SAFE BUILDING CONSTRUCTION THE MACMILLAN COMPANY NEW YORK • BOSTON • CHICAGO ATLANTA • SAN FRANCISCO MACMILLAN & CO., Limited LONDON - BOMBAY • CALCUTTA MELBOURNE THE MACMILLAN CO. OF CANADA, Lm TORONTO SAFE BUILDING CONSTRUCTION A TREATISE GIVING IN SIMPLEST FORMS POSSIBLE PRACTICAL AND THEORETICAL RULES AND FORMULiE USED IN CONSTRUCTION OF BUILDINGS AND GENERAL INSTRUCTION BY LOUIS DE COPPET BERGH, F. A. I. A. MEMBER AMERICAN SOCIETY CIVIL ENGINEERS; AUTHOR OF SAFE BUILDING, IRON CONSTRUCTION IN NEW YORK CITY, ETC. NEW'EDIl'ltnr THOAoufinrLVRBvisED THRsvanouWt* * *. THE MACMILLAN COMPANY 1908 All rights reserved Copyright, 1886, 1887, 1888, i88g, 1890, i8g2 and 1893, by TICKNOR AND COMPANY Copyright, 1908, By the MACMILLAN COMPANY First Published elsewhere. New Edition, Revised, Published April, 1908 THE MASON-HENRY PRESS SYRACUSE, NEW YORK TH€ GETTY CENTER LIBRARY TABLE OF CONTENTS PAGE LIST OF TABLES ix,x,xi LIST OF FOEMULAE xii, xiii, xiv, xv, xvi PREFACE 1 CHAPTEE I. Strength op Materials 2-84 CHAPTEE II. Foundations 85-99 CHAPTEE III. Cellar and Retaining Walls 100-120 CHAPTEE IV. Walls and Piers 121-156 CHAPTEE V. Arches 157-182 CHAPTEE VL Floor-Beams and Girders 183-238 CHAPTEE VII. * Graphical Analysis of Transverse Strains 239-269 CHAPTEE VIIL Eeinforced Concrete Construction 271-275 CHAPTEE IX. EiVETS, Eiveting and Pins 276-325 vii 18311 viii TABLE OF CONTENTS CHAPTER X. PAGE Plate and Box Girders • 326-355 CHAPTER XI. Graphical Analysis of Strains in Trusses 356-398 CHAPTER XII. Wooden and Iron Trusses 399-419 CHAPTER XIII. COLUMNS 420-431 TABLES 432-436 « LIST OF TABLES PAGE Table I. — Distance of extreme fibres, i, r, a, for different sections 12-21 Table II. — Value of n in compression formula 23 Table III. — Wrinkling strains 26 Table IV. — Strength and weights of materials 37-42 Table V. — Strengths and weights of stones, bricks, and cements 43-45 Table VI. — Weights of other materials 46 Table VII. — Bending moments, shearing and deflection 58, 59 Table VIII. — Strength of 1" wooden beams 62 Table IX. — When to calculate rupture or deflection . . 63 Table X. — Angles of friction of soils, etc 103 Table XI. — Values of p and slope for retaining walls 105 Table XV. — Iron I-beam girders 203 Table XVI. — Value of y in formula for unbraced beams 209 Table XVII. — Continuous girders 218-19 Table XVIII.— Trussed beams 220-21 Table XII. — Wooden floor -beams .at end of text. Table XIII. — ^Wooden girders " Table XIV. — Iron I beams for floors. . " Table XIX.— Explaining Tables XX to XXV. . . Table XX. — List of iron and steel I beams .... * ' ix X LIST OF TABLES PAGE Table XXI. — List of iron and steel channels., .at end of text. Table XXII. — List of iron and steel even-legged angles * * Table XXIII. — List of iron and steel uneven- legged angles " Table XXIV. — List of iron and steel tees " Table XXV. — List of iron and steel decks, ha^f- decks, and zees " Table XXVI. — Analysis of cast-irons ** Table XXVII. — Densities and vreights of cast- irons * * Table XXVIII. — Classification of irons and steels ** Table XXIX. — Classification of structural irons and steels ** Table XXX. — Amount of extension and con- traction, in inches, of cast and wrought iron bars, 100 feet long, under different strains... ** Table XXXI. — Length of cast or wrought iron bars, in feet, that will stretch or contract exactly one inch under different strains ** Table XXXII. — Ultimate breaking strength of materials under different kinds of strains (dead, rolling, inter- mittent, etc.) ** Table XXXIII. — Proportion for screw-threads, nuts and bolt heads " Table XXXIV. — Eecent tests of irons and steels. . " Table XXXV. — Bearing values for iron and steel rivets and pins ** Table XXXVl. — Bearing values for iron and steel pins and bolts, 1" to 3" diameter ** LIST OF TABLES xi PAGE Table XXXVII. — Bearing value for iron and steel pins and bolts, 3" to 6" di- ameter at end of text. Table XXXVIII. — Shearing, bending and tensional values for iron and steel rivets, pins and rods of i^" to 1" in diameter ** Table XXXIX. — Shearing, bending and tensional values for iron and steel pins and rods 1" to 3" in diameter Table XL. — Shearing, bending and tensional * values for iron and steel pins and rods 3" to 6" in diameter ** Table XLI. — Wrought-iron riveted girders: strength of webs ** Table XLII. — Wrought-iron riveted girders: strength of the angle bars... ** Table XLIII. — Wrought-iron riveted girders: strength of flanges ** Table XLIV. — Table of wind-pressures on roofs.... 360 Table XLV. — Strength of hollow cylindrical columns of cast-iron (3" to 7" diameter) . . . 432 Table XL VI. — Strength of hollow cylindrical columns of cast-iron (8" to 10" diameter) ... 436 Table XL VII. — Strength of hollow cylindrical columns of cast-iron (11" to 13" diameter).. 434 Table XL VIII. — Strength of hollow cylindrical columns of cast-iron (14" to 16" diameter).. 485 Table XLIX. — Safe compressive loads per square inch on wrought-iron colunms 436 Table L. — Reinforced concrete forms 270 LIST OF FORMULAE NO. OF FORMULA PAGE 1. — Fundamental formula — stress and strain 21 2. — Compression, short columns 22 3. — " long columns 23 4. — Wrinkling strains 25 5. — Lateral flexure 28 6. — Tension 29 7. — Shearing, across grain 32 8. — " along grain 32 9. — " at left reaction 33 10. — " at right reaction 33 11. — Vertical shearing strain, any point of beam 34 12. — " " " " cantilever 35 13. — ^Longitudinal shearing 35 14. — Amount of left reaction (single load) . 47 15. — " right " " 47 16. — ' ' left ' ' (two loads) 47 17. — ' ' right " " 47 18. — Transverse strength, uniform cross-section , 50 19. — " " upper fibres , 50 20. — " " lower " 50 21. — Central bending moment on beam, uniform load 52 22. — " " " centre load ...... 52 23. — Bending moment, any point, several loads , 53 24. — " " " " check to No. 23 53 25. — " " cantilever, uniform load . 53 26. — " " " end load 53 27. — " " any loading 54 28. — Safe deflection, beams, not to crack plastering. ....... 56 zii LIST OF FORMULAE xiii NO. OF FORMULA PAGE 29. — Safe deflection, cantilevers or uneven loads on beams, not to crack plastering 57 30. — Comparative formula — beams, strength, different cross- sections, 60 31. — Comparative formula — ^beams, deflection, different cross- sections 61 32. — Comparative formula — ^beams, strength, different lengths and sections 62 33. — Comparative formula — beams, deflection, different sec- tions 62 34. — Comparative formula — columns, different lengths 65 35. — " " " different sections 65 36. — Approximate formula for plate girders 65 37. — Deflection, cantilever, uniform load 66 38. — " " end load 66 39. — " beam, uniform load 66 40. — " " centre load 66 41. — " " any load 67 42. — " cantilever, 67 43. — Point of greatest deflection of beam 67 44. — Strain on arch edge nearest to line of pressure 78 45. — " " farthest from " 79 46. — Safe load on long piles 93 47. — Amoimt of pressure against retaining walls, with higher backing 102 48. — Amount of pressure against retaining walls, with level backing 102 49. — Formula 47 simplifiea 103 50. — " 48 " 104 51. — Pressures against cellar walls, general case 105 52. — " " " usual case 105 53. — Eeservoir walls, fresh water 107 54. — " " salt water 107 xiv LIST OF FORMULAE NO. OP FORMULA PAGE 55. — Eetaining wall, backing loaded 107 56. — " " " centre of pressure 108 57. — " " graphical method 109 58. — Buttressed wall, average weight Ill 59. — Piers, chimney and tower walls, in inches 139 60. — " " " in feet : . 139 61. — Flue area of chimneys 143 62. — Thickness of walls, in inches 146 63. — " " in feet 146 64. — Pressure on walls from barrels 152 65. — Metal bands to domes 177 66. — Height of courses in domes 178 67. — ^Approximate depth at crown for arches 182 68. — Approximate rule for arches 182 69. — Width of stirrup irons 197 70. — Thickness of stirrup irons 197 71. — " " for wrought-iron 197 72. — Eectangular beams, transverse strength, uniform load . . 198 73. — " " " centre load ... 198 74. — " " " any load 199 75. — ' ' cantilever, ' ' imif orm load . . 199 76. — " " " end load 199 77. — " " " any load 199 78. — Lateral flexure for beams 208 79. — Approximate deflection, iron (also steel) beams and girders, uniform load 217 80. — Approximate deflection, iron (also steel) beams and girders, centre load 222 81. — Safe length not to crack plastering, iron (also steel) beams and girders, uniform load 222 82. — Safe length not to crack plastering, iron (also steel) beams and girders, centre load 222 LIST OF FORMULAE XV NO. OF FOEMULA p^Qj, 83. — Average strain on chords, uniform cross-section and load 223 84. — " top chord " << 224 85. — " bottom chord " " 224 86. — " top chord, uniform section, centre load 225 87. — Average strain on bottom chord, uniform section, centre load 225 88. — Expansion or contraction from strain 225 89. — Deflection of girder, parallel flanges, uniform section. . 226 " " varying section.. 227 91. — Safe length not to crack plastering, parallel flanges, varying section 228 92. — Graphical method — moment of resistance 243 93. — " bending moment 243 94. — cross-shearing 244 95. — " deflection with i 246 96. — ** deflection with v 246 97. * * deflection, random pole distance . . . 246 " thickness of plate girder flange... 268 ^9- " value of angles in diminishing flanges 268 107. — Greatest pitch for rivets 285 108. — Least pitch for rivets 285 109. — Number of rivets required at lap and butt joints to resist bearing 288 110. — Number of rivets required at lap and butt joints to resist single shearing 288 111. — Number of rivets required at lap and butt joints to resist double shearing 288 112. — Number of rivets required at butt joints to resist bend- ing or cross-breaking 288 113. — Number of rivets required at butt joints to resist bend- ing or cross-breaking 288 xvi LIST OF FORMULAE NO. OF FORMULA PAGE 114. — Thickness of plate required to resist rivets compress- ing or tearing same 289 115. — Length of plate needed between rivet and edge of plate 289 116. — Load on cantilever only, or on overhanging pin end, nearer reaction 307 117. — Same as 116, farther reaction 307 118. — Load on both beam and cantilever, or on pin and over- hanging end, nearer reaction 308 119. — Same as 118, farther reaction 308 120. — Eequired thickness of eye-bar at pin 309 121. — Horizontal flange strain on rivets, in plate girders... 328 122. — Number of rivets required in ends of each layer of flange plates 331 123. — Horizontal slide tendency in plates of riveted girders 332 124. — Required thickness of web in plate girders 335 125. — Total compression strain on end stiffeners of plate girder 336 126. — Total compression strain on intermediate stiffeners of plate girders 336 127. — Where stiffeners are needed in plate girders 336 128. — Average fibre stress in flanges of plate girders 354 129. — Increase in stress for curved compression (struts) or tension (tie) members 380 130. — Amount of reaction (left side) due to wind 387 131. — Amount of reaction (right side) due to wind 387 132. — Safe load on each roller under end of truss 418 Safe Building Construction. PREFACE. Safe Building Construction is the logical sequence of my former work on Construction, ' ' Safe Building. ' ' Methods of construction and building materials have changed so radically that this new book seems called for. Much of my former work has been retained, though my object has been to condense — saving the student's time in these busy days — while still following the original plan which was: — "to furnish to any earnest student the opportunity to acquire, so far as books will teach, the knowledge necessary to erect safely any building. While of course, the work will be based strictly on the science of mechanics, all useless theory will be avoided. The object will be to make the articles simply practical. To follow any of the mathematical demonstrations, arithmetic, a rudimentary knowledge of algebra and plane geometry will be sufficient." Special effort has been made to introduce in this new work a general survey and fair knowledge of Concrete Construction, in- cluding concrete foundations, grillage, piles, piers, and other sub- soil as well as above-soil constructions, such as floors, columns, girders, walls, etc. For the many flattering letters and comments received for ' ' Safe Building ' ' during the last two decades, some quite recently, I thank my professional friends, young and old, and hope this newer work — Safe Building Construction — may be equally acceptable. CHAPTER L STRENGTH OF MATERIALS. (German, Festigkeit ; French, Resistance des materiaux.) All solid bodies or materials are made up of an infinite number oi atoms, fibres or molecules. These adhere to each other and resist separation with more or less tenacity, varying in different materials. This tenacity or tendency of the fibres to resume their former rela- tion to each other after the strain is removed is called the elasticity of the material. It is when this elasticity is overcome that the fibres separate, and the material breaks and gives way. There are to be considered in calculating strengths of materials two kinds of forces, viz., the external (or applied) forces and the inter- nal (or resisting) forces. The external forces are any kind of forces applied to a material and tending to disrupt or force the fibres apart. Thus a load lying apparently perfectly tranquil on a beam is really a very active force ; for the earth is constantly attracting the load, which tends to force its way downwards by gravitation and push aside the fibres of the beam under it. These latter, however, resist separation from each other, and the amount of the elasticity of all these fibres being greater than the attraction of the earth, the load is unable to force its way downwards and remains apparently at rest. Strain. The amount of this tendency to disrupt the fibres (produced by the external forces) at any point is called the " strain " at that point. Stress. The amount of the resistance against disruption of the fibres at such point is called the " stress " at that point. External (or applied) forces, then, produce strains. Internal (or resisting) forces produce stresses. This difference must be well understood and constantly borne in mind, as strains and stresses are the opposing forces in the battle of all materials against their destruction. Ultimate When the strain at every point of the material just Stress, equals the stress, the material remains in equilibrium. The greatest stress, at any point of a material that it is capable of ex- erting is the ultimate stress (that is, the ultimate strength of resist- ance) at that point. Were the strain to exactly equal that ultimate stress, the material, though on the point of breaking, would still be STRENGTH OF MATERIALS. 8 safe theoretically. But it is impossible for us to cal- pactgr-of- culate so closely. Besides we can never determine Safety, accurately the actual ultimate stress, for different pieces of the same material vary in practice very greatly, as has been often proved by experiment. Therefore the actual ultimate stress might he very much less than that calculated. Again, it is impossible to calculate the exact strain that will always take place at a certain point ; the applied forces or some other con- ditions might vary. Therefore, to provide for all possible emergen- cies, we must niake our material strong enough to be surely safe ; that is, we must calculate (allow) for a considerably greater ultimate stress at every point than there is ever likely to be strain at that point. The amount of extra allowance of stress varies greatly, according to circumstances and material. The number of times that we calcu- late the ultimate stress to be greater than the strain is called the fac- tor-of-safety (that is, the ratio between stress and strain). If the elasticity of different pieces of a given material is practi- cally uniform, and if we can calculate the strain very closely in a given case, and further, if this strain is not apt to ever vary greatly, or the material to decay or deteriorate, we can of course take a low or small factor-of-safety ; that is, the ultimate stress need not exceed many times the probable greatest strain. On the other hand, if the elasticity of different pieces of a given material is very apt to vary greatly, or if we cannot calculate the strain very closely, or if the strain is apt to vary greatly at times, or the material is apt to decay or to deteriorate, we must take a very high or large factor-of-safety, that is, the ultimate stress must exceed many times the probable greatest strain. Factors-of-safety are entirely a matter of practice, experience, and circumstances. In general, we might use for stationary loads : A factor of safety of 3 to 4 for wrought-iron and steel, " " " 6 for cast-iron, «' " « 4 to 10 for wood, " " " 10 for brick and stone. For moving-loads, such as people dancing, machinery vibrating, dumping of heavy loads, etc., the factor-of-safety should be one- half larger, or if the shocks are often repeated and severe, at least double of the above amounts. Where the constants to be used in formulas are of doubtful authority (as is the case with most of them for woods and stones), the factor-of-safety chosen should be the high est one. S.\1K liL'I l.niNti. In building-materials we meet with four kind of strains, and, o} course, with the four corresponding stresses resisting them, viz. : — STRAINS. Compression, or crushing strains, Tension, or pulling strains, Shearing, or sliding strains, and Transverse, or cross-breaking strains. STRESSES. The resistance to Compression, or crushing-stress, The resistance to Tension, or pulling-stress. The resistance to Shearing, or sliding-stress, and The resistance to Transverse strains, or cross-breaking stress. Materials yield to Compression in three different ways : — 1. By direct crushing or crumbling of the material, or 2. By gradual bending of the piece sideways and ultimate ruptu? e, or 3. By buckling or wrinkling (corrugating) of the material length- wise. Materials yield to Tension, 1. By gradually elongating (stretching), thereby reducing the size of the cross-section, and then, 2. By direct tearing apart. Materials yield to Shearing by the fibres sliding past each other in two different ways, either 1. Across the grain, or 2. Lengthwise of the grain. Materials yield to Transverse strains, 1. By deflecting or bending down under the load, and (when this passes beyond the limit of elasticity), 2. By breaking across transversely. In calculating strains and stresses, there are certain rules, expres- sions, and formulae which it is necessary for the student to under- stand or know, and which will be here given without attempting elab- orate explanations or proofs. For the sake of clearness and simplic- ity, it is essential that in all formulae the same letters should always represent the same value or meaning ; this will enable the student to read every formula off-hand, without the necessity of an explanatory key to each one. The writer has further made it a habit to express, in all (iases, his formulaa in pounds and inches (rarely using tons or feet) ; this will frequently make the calculation a little more elabo- OLDSSA m' 1)1' SYMnoLS. fi rate, but it will be found to greatly simplify the formulae, and to make their understanding and retention more easy. In the following articles, then, a capital letter, if it were used, would invariably express a quantity (respectively), either in tons or feet, while a small letter invariably expresses a quantity (respec- tively), either in pounds or inches. The following letters, in all cases, will be found to express the same meaning, unless distinctly otherwise stated, viz. : — a signifies area, in square inches. b " breadth, in inches. c " constant for ultimate resistance to compression, in pounds, per square inch. (See Tables IV and V.) d signifies depth, in inches. e " constant for modulus of elasticity, in pounds-inch, that is, pounds per square inch. (See Table IV.) / " factor of safety. g " constant for ultimate resistance to shearing, per square inch, across the grain. (See Tables IV and V.) g, " constant for ultimate resistance to shearing, per scjuare inch, lengthwise of the grain. (See Table IV.) h " height, in inches. t " moment of inertia, in inches. (See Table I.) k " ultimate modulus of rupture, in pounds, per square inch. (See Tables IV and V.) / " length, in inches. m " TwomenZ or Jenrfm^rmomeni, in pounds-inch. (See Table IX.) n '* constant in Rankine's formula for 'compression of long pillars. (See Table 11.) o " the centre. p " the amount of the left-hand re-action (or support) of beams, in pounds. q " the amount of the right-hand re-action (or support) of beams, in pounds, r " moment of resistance, in inches. (See Table I.) s " strain, in pounds. t " constant for ultimate resistance to tension, in pounds, per square inch. (See Tables IV and V.) M " uniform load, in pounds, i; " stress, in pounds. w " load at centre, in pounds. r, y, and z signify unknown quantities, cither in pounds or inches. 6 SAFE BUILDING. 6 signifies total deflection, in inches. p' " square of the radius of gyration^ in inches. 5 " diameter, in inches. r " radius, in inches. IT = 3.14159, or, say, 3-1 signifies the ratio of the circumference and diameter of a circle. If there are more than one of each kind, the second, third, etc., are, indicated with Roman numerals, as for instance, a, a„ a„, a„„ etc., or 6, 6„ 6„, 6„„ etc. In taking moments, or bending moments, strains, stresses, etc., to signify at what point they are taken, the letter signifying that point is added, as for instance : — m signifies moment or bending moment at centre, mj, " " " " point A. tHb " " " " point B. mjt ** " «' " point X. s ** strain at centre. *' " point B. Sx " " " X. V ** stress at centre. r» " " point D. vx '« " X to signifies load at centre. Wx " " " point A. CENTRE OF GRAVITY. (German, Schwerpunkt; French, Centre de gravite.) The centre of gravity of a figure, or body, is that centre or point upon which the figure, or body, will balance Gravity, itself in whatever position the figure or body may be placed, provided no other force than gravity acts upon the figure or body. To find the centre of gravity of a plane figure, find two neutral axes, in different directions, and their point of intersection will be the centre of gravity required. NEUTRAL AXIS. (Grerman, Neutrale Achse; French, Axe neutre.) The neutral axis of a body, or figure, is an imagin- Neutral Axis, ary line upon which the body, or figure, will always balance, provided the body, or figure, is acted on by no other force than gravity. The neutral axis always passes through the centre of gravity, and may run in any direction. In calculating transverse strains, the neutral axin NEUTRAL AXIH. 7 designates an imaginary line of the body, or of the cross-section of the body, at which the forces of compression and tension meet. Tin- strain on the fibres at the neutral axis is always naught. Extreme Fibres. On the upper side of the neutral axis the fibres are compressed, while those on the lower side are elongated. The amount of compression or elongation of the fibres increases directly as their distance from the neutral axis ; the greatest strain, tlierefore, being in the fibres along the upper and lower edges, these being farthest from the neutral axis, and therefore called the extreme fibres. It is necessary to calculate only the ultimate resistance of these extreme fibres, as, if they will stand the strain, certainly all the other fibres will, they all being nearer the neutral axis, and consefinf ntly less strained. Where the ultimate resistances to compression and tension of a material vary greatly, it is necessary to so design the cross-section of the body, that the "extreme fibres" (farthest edge) on the side offering the weakest resistance, shall be nearer to the neutral axis than the " extreme fibres " (farthest edge), on the side offering the greatest resistance, the distance of the " extreme fibres " from the neutral axis being on each side in direct proportion to their respective capacities for resistance. Thus, in cast-iron the resistance of the fibres to compression is about six times greater than their resistance to tension ; we must therefore so design the cross-section, that the distance of the neutral axis from the top-edge will be six- sevenths of the total depth, and its distance from the lower edge one-seventh of the total depth. To find the neutral axis of any plane-figure, some writers recom- mend cutting, in ~T~ stiff card-board, Neutral Axis. k. orec iV£^ I T d 1 oi-eq-QH-- ill a duplicate of the figure (of which the neutral axis is sought), then K to experiment until it balances on the edge of a knife, the line on which it balances being, of course, the neutral axis. This Fig. I. is an awkward and unscientific method of procedure, though there may be some cases where it will recommend itself as saving time and trouble. The following general formula, however, covers every case : To find the neutral axis M-N in any desired direction, draw a line X-Y at random, but parallel to the desired direction. Divide the figure into any number of simple figures, of which the areas and cen* 8 SAFE BUILDING. tres of gravity can be readily found, then tlie distance of the neutral axis ilf-iVfrom the line X-Ywi\l be equal to the sum of the products ^ of each of the small areas, multiplied by the distance of the centre of gravity of each area from X-Y, the whole to be divided by the entire area of the whole figure. An Example will make this more clear. Find the horizontal neutral axis of the cross-section of a deck-beam, standing vertically on its bottom-flange. Draw a line (X-Y) horizontally (Fig. 1), then letc?,, d,„ rej)- resent the respective distances from ^-Fof the centres of gravity of the small subdivided simple areas a„ a„, a,„, then let o stand for the whole area of section, that is : — Oi + an + a,,, =a, then the required distance (d) of the neutral axis Af-iV from X-Y, will be a To 6nd the centre of gravity of the figure, we might find another neutral axis, but in a different direction, the point of intersection of the two being the required centre of gravity. But as the figure is uniform, we readily see that the centre of gravity of the whole figure must be lialf-way between points A and B. Centres of "^^^ centre of gravity of a circle is always its cen- Cravlty. t^e. The centre of gravity of a parallelograiSi is al- ways the point of intersection of its two diagonals. The centre of gravity of a triangle is found by bi-secting two sides, and connecting these points each with its respective opposite apex of the triangle, the point of intersection of the two lines being the required centre of gravity, and which is always at a distance from each base equal to one-third of the respective height of the triangle. Any line drawn through either centre of gravity is a neutral axis. MOMENT OF INERTIA. (German, Tr&gheitsmoment ; French, Moment d'inertie, or Moment de giration.) Moment of iner- "^^^ moment of inertia, sometimes called the mo- t i a. > See Table I,) ment of gyration, is the formula representing the inactivity (or state of rest) of any body rotating around any axis. The reason of the connection of this formula with the calculation of strains and the manner of obtaining it cannot be gone into here, as it would be quite beyond the scope of these articles. The moment of ' If line X-Y is inside of (bisects) figure, take sum of products on one side only and deduct sum of products on other side. TIIK CENTRE OF GYRATION AND RADIUS OP GYRATION. 9 inertia of any body or figure is the sum of the products of each par- ticle of the body or figure multiplied by the square of its distance from tlie axis around which the body or figure is rotating. A table of moments of inertia, of various sections, will be given later on and will be all the student will need for practical purposes. THE CENTRE OF GYRATION AND RADIUS OF GYRATION. (German, Trdgheitsmiltelpunkt ; French Centre de giration.) The centre of gyration "is that point at which, if ^ ^, r , . • Square Of Radi- tlie whole mass of a body rotating around an axis us of Gyration. . , c ■ 11 ^ J • f (See Table I.) or point of suspension were collected, a given force apjilied would produce the same angular velocity as it would if ap- plic d at the same point to the body itself." The distance of this cen- tre of gyration from the axis of rotation is called the radius of gyra- tion (German, Tragheitshalhmesser ; French, Rayon de giratiori) ; this latter is used in the calculation of strains, and is found by dividing the moment of inertia of the body by the area or mass of the body, and extracting the scjuare root of the quotient, or, 9=\/l, or S — a- A table will be given, later on, of the "squares of the radius of gyra- tion " (German, Quadrat des Trdgheitshalbmessers ; French, Carri du rayon de giration}. THE MOMENT OF RP:SI8TANCE. (German, Widerstandsmonient ; French, Moment de resistance). The moment of resistance of any fibre of a body, _ . . , Moment of Re* revolving around an axis, is equal to the moment of sistance. THE MOMENT OF INERTIA OF ANY CROSS-SECTION. 11 Therefore ilit* inomrnt of inertia, t, of the whole deck-beam would be: — But a,=^x,^ Further, a„ = b^, h„. And a,„ = h,„ A„„ which, inserted above, gives for The following table (T) gives the values for the moment of inertia (i), moment of resistance (r), area (d), square of radius of gyration etc., for nearly every cross-section likely to be used in building. Those not given can be found from Table I by dividing the cross- section into several simpler parts, for which examples can be found in the table. Note, that it makes a great difference whether the neu- tral axis is located through the centre of gravity (of the part), or elsewhere. When making calculations we must, of course, insert in the different formulae in place of i, r, a, Q^, their values (for the re- spective cross-section), as given in Table I. ]\/ote. — Throughout this ivork the writer uses the foreign punctuation in formula:, vis: a period or point denotes the sign of multiplication; while a comma in connection with numerals denotes the decimal sign. 12 SAFE BUILDING. TABLE I. DISTANCE OF EXTREME FIBRES, MOMENTS OF INERTIA AND RESISTANCE, SQUARE OP RADIUS OF GYRATION, AND AREAS OF DIFFERENT SHAPES OF CROSS-SECTION'S. Number and Form of Section. 5 • !z! 12 12 12 12 11 6 bd^ 6 d^-d*- 6d hd^-b,d,» 6d 11 14^ CZ2 bd d^~d,^ bd-bJ, 22 TABLE I, CONTINUED. 18 14 SAFE BUILDING. Number and Form of Section. a r SI. I :+ 3=^ + a- I + TABLE I, CONTINUED. Id 16 SAFE BUILDING. Number and Form of Section. a : a> ? Iz! ? 19 m 20 M- 22 + + 5? d £,3 i= 3 3 3 3 .. O + 1 <" + 1 _5>i a. s 2 w "a TABLE I, CONTIKUED. 17 *^ b o B o B o B axis a » to a Number and Form of extr bres a o o B B a B (6 Section. r- o a O F • a B M B o Neu- CD ►1 0r CO* 23 + + I Si- "l rr a- ++ 1 I ? T + + + + i + a- + 18 SAFE BUILDING. TABLE I, CONTINUED. 19 20 SAFE BUILDING. TABLE I, COXTINTJED. 21 Number and Form of Section. a s. Corrugated Iron. T" Breadth of each sheet = 6 224 112 15 11, 14 hd CALCULATION OF STRAINS AND STRESSES. As we have already noticed, the stress should exceed the strain as many times as the adopted factor-of-safety, or : — Stress r ^ e e ^ —= ractor-oi-saiety. btrain Or, stress = strain X factor-of-safety. This holds good for all calculations, and can be expressed by the following simple fundamental formula : — ■ - Fundamental v = s.f (1) Formula. Where r = the ultimate stress in pounds. " « = " strain in pounds. 22 SAFE BUILDING. And where f = the f ac.tor-of-saf ety. COMPRESSION. Compression, In pieces under compression the load is directly ' applied to the material. In short pieces, therefore, which cannot give sideways, the strain will just equal the load, or we have : — s = w. Where s — the strain in pounds. And where w — the load in pounds. The stress will be equal to the area of cross-section of the piece being compressed, multiplied by the amount of resistance to com- pression its fibres are capable of.^ This amount of resistance to compression which its fibres are capable of is found by tests, and is given for each square inch cross-section of a material. A table of constants for the resistance to crushing of different materials will be given later on. (See table IV.) In all the formulae these constants are represented by the letter c. We have, then, for the stress of short pieces under compression : — V = a.c Where v is the ultimate stress in pounds. Where a is the area of cross-section of the piece in inches. And where c is the ultimate resistance to compression in pounds per square inch. Inserting this value for v, and w for s in the fundamental formula (1), we have for short pieces under compression, which cannot yield sideways : — a.c = w.f, or : — Short Columns. Where w = the safe total load in pounds. Where a = the area of cross-section in inches. Example. What is the safe load which the granite cap of a 12" x 12" pier will carry, the cap being twelve inches thick f The cap being only twelve inches high, and as wide and broad as 1 This is not theoretically correct, as there is in every case a tendency for the material under compression to spread; but it is near enough for all practical purposes. (2) And where =: the safe resistance to crushing per square inch. COMPRESSION. 23 high, is evidently a short piece under compression, therefore the above formula (2) applies. The area is, of course: a = 12.12= 144 square inches. The ultimate resistance of granite to crushing per square inch is, say, fifteen thousand pounds, and using a factor-of-safety of ten, we have for the safe resistance : — y = - = 1500 lbs. Therefore the safe load w on the block would be : — w=144. 1500 = 216000 pounds. Where long pieces (pillars) are under compres- Compression, . , • ' ^^• -1 Long Columns. sion, and are not secured agamst yielding sideways, it is evident they would be liable to bend before breaking. To ascer- tain the exact strain in such pieces is probably one of the most dif- ficult calculations in strains, and in consequence many authors have advanced different theories and formulae. The writer has always preferred to use Rankine's formula, as in his opinion it is the most reliable. According to this, the greatest strain would be at the cen- tre of the length of the pillar, and would be equal to the load, plus an amount equal to the load multiplied by the square of the length in inches, and again multiplied by a certain constant, n, the whole divided by the " square of the radius of gyration " of the cross-sec- tion of the pillar. We have therefore for the total strain at the cen- tre of long pillars : — Where s = the strain in pounds. " w = the total load in pounds. " Z=the length in inches. " p2 — - tiie square of the radius of gyration of the cross-section. " n = a constant, as follows : — TABLE II. VALUE OF n IN FORMULA FOB COMPRESSION. Material of pillar. Both ends of pillar smooth (turned or planed.) One end smooth (turned or planed) other end a pin end. Both ends pin ends. Cast-iron O.0O03 0.0004 0.00057 "Wrought-iron 0.000025 0.000033 0.00005 Steel 0.00002 0.000025 0.000033 Wood 0.00033 0.00044 0.00067 Stone 0.002 Brick 0.0033 24 BAFB BUILDING. The stress of course will be as before : — v=za. c. Where w = the ultimate stress in pounds. " a = the area of cross-section in inches. " c = the ultimate resistance to crushing per square inch. Inserting the values for strain, s, and stress, r, in the fundamental formula (1) we have: — "(7) Long Columns. 1 | „ Where w = the safe total load on the pillar. ** a = the area of cross-section in inches. " = the square of the radius of gyration of the cross-section. " 1 = the length in inches. " ~Y~~ ^^/^ resistance to crushing per square inch, n = a constant as given in Table II. Example. What safe load loilla 12" x 12" brick pier carry, if the pier is ten feet long, and of good masonry f The area of cross-section will be : — a = 12.12 = 144 square inches. .The square of the radius of gyration according to Section No. 1 in Table I would be : — and as d=12, we have Q ^IHI— 1 9 12 ^ 12 For the safe resistance to crushing per square inch, we have, using a factor-of-safety of ten, and considering the ultimate resistance to be 2,000 pounds per square inch, / c \ 2000 ,^ (-^)=^=200lbs. The length will be ten feet, or one hundred and twenty inches ; therefore : — P = 14400 For n we must use (according to Table II), for brickwork : — n z= 0,0033; COMPRESSION. 26 Therefore the safe total load on the pier would be : — 144. 200 28800 = — j — — = 5806 lbs. «>= I 14400. 0,0033 l-}-3,96 12 In all formulae where constants and factors of safety are used, it will be found simpler and avoiding confusion to immediately reduce the constant by dividing it by the factor-of-safety, and then using only the reduced or safe constant. Thus if c= 48000 pounds, and if y=:4, do not write into your formula for ^-^^ = 1^^^, but use at once for ^-^ ^ = 12000. Materials in compression that have an even bearing on all parts of the bed will stand very much more compression to the square inch than materials with rough, uneven or rounded beds, or where the bearing is on part of the cross-section only, as in the case of pins (in trusses) bearing on eye-bars. It is usual in calculating to make allowance for this. Columns with perfectly even bearing on all parts of the bed (planed or turned iron or dressed stone) will stand the largest amount of compression. Columns with rough, rounded or uneven ends are calculated the same as for pin-ends of eye-bars. In the table (II) giving the values for n of Rankine's formula for com- pression, the different values for smooth and also for pin ends are given. WRIKKLING STRAINS. Thin pieces of wrought-iron under compression endwise may neither crush nor deflect (bend), but give way by wrinkling, that is, buckling or corrugating, provided there are no stiffening-ribs length- wise. Thus a square, tubular column, if the sides are very' thin might give way, as shown in Figure 2, which is called wrinkling. Or, in a similar way, the top plate of a boxed Fig- 2. girder, if very thin, might wrinkle, as shown in Figure 3, under heavy compressive strains. To calculate this strain use the following formula : Where w = the amount of ultimate compression in ^ul rial. pounds per square inch, which will wrinkle the mate- «?,. = a constant, as given in Table III, d ■= the thickness of plate in inches. SAFE BUILDING. b = the unstiffened breadth of plate in inches. If a plate has stiffening ribs along both edges, use for b the actual breadth between the stiffening ribs ; if the plate is stiffened along one edge only, use 46, in place of b. Thus, in the case of the boxed girder, Figure 3, if we were considering the part of top plate between the webs, we should use for b in the formula, the actual breadth of b in inches ; while, if we were considering the overhang- ing part b, of top plate, we should use 46, in place of 6 in formula. For rectangular columns use 160,000 pounds for Wr', for tubular beams, top plates of girders, and single plates use 200,000 pounds for Wf. With a factor-of-safety of 8, we should have 160000 3 53000 pounds for rectangular columns, and ^^^^^^ = 66000 pounds for 3 tubular beams, top plates of riveted girders and single plates. For to we shall use, of course. 36000 = 12000 pounds, which is the safe allowable compressive strain. This would give the following table for safe unstiffened breadth of wrought-iron plates, to prevent wrinkUng of plates. TABLE III. Thickness of Plate In inches. Safe breadth iu inches of Plate Btiifened along both edges, (use 6.) Rectan- gular Col- umns 6 O 7 41 1 H 1 f 14| i 1 u 24 f H If 34^ 2 39 Tubular Beams, riveted Girders, and single Plates. 111 15J 18i 22H 26^ ^ * 1 6 45f 52^ 60^ Safe breadth in inches of Plate stiifened along one edge only, (use 46i) Rectangular Columns. 2tV 3 Riveted Girders and single Plates. H m 3| 15^ The above table will cover every case likely to arise in buildings. WRINKLING. Two facts should be noticed in connection with wrinkling : 1. That the length of plate does not in any way affect the resi9^ aiice to wrinkling, which is dependent only on the breadth and thick- ness of tiie part of plate unstifFened, and 2. That the resistance of plates to wrinkling being dependent on their breadth and thickness only, to obtain equal resistance to wrink- ling at all points (in rectangular columns with uneven sides), the thickness of each side should be in proportion to its breadth. Thus, if we have a rectangular column 30" X 15" in cross section and the 30" side is 1" thick, we should make the 15" side but J" thick, for as 30" : 1" : : 15" : i". Of course, we must also calculate the column for direct crushing and flexure, and in the case of beams for rupture and deflection, as well as for wrinkling. Example of Wrinkling. It is desired to make the top plate of a boxed girder as wide as pos- sible, the top flange is to be 1^" thick, and is to be subjected to the full amount of the safe compressive strain, viz: 12,000 pounds per square inch ; how wide apart should the webs be placed, and how much can the plate overhang the angles without danger of wntikling f Each web to be y thick, and the angles 4" X 4" eachf For the distance between webs we use b in Formula (4). which is the safe width between webs to avoid wrinkling For the overhanging part of top plate we must use 46, in place of b in Formula (4). / 66000 \2 46. = 1 i ^ ^2000 / = ^ 'if ' therefore, 37iA ^1= = 9,453, or say, 6, = 9^". The total width of top plate will be, therefore, including 1" for two webs and 8" for the two angles, or 9", and remembering that there is an overhanging part, 6„ each side, 9" + 6 + 6,-f 6. = 9+37|f + 9^ + 9,'5 ' = 6511". By referring to Table ITT, we should have obtained the same result, with- Fi£. 4 out the necessity of any calculation. Figure 4 will make the above still more clear. SAFE BUILDING. LATERAL FLEXURE IN TOP FLANGP:8 OF BKAMS, GIRDERS, OR TRUSSES, DUE TO COMPRESSION. The usual formulae for rupture and deflection assume the beam, girder or truss to be supported against possible lateral flexure (bending sideways). Now, if the top chord of a truss or beam is comparatively narrow and not supported sideways, the heavy, com- pressive strains caused in same may bend it sideways. To calculate this lateral flexure, use the formula given for long columns in com- pression, but in place of I use only two-thirds of the span of the beam, girder or truss, that is |Z, and for w use one-third of tlie great- est compressive strain in top chord, which is usually at the centre. Inserting this in Formula (3) we have : a(f) ^ij) Y ~ 4Pn *i"ansposing, we have, w = (5) where a the area of the cross-section of the top chord in inches, = a.(-^) (9) And at the right-hand end of beam : — Where /) = the amount of load, in pounds, carried on the left- hand support. Where q = the amount of load, in pounds, carried on the right- hand support. Where a = the area of cross-section, in inches, at the respective support. Where (^y^ = the safe resistance, per square inch, to cross-shear- ing. Example. A spruce beam of 5' clear span is 24" deep and 3" wide ; how much uniform load will it carry safely to avoid the danger of shearing off at either point of support ? The beam being uniformly loaded, the supports will each carry one-half of the load ; if, therefore, we find the safe resistance to shear- ing at eiiher support, we need only double it to get the safe load (in- stead of calculating for the other support, too, and adding the results). Let us take the left-hand support. From Formula (9) we have : — p = a.{j.) Now, we know that a = 24. 3 = 72 square inches. 84 SAFE BUILDING. The ultimate resistance of spruce to cross-shearing is about thirty- six hundred pounds per square inch ; using a factor-of-safety of ten, we have for the safe resistance per square inch : — (^j^'j==^_^^3Q0 pounds. We have, now : — p = 72.360 = 25920 pounds. Similarly, wc should have found for the right-hand support: — ^ = 25920 pounds. And as: — u =p -\-q = 51840 pounds, that will, of course, be the safe uniform load, so far as danger of shear- ing is concerned. The beam must also be calculated for transverse strength, deflec- tion and lateral flexure, before we can consider it entirely safe. These will be taken up later on. Should it be desired to find the amount of vertical shearing strain X at any point of a beam, other than at the points of support, use : — (11) Where x = the amount of vertical shearing strain, in pounds, at any point of a beam. C p'\ the reaction, in pounds, (that is, the share of the . Where ^ or >• = total loads carried) at the nearer support to the \q) point. Where S = the sum of all loads, in pounds, between said nearer support and the point. When X is found, insert it in place of w, in Formula (7), in order to calculate the strength of beam necessary at that point to resist the shearing. Example. A spruce beam, 20' long, and 8" deep, carries a uniform load of one hundred pounds per running foot. What should be the thickness of beam 5' from either support, to resist safely vertical shearing? Each support will carry one-half the total load ; that is, one thou- sand pounds; so that we have for Formula (11) : — 1t1= = 1000 pounds. The sum of all loads between the nearer support and a point 5' from support will be : — S = 5. 100 = 600 pounds. HORIZONTAL SHEARING IN BEAMS. S5 Therefore, the amount of shearing at the point 5' from support will be: — X = 1000 — 500 = 500 pounds. Inserting this in Formula (7) we have : — 500 = a. (^-^j, or, a = ^g^^ We have just found that for spruce, (5) ^ ^ — 360 pounds. Therefore, a = |^ = 1,39 square inches. ooO And, as b.d = a, or b=2-, we have, b = M£ = 11' c? '86 This is such a small amount that it can be entirely neglected in an 8" wooden beam. To find the amount of vertical shearing at any point of a canti- lever, other than at the point where it is built in, use : — x='S, w (12) Where x the amount of vertical shearing strain, in pounds, at any point of canti-lever. Where S w the sum of all loads between the free end and said point. To find the strength of beam at said point necessary to resist the shearing, insert x for w in Formula (7). In transverse strains there is also a horizontal shearing along the entire neutral axis of the piece. This stands to reason, as the fibres above the neutral axis are in compression, while those below are in tension, and, of course, the result along the neutral line is a tendency of the fibres just above and just below it, to slide past each other or to shear off along the grain. We can calculate the intensity (not amc-unt) of this horizontal shear- ing at any point of the piece under transverse strain. If X represents the amount of vertical shearing at the point, then the intensity of horizontal shearing at the point is = ^ — -. If this intensity of shearing does not exceed the safe-constant ( ^) for shearing along the fibres, the piece is safe, or : — 86 SAFE BUlLDINfi. Where x is found by formulae (11) or (12) for any point of beam or, Where x ^or\ amount of supporting foroe, in pounds, for either point of support. Where a = the area of cross-section in square incher. Where ^^^ = the amount of safe resistance, per square inch, to shearing along fibres. Example. Take the same beam as before. The amount of vertical shearing 5' from support we found to be five hundred pounds, or : — a; = 500. The area was 8" multiplied by thickness of beam, or : — a = 8l. The ultimate shearing along the fibres of spruce is about four hun- dred pounds per square inch, and with a factor-of-safety of ten, we should have : — V// 10 Inserting this in Formula (13) • tt^ = 4:0 or 6 = = 2,"34. 16.40 The beam should, therefore, be at least 2^" thick, to avoid danger of longitudinal shearing at this point. At either point of support the vertical shearing will be equal to the amount supported there; that is, one-half the load, or one thousand pounds. Substituting this for X in Formula (13), we have: — 1. 12^^40, or =4,"68. 2 8b 16.40 ' The beam would, therefore, have to be 4-|" thick at the points of support, to avoid danger of longitudinal shearing. The beam, as it is, is much too shallow for one of such span, a fact we would soon dis- cover, if calculating the transverse strength or deflection of beam, which will be taken up later on. It will also be found that the greater the depth of the beam, the smaller will be the danger from longitu- dinal shearing, and, consequently, to use thinner beams, it would be necessary to make them deeper. 88 SAFE BUILDING. IBS I IS IS I 1^ |S8 |§§ 13 |« |Sa I |35:S |g I |.sii||i|||Sil|i|iili|i|iil|Siili 7^ 3 3. I- iilllliliilliliilililiilliiili i||^s-|i||||Si|ii||''rsi||iiisi iiliillipiilliiPi illiS nil II nil II II iiiMiiiii II II II II II ^iiiiisiiiiiiisiiiii§iiiisiiPii iiiiPiiiilHiiiiiiipMPi I p I jispsi I |iS|iiiiiiiSi|iiiiii iillPliiillliiiiiiiiiiiliiiiii TABLE IV, CONTINUED. 99 9^ S CO I 05 CM S SJ S3 S' r~> 00 I CO c ISS JCOOO 10 , a ^ SggSgS882S I lii ISS Cft ^ C5 ^ 00 a> a> o> O Q lO 00 ic Q I 00 35 loi-i t- o < i|2 i 1-1 ^ oooooooo lO i-O lO lO o o OOOOIOOOOOO OOOlOt-i-IOOi-IO kO tH CO 00cnH t-rt 00 I ai |iQ>«2« I 00 bO O !M 1 o o O Q o o © o o o o 1003 1 OCrO -t< a 5 oo o p s s a li^liiii^iiliiiiiii §|8||S||? •3 <« 22 MiS oja c »s o a a Sd OS o o .^-T >H 1- so £ «3 m . . . _ -■3)9 ,•3 9 a s - ^ Ah c« ■a a ® s , o a 14 SAFE BUILDING. 1^ i|iiliii!ll||i|i||Pllii|i|i|Si|} 31 11 lipi||||§llll|i||i§|§l§lip| it i|i||i|iiiiil|i||8|||8S||i|||M| llilliliiiililllillliillilllllj 8iii8igiSiiiiiiiS8iHpiiiiiiiii|i 1i i Jill! Illf TABLE V, CONTINUED. 45 SJ^K^ |S50i-ieqo« «o ^ 05 oo oo o« os «d o 1^ ,^ lOOOlOl^OJOOC ooS CO— • -J _ „ 5a CO 000 110000 I Mso 000 Ic-jooo I o IS b« 10 CO >n o 1 o o — 1 CO o o us in o I 3 lOQOOOOOQOOQ w O O O CO I- o 10 o o o _ OOOOOiOOQOOO ■0 g (o 50 o 23 to >-< »H »-< "-I 10 M «-! 000 c" 01000000 «0<0'*10t-0'«J E-B n c« fl h O eS 46 SAFE BUILDING. The amounts given in Table IV for compression, tension and shear ing are along fibres, except where marked across. It will also be noticed that the factors -of-safety chosen are very ( if- ferent ; the reason being that where figures seemed reliable the fac- tor chosen was low, and became higher in proportion to the unre- liability of the fi;Tures. The tables, as they are, are extremely un. satisfactory and unreliable, though the writer has spent much lime in their construction. Any one, who will devote to the subject even the slightest research, will find that there are hardly any two origi- nal experimenters who agree, and in most cases, the experiments are so carelessly made or recorded that they are of but little value. TABLE VI. WEIGHT PER CUBIC FOOT OF MATERIALS. (Not included in Tables IV and V.) Material. Ashes Asphalt Butler Camphor Charcoal Coal, solid " loose Coke Cork Cotton in bah s Fat Gunpowder Hay in bales Isinglass Lead, red Paper Weight. 59 150 60 63 23 93 54 50 15 20 56 17 70 660 56 Material. Peat Petrified wood Pitch Plumbago Punlice-^tone Resiu lioi k crystal Uubber Salt Saltpetre Snow, fresh fallen " soli, I Sugar Su.phur Tiles Water REACTION OF SUPPORTS. 47 TRANSVERSE STRENGTH. — RUPTURE. If a beam is supported at two ends, and loads are applied to the beam, it is evident : — 1st, that the beam will bend under the load, or deflect. 2d, that if the loading continues, the beam will eventually break, or be ruptured. Deflection when ^he methods of calculating deflection and rup- non-important. ture diff'jr very greatly. In some cases, where de- flection in a beam would do no damage — such as cracking plaster, lowering a column, making a floor too uneven for macliinery, etc., — or where it would not look unsightly, we can leave deflection out of the question, and calculate for rupture only. Wliere, however, it is im- portant to guard against deflection, we must calculate for both. REACTION OF SUPPORTS. If we imagine the loaded beam supported at both ends bv two giants, it is evident that each giant would have to exert a certain amount of force upwards to keep his end of the beam from tipping. We can therefore imagine in all cases the supports to be resisting Amount of Reac- reacting with force sufficient to uphold their re- spective ends. The amount of this reaction for either support is equal to the load multiplied by its distance from the further support, the whole divided by the length, or (see Pig. 7) w. n P — I (14) Where jt>=the amount of the lefl hand reaction or supporting force. and y = !£l^ (15) Where 5 = the amount of the right ^'g' hand reaction or supporting force. If there are several loads the same law holds good for each, the reaction being the sum of the products, or (see Fig. 8) w„ s T" (16) „ w^m , w..r and9 = -Y- + -|- (17) As a check add the two reactions together and their sum mast equal the whole load, that is, p-\-q = «',-!-«;„ 48 SAFE BUILDING. Example. A beam 9' 2" long between hearings carries two loads, one of 200 lbs. A' 2" from the left-hand support, and the other of 300 lbs. 3' A" from the right-hand support. 5 jrom me rigia-iiana support. * ^ What are the right-hand(^_ Vi n " Wj /r|_. Viin , VVgr and left-hand reactions f "X-^ ^^-^1 Referring to Figure 8 Fig. 8. we should have = 200 lbs., and 7tf„ = 300 lbs., further Z= 110"; m=50"; n = 60"; s = 40", and r=70", therefore the left-hand reaction would be : — 200. 60 , 300. 40 p: 110 110 = 218^ pounds. and the right>-hand reaction would be : — ^ 200. 60 , 300. 70 « , q = + = 281 ^ pounds. As a check add p and 5 together, and they should equal the whole load of 600 lbs., and we have in effect : — p.\-q = 218^j- + 281y\ = 500 pounds. If the load on a beam is uniformly distributed, or is concentrated at the centre of the beam, or is concentrated at several points along the beam, each half of beam being loaded similarly, then each sup- port will react just one half of the total load. THE PRINCIPLE OF MOMENTS. Law of Lever. The law of the lever is well known. The distance of a force from its fulcrum or point where it takes effect is called its leverage. The effect of the force at such point is equal to the amount of the force multiplied by its leverage. Moment of a The effect of a force (or load) at any point of a beam is called the moment of the force (or load) at said point, and is equal to the amount of the force (or load) multi- plied by the distance of the force (or load) from said point, the distance measured at right angles to the line of the force. If therefore we find the moments — for all of the forces acting on a beam — at any single point of the beam we know the total moment at said point, and this is called the bending-moment at said point. Of course, forces BENDING MOMENT. 49 Bending acting in opposite directions will give opposite mo- moment, ments, and will counteract each other; to find the bending-moment, therefore, for any single point of a beam take the difference between the sums of the opposing moments of all forces acting at that point of the beam. Now on any loaded beam we have two kinds of forces, the loads which are pressing downwards, and the supports which are resisting upwards (theoretically forcing upwards). Again, if we imagine that the beam will break at any certain point, and imagine one side of the beam to be rigid, while the other side is tending to break away from the rigid side, it is evident that the effect at the point of rupture will be from one side only ; therefore we must take the forces on one side of the point only. It will be found in practice that no matter for what point of a beam the bending moment is sought, the bending moment will be found to be the same, whether we take the forces to the right side or left side of the point. This gives an ex- cellent check on all calculations, as we can calculate the bending moment from the forces on each side, and the results of course should be the same. Now to find the actual strain on the fibres of any cross-section of the beam, we must find the bending moment at the point where the cross-section is taken, and divide it by the moment of resistance of the fibre, or, Where m = the bending moment in lbs. inch. Where r = the moment of resistance of the fibre in inches. Where s = the strain. The stress, of course, will be equal to the resistance to cross-break- ing the fibres are capable of. In the case of beams which are of uni- form cross-section above and below the neutral axis, this resistance is called the Modulus of Rupture (k). It is found by experiments and tests for each material, and will be found in Tables IV and V. We have, then, for uniform cross-sections : — v=k Where v = the ultimate stress per square inch. Where the modulus of rupture per square inch. Inserting this and tiie above in the fundamental formula (1), viz.: v=.sf., we have : — , 771/. ^=_./,or 50 SAFE BUILDINQ. Transverse tn /t b\ strength unl- . = f K*^^) form cross-sec- i _ i tion. V// Where 7w = the bending moment in lbs. inch at a given point of beam. Where r = the moment of resistance in inches of the fibres at said point. Where (^y^ =the safe modulus of rupture of the material, per square inch. If the cross-section is not uniform above and be- "'"stre ngtl? * s e c- loathe neutral axis, we must make two distinct tion not uni- calculations, one for the fibres above the neutral axis, the other for the fibres below ; in the former case the fibres would be under compression, in the latter under tension. Therefore, for the fibres above the neutral axis, the ultimate stress would be ec^ual to the ultimate resistance of the fibres to com- pression, or v = c. Inserting this in the fundamental formula (1), we have: — c =~'f, or (19) Upper fibres. Where m = the bending moment in lbs. inch, at a given point of beam. Where r = the moment of resistance in inches of the fibres at said point. Where ^y^=the safe resistance to crushing of the material, per square inch. For the fibres below the neutral axis, the ultimate stress wo\ild be equal to the u timate resistance of the fibres to tension, or, v= t. Inserting this in the fundamental formula (1) we have : — t = ^.f, or r = r (20) Lower fibres. (^—^ Where m = the bending moment in lbs. inch at a given point of beam. GREATEST BENDING MOMENT. 51 Where r = the moment of resistance in inches of the fibres at said point. Where (^y^ =the safe resistance to tension of the material, per square inch. The same formuIaB apply to cantilevers as well as beams. The moment of resistance r of any fibre is equal to the moment of inertia of the whole cross-section, divided by the distance of the fibre from the neutral axis of the cross-section. The greatest strains are along the upper and Greatest strains lower edges of the b,eam (the extreme fibres') : we. on extreme fi- xi, r i , \ , . y 7 > bres. tneretore, only need to calculate their resistances, as all the intermediate fibres are nearer to the neutral axis, and, consequently, less strained. The distance of fibres chosen in calculating the moment of resistance is, therefore, the dis- tance from the neutral axis of either the upper or lower edges, as the case may be. The moments of resistance given in the fourth column, of Table I, are for the upper and lower edges (the extreme fibres), and should be inserted in place of r, in all the above formulse. To find at what point of a beam the greatest bend- Point of great- ing moment takes place (and, consequently, the moment*"'*'"^ greatest fibre strains, also), begin at either support and move along the beam towards the other sup- port, passing by load after load, until the amount of loads that have been passed is equal to the amount of the reaction of 'the support (point of start) ; the point of the beam where this amount is reached IS the point of greatest bending moment. In cantilevers (beams built in solidly at one end and free at the other end), the point of greatest bending moment is always at the point of the support (where the beam is built in). In light beams and short spans tlie weight of the beam itself can be neglected, but in heavy or long beams the weight of the beam should bo considered as an independent uniform load. RULES FOR CALCULATING TRANSVERSE STRAINS. 1. Find Reaction of each Support. Summary oir^^^^ If the loads on a girder are uniformly or sym- metrically distributed, each support carries or re- acts with a force equal to one-half of the total load. If the weights are unevenly distributed, each support carries, or the reaction of each support is equal to, the sum of the products of each load into its 6S SAFK BUILDING. distance from the other sup[)Ort, divided by tlie whole length of span, See Formulae (14), (15), (16), and ( 1 7). 2. Find Point of Greatest Bending Moment The greatest bending moment of a uniformly or pymmetrically distributed loiid is always at the centre. To find the point of great- est bending moment, when the loads are unevenly distributed, begin at either support and pass over load after load until an amount of loads has been passed equal to the amount of reaction at the support from which the start was made, and this is the desired point. In a cantilever the point of greatest bending moment is always at the wall. 3. Find the Amount of the Greatest Bending Moment. In a beam (supported at both ends) the createst bending moment is at the centre of the beam, provided the load is uniform, and this moment is equal to the product of the whole load into one-eigiith of the length of span, or m=^ (21) Where m = the greatest bending moment (at centre), in Ib.'^. inch, of a uniformly-loaded beam supported at both ends. Where u = the total amount of uniform load in pounds. Where I — the length of si)an in inches. If the above beam carried a central load, in place of a uniform load, the greatest bending moment would still be at the centre, liut would be equal to the product of the load into one-quarter of the length of span, or W.l , e)e)\ 4 Where m = the greatest bending moment (at centre), in lbs. inch, of a beam with concentrated load at centre, and supported at both ends Where w = the amount of load in pounds. Where I = the length of span in inches. To find the greatest bending moment of a beam, supported at both ends, with loads unevenly distributed, imagine the girder cut at the point (previously found) where the greatest bending moment is known to exist ; then the amount of the bending moment at that point will be equal to the product of the reaction (of either support) into its distance from said point, less the sum of the products of all the loads on the same side into their respective distances from said point, i. e., the point where the beam is supposed to be cut. To check the whole calculation, try the reaction and loads of the discarded side of the beam, and the same result should be obtained. KULH8 TRANSVERSE STRAINS. 6; To put the above in a formula, we sliould have : — ruj, = ]).x- S (?/;, -j- w„ x,, -f- «t',„ ar„„ etc.) (23) Amount of 114 u greatest bend- '■'"^^^ ^"^ove : Ins moment. = 9.(Z-x)-S («;.„, a:.,,, + + Wv, a:,,) etc. (24; Where A = h the point of greatest bending moment. Where = is the amount of bending moment, in lbs. inches, at A Where JO = is the left-hand reaction, in pounds. Where 7 = is the right-hand reaction, in pounds. Where x and (J-x) = tlie respective dis- tances in inches, of the left and right reac- tions from A. Where x„ a:,,,, f ^ etc., = the respective ^ — ' distances, in inches, r "Xir- - j — Xv - ...X. -^.^jt)- Fig. 9. from A, of the loads w„ w„, etc. Where «;„ w„, w„„ etc., = the loads, in pounds. Where S = the sign of summation. The same formula, of course, would hold good for any point of beam. In a cantilever (supported and built in at one end only), the great- est bending moment is always at the point of support. For a uniform load, it is equal to the product of the whole load into one-half of the length of the free end of cantilever, or m = 'tL (25) 2 Where m =z the amount, in lbs. inch, of the greatest bending moment (at point of support). Where u = the amount of the whole uniform load, in pounds. Where I = the length, in inches, of the free end of cantilever. For a load concentrated at the free end of a cantilever, the great- est bending moment is at the point of support, and is equal to the product of the load into the length of the free end of canti- lever, or m = w.l (26) Where m = the amount, in lbs. inch, of the greatest bending moment (at point of support). 64 SAFE BUILDING. Where w = the load, in pounds, concentrated at free end. Where I = the length, in inches, of free end of cantilever. For a load concentrated at any point of a cantilever, the greatest bending moment is at the point of support, and is equal to the pro duct of the load into its distance from the point of support, or m = w. X (27) Where m = the amount, in lbs. inch, of the greatest bending moment (at point of support). Where w = the load, in pounds, at any point. Where x = the distance, in inches, from load to point of support of cantilever. Note, that in all cases, when measuring the distance of a load, we must take the shortest distance (at right angles) of the vertical neutral axis of the load, (that is, of a vertical line through the centre of gravity of the load.) 4. Find the Required Cross-section. To do this it is necessary first to find what will be the required moment of resistance. If the cross-section of the beam is uniform above and below the neutral axis, we use Formula (18), viz. : — If the cross-section is unsymmetrical, that is, not uniform above and below the neutral axis, we use for the fibres above the neutral axis, formula (19), viz.: — and for the fibres below the neutral axis, Formula (20), viz. : 111 the latter two cases, for economy, the cross-section should be so designed that the respective distances of the upper and lower edges (extreme fibres), from the neutral axis, should be proportioned to their respective stresses or capacities to resist compression and ten- sion. This will be more fully explained under cast-iron lintels. A simple example will more fully explain all of the above rules. EXAMPLE TRANSVERSE STRAINS '66 210 210 Example. Three weights of respectively 600 lbs., 1000 lbs., and 1600 lbs., are placed on a beam of IV 6" (or 210") clear span, 2' 6" (or 30"), 7' 6" (or 90"), and 10' 0" (or 120") /rowi iAe left-hand support. The mod- ulus of rupture of the material is 2800 lbs. per square inch. The fac- tor-ofsafety to be used is 4. The beam to be of uniform cross-section. What size of beam should be used? 1. Find Reactions (see formulce IQ and 17). Reaction p will be in pounds, = ^^0080 . 1000.120 . '1500.90 ^ ^ ' 210 1642^ pounds. Reaction q will be in pounds, = + IMli^ + ^500-120 ^ ^ ^ ' 210 ~ 210 ' 210 1357| pounds. Check, p-\-q must equal whole load, and we have in effect: — jt>-}-9=1642f-{-1357| = 3000, which being equal to the sum of the loads is correctj for : — 500 -f 1000 + 1500 — 3000. 2. Find Point of Greatest Bending Moment. Begin at p, pass over load 500, plus load 1000, and' we still need to pass 142f pounds of (looo) K/ y load to make up amount of i*eaction p (1642^ lbs.) ; therefore, the greatest bending mo- ment must be at load 1600 ; check, begin at q and we arrive only at the first load (1500) be- fore passing amount of reaction q (1357 J' lbs.), ■\'2o- 90- • - 2io' - ^ - Z, Fig. 10. therefore, at load 1500 is the point sought. 3. Find Amount of Greatest Bending Moment. Suppose the beam cut at load 1500, then take the left-hand side of beam, and we have for the bending moment at the point where the beam is cut. m = 1642f 120 — (500.90 + 1000.30 -f 1500.0) = 197143— (45000-|-30000-f0) = 197143 — 75000 = 122143 lbs. inch. 6C SAFE BUILDING. As a check on the calculation, take the right-hand side of beam and we should have : — m = 1357f 90 — 1500.0 = 122143 — 0 = 122143 lbs. inch, which, of course, proves the correctness of former calculation. 4. Find the Required Cross-section of Beam. We jaust first find the required moment of resistance, and as the cross-section is to be uniform, we use formula (18), viz. : — (7) Now, m = 122143, and 4 = = 700, therefore, / 4 122143 700 174,49 or say = 174,5 Consulting Table I, fourth column, for section No. 2, we find r = we have, therefore, 6 ?^= 174,5 or 1047. If the size of either b or d is fixed by local conditions, we can, of course, find the other size (d or 6) very simply ; for instance, if for certain reasons of design we did not want the beam to be more than 4" wide, we should have 6 = 4, therefore, 4.rf2 =1047, and ]Ml _ 2g2^ therefore, rf= (about) 16", or, if we did not want the beam to be over 12" deep, we should have c? = 12, and = 12.12 = 144, therefore, 6.144 = 1047, and 6 = ^~ = 7,2" or say 7f . The deepest One thing is very important and, must be remem- beam the most ^ , , . , economical. berea, that the deeper the beam is, the more eco- nomical, and the stififer will it be. If the beam is too shallow, it might deflect so as to be utterly unserviceable, besides using very much more material. As a rule, it will therefore be necessary to calculate the beam for deflection as well as for its transverse strength. The deflection should not exceed 0/'03 that is, Safe deflection, jj^j.^^ one-hundredths of an inch for each foot of span, or else the plastering would be apt to crack, we have then the formula : — 8 = £. 0,03 (28) DEFLECTION. 67 Where 6 = the greatest allowable total deflection, in inclies, at centre of beam, to prevent plaster cracking. Where L = the length of span, in feet. In case the beam is so unevenly loaded that the greatest deflection will not be at the centre, but at some other point, use : — 8 = X 0,06 (29) Where d = the greatest allowable total deflection, in inches, at point of greatest deflection. W^liere X = the distance, in feet, to nearer support from point of greatest deflection. If the beam is not stiffened sideways, it should also be calculated for lateral flexure. These matters will be more fully explained when treating of beams and girders. COMPARATIVE STRENGTH AND STIFFNESS OF BEAMS AND CANTILEVERS. (1) If abeam supported at both ends and loaded uniformly will safely carry an amount of load = u; then will the same beam: (2) if both ends are built in solidly and load uniformly distributed, carry 1^. u, (3) if one end is supported and other built in solidly and load uni- formly distributed, carry 1. u, (4) if both ends are built in solidly and load applied in centre, carry 1. u, (5) if one end is supported and other built in solidly and load ap- plied in centre, carry §. «, (G) if both ends are supported and load applied in centre, carry ^. «, (7) if one end is built in solidly and other end free (cantilever) and load uniformly distributed, carry ^. u, (8) if one end is built in solidly and other end free (cantilever) and load applied at free end, carry ^. u. That is, in cases (1), (3) and (4) the effect would be the same with the same amount of load ; in case (2) the beam could safely carry 1^ times as much load as in case (1) ; in case (5) the beam could safely carry only § as much as in case (1), etc., provided that the length of span is (he same in each case.^ If the amount of deflection in case (1) were then would the amount of deflection in the other cases be as follows : Case (2) d„ =f <5, Case (4) d,v = |. <5, Case (7) dv„ = 9-|- ^, Case (3) (J,„ = |. (J, Case (5) 6y = f . d, Case (8) <5v,„ = 25|. d. Case (6) «5v, = If-*'. 2 To count on the end of a beam being built in solidly would be very bad prac- tice in most cases of building construction; as, for instance, a wooden beam with end built in solidly culd not fall out in caseof Are, and would be apt to throw the wall. Even where practicable, it would require very careful supervicion to get the beam built in properly ; then, too, it causes upward strains which must be overcome, complicating the calculations unnecessarily. In most cases where it is necessary to " build in " beam ends, the additional strength and diminished deflection thereby secured had better be credited as an additional margin of safety. The above rules for deflection do not hold good if the beam is not of uniform cro8s-section throughout ; the deflection being greater as the variation in oross-section is greater. fiS SAFE BUILDING. TABLE BENDING-MOMENT (m) AND AMOUNT OF SHEARING-3TKAIN («) A .. i Load at any point of Cantilever. T. Load at free end of cantilever. smallex than4p: greater than-^i If y I Co 2 09 II s II II II II s i Uniform load on cantilever. If X smaller If x, greater If a; = i than 2/ use: than ^ use: use: CO S For a; use: -At free end use: _, At free For X use: ^nd use: II ei5| I— in s ?> ci a 3 0< Cyy 05 1— » 1 ^ o 1 00 » p. Is oo re I e 60 SAFE BUILDING. Strength of The comparative transverse strength of two or tere^t"cro8s-sec^ rectangular beams or cantilevers is directly tions. as the product of their breadth into the square of their depth, provided the span, material and manner of supporting and loading are the same, or x=:bd^ (30) Where x = a. figure for comparing strength of beams of equal ppans. Where b = the breadth of beam, in inches. Where Tlie point of greatest deflection can never be further from the centre of beam than 2-25 of the entire length of span. It can as a rule, therefore, be safely as- Bumed to be at the ce ntre. If it is desired to find its exact location, use where n — the distance from weight to nearer Bupport; x = the distance of point of greatest deflection from farther support; and / = the length of span; x, I and m should all be expressed either in feet or inches. > Formula (i2,) is approximate only, but sufficiently exact for practical use. 68 8AFK BUILDING. feet long at 10" Fahrenheit, would gain in length (if the tempera 90.145 9 ture advanced to 100° Fahrenheit), = of a foot, or. 145000 100 pay, 1^ inches, so that at 100° Fahrenheit the truss would be 145 feet and 1^ inches long; this amount of expansion would necessitate roll- ers under one end. Of course the contraction would be in the same proportion. The approximate expansion of other materials for each additional degree Fahrenheit would be (in parts of their lengths), as follows : Expansion and„ , ^ . l contraction of Wrought-iron j^gQQQ materials. Cast-iron , Steel Antimony Gold, annealed. Bisniutli Copper Braes Silver Qon metal Tin Lead Solder 162000 1 151000 1 166000 1 123000 t 130000 1 104000 1 95000 1 95000 90000 1 87000 1 • 63000 1 • 70000 Pewter Platina Zinc Glass Granite Fire Brick Hard Brick.... White Marble. Slate Sandstone White pine.. . . Cement , 1 • 78000 1 208u00 1 62000 1 210000 1 208000 1 365000 1 600000 1 173000 1 173000 1 103000 1 440000 1 120000 The tension due to each additional degree of Fahrenheit would be equal to the modulus of elasticity of any material multiplied by the above fraction ; or about 186 pounds per square inch of cross-section, for wrougbt-iron. Above figures are for linear dimensions, the area of the superficial extension would be equal to the square of the linear, while the contents of the cubical extension would be equal to the cube of the linear. Water is at its maximum density at about 39° Fahrenheit ; above tnat it expands by additional heat, and below that point it expands by less heat. At 32° Fahrenheit water freezes, and in so doing ex- pands nearly ^ part of its bulk, this strain equal to about 30000 lbs. per square inch will burst iron or other pipes not sufficiently strong to resist such a pressure. The above table of expansions might be useful in many calculations of expansions in buildings ; for instance, were GRAPHICAL ANALYSIS. 69 Fif.lS. we to make the sandstone copings of a building in 10-foot lengths, and assume the rariation of temperature from summer sun to winter cold would be about 150«> Fahrenheit, each stone would expand 150.10 __ 1 103000 — 68 * ^^y' about J inches, quite sufficient to open the mortar joint and let the water in. The stones should, therefore, be much shorter. GRAPHICAL METHOD OF CALCULATING STRAINS. — NOTATIOW. Notation. The calculation of strains in trusses and arches is based on the law &nown as the "Parallelogram of Forces." Be- fore going in- g to same it will ' be necessary to explain the 1 notation used. If Fig. 15 rep- resents a truss, and the arrows the loads, and the two reactions (or supporting forces), we should call the left reac- tion O A and the right reaction F O. The loads would be, taking them in their order, A B, B C, C D, D E and E F. The foot, or lower half, of left rafter would be called B K, the upper half C I, while the respective parts of right rafter would be G E and H D. The King-post (tie) is I H, and the struts K I and H G, while the lower ties are K O and O G. In the strain diagram, Fig. 16 (which will be explained presently), the notation is as usual ; that is, loads A B, B C, C D, etc., are rep- resented in the strain diagram by the lines ab, be, cd, etc. Rafter pieces B K, C I, D H and E G are in the strain diagram b k, c t, dh and eg (^g and k falling on the same point). I H in Fig 15 becomes i h in strain diagram. K I becomes ^ i, H G becomes h g. O K becomes o ^, G O becomes o, O A becomes 0 a and F O becomes f o. Or, in the drawing of the truss itself the lines are called, not by letters placed at the ends of the lines, but by letters placed each side of the lines, the lines being between ; it is also usual to put these Fig. ic. 70 SAFE BUILDING. letters in capitals to distinguish them from the letters representing the strain diagram, which are, as usual, at each end of the line thej represent. One thing is very important, however, and that is, always to read the pieces off in the correct direction and in their proper order. For instance, if we were examining the joint at middle of left rafter we must read off the pieces in their proper order, as B C, C I, I K, K B, and not jump, as B C, I K, C I, etc., as this would lead to error. Still more important is it to read around the joint in one direction, as from left to right (Fig. 17), that is, in the direction of the arrow. If we were to re- verse the reading of the pieces, we should find the direction of the strain or stress re- versed in the strain diagram. For instance, if we read K I and then find its correspond- ing line k i in the strain diagram, we find its direction downward, that is, pulling joint, which would make K I a tie-rod, which, of course, is wrong, as we know it is a strut. If, however, we had read correctly i k it would be pushing upwards, which, of course, is correct and is the action of a strut. When we come to examine the joint at O, however, we reverse the above and here have to read k i, which is in the same relative direc- tion for the point O, as was t k for the point at centre of left rafter. I Re. 17. away from the V ^ = Pi ■V o ' Fig. 18. The arrows in the accompanying figure (18) show how each joint must be read, and remember always to read the pieces in tlieir proper succession. It makes no difference with which joint or with which piece of the joint we begin, so long as we read in correct succession and direc- tion, thus : for joint No 1 we can read A B, B K, K O and O A or K O, O A, A B and B K, PARALLELOGRAM OF FORCES. 71 or B K, K O, O A and A B, etc. In the strain sheet of course we read in the same succession, and it will be found that the lines, as read, point always in the correct direction of the strain or stress. PARALLKLOORAM OF FORCES. Parallelogram If a ball lying at the point A, Fig. 19, is propelled of Forces, by a power sufficient to drive it in the direction of B, and as far as B in one minute, and at B is again propelled by a power sufficient to drive it in the direction of and as far as the point C in another minute, it will, of course, arrive at C at the end of two min- utes, and by the route ABC. If, on the Other hand, both powers had been applied to the ball simultaneously, while lying at A, Fig. 20, it stands to reason that the ball would have reached C, but in one minute and by the route AC. A C (or E D), is, there- fore, called the resultant of the forces A E and D A. If, now, we were to apply to tlie ball, while at A, simultaneously with the forces D A and A E, a third force (E D) sufficient to force the ball in the oppo- site direction to A C (that is, in the direc- tion of C A), a distance equal to C A in one minute it stands to reason that the ball would remain perfectly motionless at A, as C A being the result- ant (that is, the result) of the other two forces, if we oppose them with a power just equal to their own result, it stands to reason that they are com- pletely neutralized. Now, applying this to a more practical case, if we had two sticks lying on A E and D A, Fig. 21, and holding the ball in place, and we apply to the ball a force E D = C A and in the direction C A, we can easily find how much each stick must resist or push against the ball. Draw a line e d, Fig. 22, parallel to E D, and of a length at any convenient scale equal in amount to-''' force E D ; through e, Fig. 22, draw a e par- allel to A E, and through d draw d a parallel to D A, then the tri- angle eda (not eac?) is the strain diagram for the Fig. 21, and d a, Fig. 19. Fig. 20. 72 SAFE BUILDIXO. measured by the same scale as e d, is the amount of force Tt C_. quired for the stick D A to exert, while a /^^^ measured by the same scale, is the amount of / ""s force required for the stick A E to exert. If / \ in place of the force E D we had had a load, the same a< FIgi. 21 and 22. truths would hold good, but we should represent the load by a force acting downward in a vertical and plumb line. Thus, if two sticks, B A and A C, Fig. 23, are supporting a load of ten pounds at their summit, and the inclination of each stick from a horizontal line is 45°, we proceed in the same manner. Draw c b, Fipc- 24, at any scale equal to ten units, through 6 and c draw b a and a c at angles of 45° each, with c b, then meas- ure the number of (scale measure) units in 6 a and a c, which, of course, we find to be a little over seven. Therefore, each stick must resist with a force equal to a little over seven pounds. Now, to find the di- rection of the forces. In Fig. 23 we read CB,BAand AC,the corresponding parts in the strain diagram. Fig. 24, are c b,b a and a c. Now the direction of c 6 is downwards, therefore C B acts downwards, which is, of course, the effect of a weight. The direction, however, of h a and a c is upwards, therefore B A andS A C must be pushing upwards, or towards the weight, and therefore they are in compression. The same truths hold good no matter how many forces we have acting at any point; that is, if the point remains in equilibrium (all the forces neutralizing each other), we can con- struct a strain diagram which will always be a closed polygon with as many sides as there are forces, and each side equal and parallel to one of the forces, and tlie sides being in the same succession Figs. 23 and 24. ANALYSIS OF TRUSS. 73 Fig. 25. to each other as the forces are. We can now proceed to dissect a Roof Trusses, simple truss. Take a roof truss with two rafters I and a single tie-beam. ^-{.^ The rafters are sup- posed to be loaded uni- formly, and to be strong enough not to give way transversely, but to transfer safely one-half flP^3,' *'of the load on each rafter to be supported on each joint at the ends of the rafter. We consider each joint separately. 1 ake joint No. 1, Fig. 25. We have four forces, one O A (the left-hand reaction), being equal to half the load on the whole truss ; next, A B, equal to half the load on the rafter B E. Then we have the force acting along B E, of which we do not as yet know amount or direction (up or down), but only know that it is parallel to B E ; the same is all we know, as yet, of the force E O. Now draw, at any scale. Fig. 26, No. 1, o a =:and parallel to O A, then from a draw a 6 = and parallel to A B (a 6 will, of course, lap over part of 0 a, but this does not affect anything). Then from b draw I e parallel to B E, and through o draw e o par- ^ allel to E O. Now, in read- ing off strains, begin at O A, then pass in succession to A B, B E and E O. Follow on the strain diagram Fig. 26, No. 1, the direction as read off, with the finger (that is, 0 a, ab,b e and e o), and we have the actual direc- tions of the strains. Thus o a is up, therefore pushing up ; a 6 is down, therefore pushing down; 6 e is downwards, therefore pushing against joint No. 1 (and we know it is compression); lastly, c o is 74 SAFE BUILDING. pushing to the right, therefore pulling away from the joint No. 1, and we know it is a tie-rod. In a similar manner we examine the joints 2 and 3, getting the strain diagrams No. 2 and No. 3 of Fig. 26. In Fig. 27, we get the same results exactly as in the above three diagrams of Fig. 26, only for simplicity they are combined into one diagram. If the single (combination) diagram, Fig. 27, should prove confusing to the student, let him make a separate diagram for each joint, if he will, as in Fig. 26. The above gives the principle jQ of calculating the strength of trusses, graphically, and will be more fully used later on in practical examples. Should the student desire a fuller knowledge of the subject, let him refer to " Greene's Analysis of Roof Trusses," which is simple, short, and one of the best manual on the subject. In roof and other trusses the line Line of Pressure „ . .„ , Central. of pressure or tension will always be co-incident with the central line or longitudinal axis of each piece. Each joint should, therefore, be so designed that the central lines or axes of all Fig. 27. the pieces will go through one point. Thus, for in- stance, the foot of a king post should be designed as per Fig. 28. In roof-trusses where the rafters support purlins, the rafters must not only be made strong enough to resist the compressive strain on them, but in addition to this enough material must be added to stand the transverse strain. Each part of the rafter is treated as a separate beam, supported at each joint, and the amount of reaction at each joint must be taken as the load j at the joint. The same holds — good of the tie-beam, when it has a ceiling or other weights suspended from it ; of course these weights must all be shown by arrows on the drawing of the truss, so as to get their full allowance in the strain diagram. Strains in opposite directions, of course, counteract each other; the stress, therefore, to be exerted by the material need only be equal to the difference between the amounts of the opposing strains, and, of course, this stress will be directed against the larger slirain. Fig. 28. ANALYSIS OF ARCH. 75 We consider an arch as a truss with a succes sion of straight pieces; w e can calculate it graphically the same as any other "1^°-^ truss, only we will find that the ab- sence of central or inner members (struts and ties) will force the line of pressure, as a rule, faraway from the central axis. Thus, if in Fig. 29, we con- sider A B C D as a loaded half-arch, we know that it is held in place by three forces, viz.' : 1. The load B C L M which acts through its centre of gravity as indicated by arrow No. 1. 2. A horizontal force No 2 at the crown C D, which keeps the arch fronuspreading to the right. 3. A force at the base B A (indicated by the arrow No. 3), which keeps the arch from spreading at the base. Now we know the direc- tion and amount of No. 1, and can easily find Nos. 2 and 8. In an arch lightly loaded, No. 2 is always assumed to act at two-thirds way down C D, that is at F (where C E = EF = FD = JC D). In an arch heavily loaded, No. 2 is always assumed to act one-third way down C D, that is at E; further the force No. 3 is always assumed to act through a point two-thirds way down B A, that is at H (where ^ 0 = GH = HA = ^B A). The reason for these assumptions need not be gone into here. Therefore to find forces Nos. 2 and 3 proceed as follows : If the arch is heavily loaded, draw No. 2 hori- zontally through E (C E being equal to ^ C D), prolong No 2 till it intersects No. 1 at O, then draw O H (H A being equal to i B A), which gives the direction of the resistance No. 3. We now have the three forces acting on the arch concentrated at the point O, and can easily find the amounts of each by using the parallelogram of forces. Make O I vertical and (at any scale) equal to whole load (or No. 1), draw I K horizontally, till it intersects O H at K ; then scale I K, 76 SAFE BUILDING. and K O (at same scale as O I), which will give the amount of the forces Nos. 2 and 3. The line of pressure of this arch A B C D, is Line of pressure therefore not through the central axis, but along not central. E O H (a curve drawn through E and H with the lines Nos. 2 and 3 as tangents is the real line of pressure). Now let in Fig. 30, A B C E F D A represent a half-arch. We can examine A B C D same as before, and obtain I K = to force same scale as 0 1. Fig. 30. No. 2; K 0 = to resistance (and direction of same) at U, where U C=J C D;OI being equal to the load on D A. Now if we consider the whole arch from A to F, we proceed similar- ly. L G is the neutral axis of the whole load from A to F, and is equal to the whole load, at That L G passes through D is accidental. Make E M = | E F and draw L M; also G H horizontally till it intersects L M at H, then is G H the horizontal force or No. 2. We now have two different quantities for force No. 2, viz. : I K and G H, I K in this case being the larger. It is evident that if the whole half- arch is one homogeneous mass, that the greatest horizontal thrust of any one part, will be the horizontal thrust of the whole, we select therefore the larger force or I K as the amount of the horizontal thrust. Now make S P = to I K and P Q = to No. 1, or load on A D and P R = to L G or whole load on F A, at any scale, then draw Q S and R S. Now at O we have the three forces concentrated, which act on the part of arch A B C D, viz. : Load No. 1 (= P Q), horizontal force No 2 (= S P) and resistance K O (= Q S). Now let No. 8 repre- sent the vertical neutral axis of the part of whole load on F D, then LINE OF PRESSURE. 77 prolong K O until it intersects No. 3 at T; then at T we have the three forces acting on the part of arch E C D F, viz. : The load No. 3 (= Q R), the thrust from A B C D, viz. : O T (= S Q), and the resistance N T (= R S). To obtain N T draw through T a line par- allel to R S, of course R S giving not only the direction, but also the amount of the resistance N T. The line of pressure of this arch therefore passes along P O, O T, T N. A curve drawn through points P, U and N— (that is, where the former lines intersect the joints A B, D C, F E) — and with lines P O, O T and T N as tan- gents is the real line of pressure. Of course the more parts we divide the arch into, the more points and tangents will we have, and the nearer will our line of pressure approach the real curve. Now if this line of pressure would always pass through the exact centre or axis of the arch, the compression on each joint would of course be uniformly spread over the whole joint, and the amount of this compression on each square-inch of the joint would be equal to the amount of (line of) pressure at said joint, divided by the area of the cross-section of the arch in square inches, at the joint, but this rarely occurs, and as the position of the line of pressure varies from the central axis so will the strains on the cross section vary also. Stress at Intra- Let the line A B in all the followino- fio-ures renre- dos and extra- ^ . . , . , ° ° ic^^ic dos. sent the section of any joint of an arch (the thick- ness of arch being overlooked) C D the amount and actual position of line of pressure at said joint and the small arrows the stress or resistance of arch at the joint. We see then that when C D is in the centre of A B, Fig. 31, the stress is uniform, tliat is the joint is, uniformly compressed, the ^ — V amount of compression (CJ being equal to the aver- 1^ age as above. As the ! line of pressure C D ap- A IP • 1 K side, Fig. 1 1 1 f f [ ITf'iT'fTf t ffl if'ff 1 1 1 1 1 I ' amount of com- Flg. pression on that side in- creases, while on the fur- {Cj ther side it decreases, until the line of pressure . |\| CD, reaches one-third Fif. SSt see there is no compres- 78 SAFE BUILDING. A' • ^''♦tttt I'll (c) t sion at A, but at B the compression is equal to just double the average as it was in Fig. 31. Now, as C D passes beyond the central third of A B, Fig. 34, the com- pression at the nearer side increases still fur- ther, while the further side begins to be sub- jected to stress in the opposite direction or ten- sion, this action increas- ing of course the further C D is moved from the central third. This means that the edge of arch section at B would be subject to very severe crushing, while the other When the line C D Fig. 33. aU Fig. 34. edge (at A) would tend to separate or open, passes on to the edge B, the nearer two-thirds of arch joint will be in compression, and the further third in tension. As the line passes out of joint, and further and further away from B, less and less of the joint is in compression, while more and more is in tension, until the line of pressure C D gets so far away from the joint finally, that one-half of the joint would be in tension, and the other half in compression. Tension means that the joint is tending to open upwards, and as arches are manifestly more fit to resist crushing of the joints than opening, it becomes apparent why it is dangerous to have the line of pressure far from the central axis. Still, too severe crushing strains must be avoided also, and hence the desirability of trying to get the Une of pressure into the inner third of arch ring, if possible. But the fact of the line of pressure coming outside of the inner third of arch ring, or even entirely outside of the arch, does not necessarily mean that the arch is unstable ; in these cases, however, we must cal- culate the exact strains on the extreme fibres of the joint at both the inner and outer edges of the arch (intrados and extrados), and see to it that these strains do not exceed the safe stress for the material. The formulas to be used, are : For the fibres at the edge nearest to the line of pressure „^P_M6.^. (44) a a.d And for the fibres at the edge furthest from the line of pressure PRE8SUKK ON JOINT. 79 ' = ^-^•^1 (45) Where r = the stress in lbs., required to be exerted by the extreme edge fibres (at intrados or extrados). Where x = the distance of line of pressure from centre of joint in inches. Where a = the area of cross section of arch at the joint, in square inches. Where p = the total amount of pressure at the joint in lbs. Where rf = the depth of arch ring at the joint in inches, measured from intrados to extrados. When the result of the formulae (44) and (45) is a positive quan- tity the stress v should not exceed (y), that is the safe compressive stress of the material. When, however, the result of the formula (45) yields a negative quantity, the stress » should not exceed f-A that is the safe tensile stress of the material, or mortar. The whole subject of arches wUl be treated much more fully later on in the chapter on arches. PIC.SS. 80 SAFE BUILDING. TO ASCERTAIN AMOUNT OF LOADS. Let A B C D be a floor plan of a building, A B and C D are the walls, E and F the columns, with a girder between, the other lines being floor beams, all 12" between centres; on the left side a well- hole is framed 2' x 2'. Let the load assumed be 100 pounds per square foot of floor, which includes the weight of construction. Each Load on right-hand beams, also the three left-hand Beams, beams E L, K P and F Q will each carry, of course, ten square feet of floor, or 10.100= 1000 pounds each uniform load. Each will transfer one- half of this load to tlie girder and the other half to the wall. The tail beam S N will carry 8 square feet of floor, or 8.100 = 800 pounds uniform load. One-half of this load will be transferred to the wall, the other half to the header R T, which will therefore carry a load of 400 pounds at its centre, one-half of which will be transferred to each trimmer. The trimmer beam G M carries a uniform load, one-half foot wide, its entire length, or fifty pounds a foot (on the o£f-side from well- hole), or 50.10 = 500 pounds uniform load, one-half of which is transferred to the girder and the other half to the wall. The trimmer also carries a similar load of fifty pounds a foot on the well-hole side, but only between M and R, which is eight feet long, or 50.8 = 400 pounds, the centre of this load is located, of course, half way between M and R, or four feet from support M, and six feet from support G, therefore M will carry (react) - •^^^ = 240 pounds and G will carry 1^= 160 pounds. See FormultB (14) and (15). We also have a load of 200 pounds at R, transferred from the header on to the trimmer ; as R is two feet from G, and eight feet from M, we will find by the same formulae, that G carries 8.200 10 2.200 160 pounds and M carries : 40 pounds. 10 So that we find the loads which the trimmer transfers to G and M as follows : At M = 250 -f 240 -}- 40 = 530 pounds. « G = 250 + 160 4-160 = 570 pounds. DISTRIBUTION OF LOADS. 81 The loads wliich trimmer O I transfers to wall and girder will, of Load on course, be similar. We therefore find the total load- ing, as follows: On the wall A B : At L = 500 pounds. « M= 530 pounds. " N = 400 pounds. »« O = 530 [)ounds. " P = 500 pounds. " Q = _500 pounds. Total on wall A B = 29G0 pounds. On the wall C D we have «ix equal loads of 500 pounds each, a l-oad on Girder, total of SOOO pounds. O o 9 II M o o I I O 0 II o o o o II - -2-c>--4- ,'-3:0'-!- I 3 o- — —z-p" ^; ---4-o' — - i-o- Fig. 36. On the girder E F, we have : At E from the left side 500 pounds, from the right 500 pounds. Total 1000 pounds. At G f rom the left side 570 pounds, from the right 500 pounds. Total 1070 pounds At H from the left side nothing, from the right 500 pounds. Total 600 pounds At I from the left side 570 pounds, from the right 500 pounds. Total 1070 pounds. At K from the left side 500 pounds, from the right 500 pounds. Total 1000 pounds. 82 SAFE BUILDING. At F from the left side 500 pounds, from the riaht 500 pounds. Total 1000 pounds. ° Total on girder 5640 pounds. As the frirder is neither uniformly nor symmetrically loaded, we must calcufate by Formulae (16) and (17), the amount of each reac- tion, which will, of course, give the load coming on the columns E and F. (These columns will, of course carry additional loads, from the girders on opposite side, further, the weight of the column should be added, also whatever load comes on the column at floor above.) Girder E F then transfers to columns. At E = 1000 + (f 1070) + (|. 500) + (|. 1070) + (f 1000) + (0. 1000) = 2784 pounds. At F = 1000 + (f 1000) + (f. 1070) + (f 500) + (f 1070) + (0. 1000) = 2856 pounds. As a check the loads at E and Fmust equal the whole load on the girder, and we have, in effect, 2784 + 2856 = 6640. Now as a check on the whole calculation the load on the two col- umns and two walls should equal the whole load. The whole load being 20' X 6' X 100 pounds minus the weU-hole 2'x 2'x 100 pounds, or 12000 — 400= 11600 pounds. And we have in effect, Load on A B = 2960 pounds. « C D = 3000 pounds. «< two columns = 5640 pounds. Total loads =11600 pounds. We therefore can calculate the strength of all the beams, headers and trimmers and girders, with loads on, as above given.^ For the columns and walls, we must however add, the weight of walls and columns above, including all the loads coming on walls and columns above tlie point we are calculating for, also what- ever load comes on the columns from the other sides. If there are openings in a wall, one-half the openings. load over each opening goes to the pier each side of the opening, Tncluding, of course, all loads on the wall ^.^^ above the opening. . ^ i Thus in Figure 37, the weight of walls would be distributed, as in making any modifications of floor loads on columns. WIND AND SNOW. 8S indicated by etched lines ; where, however, the opening in the wall is very small compared to the mass of wall-space over, it would, of course, be absurd to consider all this load as on the arch, and pract- ically, after the mortar has set, it would not be, but only an amount about equal to the part enclosed by dotted lines in Figure 88, the inclined lines being at an angle of 60° with the horizon. Where only part of the wall is calculated to be carried on the opening, the wooden centre should be left in until the mortar of the entire wall has set. In case of beams or lintels the wall should be built up until the intended amount of load is on them, leav- ing them free underneath ; after the intended load is on them, they ehould be shored up, until the rest of wall is built and thoroughly set. Wind Pressure Wind-pressure on a roof is and Snow, generally assumed at a certain load per square foot superficial measurement of roof, and added to the actual (dead) weight of roof ; except in large roofs, or where one foot of truss rests on rollers, when it is im- portant to assume the wind as a separate force, acting at right angles to incline of rafter. ^'s- 38. The load of snow on roofs is generally omitted, when wind ia allowed for, as, if the roof is very steep snow will not remain on it, while the wind pressure will be very severe ; while, if the roof is flat there will be no wind pressure, the allowance for which will, of course, offset the load of snow. If the roof should not be steep enough for enow to slide off, a heavy wind would probably blow the snow off. In case of " continuous girders," that is, beams or girders sup- ported at three or more points and passing over the intermediate supports without being broken, it is usual to allow more load on the central supports, than the formulas (14) to (17) would give. This subject will be more fully dealt with vn the chapter on beams and girders. FATIGUE. If a load or strain is applied to a material and then removed, the material is supposed to recover its first condition (provided it has not been strained beyond the limit of elasticity). This practically, however, is not the case, and it is found that a small load or strain often applied and removed will do more damage (fatigue the mate- rial more) than a larger one left on steadily. Most loads in buildingi 84 SAFE BUILDING. are stationary or "dead" loads. But where there are "moving" Movin loads, such as people moving, dancing, marching, °*'Loads. etc., or machinery vibrating, goods being carted and dumped, etc., it is usual to assume larger loads than will ever be imposed; sometimes going so far as to double the actual intended load, or what amounts to the same thing, doubling (or increasing) the factor-of-safety, in that case retaining, of course, the actual intended load in the calculations. This is a matter in which the architect must exercise his judgment in each individual case. Dead Wei ht "^^^ vertical effect of wind at right angles to a of^Wind vertical wall, is generally assumed as equal to 15 wlfls" pounds vertical load for each square foot of out- side vertical wall surface exposed to wind. Thus on a wall 100 feet high, there would be (for wind pressure) an additional vertical load to be added to it, per running foot horizontal, of 150 lbs. ten feet below roof, 300 lbs. twenty feet below roof, etc., and of 1500 lbs. per running foot at ground level. This load is supposed to act at the centre of thickness of wall. CHAPTER II. FOUNDATIONS. Nature Of T!^^ nature of the soils usually met with on SolISi building sites are: rock, gravel, sand, clay, loamy earth, "made" ground and marsh (soft wet soil). If the soil is hard and practically non-compressible, it is a good foundation and needs no treatment; otherwise it must be carefully prepared to resist the weight to be superimposed. Stepping base-courses of all foundation walls must be Courses, spread (or stepped out) suflS.ciently to so distribute the weight that there may be no appreciable settlement (compres- sion) in the soil. Two important laws must be observed: — 1. All base-courses must be so proportioned as to produce exactly the same pressure per square inch on the soil under all parts of building where the soil is the same. Where in the same building we meet with different kinds of soils, the base-courses must be so proportioned as to produce the same relative pressure per square inch on the different soils, as will produce an equal settlement (compression) in each. 2. Whenever possible, the base-course should be so spread that its neutral axis will correspond with the neutral axis of the super- imposed weight; otherwise there will be danger of the foundation walls settling unevenly and tipping the walls above, producing unsightly or even dangerous cracks. Example. In a churcJi the gahle wall is 1' 6" tMcTc, and is loaded (includ- ing weight of all walls, floors and roofs coming on same) at the rate of 52 lis. per square inch. The small piers are 12" x 12" and 5' high, and carry a floor space equal to 14' x 10'. What should he the sise of base-courses, it being assumed that the soil will safely stand a pressure of 30 Ihs. per square inch? If we were to consider the wall only, we should have the total pressure on the soil per running inch of wall, 18.52=936 lbs. Dividing this by 30 lbs., the safe pressure, we should need = 31,2" or say 32" width of foundation, or we should step out each side of foundation wall an amount — 7" each side. 2 Now the load on pier, assuming the floor at 100 lbs. per square 86 SAFE BUILDING. foot, would be 14 x 10 X 100 = 14000 lbs. To this must be added the weight of the pier itself. There are 5 cubic feet of brickwork (weighing 112 pounds per foot) =5. 112=560 lbs., or, including base-course, a total load of say 15000 lbs. This is distributed over an average of 144 square inches; therefore pressure per square inch under pier. ^^"^^ = 104 or, say, 100 lbs. 144 We must therefore make the foundation under pier very much wider, in order to avoid unequal settlements. The safe pressure per square inch we assumed to be 30 lbs. ; therefore the area required would be =^^|^ = 500 square inches, or a square about 22"x22", We therefore shall have to step out each side of the pier an amount — — = 5". 2 The safe compressions for different soils are given in Table V, but in most cases it is a matter for experienced judgment or else experiment. ' Testing "^^^^ ^^'^^ ^"^^^ ^* intervals, considerably Soils, deeper than the walls are intended to go, at some spot where no pressure is to take place, thus enabling the architect to judge somewhat of the nature of the soil. If this is not sufficient, he takes a crowbar, and, running it down, his experienced touch should be able to tell whether the soil is solid or not. If this is not sufficient, a small pumping or tube testing-machine should be obtained, and samples of the soil, at different points of the lot, Dottled for every one or two feet in depth. These can be taken to the office and examined at leisure. The boring should be continued if possible, until hard bottom is struck. If the ground is soft, new made, or easily compressible, experi- ment as follows: Level the ground off, and lay down four blacks each, say 3"x3" ; on these lay a stout platform. Alongside of plat- form plant a stick, with top level of platform marked on same. Now pile weight onto platform gradually, and let same stand. As soon as platform begins to sink appreciably below the mark on stick, you have the practical ultimate resistance of the foundation ; this divided by 36 gives the ultimate resistance of the foundation per square inch. One-tenth of this only should be considered ais a safe load for a permanent building. 1 In most large cities the building laws cover the amount of safe com- pressions to be allowed on various soils. WET FOUNDATIONS. 87 Drainage of Drainage is essential to make a building healthy, Soil. but can hardly be gone into in these articles. Some- times it is also necessary to keep the foundations from being undermined. It is usual to lead off all surface or spring water by means of blind drains, built underground with stone, gravel, loose tile, agri- cultural tile, half-tile, etc. To keep dampness out, walls are ce- mented and then asphalted, both on the outsides. If the wall is of Damp-proof- cementing can be omitted. Damp-courses ing. of slate or asphalt are built into walls horizontally, to keep dampness from rising by capillary attraction. ^ Cellar bottoms are concreted and then asphalted; where there is pressure 1 , ... Fig. 39. Fig. 40. of water from underneath, such as springs, tide-water, etc., the asphalt has to he sufficiently weighted down to resist same, either with brick paving or concrete. Frequently courses of tarred paper between concrete layers are used for damp-proofing; these are carried up walls, around and up piers, pipes, etc., being careful to make all tight, to well above water level. Where there are water-courses they should be diverted from the foundations, but never dammed up. They can often be led into iron or other wells sunk for this purpose, and from there pumped into the building to be used to flush water-closets, or for manu- facturing or other purposes. Clay, particularly in vertical or in- clined layers, and sand are the foundations most dangerously affected by water as they are apt to be washed out. Where a very wide base-course is required by the nature of the soil, it is usual to step out the wall above gradually; the angle of stepping should never be more acute than 60°, or, as shown in Figure 39. Care must also be taken that the stepped-out courses are sufficiently wide to project well in under each other and wall, to prevent same breaking through foundation, as indicated in Figure 40. Where, on account of party lines or other buildings, the stepping ' When it is preferred to use liquid asphalt in place of a prepared asphalt damp-course, a key should be built along the entire length of the wall to keep the wall above from slipping. This key consists of a course of brick, 8 in. wide, laid about centre of wall, and top and sides of key must be asphalted, too. 88 SAFE BUILDING. mnda- / \^/^\/\ out of a foundation wall has to be done entirely at one side, the step- ping should be even steeper than 60°, if possible ; and particular at- tention must be paid to anchoring the walls together as soon and as thoroughly as possible, in order to avoid all danger of the foundation wall tipping outwardly. • , , , Where a front or other wall is ij |^ composed of isolated piers, well to combine all their founds tions into one,and to step the piers down for this purpose, as shown in Figure 4 1 . Where there is not u • * i . sufficient depth for this purpose, inverted arches must be resorted to. The manner of calculating the strength of inverted arches will be given under the article on arches. Inverted arches are not recom- mended, however (except where the foundation wall is Inverted , , „ ^ . • ^ arches, by necessity very shallow), as it requires great care and good mechanics to build them well. Two things must particularly be looked out for : 1, That the end arch has suflicient pier or other abut- ment; otherwise it will throw the pier out, as indicated in Figure 42. (This will form part of calculation of Fig. 42. strength of arch.) Where there is danger of this, ironwork should be resorted to, to tie 8" e of Skew- ^^^^ ^^^^ pier. back, 2. The skew-back of the arch should be sufEciently wide to take its proportionate share of load from the pier (that is, amount of the two skew-backs should be proportioned to balance of pier or centre part of pier, as the width of opening is to width of pier) ; otherwise the pier would be apt to crack and settle past arch, as shown in Figure 43. An easy way of getting the graphically is given below. In Figure 44, width of Fig. 43. skew-back INVERTED ARCHES. 89 draw A B horizontally at springing-line of inverted arch ; bisect A C at F, and C B at E. Draw E O at random to vertical through F ; then draw O C, and parallel to O C draw G D ; then is C D the required skew-back.* A good way to do is to give the arches wide skew-backs, and then to introduce a thick granite or blue, stone pier stone over them, as shown in Figure 45. This will force all down evenly and avoid cracks. The stone must be Fig. 44. thick enough not to break at dotted lines, and should be carefully bedded. Example. A foundation pier carrying 150000 lbs. is 5' wide q.nd 3' broad The inverted arches are each 24" deep. What thickness should the granite block have ? W e have here virtual- ly a granite beam, 60" long and 36" broad, s u p - ported at two points (the centre lines of s k e w - backs) 36" apart. The \ Fig. 45. load is a uniform load of 150000 lbs. The safe modulus of rupture, according to Table V, for average granite is ^ = 180 lbs. » In reality C D should be somewhat larger than the amonnt thus obtained; but this can be overlooked, except in cases where the pier approaches in width the width of opening. In such cases, however, stepping can generally be resorted to in place of inverted arches. Then, too, if the opening were very wide and the line of pressure came very much outside of central third of C D, it might be necessary to still further increase the width of skew-back, C D. 90 SAFE . BUILDING The bending moment on this beam, according to Formula (21) is u.l 150 000.36 = 675 000 The moment of resistance, r, is, from Formula (18) m 675 000 180 3 750 From Table I, No. 3, we find r— -g-, therefore 6 -^ = 3750 ; now, as 6 = 36, transpose and we have 3750.6 36 625. Therefore rf= 25" or say 24". nave to be 5' x 3' x 2'. Fig. 46. The size of grapite block would As this would be a very un- wieldy block, it might be split in two lengthwise of pier; that is, two stones, each 6' x 18" x 2' should be used, and clamped together. Before building piers, the arch should be allowed to get thoroughly set and hardened, to avoid any after shrinkage of the joints. A parabolic arch is best. Next in order is a pointed arch, then a semi-circular, next elliptic, and poorest of all, a segmental arch, if it is very flat. But, as before mentioned, avoid inverted arches, if pos- sible, on account of the difficulty of their proper execution. ^^^y^ A rock foundation makes an excellent one, and needs foundations, little treatment, but is apt to be troublesome because of water. Remove all rotten rock ati step off all slanting surfaces, to make Co^c^ETC Fig. 47. level beds, filling; all crevices with condl^ete, as shown in Figure 47 : In no case build a wall on a slanting foundation. Look out for springs and water in rock foundations. Where soft ROCK FOUNDATIONS. soils are met in connection with rock, try and dig down to solid rock, or, if tliis is impossible, on account of the nature of the case or ex- pense, dig as deep as possible and put in as wide a concrete base- course as possible. If the bad spot is but a smaU one, arch over frona rock to rock, as shown in Figure 48 : Fig. 48. s nd gravel ^^^^ hard sand or even quicksand makes an excel- ^nd clay, lent foundation, if it can be kept from shifting and clear of water. To accomplish this purpose it is frequently " sheath- piled " each side of the base course. Gravel and sand mixed make an excellent, if not the best founda- tion; it is practically incompressible, and the driest, most easily drained and healthiest soil to build on. Clay is a good foundation, if in horizontal layers and of sufficient thickness to bear the superimposed weight. It is, however, a very treacherous material, and apt to swell and break up with water and frost. Clay in inclined or vertical layers cannot be trusted for im- portant buUdings, neither can loamy earth, made ground or marsh. If the base-course cannot be sufficiently spread to reduce the load to a minimum, pile-driving has to be resorted to. This is done in many Short different ways. If there is a layer of hard soil not far plies, down, short piles are driven to reach down to same. These should be of sufficient diameter not to bend under their load ; they should be calculated the same as columns. The tops should be well tied together and braced, to keep them from wobbling oi spreading. Example. Georgia-pine piles of 16" diameter are driven through a layer of soft soil 15' deep, until they rest on hard bottom. What will each pile safely carry f The pile evidently is a circular column 15' long, of 16" diameter, solid, and we should say with rounded ends, as, of course, its bear- 92 SAFE BUILDING. ings cannot be perfect. From Formula (3) we find, then, that the pile will safely carry a load. At) to , now from Table I, Section No. 7, and fifth 1^ ccl'imn, we have 22 , 22.8» a=-^. r*= —J— = 201. From the same table we find, for Section No. 7, last column, 8^ 4 From Table 17 we find for Greorgia pine, along fibres, ^ Jr^ = 750, And from Table 11, for wood with rounded ends, n = 0,00067, therefore : 201.750 _ 15 0750 _ , 1802.0,00067— 2,357 ' ^ 16 or say 30 tons to each pile. g^^^ Sometimes large holes are bored to the hard soil and piles, filled with sand, making " sand piles." This, of course, 3an only be done where the intermediate ground is sufficiently firm to ceep the sand from escaping laterally. Sometimes holes are dug down and filled in with concrete, or brick piers are buUt down ; or large iron cylinders are sunk down and the space inside of them driven full of piles, or else excavated and filled in with concrete or other masonry, or even sand, well soaked and packed. If filled with sand, there should first be a layer of concrete, to keep the sand from possibly escaping at the bottom. Where no hard soil can be struck, piles are driven over a large area, and numerous enough to consolidate the ground ; they should not be closer than two feet in the clear each way, or they will cut up the ground too much. The danger here is that they may press the ground out laterally, or cause it to rise where not weighted. Some- tunes, by sheath-piling each side, the ground can be sufficiently com- pressed between the piles, thereby being kept from escaping laterally. But by far the most usual way of driving piles is where *"*pRe8. they resist the load by means of the friction of their sides against the ground. In such cases it is usual to drive experimental piles, to ascertain just how much the pile descends at the last blow of the hammer or ram ; also the amount of fall and weight of ram, and PILES. 93 then to compute the load the pile is capable of resisting : one-tenth of this might be considered safe. The formula then is : — Where w = the safe load on each pile, in lbs. " r = the weight of ram used, in lbs. " /= the distance the ram falls, in inches. " s = the set, in inches, or distance the pile is driven at the last blow. Where there is the least doubt about the stability of the pile, use three-fourths w, and if the piles drive very unevenly, use only one- half w. Some engineers prefer to assume a fixed rule for all piles. Profes- sor Rankine allows 200 lbs. per square inch of area of head of pile. French engineers allow a pUe to carry 50000 lbs., provided it does not sink perceptibly under a ram falling 4' and weighing 1350 lbs., or does not sink half an inch under tliirty blows. There are many other such rules, but the writer would recommend the use of the above formula, as it is based on each individual experiment, and is therefore manifestly safer. Example. All experimental pile is found to sink one-half inch under the last blow of a ram weighing 1500 lbs., and falling 12'. What will each pile safely carry f According to formula (46), the safe load w would be w = —-yfe have, 10. s r= 1500 lbs. /= 12.12 = 144", and s — therefore 1500.144 ,„„«^„ w = -j^y— =43200 lbs. K several other piles should give about the same result, we would take the average of all, or else allow say 20 tons on each pile. If, however, some piles were found to sink considerably more than others, it would be better to allow but 10 tons or 15 tons, according to the amount of irregularity of the soil. All cases of pile-driving require experience, judgment, and more or less experiment ; in fact all foundations do. All piles should be straight, solid timbers, free from projecting 94 SAFE BUILDING. branches or large knots. They can be of hemlock, spruce or white pine, but preferably, of course, of yellow pine or oak. There is danger, where they are near the seashore, of their being destroyed by worms. To guard against this, the bark is sometimes shrunken on ; that is, the tree is girdled (the bark cut all around near the root) before the tree is felled, and the sap ceasing to flow, the bark shrinks on very tightly. Others prefer piles without bark, and char the piles, coat them with asphalt, or fill the pores with creosote. Copper sheets are the best (and the most expensive) covering. Piles should be of sufficient size not to break in driving, and should, as a rule, be about 80' long, and say 15 " to 18" diameter at the top. They should not be driven closer than about 2' 6" in the clear, or they will be apt to break the ground all up. The feet should be shod with wrought-iron shoes, pointed, and the heads pro- tected with wrought-iron bands, to keep them from sphtting under the blows of the ram. Sheath ^ sheath-piling it is usual to take boards (hemlock, piling, spruce, white pine, yellow pine or oak) from 2"to6"thick. Guide-piles are driven and cross pieces bolted to the insides of them. The intermediate piles are then driven between the guide piles, mak- ing a solid wooden wall each oide, from 2" to 6" thick. Sometimes the sheath-piles are tongued and grooved. The feet of the piles are cut to a point, so as to drive more easily. The tops are covered with wrought-iron caps, which sUp over them and are removed after the piles are driven. 1^^^ Piles are sometimes made of iron ; cast-iron being pref- plles. erable, as it will stand longer under water. Screw-piles are made of iron, with large, screw-shaped flanges attached to the foot, and they are screwed down into the ground Uke a gimlet. Sheath-piling is sometimes made of cast-iron plates with vertical strengthening ribs, and sometimes of steel plates, channels or other metal sections. Where piles are driven under water, great care must be taken that they are entirely immersed, and at all times so. They should be cut off to a uniform level, below the lowest low-water mark. If they are alternately wet and dry, they will boon be destroyed by decay. After the piles are cut to a level, tenons are often cut ^^ove^°p"ris.^ on their tops, and these are made to fit mortises in heav^ wooden girders which go over them, and on which the superstructure The sizes given on this page are for heavy buildings ; for very light work use piles of 8" to 9" diameter at tops, about 20 feet long aud IG" apart. CONCRETE PILES 86 rests. This is usual for docks, ferry-houses, etc. For other build- ings we frequently see concrete packed between and over their tops; this, however, is a very bad practice, as the concrete sur- rounding the tops is apt to decay them. It is better to cover the piles with 3" x 12" or similar planks (well lag-screwed to piles, where it is necessary to steady the latter) and then to build the concrete base course on these planks.^ Better yet, and the best method, is to get large-sized building- stones, with levelled beds, and to rest these directly on the piles. In this case care must be taken that piles come at least under each corner of the stone, or oftener, to keep it from tipping, and that the stone has a full bearing on each pile-head. On top of stone build the usual base-courses. Now-a-days the concrete is frequently reinforced and strengthened by steel grillage beams, where stone courses would be too expensive. Piles should be as nearly uniform as possible (particularly in the case of short piles resting on hard ground), for otherwise their respective powers of resistance will vary very much, gllp. It is well to connect all very heavy parts of build- JolntSi ings (such as towers, chimneys, etc.) by vertical slip-joints with rest of building. The slip-joint should be carried through the foundation-walls and base-courses, as wejl as above. Where there are very high chimneys or towers, or unbraced walls, the foundation must be spread sufl&ciently to overcome the leverage produced by win'd. These points will be more fully explained in the chapter on "Walls and Piers." Action of base-courses should be carried low enough to Frost, be below frost, which will penetrate from three to five feet deep in our latitudes. The reason of this is that the frost tends to swell or expand the ground (on aecoimt of its dampness) in all directions, and does it with so much force that it would be apt to lift the base-course bodily, causing cracks and possible failure above. Concrete -^^^ years concrete piles have begun to Piles. supersede wooden piles. The latter owing to the cutting of our forests, are becoming scarce, difficult to obtain of good quality and are very expensive. The concrete pile is made in a mold of concrete reinforced with longitudinal iron wires or small rods, generally wound with wiring. They are of all shapes, many being polygonal (many sided) and some have flutings in their sides running the entire length, as a 1 The strength of planks is calculated the same as for beams laid on their nat sides. 96 SAFE BUILDING. rule they are iron shod at the foot, and some iron capping device — which is later withdrawn — protects their heads from damage by the blow of the hammer, the latter being manipulated by hand, steam, or other power. These piles are vastly superior to wooden piles in every way, and in most cases to iron piles. They are not liable to decay, etc., and they do not rust. Such iron as they contain is perfectly protected by the surrounding con- crete, which is now-a-days acknowledged to be the best protection to a once well cleansed and properly coated piece of iron. Even should the pile crack, which is very infrequent, the parts would still hold and act together, owing to the wiring. Where the foundations have to go very deep to hard pan, or into deep water, concrete piles are the most economical solution. They avoid the great expense of pumping, nearly all excavating, blasting or levelling of bottom, and the weight and expense of solid deep masonry or concrete piers. One great advantage of reinforced concrete piles is: that they need not be cut off below water level, but can stick up through the ground to the underside of base courses, just below the cellar bottom. The most curious adaptation of reinforced concrete piles is their use as telegraph poles. The writer believes that before long they will be used as sleepers for ties for E. E. tracks to avoid noise and shock of the present system of iron ties or plates against rails. Sometimes a solid concrete wall is built all around the building between continuous inner and outer caissons, which are built inside and outside (all around) thus forming a continuous open well to receive the solid continuous concrete foundation wall. This has the advantage that the caisson can be built large enough to allow of waterproofing the outside of the concrete, and then excavating the entire interior surface of the lot down to rockr, for cellars and sub-cellars, which will be waterproof against side pres- sure of water. If there should be springs or fissures in rock from which water rises, the entire bottom should be stoned up and drained to certain basins (sump tanks) to be built below bottom, which must of course, be jiumped out continuously. The stone is then covered with a five or six-inch layer of concrete, then waterproofed, as previously described, and then sufficient weight of concrete put over this to hold down the water. Such a continuous foundation would be very expensive. It is customary therefore to put down piers only, under the outer GEILIiAGE FOUNDATIONS. 9f (or long inner) walls, and near the cellar bottom to span from pier to pier with steel beams, or riveted girders. The foundations proper begin near the cellar level with grillage beams and concrete. Grillage Where buildings are very heavy and not carried to Foundations, rock, it becomes necessary to spread the foundation to such an extent that the cost would become excessive, were step- ping up in mason work attempted. In such cases grillage is resorted to. A thin layer of concrete, ^say one foot thick, is spread over the necessary surface. On this steel beams are laid parallel to each other, at right angles to wall above, and close enough together for each to bear on as much soil as its share of the superimposed load will be: for instance, if the load per running foot is 100000 pounds, (including an estimated allowance for grillage and other foundation work) and the ground is capable of carrying safely 35 pounds per square inch or say 5000 pounds per square foot, the foundation would have to be 20 feet wide, now if we place our grillage beams 1 foot between centres, each beam will be 20 feet long, will bear on 20 square feet of con- crete, and the latter on 20 square feet of ground. The steel beams should be calculated upside down, that is, as supported in the center as a double lever, each arm projecting 10 feet and bearing a uniform load of 100000 pounds, that is, 50000 pounds uniform load on each lever arm. After the beams are in place, they are filled in between solidly with concrete and another layer of concrete placed over them to protect them. Beams should of course, be treated same as for other parts of building, thoroughly cleansed of rust, dirt, sand, scales, etc., then oiled, and then painted two or more coats. On top of the upper layer of concrete the wall can then be begun without any stepping courses. If the wall is on a party line, and therefore at the end of the grillage beams, struts from above or truss work above the grillage beams has to be resorted to, to force down the inner ends of the grillage beams uniformly with their outer ends at party lines; in this case each beam would be supported at both ends and load a uniform one. The example given above would be bad practice, except in cases where foundations must necessarily be shallow; for it makes an unnecessarily long span, and besides makes a centre support, and beams a cantilever each side of wall. To remedy this, other layers of grillage above the lower one are 98 SAFE BUILDING. resorted to, at right angles to each other; or in other words the stepping up is done in layers of concrete surrounding steel beams. The top layer of beams should always be at right angles to wall. These of course make the load a uniform pressure on the lower grillage and leave only their projecting length as levers the same as on the other layers, excepting possibly the top one, which, if its beams project much beyond the wall either side, should be calcu- lated for a central support with two lever arms. In every ease the beam should be considered reversed with the resistance of the ground as an upward pressure : and this pressure a uniform load. In each layer the space between the grillage beams is filled with concrete, but no concrete is placed over the grillage beams, the bottom flange of one layer being laid on top flange of layer below, but all exposed ends and the top layer of beams should be thorough- ly protected with a layer of concrete. Of course all the concrete should be of a good quality — Portland cement concrete, as hereinafter explained. Great care should be taken to get even bearing under all parts of grillage. Where there are columns, etc., sort of floating islands of grillage should be resorted to no matter of what shape, square, rectangular, polygonal, or even figures with acute angles, so long as its part of the load is properly carried to each square foot of the ground underneath. When the entire foundation is made of one solid mass of grillage and concrete, steel struts, riveted trusses, girders and other devices must be resorted to, to see that every square foot of the concrete mass receives its share of load to be transmitted to the ground. Otherwise the upward pressure of the latter, on the non-loaded parts, will tend to crack the concrete mass and cause the building ultimately to settle. _ „ When rock can be found at a moderate distance Open Caissons and Pumping. below the water level, open caissons should be used — that is water tight boxes built of tongued and grooved sheath piling with joints glued together with a resinous mixture. Sometimes steel sections, with plates between, are used for sheath piling. After the sheath piling is driven, excavation begins, and, as this proceeds, timbers and braces are used to keep the piling from being forced inwardly. Sheath piling is about 12 feet long, so that at about every 10 feet of depth a new box or caisson has to be driven CLOSED CAISSONS. 99 down. This is usually one foot smaller all around, (that is inside of) than the caisson above. Borings should therefore be made to ascertain depth of rock, the lower caisson made size required for lowest part of pier, and for every ten feet of height two feet added each way for width of upper caissons. This gives the required size for starting the top caisson. Pumps are used to keep the water down. As is well known, the atmospheric pressure will allow a suction theoretically of a column of water not over 32 feet high, but prac- tically no pumps can do this, so that 28 feet is about the limit that a pump can raise water by suction, as pumps in open caissons have to do. As the depth of the caisson increases it is necessary therefore to lower the pumps, which is done by allowing sufficient offset on one side to receive them at the lower levels. Where there are many caissons being sunk at once, a central pumping plant is established and connected with each. Closed When the column of water is so deep, or its pres- Caissons. sure so great as to make pumping impractical or too expensive, closed caissons are resorted to. These consist of air and water tight boxes with tops, but no bottoms, two (or more), super- imposed above each other, and called the lower and upper chambers. Each chamber has an outlet opening at the top sufficiently large to pass a bucket. The air is put under suflicient pressure in the lower chamber to keep out the water; and air is also supplied, under pressure, to the upper chambers. The outlet openings have trap doors which can be opened from below and shut air tight. The men, tools, or buckets are first entered into the upper cham- ber, the upper outlet closed and then air pressure put into the upper chamber. When this has been brought up to the same pressure as in the chamber below, the outlet to lower opening is opened. The same process of going from chamber to chamber is resorted to when leaving, except that air in upper chamber is reduced before leaving. The pressure has to be sufficient in the lower chamber to keep out water or the m^en would drown. The men in lower chamber dig out the ground under them and around under the edges of lower caisson, and weights (usually pig iron) piled on top force the whole caisson down. When rock ia reached it is cleaned off, levelled (sometimes stepped) and then concrete lowered into lower chamber to build pier. CHAPTER III. CELLAR AND RETAINING WALLS. 3P' ANALYTICAL SOLUTION. fliE architect is sometimes called upon to build retaining-walls in connection witli ter- races, ornamental bridges, city reservoirs, or similar problems. Then, too, aU cellar walls, where not adjoining other buildings, become re- taining walls ; hence the necessity to know how to ascertain their strength. Some writers dis- tinguish between " face-walls " and " retaining- walls " ; a face-wall being buUt in front of and against ground which has not been disturbed and is not likely to slide ; a retaining-wall being a wall that has a filled-in backing. On this the- ory a face-wall would have a purely ornamental duty, and would receive no thrust, care being taken during excavation and building-operations not to allow damp or frost to get into the ground so as to prevent its rotting or losing its natural tenacity, and to drain off aU surface or underground water. It seems to the writer, however, that the only walls that can safely be considered as " face-walls " are those built against rock, and that all walls built against other banks should be calculated as retaining- walls. Most EconomI- The cross-section of retaining-walls vary, accord- cal Section, ing to circumstances, but the outside surface of wall is generally built with a " batter " (slope) toioards the earth. The most economical wall is one where both the outside and back sur- faces batter towards the earth. As one or both surfaces become near- ly vertical the wall requires more material to do the same work, and 100 DETAINING WALLS. 101 the most extravagant design of all is where the back face batters away from the earth ; of course, the outside exposed surface of wall must either batter towards the earth (A B in Figure 49) or be vertical, (A C) ; it cannot batter away from the ground, otherwise the waU would overhang (as shown at A D). Where the courses of masonry are bmlt at right angles to the outside surface the wall will be stronger than where they are all horizontal. Thus, for the same amount of material in a wall, and same height, Figure 50 will do the most work, or be the strongest retaining-wall, Figure 51 the next atrongest, Figure 62 the next, Figure 53 next. Rf. 80. Fig- if' ^t' 52. Figure 54 next, and Figure 55 the weakest. In Figure 50 and Fig- ure 52 the joints axe at right angles to the outside surface; in the Fie. SS. ng. 54. Fig. 55. other figures they are all horizontal. For reservoirs, however, the shapes of Figures:52 or 55 arft'oftenTcfeipJeyed.... • ."• 102 SAFE BUILDING. To calculate the resistance of a retaining-wall proceed as follows ■. Height of Line The of Pressure, c e n - tral line or axis of the pressure O P or/) of backing will be at one- third of the height of back surface, meas- ured from the ground lines,^ that is at O in Figure 56, where A O = 4, AB. The direction of the pressure-line (except for reservoirs) is usu- ally assumed to form an angle of 57° with the back surface of wall, or L P0B = 57o. For water it is as- sumed normal, that is, at right angles to the back surface of wall. If it is desired, how- ever, to be very exact, erect O E perpendicular to back surface, and make angle E O P, or ( Z. x) = the angle of friction of the filling-in or backing. This angle can be found from Table X. Amount of prea- The amount (p) of the pressure P O is found from the foUowing formula : If the backing is filled in higher that the wall. Backing higher than Wall. sm'' 2 sin=*. y. sin. {}J-\-x) If the backing is filled in only to the top level of wall, w.L^ sin. X p—- Backlng level with Wall. (47) (48) sin (?/-|-2 x) ^ y/cot. x-cot (?/-|- 2x) — y/cot, Xj - cot -\- 2x) ^ 1 Where the earth in front of the outside surface of wall C D is not packed very solidly below the grade line and against tlio wall, the total height of wall N B (including part underground) should be taken, in place of A li (the height above grade line). «Th»'Wjp;Bl4pe'ef«bakp:fri.cf ioa aygie. j; . . |« 1 .*. .* FKICTION OF GBOUND. 103 Where p = the total amount of pressure, in pounds, per each run- ning foot in length of wall. Where «;= the weight, in pounds, per cubic foot of backing. Where L = the height of retaining wall above ground, in feet. See foot note 1, p. 102. Where y = the angle formed by the back surface of wall with the horizon. Where t = the angle of friction of the backing as per Table X. TABLE X.» Material. Angle of friction. AVEBAGB (except water). Very compact earth Dry clay Sharp pebbles Dry loam Sharp broken stonea Dry rammed earth Dry sand Dry gravel Wet rammed earth Wet sand Wet gravel Round pebbles Wet loam Wet clay Salt water Rain water Even those who do not understand trigonometry can use the above formulae. It wiU simply be necessary to add or subtract, etc., the numbers of degrees of the angles y and x, and then find from any table of natural sines, cosines, etc., the corresponding value for the amount of the new angle. The value, so found, can then be squared, multi- plied, square root extracted, etc., same as any other arithmetical problem. Should the number of degrees of the new angle be more than 90°, subtract 90° from the angle and use the positive cosine of the difference in place of the sine of whole, or the tangent of the difference in place of the co-tangent of the whole ; in the latter case the value of the tengent will be a negative one, and should aave the negative sign prefixed. Thus, if a: = 33° and y = 50°, formula (47) would become: w.La sin." (17)^ sin.3 50.° sin. 83° 1 Above table of friction angles is taken from Klasen's " Hochbau und BrUck- ettbau- Cmstructionen." As a rule it will do to assume the angle of friction atSS' and the weight of backing at 120 lbs. per cubic foot, except in the case of water. 104 8AFE BUILDING. The values of which, found in a table of natural sines, etc., is * _ w. L2 0,29242 ^ 2 ' 0,7662. 0,9925 ^' 0,1468 = 0,734. w.lfi 2 Similarly, in formula (48), we should have for the quantity : ^cot. X - cot. 4- 2x) = y/cot. 33° - cot. 116° = y/cot. 33° - [-tg. (116° - 90°)] = y'cot. 33° + tg. 26° = ^1,5399 -f 0,4877 = ^^2,0276 = 1,424 Average Case. As already mentioned, however, the angle of frb- tion — (except for water when it is=:0°, that is, normal to the back surface of wall) — is usually assumed at 33° ; this would reduce abore formulae to a very much more convenient form, viz. : For the average angle of friction (33°) If the backing is higher than the wall : Backing higher w. L2 . (10 - n. 0,55)2. y 144 _|_ than Wall. P 2 (lO + n. 0,55). 144 Or, if the backing is level with top of wall : Backing leve! _ w. TJ V 144 -f- ( \[T7a "-0,4-11 With Wal l. P- 2 • 9 + n.l,7 • V V ' b-\-n.Q,d~ i/n_ n.0,4-ri y . V12 5 + 0,9/ ^^"^ Where p, w and L same as for formulae (47) and (48). Where n = amount of slope or batter in inches (per foot height of wall) of rear surface of wall. Thus, if the rear surface sloped towards the backing three inches (for each foot in height) we should have a positive quantity, or » = +3. If the rear surface sloped away from the backing three inches (per foot of height), n would become negative, or n = — 3. When the rear surface of wall is vertical, there would be no slope, and we would have Cellar Walls. n=0. The latter is the case generally for all cellar walls, which would still further simplify the formula, or, for cellar walls where weight of SOU or backing varies materially from 120 pounds per cubic foot. CELLAR WALLS. 105 (51) ^ly^^rirSL's'or i> = «'.L» 0,138. For cellar walls, where the weight of soil or backing can be safely assumed to weigh 120 pounds per cubic foot ^'^i'snTal'^liif- i'=16|.L=». (52) Where ;) = the total amount of pressure, in pounds, per each running foot in length of wall. Where «;=the weight, in pounds, per cubic foot of backing. Where L = the height, in feet, of ground Une above cellar bottom. For different slopes of the back surface of retaining walls (assum- ing friction angle at 33°) we should have the following table ; + denoting slope towards backing, - denoting slope away from bacBng. TABLE XI. Slope of back sur- face of wall in inches per foot of height. -j-4// --3" .-lit 0" —1" _2W —3" —4" Value of p for backings of different weights per cubic foot. />=0,072. w. L» p=rO,088. W. 1? J?— 0,098. w. L> J9=0,112. w. L» i>=:0,138. w. L» i>=0,157. w. L» iJ=0,185. w. L» p=0,205. w. L> p=0,258. w. L» Value of p for the average backing, assumed to weigh 120 lbs. per cubic foot. p= 81. L» p=\\ . Ifl p=12 . L» p=13i. L» p=16i. L» i>=19 . L» p=22i. L» i3=24§. L» j>=31 . L» Now having found the amount of pressure p from the most conven- ient formula, or from Table XI, and referring back to Figure 56, proceed as follows : To find Curve of ^'^^^ the centre of gravity G of the mass A B C D,^ Pressure. from G draw the vertical axis G H, continue P 0 till it intersects G H at F. Make F H equal to the weight in pounds of the mass A B C D (one foot thick), at any convenient scale, and at same scale make H I = ^ and paral- lel to P O, then draw I F and it is the resultant of the pressure of the earth, and the resistance of the retaining wall. Its point of intersection K with the Amount of base D A is a point of Stress at Joint, the curve of pressure.^^: To find the exact amount of pressure on the joint D A use formula (44) for the edge of joint nearest to the point K or edge D, and formula (45) for the edge of joint farthesb from the point K, or edge A. » To find the centre of gravity of a trapezoid A B C D, Fig. 57, prolong C B until BF = C I = DA and prolong D A until A E = D H = CB.draw E I andHFaud their point of intersection G is the centre of gravity of the whole. 106 SAFE BUILDING. Formula (44) was *' = — + 6.-^, a ' a.d If M be the centre of D A, that is D M = M A = ^. DA, and re- membering that the piece of wall we are calculating, is only one run- ning foot (or one foot thick), we should have For X = K M ; expressed in inches. For a = A D. 12 ; (AD expressed in inches). For = A D ; in inches, and For j9 = F I, in lbs., measured at same scale as F 11 and II I ; or, the stress at D, (the nearer edge of joint) would be r, in pounds, per square inch, Fl 6. KM.F1 AD. 12 • "■l2.ADa Remembering to measure aU parts in inches except F I, which must be measured at same scale ^ as was used to lay out F H and H I. Similarly we should obtain the stress at A in pounds per square inch. FI 6. KM.FI AD. 12 "12.AD2 V should not exceed the safe crushing strength of the ma- terial if positive ; or if » is negative, the safe tensile strength of the mortar. If we find the wall too weak, we must enlarge A D, or if too strong, we can diminish it ; in either case, finding the new centre of gravity G of the new mass A B C D and repeating the operation from that point ; the pres- sure, of course, remaining Pig. 5(. the same so long as the slope of back surface remains unaltered. If the wall is a very high one, it should be divided into several sections in height, and each section ex- amined separately, the base of each section being treated the same as if it were the joint at the ground line, and the whole mass of wall in the section and above the section being taken in each time. rf:servoir walls. 107 Thus in Figure 58, when examining the part A, B C D, we should find O, P, for the part only, using L, as its height, the point O, being at one third the height of A, B or A, O, = ^. A, B ; G, would be the centre of gravity of A, B C D„ while F, H, would be equal to the weight of its mass, one foot thick ; this gives one point of the curve of pressure at K„ with the amount of pressure = F, I„ so that we can examine the pressures on the fibres at D, and A,. Similarly when com paring the section A„ B C D„ we have the height L,„ and so find the amount of pressure 0„ P,„ applied at 0,„ where A„ 0„ = ^ A„ B ; G„ is centre of gravity of A„ B C D„ while F„ H„ is equal to the weight of A„ B C D,„ one foot thick, and F„ I„ gives us the amount of pressure on the joint, and another point K„ of curve of pressure, so that we can examine the stress on the fibres at D„ and A„. For the whole mass A B C D we, of course, proceed as before. Resenoir For reservoirs the line of pressure O P is always Walls, at right angles to the back surface of the wall, so that we can simplify formula (50) and use for rain water : p — La (53) For salt water : p = 32. LS (54) Where p = the amount of pressure, in pounds, on one running foot in length of wall, and at one-third the height of water, measured from the bottom, and p taken normal to back surface of wall. Where L = the depth of water in feet. If Backing Is Where there is a superimposed weight on the back- Loaded < ing of a retaining-wall proceed as follows : In Figure 59 draw the angle C A D = a;, ; i^isisMc^m^assm-i:^^^ the angle of friction of / I ' ' ^ the material. Then / I take the amount of / I ^^'^ load, in pounds, com- / / ing on B C and one / / ^ running foot of it in ' ^ -* f;-- thickness (at right " angles to B C), divide this by the area, in feet, of the triangle ABC and add the quotient to w, the weight of the backing per cubic foot, then i)roceed as before, inserting the sum in place of w in formulte (47) to (51) and in 2 z Table XI, when calculating p; or w, = w-^ (55) 108 SAFE BUILDING. Where w, = the amount, in pounds, to be used in all the formula (47) to (51) and in Table XI, in place of w. Where w = weight of soil or backing, in pounds, per cubic foot. Where z = the total superimposed load on backing, in pounds, per running foot in length of wall. Where B = length, in feet, of B C, as found in Figiire 59. Where L = the height of wall, in feet. Where there is a superimposed load on the backing, the central line of pressure p should be assumed as striking the back surface of wall higher than one-third its height, the point selected, being at a height X from base ; where X is found as per formula (56) L.K-§«>) ^5g. Where X = the height, in feet, from base, at which pressure is ap plied, when there is a superimposed load on the backing. Where L = the height, in feet, of wall. Where to = the weight, in pounds, per cubic foot of backing. Where «?,=is found from formula (55) When calculating the pressure against cellar walls, only the actual weight of the material of walls, floors and roof should be assumed as coming on the wall, and no addition should be made for wind nor for load on floors, as these cannot always be relied on to be on hand. The additional compression due to them should, however, be added afterwards.^ The graph- ical method of calculating re- taining-walls is easier the ana- lytical, being less liable to cause errors, and is recom- mended for office use, though the an- alytical method » Where a wall is not to be kept bracud until the superimposed wall, etc., is on It, these should of course be entirely omitted from the calculation, and the wall jiust be made heavy enough to stand alone. OKAFHICAL METHOD- X \ \,_much \ _ '''"'fthan GRAPHICAL METHOD. 109 might often serve as a check for detecting errors, when undertaMng important work. If A B C D is the section of a retaining wall and B I the top line of backing, draw angle F A M = x = the angle of friction, usually assumed at 33° (except for water) ; continue B I to its intersection at E with A M ; over B E draw a semi-circle, with B E as diameter ; make angle B A G = 2 x (usually 66°), continuing line A G till it intersects the continuation of B I at G ; draw G H tangent to semi- circle over B E ; make G I = G H ; draw I A, also I J parallel to B A ; draw J K at right angles to I A ; also B M at right angles to A E. Now for the sake of clearness we wUl make a new drawing of the wall A B C D in Figure 61. Calling BM = Z and K J = Y (both in Figure 60) makeAE = Q (Figure 61) where Q is found from formula (57) following: — L. m Where Q = the length of A E ia Figure 61, in feet, \Vliere Y = the length of K J in Figure 60, in feet,» Where Z = the length of B M in Figure 60, m feet, C lift' (57) V Figt. 61 and 62. Where s — the weight of one cubic foot of backing, in lbs. Where m = the weight of one cubic foot of wall, in lbs. Where L = the height of backing, in feet, at wall. 1 If the incline of line B I to the horizon is equal to the angle of friction, as is often the case, find A G as before and use this length in place of K J or Y, which, of course, it will be impossible to find, as A M and B I would be parallel and would haye no point of intersection ; of course, B I should never be steeper than A M, or else all of the soil steeper than the line of angle of frictiou would l» apt to slide. 110 SAFE BUILDING. Sections must Draw E B, divide the wall into any number of sec- he^g^s.^ tions of equal height, in this case we will say three sections, A A, D, D ; A, A„ D„ D, and A„ B C D„. Find the cen- tres of gravity of the different parts, viz. : G, G, and G,„ also F, F, and F„. Bisect D D, at S, also D. D„ at S, and D„ C at S„. Draw S N, S, N, and S„ N„ horizontally. Through G, G, and G„ draw vertical axes, and through F, F, and F„ horizontal axes, till they in- tersect A B at O, O, and 0„. Draw O P, 0, P, and 0„ P„ parallel to M A, where angle M A E = x = angle of friction of soil, or back- ing. In strain diagram Figure 62 make a &, = R„ N„ ; also 6, c?, ^ R, N, and s?, /, = R N. From 6„ and draw the vertical lines. Now begin at a ; draw a h parallel to M A ; make & c = S„ R„ ; draw c d parallel to M A ; make d e = S, R, ; draw e f parallel to M A and make / f? = S R. Draw a c, a d, a af and a g. Now returning to Figure 61, prolong P„ 0„ tUl it intersects the vertical axis through G„ at II„ ; draw Il„ H, parallel to a c tUl it intersects P, O, at II, ; draw II, I, parallel to a till it intersects the vertical axis through G, at I, ; draw I, II parallel to a e till it intersects P 0 at H ; draw II I parallel to a / till it intersects the vertical through G at I ; draw I K parallel to a g. Then will points K, K, and K„ be points of the curve of pressure. The amount of pressure at K„ will be a c, at K, it will be a e, and at K it will be a g, from which, of course, the strains on the edges D, D, and D„, also A, A, and A„ can be calculated by formulae (44) and (45). To obtain scale, by which Scale of strain *° measure a c, a e and a g, make g h, Figure 62 at diasram. any scale equal to the weight, in pounds, of the part of wall A A, D, D one foot thick, draw h i parallel /a, then g i meas- ured at same scale as g h, is the amount of pressure, in pounds, at K. Similarly make e k = weight of centre part, and c m = weight of upper part, draw k I parallel d a, and m n parallel b a, then is e Z the pressure at K, and c n the pressure at K„, both measured at same scale ; or, a still more simple method would be to take the weight of A A, D, D, in pounds, and one foot thick, and divide this weight by the length of ^ / in inches ; the result being the number of pounds per inch to be used, when measuring lengths, etc., in Figure 62. The above graphical method is very convenient for high walls, where it is desirable to examine many joints, but care must be taken to be sure to get the parts all of equal height, otherwise, the result would be incorrect. If backing case of a superimposed weight find w„ as di- loaded. rected in formula (55), make A T at any scale equal to w and A U = e«„ draw T E and parallel thereto U V, draw V B, BUTTRESSED WALLS. Ill parallel to E B and use V B, in place of E B, proceeding otherwise as before. The points O of application of pressure P O, will be slightly changed, particularly in the upper part, as they will be hori- zontally opposite the centres of gravity of the enlarged trapezoids, and in the upper case this point would be much higher, the figure now being a trapezoid, instead of a triangle as before. Buttressed Where a wall is made very thin and then buttressed walls, at intervals, all calculations can be made the same aa for walls of same thickness throughout, but the vertical axis through centre of gravity of wall should be sliifted so as to pass tlirough the centre of gravity of the whole mass, in- cluding buttresses ; and the weight of thin part of wall should be increased proportion- ately to the amount of buttress, thus : If a 12" wall is buttressed every 5 feet (apart) with 2' X 2' buttresses, proceed as follows : Find the centre of gravity G of the part of wall A B C D (in plan) Figure 63, also centre of gravity F of part E I H C, draw lines through F and G parallel to wall. Now make a h parallel to wall and at any scale equal to weight or area of A B C D jcaie of strains (ifc/^^^ ^ ^^^^^ t^^t of E I H C. From Fig. 63. any point o draw the lines oa,ob and o c ; now draw K L (anywhere Ijetween parallel lines F and G), but paral- lel to b o, and from L draw L M parallel to o c, and from K draw K M parallel to a o, a line through their point of intersection M drawn parallel to wall is the neutral axis of the whole mass. When p . drawing the vertical section of wall-part A B C D, Figure 64, therefore, instead of locating the neutral axis through the centre of waU it will be as far outside as M is from B C, in Figure 63 ; that is, at G H, Figure 64. When considering the weight per cubic foot of wall, we add the proportionate share of buttress; now in Figure 63'''~AtUB there are 4 cubic feet of buttress to every 7 feet of wall, '"'s- 64. so that we must add to the usual weight w per cubic foot of wall ^ ic or w. (1 -\- 4) To put this in a formula. D 1 1 f 1 gI 1 1 1 I 5 H goo loop Zooo w„ = (1 4- ^) 112 SAFE BUILDING. Where w„ = the weight per cubic foot, in pounds, to be used for buttressed walls, after finding the neutral axis of the whole mass. Where w = the actual weight, in pounds, per cubic foot of the material. ^Vhere A = the area in square feet of one buttress. Where A,= the area in square feet of wall from side of one but- tress to corresponding side of next buttress. Walls with Buttresses, however, will not be of very much value, counterforts, unless they are placed quite close together. But- tresses on the back surface of a wall are of very little value, unless thoroughly bonded and anchored to walls ; these latter are called counterforts. It is wiser and cheaper in most cases to use the additional masonry in thickening out the lower part of wall its entire length. - . Where frost is to be resisted the back Resistance to frost, part of wall should be sloped, for the^ depth frost is likely to penetrate (from 3 to 4 feet in our climate), and finished smoothly with cement, and then asphalted, to allow the frozen earth to slide upwards, see Figure 65. Example I. Cellarwall to ^ ^^'^ story and aide frame house has a 12" brick. franne awa\\\nS' foundation wall, the distance from cellar bottom to ground level being 6 feet. The angle of friction of ground to be assumed at 33° and the weight per cubic foot at 1 20 pounds. Is the wall safe f The weight of wall and superstructure must, of course, be taken at its minimum, when cal- culating its resistance to the ground, we shall, therefore, examine one of the sides on which no beams or rafters rest. The weight will con- sist, therefore, only of brick wall and frame wall over. We examine only one running foot of wall, and have 8 cubic feet of brick at 112 lbs. = 896 lb« 24 feet (high) of frame wall at 1 5 lbs. = 3 60 " Total weight resisting pressure = 1256 " The pressure itself will be according to iono- ula (52) jt) = 16|. L2=16§. 36 = 600 lbs. Now make D O = J. D C. Make angle P O C = 67° ; prolong P O tillit intersects the Jcale of SlmmJdts) i )I t4 *» » « Fig. 60 EXAMPLES. 113 vertical neutral axis of wall^ at F ; make F H = 1256 pounds, at anj scale, draw PI I parallel to P O, and make H I =p = 600 pounds at same scale. Draw I F, then is its point of intersection K (with D A) a point of the curve of pressure, and F I (measured at same scale) it the amount of pressure p to be used in formulae (44) and (45). B} careful drawing we will find that K comes ^" beyond A (outside ol A D), or 6^" from centre E of A D. F I we find measures 166C units, therefore j9 = IGGO pounds. To find the actual stress or resistance v of edge of fibres of brick- work at A use (44), viz. : „ = _L4.6.^ a a. a and as p = 16G0 and a; = E K = 6^" and a = 12. 12 = 144 inches and d = A D = 12" we have : "= w m + P"-'^» as this is a positive quantity it will be compression. The resistance of edge-fibres at D will be according to formula (45) 1660 6A. 1660 m 071 o/' 1 ,;=_-6.^^^-^ = ll^-37i=.-26 pounds as this is a negative quantity D will be subjected to tension ; that is, there is a tendency for A B C D to tip over around the point A, the point D tending to rise. The amount of tension at D is more than ordinary brickwork will safely stand, according to Table V, still, as it would only amount to 26 pounds on the extreme edge-fibres and would diminish rapidly on the fibres nearer the centre, we can con- sider the wall safe, even if of but fairly good brickwork, particularly as the first-story beams and girders and the end and possible cross- walls, will aU help to stiffen the wall. Had we taken a foot-slice of the waU under . the side carrying the beams, we should have had an additional amount of weight resisting the pressure. If the beams were 18 ft. span, we should have three floors each 9 ft. long and with load weighing, say, 90 pounds per foot; to this must be added the roof, or about 13 ft. k 50 pounds, the additional load being : Floors, 3. 9. 90 = 2430 Roof, 13. 50= 650 Total 3080 Now make I M = 3080 pounds at same scale as F H, etc., draw M F and its point of intersection N with D A would be a point of > It should really be the vertical neutral axis of the whole weight, which would be a trifle nearer to I) C thau centre of walL 114 BAFE BUILDIKG. the curve of pressure, and F M, at same scale the amount of pressure on D A for the bearing walls of house; E N we find measures 2 J", and F M measures 4600 units or pounds. The stress at A, then, would be : + 6. ^^-^^QQ = + 72 pounds. 5 144 ■ 144. 12 and the stress at D would be : ni^ = - 8 pounds. 144 144.12 ^ There will, therefore, be absolutely no doubt about the safety of bearing walls. Example II. Cellar wall A cellar wall A B C D is to he carried 15 feet he- ;o^ning'^bufiding./o«; the level of adjoining cellar; for particular rea- sons the neighboring wall cannot be underpinned. It is desirable not to make the wall A B C D over 2' 4" thick. Would this be safe t The soil is wet loam. In the first place, before excavating we must sheath- pile along line C D, then as we excavate we must secure horizontal timbers along the sheath-piling and brace these from opposite side of excavation. The j sheath-piling and horizon- tal timbers must be built in and left in wall. The braces will have to be built around and must not be removed until the whole weight is on the wall. The weight of the wall C G, per running foot of length, including floors and roofs, we find to be 13000 pounds, but to this we must add the possible loads com- ing on floors, wliich we find to be 6000 pounds addi- tional, or 19000' pounds to«aI, possible maximum load. This load will K 4a to 72 e* ■?« Jcale of Lonjrhj (mchw) Jcale of- JtromJdbJ) Fig. «7. DEEP CELLAK WALL. 115 be distributed over the area of C D E. lu calculating the weight of A B C D resisting the pressure, we must take, of course, only the min- imum weight ; that is, the actual weight of construction and omit all loads on floors, as these may not always be present. The weight of walls and unloaded floors coming on A B C D, and including the weight of A B C D itself, we find to be 21500 pounds per running foot. Now to find the pressure p, proceed as follows : Make angle E, D M = 17°, the angle of friction of wet loam (See Table X), and prolong D E, till it intersects C E„ at E. Now C E, we find, measur es 49 feet ; CD or L is 15 feet ; then, instead of using w in formula (51), we must use w„ as found from formula (55), viz. : , 2. 19000 = CE7^ w for wet loam (Table X) is 130 pounds; therefore, w, = 130 + 1^^^ = 130 + 48, 7 = 179. Inserting this value for w in formula (51) we have : p = 179. 152. 0, 138 = 5558. The height X from D at which P O is applied is found from formula 56, and is : X = 15. (179 — f . 130) _ 15. (179— 87) _ 2.179 — 130 358 — 130 ^ 6',053 = 6'0%" Make D 0 = X=: 6'0%"; draw P O parallel to E D till .it inter- sects the vertical neutral axis of wall (centre line) at F ; make F H (vertically) at any scale equal to 21500, draw H I parallel to P O and make H I=/> = 5558 pounds at same scale, draw I F, then is its point of intersection with the prolongation of A D at K a point of the curve of pressure, and F I measured at same scale as F H is the amount of pressure on joint. The distance K from centre of joint N we find is 14^", F I measures 23800 units or pounds ; the stress («;) at A, therefore, will be, formula (44) : 23800 . 14^ 23800 ^ ■ 28.12 ' 28.12.28 ~ WTiile the stress at D would be formula (45) 23800 14^23800 28.12 * 28.12.28 Or the edge at A would be subjected to a compression of 292 poimds, while the edge at D would be submitted to a tension of 150 pounds per square inch, both strains much beyond the safe limit of even the best masonry. The wall will, therefore, have to be thickened and a new calculation made. 116 SAFE BUILDING. Example III. Wall to stage-pit 30 feet deep is to he enclosed by a stone- stage pit. i^aii^ 3 Jed thick at the top and increasing 4 inches in thickness for every 5 feet of depth. The wall, etc., coming over this wall weighs 25000 pounds per running foot, but cannot be included in the calculation, as peculiar circumstances will not allow braces to be kept against the cellar wall, until the superimposed weight is on it. The surrounding ground to be taken as the average, that is, 120 pounds weight per cubic foot, and with an anale of friction of 33". Fig. 68. Find B M (= Z) and Q T (= Y) by making angles T A E = 83° and B A U = 66° and tlien proceeding as explained for Figure 60. We scale B M and Q T at same scale as height of wall A B is drawn, and find : BM = Z = 25 ft. 6" = 25 J Q T = Y r= 9 f t. 8" = 9f ; assuming each cubic foot of wall to weigh 150 pounds we find Q from Formula (57) STAGE-PIT WALL. 117 A E = Q = 6' 7" and draw B E. At equal heights, that is, every 5 feet, in this case, draw the joint lines D E, D, E„ D„ E„, etc. Find the centres of gravity F, F,, F.„ u i| hi fi di b- ^^^'^ °^ parts of A E B, [ I \ 1 Y~i^^(see foot-note, p. 101) and also the centres of gravity G, G„ G,„ etc., of the six parts of tlie wall itself, which, in the latter case, will be at the centre of each part. Horizontally, opposite the centres F, F„ F,„ etc., apply the pressures P 0, P, 0„ etc., against wall, and parallel to M A. Through centres G, G„ G„, etc., draw vertical axes. Draw the lines S N, S, N„ S„ N„, etc., at half the vertical height of each section. Now in Figure 69 make a 6, = By Nv ; 6, d, = R,y N,v ;d,f,= R,„ N„, ; /. A, = R„N„; ^,y. = R,N.; and j\ Z, = R N. Draw the vertical lines through these points. Now begin at a, make a b parallel to P O, make bc = Rv Sy ; draw c d parallel P O ; ^'s- make c? e = R,y S.y ; draw e J parallel P O, make f g = R,„ S,„, and similarly g h, i j and k I par- allel P O, and h i = R„ S„ ; _/ A; = R, S, and Z m = R S. Draw from a lines to all the points c, d, e, f, g, etc. Now in Figure 68 begin at Py Oy, prolong it till it intersects vertical axis Gy at ly, draw ly Hy parallel a c till it intersects P,y 0,y at Hy ; draw Ply I,y parallel to a d till it intersects vertical axis G,y at I,y ; draw I,y H,y parallel to a « till it intersects P„, 0,„ at H,y and similarly H,y I,„ parallel a f; I„, H,,, parallel a g ; H„, I„ parallel ah; I„ H„ parallel to a i; H„ I, parallel to a j ; I, H, parallel to a & ; H, I parallel to a I, and I K parallel to am. The points of intersection K, K„ K,„ etc., are points of the curve of pressure. To find the amount of the pressure at each point, find Jcale of= Lengths (Inches) 118 SAFE BUILDING. weight per running foot of length of any part of wall, say, the bottom part (A A, D, D) the contents are 5' high, 4' 8" wide, 1' thick = 5.4§. 1 = 23| cubic feet k 150 pounds = 3500 pounds. Divide the weight by the length of m I in inches, and we have the number of pounds per inch, by which to measure the pressures. As m I measures 56 inches, each inch will represent = 62^ lbs. Now let us examine any joint, say, A,„ D„, ; I„, H„, which intc r- sects A„, D„, at K„, is parallel to a g. Now a g scales 166 inches, therefore, pressure at K,„= 166. 62|^ = 10375 pounds. In measur- ing the distance of K,„ from centre of joint in the following, remem- ber that the width of A,„ D,„ is 44 inches, the width of masonry above joint, and not 48" (the width of masonry below). A,„ K,„ scales 38", therefore, distance x of K,„ from centre of joint is a; = 38 — 22 = 1 fi" We have, then, from Formula (44) ^ -r, 10375 ,„ 16.10375 , stress at D„. ; v = -f 6. -^^-^ = + 63 pounds. and from formula (45) ^ . 10375 ^ 16.10375 _„ , stress at A.„ ; v = — 6. -^^^^ = — 23 pounds. The joint A,„ D„„ therefore, would be more than safe. Let us try the bottom joint A D similarly. I K is parallel to a m now a m scales 480", therefore, the pressure at K is/) = 480. 62^ = 30000 pounds. Now K is distant 53 inches from centre of joint, therefore, stress . p. . 30000 , p 53.30000 , , ^ " " = 56J2- + ^' 56.12.56 = + P°"°^«- , . . A • 30000 _ 53.30000 and stress at A is » = — — 6. ,^ = — 209 pounds. 00.12 5().12.0o The wall would evidently have to be thickened at the base. If we could only brace the wall until the superimposed weight were on it, this might not be necessary. K we could do this we should lengthen 6 c an amount of inches equal to the amount of this load divided by 62^ (the number of pounds per inch), or 6 c instead of being 36 inches long would be : 36 -f = 436 inches long. While tliis lengthening of 6 c would make the lines of pressure a c, a e, a /, etc., very much longer, and consequently the actual pressure very much greater, it will also make them very much steeper and conse- quently bring this pressure so much nearer the centre of each joint, EXAMPLE-RESERVOIR. 119 that the pressure will distribute itself over the joint mflch more evenly, and the worst danger (from tension) will probably be entirely removed. Example IV. Reservoir ^ stone reservoir wall is plumb on the outside, 2 feet Wall, tcide at the top and 5 feet wide at the bottom; the wall is 21 feet high, and the possible depth of water 20 feet. Is the wall safe f Divide the wall into three parts in height ; that is, D D, = D, D„ = D„ C. Find the weight of the parts from each joint to top, per run- ning foot of length of wall, figuring the stone- work at 150 pounds per cubic foot, and we have: Weight of A„BCD„ = 2975 pounds. Weight of A. BCD, = 7140 pounds. Weight of A B C D = 12495 pounds. Find centres of grav- ity of the parts A B C D (at G), of A, B C D, (at G.) and of A„ B C D„ (at G„). Apply the pressures P O at J height of A E ; P, O, at J height of A, E and P.. 0„ at ^ height of A„ E, where E top level of water. The amount of pres- sures will be from form- ula (53). For part A„ E. P,. 0„ = 31f L2, =^ 3 If 62=1125 pounds. For part A, E ; P, O, = 31 — i\ Jeoo yjoo lO OOO Jcaleof Jtrain; Qf-f) Jtale of Lengths {)n(.t\«i) Fig. 70. 31f 13a = 5281 pounds. For part A E ; P O = 31f 1.2= si |. 20^ = 12500 pounds. The pressures P O, P, 0„ et.-;., will be applied at right angles to A E, prolong these lines, till they intersect the vertical axes through .120 RAFE BUILDING. (the centres of gravity) G, G, and G„ at F, F, and F„. Then make F„ H„= 2975, weight of upper part. F, H. = 7140, weight of A, B C D„ and F H = 12495, weight of A B C I). Draw through H, H, and H„ the lines parallel to pressure Unes making H„I„ = P„0..= 1125 H, I, = P. O, = 5281 11 I =P O =12500 Draw T„ F„, I, F, and I F, then will their lengths represent the amounts of pressure at points K,„ K, and K on the joints A„ D„, A, D, ind A D. F„ I„ measures 3300 units or pounds. F, I, « 9500 " " F I « 18800 " " By scaling we find that K„ is 10^ inches from centre of D„ A„ K, is 37 " " D. A, K is 74 « "DA The stresses to be exerted by the wall will, therefore, be at D„ at A„ at D, at A, at D at A 3300 „ 10*.3300 , , 6- ^r;rQ7r = + 2i pounds. 36.12 ' 36.12.36 3300 „ lOi.3300 , , W2-^--3fc:36 = P"""*^^' 9500 , ^ 37.9500 93 pounds. 48.12 ' 48.12.48 9500 „ 37.9500 , 48T2-^-4832^^~^^P^""^«- 18800 . „ 74.18800 , , ■ - 60J2- + 60.12.60 = + ^ 18800 „ 74.18800 , -G0J[2 60.12.60 =-167 pounds. From the above it would appear that none of the joints are subject to excessive compression : further that joint D„A„ is more than safe, but that the joints D, A, and D A are subject to such severe tension that they cannot be passed as safe. The wall should, therefore, be redesigned, making the upper joint lighter and the lower two joints much wider. CHAPTER rV. WALLS AND PIEBS.^ WALLS are usually built of brick or stone, which are sometimes, though rarely, laid up dry, but usually with mortar fiUing all the joints. The object of mortar is threefold : Object of 1. To keep out wet and changes of temperature by mortar, filling all the crevices and joints. 2. To cement the whole into one mass, keeping the several parts from separating, and, 3. To form a sort of cushion, to distribute the crushing evenly, tak- ing up any inequalities of the brick or stone, in their beds, which might fracture each other by bearing on one or two spots only. To attain the first object, " grouting " is often resorted to. That is, the material is laid up with the joints only partly filled, and liquid cement-mortar is poured on till it runs into and fills all the joints. Theoretically this is often condemned, as it is apt to lead to careless and dirty work and the overlooking of the filling of some parts ; but practically it makes the best work and is to be recommended, except, of course, in freezing weather, when as little water as possible should be used. To attain the second object, of cementing the whole into one mass, it is necessary that the mortar should adhere firmly to all parts, and this necessitates soaking thoroughly the bricks or stones, as other- wise they will absorb the dampness from the mortar, which will crumble to dust and fail to set for want of water. Then, too, the brick and stone need washing, as any dust on them is apt to keep the mortar from clinching to them. In winter, of course, all soaking must be avoided, and as the mortar will not set so quickly, a little lime is added, to keep it warm and prevent freezing. Thickness attain the third object, the mortar joint must be of Joints, made thick enough to take up any inequalities of the brick or stone. It is, therefore, impossible to set any standard for ^ See Chapter VIII on "Reinforced Concrete Construction." 121 122 SAFE BUILDING. joints, as the more irregular the beds of the brick or stone, the larger should be the joint. For general brickwork it will do to assume that the joints shall not average over one-quarter of an inch above irregu- larities. Specify, therefore, that, say, eight courses of brick laid up " in the wall " shall not exceed by more than two inches in height eight courses of brick laid up " dry." For front work it is usual to gauge the brick, to get them of exactly even width, and to lay them up with one-eighth inch joints, using, as a rule, " putty " mortar. While this makes the prettiest wall, it is the weakest, as the mortar has little strength, and the joint being so small it is impossible to bond the facing back, except every five or six courses in height. " Putty " mortar is made of lime, water and white lead, care being taken to avoid all sand or grit in the mortar or on the beds. Quality of "^^^^ ^^s* niortar consists of English Portland ce- mortars. ment and sharp, clean, coarse sand. The less sand the stronger the mortar. • Sand for all mortars should be free from earth, salt, or other impu- rities. It should be carefully screened, and for very important work should be washed. The coarser and sharper the sand the better the cement will stick to it. Enghsh Portland cement will stand as much as three or four parts of sand. Next to English come the German Portland cements, which are nearly as good. Then the American Portland, and lastly the Rosendale and Virginia cements. Good qualities of Rosendale cements will stand as much as tAvo-and-a-half of sand. Of limes, the French lime of Teil is the strongest and most expensive. Good, hard-burned lime makes a fairly good mortar. It should be thoroughly slacked, as otherwise, if it should absorb any dampness afterwards, it will begin to burn and swell again. At least forty-eight hours should be allowed the lime for slacking, and it is very desirable to strain it to avoid unslacked lumps. Lime will take more sand than cement, and can be mixed with from two to four of sand, much depending on the quality of the sand, and particularly on the " fatness " of the lime. It is better to use plenty of sand (with lime) rather than too little ; it is a matter, however, for practical judgment and experiment, and while the specification should call for but two parts of sand to one of lime, the architect should feel at lib- erty to allow more sand if thought desirabl*" Lime and Rosendale cement are often mixed in equal proportions, and from three to five parts of sand added ; that is, one of lime, one of cement, and three to five of sand. It is advisable to specify that all parts shall be actually measured in barrels, to avoid such tricks, for instance, as hiring a MORTARS. 123 decrepit laborer to shovel cement or lime, while two or three of the strongest laborers are shovelling sand, it being called one of cement to two or three of sand. A little lime should be added, even to the very best mortars, in winter, to prevent their freezing. Frozen When a wall has been frozen, it should be taken walls. down and re-built. Never build on ice, but use salt, if necessary, to thaw it ; sweep off the salt-water, which is apt to rot the mortar, and then take off a few courses of brick before continu- ing the work. Protect wallf from rain and frost in winter by using boards and tarpaulins. Some writers claim that it does no harm for a wall to freeze ; this may be so, provided all parts freeze together and are kept frozen until set, and that they do not alternately freeze and thaw, which latter will undoubtedly rot the mortar. Plaster-of-Paris makes a good mortar, but is expensive and cannot stand dampness. Cements or hmes that will set under water are called hydraulic. Quickness of setting is a very desirable point in cements. All ce- ment-mortars, therefore, must be used perfectly fresh; any that has begun to set, or has frozen, should be condemned, though many con- tractors have a trick of cutting it up and using it over with fresh mortar. To keep dampness out of cellar-walls the outside should be plastered with a mortar of some good hydraulic cement, with not more than one part of sand to one part of cement ; this cement should be scratched, roughened, and then the cement covered outside with a heavy coat of asphalt, put on hot and with the trowel. In brick walls, the coat of cement can be omitted and the joints raked out, the asphalt being applied directly against the brick. This asphalt should be made to form a tight joint, with the slate or asphalt damp- course, which is built through bottom of wall, to stop the rise of dampness from capillary attraction. In ordinary rubble stonework the mortar should be as strong as possible, as this class of work depends entirely on the mortar for its strength. For the strengths of different mortars, see Table V. Some cements are apt to swell in setting, and should be avoided. Smoke Where flues or unplastered walls are built, the joints flues, should be '•^struck," that is, scraped smooth with the trowel. No flues should be " pargetted " ; that is, plastered over, as the smoke rots the mortar, particles fall, and the soot accumulating in the crevices is apt to set fire to the chimney. Joints of chimneys are liable to be eaten out from the same reason, and the loose per- 184 SAFE BUILDING, tions fall or are scraped out when the flues are cleaned, lea\dng dangerous cracks for fire to escape through. It is best, therefore, to line up chimneys inside with burned earthenware or fire-clay pipes. If iron pipes are used, cast-iron is preferable; wrought- iron, unless very thick, will soon be eaten away. Where walls are to be plastered, the joints are left as rough as possible, to form a good clinch. ^ Outside walls are not plastered directly on the in- furrlngs. side, \mless hollow; otherwise, the dampness would strike through and the plaster not only be constantly damp, but it would ultimately fall off. Outside walls, unless hollow, are always "furred." In fireproof work, from one to four-inch thick blocks are used for this purpose. These blocks are sometimes cast of ashes, lime, etc., but are a very poor lot and not very lasting. Generally they are made of burnt clay, fire-clay or porous terra- cotta. The latter is the best, as, besides the advantages of being lighter, warmer and more damp-proof, it can be cut, sawed, nailed into, etc., and holds a nail or screw as firmly as wood. These blocks are laid up independently of the wall, but occasionally anchored to the same by iron anchors. The plastering is applied directly to the blocks. Where space is very desirable, or the eco- nomy of saving expense of fire proof furring is necessary, or it is desired to avoid the dangers of wooden furrings, viz: fire-spaces, vermin, rats, mice, etc., Antihydrine should be applied immediately to the inside surface of the brick or concrete wall, care being taken to have this surface smooth and free from cracks or holes of any kind. Antihydrine — the original wall damp-proofing material — has to- day many imitators, some good, many worse than useless. The architect, if he cannot obtain the genuine, should make thorough practical tests, as otherwise he may later have to remove all plastering. Even where fire proof furrings are used it is desirable to coat them with Antihydrine, or similar good material, to avoid subse- quent staining of plasters. Similar damp-proofing material applied to fire proof ceilings and partitions, will stop the staining and fre- quent ruin of subsequent decorations. In all cases the first coat of plaster should be put on while the Antihydrine if .still tacky. The damp-proofing material is best ap- 1 Except in cases wliere Antihydrine or similar damp-proof materials, such as tar, etc., take the place of furrings. In such cases the walls and joints should be as smooth as possible, as even a small pin-hole may cause dampness and staining of plaste::. SHRINKAGE OF WALLS. 125 plied with a whitewash brush, after thorough mixing, and should be gone over to find breaks or holes in it, before plastering. The architect should allow no thinning or adulteration, on any excuse whatever, of this or any other damp-proofing material. In cheaper and non-fireproof work, furrings are made of vertical strips of wood about two inches wide, and from one to two inches thick, according to the regularity of the backing. For very fine work, sometimes, an independent four-inch frame is built inside of the stone-wall, and only anchored to same occasionally by iron an- chors. Where there are inside blinds, a three or four inch furring is used (or a fireproof furring), and this is built on the floor beams, as far inside of the wall as the shutter-boxes demand. To the wooden furrings the lath are nailed. Turrings are set, as a rule, sixteen inches apart, the lath being four feet long; this affords four nailings to each lath. Sometimes the furrings are set twelve inches apart, affording five nailings. All ceilings are cross-furred every twelve inches, on account of stiffness, and the strips should not be less than one-and-three-eighths inches thick, to afford strength for nailing. Furring-strips take up considerable of the strain of settlements and shrinkage, and prevent cracks in plaster- ing by distributing the strain to several strips. To still further help this object, the ''heading- joints" of lath should not all be on the same strip, but should be frequently broken (say, every foot or two), and should then be on some other strip. Laths should be separated sufaciently (about three-eighths inch) to allow the plaster to be well worked through the joint and get a strong grip or ' ' clinch ' ' on the back of the laths. If a building is properly built, theoretically correct in every respect, it should not show a single crack in plastering. Practically, however, this is impossible. , , But there never need be any fear of shrinkage or Shrinkage . t ■. of Joints. settlement, m a well-constructed building, where the foundations, joints and timbers are properly proportioned. The danger is never from the amount of settlements or shrinkage, but from the inequality of same in different parts of the building. Inequalities in settlements are avoided by properly proportioning the foundations. Inequalities in shrinkage of the joints, though quite as important, are frequently overlooked by the careless archi- tect. He will build in the same building one wall of brick with many joints, another of stones of all heights and with few joints, and then put iron columns in the centre, making no allowance whatever for the difference in shrinkage. If he makes any, it is probably to call for the most exact setting of the columns, for the hardest and quickest-setting Portland cement for the stonework, 126 SAFE BUILDING. and probably be content with lime for the brickwork. To avoid uneven shrinkages, allowances should be made for same. Brick- work will shrink, according to its quality, from one-sixteenth to one-eighth inch per story, ten to twelve feet high, and according to the total height of wall. The higher the wall, the greater the weight on the joints and the greater the shrinkage. Iron columns should, therefore, be made a trifle shorter than the story requires, the beams being set out of level, lower at the column. The plan should provide for the top of lowest column to be one-sixteenth or one-eighth inch low, while the top of highest column would be as many times one-sixteenth or one-eighth inch low as there were stories ; or if there were eight stories, the top of bottom column for the very best brickwork would be, say, one-sixteenth inch low, and the top of highest column would be one-half inch low. Stone walls should have stone backings in courses as high as front stones, if possible; if not, the backing should be set in the hardest and Slip- quickest-setting cement. Stone walls should be con- joints, nected to brick walls by means of slip-joints. By this method the writer has built a city stone-front, some 150 feet high and over 50 feet wide, connected to brick walls at each side, without a single stone sill, or transom, or lintel cracking in the front. The slip-joint should carry through foundations and base courses where the pressure is not equal on all parts of the founda- tion. If for the sake of design, it is necessary to use long columns or pilasters, in connection with coursed stone backings, the columns or pilasters must either be strong enough to do the whole work of the wall, or else must be bedded in putty-mortar with generous top and bottom joints, to allow for shrinkage of the more frequent joints behind them; otherAvise, they are apt to be -shattered. Such unconstructional designs had, however, better be avoided. In no case should a wall be built of part iron uprights and part masonry; one or the other must be strong enough to do the work alone; no reliance could be placed on their acting together. In Shrinkage frame walls, care should be taken to get the amount of timber. of "cross" timbering in inner and outer walls about equal, and to have as little of it as possible. Timber will shrink "across" the grain from one-fourth to one-half inch per foot. Where the outer walls are of masonry, and inner partitions or girders are of wood, great care must be taken that the shrink- age of each floor is taken up by itself. If the shrinkage of all beams and girders is transferred to the bottom, it makes a tre- mendous strain on the building and will ruin the plastering. To STONE-WORK. 127 effect this, posts and columns should bear directly on each other, and the girders be attached to their sides or to brackets, but by no means should the girder run between the upper and lower posts or columns. If there are stud-partitions, the head pieces should be as thin as possible, and the studs to upper partitions should rest directly on the head of lower partitions. AH beds masonry, all beds should be as nearly level aa level, possible, to avoid unequal crushing. Particularly is this the case with cut stonework. If the front of the stone comes closer than the backing (which is foolishly done sometimes to make a smpll-looking joint), the face of the stone will surely split off. If the back of a joint is broken off carelessly, and small stones in- serted in the back of a joint to form a support to larger stones, they will act as wedges, and the stone will crack up the centre of joint and wall. Stones should be bedded, therefore, perfectly level and solid, except the front of joint for about three-fourths inch back from the face, which should not be bedded solid, but with Cement " putty "-mortar. Light-colored stones, particular- Stains, ly lime-stones, are apt to stain if brought in con- nection with cement-mortar. A good treatment for such stones is to coat the back, sides and beds with Antihydrine or similar damp- proofing material. This should be put on both sides, top, bottom and back of stone before setting. No space should be left for marking, but contractor be compelled to cut marks with chisel, before coating. After stones are set it is well to give an addi- tional coat on back, covering joints as well. La Farge or similar white, non-staining cements, though expensive, will still further insure immunity against unsightly stains, after cleaning down. Natural stones should be laid on their "natural beds"; bed. that is, in the same position as taken from the quarry. This will bring the layers of each stone into horizontal positions, on top of each other, and avoid the "peeling" so fre- quently seen. Ashlar should be well anchored to the backing. The joints should be filled with putty-mortar, and should be suffici- Sire of large to take up the shrinkage of the backing. Stones. Stones should not be so large as to risk the danger of their being improperly bedded and so breaking. Professor Rankine recommends for soft stones, such as sand and lime stones, which will crush with less than 5000 pounds pressure per square inch, that the length shall not exceed three times the depth, nor the breadth one-and-a-half times the depth. For hard stones, which will resist 5000 pounds compression per square inch, he 128 SAFE BUILDING. allows the length to be from four to five times the depth, and the breadth three times the depth. Stones are sometimes joined with "rebated" joints, or "dove-tail" joints, the latter particularly in circular work, such as domes or light-houses. The practice (Early Italian Eenaissance) of making stone-work in courses, of long stones, alternately thin and thick, and breaking joints •cn- trally, is very pretty in effect, but very apt to break every thin stone immediately over joints in big stones, as can be seen, for instance, in the Memorial Arch, Washington Square, New York City, and in innumerable other examples. Sills bedded -^^^ ^^^^^ either stone or brick walls should be hollow, bedded at the ends only, and the centre part left hollow until the walls are thoroughly set and settled; otherwise, as the piers go down, the part or panels between them, not being so much weighted, will refuse to set or settle equally with them, and will force up the centre of sill and break it. Where there are lintels across openings in one piece, with central mullion, the lintel should either be jointed on the mullion, or else the mullion bedded in putty at the top. Otherwise, the lintel will break; or, if it be very strong, the mullion will split; for, as the piers set or settle, the lintel tends to go down with them, and, meeting the mullion, must either force it down, or else break it, or break itself. * gllp. Walls of uneven height, even where of the same joints. material, should be connected to each other by means of a slip-joint, so as to provide for the uneven shrinkage. Slip- joints must be so designed that while they allow independent vertical movement to each part, neither can separate from the other in any other direction. Figures 71 to 73 give a few examples. Figure 71 shows the plan of a gable-wall connected with a lower wall, by means of a slip-joint. Figure 72 shows the corner of a front stone-wall connected similarly with side-walk of brick. Figure 73 the corner of a tower ' pig. 71. or chimney connected with a lower wall. The joint must be built plumb from top to bottom. Where the higher wall sets over the tongue above lower wall, one or two inches must be left- hollow over the tongue, to allow for settlement or shrinkage of the higher wall and to prevent its resting on the tongue and possibly cracking it off. Where iron anchors are used in connection with slip-joints, they should be so arranged as to allow free vertical movement. ^ Top joints of mullions in such cases had better not be filled at all until "cleaning-down," and then only with putty mortar. In meanwhile keep joint clear by sawing it out occasionally. Gable Wall. Low Vail. TIMBER IN WALLS. 129 Steiie nil cwWall Such joints and anchors must be designed with reference to each special case. In stepped-foundations (on shelving-rock, etc.), or in walls of uneven heights where slip-joints are impracticable, the foundations or walls should be built up to each successive level and be allowed to set thoroughly before building further. A hard and quick-setting cement should be used, and the joints made Brick: oidi. as small as possible. In no case should one IL_ wall of a building be carried up much ^'2* higher than the others, where slip-joints are not to be used. When building on top of old work, clean same off thoroughly or the mortar will not take hold (clinch). In summer, soak Old and t^ie old work thoroughly. Where new work has to new walls, be built against old woi-k, a slip-joint should be used, if possible, or else a straight joint should be used with slip-anchors, and after the new work has thoroughly set, bond-stones can be cut in. In such cases, the foundations should be spread as nuich as possible, to avoid serious set- tlements. In all work involving old and new walls combined, the quickest and hard- est-setting cements should be used. Some- times it is advisable not to load walls until Fig. 73. they have set, unless all walls are loaded alike, as the uneven weights on green walls are apt to crack them. All walls should be well braced, and wooden centres left in till they have set thoroughly. Timber Timber of any kind, in walls, should be avoided, in walls, if possible. It should only be used for temporary support, as it is liable to rot, slirink, burn out, or to absorb dampness and swell, in either case causing settlements or cracks, even if not endangering the waU. In ^S.^no case bond a wall with timbers. Wliere it ris necessary to nail into a wall, wooden plugs are sometimes driven into the joints ; they are very bad, however, and liable to shake the wall in driving. Wooden strips, let in, weaken the Fig. 74. wall just that much. Wooden nailing-blocks, though not much better, are frequently used. The block should be the fuU thickness of the bricks, plus the upper and iower joints. If 130 SAFE BUILDING. there is any niortar over or under the block, the nailing will jar it loose and the block fall out. The best arrangement to secure nail- ings is to build-in porous terra-cotta blocks or bricks. Beam Where ends of beams are built into the wall, they ends, should always be cut off to a slant, as shown in Fig- ure 74. The anchors should be attached to the side, so as to allow the beam to fall out in case it is burned through. If the beams were not cut to a slant, the leverage produced by their weight, when burned through, would be apt to throw the wall ; as it is, each beam can fall out easily and the wall, being corbelled over the beam-opening, remains stand- ing. It is desirable to " build-in " the ends of wooden beams as little as possible, to prevent dry-rot ; if it can be arranged to circulate air around their ends, it will help preserve them. Beams should always be levellcd-up with good-sized pieces of slate, and not ■with wood- chips, which are liable to crush. The old-fashioned way of corbelling out to receive beams, leaving the wall intact, has much tp commend it. A modern practice is to corbel out one course of brick, at each ceiling-level, just sufficient to take the projection of furring-strips; 'this will stop draughts in case of fire, also rats and mice from ascend- ing. All slots for pipes, etc., should be bricked up solid around the pipes for about one foot at each ceiling-level for the same purposes. Wooden Where wooden lintels arc used in walls, there lintels, should always be a relieving-arch over them, so ar- ranged that it would stand, even if the lintel were burned out o^ re- moved ; the lintel should have J ' 'I ' 1 — ^'T^" httle bearing as possible, and be shaved off at the ends. Figure 75 shows a wooden lintel correctly built-in. Figure 76 shows a very blundering way of building-in a wooden lintel, but one, nevertheless. Fig. 75. frequently met with. It is ob- vious, however, in the latter case, that if the lintel were removed the abutment to the arch would sink and let the arch down. The relieving-arch, after it has set, should be strong enough to carry the wall, the lintel being then used for nailing only. The rule for lin- Bonded tels is to make their depth about one-tenth of the arches best. span. Arches are built of "row-locks" (that is, " headers,") or of " stretchers," or a combination of both, according BRICK FACINGS. 131 I . I TTT to design. The strongest arch, however, is one which has a com- bination of both headers and stretchers ; that is, one which is bonded on the face, and also bonded into the backing. Straight arches and arches built in circular walls should always be bonded into the backing, or if the design does not allow of this, they should be anchored F'g- 76. back. " Straight " arches should be built with a slight " camber " up towards the centre, to -r allow for settlement -T~ and to satisfy the eye. ZT About one-eighth inch -J — rise at the centre fo — each foot of span is sufficient. Straig/it Fig. 77. arches should never be built, as shown in Figure 77, and known as the French or Datch arch, as there is absolutely no strength to them. Fireplaces ar3.1080-fl0 ^ /T"^ or the radius of flue will be 4 feet (diameter 8 feet). Now making the walls at top of chimney 8" thick and adopting the rule of an outside batter of about \" to the foot, or say 4" every 15 feet, we get a section as shown in Figure 87. Let us examine the strength of the chimney at the five levels A, B, C, D and E. The thickness of the base of each part is marked on the right-hand section, and the average thickness of the section of the part on the left-hand side. 144 SAFE BUILDING. Take the part above A ; the average area is (2^2.5^ — flue area) or, 78 — 50 = 28 square feet. This multiplied by the height of the part and the weight of one cubic foot of brickwork (112 lbs.) gives the weight of the whole, or actual load. W. = 28.30.112 = 94080 lbs., or 47 tons. The area of the base at A would be : flue ■60: area^; or A = 89- 39 square feet. The height of the part is L = 30. The sc^uare of the rar (iius of gyration, in feet, is : L2 -j- 42 . 11,11 Inserting these values in Formula (GO) the safe load at A would be : 39.200 W= 14 4- 0,046 30.30 11,11 = 440 tons, or about nine times the actual load. Now, in examining the joint B we must remem- ber to take the whole load of brickwork to the top as well as whole length L to top (or 60 feet). The load on B we find is : W, = 131 tons, while the safe load is * 1 1 r 1 1 1 -6 1 o i I I I ^8 36 -Pa. I .24 8?. [Am. _i TOPS OF CHIMNEYS. 145 W= =472 tons. 14 + 0,046.^ ' ' 13 Similarly, we should find on C the load : W,= 259 tons, while the safe load it: W= ?^ = 461tOD.. 14 + 0,046.?||» On i) we should find the load : AV, = 432 tons, while the safe load is : W= liM^^ = 448 ton«. ,.,r,(,,(. 120.120 14 + 0,046.^^ Below D the wall is considerably over three feet thick, and is solid, therefore we can use ^^^ = 300, provided good Portland cement is used and best brick, which should, of course, be the case at the base of such a high chimney. We should have then the load on E : W,= 657 tons, whUe the safe load is : W=z 1^1:^ = 690 tons. 14 + 0,046.1^ The chimney is, therefore, more than amply safe at all points, the bottom being left too strong to provide for the entrance of Hue, which will, of course, weaken it considerably. We might thin the upper parts, but the bricks saved would not amount to very much and the offsets would make very ugly spots, and be bad places for water to lodge. K the chimney had been square it would have been much stronger, though it would have taken considerably more mate- rial to build it. It is generally best to build the flue of a chimney plumb from top to bottom, and, of course, of same area throughout. Sometimes the flue is gradually enlarged towards the top for some five to ten feet in height, which is not objectionable, and the writer has obtained good results thereby ; some writers, though, claim the flue should - . «... be diminished at the top, which, however, the writer Tops of Chim- ^' ' , ; ney Flues. has never cared to try. Galvanized iron bands should be placed around the chimney at intervals, particularly around the top part, which is exposed very much to the disintegrating effects of the weather and the acids contained in the smoke. No smoko flue should ever be pargetted (plastered) inside, as the acids in the smoke will eat up the lime, crack the plaster, and cause it to fall. The 146 SAFE BUILDING. crevices will fill with soot and be liable to catch fire. , The mortar- joints of flues should be of lime and cement, or, better yet, of fire-clay, and should be carefully struck, to avoid being eaten out by the acids. Calculation of Where walls are long, without buttresses or cross- Walls.- Bulging, walls, such as gable-walls, side-walls of building^ etc., we can take a slice of the wall, one running foot in length, and consider it as forced to yield (bulge) inwardly or outwardly, so that for p2 -vve should use : Q = — : where d the thickness of wall in inches. The ^ 12 ' area or a would then be, in square inches, a = l 2.d. Inserting these values in formula (59) we have for BRICK OR STONE WALLS. « 0,0833 + 0,475. ^ f \ (62) Where w — the safe load, in lbs., on each running foot of wall {d" thick). Where d = the thickness, in inches, of the wall at any point of its height. Where L = the height, in feet, from said point to top of wall. Where ^-y ^ = the safe resistance to crushing, in lbs., per square inch, as given in Table V. (See page 135.) If it is preferred to use tons and feet, we insert in formula (60) : for A = D, where D the thickness of wall, in feet, and we have : = — ; therefore W = (63) 14 + 0,552.^^ AVTiere W = the safe load, in tons, of 2000 lbs., on each running foot of wall (D feet thick). Wliere D = the tliickness of wall, in feet, at any point of its height. Where L = the height, in feet, from said point to the top of wall. Where ^ ^ = the safe resistance to crushing of the material, in lbs., per square inch, as found in Table V. (See page 135.) Anchored Walls. Where a wall is thoroughly anchored to each tier EXAMPLE OF WALLS. 147 of floor beams, so that it cannot possibly bulge, except between floor- beams, use the height of story (that is, height between anchored f n Fig. 88. Fig. 89. Fig. 90. beams in feet) in place of L and calculate or D for the bottom of wall at each story. The load on a wall consists of the wall itself, from the point at which the thickness is being calculated to the top, plus the weight of 148 SAFE BUILDING. one foot in width by half the span of all the floors, roofs, partitions etc. Where there are openings in a wall, add to pier the proportion- ate weight which would come over opening ; that is, if we find the load per running foot on a wall to be 20000 lbs., and the wall con- sists of four-foot piers and three-foot openings alternating, the piers will, of course, carry not only 20000 lbs. per running foot, but the 60000 lbs. coming over each opening additional, and as there are four feet of pier we must add to each foot = 15000 lbs. ; we therefore calculate the pier part of wall to carry 35000 lbs. per run- ning foot. The actual load on the wall must not exceed the safe load as found by the formula (62) or (63). Example. Wall of Country A two-siory-and-attic dwelling has brick walls 12 House. inches thick ; the walls carry two tiers of beams of 20 feet span ; is the wall strong enough f The brickwork is good and laid in cement mortar. We will calculate the thickness required at first story beam level, Figure 88. The load is, pei running foot of wall : Wall =22.112 = 2464 lbs. Wind =22. 15= 330 lbs. Second floor =10.90 = 900 lbs. Attic floor =10. 70= 700 lbs. Slate roof (incl. wind and snow) = 10. 50 = 500 lbs. Total load =4894 lbs. For the quality of brick described we should take from Table V : ^-^) = 200 lbs. The height between floors is 10 feet, or L=10, therefore, using formula (62) we have : LL = 12.200 58071b. 0,0833 + 0,475.^2 0,0833 + 0,475. So that the wall is amply strong. If the wall were pierced to the extent of one-quarter with openings, the weight per running foot would be increased to 6525 lbs. Over 700 lbs. more than the safe load, still the wall, even then, would be safe enough, as we have allowed some 330 lbs. for wind, which would rarely, if ever, be so strong; and further, some 1200 lbs. for loads on floors, also a very WAEEHOUSE WALL. 149 ample allowance; and even if the two ever did exist together it would only run the compression up to 225 lbs. per inch, and for a temporary stress this can be safely allowed. The writer would state here, that the only fault he finds with for- mula; (59), (GO), (62) and (G3), is that their results are apt to give au excess of strength ; still it is better to be in fault on the safe side and be sure. Example. Walls of City The brick walls of a warehouse are 115 feet high, Warehouse, the 8 stories are each 14 feet high from floor to floor, or 1 2 feet in the clear. The load on floors per square foot, including the fire-proof construction, will average 300 lbs. What size should the walls be f The span of beams is 26 feet on an average. (See Fig. 89, page 142.) According to the New York Building Law, the required thicknesses would be: first story, 32"; second, third, and fourth stories, 28"; fifth and sixth stories, 24" ; seventh and eighth stories, 20". At the seventh story level we have a load, as follows, for each run- ning foot of wall : Wall =30.1|.112 = 5600 Wind = 30.30= 900 Roof — 13.120 = 1560 Eighth floor = 13.300 = 3900 Total =11960 lbs., or 6 tons. The safe load on a 20" wall 12 feet high, from formula (63) is : 1|.200 333 ,o,, * 14 + 0 552 ^-^-^^ ^ ^^ + ^>^^^-^^'^^ ^ ' ' " 14 + 0,o52.^-p^ 15622,lbs. If one-quarter of the wall were used up for openings, slots, flues, etc., the load on the balance would be 8 tons per running foot, which IS still safe, according to our formula. At the fifth-story level the load would be : Load above seventh floor = 11960 Wall = 28.2.112= 6272 Wind =28.30 = 840 Sixth and seventh floors = 2.13.300 = 7800 Total = 26872 lbs. or 13^ tons. 150 SAFE BUILDING. The safe load on a 24" wall, 12 feet high, from forn.ula (63) is: W— 2.200 400 M 4- n npg 12.12 — 14 + 0,552.36 = ^^'^^^ ' * 2.2 or 23618 lbs. This is about 10 per cent less than the load, and can be passed as safe, but if there were many flues, openings, etc., in wall, it should be thickened. At the second-story level the load would be : Load above fifth floor — 26872 Wall = 42. 21.112 = 10976 Wind := 42. 30 = 1260 Third, fourth and fifth floors = 3.13.300 = 11700 , 'rotal _ 5Q808, or 25 tons. The safe-load on a 28" wall, 12 feet high, from formula (63) is : W— 2|.200 4G7 _ 14 + 0,552.1^ ~ 14 -+- 0,552.26,45 " ^^'^^ or 32660 lbs. Or, the wall would be dangerously weak at the second-floor level. At the first-floor level the load would be : Load above second floor = 50808 Wall = 14. 2|.112 = 4181 Wind 14.30 = 420 Second floor = 13.300 = 3900 Total = 59309, or 29^ tons. The safe-load on a 32" wall, 12 feet high, from formula (63) is : w_ 2f.200 533 14 + 0,552.1^2-= 14 + 0,552.20,25= 2|-2f or 42338 lbs. The wall would, therefore, be weak at this point, too. Now while the conditions we have assumed, an eight-story ware- house with all floors heavily loaded, would be very unusual, it answers to show how impossible it is to cover every case by a law, not based on the conditions of load, etc. Tn reality the arrangements of walls, as required by the law, are foolish. Unnecessary weight is piled on top of the wall by making the top 20" thick, which wall has nothincr to do but to carry the roof. (If the span of beams were increased to 31 feet or more the law compels this top wall to be 24" thick, if 41 feet, it would have to be 28" thick, an evident waste of material.) It THRUST OF BARRELS 151 would be much better to make the top walls lighter, and add to the bottom ; in this case, the writer would suggest that the eighth story be 12" ; the seventh story 16" ; the sixth story, 20"; the fifth story, 24" ; the fourth story, 28" ; the third story, 32" ; the second story, 36", and the first story 40", see Figure 90. (Page 142.) This would represent but 4| cubic feet of additional brickwork for every running foot of wall; or, if we make the first-story wall 36" too, as hereafter suggested, the amount of material would be exactly the same as required by the law, and yet the wall would be much better proportioned and stronger as a whole. For we should find (for L = 12 feet), Actual load at eighth-floor level, Safe load on a 12" wall from Formula (62) Actual load at seventh-floor level. Safe load on a 16" wall from Formula (62) Actual load at sixth-floor level. Safe load on a 20" wall from Formula (62) Actual load at fifth-floor level. Safe load on a 24" wall from Formula (62) Actual load at fourth-floor level, Safe load on a 28" wall from Formula (62) Actual load at third-floor level, Safe load on a 32" wall from Formula (62) Actual load at second-floor level. Safe load on a 36" wall from Formula (62) Actual load at first-floor level. Safe load on a 40" wall from Formula (62) The first-story wall could safely be made 36" if the brickwork is good, and there are not many flues, etc., in walls, for then we could use ^y.^ = 250, which would give a safe load on a 36" wall = 65127 lbs., or more than enough. The above table shows how very closely the Formula (62) would agree with a practical and common -sense arrange- ment of exactly the same 3832 ) 4298) 10243 ) 91351 17176 ) 15729 > 24632 I 23750 I 32611 I 32825 ) 41112 I 42638 j 50136 ) 52902 I 69683 ) 63492 ^ ng.91. amount of material, as required by the law. 152 SAFE BUILDING. Thrust of Now, if tlie upper floor were laden with barrels, barrels, there might be some danger of these thrusting out the wall. We will suppose an extreme case, four layers of flour barrels packed against the wall, leaving a 5-foot aisle in the centre. We should have 20 barrels in each row (Fig. 91), weighing in all 20.19G =: 3920 lbs. These could not well be placed closer than 3 feet from end to end, or, say, 1307 lbs., per running foot of wall; of this amount only one-half will thrust against wall, or say, 650 pouuds. The dia- meter of the barrel is about 20". If Figure 92 represents three of the barrels, and we make A B =: MJ, = I the load of the flour barrels, per running foot of walls, it is evi- dent that D B will represent the \l/ \ horizontal thrust on wall, per run- ning foot. As D B is the radius, and as we know that AD=r2DBor = g2 *2 radii, we can easily find A B, for : A — D B2 A B2 or 4. D B2 — D B2 = or D B9 =-^or D B = 0,5 78.w. 3 y/3 1,73 Or, h z= 0,578. w, (64) Where h = the horizontal thrust, in lbs., against each running foot of wall, = one- half the total load, in lbs., of barrels coming on one foot of floor in width, and half the span. In our case we should have : /j = 0,5 78.650 = 3 75 lbs. Now, to find the height at which this thrust would be applied, we see, from Fig- ure 91, that at point 1 the thrust would be from one line of barrels ; at point 2, from two lines ; at point 3, from three lines, etc. ; therefore, the average thrust will be at the centre of gravity of the triangle ABC, this we know would be at one-third *L the heiiiht A B from its base A C. ''^^^^ oFz.g^^r^J ONCff^s) ^>--r — ^ I IHH l2 2A- 36 Now B C is equal to 6r or six radii of 1500 50Q0 4500 the barrels ; further, A C = 3r, therefore : Flg.93. A B2=:36.r2 — 9r2= 25.r2 and A B = 5r; therefore, — = Ifr, WIND-PBESSXJRE. 153 To this must be added the radius A D (below A) so that the central point of thrust, O in this case, would be above the beam a distance 2/ = 2f.r. Where y = the height, in inches, above floor at which the average thrust takes place. Where r = radius of barrels in inches. Our radius is 10", therefore : Now, in Figure 93 let A B C D be the 12" wall, A the floor level, G M the central axis of wall, and A O = 2G|; draw O G horizon- tally ; make G II at any scale equal to the permanent load on A D, which, in this case, would be the former load less the wind and snow allowances on wall and roof, or 3832 — (1G.30 + 13.30) = 2962, or, say 3000 lbs. Therefore, make G H = 3000 lbs., at any scale ; draw 111 = h = 375 lbs., at same scale, and draw and prolong G 1 till it intersects D A at K. The pressure at K will be p = G I = 3023. We find the distance M K measures Therefore, from formula (44) the stress at D will be : 3023 + 6 3i.3023 = -|- 56 lbs. (or compression) 144 ' 12.144 While at A the stress would be, from formula (45) : 3023 „ 31.3023 , . ii, / * • n .i, * v = 6.-3 = — 14 lbs. (or tension), so that 144 12.144 ^ ^' the wall would be safe. The writer has given this example so fully because, in a recent case, where an old building fell in New York, it was claimed that the walls had been thrust outwardly by flour barrels piled against them. Narrow Piers. Where piers between openings are narrower than they are thick, calculate them, as for isolated piers, using for d (in place of thickness of wall) the width of pier between openings ; and inP - place of L the height of opening. The load on the pier will consist, besides its own weight, of all walls, girders, floors, etc., coming on the wall above, from centre to centre of openings. Wind-pressure. To calculate wind-pressure, assume it to bu normal to the wall, then if A B C D, Figure - 94, be the section of the whole wall above ground (there being no beams or braces against wall). O DMC H Fig. 94. 154 SAFE BUILDING. Make O D = ^. A D = — ; draw G H, the vertical neutral axis of tlie whole mass of wall, make G II, at any convenient scale, equal to the whole weight of wall ; draw H I horizontally equal to the total amount of wind-pressure. This wind-pressure on vertical surfaces is usually assumed as being equal to 30 lbs. per square foot of the surface, provided the surface is Hat and normal, that is, at right angles to the wind. If the wind strikes the surface at an angle of 45° the pressure can be assumed a,t 15 lbs. per foot. It will readily be seen, therefore, that the greatest danger from wind, to rectangular, or square towers, or chimneys, is when the wind strikes at right angles to the widest side, and not at right angles to the diagonal. In the latter case the exposed surface is larger, but the pressure is much smaller, and then, too, the resistance of such a structure diagonally is much greater than directly across its smaller side. In circular structures multiply the average outside diameter by the height, to obtain the area, and assume the pressure at 15 lbs. per square foot. In the examples already given we have used 15 lbs., where the building was low, or where the allowance was made on all sides at once. Where the wall was high and supposed to be normal to wind we used 30 lbs. Referring again to Figure 94, continue by drawing G I, and prolonging it till it intersects D C, or its prolonga- tion at K. Use formulae (44) and (45) to calculate the actual pres- sure on the wall at D C, remembering that, a; = M K ; where M the centre of D C, also that, p = G 1; measured at same scale as G H. Remember to use and measure everything uniformly, that is, all feet and tons, or else all inches and pounds. The wind-pressure on an isolated chimney or tower is calculated similarly, except that the •neutral axis is central between the walls, instead of being on the wall itself; the following example will fully illustrate this.' Example. Wind -pressure chimney, Figure 87, safe against wind- on Chimney, pressure? We need examine the joints A and E only, for if these are safe the intermediate ones certainly will be safe too, where the thickening of walls is so symmetrical as it is here. The load on A we know is 47 tons, while that on E is 657 tons. ^ Many cities require an additional vertical load for wind pressure to be added to actual weight of walls when calculating lintels, girders, founda- tions, etc., carrying outside walls. See "Dead VVeight of Wind Pressure on Walls," p. 84. STRENGTH OF CCRBEI,S. 155 Now the wind-pressure down to A is : Pa = 10,30.15 = 4500 lbs., or = 2^ tons. On base joint E, the wind-pressure is : P = 12f. 150.15 = 28500 lbs., or = 141 tons. We can readily see that the wind can have no appreciable effect, but continue for the sake of illustration. Draw P^ horizontally at half the height of top part A till it intersects the central axis G, ; make G, H, at any convenient scale = 47 tons, the load of the top part; draw H. I, horizontally, and (at same scale) = 2 J tons = the wind-pressure on top part; draw G I,; then will this represent the total pressure (from load and wind) at K, on joint A A,. 11*^0 Formula (44) to get the stress at A, where x = K, M, = 9", or I feet; and p = G, I„ which we find scales but little over 47 tons ; and Formula (45) for stress at A. For d the width of joint we have of course the diameter of base, or 10| feet. Therefore Stressat A, = 1^-1-6.^4^-^ =-}- 1,71 tons, per s(juare foot, or ^'^2^^^^^ — ~l~ (compression), per square inch, and 47 47 -?• stress at A =— — 6. — — ^ = -(- 0,7 tons, per square foot, or 0,7.2000 I 1A IK / • N -1 ' — =-[-10 lbs. (compression), per square inch. To find the pressure on base E E, ; draw P G horizontally at half the whole height; make G jNI = 057 tons (or the whole load)^ and draw M K horizontally, and = 14|- tons (or the whole wind-pressure). Draw G K and (if necessary) prolong till it intersects E E, at K. From formulas (44) and (45) we get the stresses at E, and E : p being = G K = 658 tons ; and a; = M K = 20", or If feet. For tl tlie width of base we have the total diameter, or IG feet. Therefore stress at G58 I „ 658.1* , r. 7.2000 , „ E, = — -f 6. = + 7 tons per sq. ft., or = -f 9 7 lbs. (compression), per square inch, and stress at E — ^' i'm'ig ~ 1| tons (compression), per square foot, or ^— '^^^^ = -{-23 lbs. (compression) per square inch. There is, therefore, absolutely no danger from wind. Strength of Corbels carrying overhanging parts of tlie walls, Corbels, etc., should be calculated in two ways, first, to see ' Scale of weiglits. Fig, 87, applies to Gi H,, etc., but not to G M. 156 SAFE BUILDING. whether the corbel itself is strong enough. "We consider the corbel as a lever, and use either Formula (25), (26) or (27) ; according to how the overhang is distributed on tlie corbel, usually it will be (25). Secondly, to avoid crushing the wall immediately under the corbel, or possible tipping of the wall. Where there is danger of the latter, long iron beams or stone-blocks must be used on top of the back or wall side of corbel, so as to bring the weight of more of the wall to bear on the back of corbel. To avoid the former (crushing under corbel) find the neutral axis G n of the whole mass, above corbel, Figure 95 ; continue G II till it intersects A B at K, and use Formulae (44) and (45). If M be the centre of A B, then use x = Vi M, and p — weight of corbel and mass above ; remembering to use and measure all parts alike ; that is, either, all tons and feet, or all pounds and inches. CHAPTER V. ARCHRS. ITTIIE manner of laying arches has been de- *X. scribed in the previous chapter, while in the first chapter was given the theory for calcu- lating their strength ; all that will be necessary, therefore, in this chapter will be a few practical examples. Before giving these, however, it will be of great assistance if we first explain the method of obtaining graphically the neu- tral axis of several surfaces, for which the arithmetical method has already been given Neutral axis (P* '^•)- To find the vertical neu- found graphi- tral axis of two plane surfaces A B E F and B C D E, proceed as follows : Find the centres of gravity G„ of the former surface and G, of the latter surface. Through these centres draw G„ H„ and G, H, vertically. Draw a vertical line a c anywhere, and make ab at any scale equal to surface A B E F, and at same scale make 6 c = B C D E. From any point o, draw o a, ob and oc. From the point of intersection of o c with H, G„ draw g, ^r,, paral- lel 0 b till it intersects H„ G„ at g,, ; from ^r,, draw ^r,, g jiarallel with o a till it intersects o c at g; through g draw g G vertically and this is the de- sired vertical neutral axis of the whole mass. Where there are many parts the same method is used. We will assume a segmental arch divided into five equal parts. Calling part A B C D = No. I ; part D C E F = No. II, etc., and the verti- cal neutral axis of each part. No. I, No. II. etc. 158 SAFE BUILDING. Draw a f vertically and at any convenient scale, make a fe = No. I or A 13 C D 6 c = No. II or D C E F, c = No. Ill de — No. TV and e/= No. V From any jioint o draw oa, ob, oc, od, oe and of. From point of intersection 1 of verti- cal axis No. I and oa, draw 1 g parallel with o b till it intersects vertical axis No. II at g ; then draw g h parallel with 0 c till it intersects vertical axis No. Ill at h ; similarly draw h i parallel with o d to axis No. IV and i j parallel with o e to axis No. V, and, tinaUy, draw j 5 par- allel with o f till it intersects a o (or its prolongati jn) at 5. A vertical axis A Fig. 99 loo ^o o 300 ■»oo 5o » JCALe ot= pounDS. JC^Ue OK H£CT. through 5 is the ver- tical neutral axis of the whole mass. Prolong j i till it intersects o a at 4, i li till it intersects 0 a at 3, and A g till it intersects o a at 2. A vertical axis through 4 will then be the vertical neutral axis of the mass of Nos. I, II, III and IV ; a ver- tical axis through 3 will be the verti- cal neutral axis of the mass of Nos. I, II and III ; and Fig 100 through 2 the vertical neutral axis of the mass of Nos. I and II course the axis through 1 is the vertical neutral axis of No. I. We will now take a few practical examples. ARCHES. 159 Example I. In a solid brick wall an opening 3 feel wide is bricked over with an 8-i7ich arch. Is this strong enough f ^ The thickness of the wall or arch, of course, does not matter, where the wall is solid, and we need only assume the wall and arch to be one foot thick. If the wall were thicker, the arch would be correspondingly thicker and stronger, so that in all cases where a Arch In load is evenly distributed over an arch we will solid wall, consider both always as one foot thick. If a wall is hollow, or there are uneven loads, we can either take the full actual thickness of the arch, or we can proportion to one foot of thickness of the arch its proportionate share of the load. In our example we as- sume everything as one foot thick. The load coming on the half-arch B J I L Fig. 100, will be enclosed by the Voussoirs lines A L arbitrary. I A at 60° with the horizon. We divide the arch into, say, four equal voussoirs B C = CF = FG=GJ. (The manner of dividing might, of course, have been arbi- trary as well as equal, had we preferred.) Draw the radiating lines through C, F and G, and from their upper points draw the ver- . , ,. , T-w -n 1 TT ■^rAi^E OF FEET. tical hnes to D, \^ and 11. Fig. loi. Now find the weight of each slice, remembering always to include the weight of the voussoir in each slice. We have, then, approximately, No. I (A B C D) = 3' X J' X 112 pounds = 168 pounds No. ir (DCFE) = 2|'x ^'xll2 " z= 119 " No. Ill (E F G H) = ij' x x 112 " = 73 " No. IV (II G J I) = |'xj^'xll2 " " Total = 403 " 1 For convenience, most of the figures are duplicated : the first, showing man- ner of obtaining the horizontal thrust ; the second, the manner cf obtaining the line of pressure. 160 SAFE BUILDING. As the arch is evidently heavily loaded at the centre, we assume the point a at one-third the height of B L from the top, or L a = — =2§" and draw the horizontal line a 4. 3 ^ As previously explained, find the neutral axes : 1 n, of part No. I, Horizontal , tvt t i -vr tt pressures. 2 g2 parts No. I plus No. 11, 3 ga of parts Nos. I, II and III, and 4 ffi of the whole half arch. Now make at any scale 1 .(7, = 1G8 units = No. T ; similarly at same scale 2 (/g = 1G8 + 119 units = 287 = No I and No. II, and 3 (73= 287 -j- 73 units = 3G0 = No. I + No. II + No. III. 4 ^4 = 360 -f 43 units = 403 = weight of half arch and its load. Now make : C ?i = i C L, = 2§". Similarly, Fk=^FU= 2|" ; also, G 4 = ^ G Ls = 2§", And, J Z4 = H I = 2§". Through g„ g., g^ and gi draw horizontal lines, and draw the lines 1 2 li, 3 li and 4 k till they intersect the horizontal lines at 7i„ h, K and h; then will g, h, measured at same scale as 1 g, represent the horizontal thrust of A B C D ; ,72 ^'2 the horizontal thrust of A B F E ; g^ hs the horizontal thrust of A B G II and h the horizontal thrust of the half arch and its load. In this case it happens that the latter is the greatest, so that we select it as our horizontal pressure, and make (in Fig. 101) ao = gih = 620 pounds, at any convenient scale. Now (in Fig. 101) make a & = 168 pounds = No. 1. 6 c = 119 " = No. n. cd= 73 " = No. III. de= iS " = No. IV. Draw ob, oc, od and oe. Now begin at a, draw a 1 parallel with 0 a till it in- curve o<'^gggy^g_ tcrsccts axls No. I at 1 ; from 1 draw 1 1„ parallel with 0 h till it intersects axis No. II at i,; from i, draw i, h parallel with 0 c till it intersects axis No. Ill at h ; from ?2 draw u i, parallel with o d till it intersects axis No. IV at i^; from h draw {3 parallel with 0 e. A curve through the points a, K„ lu, K3 and 1^ (where the former lines intersect the voussoir joint lines) and tangent to the line a 1 ?, iih would be the real curve of pressure. The amount of pressure on joint C L, would be concentrated at K, and would be ecpial to 0 ft (measured at same scale as a b, etc.). The pressure on joint F I.3 vDuld be concentrated at and be equal to 0 c. The pressure on BRICK FLOOR-ARCH. 161 joint G L3 would be concentrated at and equal to 0 d. The pres- sure on the skew-back joint J I would be concentrated at iT, and be equal to 0 e. The latter joint evidently suffers the most, for not only has it got the greatest pressure to bear, but the curve of pressure is farther from the centre than at any other joint. We need calculate this joint only, therefore, for if it is safe, the others certainly are so, too. By scale we find that J /£( measures 2A-", or Ki Stress at ex- ^ ... 1? j j- trades and is (x =) IJ" from the centre of joint; we nnd lur- intrados. ^^^^ ^^iZit, 0 e scales 740 units, therefore {p =) 0 e = 740 pounds. The width of joint is, of course, 8", and the area = 8 x 12 = 96 square inches ; therefore, from Formula (44) Stress at edge J =^ + 6. yb 740. 96. 8 16,26 and from (45) Stress at edge I = ■ 9d 740. U 96. 8 = — 0,85 Or the edge J would be subject to the slight compression of 16^ pounds, and edge at I to a tension of a little less than one pound per square inch. The arch, therefore, is more than safe. Example II. A four-inch rowlock brick arch is built between two iron beams, oj five feet span, the radius of arch being Jive feet. The arch is loaded at the rate of 150 pounds per square foot. Is it safe f In this case we will divide the top of arch A D into five equal sec- ions and assume that each section carries 75 pounds — (which, of course, is not quite correct). We find the horizontal Brick floor- ^ , 7 . » 1 r ■, n j arch, pressures (Fig. 102) gih„gzhz, etc., as before, and find loo goo .500 4oo yoo SCALS of POUNOJ JCALE oP Feet. Fig. 10 2. again g^ h^ the largest and equal to 575 pounds. We now make (Fig. 103) at any convenient scale, 0 a = g^h — 575 poimds, and ab = 162 SAFE BUILDING. hc=cd = de = ef = 75 pounds and draw o a, o b, o c, etc. We now find the broken line a 1 i, K„ where : a 1 is parallel w^ith oa 1 z", is parallel with o h i, la " " 0 c 4 is " " od ii U " " oe « " o/ In this case again evidently the greatest stress is on the skew-back joint C D, for it not only has the greatest pressure o f, but the curve of Rg. 103. .TCAie of p-eer. pressure passes farther from the centre than at any other joint. We find that C scales 1\ inches, therefore the distance of from the centre is {x =) f". We scale o f and find it scales 690 units, or (p )= 690 pounds. The joint is 4" wide and its area = 4 x 12 = 48 square inches. From Formulae (44) and (45) we have then : Stress at C = — H- 6. = + 30,6 pounds and 48 ' 48. 4 ' ^ Stress at D = 690 690. '■■ = — 1,8 pounds. 48 48. 4 The arch, therefore, is perfectly safe. Example III. Two iron beams, five feet apart, same as before, but filed with a fitraight 7" hollow f re-clay arch. The load per foot to be assumed at 140 pounds. Is the arch safe ? Fireproof pounds on the half arch we will assume floor-arch. gQ pounds to come on each of the blocks and 30 j- D - hi g5 S5 Fig. 104. FIREPROOF FLOOR-ARCH. 163 pounds on the skew-back. We then (in Fig. 104) 5nd, as before the horizontal pressures, g, Zi„ ho, etc. Again we find the largest pres- sure to be ^5 ^5, and as it scales 2040 units, we make (in Fig. 105) at any convenient scale and place o a = gjir,= 2040 pounds. We also make a'b=.'bc = cd = de — ^(i units and e/= 30 units. Draw oa, ob, oc, etc. Drawing the lines parallel thereto, beginning at a we get the line a 1 i, 4 13 h K^, same as before Imagining a joint j?V N|'llf ' _2o+o V- jiwo J-CAl-e OF Pour di — ei-v.v 5CA1_E OF F=6ET. Fig. 105. at C D this would evidently be the joint with greatest stress, for tlie same reasons mentioned before. We find C scales 2|", and as C D scales 7^" the point is distant from the centre of joint. (a: = )3|-2|=l" as o / scales 2100 units or pounds, and the joint is H" deep with area = 7^. 12 = 87 square inches, we have : „^ . 2100 , „ 2100. 1 I , . A A Stress at C = -)- 6. = -|- 44,14 pounds and 6. -^1^^= + 4,14 pounds. The arch, therefore, would seem perfectly safe. But the blocks are not soUd ; let us assume a section tlirough the skew-back joint C D to be as per Fig. 106. We should have in Formulae (44) and (45) X, p, and the depth of joint same as before, but for the area we should use a = 3.1^.12 = 54 square inches, or only the area of solid parts of block. Therefore we should have : 164 SAFE BUILDING. Stress at C = Stress at D = 2100 54 2100 54 6. 6. 2100. 1 . 64. 7^' 2100. 1 71 pounds, and = -|- 6,71 pounds. 64. 7^ There need, therefore, be no doubt about the safety of the arch. Example IV. Over a 20-inch brick arch of 8 feet clear span is a centre pier 16' wide, carrying some two tons weight. On each side of pier is a window opening 2 J feet wide, and beyond, piers similarly loaded. Is the arch safe f Arch In front divide the half arch into five equal voussoirs. wa 1 1 concen- The amounts and neutral axes of the different vous- tratAd loads. ^^^^8, and loads coming over each, are indicated in circles and by arrows ; thus, on the top voussoir E B (Fig. 107) we have a load of 2100 pounds, another of 62 pounds, and the weight of voussoir or 228 pounds. The neutral axis of the three is the verti- cal through G, (Fig. 108). Again on voussoir E F (Fig. 107) we Fig. 107. have the load 174 pounds, and weight of voussoir 228 pounds; the vertical neutral axis of the two being through G„ (Fig. 108). Simi- larly we get the neutral axes G„„ G.v and Gv (Fig. 108) for each of the other voussoirs. Now remembering that 1 g, (Fig. 107) is the ARCH IN FRONT WALL. 165 ueutral axis of and equal to the voussoir B E and its load ; 2 the neutral axis of and equal to the sum of the voussoir s B E and E F and their loads, etc., we find the horizontal thrusts h„ g^ h^, ga h^, etc. The last g^ is again the largest, and we find it scales 7850 units or pounds. The arch being heavily loaded we selected a at one-third from the top of A B. We now make (Fig. 108) 00= 7850 pounds or units at any scale ; and at same scale make ab= 2390 pounds ; 6 c = 402 pounds; c d = 432 pounds; d e = 2956 pounds, and e /= 18G8 pounds. Draw o b, 0 c, 0 d, etc. Now draw as before a 1 parallel with 0 a to axis G, ; also 1 i, parallel with 0 6 to axis G„ ; t, {2 parallel with o c to G„„ etc. We then again have the pomts a, K„ K^, K^, /tT, O to o 2oo A»o 4^0 900 4oo 5CA1.E op foun»r O 1 g 5 ■» I i Fig. 108. and ^5 of the curve of pressure. As K^, is the point farthest from the centre of arch-ring and at the same time sustains the greatest pressure {of) we need examine but the joint C D ; for if this is safe so are the otliers. We insert, then, in Formula3 (44) and (45) for p = of= 1125U pounds, and as C measures 6^", X = 10" — 6^" = 3^" ; also as the joint is 20" wide, a = 12. 20 = 240 square inches. 166 SAFE BUILDING. Therefore Stress at 0 = 11^ + 6.11^^: 240 ~ 240.20 Stress at D = 11250 240 11250. 3^ 240. 20' -|- 96 pounds, and — 3 pounds. The arch is, therefore, safe. Example V. A 12-inch brick semi-circular arch has 12 foot span. A solid brick wall is built over the arch to a level with one foot above the keystone. The abutment piers are 5 feet high to the spring of arch and are each 3 feet wide, including, of course, the width of skew-backs. Are the arch and piers safe ? Arch in fence- As before, we will assume arch, pier, and wall ment^*^"*' o'^^^ ^^^li, each one foot thick. We will divide the load over arcli into seven equally wide slices. This will make uneven voussoirs, but this does not matter, oas our joint hnes (and voussoirs) are only imaginary anyhow, and not necessarily of the shape of the ac- tual voussoirs, which in brick would, of course, be repre- sented by each single brick. The amount of the sums of each voussoir and its load, and the vertical neu- tral axes of the differ- ent sets are given by the arrows and lines G„ Gg, Gs, etc. (in Fig. 110). When considering the safety of the abut- ment we treat it ex- actl}' the same as the voussoirs (and loads) of the arch ; that is, we THRUST ON ABUTMENT. 167 take the whole weight of the abutment, viz., C D E F I H C and find its neutral axis Gg. Keturning now to the arch, we go through the same process aa before. We find the horizontal pressures (Fig. 109) <7, A„ h^, etc. In this case we find that the last pressure ffj hj is not as large as ; therefore we adopt the latter; it scales 1423 units or pounds. We now make (Fig. 110) a o = 1425 pounds ; and a b = 251 pounds ; i c z=. 280 pounds ; c J = 373 pounds, etc. ; g his equal to the last section of arch or 1782 pounds. We continue, however, and make h i = 4600 pounds = the weight of abutment. Draw o a, o b, o c, etc., to 0 i. Then get the tangents to the curve of pressure, as be- fore, viz. : a 1 2, 4 h h h h Ki > we now continue K^, which is par- allel with 0 h till it intersects the vertical axis Gn of the abutment at and from thence we draw ij IQ parallel with o i. Thrust on ^'^^ ^'^^^ examine the base joint I II of pier. abutment. I /^^ scales IQ\", and as the pier is 36" wide, is 7|" from the centre of joint. The area is a = 12.36 = 432 and the pressure is/i = o i = 9100 pounds. Therefore, and o, . T 9100 , „ 9100.7J , .Q , Stress at I = — + 6. = -f 48 pounds, Stress at H = — 6. ^}^^'^* = — 6 pounds. 432 432.36 ^ There is, therefore, a slight tendency for the pier to revolve around the point I, raising itself at H ; still the tendency is so small, only 6 pounds per square inch, that we can safely pass the pier, so far as danger from thrust is concerned. Joint C D at the spring of the arch looks rather dangerous, how- ever, as I'e cuts it so near its edge D. Let us examine it. D measures 1^", therefore Ki is 4^'' from the centre of joint, which is 12" wide. The area is, of course, a = 12. 12 = 144 and the pressure p = 0 h = 4600 pounds. Therefore, o. . -r, 4G00 , „ 4600.4i , , Stress at D = -— — - + 6. , ^ ^ = -4- 104 pounds, 144 144.12 ' and o. . r- 4600 ^ 4600.4^ .„ , Stress at C = — — — 6 — — = — 40 pounds. 144 144.12 It is evident, therefore, that the arch itself is not safe, and it should be designed deeper ; that is, the joints should be made deeper (say, 16"), and a new calculation made. 168 SAFE BUILDING. Tie-rods to If> instead of an abutment-pier, we had used an arches, iron tie-rod, its sectional area would have to be suf- ficient to resist a tension equal to the greatest horizontal thrust o a; 1 and care should be taken to proportion the washers at each end, large UNEVEN ARCHES. 169 enough that they may have sufficient bearing-surface so as not to crush the material of the skew-backs. Thus, in Example III, Fig. 105, if we should place the iron tie-rods to the beams 5 feet apart, they would resist a tension equal to five times the horizontal thrust o a, which, of course, was calculated for 1 foot only, or t = 5. 2040 = 10200 pounds. The safe resistance of wrought-iron to tension is from Table IV, 12000 pounds per square inch ; we need, therefore : 10200 . , = 0,85 square inches 12000 ^ of area in the rod; or the rod should be l^y diameter. A 1" or even |" rod would probably be strong enough, however, as such small iron is apt to be better welded, and, consequently, stronger, and the load on the arch would probably be a " dead " one. As the end of rod will bear directly against the iron beam, the washers need have but about \" bearing all around the end of the rod, so that the nut would probably be large enough, and no washer be needed. Example VI. A pier 28" wide and 10' high supports two abutting semi-circular arches ; the right one a 20" Vrick arch of 8' span ; the left one an 8" brick arch of 3' span. The loads on the arches are indicated in the Figure 111. Is the pier safe f Uneven arches The loads are so heavy compared to the weight of with central pier, the voussoirs, that we will neglect the latter, in this case, and consider the vertical neutral axis of and the amount of each load as covering the voussoirs also ; except in the case of the lower two voussoirs, where the axes are considerably affected. We find the curve of pressure of each arch as before. For the large arch we would have the curve through a and i, for the small one through a, and ; the points i and being the inter- sections of the curves with the last vertical neutral axis of each arch. Now from i draw i x parallel with o f and from i, draw i, x (back- wards), but parallel with o,/, till the two lines intersect at x. Now make / g parallel with and = o, /„ and draw g h vertically = 2600 pounds = the weight of the pier from the springing line to the base (1' thick). Draw o g and o h. Now returning to x, draw x y paral- lel with g 0 till it intersects the neutral axis of the pier at y, and from y draw y z parallel with o h till it intersects the base joint C D of 170 SAFK BUILDING. pier (or its prolongation) at Ei. Continue also x y till it intersects the springing joint A B at iT. Now, then, to get the stresses at Fig. Ml. joint A B we know that the width of joint is 28", therefore a = 28.1 2 = 336 ; further p = o g = 19250 pounds, and as B K scales one inch, K is distant 13" from the centre of joint, therefore Stress at B = 19250 19250. 13 and Stress at A : 336 19250 336. 28 19250. 13 = -{-217 pounds. 336 336. 28 = — 102 pounds. Thrust on arch, therefore, cannot safely carry such heavy central pier, loads. The pier we shall naturally expect to find still more unsafe, and in effect have, remembering that joint D C is 28" wide, therefore, area 336 square inches, and as K, distant 54' from centre of joint, and p = o h — 21750 pounds. 21750. 54 , , = -j- 813 pounds. Stress at D = 6. 336 336. 28 INVERTED ARCHES. and Stress at = 21750 6 21750. 54 p^^^^g. 336 336. 28 » 1- # ».h The construction, therefore, must be radically Relief through , i i rr iT iron-work, changed, if the loads cannot be altered, ii tlie arches are needed as ornamental features, they should be constructed to carry their own weight only, and iron-work overhead should carry the loads, and bear cither nearer to, or directly over, the piers, as farther trials and calculations might call for. If this is done the wall should be left hollow under all but the ends of iron- To avoid ., 1 . • y cracks, work unlil it gets its "set"; that is, until it lias taken its full load and deflection; and then the wall should be pointed with soft " putty " mortar. Example VII. The foundations of a building rest on brick piers 6' wide and 18' apart. The piers are joined by 32" brick inverted arches and tied to- gether 8' above the spring of the arch. Piers and arches are 3' thick. Load on central piers is 72 tons; on end pier 60 tons. Is this con- struction safe ? Inverted We will first examine the inner or left pier. The arches, pier being 3' thick and the load 72 tons, each 1' of thickness will, of course, cirry -^^ = 24 tons. The width from centre to centre of piers is just 24', so that each running foot of wall under /IS. ^CAUa »F TonJ . » » . » » » 7 « « ■■ 'J '» I* l« '« JCAl-E rEHT. Fig. 112. arch will receive a pressure of one ton. Now all we need to do is to imagine this pressure as the load on the arch. We can either draw 172 SAFE BUILDING. the arcli upside down with a load of one ton per foot, or we can make the drawing with tlie arch in correct position and the weights press- ing upward, as shown in Fig, 112. We divide the load into five equal Strength of slices, each about 2' wide, therefore = 2 tons each. arch. We make a 6 = 2 tons ; 6 c = 2 tons, etc., and find the horizontal pressures h„ h^, etc., same as before. Again, g^, is the largest, measuring 13 units or tons. Now make o a = 13 tons, and draw ob,oc, etc. Construct the line al K^or curve of pressure same as before. Joint L M is evidently the most strained one. We find M measures 11", and as the arch is 32" (= M L), of course, Ki is 5" from the centre of joint. The area of joint is a = 32. 12 = 384. We scale of, the pressure at K^, and find that it measures 16^ units, or 33000 pounds, therefore qfrp«9 at AT — 33000 I „ 33000. 5 btress at M = -(- 6. -^^^-^ = + 168 pounds Stress at T — ?^ R 33000. 5 , ^ , btress at L _ ^ _ 6. = + 6 pounds. Central The arch, therefore, is perfectly safe. Of course, pier. the left pier is safe, for being an inner pier the re- sistance X of the adjoining arch to the left will just counter-bal- ance the thrust of our arch, or x. But at the end (right) pier this is different. We, of course, proportion the length of foundation A C, to get same pressure as on rest of wall under arch. The end pier car- 60 ries 60 tons, or y = 20 tons per foot thick. The pressure per running foot on wall under arch we found to be one ton, therefore A C should Thrust on be 20 feet long. The half-arch will take the pressure end pier, of 10 feet (from A to B) or 10 tons, and the balance (10 tons) will come on B C. This will act through its central axis G II, which, at the arch skew-back, will be half-way between the end of arch J and outside pier-line I F. This will, of course, deflect some- what the abutment (or last pressure) line K^ H of the arch. At any convenient place draw o,/ vertically equal 10 tons, and a, o, horizon- tally equal 13 tons, the already known horizontal pressure. Then Kt H is, of course, parallel with and equal o,/. Now makey; g, — 10 tons and draw o, g,. Now on the pier draw H K parallel with o, g,; H being the point of intersection of G II and K^ II. As the pier is tied back 8' above the arch, we take our joint-Hne at E F, being 8' above D. We find, by scale, that K is 58" from the centre of E°F, the latter being 72" wide. The area is a = 72. 12 = 8G4. And as 0, g, scales 24 units, the pressure is, of course, 24 tons or 48000 pounds, CIRCULAR DOME. 173 therefore : and a* * -P 48000 , „ 48000.68 i qik^„„„i. Stress at i = + 6. = -t- 315 pounds. 864 ~ 864. 72 ' ^ ^ 48000 „ 48000.58 , Stress at E = -tt^t 6. . = — 207 pounds. 864 864. 72 Us^ There is, therefore, no doubt of the insecurity of buttress, the end pier. Two courses for safety are now open. Either we can build a buttress sufficiently heavy to resist the thrust of the end arch, or we can tie the pier back. The former case is easily calculated ; we simply include the mass of the buttress in the resistance and shift the axis G H to the centre of gravity of the area of pier and buttress up to joint E F. Of course, the buttress should be carried up to the joint-Une E F,but it can taper away from there. Use of ^® back with iron we need sufficient area to tie-rods, resist a tension equal to the horizontal pressure a o, which in this case is 13 tons or 26000 pounds per foot thick of wall. As the wall is 3' thick, the total horizontal thrust is 3. 26000 = 78000 pounds. The tensional stress of wrought-iron being 12000 pounds 78000 per square inch, we need jgooo ~ square inches area, or, say, two wrought-iron straps, 4" x |", one each side of pier. By this method the inner or left arch becomes practically the end arch. For the last two piers and the right arch become one solid mass ; and not only is their entire weight thrown against the second or inner arch, but the centre of gravity of the whole mass shifts to the centre line of end arch, or in our case 9' inside of the end pier ; so that there is no pos- sible doubt of the strength of the abutment. There is one element of weakness, however, in the small bearing the pier has on the skew- back of the arch, the danger being of the pier cracking upwards and settling past and under the arch. This can be avoided, as already explained, by building a large bond-stone across the entire pier, form- ing skew-backs for the arch to bear against. Example VIII. A semi-circular dome, circular in plan, is 40' inside diameter. The shell is 5' thick at the spring and 2' at the crown. The dome is of cut- stone. Will it stand f Calculation ^'^^ ^^^^ ^ section (Fig. 1 14) of the half-dome an** of dome, treat it exactly the same as any other arch. The only difference is in the assumption of the weights. Instead of assuming the arch 1' thick, we take with each voussoir its entire weight around 174 SAFE BUILDING. one-half of the dome. Thus, in our case, we divide the section into six imaginary voussoirs. The weight on each voussoir will act through 6 ...S. 3 , .^■..^-^■^•^^•""Tr ■ ■ 1°' its centre of grav- I ,^ /'! A^-^-fT^^^^^ Now weight J I / ' I J/'V''^^^^^^ ^ ^^^^ ^® equal to the area of the top voussoir multiplied by the circumfer- ence of a semi- circle with A B as radius. Similarly, No. II will be equal to the area of the second voussoir multiplied by the circumference of a semi-circle with A C as radius; No. Ill will be equal to the area of voussoir 3, multiplied by the circumference of a semi-circle with A D as radius, etc. The vertical neu- tral axes Nos. I, II, ni, etc., act, of course, through the centres of gravity of their respective voussoirs. Now the top voussoir measures 5' 6" by an average thickness of 2' 1" or 51-. 23^ = 11,46 or, say, 11 J square feet. As A B (the radius) measures 2' 9", the circumference of its semi-circle would be 8',6. Taking the weight of the stone at 160 pounds per cubic foot, we should then have the weight of No. r =: 11,5. 8, 6. 160 = 15824 pounds, or, say, 8 tons. Similarly we should have : No. II = 12,4. 25,1. 160 = 49798 pounds, or, say, 25 tons. No. Ill = 14,2. 40. 160 = 90880 " « 45 " No. IV = 17,4. 52,7. 160 = 146716 « 73 " No. V= 21,2. 62,8. 160 = 213017 " « 106 " No. VI = 28,7. 69,1. 160 = 317307 " « 168 « STRESS AT JOINTS. 175 We now make ab = 8 tons, J c = 25 tons, Fig. 114. we find that the largest pressure is not the last one, but g^ /jj which measures 78 units; we therefore select the latter and (in Fig. 114) make a o = g^ hi= 78 tons. Draw o b, o c, o d, etc., and construct the line a 1 i, 4 /iTj, same as before. In this case we cannot tell at a glance which is the most strained voussoir joint, for at joint H I the pressure is not very great, but the line is farthest from the centre of joint. Again, whUe joint J L has not so much pressure as SAFE BUILDING. tlie bottom joint M N, the line is farther from the centre. We mast, therefore, examine all three joints. Wc will take H I first. The width of joint is 2' 5" or 29". The pressure is o c, which scales 88 tons or 17G000 pounds. The distance of K2 from the centre of joint P is 16". The area of the joint is, of course, the full area of the joint around one-half of the dome, or equal to II I multiplied by the circumference of a semi-circle with the dis- tance of P from a <7 as radius. The latter is 10' 6", therefore area = 2^^. 33 = 80 square feet or 11520 square inches, therefore : Stress at I = 1^ + 6. li^^J^ = + 65,9 11520 ' 11520. 29 ~ ' Stress at H = - 6. lI^Md^ = - 35,3. 11520 11520. 29 ' For joint J L we should have : the width of joint = 50". The pressure = 0/= 272 tons or 544000 pounds. The distance of from the centre of joint is 8", while for the area we have 50. 66§. 12 = 40000 square inches, therefore : and Gf 0+ T 544000 , „ 540000. 8 , „ , ^'^= 40000 + 40000750 = + P°""^^' Stress at L = _ 6 ^^^^QQ" « = + 0,5 pound. 40000 ^ 40000. 50 T ' 1 ^ ^• For the bottom joint M N we should have the width of joint = 60". The pressure — og = 428 tons or 856000 pounds. The distance of /Tg from the centre of joint is 4", while for the area we have 60. 70§. 12 = 50880 square inches, therefore : _ 856000 , 856000.4 Stress at M = -f 6. g^gso. 60 = + ^^'^ P'^""'^'' and - ... 856000 „ 856000.4 Stress at N = - 6. ^oggo. go = + P°"»^^- The arch, therefore, would seem perfectly safe except at the joint H I, where there is a tendency of the joint to open at H. Had we, however, remembered that this is an arch, lightly loaded, and started our line at the lower third of the crown joint, instead of at the upper third, the line would have been quite different and undoubtedly safe. The dangerous point in that case would be much nearer the centre of joint, while the other lines and joints would not vary enough to call for a new calculation, and we can safely pass the arch. One thing must be noted, however, in making the new figure, and that is that the horizontal pressures A„ ^2 Aj, etc. (in Fi{:. 11 3^, ■would have ITHRUST OF DOME. 177 to be changed, too, and would become somewhat larger than before, as the line a, 6 would now drop to the level of 03. A trial will show that the largest would again be g-, h, and would scale 80 units or tons, which should be used (in Fig. 114) in place of a 0. -Thrust of regards abutments, there usually Ib dome, are none in a dome ; it becomes neces- sary, therefore, to take up the horizontal thrust, either by metal bands around the entire dome, or by dovetailing the joints of each horizontal course. We must, of course, "2 take the horizontal thrust existing at each joint. Thus, if " ^' we were considering the second joint 11 I the horizontal thrust would be g.^ or, if we were considering the fifth joint J L the horizontal thrust would be g^ h. For the lower joint M N we might take its own horizontal thrust g^ h, which is smaller than ^3 hi, provided we take care of the joint J L separately ; if not, we should take the larger thrust. Metal bands. If we use a metal band its area manifestly should be strong enough to resist one-half the thrust, as there will be a sec- tion in tension at each end of the semi-circle, or Where a = area, in square inches, of metal bands around domes, at any joint. Where h = the horizontal thrust at joint, in pounds. Where ^-^^ = the safe resistance of the metal to tension, per square inch. In our case, then, for the two lower joints, if the bands are wrought- iron, we should have, as h = 80 tons = 160000 pounds. 1600Q0 _ ° — 2. 12000 Or we Avould use a band, say, 5" x 1^". ^. , If dovetailed dowels of stone are used, as shown in Strengtn o' . . dowels. 115, there should be one, of course, in every ver- tical joint. The dowels should be large enough not to tear apart at A D, nor to shear off at A E, nor to crush A B. Similarly, care should be taken that the area of C G + B F is sufficient to resist the tension, and of H C to resist the shearing. The strain on A D will, of course, be tension and equal to one-half the horizontal thrust; the same formula holds good, therefore, as for 178 SAFE BUILDING. metal bands, except, of course, that we use for ^^s value for what- ever stone we select. Supposing the dome to be built of average mar- ble, we should have from Table V, ^-^ ^ = 70, therefore, 160000 Now if the dome is built in courses 2' 6" or 30" high, the width of dovetail at its neck A D would need to be : , ^ 1143 „„ . , A D = -gQ- = 38 inches. As this would evidently not have sufficient area at G C and B F we must make A D smaller, and shall be compelled to use either a metal dowel, or some stronger stone. By reference to Table V we find that for bluestone, (^y) = 1^^' therefore : 160000 , ° = Xl40 = ^^^ 571 A D = -gQ- = 19 inches. This would almost do for the lower joint, which is 60" wide, for if we made : G C + B F = 38" and A D = 19" it would leave 60 — 38 + 19 = 3" or, say, IJ" splay; that is, D 11 = 1 ^". Had we used iron, or even slate, however, there would be no trouble. The shearing strain on either A E or H C will, of course, be equal to one-cpiarter of the horizontal thrust or h b = / Where h = the width of one-half the dowel, in inches (A E) Where h =.- the horizontal thrust at the joint, in pounds Where d = the height of the course, in inches. Where the safe resistance to shearing, per square inch, of either dowel or stone voussoir (whichever is weaker). When the shearing stress of a stone is not known, we can take the tension instead. Thus in oxir case as the marble is the weaker, we TILE FLOOR ARCH. 179 should use ^-^^ = 70, therefore 160000 A E = H C = = 19 inches. 4. 30. 70' The compression on A B need not be figured, of course, for while the strain is the same as on A E, the area of A B is somewhat larger, and all stones resist compression better than shearing. Cambered When a plank is "cambered," sprung up, and the plank arch, ends securely confined, it becomes much stronger transversely than when lying flat. The reason is very simple, as it now acts as an arch, and forces the abutments to do part of its work. Such a phmk can be calculated the same as any other arch. Spanish tile Quite a curiosity in construction, somewhat in the arches, above line, has been recently introduced in New York by a Spanish architect. He builds floor-arches but 3" thick, of 3 suc- cessive layers of 1 "-thick tiles, up to 20' span, and more. His arches have withstood safely test-loads of 700 pounds a square foot. The secret of the strength of his arches consists in their following closely the curve oi pressure, thus avoiding tension in the voussoirs, as far as possible. But even were this to exist, it could not open a joint without bodily tearing off several tiles and opening many joints, as shown in Fig. 116, owing to the fact that each course is thoroughly bonded and breaks joints with the course below ; besides this, each upper layer is attached to its lowei layer by Portland-cement mortar. Specimens of these tiles have been tested for the writer and were found to be as follows : Compression (12 days old) c = 2911 pounds or say (^y,j = 300 pounds. Tension (10 days old) / = 287 pounds or say ^ ^ = 30 pounds. The shearing stress of the mortar-joints is evidently greater than the tension, as samples tested tore across the tile and could not be sheared off. The modulus of rupture (about 10 days old) was 180 SAFE BUILDING. i = 91 pounds or, say, ^ = lO pounds. Of course older specimens would prove much stronger. There is only one valid objection that the writer has heard so far against these arches, and this is, that in case of any uneven settle- ment they might prove dangerous ; as, of course, the margin in which the curve of pressure can safely shift in case of changed surround- ings is very small. The writer does not think the objection very great, however, as settlements are apt to ruin any arch and should be carefully provided against in every case. The arches have some very great advantages. The principal one, of course, is their light- ness of construction and saving of weight on the floors, walls and foundations. Then, too, in most cases iron beams can be entirely dispensed with, the arches resting directly on the brick walls ; of course, there must be weight enough on the wall to resist the hori- zontal thrust, or else iron tie-rods must be resorted to. The former is calculated as already explained for retaining walls. If tie-rods are used, they are calculated as explained above for the example of the 7" flat floor arch. An example of these 3"-tile arches may be of interest. Example IX. A segmental 3" tile arch, built as explained above, has 20' clear span with a rise at the centre of 20". The floor is loaded at the rate of 150 pounds per square foot. Is the arch safp f Example of divide the arch into five parts or voussoirs, tile arch, each voussoir carrying 2' of floor = 300 pounds ; we make (Fig. 118) a5 = 300 pounds, be = 300 pounds, etc., and find * * iiii^ y* lOoe ijoo aoce aJOo Jo o» » I 2 3 -5 g 7 a g |0 Fig. I I 7. the horizontal pressures (Fig. 117) g, h„ g^ h^, etc. The last one g^h^ is the largest and scales 4500 units or pounds. We now make (Fig. DANGER OP SLIDING. 181 118) ao = gihi, = 4500 pounds and draw ob, oc, od, etc. We next construct the curve of pressure a K and find that it coincides as closely as possible with the centre line of the arch. This means that -ft? Fig. I I 8. the pressure on each joint will be uniformly distributed. That on the lower joint is, of course, the largest and is = o /, which scales 4700 pounds. The area of the joint is, of course, a = 3. 12 = 36 square inches, therefore the greatest stress per square inch will be 4700 36 = 130 pounds compression. As the tests gave us 300 pounds compression, per square inch, as safe stress of a sample only twelve days old, the arch is, of course, perfectly safe. If, however, instead of a uniform load, we had to provide for a very heavy concentrated load, or heavy-moving loads, or vibrations, it would not be advisable to use these arches. Danger of ^® hdiNO. simply considered the danger of sliding, compression or tension at the joints of an arch; there is, however, another element of danger, though one that does not arise frequently, viz. : the danger of one voussoir sliding past the other. Where strong and quick-setting cements are used this danger is, of course, not very great. But in other cases, and particularly in large arches, it must be guarded against. The angle of friction of brick against brick (or stone against stone) being generally assumed at 30°, care must be taken that the angle formed by the curve of pressure at the joint, with a normal to the jokit (at the point of in- tersection K) does not form an angle greater than 30°. If the angle be greater than 30° there is danger of sliding; if smaller, there is, of course, no danger. Thus, if in Fig. 114 we erect through Ki a nor- mal /C, X to the joint, the angle X Ki 4 should not exceed 30°. . ^, In arches with heavy-concentrated loads at single Danger of ir r. shearing- points, there might, in rare cases, be danger of the 182 SAFE BUILDING. load shearing right through the arch. The resistance to shearing would, of course, be directly as the vertical area of cross-section of the arch, and in such cases this area must be made large enouo-h to resist any tendency to shear. Depth at Arches are frequently built shallower at the crown crown, and increasing gradually in depth towards the spring, the amount being regulated, of course, by the curve of pressure and Formulas (44) and (45). To establish the first experimental thickness at the crown of an arch, many engineers_use the empyrical formula : x=y.\/^ (67) Where x = the depth of arch, at crown, in inches. Where r = the radius at crown, in inches. Where y = a constant, as follows : For cut stone, in blocks : y = o,3 For brickwork y = o,4 For rubblework y = o,45 When Portland cement is used, a somewhat lower value may be assumed for y. The depth thus established for crown is only exper- mental, of course, and must be varied by calculation of curve of pressure, etc. Approximate cases where the architect does not feel the ne- rule. cessity for such a close calculation of the arch, it will be sufficient to find the curve of pressure. If this curve of pressure comes witJiin the inner third of arch-ring, at every point, the arch is safe, provided the thrust on each joint, divided by the area of joint, does not exceed one-half o£ the safe compressive stress of the material, or : Where p = the thrust on joint, in pounds. Where a = the area of joint, in square inches. Where ^ = the safe resistance to compression, per square inch, of the material. Vaulted and Vaulted and groined arches are calculated on the Croined Arches, gg^j^g principles as ordinary arches and domes ; though, in groined arches, as a rule, the ribs do all the work, the spandrels between the ribs being filled with stone slabs or other lifht material ; in some European examples even flower-pots, plastered on the under side, have been used for this purpose. CHAPTER VL FLOOK BEAMS AND GIRDERS. n SLICE (h-l) SLICE 4* SLICE ^ Fig. I I 9. Moment of Inertia, writer has so often been asked for more in- formation as to the meaning of the term Moment of Inertia that a few more words on this subject may not be out of place. All matter, if once set in mo- tion, will continue in motion unless stopped by grav- ity, resistance of the atmosphere, friction or some other force; similarly, matter, if once at rest, will BO remain unless started into motion by some external force. Formerly it was believed, however, that all matter had a certain re- pugnance to being moved, which had to be first overcome, before a body could be moved. Probably in connection with some such theory the term arose. In reahty matter is perfectly indifferent whether it be in motion or in a state of rest, and this indifference is termed " Inertia." As used to-day, however, the term Moment of Inertia is simply a symbol or name for a certain part of the formula by which is calculated the force necessary to move a body around a certain axis with a given 184 SAFE BUILDING. velocity in a certain space of time ; or, what amounts to the same thing, the resistance necessary to stop a body so moving. In making the above calculation the " sum of the product of the weight of each particle of the body into the square of its distance from the axis " has to be taken into consideration, and is part of the formula ; and, as this sum will, of course, vary as the size of the body varies, or as the location or direction of the axis varies, it would be difficult to express it so as to cover every case, and there- fore it is called the " Moment of Inertia." Hence the general law or formula given covers every case, as it contains the Moment of Inertia, which varies, and has to be calculated for each case from the known size and weight of the body and the location and direction of the axis. In plane figures, which, of course, have no thickness or weight, the area of each particle is taken in place of its weight ; hence in all plane figures the Moment of Inertia is equal to the " sum of the prod- ucts of the area of each particle of the figure multiplied by the square of its distance from the axis." Calculation of Thus if we had a rectangular fio^ure (119') & inches IWIoment of In- .■■ i i • ■, -, ,. ertla. wide and d inches deep revolving around an axis M-N, we would divide it into many thin slices of equal height, say n slices each of a height = 2. X. The distance of the centre of gravity of the first slice from the axis M-N will, of course be = ^. 2. X. = 1. X The distance of the centre of gravity of the second sUce will be = 3. X, that of the third shce wiU be= 6. X, that of the fourth slice will be = 7. X, that of the last slice but one will be = (2 n — 3). X. and that of the last slice will be = (2 n — 1). X The area of each slice will, of course, be = 2. X. 6 ; therefore the Moment of Inertia of the whole section, around the axis M-N will be (see p. 8), t = 2. X. h. (1. X)2-|- 2. X. h. (3. X)2 4- 2. X. h. (5. X)2-|- 2. X. h. (7. X)2 + etc + 2. X. &. [(2n — 3).X]2 + 2. X. &.[(2n — l).X]2or, t = 2.X.8 S. [1 2 _}_ 32 52 _,_ etc _|_ (2n_3)a -f-(2n-l)2] now the larger n is, that is the thinner we make our slices, the nearer will the above approximate : Effect of load. 185 i==2. X«.6. [-I"-""] Moment of Resistance. = 8. X8 ««. 3 Therefore, as : 2. X. n = cf we have, by cubing, 8. X8, ra8 = rf«; inserting this in above, we have : . d\ b. b. t= — - — or 3 3 The same value as given for i in Table I, section No. 29. Of course it would be very tedious to calculate the Moment of Inertia in every case ; besides, unless the slices were assumed to be very thin, the result would be inaccurate ; the writer has therefore given in Table I, the exact Moments of Inertia of every section likely to arise in practice. The Moment of Inertia applies to the whole sec- tion, the " Moment of Resistance," however, applies only to each individual fibre, and varies for each ; it being equal to the Moment of Inertia of the whole section divided by the distance of the fibre from the axis. /Ck Now to show the connec- tion of the Moments of In- ertia and Resistance with transverse strains, let us consider the effect of a weight on a beam (sup- ported at both ends). If we consider the beam as cut in two and hinged at the point A (where the weight is applied), Fig. 1 20 ; further, if we consider a piece of rubber nailed to the bottom of each side of the beam, it is evident that the 1 H I B 1 Fig. 120. effect of the weight will as per Fig. 121. be. Effect of load on beam. Examin- ing this closer we find that the cor ners of the beams above A (or their fibres) wiH crush Fig. 121. each other, while those below A, are separated farther from each other, and the piece of rubber at B greatly stretched. It is evident, therefore, that the fibres nearest A experience the least change, and 186 SAFE BUILDING. that the amount of change of all the fibres is directly proportionate to their distance from A (as the length of all lines drawn parallel to the base of a triangle, are proportionate to their distance from the apex) ; further, that the fibres at A experience no change whatever. Now, if instead of considering the effect of a load on a hinged beam we D Fig. 122. took an unbroken beam, the effect would be similar, but, Instead of being concentrated at one point, it would be distributed along the entire beam; thus the beam A B D C (Fig. 122) which is not loaded, becomes when loaded, the slightly curved beam (A B D C) Fig. 123. A /Oi evident B that the fibres along the upper edge are com- pressed or A B is shorter than before; on the other hand the Fig- 123. fibres along CD are elongated, or in tension, and C D is longer than before ; if we now take any other layer of fibres as E F, they — being below the neutral (and central)^ axis X-Y — are evidently elongated; but not so much so, as C D : and a little thought will clearly show that their elongation is proportioned to the elongation of the fibres C D, dir- ectly as their respective distances from the neutral axis X-Y. It is further evident that the neutral axis X-Y is the same length as before, or its fibres are not strained; it is, therefore, at this point that the strain changes from one of tension to one of compres- sion. In Fif. 124 we have an isometrical view of a loaded beam. 1 As a rule the neutral axis can be safely assumed to be central, but it is not necessarily so. In materials, such as cast-iron, stone, etc., where the resistance of the fibres to compression and tension varies greatly, the axis will be far from the centre, nearer the weaker fibres. ROTATION AROUND NEUTRAL AXIS. 187 Fig. I 24. Rotation around consider an infinitesimally thin (cross) neutral axis, section of fibres A B C D in reference to their own neutral axis M-N. It is evident that if we wore to double the load on the beam, so as to bend it still more, that the fibres along A B would be com- pressed towards oi would move towards the centre of the beam ; the fibres a- long D C on the contrary would be elongated or would move away from the centre of the beam. The fibres along M-N", being neither stretched nor compressed, would remain stationary. The fibres between M-N and A B would all move towards the cen- tre of the beam, the amount of motion being proportionate to their distance from M-N ; the fibres between M-N and D C on the contrary would move away from the centre of the beam the amount of motion being proportionate to their distance from M-N ; a little thought, therefore, shows clearly that the section A B C D turns or. rotates on its neutral axis M-N, whenever additional weight is imposed on the beam. This is why we consider in the calculations the moment of Inertia or the moment of resistance of a cross-section as rotating on its neutral axis. Now let us take the additional weight off the beam and it will spring back to its former shape, and, of course, the fibres of the in- finitesimally thin section A B C D will resume their normal shape ; that is, those that were compressed will stretch themselves again, while those that were stretched will compress themselves back to their former shape and position, and those along the neutral axis will remain constant ; or, in other words, this thin layer of fibres A B C D can be con- sidered as a double wedge - shaped figure A B A. B. M N D C 188 SAFE BUILDING. D, C, (Fig. 125) the base of the wedges becoming larger or smaller as the weight on the beam is varied. Resistance of Now to proceed to the calculation of the resistance Wedge, of this wedge. It is evident that whatever may be the external strain on the beam at the section A B C D, the beam will owe whatever resistance it has at that point to the resistances of the fibres of the section or wedge to compression and tension. Now considering the right-hand side of the beam as rigid, and the section A B C D as the point of fulcrum of the external forces, we have only one external force p, tending to turn the left-hand side of the beam upwards around the section A B C D, its total tendency, effect or moment m at A B C D, we know is m=p. x (law of the lever). Now to resist this we have the opposition of the fibres in the wedge A B A, B, M N to compression and the opposition to tension of the fibres in the wedge D C D, C, M N. For the sake of conven- ience, we will still consider these wedges, as wedges but so infinites- imally thin that we can safely put down the amount of their con- tents as equal to the area of their sides, so that — if A B = i (the width of beam) and A D = (the depth of beam) — we can safely lall each wedge as equal to 6. ^- . Now as the centre of gravity of a wedge is at ^ of the height from its base, or § of the height from its apex (and as the height of each wed"-e is=:— ^ it would be = — . — = ~ from axis M-N. The ° 2 / 3 2 3 moment of a wedge at any axis M-N is equal to the contents of the wedge multiplied by the distance of its centre of gravity from the axis, the whole multiplied by the stress of the fibres, (that is their resistance to tension or compression). Now the contents of each wedge being = h. ~, the distance of centre of gravity from M-N =y ' and the stress being say = s, we have for the resistance of each wedge 2 3 b. rf2 Now if the stress on the fibres along the extreme upper or lower edges = yfc (or the modulus of rupture), it is evident that the average stress on the fibres in either wedse will = , or s = (for the FORMATION OF WOOD. 189 Stress on each fibre being directly proportionate to the distance from the neutral axis the stress on the average will be equal to half that on the base). S'ow inserting for s in the above formula, and multiplying also by 2, (as there are two wedges resisting), we have the total resistance to rupture or bending of the section A B C D (A. B. C. D.) g-. 10 Kow, by reference to Table I, section No. 2, we find that — g- — Moment of Resistance for the section A B C D ; therefore, we have proved the rule, that when the beam is at the point of rupture at any point of its length the bending moment at that point is equal to the moment of resistance of its cross-section at said point multipUed by the modulus of rupture. Where girders or beams are of wood, it becomes of the highest importance that they should be sound and perfectly dry. The for- mer that they may have sufficient strength, the latter that they may resist decay for the longest period possible. Formation of Every architect, therefore, should study thor- Wood. oughly the different kinds of timber in use in his locality, so as to be able to distinguish their different qualities. The strength of wood depends, as we know, on the resistance of its fibres to separation. It stands to reason that the young or newly formed parts of a tree will offer less resistance than the older or more thor- oughly set parts. The formation of wood in trees is in circular lay- ers, around the entire tree, just inside of the bark. As a rule one layer of wood is formed every year, and these layers are known, there- fore, as the " annular rings," which can be distinctly seen when the trunk is sawed across. These rings are formed by the (returning) sap, which, in the spring, flows upwards between the bark and wood, supplies the leaves, and returning in the fall is arrested in its altered state, between the bark and last annular ring of wood. Here it hard- ens, forming the new annular ring. As subsequent rings form around it, their tendency in hardening is to shrink or compress and harden still more the inner rings, which hardening (by compression) is also assisted by the shrinkage of the bark. In a sound tree, there- fore, the strongest wood is at the heart or centre of growth. The 190 SAFE BUILDING. heart, however, is rarely at the exact centre of the trunk, as the sap flows more freely on the side exposed to the effects of the sun and wind ; and, of course, the rings on this side are thicker, thus leaving the heart constantly, relatively, nearer to the unexposed side. Heart-Wood. From the above it will be readily seen that timber should be selected from the region of the heart, or it should be what is known as " heart-wood." The outer layers should be rejected, as they are not only softer and weaker, but, being full of sap, are liable to rapid decay. To tell whether or no the timber is " heart-wood " one need but look at the end, and see whether it contains the centre of the rings. No bark should be allowed on timber, for not only has it no strength itself, but the more recent annular rings near it, are about as valueless. Medullary Rays. In some timbers, notably oak, distinct rays are noticed, crossing the annular rings and radiating from the centre. These are the "medullary rays," and are elements of weakness. Care should be taken that they do not cross the end of the timber horizontally, as shown at A in Fig. 126, but as near vertically as possible, see B in Fig. 127. The beautiful appearance of quartered oak and other woods is obtained by cutting the planks so that their surfaces will show slanting cuts through these medullary rays. Seasoning -^^^ timber cracks more or less in seasoning, nor cracks, ^ggj these cracks cause much worry, unless they are very deep and long. They are, to a certain extent, signs of the amount of seasoning the timber has had. They should be avoided, as much as possible, near the centre of the timber, if regularly loaded, or near the point of greatest bending moment, where the c c Fig. 12 6. Fig. 127. Fig. 1 28. Fig. 1 29. Fig. I 30. Fig. 131. loads are irregular. If timber without serious cracks cannot be ol> tained, allowance should be made for these, by increasing its size. Vertical, or nearly vertical cracks (as C, Fig. 128) are not objec- tionable, and do not weaken the timber. But horizontal cracks (as D, Fig. 129), are decidedly so, and should not be allowed. SOUND TIMBER. 191 •*"ot«' Knots in timber are another element of weakness. They are the hearts, where branches grow out of the trunk. If they are of nearly the same color as the wood, and their rings gradually die out into it, they need not be seriously feared. If, however, they are very dark or black, they are sure to shrink and fall out in time, leaving, of course, a hole and weakness at that place. Dead knots, — that is, loose knots, — in a piece of timber, mean, as a rule, that the heart is decaying. Knots should be avoided at the centre of a beam, regularly loaded, and at the point of greatest bending moment, where the loads are irregular. The farther the knots (and cracks) are from these points the better. Wind-shakes. Timber with " wind-shakes " should be entirely avoided, as it has no strength. These are caused by the wind shak- ing tall trees, loosening the rings from each other, so that when the timber is sawed, the wood is full of small, ahnost separate pieces or splinters at these points. A timber with wind-shakes should be condemned as unsound. A timber with the rings at the end showing nearly vertical (E Fig. 130) will be much stronger than one showing them nearly hori- zontal. (F Fig. 131.) Signs of sound ^^^^ sound timber. Lord Bacon recommended Timber, to speak through it to a friend from end to end. If the voice is distinctly heard at the other end it is sound. If the voice comes abruptly or indistinctly it is knotty, imperfect at the heart, or decayed. More recent authorities recommend listening to the ticking of a watch at the other end, or the scratching of a pin on its surface. If, in sawing across a piece it makes a clean cut, it is neither too green nor decayed. The same if the section looks bright and smells sweet. If the section is soft or splinters up badly it is decayed. If it wets the saw it is full of sap and green. If a blow on timber rings out clearly it is sound ; if it sounds soft, subdued, or dull, it is very green or else decayed. The color at freshly-sawed spots should be uniform throughout ; timbers of darker cross-section are generally stronger than those of lighter color (of the same kind of wood.) The annular rings should be perfectly regular. The closer they are, the stronger the wood. Their direction should be parallel to the axis throughout the length of the timber, or it will surely twist in time, and is, besides, much weaker. Where the rings at both ends are not in the same direction the timber has either twisted in 192 SAFE BUILDING. growing, or has a "wandering heart," — that is, a crooked one. Such timber should be condemned. Besides looking at the rings at the end, a longitudinal cut near the heart will show whether it has grown regularly and straight, or whether it has twisted or wandered. The weight of timber is important in judging its quality. If spec- imens of a wood are much heavier than the well-known weight of that wood, when seasoned, they may be condemned as green and full of sap. If they are much lighter than thoroughly seasoned speci- mens of the same wood, they are very probably decayed. Methods of Tredgold claims that timber is " seasoned " when Seasoning. jggj. one-fifth of its original weight (when green) ; and " dry " when it has lost one-third. Some timbers, how- ever, lose nearly one-half of their original weight in drying. Many methods are used to season or dry timber quickly. The best method, however, is to stack the timber on dry ground (in as dry an atmosphere as possible) and in such a position that the air can circulate, as freely as possible around each piece. Sheds are built over the timber to protect it from the sun, rain, and also from severe winds as far as possible. Timber dried slowly, in this manner, is the best. It will crack somewhat, but not so much so as hastily dried timber. Many proc- esses are used to keep it from cracking, the most eflEective being to bore the timber from end to end, at the centre, where the loss of material does not weaken it much, while the hole greatly relieves the strain from shrinkage. Some authorities claim that two j'ears' ex- posure is sufficient, though formerly timber was kept very much longer. But even two years is rarely granted with our modern con- ditions, and most of the seasoning is done after the timber is in the building. Hence its frequent decay. There are many artificial methods for drying timber, but they arc expensive. The best known is to place it in a kiln and force a rapid current of heated air past it, this is known as " kiln-drying." It is very apt to badly " check " or crack the wood. To preserve timber, besides charring, the " creo- soting " process is mo'st effective. The timber is placed in an iron chamber, from which the air is exhausted ; after which creosote is forced in under a high pressure, filling, of course, all the pores which have been forced open by the suction of the departing air. Creo- soted wood, however, cannot be used in dwellings, as the least appli- cation of warmed air to it, causes a strong odor, and would render the building untenantable. DECAY OP TIMBER. 193 Effect of Timber properly dried can be raised to four times Drying. (400% of) the strength of its green state (notably spruce, see U. S. Dept. of Agriculture Circular 108, Aug. 26th, 1807). But as it absorbs more or less moisture from the air, from 8% to 16% of its dry weight, say 12% average, the actual gain in strength by drying is only 240%. Again wood, in dry- ing, checks and is thus weakened, so that no greater strength should be allowed for than it shows in its green state. Timber is so uncertain, and affected in so many ways by local conditions, that a large factor of safety should always be used. Now-a-days wood is fire-proofed, that is made proofing, unburnable. There are many different processes; some — treatment by electricity; others — chemical baths; again — application of fire-proof chemicals by pressure, filling the pores; or direct application or washes of chemicals. Woods so treated are very much heavier, and are very much hardened, greatly increasing the expense, not only on account of the cost of treatment, but on account of the greatly increased labor necessary to work the wood, and the destruction of edges of and of tools on account of the hardened character of the wood. . In shrinking, the distance between rings remains Manner of Shrinkage, constant, and it is for this reason that the finest floors are made from quartered stuff ; for (besides their greater beauty), the rings being all on end, no horizontal shrinkage wiU take place ; the width of boards remaining con- stant, and the shrinkage being only in their thickness ; neither will timber shrink on end or in its length. Figures 132 and 133 show how timber will shrink. The first from a quartered log, the other from one with parallel cuts. The dotted part shows the shrinkage. The side-pieces G in Fig. 133 will curl, as shown, besides shrinking. By observing the directions of the annular rings, therefore, the future behavior Fig. 132. Fig. 133. 194 SAFE BUILDING. of the timber can be readily predicted. Of course^ the figures are greatly exaggerated to show the effect more clearly. Decay of heart is not straight its entire length, the Timber, piece will twist lengthwise. Shrinkage is a serious danger, but the chief danger in the use of timber lies in its decay. All timber will decay in time, but if it is properly dried, before being built in, and all sap-wood discarded, and then so placed that no moisture can get to the timber, while fresh air has access to all parts of it, it will last for a very long time; some woods even for many centuries. In proportion as we neglect the above rules, will its life be short-lived. There are two kinds of decay, wet and dry rot. The wet rot is caused by alternating exposures to dampness and dryness; or by exposure to moisture and heat; the dry rot, by confining the timber in an air-tight place. In wet rot there is "an excess of evaporation ; " in dry rot there is an ' ' imperfect evaporation. ' ' Beams with ends built solidly into walls are apt to rot; also beams surrounded solidly with fire- proof materials; beams in damp, close, and imperfectly ventilated cellars; sleepers bedded solidly in damp mortar or concrete, and covered with impervious papers or other materials; also timbers exposed only at intervals to water or dampness, or timbers m "solid" timbered floors. Dry rot is like a contagious disease, and will gradually not only eat up the entire timber, but will attack all adjoining sound woodwork. Where rotted woodwork is removed, all adjoining woodwork, masonry, etc., should be thoroughly scraped and washed with strong acids. Ventilation Where wood has, of necessity, to be surrounded necessary, with fireproof materials, a system of pipes or other arrangements, should be made to force air to same through holes, either in the floors or ceilings, but in no case connecting two floors; the holes can then be made small enough not to allow the passage of fire. Where the air is forced in under pressure it would be advisable at times to force in disinfectants, such as steam con- taining evaporated carbolic acid, fumes of sulphur, etc. Coating woodwork with paint or other preparations will only rot the wood, unless it has been first thoroughly dried and every particle of sap removed. When wood is painted, it should be on CROSS-BRIDGING. 195 all surfaces, or the unpainted surface will absorb moisture, ulti- mately causing rot. Cross-bridging. Timber must not be used too thin, or it will be apt to twist. For this reason floor-beams should not be used thinner than three inches. To avoid twisting and curling, cross- bridging is resorted to. That is, strips usually 2" x 3" are cut between the beams, from the bottom of one to the top of the next one, the ends being cut (in a mitre-box), so as to fit ac- curately against the sides of beams, and each end nailed with at least two strong nails. The strips are always placed in double courses, across the beams, the courses crossing each other like the letter x between each pair of beams. This is known as "herring-bone" cross-bridging. Care should be taken that all the parallel pieces in each course are in the same line or plane. The lines of cross-bridging can be placed as fre- quently as desired, for the more there are, the stiffer will be the floor. About six feet between the lines is a good average. Sometimes solid blocks are used between the beams, in place of the herring-bone bridging. Cross-bridging is also of great help to a floor by relieving an individual beam from any great weight accidentally placed on it (such as one foot of a safe, or one end of a book-case), and distributing the weight to the adjoining beams. Unequal settlements of the individual beams are thus avoided. "Where a floor shows signs of weakness, or lacks stiff- Stiffening ness, or where it is desirable to force old beams, weak floors. that cannot be well removed, to do more work, two lines of slightly wedge-shaped blocks are driven tightly between the beams, in place of the cross-bridging. The beams are then bored, and an iron rod is run between the lines of wedges, from the outer beam at one end to the outer beam at the other, and, of course, at right angles to all. At one end the rod has a thread and nut, and by screwing up the latter the beams are all forced up- wards, ' ' cambered, ' ' and the entire floor arched. It will be found much stronger and stiffer; but, of course, will need leveling for both floor and ceiling. Under the head and nut at ends of rod, there must be ample washers, or the sides of end beams will be crushed in, and the effect of the rod destroyed. 196 SAFE BOTLDING. Girders, which cannot be stiffened sideways, should be, at least, half as thick as they are deep, to avoid lateral flexure. Framing of using wooden beams and girders, much fram- Beams. ing has to be resorted to. The used joints be- tween timbers are numerous, but only a very few need special mention here. Beams should not rest on girders, if it can be avoided, on account of the additional dropping caused by the sum of the shrinkage of both, where one is over the other. If framing is too expensive, bolt a wide piece to the under side of the girder, sufficiently wider than the girder to allow the beams to rest on it' each side. If this is not practicable bolt pieces onto each side of the girder, at the bottom, and notch out the beams to rest against and over these pieces. The bearing of a beam should always be as near its bottom as possible. If a beam is notched so as to bear near its centre, it will split longitudinally. Where a notch of more than one-third the height of beam, from the bottom, is necessary, a wrought-iron strap or belt should be secured around the end of beam, to keep it from splitting lengthwise. ^ If framing can be used, the best method is the "tusk and tenon" joint, as shown in Figs. 134 and 135. In the one case the tenon goes through the girder and is secured by a wooden wedge on the other side; in the ■ -' other it goes in only about a length equal to twice its depth, and is spiked from the top of girder. The Fig. 134. latter is the most used. By both methods the girder is weakened but very little, the principal cut being near its neutral axis, while the beam gets bearing near its bottom, and its tenon is thoroughly strengthened to prevent its shearing off. The dimensions given in the figures are all in parts of the height of beams. Headers and trimmers at fire-places and other openings are frequently framed together, though it would be more advisable to use "stir- rup-irons. ' ' The short tail-beams, however, can be safely tenoned into the header. 1 Where beams come against the sides of girders, the top of beams should be from one to four inches above the top of girder as these beams shrink much quicker and more than thick girders, and thus the long humps or ridges in floors over girders will be avoided. 8TIKRUP-IRONS. 197 In calculating the strength o£ framed timber, the point where the mortise, etc., are cut, should be carefully calculated by itst'lf, as the cutting frequently renders it dangerously weak, at this point, if not allowed for. For the same reason plumbers should not be allowed to cut timbers. As a rule, however, cuts near the wall are not dan- gerous, as the beam being of uniform size throushout, there is usu- ally an excess of strength near the wall. Stlrrup-lrons< Stirrup-irons are made of wrought-iron ; they are secured to one timber in order to provide a resting-place for another timber, usually at right angles to and carried by the former. They should always lap over the farther side of the carrying timber, to prevent slip- ping, as shown in Fig. 136. The iron should be sufficiently wide not to crush the beam, where resting on it ; the section of iron must be sufficient not to shear off each side of beams. The twist must not be too sudden, or it will straighten out and let the carried timber down. To put the above in formulsB we should have for the width of stirrup-iron (x) s Width of ^= T / c\ ''69) Stirrup-Irons. ( "7 /"* / Where x = the width of stirrup-iron, in inches. Where s = the shearing strain, in lbs., on end of beam, being car- ried. Where & = the width of beam being carried, in inches. Where ^ = the safe resistance, in pounds, to compression, across the fibres, of the beam, being carried. For the thickness of stirrup-iron we should have : s 2. ..(5) Which for wrought-iron (Table IV.) becomes, (7D; Thickness of ^ (71) StIrrup-iron. ^ 16000. x Where y=-the thickness of stirrup-iron, in inches. Where s = the shearing strain on end of beam, in lbs. VVhere a: = is found by formula (69). 198 SAFE BUILDING. Providing, however, that y should never be less than one-<]uarter inch thick. Example. A girder carries the end of a beam, on tvkich there is a uniform load of two thousand pounds. The beam is four inches thick, and oj Georgia pine. What size must the stirrup-iron be ? Lxample stir- '^^^^ shearing strain at each end of the beam will, rup-lrors. of course, be one thousand pounds, which will be the load on stirrup-irons. (See Table VI T). From Table IV we find for Georgia pine, across the fibres, ^-^^ = 200, we have, therefore, for the width of stirrup-iron from Formula (G9) ^ — 4.200 ~ * Therefore the thickness of iron from Formula (71) sliould be — 1000 _ 1 ff IGOOO.l^ "~ 20 we must make the iron liowever at least i" thick and therefore use a section of I5 X In calculating ordinary floor-beams the shearing strain can be overlooked, as a rule ; for, in calculating transverse strength we allow only the safe stress on the fibres of the upper and lower edges, while the intermediate fibres are less and less strained, those at the neutral axis not at all. The reserve strength of these only partially used fibres will generally be found (^uite ample to take up the shear- ing strain. Rectangular The formula for transverse strength are (luite beams, complicated, but for rectangular sections (wooden beams) they can be very much simplified provided we are calcu- lating for strength only and not taking deflection into account. Remembering that the moment of resistance of a rectangular sec- tion is (Table I) = -j- and inserting into Formula (18) the value for m according to the manner of loading and taken from (Tabk VII), we should have : For uniform load on beam. {j) (72) For centre load on beam. Transverse • ^72 / i. strength of „ . ( JL rectangu- 9.L V / lar beams. KECTANGULAK 15EAM8. 199 For load at any point ofheam. _ b.d^.h / k \ For unifoi m load on cantilever. Ftjr load concentrated at end of cantilever. b.d^. / k\ "' = T2X-(7) For load at any point of cantilever. W here u — safe uniform load, in pounds. Where w= safe centre load on beam, in pounds; or safe load at end of cantilever, in pounds. Where safe concentrated load, in pounds, at any point. Where Y= length, in feet, from wall to concenti'ated load (in can- tilever). Where M and iV = the respective lengths, in feet, from concen- trated load on beam to eacli support. Where L= the length, in feet, of span of beam, or length of canti- lever. Where h = the breadth of beam, in inches. Where d = the depth of beam, in inches. Where^-^ ^= the safe modulus of rupture, per scjuare inch, of the material of beam or cantilever (see Table IV). The above formulae are for rectangular wooden beams supported against lateral flexure (or yielding sideways). Where beams or gir- ders are not supported sideways the thickness should be equal to at least half of the depth. No allowance above formulce make no allowance for device- for deflection, tion, and except in cases, such as factories, etc., where stretigth only need be considered and not the danger of crack- ing plastering, or getting floors too uneven for machinery, are really of but little value. They are so easily understood that the simplest example will answer : Example. Take a 3" X 10" hemlock timber and 9 feet long (clear span), loaded in different ways, what will it safely carry ? taking no account of deflection. (74) (76) (76) (77) 200 SAFE BUILDING. The safe modulus of rupture ^A^for hemlock from Table IV is = 750 pounds. If both ends are supported and the load is uniformly distributed the beam will safely carry, (Formula 72) : Q 1 A2 uz=^^. 750 = 2778 pounds. If both ends are supported and the load concentrated at the centre, the beam will safely carry, (Formula 73) : m; = ?^. 750 = 1389 pounds. Io.9 If both ends are supported and the load is concentrated at a point J, distant four feet from one support (and five feet from the other) the beam will safely carry, (Formula 74) : w, = ^'^^ . 750 = 1406 pounds. 72.4.5 ^ If one end of the timber is built in and the other end free and the load uniformly distributed, the cantilever will safely carry, (Formula 75): ^_o^ . 750 = G94 pounds. 3G.9 ^ If one end is built in and the other end free, and the load concen- trated at the free end, the cantilever will safely carry, (Formula 76) : w = Y^' 750 = 347 pounds. If one end is built in and the other end free, and the load concen- trated at a point I, which is 5 feet from the built-in end, the canti- lever will safely carry, (Formula 77) : = 750 = 625 pounds. Where, however, the span of the beam, in feet, greatly exceeds the depth in inches (see Table VIII), and regard must be had to de- flection, the formulae (28) and (29), also (37) to (42) should always be used, inserting for i its value from Table 1, sectiorf No. 2, or : — . h.d« '=-12 Where b = the thickness of timber, in inches. Where d = the depth of timber, in inches. Where i = the moment of inertia of the cross-section, in inches. Table IX, however, gives a much easier method of calculating wooden beams, allowing for both rupture and deflection, and For- mula (72) to (77) have only been given here, as they are often erro- USE OF TABLES. 201 neously given in text-books, as the only calculations necessary for beams. Basis of Tables ^^^^^ further simplify to the architect the labor XII and Xllli of calculating wooden beams or girders, the writer has constructed Tables XII and XIII. Table XII is calculated for floor beams of dwellings, offices' churches, etc., at 90 pounds per square foot, including weight of construction. -The beams are supposed to be cross-bridged. Table XIII is for isolated girders, or lintels, uniformly loaded, and supported sideways. When not supported sideways decrease the load, or else use timber at least half as thick as it is deep. In no case will beams or girders (with the loads given) deflect sufficiently to crack plastering. For convenience Table XII has been divided into two parts, the first part giving beams of from 5' 0" to 15' 0" span, the second part of from 15' 0" to 29' 0" span. How to use '^^■'^ ^^'^ table is very simple and enables us Table Xil. to select the most economical beam in each case. For instance, we have say a span of 21' G". We use the second part of Table XII. The vertical dotted line between 21' 0" and 22' 0" is, of course, our line for 21' 6". We pass our finger down this line till we strike the curve. To the left opposite the point at which we struck the curve, we read : 21.6 spruce, W. P. 56—4-14-14 or: at 21' 6' span we can use spruce or white-pine floor beams, of 56 inches sectional area each, viz. : 4" thick, 14" deep and 14'' from cen- tres. Of course we can use any other beam below this point, as they are all stronger and stiffer, but we must not use any other beam above this point. Now then, is a 4" X 14" beam of spruce or Avhite pine, and 14" from centres the most economical beam. "We" pass to the columns at the right of the curve and there read in the first column 48". This means that while the sectional area of the beam is 56 square inches, it is equal to only 48 square inches per square foot of floor, as the beams are more than one foot from centres. In this column the areas are all reduced to the " area per square foot of floor," so that we can sec at a glance if there is any cheaper beam beloio our point. We find below it, in fact, many cheaper beams, the smallest area (per square foot of floor) being, of course, the most economical. The smallest area we find is 36, 0 or 86 square inches of section per square foot of floor (this we find three times, in the sixteenth, twenty-ninth and thirty-first lines from the bottom). Pass- ing to the left we find they represent, respectively, a Georgia pine 202 SAFE BUILDING. beam, 3" thick, 16" deep and 16" from centres; or a Georgia pine beam 3" thick, 14" deep and 14" from centres; or a white oak beam 3" thick, 16" deep and 16" from centres. If therefore, we do not consider depth, or distance from centres, it would simply be a ques- tion, which is cheaper, 48 inches (or four feet " board-measure ") of white pine or spruce, or 36 inches (or three feet " board-measure ") of either white oak or Georgia pine. The four other columns on the right-hand side, are for the same purpose, only the figures for each kind of wood are in a column by themselves ; so that, if we are limited to any kind of wood we can examine the figures for that wood by themselves. Take our last case and suppose we are limited to the use of hemlock ; now from the point where our verti- cal line (21' 6") first struck the curve, we pass to the right-hand side of Table, to the second column, which is headed " Hemlock." From this point we seek the smallest figure below this level, but in the same column; we find, that the first figure we strike, viz: 41, 2 is the smallest, so we use this ; passing along its level to the left we find it represents a hemlock beam of 48 square inches cross-section, or 3'' thick, 16" deep and 14" from centres. In case the size of the beam is known, its safe span can, of course, be found by reversing the above procedure, or if the depth of beam and span is settl:d, we can find the necessary thickness and distance between centres; in this way the Table, of course, covers every problem. Table XIII is calculated for wooden girders of all sizes. Any thickness not given in the table can be obtained by taking the line for a girder of same depth, but one inch thick and multiplying by the thickness. For very short spans, look out for danger of horizontal shearing (see formula 13) ; where this danger exists, pass vertical bolts through ends of girder, or bolt thin iron plates, or straps, or even boards wi Ji vertical grain, to each side of girder, at ends. How to use The use of this table is very simple. The vertical Table XIII. columns to the left give the safe uniform loads on girders (sufficiently stiff not to crack plastering) for different woods : these apply to the dotted parts of curves. The columns on the right- hand side give the same, but apply to the parts of curves drawn in full lines. If we have a 6" X 16" Georgia pine beam of 20 feet span and want to know what it will carry, we select the curve marked at its upper end 6 X 16 = 96; we follow this curve till it intersects the vertical line 20' 0" ; as this is in the part of curve drawn full, we TABLE XV. IRON I-BEAM GIRDERS, BRACED SIDEWAYS. 203 SPAN OF GIRDER IN FEET. 18. . - For Steel Beams, add one-quarter to safe uniform load on Iron Beams ; but leny;t"h J f span in feet must not exceed twice the depth of beam in inches, or deflection e too great for plastering. IRON BEAM TABLES. 205 pass horizontally to the right and find under the column marked * ' Georgia Pine, ' ' 7980, which is the safe, uniform load in pounds. Supposing, however, we had simply settled the span, say 8 feet, and load, say 7000 pounds, and wished to select the most economical girder, being, we will say, limited to the use of white pine: the span not being great we will expect to strike the dotted part of curve, and therefore select the fourth (white pine) column to the left. We pass down to the nearest figure to 7000 and then pass horizontally to the right till we meet the vertical 8 feet line; this we find is, as we expected, at the dotted part, and therefore our selection of the left column was right. We follow the curve to its upper end and find it requires a girder 4" X 12" =48 square inches. Now, can we use a cheaper girder? of course, all the lines under and to the right of our curve are stronger, so that if either has a smaller sectional area, we will use it. The next curve we find is a 6" X 10" = 60"; then comes a 4" X 14" = 56"; then an 8" X 10" = 80" ; then a 6" X 12" = 72" and so on ; as none has a smaller area we will stick to our 4" x 12" girder, provided it is braced or supported sideways. If not, to avoid twisting or lateral flexure, we must select the next cheapest section, where the thickness is at least equal to half the depth ;^ the cheapest section beyond our curve that corresponds to this, we find is the 6" X 10" girder, which we should use if not braced sideways. In the smaller sections of girders where the difference between the loads given from line to line is proportionally great, a safe load should be assumed between the two, according to the proximity to either line at which the curve cuts the vertical. The point where the curve cuts the bottom horizontal line of each part is the length of span for which the safe load opposite the line is calculated. Heavier Floors. Where a different load than 90 pounds per square foot, must be provided for, we can either increase the thick- ness of beams as found in Table XII, or decrease their distance from centres, either in proportion to the additional amount of load. Or, if we wish to be more economical, we can calculate the safe uniform load on each floor beam, and consider it as a separate girder, supported sideways, using of course. Table XIII. Basis of Tables 1"^^ Tables XIV and XV are very similar to the XIV and XV. foregoing, but calculated for wrought-iron I-beams.^ Table XIV gives the size of beams and distance from centres re- ^ The rule for calculating the exact thickness will be found later, Formula (78). ^ To save the great additional cost of this book to the student, if entirely- new tables were used for the present sizes, weights and shapes of rolled sections, the tables from "Safe Building" have been retained as they 206 SAFE BUILDING. quired to carry different loads per square foot of floor, 150 pounds per square foot of floor (including the weight of construction), however, being the usual load allowed for in churches, office-build- ings, public halls, etc., where the space between beams is filled with arched brickwork, or straight hollow-brick arches, and then covered over with concrete. A careful estimate, however, should be made of the exact weight of construction per square foot, including the ironwork, and to this should be added 70 pounds per square foot, which is the greatest load likely ever to be produced if packed solidly with people. Furniture rarely weighs as much, though heavy safes should be provided for separately. The load on roofs should be 30 pounds additional to the weight of construction, to provide for the weight of snow or wind. Look out for tanks, etc., on roofs. Plastered ceilings hanging from roofs add about 10 pounds per square foot, and slate about the same. Where a different load than given in the Table must be provided for, the distance between centres of beams can be reduced, proportionally from the next greater load; or the weight on each beam can be figured and the beam treated as a girder, supported sideways, in that case using Table XV. Both tables are calculated for the beams not to deflect sufficiently to crack plastering. How to use The use of Table XIV is very simple. Suppos- TableXIV. ing we have a span of 24 feet and a load of 150 pounds per square foot. "We pass down the vertical line 24' 0" and strike first the 12" — 96 pounds beam, which (for 150 pounds) is opposite (and three-quarters way between) 3' 0" and 3' 4" therefore 3' 3" from centres. The next beam is the 12" — 120 pounds beam 4' 0" from centres; then the 12" — 125 pounds beam 4' 1" from centres; then the 15" — 125 pounds beam 5' 0" from centres and so on. It is simply a question, therefore, which "distance from centres" is most desirable and as a rule in fire- proof buildings it is desira,ble to keep these as near alike as possible, so as not to have too many different spans of beam arches and centres. If economy is the only question, we divide the weight of beam by its distance from centres, and the curve giving the smallest result is, of course, the cheapest. Supposing, however, that we desire all distances from centres alike, say 5 feet. In that case we pass down the 150-pound column to and then along the horizontal line 5' 0" till we strike the vertical (span) line, in this case 24' 0", and then take the cheapest beam to the sufficiently teach the student the theory and method of using all sections. But little, if any, structural iron is rolled now-a-days, and the shapes of steel are varied so often, it would anyhow be imposcsible to cover all of them permanently in a book of this general character. Fortunately the steel companies all issue elaborate hand-books from time to time, and these should always be used in actual practice. USE OF STEEL. 207 right of the point of intersection. Thus, in our case the nearest beam would be 15"— 125 pounds; next comes 1214"— 170 pounds; then 15"— 150 pounds, etc. As the nearest beam is the lightest in this case, we should select it. The weight of a beam is always given per yard of length. The reason for this is that a square inch of wrought-iron, one yard long, weighs exactly 10 pounds. Therefore if we know the weight per yard in pounds we divide it by ten to obtain the exact area of cross-section in square inches; or if we know the area, we multiply by ten and obtain the exact weight per yard. How to use The use of Table XV, is very similar to that of Table XV. Table XIII, but that the safe uniform load is given (in the first column) in tons of 2000 pounds each. The continuation of the two 20" beams up to 42 feet span is given in the separate table, in the lower right-hand corner. To illustrate the Table: if we have a span of say 21 feet we pass down its verti- cal line; the first curve we strike is the lOy," — 90 pounds beam, which is three-quarters space beyond the horizontal line 5 (tons) ; therefore a lOy^"— 90 pounds beam at 21 feet span will carry safely 53^4 tons uniform load, and will not deflect sufficiently to crack plaster. (Each full horizontal space represents one ton). The next beam at 21 feet span is lOyg" — 105 pounds, which will safely carry Gy^ tons. Then comes the 12" — 96 pounds beam, which will safely carry 7 tons, and so on down to the 20" — 272 pounds beam, which will safely carry 33% tons. If we know the span (say 17 feet) and uniform load (say 7y2 tons) to be carried, we pass down the span line 17' 0" and then horizontally along the load line 7y2 till they meet, which in our case is at the 9" — 125 pounds beam; we can use this beam or any cheaper beam, whose curve is under it. We pass over the different curves under it, and find the cheapest to be the 12" — 96 pounds beam, which we, of course, use. Iron beams must be scraped clean of rust and be well painted. They should not be exposed to dampness, nor to salt air, or they will deteriorate and lose strength rapidly. The best method is to have all rolled iron or steel cleaned and inspected at the mill, then coated with a heavy coat of linseed oil before shipment, then a coat of paint or preservative on arrival, and a final coat after set- ting. Before each of the three coatings remove all scales, rust, sand and dirt. Steel beams. Steel beams are used almost exclusively to-day, as they are cheaper and stronger, though not so reliable as wrought- iron. They are cheaper to manufacture than iron beams, as they are made directly from the pig and practically in one process; while with iron beams the ore is first converted into cast-iron, then puddled into the muck-bar, re-heated, and then rolled. Steel beams, however, are not apt to be of uniform quality. Some may 208 SAFE BUILDING. be even very brittle; they are, however, very much stronger than iron (fully 25 per cent, stronger), but as their deflection is only about 7, 3 per cent, less than that of iron beams, there is not so much relative economy of material possible in their use. If steel beams are used they can be spaced one quarter distance (between centres) farther apart than given in Table XIV for iron beams; or they will safely carry one-quarter more load than given in Table XV; but in no case, where full load is allowed, must the span in feet, (of steel beams), exceed twice the depth in inches. With full safe loads the deflection of steel beams will always be greater than that of iron beams (about Vs larger). Where, therefore, it is desirable not to have a greater deflection than with iron beams, add only Ty, per cent, to the distances between centres or ' ' safe loads" as given in Tables for iron beams, instead of 25 per cent. The strength and consistency of steel beams is variable. It has been found in some cases that steel beams broke suddenly when jarred, (that is, were very brittle,) though test pieces off the ends of these same beams gave very satisfactory results. If steel is used, not only should samples of each rolling be carefully tested, for tenacity, ductility, elasticity, elongation, etc., but now and then the whole beam itself should be tested by actual loading. It will be readily seen that the expense of such tests would bar the use of steel, but no architect can afford to take any chances in such an important part of his building. Many writers even claim, that, "within the elastic limit," the additional stiffness of steel over iron does not appear; and that it is only beyond this limit that steel is somewhat stiffer than iron. Nevertheless, better or not, steel is here, and here to stay, while wrought-iron is practically unobtainable. Lateral Flexure In using iron and steel beams it is very impor- in Beams, tant that they be supported sideways, so as not to yield to lateral flexure. Where the beams are isolated and un- supported sideways, the safe load must be diminished. Just how much to diminish this load is the question. The practice amongst iron workers is to consider the top flange as a column of the full length of the span, obliged to yield sideways, and with a load equal to the greatest strain on the flange. Modifying, therefore. Formula (3) to meet this view, we should have: Beams not w (78) braced T~y7l? sidewaysi 1 -f- J2 Where ivi = the safe load, in pounds, on a beam, girder, lintel or straight arch, etc., unsupported sideways. Where w — the safe load, in pounds, on a beam, lintel or straight arch supported sideways. LATERAL FLEXURE. 209 Where L = the length of clear span, in feet, that beam, etc., is unsupported sideways. Where b = the least breadth in inches of top flange, or least thickness of beam, lintel or arch. Where y = a constant, as found in Table XVI. (In place of w we can use r = the moment of resistance of leam supported sideways, and in place of w, we use r, = the moment of resistance of beam not supported sideways.) The above practice, however, would seem to diminish the weight unnecessarily, particularly where the beam, girder, etc., is of uniform section throughout ; for while the beam in that case, would, be equally strong at all points, it would be strained to the maximum compression only at the point of greatest bending-moraent, the strain diminishing towards each support, where the compression would TABLE XVI. VALUE OF Y IN FORMULA (78). Material of beam, girder, lintel, straiglit arcli, etc. Value of y for girders, beams, etc., of varia- ble cross-sec- tions. Value of y for beame. girders, lintels, straight arches, etc., of uniform cross -sec- tion throughout. 0,5184 0,0432 0,0346 0,5702 3,4560 6,7024 0,J304 0,01'J2 0,0154 0,2534 1,5360 2,5344 cease entirely. To consider, therefore, the whole as a long column carrying a weight equal to this maximum compressive strain, seems unreasonable. Box has shown, however, that the maximum tendency to deflect laterally is when we consider the top flange (or top half in rectangular beams, lintels and straight arches) as a column ecpal to two thirds of the span (unsupported sideways) loaded with a weight equal to one-third of the greatest compressive strain at any point. This greatest compressive strain is always at the point of greatest bending moment (usually the centre of span), and is equal to the area of top flange, multiplied by ( j") ^- In case of plate girders the angle-irons and part of web between angle-irons should be included in the area. Box's theory is given in Formula (5) ; if then we take > This is not quite correct. The greatest compressive strain is really a little less, as will be explained in Vol. II. 210 SAFE BUILDING. one-third of this " maximum tendency to detiect " as safe, we should hare the same Formula as (78) but with a smaller value for y. The Use of writer would recommend using the larger value for Table XVI. where, as in plate girders, trusses, etc., the section of top flange or chord is diminished, varying according to the com- pressive strain at each point; and usiug the smaller value for y, where the section of beam, girder or top chord is uniform throughout. Thus the 10^"- 90 pounds beam at 20 feet span will safely carry (if supported sideways) a uniform load of 5,9 tons or 11800 pounds (see Table XV.) The width of flange being 4^", and this width and its thickness, of course, being uniform throughout the entire length of beam, we use the smaller value for y (second column) and have for the actual safe uniform load, if the beam is not secured against I?*^.eral flexure: 11800 11800 V) — z — '""1 4- 0,0192. 202 l-f 0,379 4^2 ' 11800 , , ^ ■ =85o7 pounds, or 4,28 tons. 1,379 Had we used the larger value for y — 0,0432 we should have had ,„ _ 11800 11800 „ocr. A o ,o . "^■- l4-0,854 "^r;854 ^'^'^ P°""^^' ^''^ which closely resembles the value (3,29) given in the Iron Com- panies' hand-books, but is an excessive reduction under the circum stances. Doubled Where two or more beams are used to carry the Beams, same load, as girders for instance, or as lintels in a wall, they should be firmly bolted together, with cas-t-iron separa- tors between. In this case use for b in Formula (78) the total width, from outside to outside of all flanges, and including in b the spaces between. The separators are made to fit exactly between the inner sides of webs and top and bottom flanges. The separator is swelled out for the bolt to pass through. Sometimes there are two bolts to each separator, but it is better (weakening the beam less) to have but one at the centre of web. The size of separators and bolts vary, of course, to suit the different sizes of beams. They should be placed apart about as frequently as twenty times the width of flange of a single beam. Where beams are placed in a floor, the floor arches usually provide the side bracing. But in order to avoid uu- Tle-rods. equal deflections, and possible cracks in the arches, Qfrom unequal or moving loads or from vibrations) and also to take up the thrust on the end beams of each floor, it is necessary to place TIE RODS. 211 lines of tie-rods across the entire line of beams. The size of these rods can be calculated as already explained in the Chapter on Arches (p. 169) ; they are usually made, however, from %" to %" diameter. Each rod extends from the outside web of one beam to the outside web of the next beam. The next rod is a little to one side of it, so that the rods do not really form one straight line, but every other rod falls in the same line. Care must be taken not to get the rods too long, or there will have to be several washers under the head and nut, making a very unsightly job, to say t]ie least. Contractors will do this, however, for the sake of the con- venience of ordering the rods all of one or two lengths. Where, therefore, the beams are not spaced evenly the contractor should be warned against this. One end of the rod has a " head " welded on, the other has a " screw-end," which need not be " up-set ;" the nut is screwed along this end, thus forcing both nut and head to bear against the beams solidly. The distance between lines of tie-rods, would depend somewhat on our calculation, if made ; the usual prac- tice, however, is to place them apart a distance equal to about twenty times the width of flange of a single beam. Flltch-plate Sometimes where wooden girders have heavier Girder, loads to carry than they are capable of doing, and yet iron girders cannot be afforded, a sheet of plate-iron is bolted be- tween two wooden girders. In this case care must be taken to so proportion the iron, that in taking its share of the load, it will deflect equally with the wooden girders, otherwise the bolts would surely shear off, or else crush and tear the wood. We consider the two wooden girders as one girder and calculate (or read from Table XII 1) their safe load, taking care not to exceed 0,03 inches of deflection per foot of span. We then, from Table VII or Formulae (37) to (41) obtain the exact amount of their deflection under this load. We now calculate the iron plate, for deflection only, inserting the above amount of deflection, and for the load the balance to be borne by the iron-work. An example will best illus- trate this : Example. 1 Flitch-plate girder of 20-foot span consists of two Georgia pine beams each 6" X 16" with a sheet of plate-iron 16" deep bolted between them. The girder carries a load of 13000 pounds at its centre; of w.'iai thickness should the plate be f The girder supports a plastered ceiling. 212 SAFE BUILDING. Strength of From Table XIII we find that a Georgia pine wooden part, beam 6" X 16" of 20-foot spaa will safely carry witLout cracking plaster 7980 pounds uniform load, or 3990 pounds at its centre (See Case (G) Table VII,) so that the two wooden beams together carry 7980 pounds of the load, leaving a balance of 5020 pounds for the iron plate to carry. The deflection of a 20-foot span Georgia pine beam 6" X 16" with 3990 pounds centre load will be, Formula (40) • , 1^ 3990. 2408 ~48 " ~e. iT e for Georgia pine (Table IV) is = 1200000 and * j2" (Table I. section No. 2), or t = ^'=2048, therefore . 1 3990. 2408 0 = — . 0 47" 48 1200000. 2048 Size of We now have a wrought-iron plate which must Iron Plate, carry 5020 pounds centre load, of a span of 20 feet, 16" deep, and must deflect under this load only 0,47". Inserting these values in Formula (40) we have : „ _ 1 6020.2408 0,47 =.-5. ^ ' 48 e. I. ^ From Table IV we have for wrought-iron 6 = 27000000 While for i, we have (Table I. Section No. 2) . 6.^/8 fe, 168 Inserting these values and transposing we have : 5020. 2408 ~ 48. 27000000. 341. 0,47 ~ ^'^^ Or the plate would have to be J" X 16". Now to make sure that this deflection does not cause too great fibre strains in the iron, we can calculate these from Formulse (18) and (22). The bendinc moment at the centre will be (22) 5020. 240 m = 1 = 301 200 The moment of resistance will be (Table T. Section No. 2) h. 0,33.162 r=-g- = -^ = 14 FLITCH PLATE GIRDEH. 213 A.nd from (18) if) — — = r, or transposing and inserting values, (7) 301200 14 = 21514 pounds. As the safe modulus of rupture of wrought-iron is only 12000 pounds (Table IV) we must increase the thickness of our plate. Let us call the plate |" X 16", we should then have 6. 162 = 26,G7 and (7)^ • 8. 6 301200. ■ 26,67 11256. So that the plate would be a trifle too strong. This would mean that both plate and beams would deflect less. The exact amount miglit be obtained by experimenting, allowing the beams to carry a little less and the plate a little more, until their deflections were the same, but such a calculation would have no practical value. We know that the deflection will be less than 0, 47" and further, that plaster- ing would not crack, unless the deflection exceeded | of an inch (Formula 28) as 20.0,03 = 0, 6" Size of Bolts. In regard to the bolts, the best position for them would, of course, be along the neutral axis, that is, at half the height of the beam. For here there would be no strain on them. But to place them with sufficient frequency along this line would tend to weaken it too much, encouraging the destruction of the beam from 0 lr\ o CM ^< *- - T- Fig. 13 7. Fig. 138. longitudinal shearing along this line. For this reason the bolts are placed, alternating, above and below the line, forming two lines of bolts, as shown in Fig. (137). The end bolts are doubled as shown; the horizontal distance, a-b, between two bolts should be about equal to the depth of the beam. If we place the bolts in our exam- Sl4 SAFE BUILDIXG. pie, say 3" above and 3" below the neutral axis, we can readily cal- culate the size required. Take a cross-section of the beam (Fig. 138) showing one of the upper bolts. Now the fibre strains along the upper edge of the girder, we know are j or 1200 pounds per square inch, for the wood, and we just found the balance of the load coming on the iron would strain this on the extreme upper edge = 11256 pounds per square inch. As the centre line of the boltls only 3" from the neutral axis or f of the distance from neutral axis to the extreme upper fibres, the strains on the fibres along this line will be, of course, on the wood f of 1200, or 450 pounds per square inch : and on the iron f of 11256 =4221 pounds per square inch. Now, supposing the bolt to be 1" in diameter. It then presses on each side against a surface of wood = 1" X 6" or = six square inches. The fibre strain being 450 pounds per square inch, the total pressure on the bolt from the wood, each side, is : 6.450 = 2700 pounds. On the iron we have a surface of 1" X |" = | square inches. And as the fibre strain at the bolt is 4221, the total strain on the bolt from the iron is = |. 4221 = 2638 pounds. Or, our bolt virtually becomes a beam of wrought-iron, circular and of 1" diameter in cross-section, supported at the points A and B, which are 6|" apart, and loaded on its centre C with a weight of 2638 pounds. Therefore we have, at centre, bending-moment (Formula 22) 2638. 6* 4 ■ = 4369. From Table I, Section No. 7, we know that for a circular section, the moment of resistance is. ^ r3=H 14 ■ ' 14 (1/= 0,098 Now for solid circular bolts, and which are acted on really along their whole length it is customary to take (^j^ the safe modulus of rupture rather higher than for beams. Where the bolts or pins have heads and nuts at their ends firmly holding together the parts acting across them they are taken at 18000 pounds for steel and at 15000 pounds for iron. We have therefore transposing (Formula 18) for the required moment of resistance 4369 ^ — 15000 Inserting this value for r in the SIZE OF BOLTS. 215 above we have for the radius of bolt, — . r3 = 0,291 and x = -\lii = 0,715 0,291 =^0,3704 Or the diameter of bolt should be 1,436" or say 1 7-16". But 1" will be quite ample, as we must remember that the strains calcu- lated will come only on the one bolt at the centre of span of beam; and that, as the beam remains of same cross section its whole length the extreme fibre strains decrease rapidly towards the sup- ports, and therefore also the strains on the bolts. The end bolts are doubled however, to resist the starting there of a tendency to longitudinal shearing. We might further calculate the danger of the bolt crushing the iron plate at its bearing against it; or crush- ing the wood each side; or the danger of the iron bolt being sheared off by the iron plate between the wooden beams; or the danger of the iron bolt shearing oflE the wood in front of it, that is tearing its way out through the wood; but the strains are so small, that we can readily see that none of these dangers exist. When girders run over three or more supports m C^rdere. one piece, that is, are not cut apart or jointed over the supports, the existing strains and reactions of ordinary girders, are very much altered. These are known as ' ' continuous girders. If we have (Fig. 143) three supports, and run a continuous girder over them in one piece and load the girder on each side it will act Fig. 143. as shown in Tig. 143; if the girder is cut it will act as shown in Fig 144 Very little thought will show that the fibres at A not being able to separate in the first case, though they want to, must causf considerable tension in the upper fibres at A. This tension, of course, takes up or counterbalances part of the compression ex- isting there, and the result is that the first or continuous girder (Fig 143) is considerably stronger, that is, it is less strained and considerably stifPer, than the sectional or jointed girder (Fig 144) Again we can readily see that the great tension and conflict of the opposite stresses at A would tend to cause more pressure on the central post in Fig. 143, than on the central post m Fig. 144, and this, in fact, is the case. 216 SAFE BUILDING. In Table XVII, pages 218 and 219, are given the various for- mute for reactions, greatest bending moments and deflections, for the most usual cases of continuous girders. The architect can, ir Fig. 144. he wishes, neglect to allow for the additional strength and stiffness of continuous girders, as both are on the safe side. But he must Variation of never overlook the fact tJiat the central reactions Reactions. are much greater, or in other words, that the end supports carry less, and the central supports carry more, than when the girders are cut. Bending moments can be figured, at any desired point along a continuous girder, as usual, subtracting from the sum of the re- actions on one side multiplied by their respective distances from the point, the sum of all the weights on the same side, multiplied by their respective distances from the point. Sometimes the result will be negative, which means a reversal of the usual stresses and strains. Otherwise the rules and formulae hold good, the same as for other girders or beams. Table XVII gives all necessary in- formation at a glance. Strength is frequently added to a girder or beam by trussing it, as shown in Table XVIII, pages 220 and 221. One or two struts are placed against the lower edge of a beam and a rod passed over them and secured to each end of the beam; by stretching this rod the beam becomes the compression chord of a truss and also a con- Trussed tinuous girder running over one or two supports. Beams. There must therefore be enough material in the beam to stand the compression, and in addition to this enough to stand the transverse strains on the continuous girder. If the loads are concentrated immediately over the braces, there will be no transverse strain whatever, but the braces will be compressed the full amount of the respective loads on each. In the case of uni- form loads, transverse strains cannot be avoided, of course, but Struts placed where loads are concentrated the struts should al- underload. ways be placed iinmediately under them. Even where loads are placed very unevenly, it is better to have the panels of the truss irregular, thus avoiding cross or transverse strains. This same rule holds good in designing trusses of any kind. IMPORTANCE OF JOINTS 217 Necessary Table XVIII shows very clearly the amount and Conditrons. kind of strains in each part of trussed beams. Where there are two struts and they are of any length care must be taken by diagonal braces or otherwise, to keep the lower ends of braces from tipping towards each other. Theoretically they cannot tip, but practically, sometimes, they do. Care must be taken that the beam IS braced sideways, or else it must be figured for its safety against lateral flexure (Formula 5.) Then it must have material enough not to shear of£ at supports, nor to crush its under side where lying on support. The ends of rods must have sufficient bearing not to crush the wood. Iron shoes are sometimes used, but if very large are apt to rot the wood. In that case it is well to have a few small holes in the shoes, to allow ventilation to end of timber. If iron straps and bolts are used at the end, care must be taken that the strap does not tear apart at bolt holes; that it does not crush itself against bolts; that it does not shear off the bolts, and that it does not crush in the end of timber. Care must also be taken to have enough bolts, so that they do not crush the wood before them, and to keep the bolts from shearing out, that is tearing out the wood Importance before them. In all trusses and trussed works the of Jornts. joints must be carefully designed to cover all these points. Many architects give tremendous sizes for timbers and rods in trusses, thus adding unnecessary weight, but when it comes to the joint, they overlook it, and then are surprised when the truss gives out. The next time they add more timber and more iron, till they learn the lesson. It must be remembered that the strength of a truss is only equal to the strength of its weakest part, be that part a member or only a part of a joint. This subject will be fully dealt with in the chapter on Trusses. Depth The deeper the truss is made, that is, the further Desirable. we separate the top and bottom chords, the stronger will it be; besides additional depth adds very much to the stiff- ness of a truss. Deflection All trussed beams, and all trusses should be ' * eam- of Girders bered up, ' ' that is, built up above their natural lines and Beams. sufficiently to allow for settling back into their cor- rect lines, when loaded. The amount of the camber should equal the calculated deflection. For all beams, girders, etc., of uniform cross- section throughout, the deflection can be calculated from Formute (37) to (42) according to the manner of loading. For wrought-iron beams and plate-girders of uniform cross-section throughout, the de- flection can be calculated from the same formula; where, however, the load is uniform and it is desired to simplify the calculation, the deflection can be quite closely calculated from the folio wing Formula : Uniform Cross- 5^ section and O — > (79) Load. I'^-d ^ 218 SAFE BUILDING. Amount of Greatest Deflections. e ^ g '"t 1 • -el- ~ ' CO - ZO ■S ri CO cT CO S • rH S • .2 CO o CO CO CO 53 , g a^ k 1 i-c; "S5 * • .8 _ ''t 8 s. - '^i .2 o o o ^ II 1 CO 53 = a, .8 ^ 8 ""t ^ 1 1 " 185. e. I Amount of Greatest Bending Moments. . + « eo 1*^ O "5 IcO ^ II g g t + 1 g w 'J . . II II Located at r > CO Amount of Keaetions. . . c a • • ico -2 X §^ 1 ^ II ^ 1 ^ ? " f - + i t 1 g s --g g 4 i;:: loo;::: IS -5 II II 1 II II g + ?5 CO |oo ec 1;^ i II n II Description. Two equal spans each carrying a central load but loads not equal. l = h Two equal spans each carrying a central load = w, loads equal. 1 = 1, Two equal spans each loaded with a Illustration. UJhS> rpr-0 ^ 'f ^ 1 I — Where ^, <'i,^ii = tho amount of deflection in inches, if girder of uniform cross- section throughout. •' e = the modulus of elasticity of the material, in pounds-inch, (see Table IV). •« izrthe moment of inertia of the cross-section, in inches, (see Table I, CONTINUOUS GIRDERS. 219 3 + t CO 3 + s" + s e Si eo o • o s + + 3 CO + + 05 eo .7* + 3 + 3" + 2 05 -IS + s" + s + >0 I'* lO |oo -3 II II 3 — 3 + 3" + 3 3 + 3" + s ■ c3 -Si c« O 5 3" - _ ^ 3 H o 3 + 3" + 3 + 3" + 1 o We draw the longitudinal neutral axes of each part, namely A B, B C and A C. The latter is so drawn that the neutral axis of the reaction, which is of course half way between end of girder and E (or 8" from E) will also pass through A. In designing trusses this should always be borne in mind, that so far as possible all the neutral axes at each joint should go through the same point. The beam A F virtually becomes a continuous girder, of two equal spans of 12 feet or 144" each, uniformly loaded with 20400 pounds each, and supported at three points A, B and F. From table XVII we know that the greatest bending moment is at B and u.l 20400.144 op^of^A 1 • = — z= = 36 7200 pounds-inch. 8 8 The modulus of rupture for Georgia pine (Table IV) is ^ = 1200, therefore moment of resistance (r) from Formula (18) and Table I, section No. 2, b.d^ 367200 r = = ■ or 6 1200 &.(/2=i836 Now we know that c?= 16, or d^= 256, therefore 1836 & = = 7,2 or say we need a beam 7i" x 16" for the 256 EXAMPLE, TRUSSED BEAM. 231 transverse strain. We must add to this however for the additional compression due to the trussing. Compression "^^^ amount of the load carried by strut C D, see in Strut. Table XVII, is = |. M from each side, or = 25500 on the strut B C, of which = 12750 from each side. If now we make at any scale a vertical line b c = half the load _ , carried at point B or = 12750 in our case, and Compression ^ ,, , ^ ^, in Beam. draw h a horizontally and a c parallel to A t , we find the strain va. B A by measuring h a = (32300 pounds) or in J. C by measuring a c = (34638 pounds) both measured at same scale as h c. We find, further, in passing around the triangle c b a c — (c b being the direction of the reaction at A), that & a is pushing to- wards A, therefore compression; and that a c is pulling away from .4, therefore tension. Using the usual signs of -|- for compression, and — for tension, we have then : AB = -{- 32300 pounds. A C = — 34638 pounds. B C=+ 25500 pounds. Had we used Table XVIII we should have had the same result for : Compression in A B = = _|- 32300 pounds and Tensioning. (7 = ^-^ = — 34638 pounds. Now the safe resistance of Georgia pine to compression along fibres (Table IV) is 750 pounds. 11 A B were very long, or the beam very shallow or very thin, we should still further reduce (^-y^ using Formulae (3), or (5). But we can readily see that the beam will not bend much by vertical fiexure due to compression, nor will it deflect laterally very much, so we can safely allow the maximum safe stress per square inch, or 750 pounds, that is, consider A B a short column. The necessary area to resist the compression. Formula (2) is : 32300 = a. 750 or 32300 ,„ . , a = — y^^ = 43 sc^uare inches. 232 SAFE BUILDING. As the beam is 16" deep, this would mean an additional thickness ji_3 — oil — 16 — ■'16 Adding this to the 7^" already found to be necessary, we have 7J. -L 211— ql 5 '4r^l6- — ''16 or the beam would need to be, say 10" x 16", Size of Strut. Now the size of B C must be made sufficient not to crush in the soft underside of the beam at B. The bearing here would be across the fibres of the beam, and we find (Table IV) that the safe compressive stress of Georgia pine across the fibres vi ^-^^ = 200 pounds. We need therefore an area 25500 ,„o • 1 a — ■ = 128 inches. 200 As the beam is only 10" wide the strut B C will have to measure, 1 28 --^ = 12| inches the other way, or we will say it could be 10" x 12". This strut itself might be made of softer wood than Georgia pine, say of spruce ; the average compression on it is 25500 , . , ^ — 212 pounds per square inch. Now spruce will stand a compression on end (Table IV) of ^y. ^ = 650 or, even if spruce is used, the actual strain would be less than one-third of the safe stress. At tiie foot of the strut B C w& put an iron plate, to prevent the rod from crushing in the wood. The rod itself must bear on the plate at least Iron Shoe to ^^^r=2,l square inches, or it would crush the Strut. 12000 ^ iron — (12000 pounds being the safe resistance of wrought-iron to crushing). Size of Tie-rod. The safe tensional stress of wro .-ght-iron being 1 2000 pounds per square inch (Table IV), we have the necessary area for tie-rod A C from Formula (6) 34638 = a. 12000 or a= ^1^^^ = 2,886 square inches. 12000 ^ From a table of areas we find that we should require a rod of 1 ^1" diameter, or say a 2" rod. , The area of a 2" rod being =3,14 square inches the actual ten- sional stress, per square inch on the rod, will be only ^3 14^ = 11312 pounds per square inch. SIZE OF ROD. 233 Size of Washer. We must now proportion the bearing of the wash- er at end of tie-rod. The amount of the crusliing coming on washer will be whichever of the two strains at A, (viz. B A and A C) is the lesser, or J3 ^ in our case, which is 32300 pounds. We must therefore have area enough to the washer not to crush the end of beam (or along its fibres), the safe resistance of which we already found to be : (-^) = "^^^ pounds per square inch ; we need there- fore 32300 .o • = 43 square inches. 750 ^ The washer therefore should be about ^" by 61" . _ The end of the rod must have an " upset " screw- end. end ; that is, the threads are raised above the end of rod all around, so that the area at the bottom of sinkage, between two adjoining threads, is still equal to the full area of rod. If the end is not " upset " the Avhole rod will have to be made enough larger to allow for the cutting of the screw at the end, which would be a wilful extravagance. It is unnecessary to calculate the size of nuts, heads, threads, etc., as, if these are made the regulation sizes, they are more than amply Central Swivel. strong. It should be remarked here that in all trussed beams, if there is not a central swivel, for tightening the rod, that there should be a nut at each end of the rod ; and not a head at one end and a nut at the other. Otherwise in tightening the rod from one side only it is apt to tip the strut or crush it into the beam on side being tightened. We must still however calculate the verti- cal shearing across the beam at the supports, which we know equals the reaction, or 20400 pounds at each end. To resist this we have 10" X 1G" = 160 square inches, less 3" x 16", cut out to allow rod end to pass, or say 112 square inches net, of Georgia pine, across the grain; and as ^-^^ = 570 pounds per square inch (see Table IV); the safe vertical shearing stress at each support would be (Formula 7) 112.570 = 63840 pounds or more than three times the actual strain. Then, too, we should see that the Bearing of , , . \ i t i ^ Beam, bearing of beam is not crushed, it bears on eacn re- action 16 inches, or has a bearing area= 16.10= 160 square inches. 234 SAFE BUILDING. ^ ^ for Georgia pine, across the fibres, Table IV, is ^-^^ = 200, therefore the beam will bear safely at each end 160.200 = 32000 pounds or about one-half more than the reaction. There will be no horizontal shearing, of course, except in that part of beam under transverse strain, and this certainly cannot amount to much. The beam is therefore amply safe. Deflection Now let us calculate the deflection. The modulus of Beam, of elasticity for Georgia pine, Table IV is : e = 1200000 pounds-inch. The average compression strain in ^ jP was 750 pounds per square inch, therefore the amount of contraction (Formula 88)i X = = 0,19 inches. 1200000 ' Now A D (in Fig. 146) will be one-half of this, or 0,095 inches. The amount of elongation in ^ C will be, remembering that we ■found the average stress to be only 11312 pounds per square inch, and that for wrought-iron e = 27000000 (Formula 88) ^^11312.163^ 27000000 ' The exact length of ^ C (Fig. 147 should be 163,41 not 163"). Therefore D F (Fig. 146) will be D F= 163,41 -|- 0,0682 = 163,4782" Z)J5 = 152 — 0,19 = 151", 81 Therefore (Fig. 146) 2 5-P'=\/ 163,4782 » — 151,81 « = 60", 655 Now B C (Fig. 147) would be = 60", deducting this from the above we should have a deflection = 0", 655. To this we must add the contraction of B C. The strut will be less than 60" long, say about 50". The average compressive stress per square inch we found = 212 pounds. The modulus of elasticity > In reality the contraction ot A F would be much less, as the part figured for transverse strain only would very materially help to resist the compress'on, on» half of it being in tension. TABLES XIX TO XXV. 236 for spruce, Table IV, is e = 850000, therefore contraction in strut (Formula 88) x=: -^=0,0125 850000 ' Adding this to the above we should have the total deflection S=: 0,655 4-0,0125 = 0,6675 This would be the amount we should have to " camber " up the learn, or say |". The safe deflection not to crack plastering, would be (Formula 28) S = L. 0,03 = 24.0,03 = 0,72 So that our trussed beam is amply stiff. Table XIX gives all the necessary data in regard of TaSres°" to the use of Tables XX, XXI, XXII, XXIII, XIX to XXV. XXIV and XXV. (See foot note p. 205). These Tables give all the necessary information in regard to all archi- tectural sections, which are used in this work. Where, after the name of the Company formerly rolling the section, there are several letters, it means that practically the same sections were . ^ ^. ^ rolled by several Companies. It should be re- I Sections not ■' _ / « • i j. i • i Economical. marked that except m the case of the simplest kind of beam work, it is cheaper to frame up plate girders, or trusses, of angles, flat-irons, tees, etc. Steel beams and sections are sold cheaper than iron, if the latter can be had at all, as they are very much cheaper to manufacture, and where their uniformity can be relied on, should be used in pre- ference, as they are much stronger and also a trifle stiffer. As a rule, however, the uniformity of steel in beams or other rolled sec- tions cannot be implicitly relied upon, but — as already stated — the architect has no option now, he must use steel, and hence should be extra vigilant. One example of an iron beam will make the application of the Tables to transverse strains clear, and help to review the subject, be- fore taking up the graphical method of calculating transverse strains. Example. Use of Tables wrougM-iron I-deam of 25-foot clear span, car- XIX to XXV. ries a uniform load of 500 pounds per foot including weight of beam; also a concentrated load of 1000 pounds 10 feet from, 236 SAFE BUILDING. the right hand support. The beam is not supported sideways. What size beam should be usedf The total uniform load u= 500.25 = 12500 pounds of which one- half or 6250 pounds will go to each reaction ; of the 1000 pounds load 180 — — or I will go to the nearer support q (Formula 15), therefore 9 = 6250 + |. 1000 = 6850 Similarly we should have (Formula 14) p = 6250 -f- f . 1000 = 6650 As a check the sum of the two loads should = 13500, and we have, in effect : 6850 4-6650 = 13500 To find the point of greatest bending moment begin at q pass to load 1000, and we will have passed over ten feet of uniform load or 5000 pounds, add to this the 1000 pounds making 6000 pounds, and we still are 850 pounds short of the reaction, we pass on therefore towards jo one foot, which leaves 350 pounds more, and pass oa another of V2o [ : 5oolb6.pey ft-.- 1250O lbs _ l6o A _ JilP . u Fig. 148. afoot (to A) which very closely makes the amount. The point of greatest bending moment therefore is at A, say 1' 8" to the left of the weight, or 140" from q : As a check begin at p and we must pass along 160" or 13' 4" of uniform load before reaching the point .4, at 500 pounds a foot this would make 13^.500. = 6666 or close enough to amount of reaction p for all practical purposes. The uniform load per inch will be = 41§ pounds. Now the bending moment at A will be, taking the right-hand side (Formula 24) mj, =6850.140— 41§.140.70— 1000.20 = 530 667 pounds-inch. As a check take the left-hand side (Formula 23) 771^ = 6650.160 — 41f. 160.80 USE OF TABLES. 237 = 530614 pounds-inch, or near enough alike for all practical purposes. Now the safe modulus of rupture for wrought-iron (Table IV) is 3=12000 pounds, therefore the required moment of resistance r from Formula (18) .530667 „ r = -=44,2 12000 Looking at the Table XX we find the nearest moment of resistance to be 4G,8 or we should use the 12" — 120 pounds per yard I-beam. But the beam is unsupported sideways. The width of top flange is b = 5y. We now use Formula (78) to find out how much extra strength we require. Reduction for In inserting value for y, we use the second column ure.^"^^' °^ Table XVI, as the beam is, of course, of uniform cross-section throughout, and have y = 0,0192. In place of w we can insert the actual value r of the beam, and see what proportion of it is left to resist the transverse strength, after the lateral flexure is attended to, or r,= , 0,0192.252 1 ^ 0,3966 _^46^__gg g ^j^^ beam would not be strong ' 1,3966 ' ° enough. The next size would be the 12}" — 125 pounds per yard beam, but as the 15" — 125 pounds per yard beam would cost no more and be much stronger we will try that. Its width of flange is b = 5" and moment of resistance r = 57,93. Inserting these values in (Formula 78) and using r in place of w we have — 57,93 57,93 , 0^0192^252— 1^48 52 = 39,14 The required moment of resistance was r = 44,2 so that this is still short of the mark, and we should have to use the next section or the 15"— 150 pounds per yard beam. The moment of resistance of this beam is r = 69,8 its width of flan"-e the same as before, therefore: 238 SAFE BUILDING. Or this beaifl would be a trifle too strong even if unsupported side- ways. We need not bother with deflection, for the length of beam is only 1| times the span, and besides not even f of the actual transverse strength of the beam is required to resist the vertical strains, and, of course, the deflection would be diminished accordingly. Safe Uniform column in Table XX headed " Transverse Load. Value," gives the safe uniform load, in pounds, if divided by the span in feet, for beams supported sideways. Of course the result should correspond with Table XV, except that the uniform load will be expressed in pounds here, while it is expressed in tons of 2000 pounds each in that table. For Tables XXI, XX 1 1, XXIII, XXIV and XXV the use of the "Transverse Value" is similar, and as more fully explained in Table XIX. CHAPTER VIL Graphical Analysis of Transverse Strains. —"EI ~S2 ■ Fig. 149. LL the (lif- erent cal- culations to ascertain the amounts of bending-mo- ments, the re- quired moments of resistance and inertia, the amounts of re- actions, vertical shearing on beam, deflec- tions, etc., can be done graph- ically, as well as arithmetical- ly. In cases of complic ated loads, or where it is desired to economize by reducinsc size of flanges, the graphical method is to be preferred, but in cases of uniform loads, or where there are but one or two concentrated loads, 239 240 SAFE BUILDING. the arithmetical method will probably save time. As a check, how- ever, in important calculations, both methods might be used to advantage. Basis of Cra- if "^^t; have three concentrated loads w, w„ and w„ phicai Method. ^ hQiim A D (Fig. 149), as represented by the arrows, we can also represent the reactions p and q by arrows in 3p posite directions, and we know that the loads and reactions all counterbalance each other. The equilibrium of these forces will not be disturbed if we add at a force = -\-7j, providing that at F we add an equal force, in the same line, but in opposite direction or = —y- We have now at E two forces, -\- y and/>. If we draw at any scale a triangle a o x (or I) where a o parallel and =p, and where o x paral- lel and = 4" 2/' ^ force a; a, which would just counterbalance them, or a x, which would be their resultant. That is, a force G E thrusting against E with an amount a x (or a:,) and parallel a x would have the same effect on E as the two forces -|- y andj9. Continuing a;, till it intersects the vertical neutral axis through load w at G, we obtain the resultant 0:3 of the two forces acting at G, namely a:, and (see triangle 6 a a; or II). Similarly we get resultant x^ at of load and x^, (see triangle c & a; or III) ; also resultant x^ at I of load w„ and ^3 (see triangle dcxov IV) ; and finally resultant -\-y,2AF of re- action of q and x^ (see triangle 0 a; or V). As this resultant i's,-\-y\t must, of course, be resisted by a force — ythat the whole may remain in equilibrium. By comparing the triangles I, II, III, IV and V, we see that they might all have been drawn in one figure (Fig. 150) for q -}-/> = io„ -\-w^-\-Wy therefore : dn-\-oa = dc-\-ch-\-ha, further both V and IV contain dx = x^ « a Y " I li ox = y t( « II " I <( ax = x. " II " III (( hx — Xi <( " III " IV « c X aTg We know further that the respective lines are parallel with each other. In Fig. 150 then, we have dc = w„ cb = io, ba = w ao-=-p and od = q POLE AND LOAD LINE. 241 The distance xy oi pole x from Imd line da being arbitrary, and the position c£ pole x the same. The figure E G H I F E (Fig 149) has many valuable quaUties. If at any point K of beam we draw a vertical Hne K L M, then L M will represent (as compared with the other vertical lines) the pro- portionate amount of bending moment at K. If we measure L M in parts of the length of A D and measure x y (the distance of pole, Fig. 150) in units of the load line da, then will the product of L M and x y represent the actual bending moment at K. That is, if we measure L M in inches and — (having laid out d c, c b, Fig. 150. etc., in pounds) — measure x ^ in pounds, the bending moment at K will be = x y. L M (in pounds-inch.) Similarly at w the bending moment would be = x y. N G (in pounds-inch.) and at v), it would he = x y. R H " " " and at M7„ it wouldbei=a; 7/. SI " " measuring, in all cases, x y in pounds and N G, R H and S I in inches. Average Strain The area of E G H I F E, divided by the length prbfes.'^^'"^ ^P^i^ i'^ inches will give the average strain for the entire length on extreme top or bottom fibres of beam, providing the beam is of uniform cross-section throughout. The area should be figured by measuring all horizontal dimensions in inches, and all vertical dimensions in parts of the longest vertical {R H'm our case), this longest vertical being considered r= ^ ^ for top, or ^-^^ for bottom fibres, or where these are practically equal = ^ The greatest bending moment on the beam will occur at the point where the longest vertical can be drawn through the figure. From this figure can also be found the shearing strains and deflection of beam, as we shall see later. Distance of If now instead of selecting arbitrarily the distance X y ol the pole from load line d a (Fig. 150) we had made this distance equal the safe modulus of rupture of the material, or X y=z^-y ^ — measuring x y in pounds at same scale as the load line da — it stands to reason that any vertical through the Figure 242 SAFE BUILDING. E GHIFE (Fig. 149) measured in inches, will represent the re- quired moment of resistance, for if LM. xy = m, we know from Fig. 151. Formula (18), that m — r. ) made x y = ^y j, we have, inserting values in above : iM.(|)=n(|)or LM=r 8EVEKAL LOADS. 243 Having thus shown the basis of the graphical methoil of analyzing transverse strains, we will now give the actual method without wasting further space on proofs. Several Concert- there are three loads w,w, and w„ on a beam _ fated Loads. A B (Fig. 151) we proceed as follows : at any conven- ient scale — to be known as the pounds-scale — lay off in pounds, rfc=M'„; alsoc&=M>, and ba=w. Let A £ = Z measured in inches - this scale being called the inch-scale. Now select pole x at random, Strain Diagram, but at a distance (measured with pounds-scale) *3' = ( y ) = tlie safe modulus of rupture of the material. Draw x d, xc,xhaniix a. Now begin at any point G of reaction q, draw G F parallel d x, till it intersects vertical w„ at F : then from F draw F E parallel ca; to vertical w,; then draw E D parallel 6 a; to vertical w ; and then D C parallel a a; to reaction p. From C draw C G, and through X draw x o parallel C G. Reactions. We now have the following results : oc? = reaction q (measured with pounds-scale.) ao= " p " " « II any vertical through figure CDEFGC, (measured with inch-scale) gives the amount of rz= required moment of resistance in inches, at point of beam where vertical is measured. The longest vertical passes through the point of greatest bending-moment in beam. Mul- tiply any vertical (in inches) with x y (in pounds) to obtain amount of bending-moment at point of beam through which vertical passes, "* R"^trstance. ^^^^^^i '' = ^ (92) Where r = the required moment of resistance, in inches, at any point of beam, provided pole distance xy=i(^Jy^. Where i; = the length (measured with inch-scale) of the vertical through upper figure CDEFGC at point of beam for which r is sought. And further : """''"r^;ment. "^ = ^^3/ (93) Where m = the bending moment at any point of beam in pounds- inch. Where « = the same value as in Formula (92) Where = the length, (measured with pound-scale) of distance of pole X from load line, in upper strain diagram xad. 244 SAFE BUILDING. If now -we draw horizontal lines through d, c, b and a ; and through 0 the horizontal line for horizontal axis ; and continue these lines until they intersect their respective load verticals w,„ w, and w, the shaded figure HIJKLMNO O, will give the vertical shearing strain along beam. Any vertical (as it, S) drawn through this figure to horizontal axis and measured with pounds-scale, gives the amount of vertical shearing at the point of beam {R) through which vertical is drawn. 0: , Vertical Cross- fnA\ shearing. * — ^" Where s = the amount of vertical shearing strain in pounds, at any point of beam. Where v,, = the length (measured with pounds-scale) of vertical through figure 0,HIJ KLM N 0 0, dropped from point of beam for which strain s is sought. We now divide G C into any number of equal parts — say twelvp in our case — and begin with a half part, or Giol = l2toC = -^^. G C ; also 1 to2z=2to3 = 3to4 = 4to5,etc. = Jj. G C and make the new lower load line g c with inch-scale so that Deflection o to I = length of vertical 1 e Diagram. ^ ° further I to II = length of vertical 2 / " II " 111= " " " 3 A " III " IV =r " " " 4 i, etc. untL " XI "c = " " " 12 A: Now select arbitrarily a pole z at any distance z j from load line g c. Now draw below the beam where convenient (say I. Fig. 151) beginning at (/„ the line g, e, parallel g z till it intersects the pro- longation of 1 e (from above) at e, ; then draw e,/, parallel I z till it intersects verticd 2 /at/,; and similarly draw / /i, parallel II z ; also A, i, parallel III z, etc., to m, parallel XI z and finally /fc, c, parallel c z. The more parts (Z,) we divide the beam into, the nearer will this line g, e,/ m, c, approach a curve. The real line to measure deflections would be a curve with the above lines as tang- ents to it ; we need not, however, bother to draw this curve for prac- tical work. Now draw c, g, and parallel thereto z o. Divide g, c, at o„ so tliat^: This would be the greatest possible deflection. If the beam were not so pro- portioned, but of uniform cross-section throughout, the deflection would be less. DEFLECTION. 247 Where §, v„ l„ e, z j, and i same value as in Formula (95). Where x y = the length of pole distance from load line in upper strain diagram, measured in pounds. The same formulae and methods could be applied to cantilevers, but for these the arithmetical calculations are so very simple that it would be taking unnecessary trouble. A few practical examples will make all of the foregoing more clear. Example I. Single concen- Georgia pine girder A B of lO-foot span carries trated Load, g load of 2000 pounds 5' 0" from right reaction B. What size should the girder be f We draw (Figure 152) ^ 5 = 240" at inch-scale, and locate w, at 60" to the left of B. Now draw a vertical line b a = 2000 pounds at pounds-scale. Select point x anywhere, but distant x y = 1200 pounds. (1200 pounds being = ^-j-^ or the safe modulus of rup- ture, per square-inch, of Georgia pine). Draw x b and x a. Draw verticals through A, w, and B. On vertical A begin at any point C, draw C E parallel x a, till it intersects verticals w, at E ; then draw E G till it intersects vertical B at G. Draw G C and o x parallel to G C. We scale o b, it scales 1500 pounds, so this, is the reaction at B. We scale a o, it scales 500 pounds and this is the reaction at A. The longest vertical through C E G is vertical w„ therefore greatest bending-moment is at w, which we know is the case. We scale E D at inch-scale, it scales 75 inches, therefore the (greatest) required moment of resistance will be at and will be Formula (92). r=75. From Table I, section No. 2, we know for rectangular beams, r = therefore ; — = 75, or 6. rf* = 450. We will suppose the girder is not braced sideways, and needs to be pretty broad ; let us try b = 5", we have then : 5.d^ = 450 or ^2^450 ^ 5 d = 9, 5" or the girder would have to be 5" x ^" or say 5" x 10". The bending-moment at 248 SAFE BUILDING. w, is, of course, Formula (93) = E D. x y = 75.1200 = 90000 (pounds-inch). Had we calculated arithmetically, we should have had, Formulae (14) and (15) : fin reaction A =—t. 2000 = 500 pounds. 240 " B= i^. 2000 = 1500 pounds. 240 Bending moment at would be (right side) Formulas (23) and (24). m^, = 1500.60 — 0.2000 = 90000 (pounds-inch) or check (left) side ??iw, = 500.1 80 — 0.2000 = 90000 (pounds-inch.) There- fore required moment of resistance. Formula (18) ^_90000_yg ^~ 1200 "~ or same result as graphically. By drawing the horizontals from b between verticals B and w, ; from a between verticals A and w, ; and from o between verticals A and B we get the etched figure for measuring vertical shearing strains. We see at a glance that the shearing to the right of load is equal to the right reaction, and is constant at all points of right side of beam ; while on the left side of load it is equal to the left reaction, and is constant at all points of the left side of beam. And this we know is the case. We need not bother with shearing, however, for we can readily see there is no danger. For even immediately to the right of the load, the weakest point in our case, we know that one- half of the dbres of cross-section are not strained at all, or we should 5 10 have one-half of area or 25 square-inches to resist 1500 pounds of shearing, or 1|^ = 60 pounds per square-inch, while the safe resistance, per square-inch, of Georgia pine to shearing across There is, however, some danger of excessive deflection; we draw, therefore, the figure c,f, g, by dividing the beam into ten equal parts, beginning and ending with half parts at the reaction, (each whole part being 24" long, or Z, = = 24") We draw the verticals through these parts and get their lengths through figure C E G. These lengths we carry down in their proper succession on the load line gjcot the lower strain diagram, begin- the grain is (Table IV) SINGLE LOAD. 249 ning at the top witli the right vertical 1, putting immediately under this the length of second vertical 2, then 3 and so on till g c = sum of lengths of all ten verticals through C E G. We now select z at random (in our case 120 inches from load line or z y=120"). We now draw lines from ztogl, II, III, etc., to c. Construct figure fl'i/i beginning at ^, drawing line parallel to zg until it intersects a > : i 1." " prolongation of first vertical 1 ; then line parallel to 3 1 till it intersects prolongation of second vertical 2, etc. We now draw z o parallel c, g^. We scale g o and find it scales 222", also c o which scales 162"; we divide c, ^, at f, so that c./: /sr. = 222: 162. / 250 SAFE BUULDING. Carrying vertical//, through figure we find it scales (v.) = 117" con- tinuing// up to beam it gives us point Fas the point of greatest de- flection, we find A F scales 138". Had we used Formula (43) we should have located Fat a distance from A or A F=^- ^^"^~^^ l = 134, 17". So that we have a sufliciently accurate result. For the amount of deflection at Fwe use Formula (95) ; we know that (Table I, Section No. 2) t = ^ = ^'= 417, further for Georgia pine ^A^:=i200 pounds. e = 1200000 (inch-nounds.) /, = 24" «,=// = 11 7" 2/= 120", therefore: g _ 117. 24. 120. 1200 1200000. 417 = 0,808" Had we calculated the deflection by Formula (41) we should have had: remembering that m = ]80" and n = 60" and /-f n = 240 + 60 = 300" ^ _ 2000. 180. 60. 300 /l80. 300 9. 240. 1200000. 41 7* A/ 3 = 0,803" Which proves the accuracy of the graphical method. For a beam of 20 feet span the deflection not to crack plastering should not exceed, Formula (28). 8 = 20. 0,03 = 0,6" Therefore, if our beam supports a plastered ceiling, it must be re- designed to be stiffer. Either made deeper, in which case it can be thinner, if braced sideways, or it can be thickened sufficiently to re- duce the deflection, see Formula (31). Example II. Single centre A hemlock girder A B {Fig. 153) of 16-foot span, carries a centre load w of 1000 pounds. What size "hould the girder be ? We make A 5 = 192" at inch scale; locate w at its centre F; make Z» a at any scale — (pounds-scale) — equal 1000 pounds. Se- lect pole X distant, xy=750 pounds, from load line b a, (as 750 CENTRE LOAD. 251 pounds =^-A^ the safe modulus of rupture per square inch of hem- lock). Draw xb and x a. Begin at G, draw G E parallel 6 a: to vertical through load, and then draw E C parallel a x. Draw C G and then x o paralleLC G, we find that o bisects 6 a or a o = o6 = 500 pounds. Each reaction is therefore one-half of the load ; this we know is the case. Greatest Une through C G E vie, find is at DE, so that greatest bending-moment is at load ; this we know is the case. D E scales 64" at inch-scale, therefore the required moment of resistance for the beam is, Formula (92) : r = 64. and the greatest bending-moment at load* Formula (93) : mw = 64. x?/= 64.750 = 48000 Had we calculated arithmetically we should have obtained the same results, for Formula (22) 1000. 192 4 and Formula (18) : 48000 = 48000 = 64 750 Now from Table I, Section No. 2, we know that for rectangular sections : r=_or 6.(^2= 64.6 = 384. If we assume the beam as 4" thick, we have then : 4. d2 = 384. rf2^?^=96 or 4 d = y/9i6 = 9,8" or we will make the beam 4" x 10". We draw the figure 0, H J K N 0 ior shearing and find it is con- stant throughout the whole length of beam and equal to length 0, H or N 0 measured at pounds scale, or 500 pounds. This is so small we need not bother with it. To obtain the deflection diagram we divide G C into eight equal 1 09 parts, each part Z, = — - = 24" and begin at each end with half 8 parts, drawing the eight verticals through C E G. We lay off their exact lengths in proper succession on the lower load line g c, beginning at the top with the right vertical. Select 252 SAFE BUILDING. pole 2 at random, in this case distant from load line zj = 180". We now di'aw the figure c, /, and find greatest deflection is at its cen- tre /yj ; for z 0 parallel c, bisects g c. We scale / /, at inch scale TWO LOADS. 253 (Table lY)e= 800000 g^ 44. 24.180. 75 0 ^ . " 800000. 33?> Had we calculated the deflectiou arithmetically from Formula (40) we should have had : g_£ 1000^»__ „ 48 800000.333 ' or practically the same result. If the beam supported plastered work the deflection should not ex- ceed, Formula (28) 8 ==16.0, 03 = 0,48" Still, unless we were very particular, the beam could be passed as practically stiff enough. Example III. Two concen- A while pine beam A B Fig. 154, of \1 foot span trated loads, carries two loads, one w, = 800 pounds, four feet from left support, the other ti'„ = 1200 pounds, two feet from right support. What size should the beam be f Make A B at inch scale = 144 inches, locate to, so that A «;, = 48", and w,, so that B w,, = 24". At any (pounds scale) make 6 c = 1200 pounds and c a = 800 pounds. Select pole x distant from b a ; z?/ = 900 pounds, the safe modulus of rupture per square inch of white pine ; draw xb,xc and x a. Construct C DE G parallel to these lines. Draw C G, and parallel to same x o, then will ao = 733 pounds be reaction at A, and o 6 = 1267 pounds be reaction at B. We scale vertical X) iV at =39" and TE at tr„ = 35", therefore greatest bending-moment is at t«, and Formula (93) TOw, = 39. 900 = 35100 Further, the required moment of resistance at w. Formula (92) will be : r = DN=39. Now from Table I, Section No. 2, h,d^ r=-—, or 6 6.(/2 = 6.39 = 234 Now if 6 = 3" we should have (Z2 = HM=78 and 3 d =y/T8 = say 9", or the beam would need to be 3" x 9". 254 SAFE BUILDING. We should have obtained practically the same results af ithmet' cally, for : Formulae (16) and (17) : ^ 800.96 , 1200.24 Reaction at^= ... h — ttt — = 733 144 144 „ . „ 800.48 , 1200.120 Reaction at 5= H t-t-. — = 1267 144 144 check : A -\- B = w, -\- w,, = 800 1200 = 2000 pounds and 733 -f 1267 = 2000 pounds. Beginning at B we have to pass over load w„ (1200 pounds) and on to w„ before passing amount of reaction B (12G7 pounds) there- fore greatest bending-moment at w,. We know from Formula (24) it would be : wi„,= 1267.96 — 72.1200 = 35232 and check from Formula (23) 7n„, = 733.48 — 0.800 = 35184 being near enough for practical purooses. From Formula (18) we should have had : r = ^= 39,09 • 900 ' We now draw the shearing diagram 0, H I J K L M 0, as shown in Figure 154, and find the amount of shearing from .4 to m;,= 0, // = 733 pounds from to w„ = / aS = 67 pounds, from w„ to B=M 0=1267 pounds. We can overlook it, for even at the weakest point of beam for re- 3 9 sisting cross-shearing we have half the area, or-^ = 13| s(iuare inches. White pine will safely resist 250 pounds per square inch in cross- shearing (Table IV) or the beam would resist. 13^. 250 = 3375 pounds at its weakest point for cross-shearing, {\iz.: at «;,) and twice as much at the reactions. To find the deflection we divide G C into eight equal parts, begin- ning with half parts (or /,=lii = 18") and draw the verticals 8 hrough C D E G. We now make the lower load line g c equal the am of these verticals, beginning at the top with the right vertical. Select z distant from g c (the load line) zj= 108". Draw zg, zc, etc., and construct g, c^f^ as before. We draw z o parallel c, g^. Now g o mear.ures 116 inches and o c TWO LOADS. 255 108 inches, therefore divide c, g, at f so that : Carrying the vertical f f^ up to point F oi beam, we find the point of greatest deflection F, where B F= &^" and ^ F=: 74^" We find / /, scales 42", remembering that (Table I, Section Xo. 2) 1 = ^ = 182, and that for white pine Table IV e= 850000 pounds we have Formula (95) : cv ^ 42. 18. 108. 900 ^ „ " 850000 .182 ' Had we attempted to get this result arithmetically by inserting the values in Formula (41) (and remembering that n is always the nearer support, or in our case respectively 48" and 24", while m respectively 96" and 120") we should realize the advantage of the graphical method, for : 800. 90.48. (144 + 48). V '^^' + + 1200. 96.48. (144 + 24). y/^^Ml^T^) 9. 144.850000.182 Tf we figure out the above tedious formula we should have 8 = 0,422' or practically the same result as we obtained graphically. The safe deflection, were the beam to carry plastering, should not exceed Formula (28) 8= 12.0,03=0,36" Our beam is therefore not nearly stiff enough, and we must make it thicker ; or else if we wish to save material, we will make it thinner, but deeper ; and then brace it sideways, see Formula (31). Example IV. Five Concentra- A spruce girder A B of 18-foot span carries five ted Loads. iQ^dg^ shown in Figure 155. What size should the girder be f We draw A 5 = 216" (inch scale) ; further 6a=2700 pounds (sura of loads at pounds scale) ; make hh — Wy = 540 pounds. he = w,v = 180 pounds', e d = w,,. = 360 pounds, dc = w^ = 720 pounds, and c a—w. = 900 pounds. 256 SAFE BUILDING. Fig. 155. FIVE CONCENTRATED LOADS. 257 Select X distant a; ?/ = 1000 pounds from & a, (as 1000 = ( j-) fo' Bpruce, see Table IV). Draw a; 6, x}i,xe, etc., and figure C D G. Draw xo parallel C G; it divides load line as follows: a 0= 1580 pounds or reaction at .4. 0 6 = 1120 pounds or reaction at B. We find longest vertical through C D G, is at load w,., therefore greatest bending-moment on beam at w„ ; now D E scales 70 J", there- fore Formula (93) : m„„=70^. 1000=70500 and Formula (92) r=70, 5 From Table I, Section No. 2 h r = — = 70, 5 and if 6 = 5, we have 5. c?2 = 6. 70, 5 or w'nu and io„ in their order. We take no notice of vertical w, as it does not fall in one of the even divisions of C G or A B into lengths We select pole 2 distant zj = 288" from load line, draw zg, zc, etc., and then figure c,f,g,. We now draw 2 o parallel c, g^ it divides g c, so that go = 295" and o c = 245", we di- vide c, g, in same proportion at f, and carry this up to F at beam, which is the point of greatest deflection of beam, and is distant 163" from B, and 197" from A. We scale / /, = 106" (inch scale) and have from Formula (9 7) tN _ 106. 60. 288. 6500 _ , „^ „ 27000000.1238 ~ ' 1238 being = t, the moment of inertia of beam as found in Table XX. The beam is therefore amply stifif even to carry plastering. Irregular Cross- The graphical method lends itself very readily to sections, finding centres of gravity and neutral axes, as ex- plained in the chapter on arches, and also for finding the moment" of inertia of difficult cross-sections. MOMENT OF INERTIA. 265 If we have an irregular figure ABODE (Figure 158) we divide it into simple parts I, IT, III and IV. We find the centres of To find Neutral gravity g,, g,„ A«is. and g,y of each part and draw their re- spective horizontal neutral axes through these. Anywhere's make a line ae=area of whole figure and divide it, so that : ab = area of I 6 c = area of II cd = area of III and de = area of IV. Select pole x at random, draw X a, xb, xc, xd, and x e. From any point of horizontal g, draw fh parallel bx till it in- sects horizontal gr„; then draw hj parallel c a; to horizontal ^f,,, ; To find Moment J ^ parallel 6? a; to last horizontal, and finally ko of Inertia, parallel xe; and fo parallel ax till they intersect at 0. A horizon- ^ tal through o is the main neu- tral axis of the ■whole.* If we multiply the area of the figure fokjh by the area of the figure CDS (both in square inches) we have the value of moment of inertia i of ABODE in inches, around its horizontal neutral axis o. » » Ti * r' Fig. 159. > The point of intersection of this line with a main neutral axis, found similarly, in any other direction, would be the centre of gravity of the whole figure. SAFE BUILDING. A simple way of obtaining the area of the figure fo k would be to To find area. draw horizontal lines through it at equal distances beginning with half distances at top and bottom, and to multiply the sura of these horizontals in length by the distance apart of any two horizontals, all measurements in inches. This will approximate quite closely both the area and moment of inertia. Of course the more parts we take in all of the processes, the closer will be our result. A practical example will more fully illustrate the above. Example VII. Rolled Deck- Find horizontal neutral axis and the corresponding beam, moment of inertia of a 1" — 55 pounds per yard deck beam, resting on its flat flange {Figure 159). We will take the roll as one part, divide the web into four equal parts, the flange into two parts, one the base which will be practi- cally rectangular, and its upper part which will be practically tri- angular. The whole area we know is for wrought-iron : 55 a = — =5, 5 square inches. The bottom rectangular part of flange will be Cv,, = 4 J. I = 1,7 square inches next triangular part a,. = iH = 0,9 The web parts a„ = a,,, = a,v= ay = — - — =0,4 square inches each. 4 Leaving for the roll at top a,= 1,3 We now make the horizontal line ah = 5,5" and divide it, so that a& = 1,3 inches bc — cd = de = ef=0,4: inches fg = 0,9 inches and gh = l,7 inches. Select X at random and draw xa, xb, xc, etc. Draw the horizontal neutral axes I, II, III, etc., through their re- spective parts. Begin anywhere on I and draw j k parallel 6 a: to line II; then kl parallel ca: to III; then Im parallel dxtolY ; tli(;n mn parallel ex to V; then np parallel /a; to VI ; then/) parallel ROLLED DECK-BEAM. 267 g X to VII; Now draw from q the line qo parallel xh, and from> Horizontal '^"^ ■'^ parallel a a; till they intersect at o. A Neutral Axis, horizontal through o is the neutral axis of whole beam. We will now make a new drawing of figure j'oq for the sake of clearness. Dra^v horizontals through it every inch in height be- ginning at both top and bottom with one-half inch. The top one scales nothing, the next then f*, then IJ", then 2|", then Area of Dia- and the bottom one the sum of all being 6^^" gram, or 6,416". This multiplied by the height of the parts, which is one inch, would give us, of course, 6,416 square inches area. Multiplying this area by the area of the cross-section of deck beam 5,5 square inches, we should have 1 = 5,5.6,416 = 35,288. Moment of In- In ^^ble XX it is given as 35, 1 so that we are ertlaof Beam, not very far out. If we had taken more parts, of course the result would have been more exact. rreducing When constructing plate girders of large size, '''^"^^^'cirders.much material can be saved by making the flanges heaviest at the point of greatest bending-moment, and gradually re- ducing the flanges towards the supports. This is accomplished by making each flange at the point of great- est bending-moment of several thicknesses or layers of iron, the outer layer being the shortest, the next a little longer, etc. Of course the angles, which form part of the flange are kept of uniform size the whole length, as it would be awkward to attempt to use different sized angles. Generally (though not necessarily) the inner or first layer of the flange plates, is also run the entire length. Of course, where the flanges are gradually reduced in this way, it becomes ne- cessary to figure the bending-moment and moment of resistance at many points along the plate girder to find where the plates can be re- duced. This would be a wearisome job. By using the graphical method, however, it can be easily accomplished. Referring back to Figure 151, we take the point of greatest bending-moment (at w,) of the beam A B. The required moment of resistance at this point, it will be remembered was the length (inch scale) of vertical S through C D E F G. We now decide what size angles we propose using and settle the necessary thickness of the flanges by Formula (36), insert- ing for the value of r, the length (inch scale) of v or vertical at E. Further a, will, of course, be the sum of the area of two angles, d the 268 SAFE BUILDING. total depth of girder in inches and b the breadth of flange, in inches, less rivet holes. The above is on the assumption that the distance xy of pole X from load line rfa was equal to the safe modulus of rup- (k\ jjoi steel or wrought-iron according to whichever material we were using, or we should have : V -J — a, Thickness or x=:z- (^^) Flanges. h Where a:=the thickness, in inches, of each flange of a plate girder at any point of its length. Where v = the length of vertical, inch scale, through upper or resistance figure, providing we have assumed the distance of pole from load line (pound scale) = of the material. Where d = the total depth, in inches, of the plate girder. Where b = the width, less rivet holes, in inches, of the flange. Where a, = the sum of the areas of cross-section, in sauare inches, of two of the angles used. We now calculate as above, the thickness x of flange at point of greatest bending- moment and then decide into how many layers or thicknesses we will divide the flanges. Say, in our case we decide to make the flange of four layers of plates, each | or one quarter x in thickness. Then make E,E„ = a,.d (99) Where E, E^ = the amount to be substracted (inch scale) from moment of resistance or vertical v and representing the work of two angles. Where a, = the sum of the area of cross-section, in square inches, of the two angles. Where d = the total depth, in inches, of the girder. Where to drop Now draw through £J„ a parallel to base of figure orr Plates, c G, divide E,, E into as many parts as we decide to use thicknesses of plates (four in our case) and draw parallel lines to base C G through these parts. Vertically over the points where these lines intersect the curve or outline of figure C DEFG will be the points at which to break off plates, as illustrated in drawing. This method, of course, is approximate, but it will be found sufiici- ently accurate for all practical purposes. It is not necessary that x KEDUCING GIKDER FLANGES. 269 or E £J„ be divided into equal parts. Had we decided to use plates of varying thicknesses we should simply divide E i?„ in proportions to correspond to thicknesses of plates in their proper order, beginning at iJ„ with plate immediately next to angles and ending at E with extreme central outside plate, ^n example, more fully illustrating the above, will be given in the chapter on plate girders. 270 SAFE BUILDING. CHAPTEE VIIL REINFORCED CONCRETE CONSTRUCTION. Modern useful concrete constructions began in France, in the beginning of the last century; but owing to the nature of the material, the constructions were necessarily confined to mostly heavy work, such as foundations, retaining walls, dams and similar Historical structures. As a matter of fact concrete was Facts. well-known to the Ancients. The dome of the Pantheon is of concrete; while during the various Gothic period's, vaultings, floors, etc., were made of concrete, and frequently walls and other parts of cities, forts and castles as well. But even further back the Aztecs of Mexico, the old Greeks and other very ancient nations used concrete. They however used lime or ground shells or similar substances. It was with the discovery of the method for making Portland cements, that concrete became an active factor in modern above- ground constructions. The French used it largely in floors, forcing between the iron beam flanges, corrugated sheet iron, in segmental arches, and fill- ing above this with concrete. By the time the concrete had set thoroughly, it mattered little whether the sheets lasted or not. The subject of concrete itself, and of concrete underground constructions has already been treated in Chapter II. In over-ground constructions the use of concrete began, (after dams, retaining walls, etc.) by the use of moulded, generally hollow, interlocking blocks, formed in moulds, of fine concrete. This system is still in vogue, and is so simple and well-known it needs no further comment here. But with the beginning of reinforced concrete, invented by the French about the time of the ending of our Civil War, a new era dawned for construction in this country, and to-day re-inforced concrete bids fair, before very long to re-place many — if not most — other systems of masonry or timber construction. ^^^^ As is well known concrete properly made of Reinforced Portland cement continues to set until it becomes Means. as hard as (almost) the hardest stones used for eonstruction purposes. It resists compression strains admirably, 272 SAFE BUILDING. and is therefore used in piers, walls and similar constructions, to carry super-imposed loads, providing it cannot give way laterally. But, where it is not braced laterally, the bending tendencies in walls or columns of great height, compared to their bases, would quickly crack and let the walls or columns down, as concrete is very poor indeed in resisting tension. Similarly in floors, the concrete would be admirable for the upper layers of floors, beams or girders, which layers are in com- pression, but the bottom layers being in tension, the bottom would tear apart, being unable to resist much tension strain, and the whole give way in consequence. Metal Bars, off-set this weakness steel square, round, Etc. rectangular, or other shaped bars are introduced into the concrete and so designed as to act together with the con- crete, the latter attending to the compression, the steel taking up and resisting the tension. The methods of doing this are innumerable. Almost every large city in the United States has several companies doing con- crete construction, each in their own way. The essential thing is to roughen the steel so that the concrete will take hold of the metal, and not allow the steel to slip past the concrete, without bearing on each other, and every part of both working together. Many companies do this by means of specially rolled shapes, with projections, indentations, ribs, etc., to form roughened sur- faces for the concrete to clinch to. Other companies use wire, others still, punched metals, patent laths, expanded metals, etc. The best thing for the architect, lacking experience, is to select the method that appeals to him and study it carefully; it would be impossible to cover the subject in a general work of this kind. These metal forms are imbedded into the wooden moulds and frequently wired together, before the concrete is poured in. It should be explained that for all floor constructions wooden moulds on centres (with proper support), are formed underneath, Removing ^^^^ ^^i^^ the concrete is poured. Moulds should Moulds. never be removed until the concrete is known to be thoroughly set. At least two weeks, or longer, is a good rule for floors; three weeks or longer for girders, columns and walls. Continuous Floors that are built over girders and beams. Floors, that is on top of them, instead of in between are much stronger, in the same ratio as the continuous girder is stronger than the girder butting over supports. But, it must be ALLOWABLE SPANS. 278 borne in mind that this very strength is obtained by a reversal of the strains, (see Figs. 143 and 144, Chap. VI) the top layers over and near the supports being in tension, the bottom in com- pression. It is necessary therefore where bars are used, to rivet them together over supports, to get the benefit of the additional strength, and where rods and similar shapes are used to bring them up from the lower layers of concrete to near the top. Plate L gives a good idea of what is meant, and shows the steel skeleton and wire construction for columns and girders, walls, etc. as well. Where floors are supported on all four sides, that is built in squares over interlacing or crossing girders, thus forming square panels, they are still stronger. Concrete roofs are built similarly to floors, but generally sloped to form pitch and to save extra weight. Domes are similarly moulded up and reinforced. In all floor constructions, care should be taken to avoid a steel that is very soft and will easily stretch; as, while it might hold, it will badly crack the whole construction. One system of construction avoids this difficulty, by fixing each rod of steel in vises at each end, so that the length between secured ends cannot change, and then twisting the steel, thus taking out the stretch to close to the elastic limit, and at the same time the twisted, cork-screw shaped rods give the concrete a good chance to take hold. No formulae, that are sufficiently scientific can be given as yet, as most of them are empirical, many even rule-of -thumb only; it is remarkable however, how much a thin floor of flat concrete slabs will hold, when properly reinforced, and what wide spans may be used. As a rule, where the floor slabs are supported only at the two ends the following will be sufficient for the architect to approxi- mate his construction. Allowable including the load of the construction, for Spans. a live load of 150 pounds per square foot, the safe thickness of slabs should not be less than: 3 * thick for spans of 7 feet or less. 3%" thick for spans of 8 feet. 4 " thick for spans of 9 feet. 41/3" thick for spans of 10 feet. 5 " thick for spans of 11 feet. 6 " thick for spans of 13 feet. 274 SAFE BUILDING. But in each case the slabs must be well supported and thoroughly reinforced. Where the slabs are square and supported on all four sides, very much greater spans may be used. The deeper the slab, the stronger of course, but the greater the expense and the weight on other parts. Girders and beams are formed similarly to floor ^^''''^Beams. slabs. The necessary moulds are built, in forms of boxes, open on the top, and thoroughly supported from below. The steel re-inforcement is then put in, and is usually in the shape of trussed bar work, with additional U bars or stirrups. These latter are spaced, similarly to stiffeners on riveted girders, closer together at supports, and further apart at centre or near point of greatest bending moment of the concrete girder. Their object is to take care of the shearing strains at supports, and prevent cracking of concrete there. Girders thus trussed ixp, with U straps, etc. are frequently called armored girders. The concrete is poured into the moulds from the top, and the frames and sup- ports of the moulds should be left in place as long as possible. For purposes of laying out his work, the architect should make the depth of the girders, in inches, not less than the span in feet. Columns and piers are built similarly as girders, ^^'ancTpiers. but the boxes are whole boxes with only the ends open. The vertical rods are placed in position and then wired together when the concrete is poured in from the top and allowed to set before removing the jacketings from mould. Walls. Walls are similarly built. Moulds are made, placed in position and thoroughly secured in place. The vertical rods and horizontal truss rods are placed inside, wired together, concrete poured in and the whole allowed to set thoroughly. A building, that is its floors, columns, girders, walls, can be made a practical monolith by attaching all ends together, both of rods, wires and concrete. The diameter or sizes of columns, piers, etc. and the thicknesses of walls in most cities is regulated by Law, or Building Depart- ment regulations. But after all is said, the greatest thing for the architect to do, is to watch, eternally watch, the construction, to see that the con- crete is properly mixed, and allowed to set sufficiently long before removing moulds. Great care must be taken too, by the architect to see that no frost gets into any part of the work. ,. The question of exterior finish of concrete hardly enters into COLOEED TEEATMENT OF FACADES. 27r a work of this kind, but the writer has discussed this quite fully in a letter, one of a series asked for by the Editors of the American Architect on the subject: "The Aesthetic Treatment of Concrete." This letter was published in the edition of May 4, 1907. (American Architect.) The main and most original suggestion made by the writer was Colored Treat- that where a colored treatment of the exterior or Facades. main facade was desired, the constructional con- crete be kept back some two or more inches, with sufficient anchor- ing wires left to anchor outside finish. That the outside finish then be cast into moulds, specially made for that purpose, with a concrete made of very small parts of either broken stone, brick or terra-cotta (or a mixture thereof), Portland cement, water and finely ground similar stone, brick or terra-cotta grains, (or a- mixture thereof) the latter to take the place of the sand. By experimenting and allowing for the effect of the cement coloring in connection with the mixtures of small broken parts and finely ground grains, as above, almost any scheme of coloring for the facade can be thus obtained, and it has the benefit of being natural to the construction, a part of it, and very durable. The reader is referred to pages 95 to 99 (incl.) for further information on reinforced concrete in sub-soil constructions. CHAPTER IX. RIVETS, RIVETIKG AND PINS. WHEN it is necessary to secure two or more pieces of iron or steel together in such a manner that they can be readily separated, bolts are used. These are iron or steel pins with solid heads at one end and threads cut on the other end, onto which the nut is screwed, thus holding the two pieces together. How closely the two pieces may be held together depends, of course, entirely upon the man who handles the wrench. Then too, bolts or pins do not completely fill the holes through which they pass, which fre- quently is a cause of great weakness, besides the danger of water getting into the spaces and rusting them. Where, therefore, it is not necessary to ever separate the pieces — and the latter are of either wrought-iron or steel — rivets should be used, which, for all practical purposes, might be considered as permanent bolts or pins. Cast^ iron, of course, always has to be bolted, as it would break if riveting were attempted on it. A rivet is a piece of metal Descrlption^of ^ (wrought-iron or steel) with a solid head at one end and a long circular shank. In its shortest descrip- tion the process of riveting consists in heating the rivet, passing its shank through the two (or more) holes, while hot, and then forging another solid head out of the projecting end of the shank. The hammering forces the heated shank to fill all parts of the holes, and the shank contracting in its length, as it cools, draws the heads nearer together, thus firmly forcing and holding the pieces together. Rivets are made of mild steel or the very best wrought-iron, the latter being the most reliable. According to some writers, the shank is made tapering in length and circular in shape, being larger at the head and smaller at the end. In this country, however, the shank is always of uniform diameter. The length of shank Length of Tall. ^^^^ head to end varies with the thickness of pieces to be riveted together, but is long enough not only to allow for pass- ing through the pieces, but has also enough additional length to SHAPE OF HEADS. 277 provide for filling of holes and forming head, the additional length being about 2^ times the diameter of hole. The rivets are manu- factured by heating rods of the diameter of rivets, which are pushed while hot into a machine, which at one operation forms the head and cuts off the shank to the desired length. The shank before heating is about one-sixteenth smaller in diame- ter than the hole, to allow for its expansion when heated, i. e., for nor lea. 1 N I" rivets, i|" holes are punched and for |", \ f holes. There are four kinds of rivets, all answer- ing the same uses, and only distinguishable by the shapes of their heads. These are the button or round headed rivet ; the conical headed rivet ; the pan or flat headed rivet ; and the countersunk rivet. The latter is only used when a smooth surface is desirable. The first is the most used shape. Figure 163 shows the different shapes, the dotted Hues indicating how the second head is finally shaped. Sometimes a rivet is counter- sunk on one end only. The exact sizes of heads, shapes, etc., vary in different mills. The diameter of head should be from 1^ to 2 times the diameter of shank,i according to shape adopted and the height of head should be about f the diameter of hole. In countersunk work, the head may extend entirely through the plate or not, its diameter being accordingly smaller or larger. Where it extends through the sharp edges will shear the rivet, countersinking therefore, should be avoided in the plates. In showing riveted work the diameter of the shank is always drawn and figured wher« the hole is to be left open and the size of rivets is spoken of accordingly, the hole is always made larger. Where the riveting is to be done at the shop or mills, the size of head is shown. The spacing of rivets will be considered later, the direct distance from centre of hole in centre of hole is known as the ' The Franklin Institute standard for button-heads (which are usually useil in the United States) is to make the head larger in diameter than 1^ times the diameter of shank. Size of Head. Pitch of Rivets. 278 SAFE BUILDING. pitch." The pitch should never be less than 2^ diameters ; nor should the centre of any hole (if possible) be nearer to any edge than 1^ diameters. By diameters is understood the diameter of shank. In riveted angle work the distance is frequently necessarily less. In thick plates it should be more. In drilled work the pitch might be reduced to 2 diameters. If rivet-heads are countersunk the pitch should be increased, according to the amount of metal cut away to make room for the rivet-head. In punching rivet-holes the position of holes are usually marked off on a wooden template and then through this marked or indented by a hand-punch on the iron plate ; the plate is then passed under a punch which 'u usually worked by steam-power, the Punchingof ^j- ^.j^g punch being adjusted to the size of Holes* ^ o J the rivet-hole wanted, the punch is usually larger than the rivet, and the die about ^" larger. The thickness of plates to be punched should not, as a rule, be greater than | of an inch; nor in any case should the thickness of plate be as large as, nor larger than, the rivet-hole, as, unless the diameter of the hole is larger than the thickness of the plate the punch is apt to break. Where holes are punched at the building, small screw or hydraulic (alcoho!) punches are used, which can be readily carried around by one or two men, the power being obtained by screwing or pumping ; or sometimes, where mechanics are not quite up to the times, a rather more clumsy lever-punch is used, the power being obtained by increasing the direct pressure by leverage. Punching makes a ragged and irregular hole, and as it gives the plate a sudden blow or shock it injures the metal considerably, unless the rivet-holes are so close, that the entire plate is practically cold-hammered. The loss in strength to the remaining fibres in a punched Loss by Punch- -yvrought-iron plate is about 15 per cent, this loss be- ing, of course, in addition to the loss of area, and it is a loss that cannot be restored. In steel plates the remaining fibres are damaged about 33 per cent, but in them the loss can be restored by annealing the plate which, however, adds considerably to the expense. In drilled-work there is no loss, and the holes are not only accu- rately located but are accurately cut. But drilled-work is very ex- pensive, as it has to be done by hand or by machine-drills, the process being slow at best and consequently meaning a very large PUNCHING AND DRILLING. 279 addition to the charge for labor. In riveted girders it would proba- bly double the expense of the girder. The advantages of drilling are, that the holes can be cut after the plates have been partially secured together, thus as- Advantages surino- a perfect fit of the holes over each other; of Drilling. ° ^ , , . ^ , i and that the holes being perfectly smooth and even bear more evenly on the rivet, and the work is less apt to fail by compression, than where the bearing of plate against rivet is ragged and uneven, as in punched work. On the other hand, the edges of drilled holes are so sharp that they promote shearing, and for this reason the edges of drilled holes in plates should be filed or reamed off. As a rule, however, the architect will find the bearing and cross-breaking strengths of rivets less than their shearing, excepting where rivets are small in comparison to thickness of plates being riveted, which is not often the case. To settle, then, whether work should be drilled or punched, is mainly a matter of expense. Drilled-work, of course, is far prefer- able as regards strength and it costs accordingly. The rule of the mills is to punch all holes, excepting for counter- sunk rivets, which, after punching, of course, have to be drilled, to obtain the slanting sides of the hole. A medium course between drilling and punching would be to punch the holes smaller than desired and then drill or ream them to accurate size when partially secured together.^ Steel should always be drilled unless annealed after punching. In most work, however, the architect will have to be satisfied with punching, and must therefore allow sufficient material to make good the damage done and to allow for inaccuracies. In riveting proper, that is, filling the holes, there are also the two methods of doing it, by hand (hand-riveting), or by Machine-Rivet- machinery (machine-riveting), but unlike the making of the hole, in this case, the machine-work is both better and cheaper. A machine-driven rivet is driven and completed by one power- ful scjueeze of the steam (or compressed-air) riveting-machine ; this squeeze not only forces the plates more closely together, but more completely fills the hole with the rivet metal, besides the great ad- 1 If this is not done the " drift pin " will be used to force all the holes into line, and this means crushing and possibly buckling of the plates. 280 SAFE BUILDING. vantage of doing the entire work while the rivet is hottest, and while it is, of course, at one temperature. In hand-riveting these advantages are lost, the power being only equal to that of the mechanic's blow, and as in hand-work the pro- cess consumes some time, the rivet changes its temperature and cools considerably. In riveting, the entire rivet, including the head, should be heated to at least a red heat. It should not be heated beyond this for fear of " burning," particularly with steel rivets. Rivets that have been heated once and allowed to cool without working should be discarded. If rivets are driven at a lower heat than a red one, they will be greatly damaged, unless riveted cold. In hand-riveting at least two men are required, one to hold the Hand-Riveting. " holder-up," the other to do the riveting ; but generally there is a boy to heat and bring the rivets, one holder-up, and two riveters, whose strokes alternate and thus accelerate the process. The riveter puts a punch or drift-pin through the holes to clear them and force them into line ; the holder-up seizes the hot rivet with his pincers and puts its shank through the hole, he then covers the head with a holding-up-iron shaped to fit it, and the riveters at the other end begin hammering down the projecting end of shank. When this is roughly shaped the use an iron (called "button set," for round-headed rivets), which is properly shaped to make the head. , Before beginning to hammer down the end of shank, the riveters should always thoroughly hammer the plates around the hole, to bring the plates closely together. Hand-riveted work can sometimes be distinguished from machine- riveted work by the many marks at the made head. In machine- work there is but one mark, and this may be a little out of the centre with the shank and so show the squeezed material around its edge. But usually the work cannot be distinguished gulsh Rivetin'g. ^^7- hand-riveting has been conscien- tiously done and by careful mechanics, it is difficult to distinguish it from machine-riveting. But the shanks of hand-riveted work, as a rule, do not fill the hole as well as those in machine- riveted work, and they can more easily be " backed out " after the head has been cut off with a chisel and hammer. The only reliable test in both methods is to hold the hand one side of the head and HAND RIVETING. 281 strike the other side with a light hammer— the hammer test — when the sound will quickly disclose loose shanks. If in machine-riveting the plate has been sprung, and the pressure is quickly removed while the rivet is still hot, the plate may settle back, lift the hot head, and so form a loose shank. In designing riveted work, whether to be hand or machine riveted, the architect should bear in mind the necessity of placing the rivets so that they can be inserted in the holes from one side and ham- mered from the other, and for machine-work, that the machine can reach them.^ Steel rivets are very seriously damaged during the process of riveting. Box gives as the average of a number of ^**^'damaged. ^^^^^ °^ ^^^^^ ^ tensional-strength of 47,84 tons (gross), which, after riveting, was equal in shearing- strength to only 23,77 tons (gross), a loss of 50 per cent as between the tensional-strength of the steel and shearing-strength of the riveted work, whereas in ordinary steel work this loss never exceeds 33^ per cent, as between tension and shearing. In wrought-iron the loss is about 15 per cent. In steel the strength of the rivets will depend greatly upon the composition and nature of the steel itself, but in order to be able to rivet, the steel will necessarily have to be of a mild character. The safe values given in Table IV, for compression and shearing of wrought-iron and steel, can therefore be used with perfect safety. When calculating bending-moments on rivets, a modulus of rupture 25 per cent greater than given in Table IV "!is of miu^reT rivet-heads answer the same purpose as nuts and heads on pins in holding together the plates which are pulling in opposite directions, and thus reducing the bending-moment by friction. In figuring the number of rivets required an architect should err on the side of liberality, rather than to stint them, as there will necessarily be more or less of them poorly driven. He should particularly do this where strains are small and the number of rivets required are few, as one weak rivet in a small lot would quickly diminish the factor-of-safety, where in a large lot it would scarcely vary it appreciably. »The minimum distance, from inside face of one leg of an angle-iron to centre of nearest rivet-hole, in other leg, should be at least IJ" for 1" diameter rivet ; H"for 3" diameter rivet ; 1" for |" diameter rivet; J" for f" diameter rivet ; 13-16" for diameter rivet ; and if possible these distances should be increased. SAFE BUILDING. Of course the more rivets there are, the more will the plates be in- jured and cut away; this loss, however, can be Chatn, zig-zag largely overcome by what is called " chain-riveting," or staggered. "zig-zag" riveting, and by making the laps or cover-plates pointed. Chain riveting consists of placing the alternate rivets on different lines instead of all on one line, see Figure 164. This again is called CHAIM EUVETED ZIG-ZAG FIG. 164. zig-zag if they alternate as shown. ^ A cross-cut through this plate at any point can only pass through two rivets if chain-riveted or through one rivet-hole if zig-zag riveted; so that the plate is only weakened by two or one rivet-hole, respectively, while it may have a large number of rivets. Where plates are lapped or joined by cover-plates, the rivets have to transfer the full strain from one plate to the other (in case of a lap) ; or from one plate to the cover-plate and from that to the other plate (in case of cover-plates). Of course, it can be Platens weak- j-g^jjiy geen that this means a large number of rivet- ened by holes. , . r i i tc holes and a very great weakening ot the plates. it it were practical to suddenly enlarge the plate at the riveting point there would be no loss, but this would be clumsy and besides it would not be practicable to roll plates with certain points enlarged or thick- ened. As the whole plate, however, will be equal only to its strength at its least cross-section the rivets should be so disposed as to weaken the plate as little as possible. This is done by pointing the plate ends in the case of lapping, or the ends of cover-plates where these are used, and are not covered in the construction. Where the plates are to be ultimately hidden out of sight, this expense is saved, the plate ends are left square, but the rivets are placed pointedly or pyramidically. Thus, in Figure 165 is shown a lapped joint with staggered rivets ; we will suppose that calculation has shown the Lapped joints, j^g^gggj^y ^j^ets to equal the tensional strain 1 Most mills use the term " staggered " In place of " zig-zag," but for conven- ience in writing the writer prefers to use the term staggered to mean zig-zag rivets placed pyramidically. RIVETED PLATE JOINTS. 283 on each plate. The under plate A is dotted, the upper plate B drawn with full lines. By arranging the rivets as shown in Figure 165, each plate is weakened only by one rivet-hole. (As already ex- plained the plates themselves need not necessarily be pointed, but can be left square, if expense must be considered and looks are no object.) For, while a section at C shows plate A weakened by two rivet-holes (Nos. 2 and 3) |_4^___ .J it must be remembered 4 j : that the strain on A is no } longer the full strain, but it i D. C. iCr. 163. been diminished by an amount equal to the work that would have come on the one rivet-hole ; for rivet No. 1 has already transferred this amount to plate B. Similarly while a cut at D shows three rivet-holes (Nos. 4, 5 and 6) the plate is really not weakened at all here, for an amount of strain equal to what would have come on the metal taken from these rivet-holes has already been transferred to plate B by rivets Nos. 1, 2 and 3. Similarly as the strain on B increases the rivet-holes in it diminish till at No. 9, plate B has got the full strain and is therefore only weakened by this one rivet-hole. The disadvantage of lapping plate ends is obvious, as the plate would not continue in the same plane. For this Butt Joint with reason ioints are generally made by butting the cover-plates. ^ . , , . , ends of the plates and covering one or both sides with cover-plates. The principle of riveting is the same as for lapped joints, but it will require a different disposition of the rivets, and twice the number of rivets, as plate A (Figure 166) has to transfer its strain to the cover-plate by one series of nine rivets, and the cover-plate transfers it to plate B by another series of nine rivets. This can be readily seen in Figure 166. In this case the disposition of the rivets requires an extra rivet each side, or 20 in all. Where it is not necessary to keep plates A and B in the same plane it would be cheaper and better of course to lap them, rather than to use one cover-plate. If, however, two cover- Two plates can be used, one on top of the plates and the other under them, the advantage is very great, as the strain will be transmitted in a direct line or plane from plate A 284 SAFE BUILDING. to plate B, and besides this the joint is greatly strengthened by the friction between the cover-plates and plates A and B. In an experi- ment made by Clark with three | inch thick plates riveted with one I inch rivet through an oblong hole, it was found that friction added about 5 tons resistance against pulling out the centre Oalnby Frlctlon.pj^j.g^ This would seem to add a safe-strength to each I inch rivet of ^ and if /=5 (or a factor-of-safety of 5 were used) it would add just one ton to the calculated strength of a | inch rivet. This increase, however, is a doubtful one, and would prob- ably be lost in time by a gradual wearing of the plates due to this very friction, or possibly from slight rusting or other causes; no no, 167. FIG- 166. allowance should therefore be made for extra strength due to friction, but it certainly is a great advantage in making joints ; and the above facts may account largely for the discrepancies in experiments on riveted joints, where no allowance is made for friction. The pitch of rivets is, of course, governed by circumstances. The rule is to try to arrange the rivets, so that the strength of the plate between them shall equal the actual strength of the rivet. In boiler-work, however, they must be located not only for strength, but must be placed close enough to make *' the joint steam-tight. For this reason, too, boiler- plates are always lapped, the joint being more easily caulked and made tight. In constructional work, however, there will be a great loss and waste of material, if the rivets are placed too closely. In plate Boiler RIvetlngi PITCH OF RIVETS. 285 girders and riveted joints in trusses the rule is not to make the pitch less than 2^ diameters for punched-work, nor more than sixteen times the thickness of the least plate at the joint, or : p = l6.t (107) Where j9=:the greatest pitch, in inches, for rivets of plati- Createst Pitch, girders or riveted trusses. Where t = the thickness of the thinnest plate, in inches. The pitch is measured from centre of hole to centre of hole on a direct (straight) line. p, = 2^.d (108) Where = the least pitch, in inches, for rivets of plate-girders Least Pitch. or riveted trusses. Where d = the diameter of rivet-holes, in inches. The exact pitch must be between these two limits ; and is, of course, calculated. Different writers have attempted to lay down exact rules for the size of rivets to be used, using for a basis for the Dlamater^of formulse the thickness of thinnest plate to be riveted. Such rules, however, generally do not agree with good practice, as they either make the rivets too small for thin plates, or too large for thick ones, or vice versa. As a rule the local circumstances must control the selection of the size of rivet ; the following, however, may serve as a general guide : For plates from \" to thick, use rivet-holes |" diameter. For plates from ^" to |" thick, use rivet-holes f " diameter. For plates from \\" to " thick, use rivet-holes |" diameter. For plates from |" to 1" thick, use rivet-holes 1" diameter. Of course, larger or smaller rivets can be used, but as a rule | inch, I inch, and | inch are most desirable. Figure 167 shows the different ways in which riveted-work will yield. This figure is made from a photograph of an actual speci- men, tested and torn apart at the Watcrtown arsenal. It is evident that the plate began yielding by all of the rivets com- pressing or crushing the plates, and finally yielded How completely by tearing apart from 2 to 1 and to left edge and tiie same from 4 to 5 and the right edjre, while rivet 3 tore its way completely out, shearing off a piece of the plate, and rivets 2 and 4 partially so. The iron plates tested were 15 inches wide, ^ inch thick, with two 286 SAFE BUILDI^G. 5} D " D 1 — ¥ £ M 1 ^ ? strain. 44410 square iron cover-plates 15 inches x inch each. The rivets were ^| incli of iron and filled 1 inch drilled holes, pitch 3 inches. The gross area of plate was 3,7G5 square inches, the net area 2,510. The total bearing-surfaces of rivets on the plates aggregated 1,255 square inches, and tlieir aggregate shearing areas, (being in double shear) was 7,854 square inches. At 11G715 pounds strain the eJges contracted and scale on the specimen began to start and the plate 3 ielded as shown at 167200 pounds This was equal to pounds tension per inch on the uncut plate, or 66610 pounds per square inch on a line at the rivet-holes (or net an a). The compression from the FIG- 170. rivets was 133230 pounds per square inch, while the shearing was only 21290 pounds per square inch. This example shows clearly how the plate yields. Besides this the joint might yield by breaking or shearing otf the rivets. We have then tlie following six manners in which a riveted-joint might yield. How Riveted 1- By crushing either the rivet or the rivet crush- Joint yields, the plate. 2. By shearing off the rivet — in sirigle shear. 3. By shearing off the rivet — in double shear. 4. By bending or cross-breaking of the rivet. 5. By tearing the plate apart or crushing it between rivet-holes. 6. By the rivets shearing out the part of the plate between them and the edge. In Figures 168, 169 and 170 are shown three kinds of joints, each with a single rivet transferring the whole strain; in Figure 168 directly from plate A to plate B; in Figure 169 transferring strain from plate A to cover-plate and thence to plate B; and in Figure HOW RIVET YIELDS. 287 170 transferring one-half of the strain A to each cover-platu and thence each half is transferred back again to plate B. It should be remarked here that cover- plates (as in Figure 169) should be at least the full width and thickness of the original plates. In practice they are made a trifle thicker (about ^ inch or more). Cover-plates, as in Figure 170, should each be the full width of original plates and at least one-half the thickness of same ; in practice they too are each made about inch (or more) thicker. The plates A and B are themselves, of course, of the same thick- ness. Now to prevent failure by the first method, compression, there must be area enough at both G H and C D, (if "^Compression. P^**^^ tension), not to crutch the rivet or the plates at these points, D E I and sA G H in Figure 170). This area is considered equal to the tliickness of either plate A or B (or of cover-plate or their sums) multiplied by the diameter of rivet-hole. To resist failure by the second method, single-shearing of rivet, the area of cross-section of each rivet must be suffi- ^^""^hearing. ^'^ shear off under the total strain on either plate A or B. It will be readily seen that only the rivets in Figures 168 and 169 are subjected to single shearing, viz: at their sections (? /). The rivets in Figure 170 have two areas resisting shearing, GD and HE, hence are subjected to double shearing ; therefore their area of cross-section need only be sufficient to resist a shearing strain equal to only one-half of the total strain on either plate A or B, in order to avoid failure by the third method. To avoid failure by the fourth method, the rivet must be suffi- Failure by ciently strong to resist the load as a lever in Figures bending. 168 and 169, and as a beam in Figure 170. In Figures 168 and 169 we can consider the part D C F G aa the built-in part of a lever, with a free end D E H G which carries a uniform load equal to the whole strain on either plate A or B. In Figure 170 we have a beam supported at C D G F and E I J H, with its span or central part G H E D loaded with a uniform load equal to the whole strain on either plate A or B, To prevent failure by the fifth method the area of cross-section 288 SAFE BUILDING. of either plate taken at right angles across same through the rivet-hole — (that is, deducting the rivet-hole from Failure of ^jjg area of cross-section) — should be sufficient to plate. . ^ resist the tension or compression. To prevent failure by the sixth method the rivets must be far enough from the edges of plates (cover and original Failure by rivets pj3^(-gg\ shear out the metal ahead of them, tearing out. ^ ^ . t-.- The rule is shown in Figure 171. Make angle A O C— 90° that is a right angle (0 being the centre of rivet-hole and C A part of its circumference), and so that the directions of 0 A and 0 C are at 45° with edge of plate D B. Then the sums of the areas A B-\- C D — (that is, A B C D multiplied by thickness of plate) — must be sufficient to resist the longitudinal shearing strain, which in this case would be the strain on either plate ^ or ^ (Figures 168 to 170). To put the above in formulae we should have : Bearing, Use X for lap joints only. Use 2 a; in place of x for butt joints with single or double cover- plates. SingleShearing.^- ^^^^^^^y^T^^ (110) Use X for lap joints only. Use 2 a; in place of x for butt joints with single cover-plate. s Double ^— ,„ / a\ (HI) Shearing. 1,5714. c?^. ^ j ^ Use 2 x or butt joints with double cover-plates. s. h. 0,1964. d«. (^y^ Use X for lap joints only. Use 2 X for butt joints with single cover-plates. s. h Bending- — / k \ (112) moment Lever. / \ v / Bending- ^ / k \ (113)' moment Beam. Tocr, jn / \ 0,7857. j) ' The fourth decimal given in formulae is not quite right, but is made to corres- pond with fractions used in Table I, FORMULA FOR RIVETS. 289 Use 2 X for butt joints witli double cover-plates. Tension on — . Plate. I, j Use ^-^^ instead of ^-^^ if plate is in compression. Shearing end — (114) /'"^ ) of Plate. 2.A.(-^) (If more than one rivet use t instead of s for distance of each rivet X from end as shown in Figure 171.) Where s = the whole load or strain, in pounds, to be trans- ferred from one side of the joint to the other. Where fZ= the diameter of rivet-hole, in inches. Where A=the thickness of plate, in inches. Where more than one plate is used, take for Ti the least aggregate sum of thick- nesses of all plates acting in one direction. (The sum of cover-plates should at least equal this aggregate in thickness and should be larger, where the net h of cover-plates is smaller than the net h of connected plates.) Where h — the net breadth of plate, in "FIG- 17 inches, that is the breadth, less rivet-holes, at the weakest section ; where more than one plate is used they should all of course be of same breadth. The net b of cover-plates will frequently be much less than that of original plates, as they lose the greatest number of rivet-holes at their centre, where they are carrying the full strain. Where x = the total number of rivets required at the joint for lap joints, and the number required each side of joint for butt joints with single or double cover-plates ; that is in the latter two cases 2 x will be the total number of rivets required. Where y='m inches, is the length A B or C D (Figure 171) from any rivet to free edge of plate ; where more than one rivet is used, insert - in formula, in place of s. It will only be necessary, of course, to calculate y for the line of rivets nearest free edge. Where — safe compression stress, per square inch, 290 SAFE BUILDING. Where ~ ^^^^ tension stress, per square inch, Where ^ ^ — safe shearing stress, per square inch. Where ^-^^ = sa£e modulus of rupture, per square inch. k ■f all in pounds, (see Table IV). Safe Stresses The writer uses the following values, as a rule for on Rivets and . ■ , . Pins. rivets ana pins. For Wrought- (^j^= 12000 pounds, per square inch. (7) ^ I — 12000 pounds, per square inch. ^, I — 8000 pounds, per square inch. (I) ^ J — 15000 pounds, per square inch. ForSteel. (^} ~ ^^^^^ pounds, per square inch. i^y^ — 15000 pounds, per square inch. ijj^ — 10000 pounds, per square inch. 18000 pounds, per square inch. Example I. Lap Joint. A wrought-iron plate, wliicli cannot he over 12" loide, is required to he so long that it has to he made up from two lengths; the joint is to be a lap joint. The plate is in tension and is strained 65000 pounds. Design the joint. We will assume that we propose to design the joint as shown in Figure 165, with staggered rivets, in that case the plate will only be weakened by one rivet-hole. We can readily see that the plate will not need to be very thick and decide to use |" rivets, (that is |" rivet-holes)^ ; we then shall have a net breadth of plate Size of plate. 6=. 12" — f" = 11^". Of course s = 65000 in this case, and we know that (7) = 12000 ; inserting these values in Formula (114) we have: 1 Before heating, in this case, the rivets would be ^ "• EXAMPLE — LAP JOINT. 291 65000__ llf 12000 ' Or we should use a plate oae-half inch thick. We next determine the number of rivets required. In the first place there must be enough for bearing, that is not to Required crush the plate or get crushed by it. We use nimiberof RIv- formula (109) inserting the values, and have: 65000 " = 1:^12000 = '^'"' Or we should need 15 rivets for bearing. Had we figured without using the formula, we should have said, the bearing area of each rivet is f" by ^" = f square inches, this at 12000 pounds, per square inch, would equal 4500 pounds safe-compression for each rivet, dividing this into 65000 pounds, the strain, would, of course, give the same result 1 4,44 or say 1 5 rivets. We next see if there is any danger from shearing. The joint be- ing a lap joint, the rivets will have, of course, only one sectional area to resist shearing, that is, will be in single shear, so that we use Formula (110) and inserting values have: 65000 0,7857. (1)2. = 18,39 Or we must use 19 rivets to prevent the shearing. Had we figured without the use of formula, we should have said, area of a f" rivet is = 0,4417 square inches. This multiplied by 8000 pounds (the safe shearing stress per square inch) =: 3533,6 pounds, or each rivet could safely assume this amount of the strain without shearing. This amount being less than the safe compression on each rivet, would, of course, require a larger number of rivets, and should therefore be used, rather than the latter. We have, in effect ^^^ = 1^'^^ or say 19 rivets, being four more than re- quired for bearing. We next take up the question of bending ; the joint being lapped the rivets will oractically become short levers. We use Formula (112) ; inserting values, we have : r- ^5000i _ 0,1964. (1)8. 15000 ' Or we should have to use at least 26 rivets to prevent bending; 292 SAFE BUILDING. which readily illustrates the great disadvantage of not transferring the strain in a direct plane, by using two cover-plates. Had we not used the formula, we should have said, we have here a I" circular lever, the free end projecting ^" and loaded uniformly with a load of 65000 pounds. From Formula (25) or Table VII we have the bending-moment 65000. A . . , m = — ^=16250 pounds-inch. and from Table I, section No 7 the moment of resistance, r — \\. (1)3 = 0,04143 From the formula on page 49, Volume I, m ^ r we have the total extreme fibre strains on all the rivets, This divided by the safe strain, or safe modulus of rupture ^-^^ = 15000 pounds, will give the number of rivets required, viz: ?^« = 26,15 15000 ' or 26 rivets as before. We still have to decide the distance y (or A B, Figure 171). We use Formula (115). As we have more than one rivet we use in place of s the strain on each rivet or which was the largest number required as above, therefore : ^_65000_2500 26 26 Inserting this in Formula (115) we have : _2500_^. „ ^ 2. ^. 8000 This, however, being less than our rule which requires 1^ diameters from centre of hole to edge, we will stick to the rule. We now design the joint. We have a plate 1 2" wide, ^" thick, lapping, and require 26 rivets. Deslgningthe They must be arranged not to weaken the plate by Joint, more than one rivet-hole. If we arrange the rivet-holes as shown in Figure 172, we will'find it the most economical arrangement. To be sure it allows for only EXAMPLE — BUTT JOINT. 293 FIG. 172 25 rivets, but that will probably be near enough, otherwise we should have to insert at least five more to keep them symmetrical. It will be readily seen that the weak- est point is at section A where one rivet-hole is lost. Section 5 is of same strength, two rivet-holes being lost, but the strain has I been reduced by an amount equal to the value of one rivet-hole. At section C we lose three rivet-holes, but the strain has been re- duced by the value of three rivet-holes, so that the plate practically has its full value here. At sections D and E the plate is stronger to resist the remaining tension than required. By figuring out E (12" wide) it will be seen that the pitch on this line is more than required by the rule, Formula (108). The pitch F G between two adjacent lines of rivets, measured on the slant from centre to centre of rivets, should be at least 2^ diam- eters, 2^. 1= If" or say 2". It will be good practice for the student to carefully lay this joint out to scale. Example II. A steel plate 10" wide has to he pieced, and for °" cover*ptatl'^^^'^"^ ?-easons this can only be done by a cover-plate on one side. The plate is subjected to a tensional strain of 135000 pounds. Design the joint. Of course, the rivets must be of steel too. We will again assume that we can stagger the rivets, so that we shall lose only one rivet-hole. The plate will evidently have to be thick and we will decide to use 1" rivets ; this would leave us a net breadth of plate Size of plate. 6 = 1 0" — 1" = 9" , From Formula (114) we have : 135000 _ „ ~ 9. 15000" Or the plate will have to be just one inch thick. The cover-plate 294 SAFE BUILDING. should be at least the same, if there were only one rivet, but as there will evidently be more than one and we propose staggering the rivets, the cover-plate will have to be considerably thicker ; we can therefore leave the cover-plate out of consideration for the present as, being thicker, or in case of one rivet equal to the plates to be joined, it will certainly be as strong, and not crush. Required Now, as for the number of rivets, from Formula number of riv- (io9) we have for bearing : * ^' _ 135000 _ g 1. 1. 15000 "~ Or nine rivets are required not to crush the plate or be crushed by it (each side of joint). From Formula (110) we have for single shearing (as there is evidently only one area to each rivet to resist shearing) : 135000 0,7857. 12. 10000 ' Or seventeen rivets are required to resist the shearing (each side of joint). The rivets will evidently be levers in this case and we have from Formula (112) 135000.1 „Q , X= s oo,l 0,1964. 18. 18000 ' Or it will require thirty-eight rivets to resist the bending-moment (each side of joint). This again shows the advantage of using both a top and bottom cover-plate and so avoiding the great leverage on the rivets. It must now be borne in mind that the joint is a butt joint, there- ^ fore, unlike the case of lap joint where the whole Designing^t^e^^ number of rivets bear on each plate, we must here use twice the number, as only one-half, or those each side of the joint bear on one plate, or we require in all 76 rivets. The plate is so narrow that we cannot get more than three rivets on a line across the plate, without infringing on the rules for pitch. We can, therefore, stagger the end rivets and make the rest either chain riveted or zig-zag riveted. The chain riveting will require a little longer cover-plate, but in the case of a plate-girder flange would have the advantage of using the rivets that are needed for the angle-irons. In that case we should not stagger the end rivets, for our plate would only be weakened by one additional hole, and, of course, the DESIGNING THE JOINT. 295 gross breadth of plate would have to be 12 inches instead of 10 inches to give same strength. Figure 173 shows this joint chain riveted, with end rivets, stag- gered, to correspond to our calculated example. We see that it takes 39 rivets each side of joint for symmetry. Figure 174 shows this joint zig-zag riveted. It has three advant- ages over the other, it is shorter, takes just the right number of 1 O O O Q O O O O o o o o ooooooooooo oooooooo o o o o o o o o o o o o o o o g o 0 FlCn 173. rivets and requires a thinner cover-plate. For in Figure 173 at the first line of rivets next to the joint (C), the cover-plate loses three rivet-holes and bears the full strain, or its clear breadth would be o o o 9 o o 0 6 o o o o o o o o o o o o o o o o o o o 0 0 0 0 o o o 0 0 0 0 0 o only 7 inches and from Formula (114) would require a cover-plate of thickness 135000 h. = 1,3 7.15000 or say 1 jY' thick. Whereas, in Figure 174, next to the hne C we lose only two rivets and have consequently a clear breadth of 8 inches and require from Formula (114) a cover-plate of thickness equal to 135000 _ ^ ^5 8.15000 ' or only 1^ inch cover-plate. 296 SAFE BUILDING. Had Tve not used the formula we should have figured out our rivets, etc, as follows : Required net or clear area of plate 135000 _ Q gquare inches. Net breadth being 9" gives, of 15000 ^ c & ' course, 1" thickness. Bearing area of each rivet = 1.1 = 1 square inch, which will safely bear 15000 pounds or we should need 135000 „ . , _ = 9 rivets. 15000 Single shearing area of each rivet (or area of a circle 1" diameter) = 0,7854 square inches, which at 10000 pounds per square inch, would give a resistance to shearing per rivet = 7854 pounds or we should need 135000 __ J y 2 rivets. For bending we should have a one 7854 ' ° inch circular lever, projecting one inch and uniformly loaded with 135000 pounds. The bending-moment would be Formula (25) 135000.1 , m= = 67600 pounds. The moment or resistance would be Table I, section No. 7 r = H- (1)3 = 0,0982 The strain s, therefore, on all the rivets will be [page 49]. - ''''' 687373 0,0982 This divided by the safe modulus of rupture for steel rivets ^-^^ = 18000 will give the required number, as before, 68!iH = 38,2 18000 Example III. Butt-joint Two Same problem as before, but two cover-plates to be Cover-plates, used, one above and one below the joint. We will again decide to stagger the rivets and losing only one rivet-hole will again require a 1" thick plate. COVER PLATES. 297 Now, for bearing we will have the same result as before, viz. : 9 rivets required, but in shearing it is evident that Required num- QQ^jy each rivet has two resisting areas or is in double ber of rivets. ° shear, so that we will need only one-half of the 17 2 previous quantity or -^^=8,6 or say, nine rivets. Had we used Formula (111) we should have obtained this result, for : _ 135000 _ g g 1,5714. 12. 10000" ' For the bending-moment we use Formula (113) and have 135000. 1 _ g ^ 0,7857.13.18000 or say ten rivets required to resist cross-breaking. This shows how much better proportioned the different strains '^^'^Tw^Tcovefrs ('^^^^*^^°S' bending and bearing) are to each other, where we use two cover-plates. Had we calculated directly without use of formula we should have obtained the same results. We have already calculated net area of plate and bearing value of rivets in Example II, also the single sliearing value, which was 7854 pounds per rivet ; as we now have two areas this would be doubled or the resistance of each rivet to shearing would be 15708 pounds and the number required ^^^^Qg — before. For bending-moment we should have a 1" circular beam with a clear span of 1", uniformily loaded with 135000 pounds. From Formula (21) we have the bending-moment 135000.1 .nor^ 1 ■ -u m = = 16875 pounds-inch 8 Moment of resistance will be as before r= 0,0982 Therefore the total strain on all the rivets 16875 ^^^^843 0,0982 This divided by ^ = 18000 gives the required number of rivets as before ^_m8i3_ 18000 Designing the We now design the joint, as before, remembering joint. stagger the rivets and to place the required num- ber each side of joint. The greatest number required was to resist cross-breaking, viz., ten. 298 SAFK BUILDING. We can design as shown in Figure 1 75 or as shown in Figure 3 76 ; both require eleven rivets each side of joint, but cover-plates in Figure 176 need only aggregate in thickness, that is, be j^g" thick each; while those in Figure 175 would have to aggregate l^^g" in thickness, or be, say, \ |" thick each. Joint shown in Figure 175 looks a little better, but otherwise there is no preference. If cover-plates are not equal in thickness each side of plate, it would require very many more rivets. Each rivet Covers of same become a thickness. double lever, with its central part built-in and a pro- jecting free arm each side, the length of arms being equal to their respective thicknesses of cover- plates. The load on each arm would be the proportion of whole strain, that the thickness of its respective cover-plate would be of the whole required (aggregate) thicknesses of cover-plates. There would be no sense in such an arrangement however. It would produce all sorts of unequal stresses, in shearing, bearing, cross-breaking, etc., and should be avoided. Riveted work at best is very theoretical, as the calculations depend entirely upon the accuracy and fit of each rivet. If a single rivet fails to do its share, it will at once disarrange all the strains and produce unequal stresses in different parts. Still if the above rules are followed, riveted work can be used with perfect safety. Where the result gives a fraction, a whole rivet should as a rule be used in place of the frac- tion. If the necessary spacing requires still more rivets, they can either be used, or, all the rivets can be reduced in size enough to bring them nearer to the allowable stresses. No account has been taken of the loss due to punching, for this will affect the plate in tension mainly, and the safe stresses allowed for tension are very low. In compression the metal would not be strained as heavily as in tension, for the rivets will not weaken the plate so much if they entirely fill the holes, thus giving full bearing on the entire plate. Then, too, the butt joint if planed and carefully TABLES XXXV TO XL. 299 made and joined, will transfer directly more or less of the compres- sion. But in all good work it is customary to place no reliance whatever on the butt, and to calculate in compression the same No reliance on as for tension, namely, sufficient net area in each plate, at its weakest section, to resist the whole compression strain. Tables XXXV to XL inclusive, have been calculated and laid out to save most of the wearisome figuring necessary in riveted work and in connection with pins. The first three give the Explanation of bearino- value of pins and rivets against eye-bars or Tables. ° , / , , , . ^ . plates, and the latter three the values in tension, single and double shearing and in cross-breaking of pins and rivets. All the tables are laid out for both steel and wrought-iron. The full heavy straight lines in Tables XXXV, XXXVI and XXXVII represent the thicknesses of plates or eye-bars against T^^^l^= vvv» which the different sizes of rivets or pins bear. XXXVI, xxxvii. The thicknesses given are from ^" to If' in iable XXXV and from i" to 2" in Tables XXXVI and XXXVII; all by ^y. For thicker plates or eye-bars it will only be necessary to increase the bearing value found, in proportion to extra thickness. The columns to the left give the diameters of pins and rivets, run- ning in Table XXXV from ^ inch diameter (by inch) to 1 inch diameter; in Table XXXVI from 1 inch diameter (by Jg inch) to 3 inches diameter ; and in Table XXXVII from 3 inches diameter (by ^ inch) to G inches diameter. The figures at the tops of these tables give the bearing values in pounds for wrought-iron, and those along the bottoms, the bearing values in pounds for average steel. The full heavy curved lines in Tables XXXVIII, XXXIX and XL give the single and double shearing values for the same sized pins and rivets as in the previous three tables. _ ^, .., The additional vertical columns to the left in these Tables xxxviiif xxxix, xl. tables give the areas of cross sections in square inches, of the different sized pins and rivets, which multiplied by ten give their weights in pounds per yard of length for wrought-iron, (for steel add 2 per cent to the weight of wrought-iron). There are also full heavy curve lines giving the strength, in tension, of tie-rods of same diameter as pins or rivets. The values selected for these curves are those always used by the writer in calculating pins, rivets or tie-rods. 300 SAFE BUILDING. It sometimes becomes desirable, in temporary work to use higher values, or in very important permanent structures For differently^ with moving loads to use lower values. But even in such cases the tables can be used, for, as all of these curves are directly dependent on the area (or double area), of cross section of the rivet or pin, they can, of course, be used interchange- ably. That is, any one who wishes to figure the safe shearing (^-y^ for steel as = 15000 pounds, instead of 10000 pounds, can take the curve marked " 15000 pounds — tension steel," in any of the three tables. Or, if he wishes to figure his iron at only 10000 pounds safe stress for tension, that is ^-^^ = 10000 pounds, he will, instead of using the curve marked " 12000 pounds — tension wrought-iron," take the curve marked " 10000 pounds — single shear steel." The writer, however, always sticks to the one set of values for ten- sion, and for pins and rivets to those given in tables (also after Formula 115), as they are certainly safe values, and yet not low enough to make the work excessively costly. Iron contractors will frequently quote off-hand opinions of celebrated engineers saying these values are much too low and thus backing up their economical tendencies in trying to add lightness (and beauty) to roof trusses and plate girders, but as a rule the opinions are either not authentic, or else it is found that the celebrated engineer, when delivering the opinion, had a similar axe to grind, as had the contractor. Good engineers as well as good architects, will attempt to save their clients all they can, but hardly at the risk of taking chances in their most important constructive works. The figures along the tops of Tables XXXVIII to XL give the values of either iron or steel, according, of course, to which curve is used. The dotted curved lines in the Tables XXXVIII to XL give the safe bending-moments for iron and steel pins and rivets, and for these the lower horizontal lines of figures are used. The use of the tables is simple, similar to the How to use, other tables with curves. Thus if we have a 4" steel Tables. . . » , rivet bearing against a steel plate, we use Table XXXV and pass horizontally to the right from the vertical figure (or TABLES CONTINUED. 301 diameter marked) |" till we strike the (fourth) slanting full bearing line, marked ^^^g^". This is one-third way between Bearing Values, ^.j^^ vertical lines marked (helow for steel) 4000 and the next vertical unmarked line to the right ; as each vertical space at the bottom (steel) evidently is one-fourth of a thousand or 250 pounds, a third space will, of course, be 83 pounds, or a |" steel rivet bearing against a ^Y' steel plate will resist safely 4083 pounds. Had we calculated arithmetically we should have had |. 15000 =4101 pounds. Had the rivet and plate been of wrought-iron we should have used the upper line of figures ; here the intersection is one-third way be- tween the second and third unmarked lines after 3000 ; as there are five spaces above between each thousand, each space evidently represents one-fifth of a 1000 or 200 pounds for iron, so that the bearing value for a |" iron rivet against a y^g" iron plate would be 3000 -|- 200 200 = 3267 pounds. By calculation we should have had |. Jg. 12000=: 3281 pounds. The single shearing value of a |" steel rivet at ^-^^ = 10000 pounds would be = 3067 pounds ; for, the horizontal line |" in Table XXXVIII strikes the single shearing curve for steel Shearing about one-third way between vertical line marked at Values* the top 3000 and the next unmarked line to the right. Each space evidently represents 200 pounds, so that we should have for the steel |" rivet in single shear 3000 -|- 1. 200 r= 3067 pounds. By arithmetical calculation we should have had the area of a |" circle multiplied by 10000 pounds or 0,3068. 10000 = 3068 pounds. The double shearing value of a f " rivet would be double this, or = 6136 pounds, and this is confirmed by the Table (XXXVIII), as the horizontal line -f" strikes the double shearing curve for steel 20000 pounds about two-thirds way between vertical line marked above 6000 and its next unmarked line to the right, or 6000 -j-f. 200=6134 pounds. For the safe bending-moment we find the horizontal line |" strikes dotted curved line marked " safe bending-moment on steel at 18000 pounds " on the second vertical unmarked line to the left of the one marked at the bottom 500 ; each space below is evidently 20 pounds, and as they increase to the left we must add the two spaces, or 302 SAFE BUILDING. 600 -|-40 = 540 pounds, which would be the safe bending-moment on a I" steel rivet or pin. Or to illustrate further take the last Example (III). We had a Bending- ^" ^^^^^ plate and 1" steel rivet. The actual bend- moment Curve, ing-moment we found was u.l 135000.1 8 8 From Table XXXVITI we see that the safe bending-moment on a 1" steel rivet is the intersection of horizontal line 1" being one-third way between third and fourth vertical unmarked lines to the left of 1 700 = 1 700 -f 3i. 20 = 1 767 pounds-inch. Dividing the actual bending-moment at the joint 16875 pounds- inch, by the safe bending-moment on each rivet, will, of course, give the number of rivets required, or 1^^ = 9,55 1767 being the same result as before, namely, 10 rivets. Take Example IT, we had the same rivet, plate and strain, but there was but one cover- plate and the rivet was practically a cantilever. The bending-moment was u.l 135000.1 , . , = 62500 pounds-inch. This divided by the safe bending-moment on each rivet (1767 pounds-inch) as found in Table XXXVIII gives the number of rivets required, or 62500 „- . , r = 35,4 or same as before. 1767 The use of Tables for pins or tie-rods, is, of course, exactly similar to use for rivets, as already described. Bearing on pins. For instance, we have a 2" pin bearing against a 1^ inch thick eye-bar or head of a tie-rod. We use Table XXXVI, the 2 inch horizontal line, strikes the full slanting line marked 1| inch exactly on one of the vertical lines. If both pin and eye-bar are wrouglit-iron — (or if either were wrought-iron we should use the smaller value) — we use the top line of figures, or we find our vertical line the third to the right of 24000 and as each space evidently represents 1000 pounds, the safe bearing value of our 2 inch pin against a 1^ inch thick eye-bar would be 27000 pounds, if one or both are of wrought-iron. Had both pin and eye-bar been of steel, we should have used the bottom line of figures, where each space PIN JOINTS. 303 evidently represents 1250 pound?, and we should have had 33750 pounds. The safe shearing value for a 2 inch -wrought-iron ^"^p'rlsf °' P^"' ^^"'"^ '^^^^'^ XXXIX 25000 pounds, and for steel 31670 pounds. By calculation we should have had for iron = 25133 pounds and for steel = 31416 pounds. If we had a 2 inch circular tie-rod its strength in Tension on^^^ tension from Table XXXIX would be, if of wrought- iron =37600 pounds and if of steel =47000 pounds. By calculation we should have had for iron = 37752 pounds and for steel =47190 pounds. Pins are calculated similarly to rivets. The same formulfe can be used for bearing area (109) h being the thickness of ^ PUis^ °' head of eye-bar, in inches, and for single shear (110) and double shear (111). For bending where there are but two eye-bars or rods, each pull- ing in opposite directions, Formula (112) could be used; using for h the thickness, in inches, of either rod ; where there are two eye- bars or rods pulling in one direction with only one between them pulling in the opposite direction Formula (113) might be used; h in this case being the thickness, in inches, of the large (central) single rod, and twice the thickness of either of the smaller rods. In all of the formul£e a; should be= 1. As a rule, however, there are several rods at each pin-connected joint, and the bending-moment has to be carefully calculated for each special case. In designing pin-joints this should be borne in mind, and the pieces with largest strains brought as closely together, as possible, to avoid excessive bending-moments, which ^^^'hTintf require very large pins, and necessitate large eyes and heads to the rods and bars, which means, of course, very great expense; pin-connections for small trusses are more expensive than riveted joints ; for large trusses they are cheaper. They are very much better in both cases, Pin-Joint best. , . ^ . ^ , more easy to transport and erect, and agree much nearer with the theoretical calculation, which assumes that all members -around a joint are free to move, and not rigid, as is the case in riveted trusses. Then, too, in case of any movement in the truss, it can be readily adjusted by means of nuts, swivel-link^, 304 SAFE BUILDING. sleeve-nuts, etc. This cannot be done in a riveted truss, as the joints not being flexible, the tightening of any one part might throw strains on some joint not able to bear it. In calculating a pin-joint, the strength of pin, as a rule, should be calculated in the several lines or directions in which it is being strained, that is first along the line of one rod, then along the line Strains another rod (or compression piece) and so on. In reduced to doing this all the other strains must be projected (and reduced) to the same line. If this is correctly done, it will be found in every case, that the sum of all the strains acting in one direction along this line will be equal to the sum of all act- ing in the opposite direction along the line. Thus in Figure 1 77 we have the end elevation of a pin A strained by three forces in tension B, C and D and one force in compression E. We will first examine the pin along the line A C. Make, at any convenient scale, the lines A C, A B, A D and A E, each in length just equal to the amount of strain they are subjected to. Project them all on to A C, then will A b represent the resultant or equiva- lent of strain A B along the line and in the direction A C. Similarly A d will be the resultant of ^ D along the line of A C, but it will be in the opposite direction ; e A will be the resultant of E A along the line A C, but it will be noticed that though on the opposite siae of the pin it is in the same direction as A C. STRAINS ON PINS. 305 Were we to examine the pin along the line A D, we should have the strain of A D in one direction and in the opposite directions the resultant strains A c, and A b,. It will be noticed that E A has no resultant along the line A D, being normal to it ; and it need not be considered, when calculating the pin along the line A D. We might still calculate the pin along the line ^ we should then have the strain E A in one direction and in addition thereto the resultant of A C ; in the opposite direction we should have the resultant of A B. The largest strain A D has no resultant along the line A E. It is evident, however, that all of these resultants will be very small as compared to those along the other lines, and therefore need not be calculated. In Figure 178 we have a plan of the pin, showing in plan the strains and resultants acting along the line A C. As arranged in this figure the pin evidently becomes a double-ended ^"oPHeadV!* cantilever, supported at the fulcrum d, and loaded on one free end with the load c and on the other free end with two loads & and e. If the strains were small, this would probably be the most econom- ical arrangement, as A D could then be made in one piece. But if the strains were heavy it would require a tremendous pin. In the latter case the arrangement shown in Figure 179 would be better. Here the pin becomes a beam, supported at both ends by ^ (or one-half of the force A D) ; the beam carries three loads c, e and b. In this case A D would be made up of two rods, separated. The forces c, e and b must be so distributed as to make the least FIG. 179. bending-moment on the beam or pin, that is, the larger ones placed nearer the supports or outer edges, and the smaller ones in the centre. Frequently it would be more economical to divide c the 306 SAFE BUILDIKG. larger force into two halves and place them immediately next to £ The small force e is frequently put on the outside, as it is not sufficient to afEect the beam or pin seriously and only enlarges the span of beam, that is, length of pin or distance between supports - thus greatly increasing the bending-moment. This arrangement is shown in Figure 180. Here we have a beam supported at two points ^with a free end carrying load e ; the beam being loaded with three loads, two each = I and one = 6. To calculate the bending-moment at any point of this pin we have to consider the end of pin farthest from e as built in '*symmetVical solidly (and after getting reactions) we should multi- Strains. ^ ^ijg forces to the right or left of the point into their distances from the point. If we select the forces on the right we deduct the sum of the moments of those (right hand forces) acting in one direction from the sum of the moments of those (right hand forces) acting in the opposite direction, the difference will be the bending-moment at the point. To check the calculation we take all the forces on the left side of the point. But the reactions will not be ^ each as shown in the table unless e were divided and one- half put at each end of pin. In every case of pin calculation, excepting where the forces form simple beams, as in Figure 179, the forces are sup- ''through Cen- posed to act through their central axes, that is tral Axes. through a line drawn at half the thickness of their heads and at right angles to the pin. The heads are frequently thickened up, that is, made thicker A^^. ^ ^ than the rod or flat bar, in order ^ ' to get the necessary bearing sur- face on the pin, but it will be readily seen that this lengthens the pin y-^ greatly, thus largely increasing the bendino;-moments. It is better, as a rule, to get the extra bearing FIG 180- surface by dividing the rod or flat bar into two or more parts and then distributing them along the ARRANGEMENT OF EYE-BARS. 307 pin symmetrically and at the most favorable points to avoid bending- monient. Forces arranged as shown in Figure 180 and arranged Un- we call the different lengths along the pin, as shown, /, I, and Z„ the reaction nearest e will be symmetrically. and the further reaction will be _c-\-h In practice, however, e would probably be so small and the con- venience so great in making the two bars | alike, that the unequal stresses caused by e in the two ^ would probably be overlooked. But in calculating the bending-moment on pin, they will have to be 2 2 y considered as unequal, otherwise, the bending- moments calculated from the right hand or left hand of any point would not agree. To put the above in formulae we J should have the following : If a pin p q IS strained by a number of forces a;,, x,,, etc., in one direction, the forces p and q in the opposite direction will equal the reactions of a beam, see P'G- '81 . Formulae (14, 15, 16 and 17). If a single force y is placed beyond one of these resisting forces (say q Figure 181) the additional reaction on the nearer force (q) will be Nearer Reaction „ _i_ ^ Force. and on the further force Further Reaction Force. ^' — (116) (117) or, we must add 7, to q and substract jo, from p to get the real strains or reactions in the tie-rods p and q. Where 7, = the strain, in pounds to be added to nearer strain q owing to force y being placed on the (5') nearer free end of pin. 305 SAFE BtTILDIXG. Where p, — the strain, in pounds, to be deducted from further strain p owing to force y being placed on the other further free end of pin. Where q = the force (reaction) at q, in pounds, resisting the forces a;,, x„, etc., (see Formul83 14, 15, 16 and 17) and to which must be added. Where p = the force (reaction) at p, in pounds, resisting the forces x,, x,,, etc., (see Formulae 14, 15, 16 and 17) and from which must be deducted. Where a;,, x,, = the forces, in pounds, acting in opposite direction to^ and q, and all projected to line^ and q as shown in Figure 177. Where I, Z„ = the respective lengths, or distances, in inches, measured along pin, from centre lines to centre lines of respective pins, as shown in Figure 181. As the forces x,, x„, etc., should always, if possible, be located along pin to make the resisting forces p and q even, which is done by putting the smallest force (a:,„) in the middle and dividing the others, we should have, where tliis is the case: Nearer Reaction Q_ 'Z.x ,l CllS") Force. ^ 2 " Z, Further S. a; _Z„ ^ Reaction Force. ^ — ~2~" 1/ ^ ^ Where S a;=the sum of all the opposing forces (a;,, a;„, a;,,,, etc.,) to and between p and q, in pounds, provided that they are divided and the halves or fractions located symmetrically with respect to p and q. Where 7; = the total resisting force (reaction), in pounds, or strain on p, the reaction furthest from free end. Where the total resisting force (reaction), in pounds, or strain on q, the reaction next to free end. Where Z, Z„ Z,„ and ?/=: same as in Formula5 (116 and 117); of course, all the forces must be projected to one line as shown in Figure 177. If one of the forces, a;,, x^„ etc., were = y and were placed to the left of p (Figure 181) the forces p and q would be equal and each = one-half of the sum of all the other opposing forces, provided always that the forces a;,, a:„, etc., are symmetrically located in respect to p and q. Where there are a number of forces on both sides of pin, the pin might be treated as a continuous girder (see Table XVII), PIN OR BEAM WITH OVERHANGING END. 309 but the calculation would become very difl&cult and the parts of all rods (or compression piece) acting in the same lines and direction would become very irregular. It is customary, therefore, to locate the forces along the pin sym- metrically, regardless of their true resistances as Locate Forces ^j^^ would be if treated as continuous girders, and Oj m lYi Gt ncai I y • ^ to consider, that each takes its proportionate load (according to the thickness of its head) of the whole load along its respective line and direction. In each case it will require special study to obtain the most economical arrangement. An example will more fully illustrate the method of calculating pins. Example IV. The Joint A of a pin-connected wrougJit-iron truss Calculation of strained hy the followinn members: The tie-rod Pin-Joint. J J J B = — 28000 pounds : the tie-rod C — — 70000 pounds ; the strut D = -\- 20000 pounds ; and the tie-rod E = — 88000 ^owrwZs. All as shown in Figure 1^2; design the j oint. We will assume that for certain reasons we wish to use a 2|" diameter pin. Now the first thing to do is to settle the thickness of the (eye) heads. These, of course, must have sufficient bea,ring against pins not to crush the pin or be crushed by it. We use there- fore the following formula : Thickness of t Heads. Where t =r the necessary thickness of eye-bar head, in inches. Where s =i the strain on each eye-bar, in pounds. Where fZ— the diameter of pin, in inches. Where (^^-^ = the safe-compression stress, per square inch, of the material of eye-bar or pin (whichever is the weaker in resisting crushing should be used.) Accordingly we should have for thickness of the different eye-bar heads in our example the following : AB= -J^-^ = 0,85 or say f " 2|. 12000 ' ^ * • ^fe-/^=2,12orsay2". 23.12000 ' •' 310 SAFE BUILDING. J. A 20000 „ „ ^^ = 2-#T2(ro-o=''' '^^^ AE= 88000 2,67 or say 2f" 2J.12000 The values for above thicknesses have been rather too broadly rounded off, but this is done to simplify the subsequent calculations. 1 \ 1 V \ \ 1 \ / ^ 2.8000, ^^^), K. I I I s I 2 s ' I I I I n I I I SCAUC orPOUHDS. FIG, 182. Had we used Table XXXVI we should have had the same results. For A B (28000 pounds) the horizontal line 2|" and """r^bie XXXVI. line 28000 (from above for iron) meet be- tween the heavy bearing lines {f " and |", for con- venience, however, we will call itf" though this should not, of course, be done in a real calculation. For A C (70000 pounds) horizontal line 2|" and vertical line 70000 from above (the second to the right of 68000) meet just be- THICKNESS OF BAR AT EYE. 311 yond the 2" heavy bearing line. It should be, therefore, a little over 2" thick, but we will call it even 2". For D A (20000 pounds) the horizontal line 2|" and vertical line 20000 from above, meet between heavy bearing lines and |", we will call it I". To find A E (88000 pounds) which is larger than A C, we shall evidently have to divide it in halves, and, of course, double the result. We find that horizontal line 2f" and vertical line 44000 from above, meet between the heavy bearing lines l^^" and If", or we will say If" ; this doubled or 2f" is the required thickness there- fore, for head A E. Of course, if we use more than one bar for either of the strains we will divide the required thickness of head accordingly. Thus, if we decide to use two bars along the line A C, each would be strained — 70000 — 35000 pounds, and the thickness of head required for 2 each would be only = = 1,06 or say 1". Now to arrange the different heads along the pin, we first lay off along each line (Figure 182) the amount of strain o^HeadTarong (measuring all at the same scale) and project these strains on to the different lines. We measure these projections and have along the line A B the strain A B = 28000 pounds pulling to the right; the projection of A C= 34500 pounds E = 75500. •D=8500 73500. D=11000. LIHE A.-B B= 14 000 C= 34-30O LTOE A-C. FIG- 164-. C= 70000 D=J18000. FIG. 183. also pulling to the right; also the projection of -D^ = 11000 pounds pushing to the right ; the projection of ..4 -E= 73500 pounds pulling to the left, thus opposing the other three.^ If our measurements are right the sum of the forces acting in one 1 These figures have been rounded off to simplify the calculations. 312 SAFE BUILDING. direction along line A B, must equal the sum of their opponents, or, AB^ A C-^DA — A E and we have in effect, 28000 + 34500 + 11000 = 73500 The strains on the pin along line A B are shown in Figure 183. Those along line A C are shown in Figure 184 and those along A E E=6aooo ■D=20000 B= 16000 f -UKE A. E. LIME A-D B-a+ooo ■p'lG-. 185 C=3600O. are shown in Figure 185 ; it will be noticed that in the latter ease D A becomes = 0. In Figure 186 are shown the strains along D A, in this case A E becomes = 0. As the largest strain in one direction (88000 pounds) is along line Design for ^ "^^ "^^^^^ ^^^^^^ Figure 185 to design the joint largest Strain and when we have settled the arrangement of heads along the pin to suit these strains, we will see how it affects the pin according to the strains along the other Hnes. The simplest arrangement of the parts would be evidently that shown in Figure 185. We will first consider the shearing. The largest shearing strain will be between C and E and will be = 64000 pounds. From Table XL we find for wrought-iron, single shear (at 8000 pounds per square inch) that we should need a diameter pin to resist 64000 pounds single shear (as each of the vertical spaces at the top evidently represent 10000 pounds), we must pass down vertically not quite half-way between the second and third lines to the right of 50000, till we strike the single shear iron curve and then pass along the horizontal Hne to the left to Shearing strain find diameter of pin, which is between 3A" and 34-" too greati <, tt or say 3^%". Had we calculated arithmetically we should have had, area of cross-section required from Formula (7) transposed 64000 o • 1 a = --— ~- = 8 square inches. oOOO By referring to a table of areas of circles, or by calculation we find SHEARING STRAIN ON PIN. 313 that for an area of cross-section of 8 square inches we require a diameter of 3 Jg". This is a larger pin than we want to use and besides seems a very large pin for the strains ; it is evident, there- fore, that our heads are badly arranged along the pin ; we will decide, therefore, to divide the rods E and C each in two parts, making each head one-half the thickness as above found, and arrange them as shown in Figure 187. L -) E. 44000 E 44OO0, I i 1 1 1 1 1 1 1 - t - 1 1 V [ ' 1 1 1 1 C.3ZOOO 1 ^C.= 32O0O 1 FIG. laz Now the safe cross-shearing on our pin (2|") for single-shear would be from Table XXXIX = 47600 pounds, we ^Table xxxix. P^®^ ^^°"S horizontal line 2J" till we strike iron single-shear curve and then pass upward about four- fifths way between the third and fourth vertical lines to the right o£ 40000 ; as each space is evidently 2000 pounds, we sliould have 40000 -|- 3|.2000 = 47600 pounds. By calculation we should have had area of 2|" circle = 5,939 square inches, therefore safe (single) cross-shearing see Formula (7) =5.939.8000 = 47512 pounds. The greatest cross-shearing strain on the pin with arrangement as shown in Figure 187 is 44000 pounds, and is between the heads E and C (or E^ and C,), so that we need not fear shearing. The shearing area of pin being all right we no\v consider the bend- 314 SAFE BUILDING. ing-moment. We have marked along the pin, Figure 187, the thicknesses of heads, the length of pin required Bending- beins 6A'', to this must be added the head and nut monfient on pin. ° ^ ' and also sufficient for strut D (|") which we remem- ber did not come into the calculation along line A E (Figure 182.) Immediately under the pin, Figure 187, we have marked the distances from centre to centre of heads, which are, of course, the distances we consider when calculating the bending-moment. Accordingly our pin becomes a circular wrought-iron beam of 2|" diameter, with a span or length of 4^" and supported at both ends by forces E and E^. The beam is loaded with the forces C, C, and B as shown in Figure 187. The greatest bending-moment will be at the centre (See p. 51, Vol. I), and will be rriB = 44000. (1 -/g + |) — 32000.| — 24000.0 = 90750 — 28000 — 0 = 62750 pounds-inch. This will be much more than the pin can stand for we have, for the safe bending-moment on a 2|" pin, moment of resistance ^ = (If)' =2,042 (Table I, Section No. 7) and from Formula (18) transposed, the safe bending-moment on pin m = 2,042 .15000 = 30G30 pounds-inch or, only about one-half of the actual bending-moment. Had we used Table XXXIX instead of calculating arithmetically rmTmeru from we should have passed along the horizontal line 2|" Table XXXIX. till we struck the dotted bending-moment curve for wrought-iron at 15000 pounds and then passed vertically to the bottom. This would be about two-fifths way between the vertical lines 30000 and the first one to its right, each space being evidently 1500 pounds, this would mean 30G00 pounds-inch safe bending- moment on a 2|" pin. It is evident, therefore, that we must re-arrange the rods, trying to get the span between loads E and E, shorter if possible. Try new ^^^^ arrangement shown in Figure arrangement. 188, placing load B at one end. The arm end B of pin now becomes a lever and we know from Formulse (118 and 119) that the reactionary forces E and E, can no longer be equal. BENDING MOMENT ON PIN. 315 From Formula (118) we have _ 32000 + 32000 , 4^^ ^ 2 ^3f = 63555 From Formida (119) we have _32000+_32000_ljV 240OO '~ 2 3f = 24445 As a check the sum of the two should equal the whole strain A E and we have in effect G3555 + 24445 = 88000. This arrangement must evidently be abandoned as a bad one, for the difference in the strains on E and -B, is altogether too great to E-4-4000. E- 4-4:00O» I L- 1 B-Z400C 1 " _ 4 ifit C= 320OO. li-'l __ Jl._-^ ^ 1 1 FIG. laa. be overlooked. Besides, we can readily see that the moments at E and C, will exceed 24504 pounds-inch, the safe bending-moment on this sized pin. For practice, however, we will figure out the bending-moments ; they are : at E (left side) 24000. 1^ — 63555.0 — 25500 pounds-inch. Check (use right side) = 32000. 2^5 + 32000. 1 — 24445.3f = 25500 316 SAFE BUILDING. At C we should have : (Leftside) G3555. 1 — 24000. 2^ = 21472 Check (use right side) 7?ic = 24445. 2Jg — 32000. 1 : - 21472 At C, we should have : (left side) wic. = 63555. 2^8^, _ 24000. 3;^ — 32000. 1 ' = 29027 Check (use right side) mc,= 24445. 1 — 32000. 0 = 29027 Had we used rule given on p. 51, Vol. T, we should have known that the greatest bending-moment was at C,. In applying the rule to this case the end load B should be deducted from the nearer reaction to B. We might next try dividing the force B in two halves of 12000 pounds each, (heads f" thick), leaving one at B and placing the other to the right of E^. This will restore equality to the forces E and E^, but it will be found that even this arrangement will not do, as the bending-moment at C or C, will still be found to be too large, namely, = 27500 pounds-inch. After these numerous failures it is evident that we cannot well arrange the heads satisfactorily along the pin, unless Dlvid^ largest^^ enlarge the pin, or else divide up the larger forces which cause us most trouble. We decide to do the latter and divide A E into four parts of 22000 pounds each, with heads each = — thick. 4 16 We now arrange the heads as shown in Figure 189. The correct way to calculate the bending-moment would, of course, be to consider the pin as a continuous girder running over four supports. This would make J?, and iJ„ much larger than E and 2?,„. Their heads would therefore have to be thickened so as not to crush the pin, or be crushed by it. It can readily be seen that were Forces will , equalize them- we to do this, the calculation would be almost inter- selves. mj„a,ble. Besides, practically, it would be expensive to use so many different sizes of rods, heads, etc. We must, there- fore, assume that all the forces E, E^, E„ and i^,„ are equal. This NEW ARRANGEMENT OF BARS. 317 they will be, too, for as soon as E, and E,, tend to take more load, they will stretch under the added tension, and this stretching will bring the pin to bear more heavily on the ends and thus the strains will even themselves up. The bearing of these heads against the pin we know are all right, also the shearing, as the greatest shearing under this arrangement E=2.zooo. Ei-22000. Ejj=aaooo. Ejj£=aaooo. I 32. _Z2 i._?:3 _ ■ 32 T 32. _Zi i £7 02 "T" 32,' 32. ^ C= 32000. B= 24-000. PIG 183. C =32000. will be a single shear, between E and C (or and C.) and equal 22000 pounds, or considerably less than half the safe single shearing on this pin, which we previously found to be 48000 pounds. The bending-moments on the pin will now be, at C : (Left side) OTc = 22000.f^ — 32000.0 =: 18563 pounds-inch. Check (use right side) me = 22000. ( ^I+i^+l^ ) _ 24000.^ — 32000.J^ = 18563 pounds-inch. Ao ii, we should have : (Left side) m^, = 22000.f| — 32000.f^ = 10125 pounds-inch. Check (use right side) m^, = 22000. (^iil^ ) - 24000.-11 - 32000.^ =rz 10125 pounds-inch. 318 SAFE BUILDING. At B we should have : (Either side) = 22000. _ 32000.ff = 18750 pounds-inch. At the ends, of course, there would be no bending-moment, for take end E we should have : = 22000. ( ^^^ + + -32000.(^-^^) - 24000.- = 0 This arrangement (Figure 189) is therefore satisfactory, so far at least as the strains along the line A E are concerned. We must now see if it will answer as well for the strains along the other lines (See Figure 182). The direction we ^''^otheMinel. ^^^^^^ ^^^^ ^i^^ be along the hne D A for force D must be placed entirely on the out- side edge of pin (not having been located yet) and being quite large, E=0. C=1j5000, E=0- C,-16000 .1^-4- J, 32 aa ~x 3a FI&. 190. D» lOOOO. ' 27 ' Z7 I Ife B= 16000. Dj?l10000. 20000 pounds, may make us some trouble. In Figure 190 we have drawn the forces arranged the same as we settled on last (in Figure 189) but have added the two forces on the extreme ends D and Z), It will be noticed that along this each = = 10000 pounds line {DA Figure 182) the forces E and Care in the same direction. We have divided D into two parts for two reasons. Had we placed it entirely to one side, say to the right of j&,„, the distance D, would have been one-quarter of |" larger or = |i" ; the leverao-e of STRAINS ON OTHER LINES. 319 27 —I- 21 D, therefore at C, would have been — — = i^" and the bending- o ^ moment mci= 1^ 20000 = 30000 pounds-inch, too much for our pin. Besides were we to calculate C and C, by Formulae (118 and 119) we should find all strain on C removed and C, more than doubled. This evidently would not do, without special provision to meet the unequal strain by increasing C„ which would lengthen the pin, so that we prefer to divide D A, making each head of half the thick- ness, or j^q" thick. The largest shearing will be between D and C (or D, and C,) and equal 10000 pounds single cross-shear or about ^ only of the safe resistance to shearing. The bending-moments will be (Figure 190) at C. (Left side) mc = f|.10000 — 18000.0 1=13437 pounds-inch. Check (right side). mc = W-10000- = 13437 pounds-inch. and at B (Either side) = ||.10000 — 1^18000 =z 9375 pounds-inch. The bearings, of course, are safe, as the thickness of head was origi- nally determined If. 16000 — ^2^^.18000 JET- 18373 Sir* — tt ■■ 3z T by the largest strain on each rod along its own line. So that we are all right with our pin for strains D» ssoo along line D A. We now take 33350 017230 B^ZOOOO C»"t7J .FIG. 131. up the strains on the pin along the hne A B, Figure 182, which will be as shown in Figure 191. The greatest shearing-strain here will be caused by B, and will be a double shear of 28000 pounds, or 14000 pounds on each area, perfectly safe on our size of pin (2|'')- The moments will be : 320 SAFE BUILDIXG. At E (Left side) Toe = ^.5500 — 18375.0 = 2750 pounds-inch. Check (right side) = 5500. V/ + 1 ^250. ( ^^ + ^^^ ) + 28000.|| — 18375. rrr2750 pounds-inch. At C: (Left side) TOc = 18375.f| — 5500.f| = 8113 pounds-inch. Check (right side) Too =18375. ( ^^ 5500.-i^3._i7250.1^'y>— 28000.f^ = 8113 pounds-inch. D=4-E50.E= 18875. 18575. 18873. 15675 P=42.50, . &7 3a T "32: ~t- C=350oo. B=-l4rOOO. C =35000. FIO. 192. At E,'. (Left side) m,. = 18375.11 — 5500.11 — 1 7250.f 1 = 4422 pounds-incli. Check (right side) = 18375.(^Ai_i^) - 28000.|f - 17250.^f - 5500.V^ = 4422 pounds-incli. BENDING-MOMENT GRAPHICALLY. 321 At 5: (Either side) 7713 = 18375. - 17250.ff - 5500.|| = 14484 pounds-incli. So that the arrangement of heads along pin is all right so far as strains along line A B (Figure 182) are concerned. We finally examine for the strains along line A C and have strains FIG. 193. SCALE OPPOUMDS- X-tTil iriHrliilii.l4iJil,lilil III SCAIXOP IHCWP.n. on the pin accordingly as shown in Figure 192. The greatest shear- ing here will be between E and C (or C, and £J„,) and will be = 4250 -{- 18875 = 23125 pounds, single shear, still, less than one-half of the safe single-shear on the pin. By calculation the moments will be found to be at the different points, as follows : 7We= 2125 pounds-inch. Too = 21G87 pounds-inch. mjs,f= 11617 pounds-inch. ma = 16648 pounds-inch. 322 SAFE BUILDING. So that this arrangement of the different bars and strut along the pin is in every way satisfactory. The graphical method of obtaining bending- mTthod more moments is frequently much more simple than the simple, arithmetical method,; in important calculations both should be used so as to check each other. All the rules for formulje given in Chapter VII for the calculation of transverse strains by the graphical method will apply equally well for pins ; the only difference will be that where there are more than two forces on each side of the pin, the base line of the polygonal figure between reactions p and q will no longer be straight, but will be composed of several lines. Thus, if we take the pin and forces shown in Figure 189, we should change the notation to that adopted for the graphical method, which would be as shown in Figure 193. That is force E (22000 pounds) of Figure 189 would be known as J. £ in Figure 193 ; again force C, (32000 pounds) of Figure 189 would be called force E F (not FE) in Figure 193. ^ I 1 w Proceeding now to the calculation, we lay off along Example solved <=> ' ^ a graphically. vertical load line a e, the following forces : ab = AB = 22000 pounds. b c = B C = 22000 pounds. c d=C D — 22000 pounds. d e = DE = 22000 pounds, and in the opposite direction, we lay off : ef=E F= 32000 pounds. fg = FG = 24000 pounds. g a= G A = 32000 pounds, which will, of course, bring us back to the starting point a, as the opposing forces must aggregate the same sum. We now select our pole o. This we remember can be arbitrarily located, or else at a distance o c — (^y^^; in our case we will make the distance, say, 12000 pounds. The distance (of pole from load line) being arbitrary we shall multiply the verticals v (inch-scale) in Figure 193, Pole stance. ^^^.^ ^^^^ distance (pounds-scale) to obtain the bending-moments at the points of pin immediately below verticals. If the pole distance from load line had been made = ^, then the GRAPHICAL METHOD. — CONTINUED. 323 length o£ verticals v measured at inch-scale would have been the required moments of resistance of the corresponding points of pin below verticals ; and each respective v multiplied by ^-^^ would be the bending-moment at each point. We will, however, make in our case the pole distance, arbitrary, viz: c o = 12000 pounds. We now begin at any point of line A B and draw A I parallel b o till it inter- , _.. sects B C Sit I : next draw I L parallel c o to intersect Strain cliagram> tion with CD at Z ; and similarly draw L E parallel do; E K parallel o e ; KJ parallel of; J H parallel o g and H A parallel oa ; the last line must intersect the first at point of starting A or some error has been made. It will be noticed that the in lividual outlines oi A I L E KJHA cover the capital letters in Figure 193, corresponding to small letters from which their respective parallel Unes started in strain diagram. Thus, for instance A I covers letter B and is parallel to ho; similarly I L covers C and is parallel to c o ; L E covers D and is parallel do ; KJ covers F and is parallel of; J H covers G and is parallel o g ; similarly we can consider E K as covering E and it is parallel o e ; and as covering A and it is parallel o a. We now measure the verticals through the figure A I L E K J H A, Curve of bend- longest, of course, will give the greatest bending- Ing-moments. moment. This happens to be the central one / /, it measures l^^g", therefore mj = 1 .12000 = 18760 which corresponds to of Figure 189. Similarly we should have : m, (formerly m^^ =:f|.12000= 10125 pounds-inch. jWh (formerly mc) == 1^.12000 = 18563 pounds-inch. Had we analyzed the strains on pin as shown in Figure 192 graphically, our verticals would have measured, at .• at C: at E,: and at B 324 SAFE BUILDING. The corresponding bending-moments would have been : at E: wij. = ^8_.i2000 = 2250 pounds-inch. : li|.12000 = 21750 pounds-inch. 1^.12000 = 11625 pounds-inch. at C: at E, : at B: = 1|.12000 = 16500 pounds-inch, which are very close to the correct moments, which we found arith- metically to be : = 2250 pounds-inch ; rwo = 21637 pounds-inch ; mE, = 11617 pounds-inch; and = 16648 pounds-inch. The simplest method of calculating pins, as a rule, will be — after calculating (or ascertaining from Tables) the safe bearing and shear- Simplest stresses of the pin, — to calculate the actual method use moment of resistance of the pin, see Table I, Section moments of -^o. 7, fourth column. Now proceed graphically, resistance, being sure to make the pole distance in every case equal to the safe modulus of rupture ^ ) material of pin. After this it will only be necessary to see that none of the verticals through the different polygonal figures (corresponding to A IL E K J H A oi Figure 193) — that none of these verticals measure at inch- scale more then the actual moment of resistance of the pin. If this is done the calculation and selection of the best arrangement becomes very simple and easy. After the final and best arrangement has been determined on, it would be well to calculate arithmetically the moments as per this final arrangement, thus checking the graphical solution. The writer has frequently been told by contractors that owing to Contractors friction due to the pressure of the nut and head, claim of no that it was impossible for any bending-moment to rnoment t^ke place on a pin. As well it might be claimed ridiculous, that owing to the pressure of the walls, there is no bending-moment on a built-in beam. The safe modulus of rupture of a built-in beam can be assumed higher same as we do for pins, but the beam will break across if too heavily strained and so will the pin. Besides it must be remembered that the heads of eye-bars liecessarily cannot fit the pins perfectly ; and even if the argument BENDING ON RIVETS. 325 were correct, (which it is not,) that friction offsets the bending- moment, the least rusting of the joints would diminish the pressure between nut and head, thus destroying its value. Again, contractors will admit that there exist bending-moments on pins, but will deny it in the case of rivets, though the cases are pre- cisely analogous; this latter argument though, if sifted, will generally lead the contractor to admit that its real basis is the large number of rivets it frequently requires ; and the less rivets he can get along with, the happier will your contractor be. CHAPTER X. PLATE AND BOX GIRDERS. WHEN it becomes neces- sary to cover floors or spaces of such, large span, or to carry loads so heavy, that rolled-beams will not answer the purpose, girders are made up of plates I and angles riveted together, ! I and are known according to ' FIGr 195 their section as " riveted plate FIG-; 19 4-. girders " with single webs (Figure 194), "or riveted box girders" with two or more webs (Figure 195). As a rule, too, riveted girders of equal strength can be more cheaply manufactured than the heavier sections of rolled beams. In the case of the former the vertical plate or web has two angle irons riveted horizontally along its entire top edge and the Single web same along its bottom edge, or four in all. In verv girdersi ° ° ' light construction the free legs of the angles might answer for the top and bottom flanges, but, as a rule, a plate is riveted to these free legs, at right angles to the web, both top and bottom, thus forming the flanges of the girder. This plate need not necessarily extend the entire length, but it usually does. Where the thickness of flanges required is very great, say one inch or more, each flange is made up of two or more thicknesses or layers of plates. In such cases only the layer nearest to the web is carried the entire length, the other layers gradually decreasing from the point of greatest bending-moment (usually the centre) towards the ei^ds. DESCRIPTION OP GIRDERS. 327 To carry all the layers to the ends in heavy work would be a great extravagance, the only advantage gained being a slight increase in stiffness, which can be very much more readily and economically gained by increasing the depth of web. In double web box girders only two angles are attached to each web, one at the top and one at the bottom, both on Box girders. ^j^^ outside surface of each web. To place angles on the inside surface is impracticable, as webs would have to be placed s^ufficiently far apart for the " holder-up " to crawl in, and the rivet- ing would not only be weak, having to be done by hand, but it would weaken tlie flange by just so many additional rivet holes. In short girders with heavy loads, where shearing is the main danger, box girders with three webs are sometimes made ; in that case the middle web has the usual four angles, but the two outside webs only two angles each. Beyond the additional stiffness sideways, in resisting lateral flexure, there is no particular advantage in using a box girder. It is more clumsy to handle and to make, and not readily painted on all exposed surfaces, and besides is more extravagant of material in proportion to its strength. Where the flange is of great breadth and it becomes desirable to have two or more webs, the writer always prefers to use two or more single web plate girders, and to secure them together with bolts and separators, or by latticing the top and bottom flanges, together. The angles need not necessarily be even-legged ; nor need the web necessarily be of same depth throughout, nor of same thickness throughout. It will, however, greatly simplify the calculation to keep the web uniform throughout, and in most cases the extra labor involved in varying the thickness of web, would more than offset the cost of the unnecessary material at the centre. Where girders are very deep, the web is made in sections or panels, as already explained. In such cases the web can readily and economically be thickened towards the ends. Whenever possible, the girder should be cambered up at the centre an amount equal to the calculated deflection. Camberln^s^^^ But as girders are usually made of straight plates, ^ and machine riveted and punched, the cambering is rarely practicable. Should the girder, however, show any bending or cambering in transportation, the architect should be sure to have the cambered side placed on top. 328 SAFE BUILDING, The calculation of riveted girders is exactly the same as for iron beams, but has the additional element of the number and location of rivets to be looked into, also the stiffness of web and overhand of flanges. The reactions, vertical shearing, bending-moments, actual and required moments of resistance, deflections, etc., can be calculated arithmetically by the rules given in Chapters I and VI ; or graphi- cally by the rules given in Chapter VII. The rules for calculating riveted work were given in the previous chapter (IX). The only new matter is to find what the strain on the rivets will be. It will be readily seen that when a plate girder f?ange*straln. loaded the tendency of the flanges and angle irons is to slide horizontally past the web (see Figures 120 to 125). This tendency to slide is called the horizontal flange strain. The rivets connecting angles to web resist this tendency and there must be sufficient rivets to do this safely. The total amount of this tendency to slide or horizontal flange strain between any selected point of girder and the nearer end of girder, is equal to the bending-moment at the selected point, divided by the depth of web of girder at the point, or d (121) Where s — the total strain, in pounds, coming on all the rivets connecting either top or bottom flange to web, between any selected point of girder and the nearer end. Where TO = the bending-moment in pounds-inch, at the selected point of girder. Where d = the total depth, in inches, of the web of girder at the selected point. The above strain s will exist in both top and bottom flanges and will be resisted by all the respective rivets in either top or bottom flange that connect the angles to the web. It should now be ascertained which is the weakest resistance of Number of ^^^^ rivet, whether it be to bearing (compression), rivets in web leg to shearing, or to bending, and this weakest resist- or angles. ^^^Q divided into strain s, as found by Formula (121) will, of course, give the number of rivets required along each edge of web between the selected point and the nearer end of girder. NUMBER OF RIVETS REQUIRED. 329 Frequently many more rivets will have to be used than are required by calculation in order not to exceed the greatest pitch for rivets given in Formula (107). To ascertain the number of rivets required (along either web edge) between any two points of girder, we will, of course, take the difference between the numbers required from each point to end of girder. In the flange leg of angles usually fewer rivets can be placed than in the web leg, though many good engineers frequently make them equal in number. But this is really unnecessary, for even if the strains on the flange rivets were the same as those on the web rivets (which they are not) we should still have two rivets in the flange to one in the web on all single plate girders. There seems to be considerable difference of opinion as to just how to figure the strain on the flange rivets. The rivets in flange best course would seem to the writer to be, to leg of angles, g^ggm^g ^jjg^^ each flange cover plate must transfer at each of its ends, by rivets, to the angle iron and parts of flange plates between it and the angle iron, an amount equal to the safe stress the plate is capable of exerting (that is, net area of cross section of the plate multiplied by either the safe compression stress ^or by the safe tensional stress ^^^^ b^). This amount should be transferred by sufficient rivets, between the end of each plate, and the point of girder at which the full thick- ness of the plate is required to make up the required moment of resistance. From this point to centre the rivets can be spaced according to the rule for greatest pitch, Formula (107), but when rivets are so spaced the pitch of the rivets immediately nearest the ends of any cover plate should be greatly decreased for a distance of three or four rivets at each end. By the above method the amount of strain on rivets can be quite accurately computed. The simplest method of locating rivets is to construct what might be called the curve of moments of resistance. This Locating flange ga^Q be done as shown in Figure 151, Chapter VII rivets. ^ (where C D E F G C is the curve of moments of resistance), or we can calculate arithmetically the required moments of resistance at several points of girder, and lay out the curve as 330 SAFE BUILDING. shown in Figure 196, wliere A B represents the length of girder, and M C, I D and N E the calculated required moments of resist- ance at points M, I and N. The curve of moments of resistance is, of course, A C D E B, and its axis or base We now make I H=a,d see Formula (99) ; a, being the net area of cross-section in square inches of two angles, and d the total depth of girder in inches. HD will now represent the total required thickness of flange plates, which we can find from Formulse (36) or (98). We will decide to divide it into three layers H G, G F and F D. We draw the lines as shown and find that plate No. 1 can stop at K, though it would be better to run it full length, it is, however, needed of full thickness at J. Again plate No. 2, can stop at /, but is n I n needed full thickness at L. The ton plate, of course, will run from L to centre. The left half of girder, will of course be similar, the loads evidently being symmetrical each side of centre. In practice the plates rarely are stopped at the exact points calculated, but are usually extended beyond these points a distance equal to from once to twice the width of plate. There must now be rivets enough between D and L to equal the efficiency of plate No. 3, between L and J to equal the efficiency of plate No. 2 and between /and K to equal the efliciency of plate No. 1. If there is not room to get them in the plates must be suffi- ciently extended to get them in, that is No. 3 must be lengthened beyond L and towards /; No. 2 must be lengthened beyond J towards K and No. 1 carried on towards end. SPACING THE RIVETS. 331 In laying out the rivets they should be as regular as possible, the best method is to lay out the total number of rivets Regularity of required from centre to end, gradually decreasing the spac ng p.^j^ towards ends, and then to make each of the plates No. 3, No. 2 and No. 1 of sufficient length beyond their respective point D (for No. 3) L (for No. 2) and / (for No. 1) to take in the number of rivets required. The length of plates may always be more than shown in Figure 196 without harm, but never less. We should have then for the bottom flange : Numberrivets a. ( —] in end of each \ f ^ flange plate. x = (122) Where x = the number of rivets required in each end of each flange plate between its ends and the nearer points to ends at which its full strength is required by the girder. Where a = net area of cross-section of the flange plate, in square inches (less rivet holes), at its weakest section. Where ^-y^ =the safe tensional stress, per square incii of the material. For top flange use — ^ and look out that rivets are not so far apart as to cause bending or wrinkling of plates. Where v = the safe stress, in pounds, or least value of each rivet. That is the bearing, shearing or cross-breaking value of the rivet, whichever is the smaller. The rivets in the flanges will, of course, be cantilevers, loaded with Value of rivets, their respective amounts of a. (^^^ or a. (^y^ respectively, a being the area as given in Formula (122). The free end of cantilever will be of a length equal to the thickness of the respective flange plate, or equal to the thickness of leg of angle iron, whichever is the smaller should be used. The bearing area will be the diameter of the rivet multiplied by the same smaller thickness. In single plate girders the shearing of rivets will rarely determine their number, this will generally be more than the bearing or bending value. Figure 197 shows clearly the way in which rivets are strained. The web-rivet. No. 3, is bearing against web on the surface E F and a<^ainst angles on the two surfaces (sum of) DE and FG 332 SAFE BUILDING. The rivet has two cross-shearing areas, at E and F. This rivet is a beam supported at D E and F G and loaded uniformily with its share of horizontal flange strain, w^ich it is transferring to web. The flange rivets, Nos. 1 c 5) G- PIGr. 197. and 2, we will suppose are connecting the flange plate A H to the angle iron. Their bearing then is against A H and in the opposite direction against B C, the lesser should be used. Their shearing area is either along the line Hov the line B according to which end is considered the cantilever, so that they have only one shearing-area practically in the calculation, instead of two as with the web rivets. Then rivets Nos. 1 and 2 are cantilevers and are built in either from H to C and loaded uniformily on the free end A H, or built in from A to B and loaded uniformly on the free end B C, whichever projection 4 if or jB C is smaller should be used. The load on the cantilever being as already explained equal to each rivet's share of an amount equal to the net-area, of top plate A H multiplied by the safe tensional or compressive stress per square inch of the material. There is, of course, a tendency of the plates HI, IB, etc., to slide past each other and past angles. This tendency will exist particularly at the centre of girder and in those parts of rivets which simply tend to hold Platetendency ^j^g plates together after the plates have once trans- ferred their strength and become a permanent part of the flange. But this tendency rarely amounts to much, unless the plates are very thick ; and if the rivets are spaced according to rules given can be overlooked. If it is desired to calculate the strain on each rivet, due to this tendency of the flange plates to slide past each other, it can be done by the following formula, which assumes that at any right angled cross-section of flange through rivets there are always two rivet holes. v = \.{x-yY.(^^^) (123) Where v = the safe value or stress on any flange rivet, in PLATE SPLICING. 333 Where = the safe modulus of rupture, in pounds, of pounds, to resist the tendency of any two flange plates (or plate and angle leg) to slide past each other. Where b — the total breadth of flange plate, in inches. Where d = the total depth of girder, in inches. Where x = the distance, in inches, from the horizontal neutral axis of girder to centre of flange plate, further from neutral axis. Where y = the distance, in inches, from the horizontal neutral axis of girder, to centre of flange plate (or centre of flange leg) immediately next to other plate, but on the neutral axis side of same. k_ ■f the material. If any part of a girder, either web or flange, is spliced, made of two parts, the number of rivets each side of splice, SpMchig^gl^der ^^^j ^^^^ amount of additional cover plates, etc., should be made sufficient to transfer the full strength of original plate across the joint. In locating the rivets of a splice care should be taken not to weaken the original plate by holes not allowed for in the original calculation of moment of resistance of the section. There is no difficulty in splicing webs, as cover plates can be put on each side, and the strains in the web are comparatively small. These (web-splice) plates and their rivets each side o/ joint should be of sufficient strength to transfer the amount of the Web-splice. vertical shearing strain at the joint from one side of (spliced) web joint to the other side of joint. In the flange, how- ever, it is more troublesome. In heavy girders, however, (the only ones usually, where it is necessary or where it pays to splice the flange plates), it is best to carry the upper or outside layers of flange plates a longer distance from the centre (or point of greatest bending- I" l3f1£6"SDllCGi moment) than required by calculation, thus gaining extra material in the flange, and more than required there by calculation, and then using this extra material to offset the loss suffered by making the additional rivet holes and by cutting or joining one of the flange plates at the point. For instance. Figure 198 repre- sents the side-view of part of the top flange of a plate girder. A B is the first flange plate running entire length of girder, A being towards end and B towards centre. 334 SAFE BUILDING. This plate has to be spliced. We have previously found that we can thin down our flange at the points F, E, D and C. We will decide to piece plate A B between D and E say at G. Of course the flange will thus be weakened at the point G by the entire loss of FIG, 19 a. plate A B and if we attempt to regain this by cover plates it will lose the additional rivet holes. But by prolonging the upper plates as shown by dotted lines this loss can be made good and without any additional rivet holes. For by the time the plate which originally ended at E has been ex. tended to G the girder is considerably stronger than needed, that is^ stronger by the amount of thickness of this extended plate, and the girder can therefore bear the loss suffered by the cutting of the lower plate. Providing, of course, the plates are of equal thickness. If there are not enough rivets between G and D to take up the strength of the spliced plate, the plate which ends at D will also have to be extended, as shown by dotted lines, till the number of rivets desired have been covered. In many cases the extending of flange plates is sufficient to form the splice, but frequently an additional cover plate over the extended flange plate may simplify and cheapen the cost. The arrangement in each case will depend upon the number of rivets required, the respective thickness of plates and other local circumstances. As a rule the angles are made in one piece from end to end, as they can easily be obtained of great length, and are Angles In^ne^ awkward to splice. Angles should be used as heavy as possible, but if very thick they are diflScult to straighten, and besides reduce greatly the value of flange rivets, owing to the bending-moment. In determining the thickness of web it has to have sufficient area DESIGNING WEB. 335 of vertical cross-section at all points to resist the vertical cross-shear- inir, and must be stiff enough not to buckle under its load, which will be equal to the vertical cross-shearing at each point of web. As this vertical cross-shearing is always greatest at one or both supports, we should have for thickness of web : Web-thickness. b= - — ^ /io/i\ '^■(^9^ (124) Where b = the required thickness of web, in inches, (should never be less than \'' thick). Where d = the depth of web, in inches, this should be the net depth d, that is depth less all rivet holes coming oh any vertical section at or near reaction. Where p = the reaction, in pounds, at either end, (Formulae 14 to 17). The larger reaction should be used, where they are unequal. If web is not to be of uniform thickness throughout, use, in place of p, the amount of vertical shearing-strain, in pounds, at the point for which thickness of web is being calculated. Where ^ ^ = the safe cross-shearing stress of the material, in pounds, per square inch. In many plate girders the web will be so thin in comparison to its depth, that there will be serious danger of the web Danger of^^^^ buckling, particularly towards the ends where the vertical cross-shearing (except in case of single con- centrated loads) is always greatest. To avoid this danger the web is stiffened by riveting upright angle irons, or T-irons to same, between the flanges. The ends of these stiffeners should always be planed and bear truly against each flange. At the very ends of girders there should always be stiffeners over the reactions, of sufficient strength, as columns, to carry the amount of reaction, less the amount of bearing of web on reaction. As the length of these columns will be only equal to the depth of girder, and the column will generally consist of the bearing Use of amount of web, plus two angles, they, the stiffeners, stIffenerSn o ? ./ ? can safely be considered as short columns and the full safe compression stress, per square inch, allowed on them. The number of rivets connecting any of the stiffeners to web, should equal the amount of cross-shearing being carried by the stiffener, 336 SAFE BUILDING. that is vertical cross-shearing at the stiffener, less the amount borne by web. This latter amount at the ends is the safe load on a column or section of web, equal to its bearing on reaction ; between reactions a section of web equal to its depth is taken as assisting or being assisted by the stiffener, that is as acting together with the stiffener. While the web really receives its load from the flanges by pin-connected ends — rivets — it is, nevertheless, assumed by most engineers to have planed ends, presumably to avoid too many stiffeners, the whole calculation, as it is, being but very theoretical anyhow. We should have, then, amount of strain on end stiffeners, _ _ 12000 . a; ®*'^'St?f?/n'y'rs. , 0,0-000-^ (125) -r ^2 and amount of strain on intermediate stiffeners. 12000.6. d ■y- Straln on in+er- „ „ mediate sti^f- ^ ^ _,_ 0.0003. rf^ (126) Where s = the total compression strain on stiffeners in ^ounds ; the stiffeners should have sufficient area of cross-section to resist this strain, considered as short columns, and sufficient rivets con- necting them to web to = s in value. Where p = the greater reaction, in pounds, where the reactions are unequal; or either reaction where they are equal, see Formulae (14 to 17). Where y = the amount of vertical cross-shearing, in pounds, at the point of girder, see Formula (11), at which stiffener is applied. Where x = the distance, in inches, that girder rests on (selected) reaction p. Where b = the thickness of web, in inches, at end or at point y, as the case may be. Where d = the depth of web, in inches, at end or point y, as the case may be. To decide whether the web needs or does not need stiffeners, and if so, at what points, use the following formula. _ 12000. b.d *^yi?fe?i;'la"rr ^ ~ ^ 0.0003. (127) Where b and d same as in Formulae (125 and 126). Should y be STIFFENING WEB. 337 larger than the greater reaction p no stiffeners are required, except at the very end. Should y be less than either reaction, stiffeners will be required up to the point of web where the vertical shearing — (as found by Formula 11) — just equals y. At this point place stiffeners a distance apart equal to the depth of web. Stiffeners should always be placed under concentrated loads. At end of web place stiffeners and again just inside of reaction, and between end and point where y equals shearing place stiffeners, not less than the depth of web apart, and gradually diminishing the distance between them towards end ; this distance should be regu- lated by the amount of increase in vertical shearing towards end. In regard to the deflection of plate girders, the same rules apply, as for beams, that is Formulae (36) to (42), Table °pVl2girders. Formula (95) to (97). It should be noted, however, that owing to more or less imper. fections in riveting, fitting of parts, etc., the plate girders will deflect very much more than if calculated by these rules, with a modulus ot elasticity same as for perfectly rolled beams. To allow for these imperfections in manufacture a lower modulus of elasticity should be used, to be varied according to the care exercised in manufacturing the girder. Experiments on riveted girders have given moduli of elasticity for steel as low as 5000000 pounds-inch. This, however, is probably an extreme case. The writer would recom- mend that the following be used, where no experiments can be made : For wrought-iron plate girders e = 18000000 Decreased pounds-inch, modulus of ^ elasticity. For mild-steel plate girders e = 20000000 pounds' inch. Where 6 = the modulus of elasticity, in pounds-inch, to be vised in calculating the deflection of plate girders. Before giving an example, Tables XLI, XLII and XLIII should be explained. They have been calculated to enable architects to lay out the required size of plate girders by their use, and without elaborate calculations. They will be found to be very accurate and valuable for preliminary estimates, quick designing of girders, and checking of final calculations. 338 SAFE BUILDING. Table XLI gives the value of the web in resisting the bending- moment. It should be remarked here, that some *'xLl°'xu^and engineers do not include the web at all; others include only one-sixth of the web at top and bottom. This is practically reducing the web to the same level as if the top and bottom flanges were merely latticed together. The writer believes, that in house-work at least, it can and should be safely included, particularly as it does not greatly affect the final result anyhow. In box girders the two webs should be considered as one web of thickness equal to the sum of the two ; except when calcu- lating for buckhng, when, of course, each web is taken separately. Table XLII gives the value of the four angle-irons, for six different sizes of angles, and Table XLIII the value of each inch of effective width of flange. In all of the Tables the horizontal column of figures at the top indicates the length of span of girder, in feet ; the vertical columns of figures to the left indicate the respective values in tons (of 2000 pounds each); while the figures on the curves indicate the depth of web of the plate girder. The tables are calculated for a safe modulus of rupture ^ ^ or extreme fibre strain of 12000 pounds per square inch, and intended, of course, for wrought-iron. For those desiring to use a smaller or greater strain it will only be necessary to increase or reduce the actual load For different (respectively) in the calculation. Thus, if it is fibre strains. ^ ^ jj , , . desired to use a fibre strain of 1 5000 pounds, this will be one-quarter more than allowed for in Tables. We therefore use but four-fifths of our load in the calculation and find by the Tables, what sized girder will safely carry four-fifths of our load with an extreme fibre strain of 12000 pounds. When we then add one-fifth of the original load (or a quarter additional to the calcu- lated load) it will, of course, add also one-quarter or 3000 pounds to the extreme fibre strain. Or, we wish to use an extreme fibre strain, of only 10000 pounds. We add one-fifth to our load making it f (or one and one-fifth) and find from Tables the size of plate girder to carry this increased load at 12000 pounds fibre strain; when now we deduct one-fifth of the original load (or one-sixth of the calculated load), we will, of course, at the same time diminish our extreme fibre strain one-sixth to 10000 pounds. TABLES XLl TO XLIII. 339 The use of the Tables is very simple and easy. For loads other than uniform, and for steel, the data at bottom of Table XLI shows their respective values as compared to those given in Tables. It should be noted that the " greatest deflection " has been calcu- lated for the most perfect work. For ordinary work this deflection will be increased, according to the quality of the workmanship, to. one-half more than for perfect work. In using the Tables, first settle the size of web, then of the angles, and finally the size of flange plates. Example I. Designing ^^^^ suppose that we have a wroughi-iron plate girders by girder of 60 feet span, which is to carry a uniform Tabies. ^^^^ ^^^^ loads of 44ii- tons each, one concentrated near each end, and one quarter span from reactions. We are to use a web 36" deep and flange 21" wide. Design the girder parts hy use of Tables. From the arrangement of loads W,,, and (at bottom of Table XLI) we see their sum is equal to a uniform load, or that is, our two concentrated loads will have the same eti!eet on the girder as a uniform load of 89f tons ; our total load on the girder, therefore, will be equal to 178f tons uniform load, which we will assume includes the weight of the girder itself. We will decide , to use four 6''x6"x|" angles, as the loads are very heavy, and |" rivets throughout. We now settle the thickness b of web, from formula given in right- hand corner of Table XLT, namely : 6:^i^=.l!y^0,621 8.rf 8.36 ' or, say, web should be |" thick. We now find the value of web in resisting the bending-nioment from Table XLT. Pass down the vertical span line Value of web. marked 60' 0" till we strike the curve marked 36", this is two-thirds way between the horizontal lines marked on the left — (in the |" thick vertical column, the second from left) — 7,5 and 10,0 respectively, or our web would safely carry about 9 tons. We next take Table XLTI, remembering that we selected the 6" X 6" X I" angles, the values for which are in the extreme left column. 340 SAFE BUILDING. We again pass down the 60' 0" vertical line till we strike the 36" curve, which is a little more than half-way between Value off our^^^ ^j^g horizontal lines marked in the extreme left — (6" X 6" X I") — column 28,2 and 32,9 respectively; or, our four angles together will take care of about 30| tons, this added to the web value (9 tons) makes 39| tons of the 178f tons to be cared for, or a balance of 139 tons to come on the flange. The flange is to be 21" wide, from this we must deduct the two |" rivet holes,^ or our effective width of flange would be = 19^", there- fore each inch must carry i|^ = 7,22 tons. We now take Table XLIII, pass down the 60' 0' vertical line till ... we strike the curve 36" and then pass to the left to Valueof flange^ find the above value 7,22 tons. We strike the curve on the fifth horizontal line from the top and passing to the left find that we cannot find any such value as 7,22 tons, in other words the flange will have to be thicker than two inches. The value of 2'' we find is 4,8 tons, leaving us 7,22 — 4,8 = 2,42 tons to care for in addition to the 2" thickness; this, we find (still on the fifth horizontal line) is under the thickness marked 1 or our flanges will have to be exactly 3" thick at the centre of girder. In regard to the web, should we decide to make a box girder, each web should be at least one-half the calculated thickness or thick. In practice it would be better to make them a little heavier, for such heavy girders, say about f " thick each. Example II. A single web riveted plate girder is of 59^,0''' span. "'hveted gl^er. ^'/^'"^ fo^^ from left support, and thence every five feet to five feet from right support it carries a concen- trated load of 19500 pounds, the third loads from each end being in- creased (by columns) to 91000 pounds. These loads include the allowance for weight of girder. The web must not be more than 36" deep. Detail the girder. This girder is one of some twenty-five used by the writer in a large public building in New York City, hence the limitation as to depth of girder. 'Many engineers deduct in addition to size of rivet, 1-16" for punching and 1-16" for reaming, which in our case would make the rivet holes 1" instead of J". 344 SAFE BUILDING. The working-drawings, as they were ♦furnished to the contractors by the writer, are given in Figures 204 to 210 inclusive.^ The only objection (on the score of cost) was to the length of some of the flange plates, but this could not be avoided, as the level of beams resting on the girders could not be disturbed, and there was not room enough between beams to get in the necessary length of cover- plates for splices. The reader will readily see that this is practically the same example as the former one (Example I), so that we need not refer to the Tables for preliminary designing. We know then from the Tables that the girder, at the centre, will need to Size by Tables. ° have a 36" x|" web, a3"x21" flange, four 6" X 6" X I" angles, and that we will use |" rivets. We will now see whether this is confirmed by calculation. We will first use the graphical method (see Chapter VII) on account of the large number of loads. In Figure 202 (p. 121) we lay off our vertical load line m a, where nil = lk=jh = hg, etc., = 19500 pounds, and kj=dc = Curve of 91000 pounds at pounds-scale. We would select the moment of ^ k resistance, ^^j^ distance a;^/ =rr^_ ^ = 12000 pounds, but that it will make the moment-of-resistance curve too deep for con- venience. We will, therefore, decide to make the distance xy = 10.^^^ = 120000 pounds at pounds-scale. We shall, therefore, have to multiply all the verticals through the moment>-of-resistance curve in Figure 199 by ten to get their actual values. We draw the moment-of-resistance curve Figure 199 (see Chapter VII) and find its base line A M. As our loads are not symmetrical on the beam, being four feet distant at one end and five feet at the other, we draw in Figure 202 X parallel A M oi Figure 199, and find our reactions y^m = q = \ 75000 pounds and a =p = 182500 pounds. The greatest bending-moment will be at load Wy, where the great- est vertical v through the moment-of-resistance curve (Figure 1 99 on p. 121) measures 276 inches (by inch-scale). Nothaving Greatest bend- / k \ ing-moment. used the pole distance x y — y-^J Figure 202 we * It is to be regretted that some of the illustrations are necessarily very small ; the reader is advised to use a magnifying glass. CALCULATING A RIVETED GIRDER. 345 must multiply this by ten to get the actual required moment of resistance whicli would be = 2760. For the same reason we cannot use Formula (98) to calculate the flange thickness and therefore refer to Formula (3G) and have for thickness of flange at centre 11^-16,4 o 7 ' = 3,02 ~ 19,25 Or we need, as found by Tables, a flange thickness of three inches at the centre of girder. In the above formula, Required flange ghould be explained, 2760 was the required moment of resistance at the pomt (that is at load t«v,) for which we were calculating flange thickness; 37 represented the approximate total depth of girder, allowing say for one half-inch plate to each end of girder ; 19,25 was the net width of flange, after deducting two rivet holes ; and 16,4 was the net area of cross-section (after deducting four rivet holes) of two 6" x 6" x f" angles. We will decide to use six half-inch thick plates in each flange and must next decide where to break them off. Accord- diminish flange ingly we use Formula (99), and have value of two thickness, ^^gjgg = 16,4 .36 = 590,4 or of the whole required moment of resistance (2760) the angles furnish an amount = 590; now, as our moment of resistance curve is only of one-tenth the required depth, we divide this by 10 and make O r,= 59 at inch-scale. We now divide T, Tinto six equal parts and draw the parallel lines to base A M, their intersections N, 0, P, Q, R, and S with the curve are the points at which the respective plates can be broken off. We shall, however, carry the first plate NO the entire length, and as this plate is spliced inside of the curve, we shall have to carrv plate No. 5 over the joint to make up for the lost section, and we shall have to carry both plates Nos. 4 and 5 suflSciently far to the left of the splice to get in the necessary number of rivets to equal the value of cut plate. It will also be necessary to prolong some of the plates to get in the necessary rivets beyond their points of contact ^fVange' rivets with curve. Thus between 0 and iV we must get required. g^Q^gi^ rivets to equal in value plate No. 1, or else prolong plate beyond N ; between P and O enough rivets to equal plate No. 2, or else prolong plate beyond 0; and so on till in the last 346 SAFE BUILDING. plate Ko. 6 we must get enough rivets between T and S to equal plate No. G. Now these plates are all of equal value = i" x = 9| square inches of cross-section, which multiplied by ^^.^ = 12000 pounds gives the real strain s on the rivets, or s = 9|.12000 = 115500 pounds. We will, therefore, lay out rivets enough, in each flange, between S and the end A to take this strain five times, Spacing flange gra(jua]iy (jgcreasing the pitch towards the end of girder. We will then carry each plate suflicientiy far beyond the length required by curve, to get its respective number of rivets. Now the value of rivets (|") in flange will be for shearing (single area) =4800 pounds each. For bearing and bend- Value of each jno- the rivets will evidently get their value from the flange rivet. ° , , . , . , . , , , , f plate, this being thinner than the angles, and we have bearing value = 5250 pounds per rivet. Either of the above can be found by calculation, or from Tables XXXV and XXXVIII. For bending we have from Table XXXVIII for a | rivet, tho safe bending-moment = 990 or say 1000 pounds-inch. The actual greatest bending-moment will be, Formula (25) u. (1) u and as the safe m = 1000 pounds-inch, we have 1000 = - or 4 u =4000 pounds. The value of the rivets against bending — (4000 pounds each) — being their least value, will control the design. Each cover-plate therefore requires 115500 — rivets, and from 5 to end we shall require 145 4000 rivets in each flange. From S to T" we require only 29 rivets, but they will have to be spaced more frequently to comply with the rule for greatest pitch, or Formula (107), accordingly the pitch of the. latter should not exceed = 1G.1= 8 inches. Figure 207 shows a plan of the top flange of left half of girder. Plate No. 6 mic^ht have stopped at S, but is carried two rivets FLANGE RIVETS. 347 further to avoid breaking under a beam, whicli rested on the girder at this point. The spUce of plate No. 1 has been Splicing the made iust to the left of R, where plate No. 5 might flange. , r i tvt have stopped ; we must, therefore, carry plates JSios. 5 and 4 at least 29 rivets beyond the splice, which has been done. Plate No. 4 might have stopped two spaces nearer E, but for the spUce ; Plate No. 3 we stop at the right number of rivets to the left of Q, and plate No. 2 to the left of P. Plate No. 1, which might stop 29 rivets to the left of O, we decide to carry to the end. The countersunk rivets shown come under beam or column ends. The blank spaces were to bolt column plates to. The blank spaces for beams were marked on later from memoranda in the con- tractor's shop. Having detailed our flange, which will be the same both for top and bottom flange, we will now consider the web. The size of web we settle from Formula (124) and have for . , . thickness b, assuming that there will be no more than Thickness of ^ six rivet-holes in any vertical section, or rfi=36 — 6.| = 30f"; ^^182500 ^ 30f.8000 or nearly f". The web, however, was made |" as the above was required onlv at the one extreme end and through its rivet-holes. The effect of decreasing the breadth of web being, of course, to raise the actual shearing per square inch at this point to a little over 9000 pounds per square inch. «... ^.«-. We next decide where stiffeners are required; Where stiffeners ^ are required, use Formula (127) and have _ 12000. |. 36. ^ ~~ , 0,0003. 36^^ ~r" (1)2 = 135000 pounds, Or we require stiffeners from the end to the point where the vertical shearing is less than 135000 pounds. By referring to Figure 200, we find this would be about fifteen feet from the end. The stiffeners, however, were placed more frequently, as shown in Figure 206, both for looks, and as there was some danger of heavier' loads being placed on the centre of girders, which would, of course, increase the vertical shearing near centre. These stiffeners were made of 6"x6"xi" angle irons, with 6"x|"x24" filler plates 348 SAFE BUILDING. behind them, so as not to bend the stiffeners. The filler plates being cheaper than would be the cost of blacksmith work involved in bend- ing these angles around the vertical legs of flange angles. Their upper and lower ends were " milled " off and made to bear firmly. Now as to value of rivets through web, we should have for bearing I". I". 12000 = 6562 pounds; for shearing, being in Value of web^^^ double shear, twice the value previously found for single shear or, 2.4800 = 9600 pounds; and for bending we have a circular beam of f " span. The safe bending- moment we previously found to be 1000 pounds-inch, the actual bending-moment is therefore ^^^=1000 and o u= 12800 pounds, Or the value against bending would be 12800 pounds. As the bear- O O O o o o _ O O (I O O OS ^^-v^ lUVETS HERE SHOWN TO BE COUNTERSUNK ON UNDER SIDE OF LOWER FLANGE SAME AT OTHER ■ENI> FIG. 20 8. ing value (6562 pounds) is the smallest we will use that in determin- ing the number of rivets in web. For end stiffeners we use Formula (125) 12000.1.16 Number of rivets « = 182500 — in end stiffeners. = 122500 pounds. * Or we should need 122500 0,0003. 36^ (1)^ 6562 :18,6 Or we should need some 19 rivets in the end stiffener,we therefore decide to use a filler plate 24" x 23|" x |" each side, which will not WEB RIVETS. 349 only help stiffen the web, but affords us the room to get in the neces- sary number of rivets, without cfttting more than six. rivet-holes on any Numberof rivets^"® vertical line. Figure 208 gives a plan of in central arrangement of web at end. We next take the stiffeners. g(.j£fgjjgj. located some 4' 6" from the entl. The verti- cal shearing here (see Figure 200) is 163000 pounds. From Formula (126) we have therefore strain on this stiffener 12000.1. 36 s = 163000 0,0003. 362 = 28000 pounds. Or we should need 28000 _ 6562 ' or say five rivets, we must, however, locate them oftener, see Formula (107). We next decide to splice the web at the point Splicing the^^ shown in Figure 206, and as shown in plan Figure 210. The vertical shearing at this point (see Figure 200) is 163000 pounds, we need, therefore, each side of joint 163000 „. „ or say 25 rivets. Including those in the angles, which, of course, help splice the joint, we have 26 each side. We next settle the size of splice plate by Formula (114). We shall have for its neat breadth &==24"— 6.| — 18|" and have for thickness of splice plates , 163000 „ ^ = 181712000^'''' As there are two plates, one each side of the web, we shall make each one-half the above, or say f" thick, remembering, however, to fill out behind the angle with an additional SECTION OF I" thick (filler) plate. ELANGE SFLTCE. ^ We must Numberof rivets connecting web next settle and angles, the number of rivets connecting the angle and the web. riG ao9. 350 SAFE BUILDING. The vertical through moment of resistance curve (Figure 199) at the centre measures 276" and the axis x y in (Figure 202) we made 120000 pounds, therefore, from Formula (93) the bending-moment at centre, or : ?Kcentre=276. 120000 = 33120000 pounds-inch. This divided by the depth will give the horizontal flange strain from centre to end of girder, see Formula (121), or 33120000 r,„r^^r^r^ A ' s - — — = 920000 pounds. This again divided by the least value of web rivets, which we FIG. 210. previously found to be 6562 pounds, gives the total number of rivets required, or 920000 = 140 6562 In reality we have placed 143 rivets from centre to end, so as not to place the central ones too far apart. Again take a point just under the column (or w,x) say fourteen feet from the left reaction. The vertical (Figure 199) measures 225", therefore bending-moment at M'lx, or Tow =225.120000 IX = 27000000 pounds-inch and horizontal flange strain 27000000 36 and required number of rivets 750000 , , , = 114 6562 750000 WEB RIVETS CONTINUED. 351 In reality there are only 113 between this point and end, but that is near enough, as it spaces more evenly so. Again, take a poirit at the first load to the left, or Wxi which is four feet from left reaction. The vertical (Figure 199) measures 75", therefore bending-moment, my, =75.120000 = 9000000 pounds-inch and horizontal flange strain to end 9000000 36 250000 Therefore number of rivets required, 250000 _ 6562 "~ In reality there are 42 rivets. It will be noticed, that in allowing for horizontal flange stress we take all the rivets to the very end of girder, this, of Length of bear- course, is rigrht; although before right through for ing on reactions. ' o ' o = o convenience we have considered the end as at the reaction. The amount the girder will run over the reaction will be determined by the crushing strength of the wall, or pier, or column it is supported by. In our case we have a bearing 21" x 16 = 336 square inches, and therefore load per square inch on masonry 182500 , = = 543 pounds 336 per square inch. This was distributed onto the brickwork by heavy, ribbed, enlarged cast-iron plates. The only thing remaining to be done now is to figure the deflection. In Figure 199 we draw the vertical lines, 1, 2, 3, 5, 6, etc., through the moment of resistance curve, the distance be- Deflection , , . . „ , r- found tween them (/,) being practically 60' , the first one graphically, jjgjjjg g, half distance. In Figure 203 we carry down these lengths in succession on line m 1, 2, 3, etc., to a. We select our pole z arbitrarily at a distance 2yr=1000". In Figure 201 we now construct the deflection curve. The longest vertical is in the centre of girder and = 243". The moment of inertia of the section of the girder at the centre will approximate very closely to 58000 (for exact amount see Table I, 352 SAFE BUILDING. section No. 14) and remembering that for built-up plate girders of wrought-iron we must use a modulus of elasticity equal to only 18000000 pounds-inch, we have the central deflection, in inches, of the girder (see Formula 97) g _ 2h. 60. 1000. 1 20000 _ „ 18000000.59000 ~ ' The safe deflection, not to crack plastering, would be, (see Formula 28) 8 = 59.0,03=1:1,77" Or, our girder is amply stiff. Were we to con- '^^''xable XLl. ^^^^^ ^^^^ equal to a uniform load of 357500 pounds we could use the approximate formula for deflection given in Table XLI, and should have had g^^92_ 75.42 ' We must add one-half to this for a modulus of elasticity of only 18000000 (the approximate formula being based on 27000000) and would have 8 = 1,105 + 0,552 = 1,657" or the same as by the graphical method. Or, we might have calculated the deflection br Deflection ^, , ° . . , . , , , found l?ormula (39) agam considering the load as a arithmetically, uniform load, and should have had jv__5^ 357500.708 3 384 18000000.58000 = 1,62" Or, practically the same result, and showing how closely the different methods agree. Had we figured the girder arithmetically we should have obtained practically the same results throughout. We should have considered our load as a uniform load of 357500 poujids, which would give us equal reactions, of 178750 pounds each, an error of hardly 2 per cent. Bending- "^^^ bending-moment at the centre would be, arith.S'eTc^Vfy. Formula (21), 357500. 708 qicqqvka i • u m = =31638750 pounds-inch. The required moment of resistance therefore, would be, Formula (18), - ^^ 31638750 ^ 12000 DEFLECTION OF GIRDER. 353 'The actual moment of resistance (by section No. 14, Table I) will be found to be at the centre r = 2740 or considerably more than required. The load, however, is not strictly equal to a uniform load, hence the discrepancy, we should, of course, use the result found graphically, which was based on the actual conditions. In figuring the girder arithmetically the required moments of resistance at different points along the girder should be ascertained ; after which the curve of moments of resistance can be laid out and the flanges, web, rivets, etc., of girder, calculated the same as already explained. If our girder were not braced sideways we should have to calculate for lateral flexure, using Formula (5). For the area a we should take the area of top flange at centre, plus two angles and the part of web between angles. For the square of the radius of gyration we should take the same parts around Lateral flexure, ^^is M. . . N a.i right angles to flange, or as top flange, shown in Figure 211. We omit rivet-hol^s in this case, for ease of calculation, and as all parts are in compression. We have then a = 3.21 + 2.9,73 + 6.| = 86,21 square inches. Now for §2 not finding the exact section in Table I, we must find the moment of inertia i and divide this by the area. We have then F fir in 3.218 12 = 2468 |.12|8 5^.2|8 12 ~ ^2 Therefore q2= 2468 ^ 55 86,21 ' und from Formula (5), I being, of course, the span of girder in inches: 354 SAFE BUILDING. _ 3.86,21 .12000 . 4. 708^^. 0.000025 ' 9.28,6 = 2597800 pounds. One-third of this would be safe, therefoit I = 865933 pounds would be the safe stress in flanges not to cause lateral flexure, or the safe stress per square inch should not exceed 865933 ^;loo94 pounds. 86,21 ^ The actual stress will, of course, equal the area 86,21 multiplied by the average fibre stress, per square inch. To find the average fibre stress, use the following Formula ; ""^'^mlkt'^n ^•'^•(/) (128) flanges. v= ^ ■ Where v = the fibre stress, per square inch, in flanges of plate girders. Where x = the distance of the centre of gravity of part of flange being strained — (flange, angles and part of web between them) — from the centre of depth of web, in inches. k_ f inch, or stress on extreme fibres, in pounds per square inch. In our case this distance x would be found by rule given on p. 7, (Vol. I), and would be (see Figure 211) ^ ^ 21.3. 19^ + 12f.l. 17^9^+ 21. 5i.l4,9g 86,21 = 18,63 Therefore the average fibre stress from Formula (128) 2. 18,63 .12000 V = 42 = 10645 pounds. The actual total compressive stress on flange will therefore be = 86,21 .10645 = 917705 pounds. This result should, of course, be the same as our horizontal flange stress, previously found, and by referring back, we see that this was practically the same (920000 pounds.) Where ^y^ = safe modulus of rupture, in pounds, per square . LATERAL FLEXURE AND WRINKLING. 355 Our actual compressive stress we see therefore is about two and one-half times larger than the safe stress to resist lateral flexure as found above (383200 pounds.) We should therefore, either brace the girder sideways — which was done by the beams in our case — or we should have to broaden the top flange. Wrinkling of ^® readily see that there is no danger of flange, wrinkling in so heavy a flange, but did we wish to calculate it, we would do so by Formula (4) or Table III. A new deep Since the publication of Table XX, the Home- beam, stead Steel Works of Pittsburgh (E) have begun rolling 24" deep steel beams from 240 to 300 pounds per yard in weight. The data in regard to these beams is as follows : Neutral axis normal to web. Neutral axis parallel lo web. Depth of beam (d) Weight per yard Width of flanges (6) Thickness of web Area of each flange Area of web Total area (a) Moment of inertia (t) Moment of resistance (r) Sq. of rad. of gyration ^ Transverse value (steel) Moment of inertia (i) Moment of resistance (r) Sq. of rad. of gyration ^. Transverse value (steel) 24" 300 7,20 0.75 6,83 16.34 30.00 2349.00 195,75 78,30 1958000 47,13 ]>.,10 1,57 131000 CHAPTER XL GRAPHICAL ANALYSIS OF STRAINS IX TRUSSES. HE same general rules which apply to beanas and girders apply- equally well to trusses ; but as the latter are made up of a large number of parts, some sustaining the loads directly, others transmitting the consequent strains and thus helping indirectly to sustain the loads, it becomes difficult and often very complex to follow out all the strains arithmetically. For this reason the graphical method is generally used, and for the architect, who has many other things to remember, besides strains and stresses, will always be found to be the most convenient. There are three steps necessary in designing a truss : 1st. Ascertaining the amounts of loads on each part, and their points of application 2d. Ascertaining the consequent strains on each member of truss. 3d. Designing the members and joints of truss. In calculating trusses it is always assumed that all the members meeting at any joint are connected by a single pin, and are, there- fore, at liberty to move around this pin, until they assume equili- brium towards each other, when of course, they will all counter- balance each other and remain stationary. All loads are, therefore, assumed to act directly on the joints, and are considered as vertical forces at these points (except where wind is allowed for separately). It will frequently happen, however, that the loads are not placed directly over the joints. For instance, the load might be uniformly distributed over the ^ entire rafter (or chord) : in that case, we should strain on have to assume one-half the load on each panel as members. gQming directly (and vertically) on the joint (that is, each joint would act as a vertical reaction, made up of several parts), and afterwards when designing the truss members, we should have to add sufficient material to the rafter (or chord) to take care of the transverse strain, due to the uniform load. DISTRIBUTING LOAD. 357 Or, again, we might have a load w, as shown in Figure 212. The amounts of this load coming on joints Nos. 2 and 3 would be figured exactly the same as reactions (Formulae 14 and 15). In these formulae p would be load Xo. 2 ; q load No. 3 ; I would be the length of rafter from No. 2 to No. 3 ; m ''°'"*feactions. ^^^"^^ be the length of C or from No. 2 to load ; and n would be the length of D or from load to No. 3. All these lengths can be measured either along the rafter, or horizontally between the vertical lines, the result will be the same. Where loads are suspended from the lower chord, or wliat amounts to the same thing — rest on same (as ceilings for instance), the load can be considered as hanging from the bottom chord as shown at zo,„ Figure 212. This, however, seriously complicates the strain diagram, it is better therefore to consider (which is also the fact) that Loads onjower ^j^^ j^^^ ^^^^ transferred directly to No. 3 by the tie-rod A B. We will therefore in making our strain diagram add the amount of to load No. 3, and must remember later to add an amount of tension (^equal to w,„) to rod A B over and above that found by strain diagram. If we had a load ii?„ placed half-way between p and w„, we should have to make the tie- beam or lower chord sufficiently strong to bear this transverse load in addition to the tension existing in it. In this case, the point and/) would be the reactions for load w„, and the rod A B would 358 SAFE BUILDING-. transfer its share up to No. 3 in addition to load w„,. But as a rule it is more economical to transfer the load w^^ up to joint No. 2 directly, by means of the tie-rod. When making the strain diagram, however, this rod should be omitted and the truss shown as at H. Otherwise it will be found that the corresponding points f and g of strain diagram would coincide. This would mean that there was no strain on F G due to the strains in truss ; and this is a fact, as the only stress in the rod is in resisting the direct tension due to the load hanging from No. 2 by the rod. When figuring the reactions p and q they should be figured by the Formulae (14) to (17) inclusive. If all the loads Reaction^at^^^ g^j,g uniformly or symmetrically placed along the truss, each reaction will be just one-half of the total load. If not, then the truss is considered the same as if it were a beam and the reactions figured by the formulse, the distances m, n, r and s are measured horizontally between verticals as shown ; the distance I is, of course, the entire (horizontal) distance from p to q. Wind can safely, as a rule, be assumed to act vertically on the truss and to simply add just so much to the calcu- Allowancefor lated dead load. The writer generally adds for this ^'"*'snow. climate (New York City) 30 pounds per square foot of surface of roof (measured on the slant, not hori- zontally). This will do for small roofs and approximate calculations of large roofs. This allowance will include the necessary allowance for snow, for, if the roof is steep, the snow will "^"^^ ' either slide off, or be blown off, and if the roof is flat the wind pressure will be very much smaller and the reduction in wind pressure will fully offset the weight of snow. Of course taking the wind as a vertical dead load involves two errors : first, the wind is never on the entire roof, as it can mani- festly act on one side only; secondly, the wind does not act vertically. Where wind pressure is calculated sep- Separate Dia- arately it is assumed to act at right angles to the wind surface of the roof and on one side only. In large trusses this should always be done, as it will frequently be found that stresses in certain members will be reversed. That is, members, which under a dead, vertical load show only tension or compression in the strain diagram may (with wind taken EFFECT OF WIND. 359 Reversal of normal to roof and on one side only) be reversed Strains, ^^^j show, respectively, compression or tension. Iron trusses, over eighty feet long, need some arrangement to allow for expansion and contraction. In roof trusses this is provided by anchoring down one end and leaving the other end free to move (horizontally) by placing it on rollers. It will readily be seen that where there are rollers the effect on the truss will be very different, according to which side the wind is blowing from. In such trusses, therefore, it will be necessary to make three strain diagrams, one for vertical dead load (including snow but no wind); one for wind only on right side; one for wind only on left side. The truss must then be designed to withstand the strains due to the dead load only ; and enough added, where necessary, to with- stand the additional or different strains due to either pressure. Where both strains are of the same nature they should be added together; where they are of opposite natures they will, of course, offset each other, but the member should be strong enough or stiff enough to withstand either separately. As the wind acts horizontally, it will on striking a roof of course cause a different pressure at right angles to the in- Wind Pressure clination of roof, than is its pressure against a ver- Dependson j surface. This pressure will therefore vary Inclination) / , rr, , • with the inclination of the roof. To determine it, it is assumed that the greatest wind pressure per square foot of roof surface will never exceed forty pounds. For steep roofs with an inclination of 60° to 90° with the horizon, this is the pressure assumed. For roofs forming smaller angles with the horizon a complicated trigonometrical formula is used. Its results are as follows : 360 SAFE BUILDING. TABLE XLIV. TABLE OF WIND PRESSURES ON ROOFS. Angle of Inclination of Rafters with Horizon, Pressure or Load, in pounds, per square foot of Koof Surface. 10° 9§ 15° 14 20° 18J 25° 22i 30= 26J 35° 30 40° 33J 45° 36 50° 38 55° 391 60o) to } 90°) 40 Approximate Rule for Wind Allowance for Snow It will be noticed that approximately the pressures in pounds are about eighty per cent or four-fifths of the number of degrees of the angle of inclination. Where the wind is taken separately the allowance for snow in the strain diagram for dead loads should be fifteen pounds per square foot of roof surface. Having once ascertained the amount of load on each joint, the strains on the different members of the truss are found by the general methods given at the end of Chapter I. (Pages 69 to 74, Vol. I.) Particular attention is again called to the method of notation, and to the necessity of reading off the pieces in their proper order, and of reading around each joint in the same direction. The writer always uses the direction in which the hands of a watch would travel around each joint. In calculating the strains each joint can be analyzed by a separate strain diagram, or all of the strain diagrams can be combined into one. As the latter method is much more convenient and less liable to error, it is the one always adopted. In laying out the strain diagram we begin with the joint with the Drawing of least number of members and this usually is at one strain diagram. Qf t]jg reactions, where we have only one strut and one tie. Having found these strains we pass to one of the joints at WIND AND SNOW ALLOWANCES. 361 their other ends, and so on. The reason for doing this is that it will be found impossible to draw the Hnes in the strain diagram representing any joint where there are more than two unknown strains. By beginning, therefore, with a joint of two members only, the strains on these can be found. We then can pass to joints of three members, containing at least one of the former strains and so on. Figures 213 and following ones give a large number of roof designs with their corresponding strain diagrams. We will analyze one or two of these and the student can puzzle out the rest. Only a few will offer any particular difficulty, and these will be taken up and explained later. In all the figures dotted lines mean that the dotted member is in tension and full lines that the member is in corn- Dotted lines in pression. The numbers in Figures 213 to 228 o-ive tension. ^ , , ° ° the amount of strain, in pounds, on each member due to one pound of load at each joint for roofs with inclination angles as shown in figures. All that is necessary therefore, where roofs are designed similar to any of these figures, and with same inclinations and angles, td CD 1 o I 3.0 O FIG. 213 ascertain the amount of load on the joints and then multiply the number or given strain on each member by the amount of load on each joint. This will give the actual amount of strain on each member. For instance, we will say we have designed a roof truss similar to Figure 213 with the principal rafter at an inclination of 26° 30'; we will say the trusses are 10 feet apart and 48 feet span, and weight of roof including snow and yrind 50 pounds per square foot 362 SAFE BUILDING. measured on the slant. By scaling the rafters we find they measure 27 feet each in length, therefore load on each joint = y .10.50 = 6750 pounds. We now refer to Figure 213 and have the strains, as follows : Compression on rafter B G = 3,35.6750 = -|- 22612 pounds. Compression on rafter C H= 2,25.6750 15187 pounds. Compression on strut GH=1,1.Q750 =-|" 7425 pounds. Tension on tie G 0= 3,0.6 750 = — 20250 pounds. Tension on tie H I = 1,0.6750 = — 6750 pounds. Had we drawn the strain diagram and made ab = ef= ^^-^ z= 33 75 pounds at any scale, and at same scale made hc — cd = de = Q 750 pounds, we should find that at the same scale the respective lines would measure : hg =22700 pounds. ch = 15200 pounds. gh= 7400 pounds. go= 20300 pounds. hi= 6 750 pounds. and to ascertain whether these strains were compression or tension we should follow the direction of each line at each joint and see whether it thrusts against or pulls away from the joint. To draw the strain diagram we first select a convenient scale, by which we will measure all the loads and strains. We now draw our load line, making in (Figure 213) a 6 = the load on A B the foot of main rafter, we then make ic = the load on joint B C, the next one on main rafter, cd = the load on apex and so on ; as we know the reactions will each be just one-half the load, we locate o half way between / and a, in other words we make fo = reaction F 0 and o a — reaction 0 A . To get the strains we begin at the joint A B G 0 A at foot of main rafter, for here there are only two unknown strains. Whereto begin. , , n ^-y i , . namely, the compression on B G and the tension on GO. In strain diagram draw b g parallel B G ; and g o parallel G 0 till they intersect at^r; then will bg be the amount of thrust on the joint or the compression in B G, and g o will be the amount of pull on the joint or tension in G 0. To make sure of these we read off the lines following the proper succession, namely, A B, B G, G 0, OA; referring now to strain diagram we read a b, this is down or a EXPLANATION OF DIAGRAMS. 363 vertical load ; next b g, this is down or towards the joint, therefore compression ; next g o, this is to the right or pulling away from the joint, therefore tension; and finally oa (which brings us back to the point of starting a) and being upward is, of course, the direction of the reaction. As our strain diagram is a closed figure abgoa, we know the joint is in equilibrium. Had we failed to get back to the point of starting a, we should have known there was some missing member to the truss at this joint. On the other hand, if we had had two letters coming onto the same point — (which would be the case with letters F and ^ere we to draw strain diagram for Figure 212) — we should know that the member or line between these two letters was superfluous so far as aiding the general truss is concerned. We now (in Figure 213) pass to the next joint with 364 SAFE BUILDING. only two unknown strains. This is evidently the joint at half the height of rafter, we have just found the amount of compression on G 5 — (note that we now read GB and not B G as before, for around this new joint the hands of a watch would travel in the direction G B) — and the only unknown strains are the com- pressions on CH and on H G. In strain diagram we draw ch parallel C H and Ti g parallel H G till they intersect ; c Ti will then be the amount of compression on C H and h g the amount of com- pression on H G. We now read off the pieces in succession B C, C E, HG and G B; and in strain diagram; be, downwards, therefore a vertical load ; c h downwards towards joint, therefore DESIGNS FOR KOOF TRUSSES. 365 compression ; h g to the left, towards joint, therefore compression ; and g b upwards towards the joint, therefore compression ; as our figure is ia closed one, we having arrived back at the point of start b, the joint is in equilibrium. We next examine similarly the joint at apex and at centre of horizontal tie. The joints to the right will be similar to the corresponding left-hand joints, as the truss is uniformly- loaded. Figures 214 and 215 are similar to 213, the only difference being in the increased pitch of the rafter. When we analyze Figure 216 we first take the joint A B at foot of rafter; next the joint BC immediately above; next the first joint along tie-rod; next the second joint along rafter and so on. Figures 217, 218 1 and 219 are similar. Figures 220, 221 and 222 will 1 Angle of rafter (or angle 2) in Figure 218 is 26° 30'. 366 SAFE BUILDING. present no difficulty ; nor will Figures 226, 227 and 228, but in the latter three we find that we cannot follow the above rule of passing from a joint along rafter to one on tie-rod; for we begin at joint AB (Figure 226) and find the strains BS and SO; we next pass to joint B C and find the two unknown strains C R and R S ; were we now to pass to the joint at the tie-rod, we should find it impossible to continue, for there would be three unknown strains, namely, R N, (i) © iVMand MO; but if we pass first to joint CD we find the two unknown strains 7)iVand N R, and then going back to the joint at tie-rod, there remain only two unknown strains NM and MO which we can now find. ROOF TRUSSES CONTINUED. 367 Figure 223, we will find, presents a difficulty at the joint C D and this truss, in effect, cannot be analyzed by the same method as the others. We begin at joint A B and find the unknown strains B L and L 0 ; we pass to the joint B C and find the strains C M and M L ; vfe now pass to first joint along tie-rod and find the strains M N and NO; we now pass to joint CD but find three '^"Figure°223. "^known strains D S, S R and R N ; we try the second joint along tie-rod, but this, too, has three unknown strains N R, R U and U 0; so have all the other joints, we are, therefore, completely stopped. We reason, however, that the duties oi L M and M N, are apparently the same as those of T S and [jj in to J CVJ CM o swayed back and forth by the hand. Hence their only service was their own dead weight. The way to put in such rods is to leave the lower end free to move Vertical tie- vertically, that is up and down, but secure it against rods in steeples. ^g^^gj.g^| movement and then to attach to the lower end a heavy weight, proportioned according to circumstances. In figuring the allowance for wind it is customary to take only one- half the usual allowance for circular or surfaces slanting to the direction of wind. This is done because they offer less resistance than a flatly opposed surface, allowing the wind to slip by more readily. In a circular steeple we should take as our area the half circumference of base multiplied by the length on the rafter line and Wind allow- ances for different surfaces. 398 SAFK BUILDING. multiply this "by only one-half the pressure for wind as given in Table XLIV. This usually would be 20 pounds per square foot. For an octagonal roof we should assume full pressure on the central surface and only half pressure on the two side surfaces. Where the octagon is a regular one, this would amount practically to assuming double pressure on one surface only. There is no danger of a steeple blowing over diagonally, as the sides would in such a case present slanting surfaces to the wind, allowing it to slide off readily ; and besides the base line, being the diagonal of the square, would be so much longer. On vertical (wall surfaces) the writer usually allows only for a maximum wind pressure of 30 pounds per square foot, for, as a rule, they (or part of them) are low down near the ground and therefore ' not exposed to the full force of the wind. CHAPTETl XTI. WOODEN AND IRUN TRUSSES. IN Figure 256 we have the design of one-half of an ordinary king- post truss, with three bays to each principal rafter. Figure 257 gives the strain diagram and Figure 258 the design of the truss in detail. The length of principal rafter is 25 feet 6 inches, therefore length of each bay 8 feet 6 inches. The trusses in the building from which this example has been taken had to be spaced Large spaces f^j. apart — 19 feet from centres — on account of extravagant. ^^.^.^^ and skylights ; this, however, is not an economical arrangement, as it increases not only the sizes of the truss members, but of all the purlins as well. The weight assumed was 55 pounds per square foot of roof surface, being made up as follows : Slate, 7 pounds. Plaster, 8 pounds. Boards and construction, 10 pounds. Wind and snow, 30 pounds. Total 55 pounds. 400 SAFE BUILDING. The slate was the ordinary roofing slate ; if the slate usvd 1 ad been called for of even thickness and ^ inch thick the allowance would have been 10 pounds. The weight of " Boards and construction " was estimated and not checked off, in this case. In important constructions and particu- larly in ironwork, this should be carefully weighed up. The load on each joint of our truss was therefore : 19.8^55 = 8882 pounds, or ^^ay 9000 pounds on each joint, excepting, of course, the joints at reactions which carry only half-loads. i FIG. 257. The loads on the truss being symmetrical, each reaction will be one-half the total load or 27000 pounds. There is no difficulty in drawing the strain diagram, or in finding the stresses in each member, the latter are marked on Figure 256, the positive + sign denoting compression, and the negative — sign tension. We now proceed to detail the truss. The principal rafter will evidently, though not necessarily, be made in one piece from foot to apex, the largest strain — at the bottom panel — Detailing will therefore determine its size. This is 58400 wooden truss. ^^^^^^ compression. The rafter is evidently a series of columns each 8 feet 6 inches long, or comparatively short columns; for the struts brace the rafter against yielding downward; SAFE BUILDING. 402 the loads and tie-rods keep it from yielding upwards ; and the purlits keep it from yielding sideways. The safe compression per square PIG. 253. inch on Georgia pine (see Table IV) is 750 pounds. Without bother- ino' with the complex Formula (3) for columns, we Principal rafter. ° ^ ^ i will assume that we can sarely use ^ = 700 pounds in our case, we require then in the main rafter an area of cross-section 58400 700 =--83,4 square inches, or say a timber 8^ inches x 10 inches. This we increase to 10 inches X 10 inches to allow for cutting away at bolt-holes. If the purlins had not been placed directly over the joints we should have to increase this size to provide for the trans- PIG. 2G0. verse strain in each panel. In such a case we should have a beam 8 feet 6 inches long, supported at both ends. The transverse load on the beam would not be the full SIZES OF TKVSS MEMBEKS. 403 vertical load, but only its resultant normal to the rafter, as shown in Figure 260. II CD were drawn at any scale equal to the vertical transverse Vertical load °° ^^^^^^ ^ ^' '^''^^ parallel rafter, and on s'a'Jjt^^g ^ C normal to same, then will E C measured at same scale as Z) C represent the transverse load on the rafter A B, the latter being considered as a beam of length A B supported at A and B. Returning to our truss; if it were not supported sideways by purlins we should have to greatly increase the size of rafter, to guard against lateral flexure, see Formula (5). The stress in tie-beam is 54000 pounds tension, the safe ^— ^ for _, ^ Georgia pine (Table IV) is 1200 pound",, therefore required area 54000 ^~"r>00 ~ square inches. In a wooden truss the principal members are naturally of same thickness, therefore we should require 4| inches x 10 inches to resist tension. To this must be added the necessary area for bolt-holes, transverse strain due to ceiling hanging to tie-beam, etc. In our case the beam was pieced, as shown at centre, and it was therefore rather heavily increased and made 10 inches x 12 inches. The struts were of spruce ; their size will, as a rule, be determined Struts. ^^^^ bearing against the principals — so as not to indent these — rather than by their length as columns, though both should be tried. By making our larger strut 6 inches x 8 inches we get a' bearing against the rafter of about 6 inches x 10 inches, or say 60 square inches, this would make a compressive stress across the grain on the Georgia pine ratter of 14000 OQQ ^ = ^oo pounds per square inch. Table IV gives 200 pounds as safe across the grain on Georgia pine, but we can pass the above as safe. It should be remembered that Nature of action of crushing stress in short blocks greatly compression resembles that of an explosion. A short block or arge ocks. ^^^i^^ ^£ g^^j^g^ j£ compressed, will fly off in the four directions normal to the four free sides, with sudden and great force, as if there had been an internal explosion. If the block is lar^e 404 SAFE BUILDING. the central fibres will tend to explode outwardly, while, of course, the fibres nearer the edge will tend to explode inwardly and as a result the central fibres, meeting with the opposition of the external fibres, will resist more compression than they would if they were free. This is confirmed by actual experiment, where it is found that large blocks of the same material will resist crushing pro- portionately to a very much greater extent than will the small blocks. Where, therefore, we are proportioning the sizes of struts - or later the sizes of washers — fur their bearing area, across the wood, we will bear the above in mind, and if the safe values for crushing across the grain, as given in Table IV, require unusual dimensions, we will increase the amount of the safe value per square inch, as our judgment may dictate. Now tryincr the strut as a column of spruce 6 inches x 8 inches, or 48 inches are°a, and 9 feet, or 108 inches long, we should have from Formula (3) the safe load : 48.650 w = ^ _|_ 1082. 0,00033 12 = 13700 pounds, which is near enough to be safe. In calculating the other strut we should find that as a column 6 inches x 6 inches and 6 feet 3 inches long it can safely carry 14400 pounds, the actual strain being 12000 pounds. For pressure across the grain against the rafter and main tie-beam we have an area of about 6 inches x 8 inches against each or 48 square inches, therefore actual compression per square inch across the grain = 250 pounds, which they can safely stand, as these parts are of Georgia pine and the crushing area quite large. We next proportion the sizes of tie-rods, which will be of wroug;htr iron. The safe tensional stress of wrought-iron Tie-rods. being 12000 pounds per square inch we require areas as follows : For the principal rod. 18000 _^ 1 square inches. "12000 ^ ^ For the side rods, — X or sav ^ square inch. 12000 - 2 1 TIE-RODS AND WASHERS. 405 The rods being circular the principal rod would need to be If inch diameter and the side rods i| inch diameter or say | inch. We also place a small | inch diameter rod in each end panel to keep the tie-beam from sagging. The rods will all Upset ends. have to have their ends upset, or else they will not have enough sectional area between the threads. In practice, however, it would be cheaper, where the rods are so small and short, to enlarge their diameter and pay for the extra material, rather than to save this and have to pay for expensive blacksmith's work. Screw ends should only be upset where the material saved fully counterbalances the labor. We next proportion the washers, they bear against Georgia pine and across its grain and should, therefore, be pro- Washers, portioned at about 200 pounds per square inch. For the principal rod we will need, therefore, a bearing area for each washer of 14000 . , = 70 square inches, 200 ^ ' or say 8 inches x 8 inches. For the side rods we will need 12000 • 1, =: 60 square inches, 200 ^ ' or about the same, we will therefore to save expense maKe them all alike. The lower washers are made to bear horizontally against both head (or nut) and tie beam, but the upper washers have to be modelled to bear against the slanting side of rafter, and horizontally against nut. It will also be noticed that the lower end of the washer is " toed-in " to the rafter to keep the washer from sliding. The wooden blocks which are screwed onto the rafter at each washer were made to allow for cutting away, to provide horizontal surfaces on which to rest the bridle irons which carried the purlins. It will be noticed that the principal tie-beam is pieced at the centre. The cut or halving was made slanting, so Halving ^ng^^^ force each half to bear on and pull against the other half ; and all sharp edges, where sudden increase in strains would take place were avoided by rounding off, as shown. Wrought-iron plates were placed over and below the cut and sufficient bolts placed each side of the cut, not to crush the wood. 4.06 SAFE BUILDING. We now must still design the foot-joints. The bearing area required must first be found ; we can safely put 200 pounds per square inch on Georgia pine, across earing a . grain, and our load at each reaction being 27000 pounds, we need 27000 square inches of bearing area, and as the timber 200 ^ => ' is 10 inches wide, it would have to rest on its bearing for a length of 13^ inches. In this case, however, we made it 24 inches, because the principal rafter bearing on the tie-beam so far from the support, it would have been apt to bend it. The left end rested on an iron column and was bolted to it as shown in Figure 258. In Figure 259 is shown the right end, which rested on and was bolted to a wall. The latter was corbelled out under the truss and capped with blue stone. We next provide at each foot an inclined bolt, to hold the rafter down to the tie-beam and keep it from jumping out Bolt at foot. ^£ ^jjg strap, which it might do, if subjected to heavy transverse strains. We next "toe-in " the rafter to the tie-beam as shown, taking care to get enough bearing area and area ahead of the toe to keep it from pushing or shearing its way out through the tie- beam, or from being sheared off itself. In our case, however, the toe is made small, the only reliance we place on it being to steady the rafter sideways. The rafter transfers all the load vertically across the tie-beam, we shall therefore need the same length of bearing area on the tie-beam, 13^ inches, as the latter needs on the walL We more than get this by means of the hardwood block, driven in tightly with its edge grain against the rafter and tie-beam and held in position by screwed on blocks. We next calculate the strap. The strains coming on it are 64000 pounds tension and resisting this 58400 com- Calculating^^^ pression. The lesser will, of course, be the one it must resist. The width of strap required not to crush the rafter must first be determined. The strap bears almost along the grain of the rafter, but not quite, we will therefore reduce slightly the safe allowance for compression along the grain as given in Table IV. and allow, say, 670 pounds per square inch, we then require ^^^^ = 80 square inches, or the rafter being 10 inches wide, 670 STRAPS AND BOLTS. a strap 8 inches wide. The thickness of strap should be sufficient not to shear off, we have two shearing areas one at each angle or 54000 __ 27000 pounds on each. The safe shearing stress on wrought-iron being 8000 pounds per square inch (see Table IV) we need an ar^a at each angle = = 3f square inches. The strap ° 8000 ^ ^ ^ being 8 inches wide would therefore need to be about ^^g^ inch thick. It was made ^ inch, however, as in this case, it was doubtful whether the ironwork would be of a high character. Now each side of the strap will have to withstand a tensional strain of 27000 pounds. While the safe tension on wrought-iron in Table IV, is 12000 pounds, in this case it was assumed at 9000 pounds only, or the strap needed a sectional area = ?-^^^^=:3 square inches. 9000 ^ The strap being ^ inch thick, would therefore need to be 6 inches wide, to which must be added 1^ inch to allow for bolt holes, making the strap 7^ inches x ^ inch. We now settle the number of bolts, which we will make 1 inch diameter. The safe shearing on a 1 inch bolt will be = 11.8000 = 6300 _. ^ , pounds. Each bolt has two shearing areas, one at Number of ^ o > bolts, each end and will therefore resist 12600 pounds. The total strain being 54000 pounds, we need ^^-^^ = 4,3 or say five ° ^ 12600 bolls to resist shearing. For bearing we have an area 1 inch diameter by ^ inch thick against the strap at each end of each bolt, or just one square inch bearing area against iron to each bolt, which will equal a safe resistance to compression of 1 2000 pounds per bolt, we need therefore ^^^^ = 4,5 or say five bolts to resist the compression of the strap. The bolts also bear against the Georgia pine tie-beam and tend to crush it along the grain, the safe resistance of the wood being 750 pounds per square inch. Each bolt bears against 1 inchx 10 inches or 10 square inches of wood, and will safely bear therefore 10.750= 7500 pounds. We need therefore, to avoid crushing the wood ^^^QQ = or say eight bolts, which is the number shown in drawings 408 SAFE BUILDING. We will now similarly analyze an iron truss. This, the same as the above wooden truss, is not intended so much as a guide in design- TRU5S 6 OTHEB IRON 1 0 3 BLOCKS 16 PLASTER 8 SLATE 10 SNOW 15 TOTAL "^LBS PER SOFT Iron truss. ing, as it is as a guide in analyzing the details of each joint. Every truss should be designed with reference to its special duty and position, and joints should be designed in each case to be made up as simply as possible. The designer should not hesitate for fear of criticism as to novelty, to make his truss or its joints of any shape that may be most convenient, bearing always in mind, that the simpler the parts and the nearer to standard sizes, the cheaper will be the exe- cution of the work. Almost any design can be made by the use of wrought-iron or steel, or of cast- iron. In the two former, care should be taken to analyze carefully the work required of each rivet. Wrought-iron is gradually taking the place of cast-iron, even for shoes and such parts of trusses, as the expense o riveting them up out of different parts, is apt to be cheaper than the cost of the patterns required for castings. nc.,262. LOADS ON IRON TRUSS. 409 Vertical load. In Figure 261, we have the design of the axial Hnes of an iron roof truss recently erected in New York City. The vertical load on this truss was assumed at 65 pounds per square foot and made up as follows : Weight of truss = 6 pounds. Weight of other iron =10 pounds. Weight of 3 inch blocks =: 16 pounds. Weight of plaster 8 pounds. Weight of slate =10 pounds. Weight of snow =15 pounds. Total (per square foot) = 65 pounds. Figure 262 gives the strain diagram for this load; the rafter panels were each 17 feet long, and the trusses placed 17 feet 6 inches from centres, so that the load on each panel was 17.17^^.65 = 19337 or say 19300 pounds. The wind-pressure was calculated separately. In Figure 263 it is supposed to blow from the left. The pressure normal to the roof WIND PRE5SURB ONLY 38,4-LBS - SQ.PT. PCR ANGLE 53° surface per square foot for a roof of this inclination (51°) will be, say, 38,4 pounds, (see Table XLIV.) We should have then on each panel : 17.17^38,4 = 11424 or say 11400 pounds. The total wind-pressure therefore was = 34200 Wind pressure, pounds. By prolonging the central axis of the wind 410 SAFE BUILDING. pressure CR till it intersects tlie horizontal ^ at i2 we obtain the reactions due to wind pressure. As A R measures 40 feet and R H 23 feet, we know that the reaction at H will be ^ = 11.34200 = 21700 pounds, and at A = |f.34200 = 12500 pounds. The strain diagram Figure 264, can now be easily constructed. We notice that there are no strains in M N nor in N P due to wind; also that the latter reverses the strains \n Q N, QO and P 0, they all having to resist compression. In Figure 265 are tabulated the strains due to both vertical load and wind pressure, the last column giving the actual result, that is, their sum or difference, as the case may be. In TABLE OF STRAINS OF ROOF. WTNDLOAD FIG. 264. Name of piece. Standing load. "Wind load. Result. B. J. -f 76000 + 8000 + 84500 G. K. + 61000 + 8500 + 69500 D. L. + 46000 + 8500 + 54500 E. M. + 46000 + 15500 + 61500 F. N. + 61000 + 15500 + 76500 G. P. + 76000 + 15500 + 91500 J. K. + 12300 + 11400 + 23700 K. L. + 12300 + 11400 + 23700 L. M. — 52000 — 15000 -^67000 M. N. + 12300 + 12300 N. P. — 12300 + 12300 R. Q. — 24000 — 12000 — 36000 G. N. — 23500 + 2000 — 21500 J. 0. — 49000 — 11000 — 60000 Q. 0. — 23500 + 6000 — 17500 P. 0. — 49000 + 7500 — 41500 Fig. 265. Figures 266 and 267 we reach exactly the same results, by combining both vertical load and wind pressure in one diagram. In designing the truss, we must remember that were the wind STRAINS AND SIZES. 411 blowing from the right, the corresponding members of each half of the truss would exchange their respective strains, we must therefore Detallingthe design each such pair of (right and left) members to truss, resist the larger strain. Figure 268 gives the drawing of truss and detail of joints. In designing the truss we find that the heaviest strains exist when the wind blows with the exception of the horizontal ^ * *'str'ain*s. ^^^^^^^ Q 0 which has its largest tension, when there is no wind ; we select accordingly, the heaviest strains for each member and proceed to design the parts. For the tie-rods we require areas, as follows : 6 7000 Size of rods. Vertical rod LM=j^^ = 5,58 square inches or a diameter of say 2| inches. 36000 Inchned rod K J2000~ ^ square inches or a diameter of say 2 inches. Lower inclined rod J 0=^^^ = 5 square inches or a diameter of say 2^ inches. Horizontal rod Q0 = ^^^^ = 2 square inches or a diameter of say 1| inches. By placing sleeve nuts (or turn buckles), as shown, we can tighten up the truss at any time. We next design the struts to, stand 23700 pounds compression each. They are 13 feet 4 inches long or 160 inches long. After several trials we decide to use two 3 inches x 3 Size of struts. inches x J inch tees, placed one inch apart, back to back. The weakest way in compression will evidently be with the neutral axis parallel to the web. From Table XXIV, we find their square of radius of gyration for this axis to be g2 _ 0^42 and their area of cross-section = 2,75 square inches each or 5^ square inches for the two.^ 1 It should be noted that no matter how many tees were placed along the same axis m the same direction we should have the 8amep2 : for while each additional one would increase the area, and the moment of inertia i, and the moment of resistance r, the square of the radius of gyration g^, being simply the quotient of the inertia divided by the area, would remain constant, no matter how manv tees were used. ' 412 SAFE BUILDING. We have then for the safe compressive stress in these tees to resist the 23700 pounds compression, (see Formula 3) : 12000.5A ooooo 1 w = - ? 23322 pounds. . 0,00005.160= ^ ^ 0,42 ■which is near enough. It should be noted that we consider the strut as a column with pin ends. The ends have got separately forged riG 266. pieces, bearing on the pins and riveted between the ends of struts with eight | inch rivets. The rivets were determined, as follows : Bearing area of each rivet | inch x 1 inch = | Rivets in ^ square inch or = 1. 12000= 7500 pounds, value per strut ends. rivet. Shearing area of each rivet= 0,3068 square inches, there being two shearing areas to each rivet, we have 0,6136 square inches resisting shearing, or shearing value of each rivet = 0,6136.8000 = 4908 pounds. For safe bending-moment on each rivet, we have from Table I, section No. 7 and from Formula(18) transposed : m = l^.(j^)8.15000 = ^1^.15000 = 360 pounds-inch. Remembering to use the larger value ^-^^ = 15000 pounds for bending-moment on pins or rivets, we could have read the same RIVETS REQUIRED. 413 results for bearing, shearing and bending moment directly from Tables XXXV and XXXVITI. The actual bending-moment on all the rivets would be Formula (21): 23700.1 , . , = 2962 pounds-inch. We require therefore, to resist bending 2962 360 8,2 or say eight rivets. Calculating the pins. This being more than required for either bearing or shearing determines the number of rivets to be used. We next decide on size of pin used. It is evident that the joints will be similar as regards arrange- ment. The largest rod will be central on pin, each side of it will be another rod and outside of this the strut. We can also readily see that the joint at foot of vertical rod L M will be the most severely taxed, and as it is usual to use the u same size pin throughout a truss we will calculate for this joint. It is, also, evident that we need calculate only for the strains in the vertical line, as these will be the heaviest. In order to reduce the bending-moment on the pin, we reduce the head or eye-part of vertical rod to 2 inches thick, this will make a 6 7000 compression of — - — = 33500 pounds per inch thickness and require 2,8 or say inches diameter, (which STEADY~T.OAD 65 LB5 PER Sp.M. WIND 38.+ LBS, PEC 50 FT. FIG. 267 , , . 38500 a diameter of pin = ^ 12000 was the nearest regular size of pin made by the mill who had the contract). The same result could have been read directly from Table 414 SAFK BUILDING. XXXVI. For the rod if Q we require a width of bearing 36000 2l|. 12000 = 1 inch. For each strut we require a width of bearing _ 23700 3. - 21^2000 = The single shearing area of pin being about 6| square inches or = 6|.8000 = 54000 pounds, there can evidently be no danger from that quarter. We now calculate the bending-moment, we first obtain the vertical resultants of all the strains, by laying off each strain along its respective line of action at a certain scale, and then measuring the length of its pro- jection along the vertical line. The vertical projection of the 36000 pounds- tension is 28000 pounds pulling downwards; the projection of the 21500 pounds-tension is 16500 pounds pulling downwards; the projection of the 23700 pounds-compression is 15000 pounds pushing downwards; the projection of the 12300 pounds-compression is 7500 pounds-compression pushing down- wards; the sum of all of these is 28000 -j- 16500 -|-i5000-(- 7500 = 67000 pound's downward. Resisting this we have the 67000 pounds of upward pull. We now lay out, Figure 269, the pin, and find that we have a double lever arrange- ment; the fulcrum being the 2 inch wide 67000 pounds strain. To the left of this we have a lever If inch wide loaded with two loads, one 28000 pounds one inch wide and one beyond 15000 pounds | inch wide. To the right we have lighter loads, the heaviest bending-moment will therefore be on the left side, or : 7^ = ^.28000+ 1|.15000 = 34625 pounds-inch. The safe-bending-moment on a 2i| inch pin is : = Ti-(1M)^-15000 = 37500 pounds-inch. The pin is therefore safe. The same result could have been read off directly from Table XXXIX. FIG. 269. Back of Foldout Not Imaged SIZE OF PINS AND PKINCIPALS. '415 We next design the main rafter; as this is to be in one length, we design it only for lower panel (91500 pounds-com- Sizeof Main pression) and the other panels will be, of course, too strong. Besides the compression we will have a transverse strain or bending-moment on rafter, per square foot, as follows : Wind = 38,4 3 inch blocks, =16 Slate =ilO Iron tees (say) = 5 Total =69,4 pounds, or allowing for weight of rafter, say, 70 pounds per foot. The rafter lengths being 1 7 feet and 1 7^ feet apart, we have total uniform load M = 17.17^.70 = 20825 or say 21000 pounds. We now draw any- where along rafter a vertical line a 6 = 21000 Transverse ***"^rafter. P°"""^ ' ^ben draw a c normal to rafter ; it scales 13000 pounds, therefore the actual transverse load on rafter is 13000 pounds. 1'he required moment of resistance to resist this load will be Formula (18) 13000.(17.12)_2 8712000 ^^'^ By reference to Table XXI, we find we require one 12^ inch 80 pounds channels to take care of the transverse strain. We also note that the area of channel is 8 square inches, which is about what We need additional for resisting the compression. We will decide then to use two 12:|- inches 80 pounds channels. The square of the radius of gyration is g^ = 21,10 The area of the two will be 16 square inches, but as just one-half of their total moment of resistance r is needed to Compression • . , , . .„ , , „ on rafter, ^^^^^t transverse strain, we will have only 8 square inches left of the total area to resist compression. The column is 1 7 feet or say 204 inches long. One end will be a pin end, the other a milled or planed end with fair bearing. We have then for safe compressive load Formula (3) : 8.12000 nnonn 204^.0,00003 -3-^^^'^ P«"°^^' 21,1 or near enough to pass as safe. If the transverse strain had reauired 416 SAFE BUILDING. say a moment of resistance of r = S5 we should have had left to resist compression of the total area 16 square inches only, V2.27,7/ = (about) 6 square inches instead of 8, which (in,such a case), would not be enough and would require a heavier channel. The thickness of our channel web we see from Table XXI, is 0,39 inches, or two webs = 0,78. The safe bearing ^^afte^rwebs on pin will therefore be = 0,78.2^|. 12000 = 27495 on pins, pounds. The web of channel will therefore need thickening at the shoe pin, but not at the two others. At the apex the pin does not bear on channels, but is connected indirectly by a 2-inch thick plate. At the shoe the channels have planed ends and rest directly on the shoe, the strain therefore against their webs, will be 60000 jiounds, due to the rod trying to tear the pin out. Reinforce -^1-,^ ^g^g ^^ke care of 28200 pounds of bearing, plates to webi , , , leaving 60000 — 28200 = 31 800 pounds, to be trans- ferred by rivets to thickening or reinforce plates. Of this each side takes care of 31800_^ggQQ pounds. The thickness required for each plate, is therefore 15900 _ Q^^2 or say ^ inch each. 2+1.12000 1 6' The number of rivets required must next be settled. We use | inch rivets: The channel being thinner than the plate determines the bearing value for each, or = 0,39.1.12000 = 3000 pounds, Rivets in rem- ' » ^ force plates, or i^i^ = 5,3 or say six rivets required for bearing. 3000 'J ^ The rivets are in single shear, their area = 0,3068 and their value 15900 „ . = 0,3068.8000 = 2454 pounds, and we require ^^^^ = 6,5 or say seven rivets. For bending-moment each rivet is a single k'ver held by the half- inch plate, projecting 0,39 inches and uniformly loaded on free end PLATES AND RIVETS. 417 with its share of the 14100 pounds carrit d by each channel web, the total bending-moment will therefore be from Formula (25). 14100.0,39 ^ = ~- =2750 pounds-inch. The safe bending-moment on a | inch rivet we previously found to be 360 pounds-inch, and require therefore 2750 „ _ ... 360" ^ig"t rivets. The disposition of rivets around the pin, however, requires nine rivets. We now go to the apex point. We use here a 2-inch plate, its Pin plates at ^^^""S pin must be all right, as we made the n p a ^^^^^^ other end of vertical rod 2 inches thick. The upper end of rod we make forked, each side 1 inch thick, and 2 inches between to admit plate. The plate is so large that there is evidently no danger of the pin shearing out. F or the rivets we can see it will require a large number and there- fore decide to use larger or say | inch rivets. Number of ^^^^ bearing value on two webs of each rivet will rivets, be = 1.2.0,39.12000 = 8190 pounds. The (double) sliearing value of each rivet will be = 2.0,6013.8000= 9620 pounds. The safe bending-moment on each rivet will be = H-(t6)''-15000 = 987 pounds-inch. If now we consider that the two rafters have planed ends and butt fairly against each other, taking up the thrust from each, the rivets need only take care of the vertical down pull 67000 pounds, all of ^^'''"^ ^ ^^se we should need for bearino- oZOOO -70-^1 ° -ggg^ — 7,2 rivets, for shearing less, and for total bending-moment, remembering that the channel backs are 2 inches apart and that the rivets are beams supported at both ends and uniformly loaded, from Formula (21) m = ^'^^^ '^ = 16750 pounds-inch. Therefore number of rivets required to resist bending-moment _ 16750 987~ "'"6 each side of joint. The plate, however, requires ten each side for even distribution. 418 SAFE BUILDING. If the channels do not butt fairly the plate will have to transfer the thrust from one to the other. This thrust will be equal to the horizontal resultant of compression on rafter at apex which is 54500 pounds. Its horizontal projection measures 34000 pounds. The rivets each side of joint must take care of this strain and the plate be large enough not to crush under it. We need not calculate any in this case, however, as this strain is just about one-half of the strain for which the total number of rivets were proportioned. We next design the shoe. It is a flat cast-iron plate 2 inches thick, and 28 inches by 24 inches with two flanges Designing ^^^j^ g inches thick to receive the pin. As the shoe-plate. , , T i channels bear directly on the plate, the only stram on the flanges will be due to the pin trying to shear its way out, the strain being 60000 pounds. Resisting this there are two areas to each of the two flanges, each area 3 inches x 3 inches, or a total area =4.3.3 = 36 square inches, the actual stress is therefore 60000 1667 pounds per square inch, which is safe. 36 The wall must carry the whole load of truss, each reaction is 57900 pounds due to vertical load. To this must be Bearing area, added the vertical resultant (or projection) of the largest wind reaction 21700 pounds. Its vertical projection measures about 14000 pounds making the total Includewind vertical reaction 71900 pounds. The area of plate reaction. ^ is 28x 24= 672 square inches, therefore compres- sion per square inch on brick-work _ 71900 _ pounds, which is safe. 672 Had this truss been of larger span, we should have had to place one shoe either on a rocking saddle, or else on rollers. Rockers or jf ^j^g latter there should be sufficient rollers, and * " they should be of large enough diameter not to indent the plate, and to roll back and forth freely. It is usual to put upward flanges all around the bottom plate and downward flanges all around the upper plate to hold the rollers in place between them. The flanges should be less than radius of rollers, so as not to meet. The foot of girder must be secured against yielding sideways. The size of rollers is determined by the following formula : Formula for to = 1.750. Jd (132) rollers. V ^ ' FOOT OF TRUSS. 419 Where w = the safe load, in pounds, on each roller. Where Z = the length, in inches, of each roller. Where c? = the diameter in inches, of each roller, if of steel or wrought-iron and rolling between cast-iron plates. If be- tween wrought-iron plates add 25 per cent to w. Rollers should be used, (under one shoe only,) where trusses in- doors are over eighty feet span, or out-doors if span is over sixty-five feet. Example. A truss of one hundred feet span has a reaction at each end of 95000 pounds. The shoe-plate is 20 inches wide the long way of rollers, and rollers are 1 inch in diameter. How many rollers are required ? Each roller will safely carry from Formula (132) w = 20.750.y/f =15000 pounds, we shall require therefore 95000 _ „ ^ ^P^^ = 6,3 or say 7 rollers. Of course, where rollers are used some arrangement must be made in the cornice to allow for the movement due to expansion and con- traction of the truss ; or if the roof is continuous a slip-joint must be provided in the roof itself, the detail of which will depend upon the local circumstances. CHAPTER XIIL COLUMNS. OG 273 FIG TTTHE Formula (3) on page 24 of Vol. I, is, of course, applicable to every kind and shape of column. The square of the radius of gyration used in this formula, can be found by any of the formula 'given in Table I; or if the shape of the column is so unusual that it is not given in Table I, the moment of inertia (i) of the cross-section of the column can be found, and To find radius ^j^; divided by the area (a) is equal to the square of of gyration. „ ^ ■ . - ^j-, y the radius of gyration, see page 9 ot Vol. 1. In Table II are given the different values of n for cast and wrought iron, steel, wood, stone and brick ; also the variations in COLUMN ENDS. 421 this value for ^^pin" that is, rounded or rough bearings at ends, and for " smooth" that is, turned, planed or smoothed off bearings at ends. These different values have been arrived at largely by experiments, but the reason why the end bearing affects the strength of column can readily be seen in Figures 270 to 274. In Figure 270 we have a column with smooth ends between two forces crushing towards each other ; as a result the column tends to bend, in this case to the right. The bending of the column will, of course, tip its ends, as these are at right angles to the (dotted) longitudinal axis of the column ; as a result the further crush- ing is taken entirely at the edges of the ends, or at points Why smooth ^and^, end stronger. , . marke a with arrows in the figure. FIG. m riG. 276. FIG. 276 A. It will be readily seen that the least pressure at A ov B tends to bring the ends back to the horizontal plane and consequently to straighten the column again. For this reason it is that columns with smooth end bearings give way the same as columns with fixed end bearings, as shown in Figure 271. If on the other hand the ends of the column are rounded, as shown in Figure 272, the effect of the crushing at A and B is to constantly increase the bending of the column, as the point of contact against the column simply slips around the circular ends, as the column bends more and more. Such 422 SAFE BUILDING. columns break as one curve instead of the triple curve shown in Figure 271. Where there is a pin bearing, as in Figure 273 the effect is, of course, the same as for rounded ends. Rough ends are considered the same as rounded ends, on account of the danger of some roughness or projection on the end of a bearing, which would greatly increase the tendency to bend, as shown at A and B in Figure 274. For this same reason wedging of joints of columns, should never be allowed. Only those columns should be considered as having smooth ends, where the entire bearing end is perfectly smooth, and forms a true and perfect plane at right angles to the longitudinal axis of the column. In iron and steel columns the ends have to be "planed" or " turned " off, both of which are done by Planing and machinery. Planing is, as its name implies, a pass- ing back and forth of a sharp metal planer which removes the surface iron, little by little. Turning is the same as planing but the motion of the planer is circular. Turning is done on circular columns, particularly where there are lugs or projections be- yond the bearing surfaces, which would be in the way of a straight planer. Cast-iron columns are usually made of circular cross-section and hollow. This is the cheapest cross-section there is for Shapes of cast ^ column, and the metal will do more work columns. •' ' , . , . per square inch, the thinner (within reason, of course,) the shell is made. Cast-iron columns are also frequently made square in cross-section, and hollow, which does not make a badly proportioned column. All other shapes, however, are bad, and should only be resorted to in unusual circumstances. Such shapes, for instance, are rectangular and hollow, one side (or diameter) being shorter than the other, in this case the neutral axis, when computing the square of the radius of gyration, should be taken parallel to the long side ; then there are H or I shapes, T or X shapes, etc., all bad and weak. Sometimes a column is made square or rectangular but with only three sides. The column will be greatly strengthened, if at intervals there can be cast on the fourth side a connecting bar. All hollow castings should be drilled, as already explained, to ascertain their thickness, also near their base to allow any water to escape, which might otherwise freeze and burst the casting. Brackets, COLUMN CAPS AND BASES. 423 riG. 277 and other projections should, by preference, be cast on the column, rather than secured to it by bolts or tap-screws. Of course, they can't be riveted to cast-iron. They should, as far as possible, be of same thickness as shell. ^ Where there are ^ heavyca pitals, m bases or other ^ mouldings or orna- mental that greatly incr ease the thickness of shell, they had better be made separately and slipped on afterwards, and then secured by tap-screws with countersunk heads. The bearing should always, if possible, be vertically under and over the shell, not flanged out as shown in Figure 275. Ends of columns are usually flanged out for bolting together, in which case the angles should be well rounded. It is also well to cast on one of the columns a lug as shown on the upper, etched column in Figure 276. To pre- vent fire spreading up t h r o u g h columns, some build- y^l^ 2J7g i n g 1 a w s require solid plates at all joints ; in such cases they should be made 424 SAFE BUILDING. as shown in etched part of Figure 276a with upper and lower lugs, and the columns bolted together through the plate. Bottom plates, which usually have to spread the weight are made, as shown in Figure 277 ; as they are often very Bottom platest j^rge, and necessarily therefore quite thick, metal can be saved by gradually reducing the thickness towards the edges, as shown. In such plates a hole should be cast in the centre, to relieve the strain on the plate, when cooling and thereby avoid warping. No bolts are needed where a bottom plate is used, as shown in Figure 277, as there is no possibility of tipping. Where the spread of base plate has to be very great, flanges are cast on same as shown in section in Figure 278 and in plan in Figure 279. Of course, the flanges can be more numerous. All parts of such a base should be of even thickness. If the casting is unusually large and unwieldy it can be made in two parts, as shown i n Figure 279, but in such cases great care should be exercised with the foundation or substructure, to avoid one-half settling away from the other half. rv S // ^■ riG.279. This can be done by thick granite stones, or additional iron plates under the main bottom plate. The calculation of hollow circular columns is very tedious, partic- ularly as one has to guess at the size and then calculate the strength, often involving several calculations for a single case. Explanation of Tables XLV, XLVI, XLVII and XL VIII, have Tables therefore been prepared by the writer to take the place of these calculations, for cast-iron hollow circular columns. HOLLOW CAST-IRON COLUMNS. 425 Across the top of the tables, in the horizontal line, are given thie lengths in feet of the columns. Down the left side in the vertical line are the safe loads in tons for each particular shape. The curved lines each represent a hollow circular section, and at the end of each curve is given first, the diameter of column, second the thickness of shell, third the area of cross-section. The latter enables one to pick out the cheapest section from any of the tables, by selecting any curve below and to the right of the one found — which has a smaller area of cross-section. Sometimes by reference to a subsequent table a still cheaper section can be found. The tables have been calculated for cast-iron, according to Formula (3) ; the columns are supposed to have smoothly dressed pounds. Any one desiring to use any other value, need only pro- portion the vertical column of safe loads in tons accordingly ; thuSj for 12000 pounds we should take \ of the loads as safe, or we could add \ to the load to be carried by the column, and find from the tables the diameter, etc., of column necessary to carry the increased load. Table XLV gives columns from 3" to 1" diameter of different thicknesses, and from 5 feet to 12 feet long. Table XL VI gives columns from 8" to 10" diameter of difFerent thicknesses, the 8" and 9" columns from 8 feet to 15 feet long, the 10" columns from 10 feet to 20 feet long. Table XLVII gives columns from 11" to 13" diameter, and Table XLVIII from 14" to 16" diameter, in both, of different thicknesses and from 10 feet to 20 feet long. An example will best illustrate the use of table. Example I. What is the safe load on a hollow, circular, cast-iron column, with turned ends, 18 feet long, 11" diameter and l^" thickf Columns of 11" diameter are given in Table XLVII to which we turn, we find the curve marked 11 — — 44,8 cuts Exampleof u^se vertical line 18 about two-fifth way down between the horizontal lines 150 and 155, our column will therefore safely carry 152 tons. Had we calculated our column by Formula (3) we should have ends and true bearings. The value used was 15000 426 SAFE BUILDING. had from Table I, section Ko. 8 for g2^:^^i^= 11,56 and for Z2 — (is. 12)^ = 2162 = 46656. Inserting the values in Formula (3) we have the safe load 44,8.15000 _ 672000 _|_ 46656.0,0003 1 +1,2108 = 303962 pounds, 11,56 or, 151,98 or say 152 tons same as found from table. Now, is this the cheapest section for that load and length of column. Its area is 44,8 square inches. We now pass to the next curve below and to the right, it is 13 — 1 — 37,7 or a 13'' diameter column, 1" thick will be cheaper, as its area is only 37,7 square inches. Then, too, it is stronger, for at 18 feet long it will carry 158 tons cutting the vertical line 18 three-fifth way between 155 and 160. By passing further along, we find also that the 12" diameter 1^" thick column of 42,1 square inches area will be stronger, carrying 161 |- tons safely at 18 feet of length. By referring to the next Table XLVIII, we find some much cheaper sections, than any of these. For the 14" diameter, |" thick column, has only 31,2 square inches area of cross-section, but we find it carries safely only 143 tons and therefore will not answer. However the 14" diameter 1" thick column, has only 40,9 square inches of area and carries 185 tons, and would answer therefore, provided of course, larger diameter is not objectionable. Or, cheapest of all would be the 15" diameter, |" thick column which has only 33,6 square inches of area and carries 161 tons. The 16" diameter, |" thick column would also be economical having only 35,9 square inches of area and carrying 180 tons at 18 feet of length. In wrought-iron construction, any number of shapes of columns are used, from plain flat bars, to the most elaborate Shapes of combinations of the different shapes rolled, most of wrought^ ron^^ which are given, with their areas, squares of radius of gyration, etc., in Tables XIX to XXV. In Figures 280 to 287 are given a few combinations, which are frequently used for columns. Besides those shown there are fre- quently used combinations of I-beams and channels, or of angles and plates formed in the shape of plate and box-girders stood on end. Figure 280 shows two channels latticed together. A plate might WROUGHT-IRON COLUMNS. 427 be used in place of latticing if the channels were placed as shown in Figure 281, but in that case the interior would not be accessible for painting. Figure 281 is more easily riveted up than 280, but is not quite as strong. Figure 282 shows an elevation with wrought-iron base, and Figure 283 the side view of base. Architects are apt to use cast bases with flG. 281. flG. 252. wrought-iron work, but as a rule a wrought-iron base can (and fihould) be designed, which will not only be better adapted to wrought-iron construction, but will be cheaper and stronger, and has the merit that it can be riveted fast to the column or other construc- tion. Figure 284 shows a column made of four Z irons, with central plate. It has the great merit of being a strong column, and though of five parts, it requires only two lines of rivets. This column adapts 428 SAFE BUILDING. itself excellently to building walls, as the horizontal wall girders, and floor girders are easily attached to it, and it readily holds the "fiUing-in walls" in place, without the use of anchors. Then too, it can be easily covered with fireproof blocks. The writer ? number of years ago erected in New York City, a fireproof office- building, ten stories high above the sidewalk, with only twelve inch thick brick walls all the way up, by using these columns combine/ with horizontal girders in all the walls. At each floor level thi vertical webs of the floor girders run between the Z irons; the vertical plates of columns butting against web plate ends both top and bottom. Joints of Z irons should not be at floor levels, and they should "break joints." When more than one thickness of plate is used vertically between the Z bars, the plates where joined should ' ' break joints. ' ' Of course, in such construction, proper precautions must be taken to prevent the building from collapsing under wind pressure. In the case above referred tc, in order to avoid cross-partitions, the FIG. 283. Pie 284. flG. 2§5. wind-bracing was all done in the front and rear walls, and in the floor levels. Figure 285 gives a combination column of two tees, riveted together with separators. Figure 286 is made of four angle bars latticed together, a very light, but strong column. Figure 287 consists also of four angle bars, which adapt them- selves more readily to fireproofing, and require less riveting, but the column is not nearly as strong as the previous one. In the last case separators are used in place of lattice bars. It would be quite impossible to give any curve tables for wrought iron construction, but Table XLIX will greatly ^Table^XLIX?' facilitate the calculation. It will be necessary in each case to find only the ratio of the length of column (in inches) divided by the radius of gyration, or the square of the length, divided by the square of the radius of gyration, and look up the value per square inch of cross-section, according lo the WROUGHT-ipON CQLUIV^NS, CONTINUED. 429 condition of ends of column, and the assumed safe value for (^~^)* The table is calculated respectively for 8000, 10000 and 12000 pounds per square inch values for (^y^* buildinn;s use the value 12000 for wrought-iron. To find the ratio look tip the value of the square of the radius of gyration in the tables, or if it is not given, fjnd the moment of inertia according to rules given in Table I, or on page 10, and divide by the area, see page 9. It should be borne in mind that where pieces are doubled or their number increased along the same neutral axis, the moment of inertia will be doubled or increased accordingly. But the square of the radius of gyration will remain constant, as it simply represents a flG. 236. ri6. 287 riG. 255 ratio, and the area and moment of inertia increasing in the same amount, their ratio, wliich gives the square of tlie radius of gyration, will, of course, remain constant. In some cases, it will be easier to find the radius of gyration, instead of its square, and in such cases the second column in Table XLIX should, of course, be used. An example will best illustrate the use of Table. Example II. A flat eye-bar of wrought-iron \\" x&" in a truss, of'Table XLIX* liable at time? to be under compressive stress, what will it safely stand f J'he bar is 5' 6" long from centre to centre of eyes. From Table I, section 2, we have ■2 CU)'_.,0 430 SAFE BUILDING. the area will be a — l^.G = 9 square inches, and The ends being eye-bars, we use, of course, in Table II, the value n for " both ends pin ends," or n = 0,00005 Inserting the values in Formula (3) we have for the safe com- pressive strain, 9.12000 _j_ 4356.0,00005 = 50322 0,19 Now had we used Table XLIX we should have had the ratio 4356^ 0,19 The nearest value to this in the first column of the Table is 22500 and under the heading " both ends pin ends " for a value of ^-y^ = 12000 pounds, we find 5655 which is the safe load per square inch on our bar, or the total safe strain,^ w = 9.5655 = 50895 pounds. Which closely approximates the above result. In our case it would have been easier to use the second column of Table XLIX, we should have had ^ M 12 4^Y2 3,464 ' and for the length ^=66 therefore the ratio g~ 0,433-^^^ The nearest value to this in the second column of Table XLIX is 150 which would give the same result as before. There are several patent wrought-iron columns ^"columns, ^^^^f of which the "Phoenix" column is un- doubtedly the best. It is made up of from four to eight segments, riv-eted together. ' (obtained by multiplying by the area of bar.) PHOENIX COLUMNS. 431 Each segment somewhat resembles a channel with the web bent to a segment of a circle, instead of being straight. Figure 288 shows one of the smaller cohmms, made up of four segments. For heavier columns each segment is rolled thicker, as shown in the figure in outline. When it is necessary to have very heavy columns flat pieces are inserted between each flange, as shown in Figure 289. These columns can be readily covered with fire proof blocks to make a circular finish in buildings, and are largely used both for this reason, and on account of their great strength, (ow- ing to all the metal being near the outer edge). When calculating the load on a column it should be borne in mind, that if the girder o^ beam is continu- ous over the column, the loads will be equal to the reactions as given in Table XVII on pages 218 and 219, Load on I f t h e Column. girders o r beams overhang columns and are built into the wall at the other end (such as gallery beams for Gallery instance) the respective loads on the column and Beams, wall and upward pressure on wall can be found from Formulae 116 to 119 inclusive. If the load on the overhang is uniform its reaction would be the same as a similar amount of load concentrated at one-half the span of overhang. If the beams or girders are inclined they can be calculated the same as already explained (in the previous chapter), when calcu- lating transverse strains on rafters; and the amount of anchoring necessary to prevent pulling out, can readily be found by obtain- ing the horizontal thrust by the graphical method, as already explained. TI6. 28a TABLE XLVI. TABLE XLVI. STRENGTH OF HOLLOW CYLINDRICAL COLUMNS OF CAST- 433 LENGTH OF COLUMN IN FEET. ThickDess of metal. Diameti Area. IX. IN. SQ. TABLE XL VIII. 435 TABLE XLViri. STRENGTH OF HOLLOW CYLINDRICAL COLUMNS OF CAST-IRON. LENGTH OF COLUMN IN FEET. 10 11 12 13 14 15 16 17 18 19 20 .2 S IN. t5iooooo»C CO OC O r-f OJ O C-V W O CO OO 1-1 "XV O CO t-- C<5 t- CO —I t-. CO 'O >0 kft otv r.- -^.^ M rH 00 CO -i" 1-1 as co o >o c-i O t- -OCOCCif5>CflO'fS--t-(C l3SXiQCOO'*-COCOCOCOCOO»00>CiO'*-*- 0-^COOCCOODt-'*>OOOiOCO-*COt>'-*<5^ieot^OOOOOOOOO< 'O'— i-*'0>OeOOCOO-*000i005TtiOO-.t>.t^t>.t^t>.t>-l>-t— COCOCOCOCOCO^lCOOiO'^-^Tj^^'^-^'^COCOG IO'-l(>»»OOC30i-(OOr-(->i(iOif5>00'-lO>OOiOO>OiCOOOOOiO< !0.-C5COt^O>-(T-li— lOOiOO^fNOt^'OS^'-IClOOt^COiOiOC CacsOOO-COiOCOlMOCJt— lOCOi— (OOCO^(NOt>.iOCO>-lXiCD'i!l*<>'OOOC irt — .-l-H.-i,-irt,-.,-lOOOOO05C505Ci05 00 00 00 00t>-l>-t-t^t-.coc t0l0a0l000'^e0t^.O0^'0t~.O.0(N>0'>-'— I' CSCiODt^t^CO-rfO^CJOCit^'O'^OOCOCO'OCO^Ot^-O'^'^'^*!^' (3iC5CSC2C:3C5C5CJOJ05 00 00 00GOOOOOl:~t^l»t-t-COeOcoCOCOCO>0>0> CSO^COl^^COiO-jJtCOt^COOt^coCOt^OOOOOOOOOOOOOt l>-->tlC5e'5COCOCiC500CO-^i-ll^COO^-^C5l(5 0lOOlOi-IC0 1MQOTj-t-l->.t^t^t>-t>-l>»t^t>-l>-COCOCOCOCOCOCO»0*0»OlOlO*0^'^'^'^" coooecoo'Mco'Oi35ireoe'siOK5C5-HOt^oioo>noo>coo< aoeooo<-lCo■^C500'-l'-l(^^.':>cooo< CSClOOOOt^COiO-^fMrtOOOCOOO^^CSt^CS-^CMOOOCO-^C^OOOCOl ^,_(^,_<,-(^.-(^i-H — i-iOOOOOCaClC50JO505Q000000000t>-t>-i CO-^.-lC005000.-ICO-«flO'»CSO'-lcOC5005lO(MOt^lOlOCC0.-IO=0t-l«;^; C5C50J05C505C505C50CiOiQOOOOOOOGOOOCOt-C-t-t-t-t-COiXiCOCO CftirtTHl:*.-i(N(M-*C5t^CiCOt^iOCO>-eot-0^000000000 oolOC>^l^~(NcoosrH(Ncococo(^^^05t^>oc^Ot^-<1^rtCClO<^lO^-■^<^^ 010i<35<»00l^C0CD>O-*C0C^.-.?-t^t-t-.c^t^t^i^cocococococococoiooooiOio>rs CO t. O 3^ fcr M .. 0»00>«Oi«0>^50>00>OOl«OlOOlOOOOlOOlOO>00>20 II<.-((M<>Je0C(S-^'^O'0C0C0t^l^0D00C5«JOOi-l>—S^<>«00C0-^'^'O OiOO>OOiOOiOO>OOiOO>OOiOO>00>«OiOOiOO>00>00 0)OC>)0(NO(MO>'-lOOOr-l(N'^C005IMCOOO >-(rHSMC-J05CQ-^-^OCOt-00(3SO-H(MC5-f'OCOXlC5'— !'» 1—ti—(f—tf— It— <*—(*— t<-HT—lG^Ceam; in case of centre load, deduct one.'ha{f the weight of beam; in case of loads at thirds of span, deduct from each load f the weight of beam; in case of loads at quarters of span, deduct from each load ^ the weight of beam. Steel sections will be slightly heavier (about 1%) than iron sections of exactlv the fi.ame dimensions. In ordering steel or iron, give either the required dimensions or the required weight of section, iievrr both. The capital letters in the first, column headed "Mills Rolling Shape" in each Table represent the following Rolling ^[ills : — A — NEW JERSEY STEEL AND IRON COMPANY, Trenton, N. J. B — PHCENIX IRON COMPANY, Philadelphia, Pa. C— PENCOYD IRON WORKS. Philadelphia, Pa. D — POTTSViLLE IRON AND STEEL COMPANY, Pottsville, Pa. E— UNION IRON MILLS and HOMESTEAD STEEL WORKS, Pittsburgh, Pa. F— PASSAIC ROLLING MILL COMPANY, Passaic. N. J. In the columns headed "Mills Rolling Sliape " the first letter on each line indic.ites the Mill which rolls the exact shape given in tlie Table; the other letters give the l\Iills which roll an approximately similar shape. Tn (jetting the areas of parts of sections, they were taken as shown below: — Back of Foldout Not Imaged Table XX. LIST OF IRON AND STEEL I BEAMS. (Foe Ixformation as to thb Use of this Tablk, skb Table XIX.) S 53 sS i & 1. * i S 1 i i t i < ■3 Web. Pa web. «-D=C,D,E . . B,A,C,D,E . . 10 75 2.63 B,A,C,D,E . . 10 1 AO A,B,C,D,E,F. -n A,BiC,D,F. . . - ■ • ■ 7^ 9 ftR A,BiC,D,E . . ^)^>C>D'E . . A,B,C,D,E . . ) 25 9'nn A 6 66 9*Rn A,BjC,D,E,F. B,A,C,D,E,F. 99 317 A9 9 91 3'545 D.C.E 31.5 2.06 0.44 B,A,C,D,E . . 4 24 2.00 0.31 A,B,D 4 16.5 1.50 0.20 F 4 13.5 1.44 0.17 A 3 25 1.S3 0.53 B,A,C,E 3 IS l.fiS 0.375 A,B,C,E 3 10 l.:jO 0.20 C 11.3 1.3>^ 0.25 C 1.09 E If 3.9 0.63 0.13 to Vel 280000 221600 200000 164960 197000 176800 134700 132800 224000 155760 117040 102400 739400 781300 02600 598800 501300 350000 277000 250300 206200 246400 221000 168400 166000 280000 194700 146300 128000 193300 241600 158400 146000 104600 74640 129100 120200 97000 89000 56480 86640 58500 62000 40240 74600 46000 27760 42720 33000 30400 20320 28000 22160 15600 12960 14400 12080 10640 5680 3840 1520 198000 182400 130700 93300 161400 150300 121000 111200 70600 108300 7S100 77400 60300 93300 78000 72300 57300 34700 63400 41200 3S000 25400 35000 27700 19500 16200 18000 15100 13300 7100 4800 1900 to Paralle "Web. 21 60 6 93 1 21 1.06 55600 69500 1.09 54720 68400 1.28 80500 100500 1.08 57200 1.26 74000 892 I'so 1.23 71360 89200 48720 60900 0.95 38000 6 5' 1 16 1.16 52500 66500 1.12 49600 0.86 29440 36800 0.73 19120 23900 5 04 2' 24 0*7' 0.755 18000 22400 0.80 25040 711 3 29 079 0.84 26320 3^900 0.62 12960 0.69 16000 255 0*71 0.80 20400 26600 0 63 14600 18100 0.70 13600 7*79 2 93 061 0.84 23440 29300 0.76 18800 23600 0.626 10500 13100 1 78 047 0 66 14240 17800 0.53 9200 11500 0.565 9100 11400 0.91 22000 4 95 20^ 047 0.73 16160 20200 0.86 18S00 23500 0.63 10800 13500 0.55 7920 9900 0.73 15840 3 53 1 69 0 48 0.756 13500 16900 0.584 7840 9800 2 54 1 46 0 56 0.76 11700 14600 0.60 7100 0.68 12240 15300 111 062 0 24 0.51 4900 6200 0.715 8800 0.470 3920 4900 0.788 11440 14300 0.662 7600 9500 0.725 9600 130 080 0 39 0.630 6400 8000 06 046 0 28 0.400 3680 4600 0.040 7520 9400 0.493 3620 4400 0.610 5440 6800 0 43 0 37 0 "^5 0.470 2928 3660 1.14 0.83 0.36 0.680 6640 8300 0.79 0.56 0.33 0.600 4512 6640 0.32 0.31 0.19 0.460 2480 3100 0.23 0.22 0.17 0.380 1760 2200 0.47 0.37 0.19 0.569 2960 3700 0.36 0.33 0.20 0.530 2616 3270 0.29 0.29 0.19 0.510 2320 2900 0.21 0.24 0 185 0.460 1920 2400 0.08 0.11 0.096 0.370 880 1100 0.014 0.032 0.036 0.200 256 360 Back of Foldout Not Imaged Table XXII. LIST OF IRON AND STEEL EVEN-LEGGED ANGLES. (Foe I.VFoaMATioN as to Use of this Table, see Table XIX.) A,B,C,D,E... A, B,C,D,E,F. B, C C B,C,F B,C A A A, C,E,F B, A,C,E,F.. , A,B,C,D,E,F. E A,B,C,E,F.. . A,B,C,D,E,F. B A, C,E,F B, A,C,D,E,F. A,B,C A,B,C,D,E. . , A, B,C,E C, B B, E,F A,B,C,D,E,F, A, B,C,D,E,F, B B, E,F A,B,C,D,E,F, A,B,C,D,E,F B F A,B,C,D,E,F, A,B,C,D,E,F C, B E,F A,B,C,D,E,F, A,B,C,D,E,F, C,E,F A, C,E,F B, A,C,E,F.. A, B,C,D,E,F B, D F A,C,E,F. . . . A,B,C,D,E,F A, C,F B, A.C,D,E,F A,B,C,D,E,F A,F A,F A,F A,F Axis Parallel to One S ide. a. £ -J Itansver Be Value •5 lit "Ill sis s" |£ |||^ 6 XG 110 0,662 1 5.00 11. 1.860 60350 76400 16 1.37 6 X6 97.3 6,375 .875 4.865 9. 73 1.820 66100 70100 13.10 1 34 6 X6 6 .500 2.875 6.75 1,685 36900 46100 7.75 1.35 6 X6 50.3 6 .438 2.515 5.03 1,580 31200 39000 6.77 1.35 5 Xo 90 5.562 1 4.60 9 1,610 39760 49700 8.67 .96 5 Xo C2 5.282 .688 3.10 6.20 1,550 31550 39400 6.07 .98 5 X5 37 .400 1.85 3.70 9 3' ''64 1,460 21100 26400 3.77 1.02 HX4J 61.9 4.812 3.095 6.19 1,396 26250 32800 4.88 .772 4}X4J 37.5 4.600 .438 1.875 3.75 7*20 094 1 92 1,286 18000 22400 2.65 .707 4 X4 54.4 4.376 .750 2.72 5.44 1,271 19800 24700 8.45 .634 4 X4 61.6 4.313 .688 2.68 5.16 1.220 18550 23200 3.01 .624 4 X4 28. 6 .375 1.43 2.86 4 36 1 52 1.138 12200 15200 1.86 .660 Six 34 61 3.876 .750 2.65 5.10 1.220 19200 24000 2.40 .470 3iX3S 43.4 3.813 .683 2.17 4.34 1,122 13920 17400 2.04 .470 3iX3j 24-8 3.600 .375 1.24 2.48 1.013 9200 11500 1.20 . 184 34X35 20.6 3.600 .313 1.025 2.06 .930 7200 9000 .95 162 3 X3 36.6 3.438 .688 1.825 3.66 .996 9040 11300 1.05 338 3 X3 28.1 3.250 .500 1.405 2.81 ,930 7680 9600 .95 336 3 X3 14.4 3 .250 .72 1.44 .842 4640 5800 .62 361 2fX-'f 27.7 .563 1.385 2.77 .887 6320 7900 .61 300 -fx-l 16.2 2.760 .313 .81 1.62 115 "59 ' 71 .802 4720 6900 .60 309 2jX2f 13.1 2. 760 .250 .665 1.31 .780 3840 4800 .39 303 24X24 23.6 2.781 .500 1.18 2.36 .770 4880 6100 .52 221 24x24 22.5 2.750 .500 1.125 2.26 1 28 66 '57 .806 5280 6600 .51 227 24x21 11.9 2.500 .250 .595 1.19 .717 3120 3900 .30 252 24x24 10.5 2.500 .219 .525 1.05 .700 2720 3400 .25 240 2^X2? 18.3 2.500 .438 .915 1.83 .740 S680 4600 .36 194 2iX2l 17.8 2.438 .438 .89 1.78 .720 3600 4500 .35 197 2|X2| 10.6 2.250 .250 .53 1.06 .654 2480 3100 .22 208 2|X2| 8 2.250 .188 .40 .80 .690 2080 2600 .17 212 2 X2 19.2 2.250 .500 .96 1.92 .730 4080 5100 .28 150 2 X2 13.6 2.156 .375 .68 1.36 .634 2400 3000 .21 154 2 X2 9.4 2 .219 .47 .94 .580 1765 2200 .13 158 2 X2 7.1 2 .188 .355 .71 .570 1520 1900 .11 160 IfXlf 15 2 .438 .75 1.80 48 '35 31 .640 2800 3500 .18 120 l}Xl| 10 1.875 .313 .60 1 .550 1600 2000 .12 120 IfXll 6.21 1.750 .183 31 .62 .507 1200 1500 .08 129 14X14 9.80 1.088 .375 .49 .98 .19 .17 .19 .510 1360 1700 .09 096 14X14 8.40 1.C25 .313 .42 ,84 .16 .14 .19 ,487 1120 1400 .07 083 14X14 7,10 1.594 ,250 .355 ,71 .14 .12 .19 ,450 960 1200 .06 081 14X14 5.27 1.500 ,183 ,265 ,53 ,11 .10 .21 ,444 800 1000 .05 094 14X14 1.500 ,09 .085 .21 081 i|xM 7.50 1.438 .313 .375 ,76 ,123 .126 .162 ,460 1010 1260 .06 070 liXlj 5.63 1.375 .250 .28 ,66 .077 .079 .138 .404 632 790 .04 071 HXH 3 1 . 250 .125 .15 .30 .044 .060 .147 .358 400 500 .02 067 1 XI 4.40 1.125 .250 .22 ,44 .037 .047 .084 .340 376 470 ,02 046 1 XI 3.60 1.063 .186 .18 ,36 .030 .040 .084 .310 320 400 ,01 040 1 XI 2.30 1 .125 .116 ,23 .022 .031 .096 .296 250 810 ,01 043 IX i 2.93 .938 .183 .146 .29 .019 .029 .066 .286 232 290 fx ! 2.03 .876 .125 .100 20 .014 .023 .070 .264 184 230 2.46 .813 .188 .125 .25 .012 .021 .048 .254 168 210 fx 1 1.72 .750 .125 .085 .17 .009 .017 .063 .233 136 170 Back of Foldout Not Imaged Table XXIII. LIST OF IRON AND STEEL UNEVEN-LEGGED ANGLES. (For IN-POKM-ITION AS TO Use of this Tabi.!!, see Table XIX.) Parallel to She. 2.40 2.53 2.39 2.55 2.59 2.40 2.43 2.56 1.66 1.49 1.56 1.69 1 64 1.61 1.25 1.26 1 . 14 l."8 1.49 1.41 1.23 1.12 1.49 1.36 Ik--' d Parallel to Long Leg. 54500 60000 549C0 32560 50000 40000 30050 27200 54300 43360 25900 32700 22000 37450 27900 18000 30550 18400 16250 29750 97200 68200 82500 68700 40700 70000 57500 38300 34000 6 7900 49200 32400 40900 27500 46300 34900 23300 38200 23000 20300 37200 14050 24040 15900 11460 24000 15700 9850 19200 11850 6200 10800 6320 5500 10400 5040 8160 4480 0960 5840 3840 4640 3280 2480 7.21 1 5.18 1 3.38 1 4.96 1 ISSIKI 30800 19900 14300 30000 19700 12300 24000 14800 13500 10200 5600 8700 2.90 1.98 6.05 4.21 2.98 2.31 1.55 1 .80 1.21 1.06 1 .10 20880 15680 25680 21600 14000 26360 21450 14720 13200 18880 14480 10100 14000 9760 24000 18480 12400 14880 9680 8560 1 1040 lOoOO 0160 148O0 9000 7040 1 2480 9120 14400 9120 5856 14400 8640 4720 5760 3440 1150 4480 2080 5680 3220 2920 2800 1840 3100 26100 19600 32100 27000 17600 31700 26800 18400 16500 23600 18100 12600 17600 12200 30000 23100 15500 18600 12100 10700 13800 12500 I14IJ0 18000 11400 7320 18000 10800 6900 7200 4300 1440 6600 2600 7100 4020 3050 3500 2300 3800 Back of Foldout Not Imaged Mills ROLLIXG Table XXIV. LIST OF IRON AND STEEL TEES. (For Infoemation as to the Use of this Table, see Table XIX.) A,C,E,F. ... E,C,F B E,C B E E,0 A,C,E,F.... 0 E,C,D E,E,C,F. .. . E A,0,E,P ... A,B,D,E,F . E,D E.F E 0 A,0,E,F.... A,B,C,D,E,F, C,E C C,E C,E A, F A F 0 C,E C,E E B, E C, E A,B,C,E,F.. .A,F E,C A,C,E I' C B A,C,F A,B,D,E,F. A,C,E A,C C F C,F A,C,D,E,F. A,E A,0,E,F... 0.600 2 0..500 2 0.500 2 0.500 2 .375 1 0.560 i 0.313 : 0.625 2 0.625 0 . 500 2 0.438 2 .470 1 .438 I 0.375 1 0.530 1 0 . 500 1 0.438 1 0.530 1 0.625 .690 1 . 500 1 0.438 1 0.438 1 0.375 1 0.375 t 0.313 0.375 1 0.250 0.344 5 0.344 0.440 0.375 0.407 1 0.375 0.313 0.313 0.281 0.250 0 . 250 0 . 250 0.188 0.313 0 . 250 0.281 0.250 .260 .20 0.250 50.250 i 0.250 ) 0.250 ) 0 . 250 ; 0.219 ) 0. 188 ; 0.156 ) 0.125 1.91 2.35 1.10 6.25 6.24 5.37 2.50 2.21 1 .50 1 .40 1.39 5.20 1.94 10.50 7.80 5.56 4.65 2.10 1.10 1.32 1 .42 1.10 1.50 1.34 1.14 1.99 1 . 73 1 .46 1 .51 1.12 2.10 1 .88 1.95 1.80 1.95 1.73 1.47 1.51 1.12 1.14 1 .67 1.25 1 .07 1.57 1.37 1.18 16560 16640 17360 17440 6880 24400 19840 16760 16620 7680 14960 11920 10560 8000 1G800 13360 10000 8560 6670 4160 2960 2590 1600 1360 5280 4080 64S0 5280 6000 4500 3860 1536 1344 1000 20700 20800 21700 11000 11100 8000 7100 8100 21800 8600 30500 24800 19700 19400 9600 6000 3500 18700 14900 13200 10000 21000 16700 12600 10700 8340 7600 8500 6000 5200 3700 3240 2000 1700 6600 5100 8100 6600 7500 5630 4830 1000 3520 970 720 2600 10600 3020 2470 1700 540 2600 1920 1680 1250 5.31 6.70 5.24 6.23 4.60 3.94 3.90 2.39 1.53 1.60 ] .30 1 .40 2.12 2.30 2.10 2.09 1.70 1 .06 1 .40 1.40 1 .31 1.61 1.10 16800 16960 18400 16800 16720 14400 12640 13600 8480 11200 11200 10480 12880 8800 8000 8320 6960 7440 6960 7360 5680 6160 5200 4000 5040 4800 3920 1640 3630 3640 2976 3680 3600 3200 3200 3680 3120 2560 2080 1850 1712 1280 1704 1120 1680 1280 1440 1120 Back of Foldout Not Imaged Table XXV. LIST OF IRON AND STEEL DECKS, HALF-DECKS, AND ZEES. (Fob iNFOHftU-TION AS TO USS OF THIS TaBLB, SEE TABLE XIX.) n-^n n-=\-^ ..^n Azia Parari to a ■ c 12 104 5.75 0.406 3.59 2.89 3.90 10.4 221.98 32.80 21.34 5.24 262400 328000 9.33 3.18 0.90 1 25440 31800 B 95 0.438 3.05 2.05 4.40 9.5 168.75 23.35 17.72 4.27 186800 233500 5.17 2.07 0..'i6 16560 20700 C a* 91 5.5 0.375 3.26 2.52 3.28 9.1 164.09 25 . 99 18.06 4.68 207900 259900 7.64 2.78 0.86 22240 27800 B 10 85 5 0.438 3.05 2.05 3.40 8.5 151.53 24.32 17.80 3.77 194600 243S00 6.16 2.06 0.61 16480 20600 C 10 80 5.26 0.375 2.87 2.19 2.96 8 118.22 20.64 14.75 4.27 166100 206400 6.13 2.31 0.76 18430 23100 B,E 9 93 6 0.593 2.60 2.30 4.50 9.3 101.08 20.39 10.96 4.03 163100 203900 4.41 1.76 0.48 14080 17600 C,B,E.... 9 72 5 0.875 2.50 2.06 2.M 7.2 84.77 17 11.83 130000 170000 4.92 0.69 15760 19700 E,B.. ,. 8 84 4 0.750 1.80 4.60 8.4 63.30 14.10 7.51 3.50 112800 141000 2.96 1.48 0.35 11800 14800 C.A.B.E.. 8 Gl 4.625 0.344 2.17 1.85 2.09 6.1 57.66 12.81 9.42 3.50 102500 128100 3.63 1 .56 0.59 12480 16600 B,E .... 7 75 0.438 2.66 2.40 7.5 53.13 12.80 7.24 2.85 102400 128000 5.34 2.14 0.72 17120 21400 C,A,B,E.. 7 52 4.25 0.344 1.86 1.55 1.80 5.2 34.40 9.05 6.60 3.20 72400 90500 2.59 1.22 0.50 9760 12200 B 6 54 4.5 0.438 2 1.40 5-4 25.16 6.97 4.75 2.39 55760 69700 3.08 1.32 0.58 10560 13200 C,B 6 42 3.75 0.313 1.5! 1.28 1.38 21.95 6.55 5.24 2.65 62400 65500 1 .64 .88 0.40 7040 8800 C,B 5 34 3.25 0.313 1.22 1.04 1.11 3.4 12.04 4.29 3.57 2.22 34320 42900 .98 .61 0.29 4880 6100 Half-Deck Beams. 0.625 0.500 0.640 0.484 0.670 0.515 0.578 0.453 0.610 0.4S4 0.515 0.360 10 11.58 9.37 8.63 7.05 6.49 5.35 4.31 3.25 119800 100300 100100 80000 92640 75000 69000 56400 51900 42800 34500 26000 149800 125400 125100 100000 115800 93700 86300 70500 64900 53500 43100 32500 16720 13360 16960 1 2800 18160 13440 11520 8720 11920 9120 7200 4800 20900 16700 21200 16000 22700 16800 14400 10900 14900 11400 9000 5.34 5.68 4.50 3.46 2.72 0.77 ».74 0.90 0.89 0.76 0.75 0.58 10 66 3 —3 0.453 Long«„n. 1 .30 1.30 6.6 87.37 17.47 13.26 5 139800 174700 6.37 2.12 .98 16960 21200 10 67 3 —3 0.375 1.12 1.12 3.46 5.7 76.88 15.37 13.47 5 123000 153700 6.60 1.37 .98 14960 18700 6 48 3J-3 3|— 3 0.422 1 .44 1.23 2.13 4.8 25.78 8.32 5.38 2.90 66600 83200 7.62 2.21 1.56 17680 22100 6 42 0.344 1.25 1 .05 1.90 4.2 22.30 7.15 5.29 2.88 67200 71.500 6.77 2.00 1.61 16000 20000 41 3 —3 0.406 1.20 1 .20 1.70 4.1 15.61 6.25 3.80 2.60 60000 62500 5.85 1.95 1.4! 15600 19500 36 3 —3 0.344 1.04 1.04 1.52 3.6 13.50 6.40 3.76 2.60 43200 54000 5.20 1.73 1 .44 13840 17300 37 0.422 1 .27 1.06 3.7 8.88 4.27 2.40 1 .92 34200 4 2700 5.10 1.76 1.37 14080 17600 33 3 — 2| 0.375 1.13 .94 1.23 3.3 7.80 3.71 2.37 1.896 29700 37100 4.28 1.48 1.30 11840 14800 Back of Foldout Not Imaged Back of Foldout Not Imaged TABLE XXVIII. CLASSIFICATION OF IRONS AND STEELS. Name. Percentage of Carbon. Properties. 1. Malleable iron. 0,26 Is not sensibly hardened by snddeu eooling. 2. Steely iron. 0,35 Can be sligUtly hardened by quenching. 3. Steel. 0,60 Gives sparks with a flint when hardened. 4. Steel. 1,00 to 1,50 Limits for steel ot inaxiniurii hardness and tenacity. 6. Steel. 1,75 Superior limit of welding steel. 6. Steel. 1,80 Very hard cast steel, forging with great diifl- cnlty. 7. Steel. 1,90 Not malleable hot. 8. Cast-iron. 2,00 Lower limits of cast-iron, cannot be ham- mered. 9. Cast-iron. 6,00 Highest carburetted compound obtainable. TABLE XXXII. ULTIMATE BREAKING STRENGTH OF MATERIALS UNDER DIFFER' ENT KINDS OF STRAINS. Dead Load (Static). Wrought Copper and Brass, also Slate, Timber, Masonry, etc. Cast metals : Iron, Copper, Brass, Lead, etc. Intermittent Loads (off-and-on continuously.) If in one direction only. Rollmg (dynamic), If in opposite directions. Wiihont Rolling shock. (dynamic). 1 6 J. 8 i I I'2 6 By combining this table with the safe stresses given in Tables IV and V, that is, taking whatever part of the safe-stress there given for (lead loads, that the nature of the load demands, we can obtain the safe-stress under any manner of loading. Where stresses in opposite directions take place, the material will yield in the direction of the weakest stress. Back of Foldout Not Imaged TABLE XXX. Amount of Extension and Contraction, in inclies, of Cast and Wrought Iron Bars, 100 ft. long, under different strains. Strain, per square inch, in pounds. CAST IRON. WROUGHT IRON. under tension. under compression. Extension, under ten- sion or contraction, under compression. 1000 0,08308 0,09155 0,0444 2000 0,17150 0,18404 0,0889 3000 0,26538 0,37747 0,1.333 4000 0,36443 0,37185 0,1778 5000 0,46890 0,46715 0 3323 6000 0,57874 0,50341 0,3667 7000 0,69393 0,66061 0,3111 8000 0,81440 0,75875 0,3.550 9000 0,94036 0,85782 0,4000 10000 1,07100 0,95784 0,4444 11000 1,30820 1,0.5880 0,4889 12000 1,35014 1,16070 0,5333 l.SOOO 1,49744 1,363.54 0,5778 14000 1,65010 1,36733 0,6232 15000 1,80810 1,47205 0,6667 IfiOOO 1,57871 0,7111 17000 1,68432 0,7556 18000 1,79186 0,8000 10000 1,90035 0,8444 20000 2,00978 0,8889 31000 2,11994 0,9383 32000 2,23145 0,9778 23000 3,34370 1,0323 24000 2,45690 1,0667 25000 3,57103 1,1111 Back of Foldout Not Imaged TABLE XXXI. Length of Cast or Wrought Iron Bars, in feet, that will stretch or contract exactly one inch under different strains. Strain, in pounds, per square inch. CAST IRON. WROUGHT IRON. Length, in feet, to extend one inch. Length, in feet, to shorten one inch. Length, in feet, to either extend or shorten one inch. 1000 1204 1094 2250 3000 583 543 1135 3000 377 360 750 4000 274 269 563 5000 213 214 450 6000 173 177 375 7000 144 151 331 8000 133 132 381 9000 106 117 350 10000 93 104 335 llOOO 83 94 204 12000 74 86 187 13000 67 79 173 14000 61 73 161 15000 55 68 150 16000 63 141 17000 59 133 18000 56 125 19000 53 118 20000 50 113 21000 47 107 23000 45 103 33000 43 98 34000 41 94 35000 40 90 Back of Foldout Not Imaged SCREW THREADS, TABLE XXXIII. \PEOP,bETioNS for NUTS, and BOLT HEADS. ^(^^ Rough Nut=one and one-half diameter of bolt -|- i. Finished Nut=one and one-half diameter of bolt + ^ liough Nut=diameter of bolt. Finished Nut=diameter of bolt — Rough] I Head=one and one-half diameter of bolt -|- ■! ^(^^ Finished Head=one and one-half diameter of bolt -f- ^ — '-^*T-' Finished head=diameter of bolt — 1*^, Back of Foldout Not Imaged Back of Foldout Not Imaged TABLE XXXV. Bearing Values for Ieon and Steel Rivets and Pins. Back of Foldout Not Imaged TABLE XXXVI. Bearing Values for Iron and Steel Pins and Bolts. Back of Foldout Not Imaged Back of Foldout Not Imaged Back of Foldout Not Imaged Back of Foldout Not Imaged TABLE XL. Sheaking, I^endtng and Tensiokal Values for Iron and Steel Pins and Rons 3 to G Inches in Diameter. CROSS SECTION, o «s 1= a ■< f? 7,069 3- 7,070 3* 8,206 8,94G 9,621 3i 10,.321 11,0 J 5 H 11, 7u;; H 12,560 4- 13,364 14,186 15,033 15,904 H 10,800 H 17,721 4| 18,665 4i 19,635 5- 20,629 5^ 21,648 5i 22,691 51 23,758 H 24,850 H 25,907 5f 27,109 5i 28,274 6- AREA OF CROSS SECTION. DIAMETER OF 1 PIN OK ROD. 1 Values (in lbs.) for Full Lines. OOOOOOOOO' oooooooo< oooooooo* - ' ' - O O O 'OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO '0000000000000000000000000 0~ ~ ' ~ " " " - - - - " - - - - - - - - - - ' 'OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO ■ o o o o OOOOOOOOOOOOOOCOOOOO-: cot>'Ooa:iO'-i(Mco-^iotc>t-x<720'-i(Mi:0"Ir^oc^ool>-cDut^■^coc^rHOCicc^-ccor^^otlC^lI-^oa5 Values (in lbs.) for Dotted Lines. Back of Foldout Not Imaged Back of Foldout Not Imaged TABLE XLII. Wrought Iron Riveted Girders.— Strength of the Angle Bars. FoK Different Manners of Loading, Deflection, Etc., Sep. Table XLI. Back of Foldout Not Imaged Back of Foldout Not Imaged "TOG 25- 2-7