while with Rankine's formula the factor of safety is applied to the ultimate strength of the material, the values of the constants in the straight line and parabolic formulas are based on a unit stress only slightly larger than the elastic limit, and therefore a factor of three for the two latter formulas is the equivalent of one of five or six applied to Rankine's. Cast iron columns are common in engineering work on account of the large compressive strength of the material. As long as the unit stress due to the bend- ing does not equal the unit compressive stress due to the load there is no tensile stress in the column. The columns are generally made hollow and round in section. Wrought iron and steel pipes are also often used for columns. The values of / or r may be found as soon as the internal and external diameters are known. Rolled steel shapes, in channel and I beams, are joined together by plates, which are riveted to them, and used as columns. Timber is used in the solid section and in the hollow box section. In the case of circular sections the value of / is the same for bending in all directions. When the column is " built " up, as is the case when the rolled steel shapes are used with the joining plates, the spacing should be such that the values of I for the " built " section will be equal about the two principal axes. The same is true of the wooden box sections. For the rolled steel shapes the value of / for the gravity axes of the rolled section, parallel and perpendicular to the web, are given in the tables, and must be transferred to parallel axes passing through the center of gravity of the built section. For the sections 142 MECHANICS OF MATERIALS where the elements are rectangular the value of the o gravity moment of inertia is — — , where d is the dimension perpendicular to the axis. EXAMINATION Explain why the formula P = AS does not give results that agree with experiment when the bar under axial com- pression is very long. Define a Long Column. What assumptions were made when the formula P = m — -— was derived ? What is meant by the expression " a column with round ends"? "a column with square ends" ? "a column with found and square ends " ? Derive Euler's formula for long columns with square ends ; with round ends. Explain why the formula as derived for columns Avith round and square ends is approximate. Explain why the use of Euler's formula for long columns does not always give satisfactory results. Give the conditions under which the use of Euler's formula for long columns will give satisfactory results. State the assumptions that were made for the deriva- tion of Rankine's formula for long columns. Derive Rankine's formula for the strength of long columns. State the meaning of each symbol and the units to be used in making substitutions in the following formulas : P JSr^TT^ , P S^ and —r = A l^ A ^ l^ 1 + ^*^ LONG COLUMNS 143 Does Rankine's formula for long columns give results that are more reliable than Euler's ? Why ? On what does the reliability of Rankine's formula depend ? Mc The formula aS^ = —~ was used in the derivation of Rankine's formula for long columns. Does this fact limit the use of the formula to materials which satisfy the con- Mc ditions for the use of the formula aS' = -— ? Show by the use of Euler's formulas that the strength of a column w^itli square ends being taken as unity, the strength of a column of the same size with round ends is ■J^ and that of a column with round and square ends is -^. Explain why the value of n in Rankine's formula is 1 for a column with square ends 4 for a column with rounds ends, and ^ for a column with round and square ends. Show that Rankine's formula applied to the solution for any column, no matter how short, will always give a safe load for that column. Is the same true for Euler's formula ? Is there any limit to the length of a column to which Rankine's formula will apply? Why is the ratio of - used as limiting the use of either formula ? If the section of a column is a rectangle having one side 4 in. and the other 6 in., find the values of I and r to be used in the formulas for the strength of long columns. Why will a hollow cylindrical form make a stronger column than a solid cylinder of the same section area ? Given the moment of inertia about an axis through the center of gravity, how can you find the moment of inertia of the section about an axis parallel to the gravity axis ? 144 MECHANICS OF MATERIALS PROBLEMS 1. A wooden column 10 ft. long is rectangular in sec- tion and has round ends. The sides of the rectangle are 6 and 8 in. respectively. What is the maximum load the column will carry? Find the safe load. (Use Euler's formula.) Solution. The formula is P = -^^, /= M! =1^1'^ M4, Z^ 12 12 ' since the least moment of inertia must be used. E may be taken as 1,500,000 and tt^ used as 10. I = 120 in. Substitutina:, P = 1.500,000 X 144 X 10 ^ ^ ^^^ 14,400 This is the greatest load that can be carried without failure by bending. The corresponding unit stress due to the axial force P id — = ^ = 3130 Ib./sq. in., or a stress about one half the ultimate ^48 / H » strength of the material. In order to carry this load the column must be straight and the load axial. As these conditions are rarely ever satisfied, a factor of safety of at least 5 should be applied to the result. Using this factor, the safe load is 30,000 lb. and produces an axial unit stress of approximately 600 Ib./sq. in., which has a fair margin for safety. 2. If the column in problem l was cast iron and had square ends, determine the maximum load it will carry and the unit stress induced by that load. Is it possible for the column to carry the load ? 3. Show that a cast iron column with square ends must have the ratio of - approximately 78 in order that the r axial unit stress corresponding to the load P in Euler's formula shall be less than the ultimate compressive strength of the material. 4. If the axial unit stress produced by a force equal to the value of P as derived by Euler's formula is equal to the elastic limit of mild steel, find the ratio of L for the ' LONG COLUMNS 145 column. Consider the ends round ; square ; and round and square. 5. A standard 12 in. I beam weighing 35 lb. /ft. is to be used as a column with round ends. The length is 10 ft. What load may be carried with a factor of safety of 5? If there should be a factor of 5 used with the formula for axial compression, is the load given by Euler's formula safe ? 6. A cast iron column 20 ft. long has a hollow cir- cular section. The external diameter is 10 in. and the internal diameter is 8 in. Determine the value of the maximum load that can be carried, if the allowable unit stress for axial compression is 4000 Ib./sq. in. What will be the factor of safety against failure by bend- ing? (Column has square ends.) 7. Solve problem 1 by the use of Rankine's formula. AS 1 Solution. The formula isP = ^-r^> ^ = 48 sq. in., S = — -. In such cases as approximate Jx solution of the problem of determining the maximum unit stress parallel to the axis is obtained by adding the unit stress due to the axial forces and the maximum similar unit stress due to the flexural forces. If Sm is the maximum unit stress parallel to the axis then Sm = S -\- Sb P Mc = — - H — , in which *S, Sb and Sm are the same kind of stress, is the approximate value of the combined unit stress. When the axial force is a tensile force, the maximum stress found in this way is too large, as the axial force on the bent beam produces a moment PA that tends to reduce the bending moment of the flexural forces. See Fig. 76. When the axial force produces compression, the moment PA of the axial force tends to increase the bending of the bar, and the approximate solution gives the resultant unit stress too small. As the error is due to the moment PA, when the deflection is small the error is also small. While the engineer as a rule desires to be as nearly accurate in his calculations as possible, when the approximate formula is simple and errs on the side of safety, it is often used in preference to the more exact and complicated formula. A bar is often designed to resist a combination of compression and bending stress by the use of the approximate formula, but when this 150 MECHANICS CF MATERIALS is done the resulting dimensions should be used in a more exact expression and the actual unit stress determined. Let Fig. 76 represent a beam under the action of flexural and axial forces. Let M be the maximum mo- ment of the flexural i ^ — — ^'A ^ forces, P an axial force which may be either tensile or compressive, and A the maximum de- flection of the beam. If Ml and *Si are the maximum bending moment and unit stress due to the sum of the moments M and PA, then Mic (M±PA)c Fig. 76. Si = (1) By analogy with beams under the action of flexural forces a SP only, the value of A may be assumed as A = — — - without *^' ^ Ec serious error. This value of A substituted in equation (1) gives the unit stress due to bending. Mc Si = /3 E (2) in which the positive sign is to be used when P produces a tensile stress and the negative sign when P is a compres- sive force. The maximum unit stress then becomes, Sm — ~T~i~ '^l A (3) COMBINED STRESSES 151 where Si has the value given in equation (2), Sm tension or compression depending on the kind of stress represented by>Si. The values of a and /3 for various kinds of beams and loadings are given in the appendix. There are no values that apply strictly to this case as the bending moment is increased by the value of PA. Since A is generally a small quantity the error made in assuming that the values of a and ^ are those found for beams under flexural forces only is very small, hence a and (3 will be determined by the kind of a beam and the nature of the loading. The percentage of the error due to the use of Sj, instead PPa of ^1, is rb— T- 100. As the amount of the error depends directly on the value of P, and inversely on ^, when P is small, and JS large, the error will be so small that it may be neglected. The error also varies directly as P; therefore it is more liable to be serious when I is large. For a timber beam 20 ft. long, 6 in. wide, and 12 in. deep, carrying a load of 8000 lb. at the middle and at the same time a compressive load of 45,000 lb., the use of the approximate formula will result in an error of approximately 8 per cent. For a steel I beam, 10 ft. long, having a moment of inertia of 122, a concentrated load of 6000 lb. at the middle and an axial compressive force of 20,000 lb., the values of S-^ and S^^ differ by less than one per cent, an error too small to be taken into account. Art. 77. Roof Rafters. The common roof rafter is an example of a beam under axial and bending forces. As a part of a truss there is some 152 MECHANICS OF MATERIALS compression in the rafter, and the weight of the roof and the probable snow load produce both bending and compression. Let Fig. 77 represent a roof rafter, length I in., carry, ing a uniform load of w lb. in. The rafter is in equi- librium under the horizontal forces H and H, V acting Fig. 77. vertically at the wall, and the uniform load. Taking mo- ments about a point at the foot of the rafter, from which HI sin <^ = — cos ^, H = -- cot 4>. The bending moment of the force H about a point in a section distant x from the upper end of the rafter is Hx sin (]), while that for the uniform loads on the left of the wx same section is — - cos (j). The unit stress due to the bending forces is a _ Mc _ (^Hx sin — ^ wx^ cos <^) g The compressive force on the same section is H cos due to the force H and wx sin , due to the uniform loads on the section. Hence the total compressive stress in the section is JP__ (.ZTcos (}) + wx sin c^) A" A COMBINED STRESSES 153 Using the approximate formula, the greatest combined unit stress parallel to the axis in any section distant x from the top of the rafter is : P ^ {H cos (f)-\-wx sm (f)) (Hx sin cl) — ^wx^ cos ^)c wl Substituting — cot for H and proceeding in the usual way the maximum value of S is found to be, _ SwP cos (f) wl cosec ^ ,w sin (f> tan (^ "*" Ihd? 2^d Wb ' In any beam carrying vertical loads and supported at points not on the same level, the maximum unit stress parallel to the axis is a combination of axial and flexural stresses. The forces supporting the beam taken with the loads form a system of forces in equilibrium. These forces may be resolved into components parallel and perpendicular to the axis of the beam. The components parallel to the axis will produce tension or compression while the per- pendicular components will induce flexural stresses, and the value of these stresses may be found when the forces acting are all known. Art. 78. Eccentric Axial Loads. When a beam is to resist axial as well as flexural forces, it is nearly always possible to make the point of applica- tion of the axial force so that its moment about a point in the neutral axis of the mid-section of the beam will be equal and opposite to the moment of the flexural forces. If P (Fig. 78) is an eccentric axial force, and ?/ is the dis- tance of its point of ■^--> application above or — Z ~ \^ P below the neutral plane, then before any bending takes <£- T T I Fig. 78. | 154 MECHANICS OF MATERIALS place Py is the moment of the axial force about a point in the center of the beam. Let M be the maximum moQient of the flexural forces ; then if the axial force is compression and the distance y measured below the neutral plane has such a value that M= Py^ the resultant bending moment will be zero. Similarly, when the axial force is tension, if 2/ is measured above the neutral surface and its value taken so that Py = M^ the resultant bending moment will also be zero. When the axial force is compressive and the maximum deflection due to the flexural forces is greater than the value of y determined as above, the value of y should be made equal to that maximum deflection. Art. 79. Shear and Axial Stress. When a bar which is subjected to an axial stress is acted on by forces at right angles with the axis, there are tensile or compressive and shearing stresses at every point in that bar. Let Fig. 79 represent an elementary cube cut from any portion of the A^V bar at which there are tensile or compressive stresses par- allel, and shear- ino^ stresses ^ _„ o Fig. 79. perpendicular to the axis. The equal tensile or compressive forces Ti, T2 act per- pendicular to opposite faces of the cube, while the equal shearing forces Vi, V2 act in the same opposite faces and are perpendicular to Ti, 7^2. H r«- r COMBINED STRESSES 155 The cube is not in equilibrium unless a pair of equal shearing forces Hi, H2 are introduced. As the cube is in equiUbrium and the arms of the couples are equal, it follows that V = H. Since the elementary block was a cube, the unit stresses due to the equal forces H and V must be equal. Hence at every point of the bar there exists a pair of equal unit shearing stresses at right angles to each other, in addition to the unit tensile or compressive stresses. The tensile or compressive and shearing unit stresses that exist at every point in the bar combine and create shearing and tensile or compressive unit stresses that are greater than the original unit stresses. Art. 80. Maximum Stresses. To determine these maximum stresses let Fig. 80 repre- sent an elementary parallelopiped cut from any portion of the bar. Let its length be dx^ height c??/, width unity, and the faces be parallel and perpendicular to the axis of the bar. The area on which the tensile or compressive forces act is c?z/ times unity, while that on which the shearing forces act, is either di/ or dx times unity. The forces that act on opposite sides may be considered to be equal since they differ by an infinitesimal quantity. Let Sg be the unit shearing stress and S the unit tensile or compressive stress. The force that acts perpendicular to the dy face is Sd^ and the shearing force in the same Fig. 80. 156 MECHANICS OF MATERIALS plane is S^dy. The shearing force in the dx face is S^dx. Let dz be the diagonal, <^ the angle between dx and dz^ S„ the unit stress perpendicular to dz, and jS^ the unit shear- ing stress along dz. Resolving the forces that act on either side of dz into components parallel and perpendicu- lar to dz we have zSidz^ S dy cos (f) -\- S^ dx cos (f) — iS^ dy sin 0, (1) S2 dz = Sg dx sin -^ S dy sin + ^S^^^ dy cos ^. (2) Divide each of these equations by dz^ make -^ = sin <^ and — = COS (f>, and equations (1) and (2) reduce to dz S\ = S sin (f> cos 4*-{- Sg (cos^ ^ — sin^ <^), (3) ^2= S sin^ ^ + 2 aS'^ sin cos . (4) Writing the equivalents of sin ^ and cos (j) in terms of 2 ^, we have q Si=^ sin 2 (/)+ /S', cos 2 c^, (5) AS'2 = f (1-cos 2 ^)+AS'sin2<^. (6) By the usual process the value of that makes *Si* a maximum and equal to >Sp is given by tan 2 = -^, while 2 Os the value of ^ that makes *S2 * a maximum and equal to &n 2 *S is given by tan 2 0= — ^. Substituting these values of 4> in equations (5) and (6), we find that the maximum values of >S2 and &\ are * The planes in which the maximum stresses Sn and *Sp are induced make angles with the axis of the bar that differ by 45 degrees, since tan 2 = oo or = 45°. (/> is the angle between the directions of the tensile or compressive unit stresses and the plane on which the maximum stress acts, hence the maximum S^ makes an angle of 45° or 135° with the direc- tion of jS. The angle that jSn makes with S under the same conditions is the tan"^ (^ = 0. This latter expression P shows that S — — will g'ive the maximum unit tensile or A compressive stress in a bar on which there are axial forces only. These formulas are general and apply to all combina- tions of . tensile and compressive with shearing stresses without regard to the nature of the force producing the stresses. When the unit stress S is induced by axial forces, the p value of aS' = — -. The stress S may also be due to a bend- ing moment, and in that case the value of S is taken from S = — -. Sg may be due to a simple shearing force or the result of a twisting moment. In the former case the value of tS^ to be used is derived from Sg— —, and in the latter case jS^ = — ^. When 158 MECHANICS OF MATERIALS the bar under axial forces is a long colamn, Rankine\s formula for long columns must be used to find the value of aS'. This formula may be written aS^=— (1 + n(j)-—L from which the value of S may be easily found. Art. 81. Horizontal Shear in Beams. There is a tensile or compressive unit stress in every tiber of a beam that is equal to aS' = -zt and at the same time a unit shear- ing stress resulting from the vertical shear. When the formula V = ASg was derived, it was assumed that the shearing stresses were uniformly distributed over the section of the beam. In the pre- vious article it was Fig. 81. shown that there was a pair of shearing stresses at right angles to each other. Therefore in any beam there is also a liorizontal unit shearing that is equal to the vertical unit shearing stress at all points of the beam. To deduce an expression for the horizontal unit shearing stress at any point of a beam imagine a parallelopipedon cut from the top half of any beam (Fig. 81). Let the length be dx^ the width 5, the distance of the lower side from the neutral axis ^/j, and the distance of the COMBINED STRESSES 159 top of the beam from the same axis e. The faces are to be taken as parallel and perpendicular to the axis of the beam. There are compressive forces acting on each end of the elementary block which vary in intensity directly as their distance from the neutral axis. Let S be the unit com- pressive stress at the upper surface, and dA a differential area at any distance ?/ from the neutral axis; then — i/dA c is the force acting at the distance ^ from the neutral axis. The sum of the horizontal forces acting on either end of the elementary block is the sum of the compressive stresses acting on the same area. Calling this sum 11=— \ ydA^ and writing — for — , we have H = -- \ yd A. Let the I c I^vi bending moments at the ends of the block be Mi and M2, and the sum of the compressive stresses on the same ends be H^ and IT^. Since the ends are separated by dx, Mi--M^= dM; then H^- H^ = ^C ydA, For I ^y^ equilibrium a force equal to H^ — H^ must be introduced, and this force must be equal to the sum of the shearing stresses on the area dx times h. This sum is a horizontal shearing force, and the unit stress is — l— ^. Therefore if Si^ is the horizontal unit rJ 71//" f*C shearing stress, Sj^ = -— I ydA. From the theory dx Ih*^yi of beams Vdx = dM ov — — = F. Substituting T^for — — in the expression for Sf^^ we have /S'^ = — - J ydA as tlie value of the unit shearing stress at a distance y^ from the neutral axis. 160 MECHANICS OF MATERIALS The parallelopipedon could have been cut from the lower half of the beam where the stress is tension and the reasoning would be equally true. The formula gives the value of the unit shearing stress at a distance y-^ from the axis, in a section for which F'is the vertical shear as defined in the chapter on beams. The width of the beam at a distance y-^ from the axis is 5, Zis the moment of inertia of the entire section, while I ydA is the static moment of the area of the section lying above the distance y-^ from the neutral axis. If e^ is taken as the distance of the center of gravity of the area above y-^ from the neutral axis ydA — a^e^, and S/i = —- a^c^. For a point y^— c from the axis, a^c-^ = 0, and therefore jS\ = at the distance e from the neutral axis. When a^ is the whole area above the axis, S^^ will be a maximum for that section. The greatest value of Sft for the beam will be found at the neutral surface in the section where the vertical shear is a maximum, since the value of jS/^ varies directly with V. Sitnilarly, there will be no horizontal unit shearing stress in the sections where V= 0. It has been proven that for any point in a section of a beam there was a horizontal unit shearing stress that was equal to the vertical unit shearing stress at the same point. The expression just derived for Sf^ shows that the hori- zontal unit shearing stress is^ a variable quantity in any section of the beam, therefore the vertical unit shearing stress must also be variable. For a rectangular section, breadth 6, depth d, the value of iSf^ for any section is - -— instead of S, = — , showing: the maximum horizontal 2bd ' bd ^ COMBINED STRESSES 161 shearing unit stress, and therefore the maximum vertical unit shearing stress is 50 per cent greater than the assump- tion of uniform distribution of stress would indicate. In determining the conditions for the safety of a beam, if the unit stress derived from the formula S — -— is a safe stress, the beam will in most cases be safely loaded. When a beam is short and deep, the horizontal unit shearing stress along the neutral surface may exceed the safe unit shearing stress. This is especially true of timber beams on account of the low value of the ultimate shearing strength of timber along the grain. Hence the value of the shearing stresses should always be investi- gated, as no beam is known to be safely loaded until the unit stresses of all kinds have been determined and found safe. Art. 82. Maximum Stresses in Beams. In the general theory of beams as presented in Chap- ter III, the' sliearing stress due to the vertical shear was assumed to be uniformly distributed over the area of the section of the beam. That this assumption was not strictly true is evident from the equality of the variable value of the horizontal unit shearing stress with the unit vertical shearing stress in the same section. The value of S/^ being a maximum or zero, as V is a maximum or zero, shows that S;^ is zero when ilf" is a maximum, or that there are no shearing stresses in the section for which M is a maximum. The value of S as derived from S = —— will, therefore, be the true unit stress for any section where Ji" is a maximum bending moment for the beam. 162 MECHANICS OF MATERIALS The shearing stresses in the fibers along the upper and lower sides of the beam are also zero, since a^c^ = 0. Hence the unit stresses in these fibers at the various sec- tions of the beam will be the value of S ss derived from — ^, when M is the bending moment for the section con- sidered. The unit tensile or compressive stress being zero along the neutral surface, the unit stress along that axis is one of shear, and its value may be found from the expression for S/^, For simple beams M is zero at the supports and the unit stress at all points of the section is simply With these exceptions the unit stress at all points in a beam is a combination of the tensile or compressive stresses with the shearing stresses. The value of the maximum unit stress at any point in the beam may be found from the expressions for the maximum values of Sj^ or Sp^ when jSf^ is substituted for S^. These maximum stresses make angles with the axis of the beam that depend for their value on the relative values of S and Sfi. The unit stress given by -— - is the true unit stress, when M is s. maximum moment, and it is easy to see by an in- spection of the expressions for the maximum values of iSji and Sp that their values can rarely ever exceed those Mo V C^ given by -— - and — I ydA, When a beam is deep vertically and carries a concentrated load at the center, the value of V is constant for all sections up to the middle of the beam. Therefore it is possible in such a beam that the maximum value of S^ may exceed the COMBINED STRESSES 163 Mc . . maximum value of S as found from . This is espe- cially true of beams having I sections, as the value of the static moment of the flanges is nearly as large as the moment of the whole area above the neutral axis. The value of S^ in the web just below the flange being very large, for a section just to the left of the load where F^is a maximum and S nearly so, the values of S.^ and aS^^ may be greater than the maximum values of S and Sj^. The shearing stress S^ at any point of the beam is a stress that is inclined at angles that vary from 0° to 45° with the axis of the beam. When the beam has an I section and is deep, these shearing stresses have a tend- ency to cause the web to buckle. To resist this tendency vertical angle irons are often riveted to the webs. When a beam of an I section is composed of angle irons riveted to a web plate, making what is known as a plate girder, the force on the rivets joining the angles to the web can be found from the values of S^ and /Sy^, and the areas over which these stresses are distributed. While the theory of beams as presented in Chapter III was defective in its assumptions regarding the distribution of the shearing stress, and its neglect of the combined stress, this discussion shows that for the majority of beams Me the formula S= -— ^iH gjye the maximum bending stress when the value of itf is a maximum. The formula may be used for the design of all beams and the dimensions checked for safety by determining the values of /S'^, /S^, and Sf^, 164 MECHANICS OF MATERIALS EXAMINATION Give some examples of bars where the forces acting produce more than one kind of stress. When a bar under bending forces also has a tensile or compressive axial force acting on it, why does not the resulting unit stress always equal the sum of the unit stresses due to each force ? Develop the expression ^ Me '^1 ~ ;u^' i± p E where S-^ is the maximum unit stress due to both the axial and bending forces acting on any bar. What assumptions are made in the development of the formula that are not strictly true ? When a beam which carries bending loads also has to resist tensile or compressive loads, how can the resulting bending moment be made practically zero ? When a bar is subjected to both axial and shearing forces, show that at every point of the bar there is a pair of equal unit shearing stresses whose directions make right angles with each other. What is the effect of the combination of shearing with unit tensile stresses ? Is the effect any different if the shearing stress is combined with an equal stress in com- pression ? Show how to determine the value of the maximum unit stress when tensile or compressive stresses are combined with shearing stress ? A bar is being acted on by tensile or compressive forces COMBINED STRESSES 165 applied in the line of the axis. Is there any shearing stress ? How may its value be determined ? If the maximum tensile or compressive unit stress is given by and the positive sign is used before the radical, what kind of stress is jS^? What is meant by the expression " horizontal shear " in beams ? Deduce an expression for the horizontal unit shearing stress at any point of a beam. Under what conditions will the unit stress given by S=—- be the maximum unit tensile or compressive stress in a beam ? Under what conditions may the horizontal shearing stress become the most dangerous stress in the beam ? A deep I beam is short and carries a load at the middle. Is it possible that at some point in the beam there may be a unit tensile or compressive stress greater than that Mc given by aS'^ — - when the value of Mis that of the maxi- mum bending moment for the beam ? When a beam is rectangular in section, where will the maximum unit shearing stress be found? Is the same true for beams of other sections ? PROBLEMS 1. A roof with two equal rafters has a span of 40 ft. and a rise of 15 ft. If the weight of the rafter is neglected, determine the size of the rafters 6 in. wide 166 MECHANICS OF MATERIALS when each rafter carries a uniform load of 50 lb. per foot. Assume the allowable unit stress as 600 lb. / sq. in. 2. Assume the rafters in problem 1 to carry a load of 1000 lb. at the middle, instead of the uniform load, and find the depth of the rafters for the same unit stress. 3. A simple wooden beam, 30 ft. long, 12 in. deep, and 4 in. wide, carries a single load of 320 lb. at the middle and an axial compressive load of 14,400 lb. Find the maximum unit stress in the beam, (^) by the approximate method, (5) by the more exact method. 4. A simple steel I beam 20 ft. long, 12 in. deep, weighing 35 lb. / ft., carries a uniformly distributed load of 500 lb. /ft. and sustains a compressive load of 40,000 lb. Is it safe if a factor of safety of 5 is needed ? 5. Find the points of application of the compressive loads in problem 4 so that the unit stress resulting from the effect of the loads will be as small as possible. 6. A steel eye bar 30 ft. long has a section area of 2 by 6 in. The eye bar is to be used horizontally, and carries a tensile load of 288,000 lb. Determine the dis- tance from the center of the eyes to the axis of the bar so that the resulting unit stress will be as small as possible. 7. A horizontal steel shaft, 5 in. in diameter, is 20 ft. between bearings, and carries a load of 1200 lb. at the center. It transmits 250 H.P. at 100 revs. /min. Required the maximum unit stress induced. 8. ' A standard 2-in. steel bolt is screwed so as to cause a unit tensile stress of 10,000 lb. /sq. in. What shearing force may be resisted if the maximum unit stress is not to exceed 15,000 lb. /sq. in.? COMBINED STRESSES 167 9. How many 1-in. standard steel bolts must be used to resist a shearing force of 300,000 lb. if the tensile stress due to screwing up is assumed to be 10,000 lb. /sq. in. and the allowable maximum unit stress is taken as 12,000 lb. /sq. in.? 10. An angle iron is bolted to the side of an I beam and supports another beam whose reaction at the sup- ported end is 50,000 lb. Select a number of 1-in. bolts to carry the load. Assume that the unit stress in each bolt due to screwing up is 12,000 Ib./sq. in., and the factor of safety" is 4. 11. Given an I-beam having flanges and web whose cross-sections are rectangles of the dimensions given in Fig. 82, used as a simple beam 20 ft. long and carrying a con- centrated load of 30,000 lb. at the middle of the span. Find the maximum horizontal unit shearing stress and the horizontal unit shearing &t ess at a. 12. Given the same beam as in problem 11, carrying the same loads, find the difference between the maximum combined compressive unit stress in a section taken just to the left of the central load and the value of Mc ^ ^ -J- at the dangerous section. 13. Given the same beam as in problem (11) and carrying the same loads, find the total horizontal shearing force acting along the neutral surface and to the left of the load. What is the value of this shearing force per Hneal foot of the beam? \( /a — — -)( 1 1 1 1 1 1 1 .ii% 1 1 1 Fig. 82. 168 MECHANICS OF MATERIALS 14. Given a beam having the same section as in problem (11) to be used as a simple beam on a span of 8 ft. and supporting a load of 160,000 lb. concentrated at the middle. Find the magnitude of the maximum combined unit stresses in tension and shear. What angle will the maximum com- bined stresses make with the axis of the bar? What will be the value of the greatest horizontal unit stresses in tension and shear? CHAPTER VIII V COMPOUND BARS AND BEAMS Art. 83. Definition. When a bar is composed of more than one kind of material it is sometimes termed a compound bar. The formulas derived in the previous chapters apply only to bars made of one material throughout. This chapter will be devoted to the investigation of a few of the simpler cases of stress in the compound bars. Art. 84. Compound Columns; alternate layers. A column or pier built with alternate layers of different materials, as in Fig. 84, will evi- dently carry only the load the weaker section will support. The unit stress in any section may be found when the area of the section is known, as each section has to support the entire load. When the modulus of elasticity of the material and the length of each section are known, then the amount of shortening for each section can be found as for simple bars. The total shortening of the entire column is the summation of the deformations of the different sections. Art. 85. Compound Columns ; longitudinal layers. When a column is composed of different materials arranged longitudinally, the column becomes a bundle 169 Fig. 84. 170 MECHANICS OF MATERIALS of simple bars. Each bar does not carry a part of the total load that is proportional to its area as in the previous case. All the bars have the same amount of deformation, and when the unit stress is within the elastic limit, the unit stress in each bar must vary directly as the modulus of elasticity. Let Fig. 85 represent a section of a hollow steel column encased in concrete carrying a compressive load which produces axial stress only. Considering each part of the compound column as a separate bar, the kind of material in that part and the necessary factor of safety determines the safe unit stress Sc This stress depends solely on the factor of safety and the material. When the two columns act together and support the load as one column, the condition that each of the columns of the different materials must sustain the same deformation, gives the relation e = flh /2^2 which may be written e = El Pih AiEi E, P2I 2^2 A2E2 (1) (2) In equations (1) and (2) /i and /2 are the unit stresses that actually occur in the different materials composing the column; P, I, A, e, and E have their usual meanings, and the subscripts 1 and 2 refer to steel and concrete respect- ively. When the lengths of the two columns are the same, equation (1) reduces to fl f2 fi El 11- = Jji or — = — ^. El E2 /2 E2 e = (3) COMPOUND BARS AND BEAMS 171 That is, the actual unit stresses induced in the different materials composing the column are in the same ratio as their respective moduli of elasticity. As the ratio of the safe unit stress — in any two materials is independent of 02 the ratio of their moduli of elasticity, it is evident that the ratio of the safe unit stresses for the materials will not necessarily be equal to the ratio of the actual unit stresses induced. Si and S2 are the limiting stresses that can not be exceeded with safety. In general, one of the safe unit stresses may be used for either /i or /2 provided the relation imposed by equation (3) gives a value for the other actual stress that is either equal to or less than the safe unit stress for that material. When the values of /i and /2 have been determined and the areas of the different parts known, the partial loads carried by each part is determined by Pi = Aifi and P2 = A2/2, the total load being given by P,= Pi + P2. When only the total load and the materials are known, the problem of finding the proper areas for the two columns is indeterminate, as any values can be assigned to Pi and P2 as long as their sum is equal to P. The same reasoning may be applied to columns of three or more materials by writing equation (1) as = /lil = /2^2 _ fsh fJn ,-v El E2 Es Eft and equation (2) as P = Pi + P2 + P3 . . . + P« (6) Art. 86. Compound Beams. When two or more beams of different materials are fastened together so that they act as one beam, the com- bination is called a compound beam. 172 MECHANICS OF MATERIALS Beams having cross-sections similar to those shown in Figs. 86a and 865 are examples of compound beams com- posed of two materials, wood and steel. The form of the section is immaterial as long as the different parts are con- nected so as to act as one beam. If the connecting bolts are removed, the safe unit stresses in each of the component parts of the compound beam Fig. 86 a. Fig. 86&. may be assigned at once, as in this case each beam acts independently and the safe unit stresses will depend on the material and the factor of safety. When the two parts of the beam are bolted together as shown in the figures, each part of the beam deflects equally and this condition gives the relation A = OL fih a f2l2^ (1) (3 Eici (3 E2C2 Which when expressed in terms of the partial loads becomes A = 1 Wlh^ 1 W2l2^ (2) i8 EJi /3 E2I2 where /i and /2 are the actual unit stresses in tension or com- pression induced in the different parts of the beam by the COMPOUND BARS AND BEAMS 173 load carried, and the other symbols have their usual meanings. Equations (1) and (2) may be expressed as and Wi EJih^ W2 E2l2ll^ (4) giving the ratio of the actual unit stresses and the ratio of the partial loads for each part of the beam. As was found to be the case when considering compound columns, the safe unit stress for each part of the compound beam will depend on the material and the factor of safety. These unit stresses are the limiting values for the unit stress for each material. They can not be exceeded with safety. The value of /i or /2 may t>e assumed as equal to the limiting value of the unit stress for that material provided the relation imposed by equation (3) does not cause the actual unit stress in the other material to exceed the safe limiting value of the unit stress for that material. When /i and /2 have been determined, the partial load carried by each of the component parts of the beam may fl be found from the formula M = — which in this case may be expressed as TFi=^ or W2 = i^. (5) llCi I2C2 The sum of the partial loads is the total load on the com- pound beam or W = Wi -\-W2 (6) 174: MECHANICS OF MATERIALS When the load to be carried by a given compound beam is known, the partial loads for the different parts may be found by the use of the equations (4) and (6) and then the actual unit stresses may be determined from equation (5). The problem of designing a compound beam to cover a given span and carry a given load is indeterminate. The values of the safe unit stress for each material will be known, but the relative values for / cannot be found until the values for I, c, and I for each of the component parts of the beam are known. c The value of the ratio — for each beam may be chosen so that the value of / for each of the component parts will be equal to the limiting stress for the material, and if at the same time the values of — are chosen so that the equation c (6) will be satisfied, the design will be satisfactory so far as unit stress is concerned, as the unit stress in each part will equal the limiting value. Writing equations (3), (4) and (6) as EiCi E2C2 EsC3 ' ' ' EnCn _ Wih^ _ W2h^ _ W.ih^ Wnlr? ,^x ~/3^l7l ~ ^E2l2 ~ ^Esls ' ' ' ^Enln W = W1 + W2 + WS . . . -\-Wn (9) it is evident that this discussion will apply equally well to compound beams with any number of different materials. EXAMINATION What is a compound bar? When a compound bar is used as a short column, will the formula for axial compression always give the true . COMPOUND BARS AND BEAMS 175 maximum unit stress? Explain your answer fully, giving reasons. When a be^m is composed of more than one kind of material, name the conditions that are used to determine the part of the load carried by each material. Is it always possible to use the safe unit stresses as determined by experiments on the materials that compose a compound bar or beam to determine the loads that may be carried? Give reasons for your answer. When the depth of the different parts of a compound beam are not the same, will the unit stresses induced be in the same ratio as the moduli of elasticity? PROBLEMS In the following problems the values of the moduli of elasticity to be used are: Steel, 30,000,000 Ib./sq. in.; wood, 1,500,000 Ib./sq. in.; cast iron, 12,000,000 Ib./sq. in.; and concrete, 2,400,000 Ib./sq. in. 1. A hollow circular cast-iron column 6 in. outside diameter, 4 in. internal diameter, was designed to carry a given load using 10,000 Ib./sq. in. as the allowable unit stress. If the column were enclosed in a square box section wooden column 10 in. outside and 6 in. inside, how much more load could be carried on the compound column than on the cast iron one? Use 1000 Ib./sq. in. as the safe allowable unit stress in the wood. 2. A load of 138,270 lb. is to be carried on a compound column formed by filling the center of a circular cast-iron column 6 in. internal diameter with concrete. Assuming the limiting unit stresses in the cast iron and concrete to be 10,000 Ib./sq. in. and 1000/sq. in. respectively, find the necessary external diameter of the column. 176 MECHANICS OF MATERIALS - 3. Given a circular cast-iron column, 8 in. external and 6 in. internal diameter filled with wood. Assuming that the limiting unit stresses in the wood and cast iron are 1000 Ib./sq. in. and 10,000 Ib./sq. in. respectively, find the load that the compound column will carry. 4. A steel pipe 6 in. external diameter and a sectional area of 5 sq. in., is to be encased in concrete 2 in. thick for fire protection. Find the load that may be carried on the steel pipe, assuming that the concrete does not carry any of the load, and compare the result with the load that may be carried when the concrete carries its share of the load. Use 15,000 Ib./sq. in. and 600 Ib./sq. in. as the limiting unit stresses in the steel and concrete. 5. A compound beam 20 ft. long is formed by bolting two standard 12-in. channel beams to the sides of a timber beam 4 in. wide and 12 in. deep. Assume that the safe unit stresses for the wood and the steel are 1000 Ib./sq. in. and 15,000 Ib./sq. in. respectively and find the uniform load that may be carried, neglecting the weight of the beam. 6. A timber beam 24 ft. long, 10 in. wide and 12 in. deep . is to be reinforced by bolting a steel plate 10 in. wide by J in. thick to each side of the beam. Find the uniform load that may be carried. Assume the hmiting unit stresses in the steel and timber to be 15,000 Ib./sq. in. and 600 Ib./sq. in. respectively. 7. A timber beam 20 ft. long, 10 in. wide and 12 in. deep is to be reinforced by bolting steel plates 12 in. wide to each side of the beam in order that a load of 14,400 lb. at the middle of the beam may be carried. Assume that the limiting unit stresses in the steel and the timber are 15,000 Ib./sq. in. and 600 Ib./sq. in. respectively and find the thickness of the steel plates. , COMPOUND BARS AND BEAMS 177 8. A wooden beam 20 ft. long, 10 in. wide and 12 in. deep has steel plates 10 in. wide and J in. thick fastened to the top of the bottom of the beam. Assume that the limiting unit stresses in the steel and the wood are 15,000 Ib./sq. in. and 600 Ib./sq. in. respectively and that there is no rela- tive motion between the steel and the wood. Find the uniform load that may be carried, neglecting the weight of the beam. 9. If the uniform load was 60,000 lb., find the total force exerted over one-half of the length of the beam that would tend to shear the bolts used to fasten the plates. CHAPTER IX REINFORCED CONCRETE Art. 87. Steel and Concrete in Combination. Assuming the safe unit compressive stress in steel and concrete to be in the ratio of 30 : 1 and the costs of the two materials per unit of volume in the ratio of 60 : 1 it follows that the cost of the steel. member to support a given com- pressive load will be about twice that for a concrete member. As concrete is approximately ten times as strong in compression as in tension, it is evident that the cost of a concrete member to carry a given tensile load will be about five times that for a steel member to carry the same load. Hence, concrete is the cheaper material when only compressive stresses are induced, and steel when the stresses are both tensile and compressive. When a combination of steel and concrete is used to resist tensile forces the full strength of the steel is not available unless the unit stress in the concrete is dis- regarded; since before the unit stress in the steel reaches a safe allowable value, the unit stress in the concrete will have reached its ultimate value and the entire load will b3 carried by the steel. Steel beams in bridges and buildings are often encased in concrete on account of its fireproofing and non-corrosive properties. In the calculations for the strength of such encased beams no account is taken of the resistance of the concrete. 178 REINFORCED CONCRETE 179 Art. 88. Reinforced Concrete Beams. When a concrete beam has steel bars embedded in the concrete on the tension side of the neutral surface, the beam is called a reinforced concrete beam. The same kinds of stresses are induced in reinforced con- crete beams as in beams of one material and in addition there is a bonding or shearing stress between the steel and the concrete. The relations between the loads, stresses of all kinds, and the dimensions of a reinforced concrete beam must be determined before the design or investigation as to the safety of a reinforced concrete beam can be under- taken. Art. 89. Tensile and Compressive Stresses Parallel to the Neutral Surface. The common theory for reinforced concrete beams assumes that the concrete above the neutral surface carries all the compressive stresses and that these stresses vary in intensity directly as their distance from the neutral surface. On the tension side of the neutral surface the tensile resistance of the concrete is neglected, and it is assumed that the steel reinforcement carries all the tensile stress, the concrete being used to keep the steel in its proper position and to resist the horizontal shearing stresses. The assumption that there is no tensile resist- ance in the concrete is undoubtedly erroneous, but it errs on the side of safety. The relation between the moment of resistance in a section of a beam for which the bending moment is a maxi- mum, and the tensile or compressive unit stresses parallel to the neutral surface is derived on the following assump- tions : (1) That the concrete carries all the compressive stresses 180 MECHANICS OF MATERIALS and that these stresses vary in intensity directly as their distance from the neutral surface. (2) That the steel reinforcement carries all the tensile stress. (3) That the unit stress in the steel is constant and depends on the distance of its center of gravity from the neutral axis. (4) That the neutral surface is so located that the tensile resistance of the steel is numerically equal to the compres- sive resistance of the concrete. Fig. 89. (5) That any plane section of the beam remains a plane during bending. (6) That there is perfect adhesion between the steel and the concrete. Let Fig. 89 represent a portion of a reinforced concrete beam, and the section X-X the dangerous section. If the value of the resisting moment for this section can be expressed in terms of the maximum unit stresses in tension REINFORCED CONCRETE 181 or compression, and the dimensions of the beam, the relation between the load on the beam and the unit stresses becomes determinate, as the expression for the maximum bending moment can always be written and equated to the expression for the moment of resistance. Let a = area of section of steel reinforcement in square inches. A = hd = area of section of concrete in square inches. d = the distance from the outer fiber in compres-. sion in inches to the center of gravity of the steel. h = the breadth of the beam in inches. I = length of the beam in inches. Es = the modulus of elasticity for steel in pounds per square inch. Ec = the modulus of elasticity for concrete in pounds per square inch. Mr = the maximum moment of resistance for the beam . in inch pounds. M = the maximum bending moment for the beam in inch pounds. fs = the maximum unit tensile stress in the steel in pounds per square inch. fc= the maximum unit compressive stress in the outer fiber of the concrete in pounds per square inch. n = the ratio Es/Eq. r = the ratio fs/fc- p = the ratio of the sectional area of the steel reinforce- ment to the area of the concrete above the center of gravity of the steel; p = ahd. k = the ratio of the distance of the neutral axis of the section to the distance of the center of gravity 182 MECHANICS OF MATERIALS of the steel, both being measured from the outer fiber in compression. j = the ratio of the distance between the resultants of the stresses in the concrete and steel to the depth of the beam. For equilibrium the resultants of the stresses in the steel and concrete must be numerically equal and the moment formed by these resultants must be equal to the maximum bending moment for the beam. Since all the unit stresses considered are within the elastic limit, under the asumptions made it follows that k E. (1 - k)d tih^l^lJl n^ h kd JcE, h ^' Ec f w Substituting r for j, n for — ^ and solving for k gives k = -^. (2) 1+- n The area over which the unit compressive stresses are distributed is hkd, while the constant tensile unit stress in the steel is distributed over an area phd, and since the total compressive force must be equal to the total tensile force, pdhfs = kdh{-' or r =^f = A_. (3) / ^ Jc ^P Substituting the value of k from equation (2) in equation (3) and solving for p, P=—T^ V- (4) 2r ('+y REINFORCED CONCRETE 183 Equation (4) shows that there is a fixed relation between the ratios p and r. When either ratio is given the value of the other is fixed. In the design of a reinforced concrete beam, the steel ratio p may be chosen so that equation (4) will be satisfied when fs and fc are respectively to the safe unit stresses for the two materials. The safe unit stress in compression for a concrete of average strength lies between 600 and 700 Ib./sq. in., while the allowable stress in steel may be anywhere between 16,000 and 30,000 Ib./sq. in. Con- sidering horizontal stresses in tension and compression only, an ideal value of p is obtained when fs and fc are assumed as equal to the limiting unit stresses for the steel and con- crete respectively. As an increase in the unit stress in the steel increases the bonding stress, the actual unit stress in the steel is generally limited to 16,000 Ib./sq. in., regard- less of the possibility of the steel being able to carry a higher unit stress with safety. When p has any value other than the ideal one, the ratio of fs to fc can not be equal to the ratio of the safe unit stresses for steel and concrete. Substituting the value of r from equation (2) in equation (4) and solving the resulting quadratic for k, /b = pnL/i+A_i (5) L \ pn J When the problem requires the determination of the safe load for an existing beam, equation (5) will determine the value of k, as the ratios p and n will be given by the data for the problem. This value for k, substituted in equation (2), will deter- fs mine the ratio ^ = t- When the value of this ratio is Jc 184 MECHANICS OF MATERIALS known and the safe limiting unit stresses are given, the maximum values for/s and fc may easily be determined. The resisting moment for the section being equal to the moment of the couple formed by the resultants of the tensile and compressive stresses, jd times either resultant gives the value of the resisting moment. Hence the value of the bending moment at the dangerous section is M = Mr =fs V^djd = ^ khdjd, (6) in which jd = d ll 1 . This theory has been developed on the assumption that M was the bending moment for the dangerous section of the beam, but it applies equally well to any section when M is taken as the bending moment for that section. Substituting for M in equation (6), the value of the maximum bending moment for the beam and for either fs or fc the maximum possible value, will give the relation between the loads on the beam and the maximum unit stresses parallel to the neutral axis that may be used to investigate the safety of any given reinforced concrete beam. From equation (6) we may obtain a value of hd^ as To design a reinforced concrete beam to resist a given maximum bending moment and considering horizontal stress in tension and compression only, the values oi fs and fc may be taken as the maximum unit stress allowable in each of these materials when used in reinforced concrete beams. REINFORCED CONCRETE 185 The values oi p, j and k may then be determined by the use of equations (4) , either (2) or (3) , and j = Substituting the proper values for p, j and fs, or k, j and fc in equation (7) the value of bd'^ may be determined. While it would appear that any values that would satisfy equation (7) might be used for b and d, other conditions limit the least value of b to ^d. Art. 90. Reinforced Concrete T-Beams. Since the concrete below the neutral surface is not con- sidered as carrying any kind of the horizontal tensile stress, any more concrete below the surface than that required to resist the horizontal shearing stresses and hold the steel reinforcement in position adds unnecessary weight to the beam. For this reason beams of the form shown in Figs. 90, called T-beams, are in common use. In the derivations of the formulas for T-beam sections the same notation applies as was used for beam of rec- tangular sections with the following exceptions and addi- tions : b = the width of the flange. 6' = width of the web. t = the thickness of the flange. z = the distance of the resultant compressive force below the top of the flange. p = the steel ratio a/bd and not a -^ area of concrete. The distance of the neutral axis below the top of the flange will depend on the relative depths of the flange and web and on the amount of steel reinforcement. As this distance may be more or less than the thickness of the flange, the neutral axis may lie either in the flange or web. When the neutral axis lies in the flange (Fig. 90a), 186 MECHANICS OF MATERIALS all the theory used with rectangular sections applies equally well for T sections, as in either case any effect the concrete below the neutral axis may have on the strength of the beam is neglected. When the axis lies in the web the conditions are changed (Fig. 906). While the unit stress in the concrete is still assumed to vary directly as the distance from that axis, the position of the resultant compressive stress is not the same as for a rectangular section. The unit compressive stresses in the web will be small, as they are near to the neutral axis, and since the width of the flange is generally much greater than the width of the web, the compressive resistance of the concrete in the I- .b'. d 1 A/eutralAxis &OB- FiG. 90a. Fig. 906. web will be small compared with the resistance of the con- crete in the flange. In the great majority of T-section beams the proportion of the compressive stresses carried in the web is so small that its neglect will not seriously affect the result, and the knowledge that any error made will be on the side of safety has led to the development of an expression for the resist- ing moment for a T-section reinforced concrete beam, in REINFORCED CONCRETE 187 which the effect of any compressive stress there may be in the web is neglected. Following the same line of argument as was used in the beams of rectangular section the value of k is found to be k = ^— . (1) r 1 + - n The average unit compressive stress in the flange is one- half of the sum of the unit compressive stresses at the top and bottom of the flange. The unit stress at the top of the flange is fc and that at the bottom is /<.( 1 — — j . The average unit stress, one-half the sum of these stresses, is This average unit stress is distributed over an area equal to ht, and since the sum of the compressive stresses must be numerically equal to the tensile resistance of the steel, f^Pbd=f{t-^^bt. (3) This may be written as Eliminating r between equations (1) and (4) and solving for k gives k = ^. (5) pn + - The resultant of the compressive stresses in the flange passes through the centroid of the part of the shaded area 188 MECHANICS OF MATERIALS lying in the flange at a distance from the top (Fig. 90) of the flange equal to a 2/c--, a (6) From Fig. 905 the arm of the couple formed by the resultants of the tensile and compressive stresses is jd = d - z. (7) Using the expression for the arm of the couple and the magnitude of the resultant forces from equation (3) the moment of resistance is given by Mr =fsVhdjd f'^'-irr^' (8) in which the values of /s and fc are such that equations (1) and (5) are satisfied. Art. 91. Shearing Stresses. Let Fig. 91 represent a differential length, dl, of a beam, C and C' the resultant compressive forces, and T and T' the resultant tensile forces acting on each end of the section. From the necessary equality of the resulting forces acting in any section, C = r,C' = T'andC -C' = T-r. The algebraic sum of the forces acting below the neutral surface is a force T — T\ applied at the center of gravity of the steel. This force tends to shear the concrete on lany horizontal section between the steel and the neutral sur- face. The stress resisting this force is distributed over an area equal to the width multipHed by the length of the Fig. 9L REINFORCED CONCRETE 189 section, hence the unit shearing stress parallel to the neutral surface between the steel and the neutral surface will be >s. =^^— ^. (1) oal For equihbrium the moment of the couple {C — C'){T — T') must be equal to Vdl, where V is the vertical shear for the section considered. The arm of the couple is jd, hence Vdl = (T - r)jd (2) , Ehminating T — T' between the equations (1) and (2) Sn=^.- (3) hjd Equation (3) holds true for T-section beams as well as for rectangular ones when h is the width of the section on the horizontal plane for which Sn is the unit shearing stress. Above the neutral surface the horizontal unit shearing stress is always less than below that surface, hence the maximum horizontal unit shearing stress occurs below the neutral axis in a section where F is a maximum, and will have the greatest possible value for that stress when h is the least width of that section. Art. 92. Bond Stress. Since the theory for reinforced beams assumes that there is a perfect bond between the steel and the concrete, the difference between the total tension in the steel at any two sections of a beam must be equal to the resistance that the concrete surrounding the steel between those sections offers to prevent any relative motion between the steel and the concrete. The value of the bond stress per square inch of horizontal rod surface may be found as follows: Let T and T^ be the 190 MECHANICS OF MATERIALS total tensions in the steel at two sections dl apart, then T — T' is the shearing or bond strength per inch of length dl of the beam. From the equality of moments (Fig. 91), Vdl = (T - T')jd. T — T' V It follows that — = — , which gives the value of di jd the bond stress per unit of length of the beam at a section of the beam where the vertical shear is V. This stress, divided by the sum of the perimeters of all the bars, is the bond stress per square inch of rod surface. If u is the bond stress/sq. in. and 2o is the sum of the perimeters, then V u = — . Zojd Art. 93. Diagonal Stress. In Chapter VII, Article 80, it was proven that in any section of a beam there was a maximum tensile stress whose value was determined by 2 S acting at an angle 6 such that tan 2 6 = —~. In reinforced concrete beams the horizontal tensile stress below the neutral axis is assumed to be zero, and since Ss = Sh, it follows that Sn, equal in magnitude to Sn, is a maximum unit tensile stress which acts at an angle of 45 degrees with the axis of the beam. Since the value of this diagonal tensile stress depends on the magnitude of the horizontal shearing stress, the greatest value of Sn will be found in a section at or near to the supports where the vertical shear is a maximum. When the failure of the beam is due to the diagonal tension, REINFORCED CONCRETE 191 the line of fracture generally starts on the plane of the reinforcement at or near to the supports and extends towards the center of the beam, making an angle of approx- imately 45 degrees with the axis of the beam. The value for Sn = Sn was obtained on the assumption that the horizontal tensile stress was zero. As there is some tensile stress in the concrete at all sections of a beam, it must always be remembered that the true value of Sn is always greater than Sn- While in some beams the value of Sn never becomes large enough to be a dangerous stress in the concrete, a great many of the failures of reinforced concrete beams are due to the neglect to provide proper reinforcement against the diagonal stress. To determine the area of the steel required to aid the concrete in resisting the diagonal tension, it will be neces- sary to make some assumptions as to the distribution of the stress between the steel and the concrete and the probable value of the vertical and horizontal components of the diagonal stress. Regarding the distribution of the stress one very common assumption considers that the steel carries f and the con- crete ^ of the diagonal stress, no steel being required until that stress exceeds 35 to 40 Ib./sq. in. Another considers that the concrete will carry 35 to 40 Ib./sq. in. and that the balance is to be carried in the steel. If the unit tensile stress in the concrete is hmited to 35 to 40 Ib./sq. in. and the maximum value of Sn for any beam, regardless of the amount of steel reinforcement, lies between 105 and 120 Ib./sq. in., each of the assumptions will give the same result when Sn has the maximum value, but for all values of Sn less than the maximum the former will require more steel than the latter. The present ten- 192 MECHANICS OF MATERIALS dency seems to be towards the use of the one that requires the most steel. Similarly while the common practice assumes that ver- tical and horizontal components of the diagonal stress are each equal in magnitude to S^, one authority gives the V V value of Sn as -— and another as -— . Since the average jbd od value of j is |, these two assumptions will give the same result if the maximum stresses in the latter case are f of those used in the former case. Consider a short length of a beam, s, lying between two sections where the vertical shears are Vi and F2. If Vi + V2 the section is short then Va = may be taken as value of a uniform vertical shear extending over the small bngth s. Taking the vertical component of the diagonal Va VaS stress as numerically equal to Sn = -rr-.j —rr is the total Jbd jd vertical stress exerted over "a section sXh in area. Of this total vertical force the steel is to carry |, therefore the force that the steel has to carry is given by ~. 3 jd When the steel reinforcement is placed with the axis of the bars perpendicular to the neutral surface of the beam, then the area of steel required in the length s is 2_VaS 3 jdfs When the axis of the bars is perpendicular to the plane of the diagonal fracture the effective area of the steel to resist vertical force becomes ^ 2^ Va cos 45°s "^ " 3 jdfs ' REINFORCED CONCRETE 193 The minimum number of bars that should be used in the distance s is given by the experimentally determined fact that the bars are not effective if the distance between any two bars is greater than f d. The number of bars in excess of this minimum is deter- mined by the bond strength of the bars, and other con- ditions depending on practice. Art. 94. Columns. A concrete column may be reinforced by steel bars placed parallel to the axis of the column, or by steel bands or wire being wound spirally around the column. In the former case both the steel and the concrete sustain the same deformation and the laws for columns of two materials hold true. Failure of the column would probably take place when the elastic limit for the concrete is reached. When the reinforcing steel is wound spirally around the column, the steel cannot carry any of the compressive load directly. When a bar sustains an axial compressive force there is an increase in the lateral dimensions. If this deformation, could be prevented the allowable compressive load would be greatly increased. Since for compressive stresses within the elastic limit of the concrete there is only a slight lateral deformation, it is evident that the elastic strength of the column is but slightly increased by the use of spiral reinforcement. The steel bands will undoubtedly aid in preventing failure of the column when the elastic strength of the column is exceeded, but the longitudinal deformation will be excessive. To be effect- ive the clear spacing between the spiral bands must not exceed one-quarter of the diameter of the enclosed column. The effective area of a reinforced concrete column is 194 MECHANICS OF MATERIALS the area enclosed by the reinforcing steel. Any concrete there may be that is outside of the steel is placed there for the protection of the steel. Since the use of spiral banding may increase the ultimate load that a column can support, r» \ — ^ ===== =^ 1ft' a", .___ ^ O n n t » 1 S-^a'^Rods tied with No. 16 Wire to Mild Steel Hoops spaced TO'cto c Spacing Bars jTHigh Carbon Steel Spiral with I'Pitch Spacing Bars _ J^'High Cart»h Steel Spiral vyith TPitch. ^^''♦Rods Beo to Spiral with No. 16 Wire at ro** Intervals Fig. 94. Spadng Bars and has but little effect on the elastic strength of the column, the allowable unit stress in the concrete in a column having both spiral and longitudinal reinforcement is higher than when the longitudinal bars are used alone. The common practice seems to tend towards the use of longitudinal reinforcing bars held in position by ample spiral banding, REINFORCED CONCRETE 195 and to take no account of the spiral reinforcement in so far as it is effective in aiding the concrete to carry the safe load. Art. 95. Deflection of Reinforced Concrete Beams. Its complex character and the uncertainty as to the dis- tribution of the stresses in the concrete, render the deriva- tion of a theoretically correct formula for the deflection of a reinforced concrete beam practically impossible. It is, however, possible to obtain an expressiou for the deflection which while theoretical in form contains an empirically determined constant. When the conditions under which the beam is to be used are approximately the same as those under which the constant was determined, the expression will give very satisfactory results. The following discussion is due to Turneaure and Maurer,* with slight change in notation. The form of the expression is the same as that for beams of one material throughout. This for a rectangular con- crete beam is ^EJ (1) To find the value of I for a reinforced beam, the area of the steel will be considered as increased n times without changing the position of its center of gravity relative to the neutral axis of the section and the steel replaced by this area of concrete. This gives an all concrete section with the same theoretical strength as the given reinforced section. The tensile stress in the concrete below the neutral axis affects the amount of deflection and should not be neglected. In the absence of any information as to the depth below the axis that the tensile stress is effect- * Principles of Reinforced Concrete, Turneaure and Maurer. 196 MECHANICS OF MATERIALS ive it is assumed that the entire area between the steel and neutral axis carries a tensile stress. This assumption changes the value of k from that given in Art. 89 to , ^ ^ + 2np "~ 2 + 2 np* Referring to Fig. 89 it will be seen that the moments of iiiertia of the areas in tension and compression relative to the neutral axis are/— (/cd)^ for concrete area in compression, 6 r p — (1 — k)d\ for the concrete area in tension, and for the changed steel area nphd[(l — k)d]^ approximately, hence the moment of inertia of the entire section is, / = i-| A;3 + (1 _ /c)3 4_ 3 pri(l - k)AhdK If the coefficient of hd^ is represented by cf), substituting Es hd^ <^ for 7, and — for Ec in equation (1), the expression n for the deflection of a rectangular reinforced concrete beam becomes, ^ = ^^rL (2) The value of n * to be used in the above formula, while theoretically the ratio between the moduli of elasticity for steel and concrete, should be chosen so that for the same beams, the calculated value of A will agree with the experi- mentally determined value of the deflection. The variation in the distribution of the tensile stress in * Turneaure and Maurer recommend that 8 or 10 be used for n in the formulas for deflection. REINFORCED CONCRETE 197 the concrete for various intensities of load on any beam renders it unlikely that any single value of n could satisfy all variations of load, hence the formula as derived should be used to determine the deflection under the maximum safe load for the beam only. Following the same line of reasoning and considering the areas of the web above and below the neutral axis as effective in carrying compression and tension respectively the maximum deflection for a reinforced concrete T-beam becomes, in which the value of d is given by, and of k by, k = , 1 \b' v/tv ,(t \n ,6' V t\ t Art. 96. Use of Formula. While the assumptions made when the relation between the external forces and the induced unit stresses in rein- forced concrete bars were derived are approximately true, none of them can be considered as exactly so. Hence the formulas derived for the various unit stresses in reinforced concrete bars should be considered as theoretical in form only. Each expression contains one or more constants of materials such as the safe unit stress modulus of elasticity, etc. 198 MECHANICS OF MATERIALS When the value of at least one of these constants in each expression is determined so that the result as cal- culated by the formula will agree with those obtained by experiments on similar bars, the expressions become empiri- cal and their use should be limited to the range of experi- mental work. Considering the complex character of con- crete, and the many conditions that may affect the strength of reinforced concrete over which the designer has no direct control, it is to be expected that there may be considerable difference in the designs offered for a reinforced concrete structure by reputable engineers, but such discrepancies will be greatly reduced as our knowledge of the action of reinforced structures under load is increased. In 1909 a joint committee from the American Society of Civil Engineers, The American Society for Testing Materials, the American Railway Engineering and Main- tenance of Way Association, and the Association of Ameri- can Portland Cement Manufacturers, after an exhaustive study of the then existing experiments on reinforced con- crete, made certain recommendations regarding the proper values for material constants to be used, and when these values are used in connection with the formulas as derived in this chapter the design will represent the average prac- tice of that day.* PROBLEMS Es ■ The values of — and the allowable unit stresses to he Ec used in the solution of the following problems will be found in the appendix under " The Recommendations of the Joint Com- mittee,^' unless otherwise stated. * An abstract is given in the appendix for the use of the student in solving the problems given in the text. REINFORCED CONCRETE 199 1. A reinforced concrete slab 6 in. deep, 4 ft. wide and 6 ft. long, has 2 sq. in. of steel reinforcement placed 2 in. from the lower surface of the beam. Find the maximum horizontal unit tensile and compressive stresses in the steel and concrete and the greatest uniform load that can be carried. Consider the weight of the slab as part of the uniform load. 2. Find the ideal sectional area of the horizontal steel reinforcement for a slab of the same dimensions as the one given in problem 1. Determine the uniform load that may be carried with the ideal amount of reinforcement. 3. Find the area of the steel reinforcement needed in a concrete beam 10 ft. long, 12 in. wide and 12 in. deep to the center of the steel reinforcement to carry a load of 6900 lb. at the middle of the span. Consider horizontal tensile and compressive stresses only. 4. Find the maximum horizontal unit shearing stress in the beam as given in problem 3. Does this beam need any reinforcement against diagonal tension? 5. A concrete beam 8 in. wide and 18 in. deep with 3 f -in. square bars placed 2 in. from the lower surface of the beam, is used to cover a span of 12 ft. and support a uniform load of 13,250 lb. Find the maximum bond stress and the greatest horizontal unit stresses in tension and compression and shear. 6. Find the sectional area of the vertical reinforcement needed in the beam as given in problem 5 in a distance of 16 in. from the end of the beam. Assume the allowable unit stress in the vertical steel as 12,000 Ib./sq. in. 7. The unit tensile stress in a steel rod J in. diameter is 16,000 Ib./sq. in. Find length that this rod should be 200 MECHANICS OF MATERIALS embedded in concrete if the bond stress is limited to 100 Ib./sq. in. of rod surface. 8. Find the maximum horizontal unit shearing stresses in the beams given in problems 1 and 2. Assume that the length of the columns in the following problems is less than 15 diameters. 9. A concrete column 16 in. effective diameter is rein- forced by eight f -in. steel rods parallel to the axis and held in place by being connected to a No. 16 wire wound spirally around the longitudinal bars, spaced 4 in. from center to center. Find the safe load for the column. 10. Determine the load that could be safely carried on the column given in problem 9, assuming that the longi- tudinal reinforcement was the same and that there were but four bands used to hold the steel in position. 11. Find the sectional area of the longitudinal reinforce- ment needed in order that a concrete column 12 in. effective diameter may safely carry a load of 102,000 lb. Consider that there is ample spiral reinforcement and length = 12 ft. 12. Find the probable amount of shortening of the column given in problem 11 when the given load is applied. 13. Find the maximum deflection for the beams given in problems 3 and 5. 14. Given a T-section concrete beam, flange 12 in. thick by 48 in. wide, web 18 in. wide, reinforced by five if-in. and seven IJ-in. diameter steel rods placed horizontally in three rows with their center of gravity 47 in. from the top of the flange. Find: (a) The maximum horizontal unit stresses in the steel and concrete in tension, compression and shear. REINFORCED CONCRETE 201 (6) The bond stress per unit of rod surface. (c) Consider a length of the beam extending from the edge of the support towards the center of the span for a distance of 2 ft. Find the sectional area of the vertical steel rods required to reinforce this part of the beam against diagonal tension, allowing a unit stress in the steel of 12,000 Ib./sq. in. (d) Find the deflection of the beam under the given load. EXPLANATION OF TABLES Table I. Fundamental Formulas. Table II. Derived Formulas. The numbers following each expression refer to the chapter and article in which the formula was derived. Table III. Prop^erties of Beams. The columns 1 and 2 give the relative strengths and stiffness of the various kinds of beams of the same length and shape. Columns 3 to 6 are the expressions for Maximum Vertical Shear, Bending Moment, Unit Stress, and deflection of the various beams under uniform and single concentrated loads. Columns 7 and 8 give the values of a. and /3 for the various beams; for a description of these symbols see: for a, Art. 48; i8. Art. 64. Table IV. Constants of Materials. This table has been compiled solely for the use of the student in solving the problems in the text. As all the constants are liable to considerable variation, it should not be used in the design of a structure that is to be built. Table V. Properties of Sections. In the rectangular sections d is the dimension in the direction of bending. In the hollow sections dx and 6i are the inside dimensions. Tables VI and VII. Properties of I and Channel Beams. — Cambria Steel. These tables have been inserted for the convenience of the student. As every engineer should own some of the trade books giving the properties of the various steel shapes, he will prefer to get his data first hand. Table VIII. A partial list of the recommendations of the Joint Committee regarding the allowable unit stresses in reinforced concrete beams and columns. 202 FUNDAMENTAL .J^ORMULAS 203 FUNDAMENTAL FORMULAS.— TABLE I Tension, Compression, and Shear (a) P = AS. Chap. I, Art. 4. Applies to all cases of uniformly distributed stress. Modulus of Elasticity for Tension and Compression (b) E = - = ^. Chap. I, Art. 11. e Ae Applies to all problems where the unit stress in tension or com- pression is within the elastic limit. Beams. — Vertical Shear (c) V=AS. Chap. Ill, Art. 39. True for all values of S. Bending Moment (d) M = ^. Chap. Ill, Art. 40. c Applies to all problems where the value of S is within the elastic limit. (/) Applies to all problems where the value of S is within the elastic limit. Equation of Elastic Curve EI^ = M. Chap. V, Art. 63. Twisting Moment in Shafts > p _SJ c Chap. IV, Art. 53. p_Sl_Ppl Oc 6J' Chap. IV, Art. 56. 204 MECHANICS OF MFATERIALS DERIVED FORMULAS. — TABLE II Strength of Bars of Uniform Strength logio ^0- Chap. II, Art. 23. w logioA = 0.434- y + Thickness op Steam and WaterJ Pipes, Cylinders, etc Thin pipes : Longitudinal ruptures. RD = 2 St. Circumferential ruptures. RD = 4: St. Thick pipes : Longitudinal ruptures. RDj^ = 2 St. Chap. II, Art. 24. Chap. II, Art. 24. Strength of Riveted Joints Tension in plate. t{p — d)St = Pf cird^ Shear on rivet. 4 5, = P., Chap. II, Art. 25. Chap. II, Art. 30. Chap. II, Art. 30. Chap. II, Art. 30. Chap. IV, Art. 58. Compression on rivet or plate. c^tdSc = P^ Horse Power of Shafts „ PpN , . SsJN H= —^ — (approx.) = \—^ ', 63,000^^^ ^ 53,000 c Shaft CouplijJ^gs Diameter of bolts. Pp = n~- SJ'h (ai)prox.). Chap. IV, Art. 59. Helical Springs Strength. Deflection, S = P = l^Ss. SD 8PZ)3 s.ttD^ Fd^ Fd Chap. IV, Art. 61. DERIVED FORMULAS.— TABLE II 205 Continuous Beams. — Three-moment Equation N^li + 2N2 (li + I2) + N^l^ = - !f!A!_±J£2^. Chap. V, Art. 67. Long Columns Round ends. Euler. P = ^^. Chap. VI, Art. 70. Rankine. P = ^^' ., . Chap. VI, Art. 73. Square ends. Euler. p = 4:^I^. Chap. VI, Art. 71. Rankine. P = -^^^' Chap. VI, Art. 73. 1 + <^1 Round and square ends. Euler. P = ^ ^^. Chap. VI, Art. 72. Rankine. P = ^^^"7^ • Chap. VI, Art. 73. 9 ^2 Combined Stresses Tension or compression with ben'ding. ^^ ^^a P/2' a 1 1 -^ ^ E P Mc or 5n, = - + -— (approx.). ^ 1 Tension or compression combined with shear. Max. shear. Sp =^ ± V,S7TT^ Chap. VII, Art. 80. Max. tension or , compression . «» = ^ « ± -^SjTlS^. Chap. VII, Art. 80. Horizontal Shearing Stress in Beams V C V 'Sfft = — I yd A = —aici. Chap. VII, Art. 81. lb J lb 206 MECHANICS OF MATERIALS Compound Coujmns Relative unit stress induced. ^=-^...=-^. Chap. VIII, Art. 85. Mix JOJ2 J^n Relation between the partial loads. Pih Pih. Pnln AiEi A2E2 AnEn Totalload = Pi + P2 . . . + Pn- Compound Beams Relative unit stress induced. « fik"^ a. fih^ a fnl\ Chap. VIII, Art. 85. |8 EiCi ^ E2C2 fi EnCn Relation between the partial loads. Wik^ ^ W2II _ Wnl'n Total load = W1+W2 . . . Wn. . Chap. VIII, Art. 86. Chap. VIII, Art. 86. Reinforced Concrete Rectangular Beams. Jc = = pn 1+4 nfc Jl+--l-l|. _M pn J 1 k Sc\ nfc/ M = fs pbdjd = —khdjd.] Wl^ n A = • -. pEsbd^ DERIVED FORMULAS.— TABLE II 207 = -[k' + (1 - /c)3 + 3 pn(l - A;)2]. Chap. IX, Art. 95. o For deflection only. 1 +2pn k = 2 +2pn T-beams. vn -f 1 ^ 2\d, k = — = . Chap. IX, Art. 90. 1 + — - pn + — nfc a i 3/c - 2 - d z = -, jd = d - z. Chap. IX, Art. 90. 2/c - - d M =fs pbdjd =f4^ - ^j^l ^h'd- Chap. IX, Art. 90. Wl^ n A =^^^3-. Chap. XI, Art. 95. Chap. IX, Art. 95. For deflection only. b b\d) '^ \d) \ — \r , • Chap. IX, Art. 95. 6 bit ^ b bd d Horizontal Shearing Stress (all beams). V Sh = — . Chap. IX, Art. 91. = r[ k ^ jbd Bond Stress (horizontal bars in beams) Columns. P = Acfc + nAsfc V u = -— . Chap. IX, Art. 92. P /c = -; — TT"- C^ap. IX, Art. 94. Ac + Agn 208 MECHANICS OF MATERIALS 02 W M O H o Ph 00 00. CO 00 00 00 CO lO CO 00 r-i (M 00 CO t- s tH (M ■ ^ 00 00 00 (M 6 Max. Deflec- tion 5 CO eo GO 00 eo 00 CO CO 5 CD 00 T— ( eo CM Oi T— 1 eo 00 CO 5 Max. Tensile or Compressive Unit Stress ^ CM ^ ^ ^ k 00 00. 00 rH 4 Max. Bending Moment s ^ CM s °° ^ CO Ss 3 Max. Vertical Shear fe fe ^ CM ^|oi 00 ^^ ^ CM 2 Eelative Stiffness 1— 1 ooico CD 00 (M O (M CO CD 00 CM l-H 1 Relative Strength ■ iH (M rtH 00 00 00 CM T— ( Q — t pi Z) l-H 'd 'd • rH B 'd CD o 1 — 1 C3 C« M a :« CD 03 rH I rH iz; o Si o S^Ico ^ o ■* 1 1 O •«»< r-( 'Q ^^ 1 ^ t^ 02 ^0 55g << y s C4 SL|C^ (M IfM 1 1 ^ 1 CO CD «* ^It-H '^It-I 1 -^I^ + rH ^ rO CO p ^ >< H S + CO CM ^1^ 1 CO 1 (M CO C5 '^1 <) ^ « w S5 t-H CO ■* 1^ ^ ■^ o K| ^ (M ^I'^l fO 3^ ^ -^ 1 '^ iz; I-o ""I ^ |tH 1 i—l ^ CO ■^ CO H s ^ CO ^ s rO V i-» IM -«. '^^ <1 fC) 5 a o '■+3 c3 (-, >» 3 o - tj ft o m a: <0 C|_l • a a .2 u ft a 1^ », II 2 n 1< a; 03 3 2 0) . CO Q ^ < H S s 4) 02 « ^ 05 d A t b I s r I' r' Ins. Lbs. Sq. in. Ins. Ins. Ins.4 Ins.3 Ins. Ins.4 Ins. B5 3 5.5 1.63 .17 2.33 2.5 1.7 1.23 .46 .53 ** " 6.5 1.91 .26 2.42 2.7 1.8 1.19 .53 .52 ** " 7.5 2.21 .36 2.52 2.9 1.9 1.15 .60 .52 B9 4 7.5 2.21 .19 2.66 6.0 3.0 1.64 .77 .59 " " 8.5 2.50 .26 2.73 6.4 3.2 1.59 .85 .58 " " 9.5 2.79 .34 2.81 6.7 3.4 1.54 .93 .58 * " 10.5 3.09 .41 2.88 7.1 3.6 1.52 1.01 .57 B13 5 9.75 2.87 .21 3.00 12.1 4.8 2.05 1.23 .65 ** " 12.25 3.60 .36 3.15 13.6 5.4 1.94 1.45 .63 ** ** 14.75 4.34 .50 3.29 15.1 6.1 1.87 1.70 .63 B17 6 12.25 3.61 .23 3.33 21.8 7.3 2.46 1.85 .72 " " 14.75 4.34 .35 3.45 24.0 8.0 2.35 2.09 .69 " " 17.25 5.07 .47 3.57 26.2 8.7 2.27 2.36 .68 B21 7 15.0 4.42 .25 3.66 36.2 10.4 2.86 2.67 .78 " *' 17.5 5.15 .35 3.76 39.2 11.2 2.76 2.94 .76 ** ** 20.0 5.88 .46 3.87 -42.2 12.1 2.68 3.24 .74 B25 8 17.75 5.33 ■■ .27 4.00 56.9 14.2 3.27 3.78 .84 " " 20.. 25 5.96 .35 4.08 60.2 15.0 3.18 4.04 .82 " " 22.75 6.69 .44 4.17 64.1 16.0 3.10 4.36 .81 ** ** 25.25 7.43 .53 4.26 68.0 17.0 3.03 4.71 .80 B29 9 21.0 6.31 .29 4.33 84.9 18.9 3.67 5.16 .90 " " 25.0 7.35 .41 4.45 91.9 20.4 3.54 5.65 .88 " " 30.0 8.82 .57 4.61 101.9 22.6 3.40 6.42 .85 ** ** 35.0 10.29 .73 4.77 111.8 24.8 3.30 7.31 .84 B33 10 25.0 7.37 .31 4.66 122.1 24.4 4.07 6.89 .97 *' " 30.0 8.82 .45 4.80 134.2 26.8 3.90 7.65 .93 ** " 35.0 10.29 .60 4.95 146.4 29.3 3.77 8.52 .91 " 40.0 11.76 .75 5.10 158.7 31.7 3.67 9.50 .90 12 31.5 9.26 .35 5.00 215.8 36.0 4.83 9.50 1.01 B 41 ' ' 35.0 10.29 .44 5,09 228.3 38.0 4.71 10.07 .99 " 40.0 11.76 .56 4.21 245.9 41.0 4.57 10.95 .96 B 53 15 42.0 12.48 .41 5.50 441.8 58.9 5.95 14.62 1.08 ** " 45.0 13 . 24 .46 5.55 455.8 60.8 5.87 15.09 1.07 ** " 50.0 14.71 .56 5.65 483.4 64.5 5.73 16.04 1.04 ** ** 55.0 16.18 .66 5.75 511.0 68.1 5.62 17.06 1.03 60.0 17.65 .75 5.84 538.6 71.8 5.52 18.17 1.01 212 MECHANICS OF MATERIALS PROPERTIES OF STANDARD CHANNELS.— TABLE VII. Axis 1-1 Passes through Center of Gravity Perpendicular to Web. Axis 2-2 Passes through Center of Gravity Parallel to Web. 1 2 3 4 5 6 7 8 9 10 11 12 13 a 'S a a 03 O o 4J o o ft d o '•+3 a) CO m a> bC c3 S o '■+3 a .2 '■+3 u I— 1 o 03 u >> . 4; Bji £3 >-> a o ja o j:i ID 03 §■2 S.2 0) 2 §.2 3.2 sO'2 -fj bc c^ Q -tj • rt X ^ ^ ■t X •S X K ^ a '53 0) 'jg TJ t< 03^ t< '§-< 2oO to P ^ < H g s rt ^ m « Q d A t b I S r I' S' r X Ins. Lbs. Sq.Ins. Ins. Ins. Ins.4 Ins.3 Ins. Ins.* Ins.3 Ins. Ins. C5 3 4.00 1.19 .17 1.41 1.6 1.1 1.17 .20 .21 .41 .44 " " 5.00 1.47 .26 1.50 1.8 1.2 1.12 .25 .24 .41 .44 " 4t 6.00 1.76 .36 1.60 2.1 1.4 1.08 .31 .27 .42 .46 C9 4 5.25 1.55 .18 1.58 3.8 1.9 1.56 .32 .29 .45 .46 " " 6.25 1.84 .25 1.65 4.2 2.1 1.51 .38 .32 .45 .46 " " 7.25 2.13 .33 1.73 4.6 2.3 1.46 .44 .35 .46 .46 C13 5 6.50 1.95 .19 1.75 7.4 3.0 1.95 .48 .38 .50 .49 " " 9.00 2.65 .33 1.89 8.9 3.5 1.83 .64 .45 .49 .48 (■ " 11.50 3.38 .48 2.04 10.4 4.2 1.75 .82 .54 .49 .51 C17 6 8.00 2.38 .20 1.92 13.0 4.3 2.34 .70 .50 .54 .52 <■ " 10.50 3.09 .32 2.04 15.1 5.0 2.21 .88 .57 .53 .50 •• " 13.00 3.82 .44 2.16 17.3 5.8 2.13 1.07 .65 .53 .52 T, All bindings are in cloth unle^ss cthert£>i^e noied. Abbott, A. V. The Electrical Transmission of Energy .Sv^o, *S5 oo A Treatise on Fuel. (Science Series No. 9.) i6mo, o 53 Testing Machines. (»Science Series No. 74.) i6nio, o 53 Adam, P. Practical Bookbinding. Trans, by T. E. Maw i2mo, *2 50 Adams, H. 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