vX f. ' 845 (flornrU Untufraftn Cibrarg BOUGHT WITH THE INCOME OF THE SAGE ENDOWMENT FUND THE GIFT OF iliMirg M. Sage 189X B-'Ari.S-^.;>^-b..- |.5.lTil-l.b. CORNELL UNIVERSITY LffiRARIES ITHACA, N. Y 14853 JSne Am Lht&^ '"" Hail Cornell University Library The mechanics of building construction, b 3 1924 015 328 374 Cornell University Library The original of tiiis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924015328374 WORKS BY HENRY ADAMS, M.Inst.CE., etc. JOINTS IN WOODWORK. Demy 8vo, 44 pp., witt large plate of 80 joints, dd. TIMBER PILING, IN FOUNDATIONS AND OTHER WORKS. Demy 8vo, 24 pp., with folded plate, 6rf. STRAINS IN IRONWORK. Crown 8vo, cloth, with 8 folded plates, containing 172 illustrations, 5^. DESIGNING IRONWORK (Second Series). Part I. Steel Box Girders. 26 pp. and folded plate, Part 2. Built-up Steel Stanchions. 59 pp. and 2 folded plates, \s. yl. Part 3. Cisterns and Tanks. 48 pp. and z folded plates, i^. Part 4. Rolled Joists, Compound Girders, and Fire-resisting Floor. 52 pp. and folded plate, IS. PRACTICAL TRIGONOMETRY FOR ENGINEERS, ARCHITECTS, AND SURVEYORS. Crown 8vo, 77 pp., illustrated, 2J, td. net, postage 30'. ENGINEERS' HANDBOOK. Royal 8vo, 576 pp., 7^. dd. net, postage M. CASSELL'S BUILDING CONSTRUCTION. Royal 8vo, 568 pp., 2284 illustrations and 12 coloured plates, Ts. dd. net, postage dd. EXAMINATION WORK IN BUILDING CONSTRUC- TION. Demy 4to, 56 pp., including 23 full-page plates, 2j. dd. net, postage 3rf. THEORY AND PRACTICE IN DESIGNING. Medium 8vo, cloth, 240 pp., illustrated, ds. net, postage 41/. ELEMENTARY GEODESY. In preparation. TO BE HAD FROM THE AUTHOR, 60, QUEEN VICTORIA STREET, LONDON, E.G. Reinforced Concrete Construction In Theory and Practice An Elementary Manual for Students and Others By Henry Adams, M.Inst.C.E., M.I.Mech.E., F.S.I., F.R.San.I., etc. ; and Ernest R. Matthew.s, Assoc. M.Inst.C.E., F.R.S. (Ed.), F.R.San.I., F.G.S., Borough Engineer and Surveyor of Bridlington. 8vo, lOJ. 6rf. net. LONGMANS, GREEN, AND CO. LONDON, NEW YORK, BOMBAY, AND CALCUTTA THE MECHANICS OF BUILDING CONSTRUCTION BY HENRY ADAMS M.lNST.C.E., M.I.Mech.E., F.S.I. , F.R.San.I., M.S.A., Etc. PAST PRESIDENT SOCIKTV OF ENGINEERS, ClVII, AND MECHANICAL ENGINEERS* SOCIETY, INSTITUTE OF SANITARY ENGINEERS, AND ASSOCIATION OF ENGINEERS-IN-CHARGE ; LATE PROFESSOR OF ENGINEERING AT THE CITY OF LONDON COLLEGE EXAMINER TO THE BOARD OF EDUCATION, THE SOCIETY OF ARCHITECTS, THE INSTITUTION OF MUNICIPAL ENGINEERS, THE INSTITUTE OF SANITARY ENGINEERS, THE ROYAL SANITAHY INSTITUTE, ETC. WITH 590 DIAGRAMS LONGMANS, GREEN, AND CO. 39 PATERNOSTER ROW, LONDQN NEW YORK, BOMBAY, AND CALCUTTA 1912 All rights reserved PREFACE During the summer of 1906 the author was requested by the Board of Education to give a course of lectures at South Kensington upon " The Mechanics of Building Construction " to a number of Science teachers selected by the Department from aU parts of the Kingdom, with the object of perfecting their knowledge of the subject and at the same time illustrating the manner in which it should be taught. The course was so well appreciated by the teachers that a repetition was asked for in 1907, and was duly given to another selected group. Those lectures form the basis of the present work ; but in preparing his notes for the press the author has made some alteration in the division of the sections and many additions to the text and drawings, and has revised the remainder where expedient. The details have been expanded, and explained more fully than they were in the first instance, so that what was originally prepared for teachers only is now available also for students, and forms a complete text-book of all the more important branches of the subject. The matter is divided up into thirty chapters so that, if used as lectures, each would occupy not more than one and a half hours in delivery, including the time taken up in sketching the diagrams on the blackboard. The want of a good general work upon the principles of Structural Engineering, which covers the whole ground that a student needs to traverse before he begins to specialise, has often been felt by young architects and engineers, although there are many books dealing with some of the elementary branches which leave little to be desired. This work is intended to supply the deficiency. It must not, however, be vi • PREFACE thought that it is merely a student's book ; the information in it is in such a form as to make it useful as a book of reference for Architects and Engineers who need at any time to look up the theory of their constructions. A glance at the list of contents will show the wide range that is covered. 60, QnsEN Victoria Stkeet, London, E.G. March, 1912. CONTENTS LECTUBE I PAQB Matter and force. Statics. Pressure and reactions. Specification of a force. Parallelogram of forces. Triangle of forces. Polygon of forces. Forces in more than one plane. Leverage. Moments. Parallel forces. Couples 1 LEOTUKE II Force of gravity. Weight. Centre of gravity. Centroid. Neutral axis. Moment of inertia. Section modulus. Bending moment .... 12 LECTURE III Moment of resistance. Modulus of rupture. Stress and strain. Permanent set. Modulus of elasticity. Deflection 18 LECTUEE IV Graphic Statics. Keciprocal diagrams. Bow's notation. Funicular poly- gons. Bending moment diagrams. Shearing forces. Vortical shear. Horizontal shear. Shear diagrams. Combined diagrams .... 23 LECTUEE V Construction of bending moment and shear diagrams, for beams with standard loading and supporting .... . ... 31 LECTUEE VI Rolled joists. Finding moment of inertia graphically. Shear stress in roUed joists. Deflection of rolled joists. Compound girders ... 36 viii CONTENTS LECTUBE VII I'AGE Plate girders. Continuous beams. Comparative strength of structures. Safe load on structures . . . . . il LECTURE VIII Bent girders for staircase. Cambered girders. Cast-iron cantilever. Gallery cantilevers. Painters' hangers 47 LECTUEE IX Gates and doors. Overhanging steps. Framed cantilevers and brackets . 54 LECTURE X Warren girders variously loaded. Lattice girders. Continuous lattice girder. Bent lattice girder 60 LECTURE XI N girders. Lattice girder with uprights. Trellis girder. Compound trussed ' girders. Trussed beams. Combined longitudinal and transverse stresses 66 LECTURE XII Struts — timber, iron, and steel. Stanchions— oast iron, roUed joist sections, built-up sections. Columns — oast-iron solid, mild steel solid, oast-iron hollow, stone solid. Piers — brick, stone ... 75 LECTURE XIII Finding stresses in roof trusses by the " Principle of Moments," and by the " Method of Sections." Allowance for wind pressure. Stresses *n roof truss according to mode of fixing. Stresses in collar-beam truss . . 82 CONTENTS ix LECTUKE XIV PAGE Stresses in queen-post truss. Effect of wind. Bending moment on tie beam. Lantern lights. Substituted members .... 91 LEOTUBE XV Hammer-beam trusses, vertical and inclined loads, rigid and yielding walls. Calculation of bending moment on principal rafter and braces. Braced collar-beam truss . . . . ... .... 99 LEOTUBE XVI Exceptional forms of braced roofs, involving addition of virtual forces, and bending moments on rafters , 107 LEOTUBE XVII Root trusses formed of bent ribs stiffened at joints, with and without tie. Calculation of joints ... ... . . . . 116 LEOTUBE XVIII Braced arch roof trusses. Arched rib truss. Comparison of stresses . . 126 LEOTUBE XIX Stability of walls. Overturning on edge. Pressure on base according to position of resultant. Safe stresses on materials, Boundary or fence walls. Wall with attached piers 1.33 LEOTUBE XX Betaining or revetment walls. Earth pressure. Eankine's theory. Other theories. Section of earth slip. Angle of repose. Line of rupture. Graphic determination of thrust on wall. Analysis of forces. Sloping back. Surcharged retaining walls. External loads at back of wall . . 139 X CONTENTS LECTURE XXI PAGE Stability of stone pinnacle. Pier at angle of porch. Railway bridge abut- ment. Church buttresses. Counterforts. Plying buttresses . . . 148 LECTUBE XXII Pressure of water. Pressure on tank and reservoir walls. Stability of masonry dam . . . • 156 LECTURE XXIII Stability of arches. Forces acting on a voussoir. Curve of thrust. Depth at crown. Primary thrusts. Analogy with girders. Catenary and parabolic curves. Virtual arches, straight arch and lintel. Curve for ordinary distributed loads. Maximum stress. Critical joint. In- complete arches. Minimum thickness of arch. Thrust in semicircular arch. Keystones not necessary 162 LECTURE XXIV Estimation of loads upon arches. Load on ogee arch. Load on culvert. Stability of arch and abutment. Arched viaduct. Stop abutments. Reservoir arched roofing. Concentrated loads. Rolling loads . . . 169 LECTUBE XXV Theory of the modern arch. Curved struts and ties. Arched ribs . . . 177 LECTURE XXVI Vaulting. Groined vaults. Bibs and panels. Finding elevation of ribs. Curve of thrust. Determination of stresses 190 LECTURE XXVII Theory of domes. Domes compared with arches. Curve of thrust in dome. Joint of rupture or point of contrary flexure , ^ jgg CONTENTS xi LECTURE XXVIII PAOX Principles of shoring. Raking, flying, dead and needle shores. Formulae and scantlings. Foundations, width and depth. Supporting power and natural slope of soils. Pile driving and supporting power of piles. Formulae .... ... 204 LECTURE XXIX Gin poles. Derrick poles. Guy ropes. Shear legs. Tripods. Cranes . . 213 LECTURE XXX Reinforced concrete. Beams. Floor slabs. Tee beanxs. Pillars. Formulas and Calculations ... . ... 224 LABORATORY WORK Testing materials. Plotting curves from results. Visit to Works in progress ... . . • . .... . 235 Index . . ..... 237 THE MECHANICS OF BUILDING CONSTEUCTION LEOTUKE I Matter and Force — Statics — Pressures and Reactions — Specification of a Force — Parallelogram of Forces — Triangle of Forces — Polygon of Forces — Forces in more than one Plane —Leverage — Moments — ^Parallel Forces — Couples. It will be useful to commence with an explanation of the terms that will be adopted in dealing with this subject. All materials come under the general term of matter, which may be defined as " the element of resistance in the sensible world." Even the greatest philosophers are unable to say what matter really is in itself, but we all know that it has certain essential properties. They are impenetrability, extension, and figure or form. Impenetrability means that two given portions of matter, or bodies, cannot occupy the same space at the same time. Extension is the fact of occupying space, expressed by the three dimensions of length, breadth, and thickness. By figure or form is meant the presentation of a definite shape at a given instant. Then there are certain accessory properties, and an illustration of each will render its nature clear. Divisibility — one grain of iodide of potassium dissolved in 480,000 grains of water, when mixed with a little starch, will tint every drop of the fluid blue on the addition of a solution of chlorine. Flexibihty — the property of permitting change of shape without disintegration or fracture, as indiarubber. Tenacity or toughness — a steel wire may bear a direct load of 100 tons per square inch without failure. Brittleness — depends largely upon molecular arrangement; steel when heated and suddenly cooled in water may be very brittle, but this condition may be removed by again heating and cooling slowly. Elasticity — ^is the property of recovering its shape after distortion ; a bent spring tends to regain its original form immediately the distorting pressure is removed. In a perfectly elastic body the force of restitution would be equal to the impressed force, but no body is perfectly elastic. Malleability — is the property of allowing the shape to be changed by forging or hammering without severance of the constituent molecules ; soft charcoal iron, gold, and lead are typical of malleable metals. Ductility — is the capability of being drawn into wire ; platinum, silver, iron, and copper are particularly ductile metals. Hardness — may be associated with brittleness, but not necessarily so ; hard wrought-iron is more tenacious than soft iron, but breaks more readily under a sudden blow ; the diamond is the hardest substance known. Then there are three condi- tions of matter — solid, liquid (including viscous or semi-liquid), and B 2 THE MECHANICS OP BUILDING CONSTRUCTION gaseous — depending upon the comparative intensity of the molecular forces of attraction and repulsion, and influenced largely by temperature. In solids, the molecules are relatively iixed, in liquids they are coherent but not fixed, iu gases they are repellent to each other. Hence, solids press downwards only, liquids press downwards and sideways, gases press in all directions. Force is as difficult to define strictly as matter is. Force is that which produces or destroys motion, or which tends to produce or destroy it ; or which alters or tends to alter its direction. By Newton's " Laws of Motion," we learn (1) that every change of state is due to external force ; (2) that every force produces its own result ; and (3) that action and reaction are equal. If only one force act upon a body, motion must ensue, but two or more forces may be in equilibrium and the body is then at rest. Forces at rest are usually called pressures or reactions ; pressure is a force balanced by a resistance. The science of forces in equihbrium is called Statics, and it is the branch specially applying to the Mechanics of Building Construction. It forms part of the larger group of Applied Mechanics, which is that branch of applied science explaining the principles upon which machines and structures are made, how they act, and how their strength and efficiency may be tested and calculated. It is necessary to obtain a clear idea of pressures and reactions. When a surface supports a load the pressure or active force is caused by the attraction of gravitation pulling down the load, but action and reaction being equal, there is a passive force, or resistance, or reaction, tending to push the load upwards with exactly the same force that gravity is pulUng downwards. This would be self- evident on a surface of indiarubber, but it is equally true for all surfaces, although the actual compression and tendency to regain its original form are not visible to the naked eye with ordinary materials. Forces may be represented graphically by straight lines whose position upon the paper gives their line of action or direction ; the length of each line to any given scale represents the magnitude of the force ; an arrow-head placed anywhere upon the line gives its sense, or the direction in which it presses ; and if a force acts upon a body, the point at which it touches is called the point of apphcation. The statement of these various particulars for any given force, pictorially or verbally, is called the specification of the force. If two forces act upon a point, Fig. 1, each tends (by the second law of motion) to produce its own result, but as the point cannot move in two directions at the same time, it takes an intermediate direction more nearly coinciding with that of the stronger force. By forming a parallelogram, of which the two forces make two adjacent sides, the diagonal of the parallelogram meeting the same point represents a single force that could be substituted for the two forces to produce exactly the same effect. This is called their resultant, and it is self-evident that an TKIANGLE OF FOKCES 8 equal and opposite force to this would exactly balance it. This latter is called the equilibrant, and a brief consideration will show that the equilibrant may be considered as a third force that will exactly balance the first two. The converse of all this is equally true, and by working backwards a single force may be resolved into two others, of which either both directions may be given, leaving the magnitude to be determined, or both magnitudes may be given, leaving the direction to be found ; or the direction and magnitude of one force may be given (within practicable limits), leaving the magnitude and direction of the other to be determined. The first proceeding by which two forces are converted into one is called the composition of forces, and the latter proceeding by which one force is converted into two is called the resolution of forces. Two given forces can have only one resultant, but one given force may be resolved into any number of forces. A force may be considered to act anywhere in its line of direction, and to compound two forces when one is acting towards the point and the other is acting away from it as in Fig. 2, one of the forces must be transferred to the opposite side of the point as shown, before the parallelo- gram can be drawn and the resultant and equilibrant found F,g.2 Force X /,q; ^ :y ipy \^. ^9-"^' ^e& ^c When three forces acting upon a point are in equilibrium they may be used to form the three sides of a triangle ; they may be taken in any order so long as the arrow-heads all run the same way round, called " circuitally." The shape of the triangle will be the same as one or other half of the parallelogram, as seen in Figs. 3 and 4, both based upon the forces shown in Fig. 1. The fact of the lines forming a perfectly closed triangle is the test of their equilibrium, and, therefore, if any two forces be used for two sides of a triangle the third side will represent the equilibrant, or reversing the arrow head will represent the resultant. This is called the triangle of 'forces. The conditions of equilibrium for three forces acting on a point, are (1) they must be in the same plane ; (2) the resultant of any two of them must be equal and opposite to the third ; (3) if lines be drawn representing the forces in position, magnitude and direction, these lines taken in order, with the senses concurrent, will form a perfect triangle. . , , , , . When several forces meet in a point they may be compounded in pairs, and then the resultants compounded to obtain the final resultant, as in Fig. 5. Or the resultant of two adjacent forces may be found and this compounded with the next force, and so on, as in Fig. 6, where F/q. J ^ - 4 THE MECHANICS OF BUILDING CONSTRUCTION the forces are the same as shown in Eig. 5. When the equilibrant is found, the ^vhole of the forces put together will form a closed polygon (Eig. 7), and if the forces be placed with the arrow heads to run circuitally, the equilibrant will be represented by the line necessary to complete the polygon. When only two forces act upon a point they are necessarily in the same plane, but when more than two forces act upon a point they may be in one or more planes. The preceding examples have all been in one plane, but cases arise in practice, as for example in connection with shear legs, where the forces are in more than one plane. In such cases a knowledge of practical geometry is necessary because each pair of forces, whether original forces or resultants, must be compounded in their own plane. The method of doing this must be left until shear legs are being dealt with. Parallel forces offer difficulties of their own, and before these are considered the principles of leverage must be studied. It is generally stated that there are three kinds of lever according to the relative position of the fulcrum, weight, and power, but there is really only one Fig -7 kind of lever with various arrangements of the components. In every case there is a force acting with a certain length of arm to balance another force acting with its length of arm. Generally there is an LEVERAGE AND MOMENTS active force called the power * and a resisting force called the weight. If the power be marked P and its arm x, the weight W and its arm y, and the fulcrum or pivot upon which the lever works F, the various arrangements will be as shown in Figs. 8, 9, and 10, and the following equation holds good for them all : in which if any one factor be an unknown quantity, it can be found by dividing the other side of the equation by its companion letter, viz. T, Ww Ww ^, Pa; Pa; a; ' P ' y ' ■^ ♦I* - y /v'g-. 8 ® W - H - ^ >/ /v^. /O Pa; is called the moment of the I "fIs. power and Wy the moment of the -^'^ 3 Jt ~ V ~ "H weight, "moment" being the name given to a force multiplied by a leverage. These equations may be called algebra, but when the values are inserted instead of the letters, it is only common arithmetic, and no student should be afraid of a formula because it has an algebraical look about it. When two parallel forces act upon a point they must be in the same straight hue, either acting together or opposed to each other. In the former case the resultant is their sum, and in the latter their difference. When the parallel forces are not in the same straight line, they may be considered as acting at different points along a lever. Suppose two parallel forces of 4 and 8 lbs. respectively be acting in the same direction at a distance of 15 in. as in Fig. 11, the resultant is equal to their sum, and must act at such a point that 4 multiplied by its arm equals 8 multiplied by its arm. Thus the sum of the forces is to the distance between them as either K - force is to the arm on the opposite side, or 12 : 15 : : 4 : 5 and 12 : 15 : : 8 : 10, or the short arm for the larger force = '^' 15 X 4 = 5, and the longer arm for the I /5 - lO // 12 smaller force = 15 X 8 12 = 10. The proof « >^ s I of accuracy is that 4 x 10 = 8 x 5, ox the. moments are equal. The same results may be obtained graphically as in Fig. 12. Add equal ' Power is hardly a proper term to use, as it really involves " the action of a force through a given distance in a given time," but custom authorizes its use in this instance, when force only should have been sufficient. 6 THE MECHANICS OF BUILDING CONSTRUCTION forces at opposite ends of the lever acting towards each other, which will not alter the balance ; complete the parallelogram and pro- duce the diagonals to intersect, which will give a point in the line of action of the resultant or equilibrant, the magnitude of which will ^ig /2 svl ■«- - /O' 5 ■*■ ' H-\- -/s' • 1 - - A \ / \ 1 / \ . / \ ' / s ' / ^ .i..j \^ s- - IS - P'ig. 13 equal the sum of the forces. If two unequal parallel forces be taken, acting in opposite directions as in Fig. 13, the equilibrant force must be equal to their difference, and must act at such a distance beyond the greater force, that the moment is equal to the moment of the first force about the point of application of the second. Thus, with parallel forces of 6 and 8 lbs. acting in opposite directions at a distance apart of 12 in., the equilibrant force must be 8 - 6 = 2 lbs., acting at a distance of — ^ — = 15 in. The two forces cannot quite be said to have a resultant, and the reason of this will appear presently. When two equal parallel forces act in opposite directions, and not in the same straight line, they form what is called a " couple," and they Si <0| - IZ Fig 14 Ho 4 U>s h- e- 6^ ->, lo F/g IS 4 Lbs. liave no single resultant, but they may be balanced by two equal and opposite forces as shown by dotted lines in Fig. 14. Or if two levers be imagined joined together at right angles as in Fig. 15, then the two forces, forming the couple, tending to turn the system in a clockwise COUPLES 7 12 •<,. Fig. 16 direction, may be balanced by two other forces giving equal moments and tending to turn the system anti-clockwise. In calculating the moments we may either multiply each force by its distance from the fulcrum, or, as many cases arise where there is no apparent fulcrum, we may multiply one of the forces of the couple by the distance between them. The latter is the usual method, the distance between the forces being called the arm of the couple. When several forces in one plane, whose lines of action do not pass through a point, act upon a body, there may of course be numberless variations of magni- tude, position, etc., but certain general principles may be applied for elucidating the re- sults. In Fig. 16 four forces are shown acting at the points of a square. Upon inspection it will be seen that two couples of 5 lbs. can be formed tending to turn the square in opposite directions, leaving two forces at right angles of 4 and 7 lbs, giving a ^ ^ resultant of 8-06 lbs. ^ "■ ^ ^ acting through one comer ^ ^ ■" ^ . of the square. Now let the four forces act at the corner of the square, but all towards the square in the direction of a side, as in Fig. 17. Then they may be combined into two resultants of 18 and 10"3 respectively. The resultant of these resul- tants will be obtained by producing them as shown, and constructing a paral- lelogram of which the diagonal gives the final resultant of 8'06 lbs. acting away from the square in the position shown. It will be evi- dent that in forming a polygon from these forces, as in Fig. 18, the value of the final resultant would be given, but there would be nothing to fix its position ; another condition must be fulfilled to prove that the forces will be in equilibrium, and that is that the funicular polygon must also close. This introduces a new subject, the consideration of which will be better postponed until Graphic Statics in f ' Re*' .i^'- ipCA r- I \ ■V 12 Fig 17 _ ^ 8 THE MECHANICS OF BUILDING CONSTRUCTION general are dealt with, but one other matter may be referred to now that moments have been explained. It is that when any forces are in equihbrium the sum of their moments about any given point is zero. Repeating in Fig. 19 the forces shown in Fig. 1 which are in equilibrium, F/g . /a ^ r/g /9 their moments round point will be found by dropping perpendiculars on to the directions of each of the forces and scaling the distances, then multiplying each force by its distance and taking the algebraical sum, clockwise moments being — and anti-clockwise +. Thus + (7 X 7-8) + (12 X 6) - (1.5-35 x 8-26) = + 54-6 + 72 - 126-6 = EXEEOISES ON LECTURE I Q. 1. Forces of 10 lbs. and 5 lbs. act upon a point with an included angle of 60 degrees. Find the resultant and equilibrant. /v. \ \ \ ) ^\\ v^ ngzo ■^ /F/g.22 Fig. 23 F/g. 2-^ For answer, see Fig. 20. EXAMPLES FOR PRA.CTIOE 9 Q. 2. A force of 12 lbs. acts upon a point. Besolve this into two forces, maUng respectively 30 degrees and 45 degrees with the given force. For answer, see Fig. 21. Q. 3. A force of 10 lbs. acts upon a point. Resolve this into two forces of 7 lbs. each. For answer, see Fig. 22. /^/gr. 25 . Q. 4. A force of 15 lbs. acts upon a point. Resolve this into a force of 10 lbs. at an angle of 30 degrees, and another force to be determined in magnitude and direction. For answer, see Fig. 23. Q. 5. A force of 8 lbs. is acting towards a given point, and another of 6 lbs. making an angle of 60 degrees with the first is acting away from the pomt. Find the resultant and equilibrant. o^ ^"3f ^^ F/p.2? For answer, see Fig. 24. Q. 6. Show a force triangle for each of the above cases. For answer, see Fig. 25. Q. 7. Forces of 5, 7, 9, and 11 lbs. acting upon a point are separated by angles of 90 degrees. Find their resultant and equilibrant, and construct the corre- sponding polygon. For answer, see Figs. 26 and 27. 10 THE MECHANICS OF BUILDING CON8TEU0TION Q. 8. Parallel forces of 5 and 7 lbs. act in the same direction at a distance apart of 17J in. Find the position and magnitude of their resultant. I ■^ -/OS - ■>-'' -73- •- N / -n-5 /^ig 28 For answer, see Pig. 28. Q. 9. Parallel forces of 4 and 9 lbs. act in opposite directions at a distance apart of 5 in. Find the magnitude and position of their equilibrant. F.c,. 29 , ^ F'llJ. 3/ ^'9 JO ,,■4 U>s. i f' <1 '- ■'o 4 Lbs For answer, see Fig. 29. V EXAMPLES FOR PRACTICE 11 Q. 10. Two parallel forces of 5 lbs. each act in opposite directions at a distance apart of 8 in. Find the balancing couple whose arm is 10 in. For answer, see Fig. 30. Q. 11. Forces are applied at the four angles of a square as follows : 4 lbs. at 45 degrees, 10 lbs. at 60 degrees, 8 lbs. at 7S degrees, and 6 lbs. at 30 degrees. Find the magnitude, line of action, and sense of the resultant. For answer, see Pig. 31. Q. 12. Three equal forces are applied at the angles of an equilateral triangle, and act anti-clockwise parallel with the opposite side. Find an equilibrant force or forces. For answer, see Fig. 32. I Note. — If two of the three given forces are combined, the resultant taken with the remaining force will form a couple whose arm is a 6, and can only be balanced by another couple of equal value acting in the opposite direction. Then taking moments about the centre of arm c, construct the parallelogram of moments a d e c. Assume the half-arm of the new couple as c /. Join a f, and parallel to this through d draw d g to meet a line through / parallel with a c. Then <; / will be the value of one of the forces of the equilibrant couple. LECTURE II Force of Gravity— Weight — Centre of Gravity — Centroid — Neutral axis — Moment of Inertia — Section Modulus — Bending Moment. SiE Isaac Newton was the first to show that all bodies attract each other with a force proportional directly to their masses and inversely to the squares of the distances between them. " The reason of these properties of gravity," he said, " I have not yet been able to deduce ; and I frame no hypotheses." As the attraction is proportional to the mass, or quantity of matter in the body, that of the earth practically overwhelms all others ; and as the attraction is towards the centre of the mass, the force of attraction, which we call weight, acts vertically downwards towards the centre of the earth. A falling body travels with an increasing velocity because the force is constantly acting upon it. It falls 16 ft. in the first second, but as it starts from rest the velocity at the end is 32 ft. per second, with which it begins the next second, and it gets another 32 ft. per second added during the second interval of time, and so on. The force of gravity, or the acceleratrix of gravity, is therefore described as 82 ft. per second per second, and is denoted by the letter ^ = 32. We are not concerned with this in statics beyond the fact that the attraction produces the effect known as weight. Mass is the quantity of matter in a body of any volume or temperature, and is constant at all heights and in all latitudes. Weight = mass x gravity, and is constant only at the same level and same latitude. The centre of the mass is called the centre of gravity ; it is that point through which the resultant of the gravities (or weights) of all its pai-ts passes, in every position the body can assume, and is such that if it be supported the whole body will be supported in eqnilibrium ; but the centre of gravity is not necessarily situated in the solid portion of a body, or enclosed by its surfaces, as, for instance, in the case of a wire bent into a semicircle. The centre of gravity of a number of bodies in a straight line may be found by taking moments about a point. Let K - -a: - - ->i ^-^ ^' - -r I «i: SI (^ - £6 -^ ^ ^ K- - - Vs - - -^ "k "- 4 M f^/g. .33 Fig. J-^ a series of bodies, W, W^ W,, etc., be situated at distances of y^ y^ y.,, etc., from a given point A, as in Fig. 33, then the mean centre of MOMEXT OF INERTIA 13 gravity .c will be foimd by the equation kx = ^ "^^V^ J^'^" ^^' VV I -f- W 2 "T* VV 3, etc. The centre of gi'avity of an in-egnlar surface is frequently spoken of, such as the section of a tee iron ; but a surface has no weight, and it is therefore more correct to call it the centroid, or centre of form ; no practical difficulty is, however, likely to arise by using the term " centre of gravity," and it may be justified by assuming a slight thickness for the section under consideration. The centre of gravity of a tee iron, as Fig. 34, may be found by the rule just given, the area of each part being taken as a weight, and the distance of its centre of gravity from the base being seen by inspection. Then (^ X 2) X ? + (^'h X I) X2; ^ ^ (4xi) + (3|xi) •5 + 3-937r. 4-487.J ...qc, ■ u .i. v r i, , = — - = - -= 1'183 m. above the base line. By formula, if a = area of flange, A — area of web, d = total depth, / = thick- ness, H = height of centre of gravity from outer edge of flange. H = li ^*\^l\^'^ ,or\i = \(d + t- ^) but it is always better to rely upon general principles than upon formulae. The neutral axis of a section is a line passing through its centre of gravity. When a beam is bent, the concave surface is compressed and the convex surface is stretched, but there is always an intermediate layer which is unaltered in length ; this is the neutral layer, and coincides with a line through the centre of gravity of every section.' The moment of inertia is the value of a section as regards its power of resistance due to its area and the disposition of its parts. It is the summation of the areas of all its individual parts multiplied by the squares of their distances from the neutral axis. This is generally expressed as I = Say^- It is necessary to divide it up into very small portions, otherwise we could take the whole area on one side of the neutral axis of a beam, as in Fig. 35, and multiply its area by the square of the distance of its centre of gravity from the neutral axis, and the same for the other half, giving 2(6 x 6 x 3^) = 648 for the moment of inertia ; this, however, would be too small a value. The result would be a little closer if smaller divisions were taken, as in Fig. 86, where 2(6 X 2 X 5^ + 6 X 2 X 3' + 6 X 2 X 1') = 840 ; but this again would be insufficient. Doubling the number of divisions, as in Fig. 37, we shouldhave2(6xl X 5-5^-f- 6 x 1 X 4-5"+ 6 xl X 3-5^ + 6x1x2-5^ + 6 X 1 X 1-5^ + 6 X 1 X '5^) = 858, which would still be too small ; but as the number has only increased from 840 to 858 by doubling the number of divisions, it is clear that we are getting much closer to the true result. If the divisions could in this way be caried to infinity, we should find the value come out at 864. It can be arrived at very readily bet' 6 X 12' in a rectangular beam by the formula I = y^ = — ^r^ — = 864 m. units. ' An indiarubber beam 12 in. X IJ in. X IJ in. was here exhibited. One face was left plain, the second wag marked off in transverse parallel lines \ in. apart, the third was covered with squares of f in. side, and the fourth with circles J in. diameter touching each other. The efEeots of compression, tension, torsion, and shearing were thus rendered visible in a very interesting manner. 14 THE MECHANICS OF BUILDING CONSTRUCTION The formulse for this and other sections will be found at p. 92 of my " Engineers' Handbook " (Cassell, 7s. ed. net). The way this formula is made up is instructive. Draw diagonals on the section, as Fig. 38 ; the horizontally shaded portions represent what is called the inertia area. The length of each horizontal line may be looked upon as the measure of the intensity of stress in that layer, the whole width being equal to the \* -A- H T - o - — t> _ /V o 1 o 1 o 1 1 / \ t f- , ^ K -e- H ^ig 35 F-fg 36 /=>?■ J7 Fig. 38 T ^ maximum stress on the outer fibres ; or upon another view, the shaded area may be looked upon as the virtual section if each component fibre were under the full stress. Let A be the area of shaded part on one side of neutral axis, G- the distance of its centre of gravity from the neutral axis, and D the depth from outer surface to neutral axis, then the moment of inertia I = 2D AG ; but D = |<^, A = -r^, G = | x ^i, therefore I = 2 x it? X-^*^ X | X i .0^' lO RECIPROCAL DIAGRAMS 25 different funicular polygon would be drawn, but the closing line would afcill give the true direction for the vector to point 5. Now take a triangular frame, as Fig. 57, draw the load line of the reciprocal diagram Fig. .'(8, then parallel to 1-i and 2-4 draw lines from points 1 and 2 to intei-sect in 4, and draw 4-8 parallel to 4-3 of the frame dia- gram. Here the reactions are obtained without a funicular polygon, but one may be drawn as in Fig. 59 to show the apphcation. No vector is required to point 4 as it is only the external forces that are dealt with by the funicular polygon. Some other useful information not yet men- tioned is given by the reciprocal diagram. The load line having been drawn to scale in Fig. 58 the length of line 1-4 gives the magnitude of the stress in bar 1-4 of the frame, line 2-4 the stress in bar 2-4, and line 3-4 the stress is bar 3-4. Besides this the nature of the stress, whether tension or compression, can be ascertained. Take the parts meeting together at the apex of the frame ; the force 1-2 is acting downwards, therefore place an arrow head in that direction upon 1-2 in the reciprocal diagram. Then, working clockwise, the stress in bar 2-4 is found by putting' an arrow head on line 2-4 in the reciprocal diagram to run circnitally, and transferring it in the same direction to the frame diagram as shown. It will be seen that this force acts towards the joint and therefore indicates compression. In the same way bar 4-1 will be found to be in compression, and it is convenient to indicate all compression bars by tluck lines. With a little practice the stresses can be equally well discovered without actually marking the arrow heads on the paper, and it is better to omit them because, in considering the other joints, it will be found that some of the same lines want the arrow heads the other way on. Take for instance the joint at B ; bar 2-4 being in compression will have the arrow head towards the joint, force 2-3 acts upwards, and putting these arrow heads on the reciprocal diagram (or imagining them there) it will be seen that, to run circuitally, 3-4 acts away from the joint and therefore indicates tension. One other fact 'to observe is that lines meeting in a point in the frame diagram form a closed figure in the reciprocal diagram, and vwe versa, lines forming a closed figure in the frame diagram radiate from a point in the reciprocal diagram. 26 THE MECHANICS OF BUILDING CONSTRUCTION Applied to beams the funicular polygon forms a bending moment diagram, but before these are further investigated it will be well to consider the shearing forces in a beam. When a beam supported at the ends carries a load there is a tendency to cut through vertically at all points, which is represented by Fig. 60. This is vertical shear. There is at the same time a tendency to cause the layers to slide upon one another horizontally as represented in Fig. 61, and as would be shown visibly if separate planks were superposed. This is horizontal shear. Horizontal shear is -only taken account of in determining the pitch of rivets, but vertical shear has often to be considered, particularly when a heavy . load is carried upon a short beam. The shearing force at any point is equal to the amount of load passing through that point to the nearest abutment ; at the abutment the shearing force is therefore equal to the reaction. For instance, in Fig. 62 with central load, if the reaction is measured up- wards from A in Fig. 63 to indicate the shearing force, it will be seen that it remains constant until the load is reached. Then it is lessened by the amount of load passed, so that at the centre it changes from ffg. 60 Fig. 62 rig. 65 + |W to — |W, and remains — fW up to the abutment B. This is a very simple mode of determining shear and applies even to the most complex cases. Now, applying these principles, a combined diagram of bending moments and shearing forces can be obtained. Let Fig. 64 represent a loaded beam to a scale of j in. to 1 ft. Upon the line of direction of the reaction B, draw the line of loads Fig. 65 to a scale of \ in. to 1 ton. Select a pole at 10 ft. distance from the line of loads, and draw vectors. Parallel to these in the space between the line of loads and the beam draw the funicular polygon Fig. 66, and parallel to the closing line draw the vector to give point 5. Now draw horizontal lines from the divisions on the load line across the lines of direction of the loads, and the complete shear diagram will be as shown in Fig. 67. The funicular polygon forms the bending moment diagram to the same horizontal scale as the beam and to a vertical scale of 4 tons to 1 in. x 10 ft. distance of pole = 40 ton-ft. to 1 in. The vertical ordinates of both diagrams give the value at intermediate points. An important point to observe is that the shear is a minimum where the bending moment is a maximum, also that the bending moment at BENDING MOMENT AND SHEAR DIAGRAMS 27 any point is equal tx) the area of the shearing force diagram from the nearest abutment to that point. As the parabola is frequently required in connection with stress diagrams and in designing girders, the simplest methods of constructing It may be given. Fig. 68 shows the_method for ordinary cases, where Ffg 64- 1 \ ' :, 1 1 — ^ 1 1 ? 1 1 g 9/ w per ft run S^ r3*^ i!£ F/g 32 no/ 93 I 1 calculations of strength, the signs merely indicate whether the forces are measured upwards or downwards from the line of origin. With two unequal concentrated loads dividing the beam into three segments x, y, z, as Fig. 97, the bending moment triangle from each load is conveniently placed on opposite sides of the line of origin and the ordinates measured right through, as Fig. 98, but it is equally correct to put them both on the same side, and add on to the outside the over- lapping parts, as Fig. 99. The shear diagram, determined as before, will be as shown in Fig. 100. When a girder is only partially loaded with a uniformly distributed load, as Fig. 101, the principle of the bending moment diagram Fig. 102 can be very easily explained. First the triangle of moments is drawn as if the load were concentrated at its centre of gravity, then the triangle being intersected by lines dropped from the ends of the load, the points of intersection are joined by a straight line, upon which a parabola is STAXDAED BENDING AND SHEAR DIAGRAMS 33 constructed as if it were the whole span of a girder. The shear diagram Fig. 103 is coustrncted upon the same principles as before, and is now a F/q. 94 5 ^ '=79' S6 r/g.95 Fyg.97 (P) (^ WJCfUtZ'i F/g.98 ^1 I I I I I II F/q 99 f 1^ F/g /oo convenient illustration of the statement that the bending moment is a maximum where the shear is a minimum, the point where the shear is ^9-^0/ ^ ^^ f^ ^^ ^ (- - 2 - -H ur per- fc run 'V F/g. /O^ f^t- ^ - 'I fc run ur /XT ft run -^ L i 1 IjlZ' r X 'V' ^1 &lllllill>v. M^yc-^-'^z) ^^./^^ F/g /03 ^ ^ pro /ofi FTg /06 -^ Y nil is opposite the ordinate to be measured for maximum bending moment. In Fig. 104 the distributed load does not reach the supports, but the method of finding the bending moment diagram, Fig. 105, and D 34 THE MECHANICS OF BUILDINa C0N8TKUCTI0N shear diagram, Fig. 106, is the same as the last case, the formulae, how- ever, come out a little more complex. EXEECISES ON LECTUEE V Q. 27. A rolled joist 12 ft. span carries concentrated loads of 1 ton at 2 ft. r,^. J07 i t con p er ft run \-zo^. -4'0'- »K^'C>*»K - 4.0' ■ 12'. O" ^/^r /09 F/'g//0 ng. //4 from A, 3 tons at 6 ft. from A and a distributed load of 1 ton per foot run for 4 ft from B. Construct the bending moment and shear diagrams. EXAMPLES FOR PKACTICE 35 For answer, see Figs. 107, 108, and 109. Q. 28. Draw a parabola with a base of 2 in. and height of 21 in. For answer, see Fig. 110. Q. 29. Draw by another method a parabola with a base of 3 in. and height of J in. For answer, see Fig. 111. Q. 30. A cantilever 8 ft. span carries a load of 1 ton distributed over the outer half ; draw the bending moment and shear diagrams. For answer, see Figs. 112, 113, and 114. Q. 31. A girder, 11 ft. span, carries a uniformly distributed load of J ton per foot run, and a concentrated load of 1 ton at 3 ft. from one end. Find the bend- ing moment and shear diagrams. For answer, see Figs. 115, 116, and 117. Q. 82, A girder of £i span supported at the ends carries a non-central load of W tons at X feet from one support ; what will be the bending moment at x feet from the other support ? For answer, see Figs. 118 and 119. LECTUEE VI Rolled Joists— Finding Moment of Inertia Graphically— Shear Stress in Rolled Joists — Deflection of Rolled Joists— Compound Girders. Although a solid rectangular beam contains the greatest possible amount of material for the dimensions, it is not all equally effective, those parts lying near the neutral axis being of very little value in resisting the bending stresses, because they have so short a leverage. It is clear, therefore, that if a considerable portion of the material could be removed from the neighbourhood of the neutral axis, and placed as far as possible away from it, a much stronger beam would be obtained with a given amount of material. This is the principle of flanged beams, and rolled steel joists represent the most efficient kind. Now that published tables of what ai'e called the " mechanical elements " of rolled joists are so readily obtainable, there is not the necessity that formerly existed for being able to calculate the section modulus from the dimensions ; approximately it is the area of one flange in square inches multiplied by the distance in inches from centre to centre of the flanges. For example, B.8.B. 21, that is, British Standard Beam No. 21, 12 ins. by 6 ins. by 44 lbs. per foot run, has a mean flange thickness of 0'717 in. and web thickness of 0*4 in., as in Fig. 120. Then the area of one flange will be 6 X 0-717 = 4'302 sq. ins., and the mean depth 12 - 0-717 = 11-283 ins., making the section modulus 4-802 X 11-283 = 48-54 inch units. The true section modulus, according to the pub- lished tables, is 52-65, and this takes the web into account. The formula for calculating the section modulus of a rolled joist made up of rectangles, as in Fig 121, is Z = — . Applied to the same joist, this gives 6 X 12" - (6 - 0-4)(12 - 2 X 0-717^ ^ 10368 - 6605-7 _ 6 X 12 72 ~ ' which is very nearly the full amount. The same section may be used to illustrate the two geometrical methods of finding the moment of inertia, and from it the section modulus. The first is shown in Fig. 122 ; it is on the same principle as the rectangular beam. Fig. 38. Every point on the outline may be considered as brought in towards the centre Una as far as the line joining the projection of the point on the upper surface with the centre of the neutral axis. The area of the shaded figure must be obtained by a planimeter, and then the piece must be cut out and suspended from two points to find the centre of gravity, which makes it rather a troublesome method. LetD = the depth from neutral axis to extreme fibres, A = area of shaded figure or inertia area, Gr = distance of centre of gravity of shaded area from neutral axis ; FINDING MOMENT OP INERTIA 87 then this being one-half of the whole section the moment of inertia will be 2DAGr, or the section modulus will be 2AG. The least moment of inertia will be found by taking the neutral axis in the opposite direc- tion and working similarly, but it is only wanted when the rolled joist is to be used as a stanchion. The other method is shown in Figs. 123, 124, and 125. Divide up the section into portions at each change of width as before, and project a horizontal line from the centre of gravity of each portion. Take the area of these portions as if they were weights or forces and number the spaces between them. Then set off distances from the point 1, Fig. 124, to any given scale to represent these areas, and draw a vertical line from 1 equal to half the length of the horizontal line, making the lower end a pole. Then draw vectors from the pole to the divisions on the horizontal line, and construct the funicular polygon, Fig. 125, by drawing lines across the spaces parallel to the vectors. Then the area of the shaded funicular polygon. A, multiplied by the whole area of the section, a, gives the moment of inertia, I, and this, divided by the distance from neutral axis to edge of section, or half the depth, gives the section modulus Z. Thus I =k.a, 7i =^. The common formula in use by architects for the strength of a 38 THE MECHAJsriCS OF BUILDING CONSTEUCTION rolled steel joist is W = ^j— , wbere W = breaking weight in tons in centre, a = sectional area of one flange in square inches plus one-sixth of the web area, d = total- depth of joist in inches, L = clear span in feet. Applied to the 12 ins. by 6 ins. by 44 lbs. r.s.j. for a span of 12 ft., ,, . ... . _- 10 X (4-302 + 1-70 X 0-4) X 12 _, „ . this will give W = ^^ ^. = 50 tons ; allowing a factor of safety of 4, the safe central load would be 12-5 tons, or safe distributed load 25 tons. By the table of safe loads in Dorman, Long & Co.'s catalogue (1906) the safe load on this section at 12 ft. span is 22 tons. With ordinary loads and spans the web of a rolled joist has usually ample strength to resist the shear stresses, but the load-carrying capacity, as measured by the moment of resistance, increases as the span reduces until a proportion is reached where the web is insufficient to resist the shear stress at the supports. It is sufficiently accurate to take two-thirds of the full depth of a rolled joist as the depth of web free to take the shear stress. In the last exam-pie the effective depth of web will be § X 12 = 8 ins., and the thickness is 0"4 in. With a distri- buted load of 22 tons the shear at the support will be 11 tons, or TT— -TT^ = 3'44 tons per sq. in. With half the span the load would be doubled to give the same bending moment, but this load would double the shear stress, and a stiffening plate should then be riveted on the side of the web. The maximum working stresses for rolled steel joists in tension and compression are 6, 7^ or 10 tons, according to circum- stances, being ^, |, and \ of the ultimate strength. The maximum safe shear stress is generally taken as 6 tons per sq. in., but it might be more satisfactory to adopt the formula s = h — \d, where d = depth of web in inches, and s = tons per sq. in. maximum safe shear stress in rolled steel joists. Examination questions that involve the determination of the size of a rolled joist are always difficult because there are 80 standard sizes, and it is impossible to carry all these in the memory. There are, however, certain approximate rules to bear in mind that will facilitate the work. The depth in inches should not be less than half the span in feet nor more than the whole span. The maximum bending moment having been found, the formula M =fad may be used, where /= maximum stress per square inch, say 7J tons, a = area of each flange in square inches, d = mean depth of joist. Or the following rules may be adopted. Let d = total depth of joist in inches, M = maximum bending moment in ton-ft., a = sectional area of one flange in square inches, h breadth of flange in inches, t = mean thickness of flange in inches, w = weight of joist in lbs. per foot run. Then approximately . "^-^J^, K - L-2o:o"- /2 tons c//str/fouteaf - - H Fig. /^S 46 THE MECHANICS OF BUILDING CONSTEUCTION Q. 37. A rolled joist 20 ft. long is supported at two points 10 ft. apart and carries a uniformly distributed load of 12 tons over the whole length. Draw the bending moment and shear diagrams. For answer, see Figs. 151 to 155. /y^. /se I-f- - - - - - /7^j"- - - - -H - - - -/3f0'- - - - >l 'Ta'pl^t^ 1/.,---" ^ ± H- - - - - - - 20 O - - - - - H Q. 38. A plate girder 20 ft. span requires a flange thioliness of 2J in. Find the length and thickness of the plates when the angle irons are equivalent to i-in. plate. Half-cover length may be taken as 12 in. For answer, see Fig. 156. LEOTUEB YIII Bent Girders for Staircases — Cambered Girders — Cast Iron Cantilever — Gallery Cantilevers— Painters' Hangers. Stone staircases are frequently supported on bent girders, either actually bent or formed of separate pieces connected together by plates at the bends. The bending moments and shear stresses are precisely the IG CNC. same as for a straight girder. Fig. 157 shows the frame diagram of one of a pair of bent girders carrying two square landings and a short flight of steps, together with the top end of a lower flight and the bottom end of an upper flight. Fig. 158 shows the bending-moment diagram, and Fig. 159 thie shear diagram. When a girder is bent with a plain camber, a very interesting problem is produced. Fig. 160 shows a girder 20 ft, span, cambered 48 THE MECHANICS OF BUILDING CONSTKUCTION with a rise of 5 ft., and carrying a concentrated load of 4 tons in the centre, the ends being rigidly fixed by bolting down or otherwise. Join the point of application of the load with the points of support, and these lines represent the direct course of the forces. Under the load draw W to scale equal to the load, and complete the parallelogram ; then H will be the horizontal thrust at each end, and T the direct thrust from load to abutment. To ascertain the effect of this thrust, draw a radial line through the centre of the arc at a to meet the line of thrust in b, draw a tangent through the intersection a parallel with the thrust, and from the radial line set off a length Ic equal to the thrust to complete the parallelogram, the area of which gives the maximum bending moment on each half of the bent girder. This may be converted graphically into a line which may be scaled to give the bending moment. To the same scale as W was drawn, take unity from J along Ic, giving point d ; join da, and parallel to da draw line ce to meet the radial line first drawn. Then eh gives the bending moment. By calculation, let W = the load in tons in centre, T = working stress in tons per square inch, say 7 for steel, T = thrust found graphically, A = area of girder in square inches, M = bending moment in ton-inches found graphically, Z = section modulus of section to be tried. Then — -i- _ = F or A^ Z ' 4-5 6 X 12 _, X + — z — ^ ^''^ ^^^ ™^^ *^° unknown quantities, but on referring to a table of sections, it appears that a B.S.B.9, 6 in. by 4i in. by 20 lbs. rolled joist would be suitable. Inserting the values for this 4"5 72 section, we liave^— + j^^ = 0-7G + C-24 = 7, as required. If the ends of the girder were not rigidly fixed, the maximum bending moment WL 4 X 20 . would be -^ = — 7 — =20 ton-ft. instead of 6 ton-ft., and would require a B.S.B.20, 12 in. by 5 in. by 32 lbs. joist. A cast-iron cantilever is an instructive subject to study. The load being distributed, the bending-moment diagram will be a semi-parabola, GALLERY CANTILEVERS 49 r/g./63 -m w ^ V-\ as Fig. 86, and with uniform width of section this would be the theo- retical elevation to give to the cantilever. Practical considerations, however, lead us to the modification of shape shown in Fig. 161. The section of the cantilever has unequal flanges like a cast-iron girder inverted, and very unlike a rolled joist with equal flanges and sym- metrical section. Cast iron being six times as strong in compression as in tension, the lower or compression flange would theoretically be only one- sixth the area of the upper or tension flange ; but practical considera- tions, again, usually limit it to one-fourth, making the section as in Pig. 162, and end view as Fig. 163. The span of a cantilever built into a brick wall should, as a rule, be taken at least 4i inches more than the external projection, to reach the centre of effective bearing surface, and at the back end the cantilever should have a stone template, care- fully fitted down on to it to resist the upward thrust. Taking the dis- tributed load on the cantilever as 2 tons, the span 6 X 12 -|- 4*5 = 76"5 in., the maximum bending moment will be 76'5 ton-inches. Taking the effective distance from centre of pressure to centre of pressure of the bearing surfaces as 14 in., the upward thrust at the back will be = 5"5 tons, and there must be this amount of load provided above, or the cantilever must be anchored down. The load on the template at face of wall will be 2 -f- 5"5 = 7'5 tons, and sufficient bearing area must be given for this load. Gallery cantilevers require very careful consideration, as the case in Fig. 164 will show. The bend- ing moment due to the load on the overhanging portion will be 6 X f = 60 ton-ft. Now, at times the front por- tion only will be loaded, as Fig. 165, therefore the back end must be anchored down sufficiently to balance the front load, or ff = 5-45 tons, making the load on the column 6 -I- 5-45 = 11-45 tons. The bendmg- moment diagram will then be as Fig. 166, and the shear diagram as Fig. 167. When the whole of the cantilever is loaded as Fig. 168 the holding down force needed at the wall end will be 60 - 4 X 5 -x.o r/g. /69 11 -= 3'63 tons, 50 THE MECHANICS OF BUILDING CONSTRUCTION and the load on the column 6 + 4 + 3'63 = 13'63 tons. The bending- moment diagram will be as Fig. 169 and the shear diagram as Fig. 170. The wrought iron hangers used by painters for support- ing the plank upon which they stand while painting the outside of a railway bridge are cantilevers. A hanger, as Fig. 171, having failed in use and killed a man, an investigation was made. The bending mo- ment at the time of the acci- dent was, according to the evidence, 168 x 9= 15121b.- ins., while the moment of re- sistance of a wrought iron bar If in. diameter would be ^ ~ „ triangle = ^ = I of large rect. = „ small „ = 10 X 4' 12 2 X 4- 12 8x4' = f = 10-66 = 14-22 78-22 Area large rect. x dist.- e.g. from N.A. = 10 x 4 x(35 - 2/ = 71-i small triangle X X = 2 X 4 X (6 - 3i)^ = 56-8 8 X 4xix(5^ Add previous result ■ 3iy =64-0 192-0 78-^ 270-2 Moment of inertia of whole figure, say 2702, and modulus of I Moment of inertia 270-2 section = -=-■ = 81. y Distance of extreme lamina from N.A. Sj Moment of resistance = modulus of section x ultimate tensile stress = 81 X 1000 = 81000 Ib.-ins. stress is in aU cases actually less than the formulse appear to give, as hefore mentioned. 56 THE MECHANICS OP BUILDINa CONSTEUCTION Allowing for weight of stone at 150 lbs. per cubic ft. the weight of each , .,, , 5(10 X 4 + 284 + 8 X 4 X i) X 150 „„,,,. step will be -^^ j-rj ^ = 333^ lbs. Then moment of load ^ (W+e.y ^ (3331+3x12x14X5x12) ^ ^^^^^ j^__.^^_^ and factor of safety = f^^ff = 3-225. The nature of the reactions produced by a cantilever can be best shown by a braced structure. Fig. 183 shows the frame diagram of a 1 2 i h %1^[% Ay / -/0 /as p srii /=>g /as- Z " Fig lee 1 S Z 1 ^%Ml^i^' /7^. /av / e 8 1 A /^/g /ae / /. / 2.4 S / 9 3 11.123 braced cantilever carrying a load at the free end and held down by a top load on the short end. Fig. 184 shows the stress diagram. Fig. 185 shows the same cantilever with the short end anchored down, for riy^ \ 6 5\ F/^. /99 W 5 3 tons and M = 57-8 ton-ins., giving a maximum stress of -~ + ^^ = 5-341 tons per sq. in., against 7 tons per sq. in. maximum safe load. By the ordinary method, taking the same distributed load and horizontal span, the maximum bending , ... , WL 50 X 15-5 n„ ^ ,^ 97 X 12 moment will be -g- = g = 97 owt.-ft. = — -— = 58 ton-ins., as in EXAMPLES FOR PRACTICE 59 Pig. 198. Then taking = 7 tons, Z = ^^ = say 8-3, which will be covered by the joist found above. It will be seen that as the maximum stress in Fig. 197 occurs at less than half way, the bending moment will be less than the maximum found in the usual way, but the difference is made up by the direct thrust which has to be added in the first method.) Q. 43. Draw a frame and stress diagram for the bracket shown in Fig. 198. For answer, see Figs. 199 and 200. Q. 44. A braced cantilever consists of three equilateral triangles making two bays on top and one on bottom. It carries an end load of 1 ton. Draw the stress diagram to a scale of 1 in. to 1 ton, and state the magnitude of the reactions. For answer, see Figs. 201 and 202, LECTURE X In 1 1 J ii 2 o 3 o 4 ^ *i *i >■ • \^, / \'/\"/ 1 V ^ £ ' \/' '° V 1 ^c 6 !( 7/ A. \^ \ \ \ /^ ' '/^ /•* ? X '-■t? yS \ 21 \/ "■\ F/g. 214^ S!: 3€ load by the centre support, thus proving practically the result arrived at by the theorem of three moments. Bent lattice girders are much used for overhead footbridges at rail- way stations. Fig. 215 shows the frame diagram of one from actual practice, and Fig. 216 the corresponding reciprocal stress diagram. The verticals 25-26 and 44-45 give a slight difficulty. The work is straight forward until point 25 is reached, then a new start must be made from the centre point 35. As this is the centre point of a symmetrical girder with symmetrical loading, it must come somewhere upon the horizontal line from point 1. Assume any given position for it and proceed with the diagram until point 26 is reached, as indicated by the dotted lines. This point will give the level at which to cut off 25-26, and point 26 being now in its correct position, the dotted figure may be reconstructed in its proper place. It is a mere accident that point 26 occurs on line 10-31, and no attention must be paid to that. 64 THE MECHANICS OF BUILDING CONSTKUCTIOX EXEECISES ON LECTURE X Q. 45. A lattice girder of double triangulation is composed of five bays with bars at 45 degrees, and loaded as in Eig. 217. Draw the frame and stress diagrams. For answer, see Figs. 218 to 222. (Note. — This extra horizontal force 11-1 required to balance the structure appears to be due to 13-14 being in tension from 3-4, while 12-14 has compres- sion from 2-3 and 0-5 of 4-5, leaving 0-5 unbalanced. It is clear that this external force does not exist, and the thrust therefore puts a cross strain on the end vertical which produces tension in the short portions of the top and bottom members making the beginning of stress diagram, as in Fig. 221, and the end of frame diagram, as in Fig. 220. The bending moment on the vertical will be as in Fig. 222. The tension added in top and bottom flanges acts throughout the whole length of girder to modify the stresses first found.) Q. 46. Show by bending-moment diagrams the comparison between (a) two girders carrying a distributed load, supported at the ends, and meeting on a central column, and (6) a continuous girder similarly loaded over the same two spans. For answer, see Pigs. 223 and 224. EXAMPLES FOR PEACTICE 65 Fig Z/7 Figr. 222 L . . ^ . - i F/of. 223 i^zy'ii J LECTURE XI N Girders — Lattice Girder with Uprights — Trellis Girder— Compound Trussed Girders — Trussed Beams — Combined Longitudinal and Transverse Stresses Lattice girders formed of inclined ties and vertical struts as Fig. 225 are the original form of the Pratt truss, although Prof. Jamieson calls them Linville or N trusses. With inclined struts and vertical ties as Fig. 226 they form the Howe truss, while Fig. 227 shows the Rg.22S^ F'ig 226 Fit}.227 modified Pratt truss, or single quadrangular truss, very largely used all over the world for railway bridges. There is no difficulty in ascertaining the stresses under a uniformly distributed load in either of these girders, but with a rolling load and single diagonals those near the centre would be in compression and tension alternately, and to avoid this the central bays are, in practice, braced both ways, so that there will always be a diagonal to take the tension. The maximum stresses under a rolling load may be found by assuming consecutive bays covered by the load and forming a new stress diagram at each step in advance. The stresses will then be collected in a table, and the maximum provided for in the design of each member. Lattice girders with verticals as Fig. 228 were formerly called Howe trusses, but by some writers are known as Whipple-Murphy trusses or Pratt trusses of single intersection. They are strictly indeterminate structures, as the stresses in the various members depend to some extent upon the workmanship. They may be considered as two N girders superposed, as Figs. 229 and 230, each carrying half the LATTICE GIEDER WITH UPRIGHTS 67 load, Figs. 231 and 232 beiag the corresponding stress diagrams, and this consideration provides a very simple solution by the summation of the stresses, as Pig. 233. It will now be found that by assuming half the load to pass down each vertical the final diagram could have been drawn direct, but the proof that there is more in the subject than meets the eye may be gathered from the fact that in Crehore's " Mechanics of the Girder " nearly 100 pages are taken up in eluci- dating the stresses in a truss of this type. Let N = number of bars in any framed structure. V = number of joints in structure, then a ',20 21 F/a/.lSZ simple stiTicture occurs when N = 2V — 3, in any other case there are redundant members and the stresses require certain assumptions. A trellis girder is one with extra diagonals, as Fig. 234, and this of course involves a little more trouble owing to the number of bars, but there is no iaherent difficulty. The stress in the end pillar may be 68 THE MECHANICS OF BUILDING CONSTEUCTION found by inspection, observing the amount of load passing down each part to the support. For instance, the load passing down 7 — 8 is equal to 3 — 4, being a part of 3 — 4 and part of the similar load at the other end of girder ; and the load passing down 7 — 10 is that coming from 7 — 8 + 2—3 + 4 — 5. No difficulty will then be found in drawing the stress diagram Fig. 235. Compound trussed girders, such as the Linville, Whipple-Murphy, or double quadrangular truss, are the most complicated girders to work out, particularly when they carry a rolling load. The confusion of names and types leads Fidler in his "Treatise on Bridge Con- struction," p. 79, to say, "Braced girders are called Linville, Pratt, Murphy and Whipple-Murphy according to variations not easily defined." An elementary case of the latter type may be taken as in Fig. 236, which is loaded on the bottom flange. This may be divided up into three component girders which are shown with their stress diagrams in Figs. 237 to 242. The stresses being marked on the frame diagram, a complete stress diagram. Fig. 243, may be formed, to prove the accuracy of the work by properly closing, but this stress diagram could not have been formed direct in the first instance. A single trussed beam as in Fig. 244, with its stress diagram. Fig. 245, presents no difficulty, but when the beam is divided into three bays as Fig. 246 it is necessary that the load should be uniformly distributed, giving the stress diagram. Fig. 247. When used as a WHIPPLE-MUEPHY GIKDER 69 support for a travelling crane on a gantry, or any similar rolling load, it is necessary to brace the rectangular bay, as in Figs. 247 and 249. If this were not done, when the load is over one strut the other would tend to rise, but when braced with tie-rods the load from one strut is transmitted to the other. To properly understand this bracing, the load must be shown over the other strut, as in Figs. 248 and 250, and the two stress diagrams, Figs. 249 and 251, compared. It would be equally correct in theory to brace with two struts, but ties are more economical than struts. In the case of trussed beams with a rolling load, when the load is midway between two points of support the beam 70 THE MECHANICS OF BUILDING CONSTEUCTION itself is under a longitudinal compression and a transverse stress at the same time. These may be readily combined by means of the formula 1 ^ 2 ^^^^^--,4 5---'^ ^ Fig 244- F,€jf 2/ w j/ 7 over one strut. As there wiU be a bending moment over the other strut due to the upward thrust the frame diagram must be as in Fig. 258, where the strut 4—6 and ties 1—4 and 1—6 are added as substituted forces to take the place of the bending moment on the beam and permit a stress diagram to be drawn This stress diagram wiU be as in Fig. 259, and gives all the direct stresses'in the original members. To ascertain precisely the value of the bending moment take away the added menibers, then resolving the tension in member 3—5 gives as in i ig. 260 a horizontal component of 2 tons and a vertical component of % ton Now the 2 tons thrust against the beam wiU be balanced by the 2 tons com' pression in the beam, and as the reaction is \ ton acting upwards, the S ton force WIU Have a balance of J ton acting downwards as in Fig. 261. At the first EXAMPLES FOR PKACTICE 73 strut there will be the upward thrust of § ton unbalanced, but at the next strut there will be the 1 ton load acting downwards, and the p, ton from strut acting upwards, leaving a balance of J ton acting downwards. At the right hand abut- _- — — "^TTg^ 1 •V 2 ^^"^-..,5 7 3""""^^ 3 r/gi 2S-a ^^-^^^ p = 4 „ 15 „ 20 1) = 3 „ , 20 „ 25 5) _ 2 25 „ 30 ?) — 1i — ■'■a " " 30 „ 35 5» _ 3 formula witl 1 a = 1 800' Stone columns may be calculated by the formula Safe W in tons = ,J — - but care should be taken that the separate stones are not bedded hoUow, or the outer edges will have all the load and spalling will take place. To prevent this there is sometimes a chisel draft 1\ in. wide left round the edge of the joint and afterwards pointed up. Soft cement mortar carefully made with sand screened through a fine sieve, may be used to get a solid joint, but the interposition of a sheet of lead as is sometimes done is apt to set up " hydraulic " pressure and burst the joint. Brick and stone piers may be calculated according to the safe load on the material, say stock brickwork in cement 6 tons per ft. sup., stock brickwork in mortar 3 tons per ft. sup. ; but when the height exceeds six times the least width it is necessary to reduce the load. A suitable formula for the purpose is W = W /24 -_r\ V 18 / where W = safe load tons per square foot on pier, W = safe load on cube of brickwork as given above, r = ratio of height to least width. When a brick pier is bonded to the end of a wall the strength may be assumed to be increased 25 per cent., and when bonded to the side of a wall, the whole thickness, including the wall, may be taken and 25 per cent, added to the strength for the adjacent bonding. These are only approximate rules to give what may be considered a fair allowance. EXAMPLES FOR PRACTICE 81 EXERCISES ON LECTURE XII of S^toM? ^^** "^^ '"^'^^'^° ^°^' °' ^"^ ^° "■ ^'^'^ ^'^^ ^^ required to carry a load (Note.— The approximate rule of f„ ton per sq. in. for a post 10 diameters long wiU not apply as this post would then be 5 in. x 5 in. and is 10 ft. high, which would make it 24 diameters long.) ^ ' Answer. Try a 7-in. x 7-in. post, allowing a factor of safety of 10. Then T,T FS 2-5 X 49. W = ^ = -^gqa = 52 tons breaking weight, "^ 260d' ■^ ""^ 250 X r or fg = 5-2 tons safe load so that this size will be sufficient. Q. 51. What is the safe thrust by Gordon's formula on a 3-in, by 3-in. by |-in. steel angle 6 ft. long, one end fixed and the other pivoted ? . / 26 Answer, p = = 26 /e X 12\ = «»y 1° *°°s per sq. in. ^ "*" mA/ ^ "^ 5 X 3250(^ 2'25 ) on an area of 2-11 sq. in., or say 21 tons breaking weight. Factor of safety = i + -Osl^j = 4 + 1-6 = 5-6, 21 giving the safe thrust as -^ = 3-75 tons. Q. 52. What load will a 12-ft. oast-iron stanchion of cruciform section carry when the four arms are 1 in, thick and project 6 in, from the axis 1 12 X 12 Answer. Ratio of length to least diameter = — = — = 16 diameters long = say 2J tons per sq. in. on an area of 23 sq. in. = 23 X 2-5 = 47'5 tons. Q. 53. A detached brick pier of stock brickwork in cement 2 ft. 3 in. by 3 ft. is 20 feet high. What will be the safe load upon it ? Answer. W = W^^*^g ~\ = el —-i:^ j = say 5 tons per sq. ft. or 5 X 3 X 2-25 = 33-75 tons total at base. The weight of the brickwork at 1 cwt. per cub. ft. will be 20 X 3 X 2-25 = 185 cwt. or 6-75 tons, making the safe load upon the pier 33-75 — 6-75 = 27 tons. LECTURE XIII Finding Stresses in Roof Trusses by the " Principle of Moments," and by the " Method of Sections " — Allowance for Wind Pressure — Stresses on Eoof Truss according to mode of Fixing — Stresses in Collar-Beam Tniss. There are so many varieties of roof trusses and elementary illustra- tions are so numerous in the various text books that only a few special cases need claim attention in this course. First draw a simple truss, Fig. 270, and determine the stresses by reciprocal diagram, Fig. 271. Then take the same truss and load and find the stresses by the principle of moments. Take first the moments on the left about point B, Fig. 272, to find the stress in AC. The forces to consider are the reaction at A acting clockwise and the load A and ^'^. 2Z2 tension in AC acting contra-clockwise. Then |W at A x ^l - ^W at A X 5Z - stress in AC x rise of truss = 0, whence jW x^l-iWx¥ rise = stress in AC. The stress in AB does not affect this ■(yorking because it acts^ through the fulcrum and does not tend to turn the bars round that point. To find the stress in AB by this method, assume the fulcrum at, say, METHOD OF SECTIONS 83 intersection of centre line with AC, marked D. Draw a perpendicular from D to E on AB, then leverage DE = '^""^^^ '" ^^• There is another method that may bs used, called the method of sections, which is based upon the prin- ciple that for equilibrium to exist all the resolved parallel forces acting in one direction must equal all those acting in the opposite direction. In Fig. 273 draw any vertical line DE between A and B, the stress in AB may be resolved into horizontal and vertical components, but the vertical components must be equal to the loads which have to be transmitted through it. The load passing through D is :^W from B, then DE : AD : : load passing through D : stress in AB, AT) or stress in AB = ^W x tj|,, which may be stated trigonometrically as, stress in AB = — - — sin DAE' AE Similarly, DE : AE : : ^W : stress in AC, or stress in AC = ^W x tytt, which is, trigonometrically, stress in AC = stress in AB x cos BAC, or stress in AC = ^ x cot DAE. An allowance of 5 cwt. per foot super of sloping surface supported by truss, taken as a vertical load distributed over the points of support, is suflBcient to include weight of truss and covering of slates, wind, and other accidental loads ; generally, however, it is desirable to take the wind separately. An allowance of 21 lbs. for slated roof and 28 lbs. for tiled roof per foot super will be sufficient for the structural load, and 28 lbs. per foot super normal to the surface will provide for the wind. The actual force of the wind is a doubtful quantity. Unwin's table of wind pressures according to the angle of the roof is generally adopted, but it is based upon inadequate experiments, and a formula that leads to the impossible conclusion that the pressure is greater against a plane raised at 70 degrees than against one at 90 degrees. The author's formula for wind pressure gives a value increasing with the height of the object above the ground and decreasing with the width to be taken. It is log^ = 1-125 + 0-32 log h - 0-12 log w, where p = ultimate wind pressure in lbs. per sq. ft. necessary to be allowed for against a plane surface normal to the wind, h = height of centre of gravity of surface considered, above ground level in feet, IV = width in feet of part taken as one surface, and when the surface is inclined at. 6 degrees to the direction of the wind, the ultimate 84 THE MECHANICS OF BUILDING CONSTRUCTION i to .^ y § § § 5 ? IS ? ^ § ^ 1- § § ^ Si is A 10 lo 8 N 8 5 $ V) lo N 5t ^ 3 5 1 It O 5 ^ ^ ^ 3 ^ s ? § •i 5 s !2 21 2 5! IB a ^ s 5 Is k Is ssa -OS + + + + + 1 1 + 1 1 1 WIND PEESSUEE ON EOOFS 85 pressure normal to the surface may be taken as 'p sin B, or its effect in the same direction as the wind as p sin^ B. The author's practice in ordinary cases is to take 28 lbs. per ft. super as vertical load over the whole truss and 28 lbs. per ft. super normal to the surface on one side only for the wind. Taking for example a simple trussed rafter roof, Fig. 274, 25 ft. span, 6 ft. 3 in. rise, camber \ in. per ft. span, trusses 10 ft. centres, dead load 28 lbs. per ft. super, normal wind pressure 28 lbs. per ft. super. The ordinary method of finding the stresses will be as Fig. 275, where the reactions are taken as parallel. If the truss is fixed at A and is free to slide without friction at B, the stress diagram will be as Fig. 276. If fixed at B and free at A the stress diagram will be as Fig. 277. If fixed both at A and B, the stress diagram will be as Fig. 278. In the latter, after the load line is drawn, a vertical is dropped from point 3 and cut bff by a horizontal from 5. The hori- zontal is bisected to give the full line 7a-6a, so that the fixing at B takes the whole of the horizontal component of force 5-6, and half the remainder equally with the fixing at A. Probably the most correct method of working is that shown in Fig. 279 and Fig. 280, where the reactions of the dead load and wind are taken separately and combined to give the total reactions, which will be found to have different incli- nations. The funicular polygon can be drawn as a check upon the work, but the reactions being found independently they can be added to the load line before the stress diagram is drawn. The difference in the stresses produced by the five methods is shown in the table, and it is noteworthy that the last method gives approximately the mean of all the others. A collar beam truss is the simplest form of truss constructively, but is a very complex one when the stresses are considered. If the walls are taken as rigid and the loading vertical, then the frame diagram will be as Fig. 281, and the stress diagram as Fig. 282. Fig. 282 The horizontal thrusts at foot of rafter being unknown, the method of working will be as follows : Set down the load line 1 to 6 ; then 3-10, 4-10 ; 10-8, 2-8, 5-8 ; 8-9, 1-9 ; and 8-7, 6-7. It will be seen that in this case all members are in compression. If the walls are taken as yielding, the nature of the stresses will be altered and a bending moment on the rafters will be induced owing to the leverage effect of the reactions at the ends of the rafters. In order to work out a stress diagram, virtual forces at right angles to the 86 THE MECHANICS OF BUILDING CONSTRUCTION rafters must be mtroduced to counteract this effect, as shown by dotted lines in Fig. 283. To obtain force 9-1, take moments about the junction of rafter and collar. Then 9-1 will equal reaction 8ar-9 multi- plied by its leverage minus 1-2 multiplied by its leverage and divided by the length of rafter from foot to junction with collar. Then as the Za,Aa Fig 28-i coUar is half way up the rafter, the force 3-3a will be equal to 9-1, and 8-8a the equiUbrant will be equal and opposite to these two forces. As the loading is symmetrical the virtual forces on the other rafter will be the same, and the stress diagram, Fig. 284, may be completed, from which it will be seen that in this case the collar is in tension. The bending moment will be found by taking moments about the junction of rafter and collar, and will equal (8a-9 X its leverage) — (1-2 X its leverage). The next case to take will be with the supports rigid and allowing for wind pressure on one side. Assuming 2 tons distributed for the dead load and 2 tons horizontally on one side for the wind pressure, the frame diagram with loads will be as in Fig. 285. The stress diagram cannot be worked straight away, and it will be necessary to divide the truss into two portions. The upper portion of the truss is shown in Fig. 286, with the resistances obtained from the reciprocal diagram, Fig. 287. The lower portion is shown in Fig. 288, with the forces resolved graphically, and it will be seen that there is a force of 1 05 tons in the collar unbalanced. In Fig. 289 the information obtained by the other diagrams is brought into one view, including the balancing force 2-2a, which is the true equilibrant of all the other forces, and a reciprocal diagram. Fig. 290, is drawn, the funicular polygon being constructed as a check upon the work. In substitution, in Fig. 291, half the amount of 2-2a must be transferred to the supports and combined with the reactions to produce the final re- sultants on the supports. In making this substitution, a bending moment diagram must be added as shown, the maximum ordinate of which is equal to the horizontal force of 0*525 ton multiplied by the vertical distance to junction of collar and rafter = 4 ft. X 0*525 ton = 2-1 ton-ft. The final case will be where the supports are not rigid and wind pressure is taken into account. Assuming the same loading as in the last case, the frame diagram with loads will be as Fig. 292. The STKESSE8 IN COLLAR-BEAM ROOF ■525 525^ Sa Azii!, ' /=ipf.Z96 Fig 297 88 THE MECHANICS OF BUILDING CONSTRUCTION combined loads at each point are shown in Fig. 293, and the reactions are obtained by vectors in Fig. 294, and funicular polygon in Fig. 293. Then the virtual force d = 1-G X a - 0-25 X i acting at right angles to the rafter at the foot. A similar and equal force will be exerted at the head of the rafter, and at the junction of collar and rafter wUl be the equilibrant of the two. A similar proceeding must be adopted for the other rafter and the whole of the forces, virtual and actual, will then be as shown in Fig. 295, from which the reciprocal diagram. Fig. 296, may be obtained. The spreading tendency will cause a bending moment, as shown in Fig. 297, the maximum bending moment being equal to 1'6 x a — 0-25 x h. BXEEOISES ON LECTURE XIII Q. 54. Pig. 298 shows the outline of an abnormal truss for which the are required, and also the bending moment and shear diagrams for the tie. Ffe^. 298 Benafincf rrhmenc a/t'agra^m /or tie ' F/'g^ SO/ I Shear- of/agram for C/e Frg. 302 EXAMPLES FOR PRACTICE 89 For answer, see Pigs. 299 to 302. (Note.— Set down load line la to 3a. Then 3-8, 3a-8 ; 8-7, 2-7 ; 7-6, la-6 ; and 6-5a, 7-5, 8-ia, giving assumed forces 4a-5 and 5-5a. Calculate 4a-4 and 5a-l and set down on load line, thus giving the assumed forces 3a-4 and 1-la. >-^--^o I / *4 Complete the diagram and test by taking pole and drawing funicular polygon. For bending moment diagram take pole P and draw funicular polygon, the closing line of which happens to nearly coincide with point 5. Shear diagram drawn from reactions to virtual forces in the usual way.) Q. 55. Pig. 303 shows a pair of rafters inclined at 39 degrees from the hori- zontal, with loads applied at their centres, and the feet supported as shown ; (a) find and measure the horizontal thrust on the walls ; (6) find and measure the thrusts in the upper and lower halves of each rafter. For answer, see Fig. 304. (Note. — The load ab in Fig. 304 may be resolved into a transverse force ac, which causes a bending moment, and a thrust ad in the lower half of the rafter. The force ac may then be resolved into the two equivalent forces ef and gh, acting /«■ 90 THE MECHANICS OP BUILDING CONSTKUCTION at the points of support. Then at point e, from the other side there wiU be a similar force e/. Combine e] and e/ to give the resultant ek, which is then resolved into the two thrusts el and era, which wiU be the thrust in the top portion of beam. This thrust will also be carried through to the lower portion, in addition ^345^Lbs ^q to the thrust ad already found. Add these two thrusts together to give jo, and combine with the force gh, giving a resultant g'p. This last resultant wiU have to be resisted by an equilibrant gsf, which may be resolved into the horizontal thrust gr and the reaction sg equal to the total load.) LEOTUKE XIV Stresses in Queen-post Truss — Effect of Wind — ^Bending Moment on Tie-beam — Lantern Lights — Substituted Members. A QUEEN-POST truss with a rectangular space in the centre is a de- formable structure as indicated in Fig. 305, the stiffness of the tie-beam alone resisting the bending moments produced by the irregular loading due to the wind on one side. To enable a stress diagram to be drawn Fig. 30S Span SOfo' Trusses /^. -j/a at the leeward support. Span 30 feet, slope 30 degrees, trusses 10 feet centres, total structural load 6 tons-, horizontal wind pressure 30 lbs. per ft. sup. Draw EXAMPLES FOE PRACTICE 97 frame, stieBS, and bending moment diagrams. Scales 8 it. to 1 in,, \ in. to 1000 lbs., J in. to 1000 Ib.-ft. For answer, see Figs. 319, 320, and 321. ^/gr. 322 a « H 98 THE MECHANICS OF BUILDING CONSTRUCTION (Note. — Order of working as follows : — Set down the load line 1 to 8, then 2—13, 7—13; 13—11 horizontal; 4—16, 5—16; 16—10 horizontal; 13—14, 3—14 ; 14—15, 16—15 ; 15—17, 6—17 ; and 17—18, 7—18.) Q. ■ 57. Pig. 322 shows roof truss for an island platform at a railway station. Span 30 feet, slope 30 degrees, trusses 12 ft. 6 in. centres, structural load 28 lbs. per ft. sup., wind pressure (normal) 28 lbs. per ft. sup. Draw frame and stress diagrams. Scales 8 ft. to 1 in., and \ in. to 10 cwt. For answer, see Figs. 323 and 324. LECTURE XV Hammer-beam Trusses, Vertical and Inclined Loads, Rigid and Yielding Walls — Calculation of Bending Moment on Principal Rafter and Braces — Braced Collar-beam Truss. A HAMMER-BEAM truss is a form frequently adopted for church work, and therefore of great importance, but unfortunately no entirely satis- factory stress diagram can be drawn, and the stresses are largely matter of conjecture, varying according to the nature of the support the trusses , 3 : 5 aJQ7 Xl6\ k ts 1 2 JrJ4 f.:^ 2lX9 /2zM 1 a \2jr F,g 3Z7 ^ 10 ng3za receive from the walls. The frame diagram of a hammer-beam truss with vertical loading only, is shown in Fig. 825, and allowing for yielding walls the stress diagram will be as in Fig. 326. Taking the walls to be moderately rigid, the frame diagram will be as in Fig. 327, and the stress diagram as in Fig. 828, the wall piece, 1-12 and 23-10, 100 THE MECHANICS OF BUILDING CONSTRUCTION or wall behind ifc, being assumed to carry one-fourth of the total load. On the assumption that the walls are perfectly buttressed and rigid, the frame diagram will be as in Fig. 329, and the corresponding stress diagram as in Fig. 330. The elevation of a common type of hammer-beam truss, 25 ft. span, 12 ft. centres, is shown in Fig. 331. A slight modification of the truss must be made for the frame diagram, by introducing additional members as in Fig. 832, 15-16, 16-17, i8-19, 22-23, 24-25, 2.0-26, in order to enable the stress diagram. Fig. 333, to be drawn out. It is also neces- sary to divide the load on the lower purlin, between the foot of rafter and junction between rafter and vertical. After the completion of the stress diagram the added members must be removed and the bending moments then produced must be ascertained and allowed for. A portion of the truss on the right-hand side where the maximum stresses occur, is shown enlarged in Fig. 334, and the stresses from the added members will be resolved into virtual forces parallel with, and perpendicular to, the various members as shown, from which the bending moments may be calculated. The diagram of bending moments on the principal rafter is shown in Fig. 335. A braced collar-beam truss 35 ft. span, 13 ft. centres, 28 lbs. per sq. ft. vertical dead load, and 28 lbs. per sq. ft. normal for wind, is shown in elevation Fig. 336, the frame diagram with loading being taken as in Fig. 337. In order to work out the stress diagram, Fig. 338, the members 14-15 and 18-19 must be added. As these members are not actually present the effect of the stresses in them must be considered. HAMMER-BEAM TRUSSES 101 The stress at junction of this added member and the principal rafter may be resolved into two forces, one at right angles to and the other parallel with the principal rafter. At the junction with the collar the Kialc/e Cbmmon natter //, '.X A.'],e;' fi/riin 8'x4'- y^boC, ** ioU 331 stress may be resolved into a force at right angles to and one parallel with the collar, as shown dotted in Fig. 339. The bending moments produced by the forces at right angles to the rafters and collar are 102 THE MECHANICS OF BUILDING CONSTRUCTION shown in Fig. 340. The stresses in the various members may now be worked out and will be as follows : King post, 2 cwt. per sq. in. ; , ' Fig SS3 Fkj. sse HAMMER-BEAM TRUSSES 103 struts, 6 cwt. per sq. in. ; collar, 29 cwt. per sq. in. ; principal rafter at junction with strut, 23 cwt. per sq. in. ; and midway between junction with collar and foot, 7"8 cwt. per sq. in. The end bolt between curved JCL?7^ .",- ncf. SJ8 PnmcipaL rnfter ■%. V 'V V ■sa-Gcrft /"mom puriin 104 THE MECHANICS OF BUILDING CONSTRUCTION brace and collar will have a pull on it of 128 cwt. = 6-4 tons. The net area is 0'42 sq. in., giving a stress of say 15 tons per sq. in. It will therefore be seen that the truss will be perilously near failure if the full loading is realised. ^232 crrC-ft. EXERCISES ON LECTURE XV Q. 57 a. Draw frame and stress diagrams for the modified hammer-beam truss shown in Pig. 341, allowing 28 lbs. per sq. ft. for wind pressure normal to slope in addition to structural load. Calculate the stress in the members. For answer, see Pigs. 342 to 347. Calculation as follows : — , 8 cwt. ,„ , Kmg post ^Q ^^^ .^ ^ = Ig cwt. per sq. m. ; P X 48 struts 88 = ,- -:- , .Z-.X-, — 4-12 cwt. per sq. in. ; 1 + \hxi) W M 2322 collar -^ + g = Jl + „g ^ ^j^;, = 0-8 + 13-5 = 14-3 cwt. per sq. in. ; W M 2287 principal at junction with strut 7+7 = ygi + ,75 — TjS = 1'^ + 13-3 = 14-9 cwt. per sq. in. EXAMPLES FOE PEACTICE 105 The maximum bending moment at the joint in the curved piece will be equal to the stress in 11-23 multiplied by its versin or leverage = 115 'cwt. X 28 in. = 3320 cwt.-in. As a check upon this method, moments may be taken, as in ^^^. . " /' ,' I I . I 106 THE MECHANICS OF BUILDING OONSTEUCTION Pig. 345, about point x where the maximum bending moment will occur. Then (80 X 54) - (15 X 73'§) = 3220 cwt.-in. as before. To find the section modulus at the joint draw out an enlarged view at this point, as shown in elevation. Pig. 346, and section on AB, Pig. 347. Then to find the position of neutral axis, take moments about CD, the iron plate being taken as having a comparative sectional area of 20 times that of the timber owing to its greater strength, but the leverage remains unaltered and will still be taken about the line CD. Then 0-375(3-75 X 0-75 X 20) + 5-25 x 7 X 3-875 + 7-25 X 11 X 17-875 1585-5 y - 56-25 + 36-75 + 79-75 ~ 172-75 = 9-2 in. from AB, or 14-55 in. from the top of rafter, giving a virtual depth of beam of 14-55 x 2 = 29-1 in. Then Z = hd' - h'(b - t) 7-25 X 29-1' - 7-l'(7-25 - 0-75) _ 6(2 6 X 29-1 1033-6, and the stress will be W M 4 + >/ ■ 180 3220 ■ 152-8 + 1038-6 = 1-17 + 3-11 = 4-28 owts. per set. ™- Q. 58. Pig. 348 shows the nave roof of a small church to a scale of J-in. to 1 ft. Draw frame, stress, and bending moment diagrams. Pot answer, see Pigs. 849, 350, and 351. LECTURE XVI Exceptional Forms of Braced Roofs, involving addition of Virtual Forces, and Bending Moments on Rafters. The outline of a roof truss sometimes used for sheds, but of very imperfect design, is shown in Fig. 352. Assuming the supports to be pei-fectly rigid the frame diagram with unity load will be as in Spon ZO O' rig. 3SZ Fig. 353, and the stress diagram as in Fig. 354. The following is the order of^working the latter, 3—11, 4—11 ; 11—10, 2—10; 11—12, 5—12 ; 10—8, 12—8 ; 8—9, 1—9 ; and 8—7, 6-7 ; thus giving the horizontal thrusts at supports. "With free bearings the stress diagrams cannot be directly worked as the tension in the two ties will produce 108 THE MECHANICS OP BUILDING CONSTRUCTION bending moments on the rafters. In order to find these bending moments an additional temporary member 9 — 10 must be added to the frame diagram as in Pig. 355, and the stress diagram for this will be as in Pig. 356. As the member 9 — 10 is non-existent, it must be replaced by virtual forces as in the frame diagram, Fig. 357, each force being equal to the amount of stress in the temporary member. The cor- responding stress diagram will then be as in Pig. 358. Now in order to obtain the measure of the bending moments on the rafter, this virtual horizontal force must be resolved into two forces, one at right angles to the rafter and the other parallel with the rafter, as in Fig. 359, and the final stress diagram, Fig. 360, may be readily drawn out. The virtual forces 3—4 and 7—8 in Fig. 359 will then give the measure of the bending moments, the rafters being taken as beams r/g. SS4 supported at both ends with a concentrated load in the centre. In this case with a distributed vertical load of only 14 lbs. per sq. ft. of sloping surface to allow for truss and covering, exclusive of wind, the cross bracing will require to be 4 in. by 2 in., and the principal rafters IMPEKFECT TEUSSES 109 10 in. by (i in. A slight modification of this truss is shown in Fig. 361, with a king rod inserted, and the corresponding stress diagram will ng. SS8 he as in Fig 362, showing how an unsatisfactory form can be easily converted into a satisfactory one. A modified ^'^-^ ^^J^'-^P wS has been used for church roofs, is shown in elevation, Fig. 36o. 110 THE MECHANICS OP BUILDING CONSTRUCTION The frame diagram with vertical loading and rigid walls will then be as in Fig. 364, the additional member, 10a — 11, being equivalent to the horizontal thrust to be taken by the supports. The stress diagram / ceiUnq boards Span 2S' (f rig. se2 presents no diificulty, and may be drawn out as in Fig. 365. If the supports are yielding, then bending moments will be produced on the rafters, and the stresses will be greatly increased. The frame diagram EXCEPTIONAL FORMS OF ROOF TRUSSES 111 will be as in Fig, 366, and in order to work out the stress diagram the bending moments mnst be replaced by virtual forces at right angles to 112 THE MECHANICS OF BUILDING CONSTEUCTION the rafters. In order to find the value of these virtual forces take moments about the junction of the tie with the rafter, then virtual £ , J _ (11Z> — 1 X its leverage) — (la — 2 X its leverage) . ~ leverage of 1 — la EXAMPLES FOR PRACTICE 113 Having obtained this force, 3 — Sa and 11 — 11& may be readily found, and the operations may be repeated for the other side ; in this case as the loading is symmetrical the virtual forces will also be symmetrical. The frame diagram, Fig. 367, may now be drawn out, and it will be seen that there is a large increase in stress in the majority of the members in addition to the bending moments produced by the virtual forces. EXERCISES ON LBOTUBE XVI Q. 59. Fig. 368 shows a skeleton outline of a roof truss for part of a brewery ; draw frame and stress diagrams, allowing 2J owt. per ft. run for the weight of the ventilator. s 1 N. ^ ^ 2 'V ?^ / / a IS / / ± _ A^ 1/ ^-^M (:r-- y"^^—^ y^-^x.f^ / ^ \ /.-;'--' A ^^^ '=''9 370 /^ 10 xii For answer, see Figs. 369, 370. Q. 60. Fig. 371 shows oak ribs and purlins for roofing a small church ; test the sufficiency of the scantlings. I 114 THE MECHANICS OF BUILDING CONSTRUCTION For answer, see Pig. 372, frame diagram with substituted bracing ; Fig. 373, stress diagram ; Fig. 374, tlirust in rafter ; and Fig. 375, bending moment diagram. (From the frame diagram it will be seen that the roof divides itself into two portions, which may be assumed to be pivoted at the apex. Therefore the loads and reactions form two couples acting in opposite directions. The loads 1 — 2, 2 —3, 3 —4, and one-half of 4 — 5 are equal to and balanced by reaction 9 — la of 48 cwts., and the moment of this couple may be obtained by multiplying 48 cwts. \a Span Z2:£' Tnusses 8'. O" centres JOctfC aa ^ by 5-625 ft., which is the mean arm of the couple, giving 48 x 5'625 = 270 owt.-ft. This is balanced by the couple consisting of the force 1 — lo, and the horizontal thrust at crown which will be the same value, acting with a leverage arm of A/^. -JTT 9 ft. Therefore force 1 — lo = 270 cwt,-ft. yfz = 30 cwt. The load line may then be set down and the stress diagram completed without difficulty. In this case, however, the stress diagram appears to be of no assistance in calculating the EXAMPLES FOE PEACTICE 115 sufficiency of the scantlings. The thrust in the rafter may be obtained as shown in Fig. 374. The bending moment diagram for the rafter is next obtained as in Fig. 375, and the maximum is found to be 720 owt.-in. Then by the formula W M 82-4 720 -r + -y the maximum stress will be -r^ + -gj- = 0-6 + 8-9 = 9"5 cwt. per sq. in. \30cfyc. -pr- I 3h^ .,(1 ? ^'9' 375 compression ; and taking oalc as crushing with 3-2 tons per sq. in. (as given by TT .V 3'2 X 20 ^'^ '' — 9^5 — ~ "'"^ factor of safety. The size of the rafter is taken as 9 in.- by 6 in. instead of 12 in. by 6 in. owing to the underside being moulded. LECTURE XVII Boof Trusses formed of Bent Bibs Stiffened at Joints, with and without Tie — Calculation of Joints. In dealing with bent-rib trusses it will be well to consider first a simple frame, as Fig. 376. It is clear that at least one additional member must be inserted to enable a stress diagram to be drawn. This would be a diagonal from one side of top to opposite side of bottom, and when riq. S7e riq 377 the wind blows to meet the diagonal it would put it in compression, but when it blows from the opposite side it would put it in tension. To avoid this change of stress it might be desirable to put two diagonals Fi^. 378 ^-© both designed for tensile stress only, so that whichever way the wind blew there would be a member to take the whole diagonal stress by tension. The diagonals may, however, be arranged to each take half the duty, being in either tension or compression according to the side BENT RIB ROOF TRUSSES 117 from which the wind was blowing. To ascertain the stresses in that case the frame might be dealt with in two stages ; first as in Fig. 377 with half the total load and the diagonal in compression, giving the stress diagram. Fig. 378 ; then as in Fig. 379, with the other half of the load and the other diagonal in tension, giving the stress diagram. Fig. 380. The combination of these two would give the frame diagram, Fig. 381, and stress diagram. Fig. 382, but it must be understood that if the diagonals are designed to resist tension only, the frame diagram would be as Fig. 379 and the stress diagram as Fig. 380, with all the stresses and reactions doubled. The calculations of the bending moments on the joints had better be left for consideration with the next example. A convenient form of roof truss for factories, involving some special considerations of stress, is shown in part elevation in Fig. 383. It consists of a rolled-joist tie forming floor girder, and double channel iron rafters with gusset plates at the angles. To obtain a stress diagram additional members must be assumed temporarily as in the frame diagram. Fig. 384, for which Fig. 385 is the corresponding stress diagram, but this diagram cannot be worked out straight away ; the frame diagram must be divided up into three parts, as in Figs. 386, 387, 118 THE MECHANICS OF BUILDING CONSTRUCTION and 388, taking one-third of the total load on each. The coiTesponding stress diagrams will then be as in Figs. 389, 390, and 391, from which Span 3l'0' Trusses 12' ff' cencr-es ScruccuraL toad IS Lbs yoer sif ft Normat nind pressure 40UIS per sif fe ru»rer) 2€0bs. per st^.f'e. (upper) ■Tm 5 X 2li X l/lis C/Krr7neis ~, R'p^ -10/ II Nf %^V/4 2/9- / \ sIsA 8 \v \^ Nl^ /v$r. 384 JO a /if: figr.^S BENT RIB ROOF TRUSSES 119 the complete stress diagram, Fig. 385, to one-third the scale of the others may be constructed. The effect of the additional members will he to avoid bending moments on the joints A, B and C, the bending moments when they are removed being equal to the stress in the member multiplied by the perpendicular distance to the neutral axis of the joint, and so far it is assumed that no stress is taken by the joints D and E. Next, assuming that the joints A, D, and E take the stress, the frame diagram will have to be as in Fig. 392, and the corresponding stress diagram as in Fig. 393. Then, as the joints B, C, D, and E will all be helping to stiffen the truss at the same time, half the bending moment found from Figs. 384 and 385 at joints B and C, may be taken together with half the bending moment found from Figs. 392 and 393 at joints D and E, making the final bending moments more nearly uniform at all the joints. Now, the maximum bending moment at A from Figs. 120 THE MECHANICS OF BUILDING CONSTRUCTION 384 and 385 will be 0-43 ton x 72 ins. = 31-2 ton-ins., at joints B and C, 2-1 tons X 52 ins. = 109-2 ton-ins., and taking half this = 109"2 — ~ = 54-6 ton-ins. The maximum bending moment at A from Figs. 392 and 393 will be 0-57 ton X 72 ins. = 41-04 ton-ins., and at joints D and E, 1-12 ton x 108 ins. = 120'96 ton-ins., and taking half = ^^^ = 60-48 ton-ins. 2 It will be desirable in such cases as this to make the section of the rafters uoiform throughout, so that after assuming a probable section the maximum stresses and moments must be allowed for by the formula W M — ± -= , the result being kept within safe limits. Similarly it will be desirable to have the gusset plates uniform ; the maximum case may. therefore, be taken and calculated as follows. In principle the gusset is as Fig. 394, the stress being a maximum at outer edge and nil at apex of joint. Taking a cross section through A — B the variation of stress will be as ordinates to a triangle (Fig. 395), the centre of effort being at two-thirds the breadth of gusset, which will therefore be the lever arm. Then if the maximum stress in compression be fixed at 5 tons per sq. in. and the thickness of gusset as ^ in., the maximum resistance per inch wide will be 5 X | = 2| tons. Then 2| reducing to nil gives an average of I5 tons per inch. Let x = breadth of gusset, then the moment of resistance will be 1| x a- x fa; = |a;^ and the greatest bending moment being 60 ton-ins., x = / ^ X b __ g.^g^^ say 9 ins., from the centre of the joint, or a total of, say, 10 ins. This truss without a tie would have the reactions in Fig. 384, combined with BENT EIB ROOF TEUSSES 121 the forces substituted for the tie and producing combined reactions as in Fig. 396, which may be converted to vertical and horizontal forces, F/g. 393 r/g S3^ n-g. S9S- V X 2 'S Cons per sq In the former giving the direct loads and the latter the overturning force on the walls. The bending moments and other stresses would be those 122 THE MECHANICS OF BUILDING CONSTRUCTION due to Fig. 384, and not to Fig. 392, because the bottom gussets would be absent. If the walls are not rigid, it is the same as removing the horizontal forces, and increasing the bending moments on the joints. Fig. 397 will give the frame diagram for this case, and Fig. 398 the stress diagram, which can be drawn direct. EXEECISES ON LECTUEE XVII Q. 61. Draw the stress diagram for the truss shown in Pig. 399. Span 25 feet, trusses 7 ft. 6 in. centres, structural load 21 lbs. per ft. sup., and horizontal wind pressure 42 lbs. per ft. sup. For answer, see Pigs. 400 to 403. (The frame diagram with loading will be as in Pig. 400. Now, as the truss EXAMPLES FOR PRACTICE 123 may be considered as pivoted at the supports and at A and B, in order to obtain a stress diagram virtual forces must be introduced at the latter points. To obtain these forces divide the truss into the two parts as in Figs. iOl and 402, taking into account the amount of load 4 — 5 coming on to A and B. The reactions on these two parts may then be found as shown, giving the two virtual forces and the reactions on the supports which take account of the reactions from the virtual Seruct(/rai. Loaa 2/ Lbs. per f^.s^up HorixonCaC mna/ pressure 42 U>s per fCsup. forces. The stress diagram, Pig. 403, may now be drawn out and completed. To find the true reactions the amounts coming on to the supports from the virtual forces must be subtracted from the reactions found by Figs. 401 and 402 . The method of doing this is shown in Fig. 404, and the true reactions will be as shown by the dotted lines in Fig. 403.) Q. 62. Fig. 405 shows the line diagram of an elliptical roof truss 20 ft. span, 7 ft. 6 in. rise, and 10 ft. centres. Draw the frame and stress diagrams, allowing for a vertical load of 42 lbs. per ft. sup. EXAMPLES FOR PRACTICE 125 For answer, see Figs. 406 and 406a. (NoTB. — The bending moment on the principal rafter is found by multiplying the stress in the member by the perpendicular distance to the curve, and the W M maximum stress is worked by the formula -:- + =-.) r/g. 404- Span 20' O' Trusses /O'O" centres Rise y'.G" ycrc. Loact 42 U>s. per fe. sup. rtg. -406 LECTUEE XVIII Braced Arch Roof Trusses— Arched Kib Truss— Comparison of Stresses. In dealing with arched ribs, or bent-rib roof trusses, it will be well to commence with a braced arch as Fig. 407. It will simplify the case if it be assumed semicircular, and the bracing should be about 45 degrees to a radial line. Owing to this example being semicircular and the resultant of wind pressure on each portion being normal to the curve, all the resultants will pass through the centre and produce an equal pressure on each support ; the structural load being vertical and symmetrical will also produce equal loading on the supports. The reactions may therefore be obtained by joining the extremities of the load line and bisecting, otherwise it would have been necessary to BKAOED ARCH KOOFS 127 calculate separate reactions for each differently sloping force, and the "reactions would then be joined continuously to give final resultant as shown at A and B, Fig. 407. The stress diagram, Fig. 408, can be set out in the usual way, but it should be worked from both ends, as otherwise there will probably be a difficulty in getting the figure to £P. close, owing to the risk of small errors creeping in, due to the shortness of the lines to which parallels have to be drawn. Jn order to compare the stresses with those obtained by treating the 128 THE MECHANICS OF BUILDING CONSTRUCTION truss as a pure arch, it will be useful to set off the stresses in the bars radially, as in Fig. 409 for outer flange, and Fig. 410 for inner flange. Now the investigation of the case as a linear arch may be proceeded with. Draw an elevation as Fig. 411, add the resultants of the structural and external loads, and set down the load line, Fig. 412. Join the extremities and bisect to give point 18, this being permissible for the reasons described above. Then draw a horizontal line both ways through point 18, and from point 2 draw a line parallel with 2-20 (Fig. 411), and from point 16 a line parallel with 16-48 (Fig. 411), bisect 20-4B in Fig. 412, giving approximate position of pole for obtaining the best curve of thrust. Draw the vectors from pole to all the points on load line, and parallel with these draw the curve of thrust in Fig. 411. Professor Goodman says in his " Mechanics Applied to Engi- neering " that with irregular loading an infinite number of curves of ARCHED RIB TRUSSES 129 thrust may be drawn, but that the true curve will be the one whose ordinates give equal areas inside and outside the linear rib. Judging by the result of the braced rib this does not appear to the author to be exactly correct, and the mode of working shown in Fig. 412 is put forward as being the best working approximation. In order to compare the maximum stresses produced in the braced rib with those given by the method now under consideration, set out on Figs. 413 and 414 the centre line of rib and radial ordinates obtained M by the formula T ± jj, where T = thrust across each space respectively, M = bending moment given by the product of the thrust T into its perpendicular distance from the centre of the space on neutral axis of rib, and D = depth of rib between centres of gravity pf flanges. Now for the stresses in the flanges, draw elevation curves as in Fig. 413 and Fig. 414, notice the position of curve of thrust in Fig. 411, and that when it passes within the line of rib the outer flange will be tension and the inner compression, and that when it passes outside the line of 24S rib the outer flange will be compression and the inner tension. This will indicate whether the + or - value in the formula should be given as the ordinate indicating the stress in the outer and inner flanges respectively. The regularity of the curves shows probable accuracy of calculation and plotting. Comparing Figs. 409 and 410 with Figs. 413 and 414, the similarity of stress will be apparent, although the actual amounts do not agree. It looks as if the line of thrust in Fig. 411 should be a trifle more eccentric to make the stresses in the two methods of working more nearly alike. E 130 THE MECHANICS OF BUILDING CONSTRUCTION la comparing tlie maxfmum stresses produced in the flanges of a braced girder and a solid web girder, it will be found that while in the WL latter they are given by the formula -^, in the former they vary according to the number of bays made by the lattice bars, being less in every case, but approaching the value of the solid web as they become moi'e numerous. In the case of the arched rib this relation does not seem to hold good, as the flange stresses are, on the whole, shown to be greater in the braced rib. EXERCISES ON LEGTUBB XVIII Q. 63. Pig. 415 shows cast liron roof principals 36 ft. span and 10 ft. rise, placed at 15 ft. centres, over a warehouse in Manchester, and carrying the roof by EXAMPLES FOR PRACTICE 131 four symmetrically placed porlins and a stmt from the ridge. Each rib is of I section, 7 in. by 4 in. by j in. thick, with pivot at each end, fixed to 14 In. by 9 in. floor beam. The load from the purlins may be taken as 3 tons vertical on the windward side and l.J tons on the other side, the ridge transmitting the mean of these or 2\ tons. Find graphically the reactions and curve of thrust and check the value of thrust and reactions by calculation. For answer, see Fig. 416 and Fig. 417. Set down the load line 1 to 8 in Pig. 417 and select any pole 0, draw vectors, and construct the funicular polygon in Fig. 416, the closing line of which will give point 9 in Fig. 417. From this point 9 draw a horizontal line, and from point 2 draw 2a parallel with the curve tangent 2a in Fig. 416, and from point 7 ^5'<'-^, draw 76 parallel with 76. Bisect ah to give point 10, and draw vectors from point 10 to the various divisions on the load line, these will then give the direc- tions for the curve of thrust crossing each space. By calculation the horizontal component of the thrust in the arch due to the central load W will be 7—, where 0.95 V 36 I = span and \> = rise, -r — ^^p = 2-025. The additional horizontal thrust at 132 THE MECHANICS OF BUILDING CONSTRUCTION foot caused by a load W at a distance x from centre will be also be the increased thrust at crown. W(; - 2x) iv This will ^(^^-^^^^■^^) . 1.6, 3(36- 2 X 13-83) ^ ^.^^^ 4 X 10 ' 4 X 10 Then from the other side 1-5(36 - 2 X 7-33) _ 1-5(36 - 2 x 13-83) iino — -^^' 4^rio — =°^^^' 2-025 + 1-6 + 0-625 + 0-8 + 0-313 = 5-363 tons total horizontal thrust on each side. For the vertical thrust or reaction on the left we have (3 X 31-88 + 8 X 25-33 + 2-25 X 18 + 1-5 X 10-67 + 1-5 X 4-17) -r- 36 = 6-5 + 1-5 = 8 tons, and on the right 3 + 3 + 2-25 + 1-5 + 1-5 - 6-5 = 4-75 + 0-75 = 5-5 tons. Also as a check upon the horizontal thrust wo have, by taking moments about the centre, (8-0 X 18 - 1-5 X 18 - 3 X 13-83 - 3 X 7-33) H- 10 = 5-352 tons. Without a tie the reactions would be as Fig. 418, but the tie being substituted for the horizontal reactions leaves only vertical reactions as in Pig. 416. LECTURE XIX Stability of Walls — Overturning on Edge — Pressure on Base according to position of Resultant — Safe Stresses on Materials— Boundary or Fence Walls — Wall -with attached Piers. In considering the stability of walls it is usual to take a length of one foot as representative of the whole. The simplest form of calculation is that for overturning and it conveniently illustrates some of the general principles. Fig. 4T9 represents an ordinary 9-in. boundary wall, or fence wall. If failure takes place by wind pressure while the mortar is green it will presumably overturn at the joint just above the ground line. In Fig. 420 the upper portion of the wall is shown to a larger scale. The total force of the wind multiplied by the distance from the centre of effort to the joint under consideration is the moment of effort. The moment of the re- sistance is given by the weight of the wall multiplied by the leverage on which it acts, that is, half the thickness. Then if p = wind pres- sure lbs. per sq. ft., to = weight of wall in lbs. per cubic ft., /* = height of wall in feet, t = thickness of wall in feef, for equilibrium, "'T.'-f.'V' W F'.g 4/9 Fi^-}20 and the parallelogram of forces will show the diagonal passing exactly » through the outer edge of joint. If we consider the wind pressure which will just overturn the wall as the measure of stability, then the stability of such a wall varies as -7- • or directly as the weight per cubic foot, inversely as the height, and directly as the square of the thickness. Failure may also take place by sliding when the ratio of thickness to height is equal to or less than the coefficient of friction. The coefficient of friction for fresh mortar is variously stated as 0'5 to 0"75. The shearing strength of old mortar is about | ton per sq. ft. The safe load in compression under ordinary conditions on — 134 THE MECHANICS OF BUILDING CONSTEUCTION 9 tons per ft. sup. 6 „ 5 3 5 3 Blue brick in cement . . Stock „ ., • • ,, „ lias mortar . „ grey lime . Cement concrete (6 to 1) Lias lime concrete (4 to 1) The safe load in tension on — Brickwork in cement (fresh) ... 3 tons per ft. sup. „ 6 months old . „ lias mortar (fresh) „ „ 6 months old „ grey lime mortar (fresh) „ „ 6 months old . , The pressure on foundations should not as a rule exceed for Gravel and compact earth 2 tons per sq. ft. Ordinary subsoil . . . 1| „ „ Made ground (6 months) f „ „ As walls are not built to be overturned the calculation of their stability should be limited to working conditions. A plain rectangular 3 tons per ft 1 A>^ 42/ f^ig*2l wall, as Fig. 421, subject only to its own weight will produce a uniform W pressure on the base = -j- , as shown by the ordinates, or where the weight per cub. ft. iv and the height h are given, the intensity of the pressure will be = wh. If, however, a horizontal force acts against DISTRIBUTION OF PEESSURE ON FOUNDATIONS 135 the wall at any point it will cause the resultant to deviate from the centre, increase the intensity of the pressure on the side toward which the resultant approaches and reduce it on the opposite side, as in Fig. 422. The sum of the ordinates, or the area of the figure contain- ing them, will remain unaltered, because no change has been made in the vertical loading. When the resultant reaches the edge of the middle third of the base, as in Fig, 423, the ordinates of pressure will fZf±±=iI- - H form a triangle having twice the depth of the parallelogram, and it is generally stated that this position is the limit of safety. It is an incorrect expression to use, the limit of the middle third only signifies that there will be no tension on the inner edge, the compression on the outer edge will be according to the load, and may readily exceed the limit of safety. On the other hand, the middle third may often be passed without exceeding the limit of safety. Suppose no tensile strength or adhesion at the base of wall, and the resultant to be pushed over to if from outer edge, as in Fig. 424, then by Prof. Crofton's theory the figure containing the ordinates will be a triangle having a base of 3 times the distance from resultant to outer edge and a maximum depth equal to the load divided by \\ times the distance. W The simplest form of the expression is p = | . -5- • Whether the resultant be shifted over by a horizontal force, or be due to unsym- metrical loading, or to a thrust at any angle, the vertical component of the resultant must be taken as W and the distance from where it cuts the base to the outer edge as A. When the wall has tensile strength, it does not come into play until the resultant passes the middle third, as in Fig. 425, and the maximum W M and minimum pressures are then found by the formula -r ± y where W = the vertical load or vertical component of the resultant, A = sectional area of base, M = bending moment, or product of the load into its distance from the centre of gravity of the base, Z = section modulus of the base. For 1 foot run of a simple wall, Z = -^ becomes \f. The same units must be retained throughout, whether tons, cwts., or lbs., and feet or inches. When the resultant reaches the outer edge, as in Fig. 426, the maximum compression wUI be four times the mean pressure under central load and the maximum tension twice the mean pressure. The same formula applies when the resultant falls beyond the base, but the stresses are greatly increased, as in Fig. 427. To find the wind pressure Q?) per square foot on face of wall to produce any given maximum pressure (K) per square foot on base W M M without tension on inner edge -j + g" = K, or K = wA + r^, but when 136 THE MECHANICS OP BUILDING CONSTRUCTION M the resultant is at edge of middle third K = ^wh, therefore n2 <= wh, or M = \whf^, also M =^A X Ih, therefore ^W = ^wM, or ph = ^wf, whence p = -^. The overturning pressure will, therefore, always be three times the pressure by the middle-third rule. When a wall is buttressed, a unit length from centre to centre of the panels must be taken as one piece in the calculations and the neutral axis and moment of inertia found as for a tee section, the line through the neutral axis being substituted for the centre line of the plain wall. A very curious fact comes to light in the course of the investigation, viz. that adding a buttress weakens the wall. The buttresses from centre to centre are usually 1^ to 2^ times the height of the wall or 15 to 20 times its thickness. Fig. 428 represents a unit length of an ordinary buttressed wall. The brickwork being taken as 112 lbs. per cub. ft., and the wind pressure as 28 lbs. per sq. ft. The neutral axis will be distant from face of buttress, _ BD" - h cP _ 12 X l-1 2;r - 10-5 x -375' _ y ~ 2(B1) - id) ~ 2(12 X l-i2r» - 10-5 x 0-;i75) ~ ^''^'^> and from plain face of wall, a; = D - ,y = 1-125 - 0-717 = 0-108. ^'^^^ ^- 12(BD-M) = (12 X 1-125 -'- 10-5 X 0-375^)^ - 4 x 1 2 x 1-125 x 10-5 x 0-.375(l-125 - 0-375)^ 12(12 X 1-1 «- 10-5 X 0-375) = 0-597. I ■5')7 I •507 - =:^^-^ = 0-8.32 =:Z„- = -.^^g= 1-46.3 = Z^ Then for buttress side W M _ 6(12 X -75 + 1-5 X •■37.5)112 12 X6 x 28 X 3 A "^ Z, ~ (12 X -75 + 1-5 X 3-7.5) "*" 0-832 = 672 + 7260 = 7941 lbs. per sq. ft. compression, and for plain side - - ,y- = 672 - 7^'^ = 672 - 41.34 = - 3462 lbs. per sq. ft. tension. A /j^ l-40.i For the plain wall without buttress Z„ and Z^ will be equal, viz. 12 X -7-5" iBD^ = ^ = 1-125, and the stresses from wind pressure will be equal on each face of wall but of opposite sign, viz. W M 6 X 9-1875 X 112 , 6048 _„,,.,„„ p^.qiu ^, A ± Z " iH875 n25 ="'2 ±53/6 = 6048 lbs. per sq/ft. compression on outer edge and 4704 lbs. per ft. tension on inner edge. Trying the effect of piers of the same width and varying projection it will be found that the resulting maximum stresses form the curves STREXGTTH OF BUTTRESSED • WALLS shown in Fig. 429. Theoretically counterforts have the same effect as buttresses, but differences arise in practice. While a buttress does not tend to separate from the main wall when under thrust a counterfort does, and is only kept in connection by the bonding, and where much projection occurs the bonding requires to be strengthened with hoop iron. The popular idea is that counterforts act chiefly by their weight while but- tresses act chiefly by extending the crushing edge further out from the line of the resultant, but this explana- tion is not supported upon analysis. The reason for putting piers on both sides is that from whichever side the wind blows there will be buttresses to resist it. When a wall varies in thickness at different heights each stage should be calculated separately. c ■^ ^ ^^ 1 ^ ^^ eon, 1 o« EXERCISES ON LECTURE XIX Q. 64. What wind force per foot super would overturn a 14-in. brick wall, 10 ft. high and weighing 108 lbs. per foot cube, neglecting the strength of the mortar ? Answer. Assuming the wall to overturn without crushing ■ph = wt^ :. P = '^ = l_2i|^' = 14.7 lbs. per ft. sup. Q. 65. Determine the height of a 14-in. brick wall which would just fail Fi^. ^SO I*- — - ur — *|* Dry sand, clay or vegetable earth 30 Loamy earth, loose shingle, clay well drained ... 40 Firm gravel, and hard dry vegetable earth .... io The working theory in designing retaining walls is that while the /v$r '^S2 natural slope is determined by the ultimate angle the soil will assume, after an unlimited period, only half the material above this slope would fall away upon the sudden failure of the wall, and therefore only half the amount requires supporting, hence the line of rupture is made to bisect the angle between the natural slope and the vertical, and the half wedge next the wall is the part considered to produce the thrust. Fig. 433 shows the ordinary mode of finding the thrust on a retaining wall and the resulting stresses at the base. This is set off as follows. Draw the proposed section of wall ABCD, which may be already given or may be assumed approximately to be height X degjn_wedge ^ f^_ ^j^j ^ ^^ ^ ^ 90 From a horizontal line through the wall at the lower ground level set off the natural slope ^ according to the material to be supported, bisect STABILITY OF RETAININa WALL 141 the angle between this line and the vertical to give the line of rupture. Find the centre of gravity of the wedge of earth between the line of rupture and the wall, and drop a vertical to cut the line of rupture in 0. Calculate the weight of the wedge of earth, and set it up to scale from as Oa. Draw a horizontal line through to give the line of thrust against the wall, and from a parallel with the line of rupture draw ab, then &0 will be the total amount of thrust from the earth, /y^'^.-fSJ ♦j^K -s'o\- H J" collected and acting in the line Ob. Now find the centre of gravity of the wall, and drop a vertical line cutting Ob produced in the point c. From c horizontally set off the thrust cd equal Ob, and from c vertically downwards set ofif ce 6qual to the weight of the wall in the same units as the weight of earth. Complete the parallelogram cefd, draw the diagonal cf, which will be the resultant, and produce it to meet the base line in g. As in this case it falls outside the actual base of wall, it is evident that the inner edge of the wall will be in tension, and care must be taken not to overstep the safe limits. The actual stresses will W _^ M _ 24 , 24(1 + -26) = 36 cwt. be given by the formula -r- + = + ^n or 1*8 tons per sq. ft. compression at outer edge, and 20 cwt. or 1 ton per sq. ft. tension at inner edge. Among the common errors in this class of work are (1) joining the centres of gravity of the wedge and wall for the direction of thrust. 142 THE MECHANICS OF BUILDING CONSTRUCTION and setting off the weights in each direction from the centre of gravity of wall to form the parallelogram ; (2) calculating the bending moment from the vertical through centre-of gravity of wall instead of from centre of base ; and (3) taking the length of the resultant to measure ^ /=/g. 43S ir/g. -^j^ the load on the base instead of its vertical component. The thrust of the earth does not increase the total load on the base, it only affects its distribution. Now to prove this mode of working, let be the point in the line of rupture at which a vertical through the centre of gravity of the wedge of earth would fall, then the forces at point must be in equilibrium. In Fig. 434 let AB represent the line of rupture, and the point at which the forces meet. Draw W vertically equal to the weight of wedge upon any suitable scale. From the upper extremity parallel with the line of rupture draw a line to intersect with a horizontal through 0, giving T the direction and magnitude of the total thrust at back of wall. There will be a certain amount of friction between the wedge of earth and the back of the wall depending upon the nature of the surface, etc., which may be approximately estimated at half the thrust, equivalent to a coefficient of friction of 0*5 ; this friction will act in direct opposition to W, as shown at F„. Deducting this from W, a parallelogram may be drawn for W — F„ and T, and the diagonal or resultant produced to the opposite side of for an equal length to form the equilibrant, which will be the resultant of the reaction of the earth behind the wedge and the friction along the line of rupture. These two directions are given, therefore the value of each force will be found by completing the parallelogram of which this equilibrant is the diagonal. Having traced the complete forces in action at point 0, a force polygon may be drawn to correspond, as in Fig. 435, which can be readily constructed without the preliminary work now that the principles are known. It must be remembered that these forces are totals, and it will be instructive to analyse them into their details as in Fig. 436. Draw the force lines as in Fig. 434, and the line of rupture AB. The triangle ABC represents by the vertical ordinates the individual pressures on AB which go to make up the weight "W. The point of application of force ANALYSIS OF PKESSURE ON RETAINING WALL 143 W being at oiie-tbird length of AB from A the distribution of pressures will be a triangle on base AB. Join de, then C/ being the perpen- dicular of the triangle ABC, set up Og = C/, draw gh parallel with de, and hi parallel with AB and cut off by the perpendicular Ki. Join j'B, then the triangle AB» represents the reactions on the line of rupture to the same scale as the wedge of eai-th. Similarly the thrust being applied at one-third the height of wall from base, the distribution of stress will be found by joining dj, setting up OJc = CB, drawing M parallel with dj to meet 0; produced in point I ; then kl produced to m in the hori- zontal line from A, and -Cm joined, gives the whole triangle of pressures AGm. The graphic method shown in Fig. 433 will equally apply to a wall supporting water as in the case of a tank, reservoir, or dam, but the angle of repose or natural slope of the water will be 0°, so that the line of assumed rupture will be 45°. Numerically the thrust in any case will be T = I ivh^ tan^ — ^-?' or, using Rankine's form, which is a little simpler to work, but not so easy to understand, T = IwW' , ~ ^!" 7 . ^ 1 -f- sm <^ When the material behind a retaining wall is waterlogged, through the absence of weep holes, the pressure is considerably increased beyond even 144 THE MECHANICS OP BUILDING CONSTRUCTION that due to water of the full depth. Sir B. Baker stated that the thrust of dry earth being taken as 1 and water ^, the thrust of a waterlogged soil would be 1^. The triangle of thrust in Fig. 433 may be turned the other way round and set off as in Fig. 437. Then the thrust for walls with back sloping out or in will be perpendicular to back and at one-third the height as in r,gr 'fsr /=;g ^8 /=/ff 439 Figs. 438 and 439, it being remembered that in Fig. 438 the weight of wedge acting will be less, and in Fig. 439 more, than in Fig. 437, because the quantity of earth varies in the wedge between back of wall and line of rupture. In the case of reservoir walls the triangle may be set off in the same way but the angle will be always 45°, and the vertical line will be the weight of the wedge from line of rupture as in the case of earth. Figs. 440 and 441 show the common methods of dealing with water pressure against walls. The pressure varies uniformly as wh from the surface to the bottom, and the whole may be represented as the ordinates to a triangle. The centre of pressure will be opposite the centre of gravity of the triangle at one-third of the height, and the total pressure assumed to be collected there will be \wW, or \wM, the same as found by calcu- lating the weight of the wedge in the previous method. SURCHARGED RETAINING WALLS 145 A surcharged retaining wall is one supporting a sloping bank of earth rising above its top. The surcharge may be limited and definite, with a flat top, or " infinite," i.e. when the angle coincides with the natural slope and extends beyond the limit of the line of rapture ; or it may be intermediate as when the slope of surcharge is less than the natural slope but yet extends for some distance. Ran- kine's graphic method applies particularly to the latter case. Let AB, Fig. 442, be the back of a wall 15 ft. high supporting a bank of earth which slopes at 1^ to 1, while the natural slope is 1 to 1. Put in the line of rupture BO, draw the angle ABD = 4>, and produce CA to D. Bisect BD in E, and from E as centre draw the semicircle BD. From A drop a perpendicular on to DB, cutting it in point F, and when r/y 443 produced cutting the semicircle in point G. From centre D with radius DG cut DB in H. Then the horizontal thrust acting at one-third the height will be T = iM;(BH)^. Let the earth in this case weigh 1 cwt. per cub. ft., then the thrust JK = 28"57 cwt., which being at 5 ft. height gives an overturning moment of 142"85 cwt.-ft. From J draw JL parallel to natural slope and KL vertical, then the thrust in direction LK scales 34'3 cwt., and its perpendicular distance from B = 4'16 ft., giving an overturning moment of 142"69 cwt.-ft., showing practically the same result whichever line is taken for thrust. From this graphic method it will be seen that when the surcharge is equal to the natural slope the horizontal thrust T = \w(h cos ^y = \wli? cos^ 4>. There is no recognised formula or means of arriving at the thrust L 146 THE MECHANICS OF BUILDING CONSTRUCTION upon a retaining wall when the earth at the back is subject to an external load, such as a st«am roller, crane, warehouse wall, stack of bricks, etc., but some such method as that shown in Fig. 443 will probably give approximate results. Find the thrust from wedge of earth as usual, then from the point of application of the external loads draw lines parallel to the line of rupture to find the point at back of wall at which thrust is to be taken. Draw triangle of weight and thnist for each and combine the thrusts in successive parallelograms. If the front wall of a building occurs behind the retaining wall, the centre of the foundation at the underside will give the point for the line parallel to the line of rupture. Any load that lies on the earth out- side the line of rupture may be ignored. EXERCISES ON LECTURE XX Q. 67. It is desired to put a J-brick wall in cement to protect a bank of earth 5 ft. high assumed to have a natural slope of 30 degrees. What should be the least batter of the wall to relieve it from all thrust ? Answer. Natural slope ^ = 30 degrees, line of rupture 90- ^_ 90-30 = 30 2 ~ 2 degrees from vertical, which will be the batter of the wall. The wall should be in hard bricks and have rubble drains at back with weep holes at intervals as in Pig. Hi. Weep hole F/g. 444 Q. 68. Pig. 445 shows a breast wall with an infinite surcharge. Investigate the stability. Answer. See Pig. 446. The maximum stresses will be as follows : — W,M 7.7X0-5 ,,,„A ,. compression and i-& cwt. per sq. ft. tension. Q. 69. Pig. 447 shows a warehouse wall at the back of a brick retaining wall. Fmd the maximum pressure at base of wall. EXAMPLES FOR PRACTICE 147 iout^s^- ^^^ ^'^' ^^^' ^^® maximum pressures at base of wall will be as W M_84 , 84 X 4-9 A Z ~ 7 I X (1 X 7^) 12 ± 50-4 = 62'4 cwt. or 3-12 tons per sq. ft. compression, and 38-4 cwt. or 1-92 tons per sq. ft. tension Noturai slope of eorch SO' tVeiqht of earch lOOlMs per cub Ft frfg 447 r/g. 448 LECTURE XXI Stability of Stone Pinnacle — Pier at angle of Porch — Railway Bridge Abutment — Church Buttresses — Counterforts — Flying Buttresses. Thekb are various interesting and special cases of stability that may now be considered. Fig. 449 shows a stone pinnacle reduced to its simplest elements. Assume it to be of stone weighing 140 lbs. per cub. ft., the shaft 3 ft. square, and the sides of apex making an angle of 75 degrees with the base. Dealing with the upper part first, the area of the side 42 lbs. per soft 3SSA l£s fieoibs I Base, s'o" square will be AB X BC 3 X 5-8 = 8-7 sq. ft., 2-2 and assuming a horizontal wind pres- sure of 42 lbs. per sq. ft., the total hori- zontal force wUl be 8-7 x 42 = 365-4 lbs. Part of this will slip off and the re- mainder will act normal to the surface, which by triangle of forces is found to be 353 lbs. Produce this force hue to cut the centre of gravity line through the pyramid and set off the value, to scale, beyond the intersection. The solid contents of the pyramid are area of base X \ vertical height 3 X 3 X 5-6 16-8 cub. ft., Figt ■443 and the weight 16-8 X 140 = 2362 lbs. Set this off below the intersection, com- plete the parallelogram, and draw the diagonal for resultant = 2460 lbs. This cuts the joint at 0-25 ft. beyond the centre and, using the vertical component of the resultant = 2450 lbs., creates a bending moment of 2450 X 0-25 = 612-5 Ib.-ft. STABILITY OF STONE POETICO 149 Then the stress on the two edges by formula W . M 2450 612-5 A ± Z = S^TS ± fx3~^^- = *°^ ^""^ ^^^ ^^'- ^' '1- ^*-' which is of course very little. Now the resultant force from the upper part must be transmitted through the prismatic block below, but it will be well to reduce the scale. The weight of the pier is 3 X 3 X 10 X 140 = 12600 lbs., set this off downwards from the centre of gravity of pyramid, complete the parallelogram and draw the diagonal for resultant. The wind pressure against this part will be 3 x 10 x 42 = 1260 lbs. acting through the centroid half-way up. Combine this with the last resuJtant to form a parallelogram, and the resultant will then be found to be 15200 lbs., cutting the base at a distance of 0-75 ft. from the centre line. The vertical component will be 15130 lbs., then W ^M. 15130^15130x0-75 ,,,, , ^ ^^„, ^ , -,«« -r iv A ± Z = 3ir^ ± i X 3 X 3- = 1681-1 ± 2521-6 = 4202-7 lbs.. or 1-875 tons per sq. ft. compression at outer edge and 840-5 lbs. or 0-375 ton per sq. ft. compression at inner edge. A porch consisting of four piers with arches between is not un- common in connection with mansions and public buildings, as in Fig. 450, elevation, and Fig. 451, plan. The thrust from each arch will be as shown in Fig. 452. The method of finding the thrust will be explained subsequently when arches are under consideration, and the accuracy of the thrust 71 cwt. must for the present be assumed. Now let Fig. 453 be the elevation of pier and thrust from front arch. Fig. 454 plan of pier and thrust from both arches. Fig. 455 diagonal elevation of pier. The next step is to find the resultant of the two thrusts in their own plane and to show this on the diagonal elevation. The order of working is shown by the small letters g, h. Fig. 456, projected from plan, hi = ca ; M = ef; Im, km, the two thrusts in the plane of the paper, mn resultant. This resultant carried back to go gives the diagonal thrust on Fig. 455, giving finally the vertical component "W = 296 cwt. acting at a distance of 0-6 ft. from outer point on the area 3 ft. square placed diagonally. The bending moment will be 296('-^ - O-e) = 296 X 1-52 = 450 cwt.-ft. The section modulus = -A- = 0-118^?^ = 0-118 X 3' = 3-186. Then the maximum stresses will be ^ + M = 29C + _i^- = 33 + 141 = 174 cwt. (say 8-7 tons) per sq. ft. A Z ® 3-I80 compression, and 108 cwt. (say 5-4 tons) per sq. ft. tension. This □ ■ 10' o - □ rig 4S2 rig 4SO Pier \ /^rch _ ^Pier Scone BO Lbs Der cutift O* ill Pier I /^S^ J ^'^'^ Porch cowereaf ivith eUxs Lead Pig. 45/ Icifcf frtm fcii- P}g -fSf RAILWAY BRIDGE ABUTMENTS 151 compression is rather high for Bath stone, but might possibly be per- mitted in first-class work with no flaws, to which Bath stone is rather subject. The tension is about the maximum that may be permitted with very carefully made Portland cement mortar, but four rows of hoop iron bond embedded in the masonry of the string course over the arch would remove all doubt as to the stability. In the case of an abutment for a railway girder bridge, as in Fig. 4:57, there are several points to note. There will usually be wing walls to support the embankment, and although these are bonded to the abutment it is not usual to con- sider them as adding any strength, because the slightest disturbance of the foundations results in a crack or fissure between the two parts. When a train is standing on the embankment just clear of the bridge the weight supported over the wedge of earth will help to increase the thrust and must be allowed for ; the train may, however, be on the bridge and embankment at the same time, the portion on the bridge adding weight to the abutment, and therefore increasing the vertical component of the thrust. But, again, the train may be on the bridge only, adding weight to the abutment but not increasing the thrust at the back. The maximum stresses will evidently be pro- , duced when the bridge and adjacent embankment are both loaded, and in the absence of special information it may be taken as 5 cwt. per foot super, or say 5 tons per foot run of its face on the abutment, and 5 cwt. per foot run on the wedge of earth at back. The method of working will be as shown, the final resultant cutting the ground line 4 ft. from centre of base of wall. Then F/q 4S7 W M A - Z 183 183 X 4 6-375 - \{\ X 6-375'0 = 28-7 ± 103 = 131-7 cwt.. or 6-58 tons per sq. ft. compression and 74-3 cwt. or 3-71 tons per sq. ft. tension. When the bridge is continuous with a brick arched viaduct, the abutment must be calculated to withstand the thrust of the adjoining arch when loaded, but no difficulty will be experienced in this. In the case of churches, school buildings, etc., with buttressed walls the buttress and portion of solid wall between the windows, forming a T section, may be taken as resisting the thrust from the roof. Pig. 458 shows elevation, and Fig. 459 plan of such a portion. The 152 THE MECHANICS OF BUILDING C0N8TEUCTI0N vertical dead load from one-half of the roof truss, amounting to 94-5 cwt., is assumed as acting on the wall plate 4J in. from the inner edge of the wall, and this is combined with the whole of the wind pressure on the roof amounting to 63 cwt. acting in a direction normal to the slope of the roof surface and striking the opposite wall where shown by the arrow in Fig. 458. The result is then combined with the weight of the wall and buttress acting at their mutual centre of gravity. Having obtained the final resultant, the maximum stress per W M square foot will be found by the formula -r- ± y, where W = vertical component of resultant in cwts., A = sectional area of wall and buttress in square feet, M = bending moment in cwt.-ft., Z = section modulus in foot units. Then W^M 404 _, 404 X 1-43 „„...„. X - Z = ifH 9^4 = ^^'^^ - ^^'^ = '^y ^^ °^*^-' or 4'25 tons per sq. ft. maximum compression on material of buttress, and 38 cwt. = 1-9 tons per sq. ft. tension at inner face of wall. In many instances there are more thrusts than one on a buttress and each must be combined with the partial weight of buttress as the line of thrust is constructed downwards. The centre of gravity of each portion should be found graphically, and the mean centre of gravity of each group permanently marked by pricking in, the lines may then be rubbed out and the force parallelograms proceeded with. The force scale may be reduced at each stage so as to keep the parallelo- grams within reasonable dimensions, but the smaller the scale the less exact will be the results, although probably near enough. At the first working out of the curve of thrust and the stresses in a buttress it may be found desirable to modify the height or projection of the various stages, but unless very great care has been taken in ascertaining the original thrusts it is useless to spend much time over the buttress. Piers, pilasters, or counterforts are the reverse of buttresses, being placed inside a wall instead of outside. The mode of calculating them is exactly similar to that of buttresses, but unless hoop-iron bond is inserted between wall and counterfort no tension should be allowed on inner edge owing to the facility with which separation in the bonding occurs between a wall and counterfort. Flying buttresses are more or less detached from the wall to which they give support. By standing further out they act more as raking shores, and can therefore be somewhat reduced in bulk, while they are also capable of architectural treatment. Take a simple case, as in Fig. 460, and divide into the portions shown by lines at A, B, and C, and find the weight and centre of gravity of each part. Combine the weight of the wall and buttress down to line A with the given thrust, and then combine the resultant thus found with the weight of the portion down to line B. The section at this line will be as in Fig. 461, and the distance of the neutral axis from inner edge of wall will be 5-5^ X 2 + 1-5Y6 - 2) 2 -^y = 1-5 ft., which coincides with the outer edge of wall, and the section modulus will be STABILITY OF BUTTRESSES 153 „ _ (6 X 5-5^ - 4 X 4°)" - 4 X 6 x 5-5 X 4 x 4(1-5^ ^^^ . . ^ - 6(6 X 5-5^ - 4 X 4^) " ^^■^*- Then the stresses will be T ± f = ^¥^ ± ^^?o"o^ ^ = 1870 ± 3302 = 5172 lbs., A Z 17 — 12'84 ' or 2'31 tons per sq. ft. compression, and 1432 lbs. or 0'63 tons per sq. ft. Fig 458 5 :7 / 'L' i^',1 ■^i. ^^' (^ /^ '^ a\ t.i l^4S/t. 1 31 B?>^ /«/9' 4-6/ tension. As the neutral axis coincides with outer face of wall, the whole of the tension comes on the wall and reduces to that extent the 154 THE MECHANICS OP BUILDING CONSTRUCTION weight of the wall below upon the foundations, aud need not be further considered. The compression which is taken by the buttress is spread over the triangle shaded by wide lines in Fig. 461, and the total amount acting at the centre of gravity of the triangle will be 2x4 X 5172 = 20688 lbs. This force may be taken as acting parallel with the last resultant, and is then combined with the weight of buttress down to line C, and the resultant thus found is combined with the weight down to ground level, giving the final resultant. The stresses produced by this re- sultant will be W M _ 3120 A - Z ~ 3-6"x 2 - 1(2 X 3-5") ± ?i^*^° ?i = 4457 ± 2547 = 7004 lbs., or 3-12 tons per sq. ft. compression, and 1910 lbs. or 0'85 ton per sq. ft. compression, and these figures are within safe limits. EXERCISES ON LECTURE XXI Q. 70. Fig. 462 shows a granite obelisk weighing 160 lbs. per cub. ft. Assuming the base to be reduced 6 in. all round by weathering, what horizontal T \'2'0'4 srfuari' \ ^ 1 1 u N< Qd (! « i O t% 1 C k \f— . TocaL '»rnd 81 1 1 ^8 [ :,; 1 I 1 ■»* h 1 ♦ ^' ' 1 '' _ I* -4'0'- H *^ k.*^ <• ' 1 ^uar'e ' ^o-s 1 F/gi 462 F/^.-aes wind pressure per sq. ft. would be required to produce a maximum compression with no tension ? Answer. See Fig. 463. EXAMPLES FOR PRA.CTIOE 155 Then 8-9 : 29866 : : 0-5 : x, whence x = 29866 X 0-5 = 1678 lbs. 8-9 4 + 2 area = ' - x 20 = 60 sq. ft. therefore wind pressure per sq. ft. = i|ja = 28 lbs. Q. 71. Find the centres of gravity of each portion of buttress shown in Fig. 464 ; trace the amount and position of combined thrust and weight at each change of section, and state the maximum stresses at the base. h^'oU -ff.V'-r /z/g. 'fe-f n'cf. -466 F/gr. <66 Answer. See diagram, Figs. 465 and 466. The final resultant is 22200 lbs., and the resultant cuts the base at the edge of middle third, or 1 ft. from the centre of base. The stress will therefore be W , M 22200 , 22200X1 ,q-„ , ,„-„ ,,„„„,, T ± Z = 6l<-2 ± 1(2X6^ = ^^^ =*= ^^^ = ^^°° ^^^- or 1-65 tons per sq. ft. compression at outer edge and nil for tension on inner edge. LECTUKE XXII Pressure of Water — Pressure on Tank and Reservoir Walls — Stability of Masonry Dam. The pressure of water against a suiiace, at any point in it, depends only upon the depth of the point below the surface of the water. That is to say, the pressure upon a square foot at any depth will be tvh, where w = the weight of a cubic foot of water = 62J lbs., and h = height or depth in feet from surface. The pressure gr^ually increases F/g. 468 therefore from the top to bottom, as ordinates to a triangle, as shown in Fig. 467. The centre of effort, or centre of pressure, will be opposite the centre of gravity of the triangle at g the height, and the total amount will be wh xh = ^wh^. The pressure at the various depths will be the same whatever the angle of the containing surface, but in the case of a sloping wall, the sloping length being greater than the vertical height, the total pressure will be greater. If h be the vertical height, and I the inclined length, the total pressure will be ^whl, as in Fig. 468. It must be remembered that water pressure always acts perpendicularly to the surface against which it presses. The stability of walls against water pressure may be worked graphi- cally precisely in the same way as for earth ; the natural slope of water being nil, the so-called line of rupture will be at 45°, and the wedge of water between this line and the back of wall, whether it is vertical or sloping in or out, is the weight producing the thrust. Then graphically in Fig. 469, at g height, set up W = weight of wedge of water, cut off the horizontal thrust line by an angle of 45° ; then TANK AND KESERVOIK WALLS 157 the amount of thrust will be given by the length of horizontal line. Let the wall slope away from the water as iu Fig. 470, then the weight of wedge will be greater, but the thrust line will be cut off at nearly the same length as before by the angle of 45°, the difference \ \ X'^ -T-\ nA A/aturaL sLooe \ Fig 459 L__.\ "^x / 1 \ ^y / V-^- 1 1 <^0\ / V-1 !\ i Fig. 470 ^-■-^ being that due to the difference between Jif A^ and ^M. By formula the thrust will be ^wh^ cosec 0. Again, let the wall slope towards the water as in Fig. 471, then the weight of wedge will be less, but the thrust will be of the same value as before if the sloping length is the same, or T = i^wh^ cosec (180 - 6). Now let us take the case of a wall subject to earth pressure on one side and water pressure on the other, such as the wall surrounding a sunk tank or small reservoir. Assume the wall to be of concrete, 158 THE MECHANICS OP BUILDING CONSTRUCTION 7 ft. high from bottom of reservoir, with a vertical hack and battering face, 1 ft. thick at top and 3 ft. at bottom, as in Fig. 472. The earth has a natural slope of 35" and finishes level with the top of wall. The highest water-level reaches to 1 ft. from top of wall. The weight per cubic foot of the earth is Ij times and that of the concrete twice that of a cubic foot of water. This should be drawn to a scale of ^ in. to 1 ft. and the stability estimated graphically as shown. Then, the figures having been scaled off, the calculation of maximum stresses at inner and outer edges of the base will be as follows : When the tank is K/.'o'H r/g.472 IVater , €25 Cbs pen , cub. ft. c.g. r i _-i-_i' -JfO"- - *\ empty the maximum stresses from the resultant of the earth pressure and • ui fiv. 11 -11 u ^^" + M 1750 . 1750 X 0-42 _ „„.„ . .„„ weight of the wall will be -j^ F ~ ~W Ml y ¥')' = 1073'3 lbs. per square foot compression at A, and 93"3 lbs. per square foot compression at B. Then combining this resultant with the water pressure and assuming no tension possible, the stress will be W 2 X 2080 ^ — =1541 lbs. per square foot compression at B. When X 0-9 W^ M tension is permissible, the stresses will be -r- ± ^ = 2080 + 2080 X 0-6 - K1X3^) = 693-3 I 832 = 1525'3 lbs. per square foot compression at B and 138'7 lbs. per square foot tension at A. The principles of designing and calculating the main stresses of a modern masonry dam are exemplified in Fig. 473, which shows a conventional section of a dam 70 ft. high. The curve of stability when the reservoir is empty will be found by first dividing the section into any convenient number of parts, as shown by the dotted lines STABILITY OF MASONRY DAM 159 A, B, C, D. Find the centre of gravity of each part, and through each centre of gravity drop a vertical line to cut the assumed base line of each part. The curve drawn through these points will be the curve of stability when the reservoir is empty. When the reservoir is full the .1 #..M Masonry ISO lbs per cub re Ky -20'0'- K - - - - - 62's'- - - »H method will be as follows : taking the part down to line A the pressures due to the head of water will be in the form of a triangle, with a base at A of 62-5 x 20 = 1250 lbs. The total pressure will then be — ^ = 12,500 lbs. acting at right angles to back of wall through the centre of gravity of the triangle. This pressure must be combined with the weight of the part considered, and where the resultant cuts the line A will be one point in the curve. The part from A to B must next be considered, and the pressures due to the head of water wiU now vary from 1250 lbs. at A to 62-5 X 40 = 2500 lbs. at B. The total pressure acting through the centre of gravity of the quadrilateral will be 1250 + 2500 j^ 20 = 37,500 lbs., and this amount must be combined with the resultant from the first portion. The resultant thus formed is then combined with the weight of wall from A to B, and another point in the curve will be given where the resultant of the last step cuts line 160 THE MECHANICS OF BUILDING CONSTRUCTION B. In a similar manner the other divisions of the wall may be worked out and other points found, through which the curve of stability when reservoir is full may be drawn. • i j • • Other calculations for shear are required in the practical desigmng of masonry dams, but these involve considerations rather beyond the present course. EXERCISES ON LECTURE XXII Q. 72. What will be the weight of water in lbs. contained in a tank 6 ft. cube when full ? and what will be the total pressure upon the sides and bottom ? h/'oM \-l'0''\ Fig 4Za Fig 475 Answer. 6 X 6 X 6 X 62-5 = 13,500 lbs. weight, \ X 62-5 X 6» X 6 X 4 + 13,800 = 40,500 lbs. pressure, or the pressure is three times the weight. Q. 73. A storage tank is built in concrete 1 ft. below the surface of the ground. The external wall is 7 ft. high inside, 1 ft. thick at top, and 2 ft. at bottom, battered inside and vertical outside. The earth, haying a natural slope of 36°, is banked up against the outside, with a 2 ft. pathway on top 1 ft. below top of wall. The water is 6 ft. deep. Find the maximum tension and compression at the base when full and empty, taking the earth \\ times weight of water, and concrete twice weight of water. Answer. See Fig. 474. When empty W I M _ 1312 _t, 1312 X 0-45 A Z 2-0 J(l X 2^) = 656 ± 885-6 = 1541-6 lbs., or 0-69 ton per aq. ft. compression at B. compression at A, and 229-6 lbs. or 0-1 ton per sq. ft. When full W , M^ A Z 1500 , 1500 X 1-08 2-0 J(l X 2') '■ : 750 ± 2431 = 3181 lbs., or 1-42 tons per B0[. ft. compression at B, and 1681 lbs. or 0-75 ton per sq. ft. tension at A. Q. 74. A division wall in the above tank is 1 ft. thick at top and 3 ft. thick at bottom. Find the maximum tension and compression when one side is full and the other empty. EXAMPLES FOR PRACTICE 161 Answer. See Pig. 475. The stresses will be W , M 1920 . 1920 X 1-11 A±S = i(l X 3') ; 640 ± 1420 = 2160 lbs., or 0*96 ton per sq. ft. compression at A, and 780 lbs., or 0-35 ton per aq. ft. tension at B. Q. 75. Fig. 476 shows the section of a concrete dam. Find the maximum -S'O SecC. area above AB = SOO scf ft. rfg. 476 r/q.477 ••5^'iSi'" V "<,-^ 4SJ l_f stresses on the base AB when the reservoir is full. The material weighs 180 lbs. per cubic foot, and the sectional area above AB is 500 sq. ft. Answer. See Fig. 477. The stresses will be ^ ± ^ = ?5 ± ^ ^ ^"* = 1-25 ± 1-68 = 2-93 tons per sq. ft. compression at B, and 0*43 ton per sq. ft. tension at A. LECTURE XXIII Stability of Arches — Forces acting on a Voussoir — Curve of Thrust — ^Depth at Crown — Primary Thrusts — Analogy with Girders — Catenary and Parabolic Curves — Virtual Arches, Straight Arch and Lintel — Curve for Ordinary Distributed Loads — Maximum Stress — Critical Joint — Incomplete Arches — Minimum Thickness of Arch — Thrust in Semicircular Arch — Key stones not necessary. The first essential for the stability of an arch is that the abutments should be rigid; if they yield in the slightest degree the arch is extremely liable to fail. This is especially likely to happen at the angle of a building where there are arches on both faces, and an example of this was given in Lecture XXI. Whether an arch is made up of separate stones, called voussoirs, or is composed of half brick rings, or is a homogeneous mass such as concrete, it may be divided up into real or imaginary portions, each of which must be in equilibrium under the forces which traverse it. Fig. 478 shows a single voussoir in the haunch of an arch, subject to a vertical force due to its own weight and the superincumbent load, a thrust trans- -gdoJ-rrrrv. -.^ mitted from the next voussoir above it, and a J'^'^'^J ,'l" ' resisting force from the voussoir below it. The Y ' ''ll.l I thrust and the weight combined into a paral- V'^V"^''']!- lelogram give the resultant, to which the equi- __^V^ j50/r 1^ librant or resisting force must be equal and vO'^1---Ts opposite. The ideal condition is when the thrust ''V^pc?^""* and resistance both act in the centre of and /va ^ZS perpendicular to the joints through which they pass, but this seldom happens. The angle is usually somewhat more or less than a right angle, and the line of action is generally out of the centre, being nearer to either the intrados or extrados. The angle 6 or its supplement must be less than the limiting angle of resistance, or sliding will occur, and the line of action must be within the middle third of the joint in order that the effect may be one of pure compression. The curve drawn through the points where the lines of thrust cut the joints will give the line of thrust or curve of thrust, or curve of equilibrium, which for perfect construction should be followed through- out its course by the centre Hne of the thickness of the arch. That is to say the curve of thrust is the theoretical shape of the arch, which requires to be clothed with material to make it practical. Other considerations of construction or sesthetics necessitate the use of a LINE OF THRUST IN ARCHES 163 simple curve such as a semicircle, segment of circle, ellipse, or gothic arch, without reference to the curve of thrust, but ifc is requisite that the curve of thrust should nowhere be so near the edge as to make the stress in excess of the safe load. In designing an arch the depth of arch ring may be assumed for trial as D = /iVR, where D = depth at crown in feet, R = radius in feet, n = constant = 0'3 for block stone, 0*4 for brickwork, 0'45 for rubble stone. In the investigation of the stability of an arch it will be found that many different curves of thrust may be drawn depending upon the points taken on the centre line and skewbacks where it is to pass through. Nature chooses that one of them which gives the minimum stresses, and we cannot be sure of selecting the right one in practice. Sometimes it is necessary to make a second trial in order to get a better curve, but usually the best result is obtained by making the curve pass through the upper edge of middle third at the centre line and lower edge of middle third at the skewbacks, except in semicircular or elliptical arches, where it is better to keep to the upper edge of middle third at the skewbacks also. With regard to the position of the line of thrust in an arch. Prof. Goodman's "Mechanics Applied to Engineering," pp. 522 to 535, is worth careful study. He says, " An infinite number of lines of thrust may be drawn in for any given distribution of load. Which of these is the right one, is a question by no means easily answered, and what- ever answer may be given, it is to a large extent a matter of opinion. For a full discussion of the question the reader should refer to I. 0. Baker's ' Treatise on Masonry ' (Wiley and Co., New York) ; and a paper by H. M. Martin, Inst. G. E. Proceedings, vol. xciii. p. 462." Having determined the guiding points for the curve of thrust, the next step is to ascertain the amount of thrust at those points, or the primary thrusts. Fig. 479 shows the elevation of half an arch with its load, a length of 1 ft. being assumed as in the case of retaining walls, and the centre of gravity of the whole is found by suspension of the figure /rv -, _y^\ , when cut out in drawing paper. A hori- >C/^' '' ^' zontal line is drawn from the point on c'^^zir^/'- \b the centre line through which the curve F/g. 479 has to pass, to cut the vertical through the centre of gravity, and from the intersection a line is drawn through the point selected on the skewback. Then the total load is ascertained by taking the whole area of the figure with a planimeter, which load is then measured downwards from the point of intersection on centre of gravity line, and the parallelogram completed to give the horizontal thrust H, and thrust at skewback T ; T will always be greater than H, and that is why some arch rings are made thicker towards the abutments. The horizontal thrust in an arch under a uniformly distributed load is determined mathematically on the same principle as the com- WL pression in a girder flange, the formula -^ applies equally to both, W 164 THE MECHANICS OF BUILDING CONSTRUCTION being the distributed load, L the span, and A the depth of girder or rise of arch. Generally speaking, the clear span of an arch between the abutments, and the rise of the arch from the springing line to the soffit at crown, will not appreciably differ from the true ratio of span and rise, which, however, ought strictly to be measured from the curve of thrust. If the load were distributed uniformly over the ring of the arch, as in an arch ring without superincumbent weight, the curve of thrust would be a catenary such as occurs in the inverted form of a suspended chain. If it were distributed uniformly over the horizontal width of span the curve would be a parabola, but owing to the increased weight towards the abutments, the curve is not a true geometrical outline, but approximates more to the catenary. A straight arch as in Fig. 480 is considered by many persons to be no true arch, but if the conditions be investigated it will be found that it virtually contains an arch ring half the depth of the straight arch. ^- "sjfrtuqi^ arp/i- - - . Flat arch f/'g. 480 LinCeL Fig ^48/ and with a rise of the same amount, as shown by the dotted lines, and these lines determine the available strength of the arch. A straight lintel follows the same law, provided that the abutments are rigid, and although there is no appearance of a skewback, there is virtually one contained in it as shown in Fig. 481, being similar to an arch in one piece with the impost stones. In the practical investigation of the stability of an arch, the outline is drawn to scale as in Fig. 482, where the part marked " filling " includes the weight of material in the superstructure, road metal, etc., for 1 ft. run, and above this is added the equivalent height of material, of the same weight, which would be equal to the external load due to the traffic, say 2 cwt. to 5 cwt. per foot super, according to circumstances. Upon one half of the diagram the centre of gravity is found as previously described, and a vertical line drawn through it. Then a horizontal through top of middle third of arch ring at crown, and from the inter- section a line through the lower side of middle third at skewback, the total load on the half arch is then set off to scale and the parallelogram completed to give the primary thrusts. The next step is to divide up the other half-span into portions, which may be equal or unequal, and may consist of actual voussoirs with the superstructure resting upon each, but the simplest division is given by equidistant vertical lines, and the result is perfectly accurate as regards the production of the line of thrust. The weight of each portion and a line through the centre of gravity is next required ; these are to be considered as force lines, and the spaces numbered as for ordinary reciprocal diagrams. LINE OF THRUST IN ARCHES 165 Then in the diagram, Fig. 483, take any pole, 0, set off the horizontal thrust, and on the load-line mark off to scale the separate loads 1-2, •2-3, 3-4, etc. Draw vectors as shown, and parallel to the vectors construct the funicular polygon upon the half elevation of arch, as shown. The curved side of the funicular polygon gives the line of thrust and the straight sides the direction of the two primary thrusts. Increasing the load does not alter the direction of the Una of thrust, but only the intensity of the thrust. If the distribution of the load is altered, the shape of the line of thrust will be altered, bulging upwards at the part where the increased load occurs. 166 THE MECHANICS OP BUILDING CONSTEUGTION The stress at any point in an arch will be given by the formula W M V- ± "2 , where "W is the thrust passing through the given part, A the sectional area or depth of arch ring, M the bending moment or thrust multiplied by the distance from line of thrust to centre of depth of arch ring, and Z the section modulus of arch ring ^bd^. It is clear that the line of thrust, being within the middle third, would be no criterion of stability, as the maximum safe load might be overpassed, but at the same time it is a much more serious matter for the line of thrust to be outside the middle third, as the chances of over-pressure on the one side or tension on the other are certainly greater. The critical joint in an arch is that joint where the line of thrust most nearly approaches the intrados or extrados. It is usually situated at from 45 to 50 degrees from the crown. If a portion of an arch be removed, and replaced by a rigid abut- ment giving the same bearing surface, the thrust in the remainder will be unaltered, and therefore if the curve of thrust be found for the com- plete side, it may be repeated as far as it applies on the incomplete side. It is a curiojus fact that there is a minimum depth of arch ring accord- ing to the span and rise of the arch, independent altogether of the load upon it. This arises from the necessity of keeping the curve of thrust suflBciently within the arch ring. A covering arch, 10 ft. span, formed of one ring of brickwork 4^ ins. thick, set in cement, was built over a tank and had to carry its own weight only. Before the centering was removed, the arch bulged at the sides, about half-way between the springing and the centre, and when the centering was removed, the arch collapsed altogether. The reason will be seen by observing the position of the curve of thrust, as shown in Fig. 484, which is con- structed from the reciprocal diagram, Fig. 485. The lettering in the illustrations shows the order of working. AB is the half elevation of the arch, which is divided up not into the actual bricks, but into con- venient portions for the method of working. Draw a vertical line through the centre of gravity of each portion, representing the direc- tion in which its weight acts. Number the spaces between these force lines and draw the line of loads CD. Select any pole, E, and draw vectors to CD. From any point, F, on line 1-2 of Fig. 484, and across space 2 draw a line parallel with the vector from 2 in Fig. 485. Now, in Fig. 485 draw DM parallel to JK, that is, horizontal, and CM parallel to KL. Join all points of CD with M, then these lines will represent the thrust throughout the arch. The " curve of thrust," N, is found as follows : From point L across space 2 draw a line parallel with M-2, then continue across space 3, parallel with M-8, and so on until B is reached. For the arch to be stable without tension on any part this curve should be everywhere within the middle third of the arch ring. If the arch be made to the same curve as the line of thrust, the arch will, of course, be under the best conditions of stability, provided that in finding the line of thrust all the circumstances, such as accidental load, wind, etc., have been taken into account. It is a great mistake to suppose, as many do, that there is no out- ward thrust from a semicircular arch. Whatever tlie horizontal thrust may be at the crown, there is a similar horizontal equivalent on each THEU8T IN SEMICIRCULAR ARCH 167 side acting outwards. An illustration of this occurs in Fig. 485, where the inclined thrust at the skewbacks may be resolved into two direc- tions, vertical and horizontal, when it will be found that the horizontal component is equal to the horizontal thrust at the crown. It is a law of nature that the line of thrust takes the shortest possible course from the load to the support, so that if an arch ring be assumed to have no weight, the thrust from a concentrated load on the centre would pass in straight lines to the skewbacks ; and where a distributed load is carried, the horizontal thrust at crown is depressed by the load it meets as it passes each joint towards the skewback. It is also a common error to suppose that a keystone is necessary for the stability of an arch ; it is purely a matter of taste, and the fact that countless thousands of brick arches exist without a keystone ought to be a sufficient answer to the holders of the idea that it is necessary. In the fronts of buildings the arches are often finished with a keystone or similarly shaped block of gauged brickwork ; but this is for the sake of appearance only. EXERCISES ON LBOTURE x:xill Q. 76. State the conditions of stability of a masonry arch. Answer. The conditions of stability are — (a) The line of resistance should everywhere fall within the middle third . of the arch ring, so that the joints are not under tension at any part. (6) The intensity of pressure should nowhere exceed the safe load upon the material. (c) The line of thrust passing through any joint must not make an angle greater than the limiting angle of resistance of the joint to sliding. (d) The abutments must be rigid to prevent the failure of the arch as a whole. Q. 77. A brick-arched railway bridge 25 ft. span, i ft. rise, 4 half-brick rings, carries a total load of 27,000 lbs., the vertical through centre of gravity of each half being situated 5 ft. horizontally from the spring of the arch : determine the amount of the thrusts at orovm and springing, and the maximum stresses. Scales, J in. to 1 ft., 1 in. to 10,000 lbs. Answer. See Fig. 486. Thrust at crown = 18,000 lbs., compression 5'3 tons per sq. ft. ; thrust at springing = 22,500 lbs., compression 6-7 tons per sq. ft. Had the line of thrust been through top of middle third at crown and bottom of middle third at springing, the compression would have been 9| tons per sq. ft. at at crown and 12J tons par sq. ft. at springing. Q. 78. A coke breeze concrete arch has a span of 8 ft., a soffit radius of 8 ft., a rise of 1 ft. J in., a depth of arch ring of 1 ft. ^ in. in centre, and radius of extradoa 168 THE MECHANICS OF BUILDING CONSTKUCTION ?^ t*'/t ' J* ^^^S^^ 120 lbs. per cubic foot, and is loaded with a uniformly distri- buted load of 10 cwt. per foot super : what is the maximum compression iu tons per square foot ? lO'.O" radius Span a'.o" Rise /' O^A- _ Answer. See Pigs. 487 and 488. TaUng the line of thrust in centre of arch ring at CTown and springing, the maximum compression wiU be at the springing, and = _ = = 90 cwt. or 4-5 tons per sq. ft. LECTUEE XXIV WALL pg 489 Estimation of Loads upon Arches — Load on Ogee Arch— Load on Culvert — Stability of Arch and Abutment— Arched Viaduct— Stop Abutments— Eeservoir Arched Eoofing— Concentrated Loads— EoUing Loads. It is very difficult to judge the amount of load that will come upon an arch in a wall. The actual load will depend upon the nature of the material, the bonding, the strength of the mortar, the age of the work and other circumstances. When the wall extends for some distance on either side of the opening it will generally be safe to assume that the load upon the arch will be the weight of brickwork, etc., included in an equi- lateral triangle having the arch for its base, but when the piers at each side of the opening are narrow, say less than one-half the span of arch, the abutments will be more liable to yield, and it will be safer to allow for the whole weight of the superstructure over the span of the arch. A curious case that arose is shown in Fig. 489. The material was a very light stone, 120 lbs. to the cubic foot, and it was assumed that the extreme load would be represented by the por- tion of the figure, of which one-half is shaded, and the arch itself. Then dividing up the arch and the load into vertical strips, drawing a line through the centre of gravity of each, and taking the lines to represent forces equivalent to the weight of each part, the mean centre of gravity of the load will be obtained by cutting out and suspending one-half. Then a horizontal through the centre of the crown to meet the mean centre of gravity line will give the point from which to draw the line to the centre of the skewback. The half total load being set oif to scale on the vertical line through the mean centre of gravity, the parallelogram may be completed, giving the horizontal thrust at the crown as 4"7 cwt. and the inclined thrust at the skewback as 15 cwt. Then the load-line being set down in Fig. 490 and the horizontal and inclined thrust drawn in, the vectors to the other 170 THE MECHANICS OF BUILDING CONSTRUCTION points will give the direction of the line of thrust across the inter- mediate portions of the arch. The line of thrust passes outside the arch at A, and the calculations will be W M A- Z 8-3 0-9 y 0'42 i.g =16-6±83-6 = +100-2cwt. = 5tonsperBq.ft. lxi-ixix(|) compression, and —67 cwt. = 3'35 tons per sq. ft. tension, which are about the extreme limits of stress for safety. At point B the calcula- tions will be „W 2 X 12-5 ■d 3 X 1-25 12 = 80 cwt. = 4 tons per sq. ft. compression. An arch of this kind is not a true arch, but only an ornament that is kept in place by the adhesion of the mortar. Where any material is interposed between the point of application of a concentrated load and the extrados of the arch, such as a bed of road material, the pressure of the load is spread in its transmission downwards, and may be considered to extend uniformly over a surface represented by the base of a cone of which the load is the apex. If the ex ■erna i h >oc/ /?a/( lei^el l" ■V _r--L r \ 1 Wt^p/' Mo'CJ'e S H'orch ILl If' /,' /' /Moore grounof \ \ /'T a J.' / / groL. no/ Bnck abuCmenC ^ ■f - lO'O- * WaCer le^eL SoOaf ground I jv^T; "^ ^ F/g. 491 interposed material is a brick wall, the pressure will extend over a surface represented by the base of a wedge of a length equal to the thickness of wall, and apex angle of 60 degrees. As against this diffused loading may be taken the concentration that might arise in the case of a culvert in a new embankment while the filling is still in a granular condition. But this state may be aggravated by an external load acting on one-half of the arch only, as in Fig. 491. It will be seen that the line of rupture is taken as the limit of the earth pressing upon the arch, and this is divided over assumed voussoirs as COMPELLING THRUST TO FOLLOW CURVE 171 shown, and equally divided along the surface line. The centre of gravity of each portion being found, as at a, Fig. 491, the weight is set up above it to scale = ab, and a line drawn through a parallel with the abutment side of the portion of load under consideration, and draw the line he at right angles to ea. Then ca is the amount and direction of the load to be taken into account in constructing the polar diagram. Fig. 492. The same being done for each portion and the vectors „ '^^ ''^■^ ./ drawn, the funicular polygon may ^ f^ be constructed on Fig. 491 to \\V\^^'^~^ // ^^ give the position and direction of \V\\ ^^r^ ^^ the mean centre of gravity line of \\\ ^\ // ^^ \ the whole load. The next step is Wi \ \ ^/ / ^ ^ ^ to draw the horizontal thrust line A\ \ V/ ~- ^^O from centre of arch at crown to ^\ \ \y / ""-^Onn meet this line, and from the inter- \ \ m\L -;V* section drawing a line to centre of \ W/ _,''^'' skewback, completing the paral- \ V / , - - ' "^ ' ' lelogram to obtain the primary \ /V ' ' ' , ^ ' thrnsts. Then the curve of thrust \ !/ ^f.^ may be drawn in as before. V-'' '^ To cause the line of thrust to ^ follow exactly the outline of a semi- circular arch the load would have to be increased to infinity over the ends ; the vertical depth at any point = a-r„ where a = depth of arch ring in centre, h = height to soffit at given point, d = rise of arch at centre, as shown by dotted lines on Fig. 493. Approximately the dis- tribution of the loading can be shown by the loads required to produce equilibrium on a series of bars whose jointed ends lie in a semicircle, as in Fig. 493. The reciprocal diagram, Fig. 494, is formed by drawing vectors from point 12 and parallel to the bars, and cutting them oflf by a vertical line, say 5 in. away, to give the amount of each load. Then set up these loads over the joints of the frame diagram, join the upper extremities, and the outline will be seen to be of the same character as that given by the material required above a semicircular arch to bend the line of thrust round the curve. To find the stability of an abutment it is not necessary to construct the whole line of thrust in the arch. For example, Fig. 495 shows half an arch in a brick wall without surcharge or added weight. Find the weight of the arch and brickwork above, mark its centre of gravity and construct the parallelogram of thrusts. Find the centre of gravity of the remaining brickwork with pier, and its weight. Draw a vertical through the centre of gravity and produce the line of thrust from the arch. Complete the parallelogram and the resultant is found to cut the base 8 ins. from the outer edge, or 1 ins. from the centre. The vertical equivalent of the thrust will be 27 + 38^ = 65^ cwt. Then by the W M formula -r + tt the maximum stresses will be A - ii r25 65;25j^ ^ _l_ gg ^jjj -14-5, X 1-^ X 1 X 3^ 65-25 3 172 THE MECHANICS OP BUILDING CONSTRUCTION that is, 58 cwt. per sq. ft. compression and li'b cwt. per sq. ft. tension. If no tension can be allowed the formula to be used for maximum com- pression will be K = |5 = ! X ^^ = 65-25 cwt. per sq. ft., or say 85 tons, which is quite safe. An arched viaduct, or series of arches, is usually made with a stop abutment at each tenth arch. Between the other arches a thin pier is placed, just sufficient to carry the dead load and the portion of thrust /^,gf 43a due to an external load on the one arch while the adjoining arch has only the structural load to carry. In the case of failure of one arch from any cause, the whole thrust from the adjoining arches would over- throw the piers, and the failure would spread the whole length of the viaduct except for the stop abutments, which are made sufficiently thick to resist the whole thrust of the full-loaded arch on one side, even when the arch on the other side has fallen. When reservoirs are covered with a series of arched roofs great care must be taken in placing any earth covering on top, as the loading of one span may cause the adjoining span to rise and collapse. The failure of an important reservoir at Madrid during construction in 1905 was probably due to this cause. In the next diagram an example is given of irregular loading. Fig. 496 is an arch under a distributed load, and a concentrated load of 3 tons over one of the haunches, due to a rolling load coming on the ROLLING LOADS ON ARCHES arch. The load of 3 tons may be assumed to be spread equally over two of the voussoirs, giving an addition of 30 cwt. to the load on each. It is necessary in a case of irregular loading such as this to find the mean centre of gravity of the whole load. This is done by means of the polar diagram in the centre of Fig. 497 on the right, and the corresponding large funicular polygon on the top of Fig. 496. The dotted closing lines of the funicular polygon give at their intersection the line of mean centre of gravity of the loads. The next step is to find the mean centre of gravity of the group of loads on either side, by means of the ^5~"~~^^/a'. soo t3^ two smaller polar diagrams in Fig. 497 and the corresponding funicular polygons in Fig. 495. Then the primary thrust parallelograms being drawn, the thrusts at abutments give the direction of the outer lines in the vector diagram on the left side of Fig. 497, the vectors of which give the parallels for the curve of thrust in the arch. Cases of irregular loading always give the greatest diificulty in finding the best position 174 THE MECHANICS OF BUILDING CONSTRUCTION for the curve of thrust. As it approaches the extrados and intrados at two points, the curve may be taken first to pass through the centre of arch ring on mean centre of gravity line, and through centre of skew- back, but if, when the curve is drawn, a better line may be found by taking other points on mean centre of gravity line and skewback, it must be drawn over again from a new parallelogram. In the present case the curve appears to be in a very good position, and no alteration is needed, but it will be seen that the curve is not quite horizontal across the mean centre of gravity line, and it will, therefore, be well to check the work by another method of construction. Draw out the arch and loads as before and proceed as in Fig. 498. Set down the load line 1-15 in Fig. 499 and select any pole ; draw vectors, and parallel to these construct the funicular polygon EFG, then Gr will give the position of the mean centre of gravity line. From draw 0-1 6a parallel with EF, giving the vertical reactions at the abutments l-16ffl and 16a-15. Next from B, the centre of the skew- back, draw a horizontal line to meet a vertical line through the centre of the arch in H. From H set up HJ = 1-16«, and HK = 16a-15. Join AJ and BK, and if produced they should intersect at L on the mean centre of gravity line. Then in Fig. 499 draw 1-16 parallel with AL, and 15-16 parallel with BL to obtain point 16. The remaining points on the load line are now drawn to point P, and the curve of thrust AMB constructed parallel with these lines. This does not pass through the arch at all, but nevertheless it is a true thrust curve, and only wants raising into its proper position by magnifying all its vertical ordinates as follows : Draw QR, Fig. 500, equal to MN, Fig. 498, and at any distance from QR set up a height ST equal to the distance from N to the centre of arch ring in Fig. 498. Join SQ and TR and pro- duce to meet in a point U. Then the ordinates to the curve AMB being marked off on QR and lines drawn from the point V through the divisions on QR will give the corresponding magnified ordinates on ST. These being set off on the verticals of Fig. 498 will give the true curve of thrust. EXERCISES ON LECTURE XXIV Q. 79. Draw to a scale of J in. to 1 ft. the half-elevation of a semicircular arch 10 ft. span, in three half-brick rings, with 12 ft. 6 in. of brickwork above the springing line and external load equivalent to 1 ft. 6 in. additional. On the other side of centre line, with the same springing line, draw the half elevation of an equilateral Gothic arch of the same thickness. The piers are 3 ft. wide and 6 ft. high. Obtain and draw the line of thrust for each arch and carry it down the pier, stating the maximum stresses on arches and piers. Answer. See Pigs. 501, 502, and 503. The calculations for the semicircular arch will be as follows : — . , W , M 60 , 60 X A Arch ;^- ± 2 = 1:125 =•= i(rx~l-125^ ^ ^^'^ ^ ^'^^'^ = ^''^'^ °^*-> °'^ ^'^^ ^°^^ per sq. ft. compression, and 65'2 owt. or 3-26 tons per sq. ft. tension. W M 111 X 1"25 A-butment -^ ± ^ = H^ ± ^^^ ^ 3^^ - = 37 ± 92-6 = 129-6 cwt., or 6-48 tons per sq. ft. compression, and 55-6 cwt. or 2-78 tons per sq. ft. tension. EXAMPLES FOE PEACTICE Fiq. SOI 175 F/g SQ2 r/gi. SOS earth SO Obs. per cut. ft. Filli'nqi F/g. S04 176 THE MECHANICS OF BUILDING CONSTRUCTION The calculations for the Gothic arch will be as follows : — ^■^''^ r ± i = rSs ± jinnw) = ''■' ± ^« = ''■' <="*■■ °' ^'^^ *°°^ ^'"^ sq. ft. compression, and 2-5 cwt. or 0-125 ton per sq. ft. tension, w M inn y 0'^ Abutment -^ ± -^ = 't^ ± i (i x 3n = ^^'^ =*= ^^'^ = ^^'^ """'•' °^ ^'^" *™^ per sq. ft, compression and no tension. Q. 80. Fig. 504 represents an abutment for a brick arch with a given thrust, find the final resultant allowing for the weight of abutment and filling and thrust of earth. Answer. See Pig. 505, the final resultant being shown by stroke-and-dot line. LECTUB,E XXV Theory of the Modern Arch — Curved Struts and Ties — Arched Ribs. An arch of brick or stone masonry is an inelastic arch ; an arch of steel or ferro-concrete is an elastic arch, or what is commonly known as an arched rib. The following description from The Builders' Journal of Jan. 17, 1906, puts the case of the modern arch very clearly : — " The ' theory ' of arch design remained in an elementary state until the last century, when engineers were led to examine critically into all structural means in order to fit conditions of modern industry. They took the masonry arch and developed it a certain way, but they went little further than the greatest of the older builders, though they raised the average knowledge of the subject. Their empirical inelastic arch theory assumed the arch stones to be rigid, and required the line of resistance within the arch ring (for safety within the middle third). In the latter part of last century, however, the theory of arch design Ijegan to advance again, particularly with continental engineers. The construction of arches in metal marked the point of departure. The theory of elastic flexure was then applied to the theory of the arch, and the arched rib came into being. Instead, then, of it being a necessity for the line of resistance to be confined to the arch ring, it was easy, at small sacrifice of economy of metal, to stiffen the arch ring against flexure. This was now required to resist combined thrust and bending (with shear as a corollary). This is the elastic flexure theory of arch design. The older inelastic theory had led to the adoption of pin and similar joints at the crown and springing so as to give greater precision in design or to afford control of the line of resistance. The pin joints simplified the modern elastic fiexure theory by allowing unknown quantities to be exactly determined, just as the fixing of one end of a roof truss and the freedom of the other simplifies the theoretical determination of stresses in such structural members. With the two-hinged arch (pin joints here being at the springing) or the three-hinged arch (a pin joint here being at the crown as well as at each springing), the analysis of stresses must be so conditioned as to make bending moments zero at the pin joints, whatever may be the condition of loading. This flexure theory has increased the range of the arch to immense spans. The Clifton Arch rib bridge at Niagara has a span of 800 ft. The arched rib, too, is a graceful form of construction, and is much superior in line to the truss and cantilever, and we may therefore look forward to the future N 178 THE MECHANICS OF BUILDING CONSTEUCTION development of engineering structures with more equanimity than formerly. The modern theory of arch design finds its architectural application to reinforced concrete, of which there are examples with and without hinges." We have already in Lecture XI. dealt with combined longitudinal and transverse stresses, in Lecture XII. with pillars or stanchions eccentrically loaded, and in Lecture XVIII. with arched rib roof trusses, and we now proceed to consider bent struts and curved members generally. These matters are all very closely allied. First let us consider the terms used in connection with curves and F/g S06 the formulae for obtaining the different elements, as it may be useful in connection with other matters. These are shown upon Fig. 506, and the formula will be as follows : — Chord of 1 arc = V(i span)^ + rise^ Eadius = ja4^5)!+risel ^ L rise J cos I rise J ^ _ , _ (chord of whole arc)^ 2 X radius^ whence /3 is obtained from table and oc = 180 — /3 Curve tangent = R cot— h = V (curve tan)^ + rad.^ - radius Rise = rad. - /y/^rad.^ - chord^ «/ for a: = Vrad.2 -x - (rad. - rise) ^ for 3 = rad. - Vrad.^ - f THRUST IN CURVED STRUTS 179 It may help to throw some light on the subject generally if we compare the action of a thrust on a curved member with the bending moment produced by a concentrated load on the centre. The end thrust on a curved strut or roof member produces a bending moment = Tv, where T = thrust, v = versin of curve. The condition of a curved strut may be illustrated by reciprocal diagram. Let AB, Fig. 507, be a curved strut, as crane jib, upper flange of bowstring girder, principal rafter of curved roof, etc., under known end thrusts. Draw chords AC, CB, and vertical force lines as shown, and number the spaces. Then for re- ciprocal diagram. Fig. 508, draw 2-3 = thrust, and draw 2-4, 3-4 ; 4-1, 2-1 ; and 1-5, 4-5. Then 1-2 will be the external load required to maintain equilibrium, or it _ will represent the virtual transverse stress on the strut resisted . "WZ , by its own stiffness, the bending moment on the strut bemg -j- the same as would be produced in a straight girder of the same span under the same load. /v'gK. sog Example, Fig. 509.— Bending moment by thrust = 3 tons x 2 ft. = 6 ton-ft. Bending moment by reciprocal diagram, 6 : 2 : : d : l, 180 THE MECHANICS OF BUILDING CONSTRUCTION the vertical reaction each end = 2 tons load, — = — -, — = 6 ton-ft. 4 4 Again, we have in Fig. 510, ^ = Tv, whence W = -j- = 2 T tan e, ,_, WZ W and i. = -r~ = ^~. a iv 2 tan Direct thrust at C and S = T Wl „ „ „ H = T sec e = ^ sec 6 „ „ from H to C and S = T sec ^ decreasing to T Wl „ „ at any point D = -j- sec cos ^ = T sec 6 cos <^ so that it is greater at H than anywhere else with load W because the thrust along the chord of half the arc is greater than the horizontal 1, , . , , , . chord of i arc thrust in the ratio 5 • J span If a' circular arch rib be resisted by reactions tangent to the curve, the concentrated load for equilibrium would be given as in Fig. 511 and Fig. 512, where the order of working is 2-3, 3-1, then 1-2 represents the balancing force, and the maximum bending moment would appear to be T y, supposing that an actual resistance occurs at the point shown for the application of W. Upon consideration it will be seen that the concentrated load in Fig. 507 does not represent the true condition of the balancing forces in a bent beam under thrust. The resistance to the thrust being dis- tributed along the strut, a distributed load of double the concentrated should really be taken, but this is not capable of being dealt with completely by the parallelogram of forces or reciprocal diagram. The nearest approach will be made by taking a series of loads equally dis- tributed along the curve and obtaining a line of thrust as for an ordinary arch, as in Fig. 513. With a suificiently large scale we should THRUST IN CUEVED STRUTS 181 find that the line of thrust produced by means of Fig. 514 would be a catenary curve, almost identical with the circular arc. The diagonal resultant of the reactions, as shown on the left, would be tangent to the catenary as shown by the full line, while the tangent to the circular curve is shown by the dotted line. The bending moment at the centre by calculation will be 2 X 6 - i(5-38 + 3-94 + 2-38 + 0-8) = 12 - 6-25 = 5-75 ton-ft. instead of 6 ton-ft., the difference being due to the splitting up of the load into concentrated portions. Under a fully distributed load the bending moment at any point x might be taken as = ^^ which v would give a parabolic outline. The bending moment is, however, only part of the stress to be resisted by the curved strut, it will have in addition the direct load, the combined effect being given by the formula -T^ ± y Now consider a 2j-in. by 2^-in. by 5-in. steel tee made up of plain rectangular sections, as Fig. 515, the sectional area will be im + (2i - m = 1-1875 sq. in. The neutral axis from the edge of web will be _ BD^ - b(P _ 25 X 2-5^ - 2-25 X 2-25° _ 1 700 • ^ ~ 2(BD -bd)~ 2(2-5 X 2-5 -2-25 X 2-25) ~ '^'^ '"•' 182 THE MECHANICS OF BUILDING CONSTRUCTION and from face of table a; = 2-5 — 1-78S = 0-717 in. The moment of inertia ^ ~ 12(BD - bd) (2-5 X 2-5^ - 2-25 X 2-25"y - 4 X 2-5 X 2-5 X 2-25 X 2-25(2-5 - 2-25)° ~ 12(2-5 X 2-5 - 2-25 X 2-25) = 0'703 in inch units. I 0'703 Then section modulus for web side Z,, = — = = 0-.394:, and for ff J. i oo I 0'703 table side Z, = - = ^ _■,„ = 0"98. Radius of gyration squared = r^ = — = = 0"592. A l*lo7o It is assumed that this tee section is to be used as a bent strut or tie 3 = ^^ - -H S i * |-t>SA « /^' 1 ^' 1 /^ig: s/s across the top of the trussed sides of a footbridge, as in Fig. 516. The chord of arc may be taken as 4J ft. and the rise 18 in. Suppose a horizontal thrust of 2 cwt. against one end, the other being rigid, what y-t' cef stresses will be induced in the bar ? By a common, although incom- plete, method of calculation we have Tv = §g x 18 = 1-8 ton-in. bend- ing moment, the moment of resistance being ZC ; then for web side (web CUEVED STRUTS AND TIES 183 1*8 upwards) = q:^ = 4-57 tons per sq. in. tension, and for table side 1"8 (web upwards) = — ^ = ISi tons per sq. in. compression. This, however, has been previously shown to be incorrect by reason of the omission of the thrust. By the formula W .M 1 ,1-8 ^ -"2 ~ 1-1875 0^ ~ ^^'^ + 1'84 = 2-68 tons persq. in. compression in table and 0"84 - 4-57 = 3-73 tons per sq. in. tension in web. Whether these stresses are permissible is generally considered to depend upon the crippling stress of the section as a column. By the Rankine-Gordon formula for mild steel struts (Alexander and Thomson's "Elementary Applied Mechanics," p. 415), , . 6i tons p tons per sq. in. = — "' 1-4-— 1—- ^ I 7000 ^ but this is for hinged ends, and to allow for fixed ends the length I may be taken as -^ of the actual length L. It may be open to question whether the full length round the curve should be taken, or the length of the chord. It will probably be suiEcient if we take the full chord length for I. 6i _ 6-5 _j_r 54^ \ 1 + 0-7035 ' 7000^0-5921/ Then p = T^IT-^ = , ^ ^-^a^^ = ^'^ ^°"« P^"^ ^1- i" safe load in compression, while the actual load is 2-68 tons per sq. in. For the next illustration of the stresses in bent beams take the case of a 3-in. by l|-in. steel channel section, as Fig. 517, bent to a semicircle 10-in. mean radius, with the flanges outwards, held at the ends and carrying a concentrated hanging load in the centre, it is required to find the ultimate crippling load W. The sectional area may be taken as IJ sq. in., the neutral axis 1 in. from edge of section, and the moment of inertia 0-25 in. units. Then R = radius = 10 in., y = distance from neutral axis to furthest edge of section = 1 in., I = moment of inertia = 0'25 in., A = sectional area = 1-5 sq. in., X = leverage of diagonal pull, or versine from line of tension to curve of neutral layer = approximately 3 in., F = ultimate tensile strength per sq. in. = 28 tons for steel, 20 tons for wrought iron. Then on the W M principle that F = -r- + 7" > instead of W will be used the diagonal pull which in a semicircular bar will be 0'7'W, and _ 0-7W , 0-7Wa; „„ 0-7W , 0-7W X 3 1 28 or 28 = 0-46W + 8-4W = 8-86W, whence W = gTgg = 3-16 tons. This 184 THE MECHANICS OF BUILDING CONSTRUCTION can be stated in another form which, however, does not show so readily the origin of the formula, viz. "W = 28 28 _^_2^ 0-7a:y O"? ~ 0-7 X 3 X 1 O"? ~ 8-4 X 0-46 886 I "^ A 0-25 ■'"1-5 = 3-16 tons. Graphically the forces at work can be found as follows : — Join ca, ch b ^ ^"- r^ T and let dc = weight W. From d draw de and df parallel to Ic and ac. Join ef, cutting cd in g. Then the vertical reactions EiBg = dg and eg ■= — . The horizontal tension 'E. = eg anAfg = -^ for semicircular beam. The diagonal force tending to straighten curved beam = e/'andice. Now take the case of an arched rib used in centering and consisting of bent 4-in. by 5-in. by J-in. steel tees, as Fig. 518, 3 ft. centre to centre, used web downwards to support the lagging of an arch of 18-ft. span and 3-ft. rise. Assuming the arch to be built on half the span before the other half is commenced, the problem is to find what will be the maximum stresses produced in the tee bars. First draw the outline as in Fig. 519, divide it up into assumed voussoirs, and find the weight of each. Suppose these to be as marked on diagram, then find the centre AECHED EIB CENTERING 185 of gravity of all the loads by polar diagram, Fig. 520, and link polygon, Fig. 521, and obtain the vertical reactions by drawing a line — I3a in Fig. 520, parallel with the closing line of the link polygon. From the centre of the horizontal line joining the tops of middle-thirds at abut- ments in Fig. 518, set up the reaction 13« — 1 and join the top point to top of middle-third at left-hand abutment. From the intersection of this last line with the mean centre of gravity line draw a line' to top of middle-third at right-hand abutment. From points 1 and 12 in Fig. 520 draw lines parallel with these two reactions, giving point 13 by their -^ F/g S20 intersection. Join 13 to the points on the load line and by parallel lines construct the curve of thrust on Fig. 519. The area of the section may be taken as 4-25 sq. ins., the distance of neutral axis from edge of web y = 3-4 in., and from face of table a; = 5 - 8-4 = 1-6 in. The moment of inertia I = 10-45 in. units. Then section modulus for web side _I_ 10-45 _ and for table side Z, = - = . „ = 6-53. X Id The stress per sq. in. may 186 THE MECHANICS OP BUILDING CONSTEUCTION now be calculated by the formula x - Z" ' '"^ ^^^^ ^^^^ ^ ~ ^^"^^^^ °* 9-13 in Fig. 520 = 47-5 cwt., A = 4-25 sq. in., M = 47-5 X 11 = 522 cwt.-in., 7i. = 6-53 and Z^ = 3-07. Then WM^47:5522^ ^ ^^^.^ A ^ Z„ 4-25 ^ S-07 or 9-05 tons per sq. in. compression in web, and W M 47-5 522 ,, , _. „. „ , A-Z;=4-^-6^=ll-l-«0=-*''-^^^*- or 3-45 tons per sq. in. tension in table ; whereas if the section had been turned up the other way, web upwards, the stresses would have been 11-1 + 80 = 91*1 cwt., or 4-55 tons per sq. in. compression in table, and 11-1 — 170 = — 158"9 cwt., or 7*94 tons per sq. in. tension in the web, which would have been a more satisfactory arrangement. An important and interesting question upon an arched rib was set in the Honours Practical Geometry, 1904, as follows: — "The form of an arched rib is a circular arc of 100-ft. span on 16 ft. rise, the supports being at the same level. It is hinged at the ends and loaded with a weight of 12 tons at a horizontal distance of 30 ft. from one end. The horizontal thrust of the arch is known to be 12'5 tons. Draw a diagram of bending moment for the arch. Indicate the places where the shearing force, thrust, and bending moment on the rib have their maximum values, and give these values." The solution of it given below is by R. E. Marsden. Looking at the loaded arched rib. Pig. 522, it can be seen that, if the ends are hinged as stated, the load tends to flatten the curve of that part of the arch beneath the load, and to increase the convexity of the opposite half upwards. To solve the problem, set out the load line, Fig. 523, and use a pole for vector polygon distant from this load line equal to 12'5 tons on the force scale. If this polar distance is set out on a horizontal line drawn 3'6 tons from the top end of the load line, it will save drawing two funicular polygons in Fig. 522. Next draw the line of thrust, as shown by chain-dotted lines in Pig. 522, this gives the form of structure required to give a horizontal thrust of 12"5 tons while supporting the given load of 12 tons. Where this line of thrust crosses the arched rib, there is a change in the bending moment from positive to negative ; that is to say, this is a point of no bending moment. To find the amount of bending moment on any point of the arch, drop a perpendicular on to the line of thrust, and measure this amount, as ac in Fig. 524 on the length scale, and multiply by the amount of thrust (13-01 tons) in the line of thrust. The same result will be obtained by dropping a vertical line ah from a point on the arched rib, and multiplying this length by the horizontal thrust 125 tons. In both cases the bending moment is given in ton-feet. The bending moments for various points on the arch have been calculated in this way, and plotted on verticals having one end on arched rib, and passing through the point at which the bending moment was calculated. In Goodman's BENDING MOMENTS ON ARCHED RIB 187 '• Mechanics Applied to Engineering " a proof is given that if a number of equidistant spaces are marked off along the line of the arched rib, and verticals drawn through these as in Fig. 525, then when the length of the ordinate up to the arch rib y. Fig. 525, is multiplied by that portion lying between the line of thrust and the arched rib z. Fig. 525, the total of all the minus or negative quantities thus obtained should equal the total of all the plus or positive quanti- ties. When the line of thrust passes below the rib, the ordinates between are minus quantities ; and when it passes above, they are positive or plus. Tested in this way. Fig. 522 is proved to be correct. The shearing force is determined by drawing a tangent to the rib through the point, and resolving a force equal and parallel to the line of thrust immediately below the point along this line, and along one perpendicular to the rib, as in Fig. 526. The shear is given by the component of thrust at right angles to the direction of the rib. The greatest bending moment is found under the load, which is also the place of greatest shear. The horizontal thrust could not have been determined had it not been given by the examiners, unless other data had been provided. The section of the arch, affecting its stiffness, etc., has something to do with the determination of the horizontal thrust, for the section may be so stiff in proportion to its length that it exerts no more thrust on the supports than does an ordinary straight girder. The condition of loading greatly affects the amount of horizontal thrust, as also the method of fixing. A good method of understanding some- of the stresses that result in a curved beam from loading is to substitute a lattice girder of the same shape, and work out the stresses in the individual members. By doing this for two or three different systems of lattice work a check could be had on the result. A discussion of , metal arches under all kinds of conditions will be found in Perry's " Applied Mechanics," Goodman's " Mechanics Applied to Engineering," Rankine's " Civil Engineering," and also in '' Statically Indeterminate Structures and the Principles of Least Work " by Martin (35 and 36, Bedford Street, Strand, London). Other works will be found in the Patent Office library. 188 THE MECHANICS OF BUILDING CONSTRUCTION EXEBGISES ON LECTURE XXV Q. 81. The curved tie-bar of a roof is composed of a standard steel tee, 4-in. table and 5-in. web, both g in. thick. The section area is 3'25 sq. in., the moment of inertia 7"7Y, neutral axis from edges of section 1-47 in. and 3'53 in. The bar is curved to a versiue of 3 in. in a chord length of 5 ft., web upwards, and the tensile stress on the chord line is 5 tons. Find the maximum intensity of the stresses in tension and compression. Would any serious difierenoe have occurred in the stresses if the web had been downwards ? Answer. The first step will be to find the section modulus for each side of the I 7-77 I 7-77 W M .,> neutral axis. Z. = - = ^ = 5-28, Z, = ,^ = gTgg = 2-2. Then / = -^ ± 2" ^'^ W M W be the formula to use, or more correctly — T"^ y' ^°^ ^^^ '^^^< ^^^ — T ^, W.M 5, 5x3 ^^^•^-A+z;,= ~3T5 + - for the table side per sq. in. maximum compression in web, and 2'2 M Z], -l-538-(-6-818=5-28tons r,^ . S2a C W M A Z^~ 5 3-25" 5x3 5-28 ~ - 1-538 - 2-841 = 4-38 tons per sq. in. maximum tension in table. EXAMPLES FOR PRACTICE 189 Web downwards the stresses would have been _ W_ M^_^ 5x3 A Z, 25 2-2 - 1-538 - 6-818 = 8-356 tons per sq. in. maximum tension (in web), and W . M 5.5X8 5^' A + Z 3-25 ■*" - 1-538 + 2-841 = 1-303 tons per sq. in. ma xim n n i compression (in table), showing an increase in the Tna.YiTrmTn stress of W^g^^ - 100 = 58 per cent. Q. 82. Pig. 527 represents a curved rib, consisting of a (B.S.B. 9) 6 in. by 4 J in. by 20 lbs. rolled steel joist, supporting the roof purlins as shown. Assuming the ribs to be 10 ft. centres and the load to be J owt. per ft. super on the roof, draw the curve of thrust to pass through centre of rib at crown and abutments, and calculate the stresses on the rib. Answer. Draw the frame diagram. Pig. 528, and the load from each purlin wiU be - - - - = 55 cwt. as shown. Set down the load line. Pig. 529, and select any pole 0, but as the loading is symmetrical is taken on a line central between 1 and 5. Draw vectors to the points on the load line and parallel with these ^5 vectors construct the curve of thrust AbcdeB. This curve does not pass through point D, another pole P must therefore be selected so that the distance P3 = 03X§J. Then, as before, draw vectors and complete the curve of thrust passing through the points ADB. The maximum stress will then be across spaces 2 and 4, where the thrust is 80 cwt. = 4 tons and the leverage 10 in., then W M _ _^ A Z 5-8 4 X 10 ' 11-54 : 0-68 ± 3-47 = 415 tons per sq. in. compression and 279 tons per sq. in. tension so that this section will be sufficient. LECTURE XXVI Vaulting — Groined Vaults — Bibs and Panels — Finding Elevation of Eibs — Curve of Thrust — Deterfliination of Stresses. In Roman vaulting, called barrel or wagon vaulting, where two cylindrical vaults intersected at right angles there was an elliptical groin, but as a rule no groin rib. The vaulting was of considerable thickness, say 2 to 3 ft., upon which it relied for stability. In Gothic vaulting the groin ribs at the intersection formed an independent diagonal arch from corner to corner, which supported the filling of the spandrils, say 6 in. thick, and transmitted the load directly to the abutments. The ribs or groins are thus a source of strength to the Gothic vaulting which is absent in the Roman. Fig. 530 shows a perspective view of simple Gothic vaulting. It was stated in a recent article in Engineering Record that " the groined arch, though used centuries ago, is a purely empirical type of structure, designed, or, speaking more accurately, drawn without satis- factory mathematical analysis and used without much knowledge of the factor of safety." It may be the fact that in many cases the modern architect is unable to make the necessary calculations, and, instead of calling in expert assistance, copies as nearly as possible what has been done before, but it is impossible to believe that the ancient and mediseval stonemasons were not fully cognisant of the essential nature and value of the stresses brought into action in their structures. Groined ceiling vaults such as occur in the crypt of a Gothic church, with a pendent stone in the centre, and on a larger scale at the junction of nave and transepts with or without an open eye in the centre, consist in the simplest cases of four ribs running diagonally to the four comer piers, with an arched wall rib or transverse rib from pier to pier, and between the ribs a thin panel filling, sometimes curved slightly in its cross section to help to carry its own weight as an arch, the weight of the panels being then entirely transmitted to the ribs. The vonssoirs of the ribs are for two or three feet up from the springing, or where they branch out separately, worked out of the solid stones with hori- zontal beds, above the capitals of the piers or columns, in order to give a solid backing and reduce the arch thrust ; this portion is called the " tas de charge." Besides arching in cross section, another method by which the failure of the panels was prevented was the introduction of intermediate ribs, reducing the width of the panels, and a later arrange- ment was the insertion of lierne ribs between the intermediate and main ribs. Fig. .531 shows the plan of a simple Gothic vault with the names of the parts marked on, and two methods of coursing the panels. The STKESSES AND GROINED VAULTING 191 A . B " alCemaCivt courauKj fbr ponei* 192 THE MECHANICS OP BUHiDING CONSTEUCTION ribs are usually rebated to receive the panelling except in a few early examples where the filling simply rests on the back of the rib. The stresses in groined vaulting follow exactly the same principles as those laid down for arches and arched bridges, but they have only their own weight to carry, except in the few cases where the haunches are loaded with chalk rabble. Fig. 532 shows a section of the groined vault corresponding with the plan. Fig. 533 shows a section of the transverse ribs, the wall rib in section being similar to the portion on one side of the centre line. Fig. 534 shows a section of the diagonal ribs. _ The dotted lines show the rectangles taken in the calculations as the virtual sections. "When the fiUing-in is flat, it compares with a section of an ordinary -111 5s 5 5 ^ « S J §, sv, ,-e a - s ^ J ' barrel vault and its stability is therefore subject to the same limitations. Assuming no ridge ribs, one spandril is kept in position by the pressure of the one opposite, but on account of its want of thickness it has not the advantage of an ordinary arch with deeper voussoirs, and therefore considerable reliance has to be placed on the mortar between the joints. In this form of vaulting very little pressure falls on the ribs from the fiUing-in, and unless the fiUing-in is carefully built, and with good mortar, it is liable to give way ; this weakness, no doubt, was the reason for the introduction of intermediate ribs. When the filling-in is arched from rib to rib its stability is increased, because it is then RIBS IN VAULTING 193 arched in both directions. Each ring forming a complete segmental arch need not necessarily be so carefully formed as when flat, and is therefore always stronger and not so liable to failure as the former. In this case the ribs are subjected to direct thrusts normal to the line of rib in plan, but inclined according to the curvature of web. It will be seen that the thrusts neutralise each other to a certain extent, and their resultant simply acts as dead weight on the rib. The half eleva- tion of a transverse rib. Fig. .533, with the loads upon it is shown in Fig 535, and the curve of thrust is found in the usual manner from Fig. 536. The maximum stresses will occur across space 7, and will be W , M 980 . 980 X 2 , A - Z = (Tl^ - |(4^rr) = 41 ± 82 = 123 lbs. per sq. m., or 7*87 tons per sq. ft. compression, and 41 lbs. per sq. in. or 2-62 tons per sq. ft. tension, which is quite safe. The curve of the diagonal rib, Fig. 534, must next be found as follows. In Fig. 537 set up the elevation of the transverse rib, and also the springing line of the diagonal rib. From any point a in the springing line of the transverse rib draw a perpendicular ab. Produce ba to d, and from d at right angles to springing line of diagonal rib set up dV equal to db. This process repeated with other points will give the curve of the diagonal rib, and the half-elevation of diagonal rib with loads wUl then be as Fig. 538. The curve of thrust will be obtained from Fig. 539, and the maximum stresses occur across space 6 ; then W.M 1620 .1620x2-5 „, ^ ^„ , ,n« = tv -r±-fr = t; — —^. ± XTT. s^^^T^ = 35 ± 67-5 = 102'5 Ibs. per sq. m., A Z 6 X 7-7y ^(6 X 7-75") ^ ^ or 6'56 tons per sq. ft. compression, and 32'5 lbs. per sq. in., or 2"08 tons per sq. ft. tension, which is satisfactory, and the vaulting will therefore be perfectly safe. BXEBCISES ON LECTURE XXVI Q. 83. Sketch the plan of one quarter of a bay, showing what is meant by " lierne rib vaulting." Name the parts shown. Answer. See Pig. 540. Q. 84. A semicircular barrel vault, 15 ft. span, intersects another of the same span at right angles. Find the section across the groin. Answer. See Pig. 541. Q. 85. In the example worked out in the Lecture find the stresses on the buttress from the vaulting ribs. Answer. There will be the thrusts from the transverse rib and two diagonal ribs on each buttress. The horizontal component of the thrust from each diagonal rib (Pig. 539) = 1240 lbs. acting at 45° to face of wall, and com- bining the two, as in Pig. 543, gives a thrust of «./2(1240)* = 1753 lbs. perpen- dicular to wall. In addition to this there will be the horizontal component of thrust from transverse rib (Pig. 586) = 720 lbs., making the total thrust perpen- dicular to wall = 2473 lbs., as shown in elevation, Pig. 542. This must now be combined with the combined vertical components of the thrusts = (2 x 2765) + 1822 = 7352 lbs., and the resultant so found combined with the weight of wall, buttress and roof = 64,040 lbs., this latter step being drawn to a reduced scale. The final resultant is then found to cut the base at 1-39 feet from inner face of wall. The calculations about the section AB are as follows : — Distance of centre of gravity of base from inner edge F'/g. 543 EXAMPLES FOE PRACTICE 195 _ (11 X Ij X 0-75) + (Ij XljX 2-2 5 ) + (1^ Xl\X 3-75 ) _ ^ ,„ ^ (11 X IJ) + (1| X IJ) + m X IJ) Z = (11 X 4^' - 9-5 X 3')' - 4 X 11 X 4^ X 9-5 X 3( 1-5)' _ . ^g . ., 6(11 X 4J» - 9-5 X 3^) ~ 71 000 X /' i-ag - 1•19 ^ or 1-61 tons per sq. ft. compression at outer edge and 3277 lbs. or 1'47 tons per sq. ft. compression at inner edge. LECTURE XXVII Theorj' of Domes — Joint of Eupture or point of Contra-flexure — Domes compared with Arches — Curve of Thrust in Dome giving preesure on Beds — Method of finding horizontal thrusts or tension and compression in Vertical Joints. Domes are not generally supposed to be reducible to the same principles as ordinary arches ; we are told that the dome differs from the arch in having no thrust at the crown and when hemispherical having no thrust at the base, but this is a mistake. A similar state- ment about no thrust at the base is made regarding a semicircular arch, but we have seen that the true arch is the line of thrust, and as this is never vertical at the springing, there is of necessity always an outward thrust. A theory of the stability of domes will be found in Rankine's " Applied Mechanics," p. 265, and this is repeated in Gwilt's " Encyclo- psedia of Architecture." The same principles are adopted in Marsh and Dunn's " Reinforced Concrete," and may be stated as follows : — Let the arc ab, Fig. 544, represent half the section through a thin hemispherical dome of radius Oa and OJ, and uniform thickness. The weight of any segment or zone is proportional to its vertical height, because the surface area = circumference x height, therefore if the vertical axis be divided into say four equal parts the horizontal planes through these points will cut off equal weights. Take any point c, and draw a tangent to meet a horizontal line through the crown in d, com- plete the parallelogram edef as shown, then hf represents the vertical load at c, be the radial thrust shown . acting outwards at c, and really being the equilibrant to balance the inward tendency, and ef the direct compression in the material at point c. The outline bef gives the triangle of the forces acting at c, which are shown separately in Fig. 545, the ordinary force triangle for which would be as in Fig. 546. The joint of rupture, as it is called, but more correctly point of contra- flexure, in a thin hemispherical dome of uniform thickness, occurs at a height of ^(Vs — 1) X radius = 0-618 radius, the point forming a centre angle with the vertical of 51° 49' as shown. The matter will perhaps be made clearer by taking out the force triangles from Fig. 540 and comparing them with each other. The triangle of forces at point c will be as Fig. 547. In a similar manner the triangle of forces at c^ is given in Fig. 544, and will be as shown separately in Fig. 548. The force fe is the thrust in the dome from point c, ff^ the weight of the segment from c to Cj, and /i« the reaction at Cj, or direct thrust in the dome at that point, showing no horizontal thrust as points c and Cj are approximately equidistant from the joint of rupture. At point c^ the polygon of forces will be as Fig. 549, where efi is the thrust from point THEOEY OF DOMES 107 Ci, J)B the horizontal thrust acting inwards, or the reverse way to its action at point c,fif., the weight of the segment, and fjb^ the reaction at point Cg. At point Cg at the base of dome, the polygon of forces will be as Fig. 550, where ^ftj is the thrust from point C2)/a/3 the weight of segment, fj) the reaction at point Cg, and S&i horizontal thrust acting ffoaf/ofi ehrusc Fi^. 544 'F/g €48 inwards. The horizontal force at c is shown to act outwards to balance the tendency of the dome to fall inwards at that point, while the hori- zontal forces at c.^ and Cj are shown to act inwards to balance the tendency of the dome to spread outwards at those points. If these forces can be applied, the line of thrust will pass down the centre of 198 THE MECHANICS OF BUILDING CONSTKUCTION the thickness of the dome, but in practice this cannot easily be done. It is only on the supposition that it is effected, that it can be said there is no thrust at the base of a hemispherical dome. The diagram of stresses may take another form, Fig. 551. Set oflf the load line as before and through A draw lines parallel to those tangential to the surface of the dome at the various points to intersect with the horizontal lines. Through these intersections draw the curve from A to B. Then we have the stress diagram giving the force triangle klc for point c, showing vertical load Kb, horizontal thrust he, and direct compression Ac. Any number of divisions may be taken on LINE OP THKUST IN DOME 199 the vertical line, and the tangents and horizontal lines drawn in. Then the actual horizontal thrust at any point will be the difference between the length of line at that point and the length of line at the point above. Above the point of contra-flexure the horizontal thrust will act inwards and below it outwards, because the upper part tends to fall in and the lower part to be thrust out. These radial thrusts being calculated upon the whole circumference require to be divided by the circumference at that point in order to VAfi 1 Hil i" n 7*11 st give the local intensity, ^^its in circumference = '"*^°^^'y- ^^^ "^°°P tension," as the circumferential stress below the point of contra-flexure „ J . . , „ . , radial thrust x radius radial thrust 18 called, IB given by the formula circumference ' = "¥2832- ' or approximately 0'16 thrust. The weight W of a thin dome, where r = mean radius of curvature of dome, v = mean rise, %v = weight per unit area of surface, is W = 241 X 277r X V. It will now be interesting to see how the stresses in a dome may be worked upon the same principles as the arch. Fig. 552 represents the half section of a dome 20 ft. mean diameter, and 1 ft. thick. Assuming the weight to be 112 lbs. per cubic foot, the weight of dome will be "W = 2<'X2i7rXi!'=lx2x 3-1416 X 10 X 10 = 630 cwt. Dividing the mean rise into 10 equal parts and drawing horizontal lines across will give rings of voussoirs having equal weights = ^ = 63 cwt., but the top voussoir may be again divided for greater accuracy in obtaining the curve of thrust. The load line may now be set down as in Fig. 553, and the curve of thrust obtained in the usual manner. The maximum stresses will occur across space 5, where the thrust = 400 cwt., and distance from curve of thrust to centre line = 0'4 ft. The mean radius at this point = 9'1 ft., therefore the circumference = 3'1416 x 9'1 X 2 = 57-2 ft., and the thrust per ft. run = f^:^ = 7 cwt. The stresses W M 7 7 X 0'4 will then be -^ ± -^ = j-^^ ± i^ ^ ^^. = 7 ± 16-8 = 23-8 cwt. or 1-19 tons per sq. ft. compression at inner edge, and 9'8 cwt. or 0'49 ton per sq. ft. tension at outer edge. The curve of thrust in this case does not allow for any tension hoops until the springing is reached, where 132 the hoop tension will be -..g^-oo = 21*04 cwt. These stresses are in the bed joints of the dome, and are independent of the stresses at the vertical joints. The latter stresses are caused directly by the tendency of the upper part of the dome to fall in, and the lower to push out. Taking the section of dome again in Fig. 554, and setting off the same divisions as before, construct a graphic diagram of thrusts. Fig. 555, on the principle of Fig. 551. Then at the centre of each division on Fig. 554 set off horizontally the difference between the successive thrust values in Fig. 555 divided by the vertical depth of the division, and join the points so found to give the horizontally shaded areas. Above the point of contra-flexure, or reversal of thrust, the joints will 200 THE MECHANICS OF BUILDING CONSTEUCTION be in compression, resisted simply by the strength of the stone. For instance, take the portion of the dome AB, Fig. 554, then the total inward horizontal thrust = -- '^^^ x i =iO cwt., and the hoop compression = ^^ = 6-36 cwt., and the area of the vertical joint between A and B being 1-9 x 1 = 1-9 sq. ft., the compression on this joint will be^ = 3-85 cwt. per sq. ft., and the curve for compression ."/-•^ /^gf. S5S on the vertical joints will be as shown. Below the point of contra-ilexure the joints will be in tension which may be resisted by a single belt or chain round the dome at the level of the centre of gravity of the outward thrusts, say at X, the courses being dowelled together in order to transmit the effect of the belt throughout the required depth. The tensile resistance necessary in the belt will be found thus : — The total STEESSES IN A DOME 201 outward thrust collected at X, or horizontally shaded area below point of contra-flexure, by taking area and mean pressure on each section, = say 188-3G cwt., and the hoop tension will therefore be 777^^7.-. = :]0 cwt. or 1'5 tons. By a more accurate method, taking the area by plani- meter = 1 sq. in. x 100 cwt. to 1 in. x 2 ft. to 1 in. = 1 x 100 X 2 = 200 cwt. total outward thrust, and hoop tension — 200 The lower radial ordinates, Fig. 554, represent the compression per DOME MeO/f rac/ius iO'.o" TA/cfcness /2'^ on ireracal jo/ncs i^re/coi Joints) ^i^ fyaw K'> m 9./. ■m y / 14 Y leslcM:] Y I 1 J J \ i ; "9 SS5 \ Ordinaees of otjororef (Tension tr) ygrticot Joirxi) square foot on the bed joint, and to obtain this the total thrust must be divided by the circumference and the thickness of dome at the point under consideration. For instance, take the horizontal joint at springing, the total thrust is 630 cwt., circumference = 2 x 10 x 3'1416 = 62"832 ft., and thickness 1 ft., therefore the compression per square ""'' = (^2W^ = ''■'' ^^^- EXERCISES ON LECTURE XXVII Q. 86. Find the curve of thrust on a concrete dome 20 ft. mean span, 7 ft. 6 ins. mean rise, 1 ft. thick, 1\ cwt. per cub. ft., and state the hoop tension collected at base. Answer. The outline of dome will be as in Pig. 556, and the weight will be iwirrv = 2 X 1-25 X 3-1416 X 10-416 X 7-5 = 615 owt. The load will then be divided up as shown and the curve of thrust obtained from Pig. 557, whence the 205 hoop tension = _ -.-„ = 32-62 cwt. 202 THE MECHANICS OF BUILDING CONSTRUCTION EXAMPLES FOE PKACTICE 203 By another method draw out the dome again as in Pig. 558, and from it set oS the curve of thrusts in Pig. 559 and also the ordinates of outward horizontal thrust aa shown by the shaded area in Pig. 558. The area of this shaded portion gives the total outward thrust = 75'5 owt., and the hoop tension wUl therefore be A „^i^ = 12 owt. 6-2832 By Marsh and Dunn's formulae, rise of tangent = -in.^iA _ 7.1; ~ ^*'^ ^^'' total radial thrust = — ^^Tj — = 180 cwt., and the hoop tension = ^.oaio = 28-64 owt. Q. 87. The circular domed roof of a furnace chamber, 20 ft. mean diameter, 2 ft. 6 ins. rise, 18 ft. radius of curvature, 13| ins. thickness of brickwork, collapsed after being heated to 450° P. and fractured the steel belt encircling it. Pind the hoop tension when cold by various methods. Answer. An approximate idea of the thrust at the base of the dome may be obtained by taking a strip as Pig. 560 1 ft. wide at base, and nothing at crown, as a half-arch, the thrust being found as in Pig. 561. Then the thrust per foot run, multiplied by the mean diameter, wiU give the total tension to be resisted by two sides of the encircling band, and half of this will give the tension in the band ; thus 7-2 20 „ „ , 20 X -2- = 3-6 tons. By Marsh and Dunn's formulae (based on Rankine). Clear span 18 ft., rise 2 ft. 6 ins., thickness 13J ins., virtual span 20 ft., radius 18 ft. Weight of dome = 36 X 3} X 2-5 X 135 = 38,187 lbs. = 17-05 tons. Rise of tangent = -.r, _ q.k ~ "'^^ ^^^*- Total radial thrust = — ^-r= — = 26-43 tons. D-45 _ ^ . 26-48 X 10 , - , Hoop tension = on x a f ~ tons. By Godard's formulae : — ^ 2WR 17-05x10 „„„, Q = -gr = —3^25-- = 22"^3 *°"'- Q 22-73 ^ = S = 2-x|=3-61t-B. By Eankine's formula : — TT * - rr A Q . 2 0-3 X 135 X 1-125 X 18^ . ,. , Hoop tension T = Q-Zwtr^ = sht?. — = 6-58 tons. ' 2240 By another formula : — Total radial thrust R = ^ = ^-^^-^ = 50914-7 lbs. = 22-73 tons. ov D X 2-5 R 22-73 Hoop tension at base T = „- = ^aoo = ^'^^ *°°^- By still another formula : — W 17-047 Radial thrust per foot run R' = g^ = g-^ s.^^fg x 2^ ~ O'^&i ton. Hoop tension = R'{J(2) = 0-362 X 10 = 3-62 tons. LECTUEE XXVIII Principles of Shoring— Baking, Plying, Dead and Needle Shores— Formula and Scantlings — Foundations, Width and Depth— Supporting Power and Natural Slope of various Soils— Pile Driving and Supporting Power of Piles— Formulse. The principles governing the stresses in shoring are very easily laid down, but the actual stresses in any given case are so difficult to deter- mine as to be practically impossible of solution, and reliance has to be placed upon personal judgment and experience. With an ordinary raking shore as A, Fig. 562, there is a certain thrust B exerted, or liable to be exerted, by the wall, and the weight of r/g S64 wall above the point of application of the shore, represented by the force 0, must be such as, compounded with B, will produce a resultant in the direction of the length of A. It does not matter if the weight of wall exceeds the required amount as only so much of it will be brought into action as may be required. The weight of wall may be taken as that part within an angle of 60 degrees above the point of the shore, together with the weight of roof that may be resting on that portion. The wedging up of the shore brings the forces into play, and usually nothing more is done than wedging up until everything appears to be tight. Then any further tendency of the wall to fall out is met by the resistance of the shore and the weight above it. The head of a raking shore is fitted as shown in Fig. 563, the lower end of the shore being wedged up or prised along the sole piece to tighten it against the PRINCIPLES OF SHORING 205 needle. In Fig. 564 let AB be the face of wall, CD the line of the shore, and DB a horizontal line. Construct the parallelogram of forces as shown, where W is the weight of the wall DB contained in an equilateral triangle having its apex at D, including roof, floors, etc., H the horizontal thrust which can be resisted by the shore, and T the Wait p/ece -//'x3" -Cleat X-- Oak rteea/ie @ /=^/^ SGJ . /o\ /o" ra/r/ngf <§r ■tVall piece J/'xS" tycrii /took -Cieae - Oak neealtje r^akinqf •s/iore thrust or compression in the shore when in full action, that is, this diagram gives the maximum figures possible. In a system of raking shores as Fig. 565 the wall piece enables the lower needle to assist the upper one when the top shore is near the top of wall. It should be observed that the supporting power of the shores is transmitted entirely through the needles to the wall, and it will often be found in practice that the needles are the weakest points in the system, particidarly in 206 THE MECHANICS OF BUILDING CONSTRUCTION heavy shoring. They should, therefore, be of oak or other hard wood. The raking shores may be calculated by Gordon's formula, or by similar means, taking the width as least diameter, but the following formulfe are sometimes adopted : a = Area of wall supported in sq. ft. t = Thickness of wall in inches. tv = Weight of wall per cubic foot, say 1 cwt. h = Height to head of shore in feet. w = Weight of shore in cwts. = Angle of shore with horizon. P = Upward vertical thrust of shore. Q = Horizontal thrust at top of raking shore in cwts. CALCULATION OF SHOEING 207 Cwts. horizontal thrust of wall at top of raking shore = Q = oeo/,- Cwts. upward vertical thrust of shore = P = Q tan 6 — ^w'. And to provide against this the distance from head of shore to top 8P 4P of wall should not he less than — r, or with roof = — ,-. Cwts. compression in raking shore = P sin ^ + Q cos 6. Safe load cwts. on shore = Ib'bd ins. ^ u ' ' In these formulae Q is the maximum horizontal thrust that the wall would stand without shoring, and the calculations are baaed upon that amount of thrust. The formulae given in " Stock on Shoring " are equivalent to these. Applying these methods to a given case, say a 9-in. wall 30 ft. high, the height to top of shore 20 ft., shore 6 ins. by 6 ins. inclined at 60 degrees. Then weight of wall coming on to shore 8-66 X 10 „ , ^ u- * . ooK * = ^ X 12 X 1 cwt. per cubic foot = 32-5 cwt., then the maximum horizontal thrust by the method shown in Fig. 564 = 32'5 X cot 60° = 18'8 cwt., and the maximum thrust in shore = 32"5 cosec 60° = 37"6 cwt. By the above formulte r8-66 X 10\ ^ gj 1 X Q = :: „ = 0-61 cwt., ^ 288 X 20 assuming the shore to weigh 35 lbs. per cubic foot. 90 V 2 P = 0-61 X /3 - i X ^ X ^^^ X ^ = 0-156 cwt.. and the compression in shore = 0-156 X -^ + 0-61 X i = 0-44 cwt. The usual scantlings for raking shores are as follows : Height of wall in feet. No. of shores in each set. Scantling of each shore. Up to 20 2 \ 20 sq. in. ~Jl " f^ g 10 to 15 ft. between ^Jl ''" >» ^" ^\ PQ/.V, opt, ^ ^5 " 40 3 Anar60°to75° '^^ 40 „ 50 4 ^°g^«'>" '0 75 gQ 50 and upwards 4/ 100 Dead shores should be calculated as posts by Gordon's formula, and needle shores as beams with concentrated loads. Flying shores are hardly subject to calculation, but an empirical formula is sometimes used for the thrust on flying shores which has no apparent scientific foundation. It is T = -kt, where T = thrust in cwts., W = weight of wall in cwts., t ~ thickness of wall, in feet at ground line, h = height in 208 THE MECHANICS OF BUILDING C0N8TKUCTI0N feet to principal stmt. The usual scantlings for flying shores are as follows : Span. Principal Strut. Raking Sirvt. Straining Pieces. Up to 15 ft. 6x4 4x4 4x2 „ io „ 9x6 6x4 6X2 „ 35 „ 9x9 9 X 4J 9x3 The maximum span is about 35 feet on account of the diificulty of getting longer timber. The sets of shores are placed from 10 to 15 ft. apart, and \t there is only one principal strut in each set it should be placed about | height of wall ; if two, then at J and | with continuous wall pieces. There are two points to particularly observe in connection with foundations — ^the width and the depth. The width depends primarily r/g. S66 * m' - -> \ ^/g.6by \ * -m' - ■> / r \ / Sr/cA/york \ ^•H V h.' ^-r f "• \ + Concrete A % t 1 ; IF V \ t rig sea ■18' ■ r/g. se9 fj •<' u 6-' ' s n e- ♦ N 1 ^ N 1 ^ 1 A 1 ' t' / ^ ' y t" t i ^ upon the load imposed by the wall or pier and the supporting power of the soil. The load is usually a simple matter of calculation, but the resist- ance is matter for judgment. The text-books give tables of safe loads for different soils, but the classification of the given soil has to be dis- covered before use can be made of the data. In the north of England and in Scotland it is not usual to put footings to walls, but it is universal in the south. The rule is, as many courses of footings as there are half bricks in the thickness of wall, projecting 2\ ins. each on each side, so that the bottom course will be twice the width of the wall. By the London Building Act and many local byelaws, there must be concrete under the footings not less than 6 ins. thick, and projecting beyond the brickwork not less than 4 ins. on each side. Architects, however, specify 6 ins. projection, because that is the width LOAD ON FOUNDATIONS 209 of excavation beyond the bottom course of footings allowed for the bricklayer to work in, and they frequently make the thickness very variable but much greater than the minimum allowance ; in fact, they seem to have no rule for thickness. The 6 ins. projection requires a minimum depth q^f say 9 ins., i.e. half as much again, to prevent it breaking off, and it is only natural to give a proportionately wider base to a thick wall, and not limit it to 6 ins. projection. The following figures will therefore be of some interest. Fig. 566 shows the ordinary method for reasonably good work. Fig. 567 shows Mitchell's sugges- tion for proportioning the size of concrete which makes it too thick. Fig. 568 shows an improved modification of Mitchell's method, and Fig. 569 shows an improvement on all of them. The depth of the bottom of concrete from the surface of the ground will naturally vary in different soils independent of the question of basement rooms or cellarage. Generally speaking, a minimum depth of 2 ft. 6 ins. should be adopted on gravel soils and 5 ft. on clay soils, but in the London suburbs one often sees the concrete started on top of the grass, pegs being put in to hold up scaffold boards edgeways to contain it. Eankine's formula for depth of foundation is based upon the same theory as his formula for the pressure of earth against a retaining wall, and assumes to give a depth that will prevent the ground rising round the foundation by being squeezed out. The formula is as follows : — Safe Load on Foundations (Eaneinb). p = maximum vertical pressure in lbs. per sq. ft. d = depth of underside of foundation below suiTOunding surface in feet. to = weight of earth in lbs. per cubic ft. 6 = angle of repose of earth. ^ \1 — sm $/ u)\l + sin ^' / l + sin eV /I - sin 6 Y yi-smo) Vl+sine/ ^=15° 30 45 2-89 9-00 33-94 + sin e^ 0-346 0-111 0-0295 Weight and Natueal Slope of Various Soils. Soil. Lbs. per cub. ft. (w). Angle of repose $. Vegetable earth . . 90 30 Sandy loam . . . . 100 34 Loamy clay . . 110 36 Firm gravel 120 40 Loose gravel . .... 110 36 Stifi olay .... 128 45 Wet clay . . 120 16 210 THE MECHANICS OF BUILDING CONSTRUCTION Safe Loads on Materials. Granite Portland and compact limestone Hard York stone . . . Limestone (ordinary) . Blue brick in cement . Stock „ „ ,, „ lias mortar „ „ grey lime mortar . Portland cement concrete (6 to 1) Lias lime concrete (3 to 1) . . Gravel and natural compact earth Hard clay .... . . Made ground rammed in layers . Ordinary load. Maximum load. Tons per ft. sup. Tons per ft. sup. 20 30 15 20 12 15 6 6 9 12 6 8 5 6 .3 4 5 10 3 5 3 3 IJ 2 i 1 Safe Load on Mortar. Stone lime mortar — 50 lbs. sq. in. or 3 tons sq. ft. compression. 25 „ „ 1^ „ tension. Lias lime mortar — 150 lbs. sq. in., or 9 tons sq. ft. compression. 50 „ „ 3 „ tension. Portland cement mortar — 200 lbs. sq. in., or 12 tons sq. ft. compression. 75 „ „ 5 „ tension. There are several very interesting points for consideration in con- nection with pile-driving and the supporting power of piles, but time will not permit us to do more than glance at the question. The best known formula is that of Major Saunders, U.S.A., but it is only a rough approximation which assumes that the safe load on a timber pile is given by ^ -r WH where L = safe load in cwts. W = weight of ram in cwts. H = height ram falls in inches. D = distance driven by last blow in inches ; in other words, the safe load is taken at one-eighth of the mean resist- ance to penetration of the last blow. The monkey is the little slip-hook that runs up and down to lift and release the ram, although there are many persons who call the ram the monkey. The weight of the ram should be not less than the weight of the pile, and not more than 1^ times that. Theoretically a 2-cwt. ram with 10-ft. fall will have the same effect as a 10-cwt. ram with 2-ft. fall, but the result will be very different. A light ram and long fall will expend all its work on the Itop of the pile and break it up, while a heavy ram with a low fall will SUPPORTING POWER OF PILES 211 do little damage, expending its energy in pushing the pile down. For a full understanding we ought to know the nature of the resistance during the penetration. " Mean resistance " gives a rectangular diagram which is an impossibility. It may in reality be triangular or parabolic ; at any rate, the resistance is small at the commencement of the movement and very great at the finish. For fuller discussion of the point see the author's paper on " Timber Piling in Foimdations and other Works," 2nd edition, 6d. Mr. A. C. Huitzig, M.Inst.C.B., made some useful investigations into the formulae for piles, and in a Paper read before the Liverpool Engineering Society in November, 1886, gave the following formula : _ ^ _ P ^ ~ P 625 where P = extreme supporting power of pile in tons. tj z= " set " of last blow in feet. X = energy of last blows in foot-tons. Let us now test these formulae. An experimental pilewas driven and the following data were obtained : Length of pile 30 feet. Scantling 12^ ins. by 12J ins. at top, 11^ ins. by 11 ins. at bottom. Weight of ram 910 lbs. Fall of last blows 5 ft. Depth driven at last blow | in. Actual load to just cause further sinking 62,500 lbs. By Saunders' formula, safe load By Hurtzig's formula, extreme load ' P 625 y = -^ _ ^, whence P' + 625P2/ = 625a; which is a quadratic equation, but y = 0-03125 and x = 2-03125, then P^ + 19-53125P = 1269-53125, and by adding the square of half the coefiicient of P to both sides, and taking the square root of both sides of the equation, P + 9-765625 = 36-94, or P = 27-17 tons, and allow- ing a factor of safety of 3, the safe load will be -g— = 9-05 tons, which is only slightly different from Saunders' result. Another experimental pile was driven on another occasion and the following data were obtained : — Pitch pine pile 12^ ins. by llj ins. Weight of ram 18 cwt. Fall of ram 8 ft. Depth driven at last blow -^ = 0-5G25 m. Loaded Witb §§-9 tons sunk i in. 212 THE MECHANICS OF BUILDING CONSTRUCTION By Saunders' formula, safe load p WA 2016X96 ,„.„„,, ,„a, P = -5-5- = . ^^ - -. -„- = 43008 lbs. = 19-2 tons. 8d 8 X 0-5625 By Hurtzig's formula, extreme load P^ + 625Py = 625a; 2/ = 0-046875, x = 7-2 P^ + 29-3P = 4500 P + 14-65 = ^4500 + 14-6.5^ .*. P = 54 tons, and with, factor of safety ?i, P = 18 tons, which is not quite so close an agreement as before, but it should be noted how closely the latter formula agrees with the actual test load. EXEECISBS FOE LEOTUEE XXVIH Q. 88. A needle beam and dead shores are required to carry a load estimated at 10 tons. What size timbers should be used ? The dead shores are to be i ft. centre to centre and 13 ft. high. Answer. W = y- toi safe distributed load owts. ; id* = 10 X 20 x 2 X 4=1600 say b = 12, then d = * / ^tct = H'Si therefore say 12 ins. by 12 ins. for beam. ^^ 1^ For dead shores try 9 ins. by 9 ins. ; then by Gordon's formula taking ends as one fixed and one rounded, safe load ^ = m/J\' = 9.H /iavi9.va = 121-5 cwts., VJV ~^ , 2-5 / 13X12 Y q[d) 1 X 250y 9 J while the actual load is 5 x 20 = 100 cwts., so that there appears to be good margin, but the bearing area of post on beam should be tested. The maximum safe load perpendicular to the grain of timber is 250 lbs. per sq. in. and the post 100 X 112 must have a sectional area of at least — ^^ — = 45 sq. ins., while it actually has 9 X 9 = 81 sq. ins., so that it will be quite safe. Q. 89. What depth is required for the foundation of a wall carrying a load of 1,J tons per sq. ft. at underside of concrete on a soil with a natural slope of 30° and weighing 100 lbs. per cubic foot ? »/ 1 - sin eV 1-5 X 2240 Answer. ^=^1 fqr^gfg j = — JqO — X 0-111 =3-729. say 3 ft. 9 ins., apart from any atmospheric considerations which would have to be taken into account. Q. 90. A 14-in. by 14-in. pile is to carry 25 tons with a factor of safety of 3. The ram weighs 30 owt. Prom what height must it fall when the last blow is to drive the pUe J in. ? Answer. By Hurtzig's formula _x _ P ^ ~ P 625 ^, , 0-25 X 25 X 3 , therefore jg = 25-^3 - 625~ ^^^"^"^ ^ = ^O'^S ft.-tons, , 10-56 ft.-tons „ ,, , , '^^^ -^-rn^-i:^^ = 7 ft, fall, 15 ton ram LECTURE XXIX Gin Poles — Derrick Poles — Guy Eopes— Shear Legs— Tripods— Cranes. A GIN pole, or derrick pole, is usually a single stick of timber, round or square, but it may be a rolled steel joist, or a solid or iioUow steel mast, which may be trussed on four sides to stiffen it. It is used as a support for lifting girders and roof trusses by means of a pulley wheel or pulley blocks attached to the top, the rope or chain being led down through a snatch block at the bottom and away to a crab winch worked by hand or steam. The top of the pole Is kept in position by two or more guy ropes made fast at their lower ends. The pole may be upright or made to rake over. in any required direction by slackening the guy ropes on one side and tightening them on the other side. Up to 20 ft. long there are often only two guy ropes, and the pole is kept steady by giving it a rake or inclination away from the guy ropes. The term " derrick " is generally used to signify something that rakes over or lifts at different distances from its foot, while a gin pole might be simply an upright pole carrying a gin- or jenny-wheel. Take now a simple case of a 9-in. by 9-in. fir pole 20 ft. long with a 5-ft. over- hang, find what thrust it will safely bear, and then what load it will carry and what size guy ropes will be needed. All derrick poles and shear legs must be calculated as if with " ends rounded." The formula for the strength of poles give results that differ very considerably ; three of these only will be considered. A/" Gordon's formula W = ■ ' H- 4) where "W = safe load in cwts., A = sectional area in square inches, /= working load in cwts. per square inch = 13 for fir, a = constant = 2I0 for both ends rounded, I = length in inches, d = least width in inches. Then W = ■' X 9 X 13 ^ 1053 ^J053 ^ g^ ^ /20 X 12 y 1 + 0-016 X 711 12-375 214 THE MECHANICS OF BUILDING CONSTRUCTION Reuleaux's formula for timber posts (fir) is W = 2000^ r where h = greatest width, d = least width, / = length, all in inches. Then W = 2000 (.go ^ \^y = 227 cwt. Eitter's formula is Gordon's formula with a diiferent constant, viz. 0'0027 instead of ^. Using this constant the calculation will be ^ 9 X 9 X 13 ^ 1053 ^ 1053 ^ ggg ^^^ 20 X 12f 1+0-0027x711 2-9 l + 0-0027(?^) The last two are so greatly in excess of the strength by the ordinary Gordon formula that we should hesitate to adopt anything like the load indicated, and if we assume a 5-ton load we shall probably be putting as much on the pole as it ought to bear. If we allow 50 per cent, additional stress for the possible surging of the load, the thrust from the a dead load should not be more than ^i = 3'33 tons. Now, assuming the guy ropes to be fixed 10 ft. back from the pole and 20 ft. apart, construct a parallelogram of forces (Fig. 570) with 3'33 tons down the pole, and on scaling off it will be found that the net load will be 2-1 tons. For stress in the guys the allowance of 3'33 tons in the pole must be taken, making the stress in the plane of the guys 1'4 tons. Projecting the real plan of the guys as shown by the dotted lines in elevation, Fig. 570, and plan, Fig. 571, the tension in each is obtained and found to be 0'75 ton. The safe load in cwts. on a good hemp rope fall is l(circumference)', therefore Kcircumference)" = 0'75 x 20, whence /0'75 X 20 X 3 circumference = ^ ^^ = 4-74 ins. If wire ropes be used the safe load in cwts. will be |(circnmference)^ X 6, whence circum- ference = ./Q^ X^Ox 8 = 1-7 in. V 6 X 7 Shear legs consist of two derrick poles and a single guy rope. Assume two poles 35 ft. long, 8 ins. diameter at top and 12 ins. at bottom, the lower ends 10 ft. apart, the overhang at top 8 ft., and the guy rope making an angle of 45 degrees with the ground. Find the stresses in each part with a dead load of 1 ton. The elevation and plan will be as shown in Fig. 572. Let AB be the position of the lower ends and AC, BC the legs laid down on the ground, each 35 ft. long. Bisect AB in D, produce CD to E, making DE equal to the overhang 8 ft. From point D with radius DC strike an arc to cut a vertical from E in point F. Join DF and this will represent the side elevation of the legs. On DC set off point G, making angle EFG = 45°, then join GF to represent the elevation of the guy rope. As the load to be lifted is 1 ton, or allowing for surging, say 1-5 ton, set off a parallelogram STRESSES IX DERRICK POLE AND GUYS 215 from point F as shown, from which it will be seen that a stress of 0'6o ton must be resisted by the guy, requiring a */ ins. circumference hemp rope, or ^/ — 0-65 X 20 X 3 = 4-il 65 X 20 X 8 6X7 1"58 in. cir- cumference wire rope. The thrust of 2 tons in the plane of the legs ng . S70 must be resolved into an equivalent force in each leg. This is done by parallelogram as shown at G, where it is found that the thrust in each leg amounts to 1'03 ton. In addition to this thrust it has a cross strain due to its own weight, which will be appreciable owing to the 216 THE MECHANICS OF BUILDING CONSTRUCTION length and overhang. Taking the weight of fir as 35 lbs. per cubic foot, the weight of one leg will be ^^(A + a+ ^~Aa) = i X gp( Q"785^X 12^+ 0-785^ X «' + Vo-7854 x 12^ X 07854 X 8^ 144 ) = 19"5 cubic ft., and 19-5 X 35 = 682 lbs., say O'Sl ton, which, allowing for the taper, will produce a bending moment of about 0-305 ton-feet found as follows. The height of centre of gravity of pole from ground 2tP 4- D^ 2(-Y 4- 1^ = '* 3#T3l? = '' X 3(|f + 3Xl- = 15-2 f^^*^' 15'2 and the diameter at this point will be 12 — 5^(12 — 8) = 10'26 ins. STKESSES IN SHEAR LEGS 21 The weight of the pole below the centre of gravity will therefore be ; X 15-2/'°l'?^* ^1^ + 0"'^854 X 10-26' + ^0-7854 X 12' + 0-7854 X 10 -26'^_a V ■ 144 " / 5_ 3240 = 0'16 ton, and the height of the centre of gravity of this part = 15-2 X ,,,„/„ = 7 ft. From Fig. 573 it will be seen 3(-^)+3xl- that the amounts of these two loads acting at right angles to the pole = ^ X 0-31 = 0-071 ton and ^ x 0-16 = 0-037 ton. The reaction at 0'071 X (B5 — 15-2^ ground = \,- ^- = 0-04 ton, therefore the bending moment at centre of gravity of pole = 0*04: x 15-2 - 0-037 X (15-2 -7) = 0-305 ton-ft. Then the maximum working strength of one leg taken as a pillar, reckoning from the mean diameter, will be 0-65 ton per sq . in. /35 X 12\ ~ 0-022 ton per sq. in., and the actual stress with 1 + ^H 10 ) the thrust of 1*03 ton and the weight of leg 0-31 ton will be W,M 1-03 + 0-31 0-305x12 ..,„,„„„, ^„.. A + Z = 0-7854x10- 26"^ + 0^0982"x~10^« = 0-016+0-034 = 0-05 ton per sq. in. This shows the poles to be insufficient in strength, although with care they might doubtless be used. Try 15 ins. bottom diameter, and 12 ins. top diameter, but not in quite such detailed calculations. The 0*8 /15^ + 12^\ .35 weight will now be 35 x Yni 5 ) ^ 9540 ~ ^"^^ *'°°' ^""^ ^^^ bending moment will be approximately -k- = 5 =0-56 ton-ft. rn. W M 1-03 + 0-56 0-56 x 12 ^ , , , , „ ^^o ^'^^^ A- +Z- 0-7854 X 13-5- + 0-0982 x 13-5^ = 0-^1^ + ^'^^^ = 0-041 ton per sq. in., whereas the maximum allowable 0-65 /35xl2y i -I- 250\^ 13.5 J = 0-04 ton per sq. in., so that this size is probably just sufficient, although the real overhang will be the diagonal distance from E to B in Fig. 672. Taking the case of a tripod, suppose we have three poles having a vertical height of 20 ft., and the bases placed at the points of an equilateral triangle of 10 ft. side, the load to be lifted being 5 tons without any additional allowance. Fig. 574 will represent the plan and Fig. 575 the elevation projected from it. Draw the parallelogram of forces from the load in the elevation, and the thrust in each pole will be as shown in DC = 1-03 ton. The proof that this will be correct for 218 THE MECHANICS OF BUILDING CONSTRUCTION each of the poles will be to project the full length of the poles AB in plan, as shown by EB, EA. Then from E, along ED, set off the thrust shown in elevation on AB-D, and completing the parallelogram it will be found that the thrust shown on EB, EA, will be equal to that /=>5r. srs A.B /v^. S74- shown on CD in the elevation. Now having a thrust of 1-03 tons to provide for in poles 20 ft. long the problem is to find the scantling. A/" By Gordon's formula W = T7T2- taking the length from DC in the elevation, Try first 8 ins. by 8 ins., then .s X .s X 13 1 + 2D0y 20-75 X 12Y 53-6 cwt. = 2-68 tons STRESSES IX CKANES 210 as the safe load, against 1-03 tons the actual thrust, but this does not allow for the overhang. The weight of each pole = 20-75 X ^ X ^io = 0-14 ton , , , ,. , WL 0-14 X 6 „,„^ , ,, and the bending moment = -^ = ^ = 0-105 ton-tt. Then W . M _ 1-03 + 0-14 , 0-105 X 12 A - Z ~ ' 6i '*' ^8X8') = 0-018 + 0-015 = 0-033 ton per sq. in., and the maximum allowable = 0-0-12 ton per sq. m. 0-65 1 + ^ / 20-75 X 12 Y 250\ 8 / Some very difficult problems occur at times in finding the stresses 220 THE MECHANICS OF BUILDING CONSTKUCTION on cranes, but these varieties would hardly come under the head of building construction. One example of a wall crane to show the nature of the stresses will be taken. Fig. 576 shows an awkward case that may arise among the smaller cranes, where the outline of the crane is shown by thick lines and the force diagrams by dotted lines, and the bearings in the top and bottom brackets are at F and E. The points of application of the thrusts and resistances not being opposite each other, bending moments are caused in addition to the direct stresses, and are shown by the shaded areas. The load of 3 tons at A is equivalent to a load of 3 X -^ = 8-643 tons at B, and taking this latter load the stresses in the members may be found by parallelogram of forces as shown, and the thrust at D and pull at C are also given = 2-6 tons each. The thrust at E and pull at F will therefore be 2-6 X =^= 2'6 X if tons each, i o o producing a bending moment at D of 2'6 X ff X 2-5 = 4-8 ton-ft., and at C, 2-6 X If X 1 = 1-9 ton-ft. The 3 tons load at A produces a bending moment of 3 x 1"5 = 4-5 ton-ft. at B as shown in Fig, 576. In the foregoing calculations it has been assumed that the guy ropes are straight and without weight, but straight guys are impossible, they will always hang in a curve, and it is the tangent to the curve at the head of the poles that fixes the virtual anchorage of the guy rope. EXERCISES ON LECTURE XXIX Q. 91. From the conditions given in Pigs. 5T7 and 578 find the stresses in the pole and ropes, and calculate the sizes required. Answer. The elevation of the pole and ropes is given in Fig. 577, and by parallelogram of forces the thrust in pole is found to be 35 owt. = 1-75 tons, and the tension in plane of ropes 13 owt. By setting off the true plan of ropes in Fig. 578 and constructing the parallelogram of forces the tension in each rope / 7'2 X 3 = 7-2 cwt. Then the size of hemp rope required = /v/ — g — = 3'3 in cir- /7'2 X 8 oumferenoe, or \/ -^-tj-yj — 1"17 in. circumference for wire rope. Assuming a 9-in. by 9-iu. pole, the weight at 35 lbs. per cubic foot will be ^^X25 X j^5B = 0'22ton, and the bending moment will be —5— = 5 = 0'18 ton-ft. The stresses o o produced will then be W M 1-75 + 0-22 0-18 X 12 A + Z ■= 9 X 9 ' + f(r>r9^ = °'°^^ + °'°^^ = °"°^^ °""*- P^"^ ^1' ™-' whereas the maximum allowable stress = °^,^T ^^'^f^ = d'035 ton per sq . in. -, , . /25 X 12 y ^ ^ ^ + ^\~^ ) This shows that the scantling is not quite enough, so a 10-in. by 10-in. pole may be EXAMPLES FOR PRACTICE 221 tried. The weight wiU ho ^^^ X 25 x i^^ = 0-27 ton, producing a hending . , WL 0-27 X 6-5 moment oi 8 ~ 8 W , M 1-75 + 0-27 = 0-22 ton-ft. The stresses produced will be X 12 = 0'02 + 0-016 = 0'036 ton per sq. in., and the A "*" Z ~ 10 X 10 "^ 1(10X10") allowable stress = ,n^ .. ,a ■ „ ■ = 0-042' ton per sq. in., so that this size 1 + 4/2S X 12 \" will be quite sufficient. g. 92. A derrick consisting of two poles 15 ft. apart at the ground, joined together at a height of 25 ft. from the ground and raking over 8 ft., has a single F>cf 578 guy rope made fast at ground level 50 ft. in direct line from the head of derrick. Assuming the sag in the rope to give a versine of 2 ft. 6 ins., draw an elevation and plan and figure the parallelograms of forces at head of derrick for a load of 30 owt. 222 THE MECHANICS OF BUILDING CONSTRUCTION and also for no load, making the size of rope suitable for the fuU load, and show for the given weight and position of rope what puU it will give on head of derrick. Discuss the relative bearing of the three results. Answer. The elevation of the poles and rope will be as in Pig. 579 and the plan as Pig. 580. Draw ab 'tangent to the curve of rope at head of poles and ed tangent to curve at ground. Taking ab as the line of the rope, draw the parallelogram of for ces for th e full load, giving 18-5 cwt. tension in the rope, requiring a rope . / o w it = 1'87 in. circumference. The 44 cwt. compression in the plane of the legs must be divided over the two legs as shown in Pig. 580i /v'jr sso giving 23 owt. = ]'15 ton compression in each leg. Assume a pole 13 ins. diameter at base and 11 ins. at top, then the weight wiU be approximately „„„„ 0-8/13'i + 112\ jljj = 0'34 ton, and the bending moment WL 0-34 X 11 8 Then W M 1-15 + 0-34 A + Z 0-7854 X 122 + whereas the maximum allowable = 0-47 ton-ft. 0'47 X 12 0-0982 X 12' = °'°^^ + °'°^* = °"°*'^ *°° P^' ^^- ™-' 0-65 1 + 27-25 X 12\'' = °*°^ '°'^ P®"^ ^1- '"• , /27-25 X 12\' '^\ 12 ) Half the weight of the poles may be set ofi as load in Pig. 579, and by parallelogram of forces it is found to produce 38 cwt. tension in the rope. The steel rope will weigh = ^y= -■ ■■ 0-234 owt., which by parallelogram EXAMPLES FOR PRACTICE 22?, of forces at d in Fig. 579 produces a tension of 0-575 cwt. in the rope. It is evident that with a derrick of this kind the angle of the legs and the sag of the rope will vary with the load. With the sag shown the rope only gives a puU of 0'575 cwt., while without a load the weight of legs alone gives a pull of 38 ovrt., so that the rope would yield and show a smaller versin. Still more will the full load tighten the rope and reduce the versin. In every case the plane of the legs will make such an angle as to cause the pull of the rope and load to balance each other. LECTUKE XXX Reinforced Concrete — Beams — Floor Slabs — Tee Beams — Pillars — Formulae and Calculations. The essential feature of reinforced concrete is that the cement concrete is strengthened by steel rods in snch a manner that practically the whole of the compressive stress will be taken by the concrete, and the tensile stress by the steel. The shear is taken partly by the concrete and partly by the steel. The ordinary bending moment and shearing force diagrams for beams and girders apply equally to reinforced concrete construction, but owing to a doubt in some cases of the efficiency of the design and execution of the work certain allowances are made. For example, a beam with fixed ends and uniformly distributed load has a bending moment in the centre of -— , but to allow for possible defect in fixing the ends of a reinforced concrete beam it is taken as jr-. Also when fixed one end and supported at the other the negative bending moment at support is taken as -^ both for ordinary and reinforced beams, but the maximum positive 9 bending moment at %l from supported end is taken as .r^ wPfoi ordinary beams and j^l^ for reinforced concrete beams. In calculating the bend- ing moment the effective span should be taken, as with ordinary girders, and this should not exceed twenty-four times the effective depth. For floor slabs supported direct by the walls, the effective span will be the clear span plus the thickness of slab. When supported by intermediate beams the effective span of the slab will be from centre to centre of the beams. The bending moment across the centre of a square slab supported on four edges and reinforced in both directions, with a load uniformly distributed, may be taken as tttt ; and with the edges fixed the bending moment will be taken as — in the centre, and -^ at the edges. Other modifications will occur for other methods of fixing or difference of length and breadth, but the above will be sufficient to indicate the general result. The modulus of elasticity for 1 : 2 : 4 cement concrete (i.e. 1 part British Standard Portland cement, 90 lbs. to a cubic foot, 2 parts clean EEINFORCED CONCRETE DATA ■i-lb sharp sand, 4 parts broken stone or gravel, graded from \ in. to | in.) averages one-fifteenth of that of mild steel or , lbs. per sq. in. For concrete E = 2,000,000 „ mild steel E = 30,000,000 F Modular ratio m = =^ = 15 E, The following are the allowable working stresses : — lbs. per sq. in. Concrete in compression in beams 600 „ „ columns under simple compression . 500 „ „ shear in beams 60 Adhesion of concrete to metal 100 Steel in tension 16,000 „ „ shear 12,000 The grip length of a reinforcing bar, round or square, to prevent pulling out, is 2500f(^, where t = tensile stress in tons and d = diameter in inches. The reinforcement should not be nearer to the face than I in. in slabs, 1 in. in cross beams, \\ in. in main beams, and 2 ins. in pillars. There should be at least 1 in. between reinforcing bars horizontally and | in. vertically. The formula for reinforced concrete work are rather complicated, and in general require the neutral axis to be first determined. When the subject was first before the public the writer used an approximate empirical formula which has saved much time in dealing with simple beams and plain floor slabs. It is W = (0-37p + 0-214) liV- "'g sgi 4- where W = safe load in cwts. distributed including weight of beam itself, 2) = percentage of reinforcement (O to 2|) excluding 1^ in. of concrete below reinforcement, 6 = breadth of concrete in inches, d = depth of concrete in inches to centre of rein- forcement, L = clear span in feet. Nowadays more precise calculations have to be made, and f uU allowance given for the eifect of the floor slab in converting the beams into tee beams. The best method of explaining the work will be to take the case of a floor as in Fig. 581 to carry an external load of 1 cwt. per ft. sup. The calculations may be based upon the following formulae, where y = distance from neutral axis to extreme compressed fibre, y" = distance from neutral axis to extreme fibre in tension, m = modular ratio = 15, A = area of tensile reinforcement in square inches, Q -+- *-" -, /ff'O'- CP.OSS BE< -/fW - » J_ 226 THE MECHANICS OF BUILDINa CONSTRUCTION h = breadth in inches of tee beam, d = depth in inches from upper surface to centre of reinforcement, e = total depth of floor slab, I = moment of inertia in inch units, c = compression on the concrete in lbs. per sq. in., t = tension in the steel in lbs. per sq. in., M = bending moment in lb. -inches. Then for floor slabs taking 1 ft. width, tj may be obtained from the formula (/y"- + mkty — mA,d = 0, I = i?/ + mA,(d — yf, , = ^and^ = '"^. For tee beams when the neutral axis falls within the depth of slab, hf + 2«A,«/ - 2mA,d = 0, l = -f + niA,(d - yf, c and t as before. ^p rvin/brce/nr^r aiternalx bars -/ 3 s i^ 3' centres Fig SBZ J- / 'rods ~ ^^ ^j ^/^' K«■*rf^ "^ CROSS BEAM ' ? - - ■ < -. : . ^ . L .. ,.^4.^ .^ ■> -^iCernctce r-ocfy Aenc ci/o or f'j' frtyn centre of These formulaj may also be used for rectangular beams, taking 6 as breadth of beam. For tee beams when the neutral axis falls outside the depth of slab. _ "imkid + 'bf_ y ~ 2{mkt + bey , be' - yy- . -j- tnAt(d - t/f. It will be on the safe side and not cause any appreciable waste of strength if the whole of the slab panels and the beams be assumed to be simply supported at the ends, giving for distributed load a bending moment of ^wP. Taking first the floor slab, assuming the sizes shown in Fig. 582, the load per ft. run = 112 + (jf X 150^) = 180 lbs. and M = -o- = W (8 X 12)= = 17280 Ib.-ins. then ijy- + mAiy — niAtd = or V,y- + 15 X 0-262/ = 15 X 0-26 X 4^ or y= + 0-65«/ = 2-92 o.,. + 0-65, + (<^7=2-02 + (^7 or 2/ + 0-325 = r7?.5 whence y = 1-41 REINFOKCED CONCRETE FLOOR SLABS 227 I = -if + mAt(d — yy = 4 X 1'4P + 15 X 0-26(4-5 - l-ilj = 4:8'44 inch-units My 17280 X 1-41 _ ,^ ^ = -]- = jgq^ = 503 lbs. per sq. in. , mMy' 15 X 17280(4-5 - 1-41) ,„.,„,^ t = — J— = ■„.. ■ = lGo40 lbs. per sq. in. so tliat this size will just do.- To resist what are known as temperature cracks light rods are laid at right angles to the main reinforcement of the slab, say, |-in. rods at 8-in. centres. For the cross beams, as shown L s' o' ^o otJffff/T rods bene up ■i J '■. , ^=^ -,: — y r' / - - i' 10'- *f* -/'l'- n* -/'/'- *H- /'o' M*- /o' *\^- s' *^4U^ CROSS BEAM -,, in Figs. 582 and 583, the width of floor slab to be taken into calcula- tions is 12 times the thickness = 12 X 5"5 = 5 ft. 6 ins. Weight per ft. run = f ^^^.^^ + ^^^150 = 429 lbs Note. — The figures 66 given in the bracket after weight per foot run at middle of page should be 96, a consequent error runs through the remainder of the chapter. 0. ,3 + 1.06, +(t06J^ 12-72 + (^1-7 oxy + 0-53 = 3-6 whence y = 3-07 I = *f ' -I- mkt{d - yf 66 X 3-07' + 15 X 2-35(12 - 3-07)' ;; = = 3448 inch-units _ M?/ _ 401280 X ~ I~ ~ mM.?/ i-07 = 357 lbs. per sq. in. = 15590 lbs. per sq. in. 3448 15 X 401280 X (12 - 3-07) 3448 228 THE MECHANICS OF BUILDING CONSTRUCTION The shear reinforcement may be calculated by the formula M 6-5 X 12 , ^„^ . ''^ = 48 = ^ = ^'^^^ ^^- ^°- say 8 stirrups in half span = — ^ — = 0'203 sq. in. each, or 6 stirrup bars = — ^ — = 0'034 sq. in. each, therefore say i-in. diameter rods. For the main beam, as shown in Figs. 583 and 584, the total concentrated load from each cross beam = 16 X 1045 = 16720 lbs., producing a bending moment in the centre of (25080 X 16) - (16720 X 8) = 267520 Ib.-ft., or 3210240 Ib.-inches. 21 X 4-8 The weight of beam = —^rr~ X 150 x 32 = 33600 lbs., producing T, A- . , 33600 x (32 X 12) ,^,,-,o,.a,u • i- a bendmg moment of ^^ = 1612800 Ib.-ms., makmg a 8 total of 3210240 + 1612800 = 4823040 Ib.-ins., as in Fig. 585. Then by^ + 2mKty = •2mA.td or nf + 2 X 15 X 12-3y = 2 X 15 X 12-3 X 45 OTf+ 17-62/ = 792 • ory^+17%+(^;=702 + (l^^:; or tj+ 8-8 = 29-5 whence y = 20"7 1= f + mHd - yy ■>1 V 9()-7^ = -^ Y ^ +15xl2'3(24-3)-^ = 171000 inch-units My 4823040 X 20-7 ,^, ,, " " X "^ 171000 "^ ^^ P^^' ^^- '°- , niMy' 15 X 4823040 X 24-3 , ,^„o,, ,, t = —JL = _— __ = 10280 lbs. per sq. in. The shear reinforcement as =-t^= — j^ — = 20 sq. ins., say 16 rows of stirrups = f| = 1"25 sq. in. in each row, 1'25 10 stirrup bars = -r^ = 0'125 sq. in. area each, say ^ in. square bars. The points at which the rods may be bent up can be ascertained as follows. For instance, take the main beam, the bending moment diagram must be drawn out as in Fig. 585, then the upper middle rod may be turned up at the points where the bending moment is one-tenth less, as at ab, or 12 ft. from either end. The two upper inner rods may be turned up at the points where the bending moment is Yo less, .is at cd, or 7 ft. 6 ins. from either end. The two upper outer rods may be turned up where the bending moment is reduced to ^, or half the total, as at ef, or 5 ft. from either end. REINFORCED CONCRETE BEAMS 229 230 THE MECHANICS OP BUILDING CONSTRUCTION As a rough check upon the section adopted it may be useful to note that approximately the bending moment in cwt.-ins. on the simple reinforced concrete beam = hf. In this case M = iSfaMo _ ^gQg3 cwt.-ins., as already found, and M^ = 21 x 45'^ = 42525, showing a very close agreement. In calculating the loads on columns in buildings with a height of more than two stories, with the exception of warehouses, the super- imposed loads from roof and top story must be calculated in full, but for each story below a reduction of 10 per cent, of the load may be made up to a maximum of 50 per cent. The maximum permissible stress on the concrete in columns under simple compression should not exceed 500 lbs. per sq. in., and in the steel 12,000 lbs. per sq. in. The vertical rods in pillars should not be placed nearer than 1 J in. to the face of the column, or 2 ins. in important cases, and if hooping is employed there must be four lines of vertical reinforcement with straight laterals and six lines with curved laterals. The straight laterals must not be less than ^ in., and curved laterals \ in. in diameter, and the pitch must not exceed ^ of the effective diameter of the pillars, that is, the diameter to the outside of the ver- In rectangular pillars, where the ratio of the or square pillars with more F,g SSS n. tical reinforcement greater and lesser diameters exceeds than four vertical rods,' the cross-section of the pillar must be sub divided by cross ties, "and the distance between the rods on the longer side must not exceed that between the rods on the shorter side. The total area of the vertical reinforcement in pillars must not be less than 0'8 per cent, of the hooped core, and the volume of lateral reinforcement must not be less than 0-5 per cent, of the volume of the hooped core. When the column has fixed ends, the following stresses may be allowed according to the ratio of length to effective diameter : — I d Stress 18 22 0-8 of 24 0-6 „ 27 0-4 „ SO 0-2 „ Full stress If the pillar has one end fixed and one hinged allow J-P, both ends hinged ^P, and one end fixed and one free ^P. The stress on the steel REINFORCED CONCRETE PILLARS 231 mW ^aud on the concrete may be worked by the formula i! = ~. , -» / n i A + A„(m - 1) by the formula c = r -- ^ where t = the stress in the steel A + A„(m — 1) in lbs. per sq. in., c = the stress on the concrete in lbs. per sq. in., W = total axial load on the pillar in lbs., A = effective area of pillar in square inches, i.e. the area bounded by the lateral reinforcement, A„ = area of vertical reinforcement in square inches, m = modular ratio = 1.5. The stress thus found should not exceed the safe stress given by the formula kC(l +fsr) where C = the ultimate crushing resistance of concrete at 3 months, S = the safety factor = say 4, Ic = the reciprocal of the safety factor = ^,/= form factor, depending upon the form or type of laterals (see table), s = spacing factor, depend- ing upon the spacing or pitch of the laterals (see table), r = ratio of the volume of lateral reinforcement to the volume of the hooped core in any given length of pillar. Form of lateral „ *,^_ j, reinforcement. F"'™ ^^"^^ =/• •Spacing of laterals in terms of diameter of hooped core. Spacing factf)r = s. Diameter. Helical . . 1-0 0'2 32 )» 1-0 0-3 24 )) • • . ] 1-0 0'4 16 Circular hoops . i 0-75 0-2 32 Ji ii . I 0'75 0-3 24 t) )> . i 0-75 0-4 16 Rectilinear . 0-5 0-2 32 )» . 1 0-5 0-3 24 . 1 0-5 0-4 , 16 )» 0-5 0-5 1 8 . : 0-5 0-6 1 When the pillar is eccentrically loaded the maximum stress will be W Wx given by the formula /= -r- + -^, where /= maximum stress in lbs. per sq. in. at edge of section, W = total load from all sources in lbs. A = equivalent area in square inches = A + (m — 1)A„ W' = eccentric, load in lbs., x = distance in inches from neutral axis of pillar to centre of application, Z =» section modulus of pillar in inch units, and the stress so found should not exceed the allowable stress given by the previous formula. h^ Tor a rectangular reinforced concrete pillar, Z=iAcA+^(m-l)A -,p where A is the whole diameter of the pillar at right angles to the neutral axis, and A, is the diameter from centre to centre of reinforcement. For a circular pillar reinforced with four bars, Z = iAcA + i(M - 1)A; Jh' 232 THE MECHANICS OF BUILDING CONSTEUCTION and for a circular pillar reinforced with bars arranged in a circle Z = lAcA + i(m - 1)A„|' The following limits of stress should be observed in pillars : — (1) The stress on the steel should not exceed one-fourth of the ultimate tensile strength, or O'o of the elastic limit, or the value of me. (2) The working stress on the concrete of pillars must not exceed 0-5O with rectilinear laterals, 0'58C with independent circular hoops, and 0'66C with helical reinforcement. EXEEOISES ON LECTURE XXX Q. 93. Calculate and design the cross section and elevation of one of the main beams of the given floor if supported by a reinforced concrete pillar in the centre. Also calculate and give a cross section of the pillar assuming the height to be 14 ft. from floor to floor. Answer. Assuming the general dimensions of the beam to be as shown in Pigs. 586 and 587, io V 9fi the weight of beam = — -ij=^ x 150 X 16 = 5200 lbs. 144 and M = 5200 X (16 X_12) ^ ^^^^ ^^ _.^^_ O ivTf 1 J Tiin 3X16720X16X12 „niQOn ly, ir,= M from load = f^WJ = r^ = 601920 Ib.-ins. making a total of 726,720 Ib.-ins. Then 12y' + 2 x 15 X 2-4 X 2/ = 2 X 15 X 2-4 X 24 or 3/' + 6j/ = 144 or y'' + &y+ (i)» = 144 + %' or 2/ + 3 = 12-4 whence y = 9'4 19 V Q'43 I = -^^ ^ "^ + 15 X 2'4(14-6)« = 10998 e = !?«^^^ = 6201bs.persq.in. The points at which to turn the rods up and to provide for reverse stresses may be found from the bending moment diagrams, Figs. 588 and 589. As the loads in this case are concentrated the shear stirrups should he equally spaced, the shear at left-hand abutment will be ^ X 16720 + i X 5200 = 7175 lbs., and at 5 tons per sq. in. this requires g ''^H^q = 0-64 sq. in. per ft. run, say stirrups 8 in. centres, then area of each = jj — g = 0-05 sq. in., say J-in. rods. At the column the shear will be {J x 16720 + 1 X 5200 = 14745 lbs., and at 5 tons per sq. in., requires _ ^ „_ ., = 1-32 sq. in. per ft. run, say stirrups 4-in. centres, then area 1'32 of each = ^^ 5 = 0'05 sq. in., say J-in. rods. The load on pillar wiU be | of total distributed load on both spans and jj of central concentrated loads, to which must be added the direct load over the pillar = 1(2 X 5200) + |J(2 X 16720) + 16720 = 46210 lbs. Assume the pillar to EXAMPLES FOR PRACTICE 233 234 THE MECHANICS OF BUILDING CONSTRUCTION bel2_ins. square with four J-in. vertical rods, bound together spirally at intervals of i ins. with ^-in. diameter wires, as in Fig. 590. The ratio of length to effective diameter = ^ - _ ' = 15-8, so that the full stress may be allowed on the pillar. Then stress in concrete c = W 46210 A+A„(ot-1) "81+1-76(15- 1)' : 440 Ibp. per sq. in. Then taking ftC = 500 lbs. per sq. in., the maximum allowable = 7i;c(l + /sr) = 500(1 H- 1 X 16 X 0-003) = 525 lbs. per sq. iu. wiW 15 X 46910 The stress in the steel = £:fj^^;^^~i) = 81+1-76(15-^) "= ^^'^ '^^- P®' '^^^ "■ EXAMPLES FOR PRACTICE 235 LABORATORY WORK. The laboratory work accompanying this course of instruction consisted of testing rectangular beams of white pine to breaking point by transverse stress, noting the deflections under given loads, plotting the results and calculating the numerical coef&oients in the formulse ; also making up and testing Portland cement pats and briquettes, neat and in various combinations, and at different ages. A visit was made to a large building in course of construction, and various matters of interest therein were pointed out. INDEX A. C. Abutments, stability of, 151, 171, 173, 175. Acoeleratrix of gravity, 12. Action and reaction, 2. Algebra and arithmetic contrasted, 5. Analysis of pressure on retaining vpall, 143. Angle of repose, 140. Anti-clockwise, 7. Arched girder, 48. ribs, 182. Arches, estimation of loads upon, 169. hinged, 177. line of thrust in, 165. linear, 48. semicircular, 165, 171, 175. stability of, 162. theory of, 177. Attraction of gravity, 12. B. Battering walls, 146. Beam, stresses in, 32, 33. Beams, continuous, 43, 62. reinforced concrete, 226. trussed, 70, 73. Beams with fixed ends, 42. Bending moment, definition of, 15. and shear diagrams, 27. for standard loading, 31. in roof trusses, 85, 94, 96. on stanchion, 78. Bent girders, 47, 48, 63, 182. rib trusses, 116, 118, 123, 130. Bow's notation, 28. Braced arch roof trusses, 126. cantilevers, 56, 58. girders, 60. Brackets, to find strength of, 56. Brewery roof truss, 112. Brick piers, 80. Bridge abutments, 151. Brittleness, illustration of, 1, Buckling, definition of, 20. Built-up stanchions, 79. Buttressed wall, 136. Buttresses, stability of, 152, 155. Camber, definition of, 21. Canopy for hotel entrance, 52. CantUever, cast iron, 49. stresses in, 31, 49, 58, 59. Cantilevers, gallery, 49. Cast-iron cantilever, 49. columns, 80. Ceiling vaults, 190. Cement, strength of, 210. Centre of gravity, 12. of an^e bars, 16. channel bar, 16. Centroid, 13. Circular curves, elements of, 178. Clockwise, 6. Closing force, 28. Coefficient of transverse strength, 18. Collapse of floor, 20. Collar beam truss, 85. Columns, cast-iron, 80. stone, 80. Component forces, 4. Compound girders, 39. Compression, definition of, 20. strength of fir in, 75. Concrete dam, 161. under footings, 208. Conditions of matter, 1. Continuous beams, 43, 62. Couples, 6. Coverplates to joints, 41. Crane, stresses in, 219. Culvert, loads on a, 170. Curve of thrust in arches, 165. Curved struts, 178. Curves, formulae for, 178. D. Deflection, coefficients of, 21. defiinition of, 21. formula for, 21. Derrick poles, 213, 220. Detrusion, definition of, 20. Diagrams for beams, 32, 33. with fixed ends, 42, 48. bent girders, 47. 2-68 INDEX Diagrams for cantilever, 31. continuous teams, 43. fixed beams, 42, 43. Distribution of pressure on base of wall, 184. Divisibility, illustration of, 1. Domes, theory of, 196. stresses in, 196. thrust in, 196. Door, stresses on hinges of, 57. Ductility, illustration of, 1. E. Earth pressure, 139. slip, shape of, 140. Eccentric loads, 78, 134, 231. Elasticity, illustration of, 1. Elastic limit, definition of, 20. Empirical formula, 19. Bquilibrant, 3. Equilibrium of forces, 3. Extension, definition of, 1. Extreme fibre stress, 18, 54. P. Factor of safety for rolled joists, 39. Fcrro-conorete, 224. Figure, definition of, 1. Fir beam, strength of, 19. Fixed ends, 39. Flexibility, illustration of, 1 . Floors, reinforced concrete, 225. Flying buttresses, 153. shores, scantlings for, 208. Footings to walls, 208. Force, definition of, 2. direction of a, 2. magnitude of a, 2. sense of a, 2. Force polygon, 4. Forces, equilibrium of, 3. graphic representation of, 2. parallelogram of, 2. polygon of, 4. specification of, 2. triangle of, 3. Forces in equilibrium, 2. Form, definition of, 1. Foundations, safe load on, 13d, 209. Fulcrum of lever, 5. Funicular polygon, 24. G. Gallery cantilevers, 49, 53. stress diagrams for, 50. Gate, stresses in a, 54. Gin poles, strength of, 213. Girder plates, length of, 41. Girders, bent, 47. braced, 60. compound, 39. continuous, 62, 65. formula for weight of, 41. lattice, 60. Linville, 66. plate, 41. trellis, 67, 68, 71. triangulated, 60. trussed, 66. Warren, 60. Whipple-Murphy, 66, 69. Gordon's formula, 75, 76. Gothic vaulting, I90. Graphic statics, definition of, 23. Gravity, centre of, 12. force of, 12. Groined vaults, 190. Gusset plates in roof, 118. Guy ropes, 214. H. Hammer-beam truss, scantlings for, 101. trusses, 99. Hardness, illustration of, 1. Hinged arches, 177. Hooke's modulus, 21. Hooks and hangers, painters', 51. Hoop tension in domes, 199, 203. Horizontal shear, 26. Howe truss, 66. I. Impenetrability, definition of, 1. Inclined joists, strength of, 57. Indiarubber beam, 13. Inertia, moment of, 13. Inertia area, 36, 37. construction of, 37. Island platform roof truss, 98. Joists, strength of fir, 19. L. Laboratory work, 235. Lantern light, 92 Lattice girders, 60. bent, 63. Leverage moments, 5. Leverage, orders of, 5. Levers, 5. Lierne rib vaulting, 193. Limit of elasticity, definition of, 20. INDEX i-d'J Lino of thrust in arches, 163. domes, 198. Linear arches, 48, 128, 187. Link polygon, 23. Linville girders, 66. Load, definition of, 20. M. Magnitude of a force, 2. Malleability, illustration of, 1. Masonry dam, 159. Mass, definition of, 12. Matter, conditions of, 1. definition of, 1. properties of, 1. Mean centre of gravity, 12, Mechanical elements of rolled joists, 36. Method of sections, 83. Moduli of elasticity for various ma- terials, 21. Modulus, definition of, 14. Modulus of elasticity, 20. rupture, 18. for verought iron, 52. section, 14, 18. Moment, definition of, 5. Moment of flexure, 14. resistance, 18. inertia, 13, 87, 55, Bankinc's rules for, 55. to find, 37. Mortar, strength of, 210, Motion, laws of, 2. N. N girders, 66. Natural slope, 140, 20'J. Needles in shoring, 205. Neutral axis, 13. layer, 13, Newton's Laws of Motion, 2, 0, Obelisk, stability of, 154. Ogee arch, 169. Open timber truss, 99, 101,106, 113. Painters' hangers, 51. Parabola, construction of, 28, Parallel forces, 5, 24. Parallelogram of forces, 2. Permanent set, definition of, 20, Piers in walls, effect of, 136. Pile driving, 210. Piles, formula; for, 211. Piles, supporting power of, 211. Pillars in reinforced concrete, 230. Plate girders, 41. Poles, strength of, 213. Polygon of forces, 4. Posts, strength of wood, 75, 81 Power, definition of, 5. Pratt ti-uss, 66. Pressure on foundations, 134. Pressures, 2. Principle of moments, 82. Prof. Crofton's theory, 135, Queen post truss, 91, 90. K. Radial thrust in domes, 199, 203. liadius of gyration, 77, Raking shores, 205. formula: for, 207, scantlings for, 207, Rankine-Gordon formula, 77. Bational formula, 19. Reaction, 2. Reciprocal diagrams, 23. Reinforced concrete, 224. beams, 226. pillars, 230. Reservoir walls, 156. Resistance moment, 18, Resultant, 2. Retaining walls, 139, 141. loads at back of, 145. surcharged, 144. Rolled joist, deflection of, 38. section modulus of, 36. strength of, 38. used as stanchion, 77. Rolling load on arch, 172. Roof truss, exceptional forms of, 107. stresses in, vary with fixing, 84. Ropes, strength of, 214. Rupture, modulus of, 18. S, Safe load on brickwork, 134, floors, 44, foundations, 134, materials, 44, 134, 210, reinforced concrete, 225, Section modulus, 14, 18, 37, Sense of a force, 2, Shear diagram, 26, in webs, 38, legs, 214, 222, stirrups in reinforced concrete, 229. Shearing force, nature of, 26, Mi) INDEX Shoring, principles of, 204. Soils, natural slope of, 140, 209. weight of, 209. Stability of arches, 162. walls, 133. Staircase girders, 47. Stanchions, 77. Statics, definition of, 2. Stifieners, 41. Stone columns, 80. piers, 80. pinnacle, stability of, 148. porch, stability of, 149. steps, strength of, 55. Strain, definition of, 20. Strength, equations of, 18. measure of, 20. Strength of fir beam, 19. rolled joist, 38. stone steps, 55. structure compared with model, 44. Stress, definition of, 20. Stress-strain modulus, 20. Stresses, to find nature of, 25. Stresses in a frame, 25. Strut, definition of, 75. Struts, curved, 178. shape of, 76. strength of, 75, 81. Substituted members, 95, 101, 105, 106, 114. Surcharged retaining wall, 144. T. Tank walls, 160. Tee, section modulus of, 182. Tee beams, reinforced concrete, 226. Tenacity, illustration of, 1. Tension, definition of, 20. Theorem of three moments, 43, 63. Theory of arches, 177. domes, 196. earth pressure, 139. Thrust of arches, 150. in domes, 196. Timber, strength of, 19. in compression, 75. Torsion, definition of, 20. Toughness, illustration of, 1. Transverse stress, definition of, 20. strength, coefficient of, 18. Trellis girder, 67, 68, 71. Triangle of forces, 3. Triangulated girders, 60. Trussed beams, 70, 73. girders, 66. Vaulting, 190. Vectors, 23. Vertical shear, 26. Virtual arch in lintel, 164. Virtual forces, 86, 91, 111. Voussoirs, pressures on, 162. W. Wall hook, 205. Wall-piece, 205. Warren girders, 60. Water pressure on walls, 144, 156. Weight of soils, 209. Whipple-Murphy girders, 66, 69. Wind pressure, allowance for, 83. formula for, 83. Wrought iron, modulus of rupture of, 52. Y. Young's modulus, 20. THE EOT) PUINTKD By WILLIASI CLOWES AND SOWS, LIUITUD, LONDON AND BKCCLES.