ACCESS ^ ^ 3 y VOL. FRANKLIN INSTITUTE LIBRARY PHILADELPHIA, PA. REINFORCED CONCRETE CONSTRUCTION VOLUME I FUNDAMENTAL PRINCIPLES McGraw-Hill BookCompai^/^ Puj6Cis/iers of 3oo£§ /or ElGCtrical World TheEngineei-in^ and Mining Journal En^iriGGriiig Record Engineering News Railw^A^e Gazett(? American Machinist Signal EnginQor American Engineer Electric Railway Journal Coal Age Metallurgical and Chemical Lnt^ineerin^ Power ENGINEERING EDUCATION SERIES EEINFOECED CONCEETE CONSTEUCTIOI VOLUME I. FUNDAMENTAL PRINCIPLES INCLUDING NUMEROUS TABLES AND DIAGRAMS TO FACILITATE THE CALCULATION AND DESIGN OF REIN- FORCED CONCRETE STRUCTURES PREPARED IN THE EXTENSION DIVISION OF THE UNIVERSITY OF WISCONSIN BY GEORGE A. HOOL, S. B. ASSOCIATE PROFESSOR OF STRUCTURAL ENGINEERING THE UNIVERSITY OF WISCONSIN First Edition Fifth Impression — Corrected Total Issue, 6,000 McGRAW-HILL BOOK COMPANY, Inc. 239 WEST 39TH STREET, NEW YORK 6 BOUVERIE STREET, LONDON, E. C. 1912 Copyright, 1912^ by the McGraw-Hill Book Company THE. MAPLE. PRESS-TORK-PA PREFACE This volume forms the first part of the regular course on Rein- forced Concrete Construction offered by the Extension Division of The University of Wisconsin and, in common with a number of the other structural engineering courses offered, presupposes a knowledge of the elements of structures. It has been written primarily to meet the needs of those who desire to take up the study of this subject by correspondence, but the author sees no reason why a text of this nature may not be employed for other purposes. The complete text for the course in Reinforced Concrete Construction is in three volumes: one on the fundamentals, one on the design and construction of retaining walls and buildings, and one on the design and construction of bridges and miscellan- eous structures. The present volume on fundamentals omits, for simplicity, the flat-slab type of floor construction — a subject reserved for thorough treatment under the heading of concrete floors in Volume II. The text is intended to be supplemented with such material as is suited to the special needs of the indi- vidual student. Information on the subject has been drawn from many sources but it should be stated that the text-books, "Principles of Rein- forced Concrete Construction" by Turneaure and Maurer (Copyright, 1907, 1909 by F. E. Turneaure and E. R. Maurer) and "Concrete Plain and Reinforced" by Taylor and Thompson (Copyright, 1905, 1909 by Frederick W. Taylor) have been referred to constantly. The author wishes to express his indebtedness to Mr. F. C. Thiessen for his excellent work in preparing the illustrations and for his help with the computations. The photographs of the beam and column tests were kindly lent by Mr. M. O. Withey, Assistant Professor of Mechanics in The University of Wisconsin. G. A. H. The University of Wisconsin, Madison, Wisconsin, June 1, 1912. v 188G7 ViiiM"iiiiiiiir^irf-riiiiTii^^^ MUUf^ TABLE OF CONTENTS PART I Properties of the Material CHAPTER I Concrete Article Page 1. General requirements 1 2. Cement 2 3. Sand 2 4. Stone 4 5. Consistency ■ ^ 6. Unit for proportioning 6 7. Theory of proportions 7 8. Proportioning by mechanical analysis 7 9. Quantities required per cubic yard 14 10. Compressive strength 15 11. Tensile strength 1^ 12. Shearing strength 19 13. Contraction and expansion 19 14. Fireproofing qualities 20 15. Waterproofing qualities 21 16. Modulus of elasticity 22 17. Weight of concrete 22 CHAPTER II Steel 18. General requirements 22 19. Tensile strength 24 20. Coefficient of expansion' 24 21. Modulus of elasticity 24 22. Bending test for steel 24 CHAPTER III Concrete and Steel in Combination 23. Advantages of the combination 26 24. Bond between concrete and steel 27 25. Ratio of the moduli of elasticity 29 26. Behavior of reinforced concrete under tension 35 vii ■ viii TABLE OF CONTENTS. Article p^^j, 27. Shrinkage and temperature stresses 36 28. Repetition of stress 37 PART II The Theory and Design of Slabs, Beams, and Columns CHAPTER IV Rectangular Beams 29. Inner forces in a homogeneous beam 39 30. Assumptions in common theory of beams 47 31. Plain concrete beams 49 32. Flexure formulas for reinforced concrete beams 52 33. For working loads 53 34. For ultimate loads 5g 35. Shearing stresses q3 36. Inclined tensile stresses 65 37. Methods of web reinforcement gg 38. Bond stress 71 39. Tests 74 40. Working stresses 84 41. Vertical and inclined reinforcement 90 42. Vertical stirrups 93 43. Horizontal bars bent up for web reinforcement 99 44. Vertical stirrups and bent rods comqined 103 45. Points to bend horizontal reinforcement 105 46. Transverse spacing of reinforcement 107 • 47. Depth of concrete below rods 109 48. Ratio of length to depth of beam for equal strength in moment and shear , IO9 49. Notation lU 50. Formulas 112 51. Deflection of beams ' 124 52. Economical proportions *. . 128 53. Restrained beams 129 54. Continuous beams 130 CHAPTER V Slabs, Cross-beams, and Girders 55. Slabs 135 56. Distribution of slab load to cross-beams 140 57. Distribution of beam and slab loads to girders 141 58. Arrangement of beams and girders 142 59. Design of beams and girders — T-beams 143 60. Economical proportions of T-beams 151 TABLE OF CONTENTS. ix Article ^^^^ 61. Conditions met with in design of T-beams 152 62. Beams with steel in top and bottom 157 63. Design of a continuous beam at the supports 159 CHAPTER VI Columns 64. Plain concrete columns 167 65. Columns with longitudinal reinforcement 169 66. Columns with hooped reinforcement 171 67. Columns with hooped and longitudinal reinforcement 173 68. Columns reinforced with structural steel shapes 174 69. Tests on plain and reinforced concrete columns 175 70. Working stresses 71. Value of longitudinal reinforcement of columns 185 CHAPTER VII Slab, Beam, and Column Tables 72. Illustrative problems 187 CHAPTER VIII Slab, Beam, and Column Diagrams 73. Illustrative problems 215 CHAPTER IX Bending and Direct Stress 74. Theory in general 241 75. Case I. — Compression over the whole section 244 76. Case II. — Tension over part of section 246 Tables NUMBEB 1. Areas, perimeters, and weights of rods 193 2. Data for design of rectangular beams 194 3. Data for reviewing rectangular beams 195 4. Spacing of round rods in slabs 196 5. Spacing of square rods in slabs 196 6. Use for designing slabs 197 X TABLE OF CONTENTS. ' Number Page 7. Use for reviewing slab designs 198 8. Use for continuous rectangular beams 200 9. Use for T-beams 202 10. Use for T-beams 207 11. Use for rectangular beams with steel in top and bottom 208 12. Use for columns 210 13. Number of rods and sectional area in square inches for beam and column reinforcement 211 14. Hooped column reinforcement 212 15. Maximum diameter of round or square stirrups 213 16. Minimum length of embedment of inclined rods 213 Diagrams 1. Use for design of rectangular beams 223 2. Curves for j and k for rectangular beams 224 3. Spacing of round rods in slabs 225 4. Spacing of square rods in slabs 226 5. Bending moments for uniformly distributed loads 227 6. Use for designing slabs 228 7. Use for rectangular beams 230 8. Use for T-beams 233 9. Use for T-beams 234 10. Use for rectangular beams with steel in top and bottom 235 11. Use for rectangular beams with steel in top and bottom 236 12. Use for columns 237 13. Bending and direct stress — compression over whole section . . . . 238 14. Bending and direct stress — tension over part of section . . . . ■ . 239 15. Bending and direct stress — tension over part of section 240 REINFORCED CONCRETE CONSTRUCTION PART I PROPERTIES OF THE MATERIAL Reinforced Concrete is concrete which is strengthened by having embedded in it some metal, usually steel. The component materials should separately possess certain properties, if satisfactory strength and durability are to be obtained in the structures having these materials in combination. The properties of each material will now be discussed, and those properties in particular will be emphasized which have the most to do with the safe and economic designing of structures. CHAPTER I CONCRETE 1. General Requirements. — Concrete used in reinforced con- crete construction should be strong, of uniform quality, free from voids, and thoroughly sound. These qualities are required even more than in massive concrete, as the sections in reinforced concrete structures are comparatively small and the stability of a given structure depends upon the strength and durability of every part. The proportions commonly used in American practice vary from about 1:1 1/2:3 to 1:3:6, using either crushed stone or gravel. The rich mixture is usually required in structural parts subjected to high stresses or where exceptional water- tightness is desired. On the other hand, the use of a 1:3:6 concrete requires careful grading of the materials to produce satisfactory results, even for ordinary work. 2 REINFORCED CONCRETE CONSTRUCTION 2. Cement. — The cement employed in reinforced concrete construction should be of high grade; only portland cement should be used, and the brand selected should conform to the specifications of the American Society for Testing Materials — for these specifications are now accepted as the American standard, 3. Sand. — The sand employed should be free from clay, vegetable loam, sticks, and organic matter and should be of hard, dense, tough material. Siliceous quartz sands are the best, although sands from any durable rock will answer. Sharp sand was formally a requirement in all important con- struction, but this property is by no means essential. To be sure, by the use of sharp sand there is a slight tendency toward a concrete of greater crushing strength than when sand of rounded grains is employed, but this influence on the result is of less importance than the size of grain, or granulometric composi- tion. Moreover, the sharper the sand employed — the relative sizes of the grains remaining the same — the greater the percent- age of voids, and consequently the greater the amount of cement required to produce a given density. (The term density is here used to express the ratio of the volume of the solid particles to the total volume of the concrete.) It is now generally conceded that the requirement of sharpness of sand should be omitted from concrete specifications. Tests of mortar and concrete show that strength and water- tightness increase with density, and so the best sand as to size is one which will produce the smallest volume of mortar of stand- ard consistency when mixed with the given cement in the required proportions. To put it somewhat differently, — the best sand for strength, for water-tightness, and also for economy (as will be seen later) is one which is so graded from fine to coarse that the percentage of voids in the resulting mortar is reduced to a minimum. Such a sand has a very coarse appear- ance as the amount of fine material required is small. It has been found that the densest mixture occurs with par- ticles of different sizes and also that the least density occurs when the grains are all of the same size. Coarse and fine sands are thus inferior to graded sands for concrete, but of the two extremes the coarse -sand is preferable. The reason for this is due to the fact that the coarse sand has a less total grain surface in a unit volume, even when the sands considered contain the same pro- CONCRETE 3 portion of solid matter and voids. Less total grain surface means less cement and water to coat the grains, and less labor required in mixing. The additional amount of cement and water required in the case of the fine sand reduces the density of the resulting mortar and likewise its strength. (The density of neat cement ranges between 0.49 and 0.59, while the density of a sand mortar ranges from 0.60 for a fine sand to 0.75 for a coarse sand or a well-graded sand.) A fine sand is one containing more than 30 per cent of par- ticles that will pass a No. 40 sieve (diameter of hole = 0.015 in.). 8 YiG, 1.— Typical mechanical analyses of fine, medium and coarse sands. The finer the sand, the more nearly uniform the size of the grains, and consequently the greater the proportion of voids. Fine sand is seldom satisfactory and should not be used unless a coarse sand is not available. Even in such cases, tests of strength should be made with the idea of determining what extra cost may be justified in securing a coarser material. The most accurate method of determining the value of a sand with reference to its size is by means of a sieve analysis. This consists of sifting the sand through several different sieves, and then plotting upon a diagram the percentage by weight which is passed (or retained) by each sieve— abscissae representing size and ordinates representing percentage. Fig. 1 represents the analyses of three natural sands— a fine, a medium, and a coarse well-graded sand. Uniform grading is indicated by an approach to a straight line. A standard size of sieve is 8 in. in diameter and 2 1/4 in. high. Woven brass wire sieves are employed for 4 REINFORCED CONCRETE CONSTRUCTION openings less than 1/10 in. in diameter; while for larger openings sheet brass is used, having circular openings drilled to the required dimensions. The woven brass wire sieves are given commercial numbers which approximately coincide with the number of meshes to the linear inch. The actual size of hole, however, varies with the gauge of wire used by different manu- facturers and every set of sieves must be calibrated separately. A common defect in sieves is the displacing of the wires so that they are not perpendicular to each other; such sieves should be discarded. Sieves are made to fit together in nests, so that when a sample of sand is placed in the upper (or coarsest) sieve and the nest of sieves is thoroughly shaken, the quantity caught on each sieve can be determined at once. For analyzing sand the following sizes ^ are desirable: Commercial No 10 20 30 40 50 80 100 200 Approximate size of hole in inches 0.073 .034 .022 .015 .011 .007 .0055 .0026 A screen with 1/4-in. openings is generally employed for separat- ing out large material from sand. Specifications should limit the maximum amount of loam or clay to be allowed in any given work. Loam should never be permitted, but clay to the amount of 5 to 10 per cent, if evenly divided, is often beneficial in lean mortars. In rich mortars the strength and density is decreased by even slight additions of clay; but in lean mortars the clay helps to fill the voids of the sand, and causes the cementing material to coat the grains better and to bind them together more strongly. Broken stone screenings have a small percentage of voids and, when free from clay, usually make excellent sand. These screenings ordinarily give a stronger mortar than natural sand but are likely to contain an undue amount of dust, especially when obtained from soft stone; in such a case the mass should be screened before being used in mixing mortar. Gravel screen- ings also constitute a good material in place of sand. All material passing a 1/4-in. screen is generally considered as sand, or fine aggregate; while all material larger than this size is classed as coarse aggregate. 4. Stone.— For the coarse aggregate, either crushed stone or gravel Is generally used. Any stone is suitable which is clean and durable and which has sufficient strength to prevent the strength of the concrete from being limited by the strength of ' From American Civil Engineers' Pocket Book, 1st edition, page 415. CONCRETE 5 the stone. Traps, granites, limestones, and the more compact sandstones are generally employed. Aggregates containing soft, flat, or elongated particles should never be used. All that has been said concerning voids in sand applies with equal force to the coarse aggregate. Fig. 2 illustrates the analy- sis of a bank gravel and of a crushed stone. Screens varying by a quarter of an inch from 1/4 in. up are desirable, but a very useful analysis may be made with fewer screens. A uniform size of stone filled with mortar does not make as dense or as 0 y IJ-^ I I I I I I I M I I I M I I I I I I I I I I I linn 0 0.25 0.50 0.75 l.OO 1.25 1.50 Diameter of- Particle m Inches. Fig. 2. — Typical mechanical analyses of bank gravel and crushed stone. strong a concrete as one in which the coarse aggregate is well graded — that is, where the small stones partly fill the larger interstices. A straight line on a mechanical-analysis diagram indicates a uniform grading of size. Other things being equal, the larger the stone, the stronger and denser the concrete. Experience has shown that for rein- forced concrete the maximum size should not be more than about 1 in. to 1 1/2 in., in order that the concrete may fit itself closely around the reinforcing metal. The smaller the stone, the greater the surface to be coated, and the greater the amount of cement required. Most gravels are sufficiently durable for use in concrete. They should be at least reasonably clean, although a quantity of finely divided clay equal to 5 to 10 per cent of the gravel may add to G REINFORCED CONCRETE CONSTRUCTION the strength of the concrete, if the cement paste does not entirely fill the voids. The presence of clay requires very thorough mixing. When gravel is used, it should be screened to separate the sand and then be remixed in order that the proportions may be definite. Cinders make good fireproof concrete, but are not recommended by the best authorities for reinforced work. The allowable stress is too low for economical use and, unless great care is taken in having a wet mix and in thorough mixing, there is danger of corrosion of the embedded steel due to porosity. Cinders for use in concrete should not contain many, if any, fine ashes and should consist of hard, clean,, vitreous clinker, free from any unburned coal. Concrete containing this aggregate can safely be employed for filling between steel beams and for fireproofing steel or iron columns, and for a concrete fill on top of reinforced concrete floors and roof slabs. 5. Consistency. — Opinion differs as to the quantity of water that should be employed in mixing, but it is safe to say that a somewhat wet or mushy mixture should be used in reinforced concrete construction. Such a mixture flows easily under and around the metal reinforcement and ensures its preservation. It also conforms readily to the molds and gives a smooth surface. Experiments show that while dry concrete carefully mixed and rammed is stronger at the earlier ages than wet concrete, in six months' time but little difference in strength is found. Moreover, with dry mixtures there is difficulty in obtaining a uniform consistency — occasional batches being too dry. The water used in mixing concrete should be free from oil, acid, alkalies, or vegetable matter. 6. Unit for Proportioning. — When proportions of the ingre- dients of a concrete are specified, the specifications should state whether the cement shall be measured loose, or as packed in bags and barrels. The reason for this is clear when it is con- sidered that loose cement occupies about 30 per cent more volume than packed cement. The usual method is to specify the barrel of packed cement as the unit, and to assign it some definite volume — the sand and stone to be measured loose. A barrel of Portland cement weighs 376 lb., not including the barrel, and a bag of Portland cement weighs 94 lb.; in other words, there are four bags to a barrel. The cement as packed in a barrel occupies, on an average, a volume of about 3.2 cu. ft., but as the unit adopted is an arbitrary one in any case, 3.8 cu. ft. CONCRETE 7 to the barrel is generally taken as the standard. The value 3.8 has been selected for convenience since 100 lb. of cement can thus be considered as 1 cu. ft. 7. Theory of Proportions. — Two well-established laws govern the theory of proper proportioning, namely: 1. With the same percentage of cement in a unit volume of concrete, the strongest and most impermeable concrete is that which has the greatest density. 2. If the sand and stone remain the same, the strongest and most impermeable concrete is that containing the greatest percentage of cement in a unit volume. The first law is extremely important. Another way of express- ing it is to say that, to obtain the greatest strength and imper- meability, the cement should fill the voids of the sand and the resulting mortar should fill the voids of the stone. The second law means that with the same aggregates the strength and water-tightness increases with the amount of cement used — provided, however, that in some cases this amount be not in excess of the voids in the sand, and that the amount of mortar used in each case be the same. If the cement more than fills the voids of the sand, or if the mortar more than fills the voids of the stone, the concrete will be less dense than if the voids were just filled (ordinary concrete has a density between 0.80 and 0.88 and hence is denser than either neat cement or cement mortar) ; and thus the strength due to increase of cement may be offset by the decrease in density. 8. Proportioning by Mechanical Analysis. — Certain standard proportions, such as 1:2:4 and 1:2 1/2:5, are commonly em- ployed in practice; but better results with greater economy can often be secured by the use of mechanical-analysis curves. These curves make it possible to find the best proportions of different aggregates, and they also afford means of finding the best proportions attainable by screening the sand and the stone, and by making artificial combinations of the several portions. In proportioning by mechanical analysis, the object to be aimed at is to grade the fine and coarse aggregate so that the densest concrete will result from the use of a given amount of cement. This means that the object to be kept in view while grading should be a minimum percentage of voids; however, the use of much very fine material should be avoided for the reason cited in Art. 3. After this grading is accomplished, an amount of 2 8 REINFORCED CONCRETE CONSTRUCTION cement should be used which will give the requisite strength or degree of imperviousness. The curve of maximum density, or ideal curve, for the com- bination of sand and crushed stone with a maximum size of 1 in. is shown in Fig. 3. This curve was plotted from a set of rules determined by William B. Fuller and Sanford E. Thompson in an extended series of tests at Jerome Park Reservoir, New York, in 1903 and 1904\ These tests, from which the rules 1 Transactions of the American Society of Civil Engineers, Vol. 59, 1907. CONCRETE 9 were deduced, comprised the screening of crusher-run stone and bank gravel into twenty-one sizes ranging from 3 in. down to material passing a No. 100 sieve, and then re-combining these sized materials in a predetermined mechanical-analysis curve by weighing out the necessary quantities of each size. Over 400 different mechanical-analysis curves were made. The different mixtures were thoroughly mixed with a given weight of cement to a given consistency; then they were tamped into a strong cylinder, and their volume determined. These tests led to the determination of valuable rules for the plotting of ideal curves for density. Many of the mixtures were also made up into prisms and beams, and the strength of each was determined by breaking tests. These tests substantiated the two laws of the theory of proportions given in the preceding article. The maximum density curve was found to be of substantially the same form for different materials, whatever the maximum size of stone. The curve in all cases may be taken as a combina- tion of an ellipse and a straight line. First a straight line should be drawn from the point where the largest diameter stone reaches the 100 per cent line, to that point on the vertical ordinate at zero diameter which is given in column (1) in the following table: DATA FOR PLOTTING CURVES OF MAXIMUM DENSITY Intersection of Height of Axes of ellipse tangent with tangent Materials vertical at zero point diameter a (6 + 7) (1) (2) (3) (4) Crushed stone and sand 28.5 35.7 0.150D 37.4 26.0 33.4 0.16.4D 35.6 Crushed stone and screenings 29.0 36.1 0.147D 37.8 In this table, D = the maximum diameter of the stone, in inches. Next mark the tangent point on this line — namely, where this line is intersected by the vertical ordinate for one-tenth the maximum stone. This mark should check with the values given in column (2) of the above table. Then plot the location of the axes of the ellipse from the values of a and (6 + 7) given in columns (3) and (4) respectively in the above table. The major axis of the ellipse should be placed on the 7 per cent line of percentages. (Fig. 3.) Now to plot the ellipse: take a strip of __1 10 REINFORCED CONCRETE CONSTRUCTION paper and mark off the lengths of the semi-major and semi- minor axes upon it; each of these lengths should be laid off in the same direction from a common point, which we shall call K; denote the length of the semi-major axis as KA and the length of the semi-minor axis as KB; now swing the strip of paper little by- little so that the outline of the curve may be marked off by the point K while the points A and B are kept at all times upon the axes h and a respectively. The principles by which sand and stone curves are combined into a single curve, with the object in view of approaching closely to the curve of maximum density, may best be explained by some typical problems as developed by Messrs. Taylor and Thompson. ^ Suppose that we have for concrete the fine sand OA and the crushed stone DE of Fig. 4, and suppose the problem is to find what proportion of each material should be employed. The curves of the two materials are plotted to the same scale and the ideal curve is drawn by the method previously described. Experiments have shown that where the materials to be mixed are represented by only two curves, the best results are obtained when the combined curve intersects the ideal curve approxi- mately on the 40 per cent line at F, and when the finer material is assumed to include the cement. The sand and stone curves in this case do not overlap, and hence for the best proportions, 60 per cent by weight should be stone, and 40 per cent by weight should be sand plus cement. The combined curve is not drawn for this simple case. Now the proportion of cement to be used to give the required strength of concrete must always be assumed ; and in this case one part by weight of cement to six parts by weight of dry aggregate (measured before the sand and stone are mixed together) will be considered as satisfactory. This will make the cement 1/7, or 14.3 per cent, of the total materials. Deducting this from the percentage of cement plus sand, we have 40% -14.3% =25.7% sand. The best proportions, then, for a 1:6 mixture by weight are 14.3 parts cement: 25.7 parts sand: 60 parts stone, or a 1:1.8:4.2 concrete. In determining the corresponding proportions by volume, the weights of the sand and stone per cubic foot should be considered. Consider now the proportioning of the medium sand OB with the stone DE as before. The curve OB may be transformed so that it will pass through F, by changing the distances from the bottom line of the diagram to the curve OB in the proportion ' From Taylor and Thompson's "ConoreJ,e, Plain and Reinforced," 2nd edition, pages 784 to 788 inclusive. Copyright, 1905, 1909, by Frederick W. Taylor. CONCRETE 11 HF 40 = — = 43 %, which means that 43 per cent of the dry materials by weight should be cement plus sand, and 57 per cent stone. About 3 per cent of the stone is overlapped by the sand, but this is so slight it need not be considered here in determining the 3A3IS UO P3U1104.3J 36lD4-U30-l3d ■^8 combined curve. Theoretically, no overlapping in this case should occur, but jt is practically impossible in a revolving screen to prevent some fine material being carried over to the openings of larger size. The best proportions of the cement, 12 REINFORCED CONCRETE CONSTRUCTION sand, and stone should be determined in a similar manner to the previous problem. The combined curve is shown as ONFRE. The part of this curve to the left of F very nearly coincides with the ideal curve. For example, 43% of LM = 0.43 X 83 =35.7%, or almost exactly LN. From F to the right, the combined curve varies somewhat from the ideal. For example, the per- centage of particles larger than 4/10 of an inch =57% of PS = 0.57X82.5=47%, or PR. Now suppose that we consider two aggregates, the curves of which overlap — namely, OB and OKE of Fig. 4. The proportion J 1 . . • 35 ^ GF 53 of sand plus cement to use is v?7^= ^=40 %, with 7^= ^ = 60% stone. The combination curve, with the exception of the overlap, may be drawn in the same manner as before. The method of finding a point such as J" is as follows: DV(Om) +DT1^(0.40) =DT = iS%. The student should understand that the location given for the point F in the above problems can be only approximate. If possible in any given case, it would be advisable to vary the proportions somewhat each way from those obtained in the above manner, and determine the corresponding densities by volumetric tests. It should also be understood in this connection that the curve of maximum density may vary slightly from the so-called ideal curve; but the variation is never great, and there is the advantage of being able to plot, by means of simple rules, a curve which will lie at least very close to the maximum density curve for the given aggregates. It should be clear that plotting mechanical-analysis curves shows approximately not only the best proportions for given materials, but also shows how the materials may be improved by adding or subtracting some particular size. The most valu- able use, however, of the method of proportioning by mechanical analysis is in the kind of work which warrants employing several grades or sizes of sand and stone. The process of determining the percentage of each material varies for different cases and is more complicated than where but two aggregates are used. The following problem will give a fair idea of the method of solution for any given case. The two sizes of crushed stone and the sand shown in Fig. 5 will be considered. The crushed stone curves show the sizes of stone which ordinarily pass through crusher screens of given CONCRETE 13 diameter of hole, and also illustrate how inefficient the screening process may be. For example, if the sizes of the particles had corresponded exactly to the diameters of the holes and the screening had been more perfectly done, the curves would have 8u> o in ~ 3A3I9 BuiSSOd 3D04-USDJ3d had more nearly the direction and location of dotted lines No. la and No. 2a. In combining these curves so as to obtain the strongest and most impermeable concrete, it is quite clear that the grains smaller than 0.10 in. diameter must be supplied wholly from the portion Oa of the sand curve; while the larger 14 REINFORCED CONCRETE CONSTRUCTION grains of the sand, represented by the portion ab, are found also dc 36 in the No. 1 curve. The percentage of sand to use ^ ~ 35 ~ 42%, since this percentage transforms the sand curve so that it fits very nearly the ideal curve from 0 to C. The percentage of No. 2 stone required is^= ^^ = ^S%. This leaves 100 — ^ fm 100 (42+43) =15% to be furnished by the No. 1 stone. Suppose now that a 1:7 concrete is desired. Then, since for more than two aggregates the finer material is not assumed to include the cement, there will be = 14.3 parts cement, and the propor- tions will be 14:42:15:43 or 1 :3.0: 1.1 :3.1— the parts being by weight. An ordinate such as rs may be found as follows: (0.43) (100) + (m) (0.15) + (rt) (0.42) =57% With three or more sizes of stone, it is often necessary to assume the required percentages of the intermediate sizes, and make many trial plottings, before we are able to determine the proper proportions to give the best combined curve. The following statement by William B. Fuller illustrates the possibilities of mechanical analysis:^ ''The ordinary mixture for water-tight concrete is about 1:2:4, which requires 1.57 barrels of cement per cubic yard of concrete. By carefully grading the materials by methods of mechanical analysis, the writer (Fuller) has obtained water-tight work with a mixture of about 1:3:7, thus using only 1.01 barrels of cement per cubic yard of concrete. This saving of 0.56 barrels is equivalent, with Portland cement at $1.60 per barrel, to $0.89 per cubic yard of concrete. The added cost of labor for proportioning and mixing the concrete because of the use of five grades of aggregate instead of two was about $0.15 per cubic yard, thus effecting a net saving of $0.74 per cubic yard. On a piece of work involving, say, 20,000 cu. yd. of concrete such a saving would amount of $14,800.00, an amount well worth considerable study and effort on the part of those in responsible charge." The above statement does not take into account the cost of the sand and stone which would be needed for the approximate 1:3:7 concrete over and above that required for the 1:2:4 mix- ture, but this amount would be small compared with the total saving. 9. Quantities Required per Cubic Yard. — The following approxi- 1 From Taylor and Thompson's "Concrete, Plain and Reinforced," 2nd edition, page 183. Copyright, 1905, 1909, by Frederick W. Taylor. CONCRETE 15 mate rule devised by William B. Fuller gives the quantities of packed cement, loose sand, and loose stone required, to make a cubic yard of concrete. Let c, s, and g be the number of parts by volume of cement, sand, and stone, respectively. Also, let C, S, and G be the required number of barrels of packed cement, the required number of cubic yards of loose sand, and the required number of cubic yards of loose stone, respectively. Then c+s+g *S = C X S X ^ ^ 3.8 G = CxgX-^ If the stone is of nearly uniform size, about 5 per cent should be added to all quantities computed by the above rule and, on the other hand, if the stone is well-graded, about 5 per cent should be deducted. Although stone is sometimes screened to approxi- mately one size, it is only a waste of labor and material, for the screened stone makes a weaker concrete, and requires more cement. 10. Compressive Strength. — The compressive strength of concrete varies within wide limits, due to the fact that there are so many reasons for variation. The principal factors which affect compressive strength are: (1) the quality of cement used; (2) the quantity of cement in a unit volume of the concrete; (3) the character and size of the aggregates; (4) the density of the concrete; (5) the care taken in mixing; (6) the age of the mixture; and (7) the conditions under which the concrete seasons. Because of the different conditions met with in practice, it is somewhat misleading to present average values for the compres- sive strength of concrete. Where possible, the strength of any given concrete should be determined by actual tests and, if the results are too low, the ingredients or proportions should be changed until a satisfactory result is obtained. The two chief factors which determine the compressive strength of concrete, considering the materials as satisfactory, are age and proportions of ingredients. The relative amount of increase in strength of concrete, from 7 days to 6 months, for two common mixtures, is shown approximately in the table which follows. The table gives average values of the compressive strength in pounds per square inch based on tests made for the Boston 16 REINFORCED CONCRETE CONSTRUCTION Elevated Railway Company at the Watertown Arsenal in 1899. The test pieces were 12-in. cubes. Mixture 7 Days 1 Month 3 Months 6 Months 1:2:4 1565 2399 2896 3826 1:3:6 1311 2164 2522 3088 The theoretij3al angle of rupture in crushing is about 60 degrees with the horizontal, and this theoretical conclusion is borne out by actual tests. For example, cubes of concrete will leave after crushing, pyramids whose surfaces are at an angle of about 60 degrees with the base. Thus, it should be clear that to, develop simply the normal compressive strength, the height of a speci- men should be at least one and one-half times its least lateral dimension. Conclusive evidence of the increased strength of cubes as compared with cylinders, due to the reason above given, is shown by the United States Government tests at St. Louis, the results of which are discussed in the U, S. Geological Survey Bulletin No. 344, 1908. Computations from these tests give a ratio of strength of 8 in. Xl6 in. cylinders to 6-in. cubes, at ages of thirteen and twenty-six weeks, as 0.88. This value coincides almost exactly with the empirical formula evolved by Prof. Johnson from results on sandstone and cast-iron prisms. The formula follows: strength of cylinder _q 778+0 222 cylinder strength of cube " " height of cylinder A study of a number of tests on concrete tend to show that this formula may be applied with sufficient accuracy, considering the variability of the material, when comparing all sizes of concrete cylinders and cubes. Tests were formerly made on specimens of cube form, but recently the prismatic or cylindrical form of a height of 2 to 3 diameters has been more commonly employed. Since there is a greater freedom for shearing action in the cylindrical specimen, it is generally used in studying the results of tests on columns. The cube form, however, is useful for comparison with the compressive strength of concrete in a beam. Tests on cylindrical specimens under different conditions show CONCRETE 17 a variation in compressive strength at the end of 1 month, for concrete of ordinary proportions, from 1500 to 4000 lb. per square inch. Under reasonably good conditions as to character of material and workmanship, an average strength of 2000 lb. per square inch may be expected of a 1:2:4 concrete at the end of 1 month; and for a 1:3:6 mixture, a strength of 1600 lb. per square inch. Taylor and Thompson present a practical working formula of sufficient accuracy to compare the compressive strength of mixtures of the same materials in different proportions. The values in the following table^have been obtained from this for- mula based on cube specimens and medium consistency: Age, one month Age, six months Proportions Voids in crushed stone or gravel Voids in crushed stone or gravel 50% 2 45% 3 40%'' 30% 5 20% 5 50% 2 45% 3 40% 4 30% 5 20% 5 a a a lb. per lb. per lb. per lb. per lb. per lb. per lb. per lb. per lb. per lb. per s a S3 o -p sq. in. sq. in. sq. in. sq. in. sq. in. sq. in. sq. in. sq. in. sq. in. sq. in. m m U 2 2880 2860 2840 2800 2760 3890 3870 3840 3780 3730 3 2780 2750 2720 2670 2610 3750 3710 3680 3600 3530 u 4 2680 2650 2610 2540 2460 3620 3570 3520 3430 3330 2 3 2560 2540 2510 2460 2410 3460 3420 3390 3320 3250 2 4 2480 2440 2410 2350 2290 3340 3300 3250 3170 3090 2 5 2400 2350 2310 2230 2170 3230 3180 3120 3010 2930 2 6 2320 2260 2230 2140 2060 3130 3060 3010 2890 2780 2i 3 2370 2340 2320 2270 2230 3200 3160 3130 3070 3020 2h 4 2290 2260 2230 2180 2110 3090 3050 3010 2940 2850 2h 5 2210 2180 2130 2070 2000 2980 2940 2880 2790 2700 2i 6 2140 2100 2060 1980 1910 2890 2830 2780 2670 2570 3 4 2120 2090 2060 2020 1970 2860 2830 2780 2720 2660 3 5 2060 2030 1990 1930 1870 2780 2740 2690 2610 2530 3 6 1990 1950 1910 1840 1770 2680 2630 2580 2480 2390 3 8 1860 1810 1770 1680 1600 2510 2440 2390 2280 2160 Note. — Proportions are based on a barrel of 3.8 cu. ft. Values are for average ultimate strength, which must be divided by a factor of safety for working loads. Quality of mate- rials and methods of mixing may affect the strength by 25 per cent in either direction, while the relative values for different proportions are not materially changed. 1 From Taylor and Thompson's "Concrete, Plain and Reinforced," 2nd edition, page 360. 2 Use 50 per cent columns for crushed stone screened to uniform size. ' Use 45 per cent columns for average conditions and for crushed stone with dust screened out. * Use 40 per cent columns for gravel or mixed stone and gravel. ' Use these columns for graded mixtures. 18 REINFORCED CONCRETE CONSTRUCTION It should be noticed in the table that the stone with the smaller percentage of voids gives the lower strength. This seeming irregularity is due to the fact that this stone measured loose has more solid material per cubic foot than the stone with the higher percentage of voids, and hence with the same proportions by volume, this stone gives a greater bulk of concrete, and less cement per unit volume. A smaller amount of cement shows here to have more influence in decreasing the strength than the greater density has in increasing it. It must not be inferred from this that the aggregate with the largest percentage of voids is best to use. Such an aggregate requires more cement to a given volume of concrete, and although the aggregate with fewer voids is sometimes slightly inferior in strength, the latter is the more economical. All the preceding results for the crushing strength refer to a compressive force applied over the entire upper surface of the specimen. Experiments show that if the load is applied upon only the central portion of the upper surface, the specimen will bear a greater unit load, due to the fact that the outer portions will aid the interior portion in resisting stress. In connection with the designing of concrete footings for the Boston Elevated Railway, thirty-six 12-in. cubes of 1:0:2 and 1:2:4 concrete were crushed at different ages by applying the load over the entire upper surface of the cube; then the same number of similar cubes were crushed by applying the stress over an area of 10 by 10 in. and then a third set by applying the stress over an area of 8 by 8 1/4 in. The third series gave a strength 128 per cent of the first, and the second series gave 112 per cent. Age of the concrete and proportions of ingredients did not seem to have any influence on the above results. 11. Tensile Strength. — Concrete is frequently subjected to tension in reinforced concrete construction; not directly, how- ever, but as the result of bending as in beams, girders, and slabs. The value of this transverse strength is of little impor- tance; because, on account of the brittleness of concrete in tension and its liability to crack from shrinkage or from the shock of some of the applied loads, it is unsafe to depend upon the tensile strength of concrete in reinforced concrete design. In certain computations, however, the tensile strength must be considered. The character of the sand and aggregate, and poor workman- ship, have a greater influence on the tensile strength than upon CONCRETE 19 the compressive. Tests seem to show that reasonable values for ultimate strength in direct tension at the end of one month are about as follows: 1:2:4 mixture , 1:3:6 mixture . 160-200 lb. per square inch 100-125 lb. per square inch The tensile strength of concrete at the place of greatest stress, that is, at the fiber most remote from the neutral axis, limits the strength of unreinforced concrete beams. Transverse tests of plain concrete should therefore show about the same relative results as tensile tests, and, in fact, they are quite as significant in this connection. The value of the ratio of modulus of rupture to tensile strength will ordinarily range from 1.8 to 2.0. 12. Shearing Strength. — A true shearing failure can only occur when the external forces producing the failure act in opposite directions and close together. The actual strength of concrete in direct shear is greater than was formerly supposed because, in the earlier experiments, shear was confused with the diagonal tension which occurs in the web of a beam. Diagonal tension is explained at length in Chapter IV. At the present time it is sufficient for the student to realize that, in this article, shear is considered to be the strength of the material against a sliding failure, when tested as a rivet or bolt would be tested for shear. Very few tests have been made to determine the shearing strength of concrete since it is difficult to arrange an experiment to determine this strength without involving some bending stress. Perhaps the best set of tests was made at the Massachu- setts Institute of Technology in 1904 and 1905 under the direction of Prof. Charles M. Spoff ord. Three grades of concrete were used • — 1:2:4, 1:3:5, and 1:3:6 — and the specimens were stored in both air and water from 24 to 32 days. The specimens tested were cylinders 5 in. in diameter and 18 in. long, and the end thirds were held rigidly by cast-iron yokes. The pressure was applied through a cast-iron, half-cylinder bearing which fitted between the two yokes and caused the shearing of the concrete across two planes. Six extra cylinders of the same concrete were tested at the same time for compression. The tests gave a shearing strength ranging in general from 60 to 80 per cent of the compressive strength of the concrete. 13. Contraction and Expansion. — The contraction and expan- I 20 REINFORCED CONCRETE CONSTRUCTION sion of concrete occurs from two causes: (1) temperature, and (2) hardening action. Like most building material, concrete contracts as the temperature falls, and expands as it rises. Its volume is also affected when it hardens, a shrinking action taking place when hardened in air and some expansion when hardened in water. Experiments show an average value of the coefficient of expan- sion of concrete, due to temperature changes, to be about 0.000006 per degree Fahr. which, it should be noted, is very nearly that of steel. Thus, when steel is embedded in concrete, the two materials will be but slightly stressed because of any difference in their rates of expansion. ■» The amount of change in volume of concrete due to hardening increases with the proportion of cement used in the mixture. When hardened in air, concrete contracts about 0.0005 of its length; while the expansion, when hardening takes place under water, is only about 0.0001 of the length. The whole change is usually not accomplished for some months, but about half the whole is effected in about a week. 14. Fireproofing Qualities.— The value of concrete as a material to be used in the construction of the walls, columns, and floors of buildings is largely dependent upon its fireproofing qualities. Concrete has a low conductivity of heat due to two causes: (1) the presence of combined water, and (2) porosity. The water of crystallization, being chemically combined, is not given off at the boiling-point. A part of this water goes off at about 500° Fahr., but the dehydration is not complete until 900° Fahr. is reached. This vaporization of the water at the surface absorbs heat and protects the interior. The layer of changed material is then a better nonconductor than before, so the process goes on very slowly. The porosity of concrete also offers considerable resistance to the passage of heat since it is well known that air spaces are a most efficient protection against conduction. Steel is said to lose 10 per cent of its strength at 600° Fahr. and 50 per cent at about 750° Fahr. A fire in a building may reach a temperature of 2000° to 2300° Fahr., and the importance of sufficiently protecting the reinforcing steel from loss of strength due to overheating is therefore evident. The surface of a mass of concrete exposed to the action of flames for some time may be CONCRETE 21 ruined and may be flaked off by the application of a stream of water from a hose; but fortunately, the depth to which the heat penetrates is very limited. The conclusions to be drawn from experimental tests, and also from observations of the results of large fires, are that sharp corners of beams and columns are more susceptible to attack than wide, flat surfaces, such as slabs. Sufficient protection seems to be afforded the reinforcing metal by a 1 1/2 to 2 1/2 in. covering in beams and columns, and a 3/4 to 1 in. covering in slabs. Little difference was observed between stone and cinder concrete — the burning of the bits of coal in poor cinder concrete being often balanced by the splitting of the stones in the stone concrete. It seems probable, from the composition of the rock, that hard trap or gravel may be preferable to hard limestone, slate, or conglomerate, as fire-resisting material, although further experiments are needed to determine their relative durability in this respect. Soft limestone as the aggre- gate should never be employed, as under the action of fire and water it disintegrates. Prof. C. L. Norton, in his report on the Baltimore fire to the Insurance Engineering Experiment Station, says: "Where con- crete floor arches and concrete-steel construction receive the full force of the fire, it appears to have stood well, distinctly better than the terra-cotta." The reason for this he considers to be due to the fact that terra-cotta expands about twice as much as steel, while concrete expands about the same. 15. Waterproofing Qualities of Concrete. — Most of the tests and observations on the waterproofing qualities of concrete have been made with the idea of determining whether or not embedded steel is subject to corrosion from moisture and other causes. In general, the rusting of iron occurs only under the combined action of moisture and carbon dioxide. A coating of Portland cement has long been known to not only exclude moisture and carbon dioxide, but in hardening to absorb any carbon dioxide which may be present. Concrete as actually deposited around reinforcing steel cannot be said to be as effective as an unbroken coating of cement, but experiments show that this is practically accomplished if the concrete is mixed quite wet, so as to furnish a thin coating on the steel, and if it is free from voids and cracks. The preserving quality of concrete is thus dependent upon its consistency and the proportions employed, but for ordinary reinforced work it may be concluded from the results of tests 22 REINFORCED CONCRETE CONSTRUCTION and observations that, when well placed, the concrete affords complete protection of the steel against corrosion. 16. Modulus of Elasticity. — To gain clearness in this study, modulus of elasticity will be treated in Art. 25. 17. Weight of Concrete. — A dense, well made, 1:2:4 concrete, when dry, will weigh approximately 155, 152, 148, and 143 lb. per cubic foot, according as the aggregate is trap rock, gravel, limestone, or sandstone. These figures represent average values although the weight is affected but little by any ordinary varia- tion of proportions. It is found that a wet concrete when dried out will weigh less than a well compacted concrete containing originally less water. The addition of reinforcing steel in the usual proportions will add from 3 to 5 lb. Cinder concrete has an average weight of 112 lb. per cubic foot. The assumed weight of reinforced concrete is usually 150 lb. per cubic foot. CHAPTER II STEEL 18. General Requirements. — The reinforcing steel in reinforced concrete construction is mostly in the form of rods, or bars, of round or square cross-section. These vary in size from about 1/4 to 3/8 in: for light floor slabs, up to 1 1/2 in. as a maximum size for heavy beams. Somewhat larger sizes are sometimes used for columns. Rods can be procured varying by 1/16 in. increments from 1/4 in. to 1 in., and then by 1/8 in. increments to 2 in. Above 2 in., rods to the nearest 1/4 in. should be se- lected. Stock lists of steel companies should be consulted as these show the normal stock in tons of the different sizes. Various bars of deformed cross-section are widely advertised, some of the more common types being shown in Fig. 6. Woven wire and thin punched plates are sometimes used in thin slabs with good effect. Authorities differ as to the quality of steel to be used for re- inforcement — medium and high carbon steel being used by different engineers. Medium steel is better fitted for most classes of structures than high steel, while in some structures high steel will answer just as well although probably with very little gain in economy. No matter what kind of steel is used in beams, low unit stresses are much to be preferred on account of the large distortions involved. STEEL 23 Brittleness is to be feared in high steel, although this quality is not so dangerous when the metal is used in heavy reinforced concrete members — for example, in heavy beams or slabs — Ransome Bar Johnson Bar Thacher Bar Diamona Bar Cup Bar Cold Twisted Lug Bar Fig. 6. — Deformed bars. as the concrete to a large extent absorbs the shocks and protects the steel. Medium steel is manufactured and sold under stand- ard conditions in the open market. This steel is of excellent quality, and an engineer can feel sure of the safety of his structure without the expense of exhaustive tests. As a rule, a satis- 3 24 REINFORCED CONCRETE CONSTRUCTION factory high elastic limit steel cannot be obtained in this way. To prevent brittleness a careful inspection of high steel should be made during manufacture. The expense of doing this may so increase the cost of the material that little or no economy will result from the use of high steel, even though it be considered permissible to employ a greater working stress than in the case of medium steel. Open hearth steel is preferable to Bessemer steel, as it is more uniform in quality and does not possess the brittleness sometimes met with in Bessemer steel. An important test in the specifications for reinforcing steel is the bending test, and no steel which fails to pass this test should be used under any circumstances. In case a lot of steel has been delivered without previous test by the purchaser, one bar selected at random in every 100 should be subjected to this test and, if it fails to pass, the portion from which it is taken should be rejected. 19. Tensile Strength. — The elastic limit and ultimate strength of the two general grades of steel above mentioned will range about as follows: Medium High Elastic limit, lb. per sq. in. . . . 35 to 40,000 50 to 60,000 Ultimate strength, lb. per sq. in. 60 to 70,000 80 to 100,000 20. Coefficient of Expansion. — The coefficient of expansion of steel may be taken at 0.0000065 per degree Fahr. 21. Modulus of Elasticity. — The modulus of elasticity of all grades of steel is very nearly the same and is usually taken at" 30,000,000. (Art. 25.) If steel could be made with a high modulus of elasticity, it would be especially fitted for reinforced concrete work, because the higher the modulus the less the deformation under any given loading. Unfortunately, however, the modulus of elasticity is a constant for all the different kinds of steel. 22. Bending Test for Steel. — Medium Steel: Full-sized mate- rial 1 in. thick, and over, should bend cold 180 degrees around a pin, the diameter of which is equal to twice the thickness of the bar, without fracture on the outside of bend. High Elastic Limit Steel: Specimens for bending should bend cold under the following conditions without fracture on the outside of the bent portion: Around twice their own diameter Around their own diameter 1 in. diameter, 80 degrees. \ in. diameter, 130 degrees. I in. diameter, 90 degrees. in. diameter, 140 degrees. I in. diameter, 110 degrees. -j in. diameter or less, 180 degrees. STEEL 25 Test pieces of steel wire used for reinforcement should bend ISO degrees around their own diameter without fracture. Peoblems 1. A concrete as proportioned by mechanical analysis is a 1 : 2.1 : 4.9 mixture by weight. If both the sand and stone weigh 100 lb. per cubic foot, what are the corresponding proportions by volume considering 100 lb. of cement as 1 cu. ft.? 2. What would be the correct proportions by volume of the concrete of Problem 1, if the sand weighed 95 lb. per cubic foot and the stone 90 lb. per cubic foot? 3. Consider a 1 : 2 : 4 concrete, and let a cubic yard be composed of 1.57 bbl. of cement, 0.44 cu. yd. of sand, and 0.88 cu. yd. of stone. The cement costs $1.50 per barrel and the stone $1.20 per cubic yard. Two sands are available, one at 50^5 and the other at $1.00 per cubic yard, which give a corresponding compressive strength of concrete of 2000 and 2200 lb. per square inch respectively. Which sand is the more economi- cal if the concrete is to be used in members sustaining purely compressive stresses? 4. By carefully grading the materials in Problem 3 using the 50^5 sand, it was found that practically a 1 : 3 : 6 mixture having 1.11 bbl. cement, 0.47 cu. yd. sand, and 0.94 cu. yd. stone, per cubic yard of concrete, could be used to give the same strength that the $1.00 sand gave in Problem 3. , The added cost of labor for proportioning and mixing the concrete, due to the use of graded materials, was 10^ per cubic yard. What would be the saving on a piece of work involving 10,000 cu. yd. of concrete? 5. The sieve analysis of a bank gravel is as follows: Sieve number Size of mesh in inches Percentage passing 1-in. 1.00 100.0 3/4-in. 0.75 92.5 1/2-in. 0.50 80.0 1/4-in. 0.25 71.0 .15-in. 0.15 64.4 10 0.073 50.0 20 0.034 34.0 30 0.022 25.7 40 0.015 20.8 50 0.011 17.5 80 0.007 11.0 100 0.0055 9.5 Show by mechanical analysis curves how you would screen the gravel into two sizes and combine to secure a concrete of greater density an(i strength. CHAPTER III CONCRETE AND STEEL IN COMBINATION 23. Advantages of the Combination— The highest success in the use of concrete and steel in combination is attained only when maximum strength is secured at minimum cost.^ This is accomplished when the steel and concrete are placed in such a manner as to derive their greatest strength, and when economical proportions of these materials are employed. Steel can be put into a form to resist a given tensile stress much more cheaply than it can be put into a form to resist a cor- responding compressive stress. This should be clear, when it is considered that the solid bar is well adapted to take tensile stresses, while for compressive stresses the steel must be made into forms of more extended cross-section, in order to provide the necessary lateral rigidity. There is a serious objection to the use of steel in many locations due to its lack of durability and its failure to stand up under a high heat. Steel is also a relatively expensive building material. Concrete, on the other hand, cannot be used in tension except to a very limited extent, but its compressive strength is fairly high. It is also a good fireproof material and has great durabil- ity. In addition, concrete is a comparatively cheap material; is readily available in almost any locality, and tests and the results of observations show^ that it thoroughly protects em- bedded steel from corrosion. From the above considerations, it follows that reinforced concrete construction is advantageous to varying degrees in different types of structures. In structural forms subjected to both tension and compression, such as beams, the proper com- bination of the two materials meets with the best success. Steel rods embedded in the lower side of the beam carry the tensile stresses while the compressive stresses are carried by the concrete. Here, then, the steel is used in its cheapest form and the whole structure may be made strong, economical, and very durable. For columns, also, a combination of the two materials is quite advantageous although to a varying degree and, in any case, not to such a large extent as in beams. In this type of 26 CONCRETE AND STEEL IN COMBINATION 27 structure, steel may be used with concrete in the form of small rods to reinforce the concrete; or it may consist of structural steel shapes in the form of a core, simply surrounded and held rigidly in place by the concrete — most of the load being carried by the steel; or finally, a steel column may be used and the concrete employed merely for the purpose of adding symmetry and providing fire protection. This last case does not come under the head of reinforced concrete and will not be considered in this course. The use of a moderate amount of steel with concrete so as to produce a material with a reliable tensile and bending resistance has opened the way for the use of this combination in a great variety of structures and in special individual forms. 24. Bond Between Concrete and Steel. — Usually the entire stress which is carried by the steel of a reinforced concrete beam is transmitted to the reinforcing bars by the bond or adhesion between the concrete and the steel. Stress is also conveyed to the steel of a reinforced concrete column in like manner. Experience has shown the bond to be reliable and per- manent, and that plain bars may be used in most structures with success. Deformed and twisted rods are used to a large extent in structures where the stress between the steel and concrete exceeds the safe working adhesion of the plain rod. The in- dented surfaces of deformed bars produce a mechanical bond in addition to the adhesion already mentioned. The values derived from different experiments for the adhesion in pounds per square inch of contact surface vary quite widely. With plain rods, results vary from about 200 to 750 lb. per square inch. The quality of the concrete and the manner of making the tests are important factors. Tests to determine the strength of bond are usually made either by pulling out a rod that has been embedded in a block of concrete or by forcing the steel out by compression. In either case the concrete is in compression, and there is the tendency to increase the friction by adding to the pressure at the surfaces in contact. In an important series of tests made at The Univer- sity of Wisconsin, test beams were arranged as shown in Fig. 7 — the reinforcing bars being embedded only a short distance from each end, leaving the middle portion exposed. The stress in the rods was computed from the observed deformations. The conditions were quite similar to those which exist in an ordinary beam, but the beam was prevented from failing, in the early 28 REINFORCED CONCRETE CONSTRUCTION stages of the tests, by an upper set of auxiliary rods. Failure finally occurred by the pulling out of the lower rods, as intended. Tests indicate that the bond per square inch of surface is not affected by the size of rod. A study of the results of many experiments leads to the conclusion that for ordinary round or square bars, not too smooth, the adhesive strength may be taken at from 200 to 300 lb. per square inch. After bond is once broken, plain rods give a frictional resistance of about two- thirds of the original bond resistance; in other words, the rein- forcement may be moved slightly with reference to the concrete, as by a sharp blow, and still leave about two-thirds of the bond strength effective. A rough surface gives a higher adhesive value than a smooth surface, consequently a thin film of rust on the reinforcing metal should not cause its rejection. Loose or scaly rust should not be allowed, however. Bars in this state of corrosion may be used, if they are first cleansed with a stiff wire brush or given a pickling bath of sulphuric acid solution (consisting of 1 part acid to 6 parts of water) , and then dipped in clean water. Oiling and paint- ing of reinforcing steel should not be permitted, as the adhesion is greatly reduced thereby. Round bars show the greatest ad- hesion — flat bars the least. Let fs be the working tensile strength of the steel, as the area of bar, 0 the circumference of bar, d the diameter or thickness of bar, u the working unit bond strength, and x the required length of embedment (or grip) for the above values of fs and u. Then, to develop the strength of the steel, using either round or square bars, Fig. 7. xou=asfs=-^- fs CONCRETE AND STEEL IN COMBINATION 29 The initial movement in the case of indented bars is probably due to a slight crushing of the concrete under the projections. The strength mechanically obtained from the indented surfaces, however, depends upon the shearing strength of the concrete. The bars cannot be pulled through the concrete without shearing off an area equal to the total area of the in- dented portion and, in addition, overcoming adhesion along the remaining surface. In tests of such bars, failures have usually occurred by the splitting of the specimen or by the breaking of the bar, but the results indicate that at least 500 to 600 lb. per square inch in bond may be expected. Fig. 8. It is a common practice to bend the ends of reinforcing rods into hooks consisting of either curved or right angle bends. Experiments, on the efficiency of such hooks have shown that the ultimate bond strength was greatly increased. It was found that with an embedment in concrete in all directions equal to 8 diameters of the bar, a hook of 5 diameters may be assumed to develop the elastic limit of the steel. When curved ends are used, they should consist of bends through 180 degrees with a short length of straight rod beyond the bend, as shown in Fig. 8. Short square hooks upon the ends of bars are not of great value. 25. Ratio of the Moduli of Elasticity. — In order to treat the subject clearly, it will be necessary to give a somewhat extended discussion of the term modvlus of elasticity. The modulus of elasticity of a material such as steel is the ratio of the unit stress to the corresponding unit deformation, provided the elastic limit of the material is not exceeded. This ratio is usually denoted by E, or in which / denotes the stress in pounds per square inch and d the corresponding deformation per unit of length. 30 REINFORCED CONCRETE CONSTRUCTION Thus, if an external load of P lb. be applied to a steel bar whose length is I in. and whose cross-section is A sq. in., and if the compression under this load be d in., we should have d d/l Ad Suppose an external force 'of 30,000 lb. be applied to a bar whose length is 10 in. and whose cross-section is 1 sq. in., and suppose the extension under this force is 0.01 in. We should have PZ_ (30,000) (10) _ ^-Ad- (1)(0.01) 30,000,000 If the external force, or load, in pounds per square inch be represented by ordinates and the corresponding elongations or Deformation ( Unit Elongations, Inches per inch ) Fig. 9. compressions be represented by abscisses, then the action of the specimen under test may be indicated by what is known as a stress-deformation diagram. In Fig. 9 a typical stress-deforma- tion diagram is shown for medium steel in tension. There are three significant points which need to be noted. These are: the elastic limit, the ultimate strength, and the hreaking-'point. These three points are indicated by the letters A, B, and C, respectively. CONCRETE AND STEEL IN COMBINATION 31 Steel has a well-defined elastic limit in tension, as is readily seen by the diagram, and which is at least one-half the ultimate strength. It is represented by the ordinate of the point of tangency between the straight line and the rest of the curve. For medium steel in compression, the transition from the straight line to the curve is more gradual, and the point of tangency is more difficult to locate. If loads are applied which cause stress less than the elastic limit, no appreciable permanent deformation will exist when the loads are removed. Loads causing stress in the steel above this limit will, however, leave a permanent set on their removal. For example, suppose a steel bar is stretched with a force of 50,000 lb. per square inch of cross-section. The point X will be reached in Fig. 9. The elongation of the bar will be about 0.07 in. per inch of length. If the load is removed, the diagram will go back along some line such as xx'. The perma- nent, set is ox\ If the bar is stretched so that the resulting stress per square inch is less than the elastic limit, then the diagram will go back along the line OA, and no deformation results. Since the line OA is straight, it is clear that below the elastic limit, stress is proportional to deformation. The above illustration could be applied equally well to steel in compression. Above the elastic limit the elongation of steel increases at a rate which becomes more and more rapid until finally the condi- tion of perfect plasticity is reached, and the body elongates under a constant force, while the lateral dimensions reduce more and more rapidly. The point B is reached when this action occurs. The piece at this point draws out rapidly, the breaking point C is soon reached, and the piece breaks under a reduced load. For a piece tested in compression, the points B and C are almost identical. The modulus of elasticity may be figured theoretically by any stress and its corresponding deformation from 0 to A, but is generally figured near the elastic limit stress since practically a more accurate result can be thus obtained. The modulus of elasticity for steel in both tension and compression is about 30,000,000. This value may be taken the same for all grades of steel, the variation being so slight. For medium steel, the elas- tic limit is about 36,000 lb. per square inch for both tension and compression. In the design of reinforced concrete structures, it is necessary to know the relative stresses in the concrete and steel of com- 32 REINFORCED CONCRETE CONSTRUCTION pression members under like distortions; and in beams, it is necessary to find the position of the neutral axis and the required percentage of steel. These may easily be determined if the modulus of elasticity is known for both the concrete and the steel. The modulus of elasticity for steel we already know. Let us now discuss the method of determining the modulus for concrete.^ With concrete we have an entirely different material to deal with as regards elasticity. Fig. 10 shows a typical stress- deformation diagram for concrete in compression. As you can -y see, the compression curve is slightly ^ curved almost from the beginning, the curvature gradually increasing toward a. the end in approximately the shape of a parabola. A release of load at even moderate stresses will show some permanent set, as OF, indicating im- perfect elasticity. After a few repe- ° ^ ^ titions of such loads, however, there DeTormaTion . i i •p^^ will be no further set, and the stress deformation diagram line will become straight up to the load applied, as Y A, for example. There is a limit of stress beyond which repeated applications of the same load will continue to add to the permanent deformation, and the specimen will ultimately fail. This limit is found to be from one-half to two-thirds the ultimate strength and is called the elastic limit. Point & is the breaking-point and ultimate strength. For very low stresses, up to perhaps 300 to 400 lb. per square inch, the variation of the .curve from a straight line is so small that it may be considered straight, and an average straight line may be drawn, as 02", and its slope taken as the modulus of elasticity. For higher stresses, the total deformation is OB and the modulus elasticity becomes ^ = 0B This ratio will be less than the slope of the line OT. Later in the course when referring to these two methods of calculating the modulus, the slope of the tangent OT will be called the initial modulus and the slope of the line OA, the secant modulus. 1 Discussion as given in Turneaure and Maurer's "Principles of Reinforced Concrete Con- struction," 2nd edition, page 20. CONCRETE AND STEEL IN COMBINATION 33 Suppose a load passes frequently over a reinforced concrete structure, and let us consider a certain member in compression. The first application, let us say, causes d deformation with a permanent set d\ Then it should be clear, that the second appli- cation causes an additional deformation of only d-d\ which is the same as saying that both the concrete and the steel did not return to their original position before the second application. There must have been some compression still left in the steel, with an equal amount of tension in the concrete, when the second load was applied; in other words, the total deformation fixes the stress in the steel. Thus, for the determination of the relative stresses in the two materials for a working stress in the concrete, AB the modulus for the concrete should be expressed by ^-g (Fig. 10) AB and not by the ratio ^f^- Let us take the case of a beam. The compressive stresses in the concrete at any section will vary from zero at the neutral axis to the value AB, for example, at the extreme fiber. At inter- mediate points, the stresses and corresponding deformations follow approximately the law of the curve OA. In this case the slope of the chord OA does not exactly represent the facts, but for working loads the error is small. Tests on prisms show that the modulus of elasticity as desig- nated above for ordinary concrete, 30 days old and under work- ing loads, ranges from 2,500,000 to 3,500,000 lb. per square inch. As a rule, the denser and older the concrete, the higher the modulus. Now that the meaning of the term modulus of elasticity has been made clear with reference to both steel and concrete, the reason for the occurrence of a constant ratio of the moduli in specifications and building codes will be discussed. Consider the stresses in the steel and concrete of a reinforced concrete column, reinforced with longitudinal rods only. Let fs =unit stress in steel. fc =unit stress in concrete. Es = modulus of elasticity of steel. Ec = modulus of elasticity of concrete. Since the modulus of elasticity of a material is the ratio of stress to deformation, it follows that, for equal deformations, the 34 REINFORCED CONCRETE CONSTRUCTION stresses in the steel and concrete of a reinforced concrete column will be as their moduli of elasticity. Thus, fc Ec This ratio of the moduli is generally denoted by the letter n, or fs = nfc The term n, then, is the subject of this discussion and is the value referred to in the various specifications and building codes. The equation just given shows that if the stress in either the steel or concrete of a concrete column is known, the stress in the other material can be found, and this relation is made use of in the derivation of column formulas. Fig. 10 shows that the modulus of elasticity of concrete in compression is less for the greater loads, and hence the value of n is greater. Thus, it is plain that with increasing loads in concrete columns the steel receives a greater proportionate stress, the variation in the amount carried by the steel depending on the variation in the value of n. In order to take account of the fact that under increasing loads the steel receives an increasing proportion, it is desirable to use a value of n in the computations for design somewhat larger than that which is obtained by taking a value of E^ corresponding to working loads on small prisms (about 10). A value of 15 for n may well be used for all ordinary mixtures and for all types of columns. The value of 12 sometimes specified is now considered unduly low. In concrete beams, experiments show that the tension which remains in the concrete just below the neutral axis, and properly not allowed for in the derivation of the beam formulas, has its effect in the position of the neutral axis and the strength of the beam. It is found that a value of 15 for n — the same as is used for columns — is not too large for calculations of strength of beams under the usual assumptions, although great accuracy in this respect is not necessary. This value of 15 for n is the one most generally used, but a value of 12 is also frequently employed. The value of 15 corresponds to a value of E^ of 2,000,000 which is somewhat low as determined by compressive tests. Comparatively few tests have been made on the elasticity of concrete in tension, but these seem to indicate that for small stresses, it is practically the same as in compression, although probably slightly less. CONCRETE AND STEEL IN COMBINATION 35 26. Behavior of Reinforced Concrete Under Tension— The be- havior of reinforced concrete under tension, and especially when constituting the tensile side of a beam, has been the result of much study by many experimenters. Early tests indicated that the ultimate strength of reinforced concrete is as much as ten times that of plain concrete, but such results were due to the fact that it was found extremely difficult to determine just when the concrete begins to crack. Cracks do not become noticeable, even on very close examination, until a stretching occurs corresponding to a tensile stress much beyond the ultimate strength of the concrete. The steel forces the concrete to elongate uniformly throughout, so that a crack will open up very slowly and will remain invisible for some time. A method of detecting minute cracks was accidentally discov- ered in 1901-2 in some experiments made at the University of Wisconsin. It was found that when beams were hardened in water and only partially dried before testing, very fine hair-cracks became noticeable at a moderate load. Before these cracks occurred, however, dark wet lines appeared across the beam, and it was observed that each of these lines was later followed by a very fine crack. That these watermarks were incipient cracks was determined by sawing out a strip of concrete along the outer part of the beam. Careful measurements of extension showed that these streaks, or watermarks, occurred at practically the same deformation at which the concrete ruptured when not reinforced. This same phenomenon has since been observed by many careful experimenters, and the fact is now generally established that concrete, reinforced with steel, does not elongate under tensile stress to any greater extent before cracking than plain concrete. A reinforced concrete beam for working loads is usually more heavily stressed on the tension side than the ultimate tensile strength of plain concrete— enough steel being usually embedded near the lower face to permit the full allowable compressive strength of the concrete to be utilized. The presence, then, of the cracks above referred to, occurring long before a reinforced con- crete beam has obtained its working load, must seriously affect the tensile strength of the concrete. The formulas now in most general use for the design of reinforced concrete beams neglect entirely the tensile strength of the concrete. The important question which arises is how far concrete may 36 REINFORCED CONCRETE CONSTRUCTION be cracked without exposing the steel to corrosion. Concrete, fortunately, contains a large proportion of lime which readily absorbs carbon dioxide. For this reason, steel is effectually protected by concrete if it is covered with even a thin film. Experiments tend to show that concrete when well placed and mixed wet, completely protects the steel in the tensile side of a beam from corrosion, even when the unit stress in the steel approaches the elastic limit. 27. Shrinkage and Temperature Stresses. — In reinforced con- crete structures which are free to contract and expand, the stresses occurring from temperature changes and from shrinkage in hard- ening are due wholly to the mutual action of the steel and con- crete. Of the stresses produced from these two causes, those which result from hardening are the greater, but experiments show that even these are not sufficient to be of practical importance. In regard to the temperature stresses, they are negligible by reason of the nearly equal rates of expansion of the two materials. On the other hand, if reinforced concrete structures are re- strained by outside forces, or if they are of such dimensions that they cannot be considered as sufficiently well bonded to act as a unit — such as long retaining walls — then the stresses resulting are much greater, and the tensile strength of the con- crete will be reached (this will occur with a drop in temperature somewhere between 10 and 20° Fahr.), thus producing cracks, called contraction cracks. To prevent plainly noticeable cracks due to shrinkage and lowering of the temperature, steel should be inserted — the amount used varying in practice from 0.2 of 1 per cent to 0.4 of 1 per cent, based on the cross-section of the concrete. This is less than the amount required theoretically, but experience shows this amount to give very satisfactory re- sults where the foundations are stable. If the structure is fixed in two directions, the reinforcement must be placed accordingly. No amount of reinforcement can entirely prevent contraction cracks. The steel can, however, if of small diameter and placed close to the surface, force the cracks to take place at such fre- quent intervals that the required deformation occurs without any one crack becoming large. No cracks will open up to be plainly noticeable until the steel is stressed beyond its elastic limit. The amount of steel should be such, then, that without being stressed beyond its elastic limit, it will withstand the tensile stress resulting from the maximum fall of temperature (usually CONCRETE AND STEEL IN COMBINATION 37 considered to be 50°) in the steel itself plus the tensile stress necessary to crack the concrete. A high elastic-limit steel is thus advantageous. The size and spacing of the cracks will also depend upon the bond strength of the reinforcing rods. The distance between cracks in any given case will be the length required to develop a bond strength equal to the tensile strength of the concrete. Thus, bars with irregular surfaces which provide a mechanical bond with the concrete are in general more effective than smooth bars. 28. Repetition of Stress. — Experiments on cubes of neat cement and concrete for repeated loads have shown that the limit of ,00| , 1 1 1 1 1 1 1 1 c "s: 90 (3 C O 9> 80 peti i mi 1 70 ^§ %^ 60 D) 0 ^ > ITT^iT^ K 5 1 (' Fig. 17. also the angle K it makes with the horizontal. The stress t, Fig. 17, produces the greatest intensity upon a plane perpendicular or normal to its line of action — namely, upon the plane be. In the same manner considering point A', it is a comparatively simple matter to derive the magnitude of f and the angle K' it makes with the horizontal. The stress t' causes the greatest stress on the inclined plane 6'c'. RECTANGULAR BEAMS 45 Theoretically, a plain concrete beam will fail by cracks opening up along the zigzag lines which are shown in Fig. 16. This is found to be true in tests on this type of beam. A plain concrete beam will always fail in tension due to the low ultimate strength of concrete in tension as compared to its strength in compression. The exact direction of the maximum tensile stress at any point in a beam depends only upon the relation between shear and bending moment. Hence by means of the formulas given above, the student should be able to trace the general direction of the stresses in a beam with the loading other than uniform. The introduction of steel into a concrete beam, however, changes the direction of the stress lines somewhat. The exact effect of adding steel will be taken up later. The student should have some idea of the variation of the Compression A Fig. 18. maximum tensile stress throughout a vertical cross-section. Of course, in studying this variation of stress, it should be remem- bered that the direction of the stress is not normal to the ver- tical section at the different points. Maximum tensile stresses, as we have already shown, are normal to a vertical section only when the section is taken where the shear is zero. In other sections these stresses are normal only along the lower fibers of ' the beam. Fig. IS^shows the variation in the maximum tensile stress on a section such as XX', Fig. 17. The variation of the normal and shearing stresses is also shown. It should now be clear that the steel reinforcement in a con- crete beam for uniform loading should, have the general direc- tions shown in Fig. 19 in order to take the tension in the beam and prevent the cracks starting along the lines indicated. Fig. 20 is the simplest method of reinforcement and quite often used for light "loads. In beams highly stressed, the student can appreciate the reason for curved reinforcement in addition to the horizontal rods. The most common method is to use several rods for the horizontal reinforcement and then to bend I'From Turneaure and Maurer's "Principles of Reinforced Concrete Construction," 2nd edition, page 50. 46 REINFORCED CONCRETE CONSTRUCTION a part of these upward as they approach the end, where they are not needed to resist bending stresses. This matter will be taken up in detail later. The concrete is depended upon for the compressive and shearing stresses, its resistance to such stresses being large. It cannot truly be said that the above remarks refer directly to homogeneous beams of concrete. As we have already seen, Fig. 19. the elastic property of concrete is very much inferior to that of My steel, consequently the formula / = does not exactly apply except for very low stresses. In fact, in the previous inves- tigation we have erred slightly by representing the direction of the maximum internal stresses in a concrete beam to be iden- tical with the direction of the maximum stresses in a steel beam. The difference, however, is inappreciable, especially for low stresses. The student should appreciate the fact that the Fig. 20. directions of the internal stresses at. the principal points in con- crete beams are the same as in steel beams — namely, at the upper and lower fibers and at the neutral plane. This statement should be understood when it is considered that the direction 2u at these points, as determined by the formula tan 2K = ~, depends only upon a zero value for either v or f. For example, at the lower and upper fibers, the value of v becomes zero and the magnitude of / does not control. Likewise, at the neutral plane, / becomes zero and the value of the angle K is not affected by the magnitude -of v. RECTANGULAR BEAMS 47 30. Assumptions in Common Theory of Beams. — In order to derive a formula by which we can design concrete beams rein- forced with steel, it will first be necessary to overhaul the assump- tions on which the common theory of flexure is founded and see if we can derive a formula, or formulas, by which we can deter- mine stresses in both the steel and concrete of such beams. The two main assumptions in the common theory of beams may be stated as follows: 1. If, when a beam is not loaded, a plane cross-section be made, this cross-section will still be a plane after the load is put on and bending takes place. (Navier's hypothesis.) 2. The stress is proportional to the deformation — namely, to the elongation or compression per unit of length. (Hooke's Law.) From the first assumption the following principle is obtained : the unit deformations of the fibers at any section of a beam are proportional to their distances from the neutral axis. From the second assumption: the unit stresses in the fibers at any section of a beam are also proportional to the distances of the fibers from the neutral axis. It may be well to explain more at length what is meant by the two assumptions and the results derived from them. Assumption 1. — Imagine two originally parallel cross-sections, ED and GH, Fig. 21a or Fig. 21b, so near to each other that the curve taken after bending by that part of the neutral plane between these sections may, without appreciable error, be accounted circu- lar. Let ED and GH, Fig. 22a or 22b, be the lines in the loaded beam in which these cross-sections cut the plane of the paper, and let 0 be the point of intersection of the lines ED and GH. Let OF^ r, FL^y, FK = l, LM = l-\-al, in which a is the elongation per unit of length of a fiber at a distance y from the neutral axis, y being variable; then, because FK and LM are concentric arcs 48 REINFORCED CONCRETE CONSTRUCTION subtending the same angle at the center, we shall have the proportion I / A i • Weight ^0 Fig. 22a. \\ \\ Wei9ht Fig. 22b. but, as y varies for different points in any given cross-section while r remains the same for the same section, it follows that if a certain cross-section be assumed, the deformation of any fiber RECTANGULAR BEAMS 49 at the point where it cuts this cross-section is proportional directly to the distance of this fiber from the neutral axis of the cross-section. Assumption 2. — As to the evidence in favor of this law, experi- ment shows that as long as a material such as steel is not strained beyond safe limits, this law holds. However, wrought iron and steel are the only important structural materials which closely follow this law, and they only within their elastic limits. But under working conditions, these materials are not stressed beyond their elastic limit and so the formulas ordinarily hold. Timber, stone, and cast iron can hardly be said to obey Hooke's law, yet for working conditions the common flexure formulas for these materials are roughly correct and they are in general use. The important question to be decided is — does concrete follow the laws stated above? In regard to the first assumption, it can be said that careful measurements show some deviation from a plane, but in general this assumption seems to be war- ranted by the results of observed deformations under working loads. Concerning the second law, as regards the concrete in a reinforced concrete beam, it should be clear that the assumption will not strictly apply except for low stresses. 31. Plain Concrete Beams. — OS in Fig. 23 is the stress-defor- mation diagram for concrete in compression, with which the student is already familiar. The curve shown here is identical with Fig. 10, the only difference being that the deformations are represented vertically instead of horizontally. This change has been made in order that we may apply the curve directly to the cross-section of a beam. The curve OT" is the stress-deformation diagram for concrete in tension and is of use only in the analysis of plain concrete beams. We have just seen that Assumption 1 may be applied to beams of concrete. This assumption leads us to the conclusion that deformations of the fibers are proportional to the distances of the fibers from the neutral axis. Figs. 24 and 25 show the curves applied to the vertical cross-section AX of a beam. For example, suppose the deformation at point a, Fig. 25, is the same as rep- resented by Oa in Fig. 23. The corresponding stress given by the compression curve in Fig. 23 is aa\ Lay off the distance aa' in Fig. 25 to represent that stress. Proceeding similarly for all points and connecting, we have the stress curve A"OX", which is nothing more than a portion of the stress-deformation diagrams in Fig. 23 plotted to a different scale. 50 REINFORCED CONCRETE CONSTRUCTION . Turning our attention to the principles of mechanics, in order to determine which ones may bo used to find the resisting moment in a plain concrete beam, we find we have three, as follows: 1. For beams rectangular in section, the average unit tensile and compressive fiber stresses on any cross-section are represented .S a Deformatior Tension 0 Compression T Fig. 23. by the average abscissae in the tensile and compressive parts of the combined stress-deformation diagram respectively. Also, the total tension T, Fig. 26, and the total compression C on a cross-section are proportional respectively to the areas XOX" Fig. 24. and AOA"; hence, according to some scale, the areas represent T and C respectively. 2. The resultant tension T and resultant compression C act through the centers of gravity of the tensile and compressive areas in the combined stress-deformation diagram. 3. When all the forces (loads and reactions) applied to the beam act at right angles to it, then the resultant tension T equals RECTANGULAR BEAMS 51 the resultant compression C; hence, the two stresses constitute a couple which is the resisting couple. (This principle comes from one of the three conditions of equilibrium; namely, IH = 0.) In order to introduce important ideas concerning concrete beams, the perfectly general method of figuring the ultimate resisting moment at any cross-section will be explained. Fig. X" X Fig. 25. 27 shows the combined stress-deformation diagram for a given concrete. A plain concrete beam will break in tension because of the low strength of concrete in tension as compared to its strength in compression. BB' then gives this breaking tensile stress and T is represented by the area BOB'. Likewise, C \ ] ' Neutral plane v" 0 — c Fig. 26. is represented by the area AOA' — this area being determined from the principle that BOB' and AOA' are equal. The next step is to locate the center of gravity of each area and scale the vertical distance between them. The resisting couple is jT or C multiplied by this scaled distance. Partly to test the correctness of the theory of flexure of concrete beams, Prof essor Morsch* made three beams and several tension and * Of the Zurich Polytechnic, Zurich, Switzerland. 52 REINFORCED CONCRETE CONSTRUCTION compression specimens of the same mix of concrete.^ From tests on the specimens, he obtained a combined stress-deformation diagram from which he computed the probable resisting moment of the beams. The average of the actual resisting moments of the beams and the moment computed by the theory of flexure agreed very closely. This general theory of the flexure of concrete beams may thus be considered highly satisfactory. It should be clear to the student by this time that a great deal of compressive strength cannot be made use of in a plain concrete beam. If concrete were only stronger in tension, then the plain con- crete beam might be of some structural value. In order to offset this disadvan- tage of plain concrete, steel is used. In the discussions which follow it will be assumed that the concrete and steel adhere perfectly and therefore deform equally. Many tests show that, under proper design, this is true for all prac- tical purposes. 32. Flexure Formulas for Reinforced Concrete Beams. — A great many formu- las have been proposed from time to time to be used in the design of reinforced concrete beams. Of course, the object has been in each case to obtain a formula whereby the resisting moment may be obtained at any cross-section in a sufficiently accurate manner, but at the same time with a view to the ease with which it may be used in practice. As might be expected, many of the earlier formulas considered the concrete to carry its share of the tension which we know now cannot be done with absolute safety. By means of some formulas the ultimate resisting moment of a beam may be computed and the ultimate loads at once determined which will produce this moment. A factor of safety is then applied and the safe load obtained. In other formulas the working loads are found directly. When the ultimate load is obtained from a formula, the utimate strength of the concrete and the elastic limit of the steel (the student will find later that the elastic limit of the steel may determine ultimate strength) must be used. Likewise, when deriving safe loads, the working strengths should be substituted. Unlike steel beams, reinforced concrete beams require a prelim- ' From Turneaure and Maurer's "Principles of Reinforced Concrete Construction," 2nd edition, page 55. Fig. 27. RECTANGULAR BEAMS 53 inary formula to be solved before the formula for resisting moment may be employed. Solving this preliminary formula locates the position of the neutral axis which is in the same position only for beams of a given concrete and of a given percentage of steel reinforcement. The two most general varieties of flexure formulas in practical use will be taken up in Arts. 33 and 34. In each of these two classes of formulas, tension in the concrete is neglected. For purposes of discussion, the subject of beams will first be treated with reference only to the horizontal reinforcement. The in- clined tensile stresses will be considered separately. Analysis and results of experiments discussed in succeeding assignments prove beyond any doubt that the rational formulas which we will now develop may be used safely and economically in design- ing. No empirical formula is needed. The student should realize that the following assumptions must be made in order to derive working formulas: (1) the union between the steel and the concrete is sufficient to cause the two materials to act as one material; (2) no initial stresses are con- sidered in either the concrete or the steel due to temperature or shrinkage; (3) the applied forces are parallel to each other and perpendicular to the neutral surface of the beam before bending; (4) sectional planes before bending remain plane surfaces after bending within the elastic limit of the steel. 33. Flexure Formulas for Working Loads Based on Rectilinear Variation of Stress in Concrete. — The loads being working loads, the unit stress in the steel is within the elastic limit, and the unit stresses in the concrete may be considered without material error to vary as the ordinates to a straight line. The following notation will be employed referring to Fig. 28. Let /c = maximum intensity of compressive stress in the concrete under a given load. It is represented by the distance AA" . /s = maximum intensity of tensile stress in the metal under the same load (the area of reinforcement is assumed to be so small with reference to the total area of cross-section of the beam that the stress in the metal is practically uniform) . AA' represent the deformation, or deformation per unit length, in the concrete which is stressed to the amount /c. 54 REINFORCED CONCRETE CONSTRUCTION C'C represent the unit deformation, or deformation per unit length, in the metal which is stressed to the amount /s. C — total compression in concrete at a section of the beam. T =: total tension in steel at a section of the beam. Ec represent the modulus of elasticity of concrete in compression. Es represent the modulus of elasticity of steel in tension. n = ratio " . d = distance from compression surface to axis of reinforcement. ]c = proportionate depth of neutral axis from below the compression surface, tts =area of cross-section of steel. h = breadth of a rectangular beam. p = "steel ratio" =the ratio of the area of steel to area of concrete =r^,' od Mc = resisting moment as determined by concrete. Ms = resisting moment as determined by steel. M = bending moment or resisting moment in general. Now the total compressive resistance is equal to the area of the triangular figure AO A" multiplied by h, the breadth of the beam. But the area AOA" = 1/2 {AA")kd = l/2fckd. Hence, the total compressive resistance C is equal to 1/2 fckbd. The total tensile resistance T is evidently the cross-sectional area of the metal multiplied by the uniform intensity of stress thereon =as/s. Since the total compressive resistance above the neutral axis must be equal to the total tensile resistance below the same, we have l/2fckbd = asfs (1) From the assumption that deformations vary as the distances of the fibers from the neutral axis, A A' CC A A' CC or OA OC kd d{l-k) But by definition of the values represented by AA' and CC , we have ^^'=^and CC = ^ tjc Es RECTANGULAR BEAMS 55 Substituting fc _ fs EM E,dil-k) which reduces to //. 1 ~k ^-fcn-r- (2) The stress-deformation diagram of the concrete being straight, the center of action of the compressive stresses is at a point 2/3 of the height of OA from 0. The lever arm for the resisting Deformation Stress Cross Diagram Diagram section Fig. 28. moment of the summation of the compressive stresses with respect to the neutral axis, is therefore represented by2/3A;d. The center of action of the tensile stresses is at a point distant OC from 0. The lever arm for the resisting moment of the sum- mation of the tensile stresses with respect to the neutral axis, is therefore represented hj d{l — k). The total resisting moment of the beam is the sum of the moments of the total compressive stresses and of the total tensile stresses about the neutral axis. Therefore, we have 2/3 M(l/2 f,khd) + d{l-k)asfs = 1 /3 fckW +asfsd (1 -k) (3) Elimmating k between equations (1) and (2), the following formula for steel ratio results 1/ 2 fc \nfc I This formula shows that for a given concrete and ratio of working stresses, p has the same value for all sizes of beams. Introducing the value of from equation (2) into equation (1), we have l/2F6(^-asn(l-A;)=0 or lJ2k^b-pbn{l- k) =0 5 56 REINFORCED CONCRETE CONSTRUCTION from which k = \/2pn+ {pny-pn (5)^ Sustituting the value of asfs from (1) into (3) , we get Mc^ll2fcka-ll3k)hd' (6) Substituting the value of /c from (1) into (3) and remembering that as=pbd, Ms=pfs{l-ll^k)hd' (7) Solving equation (1) for fc, fc = ^ (8) It will be noted that equation (6) gives the resisting moment when the maximum allowable value of fc is introduced as the limiting factor, and that equation (7) gives the resisting moment when the maximum allowable value of fs is the limiting factor. The lesser of these two resisting moments, when proper working values are assigned to fc and/^, is the allowable resisting moment of the beam in question. The distance from C to T is in most formulas denoted by jd; namely, j denotes the ratio of this distance to the total depth of the beam. Since jd = d-l/Skd, then j =1-1 /3k and the above formulas may be simplified by substituting / for the quantity (1 — 1 /3A;.) The formulas which we shall need to use in the design of re- inforced concrete beams are collected for convenience as follows: P^bd 1/2 fc\nfc / k= V2pn-\-ipn)^—pn (5) M Mc = ll2fckj(hd^) or ^d'=Y/2fck] M Ms = pfsj(bd^) or bd^--^- (7) , ' (8) k Illustrative Problem.— What will be the resisting moment (M) for a beam whose breadth (b) is 8 in. with a distance from the center of the rein- forcement to the compression surface (d) of 12 in., the area of steel section RECTANGULAR BEAMS 57 being 0.96 sq. in.? = 30,000,000. jE;c = 2,500,000. /c = 500 lb per square inch. = 16,000 lb. per square inch. ^ 30,000,000 as 0.96 ~2:mm~ ^ = 65= (8)112) From equation (5) k = \/ (2)(0.01)(12) + (0.01)2(12)2^ - (0.01) (12) = 0.384 From equation (6) Mc = 1/2(500) (0.384) (0.872) (8) (12)=' = 96,500 in.-Ib. From equation (7) Ms = (0.01) (16,000) (0.872) (8) (12)2 = 160,700 in.-lb. is the lesser of the two resisting moments and hence controls in the design =96,500 in.-lb. _ Illustrative Problem.— Suppose that the beam of the preceding problem is 14 m. deep and is subjected to a bending moment of 130,000 in.-lb. Compute the greatest unit stresses in the steel and concrete. ^=S=(Iot=^-^°^' From equation (5) k = V{2) (00086) (12) +(0.0086)2 (12) 2- (0.0086) (12) = 0.363 From equation (6) 130,000= (0.363) (0.879) (8) (14) 2 /c = 520 lb. per square inch From equation (7) 130,000 = (0.0086) (/,) (0.879) (8) (14)2 fs = 11,000 lb. per square inch Illustrative Problem.— A beam is to be figured to withstand a bending moment of 300,000 in.-lb. A 1:2:4 concrete will be used with E^ = 2,000,000 and/, = 600 lb. per square inch. The pull in the steel is to be limited to 14,000 lb. per square inch. Its modulus of elasticity is 30,000,000. ^ * From equation (4) p = 0.0084 With this value of p, Eq. (5) gives A; = 0.391 7=0.870 Either equation (6) or (7) may now be used in determining b and d since 58 REINFORCED CONCRETE CONSTRUCTION the amount of steel to be employed will cause simultaneous maximum working stresses. From equation (7) , ,3 300.000 ^ g ~ (0.0084) (14,000) (0.870) Many different values of h and d will satisfy the last equation. If b is taken as 10 in., then ^2=29^ =293, ord = 17 1/4 in. Finally as = (0.0084) (10X17.25) = 1.45 sq. in. If 1 3/4 in. is allowed between the tension surface of the concrete and the center of the steel, the entire depth of the beam should be 19 in. Problems 16. What will be the resisting moment for a beam with 6 = 12 in. and d = W in., the area of steel section being 3.5 sq. in.? =30,000,000. Ec = 3,000,000. fc = 600 lb. per square inch, fg = 16,000 lb. per square inch. 17. Suppose the beam of the preceding problem is 20 in. deep (d) and is subjected to a bending moment of 550,000 in.-lb. Compute the maximum unit stresses in the concrete and steel. 18. Design a beam of 12.5 ft. span to carry a load of 1000 lb. per linear foot (including weight of beam), using a 1 : 2 : 4 Portland cement concrete having an allowable working strength of 650 lb. per square inch. A mild steel, with an allowable working stress of 14,000 lb. per square inch is to be used. = 30,000,000. Ec = 2,500,000. . 2v 19. By means of the formulas < = l/2/±V 1/4/2 + ^2 and tan 2X = y, prove the following statements concerning homogeneous beams: (a) Wherever shearing stress exists, the maximum stress will not be on a vertical section, but on an incHned one. (b) At all points in a beam where the shear is zero, the direction of the maximum tension is horizontal, as at points of maximum bending moment and along the outer fibers of the beam. (c) Wherever the horizontal fiber stress is zero (at the neutral plane and at all sections of zero bending moment), the direction of the maximum tension is inclined 45 degrees to the horizontal, and its intensity is equal to the vertical shearing stress at the same place. 34. Flexure Formulas for Ultimate Loads, Based on Parabolic Variation of Stress in Concrete. — The stress-deformation curve for concrete in compression agrees very closely with the parabola; in fact, so nearly so that the parabola is often used in theoretical analysis to represent the distribution of compressive stress in a RECTANGULAR BEAMS 59 reinforced concrete beam. The form of parabola used has its axis vertical in the stress-deformation diagram, Fig. 10, and its vertex at the point S of the curve representing the ultimate strength. The assumption is made in deriving formulas for ultimate loads, that the amount of reinforcement is sufficient to develop the full compressive strength of the concrete without stressing the steel beyond its yield point. Failure under such conditions will occur by crushing the concrete, with the yield point of the steel not exceeded. Then, the parabola representing the variation of com- pression is a full parabola, the upper end being the vertex and the axis horizontal. (Fig. 29.) Averaqe compressive stress = Total compressive Stress = i fr bKd . 3 ^ Fig. 29. When ultimate loads are considered, the secant modulus for the concrete should not be used. The initial modulus should be employed and will be denoted in this article by Ec. Considering Fig. 29 as drawn to scale, ^'~A0 It is a well-known property of the parabola that Since AO represents the deformation A A' (Fig. 28). AA :) or AA' = Ec In the present connection, the two following properties of a parabola (Fig. 29) are useful: (1) the average abscissa of the parabolic arc equals two-thirds the greatest (/c) ; (2) the distance from the center of gravity of the parabolic area to its top equals three-eighths the total height kd. Following the same procedure as employed in deriving formulas for a rectangular variation of stress in the concrete and using 60 REINFORCED CONCRETE CONSTRUCTION the value of AA^ given above, the following equations may be obtained: as V bd 2/3 III A^j^l fc\2nfc y=l-3/8/c M Me=-2/3fckiibd^) or bd^ = Ms=pfsj{bd^) or bd^ = 7< 2J3fcki M Vfsj 2k In the above formulas, fs = elastic limit of the steel, and fc = ultimate compressive strength of concrete. When using the above formulas, it should be remembered that it was assumed at the outset that the amount of steel in the beam is sufficient to cause the ultimate resisting moment to be due to the concrete. Thus, the resisting moment of the beam may be figured by using the formula for Mc. If an amount of steel is used such that the ultimate strength of the concrete and the elastic limit of the steel would be reached simultaneously, either Mc or Ms may be used to determine the ultimate resisting moment. If a less amount of steel is used than the amount just mentioned, the conditions of the assumption do not hold, and the formulas given above cannot be used. When this happens the ultimate moment may be figured by means of formulas based on a parabolic variation of compression in the concrete and applicable for any load up to the ultimate. The parabola for such a case is not a full one and the formulas are cumbersome to use and not at all fitted for practical use. The formulas for ultimate loads, however, can easily be em- ployed in designing. The method is to find the amount of steel to give equal strength in tension and compression. Then either ^ 2l3fckj RECTANGULAR BEAMS 61 or Pfs] may be used to determine the size of beam necessary. Illustrative Problem. — A beam is to be figured to safely withstand a bending moment of 200,000 in.-lb., the ultimate compressive strength of the concrete being taken at 2000 lb. per square inch and the elastic limit of the steel at 40,000 lb. per square inch. n = 15. =0.02 20 A;=-yj3(0.02) (15)+ (|)\0.02)'(15)' (0-02) (15) = 0.598 /=1_3/8A! =0.775. With a factor of safety of 3, the ultimate bending moment is 600,000 in.-lb. and , 600,000 2 ~ (2/3) (2,000) (0.598) (0.775) ' With b =8 in., then ^2 = 972^ ^21,5, orcf =11 in. 8 Also, = (0.02) (8) (11) = 1.76 sq. in. Some designers consider the stress-deformation curve to be a full parabola even at working loads and use working formulas similar to those derived for a rectilinear variation of stress. The results obtained in designing under such an assumption are not as much on the side of safety as those obtained by the straight line formulas. This method of figuring beams is more likely to be found in practice where the allowable unit stress on the con- crete as specified by a building code is considered lower than conservative practice requires. Formulas for ultimate loads are open to the objection that when a factor of safety is applied which will bring the stress in the concrete to about a good working stress, the stress in the steel becomes unduly low from a standpoint of economy. A factor of safety of 3 or 4 as is usually taken leaves a high stress in the concrete with the stress in the steel far below what is 62 REINFORCED CONCRETE CONSTRUCTION usually considered a safe stress. Beams designed by the ultimate load formulas will generally be of smaller cross-sectional dimensions than when the straight-line formulas are employed, but, on the other hand, a larger amount of steel is required. Experienced designers, however, will arrive at satisfactory results by either the "factor of safety" (referred to ultimate strengths) or the "working stress" methods, but there seems to be no good reason why the simple formulas based on the straight line stress variation should not be used for purposes of design, — safe working stresses being employed. It may be well to state here, that the straight line theory of stress distribution will be assumed in all the discussions which follow. Problems 20. Solve Problem 16 for the ultimate resisting moment, assuming a 2000-lb. concrete and assuming the elastic limit of the steel equal to 40,000 lb. per square inch. Consider the given value of Eg to be the initial modulus. 21 Solve Problem 18 by the ultimate load formula, assuming a 2400 lb. con- crete and assuming the elastic limit of the steel equal to 35,000 lb. per square inch. Use a factor of safety of 4. Consider the given value of E(. to be the initial modulus. [In the preceding paragraphs have been shown the two usual methods of calculating the maximum fiber stresses in the concrete and steel of a reinforced concrete beam. The method of pro- cedure is to determine the vertical section of the beam where the moment is a maximum and apply the formulas at that section. The formula for p, containing the values of /c and/s, determines the amount of steel reinforcement which is needed to cause the beam to be of equal strength in tension and compression. The formulas for resisting moment determine the bending moment which a beam will safely withstand (for an existing structure) or the size of the beam needed to resist a given bending moment (for a proposed structure). In steel I-beams the above mentioned calculations, which for convenience we shall call "moment calculations," are the only ones needed except in the case of short beams heavily loaded, when the matter of diagonal compression must be investigated. Steel beams are strong in tension but the thinness of the web RECTANGULAR BEAMS 63 makes some investigation necessary in the design of short deep beams to prevent crippling of the web due to column action. In reinforced concrete beams, on the other hand, the concrete is strong in compression but exceedingly weak in tension and usually the diagonal tensile stresses become fully as important as the maximum fiber stresses. An investigation will now be made of the distribution of tension and shear in the concrete to deter- mine what effect such stresses have on the diagonal tensile stresses, and then we can decide on the method of providing for these stresses. The variation of the bond stress between the concrete and steel will also be considered.] 35. Shearing Stresses. — As previously mentioned, the intro- -X --->j A 0' ( Fig. 30. duction of steel into a concrete beam affects the direction of the diagonal tension lines to a certain extent, by reason of the large shearing stresses which are brought into existence immediately above the steel. First of all, however, let us see the method employed in determining this shear. In Fig. 30 is represented a short portion of a beam which we will assume so short that no part of the load need be directly considered. The total vertical shear will be denoted by V. The student has already learned that the horizontal and vertical shearing stresses per unit area at any point in a beam are the same. Let v = intensity of either stress at the neutral axis. The tension area of the concrete may be neglected. Then, C = T' and C=T. The total shearing stress on any horizontal plane between the steel and the neutral plane will be equal to 64 REINFORCED CONCRETE CONSTRUCTION T' - T, and, if h is used to denote the breadth of beam, the stress per unit area (1) V — T'-T hx The forces acting on this portion of the beam must be in equilib- rium, hence the couples must balance. Thus, Vx=ir-T)jd Vx {T'-T) or id (2) Substituting in equation (1) , we have V The value of / for working loads varies within narrow limits and V will change but slightly if the different values of / are Neutral plane Steel 8 Y. 7bo( Fig. 31. inserted in equation (2). The average value of / for beams in ordinary construction is 7/8. Using this value, equation (2) reduces to 8 V (3) v = 7'bd It is clear that the shearing stress is the same at all points between the neutral axis and the steel; above the neutral axis the shear follows the parabolic law as in the plain concrete beam. Fig. 31 represents the distribution of shearing stress (both horizontally and vertically) on a vertical cross-section where the bending moment is not a maximum. It is quite as satisfactory for purposes of comparison to use the average value of the shearing stress, or V (4) RECTANGULAR BEAMS 65 It should be remembered that the true maximum intensity of shear will generally be from 10 to 15 per cent higher than the value thus determined. The longitudinal tension in the concrete near the end of beam modifies the distribution of the shear, increasing the shearing stress somewhat at the neutral axis and decreasing it at the level of the reinforcement. Equation (3), however, gives results which are sufficiently accurate and are derived for beams having the horizontal bars straight throughout. When any web rein- forcement is used, the distribution and the amount of the shearing stresses at the end of a simply supported beam are materially different from the foregoing. The analysis of the stresses become more complex and a determination of their value impracticable. Even here, however, the above formula serves a useful purpose. It is found that shear is the chief factor in the failure of a beam by diagonal tension and either formula (3) or formula (4) may be used in design if properly controlled by the results of experiments. 36. Inclined Tensile Stresses. — To determine the approximate amount and direction of the diagonal tensile stresses in the concrete of reinforced concrete beams, the two equations given in Art. 29 will apply. The equations there given are as follows: « = 1/2/±v/T/47M^ (1) tan2Z = ^ (2) Although we properly neglected any tension which may exist in the concrete, when making the "moment calculations" and when deriving the formula for horizontal and vertical shear, still when it comes to the consideration of diagonal tensile stresses this tension must be taken into account. As already explained, when the maximum fiber stresses in a beam at the section of maximum bending moment have reached their allow- able working values, the cracks near this section, across the bottom of the beam, show that tension in the concrete no longer exists. This is the reason why the concrete is not considered to take any tension in the moment calculations. But it must be understood that cracks due to the same cause do not exist near the ends of the beam where the bending moment is small and where the diagonal tension cracks are likely to occur. Also, maximum vertical and maximum horizontal shear will be found at the ends of a beam where the tension in the. concrete exists, 66 REINFORCED CONCRETE CONSTRUCTION but in the preceding equations for shear we have neglected the tension existing in the concrete, because the effect of such tension on the distribution of the shear is very small and need not be taken into account. When it comes to the matter of diagonal tensile stresses, however, the tension in the concrete must be considered. As already explained, it is impossible at any point to determine how much tension still remains in tne concrete and, because of this, our formulas for direction and magnitude of the diagonal tensile stresses cannot be used to give us accurate results nor can they be used in design. By means of them, however, we can arrive at some important conclusions. Suppose, for example, at some point between the middle and end of a beam, the stress in the steel is 3000 lb., due to a smaller bending moment than the maximum. Consider J^c = 2,000,000 and =30,000,000. The steel and concrete at the bottom of the beam will have the same deformation due to plane sections remaining plane. We know, also, by definition that tensile stress in concrete Deformation = ^ = ^ f E or tensile stress in concrete ="^^= 200 lb. per square , inch. Es This is about the ultimate tensile strength of a good concrete and so between this point and the end of the beam, tensile stress will exist in the concrete, even along the bottom of the beam. Between this point and the center, tension will also exist in the concrete but the lower limit of the stress will gradually approach the neutral plane until the amount is approximately zero near the section of maximum moment. The resulting diagonal tension at the point of the beam referred to above, assuming a reasonable shearing stress in the lower part of the beam, say 80 lb. per square inch, will be by equation (1) f = 1/2(200) + \/l / 4 (200) 2 + 80^ = 228 lb. per square inch. and it will have a direction inclined about 19 1/4 degrees from the horizontal. This stress may exceed the ultimate strength of the concrete and the result will be an inclined crack. The above discussion shows us that the maximum tensile stresses become considerably inclined immediately above the line of the steel. From equation (2) it is plain that this inclina- tion is greater, the greater the shear, and the less the horizontal RECTANGULAR BEAMS 67 tension. It will, therefore, increase toward the end of the beam. At points nearer the neutral plane, the horizontal tensile stresses become less and the inclined tension approaches the value of the shearing stress, while its inclination approaches 45 degrees. The student through all this discussion should keep in mind that these diagonal tensile stresses can only occur where the concrete still takes its proportion of the tensile stress. Fig. 32 is an attempt to represent roughly the general direction of the inclined tensile stresses in a uniformly loaded beam with hori- zontal reinforcement. From equation (1) it is evident that the intensity of the diagonal tensile stress at any point depends upon the shear and horizontal tension in the concrete. The percentage of reinforcement is also a factor to be considered, since a large percentage reduces the horizontal deformation and consequently the tension in the concrete, and tends to strengthen the beam as regards failure in diagonal tension. Remembering that for a given percentage of reinforcement, the horizontal tension in the concrete depends entirely upon the bending moment, we may say that the strength of a beam as regards diagonal tension failure depends upon the relation between shear and bending moment and upon the amount of reinforcement; shear, however, is the chief factor. From the preceding considerations the student should now see clearly that the character of the loading influences the strength of a beam as regards diagonal tension, the amount of reinforce- ment remaining the same. For example. Fig. 33 represents the variation in moment and shear in a beam with a concentrated load at the center; Fig. 34 represents the' variation of these functions in a beam loaded at the third points; while Fig. 35 shows similar curves for a uniformly loaded beam. In the first Diagonal -tension cracKs UKely to occur -Very little tension in the concrete here (if any) on account of concrete cracking across the tension face Fig. 32. 68 REINFORCED CONCRETE CONSTRUCTION and second cases maximum shear occurs where maximum mo- ment exists, while in the latter case it occurs at the point of zero moment. Conditions are thus seen to be somewhat more favor- able in the continuously loaded beam. The next question to be decided is this : How may we prevent diagonal tensile stresses which are excessive? Obviously, they will be reduced by keeping the horizontal tension small through the use of considerable horizontal steel at points of heavy shear, by avoiding heavy shearing stresses, and by providing some type Fig. 33. of web reinforcement. The different methods of reinforcing for diagonal tension failure will now be taken up. From the pre- ceding discussion it is sufficient to bear in mind that the shear is the principal factor to be considered in the matter of diagonal tension and is a convenient measure of such stress. Also, that the shearing strength of the steel cannot be considered to aid the strength of a beam with respect to diagonal tension. 37. Methods of Web Reinforcement. — There are in use many methods of placing steel in the web of a reinforced concrete beam in order to reinforce it against diagonal tension failure. The various methods may, for convenience, be divided into three RECTANGULAR BEAMS 69 groups: (1) reinforcing metal placed at an inclination; (2) rein- forcing metal placed vertically; (3) miscellaneous methods. i i I Fi( J. 34. y//////////////// V/////////////////. > < iiiiiiiT^ Fig. 35. Referring to Fig. 32, it is evident that the ideal web reinforce- ment for uniform loading would be a system of rods arranged 70 REINFORCED CONCRETE CONSTRUCTION approximately as shown in Fig. 36. In any given case the exact arrangement must depend upon the nature of the loading, concentrated loads tending to extend the region of large shear to greater distances from the supports. Some horizontal rods should be carried to the end of the beam in order to keep the tension in the concrete low and reduce the tendency to the forma- tion of inclined cracks. Also, if some of the horizontal rods are bent up to take the place of inclined stirrups, the bends should be made somewhat beyond the theoretical points required for bending moment, so that the actual working stresses in the horizontal steel near the end of the beam will be low. The method of reinforcement indicated in Fig. 36 cannot, how- ever, be conveniently used in practice. Often the horizontal Fig. 36. rods are too few in number to bend up at the required number of points for thorough web reinforcement, and besides it is not convenient to handle rods with various inclinations at their ends. The common practice is to use bent rods (all bent at the same angle) combined with vertical stirrups. It should be clear that rods bent at a moderate angle are well suited for sections toward the center of beam, and vertical stirrups for sections near the end where the steel must be spaced closer together and at greater inclinations. Sometimes separate inclined reinforcement ( > 1 1 1 1 1^ 1 1 1 1 1 1 A- 1 i 1 ! > > 1 Fig. 43. appearance, final failure occurring at crack No. 4 and being due to inadequate web reinforcement. The stirrups were stressed beyond their yield point. It appears from tests of beams in which bent rods were em- ployed with a good anchorage at their ends, that considerable arch action is developed, and that the anchorage is quite advantageous in increasing web resistance. This form of construction is also found to be an insurance against failure at low loads through de- fective concrete or insufficient bond. The results of experiments show that the ultimate compressive strength of concrete in a beam is at least equal to its crushing strength as determined by tests on cubes hardened under similar conditions; also, that the yield-point of the steel should be regarded as ultimate strength as far as reinforced beams are con- cerned. When the steel reaches its yield point, the beam de- RECTANGULAR BEAMS 79 fleets, and failure soon occurs by the crushing of the concrete. The ultimate shearing strength of a beam having no web reinforcement may be taken at 100 to 150 lb. per square inch, for a 1:2:4 concrete, calculated as average shearing stress on the cross-section. The stresses here considered relate to shearing stresses involving large diagonal tensile stresses. Where such tensile stresses are not developed^^ to any extent, as in punching shear, a value of at least one-half the compressive strength (as previously mentioned) may be employed. However, it is almost impossible in practice to avoid altogether such tensile stresses, and it is not advisable to raise the first value given any consider- able amount. The student must continually bear in mind that the kind of failure denoted as a shear failure is so called for con- venience; they are diagonal tension failures brought about by large shearing stresses and hence may be measured by the shear- ing forces present. The average shearing stress on a vertical section at failure is the value always given. While the maximum shearing stress is 12 to 15 per cent greater than this, the average stress is practically as good a standard of measure and is much more readily calculated. The following is taken from "Principles of Reinforced Con- crete Construction" by Turneaure and Maurer: ''As already stated the amount of horizontal steel has a direct bearing on shear failures for the reason that large areas of steel with low unit stresses permit less extension of the concrete than small areas with high working stresses. This effect is shown in a marked manner in a series of tests made at The University of Wisconsin on small mortar beams of 1:3 mixture. The beams were 3 in. X4 1/2 in. in cross-section and 4 ft. span length. Loads were applied at two .points a varying distance apart. Only straight reinforcement was used, amounting to 1.41 per cent. The tensile strength of the material was high, being 490 lb. per square inch. The results were as follows: Distance apart of loads Average shearing stress Inches Center load Lb. per sq. in. 12 24 32 36 40 44 177 200 220 316 512 850 1036 80 REINFORCED CONCRETE CONSTRUCTION RECTANGULAR BEAMS 82 REINFORCED CONCRETE CONSTRUCTION 84 REINFORCED CONCRETE CONSTRUCTION The increase in strength as the loads approach the supports must be due largely to the decrease in moment stress and consequent distortion which is essentially what occurs when large areas of steel and low working stresses are used." Tests on beams with web reinforcement show that the ultimate average shearing strength may reach 300 to 400 lb. per square inch. The latter figure may, from our present knowledge, be taken as about the maximum value with ordinary, closely spaced, web reinforcement. Diagonal tension failure in a beam occurs suddenly. It is the failure to be most feared and therefore should be most care- fully guarded against. 40. Working Stresses. — The following working stresses were recommended by the Joint Committee on Concrete and Rein- forced Concrete in its first Progress Report presented early in 1909. This Joint Committee is composed of members selected from the American Society of Civil Engineers, the American Society for Testing Materials, the American Railway Engineering and Maintenance of Way Association, and the Association of American Portland Cement Manufacturers, and therefore represents the highest authority in the United States. The report on working stresses will be given in full.^ The student need not concern himself with the stresses in columns for the present. " General Assumptions. — The following working stresses are recommended for static loads. Proper allowances for vibration and impact are to be added to live loads where necessary to produce an equivalent static load before apply- ing the unit stresses in proportioning parts, "In selecting the permissible working stress to be allowed on concrete, we should be guided by the working stresses usually allowed for other materials of construction, so that all structures of the same class but composed of different materials may have approximately the same degree of safety. ''The stresses for concrete are proposed for concrete com- posed of one part Portland cement and six parts aggregate, capable of developing an average compressive strength of 2000 lb. per square inch at 28 days when tested in cylinders 1 The form given corresponds essentially with the 1909 Report of the Reinforced Concrete Committee of the American Society of Civil Engineers. RECTANGULAR BEAMS 85 8 in. in diameter and 16 in. long, under laboratory condi- tions of manufacture and storage, using the same con- ■ sistency as is used in the field. In considering the factors recommended with relation to this strength, it is to be borne in mind that the strength at 28 days is by no means the ultimate which will be developed at a longer period, and therefore, they do not correspond with the real factors of safety. On concrete in which the material of the aggregate is inferior, all stresses should be proportionally reduced, and similar reductions should be made when leaner mixtures are to be employed. On the other hand, if, with the be^t quality of aggregates, the richness is increased, an increase may be made in all working stresses proportional to the increase in compressive strength at 28 days, but this increase shall not exceed 25 per cent. "Bearing.^ — When compression is applied to a surface of concrete larger than the loaded area, a stress of 32.5 per cent of the compressive strength at 28 days, or 650 lb. per square inch on the above-described concrete may be allowed. This pressure is probably unnecessarily low when the ratio of the stressed area to the whole area of the concrete is much below unity, but is recommended for general use rather than a variable unit based upon this ratio. " Axial Compression. — (a) For concentric compression on a plain concrete column or pier, the length of which does not exceed 12 diameters, 22.5 per cent of the compressive strength at 28 days, or 450 lb. per square inch on 2000 lb. concrete, may be allowed. " (b) Columns with longitudinal reinforcement only, the same unit stress as recommended for (a) . (It is recommended that the ratio of the unsupported length of a column to its least width should be limited to 15 in any type of reinforced concrete column.) " (c) Columns with reinforcement of bands or hoops, ^ stresses 20 per cent higher than given for (b) , or 540 lb. per square inch on 2000 lb. concrete. " (d) Columns reinforced with not less than 1 per cent ' For beams built into pockets in concrete walls, the lower compressive stress of 450 lb. per square inch should not be exceeded. 2 Where bands or hoops are used, the total amount of such reinforcement shall not be less than 1 per cent of the volume of the column enclosed. The clear spacing of such bauds or hoops shall not be greater than one-fourth the diameter of the enclosed column. REINFORCED CONCRETE CONSTRUCTION and not more than 4 per cent of longitudinal bars and with bands or hoops, stresses 45 per cent higher than given for (b), or 650 lb. per square inch on 2000 lb. concrete. " (e) Columns reinforced with structural steel column units which thoroughly encase the concrete core, stresses 45 per cent higher than given for (b), or 650 lb. per square inch on 2000 lb. concrete. ''Compression on Extreme Fiber. — The extreme fiber stress of a beam, calculated on the assumption of a constant modulus of elasticity for concrete under working stresses, may be allowed to reach 32.5 per cent of the compressive strength at 28 days, or 650 lb. per square inch for 2000 lb. concrete. Adjacent to the support of continuous beams, stresses 15 per cent higher may be used. "Shear and Diagonal Tension. — Where pure shearing stress occurs, that is, uncombined with compression normal to the shearing surface, and with all tension normal to the shearing plane provided for by reinforcement, a shearing stress of 6 per cent of the compressive strength at 28 days, or 120 lb. per square inch on 2000 lb, concrete, may be allowed. Where the shear is combined with an equal compression, as on a section of a column at 45 degrees with the axis, the stress may equal one-half the compressive stress allowed. For ratios of compressive stress to shear intermediate between 0 and 1, proportionate shearing stresses shall be used. " In calculations on beams in which diagonal tension is considered to be taken by the concrete, the vertical shearing stresses should not exceed 2 per cent of the compressive strength at 28 days, or 40 lb. per square inch for 2000 lb. concrete. "Bond. — The bonding stress between concrete and plain reinforcing bars may be assumed at 4 per cent of the com- pressive strength at 28 days, or 80 lb. per square inch for 2000 lb. concrete; in the case of drawn wire, 2 per cent or 40 lb. on 2000 lb. concrete. " Reinforcement. — The tensile stress in steel should not ex- ceed 16,000 lb. per square inch. The compressive stress in reinforcing steel should not exceed 16,000 lb. per square inch, or 15 times the working compressive stress in the concrete. "Modulus of Elasticity. — The value of the modulus of RECTANGULAR BEAMS 87 elasticity of concrete has a wide range, depending upon the materials used, the age, the range of stresses between which it is considered, as well as other conditions. It is recom- mended that in all computations it be assumed as one- fifteenth that of steel, as, while not rigorously accurate, this assumption will give safe results." The yield-point of ordinary mild steel purchased in the open market cannot safely be fixed at a higher value than from 30,000 to 32,000 lb. per square inch, although frequently, and in fact in the majority of cases, a value of at least 35,000 to 40,000 lb. per square inch will be found. Since the yield-point of steel is regarded as its ultimate strength when used in reinforced concrete beams, the maximum working stress of 16,000 lb. per square inch recommended by the Joint Committee, gives, it would seem, a factor of safety of at least about 2 with respect to a steel-tension failure. In well-designed beams, however, where the concrete used is of high grade, the steel stress at failure will considerably exceed its elastic limit, the high crushing strength of the concrete enabling the steel to elongate very considerably before final failure occurs through the crushing of the concrete. For the above reason, the working stress of 16,000 will be found to give a much greater factor of safety than 2— generally somewhere between 2 1/2 and 3. The elastic limit of good concrete is between one-half and two- thirds its ultimate strength. Considering 2000 lb. per square inch as the ultimate strength in concrete, the elastic limit will be about 1000 to 1300 lb. per square inch. The working stress of 650 lb. per square inch gives the beam, then, a factor of safety as regards elastic limit of concrete of about 2, and as regards the ultimate strength of concrete, from 4 to 5 (this value is greater • than ' since above the elastic limit the stress in concrete does not vary even approximately in the same ratio as the increase in the loading) . Thus, the elastic limit of a reinforced concrete beam is deter- mined by the concrete (minimum factor of safety about 2) and its point of failure by the steel (minimum factor of safety between 2 1/2 and 3), which may be regarded as satisfactory conditions. The greater uniformity and reliability of the steel, as compared to the concrete, should be noted in this connection. 7 88 REINFORCED CONCRETE CONSTRUCTION It must be plain to the student that with the values recom- mended by the Joint Committee for the compression in concrete and the tension in steel, that a beam will reach its elastic limit (determined by the concrete) before it fails (determined by the steel). Of course, the preceding statement applies only when the beam is adequately reinforced against diagonal tension and bond stress. In choosing the above factors, the margin of safety between the elastic limit and ultimate strength has received consideration. The student should note that a working tensile stress of 16,000 lb. per square inch is recommended by the Joint Committee for all grades of steel. High working stresses in the steel involve large distortions in the concrete, not only at the center of beam but also diagonally near the end of beam. Low unit stresses in the steel are greatly to be preferred on this account. It can also be shown that very little is to be gained in economy by using high stresses such as is often done with high elastic limit steel. In determining the relative working stresses in steel and concrete some attention should be given to the question of repeated loads. If the live load is a large percentage of the total load and subject to frequent repetitions, a relatively low working stress in the con- crete may well be employed in order to maintain elastic condi- tions. As regards the steel, more perfect elasticity exists up to a definite point, and hence repetition of load need not be con- sidered in the selection of its working stresses. Low unit stresses in the. steel, are to be preferred, in order to prevent excessive deformations and reduce the liability of diagonal tension cracks. The working stress of 16,000 should be the maximum for medium steel. The "General Assumptions" of the Joint Committee's report should be studied until the student feels that he can use his judgment correctly in the selection of working stresses for any given concrete and steel. The table of crushing strengths of concrete given in Art. 10 should be employed in this connection. A 1:2:4 concrete is considered to have an average ultimate strength of 2000 lb. per square inch, and the strengths of other concretes should be proportional to this. If high carbon steel is used a stress of 20,000 lb. is frequently permitted. Age of concrete is another point frequently overlooked when choosing working stresses. The stresses recommended by the Joint Committee refer to the strength at the end of one month, RECTANGULAR BEAMS 89 so that the factors of safety already mentioned with regard to concrete are by no means the ultimate. The standard compression specimen adopted by the Joint Committee is a cylinder 8 in. in diameter by 16 in. long. Most of the important tests of concrete up to the present time have been made on either 6 by 6 in. or 12 by 12 in. cubes. It is often important to compare the strength of cylinders and cubes and this can be done by means of the formula given in Art. 10. Problems 22. A simply-supported reinforced concrete beam having 6 = 8 in. and d = 12 in. has a span of 15 ft. and sustains a uniform load (live plus dead) of 350 lb. per foot. The beam contains four 1/2-in. square rods straight throughout and properly spaced. What is the maximum bond stress on these rods in pounds per square inch? 23. Determine whether or not the beam of Problem 22 is of equal strength * in tension and compression, assuming the allowable working stresses to be /« = 16,000 lb. per square inch and /c = 650 lb. per square inch. Take n = 15. 24. (a) Wha t would be the maximum bond stress in Problem 22 if two 3 /4-in. square rods were employed? (b) If one 1-in. square rod was used? (c) What fact is brought out by (a) and (b)? 25. What is the maximum shearing stress in the beam of Problem 22? Where does this maximum occur? 26. Is web reinforcement needed in the beam of Problem 22 according to the recommendations of the Joint Committee? 27. If two of the rods in Problem 22 were bent up near the end of beam, approximately what would be the maximum bond stress along the two remaining rods? 28. If you were designing a beam under the conditions of Problem 22 and the depth (d) was limited to 12 in., what breadth of beam and amount of steel would you consider satisfactory? 29. A reinforced concrete structure is subjected to a constant repetition of a live load which is large in comparison with the dead load. Explain what you know pertaining to the proper working stresses for the same. 30. A 1 : 2 1/2 : 5 concrete (see table on page 17) is to be used under average conditions in a reinforced concrete beam. Medium steel. Determine the working stresses to use. 31. A 12-in. by 12-in. cube of a concrete, which is to be used in a reinforced beam, has an ultimate strength of 3000 lb. per square inch. High car- bon steel. Average conditions. What working stresses would you employ in design? 32. A simply-supported beam of 10-ft. span and 30-in. depth is loaded with a heavy concentrated load at a point one-fourth the span length from the left support. Considering horizontal steel, show the general directions of the diagonal tensile stresses throughout the beam. 90 REINFORCED CONCRETE CONSTRUCTION 41. Vertical and Inclined Reinforcement. — Thus far in the course the student has been brought to realize the effectiveness of inclined and vertical bars in preventing diagonal tensile cracks in reinforced concrete beams. This type of failure we have seen is the one to be most feared. The inclined reinforcement may- be separate members firmly connected with the horizontal reinforcement to prevent slipping, or some of the horizontal bars may be bent up near the ends of the beam where they are not needed to resist bending. The vertical reinforcement may be used separately or in combination with inclined reinforcement, depending upon the preference of the designer and upon the amount of diagonal tension to be provided for. When vertical stirrups are used, they should be looped around the horizontal bars so as to be firmly anchored at their lower ends where the stress is a maximum. The value of a stirrup unless looped or hooked at the top is limited by its strength of bond and, as its length is not great, this point may need consideration. The proportioning of web reinforcement cannot be done with any degree of exactness since very little experimental work has been performed along this line. However, rough determinations of what is required may be obtained on rational grounds. The only information given us by tests, is the value of the shearing stress which measures diagonal tension failure — (1) for beams with horizontal bars only, and (2) for beams having an effective system of web reinforcement. Also, tests on beams, with and without web reinforcement, show that when reinforcement is provided for diagonal tension, the concrete may be assumed to carry its full value of the shear and the steel the remainder. This action of the concrete and web reinforcement is found to exist at ultimate loads — namely, when the beam has failed by the slipping of the bars. Since this is found true at ultimate loads, it undoubtedly would be even more certain at working loads when the concrete at the most is only slightly cracked. The Joint Committee limits the shearing unit stress in a beam with an effective web reinforcement to 120 lb. per square inch for a 2000 lb. concrete — namely, three times the allowable shearing stress in a 2000 lb. concrete without reinforcement. In the dis- cussion which follows, we shall assume the concrete to take one- third of the total shear and the stirrups the remaining two-thirds. The point in the beam beyond which web reinforcement is unnecessary will, however, be determined by the maximum allowable shear in the concrete. RECTANGULAR BEAMS 91 When the unit shear, which equals ^> is found to only slightly exceed 40 lb. per square inch, it would seem that the beam de- signed according to the above assumption would be unnecessarily strong as regards web reinforcement. This is true, but on ac- count of the small amount of reinforcement which is usually needed for such cases and the importance of such reinforcement 2 in reinforced concrete beams, the value ^ is proposed for all ordinary cases. The student should always remember this element of safety when using the following formulas in design. 2 If desired, the coefficient ^ in the formulas may be changed o as required by the problem under consideration. Using Taylor and Thompson's method of analysis/ let V — total shear. s = horizontal spacing of stirrups. tts = area of section of stirrup. (In a U-shaped stirrup, is the sum of the areas of the two legs.) fs = unit stress in steel. The maximum intensity of vertical shear -y at a vertical section A A, Fig. 44, is The maximum intensity of horizontal shear _J1 at a section A A is equal to the maximum intensity of vertical shear v at the same section. Multiplying this shear by b, the breadth of the beam, gives the maximum horizontal shear per unit of length of the beam = jd Consider now a horizontal plane of length s at, or below, the neutral plane. The mtensity ' From Taylor and Thompson's "Concrete, Plain and Reinforced," 2nd edition, page 448. 92 REINFORCED CONCRETE CONSTRUCTION of the horizontal unit shear on this plane varies proportionally V V with the total shear, and is equal to -~ at section A A and to— v ]d ]d at section CC. Hence, the total amount of shear on this plane, . . V -\-V considering uniform variation of shear, is equal to ^ (s), 2ja V or, when V at BB is the average shear, to -r^s. This is the ver- jd tical component of the diagonal tensile stress in a length of beam s. When a stirrup is placed at B, Fig. 44, and the distance between stirrups is s, then the vertical component of the diagonal tension 2 7s to be taken by this stirrup is j^- This stress causes tension in the stirrup, the corresponding horizontal component of the diagonal tension being carried by the horizontal rods. The stress in a vertical stirrup is asfs, hence the cross-sectional area of vertical steel required in a length of s may be taken approximately as 2 7s . , . as=o*i^ (Vertical stirrups) (1) For stirrups or bars incHned at 45 degrees, the lines on a beam representing the direction in which the diagonal tensile cracks are likely to occur, are crossed more times per unit of length for a given horizontal spacing than would be the case if vertical stirrups were employed; that is, a given amount of incHned steel is much more effective in taking diagonal tension than the same amount of vertical steel. It may be assumed, then, that the stress in the inclined bars is approximately .^^"{t^ and the re- sm 45° quired area of steel, assuming the steel to take two-thirds of the shear, is cis ^'fjj^^ (Inclined bars 45 degree angle) (2) For other angles of inclination K, it may be assumed as approxi- mately correct to use the formula 2 sin K (Vs) , , a* ~3' ■ — (Inclined bars any angle) (3) The recommendations of the Joint Committee for the allowable shearing stresses in beams, both with and without effective web RECTANGULAR BEAMS 93 reinforcement, apply to the maximum unit shear at any point V of a vertical section, which we have previously found to be This maximum extends from the horizontal steel up to the neutral axis. The stress in the stirrups will consequently be the greatest along a plane located between the neutral plane and the horizontal rods. Assuming that the distance between centers of compression and tension {jd) is approximately 7/8d, it is clear that stirrups are V . required with horizontal bars, only when ^ is greater than 7/8(40) V = 35 lb. per square inch. Also, ^ should not be greater than 7/8(120) =105 lb. per square inch with the most effective web V reinforcement. ^ has been previously denoted by Vo, and we shall represent it in this manner hereafter. The manner of providing for diagonal tension in beams cannot be impressed too strongly upon the student. First of all, notice the most unfavorable part of the given beam as regards diagonal tensile stresses. This will be at points of excessive shear com- bined with considerable bending moment. Also, be careful to extend sufficient horizontal steel to the ends of the beam to provide for bending with low unit stresses in the steel. A great many times it will be impracticable to provide as much reinforce- ment as desired by means of bent-up rods, and some vertical stirrups will be needed. In small beams, vertical stirrups only are often employed throughout, but in large beams under heavy shearing stresses, both should be used. In detail, stirrups may be made in various forms, as indicated in Fig. 45. Special attention is called to the many excellent features of the continu- ous stirrup. Woven wire bent around the rods is a satisfactory and very effective reinforcement. The Joint Committee recom- mends that the longitudinal spacing of stirrups or bent rods should not exceed three-fourths the depth of the beam. 42. Vertical Stirrups. — We have seen that the required area of cross-section of a vertical stirrup may be determined by the formula _2 Vs_ 94 REINFORCED CONCRETE CONSTRUCTION With a given stirrup, this formula may be solved to give the spacing required, or 3 asfsl'd The value of V should be taken at the section \yhere the spacing is desired. Tests-have shown that little or no value is derived from stirrups spaced a distance apart about equal to d. Thus, the spacing of IN A -••••Continuous stirrup Fig. 45. stirrups must not exceed the depth of the beam, and, as we have seen, the practical limit suggested by the Joint Committee is three-fourths the depth. Many constructors advise the insertion of occasional stirrups throughout the entire length of the beam, especially T-shaped beams, even if they are not theoretically necessary. Where only a small amount of web reinforcement is needed, stirrups are occasionally spaced uniformly for convenience. When this is done, only the minimum value of s needs to be figured, and this can be obtained by substituting for V in the above equation, the total shear at the end of beam. RECTANGULAR BEAMS 95 The distance from the support to the point where no stirrups are required is helpful in design. This distance for uniform loading may easily be expressed by means of a formiila. Let V = unit shear. = unit, working shear. (F' = corresponding total shear). = distance in feet from left support to point beyond which stirrups are unnecessary. I = span of beam in feet. w = uniform load in pounds per foot. Then V Stirrups become unnecessary at section where v = v'. Thus, at the required section But -2-— w^i Substituting and solving for x^, we have I v'bjd Suppose a 10-ft. beam (& = 10 in. and d = 20 in.) is uniformly loaded with 2900 lb. per foot, and assume v' = AO lb. per square inch according to recommendation of Joint Committee for 2000 lb. concrete. Also assume jd=^7/8d. Then, 10 (40)(10)(17.5) 2 2900 = 2.59 ft. The following graphical method may be employed to deter- mine the spacing of stirrups in large and important beams: Lay off one-half the span to any convenient scale as shown in Fig. 46. Compute the values of s at a number of points (1, 2 and 3, for example) and lay off these values on the perpendicu- lars erected at the respective points to the same scale as the span. Draw a smooth curve through the upper ends of the perpendicu- lars. From point a on the curve directly above the point where the first stirrup will be placed, draw a line at 45 degrees to inter- sect with the horizontal line and erect at the point of intersection 96 REINFORCED CONCRETE CONSTRUCTION B a perpendicular to cut the curve in point b. A line drawn from b at 45 degrees will intersect the horizontal in point C, where the above process is repeated. The points A, B, C, D, E, thus ^obtained, are the points at which stirrups are required. In the case of uniformly loaded beams, it is only necessary to compute the minimum spacing of stirrups — that is, at the support. The spacing at two other points may be obtained from the fact that the spacing for a distance from the support ^ is a 1 1 J, Fig. 47. four-thirds the minimum, and for a distance \ is twice the 4 minimum. At the center, the spacing is infinity. The diagram of maximum shear intensity along a uniformly loaded beam is shown in Fig. 47. If the shear at the end of beam RECTANGULAR BEAMS 97 is the maximum allowable, the triangle ahc represents the total stress to be taken by the stirrups at each end of beam. This may be proved as follows: Formula (1), Art. 41, shows the total stress for a distance s along the beam (where V is the average 2 Vs shear) to be agfs = q- "vT* in the case at hand s=Xi and, if v is con- 6 ja ' 2Y sidered the unit shear at the end of beam = r^^, the total stress for the distance is ^- ^ = |- vbx^, which is the area of triangle ahc. The ordinate ah represents two-thirds of the shear at the support per 1-in. length of beam. Some attention must be paid to the diameter of stirrup which it will be possible to employ in any given case. ^ Of course, the diameter should not be so small that the stirrups will be placed too close together for convenience in construction, nor yet so far apart that the. limiting value 3 /Ad is exceeded. But, in addition to such consideration, the bond strength of the stirrup must be carefully investigated since the danger of slipping determines the maximum diameter which may be employed. Let us derive a formula for the maximum diameter to be used in any given case. We shall consider straight stirrups only. Let i — diameter of stirrup. as = area of stirrup. o = circumference of stirrup. u = allowable bond stress per unit of surface of bar. The distribution of bond stresses developed on the surface of the stirrups is indeterminate. Evidently it must not be expected that tension will be transferred to the concrete until the com- pression area of the beam is reached, or until a point but little below is reached. Experiments show that it is safe to assume the grip of a stirrup to be 0.6 the depth of beam. /stts =0.6 dou or "^ = 0.6^d 0 Js But, for round or square stirrups, 0 m 98 REINFORCED CONCRETE CONSTRUCTION Then U i = ) d u The table below gives the values of 2.4 7r for different working Js values of tension and bond, as developed by Messrs. Taylor and Thompson in "Concrete, Plain and Reinforced."^ Allowable unit bond Vertical bars stress in pounds Allowable unit tension in stirrups in pounds per square inch (m) per square inch (/») 12,000 14,000 15,000 10,000 20,000 80 0.016 0.014 0.013 0.012 0.010 100 0.020 0.017 0.016 0.015 0.012 120 0.024 0.020 0.019 0.018 0.014 150 0.030 0.026 0.024 0.022 0.018 Suppose that for a given vertical stirrup the allowable fs = 16,000 lb. per square inch and the allowable u = SO lb. per square inch. Also, let the depth of the beam in question be 20 in. (d). Then i = 0.24 in. — practically 1/4 in. For deformed bars the bond may be increased to 100, or even to 150 lb. per square inch, varying with the character of the bar. Using 150 for u and 16,000 for /s, a beam 20 in. deep to center of steel, making no allowance for the value of a bent end (called prong or hook) , would require stirrups not to exceed practically 1/2 in. in diameter. Deformed bars are therefore useful for stirrups to permit larger diameters, although the total quantity of stirrup steel required throughout a beam is not changed except where the spacing would tend to exceed the allowable. In other words, with de- formed bars the stirrups do not need to be spaced as closely as with plain bars. Recent tests show that either a right angle or a semi-circular bend of 5 diameters is sufficient to stress the steel to its elastic limit, provided the hook is well embedded in the concrete so that it cannot kick out. To rely upon such increase in strength over the straight stirrup, an embedment is required in all directions equal to 8 diameters of the bar. 1 From Taylor and Thompson's "Concrete, Plain and Reinforced," 2nd edition, page 454, Copyright, 1905, 1909, by Frederick W. Taylor. RECTANGULAR BEAMS 99 Illustrative Problem. — A concrete beam is 9 in. X 16 in. in cross-section and the tension reinforcement is 2 in. above the lower face of the beam. Span of the beam is 8.5 ft. Uniform load of 1800 lb. per foot. If necessary, the web is to be reinforced against diagonal tension using vertical stirrups. The working stresses recommended by the Joint Committee for a 2000 lb. con- crete will be used. Z = 765 0_^ square inch. bd (9) (14) ^ The allowable average shear Vo is 35 lb. per square inch, hence stirrups are necessary. The diameter of a stirrup without any prong or hook should not exceed (see table) i = (0.012) (14) = 0.17 in. We will use stirrups bent at the upper end and because of this bending 1/4- in round bars may be considered secure against slipping. Stirrups are unnecessary at a distance from support (assuming j = 7/8) equal to _ 8^ _ (40) (9) (7/8) (14) ^ 1 go f^. 2 1800 The minimum spacing of stirrups (U-shape) will occur at the supports and will be equal to _3 (2) (0.049) (16,000) (7/8) (14) ^^^g.^ ^~2" " 7650 Let X = distance from the left support. Then for x = g- = 1.1 ft., the spacing is (4/3) (3.75) =5.0 in.; and for x = ^ = 2.1 ft., the spacing is (2) (3.75) = 7.5 in. A smooth curve drawn through the points determines the spacing at any part of the beam. In monolithic construction, the first stirrup may be placed one-half the minimum spacing from the edge of the support, and the last stirrup should not be farther distant from the Hmiting point, where stirrups are unnecessary, than half of the distance between the last two stirrups. In beams simply supported, the first stirrup should be placed one-half the minimum distance from the center of support. 43. Horizontal Bars Bent Up for Web Reinforcement— The ends of the horizontal bars in a reinforced concrete beam may often be bent up to assist in providing for the diagonal tension. In some cases these bent rods may take all such stresses, and vertical stirrups then are not theoretically needed— though they are desirable, as shown by tests. Plain rods bent up to provide web reinforcement often lack sufficient bond strength to render them fully effective. Where bent up at a considerable angle they should be turned again horizontally and extend some distance along the upper part of 100 REINFORCED CONCRETE CONSTRUCTION the beam, as shown in Fig. 48. In heavy construction the ends of all bars should be bent into a hook. The most convenient method of using reinforcement is to bend up two rods at a time and make all the bars inclined at an angle of 45 degrees with the horizontal. The bars bent should theoretically be such as to keep the center of gravity of the beam cross-section in the line drawn vertically through the center of the section. An excep- tion occurs to the bending of two rods at a time, in the case of an odd number of horizontal rods. Here, one of the bends may consist of either one or three rods. If bent rods are not required to provide for diagonal tension, then the horizontal rods may be dispensed with at the points, Fig. 48. beyond which they are not needed to provide for tension due to bending. This method of stopping off the horizontal rods is not desirable, however, as the bond in the concrete near the middle of the beam is not as good as would be the case near the end where the moments are smaller. Also, when a bar is discon- tinued, the stress in those which remain is immediately increased tending still further to impair the bond between the steel and the concrete. This is true whether or not a hook is employed on the discontinued rods. With bent up rods a more ideal condition exists. The horizontal components of the upturned bars act with the bars unbent in taking the tension due to bending, and RECTANGULAR BEAMS 101 so the tension in the horizontal rods decreases gradually toward the end of beam as it should. At the bend in a horizontal rod, the stress is transferred gradually to the concrete as compression, and, if the bend is too abrupt, the unit stress in the concrete in this vicinity may be- come excessive. Theory tends to indicate that a radius of bend equal to 12 diameters is satisfactory. Cracks following the line of the rod are seldom seen in tests of beams having even sharper bends in the reinforcement. The distance from the support to the point where web rein- forcement is not needed is determined in the same manner as for vertical stirrups. The bent rods, if of the same diameter, should be so arranged that each rod will take an equal part of the diagonal tension — that is, if they can be bent in this way and still provide satisfactorily for the horizontal tension. If the rods cannot be bent at the desired points, vertical stirrups must be used to provide for the diagonal tension either toward the center, or toward the end of the beam. First of all, we shall assume that the rods can be bent up con- veniently to provide for all the diagonal tension. Consider uniform loading. The maximum shear V should be calculated at the support, point D, Fig. 48. One-third of V will be taken by the concrete and two-thirds by the bent bars. Point A should also be determined — namely, the point where the web reinforce- ment is not needed. From this point to the left support, the shear to be taken by the bent bars increases from zero to its maximum value of 2/37 at the support, and may be represented by the triangle ABC. To construct this triangle draw a line AB from point A at 45 degrees with the horizontal, and from point D draw a line DB perpendicular to the line AB. Then, consider the maximum shear per inch length of beam to be 213V represented by BC; in other words, BC= 2/ 3vh^-~- Now, suppose we intend to bend two rods at a time and to bend in all 8 rods, all of the same diameter. Then each of them will take an equal part of the diagonal tension. Divide the area of the tri- angle into four equal parts, find centers of gravity of each part, and from these centers of gravity draw lines to represent the location of points to bend up the bars in the beam. The method of division of the triangle into an equal number of parts is clearly shown in the drawing, where the line AB is divided into equal 102 REINFORCED CONCRETE CONSTRUCTION parts and dotted arcs of circles are drawn with centers at The reason for projecting the neutral axis AD upon a plane making 45 degrees with the horizontal may need some explana- tion. Our formula for the required total area of steel in each bend is _2 0.77s fsjd Hence the sum of the stresses in the inclined bars at each bend is given by the following formula . 2 0.7Vs But any ordinate as BC=^l l^^- Also, xy = l. and 5a: = 0.7s. Then, the 2 F(at No. 1 bars)0.7s area BC yx = - — (the bars being practically half way between x and B), and represents the sum of the stresses in the No. 1 bars. Similarly, the area xyy'x' represents the sum of the stresses in the No. 2 bars. In practice the line AC need not be drawn. For example, no matter what angle the line AC makes with the line AB, the points 1, 2, 3, and 4 will be the same; the points 1, 2, and 3 will remain practically half way between B and x, x and x' , and x' and x", respectively, while A4l will remain as 2/3 Ax ". The bond strength in these inclined bars must now be in- vestigated. This strength should be provided in the upper portion of the beam. As with vertical stirrups, we shall arbitra- rily assume that no stress is transmitted from the steel to the concrete below a point which is 0.6 d below the upper surface of the beam. We will assume that the stress in an inclined bar is its working stress. This gives the maximum condition. Using the notation of the preceding article and V for length, l'ou = asfs and for round or square bars, 4 or , / . I = ~ diameters. 4w 1 From Taylor and Thompson's "Concrete, Plain and Reinforced," 2nd edition, page 475. Copyright, 1905, 1909, by Frederick W. Taylor. RECTANGULAR BEAMS 103 The following table gives the values of ^ for different working values of tension and bond, as developed by Messrs. Taylor and Thompson in "Concrete, Plain and Reinforced."^ Allowable unit bond stress in pounds per square inch (u) Allowable unit tension in inclined bar, in pounds per square inch (fg) 12,000 14,000 15,000 16,000 20,000 80 37 44 47 50 62 100 30 35 38 40 50 120 25 29 31 33 41 150 20 23 25 27 33 The length of embedment may be obtained by multiplying the value selected from this table by the diameter of the bar. The bond of deformed bars may be figured the same as the bond of plain bars except using for their diameter, the diameter of a cylinder based on the longest projections — that is, of a cylinder which would be sheared out by the deformed bar. For smooth metal of the nature of tool steel not over 40 lb. per square inch should be permitted for allowable bond strength. Flat steel should be given a similarly low value per square inch. If the allowable /« = 16,000 lb. per square inch and the allow- able w = 80 lb. per square inch, then V (as shown at No. 4 bars in Fig. 48) should equal 50 diameters by the above table. Suppose a deformed bar is used with u =150 lb. per square inch. The length r should then equal 27 diameters, or 1 ft. 8 1 /4 in. for a 3/4 in. rod. Hooks should be provided at the ends of the bars to provide additional safety. 44. Vertical Stirrups and Bent Rods Combined. — Usually the diagonal tension in a simply supported beam of rectangular section can all be provided for by bent up rods and the theoretical combination of stirrups with bent up rods need not be con- sidered. This combination, however, is a common one in T-beam design and may just as well be treated at this time. Consider uniform loading and, in Fig. 49, let ABC be the diagonal tension triangle. Assume that four rods may be bent near the end of beam, but not so that they can take any diagonal 1 From Taylor and Thompson's "Concrete, Plain and Reinforced," 2nd edition, page 454. Copyright, 1905, 1909, by Frederick W. Taylor. 104 REINFORCED CONCRETE CONSTRUCTION tension near to the point A. It will be considered feasible that the bent rods should be placed to take as much diagonal tension as possible from D toward A. The area BCEF should be made equal to the allowable tensile stress in the two No. 1 bars; likewise area EFGH should represent the tensile strength of the two No. 2 bars. This would not be true, however, if the distance s should turn out to be greater than 3/4cZ. For such a case, the No. 2 bars should be placed the limiting distance from the No. 1 bars and the area toward the center of beam (representing the diagonal tension which the No. 2 bars may be assumed to take in that direction) should be made equal to rtGH. Stirrups may also be supplied to take the stress between e and A. The spacing near the point e would be obtained by the formula _3 Osfsjd V in which V is the total vertical shear at the point where the Fig, 49. stirrup is placed. It will be well on the safe side to place the first stirrup one-half the computed spacing to the right of point e. If possible, when bent up rods cannot take all the diagonal tension, it would be better design to bend up the rods as far from the end of beam as possible so that stirrups may be employed toward the end of beam where the diagonal tensile stresses are greatly inclined, rather than toward the center of beam as in Fig. 49. This method would be followed if the distance a were RECTANGULAR BEAMS 105 considerable. Even in the case assumed in Fig. 49, at least one stirrup should be employed in this distance, as shown. 45. Points to Bend Horizontal Reinforcement. — If some of the horizontal bars are bent up at a given point to provide for diagonal tension, those remaining should have sufficient sectional area to carry the tension beyond this point. '■-Parabola M 4. 4 2 T 2 i \ ^ 1 Fig. 50. In determining the horizontal length of the various bars neces- sary to resist the bending moment, the same method, as developed by Messrs. Taylor and Thompson in ''Concrete, Plain and Rein- forced,"^ may be used as in the design of plate girder flanges. Consider uniform loading. Let m = number of bars at the center of the beam. TO2 = number of bars to be bent. I = span of beam in feet. M = maximum moment -^7 in which o X2 - distance from support to point where bars may be bent up leaving sufficient steel to carry the tension. The ratio of stress at the center of the beam to that at the point under consideration equals the ratio of moments at these points. Thus, if the steel is stressed equally at both points, ilfa;,.-M=(m-m2)(area of one bar) :m(area of one bar). iSubstituting M=-^ andM.,= — i-- 1 From Taylor and Thompson's "Concrete, Plain and Reinforced," 2nd edition, page 458. Copyright, 1905, 1909, by Frederick W. Taylor. 106 REINFORCED CONCRETE CONSTRUCTION and solving for (see Fig. 50), we have wP For any bending moment M =— > (the meaning of this formula will be clear after studying Art. 54). If it is desired to bend up a number of rods two or more at a time, then should be determined for each bend. After this is done, the remaining horizontal bars should be secure against slipping. For concentrated and unsymmetrical loading, the maximum moments at various sections will need to be determined, in order Fig. 51. to ascertain the points where the horizontal bars may be bent up. From these maximum moments obtain the required area of horizontal rods at the different points (1, 2, 3, and 4, Fig. 51). Plot a curve to scale, as shown. Thus, ah represents the area required at the point a. On the center ordinate lay off the required areas of the rods, and draw horizontals as shown. The rods may be bent up where these horizontals cut the curve but it would be better, however, to carry them a short distance beyond the theoretical points. Problems (Working stresses recommended by the Joint Committee to be used throughout.) 33. A simply supported beam of 12 ft. span (c. to c. of supports) and 18 in. breadth is to carry a uniform load (Uve plus dead) of 6000 lb. per foot and rest upon concrete supports. Considering the reactions as uniformly RECTANGULAR BEAMS 107 distributed over the bearing surface, how far should the beam extend beyond the edge of the supports, taking the allowable crushing strength of the concrete at 450 lb. per square inch? 34. In the beam of Problem 33, what is the distance from the left reaction to the point beyond which stirrups are unnecessary, assuming the moment calculations to give 6 = 18 in. and d!=26 in. 35. (a) Could four 1-in. plain square bars be used in the beam of Problem 34? Give reasons, (b) If eight 3/4-in. square bars were used, how many could be bent up to take diagonal tension (allow about 30 per cent more bond stress on the horizontal rods than would be considered safe by formula if three rods can be bent up and at least two bends made at each end of beam)? 36. In the beam of the preceding problems, at what points may the horizontal rods be bent up if eight 3/4-in. square rods are used? 37. At what points in the beam of Problem 36 should the horizontal rods be bent to provide for the diagonal tension in the best possible manner, assuming the resultant reactions 3 in. from the edge of supports (ends of beam 6 in. from edge)? Submit sketch neatly drawn. 38. In Problem 36 if the horizontal bars were run straight to the end of beam what size and spacing of vertical stirrups would be required to provide thoroughly for diagonal tension? Employ double-looped, square stirrups. Graphical work to be submitted. 39. (a) What should be the grip of the inclined rods of Problem 37 consider- ing the rods stressed in tension to the allowable value of 16,000 lb. per square inch? (b) What should be the grip of these rods for the stress they are actually called upon to withstand? 46. Transverse Spacing of Reinforcement. — The amount of concrete between the horizontal bars in a beam should be suffi- cient to transmit to the upper part of the beam the stress which the bars give over to the concrete below them; in other words, the shearing stress along ah, Fig. 52, should equal the amount of stress transmitted by bond along bed. If bond and shearing strengths were equal, ah should equal hcd, and the clear space be- tween bars should be — diameters ..b 1.57 diameters. But shearing strength yig. 52. here employed is controlled by the diagonal 120 3 tension and is approximately = the bond stress, using the values recommended by the Joint Committee. Hence, ah should be (2/3) (1.57) diameters = 1.05 diameters. That is, the mini- mum net distance in the clear between rods is approximately 108 REINFORCED CONCRETE CONSTRUCTION equal to the diameter of the rod. There is likely to be more or less tension in the concrete surrounding the rods and, besides, since the concrete is not easily placed between the rods, it may have a lower strength in that vicinity. A clear spacing of 1 1/2 diameters is advisable unless it is determined by computation that the bond stress is very much lower than the bond strength allowed. In the above discussion plain rods only have been considered. Deformed bars, if stressed to their full bond value, should be spaced farther apart than plain bars. For a beam uniformly loaded, the bond stress near the center of the beam is low and the bars may be placed as closely together as the proper placing of the concrete between them will permit. Near the end of beam the bond stress should be calculated if it is desired to space the rods so as to obtain the minimum width of beam possible. In beams having the horizontal rods bent up, the bond stress at the bending points should be considered. This stress may be less than the maximum and, if so, the corre- sponding spacing may be made less than 11/2 diameters in the clear. With rods bent up, more liberal spacing can readily be made toward the end of beam. The Joint Committee recommends that "the lateral spacing of parallel bars should not be less than 2 1/2 diameters, center to center, nor should the distance from the side of the beam to the center of the nearest bar be less than 2 diameters." In order that concrete may be readily placed between the rods and also give sufficient concrete on the sides of the beam for fire protection, it is also advisable to require that the spacing of rods be not less than 1 in, in the clear (if the maximum size of aggregate does not exceed 1 in.) and that 1 in. in the clear be also considered the minimum distance of the rods from the sides of the beam. Thus, the least width of beam should be the greater of the two values determined from the following formulas: 6 = [2.5(n-l)+4]di h = ag(n-\-l) +ndi in which 6= least width of beam. c?i = thickness of the rods. w = maximum number of rods which occurs in a hori- zontal layer. aj, = maximum size of aggregate in inches. It should be clear that, for a 1-in. maximum aggregate, the width RECTANGULAR BEAMS 109 of beam for all rods greater than 5/8 in. in diameter will be governed by the first formula and for 5/8-in. rods and less, by the second formula. Where two layers of rods are used, there is less danger of vertical splitting if the rods are placed directly over each other, and with sufficient space simply to permit the mortar to run between them. The Joint Committee specifies a limiting clear space of 1/2 in. 47. Depth of Concrete Below Rods. — ^Prof. Charles L. Norton of the Insurance Engineering Experiment Station has made a careful study of the thickness of concrete which is essential to thoroughly protect embedded steel from the direct action of - flames, and recommends 2 in. for maximum conditions. An excessive thickness of concrete, however, adds to the danger of cracking, because the tension in the concrete increases with the depth below the steel and with but slight corresponding gain in strength to the beam. Also, flat slabs are found to be affected to a less depth than projecting members such as beams and columns. The Joint Committee suggests that "the metal in girders and columns be protected by a minimum of 2 in. of con- crete; that the metal in beams be protected by a minimum of 1 1/2 in. of concrete; and that the metal in floor slabs be pro- tected by a minimum of 1 in. of concrete." The following depths of concrete below the steel may be em- ployed under ordinary conditions, but wherever conditions are especially hazardous, the recommendations of the Joint Com- mittee should be followed: Slabs Depth to steel (d) Depth below center of steel 3i in. and under i in. Between 3^ in. and 4^ in 1 in. 4f in. and over 1^ in. Beams and Girders Depth to steel (d) Depth in the clear below steel 10 in. and under 1 in. Between 10 in. and 20 in IJ in. 20 in. and over 2 in. ■ 48. Ratio of Length to Depth of Beam for Equal Strength in Moment and Shear. — With given working stresses in concrete and steel, there is a definite ratio of length to depth of beam which will give equal strength in moment and shear. Let us first consider beams simply supported. 110 REINFORCED CONCRETE CONSTRUCTION In the case of a single concentrated load at the center of span, the shear V, due to a given load "PF, is 1/2W, and the moment M is 1/4TFL Hence, But from preceding formulae, we have V = v'bjd and Ms =pfsjhd^, in which v' = allowable shearing stress and fa = working stress in steel. Substituting, we have from which I ^ 2fap d v' For a uniformly distributed load, a similar process gives the ratio d v' For beams loaded with equal loads at the third points, d Taking for example, v'=40 lb. per square inch, /« = 16,000, /c = 650, n = 15, and, using an average value of 7/8 for /, we have the following ratios for ^• For concentrated load at center of span J= 6.16 For uniformly distributed load ^ = 12.32 For equal loads at the third points ^= 9.24 It should be clear that the strength of beams of greater relative length than obtained by the formulas will be determined by their moment of resistance, while that of shorter beams by their shearing resistance. In the case of continuous beams the above formulas will apply if I is taken as the length between points of inflection. It is often convenient to know the extreme limit in design. The Joint Committee recommends 120 lb. per square inch for the shearing strength of concrete when adequately reinforced against diagonal tension. This is a low figure but is adopted in order to prevent any likelihood of cracks opening up in the concrete. Suppose then, it is required to know the minimum value of J for a given RECTANGULAR BEAMS 111 beam, uniformly loaded. Our formula readily gives us the result, using the working stresses given above. Z ^ (4)(16,000)(0.0077) ^ d~ 120 • -^-^^ At the same time that the ratio of length to depth is being investigated for moment and shear, there are other conditions which must be considered. For instance, the ratio of length to breadth of beam should not exceed a value of about 25 if the beam is not supported laterally. The reason for this is found in the fact that the upper part of the beam is a column, and to prevent additional stress due to side bending the length should not exceed about 25 times the width. On the other hand, the best shaped beam is one in which h lies between l/2d and 3/4d. In any given case, to satisfy all requirements and arrive at a satisfactory design, two or three trials may be required. 49. Notation. — The notation used in this course in the design of rectangular reinforced concrete beams is summarized as follows : fc = unit compressive stress in outside fiber of concrete. fs = unit tensile stress in steel. n = ratio of modulus of elasticity of steel in tension to modulus of elasticity of concrete in compression. Us = area of cross-section of steel. b = breadth of beam. d — distance from compression surface to axis of rein- forcement. Mc ~ resisting moment as determined by concrete. Ms = resisting moment as determined by steel. M = bending moment or resisting moment in general. p = steel ratio = bd k = ratio of depth of neutral axis to depth of steel. / = ratio of lever arm of resisting couple to depth of steel. V = total shear. V = unit shear. v' = unit working shear. V i^o = ^ = average unit shear. u = unit bond. 0 = circumference of one bar. 112 REINFORCED CONCRETE CONSTRUCTION lo = total circumference of bars, s = horizontal spacing of stirrups. i = diameter of stirrup bar. w = uniform load in pounds per foot. I = span of beam in feet. Xi = distance in feet from support to point beyond which stirrups are unnecessary. V = required length of bent bars for bond strength (above a point 0.6 19 in. In the best shaped beam b lies between 1/2 d and 3/4 Assume 6 = 18 in. d2 = lM22 = 745^ or d = 27.S in., say 27 1/2 in. 18 V 12,000 = 24 lb. per square inch. ° bd (18) (27.5) Web reinforcement is not theoretically needed and we will take 6 = 18 in. and d = 27 1/2 in. Area of cross-section, fed = (18)(27.5) =495 sq. in. as = (495) (0.0077) =3.81 sq. in. We shall select four 1 1/8-in. round rods = 3.98 sq. in. (See following table.) Round rods Square rods Area Weight Area Weight Size square Perimeter per foot Square Perimeter per foot inches inches inches pounds inches inches pounds i .0491 785 17 .0625 1 000 .21 .0767 982 26 .0977 1 25 .33 * .1104 1 178 38 .1406 1 50 .48 .1503 1 374 51 .1914 1 75 .65 .1963 1 571 67 .2500 2 00 .85 .2485 1 767 85 .3164 2 25 1.08 i ..3068 1 964 1 04 .3906 2 50 1.33 U .3712 2 160 1 26 ,4727 2 75 1.61 i .4418 2 356 1 50 .5625 3 00 1.91 a .5185 2 553 1 76 .6602 I 25 2.25 i .6013 2 749 2 04 .7656 50 2.60 11 .6903 2 945 2 35 .8789 3 75 2.99 1 .7854 3 142 2 67 1.0000 4 00 3.40 u .9940 3 534 3 38 1.2656 4 50 4.30 li 1.2272 3 927 4 17 1.5625 5 00 5.31 If 1.4849 4 320 5 05 1.8906 5 50 6.43 li 1.7671 , 4 712 6 01 2.2500 6 00 7.65 The spacing adopted is shown in Fig. 53. V 12,000 The maximum = 35 lb. per square inch. lojd (4) (3.534) (7/8) (27.5) Hooks will be provided at the ends of the rods for additional safety. Spac- ing of bars is greater than necessary. Depth of concrete below the rods is practically 2 in. Minimum distance from the end of beam to edge of support is ^g^^^Q^ = RECTANGULAR BEAMS 115 3.0 in., allowing 100 per cent for unequal distribution of the reactions over the bearing surface. A practical limit is about 6 in. on concrete supports. Let us now review the beam we have designed. It is not necessary to do this unless the values of bd^ and as are made considerably different than the Fig. 53. calculated values. The method will be shown here, however, for the benefit of the student. p = ^=|^ = 0.0080 ^ bd 495 A; = \/(2)(0.0080)(15) + (0.0080)2(15)'- (0.0080)(15) = 0.384 / = 1-1/3(0.384) =0.872 /•°7 3.98)'(o'8°C27.5) ='°-^°° "i'^"' (2)(15,100)(0.0080)_ ,^ •"^ 0.384 In long beams, it frequently occurs that stock bars are too short to extend the full length of the beam. In such cases welding may be resorted to. Since the maximum moment is at or near the middle of the span in most beams, the welding should be near the ends and not near the middle of the beam. This will require two welds instead of one, but welding near the section of maximum moment should never be permitted. Illustrative Problem. — It is not always possible to commence the design of a beam with an accurate enough idea of its weight to proceed as in the previous problem. The usual way is to make an intelligent guess of the 116 REINFORCED CONCRETE CONSTRUCTION weight, design the beam, compute the weight, and if necessary revise the design. Suppose a beam is to span 10 ft. and is to support 600 lb. per foot (not including weight). We shall assume the weight of beam at 85 lb. per foot. p = 0.0077 A; = 0.378 7 = 0.874 M = f=i5MMM=io3,OOOin.lb. , ,2 103,000 (0.0077) (16,000) (0.874) Assume b = 7 in. d^=^~=^137, ordf=11.7in. We will take 6 = 7 in., and d = ll 1/2 in. Thus, web reinforcement will be needed. V (685) (5) . , 1)0 = ^= ^yy^jj-^y = 43 lb. per square mch. Allowable average shear is 35 lb. per square inch. We shall provide the web reinforcement by means of vertical stirrups. Area of cross-section, bd = (7)(11.5)=80.5 sq. in. as = (80.5) (0.0077) =0.62 sq. in. We shall select three 1/2-in. square bars = (3) (0.250) =0.750 sq. in. No difficulty will be encountered in the spacing of these bars with a 7-in. width. Weight of designed beam per linear foot is (total depth 13 in.) The assumed and calculated weights are close enough so that the beam need not be redesigned. We shall, however, consider the loading as 695 lb. per foot in the following computations. The maximum (5) (695) ^„ ^^(3)74)(0:5 ) (7/8) (11.5) ^^^ The bond stress is satisfactory. Minimum distance from end of beam to edge of support is 2. 2 in., allowing 100 per cent for unequal distribution of reactions over support- ing surface. A distance of 6 in. will be taken. Reviewing the beam we have designed, 0 Ti ^ = (7)(1T:5)=0-0093 A; = 0.406 7=0.865 ^ (695) (10) (10) (12) ..nnmu • •^- = (8)(0.75j(0.865)(11.5) =^^'°°° ^^"^'"^ , (2) (14,000 ) (0.0093) _ . , Jc = — o"406 ^ = 640 lb. per square mch. which is satisfactory. i = (0.012)(11.5) =0.138 in. We will use U-shaped stirrups bent at the upper end and 1/4-in. round rods RECTANGULAR BEAMS 117 may be considered secure against slipping. Stirrups are unnecessary at a distance from the center of support equal to 10 (40) (7) (0.865) (11 .5) 2 695 = 5-4.0=1.0 ft. = 12 in. The minimum spacing of stirrups will occur at the supports and will be equal to (3)(2)(0.049)(16,000)(0.865)(11.5) ^ ^ . *~ (2) (5) (695) -b./m. The stirrups are needed for such a short distance from each end of the beam that two stirrups near the ends will suffice, as shown in Fig. 54. Perhaps we may be able to provide web reinforcement by means of bent up rods. Let us investigate. The total stress to be taken by the inclined rods is represented by the Center Line of Support 5'">4" , k-->K--><^> -JO'O' -5" Fig. 54. triangle ABC, Fig. 55. BC represents two-thirds of the horizontal shear at the support per 1-in. length of beam. 2 Vb_2V_2 (695)(5) BC = - "3 bjd~3 jd~3 (0.865) (11.5) r = (0.7)(AZ>) = (0.7)(12)=8.4 in. Hence total stress to be taken by the rods (233) =233 lb. BC. - -(8.4) =980 lb. But the area of one rod multiplied by 16,000 gives its tensile value, or tensile value of one rod = (0.250) (16,000) =4000 lb. Thus, only one bent up rod is required. For the two straight rods the maximum bond stress (695) (5) factory. (2) (4) (0.5) (7/8) (11. 5j = 86 lb. per square inch, which is satis- or < = 2.12 ft. 118 REINFORCED CONCRETE CONSTRUCTION The distance from the support to the point where web reinforcement is unnecessary is only 12 in. The rod can thus be bent up at the required place between these two points. Fig. 55 shows the construction necessary to locate the place where the rod should be bent. If the inclined rod received a stress of 16,000, the length V should equal 50 diameters (from table) or Z' = (50)(l/2)=25 in. but this rod receives only 980 0.25 = 3920 lb. per square inch. and the length I' necessary is 3920 1/2 = 6 1/4 in. 4:u^'' (4) (80) In this case this length can easily be obtained. It should be clear from the sketch that, if a greater length than about 9 in. were required for I', some kind of anchorage at the end of rod should be Nut.- and washer Fig. 55. provided. A possible method would be to use a large nut and washer on the end of the rod, as shown. This would perhaps give sufficient bond but the work of construction would be troublesome. A hook as shown at the end of the horizontal bars is generally considered more satisfactory. Tests seem to indicate (Art. 39) that too much reliance should not be placed upon one or two bent rods. For this reason, in the preceding prob- lem it would be good design to use also the two stirrups as shown. Illustrative Problem, — Design a beam to span 10 ft. and to support a load of 4900 lb. per foot (not including weight). We shall assume the weight of beam at 400 lb. per foot. p = 0.0077 A; = 0.378 / = 0.874 RECTANGULAR BEAMS 119 M=t= '^^'"')"°»"""^' =795,000 ia.-lb. hrJ^ = 795,000 ^ (0.0077) (16,000) (0.874) Assume 6 = 14 in. ^2^7380^ or J=23in. 14 ' We shall take 6 = 14 in., and d = 23 in. Vo =^= ^^^^^^^2 lb. per square inch. (Allowable 35.) Thus web reinforcement is needed. Area of cross-section, fed = (14) (23) = 322 sq. in. as = (322) (0.0077) =2.48 sq. in. We shall select ten 9/16-in. round rods =2.485 sq. in. If all the rods extended straight to end of beam, the maximum bond would be (26,500) ^ lu -1, '*=(T77y(IO)(778)M^ "'^''^"^ The spacing of the rods is shown in Fig. 56. Weight of designed beam per linear foot is »^)=37.,b.,„rMii).(0.85,(I0,=375^ Minimum distance from end of beam to edge of support is (^280) (5) (1 .5) ^ (14) (450) 6.3 in., allowing 50 per cent for unequal distribution of reactions over bearing surface. A distance of 6 in. will be taken. Reviewing the beam we have designed, . (5280)(10)(10)(12) ,.onoiK • v, •^-^ (8)(2.485)(0.874)(2 3-) = '"^""^'^ , (2) (15,800) (0.0077) . , fc=^^^ — Q = 647 lb. per square mch. which is perfectly satisfactory. Web reinforcement is unnecessary at a distance from support _10 (40) (14) (0.874) (23) 2 5280 = 5-2.13 = 2.87 ft. or 34.4 in. If inclined rods are employed, the total stress to be taken is represented by triangle ABC, Fig. 57. BC represents two-thirds of the horizontal shear at the support per 1 in. length of beam. BC-^ g.o (5278) (5) ^^-3 bjd^^^ 0.874 (23) r = (0.7)(AD) = (0.7)(34.4)=24.1 in. Hence total stress to be taken by the rods RC 9.77 = ^ (r) = ^ (24. 1) = 10, 600 lb. But the area of one rod multiplied by 16,000 gives its tensile value, or tensile value of one rod = (0.2485) (16,000) =3980 lb. Thus, only three rods are 9 120 REINFORCED CONCRETE CONSTRUCTION required if they can be bent up at the proper points. The unit bond along the seven rods at end of beam will be but 33 per cent greater than the allowable, which seems reasonable for a beam with this number of rods bent up in two planes. The rods may be bent up in the following order. ..= or<^»(l-Aji)(12)=41in. »..-or<^(l-Aj^)a2)-.27 1in. Fig. 57 shows the construction necessary to locate the points where the rods may be bent. The drawing shows that the rods can be bent at the desired points to take all the diagonal tension and also that the longitudinal spacing wmmi ■ilO-|'Vrods'.5> 14" Fig. 56. of these rods does not exceed three-fourths the depth of the beam. The length I' should equal nearly 50 diameters (from table) or i' = (50)(9/16)=28 in, This length cannot be obtained with the bars bent up nearest the support. These bars will be provided with a hook as shown. It would be good design to add one or two stirrups, as shown, to aid diagonal tension at the end of beam where the stresses have a greater inclination with the horizontal. If stirrups were employed, their diameter and spacing would be deter- mined by the method of Art. 42. RECTANGULAR BEAMS 121 Where web reinforcement is used, it is a good plan to require the rods to be made up into frames and have them placed in the forms as such. If loose rods must be used, the reinforcing elements should be wired together to insure their being in place. A good plan is to have the main bars held in place by metal spacers. Illustrative Problem.— Design a beam to span 15 ft. and to support the ads shown in Fig. 58. The reactions are readily found and are given in the etch. The maximum moment occurs at the center load since the shear Fig. 57. passes through the value zero at this point. We shall assume the weight of beam included in the uniform load of 1000 lb. per foot. M = (25,500) (8) (12) - (23,000) (4) (12) = 1,344,000 in. lb 1,344,000 (0.0077) (16,000) (0.874) ~ ^2,500. Assume 6 = 16 in. ,2 12500 <^ =-jg- = 782, orc?=28in. We shall take 6 = 16 in., and rf = 28 in. _ 7 _ 25,500 „ „ °~hd ~ (16) (28) " square inch. 122 REINFORCED CONCRETE CONSTRUCTION Thus web reinforcement is needed. as = (16) (28) (0.0077) =3.45 sq. in. We shall select eight 3/4-in. round rods = 3.53 sq. in. Bond for one rod at the left end of beam is u = 25^00 — .^443 square inch. ^ (2.356) (7/8) (28) ^ ^ For plain rods, the number which must extend straight to the left end of beam is 443 (1.5)(80)~*' (See page 73.) Thus, at this end of beam four rods may be bent up. In a similar manner it will be found that four rods may be bent up at the right end. Figs. 58, 59. The concrete will be found to take care of any diagonal tension between the concentrated loads. Horizontal shear (which measures diagonal tension) at the support is and to the left of the adjacent concentrated load, it is Ji^-SSOlb.perUnearinch. The total diagonal tension is represented by a trapezoid (Fig. 59), the parallel sides of which are 1040 lb. and 880 lb. and the length 4 ft. Hence, RECTANGULAR BEAMS - 123 total stress in this part of beam to be taken by the concrete and by the web reinforcement is 1^+^X4x12 = 46,080 lb. One-third of the shear, or 15,360 lb., is assumed to be taken by the concrete, hence the amount to be taken by the web reinforcement is 30,720 lb. Figs. 60, 61. Since four rods are to be bent, their comparative tensile value is M0^4^ClM00)=40,400 1b. Thus, the tensile value of the rods is in excess of the stress to be provided for. Now shear is nearly uniform between the support and the adjacent concentrated load and, as regards diagonal tension, it would be sufficiently 124 REINFORCED CONCRETE CONSTRUCTION accurate to give nearly equal spacing to the inclined rods — closer spacing, however, being favored toward the support. The four rods will be bent up two at a time. To investigate for diagonal tension, the line AB, Fig. 60, should be divided into four equal spaces and the position of two of the rods determined by a line projected up at an inclination of 45 degrees vdth the horizontal, and beginning at about the point marked 1. The place to bend up the other two rods should be determined by a parallel line passing prac- tically through the point 3. The points to bend up the rods between the right support and the adjacent concentrated load, to provide for diagonal tension, is determined in a similar manner to the above. An investigation must now be made to see whether the tensile stresses in the bottom of the beam will permit the bending of the rods as required for diagonal tension. Fig. 61 shows the bending moment curve plotted to scale, and the points where the rods may be bent up are determined by the method described in Art. 45. It is clear that the rods cannot be bent up as desired to provide tho oughly for diagonal tension. The points where the rods are actually bent up are about 2 in. beyond the theoretical points as determined by moment. The area ABNM, Fig. 60, represents the amount of diagonal tension to be taken by incUned rods. We have just found, however, that the rods cannot be bent up at just the proper points to take all of this. If each bent-up rod is assumed to take diagonal tension to the amount of one-half its tensile value on each side of itself, then the area bBNc remains unprovided for. Stirrups will be provided to take diagonal tension between the point t, where the line be produced meets the neutral Une, and the adjacent load. Only two stirrups are required, but it would seem advisable in a design of this kind to also place stirrups at the positions indicated by the dotted lines. The spacing of the stirrups is determined by the formula and the student should be familiar with the method of procedure. The length of embedment of the inclined rods should be 50 diameters, or 37 1/2 in. In the above problem the web reinforcement could be designed to take considerably less than 2/3 the diagonal tension with safety. The bent-up rods would not change but the number of stirrups theoretically required might be reduced. In a problem of this kind is found the maximum deviation from the assumption of Art. 41 pertaining to the coefficient 2/3 in the formulas for the design of web reinforcement due, of course, to the trapezoidal shape of the shear diagram. 51. Deflection of Beams. — Very little has been accomplished with respect to the determination of formulas for the deflection of reinforced concrete beams. The difficulty which has been experienced has been due to the fact that the beam action is complicated as far as it pertains to deflection. During the early loading, the concrete assists the steel in taking tension and the RECTANGULAR BEAMS 125 deflection is quite uniform during this period. But as the concrete fails in tension near the center of the beam, there occurs a second or readjusting stage: the steel carries more and more of the tensile stresses and the deflection diagram is a curve. From this point on till the steel reaches its yield point, the de- flection is again quite uniform. Since deflection depends on the stress at all sections, the deflection formulas for homogeneous beams cannot be used for reinforced concrete without modifica- tion, because of the variable action of the concrete in tension throughout the length of the beam. Fig. 62 gives the general form of a deflection diagram for a reinforced concrete beam. The portion AB shows the deflection before the concrete has begun to fail in tension, BC shows the deflection during the readjusting stage, and CD the deflection with the steel near the center of beam carrying practically all the tension. The deflection formulas presented in "Prin- ciples of Reinforced Concrete Construction" by Turneaure and Maurer yield results in fair agreement with actual measured deflections and undoubtedly are the best so far proposed. These formulas have been obtained semi- rationally from the deflection formulas for homogeneous beams and thus assume the material of the beam to obey Hooke's law (stress is proportional to deformation) . Concrete in compression obeys the law very closely up to working stresses, but for concrete in tension the assumption is far from the actual conditions. Since tension in the concrete is considered in the above mentioned deflection formulas, the assumption of a linear stress-deformation relation, which is made for simplicity, should be regarded as a rough ap- proximation. The modulus of elasticity of the concrete in the type of formulas mentioned above should be taken as about the average or secant modulus up to the working compressive stress in the same. Although initial moduli of concrete for compression and tension are nearly equal, the deflection of a beam depends on the elonga- tions and shortenings of all the fibers, and hence not upon the initial modulus but on some sort of a mean value. The resulting value of n should also be governed by the value as chosen from Deflection Fig. 62. 126 REINFORCED CONCRETE CONSTRUCTION experiments on actual deflection, so that this term becomes to some extent a sort of empirical coefficient-making correction for various errors in the deduction of the deflection formulas. Turneaure and Maurer^recommend that 8 to 10 be used for n in the formulas which they have derived, and which are given below. They also state that the formulas presented are the result of modifying the deflection formulas for homogeneous beams in accordance with the following assumptions: 1. The representative or mean section has a depth equal to the distance from the top of the beam to the center of the steel. 2. It sustains tension as well as compression, both following the linear law. 3. The proper mean modulus of elasticity of the concrete equals the average or secant modulus up to the working com- pressive stress. 4. The allowance for steel in computing the moment of inertia of the mean section sljould be based on the amount of steel in the mid-sections, since stirrups and bent-up rods do not affect stiffness materially for working loads. The following are the deflection formulas for rectangular rein- forced concrete beams: Wl^ n E, hd^ a ^ ^ a-l/3[F+(l-/c)3 + 3np(l-A;)2] (2) h^]^ (3) From equations (2) and (3) , the value of a for any values of p and n may be computed, and then the deflection from equation (1). The notation employed in the above formulas is as follows: D = maximum deflection (if desired in inches, the units specified below should be used) . h = breadth of the beam (inches). d = depth of the beam to the center of the steel (inches) . W = total load (pounds) . I — span (inches). p = steel ratio. Es = modulus of elasticity of the reinforcing steel (pounds per square inch). n = ratio of the moduli of elasticity of steel and concrete. 1 In Turneaure and Maurer'a "Principles of Reinforced Concrete Construction," 2nd edition, pages 116 to 123. RECTANGULAR BEAMS 127 a = a numerical coefficient depending on p and n. k = proportionate depth of the neutral axis. Ci = the numerical coefficient in the formula for deflec- tion of homogeneous beams, '^I'^r^' depending on the loading and support. For example, for a cantilever loaded at the end, for a cantilever uniformly loaded, o for a simple beam loaded at center, Ci=^ 5 for a simple beam uniformly loaded, Cj for a beam with fixed ends, load at the center, ^^""1^ for a beam with fixed ends, uniformly loaded, c.=^^ ' ^ 384 The following are the deflection formulas for reinforced concrete T-beams (referred to later) : _ WP n El' h¥' ^ i3 = l/3[/c^-(^l-~)(^A;-^y+|-a-/o)^+3pn(l-fc)2] np+1/2 k = - ¥ bWt\ t in which /? is a coefficient depending upon the steel ratio and w, and other synibols as before. Illustrative Problem. — A rectangular reinforced concrete beam with 6 = 18 in. and d = 27 1/2 in. has a span of 40 ft. and is reinforced with four 11/8 in. round rods (and stirrups) extending along the whole length. What is its probable deflection when sustaining a uniform load of 600 lb. per foot, 128 REINFORCED CONCRETE CONSTRUCTION including its own weight, if the beam is considered as simply supported? A value of 8 will be taken for n. 1 + 2np ^ 1 + (2) (8) (0.008) 2 + 2np 2 + (2) (8) (0.008) a = 0.098 2) = A . (24.000) (480) 3 . _A__0 25in (384) (30,000,000) (18) (27.5) ^ 0.098 52. Economical Proportions. — In the designing of a reinforced concrete beam, the expression bd^ appears. If one of the dimen- sions represented in this expression is assumed, then the other is also determined. Now the cost of a beam to resist a given bending moment will vary with the proportions adopted for breadth and depth, and it is useful to investigate this variation in cost for different conditions. Without taking the matter of shear or web reinforcement into consideration, it is found from a mathematical study along this line that the cost of a rectangular reinforced concrete beam to support a given moment varies inversely with the depth, directly with the square root of the breadth, and also directly with the cube root of the ratio of breadth to depth. The depth will, however, be limited in many ways. It may be limited by the shearing stress fixing the value of bd, or it may be limited by the head room required, or it may need to be so chosen as to give a beam of satisfactory pro- portions. In the case of floor slabs, the breadth (and like- wise the depth) becomes fixed, as the breadth of beam to carry the load coming upon a strip one foot wide is also one foot. In the case where the cross-section of beam is determined by shear, the maximum depth theoretically permissible is that for which bd is just large enough to carry the shear. With a beam designed for moment alone, the cost decreases as the depth in- creases, but the area of the cross-section becomes less. A point must be reached when the beam will be of just the required strength in moment and shear. (Art. 48.) The question which now arises is whether or not a still greater depth will result in greater economy. The quantity bd must now remain constant for the greater depths. But bd^, on the other hand, is increased and the con- crete stress (/c) decreased. A smaller value for fc permits the RECTANGULAR BEAMS 129 use of a smaller percentage of steel, and the cost is still further reduced. Thus it should be clear that the proportions of a beam will not be determined by shear excepting as to minimum cross- section — an increase in depth always resulting in a gain in economy. It should be noted in this connection, however, that although deep beams are economical of concrete, the wooden forms cost more than they do for shallow beams. 53. Restrained Beams. — The discussion thus far in the course has related mainly to rectangular reinforced concrete beams, reinforced for tension and shear, and simply supported. It must be known by the student, however, that the ends of a reinforced concrete beam are often fixed, as in all-concrete construction. Such beams sometimes span only one opening, but more often they are continuous over several supports. It is sometimes difficult to provide a sufficient amount of restraint at the ends of such beams to be able to consider the ends entirely fixed, and good judgment should be used in the design for all such cases. Restrained beams of one span only will be considered under this heading, leaving the treatment of continuous beams, both fixed and supported ends, for a later discussion. A smaller beam may be used when the beam is considered restrained than when considered simply supported, so that a beam computed as simply supported is always on the safe side. For example, the moment at the center of a restrained beam of one span, with both ends fixed, due to a uniform load over the entire beam, is 1/24 wP, where w is the load per linear foot of beam (maximum shear 0.5 wl) . Also, a restrained beam with one end fixed and the other free, and under the same conditions as above, has a moment of 9/128 (maximum shear 5/8 wl). A simply supported beam, on the other hand, has the much greater moment at the center of 1/8 wP. Before one can appreciate the design of beams in all-concrete construction, it will be necessary to have a clear understanding of what is meant by a restrained beam, and the amount of re- straint which is necessary to make the beam fixed. A restrained beam may be defined as a beam fastened at one or both ends in such a manner that the beam is not free to deflect at these points. A beam cannot be considered entirely fixed at either point, however, unless the restraint is sufficient to cause the neutral surface at that point to be horizontal. An example of a beam completely fixed at the ends is shown 130 REINFORCED CONCRETE CONSTRUCTION in Fig. 63. The fixing at the ends is effected in this case by build- ing the beam for some distance into the wall. The same result, as far as the effect on the beam is concerned, might be effected as follows: having merely supported the beam and placed upon it the loads it has to bear, load the ends outside of the supports just enough to make the tangents at the supports horizontal. In reinforced concrete construction the designer must use his judg- ment of how near any particular case approaches the case above mentioned, the bending moment 'formulas for fixed beams being based upon the direction of the neutral surface as just described. With restrained beams, a negative bending moment occurs over a fixed end; that is, at such a point the upper surface of the beam is in tension, and the bottom is in compression. The negative moment at each support, for a beam uniformly loaded w Fig. 63. and fixed at both ends, is -1/12 wP. For one end free, the negative moment at the fixed end is —l/8wP. If a beam in all-concrete construction runs into a column or heavy wall girder, the end in question is practically fixed. Top reinforcement will be required for the negative moment, and rods running into the support must be bent or otherwise anchored. 54. Continuous Beams. — The bending moments and shears on beams which are continuous over one or more intermediate supports cannot be derived from the principles of statics, since the unknown conditions in the case of the reactions are greater in number than the statical conditions of equilibrium. To find the reactions for continuous beams, an additional condition is to be introduced from the theory of elasticity, by means of the equation of the elastic curve. After the reactions are computed, the shears and moments due to given loads are readily deter- mined, either analytically or graphically. Only the results of such study will be stated here, since the study itself is not important in this connection. In general practice, such as building construction, the live RECTANGULAR BEAMS 131 loading is usually indefinite, and is generally considered uniformly distributed except where the exact conditions of panel loading or concentrated loads are known in advance. Fig. 64 shows the variation in shear and moment along a continuous beam due to a uniformly distributed load. The diagram shows that if a beam is built continuously, pull or tension is bound to occur over each intermediate support, with compression at these points at the bottom of the beam. This is shown by the bending moment T T 3 Moment Shear Fig. 64. being negative at these supports. The same is also true at the ends of such beams when fixed. Many of the foremost authorities have reached the conclusion that for a continuous beam of any number of spans, the maximum positive moment in the middle of all but the end spans and the negative bending moment over the supports (with the exception of the second support from each end) may be taken as numeric- ally equal to each other and represented by the formula T2 The center of the end span and the adjoining support between this and the next span should be designed for a moment To M = - 132 REINFORCED CONCRETE CONSTRUCTION A special case, however, occurs with continuous beams of two spans. Such beams should be designed for a positive bending moment of in the middle of each span and a negative moment of over the center support. The shear at each support of continuous beams with fixed ends may be taken at one-half the span load. If the ends are simply supported the shear in the end spans near the second support will be approximately six-tenths of a span load. Many of the building laws in the United States, to provide for the possibility of poor construction and other conditions, give the more conservative figure — to be used throughout. It must be understood that when — is used, it is absolutely neces- sary that the beam be really continuous both in design and construction. In applying the values for bending moment given above to the various cases in all-concrete construction, the assumption is made that the moment of inertia of the beam is constant through- out its length. While this is not strictly true, extensive studies of various cases in reinforced concrete show that a large change in the moment of inertia makes a very small change in the bend- ing moment, so that the relations are substantially correct. Some engineers consider that the use of more steel is necessary between supports than the amount just sufficient to resist the bendmg moment j^"" They consider that by figuring for a moment of — at the center of span, the stresses over the sup- ports are relieved, and the design is more economical. It is true that the negative bending moment over the support de- creases with the increase of steel at the center of span, but it is also true that the consequent decrease of steel over the supports does not offset the increase in steel needed throughout the beam. Mr. Sanford E. Thompson in an article published in the issue of Engineering News of January 13, 1910, under the title ''Continuity in Reinforced Concrete Beams," shows that a beam designed for a moment ^ in the middle of span requires 25 RECTANGULAR BEAMS 133 per cent more steel than the beam designed with — ' for both positive and negative moments. Building laws, however, often require the beams and girders to be figured as simply supported. In such cases to obtain a safe structure and at the same time one that is as economical as possible under the conditions, the beam over the supports should have about six-tenths of the amount of steel that is employed in the middle of the beam. The spans of a continuous beam are generally taken as the distance between the centers of supports. This is the simplest plan to follow in the majority of cases and is always on the side of safety. If the support is exceptionally wide, an arbitrary length of span may be taken which should not be less than the net span between supports plus the total depth of the member which is being designed. For occasional concentrated loads which act in connection with uniform live and dead loads, and for loads produced by beams running into girders, the maximum moment may be computed with sufficient accuracy by considering the beam or girder simply ^>- — '—• y ^ A \ ^ / \ t %■ Fig. 65. supported, and then reducing this maximum moment by the same ratio used in the distributed loading. For example, sup- pose the maximum moment due to given concentrated loads is K (considering the beam supported), then if 1/12 wl"^ is used in distributed loading instead of 1/8 wZ^ required for the supported beam, 8/12 of K, or 2/3K, may be used for the concentrated loads. The negative bending moment with concentrated loading may be taken the same as the maximum positive moment due to concentrated loading, reduced as already explained. If the principal live loads on a beam are concentrated as, for example, upon a girder bridge, the moments and shears at all points must 134 REINFORCED CONCRETE CONSTRUCTION be specially computed. We will not consider such loading for the present. The negative moment at the supports of continuous beams is provided for by bending up a sufficient amount of horizontal steel in the beams on each side of the support, and carrying it across over the support to about the third point of the span. The rods that are bent up for negative moment can be figured to take diagonal tension. The moment at the supports being negative, the tendency is for diagonal cracks to start at the top while farther along the cracks tend to start at the bottom, as shown in Fig. 65. Stirrups at points of negative moment should loop about the upper bars, and at points of positive moment should loop about the lower bars. The student should satisfy himself with regard to the direction of the diagonal stress lines in such beams. Problems Make the best design of beam possible for the conditions given below. Use /c = 600, /» = 16,000, and the allowable unit stresses for shear and bond as recommended by the Joint Committee. Beams are to be considered as simply supported. p = 0.00675. A; = 0.360. j =0.880. 40. Span 30 ft., load 800 lb. per ft. (including wt. of beam). Take 6= 16 in. Use 6 plain round rods. Determine also the maxi- mum deflection. 41. Span 20 ft., load 1800 lb. per ft. (including wt. of beam). Take 6 = 16 in. Use 6 plain round rods. 42. Span 15 ft., load 3200 lb. per ft. (including wt. of beam). Take 6= 16 in. Use 7 plain square rods. 43. Span 12 ft., load 5000 Ib.^er ft. (including wt. of beam). Take 6= 16 in. Use 10 plain round rods in two rows. 44. Span 10 ft., load 8000 lb. per ft. {not including wt. of beam). Take b= 16 in. Use 14 plain round rods and bend up six. 45. Change the concentrated loads shown in Fig. 58 to 25,000 lb. each, and design a beam for the conditions shown, but with the allowable unit stresses as specified above. Consider the concrete to take its allowable value of shear. Take 6 = 20 in. Include weight of beam in given loading. Note. — In Problem 44 given above and in all slab and beam problems which follow in this course, the student is required to make his designs so that the assumed weight of beam checks within 10 per cent of the actual weight. All designs must be submitted neatly drawn in ink with the ac- companying computations. CHAPTER V SLABS, CROSS-BEAMS, AND GIRDERS 55. Slabs. — Only the style of slab occurring in all-concrete construction will be considered at this time, in order to bring forward the treatment of T-beams clearly. Fig. 66 shows a finished floor slab in this type of construction in a building. A somewhat similar construction will be found in beam and girder bridges. The beams, girders, and slabs are continuous. The 'Girder Fig. 66. girders form intermediate supports for the beams, and the col- umns for the girders. The beam BB, for example, is supported by the girder CC, between the points a and b. Fig. 67 shows a cross-section along the line hk, Fig. 66. The slab constitutes part of the beams and girders, being built simultaneously with them. However, for a slab to be considered a part of a beam in 10 135 136 REINFORCED CONCRETE CONSTRUCTION figuring stresses, the beam and slab must be well tied together by the steel reinforcement. With cross-beam and girder construction, the span of the slab will usually range from 4 to 6 ft. In such short spans, the slabs become rigidly supported if they are well bonded to the beams, by reason of the action of the adjoining floor-panels. Even where the slabs are simply supported on the tops of steel beams, the adjoining slabs prevent any lateral motion and render the slab partially continuous. With supports in monolithic con- struction, therefore, the strengthening effect is especially great and reinforcement against negative moment is hardly necessary. For such short spans, a reinforcement of rods near the bottom will be effective. They should be laid with lapped and broken joints to give continuity and to prevent the localization of contraction cracks. (Fig. 67.) i Fig. 67. In the case of spans longer than 5 or 6 ft. it becomes necessary to reinforce against negative moment. This may be done in the same manner as already suggested — by bending up a part of the rods and extending the bent ends beyond the beam. The bend in the bars should be near the 1/4 points in the span, and usually at an angle of 30 degrees with the horizontal. Too sharp an angle may tend to crack the slab. Fig. 68 illustrates two arrangements of rods. For clearness the rods are shown one above the other but, with the exception of the bends, they are actually in the same horizontal plane at the different points and are spaced alternately. Bars may extend over several spans if the spans are short, but in such cases they should be arranged to break joints so as to accomplish the same results as shown in Fig. 68. Fig. 69 is another arrangement to accomplish the same object, using separate straight rods. Either arrangement in Fig. 68 is preferable to that shown in Fig. 69 for very heavy loads, in order to provide against shearing failures. Shearing failures are not usually important in slabs, but in special cases of heavy loading the same care should be used as in the design of large beams. SLABS, CROSS-BEAMS, AND GIRDERS 137 When placing slab reinforcement in long spans, a top reinforce- ment at least to the third point will be desirable. In ordinary spans the steel should at least be lapped a sufficient distance over supports to provide adequate bond strength. The use of a webbing for reinforcing the slab is to be preferred, since the spacing of the bars is not likely to be deranged, as occurs when separate rods are used. Fig. 68. A slab should be figured in the same manner as a rectangular beam, and the depth and percentage of steel are therefore obtained by the formulas given in Art. 33 — the bending moment being figured for a width of slab equal to one foot. For short uniformly loaded spans, fully continuous over two or more intermedfate supports, a moment of 1/12 wP may be used both in the centers of all spans and also over all supports, for both Fig. 69. dead and live loads. For slabs of long span, and for all slabs over two bays only, the moments should be taken as for continu- ous beams. (Art. 54.) The ratio of steel in a slab is most readily found by dividing the cross-section of one bar by the area between two bars, this area being the spacing of the bars multiplied by the depth of steel below the top of slab. Slab bars should not be spaced too far apart to properly take stress directly nor yet should they be spaced so close that 138 REINFORCED CONCRETE CONSTRUCTION the concrete cannot properly be placed between them. The reinforced concrete regulations of New York City adopted Dec. 28, 1911, require that the reinforcement shall not be spaced farther apart than two and one-half times the thickness of the slab. The minimum limit should be about the same as in beams. When a floor panel is square, or nearly so, the slab may advantageously be reinforced in both directions. Exact analy- sis of stresses in such a case is impossible, but some important facts have been brought out by approximate solutions for uniform loading. The theory ^ applied in such an analysis depends upon the fact that the load at any point on the slab is distributed to the two systems of reinforcing bars at that point, in proportion to the stiffness of the beam elements lying in those directions. The A b d B Ti n I J, Fig. 70. distribution of load on each system of reinforcement of a square slab, either along aa' or bb' , Fig. 70, has been found to approxi- mate the curve of a parabola aOa'. The center bending moment along aa' or bV will be found by this assumption to have a slightly greater value than a slab reinforced in only one direction and carrying one-half the uniform load. Also the distribution of load along cc' (or dd') is shown by the curve aOa', and when AB (or BD) is reached by an element parallel to aa' (or 66'), the load supported by the reinforcement becomes zero. The assumption usually made is that one-half the load is car- ried uniformly by each system of reinforcement, and the rods consequently for such an assumption have an equal spacing throughout the slab. The center bending moment resulting from an analysis, as above described, is so near the center bending moment for uniform distribution of load to each system, that 1 From Turneaure and Maurer's "Principles of Reinforced Concrete Construction," 2nd edition, page 309. SLABS, CROSS-BEAMS, AND GIRDERS 139 practically it is accurate enough to consider these bending moments identical. A greater economy, however, may be obtained from the spacing of rods than is generally obtained in practice. The spacing at the center — namely, at aa' or bh' — may be determined on the equally distributed basis but at points intermediate between the center and the edge, the rods might well be spaced so that the number per foot would vary from the required number at the center to zero at the edge, following the law of the parabola. Or, the spacing for the center half of the slab may be the same and then gradually reduce the number of rods per foot to the edge of the slab, using one-half as many rods for the remaining two quarters. From a similar approximate analysis regarding the stresses in rectangular slabs of greater length than breadth and rein- forced in both directions, it seems proper to vary the spacing of the reinforcement for each system as already described, provided the panels are not far from square. As a slab becomes oblong in form, however, the relative amount of load carried by the longi- tudinal system becomes rapidly less. r S +■ ll li S' i Fig. 71. Consider uniform load over the slab represented in Fig. 71. It is required to find and w^, the parts of the whole unit load w that is carried by the reinforcement parallel to the strips rr' and ss\ The deflections of these strips of unit width at k are the same, and are proportional to the fourth power of the length of the strip. Thus, wj3^ = w^^ and, since w^-\-W2 = w w — w^ 6* w ~b*+l* 140 REINFORCED CONCRETE CONSTRUCTION which represents the proportion of load carried by the reinforce- ment parallel to the shorter axis. The proportion of the total load carried by the shorter system for various ratios of length (I) to breadth (b) is approximately as follows: Ratio ^ b 1 1.1 1.2 1.3 1.4 1.5 2.0 w 0.50 0.59 0.67 0.75 0.80 0.83 0.89 When the length of a floor panel is large compared to its breadth, the longitudinal reinforcement (that is, reinforcement parallel with the length) is of little value in carrying loads, but a small amount is nevertheless generally desirable in preventing shrinkage and temperature cracks and in binding the entire structure together. It is more important for wide beam spacing than when the beams are closely spaced. The amount of steel to use is usually selected somewhat arbitrarily, and 1/4-in. or 3/8-in. rods spaced 18 to 24 in. apart is common practice. The top of the slab over a girder should be reinforced transversely, not only for stiffening the girder, but also to provide for the negative bending moment produced with the bending of the slab at right angles to the direction of the principal slab steel. 56. Distribution of Slab Load to Cross-beams. — If a floor slab is reinforced in one direction only, the load will practically all be transmitted to the beams at right angles to the direction of the reinforcing rods. A small part, however, will be transferred directly to the girders at the sides of the panels, but this may well be neglected in the calculations for cross-beams. In fact, even with reinforcement in two directions, the load should be assumed as all transferred to the cross-beams unless the panel is nearly square. If panels, nearly square, are reinforced in both directions, the loads carried to the cross-beams and girders will not be uni- formly distributed over the length of such beams and girders, but may be assumed to vary in accordance with the ordinates of a triangle. This assumption is surely on the safe side in re- gard to moment, if the area of the triangle is made equal to that part of the total load on the panel which is transmitted to the beam in question — as determined by the table of the preceding SLABS, CROSS-BEAMS, AND GIRDERS 141 article. Assumptions of this load being either uniformly dis- tributed, or varying as the ordinates of a parabola, give a lower resulting moment than the triangle method and for this reason are not so much in use. Let w be the uniform load per unit of area on the slab, and and the parts of this unit load that go to the shorter and longer beams respectively. Applying the loads in the form of a triangle having its apex at the middle of the beam, the maximum moment will be for the longer beam, and, if this beam is considered simply supported and as carrying the load from one panel only, and for the shorter beam M= ^ w^hH w If the slab is square, is - , and ikT = 1 / 24 ivl^. For beams made continuous, the bending moment may be multiplied by the proper coefficient as already described. (Art. 54.) With beams or girders common to two panels, the bending moment should be multiplied by 2. Unless the panel is nearly square, floor slabs should not be reinforced in two directions, as it is evident that no economy results from double reinforcement when the ratio of length to breadth of panel is greater than about 1.2. If the length of the slab exceeds 1.5 times its breadth, the entire load should surely be carried by the transverse reinforcement. 57. Distribution of Beam and Slab Loads to Girders. — Fig. 66 shows cross-beams running into girders, with the one-way type of reinforcement. The load upon the girders in such construc- tion consists of the concentrated live and dead loads from the cross-beams acting at their points of intersection with the girder, the uniformly distributed weight of the girder itself, and the unsymmetrically distributed weight of a small portion of the floor slab with its live load which bears directly upon the girder. (Fig. 72.) In order to avoid computing several moments, it has been found^ that the maximum bending moment on a girder sup- porting cross-beams ^may be obtained without appreciable error by considering, as a uniformly distributed load, the weight of the stems of the girder and cross-beams plus the weight of the slab I Taylor and Thompson's "Concrete, Plain and Reinforced," 2nd edition, page 432. 1 12 1 142 REINFORCED CONCRETE CONSTRUCTION and its live load, for an area whose length is the length of the girder and whose width is the average length of the beams run- ning from each side into the girder. The sum of these loads divided by the length of the girder gives a uniformly distributed load for which the ordinary formula for bending moment may be used. Thus, in Fig. 72, instead of computing the moment on the girder as the sum of the moments produced by loads of the tri- angles abc plus the concentrated loads from the beams at c plus the weight of the girder, the entire live and dead load of the area dddd may be considered as uniformly distributed over the girder in the length aa. Cross beam Cross Cross ■a Cross beam beam b/ \ N beam -a 'c Crosslbeam ^\_^.--A55umed 45°- ^b i < d ■a Crosslbeam a Fig. 72. An exception occurs, however, in the case where two beams run into a girder at the one-third points. Here the maximum moment obtained by the uniformly distributed method gives slightly too conservative results, and may be reduced by 10 per cent. Moments in a girder other than the maximum must be computed for individual conditions. 58. Arrangement of Beams and Girders. — The spacing of columns and girders will be determined largely by architectural considerations and consequently the best spacing of cross-beams will vary for different cases. The designer must decide whether to omit all cross-beams, to insert them only at columns so as to form a square or nearly square panel, or to space them at much closer intervals using two or more to a girder panel, as shown in Fig. 66. For economy, slabs with square or nearly square panels should have both transverse and longitudinal reinforcement. For the sake of lateral stiffness, it is generally desirable to place cross-beams at columns. In some cases, however, where such stiffness is not needed, the cross-beams may be entirely SLABS, CROSS-BEAMS, AND GIRDERS 143 omitted. The amount of concrete will be more, but the amount of steel required will be less and little saving, if any, is effected by using cross-beams when their use is in doubt. Heavy loads and low stresses, however, call for large weights of concrete, and cross-beams will undoubtedly be needed under such conditions, as the deeper a beam, the greater its moment of resistance for a given volume. The economical spacing of cross-beams will vary between 4 or 5 ft. to 10 or 12 ft. Architectural considera- tions will generally play an important part, but frequently designs must conform to building regulations relative to ratio of span to depth. 59. Design of Beams and Girders. T -Beams. — When a slab and beam (or girder) are built at the same time and thoroughly tied together, a part of the slab may be considered to act with the upper part of the beam in compression. Cross-beam A and a portion of the slab which may be considered to aid in resisting stress (Fig. 66) is shown cross-hatched; a like representation is given for girder C. This form of beam is called a T-beam, and the extra amount of concrete in the compressive part of such a beam makes possible a considerable saving over the rectangular form. The thickness of the flange is fixed by the thickness of slab required to support its load, but the width of slab which can be taken as effective flange width must be estimated. This width should not be too great in proportion to the slab thickness, otherwise the shearing stresses on the vertical sections through the slab at the edge of beam will be excessive and greater than those on the horizontal section between stem and flange. Experiments show that it would be difficult to crush a flange as much as four times the width of web without failure taking place by the excessive shearing stresses in the web. The arbitrary rules which have been adopted in the assumption of the flange width may be divided into two classes. The first class makes the flange width a factor of the width of stem or thickness of slab, or else of the distance between adjacent beams. The second class makes the flange width a factor of the length of span of the beam itself. The Joint Committee, however, com- bines the two and has recommended a width not exceeding one- fourth the span length of the beams and, in addition, limits the width to use on either side of the web to four times the thickness of the slab. In no case should the breadth of the flange be taken greater than the distance between beams. 144 REINFORCED CONCRETE CONSTRUCTION The web and flange of a T-beam can be considered well tied together when slab reinforcement crosses the beam and when the web reinforcement extends well up into the slab. The bonding should be especially well looked after near the end of beam, and this is generally accomplished by means of the bent rods brought up as high as possible, in addition to the slab reinforcement (as mentioned) acting at right angles to the length of the beam. Along the center of the beam the differential stresses between the beam and slab are not large, but it is better to insert vertical stirrups extending up into the slab at occasional intervals since shrinkage of the concrete is apt to part the slab from the beam if there is not some means to hold the two together mechanically. The thinner the sections, the more thorough should be the bonding. r< k- b'-: Cross section .■■Neutlral plane Stress Diagram Fig. 73. When frames are not used for beams and girders, successful results can be obtained by wiring the main rods together and employing U-shaped stirrups with hook ends to act as hangers by which to suspend the rods in the form. The length of the hook should be sufficient to permit the stirrup to rest on the top of the slab form. The hook has the further advantage of increas- ing the bond between the beam and slab. With a T-beam it is necessary to distinguish two cases when applying formulas for design; namely, (1) the neutral axis in the flange, and (2) the neutral axis in the web. Case I. The Neutral Axis in the Flange.— All formulas for "moment calculations" of rectangular beams apply to this case. It should be remembered, however, that b of the formulas de- SLABS, CROSS-BEAMS, AND GIRDERS 145 notes flange-width, not web-width, and p (the steel ratio) is not (Fig. 73.) Case II. The Neutral Axis in the Web. — The amount of com- pression in the web (aaaa. Fig. 73) is commonly small compared with that in the flange, and is generally neglected. The method of procedure in determining formulas for " moment calculations" of T-beams is identical with that for rectangular beams and the resulting equations only will be given. The formulas to use, assuming a straight line variation of stress and neglecting the compression in the web, are: nfc _ 2ndas +bi- (2) ~ 2nas +2bt 1 /A 2 ^^+2^ (3) _3 kd-2t t^ (4) ^~2kd-t ' 3 jd = d—z (5) J= — (6) 6-3- d (7) Ms = fsttajd J f =il asjd . . fsk (8) na-k) Approximate formulas can also be obtained. From the stress diagram. Fig. 73, it is clear that the arm of the resisting couple is never as small as d — 1/2^, and that the average unit compressive stress is never as small as 1/2/c, except when the neutral axis is at 146 REINFORCED CONCRETE CONSTRUCTION the top of the web. Using these limiting values as approxi- mations for the true ones, Mc=ll2fMd-l/2t) (a) M.=a./.(^-l/20,ora. = ^^^^^^^ (b) The errors involved in these approximations are on the side of safety. Since a T-beam will usually have ample strength in compression for any ordinary depth of beam likely to be selected, the design of the stem of the T, or the beam below the slab, is therefore largely a question of providing sufficient concrete to take care of the shearing stresses and to give a good layout of the tension rods. The manner of providing reinforcement for shearing stresses in T-beams is similar to that already described for rectangular beams. In T-beams, however, the reinforcement for shear should run well up into the slab in order to tie the beam and slab together. Results from tests for both shear and moment in T-beams agree quite closely with theoretical results. It also appears that the shearing strength of a T-beam is about the same as that of a rectangular beam of the same depth and a width equal to the width of the stem of the T. All re-entrant angles in concrete are points of weakness and such angles should, therefore, be avoided. Also, T-beams should not be made too deep in proportion to the width of stem as such forms are relatively weak at the junction of stem and flange. The width should preferably be from one-third to one- half the depth in ordinary cases. For large beams the width may be made one-third to one-fourth the depth. Deflection formulas for T-beams are given in Art. 51. In the following problems and also in those problems to be worked out by the student, the working stresses recommended by the Joint Committee for a 2000-Ib. concrete will be assumed; namely, /c = 650, /s = 16,000, n = 15. In the illustrative problems following Art. 50 in which the above-mentioned allowable stresses were used, the following values were determined: p = 0.0077, A = 0.378, / = 0.874. SLABS, CROSS-BEAMS, AND GIRDERS 147 It will be well to bear these results in mind when following through the computations in slab design which follow. The rule is followed that for a slab of 4 in. (total depth) and less, the depth is taken to the nearest 1/4 in. The depth of slabs over 4 in. are taken to the nearest 1/2 in. Illustrative Problem. — What safe load per square foot (including dead weight) can be supported by a slab 6 in. deep (d =4 3/4 in.) and 10 ft. span, reinforced with 1/2-in. round rods placed 8 in. apart? The slab is simply- supported and reinforced in only one direction. 0.1963 _ P = (8)(4J5) = 0-°°'2 k = V{2) (0.0052)(15) + (0.0052)='(15)^- (0.0052)(15) = 0.325 The percentage of steel is less than 0.0077 and, consequently, the resulting moment is determined by the steel. 1 / 8 («)) (100) (1 2) = (0 . 0052) (16,000) (0 . 892) (1 2) (4 . 75) =» it> = 134 lb. per square foot, safe load. Illustrative Problem.— Design a slab to span 6 ft. and to carry a live load of 250 lb. per square foot. Slab is to be fully continuous and reinforced in only one direction. We shall assume the dead load at 50 lb. per square foot., making a total loading of 300 lb. per square foot. M="|',thenM = «(|^-10,800 m.-lb. 12 1^ Also, M = (0.0077) (16,000) (0.874) (12) (d^) = 10,800 or d = 2.9 in. — take 3 in. Taking 3/4-in. concrete below steel, thickness of slab is 3 3/4 in. Area of steel per foot of breadth = (3.0) (12) (0.0077) =0.277 sq. in. We shall use 3 /8-in. round rods for tensile reinforcement. Area of 3/8-in. round rod = 0.1104 sq. in. Required spacing =^^(12) = 4 3/4 in. on centers. (3.75) (12), V square foot. (12) (12) ^ ^ Thus the assumed and calculated dead weights are close enough, and the slab need not be redesigned. The slab should be reinforced against negative moment at the supports. The slab should also be reinforced transversely in order to prevent shrinkage and temperature cracks. Shear at ends of slab in direction of reinforcement is (300) (3) =900 lb. per foot of breadth. Allowable shear = (12) (3) (40) = 1440. Thus no web reinforcement is needed, as is usually the case except for excessive loading. Illustrative Problem.— Design a slab for a 10 ft. by 10 ft. panel to carry a Hve load of 250 lb. per square foot. Slab is to be fully continuous and rein- forced in both directions. The dead load will be assumed at 60 lb. per square foot. 148 REINFORCED CONCRETE CONSTRUCTION For a strip one foot wide at the center, M = 2j' then M = 15,500 in. -lb. Also, M = (0.0077) (16,000) (0.874) (12) (d^) = 15,500. or d = 3.5 in. Area of steel (at the center) per foot of breadth in each direction = (3.5)(12)(0.0077) =0.323 sq. in. We shall use 3/8-in. round rods for tensile reinforcement. Area of 3/8-in. round rod = 0.1104 sq. in. The required spacing for center half of slab = jj^(12) =4 in. on centers. The spacing of rods should be gradually increased to the edge of the slab, using one-half as many rods for the remaining two-quarters. The slab should be reinforced against negative moment at the supports. The depth of the slab should be made 5 in. in order to have the upper reinforcement system at the minimum distance 3 1/2 in. from the surface of the slab. The lower system will then be slightly stronger than necessary The dead weight of slab is 62 1/2 lb. per square foot, or approximately that assumed. For safety in construction, it is preferable to require the two systems -of reinforcement to be fastened together at frequent intervals. Web reinforcement is not necessary. Illustrative Problem.— A floor panel is to be 12 ft. by 13.2 ft. and the slab IS to be fully continuous and reinforced in both directions. Design such a slab to carry a hve load of 250 lb. per square foot. We shall assume the dead load at 75 lb. per square foot making a total load of 325 lb. per square foot. ^ = ^ = 11 b 12 From table, Art. 55, the 12-ft. span reinforcement should be designed to carry 0.59 w per foot, leaving 0.41 w to be carried by the longitudinal rein- forcement. The resisting moment and consequently the depth of the slab is always determined by the shorter span. ^ ^ (a59)(325) 02^12) ^ 2^ Also, M = (0 . 0077) (16,000) (0 . 874) (12) (d^-) = 27,600 or rf = 4.6— say4 3/4in. Taking 1 1/4-in. concrete below center of steel, the thickness of slab is 0 in. Area of steel at the center per foot of breadth in direction of short length = (4.6) (12) (0.0077) =0.425 sq. in. We shall use 1/2-in. round rods. Area of a 1/2-in. round rod = 0.1963 sq. in. Then, required spacing for center half of slab =--^(12) =5 1/2 in. on centers. Let us see if the long span system of reinforcement may not be placed above the short span reinforcement and still carry its load with safety. (d = 4: 1/4 in.) Using the same spacing and size of rods 'bd (4.25) TSTS) = \/r2) (0.0084) (15) + (0.0084)^^(15?- (0.0084) (15) = 0.392 / = 0.869 SLABS, CROSS-BEAMS, AND GIRDERS 149 The percentage of steel is greater than 0.0077 and consequently the resulting moment is determined by the concrete. M = 1/2(650) (0.392) (0.869) (12) (4.25)^ = 24,000 in. -lb. It is assumed to carry a load of (O^^li^^l^ffl =23,200 in. -lb. 12 Thus, this arrangement of the two reinforcing systems is satisfactory. The assumed and calculated weights are identical and the slab need not be redesigned. In practice it is always more convenient to use the same sized rods and to space them the same in both directions and, if the floor spans are nearly square, this should be done. In the construction of the floor slab, the spacing of the rods should vary as described for a square panel. It will be found that web reinforcement is not necessary. Illustrative Problem. — Design the center cross-section of the 12-ft. supporting beam for the slab of the preceding problem. Assume a center span floor panel so that M = ^^^ may be used throughout. (Art. 54.) The beam receives its load from two floor panels. The triangular load on the beam due to the live load plus dead weight of slab is equal to w,bl (Art. 56) = (0.41) (325) (12) (13.2) =21,100 lb. Assume dead weight of stem of beam at 110 lb. per linear foot. Then maximum shear occurs at either support and equals — h (110) (6) =12,200 lb. The only purpose of the concrete below the neutral axis is to bind together the tension and compression flanges, and consequently its section is deter- mined by the shearing stresses involved and space for the necessary rods. ~ Tlie shearing stress v should not be greater than 120. The area b'd (unless the value of / should turn out to be less than 7/8) should not be less than 12,200 (7/8KT20)^''''^-'"- Some rough calculations show that if four rods are to be used and all in one row, the breadth of stem necessary for the rods controls. A breadth h' of 9 in. and a depth d of 16 in. (total depth 18 in.) will be tried. The bending moment due to the hve load plus weight of slab is given by the formula M = 1/12 wJ)H (Art. 57) for a simply supported beam the load from one panel only. For a beam fully continuous and carrying the load from two panels, M = l/12(0.41)(325)(144)(13.2)(2/3)(2)(12) = 340,000 in.-lb. The moment due to the dead weight of stem (113 lb.) is 16,300 in. -lb. or a total maximum moment of 356,300 in.-lb. The flange of the T-beam is 6 in. thick and its breadth is controlled by one-fourth the span, or 36 in. Using approximate formula (b) of Art. 59, M 356 ,3 00 ^"~{fs)(d-l/2t) ~ (16,000)(13.0) ^'^ ^=(36)ii))=^-«0^^- 150 REINFORCED CONCRETE CONSTRUCTION Supposing this beam to fall under Case I, we find k = \/2(0.0030)(15) + (0.0030)2(15)2 - (0.0030) (15) = 0.258 y = 0.914 Hence, M = (0.258) (16.0) =4.1 in., and the neutral axis is in the flange; that is, the case was correctly guessed. The corrected value of a^, is _ M _ 356,300 (.fsKid) (16,000)(0.914)(16.0) ^-^"^ 1 CQ "'(mm'"-'"'' _ Hence, k = V2(0.0027)(15) + (0.0027)2(15)2- (0.0027) (15) =0.247 The stress in the concrete is then found to be . 2fsp (2)(16.000)(0.0027) „.„ . , Jc = -j^ — = Q ^ = 350 lb. per square mch which is satisfactory. Four 3/4-in. round rods will give the required steel area. It is quite evident in the above problem that the compressive stress in the concrete could no.t be a determining factor in the design, but the com- putations are all given to show the student the method to be followed in the design of T-beams of larger size. Illustrative Problem. — The flange of a T-beam is 24 in. wide and 4 in. thick. The beam is to sustain a bending moment of 480,000 in.-lb. What depth of beam and amount of steel is necessary? TrycZ = 18in. Approximately, /rf = 16in. Then formula (8) or formula (b), Art. 59, gives: _ M 480,00 0 . ~/«/(^~ (16, 000)06) ~ ^-^^ ^"l- P = (2i^f8)=«-°04^ Then equation (6) gives / = 0.910 and the corrected value of a, is 480,000 _ (16,000X0.910)(18) Then equation (2) gives M = 5.61 in., or fc = 0.312 and shows that the beam falls under Case II, as assumed. The stress in the concrete is found from equation (8) to be ^ (16,000)(0.312) „ • u u- u- ■ -u, "(15)(1 — 0 312) square inch, which is permissible. Problems 46. A floor slab is simply supported and reinforced in only one direction. The span in the direction of the reinforcement is 8 ft., d = 4 in., and 1/2-in. round bars are used, placed 7 in. center to center. Determine the safe load per square foot including the weight of slab. 47. What thickness of slab and what sizes and spacing of bars will give a resistance of 4900 ft. -lb. per foot of width? SLABS, CROSS-BEAMS, AND GIRDERS 151 48. (a) Design a center span slab to carry a live load of 250 lb. por square foot. Slab is to span 8 ft. and is to be fully continuous, with the one- way type of reinforcement, (b) Design section of continuous cross- beams with span of 12 ft. Take 6' = 8 in. 49. In a square panel, 14 ft. X 14 ft., of the two-way type of reinforcement, what thickness of slab and what size and spacing of tension rods will be required to carry a live load of 400 lb. per square foot; slab is to be in a center span and fully continuous? 50. A center span floor panel is to be 12 ft. X 14.4 ft., and the slab is to be fully continuous and reinforced in both directions. Design such a slab to carry a live load of 400 lb. per square foot. 51. Design the center cross-section of the 12-ft. supporting beams for the floor slab in the preceding problem. The beam receives its load from two floor panels. Consider the depth of beam (d) fixed at 18 in. (total depth 20 in.) and use four plain round rods. 52. The flange of a T-beam is 20 in. wide, and 5 in. thick. The beam is to sustain a bending moment of 550,000 in. -lb. What amount of steel is necessary? Take d=20 in. 60. Economical Proportions of T-Beams. — When a floor-slab forms the flange of a T-beam, it is possible to determine eco- nomical proportions for the stem. Consider a portion of a rectangular beam one unit in length. Let c = cost of concrete per unit volume; r = ratio of cost of steel to cost of concrete per unit volume; C = cost of beam per unit length; d' — depth of beam below slab. Then C=c 7/ '^^ fM + ll2t)j using the approximate formula (6) of the preceding article. When d' is fixed by the head room available, the cost will be a minimum when 6' is made as small as possible, and its value will then be determined by the shearing stress or by the space required for the rods. The expression also shows that the cost will decrease with increased values of fs, and that with a fixed value of b'd' the cost decreases with increase in depth. If the value of b' is assumed as fixed, then there is a definite value of d which will give minimum cost. By calculus the following expression has been deduced from the preceding equation and will give the value of d for minimum cost when the value of b' is fixed: frM t u 152 REINFORCED CONCRETE CONSTRUCTION From this expression the best depths for various assumed widths may readily be determined and the desirable proportions finally selected. In the design of beams, and especially T-beams, it is important for the student to note that liberal spacing favors large rods and few in number, while good bond strength without waste of material favors small rods. Also, if bent rods are to be used for web reinforcement, then numerous -small rods are also advantageous. 61. Conditions Met With in Design of T-beams. — In practice the design of T-beams will take one of the following forms: (1) The dimensions may be given, to find the safe resisting moment of the beam or the stresses in the steel and concrete under a given load; (2) the flange of the T-beam may form a portion of a floor slab which is already designed, in which case the dimensions of the flange are given — also the loading and specified working stresses — and the design comprises the determination of suitable web dimensions and steel area; (3) the loading and working stresses may be given, to determine suitable proportions for the entire beam— that is, the flange does not form a part of a floor system already determined, (1) The values of k and j may be found from Eqs. (3) and (6), or from Eqs. (2), (4) and (5), of Art. 59, and then the values of the moment of resistance from Eqs. (7), or the fiber stresses from Eqs. (8). If the value of k is found to be less than J, then the problem falls under Case I and the formulas for rectangular beams apply. (2) Depth and width of beam should be selected with reference to shearing strength, space for necessary rods, and other con- siderations. The depth having been selected, the amount of steel may be approximately determined by either Eq. (a) or Eq. (b). The amount of steel being known, the value of / may be determined by Eq. (7) , The value of k should also be found from Eqs. (2) or (3) in order to ascertain if the beam falls under Case I or Case II. The stress in the concrete, corresponding to the allowable working stress in the steel, is then found from Eq. (8) . (This method has been explained by illustrative problems following Art. 59.) (3) First, select suitable proportions for the web. A flange thickness is then assumed such as to give satisfactory proportions SLABS, CROSS-BEAMS, AND GIRDERS 153 between t and d. (See Art. 59 for suitable proportions.) The value of is then known and k and i can be determined from d Eqs. (1), (4), and (5). The area of steel and the breadth of flange are then found from Eq. (7) . (Note. — When making approximate computations for shear or bond stress along the horizontal tension rods, an average value of / = 7/8 may be assumed, as for rectangular beams.) Illustrative Problem. — A T-beam in a floor system is subjected to a maximum shear of 19,000 lb. and a maximum moment of 722,000 in. -lb. Determine the economical depth of beam. Use working stresses recom- mended by Joint Committee. Assume the ratio of unit cost of steel to cost of concrete = 70. Thickness of floor slab is 3 3/4 in. Cross-section of web as determined by shear b'd = ^^'!^^^ = 181 From formula of Art. 60, for 6'= 8 in. (^ = 21.7 in. b'= 9 in. d = 20.6 in. 6' = 10 in. d = 19.7in. If the object is to obtain suitable proportions, either of the last twQ breadths with corresponding depths may be selected but, for convenience in placing steel, the breadth of 10 in. and depth of 19.7 in. will most hkely give the best design. Illustrative Problem. — Design a T-beam with span of 40 ft. Assume dead load = 1400 lb. per foot. Live load = 3000 lb. per foot. The beam is to be simp y supported at the ends and the flange is to be pro- portioned as well as the web; that is, the flange does not form a part of a floor system already determined. Use working stresses recommended by the Joint Committee. This problem comes under Case (3) in the design of T-beams and the method previously suggested will be followed. The total bending moment, M=M2ffi) =10,660.000 in.-lb. The maximum shear F = (4400) (20) =88,000 lb. The required net web area 88,000 - b'd = = 838 sq. m. This area can be supphed by a web 6' = 16 in. (Z = 53in. 6' = 17 in. d = 50m. 6' = 18in. £? = 47in. 6' = 19 in. d = 4:4: in. 6' = 20 in. d = 42m. 154 REINFORCED CONCRETE CONSTRUCTION To give good space for the steel and still obtain satisfactory proportions for the cross-section, we will select b' = lS in. and d = 47 in. for a preliminary value. A thickness of flange of 12 in. will be tried. For this thickness i 12 4-=— =0.256, Then, by means of formulas (1), (4), and (5) of Art. 59, a 47 we find that A; = 0.379 and /d = 42.0 in. Also, from formulas (7), a8, = 15.7 sq. in. and 6 = 48 in. To illustrate the effect of varying proportions, calculations will also be made for a flange thickness of 8 in., 10 in., 14 in., and 16 in. The results are as follows: t b id ag Overhanging width of flange Area of flango outside of web 8 in. 61 in. 43.3 in. 15.2 sq. in. 21.5 in. 344 sq. in. 10 in. 52 in. 42.8 in. 15.4 sq. in. 17.0 in. 340 sq. in. 12 in. 48 in. 42.0 in. 15.7 sq. in. 15.0 in. 360 sq. in. 14 in. 46 in. 41.5 in. 15.9 sq. in. 14.0 in. 392 sq. in. 16 in. 44 in. 41.2 in. 16.0 sq. in. 13.0 in. 416 sq. in. It should be observed that the amount of concrete is less the thinner the slab and that the effect of variation of i upon the amount of steel is very small. However, a relatively thick flange is desirable considering the fact that the girder is not a part of a floor system and that the flanges Fig. 74. are unsupported at their outer edges. The 12-in. flange will be adopted. The dead load assumed is considerably on the safe side, but the design need not be changed. The steel area required is 15.7 sq. in. Five round rods 1 3/8 in. diameter and seven round rods 11/4 in. diameter, giving a total area of 16.0 sq. in., will be used. To provide sufficient spacing between rods, the rods will be placed in three rows as shown in Fig. 74. The five 1 3/8-in. rods will SLABS, CROSS-BEAMS, AND GIRDERS 155 be placed in the lower row, five 1 1/4-ln. rods above, and two 1 1/4-in. rods in the " third row. Taking moments of areas about the center of the lowest row, the center of gravity of the group is found to be (5)(1.23)(2) + (2)(1.23)(4) _ 16 above this row. Hence, the lower row should be placed about 48 1/2 in. below the top of the beam, thus giving a total depth, including the protective covering, of 51 in. Bond for one rod (1 3/8 in.) at the end is V 88,000 . , ^ = J^= (4.3197) (42.0) = '^^ P^'- ^^""^^ For plain round rods, the number which must extend to the end of the beam is 485 (1.5) (80)"* Six will be run to the end — namely, the entire lower row of 1 3/8-in. rods and also the center 1 1/4-in. rod of the second row. Web reinforcement will be provided by means of bent rods and vertical stirrups. In bending up the rods, the two uppermost rods will be bent up nearest the center. Two rods will be bent up at a time since bending ■20'- 0^' > Neutral plane Fig. 75. the rods singly greatly complicates the handling and prevents a symmetrical arrangement. The bends, also, will all be made at 45 degrees and the bent ends extended far enough to give ample strength of bond. Web reinforcement is unnecessary at a distance from support equal to 40 (40)(18)(42.0) The total stress to be taken by the inclined rods (and vertical stirrups, if necessary) is represented by triangle ABC, Fig. 75. BC represents two- thirds of the horizontal shear at the support per 1 in. length of beam. 2 (88,000) 3 13951b. (42.0) r= (0.7)(AD) = (0.7)(13.1)(12) = 110 in. 156 REINFORCED CONCRETE CONSTRUCTION , 40 3^2= or < "2 Hence, total stress to be taken by the rods = ^2^(r) =-f^(110) = 76,700 lb. But the area of one rod multiplied by 16,000 gives its tensile value, or tensile value of one rod = (1.2272)(16,000) = 19,620 lb. Thus, only -^1^1^ = 4 rods are required if they can be bent up at the proper points. Six rods, however, will be bent up as a trial. The rods may be bent up at two at a time in the following order: (Note. — Areas are used in the three preceding equations instead of the number of rods, since rods are of two different sizes. The location of the neutral plane, by formula (2) of Art. 59, is A;d = (0.38) (47) = 17.9 in. below the top of the beam.) The length I' should equal 50 diameters (from table, Art. 43) or (50)(1.25)=62.5 in.=5.2 ft. Fig. 75 shows the construction necessary to locate the points where the rods should be bent to provide for diagonal tension. It is evident that the rods can be bent at the desired points, but notice that the spacing between two bends of the bent rods is greater than 3/4 d. Computations given above show that only 4 rods are required at the end of beam and the bending up of 8 rods should be tried unless it is intended to reinforce with vertical stirrups as well. In the design at hand, with only six rods bent, the spacing of the bent rods can either be made not greater than 3/4 and the Y resulting bond per square inch of the surface is -j^- The shear per linear inch between the compression rods and the concrete will, therefore, very closely equal V Q of e quivalent compressive steel area ]d Total Q of compression area (By the term equivalent steel area is meant the area of the steel multiplied by n and considered equivalent to concrete at the same horizontal plane.) If this quantity be divided by lo', considering o' the circumference of a rod in compression, then the result will be the bond per square inch of surface along the compression rods. Now the total statical moment of the compression area about the neutral axis equals the total moment of the tension area, which is nas{l-k)d. The moment of the compressive steel is equal to na' s{hd-d'). Hence, for bond stress of compressive steel per square inch V_ ofj{kd-d!)^ ^~ lo'jd^ as{l-k)d If now we let 4 = diameter of compression rods, rf< = diameter of tension rods, and 2'o = total circumference of the tension rods, then the above formula reduces to _ V dc(kd -d') dta-k)d This formula shows that the bond stress per square inch for the tension and compression rods will be proportional to the product 164 REINFORCED CONCRETE CONSTRUCTION of the diameters by their distances from the neutral axis. Since the compressive steel will generally be nearer the neutral axis than the tensile steel, it follows that, if the compression bars are no larger in diameter than the tension bars, the bond stress per square inch will always be less than that of the tension bars and the above formula need not be used. The formula derived above may be used in any special case in designing, but generally it will be sufficient to consider simply the compressive stress in the steel and provide a sufficient length from this point to the end of the bar to transmit this stress. The working strength of the steel in compression cannot be reached without exceeding the compressive strength of the concrete in which it is embedded. Consequently, in common design it will be satisfactory to provide a lap beyond the center of support sufficient to take care of compressive stress in the steel equal to the maximum as determined by the concrete. Illustrative Problem.— A continuous T-beam, uniformly loaded, has a bending moment at the center of each span of 358,000 in.-lb. Negative bending moment at the supports and the positive bending moment at the center of span are figured by the formula™. The tensile steel at the center of span consists of four 3/4-in. round rods, b' =9 in. d = 15.5 in. Design the supports using working stresses recommended by the Joint Committee. At the supports the flange of the T-beam, being in tension, is negligible and the T-beam changes into a rectangular beam with steel in top and bottom. Two of the tensile rods on each side of the supports will be bent up and made to lap over the top of the supports, while the other two rods on each side will be continued straight and lapped over supports at the bottom of beam. It will be assumed that the steel at the center of span has already been chosen so that this may be done. The ratios of steel in tension and compression are the same, and are respectively: ^ ^ (9) (15.5) "-"^^ With these values of p and p', and n = 15, also assuming ^ =0.1 (this value is assumed simply to make it easier for the student to check up the com- putations—in practice, the correct ratio should be taken) we obtain the following from equations (3), (5), (7), and (9), of Art. 62: L = 0.291, K = 0.01 1 5, and U = 0.0266 Maximum pressure in concrete is •^'^ (9) (15^5)^(0^291) ^ square inch, by formula (4) SLABS, CROSS-BEAMS, AND GIRDERS 165 Also, fs = (9)(15 5^'^^o"oil5) ^^^^'^^ ^"^^ formula (6) Allowable compression in concrete at the support is thus satisfactory, and no haunch is necessary. The value of fs can always be determined as above, but it is seldom desired as the working strength of the steel in compression cannot be reached without exceeding the compressive strength of the concrete in which it is embedded. With n = 15, the allowable stress in the steel cannot exceed 15 times the compressive strength of the concrete. Illustrative Problems. — A continuous T-beam with 6' = 14 in., and d = 26.5 in. has a negative bending moment at the supports of 2,000,000 in.-lb., and has equal spans of 18 ft. Reinforcement at supports consists of 7/8-in. round rods. Eight bars are in tensile and four in compressive part of beam. Hence, 4.81 Ratio tension steel, ?> = 7T^/-r,r.-EV = 0-0130 ^ (14) (26.5) Ratio compression steel, p' =^^^ = 0.0065 d' With these values of p and p' and w = 15, also assuming ^- = 0.1, we obtain from equations (3) and (5), L =0.243, and from equation (4). 2,000,000 „„„ . , •^^=(W(k5)"^(o:243) ^^"^^^ which is excessive, only 750 lb. per square inch being allowed, or a stress 15 per cent greater than at the center of span. d' For depth of haunch assume ^ =0.1 and try (i = 29 in. For this depth of beam the ratios of steel in tension and compression change to p = 0.0130 =0.0119 ^. = 0:^=0.0059 The corresponding value of L = 0.233 and if = 0.0104. The maximum com- pression in the concrete , 2,000,000 r,cn^u ■ u •^'== (14)W(0:233) =729 lb. per square mch and maximum tension in steel , 2,000,000 iconmu • u = (14)(29)-(0.0104 ) = 1^'^^° P^'" '^"^'■^ ^"'^ This stress is allowable and the depth of haunch from top of beam of 29 in. will be accepted. Using the formula (4) Ml = (750) (14) (26.5)2(0.242) =1,780,000 in.-lb. ^2 = 2,000,000 in.-lb. 166 REINFORCED CONCRETE CONSTRUCTION Hence, from formula in Art. 63, the length of haunch 18^ 220,000 , . ^^T>^ 2:000:000^12^ =4-^^ ^'^^ Since maximum negative moment occurs in middle of column and necessary length of haunch is only 4.75 in., no haunch will be introduced outside of the column. Problems 53. Determine the economical depth of a T-beam in a floor system. Thick- ness of floor slab is 4 in. Maximum shear = 16,000 lb. Maximum moment = 600,000 in.-lb. Assume the ratio of unit cost of steel to cost of concrete = 60. Use working stresses recommended by the Joint Committee. 54. A T-beam has the following dimensions; b=48 in., t = G in,, d = 22 in., and 6' = 10 in. — the steel consisting of six 3/4-in. round rods. Take /s= 15,000, /c = 600, and /i = 15. What is the safe resisting moment of the beam? 55. Change t of Problem 54 to 4 in. and find the safe resisting moment. 56. Make a complete design of a T-beam with span of 30 ft. Live load = 3800 lb. per foot. Assume dead load = 1000 lb. per foot. The beam is simply supported and the flange does not form a part of a floor system already determined. Use working stresses recommended by the Joint Committee. Take b' = 17 in. and t = 10 in. Select required steel area from 1 1/8-in. and 1-in. round rods. 57. A beam of which 6 = 14 in. and d = 20 in. has 2 per cent of tensile steel and 1 per cent of compressive. What is the safe resisting moment d' of the beam? Assume ^=0.1 and use stresses recommended by the Joint Committee. 58. Suppose the beam in Problem 57 were subjected to a bending moment of 1,500,000 in-lb.; what would be the resulting fiber stresses, fc, A, and/%? 59. Suppose the conditions met in Problem 58 occurred at the supports of a continuous beam with equal spans of 30 ft. Design the necessary haunch. Take d = 23 in. CHAPTER VI COLUMNS Concrete columns may be divided into five classes, as follows: 1. Plain concrete columns. 2. Columns reinforced with longitudinal rods only. 3. Columns reinforced with bands or spirally wound metal, called hooped columns. 4. Columns reinforced with both hoops and longitudinal rods. 5. Columns reinforced with structural steel shapes. It is generally conceded that long columns should be avoided because of the fact that any variation in the quality of concrete affects the strength more seriously than in any other structural form, and also because of the fact that for such columns very little data is available from tests. The use of reinforced columns should be limited to a length of about 15 diameters (see recom- mendations of the Joint Committee, Art. 40) and for this reason the strength of longer columns will not be considered in the discussion which follows. 64. Plain Concrete Columns. — In the testing of plain concrete columns, two distinct forms of failure have been observed: (1) a diagonal shearing failure, and (2) a failure by gradual crushing. The failure by diagonal shearing is sudden, the columns breaking suddenly with a loud report and without warning. The failure by gradual crushing may be considered as due to simple com- pression, the breaking down not being discovered until the weighed load begins to decrease and the point of failure not being determined until the machine produces a further shortening of the column. For both forms of failure, however, the ultimate strength of a column could be predicted if a stress-deformation diagram were developed with the progress of the test. It has been found that rich mixtures tend to give the diagonal shearing type of failure, and columns with lean mixtures generally give a failure of the simple compression type. But few comparative tests of cylinders and columns are avail- able, but these indicate that the strength of the column and of the cylinder very nearly agree. (For the relation of crushing 12 167 168 REINFORCED CONCRETE CONSTRUCTION strength of cylinders to crushing strength of cubes, see Art. 10). Strength of plain concrete columns of an average 1:2:4 mixture at 60 days may be taken at about 1800 lb. per square inch al- though tests made at the University of Wisconsin in 1908 indicate that, with careful workmanship and testing, an average value of about 2000 lb. per square inch can be obtained. Plain concrete is entirely suitable for short columns up to lengths of 6 to 10 diameters, but for more slender proportions hooped or banded columns are much to be preferred. The Joint Committee recommends that the length of a plain concrete column be limited to 12 diameters. The compressive strength of concrete is approximately pro- portional to the amount of cement which it contains, so that ..t < ^ increasing the richness of the concrete in .'•bl...*'.. either a plain or reinforced column is an effec- tive means of strengthening the column to permit smaller section. Tests show that for a 1:2:4 plain concrete column there is no excessive variation in individual tests. For a weaker mixture than this, however, the individual tests are much more at variance, indicating greater unreliability. A rich mor- tar for the above reasons may often prove to be the more economical in column construc- tion. Bending stresses in columns due to ec- centric loads must be provided for by increas- ing the section until the maximum stress does not exceed the allowable. A formula for homogeneous columns follows, and formulas applicable to reinforced concrete will be given under " Bending and Direct Stress." The ordinary formula for the compressive fiber stress due to eccentric loading upon solid rectangular columns of homogeneous materials (Fig. 78), is as follows: i2 = total load. A = area of column.* Xo = eccentricity. i = breadth of column. fc = total unit pressure on outer fiber nearest to line of vertical pressure. Fig. 78. COLUMNS 169 Then R , ^Xo\ and the additional intensity of compressive stress due to eccentric loading is seen to be equal to -j- The above formula may be used for columns of plain concrete. 65. Columns with Longitudinal Reinforcement. — Since the modulus of elasticity of a material is the ratio of stress to de- formation, it follows that for equal deformations the stresses in the steel and concrete of a concrete column will be as their moduli of elasticity. Thus, Let A denote total cross-section of column. Ac denote cross-section of concrete. As denote cross-section of steel. p denote steel ratio = fc denote stress in concrete, /s denote stress in steel. n denote P' denote total strength of a plain column for the stress /c. P denote total strength of a reinforced column - for the stress fc- Then, P' =fcA or P ^fcAc +fsAs =fc{A - pA) +fcnpA Thus, ^ P=fcA[l + (n-l)p] (1) from which also p=l + (n-l)7> The relative increase in strength caused by the reinforcement is P-P' ... (3) Tests of columns made at the Massachusetts Institute of Technology, the Watertown Arsenal, the University of Wisconsin, and the University of Illinois, on columns with vertical steel bar reinforcement, indicate that the steel may be counted upon in design to take its portion of the loading as computed from equa- 170 REINFORCED CONCRETE CONSTRUCTION tion (1). Also the value of 15 for n (see Art. 25) as recommended by the Joint Committee, is found to give very conservative results when employed in the above formulas. In this form of column the concrete fails suddenly and in a manner similar to the diagonal shearing failure of a plain concrete column. The density and rigidity of the concrete when steel is employed is apt to be less than in the plain column, so that for small percentages of longitudinal reinforcement the gain in strength is small. On the other hand, it is undesirable to use high per- centages of reinforcement as the strength of such columns is not well determined. If great strength is desired, an increase in the proportion of cement is preferable to a high percentage of steel. The riveted column unit has advantages over the separate rod reinforcement if large amounts of steel are required. Where the concrete is depended upon to fireproof the steel in a column, a certain thickness should be deducted in calculating strength. As already mentioned, the necessary thickness for fireproofing is about 2 in., but if 1 1/2 in. be deducted all around in calculating strength this will amply provide for the weakening effect of fires. A less thickness than this should be sufficient where the contents of a building are not especially inflammable. In the construction of columns the reinforcing bars should be straight, and great care should be exercised in keeping them in place during the pouring of the concrete. Assuming a load uniformly distributed over the cross-section of the column, the reinforcing rods should be arranged symmetrically over this area. Ultimate failure is sometimes due to the buckling of the reinforcement causing the outside concrete to scale and, for this reason, at least 1 1/2 in. of concrete should cover the rods on the outside. The rods should also be held securely in place by wiring or banding the rods together at intervals of about a foot, but such banding cannot be considered as hooping in the sense usually employed. As will be seen later, hooping to be effective must be spaced relatively close. Vertical reinforcement held in place by hoops is shown in Fig. 79. In splicing columns, large rods (or structural shapes) should have their ends planed true and well spliced. The splicing of the large rods may be effected by placing a pipe-sleeve over the upper end of the lower bar and projecting above it, and then setting the lower end of the upper bar within the pipe and resting upon the lower bar. The rods should either be grouted in place COLUMNS 171 or else the diameter of the sleeve should not be over 1/16 in. larger than the diameter of the rod. All such splices should be made above the floor level but not more than 12 in. above the same. When small diameter rods are used, say up to 1 1/4 in., joints in the vertical steel may be provided for by over- lapping a sufficient distance to develop the requisite bond strength, and the lapped rods should be securely wired or bolted to each other. It is much better to avoid splices in a column between lateral supports. In footings where the length of embedment is not sufficient to take all the stress, large rods (or shapes) should rest upon suitable base plates in the foun- dation concrete. The economy of steel reinforcement is dependent upon the working stresses permissible in the concrete and the value of n, since the stress in the steel =/cn. The following table gives p the various values of -j and/, for different stresses in the concrete and different moduli of elasticity: fc p Values of for A n p = 0.01 p = 0.02 p = 0.03 p = 0.04 450 10 490 531 571 612 4500 15 513 576 639 702 6750 20 535 621 706 792 9000 550 10 599 649 698 748 5500 15 627 704 781 858 8250 20 654 759 863 968 11000 650 10 708 767 825 884 6500 15 741 832 923 1014 9750 20 773 897 1020 1144 13000 750 IG 817 885 952 1020 7500 15 855 960 1065 1170 11250 20 892 1035 1177 1320 15000 It should be noted in the table that the stresses in the steel will be relatively low except in the unusual combination of high working stresses in the concrete with large value of n. 66. Columns with Hooped Reinforcement. — Whenever a material is subjected to compression along one axis, then, as a consequence, there will be an expansion of the material along 172 REINFORCED CONCRETE CONSTRUCTION axes which are perpendicular to the one first considered. Thus, if the material of a column is held laterally, then lateral compressive stresses are developed which tend to neutralize the effect of the longitudinal compressive stresses and thus to increase the resistance against failure. This is the principle involved in the spiral or banded column. Fig. 80 shows a spiral form of Fig. 79. Fig. 80. Fig. 81. reinforcement. Longitudinal crimped spacing bars serve to maintain a uniform pitch. Banded reinforcement is similar to that shown in Fig. 79 with the exception that the clear spacing of the bands to be effective must not be greater than about one-fourth the diameter of the enclosed column. The same limit as regards spacing applies to spirals. The yield-point is taken to be the point on the stress-defor- COLUMNS 173 mation diagram where the rate of deformation increases rapidly. It is very nearly the deformation at which plain concrete would fail. Theoretically it can be shown that within this limit of elasticity the hooped reinforcement is much less effective than longitudinal reinforcement, but that such reinforcement may be quite effective in increasing the ultimate strength of the column. The results of tests verify the theoretical conclusion. The ultimate strength is found increased, but the shortening of the column is so great at a comparatively early period in the loading, that the safe strength cannot be based directly on the ultimate strength. Concrete expands laterally only a very small percentage of its vertical deformation or shortening, so that the hoops do not come much into play until the concrete has shortened to an extent such that its elastic limit has been passed. Under further loading, however, the concrete is prevented by the hooping from actual failure, but continues to expand laterally until final failure occurs by the breaking of the wire or by its excessive stretching. It is also found that the shell of concrete outside of the hooping, which is necessary for fireproofing and for the protection of the steel, begins to crack and peel off at about the same load as that which causes complete failure in plain concrete columns. If hooping is not continuous or rigid, it will also peel off with the surface concrete so that the effective strength of the column will be no greater than a similar one of plain concrete. 67. Columns with Hooped and Longitudinal Reinforcement. — ■ The addition of bands or spirals to columns having longitudinal reinforcement does not have much effect upon the deformation of such columns up to the point of failure without hooping. In fact the elastic limit and rigidity of the column appears to be decreased if anything. The effect of such hooping, however, raises slightly the ultimate strength and increases the capacity of the column to deform at loads beyond the elastic limit, so that a somewhat higher working stress may be employed on the concrete than for plain concrete columns. Tests show that about 1 per cent of a closely spaced spiral of high carbon steel is sufficient to prevent the longitudinal rods from bulging outward and will provide a satisfactory amount of toughness with a corresponding raising of the ultimate strength beyond the elastic limit. Vertical reinforcement combined with spiral hooping is shown in Fig. 81. 174 REINFORCED CONCRETE CONSTRUCTION 68. Columns Reinforced with Structural Steel Shapes. — If a structural steel column is designed to take all the load and then is simply fire-proofed with a covering of concrete, it cannot properly be called a reinforced concrete column. To be classed under this heading the steel must be designed so that it takes a Front Side 3" J6" \ 8 3" T End View Fig. 82. load in combination with the concrete; that is, the steel must be figured in the same way as vertical rods and the stresses deter- mined by the formulas previously given. Structural steel reinforcement is sometimes in the form of a cross in the center of the column or more often angles are COLUMNS 175 employed connected by riveted latticing. (Fig. 82.) Tests of columns of this character generaUy show lower ultimate strength than similar columns reinforced with the same quantity of steel in the form of vertical rods. This is most likely due to the difficulty of properly placing the concrete around the steel, and, furthermore, to the fact that the adhesion of concrete to steel where the latter presents broad flat surfaces is not good. It is often advantageous to employ steel columns for reinforce- ment and arrange them to carry the false work and dead load of two or more floors, thus enabling the placing of concrete to proceed simultaneously on several floors. The initial dead load stress applied to the steel of the column in this way enables higher steel stresses to be used. In other words, this initial stress need not be counted with respect to the stress in the concrete, but the total stress in the steel must be looked to in order to make sure that the allowable stress per square inch is not exceeded. To be able to count upon the concrete in columns reinforced with structural forms, the concrete should be well enclosed either by the steel form itself or by means of bands or hooping. However when the amount of steel becomes very large, the relative value of the concrete becomes more uncertain, and it would be good design to neglect its element of strength. 69. Tests on Plain and Reinforced Concrete Columns. — Important tests have been made on plain and reinforced concrete columns at the Watertown Arsenal, the Massachusetts Institute of Technology, the University of Illinois, and the University of Wisconsin. Similar conclusions have been reached in each case. Some of the tests made at The University of Wisconsin will be described as indicating in general what results may be expected from any series of tests. All columns of the series to be described were 10 ft. long with a 1:2:4 concrete mix throughout. Columns were reinforced (Fig. 83) as follows: Ai, A2, A3, A4 Bi, B2, S3, B^. D„ D,, D„ D, E„ E„ E,. . . none. reinforcement shown in Fig. 82. reinforcement shown in Fig. 80. reinforcement shown in Fig. 81. reinforcement shown in Fig. 79. 176 REINFORCED CONCRETE CONSTRUCTION Columns to were squares 12 in. on a side, and C^ to Eg were octagons 12 in. on short diameters. The reinforced columns of this series were all made with a protective shell 2 in. thick. In order to study the behavior of this shell and its effect on the strength of the test piece, it was removed from columns J5i, B^, and Cg before testing. Thus B^ and B^ with the outside shell knocked off had a cross-section 8x8 in., and Cg when stripped of its outside shell before testing had a cross-section in the form of a circle 10 in. in diameter. Columns 5^ and B^ exhibited considerably more toughness than did the plain columns. These columns, outside shell removed, sustained an average ultimate load of 239,50 lb, (3740 lb. per square inch). A plain concrete column of the same size, namely, 8X8 in. and composed of the same concrete. 3500 0 .001 0 .001 .002 .003 .004. 0 .001 0 .001 .002 Deformation per unit length Fig. 83. sustained 120,600 lb. (or 1880 lb. per square inch). Also, a steel column similar to the reinforcement in these columns failed under a load of 138,000 lb. To quote from Prof. M. 0. Withey's paper before the American Society for Testing Materials, Volume IX, 1909: "A comparison of these figures indicates the value of the reinforcement and also shows a means of increasing the strengths of hollow steel columns by filling the inside with concrete. In the tests of columns B^ and B^, the area of the outside shell was so large in proportion to the area of the core that the value of this form of reinforcement COLUMNS 177 was obscured. The results of tests of these two columns, together with those of columns Bi and B^, indicate that the protective shell does not carry its share of the load in this type of column, especially when there is relatively little direct bond between it and the concrete core A comparison of the stresses sustained by the column before and after the shell was removed shows that the shell sustained only about 40 per cent of the average unit stress carried by the entire cross-section. However tests of these two columns revealed that the protective shell remained intact until the yield point of the steel was passed. The point on the stress-deformation curve at which this outside coating began to crack is indicated by a cross in Fig. 83. These tests demonsl rate that this type of column possesses considerable toughness and a high load-carrying capacity; that the steel and concrete act together in carrying the load and in resisting deformation; and that the protective coating, although it remains intact until the yield point of the steel is passed, should not be counted upon to resist deformations or stresses. "Columns to exhibited considerable stiffness when tested. No cracks of any consequence appeared until the deformations indicated that the yield point of the steel had been reached. At this point, however, owing to insufficient lateral support offered by the widely separated 1/4-in. ties, they failed very suddenly. The increase in strength for each per cent of longitudinal reinforcement for these columns, over the strength of the plain columns Ai to ^4, is approximately 156 lb. per square inch. These columns were lacking in toughness and did not have a high ultimate strength. ''The preliminary tests of the spiral columns CitoD^ demon- strated that this type of column has great toughness and a high ultimate strength, which is accompanied by large deformations and deflections. By comparing the test records and the stress- deformation curves for columns to D^, which had vertical reinforcement, it was observed that the outside shell of concrete cracked at about the same time that the stress in the steel reached the yield point. As may be seen in Fig. 83, the stress- deformation curves for columns to C4 are practically the same up to a stress of 1600 lb. per square inch. As the shell concrete in columns C^, C^, and cracked at about this stress, it seems evident that the shell and core act in unison up to this point in this type of column. However as the shell on a column in 178 REINFORCED CONCRETE CONSTRUCTION a building is liable to be greatly weakened by a fire, its strengthen- ing and stiffening value should be neglected in such design." In the experiments above referred to, it was also found that the yield point of columns reinforced with both spirals and longitudinal rods is closely given by the formula ^ = (1-P)fc + Pfs and the ultimate strength approximately by the formula ^= (1 -p)fc+pfs+o.i2fyp' In these formulas, Pi=load at yield point, P = maximum load, A = area of column inside of spiral, p = percentage of lateral steel, fc = ultimate compressive strength of concrete, /« = yield point of longitudinal steel, and = yield point of lateral steel. The spiral steel is not included in the first formula as its effect on the yield point is very small. From the above and other tests made at The University of Wisconsin during the year 1909, Prof. AVithey draws the follow- ing conclusions: "1. A small amount, 0.5 to 1 per cent of closely spaced lateral reinforcement, such as the spirals used, will greatly increase the toughness and ultimate strength of a concrete column, but does not materially affect the yield point. More than 1 per cent of lateral reinforcement does not appear to be necessary. The use of lateral reinforcement alone does not seem advisable. "2. Vertical steel in combination with such lateral reinforce- ment raises the yield point and ultimate strength of the column and increases its stiffness. Columns reinforced with vertical steel only, are brittle, and fail suddenly when the yield point of the steel is reached, but are considerably stronger than plain columns made from the same grade of concrete. ''3. Increasing the amount of cement in a spirally reinforced column increases the strength and stiffness of the column. A column made of rich concrete or mortar and containing small percentages of longitudinal and lateral reinforcement, is without doubt fully as stiff and strong and more economical than one made from a leaner mix reinforced with considerably more steel. In these texts, doubling the amount of cement increased the yield point and ultimate strength of the columns without vertical steel about 100 per cent, and added about 50 per cent to the strength of those reinforced with 6.1 per cent vertical steel. COLUMNS 179 "4. From the behavior under test of the columns reinforced with spirals and vertical steel and the results computed, it would seem that a static load equal to from 35 to 40 per cent of the yield point would be a safe working load. "5. The results obtained from tests of columns reinforced with structural steel indicate that such columns have considerable strength and toughness, and that the steel and concrete core act in unison up to the yield point of the former. The shell concrete will remain intact until the yield point of the steel is reached, but no allowance should be made for its strength or stiffness. ''6. As many of the blotters on the tops and bottoms of columns bore imprints of the vertical steel after failure, it would seem a safe precaution to use bed plates at the foundations for such columns, and thus prevent any possiblity of the steel punching through the concrete under an excessive load." Further experiments made in 1910 at The University of Wis- consin were for the purpose of making a detailed study of ^ the strength and elastic properties of columns reinforced with spirals and longitudinal rods. To quote from Bulletin No. 466 of the University written by Prof. Withey : "The tests were made to obtain some data relating to 1, the effect of varying the percentage of spiral reinforcement; 2, the effect of varying the percentage of longitudinal reinforcement; 3, the effect of varying the richness of the mixture; '4, the effect of a small number of repeated loadings; 5, the effect of maintaining a constant load for different time intervals; 6, the behavior of columns eccentrically loaded; 7, the relative value of plain and deformed bars for longi- tudinal' reinforcement; 8, the effects of differences in end condi- tions. All told, 66 columns of commercial size were made and tested. "It is customary in good practice to encase all column rein- forcement in a shell of concrete about 2 in. thick. This shell is provided to protect the reinforcement against fire and rusting. Since this coating is liable to be injured or destroyed in various ways its strength should not be considered in structural design. Consequently, the core, or portion of the column inside the spiral, should be proportioned to carry the entire load. It, therefore, seemed desirable to make the protective shell on these columns as thin as possible in order that the properties of the core might be more readily observed." 180 REINFORCED CONCRETE CONSTRUCTION Prof. Withey draws the following general conclusions from the results of these tests: "1. If materials can be obtained at average unit prices, rich mixtures are more economical than lean ones. Considering materials similar to those employed herein, the more economical mixtures will be produced if the proportion of cement to aggre- gate, by weight lies between 0.2 and 0.7. "2. Although the yield point of a reinforced concrete column is practically independent of the percentage of spiral reinforce- ment, the ultimate strength and the toughness are directly affected by it. On account of the excessive deformations accom- panying loads beyond the yield point, on account of the prob- ability that both yield point and ultimate strength are less in repeated or long time load tests than in the progressive load tests ordinarily made in the testing machine, and on account of the uncertainties which always surround the hypothesis adopted in designing, good practice demands that only a portion of the stress producing disintegration of the outside shell be used as a working stress. Consequently, only enough lateral reinforce- ment is needed to prevent the longitudinal rods from bulging outward, and to provide an additional factor of safety against an overload by increasing the toughness and raising the ultimate strength somewhat above the yield point. From these tests 1 per cent of a closely spaced spiral of high carbon steel seems to be sufficient for this purpose. "3. By the addition of longitudinal steel the yield point, ultimate strength, and stiffness of a spirally reinforced column can be considerably increased. If maximum economy in floor space is desired, if a column is so long or is so eccentrically loaded that tension exists on a portion of the cross-section, or if a large dead load must be sustained by the column while the concrete is green, a high percentage of longitudinal reinforcement may often be advantageously employed. Such reinforcement is also a valuable safeguard against failure due to flaws in the concrete. If the cost of cement is extremely high, it may be economical to use a leaner mixture than suggested in (1) and considerable longitudinal steel to increase the stiffness and strength; columns like those of Series 1 may profitably be used. In general, how- ever, cement is a more economical reinforcement than steel. Therefore, for ordinary constructions it does not seem advan- COLUMNS 181 tageous to use in combination with a rich concrete more than 2 or 3 per cent of longitudinal steel. "4. The amount of data presented on tests of columns sub- jected to repeated or time loadings is far too small to warrant drawing definite conclusions as to the limiting stress for repeated loadings which will hold true for all kinds of columns and for an infinite number of repetitions, or for a prolonged loading. However, it does appear from the results presented that there is practically no increase in set or deformation after a few repetitions of loads equal to 40 to 50 per cent of the yield points of the columns tested. The results of the repeated load tests also plainly indicate that there is considerable additional strength and toughness afforded by the spiral after the yield point of the longitudinal steel has been passed. That all of this additional strength may not be permanent is suggested by the slopes of the deformation-time curves for two of the columns. "5. The close agreement between theoretical values and values derived from test data shows that the formula commonly used in designing short homogeneous columns eccentrically loaded, may be applied to reinforced concrete columns, provided suitable allowance be made for the steel. "6. The strength of a column resting upon a footing will be about as great as when bedded on a metal plate, provided con- siderable lengths of the longitudinal reinforcing bars are bent outward into the slab or footing. The results given show that the use of metal base plates and longitudinal steel milled to their required length leads to greater uniformity and strength. "7. Although only a few test pieces were reinforced with corru- gated bars of high carbon steel, the results were so uniform and the strength so high that attention should be called to this type of reinforcement. It is quite evident that, with certain ratios of unit prices, the use of deformed bars with high elastic limits will be more economical than the use of plain round bars of mild steel. ''8. Briefly summarizing the foregoing, it seems economical to use for reinforced concrete columns a very rich mixture, and advantageous to employ about 1 per cent of closely spaced high carbon steel lateral reinforcement combined with two or three per cent of longitudinal reinforcement. From the test data pre- sented it seems apparent that such columns, centrally loaded, may be subjected to a static working stress equal to one-third of the stress at yield point." 182 REINFORCED CONCRETE CONSTRUCTION The column A4, shown above, illustrates a diagonal shearing failure in a plain concrete column. The failure was sudden — the column breaking with a loud report. COLUMNS 183 As soon as the yield point had been reached in the vertical steel of column E2, shown above, failure occurred suddenly owing to the insufficient lateral support offered by the widely separated bands. The concrete failed in the same way as in some plain columns. The reinforcement, of course, buckled between the 1/4-in. ties. The above illustration (column J2) shows the great ductility of a column reinforced with both spiral and longitudinal steel. 13 184 REINFORCED CONCRETE CONSTRUCTION 70. Working Stresses. — Considering the sudden manner of failure of plain concrete columns, the factor of safety should be relatively large. The Joint Committee recommend a working stress of 22 1/2 per cent of the compressive strength at 28 days, or 450 lb. per square inch on 2000 lb. concrete tested in cylin- drical form. For richer and stronger mixtures the working stress may be increased accordingly. In a column reinforced with longitudinal steel only, the con- crete fails in a manner similar to the failure of an unreinforced column. Failure takes place suddenly at approximately the same stress per square inch on the concrete as in a plain concrete column. Thus, the factor of safety employed for unreinforced columns might well be used with respect to the concrete in columns of this type. The value of 15 for n is also recommended. (See Art. 25.) With n= 15, the corresponding stress in the steel, for a working stress of 450 lb. per square inch on the concrete, is 6750 lb. per square inch. The proper working stress for hooped columns should be selected mainly with reference to the elastic limit, but the greater toughness of the hooped column as compared with other types, insures a much larger and more certain margin of safety; in other words, the hooped column should permit the use of a somewhat higher working stress in the concrete than for similar columns without hoops. The Joint Committee recommends a working stress 20 per cent higher than for a plain concrete column, or 540 lb. per square inch on 2000 lb. concrete at the age of 28 days, when the amount of band or spiral reinforcement is not less than 1 per cent of the volume of the column enclosed. It is also recommended that the clear spacing of such bands or hoops shall not be greater than one-fourth the diameter of the enclosed column. The working stress allowed is to be applied to the concrete alone, and the hooping is not to be taken directly into account. With w= 15, the corresponding stress in the steel, for a working stress of 540 lb. per square inch on the concrete, is 8100 lb. per square inch. For hooped columns containing an ordinary amount of longi- tudinal reinforcement, the elastic limit of the column tends to approach a point corresponding to the elastic limit of the longi- tudinal steel. One per cent of hooping appears from the results of tests to be sufficient to make possible very high working stresses. The Joint Committee recommends that columns reinforced with COLUMNS 186 not less than 1 per cent and not more than 4 per cent of longi- tudinal steel and with bands or spirals spaced not greater than one-fourth the diameter of the enclosed column, shall have an allowable stress on the concrete 45 per cent higher than given for a plain concrete column, or 650 lb. per square inch on 2000 lb. concrete at the age of 28 days. With n = 15, the corresponding stress in the steel is 9750 lb. per square inch. When columns reinforced with structural steel are designed in accordance with the principles already stated, and the steel and concrete receive their load simultaneously, the working stresses may be taken the same as for hooped columns with vertical steel rods. This is also in accordance with the Joint Committee's recommendations. 71. Value of Longitudinal Reinforcement in Columns. — When a column is of any considerable length something more than plain concrete is desired in practice. Bending moments in such col- umns are apt to occur from unforeseen conditions, and tensile stresses may be produced which would rupture a column if without reinforcement. In very large and relatively short columns, little is to be feared from bending stresses, as in such a case no resultant stress is likely to occur, but in small sections where the danger of imperfect spots in the concrete is greatest, steel reinforcement is of great value in producing a more reliable structure. It should be understood, however, that from a theoretical standpoint the use of steel in columns is not economical. From equation (1) previously given, we see that with a value of n = 15, the use of each 1 per cent of longitudinal steel adds 14 per cent to the strength of a column, while with 50 as the ratio of cost of steel to cost of concrete per unit volume, the increased cost of a column with 1 per cent of steel will be 50 X 1 = 50 per cent. That is, the relative economy of the reinforced column is ordinarily 114 only th&t of the concrete column without reinforcement. Illustrative Problem. — The effective area of a column is 144 sq. in. ; load to be carried is 80,000 lb.; and working stress on the concrete is 450 lb. per square inch. What percentage of longitudinal bars without hooping will be required? Taken = 15. The safe strength of a plain concrete column would be 144X450 = 64,800 lb. 186 REINFORCED CONCRETE CONSTRUCTION Hence, from equation (2), &=^='+('^-«^ p = 1.7 per cent Illustrative Problem. — What sectional area of vertical steel will be required for a round column limited to 35 in. diameter, which has to bear 750,000 lb.; working stresses as recommended by Joint Committee? Using 1 per cent of bands or spirals, a pressure of 650 lb. per square inch may be allowed on the concrete and 9750 lb. per square inch on the steel. One and one-half inch will be considered ~a sufficient protective covering for the hooped reinforcement. Taking the area within the hooping as effective, the safe strength of a corresponding plain concrete column would be ^'(650) =^M!(650) = 523,000 lb. Hence, from equation (2), P 750 , , F-523 = 1 + ^1^-^)P p = 3.1 per cent Problems 60. What will be the safe strength of a column with a 12-in. X 12-in. effective cross-section, which is reinforced with 1.3 per cent of vertical steel, — the working stress in the concrete being 450 lb. per square inch? Taken = 15. 61. The effective area of a column is 196 sq. in. and the load to be carried is 120,000 lb. What percentage of vertical steel is required if stresses as recommended by the Joint Committee for a 2000 concrete are employed? 62. Determine the size of square column, reinforced with 2 per cent of longitudinal steel, which will be required to support a load of 100,000 lb. Provide protective covering for steel, and use the working stresses as recommended by the Joint Committee. 63. Find the diameter of a round column, reinforced by 1 per cent of hooping, designed to support a load of 150,000 lb.; stresses as recom- mended by the Joint Committee. 64. Determine the required area of a column 15 ft. high, supporting 800,000 lb., and reinforced with 3 per cent of longitudinal steel and 1 per cent of hooping; working stresses as recommended by the Joint Committee. 65. What sectional area of vertical steel will be required for a round column limited to 30 in. diameter, which has to bear 500,000 lb.? Use the working stresses recommended by the Joint Committee. CHAPTER VII SLAB, BEAM, AND COLUMN TABLES 72. Illustrative Problems. — Some of the illustrative problems of preceding articles will be partly worked out under this heading to show the amount of numerical computations which may be avoided by the use of tables. Working stresses recommended by the Joint Committee will be employed throughout. 1, Design a rectangular beam to span 40 ft. and to support 600 lb. per foot (including weight of beam). Beam is assumed to be simply supported. From Table 2, for n = 15, /3 = 16,000, and/c = 650. K = 107.4 ^^wP jm(m^ = i,U0,000 in.-lb. 107.4 Assume 6 = 18 in. ^2 = ^^1^ = 745, or d = 27 1/2 in. 18 Area of cross-section, 6d = (18)(27.5) =495 sq. in. as = (495) (0.0077) =3.81 sq. in. We shall select four H-in. round rods = 3.98 sq.*in. (See Table 1 or Table 13.) To make clear the method of reviewing beams, let us review the beam we have designed. From table 3, for this value of p, A; = 0.384 7=0.872 Then, 1,440,000 ir inniu -1, fa=-. / =15,100 lb. per square mch •'^ (3.98) (0.872) (27.5) ' ^ ^ , (2) (15, 100) (0.0080) . , fc = ^^ — ' J ^ =630 lb. per square mch O.o84- Column 4 of Table 3 shows that/^ =15,600 when/^ =650. Table 8 could also be used in the design of above beam. Table 3 may be employed to find minimum allowable depth of beam for a given percentage of steel and various assumed widths. Also, this table may be employed to determine the amount of steel for a beam with given loading, the stress in the concrete being hmited to 650 lb. per square inch and the stress in the steel to 16,000 lb. per square inch. 14 187 188 REINFORCED CONCRETE CONSTRUCTION To find the depth of beam for a given percentage of steel, select the lower value of K on Hne with the given percentage. This substituted in formula M = Khd\ ord=\R gives the smallest permissible depth for various assumed widths. To find the amount of steel for a given beam under the conditions above stated, compute the value of K from formula M = Kbd^. Locate this value either in column (5) or column (7), whichever satisfies the allowed stresses. Thus, if K =80, it must be located in column (7) instead of column (5), because the latter would give a higher stress in the steel than is allowable. The desired ratio, therefore, is 0.0056. If K =112.0, it must be located in column (5) because column (7) would give too high a stress in the concrete. Table 3 may also be employed to determine the safe resisting moment of a given beam. The preceding discussion should make clear the method of procedure. 2. Design a rectangular beam to span 10 ft. and to support a load of 4900 lb. per foot. Beam is assumed to be simply supported. We shall use Table 8 (based on M = ~) in the design of this beam. The weight of beam will be assumed at 400 lb. per foot. Assuming a width of beam of 14 in., the total load per inch of width is (14) (0 667) ^^^^ hnear foot. Referring directly to the table, we find that the depth (d), corresponding to a 10-ft. beam with this load, is 23 in. The area of reinforcement (a^) is shown to be equal to (0.177) (14) =2.48 sq. in. We shall select ten 9/ 16-in. round rods =2.485 sq. in. The spacing of the rods at the center of beam is shown in Fig. 56. At the bottom of Table 8 is given the weight of beam 1 in. wide per Hnear foot for two rows of steel with a total depth of 26 in., and the corresponding weight is (27.1) (14) =379 lb. The assumed and calculated weights do not differ materially and the beam as designed will be considered satisfactory. 3. What safe load per square foot (including dead weight) can be sup- ported by a slab 6 in. deep (d = 4 3/4 in.) and 10 ft. span, reinforced with 1/2-in. round rods placed 8 in. apart? The slab is simply supported and reinforced in only one direction. From Table 7, for p = 0.006 -safe load for slab 6 in. deep (d = 4 3/4 in.) =193 lb. per sq. ft. for p = 0.004 -safe load for slab 6 in. deep (d = 4 3/4 in.) =^ lb. per sq. ft. ^(62)= 37.2 131.0 168.2 safe load, based on M = - 10 (168.2) (0.80) = 135 lb. per square foot, safe load for slab simply supported. 4. Design a slab to span 6 ft. and to carry a H v^e load of 250 lb. per square foot. Slab is to be fully continuous and reinforced in only one direction. From Table 6 for a span of 6 ft., a 3 3/4-in. (d = 3 in.) slab, fully continuous, is seen to sustain a load of (269) (1.2) =323 lb. per square foot. Corre- t SLAB, BEAM, AND COLUMN TABLES 189 spending weight of slab is 47 lb. per square foot and the total load for the slab to carry is thus 297 lb. This depth of slab will be considered satis- factory and is on the side of safety. The area of steel per foot of breadth may be taken directly from Table 6 and, for the depth of slab chosen above, ag =0.277 sq. in. From Table 4 we may use 3/8 round rods spaced 4 3/4 in. on centers, or 7/16-in. rods spaced 6 1/2 in., or 1/2-in. rods spaced 8 1/2 in., etc. 5. Design the center cross-section of a T-beam in a floor system; the beam is to have a span of 12 ft. and be fully continuous. Maximum shear (live plus dead) is closely equal to 12,200 lb. Maximum moment (live plus dead) =356,300 in. -lb. Supported slab is 6 in. thick. The breadth of the flange is controlled by one-fourth the span, or 36 in. Using approximate formula (b) of Art. 59, assuming a depth (d) of 16 in. M 356,300 (/«)(d-l/2/)~ (16,000) (13.0) Referring to Table 9, Part 1, it is seen at once that this beam falls under Case I; that is, the neutral axis is in the flange. For the above value of p, Table 3 gives / =0.914. The corrected value of is _ M _ 356,300 _ """'Tsfd- (16,000) (0.914)06X1) '^'^'^ From Table 3 the stress in the concrete is found to be 350 lb. per square inch. The beam as designed is thus satisfactory. 6. The flange of a T-beam is 24 in. wide and 4 in. thick. The beam is to sustain a bending moment of 480,000 in.-lb. What depth of beam and amount of steel is necessary? Try d = 18 in. Approximately, jd =16 in. Then formula (8) or formula (b), Art. 59, gives: M 480,000 , "-=7^- (16,000) (16) also, t ^_f\ooo _____ 0.222 Referring to Table 9, Part 2, forp = 0.004 and ^ = 0.222 {}gl456^ for p = 0.006 and ^ = 0.222 f}c = 619^^ Thus, The corrected value of a. is for p = 0.0043 and ^ = 0.222 |}^ = 480^ 480,000 (16.000) (0.910) (18) ^' 190 REINFORCED CONCRETE CONSTRUCTION Another and perhaps better method of arriving at the same results is as follows: M ^ 4 80,000 For this value of and for - = 0.222, Table 9, Part 2, shows p to have a value between 0.004 and 0.006. 86.9 61.6 o o 58.4 58.4 p = 0.004 + (0.002) = 0.0042 28.5 3.2 ^ 3 2 /c = 456 +7— (163) =475 lb. per square inch. as = (24) (18) (0.0042) =1.82 sq. in. Suppose that the flange of the above beam had been made 5 in. thick and that the depth (d) had been taken at 16 1/2 in. Then, 480,000 " — =2.14 sq. m. (16,000) (14) ^ 2.1 4 (24) (16 also, ^ _ ^ and 2.14 P= (24)116:5) =^-^°^^ d-16.5-^-^^ Referring to Table 9, Part 2, for p = 0.0043 and ^ = 0.30 I | d 1 /c = 457 Thus, for p = 0.0060 and ^ = 0.30 / / a [Jc= 565 for p = 0.O054 and ^ = 0.30 ... .- { i " ^'^ll a [J<^~ 527 The corrected value of is 480,000 (16,000) (0.893) (16.5) ^-"^ 7. Design a T-beam with span of 40 ft. Assume dead load =1400 lb. per foot. Live load =3000 lb. per foot. The beam is to be simply sup- ported at the ends and the flange is to be proportioned as well as the web ; that is, the flange does not form a part of a floor system already determined. From Art. 61, b' =18 in. and d =47 in. are suitable dimensions and a thickness of flange of 12 in. is tried. The totial bending moment on the beam is 10,560,000 in.-lb. / 12 5 = ^^ = 0.256 From Table 10. 1 = 0.414 z=4.97 yd = 42.0 From Formula (7), Art. 59. 10,560,000 . ^«=(16;0W(4"2^^ The detailed design of this beam has been given at the end of Art. 61. SLAB, BEAM, AND COLUMN TABLES 191 8. A continuous T-beam, uniformly loaded, has a bending moment at the center of each span of 358,000 in.-lb. Negative bending moment at the supports and the positive bending moment at the center of span are figured by the formula, M =— • The tensile steel at the center of span consists of four 3/4 in. round rods, b' =9 in. d =15.5 in. Design the supports. Two of the tensile rods on each side of the supports will be bent up and made to lap over the top of the supports, while the other two rods on each side will be continued straight and lapped over supports at the bottom of beam. The ratios of steel in tension and compression are the same, and are respectively: (1.77 in above computations taken from Table 13.) We will assume d' , =0.1 as before. d From Table 11, Part 2, knowing p' = p, we obtain for o =0 010 / ^ =0.250 lor p u.uiu. j K = 0.0089 fern — nOTi ^ L =0.318 ■ lor p -U.ULb ^ i(: = 0.0133 Thus, Maximum pressure in concrete is Also, for » -0 013 / L =0.291 tor p -V.UL6 I if = 0.0115 . 358,000 . , , 358, 000 i..nniK • u •^^^ (9) (15.5) ^(0.01 15) ^ ' square mch. 9. The effective area of a column is 144 sq. in. ; load to be carried is 80,000 b. ; and working stress on the concrete is 450 lb. per square inch. What percentage of longitudinal bars without hooping will be required? Take n =15. The safe strength of a plain concrete column would be 144X450 = 64,800 lb. Hence, ^ = J0-=1235 From Table 12, for n =15, and p =0.017 ^, = 1.238 Thus, 1.7 per cent of steel is required, and As =(144)(0.017)=2.45 sq. in. From Table 13, four 7/8-in. round rods will be seen to have about the required area. 192 REINFORCED CONCRETE CONSTRUCTION When bands or spirals are used, Table 14 gives the sectional area of hooping for a maximum pitch of one-fourth the diameter of the enclosed concrete — also for a pitch somewhat less than this but which varies for the different column diameters. The amount of hooped reinforcement taken is 1 per cent of the volume of the column. This conforms to the recom- mendations of the Joint Committee. Let D = diameter of enclosed concrete. Ah= sectional area of one strand of hooping for given pitch. P= pitch allowed. Ih = length of hooping in 1 ft. in height of column. Then, for banded reinforcement A ,1 = 0.0025 PD Also, , S7.7D Results for banded and spiral reinforcement will not differ appreciably and Table 14 may be used for both bands and spirals. Problems In a certain building, it was necessary to space the columns 12 ft. centers and to have beams in one direction only. The specifications called for the following items: For slabs, M = \l\2wl^; beams, M = l/10w;F^ Loadings: Dead load, actual weight, concrete weighing 150 lb. per cubic foot. Live load, 250 lb. per square foot. Unit stresses as recommended by Joint Committee. 66. (a) What thickness (total) of floor slabs required? (b) What will be the spacing of the steel reinforcement if 1/2 in. round bars are used? 67. Design the beams to support the floor, making 6' = 10 in. and assuming weight of beam below slab = 210 lb. per hnear foot. Make h'd just large enough to resist shear due to load, considering the web properly reinforced. Required: h, b', d, as, and the number of 5/8 in. round rods required. 68. What will be the stress in the concrete at the bottom of the beam at the column if one-half of the rods from each side are bent up and the horizontal rods are lapped at the support? Use eight 5/8-in. round rods at center of span. 69. A beam 12 in. X22 in. in cross-section has four 7/8-in. round rods placed 2 in. above the lower face of the beam. If the allowable stresses as recommended by the Joint Committee are employed (a) What uniform load in pounds per linear foot could be applied to the span? Span = 16 ft. M = l/SwP. (b) What is the maximum stress developed in the steel? SLAB, BEAM, AND COLUMN TABLES 193 70. Given a column load of 80 tons, what size of square column and what total area of steel cross-section is required if 3 per cent of longitudinal reinforcement is employed? 71 Given a column whose effective dimensions are 20 in. X20 in. and whose cross-sectional area of steel amounts to 14 sq. in. What load will the column sustain? 72. Given a column load of 200 tons, what will be the design of hooped column required, if the area of longitudinal rods is taken equal to 3 per cent of the sectional area of the enclosed concrete? 73. (a) Using Tables 2 and 3 design a rectangular beam to span 15 ft. and to carry a live load of 1500 lb. per linear foot. Beam is to be simply supported. (b) Design this beam using Table 8. 74. The flange of a T-beam is 48 in. wide and 4 in. thick. The beam is to sustain a bending moment of 800,000 in.-lb. Assume the ratio of unit cost of steel to cost of concrete = 60. Take &' = 9 in. and determine the economical depth of beam, also the amount of steel necessary. Consider the computed cross-section as sufficient to resist shear. TABLE 1.— AREAS, PERIMETERS, AND WEIGHTS OF RODS Round rods Square rods Size inches Area square inches Perimeter inches Weight per foot pounds Area square inches Perimeter inches Weight per foot pounds i f 15 .0491 .0767 .1104 .1503 .785 .982 1.178 1.374 .17 .26 .38 .51 .0625 .0977 .1406 .1914 1.000 1.25 1.50 1.75 .21 .33 .48 .65 16 1 n .1963 .2485 .3068 .3712 1.571 1.767 1.964 2.160 .67 .85 1.04 1.26 .2500 .3164 .3906 .4727 2.00 2.25 2.50 2.75 .85 1.08 1.33 1.61 i ig i u .4418 .5185 .6013 .6903 2.356 2.553 2.749 2.945 1.50 1.76 2.04 2.35 .5625 .6602 .7656 .8789 3.00 3.25 3.50 3.75 1.91 2.25 2.60 2.99 1 u li 11 .7854 .9940 1.2272 1.4849 3.142 3.534 3.927 4.320 2.67 3.38 4.17 5.05 1.0000 1.2656 1 . 5625 1.8906 4.00 4.50 5.00 5.50 3.40 4.30 5.31 6.43 li 11 li IS 1.7671 2.0739 2.4053 2.7612 4.712 5.105 5.498 5.891 6.01 7.05 8.18 9.39 2.2500 2.6406 3.0625 3.5156 6.00 6.50 7.00 7.50 7.65 9.98 10.41 11.95 2 2i 2i 2} 3.1416 3.9761 4.9087 5.9396 6.283 7.069 7.854 8.639 10.68 13.52 16.69 20.20 4.0000 5.0625 6 . 2500 7.5625 8.00 9.00 10.00 11.00 13.60 17.22 21.25 25.72 3 7.0686 9.425 1 24.03 9.0000 12.00 30.09 194 REINFORCED CONCRETE CONSTRUCTION TABLE 2.— DATA FOR DESIGN OF RECTANGULAR BEAMS^ Formulas needed: Formulas used in preparing table: M = Khd\ or bd^=^ p .- k = y/2pn+ (pn)^—pn j = l — K = pf8j, or l/2fckj (from formula M=Kbd') Working stresses Ratio of Moduli w = 12 Ratio of Moduli n = 15 /« /c k J P K k 3 P K 12,000 500 0 332 0 889 0 0069 73 6 0 384 0 872 0 0080 83 7 550 0 354 0 882 0 0081 85 7 0 407 0 864 0 0093 96 4 600 0 375 0 875 0 0094 98 4 0 428 0 857 0 0107 110 0 650 0 394 0 869 0 0107 111 3 0 448 0 851 0 0121 123 6 700 0 412 0 863 0 0120 124 4 0 467 0 844 0 0136 138 0 750 0 429 0 857 0 0134 137 8 0 484 0 839 0 0151 152 0 800 0 444 0 852 0 0148 151 3 0 501 0 833 0 0167* 166 9 14,000 500 0 300 0 900 0 0054 67 5 0 348 0 884 0 0062 76 7 550 0 320 0 893 0 0063 78 6 0 372 0 876 0 0073 89 5 600 0 340 0 888 0 0073 90 g 0 391 0 870 0 0084 102 0 650 0 358 0 881 0 0083 102 5 0 410 0 863 0 0095 114 8 700 0 375 0 875 0 0094 114 8 0 428 0 857 0 0107 128 3 750 0 391 0 870 0 0105 127 6 0 446 0 851 0 0120 142 3 800 0 407 0 864 0 0116 140 4 0 462 0 846 0 0132 156 3 15,000 500 0 286 0 905 0 0048 64 7 0 334 0 889 0 0056 74 1 550 0 306 0 898 0 0056 75 4 0 355 0 882 0 0065 86 1 600 0 325 0 892 0 0065 86 7 0 375 0 875 0 0075 98 3 650 0 343 0 886 0 0074 98 4 0 393 0 869 0 0085 111 3 700 0 360 0 880 0 0084 110 3 0 411 0 863 0 0096 124 2 750 0 376 0 875 0 0094 123 1 0 429 0 857 0 0107 137 9 800 0 391 0 870 0 0105 135 7 0 445 0 852 0 0118 151 2 16,000 500 0 273 0 909 0 0043 62 0 0 319 0 894 0 0050 71 3 550 0 292 0 903 0 0050 72 2 0 339 0 887 0 0058 82 3 600 0 310 0 897 0 0058 83 2 0 358 0 881 0 0067 94 4 650 0 328 0 891 0 0067 95 0 0 378 0 874 0 0077 107 4 700 0 344 0 885 0 0075 106 2 0 397 0 868 0 0087 120 6 750 0 361 0 880 0 0085 119 1 0 414 0 862 0 0097 133 8 800 0 375 0 875 0 0094 131 3 0 429 0 857 0 0107 146 7 20,000 500 0 230 0 923 0 0029 53 1 0 272 0 909 0 0034 61 8 550 0 248 0 917 0 0034 62 4 0 292 0 903 0 0040 72 2 600 0 264 0 912 0 0040 72 2 0 311 0 897 0 0047 83 7 650 0 280 0 907 0 0046 82 4 0 328 0 891 0 0053 94 4 700 0 295 0 902 0 0052 93 3 0 344 0 885 0 0060 106 2 750 0 309 0 897 0 0058 103 9 0 359 0 880 0 0067 117 9 800 0 324 0 892 0 0065 115 6 0 374 0 875 0. 0075 130 9 1 From Taylor and Thompson's " Concrete, Plain and Reinforced," 2nd edition, page 519. SLAB, BEAM, AND COLUMN TABLES 195 TABLE 3.— DATA FOR REVIEWING RECTANGULAR BEAMS n = 15 Formulas needed: Formulas used in preparing table: _a8 k =\/2pn+{pn)^—pn M „ K = p fsj, or 1 / 2 fckj (from formula M = Kbd') p k ] /c = 650 lb. per square inch /« = 16,000 lb. per square inch. fs K fc 1 K (1) (2) (3) (4) (5) (6) (7) .0020 0.217 0.928 35300 65.5 295 29.7 .0022 0.226 0.925 33400 68.0 310 32.4 . .0024 0.235 0.922 31800 70.4 330 35.8 .0026 0.243 0.919 30400 72.6 340 38.0 .0028 0.251 0.916 29100 74.6 360 41.4 .0030 0.258 0.914 28000 76.8 370 43.6 .0032 0.265 0.912 26900 78.5 390 47.1 .0034 0.273 0.909 26100 80.7 400 49 .6 .0036 0.279 0.907 25200 82.1 410 51 .9 .0038 0.285 0.905 24400 83.9 430 55.1 .0040 0.292 0.903 23700 85.6 440 58.2 .0042 0.298 0.901 23100 87.4 450 60.4 .0044 0.303 0.899 22400 88.6 465 63.3 .0046 0.309 0.897 21800 89.8 480 66.5 .0048 0.314 0.895 21100 90.6 490 68.8 .0050 0.320 0.893 20800 92.8 500 71 .4 .0052 0.325 0.892 20300 94.2 510 73 . 9 .0054 0.330 0.890 19900 95.6 520 76 .4 .0056 0.334 0.889 19400 96.6 540 80 . 1 .0058 0.339 0.887 19000 97.7 550 82 .7 .0060 0.344 0.885 18600 98.8 560 85.2 .0062 0.348 0.884 18300 100.3 570 87.7 .0064 0.353 0.882 18000 101.6 580 90.6 .0066 0.357 0.881 17600 102.3 590 92.8 .0068 0.361 0.880 17300 103.5 600 95.3 .0070 0.365 0.878 16900 103.9 610 97.7 .0072 0.369 0.877 16700 105.4 625 101.1 .0074 0.373 0.876 16400 106.3 635 103.7 .0076 0.377 0.875 16100 107.1 645 106.4 .0078 0.381 0.873 15900 108.3 655 108.1 .0080 0.384 0.872 15600 108.8 670 112.2 .0082 0.388 0.871 15400 110.0 680 114.9 .0084 0.392 0.869 15200 110.9 690 117.5 .0086 0.395 0.868 15000 112.0 700 120.0 .0088 0.398 0.867 14700 112.2 710 122.5 .0090 0.402 0.866 14500 113.0 720 125.3 .009^ 0.405 0.865 14300 113.8 725 127.0 .0094 0.408 0.864 14100 114.5 740 130.4 .0096 0.411 0.863 13900 115.2 750 133.0 .0098 0.415 0.862 13800 115.6 760 135.9 .010 0.418 0.861 13600 117.1 770 138.6 .012 0.446 0.851 12100 123.6 860 163.2 .014 0.471 0.843 11000 129.8 950 188.6 .016 0.493 0.836 10000 133.8 1040 214.3 o018 0.513 0.829 9300 138.8 1120 238.1 .020 0.531 0.823 8600 141.6 1210 264.4 196 REINFORCED CONCRETE CONSTRUCTION TABLE 4.— SPACING OF ROUND RODS IN SLABS Diameter inches Sectional area of steel per foot of slab when spaced as follows: 2i' 3" 3r 4" 4i" 5i' 10' i .29 .23 .20 .17 .15 . lo .12 Iff .46 .36 .31 26 .23 .20 .18 .17 .15 .13 i .66 .53 .44 38 .33 .29 .26 .24 .22 .19 .17 .15 .13 .90 .72 .60 51 .45 .40 .36 .33 .30 .26 .23 .20 .18 .15 i 1.18 .94 .78 67 .59 .52 .47 .43 .39 .34 .29 .26 .24 .20 1.49 1.19 .99 85 .75 .66 .60 .54 .50 .43 .37 .33 .30 .25 i 1,84 1.47 1.23 1 05 .92 .82 .74 .67 .61 .53 .46 .41 .37 .31 u 2.23 1.78 1.48 1 27 1.11 .99 .89 .81 .74 .64 .56 .49 .45 .37 i 2.65 2.12 1.77 1 51 1.32 1.18 1.06 .96 .88 .76 .66 .59 .53 .44 ii 3.11 2.48 2.07 1.78 1.56 1.38 1.24 1 .13 1.04 .89 .78 .69 .62 .52 i 3.61 2.88 2.40 2 06 1.80 1.60 1.44 1 .31 1.20 1.03 .90 .80 .72 .60 u 4.14 3.31 2.76 2 37 2.07 1.84 1.66 1 .51 1.38 1 18 1.03 .92, .83 69 1 4.71 3.77 3.14 2 69 2.36 2.09 1.88 1 .71 1.57 1 35 1.18 1.05 .94 .78 li 4.77 3.98 3. 41 2.98 2.65 2.39 2 .17 1.99 1 70 1.49 1.33 1 . 19| .99 li 4.91 4 21 3.68 3.27 2.95 2 .68j2.45 10 1.84 1.64 1.47 1 23 2 If 5 09 4.45 3.96 3.56 3 .24 2.97 2 55 2.23 1.78 48 U 1.98 1 5.30 4.71 4.24 3 .86 3.53 3 03 2.65 2.36 2.12 1 77 TABLE 5.— SPACING OF SQUARE RODS IN SLABS Sectional area of steel per foot of slab when spaced as follows: 2i" 1.15 1.50 1.90 2.34 2.84 3.37 3.96 4.59 5.27 6.00 .30 .47 .67 .92 1.20 1.52 1.87 2.27 2.70 3.17 3.67 4.22 4.80 6.08 .25 .39 .56 .77 3i" 1.00 1.27 1.56 1.99 2.25 2.64 3.06 3.52 4.00 5.06 6.25 4i" 1.08 1.34 1.62 1.93 2.26 2.62 3.01 3.43 4.34 5.36 6.48 .19 .29 .42 .57 .75 .95 1.17 1.42 1.69 1.98 2.30 2.64 3.00 3.80 4.69 5.67 6.75 .17 .26 .37 .51 .67 .84 1.04 1.33 1.50 1.76 .15 .23 .34 .46 .60 .76 .94 5i" 1.13 1.35 1.58 1.84 2.04 2.34 2.67 3.37 4.17 |3.75 5.04 2.11 2.40 3.04 6.00 4.54 5.40 .13 .21 .31 .42 .55 .69 .85 1.03 1.23 1.44 1.67 1.9211 2.182 2.76 2 3.41 13 4.12 4.9114 .12 . .19 .28 .38 .50 .63 .78 .94 .12 .32 1 .53 1, .76|l. !ooli, .53^ .12|2. 7813. 50I3. .17 .24 .33 ,43 54 67 81 96 Ts 31^ 51 1 7lfl 17 10" 12 .13 .19 .25 .33 .42 .52 .66 .75 1.02 1.17jl 1.33 1 1.6911 |l 2.0 2 . 52 2 3.00 .17 .23 .30 .38 .47 .57 .67 .79 .92 "05] .20|1, .52 1, 87,1 .27[l_ 7012" .14 .19 .25 .32 ,39 ,47 56 66 ,77 88 00 27 56 89 SLAB, BEAM, AND COLUMN TABLES 197 TABLE 6.— USE FOR DESIGNING SLABS Based on M=^- For supported ends, = deduct 20 per cent from load. wl For fully continuous, {M=^J , add 20 per cent to load. /c = G50 lb. per square inch. = 16,000 lb. per square inch. n = 15. p = 0.0077. A; = 0.378. 6 = 12 in. M = pfsjbd^ = 1290 d2(safe moment of resistance) as = bpd = 0.0924 d (steel area) ^'(12) =pfsibd^, orw = 1074y' (safe load) 10 ^ II 2 2i 2i 21 3 3i 3i 4 4i 41 5i 51 6i 61 7i 71 8i 81 Total safe load (w) per square foot including weight of slab For safe live load deduct weight of slab Span in feet (l) 4 5 6 7 2i 205 2i 268 10 339 419 507 605 709 822 1074 1360 1520 1850 2220 2620 3070 3540 4040 4580 5150 132 172 217 268 325 387 454 527 871 971 1190 1420 1680 1960 2260 2580 2930 3300 91 120 151 i87 226 269 315 366 478 605 675 825 990 1170 1360 1570 1800 2030 2290 67 111 137 165 197 232 269 351 444 496 606 726 858 1000 1150 1320 1490 61 67 85 105 127 177 206 269 340 380 463 556 657 766 884 1010 1140 1680 1290 9 10 11 12 13 14 15 43 67 54 83 100 151 119 140 162 212 269 300 366 438 519 605 698 798 904 67 81 97 113 132 172 218 243 297 356 420 490 566 646 733 1020' 824 35 45 55 67 80 94 109 142 180 201 245 294 347 405 467 534 605 680 119 102 15i:i29 169 144 206,175 247 210 292 248 340 290 392 335 449 382 509 1 434 572 487 lb. sq. in in.-lb. 49 111 420 31 34 38 41 44 47 50 56 62 69 75 81 87 94 100 106 113 119 0.162 0.185 0.208 0.231 0.254 0.277 300 323 370 3950 5160 6530 8060 9750 11600 13600 15800 20600 416 26100 438 29100 485135600 366 125 624 .670 .716 .762 42700 50500 58900 67900 77600 87900 0.808 98800 198 REINFORCED CONCRETE CONSTRUCTION TABLE 7.— USE FOR REVIEWING SLAB DESIGNS Based on M = To* /c = or < 650 lb. per square inch. n = 15. /s =or < 16,000 lb. per square inch. For supported ends (Af = -J-), deduct o 20 per cent from loads. For fully continuous, (ikf = '^|), add 20 per cent to loads. M = lesser of ^*_^^^^^.^^2 (safe moment of resistance) as = bpd=12pd (steel area) ~{12) = pf,ibd^ ] W2 f and using Table 3 ^(12) = if^k}bd2 I , d^ For p= 0.002, M = 297 (safe load) For p = 0.004, w= 582 (safe load) For jj = 0.006, w=852 (safe load) For 2) = 0.008, w=1088y' (safe load) For p = 0.010, 1171^- (safe load) p 1 to steel 1 below steel depth of slab Total safe load (w) per square foot including weight of slab For safe live load deduct weight of slab It of slab (uare foot a. in a section 1 ft. wide moment of sistance ft Q ft Q "3 +3 o Span in feet (Z) •SP " 'S (H ^ ft ^ 02 (U ^^ m in. in in 4 5 6 7 8 9 10 11 13 14 lb. ^ O m sq. in. in.-lb. '2i i 3 94 60 41 38 50 62 0.054 0.078 0.096 3i i 4 196 125 87 64 49 1800 3760 5700 4 1 5 297 190 132 97 74 59 47 0.002 4f li 6 420 268 186 137 105 83 67 55 47 75 87 100 0.114 0.138 0.162 8060 11800 16300 5f 6i li li 7 8 615 847 393 542 273 376 200 276 153 212 121 167 98 135 81 112 68 94 58 80 50 69 44 60 7i 8f li li 9 10 1120 1420 715 910 496 632 364 464 279 356 220 281 179 228 147 188 124 158 106 135 91 116 79 101 113 125 0.186 0.210 21400 27300 '2i i 3 184 118 82 60 46 38 50 62 0.108 0.156 0.192 3540 7380 11200 3i i 4 384 246 170 125 96 76 61 51 4 1 5 582 372 258 190 145 115 93 77 65 55 47 0.004 ■ 4i 5i 6i li li li 6 7 8 823 1200 1660 526 770 1060 366 535 738 268 393 542 206 301 416 162 238 328 131 193 267 109 159 219 91 134 184 78 114 157 67 98 135 58 86 118 75 87 100 0.228 0.276 0.324 15800 23100 31800 7i 8i li li 9 10 2190 2790 1400 1780 972 1240j 714 910 547 697 432 551 350 446| 289 368 243 310 207 264 178 228j 155 198 113 125 0.372 0.420 42000 53500 SLAB, BEAM, AND COLUMN TABLES 199 TABLE 1. —Continued. p g" Depth to steel | Depth below steel g- Total depth of slab Total safe load {w) per square foot including weight of slab For safe live load deduct weight of slab ^ Weight of slab per square foot •? Steel area in a section p" of slab 1 ft. wide 5' Safe moment of g: resistance Span in feet (/) 4 5 6 7 8 9 10 11 12 13 14 15 ' 2i f 3 270 173 120 88 67 53 43 38 0.162 5160 3i i 4 562 360 250 loo 111 90 74 62 53 50 0.234 10800 4 1 5 852 545 379 278 213 168 136 113 95 81 70 62 0.288 16300 4i li 6 1200 771 535 393 301 238 193 159 134 114 98 86 75 0.342 23100 0.006 5i li 7 1760 1130 784 575 441 233 196 167 144 125 87 0.414 33800 6i li 8 2430 1550 1080 793 607 480 388 321 270 230 198 173 100 0.486 46600 7J li 9 3200 2050 1420 1040 800 632 512 423 356 303 261 228 113 0.558 61500 ,81 li 10 4080 2610 1810 LOOK) 1020 806 653 540 453 386 333 290 125 0.630 78300 '2\ f 3 344 220 153 112 86 68 55 45 38 0.216 6610 3i i 4 718 459 319 234 179 142 115 95 80 68 59 50 0.312 13800 4 1 5 1090 696 483 355 272 215 174 144 121 103 89 77 62 0.384 20900 4i li 6 1540 984 882 502 333 303 246 203 171 145 125 109 75 0.456 29500 0.008 • 51 li 7 2250 1440 1000 735 561 444 360 297 250 213 184 160 87 0.552 43200 6i li 8 3100 1990 1380 1010 774 612 496 409 344 293 253 220 100 0.648 59500 7} li 9 4080 2620 1820 1330 1020 807 654 540 454 387 334 291 113 0.743 78500 .8i li 10 5210 3340 2320 1700 1300 1030 834 688 579 493 426 370 125 0.840 100000 ■2i i 3 370 237 165 121 93 73 59 49 38 0.270 7100 i 4 773 495 344 252 193 153 124 102 86 73 63 55 50 0.390 14800 4 1 5 1170 750 520 383 293 231 187 155 130 111 96 83 62 0.480 22500 4J li 6 1650 1060 734 539 414 326 264 218 184 157 135 118 75 0.570 31800 0.010 • 5i li 7 2420 1550 1070 790 606 478 388 320 269 229 198 172 87 0.690 46500 6i li 8 3340 2140 1480 1090 835 659 534 441 371 316 272 237 100 0.810 64100 7i li 9 4400 2820 1950 1430 1100 869 704 581 489 417 360 313 113 0.930 84500 .8* li 10 56103590 2490 1830 1400 1110 897 741 623 531 459 399 125 1 1.050 107800 200 REINFORCED CONCRETE CONSTRUCTION I? o o o o Tt< CD 00 "O (N O CO ^ lO «0 O O O O I> CO O OS 00 l> CO CO t> OO 05 o o o o O O OO O (N O iH CO a ^, CO O ■* 00 lO 10 lO O O O O d d o o o o o o d d d d o o o o d d d d 3 O rH 00 O (N O CO Tt< 00 "O CO CD 00 O IN --H iH (N IN ~05 c5 iS o~ ■-H Tt< CD 05 (N IN IN — ICOINOJ O0O5.-I-* 05 OC CO 00 05 lo CO o 00 05 O IN tH CD rH O IN CO lO O >0 O lO 00 00 05 05 O lO O lO d d ^' rn' SLAB, BEAM, AND COLUMN TABLES 201 o o o o >0 CD ■ OO 05 o o o o lO CO CO o O lO rH 00 0 o o o 01 •* OS TlH (M O 00 0> .-H (N O O O O O O O O CO OO CO O O CD 00 OJ 03 OO CO (M CD TfH 00 OS (M CO O OS OS o o ^ o o CO tH lO (M C<3 CO CO ^ _ CO (M (M (M o o o o o o o o o o o o o o o o o o o (N (M CO CO CD OS (N lO CO CO Tt( OS (M ^ (N CO CO CO 2 CO CO CO TJ( lO >o "O O O CD CD OO CO t> OS O (N ^ CO CO >-l 05 1> OD O CO CO 00 OS tH CO CO 00 00 O (N lO t~ OS C^l CD O -Jtf CO CD t> t> OS t> 00 O O OS I> OS >H CO OS 1-1 <3S CO 00 (N CO t> t- 00 i-l !-( i-H 00 00 OS I> (M O i-H O CO 00 CO 00 iO lO CO CD lO ,-1 OS O O Tj( ,_, ^ T)H rH lO OS CO (N (N (N CO —I OO O T-H IM C) CO 00 CO (N Tt* 05 00 CO CO O TtH OO t— rH rH (N (N (M CO CO ^1 lO rH rH lO CO OO IC OS CO C rH CO rH oouocornoosos rHrH(N(N IMIM0 OO IM CD "O CO 00 OS O CD IM 00 rH CI Tl< lO (M N N C^l O OS OS rH (M IM CO 00 00 O O OS CO (M I> Tj< CO N rH (N CD CO Tt< »C CD t> 00 OS ^ rH ■* (N O OS 00 00 CD l> OS rH M (N (N CO CO CO CO O O Tti l> OS CO CO CO ^ CD OS CO OS lO CO OO O rH IM (N ^ Tt( CD 00 0 CO IM O OS rH Tf OS CO CO CO CO O rH CO CO CO CO OS IM Tt( >o rH CD 'J* CO OS CO lO lO CO CO O r> O (M rH i> rH CO CD OO rH 1> 00 O C^l 1^ CO CD O Tt< ^ lO lO lO 00 rH CD OO (N l> rH lO CD CD Tt< (M CO Ttl CD rH CD rH t> OO 00 OS O *3 O "O (M IM CO CO O lO O lO O lO O lO o o o o o o o Tt< Tt( lO lO CD CO rH IM rH 3 § f .2 O Q ^^ 202 REINFORCED CONCRETE CONSTRUCTION TABLE 9, PART 1.— USE FOR T-BEAMS n = 15. /8 = or< 16,000 lb. per square inch. /c = or<650 lb. per square inch. 1. The dimensions given, to find the safe resisting moment or the stresses in" the steel and concrete under a given load. 2, When flange of T-beam forms a portion of floor slab already designed — to .determine suitable web dimensions and steel area. pn+1/2 U Mc=fc[l- fc ■ k = safe resisting moment = lesser of Mc and Mg p = 0.002 5 S S ^ T hd- p = 0.004 S S o ■a » 1 s .s ^ « o ci S Mr bd^ 269 257 248 240 234 229 225 222 220 218 0.217 0.217 0.217 0.954 0.950 0.946 0.943 0.940 0.937 0.935 0.933 0.931 0.929 0.928 0.928 0.927 26500 28200 29500 30800 31900 32800 33600 34200 34600 35000 35200 35200 35200 392 369 352 337 326 316 310 304 301 297 296 296 296 Neutral axis in flange for greater values of 30.5 30.4 30.3 30.2 30.1 30.0 29.9 29.9 29.8 29.7 29.7 29.7 29.7 t_ d 406 388 373 360 349 339 331 323 317 312 0.954 0.948 0.944 0.940 0.936 0.932 0.928 0.925 0.921 0.918 0.308 0.915 0.304 0.914 0.300 0.912 14200 15400 16400 17300 18200 19000 19700 20400 21000 21500 21900 22300 22700 730 675 634 600 572 546 528 509 495 484 475 465 457 Values of p for neutral axis at lower edge of slab Values below this line correspond to values of p in first column .0023 .0025 .0028 .0030 .0033 .0036 .0039 .23 0.230 0.240 0.250 0.260 0.270 0.280 0.290 0.924 32600 319 0.918 30800 337 0.916 29300 356 0.913 27800 375 0.911 26500 395 0.908 25000 415 0.904 23800 436 34.0 36.8 41.1 43.8 48.2 52.3 56.5 .23 .24 .25 .26 .27 .28 .29, t Neutral axis in flange for greater values of — ■ 54.3 58.2 60.4 60.2 59.9 59.7 59.4 59.2 59.0 58.8 58.6 58.5 58.4 0.298 0.909 22900 45? 448 58.2 0.296 0.907 23200 58.0 0.294 0.906 23400 444 58.0 0.293 0.904 23500 442 57.9 0.292 0.903 23600 440 57.8 0.291 0.903 23800 438 57.8 0.291 0.903 23800 438 57.8 SLAB, BEAM, AND COLUMN TABLES 203 TABLE 9, PART 2.— USE FOR T-BEAMS. n=15. /8 = or< 16,000 lb. per square inch. =or< 650 lb. per square inch p = 0.004 p=0.006 I i 11^ _d u Oh ^ -2 P. I'll u H » (3 & ^ 0.406 0.388 0.373 0.360 0.349 0.339 0.331 0.323 0.317 0.312 0.308 0.304 0.300 0.298 0.296 0.294 0.293 0.292 0.291 0.291 0.954 0.948 0.944 0.940 0.936 0.932 0.928 0.925 0.921 0.918 0.915 0.914 0.912 0.909 0.907 0.906 0.904 0.903 0.903 0.903 14200 15400 16400 17300 18200 19000 19700 20400 21000 21500 730 675 634 600 572 546 528 509 495 484 54.3 58.2 60.4 60.2 59.9 59.7 59.4 59.2 59.0 58.8 Neutral axis in flangi 58.6 58.5 58.4 58.2 58.0 58.0 57.9 57.8 57.8 57.8 ;e for greater values of ^ 21900 22300 22700 22900 23200 23400 23500 23600 23800 23800 475 465 457 452 448 444 442 440 438 438 Values below this line correspond to values of p in first column 0.300 0.310 0.320 0.330 0.340 0.901 0.897 0.893 0.890 0.886 22700 21700 20700 19800 18900 457 479 502 525 550 62.1 66.1 71.5 76.9 82.3 .10 0. .11 0. 500 0.952 480|0.947 462 0.943 447 0.939 434 0.934 422 0.930 411 0.926 402 0.923 0.919 0.915 394 386 0.380 0.373 0.368 0.364 0.360 0.357 0.354 0.351 0.349 0.348 0.912 0.909 0.906 0.903 0.900 0.897 0.895 0.893 0.891 0.889 9750 10600 11300 12000 12700 13400 14000 14500 15000 15500 15900 16400 16700 17000 17300 17500 17800 18000 18200 18200 0.346 0.888 0.345 0.887 0.344 0.344 0.344 0.886 0.886 0.885 1067 984 917 862 818 779 745 716 693 670 654 635 621 610 600 592 584 577 572 569 18400 565 85 2 18500 562 85 2 18600 559 85 1 18600 559 85 1 18600 559 85 0 t Neutral axis in flange for greater values of— - a 15 204 REINFORCED CONCRETE CONSTRUCTION TABLE 9, PART 3.— USE FOR T-BEAMS n = 15. /s = or < 16,000 lb. per square inch. /c = or < 650 lb. per square inch. p=0.006 p = 0.008 a:: 2 S S !? I- a 1 f£ •Mil P, CD .10 0 500 0. 952 9750 1067 55 7 .10 0 568 0 952 7420 1400 £6 4 .11 0 480 0 947 10600 984 59 9 .11 0 547 0 947 8070 1287 60 9 .12 0 462 0 943 11300 917 64 0 .12 0 530 0 943 8650 1200 65 2 .13 0 447 0 939 12000 862 67 8 .13 0 514 0 938 9230 1125 69 2 .14 0 434 0 934 12700 818 71 3 .14 0 499 0 934 9800 1060 73 1 .15 0 422 0 930 13400 779 74 5 .15 0 486 0 930 10300 1008 76 7 .16 0 411 0 926 14000 745 77 5 . 16 0 474 0 925 10800 960 70 /y Q y .17 0 402 0 923 14500 716 80 4 .17 0 463 0 921 11300 920 83 0 .18 0 394 0 919 15000 693 83 0 .18 0 453 0 917 11800 884 85 9 .19 0 386 0 915 15500 670 85 6 .19 0 445 0 914 12100 855 88 8 .20 0 380 0 912 15900 654 87 4 .20 0 437 0 910 12500 827 91 ^ .21 0 373 0 909 16400 635 87 3 .21 0 430 0 906 12900 805 93 5 .22 0 368 0 906 16700 621 87 0 .22 0 424 0 903 13200 784 95 6 .23 0 364 0 903 17000 610 86 7 .23 0 418 0 899 13600 766 97 4 .24 0 360 0 900 17300 600 86 4 .24 0 413 0 896 13800 750 99 2 .25 0 357 0 897 17500 592 86 1 .25 0 409 0 893 14100 738 100 7 .26 0 354 0 895 17800 584 85 9 .26 0 405 0 890 14300 726 102 1 .27 0 351 0 893 18000 577 85 7 .27 0 401 0 888 14500 715 103 3 .28 0 349 0 891 18200 572 85 5 .28 0 398 0 885 14700 705 104 4 .29 0 348 0 889 18200 569 85 3 .29 0 395 0 883 14900 696 105 4 .30 0 346 0 888 18400 565 85 2 .30 0 393 0 881 15100 690 106 2 .31 0 345 0 887 18500 562 85 2 .31 0 391 0 879 15200 685 107 0 .32 0 344 0 886 18600 559 85 1 .32 0 389 0 877 15300 678 107 4 .33 0 344 0 886 18600 559 85 .1 .33 0 388 0 876 15400 676 108 0 .34 0 344 0 885 18600 559 85 .0 .34 0 386 0 874 15500 670 108 2 Neutral axis in lange for greater values of i d Values of p for neu- tral axis Values below this line correspond to at lower values of p in first column. edge of slab. 0.0063 .35 0 .350 0 .883 18100 575 89 .0 .35 0 .386 0 873 15500 670 108 6 0.0068 .36 0 .360 0 .879 17300 601 95 .6 .36 0 .385 0 .873 15500 668 108 7 0.0072 .37 0 .370 0 .878 16600 627 101 .2 .37 0 .383 0 .873 15700 662 108 7 0.0078 .38 0 .380 0 .873 15900 654 107.9 .38 0 .380 0 .873 15900 654 107 9 Neutral axis in flange for greater values of t ~d SLAB, BEAM, AND COLUMN TABLES 205 TABLE 9, PART 4.— USE FOR T-BEAMS n = 15. /g = or < 16,000 lb. per square inch, /c = or < 650 lb. per square inch. p 0.008 v = 0.010 t ~d k J Maximum fiber stress in steel correspond- ing to /c = 650 lb. per sq. in. cimum fiber stress concrete corre- jnding to /s = ,000 lb. per sq. in. IVl M r ~2 t d k 3 Maximum fiber stress in steel correspond- ing to /c = 650 lb. per sq. in. cimum fiber stress concrete corre- Dnding to fs = ,000 lb. per sq. in. M bd r 2 oj Ci ft CD cS fl ft CD . 10 0 568 0 952 7420 1400 56 4 .10 0 620 0 951 5980 1740 56 8 . 11 0 547 0 947 8070 1287 60 9 .11 0 600 0 947 6500 1600 61 5 .12 0 530 0 943 8650 1200 65 2 .12 0 582 0 942 7000 1484 65 9 . 13 0 514 0 938 9230 1125 69 2 .13 0 566 0 938 7470 1390 70 1 . 14 0 499 0 934 9800 1060 73 1 .14 0 551 0 933 7940 1310 74 1 . 15 0 486 0 930 10300 1008 76 7 .15 0 5380 929 8360 1240 77 9 .16 0 474 0 925 10800 960 79 9 .16 0 525 0 925 8820 1179 81 6 .17 0 463 0 921 11300 920 83 0 .17 0 513 0 921 9260 1121 84 9 .18 0 453 0 917 11800 884 85 9 .18 0 502 0 916 9670 1075 88 0 .19 0 445 0 914 12100 855 88 8 .19 0 494 0 913 10000 1040 91 0 .20 0 437 0 910 12500 827 91 2 .20 0 486 0 909 10300 1010 93 8 .21 0 430 0 906 12900 805 93 5 .21 0 477 0 905 10700 973 96 4 .22 0 424 0 903 13200 784 95 6 .22 0 470 0 901 11000 944 97 7 .23 0 418 0 899 13600 766 97 4 .23 0 464 0 898 11300 922 101 0 .24 0 413 0 896 13800 750 99 2 .24 0 459 0 894 11500 903 103 1 OK n u '±uy 0 893 14100 738 100 7 .25 0 453 0 891 11800 883 104 8 .26 0 405 0 890 14300 726 102 1 .26 0 448 0 888 12000 866 106 5 .27 0 401 0 888 14500 715 103 3 .27 0 444 0 885 12200 851 108 1 .28 0 398 0 885 14700 705 104 4 .28 0 440 0 882 12400 839 109 5 .29 0 395 0 883 14900 ■696 105 4 .29 0 436 0 879 12600 825 110 5 .30 0 393 0 881 15100 690 106 2 .30 0 433 0 876 12800 815 111 7 .31 0 391 0 879 15200 685 107 0 .31 0 430 0 874 12900 805 112 7 .32 0 389 0 877 15300 678 107 4 .32 0 428 0 872 13000 799 113 5 .33 0 388 0 876 15400 676 108 0 .33 0 426 0 870 13100 792 114 3 .34 0 386 0 874 15500 670 108 2 .34 0 424 0 868 13200 784 114 9 .35 0 386 0 873 15500 670 108 6 .35 0 422 0 866 13300 779 115 2 .36 0 385 0 873 15500 668 108 7 .36 0 421 0 865 13400 776 115 8 .37 0 383 0 873 15700 662 108 7 .37 0 420 0 864 13400 773 116 4 .38 0 380 0 873 15900 654 107 9 .38 0 419 0 862 13500 769 116 3 Neutral axis in flange for greater values of t d Values of p for neu- tral axis Values below this line correspond to at lower values of p in first column edge of slab. .0083 .39 0 390 0 870 15200 682 110 4 .39 0 419 0 862 13500 769 117 0 .0089 .40 0 400 0 867 14600 712 112 8 .40 0 418 0 .861 13600 766 117 0 .0095 .41 0 410 0 864 14000 742 115 2 .41 0 418 0 .861 13600 766 117 .0 .0100 .42 0 420 0.861 13500 774 117.5 .42 0 .418 0 .861 13600 766 117.0 Neutral axis in fl inge for greater values of t d 206 REINFORCED CONCRETE CONSTRUCTION TABLE 9, PART 5.— USE FOR T-BEAMS n = 15. /8 = or< 16,000 lb. per square inch. /c = or< 650 lb. per square inch. p = 0.010 p=0.012 Mr 2 S 2 o .a ft ,620 .600 ,582 ,566 ,551 ,538 ,525 ,513 ,502 ,494 477 470 464 459 453 448 444 440 436 433 430 428 426 424 422 421 420 419 419 0.418 0.418 0.418 0.951 0.947 0.942 0.938 0.933 0.929 0.925 0.921 0.916 0.913 0.909 0.905 0.901 0.898 0.894 0.891 0.888 0.885 0.882 0.879 0.876 0.874 0.872 0.870 0.868 0.866 0.865 0.864 0.862 0.862 0.861 0.861 0.861 5980 6500 7000 7470 7940 8360 8820 9260 9670 10000 10300 10700 11000 11300 11500 11800 12000 12200 12400 12600 12800 12900 13000 13100 13200 13300 13400 13400 13.500 13500 13600 13600 13600 1740 1600 1484 1390 1310 1240 1179 1121 1075 1040 1010 973 944 922 903 883 866 851 839 825 815 805 799 792 784 779 776 773 769 769 766 766 766 Neutral axis in flange for greater values of 56.8 61.5 65.9 70.1 74.1 77.9 81.6 84.9 88.0 91.0 93.8 96.4 97.7 101.0 103.1 104.8 106.5 108.1 109.5 110.5 111.7 112.7 113.5 114.3 114.9 115.2 115.8 116.4 116.3 117.0 117.0 117.0 117.0 d Values below this line correspond to values of p in first column. 0.430 0.857 0.440 0.855 12900 12400 805 839 119.8 122.1 660 641 624 608 592 579 566 555 544 535 527 518 511 504 497 490 486 481 476 472 0.469 0.465 0.462 0.460 0.457 0.455 0.453 0.452 0.450 0.449 0.448 0.447 0.447 0.951 0.947 0.942 0.938 0.933 0.929 0.924 0.920 0.916 0.912 0.908 0.904 0.900 0.896 0.893 0.889 0.886 0.883 0.879 0.876 0.874 0.871 0.868 0.866 0.863 0.861 0.860 0.858 0.856 0.855 0.854 0.853 0.852 5020 5450 5880 6290 6720 7100 7480 7810 8180 8470 8750 9080 9310 9600 9870 10100 10300 10500 10700 10900 11000 11200 11300 11400 11600 11700 11800 11800 11900 11900 12000 12000 12000 .43 0.447 0.852 .44 0.446 0.851 12000 12100 Neutral axis in flange for greater values of ^ a SLAB, BEAM, AND COLUMN TABLES 207 TABLE 10.— USE FOR T-BEAMS 3. Loading and working stresses given — to determine suitable pro- portions for entire beam. /c==650 lb. per square inch. = 16,000 lb. per square inch. n = 15. t A; = — ^- = 0.379 -= A ]d==d-z nfc t 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20 0.21 z t 0.475 0.471 0.468 0.465 0.462 0.458 0.455 0.451 0.448 0.444 0.440 0.436 t 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.30 0.31 0.32 0.33 z i 0.431 0.427 0.422 0.417 0.412 0.407 0.402 0.396 0.391 0.384 0.378 0.371 t ~d 0.34 0.35 0.36 0.37 0.38 { Neutral axis in flange when— = A: = 0.379 a z t 0.364 0.356 0.349 0.341 0.332 208 REINFORCED CONCRETE CONSTRUCTION TABLE 11, PART 1.— USE FOR RECTANGULAR BEAMS WITH STEEL IN TOP AND BOTTOM n = 15 Mc = M%L and/c = Ms = bd-faK and fs k = ^2n(^p + p'j^ +n^(p + p'y-n(p + p') 1 k d ^ > in which if = p fl — ^~ , , 2n(l-^)\3 d p' = 0.25 p p P' k L K 0 005 0 00125 0 307 0 153 0 0045 0 01 0 0025 0 394 0 202 0 0088 d' — = 0 05 0 015 0 00375 0.451 0 239 0 0130 0 02 0 005 0 490 0 269 0 0172 0 025 0 00625 0.521 0 296 0 0214 0 03 0 0075 0.546 0 321 0 0256 0 005 0 00125 0 309 0 160 0 0045 0 01 0 0025 0 398 0 198 0 0087 = 0 10 0 015 0 00375 0 454 0 232 0 0126 d 0 02 0 005 0 495 0 260 0 0170 0 025 0 00625 0 525 0 285 0 0206 0 03 0 0075 0 551 0 308 0 0261 0 005 0 00125 0 312 0 148 0 0044 0 01 0 0025 0 402 0 194 0 0086 15 0 015 0 00375 0 458 0 226 0 0127 d 0 02 0 005 0 499 0 253 0 0167 0 025 0 00625 0 530 0 275 0 0207 0 03 0 0075 0 555 0 296 0 0247 0 005 0 00125 0 314 0 146 0 0044 0 01 0 0025 0 404 0 190 0 0086 20 0 015 0 00375 0 462 0 221 0 0126 0 02 0 005 0 503 0 245 0 0165 0 025 0 00625 0 534 0 266 0 0204 0 03 0 0075 0 560 0 285 0 0243 0 005 0 00125 0 316 0 144 0 0045 0 01 0 0025 0 408 0 187 0 0086 25 0 015 0 00375 0 465 0 216 0 0125 d 0 02 0 005 0 507 0 239 0 0164 0 025 0 00625 0.639 0 259 0 0202 0 03 0 00725 0 665 0 276 0 0240 p' = 0.5 p = 0.05 = 0.10 = 0.15 = 0.20 = 0.25 V P' k L K 0 005 0 0025 0 296 0 i 163 0 0046 0 01 0 005 0 373 0 225!o 0090 0 015 0 0075 0 421 0 275 0 0134 0 02 0 01 0 454 0 320 0 0178 0 025 0 0125 0 479 0 361 0 0222 0 03 0 016 0 499 0 400 0 0266 0 005 0 0026 0 300 0 158 0 0045 0 01 0 006 0 381 0 216 0 0088 0 015 0 0075 0 428 0 261 0 0131 0 02 0 01 0 462 0 302 0 0174 0 025 0 0125 0 488 0 338 0 0215 0 03 0 016 0 509 0 374 0 0258 '0 005 0 0025 0 305 0 153 0 0044 0 01 0 006 0 386 0 207 0 0087 0 015 0 0075 0 436 0 249 0 0128 0 02 0 01 0 471 0 285 0 0169 0 025 0 0126 0 498 0 319 0 0210 0 03 0 015 0 518 0 350 0 0251 0 005 0 0025 0 309 0 149 0 0044 0 01 0 006 0 392 0 199 0 0086 0 016 0 0076 0 442 0 238 0 0126 0 02 0 01 0 479 0 271 0 0166 0 026 0 0125 0 606 0 301 0 0206 0 03 0 015 0 527 0 329 0 0245 fo 006 0 0026 0 314 0 146 0 0044 0 01 0 005 0 398 0 194 0 0085 0 016 0 0075 0 450 0 229 0 0125 0 02 0 01 0 486 0 258 0 0164 0 026 0 0125 0 514 0 287 0 0202 0 03 0 015 0 537 0 311 0 0240 SLAB, BEAM, AND COLUMN TABLES 209 TABLE 11, PART 2.— USE FOR RECTANGULAR BEAMS WITH STEEL IN TOP AND BOTTOM p P' k L K P P' A; L K ■ 0 005 0 005 0.274 0 183 0 .0046 ■ 0 005 0 0075 0 256 0 203 0046 0 01 0 01 0.336 0 271 0 .0092 0 01 0 .015 0 305 0 316 .0093 d' = 0 05 0 015 0 015 0.372 0 348 0 .0138 d' 05 0 015 0 0225 0 332 0 420 0140 d 0 02 0 02 0.395 0 420 0 0184 d = 0 0 02 0 03 0 349 0 521 0 0186 0 025 0 025 0.412 0 491 0 0230 0 025 0 0375 0 361 0 619 0 0232 [o 03 0 03 0.425 0 561 0 0275 , 0 03 0 045 0 369 0 716 0 0280 0 005 0 005 0.284 0 172 0 0045 ■ 0 005 0 0075 0 268 0 ise'o 0045 0 01 0 01 0.349 0 250 0 0089 0 01 0 015 0 322 0 284 10 0090 d' = 0 10 0 015 0 015 0.386 0 318 0 0133 d' = 0 10 0 015 0 0225 0 351 0 374 0 0134 d 0 02 0 02 0.410 0 381 0 0177 d 0 02 0 03 0 369 0 458 0 0178 0 025 0 025 0.428 0 442 0 0221 0 025 0 0375 0 382 0 541 0 0222 , 0 03 0 03 0.442 0 503 0 0265 , 0 03 0 045 0 392 0 623 0 0267 0 005 0 005 0.292 0 163 0 0044 0 005 0 0075 0 280 0 171 0 0045 0 01 0 01 0.360 0 233 0 0087 0 01 0 015 0 338 0 256 0 0087 d' = 0 15 0 015 0 015 0.399 0 292 0 0129 d' = 0 0 015 0 0225 0 369 0 333 0 0129 d 0 02 0 02 0 425 0 347 0 0171 d 15 0 02 0 03 0 389 0 404 0 0171 0 025 0 025 0.444 0 400 0 0213 0 025 0 0375 0 403 0 475 0 0213 ^ 0 03 0 03 0.458 0 451 0 0255 , 0 03 0 045 0 414 0 544 0 0258 0 005 0 005 0.299 0 154 0 0044 0 005 0 0075 0 292 0 160 0 0044 0 01 0 01 0.371 0 218 0 0086 0 01 0 015 0 353 0 2.34 0 0086 d' = 0 20 0 015 0 015 0.411 0 270 0 0126 d' = 0 20 0 015 0 0225 0 386 0 298 0 0126 d 0 02 0 02 0.439 0 318 0 0166 ~d 0 02 0 03 0 409 0 360 0 0168 0 025 0 025 0.460 0 364 0 0206 0 025 0 0375 0 424 0 420 0 0206 , 0 03 0 03 0.475 0 408 0 0246 , 0 03 0 045 0.436 0 478 0 0248 0 005 0 005 0.308 0 149 0 0045 0 005 0 0075 0.304 0 152 0 0044 0 01 0 01 0.382 0 206 0 0085 0 01 0 015 0 369 0 216 0 0084 d' = 0 25' 0 015 0 015 0.425 0 252 0 0124 d' = 0 25 0 015 0 0225 0 404 0 271 0 0123 J 0 02 0 02 0.454 0 294 0 0162 d 0 02 0 03 0.428 0 325 0 0161 0 025 0 025 0.475 0 333 0 0200 0 025 0 0375 0 444 0 373 0 0199 0 03 0 03 0.491 0 371 0 0238 0 03 0 045 0 457 0 423 0 0237 p' = 1 .5 p 210 REINFORCED CONCRETE CONSTRUCTION TABLE 12.— USE FOR COLUMNS -ll = 1 -f (tj, — l)p P = total strength of reinforced column, for stress fc. P' P' = total strength of plain column, for stress /c. p n=10 n=12 n= 15 n = 20 Values of — P' 0.005 1 045 1 055 1 070 1 095 0.006 1 054 1 066 1 084 1 114 0.007 1 063 1 077 098 1 133 0.008 1 072 1 088 1 112 1 152 0.009 1 081 1 099 1 126 1 171 0.010 1 090 1 110 140 1 190 0.011 1 099 1 121 1 154 1 209 0.012 1 108 1 132 1 168 1 228 0.013 1 117 1 143 182 247 0.014 1 126 1 154 1 196 1 266 0.015 1 135 1 165 1 210 1 285 0.016 1 144 1 176 224 \ 304 0.017 1 153 1 187 1 238 1 323 0.018 1 162 1 198 1 252 1 342 C.019 1 171 1 209 1 266 361 0.020 1 180 1 220 1 280 1 380 0.021 1 189 1 231 1 294 1 399 0.022 1 198 1 242 1 308 1 418 0.023 1 207 1 253 1 322 1 437 0.024 1 216 1 264' 1 336 1 456 0.025 1 225 1 275 1 350 1 475 0.026 1 234 1 286 1 364 1 494 0.027 1 243 1 297 1 378 1 513 0.028 1 252 1 308 1 392 1 532 0.029 1 261 1 319 1 406 1 551 0.030 1 270 1 330 1 420 , 1 570 0.031 1 279 1 341 1 434 ' 1 589 0.032 1 288 1 352 1 448 1 608 0.033 1 297 1 363 1 462 1 627 0.034 1 306 1 374 1 476 1 646 0.035 1 315 1 385 1 490 1 665 0.036 1 324 1 396 1 504 1 684 0.037 1 333 1 407 1 518 1 703 0.038 1 342 1 418 1 532 1 722 0.039 1 351 1 429 1 546 1 741 0.040 1 360 1 440 1 560 1 760 SLAB, BEAM, AND COLUMN TABLES 211 TABLE 13.— NUMBER OF RODS AND SECTIONAL AREA IN SQUARE INCHES FOR BEAM AND COLUMN REINFORCEMENT Size of rod o Q g g 10 11 12 13 14 15 16 r \ round square 0.78 1.00 0.98 1.25 1.18 1.50 1.37 1.75 1.57 2.00 1.77 2.25 1.96 2.50 2.16 2.75 2.36 3.00 2.55 3.25 2.75 3.50 2.94 3.75 3.14 4.00 round square 0.99 1.27 1.24 1.58 1.49 1.90 1.74 2.21 1.99 2.53 2.24 2.85 2.48 3.16 2.73 3.48 2.98 3.80 3.23 4.11 3.48 4.43 3.73 4.75 3.98 5.06 round square 1.23 1.56 1.53 1.95 1.84 2.34 2.15 2.73 2.45 3.12 2.76 3.52 3.07 3.91 3.37 4.30 3.68 4.69 3.99 5.08 4.30 5.47 4.60 5.86 4.91 6.25 round square 1.48 1.89 1.86 2.36 2.23 2.84 2.60 3.31 2.97 3.78 3.34 4.25 3.71 4.73 4.08 5.20 4.45 5.67 4.83 6.15 5.20 6.62 5.57 7.09 5.94 7.56 ,„ J round \ square 1.77 2.25 2.21 2.81 2.65 3.38 3.09 3.94 3.53 4.50 3.98 5.06 4.42 5.62 4.86 6.19 5.30 6.75 5.74 7.31 6.19 7.88 6.63 8.44 7.07 9.00 U" \ round square 2.07 2.64 2.59 3.30 3.11 3.96 3.63 4.62 4.15 5.28 4.67 5.94 5.18 6.60 5.70 7.26 6.22 7.92 6.74 8.58 7.26 9.24 7.78 9.90 8.30 10.56 t" ] round square 2.41 3.06 3.01 3.83 3.61 4.59 4.21 5.36 4.81 6.12 5.41 6.89 6.01 7.66 6.61 8.42 7.22 9.19 7.82 9.95 8.42 10.72 9.02 11.48 9.62 12.25 If, It / round ^° \ square 2.76 3.52 3.45 4.39 4.14 5.27 4.83 6.15 5.52 7.03 6.21 7.91 6.90 8.79 7.59 9.67 8.28 10.55 8.97 11.43 9.66 12.30 10.35 13.18 11.04 14.06 „ / round \ square 3.14 4.00 3.93 5.00 4.71 6.00 5.50 7.00 6.28 8.00 7.07 9.00 7.85 10.00 8.64 11.00 9.43 12.00 10.21 13.00 11.00 14.00 11.78 15.00 12.57 16.00 round square 3.98 5.06 4.97 6.33 5.96 7.59 6.96 8.86 7.95 10.12 8.95 11.39 9.94 12.66 10.94 13.92 11.93 15.19 12.92 16.45 13.92 17.72 14.91 18.98 15.90 20.25 u» i round square 4.91 6.25 6.14 7.81 7.36 9.37 8.59 10.94 9.82 12.50 11.04 14.06 12.27 15.62 13.50 17.19 14.73 18.75 15.95 20.31 17.18 21.87 18.41 23.44 19.64 25.00 round square 5.94 7.56 7.42 9.45 8.91 11.34 10.39 13.23 11.88 15.12 13.36 17.01 14.85 18.91 16.33 20.80 17.82 22.69 19.30 24.58 20.79 26.47 22.27 28.36 23.76 30.25 U" \ round square 7.07 9.00 8.84 11.25 10.60 13.50 12.37 15.75 14.14 18.00 15.90 20.25 17.67 22.50 19.44 24.75 21.21 27. OC 22.97 29.25 24.74 31.50 26.51 33.75 28.27 36.00 .„ j round \ square 8.30 10.56 10.37 13.20 12.44 15.84 14.52 18.48 16.59 21.12 18.67 23.77 20.74 26.41 22.81 29.05 24.89 31.69 26.96 34.33 29.03 36.97 31.11 39.61 33.18 42.25 ^„ / round " 1^ square 9.62 12.25 12.03 15.31 14.43 18.37 16.84 21.44 19.24 24.50 21.65 27.56 24.05 30.62 26.46 33.69 28.86 36.75 31 .27 39.81 33.67 42.87 36.08 45.94 38.48 49.00 . ^ „ / round ' ^ square 11.04 14.06 13.81 17.58 16.57 21.09 19.33 24.61 22.09 28.12 24.85 31.64 27.61 35.16 30.37 38.67 33.13 42.19 35.90 45.70 38.66 49.22 41.42 52.73 44.18 56.25 2" < round square 12.57 16.00 15.71 20.00 18.85 24.00 21.99 28.00 25.13 32.00 28.27 36.00 31.42 40.00 34.56 44.00 37.70 48.00 40.84 52.00 43.98 56.00 47.12 60.00 50.27 64.00 2i" J round square 15.90 20.25 19.88 25.31 23.86 30.38 27.83 35.44 31.81 40.50 35.78 45.56 39.76 50.62 43.74 55.69 47.71 60.75 51.69 65.81 55.67 70.87 59.64 75.94 63.62 81.00 2i" round square 19.63 25.00 24.54 31.25 29.45 37.50 34.36 43.75 39.27 50.00 44.18 56.25 49.09 63.50 54.00 69.75 58.90 75.00 63.81 81.25 68.72 87.50 73.63 78.54 93.75|100.00 16 212 REINFORCED CONCRETE CONSTRUCTION TABLE 14.— HOOPED COLUMN REINFORCEMENT Diameter of en- Pitch Sectional area of Length of hooping in closed concrete^ (inches) hooping (square 1 ft. in height (inches) inches) (inches) 6 H 0.0187 181 li max. 0 .0225 151 7 U 0 .0262 176 1| max. 0 .0306 151 8 li 0 .0350 172 2 max. 0 .0400 151 9 2 0 .0450 170 2i max. 0 0506 151 10 2i 0 0562 167 2i max. 0 0625 151 11 0.0687 166 2| max. 0 0756 151 12 2} 0 0825 165 3 max. 0 0900 151 13 3 0.0975 163 3i max. 0 1056 151 14 3 0 1050 176 3i max. 0 1225 151 15 3i 0 1219 174 3| max. 0 1406 151 16 3i 0 1400 172 4 max. 0 1600 151 17 3i 0 1594 171 4J max. 0 1806 151 18 4 0 1800 170 4i max. 0 2025 151 19 4i 0 2019 169 4i max. 0 2256 151 20 4i 0 2250 168 5 max. 0 2500 151 21 4i 0 2362 176 5i max. 0 2756 151 22 4i 0 2612 175 5i max. 0.3025 151 23 5 0 2875 173 5i max. 0 3306 151 24 5i 0 3150 172 6 max. 0 3600 151 25 - 0 3437 172 6i max. 0 3906 151 26 5i 0 3737 170 6i max. 0 4225 151 27 6 0 4050 170 6i max. 0 4556 151 28 6i 0 4375 169 7 max. 0 4900 151 29 6i 0. 4531 175 7J max. 0. 5256 151 30 6i 0. 4875 174 7i max. 0. 5625 151 SLAB, BEAM, AND COLUMN TABLES 213 TABLE 15.— MAXIMUM DIAMETER OF ROUND OR SQUARE STIRRUPS (From Taylor and Thompson's "Concrete, Plain and Reinforced" i) J a 11 Table gives values of 2.4 ^ for different values of tension and bond. J 8 Allowable unit Allowable unit tension in stirrups (fa) bond stress (m) (lb. per sq. in.) Lb. per sq. in. 12,000 14,000 15,000 16,000 20,000 80 0.016 0.014 0.013 0.012 0.010 100 0.020 0.017 0.016 0.015 0.012 120 0.024 0.020 0.019 0.018 0.014 150 0.030 0.026 0.024 0.022 0.018 TABLE 16.— MINIMUM LENGTH OF EMBEDMENT OP INCLINED RODS (From Taylor and Thompson's "Concrete, Plain and Reinforced"^) (Round or square) f Z' =~ diameters. Table gives values of for different values of tension and bond. Allowable unit Allowable unit stress in inclined rod (/"g) bond stress (u) (lb. per sq. in.) Lb. per sq. in. 12,000 14,000 15,000 16,000 20,000 80 37 44 47 50 62 100 30 35 38 40 50 120 25 29 31 33 41 150 20 23 25 27 33 Transverse Spacing of Reinforcement. Least width of beam should be the greater of the two values determined from the following formulas: b = [2.5(n-l)+4:]d^ b = ag(n + l) +ndi in which b = least width of beam. = thickness of the rods. n = maximum number of rods which occurs in a horizontal layer. ttg = maximum size of aggregate in inches. ' From Taylor and Thompson's "Concrete, Plain and Reinforced," 2nd edition, page 454. Copyright, 1905, 1909, by Frederick W. Taylor. 214 REINFORCED CONCRETE CONSTRUCTION Depth of Concrete Below Rods. Slabs Depth to steel (d) Depth below center of steel Bl in. and under | in. Between 3J in. and 4| in. 1 in. 4| in. and over li in. Beams and Girders Depth to steel (d) Depth in the clear below steel 10 in. and under 1 in. Between 10 in. and 20 in. IJ in. 20 in. and over. 2 in. Formulas for Shear. _V_ _7 b]d '"''~hd Formula for Bond of Tension Rods. V lojd Formulas for Vertical Stirrups. s = —2V~ " ^/Jrf ^1 ^ 2 ^ (uniform loadmg only) u ' i= or < 2.4-c^ Formulas for Inclined Rods. Z' = ~*^dianieters or < ^(^l — ^^^j (uniform loading only) (^s = §-7-^^(45 degrees incHnation) js]d CHAPTER VIII SLAB, BEAM, AND COLUMN DIAGRAMS 73. Illustrative Problems. — In this article the same illustrative problems will be worked out as in Art, 72. It is thought by so doing a good comparison can be made between tables and dia- grams as to their advantages and limitations. The working stresses recommended by the Joint Committee will be employed throughout, 1. Design a beam to span 40 ft. and to support 600 lb. per foot (includ- ing weight of beam). Beam is assumed to be simply supported. In Diagram 1, the intersection of the curves /c = 650 and/s = 16,000 is first found. Tracing down, p is found to be 0.0077, and tracing horizontally K r-.. is found to be 107.3 ^^_«|_(600)M!M_i,440,000 m.-lb. 6 = 18 in. and d = 27 1/2 in. will be satisfactory. Area of cross-section, 6d = (18)(27.5)=495 sq. in. as = (495) (0.0077) =3.81 sq. in. We shall select four 1 1/8-in. round rods = 3.98 sq. in. Diagram 1 may be also employed to determine the safe resisting moment of a given beam and the greatest unit stresses in the steel and concrete due to a given bending moment. To determine the safe resisting moment of a given beam, the value of p should be computed. After finding this value on the lower margin, trace vertically, stopping at the first of the two curves /c = 650 and /» = 16,000 (assuming the allowable stresses as recommended by the Joint Committee). Now trace horizontally to either side margin and the value of K is found. Then, M = Kbd^. Consider a beam of the above dimensions to have 1 per cent of steel. Tracing vertically from this value on the lower margin, the 650 is the first curve to be reached and at a value of if = 117.0 M == (117)(18)(27.5)2 = 1,593,000 in.-lb. To determine the greatest unit stresses in the steel and concrete of a given beam due to a given bending moment, the value of p should be computed M as before. Also, K should be computed from the formula K = With these values of p and K, find the intersection of the vertical and horizontal lines through these values respectively, and from the adjacent steel and con- crete curves the values of fc and /g may be estimated. Consider a beam of 17 215 216 REINFORCED CONCRETE CONSTRUCTION the above dimensions and with 0.7 per cent of steel, to be subjected to a . „ 1,200,000 „ m, • i bending moment of 1,200,000 m.-lb., or X = ^j|y^^--g^ = 88.2 Themter- section of the vertical and horizontal hnes through these values respectively, gives /c = 550 and/s = 14,400. This procedure is followed in reviewing beam design. Diagram 1 may be also employed to find minimum allowable depth of beam for a given percentage of steel and various assumed widths, also to find the amount of steel for a beam with given loading. The preceding discussion and the discussion under Problem 1 of the preceding article should make clear the method of procedure. 2. Design a beam to span 10 ft. and to support a load of 4900 lb. per foot. Beam is assumed to be simply supported. We shall use Diagram 7 in the design of this beam. The weight of beam will be assumed at 400 lb. per foot. ^^^^^^(5300)(10)!(12) ^795 000 in.-lb. ^. 8 8 Assuming a width of beam of 14 in., the bending moment to use for one inch in width is 795,000 _ „„„ . — -^^ — =56,800 m.-lb. 14 Selecting this value on the left-hand margin of Diagram 7, Part 3, and follow- ing the horizontal hne to the right, a depth (d) of 23 in. will give maximum efficiency. This is shown by the fact that the horizontal hne mentioned above meets a hne half way between d = 22 in. and d = 24 in. much closer to the break in the curves than it does any other line representing depth. The area of steel required, shown by the curved hnes crossing the d hnes, is found to be 0.178 sq. in. per inch width or (0.178) (14) =2.49 sq. in. for the beam. We shall select ten 9/16-in. round rods = 2.485 sq. in. The spacing of the rods at the center of beam is shown in Fig. 56. This beam as designed con- tains two rows of steel and the computed weight per foot is WMl^ = 379 lb. 144 The assumed and calculated weights do not differ materially and the beam as designed will be considered satisfactory. It should be noted that Diagram 7 may be employed for all cases cited under Problem 1 except the case of finding the greatest unit stresses in the steel and concrete due to a given bending moment. The student should have no difficulty in determining the method of procedure. Diagram 1 may also be used in the above design. 3. What safe load per square foot (including dead weight) can be sup- ported by a slab 6 in. deep (d! = 4 3/4 in.) and 10-ft. span reinforced with 1/2-in. round rods placed 8 in. apart? The slab is simply supported and reinforced in only one direction. 0.1963 „ p=r8)(4r75)=°-°^'' Referring to Diagram 6, Part 2, and tracing vertically from this value of p on the lower margin to an intersection with the curve of d = 4 3/4 in., and SLAB, BEAM, AND COLUMN DIAGRAMS 217 then tracing horizontally to the left-hand margin, a bending moment of 20,100 in.-lb. is found. Select this value of the bending moment on the left-hand margin of Dia- gram 5 and trace horizontally to the right to an intersection with a vertical hne through 10, denoting span length. The safe load, based on M = ^ can now be estimated directly by means of the curved hues and is found to be 168 lb. per square foot. ( 1 68) (0 . 80) = 1 34 1 / 2 lb . per square foot, safe load for slab simply supported. 4. Design a slab to span 6 ft. and to carry a live load of 250 lb. per square foot. Slab is to be fully continuous and reinforced in only one direction. Assume the weight of slab at 50 lb. per square foot. Total load for slab is thus 300 lb. per square foot. From Diagram 5 for this span length and load per square foot, a bending moment of 11,000 in.-lb. is found, based on — . Diagram 6, Part 1, shows that a depth (d) of 3 in. will be ample — total depth 3 3/4 in. Also, a8 = 0.275 sq. in. From Diagram 3, we may use 3/8-in. round rods spaced 43/4 in. on centers. 5. Design the center cross-section of a T-beam in a floor system; the beam is to have a span of 12 ft. and be fully continuous. Maximum shear (live plus dead) is closely equal to 12,200 lb. Maximum moment (live plus dead) = 356,300 in.-lb. Supported slab is 6 in. thick. Diagram 8 cannot be employed to solve for the resisting moment of a given beam but is useful in designing. Formula (7), Art. 59, may be put in the following form k and / in this equation are functions of/c and /,„ and hence the variables are fc, fa, and the ratio J. The curves at the left in Diagram 8 are plotted from this equation with a fixed value of fs = 16,000 lb, per square inch. Values of fc may be determined for various values of ^ and 4, or values of bd^ d hd^ may be determined for various values of fc and It must not be over- a looked, however, that this diagram will apply only when the amount of steel is such that fs = 16,000 lb. per sq. in. This amount of steel may be easily determined when the corresponding / is found from the curves at the right of the diagram. Suppose ^ = 80 and ^ = 0.2, then the intersection of horizontal and vertical lines through these values respectively shows fc to equal 600, and then tracing from this intersection horizontally to the right until the vertical line is reached indicating /c = 600 (at the right-hand side of the diagram), we find / equal to 0.91. Finally, as =^„ in which ? =0.91, fajd fs = 16,000, and M and d are known. Diagram 8 will now be employed in working out the problem stated at the beginning of this discussion. 218 REINFORCED CONCRETE CONSTRUCTION The breadth of the flange is controlled by one-fourth the span, or 36 in. Assuming a depth (d) of 16 in. M 356,300 _ bd^ (36) (16)=^ For this value of ^ and = ^^^^ ^^^^ diagram that this beam falls under Case I; that is, the neutral axis is in the flange. Diagram 1 may be used for T-beams under Case I when the problem falls within the limits of the curves. In the problem at hand, a horizontal line through the value 38.7 for K would intersect the obhque line/g = 16,000 some distance below the diagram. It should be noted that the corresponding value of fc is considerably below 400. Diagram 7, Part 2, may be used to determine the amount of steel required. The bending moment to use for one inch width of beam is — — = 9900 in.-lb. The intersection of a horizontal line through this value and the curve for d = 16 in. shows p to be 0.0028, or as = (0.0028)(36)(16) = 1.6 sq. in. This diagram also shows that the stress in the concrete is far below the allowable. 6. The flange of a T-beam is 24 in. wide and 4 in. thick. The beam is to sustain a bending moment of 480,000 in.-lb. What depth of beam and amount of steel are necessary? We will try d = 18 in. M 480,000 _ bd^ (24) (18) 2 For this value of and for -^ = ^^ = 0.222, we find from Diagram 8, /c = bd" d lo 485 lb. per square inch and j =0.910. Then, 480,000 , '^« = a6.000)(0.910)(18)=^-^^ The stress in the concrete of 485 is permissible and the beam as designed is satisfactory. Suppose 2.0 sq. in. of steel were inserted in a beam of the above dimensions, and suppose that the safe resisting moment is desired. Diagram 9 must be used for this case. - = 0.222 as before. Tracing vertically from this value on the lower margin d to a value of p = .0046 and then tracing horizontally to the left margm, we find a value of A; = 0.32 Now tracing horizontally to the right until the vertical Une is reached indicating p = .0046 (at the right-hand side of the diagram), we find / equal to 0.91 Ms -fsasjd = (16,000) (2.0) (0.91) (18) = 525,000 in.-lb. / (16.000)(0.32)_ „ •^'==;^(r^=(r5)(i-o.32y-^^^ fc is less than 650; hence, the resisting moment depends upon the steel, or Ms = 525,000 in.-lb. SLAB, BEAM, AND COLUMN DIAGRAMS 219 7. Design a T-beam with span of 40 ft. Assume dead load = 1400 lb. per foot. Live load = 3000 lb. per foot. The beam is to be simply supported at the ends and the flange is to be proportioned as well as the web ; that is, the flange does not form a part of a floor system already determined. From Art. 61, 6' = 18 in. and (i=47 in. are suitable dimensions and a thickness of flange of 12 in. is tried. The total bending moment on the beam is 10,560,000 in.-lb. i.l|=0.266 By means of Diagram 8, we find that with /c = 650 and ^ = 0.256, the value of = 99.2 Then, , 10,560,000 , Also, from the diagram, / = 0.892, Then, from Formula (7), Art. 59, 10,560,000 "(16,000) (.892) (47) "-^^-^ ^"l- The detailed design of this beam has been given at the end of Art. 61. 8. A continuous T-beam, uniformly loaded, has a bending moment at the center of each span of 358,000 in.-lb. Negative bending moment at the supports and the positive bending moment at the center of span are figured by the formula, M = ~. The tensile steel at the center of span consists of four 3/4-in. round rods, b' = 9 in. d = 15.5 in. Design the supports. Diagrams 10 and 11 have been prepared to solve problems involving the determination of stresses for rectangular beams with steel in top and bottom. The formulas used in constructing these diagrams are radically different from those previously employed and an explanation of them will now be given. Compressive reinforcement needs to be used only when the compressive concrete, if unreinforced, would be stressed too high. The question then arises of how much compressive reinforcement is needed to reduce the streso in the concrete to within the working limit. Let the following notation be used: With no compressive reinforcement fc,fs,k,j. With compressive reinforcement . . fc',fs', k', j'. Formula 2, Art. 33, may be written as follows: fc n{l-k) We also have ^ asjd • / = /sfc Mk (1) n{jL-k) jdaanil-k) Referring to Formula 1, Art. 62, and knowing that the fiber stress in the tensile steel of double-reinforced beams may be expressed by/s=^^^ 220 REINFORCED CONCRETE CONSTRUCTION (using the proper value of /) the same as in rectangular and T-beams, we obtain r ■ « and/' . _ f'J"' ^ Mk' (2) • " n(l-k') i'daMl-k') From these equations (1) and (2), the relative reduction in fc due to the addition of compressive steel is found to be /c-A _i / ^ (3) /c 2'' k' l-k' Since / and k depend on and j' and k' on p', equation (3) may be employed to show the relative reduction of fc for different percentages of tensile and compressive steel due to the addition of the steel in the compression side of the beam. Diagram 10 gives concrete curves for various values of ^ • It was found when constructing the diagram that, in some instances, one concrete curve may be employed with sufficient accuracy to represent two and sometimes three percentages of tensile steel. Adding compressive steel also reduces the stress in the tensile steel. Using the same notation as above The tensile steel curves, Diagram 11, give this relative reduction for different percentages of tensile and compressive steel. Diagrams 10 and 11 vnll now be employed in working out the problem stated at the beginning of this discussion. Two of the tensile rods on each side of the supports will be bent up and made to lap over the top of the supports, while the other two rods on each side will be continued straight and lapped over supports at the bottom of beam. The ratios of steel in tension and compression are the same, and are respectively: d' We will assume -; =0.1 as before. d The stress in the steel and concrete at the support must first be found with no compressive reinforcement. Diagram 1 may be employed when the values sought come within the limits of the curves. In this problem it will be necessary to employ Diagram 2 and do some computing. For p = 0.013, / = 0.847 and A; = 0.452 Then 358,000 1 K iu ' \. ^-° (0.013)(0.847)(9)(15.5)^ ^^'-Q'Q ^^^"^^ ^^^(2)a5,050K0.013)^3eg ^^^^^^ SLAB, BEAM, AND COLUMN DIAGRAMS 221 The question now arises, — is the stress in the concrete brought down to a value 750 lb. per square inch (or less) by the introduction of 1.3 per cent of compressive steel. Also, the corresponding value of fg should usually be determined. d' Using Diagram 10 for -^ = 0.10 and for p' = 0.013 and p = 0.013, the relative reduction in the stress in the concrete is found to be 33.3 per cent, or the resulting stress equals 865 — (.333) (865) =577 lb. , per square inch. Using Diagram 11, the relative reduction in the stress in the tensile steel equals 4.4 per cent; that is, the maximum tension in the steel is 14,390 lb. per square inch. The stresses in the concrete and steel are within the allowable and no haunch or additional steel are necessary. 9. The effective area of a column is 144 sq. in. ; load to be carried is 80,000 lb.; and working stress on the concrete is 450 lb. per square inch. What percentage of longitudinal bars without hooping will be required? Take n = 15. The safe strength of a plain concrete column would be 144X450 = 64,800 lb. Hence, ^=-^-1235 P' 64.8"^-^'^^ From Diagram 12, for n = 15, and p = 0.017 ^ = 1.238 Thus, 1.7 per cent of steel is required, and = (144) (0.017) =2.45 sq. in. Problems Unit stresses recommended by Joint Committee are to be used in all the following problems. Solve by using Diagrams. 75. Determine the cross-section and number of 1-in. round rods required for a rectangular girder to span 20 ft. and to sustain a live load of 2000 lb. per foot. The bending moment is to be figured by the formula M= Design this girder by both Diagrams 1 and 7. 76. Design a slab to span 4.5 ft. and to carry a live load of 300 lb. per square foot. Slab is to be fully continuous and reinforced in only one direction. 77. Design the center cross-section of a T-beam, fully continuous, having a span of 16 ft. Maximum shear (not including dead weight of stem) is 13,000 lb. Maximum moment (not including dead weight of stem) = 628,000 in.-lb. Supported slab is 5 in. thick. The depth of beam (d) is fixed at 18 in. 78. Solve Problem 77 considering the slab 3 3/4 in. thick. 79. A continuous T-beam, uniformly loaded, has a bending moment at the center of span of 1,400,000 in.-lb. Negative bending moment at the 222 REINFORCED CONCRETE CONSTRUCTION supports and the positive bending moment at the center of span are figured by the formula — • The tensile steel at the center of span consists of four 1-in. round bars. 6' = 10 in. d = 25 in. Design the supports assuming ^ =0.15. Assume two of the rods to be bent up and lap over the top of the supports. Consider the other two rods to lap at the bottom of beam. SLAB, BEAM, AND COLUMN DIAGRAMS 223 • DIAGRAM 1 • USE FOR DESIGN OF RECTANGULAR BEAMS Based on n = i5 Percentage Reinforcemeni- 224 REINFORCED CONCRETE CONSTRUCTION SLAB, BEAM, AND COLUMN DIAGRAMS 225 • DIAGRAM 3 • SPACING OF ROUND RODS IN SLABS 0 .Z .4 .6 .8 1.0 1.2 1.4 1.6 1.8 2.0 .2 4 .6 .8 1.0 1.2 1.4 1.6 1.8 2.0 SecTional Area of Sreel per foot of Slab 226 REINFORCED CONCRETE CONSTRUCTION • DIAGRAM 4 • SPACING or SQUARE RODS IN SLABS 2 4 6 8 1.0 1.2 14 16 1.8 2.0 Sectional Area of Steel per -foot of Slab in Inches 228 REINFORCED CONCRETE CONSTRUCTION • DIAGRAM 6- PART 1 • USE FOR DESIGNING $i.AB5 Based on -f^=i6ooo; -f^ = 650; n = I5 SLAB, BEAM, AND COLUMN DIAGRAMS 229 • DIAGRAM 6- PART 2 • USE FOR DESIGNING SLABS. Based onfs=lGOOO. •f^=S50; n = 15 •020 .022 .5% 1.0% 1,5% 2.0% [p = .005] [p = ,OI0j [p = .OI5] [p = .02O] Percentage Reinforcement 230 REINFORCED CONCRETE CONSTRUCTION ' DIAGRAM 7- PART 1 USE FOR RECTANGULAR BEAMS DEPTH AND AREA OF STEEL FOR BEAMS ONE INCH IN WIDTH Based on 1^=16000; -1^=650, n = i5 0 .002 .004. .OOS .008 .010 .012 .014 .016 .018 .020 .022 .5% 1.0% 15% 2.0% [p=005] [p=,0i0] [p-015] [p = .020] Percentage Reinforcement SLAB, BEAM, AND COLUMN DIAGRAMS 231 DIAGRAM 7- PART 2 USE. FOR RECTANGULAR BEAMS DEPTH AND AREA OF STEEL FOR BEAMS ONE INCH IN WIDTH Based on 16000; •f^;=650 , n = l5 0 .002 .004- .006 .008 .010 . 012 .014- 016 . 018 020 .022 0 5% 10% 15% 20% [p= 005j [p=OIO] [p = -OI5] [p = 020 Percentage Reinforcement 18 232 REINFORCED CONCRETE CONSTRUCTION DIAGRAM 7 PART 3 • USE FOR RECTANGULAR BEAMS DEPTH AND AREA OF STEEL FOR BEAMS ONE INCH IN WIDTH Based on -f^ = l6000. "fb = 650 , n=l5 .002 .004 .006 008 .010 .012 .014 .016 .018 .020 022 .5% \.0% 1.5% 2.07» [p=005] [p= 010] [p=015] [p=.020i PercenTage ReinforcemenT SLAB, BEAM, AND COLUMN DIAGRAMS 233 234 REINFORCED CONCRETE CONSTRUCTION SLAB, BEAM, AND COLUMN DIAGRAMS 235 ..5% 1.0% 1.5% Percentage cf Compressive Steel z.0% 236 REINFORCED CONCRETE CONSTRUCTION • DIAGRAM 1 1 • USE FOR RECTANGULAR BEAM5 WITH STEEL IN TOP AND BOTTOM TENSILE STEEL CURVES Bosed on n=ife .5% 1.0% 1.5% 2.0% Percentage of Compressive Steel SLAB, BEAM, AND COLUMN DIAGRAMS 237 238 REINFORCED CONCRETE CONSTRUCTION ' DIAGRAM 13 • • BENDING AND DIRECT STRESS • CASE I • COMPRESSION OVER WHOLE SECTION 0 0 02 0.04 006 0 08 010 0.12 014 016 018 020 0.22 Values of — » 1 From Taylor and Thompson's, "Concrete, Plain and Reinforced," 2nd edition, page 569. SLAB, BEAM, AND COLUMN DIAGRAMS 23.9 240 REINFORCED CONCRETE CONSTRUCTION CHAPTER IX BENDING AND DIRECT STRESS 74. Theory in General. — Consider any given section of a beam or column and locate its center of gravity. Now, if the resultant of the forces on one side of the section does not pass through this center of gravity, bending will result, and the distribution of stress normal to the section will not be uniform. Bending may result from lateral pressure; or it may be due to the eccentric application of a force in the direction of the axis; or < it may be due to the two combined. If the structure considered is a beam and is acted upon by forces which are all normal to its length, then the stresses resulting are due to simple bending and the formulas already deduced may be employed. If, however, any of the forces acting throughout the length of a beam be inclined, or if additional forces be applied at the ends, then our beam formulas for simple bending will not apply. Likewise, in columns, if the load be eccentrically applied or if lateral pressure be exerted, both bending and direct stresses will result and the ordinary column formulas cannot be used. The same combination of stresses occurs also in arch rings and may occur in special cases. The formulas to be derived can be employed in any type of reinforced concrete structure provided the normal component of the resultant thrust on the given section acts with a lever arm about the center of gravity of the section. In long beams and columns, the deflection resulting from flexure should be given consideration when determining the eccentricity of the axial and inclined forces. Let us first consider structures of homogeneous material, as structures of plain concrete. The distribution of pressure on any section due to a resultant pressure acting at different points will be explained. Consider a section represented in projection by EF, Fig. 84. When the resultant, R, acts at the center of gravity, 0, the intensity of stress is uniform over the section and is equal to the vertical component of R divided by the area W . . of section, or -j- If R acts at any other pomt, as A^, and if 241 242 REINFORCED CONCRETE CONSTRUCTION the projection of the section is taken such that the distance Xo represents the true lever arm of W about the center of gravity, then the force W is equivalent to an equal W at 0 and a couple whose moment is Wxq. The intensity of the uniformly varying stress due to this bending moment at a distance x from 0 is ^VxqX (by the common flexure formula for homogeneous beams) — in which / is the moment of inertia of the section about an axis through 0 at right angles to the plane of the paper. At the edges E and F this intensity = - Regarding compressive and tensile stresses as positive and negative respectively, the intensity of stress at edge E is At edge F it is , W Wxox^ _W WxqX^ J'- A I If the stress fc comes out minus, the value obtained is the maximum tension as shown in Fig. 85. In plain concrete con- struction a greater tension than about 50 lb. per square inch should not be allowed or else cracks may be expected on the tension side. When we come to reinforced concrete, which is composed of two materials (concrete and steel) with different values of E, BENDING AND DIRECT STEESS 243 then the steel area at any given cross-section may be replaced by an area of concrete equal to n times the area of the steel, placed in the plane of the steel reinforcement. This section may be called the transformed section, or section of concrete < J > Diagram 13 gives K = 1.70 and shows that the problem falls under Case I. Then by formula (6) ^ ^ ^ ^^^'(9)^20) '^"^ ^ ^'^^ square inch. Illustrative Problem. — Change the eccentricity of the preceding problem to 6 in. and solve. = 1 = 0.30 For p„= 0.0087 and y = 0.30, Diagram 13 shows that is too great for the problem to come under Case I. The method of procedure for Case II must then be followed. Diagram 14 gives A; = 0.73 for the values of pg and ^ given above. With yt = 0.73 and po = 0.0087, Diagram 15 shows L to be 0.122. Solving equa- tion (11) . M (60,000) (6) „„„„ . , •^'== L6^ = (022) (9)W^^^^ "'1"'''' Using the formula (8) gives fs-nf, (^^-1^ =(15) (820) (o;7^-2o-l) =2830 lb. per square inch. The stress f's may be found by formula 7 but is always less than n X fc- Illustrative Problem. — An arch is 20 in. deep and is reinforced with 3 rods 3/4 in. in diameter to each foot of width, both above and below. If the rods are embedded to a depth of 2 in. and the normal component of the resultant thrust on a section is 100,000 lb. for 1-ft. width of arch with an eccentricity of 3.4 in., determine the maximum intensity of compressive stress on the concrete. Assume n = 15. ^(6)(0,4A18)^ (12) (20) Diagram 13 gives if = 1.63 and the problem comes under Case I. Then by formula (6) . WK (100,000) (1.63) . , ^"^-jr - (12) (2 0) — ^^^^ '^"^'^ Problems 80. A rectangular beam is 8 in. wide and 20 in. deep, and contains four 1 /2 in. round rods both at the top and bottom. At a given section the nor- mal component of the resultant force is 77,000 lb. and acts at a distance of 2 in. from the gravity axis. Assume n = 15. (a) Compute the maxi- mum unit compressive stress in the concrete, (b) How much steel reinforcement would be needed to make the stress in the concrete 600 lb. per square inch? 81. Change the eccentricity in part (a) of the preceding problem to 5 in. and solve. INDEX Adhesion, of concrete and steel, 27-29 working values of, 86 Bars (see steel) Beams, arrangement of, 142, 143 assumptions in common theory, 47-49 bending and direct stress in, 241-249 bond stress, 71, 74 compression reinforcement in, 157-159 continuous, ' 1 30-1 34 deflection of, 124-128 depth of concrete below rods, 109 design of a continuous beam at the supports, 159-164 diagonal tension failures of, 74, 77-78 diagrams (see Table of Contents) distribution of beam and slab load to girders, 141, 142 distribution of slab load to cross-beams, 140, 141 double reinforced, 157-159 economical proportions, rectangular beams, 128, 129 T-beams, 151, 152 flexure and direct stress in, 241-249 flexure formulas for reinforced concrete, 52-62 formulas, 112, 113 horizontal bars bent up for web remforcement, 99-103 inclined tensile stresses, 65-68 inner forces in a homogeneous beam, 39-46 methods of web reinforcement, 68-71 notation, 111, 112, 126, 127, 151, 158, 160 plain concrete, 49-52 points to bend horizontal reinforcement, 105, 106 ratio of length to depth for equal strength in moment and shear, 109, 111 rectangular, 39, 134 restrained, 129, 130 shear failures of, 74, 77, 78 shearing stresses, 63-65 shear reinforcement for, 68-71 steel in top and bottom, 157-159 stirrups (see vertical stirrups) tables (see Table of Contents) T-beams (see T-beams) tests, 74-84 transverse spacing of reinforcement, 107-109 vertical and incHned reinforcement, 90-93 vertical stirrups, 93 vertical stirrups and bent rods combined, 103-105 web reinforcement in, 68-71 working stresses, 84-89 . o^c Bending and direct stress, compression over whole section, 244, 24b tension over part of section, 246-249 theory in general, 241 Bending test for steel, 24 251 252 INDEX Bent rods, 99-103 combined with vertical stirrups, 103-105 points to bend, 105, 106 Bond, 27-29, 71-74, 86 Broken stone (see stone) Broken stone screenings, 4 Cement, general requirements, 2 Cinder concrete, fireproofing, 6 weight, 22 Coefficient of expansion, concrete, 20 steel, 24 Columns, diagrams, 237 plain concrete, 167-169 hooped reinforcement, 171-173 hooped and longitudinal reinforcement, 173 longitudinal reinforcement, 169-171 structural steel shapes, 174, 175 tables, 210, 211, 212 tests, 175-184 value of longitudinal reinforcement, 185 working stresses, 184, 185 Concrete, advantage of combining with steel, 26, 27 bond with steel, 27-29 cinder, 6, 22 compressive strength, 15-18 consistency, 6 contraction and expansion, 19, 20 dry vs. wet, 6 fireproofing qualities, 20, 21 general requirements, 1 modulus of elasticity, 32, 33 proportioning by mechanical analysis, 7-14 proportions commonly used, 1 quantities of ingredients required per cubic yard, 14, 1 shearing strength, 19 table of compressive strengths, 17 tensile strength, 18, 19 theory of proportions, 7 unit for proportioning, 6, 7 waterproofing qualities, 21, 22 weight, 22 Continuous beam, design at supports, 159, 164 moments in, 130-134 Cross-beams (see beams) Deflection of beams, 124-128 Deformed bars, 22, 23, 27 Diagonal tension, 65-68 (see also shear) Diagrams (see Table of Contents) Double-reinforced beams, 157-159 Economical proportions of beams, rectangular, 128, 129 T-beams, 151, 152 Fireproofing qualities, 20, 21 Flexure and direct stress, 241-249 Formulas, bond, 73 columns, 169 continuous beams for moment, 131, 132 INDEX 253 deflection, 126, 127 economical depth of T-beams, 151 flexure, ultimate loads, 58-61 working loads, 53-56 homogeneous beams, 39-41 incUned rods, 102, 104-106 rectangular reinforced concrete beams, 112, 113 shear, 64 T-beams, 145, 146 transverse spacing of reinforcement, 107 vertical stirrups, 93, 94, 95, 98, 104 Girders (see beams) Gravel, 5, 6 screenings, 4 Hooks, 29 Horizontal bars bent up for web remforcement, 99, 103 Inclined tensile stresses, 65-68 ■ ■ Inner forces in a homogeneous beam, 39-46 Mechanical analysis, 7-14 Modulus of elasticity, defined, 29 method of determining, 29-33 Notation, 111, 112, 126, 127, 151, 158, 160 Plain concrete beams, 49-52 Properties of the material, 1-38 _ . Proportioning concrete, by mechanical analysis, 7-14 theory of, 7 unit for, 6 Ratio of the moduli of elasticity, 29-34 Rectangular beams (see beams) , . . . . j . i Reinforced concrete, advantages in the combination of concrete and steel, 26, 27 arrangement of beams and girders, 142, 143 beams, 26, 34, 39-166 beams with steel in top and bottom, 59-159 behavior under tension, 35, 36 bending and direct stress, 241-249 bond between concrete and steel, 27-29 columns, 26, 27, 34, 167-186, 210, 211, 212, 237 compared with terra-cotta for fireproofing, 21 diagrams, 223-240 double-reinforced beams, 59-159 fireproofing qualities, 20, 21 ratio of the moduli of elasticity, 29-34 rectangular beams, 39-134 repetition of stress, 37 shrinkage and temperature stresses, 36, 37 slabs, 135-140, 197-199, 205-229 tables, 193-213 T-beams, 143-146, 202-207, 233, 234 waterproofing qualities, 21, 22 Reinforcement (see steel) Repetition of stress, 37 Restrained beams, 129, 130 254 INDEX Rods (see steel) Sand, broken stone screenings, 4 clay in, 4 gravel screenings, 4 loam in, 4 requirements, 2 sharp, 2 sieve analysis of, 3, 4 size of grains, 2, 3 specifications, 4 with rounded grains, 2 Shear, in homogeneous beams, 39-41 in reinforced concrete beams, 63, 65, 75, 86, (see also diagonal tension) Shrinkage and temperature stresses, 36, 37 Slabs, 135-140 Steel, advantage of combining with concrete, 26, 27 bending test, 24 bond with concrete, 27-29 coefficient of expansion, 24 concrete as a protector against corrosion, 21, 22 deformed bars, 22, 23, 27 general requirements, 22 losis of strength from overheating, 20, 26 medium vs. high, 22-24 modulus of elasticity, 24, 29-31 open-hearth vs. Bessemer, 24 tensile strength, 24 transverse spacing of reinforcement, 107-109 wire, 22, 25 Stirrups, 93-98 combined with bent rods, 103-105 Stone, crushed, 4 gravel, 5, 6 maximum size of, 5 quality required, 4, 5 , sieve analysis of, 5 Tables (see Table of Contents) Tests, beam, 74-84, 146, 157 column, 175-183 T-beams, conditions met with in design, 152, 153 design of, 143-146 diagrams, 233, 234 economical proportions of, 151, 152 tables, 202-207 tests, 146 Vertical and inclined reinforcement, 90-93 Vertical stirrups (see stirrups) ^ Waterproofing qualities, 20, 21 Web, reinforcement, 68-71 (see also stirrups and bent rods) Wire, 22, 25 Working stresses, 84-89, 184, 185 1 GETTY CENTFR LIBRARY r ) 3125 00060 4567