TA 
 
 Southern Branch 
 of the 
 
 University of California 
 
 Los Angeles 
 
 Form L 1 
 
 TA405 
 
 M44s 
 cop.l
 
 This book is DUE on the last date stamped below 
 
 MAR 26 1928 
 3 
 
 APR 3 1931 
 MAR z 1.1932
 
 STRENGTH OF MATERIALS 
 
 A COMPREHENSIVE PRESENTATION OF SCIENTIFIC METHODS 
 
 OF LOCATING AND DETERMINING STRESSES AND 
 
 CALCULATING THE REQUIRED STRENGTH 
 
 AND DIMENSIONS OF BUILDING 
 
 MATERIALS 
 
 EDWARD R. MAURER, B.C.E. 
 
 PROFESSOR OF MECHANICS, UNIVERSITY OF WISCONSIN 
 
 ILLUSTRATED 
 
 46751 
 
 AMERICAN TECHNICAL SOCIETY 
 
 CHICAGO 
 1919
 
 COPYRIGHT. 1907. 1914. 1917, 1919. BY 
 AMERICAN TECHNICAL SOCIETY 
 
 COPYRIGHTED IN GREAT BRITAIN 
 A'.,. BIGHTS RESERVED
 
 INTRODUCTION 
 
 Q.oto,J 
 
 V> r'VERY layman is fascinated by a great engineering work a 
 I_j large sewage system, a Keokuk dam, a twenty-story steel 
 structure and wishes he might be able to construct such work 
 and carry it to completion. And yet he hardly appreciates the 
 f knowledge, experience, and judgment necessary to bring such a 
 M work to a satisfactory close. Time was when all of the details of 
 5J such structures were determined by guesswork, but the develop- 
 ^ ments in science and mathematics have changed all that. For- 
 mulas for the various types of stresses have been worked out; 
 52 constants for every known material have been collected; and a 
 multitude of diagrams and tables contribute to making the engi- 
 ' neer's work as precise as a bookkeeper's balance. The strength 
 QVJ and size of every rivet and the length and cross-section of every 
 girder in a steel structure are figured so they will bear the strains 
 put upon them. The design of a masonry dam, the strength of 
 the concrete mixture, and the amount of steel reinforcement are 
 all mathematically determined in order to safely restrain the given 
 volume of water behind it. 
 
 <I The final judgment, therefore, as to the size of every part of 
 the structure must depend upon the designer's knowledge of 
 "Strength of Materials." The treatment of Strength of Materials, 
 as found in standard textbooks, is so clothed in abstruse mathe- 
 matics that it is impossible for the average trained man to obtain 
 a working knowledge of the subject. As the subject is one of the 
 most important in any of the engineering branches, the author 
 has thought it wise to present it in simple form, culling out all 
 material that can not be used in practical design and analyzing 
 the subject from the theory through to the practical formulas 
 without the use of higher mathematics. In fact, he has used only 
 such mathematics as may be easily understood. While the author 
 has designed this work especially for home study purposes, the 
 material is valuable to the college trained man as well, as it gives 
 in clear, concise form the principles which are most used in engi- 
 neering and architectural work. It is the hope of the publishers 
 that the book will fill a place among the useful reference works in 
 the field of engineering.
 
 If. * 
 
 u* i 
 
 < J, V
 
 CONTENTS 
 
 PAGE 
 
 Simple stress 1 
 
 Kinds of stress 2 
 
 Tension, compression, shear 2 
 
 Unit-stress 3 
 
 Deformation 4 
 
 Elasticity 4 
 
 Hooke's law, and elastic limit 5 
 
 Ultimate strength 6 
 
 Stress-deformation diagram 6 
 
 Working stress and strength 8 
 
 Factor of safety 8 
 
 Strength of materials in tension 11 
 
 Strength of materials in compression 13 
 
 Strength of materials in shear 15 
 
 Reaction of supports 16 
 
 Moment of a force 16 
 
 Principle of moments 17 
 
 Kinds of beams 19 
 
 Determination of reactions on beams 19 
 
 External shear and bending moment 23 
 
 External shear 23 
 
 Rule of signs 23 
 
 Units for shears 24 
 
 Shear diagrams 27 
 
 Maximum shear 31 
 
 Bending moment 32 
 
 Moment diagrams 36 
 
 Maximum bending moment 40 
 
 Table of maximum shears, moments, etc 41 
 
 Center of gravity and moment of inertia 41 
 
 Center of gravity of an area 41 
 
 Principle of moments applied to areas 42 
 
 Center of gravity of built-up sections 44 
 
 Moment of inertia 46 
 
 Table of centers of gravity and moments of inertia 52 
 
 Strength of beams 52 
 
 Kinds of loads 52 
 
 Transverse, longitudinal, inclined forces 52 
 
 Neutral surface 54 
 
 Neutral line 64 
 
 Neutral axis 64 
 
 Stress at cross-section 65 
 
 Fibre stress. . . . . 57
 
 CONTENTS 
 
 Strength of beams (Continued) *AGE 
 
 Value of the resisting moment 58 
 
 First beam formula 59 
 
 Applications of the first beam formula 59 
 
 Laws of strength of beams 71 
 
 Modulus of rupture 72 
 
 Resisting shear 73 
 
 Second beam formula 73 
 
 Horizontal shear 75 
 
 Design of timber beams 76 
 
 Kinds of loads and beams 79 
 
 Flexure and tension . 79 
 
 Flexure and compression ' 81 
 
 Combined flexural and direct stress 83 
 
 Strength of columns 85 
 
 End conditions 85 
 
 Classes of columns 86 
 
 Cross sections of columns 86 
 
 Radius of gyration 86 
 
 Kinds of column loads 88 
 
 Rankine's column formula 88 
 
 Graphical representation of column formulas 93 
 
 Combination column formulas 94 
 
 Straight-line and Euler formulas 94 
 
 Parabola-Euler formulas 98 
 
 Broken straight-line formula 101 
 
 Design of columns 102 
 
 Strength of shafts 105 
 
 Twisting moment 105 
 
 Torsional stress 106 
 
 Resisting moment 107 
 
 Formula for the strength of a shaft 107 
 
 Formula for the power which a shaft can transmit 109 
 
 Stiffness of rods, beams, and shafts 110 
 
 Coefficient of elasticity Ill 
 
 Temperature stresses 113 
 
 Deflection of beams 114 
 
 Twist of shafts 116 
 
 Non-elastic deformation 117 
 
 Riveted joints 118 
 
 Kinds of joints 118 
 
 Shearing strength, or shearing value, of a rivet 118 
 
 Bearing strength, or bearing value, of a plate 119 
 
 Frictional strength of a joint 119 
 
 Tensile and compressive strength of riveted plates 120 
 
 Computation of the strength of a joint 120 
 
 Efficiency of a joint 122
 
 STRENGTH OF MATERIALS. 
 
 PART I. 
 
 SIMPLE STRESS. 
 
 i. Stress. When forces are applied to a body they tend in a 
 greater or less degree to break it. Preventing or tending to pre- 
 vent the rupture, there arise, generally, forces between every two 
 adjacent parts of the body. Thus, when a 
 load is suspended by means of an iron rod, 
 the rod is subjected to a downward pull at 
 its lower end and to an upward pull at its 
 upper end, and these two forces tend to pull 
 it apart. At any cross -section of the rod 
 the iron on either side "holds fast" to that on 
 the other, and these forces which the parts 
 of the rod exert upon each other prevent 
 the tearing of the rod. For example, in Fig. 
 1, let a represent the rod and its suspended 
 load, 1,000 pounds; then the pull on the 
 lower end equals 1,000 pounds. If we neg- 
 lect the weight of the rod, the pull on the 
 upper end is also 1,000 pounds, as shown in 
 Fig. 1 (5) ; and the upper part A exerts 
 on the lower part B an upward pull Q equal 
 to 1,000 pounds, while the lower part exerts 
 on the upper a force P also equal to 1,000 pounds. These two 
 forces, P and Q, prevent rupture of the rod at the "section" C; at 
 any other section there are two forces like P and Q preventing 
 rupture at that section. 
 
 By stress at a section of a body is meant the force which the 
 part of the body on either side of the section exerts on the other. 
 Thus, the stress at the section C (Fig. 1) is P (or Q), and it equals 
 1,000 pounds. 
 
 a. Stresses are usually expressed (in America) in pounds, 
 sometimes in tons. Thus the stress P in the preceding article is 
 
 9 
 
 Fig. 1.
 
 STRENGTH OF MATERIALS 
 
 1,000 pounds, or \ ton. Notice that this value has nothing to do 
 with the size of the cross-section on which the stress acts. 
 
 3. Kinds of Stress, (a) ' When the forces acting on a body 
 (as a rope or rod) are such that they tend to tear it, the stress at 
 any cross-section is called a tension or a tensile stress. The 
 stresses P and Q, of Fig. 1, are tensile stresses. Stretched ropes, 
 loaded "tie rods" of roofs and bridges, etc., are under tensile stress. 
 (b.) When the forces acting on a body (as a short post, brick, 
 etc.) are such that they tend to 
 crush it, the stress at any sec- 
 tion at right angles to the di- 
 rection of the crushing forces is 
 called a pressure or a compres- 
 sive stress. Fig. 2 (a) repre- 
 sents a loaded post, and Fig. 2 
 (5) the upper and lower parts. 
 The upper part presses down on 
 B, and the lower part presses up 
 on A, as shown. P or Q is 
 the compressive stress in the 
 post at section C. Loaded posts, 
 or struts, piers, etc., are under 
 compressive stress. 
 
 (c.) "When the forces acting 
 on a body (as a rivet in a bridge 
 joint) are such that they tend to cut or " shear " it across, the stress 
 at a section along which there is a tendency to cut is called a shear 
 or a shearing stress. This kind of stress takes its name from the 
 act of cutting with a pair of shears. In a material which is being 
 cut in this way, the stresses that are being " overcome " are shear- 
 ing stresses. Fig. 3 (a) represents a riveted joint, and Fig. 3 (J) 
 two parts of the rivet. The forces applied to the joint are such 
 that A tends to slide to the left, and B to the right; then B exerts 
 on A a force P toward the right, and A on B a force Q toward the 
 left as shown. P or Q is the shearing stress in the rivet. 
 
 Tensions, Compressions and Shears are called simple stresses. 
 forces may act upon a body so as to produce a combination of simple 
 stresses on some section ; such a combination is called a complex 
 
 r 
 
 f 
 
 
 ^ 
 
 A 
 
 
 
 
 C 
 
 T P 
 GU 
 
 B 
 b 
 
 ////w//////////////// 
 
 fr. 
 
 Fig. 2.
 
 STRENGTH OF MATERIALS 3 
 
 stress. The stresses in beams are usually complex. There are other 
 terms used to describe stress; they will be defined farther on. 
 
 4. Unit-Stress. It is often necessary to specify not merely 
 the amount of the entire stress which acts on an area, but also the 
 amount which acts on each unit of area (square inch for example). 
 By unit-stress is meant stress per unit area. 
 
 To find the value of a unit-stress: Divide the whole stress by 
 the whole area of the section on which it acts, or over which it is 
 distributed. Thus, let 
 
 P denote the value of the whole stress, 
 
 A the area on which it acts, and 
 
 S the value of the unit-stress; then 
 
 S=^-,alsoP = AS. (I) 
 
 Strictly these formulas apply only when the stress P is uniform, 
 
 Fig. 3. 
 
 that is, when it is uniformly distributed over the area, each square 
 inch for example sustaining the same amount of stress. When 
 the stress is not uniform, that is, when the stresses on different 
 square inches are not equal, then P-5-A equals the average value 
 of the unit-stress. 
 
 5. Unit-stresses are usually expressed (in America) in 
 pounds per square inch, sometimes in tons per square inch. If 
 P and A in equation 1 are expressed in pounds and square 
 inches respectively, then S will be in pounds per square inch ; and 
 if P and A are expressed in tons and square inches, S will be in 
 tons per square inch. 
 
 Examples. 1. Suppose that the rod sustaining the load in 
 Fig. 1 is 2 square inches in cross-section, and that the load weighs 
 1,000 pounds. What is the value of the unit-stress ?
 
 4 STRENGTH OF MATERIALS 
 
 Here P = 1,000 pounds, A= 2 square inches; hence. 
 
 1 000 
 S = ' " = 500 pounds per square inch. 
 
 2. Suppose that the rod is one-half square inch in cross-sec, 
 tion. What is the value of the unit-stress ? 
 
 A = -ysquare inch, and, as before, P= 1,000 pounds; hence 
 
 S = 1,000-; g- = 2,000 pounds per square inch. 
 
 Notice that one must always divide the whole stress by the area to get 
 the unit-stress, whether the area is greater or less than one. 
 
 6. Deformation. Whenever forces are applied to a body it 
 changes in size, and usually in shape also. This change of size 
 and shape is called deformation. Deformations are usually meas- 
 ured in inches; thus, if a rod is stretched 2 inches, the "elonga- 
 tion"= 2 inches. 
 
 7. Unit-Deformation. It is sometimes necessary to specify 
 not merely the value of a total deformation but its amount per 
 unit length of the deformed body. Deformation per unit length 
 of the deformed body is called unit-deformation. 
 
 To find the value of a unit-deformation: Divide the whole 
 deformation by the length over which it is distributed. Thus, if 
 D denotes the value of a deformation, 
 I the length, 
 * the unit-deformation, then 
 
 =-^, also D=fc. (2) 
 
 Both D and I should always be expressed in the same unit. 
 
 Example. Suppose that a 4-foot rod is elongated inch. 
 What is the value of the unit-deformation? 
 
 Here D=| inch, and Z=4 feet =48 inches; 
 hence *=|-=-48=-^ 8 - inch per inch. 
 That is, each inch \& elongated -^ inch. 
 
 Unit-elongations are sometimes expressed in per cent. To 
 express s^ elongation ia per cent: Divide the elongation in inches 
 by the original length in inches, and multiply ly 1QO. 
 
 8. Elasticity. Most solid bodies when deformed will regain 
 more or less completely their natural size and shape when the de.
 
 STRENGTH OF MATERIALS 5 
 
 forming forces cease to act. This property of regaining size and 
 shape is called elasticity. 
 
 "We may classify bodies into kinds depending on the degree 
 of elasticity which they have, thus : 
 
 1. Perfectly elastic bodies; these will regain their orig- 
 inal form and size no matter how large the applied forces are if 
 less than breaking values. Strictly there are no such materials, 
 but rubber, practically, is perfectly elastic. 
 
 2. Imperfectly elastic bodies; these will fully regain their 
 original form and size if the applied forces are not too large, and 
 practically even if the loads are large but less than the breaking 
 value. Most of the constructive materials belong to this class. 
 
 3. Inelastic or plastic bodies; these will not regain in the 
 least their original form when the applied forces cease to act. Clay 
 and putty are good examples of this class. 
 
 9. Hooke's Law, and Elastic Limit. If a gradually increas- 
 ing force is applied to a perfectly elastic material, the deformation 
 increases proportionally to the force; that is, if P and P' denote 
 two values of the force (or stress), and D and D' the values of the 
 deformation produced by the force, 
 
 thenP:P'::D:D'. 
 
 This relation is also true for imperfectly elastic materials, 
 provided that the loads P and P' do not exceed a certain limit depend- 
 ing on the material. Beyond this limit, the deformation increases 
 much faster than the load; that is, if within the limit an addition 
 of 1,000 pounds to the load produces a stretch of 0.01 inch, beyond 
 the limit an equal addition produces a stretch larger and usually 
 much larger than 0.01 inch. 
 
 Beyond this limit of proportionality a part of the deformation 
 is permanent; that is, if the load is removed the body only partially 
 recovers its form and size. The permanent part of a deformation 
 is called set. 
 
 The fact that, for most materials the deformation is propor- 
 tional to the load within certain limits, is known as Hooke's Law. 
 The unit-stress within which Hooke's Jaw holds, or above which 
 the deformation is not proportional to the load or stress, is called 
 elastic limit.
 
 6 STRENGTH OF MATERIALS 
 
 10. Ultimate Strength. By ultimate tensile, compressive, 
 or shearing strength of a material is meant the greatest tensile, 
 compressive, or shearing unit-stress which it can withstand. 
 
 As before mentioned, when a material is subjected to an in- 
 creasing load the deformation increases faster than the load beyond 
 the elastic limit, and much faster near the stage of rupture. Not 
 only do tension bars and compression blocks elongate and shorten 
 respectively, but their cross-sectional areas change also; tension 
 bars thin down and compression blocks "swell out" more or less. 
 The value of the ultimate strength for any material is ascertained 
 by subjecting a specimen to a gradually increasing tensile, com- 
 pressive, or shearing stress, as the case may be, until rupture oc- 
 curs, and measuring the greatest load. The "breaking load divided 
 1y the area of the original cross-section sustaining the stress, is the 
 value of the ultimate strength. 
 
 Example. Suppose that in a tension test of a wrought-iron 
 rod inch in diameter the greatest load was 12,540 pounds. What 
 is the value of the ultimate strength of that grade of wrought iron? 
 
 The original area of the cross-section of the rod was 
 0.7854 (diameter) 2 =0.7854X J=0.1964 square inches; hence 
 the ultimate strength equals 
 
 12,540-^-0.1964=63,850 pounds per square inch, (nearly). 
 ii. Stress-Deformation Diagram. A "test" to determine 
 the elastic limit, ultimate strength, and other information in re- 
 gard to a material is conducted by applying a gradually increasing 
 load until the specimen is broken, and noting the deformation cor- 
 responding to many values of the load. The first and second col- 
 umns of the following table are a record of a tension test on a steel 
 rod one inch in diameter. The numbers in the first column are 
 the values of the pull, or the loads, at which the elongation of 
 the specimen was measured. The elongations are given in the sec- 
 ond column. The numbers in the third and fourth columns are 
 the values of the unit-stress and unit-elongation corresponding to 
 the values of the load opposite to them. % The numbers in the 
 third column were obtained from those in the first by dividing 
 the latter by the area of the cross-section of the rod, 0.7854 
 square inches. Thus, 
 
 3,930-^0.7854=5,000 
 7,850-^0.7854=10,000, etc.
 
 STRENGTH OF MATERIALS 
 
 Total Pull 
 in pounds, P 
 
 Deformation 
 in inches, D 
 
 Unit-Stress in 
 pounds per 
 square inch, S 
 
 Unit- 
 Deformation, 
 s 
 
 3930 
 
 0.00136 
 
 5000 
 
 0.00017 
 
 7850 
 
 .00280 
 
 10000 
 
 .00035 
 
 11780 
 
 .00404 
 
 15000 
 
 .00050 
 
 15710 
 
 .00538 
 
 20000 
 
 .00067 
 
 19635 
 
 .00672 
 
 25000 
 
 .00084 
 
 23560 
 
 .00805 
 
 30000 
 
 .00101 
 
 27490 
 
 .00942 
 
 35000 
 
 .00118 
 
 31415 
 
 .01080 
 
 40000 
 
 .00135 
 
 35345 
 
 .01221 
 
 45000 
 
 .00153 
 
 39270 
 
 .0144 
 
 50000 
 
 .00180 
 
 43200 
 
 .0800 
 
 55000 
 
 .0100 
 
 47125 
 
 .1622 
 
 60000 
 
 .0202 
 
 51050 
 
 .201 
 
 65000 
 
 .0251 
 
 54980 
 
 .281 
 
 70000 
 
 .0351 
 
 58910 
 
 .384 
 
 75000 
 
 .048 
 
 62832 
 
 .560 
 
 80000 
 
 .070 
 
 65200 
 
 1.600 
 
 83000 
 
 .200 
 
 The numbers in the fourth column were obtained by dividing 
 those in the second by the length of the specimen (or rather the 
 length of that part whose elongation was measured), 8 inches - 
 Thus, 
 
 0.00136+8 = 0.00017, 
 .00280+8 = .00035, etc. 
 
 Looking at the first two columns it will be seen that the elonga- 
 tions are practically proportional to the loads up to the ninth load, 
 the increase of stretch for each increase in load being about 0.00135 
 inch; but beyond the ninth load the increases of stretch are much 
 greater. Hence the elastic limit was reached at about the ninth 
 load, and its value is about 45,000 pounds per square inch. The 
 greatest load was 65,200 pounds, and the corresponding unit-stress, 
 83,000 pounds per square inch, is the ultimate strength. 
 
 Nearly all the information revealed by such a test can be 
 well represented in a diagram called a stress-deformation, diagram. 
 It is made as follows: Lay off the values of the unit-deformation 
 (fourth column) along a horizontal line, according to some con- 
 venient scale, from some fixed point in the line. At the points on 
 the horizontal line representing the various unit-elongations, lay 
 off perpendicular distances equal to the corresponding unit-stresses. 
 Then connect .by a smooth curve the upper ends of all those dis- 
 tances, last distances laid off. Thus, for instance, the highest unit-
 
 8 STRENGTH OF MATERIALS 
 
 elongation (0.20) laid off from o (Fig. 4) fixes the point a, and a 
 perpendicular distance to represent the highest unit-stress (83,000) 
 fixes the point J. All the points so laid off give the curve ocb. The 
 part oc, within the elastic limit, is straight and nearly vertical 
 while the remainder is curved and more or less horizontal, especially 
 toward the point of rupture I. Fig. 5 is a typical stress-defor- 
 mation diagram for timber, cast iron, wrought iron, soft and hard 
 steel, in tension and compression. 
 
 12. Working Stress and Strength, and Factor of Safety. 
 The greatest unit-stress in any part of a structure when it is sus- 
 taining its loads is called the 
 working stress of that part. If 
 it is under tension, compression 
 and shearing stresses, then the 
 corresponding highest unit- 
 stresses in it are called its work- 
 ing stress in tension, in com- 
 pression, and in shear respect- 
 ively; that is, we speak of as 
 " many working stresses as it has 
 
 kinds of stress. 
 
 By working strength of a material to be used for a certain 
 purpose is meant the highest unit-stress to which the material 
 ought to be subjected when so used. Each material has a working 
 strength for tension, for compression, and for shear, and they are in 
 general different. 
 
 By factor of safety is meant the ratio of the ultimate strength 
 of a material to its working stress or strength. Thus, if 
 S u denotes ultimate strength, 
 S w denotes working stress or strength, and 
 f denotes factor of safety, then 
 
 / = ^;al 8 oS. = |. (3) 
 
 When a structure which has to stand certain loads is about 
 to be designed, it is necessary to select working strengths or fac- 
 tors of safety for the materials to be used. Often the selection is 
 a matter of great importance, and can be wisely performed only 
 by an experienced engineer, for this is a matter where hard-and-
 
 STRENGTH OF MATERIALS 
 
 9 
 
 fast rules should not govern but rather the judgment of the expert. 
 But there are certain principles to be used as guides in making a 
 selection, chief among which are: 
 
 1. The working strength should be considerably below the 
 elastic limit. (Then the deformations will bs small and not per- 
 manent.) 
 
 Pig. 5. (After Johnson.) 
 
 2. The working strength should be smaller for parts of a 
 structure sustaining varying loads than for those whose loads are 
 steady. (Actual experiments have disclosed the fact that the 
 strength of a specimen depends on the kind of load put upon it, 
 and that in a general way it is less the less steady the load is.) 
 
 3. The working strength must be taken low for non-uniform 
 material, where poor workmanship may be expected, when the
 
 10 
 
 STRENGTH OF MATERIALS 
 
 loads are uncertain, etc. Principles 1 and 2 have been reduced 
 to figures or formulas for many particular cases, but the third must 
 remain a subject for display of judgment, and even good guessing 
 in many cases. 
 
 The following is a table of factors of safety* which will be 
 used in the problems: 
 
 Factors of Safety. 
 
 Materials. 
 
 For steady 
 stress. 
 (Buildings.) 
 
 For varying 
 stress. 
 (Bridges.) 
 
 For shocks. 
 (Machines.) 
 
 Timber 
 Brick and stone 
 Cast iron 
 Wrought iron 
 Steel 
 
 8 
 15 
 
 6 
 4 
 5 
 
 10 
 25 
 15 
 6 
 
 7 
 
 15 
 30 
 20 
 10 
 15 
 
 They must be regarded as average values and are not to be 
 adopted in every case in practice. 
 
 Examples. 1. A wrought-iron rod 1 inch in diameter sus- 
 tains a load of 30,000 pounds. What is its working stress? If 
 its ultimate strength is 50 r OOO pounds per square inch, what is 
 its factor of safety ? 
 
 The area of the cross -section of the rod equals 0.7854 X (diam- 
 eter) 2 =0.7854 X 1 2 =0.7854 square inches. Since the whole stress 
 on the cross-section is 30,000 pounds, equation 1 gives for the 
 unit working stress 
 
 30 000 
 S = Q ' . = 38,197 pounds per square inch. 
 
 Equation 3 gives for factor of safety 
 -_ 60,000 
 '""88,197 ' 
 
 2. How large a steel bar or rod is needed to sustain a steady 
 pull of 100,000 pounds if the ultimate strength of the material is 
 65,000 pounds ? 
 
 The load being steady, we use a factor of safety of 5 (see table 
 above); hence the working strength to be used (see equation 3) is 
 
 fip\ /~vrv/\ 
 
 S = ^ = 13,000 pounds per square inch. 
 
 The proper area of the cross -section of the rod can now be com- 
 puted from equation 1 thus: 
 *Taken from Merriman's "Mechanics of Materials."
 
 STRENGTH OF MATERIALS 11 
 
 P 100 000 
 A = S = = 7 * 692 ""** inche8 ' 
 
 A bar 2x4 inches in cross-section would be a little stronger 
 than necessary. To find the diameter (<#) of a round rod of suffi- 
 cient strength, we write 0.7854: d? = 7.692, and solve the equation 
 fore?/ thus: 
 
 7 692 
 
 9 ' 794 ' r rf = 3 - 129 inches - 
 
 3. How large a steady load can a short timber post safely sus- 
 tain if it is 10x10 inches in cross-section and its ultimate com- 
 pressive strength is 10,000 pounds per square inch ? 
 
 According to the table (page 10) the proper factor of safety is 
 8, and hence the working strength according to equation 3 is 
 
 10,000 
 S = H = 1,250 pounds per square inch. 
 
 The area of the cross-section is 100 square inches; hence the safe 
 load (see equation 1) is 
 
 p = 100 X 1,250 = 125,000 pounds. 
 
 4. When a hole is punched through a plate the shearing 
 strength of the material has to be overcome. If the ultimate shear- 
 ing strength is 50,000 pounds per square inch, the thickness of the 
 plate -| inch, and the diameter of the hole | inch, what is the value 
 of the force to be overcome ? 
 
 The area shorn is that of the cylindrical surface of the hole 
 or the metal punched out; that is 
 
 3.1416 X diameter X thickness = 3.1416 X f X = 1.178 sq. in. 
 Hence, by equation 1, the total shearing strength or resistance 
 to punching is 
 
 P = 1.178 X 50,000= 58,900 pounds. 
 
 STRENGTH OF MATERIALS UNDER SIMPLE STRESS. 
 
 13. Materials in Tension. Practically the only materials 
 used extensively under tension are timber, wrought iron and steel, 
 and to some extent cast iron. 
 
 14. Timber. A successful tension test of wood is difficult, 
 as the specimen usually crushes at the ends when held in the test- 
 ing machine, splits, or fails otherwise than as desired. Hence the
 
 12 STRENGTH OF MATERIALS 
 
 tensile strengths of woods are not well known, but the following 
 may be taken as approximate average values of the ultimate 
 strengths of the woods named, when "dry out of doors." 
 
 Hemlock, 7,000 pounds per square inch. 
 
 White pine, 8,000 
 
 Yellow pine, long leaf, 12,000 
 " , short leaf, 10,000 
 
 Douglas spruce, 10,000 
 
 White oak, 12,000 
 
 Red oak, 9,000 
 
 15. Wrought Iron. The process of the manufacture of 
 wrought iron gives it a "grain," and its tensile strengths along and 
 across the grain are unequal, the latter being about three-fourths 
 of the former. The ultimate tensile strength of wrought iron 
 along the grain varies from 45,000 to 55,000 pounds per square 
 inch. Strength along the grain is meant when not otherwise 
 stated. 
 
 The strength depends on the size of the piece, it being greater 
 for small than for large rods or bars, and also for thin than for 
 thick plates. The elastic limit varies from 25,000 to 40,000 
 pounds per square inch, depending on the size of the bar or plate 
 even more than the ultimate strength. Wrought iron is very 
 ductile, a specimen tested in tension to destruction elongating from 
 5 to 25 per cent of its length. 
 
 16. Steel. Steel has more or less of a grain but is practically 
 of the same strength in all directions. To suit different purposes, 
 steel is made of various grades, chief among which may be men- 
 tioned rivet steel, sheet steel (for boilers), medium steel (for 
 bridges and buildings) , rail steel, tool and spring steel. In general, 
 these grades of steel are hard and strong in the order named, the 
 ultimate tensile strength ranging from about 50,000 to 160,000 
 pounds per square inch. 
 
 There are several grades of structural steel, which may be 
 described as follows:* 
 1. Rivet steel: 
 
 Ultimate tensile strength, 48,000 to 58,000 pounds per square inch. 
 Elastic limit, not less than one-half the ultimate strength. 
 Elongation, 26 per cent. 
 Bends 180 degrees flat on itself without fracture. 
 
 *Taken from " Manufacturer's Standard Specifications."
 
 STRENGTH OF MATERIALS 13 
 
 2. Sott steel: 
 
 Ultimate tensile strength, 52,000 to 62,000 pounds per square inch. 
 Elastic limit, not less than one-half the ultimate strength. 
 Elongation, 25 per cent. 
 Bends 180 degrees flat on itself. 
 
 3. Medium steel: 
 
 Ultimate tensile strength, 60,000 to 70,000 pounds per square inch. 
 Elastic limit, not less than one-half the ultimate strength. 
 Elongation, 22 per cant. 
 
 Bends 180 degrees to a diameter equal to the thickness of the 
 specimen without fracture. 
 
 17. Cast Iron. As in the case of steel, there are many 
 grades of cast iron. The grades are not the same for all localities 
 or districts, but they are based on the appearance of the fractures, 
 which vary from coarse dark grey to fine silvery white. 
 
 The ultimate tensile strength does not vary uniformly with 
 the grades but depends for the most part on the percentage of 
 "combined carbon" present in the iron. This strength varies from 
 15,000 to 35,000 pounds per square inch, 20,000 being a fair 
 average. 
 
 Cast iron has no well-defined elastic limit (see curve for cast 
 iron, Fig. 5). Its ultimate elongation is about one per cent. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. A steel wire is one-eighth inch in diameter, and the ulti- 
 mate tensile strength of the material is 150,000 pounds per square 
 inch. How large is its breaking load ? Ans. 1,845 pounds. 
 
 2. A wrought-iron rod (ultimate tensile strength 50,000 
 pounds per square inch) is 2 inches in diameter. How large a 
 steady pull can it safely bear ? Ans. 39,270 pounds. 
 
 18. Materials in Compression. Unlike the tensile, the 
 compressive strength of a specimen or structural part depends on 
 its dimension in the direction in which the load is applied, for, 
 in compression, a long bar or rod is weaker than a short one. At 
 present we refer only to the strength of short pieces such as do 
 not bend under the load, the longer ones (columns) being dis- 
 cussed farther on. 
 
 Different materials break or fail under compression, in two 
 very different ways: 
 
 1. Ductile materials (structural steel, wrought iron, etc.),
 
 14 STRENGTH OF MATERIALS 
 
 and wood compressed across the grain, do not fail by breaking Into 
 two distinct parts as in tension, but the former bulge out and 
 flatten under great loads, while wood splits and mashes down. 
 There is no particular point or instant of failure under increasing 
 loads, and such materials have no definite ultimate strength in 
 compression. 
 
 2. Brittle materials (brick, stone, hard steel, cast iron, etc. ) , 
 and wood compressed along the grain, do not mash gradually, but 
 fail suddenly and have a definite ultimate strength in compression. 
 Although the surfaces of fracture are always much inclined to the 
 direction in which the load ia applied (about 45 degrees), the ulti. 
 mate strength is computed by dividing the total breaking load by 
 the cross-sectional area of the specimen. 
 
 The principal materials used under compression in structural 
 work are timber, wrought iron, steel, cast iron, brick and stone. 
 
 19. Timber. As before noted, timber has no definite ulti- 
 mate compressive strength across the grain. The U. S. Forestry 
 Division has adopted certain amounts of compressive deformation 
 as marking stages of failure. Three per cent compression is 
 regarded as "a working limit allowable," and fifteen per cent as 
 "an extreme limit, or as failure." The following (except the first) 
 are values for compressive strength from the Forestry Division 
 Reports, all in pounds per square inch: 
 
 Ultimate strength 3# Compression 
 
 along the grain. across the grain 
 
 Hemlock 6,000 
 
 White pine 5,400 700 
 
 Long-leaf yellow pine 8,000 1,260 
 
 Short-leaf yellow pine 6, 500 1 , 050 
 
 Douglas spruce 5, 700 800 
 
 White oak 8,500 2,200 
 
 Red oak 7,200 2,300 
 
 20. Wrought Iron. The elastic limit of wrought iron, as be- 
 fore noted, depends very much upon the size of the bars or plate, it 
 being greater for small bars and thin plates. Its value for com- 
 pression is practically the same as for tension, 25,000 to 40,000 
 pounds per square inch. 
 
 21. Steel. The hard steels have the highest compressive 
 strength; there is a recorded value of nearly 400,000 pounds per 
 square inch, but 150,000 is probably a fair average.
 
 STRENGTH OF MATERIALS 15 
 
 The elastic limit in compression is practically the same as in 
 tension, which is about 60 per cent of the ultimate tensile strength, 
 or, for structural steel, about 25,000 to 42,000 pounds per square 
 inch. 
 
 22. Cast Iron. This is a very strong material in compres- 
 sion, in which way, principally, it is used structurally. Its ulti. 
 mate strength depends much on the proportion of "combined car- 
 bon" and silicon present, and varies from 50,000 to 200,000 pounds 
 per square inch, 90,000 being a fair average. As in tension, 
 there is no well-defined elastic limit in compression (see curve for 
 cast iron, Fig. 5). 
 
 23. Brick. The ultimate strengths are as various as the 
 kinds and makes of brick. For soft brick, the ultimate strength 
 is as low as 500 pounds per square inch, and for pressed brick it 
 varies from 4,000 to 20,000 pounds per square inch, 8,000 to 
 10,000 being a fair average. The ultimate strength of good pav- 
 ing brick is still higher, its average value being from 12,000 to 
 15,000 pounds per square inch. 
 
 24. Stone. Sandstone, limestone and granite are the 
 principal building stones. Their ultimate strengths in pounds 
 per square inch are about as follows: 
 
 Sandstone,* 5,000 to 16,000, average 8,000. 
 Limestone,* 8,000 " 16,000, " 10,000. 
 Granite, 14,000 " 24,000, " 16,000. 
 *Compression at right angles to the "bed" of the stone. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. A limestone 12x12 inches on its bed is used as a pier 
 cap, and bears a load of 120,000 pounds. What is its factor of 
 safety ? Ans. 12. 
 
 2. How large a post (short) is needed to sustain a steady 
 load of 100,000 pounds if the ultimate compressive strength of 
 the wood is 10,000 pounds per square inch ? Ans. 8X10 inches. 
 
 25. Materials in Shear. The principal materials used under 
 shearing stress are timber, wrought iron, steel and cast iron. 
 Partly on account of the difficulty of determining shearing 
 strengths, these are not well known. 
 
 26. Timber. The ultimate shearing strengths of the more 
 important woods along the grain are about as follows:
 
 16 STRENGTH OF MATERIALS 
 
 Hemlock, 300 pounds per square incK 
 
 White pine, 400 
 
 Long-leaf yellow pine, 850 
 
 Short-leaf " " 775 
 
 Douglas spruce, 500 
 
 White oak, 1,000 
 
 Red oak, 1,100 
 Wood rarely fails by shearing across the grain. Its ultimate 
 
 Fig. 6 a. Fig. 6 *. 
 
 shearing strength in that direction is probably four or five timss 
 the values above given. 
 
 27. Metals. The ultimate shearing strength of wrought 
 iron, steel, and cast iron is about 80 per cent of their respective 
 ultimate tensile strengths. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. How large a pressure P (Fig. 6 ) exerted on the shaded 
 area can the timber stand before it will shear off on the surface 
 abed, if ab = 6 inches and Ic = 10 inches, and the ultimate shear- 
 ing strength of the timber is 400 pounds per square inch ? 
 
 Ans. 24,000 pounds. 
 
 2. When a bolt is under tension, there is a tendency to tear 
 the bolt and to "strip" or shear off the head. The shorn area 
 would be the surface of the cylindrical hole left in the head. 
 Compute the tensile and shearing unit-stresses when P (Fig. 6 ) 
 equals 30,000 pounds, d = 2 inches, and t = 3 inches. 
 
 j Tensile unit-stress, 9,550 pounds per square inch. 
 ' ( Shearing unit-stress, 1,591 pounds per square inch. 
 
 REACTIONS OF SUPPORTS. 
 
 28. Moment of a Force. By moment of a force with re- 
 spect to a point is meant its tendency to produce rotation about 
 that point. Evidently the tendency depends on the magnitude of 
 the force and on the perpendicular distance of the line of action 
 of the force from the point : the greater the force and the per- 
 pendicular distance, the greater the tendency; hence the moment
 
 STRENGTH OF MATERIALS 
 
 17 
 
 of a force with respect to a point equals the product of the force 
 and the perpendicular distance from the force to the point. 
 
 The point with respect to which the moment of one or more 
 forces is taken is called an origin or center of moments, and the 
 perpendicular distance from an origin of moments to the line of 
 action of a force is called the arm of the force with respect to 
 that origin. Thus, if F t and F 2 (Fig. 7) are forces, their arms 
 with respect to O' are af and a 2 r respectively, and their moments 
 are F,'j and F 2 ' 2 . With respect to O" their arms are a" and 2 " 
 respectively, and their moments are F^" and F 2 2 ". 
 
 If the force is expressed in pounds and its arm in feet, the 
 moment is in foot-pounds; if the force is in pounds and the arm 
 in inches, the moment is in inch-pounds. 
 
 29. A sign is given to the moment of a force for conven- 
 ience; the rule used herein is as follows: The moment of a 
 force about a point is positive or negative according as it tends 
 to turn the body about that point in the clockwise or counter- 
 clockwise direction*. 
 
 Thus the moment (Fig. 7) 
 
 of F, about O' is negative, about O" positive; 
 F a Q' " , about O" negative. 
 
 30. Principle of Moments. In general, a single force of 
 proper magnitude and line of ac- 
 tion can balance any number of 
 
 forces. That single force is called 
 the equilibrant of the forces, and 
 the single force that would balance 
 the equilibrant is called the result- 
 ant of the forces. Or, otherwise 
 stated, the resultant of any num- 
 ber of forces is a force which pro- 
 duces the same effect. It can be 
 proved that The algebraic sum 
 of the moments of any number 
 of forces with respect to a point, 
 equals the moment of their re- Pig. 7. 
 
 sultant about that point. 
 
 *By clockwise direction is meant that in which the hands of a clock 
 rotate; and by counter-clockwise, the opposite direction.
 
 18 STRENGTH OF MATERIALS 
 
 This is a useful principle and is called "principle of moments." 
 
 31. All the forces acting upon a body which is at rest are 
 
 said to be balanced or in equilibrium. No force is required to 
 
 balance such forces and hence their equilibrant and resultant are 
 
 zero. 
 
 Since their resultant is zero, the algebraic sum of the mom- 
 
 oooft 
 
 IT 
 
 )s. aooolbs. eooollos. i 
 
 oooltos. 
 
 4 
 
 cT t fc u E' 
 
 : 2 i 
 
 A 
 
 fi. i 
 
 sooolfos. aooolbs. 
 
 B 
 
 Fig. 8. 
 
 ents of any number of forces which are balanced or in equilib- 
 rium equals zero. 
 
 This is known as the principle of moments for forces in 
 equilibrium; for brevity we shall call it also "the principle of 
 moments." 
 
 The principle is easily verified in a simple case. Thus, let 
 AB (Fig. 8) be a beam resting on supports at C and F. It is 
 evident from the symmetry of the loading that each reaction 
 equals one-half of the whole load, that is, of 6,000=3,000 
 pounds. (We neglect the weight of the beam for simplicity.) 
 
 With respect to C, for example, the momenta of the forces 
 are, taking them in order from the left: 
 
 1,000 X 4 = 4,000 foot-pounds 
 3,000 X 0= 
 
 2,000 x 2 = 4,000 
 2,000 X 14 = 28,000 
 3,000 X 16 = 48,000 
 1,000 X 20 = 20,000 
 The algebraic sum of these moments is seen to equal zero. 
 Again, with respect to B the moments are: 
 
 1,000 X 24 = 24,000 foot-pounds 
 
 3,000 X 20 = 60,000 
 2,000 X 18 = 36,000 
 2,000 X' 6 = 12,000 
 
 3,000 X 4= 12,000 
 
 1,000 X 0= 
 
 The sum of these moments also equals zero. In fact, no matter
 
 STRENGTH OF MATERIALS 19 
 
 where the center of moments is taken, it will be found in this and 
 any other balanced system of forces that the algebraic sum of their 
 moments equals zero. The chief use that we shall make of this 
 principle is in finding the supporting forces of loaded beams. 
 
 32. Kinds of Beams. A cantilever beam is one resting on 
 one support or fixed at one end, as in a wall, the other end being 
 free. 
 
 A simple ~beam is one resting on two supports. 
 
 A restrained beam is one fixed at both ends; a beam fixed at 
 one end and resting on a support at the other is said to be re- 
 strained at the fixed end and simply supported at the other. 
 
 A continuous beam; is one resting on more than two supports. 
 
 33. Determination of Reactions on Beams. The forces which 
 the supports exert on a beam, that is, the "supporting forces," are 
 called reactions. We shall deal chiefly with simple beams. The 
 reaction on a cantilever beam supported at one point evidently 
 equals the total load on the beam. 
 
 When the loads on a horizontal beam are all vertical (and 
 
 1000 Ibs. 2000 Ibs., 
 
 ~' s ' 
 
 Fig. 9. 
 
 this is the usual case), the supporting forces are also vertical and 
 the sum of the reactions equals the sum of the loads. This prin- 
 ciple is sometimes useful in determining reactions, but in the case 
 of simple beams the principle of moments is sufficient. The gen- 
 eral method of determining reactions is as follows: 
 
 1. Write out two equations of moments for all the forces 
 (loads and reactions) acting on the beam with origins of moments 
 at the supports. 
 
 2. Solve the equations for the reactions. 
 
 3* As a check, try if the sum of the reactions equals the 
 sum of the loads. 
 
 Examples. 1. Fig. 9 represents a beam supported at its 
 ends and sustaining three loads. We wish to find the reactions 
 due to these loads.
 
 20 STRENGTH OF MATERIALS 
 
 Let the reactions be denoted by R t and R 2 as shown; then 
 the moment equations are: 
 For origin at A, 
 
 1,000 X 1 + 2,000 X 6 + 3,000 X 8 R a x 10 = 0. 
 For origin at E, 
 
 zioolbs. 
 
 H- 
 
 IB 
 
 3600 ItoS. 
 
 
 
 leoolbs. 
 J,E 
 
 c Ic D 
 
 ID 
 
 
 K^ .. ^5 
 
 Fig. 10. 
 
 R, X 101,000 X 92,000 X 43,000 X 2 = 0. 
 The first equation reduces to 
 
 10 R 2 = 1,000+12,000+24,000 - 37,000; or 
 R 2= = 3,700 pounds. 
 The second equation reduces to 
 
 10 R 1= 9,000+8,000+6,000 = 23,000; or 
 B,= 2,300 pounds. 
 
 The sum of the loads is 6,000 pounds and the sum of the reactions 
 is the same; hence the computation is correct. 
 
 2. Fig. 10 represents a beam supported at B and D (that is, 
 it has overhanging ends) and sustaining three loads as shown. "We 
 wish to determine the reactions due to the loads. 
 
 Let R, and R 3 denote the reactions as shown; then the moment 
 equations are: 
 For origin at B, 
 
 -2,100x2+0+3,600x6 B, x 14+1,600 x 18 = 0. 
 For origin at D, 
 
 -2,100x16+^x143,600x8 + 0+1,600x4 = 0. 
 The first equation reduces to 
 
 14 R 2 = -4,200+ 21,600+ 28,800 = 46,200; or 
 R 2 = 3,300 pounds. 
 The second equation reduces to 
 
 14 B,= 33,600+28,800-6,400 = 56,000; or 
 R!= 4,000 pounds. 
 
 The sum of the loads equals 7,300 pounds and the sum of the 
 reactions is the same; hence the computation checks. 
 
 3. What are the total reactions in example 1 if the beam 
 weighs 400 pounds ?
 
 STRENGTH OF MATERIALS 21 
 
 (1.) Since we already know the reactions due to the loads 
 (2,300 and 3,700 pounds at the left and right ends respectively 
 (see illustration 1 above), we need only to compute the reactions 
 due to the weight of the beam and add. Evidently the reactions 
 due to the weight equal 200 pounds each; hence the 
 
 left reaction =2,300+200=2,500 pounds, and the 
 right -=3,700+200=3,900 . 
 
 (2.) Or, we might compute the reactions due to the loads 
 and weight of the beam together and directly. In figuring the 
 moment due to the weight of the beam, we imagine the weight 
 as concentrated at the middle of the beam; then its moments with 
 respect to the left and right supports are (400 X 5) and (400 X 5) 
 respectively. The moment equations for origins at A and E are 
 like those of illustration 1 except that they contain one more 
 term, the moment due to the weight; thus they are respectively: 
 
 1,000x1 + 2,000x6 + 3,000x8 R 2 X 10+400x5=0, 
 
 R! X 101,000 X 92,000 X 43,000 X 2400 X 5=0. 
 The first one reduces to 
 
 10 R 2 = 39,000, or R, = 3,900 pounds; 
 and the second to 
 
 10 R,= 25,000, or R 1= 2,500 pounds. 
 
 4. What are the total reactions in example 2 if the beam 
 weighs 42 pounds per foot ? 
 
 As in example 3, we might compute the reactions due to the 
 weight and then add them to the corresponding reactions due to 
 the loads (already found in example 2), but we shall determine 
 the total reactions due to load and weight directly. 
 
 The beam being 20 feet long, its weight is 42 X 20, or 840 
 pounds. Since the middle of the beam is 8 feet from the left and 
 6 feet from the right support, the moments of the weight with 
 to the left and right supports are respectively: 
 
 840x8 = 6,720, and 840x6 = 5,040 foot-pounds. 
 The moment equations for all the forces applied to the beam 
 for origins at B and ~D are like those in example 2, with an addi- 
 tional term, the moment of the weight; they are respectively: 
 2,100x2 + 0+3,600x6 R 2 x 14+1,600x18+6,720 = 0, 
 2,100x16+^x143,600x8+0+1,600x45,040 = 0.
 
 22 STRENGTH OF MATERIALS 
 
 The first equation reduces to 
 
 14 R 2 :=52,920, or R 2 =3,780 pounds, 
 and the second to 
 
 14 R,= 61,040, or R,= 4,360 pounds. 
 
 The sum of the loads and weight of beam is 8,140 pounds; 
 and since the sum of the reactions is the same, the computation 
 
 checks. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. AB (Fig. 11) represents a simple beam supported at its 
 
 ends. Compute the reactions, neglecting the weight of the beam, 
 
 ( Right reaction = 1,443.75 pounds. 
 
 Ans> ] Left reaction = 1,556.25 pounds. 
 
 eoolbs. soolbs. soolbs. looolbs. 
 
 4' 
 
 Fig. 11. 
 
 2. Solve example 1 taking into account the weight of the 
 beam, which suppose to be 400 pounds. 
 
 . ( Right reaction = 1,643.75 pounds. 
 3 ' } Left reaction = 1,756.25 pounds. 
 
 3. Fig. 12 represents a simple beam weighing 800 pounds 
 supported at A and B, and sustaining three loads as shown. 
 What are the reactions ? 
 
 A j Right reaction = 2,014.28 pounds. 
 v Left reaction = 4,785.72 pounds. 
 
 3OOO Ib3. 
 
 Fig. 12. 
 
 4. Suppose that in example 3 the beam also sustains a uni- 
 formly distributed load (as a floor) over its entire length, of 500 
 pounds per foot. Compute the reactions due to all the loads and 
 the weight of the beam. 
 
 Ana J Right reaction = 4,871.43 pounds. 
 | Left reaction = 11,928.57 pounds.
 
 STRENGTH OF MATERIALS 23 
 
 EXTERNAL SHEAR AND BENDING MOMENT. 
 
 On almost every cross-section of a loaded beam there are 
 three kinds of stress, namely tension, compression and shear. The 
 first two are often called fibre stresses because they act along the 
 real fibres of a wooden beam or the imaginary ones of which we 
 may suppose iron and steel beams composed. Before taking up 
 the subject of these stresses in beams it is desirable to study certain 
 quantities relating to the loads, and on which the stresses in a 
 beam depend. These quantities are called external shear and 
 bending moment, and will now be discussed. 
 
 34. External Shear. By external shear at (or for) any sec- 
 tion of a loaded beam is meant the algebraic sum of all the loads 
 (including weight of beam) and reactions on either side of the 
 section. This sum is called external shear because, as is shown 
 later, it equals the shearing stress (internal) at the section. For 
 brevity, we shall often say simply "shear" when external shear is 
 meant. 
 
 35. Rule of Signs. In computing external shears, it is cus- 
 tomary to give the plus sign to the reactions and the minus sign 
 to the loads. But in order to get the same sign for the external 
 shear whether computed from the right or left, we change the sign 
 of the sum when computed from the loads and reactions to the 
 right. Thus for section a of the beam in Fig. 8 the algebraic sum is, 
 when computed from the left, 
 
 -1,000+3,000= +2,000 pounds; 
 and when computed from the right, 
 
 -1,000 + 3,000-2,000-2,000 = -2,000 pounds. 
 The external shear at section a is +2,000 pounds. 
 
 Again, for section b the algebraic sum is, 
 when computed from the left, 
 
 -1,000+3,000-2,000-2,000+3,000 = + 1,000 pounds; 
 and when computed from the right, -1,000 pounds. 
 
 The external shear at the section is +1,000 pounds. 
 
 It is usually convenient to compute the shear at a section 
 from the forces to the right or left according as there are fewer 
 forces (loads and reactions) on the right or left sides of the 
 section.
 
 24 STRENGTH OF MATERIALS 
 
 36. Units foi^ Shears. It is customary to express external 
 shears in pounds, but any other unit for expressing force and 
 weight (as the ton) may be used. 
 
 37. Notation. We shall use Y to stand for external shear at 
 any section, and the shear at a particular section will be denoted 
 by that letter subscripted; thus V 15 V a , etc., stand for the shears 
 at sections one, two, etc., feet from the left end of a beam. 
 
 The shear has different values just to the left and right of a 
 support or concentrated load. We shall denote such values by V 
 and V"; thus Y 5 ' and Y 5 " denote the values of the shear at sec- 
 tions a little less and a little more than 5 feet from the left end 
 respectively. 
 
 Examples. 1. Compute the shears for sections one foot 
 apart in the beam represented in Fig. 9, neglecting the weight of 
 the beam. (The right and left reactions are 3,700 and 2,300 
 pounds respectively; see example 1, Art. 33.) 
 
 All the following values of the shear are computed from the 
 left. The shear just to the right of the left support is denoted by 
 Y ", and Y " = 2,300 pounds. The shear just to the left of B is 
 denoted by Y/, and since the only force to the left of the section 
 is the left reaction, Y/ = 2,300 pounds. The shear just to the 
 right of B is denoted by Y/', and since the only forces to the left 
 of this section are the left reaction and the 1,000-pound load, 
 V/' = 2,300 - 1,000 =1,300 pounds. To the left of all sections 
 between B and C, there are but two forces, the left reaction and 
 the 1,000-pound load; hence the shear at any of those sections 
 equals 2,300 - 1,000 = 1,300 pounds, or 
 
 Y 2 = Y 3 = Y 4 = Y 5 =, Y 6 '= 1,300 pounds. 
 
 The shear just to the right of C is denoted by Y 6 "; and since the 
 forces to the left of that section are the left reaction and the 
 1,000- and 2,000-pound loads, 
 
 Y 6 " = 2,300-1,000-2,001 -=- 700 pounds. 
 
 Without further explanation, thb student should understand 
 that 
 
 Y 7 = = + 2,300 - 1,000 - 2,000 = - 700 pounds, 
 
 Y; =-700, 
 
 Y 8 " = +2,300 - 1,000-2,000 - 3,000 =- 3,700, 
 
 V 9 =Y 10 '=- 3,700, 
 
 Y w "= 4-2,300-1,000-2,000 - 3,000 -{- 3,700=
 
 STRENGTH OF MATERIALS 2& 
 
 2. A simple beam 10 feet long, and supported at each end, 
 weighs 400 pounds, and bears a uniformly distributed load of 
 1,600 pounds. Compute the shears for sections two feet apart. 
 
 Evidently each reaction equals one-half the sum of the load 
 and weight of the beam, that is, \ (1,600+400) =1,000 pounds. 
 To the left of a section 2 feet from the left end, the forces acting 
 on the beam consist of the left reaction, the load on that part of 
 the beam, and the weight of that part ; then since the load and 
 weight of the beam per foot equal 200 pounds, 
 
 V 2 = 1,000-200 X 2 = 600 pounds. 
 
 To the left of a section four feet from the left end, the forces 
 are the left reaction, the load on that part of the beam, and the 
 weight ; hence 
 
 V 4 = 1,000-200 X 4 = 200 pounds. 
 
 Without further explanation, the student should see that 
 V 6 = 1,000-200 X 6 =-200 pounds, 
 V 8 = 1,000-200 X 8 = -6QO pounds, 
 V 10 ' = 1,000-200 X 10 === -1,000 pounds, 
 V IO "= 1,000-200x10+1,000 = 0. 
 
 3. Compute the values of the shear in example 1, taking 
 into account the weight of the beam (400 pounds). (The right 
 and left reactions are then 3,900 and 2,500 pounds respectively; 
 see example 3, Art. 33.) 
 
 We proceed just as in example 1, except that in each compu- 
 tation we include the weight of the beam to the left of the section 
 (or to the right when computing from forces to the right). The 
 weight of the beam being 40 pounds per foot, then (computing 
 from the left) 
 
 V "=+ 2,500 pounds, 
 
 V/ =+2,500-40 =+2,460, 
 
 V," =+2,500-40-1,000= + 1,460, 
 
 V 2 =+2,500-1,000-40x2 =+1,420, 
 
 V 3 + 2,500-1,000-40x3 =+1,380, 
 
 V 4 =+ 2,500-1,000-40 X 4 = + 1,340, 
 
 V 5 =+2,500-1,000-40x5 = +1,300, 
 
 V; =+2,500-1,000-40x6 = +1,260, 
 
 V 6 " =+2,500-1,000-40 X 6-2,000 = -740, 
 
 V 7 =+2,500-1,000-2,000-40x7 = -780,
 
 26 STRENGTH OF MATERIALS 
 
 V; = + 2,500-1,000-2,000-40 X 8 = -820, 
 y s " = + 2,500-1,000-2,000-40x8-3,000 =-3,820, 
 V 9 = + 2,500-1,000-2,000-3,000-40 X 9 = -3,860, 
 V 10 = + 2,500-1,000-2,000-3,000-40 X 10 = -3,900, 
 Y" 10 = + 2,500-1,000-2,000-3,000-40x10 + 3,900=0. 
 
 Computing from the right, we find, as before, that 
 
 V 7 =-(3,900-3,000-40x3) =-780 pounds, 
 Y/ =-(3.900-3,000-40 X 2) =-8 20, 
 Y 8 " =-(3,900-40 X2)=-3,820, 
 etc., etc. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Compute the values of the shear for sections of the beam 
 represented in Fig. 10, neglecting the weight of the beam. (The 
 right and left reactions are 3,300 and 4,000 pounds respectively; 
 see example 2, Art. 33.) 
 
 rV, =Y '=-2,iOO pounds, 
 Ans J Y " =V 3 "=Y 4 =Y 5 =Y 6 =Y 7 =Y' 8 = + 1,900, 
 
 1 V 8 " =Y 9 =Y 10 =Y U =Y 12 =Y 13 =Y U =Y 15 =Y' 16 =-1,700, 
 I V" V V Y V' 4-1 fiOO 
 
 ^ V 16 V 17 V 18 ' V 19 V 20 I ^"VU. 
 
 2. Solve the preceding example, taking into accoun-t the 
 weight of the beam, 42 pounds per foot. (The right and left 
 reactions are 3,780 and 4,360 pounds respectively; see example 4, 
 Art. 33.) 
 
 Y " = - 2,100 Ibs. Y 7 = + 1,966 Ibs. V 14 = - 1,928 Ibs. 
 Y, -- -2,142 Y 8 ' = + 1,924 Y 15 =- 1,970 
 Y 2 ' = - 2,184 Y 8 " = - 1,676 Y 16 ' = - 2,012 
 
 Ans. - 
 
 Y 2 "= + 2,176 Y 9 =-1,718 Y 16 "= + 1,768 
 
 Y 3 = + 2,134 Y 10 =- 1,760 Y 17 = + 1,726 
 
 Y 4 = + 2,092 Y u =-1,802 V 18 = + 1,684 
 
 Y 5 = + 2,050 Y 12 =- 1,844 Y 19 = + 1,642 
 
 I Y 6 = + 2,008 Y I3 =- 1,886 Y^' = + 1,600 
 
 3. Compute the values of the shear at sections one foot apart 
 in the beam of Fig. 11, neglecting the weight. (The right and 
 left reactions are 1,444 and 1,556 pounds respectively; see example 
 1, Art. 33.)
 
 STRENGTH OF MATERIALS 27 
 
 "V o " = v i =V/= + 1,556 pounds, 
 
 Ans.-! V 6 " =Y/= + 56, 
 
 V " V V V V V V ' 4-44- 
 
 V 7 V 8 V 9 V 10 V 11 V 12 V 13 -*44, 
 
 4. Compute the vertical shear at sections one foot apart in 
 the beam of Fig. 12, taking into account the weight of the beam, 
 800 pounds, and a distributed load of 500 pounds per foot. (The 
 right and left reactions are 4,870 and 11,930 pounds respectively; 
 see examples 3 and 4, Art. 33.) 
 
 V = V 7 = + 6,150 Ibs. V 15 =+ 830 Ibs 
 
 Vj' = - 540 Ibs. V 8 ' = + 5,610 V 18 =+ 290 
 VY'=- 2,540 V 8 " = + 4,610 V 17 ' = - 250 
 V 2 = - 3,080 V 9 =+4,070 V r /'== -3,250 
 
 Ans. -< 
 
 Y 3 = - 3,620 V 10 = + 3,530 Y 18 = - 3,790 
 
 V 4 = - 4,160 V n = +2,990 V M = - 4,330 
 
 V 5 = - 4,700 V u = +2,450 V w ' = - 4,870 
 
 V 6 ' = - 5,240 V u = + 1,910 V 20 "= 
 
 V 6 " = + 6,690 V I4 = + 1,370 
 
 38. Shear Diagrams. The way in which the external shear 
 varies from section to section in a beam can be well represented 
 by means of a diagram called a shear diagram. To construct 
 such a diagram for any loaded beam, 
 
 1. Lay off a line equal (by some scale) to the length of 
 the beam, and mark the positions of the supports and the loads. 
 (This is called a "base-line.") 
 
 2. Draw a line such that the distance of any point of it 
 from the base equals (by some scale) the shear at the correspond- 
 ing section of the beam, and so that the line is above the base 
 where the shear is positive, and below it where negative. (This is 
 called a shear line, and the distance from a point of it to the 
 base is called the "ordinate" from the base to the shear line at 
 that point.) 
 
 We shall explain these diagrams further by means of illus- 
 trative examples. 
 
 Examples. 1. It is required to construct the shear diagram 
 for the beam represented in Fig. 13, a (a copy of Fig. 9).
 
 28 
 
 STRENGTH OF MATERIALS 
 
 Lay off A'E' (Fig. 13, b) to represent the beam, and mark the 
 positions of the loads B', G' and D'. In example 1, Art. 37, we 
 computed the values of the shear at sections one foot apart; hence 
 we lay off ordinates at points on A'E' one foot apart, to represent 
 those shears. 
 
 Use a scale of 4,000 pounds to one inch. Since the shear for 
 any section in AB is 2,300 pounds, we draw a line db parallel 
 to the base 0.575 inch (2,300-^- 4,000) therefrom; this is the shear 
 line for the portion AB. Since the shear for any section in BC 
 equals 1,300 pounds, we draw a line Vc parallel to the base and 
 
 looollos. 
 
 aooplbs. aooolbs. 
 a' >k 2' 
 
 A B 
 
 ScaJe: a" AOOO Ibs. 
 
 Fig. 13. 
 
 0.325 inch (1,300-^4,000) therefrom; this is the shear line for the 
 portion BC. Since the shear for any section in CD is -70(1 
 pounds, we draw a line c'd below the base and 0.175 inch 
 (700 -=-4,000) therefrom; this is the shear line for the portion 
 CD. Since the shear for any section in DE equals -3,700 Ibs., we 
 draw a lined'e below the base and 0.925 inch (3,700-f-4,000) there- 
 from; this is the shear line for the portion DE. Fig. 13, 5, is the 
 required shear diagram. 
 
 2. It is required to construct the shear diagram for the 
 beam of Fig. 14, a (a copy of Fig. 9), taking into account the 
 weight of the beam, 400 pounds. 
 
 The values of the shear for sections one foot apart were com- 
 puted in example 3, Art. 37, so we have only to erect ordinates at 
 the various points on a base line A'E' (Fig. 14, b), equal to those
 
 STRENGTH OF MATERIALS 
 
 29 
 
 values. We shall use the same scale as in the preceding illustra- 
 tion, 4,000 pounds to an inch. Then the lengths of the ordinates 
 corresponding to the values of the shear (see example 3, Art. 37) 
 are respectively: 
 
 2,500-^4,000=0.625 inch 
 
 2,460-^-4,000=0.615 
 
 1,460-^-4,000=0.365 
 
 etc. etc. 
 
 Laying these ordinates off from the base (upwards or downwards 
 according as they correspond to positive or negative shears), we 
 get al>, I'c, c'd, and d'e as the shear lines. 
 
 ioooltas. aoooltos. aoooltas. 
 
 3. It is required to construct the shear diagram for the 
 cantilever beam represented in Fig. 15, <z, neglecting the weight 
 of the beam. 
 
 The value of the shear for any section in AB is 500 pounds ; 
 for any section in BC, -1,500 pounds; and for any section in 
 CD, - 3,500 pounds. Hence the shear lines are ab t Vc, c'd. The 
 scale being 5,000 pounds to an inch, 
 
 A.'a = 500^-5,000 = 0.1 inch, 
 
 B'J' = l,500-s-5,000 == 0.3 
 
 C'c' = 3,500-^5,000 = 0.7 
 
 Xhe shear lines are all below the base because all the values of th 
 shear are negative.
 
 30 
 
 STRENGTH OF MATERIALS 
 
 4. Suppose that the cantilever of the preceding illustration 
 sustains also a uniform load of 200 pounds per foot (see Fig. 16, a). 
 Construct a shear diagram. 
 
 Soolbs. loqolbs. sooolbs. 
 A* : ^ 
 
 First, we compute the values of the shear at several sections. 
 
 Thus V " - 500 pounds, 
 
 V, =- 500 - 200= - 700, 
 
 V 2 ' =-500 - 200X2=-900, 
 
 Y 2 " =_ 500 - 200X2 - 1,000=-1,900, 
 
 V 3 =-500 -1,000- 200 X3=-2,100, 
 
 V 4 =- 500 - 1,000 - 200 X 4=-2,300, 
 
 V 5 ' =-500 - 1,000 - 200X5=-2,500, 
 
 V 5 " =- 500 - 1,000 - 200X5 - 2,000=-4,500, 
 
 V 6 =- 500 - 1,000 - 2,000 - 200x6=-4,700, 
 
 V 7 =- 500 - 1,000 - 2,000 - 200 X 7=-4,900, 
 
 V 8 =- 500 - 1,000 - 2,000 - 200X8=-5,100, 
 
 Y 9 =- 500 - 1,000 - 2,000 - 200 X 9=-5,300. 
 
 The values, being negative, should be plotted downward. To a 
 
 scale of 5,000 pounds to the inch they give the shear lines ob, b'c, 
 
 <fd (Fig. 16, 5). 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Construct a shear diagram for the beam represented in 
 Fig. 10, neglecting the weight of the beam (see example 1, Art. 37). 
 
 2. Construct the shear diagram for the beam represented w 
 Fig. 11, neglecting the weight of the beam (see example 3, 
 Art. 37).
 
 STRENGTH OF MATERIALS 
 
 31 
 
 3. Construct the shear diagram for the beam of Fig. 12 
 when it sustains, in addition to the loads represented, its own 
 weight, 800 pounds, and a uniform load of 500 pounds per foot 
 (see example 4, Art. 37). 
 
 4. Figs, a, cases 1 and 2, Table B, represent two cantilever 
 beams, the first bearing a concentrated load P at the free end, 
 and the second a uniform load W. Figs, b are the corresponding 
 shear diagrams. Take P and W equal to 1,000 pounds, and satisfy 
 yourself that the diagrams are correct. 
 
 and satisfy yourself that the diagrams are correct. 
 
 5. Figs. , cases 3 and 4, same table, represent simple 
 beams supported at their ends, the first bearing a concentrated 
 
 aooolbs. 
 
 r - T ; 
 
 I 
 
 'A IB 
 i i 
 < i 
 
 !A' ,B' 
 
 C Df 
 C' D'| 
 
 gJuUIUliD 
 
 mil 
 
 i'50oolbs. d 
 
 Fig. 16. 
 
 load P at the middle, and the second a uniform load W. Figs. 
 b are the corresponding shear diagrams. Take P and W equal 
 to 1,000 pounds, and satisfy yourself that they are correct. 
 
 39. Maximum Shear. It is sometimes desirable to know 
 the greatest or maximum value of the shear in a given case. This 
 value can always be found with certainty by constructing the shear 
 diagram, from which the maximum value of the shear is evident at 
 a glance. In any case it can most readily be computed if one 
 knows the section for which the shear is a maximum. The stu- 
 dent should examine all the shear diagrams in the preceding 
 articles and those that he has drawn, and see that 
 
 1. In cantilevers fixed in a wall % the maadmum, shear 
 occurs at the wall.
 
 32 STRENGTH OF MATERIALS 
 
 2. In simple beams, the maximum shear occurs at a sec- 
 tion next to one of the supports. 
 
 By the use of these propositions one can determine the value 
 of the maximum shear without constructing the whole shear 
 diagram. Thus, it is easily seen (referring to the diagrams, page 
 53) that for a 
 
 Cantilever, end load P, maximum shear=P 
 
 " , uniform load W, " " =W 
 
 Simple beam, middle load P, " " =|P 
 
 " " , uniform " W, " " =i w 
 
 40. Bending rioment. By bending moment at (or for) a 
 section of a loaded beam, is meant the algebraic sum of the mo- 
 ments of all the loads (including weight of beam) and reactions 
 to the left or right of the section with respect to any point in the 
 section. 
 
 41. Rule of Signs. We follow the rule of signs previously 
 stated (Art. 29) that the moment of a force which tends to pro- 
 duce clockwise rotation is plus, and that of a force which tends to 
 produce counter-clockwise rotation is minus; but in order to get 
 the same sign for the bending moment whether computed from 
 the right or left, we change the sign of the sum of the moments 
 when computed from the loads and reactions on the right. Thus 
 for section a, Fig. 8, the algebraic sums of the moments of the 
 forces are : 
 
 when computed from tne left, 
 
 -1,000 X 5 + 3,000 X 1 =-2,000 foot-pounds ; 
 and when computed from the right, 
 
 1,000 X 19-3,000 X 15 + 2,000 X 13 + 2,000 X 1 = + 2,000 foot- 
 
 pounds. 
 The bending moment at section a is -2,000 foot-pounds. 
 
 Again, for section J, the algebraic sums of the moments of the 
 forces are: 
 when computed from the left, 
 
 -1,000 X 22+3,000 X 18-2,000 X 16-2,000 X 4+ 3,000 X 2= 
 
 -2,000 foot-pounds ; 
 and when computed from the right, 
 
 1,000x2= + 2,000 foot-pounds. 
 The bending moment at the section is -2,000 foot-pounds.
 
 STRENGTH OF MATERIALS 33 
 
 It is usually convenient to compute the bending moment for 
 a section from the forces to the right or left according as there 
 are fewer forces (loads and reactions) on the right or left side 
 of the section. 
 
 42. Units. It is customary to express bending moments in 
 inch -pounds, but often the foot-pound unit is more convenient. 
 To reduce foot-pounds to inch-pounds, multiply l>y twelve. 
 
 43. Notation. We shall use M to denote bending moment at 
 any section, and the bending moment at a particular section will 
 be denoted by that letter subscripted; thus M,, M 2 , etc., denote 
 values of the bending moment for sections one, two, etc., feet 
 from the left end of the beam. 
 
 Examples. 1. Compute the bending moments for sections 
 one foot apart in the beam represented in Fig. 9, neglecting the 
 weight of the beam. (The right and left reactions are 3,700 and 
 2,300 pounds respectively. See example 1, Art. 33.) 
 
 Since there are no forces acting on the beam to the left of the 
 left support, M (p =0. To the left of the section one foot from the 
 left end there is but one force, the left reaction, and its arm is one 
 foot; hence M,= + 2,300x1 =2,300 foot-pounds. To the left of 
 a section two feet from the left end there are two forces, 2,300 and 
 1,000 pounds, and their arms are 2 feet and 1 foot respectively; 
 hence M 2 = + 2,300x2-1,000x1=3,600 foot-pounds. At the 
 left of all sections between B and C there are only two forces, 
 2,300 and 1,000 pounds; hence 
 
 M 3 = + 2,300 X 3-1,000 X 2= + 4,900 foot-pounds, 
 
 M 4 = + 2,300 X 4-1,000 X 3= + 6,200 
 
 M 5 = + 2,300 X 5-1,000 X 4= + 7,500 
 
 M 6 = + 2,300 X 6-l,OOOx 5= + 8,800 
 
 To the right of a section seven feet from the left end there 
 are two forces, the 3,000-pound load and the right reaction 
 (3,700 pounds), and their arms with respect to an origin in that 
 section are respectively one foot and three feet; hence 
 
 M 7 =-(-3,700 X 3 + 3,000 X 1) = + 8,100 foot-pounds. 
 
 To the right of any section between E and D there is only one 
 force, the right reaction ; hence
 
 34 STRENGTH OF MATERIALS 
 
 M 8 =-(-3,700x 2)=7,400 foot-pounds, 
 M 9 =-(-3,700 X 1)=3,700 
 
 Clearly M 10 =0. 
 
 2. A simple beam 10 feet long and supported at its ends 
 weighs 400 pounds, and bears a uniformly distributed load of 1,600 
 pounds. Compute the bending moments for sections two feet 
 apart. 
 
 Each reaction equals one-half the whole load, that is, -| of 
 (1,600+400) =1,000 pounds, and the load per foot including 
 weight of the beam is 200 pounds. The forces acting on the 
 beam to the left of the first section, two feet from the left end, are 
 the left reaction (1,000 pounds) and the load (including weight) 
 on the part of the beam to the left of the section (400 pounds). 
 The arm of the reaction is 2 feet and that of the 400-pound force 
 is 1 foot (the distance from the middle of the 400-pound load to 
 the section). Hence 
 
 M 2 = + 1,000 X 2-400 X 1 = + 1,600 foot-pounds. 
 
 The forces to the left of the next section, 4 feet from the left 
 end, are the left reaction and all the load (including weight of 
 beam) to the left (800 pounds). The arm of the reaction is 4 feet, 
 and that of the 800-pound force is 2 feet; hence 
 
 M 4 = + 1,000 X 4-800 X 2= + 2,400 foot-pounds. 
 Without further explanation the student should see that 
 
 M u = + 1,000 X 6-1,200 X 3= + 2,400 foot-pounds, 
 M 8 = + 1,000 X 8-1,600 X 4= + 1,600 
 Evidently M =M 10 =0. 
 
 3. Compute the values of the bending moment in example 
 1, taking into account the weight of the beam, 400 pounds. (The 
 right and left reactions are respectively 3,900 and 2,500 pounds; 
 see example 3, Art. 33.) 
 
 We proceed as in example 1, except that the moment 
 of the weight of the beam to the left of each section (or to 
 the right when computing from forces to the right) must be 
 included in the respective moment equations. Thus, computing 
 from the left,
 
 STRENGTH OF MATERIALS 35 
 
 M =0 
 
 M, = + 2,500 X 1-40 X |= + 2,480 foot-pounds, 
 M 2 = + 2,500x2-l,OOOxl-80xl= + 3,920, 
 M 3 = + 2,500 X 3-1,000 X 2-120 X 1|= + 5,320, 
 H 4 = + 2,500x4-l,OOOx3-160x2=+6,680, 
 M 5 = + 2,500 X 5-1,000 X 4-200 X 2|= + 8,000, 
 M 6 = + 2,500 X 6-1,000 X 5-240 X 3= + 9,280. 
 
 Computing from the right, 
 
 M 7 =-(-3,900x3+3,OOOxl + 120xl|)=: + 8,520, 
 M 8 =-(-3,900x2 + 80xl) = + 7,720, 
 M 9 =-(-3,900xl + 40x|)=+3,880, 
 
 M 10 = 0. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Compute the values of the bending moment for sections 
 one foot apart, beginning one foot from the left end of the 
 beam represented in Fig. 10, neglecting the weight of the beam. 
 (The right and left reactions are 3,300 and 4,000 pounds respec- 
 tively; see example 2, Art. 33.) 
 
 (M,= - 2,100 M 6 = + 3,400 M u = + 2,100 M 16 =-6,400 
 M 2 = - 4,200 M 7 = + 5,300 M 12 = + 400 M 1T =-4,800 
 M 3 = - 2,300 M. = + 7,200 M w = - 1,300 M 18 =-3,200 
 M 4 = - 400 M 9 = + 5,500 M M = - 3,000 M 19 =-l,600 
 M 5 = + 1,500 M 10 = + 3,800 M 15 = - 4,700 M 20 = 
 
 2. Solve the preceding example, taking into account the 
 weight of the beam, 42 pounds per foot. (The right and left 
 reactions are 3,780 and 4,360 pounds respectively; see example 4, 
 Art. 33.) 
 
 ~ t = - 2,121 M 6 =+4,084 M u = + 2,799 M,.= - 6,736 
 
 Ans. 
 (in foot- -< 
 pounds) 
 
 M 2 =- 4,284 M 7 = + 6,071 M 12 =+ 976 M 17 = - 4,989 
 M 3 = - 2,129 M 8 = + 8,016 M 13 = - 889 M 18 = - 3,284 
 
 4 = - 16 M 9 = + 6,319 M 14 = - 2,796 M 19 = - 1,621 
 M 5 = + 2,055 M 10 = + 4,580 M 15 = - 4,745 M M = 
 
 3. Compute the bending moments for sections one foot 
 apart, of the beam represented in Fig. 11, neglecting the weight. 
 (The right and left reactions are 1,444 and 1,556 pounds respect- 
 ively; Bee example 1, Art. 33.)
 
 36 
 
 STRENGTH OF MATERIALS 
 
 fM^H- 1,556 M 5 = + 5,980 M 9 -= + 6,104 M u = +4,328 
 An8 ' I M.^+8,112 M 6 =+6,936 M 10 =+5,660 M u =+2,884 
 
 pounds^ M 3 =+4,068 M 7 = + 6,992 M u = + 5,216 M 15 = + 1,440 
 ' [M 4 = + 5,024 M 8 = + 6,548 M 12 = + 4,772 M 16 = 
 
 4 Compute the bending moments at sections one foot apart 
 in the beam of Fig. 12, taking into account the weight of the beam, 
 800 pounds, and a uniform load of 500 pounds per foot. (The 
 right and left reactions are 4,870 and 11,930 pounds respectively; 
 see Exs. 3 and 4, Art. 33.) 
 
 ~M,=- 270 M 6 =-19,720 M u =+ 3,980 M 16 =12,180 
 
 Ans. M 2 = - 3,080 M 7 = -13,300 M 12 + 6,700 M n = 12,200 
 
 (in foot J M,= - 6,430 M 8 =- 7,420 M 13 = + 8,880 M 18 = 8,680 
 
 pounds) M 4 = -10,320 M 9 =- 3,080 M M = + 10,520 M 19 ^ 4,620 
 
 M 8 = -14,750 M M = + 720 M 15 = + 11,620 M 20 = 
 
 44. Moment Diagrams. The way in which the bending 
 moment varies from section to section in a loaded beam can be 
 well represented by means of a diagram called a moment diagram,. 
 To construct such a diagram for any loaded beam, 
 
 1000 Ibs. 
 
 2 ooo Ibs. 3000 Ibs. 
 
 2' * 2'- 
 
 A B' C 1 
 
 ScaJc;i 10600 ft.- Ibs 
 Fig. 17. 
 
 1. Lay off a base-line just as for a shear diagram (see 
 Art. 38). 
 
 2. Draw a line such that the distance from any point of it 
 to the base-line equals (by some scale) the value of the bending 
 moment at the corresponding section of the beam, and so that the 
 line is above the base where the bending moment is positive and 
 below it where it is negative. (This line is called a "moment
 
 STRENGTH OF MATERIALS 37 
 
 Examples. 1. It is required to construct a moment dia- 
 gram for the beam of Fig. 17, a (a copy of Fig. 9), loaded as 
 there shown. 
 
 Layoff A'E' (Fig. 17, 3) as a base. In example 1, Art. 43, 
 we computed the values of the bending moment for sections one 
 foot apart, so we erect ordinates at points of A'E' one foot apart, 
 to represent the bending moments. 
 
 We shall use a scale of 10,000 foot-pounds to the inch; then 
 <he ordinates (see example 1, Art. 43, for values of M) will be: 
 One foot from left end, 2,300-=- 10,000 = 0.23 inch, 
 Two feet " 3,600-^10,000 = 0.36 
 Three" " 4,900-7-10,000 = 0.49 
 Four " 6,200-7-10,000 = 0.62 
 etc., etc. 
 
 1C 
 
 U i' 
 
 loolbs- 
 
 1- 3 1 
 
 zooolbs. 
 
 3000 IbS. 
 
 l 1 
 
 j 
 
 ,' 
 
 | 3 
 
 i 
 
 - J 
 
 l 
 
 j 
 
 IB 
 
 iC 
 1 
 
 JD 
 
 1 
 
 i 
 
 C' D 1 E' 
 
 ScoUe: l"- 10000 ft.-ltos. 
 Fig. 18. 
 
 Laying these ordinates off, and joining their ends in succession, 
 we get the line A' WE', which is the bending moment line. 
 Fig. 17, , is the moment diagram. 
 
 2. It is required to construct the moment diagram for the 
 beam, Fig. 18, a (a copy of Fig. 9), taking into account the weight 
 of the beam, 400 pounds. 
 
 The values of the bending moment for sections one foot apart 
 were computed in example 3, Art. 43. So we have only to lay off 
 ordinates equal to those values, one foot apart, on the base A'E' 
 (Fig. 18, b). 
 
 To a scale of 10,000 foot-pounds to the inch the ordinates 
 (see example 3, Art. 43, for values of M) are: 
 
 46751
 
 38 
 
 STRENGTH OF MATERIALS 
 
 At left end, 
 
 One foot from left end, 2,480-^10,000=0.248 inch 
 Two feet " " 3,920-^-10,000=0.392 
 Three " " " 5,320^10,000=0.532 
 Four " " " 6,680^-10,000=0.668 " 
 Laying these ordinates off at the proper points, we get 
 as the moment line. 
 
 3. It is required to construct the moment diagram for the 
 cantilever beam represented in Fig. 19, a, neglecting the weight 
 of the beam. The bending moment at B equals 
 
 -500x2=-l,000 foot-pounds; 
 at C, 
 
 -500 X 5-1,000 X 3=-5,500 ; 
 and at D, 
 
 -500 X 9-l,OOOX 7-2,000 X 4=-19,500. 
 
 500 Ibs. looolbs. 2060 lias. 
 
 ScaJc:i= aooooft.- 
 
 Fig. 19. 
 
 Using a scate of 20,000 foot-pounds to one inch, the ordinatea 
 in the bending moment diagram are: 
 
 AtB, 1,000-*- 20,000=0.05 inch, 
 C, 5,500--20,OCO=0.275 
 D, 19,500^-20,000=0.975 
 
 Hence we lay these ordinates off, and downward because the bend- 
 ing moments are negative, thus fixing the points 5, c and d. The 
 bending moment at A is zero; hence the moment line connects A 
 b y c and d. Further, the portions A5, be and cd are straight, as 
 can be shown by computing values of the bending moment for 
 sections in AB, BC and CD, and laying off the corresponding 
 ordinates in the moment diagram.
 
 STRENGTH OF MATERIALS 
 
 39 
 
 4. Suppose that the cantilever of the preceding illustration 
 sustains also a uniform load of 100 pounds per foot (see Fig. 20, a). 
 Construct a moment diagram. 
 
 First, we compute the values of the bending moment at sev- 
 eral sections; thus, 
 
 M 1= -500 X 1-100 X J=-550 foot-pounds, 
 M 2 =-500 X 2-200 X 1 =-1,200, 
 M 3 =-500 X 3-1,000 X 1-300 X 1 J=-2,950, 
 M 4 =-500 X 4-1,000 X 2-400 X 2=-4,800, 
 M 5 =-500 X 5-1,000 X 3-500 X 2 J=- 6,750, 
 M 6 =-500 X 6-1,000 X 4-2,000 X 1-600 X 3=-10,800, 
 M 7 =-500 X 7-1,000 x 5-2,000 X 2-700 X 3J=-14,950, 
 M,=-500 X 8-1,000 X 6-2,000 X 3-800 X 4=-19,200, 
 M,=-600 X 9-1,000 X 7-2,000 X 4-900 X 4J=-23,550. 
 
 500 Ibs. looo Ibs. 
 
 L_ 2 - 
 A c= 
 
 2000 Ibs. 
 
 ScaJe: f aoooof t-lba. 
 
 Fig. 20. 
 
 These values all being negative, the ordinates are all laid off 
 downwards. To a scale of 20,000 foot-pounds to one inch, they 
 fix the moment line A.'bcd. 
 
 EXAflPLES FOR PRACTICE. 
 
 1. Construct a moment diagram for the beam represented in 
 Fig. 10, neglecting the weight of the beam. (See example 1, 
 Art. 43). 
 
 2. Construct a moment diagram for the beam represented 
 in Fig. 11, neglecting the weight of the beam. (See example 3, 
 Art. 43). 
 
 3. Construct the moment diagram for the beam of Fig. 12
 
 40 STRENGTH OF MATERIALS 
 
 when it sustains, in addition to the loads represented and its own 
 weight (800 pounds), a uniform load of 500 pounds per foot. 
 (See example 4, Art. 43.) 
 
 4. Figs, a, cases 1 and 2, page 53, represent two cantilever 
 beams, the first bearing a load P at the free end, and the second 
 a uniform load W. Figs, c are the corresponding moment 
 diagrams. Take P and W equal to 1,000 pounds, and I equal to 
 10 feet, and satisfy yourself that the diagrams are correct. 
 
 5. Figs. #, cases 3 and 4, page 53, represent simple beams 
 on end supports, the first bearing a middle load P, and the other a 
 uniform load W. Figs, c are the corresponding moment dia- 
 grams. Take P and "W" equal to 1,000 pounds, and I equal to 
 10 feet, and satisfy yourself that the diagrams are correct. 
 
 45. Maximum Bending Moment. It is sometimes desirable 
 to know the greatest or maximum value of the bending moment 
 in a given case. This value can always be found with certainty 
 by constructing the moment diagram, from which the maximum 
 value of the bending moment is evident at a glance. But in any 
 case, it can be most readily computed if one knows the section for 
 which the bending moment is greatest. If the student will com. 
 pare the corresponding shear and moment diagrams which have 
 been constructed in foregoing articles (Figs. 13 and 17, 14 and 
 18, 15 and 19, 16 and 20), and those which he has drawn, he will 
 see that The maximum bending moment in a beam occurs 
 where the shear changes sign. 
 
 By the help of the foregoing principle we can readily com- 
 pute the maximum moment in a given case. We have only to 
 construct the shear line, and observe from it where the shear 
 changes sign ; then compute the bending moment for that section. 
 If a simple beam has one or more overhanging ends, then the shear 
 changes sign more than once twice if there is one overhanging 
 end, and three times if two. In such cases we compute the 
 bending moment for each section where the shear changes sign; 
 the largest of the values of these bending moments is the maxi- 
 mum for the beam. 
 
 The section of maximum bending moment in a cantilever 
 fixed at one end (as when built into a wall) is always at the wall.
 
 STRENGTH OF MATERIALS 41 
 
 Thus, without reference to the moment diagrams, it is readily seen 
 foat, 
 
 for a cantilever whose length is l y 
 
 with an end load P, the maximum moment is PZ, 
 " a uniform " W, " " " J ~W7. 
 
 Also by the principle, it is seen that, 
 
 for a beam whose length is Z, on end supports, 
 
 with a middle load P, the maximum moment is J PZ, 
 " uniform " W, " " " % "W7. 
 
 46. Table of Maximum Shears, Moments, etc. Table B 
 on page 53 shows the shear and moment diagrams for eight 
 simple cases of beams. The first two cases are built-in cantilevers: 
 the next four, simple beams on end supports; and the last two, 
 restrained beams built in walls at each end. In each case I 
 denotes the length. 
 
 CENTER OF GRAVITY AND HOMENT OF INERTIA. 
 
 It will be shown later that the strength of a beam depends 
 partly on the form of its cross-section. The following discussion 
 relates principally to cross -sections of beams, and the results 
 reached (like shear and bending moment) will be made use of 
 later in the subject of strength of beams. 
 
 47. Center of Gravity of an Area. The student probably 
 knows what is meant by, and how to find, the center of gravity of 
 any flat disk, as a piece of tin. Probably his way is to balance 
 the piece of tin on a pencil point, the point of the tin at which it so 
 balances being the center of gravity. (Really it is midway between 
 the surfaces of the tin and over the balancing point.) The center 
 of gravity of the piece of tin, is also that point of it through which 
 the resultant force of gravity on the tin (that is, the weight of the 
 piece) acts. 
 
 By "center of gravity" of a plane area of any shape we mean 
 that point of it which corresponds to the center of gravity of a 
 piece of tin when the latter is cut out in the shape of the area. 
 The center of gravity of a quite irregular area can be found most 
 readily by balancing a piece of tin or stiff paper cut in the shape 
 of the area. But when an area is simple in shape, or consists of 
 parts which are simple, the_center of gravity of the whole can be
 
 42 STRENGTH OF MATERIALS 
 
 found readily by computation, and such a method will now be 
 described. 
 
 48. Principle of moments Applied to Areas. Let Fig. 21 
 represent a piece of tin which has been divided off into any num- 
 ber of parts in any way, the weight of the whole being W, and 
 that of the parts W,, W 2 , "W 3 , etc. Let C,, C 2 , C,, etc., be the 
 centers of gravity of the parts, C that of the whole, and c u e n 
 etc., and o the distances from those centers of gravity respectively 
 to some line (L L) in the plane 
 of the sheet of tin. When the 
 tin is lying in a horizontal posi- 
 tion, the moment of the weight 
 of the entire piece about L L is 
 We, and the moments of the 
 parts are W^, W 2 c 2 , etc. Since 
 the weight of the whole is the 
 resultant of the weights of the 21 
 
 parts, the moment of the weight 
 of the whole equals the sum of the moments of the weights of the 
 parts ; that is, 
 
 Wc=W 1 c 1 +W a c a +etc 
 
 Now let AU A 2 , etc. denote the areas of the parts of the pieces 
 of tin, and A the area of the whole; then since the weights are 
 proportional to the areas, we can replace the W's in the preceding 
 equation by corresponding A's, thus: 
 
 Ac=A lCl + A 2 c 2 -f etc (4) 
 
 If we call the product of an area and the distance of its 
 center of gravity from some line in its plane, the "moment" of the 
 area with respect to that line, then the preceding equation may be 
 stated in words thus: 
 
 The moment of an area with respect to any line equals the 
 algebraic sum of the moments of the parts of the area. 
 
 If all the centers of gravity are on one side of the line with 
 respect to which moments are taken, then all the moments should be 
 given the plus sign; but if some centers of gravity are on one side 
 and some on the other side of the line, then the moments of the 
 areas whose centers of gravity are on one side should be given the
 
 STRENGTH OF MATERIALS 43 
 
 same sign, and the moments of the others the opposite sign. The 
 foregoing is the principle of moments for areas, and it is the basis 
 of all rules for finding the center of gravity of an area. 
 
 To find the center of gravity of an area which can be divided 
 np into simple parts, M*e write the principle in forms of equations 
 for two different lines as "axes of moments," and then solve the 
 equations for the unknown distances of the center of gravity of the 
 whole from the two lines. We explain further by means of specific 
 examples. 
 
 Examples. 1. It is required to find the center of gravity 
 of Fig. 22, a, the width being uniformly one inch. 
 
 The area can be divided into two rectangles. Let C t and 
 
 .C 
 
 T^n 
 .1: 
 
 1 
 1 
 
 OH a H 
 
 O. To 
 
 Fig. 22. 
 
 C 2 be the centers of gravity of two such parts, and C the center of 
 gravity of the whole. Also let a and I denote the distances of 
 from the two lines OL' and OL" respectively. 
 
 The areas of the parts are 6 and 3 square inches, and their 
 arms with respect to OL' are 4 inches and -| inch respectively, and 
 with respect to OL" -| inch and 1^ inches. Hence the equations of 
 momenta with respect to OL' and OL" (the whole area being 9 
 square inches) are: 
 
 Hence, a 25.5-S-9 = 2.83 inches, 
 
 1 = 7.5^-9 = 0.83 . 
 
 2. It is required to locate the center of gravity of Fig. 22, 
 the width being uniformly one inch.
 
 44 
 
 STRENGTH OF MATERIALS 
 
 The figure can be divided up into three rectangles. Let C 1? C 2 
 and C 3 be the centers of gravity of such parts, C the center of 
 gravity of the whole; and let a denote the (unknown) distance of 
 from the base. The areas of the parts are 4, 10 and 4 square 
 inches, and their " arms " with respect to 
 the base are 2, -| and 2 inches respectively; 
 hence the equation of moments with re- 
 spect to the base (the en tire area being 18 
 square inches) is: 
 
 18 X= 4X2+10X4-HX2 = 21. 
 Hence, a 21-^-18 = 1.17 inches. 
 
 From the symmetry of the area it is plain 
 that the center of gravity is midway be- 
 tween the sides. 
 
 EXAMPLE FOR PRACTICE. 
 
 1. Locate the center of gravity of 
 Fig. 23. 
 
 Fig. 23. 
 
 Ans. 2.3 inches above the base. 
 49. Center of Gravity of Built-up Sections. In Fig. 24 
 there are represented cross-sections of various kinds of rolled steel, 
 called "shape steel," which is used extensively in steel construction. 
 Manufacturers of this material publish "handbooks" giving full 
 information in regard thereto, among other things, the position of 
 the center of gravity of each cross section. With such a handbook 
 
 I-toewn 
 
 Cha.nnl 
 
 Angje 
 
 Z-b^r 
 
 Pig. 24, 
 
 available, it is therefore not necessary actually to compute the posi- 
 tion of the center of gravity of any section, as we did in the pre- 
 ceding article; but sometimes several shapes are riveted together to
 
 STRENGTH OF MATERIALS 
 
 45 
 
 make a "built-up" section (see Fig. 25), and then it may be neces- 
 sary to compute the position of the center of gravity of the section. 
 Example. It is desired to locate the center of gravity of the 
 section of a built-up beam represented in Fig. 25. The beam con- 
 
 14" 
 
 <\J 
 
 Fig. 25. 
 
 sists of two channels and a plate, the area of the cross -section of a 
 
 channel being 6.03 square inches. 
 
 Evidently the center of gravity of each channel section is 6 
 
 inches, and that of the plate section is 12^ inches, from the bottom. 
 
 Let c denote the dis- 
 tance of the center of 
 gravity of the whole 
 section from the bot- 
 tom; then since the 
 area of the plate section 
 is 7 square inches, and 
 that of the whole sec- 
 tion is 19.06, 
 
 * 
 
 ite 
 
 19.06x^ = 6.03 X6 + 
 
 - 26 ' 
 
 158.11. 
 
 Hence, 0=158.11-^19.06=8.30 inches, (about). 
 
 EXAMPLES FOR PRACTICE. 
 1. Locate the center of gravity of the built-up section of
 
 46 STRENGTH OF MATERIALS 
 
 Fig. 26, a, the area of each "angle" being 5.06 square inches, and 
 the center of gravity of each being as shown in Fig. 26, b. 
 
 Ans. Distance from top, 3.08 inches. 
 
 2. Omit the left-hand angle in Fig. 26, a, and locate the 
 center of gravity of the remainder. 
 
 ( Distance from top, 3.65 inches, 
 9> \ left side, 1.19 inches. 
 
 50. rioment of Inertia. If a plane area be divided into an 
 infinite number of infinitesimal parts, then the sum of the prod- 
 ucts obtained by multiplying the area of each part by the square 
 of its distance from some line is ealled the moment of inertia of the 
 area with respect to the line. The line to which the distances are 
 measured is called the inertia-axis; it may be taken anywhere in 
 the plane of the area. In the subject of beams (where we have 
 sometimes to compute the moment of inertia of the cross-section 
 of a beam), the inertia-axis is taken through the center of gravity 
 of the section and horizontal. 
 
 An approximate value of the moment of inertia of an area 
 can be obtained by dividing the area into small parts (not infini- 
 tesimal), and adding the products obtained by multiplying the 
 area of each part by the square of the distance from its center to 
 the inertia-axis. 
 
 Example. If the rectangle of Fig. 27, , is divided into 8 
 parts as shown, the area of each is one square inch, and the dis- 
 tances from the axis to the centers of gravity of the parts are \ 
 and \\ inches. For the four parts lying nearest the axis the 
 product (area times distance squared) is: 
 
 lX( i) 2 =J; and for the other parts it is 
 
 Hence the approximate value of the moment of inertia of the area 
 with respect to the axis, is 
 
 If the area is divided into 32 parts, as shown in Fig. 27, b, 
 the area of each part is | square inch. For the eight of the little 
 squares farthest away from the axis, the distance from their centers 
 of gravity to the axis is 1| inches; for the next eight it is 1J; 
 for the next eight |; and for the remainder J inch. Hence an
 
 STRENGTH OF MATERIALS 
 
 47 
 
 approximate value of the moment of inertia of the rectangle with 
 respect to the axis is : 
 
 Fig. 27. 
 
 If we divide the rectangle into still smaller parts and form 
 the products 
 
 (small area) X (distance) 2 , 
 
 and add the products just as we have done, we shall get a larger 
 answer than 10|. The smaller the parts into which the rec- 
 tangle is divided, the larger will be the answer, but it will never 
 be larger than 10|. This 10 is the sum corresponding to a 
 
 division of the rectangle into an 
 infinitely large number of parts 
 (infinitely small) and it is the 
 exact value of the moment of 
 &Q 3 inertia of the rectangle with re- 
 spect to the axis selected. 
 
 There are short methods of 
 computing the exact values of the 
 moments of inertia of simple fig- 
 ures (rectangles, circles, etc.,), 
 but they cannot be given here since they involve the use of difficult 
 mathematics. The foregoing method to obtain approximate val- 
 ues of moments of inertia is used especially when the area is quite 
 irregular in shape, but it is given here to explain to the student 
 the meaning of the moment of inertia of an area. He should 
 understand now that the moment of inertia of an area is sim- 
 ply a name for such sums as we have just computed. The name 
 is not a fitting one, since the sum has nothing whatever to do with 
 inertia. It was first used in this connection because the sum is 
 very similar to certain other sums which had previously been 
 called moments of inertia. 
 
 51. Unit of Moment of Inertia. The product (area X dis- 
 tance 2 ) is really the product of four lengths, two in each factor; 
 and since a moment of inertia is the sum of such products, a 
 moment of inertia is also the product of four lengths. Now the 
 product of two lengths is an area, the product of three is a vol- 
 ume, and the product of four is moment of inertia unthinkable in
 
 48 STRENGTH OF MATERIALS 
 
 the way in which we can think of an area or volume, and there- 
 fore the source of much difficulty to the student. The units of 
 these quantities (area, volume, and moment of inertia) are respec- 
 
 tively: 
 
 the square inch, square foot, etc., 
 " cubic " , cubic " " , 
 " biquadratic inch, biquadratic foot, etc.; 
 
 but the biquadratic inch is almost exclusively used in this connec- 
 tion; that is, the inch is used to compute 
 values of moments of inertia, as in the pre- 
 ceding illustration. It is often written 
 thus: Inches 4 . 
 
 52. Moment of Inertia of a Rectangle. 
 _ Axis _ i, Let b denote the base of a rectangle, and a 
 
 p- 28 its altitude; then by higher mathematics it 
 
 can be shown that the moment of inertia 
 
 of the rectangle with respect to a line through its center of gravity 
 and parallel to its base, is -^ J 3 . 
 
 Example. Compute the value of the moment of inertia of 
 a rectangle 4x12 inches with respect to a line through its center 
 of gravity and parallel to the long side. 
 
 Here 5=12, and a = 4 inches ; hence the moment of inertia 
 desired equals 
 
 =64 inches 4 . 
 
 EXAflPLE FOR PRACTICE. 
 
 1. Compute the moment of inertia of a rectangle 4x12 
 inches with respect to a line through its center of gravity and 
 parallel to the short side. Ans. 576 inches 4 . 
 
 53. Reduction Formula. In the previously mentioned 
 "handbooks" there can be found tables of moments of inertia of 
 all the cross-sections of the kinds and sizes of rolled shapes made. 
 The inertia-axes in those tables are always taken through the cen- 
 ter of gravity of the section, and usually parallel to some edge of 
 the section. Sometimes it is necessary to compute the moment of 
 inertia of a "rolled section" with respect to some other axis, and 
 if the two axes (that is, the one given in the tables, and the other) 
 are parallel, then the desired moment of inertia can be easily com- 
 puted from the one given in the tables by the following rule:
 
 STRENGTH OF MATERIALS 
 
 49 
 
 The moment of inertia of an area with respect to any axis 
 equals the moment of inertia with respect to a parallel axis 
 through the center of gravity r , plus the product of the area and 
 the square of the distance between the axes. 
 
 Or. if I denotes the moment of inertia with respect to any axis; 
 I the moment of inertia with respect to a parallel axis through 
 the center of gravity; A the area; and d the d ; stance between the 
 axes, then 
 
 Example. It is required to compute the moment of inertia 
 of a rectangle 2x8 inches with respect to a line parallel to the 
 long side and 4 inches from the center of gravity. 
 
 Let I denote the moment of inertia sought, and I the moment 
 of inertia of the rectangle with respect 
 to a line parallel to the long side and 
 through the center of gravity (see Fig. 
 28). Then 
 
 1 =Tzfa 3 (see Art. 52) ; and, 
 since 1=8 inches and a=2 inches, 
 I =yV ( 8 X 2 3 ) = 5^ biquadratic inches. 
 
 The distance between the two inertia- 
 axes is 4 inches, and the area of the 
 rectangle is 16 square inches, hence 
 equation 5 becomes Fi S- 29t 
 
 I=5J-f 16X4 2 =261^ biquadratic inches. 
 
 EXAMPLE FOR PRACTICE. 
 
 1. The moment of inertia of an "angle" 2JX2xJ inches 
 (lengths of sides and width respectively) with respect to a line 
 through the center of gravity and parallel to the long side, is 0.64 
 inches*. The area of the section is 2 square inches, and the dis- 
 tance from the center of gravity to the long side is 0.63 inches. 
 (These values are taken from a "handbook".) It is required to 
 compute the moment of inertia of the section with respect to a 
 line parallel to the long side and 4 inches from the center of 
 gravity. Ans. 32.64 inches 4 . 
 
 54. Moment of Inertia of Built-up Sections. As before 
 stated, beams are sometimes "built up" of rolled shapes (angles,
 
 50 
 
 STRENGTH OF MATERIALS 
 
 -a? 
 
 channels, etc.). The moment of inertia of such a section with 
 respect to a definite axis is computed by adding the moments of 
 inertia of the parts, all with respect to that same axis. This is the 
 method for computing the moment of any area which can be 
 divided into simple parts. 
 
 The moment of inertia of an area which may be regarded as 
 consisting of a larger area minus other areas, is computed by sub- 
 tracting from the moment of inertia of the large area those of the 
 "minus areas." 
 
 Examples. 1. Compute the moment of inertia of the built- 
 up section represented in Fig. 30 (in part same as Fig. 25) with 
 respect to a horizontal axis 
 
 passing through the center i< 1' " 
 
 of gravity, it being given 
 that the moment of inertia 
 of each channel section 
 with respect to a horizontal 
 axis through its center of 
 gravity is 128.1 inches 4 , 
 and its area 6.03 square 
 inches. 
 
 The center of gravity of 
 the whole section was found 
 
 in the example of Art. 49 to be 8.30 inches from the bottom of 
 the section; hence the distances from the inertia-axis to the 
 centers of gravity of the channel section and the plate are 2.30 
 and 3.95 inches respectively (see Fig. 30). 
 
 The moment of inertia of one channel section with respect to 
 the axis A A (see equation 5, Art. 53) is: 
 
 128.1 + 6.03 X2.30 2 = 160.00 inches 4 . 
 
 The moment of inertia of the plate section (rectangle) with re- 
 spect to the line a" a" (see Art. 52) is: 
 
 Jg. ba 3 = I J 2 -[14x (^) 3 J=0.15 inches 4 ; 
 
 and with respect to the axis A A (the area being 7 square inches) 
 it is: 
 
 0.15+7X3.95 2 =109.37 inches 4 . 
 
 Therefore the moment of inertia of the whole section with re- 
 apect to AA is : 
 
 2 X 160.00+ 109.37=429.37 inches*. 
 
 Fig. 30.
 
 STRENGTH OF MATERIALS 
 
 51 
 
 2. It is required to compute the moment of inertia of the 
 " hollow rectangle " of Fig. 29 with respect to a line through the 
 center of gravity and parallel to the short side. 
 
 The amount of inertia of the large rectangle with respect to 
 the named axis (see Art. 52) is: 
 
 T V(5X10 3 ) = 
 
 * 
 
 ""~ 
 
 Ito 
 
 Fig. 31. 
 
 and the moment of inertia of the smaller one with respect to the 
 same axis is: 
 
 1 V(4X8 8 )=170|; 
 
 hence the moment of inertia of the hollow section with respect 
 to the axis is: 
 
 416 - 170 = 246 inches 4 . 
 EXAMPLES FOR PRACTICE. 
 
 1. Compute the moment of inertia of the section repre- 
 sented in Fig. 31, a, about the axis AA, it being 3.08 inches 
 from the top. Given also that the area of one angle section is 
 5.06 square inches, its center of gravity C (Fig. 31, b) 1.66 inches 
 from the top, and its moment of inertia with respect to the axis aa 
 17.68 inches 4 . Ans. 145.8 inches 4 . 
 
 2. Compute the moment of inertia of the section of Fig. 31, a,
 
 52 
 
 STRENGTH OF MATERIALS 
 
 with respect to the axis BB. Given that distance of the center 
 of gravity of one angle from one side is 1.66 inches (see Fig. 31, J), 
 and its moment of inertia with respect to bb 17.68 inches. 
 
 Ans. 77.618 inches 4 . 
 
 55. Table of Centers of Gravity and floments of Inertia. 
 Column 2 in Table A below gives the formula for moment of 
 inertia with respect to the horizontal line through the center of 
 gravity. The numbers in the third column are explained in Art. 
 62; and those in the fourth, in Art. 80. 
 
 TABLE A. 
 
 Moments of Inertia, Section Moduli, and Radii of Gyration. 
 
 In each case the axis is horizontal and passes through the center of gravity. 
 
 Section. 
 
 Moment of 
 Inertia. 
 
 Section 
 Modulus. 
 
 Radius of 
 Gyration. 
 
 a 4 - 
 
 baf 
 12 
 
 0.049d 4 
 
 0.049 (d 4 -d/) 
 
 ba* 
 
 12 
 
 a 
 
 I/ "12" 
 
 >! 
 
 0.098d8 
 
 - 
 
 0.098 -r-J- 
 d 
 
 STRENGTH OF BEAMS. 
 
 56. Kinds of Loads Considered. The loads that are applied 
 to a horizontal beam are usually vertical, but sometimes forces are 
 applied otherwise than at right angles to beams. Forces acting on 
 beams at right angles are called transverse forces ; those applied
 
 STRENGTH OF MATERIALS 
 TABLE B. 
 
 53 
 
 Shear Diagrams (b) and Moment Diagrams (c) for Eight Different Cases (a). 
 Also Values of Maximum Shear (V), Bending Homent (M), and Deflection (d). 
 
 V=P, M=P1, d=Pl s -f-3EI. 
 
 , d= P13--48EI. 
 
 -J? 
 
 to 1 
 
 W-uniform loa-d 
 
 V=W, 
 
 W-onlform loa-d 
 
 V=P, M=Pa,
 
 54 
 
 STRENGTH OF MATERIALS 
 
 parallel to a beam are called longitudinal forces ; and others are 
 called inclined forces. For the present we deal only with beams 
 subjected to transverse forces (loads and reactions). 
 
 57. Neutral Surface, Neutral Line, and Neutral Axis. When 
 a beam is loaded it may be wholly convex up (concave down), as a 
 cantilever; wholly convex down (concave up), as a simple beam 
 on end supports; or partly convex up and partly convex down, as 
 a simple beam with overhanging ends, a restrained beam, or a con- 
 
 Fig. 32. 
 
 tinuous beam. Two vertical parallel lines drawn close together on 
 the side of a beam before it is loaded will not be parallel after it 
 is loaded and bent. If they are on a convex-down portion of a 
 beam, they will be closer at the top and farther apart below than 
 when drawn (Fig. 32), and if they are on a convex-up portion, 
 they will be closer below and farther apart above than when drawn 
 (Fig. 326). 
 
 The " fibres " on the convex side of a beam are stretched and 
 therefore under tension, while those on the concave side are short- 
 ened and therefore under compression. Obviously there must be 
 some intermediate fibres which are neither stretched nor shortened, 
 i. ., under neither tension nor compression. These make up 
 a sheet of fibres and define a surface in the beam, which surface is 
 called the neutral surface of the beam. The intersection of the 
 neutral surface with either side of the beam is called the neutral 
 line, and its intersection with any cross-section of the beam is 
 called the neutral axis of that section. Thus, if db is a fibre that 
 has been neither lengthened nor shortened with the bending of the 
 beam, then nn is a portion of the neutral line of the beam; and, 
 if Fig. 32c be taken to represent a cross-section of the beam, NN 
 is the neutral axis of the section. 
 
 It can be proved that the neutral axis of any cross-section of
 
 STRENGTH OF MATERIALS 
 
 55 
 
 a loaded beam passes through the center of gravity of that section, 
 provided that all the forces applied to the beam are transverse, and 
 that the tensile and compressive stresses at the cross-section are 
 all within the elastic limit of the material of the beam. 
 
 58. Kinds of Stress at a Cross-section of a Beam. It has 
 already been explained in the preceding article that there are ten- 
 sile and compressive stresses in a beam, and that the tensions are 
 on the convex side of the beam and the compressions on the con- 
 cave (see Fig. 33). The forces T and C are exerted upon the 
 portion of the beam represented by the adjoining portion to the 
 
 P" C 
 
 ^ T 
 
 - C 
 
 Fig. 33. 
 
 right (not shown). These, the student is reminded, are often called 
 fibre stresses. 
 
 Besides the fibre stresses there is, in general, a shearing stress 
 at every cross -section of a beam. This may be proved as follows: 
 
 Fig. 34 represents a simple beam on end supports which has 
 actually been cut into two parts as shown. The two parts can 
 maintain loads when in a horizontal position, if forces are applied 
 at the cut ends equivalent to the forces that would act there if the 
 beam were not cut. Evidently in the solid beam there are at the 
 section a compression above and a tension below, and such forces 
 can be applied in the cut beam by means of a short block C and a 
 chain or cord T, as shown. The block furnishes the compressive 
 forces and the chain the tensile forces. At first sight it appears as 
 if the beam would stand up under its load after the block and 
 chain have been put into place. Except in certain cases*, how- 
 ever, it would not remain in a horizontal position, as would the 
 
 * When the external shear for the section is zero.
 
 56 
 
 STRENGTH OF MATERIALS 
 
 solid beam. This shows that the forces exerted by the block and 
 chain ( horizontal compression and tension ) are not equivalent to 
 the actual stresses in the solid beam. What is needed is a vertical 
 force at each cut end. 
 
 Suppose that Rj is less than Lj +L,+ weight of A, i. e., that 
 the external shear for the section is negative; then, if vertical pulls 
 be applied at the cut ends, upward on A and downward on B, the 
 beam will stand under its load and in a horizontal position, just as 
 a solid beam. These pulls can be supplied, in the model of the 
 beam, by means of a cord S tied to two brackets fastened on A and 
 
 Fig. 34. 
 
 Fig. 35, 
 
 Ib fa crx L 3 
 
 A 
 
 TTl 
 
 <w//s RI 
 
 K k i 
 
 ^2///-^ 
 
 S 
 
 \v 
 
 B, as shown. In the solid beam the two parts act upon each 
 other directly, and the vertical forces are shearing stresses, since 
 they act in the plane of the surfaces to which they are applied. 
 
 59. Relation Between the Stress at a Section and the Loads 
 and Reactions on Either Side of It. Let Fig. 35 represent the 
 portion of a beam on the left of a section; and let Rj denote the 
 left reaction; Lj and L 2 the loads; W the weight of the left part; 
 
 C, T, and S the compression, tension, and shear respectively which 
 the right part exerts upon the left. 
 
 Since the part of the beam here represented is at rest, all the 
 forces exerted upon it are balanced; and when a number of hori- 
 zontal and vertical forces are balanced, then 
 
 1. The algebraic sum of the horizontal forces equals zero. 
 
 2. " " ' " " " vertical " " " 
 
 3. " " " " " moments of all the forces with respect to 
 any point equals zero. 
 
 To satisfy condition 1, since the tension and compression are 
 
 the only horizontal forces, the tension must equal the compression. 
 
 To satisfy condition 2, S (the internal shear) must equal the
 
 STRENGTH OF MATERIALS 57 
 
 algebraic sum of all the other vertical forces on the portion, that 
 is, must equal the external shear for the section; also, S must act 
 up or down according as the external shear is negative or positive. 
 In other words, briefly expressed, the internal and external shears 
 at a section are equal and opposite. 
 
 To satisfy condition 3, the algebraic sum of the moments of 
 the fibre stresses about the neutral axis must be equal to the sum 
 of the moments of all the other forces acting on the portion about 
 the same line, and the signs of those sums must be opposite. (The 
 moment of the shear about the neutral axis is zero.) Now, the 
 sum of the moments of the loads and reactions is called the bend- 
 ing moment at the section, and if we use the term resisting mo- 
 ment to signify the sum of the moments of the fibre stresses (ten- 
 sions and compressions ) about the neutral axis, then we may say 
 briefly that the resisting and the bending moments at a section are 
 equal, and the two moments are opposite in sign. 
 
 60. The Fibre Stress. As before stated, the fibre stress is 
 not a uniform one, that is, it is not uniformly distributed over the 
 section on which it acts. At any section, the compression is most 
 " intense " (or the unit-compressive stress is greatest) on the con- 
 cave side; the tension is most intense (or the unit-tensile stress is 
 greatest) on the convex side; and the unit-compressive and unit- 
 tensile stresses decrease toward the neutral axis, at which place the 
 unit-fibre stress is zero. 
 
 If the fibre stresses are within the elastic limit, then the two 
 straight lines on the side of a beam referred to in Art. 57 will still 
 be straight after the beam is bent; hence the elongations and short- 
 enings of the fibres vary directly as their distance from the neutral 
 axis. Since the stresses (if within the elastic limit) and deforma- 
 tions in a given material are proportional, the unit-fibre stress 
 varies as the distance from the neutral axis. 
 
 Let Fig. 36# represent a portion of a bent beam, 36 b its cross- 
 section, nn the neutral line, and NN the neutral axis. The way 
 in which the unit-fibre stress on the section varies can be rep- 
 resented graphically as follows: Lay off ac, by some scale, to 
 represent the unit-fibre stress in the top fibre, and join c and n t 
 extending the line to the lower side of the beam ; also make bc f equal 
 to be" and draw no'. Then the arrows represent the unit-fibre 
 stresses, for their lengths vary as their distances from the neutral axis.
 
 58 
 
 STRENGTH OF MATERIALS 
 
 61. Value of the Resisting Moment. If S denotes the unit- 
 fibre stress in the fibre farthest from the neutral axis (the greatest 
 unit-fibre stress on the cross -section), and c the distance from the 
 neutral axis to the remotest fibre, while S,, S 2 , S,, etc., denote the 
 unit-fibre stresses at points whose distances from the neutral axis 
 
 are, respectively, y n y n y n etc. (see Fig. 36 J), then 
 
 c 
 
 S : S, :: c : y,\ or S, = y r 
 
 S 2 = 2/ 2 ; S 3 = 3/3, etc. 
 
 Also, 
 
 Let # a a , a n etc., be the areas of the cross-sections of the fibres 
 
 A. S C 
 
 Fig. 36. 
 
 whose distances from the neutral axis are, respectively, y^ j/ 2 , y,, 
 etc. Then the stresses on those fibres are, respectively, 
 S, S a 2 , S 3 3 , etc.; 
 
 S S S 
 
 or, y\ fl n yfl'v ^3*35 etc. 
 
 o c ' c 
 
 The arms of these forces or stresses with respect to the neutrai axis 
 are, respectively, y^ y^ y^ etc.; hence their moments are 
 
 s 2 ^s 2 s 2 
 
 c c c ' 
 
 and the sum of the moments (that is, the resisting moment) is 
 
 s 2 s s, 
 
 a i y\ H #9 iA + etc. = (a, v\ -4- a., v\ -4- etc.) 
 
 c c J c v ' 
 
 Now a, y\ + a a y\ -f etc. is the sum of the products obtained by 
 multiplying each infinitesimal part of the area of the cross-section 
 by the square of its distance from the neutral axis; hence, it is the 
 moment of inertia of the cross-section with respect to the neutral 
 axis. If this moment is denoted by I, then the value of the resist- 
 
 OT 
 
 ing moment is m
 
 STRENGTH OF MATERIALS, 
 
 PART II. 
 
 STRENGTH OF BE AHS (Concluded). 
 
 62. First Beam Formula. As shown in the preceding 
 article, the resisting and bending moments for any section of a 
 beam are equal; hence 
 
 all the symbols referring to the same section. This is the most 
 important formula relating to beams, and will be called the " first 
 beam formula." 
 
 The ratio I -r- c is now quite generally called the section 
 modulus. Observe that for a given beam it depends only on the 
 dimensions of the cross -section, and not on the material or any- 
 thing else. Since I is the product of four lengths (see article 51), 
 I -r- c is the product of three; and hence a section modulus can be 
 expressed in units of volume. The cubic inch is practically always 
 used; and in this connection it is written thus, inches 3 . See Table 
 A, page 52, for values of the section moduli of a few simple sections. 
 
 63. Applications of the First Beam Formula. There are 
 three principal applications of equation 6, which will now be ex- 
 plained and illustrated. 
 
 64. First Application. The dimensions of a beam and its 
 manner of loading and support are given, and it is required to 
 compute the greatest unit- tensile and compressive stresses in the 
 beam. 
 
 This problem can be solved by means of equation 6, written 
 in this form, 
 
 Me M 
 
 S= T r fT~ c ( 6 ) 
 
 Unless otherwise stated, we assume that the beams are uniform 
 in cross-section, as they usually are; then the section modulus 
 (I-r-c) is the same for all sections, and S (the unit-fibre stress on
 
 60 STRENGTH OF MATERIALS 
 
 the remotest fibre) varies just as H varies, and is therefore greatest 
 where M is a maximum.* Hence, to compute the value of the 
 greatest unit-fibre stress in a given case, substitute the values of 
 the section modulus and the maximum bending moment in the 
 preceding equation, and reduce. 
 
 If the neutral axis is equally distant from the highest and low- 
 est fibres, then the greatest tensile and compressive unit-stresses 
 are equal, and their value is S. If the neutral axis is unequally 
 distant from the highest and lowest fibres, let c denote its distance 
 from the nearer of the two, and S' the unit-fibre stress there. 
 Then, since the unit-stresses in a cross-section are proportional to 
 the distances from the neutral axis, 
 
 If the remotest fibre is on the convex side of the beam, S is tensile 
 and S' compressive; if the remotest fibre is on the concave side, S 
 is compressive and S' tensile. 
 
 Examples. 1. A beam 10 feet long is supported at its ends, 
 and sustains a load of 4,000 pounds two feet from the left end 
 (Fig. 37, a]. If the beam is 4 X 12 inches in cross-section (the 
 long side vertical as usual), compute the maximum tensile and 
 compressive unit-stresses. 
 
 The section modulus of a rectangle whose base and altitude 
 are b and a respectively (see Table A, page 52), is -J&a 2 ; hence, 
 for the beam under consideration, the modulus is 
 
 y X 4 X 12 2 = 96 inches 3 . 
 
 To compute the maximum bending moment, we have, first, to find 
 the dangerous section. This section is where the shear changes 
 sign (see article 45); hence, we have to construct the shear dia- 
 gram, or as much thereof as is needed to find where the change of 
 sign occurs. Therefore we need the values of the reaction. 
 Neglecting the weight of the beam, the moment equation with 
 origin at C (Fig. 37, a] is 
 
 R, X 10 - 4,000 X 8 = 0, or R, = 3,200 pounds 
 
 * NOTE. Because S is greatest in the section where M is maximum, this 
 section is usually called the " dangerous section" of the beam.
 
 STRENGTH OF MATERIALS 
 
 61 
 
 Then, constructing the shear diagram, we see (Fig. 37, 5) that the 
 change of sign of the shear (also the dangerous section) is at the 
 load. The value of the bending moment there is 
 
 3,200 X 2 = 6,400 foot-pounds, 
 or 6,400 X 12 == 76,800 inch-pounds. 
 
 Substituting in equation 6', we find that 
 
 76,800 
 
 Q 
 
 ~96~ 
 
 = 800 pounds per square inch. 
 
 4ooolbs. 
 
 ic(a) 
 
 c'(b) 
 
 c"(c) 
 
 Fig. 37. 
 
 2. It is desired to take into account the weight of the beam 
 in the preceding example, supposing the beam to be wooden. 
 The volume of the beam is 
 
 1^1? x 10 = 3 cubic feet; 
 
 and supposing the timber to weigh 45 pounds per cubic foot, the 
 beam weighs 150 pounds (insignificant compared to the load). 
 The left reaction, therefore, is 
 
 3 s 200-f(-^-X 150) = 3,375;
 
 62 STRENGTH OF MATERIALS 
 
 and the shear diagram looks like Fig. 37, c, the shear changing 
 sign at the load as before. The weight of the beam to the left of 
 the dangerous section is 30 pounds; hence the maximum bending 
 moment equals 
 
 3,275 X 2 - 30 X 1 = 6,520 foot-pounds, 
 or 6,520 X 12 = 78,240 inch-pounds. 
 
 Substituting in equation 6', we find that 
 
 78 240 
 S = <Tp = 815 pounds per square inch. 
 
 The weight of the beam therefore increases the unit-stress pro- 
 duced by the load at the dangerous section by 15 pounds per 
 square inch. 
 
 3. A T-bar (see Fig. 38) 8 feet long and supported at each 
 "~Hr~ end, bears a uniform load of 1,200 
 
 d_pq pounds. The moment of inertia of its 
 
 cross -section with respect to the neu- 
 g tral axis being 2.42 inches 4 , compute 
 w the maximum tensile and compressive 
 
 *- unit-stresses in the beam 
 
 Evidently the dangerous section 
 
 is in the middle, and the value of the maximum bending moment 
 (see Table B, page 53, Part I) is J WZ, W and I denoting the load 
 and length respectively. Here 
 
 -g- Wl =-g- X 1,200 X 8 = 1,200 foot-pounds, 
 or 1,200 X 12 = 14,400 inch-pounds. 
 
 The section modulus equals 2.42 -*- 2.28 = 1.06; hence 
 
 _ 14,400 
 
 - * QQ - = 13,585 pounds per square inch. 
 
 This is the unit-fibre stress on the lowest fibre at the middle sec- 
 tion, and hence is tensile. On the highest fibre at the middle 
 section the unit-stress is compressive, and equals (see page 60) : 
 
 ~ g";^ X 13,585 = 4,290 pounds per square inch.
 
 STRENGTH OF MATERIALS 63 
 
 EXAMPLES FOR PRACTICE. 
 
 1. A beam 12 feet long and 6 X 12 inches in cross-section 
 rests on end supports, and sustains a load of 3,000 pounds in the 
 middle. Compute the greatest tensile and compressive unit- 
 stresses in the beam, neglecting the weight of the beam. 
 
 Ans. 750 pounds per square inch. 
 
 2. Solve the preceding example taking into account the 
 weight of the beam, 800 pounds 
 
 Ans. 787.5 pounds per square inch. 
 
 3. Suppose that a built-in cantilever projects 5 feet from the 
 wall and sustains an end load of 250 pounds. The cross -section of 
 the cantilever being represented in Fig. 38, compute the greatest 
 tensile and compressive unit-stresses, and tell at what places they 
 occur. (Neglect the weight.) 
 
 ( Tensile, 4,471 pounds per square inch. 
 
 n8 ' I Compressive, 14,150 " 
 
 4. Compute the greatest tensile and compressive unit-stresses 
 in the beam of Fig. 18, a, due to the loads and the weight of beam 
 (400 pounds). (A moment diagram is represented in Fig. 18, J; 
 for description see example 2, Art. 44, p. 39.) The section of 
 the beam is a rectangle 8 X 12 inches. 
 
 Ans. 580 pounds per square inch. 
 
 5. Compute the greatest tensile and compressive unit-stresses 
 in the cantilever beam of Fig. 19, a, it being a steel I-beam whose 
 section modulus is 20.4 inches 3 . (A bending moment diagram for 
 it is represented in Fig. 19, b; for description, see Ex. 3, Art. 44.) 
 
 Ans. 11,470 pounds per square inch. 
 
 6. Compute the greatest tensile and compressive unit-stresses 
 in the beam of Fig. 10, neglecting its weight, the cross -sections 
 being rectangular 6 X 12 inches. (See example for practice 1, 
 Art. 43.) 
 
 Ans. 600 pounds per square inch. 
 
 65. /Second Application. The dimensions and the work- 
 ing strengths of a beam are given, and it is required to determine 
 its safe load (the manner of application being given). 
 
 This problem can be solved by means of equation 6 written 
 in this form, 
 
 * = (6")
 
 64 STRENGTH OF MATERIALS 
 
 We substitute for S the given working strength for the ma- 
 terial of the beam, and for I and c their values as computed from 
 the given dimensions of the cross -section; then reduce, thus 
 obtaining the value of the safe resisting moment of the beam, 
 which equals the greatest safe bending moment that the beam can 
 stand. We next compute the value of the maximum bending 
 moment in terms of the unknown load; equate this to the value 
 of the resisting moment previously found; and solve for the 
 unknown load. 
 
 In cast iron, the tensile and compressive strengths are very 
 different; and the smaller (the tensile) should always be used if 
 the neutral surface of the beam is midway between the top and 
 bottom of the beam; but if it is unequally distant from the top 
 and bottom, proceed as in example 4, following. 
 
 Examples. 1. A wooden beam 12 feet long and 6 X 12 
 inches in cross -section rests on end supports. If its working 
 strength is 800 pounds per square inch, how large a load uniformly 
 distributed can it sustain ? 
 
 The section modulus is |J 2 , 5 and a denoting the base and 
 altitude of the section (see Table A, page 52) ; and here 
 
 1 ba? =r ~ X 6 X 12 2 = 144 inches 3 . 
 
 Hence S -L = 800 X 144 == 115,200 inch-pounds. 
 
 For a beam on end supports and sustaining a uniform load, the 
 maximum bending moment equals |WZ (see Table B, page 55), 
 W denoting the sum of the load and weight of beam, and I the 
 length. If W is expressed in pounds, then 
 
 g- W Z = - W X 12 foot-pounds = _ W X 144 inch-pounds. 
 
 Hence, equating tbe two values of maximum bending moment 
 and the safe resisting moment, we get 
 
 W X 144 = 115,200; 
 
 = 6,400 pound..
 
 STRENGTH OF MATERIALS 65 
 
 The safe load for the beam is 6,400 pounds minus the weight of 
 the beam. 
 
 2. A steel I-beam whose section modulus is 20.4 inches 8 
 rests on end supports 15 feet apart. Neglecting the weight of the 
 beam, how large a load may be placed upon it 5 feet from one end, 
 if the working strength is 16,000 pounds per square inch? 
 
 The safe resisting moment is 
 
 * = 16,000 X 20.4 = 326,400 inch-pounds; 
 
 hence the bending moment must not exceed that value. The 
 dangerous section is under the load; and if P denotes the unknown 
 value of the load in pounds, the maximum moment (see Table B, 
 page 53, Part I) equals P X 5 foot-pounds, or P X 60 inch- 
 pounds. Equating values of bending and resisting moments, 
 we get 
 
 | P x 60 = 326,400; 
 
 326,400 X 3 
 or, P= gL____ = 8,160 pounds. 
 
 3. In the preceding example, it is required to take into 
 account the weight of the beam. 375 pounds. 
 
 U -i 1 
 
 l p 
 
 JLf -1 fyl 
 
 1 
 
 i c 10 j 
 
 K 
 
 W= 375 DOS. 
 Fig. 39. 
 
 As we do not know the value of the safe load, we cannot con- 
 struct the shear diagram and thus determine where the dangerous 
 section is. But in cases like this, where the distributed load (the 
 weight) is small compared with the concentrated load, the dan- 
 gerous section is practically always where it is under the concen- 
 trated load alone; in this case, at the load. The reactions due to 
 the weight equal X 375 187.5; and the reactions due to the 
 load equal ^ P and P, P denoting the value of the load. The 
 larger reaction R, (Fig. 39) hence equals P -f 187.5. Since
 
 66 STRENGTH OF MATERIALS 
 
 the weight of the beam per foot is 375 -s- 15 = 25 pounds, the 
 maximum bending moment (at the load) equals 
 
 ( | P + 187.5) 5 - (25 X 5) 2| = 
 
 o 
 
 ^ p 4. 937.5 - 312.5 = ^ p + 625. 
 3 o 
 
 This is in foot-pounds if P is in pounds. 
 
 The safe resisting moment is the same as in the preceding 
 illustration, 326,400 inch-pounds; hence 
 
 (4~ P + 625) 12 = 326,400. 
 Solving for P, we have 
 
 ^ P + MB _0. 
 
 10 P = 79,725; 
 or, P = 7,972.5 pounds. 
 
 It remains to test our assumption that the dangerous section 
 is at the load. This can be done by computing R t (with P = 
 7,972.5), constructing the shear diagram, and noting where the 
 shear changes sign. It will be found that the shear changes sign 
 at the load, thus verifying the assumption. 
 
 4. A cast-iron built-in cantilever beam projects 8 feet from 
 the wall. Its cross-section is represented in Fig. 40, and the 
 
 moment of inertia with respect to 
 
 I . ' ^ the neutral axis is 50 inches*; the 
 
 N 1- -I *-N working strengths in tension and 
 
 JJ compression are 2,000 and 9,000 
 
 I | if pounds per square inch respect- 
 
 1 ; ' J k ively. Compute the safe uniform 
 
 load which the beam can sustain, 
 neglecting the weight of the beam. 
 
 The beam being convex up, the upper fibres are in tension 
 and the lower in compression. The resisting moment (SI -s- c), 
 as determined by the compressive strength, is
 
 STRENGTH OF MATERIALS 67 
 
 X 50 
 
 and the resisting moment, as determined by the tensile strength, is 
 ~-^ -- 40.000 inch-pounds. 
 
 Hence the safe resisting moment is the lesser of these two, or 
 40,000 inch-pounds. The dangerous section is at the wall (see 
 Table B, page 53), and the value of the maximum bending 
 moment is "W7, W denoting the load and I the length. If W is 
 in pounds, then 
 
 M = | W X 8 foot-pounds = * W X 96 inch-pounds. 
 Equating bending and resisting moments, we have 
 
 -L-W X 96 = 40,000; 
 
 EXAMPLES FOR PRACTICE. 
 
 1. An 8 X 8 -inch timber projects 8 feet from a wall. If its 
 working strength is 1,000 pounds per square inch, how large an 
 end load can it safely sustain ? 
 
 Ans. 890 pounds. 
 
 2. A beam 12 feet long and 8 X 16 inches in cross-section, 
 on end supports, sustains two loads P, each 3 feet from its ends 
 respectively. The working strength being 1,000 pounds per square 
 inch, compute P (see Table B, page 53). 
 
 Ans. 9,480 pounds. 
 
 3. An I-beam weighing 25 pounds per foot rests on end 
 supports 20 feet apart. Its section modulus is 20.4 inches 3 , and 
 its working strength 16,000 pounds per square inch. Compute 
 the safe uniform load which it can sustain. 
 
 Ans. 10,880 pounds- 
 
 66. Third Application. The loads, manner of support, 
 and working strength of beam are given, and it is required to de- 
 termine the size of cross-section necessary to sustain the load 
 safely, that is, to " design the beam."
 
 68 STRENGTH OF MATERIALS 
 
 To solve this problem, we use the first beam formula (equation 
 6), written in this form, 
 
 I _M. (6'") 
 
 T ~~S~ 
 
 We first determine the maximum bending moment, and then sub- 
 stitute its value for M, and the working strength for S. Then we 
 have the value of the section modulus (I -4- c) of the required 
 beam. Many cross-sections can be designed, all having a given 
 section modulus. Which one is to be selected as most suitable will 
 depend on the circumstances attending the use of the beam and 
 on considerations of economy. 
 
 Examples. 1. A timber beam is to be used for sustaining 
 a uniform load of 1,500 pounds, the distance between the supports 
 being 20 feet. If the working strength of the timber is 1,000 pounds 
 per square inch, what is the necessary size of cross -section ? 
 
 The dangerous section is at the middle of the beam ; and the 
 maximum bending moment (see Table B, page 53) is 
 
 -g-WZ = -g- X 1,500 X 20 = 3,750 foot-pounds, 
 
 or 3,750 X 12 = 45,000 inch-pounds. 
 
 I 45,000 
 
 -' T^OOO = 45 inches'. 
 
 Now the section modulus of a rectangle is \ba? (see Table A, 
 page 54, Part I); therefore, \bc? = 45, or ba? = 270. 
 
 Any wooden beam (safe strength 1,000 pounds per square 
 inch) whose breadth times its depth square equals or exceeds 270, 
 is strong enough to sustain the load specified, 1,500 pounds. 
 
 To determine a size, we may choose any value for or a, and 
 solve the last equation for the unknown dimension. It is best, 
 however, to select a value of the breadth, as 1, 2, 3, or 4 inches, 
 and solve for a. Thus, if we try b = 1 inch, we have 
 
 a? 270, or a = 16.43 inches. 
 
 This would mean a board 1 X 18 inches, which, if used, would 
 have to be supported sidewise so as to prevent it from tipping or 
 " buckling." Ordinarily, this would not be a good size. 
 
 Next try I = 2 inches; we have 
 
 2 X a? = 270; or a = 1/370 -*- 2 = 11.62 inches. 
 This would require a plank 2 X 12, a better proportion than the 
 first. Trying I = 3 inches, we have
 
 STRENGTH OF MATERIALS 69 
 
 3 X # 2 = 270; or a = 1/270 -5- 3 = 9.49 inches. 
 This would require a plank 3 X 10 inches; and a choice between 
 a 2 X 12 and a 3 X 10 plank would be governed by circumstances 
 in the case of an actual construction. 
 
 It will be noticed that we have neglected the weight of the 
 beam. Since the dimensions of wooden beams are not fractional, 
 and we have to select a commercial size next larger than the one 
 computed (12 inches instead of 11.62 inches, for example), the 
 additional depth is usually sufficient to provide strength for the 
 weight of the beam. If there is any doubt in the matter, we can 
 settle it by computing the maximum bending moment including 
 the weight of the beam, and then computing the greatest unit-fibre 
 stress due to load and weight. If this is less than the safe strength, 
 the section is large enough; if greater, the section is too small. 
 
 Thus, let us determine whether the 2 X 12-inch plank is 
 strong enough to sustain the load and its own weight. The plank 
 will weigh about 120 pounds, making a total load of 
 
 1,500 + 120 == 1,620 pounds. 
 Hence the maximum bending moment is 
 
 J_WZ = -i-1,620 X 20 X 12 = 48,600 inch-pounds. 
 
 Since = -rr- t>a 2 = -77- X 2 X 12 2 = 48, and S = , , 
 
 c 6 b I-f-c' 
 
 48 600 
 S = jg- = 1,013 pounds per square inch. 
 
 Strictly, therefore, the 2 X 12-inch plank is not large enough; but 
 as the greatest unit-stress in it would be only 13 pounds per square 
 inch too large, its use would be permissible. 
 
 2. What size of steel I-beam is needed to sustain safely the 
 loading of Fig. 9 if the safe strength of the steel is 16,000 pounds 
 per square inch ? 
 
 The maximum bending moment due to the loads was found 
 in example 1, Art. 43, to be 8,800 foot-pounds, or 8,800 X 12 = 
 105,600 inch-pounds. 
 
 I 105,600 
 
 -= T^000= 6 - 6lnche88 ' 
 
 That is, an I-beam is needed whose section modulus is a little 
 larger than 6.6, to provide strength for its own weight.
 
 70 
 
 STRENGTH OF MATERIALS 
 
 To select a size, we need a descriptive table of I-beams, such 
 as is published in handbooks on structural steel. 
 
 Below is an abridged copy of such a table. (The last two columns con- 
 tain information for use later.) The figure illustrates a cross-section of an 
 I-beam, and shows the axes referred to in the table. 
 
 It will be noticed that two sizes are given for each depth; 
 these are the lightest and heaviest of each size that are made, but 
 intermediate sizes can be secured. In column 5 we find 7.3 as the 
 next larger section modulus than the one required (6.6); and this 
 corresponds to a 12^-pound 6-inch I-beam, which is probably the 
 proper size. To ascertain whether the excess (7.3-6.6 = 0.70) 
 in the section modulus is sufficient to provide for the weight of the 
 beam, we might proceed as in example 1. In this case, however, 
 the excess is quite large, and the beam selected is doubtless safe. 
 
 TABLE C. 
 Properties ot Standard I-Beams 
 
 Section of beam, showing axes 1-1 and 2-2. 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 Depth of 
 Beam, 
 in inches. 
 
 Weight 
 per foot, 
 in pounds. 
 
 Area of cross- 
 section, in 
 square inches. 
 
 Moment of 
 inertia, 
 axis 11. 
 
 Section 
 modulus, 
 axis 11. 
 
 Moment of 
 inertia, 
 axis 22. 
 
 3 
 
 5.50 
 
 1.63 
 
 2.5 
 
 1.7 
 
 0.46 
 
 3 
 
 7.50 
 
 2.21 
 
 2.9 
 
 1.9 
 
 .60 
 
 4 
 
 7.50 
 
 2.21 
 
 6.0 
 
 3.0 
 
 .77 
 
 4 
 
 10.50 
 
 3.09 
 
 7.1 
 
 3.6 
 
 1.01 
 
 5 
 
 9.75 
 
 2.87 
 
 12.1 
 
 4.8 
 
 1.23 
 
 5 
 
 14.75 
 
 4.34 
 
 15.1 
 
 6.1 
 
 1.70 
 
 6 
 
 12.25 
 
 3.61 
 
 21.8 
 
 7.3 
 
 1.85 
 
 6 
 
 17.25 
 
 5.07 
 
 26.2 
 
 8.7 
 
 2.36 
 
 7 
 
 15.00 
 
 4.42 
 
 36.2 
 
 10.4 
 
 2.67 
 
 7 
 
 20.00 
 
 5.88 
 
 42.2 
 
 12.1 
 
 3.24 
 
 8 
 
 18.00 
 
 5.33 
 
 56.9 
 
 14.2 
 
 3.78 
 
 8 
 
 25.25 
 
 7.43 
 
 68.0 
 
 17.0 
 
 4.71 
 
 9 
 
 21.00 
 
 6.31 
 
 84.9 
 
 18.9 
 
 5.16 
 
 9 
 
 35.00 
 
 10.29 
 
 111.8 
 
 24.8 
 
 7.31 
 
 10 
 
 25.00 
 
 7.37 
 
 122.1 
 
 24.4 
 
 6.89 
 
 10 
 
 40.00 
 
 11.76 
 
 158.7 
 
 31.7 
 
 9.50 
 
 12 
 
 31.50 
 
 9.26 
 
 215.8 
 
 36.0 
 
 9.50 
 
 12 
 
 40.00 
 
 11.76 
 
 245.9 
 
 41.0 
 
 10.95 
 
 15 
 
 42.00 
 
 12.48 
 
 441.8 
 
 58.9 
 
 14.62 
 
 15 
 
 60.00 
 
 17.65 
 
 538.6 
 
 71.8 
 
 18.17 
 
 18 
 
 55.00 
 
 15.93 
 
 795.6 
 
 88.4 
 
 21.19 
 
 18 
 
 70.00 
 
 20.59 
 
 921.2 
 
 102.4 
 
 24.62 
 
 20 
 
 65.00 
 
 19.08 
 
 1,169.5 
 
 117.0 
 
 27.86 
 
 20 
 
 75.00 
 
 22.06 
 
 1,268.8 
 
 126.9 
 
 30.25 
 
 24 
 
 80.00 
 
 23.32 
 
 2,087.2 
 
 173.9 
 
 42.86 
 
 24 
 
 100.00 
 
 29.41 
 
 2,379.6 
 
 198.3 
 
 48.55
 
 STRENGTH OF MATERIALS 71 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Determine the size of a wooden beam which can safely 
 sustain a middle load of 2,000 pounds, if the beam rests on end 
 supports 16 feet apart, and its working strength is 1,000 pounds 
 per square inch. Assume width 6 inches. 
 
 Ans. 6 X 10 inches. 
 
 2. What sized steel I-beam is needed to sustain safely a 
 uniform load of 200,000 pounds, if it rests on end supports 10 
 feet apart, and its working strength is 16,000 pounds per square 
 inch? 
 
 Ans. 95-pound 24-inch. 
 
 3. What sized steel I-beam is needed to sustain safely the 
 loading of Fig. 10, if its working strength is 16,000 pounds per 
 square inch ? 
 
 Ans. 14.75-pound 5-inch. 
 
 67. Laws of Strength of Beams. The strength of a beam is 
 measured by the bending moment that it can safely withstand; or, 
 since bending and resisting moments are equal, by its safe resist. 
 ing moment (SI -^ c). Hence the safe strength of a beam varies 
 (1) directly as the working fibre strength of its material, and (2) 
 directly as the section modulus of its cross-section. For beams 
 rectangular in cross -section (as wooden beams), the section modu- 
 lus is -J-J& 2 , l> and a denoting the breadth and altitude of the 
 rectangle. Hence the strength of such beams varies also directly 
 as the breadth, and as the square of the depth. Thus, doubling 
 the breadth of the section for a rectangular beam doubles the 
 strength, but doubling the depth quadruples the strength. 
 
 The safe load that a beam can sustain varies directly as its 
 resisting moment, and depends on the way in which the load is 
 distributed and how the beam is supported. Thus, in the first 
 four and last two cases of the table on page 55, 
 
 M = PZ, hence P = SI -*- lc, 
 
 M = $ WZ, 
 
 W= 2SI 
 
 + lc, 
 
 M = 1 PI, 
 
 P = 4SI 
 
 -r-lc, 
 
 M = -|- Wl, ' " 
 
 W= SSI 
 
 + lc, 
 
 M = 1 PZ, 
 
 P= 881 
 
 + lc, 
 
 M = T V Wl, 
 
 W = 12SI 
 
 -*- ^,
 
 72 
 
 STRENGTH OF MATERIALS 
 
 Therefore the safe load in all cases varies inversely with the 
 length; and for the different cases the safe loads are as 1, 2, 4, 8, 
 8, and 12 respectively. 
 
 Example. What is the ratio of the strengths of a plank 2 X 
 10 inches when placed edgewise and when placed flatwise on its 
 supports ? 
 
 When placed edgewise, the section modulus of the plank is 
 \ X 2 X 10 2 = 33|, and when placed flatwise it is \ X 10 X 2 2 == 
 6|; hence its strengths in the two positions are as 33 to 6-| 
 respectively, or as 5 to 1. 
 
 EXAMPLE FOR PRACTICE. 
 
 What is the ratio of the safe loads for two beams of wood, 
 one being 10 feet long, 3x12 inches in section, and having its load 
 in the middle; and the other 8 feet long and 2x8 inches in section, 
 with its load uniformly distributed. 
 
 Ans. As 28.8 to 21.3 
 
 68. Modulus of Rupture. If a beam is loaded to destruction, 
 and the value of the bending moment for the rupture stage is 
 computed and substituted for M in the formula SI -f- c = M, then 
 the value of S computed from the equation is the modulus of 
 rupture for the material of the beam. Many experiments have 
 been performed to ascertain the moduli of rupture for different 
 materials and for different grades of the same material. The fol 
 owing are fair values, all in pounds per square inch : 
 
 TABLE D. 
 
 Moduli of Rupture. 
 
 Timber: 
 
 Spruce 
 
 Hemlock 
 
 White pine 
 
 Long-leaf pine. . . 
 Short-leaf pine . . 
 Douglas spruce.. 
 White oak. 
 
 4,000 7,000, average 5,000 
 
 3,500 7,000, " 4,500 
 
 5,500 10,500, " 8,000 
 
 10,000 16,000, " 12,500 
 
 8,000 14,000, " 10,000 
 
 4,000 12,000, " 8,000 
 
 Red oak... 
 
 .. 
 
 
 9,000 15,000, " 11,500 
 
 Stone: 
 Sandstone. 
 Limestone. 
 Granite. . . 
 
 
 
 400 1,200, 
 400 1,000. 
 800 1,400. 
 
 Cast iron : 
 
 
 
 One and one-half to two and 
 one-quarter times its ulti- 
 mate tensile strength. 
 
 Hard steel: 
 
 
 
 Varies from 100,000 to 150,000
 
 STRENGTH OF MATERIALS 73 
 
 Wrought iron and structural steels have no modulus of rup- 
 ture, as specimens of those materials will " bend double," but not 
 break. The modulus of rupture of a material is used principally 
 as a basis for determining its working strength. The factor of 
 safety of a loaded learn is computed by dividing the modulus 
 of rupture of its material by the greatest unit-fibre stress in 
 the beam. 
 
 69. The Resisting Shear. The shearing stress on a cross- 
 section of a loaded beam is not a uniform stress; that is, it is not 
 uniformly distributed over the section. In fact the intensity or 
 unit-stress is actually zero on the highest and lowest fibres of a 
 cross-section, and is greatest, in such beams as are used in prac- 
 tice, on fibres at the neutral axis. In the following article we 
 explain how to find the maximum value in two cases cases which 
 are practically important. 
 
 70. Second Beam Formula. Let S s denote the average 
 value of the unit-shearing stress on a cross-section of a loaded 
 beam, and A the area of the cross-section. Then the value of the 
 whole shearing stress on the section is : 
 
 Resisting shear = S s A. 
 
 Since the resisting shear and the external shear at any section of a 
 beam are equal (see Art. 59), 
 
 S s A = Y. (7) 
 
 This is called the " second beam formula " It is used to investi- 
 gate and to design for shear in beams. 
 
 In beams uniform in cross -section, A is constant, and S s is 
 greatest in the section for which V is greatest. Hence the great- 
 est unit-shearing stress in a loaded beam is at the neutral axis of 
 the section at which the external shear is a maximum. There ia 
 a formula for computing this maximum value in any case, but it 
 is not simple, and we give a simpler method for computing the 
 value in the two practically important cases: 
 
 1. In wooden beams (rectangular or square in cross-section), the 
 greatest unit-shearing stress in a section is 50 per cent larger than the average 
 value 8 g . 
 
 2. In I-beams, and in others with a thin vertical web, the greatest 
 unit-shearing stress in a section practically equals S g , as given by equation 7, 
 if the area of the web is substituted for A.
 
 74 STRENGTH OF MATERIALS 
 
 Examples. 1. What is the greatest value of the unit- 
 shearing stress in a wooden beam 12 feet long and 6 X 12 inches in 
 cross-section when resting on end supports and sustaining a uni- 
 form load of 6,400 pounds ? (This is the safe load as determined 
 by working fibre stress; see example 1, Art. 65.) 
 
 The maximum external shear equals one-half the load (see 
 Table B, page 53), and comes on the sections near the supports. 
 
 Since A = 6 X 12 72 square inches; 
 
 3,200 
 S s = ' = 44 pounds per square inch, 
 
 and the greatest unit-shearing stress equals 
 
 3 3 
 
 ~2~ ^s -9- 44 = 66 pounds per square inch. 
 
 Apparently this is very insignificant; but it is not negligible, as 
 is explained in the next article. 
 
 2. A steel I-beam resting on end supports 15 feet apart 
 sustains a load of 8,000 pounds 5 feet from one end. The weight 
 of the beam is 375 pounds, and the area of its web section is 3.2 
 square inches. (This is the beam and load described in examples 
 2 and 3, Art. 65.) What is the greatest unit-shearing stress ? 
 
 The maximum external shear occurs near the support where 
 the reaction is the greater, and its value equals that reaction. 
 Calling that reaction R, and taking moments about the other end 
 of the beam, we have 
 
 R X 15 - 375 x 7-|~ ~ 8 ' 000 X 10 = 0; 
 therefore 15 R = 80,000 + 2,812.5 = 82,812.5; 
 or, R = 5,520.8 pounds. 
 
 Hence S, = ' ' = 1,725 pounds per square inch. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. A wooden beam 10 feet long and 2 X 10 inches in cross - 
 section sustains a middle load of 1,000 pounds. Neglecting the 
 weight of the beam, compute the value of the greatest unit-shearing 
 stress. 
 
 Ans. 37.5 pounds per square inch.
 
 STRENGTH OF MATERIALS 75 
 
 2. Solve the preceding example taking into account the 
 weight of the beam, 60 pounds. 
 
 Ans. 40 pounds per square inch. 
 
 3. A wooden beam 12 feet long and 4 X 12 inches in cross- 
 section sustains a load of 8,000 pounds 4 feet from one end. 
 Neglecting the weight of the beam, compute the value of the 
 greatest shearing unit-stress. 
 
 Ans. 62. 5 pounds per square inch. 
 
 71. Horizontal Shear. It can be proved that there is a 
 shearing stress on every horizontal section of a loaded beam. An 
 experimental explanation will have to suffice here. Imagine a 
 pile of six boards of equal length supported so that they do not 
 bend. If the intermediate supports are removed, they will bend 
 and their ends will not be flush but somewhat as represented in 
 Fig. 41. This indicates that the boards slid over each other during 
 the bending, and hence there was a rubbing and a frictional re- 
 sistance exerted by the boards upon each other. Now, when a 
 solid beam is being bent, there is an exactly similar tendency for 
 the horizontal layers to slide over each other; and, instead of a 
 frictional resistance, there exists shearing stress on all horizontal 
 sections of the beam. 
 
 In the pile of boards the amount of slipping is different at 
 different places between any two boards, being greatest near the 
 supports and zero midway between them. Also, in any cross- 
 section the slippage is least between the upper two and lower two 
 boards, and is greatest between the middle two. These facts indi- 
 cate that the shearing unit-stress on horizontal sections of a solid 
 beam is greatest in the neutral surface at the supports. 
 
 It can be proved that at any place in a beam the shearing 
 unit-stresses on a horizontal and on a vertical section are equal. 
 
 :==? 
 
 Fig. 41. Fig. 42. 
 
 It follows that the horizontal shearing unit-stress is greatest at the 
 neutral axis of the section for which the external shear (Y) is a 
 maximum. Wood being very weak in shear along the grain, 
 timber beams sometimes fail under shear, the " rupture " being
 
 76 . STRENGTH OF MATERIALS 
 
 two horizontal cracks along the neutral surface somewhat as rep- 
 resented in Fig. 42. It is therefore necessary, when dealing with 
 timber beams, to give due attention to their strength as determined 
 by the working strength of the material in shear along the grain. 
 
 Example. A wooden beam 3 X 10 inches in cross-section 
 rests on end supports and sustains a uniform load of 4,000 pounds 
 Compute the greatest horizontal unit-stress in the beam. 
 
 The maximum shear equals one-half the load (see Table B, 
 page 55), or 2,000 pounds. Hence, by equation 7, since A = 
 3 x 10 = 30 square inches, 
 
 2 000 2 
 
 g g ' = 66-;r- pounds per square inch. 
 
 This is the average shearing unit-stress on the cross -sections near 
 the supports; and the greatest value equals 
 
 3 2 
 
 - X 66 = 100 pounds per square inch. 
 
 A o 
 
 According to the foregoing, this is also fehe value of the 
 greatest horizontal shearing unit-stress. (If of white pine, for 
 example, the beam would not be regarded as safe, since the ulti- 
 mate shearing strength along the grain of selected pine is only 
 about 400 pounds per square inch.) 
 
 72. Design of Timber Beams. In any case we may pro- 
 ceed as follows: (1) Determine the dimensions of the cross- 
 section of the beam from a consideration of the fibre stresses as 
 explained in Art. 66. (2) "With dimensions thus determined, com- 
 pute the value of the greatest shearing unit-stress from the formula, 
 
 g 
 Greatest shearing unit-stress = -^ V -*- <z5, 
 
 where V denotes the maximum external shear in the beam, and 
 b and a the breadth and depth of the cross -section. 
 
 If the value of the greatest shearing unit -stress so computed 
 does not exceed the working strength in shear along the grain, 
 then the dimensions are large enough ; but if it exceeds that value, 
 then a or 5, or both, should be increased until f V -=- ab is less 
 than the working strength. Because timber beams are very often 
 "season checked" (cracked) along the neutral surface, it is advia-
 
 STRENGTH OF MATERIALS 77 
 
 able to take the working strength of wooden beams, in shear along 
 the grain, quite low. One-twentieth of the working fibre strength 
 has been recommended* for all pine beams. 
 
 If the working strength in shear is taken equal to one- 
 iwentieth the working fibre strength, then it can be shown that, 
 
 1. For a beam on end supports loaded in the middle, the safe load de- 
 pends on the shearing or fibre strength according as the ratio of length to 
 depth (I -s- a) is less or greater than 10. 
 
 2. For a beam on end supports uniformly loaded, the safe load depends 
 on the shearing or fibre strength according as I * a is less or greater than 20. 
 
 Examples. 1. It is required to design a timber beam to sus- 
 tain loads as represented in Fig. 11, the working fibre strength 
 being 550 pounds and the working shearing strength 50 pounds 
 per square inch. 
 
 The maximum bending moment (see example for practice 3, 
 Art. 43; and example for practice 2, Art. 44) equals practically 
 7,000 foot-pounds or, 7,000 X 12 = 84,000 inch-pounds. 
 Hence, according to equation 6'", 
 
 = 152.7 inches'. 
 Since for a rectangle 
 
 -g- ba 2 = 152.7, or ba? = 916.2. 
 
 Now, if we let 5 = 4, then a? = 229; 
 
 or, a = 15.1 (practically 16) inches. 
 
 If, again, we let 6 = 6, then a? = 152.7; 
 
 or a = 12.4 (practically 1'4) inches. 
 
 Either of these sizes will answer so far as fibre stress is concerned, 
 but there is more " timber " in the second. 
 
 The maximum external shear in the beam equals 1,556 
 pounds, neglecting the weight of the beam (see example 3, Art. 
 87; and example 2, Art. 38). Therefore, for a 4 X 16-inch beam, 
 
 * See " Materials of Construction." JOHNSON. Page 55.
 
 78 STRENGTH OF MATERIALS 
 
 3 ., 1,556 
 Greatest shearing unit-stress =-g- X ^ ^ 
 
 = 36.5 pounds per square inch; 
 and for a 6 X 14-inch beam, it equals 
 
 X ' _ = 27.7 pounds per square inch. 
 2 o X 14 
 
 Since these values are less than the working strength in shear, 
 either size of beam is safe as regards shear. 
 
 If it is desired to allow for weight of beam, one of the sizes 
 should be selected. First, its weight should be computed, then 
 the new reactions, and then the unit-fibre stress may be com- 
 puted as in Art. 64, and the greatest shearing unit-stress as in the 
 foregoing. If these values are within the working values, then 
 the size is large enough to sustain safely the load and the weight 
 of the beam. 
 
 2. What is the safe load for a white pine beam 9 feet long 
 and 2 X 12 inches in cross-section, if the beam rests on end supports 
 and the load is at the middle of the beam, the working fibre 
 strength being 1,000 pounds and the shearing strength 50 pounds 
 per square inch. 
 
 The ratio of the length to the depth is less than 10; hence 
 the safe load depends on the shearing strength of the material 
 Calling the load P, the maximum external shear (see Table B, 
 page 53) equals ^ P, and the formula for greatest shearing unit 
 stress becomes 
 
 3 l P 
 50 = --X 2; or P = 1 ' 600 
 
 EXAMPLES FOR PRACTICE. 
 
 1. What size of wooden beam can safely sustain loads as in 
 Fig. 12, with shearing and fibre working strength equal to 50 and 
 1,000 pounds per square inch respectively ? 
 
 Ans. 6 X 12 inches 
 
 2. What is the safe load for a wooden beam 4 X 14 inches, 
 and 18 feet long, if the beam rests on end supports and the load 
 is uniformly distributed, with working strengths as in example 1 ? 
 
 Ans. 3,730 pounds
 
 STRENGTH OF MATERIALS 79 
 
 73. Kinds of Loads and Beams. We shall now discuss the 
 strength of beams under longitudinal forces (acting parallel to 
 the beam) and transverse loads. The longitudinal forces are 
 supposed to be applied at the ends of the beams and along the axis* 
 of the beam in each case. We consider only beams resting on 
 end supports. 
 
 The transverse forces produce bending or flexure, and the 
 longitudinal or end forces, if pulls, produce tension in the beam; 
 if pushes, they produce compression. Hence the cases to be con- 
 sidered may be called "Combined Flexure and Tension" and 
 " Combined Flexure and Compression." 
 
 74. Flexure and Tension. Let Fig. 43, a, represent a beam 
 subjected to the transverse loads L,, L 2 and L.,, and to two equal 
 end pulls P and P. The reactions R, and R 2 are due to the trans- 
 verse loads and can be computed by the methods of moments just 
 as though there were no end pulls. To find the stresses at any 
 cross- section, we determine those due to the transverse forces 
 (L,, L 2 , L 3 , Rj and R 2 ) and those due to the longitudinal; then 
 combine these stresses to get the total effect of all the applied 
 forces. 
 
 The stress due to the transverse forces consists of a shearing 
 stress and a fibre stress; it will be called the flexural stress. The 
 fibre stress is compressive above and tensile below. Let M denote 
 the value of the bending moment at the section considered; <?, and 
 c t the distances from the neutral axis to the highest and the low- 
 est fibre in the section; and S, and S 2 the corresponding unit-fibre 
 stresses due to the transverse loads. Then 
 
 Mr, Mc 2 
 
 IS, . ; and B a = j . 
 
 The stress due to the end pulls is a simple tension, and it equals 
 P; this is sometimes called the direct stress. Let S denote the 
 unit-tension due to P, and A the area of the cross -section; then 
 
 Both systems of loads to the left of a section between L, and 
 
 * NOTE. By " axis of a beam " is meant the line through the centers of 
 gravity of all the cross-sections.
 
 80 STRENGTH OF MATERIALS 
 
 L 2 are represented in Fig. 43, &; also the stresses caused by them 
 at' that section. Clearly the effect of the end pulls is to increase the 
 tensile stress (on the lower 
 fibres) and to decrease the 
 compressive stress (on the 
 upper fibres) due to the flex- 
 ure. Let S c denote the total 
 (resultant) unit-stress on the 
 upper fibre, and S t that on 
 the lower fibre, due to all 
 the forces acting on the beam. 
 In combining the stresses 
 there are two cases to con- 
 sider: Fig. 43. 
 
 (1) The flexural compressive unit-stress on the upper fibre is 
 greater than the direct unit-stress; that is, S, is greater than S . 
 The resultant stress on the upper fibre is 
 
 S c = Sj - S (compressive) ; 
 and that on the lower fibre is 
 
 S t = S 2 + S (tensile). 
 
 The combined stress is as represented in Fig. 43, <?, part tensile 
 and part compressive. ' 
 
 (2) The flexural compressive unit-stress is less than the 
 direct unit-stress; that is, S, is less than S . Then the combined 
 unit- stress on the upper fibre is 
 
 S C = S -S, (tensile); 
 and that on the lower fibre is 
 
 S t = S 2 + S (tensile). 
 
 The combined stress is represented by Fig. 43, d, and is all 
 tensile. 
 
 Example. A steel bar 2x6 inches, and 12 feet long, is sub- 
 jected to end pulls of 45,000 pounds. It is supported at each 
 end, and sustains, as a beam, a uniform load of 6,000 pounds. 
 It is required to compute the combined unit-fibre stresses. 
 
 Evidently the dangerous section is at the middle, and M = 
 \ WZ; that is,
 
 STRENGTH OF MATERIALS 81 
 
 M = -1 x 6 > 000 X 12 = 9,000 foot-pounds, 
 
 o 
 
 or 9,000 x 12 = 108,000 inch-pounds. 
 
 The bar being placed with the six-inch side vertical, 
 c l = c 2 = 3 inches, and 
 I=jL X 2x6 3 = 36 inches 4 . (See Art. 52.) 
 
 Hence S, = S 2 = 108 > 000 x 3 = 9,000 pounds per square inch. 
 Since A = 2 X 6 = 12 square inches, 
 
 S = ~ = 3,750 pounds per square inch. 
 
 The greatest value of the combined compressive stress is 
 
 S, - S = 9,000 - 3,750 = 5,250 pounds per square inch, 
 and it occurs on the upper fibres of the middle section. The great- 
 est value of the combined tensile stress is 
 
 S 2 + S = 9,000 + 3,750 = 12,750 pounds per square inch, 
 and it occurs on the lowest fibres of the middle section. 
 
 EXAflPLE FOR PRACTICE. 
 
 Change the load in the preceding illustration to one of 6,000 
 pounds placed in the middle, and then solve. 
 
 ( S c 14,250 pounds per square inch. 
 n8 ' \ S t = 21,750 
 
 75. Flexure and Compression. Imagine the arrowheads on 
 P reversed; then Fig. 43, #, will represent a beam under com- 
 bined flexural and compressive stresses. The flexural unit-stresses 
 are computed as in the preceding article. The direct stress is a 
 compression equal to P, and the unit-stress due to P is computed 
 as in the preceding article. Evidently the effect of P is to increase 
 the compressive stress and decrease the tensile stress due to the 
 flexure. In combining, we have two cases as before: 
 
 (1) The flexural tensile unit-stress is greater than the 
 direct unit-stress; that is, S 2 is greater than S . Then the com- 
 bined unit-stress on the lower fibre is
 
 82 STRENGTH OF MATERIALS 
 
 S t = S 2 - S (tensile) ; 
 and that on the upper fibre is 
 
 S c = S, + S (compressive). 
 
 The combined fibre stress is represented by Fig. 44, #, and is part 
 tensile and part compressive. 
 
 (2) The flexural unit- stress on the lower fibre is less than 
 the direct unit-stress; that is, S 2 is less than S . Then the com- 
 bined unit-stress on the lower fibre is 
 
 S t = S - S 2 (compressive); 
 and that on the upper fibre is 
 
 S c = S + S, (compressive). 
 The combined fibre stress is represented by 
 Fig. 44, J, and is all compressive. 
 
 Example. A piece of timber 6x6 
 inches, and 10 feet long, is subjected to end 
 pushes of 9,000 pounds. It is supported in 
 a horizontal position at its ends, and sustains 
 a middle load of 400 pounds. Compute the 
 combined fibre stresses. 
 
 Evidently the dangerous section is at the 
 middle, and M = J P?; that is, Fig. 44. 
 
 M = X 400 X 10 = 1,000 foot-pounds, 
 
 or 1,000 X 12 = 12,000 inch-pounds. 
 
 Since c l = c 3 = 3 inches, and 
 
 1 1 
 
 I = -jo ^ a> = To X 6 X 6 3 = 108 inches 4 , 
 
 _ 12,000 X 3 1 
 
 D t : D a ~- jQg rfrfrf ~ir pounds per square inch, 
 
 Kince A = 6 X 6 = 36 square inches, 
 
 9,000 
 = 36 = 250 pounds per square inch. 
 
 Hence the greatest value of the combined compressive stress is 
 
 1 i 
 
 S + Sj 333 -g- -f 250 = 583-15- pounds per square inch.
 
 STRENGTH OF MATERIALS 83 
 
 It occurs on the upper fibres of the middle section. The greatest 
 value of the combined tensile stress is 
 
 S 2 - S == 333^- - 250 = 83-g- pounds per square inch. 
 It occurs on the lowest fibres of the middle section. 
 
 EXAMPLE FOR PRACTICE. 
 
 Change the load of the preceding illustration to a uniform 
 load and solve. 
 
 ( S c ' = 417 pounds per square inch. 
 Ans. < o oo r , 
 
 (compression). 
 
 76. Combined Flexural and Direct Stress by flore Exact 
 Formulas. The results in the preceding articles are only approxi- 
 mately correct. Imagine the 
 beam represented in Fig. 45, a, 
 to be first loaded with the trans- 
 verse loads alone. They cause 
 the beam to bend more or less, 
 p LJ lLi_^--vP an( ^ produce certain flexural 
 
 :ic-^HEZ! ^-Ez^'t stresses at each section of the 
 
 t> beam. Now, if end pulls are 
 
 Fig. 45. 
 
 applied they tend to straighten 
 
 the beam and hence diminish the flexural stresses. This effect 
 of the end pulls was omitted in the discussion of Art. 74, and 
 the results there given are therefore only approximate, the 
 value of the greatest combined fibre unit-stress (S t ) being too 
 large. On the other hand, if the end forces are pushes, they in- 
 crease the bending, and therefore increase the flexural fibre stresses 
 already caused by the transverse forces (see Fig. 45, b). The 
 results indicated in Art. 75 must therefore in this case also be 
 regarded as only approximate, the value of the greatest unit- 
 fibre stress (S c ) being too small. 
 
 For beams loaded in the middle or with a uniform load, the 
 following formulas, which take into account the flexural effect of 
 the end forces, may be used : 
 
 M denotes bending moment at the middle section of the beam; 
 
 I denotes the moment of inertia of the middle section with 
 ^eapect to the neutral axis;
 
 84 STRENGTH OF MATERIALS 
 
 S lf S 2 , c l and c 2 have the same meanings as in Arts. 74 and 
 75, but refer always to the middle section ; 
 
 Z denotes length of the beam ; 
 
 E is a number depending on the stiffness of the material, the 
 average values of which are, for timber, 1,500,000; and for struc- 
 tural steel 30,000,000.* 
 
 c Me. , M<? 2 
 
 S -7T~FF' andS ^ : TT^:' (8) 
 ~TOE~ 10E 
 
 The plus sign is to be used when the end forces P are pulls, and 
 the minus sign when they are pushes. 
 
 It must be remembered that S t and S 2 are flexural unit- 
 stresses. The combination of these and the direct unit-stress is 
 made exactly as in articles 74 and 75. 
 
 Examples. 1. It is required to apply the formulas of this 
 article to the example of article 74. 
 
 As explained in the example referred to, M = 108,000 inch- 
 pounds; c t = c 2 = 3 inches; and I = 36 inches 4 . 
 Now, since I = 12 feet 144 inches, 
 
 108,000 X 3 324,000 
 
 > > 45,000 X I 36+3 ' 
 
 + 10 X 30,000,000 
 per square inch, as compared with 9,000 pounds per square inch, 
 the result reached by the u&e of the approximate formula. 
 As before, S = 3,750 pounds per square inch; hence 
 
 S c = 8,284- 3,750 = 4,534 pounds per square inch; 
 and S t = 8,284 + 3,750 = 12,034 
 
 2. It is required to apply the formulas of this article to the 
 example of article 75. 
 
 As explained in that example, 
 
 M = 12,000 inch-pounds; 
 c \ = C 2 = 3 inches, and I = 108 inches 4 . 
 Now, since I = 120 inches, 
 
 12 > QQO X 3 36,000 
 
 ' =: 2 = "~ X 120- = 108 - 8.64 = 362 
 
 10 X 1,500,000 
 * NOTE. This quantity " E " is more fully explained in Article 95.
 
 STRENGTH OF MATERIALS 85 
 
 per square inch, as compared with 333| pounds per square inch, 
 the result reached by use of the approximate method. 
 As before, S = 250 pounds per square inch ; hence 
 
 S c =' 362 + 250 = 612 pounds per square inch; and 
 S t = 362 - 250 = 112 " " " . 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Solve the example for practice of Art. 74 by the formulas 
 of this article. 
 
 A ( S c = 12,820 pounds per square inch. 
 ns ' \ S t = 20,320 
 
 2. Solve the example for practice of Art. 75 by the formulas 
 of this article. 
 
 A j S c = 430 pounds per square inch. 
 
 1 \ S t = 70 (compression). 
 
 STRENGTH OF COLUflNS. 
 
 A stick of timber, a bar of iron, etc., when used to sustain 
 end loads which act lengthwise of the pieces, are called columns, 
 posts, or struts if they are so long that they would bend before 
 breaking. When they are so short that they would not bend 
 before breaking, they are called short blocks, and their compres- 
 sive strengths are computed by means of equation 1. The strengths 
 of columns cannot, however, be so simply determined, and we now 
 proceed to explain the method of computing them. 
 
 77. End Conditions. The strength of a column depends in 
 part on the way in which its ends bear, or are joined to other 
 parts of a structure, that is, on its " end conditions." There are 
 practically but three kinds of end conditions, namely: 
 
 1. " Hinge " or " pin " ends, 
 
 2. "Flat" or "square" ends, and 
 
 3. "Fixed" ends. 
 
 (1) When a column is fastened to its support at one end by 
 means of a pin about which the column could rotate if the other 
 end were free, it is said to be "hinged" or "pinned" at the 
 former end. Bridge posts or columns are often hinged at the ends. 
 
 (2) A column either end of which is flat and perpendicular 
 to its axis and bears on other parts of the structure at that surface, 
 is said to be " flat " or " square" at that end.
 
 STRENGTH OF MATERIALS 
 
 (3) Columns are sometimes riveted near their ends directly 
 to other parts of the structure and do not bear directly on their 
 ends; such are called " fixed ended." A column which bears on its 
 flat ends is often fastened near the ends to other parts of the struc- 
 ture, and such an end is also said to be " fixed." The fixing of an 
 end of a column stiffens and therefore strengthens it more or less, 
 but the strength of a column with fixed ends is computed as 
 though its ends were flat. Accordingly we have, so far as strength 
 is concerned, the following classes of columns : 
 
 78. Classes of Columns. (1) Both ends hinged or pinned; 
 (2) one end hinged and one flat; (3) both ends flat. 
 
 Other things being the same, columns of these three classes 
 are unequal in strength. Columns of the first class are the 
 weakest, and those of the third class are the strongest. 
 
 
 
 B 
 
 
 _A_ 
 
 
 ._. 
 
 A. 
 
 
 ' 
 
 B 
 
 D 
 
 
 Fig. 46. 
 
 79. Cross-sections of Columns. Wooden columns are usu- 
 ally solid, square, rectangular, or round in section; but sometimes 
 they are "built up" hollow. Cast-iron columns are practically 
 always made hollow, and rectangular or round in section. Steel 
 columns are made of single rolled shapes angles, zees, channels, 
 etc.; but the larger ones are usually " built up" of several shapes. 
 Fig. 46, a, for example, represents a cross -section of a "Z-bar" 
 column; and Fig. 46, fl, that of a "channel" column. 
 
 80. Radius of Gyration. There is a quantity appearing in 
 almost all formulas for the strength of columns, which is called 
 " radius of gyration." It depends on the form and extent of the 
 cross-section of the column, and may be defined as follows:
 
 STRENGTH OF MATERIALS 87 
 
 The radius of gyration of any plane figure (as the section of a column) 
 with respect to any line, is such a length that the square of this length mul- 
 tiplied by the area of the figure equals the moment of inertia of the figure 
 with respect to the given line. 
 
 Thus, if A denotes the area of a figure; I, its moment of in- 
 ertia with respect to some line; and r, the radius of gyration 
 with respect to that line; then 
 
 r*A. = I; or r = I/I -s- A. (9) 
 
 In the column formulas, the radius of gyration always refers to an 
 axis through the center of gravity of the cross-section, and usually 
 to that axis with respect to which the radius of gyration (and mo. 
 ment of inertia) is least. (For an exception, see example 3, 
 Art. 83.) Hence the radius of gyration in this connection is often 
 called for brevity the " least radius of gyration," or simply the 
 least radius." 
 
 Examples. 1. Show that the value of the radius of gyration 
 given for the square in Table A, page 52, is correct. 
 
 The moment of inertia of the square with respect to the axis 
 is T V# 4 - Since A = a 2 , then, by formula 9 -above, 
 
 2. Prove that the value of the radius of gyration given for 
 the hollow square in Table A, page 54, is correct. 
 
 The value of the moment of inertia of the square with respect 
 to the axis is -J^ (a* - a'}. Since A = a 2 a*, 
 
 EXAHPLE FOR ^PRACTICE. 
 
 Prove that the values of the radii of gyration of the other fig- 
 ures given in Table A, page 52, are correct. The axis in each 
 case is indicated by the line through the center of gravity. 
 
 81. Radius of Gyration of Built-up Sections. The radius of 
 gyration of a built-up section is computed similarly to that of any 
 other figure. First, we have to compute the moment of inertia of
 
 88 STRENGTH OF MATERIALS 
 
 the section, ab explained in Art. 54; and then we use formula 9,aa 
 in the examples of the preceding article. 
 
 Example. It is required to compute the radius of gyration 
 of the section represented in Fig. 30 (page 52) with respect to the 
 axis AA. 
 
 In example 1, Art. 54, it is shown that the moment of inertia 
 of the section with respect to the axis AA is 429 inches 4 . The 
 area of the whole section is 
 
 2 X 6.03 + 7 = 19.06; 
 hence the radius of gyration r is 
 
 = J 429 = 4.74 inches. 
 \ 19.06 
 
 EXAMPLE FOR PRACTICE. 
 
 Compute the radii of gyration of the section represented in 
 Fig. 31, a, with respect to the axes AA and BB. (See examples 
 
 for practice 1 and 2, Art. 54. ) 
 
 . ( 2.87 inches. 
 
 AnS ' J2.09 
 
 82. Kinds of Column Loads. When the loads applied to a 
 column are such that their resultant acts through the center of 
 gravity of the top section and along the axis of the column, the 
 column is said to be centrally loaded. When the resultant of the 
 loads does not act through the center of gravity of the top 
 section, the column is said to be eccentrically loaded. All the 
 following formulas refer to columns centrally loaded. 
 
 83. Rankine's Column Formula. When a perfectly straight 
 column is centrally loaded, then, if the column does not bend and 
 if it is homogeneous, the stress on every cross-section is a uniform 
 compression. If P denotes the load and A the area of the cross- 
 Bection, the value of the unit-compression is P -r- A. 
 
 On account of lack of straightness or lack of uniformity in 
 material, or failure to secure exact central application of the load, 
 the load P has what is known as an " arm " or " leverage " and 
 bends the column more or less. There is therefore in such a 
 lolumn a bending or flexural stress in addition to the direct com- 
 ^ressive stress above mentioned ; this bending stress is compressive
 
 STRENGTH OF MATERIALS 89 
 
 on the concave side and tensile on the convex. The value of the 
 stress per unit-area (unit-stress) on the fibre at the concave side, 
 according to equation 6, is Me -r- I, where M denotes the bending 
 moment at the section (due to the load on the column), c the 
 distance from the neutral axis to the concave side, and I the 
 moment of inertia of the cross-section with respect to the neutral 
 axis. (Notice that this axis is perpendicular to the plane in 
 which the column bends.) 
 
 The upper set of arrows (Fig. 47) represents the direct com- 
 pressive stress; and the second set the bending stress if the load 
 is not excessive, so that the stresses are within the elastic limit of 
 the material. The third set represents the combined stress that 
 actually exists on the cross -section. The greatest combined unit- 
 stress evidently occurs on the fibre at the concave side and where 
 the deflection of the column is greatest. The 
 stress is compressive, and its value S per unit- 
 area is given by the formula, 
 
 8= P +M?. 
 
 Now, the bending moment at the place of 
 greatest deflection equals the product of the 
 load P and its arm (that is, the deflection). 
 ' * Calling the deflection d, we have M = Pd', and 
 this value of M, substituted in the last equa 
 tion, gives 
 
 IlPnn B __P + M,. 
 
 Fig. 47. A I 
 
 Let r denote the radius of gyration of the cross-section with respec 
 to the neutral axis. Then I = Ar 2 (see equation 9); and this 
 value, substituted in the last equation, gives 
 
 P Pde P do 
 
 = X + A? = X ( l + -?)' 
 
 According to the theory of the stiffness of beams on end sup- 
 ports, deflections vary directly as the square of the length Z, and in- 
 versely as the distance c from the neutral axis to the remotest fibre 
 of the cross -section. Assuming that the deflections of columns
 
 90 
 
 STRENGTH OF MATERIALS 
 
 follow the same laws, we may write d = k (I 2 -5- c), where k ia 
 some constant depending on the material of the column and on the 
 end conditions. Substituting this value for d in the last equation, 
 we find that 
 
 s =-j(i 
 
 P _ S 
 
 x- 
 
 and 
 
 P = 
 
 SA 
 
 (10) 
 
 Each of these (usually the last) is known as " Rankme's formula." 
 
 For mild-steel columns a certain large steel company uses S = 50,000 
 pounds per square inch, and the following values of k: 
 
 1. Columns with two pin ends, k 1 -H 18,000. 
 
 2. " " one flat and one pin end, k = 1 -=- 24,000. 
 
 3. " " both ends flat, k = 1 -*- 36,000. 
 
 With these values of S and k, P of the formula means the ultimate load, 
 that is, the load causing failure. The safe load equals P divided by the 
 selected factor of safety a factor of 4 for steady loads, and 5 for moving 
 loads, being recommended by the company referred to. The same unit is to 
 be used for I and r. 
 
 Cast-iron columns are practically always made hollow with 
 comparatively thin walls, and are usually circular or rectangular 
 in cross -section. The following modifications of Rankine's formula 
 are sometimes used: 
 
 For circular sections, 
 For rectangular sections, 
 
 80,000 
 
 1 4 
 
 ' 
 
 80,000 
 
 (ID") 
 
 In these formulas d denotes the outside diameter of the circular sec- 
 tions or the length of the lesser side of the rectangular sections. The same 
 unit is to be used for I and d. 
 
 Examples. 1. A 40-pound 10-inch steel I-beam 8 feet 
 long is used as a flat-ended column. Its load being 100,000 
 pounds, what is its factor of safety ? 
 
 Obviously the column tends to bend in a plane perpendicular 
 to its web. Hence the radius of gyration to be used is the one
 
 STRENGTH OF MATERIALS 91 
 
 with respect to that central axis of the cross-section which is in 
 the web, that is, axis 2 2 (see figure accompanying table, page 72). 
 The moment of inertia of the section with respect to that axis, 
 according to the table, is 9.50 inches 4 ; and since the area of the 
 section is 11.76 square inches, 
 
 -=0.30, 
 
 Now, I = 8 feet = 96 inches; and since k = 1 -+- 36,000, and S = 
 50,000, the breaking load for this column, according to Rankine'a 
 formula, is 
 
 P = 50.000X1176 _ _ 
 
 yb 
 1 + 86,000 X 0.808. 
 
 Since the factor of safety equals the ratio of the breaking load to 
 the actual load on the column, the factor of safety in this case is 
 
 = 4.5 (appro*.). 
 
 2. What is the safe load for a cast-iron column 10 feet long 
 with square ends and a hollow rectangular section, the outside 
 dimensions being 5X8 inches; the inner, 4x7 inches; and the 
 factor of safety, 6 ? 
 
 In this case I = 10 feet = 120 inches; A = 5x8-4x7 
 = 12 square inches; and d= 5 inches. Hence, according to 
 formula 10', for rectangular sections, the breaking load is 
 
 P = 8 ' 000 * 12 =610,000 pounds. 
 
 1 4- 
 
 r 1,000 X 5' 
 
 Since the safe load equals the breaking load divided by the factor 
 of safety, in this case the safe load equals 
 
 5^992=, 101,700 pounds. 
 
 3. A channel column (see Fig. 46, b) is pin -ended, the pins 
 being perpendicular to the webs of the channel (represented by 
 AA in the figure), and its length is 16 feet (distance between axes
 
 92 STRENGTH OF MATERIALS 
 
 of the pins). If the sectional area is 23.5 square inches, and the 
 moment of inertia with respect to AA is 386 inches 4 and with 
 respect to BB 214 inches 4 , what is the safe load with a factor of 
 safety of 4 ? 
 
 The column is liable to bend in one of two ways, namely, in 
 the plane perpendicular to the axes of the two pins, or in the plane 
 containing those axes. 
 
 (1) For bending in the first plane, the strength of the col- 
 umn is to be computed from the formula for a pin -ended column. 
 Hence, for this case, ?' 2 = 386 -r- 23.5 = 16; and the breaking 
 load is 
 
 50,000 X 23.5 ... _ n 
 
 P= (16X12)- -W^OO pounds. 
 
 f 18,000 X 16 
 The safe load for this case equals -- - -- = 260,400 pounds. 
 
 (2) If the supports of the pins are rigid, then the pina 
 stiffen the column as to bending in the plane of their axes, and the 
 strength of the column for bending in that plane should be com- 
 puted from the formula for the strength of columns with flat ends. 
 Hence, r 2 = 214 H- 23.5 = 9.11, and thebreaking load is 
 
 50,000 X 23.5 
 
 , 16 x 12 )2 = 1,056,000 pounds. 
 
 36,000 X 9.11 
 The safe load for this case equals - - = 264,000 ponnds. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. A 40-pound 12-inch steel I-beam 10 feet long is used as 
 a column with flat ends sustaining a load of 100,000 pounds. 
 What is its factor of safety? 
 
 Ana. 4.1 
 
 2. A cast-iron column 15 feet long sustains a load of 
 150,000 pounds. Its section being a hollow circle, 9 inches out- 
 side and 7 inches inside diameter, what is the factor of safety? 
 
 Ans. 8.9 
 
 3. A steel Z-bar column (see Fig. 46, a) is 24 feet long and 
 has square ends; the least radius of gyration of its cross -section is
 
 STRENGTH OF MATERIALS 
 
 93 
 
 3.1 inches; and the area of the cross-section is 24.5 square inches. 
 What is the safe load for the column with a factor of safety of 4 ? 
 
 Ans. 247,000 pounds. 
 
 4. A cast-iron column 13 feet long has a hollow circular 
 cross-section 7 inches outside and 5^ inches inside diameter. 
 What is its safe load with a factor of safety of 6? 
 
 Ans. 121,142 pounds. 
 
 5. Compute the safe load for a 40-pound 12-inch steel 
 I-beam used as a column with flat ends, its length being 17 feet. 
 Use a factor of safety of 5. 
 
 Ans. 52,470 pounds. 
 
 84. Graphical Representation of Column Formulas. Col- 
 umn (and most other engineering) formulas can be represented 
 graphically. To represent Rankine's formula for flat-ended mild- 
 steel columns, 
 
 P 50,000 
 
 we first substitute different values of I -z- r in the formula, and 
 solve for P -f- A. Thus we find, when 
 
 I -i- r - 40, P *- A = 47,900 ; 
 l]^ r = 80, P -* A = 42,500 ; 
 I + r = 120, P - A = 35,750 ; 
 etc., etc. 
 
 Now, if these values of I -=- r be laid off by some scale on a line 
 from O, Fig. 48, and the corresponding values of P -f- A be laid 
 
 5000O 
 4OOOO] 
 30000 
 2000CH 
 
 1OOOO 
 
 ,1-H" 
 
 Fig. 48. 
 
 off vertically from the points on the line, we get a series of points 
 as #, #, <?, etc.; and a smooth curve through the points a, b, c,
 
 94 STRENGTH OF MATERIALS 
 
 etc., represents the formula. Such a curve, besides representing 
 the formula to one's eye, can be used for finding the value of 
 P ~- A for any value of I -+- r; or the value of / -r- r for any value 
 of P -T- A. The use herein made is in explaining other column 
 formulas in succeeding articles. 
 
 85. Combination Column Formulas. Many columns have 
 been tested to destruction in order to discover in a practical way 
 the laws relating to the strength of columns of different kinds. 
 The results of such tests can be most satisfactorily represented 
 graphically by plotting a point in a diagram for each test. Thus, 
 suppose that a column whose I -4- r was 80 failed under a load of 
 276,000 pounds, and that the area of its cross-section was 7.12 
 square inches. This test would be represented by laying off O, 
 Fig. 49, equal to 80, according to some scale; and then ab equal to 
 276,000 -H 7.12 (P -*- A), according to some other convenient 
 scale. The point b would then represent the result of this par- 
 ticular test. All the dots in the figure represent the way in which 
 the results of a series of tests appear when plotted. 
 
 It will be observed at once that the dots do not fall upon any 
 one curve, as the curve of Rankine's formula. Straight lines and 
 
 soooo 
 
 40000 
 
 "f 
 
 
 
 30000 
 20000 
 
 b 
 
 l-s-r 
 
 100 200 300 
 
 Fig. 49. 
 
 curves simpler than the curve of Rankine's formula have been 
 fitted to represent the average positions of the dots as determined 
 by actual tests, and the formulas corresponding to such lines have 
 been deduced as column formulas. These are explained in the 
 following articles. 
 
 86. Straight-Line and Euler Formulas. It occurred to Mr. 
 T. H. Johnson that most of the dots corresponding to ordinary
 
 STRENGTH OF MATERIALS 
 
 95 
 
 lengths of columns agree with a straight line just as well as with 
 a curve. He therefore, in 1886, made a number of such plats or 
 diagrams as Fig. 49, fitted straight lines to them, and deduced the 
 formula corresponding to each line. These have become known 
 as "straight-line formulas," and their general form is as follows: 
 
 P - l 
 
 A r ' 
 
 P, A, Z, and r having meanings as in Rankine's formula (Art. 83), 
 and S and m being constants whose values according to Johnson 
 are given in Table E below. 
 
 For the slender columns, another formula (Euler's, long since 
 deduced) was used by Johnson. Its general form is 
 
 P n 
 
 X = (TTlf () 
 
 n being a constant whose values, according to Johnson, are given 
 in the following table: 
 
 TABLE E. 
 
 Data for Mild-Steel Columns. 
 
 
 S 
 
 m 
 
 Limit (I -s- r) 
 
 n 
 
 Hinged ends ...... 
 Flat ends 
 
 52,500 
 52,500 
 
 220 
 
 180 
 
 160 
 
 195 
 
 444,000,000 
 666,000,000 
 
 The numbers in the -fourth column of the table mark the point of divi- 
 sion between columns of ordinary length and slender columns. For the 
 former kind, the straight-line formula applies; and for the second, Euler's. 
 That is, if the ratio I -*- r for a steel column with hinged end, for example, is 
 less than 160, we must use the straight-line formula to compute its safe load, 
 factor of safety, etc.; but if the ratio is greater than 160, we must use Euler's 
 formula. 
 
 For cast-iron columns with flat ends, S = 34,000, and m 88; and since 
 they should never be used " slender," there is no use of Euler's formula for 
 cast-iron columns. 
 
 The line AB, Fig. 50, represents Johnson's straight-line for- 
 mula; and BC, Euler's formula. It will be noticed that the two 
 lines are tangent; the point of tangency corresponds to the "lim- 
 iting value " I -r- r, as indicated in the table. 
 
 Examples. 1. A 40-pound 10-inch steel I-beam column 8
 
 STRENGTH OF MATERIALS 
 
 feet long sustains a load of 100,000 pounds, and the ends are flu 
 Compute its factor of safety according to the methods of tbi- 
 article. 
 
 The first thing to do is to compute the ratio I -r- r for the 
 column, to ascertain whether the straight-line formula or Euler's 
 
 F-J-A 
 
 50000 
 
 AOOOO- 
 
 30000 
 
 2OOOO- 
 
 10000 
 
 100 
 
 200 
 
 30O 
 
 Fig. 50. 
 
 formula should be used. From Table C, on page 70, we find that 
 the moment of inertia of the column about the neutral axis of 
 its cross-section is 9.50 inches 4 , and the area of the section is 
 11.76 square inches. Hence 
 
 9.50 
 IL76" 
 
 = 0.81; or r = 0.9 inch. 
 
 Since I = 8 feet = 96 inches, 
 I 96 
 
 106 T 
 
 This value of I -*- r is less than the limiting value (195) indicated 
 by the table for steel columns with flat ends (Table E, p. 97), and 
 we should therefore use the straight-line formula; hence 
 
 = 52,500 - 180 X 106 
 
 - 
 
 or, P = 11.76 (52,500 - 180 X 106 ) = 391,600 pounds. 
 
 This is the breaking load for the column according to the straight- 
 line formula; hence the factor of safety is 
 
 391,600 
 100,000 ~ : d * y
 
 STRENGTH OF MATERIALS 97 
 
 2. Suppose that the length of the column described in the 
 preceding example were 16 feet. "What would its factor of safety be? 
 
 Since I = 16 feet = 192 inches; and, as before, r = 0.9 
 inch, -!-?' = 218J. This value is greater than the limiting 
 value (195) indicated by Table E (p. 97) for flat-ended steel col- 
 umns; hence Euler's formula is to be used. Thus 
 
 P _ 666,000,000 1 
 IL76 " 
 
 11.70 X 006,000,000 
 P = - ~ = 172,100 pounds. 
 
 This is the breaking load; hence the factor of safety is 
 
 172,100 _ 1 
 100,000 ~ 
 
 3. What is the safe load for a cast-iron column 10 feet long 
 with square ends and hollow rectangular section, the outside 
 dimensions being 5x8 inches and the inside 4x7 inches, with a 
 factor of safety of 6 ? 
 
 Substituting in the formula for the radius of gyration given 
 in Table A, page 52, we get 
 
 8 x 5 a -7x 4 3 
 12 (8 X 5 - 7 X 4) 
 
 Since I 10 feet = 120 inches, 
 
 According to the straight-line formula for cast iron, A being 
 equal to 12 square inches, 
 
 ^ = 34,000 - 88 X 61.22; 
 
 or, P = 12 (34,000 - 88 X 61.22) = 343,360 pounds. 
 
 This being the breaking load, the safe load is 
 
 B43 ' 860 = 57,227 pounds.
 
 98 STRENGTH OF MATERIALS 
 
 EXAMPLES FOR PRACTICE. 
 
 1. A 40-pound 12-inch steel I-beam 10 feet long is used as 
 a flat-ended column. Its load being 100,000 pounds, compute 
 the factor of safety by the formulas of this article. 
 
 Ans. 3.5 
 
 2. A cast-iron column 15 feet long sustains a load of 
 150,000 pounds. Its section being a hollow circle of 9 inches 
 outside and 7 inches inside diameter, compute the factor of safety 
 by the straight-line formula. 
 
 Ana. 4.8 
 
 3. A steel Z-bar column (see Fig. 46, a] is 24 feet long 
 and has square ends; the least radius of gyration of its cross- 
 eection is 3.1 inches; and the area of the cross-section is 24.5 
 square inches. Compute the safe load for the column by the 
 formulas of this article, using a factor of safety of 4. 
 
 Ans. 219,000 pounds. 
 
 4. A hollow cast-iron column 13 feet long has a circular 
 cross-section, and is 7 inches outside and 5| inches inside in 
 diameter. Compute its safe load by the formulas of this article, 
 using a factor of safety of 6. 
 
 Ans. 68,300 pounds. 
 
 5. Compute by the methods of this article the safe load for 
 a 40-pound 12-inch steel I-beam used as a column with flat ends, 
 if the length is 17 feet and the factor of safety 5. 
 
 Ans. 35,050 pounds. 
 
 87. Parabola-Euler Formulas. As better fitting the results 
 of tests of the strength of columns of " ordinary lengths," Prof. 
 J. B. Johnson proposed (1892) to use parabolas instead of straight 
 lines. The general form of the parabola formula " is 
 
 P, A, I and r having the same meanings as in Rankine's formula, 
 Art. 83; and S and m denoting constants whose values, according 
 to Professor Johnson, are given in Table F below. 
 
 Like the straight-line formula, the parabola formula should 
 not be used for slender columns, but the following (Euler's) is 
 applicable:
 
 STRENGTH OF MATERIALS 
 
 (14) 
 the values of n (Johnson) being given in the following table: 
 
 TABLE F. 
 
 Data for find Steel Columns. 
 
 
 s 
 
 m 
 
 Limit (I -t- r) 
 
 n 
 
 Hinged ends 
 Flat ends 
 
 42,000 
 42,000 
 
 0.97 
 0.62 
 
 150 
 190 
 
 456,000,000 
 712,000,000 
 
 The point of division between columns of ordinary length and slender 
 columns is given in the fourth column of the table. That is, if the ratio Z-r-r 
 for a column with hinged ends, for example, is less than 150, the parabola 
 formula should be used to compute the safe load, factor of safety, etc.; but 
 if the ratio is greater than 150, then Euler's formula should be used. 
 
 The line AB, Fig. 51, represents the parabola formula; and the line 
 BC, Euler's formula. The two lines are tangent, and the point of tangency 
 corresponds to the "limiting value" ZH-r of the table. 
 
 For wooden columns square in cross-section, it is convenient to replace 
 r by d, the latter denoting the length of the sides of the square. The formula 
 becomes 
 
 S and m for flat-ended columns of various kinds of wood having the follow- 
 ing values according to Professor Johnson: 
 
 For White pine, 8=2,500, m = 0.6; 
 
 " Short-leaf yellow pine, 8=3,300, m = 0.7; 
 
 " Long-leaf yellow pine, 8=4,000, m = 0.8; 
 
 " White oak, 8=3,500, m = 0.8. 
 
 The preceding formula applies to any wooden column whose ratio, l-r-d t 
 
 is less than 60, within which limit columns of practice are included. 
 
 soooo- 
 A 
 
 4OOOO- 
 3OOOO- 
 20000- 
 1OOOO- 
 
 , l+r 
 
 1OO 2OO 
 
 . Fig. 51. 
 
 300 
 
 Examples. 1. A 40-pound 10-inch steel I-beam column
 
 100 STRENGTH OF MATERIALS 
 
 8 feet long sustains a load of 100,000 pounds, and its ends are flat. 
 Compute its factor of safety according to the methods of this 
 article. 
 
 The first thing to do is to compute the ratio I -f- r for the 
 column, to ascertain whether the parabola formula or Euler's for- 
 mula should be used. As shown in example 1 of the preceding 
 article, I -f- r = 106|. This ratio being less than the limiting 
 value, 190, of the table, we should use the parabola formula. 
 Hence, since the area of the cross-section is 11.76 square inches 
 (see Table C, page 70), 
 
 P = 42,000 - 0.62 (106)'; 
 
 or, P = 11.76 [42,000 - 0.62 (106|) 2 ] = 410,970 pounds. 
 
 This is the breaking load according to the parabola formula; hence 
 the factor of safety is 
 
 410,970 ^ 41 
 
 100,000 
 
 2. A white pine column 10 X 10 inches in cross-section and 
 18 feet long sustains a load of 40,000 pounds. What is its factor 
 of safety ? 
 
 The length is 18 feet or 216 inches ; hence the ratio I ~d = 
 21.6, and the parabola formula is to be applied. 
 Now, since A = 10 X 10 = 100 square inches, 
 
 P 
 
 JQQ = 2,500 - 0.6 X 21.6 2 ; 
 
 or, P = 100 (2,500 - 0.6 X 21.6 2 ) = 222,000 pounds 
 
 This being the breaking load according to the parabola formula, 
 the factor of safety is 
 
 222,000 
 
 1poo- = 
 
 3. What is the safe load for a long-leaf yellow pine column 
 12 X 12 inches square and 30 feet long, the factor of safety 
 being 5 ? 
 
 The length being 30 feet or 360 inches, 
 
 * 360 
 d = 18" = 3 '>
 
 STRENGTH OF MATERIALS 101 
 
 hence the parabola formula should be used. Since A = 12 X 12 
 = 144 square inches, 
 
 JL = 4,000 - 0.8 X 30'; 
 
 or, P = 144 (4,000 - 0.8 X 30 2 ) = 472,320 pounds. 
 
 This being the breaking load according to the parabola formula, 
 the safe load is 
 
 = 94,466 pound,. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. A 40-pound 12-inch steel I-beam 10 feet long is used as 
 a flat-ended column. Its load being 100,000 pounds, compute its 
 factor of safety by the formulas of this article. 
 
 Ans. 3.8 
 
 2. A white oak column 15 feet long sustains a load of 
 30,000 pounds. Its section being 8x8 inches, compute the 
 factor of safety by the parabola formula. 
 
 Ans. 6.6 
 
 3. A steel Z-bar column (see Fig. 46, a) is 24 feet long and 
 has square ends; the least radius of gyration of its cross-section 
 is 3.1 inches; and the area of its cross-section is 24.5 square 
 inches. Compute the safe load for the column by the formulas 
 of this article, using a factor of safety of 4. 
 
 Ans. 224,500 pounds. 
 
 4. A short-leaf yellow pine column 14 X 14 inches in sec- 
 tion is 20 feet long. What load can it sustain, with a factor of 
 safety of 6 ? 
 
 Ans. 101,100 pounds. 
 
 88. " Broken Straight-Line " Formula. A large steel com- 
 pany computes the strength of its flat-ended steel columns by two 
 formulas represented by two straight lines AB and BC, Fig. 52. 
 The formulas are 
 
 - = 48,000, 
 -^ = 68,400 
 
 <A. 
 
 P, A, Z, and r having the same meanings as in Art. 83. 
 
 and -^ = 68,400 - 228 -, 
 
 *
 
 102 
 
 STRENGTH OF MATERIALS 
 
 The point B corresponds very nearly to the ratio I -*- r = 90. 
 Hence, for columns for which the ratio I -f- r is less than 90, the 
 first formula applies; and for columns for which the ratio is 
 greater than 90, the second one applies. The point C corre- 
 sponds to the ratio I -t- r = 200, and the second formula does not 
 apply to a column for which I H- r is greater than that limit. 
 
 10000 
 
 The ratio I -r- r for steel columns of practice rarely exceeds 150, 
 and is usually less than 100. 
 
 Fig. 53 is a combination of Figs. 49, 50, 51 and 52, and 
 represents graphically a comparison of the Rankine, straight-line, 
 Euler, parabola -Euler, and broken straight-line formulas for flat- 
 ended niild-steel columns, It well illustrates the fact that our 
 knowledge of the strength of columns is not so exact as that, for 
 example, of the strength of beams. 
 
 P+A 
 
 50000 
 
 40000: 
 
 3OOOO 
 20OOO 
 
 10000 
 
 100 
 
 Fig. 53. 
 
 200 
 
 300 
 
 89. Design of Columns. All the preceding examples relat- 
 ing to columns were on either (1) computing the factor of safety
 
 STRENGTH OF MATERIALS 103 
 
 of a given loaded column, or (2) computing the safe load for a 
 given column. A more important problem is to design a column 
 to sustain a given load under given conditions. A complete dis- 
 cussion of this problem is given in a later paper on design. "We 
 show here merely how to compute the dimensions of the cross- 
 section of the column after the form of the cross-section has been 
 decided upon. 
 
 In only a few cases can the dimensions be computed directly 
 (see example 1 following), but usually, when a column formula is 
 applied to a certain case, there will be two unknown quantities in 
 it, A and r or d. Such cases can best be solved by trial (see 
 examples 2 and 3 below). 
 
 Example. 1. What is the proper size of white pine column 
 to sustain a load of 80,000 pounds with a factor of safety of 5, 
 when the length of the column is 22 feet ? 
 
 We use the parabola formula (equation 13). Since the safe 
 load is 80,000 pounds and the factor of safety is 5, the breaking 
 load P is 
 
 80,000 X 5 = 400,000 pounds. 
 
 The unknown side of the (square) cross-section being denoted by 
 d, the area A is d 2 . Hence, substituting in the formula, since I 
 == 22 feet = 264 inches, we have 
 
 *W = 2,500- 0.6 ?. 
 
 Multiplying both sides by d 2 gives 
 
 400,000 = 2,500 d 2 - 0.6 X 264 a , 
 
 or 2,500 d* = 400,000 + 0.6 x 264' = 441,817.6. 
 
 Hence d 2 = 176.73, or d = 13.3 inches. 
 
 2. What size of cast-iron column is needed to sustain a load 
 of 100,000 pounds with a factor of safety of 10, the length of the 
 column being 14 feet ? 
 
 We shall suppose that it has been decided to make the cross- 
 section circular, and shall compute by Rankine's formula modified 
 for cast-iron columns (equation 10'). The breaking load for the 
 column would be
 
 104 STRENGTH OF MATERIALS 
 
 100,000 X 10 = 1,000,000 pounds. 
 The length is 14 feet or 168 inches; hence the formula oecomes 
 
 1,000,000 80,000 t 
 A 168 2 ' 
 
 or, reducing by dividing both sides of the equation by 10,000, and 
 then clearing of fractions, we have 
 
 1 Afi2 
 
 100 [1 + ^^-^1 = 8A. 
 
 There are two unknown quantities in this equation, d and A, and 
 we cannot solve directly for them. Probably the best way to pro- 
 ceed is to assume or guess at a practical value of d y then solve for 
 A, and finally compute the thickness or inner diameter. Thus, let 
 us try d equal to 7 inches, first solving the equation for A as far 
 as possible. Dividing both sides by 8 we have 
 
 100 f 1682 
 : 8 I + 800<Z 2 J' 
 and, combining, 
 
 4.4.1 
 
 A = 12.5 + ^. 
 
 Now, substituting 7 for d, we have 
 
 441 
 A = 12.5 + -JQ = 21.5 square inches. 
 
 The area of a hollow circle whose outer and inner diameters are 
 d and d l respectively, is 0.7854 (d 2 - d*}. Hence, to find the inner 
 diameter of the column, we substitute 7 for d in the last expres- 
 sion, equate it to the value of A just found, and solve for d t . Thus, 
 
 0.7854 (49-^ 2 ) = 21.5- 
 hence 
 
 and d* = 49 - 27.37 = 21.63 or d l = 4.65. 
 
 This value of d makes the thickness equal to 
 
 |(7- 4.65) = 1.175 inches,
 
 STRENGTH OF MATERIALS 105 
 
 which is safe. It might be advisable in an actual case to try 
 d equal to 8 repeating the computation.* 
 
 EXAMPLE FOR PRACTICE. 
 
 1. What size of white oak column is needed to sustain a load 
 of 45,000 pounds with a factor of safety of 6, the length of the 
 column being 12 feet. 
 
 Ans. d = 9.05, practically a 10 X 10-inch section 
 
 STRENGTH OF SHAFTS. 
 
 A shaft is a part of a machine or system of machines, and is 
 used to transmit power by virtue of its torsional strength, or resist- 
 ance to twisting. Shafts are almost always made of metal and are 
 usually circular in cross -section, being sometimes made hollow. 
 
 90. Twisting Moment. Let AF, Fig. 54, represent a shaft 
 with four pulleys on it. Suppose that D is the driving pulley 
 and that B, C and E are pulleys from which power is taken off to 
 drive machines. The portions of the shafts between the pulleys 
 
 P 4 
 
 Pig. 54. 
 
 are twisted when it is transmitting power; and by the twisting 
 moment at any cross- section of the shaft is meant the algebraic 
 sum of the moments of all the forces acting on the shaft on either 
 
 *NOTE. The structural steel handbooks contain extensive tables by 
 means of which the design of columns of steel or cast iron is much facilitated. 
 The difficulties encountered in the use of formula are well illustrated in this 
 example.
 
 106 STRENGTH OF MATERIALS 
 
 side of the section, the moments being taken with respect to th 
 axis of the shaft. Thus, if the forces acting on the shaft (at the 
 pulleys) are P,, P 2 , P 3 , and P 4 as shown, and if the arms of the 
 forces or radii of the pulleys are a v a z , a,, and a t respectively, then 
 the twisting moment at any section in 
 
 BC is P, a n 
 
 CD is P x a, + P 2 a v 
 
 DE is P, a, + P 2 a 2 - P 3 a y 
 
 Like bending moments, twisting moments are usually ex- 
 pressed in inch -pounds. 
 
 Example. Let a t = a t = 4 = 15 inches, a 3 = 30 inches, 
 P, = 400 pounds, P 2 = 500 pounds, P 8 = 750 pounds, and P 4 = 
 600 pounds.* What is the value of the greatest twisting moment 
 in the shaft? 
 
 At any section between the first and second pulleys, the 
 twisting moment is 
 
 400 X 15 = 6,000 inch-pounds; 
 
 at any section between the second and third it is 
 
 400 X 15 + 500 X 15 = 13,500 inch-pounds; and 
 at any section between the third and fourth it is 
 
 400 X 15 + 500 X 15 - 750 X 30 = - 9,000 inch-pounds. 
 Hence the greatest value is 13,500 inch-pounds. 
 
 91. Torsional Stress. The stresses in a twisted shaft are 
 called "torsional" stresses. The torsional stress on a cross -section 
 of a shaft is a shearing stress, as in the case illustrated by Fig. 55, 
 which represents a flange coupling in a shaft. Were it not for 
 the bolts, one flange would slip over the other when either part 
 of the shaft is turned; but the bolts prevent the slipping. Obvi- 
 ously there is a tendency to shear the bolts off unless they are 
 screwed up very tight; that is, the material of the bolts is sub- 
 jected to shearing stress. 
 
 Just so, at any section of the solid shaft there is a tendency 
 for one part to slip past the other, and to prevent the slipping or 
 
 * Note. These numbers were so chosen that the moment of P (driving 
 moment) equals the sum of the moments of the other forces. This is always 
 the case in a shaft rotating at constant speed; that is, the power given the 
 shaft equals the power taken off.
 
 STRENGTH OF MATERIALS 
 
 107 
 
 shearing of the shaft, there arise shearing stresses at all parts of 
 the cross-section. The shearing stress on the cross-section of a 
 shaft is not a uniform stress, its value per unit-area being zero at 
 the center of the section, and increasing toward the circumference. 
 In circular sections, solid or hollow, the shearing stress per unit- 
 area (unit-stress) varies directly as the distance from the center 
 of the section, provided the elastic limit is not exceeded. Thus, 
 if the shearing unit-stress at the circumference of a section is 
 
 Fig. 55. 
 
 1,000 pounds per square inch, and the diameter of the shaft is 
 2 inches, then, at ^ inch from the center, the unit-stress is 500 
 pounds per square inch; and at ^ inch from the center it is 250 
 pounds per square inch. In Fig. 55 the arrows indicate the 
 values and the directions of the shearing stresses on very small 
 portions of the cross-section of a shaft there represented. 
 
 92. Resisting Moment. By "resisting moment" at a sec- 
 tion of a shaft is meant the sum of the moments of the shearing 
 
 O 
 
 stresses on the cross- section about the axis of the shaft. 
 
 Let S_ denote the value of the shearing stress per unit-area 
 
 S O A 
 
 (unit-stress) at the outer points of a section of a shaft; d the 
 diameter of the section (outside diameter if the shaft is hollow); 
 and d^ the inside diameter. Then it can be shown that the re- 
 sisting moment is: 
 
 For a solid section, 0.1963 S s <f; 
 
 0.1963 S s (d 4 - d. 1 ) 
 For a hollow section, ^ 
 
 93. Formula for the Strength of a Shaft. As in the case
 
 108 STRENGTH OF MATERIALS 
 
 of beams, the resisting moment equals the twisting moment at 
 any section. If T be used to denote twisting moment, then we 
 have the formulas : 
 
 For solid circular shafts, 0.1963 S s d 3 = T; 
 
 , , , 0.1963 S s <Y7 4 -r7, 4 ) r 
 For hollow circular shafts, - -^ - I^= 
 
 In any portion of a shaft of constant diameter, the unit- 
 shearing stress S s is greatest where the twisting moment is greatest. 
 Hence, to compute the greatest unit-shearing stress in a shaft, 
 we first determine the value of the greatest twisting moment, 
 substitute its value in the first or second equation above, as the 
 case may be, and solve for S s . It is customary to express T in 
 inch-pounds and the diameter in inches, S s then being in pounds 
 per square inch. 
 
 Examples. 1. Compute the value of the greatest shearing 
 unit-stress in the portion of the shaft between the first and second 
 pulleys represented in Fig. 54, assuming values of the forces and 
 pulley radii as given in the example of article 90. Suppose also 
 that the shaft is solid, its diameter being 2 inches. 
 
 The twisting moment T at any section of the portion between 
 the first and second pulleys is 6,000 inch-pounds, as shown in the 
 example referred to. Hence, substituting in the first of the two 
 formulas 15 above, we have 
 
 0.1963 S s X 2 3 = 6,000; 
 
 6 000 
 or, S 8 = Q i9fi3 x g = 3,820 pounds per square inch. 
 
 This is the value of the unit-stress at the outside portions of all 
 sections between the first and second pulleys. 
 
 2. A hollow shaft is circular in cross.section, and its outer 
 and inner diameters are 16 and 8 inches respectively. If the 
 working strength of the material in shear is 10,000 pounds per 
 square inch, what twisting moment can the shaft safely sustain ? 
 
 The problem requires that we merely substitute the values of 
 S s , <7, and d^ in the second of the above formulas 15, and solve for 
 T. Thus, 
 
 T - -1963 X 10,000 (16* - 8*) 7 Q9n . 
 
 - _ - i - L = 7,od7,920 inch-pounds.
 
 STRENGTH OF MATERIALS 109 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Compute the greatest value of the shearing unit-stress in 
 the shaft represented in Fig. 54, using the values of the forces 
 and pulley radii given in the example of article 90, the diameter 
 of the shaft being 2 inches. 
 
 Ans. 8,595 pounds per square inch 
 
 2. A solid shaft is circular in cross-section and is 9.6 inches 
 in diameter. If the working strength of the material in shear is 
 10,000 pounds per square inch, how large a twisting moment can 
 the shaft safely sustain? (The area of the cross -section is practically 
 the same as that of the hollow shaft of example 2 preceding.) 
 
 Ans. 1,736,736 inch-pounds. 
 
 94. Formula for the Power Which a Shaft Can Transmit. 
 The power that a shaft can safely transmit depends on the shear- 
 ing working strength of the material of the shaft, on the size of 
 the cross -section, and on the speed at which the shaft rotates. 
 
 Let H denote the amount of horse-power; S s the shearing 
 working strength in pounds per square inch; d the diameter 
 (outside diameter if the shaft is hollow) in inches; d l the inside 
 diameter in inches if the shaft is hollow; and n the number of 
 revolutions of the shaft per minute. Then the relation between 
 power transmitted, unit-stress, etc., is: 
 
 For solid shafts, 11= Ss ^^ ' 
 
 For hollow shafts, H = 
 
 Examples. 1. What horse-power can a hollow shaft 16 
 inches and 8 inches in diameter safely transmit at 50 revolutions 
 per minute, if the shearing working strength of the material is 
 10,000 pounds per square inch? 
 
 We have merely to substitute in the second of the two for- 
 mulas 16 above, and reduce. Thus, 
 
 H = = 6,000 horse-power (nearfy). 
 
 2. What size of solid shaft is needed to transmit 6,000 horse- 
 power at 50 revolutions per minute if the shearing working 
 strength of the material is 10,000 pounds per square inch?
 
 HO STRENGTH OF MATERIALS 
 
 We have merely to substitute in the first of the two formulas 
 16, and solve for d. Thus, 
 
 10,000 X d* X 50 
 6 ' " 321,000 
 
 6,000X32^000 _ 
 therefore d? = io,000 X 50 
 
 OT) d ^3^852 = 15.68 inches. 
 
 (A solid shaft of this diameter contains over 25% more material than 
 the hollow shaft of example 1 preceding. There is therefore considerable 
 caving of material in the hollow shaft.) 
 
 3. A solid shaft 4 inches in diameter transmits 200 horse- 
 power while rotating at 200 revolutions per minute. What is the 
 greatest shearing unit-stress in the shaft? 
 
 We have merely to substitute in the first of the equations 16, 
 and solve for S. Thus, 
 
 200 - 
 
 321,000 
 
 200 X 321,000 
 S = 3 = 5,016 pounds per square inch. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. What horse-power can a solid shaft 9.6 inches in diameter 
 safely transmit at 50 revolutions per minute, if its shearing work- 
 ing strength is 10,000 pounds per square inch ? 
 
 Ans. 1,378 horse-power. 
 
 2. What size of solid shaft is required to transmit 500 horse- 
 power at 150 revolutions per minute, the shearing working strength 
 of the material being 8,000 pounds per square inch. 
 
 Ans. 5.1 inches. 
 
 3. A hollow shaft whose outer diameter is 14 and inner 6.7 
 inches transmits 5,000 horse-power at 60 revolutions per minute. 
 What is the value of the greatest shearing unit- stress in the shaft? 
 
 Ans. 10,273 pounds per square inch. 
 
 STIFFNESS OF RODS, BEAMS, AND SHAFTS. 
 
 The preceding discussions have related to the strength of
 
 STRENGTH OF MATERIALS 111 
 
 .;-j}-?T,-. 
 
 materials. "We shall now consider principally the elongation of 
 rods, deflection of beams, and twist of shafts. 
 
 95. Coefficient of Elasticity. According to Hooke's Law 
 (Art. 9, p. 7), the elongations of a rod subjected to an increasing 
 pull are proportional to the pull, provided that the stresses due to 
 the pull do not exceed the elastic limit of the material. Within 
 the elastic limit, then, the ratio of the pull and the elongation is 
 constant; hence the ratio of the unit-stress (due to the pull) to the 
 unit-elongation is also constant. This last-named ratio is called 
 " coefficient of elasticity." If E denotes this coefficient, S the 
 unit-stress, and s the unit-deformation, then 
 
 Coefficients of elasticity are usually expressed in pounds per square inch. 
 
 The preceding remarks, definition, and formula apply also to 
 a case of compression, provided that the material being compressed 
 does not bend, but simply shortens in the direction of the com- 
 pressing forces. The following table gives the average values of 
 the coefficient of elasticity for various materials of construction: 
 
 Coe 
 
 TABLE Q. 
 
 fficients of Elasticity. 
 
 
 Material. 
 
 Average Coefficient of 
 
 Elasticity. 
 
 Steel 
 
 30,000,000 pounds per 
 
 square inch. 
 
 Wrought iron 
 Cast iron 
 
 27,500,000 
 15,000,000 " " 
 
 
 Timber 
 
 1 800 000 " " 
 
 (i <( 
 
 
 
 
 The coefficients of elasticity for steel and wrought iron, for different 
 grades of those materials, are remarkably constant; but for different grades 
 of cast iron the coefficients range from about 10,000,000 to 30,000,000 pounds 
 per square inch. Naturally the coefficient has not the same value for the 
 different kinds of wood; for the principal woods it ranges from 1,600,000 
 (for spruce) to 2,100,000 (for white oak). 
 
 Formula 17 can be put in a form more convenient for use, as 
 follows : 
 
 Let P denote the force producing the deformation ; A the 
 area of the cross -section of the piece on which P acts ; I the length 
 of the piece ; and D the deformation (elongation or shortening).
 
 112 STRENGTH OF MATERIALS 
 
 Then 
 
 S = P -5- A (see equation 1), 
 and s = D -*- I (see equation 2). 
 
 Hence, substituting these values in equation 17, we have 
 
 PZ PI 
 
 E = AD' orD = AE' < I7) 
 
 The first of these two equations is used for computing the value of 
 the coefficient of elasticity from measurements of a " test," and 
 the second for computing the elongation or shortening of a given 
 rod or bar for which the coefficient is known. 
 
 Examples. 1. It is required to compute the coefficient of 
 elasticity of the material the record of a test of which is given on 
 page 9. 
 
 Since the unit-stress S and unit-elongation s are already 
 computed in that table, we can use equation 17 instead of the first 
 of equations 17'. The elastic limit being between 40,000 and 
 45,000 pounds per square inch, we may use any value of the 
 unit-stress less than that, and the corresponding unit-elongation. 
 
 Thus, with the first values given, 
 
 5,000 
 
 "With the second, 
 
 This lack of constancy in the value of E as computed from different 
 loads in a test of a given material, is in part due to errors in measuring the 
 deformation, a measurement difficult to make. The value of the coefficient 
 adopted from such a test, is the average of all the values of E which can be 
 computed from the record. 
 
 2. How much will a pull of 5,000 pounds stretch a round 
 steel rod 10 feet long and 1 inch in diameter ? 
 
 "We use the second of the two formulas 17'. Since A = 
 0.7854 X I 2 = 0.7854 square inches, I = 120 inches, and E = 
 30,000,000 pounds per square inch, the stretch is: 
 
 n _ 5,000 X 120 
 
 = 0.7854 X 30,000,000
 
 STRENGTH OF MATERIALS 113 
 
 EXAMPLES FOR PRACTICE. 
 
 1. What is the coefficient of elasticity of a material if a pull 
 of 20,000 pounds will stretch a rod 1 inch in diameter and 4 feet 
 
 long 0.045 inch ? 
 
 Ans. 27,000,000 pounds per square inch. 
 
 2. How much will a pull of 15,000 pounds elongate a round 
 cast-iron rod 10 feet long and 1 inch in diameter ? 
 
 Ans. 0.152 inch. 
 
 96. Temperature Stresses. In the case of most materials, 
 when a bar or rod is heated, it lengthens; and when cooled, it 
 shortens if it is free to do so. The coefficient of linear expansion 
 of a material is the ratio which the elongation caused in a rod or 
 bar of the material by a change of one degree in temperature bears 
 to the length of the rod or bar. Its values for Fahrenheit degrees 
 are about as follows: 
 
 For Steel, 0.0000065. 
 
 For Wrought iron, .0000067. 
 For Cast iron, .0000062. 
 
 Let K be used to denote this coefficient; t a change of tem- 
 perature, in degrees Fahrenheit; I the length of a rod or bar; 
 and D the change in length due to the change of temperature. 
 Then 
 
 D == K tl. (l8) 
 
 D and Z are expressed in the same unit. 
 
 If a rod or bar is confined or restrained so that it cannot 
 change its length when it is heated or cooled, then any change in 
 its temperature produces a stress in the rod; such are called tem- 
 perature stresses. 
 
 Examples. 1. A steel rod connects two solid walls and is 
 screwed up so that the unit-stress in it is 10,000 pounds per 
 square inch. Its temperature falls 10 degrees, and it is observed 
 that the walls have not been drawn together. What is the temper- 
 ature stress produced by the change of temperature, and what is 
 the actual unit-stress in the rod at the new temperature ? 
 
 Let I denote the length of the rod. Then the change in 
 length which would occur if the rod were free, is given by formula 
 18, above, thus: 
 
 D = 0.0000065 X 10 X I = 0.000065 I.
 
 114 STRENGTH OF MATERIALS 
 
 Now, since the rod could not shorten, it has a greater than normal 
 length at the new temperature; that is, the fall in temperature has 
 produced an effect equivalent to an elongation in the rod amount- 
 ing to D, and hence a tensile stress. This tensile stress can be 
 computed from the elongation D by means of formula 17. Thus, 
 
 S = Es; 
 and since , the unit-elongation, equals 
 
 D .0000065 I 
 j = .0000065, 
 
 S = 30,000,000 X .0000065 = 1950 pounds per square inch. 
 This is the value of the temperature stress; and the new unit- 
 stress equals 
 
 10,000 + 1950 = 11,950 pounds per square inch. 
 
 Notice that the unit temperature stresses are independent of the length 
 of the rod and the area of its cross-section. 
 
 2. Suppose that the change of temperature in the preceding 
 example is a rise instead of a fall. What are the values of the 
 temperature stress due to the change, and of the new unit-stress in 
 the rod ? 
 
 The temperature stress is the same as in example 1, that is, 
 1,950 pounds per square inch ; but the rise in temperature 
 releases, as it were, the stress in the rod due to its being screwed 
 up, and the final unit stress is 
 
 10,000 - 1,950 = 8,050 pounds per square inch. 
 EXAflPLE FOR PRACTICE. 
 
 1. The ends of a wrought-iron rod 1 inch in diameter are 
 fastened to two heavy bodies which are to be drawn together, the 
 temperature of the rod being 200 degrees when fastened to the ob- 
 jects. A fall of 120 degrees is observed not to move them. 
 What is the temperature stress, and what is the pull exerted by 
 the rod on each object ? 
 
 Ans \ Tem P eratlire stre ss, 22,110 pounds per square inch. 
 
 ' ( Pull, 17,365 pounds. 
 
 97- Deflection of Beams. Sometimes it is desirable to know 
 how much a given beam will deflect under a given load, or to design
 
 STRENGTH OF MATERIALS 115 
 
 a beam which will not deflect more than a certain amount under a 
 given load. In Table B, page 53, Part I, are given formulas for 
 deflection in certain cases of beams and different kinds of loading. 
 
 In those formulas, d denotes deflection; I the moment of inertia of the 
 cross-section of the beam with respect to the neutral axis, as in equation 6; 
 and E the coefficient of elasticity of the material of the beam (for values, see 
 Art. 95). 
 
 In each case, the load should be expressed in pounds, the length in 
 inches, and the moment of inertia in biquadratic inches; then the deflection 
 will be in inches. 
 
 According to the formulas for 6?, the deflection of a beam 
 varies inversely as the coefficient of its material (E) and the mo- 
 ment of inertia of its cross-section (I) ; also, in the first four and 
 last two cases of the table, the deflection varies directly as the cube 
 of the length (I 3 ). 
 
 Example. What deflection is caused by a uniform load of 
 6,400 pounds (including weight of the beam) in a wooden beam 
 on end supports, which is 12 feet long and 6 X 12 inches in 
 cross -section ? (This is the safe load for the beam ; see example 
 1, Art. 65.) 
 
 The formula for this case (see Table B, page 53) is 
 
 ___ 
 ~ 384 El ' 
 
 Here W = 6,400 pounds ; I = 144 inches ; E = 1,800,000 
 pounds per square inch ; and 
 
 I = -^ ba 3 = i 6 X 12 3 = 864 inches 4 . 
 
 Hence the deflection is 
 
 5 X 6,400 X 144 3 
 ~ 384 X 1,800,000 X 864 = 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Compute the deflection of a timber built-in cantilever 
 8x8 inches which projects 8 feet from the wall and bears an 
 end load of 900 pounds. (This is the safe load for the cantilever, 
 see example 1, Art. 65.) 
 
 Ans. 0.43 inch. 
 
 2. Compute the deflection caused by a uniform load of 40,000
 
 116 
 
 STRENGTH OF MATERIALS 
 
 pounds on a 42-pound 15-inch steel I-beam which is 16 feet long 
 and rests on end supports. 
 
 Ans. 0.28 inch. 
 
 98. Twist of Shafts. Let Fig. 57 represent a portion of a 
 shaft, and suppose that the part represented lies wholly between 
 
 Fig. 57. 
 
 two adjacent pulleys on a shaft to which twisting forces are applied 
 (see Fig. 54). Imagine two radii ma and nb in the ends of the 
 portion, they being parallel as shown when the shaft is not twisted. 
 After the shaft is twisted they will not be parallel, ma having 
 moved to ma', and nb to nb'. The angle between the two lines in 
 their twisted positions (ma 1 and nb') is called the angle of twist, 
 or angle of torsion, for the length I. If a' a" is parallel to ab, then 
 the angle a"nb' equals the angle of torsion. 
 
 If the stresses in the portion of the shaft considered do not 
 exceed the elastic limit, and if the twisting moment is the same 
 for all sections of the portion, then the angle of torsion a (in 
 degrees) can be computed from the following: 
 
 For solid circular shafts, 
 
 584 TZ 36,800,000 HZ 
 
 For hollow circular shafts, 
 
 584 TO 36,800,000 
 
 (19) 
 
 "E 1 (ffi-dfin 
 
 Here T, Z, eZ, d lt H, and n have the same meanings as in Arts. 93 
 and 94, and should be expressed in the units there used. The 
 letter E 1 stands for a quantity called coefficient of elasticity for 
 shear; it ia analogous to the coefficient of elasticity for tension and 
 compression (E), Art. 95. The values of E 1 for a few materiala 
 average about as follows (roughly E 1 = | E) :
 
 STRENGTH OF MATERIALS 117 
 
 For Steel, 11,000,000 pounds per square inch. 
 
 For Wrought iron, 10,000,000 " " " " 
 For Cast iron, 6,000,000 " " " " 
 
 Example. What is the value of the angle of torsion of a 
 steel shaft 60 feet long when transmitting 6,000 horse-power at 
 50 revolutions per minute, if the shaft is hollow and its outer and 
 inner diameters are 16 and 8 inches respectively ? 
 
 Here I = 720 inches; hence, substituting in the appropriate 
 formula (19), we find that 
 
 36,800,000 X 6,000 X 720 
 = 11,000,000 X (16- -8-) 50 = 
 
 EXAMPLE FOR PRACTICE. 
 
 Suppose that the first two pulleys in Fig. 54 are 12 feet 
 apart; that the diameter of the shaft is 2 inches; and that P t = 400 
 pounds, and , = 15 inches. If the shaft is of wrought iron, 
 what is the value of the angle of torsion for the portion between 
 the first two pulleys ? 
 
 Ans. 3.15 degrees. 
 
 99. Non-elastic Deformation. The preceding formulas for 
 elongation, deflection, and twist hold only so long as the greatest 
 unit-stress does not exceed the elastic limit. There is no theory, 
 and no formula, for non-elastic deformations, those corresponding 
 to stresses which exceed the elastic limit. It is well known, how- 
 ever, that non -elastic deformations are not proportional to the 
 forces producing them, but increase much faster than the loads. 
 The value of the ultimate elongation of a rod or bar (that is, the 
 amount of elongation at rupture), is quite well known for many 
 materials. This elongation, for eight-inch specimens of various 
 materials (see Art. 16), is : 
 
 For Cast iron, about 1 per cent. 
 For Wrought iron (plates), 12 - 15 per cent. 
 For " " (bars), 20-25 " " . 
 
 For Structural steel, 22 - 26 " " . 
 
 Specimens of ductile materials (such as wrought iron and 
 structural steel), when pulled to destruction, neck down, that is, 
 diminish very considerably in cross-section at some place along 
 the length of the specimen. The decrease in cross -sectional area
 
 118 
 
 STRENGTH OF MATERIALS 
 
 is known as reduction of area, and its value for wrought iron and 
 steel may be as much as 50 per cent. 
 
 RIVETED JOINTS. 
 
 100. Kinds of Joints. A lap joint is one in which the 
 plates or bars joined overlap each other, as in Fig. 58, a. A butt 
 Joint is one in which the plates or bars that are joined butt against 
 each other, as in Fig. 58, J. The thin side plates on butt joints 
 
 Fig. 58. 
 
 are called cover-plates ; the thickness of each is always made not 
 less than one-half the thickness of the main plates, that is, the 
 plates or bars that are joined. Sometimes butt joints are made 
 with only one cover-plate; in such a case the thickness of the 
 cover-plate is made not less than that of the main plate. 
 
 When wide bars or plates are riveted together, the rivets are 
 placed in rows, always parallel to the " seam " and sometimes also 
 perpendicular to the seam; but when we speak of a row of rivets, 
 we mean a row parallel to the seam. A lap joint with a single 
 row of rivets is said to be single-riveted ; and one with two rows 
 of rivets is said to be double-riveted. A butt joint with two rows 
 of rivets (one on each side of the joint) is called " single-riveted," 
 and one with four rows (two on each side) is said to be "double- 
 riveted." 
 
 The distance between the centers of consecutive holes in a 
 row of rivets is called pitch. 
 
 101. Shearing Strength, or Shearing Value, of a Rivet. 
 When a lap joint is subjected to tension (that is, when P, Fig. 58, 
 a, is a pull), and when the joint is subjected to compression (when 
 P is a push), there is a tendency to cut or shear each rivet along 
 the surface between the two plates. In butt joints with two cover-
 
 STRENGTH OF MATERIALS 119 
 
 plates, there is a tendency to cut or shear each rivet on two sur- 
 faces (see Fig. 58, 2>). Therefore the rivets in the lap joint are 
 said to be in single shear ; and those in the butt joint (two covers) 
 are said to be in double shear. 
 
 The " shearing value " of a rivet means the resistance which 
 it can safely offer to forces tending to shear it on its cross-section. 
 This value depends on the area of the cross-section and on the work- 
 ing strength of the material. Let d denote the diameter of the 
 cross-section, and S, the shearing working strength. Then, since 
 the area of the cross-section equals 0.7854 d 2 , the shearing strength 
 of one rivet is : 
 
 For single shear, 0.7854 d 2 S, . 
 
 For double shear, 1.5708 d 2 S. . 
 
 102. Bearing Strength, or Bearing Value, of a Plate. When 
 
 a joint is subjected to tension or compression, each rivet presses 
 against a part of the sides of the holes through which it passes. 
 By " bearing value " of a plate (in this connection) is meant the 
 pressure, exerted by a rivet against the side of a hole in the plate, 
 which the plate can safely stand. This value depends on the 
 thickness of the plate, on the diameter of the rivet, and on the 
 compressive working strength of the plate. Exactly how it 
 depends on these three qualities is not known; but the bearing 
 value is always computed from the expression t d S c , wherein t 
 denotes the thickness of the plate; d, the diameter of the rivet or 
 hole; and S c , the working strength of the plate. 
 
 103. Frictional Strength of a Joint. When a joint is sub- 
 jected to tension or compression, there is a tendency to slippage 
 between the faces of the plates of the joint. This tendency is 
 overcome wholly or in part by frictional resistance between the 
 plates. The frictional resistance in a well-made joint may be 
 very large, for rivets are put into a joint hot, and are headed or 
 capped before being cooled. In cooling they contract, drawing the 
 plates of the joint tightly against each other, and producing a 
 great pressure between them, which gives the joint a correspond- 
 ingly large frictional strength. It is the opinion of some that 
 all well-made joints perform their service by means of their 
 frictional strength ; that is to say, the rivets .act only by pressing 
 the plates together and are not under shearing stress, nor
 
 120 STRENGTH OF MATERIALS 
 
 are the plates under compression at the sides of their holes. The 
 " frictional strength " of a joint, however, is usually regarded as 
 uncertain, and generally no allowance is made for friction in com- 
 putations on the strength of riveted joints. 
 
 104. Tensile and Compressive Strength of Riveted Plates. 
 The holes punched or drilled in a plate or bar weaken its tensile 
 strength, and to compute that strength it is necessary to allow for 
 the holes. By net section, in this connection, is meant the small- 
 est cross-section of the plate or bar ; this is always a section along 
 a line of rivet holes. 
 
 If, as in the foregoing article, t denotes the thickness of the 
 plates joined ; d, the diameter of the holes; n^ the number of riv- 
 ets in a row ; and w, the width of the plate or bar; then the net 
 section = (w - nfy t. 
 
 Let S t denote the tensile working strength of the plate ; then 
 the strength of the unriveted plate is wt& a and the reduced tensile 
 strength is (w - n^ t S t . 
 
 The compressive strength of a plate is also lessened by the 
 presence of holes ; but when they are again filled up, as in a joint, 
 the metal is replaced, as it were, and the compressive strength of 
 the plate is restored. No allowance is therefore made for holes in 
 figuring the compressive strength of a plate. 
 
 105. Computation of the Strength of a Joint. The strength 
 of a joint is determined by either (1) the shearing value of the 
 rivets; (2) the bearing value of the plate; or (3) the tensile 
 strength of the riveted plate if the joint is in tension. Let P. de- 
 note the strength of the joint as computed from the shearing 
 values of the rivets ; P c , that computed from the bearing value of 
 the plates ; and P t , the tensile strength of the riveted plates. 
 Then, as before explained, 
 
 P s = n 2 0.7854 ?S B ; and t (20) 
 
 n, denoting the total number of rivets in the joint ; and n 3 denot- 
 ing the total number of rivets in a lap joint, and one-half the 
 number of rivets in a butt joint. 
 
 Examples. 1. Two half -inch plates 7| inches wide are con-
 
 STRENGTH OF MATERIALS 121 
 
 nected by a single lap joint double-riveted, six rivets in two rows. 
 If the diameter of the rivets is | inch, and the working strengths 
 are as follows : S t = 12,000, S.= 7,500, and S c = 15,000 pounds 
 per square inch, what is the safe tension which the joint can 
 transmit ? 
 
 Here n 1 = 3, n a = 6, and n a = 6 ; hence 
 
 1 31 
 
 P t = (7-g-- 3 X --) X -g- X 12,000 = 31,500 pounds; 
 
 P 8 = 6 X 0.7854 X (-|-) a X 7,500 = 19,880 pounds ; 
 
 p c== 6x-^-X-|-X 15,000 = 33,750 pounds. 
 
 Since P, is the least of these three values, the strength of the 
 joint depends on the shearing value of its rivets, and it equals 
 19,880 pounds. 
 
 2. Suppose that the plates described in the preceding example 
 are joined by means of a butt joint (two cover-plates), and 12 
 rivets are used, being spaced as before. What is the safe tension 
 which the joint can bear ? 
 
 Here n t = 3, n 2 = 12, and n a = 6; hence, as in the preced- 
 ing example, 
 
 P t = 31,500; and P c = 33,750 pounds; but 
 
 P g = 12 x 0.7854 X (^-Y X 7,500 = 39,760 pounda 
 
 The strength equals 31,500 pounds, and the joint is stronger than 
 the first. 
 
 3. Suppose that in the preceding example the rivets are 
 arranged in rows of two. What is the tensile strength of the* 
 joint ? 
 
 Here n l = 2. n a = 12, and n 3 = 6; hence, as in the preced- 
 ing example, 
 
 P s = 39,760; and P c ='33,750 pounds; but 
 
 P t = (7 -jL_ 2 x -|-) -i X 12,000 = 36,000 pounds. 
 
 The strength equals 33,750 pounds, and this joint is stronger than 
 either of the first two.
 
 /22 STRENGTH OF MATERIALS 
 
 EXAMPLES FOR PRACTICE. 
 
 Note. Use working strengths as in example 1, above. 
 
 S t = 12,000, S. = 7,500, and S c = 15,000 pounds per square inch. 
 
 1. Two half-inch plates 5 inches wide are connected by a 
 lap joint, with two |-inch rivets in a row. What is the safe 
 strength of the joint ? 
 
 Ans. 6,625 pounds. 
 
 2. Solve the preceding example supposing that four |-inch 
 rivets are used, in two rows. 
 
 Ans. 13,250 pounds. 
 
 3. Solve example 1 supposing that three 1-inch rivets are 
 used, placed in a row lengthwise of the joint. 
 
 Ans. 17,670 pounds. 
 
 4. Two half-inch plates 5 inches wide are connected by a 
 butt joint (two cover-plates), and four |-inch rivets are used, in 
 two rows. What is the strength of the joint ? 
 
 Ans. 11,250 pounds. 
 
 106. Efficiency of a Joint. The ratio of the strength of a 
 joint to that of the solid plate is called the " efficiency of the 
 joint." If ultimate strengths are used in computing the ratio, 
 then the efficiency is called ultimate efficiency; and if working 
 strengths are used, then it is called working efficiency. In the 
 following, we refer to the latter. An efficiency is sometimes ex- 
 pressed as a per cent. To express it thus, multiply the ratio 
 strength of joint -5- strength of solid plate, by 100. 
 
 Example. It is required to compute the efficiencies of the 
 joints described in the examples worked out in the preceding article. 
 
 In each case the plate is j inch thick and 7 inches wide; 
 hence the tensile working strength of the solid plate is 
 
 7 ~~ X T" X 12 ' 000 = 45,000 pounds. 
 Therefore the efficiencies of the joints are : 
 
 V A ; 
 
 (2) 
 (3) 
 
 45,000 
 31,500 
 
 v.-x-x, 
 
 0.70, 
 0.75, 
 
 Ul 
 
 or 
 or 
 
 ttrt 
 
 70 
 
 75 
 
 UCI CC11 I , 
 
 per cent; 
 per cent, 
 
 45,000 
 33,750 
 45,000 ~
 
 INDEX 
 
 Page 
 
 Angle of torsion 116 
 
 Angle of twist 116 
 
 Applications of first beam formula 59 
 
 Beam formula 
 
 first 59 
 
 second 73 
 
 Beams 19 
 
 deflection of 114 
 
 stiffness of 110 
 
 Bearing strength of a plate 119 
 
 Bending moment 23, 32 
 
 maximum 40 
 
 Brick, ultimate strength of 15 
 
 Broken straight-line formula 101 
 
 Built-up sections, moment of inertia of 49 
 
 Butt joint 118 
 
 Cast iron 
 
 in tension 13 
 
 in compression 15 
 
 Center of gravity of an area 41 
 
 Center of gravity of built-up sections 44 
 
 Centrally loaded column 88 
 
 Coefficient of elasticity Ill 
 
 for shear 116 
 
 Coefficient of linear expansion 113 
 
 Column formulas 
 
 graphical representation of 93 
 
 Rankine's 88 
 
 Column loads, kinds of 88 
 
 Columns 
 
 classes of ' 86 
 
 cross-sections of 86 
 
 design of 102 
 
 end conditions of 85 
 
 strength of 85 
 
 Combination column formulas 94 
 
 Combined flexural and direct stress 83 
 
 Compression, materials in 13 
 
 Compressive strength of riveted plates 120 
 
 Continuous beam 19
 
 124 INDEX 
 
 Page 
 
 Goyer-plates H8 
 
 Deflection of beams 114 
 
 Deformation 4 
 
 columns 102 
 
 timber beams 76 
 
 Direct stress 79 
 
 Double shear 119 
 
 Eccentrically loaded 88 
 
 Efficiency of a joint 122 
 
 Elastic limit 5 
 
 Elasticity 4 
 
 End conditions of column 85 
 
 Euler formulas 94 
 
 External shear 23 
 
 Factor of safety 8 
 
 Fibre stresses 55, 57 
 
 First beam formula 59 
 
 Flexural stress 79 
 
 Flexure and compression 81 
 
 Flexure and tension 79 
 
 Formula for power which shaft can transmit 109 
 
 Formula for strength of shaft 107 
 
 Frictional strength of joint 119 
 
 Graphical representation of column formulas 93 
 
 Hooke's law 5 
 
 Horizontal shear 75 
 
 Inclined forces 54 
 
 Joint 
 
 efficiency of 122 
 
 frictional strength of 119 
 
 Kinds of loads and beams 52, 79 
 
 Kinds of stress 2, 55 
 
 Lap joint 118 
 
 Laws of strength of beams 71 
 
 Linear expansion, coefficient of 113 
 
 Longitudinal forces 54 
 
 Main plates 118 
 
 Materials 
 
 in compression 13 
 
 in shear 15 
 
 in tension 11 
 
 Maximum bending moment 40 
 
 Maximum shear 31 
 
 Metala in shear 16 
 
 Modulus of rupture 72 
 
 Moment diagrams 36 
 
 Momentof a force 16
 
 INDEX 125 
 
 Page 
 
 Moment of inertia .' 46 
 
 of built-up sections 49 
 
 of a rectangle 48 
 
 unit of 47 
 
 Moments, principle of 17 
 
 Neutral axis 54 
 
 Neutral line 54 
 
 Neutral surface 54 
 
 Non-elastic deformation 117 
 
 Notation 24, 33 
 
 Parabola-Euler formulas 98 
 
 Posts 85 
 
 Principle of moments 17 
 
 applied to areas 42 
 
 Radius of gyration 86 
 
 Rankine's column formula 88 
 
 Reactions of supports 16 
 
 Rectangle, moment of inertia of 48 
 
 Reduction formula 48 
 
 Resisting moment 57, 107 
 
 value of 58 
 
 Resisting shear 73 
 
 Restrained beam 19 
 
 Rivet, shearing strength of 118 
 
 Riveted joints 118 
 
 Rods, stiffness of 110 
 
 Rule of signs 23, 32 
 
 Safe load of a beam 71 
 
 Safe strength of a beam 71 
 
 Second beam formula 73 
 
 Section modulus 59 
 
 Shafts 
 
 stiffness of 110 
 
 strength of 105 
 
 twist of 116 
 
 twisting moment of 105 
 
 Shear diagrams 27 
 
 Shear stress 2 
 
 Shearing strength of rivet 118 
 
 Shears, unit for 24 
 
 Signs, rule of 23, 32 
 
 Simple beam 19 
 
 Simple stress 1 
 
 Single shear 119 
 
 Steel in compression 14 
 
 Steel in tension 12 
 
 Stiffness of rods, beams, and shafts 110 
 
 Stone, ultimate strength of . 15
 
 126 INDEX 
 
 Page 
 
 Straight-line formulas 94 
 
 Strength of beams 52 
 
 laws of 71 
 
 Strength of columns 
 
 Strength of joint, computation of 120 
 
 Strength of shafts 105 
 
 Stress, kinds of 2 
 
 Stress-deformation diagram 6 
 
 Struts 85 
 
 Tables 
 
 bending moment 53 
 
 coefficients of elasticity HI 
 
 deflection 53 
 
 factors of safety 10 
 
 maximum shear, values of 53 
 
 mild steel columns, data for 95, 99 
 
 moduli of rupture 72 
 
 moment diagrams 53 
 
 moments of inertia 52 
 
 properties of standard I-beams 70 
 
 radii of gyration 52 
 
 section moduli 52 
 
 shear diagrams 53 
 
 Temperature stresses 113 
 
 Tensile strength of riveted plates 120 
 
 Tension, materials in 11 
 
 Timber 
 
 in compression 14 
 
 in shear 15 
 
 in tension 11 
 
 Timber beams, design of 76 
 
 Torsional stress 106 
 
 Transverse forces 52 
 
 Twist of shafts 116 
 
 Twisting moment of shaft 105 
 
 Ultimate strength [[[[ 6 
 
 Unit-deformation 4 
 
 Unit of moment of inertia 47 
 
 Unit-stress 3 
 
 Units 33 
 
 Units for shears 24 
 
 Value of resisting moment 58 
 
 Wrought-iron 
 
 in compression 14 
 
 in tension ^2 
 
 Working strength g 
 
 Working stress
 
 SOUTHERN 
 
 UNIVERSITY OF CALIFORNIA, 
 
 LIBRARY, 
 
 'LOS ANGELES, CALIF.