LIBRARY UNIVERSITY OF CALIFORNIA. Class UftfY^RSITY OF CALIFORNIA Accessions No.. MECHANICS APPLIED TO ENGINEERING MECHANICS APPLIED TO ENGINEERING BY JOHN OODMAN \V H . SCH., M.I.C.E., M.I.M.E. PBOFESSOR OF ENGINEERING IN THE UNIVERSITY OF LEEDS With 714 Illustrations and Numerous Examples UNIVERSITY \sixTff EDITION LONGMANS, GREEN, AND CO. 39 PATERNOSTER ROW, LONDON NEW YORK, BOMBAY, AND CALCUTTA 1908 All rights reserved o PREFACE THIS book has been written especially for Engineers and Students who already possess a fair knowledge of Elementary Mathematics and Theoretical Mechanics ; it is intended to assist them to apply their knowledge to practical engineering problems. Considerable pains have been taken to make each point clear without being unduly diffuse. However, while always aiming at concis_eness, the short-cut methods in common use have often and intentionally been avoided, because they appeal less forcibly to the student, and do not bring home to him the principles involved so well as do the methods here adopted. Some of the critics of the first edition expressed the opinion that Chapters I., II., III. might have been omitted or else con- siderably curtailed ; others, however, commended the innovation of introducing Mensuration and Moment work into a book on Applied Mechanics, and this opinion has been endorsed by readers both in this country and in the United States. In addition to the value of the tables in these chapters for reference purposes, the worked-out results afford the student an oppor^ cunity. of reviewing the methods adopted. The Calculus has been introduced but sparingly, and then only in its most elementary form. That its application does not demand high mathematical skill is evident from the working out of the examples in the Mensuration and Moment chapters. For the benefit of the beginner, a very elementary sketch of the subject has been given in the Appendix; it is hoped that he will follow up this introduction by studying such works as those by Barker, Perry, Smith, Wansbrough, or others. For the assistance of the occasional reader, all the symbols employed in the book have been separately indexed, with the exception of certain ones wliich only refer to the illustrations in their respective accompanying paragraphs. 1 731.98 vi Preface. In this (fourth) edition, some chapters have been con- siderably enlarged, viz. Mechanics ; Dynamics of Machinery ; Friction ; Stress, Strain, and Elasticity ; Hydraulic Motors and Machines ; and Pumps. Several pages have also been added to many of the other chapters. A most gratifying feature in connection with the publication of this book has been the number of complimentary letters received from all parts of the world, expressive of the help it has been to the writers ; this opportunity is taken of thanking all correspondents both for their kind words and also for their trouble in pointing out errors and misprints. It is believed that the book is now fairly free from such imperfections, but the author will always be glad to have any pointed out that have escaped his notice, also to receive further suggestions. While remarking that the sale of the book has been very gratifying, he would particularly express his pleasure at its reception in the United States, where its success has been a matter of agreeable surprise. The author would again express his indebtedness to all who kindly rendered him assistance with the earlier editions, notably Professor Hele-Shaw, F.R.S., Mr. A. H. Barker, B.Sc., Mr. Andrew Forbes, Mr. E. R. Verity, and Mr. J. W. Jukes. In preparing this edition, the author wishes to thank his old friend Mr. H. Rolfe for many suggestions and much help ; also his assistant, Mr. R. H. Duncan, for the great care and pains he has taken in reading the proofs ; and, lastly, the numerous correspondents (most of them personally unknown to him) who have sent in useful suggestions, but especially would he thank Professor Oliver B. Zimmerman, M.E., of the University of Wisconsin, for the " gearing " conception employed in the treatment of certain velocity problems in the chapter on " Mechanisms." JOHN GOODMAN. THE UNIVERSITY OF LEEDS, August, 1904. CONTENTS CHAP. PAGE I INTRODUCTORY i II. MENSURATION . 20 III. MOMENTS 50 IV. RESOLUTION OF FORCES 106 V. MECHANISMS 119 VI. DYNAMICS OF THE STEAM-ENGINE 160 VII. FRICTION 227 VIII. STRESS, STRAIN, AND ELASTICITY . . .... 290 IX. BEAMS 353 X. BENDING MOMENTS AND SHEAR FORCES 392 XL DEFLECTION OF BEAMS 424 XII. COMBINED BENDING AND DIRECT STRESSES ... 452 XIII. STRUTS 461 XIV. TORSION. GENERAL THEORY , . ..... , 479 XV. STRUCTURES 501 XVI. HYDRAULICS 54i XVII. HYDRAULIC MOTORS AND MACHINES 586 XVIII. PUMPS 6 3i APPENDIX . . . <.>; . 6 7 r EXAMPLES . .'--. t .. .' . *. . INDEX * Y . ." 721 MECHANICS APPLIED TO ENGINEERING CHAPTER I. INTRODUCTORY. THE province of science is to ascertain truth from sources far and wide, to classify the observations made, and finally to embody the whole in some brief statement or formula. If some branches of truth have been left untouched or unclassi- fied, the formula will only represent a part of the truth ; such is the cause of discrepancies between theory and practice. A scientific treatment of a subject is only possible when our statements with regard to the facts and observations are made in definite terms ; hence, in an attempt to treat such a subject as Applied Mechanics from a scientific standpoint, we must at the outset have some means of making definite state- ments as to quantity. This we shall do by simply stating how many arbitrarily chosen units are required to make up the quantity in question. We shall find that some of our units will be of a very com- plex character, but in every instance we shall be able to express them in terms of three fundamental units, viz. those of time^ mass, and space. The complex units are usually termed " derived " units. Units. Time (f). Unless otherwise stated, we shall take one second as the unit of time, but sometimes we shall find it convenient to take minutes and hours. Mass (M). Unit, one pound ; occasionally hundredweights and tons. i pound (Ib. ) = 0*454 kilogramme. i kilogramme = 2 '2 Ibs. I hundredweight (cwt.) = 50*8 kilos. I ton = 1016 ,, (tonneau or Millier). i tonneau or Millier = o'984 ton. B 2 Mechanics applied to Engineering. Space (s). Unit, one foot ; occasionally inches, yards, and miles. Such terms as distance, length, breadth', width, thickness, are fre- quently used to denote space in various directions. i sq. foot = 0*0929 sq. metre, i sq. metre = 10*764 sq. feet. i sq. inch = 6*451 sq. cms. i sq. mm. = 0*00155 sq. inch, i kilogramme = 2*2046 Ibs. no. per sq. I _ ^. 22 ^ ^ & ^ j nc j 1> I Ib. sq. inch = 0*0703 kilo. sq. cm. I cubic inch = 16*387 c. cms. I cubic foot = 0*0283 cubic metre. foot = 0*305 metre, metre =3*28 feet, inch = 25*4 millimetres, millimetre = 0*0394 inch, yard = 0^914 metre, metre = I '094 yards, mile = 1609*3 metres, kilometre = 1093 '63 yards. = 0*621 mile. Dimensions. The relation which exists between any given complex unit and the fundamental units is termed the dimensions of the unit. As an example, see p. 20, Chapter II. Speed. When a body changes its position relatively to surrounding objects, it is said to be in motion. The rate at which a body changes its position is termed the speed of the body. Uniform Speed. A body is said to have uniform speed when it traverses equal spaces in equal intervals of time. The O 1 2 3 * S Time ~in seconds Uniform/ sjieetL FIG. i. body is said to have unit speed when it traverses unit space in unit time. , /. r . ,v space traversed (feet) s bpeed (in feet per second) = : ; - - = - time (seconds) / Introductory. 3 Varying Speed. When a body does not traverse equal spaces in equal intervals of time, it is said to have a varying speed. The speed at any instant is the space traversed in an exceedingly short interval of time divided by that interval; the shorter the interval taken, the more nearly will the true speed be arrived at. In Fig. i we have a diagram representing the distance travelled by a body moving with uniform speed, and in Fig. 2, varying speed. The speed at any instant, a, can be found by drawing a tangent to the curve as shown. From the slope of this tangent we see that, if the speed had been O 1 Z 3 * 5 Time in seconds Varying sjieed* FIG. 2. uniform, a space of 4-9 1*4 = 3-5 ft. would have been traversed in 2 sees., hence the speed at a is = 175 ft. per 2 second. Similarly, at b we find that 9 ft. would have been traversed in 5-2 2*3 = 2-9 sees., or the speed at b is -^- = 3'i ft. per second. The same result will be obtained by taking any point on the tangent. For a fuller discussion of variable quantities, the reader is referred to either Perry's or Barker's Calculus. Velocity (v). The velocity of a body is the magnitude of its speed in any given direction ; thus the velocity of a body may be changed by altering the speed with which it is moving, or by altering the direction in which it is moving. It does not 4 Mechanics applied to Engineering. follow that if the speed of a body be uniform the velocity will be also. The idea of velocity embodies direction of motion, that of speed does not. The speed of a point on a uniformly revolving wheel is constant, but the velocity is changing at every instant. Velocity and speed, however, have the same dimensions. The unit of velocity is usually taken as i foot per second. Velocity in feet 1 _ space (feet) traversed in a given direction per second j " time (seconds) v = -, or s = vt i ft. per second = 0*305 metre per second = 0*682 mile per hour = i -i kilo. . i metre per second = 3*28 ft. per second i mile per hour { I m ^ p r 0-278 metre Ikil - {I " Angular Velocity (>), or Velocity of Spin. Suppose a body to be spinning about an axis. The rate at which an angle is described by any line perpendicular to the axis is termed the angular velocity of the line or body, or the velocity of spin ; the direction of spin must also be specified, as in the case of linear velocity. When a body spins round in the direction of the hands of a watch, it is termed a + or positive spin ; and in the reverse direction, a or negative spin. As in the case of linear velocity, angular velocity may be uniform or varying. The unit of angular measure is a " radian ; " that is, an angle subtending an arc equal in length to the radius. The length of G a circular arc subtending an angle 6 is 2-n-r X -^-5, where TT 300 is the ratio of the circumference to the diameter (zr) of a circle, and is the angle subtended (see p. 22). Then, when the arc is equal to the radius, we have 2irrO _ 360 practically 57*3 Introductory. 5 Thus, if a body be spinning in such a manner that a radius describes 100 degrees per second, its angular velocity is TOO i- i r = z>, or to = -. r Angular velocity in radians per sec. = linear velocity (ft per sec.) radius (ft.) The radius is a space quantity, hence _ s _ i ~ Ts~ 1 Thus an angular velocity is not affected by the unit of space adopted, and only depends on the time unit, but the time unit is one second in all systems of measurement, hence all angular measurements are the same for all systems of units an important point in favour of using angular measure. Acceleration (f a ) is the rate at which the velocity of a body increases in unit time that is, if we take feet and seconds units, the acceleration is the number of feet per second that the velocity increases in one second ; thus, unit acceleration is an increase of velocity of one foot per second per second. It should be noted that acceleration is the rate of change of velocity, and not merely change of speed. The speed of a body in certain cases does not change, yet there is an acceleration due to the change of direction (see p. 18). As in the case of speed and velocity, acceleration may be either uniform or varying. Uniform ac- celeration _ increase of velocity in ft. per sec. in a given time in feet per ~ time in seconds sec. per sec. /= JL y- 1 = j hence v =/./, or z/ 2 - v l =f a t . . . . (i.) 6 Mechanics applied to Engineering. where z> 2 is the velocity at the end of the interval of time, and #1 at the beginning, and z> is the increase of velocity. In Fig. 3, the vertical distance of any point on any line ab from the base line shows the velo- city of a body at the corre- sponding instant : it is straight because the acceleration is as- sumed constant, and therefore the velocity increases directly as the time. If the body start from rest, when v is zero, the mean velocity over any inter- FIG. 3. val of time will be , and the space traversed in the interval will be the mean velocity X time, or Time in, seconds s = / *?&- (see equation i.) Acceleration in feet per sec. per sec. = constant X space (in/*/) (time) 2 (in seconds) When the body has an initial velocity z/ 1 , the mean velocity during the time / is the mean height of the figure oabc. Mean velocity = v * + v * = v + 2Z/1 = v, + /a/ (ii.) 22 (see equation i.) The space traversed in the time / s = JJ*L \t On.) which is represented in the diagram by the area of the diagram oabc. By multiplying equations i. and ii., we get Substituting from iii., we get }, 2 7 ; 2 1 - VS = 2f a S or v? = z>! 2 + 2/ a s Introductory. 7 When a body falls freely due to gravity,^ = g = 32*2 ft. per second per second, it is then usual to use the letter h, the height through which the body has fallen, instead of s. When the body starts from rest, we have v l - o, and v z = v ; then by substitution from above, we have v = *J 2gh = 8*02 *J h .... (iv.) Momentum (M ). If a body of mass M l move with a velocity v, the moving mass is said to possess momentum, or quantity of motion, Mz/. Unit momentum is that of unit mass moving with unit velocity ,, ,, M.S M = Mz> = Impulse. Consider a ball of mass M travelling through the air with a velocity z/ 1} and let it receive a fair blow in the line of motion (without causing it to spin) as it travels along, in such a manner that its velocity is suddenly increased from V L to v* The momentum before the blow = Mz'j after = M# 2 The change of momentum due to the blow = M(v 2 2^) The effect of the blow is termed an impulse, and is measured by the change of momentum. Impulse = change of momentum = M(z> 2 #1) Force (F). If the ball in the paragraph above had received a very large number of very small impulses instead of a single blow, its velocity would have been gradually changed, and we should have had The whole impulse per second = the change of momentum per second When the impulses become infinitely rapid, the whole impulse per second is termed the force acting on the body. Hence the momentum may be changed gradually from M^ to M z v 2 by a force acting for / seconds. Then 1 For a rational definition of mass, the reader is referred to Prof. Karl Pearson's " Grammar of Science," p. 357. Mechanics applied to Engineering. F/ = M(z; a PI) , -p _ total change of momentum tme But Z ' 2 ~ Vl =/ B (acceleration) (see p. 5) hence F = M/ = Hence the dimensions of this unit are Force = mass X acceleration Unit force = unit mass X unit acceleration Thus unit force is that force which, when acting on a mass of one |P oun J for one second, will change its velocity by We are now in a position to appreciate the words of Newton Change of momentum is proportional to the impressed force, and takes place in the direction of the force ; . . . also, a body will remain at rest, or, if in motion, will move with a uniform velocity in a straight line unless acted upon by some external force. Force simply describes how motion takes place, not why it takes place. It does not follow, because the velocity of a body is not changing, or because it is at rest, that no forces are acting upon it; for suppose the ball mentioned above had been acted upon by two equal and opposite forces at the same instant, the one would have tended to accelerate the body backwards (termed a negative acceleration, or retardation) just as much as the other tended to accelerate it forwards, with the result that the one would have just neutralized the other, and the velocity, and consequently the momentum, would have remained un- changed. We say then, in this case, that the positive acceleration is equal and opposite to the negative acceleration. If a railway train be running at a constant velocity, it must not be imagined that no force is required to draw it ; the force exerted by the engine produces a positive acceleration, while 1 The poundal unit is never used by engineers. Introductory. 9 the friction on the axles, tyres, etc., produces an equal and opposite negative acceleration. If the velocity of the train be constant, the whole effort exerted by the engine is expended in overcoming the frictional resistance, or the negative accelera- tion. If the positive acceleration at any time exceeds the negative acceleration due to the friction, the positive or forward force exerted by the engine will still be equal to the negative or backward force or the total resistance overcome ; but the resistance now consists partly of the frictional resistance, and partly the resistance of the train to having its velocity increased. The work done by the engine over and above that expended in overcoming friction is stored up in the moving mass of the train as energy of motion, or kinetic energy (see p. 14). UNITS OF FORCE. Force. Mass. Acceleration. Poundal. One pound. One foot per second per second. Dyne. One gram. One centimetre per second per second. i poundal = 13,825 dynes. I pound = 445,000 dynes. Weight (W). The weight of a body is the force that gravity exerts on that body. It depends (i) on the mass of the body ; (2) on the acceleration of gravity (g), which varies inversely as the square of the distance from the centre of the earth, hence the weight of a body depends upon its position as regards the centre of the earth. The distance, however, of all inhabited places on the earth from the centre is so nearly constant, that for all practical purposes we assume that the acceleration of gravity is constant (the extreme variation is about one-third of one per cent.). Consequently for practical purposes we compare masses by their weights. Weight = mass X acceleration of gravity We have shown above that Force = mass X acceleration l 1 Expressing this in absolute units, we have Weight or force (poundals) = mass (pounds) x acceleration (feet per second per second) Then- Force of gravity on a mass of one pound = I X 32-2 = 32^2 poundals But, as poundals are exceedingly inconvenient units to use for practical 5 io Mechanics applied to Engineering. hence we speak of forces as being equal to the weight of so many pounds; but for convenience of expression we shall speak of forces of so many pounds, or of so many tons, as the case may be. VALUES OF g. 1 In centimetre- In foot-pounds, sees. grammes, sees. The equator ...... 32*091 ... 978*10 London* ......... 32*191 ... 981*17 The pole ...... 32*255 ... 983*11 Work. When a body is moved so as to overcome a resist- ance, we know that it must have been acted upon by a force acting in the direction of the displacement. The force is then said to perform work, and the measure of the work done is the product of the force and the displacement. The absolute unit of work is unit force (one poundal) acting through unit dis- placement (foot), or one foot-poundal. Such a unit of work is, however, never used by engineers ; the unit nearly always used in England is the "foot-pound," i.e. one pound weight lifted one foot high. Work = force X displacement = FS The dimensions of the unit of work are therefore purposes, we shall adopt the engineer's unit of one pound, i.e. a unit 32*2 times as great ; then, in order that the fundamental equation may hold for this unit, viz. Weight or force (pounds) = mass X acceleration we must divide our weight or force expressed in poundals by 32*2, and we get Weight or force (pounds) = weight or force (poundals) mass X acceleration 32*2 32-2 or mass in pounds weight or force (pounds) = -- - X acceleration in ft. -sec. per sec. Thus we must take our new unit of mass as 32*2 times as great as the absolute unit of mass. Readers who do not see the point in the above had better leave it alone at any rate, for the present, as it will not affect any question we shall have to deal with. As a matter of fact, engineers always do (probably unconsciously) make the assumption, but do not explicitly state it. 1 Hicks's " Elementary Dynamics," p. 45. Introductory. 1 1 Frequently we shall have to deal with a variable force acting through a given displacement; the work done is then the average 1 force multiplied by the displacement. Methods of finding such averages will be discussed later on. In certain cases it will be convenient to remember that the work done in lifting a body is the weight of the body multiplied by the height through which the centre of gravity of the body is lifted. UNITS OF WORK. Force. Displacement. Unit of work. Pound. Foot. Foot-pound. Kilogram. Metre. Kilogrammetre. Dyne. Centimetre. Erg. i foot-pound = 32*2 foot-poundals. = 13,560,000 ergs. Power. Power is the rate of doing work. Unit power is unit work done in unit time, or one foot-pound per second. Power = total work done time taken to do it / The dimensions of the unit of power are therefore ^-. The unit of power commonly used by engineers is an arbitrary unit established by James Watt, viz. a horse-power, which is 33,000 foot-pounds of work done per minute. Horse-power __ foot-pounds of work done in a given time ~~ time (in minutes) occupied in doing the work X 33,000 i horse-power = 33,000 foot-pounds per minute = 7*46 X io 9 ergs per minute. i French horse-power = 32,500 foot-pounds per minute = 7*36 x io 9 ergs per second. i watt =746 foot-pounds per minute = io 7 ergs per second. Couples. When forces act upon a body in such a manner as to tend to give it a spin or a rotation about an axis without any tendency to shift its c. of g., the body is said to be acted 1 Space-average. 12 Mechanics applied to Engineering. upon by a couple. Thus, in the figure the force F tends to turn the body round about the point O, its c. of g. If, however, this were the only force acting on the body, it would have a motion of translation in the direction of the force as well as a spin round the axis ; in order to prevent this motion of translation, another force, F^ equal and parallel but opposite in direc- tion to F, must be applied to the body in the same plane. Thus, a couple is said to consist of two parallel forces of equal magnitude acting in opposite directions, but not in the same straight line. 77c. 4. The perpendicular distance x between the forces is termed the arm of the couple. The tendency of a couple is to turn the body to which it is applied in the plane of the couple. When it tends to turn it in the direction of the hands of a watch, it is termed a clockwise, or positive (+) couple, and in the contrary direction, a contra-clockwise, or negative ( ) couple. It is readily proved * that not only may a couple be shifted anywhere in its own plane, but its arm may be altered (as long as its moment is kept the same) without affecting the equili- brium of the body. Moments. The moment of a couple is the product of one of the forces and the length of the arm. It is usual to speak of the moment of a force about a given point that is, the product of the force and the perpendicular distance from its line of action to the point in question. As in the case of couples, moments are spoken of as clock- wise, contra-clockwise, etc. If a rigid body be in equilibrium under any given system of moments, the algebraic sum of all the moments in any given plane must be zero, or the clockwise moments must be equal to the contra-clockwise moments in any given plane. Moment = force X arm The dimensions of a moment are therefore . 1 See Hicks's " Elementary Mechanics." Introductory. 1 3 Centre of Gravity (c. of g.). The gravitation forces acting on the several particles of a body may be considered to act parallel to one another. If a point be so chosen in a body that the sum of the moments of all the gravitation forces acting on the several particles about the one side of any straight line passing through that point be equal to the sum of the moments on the other side of the line, that point is termed the centre of gravity of the body. . Thus, the resultant of all the gravitation forces acting on a body passes through its centre of gravity, however the body may be tilted about. Centroid. In certain cases in which parallel systems of forces are concerned, the point referred to in the last paragraph is frequently termed the centroid ; such cases are fully dealt with in Chapter III. Energy. Capacity for doing work is termed energy. Conservation of Energy. Experience shows us that energy cannot be created or destroyed ; it may be dissipated, or it may be transformed from any one form to any other, hence the whole of the work supplied to any machine must be equal to the work got out of the machine, together with the work converted into heat, 1 either by the friction or the impact of the parts one on the other. Mechanical Equivalent of Heat. It was experiment- ally shown by Joule that in the conversion of mechanical into heat energy, 2 772 foot-lbs. of work have to be expended in order to generate one thermal unit. Efficiency of a Machine. The efficiency of a machine is the ratio of the useful work got out of the machine to the gross work supplied to the machine. r^rc . work got out of the machine Efficiency = - $ ^, : . work supplied to the machine This ratio is necessarily less than unity. The counter-efficiency is the reciprocal of the efficiency, and is always greater than unity. Counter-efficiency = ^ supplied to the machine work got out of the machine 1 To be strictly accurate, we should also say light, sound, electricity, etc. 2 By farTthe most accurate determination is that recently made by Pro- fessor Osborne Reynolds and Mr. W. H. Moorby, who obtained the value 776-94 (see Phil. Trans., vol. 190, pp. 301-422) from 32 F. to 212 F., which is equivalent to about 773 at 39 F. 14 Mechanics applied to Engineering. Kinetic Energy. From the principle of the conservation of energy, we know that when a body falls freely by gravity, the work done on the falling body must be equal to the energy of motion stored in the body (neglecting friction). The work done by gravity on a weight of W pounds in falling through a height h ft. = WA foot-lbs. But we have shown above that h = , where v is the velocity after falling through a height h ; whence 2< This quantity, , is known as the kinetic energy of the 2 ^ body, or the energy due to its motion. Inertia. Since energy has to be expended when the velocity of a body is increased, a body may be said to offer a resistance to having its velocity increased, this resistance is known as the inertia of the body. Inertia is sometimes denned as the capacity of a body to possess momentum. 1 Moment of Inertia (I). We have denned inertia as the capacity of a body to possess momentum, and momentum as the product of mass and velocity (Mv). If we have a very small body of mass M rotating about an axis at a radius r, with an angular velocity p and the momentum will beMz/. But if the body be shifted further from the axis of rotation, and r be thereby in- creased, the momen- F IG . 5 . turn will also be in- creased in the same ratio. Hence, when we are dealing with a rotating body, we have not only to deal with its mass, but with the arrangement of the body about the axis of rotation, i.e. with its moment about the axis. Let the body be acted upon by a twisting moment, Pr = T, 1 Hicks's " Elementary Dynamics," p. 28. Grooved jutUey considered, as TjueiqTvtless. Introductory. 1 5 then, as the force P acts at the same radius as that of the body, it may be regarded as acting on the body itself. The force P acting at a radius r will produce the same effect as a force ?iP acting at a radius . The force P acting on the mass M gives it a linear acceleration f M where P = M/" a , or P i f a = _ . The angular velocity o> is - times the linear velocity, hence the angular acceleration is - times the linear accelera- tion. Let A = the angular acceleration ; then A -/= JL-- _?L---!L " r Mr Mr 2 Mr 1 or angular acceleration = twisting moment 1 mass X (radius) 2 In the case we have just dealt with, the mass M is supposed to be exceedingly small, and every part of it at a distance r from the axis. When the body is great, it may be considered to be made up of a large number of small masses, M 15 M 2 , etc., at radii r i> r ?n etc -> respectively ; then the above expression becomes , (M^ 2 -f- M 2 r 2 2 + M 3 r 3 2 +, etc.) The quantity in the denominator is termed the " moment of inertia " of the body. We stated above that the capacity of a body to possess momentum is termed the " inertia of the body." Now, in a case in which the capacity of the body to possess angular momentum depends upon the moment of the several portions of the body about a given axis, we see why the capacity of a rotating body to possess momentum should be termed the " moment of inertia." Let M = mass of the whole body, then M = Mj+Mg+Mg, etc. ; then the moment of inertia of the body, I, = M* 2 = (M^! 2 + M 2 ; 2 2 , etc.). Radius of Gyration (K). The K in the paragraph above is known as the radius of gyration of the body. Thus, if we could condense the whole body into a single particle at a distance K from the axis of rotation, the body would still have 1 The reader is advised to turn back to the paragraph on "couples," so that he may not lose sight of the fact that a couple involves two forces. Mechanics applied to Engineering. the same capacity for possessing energy, due to rotation about that axis. Representation of Displacements, Velocities, 1 Accelerations, Forces by Straight Lines. Any displacement cekration * s ^ u ^ represented when we state its magni- force tude and its direction, and, in the case of force, its point of application. Hence a straight line may be used to represent any (displacement J velocity I acceleration , the length of which represents its magni- vforce tude, and the direction of the line the direction in which the force, etc., acts, displacements velocities accelerations forces be replaced by one force, etc,, passing through the same point, which is termed the resultant force, etc, displacements Two or more , meeting at a point, may If two velocities accelerations forces , not in the same straight line, meeting at a point , and let its velocity at A be represented by the radius OP, the inner circle being the hodograph of A. Now let A move through an extremely small space to A 1} and the corresponding radius vector to OPi ; then the line PP a FlG g represents the change in velocity of A while it was moving to Aj. (The reader should never lose sight of the fact that change of velocity involves change of direction as well as change of speed, and as the speed is constant in this case, the change of velocity is wholly a change of direction.) As the distance AA : becomes smaller, PPj becomes more nearly perpendicular to OP, and in the limit it does become perpendicular, and parallel to OA ; thus the change of velocity is radial and towards the centre. We have shown on p. 17 that the velocity of P represents the acceleration of the point A ; then, as both circles are de- scribed in the same time velocity of P _ OP velocity of A ~~ OA But OP was made equal to the velocity of A, viz. v t and OA is the radius of the circle described by the body. Let OA = R; then- velocity of P v ~~ or velocity of P = -fr 1 For another method of treatment, see Barker's " Graphic Methods of Engine Design." Introductory. 19 and acceleration of A = JT and since force = mass x acceleration we have centrifugal force C = -r- or in gravitational units, C = ^ This force acts radially outwards from the centre. Sometimes it is convenient to have the centrifugal force expressed in terms of the angular velocity of the body. We have v wR hence C = Mo 2 R Wo> 2 R ~~~ Change of Units. It frequently happens that we wish to change the units in a given expression to some other units more convenient for our immediate purpose ; such an alteration in units is very simple, provided we set about it in systematic fashion. The expression must first be reduced to its funda- mental units; then each unit must be multiplied by the required constant to convert it into the new unit. For example, suppose we wish to convert foot-pounds of work to ergs, then The dimensions of. work are , / r(L , , pounds x (feet) 2 work (in ft.-poundals = - ; - i ' (seconds) 2 work in ergs = grams x (centimetres) (seconds) 2 i pound = 453*6 grams i foot = 30-48 centimetres Hence T foot-poundal = 453^6 X 3o'48 2 = 421,390 ergs and i foot-pound = 32*2 foot-poundals = 32-2 X 421)390 = 13^60,000 ergs CHAPTER II. MENSURATION. MENSURATION consists of the measurement of lengths, areas, and volumes, and the expression of such measurements in terms of a simple unit of length. Length. If a point be shifted through any given distance, it traces out a line in space, and the length of the line is the distance the point has been shifted. A simple statement in units of length of this one shift completely expresses its only dimension, length; hence a line is said to have but one dimension, and when we speak of a line of length /, we mean a line con- taining / length units. Area. If a straight line be given a side shift in any given plane, the line sweeps out a surface in space. The area of the surface swept out is dependent upon two distinct shifts of the generating point: (i) on the length of the original shift of the point, i.e. on the length of the gene- rating line (/) ; (2) on the length of the side shift of the generating line (d). Thus a statement of the area of a given surface must involve two length quantities, / and d, both expressed in the same units of length. Hence a surface is said to have two dimensions, and the area of a surface Id must always be expressed as the product of two lengths, each containing so many length units, viz. Area = length units X length units = (length units) 2 Volume. If a plane surface be given a side shift to bring it into another plane, the surface sweeps out a volume in space. Mensuration. 21 i The volume of the space swept out is dependent upon three distinct shifts of the generating point : (i) on the length of the original shift of the generating point, i.e. on the length of the generating line /; (2) on the length of the side shift of the generating line d- y (3) on the side shift of the generating surface /. Thus the state- ment of the volume of a given body or space must involve three length quantities, /, d, t, all expressed in the same units of length. Hence a volume is said to have three dimensions, and the volume of a body must always be expressed as the product of three lengths, each containing so many length units, viz. Volume = length units X length units X length units = (length units)' FIG. 10. 22 Mechanics applied to Engineering. Lengths. Straight line. Circumference of circle. Length of circumference = ied = 3-1416^ or = The last two decimals above may usually be neglected ; the error will be less than in. on a lo-ft. circle. FIG. xi. FIG. Ceradinfs Approximate Gra- phical Method. Draw diameter CD and tangent AB; with a 30 set square, set off OA. Make AB = y. Join BC. Then BC = TJT, i.e. the semicircumference (nearly). Arc of circle. Length of arc = or = 2-irrB f 360 rO 57-3 FIG. 13. For an arc less than a semicircle o/~ _ /~ Length = - a approximately Mensuration. 2 3 The length of lines can be measured to within in. with a scale divided into either tenths or twentieths of an inch. With special appliances lengths can be measured to within looiooo in - if necessary. The mathematical process by which the value of TT is deter- mined is too long for insertion here. One method consists of calculating the perimeter of a many-sided polygon described about a circle, also of one inscribed in a circle. The perimeter of the outer polygon is greater, and that of the inner less, than the perimeter of the circle. The greater the number of sides the smaller is the difference. The value of TT has been found to 750 places of decimals, but it is rarely required for practical purposes beyond three or four places. For a simple method of finding the value of TT, see " Longmans' School Mensura- tion," p. 48. In the figure we have ~DC = zr, AB = $r (by construction), DA = -7= (Euc. I. 47)- V3 DB = AB - DA = 3^ - --.-= = 2-4237- BC = VDB 2 + DC 2 = V(2- 4 2 3 7') 2 + (2rf = ^9 BC = 3' 1416;', i.e. the semicircumference (nearly). The length of the arc is less than the length of the A circumference in the ratio - . 360 T . f , v 6 irdO Length of arc = tea X = 360 360 The approximate formula given is extremely near when h is not great compared with C a ; even for a semicircle the error is only about i in 80. The proof is given in Lodge's " Mensura- tion for Senior Students " (Longmans). Mechanics applied to Engineering. Arc of circle. Rankings Approximate Method for Short Arcs. Produce chord ab. Make be = ; describe arc ae from centre c 2 and radius ac draw be perpendicular to ob. Length of arc adb = be. Irregular curved line abc. Set off tangent cd. With pair of dividers start from a, making small steps till the point c is reached, or nearly so. Count number of steps, and step off - < same number along tangent. FIG. 15. Areas. Parallelograms. "A-- Area of figure = Ih FIG. 16. Triangles. A r Area of figure = FIG. 17. Equilateral triangle. Area of figure = P - 0*433^ Mensuration. 2 5 For short arcs this method is very accurate; the error is only about i in 1000 when adb = bo, but it increases some- what rapidly as the arc gets greater than the radius (see Rankine's " Machinery and Millwork," p. 28). The stepping should be commenced at the end remote from the tangent ; then if the last step does not exactly coincide with c, the backward stepping can be commenced from the last point without causing any appreciable error. The greater the accuracy required, the greater must be the number of steps. Areas. See Euc. I. 35. See Euc. I. 41. bh J$P ,o area - = *- = 0*433^- 26 Mechanics applied to Engineering. Triangle. ^5** a + b + c ^ Area of figure = V s(s a) (s />) (s c) FIG. 19. Quadrilateral. Area of figure = Fro. 2 rapez.it m. Area of figure = Irregular straight-lined figure. a Area of figure = area abdef area &a? or area of triangles (acb+aef+efe+ceefy FIG. 22. Menstiration. 27 The proof is somewhat lengthy, but perfectly simple (see " Longmans' Mensuration," p. 18). Area of upper triangle = - 1 lower triangle = - both triangles = b ( h * + h * \= bh Area of parallelogram = bji Area of triangle = & Area of whole figure = ^ Simple case of addition and subtraction of areas. 28 Mechanics applied to Engineering. Circle. Area of figure = Trr 2 = 3'i4i6/- a or -- = 0785 / 4 Fio. 23. Sector of circle. Area of figure = - 360 FIG. 24. Segment of circle. Area of figure = f C a / when h is small = A(6C a + 8Q) nearly FIG. 25. Hollow circle. Area of figure = area of outer circle area of inner circle or = 07854* or = f /. ft mean cir- cumf. X thickness Mensuration. 29 The circle may be conceived to be made up of a great number of tiny triangles, such as the one shown, the base of each little triangle being b units, then the area of each triangle is ; but the sum of all the bases equals the circumference, or ^b = 27ir, hence the area of all the triangles put together, i.e. the area of the circle, = - = T/*. The area of the sector is less than the area of the circle in 6 2/3 the ratio -7-, hence the area of the sector = 7- ; if 8 be the 300 360 angle expressed in circular measure, then the above ratio R becomes 27T The area = - When h is less than -, the arc of the circle very nearly coincides with a parabolic arc (see p. 31). For proof of second formula, see Lodge's " Mensuration for Senior Students " (Longmans). Simple Case of Subtraction of Areas. The substitution of /y 2 for r 2 2 r? follows from the properties of the right-angled triangle (Euc. I. 47). The mean circumference X thickness is a very convenient form of expression ; it is arrived at thus Mean circumference = 1-1 thickness = 3O MecJtanics applied to Engineering. Ellipse. -^ Area of figure = irr or = - FIG. 27. Parabolic segments. ' H Area of figure = |BH I i.e. J(area of circumscribing rectangle) FIG. a8. FIG. 39. Area of figure = -f area of A abc a, e FIG. 30. Make dc = \ce area of figure = area of A Mensuration. An ellipse may be regarded as a flattened or an elon- gated circle ; hence the area of an ellipse is | j than the area of a circle whose diameter is the ellipse {j}, in the ratio } j J axis of an From the properties of the parabola, we have ^! =^ r H 2 B V B B 1 area of strip = h . db = ( - \ B ' TT area of whole figure = - db The area fl^has been shown to be HB. Take from each the area of the A ahg, then the re- mainder abc = J the A abe ; but, from the properties of the para- bola, we have ed = fyb, hence the area abc = area of the cir- cumscribing A abd. From the properties of the parabola, we also have the height of the A abd = 2 (height of the A abc) ; hence the area of the A abd = 2 (area of A abc\ and the area of the parabolic segment = 2 x area A abc = J area A abc. FlG - By increasing the height of the A abc to f its original height, we increase its area in the same ratio, and consequently make it equal to the area of the parabolic segment. 32 Mechanics applied to Engineering. d Area of shaded figure = \ area of A Surfaces bounded by an irregular curve. v c Area of figure = areas of para- bolic segments (a b + c + d -\- e) + areas of triangles (g -f- h). FIG. 32. Mean ordinate method. Area of figure = (h + 7/j 4- h. 2 .+ h z +, etc.)* FIG. 33. Mensuration. 33 The area abc = J area of triangle abd, hence the remainder = \ of triangle abd. Simply a case of addition and subtraction of areas. It is a somewhat clumsy and tedious method, and is not recom- mended for general work. One of the following methods is considered to be better. This is a fairly accurate method if a large number of ordi- nates are taken. The value of h 4- h\ + ^2 + As> etc., is most ^2 ' ^3 ' and so on. FIG. 33. easily found by marking them off continuously on a strip of paper. The value of x must be accurately found ; thus, If n be the number of ordinates, then x = -. n The method assumes that the areas a, a cut off are equal to the areas a^ a^ put on. D 34 Mechanics applied to Engineering. Simpson's Method.- FIG. 34. -(o + Area of figure -(h 4- 4^ -f N.B. In this case, the ordi- nate h being a tangent to the curve, its height is of course = o ; it should be written down, however, to avoid slips, thus +, and so on) Any odd number of ordinates may be taken ; the greater the number the greater will be the accuracy. Mensuration. 35 This is by far the most accurate and useful of all methods of measuring such areas. The proof is as follows : The curve gfedc is assumed to be a parabolic arc. Area a/ = . (i.) (i.+ii.) Area of A gee = (i.) + (ii.) - = f( "f ( &> FJG. 34 a. -hi-hj) (i v .) Area of parabolic ) _ 4/- x __ 2x segment gcdef \ ~ ^ 1V '/ " -j( 2 ^ - * - ^) - (v.) Whole figure = (iii.) -f (v.) = ^(^ -f h^ -\ (2^ h /^ 8 ) 3 If two more slices were added to the figure, the added area x would be as above = -(/$ 3 + 4^4 + ^ 5 ), and when the two are added they become =-(h^ + 2^3 4- 4^4 4- ^e)- Mechanics applied to Engineering. Surfaces of revolution. Pappus' or GuldinuJ Method. Area of surface swept out by ~\ the revolution of the line > = L x 2irp def about the axis ab ) Length of line =- L Radius of c. of g. of line def ) _ considered as a fine wire j ~~ ^ This method also holds for any part of a revolution as well as for a complete revolution. The area of such figures as circles, hollow circles, sectors, parallelo- grams (p = cc ), can also be found by this method. Surface of sphere. Area of surface of sphere = The surface of a sphere is the same as the curved surface of a cylinder of same diameter and length = d. FIG. 36. Surface of cone. Area of surface of cone FIG. 37- Mensuration. 37 The area of the surface traced out by a narrow strip of Inland so on. Area of whole surface = 27r(/ p + /!/>!+, etc.) = 27r(each elemental length of wire X its distance from axis of revolution) = 27r(total length of revolving wire X distance of c. of g. from axis of revolution) (see p. 58) = (total length of revolving wire X length of path described by its centre of gravity) = L27T/3 N.B. The revolving wire must lie wholly on one side of the axis of revolution and in the same plane. The distance of the c. of g. of any circular arc, or wire bent to a circular arc, from the centre of the circle is y = = p, where r = radius of circle, c chord of arc, a length of arc (see p. 64). T 2r 2 2 In the spherical surface a = L = TIT, c = 2r, p - 7T Surface of sphere = TFT . sir . = Length of revolving wire = L = h radius of c. of g. = p = - . . surface of cone = - = irrh Mechanics applied to Engineering. Hyperbola. Area of figure = XY log e r log, = 2*31 x ordinary log ~X:> _ x - -> FIG. 38. Area of figure XY - n - i FIG. 39. UNIVERSITY or Mensuration. 39 In the hyperbola we have XY = XiY, = xy , XY hence y = - oc area of strip = y . dx = XY- Area of ") whole > = XY figure J = XY (log. X x - log. X) = XY 10g e *?- Using the figure above, in this case we have YX n = YjXj" = yx n y X n hence y = YX n area of strip = y . dx = -^-dx = YX W ^~ VJK 'Xi 1 -3P area of whole figure = YX J^ = ! ^-V^ = x = X i n But YiXj" = YX M Multiply both sides by Xj 1 -" Substituting, we have Area of whole figure = Y i x i - YX i = YX - YA ;/ i Mechanics applied to Engineering. Irregular areas. Irregular areas of every description are most easily and accurately measured by a planimeter, such as Amsler's or Goodman's. A very convenient method is to cut out a piece of thin cardboard or sheet metal to the exact dimensions of the area ; weigh it, and compare with a known area (such as a circle or square) cut from the same cardboard or metal. A convenient method of weighing is shown on the opposite page, and gives very accurate results if reasonable care be taken. Prisms. FIG. 41. Volumes. Let A = area of the end of prism ; / = length of prism. Volume = /A Parallelepiped. Volume = Idt Hexagonal prism. \\ Volume = 2- Cylinder. Volume = n/V FIG. 43 Mensuration. Suspend a knitting-needle or a straight piece of wood by a piece of cotton, and accurately balance by shifting the cotton. Then suspend the two pieces of cardboard by pieces of cotton or silk ; shift them till theybalance ; then mea- sure the distances x and y. Then A* = B>> wre or -oc, or B = FIG. 40. The area of A should not differ very greatly from the area of B, or one arm becomes very short, and error is more likely to occur. Area of end = td volume = ltd Area of hexagon = area of six equilateral triangles = 6 x o-4 33 S 2 (see Fig. 18) volume = 2'598S 2 / or say 2'6S 2 / Area of circular end = d* 4 volume Mechanics applied to Engineering. Prismoid. Simpson's Method. $ 3 +, etc.) (see p. 34), ons. i etc.) 4 A 2 + 2 A 3 +,etc.) _ a The above proof assumes that the sections are parallelo- grams, i.e. the solid is flat-topped along its length. We shall later on show that the formula is accurate for many solids having surfaces curved in all directions, such as a sphere, ellipsoid, paraboloid, hyperboloid. If the number of sections be even, calculate the volume of the greater portion by this method, and treat the volume of the remainder as a paraboloid of revolution or as a prism. Let the area be revolved around the axis then The volume swept out by an\ elemental area , when re- 1 _ volving round the axis at a distance p Ditto ditto 0! and p l = and so on. Whole volume swept out by all> the elemental areas, /z , a l} etc., when revolving round the axis V = 27r(a p + at their respective distances, p , Pl , etc. = 27r(each elemental area, # , a^ etc. X their respective distances, p , p lt from the axis of revolution) = 27r(sum of elemental areas, or whole area X distance of c. of g. of whole area from the axis of revolution) (see p. 58) = A X 27rp = 2?rpA But 27r/o is the distance the c. of g. has moved through, or the length of the path of the c. of g. ; hence Whole volume = area of generating surface X the length of the path of the c. of g. of the area This proof holds for any part of a revolution, and for any value of p; when p becomes infinite, the path becomes a straight line, in such a case as a prism. 44 Mechanics applied to Engineering. Sphere. Volume of sphere = , or Volume of sphere = volume of circum- scribing cylinder Hollow sphere. c _....*" .;.* Volume of \ __ (volume of outer sphere hollow sphere/ ~ I volume of inner sphere External diameter = a e Internal diameter = d\ FIG. 48. = Slice of sphere. Volume of slice = - N.E. The slice must be taken wholly / on one side of the diameter; if the slice includes the diameter, it must be treated as two of the following slices. FIG. 49. \ Special case in which Y 2 Volume of slice = -(2R 3 o R. + FIG, 50. Mensuration. 45 2 Sphere. The revolving area is a semicircle of area The distance of the c. of g. ) v (see fifi) from the diameter { ^ v volume swept out = 2ir x X - - = -f^ir 3 = - or by Simpson's rule Volume of sphere = - (o + 47r;- 2 -f o) = o Volume of elemental slice = 7r{R 2 - (R 2 Volume of'i ry = Y 2 whole I = TT (2Ry~y 2 )_ __ h_ R? H j*2 TrR'/fc Volume of slice = icr*dh = volume of solid TrR 2 f H "-"/.* H ' dh 2H Volume of slice = vr*dh = volume of cone ^T?2 rH -vj J o tfdh dh TrR 2 x H 3 = 7rR 2 H H 2 3 3 FIG. 48 Mechanics applied to Engineering. Pyramid. Volume of pyramid = - H8 u T, T, , when B 1 = B = H = % volume circumscrib- ing solid FIG. 54. Slightly tapered body. Mean Areas Method. Volume of body = (mean area)/ (approx.) FIG. 55. Ring. Volume of ring = X ?rD = 2-47^0 4 FIG. 56. WEIGHT OF MATERIALS. Aluminium Brass and bronze Copper Iron cast ,, wrought Steel Lead Brickwork Stone 0*093 lb- P er cubic inch. 0-30 0-32 0-26 0-278 0-283 0-412 loo to 140 Ibs. per cubic foot. 150 to 180 Mensuration. This may be proved in precisely the same manner as the cone, or thus by Simpson's method Volume = 49 H BB.H - (2BBi) ~ This method is only approximately true when the taper is very slight. For such a body as a pyramid it would be seriously in error ; the volume obtained by this method would be j^H 3 instead of T %H 3 . The diameter D is measured from centre to centre of the sections of the ring, i.e. their centres of gravity Volume = area of surface of revolution X length of path of c. of g. of section CHAPTER III. MOMENTS. THAT branch of applied mechanics which deals with moments is of the utmost importance to the engineer, and yet perhaps it gives the beginner more trouble than any other part of the subject. The following simple illustrations may possibly help to make the matter clear. We have already (see p. 12) explained the meaning of the terms " clockwise " and " contra- clockwise " moments. In the figures that follow, the two pulleys of radii R and R x are attached to the same shaft, so that they rotate together. We shall assume that there is no friction on the axle. FIG. 57- FIG. 59. Let a cord be wound round each pulley in such a manner that when a force P is applied to one cord, the weight W will be lifted by the other. Now let the cord be pulled through a sufficient distance to cause the pulleys to make one complete revolution ; we shall then have Moments. 5 1 The work done by pulling the cord = P x ,, in lifting the weight = W X These must be equal, as it is assumed that no work is wasted in friction ; hence P27TR = W27TRJ or PR = WRj or the contra-clockwise moment = the clockwise moment It is clear that this relation will hold for any portion of a revolution, however small ; also for any size of pulleys. The levers shown in the same figures may be regarded as small portions of the pulleys ; hence the same relations hold in their case. \ It may be stated as a general principle that if a rigid body be in equilibrium under any given system of moments, the algebraic sum of all the moments in any given plane must be zero, or the clockwise moments must be equal to the contra- clockwise moments. ( force (/) ) First Moments. The product of a < J^p > by (^ volume (v) ) the length of its arm /, viz. < m * v, is termed \htfirst moment I al ( force ^ of the < ma ^ S L or sometimes simply the moment. \ area c ( volume \ ( force ) A statement of the first moment of a < " > must j area ( volume {force units X length units. mass uriits x length units. area units X length units. volume units X length units. In speaking of moments, we shall always put the units of force, etc., first, and the length units afterwards. For example, we shall speak of a moment as so many pounds-feet or tons- inches, to avoid confusion with work units. Mechanics applied to Engineering. force (/) Second Moments. The product of a D 7 volume (v) the square or second power of the length (/) of its arm, viz. f force I , is termed the second moment of the J mass L The j area i V volume ) second moment of a volume or an area is sometimes termed the "moment of inertia" (see p. 78) of the volume or area. Strictly, this term should only be used when dealing with questions involving the inertia of bodies ; but in other cases, where the second moment has nothing whatever to do with inertia, the term " second moment " is preferable. force must volume ( force units X (length units) 2 . _ 1 mass units X (length units) 2 . consist of the product of < area units x (length units)2> f volume units x (length units) 2 . First Moments. Clockwise moments Contra-clockwise moments Levers. y ^ about the point . about the point a. 1 * I U 0/2/2 + ^3 I + ^4/4 1 = A i FIG. 60. 8- 1 . 1 1 1 dr 2 w * + * = Wi/i FIG. 61. Moments. 53 Reaction R at fulcrum a, i.e. the resultant of all the forces acting on lever. REMARKS. To save confusion in the diagrams, the / has in some cases been omitted. In every case the suffix of / indicates the distance of the weight w bearing the same suffix from the fulcrum. 54 S3 . J> jf Mechanics a} ' //*>^ /# Engim Clockwise moments about the point a. wring. Contra-clockwise moments r about the point a. 1 T i FIG. 62. w^ -j- wj O'l/l + W 3/3 + 0/5/5 , ^ ? 2 2 If w = dis- tributed load per unit length, //= W = Zf/j/j omooo i i ro y? 2 " FIG. 63. 2 2 = Wj -f~ 0/1/1 or^ 4. w I 2 -r^oooo tlf )- X _ H(s + 2Si) /S + SA H " 3(S + SJ \ 2 / < -5 * H! _ 2S + Sj _ 1 I M HC i G S 2 2^! -f- g _j_ _ 2 bee C 2 cde C 3 whole fig- ure C 1AS . FIG. 76. Trapezium and triangles. Intersection of line joining c. of g. of triangle and c. of g. of trapezium, viz. ab and cd, where etc = area of trapezium, and db area of triangle, ac \ is parallel to bd. FIG. 77- Lamina with hole. Let A = area abcde', H = height of its c. of g. from ed; Hj = height of its c. of g. from df drawn at right angles to ed; a = area of hole gji; h = height of its c. of g. from td\ h^ = height of its c. of g. from FIG. 78. Then Hr = height of c. of g. of whole figure from ed HV = height of c. of g. of whole figure from df TT __ AH ah A-a H' - AH i ~ ah A-a Moments. The principle of these graphic methods is as follows Let the centres of gravity of two areas, A 1 and situated at points Q and C 2 e respectively, and let the common centre of gravity be situated at c, distant x from C l5 and x 2 from C 2 ; then we shall have A^ =A^x z . From C 2 set off a line i and CiCa is the common centre of gravity c. FlG - 7 6tf - The two triangles are similar, therefore ! X* A" 2 = ^' r lXl = be 'A* N.B. The lines C^i and C 2 2 are set off on opposite sides of C lt C 2 , and at opposite ends to their respective areas, at any convenient angle ; but it is undesirable to have a very acute angle at l and c^b z must be set off on the same side of the line, thus Then A, x-2 A" = V' 03 " A l^l A 2 X l FIG. 7 63. 64 Mechanics applied to Engineering. POSITION OF CENTRE OF GRAVITY, OR CENTROID. Graphical method. Lamina with hole. FIG. 79. NOTE. The lines rjK,, ,, but the line K 2 K, should not cut it at a very acute angle. Tr , , r If ^ be the c. of g. of . Join a are similar ; therefore = or -' = * y l Je ab likewise y^ = -, and so on Let Y = distance of c. of g. of portion of polygon from the centre O ; w weight of each side of the polygon. Then Y = > etc - wn where n = number of sides. The w cancels top and bottom. Substituting the values of y^y^ etc., found above, we have Y = *fe + 4 +, etc.) (Proof concluded on p. 67.) 66 Mechanics applied to Engineering. POSITION OF CENTRE OF GRAVITY OR CENTROID. Semicircular arc or wire. Circular sector considered as a thin sheet. ...A 2RC 3A FIG. 82. Semicircular lamina or sheet. 3*- 3*- -- FIG. 83. Parabolic segment. where Y = distance of c. of g. from apex. The figure being symmetrical, the c. of g. lies on the axis. FIG. 84. but ^ 2 V ^+^ 2 +^ 8 +^ 4 +, etc. ) where Y = distance of c. of g. from line CD ; w' = width of strips ; H = mean height of the strips. FIG. 87. Moments. 7 1 By the principle of moments, we have Distance of c. of g. of figure from axis / f v /dist. of its c. of g.\ _ /area of para. \ v /dist. of its c. of g. \ i area of rect ' X \ from axis / \ segment | X \ from axis j ^ ^ ~o area of figure BH X - - f BH x |B 2 Likewise IBH = |B BH X - - f BH X |H BH This is a simple case of moments, in which we have Distance of c. of g. 1 _ moment of each strip about AB from line AB j area of whole figure Moment of first strip = second = third ^ + wh* -h wh 9 +, etc. w, The area of the first strip =w^ second =w/i 2 third =wh z and so on. Area of whole figure = wh w 3^e/ T 50 Distance of c. wh^ X + w/h X - + ^^3 X of g. from Y 2 2 2 line AB whi -f w&2 + win -J-, etc. one w cancels out top and bottom, and we have v __ wfk + 3^2 + 5^3 + 7^4 +, etc. and similarly with Y . The division of the figure may be done thus : Draw a line, xy, at any angle, and set off equal parts as shown; project the first, third, fifth, etc., on to xz drawn normal to AB. Mechanics applied to Engineering. POSITION OF CENTRE OF GRAVITY OR CENTROID. Wedge. On a plane midway between the ends, TT and at a height from base. For frustum of wedge, see Trapezium. IG. 88. Pyramid or cone. H FIG. 89. On a line drawn from the middle point of the base to the apex, and at a distance fH from the apex. Frustum of pyramid or cone. On a line drawn from the middle point of the base to the apex, and at a height n ^ from the apex, where n = -i xi fH(- 3 } Moments. 73 A wedge may be considered as a large number of triangular laminae placed side by side, the c. of g. of each being situated TT at a height from the base. o Volume of layer = d 2 . dh j, moment of layer about apex = 2 . h . dh \ r b__^_ & ' h~~ H , h. B f""ir /* moment of layer about apex = Tj 2 dh moment of the whole pyramid) _ B 2 about apex j H 2 'A volume of pyramid = B 2 H 2 distance of c. of g. from apex = fr B FIG. 8ga. .dh 4H' In the case above, instead of integrating between the limits of H and o for the moment about the apex, we must integrate between the limits H and Hi. ; thus Moment of frustum of pyramid) _ about the (imaginary) apex ) ~~ volume of frustum = .dh B 2 H T3 TT substituting the value -^- = -^ = n then the distance of the c. from the apex 74 FIG. 91 Mechanics applied to Engineering. .POSITION OF CENTRE OF GRAVITY OR CENTROID. Locomotive or other symmetrical body. The height of the c. of g. above the rails can be found graphically, after calculating x, by erecting a perpendicular to cut the centre line. W, the weight of the engine, is found by weighing it in the ordinary way. W 2 is found by tilting the engine as shown, with one set of wheels resting on blocks on the platform of a weighing machine, and the other set resting on the ground. Let h^ be the height of the c. of g. above the rails. FIG. 92. Irregular surfaces. Also see Barker's "Graphical Calculus," p. 179, for a graphical integration of irregular surfaces. Moments. By taking moments about the lower rail, we hav< W* = 75 But - = L PI G whence ~i G _W 2 G * x " ~W G _^ _ G _ W 2 G y " 2 ' l 2 " W By similar triangl y G(- W * V" W 3 (These symbols refer to Fig. 92 only.) The c. of g. is easily found by balancing methods ; thus, if the c. of g. of an irregular surface be required, cut out the required figure in thin sheet metal or cardboard, and balance on the edge of a steel straight-edge, thus : The points a, a and , b are marked and afterwards joined : the point where they cut is the c. of g. As a check on the result, it is well to balance about a third line cc\ the three lines should intersect at one point, and not form a small triangle. The c. of g. of many solids can also be found in a similar manner, or by suspending them by means of a wire, and dropping a perpendicular through the points of suspension. Second Moments Moments of Inertia. A definition of a second moment has been given on p. 52. In every case we shall find the second moment by summing up FIG. 93. Mechanics applied to Engineering. or integrating the product of every element of the body or surface by the square of its distance from the axis in question. In some cases we shall find it convenient to make use of the following theorems : Let I = the second moment, or moment of inertia, of any surface (treated as a thin lamina) or body about a given axis ; I = the second moment, or moment of inertia, of any surface (treated as a thin lamina) or body about a parallel axis passing through the c. of g. ; M = mass of the body ; A = area of the surface j R = the perpendicular distance between the two axes. Then I = I 4- MR 2 , or I + AR,, 2 Let xy be the axis passing through the c. of g. Let x$i be the axis of revolution, parallel to xy and in the plane of the surface or lamina. Let the elemental areas, 0i, 02, 0.3, etc., be situated at distances r lt r zj r^ etc., from xy. Then we have I = ^(R 4- r,Y 4- 2 (Ro + ;- 2 ) 2 +, etc. = ^(Ro 2 4- rf + 2 ROT,) + , etc. = a \ r \ 4- 02^2 2 4-, etc. 2/1 If i 2R62*- 2 4- air 4-, etc.) But as xy passes through the c. of g. of the section, we have 0^ 4- 2 r 2 4-, etc. = o (see p. 58), for some r's are positive and some negative ; hence the latter term vanishes. The second term, X 4- 2 4-, etc. = the whole area (A); whence it becomes R 2 A. In the first term, we have simply the second moment, or moment of inertia, about the axis passing through the c. of g. = I j hence we get I = I + R 2 A Moments. 77 We may, of course, substitute m lt m^ etc., for the elemental masses, and M for the mass of the body instead of A. When a body or surface (treated as a thin lamina) revolves about an axis or pole perpendicular to its plane of revolution, the second moment, or moment of in- ertia, is termed the second polar mo- ment, or polar mo- ment of inertia. The second polar moment of any sur- face is the sum of the second moments about any two rect- angular axes in its own plane passing through the axis of revolution, or Consider any ele- mental area a, distant r from the pole. I, about ox ay* oy = ax* pole = or 2 But r*=, and ar 2 =ax 2 +ay* hence L,= L+L In a similar way, it may be proved for every element of the surface. When finding the position of the c. of g., we had the following relation : Distance of c. of g. from ) _ first moment of surface about xy the axis xy (K) f " area of surface _ first moment of body about xy volume of body Mechanics applied to Engineering. Now, when dealing with the second moment, we have a corresponding centre, termed the centre of gyration, at which the whole of a moving body or surface may be considered to be concentrated ; the distance of the centre of gyration from the axis of revolution is termed the " radius of gyration." When finding its value, we have the following relation : Radius of gyration^ 8 second moment of surface about xy about the axis xy (ic) J ~ area of surface second moment of body about xy volume of body 'Zar 2 IT* ic zzz = "~A") or L A/i SECOND MOMENT, OR MOMENT OF INERTIA (I). Parallelogram treated as a thin lamina about its extreme end. BH 3 H If the figure be a square B = H = S we have I n = FIG. 96. Radius of gyration K. H VI Moments. Area of elemental strip = B . dh second moment of strip = B . W . dh O second moment of whole surface area of whole ) ^u surface } " square of radius j _,BH 3 __ H a of gyration J ~" 3 BH ~ "3" TT radius of gyration = p= FIG. g6a. It will be seen that the above reasoning holds, however the parallelogram may be distorted sideways, as shown. Mechanics applied to Engineering. SECOND MOMENT, OR MOMENT OF INERTIA (I). Parallelogram treated as a thin lamina about\ its central axis. 12 12 Radius of gyration H v/Ii " Parallelogram treated as a thin lamina about an axis distant 'R from its ofg- O Moments. 81 This is simply a case of two parallelograms such as the above put together axis to axis, each of length ; I ~R"H" 3 Then the second moment of each =- 3 8x3 then the second moment \ _ /BH 3 \ __ of the two together ~ \ 24 / 12 area of whole surface = BH radius of gyration = From the theorem given above (p. 76), we have T>TT3 I = I + R 2 A J = 17 ' T)TT3 =--R 2 BH A = BH !L + R 2 ) when R = o, I = I 82 Mechanics applied to Engineering. SECOND MOMENT, OR MOMENT OF INERTIA (I). FIG. 99. Hollow parallelogram. Let I, = second moment of external figure ; It = second moment of internal figure ; I = second moment of hollow figure ; Io = I. - I*. Radius of gyration Triangle ' about an axis parallel to the base passing through the apex. BH 3 4 FIG. ioo. Triangle about an axis parallel to the base passing through the c. of g. 1 = BH ; 36 Moments. The I for the hollow parallelogram is simply the difference between the I a for the external, and the I, for the internal parallelogram. Area of strip = b . dh ; but b = -g H ' O T> second moment of strip = -h* . dh rl triangle = g ' h* .dh BH 4 BH 3 11 area of triangle = radius of gyration = " 4 H 4 BH FIG. 'BH 2 .JL = V /H'=H -QTT V ft ./ Brl z v 2 \ From the theorem on p. 76, we hav< I = I + R 2 A I = I - R 2 A T BH 3 4H 2 v BH I x 492 BH 3 BH 3 Io = A, 2 4|P 9 I = radius of gyration = 84 Mechanics applied to Engineering. SECOND MOMENT, OR MOMENT OF INERTIA (I). FIG. 102. O Triangle about an axis at the base. BH 8 12 Radius of gyration H V6 Trapezium abotit an axis coinciding with its short base. < H FIG. 103. I= (3B+jy 12 T ^4- T) .._ -,."D J-jCL J3j /Zi3t H Trapezium about an axis coinciding with its long base. or BH 3 FIG. 104. v Moments. From the theorem quoted above, we have I = I + R 2 A R 2 = EH 3 H 2 EH Io EH 3 12 Radius of gyration obtained as in the last case. This figure may be treated as a parallelogram and a triangle about an axis passing through the apex. "R TT* For parallelogram, I = - for triangle, I = for trapezium, I = (B-B,)H 12 FIG. When the axis coincides with the long base, the I for the (B - B k )H 8 triangle = 12 ; then, adding the I for the parallelogram as above, we get the result as given. When n = i, the figures become parallelograms, and BH 3 I = , as found above. When n = o, the figures become triangles, and the I = *r for the first case, as found for the triangle about its apex ; and T>TT3 I = ^ for the second case, as found for the triangle about its base. 86 Mechanics applied to Engineering. SECOND MOMENT, OR MOMENT OF INERTIA (I). Trapezium about an axis passing through its c. of g. and parallel with the base. B 2 )H 3 36 2 + B) + n+i H--* O FIG. 105. For a close approximation, see next figure. Radius of gyration x. Approximate method for trapezium about axis passing through c. of g. The I for dotted rectangle about an axis passing through its c. of g., is approximately the same as the I for trapezium. For dotted rectangle f' l _ (B + BJBP I 24 or if Bj = B d FIG. 106. Moments. From the theorem on p. 76, we have T "P 2A T - ~~ J-Q JX0 \ J.Q 12 base) B) , , lon A=(I=)H Substituting the values in the above equation and simplify- ing, we get the result as given. The working out is simple algebra, but too lengthy to give here. TVTT3 The I for a rectangle is (see p. 80). Putting in the value - = B', we get __ (B + 12 24 The following table shows the error involved in the above assumption ; it will be seen that the error becomes serious when n < 0*5 : Value of. Appro*, method, the correct value being i. 0'9 ooi 0-8 005 07 on 0-6 '021 0-4 065 0*3 107 0'2 174 The approximate method always gives too high results. Mechanics appliea to Engineering. SECOND MOMENT, OR MOMENT OF INERTIA (I). Sqiiare about its diagonal. O 1 = 12 Radius of gyration K. Circle about a diameter. 64 Hollow circle about a dia- meter. Moments. gg This may be taken as two triangles about their bases (see p. 84). In this case, B = 1 = = 2 BH 12 = sj 12 area of figure = S 2 r&~ s radius of gyration = A/ ^32 = / From the theorem on p. 77, we have I P = I X + I,; in the circle, I x = I v . Then l p = 2\ x e =^=^-=^(seeFig.n8). The I for the hollow circle is simply the difference between the I for the outer and inner circles. go Mechanics applied to Engineering. SECOND MOMENT, OR MOMENT OF INERTIA (I). Hollow eccentric circle about a line normal to the line joining the two centres^ and passing through the c. of g. of the figure. where x is the eccentricity. NOTE. When the eccentricity is zero, i.e. when the outer and inner circles are concentric, the latter term in the above expression vanishes, and the value of I is the . same as in the case given above for the hollow circle. Radius of gyration Ellipse about minor axis. o DI 4 Moments. The axis OO passes through the c. of g. of the figure, and is at a distance b from the centre of the outer circle, and a from the centre of the inner circle. From the principle of moments, we have *)-?. D, 2 * whence b D e 2 - D,* 7T / \ 7T 4 4 whence #3> a*, etc., respectively ; and their mean distances from the axis be r^ r 2l r 3) r^ etc., respectively. Then I = op? + apf + ay} +, etc. But ^i = wb-H and a^ wb^ and so on , w -\w $w and TI -, r 2 = , r 3 = y and so on hence I, = 4^ + >J " 9) circle = 27r | r 3 . dr J o 27rR 4 _ TrR 4 ~T~ 2 v4 < 2 x 16 32 FIG. 98 Mechanics applied to Engineering. SECOND POLAR MOMENT, OR POLAR MOMENT OF INERTIA. Hollow circle about a pole passing through the c. f g- = ~(D e 4 - IV) Radius of gyration /D. a + D,* V T~ or /R/- + R,' V ~ Second Polar Moments of Solids. Bar of rectangular section about a pole passing through its c. of g. : H &\ For a circular bar of radius R 12 FIG. 12 v 7 ^ 3 R 2 Cylinder about a pole pas sing through its centre. o \ I Di D A/8 <- R FIG. 121. Moments. 99 The I p for the hollow circle is simply the difference between the l p for the outer and the l p for the inner circles. The bar may be regarded as being made up of a great number of thin laminae of rectangular form, of length L and breadth B, revolving about their polar axis, the radius of V'^2 I g2 - (see Fig. 117), which is the radius of gyration of the bar. The second moment of the bar will then be K 2 (volume of bar), or -l+J or LB _?(L 2 + B 2 ) The cylinder may be regarded as being made up of a great number of thin circular laminae revolving about a pole passing through their centre, the radius of gyration of each being D 2 K = The second moment of cylinder = K 2 (volume of cylinder) " "8 X 4 ~3^~ ioo Mechanics applied to Engineering. SECOND POLAR MOMENT, OR POLAR MOMENT OF INERTIA. Hollow cylinder about Radius of gyratk a pole pas sing through its axis. xA-g-- or /R7+-R1 V ' ~T~ 32 or ~? (R/ - Rfl FIG. 122. Disc flywheel. Treat each part sepa- rately as hollow cylinders, and add the results. FIG. 123. Moments. 101 The Ip for the hollow cylinder is simply the difference between the l p for the outer and the inner cylinders. It must be particularly noticed that the radius of gyration of a solid body, such as a cylinder, flywheel, etc., is not the radius of gyration of a plane section ; the radius of gyration of a plane f ' V section is that of a thin lamina of uniform thick- ness, while the radius of gyration of a solid is that of a thin wedge. The radius of gyration of a solid may be found by correcting the section in this manner, and finding the I for the shaded figure FlG . 123 I42< Thus, when the link OA rotates about O, we have velocity of A _Va__ r a velocity of B ~~ V & ~~ r b If the link be rotating with an angular velocity o> radians per second (see p v 4), then the linear velocity of #, viz. V a = wr a , and of , V 6 = o>;- 6 , but the angular velocity of every point in the link is the same. As the link rotates, every point in it moves at any given instant in a direction normal to the line drawn to the centre of rotation, hence at each instant the point is moving in the direction of the tangent to the path of the point, and the centre about which the point is rotating lies on a line drawn normal to the tangent of the curve at that point. This property will enable us to find the centre about which a body having plane motion is rotating. The plane motion of a body is completely known when we know the motion of any two points in the body. If the paths of the points be circular and concentric, then the centre of rotation will be the same for all positions of 122 Mechanics applied to Engineering. the body. Such a centre is termed a " permanent " or " fixed " centre; but when the centre shifts as the body shifts, its centre at any given instant is termed its " instantaneous " or " virtual " centre. Instantaneous or Virtual Centre. Complex plane motions of a body can always be reduced to one very simply expressed by utilizing the principle of the virtual centre. For example, let the link db be part of a mechanism having a complex motion. The paths of the two end points, a and , are known, and are shown dotted. In order to find the relative velocities of the two points, we draw tangents to the paths at a and , which ^ve us the directions in which each is moving at the instant. From the points #, b draw normals ad and btf to the tangents j then the centre about which a is moving at the instant lies somewhere on the line aa', likewise with bti hence the centre about which both points are revolving at the instant, must be at the intersection of the two lines, viz. at O. This FIG. 143. point is termed the virtual or instantaneous centre, and the whole motion of the link at the instant is the same as if it were attached by rods to the centre O. As the link has thickness normal to the plane of the paper, it would be more correct to speak of O as the plan of the virtual axis. If the bar had an arm projecting as shown in Fig. 144, the path of the point C could easily be determined, for every point in the body, at the instant, is describing an arc of a circle round the centre O ; thus, in order to determine the path of the point C, all we have to do is to describe a small arc of a circle passing through C, struck from the centre O with the radius OC. The radii OA, OB, OC are known as the virtual radii of the several points. If the tangents to the point-paths at A and B had been parallel, the radii would not meet, except at infinity. In that Mechanisms. 123 case, the points may be considered to be describing arcs of circles of infinite radius, i.e. their point-paths are straight parallel lines. If the link AB had yet another arm projecting as shown in the figure, the end point of which coincided with the virtual centre O, it would, at the in- stant, have no motion at all relatively to the plane, i.e. it is a fixed point. Hence there is no reason why we should not regard the virtual centre as a point in the moving body itself. It is evident that there can- not be more than one of such fixed points, or the bar as a whole would be fixed, and then it could not rotate about the centre O. It is clear, from what we have said on relative motion, that if we fixed the bar, which we will term m (Fig. 146), and move the plane, which we will term , the relative motion of the two would be precisely the same. We shall term the virtual centre of the bar m relatively to the plane n, Omn. Centrode and Axode. As the link m moves in such a manner that its end joints a and b follow the point-paths, the virtual centre Omn also shifts relatively to the plane, and traces out the curve as shown in Fig. 146. This curve is simply the point-path of the virtual centre, or the virtual axis. This curve is known as the centrode, or axode. Now, if we fix the link m, and move the plane n relatively to it, we shall, at any instant, obtain the same relative motion, therefore the position of the virtual centre will be the same in both cases. The centrodes, however, will not be the same, but as they have one point in common, viz. the virtual centre, they will always touch at this point, and as the motions of the two bodies continue, the two centrodes will roll on one another. This rolling action can be very clearly seen in the simple four-bar mechanism shown in Fig. 147. The point A moves in the arc of a circle struck from the centre D, hence AD is normal to the tangent to the point-path of A ; hence the virtual centre lies somewhere on the line AD. For a similar reason, it lies somewhere on the line BC ; the only point common to the two is their intersection O, which is therefore their virtual centre. If the virtual centre, i.e. the intersection of the two 124 Mechanics applied to Engineering. bars, be found for several positions of the mechanism, the centrodes will be found to be ellipses. As the mechanism revolves, the two ellipses will be found to FIG. 146. roll on one another. That such is the case can easily be proved experimentally, by a model consisting of two ellipses cut out of suitable material and joined by cross-bars AD and BC ; it will be found that they will roll on one another perfectly. Hence we see that, if we have given a pair of centrodes for two bodies, we can, by making the one centrode roll on the other, completely determine the relative motion of the two bodies. Position of Virtual Centre. We have shown above that when two point-paths of any body are known, we can Mechanisms, 125 readily find the position of the virtual centre. In the case of most mechanisms, however, we can determine the virtual centres without first constructing the point-paths. We will show this by taking one or two simple cases. In the four-bar mechanism shown in Fig. 148, it is evident that if we consider d as stationary, the virtual centre Oad will be at the joint of a and d^ and the velocity of any point in a relatively to any point in d will be propor- tional to the distance from this joint ; likewise with Ode. Then, if we consider b as fixed, the virtual centre of a and b will also be at their joint. By similar reasoning, we have the virtual centre Qbc. Again, let d be fixed, and consider the motion of b relatively to d. The point- path of one end of , viz. Qab, describes the arc of a circle about Oad t therefore the virtual centre lies on a produced ; for a similar reason, the virtual centre lies on c produced, hence it must be at Obd^ the meet of the two lines. .Old FIG. 147- FIG. 148. In a similar manner, consider the link c as fixed ; then, for 126 Mechanics applied to Engineering. the same reason as was given above for b and d, the virtual centre of a and c lies at the meet of the two lines b and d, viz. Oac. If the mechanism be slightly altered, as shown in Fig. 149, we shall get one of the virtual centres at infinity, viz. Qcd. Ocd> \ Oao (fed Ocuc Oad, Fic. 149. a/ FIG. 150. The mechanism shown in Fig. 149 is, as far as its action goes, precisely similar to the mechanism in Fig. 150. Instead of c sliding to and fro in guides, a link of infinite length has been substituted, and the fixed link d has been carried to infinity in order to provide a centre from which c shall swing. Then it is evident that the joint Qbc moves in the arc of a circle of infinite radius, i.e. it moves in a straight line in precisely the same manner as the sliding link c in Fig. 149. The only virtual centre that may present any difficulty in finding is Qac. Consider the link c as fixed, then the bar d swings about a centre at an infinite distance away ; hence every point in it moves in a straight path at right angles to c, i.e. in the same straight line or parallel to d (Fig. 150). Hence the virtual centre lies on a line normal to d ; also, for reasons given below, it lies on the prolongation of the bar fr, viz. Qac. Three Virtual Centres on a Line. By referring to the figures above, it will be seen that there are always three virtual centres on each line. In Figs. 149, 150, it must be remembered that the three virtual centres Qad, Qac, Ocd are on one line ; also O&r, Qbd, Ocd. The proof that the three virtual centres corresponding to the three contiguous links must lie on one line is quite simple, and as this property is of very great value in determining the positions of the virtual centres for complex mechanisms, we will give it here. Let b (Fig. 151) be a body moving relatively to a, and let the virtual centre of its motion relative to abe Oab likewise let Qac be the virtual centre of ^'s motion relative to a. If we Mechanisms. 127 want to find the velocity of a point in b relatively to a point in a , and the linear velocity of the point 8 as V 8 , etc., we have , (Up/34 <*>a/3 = w c /4> ana w :t - Ps also V 8 = w a p 8 V fl = o) t .p 9 Substituting the value of w a , we have Mechanisms. \ 20 Similarly, if we require the velocity of the point 6 relatively to that of point 5 6 _ 6 y _ s V^~Pa 6 ~"pT V5= V - v ^ 5 V 9 P9 '7T whenrp VG - V * W9 - WW - ^6 wiiciiUC ^y =y V 5 V 9 /Japs p 3 p9p8P5 P3P5 or this might have been arrived at more directly, thus Likewise velocity 7 _ R 7 velocity^ "~ R^ hence velocit y 6 = Pe X velocity 8 X R 8 velocity 7 p 8 x R 7 X velocity 8 FIG. 154. This can be arrived at much more readily by a graphical process; thus (Fig. 154): With Qad as centre, and p s as K 130 Mechanics applied to Engineering. radius, set off Oad.h = /o 6 along the line joining Qad to 8 ; set off a line hi to a convenient scale in any direction to represent the velocity of 6. From Qad draw a line through /', and from 8 draw a line Se parallel to hi ; this line will then represent the velocity of the point 8 to the same scale as 6, for the two triangles Qad.&.e and Qad.h.i are similar ; therefore &_ Oad.S ^ p.S __ velocity 8 hi Qad.h p. 6 velocity 6 From the centre Qbd and radius R 7 , set off Obd.f = R 7 ; draw f.g parallel to e.S, and from Qbd draw a line through e to meet this line in g ; then fg = V 7 , for the two triangles Qbd.f.g and Qbd&.e are similar ; therefore velocit y 7 Be Obd.8 R 8 velocity 8 and velocit y 6 -^ velocity 7 fg The same graphical process can be readily applied to all cases of velocities in mechanisms. Relative Angular Velocities of Bars in Mechan- isms. Every point in a rotating bar has. the same angular velocity. Let a bar be turning about a point O in the bar with an angular velocity w ; then the linear velocity V a of a point A situated at a radius r a is V V a = wr a) and w = - ^a In order to find the relative angular velocity of any two links, let the point A (Fig. 155) be first regarded as a point in the bar a, and let its radius about Qad be i\. When the point A is regarded as a point in the bar ^, we shall term it B, and its radius about Q&J, r B . Let the linear velocity of A be V^ and that of B be V B , and the angular velocity of A be W A , and of B be B r A Mechanisms. % 131 This may be very easily obtained graphically thus : Set off a line Ae in any direction from A, whose length on some given scale is equal to W A ; join e.Qbd-, from Oad draw Qad.f parallel FIG. 155. FIG. 156. to e.Qbd. Then A/= W B , because the two triangles A.f.Qad and A.e.Qbd are similar. Hence Ae = (CW)B _ o,^ A/ (Oad)A W B In Fig. 156, the distance A^ has been made equal to A.Qad, and g/is drawn parallel to e.Obd. The proof is the same as in the last case. When a is parallel to <:, the virtual centre is at infinity, and the angular velocity of b becomes zero. When finding the relative angular velocity of two non- adjacent links, such as a and c, we proceed thus : For con- venience we have numbered the various points instead of using the more cumbersome virtual centre nomenclature. The radius 1.6 we shall term r 1-6 , and so on. Then, considering points i and 2 as points of the bar &, we have V, V 2 r M 132 Mechanics applied to Engineering. Then, regarding point i as a point in bar ^1.6 X 4-7 4-7 5-4 Thus, if the length f\ s represents the angular velocity of c, and a line be drawn from 4 to meet the opposite side in 7, 4.7 represents on the same scale the angular velocity of a. Or it may conveniently be done graphically thus : Set off from 3 a line in any direction whose length 3.8 represents the angular velocity of c; from 4 draw a line parallel to 3.8; from 5 draw a line through 8 to meet the line from 4 in 9. Then 4.9 repre- sents the angular velocity of a, the proof of which will be perfectly obvious from what has been shown above. When b is parallel to d, the virtual centre Qac is at infinity, and the angular velocity of a is then equal to the angular velocity of c. Steam-engine Mechanism. On p. 126 we showed how a four-bar mechanism may be developed into the ordinary steam-engine mechanism, which is then often called the " slider-crank chain." This mechanism appears in many forms in practice, but some of them are so disguised that they are not readily recognized. We will proceed to examine it first in -its // most familiar form, viz. f the ordinary steam- engine mechanism. \ /' Having given the FIG. 159- speed of the engine in revolutions per minute, and the radius of the crank, the velocity of the crank-pin is known, and the velocity of the cross-head at any instant is readily found by means of the principles laid down above. We have shown that velocity P _ O&/.P velocity 1C = (W.X FIG. 160 134 Mechanics applied to Engineering. From O draw a line parallel to the connecting-rod, and from P drop a perpendicular to meet it in e. Then the triangle OP 7' c =250 feet per minute. in round numbers FIG. 161. (2) We must solve this part of the problem by finding the relative angular velocity of the connecting-rod and the crank. Knowing the angular velocity of a relatively to d^ we obtain the angular velocity of b relatively to d thus : The virtual centre Qab may be regarded as a part of the bar a pivoted at Qad, also as a part of the bar b rotating for the instant about the virtual centre Qbd] then, by the gearing conception -already explained, we have * = .<* = TT = J- w a Kd -K 6 JK 6 When the crank-arm and the connecting-rod are rotating in the opposite sense, the rubbing velocity V, = r> a + co d ) =r j ,o> a (i + |? n This has its maximum value when -= is greatest, i.e. when 136 Mechanics applied to Engineering. R 6 is least and is equal to the length ot the connecting-rod, /. d = \/ Taking the circle mn to represent the constant angular velocity of the crank, the polar curves op, qr represent to the same scale the angular velocity of the oscillating link c for corre- ^ spending positions of the crank. From these diagrams it will be seen that the swing to and fro of the cylinder is not accomplished in equal times. The in- equality so apparent to an observer of the oscillating engine is usefully applied as a quick-return motion on shaping machines. The cutting stroke takes place during the slow swing of c, i.e. when the crank-pin is traversing the upper portion of its arc, and the return stroke is quickly effected while the pin is in its lower position. The ratio of the mean time occupied in the cutting stroke to that of the return stroke is termed the " ratio oC the gear," which is readily determined. The link c is in its extreme position when the link a is at right angles to it ; the cutting angle is 360 0, and the return angle (Fig. Connecting 'tvd FIG. 161*. 138 Mechanics applied to Engineering. The ratio of the gear R = 3 or 0(R + i) = 360 and a = b cos - 2 LetR=2; 0=120; b = 20. R = 3- (9 = 90; b= 1-420. Another form of quick-return motion is obtained by fixing the link a. When the link d is driven at a constant velocity, the link b rotates rapidly during one part of its revolution and slowly during the other part. The exact speed at any instant can be found by the method already given for the oscil- lating cylinder engine. The ratio R has the same value as before, but in this case we have A FIG. i6xc. a = d cos - ; therefore d must be made equal to za for a ratio of 2, and i'4za for a ratio of 3. This mechanism has also been used for a steam-engine, but it is best known as Rigg's hydraulic engine (Fig. i6i he is measured in feet. 142 Mechanics applied to Engineering curve drawn through the points so obtained is the acceleration curve for the point f. This method has been adopted until point 7 has been reached; then another method has been used. This will perhaps be better explained by a separate diagram, thus Let the curve represent the velocity of any point as it moves through space. Let the time-interval between the two dotted lines be dt^ and the change of velocity of the point while passing through that space be dv. Then the acceleration . dv change of velocity during the interval is ^, ,.,. -gJL-jj-j^-', O r the change of velocity in the given interval. By similar triangles, we have -rr = dt xy When dt = i second, dv = acceleration. Hence, make xy^ = i on the time scale, then s^ the subnormal, gives us the accele- ration measured on the same scale as the velocity. The sub-normals to the curve above have been plotted in this way to give the accelera- tion curve from 7 to 16. The scale of the acceleration curve will be the same as that of the velocity curve. The reader is recommended to refer to Barker's " Graphical Calculus " for other curves of this character. Velocity Diagrams for Mechanisms. Force and reciprocal diagrams are in common use by engineers for find- ing the forces acting on the various members of a structure, but it is rare to find such diagrams used for finding the velocities of points and bars in mechanisms. We are indebted to Pro- fessor R. H. Smith for the method. (For fuller details, readers should refer to his own treatise on the subject. 1 ) Let ABC represent a rigid body having motion parallel to the plane of the paper ; the point A of which is moving with a known velocity V A as shown by the arrow ; the angular velocity (o of the body must also be known. If w be zero, then every 1 "Graphics," by R. H. Smith, Bk. I. chap. ix. ; or Machines," by R. J. Durley {]. Wiley & Sons). Kinematics of Mechanisms. H3 FIG. 164^. point in the body will move with the same velocity as V A . From A draw a line at right angles to the direction of motion as indicated by V A , then the body is moving about a centre situated somewhere on this line, but since we know V A and CD, we can find the virtual centre P, since wR A = V A . Join PB and , PC, which are virtual radii, and from which we know the direction and velocities of B and C, because each point moves in a path at right angles to its radius, and its velocity is proportional to the length of the radius ; thus V B = w.PB, and V c = w.PC The same result can be arrived at by a purely graphical process; thus From any pole p draw (i) the ray pa to represent V A ; (2) a ray at right angles to PB ; (3) a ray at right angles to PC. These rays give the directions in which the points are moving. We must now proceed to find the magnitude of the velocities. From a draw db at right angles to AB ; then pb gives the velocity of the point B. Likewise from b draw be at right angles to BC, or from a draw ac at right angles to AC ; then pc gives the velocity of the point C. The reason for this construction is that the rays pa, pb, pc are drawn respectively at right angles to PA, PB, and PC, i.e. at right angles to the virtual radii ; therefore the rays indicate the directions in Which the several points move. The ray- lengths, too, are proportional to their several velocities, since the motion of B may be regarded as being compounded of a translation in the direction of V A and a spin, in virtue of which the point moves in a direction at right angles to AB; the com- ponent pa represents its motion in the direction V A , and ab its motion at right angles to AB, whence pb represents the velocity of B in magnitude and direction. Similarly, the point C par- takes of the general motion pa in the direction V A , and, due to the spin of the body, it moves in a direction at right angles to AC, viz. ac, whence pc represents the velocity of C. The point c can be equally well obtained by drawing be at right angles toBC. 144 Mechanics applied to Engineering. The triangular connecting-rod of the Musgrave engine can be readily treated by this construction. A is the crank-pin, whose velocity and direction of motion are known. The pistons are attached to the corners of the triangle by means of short connecting-rods, a suspension link DE serves to keep the connecting-rod in position, the direction in which D moves is at right angles to DE. Produce DE and AF to meet in P, which is the virtual centre of AD and FE. Join PB and PC. From the pole/ draw the ray/# to represent the velocity of the point A, also draw rays at right angles to PB, PC, PD. From a draw a line at right angles to AD, to meet the ray at FIG. if>A.b. right angles to PD in d, also a line from a at right angles to AB, to meet the corresponding ray in b. Similarly, a line from a at right angles to AC, to meet the ray in c. Then pb, pc, pd give the velocities of the points B, C, D respectively. The velocity of the pistons themselves is obtained in the same manner. As a check, we will proceed to find the velocities by another method. The mechanism ADEF is simply the four-bar mechanism previously treated ; find the virtual centre of DE and AF, viz. Q; then (see Fig. 153) V A _AF.EQ V D ED . QF Mechanisms. 145 The points C and B may be regarded as points on the bar AD, whence PC 'PD' PB RG V c = V D , and V B = V D5 and V = V B - Let the radius of the crank be 16'', and the revolutions per minute be 80 ; then = 3-14 X 2 X 16 X 80 = 12 Then we get V B = 250 feet per minute V c = 79 ii i> V D = 515 V G = 246 V H = 79 2 ...... A \talverod By taking one more example, we shall probably cover most of the points that are likely to arise in practice. We have selected that of a link-motion. Having given the speed of the engine and the dimensions of the valve-gear, we proceed to find the velocity of the slide-valve. In the diagram A and B represent the centres of the eccentrics ; the link is suspended from S. The velocity of the points A and B is known from the speed of the engine, and the direction of motion is also known of A, B, and S. Choose a pole p, and draw a ray pa parallel to the direction of the motion of A, and make its length equal on some given scale to L 1 4 6 Mechanics applied to Engineering. the velocity of A ; likewise draw/ for the motion of B. Draw a ray from p parallel to the direction of motion of S, i.e. at right angles to US ; through a draw a line at right angles to AS, to cut this ray in the point s. From s draw a line at right angles to ST ; from b draw a line at right angles to BT ; where this line cuts the last gives us the point /. Join / /, p s, which give respectively the velocity of T and S. From/ draw a line at right angles to TV, and from s a line at right angles to SV ; they meet in ^ ; then pv^ is the velocity of a point on the link in the position of V. But since V is guided to move in a straight line, from ^ draw a line parallel to a tangent at V, and from p a line parallel to the valve-rod, meeting in z>; then/f is the required velocity of the valve, and vv 1 is the velocity of " slip " of the die in the link. Toothed Gearing. The general principles that have to be considered when deciding upon the form that shall be given to the teeth of wheels, follow directly from the work that we have con- sidered above when dealing with virtual centres and the velocities of different points in mechanisms. In the figure, let two shafts A and B be provided with circular discs, a and b, as shown ; let there be sufficient friction at the line of contact to make the one revolve when the other is turned round ; then the linear velocity v of the two rims will be the same. Let the radius of a be r M of b be r b ; the angular velocity of a be 6 = 2?rN 6 _ N,, Mechanisms. 147 or the revolutions of each wheel are inversely proportional to their respective radii. The virtual centre of a and c, likewise b and t, is evidently at their permanent centres, and as the three virtual centres must lie in one line (see p. 127), the virtual centre Qab must lie on the line joining the centres of a and b, and must be a point (or axis) common to each. The only point which fulfils these conditions is Qab, the point of contact of the two discs. The two circles representing the rims of the discs are known as the pitch circles of the gearing. If a large amount of force be transmitted from the one disc to the other, slipping will occur, and the velocity ratios will no longer remain constant ; in order to prevent slipping, projections or teeth must be formed on the one wheel to fit into recesses in the other, and they must be of such a form that the velocity ratio at every instant shall be constant, i.e. the virtual centre Qab must always be in its present position. We have seen above that the direction of motion of any point in a body moving relatively to another body is normal to the virtual radius ; hence, if we make a pro- jection or a tooth, say, on a (Fig. 166), the direction of motion of any point d relatively to b, will be a normal to the line drawn from d through the virtual centre Qab. Likewise with any point in b relatively to a. Hence, if a projection on the one wheel is required to fit into a recess in the other, a normal to their surfaces at the point of contact must pass through the virtual centre Qab. If such a normal do not pass through Oab, the velocity ratio will be altered^ and if Qab shifts about as the one wheel moves relatively to the other, the motion will be jerky. FlG - l66 ' Hence, in designing the teeth of wheels, we must so form them that they fulfil the condition that the normal to their surfaces at the point of contact must pass through the virtual centre of the one wheel relatively to the other, i.e. the point where the two pitch circles touch one another, or the point where the pitch circles cut the line joining their centres. An infinite number of forms might be designed to fulfil this con- dition; but some forms are more easily constructed than others, and for this reason they are chosen. 148 Mechanics applied to Engineering. The forms usually adopted for the teeth of wheels are the cycloid and the involute, both of which are easily constructed and fulfil the necessary conditions. If the circle e rolls on either the straight line or the arc of Qef FIG. 107. FIG. 168. a circle/, it is evident that the virtual centre is at their point of contact, viz. Oef; and the path of any point d in the circle moves in a direction normal to the line joining d to Oef, or normal to the virtual radius. When the circle rolls on a straight line, the curve traced out is termed a cycloid (Fig. 167); when on the outside of a circle, the curve traced out is termed an epicycloid (Fig. 168); when on the inside of a circle, the curve traced out is termed a hypocycloid (Fig. 169). If a straight line / (Fig. 170) be rolled without slipping on the arc of a circle, it is evident that the virtual centre is at their point of contact, viz. Oef, and the path of any point d in the line moves in a direction normal to the line *' f t or normal to the virtual FIG. 170. radius. The curve traced out by d is an involute. It may be described by wrapping a piece of string round a circular disc and attaching a pencil at d; as the string is unwound d moves in an involute. Mechanisms. 149 When setting out cycloidal teeth, only small portions of the cycloids are actually used. The cycloidal portions can be obtained by construction or by rolling a circular disc on the JHJ~ KolUng cuvle g pitch circle. By reference to Fig. 171, which represents a model used to demonstrate the theory of cycloidal teeth, the reason why such teeth gear together smoothly will be evident. A and B 150 Mechanics applied to Engineering. are parts of two circular discs of the same diameter as the pitch circles ; they are arranged on spindles, so that when the one revolves the other turns by the friction at the line of contact. Two small discs or rolling circles are provided with double-pointed pencils attached to their rims ; they are pressed against the large discs, and turn as they turn. Each of the large discs, A and B, is provided with a flange as shown. Then, when these discs and the rolling circles all turn together, the pencil-point i traces an epicycloid on the inside of the flange of A, due to the rolling of the rolling circle on A, in exactly the same manner as in Fig. 168 ; at the same time the pencil-point 2 traces a hypocycloid on the side of the disc B, as in Fig. 169. Then, if these two curves be used for the pro- files of teeth on the two wheels, the teeth will work smoothly together, for both curves have been drawn by the same pencil when the wheels have been revolving smoothly. The curve traced on the flange of A by the point i is shown on the lower figure, viz. i.i.i ; likewise that traced on the disc B by the point 2 is shown, viz. 2.2.2. In a similar manner, the curves 3.3.3, 4.4.4, have been obtained. The full-lined curves are those actually drawn by the pencils, the remainder of the teeth are dotted in by copying the full-lined curves. In the model, when the curves have been drawn, the discs are taken apart and the flanges pushed down flush with the inner faces of the discs, then the upper and lower parts of the curves fit together, viz. the curve drawn by 3 joins the part drawn by 2 ; likewise with i and 4. From this figure it will also be clear that the point of contact of the teeth always lies on the rolling circles, and that contact begins at C and ends at D. The double arc from C to D is termed the "arc of contact" of the teeth. In order that two pairs of teeth may always be in contact at any one time, the arc CD must not be less than the pitch. The direction of pressure between the teeth is evidently in the direction of a tangent to this arc at the point of contact. Hence, the greater the angle the tangent makes to a line EF (drawn normal to the line joining the centres of the wheels), the greater will be the pressure pushing the two wheels apart, and the greater the friction on the bearings ; for this reason the angle is rarely allowed to be more than 30. In order to keep this angle small, a large rolling circle must be used, but it is rarely more than half the diameter of the smallest wheel in the train. The same rolling circle should be used for all wheels required to gear together. Mechanisms. The model for illustrating the principle of involute teeth is shown in Fig. 172. Here again A and B are parts of two circular discs con- nected together with a thin cross-band which rolls off one disc on to the other, and as the one disc turns it makes the other revolve in the opposite direction. The band is provided with a double-pointed pencil, which is pressed against two flanges on the discs ; then when the discs turn, the pencil-points describe involutes on the two flanges, in exactly the same manner as that described on p. ISO- Then, from what has been said on cycloidal teethj it is evident that if such curves be used as profiles for teeth, the two wheels will gear smoothly together, for they have been drawn by the same pencil as the wheels revolved smoothly together. The point of contact of the teeth in this case always lies on the band ; contact begins at C, and ends at D. The arc of contact here becomes the straight line CD. In order to prevent too great pressure on the /. axles of the wheels, FIG. 172. the angle DEF seldom 152 Mechanics applied to Engineering. exceeds i5-| ; this gives a base circle -ff of the pitch circle (Unwin). A simple arrangement for drawing the form of tooth to work with one of any given form has been devised by Professor Hele-Shaw, which indicates clearly the actual manner in which one tooth comes into contact with another. This method has been extended by him for drawing cycloidal and involute curves, so as to give any required form of wheel tooth. His paper is printed in full, with illustrations, in the Report of the British Association for 1898. It is beyond the scope of the present work to go beyond the principles of the forms of wheel-teeth. Readers requiring details as to the proportions and strength of teeth, and various other matters regarding the design of toothed gearing, cannot do better than refer to Unwin's " Elements of Machine Design," or Anthony's "Essentials of Gearing," D. C. Heath & Co., Boston, U.S.A. Velocity Ratio of Wheel Trains. In most cases the problem of finding the velocity ratio of wheel trains is FIG. 172*1. FIG. 1723, easily solved, but there are special cases in which difficulties may arise. The velocity ratio may have a positive or a negative value, according to inform of the wheels used; thus if a in Fig. 1720 have a clockwise or + rotation, b will have an anti-clockwise or rotation; but in Fig. 172^ both wheels rotate in the same sense, since an annular wheel, i.e. one with internal teeth, rotates in the reverse direction to that of a wheel with external teeth. In both cases the velocity ratio is V = 5? = Z? = ab Qfo _ N r ~R fl T a Qab.Oac N 6 where T a is the number of teeth in a, and T 6 in ft, and N a is the number of revolutions per minute of a and N 6 of b. In the case of the three simple wheels in Fig. 172*:, we have the same peripheral velocity for all of them ; hence Mechanisms. G> a R a = WfiRft = W c 153 "c R T a N e and the first and last wheels rotate in the same sense. The same velocity ratio could be obtained with two wheels only, but then we should have the sense of rotation reversed, snce Thus the second or " idle " wheel simply reverses the sense of rotation, and does not affect the velocity ratio. The velocity ratio is the same in Figs. 1720, 172^, 172^, but in the second and third cases the sense of the last wheel is the same as that of the first. When the radius line R c of the last wheel falls on FlG 1?2< . t the same side of the axle frame as that of the first wheel, the two rotate in the same sense ; but if they fall on opposite sides, the wheels rotate in opposite senses. When the second wheel is compound, i.e. when two wheels of different sizes are fixed to one another and revolve together, it is no longer an idle wheel, but the sense of rotatipn is not altered. If it is desired to get the same velocity ratio with an idle wheel in the train as with a compound wheel, the wheel c must be altered FIG. in the proportion T, where b is the driven and & the driver. The velocity ratio V r of this train is obtained thus and and = rp-, when there is an idle wheel between, or v re = T 6 .T C /-IT rr\ I when there is a compound wheel. In the last figure the wheel c is mounted loosely on the same axle as a and d. In this arrangement neither the velocity ratio nor the sense is altered. The general action of simple, i.e. not bevil, epicyclic trains may be summed up thus : The number of revolutions of any wheel of the train for one revolution of the arm is the number of revolutions that the wheel would make if the arm were fixed, and the first wheel were turned through one revolution, + 1 for wheels that rotate in the same sense as the arm, and i for wheels that rotate in the opposite sense to the arm. It should be remembered that wheels on the nth axle rotate in the same sense as the arm when n is an even number, and in the opposite sense to that of the arm when n is an odd number, counting the axle of the fixed wheel as " one." Thus the revolutions of the nth wheel for one revolution of the arm is (V r + i) when n is even, and V r i when n is odd, where V r is the velocity ratio of the train up to the nth wheel. Epicyclic Bevil Trains. When dealing with bevil trains, it should be noticed t * m < "'"tooMn(/7/8a tnat the' wheel on the odd axis does not rotate in the same sense as the first wheel ; but as regards the velocity ratio of a and c, the wheel b simply acts as an idle wheel. When D rotates, it is equivalent to the rotation of the arm D (Fig. 1722) of a simple epicyclic train. When the FlG - I 7 a/ - wheels and the arm are all rotating, some troublesome problems are liable to arise. For the sake of dealing with such cases, the following table may be of some assistance. Symbols without a dash indicate relative velocities, i.e. revolutions per min. of the wheels on their own axles ; and symbols with a dash indicate absolute velocities of the wheels : Mechanisms. 157 Revolutions per minute of D a (Upon their own axles) Absolute velocities in revolutions per minute of D N c '=-N c +N d When a is at rest (4) When a is rotating From line 2 we have By substitution of N a of line 5 in line 2 (5) (6) Let T a = T c . If D rotates at the same speed and sense as a, the wheel c will do the same, and the wheel b will not turn at all on its own axis ; but if D rotates at the same speed and in the opposite sense to that of a, c will rotate at twice the speed of a, but in the opposite sense. By varying the speed of D, a very wide range of speed can be secured for c. Let N a ' = 100 ; then we get Revolutions per minute of a D C N a ' =100 N d = 5 20 N; = - 9 o - 60 50 80 o 60 100 100 N; = O 5 10 20 40 50 IOO 1 5 8 Mechanics applied to Engineering. Humpage's Gear. This compound epicyclic bevil train is used by Messrs. Humpage, Jacques, and Pedersen, of Bristol, as a variable-speed gear for machine tools (see The Engineer, December 30, 1898). The number of teeth in the wheels are : A = 46, B = 40, B a = 1 6, C = 12, E = 34. The wheels A and C are loose on the shaft F, but E is keyed. The wheel A is rigidly attached to the frame of the machine, and C is driven by a stepped pulley; the arm d rotates on the shaft F; the two wheels B and Bj are fixed together. Let d make one complete clockwise revolution ; then the other wheels will make T Revs, of B on own axle = - = % = nc C absolute = = + i = f + i = 4*83 * T E = revs.ofB X -~-+ i= - 1-15 X-^-fi -* e = -0-54I + I = Q'459 Whence for one revolution of E, C makes -1-J5- = 10*53 o-459 revolutions. The + sign in the expressions for the speed of C and E is ^ Observer looking this way FIG. 1727. on account of these wheels of the epicyclic train rotating in the same sense as that of the arm d. . As stated above, the ^th wheel in a bevil train rotates in the same sense as the arm when n is odd, and in the opposite sense when n is even ; hence the sign is -j- for odd axes, and for even axes, always counting the first as " one." Mechanisms. 1 59 It may Help some readers to grasp the solution of this problem more clearly if we work it out by another method. Let A be free, and let d be prevented from rotating ; turn A through one revolution; then Revs.) _ product of teeth in drivers _ _ T X T 6i ^ of E j product of teeth in driven wheels T 6 xT a ~ Revs.) T a _ ; Hence, when A is fixed by clamping the split bearing G, and d is rotated, the train becomes epicyclic, and since C and E are on odd axes of bevil trains, they rotate in the same sense as the arm ; consequently, for reasons already given, we have Revs. E _ N e __ 0*541 + i ^ i Revs. C " N 3-83 + i 10-53 Particular attention must be paid to the sense of rotation. Bevii gears are more troublesome to follow than plain gears \ hence it is well to put an arrow on the drawing, showing the direction in which the observer is supposed to be looking. CHAPTER VI. DYNAMICS OF THE STEAM-ENGINE. Reciprocating Parts. On p. 133 we gave the construction for a diagram to show the velocity of the piston at each part of the stroke when the velocity of the crank-pin was assumed to be constant. We there showed that, for an infinitely long connecting-rod or a slotted cross-head (see Fig. 159), such a diagram is a semicircle when the ordinates represent the velocity of the piston, and the abscissae the distance it has moved through. The radius of the semicircle represents the constant velocity of the crank-pin. We see from such a dia- gram that the velocity of the reciprocating parts is zero at each end of the stroke, and is a maximum at the middle ; hence during the first half of the stroke the velocity is increased, or the reciprocating parts are accelerated, for which purpose energy has to be expended ; and during the second half of the stroke the velocity is decreased, or the reciprocating parts are retarded, and the energy expended during the first half of the stroke is given back. This alternate expenditure and paying back of energy very materially affects the smoothness of run- ning of high-speed engines, unless some means are adopted for counteracting these disturbing effects. We will first consider the case of an infinitely long con- necting-rod, and see how to calculate the pressure at any part of the stroke required to accelerate and retard the reciprocating parts. The velocity diagram for this case is given in Fig. 173 (see p. 133). Let *- -X-2~-* V = the linear velocity of the crank-pin, FIG. 173. assumed constant ; then the ordinates Yi, V 2 represent to the same scale the velocity of the piston when it is at the positions A,, A 2 respectively, and Vj = V sin 0j. Dynamics of the Steam- Engine. 161 Let the total weight of the reciprocating parts = W. Then The kinetic energy of the re-) _ W V^ _ WV 2 sin 2 6 l ciprocating parts at A 1 } 2 g 2P* WV 2 V 2 Likewise at A 2 = W The increase of kinetic energy) __ WV 2 / v 2 _ v 2 v during the interval A^ } ~ 2^R This energy must have come from the steam or other motive fluid in the cylinder. Let P = the pressure on the piston required to accelerate the moving parts. Work done on the piston in accelerating the) _ p/ \ moving parts during the interval j '^ 2 ~" Xl ' But xf + Vi 2 = x.? + V 2 2 = R 2 hence V^ - V 2 2 = x? - x? and Increase of kinetic energy of the! reciprocating parts during the > = -- 9 (^ 2 2 x interval then Pfe - *,) = - 2 (^ 2 - x?) and P where x is the mean distance - 1 ** of the piston from the 1 62 Mechanics applied to Engineering. middle of the stroke ; and when x = R at the beginning and end of the stroke, we have p _WV 2 We shall term P the " acceleration pressure." Thus with an infinitely long connecting-rod the pressure at the end of the stroke required to accelerate or retard the reciprocating parts is equal to the centrifugal force (see p. 19), assuming the parts to be concentrated at the crank-pin, and at any other part of the stroke distant x from the middle the pressure is less in the ratio |. Another simple way of arriving at the result given above is as follows : If the connecting-rod be infinitely long, then it . always remains parallel x, I to the centre line of |\ the engine ; hence the \ action is the same as " / if the connecting-rod / were rigidly attached to the cross-head and FlG - T 74- piston, and the whole rotated together as one solid body, then each- point in the body would describe the arc of a circle, and would be subjected to WV 2 the centrifugal force C = ^- , but we are only concerned with the component along the centre line of the piston, marked P in the diagram. It will be seen that P vanishes in the middle of the stroke, and increases directly as the distance from the middle, becoming equal to C at the ends of the stroke. When the piston is travelling towards the middle of the stroke the pressure P is positive, yM^___- -~^^ and when travelling away from the middle it is negative. Thus, in constructing a diagram to show the pressure exerted at all parts of the stroke, we put the .^wvj 2 first half above, and the second half below the base-line. We FlG . I75> show such a diagram in Fig. 175. The height of any point in the sloping line ab above the base-line represents the pressure Dynamics of the Steam- Engine. 163 at that part of the stroke required to accelerate or retard the moving parts. It is generally more convenient to express the pressure in pounds per square inch, /, rather than the total p pressure P ; then = /, where A = the area of the piston. We A W will also put w =r ' where w is the weight of the reciprocating A parts per square inch of piston. It is more usual to speak of the speed of an engine in revolutions per minute N, than of the velocity of the crank-pin V in feet per second. v _ 27TRN 60 then, substituting these values of P, W, and V, we have W2 2 7T 2 R 2 N 2 P ~ /: 2 p - = o'ooo34/RN 2 N.B. The radius of the crank R is measured in feet. The quantity w varies between 2 and 6 Ibs. per square inch, and occasionally values outside these limits are met with. In the absence of more accurate data, it is usual to take w 3 Ibs. per square inch. Then the above equation becomes / = o-ooiRN 2 1000 which is a very convenient form of the expression for com- mitting to memory ; even if w be not equal to 3 Ibs., the above expression is readily corrected by multiplying by the w ratio 3 When working from the mean velocity of the piston instead of the velocity of the crank-pin V, it should be remembered that V is greater than the piston velocity V p in the ratio semicircumference = TT = Qr y = diameter 2 Influence of Short Connecting-rods. Referring to Fig. 159, we there showed how to construct a velocity diagram 1 64 Mechanics applied to Engineering, for a short connecting-rod ; reproducing a part of the figure, we have "OulEnd FIG. 176. cross-head velocity __ OX crank-pin velocity OQ Infinite rod cross-head velocity _ crank-pin velocity velocity of cross-head with short rod _ OX velocity of cross-head with long rod OXj But at the " in " end of the stroke -- = - = t I + R and at the " out " end of the stroke = i L Thus if the connecting-rod is n cranks long, the pressure at the " in " end is - greater, and at the " out " end - less, than if the rod were infinitely long. The value of/ at each end of the stroke then becomes p o'ooo34?'RN 2 ( i + -) for the "in" end -\ for the "out" end Dynamics of the Steam-Engine. 165 The line ab is found by the method described on p. 162. Set off aa = a ^ also bb = . The acceleration is zero where n n the slope of the velocity curve is zero, i.e. where a tangent to it is horizontal. Draw a horizontal line to touch the curve, viz. at g 1 . As a check on the accuracy of the work, it should be noticed that this point very nearly indeed corresponds to the position in which the connecting-rod is at right angles to the crank; l the cross-head is then at a distance R(V 2 + i ), n or very nearly , from the middle of the stroke. The point g 1 2n having been found, the corresponding position of the cross- head g is then put in. At the instant when the slope of the short-rod velocity curve is the same as that of the long-rod velocity curve, viz. the semicircle (see p. 134), the accelera- tions will be the same in both cases. In order to find where the accelerations are the same, draw arcs of circles from C as centre to touch the short-rod curve, and from the points where they touch erect perpendiculars to cut the circle at the points H and i\ which occurs when = 45 and 135 (see p. 1 66). The corresponding positions of the cross-head are shown at h and i respectively. In these positions the accelera- tion curves cross one another, viz. at h and i . We now have five points on the short-rod acceleration curve through which a smooth curve may be drawn. The acceleration pressure at each instant may also be arrived at thus Let = the angle turned through by the crank starting from the " in " end ; V = the linear velocity of the crank-pin, assumed constant and represented by R ; v = the linear velocity of the cross-head. Then B 8in(0 + a) ( seeFi 6) V sin (90- a) v ,, /sin cos a + cos sin a\ v = VI - - I \ COS a / In all cases in practice the angle a is small, consequently 1 Engineering, July 15, 1892, p. 83 ; also June 2, 1899. 1 66 Mechanics applied to Engineering. cos a is very neaciy equal to unity, and may therefore be neglected ; even with a very short rod the average error in the final result is well within one per cent. Hence v = V(sin + cos sin a) nearly We also have L = n = sin(9 R sin a Substituting this value v = V(sin + cos The acceleration of the cross-head/, = = . (The angle 9 is measured in circular measure.) Q _ the space traversed by the crank-pin in the time S/ radius of the crank V . 8/ . 86 V dv V .dt dv V whence in the limit /.= -- = -. R Substituting the value of v found above, we have and the acceleration pressure, when the crank has passed through the angle from the " in " end of the stroke, is a/VV , , cos 26 or/ = o-000340/RN 2 ("cos + 22i 2 ^\ Dynamics . of the Steam- Engine. It should be noticed that at the beginning and end of the stroke, i.e. when 6 = o, and 6 = 180, this expression becomes that previously found. It is also identical for the case in which the length of the rod is infinite. In arriving at the value of w it is usual to take as reciprocating parts the piston-head, piston-rod, tail-rod (if any), cross-head, small end of connecting-rod and half the plain part of the rod. When air-pumps or other connections are attached to the cross-head, they may approximately be taken into account in calculating the weight of the reciprocating parts ; thus FIG. 177. weight of | piston + piston and tail-rods + both cross-heads -f small area of endofcon.rod+ plainpart + air- pump i/ 4 2 piston The kinetic energy of the parts varies as the square of the velocity; hence the ( j Correction of Indicator Diagram for Acceleration Pressure. An indicator diagram only shows the pressure of the working fluid in the cylinder; it does not show the real pressure transmitted to the crank-pin because some of the energy is absorbed in accelerating the reciprocating parts during the first part of the stroke, and is therefore not available for driving the crank, whereas, during the latter part of the stroke, energy is given back from the reciprocating parts, and there is excess energy over that supplied from the working fluid. But, apart from these effects, a single indicator diagram 1 68 Mechanics applied to Engineering. does not show the impelling pressure on a piston at every portion of the stroke. The impelling pressure is really the difference between the two pressures on both sides of the piston at any one instant, hence the impelling pressure must be measured from the top line of one diagram and the bottom line of the other, as shown in full lines in Fig. 178. This diagram, set out FlG I?8> to a straight base-line, is shown in Fig. 179, aaa. On the same base-line we have set off the acceleration pressure diagram, bbb ; then, setting down on the base-line the difference between the two curves, we get the line ccc, giving the real pressure transmitted to the crank-pin at each instant. The area of this latter curve is, of course, equal to the area of the indicator diagram. If the engine were of the vertical type the weight of the moving parts w must be added or subtracted, accord- ing as it is making the up or down stroke; for the upstroke the base-line be- comes dd, and for the down- stroke ee. When dealing with en- gines having more than one cylinder, the question of scales must be carefully attended to; that is, the heights of the diagrams must be corrected in such a manner that the mean height of each shall be proportional to the total effort exerted on the piston. Let the original indicator diagrams be taken with springs of the following scales, H.P. -, I.P. I, L.P. -. Let the areas of x y z the pistons (allowing for rods) be H.P. X, I.P. Y, L.P. Z. Let all the pistons have the same stroke. Suppose we find that the H.P. diagram is of a convenient size, we then reduce all the others to correspond with it. If, say, the intermediate piston were of the same size as the high-pressure piston, we should simply have to alter the height of the intermediate diagram in the ratio of the springs ; thus FIG. 179. The curve aaa is the shaded area shown above to a straight base. Dynamics of the Steam- Engine. 169 Correctedheightof I.P. diagram) ; e x if pistons were of same size i } diagram J 1 X - actual height X ^- MT But as the cylinders are not of the same size, the height of the diagram must be multiplied by the ratio of the two areas ; thus Height of intermediate diagram \ corrected for scales of springs > and for areas of pistons J actual height of \ y intermediate \ X ^ X - diagram J = actual height X 2 Similarly for the L.P. diagram Height of L.P. diagram corrected > Jactual hd h{ of > sZ for scales of springs and forj - { Lp dia s X ^ areas of pistons ) It is probably best to make this correction for scale and area after having reduced the diagrams to the form given in line ccc in Fig. 179. Pressure on the Crank-pin. The diagram given in FIG. 180. Fig. 179 represents the pressure transmitted to the crank-pin at all parts of the stroke. The ideal diagram would be one in which the pressure gradually fell to zero at each end of the stroke, and was constant during the rest of the stroke, such as a, Fig. 180. The curve b shows that there is too much compression resulting in a negative pressure p at the end of the stroke ; at the point x the pressure on the pin would be reversed, and, if 170 Mechanics applied to Engineering. there were any "slack" in the rod-ends, there would be a knock at that point, and again at the end of the stroke, when the pressure on the pin is suddenly changed from p to +/. These defects could be remedied by reducing the amount of compression and the initial pressure, or by running the engine at a higher speed. The curve (c) shows that there is a deficiency of pressure at the beginning of the stroke, and an excess at the end. The defects could be remedied by increasing the initial pressure and the compression, or by running the engine at a lower speed. For many interesting examples of these diagrams, the reader is referred to Rigg's " Practical Treatise on the Steam Engine;" also a paper by the same author, read before the Society of Engineers. Cushioning for Acceleration Pressures. In order to counteract the effects due to the acceleration pressure, it is usual in steam-engines to close the exhaust port before the end of the stroke, and thus cause the piston to compress the exhaust steam that remains in the cylinder. By choosing the point at which the exhaust port closes, the desired amount of compression can be obtained which will just counteract the acceleration pressure. In certain types of vertical single-acting high-speed engines, the steam is only admitted on the downstroke ; hence on the upstroke some other method of cushioning the reciprocating parts has to be adopted. In the well-known Willans engine an air-cushion cylinder is used ; the required amount of cushion at the top of the cylinder is obtained by carefully regulating the volume of the clearance space. The pistons of such cylinders are usually of the trunk form ; the outside pressure of the atmosphere, therefore, acts on the full area of the underside, and the compressed air cushion on the annular top side. Let A = area of the underside of the piston in square inches ; A a = area of the annular top side in square inches ; W = total weight of the reciprocating parts in Ibs. ; c = clearance in feet at top of stroke. At the top, i.e. at the " in end," of the stroke we have P = o-ooo34\V.RN 2 ( i + -} - W + 14'yA \ / Assuming isothermal compression of the air. and taking the pressure to be atmospheric at the bottom of the stroke, we have Dynamics of the Steam -Engine. i4'7A,,(2R + c) = PC whence c = "94 a P - 147 A fl Or for adiabatic compression i47A tt (2R + *) r41 = P^ r41 c 171 2R p in the expression for c given above. The problem of balancing the reciprocating parts of gas and oil engines is one that presents much greater difficulties than in the steam-engine, partly because the ordinary cushioning method cannot be adopted, and further because the effective pressure on the piston is different for each stroke in the cycle. Such engines can, however, be partially balanced by means of helical springs attached either to the cross-head or to a tail-rod, arranged in such a manner that they are under no stress when the piston is at the middle of the stroke, and are under their maximum compression at the ends of the stroke. The weight of such springs is, however, a great drawback ; in one instance known to the author the reciprocating parts weighed about 1000 Ibs. and the springs 800 Ibs. Polar Twisting-Moment Diagrams. From the diagrams of real pressures transmitted to the crank-pin that FIG. 181. we have just constructed, we can readily determine the twisting moment on the crank-shaft at each part of the revolution. In Fig. T 8 1, let/ be the horizontal pressure taken from such a diagram as Fig. 179. Then fa is the pressure transmitted 1/2 Mechanics applied to Engineering. along the rod to the crank-pin. This may be resolved in a direction parallel to the crank and normal to it (/) ; we need not here concern ourselves with the pressure acting along the crank, as that will have no turning effect. The twisting moment on the shaft is then/ n R ; R, however, is constant, therefore the twisting moment is proportional to p n . By setting off values of p n radially from the crank-circle we get a diagram showing the twisting moment at each part of the revolution. / is measured on the same scale, say -, as the indicator diagram ; then, if A x be the area of the piston in square inches, the twisting moment in pounds feet = / n #AR, where p n is measured in inches, and the radius of the crank R is expressed in feet. When the curve falls inside the circle it simply indicates Bottom, of H.PGfUnder Bottom, I NT. P. Cylinder Top of I NT. P Cylinder FIG. 182. that there is a deficiency of driving effort at that place, or, in other words, that the crank-shaft is driving the piston. In Fig. 182 we have a similar diagram, worked out fully for a vertical triple-expansion engine made by Messrs. McLaren of Leeds, and by whose courtesy the author now gives it. The dimensions of the engine were as follows : Dynamics of the Steam-Engine. 173 Diameter of cylinders High pressure Intermediate Low pressure Stroke 9'OI inches. 14*25 , 22-47 2 feet. 1/4 Mechanics applied to Engineering. The details of reducing the indicator diagrams have been omitted for the sake of clearness ; the method of reducing them has been fully described. Twisting Moment on a Crank-shaft. In some instances it is more convenient to calculate the twisting moment on the crank- shaft when the crank has passed through the angle from the inner FIG iga j dead centre than to construct a diagram. Let P = the effort on the piston-rod due to the working fluid and to the inertia of the moving parts ; Pj = the component of the effort acting along the connecting-rod ; sin B P = P x cos a, and sin a = -^ from which a can be obtained, since and n are given. The tangential component T = P! COS and = 90 (0 -}- a) whence T= -A-cos (90- (6 + a)} = *.V*l COS a COS a Flywheels. The twisting-moment diagram we have just constructed shows very clearly that the turning effort on the crank-shaft is far from being constant hence, if the moment of resistance be constant, the angular velocity cannot be constant. In fact, the irregularity is so great in a single-cylinder engine, that if it were not for the flywheel the engine would come to a standstill at the dead centre. A flywheel is put on a crank-shaft with the object of storing energy while the turning effort is greater than the mean, and giving it back when the effort sinks below the mean, thus making the combined effort, due to both the steam and the flywheel, much more constant than it would otherwise be, and thereby making the velocity of rotation more nearly constant. But, however large a flywheel may be, there must always be some variation in the velocity ; but it may be reduced to as small an amount as we please by using a suitable flywheel. In order to find the dimensions of a flywheel necessary for keeping the cyclical velocity within certain limits, we shall make use of the twisting-moment diagram, plotted for convenience Dynamics of the Steam -Engine. 175 to a straight instead of a circular base-line, the length of the base being equal to the semicircumference of the crank- pin circle. Such a diagram we give in Fig. I83. 1 Its area is, of course, equal to the area of the original indicator diagram, from which it was constructed; this check should, indeed, always be applied, to see whether the workmanship is accurate. The length of its base is greater than the length of the original indicator diagram in the ratio of ir to 2 ; the mean height is, FIG. 183. consequently, less in the ratio 2 to IT. The resistance line, which for the present we shall assume to be straight, is shown dotted ; the diagonally shaded portions below the mean line are together equal to the horizontally shaded area above. During the period AC the effort acting on the crank-pin is less than the mean, and the velocity of rotation of the crank- pin is consequently reduced, becoming a minimum at C. During the period CE the effort is greater than the mean, and the velocity of rotation is consequently increased, becoming a maximum at E. Let V = mean velocity of a point on the rim at a radius equal to the radius of gyration of the wheel, in feet per second usually taken for practical purposes as the velocity of the surface of the rim : V c = minimum velocity at C (Fig. 183) ; V e == maximum E ; W = weight of the flywheel in pounds, usually taken for practical purposes as the weight of the rim ; R M == radius of gyration in feet of the flywheel rim, usually taken as the external radius for practical purposes. WV 2 Then the energy stored in the flywheel at C = - WV 2 TT e 2 g W The increase of energy during CE - (V e a - V c 2 ) (i.) 1 Figures 178, 179, 181, 183 are all constructed from the same indicator diagram. 176 Mechanics applied to Engineering. This increase of energy must have been derived from the steam or other source of energy ; therefore it must be equal to the work represented by the horizontally shaded area CDE = E n (Fig. 183). Let E n = m x average work done per stroke. Then the area CDE is m times the work done per stroke, or m times the whole area BCDEF. Or F _ m x indicated horse-power of engine x 33000 ,.. n ~ 2 N where N is the number of revolutions of the engine per minute in a double-acting engine. Whence, from (i.) and (ii.), we have W /V2 v ,v __m X I.H.P. X 33000 V 2N But * "^" c = V (approximately) or also V * ~ Vc = K, " the coefficient of speed fluctuation " and V e - V c = KV V, 2 - V c 2 = 2KV 2 Substituting this value in (iii.) -(2KV 2 ) = m X LH - ? - X 33 = E V ' 2N and W = The proportional fluctuation of velocity K is the fluctuation of velocity on either side of the mean ; thus, when K = 0*02 it is a fluctuation of i per cent, on either side of the mean. The following are suitable values for K : K = o'oi to 0*02 for ordinary electric-lighting engines, but for public lighting and traction stations it often gets as low as 0*0016 to 0*0025 to allow for very sudden and large changes in the load ; the weight of all the rotating parts, each multiplied by its own radius of gyration, is to be included in the flywheel ; = o'o2 to 0*04 for factory engines; = 0*06 to o'io for rough engines. Dynamics of the Steam-Engine. 177 When designing flywheels for public lighting and traction stations where great variations in the load may occur, it is common to allow from 2*4 to 4*5 foot-tons (including rotor) of energy stored per I.H.P. The calculations necessary for arriving at the value of E n for any proposed flywheel are somewhat long, and the result when obtained has an element of uncertainty about it, because the indicator diagram must be assumed, as the engine so far only exists on paper. The errors involved in the diagram may not be serious, but the desired result may be arrived at within the same limits of error by the following simpler process. The table of constants given below has been arrived at by con- structing such diagrams as that given in Fig. 183 for a large number of cases. They must be taken as fair average values. The length of the connecting-rod, and the amount of pressure required to accelerate and retard the moving parts, affect the result. The following table gives approximate values of m. In arriving at these figures it was found that if n = number of cranks, then m varies as -^ approximately. APPROXIMATE VALUES OF m FOR DOUBLE-ACTING STEAM-ENGINES.' Cut-off. Single cylinder. Two cylinders. Cranks at right angles. Three cylinders. Cranks at 120. O'l o'35 0-088 0-040 0-2 o'33 0'082 0-037 0'4 0-31 0-078 0-034 0-6 0-29 0-072 0*032 0-8 0-28 0'070 0-031 End of stroke 6-27 0-068 0-030 i m FOR GAS- AND OIL-ENGINES. Single cylinder. Two cylinders. Cranks at right angles. Exploding at every 4th stroke ,, ,, 8th ,, i.e. ? missing every alternate charge 37 to 4-5 8-5 to 9-8 I'O to I 'I 2'I tO 2'5 1 The values of m vary much more in the case of two- and three- cylinder engines than in single-cylinder engines. Sometimes the value of m is twice as great as those given, which are fair averages. N Mechanics applied to Engineering. Relation between the Work stored in a Flywheel and the Work done per Stroke. For many purposes it is convenient to express the work stored in the flywheel in terms of the work done per stroke. The energy stored in the wheel = - Then from equation (iv.), we also have The energy stored in thei __ E M wheel I 2K and the average work done^ _ E w per stroke I'm = I-H.P. X 33000^ | (for a double- 2 N A acting engine the number of average^ _ ^K _ m strokes stored in flywheel/ ~~ E n ~~ 2K In the -following table we give the number of strokes that must be stored in the flywheel in order to allow a total fluctua- tion of speed of i per cent., />. ~ per cent, on either side of the mean. If a greater variation be permissible in any given case, the number of strokes must be divided by the per- missible percentage of fluctuation. Thus, if 4 per cent., i.e. K = 0*04, be permitted, the numbers given below must be divided by 4. NUMBER OF STROKES STORED IN A FLYWHEEL FOR DOUBLE-ACTING STEAM-ENGINES. 1 Cut-off. Single cylinder. Two cylinders. Cranks at right angles. Three cylinders. Cranks at 120. O'l 18 4 '4 2'0 0'2 17 4'i I 9 0'4 16 3'9 r-8 0-6 15 3'6 17 0-8 H 3*5 1-6 End of stroke 13 3'4 *'$ 1 See note at foot of p. 177. Dynamics of the Steam-Engine. GAS-ENGINES (MEAN STROKES). 1/9 Single cylinder. Two cylinders. Cranks at right angles. Exploding at every 4th stroke 8th 185 to 225 425 to 490 46 to 56 112 to 122 Shearing, Punching, and Slotting Machines (K not known). It is usual to store energy in the flywheel equal to the gross work done in two working strokes of the shear, punch, or slotter, amounting to about 15 inch- tons per square inch of metal sheared or punched through. Gas-Engine Flywheels. The value of m for a gas- engine can be roughly arrived at by the following method. FIG. 184. The work done in one explosion is spread over four strokes when the mixture explodes at every cycle. Hence the mean effort is only one-fourth of the explosion-stroke effort, and the excess energy is therefore approximately three-fourths of the whole explosion-stroke effort, or three times the mean : hence m = 3. Similarly, when every alternate explosion is missed, m = 7. By referring to the table, it will be seen that both of these values are too low. The diagram for a 4-stroke case is given in Fig. 184. It has I So Mechanics applied to Engineering. been constructed in precisely the same manner as Figs. 179, 181, and 183. When gas-engines are used for driving dynamos, a small flywheel is often attached to the dynamo direct, and runs at a very much higher peripheral speed than the engine flywheel. Hence, for a given weight of metal, the small high- speed flywheel stores a much larger amount of energy than the same weight of metal in the engine flywheel. The peripheral speed of large cast-iron flywheels has to be kept below a mile a minute (see p. 181), on account of their danger of bursting. The small disc flywheels, such as are used on dynamos, are hooped with a steel ring, shrunk on the rim, which allows them to be safely run at much higher speeds than the flywheel on the engine. The flywheel power of such an arrangement is then the sum of the energy stored in the two wheels. The author's experience leads him to the conclusion that there is no perceptible flicker in the lights when about forty impulse strokes, or 160 average strokes (when exploding at every cycle, and twice this number when missing alternate cycles), are stored in the flywheels. Case in which the Resistance varies. In all the above cases we have assumed that the resistance overcome by the engine is constant. This, however, is not always the case ; when the resistance varies, the value of E M is found thus : The line aaa is the engine curve as described above, the line bbb the resistance to be overcome, the horizontal shading indicates excess energy, and the vertical deficiency of energy. The excess areas are, of course, equal to the deficiency areas over any complete cycle. The resistance cycle may extend over several engine cycles; an inspection or a measurement will reveal the points of maximum and minimum velocity. The value of m is the ratio of the horizontal shaded areas to the whole area under the line aaa described during the complete cycle of operations. For a full treatment of this, the reader is referred to an article by Prof. R. H. Smith in the Engineer of January 9, 1885. Stress in Flywheel Rims. If we neglect the effects of Dynamics of the Steam-Engine. 181 the arms, the stress in the rim of a flywheel may be treated in the same manner as the stresses in a boiler-shell or, more strictly, a thick cylinder (see p. 348), in which we have the relation or P r R w =/, when / = i inch The P r in this instance is the pressure on each unit length of rim due to centrifugal force. We shall find it convenient to take the unit of length as i foot, because we take the velocity of the rim \\\feet per second. Then W V ' 2 W V 2 FIG. 186. where W r = the weight of i foot length of rim, i square inch in section = 3 'i Ibs. for cast iron We take i sq. inch in section, because the stress is expressed in pounds per square inch. Then substituting the value of W P in the above equations, we have ' % - -/ 32-2 /= 0-096 VJ 2 V 2 or/ = (very nearly) 10 For a fuller treatment, taking into account the effect of the arms, etc., the reader is referred to Unwin's " Elements of Machine Design," Part II. In English practice V w is rarely allowed to exceed 100 feet per second, but in American practice much higher speeds are often used, probably due to the fact that American cast iron is much tougher and stronger than the average metal used in England. An old millwright's rule was to limit the speed to a mile a minute, i.e. 88 feet per second, corresponding to a stress of about 800 Ibs. per square inch. In this connection it must be remembered that the internal cooling stresses in cast-iron wheel-rims are liable to be very serious, and they may increase the real stress in the metal to an 182 Mechanics applied to Engineering. amount far above that due to centrifugal force ; to this unknown factor one may safely attribute many of the disastrous bursting accidents that so frequently occur even with wheels running at moderate speeds, and when a wheel-rim gets overheated, due to a friction dynamometer brake, the risk of bursting is still greater. To reduce this risk large wheels should invariably be made with split bosses. Many firms either build up their wheels entirely of wrought-iron or steel sections and plates, or wind channel-shaped rims with hard-drawn steel wire. In some cases a cast-iron rim is attached by a thin plate-web to the boss of the wheel; such wheels are far safer than those made entirely of cast iron, and, moreover, the plate-web wheel has the additional advantage of offering much less air resistance than a wheel with arms. In gas- or oil-engine driven electrical installations it is common to increase the store of flywheel energy by putting a small disc flywheel on the dynamo shaft. The rim velocity in such cases often reaches 300 feet per second, but they are always hooped with a shrunk wrought-iron or steel ring, to reduce the possibility of bursting. With this arrangement the belt should always be left slacker than usual, in order to allow a small amount of slip at every explosion. Experimental Determination of the Bursting Speed of Flywheels. Professor C. H. Benjamin, of the Case School, Cleveland, Ohio, has done some excellent research work on the actual bursting speed of flywheels, which well corroborates the general accuracy of the theory. The results he obtained are given below, but the original paper read by him before the American Society of Mechanical Engineers in 1899 should be consulted by those interested in the matter. Bursting speed in feet per sec. - Thickness of rim. Remarks. Ibs. sq. in. inch. 43 18,500 0-68 Solid rim, 6 arms, 15 ins. diam. 388 15,000 0*56 J J ' j 192 Jointed rim, ,, , 381 14,500 o'6<> Solid rim, 3 arms, , 363 385 13,200 14,800 0-38 it ,, 6 arms, 24 ins. diam. Two internal I 9 305 3,610 9,300 075 flanged joints, ,, Linked joints ,, ' Dynamics of the Steam- Engine. 183 From these and other tests, Professor Benjamin con- cludes that solid rims are by far the safest for wheels of moderate size. The strength is not much affected by bolting the arms to the rim, but joints in the rims are the chief sources of weakness, especially when the joints are near the arms. Thin rims, due to the bending action between the arms, are somewhat weaker than thick rims. Some interesting work on the bending of rims has been done by Mr. Barraclough (see I.C.E. Proceedings, vol. cl.). For practical details of the construction of flywheels, readers are referred to a paper by Mr. Sharpe on " Flywheels," read before the Manchester Association of Engineers in October, 1900. Bending Stresses in Locomotive Coupling-rods. Each point in the rod describes a circle (relatively to the engine) as the wheels revolve; hence each particle of the rod is subjected to an upward and downward force equal to the centrifugal force when the rod is in its top and bottom positions. As we shall express the stress in the rod in pounds per square inch, we must express the bending moment on the FIG. 187. rod in pound-inches ; hence we take the length of the rod / in inches. In the expression for centrifugal force we have feet units, hence the radius of the coupling crank must be in feet. The centrifugal force actingj = c = on the rod per inch run / where w is the weight of the rod in pounds per inch run, or w = 0*2 8 A pounds, where A is the sectional area of the rod. The centrifugal force is an evenly distributed load all along the rod ; then, assuming the rod to be parallel, we have The maximum bending moment j = C/ 2 _ /AK a in the middle of the rod, M / 8 y (see Chapters IX. and X.) 184 Mechanics applied to Engineering. where /c 2 = the square of the radius of gyration (inch units) about a horizontal axis through the c. of g. ; y = the half-depth of the section (inches). Then, substituting the value of C, we have , = 0-00034 X 0*28 xAxRcX^X^Xj /= 8 X A X , _ o-ooooi2R c N 2 / 2 j ~~ ~ The value of /c 2 can be obtained from Chapter III. For a angu BH 3 - ; rectangular section, K 2 = ; and for an I section, K 2 - i 2 (BH - bh)' It should be noticed that the stress is independent of the sectional area of the rod, but that it varies inversely as the square of the radius of gyration of the section ; hence the im- portance of making rods of I section, in which the metal is placed as far from the neutral axis as possible. If the stress be calculated for a rectangular rod, and then for the same rod which has been fluted by milling out the sides, it will be found that the fluting very materially strengthens the rod. The bending stress can be still further reduced by removing superfluous metal from the ends of the rod, i.e. by proportioning each section to the corresponding bending moment, which is a maximum in the middle and diminishes towards the ends. The " bellying " of rods in this manner is a common practice on many railways. In addition to the bending stress in a vertical plane, there is also a direct stress of uniform intensity acting over the section of the rod, sometimes in tension and sometimes in compression. The uniform stress is due to the driving effort transmitted through the rod from the driving to the coupled wheel, but it is impossible to say what this effect may amount to. It is usual to assume that it amounts to one-half of the total pressure on the piston, but it may be greater ; but whatever stress is assumed, it must be added to the bending stress found above. Professor Perry, however, does not consider that this is a complete treatment. He also allows for the fact that the rod bends, and therefore treats it as a strut loaded out of the centre (see Perry's "Applied Mechanics," p. 470). Dynamics of the Steam- Engine. Bending Stress in Connecting-rods. In the , case of a coupling-rod of uniform section, in wh : ch each particle describes a circle of the same radius as the coupling-crank pin, the centrifugal force produces an evenly distributed load ; but in the case of a connecting-rod the swing, and therefore the centrifugal force at any section, varies from a maximum at the crank-pin to zero at the gudgeon-pin. The centrifugal force acting on any small mass distant x from the gudgeon- pin is ---, where C is the centrifugal force acting on a mass rotating in a circular path of radius R, i.e. the radius of the crank, and / is the length of the connecting- rod. Let the rod be in its extreme upper or lower position, and let the reaction at the gudgeon-pin, due to the centrifugal force acting on the rod, be R^. Then, since the centrifugal force varies directly as the distance x from the gudgeon-pin, the load distribution diagram is a triangle, and R/ _C/ / KJ = X 2 3 FIG. 187*. C/ 6 The shear at a section distant ^^__ from the gudgeon-pin ) a The shear is zero when = or when x = -r=- But the bending moment is a maximum at the section where the shear is zero (see p. 401). The bending moment at a section y distant x from R,, M y = - X X - which, by substitution of the values of x and reduction, gives and by 1 86 Mechanics applied to Engineering. M = ""-max. and the bending stress/ = 164000^ which is about one-half as great as the stress in a coupling-rod working under the same conditions. Readers who wish to go very thoroughly into this question should refer to a series of articles in the Engineer, March, 1903. Balancing Revolving Axles. CASE I. " Standing Balance" If an unbalanced pulley or wheel be mounted on a shaft and the shaft be laid across two levelled straight-edges, the shaft will roll until the heavy side of the wheel comes to the bottom. If the same shaft and wheel are mounted in bearings and rotated rapidly, the centrifugal force acting on the unbalanced portion would cause a pressure on the bearings acting always in the direction of the unbalanced portion ; if the bearings were very slack and the shaft light, it would lift bodily at every revolution. In order to prevent this action, a balance weight or weights must be attached to the wheel in its own plane of rotation, with the centre of gravity diametrically opposite to the unbalanced portion. Let W = the weight of the unbalanced portion ; - Wi = balance weight ; R = the radius of the c. of g. of the unbalanced portion ; Rj = the radius of the c. of g. of the balance weight. Then, in order that the centrifugal force acting on the balance weight may exactly counteract the centrifugal force acting on the unbalanced portion, we must have 0-00034WRN 2 = o-ooo34W 1 R 1 N 2 or WR = W x Ri or WR - WjRj = o that is to say, the algebraic sum of the moments of the rotating weights about the axis of rotation must be zero, which is equivalent to saying that the centre of gravity of all the rotating weights must coincide with the axis of rotation. When this is the case, the shaft will not tend to roll on levelled straight- edges, and therefore the shaft is said to have "standing balance." Dynamics of the Steam-Engine. 187 When a shaft has standing balance, it will also be perfectly balanced at all speeds, provided that all the weights rotate in the same plane. We must now consider the case in which all the weights do not rotate in the same plane. CASE II. Running Balance. If we have two or more weights attached to a shaft which fulfil the conditions for standing balance, but yet do not rotate in the same plane, the shaft will no longer tend to lift bodily at each revolution ; but it will tend to wobble, that is, it will tend to turn about an axis perpendicular to its own when it rotates rapidly. If the bearings were very slack, it would trace out the surface of a double cone in space as indicated by the dotted lines, and the axis would be con- stantly shifting its position, i.e. it would not be permanent. The reason for this is, that the two centrifugal forces c and ^ form a couple, tending to turn the shaft about some point A between them. In order to FIG. 188. FlG. counteract this turning action, an equal and opposite couple must be introduced by placing balance weights diametrically opposite, which fulfil the conditions for " standing balance," and r88 Mechanics applied to Engineering. moreover their centrifugal moments about any point in the axis of rotation must be equal and opposite in effect to those of the original weights. Then, of course, the algebraic sum of all the centrifugal moments is zero, and the shaft will have no tendency to wobble, and the axis of rotation will be permanent. In the figure, let the weights W and W a be the original weights, balanced as regards " standing balance," but when rotating they exert a centrifugal couple tending to alter the direction of the axis of rotation. Let the balance weights VV 2 and W 3 be attached to the shaft in the same plane as Wj and W, i.e. diametrically opposite to them, also having "standing balance." Then, in order that the axis may be permanent, the following condition must be fulfilled : o-ooo 34 N 2 (WR>;+W 1 R 1 ;; 1 ) = or WRy + W^^ - W 2 R 2 j. 2 - W 3 R 3 j 3 = o The point A, about which the moments are taken, may be chosen anywhere along the axis of the shaft without affecting the results in the slightest degree. Great care must be taken with the signs, viz. a + sign for a clockwise moment, and a sign for a contra-clockwise moment. The condition for standing balance in this case is WR - WA - W 2 R 2 + W 3 R 3 = o So far we have only dealt with the case in which the balance weights are placed diametrically opposite to the weight to be balanced. In some cases this may lead to more than one balance weight in a plane of rotation ; the reduction to one equivalent weight is a simple matter, and will be dealt with shortly. Then, remembering this condition, the only other conditions for securing a permanent axis of rotation, or a " running balance," are 2WR = o and 2WRy = o where SWR is the algebraic sum of the moments of all the rotating weights about the axis of rotation, and y is the distance, measured parallel to the shaft, of the plane of rotation of each weight from some given point in the axis of rotation. Thus the c. of g. of all the weights must lie in the axis of rotation. Dynamics of the Steam-Engine. 189 Graphic Treatment of Balance Weights. Such a problem as the one just dealt with can be very readily treated graphically. For the sake, however, of giving a more general application of the method, we will take a case in which the FIG. 189*. weights are not placed diametrically opposite, but are as shown in the figure. Let all the quantities be given except the position and weight of W 4 , and the arm j>- 4 , which we shall proceed to find by construction. Standing balance, There must be no tendency for the axis to lift bodily ; hence the vector sum of the forces Cj, C 2 , C 3 , C 4 , must be zero, i.e. they must form a closed polygon. Since C is pro- portional to WR, set off WiR lf W,R/ W 2 R 2 , W 3 R 3 , to some suitable scale and in their respective directions ; then the closing line of the force polygon gives us W 4 R 4 in direction, magnitude, and sense. The radius R 4 is given, whence W 4 is found by dividing by R 4 . Running balance. There must be no tendency for the axis to wobble ; hence the vector sum of the moments C,^/,, etc., about a given plane must be zero, i.e. they, like the forces, must form a closed polygon. We adopt Pro- fessor Dalby's method of taking the plane of one of the rotating masses, viz. W 2 for our plane of reference ; then the force C 2 has no moment about the plane. Constructing the triangle of mo- ments, we get the value of W 4 R 4 ^ 4 from the closing line of the triangle. Then dividing by W 4 R 4 , we get the value of jj/ 4 . 190 Mechanics applied to Engineering. FIG. 1893. Provided the above-mentioned conditions are fulfilled, the axle will be perfectly balanced at all speeds. It should be noted that the second condition cannot be fulfilled if the number of rotating masses be less than four. Balancing of Stationary Steam-Engines. Let the sketch represent the scheme of a two-cylinder vertical steam- engine with cranks at right angles. Consider the moments of the unbalanced forces p a and/ 6 about the point O. When the piston A is at the bottom of its stroke, there is a contra-clockwise moment, p a y a , due to the acceleration pressure p a tend- ing to turn the whole engine round in a contra-clockwise direction about the point O. The force p b is zero in this position (neglecting the effect of the obliquity of the rod). When, however, A gets to the top of its stroke, there is a moment, p a y ay tending to turn the whole engine in a contrary direction about the point O. Likewise with B ; hence there is a constant tendency for the engine to lift first at O, then at P, which has to be counteracted by the holding-down bolts, and which may give rise to very serious vibrations unless the foundations be very massive. It must be clearly understood that the cushioning of- the steam men- tioned on p. 170 in no way tends to reduce this effect ; balance weights on the cranks will partially remedy the evil, but it is quite impossible to entirely eliminate it in such an engine as this. A two-cylinder engine can, however, be arranged so that the balance is perfect in every respect. Such a one is found in the Barker engine. In this engine the two cylinders are in line, and the connecting-rod of the A piston is forked, while that of the B piston is coupled to a central crank ; thus any forces that may act on either of the two rods are equally distributed between the two main bearings of the bed-plate, and consequently no disturbing moments are set up. Then if the mass of A and its attachments is equal to that of B, also if the moments of inertia of the two connecting-rods about the gudgeon-pins are the same, the disturbing effect of the obliquity of the rods will be entirely eliminated. FIG. 189*. Dynamics of the Steam-Engine. 191 A three-cylinder vertical engine having cranks at 120, and having equal reciprocating and rotating masses for each cylinder, can be entirely balanced along the centre line of the engine. The truth of this statement can be readily demonstrated by taking the sum of the three inertia forces as given by the equation on p. 166 for the angles 0, 0+120, and # + 240, which will be found to be zero. The proof was first given by M. Normand of Havre. There will, however, be small unbalanced forces acting at I w, -JB- l FIG. / right angles to the centre line, tending to make the engine rock about an axis parallel to the centre line of the crank-shaft. A four-cylinder engine, how- ever, apart from a small error due to the obliquity of the rods, can be perfectly balanced ; thus Let the reciprocating masses be W 1? W 2 , etc. ; the radii of the cranks be R 1} R 2 , etc. ; the distance from the plane of reference taken through the first crank be y 2 , y 3) etc. Then the acceleration pressure, neglecting the obliquity of the rods at each end of the stroke, will be o'ooo34WRN 2 , with the corresponding suffixes for each cylinder. Since the 192 Mechanics applied to Engineering. speed of all of them is the same, the acceleration pressure will be proportional to WR. It will be convenient to tabulate the various quantities, thus Cylinder. Weight of reciprocating parts. Ibs. Radius of crank. Proportional acceleration force. Distance of centre line from plane of reference. Proportional acceleration force moment. I w,= 750 R, = 12" WjR,= 9,000 2 W 2 =iooo R 2 =H" W 2 R 2 =i4,ooo J> 2 = 40" W S R^ 8 = 560,000 3 W 3 =I200 R 3 = i4" W 3 R 3 = 16,800 }' 3 = So" WsR^ =1,344, ooo 4 W 4 =I2 3 R 4 = I2" W 4 R<= 14,800 }> 4 =II2" W 4 R 4 ;' 4 =1,653,000 The vector sum of both the forces and the moments of the forces must be zero to secure perfect balance, i.e. they must form closed polygons ; such polygons are drawn to show how the cranks must be arranged and the weights distributed. The method is due to Professor Dalby, who treats the whole question of balancing very thoroughly in his " Balancing of Engines." The reader is recommended to consult this book for further details. Balancing Locomotives. In order that a locomotive may run steadily at high speeds, the rotating and reciprocating parts must be very carefully balanced. If the rotating parts be left unbalanced, there will be a serious blow on the rails every time the unbalanced portion gets to the bottom; this is known as the "hammer blow." If the reciprocating parts be left unbalanced, the engine will oscillate to and fro at every revolution about a vertical axis situated near the middle of the crank-shaft ; this is known as the " elbowing action." By balancing the rotating parts, the hammer blow may be overcome, but then the engine will elbow ; if, in addition, the reciprocating parts be entirely balanced, the engine will be overbalanced vertically hence we have to compromise matters by only partially balancing the reciprocating parts. Then, again, the obliquity of the connecting-rod causes the pressure due to the inertia of the reciprocating parts to be greater at one end of the stroke than at the other, a variation which cannot be compensated for by balance weights rotating at a constant radius. Thus we see that it is absolutely impossible to perfectly balance a locomotive of ordinary design, and the compromise we adopt must be based on experience. OF THE {' UNIVERSITY OF of the Steam- Engine. 193 The following symbols will be used in the paragraphs on locomotive balancing : W r , for rotating weights (pounds) to be balanced. W p , for reciprocating weights (pounds) to be balanced. W B , for balance weights ; if with a suffix p, as W Bp , it will indicate the balance weight for the reciprocating parts, and so on with other suffixes. R, for radius of crank (feet). RCI i, coupling-crank. R B , balance weights. Rotating Parts of Locomotive. The balancing of the rotating parts is effected in the manner described in the paragraph on standing balance, p. 186, which gives us W r R = W Br R B and W Br = The weights included in the W r vary in different types of engines ; we shall consider each as we come to it. Reciprocating Parts of Locomotive. We have already shown (p. 162) that the acceleration pressure at the ^%%^^%: , _ ^ ^ FIG. 190. end of the stroke due to the reciprocating parts is equal to the centrifugal force, assuming them to be concentrated at the crank-pin, and neglecting the obliquity of the connecting-rod. Then, for the present, assuming the balance weight to rotate in the plane of the crank-pin, in order that the recipro- cating parts may be balanced, we must have C = C o-ooo34W Bp . R B . N 2 = o'ooo34W p . R . N 2 W Bp . R B = W p . R and W BC = ^-?- fi.) IQI Mechanics applied to Engineering. On comparing this with the result obtained for rotating parts, we see that reciprocating parts, when the obliquity of the connecting-rod is neglected, may for every purpose be regarded as though their weight were concentrated in a heavy ring round the crank-pin. Now we come to a much-discussed point. We showed above that with a short connecting-rod of n cranks long, the acceleration pressure was - greater at one end and - less at the other end of the stroke than the pressure with an infinitely long rod; hence if we make W Bp - greater to allow for the 2 obliquity of the rod at one end, it will be - too great at the other end of the stroke. Thus we really do mischief by attempting to compensate for the obliquity of the rod at either end ; we shall therefore proceed as though the rod were of infinite length. If the reader wishes to follow the effect of the obliquity of the rod at all parts of the stroke, he should consult a paper by Mr. Hill, in the Proceedings of the Institute of Civil Engineer s, vol. civ. ; or Barker's " Graphic Methods of Engine Design ; " also Dalby's " Balancing of Engines." There is yet another point upon which there is a great difference of opinion, viz. what proportion of the connecting- rod should be regarded as rotating and what proportion as re- ciprocating. As a matter of fact, there is no room for difference of opinion here, for an exact solution of the problem is possible, though rather long. Some writers on this subject evidently find much pleasure in indulging in pages of abstruse mathematics on this point ; but their labour is in vain, for, do what we may, we cannot perfectly balance an engine as ordinarily built; and as we have to arbitrarily decide upon some proportion of the reciprocating parts that we will balance, viz. about two- thirds, it is folly to bother about a matter which may affect the result to i or 2 per cent, while we decide to leave unbalanced about 33 per cent, in wholesale fashion. In this connection, we shall assume that the big end of the connecting-rod and half the plain part rotates, while the small end and the other half of the rod reciprocates. Inside-cylinder Engine (uncoupled). In this case we have W^ = weight of (piston -j- piston-rod + cross-head + small end pf connecting-rod + J plain part of rod) ; Dynamics of the Steam- Engine. 195 W r = weight of (crank-pin -f- crank-webs * + big end of connecting-rod + ^ plain part of rod). If we could arrange balance weights to rotate in the same plane as the crank-pins, the weight of each would be W Br + W Bp , placed at the radius R B , and if we only counter- balance two-thirds of the reciprocating parts, we should get each balance weight ~~ R(|W p Balance weights cannot, however, be arranged to rotate in the same planes as the crank-pins. They might, of course, be placed opposite the crank-webs, but fo* many reasons such a position would be inconvenient ; they are therefore distributed over the wheels in such a manner that their centrifugal moments about the plane of rotation of the crank-pin is zero. If W be one weight, and Wj the other, distant y' and y-l from the plane of the crank, then WR^' = or wy = which is equivalent to saying that the centre of gravity of the two weights lies in the plane of rotation of the crank. The object of this particular arrangement is to keep the axis of Wheel i IT* . Wheel Sedf~ On * off wheel opposite "far" 1 crct-nJc // / crvcrvk. FIG. 191. rotation permanent. Then, considering the vertical crank shown in Fig. 191, by taking moments, we get the equivalent weights at the wheel centres as given in the figure. 1 See p. 203. 196 Mechanics applied to Engineering. We have, from the figure 22 2 Substituting these values, we get ??( y c) = W B1 , as the proportion of the balance weight 2y on the " off" wheel opposite the far crank and ^(y -\- 1) = W B2 , as the proportion of the balance weight on the " near " wheel opposite near crank Exactly similar balance weights are required for the other crank. Thus on each wheel we get one large balance weight W B2 at N (Fig. 192), opposite the near crank, and one small one W B1 at F, opposite the far crank. Such an arrangement would, however, be very clumsy, so we shall combine the two balance weights by the parallelogram of forces as shown, and for them substitute the large weight W B at M. Then W B = VW B1 2 + W, On substituting the values given above for W B1 and we have, when simplified W = In English practice y = 2*5*: (approximately) On substitution, we get W B = 076WBO Substituting from ii., we have __ - 7 6R(|W p + W r ) " Dynamics of the Steam-Engine. 197 Let the angle between the final balance weight and the near crank be a, and the far crank + 90. Then a= 180 - and tan 6 = = Substituting the value of y for English practice, we get tan 6 - = 0*429 6 = *3 Now, = - very nearly ; hence, for English practice, if the quadrant opposite the crank quadrant be divided into FIG. 193. four equal parts, the balance weight must be placed on the first of these, counting from the line opposite the near crank. Outside-cylinder Engine (uncoupled). \V P and W r are the same as in the last paragraph. If the plane of rotation of the crank-pin nearly coincides, as it frequently does, with the plane of rotation of the balance weight, we have w w P W r ) W B = WBO = ^ 3 - nearly J^B and the balance weight is placed diametrically opposite the crank. When the planes do not approximately coincide Let y = the distance between the wheel centres ; c = cylinder centres ; x= cylinder centre line and the " near " wheel ; z = cylinder centre line and the "off" wheel. c-y x = - - z 2 198 Mechanics applied to Engineering. The balance weight required! on the "off" wheel opposite} the "far "crank The balance weight required. on the "near" wheel oppo- 1= ^5? = II(*+j,) = w^ w site the "near "crank J7T? Then W B = which is precisely the same expression as we obtained for inside-cylinder engines, but in this case_y = o'&c to 0*9*:. On substitution, we get W B = i'i3W BO to i -o5W BO , and = 6 to 3. The same reasoning applies to the coupling-rod balance weights WBC in the next paragraphs. Inside-cylinder Engine (coupled). In this case we have W p the same as in the previous cases. W c = the weight of coupling crank-web and pin 1 4- coupling rod from a to b, or c to ;' governor more stable, the ' ) points d, d are brought in nearer the axis. The virtual centre of the arms is at their intersection ; hence the height F IG . 208. of the governor is H, which is approximately constant. The equivalent height can be raised by adding a central weight as in a Porter governor. It, of course, does not affect the sensitiveness, but it increases the power of the governor to overcome resistances. The speed at which a crossed-arm governor lifts depends upon the height in precisely the same manner as in the simple Watt governor. Astronomical Clock Governor. A beautiful applical tion of the crossed-arm principle as applied to isochronous governors is found in the governors used on the astronomica- clocks made by Messrs. Warner and Swazey of Cleveland, Ohio. Such a clock is used for turning an equatorial telescope Dynamics of ttie Steam- Engine. 211 on its axis at such a speed that the telescope shall keep exactly focussed on a star for many hours together, usually for the purpose of taking a photograph of that portion of the heavens immediately surrounding the star. If the telescope failed to move in the desired direction, and at the exact apparent speed of the star, the relative motion of the telescope and star would not be zero, and a blurred image would be produced ; hence an extreme degree of accuracy in driving is required. The results obtained with this governor are so perfect that no ordinary means of measuring time are sufficiently accurate to detect any error. The spindle A and cradle are driven by the clock, whose speed has to be controlled. A short link, c, is pivoted to arms on the driving spindle at b ; the governor weights are suspended by links from the point d; a brake shoe, affect the isochronous cha- racter of the governor. For example, the weight of the ball, except when its arm is vertical, has a moment about the pivot. Then, except when the spring arm is horizontal, the centrifugal force acting on the spring arm tends to make the ball fly in or out according as the arm is above or below the horizontal. We shall shortly show how the sensitiveness can be varied by altering the compression on the spring. Dynamics of the Steam- Engine. 213 Weight of Governor Arms. Up to the present we have neglected the weight of the governor arms and links, and have simply dealt with the weight of the balls themselves ; but with some forms of governors such an approximate treatment would give results very far from the truth. By similar reasoning to that given on p. 1 8 6, it will be seen that the centri- fugal force acting on any governor arm, is, approximately, equal to the centrifugal force acting on a mass equal to the mass of the arm con- centrated at the centre of gravity of the arm itself. When dealing with such a governor as that shown in the figure, it is con- venient to treat it as a governor having balls of greater weight concentrated at b, b. FIG. Let the weight of the top link = w^ bottom ,, = / a each ball = w c Let the weight of each equivalent ball at b = w b Let the radius of the c. of g. of the top link = ^ bottom =/ balls = r c equivalent = r b Then The height of this governor is H e , not 7f a ; i.e. the height is measured from the virtual centre at the apex. A governor having arms suspended in this manner is very much more stable and sluggish than when the arms are sus- pended from a central pin, and still more so than when the arms are crossed. In Fig. 209$ we show the governor used on the De Laval steam turbine. The ball weights in this case consist of two halves of a hollow cylinder mounted on knife-edges to reduce the 214 Mechanics applied to Engineering. friction ; the equivalent balls acting at the centre of gravity of the weights are shown in broken lines. These governors work exceedingly well, and keep the speed within very narrow limits. The figure is not drawn to scale. FIG. 209*. Crank-shaft Governors. The governing of steam- engines is often effected by varying the point at which the steam is cut off in the cylinder. Any of the forms of governor that we have considered can be adapted to this method, but the one which lends itself most readily to it is the crank-shaft governor, which alters the cut-off by altering the throw of the eccentric. We will consider one typical instance only, the McLaren governor, chosen because it contains so many good points, and, moreover, has a great reputation for governing within extremely fine limits (Fig. 210). The eccentric E is attached to a plate pivoted at A, and suspended by spherical-ended rods at B and C. A curved cam, DD, [attached to this plate, fits in a groove in the governor weight W in such a manner that, as the weight flies outwards due to centrifugal force, it causes the eccentric plate to tilt, and so bring the centre of the eccentric nearer to the centre of the shaft, or, in other words, to reduce its eccentricity, and consequently the travel of the valve, thus causing the steam to be cut off earlier in the stroke. The cam DD is so arranged that when the weight W is right in, the cut-off is as late as the slide-valve will allow it to be. Then, when the weight is right out, the travel of the valve is so reduced that no steam is admitted to the cylinder. A spring, SS, is attached to the weight arm to supply the necessary centripetal force. The speed of the engine is regulated by the tension on this spring. In order to alter the speed while the engine is running, the lower end of the spring is attached to a screwed hook, F. The nut G is in the form of a worm wheel ; the worm spindle is provided with a small milled wheel, H. If it be desired to Dynamics of the Steam-Engine. 2T5 alter the speed when running, a leather-covered lever is pushed into gear, so that the rim of the wheel H comes in contact with it at each revolution, and is thereby turned through a small amount, thus tightening or loosening the spring as the case may be. If the lever bears on the one edge of the wheel H, the spring is tightened and the speed of the engine increased, and if on the other edge the re- verse. The spring S is attached to the weight arm as near its centre of gravity as possible, in order to eliminate friction on the pin J when the engine is running. The governor is designed to be ex- tremely sensitive, and, in order to prevent hunting, a dashpot K is attached to the weight arm. In the actual governor two weights FIG. 210. are used, coupled together by rods running across the wheel. The figure must be regarded as purely diagrammatic. It will be seen that this governor is practically isochronous, for the load on the spring increases as the radius of the weight, and therefore, as explained in the Hartnell governor, as the centrifugal force. The sensitiveness can be varied by altering the position of suspension, J. In order to be isochronous, the path of the weight must as nearly as possible coincide with a radial line drawn from O, and the direction of S must be parallel to this radial line. Inertia Effects on Governors. Many governors rely entirely on the inertia of their weights or balls for regulating the supply of steam to the engine when a change of speed occurs, while in other cases the inertia effect on the weights is 216 Mechanics applied to Engineering. so small that it is often neglected ; it is, however, well when designing a governor to arrange the mechanism in such a manner that the inertia effects shall act with rather than against the centrifugal effects. In all cases of governors the weights or balls tend to fly out radially under the action of the centrifugal force, but in the case of crank-shaft governors, in which the balls rotate in one plane, they are subjected to another force acting at right angles to the centrifugal whenever a change of speed takes place ; the latter force, therefore, acts tangentially, and is due to the tangential acceleration of the weights. For convenience of expression we shall term the latter the " inertia force." The precise effect of this inertia force on the governor entirely depends upon the sign of its moment about the point of suspension of the ball arm : if the moment of the inertia force be of the same sign as that of the centrifugal force about the pivot, the inertia effects will assist the governor in causing it to act more promptly ; but if the two be of opposite sign, they will tend to neutralize one another, and will make the governor sluggish in its action. Since the inertia of a body is the resistance it offers to having its velocity increased, it will be evident that the inertia force acts in an opposite sense to that of the rotation. In the figures and table given below we have only stated the case in which the speed of rotation is increased ; when it is decreased the effect on the governor is the same as before, in as far as the moments act together or against one another. In the case of a governor in which the inertia moment assists the centrifugal, if the speed be suddenly increased, both the centrifugal and the inertia moments tend to make the balls fly out, and thereby to partially or wholly shut off the supply of steam, the resulting moment is therefore the sum of the two, and a prompt action is secured ; but if, on the other hand, the inertia moment acts against the centrifugal, the resulting moment is the difference of the two, and a sluggish action results. If, as is quite possible, the inertia moment were greater than the centrifugal, and of opposite sign, a sudden increase of speed would cause the governor balls to close in and to admit more steam, thus producing serious disturbances. The table given below will serve to show the effect of the two moments on the governor shown in Fig. 2100. In every case c is the centrifugal force, and T the inertia force. T = M , or = , where W is the weight at gat Dynamics of the Steam-Engine. 217 of the ball, and the acceleration in feet per second per at second. Sense of rotation. Position of ball. Centrifugal moment. Inertia moment for an increase of speed. Effect of inertia on governor. A B A B + I -'1*1 '1*1 TU/J -T/^x Retards its action Jf. 2 ^2*2 ^2*2 No effect + 3 -'3*3 C 3 X Z T 3 y a Tz'y* Assists its action i ~ c \ x \ '1*1 TI_J/! T-i'y-i 55 55 _ 2 CXn o No effect ~ 3 -'3*3 ** T 3 ^ 3 -T s > 3 Retards its action Sensitiveness of Governors. The sensitiveness and behaviour of a governor when running can be very conveniently studied by means of a diagram showing the rate of increase of FIG. 2io. the centrifugal and centripetal moments as the governor balls fly outwards. These diagrams are the invention of Mr. Wilson Hartnell, who first described them in a paper read before the Institute of Mechanical Engineers in 1882. In Fig. 211 we give such a diagram for a simple Watt governor, neglecting the weight of the sleeve, etc. The axis OO' is the axis of rotation. The ball is shown in its two extreme positions. The ball is under the action of two moments the centrifugal moment CH and the centripetal moment WR, which of course must be equal for all positions of the ball, unless the ball is being accelerated or retarded. The centrifugal moment is tending to carry the ball outwards, 218 Mechanics applied to Engineering. the centripetal to bring it back. The four numbered curves show the relation between the moment tending to make the balls fly out (ordinates) and the position of the balls. The centripetal moment line shows the relation between the O f moment tending to bring the balls back and the position of the balls, which is independent of the speed. We have Scale / / CH = o'ooo 3 4WRN 2 H 1*-0>22feet / I =o -ooo34WN 2 (RH) = KRH The quantity '00034 WN 2 is constant for any given ball running at any given speed. Values of KRH have been calcu- lated for various positions and speeds, and the curves plotted. The value of WR varies, of course, directly as the radius ; hence the centripetal line is straight, and passes through the origin O. From this we see that the governor begins to lift at a speed of about 82 revolutions per minute, but gets to a speed of about 94 before the governor lifts to its extreme position. Hence, if it were intended to run at a mean speed of 88 revolutions per minute, it would, if free from friction, vary about 9 per cent, on either side of the mean, and when retarded by friction it will be far worse. If the centrifugal and centripetal curves coincided, the governor would be isochronous. If the slope of the centrifugal curve be less than the centripetal, the governor is too stable, i.e. not sufficiently sensitive; but if, on the other hand, the FIG. 211. Dynamics of the Steam- Engine. 219 slope of the centrifugal curve be greater than the centripetal, the governor is too sensitive, for as soon as the governor begins to lift, the centrifugal moment, tending to make the balls fly out, increases more rapidly than the centripetal moment, tending to keep the . balk in consequently the balls are accelerated, and fly out to their extreme position, completely closing the governing valve, which immediately causes the engine to slow down. But as soon as this occurs, the balls close right in and fully open the governing valve, thus causing the engine to race and the balls to fly out again, and so on. This alternate racing and slowing down is known as hunting, and is the most common defect of governors intended to be sensitive. It will be seen that this action cannot possibly occur with a simple Watt governor unless there is some disturbing action. Friction of Governors. So far, we have neglected the effect of friction on the sensitiveness, but it is in reality one of the most important factors to be considered in connection with sensitive governors. Many a governor is practically perfect on paper friction neglected but is to all intents and purposes useless in the material form on an engine, on account of retarda- tion due to friction. The friction is not merely due to the pins, etc., of the governor itself, but to the, moving of the governing valve or its equivalent and its connections. In Fig. 212 we show how friction affects the sensitiveness of a governor. The vertical height of the shaded portion represents the friction moment that the governor has to overcome. Instead of the governor lifting at 80 revolutions per minute, the speed at which it should lift if there were no friction, it does not lift till the speed gets to about 92 revolutions per minute ; likewise on fall- ing, the speed falls to 64 revolutions per minute. Thus with friction the speed varies about 22 per cent, above and below the mean. Unfortunately, very little experimental data exists FIG. 212. 220 Mechanics applied to Engineering. on the friction of governors and their attachments ; l but a designer cannot err by doing his utmost to reduce it even to the extent of fitting all joints, etc., with ball bearings or with knife-edges see Fig. 209^). In the well-known Pickering governor, the friction of the governor itself is reduced to a minimum by mount- ing the balls on a number of thin band springs in- stead of arms moving on pins. The attachment of the spring at the c. of g. of the weight and arm, as in the McLaren governor, is a point also worthy of attention. We will now examine in detail several types of governor by the method just described. Porter Governor Diagram. In this case the centripetal force is greatly increased, while the centrifugal is un- affected. The central weight W w rises twice as fast as the balls (Fig. 204) ; hence we get the weight W w acting on each ball in the manner shown in Fig. 213. Resolve W w in the directions of the two arms as shown : it is evident that ab, acting along the upper arm, has no moment about O, but bd = W FIG. 213. has a centripetal moment, W R ; then we have CH = WR -f- W R 1 See Paper by Ransome, Proc. Inst. C.., vol. cxiii. ; also the question has recently been investigated by one of the author's students, Mr. Eurich, who finds that when oiled a Watt governor lags behind to the extent of 7*5 per cent., and when unoiled 17*5 per cent. Dynamics of the Steam-Engine. 221 Values of each have been calculated and plotted as before in Fig. 211. In the central spring governor W w varies as the balls lift ; in other respects the construction is the same. It should be noticed that the centripetal and centrifugal moment curves much more closely coincide as the height of the governor increases; thus the sensitiveness increases with the height a conclusion we have already come to by another process of reasoning. Crossed- arm Governor Diagram. In this governor H is constant, and as C varies directly as the radius for any given speed, it is evident that the centripetal and centrifugal lines are both straight and coincident, hence the governor is isochronous. Wilson Hartnell Governor Diagram. In constructing this diagram (Fig. 214) we have neg- lected the moment of the ball weight on either side of the sus- pension pin, and the other slight irregularities ; but in a big governor they are of importance, and should be taken into account. We have shown that cr^ - pr, and that c varies as R, likewise p varies as R ; hence both the cen- trifugal and the centripetal moment curves are straight, and when the governor is isochronous, both pass through the origin. When a spring is loaded in either tension or com- pression, the strain is proportional to the load applied ; hence, if an initial load be put on the governor spring, the spring pressure curves will always be parallel to one another as shown in dotted lines. If the initial load be too small, the governor will be too stable, and if too great, too sensitive, i.e. it will cause the engine to hunt. The position for the governor to be isochronous is when the centripetal and centrifugal curves are coincident, i.e. when both pass through the origin O. It may happen, however, that when the spring is adjusted to make the governor isochronous, the speed is not that which is desired, and in order to obtain it, either a weaker or stiffer spring will FIG. 214. 222 Mechanics applied to Engineering. be required. Instead, however, of getting a new spring, the stiffness can be readily varied by altering the effective length of the spring. This is most conveniently done by casting a gun-metal nut round the coils of the spring, like a cork round a corkscrew. If the spring be previously warmed and dipped in loam and water so as to coat the spring, the nut will not bind when cold; then by screwing this nut up or down the effective length of the spring can be varied at will, and the exact stiffness obtained. Instead of altering the spring for adjusting the speed, some makers prefer to leave a hollow space in the balls for the insertion of lead until the exact weight and speed are obtained ; but when this method is adopted care must be taken to prevent the lead from flying out. It is usually accomplished by making the hollow spaces on the inside edge of the; ball, then the centrifugal force tends to keep the lead in position. The sensitiveness of the Wilson Hartnell governor may also be varied at will by a simple method devised by the author some years ago, which has been successfully applied to several forms of governor. In general, if a governor tends to hunt, it can be corrected by making the centripetal moment increase more rapidly, or, if it be too sluggish, by making it increase less rapidly as the centrifugal moment of the balls increases. The governor which is shown in Fig. 2140, is of the four-ball horizontal type ; it originally hunted very badly, and in order to correct it the conical washer A was fitted to the ball path, which was previously flat. It will be seen that as the balls fly out the inclined ball path causes the spring to be compressed more rapidly than if the path were flat, and consequently the rate of increase of the centripetal moment is increased, and with it the stability of the governor. In constructing the diagram it was found convenient to make use of the virtual centre of the ball arm in each position ; after finding it, the method of procedure is similar to that already given for other cases. In order to show the effect of the conical f washer, a second centripetal curve is shown by a broken line for a flat plate. With the conical washer, neglect- ing friction, the diagram shows that the governor lifts at 430 revolutions, and reaches 490 revolutions at its extreme range ; by experiment it was found that it began to lift at 440, and rose to 500, when the balls were lifting, and it began to fall at 480, getting down to 415 before the balls finally closed in. The conical washer A in this case is rather too steep for accurate governing. 224 Mechanics applied to Engineering. The centrifugal moment at any instant is 4 X o-ooo 3 4WRN 2 H where W is the weight of one ball. And the centripetal moment is Load on spring X R, See Fig. 2140 for the meaning of R,, viz. the distance of the virtual centre from the point of suspension of the arm. Taking position 4, we have for the centrifugal moment at 450 revolutions per minute 4 X 0*00034 X 2*5 x ^y- X 45 o 2 x 1*58 = 260 pound-inches and for the centripetal moment The load on the spring = in Ibs. ; and R a = 2*78 centripetal moment =111X278 = 310 pound-inches McLaren's Crank-shaft Governor. In this governor we have CR = SR, ; but C varies as R, hence if there be no FIG. 215. tension on the spring when R is zero, it will be evident that S will vary directly as R ; but C also varies in the same manner, hence the centrifugal and centripetal moment lines are nearly Dynamics of the Steam- Engine. 225 straight and coincident. The centrifugal lines are not abso- lutely straight, because the weight does not move exactly on a radial line from the centre of the crank-shaft. Governor Dashpots. A dashpot consists essentially of a cylinder with a leaky piston, around which oil, air, or other fluid has to leak. An extremely small force will move the piston slowly, but very great resistance is offered by the fluid if a rapid movement be attempted. Very sensitive governors are therefore generally fitted with dashpots, to prevent them from suddenly flying in or out, and thus causing the engine to hunt. If a governor be required to work over a very wide range of power, such as all the load suddenly thrown off, a sensi- tive, almost isochronous governor with dashpot gives the best result; but if very fine governing be required over small variations of load, a slightly less sensitive governor without a dashpot will be the best. However good a governor may be, it cannot possibly govern well unless the engine be provided with sufficient fly- wheel power. If an engine have, say, a 2-per-cent. cyclical variation and a very sensitive governor, the balls will be constantly fluctuating in and out during every stroke. Power of Governors. The "power" of a governor is its capacity for overcoming external resistances. The greater the power, the greater the external resistance it will overcome with a given alteration in speed. Nearly all governor failures are due to their lack of power. The useful energy stored in a governor is readily found thus, approximately : Simple Watt governor, crossed-arm and others of a similar type Energy = weight of both balls X vertical rise of balls Porter and other loaded governors Energy = weight of both balls x vertical rise of balls -f- weight of central weight X its vertical rise Spring governors Energy = weight of both balls X vertical rise (if any) of balls , /max. load on spring + niin. load on spring *\- T - X the stretch or compression of spring 226 Mechanics applied to Engineering. where n = the number of springs employed ; express weights in pounds, and distances in feet. The following may be taken as a rough guide as to the energy that should be stored in a governor to get good results : it is always better to store too much rather than too little energy in a governor : Foot-pounds of energy Type of governor. stored per inch diameter of cylinder. For trip gears and where small resistances have to be overcome '5~'75 For fairly well balanced throttle-valves ... ... 0*75-1 In the earlier editions of this book values were given for automatic expansion gears, which were based on the only data available to the author at the time; but since collecting a considerable amount of information, he fears that no definite values can be given in this form. For example, in the case of governors acting through reversible mechanisms on well- balanced slide-valves, about 100 foot-pounds of energy per inch diameter of the high-pressure cylinder is found to give good results ; but in other cases, with unbalanced slide-valves, five times that amount of energy stored is found to be insufficient. If the driving mechanism of the governor be non-reversible, only about one-half of this amount of energy will be required. A better method of dealing with this question is to calculate, by such diagrams as those given in the " Mechanisms " chapter, the actual effort that the governor is capable of exerting on the valve rod, and ensuring that this effort shall be greatly in excess of that required to drive the slide-valve. Experiments show that the latter amounts to about one-fifth to one-sixth of the total pressure on the back of a slide-valve (/. } where is the angle the plane makes to the horizontal when sliding just commences. The angle is termed the ""friction angle," or "angle of repose." The body will not slide if the plane be tilted at an angle less than the friction angle, and a force F (Fig. 218) A FlG. 2TQ. FIG. 220. will have to be applied parallel to the plane in order to make it slide. Whereas, if the angle be greater than , the body will be accelerated due to the force F x (Fig. 219). There is yet another way of looking at this problem. Let the body rest on a horizontal plane, and let a force P be applied at an angle to the normal ; the body will not begin to slide until the angle becomes equal to the angle <, the angle of friction. If the line representing P be revolved round the normal, it will describe the surface of a cone in space, the apex angle being 2<; this cone is known as the "friction cone." Friction. 229 If the angle with the normal be less than , the block will not slide, and if greater the block will be accelerated, due to the force Fj. In this case the weight of the block is neglected. If P be very great compared with the area of the surfaces in contact, the surfaces will seize or cling to one another, and if continued the surfaces will be torn or abraded. Friction of Dry Surfaces. The experiments usually quoted on the friction of dry surfaces are those made by Morin and Coulomb ; they were made under very small pressures and speeds, hence the laws deduced from them only hold very imperfectly for the pressures and speeds usually met with in practice. They are as follows : 1. The friction is directly proportional to the normal pressure between the two surfaces. 2. The friction is independent of the area of the surfaces in contact for any given normal pressure, i.e. it is independent of the intensity of the normal pressure. 3. The friction is independent of the velocity of rubbing. 4. The friction between two surfaces at rest is greater than when they are in motion, or the friction of rest l is greater than the friction of motion. 5. The friction depends upon the nature of the surfaces in contact. We will now see how the above laws agree with experiments made on a larger scale. The first two laws are based on the assumption that the coefficient of friction is constant for all pressures; this, however, is not the case. The curves in Fig. 221 show approximately the difference between Coulomb's law and actual experiments carried to high pressures. At the low pressure at which the early workers worked, the two curves practically agree, but at higher pressures the friction falls off, and then rises until seizing takes place. Instead of the frictional resistance being F = /xN it is more nearly given by F = /xN ' 97 , or F = /* r VN where /x has the following values : 1 The friction of rest has been very aptly termed the "sticktion." 230 Mechanics applied to Engineering. Wood on wood . Metal Metal on metal . Leather on wood ,, metal Stone on stone . o'25 to o - 5o O'2O to O'6O 015 to 0*30 0*25 to 0-50 o'3o to o'6o 0*40 to 0-65 These coefficients must always be taken with caution ; they vary very greatly indeed with the state of the surfaces in contact. The third law given above is far from representing facts ; in Intensity of pressure FIG. 221. the limit the fourth law becomes a special case of the third. If the surfaces were perfectly clean, and there were no film of air between, this law would probably be strictly accurate, but all experiments show that the friction decreases with velocity of rubbing. The following empirical formula fairly well agrees with experiments : Let fji coefficient of friction ; K = a constant to be determined by experiment ; V = the velocity of sliding. Then . = - Friction, 231 The following experiments by Westinghouse and Galton, on steel tyres on steel rails will serve to illustrate this point : Speed miles per hour Coefficient of friction 10 O'HO 15 0-087 25 o'o8o 38 0-051 45 0-047 50 o 040 For other instances, see Proc. Inst. M.E.j 1883, p. 660. The fourth law has been observed by nearly every experi- menter on friction. The following figures by Morin and others will suffice to make this clear : Coefficient of friction. Materials. Rest. Velocity 3 to 5 ft.-sec. \Vood on wood 0*46 0*69 Metal on metal ... ... '34 O'26 Stone on stone ..-. 0-74 0-63 Leather on iron ... o-59 0-52 c The figures already quoted quite clearly demonstrate the truth of the fifth law given above. Special Cases of Sliding Bodies. In the cases we are about to consider, we shall, for sake of simplicity, assume that Coulomb's laws hold good. Oblique Force re- quired to make a Body slide on a Horizontal Plane. If an oblique force P act upon a block of weight W, making an angle 6 with the direction of sliding, we can find the magnitude of P required to make the block slide ; the total normal pres- sure on the plane is the normal component of P, viz. ;/, together with W. From a draw a line making an angle (the friction angle) with W, cutting P in 232 Mechanics applied to Engineering. the point b ; then be, measured to the same scale as W, is the magnitude of the force P required to make the body slide. The frictional resistance is F, and the total normal pressure n + W; hence F = /i( -f W). When = o, P = cd = /xW. When is negative, it simply indicates that P : is pulling away from the plane : the magnitude is given by ce. From the figure it is clear that the least value of P is when its direction is normal to ab> i.e. when 6 = ; then fmin. = 1 o COS = /X\V COS < = tan sin 6 W sin < ~ cos ( + 6) When P acts upwards away from the plane, the -f sign is to be used in the denominator ; and for the minimum value of P, < = 0; then the denominator is unity, and P = W sin <, the result given above, but arrived at by a different process. Thus, in order to drag a load, whether sliding or on wheels, along a plane, the line of pull should be upwards, making an angle with the plane equal to the friction angle. Force required to make a Body slide on an Inclined Plane. A special case of the above is that in which the plane is inclined to the horizontal at an angle a. Let the block of weight W rest on the inclined plane as shown. In order to make it slide up the plane, work must be done in lifting the block as well as overcoming the friction. The pull required to raise the block is readily obtained thus : Set down a line be to represent the weight W, and from c draw a line cd> making Friction. 233 an angle a with it ; then, if from b a line be drawn parallel to the direction of pull P,, the line bd l represents to the same scale as W the required pull if there were no friction. An examination of the diagram will at once show that bd^c is simply the triangle of forces acting on the block ; the line cd is, of course, normal to the plane. When friction is taken into account, draw the line ce, making an angle (the friction angle) with cd\ then be^ gives the pull P! required to drag the block up the FIG. 223. plane including friction. For it will be seen that the normal pressure on the plane is cdi, and that the friction parallel to the direction of sliding, viz. normal to cd, is = tan ~ = ~ X Then resolving with cd. FIG. 224. Then P min = W sin (< + a) . 234 Mechanics applied to Engineering. Horizontal Pull. When the body is raised by a hori- zontal pull, we have (Fig. 225) , = W tan (< -f FIG. 225. FIG. 226. Thus, when the pull is horizontal, the effect of friction is equivalent to making the slope steeper by an amount equal to the friction angle. \H Parallel Pull. When the body is raised by a pull parallel - to the plane, we have (Fig. 226) FIG. 227. P p = But ed = dc tan = pdc and dc = W . cos a therefore ed = /JL . W . cos a and db = W sin a hence P p = W(/x, . cos a -f- sin a) This may be expressed thus (see Fig. 227) or, Work done in dragging a body of weight W up a plane, by a force acting parallel to the plane or PjJ, = W/xB + WH work done in dragging \ the body through the ("work done same distance on a > + \ in lifting horizontal plane ' the body against friction. Friction. 235 General Case. When the body is raised by a pull making an angle with the plane p j-j -* min. ~ cos (0 ) Substituting the value of Pmin. from equation (i.) - cos (B - <) When the line of action of P is towards the plane, as in P , the 6 becomes minus, and we get W sin (< + a) 1 ~ cos ( - - <) All the above expressions may be obtained from this. When the direction of pull is parallel to ec^ viz. P , it will only meet ec at infinity that is, an infinitely great force would be required to make it slide ; but this is impossible, hence the direction of pull must make an angle to the plane < (90 <) in order that sliding may take place. We must now consider the case in which a body is dragged down a plane, or simply allowed to slide down. If the angle a be less than <, the body must be dragged down, and if a be FIG FIG. 229. NOTE. The friction now assists the lowering, hence ce is set off to the right of cd. 236 Mechanics applied to Engineering. greater, a force must be applied to prevent it from rushing down and being accelerated. Least pull when body is lowered, < a (Fig. 229). P mi n. = be = W sin (a - <) and 6> = (v.) When <>a, be l is the least force required to make the body slide down the plane. = sn - a) . (vi.) when < = a, P min . is of course zero. The remaining cases are arrived at in a similar manner ; we will therefore simply state them. *<*. >a. Least pull Parallel pull Horizontal pull General case W sin (o 0) W (sin o fj. cos a) W tan (o ) W sin (a ) W sin (4, - a) W (ft cos a sin o) W tan (<(> a) W sin (< a) cos (0 + J>) cos (0 ) NOTE. If the line of pull comes below the plane, the angle 6 takes the sign. In the case of the parallel pull, it is worth noting that when < < a, we have Total work done = work done in lowering the body work done in dragging the body through the horizontal distance against friction and when <>a we have the same relation, but the work done is negative, that is, the body has to be retarded. It should be noticed that the effect of friction on an inclined plane is to increase the steepness when the block is being hauled up the plane, and to decrease it when hauling it down the plane by an amount equal to the friction angle. Efficiency of Inclined Planes. If an inclined plane be used as a machine for raising or lowering weights, we have Em" ' wor k done (i.e. without friction) actual work done (with friction) Inclined Plane when raising a Load. The maximum efficiency occurs when the pull is least, i.e. when 6 = <. The useful work done without friction is when < = o; then Friction. 237 The work done without friction = - - from (iv.) cos with = LW sin (< + ) from (i.) where L = the distance through which the body is dragged ; a = the inclination of the plane to the horizon ; = the inclination of the force to the plane ; < = the friction angle. LW sin a Then maximum \ _ CQ $ _. sin a , .. , efficiency [ LW sin ( + a) cos 9 sin (< -f a) When the pull is horizontal, - a, and sin a tan a / \ Efficiency = --- 7-7 r = - - 7-7 x ( v111 -) cos a . tan ( + a) tan (< + ) when the pull is parallel, = o, and cos 9 = i ; -r^ff. - sin a . cos Efficiency = - . . . sin (a + ) General case, when the line of pull makes an angle 6 with the direction of sliding sin a . cos (9 <) / x Friction of Wedge. This is simply a special case of the inclined plane in which the pull is horizontal, or when it acts normal to W. We then have \ from equation (ii.) P = W tan 4- a) for a single inclined W W plane; but here we have two inclined planes, each at an angle a, hence W moves twice as far , ,. r 11 FiG. 231. for any given movement 01 as in the single inclined plane ; hence P = 2W tan (< + a) for a wedge The wedge will not hold itself in position, but will spring back, if the angle a be greater than the friction angle <. From the table on p. 236 we have the pull required to withdraw the wedge -P = 2\V tan (a - 0) 238 Mechanics applied to Engineering. The efficiency of the wedge is the same as that of the inclined plane, viz. Efficiency = - . , . when overcoming a resistance (xi.) reversed ) tan (a cf>) rf, > = when withdrawing from a resistance efficiency/ tana (see p . Efficiency of Screws and Worms Square Thread. A screw thread is in effect a narrow inclined plane wound round a cylinder ; hence the efficiency is the same as that of an inclined plane. We shall, however, work it out by another method. FIG. 933. Let / = the pitch of the screw ; ) the thread required to raise nut f ~ ( (see equation ii.) The work done in turning the nuti _ through one complete revolution f = WTT^O tan (a + (f>) WTT^/O tan a efficiency when raising the weight = WWo tan (a H T^T tan a ~ tan" (a + ) This result will be found to be the same as that obtained for the inclined plane (equation xi.). The expression - -. \ has its maximum value when a = 45 , and its value is then Friction. 239 maximum efficiency = The proof of this is somewhat long; perhaps the easiest way of arriving at it is to calculate several values, plot a curve, and from it find a maximum. In addition to the friction on the threads, the friction on the thrust collar of the screw must be taken into account. The thrust collar may be assumed to be of the same diameter as the thread ; then the efficiency of screw thread ") _ tan a and thrust collar / ~ ^T( a + 2 j) ( a PPx.) In the case of a nut the radius at which the friction acts will be about i J times that of the threads ; we may then say efficiency of screw thread and nut ) _ tan a bedding on a flat surface j ~~ tan (a -f 2-5$) If the angle of the thread be very steep, the screw will be reversible, that is, the nut will drive the screw. By similar reasoning to that given above, we have , rr- tan (a ) , reversed efficiency = (see p. 267) Such an instance is found in the Archimedian drill brace, and another in the twisted hydraulic crane-post used largely on board ship. By raising and lowering the twisted crane-post, the crane, which is in reality a part of a huge nut, is slewed round as desired. Triangular Thread. In the triangular thread the normal pressure on the nut is greater than in the square-threaded screw, in the ratio of ~ = -i-*, and W = , where is cos - cos - 2 2 the angle of the thread. In the Whitworth thread the angle is 55, hence W = 1-13 W. In the Sellers thread = 60 and W = 1*15 W; then, taking a mean value of W = 1*14 W, we have ~- . tan a efficiency = tan (a + i 'i4) 2 4 Mechanics applied to Engineering. In the case of an ordinary bolt and nut, the radius at which the friction acts between the nut and the washer is about ij times that of the thread, and, taking the same coefficient of friction for both, we have efficiency of a bolt \ _ and nut tan (a + 2-640) tan a (approx.) The following table may be useful in showing roughly the efficiency of screws. In several cases they have been checked FIG. 233. by experiments, and found to be fair average values ; the efficiency varies greatly with the amount of lubrication : TABLE OF APPROXIMATE EFFICIENCIES OF SCREW THREADS. Angle of thread a. Efficiency per cent, when no friction between nut and washer or a thrust Efficiency per cent, allow- ing for friction between nut and washer or a collar. thrust collar. Sq. thread. V-thread. Sq. thread. V-thread. 2 19 17 II 8 3 26 23 H 12 4 32 28 17 16 5 36 33 21 20 10 20 15 11 36 4 8 29 42 -* 79 75 52 44 In the above table cf> has been taken as 8'5, or fj. = 0*15. Rolling Friction. When a wheel rolls on a yielding material that readily takes a permanent deformation, the resistance is due to the fact that the wheel sinks in and makes a rut. The greater the weight W carried by the wheel, the deeper will be the rut, and consequently the greater will be the resistance to rolling. When the wheel is pulled along, it is equivalent to con- stantly mounting an obstacle at A ; then we have Friction. 241 P . BA = W . AC W.AC orP = Let AC = K ; Then P = BA W.K But h is usually small compared with R ; hence we may write ^ (nearly) P and W, also K and R, must be measured in the same units, or the value of K corrected accordingly. The above relation holds fairly well in practice, but there is much need for further research in this branch of friction. VALUES OF K. Iron or steel wheels on iron or steel rails ,, wood... ... ... ,, ,, macadam ,, ,, soft ground Pneumatic tyres on good road or asphalte ... ,, ,, heavy mud - ... ... Solid indiarubber tyres on good road or asphalte ,, ,, heavy mud ... K (inches). O'oo7-o'oi5 O'o6-o'io 0-05-0-20 3-5 o f o2-o'O22 o'O4-o - o6 0^04 Some years ago Professor Osborne Reynolds investigated the action of rollers passing over elastic materials, and showed clearly that when a wheel rolls on, say, an indiarubber road, it sinks in and com- presses the rubber imme- diately under it, but forces out the rubber in front and behind it, as shown in the sketch. That forced up in the front slides on the surface of the wheel in just the in reverse direction to the mo- FIG. 235. tion of the wheel, and so hinders its progress. Likewise, as the wheel leaves the heap behind it, the rubber returns to its original 242 Mechanics applied to Engineering. place, and again slips on the wheel in the reverse direction to its motion. Thus the resistance to rolling is in reality due to the sliding of the two surfaces. On account of the stretch- ing of the path over which /\ \ the wheel rolls, the actual distance rolled over is greater than the horizontal distance travelled by the wheel. Antifriction Wheels. In order to reduce the friction on an axle it is sometimes mounted on antifriction wheels, as FlG . 236 . shown. A is the axle in question, B and C are the antifriction wheels. If W be the load on the axle, the load on each antifriction wheel bearing will be W = W ., and the load on both -^_ 2 cos cos Let R a = the radius of the main axle ; R = antifriction wheel ; r = axle of the antifriction wheel. The rolling resistance on the surface! _. WK of the wheels / R cos The frictional resistance referred to^j ^y the surface of the antifriction wheels, > = =r*- or the surface of the main axle J The total resistance = W R cos -(K If the main axle were running in plain bearings, the resistance would be /xW ; hence friction with plain bearings _ /xR cos friction with antifriction wheels K + u.r Of course, two such sets of wheels are required for mount- ing an axle, and unless the wheel axles are perfectly parallel, Friction. 243 and the wheels true and of the same size, a great deal of trouble will be experienced with the main axle travelling sideways. The author has had to use ball thrust bearings to prevent this lateral motion. Roller Bearings. In the roller bearing shown in Fig. 237, the shaft rolls on hard steel rollers, which are kept in position by gun-metal cages shown in section. The outer casing is of cast-iron. In other bearings of a somewhat similar design the shaft is fitted with a hardened and ground steel sleeve, and the casing with a similar liner both with the object of providing a perfectly smooth and hard path for the rollers. Trouble is often experienced with bearings of this type owing to the serious "end thrust" of the rollers. If the rollers are not J/alf Cross Sectuai on Centre Ling FIG. 237. absolutely true, i.e. parallel with the shaft, they tend to roll in a helical path, but since the cage and casing prevents them from doing so, end slip has to occur, which may set up end thrust even to the extent of one-tenth of the load on the bearing. The author has two machines, which have been constantly running for the past six years, testing roller and ball bearings, and up to the present has not found any roller bearing entirely free from " end thrust." Some bearings are much worse than others in this respect, but some of the worst improve in this respect as they wear, while others that are tolerably good to start with get worse as time goes on. The " end thrust " and the friction of the bearing diminish as the speed increases. The following table gives fair average results for a friction test of an ordinary roller bearing : 244 Mechanics applied to Engineering 40 revolutions per minute. 400 revolutions per minute. Total load in Ibs. M End thrust in Ibs. M End thrust in Ibs. 2000 OOI3I 82 0-0053 51 4000 0*0094 M7 0-0035 8 9 6000 0'0082 212 0*0029 128 8000 0*0076 2 7 6 0-0026 1 66 10,000 O'OO72 340 0-0024 205 Generally speaking, a low end thrust in a roller bearing is accompanied by a small amount of friction of rotation and wear. It is much to be regretted that no one has succeeded in eliminating the end thrust on the rollers, because, as regards friction and wear, they are far better than ordinary bearings, and will support much greater loads than ball bearings of the same length of journal. The remarks as to the variation of the friction of ball bearings in the next paragraph may be taken in general terms to refer to roller bearings as well. Ball Bearings. The "end-thrust" troubles that are experienced with roller bearings can be entirely avoided by the substitution of balls for rollers. The form of the ball path, how- ever, requires careful consideration. If the rollers be removed from such a roller bearing as the one shown in Fig. 237, and balls of the same diameter be packed in the roller cages, the bearing will run perfectly for a time, but the shaft and casing will suffer by wear; if, however, a hardened and ground cylindrical steel liner and sleeve be used, the wear will be exceedingly small, even under moderately high loads and speeds, and the friction will be considerably lower than with the rollers ; but it will not safely carry so great a load as when fitted with rollers. Such a bearing would be known as a " two- point " bearing, because the ball only bears on two points at any one time. If the ball paths be made in the form of slightly hollow grooves in both l he sleeve and liner, the bearing will sup- port a somewhat higher load than when plain cylindrical paths are adopted. A large number of tests by the author, of nearly every bearing on the market, lead him to believe that the advantages claimed for special forms of grooves by certain makers } also by writers on the theory of ball bearings, are very much exaggerated. It used to be contended that the correct form of race for a ball bearing was that shown in Fig. 238, the argument being Friction. 245 that, if the sides of the ball races were made of the same slope, the balls would grind, because the circumference of the ball race at a moves faster than that of b. In order that there may be no grinding, the circle on which a rolls must be greater than that on which b rolls, in the ratio of their distances from the centre of the shaft, or aa R0 i i TL = HTJ wm ch is secured by oo R0 the construction shown, since the triangles Oaa and Obb are similar. Although this form of bearing is right in principle, it is not found to work well in practice, probably because the exact con- ditions are upset when any wear or change of load takes place. A series of tests of some bearings of this type showed that the balls began to " peel " and score, and the races to grind at very low loads and speeds. The best results have been obtained with either slightly hollow or flat races, both of which are shown in Fig. 238^. The balls should be kept apart by a light aluminium or gun-metal cage, as shown. The wear FIG. 238. Cage. on such cages is practically nil if the bearing has been properly made. In order to ensure that all the balls take a fair pro- portion of the load, the Hoffmann Manufacturing Company, of Chelmsford, allow one of the ball-race collars to swivel, as shown in Fig. 239. This is a wise precaution, and certainly tends to increase the life of the bearing. For very heavy loads it is better to increase the size of the balls rather than use several rows of smaller balls, mainly on account of the difficulty of distributing the load evenly on all 246 Mechanics applied to Engineering. the balls; but if for any reason several concentric rings of SPHERICAL SEAT BALL CAGE EXPANDING LOCK NUT BRASS CLAMPING SLEEVE CLAMPING NUT 002 FIG. 239. balls must be used, each ball race should be made as a separate ring, backed with a leather or other soft washer, to better distribute the load on each ring. In the case of journal bearings either plain or grooved races may be used. In the latter case the Wkae Metal balls can be inserted through a hole at the side of the bearing, which is afterwards plugged. For light loads and low speeds such bearings will run well without a cage, but the space between the sleeve and liner should be completely filled with balls. For heavy loads and high speeds, however, a light cage is necessary. The safe load for a ball bearing varies as the number and size of the balls, and inversely 0-01 Same loads tn all cases. all Bearings / 1 2 3 4 Revolutions of S/ia/t FIG. 2390;. Friction. 247 as the speed. Any statement as to the safe load per ball, without stipulating the speed, is not worthy of a moment's consideration. The author is about to publish a paper upon this subject, and is therefore unable to give further details at present. The friction of a ball bearing varies (1) Directly as the load ; (2) Is independent of the speed ; (3) Is independent of the temperature ; (4) The friction of rest is but very slightly greater than the friction of motion ; (5) Is not reduced by lubrication in a clean well-designed bearing. The curves given in Fig. 2390 were obtained by an auto- graphic recorder in the author's laboratory. They show clearly that the friction of rest in the case of a ball bearing is practically the same as the friction of motion, and that it is very much less than that of an ordinary bearing. The reader should refer to the following articles in Engineering on the question of "Ball Bearings:" April 12, 1901; December 26, 1902; February 20, 1903. Friction of Lubricated Surfaces. The laws which appear to express the behaviour of well-lubricated surfaces are almost the reverse of those of dry surfaces. For the sake of comparison, we tabulate them below side by side Dry Surfaces. Lubricated Surfaces. 1. The frictional resistance is I. The frictional resistance is nearly proportional to the normal almost independent of the pressure pressure between the two surfaces. with bath lubrication, and ap- proaches the behaviour of dry sur- faces as the lubrication becomes more meagre. 2. The frictional resistance is 2. The frictional resistance varies nearly independent of the speed for directly as the speed for low pres- low pressures. For high pressures sures. But for high pressures the it tends to decrease as the speed friction is very great at low veloci- increases. ties, becoming a minimum at about 100 ft. per minute, and afterwards increases approximately as the square root of the speed. N.B. The friction of liquids varies as the square, not as the square root, of the speed ; hence the friction of a well-lubricated bearing is not merely that of the lubricant. 248 Mechanics applied to Engineering. Dry Surfaces. 3. The friclional resistance is not greatly affected by the temperature. 4. The frictional resistance de- pends largely upon the nature of the material of which the rubbing surfaces are composed. 5. The friction of rest is slightly greater than the friction of motion. 6. When the pressures between the surfaces become excessive, seizing occurs. Lubricated Surfaces. 3. The frictional resistance de- pends more upon the temperature than on any other condition partly due to the variation in the viscosity of the oil, and partly to the fact that the diameter of the bearing increases with a rise of temperature more rapidly than the diameter of the shaft, and thereby relieves the bearing of side pressure. 4. The frictional resistance with a flooded bearing depends but slightly upon the nature of the material of which the surfaces are composed, but as the lubrication becomes meagre, the friction follows much the same laws as in the case of dry surfaces. 5. The friction of rest is enor- mously greater than the friction of motion, especially if thin lubricants be used, probably due to their being squeezed out when standing. 6. When the pressures between the surfaces become excessive, which is at a much higher pressure than with dry surfaces, the lubri- cant is squeezed out and seizing occurs. The pressure at which this occurs depends upon the viscosity of the lubricant. 7. The frictional resistance is least at first, and rapidly increases with the time after the two surfaces are brought together, probably due to the partial squeezing out of the lubricant. 8. Same as in the case of dry surfaces. 7. The frictional resistance is greatest at first, and rapidly de- creases with the time after the two surfaces are brought together, pro- bably due to the polishing of the surfaces. 8. The frictional resistance is always greater immediately after reversal of direction of sliding. We will now give the results of a few experiments in support of the laws of lubricated surfaces given above. i. The curves given in Fig. 240 were taken from diagrams autographically drawn by the author's own friction-testing machine. They speak for themselves. The weight of pad used in each case was 0*023 lb. It is interesting to note that the friction of a dry bearing is actually less at very low loads than the lubricated bearings, on account of the viscosity of the oil being approximately Friction. 249 constant at all loads. After about 400 Ibs. square inch the oil appears to get squeezed out. 2. The manner in which the friction of a bearing depends 46OO f}OO 12OO 1OOO SOO 6OO 4OO LO0DS IN POUNDS SO INCH FlG. 240. ZOO 300 *OO Velocity rf nMring in. feet* per minute. FIG. 241. upon the velocity of rubbing is shown in Fig. 241 ; the friction of water in pipes is also shown for comparison. 250 Mechanics applied to Engineering. 3. The two typical curves selected will serve to make this point clear. 4. Mr. Tower has shown that in the case of a flooded bearing there is no metallic contact between the shaft and bearing; it is therefore quite evident that under such circumstances the material of which the bearing is composed makes no difference to the friction. When the author first began to experiment on' the relative friction of antifriction metals, he used profuse lubrication, and was quite unable to 70 80 90 JOO Temperature' FIG. 242. ~i 1 120 Fah/ detect the slightest difference in the friction ; but on using the smallest amount of oil consistent with security against seizing, he was able to detect a very great deal of difference in the friction. In the table below, the two metals A and B only differed in composition by changing one ingredient, amounting to 0*23 per cent, of the whole. Load in Ibs. sq. inch 150 250 35 45 75o 95o Coefficient of * A 0-0143 0-OII2 0-0091 0-0082 0-0075 0-0083 friction \ B 0-0083 O*OO62 0-0054 0*0050 0-0045 0-0047 Friction. 251 5. The following tests by Thurston will show how much greater is the friction of rest than of motion : Load in Ibs. sq. inch. 50 IOO 250 500 7So IOOO Coefficient of 1 ^'ns^M 0-013 0-008 0-005 0-004 0-0043 0-009 friction | of starting) 0*07 o-i35 0*14 0-15 0-185 0-18 Oil used) sperm. 6. Experiments by Tower and others show that a steel shaft in a gun-metal bearing seizes at about 600 Ibs. square inch under steady running, whereas when dry the same materials seize at about 80 Ibs. square inch. The author finds that the seizing pressure increases as the viscosity of the oil increases. tfo load. Load <50O lb& s TIME, 6"M, Second f FIG. 243. 7. This diagram is one of many drawn autographically on the author's machine. The lever which applies the load on the bearing was lifted, and the machine allowed to run with only the weight of the bearing itself upon it ; the lever was then suddenly dropped, the friction being recorded automatically. An indirect proof of this state- ment is to be found in the case of connecting-rod ends, and on pins on which the load is constantly reversed ; at each stroke the oil is squeezed away from the pressure side of the pin to the other side. Then, when the pressure is reversed, there is a large supply of oil between the bearing and the pin, which gradually flows to the other side. Hence at first the bearing is floating on oil, and the friction is consequently very small ; as the oil flows away, the friction increases. This is the reason why a much higher FIG. 244. 252 Mechanics applied to Engineering bearing pressure may be allowed in the case of a connecting- rod end than in a constantly revolving bearing. 8. In friction-testing machines it is always found that the temperature and the friction of a bearing is higher after reversal of direction, but in the course of a few hours it gets back to the normal again. Some metals, however, appear to have a grain, as the friction is always much greater when running one way than when running the other way. Nominal Area of Bearing. The pressure on a cylin- drical bearing varies from point to point; it is a maximum at the crown, and is least at the two sides. Calculations as to the distribution of pressure are pos- sible, but the assumptions usually made are most unwarrantable. For all practical purposes, the pressure is assumed to be evenly distributed over the projected area of the bearing. Thus, if w be the width of the bearing across the chord, and / the length of the bearing, the nominal area is wl y and the nominal pressure per W square inch is , where W is the total load on the bearing. Beauchamp Tower's Experiments. These experi- Centre FIG. 245. FIG. 246. ments were carried out for a research committee of the Institution of Mechanical Engineers, and deservedly hold the highest place amongst friction experiments as regards accuracy. The reader is referred to the Reports for full details in the Institution Proceedings, 1885. Friction. 253 Most of the experiments were carried out with oil-bath lubrication, on account of the difficulty of getting regular lubrication by any other system. It was found that the bearing was completely oil-borne, and that the oil pressure varied as shown in the curves, the pressure being greatest on the "off" side. In this connection Mr. Tower shows that it is useless ay, worse than useless to drill an oil-hole on the resultant line of pressure of a bearing, for not only is it impossible for oil to be fed to the bearing by such means, but oil is also collected from other sources and forced out of the hole (Fig. 247), thus robbing the bearings of oil at exactly the spot where it is most required. If oil-holes are used, they must communicate with a part of the bearing where there is little or no pressure (Fig. 248). FIG. 247. FIG. 248. A general summary of the results obtained by Mr. Tower are given in the following table. The oil used was rape ; the speed of rubbing 150 feet per minute; and the temperature about 90 F. : Form of bear- ing. Load at which j] seizing occur- red, in Ibs. sq. inch Coefficient of h friction } 0<01 O'OI 370 0-006 550 0-006 600 O'OOI 90 001 Other of Mr. Tower's experiments are referred to in preceding and succeeding paragraphs. 254 Mechanics applied to Engineering. Professor Osborne Reynolds' Investigations. A theoretical treatment of the friction of a flooded bearing has been investigated by Professor Osborne Reynolds, a full account of which will be found in the Philosophical Transac- tions. In this investigation, he has shown a complete agree- ment between theory and experiment as regards the total frictional resistance of a flooded bearing, the distribution of oil pressure, and the thickness of the oil film, besides many other points of the greatest interest. Professor Petroff, of St. Petersburg, has also done very similar work, and has, moreover, shown both experimentally and theoretically how the thickness of the oil film varies with the viscosity of the oil and with the pressure on the bearing. Goodman's Experiments. The author, shortly after the results of Mr. Tower's experiments were published, repeated his experiments on a much larger machine belonging to the L. B. & S. C. Railway Company; he further found that the oil pressure could only be registered when the bearing was flooded ; if a sponge saturated with oil were applied to the bearing, the pressure was immediately shown on the gauge, but as the oil ran away and the supply fell off, so the pressure fell. In another case a bearing was provided with an oil-hole on the resultant line of pressure, to which a screw-down valve was attached. When the oil-hole was open the friction on the bearing was very nearly 25 per cent, greater than when it was closed and the oil thereby prevented from escaping. Another bearing was fitted with a micrometer screw for the purpose of measuring the thickness of the oil film ; in one instance, in which the conditions were similar to those assumed by Professor Reynolds, the FlG> 249 . thickness by measurement was found to be 0*0004 inch, and by his calcu- lation o'ooo6 inch. By the same appliance the author found that the thickness was greater on the " on " side than on the " off" side of the bearing. The wear always takes place where the film is thinnest, i.e. on the " off" side of the bearing exactly the reverse of what would be expected if the shaft were regarded as a roller, and the bearing as being rolled forwards. When white metal bearings are tested to destruction, the metal Friction. 255 always begins to fuse on the "off" side first, and bearings originally i inch thick are frequently only 0*9 inch thick at a after a severe test. The figure shows roughly how the bearing wears. The side on which the wear takes place depends, however, upon the arc of the bearing in contact with the shaft. When the arc subtends an angle greater than about 90, the wear is on the "off" side; if less than 90, on the "on" side. This FIG. 250. 0-25 0-J 0-73 i-0 Chords in, corvtcuct FIG. 251. wear was measured thus : The four screws a, tj>/r 2 . dr M __ 27T/./R3 3 Substituting the value of/ from above M,=f/*WR work done per minute in foot-pounds = ^ AtWDN horse-power absorbed = -^ 189,000 This result might have been arrived at thus : Assuming the load evenly distributed, the triangle (Fig. 258) shows the distribution of pressure, and consequently the distribution of the friction. The centre of gravity of the triangle is then the position of the resultant friction, which therefore acts at a radius equal to | radius of the pivot. If it be assumed that the unequal wear of the pivot causes the pressure to be unevenly distributed in such a manner that the product of the normal pressure / and the velocity of rubbing V be a constant, we get a different value for M 7 ; the f becomes \. It is very uncertain, however, which is the true value. The same remark also applies to the two following paragraphs. Collar Bearing (Fig. 260). By similar reasoning to that given above, we get FIG. 257. Moment of friction 1 on collar } = r = R 2 FIG. 258. Conical Pivot. The intensity of pressure p all over the surface is the same, whatever may be the angle a. Let P be the pressure acting on one half of the cone 2 sm a Friction. The area of half the surface of the cone is 263 2 2 sm a , ^_ PQ _ W . 2 sin a _ W __ weight A 2 sin a . TrR 2 TrR 2 projected area W FIG. 259. Total normal pressure on any elementary ring = 2-71-;^ . dl moment of friction on elementary ring = mr^pp . dl but *=- sin sm a moment of friction on whole surface = ^ I r 2 . dr sin a J 2TT/X/R 3 ' 3 sin a Substituting the value of/, we have M/-= : T 3 sin a The angle a becomes 90, and sin a = i when the pivot becomes flat. By similar reasoning, we get for a truncated conical pivot (Fig. 261)- ! 3 - R 2 3 ) 3 sin 264 Mechanics applied to Engineering. Schiele's Pivot and Onion Bearing (Figs. 262, 263). Conical and flat pivots often give trouble through heating, pro- bably due to the fact that the wear is uneven, and therefore the contact between the pivot and step is imperfect, thereby giving rise to intense local pressure. The object sought in the Schiele pivot is to secure even wear all over the pivot. As the footstep wears, every point in the pivot will sink a vertical dis- tance h, and the point a sinks to lf where aa^ = h. Draw ab normal to the curve at a, and ac normal to the axis. Also draw ba^ tangential to the dotted curve at b t and ad to the full-lined curve at a ; then, if h be taken as very small, ba l will be practically parallel to ad, and the two triangles aba and acd will be practically similar, and FIG. 261. ac X aa ac or ad = ba r.h ba ba But ba is the wear of the footstep normal to the pivot, which is usually assumed to be proportional to the friction F between the surfaces, and to the velocity V of rubbing ; hence ba oo FV oo or ba = where K is a constant for any given speed and rate of wear; hence r.h ad = But h is constant by hypothesis, and //. is assumed to be constant all over the pivot; p we have already proved to be constant (last paragraph) ; hence ad t the length of the tangent to the curve, is constant; thus, if the profile of a pivot be so con- structed that the length of the tangent ad = /be constant, the wear will be (nearly) even all over the pivot. Although our Friction. 265 assumptions are not entirely justified, experience shows that such pivots do work very smoothly and well. The calculation of the friction moment is very similar to that of the conical pivot. FIG. 262. FIG. 263. The normal pressure at every point is weight _ W . _ ~ projected area R^ - R 2 2 ) By similar reasoning to that given for the conical pivot, we have Moment of friction on an elementary \ __ z-rrr^^pdr ring of radius r } slrTcT" (but -I = t ) = 2irtu.prdr \ sin a / and moment of friction for whole pivot = zirt r . dr M, = 2* Substituting the value of/, M, = 266 Mechanics applied to Engineering. The onion bearing shown in the figure is simply a Schiele pivot with the load suspended from below. Friction of Cup Leathers. The resistance of a hydraulic plunger sliding through a cup leather has been investigated by Hick, Tuit, and others. The formula proposed by Hick for the friction of cup leathers does not agree well with experiments ; the author has therefore recently tabulated the results of published experiments and others made in his laboratory, and finds that the following formulae much more nearly agree with experiment : Let F = frictional resistance of a leather in pounds per square inch of water-pressure ; d = diameter of plunger in inches ; p water-pressure in pounds per square inch. Then F = o*o8/ + when in good condition F = o-o8/ + ~ bad Efficiency of Machines. In all cases of machines, the work supplied is expended in overcoming the useful resistances for which the machine is intended, in addition to the useless or frictional resistances. Hence the work supplied must always be greater than the useful work done by the machine. Let the work supplied to the machine be equivalent to lowering a weight W through a height h ; the useful work done by the machine be equivalent to raising a weight W M through a height h u ; the work done in overcoming friction be equivalent to raising a weight W, through a height h f Then, if there were no friction Supply of energy = useful work done or mechanical advantage = velocity ratio When there is friction, we have Supply of energy = useful work done + work wasted in friction W/fc = W A + Friction. 267 and , . , ^ useful work done the mechanical efficiency = - total work done _ the work got out the work put in Let TI = the mechanical efficiency ; then = ^ or W ' v) is, of course, always less than unity. The " counter- efficiency " is -, and is always greater than unity. *? Reversed Efficiency. When a machine is reversed, for example, when a load is being lowered by lifting- tackle, the original resistance becomes the driver, and the original driver becomes the resistance ; then Reversed efficiency = useful work done in liftin S W throu S h h total work done in lowering W tt through h u __ W When W acts in the same direction as W M , i.e. when the machine has to be assisted to lower its load, rj r takes the negative sign. In an experiment with a two-sheaved pulley block, the pull on the rope was 170 Ibs. when lifting a weight of 500 Ibs.; the velocity ratio in this case R = -~ = ^. Then * = W == ^T4 = ' 73S The friction work in this case W f fi f was 170X4^500X1 = 1 80 foot-lbs. Hence the reversed efficiency -rj r = - = 0*64, and in order to lower the 500 Ibs. weight gently, the backward pull on the rope must be S? x 0-64 = 80 Ibs. 4 If the 80 Ibs. had been found by experiments, the reversed efficiency would have been found thus 80 X 4 .( t] r = = O 64 500 X 1 268 Mechanics applied to Engineering. The reversed efficiency must always be less than unity, and may even become negative when the frictional resistance of the machine is greater than the useful resistance. In order to lower the load with such a machine, an additional force acting in the same sense as the load has to be applied ; hence such a machine is self-sustaining, i.e. it will not run back when left to itself. The least frictional resistance necessary to ensure that it shall be self-sustaining is when W^ = W u /t u ; then, substituting this value in the efficiency expression for forward motion, we have Thus, in order that a machine may be self-supporting, its efficiency cannot be over 50 per cent. This statement is not strictly accurate, because the frictional resistance varies some- what with the forces transmitted, and consequently is smaller when lowering than when raising the load ; the error is, how- ever, rarely taken into account in practical considerations of efficiency (see Appendix). This self-supporting property of a machine is, for many purposes, highly convenient, especially in hand-lifting tackle, such as screw-jacks, Weston pulley blocks, etc. Combined Efficiency of a Series of Mechanisms. If in any machine the power is transmitted through a series of simple mechanisms, the efficiency of each being ^, %, 173, etc., the efficiency of the whole machine will be ^ = ^ X % X r] 3 , etc. If the power be transmitted through n mechan- isms of the same kind, each having an efficiency ^ the efficiency of the whole series will be approxi- mately VJ = TjS tn Hence, knowing the efficiency of various simple mechanisms, it becomes a simple matter to calculate with a fair degree of accuracy the efficiency of any complex machine. _ Efficiency of Various Machine Elements. "FIG. 264. Pulleys. In the case of a rope or chain pass- ing over a simple pulley, the frictional resistances are due to (i) the resistance of the rope or chain to bending; (2) the friction on the axle. The first varies with the make, Friction. 269 size, and newness of rope; the second with the lubrication. The following table gives a fairly good idea of the total efficiency at or near full load of single pulleys ; it includes both resistances i and 2 : Diameter of rope Ito. fin. i in. ii in. chain. Maximum efficiency Clean and well oiled Dirty Clean and well oiled, \ 96 94 93 91 9i 89 88 86 95-97 93-96 per cent. with stiff new rope) 9 1 These figures are fair averages of a large number of experiments. The diameter of the pulley varied from 8 to 12 times the diameter of the rope, and the diameter of the pins from \ inch to i\ inch. It is useless to attempt to calculate the efficiency with any great degree of accuracy. Pulley Blocks.' When a number of pulleys are combined for hoisting tackle the efficiency 01 the whole' may be calculated approxi- mately from the known efficiency of the single pulley. The efficiency of a single pulley does not vary greatly with the load unless it is absurdly low ; hence we may assume that the efficiency of each is the same. Then, if the rope passes over n pulleys, each having an efficiency ift, we have the efficiency of the whole FIG. 265. The following table of efficiencies has been com- piled by plotting curves for experiments made at the Yorkshire College laboratory : 2/O Mechanics applied to Engineering. Single pulley. Two-sheaved. Three-sheaved. T A " pounds. Old f-in. New |-in. Old *-in. New *-in. Old |-in. New |-in. rope. rope. rope. rope. rope. rope. 14 94 90 __ T __ 28 94'5 90-5 80 75 30 24 56 112 95 96 91 92 84 86 78-5 9i'5 g 35 4i 168 87-5 93 65 44 224 89 93 6 9 47 280 90 94 72 50 336 74 53 448 ~ ~ 78 56 Weston Pulley Block. This is a modification of the old Chinese windlass ; the two upper pulleys are rigidly attached ; the radius of the smaller one is r, and of the larger R. Then, neglecting fric- tion for the present, and taking moments about the axle of the pulleys, we have (R - r) = PR w and the velocity ratio 2R The pulleys are so chosen that the velocity ratio is from 30 to 40. The efficiency of these blocks is always under 50 per cent., consequently they will not run back when left alone. From a knowledge of the efficiency of a single-chain pulley, one can make a rough estimate of the relative sizes of pulleys required to prevent such blocks from running back. Taking the efficiency of each pulley as 97 per cent, when the weight is just on the point of running back, the tension in the right-hand chain will be 97 per cent, of that in the left-hand chain due to the friction Friction. 271 on the lower pulley; but due to the friction on the upper pulley only 97 per cent, of the effort on the right-hand chain can be transmitted to the left-hand chain, whence for equi- librium, when P = o, we have W W p r = 0-97 X 0-97 X -R or ; = and the velocity ratio = R - = 33 which is about the value commonly adopted. The above treatment is only approximate, but it will serve to show the relation between the efficiency and the ratio between the pulleys. General Efficiency Law. A simple law can be found to represent tolerably accurately the friction of any machine when working under any load it may be capable of dealing with. It can be stated thus : " The total effort F that must be exerted on a machine is a constant quantity K, plus a simple function of the resistance W to be overcome by the machine." The quantity K is the effort required to overcome the friction of the machine itself apart from any useful work. The law may be expressed thus F = K + Wx The value of K depends upon the type of machine under consideration, and the value of x upon the velocity ratio z/ r of the machine. From Fig. 2 65 it will be seen how largely the efficiency is dependent upon the value of K. The broken and 2/2 Mechanics applied to Engineering. the full-line efficiency curves are for the same machine, with a large and a small initial resistance. W The mechanical efficiency ij - - W (K + Thus we see that the efficiency increases as the load W j increases. Under very heavy loads ^ may become negligible ; hence the efficiency may approach, but can never exceed -IS. The .following values give results agreeing well with experiments : X K - Rope pulley blocks Chain blocks of the| Weston type / i 4- o'oszv 3lbs. 3lbs. W V r I 4-0'17/r W V r W(i+o W ) + K* Levers. The efficiency of a simple lever (when used at any other than very low loads) with two pin joints varies from 94 to 97 per cent., the lower value for a short and the higher for a long lever. When mounted on well-formed knife-edges, the efficiency is practically TOO per cent. Toothed Gearing. The efficiency of toothed gearing depends on the smoothness and form of the teeth, and whether lubricated or not. Knowing the pressure on the teeth and the distance through which rubbing takes place (see p. 149), also the /A, the efficiency is readily arrived at; but the latter varies so much, even in the same pair of wheels, that it is very difficult to repeat experiments within 2 or 3 per cent. ; hence calculated values depending on an arbitrary choice of /* cannot have any Friction. 273 pretence to accuracy. The following empirical formula fairly well represents average values of experiments : For one pair of machine-cut toothed wheels, including the friction on the axles - 0*96 for rough unfinished teeth = 0*90 Where N is the number of teeth in the smallest wheel. When there are several wheels in one train, let n = the number of pairs of wheels in gear ; Efficiency of train ^ = 17" The efficiency increases slightly with the velocity of the pitch lines (see Engineering, vol. xli. pp. 285, 363, 581; also Kennedy's "Mechanics of Machinery," p. 579). Velocity of pitch line in ) feet per minute . . . \ Efficiency 10 0*940 50 0-972 100 980 150 0-984 200 0-986 Screw and Worm Gearing. We have already shown FIG. 266. how to arrive at the efficiency of screws and worms when the coefficient of friction is known. The following table is taken from the source mentioned above : 2/4 Mechanics applied to Engineering. Velocity of pitch line in feet per ) minute .. ... ... 10 50 100 150 200 Efficiency per cent. Angle of thread a, 45 87 94 95 96 97 30 82 90 93 94 9S 20 75 86 90 92 92 rc >i *j 70 82 7 9 90 ir i > 62 76 82 S 86 7 > / 53 69 76 80 81 C > D 45 62 70 74 76 The figure shows an ordinary single worm and wheel. As the angle a increases, the worm is made with more than one thread ; the worm and wheel is then known as screw gearing. For details, the reader should refer to Unwin's " Elements of Machine Design." Friction of Slides. A slide is generally proportioned so that its area bears some relation to the load ; hence when the load and coefficient of friction are unknown, the resistance to sliding may be assumed to be proportional to the area ; when not unduly tightened, the resistance may be taken as about 3 Ibs. per square inch. Friction of Shafting. A 2-inch diameter shaft running at 100 revolutions per minute requires about i horse-power per 100 feet when all the belts are on the pulleys. The horse- power increases directly as the speed and approximately as the cube of the diameter. This may be expressed thus Let D = diameter of the shafting in inches ; N = number of revolutions per minute ; L = length of the shafting in feet ; F = the friction horse-power of the shafting. ^ NLD 3 Then F = 80,000 The horse-power that can be transmitted by a shaft is accoiding to the working stress. Friction. 275 Hence the efficiency of line shafting on which there are numerous pulleys is _ horse-power transmitted friction horse-power horse-power transmitted ND 3 _ LND 3 _ 64 jJo,ooo 64 = I - - f or a working stress of 5000 Ibs. sq. inch 2000 " " 8oo L Thus it will be seen that ordinary line shafting may be extremely wasteful in power transmission. The author knows of several instances in which more than one -half the power of the engine is wasted in driving the shafting in engineers' shops ; but it must not be assumed from this that shafting is necessarily a wasteful method of transmitting power. Most of the losses in line shafting are due to bending the belts to and fro over the pulleys (see p. 279), and to the extra pressure on the bearings due to the pull on the belts and the weight of the pulleys. In an ordinary machine shop one may assume that there is, on an average, a pulley and a 3-inch belt at every 5 feet. The load on the bearings due to this belt, together with the weight of the shaft and pulley, will be in the neighbourhood of 500 Ibs. The Iqad on the countershaft bearings may be taken at about the same amount or, say, a load on the bearings of 1000 Ibs. in all. Let the diameter of the shafting be 3 inches; the 5 feet length will weigh about 120 Ibs., hence the load on the bearings due to the pulleys, belts, etc., will be about eight times as great as that of the shaft itself and considering the poor lubrication that shafting usually gets, one may take the relative friction in the two cases as being roughly in this proportion. Over and above this, there is considerable loss due to the work done in bending the belt to and fro. We shall now proceed to find the efficiency of shafting, which receives its power at one end and transmits it to a distant point at its other end, i.e. without any intermediate pulleys. 276 Mechanics applied to Engineering. Consider first the case of a shaft of the same diametre throughout its entire length. Let L = the length of the shaft in feet; R = the radius of the shaft in inches ; W = the weight of the shaft i square inch in section and i foot long ; //. = the coefficient of friction ; rj = the efficiency of transmission ; f, = the torsional skin stress on the shaft per square inch; the maximum twisting mo- ment at the motoi end of the shaft /"irR 3 SS-CL L inch-lbs. (see p. 484) of the^rTn y 1 = the effective twisting moment at the far end mission -n \ t ^ ie tw i stm S moment at the motor end _ maximum twisting moment friction moment maximum twisting moment __ _ friction moment _ maximum twisting moment _ _ 2/xWL For a hollow shaft in which the inner radius is - of the n outer, this becomes Now consider the case in which the shaft is reduced in diameter in order to keep the skin stress constant throughout its length. Let the maximum twisting moment at the motor end of the shaft = T 1 ; Friction. 277 Let the useful twisting moment at the far end of the shaft = T 2 . Then the increase of twisting moment dt due to the friction on an elemental length dl /xWTrRV/ = dt. For the twisting moment / we may substitute (see p. 484) by substitution, we get dt and T! = where e = 272, the base of the system of natural logarithms. The efficiency rf = ? e 2/iWL ~ V = and for a hollow shaft, such as a series of drawn tubes, which are reduced in size at convenient intervals The following table shows the distance L to which power may be transmitted with an efficiency of 80 per cent. For ordinary bearings we have assumed a high coefficient of friction, viz. 0^04, to allow for poor lubrication and want of accurate alignment of the bearings. For ball bearings we also 2 7 3 Mechanics applied to Engineering. take a high value, viz. 0*002. Let the skin stress f, on the shaft be 8000 Ibs. sq. inch, and let n = 1*25. Form of shafting Parallel. Taper. Kind of bearings Ordinary. Ball. Ordinary. Ball. Solid shaft with belts ,, ,, without belts Hollow Feet. 400 6000 9840 Feet. 120,000 197,000 Feet. 76,600 125,550 Feet. 1,530,000 2,505,000 These figures at first sight appear to be extraordinarily high, and every engineer will be tempted to say at once that they are absurd. The author would be the last to contend that power can practically be transmitted through such distances with such an efficiency, mainly on account of the impossibility of getting perfectly straight lines of shafting for such distances, and the prohibitive costs \ but at least the figures show that very economical transmission may, under convenient circum- stances, be accomplished by shafting and when straight lengths of shafting could be put in they would unquestionably Driuer 3Q2L_ (0) (0; E 72,= / Tl/ = 2 Tt3 TL- FIG. 267. be far more economical in transmitting power than could be accomplished by converting the mechanical energy into electrical by means of a dynamo, losing a certain amount of the energy in the mains, and finally reconverting the electrical energy into mechanical by means of a motor ; but, of course, in most cases the latter method is the most convenient and the cheapest, on account of the ease of carrying the mains as against that of shafting. The possibility of transmitting power very economically by shafting was first pointed out by Professor 1 Apart from the loss in bending the belts to and fro as they pass over the pulleys. Friction. 279 Osborne Reynolds, F.R.S., in a series of Cantor Lectures on the transmission of power. Belt and Rope Transmission. The efficiency of belt and rope transmission for each pair of pulleys is from 95 to 96 per cent., including the friction on the bearings ; hence, if there are n sets of ropes or belts each having an efficiency 17, the efficiency of the whole will be, approximately Experiments by the author on a large number of belts show that the work wasted by belts due to resistance to bending over pulleys, creeping, etc., varies from 16 to 21 foot-lbs. per square foot of belt passed over the pulleys. Mechanical Efficiency of Steam-engines. The work absorbed in overcoming the friction of a steam-engine is roughly constant at all powers ; it increases slightly as the power increases. A full investigation of the question has been made by Professor Thurston, who finds that the friction is distributed as follows : Main bearings ............... 35~47 P er cent. Piston and rod ............... 21-33 a Crank-pin .................. 5-7 Cross-head and gudgeon-pin ......... 4-5 Valve and rod ... 2 '5 balanced, 22 unbalanced Eccentric strap ............... 4-5 Link and eccentric ............ 9 The following instances may be of interest in illustrating the approximate constancy of the friction at 'all powers : EXPERIMENTAL ENGINE, UNIVERSITY COLLEGE, LONDON. Syphon Lubrication. I.H.P 2'7S 9*25 10-23 11-14 12-34 13*95 14-29 B.H.P 0*0 5-63 7-50 7-66 9-09 11-09 11-25 Friction H.P. ... 275 3-62 2'73 3'4 3'2S 2-86 3 '04 EXPERIMENTAL ENGINE, YORKSHIRE COLLEGE, LEEDS. Syphon and Pad Lubrication. I.H.P. 2-48 Vi6 6-83 8-30 11*50 I3'84 17*02 22-30 B.H.P. ... o o 2'35 3 '94 S'6i 8-70 10-82 13-89 19-09 Friction H.P. 2-48 2-81 2-89 2*69 2-80 3-02 3'i3 3-21 280 Mechanics applied to Engineering. BELLISS ENGINE, BATH (FORCED) LUBRICATION. (See Proc. I.M.E., 1897.) I.H.P 49-8 1027 147-1 193-6 217-5 B.H.P 44'5 97 -o 140-6 186-0 209-5 Friction H.P. 5*3 57 6'5 7-6 8-0 Friction Pressure. The friction of an engine can be conveniently expressed by stating the pressure in the working cylinder required to drive the engine when running light. In ordinary steam-engines in good condition the friction pressure amounts to 2^ to 3^ Ibs. square inch, but in certain bad cases it may amount to 5 Ibs. square inch. It has about the same value in gas and oil engines per stroke, or say from 10 to 14 Ibs. square inch, reckoned on the impulse strokes when exploding at every cycle, or twice that amount when missing every alternate explosion. Thus, if the mean effective pressure in a steam-engine cylinder were 50 Ibs. square inch, and the friction pressure 3 Ibs. square inch, the mechanical efficiency of the engine would be 5 "" 3 = 94 per cent if double-acting, and 5 ~ 6 5 . 50 = 88 per cent, if single-acting. The mean effective pressure in a gas-engine cylinder seldom exceeds 75 Ibs. square inch. Thus the mechanical efficiency is from 81 to 87 per cent. The friction horse-power, as given in the above tables, can also be obtained in this manner. MECHANICAL EFFICIENCY PER CENT. OF VARIOUS MACHINES. (From experiments in all cases with more than quarter full load.) Weston pulley block ( ton) 20-25 Epicycloidal pulley block 40-45 One-ton steam hoists or windlasses 50-70 Hydraulic windlass 60-80 jack 80-90 Cranes (steam) 60-70 Travelling overhead cranes -. ... 30-50 , . draw bar H.P. Locomotives j H p 5 ~ 75 Two-ton testing-machine, worm and wheel, screw and nut, slide, two collars 2-3 Friction. 28 1 Screw displacer hydraulic pump and testing-machine, two cup leathers, toothed-gearing four contacts, three shafts (bearing area, 48 sq. inches), area of flat slides, 1 8 sq. inches, two screws and nuts 2-3 / About 1000 H.P. engines, Lancashire Cotton Mills f spur-gearing, and engine (see Proceedings of the I friction 74 Manchester Association ( Rope drives 70 of Engineers, 1892) Belt ,, 71 ^ Direct (4OO-H.P. engines) ... 76 For much valuable information on the subject of friction, the reader is referred to the Cantor Lectures on Friction, delivered at the Society of Arts by Professor Hele-Shaw, LL.D. Belts. Coil Friction. Let the pulley in Fig. 268 be fixed, and a belt or rope pass round a portion of it as shown. The weight W produces a tension Tj ; in order to raise the weight W, the tension T 2 must be greater than Tj by the amount of friction between the belt and the pulley. Let F = frictional resistance of the belt ; / = normal pressure between belt and pulley at any point. Then, if /* = coefficient of friction F = T 2 - T X = :% Let the angle a embraced by the belt be divided into a great number, say n, parts, so that ^ is very small; then the tension on both sides of this very small angle is nearly the same. Let the mean tension be T; then, expressing a in circular measure, we have The friction at any point is (neglecting the stiffness of the belt) But we may write - as So, ; also T 2 ' - T/ as ST. Then a = ST 282 Mechanics applied to Engineering. which in the limit becomes 7*T .da. = dT d^ , -Tr = da. We now require the sum of all these small tensions ex- pressed in terms of the angle embraced by the belt : where ^ = 272, the base of the system of natural logarithms, and log e = 0-4343. When W is being raised, the + sign is used in the index, and when lowered, the sign. The value of /A for leather, cotton, or hemp rope on cast iron is from 0*2 to o '4, and for wire rope 0*5. If a wide belt or plaited rope be used as an absorption dynamometer, and be thoroughly smeared with tallow or other thick grease, the resistance will be greatly increased, due to the shearing of the film of grease between the wheel and the rope. By this means the author has frequently obtained an apparent value of //, of over i a result, of course, quite impossible with perfectly clean surfaces. Power transmitted by Belts. Generally speaking, the power that can be transmitted by a belt is limited by the friction between the belt and the pulley. When excessively loaded, a belt usually slips rather than breaks, hence the FIG. a68. Friction. 283 friction is a very important factor in deciding upon the power that can be transmitted. When the belt is just on the point of slipping, we have Horse-power transmitted = FV = ( T 2 - T i)V 33> 000 33,000 where the friction F is expressed in pounds, and V = velocity in feet per minute. Substituting the value of V 25 square feet per minute for double-ply 600 284 Mechanics applied to Engineering. This will be found to be an extremely convenient expression for committal to memory. Centrifugal Action on Belts. In Chapter VI. we showed that the two halves of a flywheel rim tended to fly apart due to the centrifugal force acting on them ; in precisely the same manner a tension is set up in that portion of a belt wrapped round a pulley. On p. 181 we showed that the stress due to centrifugal force was where W r is the weight of i foot of belting i square inch in section. W r = 0*43 lb., and V w = the velocity in feet per second; hence 32'2 75 hence the tension T 2 in a belt is increased by centrifugal force to V 2 or, putting the velocity in feet per minute V, the total tension is 270,000 and the effective tension for the transmission of power is T ^_ v 2 270,000 The usual thickness of single-ply belting is about 0*22 inch, and taking the maximum tension as 80 Ibs.per inch of width, this gives = 364 Ibs. per square inch of belt, and the power transmitted per square inch of belt section is ,. 72 ( 3 6 4 Y!-)V \ 270,0007 33,000 7 V V 3 891 12,400,000,000 Friction. 285 _7_ 891 3V 2 12,400,000,000 V 2 which has its maximum value when 7 = ? 4,633,000 or when V = 5700 feet per minute Substituting this value in the equation given above, we have the maximum horse-power that can be transmitted per square inch of belt for the given stress limit. H.P. max. = 29*9 For ropes we have taken the weight per foot run as 0*35 Ib. Transmitted perSq In .of Section. 5 cT! S g /" ^ \ * / \ / P^ ^ iK, \ / x \ 1 5 / r \ 2 \ 1000 2000 .3000 4000 5000 6000 7000 8000 Velocity in Feet per Minute. FIG. 268/1. per square inch of section, and the maximum permissible stress as 200 Ibs. per square inch. On this basis we get the maximum horse-power transmitted when V = 4700 feet per minute, and the maximum horse-power per square inch of rope = 17*1. The curves in Fig. 268^ show how the horse-power trans- mitted varies with the speed. The accompanying figure (Fig. 269), showing the stretch of a belt due to centrifugal tension, is from a photograph of an indiarubber belt running at a very high speed ; for comparison 286 Mechanics applied to Engineering. the belt is also shown stationary. The author is indebted to his colleague Dr. Stroud for the photograph, taken in the physics laboratory at the Yorkshire College. Creeping of Belts. The material on the tight side of a belt is necessarily stretched more than that on the slack side, hence a driving pulley always receives a greater length of belt than it gives out ; in order to compensate for this, the belt creeps as it passes over the pulley. Let / = unstretched length of belt passing over the pulleys in feet per minute ; / 2 = stretched length on the T 2 side ; 7 - T n ~ j 5? *-i 55 N, = revolutions per minute of driven pulley ; N 2 = driving d-i = diameter of driven pulley ) measured to the middle // 2 = 55 driving ) of the belt ; x = stretch of belt in feet ; E = Young's modulus ; /! and^ = stresses corresponding to Tj and T 2 in Ibs. square inch. Then x =- E Nl _ N 2 (E +/. If there were no creeping, we should have N 1 = 4 N 2 d, E = from 8,000 to 10,000 Ibs. per square inch. Taking Friction. 287 T 2 = 80 Ibs. per inch of width, and the thickness as 0*22 inch, we have when a = 3-14 f 2 = --- = 364 Ibs. per square inch 0*22 and T! = ___ ___ = = 23 Ibs. per inch width 2'72' 4 3*5 = -^- = 104 Ibs. per square inch 0'22 Hence FIG. 269. or the belt under these conditions creeps or slips 2*5 per cent. When a belt transmits power, however small, there must be some slip or creep. 288 Mechanics applied to Engineering. When calculating the speed of pulleys the diameter of the pulley should always be measured to the centre of the belt ; thus the effective diameter of each pulley is D + A where / is the thickness of the belt. In many instances this refinement is of little importance, but when small pulleys are used and great accuracy is required, it is of importance. For example, the driving pulley on an engine is 6 feet diameter, the driven pulley on the countershaft is 13 inches, the driving pulley on which is 3 feet 7 inches diameter, and the driven pulley on a dynamo is 8 inches diameter; the thickness of the belt is 0-22 inches; the creep of each belt is 2-5 per cent.; the engine runs at 140 revolutions per minute : find the speed of the dynamo. By the common method of finding the speed of the dynamo, we should get 140 X 72 X 43 = 4l68 revolutions per minute 13 X o But the true speed would be much more nearly 140 x 72-22 X 43*22 X o'975 X o'975 5 212 -- Z12.SS 3822 revs, per minute I3'22 X 8'22 Thus the common method is in error in this case to the extent of 9 per cent. Chain Driving. In cases in which it is important to prevent slip, chain drives should be used. They moreover possess many advantages over ordinary belt driving if they are properly designed. For the scientific designing of chains and sprocket wheels, the reader is referred to a pamphlet on the subject by Mr. Hans Renold, of Manchester. FIG. 270. Rope Driving. When a rope does not bottom in a grooved pulley, it wedges itself in, and the normal pressure is thereby increased to sm - 2 The angle 6 is usually about 45; hence Pj = 2'6P. The most convenient way of dealing with this increased pressure is to use a false coefficient 2 '6 times its true value. Taking /u, - o'3 for a rope on cast iron, the false /A for a grooved pulley becomes 2 '6 X 0*3 = 073. Friction. 289 The value of e^ a now becomes io'i when the rope embraces half the pulley. The factor of safety on driving-ropes is very large, often amounting to about 80, to allow for defective splicing, and to prevent undue stretching. The working strength in pounds may be taken from io; ^ ^ Gun-metal bearings { Plain cylindrical journals^ tested by Mr. with bath lubrication j Beauchamp Tower J Plain cylindrical journals^ for the Institution I with ordinary lubrication/ of Mechanical | Thrust or collar bearing | Engineers ( well lubricated / Good white metal (author) with very meagre] lubrication / Poor white metal under same conditions 0*008 CHAPTER VIII. STXJSSS, STRAIN, AND ELASTICITY. Stress. If, on any number of sections being made in a body, it is found that there is no tendency for any one part of it to move rektively to any other part, that body is said to be in a state of ease ; but when one part tends to move relatively to the other parts, we know that the body is acted upon by a system of equal and opposite forces, and the body is said to be in a state of stress. Thus, if, on making a series of saw-cuts in a plate of metal, the cuts were found to open or close before the saw had got right through, we should know that the plate was in a state of stress, because the one part tends to move relatively to the other. The stress might be due either to external forces acting on the plate, or to internal initial stresses in the material, such as is often found in badly designed castings. Intensity of Stress. The intensity of direct stress on any given section of a body is the total force, acting normal to the section divided by the area of the section over which it is distributed ; or, in other words, it is the amount of force per unit area. Intensity of stress in) _ the given force in pounds pounds per sq. inch \ ~ area of the section over which the force acts in sq. inches For brevity the word " stress " is generally used for the term " intensity of stress." The conditions which have to be fulfilled in order that the intensity of stress may be the same at all parts of the section are dealt with in Chapter XIII. Strain. The strain of a body is the change of form or dimensions that it undergoes when placed in a state of stress. No bodies are absolutely rigid ; they all yield, or are strained more or less, when subjected to stress, however small in amount. The various kinds of stresses and strains that we shall consider are given below in tabular form. Stress, Strain, and Elasticity. 291 ii strain." Hi- ll u Hi- u he ratios 1 3 I'g jt ll * " || > r 1 S rt &. c ll afe J|o .Is 1] i 1 1 4-* S| || <> 3 .8 ol 2 3 c *.| o , . f^ 8 * = A o c 3 "*3 .2 g T3 Kind of 52 !1 ii O 4J w ^ s*cs o o "T/i o gl ss Is C Q 3.S |1 ^ U A ^ .. ^ N! ; j!L* 292 Mechanics applied to Engineering. Elasticity. A body is said to be elastic when the strain entirely disappears on the removal of the stress that produced it. Very few materials can be said to be perfectly elastic except for very low stresses, but a great many are approximately so over a wide range of stress. Permanent Set. That part of the strain that does not entirely disappear on the removal of the stress is termed " permanent set." Elastic Limit. The stress at which a marked permanent set occurs is termed the elastic limit of the material. We use the word marked because, if very delicate measuring instruments be used, very slight sets can be detected with much lower stresses than those usually associated with the elastic limit. In elastic materials the strain is usually proportional to the stress ; but this is not the case in all materials that fulfil the conditions of elasticity laid down above. Hence there is an objection to the definition that the elastic limit is that point at which the strain ceases to be proportional to the stress. Plasticity. If none of the strain disappears on the removal of the stress, the body is said to be plastic. Such bodies as soft clay and wax are almost perfectly plastic. Ductility. If only a small part of the strain be elastic, but the greater part be permanent after the removal of the stress, the material is said to be ductile. Soft wrought iron, mild steel, copper, and other materials, pass' through such a stage before becoming plastic. Brittleness. When a material breaks with a very low stress and deforms but a very small amount before fracture, it is termed a brittle material. Behaviour of Materials subjected to Tension. Duct He Materials. If a bar of ductile metal, such as wrought iron or mild steel, be subjected to a low tensile stress, it will stretch a certain amount, depending on the material ; and if the stress be doubled, the stretch will also be doubled, or the stretch will be proportional to the stress (within very narrow limits). Up to this point, if the bar be relieved of stress, it will return to its original length, i.e. the bar is elastic; but if the stress be gradually increased, a point will be reached when the stretch will increase much more rapidly than the stress ; and if the bar be relieved of stress, it will not return to its original length in other words, it has taken a " permanent set." The stress at which this occurs is, as will be seen from our definition above, the elastic limit of the material. Let the stress be still further increased. Very shortly a Stress, Strain, and Elasticity. 293 point will be reached when the strain will (in good wrought iron and mild steel) suddenly increase to 10 cr 20 times its previous amount. This point is termed the yield point of the material, and is always quite near the elastic limit. For all commercial purposes, the elastic limit is taken as being the same as the yield point. Just before the elastic limit was reached, while the bar was still elastic, the stretch would only be about YoVo" f tne l en gth of the bar ; but when the yield point is reached, the stretch would amount to ~Q, or ~ of the length of the bar. The elastic extensions of specimens cannot be taken by direct measurements unless the specimens are very long indeed ; they are usually measured by some form of exten- someter. That shown in Fig. 2 7 5^ was designed by the author FIG. 275*1. some years ago, and gives entirely satisfactory results ; it reads to TO"O o o" f an ^h \ i* i s simple in construction, and does not get out of order with ordinary use. It consists of suitable clips for attachment to the specimen, from which a graduated scale is supported j the relative movement of the clips is read on the scale by means of a pointer on the end of a 100 to i lever. In Fig. 275^ several elastic curves are given. In the case of wrought iron and steel, the elastic lines are practically straight, but they rapidly bend off at the elastic limit. In the case of cast iron the elastic line is never straight ; the strains always increase more rapidly than the stresses, hence Young's modulus is not constant. Such a material as copper takes a " permanent set " at very low loads ; it is almost impossible to say exactly where the elastic limit occurs. 294 Mechanics applied to Engineering. As the stress is increased beyond the yield point, the strain continues to increase much more rapidly than before, and the material becomes more and more ductile; and if the stress be now removed, almost the whole of the strain will be found to be permanent. But still a careful measurement will show that a very small amount of the strain is still elastic. 4 6 8 10 Stress in Ions per S, Ordinary state Turned shank lin. 8 14 ins. H 2*15 inch- tons I6'8 Ordinary state lurned "shank Ihl. \ 14 ins. 14 175 inch-tons 7 '4 . Ordinary state Turned shank In this connection it may be useful to remember that the sectional area at the bottom of the thread is - '- sq. inches (very nearly) 100 where d is the diameter of the bolt expressed in eighths of an inch. The author is indebted to one of his former students, Mr. W. Stevenson, for this very convenient expression. Behaviour of Materials subjected to Compression. Aluminium. Original form. Gun metal. Cast iron. FIG. 287. Soft brass. Cast iron. Ductile Materials. In the chapter on columns it is shown that the length very materially affects the strength of a piece of material when compressed, and for getting the true compressive strength, very short specimens have to be used in order to Stress, Strain, and Elasticity. 309 fJto Li tmtt FIG. 288. prevent buckling. Such short specimens, however, are incon- venient for measuring accurately the relations between the stress and the strain. Up to the elastic limit, ductile materials be- have in much the same way as they do in tension, viz. the strain is proportional to the stress. At the yield point the strain does not increase so suddenly as in tension, and when the plastic stage is reached, the sectional area gradu- ally increases and the metal spreads. With very soft homo- geneous materials, this spreading goes on until the metal is squeezed to a flat disc without fracture. Such materials are soft copper, or aluminium, or lead. In fibrous mate- rials, such as wrought iron and wood, in which the strength across the grain is much lower than with the grain, the material fails by splitting side- ways, due to the lateral tension. The usual form of the stress - strain curve for a ductile material is somewhat as shown in Fig. 288. If the material reached a perfectly plastic stage, the real stress, i.e. FIG. 389. load at any instant (W) J sectional area at that instant (A,) Mechanics applied to Engineering. would be constant, however much the material was compressed ; then, using the same notation as before A - /A Al ~7 w and from above, - = constant Substituting the value of A l from above, we have W7 1 = constant ; /A But /A, the volume of the bar, is constant ; hence W/ : = constant or the stress-strain curve during the plastic period is a hyperbola. The material never is perfectly plastic, and therefore never perfectly complies with this, but in some materials it very nearly approaches it. For example, copper and aluminium (author's recorder) (Fig. 289). The constancy of the real stress will be apparent when we draw the real stress curves. Brittle Materials. Brittle materials in compression, as in tension, have no marked elastic limit or plastic stage. When crushed they either split up into prisms or, if of cubical form* into pyramids, and sometimes by the one- half of the specimen shearing over the other at an angle of about 45. Such a fracture is shown in Fig. 287 (cast iron). The shearing fracture is quite what one might expect from purely theoretical reasoning. In Fig. 291 let the sectional area = A ; W then the stress on the cross-section S = A FIG. 290. Asphaltc. Stress, Strain, and Elasticity. 311 and the stress on an oblique section aa, making an angle a FIG. 2Q3. Portland cement. with the cross-section, may be found thus : resolve W into components normal N = W cos a, and tangential T = W sin a. 312 Mechanics applied to Engineering. The area of the oblique section aa = A = cos a . N Wcosa Wcos 2 a Q 2 the normal stress = = = = S . cos^ a A A A COS a tangential or shear- \ _ T _ W sin a __ W cos a sin a ing stress J A~ ~ A cos a = S COS a sin a If we take a section at right angles to aa, T becomes the normal component, and N the tangential, and it makes an angle of 90 a with the cross-section ; then, by similar reason- ing to the above, we have A A The area of the oblique section = A ' = = - COS (90 a) sin a normal stress = S sin 2 a tangential stress or shearing stress = S sin a cos a So that the tangential stress is the same on two oblique sections at right angles, and is greatest when a = 45; it is then = S X 071 x 071 = 0-5 S. From this reasoning, we should expect compression speci- mens to fail by shearing along planes at right angles to one another, and in a cylindrical specimen to form two cones top and bottom, and the sides to break away in triangular sections, which in the cube become six pyramids. That this theory is fairly correct is shown by the illustrations in Figs. 290-292. Real and Nominal Stress in Compression (Fig. 293). In the paragraph on real and nominal stress in tension, we showed how to construct the curve of real stress from the ordinary load-strain diagram. Then, assuming that the com- pression specimen remains parallel (which is not quite true, as the specimens always become barrel-shaped), the same method of constructing the real stress curve serves for compression. As in the tension curve, it is evident that (see Fig. 280) S L 3 '=% Stress, Strain, and Elasticity. 313 Behaviour of Materials subjected to Shear. Nature of Shear Stress. If an originally square plate or block be acted upon by forces P parallel to two of its opposite sides, the square will be distorted into a rhombus, as shown in Fig. 294, FIG. fs I I a, b d C i i but these must be equal if the plate be in equilibrium ; then /, . ab . ad = // .ad.ab i.e. the intensity of stress on the two sides of the plate is the same. Now, for convenience we will return to our square plate. The forces acting on the two sides P and Pj may be resolved into forces R and Rj acting along the diagonals as shown in Fig. 296. The effect of these forces will be to distort the square into a rhombus exactly as before. (N.B. The rhombus V a ft, FIG. 296. FIG. in Fig. 294 is drawn in a wrong position for simplicity.) These two forces act at right angles to one another ; hence we see that a shear stress consists of two equal and opposite stresses, a tension and a compression, acting at right angles to one another. In Fig. 297 it will be seen that there is a tensile stress acting normal to one diagonal, and a compressive stress normal to the other. The one set of resultants, R, tend to pull the two triangles abc, acd apart, and the other resultants to push the two triangles abd, bdc together. Let/ = the stress normal to the diagonal. Then/o^ =_f^ 2ab, Qif^/2bc = R But -/iP, or J~2f t .ab, or J2/ t . be = R hence / =/, =// Stress, Strain, and Elasticity. 315 Thus the intensity of shear stress is equal on all the four edges and on the two diagonals of a rectangular plate subjected to shear. Materials in Shear. When ductile materials are sheared, they pass through an elastic stage similar to that in tension and compression. If an element be slightly distorted, it will return to its original form on the removal of the stress, and during this period the strain is proportional to the stress; but after the elastic limit has been reached, the plate becomes perma- nently deformed, but has not any point of sudden alteration as in tension. On continuing to increase the stress, a ductile and plastic stage is reached, but as there is no alteration of area under shear, there is no stage corresponding with the stricture stage in tension. The shearing strength of ductile materials, both at the elastic limit and at the maximum stress, is about \ of their tensile strength (see p. 327). Ductile Materials in Shear. The following results, obtained in a double-shear shearing tackle, will give some idea of the relative strengths of the same bars when tested in tension and in shear ; they are averages of a large number of tests : Work done M.3,tcricU. Nominal tensile Shearing Shearing strength. per sq. in. strength. strength. Tensile strength. of metal sheared through. Cast iron io*9 I2'9 1-18 _ Best wrought iron 22-0 18-1 0-82 57 Mild steel 26-6 20*9 079 6-9 Hard steel 48-0 34'o 071 37 Gun metal I3'S 15-2 1-13 0-9 Copper . Aluminium I5-0 8'8 iro 57 073 0-65 4'5 I'D From autographic shearing and punching diagrams, it is found that the maximum force required occurs when the shearing tackle is about \ of the way through the bar, and when the punch is about \ of the way through the plate. From a series of punching tests it was found that where /,= load on punch circumference of hole X thickness of plate 316 Mechanics applied to Engineering. f ft Ratio. Wrought iron . 19-8 2 4 -8 0'8o Mild steel . . . 22-2 28-4 078 Copper. . . . I0'4 147 C7I Brittle materials in shear are elastic, although somewhat imperfectly in some cases, right up to the point of fracture ; they have no marked elastic limit. It is generally stated in text-books that the shearing strength of brittle materials is much below ~ of the tensile strength, but this is certainly an error, and has probably come about through the use of imperfect shearing tackle, which has caused double shear specimens to shear first through one section, and then through the other. In a large number of tests made in the author's laboratory, the shearing strength of cast iron has come out rather higher than the tensile stress in the ratio of n to i. Shear combined with Tension or Compression. We have shown above that when a block or plate, such as abed, is subjected to a shear, there will be a direct stress acting normally to the diagonal bd. Likewise if the two sides ad, be are subjected to a normal stress, there will be a direct stress acting normally to the section ef\ but when the block is sub- jected to both a direct stress and a shear, there will be a direct stress acting normally to a section occupying an intermediate position, such as gh. Consider the stresses acting on the triangular element shown, which is reproduced from the figure above for clearness. The intensity of the shear stress on the two edges will be equal (see p. 314). Hence The total shear stress on the face gi =f t .gi = Pj ///=/,. /fc* = P direct = /, . n = T Let the resultant stress on the face gh, which we are about to find in terms of the other stresses, be f, t . Then the total direct stress acting normal to the face gh = f it . gh = P . Stress, Strain, and Elasticity. 317 Now consider the two horizontal forces acting on the FIG. 298. element, viz. T and P, and resolve them normally to the face gh as shown, we get T n and P n . 318 Mechanics applied to Engineering. But P n + T n = P + T -f.-M+ft. cos cos 6 alsoP B + T n = P =/^ hence, substituting the above values, we have Next consider the vertical force acting on the element, viz. P 1} and when resolved normally to the face gh, we get P . hav But P! = P sin =f*gh sin =f tt ht or/.=/^i and f -'f = M fa Substituting this value of hi in the equation above, we Completing the square The minus sign would be retained for finding the stress normal to the other diagonal section ij if the rectangle were completed, viz. gihj. Young's Modulus of Elasticity (E). We have already Stated that experiments show that the strain of an elastic body is proportional to the stress. In some elastic materials the Stress, Strain, and Elasticity. 319 strain is much greater than in others for the same intensity of stress, hence we need some means of concisely expressing the amount of strain that a body undergoes when subjected to a given stress. The usual method of doing this is to state the intensity of stress required to strain the bar by an amount equal to its own length, assuming the material to remain Perfectly elastic. This stress is known as Young's modulus of (or measure of) elasticity. We shall give another definition of it shortly. FIG. 299. In the diagram in Fig. 299 we have shown a test-bar of length / between the datum points. The lower end is supposed to be rigidly fixed, and the upper end to be pulled : let a stress- strain diagram be plotted, showing the strain along the vertical and the stress along the horizontal. As the test proceeds we shall get a diagram abed as shown, similar to the diagrams shown on p. 244. Produce the elastic line onward as shown (we have had to break it in order to get it on the page) until the elastic strain is equal to /; then, if x be the elastic strain at any point along the elastic line of the diagram corresponding to a stress/", we have by similar triangles 320 Mechanics applied to Engineering. The stress E is termed " Young's modulus of elasticity," and sometimes briefly " The modulus of elasticity." Thus in tension we might have defined the modulus of elasticity as The stress required to stretch a bar to twice its original length , assuming the material to remain perfectly elastic. It need hardly be pointed out that no constructive materials used by engineers do remain perfectly elastic when pulled out to twice their original length ; in fact, very few materials will stretch much more than the one-thousandth of their length and remain elastic. It is of the highest importance that the elastic stretch should not be confused with the stretch beyond the elastic limit. It will be seen in the diagram above that the part bed has nothing whatever to do with the modulus of elasticity. We may write the above expression thus : Then, if we reckon the strain per unit length as on p. 291, we have j = strain, and we may write the above relation thus : Young's modulus of elasticity = strain Thus Young's modulus is often defined as the ratio of the stress to the strain while the material is perfectly elastic, or we may say that it is that stress at which the strain becomes unity, assuming the material to remain perfectly elastic. The first definition we gave above is, however, by far the clearest and most easily followed. For compression the diagram must be slightly altered, as in Fig. 300. In this case the lower part of the specimen is fixed and the upper end pushed down ; in other respects the description of the tension figure applies to this diagram, and here, as before, we have For most materials the value of E is the same for both tension and compression ; the actual values are given in tabular form on p. 352. Stress, Strain, and Elasticity. 321 Occasionally in structures we find the combination of two or more materials having very different coefficients of elasticity; the problem then arises, what proportion of the total load is FIG. 300. borne by each? Take the case of a compound tension member. Let Ej = Young's modulus for material i ; E 2 = ,, 2 ; Aj = the sectional area of i ; "-2 jj )> 2 ' } fi = the tensile stress in i : / 2 = 2; W = the total load on the bar. Then : W, + Wo since the components of the member are attached together at both ends, and therefore the proportional strain is the same in both ; A,/, AA W, and T ? = -~w = ^f A a E a W y 322 Mechanics applied to Engineering. which gives us the proportion of the load borne by each of the component members ; and W a = A 2 E 2 By similar reasoning the load in each component of a bar containing three different materials can be found. The Modulus of Transverse Elasticity, or the Coefficient of Rigidity (G). The strain or distortion of an element subjected to shear is measured by the slide, x (see p. FIG. 301. 291). The shear stress required to make the slide x equal to the length / is termed the modulus of transverse elasticity, or the coefficient of rigidity, G. Assuming, as before, that the material remains perfectly elastic, we can also represent this graphically by a diagram similar to those given for direct elasticity. In this case the base of the square element in shear is rigidly fixed, and the outer end sheared, as shown. From similar triangles, we have / G _ /_ stress # strain Stress, Strain, and Elasticity. 323 Relation between the Moduli of Direct and Trans- verse Elasticity. Let abcd\>z a square element in a perfectly elastic material which is to be subjected to 2l ., (i) Tensile stress equal to the modulus stress ; then the length / of the line ab will be stretched to 2/, viz. ab-H and the strain reckoned on unit length will be (2) Shearing stress also equal to the modulus stress ; then the length / of the line ab will be stretched to V/ 2 4- / 2 = >/2/ = i'4i/, and the strain reckoned on unit length will be FIG. 302. = 0-41. Thus, when the modulus stress is reached in shear the strain is 0*41 of the strain when the modulus stress is reached in tension ; but the stress is proportional to the strain, therefore the modulus of transverse elasticity is 0*41, or f nearly, of the modulus of direct elasticity. The above proof must be regarded rather as a popular demonstration of this relation than a scientific treatment. The orthodox treatment will be given shortly. Poisson's Ratio. When a bar is stretched longitudinally, it contracts laterally ; likewise when it is compressed longi- tudinally, it bulges or spreads out laterally. Then, terming stretches or spreads as positive (+) strains, and compressions or contractions as negative ( ) strains, we may say that when the longitudinal strain is positive (+), the lateral strain is negative ( ). Let the lateral strain be - of the longitudinal strain. The n fraction - is generally known as Poisson's ratio, although in n reality Poisson's ratio is but a special value of the fraction, viz. \. Strains resulting from Three Direct Stresses acting at Right Angles. In the following paragraph it will 324 Mechanics applied to Engineering. be convenient to use suffixes to denote the directions in which the forces act and in which the strains take place. Thus any force P which acts, say, normal to the face x will be termed P z , and the strain per unit length 4 will be termed S^, and the stress on the facey,.; then Every applied force which produces a stretch or a 4- strain in its own FIG. 303. direction will be termed + , and vice versd. The strains produced by forces acting in the various directions are shown in tabulated form below. FIG. 304- Force acting on face of cube. Strain in direction x. S*. Strain in directions. Sy. Strain in direction . St. P* 1 /* ~E^ A E P, fy En /, E _/* En P, /, E / En 4 Stress, Strain, and Elasticity. 325 Thtese equations give us the strains in any direction due to the stresses / z ,/ y ,/ r acting alone; if two or more act together, the resulting strain can be found by adding the separate strains, due attention being paid to the signs. Shear. We showed above (p. 314) that a shear consists of two equal stresses of opposite sign acting at right angles to one another. The resulting strain can be obtained by adding the strains given in the table above due to the stresses /i andy^, which are of opposite sign and act at right angles to one another. The strains are 4- A = /( i + I s ) in the direction (i) L, nhi hi\ n/ <> Thus the strain in two directions has been increased by - due to the superposition of the two stresses, and has been n reduced to zero in the third direction. If a square abed had been drawn on the side of the element, it would have become the rhombus a'&Jd' after the strain, the FIG. 305. FIG. 306. long diagonal of the rhombus being to the diagonal of the square as i + S to i. The two superposed are shown in Fig. 306. Then we have or 2/ 2 + 2/x + x* = i + 28 + S a 326 Mechanics applied to Engineering. But as the diagonal of the square = i, we have 2/ 2 - I , X And let y = S ; x = /S ; then by substitution we have 2 / 2 + 2 / 2 S + /% 2 = i + 28 + S 2 and i + S + y = i + 28 + S 2 Both S and S are exceedingly small fractions, never more than about YoVo an d their squares will be still smaller, and therefore negligible. Hence we may write the above ! + S = i + 28 But G = - ' ~ ~, - , E 2G 2G(^ + I) hence = ^ FJ, and E = J Hi 2(j n When = 5, G = -^-E = o'42E. n = 4, G = f E = o'4oE. = 3} G = |E = o'38E. n = 2, G = |E = o- 33 E. Some values of n will be given shortly. We have shown above that the maximum strain in an element subject to shear is but the maximum strain in an element subject to a direct stress in tension is s-^ b '~E /(r 4- -^ S EV 1 '^ n) ( T\ hence -^_ T _ = ( I+ -) E / orS = <'+;) Stress, Strain, and Elasticity. 327 S Safe shear stress s safe tensile stress When n 5 i 1 ) > = 4 1 1 ii = 3 4 1 = 2 i Taking = 4, we see that the same material will take a permanent set, or will pass the elastic limit in shear with -| of the stress that it will take in tension ; or, in other words, the shearing strength of a material is only |- of the tensile strength. Although this proof only holds while the material is elastic, yet the ratio is approximately correct for the ultimate strength. Bar under Longitudinal Stress, but with Lateral Strain prevented in one Direction. Let a bar be subjected to longitudinal stress, f m in the direction x, and be free in the direction y t but be held in the direction z. Referring to the table on p. 324 The strain due to f x when = the bar is free laterally ^ in the direction x !/. : FIG. 306^. But since the strain in the direction z is zero, we must impose a stress, f z = - 1 on the face z t sufficient to bring it back to its original position. f f - ^~ in the direction x The strain due to A, ^ or E ^ Ea _ _ fx The result- ing strains due to f x and/ z = 4 ~ sh = i'( ' ~ f,) = T4 ^ diction * _/--/ - -S*{1_L I \= JL/ R5' + W 18 E 328 Mechanics applied to Engineering. Thus the longitudinal strain of a bar held in this manner is only as great as when the bar is free. Bar under Longitudinal Stress with Lateral Strain prevented in both Directions. Let the bar be placed in a close-fitting cylinder, and be loaded axially. The strain when the bar is free in the direction y = E En En . /*,/*_/ En~r E En En But when the bar is held the strain in the directions y and z is zero, and putting/, =/, we have E ~ En or/, = / -/, =/( - i) whence in the manner adopted in the previous case on im- posing stresses/, = ^-^-j, and/ = -^-j, we get for the strain in the direction / FIG. 306*. corresponding to L_ J* M* - n ~ 2 N - i) En(n - i) " E\ n 2 - n ) ** 25* Or the longitudinal compression of bar held in this manner is only as great as when the bar is free. Anticlastic Curvature. When a beam is bent into a circular arc some of the fibres are stretched and some com- pressed (see the chapter on beams), the amount depending upon their distance from the N. A., viz. LL in the figure ; due to the extension of the fibres, the tension side of the beam section contracts, and the compression side extends. Then, the strain at EE on the beam profile, we Stress, Strain y and Elasticity. 329 have - as much at E'E' on the section ; also, the proportional strain , and LL p E'E' - L'L' y -- -_ --- = LL Pi n p hence p x = np pi = 4p This relation only holds when all the fibres are free laterally, which is very nearly the case in deep narrow sections ; but if the section be shallow and wide, as in a flat plate, the layers which would contract sideways are so near to those which would extend sideways, that they are to a large extent prevented from moving laterally j hence the material in a flat wide beam is nearly in the state of a bar prevented from contracting laterally in one direction. Hence the beam is stiffer in the ratio of -s = 4-f than if the section were narrow. Boiler Shell. On p. 346 we show that P. = / _ 2 f . f -f* I* ~~ ~Jx ) Jx FIG. 307. Strains. Let n = 4. FIG. 308. S =- A=//I_lW^- x E En E\a n) T E 33 Mechanics applied to Engineering. Sy = " E^ "" Wn = " E\w + J " ~~ * S = - +*-//!-. ' En ^ E " EV Thus the maximum strain is in the direction S,. By the thin-cylinder theory we have the maximum strain = ^ ; thus the real strain is only f as great, or a cylindrical boiler shell will stand -J- = 1-14 or 14 per cent, more pressure before the elastic limit is reached than is given by ordinary ring theory. Thin Sphere subjected to Internal Fluid Pressure. In the case of the sphere, we have P, = P z ; f x = f t . Strains. *E EnE\ S = _ A-A=-/-(l + r\ s= _:L/L En En E\x n) 2 E c _ /. . /. -// T _ *\ - sf* ~&i E~E\ n)~*E ^ in this case =-^in the case given above;. .'. S z in this case = f X ^ = |^ Maximum strain in sphere -_3_ _ s. maximum strain in boiler shell -| 7 Hence, in order that the hemispherical ends of boilers should enlarge to the same extent as the cylindrical shells when under pressure, the plates in the ends should be f thickness of the plates in cylindrical portion. If the proportion be not adhered to, bending will be set up at the junction of the ends and the cylindrical part. Alteration of Volume due to Stress. If a body were placed in water or other fluid, and were subjected to pressure, its volume would be diminished in proportion to the pressure. Let V = original volume of body ; v - change of volume due to change of pressure ; 8/ pressure. Stress, Strain, and Elasticity. 331 Then K = = change of pressure _ z> change of volume per cubic unit of the body K is termed the coefficient of elasticity of volume. The change of volume is the algebraic sum of all the strains produced. Then, putting / = f x = f y = / z for a fluid pressure, we have, from the table on p. 324, the resulting strains ___ E En E E* E En E E K = EG hence K = v zpn - 6 3#-6 90- 3E 3 K- 2 G The following table gives values of K in tons per square inch also of n : Material. K. n. Water... 140 _ Cast iron 6,000 3-0 to 47 Wrought iron 8,800 Steel ... 11,000 3-6 to 4-6 Brass ... 6,400 3'i to 3-3 Copper Flint glass Indiarubber 10,500 2,400 2-9 to 3-0 3'9 2'O The n given above has not been calculated by the above formula, but is the mean of the most reliable published experiments. --W-* 332 Mechanics applied to Engineering. Riveted Joints. Strength of a Perforated Strip. If a perforated strip of width w be pulled apart in the testing-machine, it will break through the hole, and if the material be only very slightly ductile, the breaking load will be (approximately) 1W = f t (w d)t, where f t is the tensile strength of the metal, and / the thickness of the plate. If *\ the metal be very ductile the breaking load will be higher than this, due to the fact that the tensile strength is always reckoned on the original area of the test bar, and not on the final area at the point of O fracture. The difference between the real and the nominal tensile strength, therefore, depends upon the reduction in area. If we could prevent the area from contracting, we should raise the nominal tensile strength. In a perforated bar the minimum area of the section through the hole is surrounded by W metal not so highly stressed, hence the reduction in FIG. 309. area is less and the nominal tensile strength is greater than that of a plain bar. This apparent increase in strength does not occur until the stress is well past the elastic limit, hence we have no need to take it into account in the design of riveted joints. a Strength of an Elementary Pin Joint. If a bar, perforated at both ends as shown, were pulled apart through the medium of pins, --M/--4 failure might occur through the tearing of the plates, as shown at aa or bb, or by the shearing of the pins themselves. Let d = the diameter of the pins ; c = the clearance in the holes. FIG. 310. Then the diameter of the holes = d + c i = the shearing strength of the material in the pins. For simplicity at the present stage, we will assume the pins to be in single shear. Then, for equal strength of plate and pins, we have Stress, Strain, and Elasticity. 333 If the holes had been punched instead of drilled, a thin ring of metal all round the hole would have been damaged by the rough treatment of the punch. This damaging action can be very clearly seen by ex- amining the plate under a microscope, and its effect demonstrated by testing two similar strips, in one of which the holes are drilled, and in the other punched, the latter breaking at a lower load than the former. J Let the thickness of this damaged ring be ; then the equivalent diameter of the hole will be 5o tons per sq. inch n(d + c) The bearing pressure has been worked out for the joints Stress, Strain, and Elasticity. 341 given in the table above, and in those instances in which it is excessive they have been crossed out with a diagonal line, and the greatest permissible pitch has been inserted. Efficiency of Joints. The efficiency \ st th of joint of a riveted J = - . J , joint j strength of plate _ effective width of metal between rivet-holes pitch of rivets w d c w d ci = , or W ' W The table on the following page gives the efficiency of joints corresponding to the table of pitches given above. All the values are per cent. Zigzag Riveting, In zigzag riveting, if the two rows are placed too near together, the plates tear across in zigzag fashion. If the material of the plate were equally strong with and across the grain, then x^ >Q Q the zigzag distance between the two holes, v /' \ / 5 \ should be -. The plate is, however, ^--^-j 3 O weaker along than across the grain, con- FIG. 328. sequently when it tears from one row to the other it partially follows the grain, and therefore tears more readily. The joint is found to be equally strong in both directions when x^ . 3 Riveted Tie-bar Joints. When riveting a tie bar, a very high efficiency can be obtained by properly arranging the rivets. ^ ^_ , - - w The arrangement shown in Fig. 329 is radically wrong for FIG. 329. tension joints. The strength of such a joint is, neglecting the clearance in the holes, and damage done by punching (w 4^)./t/ at aa 342 Mechanics applied to Engineering. EFFICIENCY OF RIVETED JOINTS. Thick- Iron plates and rivets. Steel plates and rivets. Type of joint. ness of plate. Punched. Drilled. Punched. Drilled. in. i 55 60 52 57 t 52 57 49 54 A. 1 50 54 47 51 * 48 52 45 49 111 foj 1 1 M (a) 66 71 (a) 70 75 <>|3 (a) 64 73 H (a) 66 69 73 ()59 66 (a) 64 70 ! 6 7 70 f()59 I 64 (a) 64 68 J 65 68 ( (a} 62 (a) 64 66 I 63 67 60 64 62 66 59 62 C. 1 , 79 82 75 " 78 | 79 82 75 78 | 79 81 75 78 I 78 So 75 78 I 77 79 74 77 II 76 78 73 76 * 3 90 . . . ' 8 90 89 ';'; M 88 - i 78 1 i 76 3 74 I 73 i 71 4 70 Stress, Strain, and Elasticity. 343 whereas with the arrangement shown in Fig. 330 the strength is (w d)f t t at aa (tearing only) (w id}f t t 4- -if 2 /, a.t bb (tearing and shearing one rivet) 4 (w - *% at cc . at dd ( three rivets) six By assuming some dimensions and working out the strength at each place, the weakest section may be found, which will be far greater than that of the joint shown in Fig. 329. The joint in Fig. 330 will be found to be of approximately equal strength at all the sections, hence for simplicity of calculation it may be taken as being (w - d}f t t In the above expressions the constants c and K have been omitted; they are not usually taken into account in such joints. The working bearing pressure on the rivets should not exceed 8 tons per square inch, and where there is much vibra- tion the bearing pressure should not exceed 6 tons per square inch. For bridge, roof, ship plates, and riveted work other than boilers, it is usual to use smaller rivets for the same thickness of plate. Unwin gives d= riAT in. in. in. in. in. in. in. Thickness of plate I | 1 1 2 I M Diameter of rivet 1 1 7 II i Ij I* Groups of Rivets In the chapter on combined bending and direct stress, it is shown that the stress may be very seriously increased by loading a bar in such a manner that the line of pull or thrust does not coincide with the centre of gravity of the section of the bar. Hence, in order to get the stress evenly distributed over a bar, the centre of gravity of the 344 Mechanics applied to Engineering. group of rivets must lie on the line drawn through the centre of gravity of the cross-section of the bar. And when two bars FIG. 331. not in line are riveted, as in the case of the bracing and the boom of a bridge, the centre of gravity of the group must lie FIG. 33 1. on the intersection of the lines drawn through the centres of gravity of the cross-sections of the two bars. Stress, Strain, and Elasticity. 345 In other words, the rivets must be arranged symmetrically about the two centre lines (Fig. 331). Punching Effects. Although the strength of a punched plate may not be very much less than that of a similar drilled plate, yet it must not be imagined that the effect of the punch- ing is not evident beyond the imaginary ring of y^ inch in thickness. When doing some research work upon this question the author found in some instances that a i-inch hole punched in a mild steel bar 6 inches wide caused the whole of the fracture on both sides of the hole to be crystalline, whereas the same material gave a clean silky fracture in the unpunched bars. Fig. 33 la shows some of the fractures obtained. The punching also very seriously reduces the ductility of the bar; in many instances the reduction of area at fracture in the punched bars was not more than one-tenth as great as in the un- punched bars. These are not isolated cases, but may be taken as the general result of a series of tests on about 150 bars of mild steel of various thicknesses, widths, and diameters of hole. Strength of Cylinders subjected to Internal Pressure. Thin Cylinders. Consider a short cylindrical ring i inch in length, subjected to an internal pressure of / Ibs. square inch. Then the total pressure tending to burst the cylinder and tear the plates at aa and bb is /D, where D is the internal diameter in inches. This bursting pressure has to be resisted by the stress in the ring of metal, which is 2//, where f t is the tensile stress in the material. Then/D = 2// FIG. 33*. or/R =f t t When the cylinder is riveted with a joint having an efficiency 7, we have In addition to the cylinder tending to burst by tearing the plates longitudinally, there is also a tendency to burst 346 Mechanics applied to Engineering. circumferentially. The total bursting pressure in this direction is/7rR 2 , and the resistance of the metal is 27rR# where f t is the tensile stress in the material. Then/rrR 2 = or/R = 2tf t and when riveted /R = Thus the stress on a circumferential section is one-half as great as on a longitudinal section. On p. 329 a method is given for combining these two stresses. The above relations only hold when the plates are very thin; with thick-sided vessels the stress is greater than the value obtained by the thin cylinder formula. Thick Cylinders. Barlow's Theory. When the cylinder is exposed to an internal pressure (/), the radii will be increased, due to the stretching of the metal. When under pressure Let r< be strained to r t + n^\ = r^i + <) + n e r e = r e (i + n.) where n is a small fraction indicating the elastic strain of the metal, which never exceeds yoVo for safe working stresses (see p. 293). The sectional area of the cylinder will be the same (to all intents and purposes) before and after the application of the pressure; hence FIG. 333. which on reduction becomes r?(n* + 2 4 ) n being a very small fraction, its square is still smaller and may be neglected, and the expression may be written _ r " n t Stress, Strain, and Elasticity. 347 The material being elastic, the stress /will be proportional to the strain n ; hence we may write that is, the stress on any thin ring varies inversely as the square of the radius of the ring. Consider the stress / in any ring of radius r and thickness dr and of unit width. The total stress on any section of the ) _ / , elementary ring I - ,. x The total stress on the whole section) _ - 2 j r * _ 2 . of one side of the cylinder \ ~ '?* I r I (Substituting the value of/fa 2 from above) i This total stress on the section of the cylinder is due to the total pressure/;;; hence Substituting the value of/ w we have Dividing by r and reducing, we have pr. =ff, -/* 348 Mechanics applied to Engineering. For a thin cylinder, we have Thus a thick cylinder may be dealt with by the same form of expression as a thin cylinder, taking the pressure to act on the external instead of on the internal radius. The diagram (Fig. 333 abed) shows the distribution of stress on the section of the cylinder, ad representing the stress at the interior, and be at the exterior. The curve dc is/r 2 = constant. Lame's Theory. All theories of thick cylinders indicate that the stress on the inner skin is greater than that on the outer when the cylinder is exposed to an internal pres- sure, but they do not all agree as to the exact distribution of the stress. Consider a thin ring i inch long and of internal radius r t of thickness 8;-. Let the radial pressure on the inner surface of the ring be /, and on the outer surface (p 8fi), when the fluid pressure is internal. In addition to these pressures the ring itself is sub- jected to a hoop stress/, as in the thin cylinder. Each element of the ring is subjected to the stresses as shown in the figure. The force tending to burst this thin ring = p x 2r . . (i.) - P-ent bursting =| These two must be equal = 2pr -f 2/Br (in.) We can find another relation between / and /, which will enable us to solve this equation. The radial stress p tends to squeeze the element into a thinner slice, and thereby to cause it to spread transversely ; the stress f tends to stretch the element in its own direction, and causes it to contract in thickness and normal to the plane of the paper. Consider the strain of the element normal to the paper ; due to p it is ~ Stress, Strain, and Elasticity. 349 (see p. 324), and due to /it is ~, and the total longitudinal strain normal to the paper is -=4 - -%. Both /and/, however, n\L nhi diminish towards the outer skin, and the one stress depends upon the other ; hence we shall assume that the longitudinal strain is constant over the section, or that originally plane sections remain plane. In Lame's complete treatment it is proved that such is the case. Now, as regards the strain in the direction/ both pressures /and/ act together, and on the assumption just made /-}-/ = a constant = 20, The 2 is inserted for convenience of integration. From (iii.) we have p 20, 2p dp . .. .^ f = - = -f in the limit $r r r dr Integrating (see Perry's " Calculus for Engineers," p. 89), we get where a and b are constants. Let the internal pressure above the atmosphere be /<, and the external pressure p e = o, we have the stress on the inner skin Reducing Barlow's formula to the same form of expression, we get Experimental Determination of the Distribution of Stress in Thick Cylinders. In order to ascertain whether there was any great difference between the distribution of stress as indicated by Barlow's and by Lamp's theories, the 350 Mechanics applied to Engineering. author, assisted by Messrs. Wales, Day, and Duncan, students, made a series of experiments on two cast-iron cylinders with open ends. Well-fitting plugs were inserted in each end, and the cylinder was filled with paraffin wax ; the plugs were then forced in by the loo-ton testing machine, a delicate extenso- meter, capable of reading to Yo^oTo f an mc ^> was pressure pressure By experiment + Lame's theory Barlow's theory * Cylinder W&tt FIG. 333*. the outside; a series of readings were then taken at various pressures. Two small holes were then drilled diametrically into the cylinder to a depth of 0*5 inch; pointed pins were loosely inserted, and the extensometer was applied to their outer ends, and a second set of observations were taken. The holes were then drilled deeper, and similar observations taken Stress, Strain, and Elasticity. 351 at various depths. From these observations it is .a simple matter to deduce the proportional strain and the stress. The results obtained are shown in Fig. 333$, and for purposes of comparison, Lame's and Barlow's curves are inserted. Built-up Cylinders. In order to equalize the stress over the section of a cylinder or a gun, various devices are adopted. In the early days of high pressures, cast-iron guns were cast round chills, so that the metal at the interior was immediately cooled ; then when the outside hot metal contracted, it brought the interior metal into com- pression. Thus the initial stress in a section of the gun was somewhat as shown by the line ab, ag being compression, and bh tension. Then, when subjected to pressure, the curve of stress would have been dc FlG - 334- as before, but when combined with db the resulting stress on the section is represented by ef, thus showing a much more even distribution of stress than before. This equalizing process is effected in modern guns by either shrinking rings on one another in such a manner that the internal rings are initially in compression and the external rings in tension, or by winding wire round an internal tube to produce the same effect. The exact tension on the wire re- quired to produce the desired effect is regulated by drawing the wire through friction dies mounted on a pivoted arm in effect, a friction brake. 352 Mechanics applied to Engineering. STRENGTH AND COEFFICIENTS OF ELASTICITY OF MATERIALS IN TONS SQUARE INCH. Elastic limit. Breaking strength. t: tl Si . . E. jj^. c Material. .2 | Tension or com- G. Shear. go c o 1 s . .2 I . pression. c ti 3 I I 1 1 J 1 Si & 3 Wrought - iron) bars > 12-15 _ 10-12 a 1-24 17-19 f 1 1,000- (13,000 5,000) 6, ooo j 10-30 X5-40 Plates with grain 13-15 20-22 5-10 7-12 across 11-13 . 1 8-20 __ ^ _ __ 2-6 3~7 Best Yorkshire,) with grain ... J 12-14 20-23 17-19 12,000 - 15-30 40-50 Best Yorkshire,) across grain 5 1314 19-20 - - 10-12 12-20 Steel, o-x^C. ... 1314 IO II 21-22 16-17 ( 13,000 (14,000 5,000) 6, ooo 5 27-30 45-50 0'2^>C. .. 17-18 16-17 13-14 30-32 24-26 H ,, 20-23 27-32 ,. 0*5 i^c. .. 20-21 16-17 34-35 28-29 1, ,, 14-17 17-20 x'o^C. .. 28-29 22-24 21-23 50-55 42-47 I4,OOO ,, 4-5 7-8 Rivet steel 15-17 12-14 26-28 21-22 20-35 30-50 Steel castings .. Tool stee'I 10-11 15-17 35-45 (not neal ann 40-50 an- ed) ealed) 20-25 30-40 40-70 \ ^ (unh ardT | 12,000 i t0 I 12,500 (14,000 }~-\ 5-12 10-20 6-13 15-35 en ed) 1 t0 > K n i) 60-80 60-80 (hard ened) " I5,OOO ) t nil nil Cast iron (no mar lim ked) 7-ix 35-60 8-13 i 6,000 \ t0 ( 10,000 2,500) to \ 4,000 ; pra cally cti- nil Copper 2-4 10-13 anne ale7 12-15 20-25 II-X2 ( 7,000 \ to }~ 35-40 50-60 fj IO 12 hard drawn 16-20 ( 7,500 } 2-5 40-55 Gun-metal 3-4 5-6 2'5-S 9-16 8-12 C 5,000- ( 5,500 2,000) 2,500) 8-15 10-18 30-50 Brass 2-4 __ __ 7-10 5-6 4,000 2,000 10-12 12-15 Delta, bull metal, etc. Cast 5-8 12-14 14-20 60-70 ( 5,5oo 1 to } ~ 8-16 10-22 Rolled Phosphor bronze 15-25 7-9 16-22 27-34 24-20 45-6o ( 6,000 6,000 2,500 17-34 27-50 Muntz metal 20-25 (roll- 25-30 10-20 30-40 ed) Aluminium 2-7 8-10 7-10 - 5-6 C 3,000 ( S, 000 1 1,700 4-15 30-70 f with 1 Oak - 4-6 s-5 1 grain 0'2 j- 500-700 Soft woods - - - 1-3 -3 (0-4 V 450-500 CHAPTER IX. BEAMS. THE beam illustrated in Fig. 3340 is an indiarubber model used for lecture purposes. Before photographing it for this illustration, it was painted black, and some thin paper was stuck on evenly with seccotine. When it was thoroughly set the paper was slightly damped with a sponge, and a weight was placed on the free end, thus causing it to bend ; the paper on the upper edge cracked, indicating tension, and that on the lower edge buckled, indicating compression, whereas between the two a strip remained unbroken, thus indicating no longi- FIG. tudinal strain or stress. We shall see shortly that such a result is exactly what we should expect from the theory of bending. General Theory. The T lever shown in Fig. 335 is hinged at the centre on a pivot or knife-edge, around which the lever can turn. The bracket supporting the pivot simply takes the shear. For the lever to be in equilibrium, the two couples acting on it must be equal and opposite, viz. W/ = px. Replace the T lever by the model shown in Fig. 336. It 2 A 354 Mechanics af plied to Engineering, is attached to the abutment by two pieces of any convenient material, say indiarubber. The upper one is dovetailed, be- cause it is in tension, and the lo\ver is plain, because it is in FIG. 335- FIG. 336. compression. Let the sectional area of each block be a\ then, as before, we have W/ = /.ar. But/ =/0, where /= the stress in either block in either tension or compression ; Or The moment of) the external! = forces j hence W/ = fax or = 2/ay !the moment of the internal forces, or the internal moment of resistance of the beam stress on the area a x (moment of the two areas (a) about the pivot) Hence the resistance of any section to bending apart alto- gether from the strength of the material of the beam varies directly as the area a and as the distance x, or as the moment ax. Hence the quantity in brackets is termed the " measure of the strength of the section," or the " modulus of the section," and is usually denoted by the letter Z. Hence we have W/ = M =/Z, or I modulus of the section The bending moment { _ (stress on the) at any section ( "~ ( material f \ The connection between the T lever, the beam model ot Fig. 336, and an actual beam may not be apparent to some readers, so in Fig. 3360 we show a rolled joist or I section, having top and bottom flanges, which may be regarded as the two indiarubber blocks of the model, the thin vertical web serves the purpose of a pivot and bracket for taking the shear ; then the formula that we have just deduced for the model applies equally well to the joist. We shall have to slightly ESS FIG. 336^1. Beams. 355 modify this statement later on, but the form in which we have stated the case is so near the truth that it is always taken in this way for practical purposes. Now take a fresh model with four blocks instead of two. When loaded, the outer end will droop down as |^| (jy) shown by the dotted lines, pivoting about the point resting on the bracket. Then the outer blocks will be stretched and compressed, or strained, more than the inner blocks in the ratio FIG. 337. 3C V or -^ ; i.e. the strain is directly proportional to the distance from the point of the pivot. The enlarged figure shows this more distinctly, where e, e 1 show the extensions, and ^ etc -> and situated at distances y, y^ y* y*"> etc ' respectively from the pivot, which we shall now term the neutral axis, we should have, as above W/ = 2fay y y -f a,y^ 4- , etc. -f , etc.) FIG. 339. The quantity in brackets, viz. each little area (a) multiplied by the square of its distance (y" 2 ) from a given line (N.A.), is termed the second moment or moment of inertia of the upper portion of the beam section ; and as the two half-sec- tions are similar, twice the quantity in brackets is the moment of inertia (I) of the whole section of the beam. Thus we have W/ = , or y The bending moment at any section the stress on the ) ( second moment, or moment of outermost layer \ \ inertia of the section distance of the outermost layer from the neutral axis But we have shown above that the stress varies directly as the distance from the neutral axis ; hence the stress on the outer- most layer is the maximum stress on any part of the beam section, and we may say The bending moment at any section the maximum stress) J second moment, or moment of on the section ( \ inertia of the section distance of the outermost layer from the neutral axis Beams. 357 But we have also seen that W/ = /Z, and here we have W/=:/- ; J y therefore Z = - The quantity - is termed the " measure of the strength of the section," or, more briefly, the " modulus of the section ; " it is usually designated by the letter Z. Thus we get W/ = /Z, or M =/Z, or m I jthe maximum or skin) ^ I the modulus of any section! = I stress on the section > 1 the secti Assumptions of Beam Theory. To go into the question of all the assumptions made in the beam theory would occupy far too much space. We will briefly consider the most important of them. First Assumption. That originally plane sections of a beam remain plane after bending; that is to say, we assume that a solid beam acts in a similar way to our beam model in Fig. 337, in that the strain does increase directly as the distance from the neutral axis. Very delicate experiments by the author show that this assumption is true to within exceedingly narrow limits, provided the elastic limit of the material is not passed. Second Assumption. That the stress in any layer of a beam varies directly as the distance of that layer from the neutral axis. That the strain does vary in this way we have just seen. Hence the assumption really amounts to assuming that the stress is proportional to the strain. Reference to the elastic curves on p. 294 will show that the elastic line is straight, i.e. that the stress does vary as the strain. In most cast materials the line is unquestionably slightly curved, but for low (working) stresses the line is sensibly straight. Hence for working conditions of beams we are justified in our assumption. After the elastic limit has been passed, this relation entirely ceases. Hence the beam theory ceases to hold good as soon as the elastic limit has been passed. Third Assumption. That the modulus of elasticity in tension is equal to the modulus of elasticity in compression. Suppose, in the beam model, we had used soft rubber in the 358 Mechanics applied to Engineering. tension blocks and hard rubber in the compression blocks, i.e. that the modulus of elasticity of the tension blocks was less than the modulus of elasticity of the compression blocks ; then the stretch on the upper blocks would be greater than the compression on the lower blocks, with the result that the beam would tend to turn about some point other than the pivot, Fig. 340 ; and the relations given above entirely cease to hold, for the strain and the stress will not vary directly as the distance from the pivot or neutral axis, but directly as the distance from a, which later on we shall see how to calculate. For most materials there is no serious error in making this assumption ; but in some materials the error is appreciable, but still not sufficient to be of any practical importance. Neither of the above assumptions are perfectly FIG. 340. true; but they are so near the truth that for all practical purposes they may be considered to be perfectly true, but only so long as the elastic limit of the material is not passed. In other parts of this book (see Appendix), experimental proof will be given of the accuracy of the beam theory. Graphical Method of finding the Modulus of the Section (Z = -). The modulus of the section of a beam y might be found by splitting the section up into a great many 2 layers and multiplying the area of each by ~, as shown above. The process, however, would be very tedious. But in the graphic method, instead of dealing with each strip separately, we graphically find the magnitude of the resultant of all the forces, viz. fa +/!! i-fyh -f, etc. acting on each side of the neutral axis, also the position or distance apart of these resultants. The product of the two gives us the moment of the forces on each side of the neutral axis, and the sum of the two moments gives us the total amount of resistance for the beam section, viz./Z. Imagine a beam section divided up into a great number of thin layers parallel to the neutral axis, and the stress in each layer varying directly as its distance from it. Then if we con- struct a figure in which the width, and consequently the area, Beams. 359 of each layer is reduced in the ratio of the stress in that layer to the stress in the outermost layer, we shall have the intensity of stress the same in each. Thus, if the original area of the layer be a i} the reduced area of the layer will be ^/-say*; whence we havey^ = /#/ Then the sum of the forces acting over the half section, viz. 4- , etc. becomes fa +/ one side of the neutral axis The distance d of the c. of g. of the modulus! _ a y 4- figure from the neutral axis (see p. 58) / ~" a -f But W/=/Z =/X or W/ = 2f(a + a,') X -~~r = 2/ay + 2> 1 > 1 which is the same expression as we had on p. 355. The graphic method of finding Z should only be used when a convenient mathematical expression cannot be obtained. Position of Neutral Axis. We have stated above that the neutral axis in a beam section corresponds with the pivot in the beam model ; on the one side of the neutral axis the material is in tension, and on the other side in compression, and at the neutral axis, where the stress changes from tension to compression, there is, of course, no stress (except shear, which we will treat later on). In all calculations, whether graphic or otherwise, the first thing to be determined is the position of the neutral axis with regard to the section. We have already stated on p. 58 that, if a point be so chosen in a body that the sum of the moments of all the gravitational forces acting on the several particles about the one side of any straight line passing through that point, be equal to the sum of the moments on the other side of the line, that point is termed the centre of gravity of the body ; or, if the moments on the one side of the line be termed positive ( + ), and the moments on the other side of the line be termed negative ( ), the sum of the moments will be zero. We are about to show that precisely the same definition may be used for stating the position of the neutral axis ; or, in other words, we are about to prove that, accepting the assumptions given above, the neutral axis invariably passes through the centre of gravity of the section of a beam. 362 Mechanics applied to Engineering. Let the given section be divided up into a large number of strips as shown Let the areas of the strips above the neutral axis be a lt a 2 , a 3 , etc.; i ditto below the neutral axis be #/, a t y f> \ aj, etc. ; and their respective distances from the neutral axis above j' 1 ,_y 2 ,j 3 , ~ etc. ; ditto below jy/, y 2 , y.J, etc. : and the stresses in the several layers above the neutral axis be/,/2,/3, etc.; ditto below the neutral axis be//,/ 2 ',/ 3 '. FlG> 343> Then, as the stress in each layer varies directly as its distance from the neutral axis, we have i also = 4 or^ = & = , etc. = say K (a constant) * y* y\ y* y then/= Ky The total stress in any layer = fa Kay The total stress in all the j layers on one side of the > = K(ay + a^ + a^y 2 +, etc.) neutral axis J The total stress in all the ^ layers on the other side 1 = K(#'/ -f a^yf + ayl +, etc.) of the neutral axis j But as the tensions and compressions form a couple, the total amount of tension on the one side of the neutral axis must be equal to the total amount of compression on the other side; hence lL(ay -h a,y, + a 2 y 2 +, etc.) = K(0'/ -f aty + a 2 'y.J +, etc.) or ay + a 1 y l + a^y 2 +, etc. = a'y' + a^ + a 2 y 2 +, etc.) or, expressed in words, the sum of the moments of all the ele- mental areas on the one side of the neutral axis is equal to the sum of the moments on the other side of the neutral axis ; but, referring to the statement above, it will be seen that this is precisely the definition of a line which passes through the centre of gravity of the section. Hence, the neutral axis passes through the centre of gravity of the section. It should be noticed that not one word has been said in the above proof about the material of which the beam is made ; all Beams. 363 that is taken for granted in the above proof is that the modulus of elasticity in tension is equal to the modulus of elasticity in compression. The position of the neutral axis has nothing whatever to do with the relative strengths of the material in tension and compression, as is so frequently stated in text-books, and by correspondents in engineering journals. Unsymmetrical Sections. In a symmetrical section, the centre of gravity is equidistant from the skin in tension and compression ; hence the maximum stress on the material in tension is equal to the maximum stress in compression. Now, some materials, notably cast iron, are from five to six times as strong in compression as in tension; hence, if we use a symmetrical section in cast iron, the material fails on the tension side at from \ to the load that would be required to make it fail in compression. In order to make the beam equally strong in tension and compression, we make the section of cast-iron beams of such z.form that the neutral axis is about five or six times * as far from the compression flange as from the tension flange, so that the stress in compression shall be five or six times as great as the stress in tension. It should be particularly noted that the reason why the neutral axis is nearer the one flange than the other is entirely due to the form of the section, and not to the material ; the neutral axis would be in precisely the same place if the material were of wrought iron, or lead, or stone, or timber (provided assumption 3, p. 357, is true). We have shown above on p. 356 that M =/-, and that Z = -, where y is the distance of the skin from the neutral axis. y In a symmetrical section there is no difficulty in finding the value of/ it is simply the half depth of the section ; but in the unsymmetrical section y may have two values: the distance of the tension skin from the neutral axis, or the distance of the compression skin from the neutral axis. Which, then, are we to choose ? If the maximum tensile stress/, is required, the y t must be taken as the distance of the tension skin from the neutral axis ; and likewise when the maximum compressive stress f c is required, the y e must be measured from the com- pression skin. Thus we have either M = f* or/ c L yt y c 1 We shall show later on that such a great difference as 5 or 6 is undesirable for practical reasons. 3^4 Mechanics applied to Engineering. ft /c and as = , we get precisely the same value for the bending moment whichever we take. We also have two values of Z, viz.- = Z, and- = Z c ; Jt Jo andM =f&oifZ 9 We shall invariably take f t Z t when dealing with cast-iron sections, mainly because such sections are always designed in such a manner that they fail in tension. The construction of the modulus figures for such sections is a simple matter. Compression base, Vne> FIG. 344. Construction for Z c (Fig. 344). Find the centre of gravity of the section, and through it draw the neutral axis parallel with the flanges. Draw a compression base-line touching the outside of the compression flange ; set off the tension base-line parallel to the neutral axis, at the same distance from it as the compression base-line, viz. y c ; project the parts of the section down to each base-line, and join up to the central point which gives the shaded figure as shown. Find the centre of gravity of each figure (cut out in cardboard and balance). Let D = distance between them ; then Z c = shaded area above or below the neutral axis X D. Construction for Z t (Fig. 345). Proceed as above, only the tension base-line is made to touch the outside of the tension flange, and the compression base-line cuts the figure ; the parts of the section above the compression base-line have been pro- jected down on to it, and the modulus figure beyond it passes Beams. 365 outside the section ; at the base-lines the figure is of the same width as the section. The centre of gravity of the two figures is found as before, also the Z. The reason for setting the base-lines in this manner will be evident when it is remembered that the stress varies directly as the distance from the neutral axis; hence, the stress on the tension flange /, is to the stress on the compression flange f as y t is to y c . N.B. The tension base-line touches the tension flange when the figure is being drawn for the tension modulus figure Z,, and similarly for the compression. As the tensions and compressions form a couple, the total amount of tension is equal to the total amount of compression, therefore the area of the figure above the neutral axis must be equal to the area of the figure below the neutral axis, whether the section be symmetrical or otherwise; but the moment of the tension is not equal to the moment of the compression about the neutral axis in unsymmetrical sections. The accuracy of the drawing of modulus figures should be tested by measuring both areas ; if they only differ slightly (say not more than 5 per cent.), the mean of the two may be taken ; but if the error be greater than this, the figure should be drawn again. If, in any given instance, the Z c has been found, and the Z t is required or vice versa, there is no need to construct the two figures, for Z, = Z c X -^ c , or Z c = Z, x y t y c hence the one can always be obtained from the other. Most Economical Sections for Cast-iron Beams. Experiments by Hodgkinson and others show that it is un- desirable to adopt so great a difference as 5 or 6 to i between the compressive and tensile stresses. This is mainly due to the fact that, if sections were made with such a great difference, the tension flange would have to be very thick or very wide com- pared with the compression flange ; if a very thick flange were used, as the casting cooled the thin compression flange and web would cool first, and the thick flange afterwards, and set up serious initial cooling stresses in the metal. The author, when testing large cast-iron girders with very unequal flanges, has seen them break with their own weight before any external load was applied, due to this cause. Very wide flanges are undesirable, because they bend transversely when loaded, as in Fig. 346. Mechanics applied to Engineering. Experiments appear to show that the most economical section for cast iron is obtained when the proportions are roughly those given by the figures in Fig. 346. "Massing up" Beam Sections. Thin hollow beam / 9 /'S 32 f f /S . ^ FIG. 346. FIG. 347. sections are usually more convenient to deal with graphically if they are " massed up " about a centre line to form an equiva- lent solid section. " Massing up " consists of sliding in the sides of the section parallel to the neutral axis until they meet as shown in Fig. 347. The dotted lines show the original position of the sides, and the full lines the sides after sliding in. The " massing up " process in no way affects the Z, as the distance of each section from the neutral axis remains unaltered ; it is done merely for convenience in drawing the modulus figure. In the table of sections several instances are given. Section. Rectangular. Square. Examples of Modulus Figures. FIG. 348. B = H = S (the side of the square) Beams. 367 Modulus of the section Z. BH 2 6 (Square) !. 6 REMARKS. The moment of inertia (sec p. 80) = BH 3 12 _H BJP 12 BH* Also by graphic method BH: The area A = H H BH II BH 2 Z = 2A^/ = 2 X X = f- 430 368 Mechanics applied to Engineering. Section. Examples of Modulus Figures. Hollow rect- angles and girder sections. One corruga- tion of a trough flooring. Beams. 369 Modulus of the section Z. EH 3 - bh* 6H Approximate- ly, when the web is thin, as in a rolled joist B/H where t is the mean thickness of the flange. TOTTJ Moment of inertia for outer rectangle = bh* Z = , hollow H 2 EH 3 - bh* 12 H 2 12 BH 3 - bh\ 12 BH 3 - 6H This might have been obtained direct from the Z for the solid section, thus TDTT2 Z for outer section = bh* h bh* Z,, mner. = y x - = . BH 2 M 3 BH 3 -M 3 Z,, hollow,,- :___.___ The Z for the inner section was multiplied by the ratio ^j, because the stress on the interior of the flange is less than the stress on the exterior in the ratio of their distances from the neutral axis. The approximate methods neglect the strength of the web, and assume the stress evenly distributed over the two flanges. For rolled joists B/H is rather nearer the truth than B/H , where H is measured to the middle of the flanges, and is more readily obtained. For almost all practical purposes the approximate method is sufficiently accurate. N.B. The safe loads given in makers' lists for their rolled joists are usually too high. The author has tested some hundreds in the testing-machine on both long and short spans, and has rarely found that the strength was more than 75 per cent, of that stated in the list. In corrugated floorings or built-up sections, if there are rivets in the tension flanges, the area of the rivet-holes should be deducted from the BA Thus, if there are n rivets of diameter d in any one cross-section, the Z will be (B - nd)tH (approx.) 2 B 370 Mechanics applied to Engineering. Section. Square on edge. Examples of Modulus Figures. FIG. 351. Tees, angles, and LJ sections. FIG. 357. .3, I This figure j H becomes a T or -I L when massed \ up about a ver- tical line. r-J . FIG. 353. Beams. Modulus of the section Z. c-Ii8S s S 4 The moment of inertia (see p. 88) = y- B,TT, 3 +B 2 H 2 3 -/'// 3 PC 11!= o'yH approx. H 2 = o- 3 H The moment of inertia for \ __ B J V the part above the N.A.J ~~ 3 B 2 H 2 3 - bh z Ditto below = - - - B,!^ 3 + B 2 IT, 3 - Ditto for whole section = L * ' - = 7 f Z for stress at top = -^ - bh z If the position of the centre of gravity be calculated for the form of section usually used, it will be found to be approximately 0-3!! from the bottom. Rolled sections, of course, have not square corners as shown in the above sketches, but the error involved is not material if a mean thickness be taken. 3/2 Mechanics applied to Engineering. Section. T section on edge and cruciform sec- tions. Unequal flanged sec- tions. Examples of Modulus Figures. FIG. 354. **B? -- h FIG. 355. Compression base lint Tension base line FIG. 356. Modulus of the section Z. t>n 3 + B/E 3 6H Approx. Z Beams. Moment of inertia for vertical part = horizontal = ,, whole section = Z for = 373 12 B/fe 3 12 12 6H Or this result may be obtained direct from the moduli of the two parts of the section. Thus 7TT2 Z for vertical part = -7- "DLt i horizontal = x - ( see p. 369) It should be observed in the figure how very little the horizontal part of the section adds to the strength. The approximate Z neglects this part. The strength of the T section when bent in this manner is very much less than when bent as shown in the previous figure. 3H, Moment of inertia) _ ^H^ for upper part / ~~ 3 Ditto lower part = ?*!!* 3 Moment \ of inertial B 1 for whole j section J Z = j^ for the stress on the tension flange 374 Mechanics applied to Engineering. Section. Examples of Modulus Figures. This figure be- comes the same as "' the last \vhen massed H z about a centre line. -v T J , fl. ^ - ^ HZ B FIG. 357. Triangle. FIG. 358. Trapezium. B FIG. 359. Beams. 375 Modulus of the section Z. The construction given in Fig. 355 is a con- venient method of finding the centre of gravity of such sections. Where ab area of web, cd area of top flange, ef area of bottom flange, gh ab + cd. The method is fully described on p. 63. For stress at base- BH 2 12 For stress at apex- BH 8 24 The moment of inertia for aj _ BH 3 ^ g . triangle about its c. of g. / ~~ 36 H 211 y for stress at base = , at apex = - - 3 $ BH 3 Z for stress at bcse = --y "" = BH 2 12 3 BH* apex = 24 For stress at wide side BH 2 12 inertia 36 + f?) g6) 2+ for stress at wide side = H x = ( 2n j" ' ) 3 \ + i y Z for stress at wide side = BH a 36 For stress at narrow side Approximate value forZ + i + 4" + I "\ I / 2 + _^ for stress at narrow side = H 2 = I : 1 3 \ n + I / and dividing as above, we get the value given in the column. The approximate method has been described on p. 86. It must not be used if the one side is more than twice the length of the other. For error in- volved, see also p. 87. 376 Mechanics applied to Engineering. Section. Circle. Examples of Modulus Figures. FIG. 360. Hollow circle and corrugated section. DDi Beams. 377 Modulus of the section Z. TrD 3 32 D 3 or 10*2 The moment of inertia of a circle \ _ ff P 4 about a diameter (see p. 88) / ~ 64 D '=7 64 irD' = = ir 2 ir(D 4 - 320 The moment of inertia of a hollow circle ) about a diameter (see p. 88) / 7T(D 4 " Z = -- Pi 4 ) This may be obtained direct from the Z thus Z for outer circle = 32 . inner = - v W (see p. 36 hollow = 32D For corrugated sections in which the corrugations are not perfectly circular, the error involved is very slight if the diameters D and Di are mea- sured vertically. The expres- sion given is for one corruga- tion. It need hardly be pointed out that the corrugations must not be placed as in Fig. 3620. FlG - 362*. The strength, then, is simply that of a rectangular section of height H = thickness of plate. 378 Mechanics applied to Engineering. Irregular sections. Bull-headed rail. Flat-bottomed rail. Examples of Modulus Figures. FIG. 363. FIG. 364. Tram rail (distorted). FIG. 365. Irregular sections. Bulb section. Beams. Examples of Modulus Figures. 379 FIG. 366. Hobson's patent floor- ing. Fireproof flooring. One section. Four sections massed up. FIG. 367. FIG. 368. 380 Mechanics applied to Engineering. Irregular sections. Fireproof flooring. Table of hydraulic press. Examples of Modulus Figures. FIG. 369. FIG. 370. Beams. 381 Shear on Beam Sections. In the Fig. 371 the rect- angular element abed on the unstrained beam becomes a'b' . rl 313) that the shear stress along any two parallel sides of a rectangular element is equal to the shear stress along the other two parallel sides, hence the shear stress W along cd is also equal to -r FIG. 371. Fic/372. Solid timber beams The shear on vertical planes tends % to make the various parts of the ^ beam slide downwards as shown in Fig. 372, a, but the shear on the horizontal planes tends to make the '^ parts of the. beam slide as in Fig. '^ 3 7 2 , . This action may be illustrated % by bending some thin strips of wood, when it will be found that they slkle over one another in- the manner shown, often fail in this manner when tested. In the paragraph above s we assumed that the shear stress was evenly distributed over the section; this, how- ever, is far from being the case, for the shearing force * at any part of a beam section is the algebraic sum of the shearing forces acting on either side of that part of the section (see p. 397). We will now work out one or two cases to show the distribution of the shear on a beam X I ( UN OF THE UNIVERSITY 1 382 Mechanics applied to Engineering. section by a graphical method, and afterwards find a mathe- matical expression for the same. In Fig. 373 the distribution of stress is shown by the width of the modulus figure. Divide the figure up as shown into strips, and construct a figure at the side on the base-line aa, the ordinates of which represent the shear at that part of the section, i.e. the sum of the forces acting to either side of it, thus The shear at i is zero 2 is proportional to the area of the strip between i and 2 = cd on a given scale. ,, 3 is proportional to the area of the strip between i and 3 = efon a given scale. 4 is proportional to the area of the strip between i and 4 = gh on a given scale. ,, 5 is proportional to the area of the strip between i and 5 = ij on a given scale. 6 is proportional to the area of the strip between i and 6 = kl on a given scale. Let the width of the modulus figure at any point distant y from the neutral axis = b then BY by the shear at y = -- - the shear My = - - = kl - (/) in the figure */ = 5X ff, = JL(y) Thus the shear curve is a parabola, as the ordinates//!, etc., vary as y z ; hence the maximum ordinate kl = f (mean ordinate) (see p. 30), or the maximum shear on the section is f of the mean shear. In Figs. 374, 375 similar curves are constructed for a circular and for an I section. It will be observed that in the I section nearly all the shear is taken by the web ; hence it is usual, in designing plate girders of this section, to assume that the whole of the shear is taken by the web. - The outer line in Fig. 375 shows the total shear and the inner figure the intensity of shear at the different layers. The W shear at any section (Fig. 376) ab = , and the intensity of Beams. 383 W shear on the above assumption = p, where K w = the sectional W area of the web, or the intensity of shear = -. But the intensity of shear stress on aa l = the intensity of shear stress FiG. 374- FIG. 375- on ab, hence the intensity of the shear stress between the web and flanee is also = . We shall make use of this when working out the requisite spacing for the rivets in the angles between the flanges and the web of a plate girder. In all the above cases it should be noticed that the shear stress is a maximum at the neutral plane, and the total shear JVffictraZpltme I ju. 15= A d= __! FIG. there is equal to the total direct tension or compression acting above or below it. 384 Mechanics applied to Engineering. We will now get out an expression for the shear at any part of a beam section. We have shown a circular section, but the argument will be seen to apply equally to any section. Let b breadth of the section at a distance y from the neutral axis ; F = stress on the skin of the beam distant Y from the neutral axis ; / = stress at the plane b distant y from the neutral axis ; M = bending moment on the section cd ; I = moment of inertia of the section ; S = shear force on section cd. The area of the strip distant y\ _ , , from the neutral axis ) ~~ * ? the total force acting on the strip f .b .dy Substituting the value of /in the above, we have F FI F M But M = Y' or Y T ( see P< F Substituting the value of y in the above, we have M the total force acting on the strip = -^b .y . dy But M = S/ (see p. 399) Substituting in the equation above, we have -j-b.y.dy Dividing by the area of the plane, viz. b . /, we get The intensity of shearing stress on that plane = rr, \ b .y .dy y the maximum value at the neutral axis,l _ S IB Beams. 385 g the mean intensity of shear stress = -r A. where A = the area of the section. The ratio of the maximum 1 intensity to the mean J K = A b.y.dy The value of K is easily obtained by this expression for geometrical figures, but for such sections as tram-rails a graphic solution must be resorted to. Y The value of I b . y . dy is the sum of the moments of all the small areas b . dy about the N.A., between the limits of y - o, i.e. starting from the N.A., and y = Y, which is the moment of the shaded area Aj about the N.A., viz. A^, where Y c is the distance of the centre of gravity of the shaded area from the N.A. Since the neutral axis passes through the centre of gravity of the section, the above quantity will be the same whether the moments be taken above or below the N.A. Deflection due to Shear. The shearing action of a load on a beam causes the deflection to be greater than it otherwise would be. In the figure the beam is supposed to be subject to shear only, which deflects it an where K is the ratio of the maximum to the mean shear stress (see last paragraph) ; A = sectional area of beam. Hence 2 C 386 Mechanics applied to Engineering. IT" If T I KW/ for a cantilever ALr For a rectangular section, K = 1*5, A = BH and x = ^|j>, for a cantilever >riCj The deflection due to bending is (see p. 428) \W __ W/ 3 3EI 3 X |G X (see pages 326 and 80) 12 8W/ 3 8 = 3 for a cantilever 5Lr>rl The total deflection by this theory of a loaded cantilever is A = Inserting the values for 8 and x, and putting G = f E, get we similarly for a beam centrally loaded. 4 8EI V 6 B In arriving at these results we have assumed that the shear strain is due to the maximum shear stress at the neutral plane, which is only a convenient rough approximation, and which gives too high a result ; on the other hand, if we took the mean shear stress on the whole section, we should get too low a result. The true value is very nearly the arithmetical mean of the two (see Perry, "Applied Mechanics," p. 462). In solid beams the deflection due to the shear is quite a negligible quantity ; for example, in a centrally loaded beam of rectangular section, we have deflection due to shear _ x _ ^H^ deflection due to bending = S = 3 75 / J TT The value of -j may be taken as ^ which gives the shear deflection as 2*6 per cent, of the bending deflection. Beams. 387 In the case of plate web sections, rolled joists, braced girders, etc., the deflection due to the shear is by no means negligible. In text-books on bridge work it is often stated that the central deflection of a girder is always greater than that calculated by the usual bending formula, on account of the " give " in the riveted joints between the web and the flanges. It is probable that a structure may take a " permanent set " due to this cause after its first loading, but after this has once occurred, it is unreasonable to suppose that the riveted joints materially affect the deflection; indeed, if one calculates the deflection due to both bending and shear, it will be found to agree well with the observed deflection. Bridge engineers often use the ordinary deflection formula for bending, and take a lower modulus of elasticity (about 9000 to 10,000 tons square inch) to allow for the shear. EXPERIMENTS ON THE DEFLECTION OF I SECTIONS. E. Depth of Section. Span L. section H. From 3. From A. Rolled joist 28" 6" 7,I2O 12,300 55 5 36" 6" 8,760 12,300 5 5 42" 6" 9,290 12,400 >5 5 5 6" 6" 10,750 12,700 55 5 60" 6" II,2OO 12,800 Riveted girder 60' 5' 10,200 I2,2OO ' 5 60' 6' 9,000 12,200 ' 1 75' 4' 11,000 I2,6OO 1 60' I2/ 8,900 12.000 It will be seen that the value of E, as derived from the expression for A, is tolerably constant, and what would be expected from steel or iron girders, whereas the value derived from 8 is very irregular. Discrepancies between Experiment and Theory. Far too much is usually made of the slight discrepancies between experiments and the theory of beams ; it has mainly arisen through an improper application of the beam formula, and to the use of very imperfect appliances for measuring the elastic deflection of beams. The author has made a special study of this subject, and finds that the more delicate the apparatus he uses, the more nearly do experiments and theory agree. 388 Mechanics applied to Engineering. The discrepancies may be dealt with under three heads (1) Discrepancies below the elastic limit (2) at (3) after (i) The discrepancies below the elastic limit are partly due to the fact that the modulus of elasticity (E c ) in compression is not always the same as in tension (E t ). For example, suppose a piece of material to be tested by pure tension for E t , and the same piece of material to be afterwards tested as a beam for E 6 (modulus of elasticity from a bending test) ; then, by the usual beam theory, the two results should be identical, but in the case FlG - 379 * of cast materials it will pro- bably be found that the bending test will give the higher result ; for if, as is often the case, the E c is greater than the E t , the compression area A c of the modulus figure will be smaller than the tension area A, for the tensions and compressions form a couple, and A C E C = A t E. The modulus of the section will now be A, X D, or - - X y g , /1? X D ; thus the Z is 4 v &C ~r v &t increased in the ratio ._ , c .-. If the E 6 be 10 per cent. greater than E the E 6 found from the beam will be about 2 '5 per cent, greater than the E, found by pure traction. This difference in the elasticity will certainly account for considerable discrepancies, and will nearly always tend to make the E 6 greater than E r . There is also another dis- crepancy which has a similar tendency, viz. that some materials do not perfectly obey Hooke's law ; the strain increases slightly more rapidly than the stress (see p. 294). This tends to increase the size of the modulus figures, as shown exaggerated in dotted lines, and thereby to increase the value of Z, which again tends to increase its strength and stiffness, and con- sequently make the E 6 greater than E t . Beams. 389 FIG. 380. \W On the other hand, the deflection due to the shear (see p. 385) is usually neglected in calculating the value E 6 , which consequently tends to make the deflection greater than calculated, and reduces the value of E 6 . And, again, experimenters often mea- sure the deflection between the bottom of the beam and the supports as shown (Fig. 381). The supports slightly indent the beam when loaded, and they moreover spring slightly, both of which tend to make the deflection greater than it should be, and con- sequently reduce the value of E 6 . The discrepancies, however, between theory and experiment in the case of beams which are not loaded beyond the elastic limit are very, very small, far smaller than the errors usually made in estimating the loads on beams. (2) The discrepancies at the elastic limit are more imaginary than real. A beam is usually assumed to pass the elastic limit when the rate of increase of the , deflection per unit increase of load increases rapidly, i.e. when the slope of the tangent to the load-deflection diagram increases rapidly, or when a marked per- manent set is produced, but the load at which this occurs is far beyond the true elastic limit of the material. In the case of a tension bar the stress is evenly distributed over any cross-section, hence the whole section of the bar passes the elastic limit at the same instant; but in the case of a beam the stress is not evenly distributed, consequently only a very thin skin of the metal passes the elastic limit at first, while the rest of the section remains elastic, hence there cannot possibly be a sudden increase in the strain (deflection) such as is experienced in tension. When the load is removed, the elastic portion of the section restores the beam to very nearly its original form, and thus prevents any marked permanent set. Further, in the case of a tension specimen, the sudden stretch at the elastic limit occurs over the whole length of the bar, but in a beam only over a very small part of the length, viz. just where the bending moment is a maximum, hence the load at which the sudden stretch occurs is much less definitely marked in a beam than in a tension bar. I FIG. 381. 390 Mechanics applied to Engineering. FIG. 38*. In the case of a beam of, say, mild steel, the distribution of stress in a section just after passing the elastic limit is approximately that shown in the shaded modulus figure of Fig. 382, whereas if the material had remained perfectly elastic, it would have been that indicated by the triangles aob. By the methods described in the next chapter, the deflection after the elastic limit can be calculated, and thereby it can be readily shown that the rate of increase in the deflection for stresses far above the elastic limit is very gradual, hence it is practically impossible to detect the true elastic limit of a piece of material from an ordinary bending test. Results of tests will be found in the Appendix. (3) The discrepancies after the elastic limit have occurred. The word "dis- crepancy" should not be used in this connection at all, for if there is one principle above all others that is laid down in the beam theory, it is that the material is taken to be perfectly elastic, i.e. that it has not passed the elastic limit, and yet one is constantly hearing of the " error in the beam theory," because it does not hold under conditions in which the theory expressly states that it will not hold. But, for the sake of those who wish to account for the apparent error, they can do it approximately in the following way. The beam theory assumes the stress to be proportional to the distance from the neutral axis, or to vary as shown by the line ab ; under such conditions we get the usual modulus figure. When, however, the beam is loaded beyond the elastic limit, the distribution of stress in the section is shown by the line adb, hence the width of the modulus figure must be increased in the ratio of the widths of the two curves as shown, and the Z thus corrected is the shaded area X D as before. 1 This in 1 Readers should refer to Proceedings I.C.E., vol. cxlix. p. 313. It should also be remembered that when one speaks of the tensile strength of a piece of material, one always refers to the nominal tensile strength, not to the real ; the difference, of course, is due to the reduction of the section as the test proceeds. Now, no such reduction in the Z occurs in the beam, hence we must multiply this corrected Z by the ratio of the real to the nominal tensile stress at the maximum load. Beams. 391 many instances will entirely account for the so-called error. Similar figures corrected in this manner are shown below, from which it will be seen that the difference is much greater in the circle than in the rolled joist, and, for obvious reasons, it will be seen that the difference is greatest in those sections in which much material is concentrated about the neutral axis. FIG. 384. FIG. 385. But before leaving this subject the author would warn readers against such reasoning as this. The actual breaking strength of a beam is very much higher than the breaking strength calculated by the beam formula, hence much greater stresses may be allowed on beams than in the same material in tension and compression. Such reasoning is utterly misleading, for the apparent error only occurs after the elastic limit has been passed. CHAPTER X. 1 lr.. FIG. 386. BENDING MOMENTS AND SHEAR FORCES. Bending Moments. When two 1 equal and opposite couples are applied at opposite ends of a bar in such a manner as to tend to rotate it in opposite directions, the bar is said to be subjected to a bending moment. Thus, in Fig. 386, the bar ab is subjected to the two equal and opposite couples R . ac and W . bc t which tend to make the two parts of the bar rotate in opposite directions round the points; or, in other words, they tend to bend the bar, hence the term "bending moment." Likewise in Fig. 387 the couples are ^ac and Ra&r, which have the same effect as the couples in Fig. 386. The bar in Fig. 386 is termed a " canti- lever." The couple R . ac is due to the resistance of the wall into which it is built. The bar in Fig. 387 is termed a " beam." When a cantilever or beam is subjected Support Support L$a*L ^ ft bending momen t F IG< 3 88. which tends to bend it 1 If there be more than two couples, they can always be reduced to two. \ FIG. 387. Load Sit) ipcrrt Bending Moments and Shear Forces. 393 concave upwards, as in Fig. 388 (0), the bending moment will be termed positive (+), and when it tends to bend it the reverse way, as in Fig. 388 (), it will be termed negative (-). B ending-moment Diagrams. In order to show the variation of the bending moments at various parts of a beam, we frequently make use of bending-moment diagrams. The bending moment at the point c ,..,..,., in Fig. 389 is W . bc\ set down from c the ordinate cc* = W . be on some given scale. The bend- ing moment at d = W . bd ; set down from d, the ordinate dd' = W . bd on the same scale ; and so on for any number of points : then, as the bending moment at any point increases directly as the distance of that point from W, the points b, d\ c\ etc., will lie on a straight line. Join up these points as shown, then the depth of the diagram below any point in the beam represents on the given scale the bending moment at that point. This diagram is termed a " bending-moment diagram." In precisely the same manner the diagram in Fig. 390 is obtained. The ordinate dd\ represents on a given scale the bending mo- ment R^d, likewise cc l the bending moment FIG. 389- or the also bending moment The reactions R : and R 2 are easily found by the principles of moments thus. Taking moments about the point b, we have FIG. 390. 1 au In the cantilever in Fig. 389, let W = 800 Ibs., be = 675 feet, bd = 4-5 feet. The bending moment at c = W . be = 800 (Ibs.) X 675 (feet) = 5400 (Ibs.-feet) Let i inch on the bending-moment diagram = 12,000 (Ibs.- , r /iu r .\ t. i2oco Ibs.-feet feet), or a scale of 12,000 (Ibs.-feet) per inch, or - j. . 394 Mechanics applied to Engineering. Then the ordinate , = = '45 (inch) i (inch) Measuring the ordinate dd-^ we find it to be 0*3 inch. Then 0-3 (inch) x " ( lbs -- feet ) _ (3600 (Ibs.-feet) bending i (inch) \ moment at d In this instance the bending moment could have been obtained as readily by direct calculation ; but in the great majority of cases, the calculation of the bending moment is long and tedious, and can be very readily found from a diagram. In the beam (Fig. 390), let W = 1200 Ibs , ac = $ feet, be = 3 feet, ad = 2 feet. W.fc- 1200 (Ibs.) X 3 (feet) R '=^r=- 8 (feet) -=45olbs. the bending moment at c = 450 (Ibs.) X 5 (feet) = 2250 (Ibs.-feet) Let i inch on the bending-moment diagram = 4000 Ibs.- feet), or a scale of 4000 (Ibs.-feet) per inch, or - Then the ordinate i = - ,.. ' . -? = 0-56 (inch) 4000 (Ibs.-ieet) i (inch) Measuring the ordinate dd^ we find it to be 0*225 (inch). 0-225 (inch) X 4000 (Ibs.-feet) _ j 9OO (Ibs.-feet) bending i (inch) ~ } moment at d' General Case of Bending Moments. The bending moment at any section of a beam is the algebraic sum of all the moments of the external forces about the section acting either to 'the left or to the right of the section. Thus the bending moment at the section/ in Fig. 391 is, taking moments to the left off or, taking moments to the right off Bending Moments and Shear Forces. 395 That the same result is obtained in both cases is easily shown by taking a numerical example. Let W x = 30 Ibs., W 2 = 50 Ibs., W 3 = 40 Ibs. ; ac - 2 feet. cd = 2-5 feet, df=rS feet,/* = 2-2 feet, eb = 3 feet. r r f FIG. 391. We must first calculate the values of moments about ^, we have and R 2 . Taking P RI= _ 3 o(lbs.) X 9 ' .) X 7(feet) + 4 o(lbs.) X 3 (feet) ==75 5(lbs.-feet) = 6 i rs (feet) R 2 = W l + W 2 -h W, - 1 1 '5 (feet) The bending moment at/, taking moments to the left of/, = 65-65 (Ibs.) X 6-3 (feet) - 30 (Ibs.) X 4*3 (feet) - 50 (Ibs.) X i'8 (feet) = 194-6 (Ibs.-feet) The bending moment at/ taking moments to the right of/, = 54-35 (Ibs.) X 5-2 (feet) - 40 (Ibs.) X 2*2 (feet) = 194-6 (Ibs.-feet) Thus the bending moment at / is the same whether we take moments to the right or to the left of the point / The calculation of it by both ways gives an excellent check on the accuracy of the working, but generally we shall choose that side of the section that involves the least amount of calculation. Thus, in the case above, we should have taken moments to the right of the section, for that only involves the calculation of two moments, whereas if we had taken it to the left it would have involved three moments. 396 Mechanics applied to Engineering. The above method becomes very tedious when dealing with many loads. For such cases we shall adopt graphical methods. Shearing Forces. When couples are applied to a beam in the way described above, the beam is not only subjected to a bending moment, but also to a shearing action. In a long beam or cantilever, the bending is by far the most important, but a short stumpy beam or cantilever will nearly always fail by shear. Let the cantilever in Fig. 392 be loaded until it fails. It FIG. FIG. 393. will bend down slightly at the outer end, but that we may neglect for the present. The failure will be due to the outer part shearing or sliding off bodily from the built-in part of the cantilever, as shown in dotted lines. The shear on all vertical sections, such as ab or ctb\ is of the same value, and equal to W. In the case of the beam in Fig. 393, the middle part will shear or slide down relatively to the two ends, as shown in dotted lines. The shear on all vertical sections between b and c is of the same value, and equal to Hz, and on all vertical sections between a and c is equal to Rj. We have spoken above of posi- tive and negative bending moments. We shall also find it convenient to speak of positive and negative shears. Bending Moments and Shear Forces. 397 When the sheared part slides in a direction relatively to the fixed part, we term it a | positive (-f ) shear) ( negative ( ) shear j Shear Diagrams. In order to show clearly the amount of shear at various sections of a beam, we frequently make use of shear diagrams. In cases in which the shear is partly positive and partly negative, we shall invariably place the positive part of the shear diagram above the base-line, and the negative part below the base-line. Attention to this point will save endless trouble. In Fig. 392, the shear is positive and constant at all vertical sections, and equal to W. This is very simply represented graphically by constructing a diagram immediately under the beam or cantilever of the same length, and whose depth is equal to W on some given scale, then the depth of this diagram at every point represents on the same scale the shear at that point. Usually the shear diagram will not be of uniform depth. The construction for various cases will be shortly considered. It will be found that its use greatly facilitates all calculations of the shear in girders, beams, etc. In Fig. 393, the shear at all sections between a and c is constant and equal to Rj. It is also positive (-f ), because the slide takes place in a clockwise direction ; and, again, the shear at all sections between b and c is constant and equal to Ra, but it is of negative ( ) sign, because the slide takes place in a contra-clockwise direction ; hence the shear diagram between a and c will be above the base-line, and that between b and c below the line, as shown in the diagram. The shear changes sign immediately under the load, and the resultant shear at that section is R! Ra. General Case of Shear. The shear at any section of a beam or cantilever is the algebraic sum of all the forces acting to the right or to the left of that section. One example will serve to make this clear. In Fig. 395 three forces are shown acting on the canti- lever fixed at d) two acting downwards, and one acting upwards. The shear at any section between a and d= -f W due to W ,, b ** - ~r vr 2 M a 398 Mechanics applied to Engineering. Construct the diagrams separately for each shear as shown, then combine by superposing the diagram on the -f- diagram. The unshaded portion shows where the shear neutralizes the FIG. 39s 4- shear; then bringing the -f portions above the base-line and the below, we get the final figure. Bending Moments and Shear Forces. 399 Resultant shear at any section Between To the right. To the left. a and b W -W, + W 2 - (W + W 2 - WJ b and c w- w, W 2 - (W + W 2 - W,) = -(W-W 1 ) c and a W 2 - W, + W or W + W 2 - W, -(W + W.-W,) In the table above are given the algebraic sum of the forces to the right and to the left of various sections. On comparing them with the results obtained from the diagram, they will be found to be identical. In the case of the shear between the sections b and c, the diagram shows the shear as negative. The table, in reality, does the same, because W 1 in this case is greater than W. It should be noticed that when the shear is taken to the left of a section, the sign of the shear is just the reverse of what it is when taken to the right of the section. Connection between B ending-moment and Shear Diagrams. In the construction of shear diagrams, we make their depth at any section equal, on some given scale, to the shear at the section, i.e. to the algebraic sum of the forces to the right or left of that section, and the length of the diagrams equal to the distance from that section. Let any given beam be loaded thus : Loads W lf W 2 , W 3 , W 4 , W 5 at distances / 1} / 2 , / 3 , / 4 , / 6 respectively from any given section a, as shown in Fig. 396. The bending moment at a is = W 2 / 2 + W,/ 3 - W 4 / 4 or - W^ + W 5 / 6 . But W 2 / 2 is the area of the shear diagram due to W 2 between W 2 and the section a, likewise W 3 / 3 is the area between W 3 and a, also W 4 / 4 is the area between W 4 and a. The positive areas are partly neutralized by the negative areas. The parts not neutralized are shown shaded. The shaded area = W 2 / 2 + W 3 / 3 W 4 / 4 , but we have shown above that this quantity is equal to the bending moment at a. In the same manner, it can be shown that the shaded area of the shear diagram to the left of the section a is equal to 400 Mechanics applied to Engineering. Wj/! + We/j, i.e. to the bending moment at 0. Hence we get this relation The bending moment at any section of a beam is equal to the > FIG. 396. $* *' = i.c. the bending moment at that point. 406 Mechanics applied to Engineering. Cantilever irregu- larly loaded. FIG. 400. Beam supported at both ends, with a central load. FIG. 401. Bending Moments and Shear Forces. 407 Bending moment Min lbs.-inches. M = d . mn Depth of bending- moment diagram in inches. Scale of W, m Ibs. = i inch. Scale of/, ifull size. REMARKS. This is simply a case of the combina- tion of the diagrams in Figs. 398 and 399. However complex the loading may be, this method can always be adopted, although the graphic method to be described later on is generally more convenient for many loads. M.-T-' M = d . mn W/ Each support or abutment takes one- W half the weight = The only moment to the right or left . W / W/ of the section x is x - = At any other section the bending moment varies directly as the distance from the abutment ; hence the diagram is triangular in form as shown. The only force to the right or left of x is W ; hence the shear diagram is of constant depth as shown, only positive on one side of the section x, and negative on the other side. 408 Mechanics applied to Engineering. Beam supported at both ends, with one load not in the middle of the span. FIG. 401^. Beam supported at both ends, with two symmetrically placed loads. forlqft hand load B. M. for right TiancL loads B M jvr boih loads FIG. 402. Bending Moments and Shear Forces. 409 Bending moment M in Ibs. -inches. M = d . mn Depth of bending- moment diagram in inches. Scale of W, m Ibs. = i inch. Scale of 4 J full size. Imn REMARKS. Taking moments about one support, we have R/ = W/ 2 , or Rj = ^. The bending moment at x W M* = R,A = y (A/,) M, = W4 M,, = W/* M = td 2 , etc. Choose any convenient tn m point O distant O-6 from the vertical. Oh is termed the "polar distance." Join/"O, eO, etc. From any point/ on the line passing through R, draw a line/m parallel to fO ; from m draw mK parallel to eO, and so on, till the line through R 8 is reached in g. Join gj\ and draw Oa on the vector polygon parallel to this last line ; then the reaction R l =fa, and R 2 = ba. Then the vertical depth of the bending-moment diagram at any given section is proportional to the bending moment at that section. Proof. The two triangles jpm and Oa/are similar, for jm is parallel tofO, and jp to aO, and// to /a ; also^y is drawn at right angles to the base mp. Hence height of l^jpm _ base of A ii>nt height of A a f base of & Oaf or^ = * mp af A = 2^ ' D, R, For jq /, and af Rj ; and let mp D x , i.e. the depth of the bending-moment diagram at the section ;r, or RJ/J = T) x Oh = Ma; = the bending moment at x. By similar reasoning, we have Rj/2 = rtx Ok also Wj(4 - /,) = rK X OA the bending moment at y = M y = R,/ 2 W,(/ 2 /,) = OA(rt - rK) where D y = the depth of the bending-moment diagram at the section^. Thus the bending moment at any section is equal to the depth of the diagram at that section multiplied by the polar distance, both taken to the proper scales, which we will now determine. The diagram is drawn so that i inch on the load scale = m Ibs. I ,, length = n inches. Hence the measurements taken from the diagram in inches must be multiplied by mn. The bending moment expressed \ ,, ._. _... in Ibs. -inches at any section } = M .... (D . Q*) 2 E 4i8 Mechanics applied to Engineering Beam sup- ported at two points with overhanging ends and irre- gularly loaded. Bending moment Min Ibs. -inches. Bending Moments and Shear Forces. 419 where D is the depth of the diagram in inches at that section, and with a uniformly distributed load. A for cantilever A for beam W / These must be equal, as explained above wL 2 3 _ o/L^La , ^L! ~~~ I ~ .,- 326 44 2 Mechanics applied to Engineering. Let ^ = ;/L 2 . 3 2 6 2 = 3 2 + which on solving gives us n - 0732. We also have LI + L 2 = t ' 2 or i'732L 2 = t 2 L 2 = o'289L and L! = 0732 x o'289L = 0*2 nL j .. I FIG. 432. Maximum bending moment in middle of central span w\^ = w X o'289 2 L 2 = ^L 2 2 2 24 Maximum bending moment on cantilever spans WL.L, + ^L 2 = * X o-28 9 L X Q- 2 nL+ g>Xo ' 2I ' 2L2 2 2 12 Deflection of central span 3 8 4 EI Deflection of cantilevers due to distributed load ^ _ ze/(o*2iiL) 4 _ wL 4 8EI ~ 4o 3 8EI Deflection of Beams. Deflection due to half-load on central part 2 _ o/L a X L/ _ w X o'289L X o'2ii 8 L 3 60 = -lET- "sET- 443 HO5EI Total deflection in middle of central span 34EI This problem may also be treated in a similar manner to the last case. The area of the parabolic bending- moment diagram axbxc ~bf . ac t and the mean \; height ae = \bf\ whence ae, the bending moment at the ends, is FIG. 432*. 1 8 12 and bg, the bending moment in the middle, is > _ 3 8 = 24 and for the distance xx, we have Lj = o'2iiL These calculations will be sufficient to show that identical results are obtained by both methods. Thus, when the ends are built in and free to slide sideways, the maximum bending moment on a uniformly loaded beam is reduced to y^ = f , and the deflection to of what it would have been with free ends. CASE XV. Beam built in at both ends, with an irregularly distributed load. In this case not only must the area of the bending-moment diagram be zero, but in addition the centre of gravity of the 444 Mechanics applied to Engineering. pier-moment diagram, viz. abed^ must lie on the same vertical as the centre of gravity of the bending-moment diagram, since the beam is rigidly guided horizontally at each end. FIG. 4323. The slope at each! __ net area of the bending-moment diagram end = o f ~ EI~~ _A "EI . jthe area of the bending-) _ (the area of the pier- ' " I moment diagram / I moment diagram Let c = the distance of the centre of gravity of the bending- moment diagram from the middle of the span. x = the distance of the centre of gravity of the pier- moment diagram from P. Then the deflection at P = or--, Ax EI Thus the centre of gravity of the bending moment and the pier-moment diagrams are both at the same distance from the piers, or they both lie on the same vertical. From the expression for the position of the centre of gravity of a trapezium (see p. 60), we have / _ / / 2 ~ c ~ V M Deflection of Beams. 445 2 A Substituting the value of M Q = - - M P from (i.) A on reduction we get M r = ^ (/ + 6c) andM Q = ^(/- 6c) The area of the bending-moment diagram A is the mean bending moment for a freely supported beam x length of the beam, or A = M/ whence M P = M f E + - When the load is symmetrically disposed c = o, and the bending moment at the ends of the built-in beam is simply the mean bending moment for a freely supported beam under the same system of loading. And the maximum bending moment in the middle of the built-in beam is the maximum bending moment for the freely supported beam minus the mean bending moment. The reader should test the accuracy of this statement for the cases already given. Beams supported at more than Two Points. When a beam rests on three or more supports, it is termed a continuous beam. We shall only treat a few of the simplest cases in order to show the principle involved. CASE XVI. Beam resting on three supports , load evenly distributed. The proportion of the load carried by each support entirely depends upon their relative heights. If the central support or prop be so low that it only just touches the beam, the end supports will take the whole of the load. Likewise, if it be so high that the ends of the beam only just touch the end supports, the central support will take the whole of the load. The deflection of an elastic beam is strictly proportional to the load. Hence from the deflection we can readily find the load. The deflection in the middle 1 _ 5WL 3 when not propped j ' ~ " 384EI Let W x be the load on the central prop. 446 Mechanics applied to Engineering. Then the upward deflection due to W x = ^ = 48EI If the top of the three supports be in one straight line, the upward deflection due to Wj must be equal to the downward deflection due to W, the distributed load ; then we have 384EI " 48EI whence W 1 = |W Thus the central support or prop takes f of the whole X> 36 Hi FIG. 433- load j and as the load is evenly distributed, each of the end supports takes one-half of the remainder, viz. of the load. The bending moment at any point x distant /i from the end support is M, = &wI4 - 4 x The points of contrary flexure occur at the points where the bending moment is zero, i.e. when - l = o or when 3!, = 8/j or / x = f L Thus the length of the middle span is . It is readily shown, 4 by the methods used in previous paragraphs, that the maximum wl* bending moment occurs over the middle prop, and is there - 5 or \ as great as when not propped. The load on the prop may also be arrived at by another method, which is also applicable to cases of more than one Deflection of Beams. 447 prop, where both the spans and loads are uneven, and, further, where the supports are not all on a level. In Fig. 434, let abc represent to an exaggerated scale of FIG 434. deflection the form of the beam when resting on the central prop only. fL V , beam into three equal parts ......... \ ^ J (d) Irregular loading (approx. ) ......... II 3-5 VALUES OF - d Stiffness. 5353 TOT5 555 Beam, central load ............ 6 12 24 Cantilever, end load ...... ... ...1*5 3 6 Beam, evenly distributed load ...... 4'8 9-6 19*2 Cantilever, ,, ,, ... ...2 4 8 Beam, two symmetrically placed loads, as in 7 - . , Beam, irregular loading (approx.) ... ... 5*5 n 22 Cantilever, ,, ,, ...... 175 3-5 7 This table shows the relation that must be observed between the span and the depth of the section for a given stiffness. The stress can be found direct from the deflection of a given beam if the modulus of elasticity be known ; as this does not vary much for any given material, a fairly accurate estimate of the stress can be made. We have above , 2/L 2 = hence/ = The system of loading being known, the value of n can be found from the table above. The value of E must be assumed for the material in the beam. The depth of the section d can readily be measured, also 8 and L. The above method is extremely convenient for finding approximately the stress in any given beam. The error cannot well exceed TO per cent., and usually will not amount to more than 5 per cent. CHAPTER XII. W FIG. 438. COMBINED BENDING AND DIRECT STRESSES. IN the figure, let a weight Wbe supported by two bars, i and 2, whose sectional areas are respectively Aj and A 2 , and the corresponding loads on the bars R x and R 2 ; then, in order that the stress may be the same in each, W must be so placed that R x and R 2 are pro- tff ft 2 portional to the sectional areas of the bars, or ^ = . But Rj = R^, or A!# = A 2 ; hence W passes through the centre of gravity of the two bars when the stress is equal on all parts of the section. This relation holds, how- ever many bars may be taken, even if taken so close together as to form a solid section ; hence, in order to obtain a direct stress of uniform intensity all over a section, the external force must be so applied that it passes through the centre of gravity of the section. If W be not placed at the centre of gravity of the section, but at a distance x from it, we shall have and when W is at the centre of gravity R 2 (^ -f z ) = Wu Thus when W is not placed at the centre of gravity of the section, the section is subjected to a bending moment Wx in addition to the direct force W. Thus^- If an external forced acts on a section at a distance xfrom its centre of gravity, it will be subjected to a direct force W acting FIG. 439. Combined Bending and Direct Stresses. 453 uniformly all over the section and a bending moment Wx. The intensity of stress on any part of the section will be the sum of the direct stress and the stress due to bending, tension and compression being regarded as stresses of opposite sign. In the figure let the bar be subjected to both a direct stress (+), say tension, and bending stresses. The . direct stress acting uni- formly all over the section may be represented by the diagram abcd^ where ab or cd is the intensity FIG. 440. of the tensile stress (+) ; then if the intensity of tensile stress due to bending be represented by be (4-), and the compressive stress ( ) by/. y hence/> = When j a = j 6 , the expression becomes the same as we had above. Cranked Tie-bar. Occasionally tie bars and rods have FIG. 447 . to be cranked in order to give clearance or for other reasons, but they are very rarely properly designed, and therefore are a source of constant trouble. Combined Bending and Direct Stresses. 457 The normal width of the tie-bar is b\ the width in the cranked part must be greater as it is subjected to bending as well as to tension. We will calculate the width B to satisfy the condition that the maximum intensity of stress in the wide part shall not be greater than that due to direct tension in the narrow part. Let the thickness of the bar be /. Then, using the same notation as before B b B--+U-- the direct stress on the) = W_ _ /* wide part of the bar / ~~ B/ \ the bending stress on the) _ Wx wide part of the bar / "' ~z~ = B 2 / 6W /B u _ b the maximum skin stress") _ W \ 2 2 due to both / ~" B/ ~r~ " 32^ But as the stress on the wide part of the bar has to be made equal to the stress on the narrow part, we have W _ W 6W(B -f- 2u - b) bt ~ B/ + W Then dividing both sides of the equation by -T-, and solving, we get B = Both b and u are known for any given case, hence the width B is readily arrived at. If a rectangular section be retained, the stress on the inner side will be much greater than on the outer. The actual values are easily calculated by the methods given above, hence there will be a considerable waste of material. For economy of material, the section should be tapered off at the back to form a trapezium section. Such a section may be assumed, and the stresses calculated by the method given in the last paragraph ; if still unequal, the correct section can be arrived at by one or two trials. An expression can be got out to give the form of the section at once, but it is very cumbersome and more trouble in the end to use than the trial and error method. 458 Mechanics applied to Engineering. Hooks. A hook may be regarded as a special case of a cranked tie-bar, and if a rectangular section be retained, as in a railway drawbar hook, the equation given above will serve for finding the width B. Crane hooks, however, are always made on more economical lines ; the section where the bending is greatest is tapered in order to make the stress of equal intensity on the two sides. The stresses can be calculated by the formulae given on p. 456. See Engineering Oct. 18, 1901 ; Proceedings I. C.E., clxvii. j also Appendix. Inclined Beam. Many cases of inclined beams occur in practice, such as in roofs, etc. ; they are in reality members subject to combined bending and direct stresses. In Fig. 449, resolve W into two com- ponents, Wj acting normal to the beam, and P acting parallel with the beam ; then the bending moment at the section x = W^. FIG. 449- But W x and 4 = W sin a / ~~ sin a hence M x = W sin a x M, = W/ = fL W/ sm a P W cos a The tension acting all over the section = = -- W cos a W/ / cos a / hence/max. = ^ - -+ -^ = W ^ -j- -+ z W COS a W/ / COS a / and/mm. = -^ - ^ = W ( -^ - % N.B. The Z is for the section x taken normal to the beam, not a vertical section. Combined Bending and Direct Stresses. 459 Machine Frames subjected to Bending and Direct Stresses. Many machine frames which have a gap, such as punching and shearing machines, riveters, etc., are sub- ject to both bending and direct stresses. Take, for example, a punching-machine with a box- shaped section through AB. Let the load on the punch = W, and the distance of the punch from the centre of gravity of the section = X. X is at present unknown, unless a sec- FIG. 450. tion has been assumed, but if not a fairly close approximation can be obtained thus : We must first of all fix roughly upon the ratio of the compressive to the tensile stress due to bending ; the actual ratio will be somewhat less, on account of the uni- form tension all over the section, which will diminish the compression and increase the tension. Let the ratio be, say, 3 to i ; then, neglecting the strength of the web, our section will be somewhat as follows : Make A c = II then X = G a H approx. 4 rj _ H MA FIG. 451- (approx.) Z = 3A C H (for tension) But WX =/Z (/being the tensile stress) W A c = 3H/ or where n is the ratio of the compressive to the tensile stress, and A, = ;/A 460 Mechanics applied to Engineering. Having thus approximately obtained the sectional areas of the flanges, complete the section as shown in Fig. 452; and as a check on the work, calculate the stresses accurately by finding the centre of gravity of the complete section, also the Z or the I, and apply the formula given on p. 456. CHAPTER XIII. STXUTS. General Statement. The manner in which short com- pression pieces fail is shown in Chapter VIII. ; but when their length is great in proportion to their diameter, they bend laterally, unless they are initially absolutely straight, exactly centrally loaded, and of perfectly uniform material three conditions which are never fulfilled in practice. The nature of the stresses occurring in a strut is, therefore, that of a bar subjected to both bending and compressive stresses. In Chapter XII. it was shown that if the load deviated but very slightly from the centre of gravity of the section, it very greatly increased the stress in the material ; thus, in the case of a circular section, if the load only deviated by an amount equal to one-eighth diameter from the centre, the stress was doubled ; hence a very slight initial bend in a compression member very seriously affects its strength. Effects of Imperfect Loading. Even it a strut be initially straight before loading, it does not follow that it "will remain so when loaded ; either or both of the following causes may set up bending : (i) The one side of the strut may be harder and stififer than the other ; and consequently the soft side will yield most, and the strut will bend as shown in A, Fig. 453. 462 Mechanics applied to Engineering. (2) The load may not be perfectly centrally applied, either through the ends not being true as shown in B, or through the load acting on one side, as in C. Possible Discrepancies between Theory and Practice. We have shown that a very slight amount of bending makes a serious difference in the strength of struts ; hence such accidental circumstances as we have just mentioned may not only make a serious discrepancy between theory and experiment, but also between experiment and experiment. Then, again, the theoretical determination of the strength of struts does not rest on a very satisfactory basis, as in all the theories advanced somewhat questionable assumptions have to be made ; but, in spite of it, the calculated buckling loads agree fairly well with experiments. Bending of Long Struts. The bending moment at the middle of the bent strut shown in Fig. 454 is evidently WS. Then WS =/Z, using the same notation as in the W preceding chapters. If we increase the deflection we shall correspondingly increase the bending moment, and consequently the stress. From above we have W = {z or rZ, and so on Oj But as /varies with S,"i-= a constant, say K ; FIG. 454. then W = KZ But Z for any given strut does not vary when the strut bends ; hence there is only one value of W that will satisfy the equation. When the strut is thus loaded, let an external bending moment M, indicated by the arrow (Fig. 455), be applied to it until the deflection is 8 lt and its stress/ ; Then WS X + M =/Z But W8, =/Z therefore M = o that is to say, that no external bending moment M is required to keep the strut in its bent position, or the strut, when thus loaded, is in a state of neutral equilibrium, and will remain Struts. 463 when left alone in any position in which it may be placed ; this condition, of course, only holds so long as the strut is elastic, i.e. before the elastic limit is reached. This state of neutral equilibrium may be proved experimentally, if a long thin piece of elastic material be loaded as shown. Now, place a load W x less than W on the strut, say W = Wj + w, and let it again be bent by an external bending moment M till its deflection is o\ and the stress /j ; then we have, as before WA + M =/ x Z = Wo\ = WA + wSj hence M = w& FIG. 455. Thus, in order to keep the strut in its bent position with a deflection o\, we must subject it to a -f- bend- ing moment M, i.e. one which tends to bend the strut in the same direction as WA; hence, if we remove the bending moment M, the deflection will become zero, i.e. the strut will straighten itself. Now, let a load W 2 greater than W be placed on the strut, say W = W 2 w, and let it again be bent until its deflection = 6\, and the stress /j by an external bending moment M ; then we have as before WA + M =/Z = WA - A hence M = ze>6\ Thus, in order to keep the strut in its bent position with a deflection S : , we must subject it to a bending moment M, i.e. one which tends to bend the strut in the opposite direction to W 2 o\ ; hence, if we remove the bending moment M, the de- flection will go on increasing, and ere long the elastic limit will be reached when the strain will increase suddenly and much more rapidly than 'the stress, consequently the deflection will suddenly increase and the strut will buckle. Thus, the strut may be in one of three conditions Condition. When slightly bent by an ex- ternal bending moment M, on being released, the strut will- When supporting a load i. ii. iii. Remain bent Straighten itself Bend still further and ultimately buckle W. less than W. greater than W. 464 Mechanics applied to Engineering. Condition ii. is, of course, the only one in which a strut can exist for practical purposes ; how much the working load must be less than W is determined by a suitable factor of safety. Buckling Load of Long Thin Struts. Euler's Formula. The results arrived at in the paragraph above refer only to very long thin struts; we will now proceed to determine the value of W for such struts. If the deflection were entirely due to the eccentricity x of the load, then the bending moment at every section of the strut would be constant and equal to Wx, and the strut would then bend to the arc of a circle (see p. 434). For the present we will assume that struts do bend to an arc of a circle; we shall return to this point later on, and then give a more exact result. Let / = the effective length of the strut (see Fig. 458); E = Young's modulus of elasticity ; I = the least moment of inertia of a section of the strut (assumed to be of constant cross-section). FIG. 4 <6. Then for a strut loaded thus 8 M/ 2 . . WS/ 2 Fl ( see P- 425) = 8EJ- 8FI or W = -jr (first approximation) As the strut is very long and the deflection small, the length / remains practically constant, and FIG. 457. the other quantities 8, E, I are also constant for any given strut; thus, W is equal to a constant, which we have previously shown must be the case. Once the strut has begun to bend it cannot remain a circular arc, because the bending moment no longer remains constant at every section, but it will vary directly as the distance of any given section from the line of application of the load. Under these conditions assume as a second approxi- mation that it bends to a parabolic arc, then the deflection / J M/ 2 9'6EI = Struts. 465 By Euler's theory, which we must not forget is only another approximation, since he neglects the direct stress on the section, we get , T , 7r 2 EI o'87EI xoEI , W = = J ' = _ fnpirlvi /2 7l 72 \ l *V 2 (see p. 78), where p is the radius of A gyration of the section. Substituting these values in the above equation, we have TT'E^ The " effective " or " virtual " length /, shown in the diagram, is found by the methods given in Chapter XI. for finding the virtual length of built-in beams. The square-ended struts in the diagrams are shown bolted down to emphasize the importance of rigidly fixing the ends ; if the ends merely rested on flat flanges without any means of fixing, they much more nearly approximate round-ended struts. It will be observed that Euler's formula takes no account of the compressive stress on the material ; it simply aims at giving the load which will produce neutral equilibrium as regards bending in a long bar, and even this it only does imperfectly, for when a bar is subjected to both direct and bending stresses, the neutral axis no longer passes through the centre of gravity of the section. We have shown above that when the line of application of the load is shifted but one- eighth of the diameter from the centre of a round bar, the neutral axis has shifted to the outermost edge of the bar. In the case of a strut subject to bending, the neutral axis shifts away from the line of application of the load ; thus the bend- ing moment increases more rapidly than Euler's hypothesis assumes it to do, consequently his formula gives too high results; but in very long columns in which the compressive stress is small compared with the stress due to bending, the error may not be serious. But if the formula be applied to short struts, the result will be absurd. Take, for example, an iron strut of circular section, say 4 inches diameter and 40 A ~ 10 X 20000000 X i inches long; we set P = - - = 181,000 Ibs. IOOO per square inch, which is far higher than the crushing strength of a short specimen of the material, and obviously absurd. If Euler's formula be employed, it must be used exclusively Struts. 467 for long struts, whose length / is not less than 30 diameters for wrought iron and steel, or 12 for cast iron and wood. Notwithstanding the unsatisfactory basis on which it rests, many high authorities, such as Unwin, Reuleaux, Bovey, Thurston, and others, prefer it to Gordon's, which we will shortly consider. For a full discussion of the whole question of struts, the reader is referred to Todhunter and Pearson's " History of the Theory of Elasticity." Gordon's Strut Formula rationalized. Gordon's strut formula, as usually given, contained empirical constants obtained from experiments by Hodgkinson and others ; the author, however, has succeeded in arriving at these constants rationally, which will shortly be seen to agree remarkably well with the constants found by experiment. Gordon's formula certainly has this advantage, that it agrees far better with experiments on the ultimate resistance of columns than does the formula propounded by Euler; and, moreover, it is applicable to columns of any length, short or long, which, we have seen above, is not the case with Euler's formula. The elastic conditions assumed by Euler cease to hold when the elastic limit is passed, hence a long strut always fails at or possibly before that point is reached ; but in the case of a short strut, in which the bending stress is small compared with the compressive stress, it does not at all follow that the strut will fail when the elastic limit in compression is reached indeed, experiments show conclusively that such is not the case. A formula for- struts of any length must there- fore cover both cases, and be equally applicable to short struts that fail by crushing and to long struts that fail by bending. In constructing this formula we assume that the strut fails either by buckling or by crushing, 1 when the sum of the direct compressive stress and the skin stress, due to bending, are equal, to the crushing strength of the material; in using the term " crushing strength " for ductile materials, we mean the stress at which the material becomes plastic. This assumption, we know, is not strictly true, but it cannot be far from the truth, or our calculated values of the constant (a), shortly to be considered, would not agree so well with the experimental values. 1 Men's. Considere and others have found that for long columns the resistance does not vary directly as the crushing resistance of the material, but for short columns, which fail by crushing and not by bending, the re- sistance does of course entirely depend upon it, and therefore must appear in any formula professing to cover struts of all lengths. 468 Mechanics applied to Engineering. Let S = the crushing (or plastic) strength of a short specimen of the material ; C = the direct compressive stress on the section of the strut ; then, adopting our former notation, we have W , , W8 C = _and/=- then S = C +/ w w S = + ~ (the least Z of the section) A LI We have shown above that, on Euler's hypothesis, the maximum deflection of a strut is g^ M/ 2 _ /Z/ a loEI where y is the distance of the most strained skin from the centre of gravity of the section, or from the assumed position of the neutral axis. We shall assume that the same expression holds in the present case. In symmetrical sections y = -, where d is the least diameter of the strut section. By substitution, we have -ft " ' A W W If W be the buckling load, we may replace by P. Let FEZ = - Struts. 469 where r ,, which is a modification of " Gordon's Strut a Formula." P may be termed the buckling stress of the strut. The d in the above formula is the least dimension of the section, thus ft FIG. 459. It now remains to be seen how the values of the constant a agree with those found by experiment ; it, of course, depends upon the values we choose for / and E. The latter presents no difficulty, as it is well known for all materials; but the former is not so obvious at first. In equation (i.), the first term provides for the crushing resistance of the material irrespective of any stress set up by bending ; and the second term provides for the bending resistance of the strut. We have already shown that the strut buckles when the elastic limit is reached, hence we may reasonably take/ as the elastic limit of the material. It will be seen that the formula is only strictly true for the two extreme cases, viz. for a very short strut, when W = AS, and for a very long strut, in which S =f; then loEI \\ = p- or p- which is Euler's formula. It is impossible to get a rational formula for intermediate cases, because any expression for 8 only holds up to the elastic limit, and even then only when the neutral axis passes through the centre of gravity of the section, i.e. when there is pure bending and no longitudinal stress. However, the fact that our rational value for a agrees so well with the experimental value is strong evidence that our formula is trustworthy. Values of S, /, and E are given in the table below ; they 470 Mechanics applied to Engineering. must be taken as fair average values, to be used in the absence of more precise data. Material. Soft wrought iron Hard Mild steel ... Hard Cast iron ,, (hard close -grained) metal) f Pitchpine and oak Pounds per square inch. S / E 40,000 48,OOO 67,000 110,000 ( 80,000 (no \marked limit) 28,000 32,000 45,000 75,000 J 80,000 25,OOO,OOO 29,OOO,OOO 30,000,000 32,000,000 I3,OOO,OOO 130,000 8,000 130,000 8,000 22,000,000 900,000 Material. Form of section. A/ FIG. 462. FIG. 463- is full, an experienced eye immediately detects them on entering the building. The following test of a column by the author will serve to emphasize the folly of loading columns in this manner. ESTIMATED BUCKLING LOAD IF CENTRALLY LOADED, ABOUT 1000 tons. Length 10 feet, end flat, not fixed. Sectional area of metal at fracture Modulus of section at fracture Distance of point of application of load from centre of column, neglecting slight amount of deflection when loaded Breaking load applied at edge of bracket Bending moment on section when fracture occurred 1 1 14 tons-inches Compressive stress all over section when fracture occurred Skin stress on the material due to bending, assuming the bending formula to hold up to the breaking point 14 '85 34 -3 sq. inches 17 inches 65-5 tons 1*91 tons per sq. inch 47 8 Mechanics applied to Engineering. Total tensile stress on material due to combined bending and compression ! ... 12*94 tons per sq. inch Total compressive stress on material due to combined bending and compression 1676 ,, ,, Tensile strength of material as ascertained from subsequent tests 8*45 ,, ,, Compressive strength of material as ascer- tained from subsequent tests 30*4 ,, ,, Thus we see that the column failed by tension in the material on the off side, i.e. the side remote from the load. FACTOR OF SAFETY FOR STRUTS. Dead loads. Live loads. Wrought iron and steel 4 8 Cast iron ... ... 6 12 Timber 5 10 1 The discrepancy between this and the tensile strength is due to the bending formula not holding good at the breaking point, as previously explained. CHAPTER XIV. TORSION. GENERAL THEORY. LET Fig. 464 represent two pieces of shafting provided with 3 FIG. 464. disc couplings as shown, the one being driven from the other through the pin P, which is evidently in shear. Let S = the shearing resistance of the pin. Then we have W/ = y Let the area of -the pin = a, and the shear stress on the pin be/,. Then wemay write the above equation FIG. 4 6 5 . The dotted holes in the figure are supposed to represent the pin-holes in the other disc coupling. Before W was applied the pin-holes were exactly opposite one another, but after the application of W the yielding or the shear of the pins caused a Now consider the case in which there are two pins, then 480 Mechanics applied to Engineering. slight movement of the one disc relatively to the other, but shown very much exaggerated in the figure. It will be seen that the yielding or the strain varies directly as the distance from the axis of revolution (the centre of the shaft). When the material is elastic, the stress varies directly as the strain ; hence Substituting this value in the equation above, we have y Then, if a = a^ and say y l - 2- 4 Thus the inner pin, as in the beam (see p. 356), has only increased the strength by \. Now consider a similar arrange- ment with a great number of pins, such a number as to form a hollow or a solid section, the areas of each little pin or element being a, a lt <%, etc., distant y, jy l5 j> 2 etc -j respectively from the axis of revolution. Then, as before, we have W/ =(ay* + a^ + a*y} -f, etc.) But the quantity in brackets, viz. each little area multiplied by the square of its distance from the axis of revolution, is the polar moment of inertia of the section (see p. 77), which we will term L. Then The W/ is termed the twisting moment, M,. f t is the skin shear stress on the material furthest from the centre, and is therefore the maximum stress on the material, often termed the skin stress. y is the distance of the skin from the axis of revolution. Torsion. General Theory. 48: - = the modulus of the section = Z n . To prevent confusion y we shall use the suffix p to indicate that it is the polar modulus of the section, and not the modulus for bending. Thus we have M, =f s Z p or the twisting moment = the skin stress X the polar modulus of the section Shafts subject to Torsion. To return to the shaft couplings. When power is transmitted from one disc to the other, the pin will evidently be in shear, and will be distorted FIG. 466. as shown (exaggerated). Likewise, if a small square be marked on the surface of a shaft, when the shaft is twisted it will also become a rhombus, as shown dotted on the shaft below. In Chapter VIII. we showed that when an element was distorted by shear, as shown in Fig. 467 (a), it was "\ FIG. 467. \ equivalent to the element being pulled out at two opposite corners and pushed in at the others, as shown in Fig. 467, (b) and (<:), hence all along the diagonal section AB there 3 I 482 Mechanics applied to Engineering. FIG. 468. is a tension tending to pull the two triangles ADB, ACB apart ; similarly there is a compression along the diagonal CD. These diagonals make an angle of 45 with their sides. Thus, if two lines be marked on a shaft at an angle of 45 with the axis, there will be a tension normal to the one diagonal, and a compression normal to the other. That this is the case can be shown very clearly by getting a piece of thin tube and sawing a diagonal slot along it at an angle of 45. When the outer end is twisted in the direction of the arrow A, there will be compression normal to the slot, shown by a full line, and the slot will close ; but if it be twisted in the direction of the arrow B, there will be tension normal to the slot, and will cause it to open. Graphical Method of finding the Polar Modulus for a Circular Section. The method of graphically finding the polar modulus of the section is precisely similar in principle to that given for bending (see Chap. IX.), hence we- shall not do more than briefly indicate the construction of the modulus figure. It is of very limited application, as it is only true for circular sections. As in the beam modulus figure, we want to construct a figure to show the distribution of stress in the section. Consider a small piece of a circular section as shown, with two blocks equiva- lent to the pins we used in the disc couplings above. The stress on the inner block = f sl , and on the outer block = f s ; then -^ = . Then by projecting the width of the inner block on to the outer circle, and joining down to the centre of the circle, it is evident, from similar triangles, that we reduce the width and area of the inner block in the ratio , or in the ratio of -^. The reduced area of the inner block, shown shaded, we will now term aj, where =i) ="-W 2 which is the same result as we had before for W/, thus proving the correctness of the graphical method. In the figure above we have only taken a small portion of a circle ; we will now use the same method to find the Z p for a FIG. 470 circle. For convenience in working, we will set it off on a straight base thus : Draw a tangent ab to the circle, making the length = ?rD ; join the ends to the centre O ; draw a series of lines parallel to the tangent ; then their lengths intercepted between ao and bo are equal to the circumference of circles of radii O 1} O 2 , etc. Thus the triangle Qab represents the circle rolled out to a straight base. Project each of these lines on to the tangent, and join up to the centre; then the width of the line i'i', etc., represents the stress in the metal at that layer in precisely the same manner as in the beam modulus figures. Then ,, , , , f . ("area of modulus figure X distance 1 he polar modulus of > ] f f of ^ fi f the section Z, ( \ centre | f circle or Z f = Ay, 484 Mechanics applied to Engineering. The construction for a hollow circle is precisely the same as for the solid circle. It is given for the sake of graphically illustrating the very small amount that a shaft is weakened by making it hollow. This construction can be applied to any form of section, but the strengths of shafts other than circular do not vary as their polar moments of inertia or moduli of their sections ; serious errors will be involved if they are even taken to be approximately correct. The calculation of the stresses in irregular figures in torsion involves fairly high mathematical work. The results of such calculations by St. Venant and Lord Kelvin will be given in tabulated form later on in this chapter. Strength of Circular Shafts in Torsion. We have shown above that the strength of a cylindrical shaft varies as i? = Z x In Chapter III., we showed that I p = , where D y 3 2 is the diameter, and y in this case = ; hence which, it will be noticed, is just twice the value of the Z for bending. In order to recollect which is which, it should be remembered that the material in a circular shaft is in the very best form to resist torsion, but in a very bad form to resist bending ; hence the torsion modulus will be greater than the bending modulus. Fgr $ follow shaft Torsion. General Theory. 485 I p = , where D* = the internal diameter 32 TT(D* - D,*) hence Z, = jp- l6D 2 = 0-196 ( ]5~~~ L ) Hollow shafts for marine work are nearly always made with the internal diameter equal to one-half the external ; Then = o-i 9 6if(D') = o- Thus by removing one- quarter the metal from the centre of the shaft, the strength has only been reduced by one-sixteenth ; similarly, it can be shown that by removing one half the metal, the strength will be reduced by one-fourth ; or, generally, if - of the metal be removed from the centre of the shaft, we have the external area The internal- area = 486 Mechanics applied to Engineering. Strength of Shafts of Various Sections. FIG. 472- FIG. 473. FIG. 474. 3 ........ FIG. 475. Any section not containing re-en- trant angles (due to St. Venant). I z = 16 ' D 3 Z, = ,or Z, = o-i96D 3 _ D 4 - D/ 5'iD , or m =-- or = or = o-2o8S 3 Z n = 3 + I'Sm where m = - Z = where A = area of sec I^ = polar moment of in y distance of furthest edg GD _ 584M/ G(D 4 - 292 M/(^ 2 + D 2 ) GDV 3 S 4 G + B 2 ) tion; ertia of section ; e from centre of section. Torsion. General Theory. 487 Twist of Shafts. In Chapter VIII., we showed that when an element was sheared, the amount of slide x bore the follow- ing relation : 71D 7 G ' where f s is the shear stress on the material ; G is the coefficient of rigidity. In the case of a shaft, the x is measured on the curved surface. It will be more convenient if we ^ express it in terms of the angle of twist. The circumference of the shaft = The arc subtending i = FIG. 477. 7TD0 _ = ^~ Substituting the value of x in equation (i.), we have = " or fl - 360? G' TrGD X l6 hei e A - ~ for solid circular shafts. Substituting the value of Z p for a hollow shaft in the above, we get B 584M/ G(D 4 - D, 4 ) for hollow circular shafts. N.B. The stiffness of a hollow shaft is the difference of the stiffness of two solid shafts whose diameters are respectively the outer and inner diameters of the hollow shaft. When it is desired to keep the twist or spring of shafts within narrow limits, the stress has to be correspondingly 488 Mechanics applied to Engineering. reduced. Long shafts are frequently made very much stronger than they need be in order to reduce the spring. A common limit to the amount of spring is i in 20 diameters; the stress corresponding to this is arrived at thus We have above 6 = Tr But when = i, / = 2oD , TrGD G then f. = -7 - =r = - 360 X 2oD 2292 For steel, G = 13,000,000; f s = 5670 Ibs. per sq. inch Wrought iron, G = 11,000,000; f t = 4800 Cast iron, G *= 6,000,000; f = 2620 In the case of short shafts, in which the spring is of no importance, the following stresses may be allowed : Steel, /, = 10,000 Ibs. per sq. inch Wrought iron, f, 8000 Cast iron, / = 3000 Horse-power transmitted by Shafts. Let a force of P Ibs. act at a distance r inches from the centre of a shaft ; then The twisting mo- \ ment on the shaft j- = P (Ibs.) X r (inches) . in Ibs.-inches J The work done per ) p ([b } ( . h } revolution in foot- > = 2 - ' 5 - '- --- Ibs. ) The work done per) _ P (Ibs.) X r (inches) x 2?rN (revs.) minute in foot-lbs. I 12 where N = number of revolutions per minute. The horse-power transmitted = I2 x then _ 12 X 33000 X H.P. /D 3 27TN " 5'I _ 12 X 3300Q X H.P. X 5'i _ 321400 HP. 27TN/. N/. 64-3 H.P. N taking/, at 5ooo Ibs. per square inch. Torsion. General Theory. 489 3 /TT p D = 4 / y/ * ' (nearly) for 5000 Ibs. per sq. inch = 3*5 N KPT N for 7500 Ibs. per sq. inch VH.P., . , = 3^7 -jr- for 12,000 Ibs. per sq. inch Taking the value of Z p = o'i84D 3 for hollow shafts having the internal diameter equal to half the external, we get 3 /]-[ p D = 4'iA/ ' ' for 5000 Ibs. per sq. inch 7500 Ibs. per sq. inch TT p = Z'5\/ XT ' for 12,000 Ibs. per sq. inch Combined Torsion and Bending. In Fig. 478 a shaft is shown subjected to torsion only. We have previously seen **N Tension, only FIG. 478. (Chapter VIII ) that in such a case there is a tension acting Tension cnly FIG. 479. 490 Mechanics applied to Engineering. normal to a diagonal drawn at an angle of 45 with the axis of the shaft, as shown by the arrows in the figure. In Fig. 479 a shaft is shown subjected to tension only. In this case the FIG. 480. tension acts normally to a face at 90 with the axis. In Fig. 480 a shaft is shown subjected to both torsion and tension ; the face over which there is the greatest tension will therefore lie between the two faces mentioned above, and the tension on this face will be greater than the tension on either of the other faces, when acted upon only by torsion or tension. We have shown in Chapter VIII. that the stress / normal to the face gh due to combined tension and shear is If the tension be produced by bending, we have - T Likewise, if the shear be produced by twisting /. = ^ But Z p = 2Z then/ = J| Torsion. General Theory. Substituting these values in the above equation 491 f = M , / M 2 AZ = also/ s< x 2Z =f si M 2 = M + The M is termed the equivalent bending moment, the M et the equivalent twist- ing moment, that would pro- duce the same intensity of stress in the material as the combined bending and twist- ing. The construction shown in Fig. 482 is a convenient graphical method of finding M*. In Chapter VIII. we also showed that 2 + M, Met FIG. 482. =*g sn = .A cos 6 = sin L cos Or, using the notation above M, = tan From this expression we can find the angle of greatest stress 0, and therefore the angle at which fracture will probably occur. In Fig. 483 we show the fractures of two cast-iron torsion test-pieces, the one broken by pure torsion, -the other by 492 Mechanics applied to Engineering. combined torsion and bending. Around each a spiral piece of paper, cut to the theoretical angle, has been wrapped in order to show how the angle of fracture agreed with the theoretical angle ; the agreement is remarkably close. The following results of tests in the author's laboratory will show the results that are obtained when cast-iron bars are tested in combined torsion and bending as compared with pure torsion and pure bending tests. The reason why the shear stress calculated from the combined tests is greater than when obtained from pure torsion or shear, is due to the fact that neither of the formulae ought to be used for stresses up to rupture ; however, the results are interesting as a comparison. The angles of fracture, however, agree well with the calculated values. Twisting Bending Equivalent Modulus Angle of fracture. moment moment twisting of rupture M M moment fit tons per Pounds- inches. Met sq. inch. Actual. Calculated. Zero 2300 4600 25-5 777 1925 4000 267 12 11 1170 2240 475 27T H 14 1228 22 55 4820 23-1 17 15 1308 2128 4628 24-0 19 1 6 2606 2644 'III 4320 3520 20-8 16-2 33 38 " 3I o 37 3084 Zero 3084 i6'o 43 45) Mean of a Pure shear 13-0 o> large number ,, tension 'S o) of tests. Whirling of Shafts. A horizontal shaft sags between its bearings due to its own weight and that of the pulleys, hence during each revolution it is bent to and fro. If the period of this disturbing action approximates to the natural period of vibration of the shaft, the amplitude of the vibrations will continue to increase, and set up a violent whipping action known as whirling ; if the speed of rotation be increased beyond this whirling speed, the shaft will again become steady. The treatment of an unloaded shaft, i.e. one without pulleys, is tolerably simple (see Rankine's " Machinery and Millwork," p. 549), but it becomes very complex in the case of a loaded shaft. A very thorough and able investigation of Torsion. General Theory. 493 this question has been undertaken by Professor Dunkerley, of Greenwich, who has not only shown mathematically the speed at which whirling takes place, but has proved the correctness of his deductions by a series of most careful and interesting experiments. Some of the principal results obtained by him are given in the appendix, but the reader should refer to his paper on " The Whirling and Vibration of Shafts" (Phil. Trans., vol. 185 A, p . Combined torsion Pure torsion. and bending . FIG. 483. pp. 279-360), or to his paper read before the Liverpool Engineering Society, 1894-5, which is more suited to the average engineer. Helical Springs. The wire in a helical spring is, to all intents and purposes, subjected to pure torsion, hence we can readily determine the amount such a spring will stretch or compress under a given load, and the load it will safely carry. 494 Mechanics applied to Engineering. We may regard a helical spring as a long thin shaft coiled into a helix, hence we may represent our helical spring thus FIG. 484. In the figure to the left we have the wire of the helical spring straightened out into a shaft, and provided with a grooved pulley of diameter D, i.e. the mean diameter of the coils in the WD spring; hence the twisting moment upon it is . That the twisting moment on the wire when coiled into a helix is also WD - will be clear from the bottom right-hand figure. The length of wire in the spring (not including the ends and hook) is equal to /. Let n the number of coils ; then / = ?rD;/ nearly, or more accurately 1= V(^D) 2 -f L 2 , Fig. 485, a refinement which is quite unnecessary for springs as ordinarily made. When the load W is applied, the end of the shaft twists, so that a point on the surface moves through a distance*, and a point on the rim of the pulley FIG. 485. moves through a distance &, where = -, and 8 = ^. D a a ^ f< // But we have - = - x = -^ hence 8=' Torsion. General Theory. 495 VVD f ^ Also-- w I6WD = 2-55WD ~ then 8 = (frQm ^ and - Substituting the value of / = mrD G for steel = 12,000,000 8 = G for hard brass = 5,000,000 8 = 625,000^ The load a spring will support with a given deflection is ,,, G^/ 4 8 1,500,000^8 r = 8D 3 ~ = " ~D 3 -------- (circular section) , TT ,( r , W = ^~ - for brass D { Safe Load. From equation (ii.), we have VV = -y - = 4 ~ ^ or 7> 000 I DS - P er square inch . (iii.) - for 60,000 Ibs. per square inch , f y ~ for 50,000 Ibs. per square inch Experiments by Mr. Wilson Hartnell show that for steel wire the following stresses are the maxima consistent with safety : Diameter of wire. Safe stress. \ inch ...... 70,000 Ibs. per square inch ,, ...... 60,000 Ibs. ,, ,, i ,, ...... 50,000 Ibs. Taking a mean value, we may say w _ 2 W- 496 Mechanics applied to Engineering. Work stored in Springs. The work done in stretching ) W8 f = - (see or compressing a spring | 2 v (substituting the value of /) = ' r 19,500,000 .putting G = 12,000,000 Weight of Spring. Taking the weight of i cub. inch of steel = 0*28 lb., then The weight of the spring w = 0*7 85*/ 2 / X 0*28 = 0*2 Substituting the value of /, we have w = Height a Steel Spring will lift itself (h). work stored in spring weight of spring / _ . ., . . = i-62Gx o'6 9 n^D = ri 2 G ir i i f eet ' i3*4G 161,000,000 The value of h is given in the following table corresponding to various values off: f t (Ibs. per square inch) ... 30,000 60,000 90,000 120,000 150,000 h (feet) ......... 5*56 22-4 50-3 89-5 139-8 These figures are of interest in showing what a very small amount of energy can be stored in springs. All the quantities given above are for springs made of wire of circular section ; for wire of square section of side S, and taking the same value for G as before, we get Torsion General Theory. 497 2,i 35 ,oooS 4 (sted D 3 Wn 6 = 890,0008^ square section D 3 n 89o,oooS 4 8 W = - Uo - brass y Safe Load for Springs of Square Section. 20,I20S 3 . . , stress 70,000 lbs. per square inch .C3 60,000 lbs. lbs. D Taking a mean value, we have ,, 7 2t;,oooS 3 . w = (square section) work stored = , Q - (inch-lbs.) steel 24,680,000 v weight = o-88^S 2 D (steel) Height a square-section spring) /, 2 will lift itself (steel) J * " 260,600,000 f s (lbs. per square inch) ... 30,000 60,000 90,000 120.000 150,000 A (feet) 3-45 13-8 31-1 55-3 86-3 It will be observed that in no respect is a square-section spring so economical in material as a spring of circular section. Helical Spring in Torsion. When a helical spring is twisted the wire is subjected to a bending moment due to the change of curvature of the spring, which is proportional to the twisting moment. 2 K 498 Mechanics applied to Engineering. Let />! and ^ the mean radii of the spring in inches before and after twisting respectively ; n^ and n. 2 = the number of free coils before and after twisting respectively ; 6 and 2 = the angles subtended by the wire in inches before and after twisting respectively ; = 360^ and 36o;z 2 respectively; 6 = QI # 2 , or the angle twisted through by the free end of the spring in degrees ; M t = the twisting moment in pounds-inches ; I = the moment of inertia of the wire section about the neutral axis in inch units ; d = the diameter or side of the wire in inches ; L = the length of free wire in the spring in inches ; M, = Elf I - - 1 ) = Elf 2 -^! - 2 J^f \pi W V L L E = M t L _ 3 6oM t L 27rl(;/ 1 of circular section of guare section If O r be the angle of twist expressed in radians, we have LM, " W Open-coiled Helical Spring. In the treatment given above for helical springs, we took the case in which the coils were close, and assumed that the wire was subjected to torsion only; but if the coils be open, and the angle of the helix be considerable, this is no longer an admissible assumption. Instead of there being a simple twisting moment WR acting on the wire, we have a twisting moment M t = WR cos a, which twists the wire about an axis ab. Think of ab as a little shaft attached to the spring wire at a, and ei as the side view of a circular disc attached to it, then, by twisting this disc, the wire will be subjected to a torsional stress. In addition to this, let ab represent the side view of half an annular disc, suitably attached to the wire at a, and which rotates about an axis cd. Then, by Torsion. General Theory. 499 twisting this disc, the spring wire can be bent ; thereby its radius of curvature will be altered in much the same manner as that described in the article on the " Helical Spring in Torsion," the bending moment M = WR sin a. The force which pro- duces the twisting moment acts in the plane of the disc ", and L.^ WR-.-W, FIG. 4850;, that which produces the bending moment in the plane ab, i.e. normal to the respective sides of the triangle of moments abL In our expression for the twist of a shaft on p. 487, we gave the angle of twist O c in degrees; but if we take it in circular measure, we get x = rO c) where r is the radius of the wire, and and O c = rG ~ rGZ p /WR cos q GL GL Likewise due to bending we have (see p. 498) B , _ /M /WR sin a = El ~ El We must now find how these straining actions affect the axial deflection of the spring. 500 Mechanics applied to Engineering. The twisting moment about ab produces a strain be = R(9 C , which may be resolved into two components, viz. one, ef = R# c sin a, which alters the radius of curvature of the coils, and which we are not at present concerned with ; and the other, bf = R0 C cos a, which alters the axial extension of the spring /WR 2 cos 2 a by an amount r-~= -- (jL p The bending moment about cd in the plane of the imaginary disc ab produces a strain bh R# c ', of which hg alters the radius of curvature in a horizontal plane, normal to the axis of the spring ; and bg =. R0 C ' sin a alters the axial extension of the spring in the same direction as that due to twisting, by an /WR 2 sin 2 a amount -- gj -- ; whence the total axial ) _ /wp 2 / cos 2 a sin^_o \ extension 8 j " w V GI, ' El ) On substitution and reduction, we get 8D 8 W and for the case of springs in which the bending action is neglected, we get SaD'W cos a Angle. . deflection allowing for bending and torsion deflection allowing for torsion only 5 0*998 10 0-992 15 0*986 20 0-978 30 0-950 45 0'900 Thus for helix angles up to 15 there is no serious error due to the bending of the coils, and when one remembers how many other uncertain factors there are in connection with helical springs, such as finding the exact diameter of the wire and coils, the number of free coils, the variation in the value of G, it will be apparent that such a refinement as allowing for the bending of the wire is rarely, if ever, necessary. CHAPTER XV. STRUCTURES. Wind Pressures. Nearly all structures at times are exposed to wind pressure. In many instances, the pressure of the wind is the greatest force a structure ever has to withstand. Let v velocity of the wind in feet per second ; V = velocity of the wind in miles per hour ; M = mass of air delivered per square foot per second ; W = weight of air delivered per square foot per second (pounds) ; w = weight of i cubic foot of air (say 0*0807 Ib.) ; P = pressure of wind per square foot of surface exposed (pounds). Then, when a stream of air of finite cross-section impinges normally on a flat surface, whose area is much greater than that of the stream, the change of momentum per second per square foot of air stream is Uv = P g But W = wv -r, hence -- = P g or expressed in miles per hour by substituting v = i*466V, and putting in the value of w, we have p 0-0807 X i -466* X V a 32 . 2 - = 0-0054 v^ 2 If, however, the section of the air stream is much greater than the area of the flat surface on which it impinges, the change of direction of the air stream is not complete, and consequently the change of momentum is considerably less than (approximately one-half) the value just obtained. Smeaton, from experiments by Rouse, obtained the coefficient 0-005, but 502 Mechanics applied to Engineering. later experimenters have shown that, such a value is probably too high. Martin gives 0*004, Kernot 0*0033, Dines 0-0029, and the most recent experiments by Stanton (see Proceedings f.C.E.,vol. clvi.) give 0*0027 for the maximum pressure in the middle of the surface on the windward side. In all cases of wind pressure the resultant pressure on the surface is composed of the positive pressure on the windward side, and a suction or negative pressure on the leeward side. Stanton found in the case of circular plates that the ratio of the maximum pressure on the windward side to the negative pressure on the leeward side was 2'i to i in the case of circular plates, and a mean of 1-5 to i for various rectangular plates. Hence, for the re- sultant pressure on plates, Stanton's experiments give as an average P = o*oo36V 2 . Recent experiments by Eiffel, Hagen, and others show that the total pressure on a surface depends partly on the area of the surface exposed to the wind, and partly on the periphery of the surface. The total pressure P is fairly well represented by an expression of the form P = (a + ^)SV 2 where S = normal surface exposed to the wind in square feet ; / = periphery of the surface in feet. a and b are constants. When a wind blowing horizontally impinges on a flat, inclined surface, the pressure in horizontal, vertical, and normal directions may be arrived at thus the normal pressure P n = P . sin 6 the horizontal pressure P A = P w sin 9 = P sin 2 the vertical pressure P p = P w cos = P . cos . sin 6 In the above we have neg- lected the friction of the air moving over the inclined sur- face, which will largely ac- count for the discrepancy between the calculated pres- sure and that found by expe- riment. The following table will enable a comparison to be made. The experimental values have been reduced to a horizontal pressure of 40 Ibs. per Stnictiires. 503 square foot of vertical section of air stream acting on a flat vertical surface. Normal pressure. Vertical pressure. Horizontal pressure. Angle of roof. Experi- ment. 1 Psin0. Experi- ment. P cos sin0. Experi- ment. P sin" 0. 10 97 7'0 9'6 6-9 1-7 1-2 20 137 17-0 12-9 6'2 47 30 26-4 20'0 22-8 I7-3 I3-2 lO'O 40 33'3 257 25-5 197 21'4 16-5 50 38-1 30-6 24' 5 197 29*2 23'5 60 40*0 34-6 20'0 i7'3 34 'o 30-0 70 41-0 37-6 I4*O 12-9 38-5 35'4 When the wind blows upon a surface other than plane, the pressure on the projected area depends upon the form of the surface. The following table gives some idea of the relative wind- resistance of various surfaces, as found by various experiments : diameter Flat plate .. Parachute (concave surface), depth = ... i'2-2 Sphere o'36-O'4i Elongated projectile 0-5 Cylinder 0-54-0-57 Wedge (base to wind) 0-8-0-97 ,, (edge to wind), vertex angle 90 0-6-07 Cone (base to wind) 0*95 ,, (apex to wind), vertex angle 90 0-69-0*72 60 0-54 Lattice girders about O'8 The pressure and velocity of the wind increase very much as the height above the ground increases (Stevenson's experi- ments). Feet above ground Velocity in miles per hour . . , 5 I? /If 1 26 9 6 17 23 28 32 15 6 18 25 31 34 21 30 35 37 52 7'5 23 32 40 43 1 Deduced from Button's experiments by Unwin (see "Iron Roofs and Bridges"). 504 Mechanics applied to Engineering. These figures appear to show that the pressure varies roughly as the square root of the height above the ground. The wind pressure measured by small gauges is always higher than when measured with gauges offering a large surface to the wind, probably because the highest pressures are only confined to very small areas, and are much greater than the mean taken on a larger surface. And, moreover, the ratio of the periphery to the surface exposed is greater in small than in large gauges; hence, for the reasons pointed out above, we might anticipate such a result. In the experiments at the Forth Bridge, this was very clearly shown. For example Small Small Large fixed gauge, 15' X 20'. gauge. gauge. Mean. Centre. Corner. Mar. 31, 1886 26 31 19 28* 22 Jan. 25, 1890 27 24 18 23* 22 In designing structures, it is usual to allow for a pressure of 40 Ibs. per square foot. In very exposed positions this may not be excessive, but for inland structures, unless exceptionally exposed, 40 Ibs. is unquestionably far too high an estimate. For further information on this question the reader should refer to special works on the subject, such as Walmisley's " Iron Roofs " and Charnock's " Graphic Statics," published by Broadbent and Co., Huddersfield. Weight of Roof Coverings. For preliminary estimates, the weights of various coverings may be taken as Weight per sq. foot in pounds. Covering. Slates 8-9 Tiles (flat) i2-2< Corrugated iron ... ... ... ... ... ... 1^-3^ Asphalted felt Lead Copper Snow 2-4 5-8 Weight of Roof Structures. For preliminary estimates, the following formulas will give a fair idea of the probable weight of the ironwork in a roof. Structures. 505 Let W = weight of ironwork per square foot of covered area (i.e. floor area) in pounds ; D = distance apart of principals in feet ; S = span of roof in feet. Then for trusses and for arched roofs Distribution of Load on a Roof. It will often save trouble and errors if a sketch be made of the load distribution on a roof in this manner. W/ND FIG. 487. The height of the diagram shaded normal to the roof is the weight of the covering and ironwork (assumed uniformly distributed). The height of the diagonally shaded diagram represents the wind pressure on the one side. The lowest section of the diagram on each side is left unshaded, to indicate that if both ends of the structure are rigidly fixed to the supporting walls, that portion of the load may be neglected as far as the structure is concerned. But if A be on rollers, and B be fixed, then the wind-load only on the A side must be taken into account. When the slope of the roof varies, as in curved roofs, the height of the wind diagram must be altered accordingly. An instance of this will be given shortly. The whole of the covering and wind-load must be con- centrated at the joints, otherwise bending stresses will be set up in the bars. Method of Sections. Sometimes it is convenient to check the force acting on a bar by a method known as the Mechanics applied to Engineering. method of sections usually attributed to Ritter, but really due to Rankine termed the method of sections, because the structure is supposed to be cut in two, and the forces required to keep it in equilibrium are calculated by taking moments. Suppose it be re- quired to find the force acting along the bar/^. Take a section through the structure db ; then three forces, pa, qe, pq, must be applied to the cut bars to keep the structure in equili- FIG. 488. brium. Takemoments about the point O. The forces pa and qe pass through O, and therefore have no moment about it; but/ty has a moment/^ x y about O. By this method forces may often be arrived at which are difficult by other methods. Forces in Roof Structures. We have already shown in Chapter IV. how to construct force or reciprocal diagrams for simple roof structures. Space will only allow of our now dealing with one or two cases in which difficulties may arise. In the truss shown (Fig. 489), a difficulty arises after the force in the bar tu has been found. Some writers, in order to get over the difficulty, assume that the force in the bar rs is the same as in ut. This may be the case when the structure is evenly loaded, but it certainly is not so when wind is acting on one side of the structure. We have taken the simplest case of loading possible, in order to show clearly the special method of dealing with such a case. The method of drawing the reciprocal diagram has already been described. We go ahead in the ordinary way till we reach the bar st (Fig. 489). In the place of sr and rq substitute a temporary bar xy, shown dotted in the side figure. With this alteration we can now get the force ey or eq\ then qr, rs, etc., follow quite readily ; also the other half of the structure. There are other methods of solving this problem, but the one given is believed to be the best and simplest. The author Structures. 507 is indebted to Professor Barr, of Glasgow University, for this method. When the wind acts on a structure, having one side fixed and the other on rollers, the only difficulty is in finding the reactions. The method of doing this by means of a funicular polygon is shown in Fig. 490. The funicular polygon has been fully described in Chapters IV. and X., hence no further description is necessary. The direction of the reaction at the fixed support FIG. 4^9- is unknown, but as it must pass through the point where the roof is fixed, the funicular polygon should be started from this point. The direction of the reaction at the roller end is vertical, hence from/ in the vector polygon a perpendicular is dropped to meet the ray drawn parallel to the closing line of the funicular polygon. This gives us the point a ; then, joining ba, we get the direction of the fixed reaction. The reciprocal 508 Mechanics applied to Engineering. diagram is also constructed ; it presents no difficulties beyond that mentioned in the last paragraph. In the figure, the vertical forces represent the dead weight on the structure, and the inclined forces the wind. The two are combined by the parallelogram of forces. In designing a structure, a reciprocal diagram must be drawn for the structure, both when the wind is on the roller Vector polygon for finding the reactions. Reciprocal diagram. FIG. 400. and on the fixed side of the structure, and each member of the structure must be designed for the greatest load. The nature of the forces, whether compressive or tensional, must be obtained by the method described in Chapter IV. Island Station Roof. This roof presents one or two interesting problems, especially the stresses in the main post. The determination of the resultant of the wind and dead load at each joint is a simple matter. The resultant of all the forces is given by/# on the vector polygon in magnitude and direction (Fig. 491). Its position on the structure must then be deter- mined. This has been done by constructing a funicular polygon Structures. 509 in the usual way, and producing the first and last links to meet in the point Q. Through Q a line is drawn parallel to pa in Vector polygon for finding the reaction. FIG. 491. the vector polygon. This resultant cuts the post in r^ and may be resolved into its horizontal and vertical components, the horizontal component producing bending moments of different sign, thus giving the post a double curvature (Fig. 492). The bending moment _on the post is obtained by the product pa X Z, where Z is the perpendicular distance of Q from the apex of the post. When using reciprocal diagrams for determining the stresses in structures, we can only deal with direct tensions and FIG. 492< compressions. But in the present instance, where there is bending in one of the members, we must introduce an imaginary external force to prevent this bending action.. It will be convenient to assume that the structure is pivoted at the virtual joint r, and that an external horizontal force F is introduced at the apex Y to keep the structure in equilibrium. The value of F is readily found thus. Taking moments about r, we have ro Mechanics applied to Engineering. F X rY = J~a x z z is the perpendicular distance from the point Y to the resultant. On drawing the reciprocal diagram, neglecting F, it will be found that it will not close. This force is shown dotted on the reciprocal diagram / *", and on measurement will be found to be equal to F. Dead and Live Loads on Bridges. The dead loads consist of the weight of the main and cross girders, floor, ballast, etc., and, if a railway bridge, the permanent way ; and the live loads consist of the train or other traffic passing over the bridge, and the wind pressure. The determination of the amount of the dead loads and the resulting bending moment is generally quite a simple FIG. 493. matter. In order to simplify matters, it is usual to assume (in small bridges) that the dead load is evenly distributed, and consequently that the bending-moment diagram is parabolic. In arriving at the bending moment on railway bridges, an equivalent evenly distributed load is often taken to represent the actual but somewhat unevenly distributed load due to a passing train. The maximum bending moment produced by a train which covers a bridge (treated as a standing load) can be arrived at thus (Fig. 493). Take a span greater than twice the actual span, so as to get every possible combination of loads that may come on the structure. Construct a bending-moment diagram in the ordinary way, then find by trial where the greatest bending moment occurs, by fitting in a line whose length is equal to the span. A parabola may then be -drawn to enclose Structures. this diagram as shown in the lower figure ; then, if d depth ivl^ of this parabola to proper scale, we have = \ / k ez - $ C < . d e I / FIG. 498. 516 Mechanics applied to Engineering. In dealing with the second method, the forces mn, kg acting on the two end verticals are simply the reactions of "Fig. A. There is less liability to error if they are treated as two upward forces, as shown in Fig. C, than if they are left in as two vertical bars. It will be seen, from the reciprocal diagram, that the force in qs is the same as that in rt, which, of course, must be the case, as they are one and the same bar. Incomplete and Redundant Framed Structures. If a jointed structure have not sufficient bars to make it retain its original shape under all conditions of loading, it is termed an "incomplete" structure. Such a structure may, however, be used in practice for one special distribution of loading which never varies, but if the distribution should ever be altered, the structure will change its shape. The determination of the forces acting on the various members can be found by the reciprocal diagram. But if a structure have more than sufficient bars to make it retain its original shape, it is termed a " redundant " structure. Then the stress on the bars depends entirely upon their relative yielding when loaded, and cannot be obtained from a reciprocal diagram. Such structures are termed " statically indeterminate structures." Even the most superficial treatment would occupy far too much space. If the reader wishes to follow up the subject, he cannot do better than consult an excellent little book on the subject, " Statically Indeterminate Structures," by Martin, published at Engineering Office. Pin Joints. In all the above cases we have assumed that all the bars are jointed with frictionless pin joints, a condition which, of course, is never obtained in an actual structure. In American bridge practice pin joints are nearly always used, but in Europe the more rigid riveted joint finds favour. When a structure deflects under its load, its shape is slightly altered, and consequently bending stresses are set up in the bars when rigidly jointed. Generally speaking, such stresses are neglected by designers. Plate Girders. It is always assumed that the flanges of a rectangular plate girder resist the whole of the bending stresses, and that the web resists the whole of the shear stresses. That such an assumption is not far from the truth is evident from the shear diagram given on p. 383. In the case of a parabolic plate girder, the flanges take some of the shear, the amount of which is easily determined. The determination of the bending moment by means of a diagram has already been fully explained. The bending Structtires. 517 moment at any point divided by the corresponding depth of the girder gives the total stress in the flanges, and this, divided by the intensity of the stress, gives the net area of one flange. In the rectangular girder the total flange stress will be greatest in the middle, and will diminish towards the abutments, consequently the section of the flanges should correspondingly diminish. This is usually accomplished by keeping the width of the flanges the same throughout, and reducing i i , the thickness by reducing , ' j the number of plates. The NJ ; b ending-moment diagram N< lends itself very readily to the stepping of the plates. Thus suppose it were found FIG. 499. that four thicknesses of plate were required to resist the bending stresses in the flanges in the middle of the girder ; then, if the bending-moment diagram be divided into four strips of equal thickness, each strip will represent one plate. If these strips be projected up on to the flange as shown, it gives the position where the plates may be stepped. 1 The shear in the web may be conveniently obtained from the shear diagram (see Chapter X.). Then if S = shear at any point in tons, f, = permissible shear stress, usually not exceeding 3 tons per square inch of gross section of web, d w = depth of web in inches, t = thickness of plate in inches (rarely less than f inch), we have The depth is usually decided upon when scheming the girder ; it is frequently made from -g- to yV span. The thickness of web is then readily obtained. If on calculation the thickness comes out less than f inch, and it has been decided not to use a thinner web, the depth in some cases is decreased accordingly within reasonable limits. 1 It is usual to allow from 6 inches to 12 inches overlap of the plates beyond the points thus obtained. 5 i8 Mechanics applied to Engineering. The web is attached to the flanges or booms by means of angle irons arranged thus : FIG. 500. The pitch The of the rivets must be such that the bearing and shearing stresses are within the prescribed limits. On p. 314 we showed that, in the case of any rectangular element subject to shear, the shear stress is equal in two directions at right angles, i.e. the shear stress along ef shear stress along ed t which has to be taken by the rivets a, a, Fig. 500. shear per (gross) inch run of web plate, along ef FIG. 501. The shearing resistance of each rivet is (in double shear) Whence, to satisfy shearing conditions This pitch is, however, often too small to be convenient ; then two (zigzag) rows of rivets are used, and / = twice the above value. The bearing resistance of each rivet is Structures. 519 The bearing pressure f b is usually taken at about 8 tons per square inch. We get, to satisfy bearing-pressure conditions p = $d (for a single row of rivets) (approximately) p = 6d (for a double row of rivets) The joint-bearing area of the two rivets b, b attaching the angles to the booms is about twice that of a single rivet (a) through the web ; hence, as far as bearing pressure is concerned, single rows are sufficient at , b. A very common practice is to adopt a pitch of 4 inches, putting two rows in the web at a, a, and single rows at , b. The pitch of the rivets in the vertical joints of the web (with double cover plates) is the same as in the angles. The shear diminishes from the abutments to the middle of the span, hence the thickness of the web plates may be diminished accordingly. It often happens, however, that it is more convenient on the whole to keep the web plate of the same thickness throughout. The pitch/ of the rivets may then A , , , . J o 1 \ 1 o ^^ -A o o FIG. 502. be increased towards the middle. It should be remembered, however, that several changes in the pitch may in the end cost more in manufacture than keeping the pitch constant, and using more rivets. The rivets should always be arranged in such a manner that not more than two occur in any one section, in order to reduce the section of the angles as little as possible. Practical experience shows that if a deep plate girder be constructed with simply a web and two flanges, the girder will not possess sufficient lateral stiffness when loaded. In order to provide against failure from this cause, vertical tees, or angles, are riveted to the web and flanges, as shown in Fig. 502. That 520 Mechanics applied to Engineering. such stiffeners are absolutely necessary in many cases none will deny, but up to the present no one appears to have arrived at a satisfactory theory as to the dimensions or pitch required. Rankine, considering that the web was liable to buckle diagonally, due to the compression component of the shear, treated a narrow diagonal strip of the web as a strut, and proceeded to calculate the longest length permissible against buckling. Having arrived at this length, it becomes a simple matter to find the pitch of the stiffeners, but, unfortunately for this theory, there are a large number of plate girders that have been in constant use for many years which show no signs of weakness, although they ought to have buckled up under their ordinary working load if the Rankine theory were correct. The Rankine method of treatment is, however, so common that we must give our reasons for considering it to be wrong in principle. From the theory of shear, we know that a pure shear consists of two forces of equal magnitude and opposite in kind, acting at right angles to one another, each making an angle of 45 with the roadway ; hence, whenever one diagonal strip of a web is subjected to a compressive stress, the other diagonal is necessarily subjected to a tensile stress of equal intensity. Further, we know that if a long strut of length / be supported laterally (even very flimsily) in the middle, the effective length of that strut is thereby reduced to -, and in general the effective length of any strut is the length of its longest unstayed segment. But even designers who adhere to the Rankinian theory of plate webs act in accordance with this principle when designing latticed girders, in which they use thin flat bars for compression members, which are quite incapable of acting as long struts. But, as is well known, they do not buckle simply because the diagonal ties to which they are attached prevent lateral deflection, and the closer the lattice bracing the smaller is the liability to buckling; hence there is no tendency to buckle in the case in which the lattice bars become so numerous that they touch one another, or become one continuous plate, since the diagonal tension in the plate web effectually prevents buckling along the other diagonal, provided that the web is subjected to shear only. What, then, is the object of using stiffeners ? Much light has recently been thrown o*n this question by Mr. A. E. Guy (see " The Flexure of Beams : " Crosby Lockwood and Son), who has very thoroughly, both experimentally and analytically, treated the question of the twisting of deep narrow sections. It is well Structures. 521 known that for a given amount of material the deeper and narrower we make a beam of rectangular section the stronger will it be if we can only prevent it from twisting sideways. Mr. Guy has investigated this point, and has made the most important discovery that the load at which such a beam will buckle sideways is that load which would buckle the same beam if it were placed vertically, and thereby converted into a strut. If readers will refer to the published accounts (" Menai and Conway Tubular Bridges," by Sir William Fairbairn) of the original experiments made on the large models of the Menai tubular bridge by Sir William Fairbairn, they will see that failure repeatedly occurred through the twisting of the girders ; and in the later experiments two diagonals were put in in order to prevent this side twisting, and finally in the bridge itself the ends of the girders were supported in such a way as to prevent this action, and in addition substantial gusset stays were riveted into the corners for the same purpose. Some tests by the author on a series of small plate girders of 15 feet span, showed in every case that failure occurred through their twisting. The primary function, then, of web stiffeners is believed to be that of giving torsional rigidity to the girder to prevent side twisting, but the author regrets that he does not see any way of calculating the pitch of plate or tee-stiffeners to secure the necessary stiffness ; he trusts, however, that, having pointed out what he believes to be the true function of stiffeners, others may be persuaded to pursue the question further. This twisting action appears to show itself most clearly when the girder is loaded along its tension flange, i.e. when the compression flange is free to buckle. Probably if the load were evenly distributed on flooring attached to the compression flange, there would be no need for any stiffeners, because the flooring itself would prevent side twisting ; in fact, in the United States one sees a great many plate girders used without any web stiffeners at all when they are loaded in this manner. But if the flooring is attached to the bottom of the girder, leaving the top compression flange without much lateral support, stiffeners will certainly be required to keep the top flange straight and parallel with the bottom flange. The top flange in such a case tends to pivot about a vertical axis passing through its centre. For this reason the ends should be more rigidly stiffened than the middle of the girder, which is, of course, the common practice, but it is usually assigned to another cause. There are, however, other reasons for using web stiffeners. Whenever a concentrated load is applied to either flange it 522 Mechanics applied to Engineering. produces severe local stresses ; for example, when testing rolled sections and riveted girders which, owing to their shortness, do not fail by twisting, the web always locally buckles just under the point of application of the load. This local buckling is totally different from the supposed buckling propounded by Rankine. By riveting tee-stiffeners on both sides of the web, the local loading is more evenly distributed over it, and the buckling is thereby prevented. Again, when a concentrated load is locally applied to the lower flange, it tends to tear the flange and angles away from the web. Here again a tee or plate stiffener well riveted to the flange and web very effectually prevents this by distributing the load over the web. In deciding upon the necessary pitch of stifTeners / there should certainly be one at every cross girder, or other con- centrated load, and for the prevention of twisting, well-fitted plate sthTeners near the ends, pitched empirically about 2 feet 6 inches to 3 feet apart ; then alternate plate and tee stiffeners, increasing in pitch to not more than 4 feet, appear to accord with the best modern practice. Weight of Plate Girders. For preliminary estimates, the weight of a plate girder may be arrived at thus Let w weight of girder in tons per foot run ; W = total load on the girder, not including its own weight, in tons. W Then w = - roughly Arched Structures. We have already shown how to determine the forces acting on the various segments of a suspension-bridge chain. If such a chain were made of suitable form and material to resist compression, it would, when inverted, simply become an arch. The exact profile taken up by a suspension- bridge chain depends entirely upon the distribution of the load, but as the chain is in tension, and, moreover, in stable equilibrium, it immediately and automatically adjusts itself to any altered condition of loading ; but if such a chain were inverted and brought into compression, it would be in a state of unstable equilibrium, and the smallest disturbance of the load distribution would cause it to collapse immediately. Hence arched structures must be made of such a section that they will resist change of shape in profile ; in other words, they must be capable of resisting bending as well as direct stresses. Masonry Arches. In a masonry arch the permissible Structures. 523 bending stress is small in order to ensure that there may be no, or only a small amount of, tensile stress on the joints of the voussoirs, or arch stones. Assuming for the present that there may be no tension, then the resultant line of thrust must lie within the middle third of the voussoir (see p. 455). In order to secure this condition, the form of the arch must be such that under its normal system of loading the line of thrust must pass through or near the middle line of the voussoirs. Then, when under the most trying conditions of live loading, the line of thrust must not pass outside the middle third. This condition can be secured either by increasing the depth of the voussoirs, or by increasing the dead load on the arch in order to reduce the ratio of the live to the dead load. Many writers still insist on the condition that there shall be no tension in the joints of a masonry structure, but every one who has had any experience of such structures is perfectly well aware that there are very few masonry structures in which the joints do not tend to open, and yet show no signs of instability or unsafeness. There is a limit, of course, to the amount of permissible tension. If the line of thrust pass through the middle third, the maximum intensity of compres- sive stress on the edge of the voussoir is twice the mean, and if there be no adhesion between the mortar and the stones, the intensity of compressive stress is found thus Let T = the thrust on the voussoirs at any given joint per unit width ; d = the depth or thickness of the voussoirs ; x = the distance of the line of thrust from the middle of the joint; P = the maximum intensity of the compressive stress on the loaded edge of the joint. The distance of the line of thrust from the loaded edge is - x, and since the stress varies directly as the distance from 2 this edge, the diagram of stress distribution will be a triangle with its centre of gravity in the line of thrust, and the area of the triangle represents the total stress ; hence = T 2T and P = d \ i ~ *) 524 Mechanics applied to Engineering. T The mean stress on the section = d hence, when d x = , P = 2 X mean intensity of stress x = -, P = 2 x mean intensity 4 A masonry arch may fail in several ways; the most important are 1. By the crushing of the voussoirs. 2. By the sliding of one voussoir over the other. 3. By the tilting or rotation of the voussoirs. The first is avoided by making the voussoirs sufficiently deep, or of sufficient sectional area to keep the compressive stress within that considered safe for the material. This condition is fulfilled if 2't, - . ., . the permissible compressive stress The depth of the arch stones or voussoirs d in feet at the keystone can be approximately found by the following expressions : Let S = the clear span of the arch in feet ; R a = the radius of the arch in feet. Then d = ^ for brick to ^- for masonry 2'2 3 or, according to Trautwine where K = i for the best work ; i *i 2 for second-class work; i '33 for brickwork. The second-mentioned method of failure is avoided by so arranging the joints that the line of thrust never cuts the normal to the joint at an angle greater than the friction angle. Let tt be the line of thrust, then sliding will occur in the manner Structures. 525 shown by the dotted lines, if the joints be arranged as shown at bb that is, if the angle a exceeds the friction angle <. But if the joint be as shown at aa, sliding cannot occur. The third-mentioned method of failure only occurs when the line of thrust passes right outside the section j the voussoirs then tilt till the line of thrust passes through the pivoting points. An arch can never fail in this way if the line of thrust be kept inside the middle halt. FIG. 503. FIG. 504. The rise of the arch R (Fig. 505) will depend upon local conditions, and the lines of thrust for the various conditions of loading are constructed in precisely the same manner as the link-and-vector polygon. A line of thrust is first constructed for the distributed load to give the form of the arch, and if the line of thrust comes too high or too low to suit the desired rise, it is corrected by altering the polar distance. Thus, suppose the rise of the line of thrust were R , and it was required to bring it to Rj. If the original polar distance were O H, the new polar distance T> required to bring the rise to R! would be OiH = O H x ^ K-i After the median line of the arch has been constructed, other link polygons, such as the bottom right-hand figure, are drawn in for the arch loaded on one side only, for one-third of the length, on the middle only, and any other ways which are likely to throw the line of thrust away from the median line. After these lines have been put in, envelope curves parallel to the median line are drawn in to enclose these lines of thrust at every point ; this gives us the middle half of the voussoirs. The outer lines are then drawn in equidistant from the middle half lines, making the total depth of the voussoirs equal to twice the depth of the envelope curves. An infinite number of lines of thrust may be drawn in for any given distribution of load. Which of these is the right one? is a question by no means easily answered, and whatever 526 Mechanics applied to Engineering. answer may be given, it is to a large extent a matter of opinion. For a full discussion of the question, the reader should refer to I. O. Baker's " Treatise on Masonry " (Wiley and Co., New York); and a paper by H. M. Martin, I.C.E. Proceedings, vol. xciii. p. 462. K f I FIG. 505. From an examination of several successful arches, the author considers that if, by altering the polar distance OH, a line of thrust can be flattened or bent so as to fall within the middle half, it may be concluded that such a line of thrust is admissible. One portion or point of it may touch the inner and one the outer middle half lines. As a matter of fact, an Structures. 527 exact solution of the masonry arch problem in which the voussoirs rest on plane surfaces is indeterminate, and we can only say that a certain assumption is admissible if we find that arches designed on this assumption are successful. Arched Ribs. In the case of iron and steel arches, the line of thrust may pass right outside the section, for in a continuous rib capable of resisting tension as well as compres- sion the rib retains its shape by its resistance to bending. The bending moment varies as the distance of the line of thrust from the centre of gravity of the section of the rib. The determination of the position of the line of thrust is therefore important. Arched ribs are often hinged at three points at the FIG. 506. springings and at the crown. It is evident in such a case that the line of thrust must pass through the hinges, hence there is no difficulty in finding its exact position. But when the arch is rigidly held at one or both springings, and not hinged at the crown, the position of the line of thrust may be found thus : l FIG. 508. 1 See also a paper by Bell, J.C.E., vol. xxxiii. p. 68. 528 Mechanics applied to Engineering. Consider a short element of the rib mn of length /. When the rib is unequally loaded, it is strained so that mn takes up the position m'ti. Both the slope and the vertical position of the element are altered by the straining of the rib. First consider the effect of the alteration in slope; for this purpose that portion of the rib from B to A may be considered as being pivoted at B. Join BA ; then, when the rib is strained, BA becomes BAj. Thus the point A has received a horizontal displacement AD, and a vertical displacement AjD. The two triangles BAE, AAjD are similar; hence being expressed in circular measure. Likewise A]L) X A Aj ^ AA, = AB .A,D = AB * = *'* . . (u.) A similar relation holds for every other small portion of the rib, but as A does not actually move, it follows that some of the horizontal displacements of A are outwards, and some inwards. Hence the algebraic sum of all the horizontal displacements must be zero, or 20y = o .......... (iii.) Now consider the vertical movement of- the element. If the rib be pivoted at C, and free at A, when mn is moved to m'n', the rib moves through an angle C , and the point A receives a vertical displacement S . 6 C . We have previously seen that A has also a vertical displacement due to the bending of the rib ; but as the point A does not actually move vertically, some of the vertical displacements must be upwards, and some downwards. Hence the algebraic sum of all the vertical displacements is also zero, or - 4- S0 C = o By similar reasoning S0 A = o or 2(S - x)B + S0 A = o (v.) If the arch be rigidly fixed at C = o and 20* = o ,,.,,., (vi.) Structttres. 529 If it be fixed at A 0A = and 2(S - x)6 = o from v. If it be fixed at both ends, we have by addition - xB + x&) = o Then, since S is constant 20 = o ....... (vii.) Whence for the three conditions of arches we have Arch hinged at both ends, ^}>0 = o from iii. fixed 2y0, and HO - o from iii. and vii. ,, one end only, 3y0, and H^xO = o from i. and vi. We must now find an expression for 0. Let the full line represent the portion of the unstrained rib, and the dotted line the same when strained. Let the radius of curvature before strain- ing be /o , and after straining p x . Then, using the symbols of Chapter XL, we have M = 5*, and M x == Po Pi Then the bending moment on the rib due to the change of curvature when strained FIG. 509. M = M x - Mo = Elf- - -) 7 (viii.) But as / remains practically constant before and after the strain, we have 0, = L, and = - Pi Po and e i -Ot = = /- - I I Pi Po 2 M 530 Mechanics applied to Engineering. Then we have from viii. and ix. M/ = EW M/ =- If the arched rib be of constant cross-section, / El is constant ; but if it be not so, then the length / must be taken so that y is constant. The bending moment M on the rib is M = F . ab, where the lowest curved line through cb is the line of thrust, and the upper dark line the median line of the rib. Draw ac vertical at r; dc tangential at c\ ^ ab normal to dc ; ad horizontal. Let H be the horizontal thrust on the vector polygon. Then the triangles adc, dd c 1 , also adb, cda, are similar ; hence F cd T ab ac = and = H ad ad cd or a ~ = a - = 3 FIG. 510. ac Cd F or F . ab = H . ac = M but H is constant for any given case. Let ac = z. Hence the expression S0y = o may be written SMy = o, since -gy is constant or 'Sz . y = o Then, substituting in a similar manner in the equations above, we have Arch hinged at both ends, S? . y = o fixed ^ .y = o, and 2^ = o at one end only, Sz . y = o, and ^ocz = o Thus, after the median line of the arch has been drawn, a line of thrust for uneven loading is constructed; and the Structures. 531 median line is divided up into a number of parts of length /, and perpendiculars dropped from each. The horizontal dis- tances between them will, of course, not be equal; then all the values z X y must be found, some z*s being negative, and if FIG. 511. some positive, and the sum found. If the sum of the negative values are greater than the positive, the line of thrust must be raised by reducing the polar distance of the vector polygon, and vice versd if the positive are greater than the negative. The line of thrust always passes through the hinged ends. In the case of the arch with fixed ends, the sum of the z's must also FIG. 512. be zero ; this can be obtained by raising or lowering the line of thrust bodily. When one end only is fixed, the sum of all the quantities x . z must be zero as well as zy this is obtained by shifting the line of thrust bodily sideways. Having fixed on the line of thrust, the stresses in the rib are obtained thus : The compressive stress all over the rib at any section is T where T is the thrust obtained from the vector polygon, and A is the sectional area of the rib. The skin stress due to bending is , (see Fig. 510) , M T . /= =- or/ = ~ 532 Mechanics applied to Engineering. and the maximum stress in the material due to both Except in the case of very large arches, it is never worth while to spend much time in getting the exact position of the worst line of thrust ; in many instances its correct position may be detected by eye within very small limits of error. Effect of Change of Length and Temperature on Arched Ribs. Long girders are always arranged with ex- pansion rollers at one end to allow for changes in length as the temperature varies. Arched ribs, of course, cannot be so treated hence, if their length varies due to any cause, the radius of curvature is changed, and bending stresses are thereby set up. The change of curvature and the stress due to it may be arrived at by the following approximation, assuming the rib to be an arc of a circle : Let N! = N(I - - \ N 2 = R 2 + 4 FIG. sis. N 2 - N x 2 = R 2 - Substituting the value of Nj and reducing, we get But R - R! = 5 and R + R! = 2R (nearly) The fraction - is very small, viz. ?, and is rarely more than n L N 2 ; hence the quantity involving its square, viz. - 9000000' is negligible. Then we get Structures. 533 S = ^> 2 (x.) But ( - ) * = N 2 - 2 N a - R 2 = p 2 - (p - R) 2 from which we get N 2 = 2 P R Substituting in x., we have N 2 and p = -ji N, 2 also p l = -^r The bending moment on the rib due to the change of curvature is M = Elf- ) from viii. = EI( - ) \P Pi/ and the corresponding skin stress is M My M ,5 3,114,160 I4'3 -I4-3 4,163,375 H } 45,050,640 MATERIAL. Krupp's Spring Steel. Tensile strength, 57-5 tons per sq. inch. Tensile str in tons p in from ess applied er square ch to Number of repetitions before fracture. Tensile str in tons p in from 2ss applied er square ch to Number of repetitions before fracture. 4775 7-92 62,OOO . 38-20 4-77 99,700 ,, 15-92 149,800 , , 9'55 176,300 > 23-87 400,050 55 I4-33 619,600 ,, 27-83 376,700 ,, 2,135,670 ,, 31-52 19,673,000 s 19-10 35,800,000 (unbroken) (unbroken) 42-95 9'55 81,200 33'4 J 4-77 286, 100 5, 14*33 1,562,000 j) 9'55 701,800 ,, 19*10 225,300 ,, 11-94 36,600,000 ,, 23-87 1,238,900 (unbroken) II ,, 300,900 II 28-65 33,600,000 (unbroken) Structures. 539 Various theories have been advanced to account for the results obtained by Wohler, all more or less unsatisfactory, but there are one or two empirical formulas which fairly well represent the results. Of these, probably Gerber's parabolic equation with Unwin's constants member. fits the results, and is very easy of application, and is, more- over, simple to remember; but whether the assumptions of the theory are justifiable or not is quite an open question, which we Stcutzc jbrecJcing stress 1 best fits the results ; it is, however, not easy to re- The "dynamic theory" equation, however, fairly a 1-0 ft / t* W ^ ^ 9 / Mean ofto ^ /* // 7 ^* n-jt ft-2 A / Tens um, / Zert stress U"^ Cbmpn "V utMOrt/ / 0-2 O-2 O-4 O-6 O-3 t / O-4 FIG. 517. shall not discuss. We adopt it simply because it is the easiest to use, and, for all practical purposes, represents Wohler's, Spangenberg's, and Bauschinger's results. The " dynamic theory" assumes that the varying loads applied by Wohler and others were equivalent to suddenly applied loads, and conse- quently a piece of material will not break under repeated loadings if the " momentary " stress, due to sudden applications, does not exceed the statical breaking strength of the material. In Fig. 517 we show, by means of a diagram^ the results of all the 1 " Elements of Machine Design," p. 32. 540 Mechanics applied to Engineering. published experiments reduced to a common scale. In every case represented the material stood over four million repetitions before fracture. The horizontal scale is immaterial ; the vertical scale shows the ratio of the applied stress to the static breaking stress. The minimum stress on the material is plotted on the line aob, and the corresponding maximum stress, which may be repeated over four million times, is shown by the small circles above. If the dynamic theory were perfectly true, all the^ points would lie on the line marked " maximum stress ; " for then the minimum stress (taken as being due to a dead load) plus twice the range of stress (i.e. maximum stress minimum stress) taken as being due to a live load should together be equal to the statical breaking strength of the material. The results of tests of revolving axles are shown in group A; the dynamic theory demands that they should be repre- sented by a point situated 0*33 from the zero stress axis. Likewise, when the stress varies from o to a maximum, the results are shown at B ; by the dynamic theory, they should be represented by a point situated 0*5 from the zero axis. For all other cases the upper points should lie on the maximum stress line. Whether they do lie reasonably near this line must be judged from the diagram. When one considers the many accidental occurrences that may upset such experiments as these, one can hardly wonder at the points not lying regularly on the mean line. To sum up, then. When designing a member which will be subjected to both a steady load, which we will term W mil , , and a fluctuating load (W^. W min- ), the equivalent static load W = W^. + 2(W ma , - W min .) The//#j sign is used when both the loads act together, i.e. when both are tension or both compression, and the minus when they act against one another. For a fuller discussion of this question the reader is referred to Unwin's "Testing of Material;" Fidler's "Practical Treatise on Bridge Construction ; " Johnson's " Materials of Construction ; " and for the effect of very rapid reversals, a paper by Osborne Reynolds and J. H. Smith on a "Throw Testing Machine for Reversals of Mean Stress," Phil. Trans. , vol. 199, p. 265. CHAPTER XVI. HYDRAULICS. IN Chapter VIII. we stated that a body which resists a change of form when under the action of a distorting stress is termed a solid body, and if the body returns to its original form after the removal of the stress, the body is said to be an elastic solid (e.g. wrought iron, steel, etc., under small stresses) but if it retains the distorted form it assumed when under stress, it is said to be a plastic solid (e.g. putty, clay, etc.). If, on the other hand, the body does not resist a change of form when under the action of a distorting stress, it is said to be a fluid body ; if the change of form takes place immediately it comes under the action of the distorting stress, the body is said to be a perfect fluid (e.g. alcohol, ether, water, etc., are very nearly so) ; if, however, the change of form takes place gradually after it has come under the action of the distorting stress, the body is said to be a viscous fluid (e.g. tar, treacle, etc.). The viscosity is measured by the rate of change of form under a given distorting stress. In nearly all that follows in this chapter, we shall assume that water is a perfect fluid; in some instances, however, we shall have to carefully consider some points depending upon its viscosity. Weight of Water. The weight of water for all practical purposes is taken at 62*5 Ibs. per cubic foot, or 0*036 Ib. per cubic inch. It varies slightly with the temperature, as shown in the table on the following page, which is for pure distilled water. The volume corresponding to any temperature can be found very closely by the following empirical formula : Volume at absolute temperature T, taking f -p 2 _j_ OOQ the volume at 39*2 Fahr. or 500^ = ^~ absolute as i [ Pressure due to a Given Head. If a cube of water of 542 Mechanics applied to Engineering. i foot side be imagined to be composed of a series of vertical columns, each of i square inch section, and i foot high, each will weigh -?- = 0-434 Ib. Hence a column of water i foot 144 high produces a pressure of 0-434 Ib. per square inch. 32. Temp. Fahr. 39-2. 50. 1 00. 150. 200. 212. Ice. Water. Weight per cubic foot in 57'2 62-417 62-425 62-409 62-00 6l'20 60-I4 59^4 Ibs Volume of a given weight, taking water I'OpI I'OOOI I'OOOO I"OOO2 1-007 I'020 I-038 1-043 at 39-2 Fahr. as i The height of the column of water above the point in question is termed the head. Let h the head of water in feet above any surface ; / = the pressure in pounds per square inch on that surface ; w = the weight of a column of water i foot high and i square inch section ; = 0-434 Ib. Then / = wh^ or 0*434^ or h = *- = 2'305/, or say 2-31^ Thus a head of 2-31 feet of water produces a pressure of i Ib. per square inch. Taking the pressure due to the atmosphere as 147 Ibs. per square inch, we have the head of water corresponding to the pressure of the atmosphere 147 X 2-31 = 34 feet (nearly) This pressure is the same in all directions, and is entirely inde- pendent of the shape of the containing vessel. Thus in Fig. 518 Hydraulics. 543 The pressure over any unit area of surface at a = p a = 0*434^ b = A = 0-434^ and so on. The horizontal width of the triangular diagram at the side shows the pressure per square inch at any depth below the surface. Thus, if the height of the triangle be made to a scale of i inch to the foot, and the width of the base 0-434^, the width of the triangle measured iri inches will give the pressure in pounds per square inch at any point, at the same depth below the surface. Compressibility of Water. The popular notion that water is incompressible is erroneous ; the alteration of volume under such pressures as are usually used is, however, very small. Experiments show that the alteration in volume is proportional to the pressure, hence the relation between the change of volume when under pressure may be expressed in the same form as we used for Young's modulus on p. 320. Let v the diminution of volume under any given pressure p in pounds per square inch (corresponding to x on p. 320) ; V = the original volume (corresponding to / on p. 320) ; K = the modulus of elasticity of volume of water; "/ = the pressure in pounds per square inch. FIG. 518. K = from 320,000 to 300,000 Ibs. per square inch. Thus water is reduced in bulk or increased in density by i per cent, when under a pressure of 3000 Ibs. per square inch. This is quite apart from the stretch of the containing vessel. Total Pressure on an Immersed Surface. If, for any purpose, we require the total normal pressure acting on an immersed surface, we must find the mean pressure acting on the surface, and multiply it by the area of the surface. We shall show that the mean pressure acting on a surface is the pressure due to the head of water above the centre of gravity of the surface. 544 Mechanics applied to Engineering. Let Fig. 519 represent an immersed surface. Let it be divided up into a large number of horizontal strips of length /!, />, etc. , and of width b each at a depth /z,, h^ etc., respectively from the surface. Then the total pressure on each strip is A^> A4i etc., where p^ / a , etc., are the pressures corre- FlG - 519- spending to h^ h^ etc. But/ = why and l^b a^ IJ> = a^ etc. The total pressure on each strip = waji^ waji^ etc. Total pressure on whole surface = P = w(a l h l + a. 2 /t. 2 + etc.) But the sum of all the areas # 1} 2 , etc., make up the whole area of the surface A, and by the principle of the centre of gravity (p. 58) we have ajii + a^ 2 +> etc., = AH where H is the depth of the centre of gravity of the immersed surface from the surface of the water, or P = ^AH Thus the total pressure in pounds on the immersed surface is the area of the surface in square units X the pressure in pounds per square unit due to the head of water above the centre of gravity of the surface. Centre of Pressure. The centre of pressure of a plane immersed surface is the point in the surface through which the resultant of all the pressure on the surface acts. It can be found thus Let H c = the head of water above the centre of pressure ; H = the head of water above the centre of gravity of the surface ; = the angle the immersed surface makes with the surface of the water ; I = the second moment, or moment of inertia of the surface about a line lying on the surface of the water and passing through o ; 1 = the second moment of the surface about a line parallel to the above-mentioned line, and passing through the centre of gravity of the surface ; Hydraulics. 545 R = the perpendicular distance between the two axes; K 2 = the square of the radius of gyration of the surface about a horizontal axis passing through the c. of g. of the surface ; lf 2) etc. = small areas at depths h^ h^ etc., respectively below the surface and at distances * 15 # 2 , etc., from O. A = the area of the surface. Taking moments about O, we have , etc. = P* c TT -f wh&^Ci -f , etc. = sin H, sin sin 2 O^x? + a 2 xlbs ' FIG. 522. 548 Mechanics applied to Engineering. Kinetic energy of the water on j _ leaving the orifice But these two quantities must be equal, or WV 2 V 3 and V = V 2gh that is, the velocity of flow is equal to the velocity acquired by a body in falling through a height of h feet. Contraction and Friction of a Stream passing through an Orifice. The actual velocity with which water flows through an orifice is less than that due to the head, mainly on account of the friction of the stream on the sides of the orifice ; and, moreover, the stream contracts after it leaves the orifice, the reason for which will be seen from the figure. If each side of the orifice be regarded as a ledge over which a stream of water is flowing, it is evident that the path taken by the water will be the result- FIG. 523. ant of its horizontal and vertical movements, and therefore it does not fall vertically as indicated by the dotted lines, which it would have to do if the area of the stream were equal to the area of the orifice. Both the friction and the contraction can be measured experimentally, but they are usually combined in one coefficient of discharge K d , which is found experimentally. Hydraulic Coefficients. The coefficient of discharge K d may be split up into the coefficient of velocity K w viz. actual velocity of flow and the coefficient of contraction K c , viz. actual area of the stream area of the orifice Then the coefficient of discharge K d = K,K Hydraulics. 549 The coefficient of resistance K _ actual kinetic energy of the jet of water leaving the orifice kinetic energy of the jet if there were no losses The coefficient of velocity for any orifice can be found experimentally by fitting the orifice into the vertical side of a tank and allowing a jet of water to issue from it horizontally. If the jet be allowed to pass through a ring distant h feet below the centre of the orifice and h% feet horizontally from it, then any given particle of water falls h^ feet vertically while travelling h z feet horizontally, where h = and / = A v g also /# 2 = v< where v 9 = the horizontal velocity of the water on leaving the orifice. If there were no resistance in the orifice, it would have a greater velocity of efflux, viz. where h is the head of water in the tank over the centre of the orifice. Then v 9 = This coefficient can also be found by means of a Pitot tube. i.e. a small sharp-edged tube which is inserted in the jet in such a manner that the water plays axially into its sharp-edged mouth ; the other end of the tube, which is usually bent for convenience in handling, is attached to a glass water-gauge. The water rises in this gauge to a height proportional to the velocity with which the water enters the sharp-edged mouth- piece. Let this height be h^ then v 2 = tj igh^ from which the coefficient of velocity can be obtained. This method is not so good as the last mentioned, because a Pitot tube, however well constructed, has a coefficient of resistance of its own, and therefore this method tends to give too low a value for K,. 550 Mechanics applied to Engineering. The area of the stream issuing from the orifice can be measured approximately by means of sharp-pointed micro- meter screws attached to brackets on the under side of the orifice plate. The screws are adjusted to just touch the issuing stream of water usually taken at a distance of about three diameters from the orifice. On stopping the flow the distance between the screw-points is measured, which is the diameter of the jet ; but it is very difficult to thus get satisfactory results. A better way is to get it by working backwards from -the coefficient of discharge. Plain Orifice. The edges should be chamfered off as shown (Fig. 524); if not, the water dribbles down the sides and makes the coefficient variable. In this case K v = about o'97, and K c = about 0*64, giving K d = 0-62. Experiments show that the sharper the edges the smaller is the coefficient, but it rarely gets below o'6i, and sometimes reaches 0-64. As a mean value K d = 0-62. Q = 0*62 A tj 2gh We shall shortly return to a consideration of this orifice, and shall 'show how to obtain this value by a rational process. ' FIG. 524. FIG. 525. Rounded Orifice. If the orifice be rounded to the same form as a contracted jet, the contraction can be entirely avoided, hence K c = i ; but the friction is rather greater than in the plain orifice, K,, = 0-94 to 0^97, according to the curvature and the roughness of the surface. The head h and the diameter of the orifice must be measured at the bottom, i.e. at the place Hydraulics. 551 where the water leaves the orifice ; as a mean value we may take Q = 0-95 A V 2gh 1 Pipe Orifice. The length of the pipe should be not less than three times the diameter. The jet contracts after leaving the square corner, aS in the sharp-edged orifice ; it expands again lower down, and fills the tube. It is possible to get a clear jet right through, but a very slight disturb- ance will make it run as shown. In the case of the clear stream, the value of K is approximately the same as in the plain orifice. When the pipe runs full, there is a sudden change of velocity from the con- tracted to the full part of the jet, with a consequent loss of energy and velocity of discharge. Let the velocity at b = V 6 ; and the head = h b the velocity at a = V ; ,, = h a (V B V ) 2 Then the loss of head = ---- a (see p. 573) FJG ^ The velocities at the sections a and b will be inversely as the respective areas. If K e be the coefficient of contraction at V b, we have V 6 = ~ ; inserting this value in the expression given above, we get V= T The fractional part of this expression is the coefficient of velocity K,, for this particular form of orifice. The coefficient 1 The beginner will do well to leave the next four paragraphs until he has mastered pp. 565-568, also p. 573. 552 Mechanics applied to Engineering. of discharge K D is from 0*94 to 0*95 of this, on account of friction in the pipe. Then, taking a mean value Q = 0-945 K The pressure at a is atmospheric, but at b it is less (see p. 566). If the stream ran clear of the sides of the pipe into the atmosphere, the discharge would be but in this case it is Let h a = nh b . Then Q x = K D A>v/2 or the discharge is D -^ times as great as before ; hence we may write i = 5 x K d Q, = and the increased discharge in the two cases is due to an (K \ 2 ^ ) nh b , whence there must be a vacuum of this amount less h b at the throat b. If a vacuum-gauge be attached as shown, the accuracy of this theory can be demonstrated. If the length of the pipe be termed h^ we have and the vacuum head = ]( if/ "" I f^ + \v^) ^& When the pipe is horizontal n = i, and the vacuum head is Hydraulics. 553 The following results were obtained by experiment : The diameter of the pipe = 0-945 inches Length h& = 2*96 ., K = 0*612 Head k^ (inches) 16-1 I3'i lO'I 8-1 6-1 4'i 3*1 K D ... 0799 0797 0797 0-792 0-786 0-780 0773 Vacuum head by ex- periment in inches 1675 I4'37 "'95 10-50 9-00 7'45 6-35 Vacuum head by cal- culation ... 16-60 14-24 11-97 10-45 8-85 7'37 6-56 Before making this experiment the pipe must be washed out with benzene or other spirit in order to remove all grease, and care must be taken that no water lodges in the flexible pipe which couples the water-gauge to the orifice nozzle. Re-entrant Orifice or Borda's Mouthpiece. If a plain orifice in the bottom of a tank be closed by a cover or valve on the upper side, the total pressure on the bottom of the tank will be P, where P is the weight of water in the tank ; but if the orifice be opened, the pressure P will be reduced by an amount P , equal to (i.) the down- ward pressure on the valve, viz. whh. \ and (ii.) by a further amount P r , due to the flow of water over the surface of the tank all round the orifice. Then we have P = FIG. 527. In the case of Borda's mouthpiece, the orifice is so far removed from the side of the tank that the velocity of flow over the surface is practically zero ; hence no such reduction of pressure occurs, or P f is zero. Let the section of the jet be a, and the area of the orifice A. 554 Mechanics applied to Engineering. Then the total pressure due to the column \ __ , . of water over the orifice * ~ the mass of water flowing per second = the momentum of the water flowing per second = S waV 2 g The water before entering the mouthpiece was sensibly at rest, hence this expression gives us the change of momentum per second. Change of momentum) per second f = im P ulse P er second, or pressure waV 2 hence a = o'5A or K c = 0-5 If the pipe be short compared to its diameter, the value of P, will not be zero, hence the value of K can only have this low value when the pipe is long. The following experiments by the author show the effect of the length of pipe on the coefficient : Length of projecting pipe expressed in diameters o A J & f 2 K d 0*61 0- 5 6 0'55 o'54 0'53 0-52 If the mouthpiece be caused to run full, which can be accomplished by stirring the water in the neighbourhood of the mouthpiece for an instant, the coefficient of velocity will be (see " Pipe Orifice ") = - = =0-71 Experiments give values from 0-69 to 073. Hydraulics. 555 Plain Orifice in a small Approach Channel. When the area a of the stream passing through the orifice is appre- ciable as compared with the area of the approach channel A a , the value of K c varies with the proportions between the two. With a small approach channel there is an imperfect con- traction of the jet, and according to Rankine's empirical formula - r6i8 A 2 where A is the area of the orifice, and A a is the area of the approach channel. The author has, however, obtained a rational value for this coefficient (see Engineering, March n, 1904), but the article is too long for reproduction here. The value obtained is 2^ 2 -f --) where n .= the ratio of the radius of the approach channel to the radius of the orifice. The results obtained by the two formulas are . Kc Rational. K c Rankine. 2 0-665 0*672 3 0-641 0*640 4 0-634 0-631 5 0-631 0*626 6 0*629 0*624 8 0-628 0-622 10 0*627 0-620 100 0-625 0-618 IOOO 0-625 0-618 Diverging Mouthpiece. This form of mouthpiece is of great interest, in that the discharge of a pipe can be greatly 556 Mechanics applied to Engineering. increased by adding a nozzle of this form to the outlet end, because the velocity of flow in the throat a is greater than the velocity due to the head of water h above it. The pressure at b is atmospheric ; * hence the pressure at a is less than atmo- spheric (see p. 566); thus the water is discharging into a partial vacuum. If a water- gauge be attached at a, and the vacuum measured, the velocity of flow at a will be found to be due to the head of water above it phis the vacuum head. We shall shortly show that the energy of any steadily flowing stream of water in a pipe in which the diameter varies gradually is constant at all sections, neglecting friction. By Bernouilli's theorem we have (see p. 566) FIG. 528. W 2g W 2g W 2g where is the atmospheric pressure acting on the free w surface of the water. The pressure at the mouth, viz. / 6 , is also atmospheric : hence =??. w w V 2 The velocity V is zero> hence is zero. Then, assuming no loss by friction, we have H ~ ^ = ~- V, 2 or h = or V 6 = and the discharge 1 This reasoning will not hold if the mouthpiece discharges into a vacuum. Hydraulics. 557 In the case above, the mouthpiece is horizontal, but if it be placed vertically with b below, the proof given above still holds ; the h must then be measured from , i.e. the bottom of the mouthpiece, provided the conditions mentioned below are fulfilled. Thus we see that the discharge depends upon the area at b, and is independent of the area at a ; there is, however, a limit to this, for if the pressure at a be negative, the stream will not be continuous. From the above, we have A 2g A W A If becomes zero, the stream breaks up, or when =-=34 feet 2 _ V 2 V 2 hence * V * = :*> - i) = 34 *g *g or t(n 2 - i) = 34 In order that the stream may be continuous, h(n^ i) should be less than 34 feet, and the maximum discharge will occur when the term to the left is slightly less than 34 feet. The following experiments demonstrate the accuracy of the statement made above, that the discharge is due to the head of water + the vacuum head. The experiments were made by Mr. Brownlee, and are given in the Proceedings of the Shipbuilders of Scotland for 1875-6. The experiments were arranged in such a manner that, in effect, the water flowed from a tank A through a diverging mouthpiece into a tank B, a vacuum gauge being attached at the throat /. The close agreement between the experimental and the FIG. 529. 558 Mechanics applied to Engineering. calculated values as given in the last two columns, is a clear proof of the accuracy of the theory given above. Head of water in tank A. Feet. Head of water in tank B. Feet. Vacuum at throat in feet of water. Velocity of flow at throat. Feet per second. Ha. By experiment. VVHa+H,). 69-24 58'85 None 65-97 66-78 69-24 69-24 50-78 None 33'5 33-5 80-97 81-43 8i*34 81-34 12*50 8-50 37-90 38-84 I2-50 s;oo 33-5 53 98 54-43 I2-50 33'5 54-60 54-43 8-00 None 33-5 5i'67 51*70 2'00 None 8-2 24-74 25-63 0-25 None 0-52 6-66 7-04 Jet Pump or Hydraulic Injector. If the height of the column of water in the vacuum gauge at / (Fig. 529) be less than that due to the vacuum produced, the water will be sucked in and carried on with the jet. Several inventors have endeavoured to utilize an arrangement of this kind for saving water in hydraulic machinery when working below their full power. The high-pressure water enters by the pipe A ; when passing through the nozzles on its way to the machine cylinder, it sucks in a supply of water from the exhaust sump via B, and the greater volume of the combined stream at a lower pressure passes on to the cylinder. All the water thus sucked in is a direct source of gain, but the efficiency of the apparatus as usually constructed is very low, about 30 per cent. The author and Mr. R. H. Thorpe, of New York, made a long series of experiments on jet pumps, and succeeded in designing one which gave an efficiency of 72 per cent. An ordinary jet pump is shown in Fig. 530. The main trouble that occurs with such a form of pump is that the water churns round and round the suction spaces of the nozzles instead of going straight through. Each suction space between the nozzles should be in a separate chamber provided with 3 FIG. 530. Hydraulics. 559 back-pressure valve, and the spaces should gradually increase in area as the high-pressure water proceeds that is to say, the first suction space should be very small, and the next rather larger, and so on. Rectangular Notch. An orifice in a vertical plane with an open top is termed a notch, or sometimes a weir. The only two forms of notches commonly used are the rectangular and the triangular. FIG. 531. From the figure, it will be observed that the head of water immediately over the crest is less than the head measured further back, which is, however, the true head H. In calculating the quantity of water Q that flows over such a notch, we proceed thus The area of any elementary strip as shown = B . dh quantity of water passing strip per ) _ v n jh second, neglecting contraction j ~~ where V = velocity of flow in feet per second, or Hence the quantity of water passing strip I __ r per second, neglecting contraction [ ~ the whole quantity of water Q passing ~\ _ Ch = H over the notch in cubic feet per>=A/2^B second, neglecting contraction ) Q = Q = introducing a coefficient toU allow for contraction j v fadh h = o where B and H are both measured in feet; where K has values varying from 0^59 to 0-64 depending largely on the 560 Mechanics applied to Engineering. proportions of the section of the stream, i.c. the ratio of the depth to the width, and on the relative size of the notch and the section of the stream above it. In the absence of precise data it is usual to take K = 0*62. The above value for Q may also be arrived at thus : the velocity of flow at any strip at a depth h from the surface is *J zgh. Hence, if we construct a diagram of velocities we shall get the parabolic figure shown at the side, the area of which is -f-Hv^H (see p. 30), and the mean width, i.e. the mean velocity of flow, is and the total flow over the notch = area X mean velocity of flow the same result as we arrived at above. Triangular Notch. In order to avoid the uncertainty of the value of K, Professor James Thompson proposed the use of V notches; the form of the section of the stream then always remains constant however the head may vary. Experiments show that K for such a notch is very nearly con- stant. Hence, in the ab- sence of precise data, it may be used with much greater confidence than the rectangular notch. The quantity of water that passes is arrived at thus : 532. Area of elementary strip = b . dh b H - h B(H - h) ~ H "R/TT Z>\ area of elementary strip = ' dh xl velocity of water passing strip = V = *j 2gh = \ f 2g h quantity of water passingj strip per second, neglect-? ing contraction whole quantity of water Q Ch = H passing over the notch in cubic feet per second, neglecting contraction Hydraulics. 561 Introducing a coefficient for the contraction of the stream where K = 0*60 to o'6i. Rectangular Orifice in a Vertical Plane. When the vertical height of the orifice is small compared with the depth of water above it, the discharge is commonly taken to be the same as that of an orifice in a horizontal plane, the head being H, i.e. the head to the centre of the orifice. When, however, the vertical height of the orifice is not small compared with the FIG. 533- FIG. 534. depth, the discharge is obtained by precisely the same reasoning as in the two last cases ; it is Q = Kf - H a i) K, however, is a very uncertain quantity; it varies with the shape of the orifice and its depth below the surface. Drowned Orifice. When there is a head of water on both sides of an orifice, the discharge is not free ; the calculation of the flow is, however, a very simple matter. The head 2 o 562 Mechanics applied to Engineering. producing flow at any section xy (Fig. 534) is HJ H 2 = H; likewise, if any other section be taken, the head producing flow is also H. Hence the velocity of flow V = J 2*H, and the quantity discharged K varies somewhat, but is usually taken 0*62 as a mean value. Plow under a Constant Head. It is often found necessary to keep a perfectly constant head in a tank when making careful measurements of the flow of liquids, but it is often very difficult to accom- plish by keeping the supply exactly equal to the delivery. It can, however, be easily managed with the device shown in the figure. It consists of a closed tank fitted with an orifice, also a gland and sliding pipe open to the atmosphere. The vessel is filled, or nearly so, with the fluid, and the sliding pipe adjusted to give the required flow. The flow is due to the head H, and the negative pressure / above the surface of the water, for as the water sinks a partial vacuum is formed in the upper part of the vessel, and air bubbles through. Hence the pressure p is always due to the head h, and the effective head producing flow through the orifice is H ^, which is independent of the height of water in the vessel, and is constant provided the water does not sink below the bottom of the pipe. The quantity of water delivered is FIG. 535- where K has the values given above for different orifices. Velocity of Approach. If the water approaching a notch or weir have a velocity V a , the quantity of water passing will be correspondingly greater, but the exact amount will depend upon whether the velocity of the stream is uniform at every part of the cross-section, or whether it varies from point to point as in the section over the crest of a weir or notch. Let the velocity be uniform, as when approaching an orifice of area a, the area of the approach channel being A. Let v = velocity due to the head h, i.e. the head over the orifice ; V = velocity of water issuing from the orifice. Hydraulics. 563 Then V a = ~V, and V = V a + v A V = V + v A Precautions should always be taken to diminish the velocity of approach as much as possible. Time required to Lower the Water in a Tank through an Orifice. The problem of finding the time T required to lower the water in a dock or tank through a sluice- gate, or through an orifice in the bottom, is one that often arises. (i.) Tank of uniform cross-section. Let the area of the surface of the water be A a ; stream through the orifice be K A ; greater head of water above the orifice be H^ ; ,, lesser ,, ,, ,, ,, -tlaj head of water at any given instant be h. The quantity of water passing through \ _^- A / 7- ,. the orifice in the time dt \~ A ^ V 2 ^ ' Let the level of the water in the tank be lowered by an amount dh in the interval of time dt. Then the quantity in the tank is reduced by an amount h-adh, which is equal to that which has passed through the orifice in the interval, or The time required to empty the tank is 564 Mechanics applied to Engineering. It is impossible to get an exact expression for this, because the assumed conditions fail when the head becomes very small ; the expression may, however, be used for most practical purposes. (ii.) Tapered tank of uniform breadth B. In this case the quan- tity in the tank is reduced by the amount B. /.*/// m the given interval of -- FIG. 535". hence ?>J.dh = ~ ill BL hdh Integrating, we get T = -~ (iii.) Hemispherical tank. In this case r 2 = 2R// - // 2 R-h I The quantity in the tank is /7 ^ reduced by 7r(2R// - h*)dh in the interval dt. Civ.) the surface of the water falls at a uniform rate. Hydraulics. 565 In this case is constant ; at hence K d AV ' zg% -- A a X a constant But K d A ij 2g is constant in any given case, hence the area of the tank A at any height h above the orifice varies as *J h> or the tank must be a paraboloid of revolution. Plow-through Pipes of Variable Section. For the present we shall only deal with pipes running full, in which the section varies very gradually from point to point. If the varia- tion be abrupt, an entirely different action takes place. This particular case we shall deal with later on. The main point that we have to concern ourselves with at present is to show that the energy of the water at any section of the pipe is constant neglecting friction. If W Ibs. of water be raised from a given datum to a receiver at a certain height h feet above, the work done in raising the water is W/6 foot-lbs., or h foot-lbs. per pound of water. By lowering the water to the datum, W^ foot-lbs. of work will be done. Hence, when the water is in the raised position its energy is termed its energy of position, or The energy of position = W/i foot-lbs. If the water were allowed to fall freely, i.e. doing no work in its descent, it would attain a velocity V feet per V 2 second, where V = J 2g/i, or h = . Then, since no energy WV 2 is destroyed in the fall, we have WA = foot-lbs. of energy stored in the falling water when it reaches the datum, V 2 or foot-lbs. per pound of water. This energy, which is due to its velocity, is termed its kinetic energy, or energy of WV 2 The energy of motion = - If the water in the receiver descends by a pipe to the datum level for convenience we will take the pipe as one square inch area the pressure p at the foot of the pipe will be wh Ibs. 566 Mechanics applied to Engineering. per square inch. This pressure is capable of overcoming a resistance through a distance / feet, and thereby doing pi foot- Ibs. of work ; then, as no energy is destroyed in passing along W*> the pipe, we have pi = WA = - foot-lbs. of work done by the w water under pressure, or - foot-lbs. per pound of water. This is known as its pressure-energy, or The pressure-energy = Thus the energy of a given quantity of water may exist exclusively in either of the above forms, or partially in one form and partially in another, or in any combination of the three. Total energy per) __ {energy of I , I energy of I , (pressure-) pound of water) ~~ ( position [ "*" | motion j "*" \ energy f = *+X! + / 2g W This may, perhaps, be more clearly seen by referring to the figure. FIG. 536. Then, as no energy of the water is destroyed on passing through the pipe, the total energy at each section must be the same, or *, -f +^ =^ 2 + +2* = constant 2g W 2g W Hydrauhcs. 567 The quantity of water passing any given section of the pipe in a given time is the same, or Q, = Q, or A,V, = A 2 V a V 1 = A a ^ A, or the velocity of the water varies inversely as the sectional area FIG. 537. Some interesting points in this connection were given by the late Mr. Froude at the British Association in 1875. Let vertical pipes be inserted in the main pipe as shown ; then the height H, to which the water will rise in each, will be proportional to the pressure, or Hj = A and H 2 = ** V w and the total heights of the water-columns above datum w w and the differences of the heights w , - H 2 = V 2 2 Vy from the equation given above. Thus we see that, when water is steadily running through 568 Mechanics applied to Engineering. a full pipe of variable section, the pressure is greatest at the greatest section, and least at the least section. In addition to many other experiments that can be made to prove that such is the case, one has been devised by Pro- fessor Osborne Reynolds that beautifully illustrates this point. Take a piece of glass tube, say inch bore drawn down to a fine waist in the middle of, say, ^ inch diameter ; then, when water is forced through it at a high velocity, the pressure is so reduced at the waist that the water boils and hisses loudly. The pressure is atmospheric at the outlet, but very much less at the waist. The hissing in water-injectors and partially opened valves is also due to this cause. Venturi Water-meter. An interesting application of this principle is the Venturi water-meter. The water is forced through a very easy waist in a pipe, and the pressure measured at the smallest and largest section ; then, if the difference of the heads corresponding to the two pressures be H V 9 2 - V, 2 *- = H , or V 2 2 - Vf = If the area of the section of the pipe be A lt and that of the waist A 2 , we have, from above '~ A, Let Aj = A 2 ; then hence V 2 2 -= and V a = /_25f V i-i Hydraulics. 569 There is a small loss of head, due to friction, in this meter, which can readily be allowed for by a coefficient of velocity K w , which is fortunately very nearly constant If the friction varied as the square of the velocity it would be constant, but since it varies more nearly as the i '851,11 power of the velocity (see p. 581), the coefficient varies slightly from 0*97 to 0*98, accord- ing to the proportions of the meter, it has a tendency to rise with high velocities of flow. To I \tietwry Gauges FIG. 537* This meter is believed to be by far the most accurate apparatus for measuring large quantities of water. It has been very thoroughly tested by Mr. Clemens Hershall, of New York, who has got excellent results from it. The author has also tested both large and small meters of this type made by Mr. Kent, of High Holborn, and has every reason to place con- fidence in them. For tests of a large meter, readers should refer to Engineering, August 14, 1896. Radiating Currents and Free Vortex Motion. 1 Let the figure represent the section of two circular plates at a small distance apart, and let water flow up the vertical pipe and escape round the circumference of the plates. Take any small portion of the plates as shown ; the strips represent portions of rings of water moving towards the outside. Let their areas be a l3 # 2 j then, since the flow is constant, we have z> 2 #i fi or - 2 = 2 = J hence v 2 = v-^ r-2 or the velocity varies inversely as the radius. The plates being horizontal, the energy of position remains constant ; therefore 2g W 2g W 1 See Professor Unwin's article, "Hydromechanics," in " Encyclopaedia Britannica." 5/O Mechanics applied to Engineering. Substituting the value of v found above, we have i L 2 J. A _ *i V. = A 2g H' n 3 Then, substituting^ -f;- 1 =H from above, and putting FIG. 538. Then, starting with a value for h^ the h. 2 for other positions is readily calculated and set down from the line above. If a large number of radial segments were taken, they would form a complete cylinder of water, in which the water enters at the centre and escapes radially outwards. The dis- tribution of pressure will be the same as in the radial segments, and the form of the water will be a solid of revolution formed by spinning the dotted line of pressures, known as Barlow's curve, round the axis. The case in which this kind of vortex is most commonly met with is when water flows in radially to a central hole, and then escapes. Forced Vortex. If water be forced to revolve in and with a revolving vessel, the form taken up by the surface is readily found thus : Hydraulics. 571 Let the vessel be rotating n times per second. Any particle of water is acted upon by the following forces : (i.) The weight W acting vertically downwards. WV 2 (ii.) The centrifugal force act- ing horizontally, where V is its velocity in feet per second, and r its radius in feet. (Hi.) The fluid pressure, which is equal to the resultant of i. and ii. From the figure, we have WV 2 be FIG. 539- which may be written ac be But - - is constant, say C ; Then O 2 = ~ be But ac = r therefore O 2 = -?- be And for any given number of revolutions per second n* does not vary ; therefore be, the subnormal, is constant, and the curve is therefore a para- bola. If an orifice were made in the bottom of the vessel at o, the discharge would be due to the head h. Loss of Energy due to Abrupt Change of Direc- tion. If a stream of water flow down an inclined surface AB with a velocity V a feet per second, when it reaches B the direction of flow is suddenly changed from AB to BC, and the 572 Mechanics applied to Engineering. layers of water overtop one another, thus causing a breaking-up of the stream, and an eddying action which rapidly dissipates the energy of the stream by the frictional resistance of the particles of the water; this is sometimes termed the loss by shock. The velocity V 2 with which the water flows after passing the corner is given by the diagram of velocities ABD, from which we see that the component V 3 , normal to BC, is wasted in eddying, and the energy wasted per pound of water . V 8 2 V 2 sin 2 6 is = 2g 2g As the angle ABD increases the loss of energy increases, and when it becomes a right angle the whole of the energy is wasted by shock (Fig. 541). If the surface be a smooth curve (Fig. 542) in which there is no abrupt change of direction, there will be no loss due to FIG. 541. FIG. 542. shock ; hence the smooth easy curves that are adopted for the vanes of motors, etc. If the surface against which the water strikes (normally) is moving in the same direction as the jet with a velocity y , then the striking velocity will be ww Vi-- l = V n and the loss of energy per pound of water will be r _ V 'Y 1 -a) ^ Hydraulics. 573 When n i, no striking takes place, and consequently no loss of energy ; when n= oo , i.e. when the surface is stationary, V 2 the loss is , i.e. the whole energy of the jet is dissipated. Loss of Energy due to Abrupt Change of Section. When water flows along a pipe in which there is an abrupt change of section, as shown, we may regard it as a jet of water moving with a velocity Vj striking against a surface (in this case a body of water) moving in the same direction, but with a velocity -1 hence n the loss of energy per pound of water is precisely the same as in the last para- V\ 2 "~ ) The energy lost Vl , x facing P . 574). graph, viz. in this case is in eddying in the corners of the large section, as shown. As the water in the large section is moving - n as fast as in the small section, the area of the large section is n times the area of the small section. Then the loss of energy per pound of water, or the loss of head when a pipe suddenly enlarges n times, is V, 1 Or if we refer to the velocity in the large section as V,, we have the velocity in the small section V 15 and the loss of head When the water flows in the opposite direction, i.e. from the large to the small section, the loss of head is due to the abrupt change of velocity from the contracted to the full section of the small stream. The contracted section in pipes under pres- sure is, according to some experiments made in the author's laboratory, from FIG.' 544 (see also NO. 4 facing 0*62 to 0*66 hence, n 1 = from r6i p-574). to 1*5 ; then, the loss of head = 574 Mechanics applied to Engineering. Let /! = the pressure in the small part, where the velocity isV i; #2 = the pressure in the large part, where the velocity i*. 11 Then, from p. 566, we have 2g when there is a gradual change of section ; but when there is an abrupt change of section, this becomes fsv A Vi 2 _A V * / _o'3Vi a W~^~ 2g W 2g 2g Then the difference in pressure in the two sections due to the abrupt change in section is, by simple reduction We shall require to make use of this in one or two special cases. Experiments on the Character of Fluid Motion. Some very beautiful experiments, by Professor Hele-Shaw, F.R.S., on the flow of fluids, enable us to study exactly the manner in which the flow takes place in channels of various forms. He takes two sheets of glass and fits them into a suitable frame, which holds them in position at about ^~ inch apart. Through this narrow space liquid is caused to flow under pressure, and in order to demonstrate the exact manner in which the flow takes place, bands of coloured liquid are injected at the inlet end. In the narrow sections of the channel, where the velocity of flow is greatest, the bands themselves are narrowest, and they widen out in that portion of the channel where the velocity is least. The perfect manner in which the bands converge and diverge as the liquid passes through a neck or a pierced diaphragm, is in itself an elegant demonstration of the behaviour of a perfect fluid (see Diagrams i and 5). The form and behaviour of the bands, moreover, exactly correspond with mathematical demon- strations of the mode of flow of perfect fluids. The author is indebted to Professor Hele-Shaw for the illustrations given, which are reproduced from his own photographs. \Tofactp.m. FLOW OF WATER DIAGRAMS. Kindly supplied by Professor ffele-Shaw, F.R.S. Hydraulics. 575 In the majority of cases, however, that occur in practice, we are unfortunately unable to secure such perfect stream-line motions as we have just described. We usually have to deal with water flowing in sinuous fashion with very complex eddy- ings, which is much more difficult to ocularly demonstrate than true stream-line motion. Professor Hele-Shaw's method of showing the tumultuous conditions under which the water is moving, is to inject fine bubbles of air into the water, which make the disturbances within quite evident. The diagrams 2, 3, 4, and 6, also reproduced from his photographs, clearly demonstrate the breaking up of the water when it encounters sudden enlargements and contractions, as predicted by theory. A careful study of these figures, in conjunction with the theoretical treatment of the subject, is of the greatest value in getting a clear idea of the turbulent action of flowing water. Readers should refer to the original communications by Professor Hele-Shaw in the Transactions of the Naval Architects ', 1897-98, also the engineering journals at that time. Surface Friction. When a body immersed in water is caused to move, or when water flows over a body, a certain resistance to motion is experienced ; this resistance is termed the surface or fluid friction between the body and the water. At very low velocities, only a thin film of the water actually in contact with the body appears to be affected, a mere skim- ming action ; but as the velocity is increased, the moving body appears to carry more or less of the water with it, and to cause local eddying for some distance from the body. Experiments made by Professor Osborne Reynolds clearly demonstrate the difference between the two kinds of resistances .the sur- face resistance and the eddy- ing resistance. Water is caused to flow through the glass pipe AB at a given velocity ; a bent glass tube and funnel C is fixed in such a manner that a fine stream of deeply coloured dye is ejected. When the water FIG. 545. flows through at a low velocity, the stream of dye runs right through like an unbroken thread ; but as soon as the velocity is increased beyond a certain point, the thread breaks up and passes through in sinuous fashion, thus demonstrating that the water is not flowing through as a steady stream, as it did at the lower velocities. 576 Mechanics applied to Engineering. Professor Reynolds found that the change from steady to unsteady flow occurred when DV = 0*02 for a temperature of 60 Fahr., where D = diameter of pipe in feet, V = velocity of flow in feet per second. Mr. E. C. Thrupp has, however, shown that this expression only holds for very small pipes; in the case of large pipes, channels, and rivers the velocity at which the water breaks up is very much (thousands of times in some instances) greater than this expression gives. See a paper on " Hydraulics of the Resistance of Ships," read at the Engineering Congress in Glasgow, 1901 ; also Engineering, December 20, 1901. For further details of Reynolds's investigations the reader is referred to the original papers in the Philosophical Transactions for 1883, p. 943; and for 1896, p. 167. Also to Turner and Brightmore's " Waterworks Engineering," p. 67. Experiments by Mr. Froude at Torquay (see Brit. Ass. Proceedings, 1874), on the frictional resistance of long planks, towed end-on through the water at various velocities, showed that the following laws appear to hold within narrow limits : (i.) The friction varies directly as the extent of the wetted surface. (ii.) The friction varies directly as the roughness of the surface. (iii.) The friction varies directly as the square of the velocity. (iv.) The friction is independent of the pressure. For fluids other than water, we should have to add (v.) The friction varies directly as the density and viscosity of the fluid. Hence, if S = the wetted surface in square feet ; /= a coefficient depending on the roughness of the surface ; i.e. the resistance per square foot at i foot per second in pounds ; V = velocity of flow relatively to the surface in feet per second ; R = frictional resistance in pounds ; Then, neglecting any variation of the friction with the tempera- ture, we have R = S/V 2 Some have endeavoured to prove from Mr. Froude's own figures that the first of the laws given above does not even approximately hold. The basis of their argument is that the Hydraulics. 577 1 \ j , \ \ \ V \ s \ \ % ^^ C OA R* 5E 3 A* fD 4, ^ -_ ,. r "/j^ fir f a A 1 r> 1 . .'fy ^ > *f A/ f^n so as jo DISTANCE rOM CUTWATER FIG. 546. CO A 15 20 25 3O 35 DISTANCE FffOM CUTWATER Fio. 547. 2 P 578 Mechanics applied to Engineering* frictional resistance of a plank, say 50 feet in length, is not ten times as great as the resistance of a plank 5 feet in length. This effect is, however, entirely due to the fact that the first portion of the plank meets with water at rest, and, therefore, if a plank be said to be moving at a speed of 10 feet a second, it simply means that this is the relative velocity of the plank and the still water. But the moving plank imparts a considerable velocity to the surrounding water by dragging it along with it, hence the relative velocity of the rear end of the plank and the water is less than 10 feet a second, and the friction is corre- spondingly reduced. In order to make this point clear the author has plotted the curves in Figs. 546 and 547, which are deduced from Mr. Froude's own figures. It is worthy of note that planks with rough surfaces drag the water along with them to a much greater extent than is the case with planks having smooth surfaces, a result quite in accordance with what one might expect. The value of/ deduced from these experiments is Surface covered with coarse sand ...... 0*0132 Ih. ,, fine ...... 0-0096 varnish ...... 0-0043 tinfoil ......... 0-0031 Professor Unwin and others have also experimented on the friction of discs revolving in water, and have obtained results very closely in accord with those obtained by Mr. Froude. Reducing the expression for the frictional resistance to a form suitable for application to pipes, we have, for any length of pipe L feet, the pressure Pj in pounds per square foot at one end greater than the pressure P 2 at the other end, on account of the friction of the water. Then, if A be the area of the pipe in square feet, we have R = (P, - P a )A Then, putting Pj = ^W,, and P 2 = AjW^, we have R = W W A(// 1 - A.) = W,,,A// where h is the loss of head due to friction on any length of pipe L; then W W AA = S/V 2 or - = 4 / LV* i LV 2 Hydraulics. 579 The coefficient ~ has to be obtained by experiment; according to D'Arcy -- fi ^ K 32ooV " I2D/ where D is the diameter of the pipe in feet. D'Arcy's experiments were made on pipes varying in diameter from \ inch up to 20 inches; for small pipes his coefficient appears to hold tolerably well, but it is certainly incorrect for large pipes. The author has recently looked into this question, and finds that the following expression better fits the most recent published experiments for pipes of over 8 inch diameter (see a paper by Lawford, Proceedings I.C.E.^ vol. cliii. p. 297) : 4. ^ for clean cast-iron pipes 2D/ -=r j for incrusted pipes But the above expression at the best is only a rough approximation, since the value of / varies very largely for different surfaces, and the resistance does not always vary as the square of the velocity, nor simply inversely as D. The energy of motion of i Ib. of water moving with a V 2 velocity V feet per second is ; hence the whole energy of & motion of the water is dissipated in friction when V 2 LV 2 Taking K = 2400 and putting in the numerical value for g, we get L = 37D. This value 37, of course, depends on the roughness of the pipe. We shall find this method of regarding frictional resistances exceedingly convenient when dealing with the resistances of T's, elbows, etc., in pipes. Still adhering to the rough formula given above, we can calculate the discharge of any pipe thus : The quantity discharged inl __ _ _ 7rD 2 V cubic feet per second j- -= Q AV = 580 Mechanics applied to Engineering. From the same formula, we have /24 V = V~ Substituting this value, we have . Q-s Or, more conveniently Thrupp's Formula for the Flow of Water. All formulas for the flow of water are, or should be, constructed to fit experiments, and that which fits the widest range of ex- periments is of course the most reliable. Several investigators in recent years have collected together the results of published experiments, and have adjusted the older formulas or have constructed new ones to better accord with experiments. There is very little to choose between the best of recent formulas, but on the whole the author believes that this formula best fits the widest range of experiments ; others are equally as good for smaller ranges. It is a modification of Hagen's formula, and was published in a paper read before the Society of Engineers in 1887. Let V = velocity of flow in feet per second ; R = hydraulic mean radius in feet, i.e. the area of the stream divided by the wetted perimeter, and is for circular and square pipes : L = length of pipe in feet ; h = loss of head due to friction in feet ; S = cosecant of angle of slope = ; n Q = quantity of water flowing in cubic feet per second. Then V = -?1 where #, C, n are coefficients depending on the nature of the surface of the pipe or channel. Hydraulics. 581 For small values of R, more accurate results will be ob- tained by substituting for the index x the value x+y\ /?TL: \ V R In this formula the effect of a change of temperature is not taken into account. The friction varies, roughly, inversely as the absolute temperature of the water. Surface. . C. X. y- X. Wrought-iron pipes ... Riveted sheet-iron pipes i -So 1-825 0-004787 0-005674 0*65 0-677 0-018 0*07 New cast-iron pipes ... /i -85 l2'OO 0-005347 0*006752 0*67 0-63 Lead pipes 175 0-005224 0-62 Pure cement rendering /i 74 U "95 O-OO4OOO 0-006429 0*67 o - 6i Brickwork (smooth) ... 2*00 0*007746 0*61 ,, (rough) 2"OO 0-008845 0*625 0*01224 0*50 Unplaned plank Small gravel in cement 2'OO 2'00 0-008451 0*01 181 0-615 0-66 0-03349 0-03938 0-50 0*60 Large 2'00 0-01415 0705 0*07590 1*00 Hammer-dressed masonry . . . Z'OO 0-01117 0-66 0*07825 1*00 Earth (no vegetation) 2 '00 0-01536 0-72 Rough stony earth 2'00 0-02144 0-78 If we take x as 0*62, and n = 2, C = 0*0067, we g e * D 2-62 Q = 3-oiC VS which reduces to Similarly, for new cast-iron pipes LV 1 * // = 3200D 1 * taking n = 1*85, and x = 0*67. These expressions should be compared with the rougher ones given on pp. 578-580. Virtual Slope. If two reservoirs at different levels be freely connected by a main through which water is flowing, the pressure in the main will diminish from a maximum at th< upper reservoir to a minimum at the lower, and if glass pipes be inserted at intervals in the main, the height of the water in each will represent the pressure at the respective points, and 5 32 Mechanics applied to Engineering. the difference in height between any two points will represent the loss of head due to friction on that section. If a straight line be drawn from the surface of the water in the one reservoir to that in the other, it will touch the surface of the water in all the glass tubes in the case of a main of uniform diameter and roughness. The slope of this line is known as the " virtual slope " of the main. If the lower end of the main be partially closed, it will reduce the virtual slope ; and if it be closed altogether, the virtual slope will be //, or the line will be horizontal, and, of course, no water will flow. The velocity of flow is proportional to the virtual slope, the tangent of the angle of slope is the in the expressions we use for the JL/ friction in pipes. The above statement is only strictly true when there is no loss of head at entry into the main, and when the main is of uniform diameter and roughness throughout, and when there are no artificial resistances. When any such irregularities do exist, the construction of the virtual slope line offers, as a rule, no difficulties, but it is no longer straight. FIG. 547 a. In the case of the pipe shown in full lines the resistance is uniform throughout, but in the case of the pipe shown in broken line, there is a loss at entry a, due to the pipe project- ing into the top reservoir ; the virtual slope line is then parallel to the upper line until it reaches b, when it drops, due to a sudden contraction in the main ; from b to c its slope is steeper than from a to b, on account of the pipe being smaller in diameter ; at c there is a drop due to a sudden enlargement and contraction, the slope from c to d is the same as from b to (average sizes) ... ... ... ..'*). Ditto by experiment Sudden enlargement to a square-ended large area pipe, where n = ^r. small area Sudden contraction Mushroom valves (one set of experiments) Plug cock, handle 15 turned 30 Unwin through 45 Sluice and slide valves = port area TV j j' i- Pierced diaphragm n = area of opening area f pipe - - area of hole Water entering a re-entrant pipe, such as "I a Borda's mouthpiece ... ... .../ Water entering a square-ended pipe flush \ with the side of the tank ... .../ Equivalent length of straight pipe expressed in diameters, on the basis of L = 360. ( 30-40 in plain pipe \5O~9O with screwed elbow 3-5 22-30 24 16-20 12 approx. 120-150 27 200 IIOO ioo( i) 2 36(i-6- i 18 9-12 Velocity of Water in Pipes. Water is allowed to flow at about the velocities given below for the various purposes named : Pressure pipes for hydraulic purposes for long mains 3 to 4 feet per sec. Ditto for short lengths * Up to 25 ,, Ditto through valve passages 1 Up to 50 ,, Pumping mains 3 to 5 ,, Waterworks mains 2103 ,, 1 Such velocities are unfortunately common, but they should be avoided if possible. CHAPTER XVII. HYDRAULIC MOTORS AND MACHINES. THE work done by raising water from a given datum to a receiver at a higher level is recoverable by utilizing it in one of three distinct types of motor. 1. Gravity machines, in which the weight of the water is utilized. 2. Pressure machines, in which the pressure of the water is utilized. 3. Velocity machines, in which the velocity of the water is utilized. Gravity Machines. In this type of machine the weight- energy of the water is utilized by causing the water to flow into the receivers of the machine at the higher level, then to descend with the receivers in either a straight or curved path to the lower level at which it is discharged. If W Ibs. of water have descended through a height H feet, the work done = WH foot-lbs. Only a part, however, of this will be utilized by the motor, for reasons which we will now consider. FIG. 549 . FIG. 550. The illustrations, Figs. 549, 550, show various methods of Hydraulic Motors and Machines. 587 utilizing the weight-energy of water. Those shown in Fig. 549 are very rarely used, but they serve well to illustrate the principle involved. The ordinary overshot wheel shown in Fig. 550 will perhaps be the most instructive example to investigate as regards efficiency. Although we have termed all of these machines gravity machines, they are not purely such, for they all derive a small portion Sf their power from the water striking the buckets on entry. Later on we shall show that, for motors which utilize the velocity of the water, the maximum efficiency occurs when the velocity of the jet is twice the velocity of the buckets or vanes. In the case of an overshot water-wheel, it is necessary to keep down the linear velocity of the buckets, otherwise the centrifugal force acting on the water will cause much of it to be wasted by spilling over the buckets. If we decide that the inclination of the surface of the water in the buckets to the horizontal shall not exceed i in 8, we get the peripheral velocity of the wheel V w = 2/v/R, where R is the radius of the wheel in feet. Take, for example, a wheel required for a fall of 15 feet. The diameter of the wheel may be taken as a first approxima- tion as 1 2 Jeet. Then the velocity of the rim should not exceed 2^/6 = say 5 feet per second. Then the velocity of the water issuing from the sluice should be 10 feet per second ; the head h required to produce this velocity will be V 2 h = , or, introducing a coefficient to allow for the friction in the sluice, we may i'iV 2 write it h = = i'6 foot. One-half 2 of this head, we shall show later, is lost by shock. The depth of the shroud is usually from 075 to i foot; the distance from the middle of the stream to the c. of g. of the water in the bucket may be taken at about i foot, which is also a source of loss. FIG. S5 i. The next source of waste is due to the water leaving the wheel before it reaches the bottom. The exact position at which it leaves varies with the form of buckets adopted, but for our present purpose it may be taken that the mean discharge occurs at an angle of 45 as 588 Mechanics applied to Engineering. shown. Then by measurement from the diagram, or by a simple calculation, we see that this loss is 0*150. A clearance of about o'5 foot is usually allowed between the wheel and the tail water. We can now find the diameter of the wheel, remembering that H = 15 feet, and taking the height from the surface of the water to the wheel as 2 feet. This together with the 0*5 foot clearance at the bottom gives us D = 12*5 feet. Thus the losses with this wheel are Half the sluice head = o'8 foot Drop from centre of stream to buckets = i 'o Water leaving wheel too early, ) _ . 0*15 X 12-5 feet \ ~ Clearance at bottom = o - 5 4- 2 feet Hydraulic efficiency of wheel = -^ = 72 per cent. The mechanical efficiency of the axle and one toothed wheel will be about 90 per cent., thus giving a total efficiency of the wheel of 65 per cent. With greater falls this efficiency can be raised to 80 per cent. The above calculations do not profess to be a complete treatment of the overshot wheel, but they fairly indicate the sort of losses such wheels are liable to. Pressure Machines. In these machines the water at the FIG. 552. higher level descends by a pipe to the lower level, from whence it passes to a closed vessel or a cylinder, and acts on a movable Hydraulic Motors and Machines. 589 piston in precisely the same manner as in a steam-engine. The work done is the same as before, viz. WH foot-lbs. for FIG. 553. the pressure at the lower level is W^H Ibs, per square foot and the weight of water used per square foot of piston = W W L = W, FIG. 554- where L is the distance moved through by the piston in feet. Then the work done by the pressure water = W W LH = WH foot-lbs. Several examples of pressure machines are shown in Figs. 552, 553, 554, a and b. Fig. 552 is an oscillating cylinder pressure motor used largely on the continent. Fig. 553 is an ordinary hydraulic pressure riveter. Fig. 554, a, is a passenger lift, with a wire-rope multiplying arrangement. Fig. 554, b, is an ordinary ram lift. For details the reader is referred 590 Mechanics applied to Engineering to special books on hydraulic machines, such as Elaine T or Robinson. 2 The chief sources of loss in efficiency in these motors are 1. Friction of the water in the mains and passages. 2. Losses by shock through abrupt changes in velocity of water. 3. Friction of mechanism. 4. Waste of water due to the same quantity being used when running under light loads as when running with the full load. The friction and shock losses may be reduced to a minimum by careful attention to the design of the ports and passages ; re-entrant angles, abrupt changes of section of ports and passages, high velocities of flow, and other sources of loss given in the chapter on hydraulics should be carefully avoided. By far the most serious loss in most motors of this type is that mentioned in No. 4 above. Many very ingenious devices have been tried with the object of overcoming this loss. Amongst the most promising of those tried are devices for automatically regulating the length of the stroke in pro- portion to the resistance overcome by the motor. Perhaps the best known of these devices is that of the Hastie engine, a full description of which will be found in Professor Un win's article on Hydromechanics in the " Encyclopaedia Britannica." In an experiment on this engine, the following results were obtained : Weight in pounds lifted V (chain! 22 feet )\ only j 427 633 745 857 969 1081 ii93 Water used in gallons at "1 80 Ibs. per square inch / 7'5 10 H 16 17 20 21 22 Efficiency per cent, (actual) 5 1 54 56 60 58 61 6S Efficiency per cent, if stroke j were of fixed length ... J 23 34 40 46 53 59 65 The efficiency in lines 3 and 4 has been deduced from the other figures by the author, on the assumption that the motor was working full stroke at the highest load given. The great increase in the efficiency at low loads due to the compensating gear is very clear. Cranes and elevators are often fitted with two cylinders of different sizes, or one cylinder and a differential piston. When lightly loaded, the smaller cylinder is used, and the larger one 1 "Hydraulic Machinery" (Spon). * ' ' Hydraulic Power and Machinery " (Griffin). Hydraulic Motors and Machines. 591 only for full loads. The valves for changing over the con- ditions are usually worked by hand, but it is very often found that the man in charge does not take advantage of the smaller cylinder. In order to place it beyond his control, the ex- tremely ingenious device shown in Fig. 555 is sometimes used. The author is indebted to Mr. R. H. Thorp, of New York, the inventor, for the drawings and particulars from which the following account is taken. The working cylinder is shown at AB. When working at full power, the valve D is in the position shown in full lines, which allows the water from B to escape freely by means of the exhaust pipes E and K ; then the quantity of water used is given by the volume A. But when working at half-power, the valve D is in the position shown in dotted lines ; the water in B then returns via the pipe E, the valve D, and the pipe F to the A side of the piston. Under such conditions it will be seen that the quantity of high-pressure water used is the volume A minus the volume B, which is usually one-half of the former quantity. The position of the valve D, which determines the conditions of full or half power, is generally controlled by hand. The action of the automatic device shown depends upon the fact that the pressure of the water in the cylinder is proportional to the load lifted, for if the pressure were in excess of that required to steadily raise a light 592 Mechanics applied to Engineering. load, the piston would be accelerated, and the pressure would be reduced, due to the high velocity in the ports. In general, the man in charge of the crane throttles the water at the inlet valve in order to prevent any such acceleration. In Mr. Thorp's arrangement, the valve D is worked automatically. In the position shown, the crane is working at full power ; but if the crane be only lightly loaded, the piston will be accelerated and the pressure of the water will be reduced by friction in passing through the pipe C, until the total pressure on the plunger H will be less than the total full water-pressure on the plunger G, with the result that the valve D will be forced over to the right, thus establishing communication between B and A, through the pipes E and F, and thereby putting the crane at half-power. As soon as the pressure is raised in A, the valve D returns to its full-power position, due to the area of H being greater than that of G, and to the pendulum weight W. It very rarely happens that a natural supply of high-pressure water can be obtained, conse- quently a power-driven pump has to be resorted to as a means of raising the water to a sufficiently high pressure. In certain simple operations the water may be used direct from the pump, but nearly always some method of storing the power is necessary. If a tank could be conveniently placed at a sufficient height, the pump might be arranged to deliver into it, from whence the hydraulic installation would draw its supply of high-pressure water. In the absence of such a con- venience, which, however, is seldom met with, a hydraulic accumulator (Fig. 556) is used. It consists essentially of a vertical cylinder, provided with a long- stroke plunger, which is weighted to give the required pressure, usually from 700 to 1000 Ibs. per square inch. With such a means of storing energy, a very large amount of power far in excess FIG. 556. Hydraulic Motors and Machines. 593 of that of the pump may be obtained for short periods. In fact, this is one of the greatest points in favour of hydraulic methods of transmitting power. The levers shown at the side are for the purpose of automatically stopping and starting the pumps when the accumulator weights get to the top or bottom of the stroke. Energy stored in an Accumulator. If s = the stroke of the accumulator in feet ; d = the diameter of the ram in inches ; p = the pressure in pounds per square inch. Then the work stored in foot-lbs. = 0-785^/5- Work stored per cubic foot of water in ) foot-lbs. ( = Work stored per gallon of water = -77 3= 23-04^ Number of gallons required per minute 1 _ 33> 000 _ I 43 2 at the pressure/ per horse-power I ~~ 23*04^ ~~ / Number of cubic feet required per minute ) _ 33,000 229^2 at the pressure p ( ~ 1 44^ ~ p Effects of Inertia of Water in Pressure Systems. In nearly all pressure motors and machines, the inertia of the water seriously modifies the pressures actually obtained in the cylinders and mains. For this reason such machines have to be run at comparatively low piston speeds, seldom exceeding 100 feet per minute. In the case of free piston machines, such as hydraulic riveters, the pressure on the rivet due to this cause is frequently twice as great as would be given by the steady accumulator pressure. In the case of a water-pressure motor, the water in the mains moves along with the piston, and may be regarded as a part of the reciprocating parts. The pressure set up in the pipes, due to bringing it to rest, may be arrived at in the same manner as the " Inertia pressure," discussed in Chapter VI. Let w = weight of a column of water i square inch in section, whose length L in feet is that of the main along which the water is flowing to the motor = 0*434!, ; area of plunger or piston m the ratio r . . area of section of water mam 2 Q 594 Mechanics applied to Engineering. p = the pressure in pounds per square inch set up in the pipe, due to bringing the water to rest at the end of the stroke (with no air-vessel) ; N = the number of revolutions per minute of the motor ; R = the radius of the crank in feet. Then, remembering that the pressure varies directly as the velocity of the moving masses, we have, from pp. 165, 167 p = o*ooo34w(o'434L)RN 2 I i - 1 f i - } Relief valves are frequently placed on long lines of piping, in order to relieve any dangerous pressure that may be set up by this cause. Pressure due to Shock. If water flows along a long pipe with a velocity V feet per second, and a valve at the outlet end is suddenly closed, the kinetic energy of the water will be expended in compressing the water and in stretching the walls of the pipe. If the water and the pipe were both materials of an unyielding character, the whole of the water would be instantly brought to rest, and the pressure set up would be infinitely great. Both the water and the pipe, how- ever, do yield considerably under pressure. Hence, even after the valve is closed, water continues to enter at the inlet end with undiminished velocity for a period of / seconds, until the whole of the water in the pipe is compressed, thus producing a momentary pressure greater than the static pressure of the water. The compressed water then expands, and the distended pipe contracts, thus setting up a return-wave, and thereby causing the water-pressure to fall below the static pressure. Let K = the modulus of elasticity of bulk of water = 300,000 Ibs. per square inch (see p. 331); x = the amount the column of water is shortened, due to the compression of the water and to the distention of the pipe, in feet ; f = the compressive stress or pressure in pounds per square inch due to shock ; w = the weight of a unit column of water, i.e. i sq. inch section, i foot long, = 0*434 Ib. ; L = the length of the column of flowing water in feet; Hydraulic Motors and Machines. 595 d the diameter of the pipe in inches ; T = the thickness of the pipe in inches ; f t = the tensile stress in the pipe (considere4 thin) due to the increased internal pressure/"; E = Young's modulus of elasticity for the pipe material. The increase in diameter due to the) /,, = (AH ^-^ ) dV \ 24ooD/ When the power is a maximum this becomes zero ; then LV H 3 or // = hence the maximum power is transmitted when \ of the head is wasted in friction. Those not familiar with the differential method can arrive at the same result by calculating out several values of V V 3 , until a maximum is found. Whence the maximum horse-power that can be transmitted through any given pipe is H.P. obtained by inserting n = ^ in the equation above. N.B. H, D, and L are all expressed in feet. Velocity Machines. In these machines, the water, having descended from the higher to the lower level by a pipe, is allowed to flow freely and to acquire velocity due to its head. The whole of its energy then exists as energy of motion. The energy is utilized by causing the water to impinge on moving vanes, which change its direction of flow, and more or less reduce its velocity. If it left the vanes with Hydraulic Motors and Machines. 599 no velocity relative to the earth, the whole of the energy would be utilized, a condition of affairs which is never attained in practice. The velocity with which the water issues, apart from friction, is given by where H is the head of water above the outlet. When friction is taken into account where // is the head lost in friction. Relative and Absolute Velocities of Streams. We shall always use the term " absolute velocity," as the velocity relative to the earth. Let the tank shown in Fig. 558 be mounted on wheels, or otherwise arranged so that it can be moved along horizontally -zr~.>. y FIG. 558. at a velocity V in the direction indicated by the arrow, and let water issue from the various nozzles as shown. In every case let the water issue from the tank with a velocity v at an angle 6 with the direction of motion of the tank ; then we have Nozzle. Velocity rel. to tank v. Velocity rel. to ground Vo. e. Cos 9. I o cos 6 A B C D V V V V V+z/ V -v 1 80 9 o e VV 2 + v> ^/V 2 -f z> 2 + 2z/V cos e 6oo Mechanics applied to Engineering. The velocity V will be clear from the diagram. The expression for D is arrived at thus : y = ab cos = v . cos and x = v . sin 6 Substituting the values of x and jr, and remembering that cos 2 6 -f sin 2 = i, we get the expression given above. If the value of cos for A, B, and C be inserted in the general expression D, the same results will be obtained as those given. Now, suppose a jet of water to be moving, as shown by the arrow, with a velocity V relative to the ground ; also the tank to be moving with a velocity V relative to the ground ; then it is obvious that the velocity of the water relatively to the tank is given by ab or v. We shall be constantly making use of this construction when considering turbines. Pressure on a Surface due to an Impinging Jet. When a body of mass M, moving with a velocity V, receives an impulse P for a space of time /, the velocity will be increased to V 1} and the energy of motion of the body will also be increased ; but, as no other force has acted on the body during the interval, this increase of energy must be equal to the work expended on the body, or The work done onl f distance through which the body j = im P ulse x ( it is exerted = increase in kinetic energy The kinetic energy! MV 2 before the impulse J = 2~ The kinetic energy! _ MV^ after the impulse J = ~2 Increase in kinetic 1 M/v 2 v 2 ^ energy j s =~^( Vl " V ' The distance through which the impulse is exerted is hence or P/ = M(Vj - V) or impulse in time / = change of momentum in time / Hydraulic Motors and Machines. 60 1 Let a jet of water moving with a velocity V feet per second impinge on a plate, as shown. After impinging, its velocity in its original direc- tion is zero, hence its change of velocity on striking is V, and therefore Pt = MV M But -y is the mass of water delivered per FIG. 559- second. Let W = weight of water delivered per second. w _ M: wv For another method of arriving at the same result, see p. 501. It should be noticed that the pressure due to an impinging jet is just twice as great as the pressure due to the head of water corresponding to the same velocity. This can be shown thus: p = wh = *<5 where w = the weight of a unit column of water. We have W = wV. Substituting this value of WV The impinging jet corresponds to a dynamic load, and a column of water to a steady load (see p. 535). In this connection it is interesting to note that, in the case of a sea-wave, the pressure due to a wave of oscillation is approximately equal to that of a head of water of the same height as the wave, and, in the case of a wave of translation, to twice that amount. 602 Mechanics applied to Engineering. Pressure on a Moving Surface due to an Imping- ing Jet. Let the plate shown in the Fig. 560 be one of a series on which the jet impinges at very short intervals. The reason for making this stipulation will be seen shortly. Let the weight of water delivered per second be W Ibs. as before ; then, if the plates succeed one another very tapidly as in many types of water-wheels, the quantity FIG. 560. impinging on the plates will also be sensi- bly equal to W. The impinging velocity V / i \ is V , or V ( i - J ; hence the pressure in pounds' weight on the plates is WV I i - - p= r And the work done per second on the plates in foot-lbs. V P- = n ng and the energy of the jet is hence the efficiency of the jet = rA = - ( i - I The value of n for maximum efficiency can be obtained by plotting or by differentiation. 1 It will be found that n 2. The efficiency is then 50 per cent., which is the highest that can be obtained with a jet impinging on flat vanes. A common example of a motor working in this manner is the ordinary 1 Efficiency = TJ = -( i - n 2 2 r? = = 2 ' 2n~ n 2 -7- = 2;/~ 2 + 4 3 = o, when TJ is a maximum an 2 4 . -j = -,, whence n = 2 Hydraulic Motors and Machines. 603 i j , and the maximum undershot water-wheel ; but, due to leakage past the floats, axle friction, etc., the efficiency, is rarely over 30 per cent. If the jet had been impinging on only one plate instead of a large number, the quantity of water that reached the plate per second would only have been W (i - j ; then, sub- stituting this value for W in the equation above, it will be seen that the efficiency of the jet = - efficiency occurs when n - 3, and is equal to about 30 per cent. Pressure on an Oblique Surface due to an Impinging Jet. The jet impinges obliquely at an angle to the plate, and splits up into two streams. The velocity V may be resolved into V x normal and V parallel to the plate. After im- pinging, the water has no velocity normal to the plate, therefore the normal pressure WV 1 WV sin Pressure on a Smooth Curved Surface due to an ,v f FIG. 561. Impinging Jet. We will first consider the case in which the surface is stationary and the water slides on it without shock ; 604 Mechanics applied to Engineering. how to secure this latter condition we will consider shortly. We show three forms of surface (Fig. 562), to all of which the following reasoning applies. Draw ab to represent the initial velocity V of the jet in magnitude and direction ; then, neglecting friction, the final velocity of the water on leaving the surface will be V, and its direction will be tangential to the last tip of the surface. Draw ac parallel to the final direction and equal to ab, then be repre- sents the change of velocity V-, ; hence the resultant pressure on the surface in the direction of cb is s Then, reproducing the diagram of velocities above, we have y = V sin x = V cos V 2 /V 1 i > o p s When stationar ^ T L ^ i.^???^?^?*?^^^^"" -a > Jft ii, li il j=; I > -A- Hydraulic Motors and Machines. 607 Pelton or Tangent Wheel Vanes. The double vane shown in section in Fig. 566 is usually known as the Pelton Wheel Vane ; but whether Pelton should have the credit of the invention or not is a disputed point. In this type of vane the angle 6 approaches 180, then i cos 0=2, and the resultant pressure on such a vane is twice as great as that on a flat vane, and the theoretical efficiency is 100 per cent, when n = 2 ; but for various reasons such an efficiency is never reached, although it sometimes exceeds 80 per cent., including the friction of the axle. A general vie.v of such a wheel is shown in Fig. 567. It is very instructive to examine the action of the jet of water on the vanes in wheels of this type, and thereby to see why the theoretical efficiency is never reached. FIG. 567.' (1) There is always some loss of head in the nozzle itself; but this may be reduced to an exceedingly small amount by carefully proportioning the internal curves of the nozzle. (2) The vanes are usually designed to give the best effect when the jet plays fairly in the centre of the vanes ; but in other positions the effect is often very poor, and, consequently, as each vane enters and leaves the jet, serious losses by shock very frequently occur. Jn order to avoid the loss at entry, Mr. Doble, of San Francisco, after a very careful study of the matter, has shown that the shape of vane as usually used is 1 Reproduced by the kind permission of Messrs. Gilbert Gilkes and Co., Kendal. 6o8 Mechanics applied to Engineering. very faulty, since the water after striking the outer lip is abruptly changed in direction at the corners a and , where much of its energy is dissipated in eddying ; then, further, on FIG 567*. leaving the vane it strikes the back of the approaching vane, and thereby produces a back pressure on the wheel with a FIG. 567^. Dobl " Tangent Wheel" buckets. consequent loss in efficiency. This action is shown in Fig. 5670. The outer lip is not only unnecessary, but is distinctly wrong in theory and practice. In the Doble vane (Fig. 567^) Hydraulic Motors and Machines. the outer lip is dispensed with, and only the central rib retained for parting the water sideways, with the result that the efficiency of the Doble wheel is materially higher than that obtained from wheels made in the usual form. (3) The angle cannot practically be made so great as 1 80, because the water on leaving the sides of the vanes would strike the back edges of the vanes which immediately follow ; hence for clearance purposes this angle must be made somewhat less than 180, with a corresponding loss in efficiency. (4) Some of the energy of the jet is wasted in overcoming the friction of the axle. In an actual wheel the maximum efficiency does not occur 80 $ 70 60 P<50 ^40 $ 30 20 " .0 / - ^_ / ~^~~~- / / / / / 1-0 2-0 3-0 JZatw of Jet to whe&l velocity FIG. 567^. when the velocity of the jet is twice that of the vanes, but when the ratio is about 2*2. The curve shown in Fig. 567*: shows how the efficiency varies with a variation in speed ratio. The results were obtained from a small Pelton or tangent wheel in the author's laboratory ; the available water pressure is about 30 Ibs. per square inch. Probably much better results would be obtained with a higher water pressure. This form of wheel possesses so many great advantages over the ordinary type of impulse turbine that it is rapidly coming into very general use for driving electrical and other installations ; hence the question of accurately governing it is one of great importance. In cases in which a waste of water is immaterial, excellent results with small wheels can be 2 R 6io Mechanics applied to Engineering. obtained by the Cassel governor, in which the two halves of the vanes are mounted on separate wheels. When the wheel is working at its full power the two halves are kept together, and thus form an ordinary Pelton wheel ; when, however, the speed increases, the governor causes the two wheels to partially separate, and thus allows some of the water to escape between the central rib of the vanes. For much larger wheels Doble obtains the same result by affixing the jet nozzle to the end of a pivoted pipe in such a manner that the jet plays centrally on the vanes for full power, and when the speed increases, the governor deflects the nozzle to such an extent that the jet partially or fully misses the tips of the vanes, and so allows some of the water to .escape without performing any work on the wheel. But by far the most elegant and satisfactory device for regulating motors of this type is the conical expanding nozzle, which effects the desired regulation without allowing any waste of water. The nozzle is fitted with an internal cone of special construction, which can be advanced or withdrawn, and thereby it reduces or enlarges the area of the annular stream of water. Many have attempted to use a similar device, but have failed to get the jet to perfectly coalesce after it leaves the point of the cone. The cone in the Doble l arrangement is balanced as regards shifting along the axis of the nozzle ; therefore the governor only has to overcome a very small resistance in altering the area of the jet. Many other devices have been tried for varying the area of the jet in order to produce the desired regulation of speed, but not always with marked success. Another method in common use for governing and for regulating the power supplied to large wheels of this type is to employ several jets, any number of which can be brought to play on the vanes at will, but the arrangement is not altogether satisfactory, as the efficiency of the wheel decreases materially as the number of jets increases. In some tests made in California the following results were obtained : Number of jets. Total horse-power. Horse-power per jet. I 155 155 2 285 130 3 390 105 4 480 90 1 A similar device is used by Messrs. Gilbert Gilkes & Co., Kendal. Hydraulic Motors and Machines. 611 The problem of governing water-wheels of this type, even when a perfect expanding nozzle can be produced, is one of considerable difficulty, and those who have experimented upon such motors have often obtained curious results which have greatly puzzled them. The theoretical treatment which follows is believed to throw much light on many hitherto unexplained phenomena, such as (i.) It has frequently been noticed that the speed of a water motor decreases when the area of the jet is increased, the head of water, and the load on the motor, 0-1 0-2 0-3 0-4- 0-5 0-6 0-7 0-8 0-9 1-0 V3/VG Vd 've c/osed. Jtatio of Vti2ve> opening to Area of Pipe=N. fa// open, FIG. 567^. remaining the same, and vice versa, when the area of the jet is decreased the speed increases. If the area of the jet is regu- lated by means of a governor, the motor under such circum- stances will hunt in a most extraordinary manner, and the governor itself is blamed ; but, generally speaking, the fault is not in the governor at all, but in the proportions of the pipe and jet. (ii.) A governor which controls the speed admirably in the case of a given water motor when working under certain conditions, may entirely fail in the case of a similar water motor when the conditions are only slightly altered, such as an alteration in the length or diameter of the supply pipe. 612 Mechanics applied to Engineering. On p. 584 we showed that the velocity (V) of flow at any instant in a pipe is given by the expression v = L KD' In Fig. 567^ we give a series of curves to show the manner in which V varies with the ratio of the area of the jet to the area of the cross-section of the pipe, viz. n. From these curves it will be seen that the velocity of flow falls off very slowly at first, as the area of the jet is diminished, and afterwards, as the " shut " position of the nozzle is approached, the velocity falls V very rapidly. The velocity of efflux V l = of the jet itself is also shown by full lined curves. The quantity of water passing any cross-section of the main per second, or through the nozzle, is AV cubic feet per second, or 62'4AV Ibs. per second. The kinetic energy of the stream issuing from the nozzle is 6 2 - 4 AV Inserting the value of V, and reducing, we get o'76D 2 / H \ r The kinetic energy of the stream = ^ I -j^- y \i \TCn + ^2/ Curves showing how the kinetic energy of the stream varies with n are given in Fig. 567*?. Starting from a fully opened nozzle, the kinetic energy increases as the area of the jet is decreased up to a certain point, where it reaches its maximum value, and then it decreases as the area of the jet is further decreased. The increase in the kinetic energy, as the area of the jet decreases, will account for the curious action mentioned above, in which the speed of the motor was found to increase when the area of the jet was decreased, and vice versft. The speed necessarily increases when the kinetic energy increases, if the load on the motor remains constant. If, however, the area of the jet be small compared with the area of the pipe, the kinetic energy varies directly as the area of the jet, or Hydraulic Motors and Machines. nearly so. Such a state is, of course, the only one con- sistent with good governing. From a large number of curves o 0-2' & L-50' D-O-S L-/00' Va/ve 0-1 0-2 03 0-4- 0-5 0-6 0-7 0-8 0'9 W D * 05 ' l/!3/\/f2 Vfi/VP c/osed. Ratio of Value opening to Area, of Pipe^N. fu/fopen. FIG. 567?. the values of n for maximum kinetic energy has been deter- mined. The kinetic energy is a maximum when n = hence for good governing the area of the jet must be less than 614 Mechanics applied to Engineering. 4'4 (area of the cross-section of the pipe) or, Poncelot Water-wheel Vanes. In this wheel a thin stream of water having a velocity V feet per second glides up curved vanes having a velocity in the same direction as the y stream. The water moves up the vane with the velocity V -- relatively to it, and attains a height corresponding to this FIG. 568. velocity, when at its uppermost point it is at rest relatively to the / y\ vane ; it thsn falls, attaining the velocity (V J, neglecting friction ; the negative sign is used because it is in the reverse direction to which it entered. Hence, as the water is moving V forward with the wheel with a velocity , and backward / v\ relatively to the wheel with a velocity ( V ), the absolute velocity of the water on leaving the vanes is Hydraulic Motors and Mechanics 615 hence, when n = 2 the absolute velocity of discharge is zero, or all the energy of the stream is utilized. The efficiency may also be arrived at thus Let a = the angle between the direction of the entering stream and a tangent to the wheel at the point of entry. Then the component of V along the tangent is V cos a. The pressure exerted in the ) _ 2\Y / v V\ direction of the tangent f ~~ ~~JT \ ~ ~n ) The work done per second ! 2WV* / i \ on the plate - ( ~ ) And the hydraulic efficiency of the wheel The efficiency of these wheels varies from 65 to 75 per cent., including the friction of the axle. Form of Vane to prevent Shock. In order that the water may glide gently on to the vanes of any motor, the tangent to the entering tip of the vanes must be in the same direction <%- V p as the path of the water relative to the tip of the wheel; thus, in the figure, if ab represents the velocity of the entering stream, ad the velocity of the vane, then db represents the relative path of the water, and the entering tip must be parallel to it. The stream then gently glides on the vane without shock. Turbines. Turbines may be conveniently divided into two classes : (i) Those in which the whole energy of the water is converted into energy of motion in the form of free jets or streams which are delivered on to suitably shaped vanes in order to reduce the absolute velocity of the water on leaving to zero or nearly so. Such a turbine wheel receives its impulse from the direct action of impinging jets or streams ; and is known as an " impulse " turbine. When the admission only takes place over a small portion of the circumference, it is known as a " partial admission " turbine. The jets of water proceeding from the guide-blades are perfectly free, and after 6i6 Mechanics applied to Engineering, impinging on the wheel-vanes the water at once escapes into the air above the tail-race. (2) Those in which some of the energy is converted into pressure energy, and some into energy of motion. The water is therefore under pressure in both the guide-blades and in the wheel passages, consequently they must always be full, and there must always be a pressure in the clearance space between the wheel and the guides, which is not the case in impulse turbines. Such are known as "reaction" turbines, because the wheel derives its impulse from the reaction of the water as it leaves the wheel passages. There is often some little difficulty in realizing the pressure effects in reaction turbines. FIG. 570. FIG. Probably the best way of making it clear is to refer for one moment to the simple reaction wheel shown in Fig. 570, in which water runs into the central chamber and is discharged at opposite sides by two curved horizontal pipes as shown ; the reaction of the jets on the horizontal pipes causes the whole to revolve. Now, instead of allowing the central chamber to re- volve with the horizontal pipes, we may fix the central chamber, as in Fig. 571, and allow the arms only to revolve ; we shall get a crude form of a reaction turbine. It will be clear that a water- tight joint must be made between the arms and the chamber, because there is pressure in the clearance space between. It will also be seen that the admission of water must take place over Hydraulic Motors and Machines. 617 the whole circumference, and, further, that the passages must always be full of water. A typical case of such a turbine is shown in Fig. 576. These turbines may either discharge into the air above the tail- water, or the revolving wheel may dis- charge into a casing which is fitted with a long suction pipe, and a partial vacuum is thereby formed into which the water discharges. In addition to the above distinctions, turbines are termed parallel flow, inward flow, outward flow, and mixed-flow turbines, according as the water passes through the wheel parallel to the axis, from the circumference inwards towards the axis, from the axis outwards towards the circumference, or both parallel to the axis and either inwards or outwards. We may tabulate the special features of the two forms of turbine thus : IMPULSE. REACTION. All the energy of the water is Some of the energy of the water converted into kinetic energy before is converted into kinetic energy, and being utilized. some into pressure energy. The water impinges on curved The water is under pressure in wheel-vanes in free jets or streams, both the guide and wheel passages, consequently the wheel passages also in the clearance space ; hence must not be filled. the wheel passages are always full. The water is discharged freely As the wheel passages are always into the atmosphere above the tail- full, it will work equally well when water ; hence the turbine must be discharging into the atmosphere or at the foot of the fall. into water, i.e. above or below the tail-water, or into suction pipes. The turbine may be placed 30 feet above the foot of the fall. Water may be admitted on a Water must be admitted on the portion or on the whole circum- whole circumference of the wheel, ference of the wheel. Power easily regulated without Power difficult to regulate with- much loss. out loss. In any form of turbine, it is quite impossible to so arrange it that the water leaves with no velocity, otherwise the wheel would not clear itself. From 5 to 8 per cent, of the head is often required for this purpose, and is rejected in the tail-race. Form of Blades for Impulse Turbine. The form of blades required for the guide passages and wheel of a turbine are most easily arrived at by a graphical method. The main points to be borne in mind are the water must enter the guide and wheel passages without shock. To avoid losses through sudden changes of direction, the vanes must be smooth easy curves, and the changes of section of the passages gradual 6x8 Mechanics applied to Engineering. (for reaction turbines specially). The absolute velocity of the water on leaving must be as small as is consistent with making the wheel to clear itself. For simplicity we shall treat the wheel as being of infinite radius, and after designing the blades on that basis we shall, by a special construction, bend them round to the required radius. In all the diagrams given the water is represented as entering the guides in a direction normal to that in which the wheel is moving. Let the velocity of the water be reduced 6 per cent, by friction in passing over the guide-blades, and let 7 per cent, of the head be rejected in the tail-race. The water enters the guides vertically, hence the first tip of the guide-blade must be vertical as shown. In order to find the o-o GUIDES FIG. 579. direction of the final tip, we proceed thus : We have decided that the water shall enter the wheel with a velocity of 94 per cent, of that due to the head, since 6 per cent is lost in friction, whence V = also the velocity of rejection r - /** X 7 r ~V"7^ We now set down db to represent the vertical velocity with which the water passes through the turbine wheel, and from b we set off be to represent V ; then ac gives us the horizontal component of the velocity of the water, and = >y/'94 2 o*2y 2 = o'Q 1 = Y! : eb gives us the direction in which the water leaves the guide-vanes ; hence a tangent drawn to the last tip of the We omit to save constant repetition. Hydraulic Motors and Machines. 619 guide-vanes must be parallel to cb. We are now able to con- struct the guide-vanes, having given the first and last tangents by joining them up with a smooth curve as shown. Let the velocity of the wheel be one-half the horizontal velocity of the entering stream, or V w = = 0*45 ; hence the horizontal velocity of the water relative to the wheel is also 0*45. Set off a'b' as before = 0*27, and a'd horizontal and = 0-45 : we get db' ^representing the velocity of the water relative to the vane ; hence, in order that there may be no shock, a tangent drawn to the first tip of the wheel-vane must be parallel to db' ] but, as we want the water to leave the vanes with no absolute horizontal velocity, we must deflect it during its passage through the wheel, so that it has a backward velocity relative to the wheel of 0*45, and as it moves forward with it, the absolute horizontal velocity will be 0-45 -f 0^45 = o. To accomplish this, set off de = 0-45. Then Ve gives us the final velocity of the water relative to the wheel; hence the tangent to the last tip of the wheel-vane must be parallel to b'e. Then, joining up the two tangents with a smooth curve, we get the required form of vane. It will be seen that an infinite number of guides and vanes could be put in to satisfy the conditions of the initial and final tangents, such as the dotted ones shown. The guides are, for frictional reasons, usually made as short as is consistent with a smooth easy-connecting curve, in order to reduce the surface to a minimum. The wheel-vanes should be so arranged that the absolute path of the water through the wheel is a smooth curve without a sharp .bend. / 2 3 FIG- 573. The water would move along the absolute path K/ and along the path O relative to the wheel if there were no vanes 62O Mechanics applied to Engineering. to deflect it, \\ here hi is the distance moved by the vane while the water is travelling from k to * ; but the wheel-vanes deflect it through a horizontal distance hg, hence a particle of water at g has been deflected through the distance gj by the vanes, where gh = ij. The absolute paths of the water corresponding to the three vanes, i, 2, 3, are shown in the broken lines bear- ing the same numbers. In order to let the water get away very freely, and to prevent any possibility of them choking, the sides of the wheel-passages are usually provided with venti- lation holes, and the wheel is flared out. The efficiency of the turbine is readily found thus : The whole of the horizontal component of the velocity of the water has been imparted to the turbine wheel, hence the work done per pound of) Vj 2 water ) = ~^r the energy per pound of the \ _ _ V 2 water on entering / = = 2F V 2 the hydraulic efficiency = ^ = o'9 2 = 81 per cent. The losses assumed in this example are larger than is usual in well-designed turbines in which the velocity of rejection V r = o*i25V2^H; hence a higher hydraulic efficiency than that found above may readily be obtained. The total or overall efficiency is necessarily lower than the hydraulic efficiency on account of the axle friction and other losses. Under the most favourable conditions an overall efficiency of 80 per cent, may be obtained; but statements as to much higher values than this must be regarded with suspicion. In some instances an analytical method for obtaining the blade angles is more convenient than the graphical. Take the case of an outward-flow turbine, and let, say, 5 per cent, of the head be wasted in friction when passing over the guide- blades, and the velocity of flow through the guides be o' 1 25>v/2H. Then the last tip of the guide-blades will make an angle ^ with a tangent to the outer guide-blade circle, where sin 0j = =0-128, and X = 7 22'. The horizontal component of the velocity Vj = 0-125 cot 6 = 0-967. The circumferential velocity of the inner periphery of the wheel V w = 0*483. The inlet tip of the wheel-vanes makes an angle 2 with a tangent to the inner periphery of the wheel ; Hydraulic Motors and Machines. 621 or* to what is the same thing, the outer guide-blade circle, where tan 2 = ^ r = 0-259, and 2 = 14 31'. o 403 Let the velocity of flow through the wheel be reduced to 0*08 V 2*H at the outer periphery of the wheel, due to widening or flaring out the wheel-vanes, and let the outer diameter of the wheel be 1-3 times the inner diameter; then the circum- ferential velocity of the outer periphery of the wheel will be 0^483 X 1*3 = o '6 2 8, and the outlet tip of the wheel-vanes will make an angle 3 with a tangent to the outer periphery of the wheel, where tan 3 = Q . 2 g = 0*127, a d 3 = 7 16'. Pressure Turbine. Before pro- ceeding to consider the vanes for a pressure turbine, we will briefly look at its forerunner, the simple reaction wheel. Let the speed of the orifices be V ; then, if the water were simply left behind as the wheel revolved, the velocity of the water relative to the orifices would be V, and the head required to produce this velocity '- Let hi - the height of the surface of the water or the head above the orifices. ^ v Then the total head! producing flow, Hj V 2 2 y v-y FIG. 574. 622 Mechanics applied to Engineering. Let v = the relative velocity with which the water leaves the orifices. Then ^ = = V 2 + The velocity of the water relative to the ground = v V. v _ y If the jet impinged on a plate, the pressure would be per pound of water ; but the reaction on the orifices is equal to this pressure, therefore the reaction R = v- V and the work done per second ) ^ v _ V(fr V) by the jpts in foot-lbs. j g energy wasted in discharge )_ (v V) 2 water per pound in foot-lbs. ) ^ (") total work done per ) . _ # 2 V 2 pound of water j ** "* 2 g 2~V hydraulic efficiency . ' .. = As the value of V approaches v, the efficiency approaches 100 per cent., but for various reasons such a high efficiency FIG. 575- FIG. 576. can never be reached. The hydraulic efficiency may reach 65 per cent., and the total 60 per cent. The loss is due to the water leaving with a velocity of whirl v V. In order to reduce this loss, Fourneyron, by means of guide-blades, gave the water an initial whirl in the opposite direction before it entered the wheel, and thus caused the water to leave with Hydratdic Motors and Machines. 623 little or no velocity of whirl, and a corresponding increase in efficiency. The method of arranging such guides is shown in Fig. 575 ; they are simply placed in the central chamber of such a wheel as that shown in Fig. 571. Sometimes, however, the guides are outside the wheel, and sometimes above, according to the type of turbine. Form of Blades for Pressure Turbine. As in the impulse turbine, let, say, 7 per cent, of the head be rejected in the tail-race, and say 13 per cent, is wasted in friction. Then we get 20 per cent, wasted, and 80 per cent, utilized. Some of the head may be converted into pressure energy, and some into kinetic energy; the relation between them is optional as far as the efficiency is concerned, but it is con- venient to remember that the speed of the turbine increases as Y the ratio ==^ increases ; hence within the limit of the head of Vfc water at disposal any desired speed of the turbine wheel can be obtained, but for practical reasons it is not usual to make the above-mentioned ratio greater than 1*6. Care must always be taken to ensure that the pressure in the clearance space between the guides and the wheel is not below that of the atmosphere, or air may leak in and interfere with the smooth working of the turbine. In this case say one-half is converted into pressure energy, and one-half into kinetic energy. If 80 per cent, of the head be utilized, the corresponding velocity will be _ V = *2 * = -0- Thus 89 per cent, of the velocity will be utilized. To find the corresponding vertical or pressure component V r and the horizontal or velocity component V A , we draw the triangle of velocities as shown (Fig. 577), and find that each is 0*63 y^^H. We will determine the velocity of the wheel by the principle of momentum. Water enters the wheel with a horizontal velocity V A = 0*63, and leaves with no horizontal momentum. Horizontal pressure per pound of water = -* Ibs. g useful work done in foct-lbs. per second perl _ V ft V^ _ Q .gjj pound of water f g since we are going to utilize So per cent, of the head. 624 Mechanics applied to Engineering. Hence V = 0* V w = 0- We now have all the necessary data to enable us to determine the form of the vanes. Set down ab to represent the velocity of flow through the wheel, and ac the horizontal velocity. Completing the triangle, we find cb = 0*69, which gives us the direction in which the water enters the wheel or leaves the guides. The guide-blade is then put in by the method explained for the impulse wheel. ,^*>^ F 2 ? ^ Idf^ WHEEl. Jii FIG. 577. FIG. 578. To obtain the form of the wheel-vane. Set off ed to represent the horizontal velocity of the wheel, and ^"parallel to cb to represent the velocity of the water on leaving the guides ; then df represents the velocity of the water relative to the wheel, which gives us the direction of the tangent to the first tip of the wheel-vanes. Then, in order that the water shall leave with no horizontal velocity, we must deflect it during its passage through the wheel so that it has a backward velocity relative to the wheel of 0*64. Then, setting down gh = 0-27, and 7 = o 64, we get /fo'as the final velocity of the water, which gives us the direction of the tangent to the last tip of the vane. In Figs. 579, 580, 581, we show the form taken by the wheel-vanes for various proportions between the pressure and velocity energy. In the case in which V, = o, the whole of the energy is converted into kinetic energy; then V A = 0^89, and Hydraulic Motors and Machines. 625 Or the velocity of the wheel is one-half the horizontal velocity of the water, as in the impulse wheel. The form of blades in this case is precisely the same as in Fig. 572, but they are arrived at in a slightly different manner. For the sake of clearness all these diagrams are drawn with assumed losses much higher than those usually found in practice. V = 0-89 V B = o8 VA = 0-4 Vr=o'2 7 V w =i-o FIG 579. 0-8 V = 0-8$ V w = 0-5 FIG. 580 FIG. Centrifugal Head in Turbines. It was pointed out some years ago by Professor James Thompson that the centri- fugal force acting on the water which is passing through an inward-flow turbine may be utilized in securing steady running. 2 s 626 Mechanics applied to Engineering. and in making it partially self-governing, whereas in an outward- flow turbine it has just the opposite effect. Let R e = external radius of the turbine wheel ; R f = internal V e = velocity of the outer periphery of the wheel j V<= inner w = angular velocity of the wheel ; w = weight of a unit column of water. Consider a ring of rotating water of radius ; and thickness dr y moving with a velocity v. The mass of a portion of the ring of area j _ wa dr ' a measured normal to a radius ) "" g The centrifugal force acting on the mass = ^rdr wa(aP j-r.dr The centrifugal force acting on all the masses lying between the radius R i \ = ~ ~~ \ r.dr and R, or 1i The centrifugal head = - The centrifugal head = f ~ The centrifugal head per square inch ) __ V e 2 V< 2 per pound of water ) 2g This expression gives us the head which tends to produce outward radial flow of the water through the wheel due to centrifugal force it increases as the velocity of rotation increases ; hence in the case of an inward-flow turbine, when an increase of speed occurs through a reduction in the external load, the centrifugal head also increases, which thereby reduces the effective head producing flow, and thus tends to reduce the quantity of water flowing through the turbine, and thereby to keep the speed of the wheel within reasonable limits. On the other hand, the centrifugal head tends to increase the flow through the wheel in the case of an outward-flow turbine when the speed increases, and thus to still further augment the speed instead of checking it. Hydraulic Motors and Machines. 62 7 The following results of experiments made in the authors laboratory will serve to show how the speed affects the quantity of water passing through the wheel of an inward-flow turbine : GILKE'S VORTEX TURBINE. External diameter of wheel Internal ,, Static head (If,) of water above turbine... Guide blades 075 foot Q'375 24 feet Half open Revolutions per minute. Quantity of water passing in cubic feet per second. Centrifugal head He. KdAv^H, - He) O 0-68 ' 0-68 IOO 0-68 0-18 0-68 2OO o'67 072 0-67 3 00 0-66 1-6 0-66 4OO 0*64 2-9 0*64 500 0'62 4'5 0-61 600 o'59 6'5 0-58 700 o'SS 8-8 o'54 800 0*50 "'5 0-49 900 0-42 H'S o'43 The last column gives the quantity of water that will flow through the turbine due to the head H, H c . The coefficient of discharge K,, is obtained from the known quantity that passed through the turbine when the centrifugal head was zero, i.e. when the turbine was standing. Dimensions of Turbines. The general leading dimen- sions of a turbine for a given power can be arrived at thus Let B = breadth of the water-inlet passages in the turbine wheel ; R = mean radius of the inlet passage where the water enters the wheel ; V f = velocity of flow through the turbine wheel ; H = available head of water above the turbine ; 628 Mechanics applied to Engineering. Q = quantity of water passing through the turbine in cubic feet per second ; P = horse-power of the turbine; rj = the efficiency of the turbine ; N = revolutions of wheel per minute. Then, if B be made proportional to the mean radius of the water opening, say B = 0R \ and V, = b\/2gR | where a, b, and c are constants V = *vH then Q = 27TRBV, = 207rR 2 \/2 < g-H, neglecting the thickness of the blades P = g2' 550 R=, 0* V. = m\/2gH(elu.'eryfrcm. 7le/2-n FIG. 592 Where the two curves overlap, the ordinates are added. It will be observed that the fluctuation is very much greater than before. It should be noted that the second barrel begins to deliver just after the middle of the stroke. Single-acting Pump. Three barrels, Nos. i, 2, 3 cranks at 120. It will be observed that the flow is much more constant than in any of the previous cases. Similar diagrams are easily constructed for double-acting pumps with one or more barrels- it should, however, be noticed that the diagram for the return stroke should be reversed end for end, thus FIG. 594. We shall shortly see the highly injurious effects that sudden changes of flow have on a long pipe. In order to partially mitigate them and to equalize the flow, air-vessels are usually placed on the delivery pipe close to the pump. Their function 636 Mechanics applied to Engineering. on a pump is very similar to that of a flywheel on an engine. When the pump delivers more than its mean quantity of water the surplus finds its way into the air-vessel, and compresses the air in the upper portion; then when the delivery falls below the mean, the compressed air forces the stored water into the main, and thereby tends to equalize the pressure and the flow. The method of arriving at the size of air-vessel required to keep the flow within given limits is of a similar character to that adopted for determining the size of flywheel required for an engine. For three-throw pumps the ratio of the volume of the air- vessel to the volume displaced by the pump plunger per stroke is from i to 2, in duplex pumps from 1*5 to 5, and in cer- tain instances of fast-running, single-acting pumps it gets up to 30. Great care must be taken to ensure that air-vessels do contain the intended quantity of air. They are very liable to get water-logged through the absorption of the air by the water. Speed of Pumps. The term " speed of pump " is always used for the mean speed of the piston or plunger. The speed has to be kept down to moderate limits, or the resistance of the water in passing the valves becomes serious, and the shock due to the inertia of the water causes mischief by bursting the pipes or parts of the pump. The following are common maximum speeds for pumps : Large pumping engines and mining pumps 100 to 300 ft. min Exceptional cases with controlled valves ... up to 600 ,, Fire-engines 15010250 ,, Slow-running pumps 30 to 50 Force pumps on locomotives up to 900 The high speed mentioned above for the loco, force-pump would not be possible unless the pipes were very short and the valves very carefully designed; the valves in such cases are often 4-inch diameter with only |-inch lift. If a greater lift be given, the valves batter themselves to pieces in a very short time ; but in spite of all precautions of this kind, the pressures due to the inertia of the water are very excessive. One or two instances will be given shortly. Suction. We have shown that the pressure of the atmo- sphere is equivalent to a head of water of about 34 feet. No pump will, however, lift water to so great a height by suction, partly due to the leakage of air into the suction pipes, to the resistance of the suction valves, and to the fact that the water Pumps. 637 gives off vapour at very low pressures and destroys the vacuum that would otherwise be obtained. Under exceptionally favourable circumstances, a pump will lift by suction through a height of 30 feet, but it is rarely safe to reckon on more than 25 feet for pumps of average quality. Inertia Effects in Pumps. The hydraulic ramming action in reciprocating pumps due to the inertia of the water has to be treated in the same way as the inertia effects in water- pressure motors (see p. 594). In a pump the length of either the suction or the delivery pipe corresponds to the length of main, L. As an illustration of the very serious effects of the inertia of the water in pumps, the following extreme case, which came under the author's notice, may be of interest. The force-pumps on the locomotives of one of the main English lines of railway were constantly giving trouble through bursting ; an indicator was therefore attached in order to ascertain the pressures set up. After smashing more than one instrument through the extreme pressure, one was ultimately got to work successfully. The steam-pressure in the boiler was 140 Ibs., but that in the pump sometimes amounted to 3500 Ibs. per square inch ; the velocity of the water was about 28 feet per second through the pipes, and still higher through the valves ; the air-vessel was of the same capacity as the pump. After greatly increasing the areas through the valves, enlarging the pipes and air-vessels to five times the capacity of the pump, the pressure was reduced to 900 Ibs. per sq. inch, but further enlargements failed to materially reduce it below this amount. The friction through the valve pipes and passages will account for about 80 Ibs. per sq. inch, and the boiler pressure 140, or 220 Ibs. per sq. inch due to both ; the remaining 680 Ibs. per sq. inch are due to inertia of the water in the pipes, etc. Volume of Water delivered. In the case of slow-speed pumps, if there were no leakage past the valves and piston, and if the valves opened and closed instantly, the volume of water delivered would be the volume displaced by the piston. This, however, is never the case ; generally speaking, the quantity delivered is less than that displaced by the piston, sometimes it only reaches 90 per cent. No figures that will apply to all cases can possibly be given, as it varies with every pump and its speed of working. As might be expected, the leakage is generally greater at very slow than at moderate speeds, and is greater with high than with low pressures. This deficiency in the quantity delivered is termed the " slip " of the pump. 638 Mechanics applied to Engineering. With a long suction pipe and a low delivery pressure, it is often found that both small and large pumps deliver more water than the displacement of the piston will account for; such an effect is due to the momentum of the water. During the suction stroke the delivery valves are supposed to be closed, but just before the end of the stroke, when the piston is coming to rest, there may be a considerable pressure in the barrel due to the inertia of the water (see p. 639), which, if sufficiently great, will force open the delivery valves and allow the water to pass into the delivery pipes until the pressure in the barrel becomes equal to that in the uptake. When a pump is running so slowly that the inertia effects on the water are practically nil, the indicator diagram from the pump barrel is rectangular in form ; the suction line will be slightly below but practically parallel to the atmospheric line. When, however, the speed increases, the inertia effect on the water in the suction pipe begins to be felt, and the suction line of the diagram is of the same form as the inertia pressure line a ijijb in Fig. 176. Theoretical and actual pump diagrams are shown in Fig. 5940. In this case the inertia pressure at the end of the stroke was less than the delivery pressure in the uptake of the pump. If the speed be still further increased until the inertia pressure exceeds that in the delivery pipe, the pressure in the suction pipe will force open the delivery valves before the end of the suction stroke, and such diagrams as those shown in Fig. 594^ will be obtained. A comparison of these diagrams with the delivery-valve lift diagram taken at the same time is of interest in showing that the delivery valve actually does open at the instant that theory indicates. Apart from friction, the whole of the work done in accelerating the water in the suction pipe during the early portion of the stroke is given back by the retarded water during the latter portion of the stroke. In Fig. 594^ the line db' represents the distribution of pressure due to the inertia at all parts of the stroke. From d to / the pressure is negative, because the pump is accelerating the water in the suction pipe. At I-8 3 0-25 77'9 73 and 78 2 T 642 Mechanics applied to Engineering. The manner in which the "discharge coefficient" varies with the speed is clearly shown by Fig. 594^, which is one of the series of curves obtained by the author, and published in the Proceedings of the Institution of Mechanical Engineers, 1903. The dimensions of the pump were Diameter of plunger ... 4 inches. Stroke 6 Length of connecting-rod 12 ,, Length of suction and delivery pipes 63 feet each. Diameter of suction pipe 3 inches. The manner in which the water ram in the suction pipe is dependent upon the delivery pressure is shown in Fig. 595. Direct-acting Steam-pumps. The term " direct- acting " is applied to those steam-pumps which have no crank or flywheel, and in which the water-piston or plunger is on the same rod as the steam-piston ; or, in other words, the steam and water ends are in one straight line. They are usually made double acting, with two steam and two water cylinders. The relative advantages and disadvantages are perhaps best shown thus: ADVANTAGES. DISADVANTAGES. Compact. Steam cannot be used expan- Small number of working parts. sively, except with special arrange- No flywheel or crank-shaft. ments, hence Small fluctuation in the dis- Wasteful in steam. charge. Liable to run short strokes. Almost entjre avoidance of shock in the pipes. Less liability to cavitation Liable to stick when the steam troubles than flywheel pumps. pressure is low. We have seen that, when a pump is driven by a uniformly revolving crank, the velocity of the piston, and consequently the flow, varies between very wide limits, and, provided there is a heavy flywheel on the crank-shaft, the velocity of the piston (within fairly narrow limits) is not affected by the resistance it has to overcome ; hence the serious ramming effects in the pipes and pump chambers. In a direct-acting pump, however, the pistons are~free, hence their velocity depends entirely on the water-resistance to be overcome, provided the steam- pressure is constant throughout the stroke ; therefore the water is very gradually put into motion, and kept flowing much more steadily than is possible with a piston which moves practically ' ^respectively of the resistance it has to overcome. A diagram U *^-'^ : . Pumps. 643 5 3 ,3 Viuttia rv/? 63 feet long 2?t live? *V T>~r* essu, ~e D 36 70. ft /.o 2O 11 in. ^ i Vj /./I ^ z j 0-9 0Jfl 2 3 4- 5 C 7 o a 9 10 Revolutions per ~miTuie. FIG 594rf. IO 2O 3O 4-O SO 60 Revolutions per minute. FIG. 595- 70 644 Mechanics applied to Engineering. of flow from a pump of this character is shown in Fig. 596 ; it is intended to show the regularity of flow from a Worthington pump, which owes much of its smoothness of running to the fact that the piston pauses at the end of each stroke. 1 We will now-look at some of the disadvantages of direct- acting pumps, and see how they can be avoided. The reason why steam cannot be used expansively in pumps of this type is because the water-pressure in the barrel is practically constant, hence the steam-pressure must at all parts of the stroke be sufficiently high to overcome the water-pressure; thus, if the steam were used expansively, it would be too high at the beginning of the stroke and too low at the end of the stroke. The economy, however, resulting from the expansive use of steam is so great, that this feature of a direct-acting pump is considered to be a very great drawback, especially for large sizes. Many ingenious devices have been tried with FIG. 596. the object of overcoming this difficulty, and .some with marked success ; we shall consider one or two of them. Many of the devices consist of some arrangement for storing the excess energy during the first part of the stroke, and restoring it during the second part, when there is a deficiency of energy. It need hardly be pointed out that the work done in the steam- cylinder is equal to that done in the water-cylinder together with the friction work of the pump (Fig. 597). It is easy enough to see how this is accomplished in the case of a pump fitted with a crank and heavy flywheel (see Chapter VI.), since energy is stored in the wheel during the first part of the stroke, and returned during the latter part. Now, instead of using a rotating-body such as a flywheel to store the energy, a recipro- cating body such as a very heavy piston may be used ; then the excess work during the early part of the stroke is absorbed in accelerating this heavy mass, and the stored energy is given back during the latter part of the stroke while the mass is being retarded, and a very nearly even driving pressure throughout the stroke can be obtained The heavy piston is, however, not a practical success for 1 Taken from a paper on the Worthington Pump, Proc. Inst. of Civil Engineers, vol. Ixxxvi. p. 293. Pumps. 645 many reasons. A far better arrangement is D' Auria's pendulum pump, which is quite successful (Fig. 598). It is in principle Friction 597. The line of real pressure has been put in as in Fig. 179. Stecun W cuter Compensator- FIG. 599. Water the same as the heavy piston, only a much smaller reciprocating (or swinging) mass moving at a higher velocity is used. By far 646 Mechanics applied to Engineering. the most elegant arrangement of this kind 1 is D'Auria's water- compensator, shown in Fig. 599. It consists of ordinary steam and water ends, with an intermediate water-compensator cylinder, which, together with the curved pipe below, is kept full of water. The piston in this cylinder simply causes the 1 The author is indebted to Messrs. Thorp and Platt, of New York, for the particulars of this pump. Pumps. 647 water to pass to and fro through the pipe, and thus transfers the water from one end of the cylinder to the other. Its action is precisely similar to that of a heavy piston, or rather the pendulum arrangement shown in Fig. 598, for the area of the pipe is smaller than that of the cylinder, hence the water moves with a higher velocity than the piston, and consequently a smaller quantity is required. The indicator diagrams in Fig. 600 were taken from a pump of this type. The mean steam-pressure line has been added to represent the work lost in friction, and the velocity curve has been arrived at thus : Let M = the moving mass of the pistons and water in the compensator, etc., per square inch of piston, each reduced to its equivalent velocity ; V = velocity of piston in feet per second. . MV 2 Then the energy stored in the mass at any instant is ; but this work is equal to that done by the steam over and above that required to overcome friction and to pump the water, MVy 2 MV, 2 hence the shaded area px = , and likewise px a , 6 4 8 Mechanics applied to Engineering. where/ is the mean pressure acting during the interval x ; but - is a constant for any given case, hence V = \/px x con- stant. When the steam-pressure falls below the mean steam- pressure line the areas are reckoned as negative. Sufficient data for the pump in question are not known, hence no scale can be assigned. Another very ingenious device for the same purpose is the compensator cylinders of the Worthington high duty pump (see Fig. 60 1). The high and low pressure steam-cylinders are shown to the left, and the water- plunger to the right. Midway between, the compensator cylin- ders A, A are shown ; they are kept full of water, but they com- municate with a small high-pres- sure air-vessel not shown in the figure. When the steam-pistons are at the beginning of the "out" stroke, and the steam-pressure is higher than that necessary to overcome the water-pressure, the compensator cylinder is in the position i, and as the pistons move forward the water is forced from the compensator cylinder into the air-vessel, and thus by compressing the air stores up energy ; at half-stroke the com- pensator is in position 2, but immediately the half-stroke is passed, the compensator plunger is forced out by the compressed air in the vessel, thus giving back the stored energy when the steam-pressure is lower than that necessary to overcome the water-pressure ; at the end of the stroke the compensator is in position 3. A diagram of the effective compensator pressure on the piston-rods is shown to a small scale above the compensators, and complete indicator diagrams are shown in Fig. 602. No. i shows the two steam diagrams reduced to a common scale. No. 3 is the water diagram. In No. 2 the dotted curved line shows the compensator pressures ab +de etc ab'-de. a& FIG. 602. Pumps. 649 at all parts of the stroke ; the light full line gives the combined diagram due to the high and low pressure cylinders, and the dark line the combined steam and compensator pressure diagram, from which it will be seen that the pressure due to the combination is practically constant and similar to the water- pressure diagram in spite of the variation of the steam-pressure. We must not leave this question without reference to another most ingenious arrangement for using steam expansively in a pump the Davy differential pump. It is not, strictly speaking, a direct-acting pump, but on the other hand it is not a flywheel pump. We show the arrangement in Fig. 603. l The discs FIG. 603. move to and fro through an angle rather less than a right angle. In Fig. 604 we show a diagram which roughly indicates the manner in which the varying pressure in the steam-cylinder produces a tolerably uniform pressure in the water-cylinder. When the full pressure of steam is on the piston it moves slowly, and at the same time the water-piston moves rapidly ; then when the steam expands, the steam-piston moves rapidly and the water-piston slowly : thus in any given period of time the work done in each cylinder is approximately the same. We 1 Reproduced by the kind permission of the Editor of the Engineer and the makers, Messrs. Hathorn, Davy & Co., Leeds. 650 Mechanics applied to Engineering. have moved the crank-pin through equal spaces, and shaded alternate strips of the water and steam diagrams. The corre- sponding positions of the steam-piston have been found by simple construction, and the corresponding strips of the steam diagram have also been shaded ; they will be found to be equal to the corresponding water-diagram strips (neglecting friction). It will be observed that by this very simple arrange- ment the variable steam-pressure on the piston is very nearly balanced at each portion of the stroke by the constant water- pressure in the pump barrel. The pressure in the pump barrel is indicated by the horizontal width of the strips. We have neglected friction and the inertia of the moving masses, but our diagram will suffice to show the principle involved. Reversed Reaction Wheel. If a reaction wheel, such as that shown in Fig. 570, be placed with the nozzles in water, and the wheel be revolved in the opposite direction from some external source of power, the water will enter the nozzles and be forced up the vertical pipe ; the arrangement would, how- ever, be very faulty, and a very poor efficiency obtained. A far better result would be, and indeed has been, obtained by turning it upside down with the nozzles revolving in the same sense as in a motor, and with the bottom of the central chamber dipping in water. The pipes have to be primed, i.e. filled with water before starting ; then the water is delivered from the nozzles in the same manner as it would be from a motor. The arrangement is inconvenient in many respects Pumps. 65 1 A far more convenient and equally efficient form of pump will be dealt with in the next paragraph. Reversed Turbine or Centrifugal Pump. If an ordinary turbine were driven backwards from an external source of power, it would act as a pump, but the efficiency would probably be very low; if, however, the guides and blades were suitably curved, and the casing was adapted to the new conditions of flow, a highly efficient pump might be produced. Such pumps, known as turbine-pumps, are already on the market, and give considerably better results than have ever been obtained from the old type of centrifugal pump, and, moreover, they will raise water to heights hitherto considered impossible for pumps other than reciprocating. In most cases turbine-pumps are made in sections, each of which used singly is only suitable for a moderate lift ; but when it is required to deal with high lifts, several of such sections are bolted together in series. In the Parsons centrifugal augmentor pump the flow is axial, and both the wheel and guide blades are similar in appearance to the well-known Parsons steam-turbine blades; but, of course, their exact angles and form are arranged to suit the pressure and flow conditions of the water. In a test of a pump of this type made by the author, the following results were obtained : Total lift in feet. Millions of gallons pumped per day. Water horse- power. Speed in revolutions per minute. Efficiency of pump. 28l 3'53 209 2385 329 3-88 269 2664 383 3'i7 255 2625 From 52 to 56 per cent. 425 2'54 227 2555 458 2-14 206 2499 See Engineering, vol. ii. pp. 9, 32, 86 (1903). The form of centrifugal pump usually adopted is the 652 MecJianics applied to Engineering. reversed outward-flow turbine, or, more accurately, a reversed mixed-outward-flow turbine, since the water usually flows parallel to the axle before entering the "eye" of the pump, and when in the " eye " it partakes of the rotary motion of the vanes which give it a small velocity of whirl, the magnitude of which is a somewhat uncertain quantity. In order to deter- mine the most efficient form of blades a knowledge of this velocity is necessary, but unfortunately it is a very difficult quantity to measure, and practically no data exist on the question ; therefore in most rases an estimate is made, and the blades are designed accordingly; but in some special cases guide-blades are inserted to direct the water in the desired path. Such a refinement is, however, never adopted in any but very large pumps, since the loss at entry into the wheel is not, as a rule, very serious. The manner in which the shape of the blades is determined will be dealt with shortly. The water on leaving the wheel possesses a considerable amount of kinetic energy in virtue of its velocity of whirl and radial velocity. In some types of pump no attempt is made to utilize this energy, and consequently their efficiency is low; but in other types great care is taken to convert it into useful or pressure energy, and thus to materially raise the hydraulic efficiency of the pump. Almost all improvements in centri- fugal pumps consist of some method of converting the useless kinetic energy of the water on leaving the runner into useful pressure, or head energy. Form of Vanes for Centrifugal Pump. Let the figure represent a small portion of the disc and vanes, some- times termed the "runner," of a centrifugal pump. Symbols with the suffix i refer to the inner edge of the vanes. Let the water have a radial velocity of flow V r ; velocity of whirl V wl , in the "eye" of the pump just as it enters the vanes ; Let the water have a velocity of whirl V w as it leaves the vanes and enters the casing. N.I3. It should bo noted that this is not the velocity with which the water circulates in the casing around the wheel. Let the velocity of the tips of the vanes be V and V,. Then, setting off ab = V r , radially and ad = V.., tangentially, on completing the parallelogram we get ac representing the velocity of entry of the water ; then, setting off ae = V l also tangentially, and completing the figure, we get ec representing the velocity of the water relative to the disc. Hence, in order that there shall be no shock on entry, the tangent to the first Pumps. 653 tip of the vane must be parallel to ec. In a similar manner, if we decide upon the velocity of whirl V, on leaving the vanes, we can find the direction of the tangent to the outer tip of the vanes ; the angle that this makes to the tangent we term 0. In some cases we shall decide upon the angle beforehand, and FIG. 605. obtain the velocity of whirl by working backwards. Having found the first and last tangents, we connect them by a smooth curve. The radial velocity V r is usually made \c placed cJose to the water, with as little suction head as pos>il>le, e.g. a pump required to lift water through a total height of, say, 20 feet will work more efficiently with I -foot suction and 19- feet delivery head than with 5 -feet suction and I5-feet delivery head, and it will not work at all if the length of the delivery pipe is less than the suction. All high speed centrifugal pumps should have a very ample suction pipe and passages, and the entry resistances reduced to a minimum. If these precautions are not attended to, serious troubles due to cavitation in the runner will he liable to arise. Let W Ibs. of water pass through the pump j>er second, and let the runner or vanes be acted upon by a twisting moment T derived from the driving belt or other source of power. And further, let the angular velocity of the water be w. Then, adhering to the symbols used in the last paragraph The change of momentum of the water perl _ ^/\r ,_ y \ second in passing through the runner \ g ' The change in the moment of the momentum of the water per second, or, the change of angular momentum per second VV = "(V.R-V..RO This change in the angular momentum of the water is due to the twisting moment which is acting on the runner or vanes. Pumps. 655 Hence T = - V..R.) The work done per second on the) __ T _ W water by the twisting moment T/ = = ~( v Ra) v ci R i<) The useful work done by the} pump per second / ~ Hence The hydraulic efficiency \ _ of the pump rj \ ~ WH WH In some instances guide-blades are inserted in the eye of the pump to give the water an initial velocity whirl of V^, but usually the water enters the eye radially ; in that case V vl is zero, and putting Ru> = V, we get . go. W. In designing the vanes we assumed an initial velocity of whirl for the water on entering the vanes proper. This velocity is imparted to the water by the revolving arms of the runner after it has entered the eye, but before it has entered the vanes ; hence the energy expended in imparting this velocity to the water is derived from the same source as that from which the runner is driven, and therefore, as far as the efficiency of the pump is concerned, the initial velo- city of whirl is zero. Investigation of the Efficiency of Various Types of Centrifugal Pumps. CASE I. Pump with no Volute. In this case the water is dis- I/ charged from the runner at a high velocity into a FIG. 6os. chamber in which there is no provision made for utilizing its energy of motion, con- sequently nearly the whole of it is dissipated in eddies before 656 Mechanics applied to Engineering. it reaches the discharge pipe. A small portion of the velocity of whirl V w may be utilized, but it is so small that we shall neglect it. The whole of the energy due to the radial component of the velocity V r is necessarily wasted in this form of pump. In the figure the water flows through the vanes with a relative velocity v l at entry and v on leaving; then, by Bernouillis' theorem, we have the difference of pressure-head due to .any change in the section of the passages W 2g and the difference of pressure-head) _ V 2 Vj 2 . due to centrifugal force f " ~^~ If the water flows radially on entering the vanes, we have v* = V a 2 + V rl 2 and on leaving v = V r cosec The increase of pressure-head \ __ TT _ ^i 2 v* 4- V 2 Vi 2 due to both causes / ~ 2g By substitution and reduction, we find V rl 2 _ v r 2 cosec 2 6 4- V a- The total head that has to be maintained by the pump is due to the direct lift H, and to that required to generate the velocity of flow V rl in the eye of the pump. V rl 2 V rl 2 - V r 2 cosec 2 4- V 2 Whence H 4- = 2g 2g . __ V 2 - V r 2 cosec 2 andH =- W also V = ^2gH + V r 2 cosec 2 0=K tne velocity of discharge will be , and the gain of Pumps. 661 the V, 2 i \ head due to the bell mouth is - - T ). Even under 2g ri*J most favourable circumstances this is not a very large quantity, since V, seldom exceeds 15 feet a second, and is more generally about one-half that amount. Hence the gain of head due to a bell mouth cannot well be greater than 3 feet, and is therefore of no great importance, except in the case of very low lifts. The gain in efficiency is readily obtained from the expressions 900 K. ^ 500 I 400 300 200 WO Lift in, Feet including Friction, in Pipe (H) 16 20 FIG. 6n. 28 32 given above, by adding to the expression for H (Equations i. and vii.) the gain due to the bell-mouth. CASE IV. Pump fitted with a Whirlpool Chamber. If a whirlpool chamber, such as that shown in Fig. 612, be fitted to a pump, a free vortex will be formed in the chamber, and a corresponding gain in head will be effected, amounting to for a pump with no volute, "V ,"i r 4 or .!*-( i - t for a pump with a volute (see p. 570). This gain of head for well-designed pumps is usually small, 662 Mechanics applied to Engineering. and rarely amounts to more than 5 per cent, of the total head, and is often as low as ij per cent. The cost of making a pump with a whirlpool chamber is consider- ably greater than that of providing a bell mouth for the discharge pipe, hence many makers have discarded the whirlpool chamber in favour of the bell-mouth. There is, of course, no need to use both a bell-mouth and a whirlpool chamber. Some very interesting matter bearing on this question will be found in a paper by Dr. Stanton in the Proceedings of the Institution of Mechanical Engineers ) October, 1903. Skin Friction of the Rotating Discs in a Centri- fugal Pump. The skin friction of the rotating discs is a very important factor in the actual efficiency of centrifugal pumps. Let f = the coefficient of friction (see p. 576). Consider a ring of radius r feet and thickness dr rotating with a linear velocity u feet per second, relatively to the water in the casing ; then Vr = __ The skin resistance of the ring =fu' 1 2Trrdr V 2 FIG. 612. The moment of the frictional resistance) c it- f -r> 9 of the ring / R 2 The moment of the frictional resistance! __fV 2 27r j r ~ of the whole disc ) R2 . Cr = R i A dr and for the two discs 27T/VVR 6 - Rj 5 V R' 2 The moment of the frictional resistance = Pumps. The horse-power wasted in overcoming-i __ /VVR. 5 RI S \ the skin friction J~2i6\ R a / The horse-power wasted in skin friction Taking the usual proportions for the runner, viz. R x = , on substitution we get /V 3 R a 223 The practical outcome of an investigation of the disc friction is important ; it shows that the work wasted in friction varies in the first place as the cube of the peripheral velocity of the runner. The head, however, against which the pump will raise water varies as the square of the velocity, hence the wasted power varies as (head)2. Thus, as the head increases, the work done in overcoming the skin friction also increases at a more rapid rate, and thereby imposes a limit on the head against which a centrifugal pump can be economically em- ployed. Modern turbine-driven pumps are, however, made to work quite efficiently up to heads of 1000 feet. Further, the work wasted in disc skin friction varies as the square of the radius other things being equal; hence a runner of small diameter running at a very great number of revolutions per minute, wastes far less in skin friction than does a large disc running at a smaller number of revolutions with the same peripheral velocity. An excellent example of this is found in the centrifugal pumps driven by the De Laval steam turbines. The author is indebted to Messrs. Greenwood and Batley, Leeds, for the following results : Single stage. Two-stage. Diameter of wheel i '15 ft. 0*69 0-75 low, 0*24 high Width of waterway at cir- ) cumference ) 0-15 ft. 0-17 - Diameter of pipes 0-67 ft. 0*67 0*50 low, 0*33 high Quantity in cub. ft. sec. ... 2-23 3 '92 2'l8 2-52 0-83 o'75 o*54 Lift in feet 142 126 100 68 53 46 136 420 7 8t Revolutions per minute ... 1565 1540 1547 2040 1997 2000 2104 (low) 20,501 2027 jhigh 2029 Efficiency per cent. - 67-6 75-6 - 72-0 75'o not known 664 Mechanics applied to Engineering. In addition to the disc friction there is also a small amount of friction between the water and the interior of the runners, due to the outward flow of the water through the wheel passages. This, however, is negligible compared to the disc friction. | Capacity of Centrifugal Pumps. When the velocity of flow V r through the runner of a centrifugal pump is known, the quantity of water passing per second can be readily obtained by taking the product of this velocity and the circumferential area of the passages through the runner ; or, if the velocity of whirl in the volute V v be known, the quantity passing can be found from the product of this velocity and the sectional area of the volute close to the discharge pipe, which is usually made equal to the area of the discharge pipe itself. Whence from a knowledge of the dimensions of a pump both V r and V, can be readily obtained for any given discharge. The most im- portant quantity to obtain is, however, the speed of the pump V for any given discharge. Expressions have already been given for V in terms of the head H, the angle 0, the velocities V r and V,; but on comparing the value of V calculated from these expressions with actual values, a discrepancy will be discovered which is largely due to the friction of the water in the suction and discharge pipes, and in the pump itself. The latter quantity can be most readily expressed in terms of the length of pipe, which produces an equal amount of friction. A few cases that the author has examined appear to show that the internal resistance of the pump and foot valve together is equivalent to the friction on a length of about 80 diameters of the discharge pipe. This head must be added to the actual lift of the pump in order to find the real head H, against which the pump is lifting. When designing centrifugal pumps it is usual to make the velocity of flow V r about ^ decreases J f variation, i.e. \ with any given < increase or > in the length of the side ; or, ex- ( decrease J pressed in symbols, we want to know the value of , which we can obtain from the expression above thus Now, if we make L.X smaller and smaller, the fraction will become more and more nearly equal to 2jr, and ultimately (which we shall term the limit) it will be 7.x exactly. We then substitute da and dx for A# and A.r, and write it thus da dx = 7.X or, expressed in words, the area of the square varies 2x times as fast as the length of the side. This is actually and absolutely true, not a mere approximation, as we shall show later on by some examples. The fraction -r- is termed the differential coefficient of a with regard to x. Then, haying given a = ;r 2 , we are said to differentiate a with regard to x when we write y- = We will now go one step further, and consider a cube of side x ; its volume V = :r 3 . As before, let its side be in- creased by a very small amount A.T, and in consequence let the volume be increased by a corre- sponding amount AV. Then we get- V + AV = (x + A.*-) 3 = + 3-r 2 (A*) 3 FIG. 615. Subtracting our original value of V, we have AV = .r 2 Ajr The increase is shown in the figure, viz. three flat portions of 2 X 674 Appendix. area x 2 and thickness &x, three long prisms of length x and section (A.r) 2 , and one small cube of volume (A.r) 3 . From the above, we have A.T which in the limit (i.e. when toe becomes infinitely small) be- comes dV By a similar process, we can show that if s = t* Likewise, if we expand by the binomial theorem, or if we work out a lot of results and tabulate them, we shall find that the follow- ing relation holds : when x = y" ^ =>*-> dy Suppose we had such a relation as x = Ay 2 + C where A and C are constant quantities, and which of course do not vary ; then, if x increases by an amount A.T, and in consequence y increases by a corresponding amount Ay, we have X + A.T = A(j/ + Ay) 2 + C = A[> 2 + 2/Ay + (Aj) 2 ] + C Subtracting the original value of x, we have and ~ = which becomes in the limit dx r dy It should be noticed that the constant C has disappeared, while the constant A is multiplied by the differential of y 2 . Similarly for the constants in the following case. Let x = A/" + By" C/ D, where the capital letters are Appendix. 675 constants. Then by a similar process to the one just given, we get ~ l nBy*~ l C It sometimes happens that x increases to a certain value, its maximum, and then decreases to a certain value, its minimum, and possibly increases again, and so on. If at any one instant it is found to be increasing, and the next to be decreasing, it is certain that there must be a point between the two when it neither increases nor decreases ; this may occur at either the maximum or the minimum. At that instant we know that -r- = o. Then, in order to find when a given quantity has a maximum or a minimum value, we have to find the value of -j- and equate it to zero. Thus, suppose we want to divide a given number N into two parts such that the product is a maximum. Let x be one part, and N x the other, and y be the product. Then y = (N - x]x = N* - x z and -j- = N 2x = o when y is a maximum or 2x = N N and x 2 As an example, let N = 10 : N x x y 9 i 9 8 2 16 7 3 21 6 4 24 5 5 25 4 6 24 and so on. N Thus we see that_y has its maximum value when x = = 5. In some cases we may be in doubt as to whether the value we arrive at is a maximum or a minimum ; in such cases the beginner had better assume one or two numerical values near the maximum or minimum, and see whether they increase or decrease, or what is very often a convenient method to plot a diagram. This question is very clearly treated in either of the books mentioned above. The process of integration follows quite readily from that of differentiation. Let the line ae be formed by placing end to end a number of 676 Appendix. short lines ab all of the same length. When the line gets to <:, let its length be termed L c , and when it gets to *, L*. The length ce we have already said is equal to ab. Now, what is ce ? It is simply the difference between the length of the line ae and the line ac t or L. LC ; then, instead of writing in full that ce Lt is the difference in length of the two lines, we will write it A/, i.e. the difference in the length |^r~ after adding or subtracting one of the short FIG. 616. lengths. Now, however long or however short the line may be, the difference in the length after adding or subtracting one of the lengths will be A/. The whole length of the line L, is the sum of all the short lengths of which it is composed ; this we usually write briefly thus : or, the sum of (2) all the short lengths A/ between the limits when the line is of length L and of zero length is equal to L e . Sometimes the sign of summation is written /= L, S A/ = L /= o Similarly, the length LI is the sum of all the short lengths between b and c, and may be written the upper limit L c being termed the superior, and the lower limit L 6 the inferior limit : the superior limit is always the larger quantity. The lower limit of / is subtracted from the upper limit. In the case of a line, the above statements are perfectly true, however large or however small the short lengths A/ are taken, but in some cases which we shall shortly consider it will be seen that if A/ be taken large, an error will be introduced, and that the error becomes smaller as A/ becomes smaller, and it disappears when A/ becomes infinitely small ; then we substitute dl for A/, but it still means the difference in length between the two lines L e and L c , although ce has become infinitely small. We shall now use a slightly different sign of summation, or, as we shall term it, integra- tion. For the Greek letter 2 (sigma) we shall use the old English s, viz.yj and L 2 A ^ becomes I dl - L, r I J and A/ becomes / dl = L e - L& = L, I* Appendix. 677 The expression still means that the sum of all the infinitely short lengths dl between the limits of length L c and L 6 is Lj, which is, of course, perfectly evident from the diagram. Now that we have explained the meaning of the symbols, we will show a general connection between the processes of dif- ferentiation and integration. We have shown that when - dy J and dx = n*-^ But the sum of all such quantities dx, viz. fdx, = x y n (iii.) ; hence, substituting the value of dx from (ii.), we have fdx = But from (iii.) we have This operation of integrating a function of a variable y may be expressed in words thus : Add i to the index of the power of y, and divide by the index so increased, and by the differential of the variable. Thus y Similarly r n ; fmy n dy = J J J Let m = 4, n 2. Then or when m 3*2, n \ In order to illustrate this method of finding the sum of a large number of small quantities we will consider one or two simple cases. Let the triangle be formed of a number of strips all of equal length, ab or A/. The width FIG. 617. iv of each strip varies ; not only is each one of different width from the next, but its own width is not Appendix. constant. If we take the greater width of, say, the strip cefi, viz. tt/ e , the area vu e &l is too great, and if we take the smaller width a/ c , the area W C A/ will be too small. The narrower we take the strips the nearer will iv t = iv m and when the strips are infinitely narrow, w t = w c w^ and the area will be iv^H exactly, and the stepped figure gradually becomes a triangle. The area of the triangle is the sum of the areas of all the small strips iv . dl (Fig. 618). w I W/ But yr = T j or iv = -y-. Hence the area of the triangle is the sum of all the small strips of area dl, which we write FIG. 618. W WL 2 The -v- is placed outside the sign of integration, because neither W nor L varies. Now,/r0/ other sources we know that the area WL of the triangle is . Thus we have an independent proof of the accuracy of our reasoning. Suppose we wanted the area of the trapezium bounded by W and Wj, distant L and 1^ from the apex. We have But here again we know, from other sources, that the area of the trapezium is since W 1 = which again corroborates our rule for integration. By way of further illustrating the method, we will find the volume of a triangular plate of uniform thickness /. The volume of the plate is the sum of the volumes of the thin slices of thick- ness dl. Appendix. 679 Volume of a thin slice] (termed an elementary > = w . t . dl slice) J W/ a result which .we could have obtained by taking the product of the area ^- and the FIG, 619. thickness /. Similarly, the volume of the trapezoidal plate is One more case yet remains that in which / varies, as in a pyramid. Volume of an elementary slice w.t.dl w = W/ hence the volume of slice = Volume of pyramid = T/ WT j L 2 WT .dl / / 2 . dl = FIG. 620. WT \l_ L 2 x 3 WTL 3 We know, however, from other sources, that the volume of a pyramid is one-third the volume of the circumscribing solid. This is easily shown by making a solid and the corresponding pyramid of the same material, and comparing the weights or the volumes of the two ; or a cube can be so cut as to form three similar pyramids. In this case the volume of the circumscribing solid is WTL, and the volume of the pyramid - . Thus we have another proof of the accuracy of our method of integration. Similarly with the volume of the frustum of a pyramid. WT TT - L Until the reader has had an opportunity of following up the subject a little further, we must ask him to accept on faith the following results : 68o Appendix, -dx - log e x ^dx = lOge *! - ^ge *2 = lOff ^ and if j = log e x Where log* = 2-3 x the ordinary log. of the number, log e is known as the hyperbolic logarithm. The reader should also read up the elements of conic sections, especially the parabola and hyperbola. Checking Results. The more experience one gets the more one realizes how very liable one is to make slips in calculations, and how essential it is to check results. " Cultivate the habit of checking every calculation " is, perhaps, the best advice one can give a young engineer. First and foremost in importance is the habit of mentally reasoning out roughly the sort of result one would expect to get ; this will prevent huge errors such as this. In an examination paper a question was asked as to the deflection of a beam of 10 feet span under a certain load. One student gave it as 34^2 inches, whereas it should have been 0*342 inch an easy slip to make when working with a slide rule ; but if he had thought for one moment, he would have seen that it is impossible to get nearly 3 feet elastic deflection on a lo-feet beam. If an approximate method of arriving at a result is known, it should be used as a check on the more accurate method ; or if there are two different ways of arriving at a result, it should be worked out by both to see if they agree. A graphical method is often an excellent check on an analytical method, or vice versd. For example, suppose the deflection of an irregularly loaded beam has been arrived at by a graphical process ; it can be roughly checked by calculating the deflection on the assumption that the load is evenly distributed. If the load be mostly placed in the middle, the graphical process should show a rather greater deflection ; but if most of the load be near the two abutments, the graphical process should show a rather smaller deflection. In this way a very good check may be obtained. A little ingenuity will suggest ways of checking every result one obtains. In solving an equation or in simplifying a vulgar fraction with several terms, rough cancelling can often be done which will enable one by inspection to see the sort of result that should be obtained, and calculations made on the slide rule can be checked by taking the terms in a different order. Then, lastly, all very important work should be independently checked by another worker, or in some cases by two others. Real and Imaginary Accuracy. Some vainly imagine that the more significant figures they use in expressing the numerical Appendix. 68 1 value of a quantity, the greater is the accuracy of their work. This is an exceedingly foolish procedure, for if the data upon which the calculations are based are not known to within 5 per cent, then all figures professing to give results nearer than 5 per cent, are false and misleading. For example, very few steam-engine indicators, with their attendant reducing gears, etc., are accurate to within 3 per cent., and yet one often sees the I.H.P. of an engine given as, say, 2345*67 ; the possible error here is 2345*67 x 1 T) = 70*37 I.H.P. Thus we are not even certain of the 45 in the 2345 ; hence the figures that follow are not only meaningless, but liable to mislead by causing others to think that we can measure the I.H.P. of an engine much more accurately than we really can. In the above case, it is only justifiable to state the I.H.P. as 2340, or rather nearer 2350 ; that is, to give one significant figure more than we are really certain of, and adding o's after the uncertain significant figure. In some cases a large number of significant figures is justifiable ; for example, the number of revolutions made by an engine in a given time, say a day. The counter is, generally speaking, absolutely reliable, and therefore the digits are as certain as the millions. Generally speaking, engineering calculations are not reliable to anything nearer than one per cent., and not unfrequently to within 10 or even 20 per cent. The number of significant figures used should therefore vary with the probable accuracy of the available data. Experimental Proof of the Accuracy of the Beam Theory. Each year the students in the author's laboratory make a tolerably complete series of experiments on the strength and elasticity of beams, in order to show the discrepancies (if any) between theory and experiment. The following results are those made during the past session, and are not selected for any special reason, but they serve well to show that there is no material error involved in the usual assumptions made in the beam theory. The deflection due to the shear has been neglected. The gear for measuring the deflection was attached at the neutral plane of the beam just over the end supports, in order to prevent any error due to external causes. If there had been any material error in the theory, the value of Young's Modulus E, found by bending, would not have agreed so closely with the values found by the tension and compression experiments. The material was mild steel ; all the specimens were cut from one bar, and all annealed together. The section was approximately 2 inches square, but the exact dimensions are given in each case. 682 Appendix, BENDING. TENSION. COMPRESSION. Central load. Two loads dividing span into three equal parts. v WL E _WL 6 ~~ 48^1 23 WL 8 ~~ x Ax Depth of sect. (A) 2*026" Breadth ,, WZ'QIS" Depth of sect. (A) 2*020'' Breadth (^2*026" Sect. 1*066" X 2*038" Area (A) 2*172 sq. ins. Length between \ n Sect. 2*041" X 2*037' Area (A) 4*157 sq. u Length between J Span (L) . . 36*25" Span (L) . . 36*25' datum points (L) } I0 datum points (L) 5 x Load in Deflection in Load in Deflection in Load in Strain ir. Load in Strain in tons (W). inches (3). tons(W). inches (3). tons(W). inches (JT). tons (W). inches (x 0*1 0*007 0*1 0*004 , 0-0003 2 0*0003 0*2 0-012 0*2 0*009 2 0*0006 4 O*OOO6 O*3 0*017 0-3 O*OI4 3 O'OOIO 6 O'OOIO 0'4 0*023 O*OI9 4 0*0013 8 0*0013 o* c O*O28 O*5 0*023 5 0*0017 10 O-OOI7 0-6 0-033 0*6 0-028 6 O*OO2O 12 0*0021 0*7 0-037 0-7 0-032 7 0-0023 14 O*OO24 0*8 0-043 0*8 0-037 8 O*OO26 16 O*OO28 0*9 0-047 0-9 0*042 9 0-0029 18 0*003I o 0-053 *o 0-047 10 0-0033 20 0*0035 i 0*058 j 0*051 ii 0*0037 22 0-0038 2 '3 0-064 0*069 2 "3 0*056 O'o6o 12 13 0*0040 0*0044 24 26 0*0042 0*0046 '4 0-075 '4 0*065 14 0*0048 28 0-0050 o'oSo 5 0*070 15 0*005I 30 0*0053 6 0-086 *6 0*075 16 0*0054 32 0-0057 '7 0-091 '7 0*079 17 o-.oo57 34 O*OO6 1 8 0-097 1*8 0*083 18 0-0060 36 0*0065 '9 0-103 1*9 0*088 19 0-0063 38 0*0069 2'0 0-109 2*0 0*093 20 0*0068 40 0*0072 2"! 0-114 2*1 0*098 21 0*0072 42 0-0075 2-2 O'I2O 2*2 0*103 22 0-0075 44 O*OO8o 2 '3 0-126 2> 3 0*1 08 23 0*0079 46 0*0084 0-I3I O*II2 24 0-0083 48 0*0088 2'5 0*136 2*5 0*117 25 0*0087 50 0*0093 2*6 0*142 2*6 O'I22 26 0*0098 52 O-OIO9 2'7 0-148 2-7 0'I26 27 0*0130 53 0-0175 2*8 0-154 2-8 0*133 28 0*0200 54 O'O2 2*9 0*160 2*9 0*138 29 0-0300 56 O*IO 3*0 0*168 0*I42 30 0*0400 I 8 0*11 3' 1 0-176 3*1 0*147 31 0*14 60 0*13 3*2 0-183 3*2 0-153 32 0*16 62 0-15 3'3 0*194 3'3 OT57 33 o*iS 64 0-17 3*4 0*208 3'4 0*l62 34 0*20 68 0'2I g 0*236 o*57 P 0*168 0*173 36 38 0-22 0-30 72 76 0*25 0*30 Appendix. 683 Load in tons(W). Deflection in inches (3). Load in tons (W). Deflection in inches (d). Load in tons (W). Strain in inches (.ar). Load in tons (W). Strain in inches (jr). 37 0-82 37 0-177 40 0'35 80 '35 1-05 3'8 0-180 42 0-42 84 0-40 3'9 1-18 3*9 0-184 44 0-52 4-0 ^ '59 4-0 0-189 46 0*62 4' i 1-91 4' i 0*196 48 075 4-2 2*19 4-2 0-203 50 4'3 2-60 4*3 O'2IO 5 2 I-30 4*4 2-77 4*4 0-218 53 174 4'5 4-6 3-01 3 '40 4*3 4-6 0-228 0-250 5 -o 3-01 Broke 47 4-12 47 0-300 4-8 4'97 4*8 o -35 4]9 6-41 4'9 '45 8-88 0*50 5' 1 o'6o 5-2 0-80 5'3 I'lO S'4 i '45 5'5 2-50 6'o 5^5 E=i; 5,000 tons E 6 = i 2,940 tons E t = i 3,420 tons E,= i3 ,180 tons sq. inch. sq. inch. sq. inch. sq. inch. It will be seen that the E obtained by the bending experiments agrees closely with that found by tension and compression ; the mean of the former is 2*5 per cent, lower than the latter. If we had taken into account the deflection due to the shear, the discrepancy would have been even smaller. Further, the E c in this case is i '8 per cent, less than the E,, hence we should expect the E 6 to be o'5 per cent, less than the E { (see p. 388). The elastic limit of the tension bar occurs at about 27 tons, corresponding to a tensile stress of 12*43 tons per square inch ; in compression it occurs at about 53 tons, corresponding to a com- pressive stress of 1275 tons per square inch. For reasons stated on p. 389, we cannot arrive at the true elastic limit in bending. Theory of Hooks. The theory of hooks, as given on p. 458, has recently been shown to be incomplete by Professor Karl Pearson, F.R.S., and Mr. E. S. Andrews, B.Sc. of University College, London. The theory of combined bending and direct stress that we have given is only true for members which are sensibly straight ; when they are curved, terms involving the radius of curvature of the profile have to be included in the expression. The result of the investigations shows that the simple theory we have given indicates too Iowa value for the tensile stress at the intrados, 684 Appendix. and too high a value at the extrados, and in some cases the old theory is very seriously in error. The true theory is very complex, hence readers who are interested in the matter should refer to a paper by Professor Pearson and Mr. Andrews, " On a Theory of the Stresses in Crane and Coupling Hooks, with Experimental Comparison with Existing Theory," published by Dulau & Co., 37, Soho Square, W. Straight-line Strut Formula. The more experience one gets in the testing of full-sized struts and columns, the more one realizes how futile it is to attempt to calculate the buckling load with any great degree of accuracy. If the struts are of homogeneous material and have been turned or machined all over, and are, moreover, very carefully tested with special holders, which ensure dead accuracy in loading, and every possible care be taken, the results may agree within 5 per cent, of the calculated value ; but in the case of commercial struts, which are not always perfectly straight, and are not always perfectly centrally loaded, the results are frequently 10 or 15 per cent, out with calculation, even when reasonable care has been taken ; hence an approximate expression, such as one of the straight-line formulas, is good enough for many practical purposes, provided the length does not exceed that specified. An expression of this kind is P = M- N-^ a where P = the buckling load in pounds per square inch ; M = a constant depending upon the material ; N = a constant depending upon the form of the strut section ; / = the "equivalent length" of the strut ; d the least diameter of the strut. Material. Form of section. M. N. - not to d exceed Wrought iron ... 1 47,000 825 40 47,000 900 40 o 47,000 775 40 HL+ J.U 47,000 1070 30 Mild steel Ml 71,000 1570 3 71,000 1700 30 73,000 1430 30 HL + J.U 71,000 1870 30 Appendix. 685 Material. Form of section. M. N. -j not to exceed. Hard steel B II4,OOO 3200 30 II4,OOO 3130 30 o 114,000 2700 30 ML + -LU II4,OOO 3500 30 Soft cast iron ... 90,000 4100 15 90,000 4700 IS O 00,000 3900 15 HL+ J-U 90,000 5000 15 Hard cast iron ... mm 140,000 6600 IS 140,000 7000 15 I4O,OOO 6100 15 HL. + J.LJ 140,000 8000 15 Pitchpine and oak ; 8000 470 10 8000 500 10 Whirling of Shafts The following values are taken from Professor Dunkerley's paper, " On the Whirling and Vibration of Shafts," read before the Liverpool Engineering Society, Session 1-5- Let N = the number of revolutions per minute the shaft makes when whirling ; d = the diameter of the shaft in inches ; / = the length of the shaft in feet. 686 Appendix. Description of shaft. Remarks on /. N. /. Unloaded: overhanging from a long bearing which fixes its direction Length of over- hang in feet II,6 4 s VI Unloaded : resting in short or swivelled bearings at each end Distance between centres of bear- ings in feet 32,864^ V? Unloaded : supported as in last case, but one end overhanging c feet Ditto "7 d "7* For values VI of a see Table I. VI Unloaded : supported in a long bearing at one end, and a short or swivelled bearing at the other Distance between inner edge of long bearing and centre of short bearing 51,340^ Unloaded : shaft supported in three short or swivelled bearings, one at each end '=* /j = shorter span 4 = longer span in feet < For see /2= ^\/l values of a Table II. Unloaded : shaft supported in long bearings which fix its direction at each end Clear span be- tween inner edges of bearings 74,97172 32,864 V? Unloaded : long continuous shaft supported on short or swivelled bearings equi- distant Distance between centres of bear- ings in feet V? Loaded : shaft supported in short or swivelled bear- ings, single pulley of weight W pounds cen- trally placed between the bearings Distance between the centres of the bearings in feet N = 3 d* 7)35 Vw? .Loaded : long bearings which fix its direction at each end Clear span be- tween inner edges of bearings N = 74 ' 7 VW Appendix. TABLE I. 687 Value of r = l a. v. I -00 7,554 87 075 12,044 no 0-50 20,931 145 '33 28,095 168 0-25 to o'io 31,000 176 Very small 31,590 178 Zero 32,864 181 TABLE II. Value of r = l. '2 a. v. I -oo to 075 32,864 181 0-5 to 075 36,884 192 0'33 ' 41,026 203 0-25 43,289 208 0'20 tO 0-14 44,3i 2 211 0*125 to o'io 47,125 217 Very small 50^54 225 ' For hollow shafts in which d v the external diameter in inches 4 = internal substitute for d in the expressions given above the value For other cases of loaded shafts the reader must refer to the paper mentioned above. EXAMPLES CHAPTER I. 1 . Express 25 tons per square inch in kilos, per square millimetre. Ans. 39-4. 2. A train is running at 30 miles per hour. What is its speed in feet per second? Ans, 44. 3. What is the angular speed of the wheels in the last question? Diameter = 6 feet. Ans. 14$ radians per second. 4. A train running at 30 miles per hour is pulled up gradually and evenly in 100 yards by brakes. What is the negative acceleration ? Ans. 3'23 feet per second per second. 5. What is the negative acceleration, if the train in Question 4 be pulled up from 40 miles to 5 miles per hour in 80 yards? Ans. 7*06 feet per second per second. 6. If the train in Question 5 weighed 300 tons, what was the total retarding force on the brakes? Ans. 65 '8 tons. (Check your result by seeing if the work done in pulling up the train is equal to the change of kinetic energy in the train.) 7. Water at 60 Fahr. falls over a cliff 1000 feet high. What would be the temperature in the stream below if there were no disturbing causes ? Ans. 6 1 -29 Fahr. 8. A body weighing 10 Ibs. is attached to the rim of a rotating pulley of 8 feet diameter. If a force of 50 Ibs. were required to detach the body, calculate the speed at which the wheel must rotate in order to make it fly off. Ans. 61 revolutions per minute. 9. (I.C.E., October, 1897.) Taking the diameter of the earth as 8000 miles, calculate the work in foot-pounds required to remove to an infinite distance from the earth's surface a stone weighing I Ib. Ans. 21,120,000 foot-lbs. 10. (I.C.E., October, 1897.) If a man coasting on a bicycle down a uniform slope of I in 50 attains a limiting speed of 8 miles per hour, what horse-power must he exert to drive his machine up the hill at the same speed, there being no wind in either case ? The weight of man and bicycle together is 200 Ibs. Ans. o'lj. 11. Find the maximum and minimum speeds with which the body represented in Fig. 2 is moving. Express the result in feet per second and metres per minute. Ans. Maximum, 13 feet per second, or 238 metres per minute at about the fifth second ; minimum, 0*5 foot per second, or 9*15 metres per minute at about the third second. Examples. 689 CHAPTER II. 1. A locomotive wheel 8 feet 6 inches diameter slips 17 revolutions per mile. How many revolutions does it actually make per mile? Ans. 2147. 2. In Fig. 13, C = 200 feet, C T = 120 feet. Find the length of the arc by both methods. Ans. 256 feet, 253 '3 feet. 3. Find the length of a semicircular arc of 2 inches radius by means of the method shown in Fig. 15, and see whether it agrees with the actual length, viz. 6*28 inches. 4. Find the weight of a piece of f-inch boiler plate 12' 3" X 4' 6", having an elliptical manhole 15" X 10" in it. (One-inch plate weighs 40 Ibs. per square foot.) Ans. 1629 Ibs., or 073 ton. 5. Find the area of a regular hexagon inscribed in a circle of 50 feet radius. Ans. 6495 square feet. 6. Find the area in square yards of a quadrilateral ABCD. Length of AB = 200 yards, BC = 650 yards, CD = 905 yards, DA = 570 yards, AC = 800 yards. (N.B. The length BD can be calculated by working backwards when the area is known, but the work is very long. The tie-lines AC and DB in a survey are often checked by this method.) Ans. 272,560 square yards. 7. A piece of sheet iron 5 feet wide is about to be corrugated. Assuming the corrugations to be semicircular of 3-inch pitch, what will be the width when corrugated ? Ans. 3-18 feet. 8. Find the error per cent, involved in using the first formula of Fig. 25, when H = ^ for a circle 5 feet diameter, assuming the second formula to be exact. Ans. 5 '8. 9. Find the area of the shaded portion of Fig. 351. 8 = 4 inches; the outlines are parabolic. Ans. 2*67 square inches. 10. Cut an irregular-shaped figure out of a piece of thin cardboard or metal with smooth edges. Find its area by (i) the method shown in Fig- 3 2 > ( 2 ) tne niean ordinate method ; (3) Simpson's method ; (4) the planimeter ; (5) by weighing. 11. Find the area of Figs. 29, 31, 32, 33, 34 expressed in square inches. Ans. (29) 0-29, (31) o'i6, (32) 1-14, (33) 0-83, (34) 0*86. 12. Find the sectional area of a hollow circular column, 12 inches diameter outside, 9 inches diameter inside. Ans. 49*48 square inches. 13. Bend a piece of thin wire to form the quarter of an arc of a circle, balance it to find the c. of g., and then find the surface of a hemisphere by the method shown in Fig. 35. Check the result by calculation by the method given in Fig. 36. 14. Find the heating surface of a taper boiler-tube length, 3 feet ; diameter at one end, 5 inches ; at the other, 8 inches. Ans. 5 'i square feet. 15. Find the mean height of an indicator diagram in which the initial height Y is 2 inches, and cut-off occurs at \ stroke, log 4 = 0*602 (i.) with hyperbolic expansion ; (ii). with adiabatic expansion, (n = 1-4.) Ans. (i.) i '19 inch ; (ii.) 1*02 inch. 2 Y 690 Examples 16. Find the weight of a cylindrical tank 4 feet diameter inside, and 7 feet deep, when full of water. Thickness of sides jj inch ; the bottom, which is \ inch thick, is attached by an internal angle 3" X-3" X ". There are two vertical seams in the sides, having an overlap of 3 inches ; pitch of all rivets, 2\ inches ; diameter, inch. Water weighs 62-5 Ibs. per cubic foot, and the metal weighs 480 Ibs. per cubic foot. Ans. 7270 Ibs. 17. (Victoria, 1893.) A series of contours of a reservoir bed have the following areas : At water-level 4,800,000 sq. yards. 2 feet down ... 3,700,000 ,, 4 2,000,000 6 700,000 8'4 Zero Find the volume of the reservoir in cubic yards. ' Ans. 5,900,000 (approx.). 1 8. The area of a half-section of a concrete building consisting of cylindrical walls and a dome is 12,000 square feet. The c. of g. of the section is situated 70 feet from the axis of the building. Find the number of cubic yards of concrete in the structure. Ans. 195,400. 19. Find the weight of water in a spherical vessel 6 feet diameter, the depth of water being 5 feet. Ans. 6545 Ibs. 20. In the last question, how much water must be taken out in order to lower the level by 6 inches. Ans. 581 Ibs. (Roughly check the result by making a drawing of the slice, measure the mean diameter of it, then calculate the volume by multiplying the mean area by the thickness of the slice. ) 21. Calculate the number of cubic feet of water in an egg-ended boiler, diameter, 6 feet ; total length, 30 feet ; depth of water, 5*3 feet. Ans. 746. 22. Find the number of cubic feet of solid stone in a heap having a rectangular base 60' X 18', standing on level ground ; slope of sides, \\ to I foot ; flat-topped, height 4 feet ; the voids being 35 per cent. Ans. 1881. 23. Find the weight of a steel projectile with solid body and tapered- point, assumed to be a paraboloid. Material weighs 0*28 Ib. per cubic inch j diameter, 6 inches ; length over all, 24 inches ; length of tapered part, 8 inches. Ans. 158 Ibs. 24. Find the weight of a wrought-iron anchor ring, 6 inches internal, and 10 inches external diameter, section circular. Ans. 22 Ibs. CHAPTER III. 1. A uniform bar of iron, f inch square section and 6 feet long, rests on a support 18 inches from one end. Find the weight required on the short arm at a distance of 15 inches from the support in order to balance the long arm. Ans. 13*5 Ibs. 2. In the case of a lever such as that shown in Fig. 65, W, i \ ton, /! = 28 inches, / = 42*5 inches, / 2 = 50 inches, / 3 = 4 inches, w. z = I ton. Examples. 691 Find W when w 3 = o, and find w 3 when w 2 l z is clockwise and 7 2 = 12-5 feet. Ans. W = 2 tons ; w 3 = 50 tons. (N.B. In the Buckton testing-machine, w 3 is the load on the bar under test.) 3. A lever safety-valve is required to blow off at 70 Ibs. per square inch. Diameter of valve, 3 inches ; weight of valve, 3 Ibs. ; short arm of lever, 2| inches ; weight of lever, 1 1 Ibs. ; distance of c. of g. of lever from fulcrum, 15 inches. Find the distance at which a cast-iron ball 6 inches diameter must be placed from the fulcrum. The weight of the ball-hook = 0*6 Ib. Ans. 35-5 inches. ^ 4. In a system of compound levers the arms of the first lever are 3 inches and 25 inches, of the second 2*5 inches and 25*5 inches, of the third 2*1 inches and 25*9 inches, a force of 84 Ibs. is exerted on the end of the long arm of the third lever. Find the force that must be exerted on the short arm of the first lever when the system is arranged to give the greatest possible leverage. Ans. 88,060 Ibs. 5. Find the position of the c. of g. of a beam section such as that shown in Fig. 355. Top flange 3 inches wide, l inch thick. Bottom flange 15 ,, wide, if thick. Web i thick. Total height 18 Ans. 572 inches from the bottom edge. 6. Find the height of the c. of g. of a T-shaped section from the foot, the top cross-piece being 12 inches wide, 4 inches deep, the stem 3 feet deep, 3 inches wide. Ans. 24*16 inches. (Check by seeing if the moments are equal about a line passing through the section at the height found.) 7. Cut out of a piece of thin card or metal such a figure as is shown in Figs. 78 and 87. Find the c. of g. by calculation or by the graphical process, and check the result by balancing as shown on p. 75. 8. In such a figure as 77, the width of the base is 4 inches, and at the base of the top triangle 1*5 inch. The height of the trapezium = 2 inches, and the total height = 5 inches. Find the height of the c. of g. from the apex. Ans. 3*54 inches. 9. A trapezoidal wall has a vertical back and a sloping front face : width of base, 10 feet ; width of top, 7 feet ; height, 30 feet. What horizontal force must be applied at a point 20 feet from the top in order to overturn it, i.e. to make it pivot about the toe ? Width of wall, I foot ; weight of masonry in wall, 130 Ibs. per cubic foot. Ans. 18,900 Ibs. 10. Find the height of the c. of g. of a column 4 feet square and 40 feet high, resting on a tapered base forming a frustum of a square-based pyramid 10 feet high and 8 feet square at the base. Ans. 20*4 feet from base. 11. Find the position of the c. of g. of a piece of wire bent to form three-fourths of an arc of a circle of radius R. Ans. On a line drawn from the centre of the circle to a point bisecting the arc, and at a distance O'3R from the centre. 12. Find the position of the c. of g. of a balance weight having the form of a circular sector of radius R, subtending an angle of 90. Ans. o'6R from centre of circle. 692 Examples. 13. Find the. second moment (I ) of a thin door about its hinges: width, 7 feet ; height, 4 feet. Ans. 457 in feet 4 units. 14. Find the second moment (I) of a rectangular section 9 inches deep, 3 inches wide, about an axis passing through the c. of g., and parallel to the short side. Ans. 183 inch 4 units. 15. Find the second moment (T ) of a square section of 4-inch side about an axis parallel to one side and 5 inches from the nearest edge. Ans. 805*3 inch 4 units. 16. Find the second moment (I ) of a triangle 0-9 inch high, base 06 inch wide as in Fig. 100. Ans. 0-109 inc h 4 units. 17. Find the second moment (1) of a triangle 4 inches high and 3 inches base, about an axis passing through the c. of g. of the section and parallel with the base. Ans. ^ inch 4 units. Ditto (I ) about the base of the triangle. Ans. 16 inch 4 units. Ditto (I ) of a trapezium (Fig. 103) j B = 3 inches, B, = 2 inches, h 2 inches. Ans. 7*3 inch 4 units. Ditto (I ) ditto, as shown in Fig. 104. Ans. 6 inch 4 units. Ditto (I) ditto, as shown in Fig. 105. Ans. 1-64 inch 4 units. Ditto ditto, by the approximate method in Fig. 106. Ans. i '67 inch 4 units. 18. Find the second moment (I) of a square of 6 inches edge about its diagonal. Ans. 108 inch 4 units. 19. Find the second moment (I) of a circle 6 inches diameter about a diameter. Ans. 63-6 inch 4 units. 20. Find the second moment (I) of a hollow circle 6 inches external and 4 inches internal diameter about a diameter. Ans. 51 inch 4 units. 21. Find the second moment (I) of a hollow eccentric circle, as shown in Fig. no. D e = 6 inches, D* = 4 inches. The metal is \ inch thick on the one side, and \\ inch on the other side. Ans. 45*4 inch 4 units. 22. Find the second moment (I) of an ellipse about its minor axis. D 2 = 6 inches, Dj = 4 inches. Ans. 42*4 inch 4 units. Ditto, ditto about the major axis. Ans. 18*84 inch 4 units. 23. Find the second moment (I) of a parabola about its axis. H = 6 inches, B = 4 inches. Ans. 102^4 inch 4 units. Ditto, ditto (I ) about its base, as in Fig. 1 14. Ans. 263*3 inch 4 units. 24. Take an irregular figure, such as that shown in Fig. 115, and find its second moment (I ) by calculation. Check the result by the graphical method given on p. 96. 25. Find the second polar moment (I p ) of a rectangular surface as shown in Fig. 117. B = 4 inches, H = 6 inches. Ans. 104 inch 4 units. Ditto, ditto, ditto for Fig. 118. D = 6 inches. Ans. 127-2 inch 4 units. Ditto, ditto, ditto for Fig. 119. D e = 6 inches, Di = 4 inches. Ans. 102 inch 4 units. 26. Find the second polar moment (l p ) of such a bar as that shown in Fig. 1 20. B = * inches, H = 2 inches, L = 16 inches. Ans. 2 1 20 inch 5 units. 27. Find the second polar moment (I p ) for a cylinder as shown in Fig. 121. D = 16 inches, H = 2 inches. Ans 12,870 inch 5 units. Examples. 693 28. Find the second polar moment (I p ) for a hollow cylinder as shown in Fig. 122. R e = 8 inches, R,- = 5 inches, H = 2 inches. Ans. 10,903 inch 5 units. 29. Find the second polar moment (I p ) of a disc flywheel about its axis. External radius of rim = 8 inches Internal radius of rim = 6 Internal radius of web = i'5 Thickness of web =07 Internal radius of boss = 075 Thickness of boss = 3 Width of rim = 3 Ans. 14,640 inch* units. 30. Find the second polar moment (I p ) of a sphere 6 inches diameter about its diameter. Ans, 407 inch 5 units. Find the second polar moment (I 0/) ) about a line situated 12 inches from the centre of the sphere. Ans. 16,690 inqh 5 units. Find the second polar moment (I p ) of a cone about its axis. H = 12 inches, R = 2 inches. Ans. 60-3 inch 5 units. CHAPTER IV. 1. Set off on a piece of drawing-paper six lines representing forces acting on a point in various directions. Find the resultant in direction and magnitude by the two methods shown in Fig. 128. 2. Set off on a piece of drawing-paper six lines as shown in Fig. 130. Find the magnitude of the forces required to keep it in equilibrium in the position shown. 3. In the case of a suspension bridge loaded with eight equal loads of looo Ibs. each placed at equal distances apart. Find the forces acting on each link by means of the method shown in Fig. 131. 4. (I.C.E., October, 1897.) A pair of shear legs make an angle of 20 with one another, and their plane is inclined at 60 to the horizontal. The back stay is inclined at 30 to the plane of the legs. Find the force on each leg, and on the stay per ton of load carried. Ans. Each leg, o'88 ; stay, ro ton. 5. A telegraph wire T \j inch diameter is supported on poles 170 feet apart and dips 2 feet in the middle. Find the pull on the wire. Ans. 47-4 Ibs. 6. A crane has a vertical post DC as in Fig. 135, 8 feet high. The tie AC is 10 feet long, and the jib AB 14 feet long. Find the forces acting along the jib and tie when loaded with 5 tons simply suspended from the extremity of the jib. Anf. Tie, 6*3 tons ; jib, 8 '8 tons. 7. In the case of the crane in Question 6, find the radius, viz. /, by means of a diagram. If the distance x be 4 feet, find the pressures /, and/ a . Ans. fr =/ 2 = I2'2 tons. 8. Again referring to Question 6. Taking the weight of the tie as 50 Ibs., and that of the jib as 350 Ibs., and the pulley, etc., at the end of the jib as 50 Ibs. find the forces acting on the jib and tie. Ans. Jib, 8*99 ; tie, 6-39 tons. 694 Examples. 9. In the case of the crane in Question 6, find the forces when the crane chain passes down to a barrel as shown in Fig. 137. Let the sloping chain bisect the angle between the tie and the jib. Find W, if /j =5 feet ; also find the force acting along the back stay. Ans. Jib, ii'4 ; tie = 3*6 ; "W^ = 4-9 ; back stay, 5*8 tons. 10. Each leg of a pair of shear legs is 50 feet long. They are spread out 20 feet at the foot. The back stay is 75 feet long. Find the forces acting on each member when lifting a load of 20 tons at a distance of 20 feet from the foot of the shear legs, neglecting the weight of the structure. Ans. Legs, 16*7 ; back stay, 16*5 tons. 11. In the last question, find the horizontal pull on the screw and the total upward pull on the guides, also the force tending to thrust the feet of the shear legs apart. Ans. Horizontal pull, 167 ; upward pull, 875; thrust, 4*1 tons. 12. The three legs of a tripod AC, BO, CO, are respectively 28, 29, and 27 feet long. The horizontal distances apart of the feet are AB, 15 feet ; BC, 17 feet ; CA, 13 feet. A load of 12 tons is supported from the apex, find the thrust on each leg. Ans. A = 2*94 tons ; B = 2*22 tons ; C = 7*35 tons. 13. A simple triangular truss of 30 feet span and 5 feet deep supports a load of 4 tons at the apex. Find the force acting on each member. Ans. 6 tons on the tie ; 6*32 tons on the rafters. 14. (I.C.E., October, 1897.) Give a reciprocal diagram of the stress in the bars of such a roof as that shown in Fig. 139, loaded with 2 tons at each joint of the rafters ; span, 40 feet j total height, 10 feet ; depth of truss in the middle, 8 feet 7 inches. 15. The platform of a suspension foot bridge 100 feet span is 10 feet wide, and supports a load of 150 Ibs. per square foot, including its own weight. The two suspension chains have a dip of 20 feet. Find the force acting on each chain close to the tower and in the middle, assuming the chain to hang in a parabolic curve. Ans. 60,000 Ibs. close to tower ; 47,000 Ibs. in middle. CHAPTER V. 1. Construct the centrodes of O&.d and O a . c for the mechanism shown in Fig. 148, when the link d is fixed. 2. In Fig. 149, if the link b be fixed, the mechanism is that of an oscillating cylinder engine, the link c representing the cylinder, a the crank, and d the piston-rod. Construct a diagram to show the relative angular velocity of the piston and the rod for one stroke when the crank rotates uniformly. 3. Construct a curve to show the velocity of the cross-head at all parts of the stroke for a uniformly revolving crank of I foot radius length of con- necting rod = 3 feet and from this curve construct an acceleration curve, scale 2 inches = I foot. State the scale of the acceleration curve when the crank makes 120 revolutions per minute. 4. The bars in a four-bar mechanism are of the following lengths : a i'2, b 2, c i '9, d i '4. Find the angular velocity of c when normal to d* having given the angular velocity of a as 2*3 radians per second. 5. In Fig. 153, find the weight that must be suspended from the point 6 in order to keep the mechanism in equilibrium, when a weight of 30 Ibs. Examples. 695 is suspended from the point 5, neglecting the friction and the weight of the mechanism itself. 6. In Fig. 158, take the length of a = 2, b = 10, c 4, d = 1075. The interior angle at 3 ~ 156. Find the angular velocity of c when that of a is 5. Ans. 2*19. 7. Find the angular velocity of the connecting rod in the case of an engine. Radius of crank = o'5 foot j length of connecting rod = 6 cranks. Revolutions per min. 200 ; crank angle, 45. Ans. 2*5 radians per second. 8. In Fig. i6ia, the length of a = 1*5, b 5, find the angular velocity of c when the crank is in its lowest position, and when making 100 revolu- tions per minute. Ans. 0*45 radians per second. 9. Construct velocity and acceleration curves, such as those shown in Fig. 163, for the mechanism given in Question 4. 10. In a Stephenson's link motion, the throw of the eccentrics is 3^ inches. The angle of advance 16, i.e. 106 from the crank. The length of the eccentric rods is 57f inches. The rods are attached to the link in the manner shown in Fig. 164*:. The distance ST = 17! inches. The suspension link is attached at T, and is 25^ inches long. The point U is lOg inches above the centre line of the engine, and is 5 feet from the centre of the crank shaft. Find the velocity of the top and bottom pins of the link when the engine makes IOO revolutions per minute, and the link is in its lowest position, and when the crank has turned through 30 from its inner dead centre. Ans. Top pin, 2*8 feet per second ; bottom pin, 4*4 feet per second. 11. In a Humpage gear, the numbers of teeth in the wheels are : A = 42, B = 28, BI = 15, C = 24, E = 31. Find the number of revo- lutions made by C for one revolution of the shaft. Ans. 10*03. 12. Construct (i.) an epicycloidal, (ii.) a hypocycloidal, tooth for a spur wheel ; width of tooth on pitch line, I inch ; depth below pitch line, } inch ; do. above, ? inch ; diameter of pitch circle, 18 inches ; do. of rolling circle, 6 inches. Also construct an involute and a cycloidal tooth for a straight rack. CHAPTER VI. 1. Find the acceleration pressure at each end of the stroke of a vertical inverted high-speed steam-engine when running at 5 revolutions per minute ; stroke, 9 inches ; weight of reciprocating parts, no Ibs. ; diameter of cylinder, 8 inches ; length of connecting-rod, 1*5 foot. Ans. 54*8 Ibs. sq. inch at bottom ; 85*5 Ibs. sq. inch at top. 2. Find the acceleration pressure at the end of the stroke of a pump having a slotted cross-head as shown in Fig. 160; speed, 100 revolutions per minute ; stroke, 8 inches ; diameter of cylinder, 5 inches ; weight of reciprocating parts, 95 Ibs. Ans. 5*49 Ibs. sq. inch. 3. To what pressure should compression be carried in the case of a horizontal engine running at 60 revolutions per minute ; stroke, 4 feet ; weight of reciprocating parts per square inch of piston, 3*2 Ibs. ; length of connecting-rod, 9 feet. Ans. 9*57 Ibs. sq. inch at " in " end ; 6*09 Ibs. sq. inch at " out" end. 4. In a horizontal engine, such as that shown in Fig. 177, the weights of the parts were as follows : 696 Examples. Piston 54lbs. Piston and tail rods ... 40 Both cross-heads loo Small end of connecting-rod 18 Plain part of ,, 24 Air-pump plunger 18 A 3 ^et / 2 i foot w>! 20 Ibs. w z 7 Diameter of cylinder 8 inches Revolutions per minute 140 Length of connecting-rod 40 inches Length of stroke 18 ,, Find the acceleration pressure at each end of the stroke. Ans. 27-8, 17*6 Ibs. sq. inch. 5. (Victoria, 1902.) In an inverted vertical engine the radius of the crank is I foot ; the length of the connecting-rod, 4 feet ; diameter of the piston, 1 6 inches ; revolutions per minute, 200 ; weight of the reciprocating parts, 500 Ibs. Find the twisting moment on the crank shaft when the piston is descending and the crank has turned through 30 from the top centre. The effective pressure on the piston is 50 Ibs. sq. inch. Ans. 2400 Ibs. -feet. 6. (Victoria, 1903.) A weight of 50 Ibs. is attached to a long projecting pin on the ram of a slotting machine. Find the downward force acting on the pin when the ram is at the top and the bottom of its downward stroke. The speed of the machine is 60 revolutions per minute, the stroke is 12 inches, and the connecting-rod which is below the crank-pin is 24 inches long. Ans. bottom, 87-5 Ibs. ; top, 27-5 Ibs. 7. In an engine of the Willans type the annular air cushion cylinder is 12 inches and 9 inches diameter, the weight of the reciprocating parts is 1000 Ibs. The radius of the crank is 3 inches, the length of the connecting- rod is 12 inches. Revolutions per minute 450, the air pressure at the bottom of the stroke is atmospheric. Find the required clearance at the top of the stroke, assuming (i.) hyperbolic, (ii.) adiabatic compression of the air. Ans. (i.) o'2O inch, (ii.) 0^58 inch. 8. Take the indicator diagrams from a compound steam-engine (if the reader does not happen to possess any, he can frequently find some in the Engineering papers), and construct diagrams similar to those shown in Figs. 179, 181, 1820, and 183. Also find the value of m, p. 176, and the weight of flywheel required, taking the diameter of the wheel as five times the stroke. 9. Taking the average piston speed of an engine as 400 feet per minute, and the value of m as \ ; the fluctuation of velocity as I / on either side of the mean, namely K = O'O2 ; the radius of the flywheel as twice the stroke ; show that the weight of the rim may be taken as 50 Ibs. per I.H.P. for a single-cylinder engine, running at 100 revolutions per minute. 10. A two-cylinder engine with cranks at right angles indicates 120 H.P. at 40 revolutions per minute. The fluctuation of energy per stroke is 12 / (i.e. m = 0*12) ; the percentage fluctuation of velocity is 2 / on either side of the mean (i.e. K = 0*04) ; the diameter of the flywheel is 10 feet. Find the -weight of the rim. Ans. 4*87 tons. 11. Taking the average value of the piston speed of a gas-engine as Examples. 697 600 feet per minute, and the value of m as 8*5, the fluctuation of velocity as 2 % (i.e. K = 0*04) ; on either side of the mean, the radius of the fly- wheel as twice the stroke, show that the weight of the rim may be taken at about 300 Ibs. per I.H.P. for a single cylinder engine running at 200 revolutions per minute. 12. Find the weight of rim required for the flywheel of a punching machine, intended to punch holes I J-inch diameter through ij-inch plate, speed of rim 30 feet per second. Ans. 1180 Ibs. 13. (I.C.E., October, 1897.) A flywheel supported on a horizontal axle 2 inches in diameter is pulled round by a cord wound round the axle carrying a weight. It is found that a weight of 4 Ibs. is just sufficient to overcome the friction. A further weight of 16 Ibs., making 20 in all, is applied, and after two seconds starting from rest it is found that the weight has gone down 12 feet. Find the moment of inertia of the wheel. Ans. 0*014 mass feet 2 units. (N.B. For the purposes of this question, you may assume that the speed of the wheel when the weight is released is twice the mean.) 14. (I.C.E., October, 1897.) In a gas-engine, using the Otto cycle, the I.H.P. is 8, and the speed is 264 revolutions per minute. Treating each fourth single stroke as effective and the resistance as uniform, find how many foot-lbs. of energy must be stored in the flywheel in order that the speed shall not vary by more than one-fortieth above or below its mean value. Ans. 30,000 foot-lbs. 15. Find the stress due to centrifugal force in the rim of a cast-iron wheel 8 feet diameter, running at 160 revolutions per minute. Ans. 431 Ibs. sq. inch. 16. (S. & A., 1896.) A flywheel weighing 5 tons has a mean radius of gyration of 10 feet. The wheel is carried on a shaft of 12 inches diameter, and is running at 65 revolutions per minute. How many revolu- tions will the wheel make before stopping, if the coefficient of friction of the shaft in its bearing is O'o65 ? (Other resistances may be neglected.) Ans. 352. 17. Find the bending stress in the middle section of a coupling-rod of rectangular section when running thus : Radius of coupling-crank, 1 1 inches ; length of coupling-rod, 8 feet ; depth of coupling-rod, 4*5 inches ; width, 2 inches ; revolutions per minute, 200. Ans. 2*4 tons per sq. inch. 18. Find the bending stress in the rod given in the last question if the sides were fluted to make an I section ; the depth of fluting, 3 inches ; thickness of web, I inch. Ans. 1*87 ton sq. inch. 19. Assuming that one-half of the force exerted by the steam on one piston of a locomotive is transmitted through a coupling-rod, find the maximum stress in the rod mentioned in the two foregoing questions. Diameter of cylinder, 16 inches ; steam-pressure, 140 Ibs. per sq. inch. Ans. 3-09 tons sq. inch for rectangular rod; 2*92 tons sq. inch for fluted rod. 20. If the rod in Question 17 had been tapered off to each end instead of being parallel in side elevation, find the bending stress at the middle section of the rod. The rod has a straight taper of \ inch per foot. Ans. 2-22 tons sq. inch. 21. Find the skin stress due to bending in a connecting-rod when running thus : Radius of crank, 10 inches ; length of rod, 4 feet ; diameter of rod, 3 inches ; number of revolutions per minute, 120. Ans. 450 Ibs. sq. inch. 698 Examples. 22. A railway carriage wheel is found to be out of balance to the extent of 3 Ibs. at a radius of 18 inches. What will be the amount of "hammer blow" on the rails when running at 60 miles per hour? Diameter of wheels, 3 feet 6 inches. Ans. 354 Ibs. 23. In Mr. Hill's paper (I.C.E., vol. civ. p. 265) on locomotive balancing, the following problem is set : The radius of the crank =12 inches ; radius of balance- weight, 33 inches ; weight of the reciprocating parts, 500 Ibs. ; weight of the rotating parts, 680 Ibs. ; distance from centre to centre of cylinders = 24 inches ; ditto of balance- weights, 60 inches. Find the weight of the balance- weight if both the reciprocating and rotating weights are fully balanced. Ans. 325 Ibs. 24. Find the weight and position of the balance -weights of an inside cylinder locomotive working under the following conditions : \Veig it of connecting rod, big end ... ... 150 Ibs. ,, small end ... 70 ,, ,, plain part ... 180 cross-head and slide blocks ... 170 piston and rod ... ... ... 200 crank- web and pin... ... ... 330 Rad us of c. of g. of crank- web and pin ... 10 inches ,, crank (K) 12 ,, balance-weight (R 6 ) ... 30 Cylinder centres (C) 24 ,, Wheel centres (y) 72 Find the requisite balance-weight and its position for balancing the rotating and two-thirds of the reciprocating parts. Ans. 259 Ibs.; 9 = 26J. 25. An English railway company, instead of taking two-thirds of the reciprocating, and the whole of the rotating parts, take as a convenient compromise the whole of the reciprocating parts. Find the amount of each balance-weight for such a condition, taking the values given in Question 23. Ans. 138 Ibs. 26. Find the amount of balance -weight required for the conditions given in Question 24 for an uncoupled outside-cylinder engine. Weight of coupling crank and pin, So Ibs. ; radius of c. of g. of ditto, H inches; R c = 12 inches. Ans. 266 Ibs. 27. Find the amount and position of the balance-weight required for a six-wheel coupled inside-cylinder engine, where w from a to b and c to d 140 Ibs., and w from b to c 300 Ibs. ; weight of coupling-crank and pin, 80 Ibs. ; radius of c. of g. of ditto, n inches ; R c = 12 inches. The other weights may be taken from Question 2.1. Ans. 85 Ibs. on trailing and leading wheels ; 140 Ibs. on driving wheel ; 6 = 55. 28. Find the balance-weights for such an engine as that shown in Fig. 196. The weight of the coupling-rod = 250 ibs. For other details see Questions 24 and 27. Ans. 79 Ibs. on trailing wheel ; 270 Ibs. on driving wheel. 29. Find the speed at which a simple Watt governor runs when the arm makes an ani;le of 30 with the vertical. Length of arm from centre of pin to centre of ball = 18 inches. Ans. 47^5 revs, per minute. Examples. 699 30. (Victoria, 1896.) A loaded Porter's governor geared to an engine with a velocity ratio 5 has rods and links I foot long, balls weighing 2 Ibs. each, and a load of 100 Ibs. ; the valve is full open when the arms are at 30 to the vertical, and shut when at 45. Find the extreme working speeds of the engine. Ans. 92 and 83-2 revs, per minute. 31. (Victoria, 1898.) The balls of a Porter governor weigh 4 Ibs. each, and the load on it is 44 Ibs. The arms of the links are so arranged that the load is raised twice the distance that the balls rise for any given increase of speed. Calculate the height of this governor for a speed of 180 revolutions per minute. Calculate also the force required to hold the sleeve for an increase of speed of 3%. Ans. Height, 13*05 inches ; force, 2*92 Ibs. 32. Find the amount the sleeve rises in the case of a simple Watt governor when the speed is increased from 40 to 41 revolutions per minute. The sleeve rises twice as fast as the balls. Find the weight of each ball required to overcome a resistance on the sleeve of I Ib. so that the increase in speed shall not exceed the above-mentioned amount. Ans. 2*14 inches ; 20 Ibs. 33. Find the speed at which a crossed -arm governor runs when the arms make an angle of 30 with the vertical. The length of the arms from centre of pin to centre of ball = 29 inches ; the points of suspension are 7 inches apart. Ans. 43 revs, per minute. 34. In a Wilson-Hartnell governor (Fig. 209) r and r = 6 inches when the lower arm is horizontal. K = 12,000, W = 2 Ibs., n = I. Find the speed at which the governor will float. A us. 133 revs, per minute. 35. In a McLaren governor (Fig. 210), the weight W weighs 60 Ibs., the radius of W about the centre of the shaft is 8 inches, the load on the spring s = looo Ibs., the radius of the c. of g. of W about the point J is 9 inches, and the radius at which the spring is attached is 8 inches. Find the speed at which the governor will begin to act. Ans. 256 revs, per minute. 36. A simple Watt governor 1 1 inches high lags behind to the extent of 10% of its speed when just about to lift. The weight of each ball is 15 Ibs. The sleeve moves up twice as fast as the balls. Calculate the equivalent frictional resistance on the sleeve. Ans. 3*2 Ibs. 37. In the governor mentioned in Question 30, find the total amount of energy stored. For what size of engine would such a governor be suitable if controlled by a throttle valve ? Ans. 32-44 foot-lbs. About 35 I.H.P. 38. Find the energy stored in the McLaren governor mentioned in Question 35. There are two sets of weights and springs ; the weight moves out 3 inches, and the final tension on the spring is 1380 Ibs. Ans. 529 foot-lbs. CHAPTER VII. 1. (S. and A., 1896.) A weight of 5 cwt., resting on a horizontal plane, requires a horizontal force of 100 Ibs. to move it against friction. What is the coefficient of friction ? Ans. 0-179. 2. (S. and A., 1891.) The saddle of a lathe weighs 5 cwt. ; it is moved along the bed of the lathe by a rack-and-pinion arrangement. What force, applied at the end of a handle 10 inches in length, will be 7OO Examples. just capable ol moving the saddle, supposing the pinion to have twelve teeth of ijlinch pitch, and the coefficient of friction between the saddle and lathe bed to be o'l, other friction being neglected ? Ans. 13-37 Ibs. 3. A block rests on a plane which is tilted till the block commences to slide. The inclination is found to be 8-4 inches at starting, and after- wards 6 -3 inches on a horizontal length of 2 feet. Find the coefficient of friction when the block starts to slide, and after it has started. AM. 0*35 ; 0-26. 4. In the case of the block in the last question, what would be the acceleration if the plane were kept in the first-mentioned position ? Ans. 27 feet per second per second. 5. A -block of wood weighing 300 Ibs. is dragged over a horizontal metal plate. The frictional resistance is 126 Ibs. What would be the probable frictional resistance if it be dragged along when a weight of 600 Ibs. is placed on the wooden block, (i.) on the assumption that the value of n remains constant ; (ii.) that it decreases with the intensity of pressure as given on p. 230. Ans. (i.) 378 Ibs. ; (ii.) 308 Ibs. 6. In an experiment, the coefficient of friction of metal on metal was found to be 0*2 at 3 feet per second. Wnat will it probably be at 10 feet per second ? Find the value of K as given on p. 230. Ans. o* 1 1 ; K = 0-83. 7. (S. and A., 1897.) A locomotive with three pairs of wheels coupled weighs 35 tons ; the coefficient of friction between wheels and rails is O'i8. Find the greatest pull which the engine can exert in pulling itself and a train. What is the total weight of itself and train which it can draw up an incline of I in 100, if the resistance to motion is 12 Ibs. per ton on the level? Ans, 6-3 tons ; 410 tons. 8. (Victoria, 1896.) Taking the coefficient of friction to be /t, find the angle 6 at which the normal to a plane must be inclined to the vertical so that the work done by a horizontal force in sliding a weight w up the plane to a height h may be 2wh. Ans. 10 35*. 9. A body weighing 1000 Ibs. is pulled along a horizontal plane, the coefficient of friction being o - 3 ; the line of action being (i.) horizontal ; (ii.) at 45 ; (iii.) such as to give the least pull. Find the magnitude of the pull, and the normal pressure between the surfaces. Ans. Pull, (i.) 300 Ibs., (ii.) 326 Ibs., (iii.) 287 Ibs. Normal pressure, (i.) loco Ibs., (ii.) 770 Ibs., (iii.) 917 Ibs. 10. A horse drags a load weighing 35 cwt. up an incline of I in 20. The resistance on the level is 100 Ibs. per ton. Find the pull on the traces when they are (i.) horizontal, (ii.) parallel with the incline, (iii.) in the position of least pull. Ans. (i.) 3717 Ibs., (ii.) 370*67 Ibs., (iii.) 370-04. 11. A cotter, or wedge, having a taper of I in 8, is driven into a cottered joint with an estimated pressure of 600 Ibs. Taking the co- efficient of friction between the surfaces as o'2, find the force with which the two parts of the joint are drawn together, and the force required to withdraw the wedge. Ans. 1128 Ibs. ; 307 Ibs. 12. Find the mechanical efficiency of a screw-jack in which the load rotates with the head of the jack in order to eliminate collar friction. Threads per inch, 3 ; mean diameter of threads, if inch ; coefficient of friction, 0*14. Also fiud the efficiency when the load does not rotate. Ans. 30 per cent. ; 17*1 per cent. Examples. 701 13. (Victoria, 1897.) Find the turning moment necessary to raise a weight of W Ibs. by a vertical square-threaded screw having a pitch of 6 inches, the mean diameter of the thread being 4 inches, and the coefficient of friction \. Ans. 1*93 W Ibs. -inches. 14. (Victoria, 1898.) Find the tension in the shank of a bolt in terms of the twisting moment T on the shank when screwing up, (i.) when the thread is square, (ii.) when the angle of the thread is 55 ; neglecting the friction between the head of the bolt and the washer ; taking r the outside radius of the thread, a = the depth of thread, n = the number of threads to an inch,/= the coefficient of friction. . . 2T 2irT Ans. (i.) tan In the answers given it was assumed as a close approximation that (a + ) = tan a + tan . An exact solution for (i.) gives The approximate method in a given instance gave 1675 ^ s -> an d the exact method 1670 Ibs. The simpler approximate solution is therefore quite accurate enough in practice. 15. The mean diameter of the threads of a J-inch bolt is 0*45 inch, the slope of the thread 0*07, and the coefficient of friction O'i6. Find the tension on the bolt when pulled up by a force of 20 Ibs. on the end of a spanner 12 inches long. Ans. 1920 Ibs. 16. The rolling resistance of a waggon is found to be 73 Ibs. per ton on the level ; the wheels are 4 feet 6 inches diameter. Find the value of K. Ans. o'88. 17. A 4-inch axle makes 400 revolutions per minute on anti-friction wheels 30 inches diameter, which are mounted on 3-inch axles. The load on the axle is 5 tons ; p = o'l ; = 30 j K = O'OI. Find the horse- power absorbed. Ans. 1*7. 1 8. A horizontal axle 10 inches diameter has a vertical load upon it of 20 tons, and a horizontal pull of 4 tons. The coefficient of friction is o'O2. Find the heat generated per minute, and the horse-power wasted in friction, when making 50 revolutions per minute. Ans. 155 thermal units j 3-63 H.P. 19. Calculate the length required for the two necks in the case of the axle given in the last question, if placed in a tolerably cool place (t 0*3). Ans. 26 inches. 20. Calculate the horse-power of the bearing mentioned in Question 18 by the rough method given on p. 261, taking the resistance as 2 Ibs. per square inch. Ans. 4-13 H.P. 21. Calculate the horse-power absorbed by a footstep bearing 8 inches diameter when supporting a load of 4000 Ibs., and making 100 revolutions 7O2 Examples. per minute. /* = 0*03, with (i.) a flat end ; (H.) a conical pivot ; a = 30 ; (iii.) a Schiele pivot. / = R,, R, = ^1. Ans. (i.) 0-51 ; (ii.) i'O2 ; (iii.) 076. 22. The efficiency of a single-rope pulley is found to be 94%. Over how many of such pulleys must the rope pass in order to make it self- sustaining, i.e. to have an efficiency of under 50%. Ans. 12 pulleys. 23. In a three-sheaved pulley-block, the pull W on the rope was no Ibs., and the weight lifted, W, was 369 Ibs. What was the mechanical efficiency ? and if the friction were 80 % of its former value when reversed, what would be the reversed efficiency, and what resistance would have to be applied to the rope in order to allow the weight to gently lower ? Ans. 55*9 per cent. ; 36^9 per cent., 227 Ibs. (N.B. The probable friction when reversed may be roughly arrived at thus: The total load on the blocks was 369 + no = 479 Ibs., when raising the load. Then, calculating the resistance when lowering, assuming the friction to be the same, we get 369 + 227 = 3917 Ibs. Assuming the friction to vary as the load, we get the friction when lowering to be j$ = 0*82. This is only a rough approximation, but experiments on pulleys show that it holds fairly well. In the question above, the experiment gave 23*4 Ibs. resistance against 227 Ibs., and the efficiency as 38 % against 36*9 %. Many other experiments agree equally well.) 24. (S. and A., 1896.) A lifting tackle is formed of two blocks, each weighing 15 Ibs. ; the lower block is a single movable pulley, and the upper or fixed block has two sheaves. The cords are vertical, and the fast end is attached to the movable block. Sketch the arrangement, and determine what pull on the cord will support 200 Ibs. hung from the movable block, and also what will then be the pressure on the point of support of the upper block. Ans. 71$ Ibs. pull ; 301 Ibs. on support. 2$. (S. and A., 1896.) If in a Weston pulley block only 40 % of the energy expended is utilized in lifting the load, what would require to be the diameter of the smaller part of the compound pulley when the largest diameter is 8 inches in order that a pull of 50 Ibs. on the chain may raise a load of 550 Ibs. ? Ans. 7^42 inches. 26. Find the horse-power that may be transmitted through a conical friction clutch, the slope of the cone being i in 6, and the mean diameter of the bearing surfaces 18 inches. The two portions are pressed together with a force of 170 Ibs. The coefficient of friction between the surfaces is 0*15. Revolutions per minute, 200. Ans. 4*4. 27. An engine is required to drive an overhead travelling crane for lifting a load of 30 tons at 4 feet per minute. The power is transmitted by means of 2j-inch shafting, making 160 revolutions per minute. (For the purposes of this question use the expression at foot of p. 274.) The length of the shafting is 250 feet ; the power is transmitted from the shaft through two pairs of bevel wheels (efficiency 90 % each including bearings) and one worm and wheel having an efficiency of 85 % including its bearings. Taking the mechanical efficiency of the steam-engine at 80 %, calculate the required I.H.P. of the engine. Ans. 22. 28. (I.C.E., October, 1897.) When a band is slipping over a pulley, show how the ratio of the tensions on the tight and slack sides depends on the friction and on the angle in contact. Apply your result to explain why a rope exerting a great pull may be readily held by giving it two or three turns round a post. Examples. 703 29. (S. and A., 1896. ) What is the greatest load that can be supported by a rope which passes round a drum 12 inches in diameter of a crab or winch which is fitted with a strap friction brake worked by a lever, to the long arm of which a pressure of 60 Ibs. is applied? The diameter of the brake pulley is 30 inches, and the brake-handle is 3 feet in length from its fulcrum ; one end of the brake strap is immovable, being attached to the pin forming the fulcrum of the brake-handle, while the other end of the strap is attached to the shorter arm, 3 inches in length, of the brake-lever. The angle a = - 7r . The gearing of the crab is as follows : On the shaft which carries the brake-wheel is a pinion of 15 teeth, and this gears into a wheel of 50 teeth on the second shaft ; a pinion of 20 teeth on this latter shaft gears into a wheel with 60 teeth carried upon the drum or barrel shaft, fj. = o'l. Ans. 4*83 tons. 30. (S. and A., 1896.) The table of a small planing-machine, which weighs i cwt., makes six double strokes of 4$ feet each per minute. The coefficient of friction between the sliding surfaces is '07. What is the work performed in foot-pounds per minute in moving the table ? Ans. 423-3. 31. (S. and A., 1897.) A belt laps 150 round a pulley of 3 feet diameter, making 130 revolutions per minute ; the coefficient of friction is 0*35. What is the maximum pull on the belt when 20 H.P. is being transmitted and the belt is just on the point of slipping? Ans. 898 Ibs. 32. (Victoria, 1897.) Find the width of belt necessary to transmit 10 H.P. to a pulley 12 inches in diameter, so that the greatest tension may not exceed 40 Ibs. per inch of the width when the pulley makes 1500 revolutions per minute, the weight of the belt per square foot being 1*5 Ibs., taking the coefficient of friction as 0*25. Ans. 8 inches. 33. A strap is hung over a fixed pulley, and is in contact over an arc of length equal to two-thirds of the total circumference. Under these circum- stances a pull of 475 Ibs. is found to be necessary in order to raise a load of 150 Ibs. Determine the coefficient of friction between the strap and the pulley-rim. ; Ans. 0*275. 34. Power is transmitted from a pulley 5 feet in diameter, running at no revolutions per minute, to a pulley 8 inches in diameter. Thickness of belt = 0*24 inch ; modulus of elasticity of belt, 9000 Ibs. per square inch ; tension on tight side per inch of width = 60 Ibs.; ratio of tensions, 2*3 to i. Find the revolutions per minute of the small pulley. Ans. 792. 35. How many ropes, 4 inches in circumference, are required to transmit 200 H.P. from .a pulley 16 feet in diameter making 90 revolutions per minute? Ans. 10. CHAPTER VIII. I. Plot a stress-strain and a real stress diagram for the following test : Scales elastic strains, 2000 times full size ; permanent strains, twice full size ; loads, 10,000 Ibs. to an inch. Calculate the stress at the elastic limit, the maximum stress ; the per- centage of extension on 10 inches ; the reduction in -area ; the work done in fracturing the bar. Compare the calculated work with that obtained from the diagram ; the modulus of elasticity (mean up to 28,000 Ibs.). Original dimensions: Length, 10 inches; width, 1753 inch; thickness, 704 Examples, o*6n inch. Final dimensions: Length, 12*9 inches; width, 1-472 inch; thickness, 0*482 inch. Loads in pounds ... I 4000 I 8000 I 12,000 I 16,000 I 20,000 Extension in inches | 0*0009 I 0*0021 | 0*0033 | 0*0045 I ' OO 57 24,000 I 28,000 I 32,000 I 34,000 0*0069 I 0*0082 I 0*0103 I 0*016 36,000 I 40,000 0*07 0*19 I 44,000 I 48,000 I 52,000 I 56,000 I 59,780 I 54,900 | 0-30 I 0*47 I 0*75 | 1-36 I 2*5 | 2-9 Ans. Elastic limit, 15 tons square inch. Maximum stress, 24*92. Extension, 29 per cent, on 10 inches. Reduction, 34 per cent. Work (by calculation), 6*27 inch- tons per cubic inch. Modulus of elasticity, 31,000,000 Ibs. per square inch ; 13,840 tons per square inch. 2. (I.C.E., October, 1897.) Distinguish between stress and strain. An iron bar 20 feet long and 2 inches in diameter is stretched 5 l g of an inch by a load of 7 tons applied along the axis. Find the intensity of stress on a cross-section, and the coefficient of elasticity of the material (E). Ans. Stress, 2*23 tons per sq. inch ; E = 10,700 tons per sq. inch. 3. (I.C.E., October, 1897.) A bar 4" X 2" in cross-section is subjected to a longitudinal tension of 40 tons. Find the normal and shearing stresses on a section inclined at 30 to the axis of the bar. Ans. Normal, 1*25 tons per square inch; tangential, 2*16 tons per square inch. 4. A bar of steel 4" X i" is rigidly attached at each end to a bar of brass 4" X 8" ; the combined bar is then subjected to a load of 20 tons. Find the load taken by each bar. E for steel = 13,000 tons per square inch ; brass, 4000 tons per square inch. Ans. Load on steel bars, 17*93 tons ; load on brass bar, 2*07 tons. 5. The nominal tensile stress (reckoned on the original area) of a bar of steel was 32*4 tons per square inch, the reduction in area at the point of fracture was 54 %. What would be the approximate tensile strength of hard drawn wire made from such steel ? Ans. 70 tons per square inch. 6. Plot a stress-strain and a real stress diagram for the following com- pression test of a specimen of copper. Scale of loads, 5 tons per inch ; strain twice full size. Loads in tons I 13 I 14 I 15 I 16 I 18 I 20 Length of specimen in inches | 2*50 | 2*47 | 2*29 | 2*19 | 2*02 | 2*86 22 I 24 I 26 I 28 I 30 I 32 1-73 I r6i I 1*52 | 1*43 I 1*35 | 1-30 34 I 36 I 38 1-23 I 1*17 | i*ii 40 42 44 1*08 | 1*02 I 0*99 46 I 48 50 0*96 I 0*92 Original length, 2*52 inches ; diameter, 2*968 inches. 7. (S. and A., 1896.) A bar of iron is at the same time under a direct tensile stress of 5000 Ibs. per square inch, and to a shearing stress of 3500 Ibs. per square inch. What would be the resultant equivalent tensile stress in the material ? Ans. 6800 Ibs. per square inch. 8. (I.C.E., October, 1897.) Explain what is meant by Poisson's Ratio. A cube of unit length of side has two simple normal stresses p^ and p t on Examples. 705 pairs of opposite faces. Find the length of the sides of the cube when de- formed by the stresses (tensile). Ans . , + 9. Find the pitch of the rivets for a double row lap joint. Plates \ inch thick ; rivets, I inch diameter ; clearance, T ' g inch ; thickness of ring damaged by punching, T ' s inch ; ft = 23 tons per square inch ; fr = 25 tons per square inch. Ans. 4/27 inches. 10. Calculate the bearing pressure on the rivets when the above- mentioned joint is just about to fracture. Ans. 33*3 tons per square inch. 11. Calculate the pitch of rivets for a double cover plate riveted joint with diamond riveting as in Fig. 321, the one cover plate being wide enough to take three rows of rivets, l.ut the other only two rows ; thickness of plates, I inch ; drilled holes ; material steel ; the pitch of the outer row of the narrow cover plate must not exceed six times the diameter of the rivet. What is the efficiency of the joint and the bearing pressure ? Ans. 10*84 inches outer row, 5-42 inches inner row ; 887 per cent., 50*5 tons per square inch. 12. Two lengths of a flat tie-bar are connected by a lap riveted joint. The load to be transmitted is 50 tons. Taking the tensile stress in the plates at 5 tons per square inch, the shear stress in the rivets at 4 tons per square inch, and the thickness of the plates as f inch ; find the diameter and the number of rivets required, also the necessary width of bar for both the types of joint as shown in Figs. 329 and 330. What is the efficiency of each, and the working bearing pressure ? Ans. O'94 inch. 18 are sufficient, but 20 must be used for con- venience ; 17 inches and 14 J inches ; 78 and 93^4 per cent. ; 3 '6 tons per square inch. 13. Find the thickness of plates required for a boiler shell to work at a pressure of 160 Ibs. per square inch : diameter of shell, 8 feet ; efficiency of riveted joint, 89% ; stress in plates, 5 tons per square inch. Ans. 077 inch, or say | inch. 14. Find the maximum and minimum stress in the walls of a thick cylinder ; internal diameter, 8 inches ; external diameter, 14 inches ; internal fluid pressure, 2000 Ibs. per square inch. (Barlow's Theory.) Ans. 4670 Ibs. per square inch ; 1520 Ibs. per square inch. 15. A thick cylinder is built up in such a manner that the initial tensile stress on the outer skin and the compressive stress on the inner skin are both 3 tons per square inch. Calculate the resultant stress on both the outer and the inner skin when under pressure. Internal fluid pressure, 4'5 tons per square inch. (Barlow's Theory. ) Internal diameter, 6 inches; external diameter, 15 inches. Ans. 4*2 and 4*5 tons per square inch. CHAPTER IX. 1. (Victoria, 1897.) Find the greatest stress which occurs in the section of a beam resting on two supports, the beam being of rectangular section, 12 inches deep, 6 inches wide, carrying a uniform load of 5 cwt. per foot run, span 35 feet. Ans. 3*12 tons square inch. 2. Rolled joists are used to support a floor which is loaded with 150 Ibs. 2 Z 706 Examples. per square foot including its own weight. The pitch of the joists is 3 feet ; span, 20 feet ; skin stress, 5 tons per square inch. Find the required Z and a suitable section for the joists, taking the depth at not less than 5 ' 5 of the span. Ans. Z = 24' I ; say 10" X 5" X ". 3. A rectangular beam, 9 inches deep, 3 inches wide, supports a load of ton, concentrated at the middle of an 8-foot span. Find the maximum skin stress. Ans. 0-3 ton square inch. 4. A beam of circular section is loaded with an evenly distributed load of 200 Ibs. per foot run ; span, 14 feet j skin stress, 5 tons per square inch. Find the diameter. Ans. 377 inches. 5. Calculate approximately the safe central load for a simple web rivete girder, 6 feet deep; flanges, 18 inches wide, 2j inches thick. The flange is attached to the web by two 4" X 5" angles. Neglecting the strength of the web, and assuming that the section of each flange is reduced by two rivet-holes inch diameter passing through the flange and angles. Span of girder, 50 feet; stress in flanges, 5 tons per square inch. Ans. no tons. 6. Find the relative weights of beams of equal strength having the following sections : Rectangular, h $b ; Square ; circular ; rolled joist, /*, = 2, = I2/. Ans. Rectangular, roo; square, 1*44; circular, i'6i ; joist, o'54. 7. Find the safe distributed load for a cast-iron beam of the following dimensions: Top flange, 3" X i" ; bottom flange, "8" X 1-5"; web, 1-25 inch thick ; total depth, 10 inches ; with (i.) the bottom flange in tension,- (ii.) when inverted ; span, 12 feet ; skin stress, 3000 Ibs. per square inch. Ans. (i.) 57 ; (ii.) 3*2 tons. 8. In the last question, find the safe central load for a stress of.3ooo Ibs. per square inch, including the stress due to the weight of the beam. The beam is of constant cross-section. Ans. (i.) 2*65 ; (ii.) 1*4 ton. 9. (S. and A., 1896.) If a bar of cast iron, I inch square and I inch long, when secured at one end, breaks transversely with a load of 6000 Ibs. suspended at the free end, what would be the safe working pressure, em- ploying a factor of safety of 10, between the two teeth which are in contact in a pair of spur-wheels whose width of tooth is 6 inches, the depth of the tooth, measured from the point to the root, being 2 inches, and the thick- ness at the root of the tooth 1 4 inch ? (Assume that one tooth takes the whole load.) Ans. 4050 Ibs. 10. (S. and A., 1897.) Compare the resistance to bending of a wrought- iron I section when the beam is placed like this, I, and like this, -. The flanges of the beam are each 6 inches wide and I inch thick, and the web is \ inch thick and measures 8 inches between the flanges. Ans. 4*56 to I. 11. A trough section, such as that shown in Fig. 350, is used for the flooring of a bridge ; each section has to support a uniformly distributed load of 150 Ibs. per square foot, and a concentrated central load of 4 tons. Find the span for which such a section may be safely used. Skin stress = 5 tons per square inch ; pitch of corrugation, 2 feet ; depth, I foot ; width of flange (B, Fig. 350) = 8 inches ; thickness = | inch. Ans. 19 feet. 12. A 4" X 4" X " J_ section is used for a roof purlin, the load being applied on the flanges ; the span is 12 feet ; the evenly distributed load is Examples. 707 roo Ibs. per foot run. Find the skin stress at the top and the bottom of the section. Ans. Top, 4'9 tons square inch compression; bottom, 2 - i tons square inch tension. 13. A triangular knife-edge of a weighing-machine overhangs i inch, and supports a load of 2 tons (assume evenly distributed). Taking the triangle to be equilateral, find the requisite size for a tensile stress it the apex of 10 tons per square inch. Ans. S = 1-69. inch. 14. A cast-iron water main, 30 inches inside diameter and 32 inches outside, is unsupported for a length of 12 feet. Find the stress in the metal due to bending. Ans. 180 Ibs. square inch. 15. In the case of a tram-rail, the area A of one part of the modulus figure is 4' 1 2 square inches, and the distance D between the two centres of gravity is 5^55 inches ; the neutral axis is situated at a distance of 3' I inches from the skin of the bottom flange. Find the I and Z. Ans. 70*9 ; 22'86. 1 6. Find the Z for the sections given on pp. 378, 379, and 380, which ere drawn to the following scales : Figs. 363, 364, and 365, 4 inches = I foot ; Fig. 366, 3 inches = I foot ; Fig. 367, I inch = I foot ; Fig. 368, full size ; Fig. 369, - full size ; Fig. 370, inch = I foot. (You may assume that the crosses correctly indicate the c. of g. of each figure.) The areas must be measured by a plani meter or by one of the methods given in Chapter II. Ans. 363, 4-86 ; 364, 6-03 ; 365, 17-3 ; 366, 8-78 ; 367, 36*3 ; 368, !5'5; 369. 1 1 '4; 370, 460. 17. When testing a 9" X 9" timber beam, the beam split along the grain by shearing along the neutral axis under a central load of 15 tons. Calculate the shear stress. Ans. 310 Ibs. square inch. 18. Calculate the ratio of the maximum shear stress to the mean in the case of a square beam loaded with one diagonal vertical. Ans. I ; i.e. they are equal. 19. Calculate the central deflection of a tram rail due to (i.) bending, (ii.) shear, when centrally loaded on a span of 3 feet 6 inches with a load of 10 tons. E 12,300 tons square inch. I = 8o'5 inch units ; A = 10*5 square inches. G = 4900 tons square inch ; K = 4'O3. Ans. (i.) O'oi6 inch; (ii.) O'ooS inch. CHAPTER X. 1. (Victoria, 1897.) The total load on the axle of a truck is 6 tons. The wheels are 6 feet apart, and the two axle-boxes 5 feet apart. Draw the curve of bending moment on the axle, and state what it is in the centre. Ans. 18 tons-inches. 2. (Victoria, 1896.) A beam 20 feet long is loaded at four points equi-distant from each other and the ends, with equal weights of 3 tons. Find ihe bending moment at each of these points, and draw the curve of shearing force. Ans. 24, 36, 36, 24 tons-feet. 3. (I.C.E., October, 1898.) In a beam ABCDE, the length (AE) of 24 feet is divided into four equal panels of 6 feet each by the points B, C, D. Draw the diagram of moments for the following conditions of /oS Examples. loading, writing their values at each panel-point : (i.) Beam supported at A and E, loaded at D with a weight of 10 tons ; (ii.) beam supported at B and D, loaded with 10 tons at C, and with a weight of 2 tons at each end A and E ; (iii.) beam encastre from A to B, loaded with a weight of 2 tons at each of the points C, D, and E. Ans. (i.) MB = 15, M c = 30, M D = 45 tons-feet, (ii.) MB and M D = 12, MC = IS (iii.) MB = 72, M c = 36, M D = 12 4. Find the bending moment and shear at the abutment, also at a section situated 4 feet from the free end in the case of a cantilever loaded, thus : length, 12 feet ; loads, 3 tons at extreme end, \\ ton 2 feet from end, 4 tons 3 feet 6 inches from end, 8 tons 7 feet from end. Ans. M at abutment, 125 tons-feet ; ditto 4 feet from end, 17 tons-feet. Shear 16-5 tons 8-5 5. Construct bending moment and shear diagrams for a beam 20 feet long resting on supports 12 feet apart. The left-hand support is 3 feet from the end. The beam is loaded thus : 2 tons at the extreme left-hand end, I ton 2*5 feet from it, 3 tons 5 feet from it, 8 tons 12 feet from it, and 6 tons on the extreme right-hand end. Write the values of the bending moment under each load. Ans. 30 tons-feet at the right abutment ; 4^62 tons-feet under the 8-ton load ; i'42 under the 3-ton load ; 6*5 over the left abutment. 6. A beam arranged with symmetrical overhanging ends, as in Fig. 404, is loaded with three equal loads one at each end and one in the middle. What is the distance apart of the supports, in terms of the total length /, when the bending moment is equal over the supports and in the middle, and at what sections is the bending moment zero? / Ans. \l. At sections situated between the supports distant 7 " om them. 7. A square timber beam of 12 inches side and 20 feet long supports a load of 2 tons at the middle of its span. Calculate the skin stress at the middle section, allowing for its own weight. The timber weighs 50 Ibs. per cubic foot. Ans. 1037 Ibs. per square inch. 8. The two halves of a flanged coupling on a line of shafting are accidentally separated so that there is a space of 2 inches between the faces. Assuming the bolts to be a driving fit in each flange, calculate the bending stress in the bolts when transmitting a twisting moment of IO,ooo Ibs.- inches. Diameter of bolt circle, 6 inches ; diameter of bolts, inch ; number of bolts, four. Ans. 9 tons square inch. CHATTER XI. 1. (Victoria, 1897.) A round steel rod inch in diameter, resting upon supports A and B, 4 feet apart, projects I foot beyond A, and 9 inches beyond B. The extremity beyond A is loaded with a weight of 12 Ibs., and that beyond B with a weight of 16 Ibs. Neglecting the weight of the rod, investigate the curvature of the rod between the supports, and calculate the greatest deflection between A and B. Find also the greatest intensity of stress in the rod due to the two applied forces. (E = 30,000,000.) Ans. The rod bends to the arc of a circle between A and B. 5 = o'45 inch ; f = 1 1,750 Ibs. square inch. 2. A cantilever of length / is built into a wall and loaded evenlv. Find Examples. 709 the position at which it must be propped in order that the bending moment may be the least possible, and find the position of the virtual joints or points of inflection. (N.B. The solution of this is very long.) Ans. Prop must be placed O'265/ from free end. Virtual joints o - 55/, o*58/ from wall. 3. Find an expression for the maximum deflection of a beam supported at the ends, and loaded in such a manner that the bending-moment diagram is (i.) a semicircle, the diameter coinciding with the beam ; (ii.) a rect- angle ; height = $ the span (L). | 4. What is the height of the prop relative to the supports in a centrally propped beam with an evenly distributed load, when the load on the prop is equal to that on the supports? 7 WI * Ans. - - - below the end supports. 1152 El 5. (S. and A., 1897.) A uniform beam is fixed at its ends, which are 20 feet apart. A load of 5 tons in the middle ; loads of 2 tons each at 5 feet from the ends. Construct the diagram of bending moment. State what the maximum bending moment is, and where are the points of inflection. Ans. Mmax = 20 tons-feet close to built-in ends. Points of in- flection 4 '4 feet from ends. 6. Find an expression in the usual terms for the end deflection of a cantilever of uniform depth, but of variable width, the plan of the beam being a triangle with the apex loaded and the base built in. (NOTE. In this case, and in that given in Question 12, the I varies directly as the bending moment, hence the curvature is constant along the whole length, and the beam bends to the arc of a circle.) 7. Find an expression for the deflection at the free end of a cantilever of length L under a uniformly distributed load W and an upward force Wo acting at the free end. What proportion must Wo bear to the whole of the evenly distributed load in order that the end deflection may be zero ? What is the bending moment close to the built-in end ? . W W _ Wo\ w<) must be g w . M at wall> WL JL1\ O 3 / o 8. Find graphically the maximum deflection of a beam loaded thus : Span, 15 feet; load of 2 tons, 2 feet from the left-hand end, 4 tons 3 feet from the latter, I ton 2 feet ditto, 3 tons 4 feet ditto, i.e. at 4 feet^ from the right-hand end. Take I'= 242 ; E = 12,000 tons square inch. Ans. 0*32 inch. 9. (T.C.E., October, 1898.) In a rolled steel beam (symmetrical about the neutral axis), the moment of inertia of the section is 72 inch units. The beam is 8 inches deep, and is laid across an opening of 10 feet, and carries a distributed load of 9 tons. Find the maximum fibre stress, also the central deflection, taking E at 13,000 tons square inch. Ans. 7 '5 tons square inch ; 0*215 inch. 10. (I.C.E., February, 1898.) A rolled steel joist, 40 feet in length. depth 10 inches, breadth 5 inches, thickness throughout j inch, is continuous 7io Examples. over three supports, forming two spans of 20 feet each. \Vhat uniformly distributed load would produce a maximum stress of 5^ tons per square inch ? Sketch the diagrams of bending moments and shear force. Ans. 0*23 ton foot run. II. (I.C.E., October, 1898.) A horizontal beam of uniform section, whose moment of inertia is I, and whose total length is 2L, is supported at the centre, one end being anchored down to a fixed abutment. Neglect- ing the weight of the beam, suppose it to be loaded at the other end with a single weight W. Find an expression for the vertical deflection at that end below its unstrained position. _ 2WL* 12. A laminated spring of 3 feet span has 20 plates, each 0*375 inch thick and 2*95 inches wide. Calculate the deflection when centrally loaded with 5 tons. E = 11,600 tons square inch. Ans. 2*45 inches. Actual deflection, 2'2 inches when loading. ,, ,, 2*9 unloading. 13. A laminated spring of 40 inches span has 12 plates each 0*375 inch thick, and 3-40 inches wide. Calculate the deflection when centrally loaded with 4 tons. E = 1 1,600 tons square inch. Ans. 3*9 inches. Actual deflection 3*35 inches when loading. ii 4*37 . unloading. 14. A laminated spring of 75 inches span has 13 plates, each 0^39 inch thick, and 3*5 inches wide. Calculate the deflection when centrally loaded with I ton. E = 1 1, 600 tons square inch. Ans. 5*0 inches. Actual deflection 4*2 1 inches when loading. i. 4'97 . unloading. The deflection due to shear in this long spring is evidently less than in the short stumpy springs in Examples 12 and 13. The E by experiment comes out 1 2,600. Sec page 387 for the effects of shear on short beams. 15. One of the set of beams quoted in the Appendix was tested with a load applied at a distance of 12 inches from one end, total length 36 inches, /; = 1*984 inch ; b 1*973 ' nc ^ taking E = 13,000 tons square inch. Find the maximum deflection. Ans. 0^51 inch. Actual deflection = 0*53 inch. CHAPTER XII. 1. A brickwork pier, iS inches square, supports a load of 4 tons; the resultant pressure acts at a distance of 4 inches from the centre of the pier. Calculate the maximum and minimum stresses in the brickwork. Ans. 4*15 tons square foot compression. 0*59 ,, tension. 2. (I.C.E., 1897.) The total vertical pressure on a horizontal section of a wall of masonry is 100 tons per foot length of wall. The thickness of the wall is 4 feet, and the centre of stress is 6 inches from the centre of thickness of the wall. Determine the intensity of stress at the opposite edges of the horizontal joint. Ans. Outer edge, 43'75 tons square foot. Inner 6*25 ,, ,, 3. (I.C.E., October, 1898.) A hollow cylindrical tower of steel plate, having an external diameter of 3 feet, a thickness of \ inch, and a height of 60 feet, carries a central load of 50 tons, and is subjected to a horizontal wind- Examples. 7 1 1 pressure of 56 Ibs. per foot of its height. Calculate the vertical stresses at the fixed base of the tower on the windward and on the leeward side. (Allow for the weight of the tower itself.) Ans. Windward side, O'li tons square inch. Leeward 2-09 ,, 4. (Victoria, 1897.) A tension bar, 8 inches wide, \\ inch thick, is slightly curved in the plane of its width, so that the mean line of the stress passes 2 inches from the axis at the middle of the bar. Calculate the maximum and minimum stress'in the material. Total load on bar, 25 tons. Ans. Maximum, 6*25 tons square inch tension. Minimum, 1*25 ,, ,, compression. 5. (Victoria, 1896.) If the pin-holes for a bridge eye-bar were drilled out of truth sideways, and the main body of the bar were 5 inches wide and 2 inches thick, what proportion would the maximum stress bear to the mean over any cross-section of the bar at which the mean-line of force was i inch from the middle of the section. Ans. 1*15. 6. The cast-iron column of a loo-ton testing-machine has a sectional area of 133 square inches. Z< = 712 ; Z = 750. The distance from the line of loading to the c. of g. of the section is 17*5 inches. Find the maximum tensile and compressive skin stresses. Ans. 3*08 tons square inch compression. 1*71 ,, ,, tension. 7. A tube in a bicycle frame is cranked $ inch, the external diameter being 0*70 inch, and the internal 0*65 inch. Find how much the stress is increased in the cranked tube over that in a similar straight tube under the same load. Ans. Four times (nearly). 8. The section through the back of a hook is a trapezium with the wide side inwards. The narrow side is I inch, and the wide side 2 inches ; the depth of the section is 2 inches ; the line of pull is I J inch from the wide side of the section. Calculate the load on the hook that will produce a tensile skin stress of 7 tons per square inch. Ans. 3-87 tons. 9. In the case of a puriching-machine, the load on the punch is estimated to be 1 60 tons. It has a gap of 3 feet from the centre of the punch. The tension flange is 40" x 3", and the compression flange 20" x 2". There are two 2-inch webs ; the distance from centre to centre of flanges is 4 feet. Calculate the stress in the tension flange. Ans. 2*05 tons square inch. CHAPTER XIII. 1. Find the buckling load of a steel strut of 3 inches solid square section with rounded ends, by both Euler's and Gordon's formula, for lengths of 2 feet 6 inches, 9 feet, 15 feet. E = 30,000,000, / = 49,000, S = 60,000. Ans. Euler, 2,250,000, 173,600, 62,500 Ibs. Gordon, 465,500, 176,000, 79,900 Ibs. 2. Calculate the buckling load of a 2" x i'O3" rectangular section strut of hard cast iron, 16 inches long, rounded ends. Take the tabular values for a and s. Ans. 40*3 tons. (Experiment gave 41*6 tons.) 3. Calculate the buckling load of a piece of cast-iron pipe, length, 2\ inches; external diameter, 4^4 inches; internal diameter, 3*9 inches; rounded ends. Take the tabular values for a. and S. Ans. 159 tons (hard), 98 tons (soft), 128 tons (mean). (Experiment, 113*2 tons.) 712 Examples. 4. Calculate the buckling load of a cast-iron column, 9 feet long, external diameter, 3*56 inches; internal diameter, 2^91 inches ; ends flat, but not fixed ; flanges 8 inches diameter. Ans. 787 tons (hard), 48-4 (soft), 63-6 (mem). (Experiment, 6l tons.) 5. Ditto, ditto, but with pivoted encls. Ans. 28*4 tons (hard). The iron was very hard, with fine close- grained fracture. (24*9 tons by experiment.) 6. Calculate the safe load for a cast-iron column, external diameter, 3*54 inches; internal, 2'S6 inches ; pivoted ends; loaded 2 inches out of the centre ; safe tensile stress, i ton square inch. Ans. 2 tons (nearly). (N.B. Cast-iron columns loatled thus nearly always fail by tension.) 7. Calculate the buckling load of T iron struts as follows : Area. (I) S-2 I '67 sq ins. lj IT (2) 16-0 1*67 , I ff (3) r 18-0 i'95 s 1 (4) r I4'3 I'20 , 1 T (5) r 19-8 2-04 , X a (6) 26-8 T20 , d Ant. Calculated, (1)26 'I tons ;(2) 19-2 ; (3)21-5 ; (4) 14-8; (5)20-3 ;(6)8'4. By experiment, 29-0 ,, 18*5 197 15-9 19-3 12-7. 8. A hollow circular mild-steel column is required to support a load of 50 tons, length, 28 feet ; ends rigidly held; external diameter, 6 inches ; factor of safety, 4. Find the thickness of the metal. Ans. o'9 inch. 9. A solid circular-section cast-iron strut is required for a load of 20 tons, length, 15 feet ; rounded ends ; factor of safety, 6. Find the diameter. Ans. 6*25 inches. (This is most easily arrived at by trial and error, by assuming a section, calculating the buckling load, and altering the diameter until a suitable size is arrived at.) CHAPTER XIV. 1. (S. and A., 1897.) Find the diameter of a wrought-iron shaft to transmit 90 II. P. at 130 revolutions per minute. If there is a bending moment equal to the twisting moment, what ought to be the diameter t Stress = 5000 Ibs. per square inch. Ans. 3-54 inches; 4 75 inches. 2. A winding drum 20 feet diameter is used to raise a load of 5 tons. If the driving shaft were in pure torsion, find the diameter for a stress of 3 tons per square inch. Ans. io'l inches. 3. Find the diameter for the above case if the load is accelerated at the rate of 40 feet per second per second when the winding engine is first started. Ans. 13*2 inches. 4. Find the diameter of shaft for a steam-engine having an overhung crank. Diameter of cylinder, 18 inches ; steam pressure, 130 Ibs. per Examples. 7 1 3 square inch ; stroke, 2 feet 6 inches ; overhang of crank, i.e. centre of crank-pin to centre of bearing, 18 inches ; stress, 7000 Ibs. per square inch. Ans. 10 inches. 5. Find the diameter of a hollow shaft required to transmit 5000 H.P. at 70 revolutions per minute ; stress, 7500 Ibs. per square inch ; the external diameter being twice the inner; maximum twisting moment = ij times the mean. Ans. 16*9 inches. 6. A 4-inch diameter shaft, 30 feet long, is found to spring 6 '2 when transmitting power ; revolutions per minute, 130. Find the horse-power transmitted. G = 12,000,000. Ans. 187. 7. (Victoria, iSgS.) If a round bar, I inch in diameter and 40 inches between supports, deflects 0^0936 inch under a load of 100 Ibs. in the middle, and twists through an angle of 0*037 radian when subjected to a twisting moment of looo Ibs. -inches throughout its length of 40 inches, find E, G, and K ; Young's modulus, the moduli of distortion, and volume. Ans. = 29,020,000, G = 11,020,000, K = 26,250,000, all in pounds per square inch. 8. A square steel shaft is required for transmitting power to a 3O-ton overhead travelling crane. The load is lifted at the rate of 4 feet per minute. Taking the mechanical efficiency of the crane gearing as 35 %, calculate the necessary size of shaft to run at 160 revolutions per minute. The twist must not exceed i in a length equal to 30 times the side of the square. G = 13,000,000. Ans. 2 inches square. 9. A horse tram-car, weighing 5 tons, when travelling at 8 miles an hour, is pulled up by brakes. Find what weight of helical springs of solid circular section would be required to store this energy. Stress in springs, 60,000 Ibs. per square inch. Ans. 0*48 ton. 10. Calculate the amount a helical spring having the following dimensions will compress under a load of one ton. Number of coils, 20^5 ; mean diameter of spring, 2'5 inches ; coils of square section steel, 0*51 inch side. Ans. 4*61 inches. (Experiment gave 4*57 inches.) 11. Calculate the stretch of a helical spring under a load of 112 Ibs. Number of coils, 22 ; mean diameter, 0-75 inch. Diameter of wire, 0*14 inch. Ans. i'8o inch. (Experiment gave 1*82 inch.) 12. A fan shaft 3*5 inches diameter, length between swivelled bearings 7 feet, weight of fan disc 1170 Ibs. Find the speed at which whirling will commence. Ans. 725 revolutions per minute. (An actual test gave 770.) CHAPTER XV. r. Find the forces, by means of a reciprocal diagram, acting on the members of such a roof as that shown in Fig. 488, with an evenly distributed load of 1 6 Ibs. per square foot of covered area. Span, 50 feet ; distance apart of principals, 12 feet, which are fixed at both ends. Calculate the force pa by taking moments about the apex of the roof. Find the force tu by the method of sections. See how they check with the values found from the reciprocal diagram. 2. In the case given above, find the forces when one of the ends is mounted on rollers, both when the wind is acting on the roller side and on the fixed side of the structure, taking a horizontal wind-pressure of 30 Ibs. per square foot on a vertical surface. 714 Examples. 3. Construct similar diagrams for Fig. 489. 4. Construct a polygon and a reciprocal diagram for tht Island Station roof shown in Fig. 490. Check the accuracy of the work by seeing whether the force F is equal to //. 5. (Victoria, 1896.) A train of length T, weighing W tons per foot run, passes over a bridge of length / greater than T, which weighs w tons per loot run. Find an expression for the maximum shear at any part of the structure, and sketch the shear diagram. Let ^ = the distance of the part from the middle of the structure. Shear = (/ - T + 2y) + uy N.B. As far as the structure is concerned, the length of the train T is only that portion of it actually upon the structure at the time. This expression becomes that on p. 512, when the length of the train is not less than the length of the structure. 6. Find an expression for the focal length X of a bridge which may be entirely covered with a rolling load of \V tons per foot run. The dead load on the bridge = w tons per foot run. Let M = . * = /(i - 2VM 4 MM-aM) 7. A plate girder of 50 feet span, 4 feel deep in the web, supports a uniformly distributed load of 4 tons per foot run. Find the thickness of web required at the ends. Shear stress in web and rivets, 4 tons square inch. Also find the pitch of rivets, \ inch diameter, for shearing and bearing stress. Bearing pressure, 9 tons square inch. A us. Thickness, inch. Pitch for single row in shear, 2*4 inches ; better put two rows of say 4'8-inch pitch. Ditto for bearing, i '97 inch and say 4 inches. 8. Find the skin stress due to change of curvature in a two-hinged arch rib, on account of its own dead weight, which produces a mean coin- pressive stress f t of 6 tons per square inch. E = 12,000 tons square inch. Span, 550 feet; rise, 114 feel; depth of rib, 15 feet. Also find the deflection due to a stationary test load which produces a further mean compressive stress^ of I ton per square inch. Arts. Stress, O'8 ton square inch ; 8 = 079 inch. (NOTE. The above arch is that over the Niagara River. Some of the details have been assumed, but are believed to be nearly accurate. When tested, the arch deflected o - 8i inch.) 9. Bridge members are subjected to the following loads. State for what static loads you would design the members : (i.) Due to a dead load on the structure of 10 tons in tension, and a live load of 30 tons in lension. (ii.) Dead, 100 tons tension ; live, 30 tons compression. (iiL) Dead, 80 tons tension ; live, 60 ions compression, (iv.) Dead, 50 tons tension ; live, 70 tons compression. Ans. (i.) 70 tons tension, (ii.) 100 tons tension ; when under the action of the live load, the stress is diminished, (iii.) Either for 80 tons tension or 40 tons compression, whichever gave the greatest section, (iv.) 90 tons compression. 10. Find the number of 4|-inch rings of brickwork required for an arch of 40 feet clear span ; radius of arch, 30 feet. Examples. 715 Am. By approximate formula, 34-4 inches = 8 rings ; by Trautwine, 31*9 inches = 7 rings. II. (Victoria, 1902). A load ?e discharged through a short pipe 2 inches diameter leading from and flush with the bottom of a tank ? Depth of water, 25 feet. Ans. 268. 11. In the last question, what would be the discharge if the pipe were made to project 6 inches into the tank, the depth of water being the same as before? Ans. 172. 12. In an experiment with a diverging mouthpiece, the vacuum at the throat was 18*3 inches of mercury, and the head of water over the mouth- piece was 30 feet, the diameter of the throat 0*6 inch. Calculate the discharge through the mouthpiece in cubic feet per second. Ans. OTI2. 13. (I.C.E., October, 1897.) A weir is 30 feet long, and has 18 inches of head above the crest. Taking the coefficient at o'6, find the discharge in cubic feet per second. Ans. 176. 14. A rectangular weir, for discharging daily 10 million gallons of compensation water, is arranged for a normal head over the crest of 15 inches. Find the length of the weir. Take a coefficient of 0*7. Ans. 3-56 feet. 15. Find the number of cubic feet of water that will flow over a right- angled V notch per second. 1 lead of water over bottom of notch, 4 inches ; K = 0*6. Ans. 0*165. 16. (I.C.E., February, 1898.) The miner's inch is defined as the flow through an orifice in a vertical plane of I square inch in area under an average head of 6J inches. Find the water-supply per hour which this represents. Ans. 90 cubic feet per hour when K = o'6l. 17. (Victoria, 1898.) Find the discharge in gallons per hour from a circular orifice I inch in diameter under a head of 2 feet, the pipe leading to the orifice being 6 inches in diameter. Ans. 885. 18. A swimming bath 60 feet long, 30 feet wide, 6 feet 6 inches deep at one end, and 3 feet 6 inches at the other, has two 6-inch outlets, for which Krf = 0*9. Find the time required to empty the bath, assuming that all the conditions hold to the last. Ans. 29*7 minutes. 19. Find the time required to lower the water in a hemispherical l>owl of 5 feet radius from a depth of 5 feet down to 2*5 feet, through a hole 2 inches diameter in the bottom of the Ixnvl. Ans. 14^3 minutes. 20. A horizontal pipe 4 inches diameter is reduced very gradually to half an inch. The pressure in the 4-inch pipe is 50 Ibs. square inch above the atmosphere. Calculate the maximum velocity of flow in the small portion without any breaking up occurring in the stream. Ans. 98 feet per second. 21. A slanting pipe 2 inches diameter gradually enlarges to 4 inches ; the pressure at a given section in the 2 inch pipe is 25 Ibs. square inch absolute, and the velocity 8 feet per second. Calculate the pressure in the 4-inch portion at a section 14 feet 1-elow. Ana. 31*2 Ibs. square inch. 22. In the last question, calculate the pressure for a sudden enlarge- ment. Ans. 30*96 Ibs. square inch. 23. (I.C.E., October, 1898.) A pipe tapers to one-tenth its original area, and then widens oul again to its former size. Calculate the reduc- tion of pressure at the neck, of the water flowing through it, in terms of the area of the pipe ana the velocity of the water. Ans - = WAV ''A A 2 ^ r N.R. The loo is the ratio ( ' Examples. 717 24. Find the loss of head due to water passing through a socket in a pipe, the sectional area of the waterway through the socket being twice that of the pipe. Velocity in pipe, lo feet per second. Ans. 0-85 foot. 25. A Venturi water-meter is 13*90 inches diameter in the main, and 5*68 inches diameter in the waist, the difference of head in inches of mercury is 20*75 (both limbs of the mercury gauge being full of water above the mercury). Taking the coefficient of velocity K p as 0^98, find the quantity of water passing per minute. Ans. 2451 gallons. 26. Find the loss of head due to water passing through a diaphragm in a pipe, the area being one-third of that of the pipe. Velocity, 10 feet per second. Ans. 21 '3 feet. 27. A locomotive scoops up water from a trough I mile in length. Calculate the number of gallons of water that will be delivered into the tender when the train travels at 40 miles per hour. The water is lifted 8 feet, and is delivered through a 4-inch pipe. The loss of head by friction and other resistances is 30%. Ans. 2130. 28. The head of water in a tank above an orifice is 4 feet. Calculate the time required to lower the head to I foot. The area of the surface of the water is 1000 times as great as the sectional area of the jet. Ans. 250 seconds. 29. (I.C.E., February, 1898.) A canal lock with vertical sides is emptied through a sluice in the tail gates. Putting A for area of lock basin, a for area of sluice, H for the lift, find an expression for the timejpf emptying the lock. (K c = coefficient of contraction.) 2 - , = 30, neglecting losses by friction. Ans. 511 revolutions per minute. 6. What is the hydraulic efficiency, neglecting friction, of the pump in the last question ? Ans. 65 per cent. 7. In the case of the pump given in Question 5, calculate the horse- power absorbed by friction on the outside of ihe two discs. Taking the inner diameter as one-half the outer, and the resistance per square foot at 10 feet per second at O'S Ib. Ans. 1-3. INDEX OF SYMBOLS THE following symbols have been adopted throughout, except where specially noted : A, a (with or without suffixes), area; exception, p. 15, A is used for angular acceleration By used as a suffix in locomotive balancing problems, and always in connection with balance weights C, centrifugal force ; exception, p. 581, constant in Thrupp's formula, com- pressive stress in " Strut " chapter C at chord of an arc C centre of pressure D, diameter of; coils (inches) of springs, pp. 494-500 ; pipes (feet) for all questions of How of water ; shafts (inches) in torsion d = diameter of pulleys in belting questions ; rivets, pipes (inches) ; least diameter of struts (same units as length) ; diameter of wire in helical springs (inches) ; bolt at bottom of thread, p. 308 dty dvy etc. See Appendix Ey Young's Modulus of Elasticity E n . See p. 176 ey extension, percentage of, p. 307 Fy force ; exceptiorty Chap. VII., "Friction," it is used for fric- tional resistance fy stress, intensity of ; exceptiotiy in friction of water problems. See p. 57 6 f u y acceleration /, and/,. See p. 347 _/&, bearing pressure on rivets fry tensile strength of rivets f n shear stress, intensity of / tensile strength of plates in riveted joints, of walls in pipes f tt y resultant stress due to shear and direct stress. See pp. 317, 491 Gy coefficient of rigidity G a . See p. 459 g, acceleration of gravity Hy Ay height in feet, head in hy- draulics ; exceptioiiSy in beam sections the height is in inches hy loss of head in feet due to friction in pipes in all flow of water problems H e y h e . See Governors, pp. 203-226 H C y depth of immersion of centre of pressure of an immersed area H , depth of immersion of centre of gravity of an immersed area H.P.y horse-power ; exception, p. 172, high pressure /, moment of inertia, or second moment, about an axis passing through the c. of g. of a surface I y moment of inertia about an axis parallel to the above-mentioned axis IP, the polar moment of inertia, or the second polar moment I.P.y intermediate pressure (cylinder) 3 A 722 Index of Symbols. /, Joule's mechanical equivalent of heat, p. 13 J % a constant K, coefficient of elasticity of volume ; exceptions, pp. 176, 212, 241, 333, 385, 579 A a , coefficient of approach A" e , ,, contraction, p. 548 A*, discharge, p. 548 A'., velocity, p. 548 A' r , resistance, p. 549 /,, or /, length ; exception, p. 307 /... See p. 427 Af, bending moment in beam ques- tions ; exception, p. 307, moss in dynamic questions J/o, momentum Af t , twisting moment M. See p. 491 m. See pp. 176, 206 N, revolutions per minute, often with suffixes, which are explained where used ; exception, p. 311 N n revolutions per second N.A., neutral axis n, ratio of length of connecting-rod to radius of crank in engine and pump problems (see j>. 494 for helical springs, p. 511 for rolling loads, p. 568 for Venturi water- meter), p. 597 ; ratio of velociiy of the jet to that of the surface, in the pressure of jets impinging on moving surfaces O.H., polar distance of a vector polygon P, pressure ; buckling stress in Ibs. per square inch in struts, see p. 466 ; exception, see special mean- ing in " Friction" chapter p, acceleration pressure (see. p. 163) in Ibs. per square inch for engine balancing problems P t , mean effective pressure on a piston in Ibs. per square inch Q, quantity of water flowing in cubic feet per second A', radius (feet) ; hydraulic mean radius (feet) in flow of water r. See p. 469 A',, A%, reactions of beam supports ; exception, rise of arch, p. 532 A'a, radius of c. of g. of balance weight Re, radius of coupling crank of locomotive A% perpendicular distance between the two parallel axes in moment of inertia, or second moment problems A' w , radius of gyration (feet) /V, radius of crank shaft journal r a , radius of gudgeon pin /,.., radius of crank pin S, stress ; exceptions, side of square shaft (inches), p. 486 ; span of arch (feet), pp. 527-535 ; wetted surface in square feet, p. 576 j, space T , t, time in seconds ; exceptions, thickness of pipe (inches), p. 595 ; tension modulus of rupture, p. 471 ; tension in belts, pp. 281- 288 ; thickness of plate in riveted joints, girder webs, etc. '/'a, Tt, etc., number of teeth in wheels a, b, etc., respectively //. See p. 452 /', v, velocity (feet per second) ; ex- ceptions (feet per minute), p. 283, (miles per hour), pp. 501, 502 Vi and V\, used for pressure tur- bines. See p. 624, Fig. 577 y*,, velocity of rim of wheel or belt in feet per second v r , velocity ratio i' r , velocity of rejection in turbines See p. 618 \V, weight in pounds WB, W p , W r . See p. 193 W n W*. See p. 198 W r , weight of i foot of material I sq. inch in section, pp. 181, 284 IV*, weight of I cubic foot of water in Ibs. tr, weight per square inch ; excep- tions, width of belt (inches), p. Index of Symbols. 723 283 ; unit strip in riveted joints, pp. 335, 343 ; weight of unit column of water, i.e. I foot high, I sq. inch section in hydraulics x, extension or compression of a body under strain ; exceptions, eccentricity of loading in com- bined bending and direct stresses, P- 45 2 K See p. 385 y t distance of most strained skin from the neutral axis in a beam section Z, modulus of the section of a beam, moment of inertia t.e. p, polar modulus of the section of a shaft GREEK ALPHABET USED. a (alpha)) angle embraced by belt, 282 ; a constant in the strut for- mula, 468 8 (delta), deflection in every case r) (eta), efficiency in every case (t/ieta), angle, subtending arc, 22, 29 ; of balance weight, 196 ; be- tween two successive tangents on a bent beam, 426 ; between the line of force and the direction of sliding in Chap. VII., exception p. 242 C , angle expressed in circular mea- sure K (kappa), radius of gyration jt* (mu), coefficient of friction v (pi), ratio of circumference to diameter of circle p (rho), radius "5. (sigma), symbol of summation (phi), friction angle u> (omega), angular velocity GENERAL INDEX Acceleration, def., 5 ; curves, 140- 142 ; on inclined plane, 229 ; pressure due to, 161-170 Accumulator, 592 Accuracy ', 680 Air, in motion, 501 ; vessels, 636 ; weight, 501 Aluminium, strength, 300, 301, 306, 309. 315; 352 Angle of friction, 228 ; repose, 228 ; sections, 370 ; of twist in shafts, 487 Angular velocity (), def., 4 ; bars in mechanism, 130, 135 Anticlastic curvature, 328 Antifriction metals, 259 ; wheels, 242 Arc of circle, c. of g., 64, 66 ; length, 24 Arch, line of thrust, 526 ; masonry, 522 ; ribs, 527 ; temperature effects, 532 Area, bearings, 260 ; circle, 28 ; cone, 36 ; ellipse, 30 ; hyperbola, 38 ; irregular figures, 26-32 ; parallelogram, 24 ; parabolic segment, 32 ; sphere, 36 ; trape- zium, 26 ; triangle, 24-26 Asphalte, 310 Assumptions of beam theory, 357 Astronomical clock governor, 211 Axis, neutral (N.A.), 361 ; of rota- tion, 186 Axle, balancing, 186 ; friction, Chap. VII. Axoile, 122, 123 Baker, Prof. I. O., reference to, 526 Balancing, axles, 186-192 ; loco- motives, 193-203 ; weights, 202 Ball bearings, 244-247, 278, 289 Bar, moment of inertia, 90 Barker, A. H., reference to, 3, 18, 74, 142, 194 Barlow, theory of thick cylinders, 346, 350 Barr, Prof. A., reference to, 507 Beams, Chap. X. ; angles, 370 ; landing moments, 392-423 ; box, 368 ; breaking load, 388 ; built in, 422, 438-445 ; cast iron, 366 ; channel, 370 ; circular, 376 ; continuous, 445-449 ; deflection, 424-450; elastic limit, 388; X, 368 ; inclined, 458 ; irregular loads, 414-416, 444 ; irregular sections, 378 ; model, 354 ; plate, 516; propped, 434, 445. 449; rails, 378; shear, 381, 392-421 ; stiffness, 450 ; unsymmetrical sections, 363 Bearings, area, 260 ; ball, 244-247, 278, 289 ; collar, 245, 256, 262 ; conical, 263 ; lubrication, 253- 256 ; onion, 265 ; pivot, 263 ; roller, 243 ; seizing, 257 ; Schiele, 265 ; wear, 255-259 ; work ab- sorbed, 261 Bell crank lever, 56 Belts, centrifugal action on, 284 ; creeping, 286 ; friction, 281-283 ; strength, 283 ; tension on, 284 ; transmission of power by, 283- 285 Bending moment, Chap. X. ; on arched ribs, 530 ; combined with twisting moment, 490 ; on coup- ling and connecting rods, 183-185 Bends in pipes, 585 Boiler, riveted joints for, 332-345 ; shell, 329, 345 Bracing, Chap. XV. Brass, strength, 352 General Index. 725 Bridges, Chap. XV. ; arch, 522- 535 ; floors, 368, 376, 379 ; loads, 510; plate girder, 516; rolling load, 512; suspension, 108 Brit tint ess, 292, 316 Bronze, strength, 352 Buckling of struts, Chap. XIII. Bulb section, 379 Bursting pressure, 329, 345-35 1 J of flywheels, 181, 182 Cantilever, bending moment on, 393, 402-406, 422; deflection, 426-429, 434 ; stiffness, 450 Cast-iron beams (see Beams) ; columns, 471-478, and Appendix ; in compression, 308, 309, 352 ; strength, 331, 352 ; strain, 294 Catenary, 109 Cavitaiion, 640, 654 Cenient, Portland, 311 Centre of gravity (c. of g.), arc of circle, 64, 66; balance weights, 201 ; cone, 72 ; definition, 13, 58 ; irregular figures, 62-70 ; locomotive, 74 ; parabolic seg- ments, 66, 68, 70 ; parallelogram, 60 ; pyramid, 72 ; trapezium r 6o ; triangle, 60 ; wedge, 72 Centre of pressure, 544-547 Centre, virtual, 122, 126 Centrifugal force, action on belts, 284 ; definition, 17 ; on flywheel rims, 181 ; on governor balls, 204 ; reciprocating parts, 162 Centrifugal pump, 651-670 Centrode, 123, 124 Chain, pump, 631 ; stress in sus- pension, 109 Channel, friction of water in, 581 ; strength as beam, 370 Circle, arc, 22; area, 28; circum- ference, 22 ; moment of inertia, 88, 90, 96, 98 ; strength as beam section, 376 ; strength as shaft section, 485 Coefficient of friction, 227, 229, 231, 244, 250, 253, 283, 289, 578 Coil friction, 281 Collar bearing, 245, 256, 262 Columns, Chap. Xlll.jeccentricallj loaded, 475~47? Combined, bending and direct stresses, Chap. XII. ; shear or torsion and direct stresses, 316,489 Compression, cement, 311 ; strength of materials in, 308-313, 352 (see also struts, Chap. XIII.) ; of water, 543 Cone, moment of inertia of, 104 ; surface of, 36 ; volume, 46 Connecting-rods, bending stresses in, 185 ; influence of short, 163- 167 ; virtual centre, 126, 133-138 Conservation of energy, 13 Constrained motion, 119 Continuous beams, 445-450 Contraction of streams, 548-555 Copper, fracture, 297 ; strength, 352 ; stress-strain diagram, 294, 300, 301, 309 Corrugated floors, 368, 376 Couples, n Coupling-rods, stress in, 183 Crane, forces in, in, 112; hook, 458, and Appendix, 683 Crank, and connecting-rod, 126, 1 33~ I 3 6 5 P in . l6 9 shafts, 489 ; shaft governor, 214, 224; webs, 203 Cranked tie bar, 456 Creeping of belts, 286 Crossed-arm governor, 210, 221 Crushing. See Compression Cup leathers, friction of, 266 Cycloid, 148 Cylinder, moment of inertia, 98, loo; strength under pressure, 329, 346-351 ; volume of, 40 Dalby, Prof., reference to, 192, 194. De Laval steam turbine governor, 214 ; centrifugal pump, 663 D'Auria pump, 645-6 Davey pump, 649-650 Deflection of beams, Chap. XI. > due to shear, 385 ; helical springs, 494-500 Delta metal, 301, 352 Diagrams, bending moment and shear, Chap. X. ; indicator, 168, 639 ; stress-strain, 294-313 ; twisting moment, 171-174 Differential calculus. See Appen- dix, 672 Dimensions, 2, 20 Discharge of pipes, etc., 575-585 ; weirs, Chap. XVI. Discrepancies, beam theory, 387-39 1 j strut theory, 462 3 A 3 726 General Index Distribution of load on roofs, 505 Doble water-wheel, 6oS Ductile materials, compression, 308 ; shear, 315 ; tension, 297 Dunktrley, Prof., reference to, 493, and Appendix, 685-7 Durley, R. J., reference to, 142 Eccentric loading, 456, 475~478 Efficiency. Bolt and nut, 240 ; cen- trifugal pump, 654-670 ; defini- tion, 13 ; hydraulic motors, Chap. XVII. ; inclined plane, 236 ; ma- chines, 266-280 ; pulleys, 269- 272 ; riveted joints, 341 ; reversed, 267 ; screws and worms, 238, 273 ; steam engine, 279 ; table of, 280 Elastic limit, artificial raising, 301 ; definition, 292 Elasticity, belts, 286 ; definition, 292 ; modulus of, 318-322 ; table of, 352 ; transverse, 322-326 Ellipse, area, 30 ; moment of inertia, 90, 92 Energy, definition, 13 ; kinetic, 14 ; loss due to shock, 571 Engine, dynamics of, Chap. VI. " Engineer " Journal, reference to, 1 80, 649 " Engineering" Journal, reference to, 165, 273, 516, 569, 651, 670 Epicyclic trains, 155-159 Epicycloid, 148 Equivalent, mechanical, of heat, 13 Euler, strut formula, 464 Extension of test-pieces, 296 Extensometer, 293 Factor of safety, struts, 478 Fidler, Prof., reference to, 540 Flange of girder, 518 Flexure, points of contrary, 439 Flooring, strength of, 368, 376, 379, 380 Flow of water, 575-585 ; Thrupp's formula, 580 Fluctuation of energy, 175 Flywheel, 174 ; gas engine, 179 ; governor, 215; moment of inertia, 100, 102 ; shearing and punching machines, 179 ; stress in rims, 181, 182 ; weight, 176 ; work stored in, 178 Focal distance, 512 Foot, pound, II; poundal, 10 ; step, 256 Force, centrifugal, 17 ; definition, 7 ; gravity, 9 ; polygon, 106 ; pump, 633 ; in structures, Chaps. IV., XV. ; triangle, 16 Fractures of metals, 297 Framed structures, Chap. XV. Friction, angle, 228 ; on axle, 261 ; ball bearings, 244-247, 278, 289 ; belts, 281-283 ; bends, 585 ; bolt and nut, 240 ; body on horizontal plane, 227 ; ditto inclined, 228- 237 ; coefficient, 227, 229, 231, 250, 253. 283, 289, 578 ; coil, 281 ; cone, 228 ; cup leather, 266 ; dry surfaces, 231 ; gearing, 273 J governors, 219 ; levers, 272 ; lubricated surfaces, 247- 253 ; pivots, 261-265 ; pulleys, 269-272 ; rolling, 240-244 ; roller bearings, 243 ; slides and shafting, 274-278 ; temperature effects, 250 ; time effects, 25 1 ; toothed gearing, 272 ; velocity, effect of, 249 ; water, 575 ; wedge, 237 ; worms and screws, 240, 273 Froude, reference to, 567, 576 Funicular polygon, 107-8 Gallon, experiments on friction, 231 Gas-engine flywheels, 1 79 Gearing, friction, 272 ; toothed, 146-159 Girders, continuous, 446 ; modu- lus of section, 368 ; plate, 516 ; triangulated, 513 ; Warren, 515 Gordon, strut formula, 467-473 Governing of water motors, 610-613 Governors, 203 ; astronomical, 210 ; crankshaft, 214; dashpot for, 215 ; friction, 219 ; Hartnell, 212, 221- 224 ; loaded, 209 ; McLaren, 215, 224; parabolic, 210 ; Porter, 208, 220 ; power of, 225 ; sensi- tiveness, 217; Watt, 204 Gravity, centre of (c. of g.) (see Centre), acceleration of, 9 Culdinus, theorem of, 36, 42 Gun-metal, 294, 300, 309, 352 Gyration, radius of, bar, 98 ; circle, 88, 90, 96, 98 ; cone, 104 ; cylinder, 98, 100; definition, 15; ellipse, 90, 92 ; flywheel, 100, General Index. 727 1 02 ; graphic method, 96 ; ir- regular surface, 94; parabola, 92, 94 ; parallelogram, 78-82, 96 ; sphere, 102, 104 ; square, 88 ; trapezium, 84, 86 Hartnell, governor, 212, 221, 224 ; springs, 495 Has tie engine, 590 Head of water, eddies, 571 ; energy, 566 ; friction, 575-583 ; effect of sudden change, 571, 572 ; pres- sure, 542 ; velocity due to, 547, 5.58 Height of governor, 204 Hele-Shaw, Prof., reference to, 152, 281, 574 Hemp ropes, strength, 289 Hicks, Prof., reference to, 10, 12 Hill) reference to paper on loco- motives, 194 Hobsorfs patent flooring, 379 Hodgkinson, beam section, 366 Hodograph, 17 Hoffmann, ball bearing, 246 Holes, rivet, 333 Hooks, 458, and Appendix, 683 Horse-power (H.P.), 11 ; of belts, 283 ; of shafts, 488 Humpaze, gear, 158 Hydraulics. See Chaps. XVI., XVII., XVIII. Hyperbola, area, 38 Hypocycloid, 148 Impulse, definition, 7 ; strokes stored in flywheel, 179 Inclined, beam, 458 ; planes, 228- 238 Indiarubber % 331 ; tyres, 241 ; beam, 353 Indicator diagram, correction for inertia of reciprocating parts, 16^-172 Inertia^ definition, 14 ; water-press, due to, 593 ; of reciprocating parts, 160-169 Inertia, moment of, 14, 75-105 ; angles, 370 ; bar, 98 ; beam sec- tions, 356 ; channel, 370 ; circle, 88, 96, 98, 376; cone, 104; cylinder, 98-100 ; ellipse, 90-92 ; flooring, 368, 376, 379, 380; flywheel, 100, 102 ; girders, 368 ; H section, 368 ; hydraulic-press table, 380 ; irregular surface, 94 ; joists, 368 ; parabola, 92-94 ; parallelogram, 78-82, 96 ; rails, 378 ; rectangle, 366 ; sphere, 102-104 > square, 88, 370 ; tees, 370; trapezium, 84, 86, 374; triangle, 82, 84, 374 Injector, hydraulic, 558 Inside cylinder locomotive, 194 Im'olute, 151 Iron, angle, 370 ; cast, 294, 308, 39> 33 ! 352 ; ditto beams, 366 ; columns and struts, 471-478, and Appendix ; modulus of elasticity, 352 ; riveted joints, 332-344 ; strength, 352 ; stress-strain dia- gram, 294, 300, 309 j weight, 48 ; wrought, 315, 331 Jet, pressure due to, 600 Johnson, Prof., reference to, 540 Joints (see Riveted joints), 332- 344 ; in masonry arches, 525 Joist, rolled, 368 Journal, friction of, 261 Kennedy, Prof., reference to, 119, 273, 307 Kinetic energy, 14 Knees, resistance of, in water-pipes, 584 Lame's theory of thick cylinders, 348 Lap joint. See Riveted joints Lattice girders. See Structures, Chap. XV. Leather, belts, 281-288 ; cup, fric- tion of, 266 ; friction on iron, 230 Levers, 52-56 ; friction, 272 Lift, hydraulic, 589 Limit of elasticity, 292 ; table of, 352 Link-motion, 145 Link polygon, 107 Load, beams, see Chap. X. ; bridges, 510 ; live, 535 ; rolling, 512 ; roofs, 504 Locomotives, coupling-rods, 183 ; balancing, 192-203 ; centre of gravity of, 74 Lodge, Mensuration, reference to, 23, 29 Longmans, Mensuration, reference to, 27 7 28 General Index. Lubrication, 247-257 Machine, definition, 119; efficiency of, 266 ; frames, 459 ; hydraulic, Chap. XVII. Martin, H. M., reference to, 516, 526 Masonry arches, 522-526 Mass, 9 Materials, strength of, 352 Mather and Platt, centrifugal pump, 667-9 McLaren, engine, 172; governor, 215, 224 Mxhanisms, CHap. V. Mensuration, Chap. II. Metals, strength, 352 ; weight, 48 Meter, Venturi water, 569 Metric equivalents, Chap. I. Modulus, of beam sections, 357 ; circle, 376; flooring, 368, 376, 379 finder, 368 ; graphic solu- tion, 358 ; press table, 380 ; rails, 378 ; rectangular, 367 ; square on edge, 370 ; tees, angles, 370 ; trapezium, 374 ; triangle, 374 ; unsym metrical, 363 ; of shaft sections, 486 ; of elasticity, 318- 323 ; of belts, 286 Moment of inertia, or second mo- ment. See Inertia Moments, 12, Chap. III. ; bending, Chap. X.; twisting, 171, 172 Momentum, 7 ; moment of, 654 Morin, laws of friction, 229 Motion, constrained and free, 119; plane, I2O ; screw, spheric and relative, 120 Mouthpieces, hydraulic, 553 Moving loads, 511, 535 Musgrave, engine, 144 Notch, rectangular, 559 ; V, 560 Nut, friction of, 240 Oak, strength, 352 ; strut, 471-473 Obliquity of connecting-rod, 163 Oil as lubricant, 247-260 Onion bearing, 265 Orifices, flow of water through, 548-558 Oscillating cylinder, 136, 588 Outside- cylinder locomotive, 1 97- 1 99 Overlap riveted joints, 334 Pappus, surfaces, 36 ; volumes, 42 Parabola, area, 30 ; c. of g., 66-70 ; chain, 109; governor, 210 ; mo- ment of inertia, 92-94 Paraboloid, volume, 46 Parallelogram, area, 24 ; c. of g., 60; forces, 1 6 ; moment of inertia, 78-82, 96 Parsons, pump, 651 Pearson, Prol. K., reference to, 7, 467, and Appendix, 683 Pelton wheel, 607 Permanent set, 292 Perry, reference to book on Calculus, 3, 184, 349, and Appendix, 672 Phosphor bronze, 352 Pillars. See Struts Pipes, bursting, 345 ; flow of water in, 575-585 ; pressure due to water-ram, 595, 643 Pitch, circles, 146-151 ; rivets, 332- 344 Pivots, conical, 263 ; flat, 262 ; Schiele, 265 Plane, inclined, 228-237 ; motion, 120 Plasticity, 292 Hate girders, 516-522 Plate springs, 435 Poisson's ratio, 323 Polygon, forces, 106, 107 ; c. of g. of arc, 64 Ponce/ et water-wheel, 614 Porter governor, 208, 220 Power, 1 1 ; transmission, by shafting, 489 ; water-main, 597 Pressure, acceleration, 160 ; bear- ings, 261 ; bursting, 345-35 centre of, 544 ; energy, *66 ; friction, effect of, Chap. VII. ; inertia, 161 ; jets, 605 ; water, 543 ; wind, 501 Prism, volume, 40 Pulleys, friction, 269-272 Pumps, Chap. XVIII. Punchtd holes, 345 / ) w^A/^-machine, flywheel, 179 ; frame, 459 Pyramid, c. of g., 72 ; volume, 48 Quick-return mechanism, 136 Radius of gyration (K), 15. See Gyration Rail sections, 378 General Index. 729 ing^ reference to, 25, 492 Reciprocating parts of engines, 160- I7i 193 Rectangle, area, 24 ; beam, 367 ; c. of g., 60 ; moment of inertia and radius of gyration, 78, 82, 96 ; shaft section, 486 Reduction in area, 297 Relative motion, 120 Repetition of stress, 535-540 Repose, angle of, 228 Resistance, bends, knees, etc., 585 ; friction, Chap. VII. ; rolling, 240 Resolution of forces, 16, 106 Reynolds, Prof. Osborne, reference to, 13,241, 254, 568, 576 Ribs. See Arch Rigg, hydraulic engine, 138 Ring, volume of anchor, 48 Riveted joints, 332-344, 518, 519 Roller bearings, 243 Rolling load on bridges, 5 1 1 Roofs, 1 1 6, 504-509 Rofes, transmission of power by, 285, 288 Rotating parts of locomotives, 193 Running balance, 187 Schiele pivot, 265 Scoop wheel, 632 Screw, friction, 238, 273 ; motion, 1 20 Second moments, 52. See Moments of inertia Seizing of bearings, 257 Sensitiveness of governors, 217 Shaft, friction, 274 ; efficiency of, 274-278 ; horse-power, 488 ; spring of, 487 ; torsion, 481-491 Sharpe, Prof., reference to, 183 Shear, Chap. X. ; beams, 381-387, 392 ; diagrams, 400-421 ; deflec- tion of beams due to, 385-387 ; and direct stress, 316, 489 ; modulus of elasticity in, 322, 325, 352 ; nature of, 313 ; plate girders, 517; rivets, 334; shafts, 481 ; strength, 315 Shearing machine, flywheel, 179; frame, 459 Sheer legs, 113 Shock, loss of energy due to, 571 ; pressure due to, 594 Simpson's rule for areas, 34 ; volumes, 42 Slides, friction, 274 Slip of pumps, 637-643 Slope of beams, 426 Smith, Prof. R. H., reference to, 142, 180, and Appendix, 672 Speed, 2 ; flywheel rims, 181 ; pumps, 636 Sphere, moment of inertia, 102- 104 ; surface, 36 ; thin, subject to in- ternal pressure, 329 ; volume, 44 Spheric motion, 1 20 ^Springs, helical, 493-500 ; plate, - 435 safe load, 495, 497 ; square section, 497 ; work stored in, 496 ; weight, 496 Square, beam section, 370 ; moment of inertia, 88; shaft, 486; spring section, 497 Standing balance, 1 86 Stanger, W. H., reference to, 310 Stannah pendulum pump, 138 Steam-engine, balancing, 190-203 ; connecting-rod, 163-167, 185 ; crank shafts, 489 ; crank pin, 134, 169 ; friction, 279; governors, 203-226 ; mechanism, 133 ; re- ciprocating parts, 160-173 5 twist- ing-moment diagrams, 171 Steel, effect of carbon, 300 ; strength, etc., 352 ; wire, 301-306 Stiffeners, plate girders, 519-522 Stiffness of girders, 450 Strain, 290, 323-331 Stream line theory, 566 Strength of materials, 352 Stress, coupling-rods, 183 ; defini- tion, 290 ;: diagrams, 294-309; flywheel rims, 180 ; oblique, 311 ; real and nominal, 299, 312 Structures, arched, 522-535 ; dead loads on, 512 ; forces in, 505- 515 ; girders (plate), 517 ; rolling loads, 511 ; weight, 522 ; wind pressure, 501-504 Struts, bending of, 462 ; eccentric loading, 475-478 ; end holding, 465 ; Euler's formula, 464 ; Gor- don's formula, 467-474 ; straight- line formula, 474, and Appendix, 684 ; table, 473 Sudden loads, 535 Suspension bridge, 108 Table, strength of, 380 Tee section, 370, 469 730 General Index. Teeth of wheels, 146 ; friction of, 272 7'emperature, effect on arched ribs, 532 ; water, 541 Tensile strength, see table, 352 Tension, 291 ; and bending, 452 Thorpe, R. H., reference to, 558, 591, 646 Three-legs, 114 Thrupp s formula for flow of water, 576, 580 Thrust bearing, 245, 253, 256, 262- 265 Tie-bar, cranked, 456 Todhunter and Pearson, reference to, 467 Toothed gearing, 146-159; friction, 272 Torsion, Chap. X V. ; and bending, 489-492 Tower, experiments on friction, 252, 253 Trapezium, area, 26 ; c. of g., 60- 62 ; modulus of section, 86, 374 Trautwine, ref. to, 524 Triangle area, 24-26 ; c. of g., 60 ; moment of inertia, 82-84, 374 Tripod, 114 Trough flooring, 368 Turbines, 615-^630 Turner and Brightmore, reference to, 576 Twisting-moment diagram, 171 ; shafts, 481 Units, i, 19 Unwin, Prof., reference to, 109, 152, 181, 258, 274, 334, 467, 503, 539. 540 Vector polygon, 108 Velocity, approach, 562 ; curve, 140 ; points in machines, 127 ; virtual, 121 ; water machines, 611 ; water in pipes, 575-5^5 Venturi water-meter, 569 Virtual centre, 122 ; length of beams, 439 ; of struts, 465 ; ve- locity, 127 Volume, cone, 46 ; cylinder, 40 ; definition, 20; paraboloid, 46; prism, 40 ; pyramid, 48 ; ring, 48 ; Simpson's method, 42 ; slice of sphere, 44 ; solid of revolution, 42 ; sphere, 44 ; tapered body, 48 Vortex, forced, 571; free, 570; turbine, 627 Walmisley, reference to, 504 Warner and Swazry governor, 210 Warren girder, 515 Water, Brownlee's experiments, 557 ; centre of pressure, 544 ; compressibility, 543 ; flow due to head, 547 ; flow over notches, 559 ; flow under constant head, 562 ; friction, 575-585 ; hammer, 643 ; inertia, 593 ; injector, 558 ; jets, 600-606 ; motors, Chap. XVII. ; orifices, 548-555 ; pipes of variable section, 565 ; power through main, 597 ; pressure, 541 ; ditto machines, 588 ; velocity machines, 607-628 ; weight, 541 : wheels, 586-588, 614 Watt governor, 204 Wear of bearings, 255, 259 Webs of plate girder, 516-522 Wedge friction, 237 Weight, 9 ; materials, 48 Westinghouse, experiments on fric- tion, 231 Wheels, anti-friction, 242 ; rolling resistance, 240 Whirling of shafts, 492, and Appendix, 685 White metal for bearings, 259 Witksteed, reference to, 296 Wilson-Hartnell governor, 212, 221 Wind, 501 ; on roofs, 505 Wire, 301-305 Wohler experiments, 538 Work, 10, II ; in fracturing a bar, 306 ; stored in flywheel, 178 ; governors, 226 ; springs, 496 Worms, friction, 273 Worthington pump, 648 Young's Modulus of elasticity, 318 THE END. CLOWES AND SONS, LIMITED, LONDON AND BECCLES. MECHANICS, DYNAMICS, STATICS, HYDROSTATICS, ETC. WORKS BY S. DUNKERLEY, D.Sc., Assoc. 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