OSfi j ' R- * t Plf 1:7 1893 1 " Accessions No. S'O ^ Z, C/zss No. THE BEAM THE BEAM OR TECHNICAL ELEMENTS OF GIRDER CONSTRUCTION BY WILLIAM LEWIS BAKER, A.M.I.C.E. LONDON: CHAPMAN & HALL, L D 1892 [All rights reserved] ^ WESTMINSTER : PRINTED BY NICHOLS AJfD SONH, 26, PARLIAMENT STREET. PREFACE. THE subject of the following pages is treated from the foundation, with the object of making it useful to the student who may possess but little previous acquaintance with the subject itself, or with mathematics, or the theory of mechanics. No complicated algebraic formulae have been found necessary for the determination of any of the questions which arise, either in respect to the amount of stresses or strains, or to the amounts necessary to provide for them. On the other hand such elementary formulae as are required for the ordinary routine of constructive work are fully explained. It has further been attempted, by a series of extremely simple examples, to illustrate the methods which may be followed in the solution of all such cases as will be found to occur in practice, and at the same time to provide easily accessible and reliable data for reference, and for the basis of calculations. CONTENTS PAGE CHAPTER I. DEFINITIONS 1 ELEMENTARY PRINCIPLES 4 CHAPTER II. DETERMINATION OF STRESSES ...... 18 CANTILEVERS 45 CHAPTER III. BEAMS WITH CANTILEVER ENDS . . . . . .48 CHAPTER IV. CONTINUOUS BEAMS ........ 58 CHAPTER V. STRENGTH OF MATERIALS ..... WEIGHT OF MATERIALS viii CONTENTS. PAGE CHAPTER VI. SOLID BEAMS ..... .81 STRENGTH ELEMENTS IN SOLID BEAMS . . .87 CHAPTER VII. GIRDER OF THREE MEMBERS . .97 CHAPTER VIII. CAST-IRON GIRDERS .... ... 107 ,* CHAPTER IX. PLATE GIRDERS . . ... . 117 CHAPTER X. FRAMED AND LATTICE GIRDERS .... . 155 CHAPTER XI. ELASTICITY AND DEFLECTION. 203 THE BEAM. CHAPTER I. DEFINITIONS. 1 . A Tie is a bar or piece of any suitable material attacliecl at its extremities to opposite parts of a structure, in order by its tensile strength to resist any tendency such parts may have to separate. 2. A Strut is a piece, or combination of pieces, so placed that by its compressive strength it may transfer a thrust from one part of a structure to some other part, or beyond it. 3. A Beam is a single piece or structure of any material or combination of materials extending longitudinally over space to one or more points of support or resistance, for the purpose of carrying a load or pressure distributed over the whole or over a part of its length, or applied at any point or points in its length. The Deck Beams of a Vessel, the Beam of a Steam Engine, a Scale Beam, or a Weaver's Beam are examples.* * The well-known terms Tic-beam and Collar-learn are not consistent with ths above definition, and, as the office of the former is to resist tension, and the latter to resist compression, each in the direction of its length, it is submitted that Tic-la j' and Collar-bar would be preferable expressions. B THE BEAM. 4. A Girder is the name originally and still given to the main wooden beams upon which flooring joists are laid, but the term is more especially applied to any beam employed in construction for the purpose of transferring horizontally the pressure or weight of a load from one or more points or surfaces above the space which it crosses to other points upon which it is supported. The principal varieties of girders may be technically divided into the eight following classes : Top Table Bottom Table Fia. 1. (1.) The Girder of three members, con- sisting of a top and a bottom member or Table united by an intermediate vertical member called the Web. When of cast iron the upper table is usually small, and the transverse section of the girder resembles an inverted T (Fig. 1), but when the material used is either rolled steel or iron, then the transverse section resembles the letter H> placed sideways (Fig. 2). (2.) The Box Girder, usually constructed of rolled steel or iron, is so named from having two webs placed apart from each other, which, together with the tables, en- close a space. (3.) The Tubular Girder is a large structure of the Box Girder type, for railway or other bridges, and so constructed that a rolling or other load may pass through the enclosed space or rectangular tube formed by the tables and webs. (4.) The Framed Girder, in which the web is formed of a simple framework of struts and ties. (5.) The Lattice Girder, in which the web is formed of DEFINITIONS. 3 four or more series of lattice bars crossing each other diagonally, and secured by rivets or bolts at their ex- tremities to the tables, and also laterally to each other at their crossings. (6.) The Bowstring Girder, in which the upper table is arched upwards like a bow, and the lower resembles the string in tension. (7.) The Flitch Girder., usually formed of two planks of timber with an intermediate steel or iron plate, all of which are placed laterally against each other and bolted together. (8.) The Continuous Girder, extending in one piece or con- tinuous length over two or more spans. An ordinary length of railway rail extending over several chairs is essentially a continuous girder. 5. A Cantilever -is a beam or bracket projecting from a wall or other supporfcpsmtf" designed to carry a load from or by one support only, and to transfer the weight of the load to ^^r- such support. Any projecting^joortion of constructive work does duty as a cantilever in supporting its own weight, and, as the case may be, frequently a load in addition as for instance a projecting cornice, a batjjoiry, a landing, or stone steps projecting from a wall and otherwise unsupported. The " hammer beams " of medigevaT roofs are essentially cantilevers. 6. A Truss is an open triangulated framework of struts and ties, and a top and bottom member, riveted, bolted, or tenoned and morticed together, as the material used or the nature of the work may require, with a bearing at each end, for the purpose of carrying a load or resisting external weight or pressure. B2 4 THE BEAM. Trusses are adopted when considerable depth is necessary, and the form of an ordinary girder would consequently be unsuitable, as in the case of roof principals ; or when lateral rigidity, of which they have little, is either of minor import- ance or can be imparted from without as for instance by the purlins* of a roof, or by the boarding of a roof or floor. To Truss is to impart additional strength or rigidity to a beam or structure, or to any part of it, by means of one or more struts and ties so united as to constitute a truss. ELEMENTARY PRINCIPLES. ?. The Office Of the Web of a girder is to transmit the vertical pressure or weight of the load to the supports, or from one of the tables to the other ; in so doing the web will resist all vertical and diagonal stresses induced by the load. 8. The Office Of the Tables is to resist all hori- zontal stresses occasioned by the load. 9. A Stress is a force set up within the substance of a structure by the action of its own weight or of an external load or pressure. All stresses are accompanied by corresponding disturbances of material or strains. A Strain is a fractional elongation or contraction or other displacement of the particles of a body, which may either * Purlins are the longitudinal members of a roof which support the rafters at intermediate points between their feet and the ridge. Purlins arc themselves frequently constructed as trusses. Framed and lattice girders are in fact, trussed beams. ELEMENTARY PRINCIPLES. 5 cease with the cause,, or may become permanent, wholly or in part, in which case the material is said to have taken a set. Or strain may end in rupture or collapse. Thus it should not be forgotten in constructive work that the particles of all materials exist in a chronic state of more or less continual strain and consequent relative clis* integration. There are five conditions of strain in which a solid may be placed: 1 tensile, 2 compressive, 3 shearing, 4 transverse or bending, and 5 torsional or twisting. 10. Transverse Strength in a beam or girder is the resistance which it is capable of opposing to any force or forces acting upon it in any direction not parallel with its length. A beam may have to resist stresses caused by either up- ward or downward transverse pressure, or by both, as in the case of engine beams and crossheads. A beam may also be placed vertically or in any other position for the purpose of resisting transverse pressure. 11. Relative Position of Beam and Load. In the following resolutions of the elementary laws which govern strains in beams or girders, the direction of their length is assumed to be horizontal with loads causing downward verti- cal pressure, these being their usual conditions. 12. The Weight Of a Beam may be separately treated as a distributed load, and the stresses caused thereby may then be respectively added to those caused by whatever superimposed load the beam may have to carry. 13. Unit of Length, Area, Weight, Stress, or Strain. It is usual in calculating, or in what is technically called " taking out," the stresses in, or the strength of beams THE BEAM. or girders, to adopt, as may bo convenient, the foot or the inch as the unit of lineal, superficial, or cubic measurement, and the ton, cwt., or Ib. as the unit of weight or pressure. The unit of stress most commonly adopted is the pound or ton per square inch. The proper measure of strain is the amount of elongation, or contraction, or other displacement; but, so long as the strain varies with the stress, which is the case within the limits of perfect elasticity, whatever measures the one measures also the other. In an examination or solution of elementary principles, a quantity may be simply expressed as so many units of length, area, weight or pressure, stress or strain. 14. A Moment is, in statics, a technical expression, denoting the measure of the tendency of any force to turn a body or system in any direction round a given point, and this measure is the multiple of the force or pressure into the length of the perpendicular drawn from the given point to the line of the direction of the force. As an example, let B, Fig. 3, be a given point from which draw a perpendicular '' B C = 6 length units to the line of the direction of a pres- sure P=ll pressure C units acting in the direction of the arrow. Again, from the point B draw a perpendicular B A = ELEMENTARY PRINCIPLES. 7 11 length units to the line of the direction of another pres- sure P'=6 pressure units also acting in the direction of an arrow. /Then the moment of the pressure P about the point B=llx6:=66, and the moment of the pressure P' about the same point=6 x 11 = 66. Therefore the moments of the two pressures P and P' tending to turn the bent lever ABC about the point B are equal, and as the one pressure tends to turn it in an opposite direction to that of the other pres- sure, and the moment of the one has been found to be equal to that of the other, the system is consequently in a state of equilibrium. It is further evident that the result would be the same whatever may be the angle ABC contained by the perpen- diculars drawn to the lines of the directions of the pressures ; or again, if a point C'= 6 units from B in the arm A B of the lever were substituted for the arm B C ; or further, if the arm B C were in a straight line with A B and the system became the straight lever A B C", the perpendicular direction of the pressures relatively to the arms being maintained.* Thus the principle of the lever is based on the equality of moments measured from the fulcrum. * " If any number of pressures act in the sama plane and any point be taken in that plane, and perpendiculars be drawn from it upon the direction of all these pressures, produced if necessary, and if the number of units in each pressure be then multiplied by the number of units in the corresponding perpendicular, then this product is called the moment of that pressure about the point from which the perpendiculars are drawn, and these moments are said to be measured from that point," Moselcy. Thus the sum of the moments of any number of pressures acting in the same direction in the same plane round a given point is equal to the moment of their resultant. THE BEAM. It will have been seen that the forces P inversely as the lengths of the arms of the lever upon which they act. Thus if the length A B -f B C" be divided into as many units as there are units of pressure in P 4- P', then if equilibrium is to be maintained the position of the fulcrum will be inversely as those pressures measured from each. A lever may be employed under the three following condi- tions : With the fulcrum between the power and the load ; With the load between the power and the fulcrum ; or, With the power between the load and the fulcrum. But in each case the moments of P and P 7 are 3neasured from the fulcrum. 1 5. Parallelogram of Pressures. If three pres- sures acting upon a point are in equilibrium, and a line be drawn from this point in the direction of each pressure, each line being as many units in length as there are units in the relative pressure, then these lines will form the adjacent sides and diagonal of a parallelogram. j? Let B, Fig. 4, be a point, and the lines A B and C B represent the directions and amounts of two thrust pressures acting upon the point B, as in- dicated by the arrow-heads. Draw A D equal to and pa- rallel with B C, and C D equal to and parallel with A B. Then the diagonal B D of the parallelogram, when read in connection with the arrow at B, is the resultant of the two thrust pressures, in respect of both measure and direction. The same line B D, when read in connection with the arrow Pi- 4. ELEMENTARY PRINCIPLES. 9 at D, represents the measure and direction of a reverse pressure or resistance, which would exactly balance the two pressures, or place them in a state of equilibrium. If A B= C B, and A B C be a right angle, then the resultant B D will be the diagonal of a square a side of which being= 1, B D^ V 7 2= 1.4142. If A B and B C were tensile forces, acting upon the point B, then B D would be a thrust resistance and the arrow-heads should be reversed. 16. First Principles of Horizontal Strain in Beams. Let A B, Fig. 5, represent a beam equally loaded on each side of the central point of support c by the weights W W*. Now if the upper part of this beam were sawn through in any part of its length as at d, and a fulcrum assumed at /, then the ,/ e vertical pressure caused by the weight AY acting A d' through the portion A/' d of the beam, as through a bent lever, will tend to widen horizontally the space at d left there by the saw cut. If, therefore, the saw cut had not been made, the upper part of the beam at d would obviously have been in a state of horizontal tensile strain. Again, let the lower part of the beam be sufficiently divided or sawn through transversely at any part d', and a fulcrum assumed at/', then the vertical pressure caused by the gravi- tation of the weight AY 7 acting through the portion B d'f' of the beam will, through the bent lever ef'd', close up hori- zontally the space at d' left by the saw cut made there. Therefore, with or without the saw cut, the lower part of io THE BEAM. the beam at d' would evidently be in a state of horizontal compression. Now, by the law that action and reaction are equal and opposite when equilibrium is maintained, because any part d of the beam is in tension, the opposite part/ will bo in compression ; and, because a part d' is in compression, then the opposite part /'will be in a state of tension. Hence the primary vertical pressure of the load subjects the whole of the upper part of the beam to a state of hori- zontal tension, and at the same time places the whole of the lower part in a state of horizontal compression, and, equilibrium having been assumed, the moments of stress in the two parts must be equal because they are opposite. 17. The Neutral Axis and the Neutral Plane. It follows, then, that in any vertical line drawn in any vertical section of a beam, there must be a point in which horizontal compression and tension vanish and are reversed. Let such a point be assumed in the same plane in any number of vertical sections, then a line drawn through all these points will represent the neutral axis of the beam; and further, if we suppose the neutral axis to be extended laterally throughout the breadth of the i>eam, we shall then obtain what is termed the neutral plane, in which both compression and extension cease, and consequently there is no horizontal strain what- ever, 1 8. The Position of the Neutral Axis. The con- dition of the beam A B, Fig. 6, supported at each end and subjected to an intermediate pressure or Joad P, is the reverse of that shown in Fig. o ; the strains are therefore reversed, the part below the neutral plane ri n n" being in tension and the part above it in compression. Now it has been mathemati- ELEMENTARY PRINCIPLES. n cally proved that within the limits of perfect elasticity whatever the section of a beam may be, and whatever difference there may exist in the ultimate amount of resistance its particles or fibres are capable of opposing to tensile as com- pared with compressive stress, the neutral axis ordinarily passes through the centre of gravity of any given transverse section. Suppose A B, Fig. 6, to be a solid rectangular beam, the neutral axis n n n" will thus pass longitudinally along its centre. From the central point n draw two equal isosceles triangles a b n and n c d with their bases coinciding with the top and bottom of the beam. Then the base a b will represent the maximum horizontal compressive stress, and the base c d the maximum horizontal tensile stress that a transverse section has to resist, while the length of any intermediate lines as e, e' ', drawn parallel to a b and c d, will represent the pro- portional amounts of stress at those lines. Now if the two triangles were filled in with a regular series of such lines, the total number of their units of length in the one triangle would equal the total number of their units of length in the other, or the areas of the triangles being equal, the area of the triangle a b n would represent the total amount of com- pressive, and the area of the triangle n c d the total amount of tensile stress which the section undergoes. The beam would thus be deflected downwards as in the diagram, its fibres or particles above the neutral plane would be longi- tudinally compressed, while those below it would be longi- tudinally elongated. i2 THE BEAM. The moments of the resistance of a section of a beam being measured from its neutral axis, which in the case of a solid rectangular beam divides the section into two equal parts, the one part being in compression and the other in tension, the ultimate strength of the section will depend upon the half which from the nature of the material is capable of offering the least amount of resistance to either of those stresses. Thus taking cast iron, the tensile strength of which is less than the resistance it is capable of offering to compression, rupture would first commence by rending at the bottom of the solid rectangular beam (Fig. 6), whereas the reverse being the case with wrought iron, failure would first commence by crushing at the top of the beam. But whether the stress were sufficient to produce rupture or not, there would be an over- plus of strength above the neutral axis in the cast-iron beam, and an overplus of strength below the neutral axis in the wrouo-ht-iron beam. 3 These considerations, therefore, naturally led the way to the more advanced forms of section as shown in Figs 1 and 2, p. 2, in which the ultimate resistances of areas of section to com- pression and extension may be so adjusted as to be of equal efficiency, at the same time practically bringing within reach the most economical forms in the adaptation of materials to the construction of beams. 19. The Efficiency of the Depth of Beams. Until fracture takes place or is approximately near, the amount of strain and of resistance to strain, that is stress, per unit of area, in any portion of a section of a beam, whether in exten- sion or compression, will vary as its distance from the neutral axis. If we suppose the section of any two solid rectangular beams of the same width, but of unequal depths, to be divided into ELEMENTAR Y PRINCIPLES. 13 an equal number of small corresponding elements, the area of each of these elements will vary as the depth of the beam, and the distance of each such element from the neutral axis will also vary in the same proportion. The strength, there- fore, of the beam will vary as the square of the depth. 20. The Efficiency of Breadth in Beams. It will have been apparent from what has been already advanced that strain in a beam, and therefore its resistance to strain, takes effect in vertical planes, which may be represented by any lines A B, D, E F in the cross section Fig. 7, and these planes extend throughout the length of the beam. So that if a beam of uniform breadth were supposed tol)e formed of a series of plates of a given width placed vertically side by side, then, by increasing or diminishing the number of such plates, we should increase or diminish in the same ratio the strength of the section, and so the strength of a rectangular beam one inch wide would be increased five times by making it five inches wide. The strength of any transverse section of a rectangular beam varies therefore directly as its breadth. 21. Vertical or Shearing Strain in Beams. The principles which regulate the effects of this strain are the same in any form of beam. Their consideration, however, with reference to solid beams of uniform section would be quite unnecessary, for when the sectional area of such beams is sufficient to resist the horizontal stress set up by a load, it would be found to be more than sufficient to resist the direct vertical stress of the load ; and moreover, the direction of the one being at right angles to that of the other, they in no way act in conjunction or augment each other. The case of webbed girders is, however, different ; for their i 4 THE BEAM. tables, being simply horizontal plates, are practically unfitted by their form and position to resist vertical stress. Thus the tables having no appreciable vertical rigidity, this stress must devolve upon and be taken by the web But the tables nevertheless impart efficient lateral rigidity to the web. Referring back to Fig. 5 (16), the weight of each load W and W must travel through each respective half of the beam to the central support c. There is therefore a vertical stress = W at any vertical section of the beam between A and c, and as W = W there is also the same stress between B and c. In a like manner a beam, whether supported at its centre, Fig. 5 (16), or at each end, Fig. 6 (18), transmits intact the total amount of the ver- tical pressure of the load direct to the support or sup- ports. This pressure, there - fore, or in the case of a dis- tributed load a component fraction of it, will cause an equal stress at any vertical section of a beam, setting up therein what is termed shear- ing strain. Fig. 8 is given* in illus- tration of a model provided at the City and Guilds of London Technical College, Fig. 8. * By the kind permission of Professor John Perry, the author of " Practical Mechanics." Cassell and Co. ELEMENTAR Y PRINCIPLES. 1 5 for the purpose of elucidating the action of vertical or shearing strain in beams. Let ABC represent one-half of such a beam as that shown Fio;. 5 (16), but with the other half built into a wall, the O " \ * ' ' external half thus forming a cantilever with a weight "W sus- pended at its outer end A B. Let any part of this beam be removed, and in its place let a cord or link which is only capable of resisting tension be inserted at d, and a rod or strut which is only capable of transmitting thrust be inserted at e. these beino- the two horizontal stresses which would be 7 O set up by the weight "W in those parts were the beam intact. It is evident that such an arrangement will not in itself keep the portion A B d e in position without an upward force at d e equal to the weight W, plus the weight of the severed part A B d e of the beam. Thus at any vertical section as d e of a beam, besides horizontal resistance to tension and com- pression, resistance to shearing stress is an essential element of strength, and this is supplied in the model by a weight W' connected with and acting in the direction of the section d e by means of a corcl passing over a pulley. Until discharged upon the supports, shearing stress is in practice always assumed to be resisted entirely by the web of a plate girder. 22. Diagonal Strain in Beams. Another im- portant office of the web in addition to transferring the vertical pressure of the load to the points of support, is to transfer to the tables the horizontal stresses due to leverage. Now as it is impossible that vertical stress in a structure can either travel in a lateral direction, or can be resolved directly into horizontal stress, the latter being at right angles with the former, it necessarily follows that both these 1 6 THE BEAM. operations are effected by diagonal stresses set up in the web. That this is so was clearly demonstrated by the experi- ments made by Mr. Eobert Stephenson upon model girders with reference to his design for the Menai Bridge. Be- sides, the buckling of the web plates of these models was found to demonstrate unequivocally that the diagonal stresses had a tendency to take the direction of an angle of 45. It would seem that that may be their most natural direction, because they would thus bisect the right angle made by the original or vertical stress with ultimate or horizontal strain. Now, if one diagonal of a square were in tension, the other would be in compression, because a rectangular or other figure of four sides, having a tendency to lengthen in the direction of one diagonal, must also have a tendency to shorten in the direction of the other, and diagonal stresses in a plate web, that is to say not conveyed through detached bars, as in a framed or lattice girder, pervade the whole web, the tensile passing apparently at right angles to the compressive through every particle of it. The resolution of these diagonal stresses and consequent strains at any given vertical section of the web would bring us back again to vertical strain. For,, assuming the diagonal stress, either of tension or compres- sion, as acting upon the particles of the web at an angle of 45 with any vertical section, it is when compared with the vertical stress on that section as I to \/%. or 1 to 1.4142, when resolved by the parallelogram of pressures, because the ver- tical stress due to the load is split into two diagonal stresses, one of tension and the other of compression, and these in a solid beam or in a plate web may be regarded as being con- tinuously retransmitted as vertical stress. Diagonal tensile or compressive stress having been ELEMENTARY PRINCIPLES. 17 shown to be less in any one direction in a beam than vertical or shearing stress, in the proportion of 1 to 1.4142, any further notice of diagonal stress and strain becomes unnecessary until the subject of framed and lattice beams is approached. CHAPTER II. DETERMINATION OF STRESSES. THE examples given in this chapter are applicable to all solid beams and webbed single span girders. Beam with a Central Load and a Support at each End. 23. Vertical Stress. When a beam is horizontally supported at each end, and loaded at the centre of its span, the vertical pressure of the load becomes halved at the centre of the beam, and each half transmitted in an opposite direc- Span 8-u.niJ* ~ of length, JSerles MHS.= o tion through the beam to the points of support. Thus as the centre of gravity of the load is, equidistant from the supports DETERMINATION OF STRESSES. 19 there will be an equal vertical pressure of half the load upon each support. Let A B, Fig. 9, represent a beam of an uniform depth of 1 unit and a length of 8, and assume a load of 4 units at its centre, then with B as the fulcrum of the lever B A by the equality of moments 4xFB = 2xAB, therefore there must be a pressure of 2 units at A. In the same wav by taking A as the fulcrum of the lever A B there V J O is shown to be a pressure of 2 units at B, and as the stress of one-half the weight of the load passes through the medium of the beam from F to A there must necessarily be at any intermediate vertical section of the beam a vertical or shearing stress of 2 units, Series VS, Fig. 9. Thus in this example the effective* vertical stress being 2 at a support, and half the length of the beam being 4 in units of the depth, the effect of that stress upon that length = 2 x 4 = 8, or the moment of horizontal stress upon the central section of the beam as stated in the next article, or upon that of each table in the case of a plate girder. The lower diagram in the figure is appended simply to illustrate the fact that pressure and resistance are equal and opposite, and therefore convertible terms. The beam is shown suspended at each end by a chain passing over a pulley and carrying a balance weight. Now the tensile stress in each of these chains, balancing and at the same time tending to / O 3 lift the beam bodily upwards, represents and is precisely the same in amount as the resistance of each support to the pressure upon it, caused by the load upon the beam, as shown in the upper diagram. It follows that the resistance or vertical reaction of one * With a distributed load the costive vertical stress at a support is half that of the actual load upon it (27). c2 20 THE BEAM. support must meet at the central vertical section of the beam precisely the same amount of reaction set up by the other support, and these forces being equal, and acting in the same direction in parallel vertical planes, there will be no shearing strain whatever in the central vertical plane in which they meet, and unite in resisting the direct vertical stress of the load. 24. Horizontal Stresses. As the depth of the beam Fig. 9 = 1 unit, assume A F C to be a bent lever with its fulcrum at F, then the moment of horizontal stress pro- duced on C F by the reaction of 2 at A will be 4 x 2 = 8, and if the whole of the stress were at C its amount would also be 8. If the length of the beam were 4 only the moment of stress would be 4. So that with half the length of beam there is half the horizontal stress on the central section. The result will be the same if the neutral axis or any other point in the vertical section be taken as the fulcrum. For let any point /in the lower diagram be the fulcrum of a lever of three arms a/, /c,/, then the stress at any point c, c', ', or t will vary in the direct proportion of the length of the arm a/, and therefore in direct proportion to the length of the beam. In any beam loaded at its centre the amount of horizontal stress at any vertical section varies in direct proportion to its distance from the nearest point of support. For if any number of points, as F' F" F'", are assumed as fulcrums, and vertical lines drawn from these points upwards to represent the vertical arms of a series of bent levers and at the same time vertical sections of the beam, then the strain on each of these sections will be in the direct proportion of its distance from the extreme end A of the horizontal arm of DETERMINA TION OF STRESSES. 2 1 each such lover. Xow assume that in the 8-feet beam these sections are taken at 1, 2, and 3 units from A, and having 2x4- founcl the moment of stress at the central section = = 8 then the moments at the other sections will relatively = 2, 4, and 6, and in a similar beam 4 feet long with the same load the stress on the four corresponding sections would be represented by the proportionate numbers 1, 2, 3, and 4. In the same way it may be shown that horizontal stress in the other half of the beam is at each relative vertical equal and opposite, so that if stress in one-half is assumed to bo that of action, that in the other half will be stress of reaction. Fig. 9, Series MHS. It has been shown, therefore, that in any beam loaded at its centre, the horizontal stress caused by the load is greatest at the centre of the beam, that it decreases in terms of arithmetic progression towards each support, there to vanish, and that the strength of a beam of uniform section loaded at its centre is inversely as its length. Beam with a Central Support and a Load at each End. 25. Let the same beam A B, 8 units long and 1 deep, Fig. 9, be supported at its centre F, and loaded with a weight of 2 units at each end as shown Fig. 10. Under these new condi- tions what were resistances when the beam was supported at each end have now become loads, and the load itself has become a resistance. Now these terms 33 THE BEAM, are convertible because they express equal and opposite forces, as has been shown (23). The condition of the beam, therefore, Fig. 10, is nothing more nor less than that shown in Fig. 9, inverted. There is now a vertical resistance of 4 units at F 'caused by the central support, and it is apparent that the horizontal stresses in the beam under the new are equal to and coincide with those under the previous conditions, with this difference, that their direction is reversed, for the upper part of the beam, which was in compression, is now in tension, and the lower part, which was in tension, is now in compression. The vertical stresses remain also in amount precisely the same as before, but are reversed. For instance, any part A F' T of the beam has now a tendency to shear off downwards before, upwards. It follows that a beam supported at its centre will carry half the load at each end that it is capable of carrying at its centre when supported at the two ends. Beam with a Distributed Load and a Support at each End. - Let A B, Fig. 11, be a beam 8 units long and 1 deep, supported at each end A and B, and carrying a load of 16 units equally distributed over the entire length. 26. The Vertical Stress occasioned by the pressure of the load is equally divided at the centre of the span, and transmitted by the beam in opposite directions to the two points of support, at each of which it is equal to half the weight of the load. From these points vertical stress in the beam decreases in an arithmetical ratio to nothing at the centre of the span. The effect of this stress arising from a central load having been determined (23), DETERMINATION OF STRESSES. 23 we may adopt the same mode of treatment in determining its effect when caused by a distributed load. iLoaa ^ /t < '/HIS Vf iwiAifi.i* 1 ( = <--+-2--f--fi- +-2-+-: .'--f-2--t--,2-+-2 - A /4 lyj ji-'i* ^1- - -V-~ "\C |t'' !C 2 it'-'' i j [ i . i r 4 I _/j t Series VS t = J 1 ^ 1 i <5 4 \ ? i i 2 T=- = \ \" ' A ^ 1 f C = t.< 7 2 1 " 7 6 3 4 ,5 4. ,5 e 7 A 2 / O ' 1 'Series MHS = ' ) 7 12 A5 W /.I /2 7 o Fig. 11. Divide the half length of the beam V 4 C into units at points V 3 , V 2 , V 1 ,- and from the point C at the centre of the beam to the point A draw the diagonal C A. Now as half the load or 8 units is distributed between the points C and V 4 , there are two units between C and V 1 , and as the vertical stress of these has to be transmitted to the end V 4 A of the beam, it must pass through the beam in that direction, and therefore there is stress of 2 units at V 1 . Also between V 1 and V 2 , two other units of the load in like manner cause a stress of 2 units at V 2 , but the stress of the first two units has also to pass the same point, consequently there will be a vertical stress of 4 units at V 2 , and so on to 8 units at the end of the beam. Therefore if the length of the line V 4 A be taken to represent 8 units, or the stress at that perpen- dicular, the length of any other vertical line drawn from any point in the line V 4 C across the triangle A V 4 C will 24 THE BEAM, represent the vertical stress at that part of the beam. Con- sequently the stress at C, V 1 , V 2 , V 3 , V 4 = 0, 2, 4, 6, 8. It is also evident that stresses of the same amounts exist in the corresponding parts of the other half of the beam, Fig. 11, Series VS. 27. Horizontal Stress. Divide the half C C 4 of the length of the beam into units of length and depth by the dotted lines F C, F 1 C 1 , F 2 C 3 , F 3 C 3 . Now 8 units or the weight of the half of the load extending from C to C 4 may be assumed to act vertically downwards through its centre of gravity, or the line C 2 F 2 . This pressure acting through the bent lever F 2 F C would produce a horizontal tensile stress at C or a compressive stress at F the moment of which about any point in C F would be 8 x 2 = 16. But the vertical resistance of 8 at A will cause a horizontal compressive stress at C or tensile stress at F the moment of which will be 8 x 4 = 32. The result will be a compressive stress at C or tensile at F with a moment of 32 - 16 = 16. The moment of horizontal stress at each of the sections C 1 F 1 , C 2 F 2 , C 3 F 3 may be similarly determined. At C 1 F 1 , there is a stress in one direction caused by the load the moment of which is 6 x 1J = 9, and one in the opposite direc- tion from the resistance of the point of support with a moment of 8 x 3 = 24. The resulting moment of stress is therefore 24 9 = 15. Similarly at C 2 F 2 , the moment of stress is 16 - 4 = 12, and at C 3 F 3 it is 8 - 1 = 7. We thus obtain the figures of the Series MHS, 16, 15, 12, 7, and the horizontal stress at the point of support is nil. At corresponding verticals in the other half of the beam there will exist equal and opposite moments of stress which DETERMINATION OF STRESSES. 25 vanish with the leverage at each end of the beam. Fig. 11, Series MHS. Therefore although, as in the case of a beam carrying a central load (23), the vertical pressure of the load becomes equally divided at the centre of the beam and transmitted in opposite directions to the points of support and with a load of 16 units either central or distributed, there will be the weight and reaction of 8 units at the support A, the moment of stress upon the central section with the distributed load is 16, or that due to an effective weight of 4 only at A. We have therefore the following results : 1st. The horizontal stress upon the central section caused by a load equally distributed over a beam supported at each end is the same as that which would be caused by one-half of the same load placed at the centre of the same beam. There- fore if a beam of uniform section will carry a given load at its centre it will carry twice that load equally distributed over its length. 2nd. In any beam of uniform section supported at each end the horizontal stress at the centre caused by its own weight is the same as though half its weight were accumulated at the centre. 3rd, In any beam supported at each end and uniformly loaded, the ratio of the moment of stress upon the central section to that upon any other vertical section is as the square of half the length of the beam to the multiple of the segments into which the beam may be divided, That this is so may be seen by multiplying the lengths of the segments together at each point in the two Series RS, Fig. 11, the results coin- ciding with those stated Series MHS. For as in determining the horizontal stress at the central 26 THE BEAM. vertical of the beam, Fig. 11, we may regard lialf the weight of the distributed load as statically effective at that vertical, so also the proportion of one-half of the weight of the same load which would be borne at either support if the load were concentrated at any other section may be considered as effective at that section. Because when a load or pressure is assumed to act at unequal distances from the supports, then by the law of the equality of moments (14) the fraction of that pressure carried by each support will be inversely as the distance of the pressure from the support, thus 8 units at C = 4 at A anchor at B then 4 x 4 = 16 C 1 = 3 A 5 B 3 x 5 = 15 C 3 = 2 A 6 B 2 x 6 = 12 C 3 = 1 A 7 B 1 x 7 = 7 A similar ratio obtains if we divide the beam into any greater or less number of segments. As an example divide the length of the beam into 16 instead of 8 units, multiply as before, and we have the following Series MHS. 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ,(01 ' (16 15 RS. 11*1* 9876543210 MHS. 15 28 39 48 55 60 63 64 63 60 55 48 39 28 15 In practice it is frequently sufficient to determine the hori- zontal stress at the central vertical only. But for largo girders the determination of a' complete series of stresses may become necessary from the nature of the work, material, or circumstances, while to be on the safe side the higher stress should be assumed to extend down to the lower; thus a moment of stress of 16 at F should be assumed to extend to F 1 , 15 at F 1 to extend to F 2 , 12 at F 2 to extend to F 3 , and 7 at F 3 to extend to the end B of the beam. DETERMINA TION OF STRESSES. 2 7 Beam witli a Distributed Load and a Central Support. Let the beam A B, 8 units long and 1 deep, Fig. 11, be supported at its centre F, and carry the same load as before of 16 units equally distributed over its entire length as shown, Fig. 12. 28. The Vertical Stress caused by the pressure of the load under this arrangement instead of being nothing at LocuA (* 16 units oj weiqkt.- = |<-2--t--2--f-2-+-.2-+-2- + -2--|- i i i i i i tr h . r r/ | 72 iy 5 y* " "i i A > J, ' * ! * A T ~T 1 1 i i I /\ V V *' s^ra-o, i i i S 1 i 1 1 i = T' iiii S T S *?* I i i i i i i I I SerfjenAfffS* J 4 9 M 9 4 i o !< . _ 5^, a/t . ^ units of length >! Fig. 12. the centre of the beam and increasing to half the load at each end of the beam, as in the last example, is now nothing at each end from whence it increases in a direct arithmetical ratio towards the centre of the beam where the stresses caused by the two halves of the load unite in a stress equal to the whole load, and where there is no shearing but compressive stress as the weight of the load is there met by a direct and equal resistance from the point of support. Halve the length of the beam by the vertical T F, draw the diagonal V 4 F, divide the side V 4 T of the triangle V 4 T F into 28 THE BEAM. units of the depth of the beam by points at V 3 , V 2 , V 1 , and draw verticals from those points to the line V 4 F. Kow the weight of the two units of the load extending c5 & from V 4 to V 3 has to be transferred by the beam to the support at F, and therefore there must be a stress of two units at V 3 . In the same way it will be seen that the stress at V 2 and V 1 is 4 and 6, and at the centre T it will be 8 plus that due to the other half of the load, or 8 + 8 = 16 units, If therefore the length of the line T F be taken to repre- sent 8, the length of the other verticals, as for instance V 1 , V 2 , V 3 , drawn across the triangle V* T F, will represent the vertical stress in the beam at those lines, and the same stresses will exist in the corresponding parts of the other half of the beam. Fig. 12, Series VS. 29. Horizontal Stress. Divide the half of the length T T 4 of the beam into units of its depth, by the dotted lines T 1 F 1 , T 3 F 2 , T 3 F 3 , and halve the spaces F 2 F 3 and F 3 B in points d and d'. 1. The moment of stress at the central section T F is that occasioned by half the load or 8 units acting vertically through its centre of gravity and the line T 2 F 2 , as upon the end F 3 of a bent lever F 2 F T, the fulcrum being at F ; and therefore 8 x 2 = 16. The stress on the central section is therefore directly as the length of the arm F F 2 of the lever, and consequently directly as the length of the beam, as in the example already given (27). 2. The moment of stress at the section T 1 F 1 is that due to fth of the load, being the portion from T 1 to T 4 , or 6 units, acting vertically through its centre of gravity upon the end DETERMINATION OF STRESSES. 29 d of the bent lever d F 1 T 1 , the fulcrum being at F 1 ; or 6x14= 9. o. The moment of stress at the section T 3 F 2 is similarly that due to the action of Jth of the load, or 4 units, being the portion from T 2 to T 4 , at a distance of 1 unit, and is there- fore 4. 4. In the same manner the moment of stress at the section T 3 F 3 from 2 acting at d' is 2 x J = 1. The moments of horizontal stress thus determined are therefore 16, 9, 4, 1, or in direct proportion to the squares of the distances from the end B of the beam, the distances being 4, 3, 2, 1. In like manner the stresses in the other half of the beam are equal and opposite. Figure 12, Series MHS. It has therefore been shown : 1st. That in a beam supported at its centre and uniformly loaded throughout its length, horizontal stress is greatest 3 O J O at the centre where it varies directly as the length of the beam. The strength of the beam is therefore inversely as its length. 2nd. That horizontal stress is the same in amount at the centre of the beam as if the beam were supported at its ends and similarly loaded, but that it decreases towards each end of the beam in the ratio of the square of the distances from the end, and not as the multiple of the segments as in the former case. Consequently a beam of uniform section, whether supported at its centre, or at each end, will carry the same load uniformly distributed, or twice the weight that it. would be able to carry at its centre when supported at the two ends. But in the second condition the stresses right and left of the centre are greater. 3 THE BEAM. Beam with a Non-central Load and a Support at each End. Let A B, Fig. 13, be a beam of uniform section 10 units long and 1 deep, supported at each end A and B, and carrying a load of 20 units at a point C, 7 from A and 3 from B. 3O. The Vertical Pressure of the load travels in unequal fractions from the point C to each support in the inverse proportion of the distance of the support from C, the vertical of the centre of pressure of the load. Thus as Load ivxito 1 7 Al ? J 1 5 S v. 1 fi 2 i \* I T Series VS. 6 ( Series MHS. O < 1 i 6 1 1 1 3 12 i 6 < 1 I /# 2 ! < -/ 3 j <: 3 i > er/4 s i 6 ^2 y 1 fl i i /4 1 14 4 /* I Span =10 unita oj length 7 . ^ Fig. ]3. the load is 20 and the span 10, and as A is 7 distant from C, there will be a stress of 14 at B, and as B is 3 distant from C, there will be a stress of 6 at A. For let A B be a lever with the fulcrum at B, then by the equality of moments the resistance at A = -* x - = 6, and in the same way the resistance at B = 10 20 x 7 "TO" = 14. The vertical stress caused by the load therefore becomes unequally divided in the inverse ratio of the segments into which the beam is divided by the centre of pressure, thus, assuming the load to be 1, DETERMINATION OF STRESSES. 31 /> - = .3 passes from C through the beam to the support at A, and = .7 passes from C in the same way to the sup- port at B. Fig. 13, Series VS. Now the vertical shearing stress caused by the resistance at A = 6 extends from A to C, and the vertical shearing stress caused by the resistance at B = 14 extends from B to C. Two unequal shearing stresses therefore meet at C, so that the true shearing stress in the theoretical vertical plane which at C divides the beam into two segments is equal to the difference of the two, or = 14 6 = 8. 31. The Horizontal Stress at any cross section of the beam may readily be determined in proportional amounts, as shown in Article 24. Tims, the beam having been longi- tudinally divided into units, commencing with the vertical stress of 6 at A, 6 x = 0, next 6xl = G, and so on to the centre C, where 6 x 7 = 42. In the same way,, com- mencing with the vertical stress of 14 at B, 14 x = 0, next 14 x 1 = 14, then 14 x 2 = 28, while at the centre it is 14 x 3 = 42. The moment of stress of action and reaction at the section C has therefore been shown to be equal and opposite, and the same must necessarily be the case at all corresponding sections right and left of C. Fig. 13, Series MHS. It will be seen, therefore, that the horizontal stress at C, the centre of pressure of the load, is as the multiple of the seg- ments into which that point may divide tne beam, and that consequently it will be the greatest when the load is central between the supports, while in all cases it will diminish from the centre of pressure of the load in terms of arithmetic progression from a maximum at C to nothing at each support. 7 C. ~K THE BEAM. Corollary. It follows conversely that if the same beam were supported at C and B. loaded with 6 units at the end A and with 14 at the end B, Fig. 14, equilibrium would be maintained, and that the amounts and ratios of the stresses would be the same as before, but that the relative conditions of tension and com- pression would be reversed (25). so. u< Beam with tico Equidistant Loads, and a Support at each End* Let A B, Fig. 15, be a beam of uniform section 10 units long and 1 deep supported at each end A and B, and carrying 14+ fi> 20 =(34/4 i* 4 1 1 1 ? \ 1 1 1 1 1 D 1 : j 1, x~ i 1 i 1 1 I 1 j ' A Y i 1 O., ,-,* . 'i JO / / 7i if 1 Series C. 14 14 14 14+6 c 6 G 6fi ft 6* 6 1 1 1 ) I i 1 , 1 1 i 1 1 * t i i i i 1 I 1 i 1 ' . .. D 6 6 (j fi',G 6 f> 6' 14 14 14 Serifs VS 20 20 ~20 i'ofo O O Ol'^O 2O 2T> - J I 1 1 ~r~ T* T i T 1 T I 1 1 I 1 I i I 1 1 i 1 | 1 1 I | 1 j Seric* c:o 1 28 4-2 1 1 30 i i 1 G \ c \ i 1 i 1 i i i * D'o 6 1?, 18 24 30 42 2S 14 O I SerwMHS . o^ 20 =L fiO 60 C>0 60 GO "O J9_ 'Fig. 15. a load of 20 at C, 3 distant from A, and a load of 20 at D, 3 distant from B. DETERMINATION OF STRESSES, 33 32. Vertical Stress. The manner in which the direct stress of a non-central load is divided having been shov/n (30) , place the stresses due to the weight of the load at C in a Series C, Fig. 15, and those due to the weight of the load at D in a corresponding Series D. Now at any section between the verticals C and D the stress thus given is in each case 6, but when these stresses are taken in conjunction, it will be seen that there is no shearing stress at all in that part of the beam, because the two are equal and act in the same direction, and there is no vertical resistance between those points, but the load of 20 at C meets with a like resistance at A, and that at D with a like resistance at B. Therefore the two series of assumed stresses occurring between sections C and D of the beam may be subtracted the one from the other, and, as already said, the shearing stress in that part of the span is nil, while that set up by the load of 20 at C extends to A, and that due to the load of 20 at D extends to B. Fig. 15, Series VS. 33. Horizontal Stresses. The mode of determin- ing these for either the load at C or D having been given (31), it only remains to place the moments due to the load acting at C in a Series C', Fig. 15, and those due to the load acting at D in a corresponding Series D', and then to add the amounts together, placing in a final series the aggregate of the united stresses due at each section to the two loads. Thus Fin-. 15, Series MHS., sives the amounts of the combined t^ 7 / cT> moments of stress due to the two loads at each unit of length of the beam. It will be observed that from centre to centre of the two loads or between the verticals C and D the moment of stress remains constant, because, as already stated, there exists no vertical resistance between those sections. 34 THE BEAM. Corollary. Tt follows conversely that if the same beam were supported at C and D, and loaded with 20 units at each end A and B, Fig. 1(3, the amounts and ratios of all stresses would remain the same as before, but that the relative conditions of tension and compression would be reversed (25). It will also be apparent that there cannot be any vertical or shearing stress between C and D. Between these verticals there is theoretically no need of any web. In the case of a cast-iron girder, however, it would be essential from the nature of the material and its section to continue the web between C and D, and the same would also be necessary in a plate girder in order to stiffen vertically that table which has to resist compressive stress. Beam carrying Loads at various Points and supported at each End. Let A B, Fig. 17, be a beam 12 units long and 1 deep, supported at each end, A and B, and carrying six various loads at the verticals C, D, E, F, G, and H, placed 1, 4, 6, 7, 9, and 10 units from A, and = 12, 3, 2, 12, 4, and 6 units in weight respectively. 34. Vertical Stresses. Referring to 32, it will be evident that these may be readily determined if we proceed first to ascertain the total resistance each support has to present to the pressure caused by the six loads. DETERMINATION OF STRESSES. 35 Thus (30) 11 units of weight (Fig. 17, * Series vs) are transferred from C to A -f 2 from D, + 1 from E, + 5 from F, -f 1 from G, + 1 from H = a total of 21 units of weight at A. Again 1 is transferred from C to B, + 1 from s^- Span 12 unita . i % *~ jia i g f* N Series vs. J -^r- l\ 2| .3 i T^ D *! -nr- E 6, F 7i fl! G H ftl A'!| i I 1 2:1 1 1 f ,57 / 3 r.j i I *, I vs 21 21:9 9 i i 9:6 6 6:4 4'.8 8 :/^ iz'.is M tl | | | X Series C O 11 10 8 7 6 S 4 3 2 / ( D 2 468 7 & 4 3 2 / C .. E 1 234 5 6 5 4 3 2 7 < .. F 5 10 15 20 25 30 35 28 21 14 7 .. G I 2 i 4 5 6 7 8 9 6 .3 c H 1 2 i \ 4 $ 6 7 8 9 10 5 ( 8*rin MHS. 21 30 39 4S 54 60 64 56 48 36 18 c D, + 1 from E, + 7 from F, + 3 from G, + 5 from H = a total of 18 units of weight at B as given Series VS. Now having ascertained that there is a stress of 21 at A, deduct in succession the load at C, D, and E ; then the stress from A to C = 21 -- = 21, from C to D = 21 - 12 = 9, and from D to E =9 3 = 6. The aggregate thus deducted = 12 + 3 + 2 = 17, but there is a weight of 21 on the support A, so that 4 units pass from the load of 12 at F to the support A, and the stress from E to F = 4, leaving at F 12 4 = 8. In the same way having a stress of 18 at B, deduct in succession the loads at H and G, then the stress from B to H = 18 - = 18, from H to G = 18 - 6 = 12, and from G to F = 12 4 = 8. The aggregate thus deducted is 6 + 4 = 10, but as there is a weight of 18 on the support B, the 8 units we have left at F must pass D 2 36 THE BEAM. to the support B, and as already found, the stress from G to F = 8. Fig. 17, Series VS. As the units of vertical stress part company right and left in the unequal fractions T 4 and T \ at the section F, the ver- tical shearing stress at that section = 8 . 4 = 4 (30). 35. The Horizontal Stresses caused by each load are also readily determined as shown (31). Place in relative series C 5 D, E, F, G, and H, the moment of stress so given at each unit of length of the beam. Add up and place the amounts in a final Series MHS, Fig. 17, which thus gives at each unit of length of the beam the moment of horizontal stress caused by the six loads. Seam with a Concentrated Rolling Load. Let A B, Fig. 18, be a beam 10 units long and one deep, supported at each end A and B, and carrying a concentrated rolling load of 10 units. 36. Vertical Stress. It has been shown (23) that when the centre of gravity of a concentrated load coin- cides with the central vertical of the beam, the vertical stress due to the weight of the load becomes equally divided, and each half transmitted in an opposite direction to the points of support. Further (30), that when the centre of gravity of the load coincides with any vertical section of the beam other than the central one, vertical stress so caused becomes unequally divided and transmitted from that section to each support in unequal fractions in the inverse proportion of the distance of the section from the support. Now bearing in mind these premises, let the load of 10 DETERMINATION OF STRESSES. 37 units at the end A of the beam roll to the other end B. Then in successively passing the verticals 1, 2, 3, 4, 5, 6, 7, 8, 9, its effect at each will be the fractional vertical stresses given in Fig. 18, Series VS. It will furfher be seen by reading from either support towards the other, that each of these fractional stresses increases in a ratio of direct arithmetical progression until it becomes equal to the whole load at each support. Thus starting from both ends, these fractional stresses are respectively 0, 1, 2, 3, 4, and 5 at the centre of the beam, and continuing past the Q *"> } f" "*> i' V } v ^ '"^W ,'V )( X- -N ^ / /'"^ 1 Jl , "V ~\ '"~ N Depth limit { I i i i J 41 C 1 1 1 7, 1 1 ? l "i foil A 1 i I Span JO /7 J- A . 1 1 A undfr a ( Seri* VS 10 3 614 5:5 416 3'.7 < r.s IO r tizs \ A/7Y.S 1 1 9 70 & i 1 24 1 1 1 24 l I 21 i 1 i | | 1 1 1 | 1 1 i 1 | 1 1 i 1 1 1 1 under a. f t , central I ^rieit VS i 5 1 5 6 1 5 5 1 5:5 1 1 i 5 1 6 1 5 lOOAL UHS s IO 1 5 20 25 20 7.5 70 5 centre in each direction, 6, 7, 8, 9, and 10 respectively. Whereas if the same load were to remain stationary at the central vertical C, vertical stress would be represented by 5 throughout, Fig. 18, Series VS 7 . Comparing Series VS with Series VS', it will be noted that the difference implies considerable theoretic modification in the web of a girder in order that it should properly resist vertical stress caused by a concentrated rolling load. 37. Horizontal Stress. It has been shown (31) that in the case of a concentrated load carried by a beam at any point between the two supports, the moment of 38 THE BEAM. stress is directly as the multiple of the segments into which that point divides the beam. Now a concentrated rolling load, in moving from end to end of a girder, must pass every assumed vertical. If, therefore, we multiply the segments into each other, at each such vertical in succes- sion, the moment so determined for each will proportionately represent the stress caused by the load in passing, Fig. 18, Series MHS. Comparing the moments thus determined with those in the Series MHS 7 of the moments of horizontal stress at the same points when the load is stationary at the central vertical C, it will be seen that in the former case the moments are, as in the case of a distributed load, represented by the multiple of the segments giving the ordinates of a para- bola, whereas in the latter the moments increase in a direct arithmetical ratio from each support to the centre of the beam. Therefore the strength of the tables of a girder intended to carry a concentrated rolling load, passing longitudinally over it, should between its centre and supports be greater than that required for an equal central stationary load in the relative proportions of the resolved moments as given in Figure 18. Beam with Two Rolling Loads coupled. Let A B, Fig> 19, be a beam 10 units long and one deep supported at each end, A and B, and carrying two rolling loads of 10 units each, coupled together at a distance of 2 units from centre to centre of each. 38. The Vertical Stresses will in this instance become incident at the assumed respective verticals simul- DETERMINATION OF STRESSES. 39 taneously in pairs as bracketed a, Z, c, t?, e, and / in the figure. Now all that is necessary for the determination of these stresses is to find (30) the amounts of the fractions of a h c d e f I OOBOOOO 1 I \ ( 1 <7. 1 1 1 1 1 7 A '.i\ ~\ ,5| 7l ] 9\ JO 77. \ \ Span /(^ units- 1 1 1 I 1 | ^ VS. 18 26 74 72 70 5 10 72 16 1 S . \ 1 1 1 I \ 1 1 \ a. o 8 76' 74 72 70 8 6 4 2 ( > \ \ i \ i 1 i. i I 1 i I i I I 1 - 8 7 .5 4 3 2 7 i ) ) T T ~7\ T T T T T T 1 = =i r o 7 |74 7,5 72 9 6 3 ( ) S 20 \7,5 20 2,5 20 7,5 70 ,5 < j \^d. 72 C. 70 nr 30 === 30 \nr \40<- 40 -40 32 T 40 24 30 76 20 8 T 7O ( = r ) =. > MHS. 76' 28 36 40 40 40 36 2S 76 c > Fig. 19. stress which travel in the direction of one of the supports, say the support A. Assume the pair of loads to be moving from A towards B, and to be in the first positions bracketed a, then for the stress at the end A, to the load 10 at A add 8 7 or the fraction due from the load at vertical 2, and 10 + 8 = 18. Next when 40 THE BEAM. the loads are in the second position or insistent at verticals 1 and 3, the stress at vertical 1 is9 + 7 = 16 ; with the loads at 2 and 4, that at 2 is 8 + 6 = 14 ; with the loads at 3 and 5, that at 3 is 7 + 5 = 12 ; with the loads at 4 and 6, that at 4 is 6 -) L 4 = 10 ; and with the loads at 5 aiid 7, that at 5, or the central vertical of the beam, is 5 + 3 = 8. Thus, Fig. 19, Series VS, completed from C to B in reverse order, gives the vertical stress due at each assumed vertical as the coupled loads pass from end to end of the beam. 39. Horizontal Stresses. Assume as before the coupled loads to be moving from A towards B. Then at each assumed vertical resolve the moments of stress (31) due to each of the loads in passing the positions indicated by the letters a, 5, c, d, e, add the amounts due to the coupled pairs together, and place them in Series I. c, and d (Fig. 19). It will be observed that Series a does not fall within a coupled series, nor within the final result ; and that Series e can be obtained by one operation (33). Now commencing from the centre and then following the results in Series e, d, c, bj in the order just stated and in the direction indicated by the arrows, we have the maximum stress to which each assumed vertical throughout the half length of the beam has been subjected during the passage of the combined load^ from A until their centre of gravity coincided with vertical 5, or the centre C of the span. Write these down for the first half of the span, and repeat the same for the second half in reverse order, and we obtain Series MHS, or the moments of maximum horizontal stress at each assumed vertical caused by the coupled load in passing; from end to end of the beam. In this manner the stresses in a girder caused by the fore DETERMINATION OF STRESSES., 41 and aft wheels of a loaded trolley passing over it longitudi- nally may be determined, whether the loads carried by the wheels be equal or unequal. Beam with an uniformly Distributed Load advancing upon it. Let A B, Fig. 20, be a beam 8 units long and 1 unit deep supported at each end A and B, and subjected to a con- tinuous load of 64 units or eight to each single unit of the length. 40. Vertical Stresses. Divide the beam into units of length by verticals 1, 2, 3, 4, 5, 6, 7, and assume the continuous load to advance upon the beam unit by unit in length from either end, say from the end A. Assume 8 load units or a weight = 8 in respect of each unit of length to act vertically downwards through its centre of gravity. Then proceed to determine (30) the vertical stresses caused by ihe load carried by each successive unit of length. Add together at each vertical the results thus obtained until the fore end of the advancing load coincides with the central vertical 4 at C. Beyond that vertical the stress caused by any further advance must be subtracted at any point between that of advance and the corresponding point already passed on the left side of the centre of the span (32). The Series VS 1, 2, 3, 4, 5, 6, 7, 8 will then give the stress at each vertical, due to each successive fractional advance of the load, until there are 8 load units upon each unit of length. Commencing from the support A, the bracketed figures indicate the position of the leading fraction of the load for each series. Between the centre of gravity of THE BEAM. IiCad"" "- i3 1 1 i i \ C 1 1 1 4U ^4 l| 2\ 3\ 4\ 5\ flj 7\ 8 B > k Span 8 unit* - > ^ VS. 2 ( 7 5 .5) .5 .5 .5 .5 .5 .5 , 5 6 ,5 (6.5 1.5) 15 L5 1.5 1.5 1.5 7 5 2 14.0 J7o 2~0 2.0 2.0 2.0 2.0 2.0 2*0 5.5 (5.5 2.5 2.5 2.5 2.5 2.5 ., 3 79. J 72.5 7.5 4.5 4.5 4.5 4.5 4,5 4.5 4.5 4.5 4.5 (4J> z.s) 3.5 3.5 3.5 3.5 > *** 24.0 17.0 12.0 9.0 8..0 S.O S.O 8.0 8 .0 3.5 5 3.5 3.5 /(O\ 4.5) 4.5 4.5 4.5 27.5 20.5 15.5 12.5 4.5 72.5 12.5 12.5 12 .5- 2 .5 2.5 2.5/ 2.5 y2a\ 5.5) 6.5 5.5 .. <5 30.0 23.0 isa lao 2.0 70.0 18.0 18.0 18.0 Zi5 M/ /1. 5 1.5 L 5 . 1.5 (7.\ 6.5) 6.5 7 57 .5 j 16.5 8,5 .5 8.5 26. 24.5 24.5 < e f 5 5 f ( S j $} 8 1 3= as- S* ~16 8 "o" 8 ~16 v_lr\ 24 ^ c= 2 =2 MHS. 1 (?.S) 3.0 2.5 2-0 1.5 7.0 .5 6.5 9.0 7.5 6.0 4.5 3.0 1.5 .. . 2 70.0 (72.0) 10.0 S.O 6.0 4.0 2.0 i ) 5.5 77.0 12.5 10.0 7.5 5.0 2.5 ( ^ <3 JD 23.0 (22.5 ) 7S.O 13.5 9.0 4.5 4.5 9.0 13.5 74.0 10.5 7.0 3.5 4 20.0 32.0 36.0 @2.0) 24.0 16.0 S.O S.5 7.0 10.5 74.0 13.5 9.0 4.5 i ) Tf > 23.5 39.0 46.5 46.0 (37.5) 25.0 12.5 ~o J)_ 2.5 5.0 7.5 10.0 12.5 ILO 5.5 J2. 6 o 26.0 44.0 54.0 56.0 50.0 (36.0) 28.0 o JL 1.5 3^0 4.5 6.0 7.5 9.0 6.5 7 O 27.5 47.0 5S.5 62.0 57.5 45.O (24.5) ^ .6 2.0 1. 5 2.0 2.6 3.0 3.5 (j L) ,. 5 28.0 48.0 6O.O 64.0 60.0 48.0 28.0 Fig. 20. DETERMINA T2ON OF STRESSES. 43 this fraction and the support B the stress necessarily remains constant in each series. Scries VS, 8 gives the stresses when the load has advanced so as to cover the whole span, and which it is to be noted are those due to an evenly distributed load of 64 units. The greatest stress at the central vertical occurs when the load has just advanced up to it. Then at that vertical it is one quarter of the ultimate maximum stress at an abutment. Whereas with an equally distributed load covering the whole span the vertical stress at the centre of the span is nil. For let the weight of a load uniformly distributed over the whole span be 2 w, then the weight on each abutment would be w. Now the weight of the same load extending from an abutment over one-half of the same span would also be 10 with its centre of gravity at a distance of one-fourth of the span from the abutment. Therefore the weight on the further support and consequently the vertical stress at the centre of the span would be or in the example given - - = 8, Series VS4. The maximum stress which occurs at each given vertical is indicated in Series VS, Fig. 20, by two diagonal lines. From a line A B, Fig. 21, representing the half-span, set up at equal distances in relative order to an increased scale as ordinates these maximum stresses and through the extreme points thus found draw the curved line C D. Fi g . 21. Then as C A is the stress at an abutment, and D B that at the centre oF the span, following 44 THE BEAM. the curved line C D the figure C D B A proportionately repre- sents the theoretic horizontal sectional web area required for an equally distributed advancing load, and in the same way the triangle C B A proportionately represents the horizontal sectional web area required by the same load when extending over the whole span. The first of these areas is greater than the second. 41. Horizontal Stresses. Having already deter- mined as given in Series VS the vertical stress at each abutment due to each unit by unit advance of the load, com- mence from the support A, Fig. 20, and determine in succes- sion (31) the moment of horizontal stress at each given vertical as the leading end of the load becomes incident with it. Thus when the load has advanced to vertical 1, the vertical stress at B is .5, and this multiplied by leverage in units of length gives Series MHS 1. Again, when the load has advafc^ed from 1 to vertical 2, that portion of it will cause a vertical stress of 1.5 at B, and 6.5 at A, and each of these multiplied by their respective leverages gives the next or ah intermediate series, which added to Series 1 gives Series 2, or the stress at each unit of length when the load covers the portion of the beam extending from A to vertical 2. In the same way determine in succession the moment of stress at each given vertical as the leading end of the advancing load becomes incident with it until the whole beam o is covered. We shall* then have obtained Series MHS 1, 2, 3, 4, 5, 6, 7,8. The bracketed figures indicate in each series the termina- tion of the load's advance, from which point the stress CANTILEVERS. 45 diminishes in the unloaded segment of the beam in terms of arithmetic progression to nothing at the point of support. When the advance end of the load has arrived at vertical 8, the whole beam is covered with an evenly distributed load of 8 units to one of length, and the moments of stress are then directly as the multiple of the given segments (27). For instance, by dividing the length of the beam into 1G units, and then multiplying each pair of segments together, we should thus obtain the same results as those stated in Series MHS 8. It will be noted that the maximum horizontal stress at each vertical occurs when the load completely covers the beam, and that this has been shown in the last Article to be otherwise with vertical stress. CANTILEVERS. Suppose a beam instead of being supported at its centre to have one end built into a wall up to that point, then the remaining or projecting portion will be a cantilever (5). The projecting parts of cantilevers fall, therefore, under precisely the same static conditions of stress and strain, as either half of a beam supported at its centre. It is, however, desirable to note some collateral particulars attending the use of cantilevers. 42. As an example, let A B, Fig. 22, be a cantilever \\ units deep projecting 4 units from a wall at B into which it is continued for 2 units, and carrying a load of 9 units at its outer end A. Assume the resistance 46 THE BEAM. presented to the turning of the beam round a point at D by the superincumbent weight of the wall to be at C, then r-<-] C a vertical or shearing stress of i ; the load = 9 will be constant from A to B (21, 25), and as A B = 4 and B C = 2 there will I be an additional stress at D due to 9x4 Fig. 22. the resistance at C, = 18, but 2 18 9 = 9 (30) will be still the shearing stress at vertical 4, while at the same time 18 -f 9 = 27 is the total vertical pressure at D, tending to thrust down the wall in the direc- tion of the arrow. It consequently follows that as the resist- ance at C has been shown to = 18, there is an upward pressure of 18 at C as indicated by an arrow, tending to overturn the upper part of the wall about the point D. The moments of horizontal stress caused by the load are (24, 25) relatively at verticals 1, 2, 3, 4, 9 X * = 6, l.o - = 12, * -=~ = 18.- - = 24. and the moment of re- 1.5 1.5 1.5 action upon vertical 4 caused by the resistance of 18 at C = 18x2^ 24 ~T5~ 43. Assume the same cantilever, Fig. 22, to carry a load equally distributed from A to B. It will then with the same horizontal stress at vertical 4, which coincides with the assumed point of support, carry twice as much as when loaded at the end, or a load of 9 x 2 = 18 (29). 1 8 As there are -- = 4J units of load to 1 of length, the shearing stress at verticals 1, 2, 3, 4 will relatively be 4J, 9, 13J, 18 (28). Now the mean leverage of a distributed CANTILEVERS. 47 load measured from any given point is the distance of its centre of gravity from that point (27), therefore in this case the centre of gravity of the load being at vertical 2, and the fulcrum at D, it follows that the vertical resistance required at C = - - = 18, or the same as with a load of 9 at the end A of the cantilever. The total vertical pressure at D tending to thrust the wall downwards in this case is therefore 18 + 18 = 36, and the upward pressure at C tending to upset the upper part of the wall about point D will as before be 18. The moment of horizontal stress caused by the distri- buted load is (29) as the square of the distance from the end A, therefore 4 2 or 16 would represent this stress at vertical 4, but as the depth of the cantilever is 1^, the actual 18x2 stress will be - - - = 24 or, 16 x 1J at vertical 4, and i .0 therefore for the whole series of verticals 1, 2, 3, 4, they will relatively be 1J, G, 13, 24. It follows therefore that given a beam of uniform section of any length carrying a maximum central load, one-half of its length will as a cantilever carry at its outer end half that load, and if the cantilever were of the same length as the original beam, it would then carry one quarter of the original load at its outer end. In the same way, one-half of a beam will as a cantilever carry one-half of the distributed load carried by the whole beam when supported at each end, or if the cantilever were as long as the whole beam it would then carry one- quarter of that load. CHAPTER III. BEAMS WITH CANTILEVER ENDS. 44. Beam of Uniform Section placed upon Two Supports, and carrying an equal load at each end, and a load at its centre equal in weight to the two end loads. Determination of the most efficient positions for the supports. Let a c b, Fig. 23, be a platform carrying an equally distributed load of 12 units or 6 on each side of the centre, and let this be VS' \ms' o VS. 3 3 3:3 3 (-6) A///5. G 9 12 15 IS 15 12 BEAMS WITH CANTILEVER ENDS. 49 assumed to occasion a pressure of 3 units at eacli of the extremities a and b, and of 6 units at the centre c.* Let the bearers a and b be carried at the ends, and the bearer c at the centre of a beam A B. Then the beam A B will carry a load of 3 at each end, and 6 at its centre. Now the supports S, S of this beam would not be in the most efficient positions if placed at its ends, for although the beam would thus be relieved of all horizontal stress that might be caused by the loads there incident, that occasioned by the central load would be a maximum, and any advance of the supports towards the centre would not only diminish the span between them, but also convert the outer or projecting ends of the beam into cantilevers which it will be seen would further diminish the central horizontal stress. If the supports were so advanced the beam would tend to be deflected in the manner shown by the dotted line in the figure, the upper part of the beam over each of the supports S, S and the lower part under the central bearer of the plat- form being in a state of horizontal tension. The points at which this change of strain takes place are termed points of contrary flexure, x x. Fig. 23, the cur- vature which is upward on the one side of these becoming downward on the other. The horizontal stress decreases as they are approached, vanishing at them, and recommencing from nothing, again increases on the other side. It is obvious that the further the supports S, S are ad- vanced from A and B, the greater will be the horizontal tension over them, and the less that at the centre of the beam. * This will not be absolutely true unless the platform is divided at c, but under ordinary circumstances the effect of the rigidity of the platform may be disregarded. E 50 THE BEAM. The most efficient positions for the supports will therefore be those which will make the horizontal stress at each support exactly equal to that at the centre of the beam. Now the load of 3 units at each end of the beam will cause a vertical pressure of 3 upon each support S, and the load of 6 at the centre of the beam will also cause a vertical pressure of 3 at each. There will therefore be a vertical pressure of 6 upon each support, and an equal stress at the centre of the beam. But the pressure upon and the resistance of the supports are equal and opposite, and the latter may be regarded as a load acting upwards instead of downwards. The points of contrary flexure x x' may therefore be con- sidered as dividing the beam into three spans, A x, as x ', and a' B, the central span being supported at the points of contrary flexure, and each of the two outer spans by one of the supports S, S. In order, therefore, that there may be the same hori- zontal stress at the centre of each of these three spans, the loads at their centres being equal, the lengths of the spans should also be equal, and the supports should be placed at the centres of the two outer spans. Thus if the length of the beam when divided by units of its depth be 12, each span would be 4 units long, the points of contrary flexure will be at the verticals 4 and 8, and the most efficient position of the supports S, S at the verticals 2 and 10. The vertical stresses are given in Series VS, and the moments of horizontal stress in Series MHS. Fig. 23. 45. Weight of the Beam, Fig. 23, as a Plate Girder, with End Supports, as compared with BEAMS WITH CANTILEVER ENDS. 51 that with Intermediate Supports. The Scries VS' and MHS' have been appended to the diagram to show the amounts of vertical and horizontal stress which would exist if the beam were supported at each end instead of at S, S, and carried as before 6 load units at its centre. Assume the resistance of the material per unit of sectional area to be the same for all stresses. Then, as the amount of material required per unit of length would vary as the stress, add together Series VS' and MHS'. Deal in the same way with Series VS and MHS, and the totals will give the relative amount of material required for the two systems of support, which will be to each other as 144 to 72 or 2 to 1. Theoretically, therefore, two equally efficient girders when supported as shown, Fig. 23, could be made with the amount of material required for one girder of the same total length if supported at each end. Practically, however, the result for the intermediate supports would not be quite so favourable because for the sake of lateral rigidity it would not be desirable to reduce the tables of a girder to nothing at the points of contrary flexure. Now as small girders are usually made of uniform thick- ness throughout, and rolled joists necessarily so, it is for these only necessary to compare the central transverse stress due to each of the two systems of support because this the maximum will determine the weight of the beam per length unit. Thus the value at the central section, of VS + MHS =3+6 = 9 and VS' + MHS' = 3 + 18 = 21 and = ?. Therefore u O 7 girders supported as shown, Fig. 23, could practically be made under these conditions with the amount of material required for 3 girders each of the same total length and of the same efficiency, but supported at each end. E 2 5 2 THE BEAM. 46. To determine, as before, the most efficient position for the two supports, but under other conditions of loading. Let A B, Fig. 24, be a beam A. /; 2 ' 3 4 5 6 7 Si .9 10 // /2 73 /*! A5' /6 /). \ \ I k 1 > < t C 4 t i > ( ) 'i ? - ^ 6 3 < ) Fig. 24. of uniform section 1 unit deep carrying 3 load units at eacli end A and B, and 4 at its centre C. Now the two points of support S S 7 will necessarily be equidistant from C, and the central load of 4 will cause a vertical stress of 2 at each support, and consequently also the same stress at each vertical of contrary flexure x and x' (Series VS). But (44) A x may be treated as a separate beam placed on an intermediate support, and loaded in the present example with 3 units at A, and 2 at x. Therefore 3 + 2 = 5 is the pressure upon the support S. Make A x = 5 units of length, and by the equality of moments the distance of S from A and x will be inversely as the loading at those two extremities, and therefore A S = 2 and S x 3. The moment of horizontal stress in the beam at the support S will then be 3 x 2 = G. Moreover x x' when treated as a separate beam, with a vertical stress of 2 at x and at #', it follows, in order that there shall be a horizontal stress of 6 at vertical C, that x C should = x S. Therefore in this instance # C = 3, and 3 x 2 = 6, the horizontal BEAMS WITH CANTILEVER ENDS. .53 stress at C and also that at S. And the half length of the beam =2 + 3 + 3 = 8 units. Therefore, given a beam of uniform section loaded in this ratio, if we divide its length into 16 units the most efficient position for the two supports will be at 2 units from each end. Series MHS gives the moments of horizontal stress from end to end of the beam. In the same manner the most efficient positions of the supports, when assumed to be equidistant from a given intermediate load, may be readily determined for any given loads at the ends of the beam. 47. Beam of Uniform Section with Two Supports, and a Distributed Load. To find the most efficient positions for the supports. Let A B, Fig. 25, be a beam carrying an equally distributed load of 2 units to 1 of length over its entire length, and as it has been shown (44, 46) that the most efficient positions of the supports S S' cannot be at each end of the beam, let the cantilever ends A S and S' B be each one unit in length. Assume for the present that the two ends A S and S' B alone are loaded, and |< that there is no load between SanclS'. Then the moment Fi S- 26 - of horizontal stress over the supports S and S' will be 2 x .5 = 1 (43), and this stress will be continued between those verticals (33, Cor.), placing the upper part of the beam in a state of tension and the lower part in a state of compression. 54 THE BEAM. But -when the loading is continued between the points of support S, S', the beam will become deflected as indicated by the dotted line in the figure, and the horizontal stresses will be reversed in the central part of the beam, where the upper part will now be in compression and the lower in tension. It consequently follows that in order to neutralise and reverse the stress = 1 as first assumed to have been set up in that part of the beam an opposing moment = 2 will be required. Let y equal the distance between the support S and the centre C of the beam, Then 2y = the load between S and C, and %- = the leverage (27). 2 let if = 2 or the reversing moment required then y = x/2 = 1.4142. (1.) Now as A S = 1, so / C will also be 1 (44) ; there- fore the distance of each point of contrary flexure / or /' from the nearest point of support = 1.4142 1 = .4142, and the total length of the beam is 4.8284. With a given length of beam, say for instance 2 units, the relative distance of the points of contrary flexure from the nearest support will be - * 2 == .1716. An algebraic solution of this problem is given, Note (), p. 57. 48. As an example let A B, Fig. 26, be a beam 9.6568 units long, or twice the length already found (47J, and carrvinor a load of 2 to 1 of length. The length of each * O 3 r*> cantilever end will now = 2, each point of contrary flexure BEAMS WITH CANTILEVER ENDS. 55 will be .4142 x 2 = .8284 from the nearest point of support, and the distance between the points of contrary flexure will equal as before twice the length of a cantilever end or 2 x 2 = 4. A te ( i i ; i ' i & 31 4\ 5 6 r . . - | jr * 8\ it ? ^8284, ,.--.--., > 1 OOOx. ' 1 1 ~ " " 1 f 1?* ' ' --j' > 1 i *?1 - '** ^ ? ' i !<-- - ? 4 ' ^ Length units ,. VS. MHS. Fig. 26. Thus the length of the beam is 2 + .8284 + 4 + .8284 + 2 = 9.6568, and the points of contrary flexure are at verticals 3 and 7. Series VS [gives the vertical stress at each given vertical. Series MHS gives the moments of horizontal stress at the given verticals in accordance with 27 between ver- ticals 3 and 7, with 42 and 43 from verticals 2 to 3 and 7 to 8, and with 43 between A and vertical 2 and 8 and B. 49. Weight of the Beam, Pig. 26, as a Plate Girder, with End Supports, as compared with Intermediate Supports. Suppose this beam were sup- ported at each end and carried the same load of two to one of length, then the moment of horizontal stress at its central vertical 5 would be 23.3, and the maximum vertical stress would be 9.6, together 32.9, whereas, when the supports are placed as shown, Fig. 26, the maximum moment of horizontal stress is 4, and the maximum vertical stress is also 4, together 8. Therefore assuming, as in small girders, that the same transverse section is maintained throughout, four equally efficient girders could be supplied under the latter condi- 56 THE BEAM. tions from the amount of material required for one girder of the same length to carry the same load, but supported at each end. 50. Practical Examples. Paddle boards or " floats" of steam vessels, subjected to an uniformly distributed pres- sure when in action, are beams of this order when their ends project beyond the paddle wheel arms to which they are attached, the paddle arms being the supports or points of resist- Fig. 27. ance. Mr. Field's rule* for floats attached to two arms was to divide the length of the float into five equal parts, and to place the arms three apart, thus leaving a projection for each end of the floats of one part beyond the arms. Tn the case of three arms to each float, the length would be divided into eight equal parts, one arm being at the centre, and the other two each one part from the end of the float (Fig. 27). For as the conditions of stress of the two intermediate spans in the diagram B are precisely the same as those of the one span in diagram A, their relative lengths to that of the cantilever ends will remain the same. This good practical rule leaves the ends or most exposed parts of a float proportionately a small fraction shorter than theory w r ould determine in providing that the moment of stress in the float at an arm of the wheel and at the centre between the arms shall be equal, * The late eminent member of the firm of Maudslay, Sons, & Field, and formerly President of the Institution of Civil Engineers. BEAMS WITH CANTILEVER ENDS. 57 whereas by the given rule the stress at an arm and at the centre between the arms is 1 to 1J.* For if a be the length of each cantilever end of an uniformly loaded beam, and b half the length of an intermediate span, the stress at a support when compared with that at the centre of a span will be as a 2 is to IP a 2 . See 29, 47. The same considerations, both theoretical and practical, are applicable in determining the position of ledges for doors, and of hinges and fastenings for gates and valves which may have to resist a pressure of wind or water. NOTE (a.) Referring to Fig. 25 (47), let A B = 2L, S S' = 21, AS = S'B = l' t then L = I + I' and let the load upon each unit of length be regarded as the unit of load. Let//' = 2x, then the vertical stress at /occasioned by the weight of the load on//' = x, the load upon/ S = I x, and the load on the cantilever A S = V, and their respective moments about S will be (I i?Y^ I'* x (I a?), . and . 2 2 Therefore the moment of horizontal strain over the support S about S 7/0 n y,\t = or = a? (I x} + v - , and that of the horizontal strain at the x~ x" 1 centre of the beam = x"- -- = . But these must be equal /. x = V and *' = Ix - x- + ~ Ix + ~ 2i l i and I = ^/2l' L = (1 + x/2) V I = L = (v/2 - 1) L = .4142 L. That is to say, half the length of the beam being 1, the most efficient position of the supports S and S' is when their relative distance from the extremity A or B is .4142. And as x has been found = Z', or A S in Fig. 25, and 2 being the length of the beam, the distance of//' from the supports S S' = 1 .4142 X 2 = .1710 as stated, (47). * In a float when regarded as a beam, any theoretic excess of efficiency at a paddle arm would, however, be slightly diminished by the necessary bolt holes- CHAPTER IV. CONTINUOUS BEAMS. 51. The static principles which apply to beams ex- tending continuously over two or more spans are further developments of those which determine the stresses in beams with cantilever ends (44, 46, 47, 48). Of the whole load, however, upon any one of the spans, the fraction carried by each support, or the position of the points of contrary flexure are not so readily found. When this, however, has been done, a continuous beam of two spans may be regarded as virtually divided at those points into three distinct spans, 1 anc ^ a beam of three T spans into five, and A I C, I /? I | the vertical and X * k -- %'- i horizontal stresses Tig. 28. may then be easily ascertained throughout each of these spans constituting the beam. Let A B, Fig. 28, be a continuous beam supported at each end A and B, and also by an intermediate support C, thus extending over the two spans A G and C B. Now suppose the two spans each to carry either a concentrated or a distri- buted loud, they will then be deflected downwards by the CONTINUOUS BEAMS, 59 load, tlic horizontal stresses will vanish at the points of con- trary flexure / and / in consequence the reaction of the inter- mediate support C, and become reversed beyond these points in the same manner as by the central portion of the load in the case of beams with cantilever ends, so that the beam will have a tendency to take a curved form, as shown in the Figure. In the case of a beam of three continuous spans, Fig. 29, supported at each end A and B and by two intermediate TYT -i- j f . / u Fig. 29. supports C and C', there are, or may be, * four points of contrary flexure////, two of these being in the central span. The positions of the points of contrary flexure are de- pendent upon the relative lengths of the spans, the nature and extent of loading, and the fraction of it transferred to each support by deflection. These positions and the fraction of the load carried by each support cannot therefore be determined simply by the principle of the lever, but only by the solution of more complicated problems. The following simple formulas based upon such solutions are, however, applicable to various conditions of continuous beams. 52. The simplest mode of treating the four following con- ditions of continuous beams of two spans is to determine by * Not necessarily. Hi the intermediate supports arc very near together thorc will l)e no contrary flexure between them, 60 THE BEAM, formulae (2, 3, 4, and 5) the position of the point of contrary flexure in each span. When this has been clone the beam, as has just been shown, becomes virtually divided into three spans. The fraction of the load carried by each of such spans n:ay then be easily found and the strains at any given verticals ascertained by simple calculation (54, 55, 57, 58). For Two Equal Spans with Equal or Unequal Central Loads. Referring to Fig. 28 (51). Let Z = A C = C B = the length of each span. w = the load at the centre of the span A C. 10' = the load at the centre of the span C B. x = the distance of the point of contrary flexure from A , x = the distance of the point of contrary flexure from B. Then x = - 16w I For equal loads, (2.) when w = w' 1C>W ' x - x' - 1<3 S For Two Equal Spans with Equal or Unequal Distributed Loads. Let w the load per linear unit of A C. w'= the load per linear unit of C B. * Sec 54, 55. CONTINUOUS BEAMS. 61 mi LW W 7 lllCll X = - -I 8w , 7 w ' w 7 For equal loads, when w = w' / 6 7 3 x = x = I = ~ (3.) For Two Unequal Spans with Equal Central Loads. Let Z = the length of the span A C. I' = the length of the span C B. L = AC + CB = the two spans. Th SH + MJ'-H*) / 8 L Z * 8 L I' + 3 (/ 2 + Z /2 ) Unequal Spans with an Equally Distributed Load. 4 I I (5.) Then ^ = 8 it' - The two following cases of continuous beams of two spans are most simply treated by first finding the portion of the loading taken by each of the end supports A and B. This * Sec Formula (11), and 57, 58. 62 THE BEAM. having been done the load carried by the central support is known, and consequently the points of contrary flexure. With these data the stresses throughout the beam may be readily determined. For Two Unequal Spans with Unequal Central Loads. Let w = the load at the centre of the span A C. w' = the load at the centre of the span C B. Then the portion of) w _ 3 (w I 2 + w I' 2 ) the load on A J 2 " ~T(TLT~ onB = For Tiuo Unequal Spans with Unequal Distributed Loads. Let W = the total load on the span A C. W = the total load on the span C B. Then the portion ) W VV / 2 + W I' 2 of the load on A) " " 2 8 U W WZ 2 + WZ' 3 onB= ^---8Trr- (7.) CONTINUOUS BEAMS. 63 53. For Three Spans, the Two Outer Spans being equal, with an equally Distributed Load throughout. Referring to Fig. 29, (51.) Let I = the length of each of the outer spans A C andC'B. 2 I' = the length of the central span C C'. w = the load per unit of length, w '3 I 3 + 1-2 1*1'- SI' 3 - Then the load on A or B = -g j n/ wnl5Z 3 + 60Z 2 Z' + ZZ' + 24Z 3 / on or = ^ ^ ^^ r 2" -f- O t t X / q \ For Three Equal Spans, or "21' = I. (9.) 4 Then the load on A or B = wl on C or C' = -TQ w I With beams of more than three equal spans a =.28 1 may be used for the intermediate spans, and a'=.Sl for the end spans, as the differences from the true values are so small as to be practically unimportant. When the Outer Spans are equal, and the Central Span equal to the tivo, or I' = /. Then the load on A or B = r, ic I. n , 57 , (10-) on C or C = - wl. 64 THE BEAM. Jf l' = o (reducing the spans to two).* Then the load on A or B =-^ wl. 5 C 11 -) on C and C' together = rw I. A Continuous Beam of Two Spans with a Load at the Centre of each Span. 54. Vertical Stress. Let A C, Figure 30, be one span of a continuous beam of two equal spans of uniform section, each 22 units long, and one deep ; and each loaded at the centre with 16 units, C being the central support dividing the two spans. Under these conditions the vertical of contrary flexure /is Verticals, i 2 3- 4. .$ 6 7 s 9 to n 12 is u is 16 n is 19 2021 22 21 20 FR^M^4 x ^^^^^^^^^^^^^"f "T : ' i : W A i I i i i i i . ^.,.,.p.^......^. ; ; ! ! ! ! I ; i n&n a. 8 8 8 8 8 S 8 S 888 8tt S $ 8 8 8 8 8 8 S 8 818 S 8 b. 3 . 3 3 A 2 3 3. 3J3 1 1 ;/ n n n n 11 n u 11:11 u n ; =. =3 -i= =^=r = = = '= ?.' =s = i i i i i i i i i i i i i i. i i i i ' i i i l i i i i r i i i i i ' ' ' ! 1C. * 16 '24 32 40 48 5664 72 80 88 80 72 fi4 56 48 40 32 24 16 8 8 16 [d. 6 _ 12 15 1S_ 21 2J 27 SO 33 36 39 42 45 48 51 5f 57 60 a MHS. 10 75 202\W 3 40 4> 50 &> 44 3 22 22 23 M 55 6655 44 See also Formula (3.) CONTINUOUS BEAMS. 65 at 16 units from the outer support A, and consequently 6 from the central support C, Formula (2), (52). A / may therefore be treated, as indicated by hatching in the figure, as a separate beam carrying a load of 16 units at vertical 11, and supported at its two ends A and /. There would thus be a vertical stress of 5 at A and 11 at/ (30), the 11 at / being continued to C, meeting at that vertical the same amount from the other half of the beam. Now if the beam were non- continuous at C, the load of 16 would cause a vertical stress of 8 at A, and 8 also at C, and at all intermediate points between A and C (Series a). But being continuous with a vertical stress of 5 at A, and of 11 at / and at G, it follows that owing to the reaction of the central support (Series I) the vertical stress becomes 8 3 or 5 from A to vertical 11, and 8 + 3 or 11 from vertical 11 to vertical 22 at the central support C (Series YS). 55. Horizontal StreSS a Assuming the span A C to be non-continuous at the central support C, then the moments of horizontal stress due to a load of 16 at vertical 11 are, as given, Fig. 30, Series c, (24). But it has been shown that when the two spans are continuous, the reaction of the central support C relieves the support A of f of the load carried by it in the former case, and thus Series d gives the moments of horizontal stress due to that reaction. Subtract at each vertical between A and / the amount stated in Series d from that in Series c, and as at / the stresses are reversed, subtract Series c from Series d between / and the central support C. We thereby obtain Series MHS, giving the moment of horizontal stress at each relative section of the continuous beam. It will be F 66 THE BEAM, observed tliat at the vertical of contrary flexure f the value of MHS is 0, the factor values (Series c and d) which are equal, having been deducted one from the other ; also that / C has become a cantilever supported at C and carrying a load of 11 at its end/, and that the greatest horizontal stress is 66 over the central support ; whereas when the span A C was assumed to be non-continuous at the central support C, the greatest horizontal stress was 88 at its central vertical 11. The stresses will of course be the same respectively at the same relative verticals of the other equal and equally loaded span or second half of this continuous beam which is not shown in the Figure. 56. Comparative Weight of a Beam as a continuous and as two non-continuous plate girders extending over two equal spans and carrying an equal load at the centre of each Span. Firstly, the strength and consequently the weight of the web at any section will theoretically be directly as the vertical stress which it has to resist, and referring for example to Fig. 30 (54), Series a and Series VS, it will be seen that when the girder is non-continuous there is a vertical stress of 8 units throughout the whole span of 22 units, whereas for the continuous girder there is a stress of 5 throughout 11 length units and of 11 throughout the remaining eleven units. Then 22 x 8 = 11 x 5 + 11 x 11 = 176 x 2 = 352 So that the total quantities in each case are equal. Also the strength and weight of the tables at any section will likewise in either case vary directly with the horizontal stress as given per unit of length in Series c and Series MHS. CONTINUOUS BEAMS. 67 Then adding together the numbers in each of those Series we obtain for the non continuous girder, Series c, 968 x 2 = 1936 for the continuous ,, MHS, 638 x 2 = 1276 Complete the comparison in each case by adding the sum representing the weight of the tables to that representing the w T eb. Then for the non-continuous girder 1936 + 352 = 2288 for the continuous 1276 -f 352 = 1628 or as 4 to 3 nearly. Therefore the weight theoretically required for the beam when continuous is J less than that of two non -continuous beams of the same strength, covering the same spans. Practically, the weight of girders will of necessity always exceed that determined by theory, and this for a variety of reasons, as w r ill be seen by subsequent examples. A Cantinuous Beam of Two Spans with a Distributed Load. 57. Vertical Stress. Let A. B, Fig. 31, be a con- tinuous beam divided by a central support C into two equal spans, each 8 units long and 1 deep, each span carrying an equally distributed load of 16 units. Under these conditions the verticals of contrary flexure //' are at a distance of 6 units from either end of the beam or f the length of each span measured from the outer support, Formula (3), (52) and the two halves of the continuous beam may be regarded as three distinct beams A/, //', and /' B, F 2 68 THE BEAM. as indicated by hatching in the figure. The first and third span each carries a distributed load = 12 causing a vertical stress = G at A,/, /', and B. The second span //', conse- Verticals ra. I ft. vs. C. 8 2_ I I O 7 -72 4 ,5 tf 4-44 7 ,9 7 5 4 2 1 \ \ 4 2 JL JL 2 12 15 4 6 i i I i ! 2 2. 2 1 i i T* i i #J 15 > : 222 JL JLI<> :IO . I ! I 12 7 7 i I i i i i i I- U/ I ' ' I T, <7 i j i i i \ B i i i i i i i i i i I 46 S'.S 6420246 S i i i i i i i i i i i 75 1$ 75 12 7 10 8 6 4 2 _0 Fig. 31. quently carries a load = 6 at each end, and also on each arm 4 units of the distributed load, the load therefore on the central pier C = (6 + 4) x 2 = 20. Supposing the beam instead of being continuous to be separated at the central vertical 8 into two distinct beams, then Series a, Fig. 31, gives (26) the vertical stresses for each of the two beams, the stress at each of their ends being 8 units. But it has been shown that in the continuous beam the stress at A or B is 6, and that at the central pier C is 20. As in the previous example, this alteration in the stresses is effected by the reaction, Series I, caused by the additional load thrown upon the central pier, by which in this instance CONTINUOUS BEAMS. 69 Series a becomes reduced by 2 from A and B to the verticals f and f' 9 and increased by the same amount between those verticals, Series YS. 58. Horizontal Stress. As in the previous example, the moments of horizontal stress (27) in the two spans A C and C B when separated at the central vertical are given in Series c, and as it has been shown that the addi- tional amount of vertical reaction due to the central support is 2 at each end of the continuous beam, Series d gives the consequent moments of horizontal reaction (24) throughout its length. The differences between these two series con- stitute the Series MRS, or the moments of horizontal stress at each assumed vertical throughout the continuous beam. At the verticals of contrary flexure they are nil, (55). 59. Comparative Weight of a Beam as a continuous and as two non-continuous plate girders of two equal spans carrying an equally distributed load. We assume as iii Art. 56 the weight of each member of a beam of a given span to be directly as the stresses its members have to resist. Thus in this instance for the web multiply the mean of the stresses given in Series a, Fig. 31, and the mean of those in Series YS, by the length in each case over which each mean thus taken extends. Then for the web of the non-continuous beam (Series a) 4 x 8 = 32 x 2 = 64. for the web of the continuous beam (Series YS) (3 x 6) + (8 x 2) = 34 x 2 = 68. For the tables add together the numbers in Series c, and also those given at each unit of length in Series MHS. 70 THE BEAM. Then for the non-continuous beam (Series c) 84 x 2 168. for the continuous beam (Series MHS) 50 x 2 = 100. Adding together these results respectively we obtain for the non- continuous beam 168 + 64 = 232, for the continuous beam 100 -f 68 = 168, or as 4 to 3 nearly. Or as in the example given in 56 the weight required for the continuous beam is about J less than that of two similar non-continuous beams of the same strength. 60. Comparative Deflection of a Beam when non- continuous and continuous. Assume the beam, Fig. 31, to consist of a top and bottom table, connected by a web, and whether non-continuous or continuous to be of uniform depth and of uniform breadth throughout, but to have its parts so proportioned in thickness as to meet all stresses with an equal resistance per unit of section. Then the linear extension or compression caused by the load might, under each condition, be represented by 8, the number of units in the length of each actual span. But continuity has been found to divide the beam into three virtual spans, the two outer of which are each 6, and the central span 4 units long; note also that the versed sines of comparatively flat segmental curves vary as the squares of their chords. Therefore, under these conditions, the comparative deflec- tion of each actual span may be represented in the following way : j When noii-continuous 8 2 = G4 When continuous 6 s + 4 2 = 2 or as 5 to 4. CONTINUOUS BEAMS, 71 Therefore the maximum deflection will be 4-th less when tho two spans are covered by a continuous beam than it would be were the same load carried by two equally efficient but separate beams. It has been shown, therefore, that the ratios of weight and deflection are in favour of the continuous beam. In practice, however, the ratio of weight would be somewhat less favour- able, for, in order to insure lateral rigidity, the tables could not be reduced to nothing at the verticals of contrary flexure, as would be the case if proportioned to the horizontal stresses, CHAPTER V. STRENGTH OF MATERIALS. 61. Experiments on the strength of materials and their behaviour under stress of all kinds gradually increased to the moment of complete fracture, have from time to time been carefully carried out by experienced and accurate observers, assisted by the most advanced and complete appliances. The results of many series of such experiments have been carefully tabulated, averaged, and published. Moreover the process of simply testing the ultimate transverse and tensile strength of iron and steel is an every-day process, necessary for the purpose of ascertaining whether the quality of material is equal to stipulated standards of efficiency. Very many valuable data are therefore available in determining proper standards with practically sufficient accuracy. The transverse strength of any constructive material is usually experimentally determined by placing a succession of bars of given scantlings* upon or against two points of support or resistance, adjusted to a given distance apart. Pressure is then applied separately to each bar at a point midway between the points of resistance, and in a direction at right angles to its length. The pressure may be effected simply by the gradual appli- * A term used for dimensions other than length. STRENG TH OF MA TE RIALS. 7 3 cation of suspended known weights until the bar is broken. But it will be obvious that the gradual and at the same time self-registering pressure obtainable by the use of a hydraulic ram and pressure gauge is preferable for ordinary tests, especially when the condition of the bar is approaching fracture. Further, by the use of such an apparatus, tests may be made with much greater expedition. On the other hand, although the gauge if correct will give the pressure of the water within the cylinder of the ram, the actual stress to which the object is being subjected is the pressure indicated by the gauge multiplied by the area of the ram, less the friction of its hydraulic packing. The most accurate mode of testing is therefore with a lever machine,* in which friction is reduced to a minimum by the fulcrums of the levers being all fitted on the knife-edge scale beam or steelyard principle. The weight causing the trans- verse stress on the object to be tested is moved by a hand- wheel and band along a graduated" lever, so that the exact amount of the stress upon the test bar can be at once read and noted at any period of the operation. Tensile strength is ascertained by securing each end of the bar to be tested in the testing machine, by means of which it may be placed in a state of tension, and the stress gradually increased and continuously registered up to the required test, or until the bar is torn asunder. Means are also provided for accurately measuring the extension and the deflection of a bar during a test. * As cat Messrs. David Kirkaldy and Son's establishment and Museum of Experimental and Accidental Fractures. 99, Southwark Street, London, by whose machine materials may also be tested under tensile, compressive, shearing, and twisting stress. 74 THE BEAM. 62. Coefficients of Efficiency. The figures given in columns c and c ' . n and n of the following table are com- 7 O puted at not more than one -fourth to one-fifth of the average ultimate strength of steel, iron, oak, and fir, of the descrip- tions generally used in the construction of beams, girders, and other structures. These arc termed coefficients of efficiency. TABLE 1. Nature of Strain. Material. Bar 1 in.by 1 in. Square. Bar lin. Diameter. C c' Tons. Tons. I Rolled Steel 6.50 5.20 Tensile compressive and shearing < ,, Iron 4.50 3.53 Tensile and shearing Cast Iron 1.40 1.10 Compressive V 6.00 4.70 Tensile and compressive . Oak .50 .39 ... Fir .40 .31 / n n' . Rolled Steel .60 .374 Iron .40 .245 Transverse; Bars one foot between Cast Iron *.175 .112 the supports and loaded at their/ Cwt. Cwt. Oak Fir 1.00 .80 .60 .50 Landings j Hard York \ \ ) The ultimate compressive strength of wrought-iron plates and bars is when compared with their tensile strength about as * =x 3} cwt, or for small work, say, 400 Ibs., for machinery 200 Ibs. STRENG TH OF MA TE RIALS. 7 5 4 to 5. But the same co- efficient of efficiency may be used in cither case, and especially for girder work. For rivet and bolt holes diminish the efficiency of sections in tension, but not to the same extent, that of those in compression, inasmuch as the rivets or bolts fill up the holes and so resist compression. It is usual therefore to simplify and expedite such work by providing the same net sectional area for the resistance of the same amounts of stress^ whether tensile or compressive. As the tensile strength of iron plates is 10 per cent, more lengthways in the direction of the fibre, or that in which they were rolled, than the strength crossways, they should be placed lengthways when used for resisting that stress. Shearing efficiency is equal to tensile in rolled iron plates and bars. In steel compressive efficiency is fully equal to tensile, and although its ultimate shearing strength is one quarter less than its tensile, the coefficient 6-*- tons given in the table is practically a well- covered allowance of safe shearing efficiency. Engines and machinery with members in motion are subject to somewhat indefinite abnormal strains. The maximum cal- culated stresses therefore which their parts may have to resist per square inch of sectional area should in no case exceed one-half of those allowed for ordinary fixed structures, while the materials used in their construction should be of the best descriptions. 63. Crushing Resistance of Materials. The results of the various published series of experiments on the resistance offered by building materials to crushing stress vary in many instances very considerably, and moreover the strength of a pier or wall of brick or stone decreases with its 7 6 THE BEAM. height, by reason of increased tendency to bend. No abso- lutely definite coefficients of safety can therefore be laid down. But any amount of stress not exceeding that given in each case in the following Table will, under ordinary constructive conditions and requirements, be practically well within the limits of safety. TABLE 2. Safe Loads for Earth and Materials. Per Square Foot. 1 Ton. 3 Tons. For Lias Liine Concrete, 6 of Ballast, and 1 of Cement . Portland Cement Concrete. 8 of Ballast, and 1 of 6 } 7 6 in Lias Lime Mortar .... in Portland Cement .... 7 8 30 20 Hard York Stone and Landings .... 20 For important works it is usual to determine the practically effective supporting strength of the materials intended to be used, by special experiments upon samples tested under con- ditions relative to the proposed work. WEIGHT OF MATERIALS, 77 WEIGHT OF MATERIALS. 64. Iii taking out quantities from drawings either for the purpose of calculating stresses caused by the weight of a structure or for making estimates of cost, a ready and expe- ditious mode of proceeding with iron or steel is to bring castings, plates, angle, and T iron into square superficial feet measure, termed feet super, arranged according to thick- nesses.* The weights of the usual thicknesses are given in the following Tables : TABLE 3. Cast Iron. 1 Foot cube = 4 cwt., therefore 5 feet cube = 1 ton, and Thickness in inches - ,1 _ = cwt. per foot super, thus : / 3 inches thick = 1 cwt. 2 ?? = 1 i i 1 If 5? 5? - '/ J > 1 ! 5 It JJ 5? ' T2" 5? 1 ' == 3 1 5? 7 7 One foot super ' H ?? J JJ JJ = i 4 ',, 5 5 8 ? ' J5 YT ?? 1 j_ "2" ?' JJ ~ G " 3 1 8 ?? 5? 8 5? f i JJ = A U JJ _1_ * Althougli in many handbooks tables may be found of the weight of angle and T iron per foot run. 7 8 THE BEAM. As dimensions arc usually given in feet, inches, halves, quarters, and eighths, duodecimals are far preferable to decimals in taking out quantities, because with the former far fewer figures are required, and J or cannot be expressed by simple decimals. Plate Iron. One foot cube = 480 Ibs., therefore 4 feet cube = 1 ton, and 1 inch thick = 40 Ibs. per square foot. I = 30 i =20 1 K F ?? j? j> ?? In girder work 2 or 3 per cent, added to the weight for rivet heads would generally be sufficient, but it is usual to add 5 per cent. Steel Add 1 per cent, to the weight of plate iron. Timber. Per foot cube ... Oak, .5 cwt. ... Fir, .4 65. Data for estimating the Loading of Structures. TABLE 4. Weight of Materials. Sfc. Per foot cube. Brickwork in lime or cement 1.00 cwt. J? ,, Gravel or ballast . . 1.25 ,, ,, ., Portland cement concrete . 1.25 ,, ,, ,, York or Portland stone . 1.40 ,, Granite 1.50 ,, WEIGHT OF MATERIALS, tfr. 79 Per square foot 6 ins. thick York or Portland Paving . 70 cwt. 3 ins -35 Per square 10x10 =100 square feet plain Tiling 12.00 pan Tiling 8.00 Slating . 7.00 ,, ,, | -inch deal boarding . . 2.50 ,, Per square foot. A Crowd of persons * . 1.50 House floors, persons, and furniture . . 1.50 Store and Warehouse floors, loading . . . 2 to 4 , , Wind probable maximum f ) ^ pressure in the British Isles j Per foot cube fresh Water . . . .62.5 Ibs. 36 feet .,.- Iton + lOlbs. Per foot sea Water . . . . 64 Ibs. 35 feet - . . .1 ton. Per square foot. A fall of Snow per each inch j - ,, of depth before becoming consolidated . J * Careful experiments show that a crowd of men of average weight closely packed weigh 1 cwt. per square foot of the area covered. The adoption of 1-J cwt. per square foot in planning a house will therefore leave a margin of strength for further contingencies. f By a Board, of Trade regulation for the stability of viaducts and high bridges, " the work must be such as will provide for a wind pressure of 56 Ibs. on the square foot." This pressure was fully provided for in the Forth Bridge by Sir Benjamin Baker. Special trussing was provided at the Crystal Palace, Sydenham, for a wind pressure of 25 Ibs. per square foot, and even with that there would be G6 tons pressure upon the area of the semi-circular part, of an end of the large transept, which has now stood for over five-and- thirty years. The ultimate strength of this trussing is, however, four times that provided, so that a hurricane of 100 Ibs. per square foot would have to occur in order to cause anv serious failure. 8o THE BEAM. Bridges carrying public roads should be designed to carry the following Ipads : The weight of the structure platform and roadway together with a distributed load of 1J cwt. per foot super over the entire platform. Also a maximum concentrated rolling load upon any parj; of the platform, but not in addition to the distributed load. For the latter, in the case of town roads, a load of 32 tons on four wheels including thp trolley, or 8 tons on each wheel, would be a practically good allowance, assuming the wheels to be 10 feet apart from centre to centre longitudinally, and 4 feet apart from centre to centre transversely.* f Half the above weight or 4 tons on each wheel taken at the same distances a/part would be a sufficient allowance for traction engines in the case of country roads. For bridges Carrying railways, although no absolute rule has been laid down by the Board of Trade, the test rolling loads for each line or pair of rails usually adopted have been per foot run about 1-J ton up to a 50 feet span, 1 ton from 50 to 100 feet span, and 1 ton from 100 to 150 feet span. * The load adopted by Sir Joseph Bazalgette 15 years since. f Messrs. Maudslay, SODS, and Field frequently carry from 32 to 35 tons on their large four-wheel trolley. They have had 43 tons upon it, to which add 5^ tons for the trolley and 1 ton for chain, making in all 49 tons or 12f tons on each wheel. These arc, however, for roads or streets, concentrated loads of unusual weight. CHAPTER VI. SOLID BEAMS. 66. The Transverse Strength of Solid Rect- angular Beams or bars varies directly as their breadths and as the square of their depths (19, 20), and that of cylindrical as the cubes of their diameters, and for any sec- tion inversely as their lengths. The latter factor is wholly one of leverage. Thus assuming a beam to be supported at each end, and its transverse section to be uniform throughout its length, the stress at any segmental transverse section would vary in direct proportion with the length of the beam. It therefore follows that as the transverse strength of plain solid sections of constructive materials has been determined by direct tests (61), the sections of such rectangular and round beams required for given loads, and the loads which will be safely carried by beams of such sections, may be readily found by the use of the following formula}. Let b = the breadth n _ Table 1 (62), d depth I = length in feet between supports. n = coefficient for rectangular bars ri = ,, circular ,, w load at centre of span. G 82 THE BEAM. For Central Loads, 67. Required the load that a beam of oak 12 feet long between the supports and 12 inches by 12 inches section will carry at its centre. Then = x 12 2 x 1 - = 144 cwt. Required the breadth of a beam of fir 15 feet long between the supports and 10 inches deep to carry a load of 40 cwt. at its centre. Then b = i < 13 ') d~ n 40 x 15 r., i = -lonrrso = '* mches Required the depth of a bar of cast iron 4 feet long between the supports and 2i inches brqacl to carry a load of If tons at its centre. Then d = A/^ 1 ( 14 ') V b n /1| x 4 . . , = A / ~r ^TTTE 4 inches. V 2.o x .175 For Distributed Loads. 68. It has been shown (27) that if a beam will carry w at its centre, it will carry 2w evenly distributed over its span. Formula? (12), (13). (14) become therefore modified in the following manner for distributed loads : Let w = the weight of a distributed load. SOLID BEAMS. 83 To find the total safe distributed load for a beam. Then w' = 2bd 2 n (15.) I The length, depth, and loading being given to find the efficient breadth of a beam to carry a total distributed load. ,-p, w' I (16.) ihen b = - 2 dr n The length, breadth, and loading being given to find the efficient depth of a beam to carry a total distributed load. Then d = ./J" C 17 ') V 2 b n For Round Beams. Let d = the diameter. 69. To find the load that a bar of rolled iron 4 feet long between the supports and 4 inches diameter will carry at its centre. m, d 3 n' (18.) Then w = - i 4 3 x .245 = - = 3.02 tons 4 Required the diameter of a beam of cast iron 8 feet long between the supports to carry a load of 14 tons at its centre. ~ . , 3 714 x 8 ~~V -"112- For round beams with distributed loads w', substitute for d fo a 2 Equation (18) w f ^ 2 - d * n ' , and for Equation (19) d = i n 84 THE BEAM. 70. The use of the formulae given (67, 68, 69) may be readily extended by noting the following conditions : When supported at its centre a beam will carry at each end one-half of the safe central load (25), and per unit of length the same distributed load as when supported at each end (29). As a cantilever a beam of any given effective length will carry at its salient end one-fourth of the central load, and also per unit of length one-fourth of the distributed load it will carry when supported at each end (43). With a load at any given vertical in the span transverse stress will be greatest at that vertical , and will vary as the multiple of the separate lengths of the segments into which it may divide the length of the span (31). Or multiply the length of one segment by that of the other and divide the product by one-fourth the length of the whole span, and the quotient will be the effective length with the load placed at its centre. Thus if I' I" be the length of the segments, the load carried will be inversely as I the length of the beam. 4 I' I" When I = 7 71. The Nature of Transverse Strength was the subject of careful research by Mr. Robert Stephenson, by whose extensive experiments and the deductions to which they led the knowledge of it was much advanced. * The investigations of Mr. Stephenson were followed by those of Mr. W. H. Barlow, who published an account of * " The Britannia and Conway Bridges," by Edwin Clark, 1 850. SOLID BEAMS. 85 a series of elaborately accurate experiments and of the de- ductions which he based upon them.* The results of these labours have since been practically elucidated and extended by Sir Benjamin Baker, t Before Mr. Barlow's solution of this problem two elements only of transverse strength were recognised as existing in a / O rt \3 beam, the one being the resistance it is capable of offering to direct extension and the other its resistance to compression. But the efficiency of these resistances was found quite inade- quate to account for the amount of strength practically evinced by solid beams when measured by or compared with the results of experiments on the direct tensile and compressive strength of materials. In this comparison the neutral axis was assumed as passing through the centre of gravity of a beam. But in order to obtain a moment of horizontal tensile resist- ance sufficient to account for the strength of a solid cast-iron beam the neutral axis would have to be at or above the top of its transverse section. That the neutral axis passes through the centre of gravity of a beam of any form of section whatever, provided all the forces acting on the beam are applied in a direction perpen- dicular to that axis, and that the limits of perfect elasticity are not exceeded, had already been mathematically determined. The consideration of these facts led Mr. Barlow to conclude that another cause existed for the amount of transverse strength as exhibited by solid beams, that the particles of a bar of iron when under transverse stress, besides being in tension below the neutral axis and in compression above it, must be subject to a differential molecular lateral movement among themselves, * ; ' The Philosophical Transactions of the Royal Society," 1855 and 1857. f " On the Strength of Beams, Columns, an'l Arches," by B. Baker, 1870. 86 THE BEAM. and that this movement would necessarily bring into action a corresponding amount of resisting energy, which he named " the resistance of flexure." Thus referring to 18 and Fig. 6, it will be evident that O o / in a beam subjected to transverse stress, horizontal stress commences at the neutral plane and then increases in a direct ratio to a maximum at the top and bottom of the beam. As the stress increases so will the differential molecular movement of the particles of the beam and their resistance to that move- ment also increase. Now there can be no such movement among the particles of a bar under direct tensile or compressive stress, equally affecting the whole of the section strained, for its strength then entirely depends upon the direct resistance it is capable of offering to either of those stresses. But in the case of bars under transverse stress, their strength then depends not alone upon the resistance they are capable of offering to direct tension and compression, but further to that which is due to their resistance to bending or flexure. For Mr. Barlow's experiments showed that the calculated tensile stress per square inch at the lower side of cast-iron bars subjected to transverse stress exceeded, before rupture took place, that which experiments on their extension by direct tension proved them to be capable of resisting in the ratio of nearly two to one. This difference was too great to admit of the old workshop notion being any longer entertained that it is caused by the superior strength of the skin. Besides, in a direct tensile test, the resistance of the whole of the skin is brought into operation, whereas in a square bar under trans- verse stress one-half only of the skin is in tension, the remainder being in compression. But in a cast-iron bar under trans- verse stress there is a considerable excess of strength in the STRENGTH ELEMENTS IN SOLID BEAMS. 87 part in compression over that in tension quite independent of the resistance of flexure. Besides, experiments on the transverse strength of bars with the skin removed have shown that the skin itself practically possesses no superior strength. It may be open to question whether this increased resistance is or is not attributable to the lateral action of the particles, but in any case it will require to be taken into consideration in the solution of questions relating to transverse strength. * STRENGTH ELEMENTS IN SOLID BEAMS. The throe following examples illustrate the theory of the elements of transverse strength briefly delineated in the last Article. Practically, however, the efficiency of solid beams is best dealt with by formulae based upon actual strength tests (66, 67, 68, 69) in use long before any real advance had been made in the theory of transverse strength. 72. Given a Cast-iron Bar squarely strained 1 foot 6 inches long between the supports, and 1J in. x 1| in. section, to find the breaking load or weight, b w, when carried by the bar at the centre of the span. Eeferring to Table 1 (62), take b iv the ultimate trans- verse strength of cast iron as .175 x 5 = .875 ton. Then by Equation (12) (67), i 1.5 * On the subject of flexure, see an article on " Elasticity " by Sir William Thomson (Lord Kelvin) (Ency. J3rit., vol. vii.). See also an article by Professor Ewing on the " Strength of Materials " (Ibid., vol. xxii.), THE BEAM. To resolve the strength elements, let A B C D, Fig. 32, be the transverse section of the bar, the action or pressure or load being in the direc- tion of the arrow P, and let the line N N indicate the position and direction of the neutral axis passing through the centre of gravity of the section. The pres- sure P will therefore place the half of the section above the line NX in a state of compression, and the remaining half below that line in a state of tension (18), Fig. 6. Now the resistance that cast iron is capable of presenting to compression when compared with that which it is capable of offering to tension is 4J to 1. Therefore as the lower half of the section is in tension, in that part will fracture first commence. Draw the diagonals A D and B C, dividing the diagrams into triangles. Then as the horizontal resistance to stress commencing at the neutral axis N N will increase directly as the distance of any point from that line, it is evident that if the line C D at the maximum distance from N N be taken to represent the tensile resistance of the material per square inch at that line, the length of each of the other horizontal lines drawn across the bottom triangle at inter- mediate distances will also relatively represent that resist- ance at each such line. Therefore, the area of that triangle ' O multiplied by the resistance represented by the length of its base C D will be the measure (m) of the ultimate tensile resistance of the section A B C D. In the same way, the area of the top triangle will represent STRENGTH ELEMENTS IN SOLID BEAMS. 89 the measure of the resistance of the section to compression. H 2 In the example, the area of each triangle = -j- It therefore follows that, referring again to Tahle 1, and taking 1.4 x 5 = 7 tons as the ultimate tensile strength of O 3 cast iron, the ultimate tensile resistance (w) of the section _ H 3 x7 _ 39375 tons> Now the centre of tensile resistance is the centre of gravity of the bottom triangle, and the centre of resistance to com- pression is the centre of gravity of the top triangle. These two points are 1 J x f = 1 inch apart, Fig. 32, and therefore the effective depth of the section is one inch. Or for rect- angular beams, if a be the area of the section, -is the effec- tive area of resistance,- and c being the coefficient of tensile strength 5c may be taken as the ultimate tensile resistance. 5 then m = v a c - 2 and if c/ be the depth of the beam, d = the effective depth. o Having thus' determined the ultimate tensile resistance and the effective depth of the section, the multiple of these is the moment of resistance, and the next step is to find the weight of a load placed at the centre of the span which will cause an equal and opposite stress. Assume half the length of the span and the effective depth of the section to be a bent lever (24), and Let in = the ultimate tensile resistance (for the given section = 3.9375 tons) d = the effective depth ^ i 1 I in inches I = the length of the spanj w = the required load 90 THE BEAM, Then w = **. w ' 7 (20) 4 x 3.9375 x 1 = - - .87o ton lo But it has been shown by Equation (12) that given the ultimate strength of cast iron as determined by actual experi- ments, the breaking load b w is 1.969 tons. Now by Equa- tion (20), w =s .875 tons only, and therefore the resistance of flexure must equal the difference, or 1.969 - .875 = 1.094 tons. The tensile resistance being .875 and that of flexure 1.094 tons, the former is to the latter as 1 to 1J, or the average of the results of Mr, Barlow's and Mr. HodgkinsoiTs investigations. Required the safe central load for a rectangular solid beam squarely strained. Let a = the area of the section a = effective area of resistance = i a d = depth of section in inches. d' = effective depth of section = f d I = length of span . . . . | c = tensile coefficient (62, Table 1). ,' p = resistance of flexure = 1J c for cast iron. f ) ,, = ,, ,, = .70 c for rolled steel. ( ,, = ,, ,, = .56 c for rolled iron. F = c + p, w = the safe load at the centre of the span. . 4 a' d' F Ihen w 7 * See Formula (22) (76), in which a and c take the place of m. f In this and subsequent examples the values of p are derived from experi- ments and investigations made by Barlow, Eairbairn, and Hodgkinson. STRENGTH ELEMENTS IN SOLID BEAMS. 91 73. Given a Cast-iron Square Bar diagonally Strained to find the breaking load when applied at the centre of the span. Let A B D E, Fig. 33, be the section of a bar of the same dimensions as Fig. 32, namely, \\ in. x \\ in. and 18 inches in length between the supports, but strained by the load in the direction of the vertical diagonal A E as indicated O by the arrow P. The horizontal diagonal B D passing through the centre of gravity of the section at C will then coincide with the position and direction of the neutral axis. Now in this as in the last example, the lower half of the section will be in tension, and the greatest tensile stress per unit of area will be at the bottom angle E, so therefore at that point would fracture theoretically commence. Divide the half C D of the diagonal B D into any number, say 8 units of width, and the lower half C E of the diagonal A E into an equal number or 8 units. Place the units of width in Series a, and the units of depth in Series b, multiply together the relative factors, and place the results in Series c. Let E D' = C D, and D D' = C E, then assuming the section of the bar to have extended to D', E D' multiplied by half the depth E, or 8 x 8 64, will represent the tensile efficiency on the line E D', so also will 92 THE BEAM. tho numbers in Series c represent the proportionate tensile efficiency of the bar at each of the sectional widths to which they respectively refer. From tho line C E let ordinates be drawn parallel to C D, and each proportioned to it in length as the relative number in Series c is to 6'4. Draw through their extremities the curve C/E, and in like manner on CE the same curve reversed C g E. These two lower curves inclose the area of equal tensile resistance of the section when submitted to transverse diagonal stress and the two similar curves drawn in tho o upper half of the section inclose the same area of com- pressive efficiency. Now the resulting curve is a parabola, and the area of a parabola is f of that of the circumscribed rectangle as may be found by the given ordinates. Therefore taking the units of division as shown in the figure the area of the latter = (8 x 2) x 2 = 32, and the parabolic area = 32 x f = 21J. But the whole area of the bar = 8 x 8 x 2 = 128 units, therefore the resolved area of maximum tensile efficiency is to the whole area as 1 to 6, or as may be seen by the Fig. -| of J = -J- of \ \ of the whole section. Therefore as the bar is \\ x 1J in. = 2.25 in. area tho 2 25 area of ultimate resistance is -^ = .375 inches. Taking as in the last example 7 tons per square inch as the ultimate tensile strength of cast iron, .375 x 7 = 2.625, or m the measure of tensile resistance of the section. It is evident that the effective depth of the bar is the distance ff measured from centre to centre of the effective areas of resistance, or ^ * * = 1.06 inches. STRENGTH ELEMENTS IN SOLID BEAMS, 93 Therefore Equation (20), (72) to find the weight of a central load 10 which would cause a moment of stress at the centre of the span equal to the ultimate tensile moment of resist- ance, 4md 4 x 2.625 x 1.06 w = j = ,-Q =.618 tons. i lo In the case of square bars strained diagonally the ratio of their resistance to flexure to that of tension has been found to be for cast iron 1 to 1. Therefore as .618 x 1J = .927, the ultimate load on the centre will be .618 4- .927 = 1.545 ton. Now by experi- ment when squarely strained the bar broke with 1.969 ton, therefore if the ultimate strength of a cast-iron bar when squarely strained = 1, it will when diagonally strained 1-545 3 L969 :: -783, orsay-j Therefore Formula (12), (67) modified is applicable, assuming b and d to be each a side of the square bar when diagonally strained, and I the length of the span in feet. 3 I d 2 n 3 d 5 n Thus*.: -^ or .___ The area of equal resistance per unit of efficiency may be geometrically determined in the following manner. Divide the lines A C and B C into any equal number of units, say 8 as before, and from the points thus obtained in A C draw horizontal lines to the side A B of the bar. Then these lines will represent by their respective lengths either as may be required the direct tensile or the com- pressive strength of the section at each plane taken. From the points obtained in like manner in the line B C, draw 94 THE BEAM. vertical lines meeting at the side B A of the bar the hori- zontal lines already drawn. Then these vertical lines will represent the distances of the several horizontal planes of strength, measured from the neutral axis B D. Draw diagonal lines from each point taken in B C to the angle at A. Point the intersection of each horizontal line with the diagonal numbered the same in the figure, and a line drawn through the points thus obtained will be the parabolic curve defining as before the area of equal resistance. Required the safe central load w for a solid square beam diagonally strained. With the following exceptions, the symbols have the same value as those given (72). Let a! = the effective area of resistance = - a d = width of one side of the beam d' = effective depth of section = \/2 d? \ p = resistance of flexure = 1J c for cast iron. ,, = ,, ,, = .9 c for rolled steel. ,, = ,, ,, = -8 c for rolled iron. mi TC Ct Ct J? Then w = 74. Given a Boiled Bound Iron Bar 2 feet long between the supports, and 2 inches diameter to find the safe load w when carried by the bar at the centre of the span. Referring to Table 1 (62) the safe transverse strength n' of STRENGTH ELEMENTS IN SOLID BEAMS, 95 a rolled iron bar 1 inch diameter and 1 foot long = .2-45 ton, therefore by Formula (18) (69), cl? ri 2 3 x .245 w = - = - _ 6 2 i , tons. To proceed as before in resolving the strength elements of this bar, let Fig. 34 repre- sent its transverse section. Then with the load acting in the direction of the arrow P, the half of the whole section of the bar above the neutral axis N N will be in compres- sion, and the other half below that line will be in tension. The curves shown in Fig. 34 having been formed as de- scribed in the last example, inclose two areas, each of which (a) in this instance = .64 inch, the lower one being the area of equal tensile and the upper one the area of equal compressive resistance. Although the ultimate resistance of rolled iron to com- pression as compared with that which it offers to tension has been assumed to be about 4 to 5, (62) its effective resistance within limits of perfect elasticity to those two stresses may be taken as equal. Besides, this material and also rolled steel has a tendency to be " set up," and its sectional area thus increased by compression, whereas it is " drawn out" and diminished by tension. Practically therefore the efficient resistance to compression of rolled bars under transverse strain may be taken as equal to their resistance to tension. 96 THE BEAM. Now the effective depth (d') of the section is 1.2 inch, Fig. 34, and the tensile efficiency (c) per square inch of rolled iron = 4.5 tons, Table 1, (62). Therefore, in supporting a central load iv, for the transverse efficiency w' of this bar due to tensile resistance 4 a! d f c 4 x .64 x 1.2 x 4.5 w = = .576 ton. I 24 But in rolled iron round beams the resistance of flexure has been found to be .7 that of tension, therefore for the total efficiency of the bar w = .576 + (.576 x .7) = .979 ton, which corroborates the result already given by the ordinary Formula. Required the safe central load w for a solid round beam. With the following exceptions the symbols have the same value as those given (72). Let a' = the effective area of resistance = .204 a\ d = diameter of the beam ,, ? . i 4.1 r L- r 7 inches. d effective depth of section = .6 d resistance of flexure = \\c for cast iron. J? ,, = .8 c for rolled steel. = .7 c for rolled iron. CHAPTER VII. GIRDER OF THREE MEMBERS. 75. It lias been shown in the three preceding articles that the strength elements of a beam resulting from compressive and tensile resistance may be represented by the substitu- tion for its section of two equal areas of equal resistance per unit of surface, one area representing the resistance to tension and the other the resistance to compression. The difficulty, however, found with respect to solid beams, of bringing their theoretic efficiency into accordance with that shown by actual experiments on their transverse strength, led to the admission of another and very considerable strength element, the resistance of flexure. This afforded a reason for the strength evinced by solid beams, of which no satisfactory explanation had been hitherto given. Now the two equal areas of equal resistance by which the section of a solid beam has been replaced are typical of the actual top and bottom tables of a girder of three members. The resistance of flexure has, however, been found to be a comparatively small factor of strength in such girders, and in practice its value is covered by the ordinary formula} given in the two following articles. H 98 THE BEAM. The determination of vertical or shearing stress at any vertical is a very simple matter, and has been fully ex- emplified throughout Chapter 2. The effects of diagonal stress, in either the web of a cast- iron or in that of a plate girder, have for practical purposes been already disposed of (22). It only remains therefore to add some useful formulae for the strength elements of the top and bottom tables. These important members govern the efficiency of a girder ; but owing to practical exigencies of construction the scant- lings of the web will always exceed those ascertained by theory. 76. Formulae for Central Loads. The theory of the determination of the moments of horizontal stress in a beam (24, 27) will now be applied in the treatment of the strength elements of the top and bottom tables of a girder. Under any condition of loading, the stress in each of those members will be equal, but of a reverse nature, compressive in the top and tensile in the bottom table. Let a = the area of either table in inches. d = the extreme depth ) both either in I = length between supports j feet or inches, c = coefficient, Table 1 (62). w safe load at centre of span. The numerical factor 4 in each equation (21, 22) results from the resistance of each support or ^ and the leverage at the centre of the span or r . GIRDER OF THREE MEMBERS. 99 Given a plate-iron girder 20 feet long, 1.5 foot deep, with a load of 27 tons at its centre, to find the efficient area for each table at the centre of the span. CL ~j 7 " 4 d c 27 x 20 Given the area of either table at the centre of the span to find the safe central load. 4 a d c I 4 x 20 x 1.5 x 4.5 20 (22.) = 27 tons. 77. Formulae for Distributed Loads. As the centre of gravity of each half of the load is midway between the centre of the beam and a support, the effective leverage becomes ~r , and with w' as a distributed load the effect of the load on that part is w , thus 4x2 = 8 the numerical factor. With this exception, the equations will remain as before. For the area of either table at the centre of the span w'l a = (23.) 8 dc Given the area of a table at the centre of the span, the safe load ,' = 8 -^ (24.) H 2 TOO THE BEAM. Or, a = Lalf the area of table required for an equal central load, and w' ~ 2iv or twice the weight of a safe central load. It should be noted that for a cast-iron girder, c = 1 .4 ton for the bottom table and 6 tons for the top table. Having determined the sectional area of each table at the O centre of the span, the relative proportional area at any other part of the span may be readily found (23) Fig. 9, and (26) Fig. 11. But these proportional areas require in practice to be modified to suit the nature of the work and material, and small girders are for simplicity frequently made of the same section throughout. Should the depth of a girder be reduced from the centre of the span to each end, the theoretic area as determined for each table of a girder of uniform depth should at any given vertical, right and left of the centre, be increased inversely with the decrease of depth. Thus if the depth were re- duced by one-half, that area should be doubled. 78. Strength Elements. Let ABCD, Fig. 35, be the transverse section of a cast-iron girder 12 feet long between the supports, 1 foot deep with tables 9 inches wide, and 1 inch metal throughout, then by Formula (22), (76) 4: adc 4 x 9 x 1 x 1.4 . w - TIT 4 - 2 - therefore practically the safe load at the centre of the span is 4.2 tons. To determine the efficiency of the strength elements in accordance with the treatment described (72, 73), draw GIRDER OF THREE MEMBERS. lot the diagonal clotted lines A D and B C crossing each other in the neutral axis N N, at the centre of gravity of the section. Assume the central load to act in the direction of the arrow P, then the half of the section below the neutral axis will be in tension, and the hatched portions of it areas of tensile resistance equal to that existing at the bottom line C D of the section. Now, 1.4 ton being the coefficient of tensile efficiency, the sum of the two lower areas of equal resistance each multiplied by its effective leverage and by 1.4 will give the moment of the efficient tensile strength of the section, thus for the Table = (8J" x V =) 8i"arca x 11" leverage = 90.75. Web=(5x.417 = )2.085 x 6| = 13.90. 104.65 Let m = moment of efficient tensile strength. w = load at centre of girder. I = length of girder between bearings in inches. Fie:. 35. Then m = 104.65 4 and w = x 1.4 = 146.51 = 4.07. (25.) In order to determine the efficiency of the resistance of flexure due to the section multiply the load w, provided for by tensile resistance, by the coefficient p, which for cast iron = IJc, (72), then 4.07 x 1 = 5.08. And for a io2 THE BEAM. flanged girder multiply the sum thus given by the thick- ness of the web and divide by the width of the tables. Add the result to the value of iv already obtained.* rpi i)-(/o X 1 f* ft Then - - - = .56. And w now = 4.07 -f .56 = 4.63 tons. It thus appears that the usual formula which gives 4.2 tons leaves a balance of one-tenth to the credit of strength. 79. As an illustration of the error f arising from a rule which is sometimes given for determining the strength of a flanged girder by the usual formula (12), (67), based upon ex- perimental tests of the transverse strength of rectangular bars, assume A B C D, Fig. 35, to be a solid rectangular beam. Then assume the two spaces left between the rectangular outline and the section of the flanged girder to form together the section 8" x 10" of a second rectangular beam, and let each of these beams be 12 feet long between the supports. With Formula (12) determine the efficiency in cast iron of each of these solid beams, subtract the second from the first result, and the remainder will be the presumed efficiency of the flanged girder. 1. Tims , = k* = ? x * 'I 75 = 18.9 2. 7.3 tons. Now 4.2 tons has been shown to be the proper safe central load for this girder, but the last result, 7.3 tons, gives an excess of three -fourths of that load. * Experimentally demonstrated by Mr. Fairbairn, B. Baker on "Beams, Columns, and Arches." t An error which occurs in some handbooks. GIRDER OF THREE MEMBERS. lo; The same rule is wrong also for a box girder. For suppose the web to be split into two thicknesses, and a half thickness to be placed at each side, as shown by dotted lines in the figure, then all strength elements would remain precisely the same. This error has arisen mainly from the non-recognition ot the fact that a considerable proportion of the strength of a solid beam is due to the resistance of flexure. But there is also an error in assuming the stress at the top and bottom of the subtracted or inner beam when forming part of the entire solid beam to be the same as the corresponding stress in an independent beam, whereas in the example given it is only -ffths of it, and the effect of this error is partially to counteract that of the other. 80. A Circular Hollow Beam is strictly of the same order in respect to strength elements as a girder of three members. A circular hollow crane post is a beam of this kind placed upright and resisting transverse late- ral thrust. Let Fig. 36 be the cross-section of a hollow concentric cast-iron beam of 18 inches external and 15 inches internal diameter, and assume it to lie hori- zontally upon two sup- ports placed 18 feet apart, and to be subjected to a load or thrust midway between Fig. 36. io 4 THE BEAM. the direction of the arrow P. Then the horizontal line N N passing transversely through the centre of the cylinder will be the neutral axis of the section. Draw a vertical centre line a b and any convenient number of horizontal ordinates dividing the section of the beam into equal units of depth. Point off from the centre line a I upon each ordinate the thickness of the cylinder measured horizontally, and draw through those points the curved dotted line d e f. Draw the same curved line reversed upon the left side of the centre line a b. Thus the annular section of the beam has been brought into the form of a girder of three members with a top and bottom table each 10 inches wide, and a web 3 inches thick at the neutral axis of the section. Draw vertical ordinates as shown in the figure, and describe in the manner given (73, 74) the curved dotted lines c Jib I c. These will inclose the area of equal tensile resistance, and the same curved lines relatively drawn in the upper half of the section will also inclose an equal area of equal compressive resistance per unit of surface. Now in this example each of these areas = 22.5 inches; while the vertical distance between their centres of gravity = 1.07 feet, and this is the static depth g< Fig. 36, of the beam. The effective safe strength w f of this beam due to O tensile resistance is, therefore, in the terms of a central load, equation 22, (76). , 4:adc 4 x 22.5 x 1.07 x 1.4 >. , n , w = j = - - = /.49 tons. I lo But for the total safe central load w to this result must be added the resistance of flexure p, which should be determined GIRDER OF THREE MEMBERS. 105 by the rule given (78) for a flanged girder, thus the actual width of the web being 3 inches and that of each table 10 inches. 7.49 x 1J* x 3 p = ~ 1Q - = 2.80 tons. Then w = 7.49 + 2.80 = 10.29 tons = the safe central load. Now the formula which has been sometimes given for determining the efficiency of a hollow cylindrical beam for a central load w when D = the outer diameter") t 7 ,1 v r in inches. a = the inner diameter) n = a transverse strength coefficient. I = length of span in feet D 4 - # n has been w -- - which in the above example would be 18 4 - lo 4 x .112 w ' 18x18 : 18 ' 78 Whereas the true safe load has been found to be 10.29 tons only, or little more than one-half of that given by the obsolete formula incorrectly based upon the transverse strength found by experimental tests to exist in solid beams. The fourth power and the divisor D I have been adopted in this equation because the stress at the surface of the inner cylinder d when in place has been assumed to be less than that at the outer surface of the large cylinder D in the pro- portion of their diameters. * p = li c for cast iron, page 90. io6 THE BEAM. 81. The, Surface of an Area, such as one of equal static resistance, may be practically measured by drawing the curves defining the area fall size or to a large scale, and then transferring the outline of the area on to a sheet of paper ruled, say, in g- or y 1 ^ inch squares, by which means the required surface may be easily computed. Or this may be done with a piece of tracing paper so ruled and then applied over the drawing. The Centre of Gravity of any such area or of the section of a girder can be found with considerable accuracy by drawing the outline of it upon and then cutting it out of cardboard or thin plate ruled with a centre line. Suspend the section thus prepared loosely upon a pin at a point in it as far removed from the centre line as may be convenient. Plumb down from the pin with a fine thread as shown, Fig. 37, and the centre of gravity of the section will be the- point where the thread crosses the centre line. The fourth power being the square of a square is easily found by a table of squares. Thus, 13* = 13 2 x 13 2 = 169* = 28561. The fourth root being the square root of a square root can be found in the same way by a table of square roots. Thus for V V 28501; x/ 28561 = 169 and J 169 = 13. Fig. 37. CHAPTER VIII. CAST-IRON GIRDERS. 82. The Class Of Pig Iron usually chosen for girders or other castings having when in situ to resist stress is No. 3 of the eight into which such iron is divided. No. 4 is, however, suitable for heavy girders. After a furnace has been tapped and the iron run into pigs, an external inspec- tion is sufficient to enable an expert to identify and sort them into the eight classes. When pigs are broken, No. 8 shows the whitest fracture, at No. 5 the fracture becomes mottled, and the lower numbers are various greys. These classes, however, merge into each other, and varieties are caused by temperature, and even by the state of the weather at the time of tapping the furnace. Carbon is mechanically combined in pigs showing a grey fracture, and these arc softer and tougher than those in which carbon being chemically combined consequently show a white and silvery fracture when broken. Cast iron does not improve and may even deteriorate under successive remeltings, but a judicious mixture of scrap of other brands with pig iron facilitates melting, and when fused promotes fluency, the latter being a necessary condi- io8 THE BEAM. tion of iron prepared for casting, which should quickly fill the moulds to every corner. All mixtures suitable for girder work should on test show a fine evenly crystallised brightish grey fracture. But in the case of important series of castings, test bars should at in- tervals be cast simultaneously of the same metal. If the iron is of a good brand or mixture, bars 1x1 inch will when placed upon two supports stand up to a pressure of 800 Ibs. applied at the centre of a clear span of three feet. 83. Board of Trade Regulations provide that cast-iron girders must not be used for bridges carrying lines of railway, " except in the form of arched ribbed girders where the material is in compression." In a cast-iron arched bridge or in a cast-iron girder bridge carried over a railway, " the breaking weight of the girders should be not less than three times the permanent load due to the weight of the superstructure added to six times the greatest moving load that can be brought upon it." " All castings for use in railway structures should, when practicable, be cast in a similar position to that which they are intended to occupy when fixed." 84. The Depth of Cast-iron Girders. Having already given (76, 77) formulae for the strength and loading of girders generally, we may next consider certain conditions which regulate the proportions and efficiency of the leading members of a cast-iron girder. The generally received rule with respect to the depth of such a girder is, that in order to ensure a reasonable amount of rigidity it need not be more than -^ and should not be less thaiiy^g- of the span. Collateral conditions or special circum- stances may, however, sometimes necessitate to some extent CAST-IRON GIRDERS. 109 deviations from that rule. But in cast iron no advantage Fig - 1 n n i is also ^ ^ Fig. 43. Forge Test, cold, is the same as for angle iron. PLATE GIRDERS. 121 94. Board of Trade Regulations. In a wrought- iron or steel bridge the greatest load which can be brought upon it, added to the weight of the superstructure, should not produce a greater stress on any part of the material than five tons where wrought iron is used, or six tons and a half where o / steel is employed; per square inch. Steel Plates and Bars. 95. Steel Plates are readily obtained when not exceeding in any particular the following table of ordinary thicknesses and dimensions. TABLE 5 TO J.^\.JJUIll !* I ^ ^ f "I* fc "00 u= ' k S 9 I. 1 * \ I ) vl 1 < e b ^ 4 ? i * 15 S! *- *- ^ ^ C^l- ^ 8 O ^ PLATE GIRDERS. 139 for each table. Besides the two tables serve together as stiffeners to the web and thus to the girder against the stress of wind or any other cause of lateral disturbance. Assuming the width of each table to be 15 inches, and O / deducting the width of four " rivet holes, or 15 (f x 4) = 12 inches, or the effective width of the table, we then 11 2 15 5" have ^- = or a thickness of three plates, will give the sectional area required for the central part of each table. Then, bearing in mind the fraction of the whole area re- quired for each table supplied by the outer flange of the angle bars, the lengths of the table plates may be arranged as shown in diagram Fig. 49, so as to provide the propor- tional area required at every part of the span for resisting the horizontal stress set up by a distributed load (27). Let the inner line of plate of each table be in two 22-feet lengths connected by a cover plate. Now as the net transverse run of section of each plate less rivet holes is 12 inches, and the coefficient being the same for all resistances, the aggregate run of rivets in the cover plate should also be 12 inches on each side of the joint (101), and as the rivets are " diameter the number of rivets re- quired = _? = 16, or there should be four rows of four 3 rivets each on each side of the joint. This w r ith a 4-inch longitudinal pitch requires each cover plate to be 2 feet 8 inches long. Fig. 49. Let each line of 3?" x 3" x f " longitudinal angle bars be in two lengths 23' 4" and 20' 8" and cover the joint with a plate 2' 8" x 3" x " taking four outer table rivets on each side of the joint, the position of these covers being upon the 140 THE BEAM. underside of the top table, and upon the uppcrside of the bottom table. Flush rivet at each end of the bottom table a 2' 0" x I' 3" x " sole plate c, Fig. 49, for distributing the weight of half the load over a sufficient surface upon each abutment. It has been shown (22) that through diagonal action in the web the vertical stress of the load produces the horizontal stress which has to be resisted by the tables. The latter must in the present example, therefore, be imparted to each table by the rivets passing through the outer or horizontal tables of the 3"^ x 3" x f " angle bars. Now it has already been seen that three T y' plates are necessary at the centre of each table, also that a cover plate for a joint in any one of these plates requires sixteen rivets on each side of the joint. Then for conveying the horizontal stress to the table plates the number of rivets required is 16 x 3 or 48, and adding 6 for the two angle bars, 54 rivets, to properly connect the web with each table, or 27 rivets in each line of angle bars, that is 9 feet run, whereas the effective run of riveting is 20 feet." It is thus evident that the outer lines of rivets in each table do not serve for this purpose, and that these lines of J" rivets to a 4" pitch are adopted simply to bring the plates into efficient contact with each other throughout, so as to impart thereby rigidity to the whole structure. It is also evident that if each table were in one length and in one thickness of plate, the two outer lines of riveting might be dispensed with, excepting the two rivets required at the top and bottom end of each of the 62-" x 3" x -f$" vertical T bar stiffeners hereafter described. Tlie arrangement of table plates is therefore open to practical modifications in accordance with the exigencies of each case. PLATE GIRDERS. 141 106. Web Details. Theoretically the horizontal area of the section of the web for the length extending from the centre of the span to an abutment would require to be the same as that of a table (103), and therefore in this instance 13.2 inches (105). Now as half the length of the span is 240 inches, the theoretical thickness of the web throughout IS 2 would if the load were central = -- =.055 inch. But with a distributed load, as in the present example, the horizontal area of the half web would in order to resist vertical stress necessarily be the area of an acute-angled triangle extending from the centre of the span to either abutment (26), where its base in this example would = .055 x 2 = .11 inch or the extreme thickness of the web. Further .11 multiplied by the depth of the web in inches, and by the coefficient will give the shearing resistance of the web at an abutment. Thus .11 x 70 x 6-i- tons = 50 tons and 50 tons is half the weight of the distributed load. Fig. 50 is a diagram of a plan of the web with the thick- nesses drawn full size, and c being the centre of the span, the two acute-angled triangles a c b and a' c b' when measured by the scale of '' to the foot for length, and full size for width, will give the theoretical horizontal area required for the web, the base a b and a b' of each triangle being .11 inch. But in the first place it would be impossible to readily get piates of tapering thicknesses, and even if such plates could be obtained it is evident that a thickness of .11 inch tapering to nothing at the centre of the span could not be practically adopted when wear and tear and contingencies are taken into J 4 2 THE BEAM, consideration. Besides, if it should be necessary or desirable to form the web longitudinally of a series of plates, its thick- ness should then be sufficient to present an adequate bearing area to the rivets for resisting the vertical stress of the load at all vertical joints. In order to comply with conditions laid down in Table 5 (95), divide the web horizontally into four plates each 11 feet long, Figs. 47 and 50, and let suitable covers be provided for the joints. Then, as there can be no shearing strain at the central joint from an equally distributed load, proceed to deter- mine the necessary thickness of the two central plates and the number of rivets necessary for a proper rivet-bearing surface at each of the intermediate joints d, d. Now 50 tons being the load on the half span, and the length of half the span being 20 feet, the vertical stress at each of those joints = ; = 27.5 tons, and--- = 4.23 inches, or the bearing area required in the web plate for each vertical line of rivets at the joint, and as the rivets are }" diameter, the length of bearing area required = - - = 5.64 inches.. o Let these web plates be J" thick then ' 6 ^ > 4 - 22J rivets required, or say 23 rivets. By placing the centre of the end rivets 2" from the top and bottom of the plates, the intermediate rivets will coincide with a 3-inch pitch. But in any case, a requisite fractional deviation, however small, from any given regular 4 or 3 inch or other pitch, is under no circumstances a matter of any difficulty in the work. Moreover, many contractors are provided with drilling and punching Tables, or benches* upon which a plate * The invention of Mr. Field, Messrs. Maudslay, Sons, and Field. PLATE GIRDERS. 143 may bo brought under the drill or punch in a straight line of successive distances of any length of gauge or pitch. In order to give a proportionate practical increase to the remaining portions of the horizontal area of the web let the 5" plate at each end be * thick ; cover on each side of the web A// each of the three vertical joints with a 6J" x 3" x "J" bar, of the latter might, however, be added, simply as stays or steadiments, or the girder might be sup- ported at a and a. To determine the horizontal stresses in each table: as a e c' f is a parallelogram, the stress in the first bay a e of the top table and in the first bay e'/Vof the bottom table = 40 -~= 30 or 50 - = 30. Proceeding in the same way to the central vertical there will be an additional increment of 30 in each bay plus 30 at the central vertical in the case of the bottom table due to the two tie-bars which meet at c'. The latter increment will, however, be taken by the pin or rivet connecting these bars, and not by the table itself, unless each bar should have a separate attachment. The stresses as determined for each table are given Fig. 59, Series HS, HS'. These, as in the case of any ordinary beam of uniform depth carrying a central load, increase from each end to the centre of the span in terms of arithmetic pro- gression (24). M 162 THE BEAM. Load Distributed. 117. Given a Single System Square and Diagonal Framed Girder 30 feet long between the supports, 4 feet deep, with vertical and diagonal web members, and carrying at the bottom table a distributed load of 160 tons or 5^ tons per foot run, Fig. 60. For the web, let the ratio of the lengths /*, d, and I be 3, 4, and 5 (115). distributed Fig. 60. In this example the vertical lines of the diagram are tie- bars dividing the length of the girder into ten bays as in the last, and the diagonal lines are struts. Now as the span is divided into ten bays, the stress of y 1 ^ of the load =16 will become active at each vertical, and the weight of one-half of the distributed load carried by each outer bay = -^ = 8 will pass horizontally direct through the bottom table to the adjoining abutment. It is further evi- dent that one-half of the fraction of the vertical stress caused 1 ft by the distributed load = = 8 carried by the tie-bar at the central vertical will be transmitted by the members of the FRAMED AND LATTICE GIRDERS. 163 web to each support + 16 at each intermediate vertical, and + 8 at the support itself, at which the stress will thus equal 80 or half the amount of the total weight of the load, Fig. 60, Series YS. Thus, commencing from the central vertical, let s be the stress in a tie-bar, and t the thrust in the consequent strut, I 5 then s -- = t, or for the strut c/, t = 8 x j = 10, and so on for each successive strut to the abutment plus the thrust as determined for the last. Fig. 60, Series DS. For the horizontal stresses in each table, as a I' is the diagonal of the parallelogram aba b', the stress from a to b and from a' to V = s -, or in this example = 72 x - = 54 Cv ^t and so on to the central vertical, but in each case adding the o product of the last, thus for the next (56 x = 42)+54 = 96. For the central vertical, it will be remembered, that in this example s will equal 8. Series HS and HS' give respectively the horizontal stresses in each table, and these increase towards the centre of the span as the multiple of the segments into which it is divided by the verticals, as in the case of an ordinary beam of uniform depth, carrying an equally distributed load (27). The additional central increment of horizontal thrust = 6 at c, will be taken by the pin or rivet connecting the two struts which meet at that point, and not by the top table itself, unless each strut should have a separate attachment. The continuation of this girder to square ends, as shown by dotted lines in the figure, would obviously add nothing to its efficiency, but might be made useful as a steadiment in con- nection with a wall or other lateral support. 1 64 'THE BEAM. Load Central. 118. Given a Single System Diagonal Framed Girder 30 feet long between the supports, 4 feet deep, and carrying a central load P = 80 tons, Fig. 61. For the web let the ratio of the lengths 7t, d, and / be 3, 4, and 5 (115). 7 fo 7 |^- 120 M-_ Tig. 61. Thus the girder is longitudinally divided by the members of the web into 5 bays, each = 2 /*, or in this example 6 feet long, the six diagonal struts and the four diagonal tie-bars together forming the single framed web. Now vertical stress caused by the load P is halved at the central point c (23), and each half = 40 tons is transmitted in an opposite direction through the diagonals of the web to each support, becoming active at each upper and lower junction of the web diagonals and tables. Series YS, The strain in each diagonal will therefore be 40 = 40 X - - 50 tons, Cb ^t as figured in the diagram Series DS. To determine the horizontal stresses in the tables, assume in the first place the line c c" to be extended to /, and let FRAMED AND LATTICE GIRDERS. 165 f c" = 3 = a e. Then because the resistance of the support a is vertical in the direction a f and as f c" a e is a parallelo- gram of which the diagonal c" a = I and a e = A, and as there is a compressive stress of 50 in the direction of c" a, there will be a tensile stress in the bottom table from a to e, and consequently from a to the central vertical equal to 50 l = 30. Next as c" u a b is a parallelogram, the diagonal of which da I and c" c or a b = 2 A, the compressive stress from c" towards the central vertical = 50 -=- = 60, and likewise L the tensile stress from a to the centre will also be 60, but + 30 already obtained for the bay a a', and 60 + 30 = 90 the total stress in a b. For the central bay b b' of the bottom table the tensile stress will in the same way be 60, and 60 + 90 = 150 tons. It has been shown that the compressive stress due to the outer bay c" c of the top table is 60, therefore for each of the like two central bays it will be 60 + 60 = 120 tons or 30 tons less than the stress in the central bay of the bottom table. The two struts, however, meeting at c cause a stress at that point = 50 =30, and it 120 + 30 = 150. But the last increment would be resisted by the pin or bolt connecting the two struts at c, and not in any way by the table itself, unless those two members should each be separately connected with it. Fig. 61, Series HS and HS'. As a horizontal stress of 60 tons is imparted to each table at each junction of the tensile diagonals, the pin or bolt forming that junction should be so proportioned as efficiently to resist that stress, being the maximum to which it would be subjected in any one direction, for the diagonal stress has been found to be 50 and the vertical 40 tons. 1 66 THE BEAM. Load Central. 119. Given a Double System Diagonal Framed Girder 30 feet long between the supports, 4 feet deep, and carrying a central load P = 80 tons, Fig. 62. For the web let the ratio of the lengths A, f/, and I be 3, 4, and 5 (115). VS. 40 40 HS. e' /' e" c' -40 40 4040:40-40 40 40 40 40 -4- 90 -k- 120 -*-7,50 -*r-150 -4- 120 -k- 90 -*- 60 -*- 30-j 30-0"- H Fig. 62. It will be seen from the diagram reading from the centre of the span that in this example there are two quite distinct and separate systems of diagonal struts and tie-bars, and tie-bars and struts, and that without the action of the central vertical strut c c', the second system of diagonals would remain practically useless for transmitting weight from the load to the abutments. But by the addition of that strut the efficiency of each of the two systems for the transmission of stress is rendered precisely the same, so that half the stress caused by the load = 40 will be resisted by the two diagonal struts meeting at c, while the strut c c' will convey the remaining half = 40 direct to the point c', there to be taken by two diagonal tie-bars. Now with the single system the girder was longitudinally divided into five 6- feet bays, giving a stress = 50 in each FRAMED AND LATTICE GIRDERS. 167 diagonal member, but as in this example the two systems divide the length of the span into ten 3-feet bays, with two active diagonals crossing each other in each bay, it will be evident that the stress in each diagonal will now = ^ = 25 tons. Series DS. The compress! ve stress in each vertical end strut as a e 25 x 4 = - -= = 20 caused by a diagonal tie as a/, while at the o same time the strut b e also brings a vertical weight at e upon the same support = 2o * '- = 20, and 20 + 20 = 40 tons, or the weight of half the load, upon each abutment. Series VS. To determine the horizontal stress to be resisted by the top table; as a f = 5 is the diagonal of the parallelogram aa'ff /-k and a a' = 2 h = 6, the stress in a a = 25 x - = 30, in the same way as Uc is the diagonal of the parallelogram b Ue'e", the stress in b b' = 25 x ~ = 30, but as the parallelogram with the diagonal af has also been found to give a stress of 30 between b and a', and consequently onwards to the centre c, the total stress in the bay b a' of the top table = 30 + 30 = 60. So with each succeeding bay there will be an additional increment of 30, bringing the stress in the two central bays up to 150 tons, or the central stress already shown (118) to be set up in the same girder carrying the same load, but constructed with a single system web. The horizontal stress resisted by the bottom table may be determined in precisely the same way, for as b c is the diagonal of the parallelogram bb'ee', the stress in ee f r* 25 x = 30, and proceeding as with the top table there 1 68 THE BEAM. will be an addition of 30 in the bay/Y, and in every succeed- ing bay to the central vertical. Horizontal stress in each table may also be determined simply by leverage. For as there is a vertical stress of 40 at -7 = 0. d Stress in the remaining diagonals may be determined as in previous examples, and as each additional increment of vertical stress is 16, so the additional stress in each succes- sive pair of diagonals will be 10, for - = 8, and 8 = 10. Thus reading from the central vertical the series will now become 5, 15, 25, 35, 45 in either system as figured in tho diagram. By this arrangement, strain arising from any abnormal load- stress or shock occurring in any part of the span would bo divided between the two systems by the introduction of the vertical tension bars, whereas without these the resistance to stress so caused might be principally thrown upon one system only. For, assuming that the tables have no vertical strength, should such a load-stress occur at any one of the points in which the web members are connected with either table, the stress caused by it would have to be resisted by one system only. Stress in the two diagonals of each bay having been found to be equal, that in the collateral bays of each table will also be equal. In determining the horizontal stress in the collateral bays 1 86 THE BEAM. of each table as given, Series HS, Fig. 65, commence as in the last article with the left-hand end bay. Then 45- = 27 (1) i (80 ~ = 48) + 27 = 75 (2) (60 - 7i = 36) + 75 = 111 (3) (40 7 i = 24) +111 = 135 (4) (20 -* = 12) +135 = 147 (5) l ( 5 h = 3*) + 147 = 150 (6) if For Equation (2) the numeric factor is the stress in two diagonals acting at the same point d or rf', or 45 + 35 = 80, and so on with those which follow, excepting Equation (6) as one diagonal only is active at c or c in the same direction. It will be seen that diagonal and horizontal stress in each bay of the girder is now the mean of the results given in the last example. Besides the additional efficiency already stated, there is yet another advantage attending this equation of stress in collateral members, for their constructive details would then precisely correspond each with each, and thereby constructive work would be simplified and facilitated. But in girders of the type Fig. 65, this desirable arrange- ment of work could not be fairly carried out except in con- junction with an adequate provision for the series of major stresses in both systems. * This stress is taken at the top table by the pin connecting the two struts, and at the bottom table by that connecting the two tie-bars, and not in either case by the table itself unless those web members should have separate connections. FRAMED AND LATTICE GIRDERS. 187 Should the load be taken by die girder at the top instead of at the bottom table, the vertical members would then be struts, but without any alteration of the nature or amount of stress in any other member. 126. A Rolling Load and a Coupled Double System Web. The stresses caused in the web members of the girder, Fig. 66, by the same load per length unit may be determined as stated (122) and (124), and are delineated in the following tabulated Series for the system terminating with struts. Let the line A F of the Table represent the length of the span, divided into bays as in previous examples, (c) indicating the centre of the span, and let the load advance from left to right. System terminating with Struts. A a B 1) C D d E e F VS si V ,92 C' S4 D' = 88 8 888 888 = 7.2:0.86.4:1.65.6:2.44.8:3.2 4:4 3.2:4.82.4:5.01.0:6.40.8:7.2 (9 8 17 7 24 6 1) (8 + _ 1 2) 3 (7 + 1 2 ~3 3) 6 (6 + - 2 + - 2 + - 1 2 + 1 2 + - 1 2 3 3 + - 1 2 7 7 14 6 3 3 ~o 4) 3 3 3 3 3 3 3 3 4 6 (5 4 6 4 6 4 6 4 6 4 30 5 35 4 20 5 10 5 5 5 4 ~9 3 12 2 14 1 15 10 (5 10 5) 10 5 15 (?) 21 (3 10 5 15" 6 21 7) 28' (2 10 5 10 5 15" , 25 4 15 4 5 4 15 (4 15 6 39 3 42 9 44 1 29 3 19 3 1 3 11 3 21 7 28 8) 21 7 32 2 2 > - ) 2 2 2 8 2 18 2 28 8 30 9) 45 34 1 24 1 25 4 1 1 16 1 26 1 36 45 35 5 5 15 25 35 i88 THE BEAM. Summary of Results for Rolling Lead, | + 9|+ 1 + 45 - 36 - 3 -28 + 6 - 21 -10 + 15 + 15 -10 -21 + 6 + 28 -36 + i +45 The signs + and relatively denote compressive and tensile stress. As the two systems of web members are now coupled, Series si, 8%, sa } &c., relatively give the stress in each web member for each advance of the load bay by bay. The figures in black type denote either increased stresses or a reverse in their nature or both when compared with those which obtain when the load covers the whole span. In the summary of results, Series x and y relatively give the stress in each web member as the load advances from left to right and from right to left. Further, as stress in the two members, crossing each other in the same bay, has been equalised by the vertical tension-bars, so by reversing the signs + and in this summary the two series x and y will give the stresses set up by the rolling load in the .system terminating with tie-bars. Comparing, therefore, as in the last example, Series x and y with Series S, the relative increase, and also the nature of the stress set up in each web member by the rolling load, is at once .apparent. FRAMED AND LATTICE GIRDERS. 189 Load Distributed. 127. Given a Quadruple System Lattice Girder, 30 feet long between the supports, 4 feet deep, and carrvino- at the bottom table a distributed load of 160 tons or t/ ? 5J tons per foot run. Figs. 67 and 68. For the web, let the ratio of the lengths /*, c/, and I be 3, 4, and 5 (115). The length of the span is thus divided by the four systems of web members into twenty bays, and a vertical stress of 8 tons becomes active at each intermediate junction 20 " of the web members with the bottom table as indicated Fig. 68, while, as in the case of all framed and lattice girders, the weight of one-half of the distributed load carried by each outer bay,, either at the top or the bottom table, in passing direct through the medium of the relative table to the ad- joining abutment, will cause no strain in the web diagonals, or in the tables. In this example each of the four systems of web members is theoretically quite distinct and separate. Two of these four systems, Fig. 68, one carrying four and the other five of the nineteen equal fractions of the load start at two of the four extreme angles of the girder, and following the same lines as those in example Fig. 65 (123) terminate in like manner at the two extreme and opposite angles. Thus the load at each point of incidence, being now half what it then was, the stresses in the ir. embers forming these two systems will be, as stated in the diagram, Fig. 68, half those given in that example. The two remaining systems having peculiar charao THE BEAM. FRAMED AND LATTICE GIRDERS. 191 teristics, are shown for convenience of treatment in the separate diagram, Fig. 67. These starting at a point, p, in one vertical end strut, midway in the depth of the girder, terminate at a similar point, p f , in the other like and opposite strut, and being simply counterparts of each other reversed longitudinally, one of them for clearness of delineation is shown by dotted lines in the figure. Each of these systems carries five of the nineteen fractions of 8 tons each, into which the active portion of the load is divided by the quadruple-system web. But as the points of incidence of these five fractions relative to the centre of the span and the points of support are not symmetrical, the deter- mination of the stresses set up by them in the web diagonals of the system becomes a matter of special treatment. First (30), find for one system the amount of vertical stress which travels from each of the five incidental fractions of 8 tons to each abutment through the medium of the web. These will be as given Fig. 67, Series VS. Thus commencing from the left ; of the stress of 8 at A, 1.2 will travel to the right and 6.8 to the left support, and each such stress so obtained multiplied by - will give that set up relatively right or Cv left in the diagonals of the system by each incidental fraction of the load stress. For instance, from A to the right the stress in the diagonals will be 1.2 = 1.5, while to cL the left it will be 6.8 - = 8.5, Series A', and so on for each (Jb incidental load stress at B, C, D, and E, the results will be as given in the corresponding Series B', C', D', and E'. Throughout the Figure the sign + indicates compressive stress, and the sign tensile stress. When both tensile and i9 2 THE BEAM. compress! ve stress refer in the series to any one member, tlie amounts are so placed in two separate columns. Add up each of these columns, subtract the less aggregate sum from the greater, and the remainder will be the true stress either of compression or tension in the member, as given for each in the last Series S, and also relatively figured on the diagram. For the counterpart system, taking each member in reverse longitudinal order as shown by dotted lines in the diagram and indicated by the letters A, B, C, D } E, above the tabulated Series (3) and (4), page 199 it will be at once evident that the stress in each relative diagonal of the two systems will be the same as that given for the first system in Series S, Fig. 67, and figured for both systems on the diagram Fig. 68. The tabulated series of stresses, Fig. 67, gives inter- mediate results necessary for completing a determination of the maximum stresses to which the web members are subjected by a rolling load (128), but the following is a more simple method of determining: the stresses in those members caused O by an evenly distributed load when covering the wliole span. ' The five active portions of the load, Fig. 67, each = 8 being incident at distances from the right abutment of respectively 17, 13, 9, 5, and 1, twentieth parts of the span, the part of this load borne by the left abutment is (17 -f 13 + 9 + 5 + 1) x = 18, thus causing an upward vertical reaction of 18 at p. The effect of this will be a compressive stress in p a, amounting to 18 x -= + 22.5, and a vertical stress acting upwards at fl, equal to 18, causing a tensile stress in a b of - 22.5. FRAMED AND LATTICE GIRDERS. 193 In the same manner, this occasions compressive stress of 4- 22-5 in b c, from which must be deducted the tensile stress due to the portion of the load which is active at b, or 8 x | = 10. So that the resulting compressive stress " in be is 22.5 - 10 = + 12.5. In the same manner, it follows that the tensile stress in c d is 12.5. That in d e 12.5 - 10 = + 2.5 ef ....= - 2.5 fg 2.5 - 10 = - 7.5 ffk ....= + 7.5 Jc n - 7.5 - 10 = 17.5 no ....= + 17.5 op'- 17.5-10 = - 27.5 As the stress in e/is 2.5, and that would cause a stress of + 2.5 in/gr, add 10 and the result is 7.5, and so on to o p. To verify the above, we may observe that the part of the load on this system carried by the right-hand abutment is 40 - 18 = 22, and that 22 x ~ = 27.5 to which, being obviously a tensile stress owing to the direction of the member o p ', we give the negative sign and obtain 27.5 in that member as before. The stress in each diagonal member of the web having been ascertained, the stresses set up in each table may next be found (123), and then the results so obtained maybe completed in the following way. It will be observed that of the two unsymmetrical systems 194 THE BEAM. of web members meeting in the points p, and p the stress at p is + 22.5, and that atp' 27.5. Subtract independent of signs the less from the greater, and the remainder will be 5. Then 5 -^ = 3, a horizontal stress acting in the direction of the centre of the span upon each end strut midway between the two tables at the points p and p r . The effect of this stress (c) g upon each table will therefore be in this example ^- or 1-|- 2 tons. Therefore as the stress c acts in a reverse direction to the tensile stresses in the bottom table as determined in accordance with 123, so 1^ tons should be deducted from each of the stresses thus given, and because it acts in the same direction with the compressive stresses in the top table, so ! tons should be added to each of the stresses as given by each equation. For the top table commencing with the left-hand end bay, the horizontal stresses in the several bays will be as given in Series HS, Fig. 68. For the first bay, as a b c e (Fig. 68), is a parallelogram, and as the diagonal a e equals 25 or the stress in that member, then the resulting stress in a & = 25 =15, and 15 + 1-|- = 16.-5. i For the second bay produce the diagonal a' p to c', then us a ]) c' e is a parallelogram of which the diagonal a' e equals the side a' c', and the stress in each of these members is 22,5; then 22.5^ = 27, to this add 16.5 given by the L previous equation and the result will be 43.5, and so on for each bay to the centre of the span. The value of the incre- ment c having been added in the first, thus passes on in each succeeding equation. FRAMED AND LATTICE GIRDERS. 195 Thus . . . (25 \ = 15) + c = 10.5 (1) 6 (22.5 ^1 = 27) + 16.5 = 43.5 (2) it (20.0 ^ = 24) + 43.5 = 67,5 (3) i (17.5 J|A = 21) + 67.5 = 88.5 (4) t (15.0 -- = 18) H- 88.5 = 106.5 (5) V (12.5 M = 15) +106.5 = 121.5 (6) 6 (10.0 1A = 12) +121.5 = 133.5 (7) t (7.5-^ = 9) +133.5 = 142.5 (8) L (5.0-?- = 6) +142.5 = 148.5 (9) I (2.54^- = 3) +148.5 = 151.5 (10) i For the bottom table, commencing in the same way as before, the horizontal stress in any bay will be as given in Series HS', Fig. 68. For the first bay it is evident that the thrust of 20 in the diagonal b c of the parallelogram a b c e, Fig. 68, will cause a tensile stress in the bottom table = 20 ~ = 12, and this L - 1-i- = 10.5. For the second bay produce the line of the tie -bar c" p to a". Now there is a tensile stress of 27.5 acting in that bar in the direction of o" c" upon the point c' x ? and there is also o2 196 THE BEAM. a compressive stress of 17.5 acting in the direction b' c upon the same point, each of these tending to set up tensile stress in the bottom table from c" to the centre of the span. Therefore as a" c" and b" c" are diagonals of two equal parallelograms, a" a' c' c" and a' b" c" e', the two factors 27.5 4- 17.5 = 45 when thus combined cause a tensile stress in the bottom table from c" to the centre of the span = 45 j = 27. To this add the stress given by Equation (1) and 27 + 10.5 = 37.5, and so on with each succeeding bay to (9). For the tenth bay, as the tensile stress = 7.5 in the diagonal //' would cause a tensile stress, and the tensile stress of 2.5 in the diagonal //" would cause a compressive stress in the bottom table from /to the centre of the span, and 7.5 - 2.5 = 5, so (5 - = 3) + 142.5 = 145.5. The stress of 3 tons given by Equation (11) will be taken by the table when each of the two tie-bars at g has a separate connection with it. But when there is but one connection, such as a bolt or pin, between the tie-bars themselves and the table, then the stress will be taken by the connection and not by the table. Thus . . (20 | = 12) - c = 10.5 (1) (45 * = 27) + 10.5 = 37.5 (2) (40 - = 24) + 37.5 = 61.5 (3) (35 j = 21) + 61.5 = 82.5 (4) I FRAMED AND LATTICE GIRDERS. 197 (30 * = 18) + 82.5 = 100.5 (5) (25 - = 15) +100.5 = 115.5 (6) (20 - = 12) +115.5 = 127.5 (7) (15 - = 9) +127.5 = 136.5 (8) I/ (10 - = 6) +13G.5 = 142.5 (9) ii 7.5 - 2.5 = 5 and (5y = 3) +142.5 = 145.5 (10) L 5 = 3* +145.5 = 148.5 (11) The mean of the total central horizontal stresses in the top and bottom tables = *"" "** ^' J " J ^ 3 * = 150, or the stress caused in each table at the centre of the span with a continuous plate web under the same conditions of length, depth, and load. The compressive stress in each vertical end strut is 20 from a to p and 20 + 18 + 22 = 60 from p to c. For the stress at G each abutment add 16 caused by the strut b c, and ^^ caused by the fraction of load incident from c to c /x , then the total stress is GO + 16 + 4 = 80, VS, Fig. 68. 128. A Rolling Load and a Quadruple System Lattice Girder, Figs. 67 and 68. Let the * This increment of stress may or may not be taken by the bottom table (11 7, 123). 198 THE BEAM. line A F and each like horizontal line in the following- tabulated Series represent the length of the span divided into bays by vertical lines, that at (c) being the central vertical, and let the same load as before advance from either end upon the girder until the whole is covered. (1) System terminating with full-length Struts. B b C (c) D d E e O) vs { s 'as iy * 6.4 5 8 8 1.6 4.8 : 3.2 j 8.2 : 4.8 o 1.6 : 6.4 +20 | -20 +10 -10 | -10 +10 -20 | +20 Summary of Results for Rolling Load. + 8 I - 8 +20 j -20 - 2 +2 + 12 -12 - 6 I + 6 + 6 | - 6 -12 + 2 + 12 2 20 ! +20 -81+8 (2) System terminating with full-length Tie-bars. (2) / vs | s V C0 (\, 88888 7.2:0.8 I 5.6:2.4 4.0:4.0 2.4:5.6 0.8:7.2 -25 + 15 | -15 + 5 - 5 - 5 +5 -15 +15 -25 Summary of Results for Rolling Load. - 9 -25 - 1 ! + 1 + 16 I -16 -4+4 + 9 - 9 -91 + 9 + 4 | - 4 -16 +16 + 1 - 1 -25 - 9 FRAMED AND LATTICE GIRDERS. 199 (3) System commencing with a half-length Strut at p, Fig. 67, ABODE - () 88888 6.8 : 1.2 5.2 : 2.8 3.6 : 4.4 j 2.0 : 6.0 0.4 : 7.6 +22.5 -22.5 + 12.5 -12.5 + 2.5 - 2.5 - 7.5 1+ 7.5 -17.5 +17.5 -27.5 Summary of Results for Rolling Load. + 8.5 1- 8.5 +22.5J-22.5 - 1.5 !+ 1.5 + 14.0! -14.0 i - 5.0 1+ 5.0 + 7.6 j- 7.5 -10.5 j+10.5 + 3.0 j- 3.0 -18.01+18.0 - 7.01- 0.5 -27.5 - 9.5 (4) System commencing with a half-length Tie-bar at p. Fig, 67, E D C B A vs 88888 7.6:0.4 6.0:2.0 4.4:3.6 j 2.8:5.2 1.2:6.8 -27.5 + 17.5 -17.5 + 7.5 - 7.5 - 2.5 |+ 2.5 -12.5 +12.5 -22.5 |+22.5 Summary of Results for Rolling Load. - 9.5 -27.5 - 0.5 I- 7.0 + 18.01-18.0 - 3.0 !+ 3.0 + 10.5 | - 10.5 - 7.5;+ 7.5 + 5.0 1- 5.0 -14.0 +14.0 + 1.5 - 1.5 -22.51+22.5 - 8.5 j+ 8.5 It has been shown (127) that the vertical stress caused by the load at each point of incidence is 8 tons. Therefore the stresses given in the Series for (1) and (2) systems of web members are one-half of those determined (123, 124) for the like systems and tabulated, page 182. For as there are now twice the number of web systems, so there is half the vertical stress at each point of incidence caused by 200 THE BEAM. the same load per foot run, and consequently half the stress in each web member. To facilitate reference the verticals of these two systems are lettered relatively with those (124). The stresses for (3) system of web members are deductions from the Series A 7 , B', C', D', and E', Fig. 67 (127). The stresses for (4) system are those determined for (3) system reversed end for end of the girder. Series S give for each of the four systems respectively the stress in tons in each web member when the whole span is covered by the load. Series x and y give for each system the stress in tons in each web member during the advance of the load from left to right and from right to left. The signs -f and respectively denote compressive and tensile stress. Upon examination of the whole of the series it will be seen that the greatest tensile stress in a central member is 9 tons o and in an end member 27J tons, also that the greatest com- pressive stress in a central member is 6 tons, and 22 \ tons in an end member. The stresses set up by a rolling load in each of the web members of the lattice girder, Fig. 68, have thus been theoretically determined. The spans of such girders, and the distributed loads they are intended to carry, are, however, so various that the relative sections of the web members must in some measure be determined in accordance with practical requirements. See (106) and (109) on plate webs. For instance, in the treatment of small girders intended to carry light loads, the section of a member or the part of a member which would have to resist the greatest stress hav- ing been determined, that section might, in order to simplify work, be judiciously adopted throughout the like members. FRAMED AND LATTICE GIRDERS. 201 Thus in ordinary practice, the final relative proportions of the members of a structure are frequently best adjusted in accordance with the results of practical knowledge and experience. 129. The depths suitable for framed and lattice girders as arranged in the following Table were computed, like those for plate girders Table 8 (103) , from the diagram given by Sir Benjamin Baker.* In the same way as with the former table it extends from girders of 20 to 200 feet span, and from a load of 10 to 100 cwt. per foot run, thus giving for reference data covering a wide range of conditions which may have to be met in ordinary girder work. TABLE 9. DEPTHS FOR LATTICE GIRDERS. Feet span. 20 Ft. ins. 40 GO 80 100 120 140 160 180 200 Per ft. run. Ft. ins. Ft. ins. Ft. ins. Ft. ins. Ft. ins. Ft. ins. Ft. ins. Ft. ins. Ft. ins. cwt. 10 2 5 4 9 7 9 1 11 2 13 14 11 16 8 18 4 20 20 2 7 5 1 7 5 9 8 11 11 14 16 18 19 11 21 9 SO 2 9 5 4 7 9 10 2 12 6 14 10 16 10 19 21 1 23 40 2 10 5 7 8 1 10 7 13 1 15 6 17 8 20 1 22 1 24 2 50 2 11 5 9 8 5 11 13 7 16 1 18 6 20 11 23 1 25 3 GO 3 5 11 8 8 11 4 14 1 16 8 19 2 21 8 24 26 4 70 3 1 G 1 8 11 11 8 14 6 17 2 19 10 22 4 24 11 27 4 80 3 1 6 2 9 1 12 14 11 17 8 20 4 23 25 8 28 2 90 3 2 6 3 9 3 12 3 15 3 18 20 10 23 7 26 4 29 100 3 2 6 4 9 4 12 5 15 6 18 4 21 2 24 26 10 29 6 " The Strength of Beams, Columns, and Arches." B. Baker, 1870. 2O2 THE BEAM. The extreme variations in the ratio of depth to span for plate and lattice girders as given in Tables 8 and 9 are with s as the span and d the depth. Plate Lattice 200 feet span with load 10 cwt. per foot d = 20 200 20 100 10 100 d = 5.6 s w d = CHAPTER XL ELASTICITY AND DEFLECTION. 130. The Modulus Of Elasticity is the measure within the limits of perfect elasticity of the force required to produce a certain amount of extension or compression in a body in any one direction, as for instance in the length of an iron bar when subjected longitudinally to tensile or com- pressive stress. Now experiments and tests have shown that within such limits each additional unit of stress will cause an additional equal unit of elongation or compression. Thus, it is found that a tensile stress of one pound applied longitudinally will extend a cast-iron bar one square inch in section , therefore 17,000,000 pounds 1 7 000 000 would be required to extend the same bar to twice its original length, if that were possible within the limits of perfect elasticity, and E, the symbol of the modulus of elasticity = 17,000,000, while for wrought-iron it is 24,000,000.* Tables of the modulus of elasticity for materials may be found in treatises which embrace the theory of deflection. The deflection of a beam when transversely strained within the limits of perfect elasticity is (1) directly as the load or stress ; (2) as the cube of the span, because the horizontal * Hodgkinson. For different samples, however, the value of E will vary considerably. 204 THE BEAM. strain with a given load varies directly as the length, while the versed sine of a comparatively short segment of an arc of large radius varies as the square of the length ; (3) and for the same reason inversely as the cube of the depth ; (4) also inversely as the breadth and as the modulus of elasticity. The deflection of a solid rectangular beam carrying a central load is represented by the following equation : Let I = the length . . . . in inches. b breadth .... ,, d depth .... W central load . . . pounds. E modulus of elasticity 8 deflection . . . inches. This equation will apply only so long as the limits of elasticity are not exceeded. 131. The Deflection of solid beams and flanged girders may under the following conditions of load or stress be readily found by equation (28), -in which c is a coefficient determined by actual experimental tests,* / the length in feet, d the depth or diameter, and 8 the deflection in inches. = (28.) The deflection caused by a distributed load will be f ths of that due to an equal central load, and the deflection caused by a distributed load of any fraction of the breaking weight will * " Beams, Columns, and Arches." Sir Benjamin Baker, ELASTICITY AND DEFLECTION. 205 be one fourth more than that due to the same fraction of the central breaking load. Coefficients for solid beams under a central load of one- fourth the breaking weight. Section of Beam. Cast Iron.* "Wrought Iron. Steel. Rectangular .017 .019 .026 Round .019 .021 .029 Coefficients for uniformly loaded girders under tensional stresses in the bottom table of 1-^- tons,, 4J- tons, and 6^ tons for cast iron, wrought iron, and steel respectively. Form of Girder. Cast Iron.* Wrought Iron. Steel. Uniform depth and section .0192 .0135 .0156 Depth and section reduced from j the centre of span to each end \ .0228 .0162 .0185 Coefficients for girders intended to carry a distributed load under tensile proof stresses in the bottom table of 2-J- tons, 5 tons, and 8 tons for cast iron, wrought iron, and steel respectively. Form of Girder. Cast Iron.* Wrought Iron. Steel. Uniform depth and section .0264 .012 .01596 Depth and section reduced from ) the centre of span to each end J .0336 .015 .01992 * The area of the bottom table being four times that of the top table. 2 o6 THE BEAM. Should the deflection of a girder on being tested advance more rapidly than expected, and to a greater extent than that given by careful calculation, the efficiency of the material or the soundness of the structure would be doubtful. INDEX. PAGE Admiralty Tests for Iron ... 118 for Steel ... 122 Angle, Eight, a mode of setting out (Note) 158 the most efficient for Web Bars 157 of Roofs 159 Arcs, Flat, Versed Sines of ... 70 Areas, Computation of 10G Axis, Neutral, and Centre of Gravity 10 to find Position of, 106,113 Bars, Long, with Screwed Ends 124 Tie, to take Thrust ... 152 Vertical Tension, in a Framed Girder 184 Beam, Definition of the Term ... 1 Effect of Length in a 20, 29 and Load, relative position 5 Weight of, how to treat ... 5 Beams, Breadth Efficiency ... 13 Depth ... 12 Circular Hollow ... 103 Diagonal Strain in ... 15 ,, Horizontal ,, ... 9 ,, Positionof Neutral Axisiu 10 Solid, Deflection of ... 205 Formulae for 81, 83 Strength Elements 87-96 Vertical Strain in 13 PAGE Seams with Cantilever Ends with Load at each End and at Centre 48,52 with Load Distributed 53 Effect of Cantilever Ends on the Weight of a Beam ... 50, 55 Practical Examples 56 Beams, Circular Hollow 103 13 earns, Continuous ... 58-71 Formulce for Two Equal fijians (2) with Equal or Unequal Central Loads 60 (3) with Equal or Unequal Distributed Loads 60 Formula for Tivo Unequal Spans (4) with Equal Central Loads 61 (5) with Equally Distributed Loads 61 (6) with Unequal Central Loads 62 (7) with Unequal Distributed Loads 62 Formula) for Three Spans, the Two Outer equal (8) with Equally Distributed Load 63 Formula) for Three Equal Spans-*- (9) with Equally Distributed Load , 63 208 INDEX. PAGE Beams, Continuous continued. Formula; for Three Spans, tlie Tmo Outer equal, the Cen- tral Span equal to the Two (10) with Equally Distributed Load 63 Formula? if Central Span = (11) With Equally Distri- buted Load 64 Determination of Strains in Tn-o Equal Spans with Two Equal Central Loads 64 Equally Distributed Load 67 Comparative Weight of a Beam of Two Equal Spans when Con- tinuous and Non- Continuous 66, 69 Comparative Deflection of a Beam of Two Equal Spans when Con- tinuous and Non-Continuous 70 Board of Trade Tests for Railway Bridges ... 80 Regulations as to Wind (Note) ... 79 Bolts, Rods, and Nuts, Screwed ... 123 Brickwork, Strength of 76 Weight of 78 Bridge, 40-feet Span Design for a 143 Bridges, Board of Trade Regula- tions ... 79,80 Cantilever, Definition of 3 Example with Load at End 45 Example with Load Distributed... . ... 46 Carbon in Cast Iron ... 107, 122 Wrought Iron and Steel 122 PAGE Cast Iron, Coefficients for ... 74 Weight of 77 Girders, Board of Trade Regulations 108 Girders, Deflection of... 205 Depth of ... 108 Formula} for Depth of ... 109 Section for ... 110 Strength Ele- ments 100, 113 Centre of Gravity and Neutral Axis 10 of a Surface ... 106 Circular Hollow Beam 103 Classification of Pig Iron 107 Coefficients of Efficiency 74 Concentrated Rolling Load 36, 80 Continuous Beams (see Beams, Continuous) Contracts for Girders 117 Cost, Proportionate, of Iron and Steel Girders 132 Cover Plates and Joints 129 Crane Post 103 Crowd of Persons, Weight of ... 79 Crushing Resistance of Materials 75 Curves, Plat Segmental, Versed Sines of 70 Data for the Loading of Structures 78 Decimals compared with Duo- decimals 78 Deflection, Coefficients for ... 205 Formulae for 2C4 Depth of Beams, Efficiency of ... 12 Cast-iron Girders ... 108 INDEX. 209 PAGE Depth of Plate Girders 133 Table of 135 Design for a Girder Bridge ... 143 Determination of Stresses in a Beam with Load Central, Support at each End 18 with Load at each End, Support Central 21 with Load Distributed, Support at each End 22 with Load Distributed, Support Central 27 with Load Non-Central, Support at each End 30 with Load at each End, Support Non-Central 32 with Load Two Equidistant, Support at each End 32 with Load at each End, Support Two Equidistant 34 with Load Various. Support at each End 34 with Load, Single Rolling, Sup- port at each End 36 with Load, Double Rolling, Sup- port at each End 38 with Load, Distributed ad- vancing, Support at each End 41 Diagonal Strain in Beams ... 15 Diameters of Screwed Bolts, less Threads, Table of 123 Doors. Position of Hinges and Ledges... 57 Effect of Length in a Beam 20, 29 Efficiency of Breadth in Beams ... 13 ,, Coefficients of ... 74 PAGE Efficiency of Depth in Beams ... 12 Elasticity, Modulus of 203 Error in computing the Strength of Beams 102-105 Extras on Plate Iron 117 on Angle Iron and T Bars 118 Fastenings for Gates and Valves 57 Fir Timber,Coefficients forStrength 74 Weight of 78 Firm Earth, Safe Load for ... 7fi Flexure, Resistance of 84 Floats of Paddle Wheels 5f> Formula? for Solid Beams. . . 81-84 Girders ... 1)8-101 Depth of Cast-iron Girders 109 Framed Girders 155-202 Gates and Valves, Hinges opposition 57 Girder, Definition of Term ... 2 Office of Web and Tables 4 Work, Contracts for ... 117 Girders, Cast-iron, Board of Trade Regulations 108 Cast-iron, Depth of ... 108 Cast-iron, Depth of, Formula} for 109 Cast iron, Proportions for Members ... 110.113 ,, Cast-iron, Strength Ele- ments 100. 113 Cast-iron, with Load Dis^ tributed 113 Deflection, Formulae for 204 Flexure, Rule for deter- mining Resistance of 101 210 INDEX. PAGE Girders, Formula) for a Central Load 98 Formulae for a Distributed Load 99 Framed, most efficient Angle for Web Bars 157 Framed, Useful Length Ratio for Web Bars ... 158 Framed, 1 System, Load Central ... 160,163 Framed, 1 System, Load Distributed ... 161, 171 Framed, 1 System, Load Rolling 175 Framed, 2 System, Load Central 166 Framed, 2 System, Load Distributed ... 177, 184 Framed, 2 System Load Rolling ... 181, 187 ., Lattice, 4 System, Load Central 169 .. Lattice, 4 System, Load Distributed 189 ., Lattice, 4 System, Load Rolling 197 Lattice, Table of Depths for 201 Plate, Comparative Weight of Two, of the same Strength ... ... 153 Plate, Depths of ... 133-136 Table 135 .. Details of a 136-143 143-152 Plate, Deviations from Theory 143 PAGE Girders, Proportionate Cost of Iron and Steel 132 Varieties of 2 Wind Trussing for ...154 Granite,. Safe Load for 76 Weight of 78 Hand Riveting 124 Heads of Rivets 125 Hexagon Nuts 123 Hinges, Position of, for Gates ani Valves 57 Hollow Beams, Circular 103 Rectangular ... 102 Horizontal Strain, First Principles of 9 Hydraulic Riveting ... ... 124 Testing 73 Iron, Angle and T, Admiralty Tests for 120 Iron, Angle and T, Lengths of ... 118 Extras on ... 118 Bars, Admiralty Tests for -119 ,, Cast, Board of Trade Regu- lations ... ... 108 ,. Carbon in 122 Coefficients for ... 74 ,, Ultimate Transverse Strength of 108 Weight of 77 Plate, Weight of 78 Plates, Sizes and Weight of 117 Extras on 117 ,. Rivet, Tensile Strength of 126 Rolled, Coefficients for ... 74 INDEX. 211 Iron, Wrought, Board of Regulations Joints an'd Covers ... Joists Rolled PAGE Trade ... 121 ... 129 61 Lateral Ribs on Cast-iron Girders 113 Lattice Girders 169, 189 Depths for ... 201 Law of Moments ... 6 Length, Useful Ratio for Web Bars 1 58 Lever, the Principle and Varieties of 7 Loading of Structures, Data for 78 Materials, Crushing Resistance of 75 Safe Loads for ... 76 Mensuration of Surfaces 106 Modulus of Elasticity 203 Moment, Definition of ... ... 6 Municipal Bye-law for the Width of Roads and Streets (Note) 136 Neutral Axis and Centre of Gravity 10 to find Position of 106, 113 Nuts for Bolts, Proportions of ... 123 Oak Timber, Coefficients for Strength ... 74 Weight of ... 78 Office of Web and Tables 4 Paddle Wheels, Floats of ... 56 Parallelogram of Pressures 8, 155 Pig Iron. Classification of ... 107 PAGE Pitch of Riveting 128 Plate Girders, Depths ... 133-136 Table of ... 135 Examples of 136, 143 Plate Iron, Weight of 78 Plates, Iron, Sizes and Weights of 117 Extras on 117 Pressure and Resistance 19 of Wind 79 Proportionate Cost of Iron and Steel Girders ... 132 Punching and Drilling Iron and Steel 128 Punching and Drilling Tables or Benches 142 Purlin, Definition of (Note) ... 4 Railway Bridges, Board of Trade Tests 80 Regulations, Board of Trade, for Bridges (Note) 79 Regulations, Municipal, for Width of Roads and Streets (Note) ... 136 Resistance to Flexure 101 Ribs, Lateral, on Cast-iron Girders 113 Right Angle, to set out a ... 158 Rivet Heads 125 ,, Iron, Tensile Strength of ... 126 Rivets and Plate Thicknesses, Table of 127 Riveting, Hand and Hydraulic ... 124 Pitch of 128 Roads and Streets, Width of (Note) 136 Rods, Long, with Screwed Ends ... 124 Rolled Iron Joists or Beams ... 51 Roofs, Angles of 159 212 INDEX. PAGE Snow, Weight of 79 Solid Beams, Rectangular, Formula for 81 Solid Beams, Round, Formulas for 83 Strength Elements in Square ... 87, 91 Strength Elements in Round 91 Steel, Admiralty Tests for ...122 Angle and T Bars, Rolled Lengths 122 Carbon in 122 Plates, Table of Sizes and Thicknesses .... ... 121 Rivets ... 128 Weight of 78 Strain, Diagonal, in Web 15 Vertical 43 Horizontal, First Principles of 9 Transverse, Nature of ... 84 Vertical or Shearing ... 13 Strains, Determination of 18 45 Strength Elements in a Cast-iron Girder 100, 113 in a Circular Beam ... 103 Safe, of Iron, Steel, Oak, and Fir 74 Stress and Strain. Nature and Effects of 4 Strut, a Definition of the Term ... 1 Surfaces, Computation of ... 106 Suspension Bridge Chains, Nuts for 124 Table 1 , Coefficients of Efficiency 74 2, Safe Loads for Materials 76 PAGE Table 3, Weight of Cast Iron ... 77 4, Weight of Materials ... 78 5, Steel Plates, Size and Thickness ... .... 121 6, Bolts, Diameters when screwed 123 7, Rivets, and Plate Thick- nesses ... 127 8, Depths for Plate Girders 135 9, Lattice 201 Tables of Girders, Office of ... 4 T Iron, Rolled Lengths, and Extras on 118 Tests,Admiralty,for Rolled Iron 118,120 Steel 122 Threads, Screwed, Reduction of Diameter by 123 Tie, a Definiton of the Term ... 1 Bars to take Thrust ...152 Timber, Coefficients for Strength 74 Weight of 78 Transverse Strength, Nature of 5, 84 Truss, Definition of ... ... 3 Trussing to resist Wind 154 Unit of Length, Weight, Stress or Strain ... 5 Useful Length Ratio for Web Bars 158 ,, Numbers for Setting out a Right Angle 158 Valves, Position of Hinges for ... 57 Versed Sines of Segmental Curves 70 Vertical or Shearing Strain, Prin- ciples of 13 Tension Bars in a Framed Girder... ... 184 INDEX. 213 PAGE Water, Weight of 79 Web, Area of 44 Examples ... 141, 149 Office of 4 Weight of a Beam, how to treat... 5 Iron, Cast ... ... 77 Rolled 78 .. Materials, c 78 PAGE Weight of Steel 78 Wind, Board of Trade Regulations as to (Note) 79 Pressure of 79 Trussing to resist ... ... 151 York Stone, Strength of 76 Weight of . , 78 PRIXTED BY NICHOLS & SONS, 25, PARLIAMENT STREET, S.W. THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $1.OO ON THE SEVENTH DAY OVERDUE. m m LD 21-95w 7,'37