GRAPHICS AND STRUCTURAL DESIGN BY H. D. HESS, M.E. Professor of Machine Design, Sibley College, Cornell University. Formerly Designer and Computer for the Mechanical Department of the Pencoyd Iron Works and the American Bridge Co. Member of the American Society of Mechanical Engineers. FIRST EDITION FIRST THOUSAND NEW YORK JOHN WILEY & SONS, INC. LONDON: CHAPMAN & HALL, LIMITED COPYRIGHT, 1913, BY H. D. HESS Stanhope F. H. GILSON COMPANY BOSTON, U.S.A. PREFACE THIS text is intended for the author's classes in General Engineering Design in Sibley College, Cornell University. The treatment of the subject has been kept as general as possible and is that used for his students during the past four or five years. The problems chosen for discussion are those on the border line between Civil and Mechanical Engineering. To the general designer knowledge of this subject is indispensable while acquaint- ance with the methods used in determining the stresses in, and the subsequent designing of, structures is of the greatest benefit to others in all designing for strength whether of machines or structures. Although the book is called " Graphics and Structural Design" the determination of the stresses has not been confined to graphi- cal means; the other usual methods have been included. The design of a plate girder may seem out of keeping with the purpose of the book. The author formerly used a crane runway girder instead of the railway bridge, but decided that the railway girders made the more comprehensive and better problem. The design of the latter permitted the use of the moment table and acquainted the student with the usual method of treating locomotive and train loads. Incidentally, one who could design a railway plate girder was well prepared to design a runway girder, although the reverse was not so generally true. The author's practice is to use a set of problems paralleling the drawing-room work, to assign a number of these problems each week to the students upon which they recite. Although this reduces slightly the time spent over the drawing boards the work iii 271051 IV PREFACE accomplished by the students is not apparently diminished and the work seems to be generally better understood. Although intended primarily for a textbook, it is hoped that this may prove a satisfactory reference book for designers whose work is not too highly specialized. The author desires to acknowledge his indebtedness to many manufacturers, all of whom have been most generous and among whom are American Bridge Co., Pennsylvania Steel Co., Cam- bria Steel Co., Bethlehem Steel Co., McClintock-Marshall Con- struction Co., Jeffrey Mfg. Co., Best & Co., and Bucyrus Co. Among the periodicals he is especially indebted to the Engineering News. The section on Specifications has been drawn largely from the specifications of the American Railway Engineering and Main- tenance of Way Association, the specifications of Mr. C. C. Schneider in the Transactions of the American Society of Civil Engineers, and the specifications of the American Bridge Co. H. D. HESS. ITHACA, N. Y. May, 1913. CONTENTS CHAPTER I. MATERIALS. Cast Iron Steel Steel Castings Physical Properties of Metals Proper- ties of Timbers Timber Beams Timber Columns Bricks Table of Bending Moments, Deflections, Etc. Table of Properties of Sections Strength of Flat Plates Properties of Sections of Rolled Structural Steel. 1-19 CHAPTER II. GRAPHICS. Force Magnitude, Direction and Line of Action of a Force Equilibrium Couple Resultant Equilibrant Force Triangle Force Polygon - Components Composition of Forces Resolution of Forces Graphic Moments Moments of Parallel Forces Uses of Force and Equilibrium Polygons Graphical Determination of Deflections Graphical Analyses of Restrained and Continuous Beams 20-34 CHAPTER III. STRESSES IN STRUCTURES. Tension or Compression in a Member Stresses due to Moving Loads Wind- load Stresses Maximum Bending due to Moving Loads Maximum Live- load Shears 35~5 l CHAPTER IV. ALGEBRAIC DETERMINATION OF STRESSES. Method of Moments Method of Coefficients 52-60 CHAPTER V. INFLUENCE DIAGRAMS. Position of Loading for Maximum Moments Position of Loading for Maximum Shear Position of Loading for Maximum Floor-beam Reaction Moment Table for Cooper's Loading, E-6o 61-75 vi CONTENTS CHAPTER VI. TENSION PIECES, COMPRESSION PIECES AND BEAMS. Tension Pieces Compression Pieces Beams Vertical Shear Bending Moment Standard Framing Riveting Shearing and Bearing Value of Rivets Rules for Spacing Rivets 76-89 CHAPTER VII. COLUMNS. Column Formulae Combined Stresses Long Beams Unsupported Laterally. 90-100 CHAPTER VIII. GIRDERS FOR CONVEYORS. Dead and Live Loading Wind Loading Determination of Stresses Selection of Members 101-109 CHAPTER IX. TRUSSES, BENTS AND TOWERS. Trusses and Bents for carrying Pipes Towers for Electrical Transmission Lines 1 10-1 1 7 CHAPTER X. DESIGN OF STEEL MILL BUILDINGS. Loading Bracing Determination of Stresses Selection of Members .118-140 CHAPTER XI. DESIGN OF A PLATE-GIRDER RAILWAY BRIDGE. Loading Bending Moments Vertical Shears Flange Area and Selection of Sections Lengths of Flange Plates Spacing of Flange Rivets Wind Bracing End Diagonals Web Splice Flange Splice 141-161 CHAPTER XII. CRANE FRAMES. Design of a Frame for an Underbraced Jib Crane Design of a Frame for a Top-braced Jib Crane 162-173 CONTENTS Vii f CHAPTER XIII. GIRDERS FOR OVERHEAD ELECTRIC TRAVELING CRANES. Box Girders Plate Girders with Upper Flange Stiffened by a Channel Bridge Girders with Horizontal Stiffening Girders 174-190 CHAPTER XIV. REINFORCED CONCRETE. Cement Sand Stone and Gravel Physical Properties of Concrete Unit Working Stresses Rectangular Beams Approximate Formulae T Beams Beams with Double Reinforcing Web Stresses Bond Stresses Lengths of Reinforcing Rods Bent Bars Web Reinforcements Columns. 191-224 CHAPTER XV. FOUNDATIONS. Materials Allowable Pressure on Soils Machine Foundations Building Foundations Piles 225-237 CHAPTER XVI. CHIMNEYS. Kern of a Section Design of a Brick Chimney Design of a Self-sustaining Steel Chimney Design of a Reinforced-concrete Chimney 238-272 CHAPTER XVII. RETAINING WALLS. General Theory of Retaining Walls Angles of Repose of Materials Design of Concrete Retaining Walls Design of Reinforced-concrete Retaining Walls 273-288 CHAPTER XVIII. BINS. Determination of Pressures on the Sides and Bottoms of Bins Stresses in Bins Graphical Determination of the Forces acting on Bins Hopper Bins Suspension Bunkers Bin Design 289-313 viii CONTENTS CHAPTER XIX. SHOP FLOORS. Cement Concrete Wooden Blocks Asphalt Brick Wooden Upper Floors, Wooden Steel Brick Arch Reinforced Concrete Iron. 314-322 CHAPTER XX. WALLS AND ROOFS. Wooden Sides Corrugated Steel Sides Walls, Brick Stone Solid Con- crete Hollow Concrete Reinforced Concrete Windows Roof Cover- ings, Corrugated Steel Slate Clay Tiles Concrete Slag or Gravel. 323-343 CHAPTER XXI. SPECIFICATIONS. General, Materials Workmanship Painting Inspection Testing Spe- cifications for Steel Mill Buildings Specifications for a Deck Plate-girder Railway Bridge Specifications for Reinforced Concrete 344-377 CHAPTER XXII. Problems 378-413 GRAPHICS AND STRUCTURAL DESIGN CHAPTER I MATERIALS THE principal materials used in the structures considered in this book are iron castings, hard, medium and soft steels, steel castings, timber (oak, hemlock, long-leaf Southern and yellow pine, spruce and white pine) and concrete, cement, slag and stone. Cast Iron is made by remelting pig iron or pig iron and cast- iron scrap in a cupola. The quality of the product is a varying one, as the entire charge in the cupola is never melted at any one time. The quality, therefore, depends upon the time at which it was run off during the heat. For machining, cast iron should be soft, and should have an ultimate strength in tension of from 16,000 to 20,000 Ibs. per sq. in. Although cast iron has no well-defined elastic limit, it may be assumed for practical purposes at 8000 Ibs. per sq. in. Cast iron is exceedingly strong in direct compression; when no bending is introduced its ultimate strength in compression should reach 90,000 to 100,000 Ibs. per sq. in. The resilience of cast iron being very low indicates little ability to resist shock. Steel is made by refining pig iron in the Bessemer converter, also by refining pig iron or pig iron and steel scrap in the open- hearth furnace. Small quantities of special steels are also being made in electric furnaces. The refining processes remove the im- purities and the carbon, the steel being afterwards recarbonized. The ordinary merchant and structural steels are soft and medium i O r- f 2 GRAPHICS AND STRUCTURAL DESIGN steels, the carbon contents ranging from 0.08 to 0.30 of i per cent. The ultimate strength of steel varies with the percentage of carbon from 48,000 Ibs. per sq. in. for very soft or rivet steel to 70,000 Ibs. per sq. in. for medium steel. Steel, having great resilience, is peculiarly adapted to resisting shock. Steel Castings. This steel is also made in open-hearth furnaces but may be made in small converters. The steel is poured into molds as is done for iron castings. To obtain fluidity it is necessary that steel for castings have a much higher temperature than is required for cast iron. The mold frames must, therefore, be much more substantial and the molds must be thoroughly dried before the metal is poured into them. Considerable contraction, with the attendant internal stresses, results from the exceedingly high temperature and necessitates the annealing of steel castings. The ultimate strength depends upon the carbon and should be 50,000 or more pounds per square inch. Steel castings have high resilience and are, there- fore, superior to iron castings. However, both the material and the machining of steel castings cost more than do those of iron castings. TABLE OF PHYSICAL PROPERTIES OF METALS Material. Modulus of elasticity. Ultimate strength. Elastic strength. Tension compres- sion. Torsion. Tension. Shear. Tension. Shear. Cast iron (cupola) {10,700,000 15,000,000 28,000,000 30,000,000 30,000,000 31,000,000 30,600,000 4,000,000 6,000,000 11,000,000 11,800,000 11,800,000 12,100,000 11,800,000 16,000 20,000 30,000 40,000 { 47,000 \ 57,000 60,000 70,000 100,000 ( 50,000 ( 100,000 16,000 ) 20,000 ) 8,000 8,000 Wrought iron 35,000 43,000 45,000 52,000 30,000 60,000 40,000 45.000 75,ooo Steel 0.15 carbon Steel 0.25 carbon Crucible steel (high carbon) . . . Steel castings NOTE. The ultimate compressive strength of cast iron is 90,000 to 100,000 Ibs. per sq. in. Its elastic strength in compression may be taken at 25,000 Ibs. per sq. in. The ultimate compressive strengths of the other materials will approximate their ultimate tensile strengths. MATERIALS 3 Working Fiber Stresses. In the structures hereinafter de- scribed the usual working fiber stresses will be given in each problem. Ordinarily, in structures liable to little or no shock or vibration, where the stresses are fully determined, the maximum fiber stress on mild steel will range from 16,000 to 20,000 Ibs. per sq. in. Structures, such as crane frames, liable to some shock will have the maximum stress on mild steel reduced to 11,000 or 12,000 Ibs. per sq. in. In the case of high- way bridges and similar structures an addition of 25 per cent is added to the live-load stresses to allow for impact, and a unit stress of 15,000 Ibs. per sq. in. is allowed on soft steel, while a unit stress of 17,000 Ibs. per sq. in. is allowed on medium steel. In the case of railway bridges 15,000 and 17,000 Ibs. per sq. in. are used, but the impact allowance is made by a formula similar to that used in the design of the railway plate girder in Chapter XI. PROPERTIES OF TIMBER Wooden beams are designed similarly to metal ones. Being liable to fail by horizontal shearing they should be examined for this. One -fifth the ultimate shearing resistance given in the tables may be taken as the working shearing strength (lengthwise). The formula for bending is -/- M = bending moment in inch pounds. / = moment of inertia in inches. e = distance from the neutral axis to the extreme fibers in inches. / = working fiber stress, in flexure, pounds per square inch. For rectangular timber beams this becomes 6 b = width of the beam in inches. d = depth of the beam in inches. GRAPHICS AND STRUCTURAL DESIGN The maximum unit shearing stress is/, = -^ , where 2 oa f a = fiber stress in shear, pounds per square inch. R = end reaction in pounds. TIMBER COLUMNS The following formula is suggested by the United States Government reports on timber. / =F x f c = ultimate compressive strength of the column in pounds per square inch. F c = ultimate crushing strength of short timber column in pounds per square inch. c = d' where / = length of the column in inches. d = least diameter in inches. The factor of safety should vary from 5 for 18 per cent moisture when used in the open to 3^ for 10 per cent or less of moisture when used in heated buildings. Name of wood. Pounds per square inch. White oak, Southern long-leaf pine :8 Cht^ 00 . to to to tooo to oo oo 1^0 t^ H -OQQQO QQQQQ >.! 8888 t^ M coco" co O~ CO CO co M oT Name of wood. : : : : : : ^ : : : : o 'oj ' ' ' ' . . .O ' jifilPMff 6 GRAPHICS AND STRUCTURAL DESIGN weigh 125 Ibs. per cu. ft., while soft bricks will average 100 Ibs. per cu. ft. Good bricks should be sound, hard, of regular shape and size, and should have a uniform structure. When struck a sharp blow they should emit a ringing sound. They should not absorb more than T V their weight of moisture. Their crushing strength should reach 7000 to 10,000 Ibs. per sq. in. Fire Bricks. These are bricks made of clay or of a mixture of clay and sand having ability to resist high temperatures. They are used mainly for building furnaces and lining flues and some chimneys. MATERIALS BENDING MOMENTS, DEFLECTIONS, ETC., FOR BEAMS OF UNIFORM SECTION Form of support and load. End reactions A , B. Bending moment M . Relation between load " P " and mo- ment of resistance. Maximum deflection. D =: i A7=Px, ' / 1F-5?, 2 P/ 4 d=.- P /' Mmax L ' "T Pec 1 / cc l 4= = P, Between A & B, M=P-c = const. _P_/3c 78]' 8V P /c 3 . c 2 / M Pl -Mmax = 7*8' p 48 Ee GRAPHICS AND STRUCTURAL DESIGN BENDING MOMENTS, DEFLECTIONS, ETC., FOR BEAMS OF UNIFORM SECTION Form of support. End reactions A , B. Bending moment. Relation between load " P " and mo- ment of resistance. Maximum deflection. = -P, B = - P, fW P = -* w = Pl 7-794/' PI 3 d = 0.01304-^ A = B = -, Pl O 60 / _L/L 2 10 Ee M= r 3_*\ ,4 l) r 8fW ~> --- 185 / . id , 16 16 M 3P/ ikZmax 7' 16 16 fW " 3 ^ IT-A. 16 / PI 3 A = B = P , A/max == * p = ^jw i d = PI* i //* 32 Ef 5 = -, 2 = ~T- ' o P = 7-< T- 192 / 24 Ee' V * ' / MATERIALS g The bending moment is a maximum under load i when x = - 4 Pi max The letters used have the following significance and for con- venience they should be expressed in the units stated. A and B = end reactions, e = distance neutral axis to pounds. extreme fibers having P = load, pounds. fiber stress /, inches. / = extreme fiber stress, E = modulus of elasticity, pounds per square inch. pounds per square inch. / = span of beam in inches. c, Ci and x = portions of span, inches. W = resistance, = - x = left end to section where bending is wanted, inches. . 7 = moment of inertia, inches, d = deflection, inches. STRENGTH OF FLAT PLATES Nomenclature : t = thickness of plate in inches. / = maximum fiber stress in plate, pounds per square inch. E = modulus of elasticity in flexure, pounds per square inch. c factor according to Grashof or Bach. P = concentrated load in pounds, usually assumed as acting on a circle whose radius is r ins. q = uniform load acting on plate, pounds per square inch. r =. radius of circular plate, inches. b = longer side of rectangular plate, inches. a = shorter side of rectangular plate, inches. b _ ~ . a i. Circular plates, carrying uniform load q. r 2 IO GRAPHICS AND STRUCTURAL DESIGN MOMENTS OF INERTIA, RESISTANCES, CENTERS OF GRAVITY AND LEAST RADII OF GYRATION OF GEOMETRICAL SECTIONS Shape of section. Moment of inertia. Resistance. Distance base to center of gravity. Least radius of gyration. W 6 Lesser side 3-46 6 B ! bh* 36' bh* 24' The smaller A b - or -- 4.24 4.9 36(26+61) 3 2& + &J 20.38 D 3 10.19' 0.049 (D 4 ( 0.098 W z = 0.259 R 3 . 0.424 R. 0.07 2 0.7854^2. MATERIALS II Cast-iron plate supported at outer circumference, c = 0.8 to 1.2. Steel plate supported at outer circumference, c = 0.75. Steel plate bolted or riveted at outer circumference, c = 0.50. 2. Circular plates, carrying a central load P, applied on a circle of radius r . Plate supported at outer circumference. For cast iron c = 1.5. 3. Circular plates, central load similar to 2 but plate bolted or riveted at outer circumference. RECTANGULAR AND SQUARE PLATES Rectangular and square plates supported at the outer edges and carrying a uniform load q. Rectangular, V 2 Square, ca* For cast iron c = 0.75 to 1.13. For steel c 0.56 to 0.75 (max. 1.13). STRUCTURAL MATERIAL The principal rolled sections used are I beams, channels, angles, plates, flats and rounds. Manufacturers' handbooks afford the best sources of information regarding these sections and in actual design they should be freely consulted. The following data are abridged from these books. 12 GRAPHICS AND STRUCTURAL DESIGN SHEARED PLATES Width in inches. Thickness in inches. A i A 1 A i T 9 * t H 1 it 1 if Length in inches. IS 240 320 400 500 500 550 500 475 475 475 425 400 375 360 300 280 280 16 240 320 400 500 500 550 500 475 475 475 425 400 375 360 300 280 280 17 240 320 400 500 500 550 500 475 475 475 425 400 375 360 300 280 280 18 240 360 400 500 500 500 500 550 550 550 500 500 450 400 400 400 350 iQ 240 360 400 500 500 500 500 55 550 55 500 500 450 400 400 400 350 20 240 360 400 500 500 500 500 550 550 550 500 500 450 400 400 400 350 21 216 360 400 500 525 525 525 55 550 550 500 500 450 400 400 400 350 22 216 3 6o 400 500 525 525 525 550 550 550 500 500 45 400 400 400 350 23 204 3 6o 400 500 525 525 525 550 550 550 500 500 450 400 400 400 350 24 204 360 400 500 525 550 55 550 550 550 500 475 425 400 35 350 325 25* 204 360 400 500 525 550 550 55 550 550 500 475 425 400 350 350 325 26 1 80 360 400 500 525 550 550 55 550 55 500 475 425 400 350 350 325 2? 168 340 400 500 500 550 550 500 500 500 45 450 400 380 330 300 300 28 168 340 400 500 500 550 550 500 500 500 450 45 400 380 330 300 300 29 156 340 400 500 500 550 550 500 500 500 450 450 400 380 330 300 300 30- 35 320 400 500 500 550 500 475 475 475 425 400 375 360 300 280 280 36- 41 360 400 500 500 500 500 550 55 550 500 500 450 400 400 400 350 42- 47 360 400 500 525 525 525 55 55 550 500 .500 450 400 400 400 350 48- 53 360 400 500 525 550 55 55 550 550 500 475 425 400 350 350 325 54- 59 340 400 500 500 550 550 500 500 500 450 450 400 380 330 300 300 60- 65 320 400 500 500 550 500 475 475 475 425 400 375 360 300 280 280 66- 71 300 350 430 450 475 425 425 425 410 375 340 330 320 280 260 260 72- 77 260 300 400 425 450 400 400 400 390 350 320 320 300 260 240 240 78-83 240 275 380 400 420 375 375 375 37 325 300 300 300 240 220 2 2O 84- 89 200 250 35 375 385 35 350 350 350 300 280 2/5 275 230 2IO 2IO 90- 95 1 80 230 330 340 350 350 325 325 325 275 260 260 260 220 200 2OO 96-101 120 175 240 250 275 275 275 275 275 240 240 220 220 2OO 1 80 1 80 102-107 150 200 230 230 250 250 250 250 230 230 2IO 210 190 170 170 108113 1 80 1 80 200 220 22=; 22$ 22< 220 220 2OO 2OO 1 80 160 160 114119 1 80 200 *O 2IO * ^0 2IO *o 2IO 200 200 1 80 1 80 I7O I ^O I SO 120-125 I2O 150 150 1 80 1 80 i75 175 160 1 60 160 * o v 144 * o w 144 MATERIALS EDGED PLATES Thickness in inches. Width in inches. A 1 4 A 1 T% A | H 3. 4 11 I I Length in feet. 4 50 50 50 50 50 5 40 40 30 30 3 28 28 28 5 30 42 42 42 42 40 30 30 30 30 30 30 30 30 6 3 42 42 42 42 40 35 30 30 3 30 30 30 30 7 25 42 42 42 42 40 35 30 30 3 30 30 30 30 8 25 42 42 42 42 42 38 36 32 3 29 28 26 25 9 25 42 42 42 42 42 38 34 32 30 29 28 26 25 10 25 42 42 42 42 42 38 33 32 3 29 28 26 25 ii 25 42 42 42 42 42 38 33 31 29 28 27 25 24 12 25 42 42 42 42 42 37 32 30 28 27 26 24 23 13 42 42 42 42 42 37 32 30 27 25 24 22 2O 14 42 42 42 42 40 35 30 28 26 25 23 22 2O 14* 42 42 42 42 36 33 30 28 25 In the tables of angles the areas are given, and the weights in pounds per foot of section can be obtained by multiplying the areas by 3.4. The moments of inertia and location of the centers of gravity of the angles are given in the tables following. The radii of gyration are readily found from I = inertia; A = area in square inches. All data are based upon dimensions in inches pounds. and weights in GRAPHICS AND STRUCTURAL DESIGN DIMENSIONS AND AREAS OF ANGLES Uneven legs. Even legs. 1 1% 1 A I T\ i A I H I H 1 it I 4 8 X8 6 X6 *5 X5 5'o6 4-40 4.18 3-97 3-75 3-53 3-31 3-09 2.87 2.65 2-43 2.22 2.OO 1.78 1.56 1.30 7.75 5.75 5.00 4.75 4.50 4.25 4.00 3-75 3-50 3-25 3.00 2.75 2.50 2.25 2.00 8.68 6.43 5-59 5-31 5.03 4-75 4-47 4.18 3-90 3.62 3-34 3.06 2.78 9.61 7- II 6.17 5.86 5-55 5-23 4.92 4.61 4.30 3.98 3.67 3.36 10.5 7-78 6.75 6.41 6.06 5-72 5-37 5-03 4.68 4-34 4.00 3-65 II. 4 8.44 7-31 6.94 6.56 6.19 5.8i 5-44 5.06 4-69 4-31 12.3 9.09 7.87 7-47 7.06 6.65 6.25 5-84 5-43 5.03 4.62 13.2 9-74 8.42 7-99 7-55 7.11 6.67 14.1 10.4 8.97 8.50 8.03 IS.O II. 9-50 9.OO 8.50 4.36 3'6i 3-42 3-23 3-05 2.86 2.67 2.48 2.30 2. II .92 73 55 .36 17 0.99 *7 X3i 6 X4 6 X3* 5 X4 5 X3* 5 X3 *4 X3* 4 X3 3*X3 3*X2* 3 X2* 3 X2 2^X2 '56 .40 25 4 X4 '3*X3i '3'X3' *2fX2| 2*X2* *2jX2j 2 X2 ijXil i*Xii 1.44 1.31 1.19 1. 06 0.94 0.81 0.69 .09 93 -78 .62 1-47 I.3I 1. 15 I.OO 0.84 o'. 3 6 0.81 0.72 0.62 0.53 Special sections. MOMENTS OF INERTIA AND CENTERS OF GRAVITY OF ANGLES EVEN LEGS Thick- ness. 1 1 I { I * I Legs. I g / g I g I g / g 7 g / g 8 X8 6 X6 15 4 i 6 48.6 19 9 .2 7 59-4 24 2 2.2 i 7 69.7 28 2 2.3 I 8 79-6 31 9 2.3 r 8 89.0 35-5 2.4 1.9 *5 X5 8 7 I 4 II 3 13 6 I 5 4 X4 3*X3* 3 X3 I 2 o 84 4-4 2-9 I 8 i.i I.O o 89 5.6 3.6 2 2 .2 .1 o 93 6.7 4.3 2 6 1.2 I.I o 98 7-7 S.o 1-3 1.2 8.6 5-5 1.3 1.2 "2fX2f O.95 0.78 1.3 0.82 I 7 o 87 2kX2k o 70 o 72 o 98 o 76 I 2 o 81 2iX2j o 50 o 65 o 70 o 70 2 X2 O.35 0.59 0.48 0.64 O.59 o 68 Special sections. MATERIALS MOMENTS OF INERTIA AND CENTERS OF GRAVITY OF ANGLES UNEVEN LEGS Thick- ness. Axis. i 1 i 1 * 1 I Legs. / e / g *7 X3i 6 X4 6 X3* *S X4 5 X3i 5 X3 *4 X3i 4 X3 3*X3 3iX2j 3 X2* 3 X2 2JX2 i 4-9 13-5 3-3 12.9 4-7 8.1 3-2 7-8 2.0 7-4 3-0 4.2 1.9 4-0 1.9 2.7 i.i 2.6 I.O 1-7 0.54 i.S 0.51 0.91 0.94 1.94 0.79 2.0 I.O 1.5 0.86 1.6 0.70 1-7 0.96 1.2 0.78 1-3 0.83 i.i 0.66 1.2 0.71 0.96 0.54 I.O 0.58 0.83 4.4 lii 17.4 4.3 16.6 6.0 10.5 4-i 10. 2.6 9-5 3-8 53 2.4 5.1 2.3 3-5 1.4 3-2 1.3 2.1 0.67 1-9 0.64 I.I 0.78 2.5 0.09 1.99 0.83 2.1 I.I 1.6 0.91 I 7 ?:2 5 1. 00 1.2 0.83 13 0.88 i.i 0.70 1.2 0.75 I.O 0.58 I.I 0.63 0.88 5-3 30.9 7-5 21. 1 5-1 20.1 7-1 12.6 4-8 12.0 3-1 II. 4 ti 1:1 2.8 4-1 1.6 3-9 1-5 2.5 0.82 .6 .0 :!i .1 .1 .6 95 .7 .80 .8 .00 '87 4 92 .2 75 3 79 .0 6.1 36.0 8.7 24.5 5.8 23-3 0.87 2.6 i.i 2.1 0.93 2.2 6.8 40.8 9-8 27.7 6.6 26.4 0.91 2.7 i.i 2.1 0.97 2.2 7-5 45-4 10.7 30.8 7-2 29.2 0.06 2.7 1.2 2.2 I.O 2.3 56 I O 6 ? I O 13-9 3-5 13-2 1.7 0.84 1.8 15-7 3-9 14.8 1.8 0.88 1.9 'sii 3.2 4-9 1.8 4-4 '78 .8 74 .2 39 .1 0.37 0.65 0.61 i.i 0.66 0.91 0-49 0.99 0.54 0.79 0.92 1-4 0.06 1.2 0.79 1.3 ?:2 3-5 5-2 0.96 1.5 I.O 1-3 Special sections. The properties of angles of intermediate thickness can be interpolated from the tables with sufficient accuracy for most purposes. i6 GRAPHICS AND STRUCTURAL DESIGN PROPERTIES OF STANDARD CHANNELS " Weight per foot. Flange, b. Web, Gauge, m. Tan- gent, Max. bolt or rivet. Moment of in- ertia, axis i- 1. Moment of in- ertia, axis 2-2. Dis- tance base toe. ofg. Area. Radius of gyration. Axis i-i. Axis 2-2. 55-00 3-82 0.82 2.50 12.25 1 r 430 12.2 0.82 16.18 5-i6 0.8? 50.00 45-00 3-72 3-62 0.72 0.62 2.50 2.00 12.25 12.25 L 403 375 II. 2 10.3 0.80 0.79 14.71 13-24 5.23 5-32 0.87 0.88 15' 40.00 3-52 0.52 2.OO 12.25 * 1 348 9-4 0.78 11.76 5-44 0.89 35-00 3-43 0.43 2.00 12.25 320 8.5 0-79 10.29 5-57 0.91 . 33-00 3-40 0.40 2.OO 12.25 J I 312 8.2 0-79 9.90 5-62 0.91 r 40.00 35-00 3-42 3-30 0.76 0.64 2.OO 2.00 IO.OO 10.00 I J 197 179 6.6 5-9 0.72 0.69 11.76 10.29 4-09 4.17 0.75 0.76 I2< 30.00 3-17 0.51 1-75 IO.OO [ i 4 162 5-2 0.68 8.82 4-28 0-77 j 25.00 3-05 0.39 1-75 IO.OO \ \ 144 4-5 0.68 '7.35 4-43 0.78 I 20.50 2.94 0.28 1-75 10.00 J I 128 3-9 0.70 6.03 4.61 0.81 r 35-00 3-18 0.82 1-75 8.25 }r 116 4-7 0.69 10.29 3-35 0.67 30.00 3-04 0.68 1-75 8.25 103 4-0 0.65 8.82 3-42 0.67 ic n 25.00 2.89 0.63 1-75 8.25 i { 91 3-4 0.62 7-35 3-52 0.68 1 20.00 2.74 0.38 1-50 8.25 } 79 2-9 0.61 5 '88 3.66 0.70 L 15.00 2.60 0.24 1.50 8.25 [ 6? 2-3 0.64 4.46 3.8 7 0.72 r 25.00 2.81 0.61 1.50 7-25 i r 71 3-0 0.62 7-35 3.10 0.64 9 20.00 I5.OO 2.65 2.49 0.45 0.29 1-50 1-38 7.25 7-25 H 61 51 2.5 2.O 0.58 0-59 5-88 4-41 3-21 3-40 0.65 0.66 L 13-25 2.43 0.23 1-38 7-25 J I 47 1.8 0.61 3.89 3-49 0.67 r 21.25 2.62 0.58 1-50 6.25 }r 48 2-3 0-59 6.25 2.76 0.60 18.75 2.53 0.49 i-5o 6.25 44 2.O 0-57 5-51 2.82 0.60 8 i 16.25 2.44 0.40 1-50 6.25 i { 40 1.8 0.56 4-78 2.89 0.61 1 13-75 2.35 0.31 1-38 6.25 36 1.6 0.56 4-04 2.98 0.62 L 11.25 2.26 0.22 1.38 6.25 I 32 1.3 0.58 3-35 3-10 0.63 r 19-75 2.51 0.63 1-50 5-50 i r 33 1.8 0.58 5-8l -39 0.56 17-25 2.41 0.53 1-50 5-50 30 1.6 0.55 5-07 -44 0.56 H 14-75 2.30 0.42 1-50 5-50 r * i 27 1-4 0.53 4-34 -50 0.57 12.25 2. 2O 0.32 1-25 5-50 24 1.2 0.53 3.6o 59 0.57 I 9-75 2.09 0.21 1.25 5-50 J I 21 0.98 0.55 2.85 .72 0.59 r 15.50 2.28 0.56 1-25 4-50 1 r 19-5 1.3 0-55 4-56 -07 0-53 A J 13.00 2.16 0.44 1.25 4-50 I j 17-3 i.i 0.52 3-82 .13 0.53 1 10.50 2.04 0.32 1-25 4-50 i f 1 IS- 1 0.88 0.50 3-09 .21 0.53 L 8.00 1.92 O.2O 1. 13 4-50 J I 13-0 0.70 0.52 2.38 34 0.54 11.50 2.04 0.48 1.13 3-75 > c 10.4 0.82 0.51 3-38 -75 o.49 9.00 1.89 0.33 1. 13 3-75 L J 8.9 0.64 0.48 .65 -83 0.49 6.50 1-75 0.19 1. 13 3-75 J I 7-4 0.48 0.49 -95 95 0.50 f 7.25 1-73 0.33 I.OO 2.75 }r 4.6 0.44 0.46 -13 -46 0.46 4i 6.25 1.65 0.25 I.OO 75 i J 4.2 0.38 0.46 .84 Si 0.45 I 5-25 1.58 0.18 I.OO 75 I 3.8 0.32 0.46 .55 56 0.45 *{ 6.00 S.oo 4.00 i. 60 1.50 i. 41 0.36 0.26 0.17 0.88 0.88 0.88 75 75 75 H 2.1 1.8 1.6 0.31 0.25 0.2O 0.46 0.44 0.44 .76 47 19 .08 .12 17 0.42 0.41 0.41 Note. This table is taken from the handbook of the Cambria Steel Co. MATERIALS PROPERTIES OF STANDARD I BEAMS "f Depth of i beam. Weight per foot. Area of section . Thickness of web. Width of flange. Moment of inertia, axis i-i. Section modulus, axis i-i. Radius of gyration, axis i-i. Moment of inertia, axis 2-2. Radius of Ky ration, axis 2-2. i! d A t b 7 S r r r' n T Ins. Lbs. Sq. ins. In. Ins. Ins.* Ins.' Ins. Ins. In. Ins. Ins. Ins. 3 5-50 1.63 0.17 2.33 2.5 1.7 1.23 0.46 o.53l 3 6.50 91 0.26 2.42 2.7 1.8 1. 19 0.53 0-52^ i 1/8 iii 3 7-50 .21 0.36 2.52 2.9 1-9 i. IS 0.60 0.52J 4 7-50 .21 0.19 8.66 6.0 3-0 1.64 0.77 0-591 4 4 8.50 9-50 -50 79 0.26 0.34 2.73 2.8l 6.4 6.7 3-2 34 1.59 1.54 0.85 0.93 o.5Sl o.58f i ij 211 4 10.50 3-09 0.41 2.88 7-1 3-6 1.52 1. 01 0.57J S 9-75 2.87 0.21 3 oo 12. 1 4-8 2.05 1.23 o.6s1 5 12.25 3.6o 0.36 3 IS 13-6 54 1.94 1.45 0.63 !> } 1} 3i S 14-75 4-34 0.50 3.29 IS-I 6.1 1.87 1.70 o.6 3 j 000 12.25 14-75 17-25 3.6i 4-34 5-07 0.23 0.35 0.47 3 33 3-45 3-57 21.8 24.0 26.2 1:1 8.7 2.46 2.35 2.27 1.85 2.09 2.36 0.72^ 0.69^ 0.68J f 2 4A 7 7 7 I5.OO 17.50 2O.OO 4-42 13 0.25 0-35 0.46 3.66 3.76 3-87 36.2 39-2 42.2 10.4 II. 2 12. 1 2.86 2.76 2.68 2.67 2.94 3-24 0.781 0.76^ 0-74J f 4 5} 8 18.00 S-33 0.27 4.00 56.9 14.2 3-27 3-78 0.841 8 8 20.25 22.75 5-96 6.69 0.35 0.44 4.08 4-17 60.2 64.1 16.0 3.18 3.10 4.04 4.36 0.82! 0.81 f i 21 6A 8 25.25 7-43 0.53 4.26 68.0 17.0 3.03 4-71 o.8oj 9 21.00 6.31 0.29 4-33 84-9 18.9 3.6? 5.16 o.ool 9 25.0O 0.41 4-45 91.9 20.4 3-54 5-65 0.881 i , * , 9 30.00 8.82 0.57 4.61 101 .9 22.6 3-40 6.42 0.85 f t 7i 9 35.OO 10.29 0.73 4-77 in. 8 2 4 .8 3-30 7-31 0.84J 10 25.OO 7-37 0.31 4.66 122. 1 24.4 4.07 6.89 10 30.OO 8.82 0.45 4.80 134-2 26.8 3-00 7-65 . * -is 10 35-00 10.29 o.oo 4-95 146.4 29-3 3-77 8.52 * 716 10 40.00 11.76 o.75 S.io 158.7 31-7 3-67 9-50 12 31.50 9.26 0.35 5-00 215.8 36.0 4-83 9-50 i. oil 12 35-OO 10.29 0.44 5-09 228.3 38-0 4-71 10.07 0-99^ I 2\ 9! 12 40.00 11.76 0.56 5.21 245-9 41-0 4-57 10.95 0.96J IS 42.00 12.48 0.41 5-50 441-8 58-9 5-95 14.62 .081 IS 45.00 13.24 0.46 5-55 455-8 60.8 5-87 15.09 -07| IS 50.00 14-71 0.56 5.6S 483.4 64.5 5 73 16.04 .04!- I 3 12* 15 55-00 16.18 0.66 5-75 Sii.o 68.1 5-62 17.06 .03! IS 60.00 17.65 0.75 5.84 538.6 71.8 5-52 18.17 .OlJ 18 55-0 15-93 0.46 6.00 795-6 88.4 7-07 21.19 .151 0.875 18 60.0 17.65 0.56 6.10 841.8 93-5 6.91 22.38 .13 I i 18 65.0 19.12 0.64 6.18 881.5 97-9 6.79 23.47 '"I i 3i I5l 3 8 18 70.0 20.59 0.72 6.26 921.2 102.4 6.69 24.62 09J i 20 65.0 19.08 0.50 6.25 1169.5 117.0 7.83 27.86 .2l1 2O 70.0 20.59 0.58 6.33 1219.8 122. 7-70 29.04 *9 r i 3* l6J| 20 75-0 22.06 0.65 6.40 1268.8 126.9 7-58 30.25 .IJJ 24 80.0 23.32 0.50 7.00 2087.2 173-9 9.46 42.86 .361 24 85.0 25.00 0.57 7.07 2167.8 180.7 9-31 44-35 .33 24 90.0 26.47 0.63 7-13 2238.4 186.5 9.20 45-70 .31 y i 4 2oli 24 95-0 27.94 0.69 7.19 2309.0 192.4 9.09 47-10 .30 24 IOO.O 29.41 0.75 7-25 2379-6 198.3 8.99 48.55 .2 8 J Note. This table is taken from the handbook of the Cambria Steel Co. i8 GRAPHICS AND STRUCTURAL DESIGN GREY MILL SECTIONS The Bethlehem Steel Company on their Grey Mill are able to roll sections with much wider flanges than are possible with the usual methods of manufacture. These sections make better columns and can be used upon longer spans without lateral brac- ing. The following tables give some of these sections with their properties. In H columns, under each section-number, only the first three weights and the maximum weight are given although a large number of intermediate weights are rolled. The beam and girder-beam sections are given almost completely. H COLUMNS | i Axis XX. Axis FF. "5 I M 1 P. C/3 D b t r "o gj "ogri 15 jj 3 Ofl . MOW 'S *r> rj .2? 1 1 a i| IP J.S l| IP Lbs. Sq. ins. / 5 r r S' r' 83-5 24-5 I3i 13-9 .0.43 ii. i 884.9 128.7 6.01 294.5 42.3 3-47 91.0 26.8 I3J 14.0 0--47 ii. i 976.8 140.8 6.04 325.4 46.6 3-49 99-0 29.1 14 14.0 0.51 ii. i 1070.6 153-0 6.07 356.9 51.0 3-50 . 287-5 84-5 i6| 14-9 1.41 ii. i 3836.1 454-7 6.74 1226 . 7 164.7 3-81 f 64-5 19.0 nj II. 9 0.39 9-2 499-0 84.9 5-13 168.6 28.3 2.98 H 12 J HI2 71-5 21.0 i if 12. 0.43 9-2 556.6 93-7 5-15 188.2 31-5 3-00 78.0 22.9 12 12.0 0.47 9.2 615.6 102.6 5.18 208.1 34-7 3-01 I 161.0 47-3 134 12-5 0.94 9.2 1444-3 214.0 5-53 477-0 76-5 3.18 f 49-0 14-4 9i IO.O 0.36 7.7 263.5 53-4 4.28 89.1 17-9 2.49 H 10 \ 54-0 59-5 15-9 17.6 10 iof IO.O IO.O 0.39 0.43 7-7 7-7 296.8 331-9 59-4 65.6 4-32 4-35 100.4 112. 2 20. i 22.3 2.51 2.53 ( 123-5 36.3 114 10.5 0.86 7-7 790.4 137-5 4.67 259-3 49-5 2.67 f 31-5 9-2 7f 8.0 0.31 6.1 105-7 26.9 3-40 35.8 8.9 1.08 H 8 \ 34-5 10.2 8 8.0 0.31 6.1 121. 5 30.4 3.46 41- 1 10.3 2.01 39-0 ii. 5 8i 8.0 6.1 139-5 34-3 3-48 47.2 II. 7 2.03 1 90-5 26.6 94 8.5 0.78 6.1 385-3 81.1 3.8o 125.1 29.6 2.17 MATERIALS GIRDER BEAMS i 1 . |j "o 5 Neutral axis perpen- dicular to web at Neutral axis coincident with center > ' ^1 ^ "j IN w-C 8 . rH i/l center. line of web. u *s & s fc-3 8.S Q> O S s I' 2 *a QS5 a s& "Sg =1 3 X'~ "S 51 8 .a 1* g| g .3 |i la" 1 s Is? H 3J & !! It rt *> 11 Pl s - a II 00 |1 D t F I r i e /' r' n T 30 200. o 58.71 0.750 15.00 9150.6 12.48 610.0 630.2 3-28 II. OO 25.2 3 180.0 53-00 0.690 13-00 8194.5 12.43 546.3 433 3 2.86 9.00 25.2 28 180.0 52.86 0.690 14-35 7264.7 11.72 518.9 533-3 3-18 10.25 23-4 28 165.0 48.47 0.660 12.50 6562.7 11.64 468.8 371-9 2.77 8.50 23-4 26 160.0 46.91 0.630 13.60 5620.8 10.95 432.4 435-7 3-05 9-50 21.6 26 150.0 43-94 0.630 12.00 5153.9 10.83 396.5 3M.6 2.68 8.00 21.6 24 140.0 41.16 0.600 13-00 4201.4 IO.IO 350.1 346.9 2.00 ?.oo 20. o 24 120. 35.38 0.530 12.00 3607.3 IO.IO 300.6 249-4 2.66 .00 20.3 2O 140.0 41.19 0.640 12.50 2934.7 8.44 293.5 348.9 2.91 8.50 15 7 2O 112. 32.81 0.550 12.00 2342.1 8.45 234-2 239 3 2.70 8.00 16.4 18 92.0 27.12 0.480 II.SO 1591.4 7-66 176.8 182.6 2.59 7-50 14.8 15 I40.O 41.27 0.800 11-75 1592.7 6.21 212.4 331.0 2.83 7-75 10. 1 15 104.0 30.50 0.600 11.25 I22O.I 6.32 162.7 213.0 2.64 7-25 ii. i IS 73-0 21.49 0.430 10.50 883-4 6.41 117.8 123.2 2.39 6.50 12. 1 12 70.0 20.58 0.460 10. OO 538.8 5-12 89.8 II4-7 2.36 6.00 9.0 12 55. o 16.18 0.370 9.75 432.0 5-17 72.0 8l.l 2.24 6.00 9-5 10 44-0 12.95 0.310 9.00 244.2 4-34 48.8 57-3 2.10 5-50 7-8 9 38.0 11.22 0.300 8.50 I7O.9 3-90 38.0 44-1 1.98 5-25 99 8 32.5 9-54 0.290 8.00 II4-4 3-46 28.6 32.9 1.86 5.00 6.0 BEAMS 30 I2O.O 35-3 0.54 10.5 5239.0 12.2 349-0 165.0 2.2 i 6.50 26.4 28 IOS.O 30.9 0.50 10. 4014.0 II. 4 287.0 I3I.O .1 i 6.00 24-7 26 90.0 26.5 0.46 9-5 2977-0 10.6 229.0 IOI.O -9 i 5-50 23.0 24 84.0 24.8 0.46 9-3 2382.0 9.8 198.0 91.0 -9 5-25 21. 24 73-0 21.5 0.39 9-0 2091.0 99 174.0 74.0 .9 5-25 21.3 20 82.0 24.2 0.57 ' 8.9 I56o.O 8.0 156.0 80.0 .8 5.00 I7-I 20 72.0 21.4 0.43 8.7 1466.0 8.3 146.0 76.0 .9 5.00 17 I 20 69.0 20.3 0.52 8.1 1269.0 7-9 127.0 Si.o .6 4-50 17-5 2O 64.0 18.9 0.45 8.1 1222. O 8.0 122.0 50.0 .6 4-50 17-5 2O 59-0 17.4 0.38 8.0 II72.O 8.2 II7.0 48.0 7 4-50 17-5 18 59-0 17.4 0.50 7-7 883.0 7-1 98.0 39-0 .5 4-25 15-7 18 54-0 15 9 0.41 7.6 842.0 7-3 94-0 38.0 .5 4-25 15-7 18 52.0 15-2 0.38 7.6 825.0 7-4 92.0 37-0 .6 4-25 15-7 18 48.5 14.2 0.32 7-5 798.0 7-5 89.0 36.0 .6 4.25 15-7 15 71.0 20.9 0.52 7-5 796.0 6.2 106.0 61.0 7 4-25 ii. 7 15 64.0 18.8 0.60 7-2 665.0 6.0 89.0 42.0 5 4.00 12.3 IS 54-0 15-9 0.41 7-0 610.0 6.2 81.0 38.0 .6 4.00 12.3 IS 46.0 13-5 0.44 6.8 485.0 6.0 65.0 25.0 4 3-75 12.9 15 41.0 12.0 0.34 6.7 457-0 6.2 61.0 24.0 .4 3-75 12.9 IS 38.0 ii. 3 0.29 6.7 443-0 6.3 59-0 23.0 4 3-75 12.9 12 36.0 10.6 0.31 6.3 269.0 5-0 45-0 21.0 .4 3-50 9.9 12 32.0 9-4 0.33 6.2 228.0 4-9 38.0 16.0 3 3-50 10.2 12 28.5 8.4 0.25 6.1 216.0 5-1 36.0 15-0 3 3-50 10.3 10 28.5 8.3 0.39 6.0 134-0 4-0 27.0 12.0 .2 3-25 8.4 IO 23-5 6.9 0.25 5-9 123.0 4-2 25.0 II. 3 3-25 8.4 9 24.0 7-0 0.36 5-6 92.0 3.6 20.0 9.o .1 3.00 7-5 9 20. o 6.0 0.25 5-4 85.0 3.8 I9.O 8.0 .2 3.00 7-5 8 19-5 S.8 0.32 5-3 61.0 3-2 15-0 7.0 .1 2.75 6.6 8 17-5 5.2 0.25 5-2 57-0 3-3 14.0 6.0 .1 2. 75 6.6 CHAPTER II GRAPHICS Statics. Statics treats of forces at rest and therefore in equilibrium, hence the resultant force in any direction and the moments of the forces about any point must be zero. For coplanar forces the condition of equilibrium is expressed in the following equations: S horizontal forces = o, 2 vertical forces = o, 2 moment of forces about any point = o. Graphic Statics. Graphic statics relates to the solution of statical problems by geometrical constructions. Force. Force is an action upon a body tending to change its state of rest or motion. A force is completely known when its magnitude, direction, line of action and point of application are known. Magnitude. Forces may be measured by any unit of weight. It is most convenient in this work to use the pound. The magni- tude of a force can be represented graphically by the length of a line, the length being drawn to a previously chosen scale, for instance, if the scale is i inch to 1000 pounds then a line J-inch long represents a force of 500 pounds. Direction. An arrow placed on the line is used to indicate the direction of the force. Line of Action. The line of action is along the line repre- senting the force and it is along this line that the force tends to produce motion. Point of Application. The point of application is the place assumed as a point where the force acts upon the body. GRAPHICS 21 Coplanar and Noncoplanar Forces. Coplanar forces have their lines of action in a common plane, while the lines of action of noncoplanar forces do not lie in the same plane. Where not otherwise stated coplanar forces are to be understood. Concurrent and Nonconcurrent Forces. Concurrent forces have their lines of action intersect in a point, while the lines of action of nonconcurrent forces fail to meet in a common point. Equilibrium. A number of forces are in equilibrium when they produce no tendency to motion in the body upon which they act. Couple. Two equal and parallel but opposite forces con- stitute a couple. The arm of the couple is the perpendicular distance between their two lines of action. Moment of a Couple. The moment of a couple is the product of either force by the arm of the couple. Resultant. The resultant of a system of forces is the single force or simplest system that would produce the same effect as the other forces and could, therefore, replace them. Equilibrant. The equilibrant is equal in magnitude to the resultant but opposite to it in direction. It is, therefore, the single force or the simplest system that will exactly neutralize the given system of forces. Let the force FI acting on the point d be represented in direc- tion and magnitude by the line ad, and similarly the force F 2 , also act- ing on d, be represented, by dc, then completing the parallelogram and drawing the diagonal db gives R, the magnitude and line of action of the F resultant. Its direction, to produce the same effect as the two forces FI and F 2 , must be as indicated by the arrow. Force Triangle. The triangle dbc would have given the resultant force as well as the parallelogram. Suppose the two known forces are laid off so that the arrows run in the same 22 GRAPHICS AND STRUCTURAL DESIGN direction, that is either clockwise or counterclockwise; then the tine closing the triangle with its ar- row following around in the same direction as the other two arrows, here counterclockwise, will be equal to the resultant but opposite in di- rection; hence, E is the force which would hold the two forces FI and F 2 in equilibrium and is their equilibrant. Force Polygon. If more than three forces meet at a point and are in equilibrium a force polygon may be constructed by finding the resultant of two of the forces, combining this resultant with a third force to find a second resultant, and so on until all the forces have been considered and a final resultant determined. FIG. 3. FIG. 4. RI is the resultant of F 3 and F 4 ; R% is the resultant of RI and F$, and F 2 is the equilibrant of the forces FI, F 5 , F 4 and F 3 . It is evident that in the construction the drawing of the resultants RI and R% might have been dispensed with, it only being necessary to lay the forces off so that the arrows would follow in the same direction. It follows then that any number of concurrent coplanar forces will be in equilibrium if their force polygon closes and that any side of the force polygon represents the equilibrant of all the other forces. Equilibrium of Coplanar, Nonconcurrent Forces. In the case of coplanar nonconcurrent forces the closing of the force polygon is not alone sufficient proof of equilibrium. This will be seen from the following simple example. The three forces FI, F 2 and ^3 are equal and act in the same plane making an angle of GRAPHICS 1 20 degrees with each other. It is evident that, being an equilat- eral triangle, the force polygon closes. For the system to be in equilibrium, however, the equilibrant of F 2 and F z should coin- cide with E. It is evident that the equilibrant of the system would be a clockwise or negative moment FI X a. Hence, when a system of nonconcurrent forces is in equilibrium the force FIG. 5. FIG. 6. polygon must close and the sum of the moments of the forces must be zero. It follows that three nonconcurrent forces can- not be in equilibrium unless the forces are parallel and that the resultant of the group of nonconcurrent forces may be either a single force or a couple. Components. In a system of forces having a resultant each force is a component of the resultant. Hence, a force may have any number of components. Composition of Forces. Composition of forces consists in finding for a system of forces an equivalent system having a smaller number of forces. The simplest case of composition of forces is finding a single force replacing a system of forces. Resolution of Forces. Resolution of forces consists in finding for a system of forces an equivalent system having a greater number of forces. An illustration of this is where a given force is resolved into two or more forces or components. EQUILIBRIUM POLYGON A system of forces is shown in Fig. 7. Fig. 8 is the force polygon. FI and F 2 intersect and the line of action of their resultant must pass through their point of intersection and be GRAPHICS AND STRUCTURAL DESIGN parallel to a, the resultant of FI and F 2 in the force polygon. Similarly the resultant 6 of a and F 3 must pass through the point of intersection of a and F 3 in Fig. 7, and be parallel to b in the force polygon, Fig. 8. In this way we finally reach R, the result- FIG. 7. ant of F 4 , with b, the resultant of all preceding forces FI, F 2 and F 3 . In Fig. 7, R is located by the fact that it must pass through the intersection of F 4 and b. It must also be equal and parallel to R in Fig. 8. Figure TO is an equilibrium or funicular polygon. For the system to be in equilibrium a force equal and opposite to R would have to be added and its line of action would have to pass through the intersection of F 4 and b in Fig. 10. In the above system the force polygon, Fig. 9, closes but F 5 does not coincide with the resultant R of the other forces FI, F 2 , FIG. 9. FIG. 10. F 3 and F 4 , in the equilibrium polygon, hence, the system lacks equilibrium by the clockwise or negative moment F 5 X d. The system is in equilibrium for translation. The resultant GRAPHICS 25 of the system is F 5 X d. The equilibrant of the system is + F 5 X d. The method just given applies when the several forces can be conveniently made to intersect. It should be noticed that an infinite number of equilibrium polygons are possible for a given system of forces. When the forces do not intersect conveniently the equilibrium polygon is drawn as follows: Take any point O, called a pole, outside of the forces FI, F 2 and F 3 and connect the extremities of these forces with the lines called rays, a, b, c and d. The force FI will be hel'd in equi- librium by the two rays a and , acting as indicated by the arrows inside the triangle formed by the force FI and the rays a c -- FIG. ii. FIG. 12. and b. In the same way b and c hold F 2 , and c and d hold F 3 in equilibrium, also R, the resultant of FI, F 2 and F 3 , is held in equilibrium by the rays a and d. Now if two forces represented by the rays a and b hold FI in equilibrium, then in the equilibrium polygon these three forces must intersect in a point. Take any point on FI in Fig. n, and through it draw the lines called strings parallel respectively to b and a in Fig. 12. F 2 is held in equi- librium by the rays b and c\ therefore, in Fig. n, produce b until it intersects the force F 2 and through .this point of intersection draw the string c parallel to the ray c, Fig. 12. Through the point of intersection of string c and force F 3 in Fig. 1 1 draw the string d parallel to ray d in Fig. 12 and produce it until it cuts string a. Now since the resultant R, Fig. 12, is held in equilib- rium by the rays a and d, these three forces R, a and d must in- tersect in a point in Fig. 1 1 . The intersection of a and d locates R in Fig. n, its magnitude and direction being given by Fig. 12. 26 GRAPHICS AND STRUCTURAL DESIGN Figure 12 is the force polygon. Fig. n is the equilibrium polygon. The difference between these two polygons should be carefully noted. The force polygon gives the direction and magnitude of the forces while the equilibrium polygon gives the lines of action and the direction but not the magnitude. GRAPHIC MOMENTS Let Fi,F 2) F 3 and F be four forces constituting a system whose bending moment is desired about a point p. Draw the force FIG. 13. polygon (Fig. 14), take a pole 0, draw the several rays and draw the pole distance Os perpendicular to the resultant R; this dis- tance is H. In Fig. 13 draw the equilibrium polygon making the strings parallel to the rays in Fig. 14. Through the intersection of strings a and e draw the resultant R. Through the point p draw y parallel to R and limited by the string e and the string a produced. The bending moment of R about the point p is R X h. The triangles 123 and 4 5 are similar, hence R:H::y:h or Rxh = HXy. That is, the bending moment of any system of coplanar forces about any point in the plane is the product of the pole distance H, and the line y, drawn through the point p, parallel to the resultant of the forces R and limited by the two strings e and a which intersect on the resultant. GRAPHICS 2 7 Moment of Parallel Forces. The equilibrium polygon may be used to determine the moment of any or all of a system of parallel forces. The bending moment on the beam at the section y-y due to the forces R^ and FI at the left of that section is equal to 1-2, -1-6 FIG. 15. FIG. 16. the intercept in the equilibrium polygon multiplied by the. pole distance H. Expressed algebraically the bending at y-y equals M=(R 1 Xg)- (F l X d) Now triangles 135 and 680 are similar, hence 1-5 :g:: 6-8: H. (2) Also triangles 245 and 670 are similar, so that 2-5 : h :: 6-7 : Z7. (3) In equation (2), 6-8 = RI, therefore RI X g = 1-5 X H. In equation (3), 6-7 = FI, therefore FI X h = 2-5 X H, and M = (1-5 X #) - (2-5 X H) = 1-2 X #. The bending moment at any point on a beam due to a system of parallel forces is equal to the ordinate of the equilibrium polygon, cut off by a line drawn through the given point and parallel to the resultant of all the forces multiplied by the pole distance. By bending moment on any beam section is meant the algebraic sum of the moments to the left of that section. The intercept in the equilibrium polygon is a distance and should be measured to the same scale as that to which the beam is laid off. The pole distance is a force and should be measured by the scale used in the force polygon. 28 GRAPHICS AND STRUCTURAL DESIGN USES OF FORCE AND EQUILIBRIUM POLYGONS Reactions of a Beam. A supported beam of span L carries a load F a distance a from the left support. Lay off F in Fig. 18, take the pole and draw the rays a and b. Through any point 2 on line of action of load F in Fig. 17 draw the two strings parallel to the rays a and b. Through the points i and 3 where the rays a and b respectively cut the lines of the I -c-------t "VjLx^ I. FIG. 17. FIG. 1 8. reactions R\ and R 2 draw the string c. Now in Fig. 18 draw the ray c through the pole and parallel to the string c in Fig. 17. This ray c cuts the force F into two parts representing the two reactions, RI being held in equilibrium by the rays a and c and R 2 by the rays c and b. Problem. In Fig. 19 find the reactions when FI = 10 tons (20,000 Ibs.). F 2 = 2 tons (4000 Ibs.). ^iw^ze'er. RI = 9300 Ibs. R 2 = 14,700 Ibs. Problem. Draw force and equilibrium polygons for the beam in Fig. 20, finding reactions RI and R% and the maximum bending moment. Check the answers by algebraic calcula- tions. Problem. Draw force and equilibrium polygons for the beam in Fig. 21, find the reactions RI and RZ and the maximum bend- ing moment. Check the answers by algebraic calculations. In treating uniform loads graphically it is necessary to divide the load into small sections and the load corresponding to each section is considered at its center. See Figs. 22 and 23. GRAPHICS 2 9 l RI h l -T- - FIG. 19. FIG. 20. iL ^ |*--6 L ->j*-* ! >)<--5 20-- FIG. 21. Force Polygon FIG. 23. K--8-W 8^-J FIG. 24. FIG. 25. ! 3" i- -* ^ 14- FIG. 27. FIG. 26. 30 GRAPHICS AND STRUCTURAL DESIGN Problem. The beam shown in Fig. 24 carries a uniform load of 1000 Ibs. per foot of length; draw the shear and bending- moment diagrams, using the force and equilibrium polygons. Determination of the Center of Gravity. Divide Fig. 25 into convenient regular sections, in this case three rectangles. Assume forces acting through the centers of gravity of these rectangles representing their areas. Then, by means of force and equilibrium polygons (Figs. 26 and 27), locate the resultant or equilibrant of these three forces. The location of the vertical equilibrant is shown. Similar treatment of these areas taken as acting horizontally will give the horizontal equilibrant and the center of gravity of the figure will lie at the intersection of these two equilibrants. DEFLECTION or BEAMS For beams carrying irregular loadings the graphical deter- mination of deflections is convenient. Books on mechanics of materials deduce the general equation of the elastic curve of d* M If the equilibrium polygon representing the bending moment upon the beam is known the curve of its deflection may be drawn in the following way : The given bending-moment diagram is divided into sections and the areas of these sections are represented by the lines A, B, C, etc., in Figs. 28 (a) and (d). In the funicular polygon, the pole is taken a distance to represent El, E being the modulus of elasticity of the material of the beam, expressed in pounds per square inch, and / the moment of inertia of the section, in inches 4 , and referred to that axis of the beam about which the deflec- tion occurs. Fig. 28 (c) is an equilibrium polygon drawn in the usual way for the two Figs. 28 (a) and 28 (d). The intercepts y in the diagram, when properly scaled, give the deflection at that point on the beam. To determine the scale if i inch measured horizontally in Fig. 28 (a) represents s inches of span, i inch in GRAPHICS Fig. 28 (b) equals P pounds, H the pole distance in Fig. 28 (b) measured in inches, H 1 the pole distance of Fig. 28 (d) also measured in inches, further in Fig. 28 (d) i inch equals k square inches of the bending moment area of Fig. 28 (a). Then i inch of intercept in Fig. 28 (c) represents k-P H E.I inches. FIG. 28. One square inch of area of the bending moment diagram, Fig. 28 (a), represents P s 2 - H (pound inches 2 ). The correctness of this method may be shown by taking a section of the moment curve and dividing it into strips having a width dx. The area of one of these sections is M* dx. Com- paring the similar triangles Oab and the infinitesimal triangle, M dx d 2 y M d*y we have = ^ or - = ^ . This is the general equation Hi JL dX Hi L dx of the elastic curve, as well as the equation of the line in Fig. 28 (c). 32 GRAPHICS AND STRUCTURAL DESIGN Continuous Beams. Another use of the curves of deflec- tion is the solution of restrained or continuous beams with irregular loads and spans. The following is a modification of the method credited to Dr. Geo. Wilson, "Proceedings of the Royal Society," Vol. 62, Nov., 1897. Force and Vector Polygons^ (ft) FIG. 29. The method consists of plotting the bending moment and deflection curves due to the external loads assumed as carried by the outer supports. Similar curves are then drawn for the same beam excepting that now the actual loads are removed and as- sumed reactions are applied at the intermediate supports, the beam being held as before at the outside reactions. If the beam is level the deflections caused by the first loading must be equal and be opposite to those produced by the second loading. Equat- GRAPHICS 33 ing these values of the deflections at the several supports gives sufficient data for the determination of the reactions and when these are known, a revised moment diagram for the actual beam is readily made. The solution is simplified by using the same pole distance for all the force or vector polygons, and using the same scale for the forces in all diagrams. An example will show the method in detail. A continuous beam, Fig. 29 (a), carries three loads on four sup- ports. The loads are, AB = 1000 Ibs. ; BC = 800 Ibs. ; and CE = 1400 Ibs. The lengths of the spans beginning with that at A are 12 ft., 10 ft. and 8 ft. The loads are placed as shown. The curve of deflection Fig. 29 (e) is first found for a beam sup- ported at the ends and carrying the three loads, AB, BC and CD. The force polygon Fig. 29 (c) is drawn, from which the bending moment diagram readily follows in Fig. 29 (b). This polygon Fig. 29 (b) is then divided into a number of small sections of uniform width, similar to the one cross-hatched. Since they have the same widths their areas may be represented by their middle ordinates, shown heavy in the cross-hatched section and marked 5. These middle ordinates may now be laid off in the vertical line of the vector polygon Fig. 29 (d). From this vector polygon the deflection curve Fig. 29 (e) is obtained in a manner similar to that used for the bending-moment diagram Fig. 29 (b). Under the support GF the deflection in Fig. 29 (e) scales 51, while under FE it scales 40. Now removing the loads AB, BC and CD, assume a unit load GF applied below the beam and acting along the line of the reaction. As before construct a force polygon Fig. 29 (f), the equilibrium polygon of bending-moment diagram Fig. 29 (g) and finally the deflection curve Fig. 29 (i). In this polygon the deflections for the unit load at the lines of the reaction scale 27 and 19. In the same way place a unit load at EF and draw the polygons /, k and m. The deflections over the reactions are 19 and 1 8. The reactions will be some multiples of the unit loads applied. These unit loads may be any convenient one as i, 34 GRAPHICS AND STRUCTURAL DESIGN ioo, 500 or 1000 Ibs., depending upon the beam and its loading. If we call the reaction GF, p times the unit load and FE, q times the unit load, then the deflections due to GF will be 27 X p and 19 X p, respectively, while those due to FE will be 19 X q and 1 8 X q. Since the deflections at the points of support due to the reactions equal the deflections due to the loads we have in this problem, under GF (27 X p) + (19 X q) = 51 and under FE (19 X p) + (18 Xq)= 40. Solving, these equations give p = 1.26 and q = 0.92. The unit load at the reactions having been taken at 1000 pounds, the reactions become GF = 1.26 X 1000 = 1260 Ibs., while FE = 0.92 X 1000 = 920 Ibs. The reaction AG may be found by taking moments about the right support ED. Ar , (ioooX24)+(8ooXi5)+(i40oX4)-(92QX8)-(i26oXi) A(jr = ^ 3 = 385 Ibs. Since the sum of the reactions equals the sum of the loads the reaction ED equals (1000 + 800 + 1400) (385 + 1260 + 920) = 635 Ibs. It should be noted that the pole distances were all equal and the same scale of forces was used in all force polygons. When all the forces and reactions are known the bending moments may be found either graphically by constructing the force and equilib- rium polygons as usual, or the moment at any section may be found by computing the moment of the forces acting on one side of the section. CHAPTER III STRESSES IN STRUCTURES THE application of force and equilibrium polygons to the determination of stresses in structures is fairly simple espe- cially if care is taken in properly marking the structure acted on by the external^ forces and the lines of the diagram as drawn. In the simple truss, Fig. 31, having located the forces at the several points, place a letter in each triangle of the truss and a letter between each pair of external forces. The truss being Upper Chord FIG. 30. FIG. 31. FIG. 32. FIG. 33. FIG. 35. symmetrical and symmetrically loaded the reactions will be equal and since the sum of the vertical forces must be zero each re- action must equal one-half the total load, hence the reaction is BC + CD + DP' + D'C f + C'E' 2 35 AB = 36 GRAPHICS AND STRUCTURAL DESIGN To assist the explanation, numbers have been placed at some of the points but this is not usually required. Considering the forces acting at apex i, AB and BC are known and the resultant of these is held in equilibrium by forces whose lines of action are CE and EA. The values of these two forces can, therefore, be found by drawing the force polygon of the four forces acting at point i. This has been done in Fig. 32. It is important to place the directions on these forces and then indicate these directions as acting to or from their point of application. This has been done at apex i. As soon as an arrow is placed at i on CE the arrow can be placed on CE at apex 2, since it must be opposite in direction to the first arrow. Similarly, the arrow on EA can be placed at apex 3. Now considering the forces acting at apex 2, EC and CD are known and, therefore, the other two, DF and FE, being known in direction, can be found in magnitude as in Fig. 33. Again taking the forces at apex 3, we now know EF and EA from which FG and GA can be found. This has been done in Fig. 34. In this manner, force polygons could be drawn for all the apices. Generally, instead of drawing these force polygons separately as just done, they are superimposed upon each other making a diagram, Fig. 35. Where the truss is symmetrical about a ver- tical axis through the center of its span and the corresponding loads on the two sides are equal, it is only necessary to make the portion of the diagram shown in full lines. The other half is shown dotted and by rotation about AE could be super- imposed upon the full-line diagram. In this case, drawing this dotted diagram is unnecessary. TENSION OR COMPRESSION IN THE MEMBER Where the forces acting upon the structure are indicated as shown, a piece having the arrows < > shows that piece to be in compression, while < indicates tension. STRESSES IN STRUCTURES 37 NOTE. In analyzing stresses in this way the members at the apices are assumed as pin connected and the external forces of the pieces are indi- cated by the arrows as acting on the pin. That is, the forces acting on the pin and not the forces of the pin on the pieces are indicated by the arrows. Frequently trusses carry loads attached to the lower chords. The solution graphically follows the procedure just described. If the lower apex loads BC, CD, etc., are equal the reactions AB and AB' will be equal and each will be one-half the sum of the loads on the lower chord. Taking the forces at point i, we can lay off AB, knowing both its direction and magnitude, then laying off BF and FA parallel to the respective truss members F-G FIG. 37. the stresses BF and FA become known in magnitude. Now going to point 2, since AF and AG will be parallel lines passing through a common point A they will coincide and their intercept on FG will be a point or its value is zero. Hence the stress in FG due to loads carried on the lower chord is zero. The points F and G will be coincident on Fig. 37. Now going to point 3, GF, FB and BC are known, hence CH and HG are readily drawn and their magnitudes determined. The method of determining the direction of the forces is the same as previously explained and, as before, the members in the upper chord are shown to be in compression, those in the lower chord in tension. The vertical members, excepting LL' ', will be in compression while the diagonal members are in tension. Moving Loads Carried under Trusses. Not infrequently trolley or hoist runway tracks are carried by the lower chords of trusses. In this event the track is preferably fastened to the GRAPHICS AND STRUCTURAL DESIGN lower chord at the apices of the triangles, and the load may come upon any apex. It will generally be easy under these circum- stances to make a diagram for the load under each apex in one- half the lower chord, and by comparing the diagrams, which are preferably made to the same scale, the maximum stress in any member due to any of the several positions of the load is readily found. A diagram made for one position of the load will illus- trate the method. F-G FIG. 40. FIG. 42. The load is assumed at apex 3. The reactions AB and AB f must first be found. Draw the force polygon, Fig. 41, and the equi- librium polygon, Fig. 39. When the ray 3, in Fig. 41, is drawn parallel to the closing side of the equilibrium polygon, Fig. 39, the load BB f is divided into the two reactions. When it is required to find the reactions for all positions of the load, the work can be abridged by laying off the load BB f at the left reaction and taking any point Q in the other reaction and drawing BQ and B'Q; then the intercept a-a in the triangle under the load will be the left reaction for that position of the load. If BQ and B'Q are drawn parallel to the lower chord a-a will be the left reaction and a-b the right reaction. NOTE. The student should prove the truth of this last diagram, using both algebraic and graphical demonstrations. STRESSES IN STRUCTURES WIND LOAD ON TRUSSES 39 The wind load will be assumed normal to the left side of the roof and both ends of the truss will be fixed. The reactions AB and AB' will be parallel to the wind loads BC, CD, DE and EF. The sum of the reactions will equal the sum of the loads. The magnitude of the reactions can be found by the force and equilibrium polygons, Figs. 44 and 45. In Fig. 44, BB r is laid off parallel to the wind loads and equal to their sum. The resultant normal wind pressure on the roof FIG. 43- FIG. 46. equals the sum of the wind loads BC, CD, DE, EF and FB' and will act coincident with the force DE. Take any point p in the line of this resultant and through it draw the strings parallel to their respective rays i and 2 in Fig. 44. Then in Fig. 45 draw the closing line of the equilibrium polygon and in Fig. 44 draw ray 3 parallel to string 3 in Fig. 45. BB f will then be divided into two parts corresponding to the two reactions. Having found the two reactions the stress diagram follows naturally as in 40 GRAPHICS AND STRUCTURAL DESIGN Fig. 46. No difficulty is experienced until apex 3 is - reached. Here there are three forces known completely and three known in direction only. It is, therefore, impossible to determine the magnitude of the unknown forces unless more conditions are assumed. One solution of the problem is as follows. It can be shown that the forces acting in FL, LM and MA will be the same whether the truss is left as shown by the full lines JK and KL, or if these are removed and replaced by the member shown dotted,./!,. Taking moments about point 5, cutting the truss as shown by the line and equating internal and external moments we have 2 external moments = (force FL) X a. As the other two forces pass through point 5, their moments about 5 are zero. Hence, the force in FL depends only upon the external moments and the lever arm a, and is consequently inde- pendent of the truss members to the left of the cut. In a simi- lar way stresses in LM and MA can be shown to be independent of truss members to the left of the cut. Having found the tem- porary forces acting at 3, when JK is removed complete the dia- gram for the forces at 5 and 6, then remove the dotted member JL and replace the full members LK and KJ. Now, taking the forces acting at 6, EF and FL are known completely; therefore, the magnitudes of EK and KL can be found. From here taking the forces at point 7 and then returning to point 3 all the forces act- ing in the left half of the truss become known. The diagram can be closed by finally taking the forces acting at the right reaction. The stresses due to the wind in the members H'G' to L'M in the right half of the truss will be seen to be zero. It is evident that for the wind acting on the right side of the truss, the mem- bers in the right half will be stressed the same as similar members in the left half when the wind acts as shown in Fig. 46 ; hence, redrawing the diagram is unnecessary. In trusses of long span it may be necessary to have one end free to allow for expansion; this may be done by allowing the STRESSES IN STRUCTURES truss to rest upon a plate secured to the wall. The plate fastened to the lower chord of the truss has slotted holes which permit the truss to slide upon the wall plate by overcoming the friction between the plates. The friction may be greatly reduced by using rollers under the truss. The following example illustrates the method of determining the reactions and stresses due to wind loading when the truss is free at one end. The plates are placed at the right-hand end of the truss. The coefficient of friction is assumed as one- third; hence the tangent a = J or the inclination FIG. 48. FIG. 47- FIG. 49. is one in three. The line of action of the right reaction is, therefore, known. The first part of the problem requires the determination of the magnitude of the right reaction and of both the magnitude and the direction of the left reaction. It is known, however, that the left reaction must pass through point p of the truss. -First draw the force polygon, Fig. 48. Through B f draw a line parallel to the right reaction. Take any pole and draw the rays i, 2, 3 and 4. In Fig. 47, draw the cor- responding strings. Since the left reaction AB, the force BC and ray i are in equi- librium they must intersect in a common point in the equilibrium polygon; this must be point p, the only point common to BC and the left reaction AB. Hence, through p draw a string parallel to ray i. From Fig. 48 the force BC is held in equilibrium by the strings i and 2 ; hence in Fig. 47 they must have a common GRAPHICS AND STRUCTURAL DESIGN point. Through p draw also the string 2 parallel to ray 2. Force CD and rays 2 and 3 are in equilibrium and must pass through a common point in the equilibrium polygon, Fig. 47. From Fig. 48 rays 4 and 5 hold reaction AB' in equilibrium and rays i and 5 hold reaction AB in equilibrium; hence string 5 must pass through the point of intersection of string 4 with reaction AB' and it must also pass through the intersection of string i with reaction AB which is in point p. It is noticed that string i becomes merely a point in Fig. 47. This work would have been simplified by using a resultant wind pressure equal to BB', acting in the same line as CD in Fig. 47. FIG. 50. FIG. 52. The stress diagram is shown in Fig. 49, its method of construc- tion differing in no way from those previously described. It should be noted that the diagram is for the entire truss, the loading being unsymmetrical, and that the diagram checks if the force B'E' when drawn through the points B f and E' in Fig. 49 is parallel to the line B'E' of the truss, Fig. 47. The diagram should be constructed for the wind on the free side and such construction is shown in Figs. 50 to 52. When rollers are used at the free end the reaction at that end is assumed vertical. It is shown, page 54, how the stresses may be obtained algebraically by the method of moments and how the moments STRESSES IN STRUCTURES 43 due to external loads may be found graphically. A combination of these two methods may be used thus: Figure 56 is the force polygon for the external loads, with pole and the rays drawn. From this the equilibrium polygon, Fig. 55, has been drawn. Now the bending moment at any section is the product of the intercept under that section in the equilibrium polygon multiplied by the pole distance H. Hence to determine the stress in any member DF cut the truss at Z-Z and replace the cut members by the forces that would have to act in them to produce equilibrium. Now, if possible, take FIG. 53. FIG. 54. FIG. 55. FIG. 56. moments about a point common to all but one of these forces, in this case point i . Let a be the lever arm of the force DF about point i, then making the sum of the internal and external mo- ments zero we have (H X y) + (DF X a) = o, from which The Character of the Stress, whether tension or compression, can be determined as follows: Calling clockwise rotation positive and counterclockwise neg- ative find the character of the moment of the external forces H X y. This is found to be positive. Since the sum of the internal and external moments about a point is zero the moment 44 GRAPHICS AND STRUCTURAL DESIGN of the -internal forces must be of opposite character to the mo- ment of the external forces hence DF X a is negative and DF must act in theopposite direction to that assumed. This makes DF in compression. STRESSES IN STRUCTURES 45 The determination of the dead-load and snow-load stresses in the truss of a bent presents no greater difficulties than that of a truss carried upon brick walls, as there are no stresses in the knee braces due to these loads. Where the wind load is carried by the columns and truss forming the bent the magnitude of the forces acting in the truss members and of the stresses acting in the column section vary greatly with the distance from the top of the column to the foot of the knee brace and with the manner of securing the column to the foundation. The wind upon the truss may be considered as either horizontal or normal but will here be assumed as normal. The columns will first be considered as hinged top and bottom and then as fixed at the base and hinged at the top. In this latter case it can be shown that the point of contraflexure lies between the base of the column and the foot of the knee brace and between one- half and five-eighths of this distance from the base of the column. It is generally assumed as one-half for convenience, and this is sufficiently accurate. To expedite the determi- nation of the stresses the forces acting upon the column have been transferred to the truss through the extra truss mem- bers added temporarily to the columns. It should be noted that vertical sections can be taken through all members of the truss without cutting these added pieces, hence the stresses in these members will be independent of the stresses in the added pieces and the diagram drawn with their assistance will give the correct stresses in the permanent members of the truss. The wind load has been assumed as 20 Ibs. per sq. ft., and the normal pressure has been estimated by the formula p P N = X A . The slope is i in 4, the angle being approximately 45 27 degrees. rtXN P N = X 27 = 12 Ibs. 45 4 6 GRAPHICS AND STRUCTURAL DESIGN The following dimensions have also been used: span 36 ft. o ins., panel widths 16 ft., base of column to lower chord of truss 14 ft., lower chord of truss to foot of knee brace 5 ft. o ins. In Case I the columns are hinged both top and bottom, while in Case II they are fixed at the base and hinged at the top. The FIG. 60. FIG. 61. Scale fs* ' 4000^ CASE II COLUMN BASE FIXER, M" COLUMN TOP -HI NGED lettering has been made identical in both cases and, the scales being the same, the influence of the method of securing the columns upon the stresses in corresponding members can be seen at a glance by comparing the stresses in the stress diagrams, Figs. 59 and 62. The following explanation refers to Case I, but applies also to Case II. It is first necessary to estimate the apex wind loads on the sides and roof. On the side of the building the wind is assumed as acting at the base of the column, the foot of the knee brace and the top of the column. The normal STRESSES IN STRUCTURES 47 wind pressure on the roof has a resultant W N equal to the sum of apex loads EF, FG, GH and HB f , and acting centrally with this side of the roof as shown. Similarly, the horizontal wind pressure WH equals the total horizontal wind pressure between two bents and acts centrally upon the side of the building. Now these two wind forces acting on the building have a resultant acting through their point of intersection given in direction and magnitude in the force polygon, Fig. 58. Taking any point in this resultant R w in Fig. 57, an equilibrium polygon can be drawn, since it is known that the resultants AB and AB f must pass through the column bases. In Case II these resultants pass through points in the columns midway between the base of the column and the foot of the knee brace. Having drawn the strings i, 2, and 3 in Fig. 57, draw the rays i and 2 in Fig. 58, and their point of intersection Q will be the desired pole. The usual assumption is that the columns share the horizontal components of the wind forces equally. In Fig. 58 the horizontal component of the wind forces is BS, which has TU drawn perpendicular to it from its center. If, through the pole Q, ray 3 is drawn parallel to string 3 in the equilibrium polygon in Fig. 57 it will cut TU at the point of intersection of the resultants AB and AB' acting at the column bases. In Case II these resultants act on- the columns midway between the bases and the knee braces. Having found the resultant wind forces acting on the columns the drawing of the stress diagram presents no unusual difficulties. To transfer the forces from the column bases to the truss members C7, //, JD, B'l', I f J f and J'B' have been added. The stresses given by the diagram for the column AI, AT ', K'J' and KJ will not be the correct stresses for the actual structure, but the stresses of all other truss members including the knee braces are the desired ones. In the drawings of the truss the members in compression have been drawn heavy. Maximum Bending due to Moving Loads. It has been shown, page 26, how the equilibrium polygon can be used to GRAPHICS AND STRUCTURAL DESIGN determine the bending moments upon a beam. By a simple extension of the principles a diagram can be made showing the maximum bending that will occur along the beam as a system of moving loads passes over the span. The loads EC, CD, DE and EF in Fig. 63 are a constant dis- tance apart and roll across the girder ab. The several positions of the system of loads relative to the girder can either be shown by redrawing several additional positions of the loads or by redrawing the girder, moving the girder under the loads. The * H -* FIG. 64. latter method, being the easier, has been used and two additional J positions of the girder c-d and e-f have been drawn. Fig. 64 is / the force polygon for the four wheel-loads, a pole was chosen V and the rays i to 5 inclusive were drawn. Draw the equilibrium polygon for the several loads, the strings being drawn parallel to their respective rays. Strings i and 2 will intersect on the force BC, strings 2 and 3 will intersect on force CD, etc. Now draw the closing line of the first equilibrium poly- gon by drawing gh through the points of intersection of the ex- tended reactions AB and AF of the girder. Then for the first\ position of the loading upon the girder the bending moment at any J point on the girder will be the product of the intercept measured ^ in feet to the scale of the span by the pole distance H measured i by the scale of forces. In the same way move the girder to the second position c-d and locate the closing line of the second equi- librium polygon in the line io. Proceed in this way until the loads J STRESSES IN STRUCTURES 49 have been rolled across the girder. It now remains to compare the bending moments at definite points on the girder for the several positions of the loads. Suppose it is desired to know the maxi- mum bending moments at p intervals across the girder, one would begin by comparing the intercepts in the several equilibrium polygons a horizontal distance p from the left end. Several of these ordinatesjV, kl and mn have been drawn. The comparison may be assisted by drawing a curve through the points of inter- section of these ordinates with the closing lines of their respective equilibrium polygons. This is shown by the curves RR and SS. The intercepts between the curves RR and 55 and the broken line of the equilibrium polygon will give the bending moments exactly only at the vertexes of the equilibrium polygons; there- fore, when the maximum appears to be between vertexes, the ordinate should be checked by drawing an equilibrium polygon with the loads in a position to bring a vertex at the desired point. Diagram of Maximum Live-load Shears. This will be ex- plained by a loading somewhat resembling the usual locomotive and train load. In making the diagram of maximum shears, Fig. 66, lay off the forces BC, CD, etc., and take a pole distance OB equal to the girder span. Draw the rays i to 10 and beginning at O draw the strings i to 10 parallel to their respective rays and in ac- cordance with the usual method, i.e., CD in the force polygon is held in equilibrium by rays 2 and 3; hence, in the equilibrium polygon, strings 2 and 3 must intersect force CD in a common point. The side of the equilibrium polygon under the uniform load is a parabola and can be drawn tangent to the broken line. To understand the theory of the diagram draw the closing line OA , string 1 1 , of the equilibrium polygon ; this is also ray 1 1 of the force polygon and divides the line of the forces B-H into the reactions. B-A is the left reaction when the load BC is over the left support with the other loads as shown. Suppose the loads moved to the right a distance a or the same relative position of the loads 5 GRAPHICS AND STRUCTURAL DESIGN and girder is more readily obtained by moving the span a dis- tance a to the left as designated by position 2. The closing line in this case is string 12 and the reaction is MN. As the loads travel to the right the shears under load BC can be obtained by measuring from point B the distance the load BC is to the right of the left reaction and at this point measuring the ordinate from OB to the broken line 2,3- - to 10 of the equilibrium polygon. This of course must be measured by the same scale as that to which the forces are laid off. When the usual locomotive wheel loads are FIG. 65. 5pan-Positioa* J FIG. 66. preceded by a much lighter pilot wheel there may be a question as to whether the greatest shear at any section occurs when the pilot wheel, load BC, is at the section or the first driver, load CD, is there. In Fig. 66, measure the span marked position 3, produce string 2 until it cuts PQ, the line of the right reaction. From the point where string 2 cuts BH draw a heavy horizontal line to the left. To find the maximum shear a distance b from the left support draw the intercept c-dj a distance b from the line QP ; this will be the shear when the load CD is at the distance b from the left support. Compare c-d with the intercept e-f drawn between the broken line of the equilibrium polygon and the horizontal line OB and at the distance b from the line of forces STRESSES IN STRUCTURES 51 HB. In this way the maximum shear may be found for any point. This method is applicable to the determination of the maximum stresses in truss members as well as in plate girders. The following illustrates the method applied to the determi- nation of the stresses in a Pratt truss. Example. The Pratt truss, Fig. 67, spans 150 ft. It has 6 panels and height of 30 ft. The loading is Cooper's E-6o; see page 71, under Influence diagrams, for this loading. Find the stresses in BC, be and diagonal bC, locating the positions of loads for maximum stress and shear by influence diagrams and using equilibrium diagrams for determining the maximum bending moments and shears. CHAPTER IV ALGEBRAIC DETERMINATION OF STRESSES THE conditions of equilibrium used in the determination of stresses are: 1. Sum of the horizontal forces = o. 2. Sum of the vertical forces = o. 3. Sum of the moments of the forces about any point = o. The stresses in the members cut by the section a-a can be replaced by forces FI, F 2 and F 3 , equal to the respective stresses FIG. 69. in these members. These forces FI, F 2 and F 3 hold the portion of the truss to the left of the section, Fig. 68, in equilibrium and the stresses in the members may be determined by using con- ditions i, 2 and 3. Referring to conditions i and 2, they should be understood to mean that the sum of the components along any line must = o, and the sum of the components along a line at right angles to the first line must = o. Frequently the work may be abridged by using the more gen- eral scheme. Forces and components acting to the right or up- wards are positive, those acting to the left or down are negative. Clockwise moments are positive; counterclockwise moments are 52 ALGEBRAIC DETERMINATION OF STRESSES 53 negative. To find the stresses in CE and EA in Fig. 70, cut the truss at a-a t then, by condition 2, (AB - BC) + CE sin a = o, CE = - (,4 - BC) cosec a, also -CEcoso! + ,4 = o or AE = CEcos hence x = 6. In this way all the coefficients can be found and stresses com- puted by multiplying these coefficients by W sec 6 for diagonal members or by W tang 6 for horizontal chords. Live Loads. To analyze the influence of live loads at the several apexes of the lower chord, construct Fig. 77. With the load at apex 3, the left reaction is y P, the right reaction y P. The coefficients of the web members to the left of the load, apex 3, are f while those to the right are y. In the same manner from Fig. 78, which is the stress diagram for the load P at apex 5, it is seen that the coefficients of the web members to the left of apex 5 are T while to the right the coefficients are y. ALGEBRAIC DETERMINATION OF STRESSES 57 58 GRAPHICS AND STRUCTURAL DESIGN It will be seen that for maximum chord stresses the truss should be fully loaded, while for minimum chord stresses there should be no live load on the truss. In this type of truss all loads on upper or lower chords produce tension in the lower chord members and compression in the upper chord members. The maximum web stress will be produced when the longer segment from that panel to a pier is fully loaded. The minimum web stresses will be created when the shorter segment from panel to pier is fully loaded. PRATT TRUSS The analysis of the stresses in a Pratt truss can be done by the method of coefficients in the same way as the Warren truss just described. If, as is sometimes done, parts of the dead loads are assumed as applied at the apexes of the upper chord the only members whose stresses will be affected are the vertical ones. To find the shear in these verticals cut the truss diagonally through the vertical member whose stress is desired and equate the shears. (See Fig. 83.) /-/ = (C-D) - (C-B) - (B-A) - (D-E). Had there been no loads upon the upper chords, as in Figs. 81 and 82, then /-/ = (C-D) - (C-B) - (B-A). The maximum chord stresses are equal to the sum of the dead- and live-load chord stresses, while the minimum chord stresses are those due to the dead load only. The maximum and mini- mum web stresses are found by adding algebraically the corre- sponding live- and dead-load stresses. Since in this truss the diagonals can take only tension the necessity for counter diagonals to care for stresses due to un- symmetrical live loading should be noted. The stress in the counter is the same as the compression would have been in the piece for which it acts. ALGEBRAIC DETERMINATION OF STRESSES 59 w FIG. 81. FIG. 82. 6000 FIG. 84. 6o GRAPHICS AND STRUCTURAL DESIGN WARREN TRUSS Problem. Find the dead-load stresses by the method of coefficients in the following deck Warren truss, Fig. 84. Span 100 ft. Load at apexes on the upper chord, 12,000 Ibs. The secant of 45 degrees is 1.414. Tangent of 45 degrees is i. Shear in panel C = 3.5 X W = 42,00x5 Ibs. Shear in panel D = 42,000 12,000 = 30,000 Ibs. Shear in panel E = 42,000 12,000 12,000 = 18,000 Ibs. Shear in panel F = 42,000 12,000 12,000 12,000 = 6000 Ibs. Stress in EG = - 3% W sec0 = - 3.5 X 12,000 X Stress in HI = + *\ W sec 6 = + 2.5 X 12,000 X Stress in JK = i| Wsec 6 = 1.5 X 12,000 X Stress in KL = + i W sec 6 = + 0.5 X 12,000 X Stress in HA = + 3.5 W tang 6 = + 3.5 X 1 2,000 X Stress in KA = + 7.5 Wt&ngd = + 7.5 X 12,000 X .414 = 59,400 Ibs. .414 = +42, 400 Ibs. .414 = 25,500 Ibs. .414 = + 8,500 Ibs. = + 42,000 Ibs. = -f- 90,000 Ibs. Stress in DI and EJ = Stress in FL Stress in CG Stress I'D GB = 6 W tang 6 = 8 W tang 6 = o 6000 Ibs. -6X 12,000 Xi - 8 X 12,000 X i 72,000 Ibs. 96,000 Ibs. CHAPTER V INFLUENCE DIAGRAMS WHEN a system of concentrated loads moves across a girder or a trussed bridge the maximum moment or shear at a section, or stress in a given member, will be produced by a certain position of the system of loads relative to that section or member. Influence lines are used to determine this position of the loads producing maximum moments, shears or stresses. Influence Diagrams. An influence diagram shows the varia- tion of the effect at any particular point, or in any particular member, of a system of loads moving over the structure. In- fluence diagrams are commonly drawn for a load of unity. The moments, shears, or stresses for any system of loads can be com- puted from the intercepts in this diagram by multiplying them by the given loads. Influence diagrams to find the position of loading to give maxi- mum moment at a given point in a beam or girder or at a given joint on the loaded chord of a truss. In Fig. 85, SPi is the resultant of moving loads to the left of point 3. ZP 2 is the resultant of moving loads to the right of point 3. To construct the influence diagram, Fig. 86, for the bending moment at 3, compute the bending at 3 due to a unit load at this point and If p is laid off on the left reaction and (S p) upon the right reaction and their extremities are joined to the ends of the hori- 61 62 GRAPHICS AND STRUCTURAL DESIGN zontal line ea by ba and/e then the vertical intercept cd represents P P for, let this intercept cd be x then by similar triangles o eab and dac, - :. x = * - ^ and similarly for the p o o triangles dec and ae/. (S - p)p - -^-^- S S-p S or x= FIG. 85. .A FIG. 86. To find the moment at a given point 3, due to a system of moving loads, multiply the intercept under each load by that load and take the sum of these products. From Fig. 86 M = Now move the system of loads a small distance dx to the left, allowing no load, however, to pass on or off the span or across the given point 3. Then for the new position M + dM = ^P l (! Subtracting the preceding equation from this we have dM = - ZPi dyi + 2P 2 dy 2 . INFLUENCE DIAGRAMS 63 Now examining for maximum bending by placing this equation = o, we have 2Pidyi = 2P 2 dyt. (i) By similar triangles from which = dx S dx S dy* p Substituting this value in equation (i) ZPi (S - p) = SP 2 />; hence from which SPtS = 2 (P l + P 2 ) or Pi S(P!+P 2 ) S 7~ 5 This may be expressed by stating that the maximum bending will occur at a section when the average load to the left of the section equals the average load on the entire span. The influence diagram to find the position of the loading to give a maximum moment at a given joint on the unloaded chord of a truss, having either parallel or inclined chords, may be found in a similar way. Figures 87 and 88 are drawn for joint 4 on upper chord. If the system of load$ is moved a distance dx to the left the rate of change in the bending moment is = - ZPi dy, - SP 2 dy, + 2P 3 dys- (i) To be a maximum r- = o. dx I- GRAPHICS AND STRUCTURAL DESIGN But by the geometry of the construction dy\ _ S p . dyz _ _, dy 2 _ dS pc dx = S ' dx ~ S ~dx = cS FIG. 88. These values substituted in equation (i) give the position for maximum moment at joint 4. d P MAXIMUM SHEAR Reactions and Shears. By definition, shear is the algebraic sum of the vertical forces to the left of the section. When the unit load has moved a distance x to the left of the right support, x X i Fig. 90, the shear to the left of the load is - , and since this value under the load is reduced by unity the shear to the right L x of the load is - X i. INFLUENCE DIAGRAMS 65 The shear under the load for any position is given by the inter- cept in the triangle. For a system of loads the shear to the left FIG. 89. FIG. 90. of the first load is the sum of the products of the several intercepts in the triangle by their respective loads. In the case illustrated, Fig. 91, the left reaction or shear to the left of load i is P*y*. FIG. 91. If the entire system of loads is moved to the left a distance x, but no loads enter or leave the span, R increases an amount making the reaction 66 GRAPHICS AND STRUCTURAL DESIGN MAXIMUM SHEAR AT ANY POINT The vertical shear at a section a distance x from the left reaction, Fig. 92, is V Xl = 2Py. Under the load PI this is reduced by the amount PI making the shear to the right of PI, F 2 = S Py - PI. If the loads are moved to the left until P 2 is at section x from the left reaction the FIG. 92. shears, providing the same loads only are now on the span, will have increased an amount ZP > making JLt Comparing V Xl and V X2 it is seen that V Xl will be the greater so long as or L Under these circumstances the shear at x will be a maximum with p PI at x when - exceeds the sum of all the loads on the span p ^p divided by that span. If -p = - the shears will be the same L P ZP at x with either PI or P 2 at that point, while if -^ < the X/ maximum shear will occur with P 2 at x. INFLUENCE DIAGRAMS 6 7 Had another load P 6 come on the span, Fig. 93, the increase in shear, after the load PI had passed x and advanced a distance b to the left of it, would be Here SPi_5 represents the sum of the loads PI to P*> inclusive, but not the load P 6 just assumed as coming on the span, while SPi_6 is the sum of the loads from PI to P& inclusive. The dis- Fj C p a > C P 3 X P 4 1) C P b c p, ") r b 1 i ; * * L FIG. 93. tance c can only range in value from zero to c = b, hence the increase in shear will be somewhere between 2/^.5 - - P! and 2Pi_e - - PI Where the first expression is negative and the latter positive both positions of the drivers should be tried. Load PI at section x will give a maximum shear when P y IP and load PZ will give a maximum shear when Pi 2Pi- 5 In the case of a uniform load the vertical shear at x is a maxi- mum when the portion of the span to the right of x is fully loaded, while the shear will be a minimum when the portion to the left of x is fully loaded. Position of a sy stein of moving loads to give a maximum shear in any panel of a truss with parallel or inclined chords. 68 GRAPHICS AND STRUCTURAL DESIGN In Figs. 94 and 95, 2Pi is the sum of the loads to the left of the panel. 2P 2 is the sum of the loads on the panel, here panel 2-3. SPs is the sum of the loads to the right of the panel. m is the number of panels to the left of panel 2-3. n is the number of panels in the truss. p is the length of a panel. FIG. 94. FIG. 95. The influence lines for loads of unity ab and cd are drawn as for shear in a beam or girder. When point 3 is reached the in- tercept in the triangle abc is / _ _i_ * X (n m i) n Now as the load moves across panel 2-3 the shear is reduced gradually until when point 2 is reached the unit load is deducted making the shear T W ~~ Wl Wl gh = I = n n . The influence diagram for the shear in panel 2-3 is given by the INFLUENCE DIAGRAMS 69 broken line chea. The maximum positive shear in panel 2-3 is V = ^Pzyz + SP 2 }> 2 - ZPiyi. Moving the loads a small distance dx to the left, no load passing a panel point or end reaction, the shear becomes V + dV = 2P 3 (J3 + dy*) + 2P 2 (j 2 - dy z ) - 2Pi(yi - dyj. By subtraction dV = 2P 3 dy 3 - 2P 2 dy 2 + ZPidyi. (i) Comparing the differential triangles with the similar triangles cgh, hegf and fea t we have ^i = _JL; ^? = ^U and ^? = JL. d# w/> rfa; w/> dx np dV Now dividing equation (i) by dx and putting = o, to ex- amine for a maximum, we have ^_ 0=SP * 2Pl !L=j +2 p * w/> w/> and n This states that the vertical shear in any panel will be a maximum when the load in that panel equals the average panel load for the span. It is necessary that a load near the head of the train be at the panel point to the right of the panel in which the shear is sought. MAXIMUM FLOOR-BEAM REACTION Figure 96 shows two panels p\ and p z with floor beams a, b and c and accompanying stringers. To find the maximum floor- beam reaction at b construct the influence diagram, Fig. 96, making its ordinate at ef equal one. Let 2Pi be the sum of the loads in panel pi and S/> 2 the same for panel /> 2 , then for a maxi- y p y P i y P mum floor-beam load - - = - . Fig. 97 is the influence pi pi + pz line for the bending moment at b. 7 o GRAPHICS AND STRUCTURAL DESIGN Ordinates y and y\ similarly located in Figs. 96 and 97 will be to each other as the corresponding intercepts ef and hi or y yi hence a OOQO& OOOOO FIG. 96. FIG. 97. To find the maximum floor-beam reaction determine the maxi- mum bending moment on a beam whose span is / = pi + p2 at a distance pi from the left support a, and multiply it by the sum of the panels pi and p z and divide this by the product of pi and p% ; generally pi = pz, so that the maximum reaction will ordinarily equal twice the maximum bending moment divided by the panel width. 1 1 1 k j e p FIG. 98. When the stresses are determined by calculation, advantage is taken of the following principles in computing. The moment due to the loads about the section under P 5 in Fig. 98 is Mi = d^ 00 8 3 4-(D 10 M N M mNOloOlOO rn IQ * R ? fj ^ l * H " =s 2 5 S 3 ; M -* oo fO * -OOO .io>ot^ srO'p r; *t IH M_ " ~^- OMi-iONrOt^oo PO q. oq M_ 10 >q. -q. q -^POO a" 8 B 72 GRAPHICS AND STRUCTURAL DESIGN If the moment is desired under P 6 , P 6 being a distance c from P 5 , then M 2 = Pi Ol + C) + P 2 (* 2 + C) + . . . P 5 C. In the second instance if SP is the sum of all the loads on the span to the left of P 6 , then M 2 = Mi + SP X-c. In this way, starting at the head of the train, the moments of the loads to the left about a point under each successive load are calculated and noted. Having placed the loads in the position to give maximum bending at the point to be investigated, first find the reaction to the left of RI and then the bending moment at the point desired. If Ms is the bending due to loads PI to P 6 about the section under P 7 the bending moment about R 2 of loads PI to P 7 , inclu- sive, will be Mi = M 3 + 2 (Pi - P 7 ) e and the reaction is where S is the span. The bending moment at a section under P 5 then is M = RJ - ML The work of calculating stresses due to moving locomotive and train loads is facilitated by the use of the following table which has been computed in the manner just described for Cooper's E-6o loading. In this moment table the consecutive wheel loads are numbered from the left to the right, and the distances given between adjacent wheels, the sum of the distances from the train load to each wheel and the sum of the loads from load 18 to and including each wheel load are also given. In the body of the table above the heavy zigzag line are given the moments of all loads between the heavy vertical line at the right of the horizontal row of moments and any wheel load to the left, about the vertical INFLUENCE DIAGRAMS 73 line. Thus the moment of loads from 8 to 18 inclusive about the vertical line marking the beginning of the uniform train load is 8,785,500 ft. Ibs. (Both pounds and foot pounds are expressed in thousands.) Below the heavy zigzag line are given the moments of the loads to the right of the heavy vertical line in that row; thus the moment of the loads 9 to 15, inclusive, about load 9 is 3,720,000 ft. Ibs. Below the body of the table are given the sum of the distances from wheel i to the several wheel loads, and also the sum of the loads from load i to any other load including both this latter load and load i. A simple example will illustrate one use of this table. In a girder whose span is 80 ft. what is the left reaction when load i is over that support? Wheel 14 being 79 ft. from wheel i when i is on the left pier, 14 will be i ft. from the right pier. The moment of wheels i to 14 about wheel i is 14,400,000 ft. Ibs., read below the zigzag line. The right reaction is this moment divided by the span or - = 180,000 Ibs. The left reaction is the sum of the oo loads on the span minus the right reaction or 348,000 180,000 = 168,000. The reaction can also be determined by using the moments above the zigzag line. The moment of the loads from i to 14 is 13,092,000 ft. Ibs., but load 14 being i ft. from the right pier the moment of the loads as placed upon the girder about the right pier is 13,092,000 + the sum of the loads from i to 14 multiplied by i ft. M = 13,092,000 + (348,000 X i) = 1 3, 440,000 ft. Ibs. The left reaction is *3.44o. = xeS.ooolbs., 80 the same as found before. The use of this table and of the rules just derived by means 74 GRAPHICS AND STRUCTURAL DESIGN of the influence diagrams will be illustrated by the following problem. A through Pratt truss, Fig. 99, carries a train load represented by Cooper's E-6o loading. Find the stresses in the members of the second panel from the left pier. To find the maximum stress in member El determine the maximum bending at apex 4. The criterion for maximum bending at a point in a truss is that the average load to the left of the point shall be equal to or less than the average load on the entire span. 1 A 7 A 7 150' FIG. 99. W t f^ f*> ' = Average panel load on bridge. = 57,330 Ibs. Trying load 7 at apex 4: Panel load to left not including load 7 ^P = 51,500 = Trying load 8 at apex 4: 116,000 ^ = 58,000; To find reaction at the left, take moments of loads about right support. When load 7 is at apex 4, which is 50 ft. from the left pier, load 2 1 will be the last one on the bridge and it will be 3 ft. from the right pier. The moment about the right pier then is M = 24,064,000 + (344,000 X 3) = 25,096,000 ft. lbs., from which the reaction at the left is 25,096,000 p - **- - = 167, 100 lbs. INFLUENCE DIAGRAMS 75 The bending moment under apex 4 is ,, / 2 5, 006.000 X SON .. M = { v ) 2,155,000 = 6, 2 10,330 ft. Ibs. \ 15 / Here 2,155,000 is the bending due to loads i to 6, inclusive, about load 7. The stress in member El is 6,210,330 207,110 Ibs* Stress in number HI. To determine the live load stresses in HI, first find the maximum shear in panel 3-4. For the shear in any panel to be a maximum the load in that panel must be equal to or less than the average panel load on the bridge. Load in panel. Average load in panel. Load 4 at apex 4 Lbs. 50 ooo Lbs. 204. 000/6 *jo 670 Load 5 at apex 4 70,000 7Q4. OOO/6= ^0,670 The maximum shear in panel 3-4 occurs with wheel 4 at apex 4. To find the reaction under these conditions when load 19 is 4 ft. to the left of the right pier we have 17,784,000 + (304,000 X 4) = 1 9,000,000 ft. Ibs. T, ,. 10,000,000 Reaction = - - = 126,670 Ibs. 150 The reaction at 3 due to loads in panel 3-4 is D moments of loads 1-4 about 4 K. = - 25 ,, 480,000 R = - - = 19,200 Ibs. 25 Maximum shear panel 3-4 = 126,670 19,200 = 107,470 Ibs. If tried for load 3 at apex 4, the maximum shear will be found to be the same. Using this shear the live-load stress in HI is found to be HI = 107,470 X = 139,890 Ibs. CHAPTER VI TENSION, COMPRESSION PIECES AND BEAMS DESIGNING any piece of a structure requires the determination , of the resisting forces called forth in it to balance the external forces acting upon it. In the case of a purely tension piece this requires that the minimum or net section multiplied by the allowable working fiber stress shall not exceed the total force acting in the piece. Compression pieces whose lengths do not exceed five times their least diameter can be designed by assuming the total load equal to the allowable working fiber stress in compression times the area of the minimum section. In both cases care should be taken to have the load distributed over the section as improper applica- tion of the external forces to the piece may result in introducing bending or injurious local concentrations of stress in it. BEAMS Upon the following fundamental conditions of static equilib- rium determinations of the required forces or moments acting on or in the piece are made. This applies whether the deter- minations are made algebraically or graphically. The sum of all the vertical forces = o. The sum of all the horizontal forces = o. The sum of the moments of all forces about any point = o. Reactions. In a beam acted on by several vertical forces the sum of the reactions or forces at the supports must equal the sum of the loads, and the algebraic sum of the moments of all forces and reactions referred to any point must be zero. In Fig. 100 Pi + P2 + P* = #1 + #2. (i) 76 TENSION, COMPRESSION PIECES AND BEAMS 77 Taking moments about any point in the reaction R 2 we have M = Piai + P 2 a 2 + P&s = RJ = o. (2) The solution of equation (2) gives RI. R 2 will be given by sub- stitution in equation (i). Vertical Shear. The vertical shear at any section is the algebraic sum of all the external forces on the left of that section. Thus a shear diagram of the beam is shown by Fig. 101. At the left support the vertical shear equals the reaction RI, and this value of the shear continues toward the right until under the load PZ the shear becomes R\ Pa; similarly, under P 2 the vertical shear is RI -- (P 3 -f P 2 ) and under PI the shear ! - (P 8 + P 2 + Pi) = - #2. s 1 1 4 u c , ft I 2. i* i ^i FIG. 10 FIG. 101. Bending Moment. The bending moment at any section is the algebraic sum of the moments of the external forces on the left of that section referred to a point in that section. In the beam, Fig. 102, the bending at the section ab is M = Ric - P z e - P 2 d. Where rolled beams are used generally only the maximum bend- ing is required; this will occur where the vertical shear passes through zero. (See any book on Mechanics of Materials for the proof of this.) In the case illustrated in Fig. 101, the maxi- mum bending occurs under load P 2 . Having found the maximum bending moment, if the beam is supported laterally, the proper 78 GRAPHICS AND STRUCTURAL DESIGN section can usually be selected from a manufacturer's handbook. In building construction beams will commonly be supported laterally by flooring, roofing or bracing. The external bending moment induces an equal and opposite moment in the material of the beam called a resisting moment. This is expressed by equation where M = the external bending moment at a given section. 1 = the resisting moment at the same section. / = the extreme unit fiber stress, generally pounds per square inch. / = the moment of inertia of the beam section, in inches. e = the distance from the neutral axis to the extreme fibers in inches. The expression - depends entirely upon the form of the beam e section and is sometimes called the section modulus. An example will illustrate the selection of a floor beam. A floor beam with a span of 12 ft. is to carry a uniform load of 14,400 Ibs. Select a suitable beam allowing a working fiber stress of 16,000 Ibs. per sq. in. The maximum bending for a supported beam with a uniform Wl load is M = , where W is the total uniform load in pounds 8 and / is the length of the span in inches. Substituting the given values in this formula we have 1/r 14,400 X (12 X 12) . ,, M = - - j - L = 259,200 in. Ibs. 8 The handbooks usually tabulate the values of - of their sections. C/ * For the derivation of this formula see any book on " Mechanics of Materials." TENSION, COMPRESSION PIECES AND BEAMS 79 By transposition of the previously given equation - = , then by substitution 7 _ 259,200 _ , e 16,000 The lightest weight standard I beam providing a sufficient sec- tion modulus is a Q-in. I beam, weighing 21 Ibs. per ft. Its section modulus is 18.9. Standard Framing ^"Clearance Clearance FIG. 103. Not infrequently the beam must not only be amply strong but it must not deflect excessively under load. When plastered ceilings are carried under the beams this deflection is limited to -\- of the span. The formula for the deflection at the middle of a supported beam carrying a uniform load is WP , A = 34 El or 384EXA For steel E = 30,000,000 Ibs. per sq. in. If the deflection is A = ^ = IM = H k 360 360 5 the limiting inertia then will be / = = 5 X 14,400 X I44 3 = 6 384 E X A 384 X 30,000,000 X f " ''" It is, therefore, evident that the 9-in. I beam weighing 21 Ibs. per ft. demanded for strength will be amply stiff, its inertia being 84.9 or almost twice the inertia required for stiffness. 8o GRAPHICS AND STRUCTURAL DESIGN In framing beams into girders, beams or columns each manu- facturer has a standard framing. This framing is designed for the shortest span and consequently the greatest load for which the beam is likely to be used. FIG. 104. The standard framing, Fig. 104, should not be used for spans less than the following: I. Lb. Span, ft. 7. Lb. Span, ft. /. Lb. Span, ft. 7. Lb. Span, ft. 24 80.0-22.0 15 8o.o-2O.o 10 25.0-9.0 6 12.25-6.0 20 8o.O-22.O 15 60.0-15.5 9 21 .O-7.O 5 9-75-4-0 20 65.0-18.0 15 42.O-II.O 8 18.0-5.5 4 7-5 -3-o 18 55.0-14.0 12 40.0-11.5 ' 7 15.0-4.0 3 5-5 -2.0 12 31 e n o All rivets in standard framing are J in. in diameter. As this goes to press the American Bridge Company, in their specifications for steel structures, publishes a revised standard TENSION, COMPRESSION PIECES AND BEAMS 8 1 for framing, in which all beams use 4 in. X 4-in. angles from 27 ins. to 12 ins. inclusive; smaller beams use 6 in. X 4-in. angles. When work is being detailed for production in a particular shop the standards of that shop should be ascertained and adhered to. It should be noted in Fig. 103 that to facilitate erection the connecting angles extend f in. beyond the end of the web of the beam and that the distance back to back of the connecting angles is | in. less than the space into which the beam is fitted. When drawings are being made for a shop having standard framing the beam sketch may be like Fig. 103, no dimensions being placed on the standard rivet spaces. The advantages of the standard framing are: It simplifies shop work, enabling a large number of angles to be made at one time. The punching of the rivet holes both in beams and angles can be done by means of multiple punches, a group of holes being made at one stroke of the punch. The drawing-room work is also reduced. The total saving in time more than counterbalances any probable waste of material. A criticism of the method used for designing these connections is that there is twisting introduced into the rivet groups increasing the shear on the rivets. Tests made on some full-sized beams with standard framing seemed to justify the usual practice as being ample notwithstanding the above theoretical criticism. When, as is the case for very short spans or for beams with thin webs, it is necessary to design special connections the cal- culations should be based upon the shearing and bearing value of the material; see page 84. Riveting. There are two types of rivet heads: "button heads," which approximate hemispherical, and " countersunk," which are truncated cones. The button-headed rivets should always be used where clearances will permit. Button heads can be flattened slightly where additional clearance is needed and this requires less work than countersunk rivets. The following illustrations, Figs. 105 and 106, give the dimensions of various 82 GRAPHICS AND STRUCTURAL DESIGN .,* '< IT f-. I sap.IS c c TENSION, COMPRESSION PIECES AND BEAMS rivet sizes and symbols for shop and field driving, and are taken from the American Bridge Company's Book of Standards. Rivets are designated by their nominal diameter D and the length under the head before driving. This length, Figs. 107 and 1 08, is made up of the grip, the distance through the material held together plus the length called the upset or the stock required to form the head and bring the body of the rivet up to the size of the rivet hole. It is necessary for the rivet holes to be larger than the rivet stock, otherwise there would be h* Grip-- Upset H FIG. 107. difnculty in placing the rivets in their proper rivet holes and the stock would be cold before they could be driven. The holes are usually -^Q in. larger in diameter than the diameter of the rivets. The number of rivets required in any joint depends upon the forces being transmitted by the rivets, and upon the rivet values in shear and bearing. The value in shear may be that due to either single or double shear as the rivet tends to fail by shearing in one or two cross sections. The bearing value of the rivet is the rivet diameter times the lesser thickness of the materials transmitting the forces to the rivets times the allowable unit bearing value of the rivet material. The following tables give the shearing and bearing values of the usual rivet sizes. GRAPHICS AND STRUCTURAL DESIGN M ::::: 8 : : : : : ct a : : : : 8 00 " : : : : 3 j H. : : 2 : : ^ i H. " '. '. 33) 8 i i is S 1 S j || 1 I 5! : : : 5 8 . ^ 10 ' M M 03 H d *. ! M M I i M : : 8 g ; '. ro O X5 > 2 .S 1 2! : : S | I I O ci o 1 c 3 O Q Q 00 O ro rj IO & o w o. o otao i oo ^ ; : i^ d o M 5 oj a. - ; ; di M ro 10 1 IS "2 ill! 1 M 1 * : S S S o rvj different * 000 10 O to N 10 r> o" M ifferent t - * S ^ S ^ Q 1 M M < w PQ 1 -2 o o o o oo r^ \o <5 . % ^ 10 a o g -2 i tpll > . Tf 10 VO t^ *| "^ * M Q < | ri ro 4 10 vo" u 1 * ro l> O> i 2 2 O O O O O ci ro ro T? \r, a W -2 2 a O> rO o 8 00 (-; O vO 10 IO w w CO H. & 8 a^ M ri ro ro "f ! H. 10 O 10 1 10 O s q t^lio tf q N fO rojTT 10 ju rt f O O O O O 2^g & 3- 5 i Is , o o o o o o $ a s g- w -s M UJ i3 "- 1 J I 2 il ? 5 M w ro * >5 6 o o o o d C 1 < 6 d 6 d d d ii a 10 Q 10 o 10 r- o M o t^ ro 10 >o r* oo d d d d 6 : rivets, es. 1 d d d d d M 1 j g.S .2 o 1 " Q ^ TENSION, COMPRESSION PIECES AND BEAMS 85 A few examples will illustrate the application of the above principles. f c = the unit bearing value, pounds per square inch. /, = the unit shearing value, pounds per square inch. Evidently the rivets in Fig. 109 would tend to shear in but one place, along ab, and are, therefore, in single shear. Shearing value = - - X /.. Bearing value = d X /i X f c . 4 \t\ x~N /^"N i a * 1 * b c , .'A;'". " . ,f ,/ ] O O OQ *!< 2 Rivet-diam.=d. ^< tg Rivet-diam.=d. FIG. 109. FIG. no. In Fig. no the rivets are in double shear as they tend to fail along the sections ab and cd. Shearing value = --/,. Bearing value = d X k Xf c - 2 The following numerical example will illustrate the design of a joint like this one. Example. A joint of the type shown in Fig. no is to trans- mit 15,000 Ibs.; /i = | in., / 2 = f in., rivets f in. in diameter, the allowable shearing stress 10,000 Ibs. per sq. in., allowable bearing stress 20,000 Ibs. per sq. in. The rivet is in double shear while the minimum bearing value will be that of the rivet against the |-in. plate as in the other direction two plates each f in. thick act. From the tables the rivet value of f-in. rivets in double shear at 10,000 Ibs. is 4420 X 2 = 8840 Ibs. The rivet value in bearing on a ^-in. plate is 7500 Ibs. As the joint is to transmit 15,000 Ibs. the number of rivets is found by dividing 15,000 Ibs. by the smaller of the two rivet values, i.e., 15,000 = 2. 7500 86 GRAPHICS AND STRUCTURAL DESIGN Riveted joints should be designed to avoid eccentric stress if possible. Where this is unavoidable such eccentricity should be reduced to a minimum and the resultant forces acting upon the individual rivets be determined. An instance of such a joint, frequently met with, is the connection of the foot of a knee brace with the column in a light steel building frame. The force F, 24,000 Ibs., Fig. in, is transferred through the rivets i, 2 and 3 to the sketch plate and in turn from the plate by the rivets 4 to n, inclusive, to the column angles. The center of gravity of the rivet group, 4 to n, is easily located as its center of symmetry.* Rivets 4, 7, 8 and n are each yj ins. from the center of gravity and rivets 5, 6, 9 and 10 are each 6.18 ins. from the same point. FIG. in. The rivets must offer a resistance F^ equal, opposite and par- allel to FI. This resultant force F z acts through the center t)f gravity of the rivet group, and the direct force F 2 is shared equally by the rivets of the group so that each rivet carries = 3000 Ibs. The forces FI and F 2 are 4.75 ins. apart so 8 that the moment opposed by the group is 24,000 X 4.75 = 114,000 in. Ibs. The force opposed by a rivet will be proportional to its dis- tance from the center of gravity of the rivet group. The rivets * When the center of gravity of a rivet group cannot be seen by inspection it can be readily calculated. If the group has any axis of symmetry the center of gravity must lie on that axis. TENSION, COMPRESSION PIECES AND BEAMS 87 4, 7, 8 and n being equally distant from G will oppose equal forces and similarly of the rivets 5, 6, 9 and 10. Let S be the force acting at 4, 7, 8 and n, then '* X S is the force acting on rivets S, 6, 9 and 10. The moment of each of the rivets 4, 7, 8 and n will be 5 X 7.50 and the moment for each of the rivets 5, 6, 9 and 10 will be (S X -^- ) X 6.18. V 7.507 The total moment for the eight rivets is (4 X S X 7.50) + (4 X S X -^- ) = 114,000 in. Ibs. V 7.507 30 5 + 15.04 5 == 114,000 in. Ibs. 114,000 . S = - - = 2531 in. Ibs. 45-04 The force acting on each of the rivets 5, 6, 9 and 10 will be (S X ^) = 2531 X = 2085 Ibs. V 7-50/ 7-50 The forces FI and F 2 create a clockwise moment so that the resisting moment must be counterclockwise and the forces act to produce a counterclockwise moment about G. The forces act at right angles to the arms or lines joining the rivet centers with G. The resultant shear on any rivet is found by completing the triangle of forces at each rivet, and is indicated in the heavy line. This is done on the figure for the r ^ a '.. >J Y several rivets and the maximum shear comes on rivet n and is 5530 Ibs. while the minimum shear acts on rivet 4 and is 850 Ibs. 1 6 Should there be no axis of symme- . ~~f " 1 try take any two base lines, preferably *~ 'Y at right angles, and find the distance of the center of gravity from each of these base lines. Fig. 112 represents an irregular rivet group. The first base line is I 88 GRAPHICS AND STRUCTURAL DESIGN taken, passing through rivets i and 5. The rivets are assumed as being of the same diameter and the cross section of each rivet being A ; then the statical moment about axis x-x will be : Rivet. Area. Arm. Moment. I A O 2 A a A .a 3 A b A .b 4 A c A .c 5 A d A.o Total area = 5 A . Total Moment = A (a + b + c) . The distance g from the axis x-x to the center of gravity will equal the statical moment divided by the total area or = A (a + b + c) It is evident from this expression that the distance g equals the sum of the arms a, b and c, etc., divided by the number of rivets. In the same way the distance g' from the axis y-y can be found and the center of gravity is then located. In designing riveted joints the following well-established rules are used. (These do not apply to joints in boilers or cylinders where maximum efficiencies are desired.) i. PREFERABLE MINIMUM DIMENSIONS, INCHES Rivet diameters. 7 8 3 4 f i Center to center of rivets * 3 a| 2| i- Rivet center to sheared edge ll it It i Rivet center to rolled edge I 7 8 * Sometimes taken three diameters of the rivet. 2. The maximum pitch in the line of the stress for members composed of plates and shapes should be 6 ins. for f-in. rivets; 5 ins. for f-in. rivets; 4^ ins. for f-in. rivets; and 4 ins. for J-in. rivets. Where angles have two gauge lines and the rivets are staggered the maximum pitch in each gauge line should be twice the above dimensions. TENSION, COMPRESSION PIECES AND BEAMS 89 3. Where two or more plates are used in contact the rivets holding them together should not be spaced farther apart than 12 ins. in either direction. 4. The pitch of rivets in the direction of the stress should not be greater than 6 ins., nor exceed 16 times the thinnest outside plate connected, and not more than 50 times that thickness at right angles to the stress. 5. The maximum distance from any edge should be 8 times the thickness of the plate. 6. The diameter of the rivets in any angle carrying calculated stresses should not exceed one-fourth of the width of the leg in which they are driven. In minor parts rivet diameters may be | in. larger. 7. The pitch of rivets at the ends of built compression mem- bers shall not exceed four diameters of the rivets for a length equal to one and one-half times the maximum width of the member. 8. Two pieces riveted together should always be secured by at least two rivets. 9. Joints with field-driven rivets should have from 25 to 50 per cent more rivets than would have been required for shop- driven rivets. CHAPTER VII COLUMNS STRUCTURAL engineers generally use either Rankine's formula or Johnson's straight-line formula in designing columns. The latter is a modification of Rankine's formula and the results approximate those given by Rankine's formula within the usual working limits of which will range from 50 to 150. The straight-line formula is preferred as the calculations with it are simpler. The following give these formulae in their usual form : Gordon's or Rankine's formula for soft steel, *+ , 13,500 r 2 Gordon's or Rankine's formula for medium steel, 1 1 ,000 r Johnson's straight-line formula for structural steel, /= 16,000 -70-- (3) / = the allowable unit compression on gross section of col- umn in pounds per square inch. / = the effective length of the column in inches.* * The effective length of / will have the following relations to L the total length of the column: Both ends hinged or butting. . . ........... / = L Both ends fixed ......................... l = \L One end fixed and one end hinged ......... / = f Z, One end fixed and other end free .......... I = 2 L go COLUMNS 91 r = the least radius of gyration of the column section in inches A I = the least moment of inertia of the column section in inches. A = the area of the column section in square inches. Ritter's formula most nearly meets the theoretical require- ments but it is not much used as, like Rankine's formula, it is cumbersome, and the other formulae are better known. It has the advantage of being applicable wherever the modulus of elasticity and the strength of the material at the elastic limit are known. In the other formulae the constant for the material 13,500 and 11,000 in the case of Rankine's and 70 in the straight- line formula had to be determined experimentally. Ritter's formula as generally expressed is: The symbols have the same significance as just given for formulae (i), (2) and (3). In addition S c = the maximum compressive unit stress desired on the con- cave side of the column in pounds per square inch. S e = the unit strength of the material at its elastic limit in pounds per square inch. 7T 2 = approximately 10. E = modulus of elasticity of the material in pounds per square inch. If for mild steel, the following values may be assumed: E = 30,000,000, S e =30,000 and S c = 16,000, all, pounds per square inch. With these values substituted, formula (4) reduces to , 16,000 16,000 , >, ' J /1\ 1 / 1\ 9 \0/ 10X30,000,000 In this approximate form it is not especially cumbersome but it gives values slightly under those calculated by formula (3). The 92 GRAPHICS AND STRUCTURAL DESIGN straight-line formula will be used for all calculations throughout the text. The American Bridge Company have issued under date of Dec. i, 1912, new specifications for structural steel work. These specifications recommend for columns the use of two formulae, the first, / = 19,000 100 -, to be applied to values of - up to 120; the second formula applies only to secondary members, those permitted to have - values ranging from 1 20 to 200, and is / = 13,000 50- The maximum value of / is placed at 13,000 Ibs. per sq. in. NOTE. There is probably no section of Mechanics of Materials about which students have a less clear conception than that of columns. It is here assumed that the student has this fundamental conception clearly and thoroughly his own. Such a conception is essential to the intelligent design of columns and, in fact, to beams also, as will be shown later. Should the student's idea of the subject be in any way vague he is urged to review the subject in any Mechanics of Materials with which he is familiar. Problem. Plot the values of /given by formulas (i), (2), (3) and (5) upon cross-section paper for values of - between 50 and ISO- It should be noted in the several formulae that r is the least radius of gyration, hence that section will theoretically make the best column where the radius of gyration of the section is a con- stant for all axes. This requires a circular section, which, for economy of material, would generally be a hollow cylinder. On this account rolled-steel and cast-iron pipe would make good columns, but that cast iron is being but infrequently used in important structures while, owing to the difficulty of making con- nections to them by brackets or framing, round-steel columns are seldom used. COLUMNS 93 One of the simplest columns that approaches the theoretically ideal section is made by using two channels spaced so as to make the radii of gyration of the column section referred to its principal axes approximately equal. The columns, Fig. 113, are then laced with flat bars riveted to the channel flanges so that the two channels are made to act as a unit throughout the column's length. /i is the moment of inertia of one channel axis i-i. 1 2 is the moment of inertia of one channel axis 2-2. a is the area of one channel. A = 2 a. FIG. 113. The inertia of the total column section about axis i-i is L -2/1. Since r = y - it follows that if the radii of gyration are to be equal about any two axes of a section the inertias must be equal; hence I C1 = /s, from which 2 It should not be understood that columns must always ap- proach equal radii of gyration about both axes. In the case of columns of comparatively short lengths it may be cheaper, owing to the reduction in the work of manufacturing, to use a rolled section than a built-up column, even if the material is not used to such good advantage. 94 GRAPHICS AND STRUCTURAL DESIGN It therefore frequently happens that single sections, usually I beams or angles, make good columns. In some cases the base is merely an iron casting of some depth for stiffness, containing a pocket cored in it into which the section fits. When the column is brought into position some molten soft metal is poured around it to hold column and base together. In more important cases, angle and plate bases similar to those shown in Figs. 116 and 117 are used. As previously stated the channel column is one FIG. 114. FIG. 115. FIG. 116. FIG. 117. of the commonest types. These are illustrated in Fig. 113 and the bases for such a column are shown in Figs. 116 and 117. The flanges of the channels may be turned in instead of out, making a square column, but, owing to the difficulty of rivet- ing, it costs more to manufacture. With the flanges turned out as in Fig. 113 the lacing may be replaced by plates. Another design uses the plate on the outside and is laced on the inside, thus permitting of inspection and painting. The sections shown in Figs. 114 and 115 are also accessible for painting. In Fig. 115, the web may be either lacing or a solid plate. This section, when deep, makes a good column section to resist combined COLUMNS 95 compression and bending. Its strength may be increased by riveting flange plates outside the angles. A couple of the simplest forms of column bases are shown in Figs. 116 and 117. The proper distribution of the load carried by the column section to the foundation demands not only an enlargement of the foot of the column but some depth to assure its proper distribution. COMBINED STRESSES In a column the force may act parallel to the axis but eccen- tric to it. Here the column is subjected to a bending moment P (5 + A) where P is the force acting parallel to the axis and A is the maximum deflection due to such eccentric loading. The following formula by Merriman gives the maximum resulting fiber stress: A\_' ' W* 1-5 /r = maximum combined fiber stress in pounds per square inch. P = eccentric force on column in pounds. A = area of the column section in square inches. d = eccentricity of the load P in inches. e = distance from the neutral axis to the extreme fibers. r = radius of gyration of column section referred to the axis about which the bending occurs. PP PP I = length of the column in inches. E = modulus of elasticity in pounds per square inch. Where a piece is subjected to either compression or tension, together with transverse bending, the following formula, due to Johnson, is commonly used: Me ' 96 GRAPHICS AND STRUCTURAL DESIGN f b = flexural fiber stress in pounds per square inch. M transverse bending moment in inch pounds. e = the distance from the neutral axis to the extreme fibers in inches. 7 = moment of inertia of the section referred to the axis about which the bending occurs. / = length of the piece in inches. E = modulus of elasticity in pounds per square inch. The sign (+) is used when P puts the piece in tension, the sign ( ) when compression is produced. To this flexural fiber stress must be added the direct com- pression ( ) or the tension (+). The factor given as 10 varies with the character of the loads and the ends, being 9.6 for a simple beam uniformly loaded, but 1 2 for a similar beam with a central load. Owing to its simplicity this formula is frequently used instead of the first formula. When an approximation only is desired the stress due to bend- ing may be found by solving the formula/ = - for the extreme fiber stress due to bending and adding to it algebraically the direct stress due to either tension or compression. This method will be satisfactory when the longitudinal tension or compression is not large. The total fiber stress should not exceed that permitted on the piece. An example will illustrate the use of these formulae. The member of the upper chord of a bridge is 25 feet long. It weighs no Ibs. per foot, and has the following section (see Fig. 118): Top plate 22 ins. X jV in. Top angles 3 ins. X 3 ins. X f in. Side plates 16 ins. X f in. Lower angles 4 ins. X 3 ins. X T 7 e in. The moment of inertia axis i-i is 1003.8. P = 238,100 Ibs. The distance from the top of the upper plate to the center of gravity of the built-up section is 6.30 ins. COLUMNS 97 The bending moment due to the dead load is WL (no X 25) X 300 . M = - -** = 103,125 in. Ibs. o o The fiber stress due to combined bending and direct stress as given by the approximate method is / _ Me _ 103,125 X 6.30 ' ~ ~T 1003.8 Total combined stress . 238.100 650 -h - 7-^- = 8190 Ibs. 650 Ibs. 63" FIG. i i 8. FIG. 119. The following result is given by Johnson's formula, /_ Me _ 103,125 X 6.3 I - PI 2 1003.8 - 238,100 X 3QQ 2 10 X 30,000,000 = 697 Ibs. Total compression . 238.100 ., 697 + i6 = 8237 Ibs. The following example will illustrate the application of the formulae when the load is parallel to the longitudinal axis of the piece but eccentric to it. A 4 in. X 4 in. X T 7 6' m - an gl e 84 ins. long has a load of 8800 Ibs. applied to one leg. The piece is in compression. What is the total extreme fiber stress in com- pression? / = 4.97; area = 3.31 sq. ins.; radius of gyration = 1.23. 98 GRAPHICS AND STRUCTURAL DESIGN By Merriman's formula PI 2 8800 X i2o 2 48 El 48 X 30,000,000 X 4-97 __ 8800 [" , 1.16 X 1.16 i +0.0177 1 3.31 1_ i.23 2 i - (5 X 0.0177)]' fr = 5300 Ibs. By Johnson's formula The fiber stress in bending is Me /*- 10 , 8800 X 1.16 X 1.16 / 6 = - - - - = 2610 Ibs. 8800 X I20 2 4.07 -- 10 X 30,000,000 The fiber stress due to compression is 8800 pc - ~ = 2660 Ibs. 3-3i fr =fb + fc = 26lO + 2660 = 5270 Ibs. In this case the maximum stress should be kept under that allowed on the strut as given by a column formula; hence / = 16,000 (70 X ^H = 5360 Ibs. \ 0.7 LONG BEAMS UNSUPPORTED LATERALLY When a beam of long span is unsupported laterally the upper flange being in compression is liable to fail as a column by buck- ling sideways. There is no very satisfactory theoretical treat- ment of this subject. The compressive fiber stress in the upper flange is usually limited by formulae or rules that have been found safe. Mr. Christie, basing his conclusions upon tests made on full-size beams for the Pencoyd Iron Works, decided it was safe COLUMNS 99 to use a desired limiting fiber stress up to a span of twenty times the flange width and that from this point the working fiber stress should be uniformly decreased until, at a span of 70 times the flange width, the working fiber stress should be one-half the maximum desired stress. The handbook of the Cambria Steel Company suggests the following formula, 18,000 Pi = I + 3000 b 2 p= allowable compressive fiber stress in pounds per square inch. /= length of span in inches. 5= width of flange of beam in inches. Mr. C. C. Schneider in his structural specifications limits the allowable compression in flanges of girders to pi = 16,000 200 -when flange is composed of plates. pi = 16,000 1 50 r when flange is a channel. o In this case the fiber stress is given somewhat lower than for beams, which is consistent, as the shallower the beam the more the lower flange, which is in tension, reinforces the compression flange. The 1912 specification of the American Bridge Company limits the span unsupported laterally to forty times the flange width, and when such span exceeds ten flange widths the fiber stress is to be reduced to that given by the formula 19,0003007- o The following curves, Fig. 120, are based on formulae derived by the author in an attempt to explain the necessity for reducing the fiber stress in beams where the beam was un- supported laterally. The derivation of these formulae is 100 GRAPHICS AND STRUCTURAL DESIGN fully explained in the Proceedings of the Engineers' Club of Philadelphia, for April, 1909. no Hatio of Span to Mange Wldjh. 20 30 40 Curves give the 20 130 140 150 .160 >70 180 Ratio oLSpan to Badius of Gyration of. Compression Flange FIG. 120. CHAPTER VIII GIRDERS FOR CONVEYORS THE half plan and section of this girder are given in Fig. 121, the elevation in Fig. 122 and the end view in Fig. 123. Assumed Loading. - Metal at 105 Ibs. per lineal foot 5*360 Ibs. Corrugated steel covering 2,000 Ibs. Foot-walk and sheathing 2,420 Ibs. Snow 6,120 Ibs. Total dead load 15,900 Ibs. Conveyor load: Uniform load on conveyor 3,900 Ibs. Weight of conveyor 3,700 Ibs. Total moving load on conveyor . . 7,600 Ibs. Allowing 25 per cent additional for impact 1,900 Ibs. Total 9,500 Ibs. As the conveyor loading is not carried equally by both girders the maximum load on either girder must be calculated. The total maximum load on one girder 7950 + 6500 = 14,450. The apex load is ^ = 2060 Ibs. 7 Wind Load. The horizontal wind load is assumed at 20 Ibs. per sq. ft. of vertical projection. The apex wind load on the upper horizontal girder is 20 X 4 X 7^ ~ 590 Ibs. The stresses must now be determined for the truss under the given loading. This may be done graphically or the stresses may be found very readily by the method of coefficients. To illustrate the procedure both methods will be used. The method of coefficients will be used first. For the explana- tion of this method see page 55. These calculations are made 102 GRAPHICS AND STRUCTURAL DESIGN \ \ \ GIRDERS FOR CONVEYORS I0 3 FIG. 124. Wtangfl Wtangfl FIG. 125, 514 DEAD LOAD AND LIVE LOAD DIAGRAMS, J H FIG. 127. v. for Fig. 125. The graphical solution for the main truss is made in Figs. 126 and 127. The wind bracing and graphical analysis are given in Figs. 128 and 129. 104 GRAPHICS AND STRUCTURAL DESIGN = 45 Stress EF Stress EH Stress EJ Stress FD Stress IB Stress KA Stress GH Stress // 1170* tang 45 = i sec 45 = 1.41 = 3 X 2050 X 1.41 = 8500 Ibs. = 5 X 2050 X i = 10,250 Ibs. = EK = - 6 X 2050 X i = - 12,300 Ibs. = + 3 X 2050 X i = + 6150 Ibs. = stress GC. = + 5 X 2050 X i = + 10,250 Ibs. = + 6 X 2050 X i = + 12,300 Ibs. = + 2 X 2050 X 1.41 = + 5420 Ibs. = + i X 2050 X 1.41 = + 2710 Ibs. 590* FIG. 128. E-F-A FIG. 129. Selection of Members. The upper chord. Maximum compressive stress, dead and live load 12,650 Ibs. Maximum compressive stress, wind load 3,660 Ibs. Total 16,310 Ibs. Try 2-3 in. X 2 in. X i-in. angles spaced J in. back to back, with their long legs parallel. r 2 = 0.88. Their length is 88 ins. / 88 - = - - = loo. r 0.88 The allowable fiber stress according to the straight-line formula is /i = 16,000 70- = 16,000 (70 X 100) = 9000 Ibs. The load these angles will carry is 2 X 1.19 X 9000 = 21,420 Ibs. GIRDERS FOR CONVEYORS 105 End Strut BF. Its length is 120 ins. and its load 8800 Ibs. Try 1-4 in. X 4 in. angle. Its radius of gyration about its diag- onal axis is 0.70. - = -- =152. The allowable fiber stress r 0.79 according to the straight-line formula is /i = 16,000 - (70 X 152) = 5360 Ibs. The bending moment due to the eccentric loading, when the angle is assumed as T 7 g in. thick, is 8800 X i : i = 10,032 in. Ibs. The fiber stress resulting from combined bending and com- pression is given by the formula io E see page 95. 10,200 X 1.16 / = = 2600 Ibs. 8800 X I20 2 4.97 io X 30,000,000 The allowable stress according to the straight-line formula being 5360 Ibs., deducting 2600 Ibs. leaves 2760 Ibs. per sq. in. The total allowable direct stress on the angle then is 3.31 X 2760 = 91 50 Ibs.; this, exceeding the 8800 Ibs. acting on it, the angle is satisfactory. Lower Chord. The maximum stress is 16,310 Ibs. Trying 1-4 in. X 3 in. X T \ in., and placing the long leg horizontally, its inertia about its axis parallel to the short leg is 3.38; the dis- tance from the back of the short leg to the axis is 1.26 ins. The bending moment due to this eccentricity is 16,310 X 1.26 = 20,600 in. Ibs. The bending moment 16,310 X 1.26 = 20,600 in. Ibs. The fiber stress due to bending is 10 10X30,000,000 I06 GRAPHICS AND STRUCTURAL DESIGN The allowable stress on the net section is/ = 16,000 6850 = 9150 Ibs. The total allowable force on the piece then is (2.09 0.25) X 9150 = 16,850 Ibs. Vertical HI. This is a compression piece carrying 2050 Ibs. 1-3 in. X 3 in. X J-in. angle will be tried. The bending due to eccentric loading is M = 2050 X 0.84 = 1720 in. Ibs. The fiber stress due to this bending is . My 1720 X 0.84 / = T-PF = - ~20. S oX88> = I22olbs - 7 --- - 1.24 -- 10 10 X 30,000,000 The allowable compression, according to the straight-line formula, is 7 , OQ \ - / = 16,000 70- = 16,000 ( 70 X - -) = 5600 Ibs. r \ 0.597 The net compression allowed per square inch of the section is 5600 1220 = 4380 Ibs. The area of the angle multiplied by 4380 is 1.44 X 4380 = 6300 Ibs. Hence the 3 in. X 3 in. X J-in. angle is more than ample. Diagonals. The maximum tension in the diagonals is 6000 Ibs. Try i-2j in. X 2 in. X J-in. angle. Its inertia about an axis through its center of gravity and parallel to its long leg is 0.37. The distance from the back of the angle to the above axis is 0.51 in. The bending moment due to the eccentric application of the load to the angle is 6000 X 0.51 = 3060 in. Ibs. The extreme fiber stress due to this bending is - My 3060 X 0.51 0.37 , ,, 6000 = 236olbs - . 10 10X30,000,000 The allowable stress per square inch of the net section is 16,000 2360 = 13,640 Ibs. Net section X 13,648 = (1.07 0.20) X GIRDERS FOR CONVEYORS 107 13,640 = 11,850 Ibs. Tnis is satisfactory, being the smallest angle we can use. Wind Bracing. The maximum stress in the diagonals is 2550 Ibs. The smallest flat allowed being 2 X i in., and as this will carry [(2 X 0.25) 0.20] X 16,000 = 4800 Ibs. (the net area times the allowable fiber stress), it is ample. STRUT IN UPPER HORIZONTAL BRACING 57 The bending moment under the load between RI and R2 is M = 400 X 12 = 4800 Ibs. FIG. 130. FIG. 131. Trying 3 in. X 3 in. X J-in. angles, their inertia about an axis through their center of gravity and parallel to a leg is 1.24, the distance from the back of the angle to this axis is 0.84 in., the radius of gyration referred to this axis is 0.93, while the radius of gyration referred to the diagonal axis is 0.59. I QQ - for the full length = - - = 97. r 0.93 - for the length of 57 ins. = -^ = 97. r 0.59 The allowable stress / = 16,000 f 70 X -J = 16,000 (70 X 97) = 9210 Ibs. The fiber stresses due to the bending of 4800 in. Ibs. in 57-in. span are, , Me 4800 X 0.84 , 4800 X 2.16 / = -f~ ~ = 3 2 5 lbs - fc = - = 8350 Ibs. / 1.24 I.2A 108 GRAPHICS AND STRUCTURAL DESIGN The bending due to the eccentric loading as a column will be My (i 180 X 0.84) X 0.84 835 _ 6Solb5 ~_*L It24 "8QX57 8 1-23 10 X 30,000,000 . 680 X 2.16 ft = - o - = 1750 Ibs. 0.04 The direct compression = force divided by the area of the 1180 section = -- = 820 IDS. 1.44 The maximum compression will be 9170 Ibs., which, being under the 9210 Ibs. allowed the section, is satisfactory. The beams serving as struts in the lower horizontal bracing carry a total load of 1080 Ibs., -f 400 (snow load) = 1080 Ibs. R>= I08 X 57 = 685 lbs . 90 M = 685 X 33 = 22,605 in. lbs. Assuming the channel unsupported laterally for its length of 88 ins., 88 length -T- flange width = - = 45- 1.92 Allowable fiber stress about 11,000 lbs. per sq. in., M I 22,605 = - = i = 2.06. / e ii ,000 This suggests about a 5 -in. channel. The direct compression being 1 180 lbs. and applied at the upper flange the moment due to eccentric loading is If = 1180 X 2.5 = 2950 lbs. 10X30,000,000 GIRDERS FOR CONVEYORS 109 Direct stress - = 605 Ibs. i-95 The extreme compressive stress due to bending of a 5-in. U is, . Me 22,605 / = ^ = 7535 Ibs. per sq. in. Total compressive stress 7535 + 570 + 605 = 8710 Ibs., being below 11,000, is satisfactory. CHAPTER IX TRUSSES, BENTS AND TOWERS IT is frequently necessary around works to carry large gas or air mains overhead. These mains are sometimes lined with brick to prevent radiation loss from the hot gas or air. There is consequently considerable load per foot, and as long spans are preferable, thus avoiding too large a number of posts or bents FIG. 132. that would block up the yard, the pipe must be trussed. Fig. 132 represents such a trussed pipe. Care should be taken that any load transferred to or from the pipe should be distributed over the pipe by a saddle and if necessary the pipe should be reinforced at such points. In Fig. 132, AB is the reaction in the post, AC the compression in the pipe, and BC the tension in the rods. The span over which the pipe section acts as a beam is -. O Steam, air, gas and water pipes may be carried upon a light TRUSSES, BENTS AND TOWERS III trussed bridge somewhat resembling the conveyor truss (see page 102), or may be hung from a cable, as shown in Fig. 133. The stress in the wire rope depends upon the load carried and upon the sag permitted in it. As the weight of the rope will be small compared with the load, the effect of the weight of the rope may be neglected. In Fig. 133, having assumed the rope diameter, lay off R representing the maximum desired pull on the rope, then draw the horizontal and vertical components H and V. Taking the forces acting about apex i, the horizontal FIG. 133. FIG. 134. component of BC must equal the horizontal component of CA. Similarly, the horizontal components of the forces acting at W W apex 2 are equal. Now in Fig. 134, XD = -- and DA = 4 H is the horizontal component of the force BC acting in the guy rope. CA and CD give the magnitudes and directions of the rope stresses and the sag of the rope may be found by drawing the strings in Fig. 133, corresponding to these rays in Fig. 134. Should the sag prove objectionable a heavier rope with greater stress or a greater stress may be used in the first rope if permissible. W The vertical load on the post, Fig. 133, is V -\ 112 GRAPHICS AND STRUCTURAL DESIGN Bents, Fig. 135, may be used to support both pipes and wires. If the bent is an end one and guyed the vertical load will be determined as in the preceding example. If it is an intermediate bent it will not carry the vertical component of the guy rope but merely its proportion of the weight of the wires and pipes. It may also be subjected to wind load, in which event the stresses in the bracing can be determined and these pieces selected for their respective loads, see Fig. 137. The wind acting on cylin- drical surfaces is commonly assumed as -fa of the pressure that would act on the vertical projection of that surface. FIG. 135. DEAD LOAD STRESS DIAGRAM FIG. 138. As the loads are generally very light and not subjected to shock, the values of - are permitted to reach 180 or 200. As the wind load is transmitted to the truss proper by pieces subjected to bending, the diagram, Fig. 136, has been started by connecting the wind force to the truss through the dotted or substituted members. The dead-load stress diagram is shown in Fig. 138. Towers for Transmission Lines. - - There is no standard practice but the four-angle type is the most common. The TRUSSES, BENTS AND TOWERS 113 loading upon these poles consists of dead load of pole and weight of wire, also ice, which may add materially to this direct load in localities visited by sleet. A live load due to the wind acts at right angles to the dead load, and in addition there may be a direct pull along the line caused by a wire or wires breaking. The ice load is generally assumed as a |-in. coating around the wire. Wind loads assumed as acting on wires vary considerably. Mr. R. Fleming in an article in Engineering News, Nov. 28, 1912, recommends 30 Ibs. per sq. ft. of exposed surface upon the tower, 15 Ibs. per sq. ft. of projected area of the bare wire, reduced to 10 Ibs. for ice-coated wires. He further recommends that poles carrying three wires be designed to resist the unbalanced pull, due to one wire breaking, while for six wires two of them shall be considered as breaking. The unbalanced load is the force that breaks the wire and the following percentages of the ultimate strengths of the wires are recommended : the full ultimate strength up to and including a No. 4 wire; a No. 3 wire, 90 per cent; a No. 2 wire, 80 per cent; a No. i wire, 70 per cent; a No. o wire, 60 per cent; and for all larger wires, 50 per cent. The ultimate strength per square inch of hard-drawn copper wire is from 50.000 to 65,000 Ibs. The towers are designed to meet the requirements of the worst combination of these conditions. Using ordinary structural steel, Mr. Fleming recommends the following formulae for columns: / c = *~z , for the most exacting city service. I+ 18,000 r 2 r 27,000, r , f c = - , for less exacting service. 18,000 r 2 In estimating the loading upon the poles the additional load- ing due to the line running on an incline or rounding a curve should not be overlooked. This is generally provided for by 114 GRAPHICS AND STRUCTURAL DESIGN placing the poles closer together in these places rather than changing the tower design. In estimating the tension in the line and length of the wire between supports it is common practice to assume the curve of the line a parabola which differs but slightly from the catenary and is much more readily computed. S = distance between poles in feet. L = length of wire in feet, measured along the curve. d sag or drop in wire in feet, measured midway between poles. H = tension in lowest point of curve in pounds. c = tension constant. w = resultant load, measured in pounds per foot of wire. In winter the weight of wire and ice acting vertically and the wind acting horizontally. In summer the weight of wire acting vertically and the wind on the bare wire acting horizontally. (4) For a more complete mathematical discussion of this subject, see Bulletin No. 54 of the University of Illinois. "Mechani- cal Stresses in Transmission Lines," by A. Guell. Jan. 22, 1912. The length of the wire at the summer temperature will increase due to the temperature but will have less extension due to the loading as the vertical load will not include the weight of the ice and the wind will not act on so great a surface. Mr. Guell gives the following formulae: TRUSSES, BENTS AND TOWERS SECTION A-A 'Concrete.! 1% Bolt Top upset to %%' FIG. 139. Il6 GRAPHICS AND STRUCTURAL DESIGN Here LI = length of the wire along the curve, measured in feet, at the summer temperature. L = length of the wire for winter temperature and loading. a = coefficient of expansion per degree Fahr. = 0.00000956. / = temperature range, degrees Fahr. H = tension at low point in curve in winter, pounds. HI = tension at low point in curve in summer, pounds. A = area of wire, square inches. E = modulus of elasticity, pounds per square inch. For hard-drawn copper wire, 12,000,000 to 16,000,000 Ibs. per sq. in. The stresses in the wire will be lower in summer than in winter, but it is necessary to estimate the sag for the summer. It may be approximated as follows. In equation (5) estimate the value of Li, omitting the last term, as HI is not known. A preliminary estimate of HI may be made by assuming HI = Wi H w' Here Wi = resultant load per foot in summer. w = resultant load per foot in winter. This value of HI would then be substituted in equation (5) and the last term need not be discarded. Having now found LI substitute this value for L in equation (i) and solve for c. This value of c used in equation (2) gives an approximate value of d. Now in equation (4), w is known from the loading and H can be found. This is an approximation to HI required in equation (5) and the value of LI may now be revised by placing this value of HI in this equation. The trials may be repeated until the desired degree of approximation is reached. Cross arms should be designed for a minimum vertical load of from looo to 1200 Ibs., also for an unbalanced horizontal pull due to the wires breaking and estimated as previously given. TRUSSES, BENTS AND TOWERS 117 The twisting of the tower due to such unbalanced loading should not be overlooked. ! 4- Concrete FIG. 140. Figure 139 shows one type of tower and foundation and follows a design by Westinghouse, Church, Kerr & Co., as does also Fig. 140, which illustrates a type of foundation with greater base area. CHAPTER X DESIGN OF A STEEL MILL BUILDING THE design of a steel mill building as generally carried out requires considerable experience and judgment on the part of the designer. The truss stresses are usually determined for an equivalent load rather than by the separate determination of the dead-, snow- and wind-load stresses and their combination for maximum effect. The following equivalent loads may be used on spans under 80 ft. Kind of roof. Lbs. per sq. ft. Gravel or composition roofing Gravel or composition roofing Corrugated steel sheeting on boards 45-50 60 40 50 65 on 3~in. concrete. Slate on boards Slate on 3-in. concrete, flat.. . . Where no snow need be considered these figures can be reduced by 10 Ibs., excepting that no roof shall be designed for less than 40 Ibs. per sq. ft. Stiffening the Structure. The structure is stiffened first where possible by introducing a knee brace, running from a panel point in the lower chord of the truss to a point as low as convenient on the column. Where the column carries a crane runway this knee brace must generally be omitted and is usually replaced by a knuckle brace which stiffens the connection be- tween the column and the truss. The structure is also braced in the plane of the lower chords of the truss to hold the tops of the columns at constant distances apart, and in some cases to carry the wind forces acting along the side of the building to the transverse bracing at its ends. 118 DESIGN OF A STEEL MILL BUILDING 119 To prevent the trusses rotating about their lower chords, bracing is placed in the plane of the upper chords. This bracing is commonly omitted, at least in alternate bays. To resist the wind pressure on the ends of the buildings and the cumulative pressure along the roof, together with any crane thrust, lateral bracing is placed in the plane of the building columns. This bracing usually consists of an eave strut which is frequently a light latticed girder composed of four angles, the web being composed of diagonal lattice bars. Where possible the diagonal bracing should extend to the ends of the columns. Where the column height is considerable an additional strut is placed about half way up the column. Both eave strut and intermediate strut run the entire length of the building and fasten to each column. The diagonal braces are commonly placed at the end bays and then at intervals along the building as deemed necessary by the designer. Where the bracing cannot be extended to the column bases the columns must be designed to resist the bending due to the horizontal forces resulting from the wind acting on the end of the building and any crane thrust. This bending will be influenced by the way the column bases are secured and may be treated similarly to the transverse bent, page 45. The transverse bracing in the ends of the building is intended to resist whatever of the forces acting on the side of the building from wind pressure, or thrust on the columns from the crane, may be transferred to the ends by the structure. In the simpler buildings it is possible to assume certain exter- nal forces resisted by a particular bracing truss and then make the design accordingly. In the more complicated buildings the designer modifies his calculations by what has been previously found to be satisfactory. The following sketches illustrate one general plan of steel mill bracing, Figs. 141 to 143. See also Figs. 156 to 158. A number of types of roof trusses are shown in Figs. 144 to 151. 120 GRAPHICS AND STRUCTURAL DESIGN TRANSVERSE END BENT BRACING FIG. 141. ELEVATION! ^"^^'' Purlin ^_ ^-' **^~^^''' ^^-' ^-^ " ^-' ^^^ ^'" "~^^ ^^ -' ^^J'*-'" "^v^ ^x-' 1 Purlin ^^^ ^'' ^^ ^^ x ''"^^- ^^" ^v.^ .. ^ ^ '" "^^j- ^' *^^ ^^" ^ *"' "^ Eave ^.--" ^-^ ,^'""^-^ ~~-^ X Strut Strut V / \ ^X &1I ^ x / E , /\^ : / A- X \ X X T ^1^'' - N NN X X/ ~ 1 / ^^ / \ FIG. 142. Eave Strut Lower Chord Bracing \ Purlin 3rd Brae ug Purlin Eave Strut PLAN FIG. 143. DESIGN OF A STEEL MILL BUILDING 121 The number of panels into which the upper chord should be divided will depend upon the span of the truss and upon the roof covering or sheathing. In the case of corrugated-steel roofing, it is desirable that the sheets should extend over three purlins. That the roof covering be amply stiff it is necessary with the ordinary gauges to limit the distance between purlins to from FIG. 144. FIG. 145. FIG. 146. FIG. 147. FIG. 148. FIG. 149. FIG. 150. FIG. 151. 4 to 6 ft.; consequently, the purlins are generally spaced about 4 ft. In the case of other roof coverings stiffness is also re- quired, but as most of these coverings are laid upon sheathing the purlins can be spaced farther apart providing the sheathing is made correspondingly thicker. If the upper chord is not in- tended to take bending it is necessary to have a member of the truss connected to the upper chord under each purlin. 122 GRAPHICS AND STRUCTURAL DESIGN Frequently, however, the upper chord is designed of a plate and two angles forming a tee-section. In this case the section is calculated to resist bending and the purlins can, therefore, be spaced as desired, and a less complicated truss used. The pitch of the roof will vary with the character of the roof covering. The following table gives the usual pitches. Kind of roofing. Minimum pitch. Usual pitch. Corrugated, steel i 1 Slate or tile i M Shingles . .... 1 V Gravel Maximum pitch fin 12" Asphalt i fin 12" Patent compositions 1 i Design of a Steel Mill Building, Figs. 152 to 158. The truss will be designed for an equivalent load of 40 Ibs. per sq. ft. FIG. 159. of horizontal projection of the roof. The rise will be made one in four, and the trusses will be spaced 17 ft., the span being 86 ft. o ins. Total load per truss = 86 X 17 X 40 = 58,480 Ibs. Tan- 1 \ = 26 30'. In Fig. 159 the length of rafter = 47.1 ft. Distance between apices Sill 2O 3 = 9.42 ft. In this way the lengths of the several members are found to be DESIGN OF A STEEL MILL BUILDING 123 CH, DI, EK, FN and GO = 9.42 ft. m = ^ = 4.71 ft. 2 JK = 2 X 4-71 = 942 ft. LM = 3 X == 14-13 ft. NO = = 7.07 ft. 2 DESIGN OF MEMBERS. COMPRESSION PIECES Member HI. The stress in HI is 5200 Ibs. and is given by Fig. 153. The value of - = 120 or r = - = 4.71 X = 0.47. Trying a 3 in. X i\ in. X i-in. angle, which is the I2O smallest angle allowed, its net area is 1.32 sq. ins. less 0.20 sq. in. for one rivet hole or 1.12 sq. ins. - = 4.71 X - - = 106. r 0.53 The allowable unit stress on this column is /= 16,000 70 - = 16,000 (70 X 106) = 8580 Ibs. The total force allowed on the piece is 1.12 X 8580 = 9610 Ibs. This, being well above the stress in the piece, is satisfactory. Member JK. The length of JK is 9.42 ft.; the stress in it / 12 is 7800 Ibs. Limiting- to 120 as before, r = 9.42 X - - = 0.94. r 1 20 Trying two 3 in. X i\ in. X i-in. angles, their net area, after deducting for rivet holes in one leg of each angle, is (2.64 0.4) = 2.24 sq. ins. The value of r with the long legs back to back is 0.95. For this value of r we find the allowable unit stress to be / = 16,000 70- = 16,000 (70 X 120) = 7600 Ibs. 124 GRAPHICS AND STRUCTURAL DESIGN The total load these angles will carry then is 2.24 X 7600 = 17,000 Ibs. Member LM. The length of this member is 14.13 ft. Its load is 13,000 Ibs. The least radius of gyration is 14.13 X 12 r = - - = 1.41. 120 The load being very light two 4 in. X 3 in. X J-in. angles will be tried, although the radius of gyration is low. = 6 / = 16,000 - = 16,000 9520 = 6480 Ibs. r 1.25 so that - = The load on two angles, allowing one rivet hole in each angle, is load = 2 (1.69 0.20) X 6480 = 19,450 Ibs. Member NO is slightly shorter than JK and carries a lighter load, but will be made of the same angles, 3 in. X 2^ in. X i in. Members MA, JA, LA and PA are in tension and will be made alike. The maximum stress in these members is 52,400 Ibs. Trying two 3 in. X 3 in. X f-in. angles their net section is (4.22 0.60) X 16,000 = 58,000 Ibs. This is sufficiently close to the section required. Member OP. This is a tension piece carrying 17,000 Ibs. stress. Trying two 3 in. X 2^ in. X J-in. angles and allowing for one rivet hole they will carry 2 (1.32 0.20) X 16,000 = 35,800 Ibs. This is more than sufficient but two angles are preferable here and these angles are the smallest ones allowed by the specifications. The less important tension members may be made of a single angle. A 3 in. X 2^ in. X i-in. angle is the smallest permitted and it will carry (1.32 0.20) X 16,000 = 17,900 Ibs. This angle will be used for the members //, KL and MN. DESIGN OF A STEEL MILL BUILDING I2; Upper Chord. The upper chord will be made of the section shown in Fig. 160. It is first necessary to find the center of gravity, and then the moment of inertia of this section. Area, sq. ins. Moment. 2 X 2.09 = 4.18 X 0.76 = 3.18 9 X 0.313 = 2.82 X 4-50 = 2.65 x = 7 . oo 5 . 83 = 2. 26 ins. 7.00 The inertia of the combined section then follows: Angles, 2 X 1.65 = 3.30 Angles, Ah 2 , 2 X 2.09 X (2.26 0.76)2 = 9.42 Plate, g- FIG. 160. =, 9 .oo Plate, Ah 2 = 2.81 X (4.50 - 2.26)2 = 14.08 7 = 45.80 The span of CH is 9.42 ft. When two purlins are used in each 5850 span of 9.42 ft. the purlin load will be 2925 Ibs. Assum- ing the purlins centrally located on the above span the bending moment is = X 2925 X o \ 4 ' o 4 The extreme fiber stress in compression is 5I)6 ooin. Ibs. Me X 2.26 7~ 45.0 while the extreme fiber stress in tension is ft = T = 5 J >6oo X / 45.00 Combining these fiber stresses we have, Total fiber stress in compression side, - 2540 - 2 54olbs., 7600 Ibs. = - 10,940 Ibs. Total fiber stress in tension side, 7600 P ' j = 800 Ibs. \ 6.99 / 126 GRAPHICS AND STRUCTURAL DESIGN The radius of gyration of this section must now be found about the axis through the plate and parallel to the short legs of the angles. Inertia of two angles referred to axis 2-2, 2 X 3.38 = 6.76 Ah 2 of the angles, 2 X 2.09 X (1.26 + 0.155)2 = 8.36 _ / = 15-12 l~f l The radius of gyration = r = y = y- -=1.44, *b 7 ' 3 / = 16,000 70 X - = 16,000 (70 X ^^ - ) = 10,400 Ibs. r \ 1.44 / The maximum fiber stress in compression being 10,940 Ibs. the section will do. Purlins. The distance between trusses being 17 ft., the pur- lins will be considered as beams supported at the ends, having a span of 17 ft. and carrying a uniform load of 35 Ibs. per sq. ft., being 5 Ibs. less than the assumed roof load on the truss to allow for the truss weight which of course does not come upon the purlins. The distance between purlins is - = 4.71 ft. The load carried by one purlin 134.71 X 17 X 35 = 2800 Ibs. WT M = ~= = (2800 X 17 X 12)+ 8 = 71,400 in. Ibs. 8 / _ M _ 71,400 _ e ~ p ~ 16,000 This will require 6-in. channels at 8 Ibs. per ft. Columns. The height from the ground to the top of the ventilator being 48 ft., the vertical projection between adjacent columns is 48 X 17 = 816 sq. ft. Taking the horizontal wind pressure at 15 Ibs. per sq. ft. of vertical projection the total pressure on the building between adjacent columns is 816 X 15 = 12,240 Ibs. Assuming the horizontal reactions at the bases of the two columns in any transverse bent as equal, this hori- zontal reaction is I2 ' 24 = 6120 Ibs. 2 DESIGN OF A STEEL MILL BUILDING 127 The bending moment on a column, assuming that the knee braces run down the column a distance of 5 ft. making the dis- tance from the foot of the knee brace to the base of the column 15 ft, is M = 6120 X 15 X 12 = 1,101,600 in. Ibs. Trying a column made up of four 4 in. X 3 in. X iVin. angles and one 16 in. X tV m - plate we must first determine the area and then the moment of inertia of the section. The area of the section is: 4 angles, 4 X 2.09 = 8.36 sq. ins. i plate, T \ X 16 = 5.00 sq. ins. Total area 13.36 sq. ins. Moment of Inertia of Section. 4 angles, 4X1.65 = 6.60 Ah 2 of angles, 4 X 2.09 (8.250.76) = 470.00 W 3 5 i6 3 ,-, -X- = IQ 7.oo / = 583.60 The extreme fiber stress in the column due to bending then is , Me 8.25 / = = 1,101,000 X J = 15,600 Ibs. per sq. in. i 583-0 The direct load in the column due to the assumed equivalent load of 40 Ibs. per sq. ft. of horizontal projection being 29,250 Ibs., the fiber stress due to this load is 2 9' 2 5 = 2200 Ibs. per sq. in. 13-36 The total maximum fiber stress in the extreme fibers is 15,600 -f 2 200 = 17,800 Ibs. The column must now be tested by the straight-line or other column formula to see if this fiber stress is excessive. Since the radius of gyration equals y > we have 13-36 The length of the column to the knee brace divided by the radius of gyration is 15 X = 27.3, and this value substituted 128 GRAPHICS AND STRUCTURAL DESIGN in the straight-line formula gives the allowable stress per square inch as / = 1.25^16,000 - yoX- J = 1.25 [16,000 - (70 X 27.3)] = 17,600 Ibs. The factor 1.25 is used here in accordance with usual specifica- tions which allow an increase of 25 per cent in the working fiber stress when the stress is due to a combination of wind, snow and dead loads. The agreement between this latter stress and that found as existing in the column is so close that the column section tried is satisfactory. The column should also be examined for its strength about its other axis. The inertia about axis 2-2 is found to be 35.4. The total area of the column at the knee brace and below that point is i6.oosq. ins., from which r = \ = 1.49. A The unsupported height being 8 ft., - = 8 X = 64.5. r 1-49 This value of - being within the permitted limits the column is satisfactory. Where the lateral wind bracing does not extend to the base of the columns the reactions due to the wind on the end of the building should be divided among the columns and the fiber stress in the column due to the resulting bending deter- mined. The column bases may be considered fixed or hinged, depending upon the bases, anchor bolts and weight of foundation. WIND BRACING The reactions at the tops of the end columns, Fig. 161, due to wind pressure are 2X3 DESIGN OF A STEEL MILL BUILDING 129 The force in the eave strut then is 11,060 Ibs. The length of this strut is the distance between columns, or 17 ft. For an eave strut try Fig. 162, composed of four 3 in. X 2\ in. X J-in. angles latticed, the long legs of the angles being turned out, r = I7X Z^ =I45; according to the straight-line formula / = 16,000 (70 X -J X 1.25 = 7300 Ibs. per sq. in. -J- 'to ~O -SCO- FIG. 161. FIG. The total force on the four angles is 4 X 1.32 X 7300 = 38,500 Ibs. No lighter section can be used as these are the lightest allowed by the specifications, for the - value determines the section. The transverse brace between the first and second trusses will be designed as a simple truss to carry the wind loads at its several apexes. Having determined the apex loads in Fig. 163, the stresses in the truss members can be found either by the method of coefficients or by making a stress diagram, Fig. 164. The diagonals will be assumed as taking only tension. The maximum stress is in the end diagonal and is found to be 13,800 Ibs. Trying a 3 in. X 2^ in. X J-in. angle, which is the lightest allowed, and assuming one rivet hole at a section, the net area is 1.63 0.50 = 1.13 sq, ins. The total load it will carry, I 3 GRAPHICS AND STRUCTURAL DESIGN allowing a fiber stress of 16,000 Ibs. per sq. in., is 1.13 X 16,000 = 18,100 Ibs. As the stresses in the diagonals of the braces in the top chords of the roof trusses and the bracing between the columns are both lower than the stress in the above diagonal the 3 in. X 2 in. X i-in. angle will be used in all these places. The strut in the bracing in the upper chord of the trusses will be stiffened by attaching it to the purlin. Here a 3 in. X 3 in. X |-in. angle will be used. FIG. 163. F 11100 # FIG. 164. GIRTS The load on the girts 134X17X15 = 1020 Ibs. M = = 1020 x 17 X = 20,800 in. Ibs. o o 20,800 = 1.30. / e 16,000 A 5 in. X 3^ in. X -f^-in.. angle will be used. The struts in the transverse truss will be made of two 5 in. X 3^ in. X -rV m - angles, with the long legs parallel. The radius of gyration here I I2 being 1.46 the value of -is 17 X- = 140. r 1.46 In the following problem (Fig. 165) the stresses due to dead load, snow load and wind load have been tabulated and their resultants compared with the stresses produced by an equivalent dead load of 40 Ibs. per sq. ft. of horizontal projection of the roof. The building is of somewhat lighter design than the previous one. Span of truss, 86 ft. o ins.; rise, J span. Truss spacing, DESIGN OF A STEEL MILL BUILDING 132 GRAPHICS AND STRUCTURAL DESIGN 25 ft. o ins. center to center. Figures 167 and 168 are the stress diagrams for wind load and dead load. The horizontal wind pressure is assumed at 20 Ibs. per sq. ft. and the normal pres- sure on the roof is taken as that given by the formula P N = X PH = ^ X 20 = 12 Ibs. per sq. ft. 45 45 The wind is shown as acting upon the left of the truss, and the truss members in compression are represented by heavy lines. The dead-load stress diagram is given in Fig. 168 and has been made to serve for both dead load and snow load by changing the scale. The snow load has been assumed at 20 Ibs. per sq. ft. of horizontal projection. The weight of the truss is taken as 300 where W = weight of truss per square foot of building. L = span of truss, in feet. D distance center to center of trusses, in feet. P = load per square foot on truss. W 4Q X 86 300 + (6 X 86) + ^^ o The estimated total weight of one truss is = 3X25X86 = 6450 Ibs. The weight of the purlins is assumed as that given by 45 / 4 where Wi = weight of the purlins per square foot of building, in pounds. PI = load per square foot on purlins, in pounds. D distance center to center of trusses, in feet. - 1.3.25 45 4 DESIGN OF A STEEL MILL BUILDING 133 The roof covering will be No. 20 galvanized corrugated iron, weighing about 2 Ibs. per sq. ft. or 2.75 Ibs. allowing for laps, etc. The wind load per square foot of horizontal projection is the normal wind pressure per square foot multiplied by the secant of 27 degrees, the angle of the roof, and equals 12 X 1.12 = 13.44 Ibs. The snow load will be 20 Ibs. per sq. ft. of horizontal projection. The total load per square foot of horizontal projection is Truss and purlins (3 + 3.25), say 6 . 25 Roof covering 2.75 Snow load 20 . oo Wind load 13 .50 Total 42 . 50 It is commonly specified that roofs of this character shall be designed for a uniform dead load of 40 Ibs. per sq. ft. of horizontal projection. This is treated as an equivalent loading and replaces the wind, snow and dead loads. The stresses will now be determined for the estimated dead, snow and wind loadings and their resulting maximum stresses compared with stresses determined from the uniform dead load of 40 Ibs. per sq. ft. The apex dead load on the upper chord of the truss at 10 Ibs. per sq. ft. is 86 X 25 X io - = 1 790 Ibs. The apex snow load at 20 Ibs. per sq. ft. of horizontal projection is 20 X 25 X 86 = 3 580 Ibs. The apex normal wind load at 1 2 Ibs. per sq. ft. of roof surface, the distance between the panel points along the upper member of the truss being 8.05 ft., is 8.05 X 25 X 12 = 2415 Ibs. 134 GRAPHICS AND STRUCTURAL DESIGN Member. Dead load at 10 Ibs. Snow load at 20 Ibs. Wind load at 12 Ibs. (normal). Combined. Equivalent load at 40 Ibs. CJ 22,500 45,000 12,500 8o,OOO 90,000 . JK 2,OOO 4,000 2,800 8,800 8,000 LM + 3,6oo + 7,200 + 5,300 + 16,100 + 14,400 QT JA + 9.300 + 20,500 +18,600 +41,000 + IO,6oo + I7900 +38,700 + 79,400 +37,200 +82,000 AT + 11,000 + 22,000 + 5,ioo +38,100 +44,000 TU 22,500 -45,000 13,800 81,300 90,000 MN 4,800 9,600 6,700 21,100 19,200 The selection of the material for the members will now be made (see Fig. 169), the stresses given for the equivalent load being used. Lower Chord. The maximum stress occurs in the lower chord in the piece JA (Fig. 165) and is 82,000 Ibs. in tension. Allowing a working stress of 16,000 Ibs. per sq. in. and assuming f-in. diameter rivets driven in yf-in. holes, we find the net section required to be 82,000 = 5. 13 sq. ins. 16,000 Two 6 X 3^ X -rV m - an gl es having a gross area of 5.78 sq. ins. after deducting i rivet hole in each angle, that is, an area yf-in. long and yV m - wide, gives a net area of 5.78 (2 X 0.25) = 5.28 sq. ins., which is sufficiently close to the required net area. As the trusses will have to be divided for shipment the lower chord member AT will be made lighter than the members JA and MA. The stress in AT is 44,000 Ibs. tension. The net area required is 4 j"' 00 = 2.75 sq. ins. Trying two angles 16,000 3^ ins. X i\ ins. X -$ in. whose gross area is 2 X 1.78 = 3.56 sq. ins., the net section = 3.56 (2 X 0.25) = 3.06 sq. ins., which is sufficiently close to the required 2.75 sq. ins. Other Tension Members. The members NT and QT will be made of the same angles, and as the stress in QT is the greater the selection will have to be made for it. The net section 16,000 = 2.32 sq. ins. Trying two 3^ in. X 2\ in. X J-in. DESIGN OF A STEEL MILL BUILDING 135 angles whose gross area is 2 X 1.44 = 2.88 sq. ins., the net area becomes 2.88- (2 X 0.20) = 2.48 sq. ins., which will do. Eave L3H 8.L:1.L. J <3xjodk a.. 2- L . 3^'x 2^ x - FIG. 169. The members LM and M) have approximately the same stress. Taking LM, which carries 14,400 Ibs. in tension, the net area required is = 0.92 sq. n. 16,000 136 GRAPHICS AND STRUCTURAL DESIGN One angle 2\ X i\ X J in. gives a net area of 1.19 0.20 = 0.99 sq. in., which is satisfactory. Compression Members. Member MN having a length of ii ft. 6 ins. carries 19,200 Ibs. in compression. If the maximum ratio of length to least radius of gyration is 125 then the least radius of gyration is length divided by 125 or jf f = i.io. The smallest angles that will serve are 3^ ins. X 2 J ins. X i in., the long legs back to back, and separated f in. Their area is 2.88 sq. ins. According to the straight-line formula the allowable stress is p = 16,000 70- = 7250 Ibs. The total load is 7250 X 2.88 = 20,900 Ibs. This, being slightly in excess of the 19,200 Ibs. stress in the member, is satisfactory. Compression members, JK, KL and the corresponding ones on the other side of the truss have the same load of 8000 Ibs. and are approximately 84 ins. long. The least radius of gyration is y 8 2\ = 0.67. The smallest angles are i\ ins. X 2 ins. X J in., being also the smallest ordinarily permitted in this class of work. That is, no leg containing rivets shall be less than 2 ins. and no material ihall be less than \ in. thick. The minimum radius of gyration for the angle is 0.70 and the - value for this member is = 106. r 0.79 The allowable stress is p = 16,000 70 - = 8580 Ibs. The total load on the section is 2 X 1.07 X 8580 = 18,400 Ibs. Although this greatly exceeds the load on the member it is the smallest permitted by the specifications and will be used. The Upper Chord or Rafter. The distance between apices along this chord is approximately 8 feet. The piece is subjected to combined compression and flexure. It will be assumed that purlins are spaced 6 ft. o ins. center to center and that the maxi- mum bending will occur when a purlin lies midway between two adjacent apices. DESIGN OF A STEEL MILL BUILDING 137 The load carried by each purlin is 6 X 25 X 34 = 5100 Ibs. Considering the continuity and the method of securing it at the points of support it is usually safe to take the bending as five- eighths of that on a beam similarly loaded but supported at the ends. The bending on a section of the upper chord whose span is 84 ins. is M = ^X =^X (5100 X 84) -* 4 = 67,000 in. Ibs. 84^ The maximum compression along the upper chord is 90,000 Ibs. and occurs in CJ. The design will be carried out for this mem- ber. The proper section can only be determined by trial, that is, a section ' = s must be assumed and then tested by cal- l f Inr culation to see if it satisfies the condi- : j, tions. The section most frequently used is that shown in Fig. 170. The plate is i__ generally taken of sufficient width so that FlG the members joining the upper chord may be riveted directly to it. The section to be tried will be made up of two 4 in. X 3 in. X iVin. angles and one 10 X iVin. plate. It is first necessary to find the center of gravity. Area. Statical moment. Two angles 2 X 2.09 = 4.18 X 0.76 = 3.18 One plate 10 X 0.313 = 3.13 X 5-0 = I 5-^5 Total area = 7.31. Total moment = 18.83 *'= Wi 8 -= 2. 58 ins. The moment of inertia about axis i-i then readily follows: Inertia of the two angles about their axis 2 X 1.65 = 3.30 Ah* of these angles about common axis 4.18 X (2.58 0.76)* = 13 .90 bd 3 0.31 X io 3 Inertia of plate. = - = 26.00 12 12 Ah 2 of plate, 3.13 X (5-00 - 2 . 5 8) 2 = 18.30 / *= 61.50 138 GRAPHICS AND STRUCTURAL DESIGN Me The fiber stress due to bending is given by / = . f e = 67,000 X r~- = 2800 Ibs. Compressive stress. L = 67,000 X * - = 8080 Ibs. Tensile stress. 61.5 The extreme fiber stress at backs of angles is 12,300 2800 = 15,100 Ibs. The extreme fiber stress at edge of plate is 12,300 + 8080 = 3920 Ibs. The radius of gyration of the section about the axis through the center line of the plate must now be found. Since r = V/ v A the inertia must be found first. The inertia of two angles about the axis parallel to the short legs is 2 x 3.38 6.76 The AW portion is 2 X 2.09 (1.26 + o.i6) 2 = 8.36 Total inertia = 15.12 12 1.44. o = 58-5, say 60. / = 16,000 70- = 16,000 (70 X 60) = 1 1, 800 Ibs. Since wind and all other external forces have been considered this allowable fiber stress can be increased 25 per cent, making/ = 14,800 Ibs. This being within 5 per cent of the stress estimated as acting in the piece, the 15,100 Ibs. fiber stress can be allowed. COLUMNS When knee braces are omitted the building must be stiffened in some other way. This may be done by designing the columns to withstand the wind loading in each panel on the side of the building, by designing the lateral bracing to carry all wind DESIGN OF A STEEL MILL BUILDING 139 pressures to the ends or by assuming a division of this loading between the columns and the bracing. Actually the columns and bracing always share such loads, but as the proportion carried by each depends upon their relative stiffness the actual loading becomes a matter of considerable uncertainty. Just what as- sumptions are best to make will depend upon the length and height of the building, the assumed wind pressure, the manner of securing the tops of the columns to the trusses and the bases of the columns to the foundations. The assumptions for the design will also depend upon the judgment of the designer. All bracing should be given an initial stress when erected, to insure its acting promptly when the loading comes upon it. The portion of the column above the crane girder carries the roof load and will be liable to buckle about an axis parallel to the web. The load is applied concentrically to the column. There may be bending on the column about this axis from wind pressure on the end of the building, and from thrust due to the crane stopping and starting on its runway which should also be con- sidered. Below the crane girder the inside angles of the column must transfer the crane load, including impact, to the full section of the column. As this transfer will occur in a short distance the - r value may be neglected, hence the angles on this side of the f 66,000 column would require a minimum section of == 4.13 sq. ins. The two 6 in. X 3$ in. X iV m - angles used give an area of 5.78 sq. ins. and should prove ample. The roof load and the side of the building carried by the columns being both eccentric to the axes of the columns will produce bending on them. The resultant of these two loads and the crane load will most likely be eccentric to the center of gravity of the columns and will, therefore, produce a bending moment upon them. In addition to these the transverse thrust from Jhe crane and the 140 GRAPHICS AND STRUCTURAL DESIGN wind load on the side of the building will produce bending on the columns about an axis at right angles to the web. The effect of these moments will be materially reduced by the restraint at the column bases if the columns are fixed at that point. The maxi- mum fiber stress resulting from the combination of direct and flexural stresses should not exceed that permitted upon the column when the maximum allowable stress has been reduced by a suitable column formula. Considering that the stresses are LOWER CHORD BRACING X END VIEW ^---^ ~^^^ X Strut Eave Strut "XT X 3>392,ooo + (318,000 ~ 15.000) X 2 = I(5l?735lbs 68 The diagram of maximum shears checks this value of the end shear and shows that the maximum shear occurs with wheel 2 at the pier instead of with wheel i there. DESIGN OF A RAILWAY GIRDER DIAGRAMS FOR DECK PLATE GIRDER Span 68 ft. Loading Cooper's E 60 a n%'n j n K ^ V ftfr ?n p FIG. 177. Wind Bracing F 1 G I H i I &th Web plate-4.375/ Span68'0" j Lengths of Flange Plates FIG. 176. O P u v. FIG. 8925^ x ,17850^ End Bracing FIG. 179. T .1 I ' 8 ' MX* Stress Diagram / Q Kll\ / / \ / R]\ / \ / u 7 \ / \ / K/ \/ \/ C-J ' P FIG. 181. FIG. 1 80. 146 GRAPHICS AND STRUCTURAL DESIGN Dead-load Shears. The weight per foot of one girder was previously estimated at 735 Ibs. The dead-load end shear, being one-half the weight of the girder, is equal to 735 X -2 8 - = 25,000 Ibs. The dead-load shear varies uniformly across the girder changing from this value at the end of the span to zero at the middle; it will, therefore, be-** 1 - - = 12,500 Ibs. at the quarter point. The impact shear will be given by the formula that was used for impact bending, / = S I ) . This gives the impact as \L + 3007 81.5 per cent of the live -load shear when the girder is fully loaded, and the impact shear becomes 0.815 X 161,735 = 131,810 Ibs. Combining the several end shears we have: End shear, dead load 2 5>33 Ibs. End shear, live load 161,735 Ibs. End shear, impact 131,810 Ibs. 318,875 Ibs. Allowing a unit shearing stress of 10,000 Ibs. per sq. in. requires a web area of 31.9 sq. ins. A web plate 80 ins. deep and T 7 F in. - thick has an area of 35.0 sq. ins. and is ample. Flange Area and Selection of Sections. It is common practice to assume the effective girder depth, that is, the distance between the centers of gravity of the girder flanges, as approxi- mately the distance back to back of the flange angles. After the flange sections are chosen their centers of gravity are determined and this distance compared with the distance first assumed; if the error is small no correction is made, otherwise the distance is taken as the calculated distance between the centers of gravity of the flanges and new flanges are determined and a new selection of sections is made. The girder has been assumed as having an 8o-in. web plate and the angles will be set out \ in. on each side from the edge of the plate to avoid the possibility of the plate extending beyond the angles. The effective depth will be taken as 80.5 ins. The DESIGN OF A RAILWAY GIRDER 147 allowable working fiber stress being 16,000 Ibs. per sq. in. the net flange area is found by dividing the maximum bending moment in inch pounds by the product of the allowable fiber stress and the girder depth. Net flange area = 58.205,980 16,000 X 80.5 45.2 sq. ms. This area will be made up of the following sections: One-eighth web area = i X 35.0 = 4.4 sq. ins. Two 6 X 6 X f-in. angles, less 4 rivet holes 13 . 9 sq. ins. Three 14 X f-in. plates, less 2 rivet holes 27 . o sq. ins. Total area 45 . 3 sq. ins. This is sufficiently near the 45.2 sq. ins. required. The rivet holes have been assumed as i in. in diameter. As the calculated distance between the centers of gravity of the flanges exceeds that assumed by 0.20 in. the section assumed will be slightly larger than required, about } per cent. This is an unimportant difference and will be neglected. The live -load shears will be taken from the diagram of maxi- mum shears, Fig. 175, and to these shears will be added the dead- load and impact shears. Shears in pounds. End. 1 span. \ span. Dead load 24,000 12,495 o Live load 162 ooo 00,000 46,000 Impact 131,800 82,500 40,400 Total end shear 318,700 IQ3,QCK 86,400 At the quarter point, when wheel 2 is at this section, the loading extends on the girder 59 ft. from the right pier and according to the formula the percentage of live-load shear to be 300 allowed for impact is = 83.5 per cent. The impact shear + 30 is 99,000 X 0.835 = 82, 500 Ibs. In the same way the allowance for impact at the center of the span is found to be 40,400 Ibs. 148 GRAPHICS AND STRUCTURAL DESIGN Stiffening Angles. The end stiffeners act as columns, but as the load is shared by the web and as their length is short they are generally estimated by using the allowable fiber stress but not reducing it by a column formula. The area required in the end stiffeners then is Max. end shear 318.875 - = , - Lai = iQ-93 S Q- ms - 16,000 16,000 Four 5 in. X 3^ in. X rs~ m - angles, having a combined area of 4 X 5.38 = 21.52 sq. ins., will be used. The intermediate stiffeners are not usually calculated. Angles of the same dimensions but of lighter section are used. The intermediate stiffeners will be made two 5 in. X 3^ in. X f-in. angles. The four end stiffeners being over the end bearing plate, the rivets will be assumed as transferring the end shear from the web to the four angles ; that is, the load will be assumed as distributed over the entire number of rivets in the two pairs of stiffeners. The rivets will be f in. in diameter and will be in double shear and will bear on the ^g-in. we ^ plate. Allowing a shearing fiber stress of 12,000 Ibs. per sq. in. and a bearing fiber stress of. 24,000 Ibs. per sq. in. the rivet value in double shear is 14,430 Ibs. while the value in bearing is 9190 Ibs. The number of rivets required in the two pairs of end stiffeners is - L -^ = 35- 9190 Lengths of Flange Plates. The flange-plate lengths can be approximated as in Fig. 176. Lay off a line representing the span of the girder to scale and at its center erect a perpendicular representing to scale the calculated net flange area, in this case 45.3 sq. ins. Now upon the span as a base and with the 45.3 as an altitude draw a parabola. The parabola construction is shown in the dotted lines. Starting at the base line lay off in succession distances representing one-eighth the web area, 4.4 sq. ins. ; the net area of the two flange angles, 13.9 sq. ins. ; and the net area of the three flange plates, 9 sq. ins. each. The distance measured along the lower line of any section between the sides of the DESIGN OF A RAILWAY GIRDER 149 parabola gives the theoretical length on the assumption that the bending moment across the girder varies as the ordinates in the parabola. This is not exact, although approximately true. Were greater accuracy desired the bending moments could be estimated for intervals along the girder and the corresponding moment diagram drawn, which could then be used instead of the parabola. It is customary to take lengths of plates slightly exceeding the lengths determined in the way just described. The following are the scaled lengths and the lengths used. Scaled length. Actual length. Outside plate, top and bottom Ft. 3O Ft. 22 Middle plate, top and bottom ... A? 46 Inside plate, top. . C2 7O Inside plate, bottom C? C6 Flange Rivets. First consider the rivets in the vertical legs of the flange angles securing these angles to the web plate. The rivets transfer the change in horizontal shear from the flanges to the web plate. The flange forces acting at intervals along the girder can be determined and the change in this force between adjacent sections used to estimate the number of rivets required. The change in flange force divided by the rivet value will give the number of rivets between the two sections. Be- sides this change in horizontal shear the rivets also transfer the vertical loads to the web through the upper flange rivets. The maximum wheel concentrations are usually considered as dis- tributed over three ties or approximately 36 ins. along the flange. These two shears can be combined in the following manner: Find the change in flange force per inch of span by dividing the difference between the flange forces acting at two adjacent sections by the distance between these sections; this will give the average change of flange force per inch of span between the sections chosen. The vertical shear transferred from the angles 150 GRAPHICS AND STRUCTURAL DESIGN to the web plate per inch of span will be the maximum wheel concentration divided by 36 ins. The square root of the sum of the squares of these two quantities will give the resultant shear per inch of span within the given section and the rivet value divided by this resultant shear per inch will give the rivet spacing in this section. The rivet value referred to is the lower of the two values in bearing or shear. The following is the commonly used and less tedious method of determining the flange riveting. The rate of change in flange force per inch at a section is given by f v f T where V = maximum vertical shear in pounds at the section. h = the distance between rivet rows in top and bottom flanges, measured in inches. Where there are double rows of rivets in the vertical legs of the flange angles h is the average distance. Where the web is assumed as resisting bending the value / should be reduced as the rivets are not required to provide strength to secure that part of the flange to the web which is already an integral part of it, that is, the one-eighth web area. A = total net area of flange, and a = one-eighth web area, then the reduced force becomes V A-a *T? ~T~ As before the vertical shear per inch transferred by the rivets due to the maximum wheel concentration is . Max. wheel load /2= ~W and the resultant shear per inch of span at the section under consideration is DESIGN OF A RAILWAY GIRDER 151 The rivet spacing then is _ Rivet value p ' /. The rivet spacing is generally made the same in the two flanges. The rivet spacing will now be estimated for the ends of the girder. The average distance h in this case is 80.5 (3 X 2.25) = 73-75- /i = F x .4-^ ==a i8 I 8oo x (4.4 + I3-Q + 9)- 44 = 6 lbs> h A 73.75 4.4 + 13.9 + 9 /, = V 3 6 3 o 2 + 8 35 2 = 3730 Ibs. The rivet value in bearing for f-in. rivets at 24,000 Ibs. per sq. in. is 9190 and the rivet spacing then is p = f f $ = 2.46 ins. Use 2j-in. spacing. Rivet spacing a distance of one-quarter the span from the piers, fl = Z X A_I^ = 194,000 X 31.9 = ]bs * A 73-75 X 36.3 As found before, / 2 = 835 Ibs. / 3 = V 23 io 2 + 8 35 2 = 2460 Ibs. P = fUnr = 3-73 in s. Rivet spacing at the center, , V X (A -'a) 86,400 X 40-9 /1= -kXT : 73-75 X 45-3 := / 3 = ioss 2 + 8 35 2 = 1350 Ibs. P = ttf* = 6.8 ins. The rivet spacing may be calculated for the several panels by the method just shown or it will be given with sufficient accuracy by laying off, at right angles to a line representing the span, ordinates showing to scale the rivet spacing at the ends, the 152 GRAPHICS AND STRUCTURAL DESIGN quarter points and the middle of the girder and then connecting the adjacent points with lines. The spacing at any section may be found by scaling the distance from the axis to the line just drawn at the particular section. Riveting of the Flange Plates. From what has been stated concerning riveting the legs of the flange angles to the web plate it will be evident that the rivets connecting the flange plates to the angles are called upon to do much less. A common practice is to put in rivets securing the plates to the angles and have them stagger the rivets in the vertical legs. This prevents the inter- ference of the rivets and furnishes more than sufficient rivets. Where there is no liability of interference this practice need not be followed. The number of rivets required to secure a plate in place and develop its strength can be estimated by multiplying the net area of the plate by its working fiber stress and dividing this by the proper rivet value. In this girder the plates are 14 X | in. and the net area is 9 sq. ins.; at 16,000 Ibs. per sq. in. the force in the plate is 144,000 Ibs. and the rivets being f in. in diameter and in single shear their value is 7215 Ibs., which is less than the bearing value and must, therefore, be used. The number of f-in. rivets in single shear to transmit 144,000 Ibs. is 144,000 The middle plate being 46 ft. long and the top plate 33 ft. long, the middle plate extends 6.5 ft. beyond the top plate at each end and the full strength of this middle plate must be developed in 6.5 ft. The rivet spacing, considering that the rivets are in a double row, must not exceed 6.5 X 12 - = 7.8 ins. 10 The spacing ordinarily does not exceed 6 ins, and in this class of work is frequently specified not to exceed 4 ins. The rivet spacing is also commonly made less at the ends of the plates and opposite open holes left for field rivets. DESIGN OF A RAILWAY GIRDER 153 Wind Bracing.* - - The wind pressure will be assumed as a horizontal pressure of 30 Ibs. per sq. ft. acting upon the side of the girder and upon a train assumed as having an average height of 10 ft. and starting 2 ft. 6 ins. above the rail. The wind load will be assumed as transferred to the ends by bracing in the plane of the upper flanges of the truss. At the ends it will be transferred from this bracing to the piers through diagonal cross bracing. Lighter intermediate cross frames will be inserted at intervals along the span to act as spacers and to stiffen the girders, but these are not usually calculated. Fig. 177 shows a plan of this bracing in the plane of the upper flanges and Fig. 180 is the stress diagram. The apex wind loads must now be esti- mated. The depth of the girder exposed to the wind pressure is approximately 7 ft., which, with the assumed train height, makes a total of 17 ft. The wind load per foot is 17 X 30 = 510 Ibs. If the bracing is divided into 13 panels the apex load is ^ * = 2750 Ibs. The reaction is 510 X -/- = 17,850 Ibs. The diagonals will have to resist both tension and compression. The maximum stress acts in piece JK whose length is approxi- mately 8 ft. 6 ins. The stress in it is 20,400 Ibs. Allowing I 12 - = 1 20 the smallest radius of gyration permitted is 8.5 X - r 1 20 0.85. Trying a 6 in. X 4 in. X ^-in. angle the value of - = ^ = 1 16. r 0.88 The allowable stress per square inch is /= 16,000 70- =/= 16,000 (70 X 116) = 7880 Ibs. The total load carried by the 6 in. X 4 in. X J-in. angle is 4.75 X 7880 =37,500 Ibs. * Paragraph 144 of the specifications would make the loading greater than that used here. 154 GRAPHICS AND STRUCTURAL DESIGN To develop the full strength of this piece, using f-in. rivets in single shear having a rivet value of 7200 Ibs. and allowing 25 per cent extra, as the rivets are field driven, would require 37,500 X^= 7 - 720O The maximum stress in a strut will be that in the end frames which will be assumed as one-half the reaction, or '' ^ = 8925. Its length is 75 ins. If - = 120, r = -" = 0.625. This requires an angle not less than 3^ ins. X 3^ ins. and one angle 3! ins. X 3! ins. X | in. will carry, according to the formula, / = 16,000 70 - = 16,000 (70 X 108) = 8440 Ibs. The total load is 2.49 X 8440 = 21,000 Ibs. The number of f-in. rivets to develop the strength of this section as a column is 21,000 X ^^- = 4. 7200 The number of rivets between the connecting plate and the girder may be determined as in Fig. 179. Lay off on a line parallel to PQ a distance representing the number of rivets in PQ and similarly for RS, as one of these is in compression and the other in tension their components along the main girder flange will be in the same direction, and hence the number of rivets between flange and plate can be determined by projecting these lengths upon the horizontal line which here scales 10 rivets. End Diagonals. The end diagonals are 7 ft. 8 ins. long. When one is in tension the other will be assumed in compression. The load carried by the piece is 12,600 Ibs. If - is not to exceed 120 then r = T VV = 0.78. 6 in. X 4 in. X f-in. angles will meet this requirement and, the area being 3.61 sq. ins., is more than ample to carry the load. Web Splice. When the web is assumed as only resisting shear the splice is calculated to provide for the maximum shear at the DESIGN OF A RAILWAY GIRDER 155 spliced section. When, however, the web is assumed as resisting bending, that is, one-eighth of the web area is taken as acting at the flanges, then the splice plates and rivets must be designed to resist a corresponding bending moment. The depth of the splice plate will be the distance between the edges of the verti- cal legs of the flange angles. In this case, allowing ^ in. for clearance, the depth of the splice plate will be 80.5 (12 -j- ^) = 68 ins. The fiber stress at the center of gravity of the flanges was assumed as 16,000 Ibs. per sq. in. The distance from the neutral axis of the girder to the center of gravity of a flange was assumed as = 40.25 ins. The fiber stress at the top of the splice 2 plate will be (16,000 X V") + 4- 2 5 = I 3j5 ^s. per sq. in. The bending moment resisted by the web is - X depth 8 of girder X fiber stress = M = -\ 5 - X 80.5 X 16,000 = 5,635,000 in. Ibs. The area of the splice plate required to resist this moment is determined by a calculation similar to the one made for the web plate. This area is . 8 X bending moment at section, in. Ibs. depth of splice plate, ins. X extreme fiber stress, Ibs. per sq. in. 8 X 5,635,000 A = - = 49.0 sq. ins. 68 X 13,500 The thickness of each of the two splice plates then is / = = 0.36 in. 2 X68 If the rivet farthest from the neutral axis develops its full rivet value the stress upon any other rivet will be proportional to its distance from the neutral axis and the moment due to any rivet being also proportional to this distance the total resisting moment of a single row of rivets similarly arranged about the neutral axis is M = 2 X rivet value 156 GRAPHICS AND STRUCTURAL DESIGN Here M = resisting moment in inch pounds of one vertical row of rivets. R = rivet value, in pounds. (The lower of the values in bearing or shear should be used.) y = distance of the rivet from the neutral axis, in inches. yi = distance from the neutral axis to rivet in splice plate farthest from the neutral axis. Assuming the following distances of the rivets in one row above the neutral axis, the following values are calculated for Fig. 182. y 2 in FIG. 182. FIG. 183. Number of rivet. y y2 Number of rivet. y y- I o O 7 18 324 2 3 9 8 22.5 506.25 3 6 36 9 26.5 702.25 4 9 81 10 29-S 870.25 5 12 144 ii 32.5 1056.25 6 15 225 3954-00 The rivet value of a f-in. rivet bearing in a -^g-in. plate is 9190 Ibs., hence the moment of one row of rivets spaced as assumed is 2 ^^ O T OO M = - - X 3954 = 2, 240,000 in. Ibs. DESIGN OF A RAILWAY GIRDER 157 The number of rows required on this basis is 15,635,000 i 3 ' OJ ' - = 2\ rows. 2,240,000 This would take three rows or a rearrangement of the rivet spacing to increase the number of rivets in the vertical row and at the same time the moment of the rivet group. There are several other forms of splices, one of which, see Fig. 183, consists in replacing the one-eighth of the web area by straps placed on each side of the flange angles. As ordinarily calculated the area of the strap is to one-eighth web area as the distance back to back of the flange angles is to the distance a center to center of straps. If the straps are 9 ins. wide the distance center to center of straps will be 59 ins. 4.4 X 80.5 Strap area = - * = 6.0 sq. ins. 59 The number of rivets on each side of the strap will be 6.0 X 16,000 - = 10.4, say ii rivets. 9190 The central plates are designed to resist the maximum vertical shear at the section. Still another method is to splice the web where excess material in the flange makes provision for all or part of the bending as- sumed as taken by the web and then design the splice for the maximum shear and any bending not otherwise provided for occurring at the section. In the problem in hand the web splice is made 20 ft. from the piers, and the shears at this point are dead-load shear 10,300 Ibs., live-load shear 90,000 Ibs., and the impact allowance 76,000 Ibs., making a total shear of 176,300 Ibs. This would require twenty f-in. rivets to provide for the maximum shear at this point. There is more than sufficient surplus material in the flange (see Fig. 176) to care for the bending assumed as coming upon the web, so that a splice designed for shear only would prove ample. 158 GRAPHICS AND STRUCTURAL DESIGN DESIGN OF A RAILWAY GIRDER 159 Flange Splices. For the ordinary girder spans splices in the angles and plates of the flanges are not quite so likely to be required as in the web. When such splices are required the same general principles may be followed as in the case of the web splice. Splice plates may be placed where angles or plates are cut or the cut may be made where the flange plate would other- wise end and then the splice may be made by continuing the flange plate beyond the cut. This affords one of the neatest ways of splicing the angles or flange plates. Bed Plates. The area of the bed plate allowing 400 Ibs. per sq. in. on the masonry will be the reaction divided by 400 or = 797 sq - ins - In small girders carrying light loads the connection between the girder and the masonry may be made by the use of a couple of rolled plates. These should extend but a few inches beyond the flange plates as the usual depth of about i in. is not stiff enough to transmit much load to the pier beyond the edges of the flange plate. The bed plate rests upon the masonry while the sole plate lies on the bed plate and carries the girder. At one end the girder is held firmly in place by both sole and bed plates having round holes a little larger than the foundation bolts. At the other end the holes in the sole plates are slotted, permitting the movement of the sole plate across the bed plate due to the expansion and contraction of the girder. The plates at the moving end should be planed. Where heavy loads are carried upon girders of short span, greater depth of bed plates may be required; this type of cast- iron bed plate is shown in Fig. 188. In the longer spans, exceeding 75 to 80 feet, greater provision must be made for expansion and also more care used to distribute the load on the masonry. One way of doing this is shown in Fig. 189. Here both ends of each girder are carried on pins, while one end rests on friction rollers. The pin assists the distri- bution of the pressure uniformly over the rollers, if the webs of i6o GRAPHICS AND STRUCTURAL DESIGN the pedestal are of ample depth. The method of calculation would be as follows: Assuming the rollers 4 ins. in diameter the allowable load per inch of roller length is L = 1200 vQ, where L = load in pounds and d = diameter of roller in inches. L = 1 200 A/4 = 2400 Ibs. The combined lengths of all the rollers in the nest is 3 1 ^ 9 = 2400 133 ins. Making the rollers 23.5 ins. long would require -** ~ 2 3-5 6 rollers. The bearing area between pin and web is - *"? = 24,000 _o o o 6 o o o , =p= r~ o o o $ o ! o o ooo |33 e^fo-^- -}^ 5"f ^ 'Y .)^ek>- 231/3 en FIG. 189. 13.3 sq. ins. If three i-in. web plates are used the projected area is 3 ins. X I.OQ in. X 6 ins. = 18 sq. ins., and the area will be more than ample. The web plate is 45 ins. long and the load is to be distributed over 1 it. A rough approximation of the depth of the web from the bottom of the pin to the lower edge of the web plate may be made by considering it a beam section. M = ^ = / - = ^-~- ; o c 6 DESIGN OF A RAILWAY GIRDER l6l here d = depth of the web in inches, b = combined thickness of the webs in inches. M (- ,72 M = 318,790 X-.r = 16,000 X 3.00 X , from which d = 8 6 15 ins. In this calculation no account has been taken of the angles and plate secured to the lower edge of the web plates, so that the i5-in. depth used should prove ample. The rollers are kept in place by two flats held by three rods shown at the ends and the middle. Angles at the sides secured to the bed plate prevent the rollers ,f rom moving laterally. The bending on the pin is 318,700 X 1.25 * LL2 2 = 132,500 in. Ibs. 5 On the basis of an allowable extreme fiber stress in pins of 22,000 Ibs. per sq. in. a pin 6 ins. in diameter will resist a bend- ing moment of 466,500 in. Ibs., so that in this way the pin is exceedingly strong. The methods indicated illustrate a couple of the simplest methods used for securing girders in place. CHAPTER XII CRANE FRAMES FIGURE 190 represents a frame for an underbraced jib crane. The live load it is to carry is 3 tons (6000 Ibs.). To this must be added an assumed weight of block, trolley, chain, etc., and this will be taken at 500 Ibs. The total live load thus becomes 6500 Ibs. The dead load is made up of the weight of the frame and hoisting machinery and will be assumed at 2000 Ibs. The line of action of this dead load will be taken at three-tenths of the effective radius from the mast, or 11.75 X 0.3 = 3.5. The equivalent load at apex i will be the load that would produce the same amount on the mast, and is 2000 X T 3 Q- = 600 Ibs. The chain pull at different points in its length will be estimated upon an assumed efficiency of each sheave of 97 per cent, thus above the jib the chain pull will be 6500 -r- (2 X Q-97 2 ) = 3460 Ibs., while parallel to the mast it is 6500 -j- (2 X o.97 3 ) = 3560 Ibs. The pull on the racking chain will be assumed at 1000 Ibs. The maximum direct stress will occur in the bracing when the load is at its maximum radius. To determine the direct stresses graphically the apex loads must be found. To calculate the equivalent apex load at i (Fig. 191), when the load is at its maximum radius, take moments about point 3. The equivalent apex load at i = 6500 X "' ' = 7650 Ibs. 10 The upward reaction at point 3 is 7650 6500 = 1150 Ibs. The diagram (Fig. 192) gives the combined live- and dead-load stresses; these can be determined with greater accuracy if done separately but this diagram is sufficiently accurate in this instance. The jib will have to be considered for bending both when the load is at its maximum radius and when the trolley is between 162 CRANE FRAMES 163 FIG. 190. 1 Hbfeting Chain ^^\ Racking Chain "*~t ~3 FIG. 193. FIG. 194. FIG. 195. 164 GRAPHICS AND STRUCTURAL DESIGN points i and 2. Here the span is 60 ins. and the distance center to center of the trolley wheels is taken as 24 ins. Although commonly discussed as a central load the position of maximum bending will here be assumed accurately ; that is, when a wheel is one-quarter the distance between the wheels from the center of the span. Here this distance is \ 4 - = 6 ins. (see Fig. 193). The load of 6500 Ibs. being carried upon four wheels the load on two wheels or an axle is ^~ = 3 2 5 I DS - The reaction R> = to X 24) + to X 48) _ , bs DO RI = 6500 3900 = 2600 Ibs. The direct stress in the section of the jib carrying the trolley is given by Fig. 194 as 6600 Ibs. Fiber Stresses. Crane frames are liable to considerable shock. The usual method of caring for this is to take a lower working fiber stress than is ordinarily used in structures not sub- jected to shock. These fiber stresses run from 10,000 Ibs. to 12,000 Ibs. per sq. in. for mild steel, which is the material commonly used for crane frames. The allowable stress in columns is the above properly reduced by a suitable column formula. As it is difficult, if not impossible, to stiffen some parts of crane frames the jib, for instance, in the present problem the allowable stress in members acting as beams must be reduced to prevent buckling of the compression flange; see curves, page 100. Selection of Jib. The load at the maximum radius creates a direct stress in the jib of 13,900 lbs. r in tension. The hoisting- chain pull is 3640 Ibs., while the pull in the racking chain is 1000 Ibs. ; this creates a force of 1000 Ibs. acting in the jib when the trolley is pulled back and of 2000 Ibs. when the trolley is drawn forward. The average height of the two racking chains from the center of the jib is taken as 8 ins. Calling tension +, and compression , we have : Direct stresses in jib. Stress due to live and dead loads + 13,900 Ibs. Stress due to hoisting rope 3,460 Ibs. Stress due to racking rope 2,000 Ibs. CRANE FRAMES 165 The greatest stress then is the combination of the stress due to live and dead loads with the stress due to hoisting. The racking force will not be considered as it gives a lower stress than the above combination. The maximum stress is 13,900 3460 = 10,440 Ibs. in tension. The bending moment is (6500 X 23) (3460 X 14) = 101,060 in. Ibs. The stress must now be considered when the trolley is between points i and 2. The maximum .bending = RI X 24 = 2600 X 24 = 62,400 in. Ibs., due to the hook load. To this must be added the bending due to the hoisting- and rack- ing-rope pulls. The former is M 2 = 3460 X 14 = 48,400 in. Ibs. That due to the latter is MS = 2000 X 8 = 16,000 in. Ibs. The total bending is 62,400 + 48,400 + 16,000 = 126,800 in. Ibs. The direct stress is 6600 3460 2000 = 1140 Ibs. The influence of span to flange width must now be considered. Trying 8-in. channels at n| Ibs. per foot the flange width is 2.26 ins., and the span being 60 ins. it follows that Span ___ 60 ^ Flange width 2.26 On the cantilevered section, Span 23 - = *- = 10.2. Flange width 2.26 From the curve the allowable stress should not exceed 90 per cent of the maximum desired, so that allowing a maximum working fiber stress of 10,500 Ibs. per sq. in. it reduces to 10,500 X 0.90 = 9450 Ibs. per sq. in. The section modulus for an 8-in. channel at nj Ibs. is 8.1; hence, since M = f- and the bending has been calculated for two channels, we have . Me 126,800 ^, /= =- -X8.i = 7850 Ibs. The direct stress per square inch = ^-- X 3.35 = 170 Ibs. The 8-in. channels at nj Ibs. will be satisfactory. 1 66 GRAPHICS AND STRUCTURAL DESIGN Member AH. The length of this piece is 7 ft. and the force acting in it is 16,100 Ibs. This member is in compression and as it cannot be braced and its - value must be limited to 140 r the least radius of gyration will have to equal or exceed $ = 0.60. Seven-inch channels will be required and their radius of gyration is 0.59. According to the abridged form of Ritter's formula for soft or mild steel f- L = I0 '5 = IO -5 = 3550 Ibs i m 2 19,600 2 . 9 6 i-l ~ XI - 1 I ~i 10,000 \rl 10,000 and the total allowable load on two 7 -in. channels at 9! Ibs. per foot or 2.85 sq. ins. in the section is 3550 X 2 X 2.85 = 20,300 Ibs., which is satisfactory. Member AF. This is a compression piece carrying a load of 17,700 Ibs.; its length is approximately 13 ft., but, as it can be braced across, its greatest length need not exceed the length of the piece HA, and as the force acting in it is about the same as in the preceding piece it will be made of the same section, 7-in. channels at 9! Ibs. per ft. Member FG. This is also a compression piece; its length is 60 ins. and the force acting in it is 10,500 Ibs. Assuming as before that the - value shall not exceed 140, r must not be less than T 6 ? ^ = 0.43. Trying 5-in. channels at 6| Ibs., which have r = 0.50, we find- = - = 120. Substituting, as before, in the r 0.50 column formula gives 10,500 10,500 ,, = 4300 Ibs. x r 10,000 \r/ The total allowable force on the piece is 4300 X 2 X 1.95 = 16,700 Ibs., and these channels will be used. Mast. For most positions of the trolley the mast will be in tension. The position of maximum compression will require the CRANE FRAMES 167 The sever- trolley to be placed as close to the mast as possible. est stress on the mast will be due to the bending. The horizontal reaction at the upper pin can be found by taking moments about the lower pin. R X 13-75 = ( 6 5 X 11.75) + K 2000 X 11.75) * 4] = 6000 Ibs. The bending at the upper portion of the mast equals 6000 X 22 = 132,000 in. Ibs. The mast also resists bending due to the 5-in. channels which fall below the line of intersection of the other ! [tl Tin FIG. 198. pieces at apex 3. The reaction on the mast due to this force in FG, the horizontal component of which is 6800 Ibs., is RI 6800 X ii 12 = 6230 Ibs. and the bending is M = 6230 X 12 = 74,760 in. Ibs. This bending is in the opposite sense to that due to the force on the pin. The bending due to the hoisting-rope pull must now be deter- mined. The horizontal reaction at apex 3 due to the rope pull on the drum is found by taking moments about apex 5, from which l68 GRAPHICS AND STRUCTURAL DESIGN The maximum bending due to the rope pull then is M = 173 X 90 = 15,57 in - tt>s. O- If fl I M 132,000 Since M = f -> - = = -^ ^ = 12. z. e e f 10,500 This is for two channels, or 6.25 for one, and would require 7-in. channels were there no direct stress. Try 8-in. channels, the area of two such channels being 2 X 3.25 = 7.0 sq. ins. The direct stress previously found was 9100 Ibs. and the unit fiber stress is = 1300 Ibs. per sq. in. The fiber stress allowed for 7.0 bending then is 10,500 1300 = 9200 Ibs. per sq. in. It follows that I M 132,000 - e = j- -%- =I4 - 3S - This will take two 8-in. channels at nj Ibs. whose section modulus is 2 X 8.1 = 16.2. The resulting moment at any section is the intercept between the heavy lines in Fig. 198. DESIGN OF FRAME FOR TOP-BRACED JIB CRANE The capacity of the crane is to be 5 tons (10,000), at a maximum radius of 20 ft. The weight of the trolley, hook and chain will be assumed at 600 Ibs. The dead load of the frame and machinery will be taken as 3000 Ibs., and its center of gravity, or line of action, will be considered as one-fourth of the crane radius from the mast, or %- = 5 ft. The load which, acting at apex i, would produce the same moment on the mast as the frame weight is 8JHLO = 75olbs> The maximum fiber stress in tension is to be 12,000 Ibs. per sq. in. In compression the maximum fiber stress will be 12,000 Ibs. per sq. in., properly reduced by a column formula. The value of - will be limited to 140. CRANE FRAMES 169 J The skeleton of the frame is shown in Fig. 199, v while the stress diagram is represented by Fig. 201. The total equivalent load at apex i is the sum of the live load, 10,000 Ibs., the weight of the trolley, hook and chain, 600 Ibs., and the equivalent frame 1 70 GRAPHICS AND STRUCTURAL DESIGN load, 750 Ibs., making a total of 11,350 Ibs. The horizontal re- actions at points 4 and 6 can then be found by taking moments about apex 6. R = TI > 35 X 2 = 12,845 Ibs. 17-75 These several external forces can now be located at the proper points and the stress diagram drawn in accordance with the principles given in Chapter II. Selection of Members. The members CE and H B are in tension and having about the same stress acting in them will be A OOO made of the same section. The net area required is 12,000 4.50 sq. ins. Trying two 4 in. X 3 in. X yV m - angles, and assuming |-in. diameter holes for f-in. rivets, the net area afforded by these angles is Net area of two 4 in. X 3 in. X ^-in. angles = (2 X 2.88) (2 X 0.38) = 5.00 sq. ins., which is satisfactory. The member CF will also be made of the same section, as it is very short. Member GH. The maximum stress for this member occurs when the load is at apex 2. The stress, given by Fig. 202, is 2*7 OOO 27,000 Ibs. The net area required then is -- = 2.25 sq. ins. Trying 3 in.X 3 in. X i-in. angles, two of which have an area of 2 X 1.44 = 2.88 sq. ins., and allowing one f-in. diameter hole for a f-in. diameter rivet to an angle the net section for two angles is 2.88 (2 X 0.22) = 2.44 sq. ins., and these angles will do. Members A G and AH. These members will be subjected to combined compression and flexure, and as it is difficult to brace these members laterally, due consideration must be given to the reduction of the fiber stress to provide for the column action of the compression flange. The axle load is - >2 -^- =5325 Ibs. The maximum bending on the lo-ft. span will occur when one of CRANE FRAMES 171 the wheels is one-fourth the distance between the wheels from the center of the span. Finding the reactions we have R> = (53*5 X 54) + (5.^5 X 78) = ^ 120 Then RI = 10,650 5860 = 4790 Ibs. The maximum bending moment on the span then is M = RI X 54 = 4790 X 54 = 258,660 in. Ibs. The maximum stress will occur in these pieces in member HA when the load is located as shown in Fig. 204, and the direct o o FIG. 204. stress is given by Fig. 203. To determine this stress the reaction at apex i must be found when the load is in the above position. Under this loading the apex load at i is R% = 5860 Ibs. plus the equivalent frame load at i, which is 750 Ibs., these making a total of 6610 Ibs. The direct stress is found to be 31,000 Ibs. The jib will be assumed as braced in two places, so that it will be approximately braced laterally at intervals of about 8 ft. This bracing can be only a light frame, generally made of a bent angle that must clear the trolley. The total bending on the member AH is due to the vertical loading used in deter- mining the 258,660 in. Ibs. just calculated, and in addition to this there is bending due to the hoisting-rope pull and the rack- ing-rope pull, both of which act to produce bending of the same character on the beam and must therefore be added to the bend- ing moment just found. 172 GRAPHICS AND STRUCTURAL DESIGN The bending moment due to hoisting-rope pull = 5650 X 19 = 107,350 in. Ibs. The bending moment due to the racking-rope pull = 1000 X 9 = 9000 in. Ibs. The total bending moment then is 258,660 + 107,350 -f 9000 = 375,010 in. Ibs. Assuming that a section will be tried whose flange width is 3 ins., the ratio of unsupported part of span to a -j^ - 8 X 12 flange width is - =32. From the curve on page 100 the allowable fiber stress in com- pression for this ratio of length to flange width is about 85 per cent of the maximum desired. The allowable fiber stress is 12,000 X 0.85 = 10,200 Ibs. per sq. in. If we try i2-in. channels at 25 Ibs. per ft., we find their area to be 2 X 7.35 = 14.7 sq. ins. and their section modulus 2 X 24.0 = 48.0. The direct stress per square inch is " 2 = 2100 Ibs. 14-7 The stress permitted for flexure becomes 10,200 2100 = 8100 Ibs. per sq. in. The required section modulus then is I _ M_ _ 375,010 _ . ~e~~f = 8100 It is evident, therefore, that the channels tried are satisfactory. Before being accepted finally, however, these channels should be examined as columns when the load is at its maximum radius. Under this condition the jib is subjected to a direct stress of c<2 tQO 52,500 Ibs. The unit stress due to this direct load is 14.7 = 3570 Ibs. The allowable unit stress according to Ritter's formula is , 12,000 - _ 12,000 = 12,000 ' == ^ -44 io,ooo \r/ 10,000 = 4900 Ibs. per sq. in. CRANE FRAMES 173 The - value of 120 was found by dividing the unsupported length, 96 ins., by the radius of gyration of the channels about their minor axis, 0.78, or - = 120. 0.78 The i2-in. channels were found satisfactory in this respect. Mast. The direct stress as given by the diagram, Fig. 201, is 50,000 Ibs. The reactions at the pins are 11,^50 X 20 R = -^ - = ii, 120 Ibs. 20.5 The bending moment due to the above reaction is M = n, 1 20 X 21.5 = 239,080 in. Ibs. Trying two i2-in. channels at 20.5 Ibs. per ft., their area is 2 X 6.03 = 12.06 sq. ins., and the direct compression is^i 12.06 = 4150 Ibs. per sq. in. Assuming the allowable ultimate fiber stress. in bending as 12,000 Ibs. per sq. in. the balance for bending is 12,000 4150 = 7850 Ibs. The required section modulus is I = M _ 239,080 _ ~e f : 7850 An inspection of the tables of the properties of channels in a Manufacturer's handbook or on page 16 will show that either a lo-in. channel at 20 Ibs. or i2-in. channels at 20^ Ibs. will carry this load. However, since i2-in. channels are used for the member AG and as these channels give considerably greater stiffness they will be used rather than increase the number of sections demanded. As it is possible to brace the channels forming the mast at as frequent intervals as desired no account has been taken of the ratio of span to flange width. CHAPTER XIII GIRDERS FOR OVERHEAD ELECTRIC TRAVELING CRANES Specification. Design a bridge for a lo-ton O.E.T. crane whose span is 45 ft. Assume the distance center to center of trolley wheels as 4 ft. o ins., that the bridge weighs 13,000 Ibs. (two girders) , and that the trolley weight is 6000 Ibs. Design the girders to resist an additional loading laterally of one-tenth the crane capacity. Allow a working fiber stress in tension of 10,000 Ibs. per sq. in. and the same properly reduced by a column formula for compression pieces. Use f-in. rivets throughout, for which assume f-in. diameter holes. The loading will be reduced to that for a single girder or one- half the bridge. The weight of the bridge will be assumed as a uniform load and, as is usually done, the bending moment for it will be calculated for the middle of the bridge. Weight of i girder = -*"- - = 6500 Ibs. Dead-load bending = - = 6500 X 45 X =438,750 in. Ibs. 8 8 Assuming the load as hung centrally on the trolley, the load on each trolley wheel will be 1 (trolley weight -f- live load) = J (6000 + 20,000) = 6500 Ibs. The reaction _ [(6500 x 19.5) + (6500 x 23.5)] = 62IO lbs 45 The maximum bending due to live load then is M = 6210 X 21.5 X 12 = 1,602,180 in. Ibs. The total bending due to combined dead and live loads is M 438,750 + 1,602,180 = 2,040,930 in. Ibs. 174 GIRDERS FOR TRAVELING CRANES 175 The material will be determined for the two following sec- tions, the depth being taken as 3 feet. H FIG. 205. FIG. 206. FIG. 207. The box section (Fig. 207) will be considered first. If the width is made one-thirtieth the span it will be, say, 17 ins. wide. The ends of the girders resting upon the bridge trucks will be assumed 18 ins. deep. The web plates will be made \ in. thick, making the gross web area at the ends 2 X J X 18 = 9 sq. ins. FIG. 208. The maximum end shear is given by the diagram of maximum shears (Fig. 208), and is 15,800 Ibs. The unit shearing fiber stress is = 1760 Ibs. This fiber stress is very low but, 1 76 GRAPHICS AND STRUCTURAL DESIGN notwithstanding that some specifications permit i^-in. material, we will use the J-in. plates. Compression Flange. The distance between the centers of gravity of the flanges will be assumed as approximately the distance back to back of the flange angles, or 36 ins. According to the curve, page 100, the allowable fiber stress in a compression flange when the ratio of the laterally unsupported length of flange to flange width is 30 to i, is 80 per cent of the maximum desired. The allowable fiber stress becomes 10,000 X 0.80 = 8000 Ibs. per sq. in. The net flange area then becomes (see page 146), M If the web is considered as taking bending in addition to shear the web will furnish a section equal to one-eighth the web area, or f X 36 X 1 X 2 = 2.25sq. ins. In this case the flange plates will work out so light that the web plates will not be assumed as re- sisting bending. The flange angles will be tried at 3 ins. X 3 ins. X \ in. and they will be considered as having two holes, one in each leg at the same section; this makes it possible to locate the rivets at any points desired. The plate will be taken as 17 ins. wide. Net area of angles, 3 ins. X 3 ins. X i in., having two f-in. diameter holes, (2 X 1.44) (4 X 0.22) = 2.88 0.88 = 2.oosq. ins. The net area of the plate then is 7.1 2.00 = 5.10 sq. ins. The net width of the plate is 17 (2 Xf) = 15.25 ins. The thick- ness of the plate is = 0.334 in. ; use A- in. I5- 2 S Tension Flange Area. Here the allowable fiber stress is 10,000 Ibs. per sq. in. and the required net area of the flange is M 2,040,930 A = - - = ' ' vo = 5.67 sq. ins. fXh 10,000 X 36 Using two 3 in. X 3 in. X J-in. angles as before, the net area of the GIRDERS FOR TRAVELING CRANES 177 plate is 5.67 2.00 = 3.67 sq. ins. and the required thickness of the plate is -^-^ = 0.241 in., say \ in. The girders will be made approximately for uniform strength (see Fig. 209). The depths at points along the girder may be checked by drawing the diagram of maximum bending moments similar to that drawn for a railway girder, in Chapter XI, and combining the live-load bending moments thus found with the dead-load bending moments. FIG. 209. Since M = flange area X mean fiber stress X girder depth, it follows that the flange area and the mean fiber stress being made constant the depths at any section can be made to vary with the bending moments. This, of course, is approximately true only so long as the flange area times the square of its dis- tance from the neutral axis of the entire girder section at the point considered is large compared with the inertia of the flange about a parallel axis through its center of gravity. Where greater accuracy is desired the moments of inertia of the several sections may be computed and the formula M = / - used. (s The section chosen should now be tested when subjected to the lateral moment stated in the specification. This moment being produced by one-tenth of the live load it will equal one-tenth of the live-load moment, or = 160,218 in. Ibs. The web 10 plates for a distance of 12 ins. from the top will be considered as i 7 8 GRAPHICS AND STRUCTURAL DESIGN GIRDERS FOR TRAVELING CRANES 170 resisting this moment. The following calculation gives the moment of inertia about the axis B-B, Fig. 211. T 1 5 /JPlate 3x3x&L FlG. 211. -4-B One plate, i 26 . 90 Two plates, AW- = 2 X 0.25 X (12 - 0.88) X (S-^) 2 = 145-73 Two angles (about their own axis) = 2 . 48 Two angles, AW- = 2 X 1.44 X 6.09* = 106 . 82 Deducting for rivet holes (Ah 2 portion only), 2 X 0.22 X S-38 2 = 12 . 73 2 X (0.27 + 0.22) X 6.752 = 44-65 Inertia Inertia 381-93 57.38 324-5S The extreme fiber stress is then / = M -e or / 8. 160,218 X ~' J = 4200 Ibs. per sq. in. 324.6 The mean fiber stress in the compression flange is M 2,040,930 /= = 8420 Ibs. per sq. in. Axh 6.73X36 The total fiber stress is 8420 + 4200 = 12,620 Ibs. per sq. in. This fiber stress, considering that it is the maximum of both vertical and lateral bending, is satisfactory. Some designers prefer to estimate the forces acting laterally on the crane upon the assumption that the crane when carrying its maximum load is stopped from full speed in an assumed dis- i8o GRAPHICS AND STRUCTURAL DESIGN tance, the retardation being considered uniform. In the present instance we will assume a velocity of bridge travel of 300 ft. per minute, and that the crane is stopped in 5 ft. The crane velocity being ^- = 5 feet per second, the retarda- tion is / = -- , where / = acceleration or retardation in feet per second per second. v = velocity of bridge in feet per second. s = distance traveled during acceleration or retardation in feet. -2 In our case the retardation is/ = ^ = 2.5 ft. per sec. per sec. As the force equals mass times retardation, the force required to stop the bridge is F = - X 2.5 = 1000 Ibs. 32.2 This acts as a uniform load laterally along the bridge and will be resisted by both top and bottom flanges of each girder. The force to retard the trolley and live load will act upon the upper flanges of the two girders of the bridge and equals F = - 32.2 X 2.5 = 2020 Ibs. The central load on one girder the equivalent of the above forces is ^Y- - + -^ = 1260 Ibs. The bending moment due to this central load is W 'L 12 M = -- = 1260 X 45 X = 170,100 in. Ibs. 4 4 This moment differs so slightly from the moment found by the previous method that the girder dimensions should prove satis- factory viewed from either estimate. Flange Rivets. Considering that the unit fiber stress in tension is 10,000 Ibs. per sq. in. the allowable unit shearing fiber stress on the rivets, assuming about the same material, would be | X 10,000 = 7500 Ibs. per sq. in. The bearing value would be taken at twice the shearing value or 15,000 Ibs. per sq. in. GIRDERS FOR TRAVELING CRANES l8l The value of f-in. rivets in single shear and bearing in a J-in. plate is 2813 Ibs. Assuming the girder depth of the supports at 18 ins., the dis- tance center to center of rivets vertically is 18 (2 X 1.75) = 14.5 ins. The maximum end shear will be assumed as transferred to the girder in a distance equal to its depth, or 18 ins. The vertical i ^ 800 shear per inch of span is Si = - = 880 Ibs. 18 The horizontal shear per inch at this point is Vertical shear Distance center to center of rivets vertically 15.800 $2 = - iL - ~ = 1090 Ibs. 14.5 The resulting shear per inch of span is s z = Si 2 -f 5 2 2 : = 1400 Ibs. Considering that the web being double will necessitate a row of rivets in each web plate the horizontal distance center to center , . 2813 X 2 of rivets = = 4 ins. 1400 Owing to the concentration of loading at the ends the spacing will be made 3^ ins. The net area of the compression flange being 6.73 sq. ins. and the mean fiber stress 8750 Ibs. per sq. in., the number of rivets to develop the full flange strength is 6 ' 73 X 875 , about 21. 2813 These rivets placed in two rows and spaced 3^ ins. center to center would require a distance of 3.5 X n = 38.5 ins. This would indicate that for about 48 ins. from the ends of the plate the rivets should be spaced about 3! ins. center to center and at the other sections the spacing in the compression flange should not exceed from 16 to 20 times the thickness of the outside plate. The spacing in the two flanges is usually made the same. The 182 GRAPHICS AND STRUCTURAL DESIGN upper plate being T 5 g in. thick the rivet spacing should not exceed 5 ins. Angle Stiffeners. Allowing 10,000 Ibs. per sq. in. the area of .. Maximum end shear 1^,800 the stiffeners = = 1.58 sq. ms. 10,000 10,000 Using two angles the smallest permitted would be i\ ins. X i\ ins. X J in., and their section 2 X 1.19 = 2.38 sq. ins., which is more than ample. Stiffeners, however, serve a double purpose, as the opposite inside Stiffeners are riveted to a plate forming a diaphragm at the section and greatly stiffening the girder laterally. They frequently also serve to secure short channel or angle sections which pass across the flange and assist the flange plate in carry- ing the rail. These latter are commonly placed at each pair of vertical Stiffeners, while the diaphragms are generally placed at alternate panels. The web plate should be reinforced where brackets or motors are attached to it and at these points handholes should be cut in the web plate on the opposite side of the girder to facilitate bolting the brackets and motor to the girder. GIRDER DESIGN FOR FIG. 206 Using the same data the girder will now be designed for the section (Fig. 206). The maximum combined live- and dead-load bending was found to be 2,040,930 in. Ibs. The maximum end reaction pre- viously determined was 15,800 Ibs. The girder will be assumed 36 ins. deep, back to back of angles, at the center of the span, and 20 ins. deep at the bridge trucks. The web will be taken -f^ in. thick making the unit shearing fiber stress **' X 20 = 2550 0.31 Ibs. per sq. in. The effective depth will be assumed slightly less than 36 ins. ' to allow for the centers of gravity of the channels falling towards GIRDERS FOR TRAVELING CRANES 183 the center of the girder, the distance being 36 1.5 = 34.5 ins. The net flange area intension then will be M 2,040,030 A = - - = 5.02 sq. ins. fXh 10,000 X 34-5 Trying one g-in. channel at 13! Ibs. per ft., its area is 3.89 sq. ins., and two 3 in. X 3 in. X i^-in. angles, we have Sq. ins. Sq. ins. One g-in. channel at 13! Ibs., gross area ......... 3.89 Less two rivet holes in -in. material, 2 X .22 = . . .44 3.45 Two 3 X 3 X iVin. angles, 1.78 X2= .......... 3.56 Less four rivet holes in tV" 1 - plate, 4 X .27 = ... 1.08 2.48 5-93 This net area is satisfactory. In the compression flange the fiber stress must be reduced to provide for lateral strength. The ratio of span to flange width if a i5-in. channel is tried is 45 X yf = 36. Reducing the unit working fiber stress to 75 per cent of the maximum desired permits a fiber stress of 7500 Ibs. per sq. in. Again substituting in the formula = 7.88 sq. ins. fXh 7500 X 34-5 Trying two 3 in. X 3 in. X iV m - angles and one i5-in. channel at 33 Ibs. per ft., Sq. ins. Sq. ins. One i5-in. channel at 33 Ibs .................. 9 . 90 Less two rivet holes, 2 X .88 X .40 = .......... 70 9 . 20 Two 3 X 3 X rVin- angles, 1.78 X 2 = ........ 3.56 Less four rivet holes ...... .................. i . 08 2 . 48 Total area ............. n.68 This seems to be too large but before discarding it the combined stresses due to the lateral and vertical bending should be deter- mined. The flexural fiber stress with this flange is M 2.040, 11.68 X 1 84 GRAPHICS AND STRUCTURAL DESIGN In the preceding girder section the lateral bending due to stop- ping the crane in 5 ft. was estimated at 170,100 in. Ibs. The fiber stress, assuming that the bending is mainly resisted by the i5-in. channel whose section modulus about its axis i-i is 41.7, is f M*e 170,100 / = = = 4100 Ibs. per sq. in. / 41.7 The combined maximum stress then is /5ioo\ . ( ) + 4100 = 10,900 Ibs. per sq. m. This is low considering that everything of importance is taken into account, but will be used, as calculations made for i2-in. channels indicate a fiber stress undesirably high. This section of the girder may now be checked by calculating its moment of inertia and using the formula / = Flange Riveting. These calculations are similar to those made for the box girder. The rivet value in a yV m - we b plate for f-in. rivets is 3516 Ibs. As in the other case 15,800 51 = -^- ~ = 790 Ibs. 20 15,800 , ,, 52 = -~ = 960 Ibs. 16.5 The resulting shear per inch of span is 5 3 = vV + s 2 2 = V 79 o 2 +. 9 6o 2 = 1240 Ibs. and the rivet spacing of f-in. rivets is ffrl = 2.83 ins. The stiffeners can, as in the preceding problem, be made the smallest allowable, say, 2\ in. X 2| in. X |-in. angles. Bridge Girders with Horizontal Stiffening Girders. The following example is intended to illustrate this type of bridge. Span 60 ft. Girder depth f|- = 5 ft. back to back of angles. The crane capacity is to be 20 tons (40,000 Ibs.). The trolley GIRDERS FOR TRAVELING CRANES 185 wheel base is 6 ft. o ins. The bridge weight, two girders, is 30,000 Ibs. The load on each trolley wheel is 4O.OOO + I2.OOO - = 13,000 Ibs. 4 The live-load bending will be found as in the preceding cases and as follows: Rl . (13,000 X 3I..S) + (13.000 X 25.5) . lbg oo M = R! X 28.5 X 12 = 12,350 X 28.5 X 12, M = 4,223,700 in. Ibs. k^U< 25.5^ 4$- FIG. 212. In estimating the dead-load bending the girder weight will be assumed as a uniform load. The bending moment is ,, WL 15,000 X 60 X 12 M = - = 1,350,000 in. Ibs. o o The total bending is 4,223,700 + 1,350,000 = 5,573,700^1. Ibs. Allowing a mean unit fiber stress of 9000 Ibs. per sq. in. the flange area will be M . n + n) 2 - p w. (6) & ,_,_.. The fiber stresses may be found by substituting these values of k andy in equations (3) and (4). APPROXIMATE FORMULAE The curves on page 197 give the calculated relations between z, k, j, and n p. It will be noted that for the usual values of p,j approximates f and k approximates f . The following approxi- mate formulae, using these values of k andy, are sometimes used. d. (7) *. (8) The curves, Fig. 220, give the relations between the several quantities. These curves are plotted upon a base line giving the values of z; therefore when z = y =r is known, the other values k, j, j k, fc X h> s p n and p are read on the perpendicular to z. Thus f or z = 2.2 the values are read on the dotted line a-a through 2.2. The values are j = 0.895; ^ = -37> ^ "j = - 2 75J P * n = 0.072 and p for n = 15 is p = 0.0048. Had the value of p i or n = 15 been given as 0.0048 instead of the value of z = 2.2 the horizontal line bb would have been drawn through p = 0.0048 until it cut the curve of p for n = 15, and through this point of intersection the line aa would be drawn perpendicular to the z axis and upon this line aa the other values would be read as before. The following example will illustrate the use of the formulae and table. Example. A rectangular beam in which d = 22 ins. is to REINFORCED CONCRETE 197 198 GRAPHICS AND STRUCTURAL DESIGN resist a bending moment of 384,000 in. Ibs. Assume n = 15, /, = 16,000 and/ c = 650; find the width of the beam and the area of the steel. /. X E c 16,000 " f c XE 8 (6 5 oXi 5 ) = Interpolating from the curves, k = 0.379, j = 0.874, k / = 0.331 and p = 0.0077. Substituting these values in equation (4) gives , __ 2 Af c _ 2 X 384,000 _ ~f c Xk-jXd*~ 6 5 X 0.331 X 22 2 ~ 7 ' 37 ^ A = bXdXp = 7.37 X 22 X 0.0077 = 1.25 sq. ins. The solution of this problem by the approximate method gives almost identical results. If- JX^i /.-<*, . 8 M 8 X 384,000 therefore A = : : = - - = 1.25 sq. ins. 7-fs-d 7X16,000X22 M-J..t.> or - , 6 X 384,000 b = -- s2 - - = 7.94 ins. 600 X 22 2 Another example will illustrate the procedure when it is re- quired to check an existing beam. A reinforced-concrete beam has the following dimensions: b = 9 ins., d = 20 ins., A =1.50 sq. ins., n = 15, and it is desired that /. should not exceed 16,000 Ibs. per sq. in. nor/ c exceed 650 Ibs. per sq. in. p = A, = ^5^ = O . oo825 . b-d (9 X 20) From equation (5), L = A , = _ Q-390 _ 6 f e 2p (2 X 0.00825) and /, = 23.6 X 650 = 15,300 Ibs. per sq. in. This fiber stress being under the 16,000 Ibs. per sq. in. is satis- REINFORCED CONCRETE 199 factory. The bending moment corresponding to these fiber stresses can be found from either equation (3) or (4). M 8 = A Xf, Xj d = 1.5 X i5,3X 0.870 X 20 = 399,000 in. Ibs. Checking this by the other formula gives M c = (f c Xk.jXb- d 2 ) = 396,000 in. Ibs. This is about as close agreement as can be expected considering the curve readings, the difference being under i per cent. T BEAMS In reinforced-concrete design the section most frequently met is the T beam, shown in Fig. 221. The floor is usually a slab of / r r- JJ i r ::: \ \ } \ - 1 1 4Y I c 1 i i i i r T-r 1 1 1 1 1 1 1 H N k 1 j ' j L_i _ L i " r m 4 M Colum . " 1 r r rr 1 _j 1 f ^ M \ SG L ^ r r L , TT / [T-Beam ( y 1 L l~ y L r L r~ y *\ -T-Girc lc r FIG. 221. reinforced concrete laid on floor beams, corresponding in the ordinary timber construction to joists. In the case of concrete construction the floor beams and the slab over it become one piece, the horizontal shear at the junction of the two being re- sisted by the concrete and usually by reinforcing stirrups or rods. 2OO GRAPHICS AND STRUCTURAL DESIGN When the neutral axis i-i, in Fig. 222, falls in the flange or slab the formulae and table previously derived for rectangular beams i FIG. 222. are sufficient. When, however, the neutral axis falls in the stem, as is usually the case, another set of formulae are required. GENEEAL FORMULAE FOR T BEAMS REINFORCED FOR TENSION ONLY Neglecting the compression in the stem, and on the same conditions a and b that applied to rectangular beams we have as before: Jc X> -E-'a J c Also from Fig. 223 j = /i U B H 1 i k k-d > & IIU K - i -j k-d- t t A. J "T X -El -\ P I 1 -i-^3 fi J ; FIG. 223. Assuming that the fiber stress in the concrete varies as its dis- tance from the neutral axis, the distance x from the top of the beam to the resultant compression R is _/e+2/i t 3-k-d- 2-t t m * ~ f s\ j j XN j f e +fi 3 2-k-d-t 3 REINFORCED CONCRETE 2OI ~ this gives 3 * d = d ~ x = d ~ \ 2 .k .d~-t The average fiber stress in the concrete is It follows that M c = BXtXf c (i-^j.d, (10) and M t = Axf,Xj-d. (n) Equation (n) can be approximated as When the ratio of the reinforcement is known, not the fiber stresses, equation (i) cannot be used to find k as z is not known. Taking equations (10) and (n), we find, fc = _A_ i_ pd k I ~7kd \~2~&/r t 2 It follows that k = 2_- t + npd This may also be written 2 n dA + Bt 2 2nA + 2Bt To facilitate the solution of problems the following curves, Fig. 224, give the values of j and k for varying values of p and - and also the ratios of * for various values of k. These values /c have been calculated with n 15, this being a very commonly assumed value for rock concrete. For other values of n similar curves may be drawn or the formulae used. Single Reinforcement M S =A x f 8 x j.d P 202 FIG. 224. REINFORCED CONCRETE 203 To findy if - = 0.25; from 0.25, at a on the upper scale, drop d a perpendicular until it cuts the curve corresponding to the steel ratio of, say, 0.003 m ^> project the point b horizontally until it cuts the vertical scale in c. Here the value of j is given as 0.914. To find k continue the vertical a-b until it cuts the curve of k corresponding to a steel ratio of 0.003 m d. This point d pro- jected horizontally cuts the vertical scale in e or k = 0.262. To find the ratio of j produce the value of k horizontally until Jc it cuts the curve of j in /. The vertical projection of / cuts the Jc scale of ratios of '-j- in e or j = 42 .3 . Jc Jc An example will further illustrate the uses of the formulae and curves. [ / T T~ A i t f k t FIG. 225. Given n = 15, f e = 600, /. = 15,000, and the following section, Fig. 225. Area of bars = 4 X 0.766 = 3.06 sq. ins. P = -7^~: = 0.0027. 60 X 19 Consulting the curves, j = 0.918, k = 0.252, = 44.5. Jc The resisting moment then is M a = A Xf a Xj-d or M a = 3.06 X 15,000 X 0.918 X 19 = 801,000 in. Ibs. 204 GRAPHICS AND STRUCTURAL DESIGN The fiber stress in the concrete is fa I5,OOO s 7 = 44-5 = -^7 or f e = 330 Ibs. per sq. in. Jc Jc SLABS In using the formulae and table for slabs it is customary to assume a width of slab of 12 ins.; thus assuming a slab whose span is 6 ft. and whose total dead and live load is 200 Ibs. per sq. ft., taking the bending moment at Wl M= , w = 15,^. = 600 Ibs. per sq. in., p 8 12,000 Ibs. per sq. in., the thickness of the slab and area of reinforcing steel can be found by the following calculations: ,, Wl (200 X 6) X 6 X 12 Q , . M = = = 8640 in. Ibs. 10 12,000 = I -33- 15 X 600 From the table the values corresponding to z = 1.33 are k = -433>y = -855> k -j = 0.370 and p = o.on; substituting these values in equation (4) gives , J 2 M c J 2 X 8640 d = V x 600x0.37 ' ' The area of the reinforcement then is given by equation (3) or by taking b -d -p = 12 X 2.55 X o.on = 0.337 sc l- m - If f in. is allowed below the center of the reinforcing steel the slab thickness will be 2. 55 +0.753. 25 ins. BENDING MOMENTS Owing to the monolithic construction of reinforced concrete the bending moments on beams are frequently estimated as lower than those occurring on similarly loaded but merely supported beams. Thus for continuous slabs the moments are commonly Wl taken as M = , at the middle of the span. 10 REINFORCED CONCRETE 205 Square slabs reinforced in both directions and supported on four sides have the moments at the center of the span in one Wl direction taken as M = . Beams and girders although some- 20 Wl times figured as fixed or continuous beams with M = are more 10 Wl commonly taken as supported beams with M = . In the 8 above, W = the total load on the span in pounds and L = the span in inches. BEAMS WITH DOUBLE REINFORCEMENT In the preceding discussions reinforcing steel has been assumed in the tension side of the beam. The case will now be considered - -. -.^-^rzzzr: EL" ~zi~ i^i.~ ~ ZZ^TT = FIG. 226. where the reinforcement is in both flanges. This condition occurs regularly in ordinary construction where some of the reinforcing bars in the tension flange are bent up and carried over the support in the upper flange. In this case the beam resists a negative moment over the supports which creates tension in the upper flange and compression in the lower flange. In the case of T beams the lower flange width being b the width of the stem is much less than the width B of the upper flange, and unless this narrow flange is reinforced for compression the fiber stress p c might easily be excessive. The following theory applies to beams with double reinforcement. The nomenclature is that previously given. Assuming the compression in the concrete as varying propor- 206 GRAPHICS AND STRUCTURAL DESIGN tionally to the distance from the neutral axis and that the con- crete carries no tension, we have, in Fig. 226, nf c kd nf c kd Also Axf. = c.'Xc = A'f,' + $f e -kd.b, from which it follows that n(A' + A) and k = 2 (# + 0/>')+*(/> + />') 2 - n (p + /). To find the lever arm of the couple acting in the beam A Xf. Xjd = f c .kd-b(d--\ + A'f.' (d - e), f c .kd.b(d--) -v V 3/ , A'f.'(d-e). 2Af, A*f. Expressing f cj f sj f, and d in terms of k, , p and p' gives J " np (i - *) The relations between the fiber stresses are ,-.__. f ~AXjXd' ~ and In the Engineering News of August 22, 1912, Mr. Arthur G. Hayden gives the following approximate formulae for j of rein- forced-concrete beams with double reinforcements. j = 0.93 + 2 P' + ioopp f 4p o.i0 2op'. Mr. Hayden states that the maximum error does not exceed if per cent for all proportions of reinforcement between J per REINFORCED CONCRETE 207 cent and 2 per cent, and that for ordinary cases the error is in- appreciable. This formula assumes that n = 15. The following example will illustrate the use of the formulae. A T beam in which d = 21 ins. and b = 12 ins. is reinforced over the supports by two f -in. < bars in each flange. If f a = 16,000 Ibs. per sq. in., = ^ and n = 15, what moment can the beam resist at the supports and what are the fiber stresses PC and #/? 2n(A + *A')d \n(A+A'W n(A+A') = ~ f ~ _ L 12 J 12 o.28 2 fo.5o - 2iL-J -f 15 x 0.0048 (i.o - o.i) (0.28 - o.i) 15 X 0.0048 (i - 0.28) j = 0.91 and jd = 21 X 0.91 = 19.11 ins. By the approximate method jd = 18.97 ms - The bending moment, allowing /. = 16,000 Ibs., is M = A X/ 3 Xjd = 1.2 X 16,000 X 19.11 = 366,912 in. Ibs. f < = n(d M -kd) f ' = I S (21 '-5.85) X I , f k j. 0.28 o.io /, = __ , /. = - - - X 16,000 = 4000 Ibs. per sq. in. The following problems are intended to be solved by means of the theory just given and it is recommended that they be first 208 GRAPHICS AND STRUCTURAL DESIGN done by using the formulae and then checked by the assistance of the curves. Problem. The combined live and dead loads upon a floor slab are 140 Ibs. per sq. ft. ; assuming f c = 650 lbs.,/ 8 = 16,000 Ibs. per sq. in., -f = 15, and that M = , determine the thickness JtL c 10 of the slab from the top to the center of the reinforcement and the spacing of f -in. round bars if the span of the slab is 10 ft. 8 ins. Problem. Assume a continuous girder whose span is 25 ft. o ins., carries 50,000 Ibs., d = 23 ins., B ~ 100 ins., slab thickness 4^ ins., --r = 15, and M = '. Find the fiber stress in the con- h c 10 crete and steel if the steel reinforcement consists of two if -in. and two if-in. round bars. PARABOLIC VARIATION OF STRESS IN CONCRETE In the preceding discussion of reinforced-concrete beams it has been assumed that the fiber stress in the concrete above the FIG. 227. FIG. 228. neutral axis varies as the intercepts in the triangle; this is repre- sented by Fig. 227. Some designers prefer to assume that this stress varies as the intercepts in the parabola as shown in Fig. 228. The resultant flange force acts at a distance f k d from the top of the beam and the average fiber stress is f f c . This treatment is commonly limited to rectangular beams as it becomes too complicated when extended to T beams. The following formulae are derived REINFORCED CONCRETE 209 similarly to those previously determined and apply to rectan- gular beams only. The symbols are those previously used. j T / and *-'- M. = A Xf.Xjd, M c = lf c XbX kj X d\ Where the ratio of reinforcement is known the formulae are: k = V%p.n + (%p-n) 2 -lp-n and j = i - f k, M. J~ . t .' The difference between these formulae and those previously given will be evident upon working several of the problems by both methods. Considering the character of the materials used in reinforced concrete it is a question whether methods intro- ducing any greater refinements than those used in the derivation of the first formulae are warranted. Most elaborate tests of reinforced-concrete beams have been made at the Experiment Station of the University of Illinois under the direction of Prof. Arthur N. Talbot, beginning in 1904 and reported in the Bulletins of that station since then. Professor Talbot has deduced formulae and offers suggestions for reinforced- concrete design as the results of the tests ; this data can be readily obtained by reference to the above bulletins. HORIZONTAL SHEAR It is shown in the section on problems (see problems 65 and 68) that ordinarily in steel beams of usual sections horizontal shear is of minor importance, while more careful consideration must be taken of it in timber beams, particularly for those of short span. In reinforced concrete it is of still greater impor- tance and generally the webs must be carefully reinforced with 2IO GRAPHICS AND STRUCTURAL DESIGN steel to resist this shear. Taking section 2-2 in Fig. 229 the horizontal shearing force will increase from zero at the upper fibers to a maximum at the neutral axis i-i ; this force F 2 is then transferred to the horizontal reinforcing steel. Assuming that the bending at 3-3 exceeds that at 2-2 the horizontal shearing force FIG. 229. F 3 at 3-3 will exceed F 2 , and if b is the width of the beam at the neutral axis the unit shearing force on the section b X i is p p v = -r 2 . Equating the moments V X i = (F 9 - F 2 ) j d b X i or F 3 - F and v V b-j-d It has been previously shown that j d ~ f d, hence V 0.85 b -d It should be noted that the reasoning is identical with that used in determining the flange riveting in plate girders. The design is sometimes made by assuming that the horizontal shear is carried partly by the concrete and any deficiency in the concrete is supplied by the addition of steel stirrups. These are small bars or strips commonly bent into U or V shapes and placed vertically in the web. They are usually secured in the flange and are run under the horizontal reinforcing bars to hold them in place. Let v c be the allowable unit working shear in the concrete; then the deficiency per square inch is v v c . If R = the value in pounds of the stirrup in shear, S = stirrup spacing in inches at the section where the vertical shear is V. REINFORCED CONCRETE 211 If the stirrup carries the entire shear, then c 0.85 XdxR V If the stirrups take only that portion of the shear that the con- crete is unable to carry, then = 0.85 XdXR ~ 7-0.85 Xv c -b-d' It is noted from both these formulae that the spacing S will be closer the greater the value of V '. It therefore follows, as in the case of flange riveting of plate girders, page 149, that the stirrups will be spaced closer at the supports. To be effective the stirrup spacing should not exceed Graphical Determination of Stirrup Spacing. Beams or girders of a constant depth d carrying a uniform load may be /; i I ; r i ! j \ Cjn o p q A jD FlG. 231. considered as follows. In Fig. 230 lay off to some scale the horizontal line CD representing the span. At its center A , erect a perpendicular AB representing to scale the flange force acting in the reinforcing steel or in the concrete flange at the center of the span. This flange force will equal the area of the steel multi- plied by the unit fiber stress induced in it by the loading. This fiber stress can be found by the formulae previously given for beams. If in a beam A =3.06 sq. ins., and f a = 15,000 Ibs. per sq. in., the flange force then is F = 3.06 X 15,000 = 45,900 Ibs. AB equals 45,900 Ibs. and the change of flange force across 212 GRAPHICS AND STRUCTURAL DESIGN the girder may be represented by the parabola CBD. In- the case of a supported span the flange force will be zero at the sup- ports, while in a restrained beam the flange force will go through zero at some points E and F shown on the dotted line C 1 D 1 , which has been raised from the position C D by the restraint on the beam. The flange force at the supports is now C C 1 instead of zero and is of the opposite character to the force B-G at the center. The range of flange force is still A-B being CC 1 + BG. Now if AB (Fig. 231) is divided into one-half the number of parts that it is thought advisable to use stirrups in the span, AB is here divided into five equal parts, and lines parallel to CD are drawn through these points, i, j, k and /, the spaces Cn, no, op, pq and qA are the sections in each of which a stirrup must be placed. The value of each stirrup in shear must equal Ai. c \nop q FIG. 232. FIG. 233. If it is desired to allow the concrete to carry a part of the hori- zontal shear it may be allowed for by laying off A E in Fig. 232 to the same scale as AB and equal to the product of the allow- able unit shear on the concrete, the width of the beam (width of stem if it is a T beam) and one-half of the span, the first in pounds per square inch, the last two in inches. Draw the line CE and as BE represents the total shear that is to be cared for by the stirrups in the half span, divide BE into the number of equal parts that there are desired stirrups in the half span. The spaces within which the stirrups must be placed are m, , o, p and q, being located by the intersections of the parabola with the lines through i,j, k and / and drawn parallel to CE. In the best practice the stirrups are spaced a distance not exceeding-- 2 REINFORCED CONCRETE 213 Tests of reinforced-concrete beams exhibit failures along lines a-a as shown in Fig. 233. This illustrates failure by diagonal tension. If we consider any point in the body of the beam, it will be acted on by a unit horizontal tension or compression, a unit horizontal shear and a unit vertical shear. The resulting maximum tensile or compressive stress is A. = */, + ^i/, 2 + f 2 , where f tm = unit maximum diagonal tension, ft = unit horizontal tension, v = unit shear, vertical or horizontal. The direction of the maximum tension is given by 2 V tang 26 = > h where f t = o, = 45 degrees. It therefore follows that along the neutral axis and on the tension side of this axis that 6 = 45 degrees. . Where = 45 degrees, f t = o and f tm = v. It is seen then that the maximum diagonal tension acts at an angle of 45 degrees with the neutral axis and that the maximum unit diagonal tension equals the unit shear at that point. '--'-iTS' v = unit vertical shear. b = width of beam if rectangular. Use b\, the width of the stem if a T beam. V = total vertical shear on section. This formula applies for the worst condition, namely, that no tension is carried by the concrete. When it is desired to allow for the concrete the numerator of the fraction becomes V v c b d for rectangular beams and V v c b\ d f or T beams. Here v c = the allowable unit shear on the concrete. Let T = the total allowable tension in a diagonal bar, in pounds. If the bar makes an angle of 45 degrees with the hori- zontal, the spacing s along the horizontal in the first case is s = 214 GRAPHICS AND STRUCTURAL DESIGN rr\ 7 *- while, when the allowance is made for the portion of the tension carried by the concrete, = 1.41 XT Xd : V-v e -b-d' The factor 1.41 is introduced since the diagonal tension is measured along a 45 -degree line, while the spacing is laid off on a horizontal line. To be effective for web reinforcement the spacing of diagonal bars should not exceed s = d. BOND STRESS The diagram, Fig. 234, shows how the flange force varies from the supports to the middle of the span for a uniformly dis- tributed load upon a beam of constant depth. Evidently this change of force can only occur by the flange transferring force to the web, as was done in the case of the plate girder through the flange riveting. AB repre- sents the maximum flange force. Passing from AB to EF the flange force has been reduced by the amount GF which has been trans- ferred to the web. In the case Q FlG> 234< of reinforcing steel the connection between the steel and the concrete must be strong enough to carry the force GF. This may be accomplished either by the adhesion between the con- crete and smooth or plain bars, or corrugated or twisted bars may be used to increase the strength of the bonding. Considering plain round bars and letting u = unit bonding stress, pounds per square inch, d = diameter of the bar, inches / = length of bar, inches, then, u X - = X//, 24 from which- 5 2 u REINFORCED CONCRETE 215 This means that to develop the full tensile strength of a round bar its total length must be -"- times the diameter of the bar. 2 - u This relation applies also to square bars. Illustrating this with the |-in. square bars used in a former example, and allowing 60 Ibs. per sq. in. for the working bond stress, we have / /, ; 0.875 X 16,000 - = -^-or / = - - = 116.5 ins - d 2 u 2 X 60 This means that under the given conditions to develop the full strength of the bar, it must have a total length of at least 116.5 ins. and must be placed to have a length of not less than 58^ ins. on each side of the center of the span. The bond stress varies along the bars; in fact, the increment of change is that of the flange force, and, similarly to the change in horizontal shear, we have V X I = (F 3 - Fjj.d, :. F,-F 2 = 2 J -d The unit bonding stress = F,-F 2 u = surface of bars for unit length V j d (2 circumference of bars) LENGTHS OF REINFORCING RODS The lengths of the reinforcing bars besides being limited by the bond stress, as just explained, will also vary in length for the same reason the flange plates of girders do. That is, as the flange force decreases from the center towards the supports the reinforcing area can be reduced. In Fig. 235, CD is the span and AB represents the total reinforcing area, A, at the center of the span. If 4 bars are used, the line AB is divided into 4 equal parts, each division representing the area of one bar; the lengths of the several rectangles, determined by the intersections of the horizontal lines with the parabola CBD, give the respec- tive lengths of the bars. In practice the bars instead of being 2l6 GRAPHICS AND STRUCTURAL DESIGN cut off are frequently bent up, run to the top and then to the end of the beam, as in Fig. 236, thus reinforcing the upper flange of the beam over the support, where, owing to the monolithic character of a concrete beam, there is restraint and consequent tension. The bars are usually bent up in pairs. The area run over the supports generally exceeds 25 per cent of the gross area A\ the remainder of the bars are continued through the J |p ----- Span ----- FIG. 235. FIG. 236. lower flange and over the supports. If it is preferred to calculate the length of the bars the following formula may be used, where / = length of bar in lower flange between bends, inches. a = area in square inches of bars, including all shorter bars and the bar whose length is desired. In Fig. 235, if the length of bar No. 3 is desired, a = area of bars Nos. i, 2 and 3. A = area in square inches of the total reinforcing at the center of the span. L = span of beam, inches. DESIGN OF A T BEAM A T beam spans 20 ft. o ins. ; the beams are spaced 10 ft. 6 ins. and carry a uniform dead load of 30,000 Ibs., while the uniform live load is 75,000 Ibs. The slab thickness is 7.5 ins., d 29 ins., and the width of the stem 61 is 20 ins. Assume the width of slab acting as flange at REINFORCED CONCRETE 217 80 ins. The reinforcing bars are four if-in. round bars and five i^-in. round bars. The total area of the steel is Four ig-in. round bars= 4 X 0.99 = 3.96 sq. ins. Five ij-in. round bars = 5 X 1.23 = 6.15 sq. ins. Total 10. ii sq. ins. These bars to be spaced not less than i\ times their diameters. From the dimensions given we have d 29 IO.II = ao 44 - From the curves we find that k = 0.305 and that j = 0.90, also y = 34. Wl The bending moment on the beam being assumed as M 8 ,, 105,000 X 2O X 12 -11 M = **- - = 3, 150,000 in. Ibs. o K/T A f - i f >, M = A.f. -j | R s. FIG. 238. in the lower flange exceed the distances previously calculated as necessary, one-half these lengths being used as the lengths are DIAGRAM OF MAXIMUM SHEARS. FIG. 239. measured from the center line c-c. The area of the trapezoid abdc multiplied by b\ gives the load carried by the rods 3 and 4. 220 GRAPHICS AND STRUCTURAL DESIGN The mean ordinate of the trapezoid is cd -f ab _ 42 -t- 60 _ -si- ac = 19 ins. The area abdc 1351X19 = 970 Ibs. The total load on the bars 3 and 4 = 970 X &i = 970 X 20 = 19,400 Ibs. Hence the unit tension in the bars is 19400 = 9800 Ibs. per sq. in. 2 X 0.99 The tension in the bars i and 2 will be considerably less, but this is made necessary, as the spacing must be less than d, or 29 ins., and the length of these bars in the lower flange measured from the center line of the span must exceed 53 ins. The bent rods are ordinarily assumed as acting approximately through the center of gravity of the trapezoid representing the load they carry; thus bars 3 and 4 should pass approximately the center of gravity of the trapezoid cbda. Although the discussion errs on the side of safety it should be remembered that, the diagram being one of maximum shears, the shears given are not in existence at the same time. There is still a diagonal tension efg not cared for; this will be provided for with vertical stirrups and the whole beam will be strengthened by their insertion across the span. The vertical stirrups are also useful in holding the horizontal bars in place. Had the horizontal shear been assumed as being resisted by the vertical stirrups and these stirrups taken as ^-in. round bars bent as shown in Fig. 240, making four sections of the rod that would FIG. 240. na ve to be sheared, and allowing 10,000 Ibs. per sq. in. as the permissible shear fiber stress the stirrup value would be R = 4 X 0.196 X 10,000 = 7840 Ibs. The vertical shears are given by the diagram of maximum shear (Fig. 238), and have the following values at the several distances from the center of the span. REINFORCED CONCRETE 221 Distance from center Max. shear, of span, feet. pounds. 8 43,875 6 35,250 4-. 2 7,ooo 2 iS.OOO The minimum spacing will be at the supports and will be = - 8 5 dxR = 0.85X29X7840 . ns ~ 7-0.85 xv c xbiXd ~~ 52, 500 -(0.85x50x20x29) 711 The spacing 8 ft. from the center of the span is 0.85 X 29 X 7840 s = - - ~ 10 ins. 43,875 - (- 8 5 X 50 X 20 X 29) The spacing 6 ft. from the center of the span is s = 0.85 X 29 X 7840 _ ^ 35,250 - (0.85 X 50 X 20 X 29) The spacing will be made, starting at the piers and running towards the center, 2 spaces at 8 ins., 2 spaces at 10 ins., and from there to the center the spacing will be 1 2 ins. This makes all the spacing less than -, or = 14.5 ins., which would be the 2 2 maximum for the best practice. The vertical stirrups might have been made much lighter considering what the bent rods do to strengthen the web, but in this case the vertical stirrups have been figured as resisting the entire horizontal shear, the bent rods will strengthen the web and will strengthen the beam to resist bending by resisting a negative moment over the supports. COLUMNS Concrete columns are reinforced in one or both of two ways: (i) by steel rods paralleling the axis of the column and (2) by spirally wound metal bands. In the first case the metal shares the load with the concrete, while in the second case the bands strengthen the concrete by preventing lateral expansion which is assumed as occurring when the column is shortened along its axis by the load. As it is customary to tie the longitudinal 222 GRAPHICS AND STRUCTURAL DESIGN reinforcements in place with steel wires or bands it virtually amounts to combining both forms of reinforcements. The following is the usual discussion of the strength of columns having longitudinal reinforcements. The nomenclature is P = total load on column, pounds. A c area of concrete inside reinforcements, not including steel, square inches. A t = area of steel, square inches. A = A a + A c . A p = ratio, -^- If the column with longitudinal reinforcement is loaded, its length will be altered, the steel and concrete being shortened the same amount. The unit reduction is A=A = . E. E.' TJ* from which fa = f c X-=r = n X f c . &c The total load carried by the column then is P = A a ./. + A c .f e = A a . nf e + A c *f c , and P = (A. - n + A e )f e -, also P = A -f e [i + p (n - i)]. The working dimensions of the columns are generally assumed as those inclosed by the longitudinal or band reinforcements. The proportions of concrete columns ordinarily work out so large that their ratio of length to smaller dimensions will usually be less than 15, thus making them short columns. The design may be safely made upon the basis just given, even where length divided by the smaller side reaches 25. The usual coating of concrete is commonly put on the column outside the reinforcing steel to protect it and act as fireproofing. A few of the numerous forms of metal reinforcements for con- crete are shown in Figs. 241 to 247 inclusive. Fig. 241 shows a REINFORCED CONCRETE 223 FIG. 241. FIG. 242. FIG. 243- FIG. 244. FIG. 245. FIG. 246. CROSS SECTION OF BAR FIG. 247. 224 GRAPHICS AND STRUCTURAL DESIGN small section of expanded metal; this is made in varying sizes; the smaller, called metal lath, is intended to replace wooden laths. For concrete reinforcement the 3 -in. mesh expanded metal is better. It is made in weights having a cross-sectional area of from 0.06 to 0.60 sq. in. per ft. of width. The triangular-mesh steel-wire reinforcement illustrated in Fig. 242 serves about the same purposes as the expanded metal. These reinforcements have the advantage of reinforcing in both directions, that is, across the span and at right angles to it. Figure 243 is the Ransome twisted bar, Fig. 244 is the Thatcher bar and Figs. 245 and 246 are corrugated bars. These bars are designed to insure proper bonding between the steel and the concrete. Figure 247 is the Kahn trussed bar; flanges along the side" of the bar are bent up to resist the horizontal shear in the web of the beam. The Pittsburgh Steel Products Company make a reinforcing frame consisting of steel bars through the lower part of the beam. To these bars on each side of the middle of the span are welded shear bars bent up and toward the adjacent support at an angle of 45 degrees. To the upper end of these shear bars are welded horizontal bars which run back over the support. These frames are made in standard sizes and the reinforcement in the upper flange over the supports is said in all cases to be at least 25 per cent of the metal in the lower flange. These frames are built up for certain spans and are placed completed in the forms about which the concrete is then poured. CHAPTER XV FOUNDATIONS THE designer recognizes two kinds of foundations, those for structures and others for machines. The function of the former is to distribute the pressure properly upon the soil and where settlement is inevitable to design the foundations that, as far as possible, such settlement shall be uniform throughout the structure. It may also maintain the structure upright against wind or other forces tending to upset it, as in the case of a chim- ney. In machine foundations besides these functions the mass of the foundation may play an important part in absorbing shock or may even prevent the motion of a machine when acting on other machines upon foundations external to its foundation. Properly balanced rotary converters and turbo-generators re- quire only rigid support, while belt-driven machines, engines driving rolling mills, mill housings, etc., must be anchored. With modern concrete floors, foundations for most machine tools may be dispensed with, the machines being placed where desired and rag bolts used to hold the machine to the floor. The mass of a foundation to limit the vibration of an engine or other rapidly moving and imperfectly balanced machine within definite limits can only be determined when complete informa- tion of the design of the engine or machine is available. The determination of this mass therefore is properly the work of the engine or machine designer and not within the scope of this book. Numerous rules of thumb have been used in determining the minimum weights of foundations. One of these is to make the foundation weigh 1.5 times as much as the weight of the engine 225 226 GRAPHICS AND STRUCTURAL DESIGN or machine. Another rule credited to E. W. Roberts is that F = 0.21 EVN, where F = weight of the foundation, in pounds. E = weight of the engine, in pounds. N = R.P.M. The wide variations in these two rules show how unsatisfactory such formulae are. Vertical engines require heavier foundations than horizontal engines of the same capacity and speed. The foundations of machines subjected to much shock should be kept free from other foundations and the building. Although considering only pressure and settlement the best foundation bed is rock, yet vibrating machines placed directly upon the rock may result in the transfer of the vibrations through the rock for considerable distance. The usual precaution under such circumstances is to place 2 or 3 feet of sand held in a pocket in the rock; this may be either a natural pocket or it may be made of concrete. The foundation is then built upon this cushion of sand. In setting an engine or important machine the foundation is brought to within f in. to ij ins. of its final top. The machine is then lined up by driving metal wedges between the foundation and machine bed. When the machine is properly in line a dam is built around the foundation top and a thin grout poured in, filling up the space between the bed plate and the foundation. After the grout hardens it firmly secures the bed and the founda- tion regardless of the finish of the engine bed or foundation top. When the grout is properly set the wedges may be removed and the spaces left by them filled up with cement mortar. The foundation bolts are preferably located and held in place during the construction of the foundation by templates of the bed plate. The bolts may either be set right in the concrete or may be placed in pipes whose inside diameters somewhat exceed the bolt diameters and the pipes are then set in the concrete. The bolts are carried by the templates; the space between the top of the FOUNDATIONS 227 pipe and the bolt is filled with burlap, waste or paper, to prevent the concrete filling the clearance space between the pipe and the bolt. This clearance permits some adjustment of the bolts should the bed plate disagree with the template. The foundation bolts should be run well down into the foundation. Frequently the nuts or keys are placed in pockets made accessible from the out- side of the foundation. This permits of the ready replacement of a bolt should it be found to be necessary. In some cases what is known as a rust joint is made between the foundation and the bed plate. This is made by rusting cast- iron chips together with sal-ammoniac. Melted sulphur was D BOLT FIG. 248. KEY FIG. 249. WASHER FIG. 250. also much used before cement grout became so common. The foundation bolts generally have the end above the foundation threaded for a nut; the other end may either have a thread and nut or be slotted for a key. The washers in the foundation are of a great variety including scrap channels, angles or rails. Cast-iron washers may also be used but of whatever kind they should be large to distribute the pressure well over the concrete. Fig. 248 illustrates a slotted bolt, while Fig. 249 shows the usual key and Fig. 250 the cast-iron washer. In other cases the ad- justment of the bolt is provided for only at the top of the bolt for a distance of a couple of feet. This is done by using a short piece of pipe similar to the way mentioned before or by setting a wooden casing around the bolt which leaves a pocket in the finished foundation around the top of the bolt. The foundation 228 GRAPHICS AND STRUCTURAL DESIGN should be carried to proper soil and always below the frost line. The load upon the soil should be calculated to see that it does not exceed that permissible. The allowable pressure upon the soil is commonly given by the building codes where the founda- tion is within a city's limits. It may frequently be had by obtaining data about neighboring foundations. The following table from " Baker's Masonry " is also available as a general guide. Soil. Bearing power, tons per sq. ft. . Soil. Bearing power, tons per sq. ft. Rock, thick layers 200 or over Gravel and coarse sand . . 8-10 Rock, equal to best brick. Clay thick beds, dry 15-20 4-6 Sand, compact Sand, clean 4-6 24 Clay thick beds, moder- Alluvial soils P5 ately dry 2-4 Clay, soft 12 Where the engine or machine is an important one and the soil of doubtful value, piles are frequently resorted to. The subject of piles is considered under building foundations. H6le FIG. 251. Figure 251 illustrates a foundation for a machine. The bolts have been set in pipes and run through to pockets accessible from the outside so that the bolts could be replaced if necessary. This foundation shows the base extended to reduce the pressure upon the soil and a semicircular hole has been cut through the foundation to reduce its weight. FOUNDATIONS 229 In another class of foundations for machines the foundation requires weight to prevent the overturning of the machine. Fig. 252 illustrates such a foundation for a crane. The load of 33,000 Ibs. is to be carried at a radius of 33 ft. The frame and 425000* Force Polygon Distribution of Pressure on Soil FlG. 252. machinery is assumed as weighing 11,000 Ibs. and as acting at a distance of 6| ft. from the post center. The foundation is circular and has been assumed as weighing 425,000 Ibs. The total weight on the foundation then will be the sum of these three amounts, 33,000 + n,ooo + 425,000 = 469,000 Ibs. 230 GRAPHICS AND STRUCTURAL DESIGN The distance of the line of action of the resultant of these forces can be found by taking moments about the vertical axis of the post; hence _ (33 X 33,000) + (6.5 X 1 1, OOP) . x 2.40 it. 469,000 The maximum pressure should now be determined as is done for a chimney (see page 245). The foundation has been assumed as circular. The kern radius of a circle is Here then r = - 8 8 = 2.0 ft. The resultant, therefore, falls outside the kern. Re- ferring to Fig. 270, we first find then a = 0.355 an d W = <*p e D 2 , W 460.000 or p e = = = 5160 Ibs. aD 2 0.355 X i6 2 Whether or not this maximum pressure is permissible will de- pend upon the character of the soil or nature of the foundation under the section considered. The curves give < as 41 per cent and the neutral axis therefore falls 16 X 0.41 = 6.55 ft. to the left of the vertical axis, or 1.45 ft. to the right of the left edge of the foundation when the full load is in the position shown in the figure. The distribution of this pressure is indicated below the assumed foundation bottom. BUILDING FOUNDATIONS Since settlement of foundations is bound to occur, unless foundations rest on bed rock, the foundations of a structure should be designed as far as possible to lead to uniform settle- ment. The effect of unequal settlement is frequently evident where chimney foundations have been carried into wall footings. FOUNDATIONS 231 Here the settlement of the chimney may cause considerable cracking of the wall. In designing foundations it is considered better to design the areas proportionally to the dead loads, which act continuously, and then see that they are sufficiently large to carry the combined dead and live loads without bringing excessive pressure upon the soil. The dead loads acting continuously will exert a greater effect upon the settlement of the foundation than live loads which may act but infrequently and then possibly be only a small part of the assumed live loads. When the footings are proportioned for both dead and live loads only such portion of the live loads should be considered as may be assumed as acting continuously. It is also of prime importance that as far as possible the center of pressure on the soil shall coincide with the center of gravity of the footing. The building codes of the various cities give the requirements of foot- ings and these are commonly required to be a certain enlarge- ment of the wall upon them. Where these codes are not available the footings can be designed using the tables of allowable pres- sures previously given. The areas should be enlarged under pilasters carrying heavy concentrations. Foundations for building columns, in addition to resisting vertical loads, may carry the horizontal reactions of wind loads, as in the case of the bent for the steel-mill building designed on page 118. Here there will be two cases, one (Fig. 253) where the column is assumed as hinged and then the horizontal reac- tion acts at the column base; the other (Fig. 254) where the column is considered as fixed and the point of application of the horizontal wind reaction is taken midway between the foot of the knee brace and the column base. The base is commonly a rectangle. Were there no horizontal force the pressure upon the foundation would be uniformly dis- tributed. The wind pressure tends to increase the pressure upon the leeward side of the foundation and decrease it upon the windward side. 232 GRAPHICS AND STRUCTURAL DESIGN As long as the resultant pressure R cuts the bottom of the n foundation at a distance not exceeding from the center of the base there will be compression over the entire base. R is the resultant of W and H, where W = load carried by the column or wall plus the weight of the foundation and H = horizontal wind reaction on the column. Since neither masonry nor the surfaces of the foundation and soil in contact can be in tension, FIG. 254. having R pass beyond the middle third will reduce the area of the foundation under pressure and will consequently increase the pressure on the leeward side, see Fig. 254. When R falls in the middle third the extreme pressures on the edges of the foundation are given by here b = width of the base at right angles to B. FOUNDATIONS 233 When R falls beyond the middle third p e The distance x may be found graphically, as shown, or by moments, since W -H h = H ' h W In the second case (Fig. 254) the attachment of the columns to the bases must be sufficiently strong to develop the bending moment H X - at this point. 2 The more general discussion of the kern of a section and the distribution of pressure upon the soil is given under chimney foundations, page 253. Where the base of a foundation ex- tends beyond the main shaft, Fig. 255, the portion extended should be calculated as a can- tilevered beam subjected to a uniform load p e . This is com- monly stated as tons or pounds per square foot. This exten- FIG. 255. i sion e is given by e = 4 d All dimensions in inches. p = 30 to 50 Ibs. per sq. in. in concrete. p e = maximum load on soil in pounds per square foot. Figure 256 gives a foundation standard adopted by one company. Capstone i Portland cement, 2 sand, 4 blast fur- nace slag. Foundation i Portland cement, 2 sand, 6 slag, minimum bolt diameter i J in. pressure on metal at base of column not over 10 tons per sq. ft. Pressure on capstone not over 5 tons per sq. ft. Bearing pressure per square foot on gravel 2.5 tons. Bearing pressure per square foot on other soil 2 tons. 234 GRAPHICS AND STRUCTURAL DESIGN Minimum 12 to suit excava "A" so : -t B" determined "by 30 FIG. 256. FIG. 257. Where a shallow and light foundation is sufficient a grillage or reinforced-concrete footing can be used. The beams or rails used in Fig. 257 may be cal- ~*~ a ^ : ~ ^"~8 culated for the load given by a diagram similar to that of Fig. 255. The extended part is as- sumed as carrying a uniform load equal to the length e, in feet, multiplied by the maxi- mum load p e , in pounds per square foot. In the reinforced-concrete footing (Fig. 258) the rein- forcement can be only roughly estimated. One method is to consider the load upon the trapezoid abed and assume it as carried by the width be. The load W on this trapezoid FlG - 2 5 8 - is ; the distance x to the center of gravity of the trapezoid is x = 2^1 x A - BI 3 FOUNDATIONS 235 The bending moment is M = W (A x). In the formula for W, all dimensions are in feet, p e = Ibs. per sq. ft. and W is in pounds. In the formula for x all dimensions are in inches. In the equation for M all the terms are in pounds and inches. Having found the moment M the design is readily completed as explained under reinforced concrete and as illustrated by the design under a reinforced-concrete chimney. PILES Where the sustaining power of the soil is inadequate any of the preceding foundations may be placed upon piles. Piles were until a few years ago entirely of timber, preferably white oak, with a point diameter of at least 6 ins. but frequently specified as 8 to 10 ins. Their lengths will vary to suit the particular location. During the last few years extensive use has been made of concrete piles. Timber piles should be sharpened at the point, and are frequently shod with an iron band or point to protect them. The larger end is cut off square and is also some- times protected with iron caps, hoops or bands. The brooming of the butt of a pile interferes materially with driving it, so that such piles should be trimmed during the driving to facilitate this operation. The final or test blows should not be made on a broomed pile. Piles may carry their load through friction between their sides and the surrounding soil, or they may be driven to rock when the load may be carried largely by the rock. A pile may there- fore fail through crushing of the timber, or, where a pile is driven to rock or its equivalent through loose soil, it may fail as a column. Specifications limit the load on a pile to from 40,000 to 50,000 Ibs., or 600 Ibs. per sq. in. of its mean section. When a possible failure as a column is considered the load should be calculated for a column whose maximum stress does not exceed 600 Ibs. per sq. in. Timber piles when completely submerged in water and not exposed to the toredo (shipworm) are apparently preserved 236 GRAPHICS AND STRUCTURAL DESIGN indefinitely. It is therefore necessary to cut timber piles below extreme low water. A bed is then built upon the tops of the piles to carry the upper part of the foundation (see Fig. 259). The usual spacing of the piles will vary from two to four feet and upon the tops of the piles which have been sawed off to a common level is placed a grillage of timbers secured to the piles with metal pins. The masonry is then placed upon this timber. All timber must be below water. Another and later practice, shown in Fig. 260, is to incase the tops of the piles in a bed of concrete. As in the first case the piles are cut off below the water level. fe The soil around the piles is excavated for a couple of feet and a bed of sand, say, 12 ins. thick, is put in, and upon this is laid the concrete whose thickness depends upon the load to be carried. This last method firmly incases the pile tops and distributes a part of the load upon the soil. Concrete piles have an advantage over timber piles in not being attacked by the toredo and in not having to be continu- ously submerged. There are several types of patented concrete piles. The Simplex is made in place by driving a hollow tube with a pointed end the entire depth required and then filling the hole with con- crete as the tube is withdrawn. In the Raymond pile a thin tube is driven from which a collapsible case is then removed and replaced with concrete. Other forms of reinforced piles are FOUNDATIONS 237 made on the ground and when properly aged driven with a pile driver. There is no very satisfactory method of determining the bear- ing power of a pile. The best known, where the pile is not driven to refusal, is the Engineering News Pile Formula. This formula 2 -W -h , 2 -W -h f is P = - - for drop-weight hammers and P = - r for S+l S -f TO steam hammers. P = safe load in tons, 2000 Ibs. W = weight of hammer in tons, 2000 Ibs. h = drop of hammer in feet. s = penetration in inches due to the last blow. CHAPTER XVI CHIMNEYS IN the design of masonry chimneys it is usual to assume that the masonry cannot resist tensile stresses. If the weight of the column above the section i-i, Fig. 261 (a), is W pounds this load will be uniformly distributed on the section when no other forces act on the column above this section. If then the area is A sq. ins. the uniform -tr-* (> I. , . w fiber stress is Ibs. per sq. in. Assuming A. a force F acting on the right of the column, it will tend to reduce the fiber stress on the right and increase that on the left of the section. The original fiber stress / is com- pression. If the force F is increased suffi- ciently the fiber stress on the right would become zero while that on the left would be doubled, as is indicated in Fig. 261 (d). The unit stress would vary across the sec- tion as shown in Fig. 261 (d), being as- sumed as varying uniformly. The stresses would still all be compression. Under these conditions the resultant R of W and F must act through a point a distance q from the center of gravity of the section i-i. Should the resultant R act at a greater distance from the center of gravity of the section than q, which would be due to a still further increase in F, then the fiber stress on the right would pass through zero and be- 238 (fl FIG. 261. CHIMNEYS 239 come tension, should the material resist tension, and the fiber stress on the left would be increased but would continue to be compression. Should the material not be able to resist tension the pressure will adjust itself to the left until the resultant stress R coincides with the point of intersection of the resultant of W and F with the section i-i. When the extreme fiber stress on the left exceeds the strength of the material, the sec- tion i-i will fail. It is usual to assume that the resultant R must fall within the distance q from the center of gravity of the section. Had the force F been applied at all possible points around the column at the same distance y above the section i-i the points of intersection of the resultants R of W and F would have described the figure shown black in Fig. 261 (b); this is called the kern of the section. If the section i-i at no point is to be subjected to tension the resultant R must fall within the kern for all positions of the force F. The moment on the section for a given loading will be the load W times the distance from the center of gravity of the section to the point of intersection of the resultant R with the plane of the section i-i. The allowable distance q from the center of gravity of the section to the resultant R that produces zero stress at the extreme fibers of the section for that position of F is given by q = - > A X c where / = moment of inertia of the section, inches 3 . A = area of the section, square inches. e = distance from center of gravity of section to extreme fibers whose stress is to be limited to zero. The kerns for the commonest sections used for chimneys and foundations are given in Figs. 262 to 267. The minimum radii for the kerns of these several sections are, Square, Fig. 262, r = 0.118 h. b-h Rectangle, Fig. 263, r = ' 240 GRAPHICS AND STRUCTURAL DESIGN Triangle, Fig. 267, r 2 = , and n = - 12 6 Octagon, Fig. 264, r^ = 0.2256 # (R = radius of corners). Circle, Fig. 265, r =- (constant). 8 Hollow square, outer side = H, inner side = h. Similar to* Fig. 262 Circular ring, Fig. 266, outer diameter = D, inner diameter In the case of retaining walls it is necessary to consider the force F as acting in one direction; the base is then rectangular and the resultant R must not fall farther from the center of the base than one-sixth the base width, or if b = width of the base, q = - . o This is commonly expressed by saying that the resultant pressure must pass within the middle third. In chimney design in this country the wind pressure is gen- erally assumed at 30 to 50 Ibs. per sq. ft. upon flat surfaces, and at 20 to 30 Ibs. per sq. ft. on the projected area of cylindrical surfaces. The greater of these pressures considerably exceeds any pressure likely to occur in ordinary localities. Ordinary masonry chimneys are commonly either square, octagonal or round. Square sections are suitable for short and unimportant chimneys while by far the greater number of chimneys are round. Chimney sections are obtained by the use of specially shaped brick for the corners of octagonal chimneys and radial brick for round chimneys. In small chimneys the upper 25 ft. should have a thickness of 8 to 9 ins. ; this thickness should be increased about 4 to 4 J ins. L " FIG. 262. |*--R-^T R --^| 6 >j FIG. 263. \*--d >| FIG. 264. FIG. 265. FIG. 266. FIG. 267. - I i ,Top .F 30800^ a , Ground BRICK CHIMNEY FIG. 268. (241) 242 GRAPHICS AND STRUCTURAL DESIGN in each succeeding lower 25 ft. In large chimneys whose diam- eters exceed 60 ins., the thickness of the upper 25 ft. should be 12 to 13 ins. Ordinary brick dimensions are, width 4 to 4! ins. ; , length 8 to 9 ins. Chimneys are commonly provided with a fireproof lining in the lower portion extending up to from one-third to one-half the chimney height. This lining, being protected from the wind pressure by the main walls of the chimney, can be made quite thin, frequently not exceeding from 4 to 8 ins. DESIGNING A CHIMNEY The method of designing a chimney will be illustrated by making the calculations for the section d-d of the chimney in Fig. 268. The weight of the material above the section has been estimated as 545,000 Ibs., being taken at no Ibs. per cu. ft. The wind pressure has been considered as 25 Ibs. per sq. ft. of projected area for a round chimney and has been estimated as 30,800 Ibs. This wind force has been assumed as acting at the center of gravity of the trapezoid which is the projected area of the chimney above the section. The distance from the section d-d to the center of gravity is x = L+^l X d - = 14.16 + 20 x 100 X 12 = ms b + bi 3 14.16 + 10 3 b = base of trapezoid. bi = top of trapezoid. d = altitude of trapezoid. From the intersection of the chimney axis and wind force lay off hi representing the weight to scale, 545,000 Ibs., and at its lower extremity lay off ij to the same scale representing F = 30,800 Ibs. Draw the hypothenuse of this right-angled triangle and it will cut the section d-d at /. To insure the masonry at this section not being subjected to tension the point / must fall within the kern of this section. CHIMNEYS 243 For accuracy it is better to calculate the length Ik than to measure it from a drawing on so small a scale. The calculation can be made by the use of the similar triangles hlk and hij, hence hk X if 556 X 30,800 Ik = q = rr^ 1 = - - = 31.4 ms. hi 545,000 hk = distance from the section to the resultant wind force. The kern radius is given by Since 31.4 ins. is less than 35.6 ins. it is evident that the resultant falls within the kern, and the section is subjected to compression only. The weights have been calculated, using the volume of a frustum of a cone as F = ^U-H h = height of the frustum. A = area of the base. a = area of the top. The following table gives the properties of the several sections. Section above. Distance from top, ft. Diameter O.D. I.D. Mean diameter. Mean area, sq. ft. Weight above, Ibs. Fat 25 Ibs. r" q" At top \ 's'^' a-a 25' o" II' 4' 9' 4' 10' 8" 8' 8" 28 77,ooo 6,700 28.6 12.7 b-b So' o" 12' 8' 10' o' 12' o" 9' 8" 40 187,000 13,750 30.8 21.3 c-c 75' o" 14' o' 13' 4" 56 341,000 21,800 33.2 27.3 d-d 100' o" 10' 8' IS' 4' 11' 4' 10' 4 ' 14' 8 ' II' O ' 74 545,000 30,800 35-6 31-4 e-e 125' o" 16' 8' 12' o' 16' o ' n' 8 ' 94 803,000 40,700 38.o 34.8 /-/ 150' o" 18' o' 12' 8' 17' 4 ' 12' 4 ' 117 1,127,000 52,500 40.1 37-3 (Max.1 Top of foundation . . . 185' o" 18' o" n' o" 229 2,008,650 67,500 93.5, IMin.f 33-4 [66. oj f Top Base foundation. Concrete at 140 Ibs. per cu. ft.. 1 22' o", 1 Bottom 2,867,130 67,500 (Min.j 23-9 I 3o' o" 244 GRAPHICS AND STRUCTURAL DESIGN It is seen in the above table that since the resultant pressure al- ways falls within the kern the maximum pressure never exceeds twice the fiber stress due to the weight above the section assumed as uniformly distributed over that section. Considering the sec- tion at/-/, the load above that section being 1,127,000 Ibs., and the area of the section 117 sq. ft., the load per square foot is 1,127,000 , - " = 9630 Ibs. This is well below the allowable, or 144 X 150 = 21,600 Ibs. per sq. ft. Some designers prefer to use the inertia of the section in making their calculations. Making the calculations for the section 150 ft. from the top we have The section modulus of the ring = / _ TT (> 4 - hence The projected area of the chimney is - - X 150 = 2100 sq. ft. The wind pressure on this area is 2100 X 25 = 52,500 Ibs. The distance x from the section to the center of gravity is CHIMNEYS 245 The moment of the wind about the section is 52,500 X 68 = 3,570,000 ft. Ibs., since f Me 3.570,000 f . f = - - = 9100 Ibs. per sq. ft. I 393 The direct pressure upon this section is - = 9630 Ibs. per sq. ft. The maximum pressure upon the leeward side is 9630 + 9100 = 18,730 Ibs., while that on the windward side is 9630 9100 = 530 Ibs. per sq. ft., both being in compression. Some designers assume an allowable tension on the windward side of one-tenth the maximum compression on the leeward side. It is then understood that the material does not exert this tensile fiber stress but that the effect is to make the resultant com- pression pass outside of the kern and increase the maximum unit pressure. This can be understood by referring to Figs. 261 (a) to 261 (f). The determination of the maximum pressure under these circumstances presents considerable difficulties. As long as the resultant pressure at any section fell within that section the margin of security would be given by the ratio of the ultimate crushing strength of the material to the extreme fiber stress in compression at that section. Keeping the resultant within the kern affords a ready means of determining the maximum fiber stress, thus assuring safety. The maximum pressure upon the foundation will now be determined. The foundation being 30 ft. o ins. square the min- imum radius of the kern is r = 0.118 X 30 X 12 = 42.5 ins. As the resultant passes 23.9 ins. from the center line of the foundation it falls in the kern and the maximum unit compression will not exceed twice the unit uniform load on the foundation when no wind is blowing. The maximum unit pressure will be on a corner when the wind blows parallel to a diagonal. The moment on the foundation 246 GRAPHICS AND STRUCTURAL DESIGN FIG. 269. -0.60- -0.40- -0.40 -0.30- -0.30 : 0.20- J-L -0.20 0.15 =. 0.2 0.3 0.4 i i I i i i i i i i i i I i i i I i i i i i i i i i I i i i i l i FIG. 270. CHIMNEYS 247 section is the product of the total weight on the soil multiplied by q or 1/r 2,867.000 X 23.0 . M = - - = 5,720,000 ft. Ibs. 12 The resistance of a square referred to a diagonal as an axis is - = 0.118 h\ e The base being 30 ft. square and keeping the resistance in feet 3 we have, - = 0.118 X 3o e 3186. . , M-e 5,720,000 r , ,, j. Since / = = - = 1800 Ibs. per sq. ft., the direct 1 3*"^ 3190 Ibs. per sq. ft. The maximum rh . 2,867,000 pressure is - 900 compression then is 1800 -f- 3190 = 4990 Ibs. per sq. ft. The compression on the opposite corner is 3190 1800 = 1390 Ibs. per sq. ft. These loads will be satisfactory if the soil can carry 2\ tons per sq. ft. When the resultant of the wind pressure and the chimney weight passes outside of the kern of the bottom of a chimney the maximum pres- sure on the corner of a square or the circum- ference of a circular section may be found by using the curves given in Figs. 269 and 270. Fig. 269 represents a square base, Fig. 270 a circular base. D = diagonal of square and diameter of cir- cle. h = side of square. FIG. 271. q = ph = (3D = distance from center of section to resultant of wind and weight. a-a neutral axis, axis of zero pressure. 248 GRAPHICS AND STRUCTURAL DESIGN F = total wind force on chimney, acting at the center of gravity of the projected area of the exposed portion of the chimney. W = total load on the soil, pounds. hi = distance from force F to bottom of the foundation, 7T< inches. See Fig. 271. q (inches) = hi X =-. r - W SELF-SUSTAINING STEEL CHIMNEYS A self-sustaining steel chimney, Fig. 272, is held upright by the resistance to overturning offered by the chimney, its lining and foundation. The wind pressure is commonly assumed as 50 Ibs. per sq. ft., acting upon the side of a square chimney, or 25 to 30 Ibs. per sq. ft. of vertical projection upon round chimneys. Usually the thickness of the upper sheets is determined for durability rather than for strength, being made not less than T 3 g in. and frequently \ in. thick. This thickness is then in- creased by -IQ in. every 30 or 40 ft. Some designers vary by $ in. instead of by yg- in. At the lower part of each 30 or 40 ft. section the fiber stress induced in it by the bending moment due to the wind pressure can be found and the section altered in thickness or length if deemed advisable. The rivet spacing is made small to assure tightness, being not less than 2.5 times the rivet diameter nor more than 16 times the thickness of the plate. For most ring sections this gives excessive rivet strength. An assumed ring section may be checked as follows: M = bending moment in inch pounds on the section due to the assumed wind pressure above that section. D = outside diameter of the chimney in inches. h = distance from a ring section to the chimney top, inches. / = thickness of shell, inches. F = total wind pressure acting upon any portion of the chimney, pounds. FIG. 272. - H SECTION AT a-a FIG. 274. BLRA.CKET STEEL CHIMNEY (240) 250 GRAPHICS AND STRUCTURAL DESIGN The chimney above the given ring section acts as a cantilever beam, and the bending moment equals the total wind pressure upon the vertical projection of the chimney above this ring section multiplied by one-half the distance from the ring section to the chimney top. The chimney will be assumed 66 ins. O.D. and 100 ft. high. The bell section is 9 ft. high and the section to be examined is 91 ft. from the chimney top. The total wind pressure on the vertical projection of this part of the chimney is F = 5.5 X 91 X 25 = 12,510 Ibs. and M = 12,510 X - - = 6,830,460 in. Ibs. The value of - for a ring section is e I TT CD 4 - d*) ^ e 32 XD where D = outside diameter. d = inside diameter. Where /, the thickness of the shell, is small compared with D, this value may be approximated as - for the case in hand is - = - X 66.625 X -^ (66.625 - 0.9375) e e 5 16 = 1094. f Me 6,8^0,460 , / = --.= -2-- - =6250 Ibs. / 1094 This fiber stress is on the gross section. If the net section at the rivet circle is x per cent of the total section then the maximum fiber stress between rivets is f 6250 .. , 6250 f m = - -, or, if x = 80 per cent, f m =^- X .oO = 7800 Ibs. per sq. in. As these fiber stresses are satisfactory the y\-in. shell will be used. CHIMNEYS 251 RIVETING RING SEAMS 5 = pitch of rivets, inches. The force acting on the rivets per inch of circumference is / X / X /, and to this must be added the weight of the shell i in. wide from the given section to the top. The calculation of this weight can be facilitated by remem- bering that a J-in. plate 12 ins. X 12 ins. weighs 10.2 Ibs., then estimating the weight of a vertical strip 12 ins. wide and finally dividing the weight found by 12. For the section 91 ft. from the top, / X / X p = / X fV X 6250 = 1950 Ibs. 1. 30 ft. of T V m - plate, 30 X i X (J X 10.2) = 230 2. 30 ft. of J-in. plate, 30 X i X (i X 10.2) = 306 3. 30 ft. of j^-in. plate, 30 X i X (| X 10.2) = 383 Total 919 Adding 10 per cent for rivets, laps, etc. 91 1010 or - - = 84 Ibs. per in. The total force per inch of circum- ference acting on the rivets is 1950 -f- 80 = 2030 Ibs. Allowing 10,000 Ibs. per sq. in. in single shear and 20,000 Ibs. per sq. in. in bearing the value of a f-in. rivet in a -j^-in. plate is 4420 Ibs., and the rivet spacing required is - - = 2.18 ins. or, say, not 2030 exceeding i\ ins. A double row of staggered rivets will be used at this seam spaced about, but not exceeding, 2\ ins. Ordinarily the rivet spacing need be determined for only the lowest row of rivets in sheets of the same thickness. RIVETING VERTICAL SEAMS The riveting in the vertical seams must provide for the shear along the neutral axis of the chimney, and as the riveted edge may be in compression the rivets must be spaced to prevent 252 GRAPHICS AND STRUCTURAL DESIGN the buckling of the plate between them; this demands that the rivet spacing shall not exceed, 1 6 times the thickness of the plate. The unit shearing stress at any point along a cylindrical beam V -r ~T e where /, = unit shear in pounds per square inch at the section. V = shear at right angles to the neutral axis, pounds. r = radius at the section, inches. - = resistance of the section, inches 3 . e In this chimney at the section under consideration the unit shearing stress is = 8olbs . 1094 This per inch of chimney height is 380 X YG = 1 2O Mbs. The rivet spacing for shear then is ~ 37 ins. 1 20 This spacing of course is not permissible, 16 times the plate thickness limiting the spacing to 5 ins. It generally will be the buckling of the plate edge that will determine this riveting. The ring sheets are preferably of one piece, but in large chim- neys this becomes impractical both for manufacture and ship- ment. The base of the chimney is usually a bell, and is preferably a frustum of a cone, although sometimes flared to improve the appearance. The former shape is a much simpler and cheaper design. The height of the bell and the diameter at its base vary from ij to 2 times the chimney diameter. The bell sec- tion is usually made of a number of pieces, butt jointed with single outside straps as shown in the drawing. The bell sheets are usually tV~ m - thicker than the ring sheets immediately above them. CHIMNEYS 253 Where the tempera- T Lining. The method of lining varies, tures do not exceed 650 F. (and they ordinarily do not) common red brick are satisfactory; otherwise a No. 2 fire brick is required. At the top the lining should have a thickness 4! ins., which .can be increased by 4^ ins. every 30 to 40 ft. In some cases the chimney is only partially lined, say one-half the way up, and in other cases the thinnest lining is made 2\ ins.; the former, however, is the bet- ter practice. Where the breeching from the boil- ers enters the chimney above the base care should be taken not to weaken the chimney by cutting away too much, and the chimney should be sufficiently reinforced at this point. The practice of having the gases enter the chimney through flues below the steel base makes the chimney stronger and of better appearance. FOUNDATIONS The shell is maintained upright against the moment of the wind pres- sure by the weight of the foundation and chimney with lining assumed as acting about the lower edge of the foundation. The foundation is concrete and the chimney is secured to it by heavy foundation bolts. The concrete can be assumed as weighing from 125 to 150 Ibs. per cu. ft. When Fig. 277 (d) 254 GRAPHICS AND STRUCTURAL DESIGN there is no wind acting upon the chimney the combined weights of the foundation, W f , the steel chimney, W c , and the lining, Wi, are distributed uniformly over the soil through the base of the foundation, as is shown in Fig. 277 (b). Now as a wind pressure F begins to act the uniform pressure previously on the soil will be altered, the pressure being reduced on the side upon which the wind acts and increased on the opposite side. In Fig. 277 (b) 5 and W f + W c + W h which are always equal, lie in the same line. In Fig. 277 (c) the force acting upon the base from the soil, S. nas moved over, acting through the center of gravity of the trapezoid which here represents the distribu- tion of the soil pressure. When the wind pressure becomes great enough the distribu- tion of pressure on the soil will become that shown in Fig. 277 (d), being zero at the right, while the original pressure at the left is doubled. The force S will pass through the center of gravity of the triangle, or J B from the edge of the foundation. It is commonly assumed in masonry construction that the resultant pressure upon any section must pass within the kern of that section. Were the resultant S to pass to the left it would pass out of the kern. See page 238. In Fig. 277 (a) the moment due to the wind force must equal the moment of the forces W f , W c and Wi, with the arm a; hence (W f + W c + W l ) a; the limit being a = 0.118 B we have F(^ + H f \ = (W f + W c + W l } Xo.nSB. The volume of a frustum of a cone or pyramid is V = (A t + A b o where A t = area of the top of the frustum. A b = area of the bottom of the frustum. h = altitude of the frustum. CHIMNEYS 255 Maximum Pressure on the Soil. When the resultant pressure passes through the kern the maximum pressure will be less than twice the mean pressure. The maximum pressure per square foot then is P = pounds per square foot. A b = area of the base of foundation, square feet. Where the foundation is not reinforced the angle of the sides with the vertical should not exceed 30 degrees. The maximum pressure should not exceed that permitted on the soil (see page 228). In the chimney under discussion assume the base 18 ft. square and that the concrete weighs 140 Ibs. per cu. ft. Estimating the weight of the steel of the chimney, the diameter being 5 ft. 6 ins., the circumference is 17.3 ft. Sheets. Dimensions, Weights, feet. Ibs. T 8 <5 17-3 X 30 X 10.2 X | 3,980 i 17-3 X 3 X 10.2 X i 5,300 rV J 7-3 X 3 X 10.2 X 6,620 f 17.3 X 10 X 10.2 X 2,650 18,550 Ibs. Allowing 10 per cent for bolts, rivets, laps, etc., 1,850 Total 20,400 Ibs. The weight of the lining assuming the chimney to be lined to a height of 70 ft., Section. Lining, Area of lining. Volume, ins. sq. ft. cu. ft. Base to 40 ft. 9 11.2 448 40 to 70 ft. 4^ 6.05 135 Total 583 Weight of lining, 580 X 125 = 72,500 Ibs. The height of the foundation will be assumed at 10 ft. 256 GRAPHICS AND STRUCTURAL DESIGN If the resultant of wind pressure and total weight is to fall within the kern its distance from the center line of the chimney must not exceed 0.118 X B = 0.118 X 18 X 12 = 25.5 ins.; this is the distance along the diagonal. At right angles to the h 12 side of the square it would be-=i8X =36 ins. 6 6 Trying the following foundation, top 10 ft. square, bottom 1 8 ft. square and height 10 ft., the volume of this foundation is V = - (A t + A b + ^A t XA b = (io 2 + i8 2 + Vio 2 X i8 2 ) o o = 2013 cu. ft. Weight = 2013 X 140 = 281,800 Ibs. The total weight on the soil is 281,800 + 20,400 + 72,500 = 374,700 Ibs. The uniform pressure per square foot is 374>7 o 3 2 4 = 1150 Ibs. The resultant of wind pressure and total weight cuts the bottom of the foundation a distance q from the center line of the chimney. The total wind force is F = 5.5 X 100 X 30 = 16,500 Ibs. This force acts at a distance of 50 ft. from the base of the chimney or 60 ft. from the bottom of the foundation. TT 60 X 12 X 16,500 Hence q = - L2 = 31.7 ins. 374,700 The resultant evidently falls outside the kern on the diagonal but passes within it at right angles to the side. The maximum pressure on the corner can be found by the assistance of the curves, Fig. 269. From the curves for a square base when ft = 0.147, = - 22 > and since - Per sq.ft. CHIMNEYS 257 the foundation will be satisfactory if this load of 2625 Ibs. per sq. ft. is permissible upon the soil under the chimney. FOUNDATION BOLTS The chimney is assumed as tending to overturn about the axis a-a tangent to the bolt circle and differing only slightly from the lower edge of the bell. The maximum fiber stress will be f m and will occur in bolts No. 3 and No. 4 (see Fig. 278), these being farthest from a-a. The fiber stress in any other / r- FIG. 278. bolt will be f m multiplied by the ratio of its distance from a-a to the distance of the bolts farthest from a-a] thus the fiber stress in the bolts 2 and 5 is/ m X -. The resisting mo- p ment due to any bolt is the product of its area, the fiber stress acting in it and its distance from the axis a-a. The resisting moment of the entire group of bolts will be the sum of the moments of the individual bolts. Taking the case of the six bolts shown, and letting a be the area of one bolt, at the root of the thread, and R the radius of the lower bell circle, we have, No. bolts. p Fiber stress. Moment. i and 6 R (i - cos .,0) = 0.134 R fmX ^f|4 1.866 K R R (i cos 30) = 0.134 R 2 and 5 R (i cos 90) = R 3 and 4 R (i - cos 150) = 1.866 R fmXi 2-a-fm-R ^~ 2-a-f m -R-i.S66 This gives a total moment of M = 4.8 X a *f m R. In this way it may be shown that in the expression M C a *f m R, C has the following values for different numbers of bolts. c. Number of bolts. C. 3-5 4.8 12 16 9.1 12.0 6.2 20 15-0 7-7 258 GRAPHICS AND STRUCTURAL DESIGN Number of bolts. 4 6 8 10 The moment on the chimney due to the wind has been found to be 6,830,460 in. Ibs. The moment due to the weight of the* chimney which tends to hold it upright is MI = 20,400 X 2 = 918,000 in. Ibs. The net moment to be resisted by the bolts is 6,830,460 918,000 = 5,912,460 in. Ibs. Trying four bolts, M = 4.8 a .f m R, or a = 5 ' 9I2 > 46 = 4,8 X 9000 X 47 2.91 sq. ins. This corresponds to 2j-in. diameter bolts. The maximum total tension in a single bolt is 2.91 X 9000 = 26,190 Ibs. The bracket through which the bolt fastens to the bell of the chimney must be designed to resist this force. A very effective design is shown in Fig. 276. The bolts should be kept back as close as possible to the edge of the bell, and the bell sheet amply reinforced with straps at this point. The lower edge of the bell should be reinforced with a circular band. This lower edge then rests upon a cast-iron base plate, Fig. 273, that distributes the pressure over the top of the foundation. The base plate may be made in one piece for small chimneys, as shown, but is usually made of a number of sections for convenience in casting -and handling. These sections when placed in position are secured with a few J-in. diameter bolts. After erection the bell of the chimney and the foundation bolts will hold these sections in place. The foundation bolts should extend almost to the bottom of the foundation and should be fitted with a lock nut and washer or key and washer. The chimney should have a light ladder running to a platform at the chimney top. Dropping the platform a little below the top protects the handrail from the chimney gases. CHIMNEYS 259 REINFORCED-CONCRETE CHIMNEYS In a reinforced-concrete chimney the weight of the concrete above the section under consideration will produce a compression on that section similar to that produced in a masonry chimney. Take any unit area having a ratio of reinforcement of 0; under a given load the shortening of the two materials will be equal and the loads shared proportionally to their moduli of elasticity. Let/ c = compressive unit stress in concrete, pounds per square inch. j? - = n; then the fiber stress on the steel is / = n /. and c if L is the load on i sq. in. of section f c = -. The i + (n - i) load can be assumed as the weight of the concrete for the ordi- nary percentages of steel reinforcement. With wind acting on one side of a chimney the pressure upon the leeward side is increased while that on the windward side is decreased, finally becoming zero while that upon the leeward side becomes double the original direct unit pressure. The concrete not being assumed as resisting tension may however be considered as acting like a beam until the compression at the windward side becomes zero. Under these circumstances the resistance of the section may be used as in the case of a beam resisting both tension and compression. In calculating the moment of inertia of the section proper allowance must be made for the modulus of elasticity of the steel differing from that of the concrete. / = I e + /. = ^ (D* + d*) + 0.394 n A 2 fc (A -3 ; here, 7 = total inertia of concrete and steel. 7 C = inertia of concrete ring. 7, = inertia of steel. A = area of concrete ring, square inches. D = outside diameter of section, inches. 260 GRAPHICS AND STRUCTURAL DESIGN d = inside diameter of section, inches. DI= diameter of steel cylinder having same area as total reinforcement at the section. t\ thickness of equivalent cylindrical steel reinforcement. Generally h will be so small that 3^1 may be neglected, making I (D 2 + d 2 ) + 0.394 nD\t\. Where the ratio of reinforcement is known t\ = -(D d). 2 When the point has been reached where the extreme fiber stress on the concrete on the windward side is zero any further increase of the wind pressure F will be resisted by increased compression in the concrete on the leeward side and by the re- inforcing steel taking tension on the windward side. This causes the axis of zero stress to move across the section from the extreme fibers on the windward side towards the leeward side. Consider- ing now only the additional moment applied after the extreme fiber stress on the windward side becomes zero there will be created additional compression on the leeward side shown shaded, while on the windward side the fiber stress will be tensile and must be carried by the steel. Con- sidering now the fiber stress as indicated in Fig. 279, it will vary uniformly on each side of the neutral axis a-a, the stress in the steel being n times the stress in the concrete at the same distance from the axis a-a. The resultant of the flange stresses in compression, shown in the shaded por- tion in Fig. 279, must equal the resultant of the tensile stresses in the steel in the remaining part of the section, the portion not shaded. The resisting moment induced in any section equals either flange force multiplied by the distance between CHIMNEYS 261 the lines of action of the resultant flange forces on each side of the axis a-a. This distance has been found to be constant and to equal 1.56 R 8 , where R 8 = radius of the steel circle. The values of the flange forces and the moments have been calculated based upon the assumption that the stresses vary proportionally to the distances of the fibers from the neutral axis a-a, and that the fiber stress in the steel is n times that in the concrete for similar positions, the thickness of the wall RI r (Fig. 281) has been considered small compared with R, that is the flanges have been considered as lines. The flange forces are Tensile steel, F t = 2 t,f t R a X C\. Compression steel, F c = 2 t a f c R a X C 3 . Compression concrete, F co = 2/ co t co R X C 3 . The moments due to these forces are Tensile steel, M t = 2/ t t a R a 2 X C 2 . Compression steel, M c = 2f c t a R a - X C 4 . Compression concrete, M^ = 2/ co t co R 2 X C 4 . Here t a = thickness of equivalent cylindrical steel reinforcement, inches. f t = extreme fiber tension in steel, pounds per square inch. R 8 = radius of steel, inches. f c = extreme fiber compression in steel, pounds per square inch. tco = thickness of concrete, inches. f co = maximum compression in concrete at mean radius R, pounds per square inch. R = mean radius of concrete, inches. The values of these coefficients Ci, C^ Ca and C 4 have been cal- culated for the several values of k and are given in the curves, Fig. 280. The following relations are important. F t = F c + F co . The total resisting moment of a section M M t + M c + M^. 262 GRAPHICS AND STRUCTURAL DESIGN The distance between the lines of action of the resultants of the flange forces has been previously stated as 1.56^; hence the moment may be expressed as (F c + F co )i.$6R. .80 ,20 M=Fxl.5GR, s RE-INFORCED CONCRETE CHIMNEYS Scale on \ left multi >lled by 100 .30 07( Oil) FIG. 280. The table also gives the relations between the several fiber stresses and the critical percentages of reinforcements for the various values of k. k = . Here R is the radius of the flange R being considered, that is, R a for the steel and the mean radius for the concrete. The following problem will illustrate the cal- culations of a chimney as outlined. CHIMNEYS 263 Problem. A chimney is 1 50 ft. high. Its outside diameter is 10 ft. and the walls at its base are 9 ins. thick. Assume a wind pressure of 50 Ibs. per sq. ft., giving an equivalent pressure of 30 Ibs. per sq. ft. of projected area of the round chimney. The pressure on the concrete is not to exceed 500 Ibs. per sq. in., while the unit stress on the steel must not be over 15,000 Ibs. per sq. in. The chimney walls are 6 ins. thick for the upper 100 ft. and 9 ins. thick for the lower 50 ft. In Fig. 282 assume a rein- forcement of i per cent and that a bar of concrete i sq. in. in 114- FIG. 281. FIG. 282. section and 12 ins. high weighs i Ib. The direct compression in the concrete at the base of the chimney is (100 X |) + 50 117 p e = - ^ - ~ ioo Ibs. per sq. in. i +4>(n - i) 1.14 The inertia of the section with i per cent steel is / = (i20 2 + io2 2 ) + 0.394 X 15 X ii4 3 X 0.09 13,350,000 in. Ibs. ,, pi M = ^ e ioo X 13.350.000 . - - = 22, 200,000 in. Ibs. 60 The total wind force is W = 10 X 150 X 30 = 45,000 Ibs. The T 9 moment due to this wind pressure is M = 45,000 X 150 X 2 264 GRAPHICS AND STRUCTURAL DESIGN = 40,500,000 in. Ibs. The remaining moment to be resisted by the reinforced section after the stress in the extreme fibers on the windward side has become zero is 40,500,000 22,200,000 = 18,300,000 in. Ibs. The compression on the leeward side at this time is 2 X 100 = 200 Ibs. per sq. in. The permissible in- crease in the compression on the leeward side is 500 200 = 300 Ibs. per sq. in. The additional flange force is 18,300,000 18,300,000 F = ( \ ^ ^ = ~ ~ = 2o6 > 000 lbs - (1.56X12) 1.56X57 F t = 2 X t. Xfs X R 8 X Ci = 2 X 0.09 X 15,000 X 57 X 1.32 = 203,000 Ibs. This is sufficiently near the required 206,000 Ibs. From the curves, Fig. 280, the ratio of stress in the steel to that in the concrete is 57 to i; hence 15,000 p c = = 263 Ibs. per sq. in. The approximate total stress in the concrete = 200 + 263 = 463 Ibs. It should be noted that the 263 Ibs. is the flexural stress at a distance R 8 from the center of the chimney and that this can be easily corrected. The value of k from the curves corresponded to 0.58, making A = k X R 3 = 0.58 X 57 = 33 ins. The distance from the neutral axis to the steel 57 33 = 24 ins. The distance to the outside of the concrete is 60 33 = 27 ins. By similar triangles, -%- x = ^, or / max = 2 9 61bs. fc 24 The extreme fiber stress then becomes 200 -f- 296 = 496 Ibs. per sq. in. CHIMNEYS 265 DESIGN OF A REINFORCED-CONCRETE CHIMNEY In Fig. 283, assuming the chimney wall 5 ins. thick at the top, the area of this section is 1210 sq. ins. The radius of the kern This means that before any reinforcement would be required the resultant of wind force and weight could pass 18.3 ins. beyond the axis of the chimney. It follows that the weight above a section / feet from the top, divided by the wind force acting on the portion above that section equals -ins., divided by 18.3, or / 1210 X / = 2 . = 1210 X 18.3 X 2 = o r t 30X6.8 Xl 18.3' 30X6.8 X 12 . Hence the upper 18 ft. would require no reinforcement. It is usual to place a small reinforcement in this portion. Assuming a reinforcement used here of | per cent (0.005) the distance that this will serve down the chimney can be found in a similar way. From the curves the ratio of the fiber stresses for p = 0.005 ^ s 85 to i ; hence taking the fiber stress in the steel at 15,000 Ibs. per sq. in. would make the value of f e = I 5' OOQ =177 Ibs. per sq. 8 5 in. ; as this is very low the tensile steel will determine the strength of the section. The value of Ci is 1.38. The flange force in the tensile steel then is F t =2 tj t X R 8 X Ci = 2 X 0.0025 X 15,000 X 38.5 X 1.38 = 39,800 Ibs. Mi = F t X 1.56 X R 8 = 39,800 X 1.56 X 38.5 = 2, 395, ooo in. Ibs. The inertia of the section is / = A (D* + d*) + 0.394 n/Vd- .16 + 7 ' 2) + 0<394 x I5 x 7?a x Fig. 283 Fig. 284 "I Fig. 285 REINFORCED CONCRETE CHIMNEY, CHIMNEYS 267 Now M = , and when the fiber stress on the windward side e becomes zero that on the leeward side has become double the stress due to the weight above the section. If the concrete weighs 144 Ibs. per cu. ft., and the distance from the top to the section is / feet, then / = /, and we have // / X 968,406 M 2 = = - - = 23,600 /. e 41 The bending due to the wind acting on the portion above the section equals the sum of Mi and MZ. From the wind pressure and the chimney dimensions tht bending equals 2 Equating these values of M we have 1220 X I 2 = 23,600 X / + 2,395,000, from which I 2 - (19.3 X /) + (^] = 2040, or / = 54.8 ft. This reinforcement will have an area of 1210 X 0.005 = 6.05 sq. ins., and will require 14 f-in. round bars. Allowing a bond stress of 80 Ibs. per sq. in. the length of the bars at laps is where li = length of the lap in inches. /. = fiber stress in steel, pounds per square inch. f b = bond stress, pounds per square inch. d = diameter of round bar or side of square bar, inches. In the problem , _ 12,000 X d _ , d being J in. the lap should be 37.3 X 0.75 = 28.1 ins. 268 GRAPHICS AND STRUCTURAL DESIGN The reinforcement will now be determined for the section at the base. The moment due to wind at this section of the chimney is M = [(85 X 6.8 X 30) (^ + 40)] + [(40 X 8.5 X 30 X 20)] = 1,634,550 ft. Ibs. M 19,570,600 in. Ibs. D = 102 ins. d = 88 ins. The area of the section is 2085 sq. ins. The weight of the upper 85 ft. is 1210 X 85 = 103,000 Ibs. The load per square inch on the lower section is/ c = I0 3> 000 _j_ 4O 2085 = 50 + 40 = 90 Ibs. per sq. in. Trying a reinforcement of 0.0175, from the curves C\ = 1.25, j- '= 41, and k = 0.46. fa Revising f c for the steel percentage . _ 90 _ 90 _ Jc ~ i. + 00- i) ~ i + (0.0175 x 14) ~ 72< The compression on the leeward side when the tension on the windward side is zero is 2 X 72 = 144 Ibs. per sq. in. The tension in the steel corresponding to this is (500 144) 41 = 14,596 Ibs. per sq. in. F = 2 tJ t R 8 Ci = 2 X 0.1225 X 15,000 X 48 X 1.25 = 220,000 Ibs. M = F X 1.56 X R 8 = 220,000 X 1.56 X 48 = 16,500,000 in. Ibs. The inertia of the section is / = A (D 2 + d 2 ) + 0.394 X n X A . 10 = H??5 (io2 2 + 88 2 ) + 0.394 X 15 X 96 3 X 0.1225, I = 2,365,000 + 640,000 = 3,005,000, M JI_ _ 72 X 3,o5,oo . e 5 1 CHIMNEYS 269 The remaining moment to be resisted by the chimney is 19,570,000 - 4,240,000 = 15,330,600 in. Ibs. F = M -4- (1.56 X R 8 ) = 15,330,600 -T- (1.56 X 48) = 204,000 Ibs. f t = F+(2Xt a X .'XC,) = 204,000 -r- (2 X 0.1225 X 48 X 1.25) = 13,800 in. Ibs. , 11,800 fc = ^~ - = 340 Ibs. 4 1 The extreme fiber stress will be somewhat in excess of this. k = 0.46, A = k X R s = 0.46 X 48 = 22 ins. By similar triangles the extreme fiber stress is - ^ X 340 = 378 Ibs. per sq. in. The extreme fiber stress then becomes (2 X 72) + 378 = 522 Ibs. per sq. in. The reinforcing bars will require an area of 2085 X 0.0175 = 3 6 -5 sc l- ms - Using f-in. bars will demand - = 76 f-in. bars. 0.44 Similar calculations will indicate the following bars in the several sections; beginning at the bottom, we have Section. Section. First, 22 ft., 76 f-in. round bars. Fourth, 22 ft., 28 f-in. round bars. Second, 25 ft., 60 f-in. round bars. Fifth, 34 ft., 14 f-in. round bars. Third, 22 ft., 44 f-in. round bars. Horizontal reinforcing rings must be used to resist the web stresses and also to prevent cracking due to expansion from the heat. Concrete being a poor conductor of heat the inside becomes much hotter than the outside. This causes considerable circum- ferential stress in the chimney which must be resisted by hori- zontal rings, preferably placed near the outer surface of the chimney. The amount, of this circumferential reinforcement can be only roughly approximated, as the difference in temperature between the inner and outer faces of the wall must be guessed. In several chimneys this reinforcement was from J to f of i per cent. 270 GRAPHICS AND STRUCTURAL DESIGN The greater the steel ratio the higher is the compression in the concrete. These chimneys are generally lined at least one-third their height. The inner tubes, being protected from the wind, are merely called upon to resist temperature stresses. These are not usually calculated. The linings may be made of fire brick or reinforced concrete and will have horizontal ring reinforce- ment somewhat lighter than in the outer wall, say from to J of i per cent, and vertical reinforcement to resist temperature stresses of from J to ^ of i per cent. The horizontal reinforcing rings are commonly ^-in. or f-in. rounds, and are spaced from 12 ins. to 18 ins. vertically. The spacing should be closer at the point on the chimney where the lining stops, Fig. 286, and additional vertical reinforcement should also be placed at this part. Chimney Base, Fig. 288. Estimate of total weight of chimney : Section Area, Volume, Weight, sq. ft. cu. ft. Ibs. Upper iffi- X 85 = 715 X 144 = 103,000 Lower - 2 T ? Y- X 40 = 580 X 144 = 83,500 Lining ff \ X 40 = 265 X 144 = 38,200 Base 56 X 3 = 168 X 144 = 24,190 Foundation (341 + 486) X 144 = 167,000 Total 415,890 The total wind moment previously found for the base of the chimney is 1,634,550 ft. Ibs. The total force is (85 X 6.8 X 30) + (40 X 8.5 X 30) = 27,540 Ibs. The resultant wind pressure then must act- 1 ' 34> 55 = 50.5 ft. above the ground line. The 27,540 resultant of weight and wind will act a distance q from the axis of the chimney and CHIMNEYS 271 The radius of the kern for a square is r = 0.118 X h = 0.118 X 20 X 12 = 28.4 ins. From page 246, /2 2O since TJ/ r r>2 W r = ./..^, or /e = _ = _ = 0.25. The resultant pressure passes 4.4 ft. from the cen- ter so that the pressure extends a distance (5.6 X 3) = 16.8 ft. over the base. The area of the pressure on the base is 16.8 X 20 = 336 sq. ft. The total pressure equals the entire weight of the chimney, hence 336 X*~ = 415,890 and p = (415,890 X 2) -5- 336 = 2470. The reinforcement can be only approximated, the simplest way being to consider the part cdef, Fig. 288, as a can tile vered beam. The distance from the center of gravity of the portion cdef to the line cf is Area. Statical moment. + 20 X V- = ioo X -V- =333-33 -ioX |= f|X (5 + I) =166.75 166.58 75 The load on cdef approximates 75 X 2400 = 180,000 Ibs. The approximate bending moment is M = 180,000 X (5 2.22) = 500,400 ft. Ibs. The bending moment per foot along the line df is (500,400 X 12) -5- 10 = 600,480 in. Ibs. Assuming n = 15; d = 42 ins. and p '= 0.0025; j = 0.922 and kj = 0.218. Using the formula for reinforced concrete design, we have - M a _ _ 600,480 _ ** A Xjd ~ (42 X 12 X 0.0025) X 0.922 X 42 = 1 2, 300 Ibs. per sq. in. 272 GRAPHICS AND STRUCTURAL DESIGN and , 2 XM C 2 X 600,480 fc = r^ r~^ = ; = 260 Ibs. per sq. in. kj X bd 2 0.218 X 12 X 42 2 The depth of the beam being controlled by the necessity of the foundation reaching proper soil and extending below the frost line this fiber stress in the concrete is satisfactory. The spacing of i|-in. round bars will be area of bar X 12 0.994 X 12 , . spacing = = -^ = 9f ins. A 42X12 X 0.0025 CHAPTER XVII RETAINING WALLS PRESSURES ON RETAINING WALLS ACCORDING to Coulomb's theory some wedge BA C, in Fig. 289, will produce a maximum pressure E against the wall. This force will make an angle 5 with the face of the wall corresponding 1 FIG. 289. to the angle of friction between the face of the wall and the fill. The natural slope of the fill is the angle <; the angle that the face of the wedge, producing the maximum pressure against the wall, makes with the horizontal is x.' h = height of wall in feet. w = the weight of the fill in pounds per cubic foot. According to this theory the general value of the maximum pressure in pounds per foot of length of wall is sin' sin ( + (x + V/ S !" \ V sm + ;> 5 ! n <* ~ " + S)sm (0 -a 273 274 GRAPHICS AND STRUCTURAL DESIGN This is simplified for the more commonly assumed conditions as follows: For 6 = 90 degrees, and 5 = a, E = -wh 2 2 COS sin (0 + a) sin (0 a In Rankine's formula = 90, a = o, and 5 = o; substituting these values in the general equation gives: i 72 cos 2 E = ~wh 2 - - r^-r 2 - 2 (l + SH10) 2 E = \w . A 2 tan 2 (45 - 10). 2 i + sin This theory, although usually developed for retaining walls, applies also to the sides of bins. FIG. 290. Graphical Solution. The graphical method (Fig. 290) affords the neatest way of estimating the pressure E. E = %w*p.y. RETAINING WALLS 275 p and y = lengths of lines in the diagram measured to the same scale by which the wall or side is drawn. The other symbols are the same as in the preceding case. Explanation of Diagram. A B is the side of the wall or bin in contact with the fill. A I is a horizontal line through the base of the wall or side AB. AC is the line of natural slope of the fill making an angle with the horizontal. BC is the slope of the top of the fill making an angle a with the horizontal. Pro- duce AC and BC until they intersect in C. Upon AC as a diam- eter draw the semicircle ARC. Draw BG making an angle + 6) with .the wall AB. At the point of intersection G of BG and AC draw GH perpen- dicular to AC. With AH as a radius draw the arc HF and through F draw DF parallel to BG. From D drop p perpendicu- lar to AC. The proofs and discussions of the preceding equations and diagrams may be found in books on the mechanics of retaining walls and earth pressures. Figure 291 is the diagram when the angle of fill a equals the angle of natural slope . In this case since BK and AF are parallel p and y will be the same length wherever the point D is taken in the line BK. Figure 292 shows the construction when the angle of fill falls below the horizontal and also below the line BC 1 making an angle -f 6 with the back of the wall. In all the preceding cases E = | w p -y. If the fill, Fig. 293, carries a uniform load L Ibs. per sq. ft. of horizontal projection, the height of the fill may be considered as increased sufficiently to bring such loading on the soil. The diagram is given in Fig. 294. w' = w + ? and E = -\w + Distribution of Pressure on the Wall or Side. In the first cases, where the fill is not loaded, the pressure on the wall will 276 GRAPHICS AND STRUCTURAL DESIGN r i: FIG. 293. FIG. 294. FIG. 296. FIG. 297. RETAINING WALLS 277 vary as a triangle and the center of pressure will correspond to the center of gravity of the triangle, as shown in Fig. 295. Where the fill is loaded the usual assumptions regarding the distribution of the pressures behind the wall are shown in Fig. 296 and the magnitudes of these pressures are given by the follow- ing formulae. The resultant pressure E t acts through the center of gravity of the trapezoid of pressure abed. 2-L\ . , /E' + E"\ h E t = l and From which hi Center of Gravity of a Trapezoid. Fig. 297 being a trape zoid its center of gravity can be readily found as shown. WEIGHTS AND ANGLES OF REPOSE OF MATERIALS Material. Weight per cu. ft. Angle of repose, . Degrees. Clav. dry QOIIO ^04.0 Clay, damp IOO I2O 404? Clay, wet I 2O I7,< 1C 2C Gravel, wet IOO I2O 2^-40 Ashes 404.5 2540 Coke ?o 30-4? Earth, dry 80-90 2040 Earth, moist 9OIOO 7.545 Earth, wet IO5I2O 1730 Broken stone, wet. . . IOO 7 {{40 Coal, broken 56 45 5O Sand, dry 90-110 7Q 35 Sand, moist IOO-IIO 7045 Sand, wet IIC I2S 15 7Q Water 62 q o The coefficient of friction between the usual masonry materials upon themselves or upon soil will range from 0.50 to 0.75. RETAINING WALL A retaining wall may fail by being rotated about the toe A or by sliding upon the base AB. There is usually little likeli- hood of failure by sliding as the coefficient of friction between the 278 GRAPHICS AND STRUCTURAL DESIGN wall and the soil is high. The force F due to the earth pressure behind the wall creates an uneven pressure under AB as shown in Fig. 298. A tilting of the wall may result from the side at A settling faster than that at B. To improve the condition it is not necessary to enlarge the entire wall but the base may be spread as shown in Fig. 299, so that the resultant R passes through the center of the base AB. D C FIG. 298. FIG. 299. At best, the theories relating to retaining walls are unsatis- factory. This is due largely to the fact that the properties of the materials vary so widely that it is difficult to decide upon the con- ditions applicable to a particular case. The practical conditions differ greatly from the approximations necessary in the theory. The weight of the fill and its angle of repose will vary greatly as the fill is wet or dry, clean or dirty, loose or rammed, etc. Other factors, such as shock, unexpected loading and frost, cannot be taken into account in the formulae. According to Trautwine a practically vertical retaining wall sustaining a fill of sand, gravel or earth, without surcharge, with the fill loose, not rammed, the wall being of good common rubble or brick, should have a thickness at the top of the footing of four-tenths the height of the wall. The Engineering News in its issue of RETAINING WALLS 279 Sept. 26, 1912, commenting upon the failure of a retaining wall, recommends the calculation of a retaining wall upon the assump- tion that a wall with no surcharge carries a load at least equiva- lent to a mass of water behind the wall of two-thirds its depth. The following problem will illustrate the method of designing a retaining wall. Problem. Design a section for a retaining wall, height 20 ft., weight of earth no Ibs. per cu. ft., natural slope 30 degrees, maximum earth pressure not to exceed 5000 Ibs. per sq. ft. Assume that the concrete of the wall weighs 140 Ibs. per cu. ft. The center of gravity of the wall must be located with reference to the back of the wall. Dividing the wall section (see Fig. 300) into rectangles and triangles, multiplying their several areas by the distances of their centers of gravity from the back of the wall and then dividing the sum of these moments by the total area of the wall section gives the center of gravity of the entire wall section as 46.6 ins. from the back of the wall. The area of the wall section is 143 sq. ft. The weight of i ft. of length of the wall is 143 X 140 = 20,020 Ibs. The resulting earth pressure is E = %-w-p-y = | X no X 12.2 X 14 = 9394 Ibs. The resultant R passes 9.3 ins. to the left of the center line a-a of the base. Using the formula deduced for footings with vary- W ing pressures on the soil, page 232, f e = j-jr 2 (B 6#), here W = 25,000 Ibs. and is the vertical component of E and the weight of the wall, b = i foot, and the maximum pressure under the foundation is 6 Since An inspection of the wall will show that by increasing the width of the footing to 12 ft., as shown in Fig. 301, the resultant 280 GRAPHICS AND STRUCTURAL DESIGN would be brought practically through the center of the base thus making the pressure uniform across it. To prevent sliding the coefficient of friction would have to be 8100 -5- 25,000 = 0.324. As the coefficient of friction probably lies between 0.500 and 0.750 there is evidently ample margin of safety against failure in this way. 8100* FIG. 300. REINFORCED-CONCRETE RETAINING WALLS The use of reinforced concrete in retaining walls has led to some modification of the ordinary design. This results in con- siderable economy in walls exceeding 25 ft. in height. The com- mon section of such a wall is shown in Fig. 302. The vertical load upon the soil is the weight of the wall plus the weight of the soil carried directly by the wall, that is, the soil prism abed and the vertical component of the force E acting upon the back of the wall. The general lines of the reinforcement are indicated by the steel shown in Fig. 302. The buttresses B are placed at intervals of from 8 to 10 ft. along the wall. Walls under 18 ft. RETAINING WALLS 281 k |000B- >| 282 GRAPHICS AND STRUCTURAL DESIGN may have the buttresses omitted and the wall designed as a cantilever beam. Problem. Design a reinforced-concrete wall for a height of 25ft. Assume the weight of the fill as 100 Ibs. per cu. ft. The angle of repose of the material behind the wall can be taken as 30 de- grees. The maximum soil pressure is not to exceed 2 tons per sq. ft. The working unit stress in the steel is not to exceed 12,000 Ibs. per sq. in., while that in the concrete is not to be over 500 Ibs. per sq. in. The fill abed, Fig. 303, in the wall will act to prevent over- turning, so that we will first want the weight and the center of gravity of the wall, buttresses and prism abed of the fill. Section. Volume. Weight. Arm. Moment. Coping 25X140= 3>5oo 9- 33 32,670 Face of wall . . 217.5X140= 30,450 9.50 289,275 Fillet . . 7.5X140= 1,050 IO. 25 10,760 Fillet . 7.5X140= 1,050 10.83 11,372 Base 280X140= 39,200 7.00 274,400 Base 15X140= 2,100 0.75 1,575 Buttress 128.1X140= 17,934 5.55 99,534 Buttress 22 9X140= 3>2O6 8.67 27,785 Fill, prism over base I7C9 CX IOO=I7^, 9^O 4.50 791,775 Fill, prism over buttress 160.0X100= 16,000 3.00 48,000 Total for 10 ft. length 290,440 1,587,146 Distance to center of gravity of wall, x = ^ '' = 5.46 ft. 290,440 E = I (w p . y) = \ (100 X 14-43 X 1443) = 10,400 Ibs. For a length of 10 ft. this is 104,000 Ibs. The diagram made to an enlarged scale of Fig. 303 gives the line of action of the resultant of E and the total weight, 290,440 Ibs., as passing through a point 35.8 ins. to the left of the vertical line through the center of gravity. The distribution of this RETAINING WALLS 283 pressure upon the foundation is given by the formula = 14, * = = j 44 f t 12 and falls to the right of the center, hence Sliding of the Wall. The coefficient of friction between the wall and the soil will lie between 50 per cent and 75 per cent. The resistance to sliding, even assuming the coefficient at 50 per cent, would be 290,440 X 0.50 = 145,220 Ibs. This, being con- siderably above the horizontal pressure on the wall, 104,000 Ibs., indicates little possibility of failure by sliding. REINFORCEMENTS Face of the Wall. The wall will be tied into the buttresses W *L and the bending will be assumed as M = - The reinforce- 12 ments will be determined for sections down the wall and the bending moment will be calculated from the load taken from Fig. 304. As the pressure is assumed as varying proportionally to the depth, the distribution will be represented by a triangle. The area of this triangle equals 104,000 Ibs., or 104,000 = 0.5 X ac X ab or ab = (104,000 X 2) -s- 25 = 8320 Ibs. The load upon a strip i ft. wide and 10 ft. span is then given by an intercept in the triangle abc parallel to ab, the intercept being located the same distance from c that the center line of the section is from the top of the wall; thus the load on a strip 12 ins. wide whose center is 10 ft. from the top is 3450 Ibs. when scaled from Fig. 304. 284 GRAPHICS AND STRUCTURAL DESIGN W L The bending moment upon such a strip is M = - - = 3450 X 10 X xf = 34, 500 in. Ibs. Using the approximate formulae M s = J X^ Xf 8 Xd. Taking d = 10.5 ins. and/ 8 = 14,000 34,500 X 8 makes A = i2 - LLi2 = 0.296 sq. in. 7 X 14,000 X 10.5 f-in. bars would require a spacing of Area of i-in. X 12 0.196 X 12 . _ . = - - . approximately 8 ins. 0.296 If desired, this can be checked by the more accurate formula. 0.296 p = = 0.0024. 10.5 X 12 p *n 0.00235 X 15 = 0.035. From the curves j = 0.924; k-j = 0.231. Substituting these values in formulae (3) and (4) M a = A Xf 3 Xj'd or /. 0.296 X 0.924 X 10.5 AXJ-d = 12,000 Ibs. r 2 M c 2 X 34,500 and fc = 7 -^ = - T = 225 Ibs. k-jXb.d* 0.231 X 12 X io.5 2 These values are satisfactory, it not being desired to make the wall thinner. The J-in. round bars will be spaced 8 ins. center to center in the section between 8 ft. and 10 ft. from the top of the wall. The reinforcements may be determined for the other sections in a similar way. The results of such calculations are given in the table. The calculated spacing is given in the brackets, while that to be used is given outside. RETAINING WALLS 285 Depth of section, feet. Load, pounds. -^- inch pounds. Area of steel, square inches. Size of bars and spacing, inches. 4 1400 14,000 o. 109 | " spaced (21.6) 12" 7 2400 24,000 0.187 spaced (12.6) 12" IO 345 34,500 0.296 "0 spaced (8 ) 8" 12 4100 41,000 0.318 -2-"0 spaced (7.4) 7-5" 14 4800 48,000 0-373 \" spaced ( 6.3) 6" 16 5400 54,000 0.420 \" spaced (5.6) 5.5" 18 6100 61,000 0.474 \" spaced (4.96) 5.0* 20 6750 67,500 0.524 \" spaced ( 4.5) 4.5" 22 7400 74,000 0-575 \" spaced ( 4.1) 4.0" The toe or base of wall to the left of the face will carry a load given in Fig. 305 and equals = 2975 Ibs. per sq. ft., making a load per foot of wall of 2975 X 4 = 11,900 Ibs. The distance of the center of gravity of this loading from the face of the wall is x = 3350 + 2 (2600) x 48 = lbs 3350 + 2600 3 The moment on a strip 12 ins. wide then is M = 11,900 X 25 = 297,500 in. lbs. By the approximate formula M a = | X A X /. X d. A = 297,500 X 8 7 X 14,000 X 22 1.07 sq. ins. The spacing for i-in. bars is 0.785 X 12 1.07 = 8.75 ins. Rear Portion of Base. The load upon this section (see Figs. 305 and 306) will be the difference between the pressure upon the soil and the weight of the nil vertically over this portion. This amounts to 23 ft. of soil and 2 ft. of concrete, and weighs 2300 + 280 = 2600, approximately. The maximum loading is 2600 800 = 1800 Ibs. Estimating the reinforcement for maxi- mum loading the total load on a strip 12 ins. wide and 10 ft. long is 18,000 Ibs. 18,000 X io X 12 12 M = 180,000 in. Ibs. 286 GRAPHICS AND STRUCTURAL DESIGN Placing rods 3 ins. above the bottom of the slab and using the approximate formula gives, A 8 XM 8 X 180.000 A = - = - = 0.70 sq. in. jXp-d 7X14,000X21 rnu r 7 r* u MI u -6o X 12 The spacing of --in. bars will be - - = ioms., approx- 8 0.70 imately. W L W L In assuming the bending moment at If = - instead of 12 8 it becomes necessary to place reinforcement in the outer flange over the supports; this would have to equal at least 50 per cent of the reinforcement at the middle. Here short lengths of rods the same as the full-length reinforcements (see Fig. 306) will be placed over the supports and spaced the same as the other rods. Counterforts. In this design (see Fig. 302) these will resist the moment due to the thrust on the wall. The thrust on the wall from the top to the upper face of the wall is -% 5 X 23 = 87,975 Ibs. 23 X 12 The moment is 88,000 X - - = 8,096,000 in. Ibs. o The design can be made as a T beam, the wall serving as the flange. The distance from the face of the wall to the center of the reinforcement can be assumed as 108 ins. The ratio of flange thickness to depth of beam is -j 1 ^- = o.n. From the curves for T beams, j = 0.95, and substituting in the equation M = A X/ 8 Xj*d, we have M 8,096,000 A = : : = -^-^ = 5.65 sq. ins. f a Xj-d 14,000 X 0.95 X 108 This will require eight yf-in. rods. The number of rods can be reduced towards the top of the wall as the bending moment becomes less. RETAINING WALLS 287 In a beam similar to the counterfort the flange force at any distance x from the top can be approximately expressed as (F x 2 } F x = v , where h = height of wall whose flange force is F. This is not accurate, as j will vary with the depth of the beam, but may be approximated at T 9 . The flange forces will be pro- portional to the required areas. The varying flange forces can then be represented by laying off a base line equal to the height of the wall and plotting ordinates according to the formula just stated. The curve being a parabola can be laid off as in Fig. 307. ca represents the side of the wall, here 25 ft., and cd represents the side extending to the top of the slab, de is the force in the steel 23 ft. from the top of the wall, gc represents the total length of diagonal steel reinforcements, the distance df measuring 8 ft. from the wall. Eight rods taken in pairs will divide de into four equal parts and dh, hi, ij and jb each represents the area of two of the rods. To find the lengths of the shortest rods produce jk perpendicularly until it cuts the curve, from k project a horizontal line until it cuts the diag- onal line in /. The lengths of these two rods will then be given by the length gl. As this curve gives the rate of change of flange force it also gives the horizontal shear which will be the difference between two adja- cent abscissas and the stirrups can be determined as shown for beams with uniform loads, page 211. The beam being so deep the concrete alone would probably supply ample strength for horizontal shear but in practice there are generally also metal stirrups inserted. Another type of wall is the can tile vered wall. In it the 288 GRAPHICS AND STRUCTURAL DESIGN buttresses or counterforts are omitted. This type is suited to lower walls than the design just carried out. The principal reinforcements are shown in Fig. 308: The calculation of the thicknesses and reinforcing metal can be made on the same general lines as the preceding problem. The vertical wall CD is treated as a cantilever and is assumed as carried by the base AB. CHAPTER XVIII BINS PRESSURES ON BINS THE 'methods used in obtaining the pressures upon retaining walls are also applicable to the pressures acting on the sides and bottoms of bins. Bins may serve as retainers of any material but the discussions here will be confined to such bins as are ordinarily used for coal, sand, refractory materials at steel mills, and similar materials. Bins at grain elevators are generally high compared with their width; it therefore follows that the friction of the material against the sides of such bins may greatly assist in relieving the lower portion of the bin, both sides and bottom, of this accumulative pressure. The theory previously given for retaining walls does not apply in this case as the assumed plane of rupture to produce maximum pressure will not cut the surface of the grain. Several formulae have been proposed for estimating these pressures ; that of Janssen, taken from Hiitte des Ingenieurs Taschenbuch, is and L = V-k. V = vertical pressure of the grain, pounds per square foot. / = coefficient of friction of grain on the bin wall. w = weight of grain, pounds per cubic foot. R = hydraulic radius of horizontal section of bin. For circu- ,. diameter ,., , . D lar section R = -- For other sections R = 4 area of section perimeter of section 289 All dimensions in feet. 290 GRAPHICS AND STRUCTURAL DESIGN k = ratio of lateral to vertical pressure. e = base of Naperian system of logarithms. h = depth to bottom or point on side at which pressure is to be determined. For wheat, w = 50 Ibs. per cu. ft.; k = 0.60; / = 0.40. SHALLOW BINS The bins now taken up will be shallow compared with those just considered. In these the theory of retaining walls will apply. These bins will commonly be made of, (a) Timber, (b) Steel, unlined, or lined with timber or concrete, (c) Reinforced concrete. The bin should be built or lining placed to avoid pockets that might hold materials indefinitely. This is particularly neces- sary in materials in which spontaneous combustion might occur. Timber lining should be tar coated on the sides and faces against the metal, when such treatment will not affect the material held. The inclined sides of unlined steel bins frequently have wear- ing strips, flats about 3 X Jin., placed on about i2-in. centers and running at right angles to the direction of flow of the material, thus tending to retain a slight thickness of the contained material along the side or make the material slide upon itself rather than on the metal and so protect the metal from wear. STRESSES IN BESTS Although the forces acting upon the bin sides may be esti- mated by formulae, the graphical methods used on retaining walls will be largely used in the following discussions. For those who prefer formulae those given by R. W. Dull in the Engineering News of July 21, 1904, are here given, with slight modification. BINS 291 where \n + i / 2 cos 5 __ / cos \ 2 wh 2 IV = > \ + I / 2 + 5) sin cos 5 J:_MI_.J A FIG. 309. Vertical wall, no surcharge. where If _w^_ /cos^y 2 cos 5 \i+n ' ! 2 _ * /sin ( -f- 5) sin (<#> a) V cos 6 cos a a = , then w = o. FIG. 310. Vertical wall, surcharged. 2 9 2 GRAPHICS AND STRUCTURAL DESIGN E = where _ 4 /sin ( + a) V cos 5 cos a N-* E J. FIG. 311. Vertical wall, surcharge negative. where wh* 2cos08 + 5) X = cos Q3 - ^>) "I 2 ' 2 cos 03 + 5) (w + i)cos/Sj + 5) sin cos (5 +/S)cos/3 FIG. 312. Wall inclined outward. 6 < 90 + 5. Surface horizontal. BINS 293 where E = wh* 2058(5+0) X i)cos/3 _ wh 2 cos 5 ["cosQfr -/3) "I 8 2cos(S+/3) X L(w + i) cos/3 J ' sin ( + 5) sin ( a) cos (5 + 0) cos (ft a) FIG. 313. Wall inclined outward. 6 < 90 -f 5. Surcharged. IF = weight of triangle ABC = E = ( 45 <>-f) tan i E = Vw* -h Nf, _tan/3 tan 2 - /3) and T = C FIG. 314. Wall inclined outward. > 90 + 8. Surface level. 294 GRAPHICS AND STRUCTURAL DESIGN Va-cos 2 * x whj cos a + v cos 2 a cos 2 2 w = E = acts parallel to line of fill. w sin /3 cos (a /?) 2 cos 2 ft cos a FIG. 315. Wall inclined outward. 6 > 90 +5. Surcharged. where tt-tarf (-) "(*-*>, ,. #2 E = / cos0 y + i/ 2 cos 6 2 and 4 /s n=\ V sm (0 + 5a) sin cos 2 tan/3 tan ^ = JF 2 , 2 Nz sin 5 2 w (H* - h*} tan 2 (-!) Q = E cos G* /3); T 1 sin (/* j8). NOTE. The quantity Nz sin 6 2 is the friction on the vertical side and reduces the weight acting upon the side AB. FIG. 316. Hopper bin, fill level. BINS 295 cos a. Vcos 2 a cos 2 xx w (hi 2 E\ = cos a X X cos a + v cos 2 a cos 2 sing cos (<*-/?) . ( ff_ cos 2 /3 cos a 2 / cos \ 2 wh 2 4 /sin #2=1- r , where n = V ~ \ n + i / 2 cos 62 sin ( a) cos 82 cos a = jVi 2 + W 2 + 2 AW sin a. Ei is parallel to the angle of the fill. FIG. 317. Hopper bin, surcharged, unsymmetrical . where n = W = g _ cos 2 cos a 2 cos (jj. - /8) and T = E sin On - /8). FIG. 318. Hopper bin, surcharged, sym- metrical. 296 GRAPHICS AND STRUCTURAL DESIGN The following coefficients of friction are given by Mr. Dull in the article just referred to. COEFFICIENTS OF FRICTION BETWEEN MATERIALS Material. Degrees.

50 9 0.426 2 = - rf . = 0_ x 2Q 2 x _ _ 2 i + sm< 2 !'574 The side 5C is acted on by the weight of the superimposed coal and the horizontal pressure E^. It is easier to consider the triangle of coal C77, and then having found the pressure on CJ the pressure on the section CB may be scaled from the diagram. The weight of the triangle of coal CIJ is W = \ X 2O 2 X 50 = 10,000 lbs. This resultant force acts at the center of gravity of this tri- /^* 7" angle or - // from CI. E% acts a distance CK = -- from the 3 3 bottom of the bin. Combining E% and W gives the resultant R = 10,300 lbs. This force resolved parallel and at right angles to the side BC gives T 5200 lbs. and N = 8900 lbs. The normal pressure of 8900 lbs. is distributed along the line JC as the intercepts in the triangle LJC. Putting the area of the triangle LJC = 8900 lbs. and solving for the side LC gives LC = 630 lbs., which represents the pressure per square foot at the point C. The pressure at B is found, by scaling the intercept BMj to be 310 lbs. The total pressure on the side BC is the area of the trapezoid BCLM and equals 6580 lbs. The resultant of this pressure acts through the center of gravity of the trapezoid. The forces acting upon the points of the bin with supports every 12 ft. are given in Fig. 320, and are the force at A = (-|--) X 12 = 2700 lbs. The horizontal force at B = (f) X675 X 12 = 5400 lbs. The force at B normal to the side BC must be estimated as 298 GRAPHICS AND STRUCTURAL DESIGN follows. The total normal pressure on BC is 6580 Ibs. CQ = 6.5 ft. CB = 14.1 ft. The resultant at B is, therefore, (6580X6,5} = Q lbs 14.1 The total on 12 ft. of bin is 3040 X 12 = 36,480 lbs. The balance of the load represented by the trapezoid MBCL acts at C and is (6580 X 12) - 36,480 = 42,480 lbs. The vertical pressure at C is CD X DH X^Xi2=5X2oX 50 X 12 = 60,000 lbs. The load on the columns due to the fill will be the total weight of the coal in the bin. The area of the bin section is 500 sq. ft., making the load in 12 ft. 500 X 12 X 50 = 300,000 lbs. The trussed bracing will be carried to the points A, B, C, E, F and G of the bin and taking these forces together with the total column load the several forces acting in the frame may be determined. Surcharged Bins. A bin similar to the last one but sur- charged (Fig. 321) may have the forces acting upon it analyzed in much the same way. First find the pressure upon the side AB] this is done graphically in Fig. 322. The diagram gives p = 8 ft. and y = 8.5 ft., from which E l = ^XwXpXy = | X 50 X 8 X 8.5 = 1700 lbs. The normal pressure is 1620 lbs. Now finding the pressure upon IE, the axis of the bin, by using the graphical method and assuming 1 = o, we have p = y = i6ft. Ez = 0.5 X w X p X y = 0.5 X 50 X 16 X 16 = 6400 lbs. The center of gravity of the triangle IJH is at M and the weight of a prism of coal i ft. high with the base IJH is (35.4 X 20.7 X 5) = I8;300 lbs The resultant of EZ and W is R = 19,300 lbs. This is assumed as varying uniformly along the face //. The pressure at 7 is x = (19,300 X 2) = 19-3 BINS 299 The total pressure on the side BC is that represented by the trapezoid QBCS whose altitude scales 12.8 ft.; the area then is 600 + 1550 g _ 2 FIG. 321. This acts through the center of gravity O of the trapezoid which has been located graphically. The normal pressure on the side BC may now be found by laying off 13,760 Ibs. through the point and resolving it into its com- ponents parallel with and at right angles to BC. The load upon one-half of the base CD is 7210 Ibs. The forces, due to the fill, acting at FlG the points of the bin may be esti- mated as in the preceding problem and are given in Fig. 323. The distance between supports has been taken at 12 ft. The columns Column Load 232400* 3 oo GRAPHICS AND STRUCTURAL DESIGN will carry the total weight of the coal in the bin and the weight of the bin. The stresses in the internal framework of the bin may be calculated by treating these external loadings as is done for any other truss. The stress diagram may be drawn or the forces calculated by any of the methods previously given. SUSPENSION BUNKERS If the side plates of the ordinary hopper bunker are permitted to bulge but slightly they must be supported at frequent intervals by beams or other structural shapes. This added weight is avoided in the use of suspension bunkers. These bunkers are patented and are designed with the idea that the sides resist tension only. To fulfil this condition theoretically the sides would assume a shape peculiar to each possible loading. This could be illustrated by a model bin whose sides were a purely tension piece like muslin or duck. However, an actual bin differs greatly from the above illustration and any deflection due to bending in the sides when the bin is not loaded as theoretically assumed is ordinarily not sufficient to cause trouble. The lined bins will be stiffer than the unlined ones. The theoretical curve of a bin for any assumed loading may be deter- mined as follows. In Fig. 324, given L n SUSPENSION BUNKERS. FIG. 324. desired span sag, the load FIG. 325. the and in the left-hand side of the bin will not vary much from a triangle ABC. The weight of the material in the portion ABC of the bin will act through the center of gravity of this triangle ABC and lies in the vertical line DG or W, GF BF i J J 1 being 3 BINS 301 Considering the side of the bin BE the force acting at B and that at E must hold W in equilibrium; hence they must pass through a common point D. The line of action of W is known as is also the line of action of the force acting at E which must be horizontal if the bin is sym- metrical and symmetrically loaded, both of which conditions have been assumed. If then a horizontal line is drawn through E it cuts W in D and the line of action of the force acting at B must lie in the line BD passing through the point D. The loading being assumed as varying as the intercepts of a triangle we may represent this loading by the triangle FBC, and having divided it into sections of equal widths the weight of each section will be proportional to its center ordinate; these have been shown dotted. These lengths from 1-2 to 5-6 have been laid off on the load line of the force polygon, Fig. 325. The lines of action of the forces acting at B and E being known the pole is readily found by drawing Oi parallel to DB and 06 parallel to DE. The other strings may now be drawn in the force polygon. If we complete the equilibrium polygon in Fig. 324 the broken- line side will approximate the theoretical side of the bunker. On the right- hand side the line of the bunker has been drawn as it is com- monly made in practice. The portion HI is straight while the lower part HE is an arc of a circle. The sides of this bin approximate an equilateral triangle. The sag is about six-tenths of the span. For a bin which is nearly an equilateral triangle the following figures will be approximately true. Area of the bin section, fill level, = 0.40 S 2 . Area of the bin section, triangular surcharge, = 0.57 S 1 . Length of plate, no allowance for laps, = 19.8 S (inches). Force in plate per foot of length at B or / = 0.60 W (pounds). Force in plate per foot of length at = 0.30 W (pounds). Here 5 = the span in feet. W = the load in the bin for one foot of its length, pounds. GRAPHICS AND STRUCTURAL DESIGN FIG. 326. In the case of the suspension bunker with vertical sides above the points of suspension, B and 7 of the bin, Fig. 326; the cross section of the bunker will be increased by the area BCIJKL equal to S X h, and the load will be increased pro- portionally. In this case the weight of the material in one-half of the bin will act through the center of gravity of the section KLBDEK. This will be a greater distance from the cen- ter line EC than was the case in the preceding problem, and will make the bin somewhat wider below BI than at corresponding points in the former bin. W In either problem if -- is known the forces acting at B and E may be found by drawing the triangle B 1 E 1 D 1 at the side of the bin as in Fig. 326. One form of Suspension Bunker with column supports for boiler rooms is shown in Fig. 327. This bunker is intended to hold 6600 Ibs. per ft. of length. Assume coal as weighing 50 Ibs. per cu. ft., and the sag of the bunker as 60 per cent of the span. The bunker will be assumed as carrying the required load when surcharged, and the angle of repose of the coal will be taken 30 degrees. Determine the bin area by using the formula, A = 0.57 S 2 . The required area is ^-f^- = 132 sq. ft. From this 132 = 0.57 S 2 and S = 15.3 ft. If the bin is made 16 ft. wide the sag will be approximately 10 ft. Spacing the columns 20 ft. center to center makes the load of coal upon each column - = 66,000 Ibs. Adding 4000 Ibs. to this to cover 2 the weight of metal, exclusive of columns and under bracing, makes a total of 70,000 Ibs. at each column. This load laid off in Fig. 328 gives the stress in the side of the bunker per foot of BINS 303 length as - - = 4000 Ibs. The unit stress per square inch 20 in a plate - in. thick is 4000 = 1330 Ibs. - in. tmcK is / 4 (0.25 X 12) The main columns are subjected to a bending moment of 80,000 [(16 6) 6] = 320,000 in. Ibs. FIG. 328. Trying two i2-in. channels weighing 20.5 Ibs. per ft., we have Fiber stress due to the bending moment is , ' ^- = 7,400 Ibs. per sq. in. , \2 X 21. Fiber stress due to direct loading is 80,000 (2 X 6.02) Combined stress = 6,650 Ibs. per sq. in. = 14,050 Ibs. per sq. in. The direct load upon the column being 6650 Ibs. per sq. in. the 304 GRAPHICS AND STRUCTURAL DESIGN radius of gyration for an assumed length of n ft. may be found from the straight-line formula, / = 16,000 70 M = 16,000 70 70 X 1^2 or r = yf = 0.99. (16,000 6650) Two i2-inch channels are to be used back to back. The radius of gyration of a 1 2-inch channel referred to an axis at the back of the web is _ Ah = * _ v /3. 9 o + (6.02 X o.yo 2 ) T , V ~~" It therefore follows that two i2-in. channels at 20.5 Ibs. per ft. either riveted back to back or separated any distance will carry the load of 80,000 Ibs. applied axially. The horizontal strut under the bin carries a load of 40,000 Ibs. Its length is 84 ins. Trying a lo-in. channel at 15 Ibs. per ft., whose radius of gyration about an axis parallel to the web is 0.70, we find - = *- = 1 20. The allowable stress is / = 16,000 r 0.70 (70 X 120) = 7600 Ibs. per sq. in. The required area then is - = 5.26 sq. ins. The strut across the top of the bin 7600 carries the conveyor, its truss and the conveyor loading and in addition to these the direct compression due to the bin loading. Assuming the load equivalent to a central load of 3500 Ibs. the bending is M = - - = 3500 X 16 X = 168,000 in. Ibs. Trying 4 4 a i2-in. I beam at 25 Ibs. per ft., with an - value of 24.6 and an c area of 7.34 sq. ins.; we have Fiber stress due to bending moment, ' = 6800 Ibs. per 24.0 sq. in. BINS 305 Fiber stress due to direct loading, = 5450 Ibs. per sq. in. 7-34 The combined fiber stress is 6800 + 5450 = 12,250 Ibs. per sq. in. Considering that this beam is held laterally at its center by the conveyor truss this fiber stress would seem permissible. The braces occur every 10 ft. The side of the bin between columns is carried by a girder. The total load is 80,000 Ibs. or 4000 Ibs. per ft. The bending moment then is Wl 12 M = - - = 80,000 X 20 X -- = 2,400,000 in. Ibs. 8 o Assuming the distance back to back of flange angles as 3 ft., the distance between the centers of gravity of the two flanges will be about 34 ins. M = A Xf X h = 2,400,000 = A X 16,000 X 34 or A = 4.42 sq. ins. Two 5 X 3^ X TV m - an gl es having a gross area of 5.12 sq. ins. and a net section, allowing for one rivet, of 4.52 sq. ins. furnish just a little more than the area required. No allowance has here been made for the web acting as flange. This gives additional security. Figure 329 shows a bin designed for bituminous coal. The weight of the coal was taken 50 Ibs. per cu. ft., the natural slope was taken 30 degrees, while the angle of friction between the sides and the coal was assumed as 18 degrees. The diagram gives p = 1 6 ft. and y = 17 ft. E = %Xw- p. y = %X $0X16X17 = 6800 Ibs. The horizontal component of this force is 6500 Ibs. The hori- zontal pressure per square foot on the sides of the bin at the bottom is x = 6500 X -$$ = 433 Ibs. The sides are supported by I beams spaced 6 ft. center to center and horizontal rods are run across the bin at 6-ft. intervals and placed 10 ft. above the bottom of the bin. The stresses were 306 GRAPHICS AND STRUCTURAL DESIGN determined graphically as outlined for fixed and continuous beams. The force in the tie rods was 34,200 Ibs. and the maxi- mum bending moment on the I beams was found to be 507,000 in. Ibs. Bolts if ins. in diameter, having an area of 2.05 sq. ins. at the root of the threads, would have a unit stress of 34>20 = 2.05 16,500 Ibs. per sq. in. and are satisfactory. Allowing 18,000 Ibs. per sq. in. fiber stress in the beam would require an -value of - = r = S ?' 000 = 2 g 2 This W ould e e f 18,000 require a lo-in. I beam at 35 Ibs. per ft. or a i2-in. I beam at 31.5 Ibs. per ft. This bin was tied across the top by the roof framing and was braced transversely at intervals against wind pressure. BIN DESIGN Figure 330 illustrates a bin for coal storage over gas producers. The upper portion of the bin was a square of 13 ft. side. The lower part was a curve with a g-ft. sag. The coal is assumed as weighing 50 Ibs. per cu. ft. and its angle of repose has been taken 35 degrees. The angle of friction between the coal and the sides of the bin has been taken 18 degrees. Owing to the large amount of coal carried in the square portion of the bin the loading cannot be represented as a triangle; therefore an approximate outline of the bin has been assumed in Fig. 330 and the actual curve estimated from this. The left side of the bin has been divided into four strips of equal widths; the weight in each strip per foot of length of bin will be proportional to the middle ordinate, shown in full lines for each section. In the force polygon, Fig. 331, the forces may be represented by these middle ordinates or by a constant percentage of them. In the figure the scale has been taken as one-fourth of these lines. The forces A l B l , B l C l , etc., have been laid off on the right-hand side of the bin equal to and symmetrically placed with those on the left-hand side. These BINS 307 forces have been laid off on the load line of the force polygon, Fig. 331, and the vector diagram, on its left-hand side, has been used to locate the center of gravity of the load in the right-hand side of the bin. The center of gravity is found in the inter- section of the lines i and 5 in Fig. 330. R, the total weight of the coal on this side of the bin, acts through this point of inter- section. The force in the extreme low point of the bin will be horizontal, and this together with the force in the bin side acting through c must hold R in equilibrium, hence these three forces must pass through a common point b ; this gives the direction of be, which is the theoretical direction of the side of the bin at the point c. The lines for the other side of the bin have been made similar to those for the right-hand side. The vector polygon drawn on the right-hand side of the load line, Fig. 331, has been used in drawing the dotted lines in Fig. 330, thus showing the agreement between the actual and the theoretical lines of the bin. Before the stresses can be read in Fig. 331, the scale must be found. GRAPHICS AND STRUCTURAL DESIGN The mean ordinate in one side of the bin is 21 ft. The bin being 13 ft. wide makes the volume of the bin section for a length of i ft. = 21 X 13 = 273 cu. ft. This gives 273 X 50 = 13,650 Ibs. of coal per ft. of bin. The load line AE = 6825 Ibs. This makes the scale $-&- = 1700 Ibs. per in. (approximately). The force A0 l = 4.3 X 1700 = 7300 Ibs. The force E0 l = 1.6 X 1700 = 2720 Ibs. The bin plates will be made f$ in. in thickness to allow for wear and deterioration, and will have 3 X f-in. wearing strips spaced about 12 ins. center to center on the curved portion of the bin sheets and running the length of the bin. The distance center to center of columns will be assumed as 18 ft. Trans- verse braces will be placed at the columns and every 9 ft. FIG. 332. FIG. 333. The pressure against the vertical sides of the bin is determined in Fig. 332. E = % Xw p -y = | X 50 X 10.5 X n = 2890 Ibs. It should be noted that for so narrow a bin this does not give the accurate pressure but it is the limiting pressure to which this force on the side of the bin could extend. A closer approxi- mation might have been made by assuming a bin, say, 16 ft. high instead of 13 ft. but with a horizontal fill. The horizontal component of this force of 2890 Ibs. is 2700 Ibs. Two-thirds of this, or 1800 Ibs., act at the lower points e and c, while the other third act at the upper points / and g. These are each 900 Ibs. These forces act in the opposite direction to the pull from the suspended portion of the bin. BINS 309 Had the bin been level, substitution in the formula for the pressure on the side of the bin would have given 1060 Ibs. or 700 Ibs. at the points e and c; the net pressure here then would be 2700 700 = 2000 Ibs. per ft. of length of bin. The compression in the lower member of the brace due to the total load between the columns is 2000 X 18 = 36,000 Ibs. If the value of - in these braces is to be limited to 120 then r = 13 X iV\ = i-3; and this requires two 6 X 3^-in. angles having their long legs placed back to back. Before the weight is selected the possibility of additional loading from wind should be con- sidered. In exposed places this might reach 30 Ibs. per sq. ft. of vertical projection. Along the bin at the point e it would equal (6.5 + 9) X 30 = 465. This makes the total load at the column (465 X 18) + 36,000 = 44,37 lbs - The unit stress on the angle is f 16,000 f 70 X - J = 7600 lbs. per sq. in. The net area of the angles is ^* = 5.83 sq. ins. 7600 The minimum angles should therefore prove ample, unless larger angles are used to provide for corrosion and possible injury from the coal passing over them. The channel at e is acted upon by the wind and the pull from the side of the bin. Considering first the wind load, it will act as a uniform load on the span of 18 ft. Its deflection under this load will put a central load on the channel at c. If P is the load transferred to the channel at c from the channel at e, the de- r /W . /3v flection due to the wind load W is - - X ( - ) , while the de- 384 \h*lj i iP 1 3 \ flection due to the central load is -- X [ ;.) Since the 48 \Jb L / deflections of the two channels are equal, 310 GRAPHICS AND STRUCTURAL DESIGN _ __ __ 384 El ' 48 El ~ 48 El and = L or P = W = 0.31 W. 384 24 384 The maximum bending moment may now be found by adding algebraically the bending moments due to these several loads. The moment due to the wind load of 465 Ibs. per ft. is M = W - - = 465 X 18 X 18 X -V 2 - = 226,000 in. Ibs. 8 The bending moment due to the central load is M = P .- = 0.31 X 465 X 18 X 18 X -V" = 140,000 in. Ibs. 4 The bending moment due to the uniform load of 2000 Ibs. per ft. on the p-ft. span, that is between the braces, is M =W-- = 2000 X 9 X 9 X -V 2 - = 243,000 in. Ibs. 8 The resultant bending moments are given by Fig. 333, and are the intercepts between the curve abc and the lower curves, aedjc. The curve abc gives the moments due to the wind load acting on the i8-ft. span and measured from the base line ac. The moments due to the central load on the i8-ft. span are then deducted by laying off the triangle adc. The bending moments due to the uniform load on the 9-ft. span are then plotted below ad and dc, thus giving the algebraic sum of these bending mo- ments. The maximum moment scales 340,000 in. Ibs. and would , , / , ,IM 340,000 require a channel having an - value of - = r = ^- = 2I -3> e e f 16,000 which calls for a i2-in. channel at 20.5 Ibs. per ft. The fiber stress in this channel would then be ^ - - = 15,900 Ibs. per 21.4 sq. in. In a similar way the maximum bending on the upper channels is found to be approximately 139,000 in. Ibs. and requires a section modulus of -~ - =8.7. 16,000 BINS 311 The sides act as girders to transfer the loads to the columns. The uniform load due to nil is 13,650 Ibs. per ft., and assuming the weight of the bin as approximately 1000 Ibs. per ft. makes the total load per foot 14,650 Ibs. The load per foot on each girder is * 4 ' 5 = 7325 Ibs. The bending moment on the girder is M = ~l = 7325 x 18 X v = 296,660 ft. Ibs. o o If the girder is assumed as 13 ft. deep the flange force is 206,660 22,800 Ibs. J 3 When it is considered that in the lower channel the splice plate to which the curved portion of the bin is attached assists in carrying this stress it will be seen that this force of 22,800 Ibs. will add very little to the fiber stress in the i2-in. channel. If lo-in. channels at 15 Ibs. per ft. are tried for the upper channels the fiber stress due to the bending is . M-e 139,000 / = = - - = 10,400 Ibs. per sq. in. * I 3-4 The fiber stress due to the flange force of 22,800 Ibs. is 22,800 - = 5100. 4.46 The combined fiber stress is 10,400 -f 5100 = 15,500 Ibs. per sq. in. The stresses may be increased by wind acting on the ends of the bin or building and also by thrust from the traveling crane, or the forces may be provided for by additional bracing. The stiffeners on the sides ef and cq of the bin carry a triangular load of 2700 Ibs. per ft. of length of the bin; hence if the stiffeners are spaced 2 ft. 3 ins. center to center, their total load is 2700 X 2.25 = 6070 Ibs. The moment on a beam due to such a load is M = 0.128 XP-/;M= o.i28X 6o7oX 13 X 12= 121,000 in. Ibs. 312 GRAPHICS AND STRUCTURAL DESIGN = 7-6, The section modulus required is / _ M 121,000 e f 16,000 calling for 8-in. channels at uj Ibs. per ft. The maximum stress in the -^g-in. plates along the side of the bin will occur at e. Since the load of 2700 Ibs. is assumed as distributed as a triangle of pressure, it follows that if p e is the pressure per square foot at e, then 2700= -3-2s and ^ = 415 Ibs. per sq. ft. lo"Channel15lfper foot 3-6x4x^1? \ ,10'p* \ 20 Gal. steel unless otherwise specified x 18 Straps spaced every I'O "riveted to corrugated steel Straps every 4 $" - - PLAN FIG. 353. The design shown in Fig. 353 is for fixed sash for monitors; for swing monitor sash cut stops off as shown by dotted lines, and omit head stop on the inside. Make frames and sash of white pine, excepting spiking and blocking pieces, which should be spruce, hemlock or Norway pine, planed on all exposed sides. For swing sash order two trunnions for each sash, and call for lever operating device. 33 GRAPHICS AND STRUCTURAL DESIGN Cor. Steel /Purlin Tin net's Nails N-* ^_ :r-^ i /Y\ o* I "F" Muntfn Bo Top Kail m"Rafl ^Cor. Steel l-^'Wund ELEVATION -W=Glass., PLAN FIG. 354- The design shown in Fig. 354 is for continuous fixed sash in corrugated steel sides. Make sash and sill of white pine, planed on all exposed sides. WALLS AND ROOFS 331 COr ' Steel Las Screws yfkl*> r^!^P^t^ r8 - Tn 53-, ~e = Gloss, Muntins PLAN FIG. 355. The design shown in Fig. 355 is for sliding sash in corrugated steel sides. The stop used for roller track is to be hard wood, other wood the same as recommended for Fig. 353. 332 GRAPHICS AND STRUCTURAL DESIGN Glass- -- -Glass- - -Glass- *nd-* W - Glass, Muntins, and 4}"=4'o Mast **j f PLAN FIG. 356. The design shown in Fig. 356 is for a window frame with counterbalanced sash in corru- gated steel sides. Neither dimension of sash exceeds 4 feet o inches. The wood is recom- mended the same as for Fig. 353. WALLS AND ROOFS 333 |Mi rlaening 1--*-=, Purln .---6 1 [_1 5 Top Rail Muntin 1 ;i||,. rf&t Meeting Rai tfi '* M * Tn Dimem bynui on of sash dt her and eize Muniin ormined >f lighW /. f-i ^Purlin P Bottom Rail V r iJUllllll U 1 -- r *t Purlin .. _._^ imtmil . ./ . ^__ --Cor Steel window in field ELEVATION Drip/111 \,*8ill 1*-X roiuid |T H x^X 1-W Screws SECTION y^2C^^ BM ^S B jC f~ W ~Gla86,MuDtins,and 4&Wl Min. - PLAN FIG. 357. The design shown in Fig. 357 is for a window frame with counterbalanced sash in corru- gated steel sides. In this design one dimension is not less than 4 feet i inch. The wood used is similar to that specified for Fig. 353. K' ' . -Glass^VH^ 2 334 GRAPHICS AND STRUCTURAL DESIGN &xl& Lag Screw 2"x3"Block Dimensi n of sash d by number and size Mantins, and4K*=i'0*Max. PLAN FIG. 358. WALLS AND ROOFS 335 i- Flashing Purlin In'. "I* 1 T'T 13* Lag Sore w 2x3"Block s? r ----- 4 r --^p r---^ FT- -ffy =' sssaa ^ ^ yj -- - 1 - T| ii._t - i -^ Mortised Top Rail I 1 Pulley - 1 i I 1 -, Blunt in i 4" 1 j M q * i 11 ? 1 : ; ? .9 Mectii ijjRail 3 -: '^r i f! x ! r Dimensi n of sauh det rmined i * .3 by uun " -r and size < f lighta 3 1 5 - l vsi- 1 1 7. *T i i 1 1 1 1 1 III o 1 1 i 1 t 1 1 Dot to in Rail -i'4 vl^f-^ 1 itapai =- = ""- 'fijtoptff - *- Purlin " \ D^VVPrX? 1 *" 1 ** 8 " 1 _ Cor-Steel ^ roUnd # ** 8S1U ~ > ELEVATION SECTION X * I h-5;^2X' jy JiX* 2; IJi" ' J* WGla88, Muntina, and l#*l'l"Min. -*{ PLAN FIG. 359- The designs shown in Figs. 358 and 359 are for window frames with double hung weighted sash in corrugated steel siding. In Fig. 358 neither dimension of the sash exceeds 4 feet p inches, while in Fig. 359 one dimension is not less than 4 feet i inch. The sills and cas- ings should be made of white pine; the jambs and parting strips should be made of hard pine; while spiking pieces and blocks should be made of hemlock or Norway pine, planed on all exposed sides. GRAPHICS AND STRUCTURAL DESIGN CRANE CLEARANCES AND WEIGHTS I . Not less than 3 " -Not lose thin 2" Ar 3 Truck Wheel-bsatf Capac- ity, tons (2000 Ibs.). Span. A. B. Wheel base. Weight of bridge, 2 girders. Weight of trolley. Rail. Ft. Inches. Ft. Ins. Ft. Ins. Pounds. Pounds. Lbs. per yard. ( 40 8 4 9 8 8,600- 13,000 4,500 ) 5 ) 60 8 4 9 8 o 16,500- 2I,OOO to [ 40 80 8 4 9 8 o 28,OOO- 33,000 5,000 ) ( 40 8 5 o 8 6 12,300- l6,OOO 6,000 ) IO < 60 8 5 o 8 6 2O,OOO- 24,000 to 40 ( 80 8 5 o 9 o 32,000- 37,000 8,000 ) ( 40 8 5 6 9 o 14,000- 23,000 9,000 ; 15 ) 60 8 5 6 9 6 23,000- 33,000 to 50 80 9 5 6 IO O 37,000- 48,000 10,000 ) ( 4-O 6 o 23,000 28,000 10,000 ) 20 5 if-W 60 _ 6 o 30,000 37,000 to So I 80 9 6 o 40,000- 52,000 12,000 $ ( 40 9 6 6 IO O 19,000- 34,000 12,000 ) 2S] 60 9 6 6 10 6 29,000- 45,000 to f 60 i 80 9 6 6 II 45,000- 61,000 15,000 ) ( 40 9 7 o II O 34,000- 37,000 14,000 } SO] 60 9 7 o ii 6 44,000- 49,000 to [ 60 ( 80 9 7 o 12 O 58,000- 66,000 17,000 ) ( 40 9 8 o 12 6 43,000- 49,000 16,000 ^ 40 < 60 9 8 o 13 o 60,000- 63,000 to [ 80 1 80 9 8 o 13 3 70,000- 82,000 20,000 ) ( 40 9 8 6 12 6 48,000- 57,000 24,000 ) 50 60 9 8 6 13 o 66,000- 73,000 to > IOO ( 80 9 8 6 13 6 85,000- 95,000 30,000 ) ( 40 10 9 o 15 78,000 60] 60 IO 9 o 15 3 70,000- 95,000 32,000 f 80 10 9 o IS 6 100,000-126,000 IOO ( 40 10 9 6 15 6 101,000 75 60 IO 9 6 IS 6 80,000-120,000 40,000 to / 80 IO 9 6 IS 6 120,000-144,000 ( 40 12 IO O IS 6 134,000 100 < 60 12 10 IS 6 94,000-161,000 56,000 15 I 80 12 10 IS 6 125,000-187,000 WALLS AND ROOFS 337 The maximum wheel loads are approximately 25 per cent of the bridge weight plus 50 per cent of the combined weights of the trolley and crane capacity. The dimensions of windows depend upon the light required and the size of the panes used. In shop construction the greatest possible amount of light is wanted but without sun. On this account a northern light is preferred. In one-story buildings this leads to saw-tooth or similar construction. The usual sizes of common glass can be obtained in any hand- book on Building Construction. The regular sizes of glass range from 6X8 ins. to 44 X 72 ins.; the dimensions up to 16 ins. varying by inches, while those above 16 ins. vary by 2 ins. and are the even inches only. Single-thick glass is about T X g in. thick. Double-thick glass is about ^ in. thick. Wired glass has the advantage of not falling out when cracked, thus affording increased fire protection. Translucent fabric is a substitute for glass that has some advantages. It cuts out a little light but when a sufficient amount is used it affords perfect lighting. It is impervious, elastic and burns with difficulty. The subject of methods of lighting is an interesting one and should be given careful consideration in buildings of any impor- tance. According to Ketchum it is common to specify that 10 per cent of the exterior surface of ordinary mill buildings and 25 per cent of machine shops and similar buildings should be glazed, this as a rule being divided between windows and skylights. ROOF COVERINGS Corrugated-steel Roofing. This is one of the commonest kinds of roofing materials. It is made by rolling the plain sheets with corrugations. These sheets range from No. 16 to No. 28, U. S. Standard gauge (0.063 m - to 0.016 in.). The corrugations range from T 3 g in. to 2.5 ins. A few companies roll sheets with corrugations of 5, 3 and 2 ins. 338 GRAPHICS AND STRUCTURAL DESIGN The sizes most frequently used are gauge Nos. 20 and 24, with 2|-in. corrugations, f in. deep. The lengths of sheets range from 6 ft. to 12 ft. A common length is 8 ft. Roofs covered with corrugated steel should not have a slope of less than 3 ins. in 12 ins. According to Kidder's Handbook they should be supported on about the following spans: Gauge. Span in feet. Gauge. Span in feet. 24 22 and 20 2-2.5 2-3 18 16 4-5 5-6 The lap at the end of the sheet should range from 3 to 6 ins. according to the slope. Slope i to 3 Lap 3 ins. Slope i to 4 Lap 4 ins. Slope i to 8 Lap 5 ins. The side laps vary from one to two corrugations; a lap of i| corrugations makes a very good joint; see Fig. 360. FIG. 360. Corrugated roofing is secured to the building at the purlins by nails when nailing strips are used, and by hoop-iron clips or bands when secured directly to the purlins; see Figs. 350 to 352. The riveting and nailing should be done at the top of the corru- gations. The Pencoyd Handbook gives the following crippling load in pounds per square foot: ooo X / X b X d\ W /Q8, t = thickness of the metal in inches. b = center to center of corrugations in inches. d = depth of corrugations in inches. WALLS AND ROOFS 339 length of span in inches. The same handbook gives the following safe loads in pounds per square foot on corrugated steel of 2^-in. corrugations and f in. deep: Span in feet. Gauge No. 3 4 5 6 2O 59 44 36 30 22 48 36 2Q 24 24 37 28 22 19 26 3i 23 18 16 Purlins should be spaced for a roof load of not less than 30 Ibs. per sq. ft.; spacing for lighter loads results in injury to the cor- rugated-steel joints when the roof is walked upon. The sizes most frequently used are No. 20 for the roof and No. 22 for the sides. The corrugated steel should be ordered in lengths sufficient to cover two purlins. The laps should be painted before being riveted. Galvanized steel will take paint better after weathering a while. When the buildings are lined, ij by f-in. corrugated steel is frequently used. When lined the corrugated steel is better nailed to nailing strips. The greatest objection to the corrugated-steel roofing is the fact that it sweats. In heated buildings or in rooms with moist air the moisture condenses on the under side of the roof and drops to the floor, possibly striking machinery or materials it may injure. A common method of overcoming this condensation is to cover the purlins with galvanized poultry netting, running the netting from the eaves on one side, over the purlins and down to the eaves on the other side. The netting is secured to the eaves and the several widths of netting are woven together along the edges. Upon this netting are then laid one or two layers of asbestos paper, and on this one or two layers of tar or other paper impervious to moisture; finally the corrugated steel is placed on these. The netting is 60 ins. wide and weighs 10 Ibs. 340 GRAPHICS AND STRUCTURAL DESIGN per square of 100 sq. ft. The asbestos paper is about T ^ in. thick. The asbestos paper prevents the tar or other paper from taking fire, thus making an excellent fireproof roof. Weight of corrugated steel, in pounds per 100 sq. ft. Gauge. Black. Galvanized. 20 165 182 22 138 154 24 III 127 26 84 99 FIG. 361. Slate Roofing. Slate makes a very satisfactory roof cover- ing. The best slates have a somewhat metallic appearance, do not absorb water and are strong. Generally the strongest slates are the best. The sizes range from 6 X 12 ins. to 16 X 26 ins. and in thickness from \ in. to J in. The most frequently used sizes run from 6 X 12 ins. to 12 X 18 ins., and are generally T 3 ^ in. thick. In slate roofs the slope should not be less than i in 4. The usual lap is a double lap of 3 ins. ; that is, the upper 3 ins. of the first slate are covered by the third slate; see Fig. 361. As slate breaks easily under shock, the roof must be designed to deflect but slightly under loading. The slates are commonly laid on sheathing; this maybe either plain or tongued-and-grooved boards, their thickness varying with the purlin spacing and the load. The sheathing is covered with tar paper or with waterproof paper or felt. The slates may be laid on roofing laths; these are 2 to 3 ins. wide and i in. to ij ins. thick; they are spaced on the rafters to suit the nails in the slates. At gutters, valleys and ridges the slates should be laid in cement and are sometimes entirely laid that way. WALLS AND ROOFS 341 The following are examples of slate roofing: Trusses 14 ft. 6 ins. center to center. No purlins. Sheathing 3-in. yellow-pine plank. The planks were 29 ft. long and joints were broken on alternate trusses. Another roof whose purlins were 5 ft. 8 ins. center to center had 2-in. yellow-pine sheathing. Although slate is an expensive roofing material, it can reason- ably be expected, with ordinary repairs, to last from 25 to 30 years. While not resisting great heat (it cracks and disinte- grates, and then exposes the sheathing to the fire), slate itself does not ignite, and thus makes a fairly good fireproof roof. Clay tiles make a thoroughly fireproof roof. They are, however, very heavy and expensive. Several makes of tiles are also molded in glass, thus permitting skylights to be very readily placed. Their artistic effect is good. Concrete Roofs. Concrete, reinforced by expanded metal or steel rods, is also much used for roofs. The purlins may be spaced from 5 to 7 ft., or they may be omitted altogether, the concrete slabs being placed directly on the trusses or on concrete beams or rafters. When the slope is steep, slate may be nailed directly to the concrete; this should be done within a couple of weeks after placing the concrete; or if the slope is slight a stand- ard slag roof may be used. The following is an example of a reinforced-concrete roof : The roof slab was 4^ ins. thick; it was reinforced with ^-in. square bars, spaced 5 ins. center to center; there were no longitudinal beams; the transverse girders were 14 ft. center to center. Wooden nailing strips were embedded in the concrete slabs; these secured a five-ply waterproofing course upon which a stand- ard slag roof was placed. Slag or Gravel Roofing. This roof covering may be placed on either wooden sheathing or on a reinforced-concrete roof. Such a roof when on wooden sheathing is commonly made by covering the wood with dry felt paper and on this placing from three to five layers of tarred or asphaltic felt. The layers of 342 GRAPHICS AND STRUCTURAL DESIGN paper are lapped the same as slate, exposing from 6 to 10 ins. of each sheet. Another method is to lay rosin-sized sheathing paper weighing not less than 6 Ibs. per 100 sq. ft.; upon this lay two ply of tarred felt lapping 17 ins. and weighing not less than 14 Ibs. per 100 sq. ft. single thickness. These are then covered with a spread- ing of pitch, and finally three additional layers of felt are laid and similarly coated with pitch. On this layer of pitch the slag or gravel is spread; if slag, it should be crushed smaller than f in. but larger than J in. Gravel should be screened and both slag and gravel should be clean and dry. When laid on cement roofing the rosin sheathing paper should be omitted but the cement should be coated with pitch. Although a slag or gravel roof is not fire proof, tests have shown its fire-resisting qualities to be good. It protects the wooden sheathing better than slate. SLOPE OF ROOFS The slope of a roof depends upon that required by the roofing material used. The following is taken from Kidder's Handbook. Roofing material. Slope of rafter, inches per foot. Slate 8 Tiles (interlocking) 7 Tin shingles (painted) 6 Cedar shingles 6 Corrugated steel 2 Ready roofing I In the case of slate the permissible slope will vary with, the weight and size of the slate. Largest sizes of slate Slope 5 ins. per ft. Medium sizes of slate Slope 6 ins. per ft. Smallest sizes of slate Slope 8 ins. per ft. Roofs having a slope of J in. to f in. per ft. constitute what are termed flat roofs, and are generally covered with tar and gravel, asphalt, ready roofing or tin with lock and soldered joints. WALLS AND ROOFS 343 Pitch roofs should be given a slope not exceeding 2 ins. per ft. Rafters or their equivalents should be spaced not over 5 ft. when ordinary sheathing of i-in. boards is used. The following are examples of roofs : Pennsylvania Steel Co. Purlins 6 ft. center to center. Sheathing i|-in. tongued and grooved hemlock. Purlins 4 X lo-in. yellow pine. Slag roofing trusses 20 ft. center to center. A Bridge Shop Roof. Composition roofing. Purlins spaced about 7 ft. 6 ins. center to center. Sheathing 2 ins. thick. Slate Roof on Tar Paper. Purlins spaced about 4 ft. center to center. Sheathing i| in. Slate roof on 2-in. sheathing. Purlins spaced about 8 ft. center to center. Slate roof on 2|-in. plank. Purlins about 8 ft. 6 ins. center to center. Five-ply Slag Roofing. Sheathing i|-in. spruce on purlins spaced about 5 ft. 6 ins. center to center. Asphalt Roofing. Purlins spaced 5 ft. 6 ins. center to center. Sheathing if-in. tongued and grooved boards. CHAPTER XXI SPECIFICATIONS FOR STRUCTURAL STEEL WORK MATERIALS 1. Process of Manufacture. Steel may be made by either the open- hearth or the Bessemer process. NOTE. For the more important work rolled steel is preferably made by the basic open-hearth process, as this process permits the elimination of the greater portion of the phosphorus and a small percentage of the sulphur contained in the pig iron and scrap from which the steel is made. Steel for castings is commonly made by the acid open-hearth process. 2. Chemical and Physical Requirements. Chemical and physical requirements. Structural steel. Rivet steel. Steel castings. Phosphorus, maximum Sulphur, maximum Ultimate tensile strength in Ibs. per sq. in. See No. 3 .... Elongation, min. per cent in 8 0.04% 0-05% Desired 60,000 1,500,000 0.04% 0.04% Desired 50,000 1,500,000 0.05% 0-05% Not less than 65,000 ins., see Fig. i. See No. 4.... Elongation, min. per cent in 2 ins., see Fig. 2 Ult. ten. str. f>O7 Ult. ten. str. _QO/ Character of fracture 22 /Q Cftlr-r, QiTb-xr Jo/o Cold bends without fracture .... ( 180 flat \ See No. 5 180 flat ) See No. 6 f granular 90, d = 3 t NOTE. The effect of phosphorus is to make the steel cold short, brittle when cold; while the sulphur makes it hot short, brittle when hot. 3. Allowable Variations. Tensile tests of steel will be considered satisfactory if showing an ultimate strength within 5000 pounds of that desired. 4. Modifications in Elongation. For material less than T 5 ^ inch thick a deduction of 2^ per cent will be allowed in the elongation for each T a g inch the material is under T 5 ^ inch. For material more than f inch thick a de- 344 MATERIALS 345 duction of i per cent may be allowed for each | inch the material exceeds f inch. For pins and rollers exceeding 3 inches in diameter the elongation in 8 inches may be 5 per cent below that given in No. 2. 5. Bending Tests. These tests may be made either by pressure or by blows. Plates, shapes and bars less than i inch thick shall bend as specified in No. 2. Full-sized material for eyebars and other steel i inch thick and over, tested as rolled, shall bend cold 180 degrees around a pin whose diameter is twice the thickness of the bar. It must show no fracture on the outside of the bend. Angles I inch and less in thickness shall open flat, angles \ inch and less in thickness shall bend shut, cold, under blows of a hammer, without signs of fracture. This test will be made only when required by the inspector. 6. Nicked Bends. Rivet steel, when nicked and bent around a bar the same size as the rivet rod, shall give a gradual break and a fine, silky, uniform fracture. 7. In order that the ultimate strength of full-sized annealed eyebars may meet the requirements of paragraph 65 the ultimate strength in test speci- mens may be determined by the manufacturers; all other tests than those for ultimate strength shall conform to the requirements in paragraph 2. 8. The yield point, as indicated by the drop of the beam, shall be recorded in the test reports. 9. Chemical Analyses. Chemical determinations of the percentages of carbon, phosphorus, sulphur and manganese shall be made by the manu- facturer from a test ingot taken at the time of the pouring of each melt of steel and a correct copy of such analysis shall be furnished to the engineer or his inspector. Check analyses shall be made from the finished material, if called for by the purchaser, in which case an excess of 25 per cent above the required limits will be allowed. 1. About 2" About 18 * Parallel section i,, not less than 9 " t FIG. i. 10. Form of Specimens. (a) Plates, Shapes and Bars. Specimens for tensile and bending tests for plates, shapes and bars shall be made by cutting coupons from the finished product, which shall have both faces rolled and both edges milled to the form shown in Fig. i; or with both 346 GRAPHICS AND 'STRUCTURAL DESIGN edges parallel; or they may be turned to a diameter of f inch for a length of at least 9 inches, with enlarged ends. (b) Rivets. Rivet rods shall be tested as rolled. (c) Pins and Rollers. Specimens shall be cut from the finished rolled or forged bar in such a manner that the center of the specimen shall be i inch from the surface of the bar. The specimen for the tensile test shall be turned to the form shown by Fig. 2. The specimen for bending test shall be i inch by f inch in section. FIG. 2. 11. Steel Castings. The number of tests will depend on the number and importance of the castings. Specimens shall be cut cold from coupons molded and cast on some portion of one or more castings from each melt or from the sink heads, if the heads are of sufficient size. The coupon or sink head, so used, shall be annealed with the casting before it is cut off. Test specimens shall be of the form prescribed for pins and rollers. 12. Annealed and Unannealed Specimens. Material which is to be used without annealing or further treatment shall be tested in the condition in which it comes from the rolls. When material is to be annealed or otherwise treated before use, the specimens for tensile tests representing such material shall be cut from properly annealed or similarly treated short lengths of the full section of the bar. 13. Number of Tests. At least one tensile and one bending test shall be made from each melt of steel as rolled. In event of the material rolled from one melt varying in thickness by f inch or more a test shall be made from the thickest and the thinnest material rolled. 14. Finish. Finished material shall be free from injurious seams, flaws, cracks, defective edges, or other defects, and have a smooth, uniform, workmanlike finish. Plates 36 inches and under in width shall have rolled edges. 15. Stamping. Every finished piece of steel shall have the melt number and the name of the manufacturer stamped or rolled upon it. Steel for pins and rollers shall be stamped on the end. Rivet and lattice steel and MATERIALS 347 other small parts may be bundled with the above marks on an attached metal tag. 16. Defective Material. Material, which, subsequent to the above tests at the mills, and its acceptance there, develops weak spots, brittle- ness, cracks, or other imperfections, or is found to have injurious defects, will be rejected at the shop and shall be replaced by the manufacturer at his own cost. 17. Allowable Variation in Weight. A variation in cross section or weight in the finished members of more than 2\ per cent from that specified shall be sufficient cause for rejection. 1 8. Cast Iron. Except where chilled iron is specified, castings shall be made of tough gray iron, with sulphur not over o.io per cent. They shall be true to pattern, out of wind (perfectly straight or flat) and free from flaws and excessive shrinkage. If tests are demanded they shall be made on the " Arbitration Bar " of the American Society of Testing Materials, which is a round bar i \ inches in diameter and 1 5 inches long. The trans- verse test shall be on a supported length of 12 inches with the load at the middle. The minimum breaking load so applied shall be at least 2900 pounds, with a deflection of at least j ff inch before rupture. WORKMANSHIP 19. General. All parts forming a structure shall be built in accord- ance with approved drawings. The workmanship and finish shall be equal to the best practice in modern bridge works. 20. Straightening Material. Material shall be thoroughly straightened in the shop, by methods that will not injure it, before being laid off or worked in any way. 21. Finish. Shearing shall be neatly and accurately done, and all portions of the work exposed to view shall be neatly finished. 22. Rivets. The size of rivets called for in the plans shall be under- stood to mean the actual size of the cold rivets before heating. 23. Rivet Holes. When general reaming is not required, the diameter of the punch for material not over f inch thick shall be not more than Y 1 ^ inch, nor that of the die more than \ inch, larger than the diameter of the rivet. The diameter of the die shall not exceed the diameter of the punch by more than one-fourth the thickness of the material punched. Material over f inch thick, except that for minor details, and all material where general reaming is required, shall be sub-punched and reamed as per paragraphs 49, 50 and 51, or drilled from the solid. Holes in the flanges of rolled beams and channels used in the floors of railroad bridges shall be 348 GRAPHICS AND STRUCTURAL DESIGN drilled from the solid. Those in the webs of same shall be so drilled or sub- punched and reamed. NOTE. Mr. Schneider in his specifications replaces the f inch occurring in paragraph 23 by f inch. 24. Punching. Punching shall be accurately done. Slight inaccuracy in the matching of holes may be corrected with reamers. Drifting to enlarge unfair holes will not be allowed. Poor matching of holes will be cause for rejection at the option of the inspector. NOTE. Drifting is driving a taper pin through holes that fail to match properly; this distorts the hole and injures the material. 25. Assembling. Riveted members shall have all parts well pinned up and firmly drawn together with bolts before riveting is commenced. Contact surfaces shall be painted (see paragraph 52). 26. Lattice Bars. Lattice bars shall have neatly rounded ends, unless otherwise called for. 27. Web Stiffeners. Stiffeners shall fit neatly between the flanges of girders. Where tight fits are called for, the ends of the Stiffeners shall be faced and brought to a true contact bearing with the flange angles. 28. Splice Plates and Fillers. Web splice plates, and fillers under Stiffeners, shall be cut to fit within | inch of flange angles. 29. Connection Angles. Connection angles for floor girders shall be flush with each other and correct as to position and length of girder. In case milling is required after riveting, the removal of more than T V inch from their thickness shall be cause for rejection. 30. Riveting. Rivets shall be driven by pressure tools wherever possible. Pneumatic hammers shall be used in preference to hand driving. 31. Rivets. Rivets shall look neat and finished, with heads of approved shape, full and of equal size. They shall be central on shank and shall grip the assembled pieces firmly. Recupping and calking will not be allowed. Loose, burned or otherwise defective rivets shall be cut out and replaced. In cutting out rivets great care shall be taken not to injure the adjacent metal. If necessary they shall be drilled out. 31 (a). Heating Rivets. Rivets shall be heated to a light cherry red, in a gas or oil furnace. The furnace must be so constructed that it can be adjusted to the proper temperature. NOTE. Paragraph 31 (a) is inserted by Mr. Schneider but is not generally in- cluded by others. 32. Field Bolts. Wherever bolts are used in place of rivets which transmit shear, the holes shall be reamed parallel and the bolts turned to a driving fit. A washer not less than \ inch thick shall be used under the nut. MATERIALS 349 33. Members to be Straight. The several pieces forming one built-up member shall be tight and fit closely together, and finished members shall be free from twists, bends or open joints. 34. Finish of Joints. Abutting joints shall be cut or dressed true and straight and fitted closely together, especially where open to view. In compression joints, depending on contact bearing, the surfaces shall be truly faced, so as to have even bearings after they are riveted up com- plete and perfectly aligned. 35. Field Connections. Holes for floor-girder connections shall be sub- punched and reamed with twist drills to a steel template i inch thick. Unless otherwise allowed, all other field connections shall be assembled in the shop and the unfair holes reamed; when so reamed the pieces shall be match-marked before being taken apart. 36. Eyebars. Eyebars shall be straight and true to size, and shall be free from twists, folds in the neck or head, or any other defect. Heads shall be made by upsetting, rolling or forging. Welding will not be allowed. The form of the heads will be determined by the dies in use at the works where the eyebars are made, if satisfactory to the engineer, but the manufacturer shall guarantee the bars to break in the body with a silky fracture, when tested to rupture. The thickness of the head and neck shall not vary more than ^ inch from that specified. 37. Boring Eyebars. Before boring each eyebar shall be properly annealed and carefully straightened. Pinholes shall be in the center line of the bars and in the center of the heads. Bars of the same length shall be bored so accurately that, when placed together, pins fa inch smaller in diameter than the pinholes can be passed through the holes at both ends of the bars at the same time. 38. Pinholes. Pinholes shall be bored true to gauges, smooth and straight, at right angles to the axis of the member and parallel to each other, unless otherwise called for. Wherever possible, the boring shall be done after the member is riveted up. 39. Variation in Pinholes. The distance center to center of pinholes shall be correct within fa inch, and for pins up to 5 inches in diameter the diameter of the hole shall not exceed the diameter of the pin by more than fa inch; for larger pins this difference shall not exceed fa inch. 40. Pins and Rollers. Pins and rollers shall be accurately turned to gauges and shall be straight, smooth and entirely free from flaws. 41. Pilot Nuts. At least one pilot and driving nut shall be furnished for each size of pin for each structure, and field rivets 10 per cent in excess of the number of each size actually required. NOTE. Pilot and driving nuts are nuts used to guide and protect truss pins during driving on erection. 350 GRAPHICS AND STRUCTURAL DESIGN 42. Screw Threads. Screw threads shall make tight fits in the nuts and shall be U. S. standard, except above the diameter of if inches, when they shall be made with six threads per inch. 43. Annealing. Steel, except in minor details, which has been partially heated shall be properly annealed. 44. Steel Castings. All steel castings shall be annealed. 45. Welds. Welds in steel will not be allowed. 46. Bed Plates. Expansion bed plates shall be planed true and smooth. Cast wall plates shall be planed top and bottom; the cut of the planing tool shall correspond with the direction of expansion. 47. Shipping Details. Pins, nuts, bolts, rivets and other small details shall be boxed or crated. ADDITIONAL SPECIFICATIONS WHEN GENERAL REAMING AND PLANING ARE REQUIRED 48. Planing Edges. Sheared edges and ends shall be planed off at least I inch. 49. Reaming. Punched holes shall be made with a punch T 8 g inch smaller in diameter than the nominal size of the rivets and shall be reamed to a finished diameter of not more than yV inch larger than the rivet. 50. Reaming after Assembling. Wherever practicable, reaming shall be done after the pieces forming one built member have been assembled and firmly bolted together. If necessary to take the pieces apart for shipping and handling, the respective pieces reamed together shall be so marked that they may be reassembled in the same position in the final setting up. No interchange of reamed parts will be allowed. 51. Removing Burrs. The burrs on all reamed holes shall be removed by a tool countersinking about yV inch. PAINTING 51 (a). Shop Painting. Steel work, before leaving the shop, shall be thoroughly cleaned and given one good coating of pure linseed oil, or such paint as may be called for, well worked into all joints and open spaces. 52. In riveted work, the surfaces coming in contact shall each be painted before being riveted together. 53. Pieces and parts which are not accessible for painting after erection, including tops of stringers, eyebar heads, ends of posts and chords, etc., shall be given two coats of paint before leaving the shop. 54. Steel work to be entirely embedded in concrete shall not be painted. 55. Painting shall be done only when the surface of the metal is per- fectly dry. It shall not be done in wet or freezing weather, unless protected under cover. MATERIALS 351 56. Machine-finished Surfaces. These shall be coated with white lead and tallow before shipment or before being put out into the open air. 57. Field Painting. After the structure is erected, the metal work shall be painted thoroughly and evenly with an additional coat of paint, mixed with pure linseed oil, of such quality and color as may be selected. Succeeding coats of paint shall vary somewhat in color, in order that there may be no confusion as to the surfaces that have been painted. INSPECTION AND TESTING 58. Facilities for Inspection. The manufacturer shall furnish all facilities for inspecting and testing the weight, quality of material and workmanship at the shop where the material is manufactured. He shall, if required, furnish a suitable testing machine for testing full-sized members. The manufacturer shall prepare all test pieces for the machine, free of cost. 59. Access to Shop. When an inspector is furnished by the purchaser, he shall have full access, at all times, to all parts of the shop where material under his inspection is being manufactured. 60. The purchaser shall be furnished complete shop plans, and must be notified well in advance of the start of work in the shop, in order that he may have an inspector on hand to inspect material and workmanship. 61. Shipping Invoices. Complete copies of shipping invoices shall be furnished to the purchaser with each shipment. 62. Mill Orders. The purchaser shall be furnished with complete copies of mill orders, and no material shall be rolled and no work done before he has been notified as to where the orders have been placed, so that he may arrange for the inspection. 63. Inspector's Mark. The inspector shall stamp with a private mark each piece accepted. Any piece not so marked may be rejected at any tune, at any stage of the work. If the inspector, through an oversight or otherwise, has accepted material or work which is defective or contrary to specifications, this material, no matter hi what stage of completion, may be rejected by the purchaser. FULL-SIZED TESTS 64. Full-sized Tests. Full-sized parts of the structure may be tested at the option of the purchaser. Such tests on eyebars and similar members, to prove the workmanship, shall be made at the manufacturer's expense, and shall be paid for by the purchaser, at contract price, if the tests are satisfactory. If the tests are not satisfactory, the members represented by them will be rejected. The expense of testing members, to prove their design, shall be paid for by the purchaser. 352 GRAPHICS AND STRUCTURAL DESIGN 65. Eyebar Tests. In eyebar tests, the minimum ultimate strength shall be 55,000 pounds per square inch. The elongation in 10 feet, in- cluding fracture, shall be not less than 15 per cent. Bars shall break in the body and the fracture shall be silky or fine granular. The elastic limit as indicated by the drop of the mercury shall be recorded. Should a bar break in the head and develop the specified elongation, ultimate strength and character of fracture, it shall not be cause for rejection, provided not more than one-third of the total number of bars break in the head. SPECIFICATIONS FOR STEEL MILL BUILDINGS 66. Dimensions. By height shall be understood the distance from the under side of the lower chord of the roof truss to the tops of the foundations. The width and length of the building shall be measured to the outsides of the framing or sheathing. 67. Spans. The spans of trusses, beams and girders for the purpose of calculations shall be assumed as the distance center to center of bearings for supported spans, while they will be considered as the distance back to back of framing angles when the trusses, beams or girders are framed into columns. LOADS 68. Dead Loads. The dead load is the weight of all permanent con- struction and fixtures. The calculated weights shall be based upon those hereafter given. (a) Weight of Trusses. The weight of a truss may be estimated as w = where W = weight of truss per square foot of building area, L = span of truss, in feet, D = distance center to center of trusses, in feet, P load per square foot on the truss. (b) Weight of Purlins. The weight of purlins per square foot of horizontal projection of the roof may be taken as Wi= Vp i xD -l, 45 4 where Wi = weight of purlins per square foot of building, D = distance center to center of the trusses, PI = load per square foot on purlins. MATERIALS 353 (c) Weight of Roof Coverings. The weights of roof coverings not including sheathing per square foot of actual roof surface will average Tin i Ib. Corrugated steel, No. 18 3.0 Ibs. Slate, ^-inch 6.6 Ibs. Corrugated steel, No. 20. ... 2.3 Ibs. Slate, |-inch 4.4 Ibs. Felt and gravel 7-9 Ibs. Terra-cotta, i inch thick. . 6.0 Ibs. (d) Weight of Sheathing. - Wooden sheathing, per foot, board measure 3.5 ibs. Concrete, i inch thick, per square foot 10-12 Ibs. NOTE. A foot board measure is a piece 12 inches square and i inch thick. LIVE LOADS 69. (a) Live Loads. These shall not be less than the following uniform loads : Warehouses 1 20 pounds per square foot. Foundry charging floors 300 pounds per square foot. Power houses, uncovered floors . . . 200 pounds per square foot. In all cases special concentrations, such as engines, turbines, boilers, chim- neys, etc., should be considered from actual weights. (b) Crane Loads. Where available the actual weights and dimensions of traveling cranes should be used; otherwise the table of weights given on page 354 and taken from the Specifications of Mr. C. C. Schneider may be used. (c) Distribution of Wheel Loads. Wheel loads transferred to beams or girders through rails may be considered as distributed a distance equal to the depth of the girder or beam but not exceeding 30 inches. (Ostrup.) (d) Lateral Loading. Besides the vertical loading, beams and girders carrying traveling cranes shall be designed so that their upper flanges shall in addition resist a lateral bending due to one-twentieth of the crane's capacity acting at each bridge-truck wheel. (e) Impact. An addition of 25 per cent of the live-load bending mo- ments and shears shall be made to all beams, girders and columns carrying traveling cranes. 70. Flat Roofs. Flat roofs liable to be loaded with people should be designed as floors. This would subject them to a possible additional load of 40 pounds per square foot. 71. Wind Loads. The normal wind pressure on a roof whose angle of slope does not exceed 45 degrees shall be based upon a horizontal wind pressure of 30 pounds per square foot and its magnitude shall be estimated by the formula, W W h -A rV n > 45 354 GRAPHICS AND STRUCTURAL DESIGN where WH is the horizontal wind pressure in pounds per square foot, A is the angle of inclination of the roof in degrees and W n is the resulting normal wind pressure on the roof in pounds per square foot. NOTE. Some designers assume a horizontal wind pressure of from 20 to 30 pounds per square foot acting on the vertical projection of the roof. WEIGHTS AND DIMENSIONS OF TYPICAL TRAVELING CRANES Capacity. Span. Wheel base. Maximum wheel load. Side clearance. Vertical clearance. Weight of rail for Tons. Feet. Ft. Ins. Pounds. Inches. Ft. Ins. Runway girder. Beams. 5 5 40 60 8 6 9 12,000 13,000 IO IO 6 6 40 40 40 40 10 10 40 60 9 ? 9 6 19,000 2I,OOO IO IO 6 6 o 45 45 40 40 15 15 40 60 9 6 10 o 25,000 29,000 IO 10 7 o 7 o 5 50 50 50 20 20 40 60 IO O IO O 33,000 36,000 12 12 7 o 7 o 55 55 50 50 25 25 40 60 IO 10 6 40,000 44,000 12 12 8 o 8 o 60 60 50 50 30 30 40 60 10 6 II 48,000 52,000 12 12 8 o 8 o 70 70 60 60 40 40 40 60 II 12 O 64,000 70,000 12 12 9 o 9 o 80 80 60 60 50 50 40 60 II 12 72,000 80,000 14 14 9 o 9 o IOO 100 60 60 NOTE. The rail weights are in pounds per yard. The side clearance is from the center of the rail, while the vertical clearance is from the top of the rail. 72. Wind Pressure on the Sides and Ends of the Building. In build- ings not exceeding 30 feet to the eaves this pressure may be taken at 20 pounds per square foot, while it should be increased up to 30 pounds for buildings 60 feet or more to the eaves. 73. Wind Pressure on Framework. The wind pressure shall be esti- mated for the wind acting horizontally in any direction upon the total exposed surface of the framework. Such pressure shall be assumed at MATERIALS 355 30 pounds per square foot of exposed surface. (This applies during con- struction.) 74. Snow Loads. For latitudes of about 40 degrees the snow load may be assumed at 25 pounds per square foot of horizontal projection of the roof for flat roofs and those inclined up to angles of 15 degrees; above this slope the load may be decreased uniformly so that at 45 degrees it would become zero. (The slope corresponding to 15 degrees is about one in four.) These loads should be increased for higher latitudes and reduced for lower latitudes. In tropical countries snow loads may be neglected. The amount and character of the precipitation in any locality may also affect the loading which would naturally be lower in arid sections. NOTE. According to Mr. C. C. Schneider in climates corresponding to that of New York, ordinary roofs up to spans of 80 feet may be designed for the following minimum equivalent loads per square foot of horizontal projection of the roof. These loads then replace the dead loads, wind loads and snow loads given above. Lbs. Gravel or fOn boards, slope i to 6, or less 50 composition -4 On boards, slope exceeding i to 6 45 roofing [ On 3-inch flat tile or cinder concrete 60 Corrugated sheets, on boards or purlins 40 Slate, on boards or purlins 50 Slate, on 3-inch flat tile or cinder concrete 65 Tile on steel purlins 55 Glass 45 Where no snow is to be expected the above loads may be reduced 10 pounds, excepting that no roof should be designed for a load less than 40 pounds per square foot. 75. Minimum Loads on Purlins and Roof Coverings. Purlins and roof coverings shall not be designed for normal loads under 30 pounds per square foot. 76. Loads on Foundations. The areas of foundation piers shall be proportional to their respective dead loads; in no case, however, shall the combined live and dead loads on a pier divided by its base area exceed the permissible pressure on the soil. NOTE. The desire is to obtain uniform settlement. In making the foundations proportional to the dead loads it is considered that as the dead loads act continu- ously the settlement will depend more on dead loads than on the possibly very intermittent action of the live loads. 356 GRAPHICS AND STRUCTURAL DESIGN UNIT STRESSES AND PROPORTION OF PARTS SUBSTRUCTURE 77. Pressures on Soils. The pressures on the soil at the base of the foundation shall not exceed the following in tons of 2000 pounds. Clay, soft i Ordinary clay, or dry sand mixed with clay 2 Dry sand and dry clay 3 Hard clay and firm coarse sand 4 Firm gravel and coarse sand 6 Rock, according to condition 15-200 78. Compressive Stresses in Masonry. The following compressive stresses will be permitted in masonry structures: Lbs. per sq. in. Common brick, in Portland cement mortar 170 Hard-burned brick, in Portland cement mortar 200 Rubble masonry, in Portland cement mortar 150 Sandstone masonry, first class 280 Limestone masonry, first class 350 Granite masonry, first class 420 Portland cement concrete, 1-2-4 400 Portland cement concrete, 1-2-5 300 79. Wall-plate Pressure. According to Mr. C. C. Schneider, the pressure of beams, girders, wall plates, columns, etc., on masonry shall not exceed the following : Lbs. per sq. in. On brick work in cement mortar 300 On rubble masonry in cement mortar 250 On Portland cement concrete 600 On first-class sandstone, dimension stone 400 On first-class limestone 500 On first-class granite 600 80. Bearing Power of Piles. Piles shall not be spaced closer than 30 inches center to center. The maximum load on any pile shall not exceed 40,000 pounds, or 600 pounds per square inch of average cross section. When driven to rock or equivalent bearing through loose or wet soil, which gives them no lateral support, the limiting load shall be determined by reducing 600 pounds, the maximum allowable compression, by a suitable column formula. 81. Walls shall be built in accordance with the local " building code " when that is available, otherwise they may be made the thicknesses given on page 324. MATERIALS 357 82. Pillars. (a) When concentrically loaded and having a height not exceeding twelve times their least dimension, pillars may be loaded until the fiber stresses reach the figures given in paragraph 78. (b) When eccentrically loaded the resultant pressure must pass within the kern of the section and the maximum pressure should not exceed that given in paragraph 78. NOTE. For the explanation of kern of a section see page 238. UNIT STRESSES IN STEEL WORK 83. Permissible Stresses. The resulting stresses due to dead load, snow load, wind load and impact shall not exceed the following limiting values, excepting where permitted in accordance with paragraph 91. 84. Tension, net section, rolled steel, 16,000 pounds per square inch. 85. Direct Compression, rolled steel and steel castings, 16,000 pounds per square inch. 86. Bending Stresses, on extreme fibers of rolled shapes, built sections, girders and steel castings, net section, 16,000 pounds per square inch. On extreme fibers of pins, 24,000 pounds per square inch. 87. Shearing Stresses. Lbs. per sq. in. On shop rivets and pins 12,000 On bolts and field rivets 10,000 On plate girders and beams, cross section, average 10,000 88. Bearing Stresses. Lbs. per sq. in. On shop rivets and pins 24,000 On field rivets and pins 20,000 89. Pressure on Expansion Rollers. The pressure per linear inch on expansion rollers shall not exceed 600 pounds per inch of roller diameter. 90. Axial Compression in Columns. The load per square inch of gross section for columns axially loaded shall neither exceed 16,000 (C-) pounds, nor 14,000 pounds. Here / = the length of the member in inches; r the corresponding radius of gyration of the section in inches; and C has the following values: Condition of ends. C. Both ends hinged or butting 70 Both ends fixed. 2 it follows - = - ; 382 GRAPHICS AND STRUCTURAL DESIGN SELECTION or BEAM SECTIONS In selecting beams from manufacturers' handbooks it is common practice to use a factor called the section modulus which is tabulated in the hand- books. The section modulus = -or moment of inertia divided by the e distance from the neutral axis to the extreme fibers. M_ f M hence, first find by dividing the maximum bending moment in inch pounds by the allowable working fiber stress. In the following problems assume the beams sufficiently stiffened laterally. Problems where the ratio of span to flange width must be considered will be taken up later. 34. Select a beam to carry a uniform load of 1500 Ibs. per lineal foot on a span of 20 ft.; allow a working fiber stress of 15,000 Ibs. per sq. in. 35. A beam of 32-ft. span carries four loads of 5000 Ibs. each, spaced 8 ft. apart, while the first is 4 ft. from the left support. Allow 15,000 Ibs. fiber stress per sq. in. and select an economical beam section, neglecting the bending due to the beam's weight. FIG. 16. 36. In Fig. 1 6 assume also a uniform load including weight of beam as loo Ibs. per lineal foot. Select an economical beam section, allowing a working fiber stress of 12,000 Ibs. per sq. in. 37. In Fig. 17 neglect weight of beam, allow 12,000 Ibs. fiber stress and select a suitable section. 38. The loading is as shown in Fig. 18. Neglect weight of beam. Allow a working fiber stress of 12,000 Ibs. per sq. in. and select an economical beam section. 39. Select I beam for Fig. 19. Assume beam secured laterally. The moving wheel loads are 6 ft. o ins. c-c. Each wheel load is 10,000 Ibs. Use a working fiber stress of 12,000 Ibs. per sq. in. Assume dead load including weight of beam as 60 Ibs. per ft. 40. Select I beam for the span in Fig. 20. Assume beam secured later- ally. The moving wheel loads are 12 ft. o ins. c-c. Each wheel load is 1 2 ,000 Ibs. Allowable working fiber stress is 1 2 ,000 Ibs. per sq. in. Assume dead load including weight of beam as 80 Ibs. per ft. PROBLEMS DEFLECTION OF BEAMS 383 It is not proposed to review the entire discussion of deflection of beams but simply to recall the subject by one or two of the easier cases and then solve a few problems by the use of the formulae given on page 7. The general equation of the elastic curve for all beams is dx* El here M = bending moment due to the external forces at a section whose abscissa is X. I = moment of inertia of beam section. E modulus of elasticity of the material. y = ordinate along which deflection is measured. 41. A cantilever beam, Fig. 21, of uniform section carries a concentrated load W at its free end. What is its deflection if its length is L? Assume origin of coordinates at o. Here M = WX and the general equation becomes ax 2 Integrate this twice and eliminate the constants of integration by means of the following facts. At the fixed end, -^ = o and X = L, also Y = o ax when X = o. 42. Prove that the deflection of a simple beam of span L and central load W is Wl? A = 48 El Find the form of the equation of the elastic curve for this beam, Fig. 22, and loading. Integrate twice, remembering that -* = o when X = - and that X = o when y o. dx 2 43. A standard i5-in. I beam 42 Ibs. per ft. having a span of 30 ft. is to carry a uniform load; its working fiber stress is not to exceed 16,000 Ibs. 384 GRAPHICS AND STRUCTURAL DESIGN per sq. in. and the maximum deflection of its span. What load will it 360 carry? 44. A 6-in. I beam 12^ Ibs. per ft. extends 5 ft. from a wall as a cantilever beam. Assume that it is amply stiff, laterally. What load will it carry if the working fiber stress is 12,000 Ibs. per sq. in. and what will be its maximum deflection? 45. What uniform load will a i2-in. I beam 31 \ Ibs. per ft. carry on a 2o-ft. span if it is merely supported at the ends and the fiber stress is 16,000 Ibs. per sq. in.? Compare the deflection and fiber stress of this beam with a beam having fixed ends, carrying the same uniform load. NOTE. In practice one generally assumes simple beams, beams merely supported at the reactions. Fixed and continuous beams are seldom de- signed. Reinforced concrete designers make allowance for fixing and con- tinuity in their beam designs. TENSION PIECES In the usual tension piece of constant cross section the stress is uniformly distributed over the cross section and is the same for all sections. The section may not be constant, in which case the minimum or net section must be considered. 46. A 2-in. round bar is to have its end upset for U. S. Standard screw thread. What size screw must be cut if the area at the root of the thread exceeds the area at the body of the rod by 18 per cent? 47. A 4 X 4 X f-in. angle carries a load of 35,000 Ibs. in tension. Assume two if-in. diameter holes in a cross section. What is the unit stress per square inch on the net section? COLUMNS In the case of columns the stress tends to increase any initial flexure due to inaccurate workmanship or vibration and a column, when of sufficient length, fails by bending. It should be noted that tension in a piece reduces any buckling tendency. COLUMNS FAILING BY FLEXURE The derivation of column formulae is based upon the general equation of the elastic curve previously used in finding the deflection of beams. In these column formulae, of which there are several, a quantity depending entirely upon the section of the column occurs; it is , and its square root A PROBLEMS 385 is called the radius of gyration of the section referred to the same axis as 7. Here I is the moment of inertia and A the area of the section. _, 4 /Inertia of section 4 // Radius of gyration = v : = V * Area of section * A 48. Two 8-in. channels 1 1 i Ibs. per ft. are spaced such a distance back to back that the moments of inertia referred to their two principal axes are equal. What do the radii of gyration equal? 49. Two 6 X 4 X ^-in. angles are placed with their long legs parallel and separated f in. What are the radii of gyration referred to the principal axes? 50. Two 5 X 3s X i-in. angles are placed with their short legs parallel and separated f in. What are the radii of gyration referred to the two principal axes? 51. Calculate the radius of gyration of a circular section 8 ins. in diameter. Then calculate the radius of gyration of a ring section 8 ins. outside diam- eter and 6 ins. inside diameter. 52. Prove the following values of radii of gyration: Rectangle, principal axis r = Triangle, principal axis r = - . Vi8 Circle, principal axis r = - 4 The following formulae are frequently used for mild-steel column design: Ends. Fixed. Hinged. -r, , . , ,, l6,000 ,, l6,000 Rankme's / = - , / = . I" I 40,000 r 2 10,000 r 2 Straight line/' = 16,000 35 -, /' = 16,000 70 -. The most rational formula for columns is that of Ritter, /' A 386 GRAPHICS AND STRUCTURAL DESIGN /' = allowable unit working stress, pounds per square inch. / = length of column, inches. r = least radius of gyration. / = maximum desired fiber stress, occurring at the dangerous section of the column and resulting from the flexure of the column. f e = fiber stress per square inch at the elastic limit of material. E = coefficient of elasticity of the material of the column. m = a constant, depending upon the way the column ends are secured. Its value for the usual end conditions are: Ends. Both Fixed. i Fixed and i Hinged. Both Hinged m 4 2.25 i For soft or mild steel, taking E = 30,000,000, 7T 2 = 10 approximately and p e = 30,000; the formula becomes, for hinged ends, 16,000 P' = 10,000 \ r 53. Compare the allowable working stresses as given by the three for- mulae for a mild-steel hinged column, whose length is 20 ft. and least radius of gyration of the column section is r = 2. 54. Compare the allowable working stresses as given by the three for- mulae for a mild-steel fixed-end column whose length is 20 ft. and least radius of gyration of the column section is r = 2. 55. Two i5-in. channels weighing 33 Ibs. per ft. are to be made into a latticed column. Assume hinged ends, use Ritter's formula and determine the maximum load which they can be designed to carry if the column is 35 ft. long. 56. What load will a Bethlehem i2-in. H section weighing 78 Ibs. per ft. carry? Assume the column, with fixed ends, 25 ft. long and the material mild steel. Use Ritter's formula and compare the result with that given in the Bethlehem Steel Company's handbook, using the formula p' = 16,000 55-. 57. Compare the loads that can be carried by two mild steel columns, each 22 ft. long, both having fixed ends, the first being a latticed column made of two lo-in. channels 15 Ibs. per ft., spaced for maximum load, the PROBLEMS 387 other being a lo-in. column 54 Ibs. per ft., Bethlehem Steel Company's section. Use Ritter's formula in both cases. In structural design it is usual practice to limit the value of - for impor- tant work subjected to considerable shock to - = 100, while for other work the usual limit is- = 120. r 58. Would a 6 X 6 X f-in. angle used singly make a good column? Should it be used for a height of 16 ft.? Why? What load could it carry if 10 ft. long, with hinged ends? Use Ritter's formula and assume the material mild steel. 59. Would an 8-in. I beam weighing 18 Ibs. per ft. make a good column? Why? Should it be used for a height of 12 ft.? Why? What load would it carry if 8 ft. high? Use Ritter's formula. Assume fixed ends and the material mild steel. 60. Two 6 X 4 X f-in. angles are to be placed back to back, separated \ in. and used for a hinged strut. Take your data from Manufacturer's Handbook and show if it would be more economical to have the long or short legs parallel. What load would it carry if 12 ft. long? Use Ritter's formula and assume the material mild steel. In the beams in the previous problems no consideration has been taken of their lateral stiffness. In practice it is customary when using a maximum working fiber stress to limit the ratio of unsupported beam length to flange width to from 12 to 20. Where the ratio exceeds these numbers the allow- able working fiber stress should be reduced. This is made necessary by the fact that in a vertically loaded beam the upper flange, being subjected to compression, is liable to fail as a column by buckling and, therefore, to secure the same factor of safety, the working stress in this flange must be reduced. There are several ways by which allowance is made for this condi- tion, all of which are empirical and subject to criticism. Attention is called to the reduction of the working fiber stress by the use of a column formula, as in the Cambria Handbook, 3000 b f = allowable stress in pounds per square inch. / = length between lateral supports in inches. b = width of flange in inches. 388 GRAPHICS AND STRUCTURAL DESIGN This formula is for a maximum desired working stress of 16,000. For limiting values other than 16,000 Ibs. reduce the maximum fiber stresses by the percentage that the above formula reduces 16,000 Ibs. Another method is based upon the tests of long beams by the Pencoyd Iron Works and adopted by the Carnegie and Bethlehem Steel companies' handbooks. It assumes the maximum allowable fiber stress applicable to spans of 20 flange widths and a uniform reduction as the ratio of span to flange width increases until a beam whose span equals 70 flange widths is reached when the allowable working fiber stress is limited to one-half that used in the first case. Still another method is that explained in Chapter VII of this book. 61. What uniform load will a i5-in. I beam weighing 42 Ibs. per ft. carry on a 23-ft. span if unsupported laterally, and if the maximum working fiber stress for a beam whose span is 20 flange widths is limited to 16,000 Ibs. per sq. in.? Compare the results by the methods given above. 62. What central load can be carried upon a Bethlehem girder beam G-2O weighing 112 Ibs. per ft. on a 3O-ft. span? The maximum fiber stress for a span of 20 widths is 16,000 Ibs. per sq. in. Compare the results obtained by the methods given. * 63. A lo-ton crane is to be carried across a span of 30 ft. on rolled beams. The maximum wheel load is 21,000 Ibs., the wheel base 10 ft. 6 ins. The maximum desired working stress is 15,000 Ibs. per sq. in. reduced to allow for ratio of span to flange width. Select a standard Cambria beam and test its ability to resist a lateral pull of one-tenth of the live load divided between the two wheels acting on the beam. * 64. A 25-ton crane is to be carried across a 36-ft. span on rolled beams. The maximum wheel load is 40,000 Ibs.; the wheel base is 10 ft. 6 ins. All other conditions are the same as in Problem 63, excepting that a Bethlehem Grey Mill section with a wide flange is to be used. Determine the section. WEB STRESSES In addition to the points already considered, in designing beams with heavy concentrations of loads or very short spans the ability of the web to resist vertical shearing, vertical crippling and horizontal shearing must also be taken into account. In the case of plate girders the web is reinforced with stiffening angles to resist this crippling. Rolled steel beams are rarely weak in vertical or horizontal shear; timber* beams should be carefully examined for horizontal shear; while in reinforced-concrete construction * Specifications frequently call for an impact allowance of 25 per cent in the case of traveling cranes. PROBLEMS 389 careful designing to resist the web stresses, including horizontal shear and diagonal tension, is of vital importance. The horizontal shear at any section of a homogeneous beam is s = ~- Zay. Iw s = horizontal shear on section per square inch. I = moment of inertia of entire section. V = total vertical shear. w = width of web. Zay = statical moment of the area of the cross section on one side of the neutral axis, or summation of areas multiplied by their re- spective distances from the neutral axis. 65. Show that for a timber beam of width b and depth d ^ 2bd 66. If a is-in. I beam weighing 42 Ibs. per ft. is to be used on varying spans and a fiber stress of 16,000 Ibs. is allowed in tension and 12,000 Ibs. in shear, what is the minimum span for which the horizontal shear may be neglected? 67. Allowing a tensile fiber stress of 1500 Ibs. per sq. in. and a shearing fiber stress with the grain of 120 Ibs. per sq. in., what uniform load can be carried by a yellow pine timber 12 ins. deep, 3 ins. wide on a i2-ft. span? 68. Allowing a tensile fiber stress of 900 Ibs. per sq. in. and a shearing fiber stress with the grain of 100 Ibs. per sq. in., what uniform load will a spruce timber 16 ins. deep X 4 ins. wide carry on a i6-ft. span? 69. What is the maximum span upon which a yellow pine timber 12 ins. deep can be. used to carry a uniform load if its fiber stress is to be limited to 1200 Ibs. per sq. in. and its deflection to yj^ of its span? Assume its modulus of elasticity as 1,500,000 Ibs. per sq. in. 70. What central load will a white oak timber 16 ins. deep X 4 ins. wide carry on an i8-ft. span if stressed to 1500 Ibs. per sq. in.? What would be its maximum deflection if the modulus of elasticity were taken at 1,500,000 Ibs. per sq. in.? COMBINED FLEXURE AND DIRECT STRESS Sometimes beams are subjected to direct compression or tension in addi- tion to flexural stresses. It is usually sufficient to take the algebraic sum of the flexural and direct stresses at the extreme fibers of the beam. When greater accuracy than this is required, Johnson's formula, as modified by 390 GRAPHICS AND STRUCTURAL DESIGN Merriman, can be used. According to this formula the fiber stress due to bending, when direct compression or tension acts upon a piece in addition to bending, is /'= Me ~ r.nPP *m E The is for a compression force. The -}- is for a tensile force. p The combined stress is/' A P = compression or tensile force in pounds. / = span in inches. Simple beam uniform load = = . m 9.6 Simple beam central load= = -L. m 12 A = area of cross section in square inches. An exact method is given by Merriman but its application is too difficult to be generally used. 71. A yellow pine beam 10 ft. long and 12 ins. square is subjected to a compression force acting along its length of 50,000 Ibs. while carrying a uniform load of 20,000 Ibs. What is the fiber stress? E = 1,500,000. What is the maximum fiber stress, using the formula , = Me ~ 72. A horizontal steel tension bar has a cross section of 9 X i| ins. It is subjected to a unit stress in tension of 10,000 Ibs. per sq. in. The bar is 20 ft. long. Determine the maximum resulting fiber stress due to the combined action of its weight and its tensile load. The bar runs vertically with its i^-in. edge down. E = 30,000,000. jtt __ i_ m 9.6 73. In Fig. 23 the top plate is 20 X I in., the sides are i5-lb. channels at 40 Ibs. per foot each. Assume the gross cross section loaded with 7500 PROBLEMS 391 Ibs. per sq. in. Chord 25 ft. long. E = 30,000,000. Determine the maximum fiber stress due to direct loading and its own weight. Use formula E 74. In Fig. 24 the top plate is 24 X | in. The sides are, two upper angles 4 X 4 X & in., two plates 24 X }f in. and two bottom angles 6 X 4 X 1 in. Area of combined section is 82.3 sq. ins. Inertia of sec- tion is 6969. Assume chord 20 ft. long. Uniform load, including live load FIG. 23. FIG. 24. FIG. 25. and weight of section, is 4000 Ibs. per foot of span. Compression load 400,000 Ibs. acting along the center of gravity of the section. What is the total maximum fiber stress? E = 30,000,000. i 9.6 75. What is the maximum fiber stress in Fig. 25? Two angles 4 X 3 X fV in., one plate 10 X T V m - Compression due to direct load acting at center of gravity of section is 10,000 Ibs. per sq. in. on gross section. (In this problem neglect rivet holes.) Length of piece 8 ft. o ins. Neg- lect weight and assume a central load due to purlin of 1600 Ibs. What is the maximum fiber stress? /'= Me E = 30,000,000. *L JL m 12 392 GRAPHICS AND STRUCTURAL DESIGN RIVETING The following considerations are useful in designing riveted joints for structural work. 1. The rivet strength is calculated upon the nominal diameter of the rivet notwithstanding the fact that the hole in which the rivet is driven is generally T x g in. to in. greater than the rivet. 2. The unit shearing strength of steel rivets is taken about the unit tensile strength of the steel and the unit bearing strength of rivets is taken at double their unit shearing strength. 3. Rivets should not be used in tension. 4. Use table of rivet values. 76. A 24-in. I beam weighing 80 Ibs. per ft. carries a uniform load on a i9-ft. span. First find what load it will carry if the extreme fiber stress is 16,000 Ibs. per sq. in. and then find how many f-in. rivets will be required to secure two 4 X 4 X f-in. angles to the web of the beam and transfer the reaction due to the given load. Compare your result with the stand- ard framing. Allow 10,000 Ibs. per sq. in. shear. 77. Two 4 X 4 X rV m - angles are fastened back to back to a f-in. plate. How many f-in. rivets will be required in the angles if the net section of the material is stressed 12,000 Ibs. per sq. in.? Allow one rivet hole if in. diam. through angles and plate. Allow 7500 Ibs. per sq. in. in shear. 78. The long leg of a 5X 35 X f-in. angle is fastened to a f-in. plate. Assume one rivet hole it in. in diameter, in a section. Allow a fiber stress of 12,000 Ibs. per sq. in. on the net section and determine how many f-in. rivets are required. Allow 7500 Ibs. per sq. in. in shear. TIMBER COLUMNS The U. S. Dept. of Agriculture has made a large number of tests upon timbers and the following is their formula suggested for timber columns. 700+ isc + c 2 To find the safe load divide p by the required factor, where p = ultimate strength in pounds per square inch. F = crushing strength of timber. _ _L Length (inches) d small diameter (inches) 79. What load can be safely carried by a yellow pine post 10 ins. square and 20 ft. high? Use a factor of 4 and consider the ultimate crushing strength 5000 Ibs. per sq. in. PROBLEMS 393 80. A cedar post is 22 ft. high and is 10 X 12 ins. in cross section. What load can it safely carry if its ultimate crushing strength is assumed at 3500 Ibs. and a factor of safety of 5 is desired? 81. A spruce column is 18 ft. high and 9 ins. square. What load can it safely carry if its ultimate crushing strength per sq. in. is taken as 4000 Ibs. and a factor of safety of 4 is desired? PLATE GIRDERS In designing girders of greater depth than rolled sections (24 and 36 ins.) the following method is easier and is sufficiently accurate. In Fig. 26, A = area of one flange in square inches. h = distance between the centers of gravity of the two flanges in inches. Where the flange is made up of angles and plates this distance is frequently assumed as the distance back to back of the angles. / = mean fiber stress in the flange in pounds per square inch. M = external bending on the section measured in inch pounds. Then M = A X/Xh. This formula assumes the web as resisting shear only. When the web is assumed to resist bending also the formula becomes Here a is the area of the web plate in square inches. 82. Derive the two formulae for girders, M = AxfXh and M 83. Find the net flange area required at the middle of the following girder. Span 60 ft. Depth of girder back to back of flange angles is 6 ft. o ins. Uniform dead load is 600 Ibs. per ft. (this covers ties, rails and metal of girder, etc.). Uniform live load is 2250 Ibs. per lineal foot. Increase the live-load bending 80 per cent to allow for impact due to a moving train load. Allow a fiber stress of 16,000 Ibs. per sq. in. Assume that the web takes shear only. 84. A girder spans 80 ft. Its depth, back to back of flange angles, is 7 ft. o ins. Uniform dead load is 700 Ibs. per ft. (this covers ties, rails, metal of girder, etc.). Uniform live load is 2000 Ibs. per lineal foot. Increase the live-load bending 80 per cent to allow for impact due to a moving train load. Allow a fiber stress of 16,000 Ibs. per sq. in. Assume that one- eighth of the web acts as flange area and that the web is f in. thick. What 394 GRAPHICS AND STRUCTURAL DESIGN additional area is required to complete the net flange section 30 ft. from the left abutment? 85. A girder which spans 50 ft. is to be made 5 ft. o ins. deep, back to back of flange angles. It carries two wheel loads 12 ft. o ins. center to center of 80,000 Ibs. each. Assume a dead load including girder weight of 200 Ibs. per ft. Increase the live-load bending 20 per cent to cover impact. Allow a fiber stress of 16,000 Ibs. per sq. in. Assume the web T\ in. thick and that one-eighth of it is considered as being flange. Determine what additional area is required to complete the net flange area at the point of maximum bending. f^ 22-- ' 1 FIG. 26. FIG. 27. FIG. 28. 86. A bridge which spans 60 ft. is to be built for a crane. Assume that each girder weighs 15,000 Ibs. and that this forms a uniform load. The trolley wheels are 6 ft. apart and each wheel load is 16,000 Ibs. The girder at the center is 48 ins. deep, back to back of angles. On account of stiffen- ing the crane laterally the flange plates are made 22 ins. wide. Assume the section like Fig. 27. Assume the web as taking shear only. The fiber stress in the compression flange is to be 9000 Ibs., while that in the tension flange can be 12,000 Ibs. per sq. in. Determine the dimensions of the plates. SHAFTING Considering the differential area, Fig. 28, the fiber stress is proportional to its distance r from the center so that if p is the fiber stress at a distance R from the center the stress p r at this distance is Pr = The force acting on an area dA then is dF = p r X dA, The moment of this force about the center then is dM t Integrating PROBLEMS 395 It will be noticed that this is in the form of the general equation for torsion. where / is the polar moment of inertia. 87. Allowing a limiting shearing working stress of 7000 Ibs. per sq. in. upon a steel round 2 ins. in diameter, what twisting moment measured in inch pounds will the round carry? What would be the force applied at a radius of i ft.? 88. Allowing a limiting shearing working stress of 7000 Ibs. per sq. in., what twisting moment in inch pounds would be carried by a 2 in. square shaft? Acting at a radius of i ft. what force would this correspond to? 89. Find the ratio of the twisting moments carried by a round shaft and a square shaft of the same sectional area and the same limiting fiber stress. 90. What horse power will a shaft of diameter d transmit when making .V revolutions per minute and subjected to a working fiber stress of p Ibs. per sq. in. at its circumference? TORSIONAL DEFLECTION It is sometimes desirable to limit the torsional deflection of a shaft, in which case the angle of deflection in degrees is given by A= *XL. 1660 X d* Mt = twisting moment in inch pounds. L = length of shaft in feet. d = diameter of shaft in inches. 91. A shaft 2 ins. in diameter and 60 ft. long is subjected to a twisting moment which produces an extreme fiber shearing stress of 6000 Ibs. per sq. in. Through what angle is one end of the shaft twisted ahead of the other? COMBINED BENDING AND TWISTING Shafting must generally be designed for combined torsion and bending. There are several formulae but according to Guest's law the following may be used: Here M e = equivalent bending moment. M = bending moment. T = twisting moment. GRAPHICS AND STRUCTURAL DESIGN Having found the equivalent bending moment, the shaft diameter can be determined by placing these values in the equation M = ^- e I = moment of inertia referred to axis. e = distance from center of gravity to extreme fibers. 92. In Fig. 29 a force of 2000 Ibs. acts on the teeth of a pinion 5 ins. in diameter. The distance from the center line of the pinion to the center of the adjacent bearing is 6 ins. What diameter of shaft will be required if oooo Ibs. per sq. in. is permitted in flexure? ~^3\ 1 FIG. 29. Jpp^VXu^^W^ TJ^V >p:L:^ ^ FIG. 31. FIG. 32. 93. In Fig. 30 a force of 2500 Ibs. acts upon the teeth of a pinion 4% ins. in diameter. The pinion is inside the bearings as shown. Using a flexural fiber stress of 9000 Ibs. per sq. in. what diameter of shaft is required? STRESSES IN STRUCTURES DETERMINED ALGEBRAICALLY In Fig. 31 to determine the stress in any piece, as FA, cut the truss at a section containing this piece and then equate the internal and external moments about any point that most conveniently gives the desired result. Consult Chapter IV. In Fig. 32 BC = 2400 Ibs., CD = DE = EF = etc. = 4800 Ibs. 94. Calculate algebraically the stress in GC and AG in Fig. 32. 95. Determine algebraically the stresses in GH and HD in Fig. 32. 96. Calculate the stresses in HI and I A in Fig. 32. 97. Calculate the stresses in PL and LM in Fig. 32. Problems 94 to 97 should be checked graphically. In Fig. 33 BC = CD = DE = EE', etc. = 10,000 Ibs. 98. Calculate algebraically the stresses in BF and FA in Fig. 33. 99. Calculate algebraically the stresses in FG and GC in Fig. 33. 100. Calculate algebraically the stresses in GH and HI in Fig. 33. Problems 98 to 100 should be checked by method of coefficients. PROBLEMS GRAPHICAL ANALYSIS or TRUSS STRESSES 397 1 01. In using the common graphical method of determining truss stresses, what assumption is made concerning the construction of the truss? What are the three conditions necessary for equilibrium in a structure acted on by forces? M/ t \| t FIG. 33. FIG. 34. 102. What is a force triangle? A force polygon? What can you say concerning the force polygon representing a number of concurrent forces in equilibrium? What general direction will the forces in the polygon take? 103. Make a sketch of any simple truss and show how force polygons may be used to determine the magnitude and character of the stresses in the structures when the external forces are known. 104. Take Fig. 34, letter the truss and make a careful free-hand sketch of the stress diagram, then find the character of the stresses. FIG. 35. FIG. 36. FIG. 37. 105. Take Fig. 35, letter the truss and make a careful free-hand sketch of the stress diagrams, first with a live load L at i and then at 2. Deter- mine the character of the stress. * 1 06. What is an equilibrium polygon? When a number of noncon- current forces are in equilibrium what graphical conditions are fulfilled? 107. In Fig. 36 show by a free-hand sketch how the resultant of the three forces acting on the piece can be determined in direction, magnitude and located in relation to the piece. 1 08. In Fig. 37 show by a free-hand sketch how the reactions due to the three parallel forces acting on the beam can be determined. 109. In Fig. 38 a truss is acted on by wind forces. Both ends of the truss are fixed. Show how the reactions can be found. no. In Fig. 39 the truss is acted on by wind pressure. The right-hand end is on rollers while the other end is fixed. Show by a free-hand sketch 398 GRAPHICS AND STRUCTURAL DESIGN how the magnitude of the right reaction and the magnitude and direction of the left reaction can be found. Explain how an equilibrium polygon may be used as a bending- in. moment diagram and prove that the statement is correct. 112. Make a free-hand sketch showing by an equilibrium polygon how the bending can be found on the beam in Fig. 40. FIG. 38. FIG. 39. FIG. 40. The equilibrium polygon may also be used in determining the stresses as follows: In finding truss stresses algebraically the internal and external moments were equated. The number of calculations can be reduced by using an equilibrium polygon for the external bending moment, the single diagram serving for all points. . 113. In Fig. 41 show how the stresses AI, HI and HE may be deter- mined by combining both graphical and algebraic methods as suggested. 114. In Fig. 42 determine the dead-load stresses in members AJ, JK and IF. The bridge has the following dimensions: Span, 150 ft.; height, center to center chords, 30 ft.; uniform dead load, 2000 Ibs. per ft., carried at lower apex points. FIG. 41. FIG. 42. FIG. 43- 115. Show how a bending-moment diagram may be determined graphi- cally to give the maximum moments on all sections of a girder for two loads, a constant distance a apart, moving across the girder. 1 1 6. Show how a bending-moment diagram may be determined graphi- cally to give the maximum moments on all sections of a girder, due to a number of loads (use 6), a constant distance apart, moving across the girder. 117. Show how to construct a diagram giving the maximum shears at all points on a girder due to a locomotive and train load passing across it. Prove that the diagram gives the required results. PROBLEMS 399 1 1 8. In Fig. 43 show how the maximum stresses due to moving wheel loads crossing the bridge may be determined for members AH, HI and //. Use a combined graphical and algebraic method. STRESSES IN CRANE FRAMES Show how to determine the stresses in the following crane frames. Ex- plain the methods fully. 1. Find direct stresses due to live load at maximum radius. 2. Find direct stresses due to live load at minimum radius. 3. Draw bending-moment diagrams for members subjected to bending. If necessary draw a stress diagram for position of live load producing maximum bending iri any member. 4. Draw stress diagram for dead load. FIG. 44. FIG. 45. FIG. 46. 1 19. Make the necessary stress and moment diagrams for the frame given hi Fig. 44. 1 20. Make the necessary stress and moment diagrams for the frame given in Fig. 45. 121. Make the necessary stress and moment diagrams for the frame given in Fig. 46. FIG. 47. FIG. 48. 122. Make the necessary stress and moment diagrams for the frame given in Fig. 47, taking the chain or rope pull into account. 400 GRAPHICS AND STRUCTURAL DESIGN 123. Make the necessary stress and moment diagrams for the frame given in Fig. 48, taking the chain or rope pull into account. 124. In Fig. 48 show how to stiffen FB so that the carriage can travel back close to the post. Also show how channels used for EA could be stiffened to reduce the - value and permit the load chain to travel back close to the post. PLATE-GIRDER DESIGN Consult Chapter XI for information concerning these problems. 125. Assume a half dozen wheel loads and their distances apart and explain how the diagram of maximum bending moments would be drawn. Show how to include the bending due to the girder weight in this diagram. 126. A girder has a span of 65 ft. Assume that it carries a load equiva- lent to a uniform load of 2000 Ibs. per ft. Calculate the bending at the center and draw the bending-moment diagram. LENGTHS or FLANGE PLATES In plate girders the flange force varies from zero at the supports to a maximum near the center of a girder. If the girder is of approximately constant depth the bending-moment diagram can represent the flange force by changing the scale ; thus, M = A XpXh and the flange force h being constant F varies as M. This diagram can now be used to deter- mine the lengths of the several plates and angles composing the flange. 127. Assume any girder. Show how to determine the lengths of the several flange plates. FLANGE RIVETING 128. The diagram representing the forces acting in the girder flanges can be used to determine the riveting of the flange angles to the web and also the several flange plates together. Show how to do this. 129. A girder spans 60 ft. Assume the depth back to back of angles as 6 ft. The flange at the middle is composed of two 6 X 6 X f-in. angles, and two 1 5 X f-in. plates. The web is f in. thick, rivets are in. in diameter. PROBLEMS 401 Assume the bending, including dead load, live load and impact, as covered by a uniform load on the girder of 6000 Ibs. per ft. of span. By the method suggested in Problem 128, find the number of rivets required between the web and flange angles for the panel extending from 6 ft. to 12 ft. from the left pier. Allow 11,000 Ibs. shear and 22,000 Ibs. bearing on the rivets. 130. Take the same data as in Problem 129, and determine the number of rivets in the panel extending from 24 ft. to 30 ft. from the left pier. Assume that the web resists bending. 131. Explain how to construct a diagram of maximum shears for a locomotive and train load passing across a bridge. 132. Estimate the reaction from the data given in Problem 129 (this includes dead load, live load and impact) and determine the shearing fiber stress in the web plate which is 6 ft. deep and f in. thick. 133. Explain the object of web stiffeners. What is the usual spacing of web stiffeners in ordinary railroad practice? How thick must the web be, in relation to the girder depth, to permit the omission of stiffeners? 134. Show by means of a diagram of maximum shears that the maximum vertical shear at any o o o s U- t point due to a uniform load occurs when the FIG. 49 girder is loaded from this point to the more remote of the two reactions. Instead of the method previously indicated for obtaining the rivet spacing between the flange and the web, the following is the method more generally used. In Fig. 49 Rxh V s = rivet pitch, inches. R = rivet value, pounds. h = distance between gauge lines of flange angles, inches. V = vertical shear, generally taken as the mean of the panel in which the rivets are to be spaced. 135. Assume the same general data as given in Problem 129. Use the method just given and determine the rivet spacing in the lower flange for the section of the girder extending from 12 ft. to 18 ft. to the right of the left support. Assume that the web takes shear only. Assume that h = 64!. Use f-in. rivets, allowing 11,000 Ibs. in shear and 22,000 Ibs. in bearing. 136. Given f-in. web, 6 X 4 X jV m - flange angles, h = 55 ins. Rivet diameter in., shearing stress 10,000 Ibs. per sq. in., bearing stress 20,000 402 GRAPHICS AND STRUCTURAL DESIGN Ibs. per sq. in., vertical shear 75,000 Ibs. What rivet spacing is required at this section? Assume that the web takes shear only. When concentrated loads are transferred to the web plates through the flange angles the resultant of the vertical and horizontal forces must be found and the rivet spacing determined from this resultant. These con- centrations are usually assumed as distributed over a certain distance along the flange, say 36 ins. The change in the horizontal flange force per inch of flange length is The vertical force per inch of flange length is e load distance distributed T> The resultant f r = Vtf + /2 2 . Rivet spacing = . Jr 137. Solve Problem 129, taking into account the load of 6000 Ibs. per foot carried from the flange angles to the web by the rivets. 138. Given web plates f in. Angles 6 X 6 X f in. Rivets in. in diameter. Shearing stress 11,000 Ibs. Bearing stress 22,000 Ibs. Total vertical shear at middle of panel under consideration 200,000 Ibs. h = 69 ins. Assume load of 45,000 Ibs. distributed over 36 ins. of upper flange. (This includes dead load, live load and impact.) Determine the rivet spacing. Assume that the web takes shear only. The horizontal shear between the flange and the web is reduced when the web is assumed as assisting in carrying the bending. The change in this shear transferred through the rivets depends upon the ratio where A net area of flange angles and plates, a = gross web area A n and, as before, rivet spacing = Jr PROBLEMS 403 139. In Problem 138 assume that one-eighth of the web acts with the flange and that at this point the net flange area is 12 sq. ins., not including one-eighth of the web. Find the required rivet spacing. 140. Determine the rivet spacing satisfying the following conditions: Vertical shear 160,000 Ibs.; h = 64.75 ins- Net flange area, not including one-eighth of web, 20 sq. ins.; web 72 X f in. Upper flange concentration 45,000 Ibs. acting over 36 ins. Shearing stress 11,000 Ibs. and bearing stress 22,000 Ibs. per sq. in. Use rivets f-in. in diameter. * 5 x.s ol ' ojjo oo ,o FIG. 50. FIG. 51. FIG. 52. 141. A girder running over two supports is cantilevered 5 ft. beyond one of them and carries a concentrated load at its end, producing an average fiber stress in the flange angles of 15,000 Ibs. per sq. in. Assume the section given in Fig. 50 and determine the spacing in the flange angles of the canti- levered portion for f-in. rivets. Allow 10,000 Ibs. per sq. hi. in shear and double this in bearing. 142. Fig. 51 is made up of 1-8 in. LJ ii Ibs. per ft., area 3.35 sq. in. and i-i2-in. /3i Ibs. per ft., area 9.26 sq. ins. The average upper flange stress is 8000 Ibs. at the middle of a 22-ft. span, carrying a central load. What rivet spacing should be used in securing the channel to the beam if the rivets are in., the shearing fiber stress 10,000 Ibs. per sq. in. and the bearing stress double this? 143. In Problem 142 determine the rivet spacing 4 ft. from the left support if the beam carries a uniform load instead of a central one? What is the rivet spacing at this point? WEB SPLICE When the web is assumed as resisting bending it becomes necessary to rivet the splice plates so that they and the riveting shall replace the broken section in bending as well as in shear. See Chapter XI. 144. A girder is 5 ft. deep and has a f^-in. web plate; one-eighth of the web is assumed as acting as flange; the average flange stress acts 29 ins. from the neutral axis and is 12,000 Ibs. per sq. in. Using two splice plates, how 404 GRAPHICS AND STRUCTURAL DESIGN thick must they be made if they can be only 50 ins. deep? The fiber stress in the splice plates is to equal that in the web plate at the same distance from the neutral axis of the girder. 145. Assume the data the same as that given in Problem 144 and Fig. 52. How many vertical rows of rivets will be required to make the web splice properly carry its portion of the moment? Allow 10,000 Ibs. per sq. in. in shear and 20,000 Ibs. in bearing. 146. A girder 6 ft. deep has a |-in. web plate which is to be spliced at the middle. How thick must each of two splice plates be if they can be made only 58 1 ins. deep? The fiber stress at the backs of the angles is 12,000 Ibs. and the fiber stress in the splice plates is to equal the fiber stress in the cor- responding position in the web plates. 147. Take the girder in Problem 146 and Fig. 53. How many rows of rivets will be required, allowing 12,000 Ibs. per sq. in. shear and 24,000 Ibs. in bearing? Design the splice to resist bending. Use -in. diameter rivets. R=.15000* R FIG. 53. FIG. 54. 148. Assume that a girder 6 ft. deep and 6o-ft. span carries a moving load of 6000 Ibs. per ft. Determine the maximum shear at the middle and design the splice plates, assuming that the web takes shear only. How many f-in. rivets carrying 12,000 Ibs. per sq. in. shear and 24,000 Ibs. per sq. in. bearing will be required, the web being f in. thick? Depth of splice plates is 58 J ins. 149. The lateral bracing, Fig. 54, in a deck girder is placed in the plane of the upper flanges. Assume that the tension pieces take the load. The wind load is assumed as 300 Ibs. per ft., to account for wind blowing on the train, and 30 Ibs. per sq. ft. of girder. The combined load is assumed as acting upon the upper flange through the rail, (i) Determine the apex loads and (2) make a stress diagram showing how the forces acting in the wind braces may be determined. Assume vertical girders 6 ft. deep. 150. Take the data given in Problem 149 and determine the stresses in the bracing algebraically. PROBLEMS 405 151. Assume that the lateral truss in line with the upper flanges trans- mits all the wind forces in this plane to the ends of the girder, and that from here the forces are transferred to the piers by the end bracing, Fig. 55. Find the reactions from Problem 149. Assume that % R goes to 2, the other \ R through i and then to 3. Determine the forces in i, 2 and 3, and design them, assuming struts with fixed ends and that / = | the diagonal. Allow a fiber stress of 16,000 Ibs. per sq. in. FIXED AND CONTINUOUS BEAMS Fixed and continuous beams do not have much application in structural work; however, owing to the monolithic character of concrete work the bending moments are frequently taken at values between those for supported and fixed beams. 152. Fig. 56 is a beam fixed at one end and supported at the other; load is at the middle of the span. The point of inflection is at X = f /. Show how to find the bending-moment diagram graphically. w FIG. 56. FIG. 57. FIG. 58. 153. A beam fixed at one end and supported at the other carries a uniform load. The reaction at the fixed end is f the load. Determine the bending moments graphically and locate the points of inflection. 154. The beam in Fig. 57 carries a uniform load W Ibs. per ft. over its length (/ + 2x). Determine graphically the value of x in terms of / so that the central span shall fulfill the conditions of a fixed beam, i.e., WP- - 24 at the middle, and at the supports if- -HZ. 12 155. A fixed beam, Fig. 58, carries a central load W. The bending at the supports is M, while for the middle the bending is + M. (Numeri- cally equal but of opposite signs.) Determine graphically these moments and the points of inflection. 406 GRAPHICS AND STRUCTURAL DESIGN 156. If a beam carries a uniform load and is restrained at the supports so that the bending at the middle is T \j- wP, determine graphically the bend- ing moment at the supports and the points of inflection. REINFORCED-CONCRETE DESIGNING For the explanation of the theory and derivation of the formulae see Chapter XIV. Also consult page 193 for the nomenclature. 157. Derive the following formulae: A and -v/; 158. Design a rectangular beam to carry a load of 800 Ibs. per ft. of a WL 2o-ft. span. Assume M = -- Allow /, = 16,000 Ibs., f c = 500 Ibs. 10 Tf> Take d = 20 ins. and assume * = 15. Determine the width and the area c of the reinforcing steel. 159. What size rectangular beam and reinforcement will be required for a span of 16 ft. The beam ends are merely supported and the load is 1000 Ibs. per ft. of span. Given M = , * = 15, tensile stress in o H, c steel 16,000 Ibs. per. sq. in., compression in concrete 500 Ibs. per sq. in. and the width of the beam is to be 40 per cent of its effective depth. 1 60. Derive the following approximate -formulae for rectangular beams, the symbols having the significance given on page 193. M, = X A Xfs, M c = l/c X b X (P. 161. The purlins on a roof are 6 ft. center to center. How thick must a reinforced roof slab be made if the total live and dead load is 80 Ibs. per sq. ft.? Assume-* = 15; fe= 500; / = 14,000 and M = zic o 162. A reinforced-concrete floor carrying a combined live and dead load of 400 Ibs. per sq. ft. is placed over steel beams that are on 6-ft. centers. Taking |=I 5) M = ^, allowable compression in concrete 500 Ibs. per sq. in., and tension in steel 14,000 Ibs. per sq. in., find depth of slab and area of reinforcement. PROBLEMS 407 163. A rectangular tank supported on its lower edges is 5 ft. wide, 6 ft. high and 13 ft. long (inside dimensions). How thick must the floor be made WL to hold the tank full of water ? Allow M = -- Compression in concrete 10 500 Ibs. per sq. in.; tension in steel 14,000 Ibs. per sq. in. E, Find also the area of the steel reinforcement. The common practice in concrete building construction is to use rein- forced-concrete beams rather than the steel beams. In this event, owing to the monolithic character of the work, part of the floor slab acts as the compression flange of the beam. See Chapter XIV. 164. The T beams have a span of 20 ft., and are spaced 6 ft. center to center; see Fig. 59. The combined live and dead load is 250 Ibs. per sq. ft. Assuming M - ; =* = 15; f c = 500; /, = 14,000; find the thickness IO A c of the slab. 6 0- - ff< 6 0- FIG. 59. FIG. 60. 165. Take Problem 164; assume the slab 5 ins. thick and also thatd = 1 6 ins. for the T beam; find the width of slab required for the flange of the T beam and the area of the reinforcing metal. NOTE. It is usual to limit the width of the slab assumed as flange of the T beam. One specification puts this limit at eight times the slab thickness but not exceeding one-third the T-beam span. Figure 60 shows a plan of a section of a floor. The following data apply to all these problems. fc = 55 Ibs., /, = 14,000 Ibs., r = 15. Live load 150 Ibs. per sq. ft. of floor, dead load 60 Ibs. per sq. ft. of floor. 166. Design the slab, determining the thickness and area of metal. Take 10 408 GRAPHICS AND STRUCTURAL DESIGN 167. Assume the slab 5! ins. thick, and design the T beam. Assume d = 19 ins. Find width of flange and area of steel. 168. Design the girder G, assuming that it carries a T beam B, on each side of it, entering it centrally, and that it takes only 85 per cent of the live load, i.e., 150 Ibs. per sq. ft., but entire dead load of 75 Ibs. per sq. ft. As- sume the depth of the girder as d = 25 ins. It must be at least deep enough to allow its reinforcing bars to pass under the bars in the T beams. Find the area of the reinforcement and the width B of the upper flange of the beam. Assume the bending as four-fifths that due to a freely supported beam centrally loaded, i.e., M= -- 169. A beam spans 20 ft. and carries 2500 Ibs. per ft. Assume that d = 30 ins. and find the width and reinforcing area. Take/, = 12,000 Ibs., f c = iu E* WL 500 Ibs.,- 5 = 15 and M= -Cc o 170. Take the data given and found for Problem 169, and find the effect on/, and/ c of placing the steel 2 ins. above the intended position, making d = 28 ins. instead of 30 ins. 171. A slab spans 10 ft. 6 ins. and carries a load of 485 Ibs. per sq. ft. WL E Assume M = - and allow/, = 16,000 Ibs., f e = 600 Ibs. and - 15 Ibs. 12 h, c Find the thickness of the slab and the spacing of f-in. diameter round rods. Slabs should always be reinforced in both directions, the metal at right angles to the main reinforcement being used to prevent the slab from cracking. When the slabs are designed square and reinforced in both directions the maximum moment may be assumed as 20 172. A square slab reinforced in both directions has a side of 16 ft. and carries a combined live and dead load of 220 Ibs. per sq. ft. Find thickness of slab and area of reinforcement. 173. When reinforced-concrete beams are designed assuming that the WL bending at the center is - , what is the bending at the supports? Where 10 in the span is the point of zero bending and what should be the minimum reinforcing over the supports expressed as a percentage of the reinforcing at the center? In the problems thus far considered three points have not been taken into account. The horizontal shear, the vertical shear and diagonal tension. See Chapter XIV. PROBLEMS 409 174. Derive the following formula for the spacing of stirrups, o 0.85 XdxR ~ F-(o.8s Xv c Xb X<*)* 175. Derive the following formula for the length of rod required on ac- count of bonding, 2 U 176. Derive the following formula for the length of bar required to provide flange strength, / = 177. Take Problem 158 and determine the vertical reinforcements, allow- ing a shearing fiber stress of 12,000 Ibs. per sq. in. Neglect the shearing strength of the concrete. Try f-in. rounds bent into U stirrups. Would stirrups be needed if 50 Ibs. per sq. in. was allowed in shear in concrete? What about reinforcement at the ends? 178 to 181. Design a light highway bridge of reinforced concrete. Span 32 ft. o ins., fill 15 ins., roadway 16 ft. with a 4-ft. wall on each side. Assume 8 = 5^ ins. See Figs. 61 to 63. FIG. 61. C '""""""V^iiyz:'^.-.^.!. L.r.'.ul.. .. ..'. '.~~~^ t *. fl X L I i ' ~1' ' ' 5'o---i>*l 5 'o^4>l4--^'o^-4>J I GMi 1^^ J 1^^^ 40Tons on 8 Wtieels 1 - -H sV'H Car Width 8V \ : i 1 . 1 o O QiO FIG. 62. W=20 Tons FIG. 63. 178. Determine thickness of slabs under roadway. Assume live load 500 Ibs. per sq. ft., concrete 60 Ibs. per sq. ft., fill 1 20 Ibs. per sq. ft. Assume 410 GRAPHICS AND STRUCTURAL DESIGN compressive stress in concrete 650 Ibs. per sq. in., tensile stress in steel 16,000 Ibs. per sq. in. Assume that the effect of impact is 50 per cent of live-load stresses. M = and ^ = 15. IO E c Using |-in. round bars, how far apart (center to center) should they be spaced? Since the bending has been taken at M = , what reinforce- 10 ments should be placed over girders? 179. Design girders d and G 6 . Assume dead load of concrete, fill, etc., as 900 Ibs. per lineal ft. of 32-ft. span. Live load 125 Ibs. per sq. ft. of walk assumed as 5 ft. o ins. wide. z = 1.64 corresponding to f c = 650 Ibs. M = . o 1 80. Design girders G 3 and G 4 on the following basis. Each girder to carry one-half the car load or the front wheel of the roller and one-half the weight of the rear wheels, Figs. 62 and 63. Assume dead load 1350 Ibs. per ft. Take M = . Add 35 per cent to live-load stresses to allow for o impact. How many i|-in. bars will be required? Assume d = 34.5, /, = 16,000 Ibs. What width of the flange will be stressed? Make stem 14 ins. wide. 181. Draw diagram of maximum shears covering dead load and trolley load. Assume ^-in. U stirrups and find the spacing at the abutment and at 2, 4, 6 and 8 ft. from this end, allowing 60 Ibs. per sq. in. shear on the concrete and using the formula s== 0.85 XdxR V - (0.85 Xv c Xb Xd)' R = shearing value of one stirrup. 182. In a given beam d = 30 ins., four i^-in. round reinforcing bars run parallel with the lower flange through the entire span. The maximum reaction is 44,000 Ibs. What is the bond stress per square inch and is this satisfactory if the specifications allow 60 Ibs. per sq. in.? How could this be provided for? 183. In a beam d = 34 ins., three i-in. bars run parallel with the lower flange through the entire span. The maximum reaction is 19,000 Ibs. What is the bond stress per square inch and is this satisfactory if the speci- fications allow a bond stress of 60 Ibs. per sq. in.? If not satisfactory, how could it be provided for? PROBLEMS 411 REINFORCED-CONCRETE COLUMNS 184. Allowing 600 Ibs. per sq. in. on the concrete, what load can be placed on a column n ft. high, 12 X 12 ins. and reinforced by four -in. bars? When a column's height does not exceed twelve times its least dimension the influence of length upon its buckling can be neglected. Assume column dimensions inside bars as 10 X 10 ins. 185. Load 420,000 Ibs., reinforcement i per cent, height 15 ft. What size square columns allowing 600 Ibs. upon concrete will be required? How many i-in. bars will be required? What would you make the outside dimensions of the column? 1 86. Take the truss in Fig. 64, place a load L at the point indicated, show the necessity for the double lacing. Explain why this is usual in the central panels and frequently not necessary at the end panels. FIG. 64. FIG. 65 187. With Fig. 65, by means of force and equilibrium polygons find the required depth of a concrete foundation 8 ft. square, at 125 Ibs. per cu. ft. so that the moment of the foundation about the toe A shall be. twice that of the crane weight and live load about the same point. At the bottom of the foundation make a diagram showing the distribution of pressure on the soil. 1 88. Take the data given and found for Problem 187 ; turn the crane until the load lies in the line of a diagonal of the foundation. With the resultant pressure upon the soil as found in Problem 187, determine the maximum and minimum pressure on the foundation by the method given in Chapter XVI. 189. Take Fig. 66, and by means of a force and an equilibrium polygon determine the maximum load the locomotive crane can lift in the position given. Boom is 36 ft. long, and weighs 2000 Ibs. Gauge of track is 4 ft. 8^ ins. Counterweight is 15,000 Ibs. acting at center line of track. Engine, machinery, etc., weigh 18,000 Ibs., and act 12 ins. to left of center line of track. Boiler and base weigh 12,000 Ibs., and act 7 ft. 6 ins. to left of center line of track. Cab, etc., weigh 16,000 Ibs., and act at center line of track. 412 GRAPHICS AND STRUCTURAL DESIGN 190. Design a concrete footing, Fig. 67, to carry 200 tons from the column through a shoe 4 ft. square, resting on reinforced concrete. The soil can carry 3 tons per sq. ft. What spacing of f-in. round bars will be required? Assume fe = 55o; ft = 16,000 Ibs.; = 15. FIG. 66. FIG. 67. What advantages would this type of foundation have over the usual con- crete pier? Consult Chapter XV. 191. In Fig. 68 assume sag 60 per cent of width, coal (bituminous) 50 Ibs. per cu. ft. Area of surcharged bin A' = 0.57 a 2 . Assume that the load on the bin varies from the sides to the middle as the intercepts between h- FIG. 68. the sides of a triangle. Show by means of force and equilibrium polygons how to find the tension in the bottom of the bin plate and also in the sides at the point of suspension from the side girders. 192. A stack, Fig. 69, has outside diameter of 7 ft. and is 120 ft. high. Assume a uniform wind pressure of 30 Ibs, per sq. ft. acting on it. The steel PROBLEMS 413 weighs 30,000 Ibs. and the concrete base is 20 it. square. If the concrete weighs 125 Ibs. per cu. ft., how deep must the foundation be if the resultant pressure falls within the kern and what is the distribution of pressure on the soil? 193. The y-ft. stack shown in Fig. 69 is 120 ft. high, is made of T Vin. plate for 30 ft. at the top, and each succeeding 30 ft. toward the foot of the stack is made ^ in. thicker. What is the extreme fiber stress in the net section, 60 ft. from the top, assuming 20 per cent of the plate cut out for rivets? Calculate the pitch of rivets, allowing 8000 Ibs. in shear and 16,000 Ibs. in bearing and assume rivets f in. in diameter. 194. In Fig. 69 investigate the fiber stress and rivet pitch for the seam 80 ft. from the top. 195. A stack is 125 ft. high. Its outside diameter is 7 ft. 6 ins. If 12 bolts are used on a circle whose radius is 65 ins., what must their diameter be? Allowable tension is 12,000 Ibs. per sq. in. Weight of stack is 30,000 Ibs. ANSWERS TO PROBLEMS 5- 0.79 in. 45- 6. 2.01 ins. 7- 3.82 ins. 8. 3.86 ins. 14- 13-5. IS- 3.90. 4 6. 16. 1334. 47- 17. 14.6. 48. 18. 324.0. 49. 19. 241.1. 50. 20. 0.0069 ^ 4 - SI- 21. 11,340.1. 53- 24. W T R = W,M= and M r x = w?L. 54. 3 3 U 55- 27. R- W , M - 58. 2 6 59- Wr 9 Wr 3 _- KK3C ^ VV X~ MX = FT" 60. 2 3 2 61. 28. TF R\ = 5 :Vr,-,OT = O.^78 L, 63. 30. 32. 34. 35- 36. 37- 38. 39- 40. 43- 44- Ri= 7000 Ibs., Afmax= 78,ooo ft. Ibs., Mu = 75,ooo ft. Ibs. -Mmax = 446,490 ft. Ibs., MM = 443,600 ft. Ibs. i5-in. I at 45 Ibs. i5-in. I at 50 Ibs. -required 217.4 use two 2o-in. I beams at 65 Ibs. i5-in. I at 42 Ibs. i5-in. I at 42 Ibs. 2o-in. I at 65 Ibs. - = 126.9. 2o-in. I at 75 Ibs. 6 21,000 Ibs. Load 1460 Ibs., deflection 0.161 in. 66. 67. 68. 69. 70. 71- 72. 73- 74- 75- 76. 77- 78. 79- 80. Load 19,200 Ibs., fiber stress in fixed beam 10,670 Ibs. per sq. in., de- flection in supported beam 0.534 in., deflection in fixed beam 0.107 ins. 2 1 ins. diam. 15,500 Ibs. 4.92 ins. back to back, r = 3.10. r Q = 1.91, n = 1.73- r = i.oo, n = 2.53. r =2.0, r = 2.5. 6150, 7600 and 6550 Ibs. 11,400, 10,600 and n, 750 Ibs. 203,000 Ibs. 34,880 Ibs. 64,400 Ibs. 66,300 Ibs. C = 16,740 Ibs.; P = 19,100 Ibs. One i2-in. channel at 20^ Ibs. hori- zontally and on top of one 20-in. I beam at 65 Ibs. placed verti- cally. Lateral stress 5700 Ibs. per sq. in. 59.4 ins. 5760 Ibs. 6400 Ibs. 16 ft. 8 ins. 4740 Ibs.; A = 0.486 in. 1420 Ibs. 10,940 Ibs. 8125 Ibs. 8450 Ibs. compression. 1 1, 660 Ibs. compression. 7 rivets. 12 rivets. 10 rivets. 81,000 Ibs. 5 7,600 Ibs. 414 ANSWERS TO PROBLEMS 415 81. 52,500 Ibs. 83. 21.8 sq. ins. 84. 24.9 sq. ins. 85. 21.7 sq. ins. 86. Top flange two 4 X 4 X -in. angles, and one plate 22 X iV m -> lower flange two 4X4 Xf-in. angles and one plate 22 X i^-in. angles. 87. 10,990 in. Ibs., and 916 Ibs. 88. 13,400 in. Ibs. and 1117 Ibs. 89. The twisting moment of a round shaft may be 1.2 times that on a square shaft of the same weight per foot. 90. H.P. - *L 32,130 91. 21.3. 92. 2 T V ins. 93. 2\ ins. 94. GC = 37,600 Ibs. comp., AG = 33,600 Ibs. tension. 95. GH = 4300 Ibs. comp., DH = 35,460 Ibs. comp. 96. HI = 4800 Ibs. tension and AI = 28,800 Ibs. tension. 97. FL 31,180 Ibs. comp. andZJf = 14,400 Ibs. tension. 98. BF = 154,660 Ibs. comp., FA = 109,375 Ibs. tension. 99. FG = 110,470 Ibs. tension, GC = 187,500 Ibs. compression. 100. GH = 31, 250 Ibs. comp., El 66,280 Ibs. comp. 114. AL = 187,500 Ibs. comp., KE = 166,700 Ibs. tension and KL = 32,500 Ibs. tension. 129. 18 rivets. 130. 3, the spacing is commonly not allowed to exceed 4 to 6 ins. 135. 4.6 ins. 136. 4.8 ins. 137. 4.4 ins. 138. 2.28 ins. 139. 2.8 ins. 140. 2.93 ins. 141. 4.28 ins. 142. 14.67 ins. 143. 9.6 ins. 144. 0.27 in. 145. 2 rows. 146. Two plates each f in. thick. 147. 2 rows. 148. 6 rivets. 149. Check by method of coefficients. 158. b = 13.5 ins. and A = i-34sq. ins. 159. d = 24 ins., 6 = 9.5 ins., A = 1.14 sq. ins. 161. d = 2.2 ins. and A = 0.166 sq. in. 162. d = 4.9 ins. and A = 0.372 sq. in. 169. b = 20 ins. and A = 4.8 sq. ins. 170. f c = 555 Ibs. per sq. in. and /, = 1 2, 800 Ibs. 171. d = 6.85 ins. Spacing = 6.77 ins. 178. d = 4.65 ins., j-in. bars 5-5-in. centers. 179. d = 38.0 ins., five i-in. bars. 180. eight i^-in. bars or nine i-in. bars. 182. 93 Ibs. per sq. in. 183. 59 Ibs. pef sq. in. 184. 66,590 Ibs. 185. 25 ins. square and eight i-in. bars. 190. |-in. rods spaced 4.75 ins. center to center. INDEX Algebraic determination of stresses, 52. Angles of repose and weights of materials, 277. Arch floors, brick and tile, 320. Ash bins, pressures on, 290. Base for reinforced-concrete chimney, 270. Beams, design of, 76. Beams of uniform section, bending moments, deflections, etc., 7. Beams, reinforced-concrete: approximate formulae for, 196. bent bars, 217. bond stresses, 214. design of a T beam, 216. diagonal tension in, 213. forms of reinforcements, 223. horizontal shear in, 209. parabolic variation of stresses, 208. reinforcing rods, lengths of, 215. stirrup spacing in, 211. theory of beams with double reinforcement, 205. rectangular beams, 194. slabs, 204. T beams, 199. Beams unsupported laterally, 98. Bearing and shearing value of rivets, 84. Bed plates for plate girders for railway bridge, 159. Bending due to moving loads, 47. Bending moments on beams, 77. Bending moments, deflections, etc., for beams of uniform section, 7. Bending moments on a railway girder due to locomotive and train load, 141. Bents, 112. Bent bars in T beams, 217. Bethlehem rolled sections, 18. Bins: coefficients of friction between materials, 296. design of, 306. graphical determination of forces acting on, 296. 417 41 8 INDEX Bins: pressure on grain bins, 289. shallow bins, 290. stresses in bins, formulae for, 290. Bond stresses between concrete and steel, 214. Box girder for electric overhead traveling crane, 179. Bracing for steel mill buildings, 118. Bricks, 4. Bricks, fire, 4. Brick arch floors, 319. Brick chimney, design of, 242. Brick floors, 316. Brick walls, 323. Buildings, foundations for, 230. Bunkers, suspension, 300. Cast iron, i. Cement, 191. Cement concrete floors, 314. Center of gravity, graphical determination of, 30. Centers of gravity of angles, 15. channels, 16. Character of stresses, 43. Chimney, brick: design of a, 242. kern of a section, 238. Chimney, reinforced-concrete: base, 270. design of a reinforced-concrete, 259. reinforcements, 267. Chimney, steel: design of a, 248. foundation, 253. foundation bolts, 257. lining, 253. maximum pressure on the soil, 255. ring seams, 251. thickness of shell, 248. vertical seams, 251. Clay tile roofing, 341. Clearances for cranes, 336. Coal bins, pressures on, 290. Column formulae, 90. Columns for steel mill buildings, 126. Columns, 90. Columns, theory of reinforced-concrete, 221. INDEX 419 Columns, timber, 4. Combined stresses, 95. Compression pieces, design of, 76. Concrete (see also Reinforced concrete). Concrete blocks, hollow, 325. Concrete retaining wall, design of, 279. Concrete roofing, 341. Concrete walls, solid, 324. Concrete wearing surfaces, 321. Concurrent and nonconcurrent forces, 21. Continuous beams, 32. Conveyors, girders for, 101. Cooper's E-6o loading, moment table for, 71. Coplanar and noncoplanar forces, 21. Corrugated bars, 224. Corrugated steel roofing, 337. Corrugated steel sides, 323. Couple, 21. Cranes: electric overhead traveling crane, 174. top-braced jib crane, 168. underbraced jib crane, 162. Crane clearances, 336. Crane frames (see Cranes). Deflection of beams, graphical determination of, 30. Deflection, bending moments, etc., 7. Design of beams, 77. compression pieces, 76. frame for top-braced jib crane, 168. frame for underbraced jib crane, 162. railway girder, 141. roofs, 337. roof truss members, 123. steel mill building, No. i, 118. steel mill building, No. 2, 130. tension pieces, 76. Determination of stresses, algebraic, 52. method of coefficients, 55, method of moments, 54. Diagonal tension in reinforced concrete beams, 212. Diagram of maximum live-load shears, 49. Dimension of angles, 14. Bethlehem beams, 19. channels, 16. edged plates, 13. 420 INDEX Dimension of angles (continued). girder beams, 19. I beams, 17. H columns, 18. sheared plates, 12. Direction of a force, 20. Elastic limit of concrete, 192. Electric overhead traveling crane: design of box girder, 175. flange areas, 176. rivets, 180. girder with channel flanges, 182. girder with horizontal stiffening girder, 184. stiff eners, 182. specifications for, 174. Equilibrant, 21. Equilibrium, 21. polygon, 23. Expanded metal, 222. Fiber stresses in underbraced jib cranes, 164. working, 3. Flange areas for girders of E.O.T. cranes, 176. of plate girders, 146. Flange plates of plate girders for railway bridges, lengths of, 148. Flange rivets in girder for E.O.T. crane, 180. Flange rivets in plate girder for railway bridge, 149. Flat plates, strength of, 9. Floors, ground: brick, 316. cement concrete, 314. wooden, 317. wooden block, 316. Floors, upper: brick arch, 319. concrete wearing surfaces, 321. hollow tile arch, 320. iron floors, 322. reinforced-concrete, 321. steel, 319. steps, 322. wooden, 318. Floor-beam reaction, maximum, 69. Force, 20. Force and equilibrium polygons, uses of, 28. INDEX 421 Force polygon, 22. Force triangle, 21. Forces acting in bins, graphical determination of, 296. Formulae for piles, 237. for stresses in bins, 291. Foundation bolts for chimneys, 257. Foundations for buildings, 230. for machinery, 225. piles, 235. pile formulae, 237. pressure on soil, allowable, 228. Foundations for chimneys, 253. Framing, standard, 80. Friction, coefficient of, between materials, 296. Girders for conveyors, 101. Girders for E.O.T. crane, specifications for, 174. with channel flanges for E.O.T. cranes, 182. with horizontal stiffening girders for E.O.T. cranes, 184. 'Glass, 337. Grain bins, pressures on, 289. Graphic moments, 26. Graphics, 20. statics, 20. Graphical determination of forces acting in bins, 296. of stirrup spacing in reinforced-concrete beams, 211. Gravel, 191. Gravel and slag roofing, 341. Ground floors, 317. Hollow tile arch floors, 320. Horizontal shear in reinforced-concrete beams, 209. stiffening girders for E.O.T. crane girders, 184. Impact, 143. Inertias of geometrical sections, 10. angles, 14. channels, ^16. H columns, 18. I beams, 17. Influence diagrams, 61. Iron, cast, i. Iron floors, 322. Jib for underbraced crane, selection of, 164. Jib crane, underbraced, 162. top-braced, 168. ^T 422 INDEX Kahn trussed bar, 224. Kern of a section, 238. Lateral bracing for plate-girder railway bridge, 153. Line of action, 20. Lining, chimney, 253. Live-load shears, diagram of maximum, 49. stresses, 56. Long beams unsupported laterally, 98. Machinery, foundations for, 225. Magnitude of a force, 20. Masonry retaining walls, design of, 279. Mast of a top-braced jib crane, 173. Mast of an under-braced jib crane, 166. Materials, i. Maximum bending due to moving loads, 47. floor-beam reaction, 69. live-load shears, diagram of, 49. moment due to moving loads, 61. shear due to moving loads, 64. Members of frame of top-braced jib crane, 166. of underbraced jib crane, 170. Metals, physical properties of, 2. Method of coefficients for the determination of stresses, 55. moments, determination of stresses by, 54. Mixing concrete, 191. Moment of a couple, 21. table for Cooper's E-6o loading, 71. Moving loads carried under trusses, 37. maximum bending due to, 47. Physical properties of concrete, 191. metals, 2. Pile formulae, 237. Piles, 235. Plate girder for railway bridge: bed plates, 159. flange rivets, 149. flange plates, lengths of, 148. rivet spacing in flange angles, 151. stiffening angles, 148. web splice, 154. wind bracing, 153. Plate girders: dead-load shear, 146. maximum end shear, 144. INDEX 423 Plate-girder railway bridge, 141. Plates, strength of flat, 9. dimensions of edged, 9. of sheared, 9. Point of application of a force, 20. Pressures on retaining walls, 273. soils, allowable, 228. soil under chimney, maximum, 255. Properties of concrete, physical, 192. metals, 2. sections, 10. Properties of timber, 3. table of, 5. Purlins, 126. Railway girder, design of, 141. girders, impact allowance for, 143. Ransome twisted bars, 224. Rectangular beams, theory of, 194. Reinforcing frame, 224. rods, lengths of, 215. Reinforced concrete: approximate formulae for rectangular beams, 196. bond stresses, 214. cement for, 191. columns, 221. design of T beams, 216. diagonal tension, 213. elastic limit, 192. forms of reinforcements, 223. gravel, 191. horizontal shear in beams, 209. mixing, 191. parabolic variation of stress, 208. physical properties, 191. reinforcing rods, lengths of, 215. sand, 191. stone, 191. stirrup spacing in beams, 211. theory of beams with double reinforcement, 205. rectangular beams, 194. slabs, 204. T beams, 199. Reinforced-concrete chimney, 259. retaining walls, 280. 424 INDEX Reinforced-concrete walls, 325. Reinforcements, forms of, 222. Resistances of sections, 10. Resultant, 21. Retaining walls: design of a masonry, 279. reinforced-concrete, 280. distribution of pressure on, 275. graphical determination of the pressures on, 274. pressures on, 273. weights and angles of repose of materials in fills, 277. Ring seams of steel chimneys, 251. Rivet spacing in flange angles, 151. Rivet values in shear and bearing, table of, 84. Riveting, 81. Riveting, rules for, 88. Roofs, types of, 121. Roof coverings: clay tile, 341. concrete, 341. corrugated steel, 337. slag or gravel, 341. slate, 340. Sand, 191. Sections, inertias, resistances, etc., 10 properties of, 10. Shallow bins, pressures on, 290. Shear and bearing values of rivets, 84. Shear in reinforced-concrete beams, horizontal, 209. Shell thickness of steel chimney, 248. Shop floors (see Floors). Slag or gravel roofing, 341. Slate roofing, 340. Soil, allowable pressure on, 228. Specifications for girders of an E.O.T. crane, 174, Splice for plate-girder of railway bridge, web, 154. Standard framing, 79. Steel castings, 2. Steel floors, 319. Steel mill buildings: design of Problem i, 118. design of Problem 2, 130. Steps, 322. Stiffeners for girders for E.O.T. crane, 182. INDEX 425 Stifteners for plate girders of railway bridge, 148. Stiffening steel mill buildings, 118. Stirrup spacing in reinforced-concrete beams, 211. Stone, 191. walls, 324. Strength of flat plates, 9. Stresses: algebraic determination of, 52. character of, 43. Stresses in concrete, unit working, 192. structures, 35. Structural material, n. Structures, stresses in, 35. T beams: bent bars, 217. design of a, 216. web reinforcements, 218. Tension pieces, design of, 76. Tension or compression in a member, 36. Thatcher bar, 224. Theory of columns, 221. Tile arch floors, 320. Timber columns, 4. Timber, properties of, 3. table of properties of, 5. Top-braced jib crane: design of frame, 168. mast, 173. selection of members, 170. Towers, 112. Traveling crane (see Electric overhead traveling crane), 174. Triangular mesh steel reinforcement, 224. Truss, determination of the stresses using moment table, 74. Trusses, no. Trusses carrying moving loads under them, 37. wind loads on, 39. Twisted bars, Ransome, 224. Underbraced jib crane: design of frame, 162. fiber stresses in, 164. mast, 1 66. members of frame, 166. selection of jib, 164. Unit working stresses for concrete, 192. 426 INDEX Vertical seams of a steel chimney, 251. shear in beams, 77. Walls: brick, 323. corrugated steel, 323. glass, 337. hollow concrete blocks, 325. reinforced-concrete, 325. solid concrete, 324. stone, 324. wooden, 323. Wearing surfaces, 321. Web reinforcements in T beams, 209. Web splice for plate girders of railway bridge, 154. Weights and angles of repose of materials, 277. 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