# Full text of "Mechanics applied to engineering"

## See other formats

BOUGHT WITH THE INCOME FROM THE SAGE ENDOWMENT FUND THE GIFT OF fletirg W, Sage 1S91 ll;.f^MaqH^ iLimif... 3777 Cornell University Library TA 350.G65 1914 Mechanics applied to engineering. 3 1924 004 025 338 Cornell University Library The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924004025338 MECHANICS APPLIED TO ENGINEERING MECHANICS APPLIED TO ENGINEERING JOHN GOODMAN Wh. Sch., M.I.C.E., M.I.M.E. PROFESSOR OF ENGINEERING IN THE UNIVERSITY OF LEEDS With 741 Illustrations and Numerous Examples EIGHTH EDITION LONGMANS, GREEN AND CO. 39 PATERNOSTER ROW, LONDON FOURTH AVENUE & 30th STREET, NEW YORK BOMBAY, CALCUTTA, AND MADRAS I9I4 All rights reserved 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 conciseness, 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 which only refer to the illustrations in their respective accompanying paragraphs. 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. Aiidrew 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. Thk University of Leeds, August, I904' PREFACE TO EIGHTH EDITION New Chapters on " Vibration " and " Gyroscopic Action " have been added to this Edition. Over a hundred new figures and many new paragraphs have been inserted. The sections dealing with the following subjects have been added or much enlarged — Cams, Toothed Gearing, Flywheels, Governors, Ball Bearings, Roller Bearings, Lubrication. Strength of Flat Plates, Guest's Law, Effect of Longitudinal Forces on Pipes under pressure. Reinforced Concrete Beams, Deflection of Beams due to Shear, Deflection of Tapered Beams, Whirling of Shafts, Hooks, Struts, Repeated Loading. Flow of Water down Steep Slopes, Flooding of Culverts, Time of Emptying Irregular Shaped Vessels, Continuous and Sinuous flow in Pipes, Water Hammer in Pipes, Cavitation in Centri- fugal Pumps. The mode of treatment continues on the same lines as before ; simple, straightforward, easily remembered methods have been used as far as possible. A more elegant treatment might have been adopted in many instances, but unfortunately such a treatment often requires more mathematical knowledge than many readers possess, hence it is a "closed book" to the majority of engineers and draughtsmen, and even to many who have had a good mathematical training in their student days. There are comparatively few Engineering problems in which the data are known to within, say, s per cent., hence it is a sheer waste of time for the Engineer in practice to use long, complex methods when simple, close approximations can be used iii a fraction of the time. For higher branches of research work exact, rigid niethods of treatment may be, and usually are, essential, but the number of Engineers who require to make use of such methods is very small. Much of the work involved in writing and revising this viii Preface to the Eighth Edition. Edition has been performed under very great difficulties, in odd moments snatched from a very strenuous life, and but for the kind and highly valued assistance of Mr. R. H. Duncan in correcting proofs and indexing, this Edition could not have been completed in time for this Autumn's publication. JOHN GOODMAN. The University of Leeds, August, 1914. CONTENTS CHAP. JAGE I. Introductory i 11. Mensuration .20 III. Moments ... 50 IV. Resolution of Forces . . . , < 106 V. Mechanisms . .... ...... iig VI. Dynamics of the Steam-engine .... -179 VII. Vibration . . . 259 VIII. Gyroscopic Action . . 277 IK. Friction . . . ... 284 X. Stress, Strain, and Elasticity ... ... 360 XI. Beams ... 429 XII. Bending Moments and Shear Forces . . 474 XIII. Deflection of Beams 506 XIV. Combined Bending and Direct Stresses . . . 538 XV. Struts 550 XVI. Torsion. General Theory 571 XVII. Structures S93 XVIII. Hydraulics 637 XIX. Hydraulic Motors and Machines . . . .691 XX. Pumps 738 Appendix 781 Examples 794 Index ... 846 ERRATA. Pages 34 and 35, bottom line, " + " should be ^' - ." Page 79, top of page, " IX." should be " XI." „ 102, the quantity in brackets should be multiplied by " — ." „ 203, middle of page, "IX." should be "XI." St 0-2' bottom, " 45° " should be " 90°." top, " sin Ra " should be " sin 20." top, " A " should be " Ao." top, " h* " should be " h." top, " /i " should be " //„.' top, " L " is the length of the suction pipe in feel. 247, line 12 from top, should be ' 16 395. . , 10 663, , 6 »» J . 7 676, , M 747. . 5 MECHANICS APPLIED TO ENGINEERING CHAPTER I. INTRODUCTOR Y. 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. Units. Mass (M). — Unit, one pound. I pound (lb.) = 0'454 kilogramme. I kilogramme = 2"2046 lbs. I hundredweight (cwt.) = SO'8 kilos. I ton = 1016 ,, (tonneau or Millier). I tonneau or Millier = 0'984 ton. Space {s). — Unit, one foot. t foot = 0-305 metre. i mile = l6o9'3 metres. I metre = 3'28 feet. i kilometre = I093'63 yards. [ inch = 25'4 millimetres. = 0'62I mile. I millimetre = o'0394 inch. i sq. foot = 0-0929 sq. metre. I yard = 0'9I4 metre. I sq. metre = 10764 sq. feet. I metre = l'094 yards. I sq. inch = 6'45I sq. cms. B Mechanics applied to Engineers I sq. I sq. mm, cm. r sq. metre I atmosphere I lb. per sq. inch = O'00l55 sq. inch. = O'ISS sq. inch. = o'ooio76 sq. feet. = 10764 sq. feet. = I'igS sq. yards. = 760 mm. of mercury. = 29-92 inches of mercury. = 33'9o feet of water. = I4'7 lbs. per. sq. inch. = I '033 kg. per sq. cm. = 0-0703 kg. per sq. cm. = 2-307 feet of water. = 2-036 inches of mercury. = 68970 dynes per sq. cm. I lb. per sq. foot = 479 dynes per sq. cm. I kilo, per sq. cm. = 14-223 lbs. per sq. inch. I cubic inch = 16-387 c. cms. I cubic foot = 0-0283 cubic metre. I cubic yard = 0-7646 c. metre. I c. cm. = 0-06103 c. inch. I c. metre = 35-31 c. feet. (See also pp. 4, 9, 10, 11, 19.) 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 when moving in a straight line 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 body is said to have unit speed when it traverses unit space in unit time. „,,.,, ,, space traversed (feet) s Speed (m feet per second) = — — : ; f-r — ~ = - time (seconds) t X 3 * Tim& "in seconds Umforiwsp. Fig. 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 1 i. 3 4- 5 Tune in seconds Varying sjieett Fig. a. 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 h we find that 9 ft. would have been traversed in 5-2 — 2*3 = 2-9 sees., or the speed at 3 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 (z/). — 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 ) "~ time (seconds) s . V = -J OT s — vt I ft. per second = o"3o5 metre per second „ „ = o"682 mile per hour „ „ = IT kilometre per hou: I metre per second = 3'28 ft. per second J ( = o'o^28 ft. per second I cm. per second < i u ^ ( = 0*0224 miles per hour ., u f = I '467 ft. per second I mile per hour < ^ ' f , '^ ( = 0-447 metre per second I kilometre " \ = °'^'l ^^- , ( = 0-278 metre „ Angular Velocity (u), or Velodty 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 J the direction of spin must also be specified. 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 6° a circular arc subtending an angle 6° is 2irr X -^-5, where ir 360 is the ratio of the circumference to the diameter {2r) of a circle and 6 is the angle subtended (see p. 22). Then, when the arc is equal to the radius, we have — • 2irrO n ^60 ,□ —T- = >' e=i_ = 57-296° 360 2ir >" ' Introductory. 5 Thus, if a body be spinning in such a manner that a radius describes 100 degrees per second, its angular velocity is — 0) = = i*7S radians per second 57-3 It is frequently convenient to convert angular into linear velocities, and the converse. When one radian is described per second, the extremity of the radius vector describes every second a space equal to the radius, hence the space described in one second is wr = v, ox <a = —. r Angular velocity in radians per sec. = "near velocity (ft. per sec.) radius (ft.) The radius is a space quantity, hence — _ J _ I "^ ~ Js~ 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 (/,) 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 I _ increase of velocity in ft. per sec, in a given time in feet perj ~ time in seconds sec. per sec.J f _ ^2 - ^1 _ V ^'~ t ~1 hence v =fj, 0XVi-v^=fJ . . , . (j.) Mechanics applied to Engineering. where »a is the velocity at the end of the interval of time, and »! at the beginning, and v 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-i is zero, the mean velocity over any inter- val of time will be — , and the 2 spate traversed in the interval will be the mean velocity X time, or — s = —t = •'-5— (see equation i.) and/. = - Acceleration in feet per sec. per sec. = constant X space (in>/) (time)^ (in seconds) When the body has an initial velocity v^, the mean velocity during the time t is represented by the mean height of the figure oabc. t a- 3 Time irv s0conds Fig. j. Mean velocity = ■ ' = —^ — 1 = z-^ 4--ii 2 2 2 (see equation i.) The space traversed in the time t — . = (.+4^. (ii.) aii.) which is represented in the diagram by the area of the diagram oabc. From equations i. and ii., we get — v^ Substituting from iii., we get — ^" ■(;-)/•'=/•• '-* = 2/,J or v^ = z/,2 -f 2/.J 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 lei'ter A, the height through which the body has fallen, instead of s. When the body starts from rest, we have Vi = o, and z'j = » ; then by substitution from above, we have — V = ij 2gh = 8'o2 ij h .... (iv.) Momentum (Mo). — If a body of mass M * move with a velocity v, the moving mass is said to possess momentum, or quantity of motion, = Mv. Unit momentum is that of unit mass moving with unit velocity — Mo = Mv = — - Impulse. — Consider a ball of mass M travelling through space with a velocity z/j, 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^ to V2- The momentum before the blow = M»i „ after „ = Mw^ The change of momentum due to the blow = M{vz — »i) The effect of the blow is termed an impulse, and is measured by the change of momentum. Impulse = change of momentum = M(Vi — v^) 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 wq 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 \!ae. force acting on the body. Hence the momentum may be changed gradually from M.-ffl\ to MaZ/j by a force acting for t seconds. Then — ' For a rational definition of mass, the reader is referred to Prof. Kar Pearson's " Grammar of Science," p. 357. 8 Mechanics applied to Engineering. Yt = M(z/si - »,) , „ _ total change of momentum time But ^' ~ ^' =/, (acceleration) (see p. 5) hence F = M/, = -r- 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 jP"'™ \ for one second, will change its velocity by °"^ (Simetre) P" '^^°"'^' ^"'^ ^' *^™^*^ °°^ {d?ne!^^'' 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 t/ie direction of the force ; . . . zho, a body will remain at rest, or, if in motion, will move with a uniform velocity in a straight line unless acted tipon by some extei-nal 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 ' The poundal unit is nevei used by engineers. Introductory. 5 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 pf 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 (£), 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 W = M^ We have shown above that — Force = mass X acceleration ' ' Expressing this in absolute units, we have — Weight or force (poundals) = mass (pounds) x acceleration (feet pei 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 lo 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-.' In centimetre- In foot-pounds, sees. grammes, sees. The equator 32'09i ... gyS'io London 32'i9l ••. 9^i'i7 The pole 3Z'2SS — Q^S"" 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 unif 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 —5- . purposes, we shall adopt the engineer's unit of one pound weight, 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 or — , , , , mass in pounds , ... weight or force (pounds) = -— x acceleration in ft. -sec. per sec. 32 2 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 il 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. ' 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 ' 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. Kilogiam, 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. „ total work done Ff Power = -. i — 5 — r- = ■— time taken to do it / The dimensions of the unit of power are therefore -—. The unit of power commonly used by engineers i^ 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 = 33jOoo foot-pounds per minute = 7*46 X 10° ergs per second. I French horse-power = 32,500 foot-pounds per minute = 736 X 10^ ergs per second. I horse-power = 746 watts I watt =10' 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 ' 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. 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 j in order to prevent this motion of trans- lation, another force, Fu 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. P,Q ^ 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 and contra-clockwise. 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 = F« The dimensions of a moment are therefore — ^. C' ' See Hicks's " Elementary Mechanics." Introductory, 13 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. — The corresponding point in a geometrical surface which has no weight is frequently termed the centroid ; such cases are fully dealt with in Chapter III. Suergy. — 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,i 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,* 772 foot-lbs. of work have to be expended in order to generate one thermal unit. Efficiency of a Machine. — The efificiency of a machine is the ratio of the useful work got out of the machine to the gross work supplied to the machine. _„. . work got out of the machine Efificiency = — = — 2 — — 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. _ ^ „ . work supplied to the machine Counter-efficiency = -, — &£ ^-^ -^. — work got out of the machine ' To be strictly accurate, we should also say light, sound, electricity, etc. ' By far the 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. igo, pp. 301-422) from 32° F. to 212° F., which is equivalent to about 773 at 39° F. and 778 at 60° 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 m 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 — W/4 = — , or 2g 2 This quantity, , is known as the kinetic energy of the 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 defined as the " deadness of matter." Moment of Inertia (I). — We may define 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 ui, the linear velocity of the body will be z/ = ar, 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- tum 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, Yr = T, M GrooveAjUiUey considered^ 0£ -*/» Fig. 5. 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 r force n? acting at a radius . The force P actmg on the mass M gives it a linear acceleration /„ where P = M^, or P • I /, = -—. The angular velocity (o is - times the hnear velocity, M T hence the angular acceleration is - times the linear accelera- tion. Let A = the angular acceleration ; then — r Mr M/-2 M^ , , . . twisting moment ^ or angular acceleration = 5__ — _ — _ mass X (radius)" 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. Mi, M^, etc., at radii »-i, ^2, etc., respectively ; then the above expression becomes — A = (Min' + M^Ta" + Mar,^ +, 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 = M1+M2+M3, etc. ; then the moment of inertia of the body, I, = Mk^ = (Miz-i" + M^r^^ 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 ' 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 tuio forces. i6 Mechanics applied to Engineering. the same capacity for possessing energy, due to rotation about that axis. Representation of Displacements, Velocities/ Accelerations, Forces by Straight Lines. — Any I displacement] ,' ■ I is fully represented when we state its magni- force J tude and its direction, and, in the case of force, its point of application. Hence a straight line may be used to represent any Idisplacemenfj velocity r ^^ length of which represents its magni- force j tude, and the direction of the line the direction in which the force, etc., acts. I displacements! Velocities 1 • • accelerations ' "^^^^^ ^' ^ P°''''' ""^^ forces / be replaced by one force, etc., passing through the same point, which is termed the resultant force, etc. (■displacements! If two P^l°"'ies. not in the same straight line, 1 accelerations , 6 > I forces Fig. 6. meeting at a point a, be represented , by two straight lines, ab, ac, and if two other straight lines, dc, hd, be drawn parallel to them from their extremities to form a parallelogram, abdc, the diagonal of the parallelogram ad which passes through that point I displacement \ acceleration I ^ magnitude force ) and direction. Hence, if a force equal and opposite to ad act on the point in the same plane, the point will be in equilibrium. It is evident from the figure that bd is equal in every Including angular velocities or spins. Introductory. 17 respect to ac; then the three forces are represented by the three sides of the triangle ai, bd, ad. Hence we may say that if three forces act upon a point in such a manner that they are equal and parallel to the sides of a triangle, the point is in equi- librium under the action of those forces. This is known as the theorem of the " triangle of forces." Many special applications of this method will be dealt with in future chapters. The proof of the above statements will be found in all elementary books on Mechanics. Hodograph. — The motion of a body moving in a curved path may be very conveniently analyzed by means of a curve called a "hodograph." In Fig. 7, suppose a point moving along the path P, Pj, Pa, with varying velocity. If a line, op, known as a "radius vector," be drawn so that its length represents on any given scale the speed of the point at P, and the direction of the radius vector the direction in which P is moving, the line op completely represents the velocity of the point P. If other radii are drawn in the same manner, the curve traced out by their extremities is known as the "hodograph" of the point P. The change of ve- locity of the point P in pass- ing from P to Pi is represented on the hodograph by the distance ppi, consisting of a change in the length of the line, viz. q-^p-^ representing the change in speed of the point P, and/^i the change of velo- city due to change of direction, Fig. 7. if a radius vector be drawn each second ; then //i will represent the average change of velocity per second, or in the limit the rate of change of velocity of the point P, or, in other words, the acceleration (see p. s) of the point P ; thus the velocity of / represents the acceleration of the point P. • If the speed of the point P remained constant, then the length of the line op would also be constant, and the hodo- graph would become the arc of a circle, and the only change in the velocity would be the change in direction pq-^. Centrifugal Force. — If a heavy body be attached to the end of a piece of string, and the body be caused to move round 1 8 Mechanics applied to Engineering. in a circular path, the string will be put into tension,the amount of which will depend upon (i) the mass of the body, (2) the length of the string, and (3) the velocity with which the body moves. The tension in the string is equal to the centrifugal force. We will now show how the exact value of this force may be calculated in any given instance.' Let the speed with which the body describes the circle be constant; then the radius vector of the hodograph will be of constant length, and the hodograph it- self will be a circle. Let the body describe the outer of the two circles shown in the figure, with a velocity v, 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 Ai, and the corresponding radius vector to OPj; then the line PPj p,e J represents the change in velocity of A while it was moving to Ai. (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 lad of OA = R; then— But OP was made equal to the velocity of A, viz. v, and OA is the radius of the circle described by the body. Let velocity of P v V = R or velocity of P = R ' For another method of treatment, see Barker's " Graphic Methods o( Engine Pesi{rn." Introductory. 19 and acceleration of A = ^ and since force = mass x acceleration we have centrifugal force C = ^- . . , . ^ W»2 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 = <dR hence C = Mw^R W<o=R or C = g 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 — 5- r , . „ J , pounds X (feet)" work in ft.-poundals = - — ; ,^,„ '' (seconds)^ work in ergs = g^ams X (centimetres)^ (seconds)^ I pound = 453"6 grams I foot = 3o"48 centimetres Hence — I foot-poundal = 4S3'6 X 3o"48' = 421,390 ergs and I foot-pound = 32-2 foot-poundals = 32-2 X 421,390 = 13,560,000 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 sraface 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- ^T I rating line (J); (2) on the length of the U i side shift of the generating line (d). _X_ 1 Thus a statement of the area of a given — I > surface must involve two length quantities, Fio. 9. / 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)' 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 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; (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 >< length units = (length units)' /i f 1 ,'' / < 1-- FlG. 10. — * 22 Mechanics applied to Engineering. Lengths. Straight line. Circumference of circle. Length of circumference = ird /^ ^\ = 3 •14161/ / \ 6*2832r Y / The last two decimals above may usually V^ _...^ be neglected ; the error will be less than \ in. ■* dj * on a lo-ft. circle. Fis. II. Length of arc = —7- 2irr0 rO or =• ^ 360 57-3 For an arc less than a semicircle — 8C — C Fioril Length = — ^ ° approximately Arc of ellipse. d \^ Length of circumference approx. /^ A 4D - d) = ir^+ 2(D -d)- ^^ '- r,a. x3. ^ V(D+rf)(D + 2rf) Mensuration. 23 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 ' in. if necessary. 1000000 The mathematical process by which the value of ir 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 bscribed in a circle. The perimeter of the outer polygon is greater, and that of the inner less, than the perimetpr of the circle. The greater the number of sides the smaller is the difference. The value of ir 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. The length of the arc is less than the length of the Q circumference in the ratio —^. 360 Length of arc = -ira X -— - = -pr- 360 360 The approximate formula given is extremely near when A is not great compared with C„-; 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). No simple expression for the exact value of the length of an elliptic arc can be given, the value opposite is due to Mr. M. Arnold Pears, of New South Wales, see Trautwine's " Pocket-book," 18th edition, p. 189. 24 Mechanics applied to Engineering Arc of parabola. Length of arc = 2 (approximately) Fig. 14. Irregular curved line abc. Set ofT 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 -— . d same number along tangent. FrG. 15. Areas. Area of figure = Ih Triangles. Area of figure = bh Equilateral triangle. Area of figure tbv^ = 0-4., 433^ Fio. 18. Mensuration No simple expression can be given for the parabolic arc — a common approximation is that e opposite page. The error is negligible when h is pared with b, but when h is equal to b the error about 8J per cent. 25 length of a ;iven on the small corn- amounts to 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. 2 4 26 Mechanics applied to Engineering. Triangle. Let s = a ■\-h -\- c A^i Area of figure = ,Js(s — d){s-b)(s—c) FrG, ig. Quadrilateral, Area of figure = bh Fic. 9a Trapezium, -A. Area of figure = ( j li Fig. «. Irrtgular straight-lined figure. , 6 Area of figure = area ahdef — area 3frf or area of triangles (acb-^acf-\-cfe->rced) Fig. »9* Mensuration. 27 The proof is somewhat lengthy, but perfectly simple (see " Longmans' Mensuration," p. 18). Area of upper triangle = — -' 2 „ lower triangle = — - 2 both triangles = b( ^l±Jh \ = bh 2 Aiea. of parallelogram = 61/1 Area of triangle = \ ~ " 2 Area of whole figure = (^ - '^1 + ^'^■)^ = iA±m 2 2 Simple case of addition and subtraction of areas. 28 Mechanics applied to Engineering. Area of figure = izr^ = 3-1416^' or — = o-i^SAiP Sector of circle. Area of figure = — ^ 360 Fig. 34. Segment of circle. Area of figure = f C^^ when h is small = A(6C. + 8C,) nearly Fig. 25. Hollow circle. Area of figure = area of outer circle — area of inner circle = TrTj" — wTi* = T(r,» - r^) = irr^ or = cySSifj' or = '' ^V , i.e. mean cir- 2 cumf. X thickness Fig. 36. 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 . br IS — J but the sum of all the bases equals the circumference, or %b = 2irr, hence the area of all the triangles put together, . , 2irr . r , ue. the area of the circle, = = '^r'- The area of the sector is less than the area of the circle in 6 Trr^d the ratio —r-, hence the area of the sector = — j- ; if fl be the 300 360 '^ angle expressed in circular measure, then the above ratio becomes — • The area = — ^ 2 When k 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 r^ for r^ — 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 — Ttid^ + (fi) Mean circumference = — ' ■ 2 thickness = — ^ 2 product = ^-^^^ X ^ = ^(4= - d^) 30 Mechanics applied to Engineering. Ellipse. Area of figure = Trr^r^ or = —d-id^ 4 Fig. 37. Parabolic segments. '^ Area of figure = |BH I i.e. f (area of circumscribing rectangle) Fig. 29. Area of figure = f area of A aie Make de = ^e area of figure = area of A aid Mensuration. 31 An ellipse may be regarded as a flattened or an elon- gated circle j hence the area of an ellipse is |^®* \ than igreaterJ the area of a circle whose diameter is the /™?J°'^1 a^is of an Immorf ellipse {J}, in the ratio } J ;° J- Area = ^\ 4' = -'^i<^2, or !^' X ^ = ""-M^ 4 "2 4 4 t^i 4 From the properties of the parabola, we have — H= B V B B* area of strip = h . db = ( — - ) V B* ^ r- rfS-.^-^ X k ¥ k -«— . i s/ db Tig. 28a. b = B area of whole figure = |HB l>\db = ^X?r B* t The area ac^^has been shown to be fHB. Take from each the area of the A «^^, then the re- mainder abt: = 5 the A i^be ; but, from the properties of the para- bola, we have ed = \eb, hence the area abc = § area of the cir- FlG. 29a. 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 = 2X1 area A abc = | area A abc. By increasing the height of the A abc to g 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 o.bd Surfaces bounded by an irregular curve. Area of figure = areas of para- bolic segments (a — b-\rC-\-d-\-e) + areas of triangles {g + K). Fig. 32. Mean ordinate method. Area of figure = (/« + Aj + ^ + hi +, etc.)a: Fio. 33. Mensuration. 33 The area abc = f 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 ^ + ^1 -)- ^2 + h^, etc., is most ^ ' ^1 I K^ /ij I and so on. Fig. 33fl. 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. Area of figure = -{k-\- 4^1 ■\-2hs, + 4/^3 + 2hi + 4/^6 + 2-4g + 4/4, + 2,48 + A^h +^a) The end ordinates should be obtained by drawing the mean lines (shown broken). If they Fig. 34. are taken as zero the expres- sion gives too low a result. Any odd number of ordinates may be taken ; the greater the number the greater will be the accuracy. Curved surface of a spherical indentation. I Curved surface Fig. 34*. 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 aieg = j/ '' "*" ' -' j . . . . (i,) „ .3« = 4^) .... (ii.) „ abceg= ^(Ai + 2^^ + ^^) . (i.+ii.) ^/f-' „ abcjg= 2:c(^L±i5)=;c(/5, + >^3)(iii.) I*' Area of A g'^ = (i-) + ("•) — (iii.) X 2 2 = -{K-V'iK-'rh^ — x{h^-\-h^ Fio. 34a. (2/^ - h^- h^ (iv.) Area of parabolic ) _ 4/- v 2X, segment gcdef \ " 3^'^-/ " y (2'^2 - fh- 4) ■ (v.) Whole figure = (iii.) + (v.) = x{hy, + h^) ->r^{2fh - fh- h^ = ^(/i, + 4/i, + /g If two more slices were added to the figure, the added area X would be as above = -{h^ + 4/^4 + h^, and when the two are X added they become = -(/^i + \li^ + 2^3 + 4/^4 + h^. The curved surface of the slice = ^irr^s By similar triangles we have S _ R ly- r^ Substituting the value of S we have Curved surface of slice = ziiRZy „ ,, indentation = zttRY Expressing Y in terms of R and d we have H-Y ^ Fig. 34^. tRY = 2!rR(^R +/^R^ - -) 36 Mechanics applied to Engineering. Surfaces of revolution. Pappus^ or Guldiwis' Method. — Area of surface swept out by ^ the revolution of the line > = L X zirp defaboMt the axis ab ) Length of line =• L Radius of c. of g. of line defi _ considered as a fine wire y ~ ^ 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 = 47rr^ 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. I .1 Area of curved surface of cone = wrh Fig. 37- Mensuration. 37 The area of the surface traced out by a narrow strip of i<.„„i.i, /4 and radius po = 2t/|,po\ , length \, f^" T"], and so on. Area of whole surface 1 = 2t(/oPo + kp\ +, etc.) -^ 1 ^ = 2ff(each elemental length of u/^ I ^x wire X its distance from axis 'F^' y-~-'-'y-~---~-'-\ of revolution) | ° _a ^ = 2'7r(total length of revolving wire I '"* X distance of c. of g. from j axis of revolution) (see p. | 58) Fig. 35a. = (total length of revolving wire X length of path described by its centre of gravity) = Lzirp 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 re to a circular arc, from the centre of the circle is y = — = Pi a where r = radius of circle, t: chord of arc, a length of arc (see p. 64). In the spherical surface a = L = irr, c = 2r, p = — = — 2r Surface of sphere = irr . 2t . — = 47ir* Length of revolving wire = \, = h radius of c. of g. „ „ = p = - hlirr surface of cone = = m-h 38 Mechanics applied to Engineering. Hyperbola. Area of figure = XY log.r log, = 2-31 X ordinary log Fig. 3» Area of figure = XY - X.Yi Mensuration. In the hyperbola we have — XY = X,Yi = xy u XY hence v = — X dx area of strip = y . dx = XY — Area whole figure °n r lie > = XY re j j « = Xi dx X x= X = XY (log. X, - log. X) I = XY log. Xi Using the figure above, in this case we have — YX» = YiXi" = yxT YX» hence y = — rr- YX- area of strip = y .dx = —^dx = YX"x-"dx J* = Xi ,-v 1 - n _ Vl - «\ x-"dx = YXH \_„ ) jc = X • _ YX"Xi'-"- YX I — n But Y,Xi" = YX» Multiply both sides by Xj'"" then Y,X, = YX"X,'-- Substituting, we have — Area of whole figure = iii^=i lii I — « YX - YiX, or = 1 — 1. 40 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. ^ I Fin. II. Ur Volumes. Let A = area of the end of prism ; / = length of prism. Volume = /A Parallelopiped. Volume = Idt Hexagonal prism. Volume = 2'598.fV Cylinder. Volume = "^ — = o-yS^aTV 4 ^ Mensuration. 41 Suspend a knitting-needle or a straight piece of wire or 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 tilltheybalance ; then mea- sure the distances x and^. Then A^ = '&y or B=: — y 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 Fig. 40. Area of hexagon = area of six equilateral triangles = 6 X o*433S'' (see Fig. 18) volume = 2'598SV or say 2"6SV Area of circular end = -d^ 4 ■KdH volume = 4 42 Mechanics applied to Engineering. Prismoid. Simpsoris Method. — Volume = '(A1+4A2+2A8+4A44-AJ) 3 and so on for any odd number of sections. Contoured volume. N.B. — Each area is to be taken as in- cluding those within it, not the area between the two contours. A3 is shaded over to make this clear. &ddntorCber of £yuidCstant slices FjR. 45- Solids 0/ revolution. Method of Pappus or Guldinus.— Let A = area of full-lined surface ; p = radius of c. of g. of surface. Volume of solid of revolution = 2irpA N.B. — The surface must lie wholly on one side of the axis of revolution, and in the same plane. This method is applicable to a great number of problems, spheres, cones, rings, etc. Fig. 47. Mensuration. 43 Area of end (or side) = -(h^ 4. 4^^ + 2,^3 +, etc.) (see p. 34), where h^^ k^, etc., are the heights of the sections. X Volume = -{hJ-\- iji4 -V 2/^3/+, etc.) = -(Ai + 4A2 + 2A3 +, etc.) 3 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 _ „ ^^o ( volving round the axis at aj " / f* distance p, ' \ ^' Ditto ditto a^ and pi = fli X 'iirp^ \ < and so on. V Whole volume swept out by all^ the elemental areas, a^, a^, etc., when revolving round the axis at their respective distances, poj Pi, etc. = 2ir(each elemental area, a^, a^, etc. x their respective distances, po, Pi, from the axis of revolution) = 2ir(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 2irp = 27rpA But 27rp 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. Fn. 46a. = 27r(a„po + aipi +, etc.) 44 Mechanics applied to Engineering. Sphere. Volume of sphere = — , or ^irr^ Volume of sphere = | volume of circum- scribing cylinder Hollow sphere. Volume of ■> _ f volume of outer sphere — External diameter = a. hoUow Sphere/ " \ volume of inner sphere Internal diameter :: Fig. 48. _ -n-d.^ _ ltd? 6 6 = JW d?) Slice of sphere. /T "^ f "^ ..^ — '■■^., ,/■' Volume of slice = -{3R(Yj,''-Yi'')-Y,»-l-Y,n 3 N.B. — 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. I .--ttT---,. V ; i -^^ ! A^ Special case in which Y, = R. / Volume of slice = -(2R» - 3RY1' + Yi') Fig. 50. Mensuration. 45 Sphere. — The revolving area is a semicircle of area The distance of the c. of e. 1 Ar , , ,, from the diameter f = '' = ^ ('^^ P- ^^) Ar ■K(P volume swept out = 27r X ^L, x — - = \iti^ = _ 3T 2 ^ 6 or by Simpson's rule — ' Volume of sphere = - (o + 47r;i + o) == iwr" Volume of elemental slice = icc^dy = 7r{R2 - (R2 +/ - '2.^y))dy — ir(zRy — y'^)dy Volume of whole 1- = slice ry=~-i (2Rj Ky-f')dy -t 2R/ _/ y=Y, y=Y, _s_ ±rJi_ = J zRC^iri^) _ Y,3-Y.n Fic. 4g<z. The same result can be obtained by Simpson's method — Volume = -(irCj2 + /[ttC' + ■rCi') \4i For X substitute (^ „ (2 R;/ —y) with the proper suflSxes. The algebraic work is long, but the results by the two methods will be found to be identical. 46 Mechanics applied to Engineering. *- /? Special case in which Yj = o. Volume of slice = -(sRY^^ - Y/) When Ya = R, and Yj = o, the slice becomes a hemisphere, and the^ Volume of hemisphere = -(2R') Fig. 51. which is one-half the volume of the sphere found by the other method. Paraboloid. Volume of\_^„2„ ^a paraboloid /- 2^ "■"'^ 8° " = iSyR'H, or o-39D''H 's \ volume circumscrib- ing cylinder Fig. 51. Cone. Volume of cone = -R'H 3 =J!:d»h 12 = ^ volume circumscrib- ing cylinder Fig. sj. Mensuration. {Continued from page 45.) For the hemisphere it comes out very easily, thus— R R «»= R»- R' Volume = ^^{o+,r(4R2 - R^) + ,rR^} 6 = |7rR^ Fio. 51, 47 From the properties of the parabola, we have — R" H H Volume of slice = volume of solid = -J, irRV/ r R Volume of slice = irT^dh = volume of cone = # 48 Mechanics applied to Engineering. Pyramid. ., B,BH Volume of pyramid = — — = — , whenB,=B=H 3 = i volume circumscrib- «^--— ^f- ->'■' ing solid F:g. 54- Slightly tapered body. 'a' 'I™ Mean Areas Method. — ^ ''?..V.V.:^/;:.':1| volume of body=(^^t^^^t^)/(approx.) ' ill = (mean area)/ SI' Fig. ss- Ring. wd' Volume of ring = — X tD = 2'^i(PY) 4 Fig. s6. Weight or Materials. Aluminium .. 0'093 lb. per cubic inch, | Brass and bronze • o'3o ,, i Copper • 0-32 Iron — cast ., 0-26 „ „ wrougtit ■ 0-278 ,, Steel .. 0-283 Lead .. 0-412 „ Brickwork .. 100 to 140 lbs .per cubic foot. Stone .. 150 to 180 n i> Mensuration. 49 This may be proved in precisely the same manner as the cone, or thus by Simpson's method — Volume=^jO+4(?X?^)+BxB. 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 T^HMnstead of ,^H3. 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 Weight of Materials. Concrete Pine and larch Pitch pine and oak Teak Greenheart ... 130 to 150 lbs. per cubic foot. 301040 „ 40 to 60 ,, ,, 4StoS5 65 to 75 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 Rj are attached to the same shaft, so that they rotate together. We shall assume that there is no friction on the axle. Fio. 57. n^ -R. — ' J 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 j we shall then have — ■ Moments. 5 1 The work done by pulling the cord = P x 2irR „ „ in lifting the weight = W X zttRj These must be equal, as it is assumed that no work is wasted m friction; hence — PairR = W2irR, or PR = WRi 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 De 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. r force (/) \ rirst Moments.— The product oi & < mass («;) f \ volume {v) ) the length of its arm /, viz. <^ ^/ ^> is termed ihe first moment force "^ of the < „ >>, or sometimes simply the moment. volume \ i force A statement of the first moment of a -s „__„ \- must I area \ volume f force units X length units. consist of the product of \ "^^^^ "'?[*« ></^'^g* "'?''^- '^ I 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. 52 Mechanics applied to Engineering. /force (/) \ Second Moments. — The product of a -; ^g^ (j^ \ ^^ (^volume (v)) the squarq or second power of the length {I) of its arm, viz. (fl\~\ (force I ^^f ( , is termed the second moment of the } ^^^* \ . The ^vP J (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. C force \ A statement of the second moment of a < ™*®^ > must 1 area ( I volume I ( force units X (length units)'! ,1 mass units X (length units)*, consist of the product of < ^rea units X (length units)". \ volume imits x (length units)'. First Moments. Levers. <r-ljr->^ — ^- ■ *S «5 -"i Fig. 60. T' Fig. St. Cloclcwise moments about the point a. Contra-clockwise moments about the point a. lUjt + wj. = a'iA = a/,4 Moments. 53 Reactidh R at fulcrum tf, z.f. the resultant of all the forces acting on lever. Remarks. o'l+w.+a'a+a'. To save confusion in the diagrams, the / has in some cases been omitted. In every case the sufBx of / indicates the distance of the weight w bearing the same suffix from the fulcrum. w^—w^—w. i. (_ 54 Mechanics applied to Engineering. 1(3 ITS rr Ai^ Fig. 62. <■—- ij"»* -- ? -» -«? i^-f-.: li'lG. 63. rSnnnnnoo Clockwise moments about the point a. W-or 2 2 If w = dis- tributed load per unit length, w/= W W-or — 2 2 W/ W = weight of long arm of lever W, = weight of Contra-clockwise moments r about the point a. =wJi+w.J.i+wJi = w^l^ = Wi^ + wA or — i — I- Wi/i 2 = W,/,+w/„-|-a:'3/, /= distance of c. of g. of long arm from a /i = distance of c. of g. of short arm from a = P/ /i = distance of c. of g. of lever from a orl P^J_ is Reaction R at fulcrum a, i.e. the resultant of all the forces acting on lever. Moments. 55 Remarks. TO, + K'j — JOj + TOj — 01/5 H/, + W N.B.— The unit length for w must be of the same kind as the length units of the lever. Wi+W.+W Wa + W.+a/j+W This is the arrangement of the lever of the Buckton testing machine. Instead of using a huge balance weight on the short arm, the travelling weight OTj has a contra-clockwise moment when the lever is balanced, and the load on the specimen, viz. KI3, is zero. As w, moves along its moment is decreased, and consequently the load w, is increased. When a/, passes over the fulcrum, its moment is clockwise ; then we have W/ x wj„ = W,/, + w,l,. W+a;,-P This is the arrangement of an ordinary lever safety-valve, where P is the pressure on the valve. The weight of the levers may be taken into account by the method already shown. S6 Medianics applied to Engineering. Fic. 68. Clockwise moments about the point a. ii'Ji Contra- clockwise moni«?nL5 r about the point a. = Wji + Wji ■w4i + w/s = Wj/i + w/. W/ + w4^ W = weight of horizon- tal arm / = distance of c. of g. from a Fig. 70. tia a W2= weight of long curved arm of lever Wi = ditto short arm = W.A + uuU 12= distance of c. of g. of long arm from a li = ditto short arm /<= perpendicular distance of the line of w^ from a W = weight of body I = perpendicular distance of force W from a Moments. Reaction R. Remarks. 57 ■ In all these cases it must be found by the paral- lelogram of forces. It should be noticed that the direction of the resultant R varies with the position of the weights ; hence, if a bell-crank lever be fitted with a knife- edge, and the weights travel along, as in some types of testing-machines, the resultant passes through the knife-edge, but not always normal to the seating, thus causing it to chimble away, or to damage its fine edge. Fig. 68a. The shape of the lever makes no difference whatever to the ! leverage. Consider each force as acting through a cord wrapped round the pulleys as shown, then it will be seen that the moment of each force is the product of the force and the radius of the pulley from which the cord proceeds, i.e. the perpendicular distance of the line of action of the force from the fulcrum. 58 Mechanics applied to Engineering. Centres of Gravity, and Centroids. — We have already given the following definition of the centre of gravity (see p. 13). 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 ; 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. From this definition it will be seen that, as the particles of any body are acted upon by a system of parallel forces, viz. .1. "■^S- ■Jiof. w. <> gravity acting upon each, the algebraic sum of the moments of these forces about a line must be zero when that line passes through the c. of g. of the body. Let the weights Wi, Wj, ^'G' 72- be attached, as shown, to a balanced rod — we need not consider the rod itself, as it is balanced — then, by our definition of the c. of g., we have W,L, = W,Lj. In finding the position of the c. of g., it will be more convenient to take moments about another point, say x, distant /j and ^ from Wj and Wj respectively, and distant /, (at present unknown) from the c. of g. Wi4 + WaLa= R/, = (W, + W,)/, . _ W/i + Wa4 ' W, + W, If we are dealing with a thin sheet of uniform thickness and weighing K pounds per unit of area, the weight of any given portion will be K« pounds. Then we may put Wi = Kai, and W, = Kfflj ; J / _ K(gi/i + aa4) _ g/i + aj^ '"'^ - K(«. + a,) A— or, expressed in words — distance of c. of g. from the point x the sum of the moments of all elemental su rfaces about x area of surface or = Moments. 59 the moment of surface about x area of surface where A = «i + a^ = whole area. In an actual case there will, of course, be a great number of elemental areas, a, a^, a^, a^, etc., with their corresponding arms, /, /j, 4> 4> etc. Only two have been taken above, they being sufficient to show the principle involved. When dealing with a body at rest, we may consider its whole mass as being concentrated at its centre of gravity. When speaking of the c. of g. of a thin weightless lamina or a geometrical surface, it is better to use the term " centroid " instead of centre of gravity. Position of Centre of Gravity, or Centroid. Parallelograms. Intersection of diagonals. Height above base ab = — 2 In a symmetrical figure it is evident that the c. of g. lies on the axis of symmetry. A parallelogram has two axes of symmetry, viz. the diagonals ; hence the c. of g. lies on each, and therefore at their intersection, and as they bisect one another, the intersection is at a height — from the base. F a- i IG. 73. 6o Mechanics applied to Engineering. Triangle. H I Fig. 74. Intersection of ae and bd, where d and e are the middle points of ac and be respectively. Height above base be = — ,. • . V. 2H height above apex a = ^ — Triangle. a Distance of c. of g. from b « + S = ^/=: Ditto from e = ef = 3 z + S Fig, 74a. Trapezium. <?<-;:v5r;;.::^ * Intersection of a3 and ed, where a and b are the middle points of S and Si, and ed = Si, /^ = S. J Height above) _ H(2S + SQ base Hi 5 3(5 + sj depth below) _ H(S + 2S1) top H, ] 3(s + Si) 'I Moments. 6i Conceive the triangle divided up into a great number of very narrow strips parallel to one of the sides, viz. be. It is evident that the c. of g. of each strip will be at the middle points of each, and therefore will lie on a line drawn from the opposite angle point a to the middle point of the side e, i.e. on ae; likewise it will lie on bd; therefore the c. of g. is at the intersection of ae and bd, viz. g. Join de. Then by construction ad= dc = — , and be — ec 2 = - ; hence the triangles acb and dee are similar, and therefore 2 de = —. The triangles agb and dge are also similar, hence ag ae eg= ^=—. ^. 3 Since g is situated at 5 height from the base, we have 2/S jr = - t.e. = -( x] 3 3\2 / t/+x = b/= Draw the dotted line parallel to the sloping side of the trapezium in Fig. 75*. Height of c. of g. of figure from base _TT _ area of parallg. x ht. of its c. of g. + area of A x ht. of its c. of g. ' area of whole figure SHxH , ,„ „>H H /S + S,w ■ 3(S + S,) SH X H . /„ (j,H 2H -^— + (b. - S)- X — ^ ^^g ^ ^g^ ^ (S±S)„ 3(S + S,) ^1 = gS + Si ^ I H, 2S, + S s, +§ < — s \k H V ■ ' \ V or^=— ' *-^'-'- ga ad ■ ^'o- 7S«. 62 Mechanics applied to Engineering. Position of Centre of Gravity, or Ckntroid. Ti \i ,..->Ij:;^-:;;-- Area dbe = Ai ice = Aj c^e = A, c. of g. of area a&e Q J) JJ i^CC U2 „ „ whole fig- ure C1.J.S. Trapezium and triangles. Intersection of line joining c. of g. „.(; of triangle and c of g. of trapeziunij viz. ab and cd, where ac = area of trapezium, and db area of triangle, ac \ is parallel to bd. Fig. 77. r Lamina with hole. Let A = area abcde; H = height of its c. of g. from ed; Hi = height of its c. of g. from <Ci'' drawn at right angles to ed; a = area of hole g/i; h = height of its c. of g. from ed; hi = height of its c. of g. from d/. Then— Hf = height of c. of g. of whole figure from ed K'c = height of c. of g. of whole figure from df Tjr AH — ah Kc=—r- A — a HV = • '^■^1 ~ ^^ Fig. 78. Moments. 63 The principle of these graphic methods is as follows : — Let the centres of gravity of two areas, Aj and Aj, be situated at points Q and Ca ^ respectively, and let the common centre of gravity be situated at c, distant «, from Ci, and X2 from •■^■■ 'A/ Caj then we shall have Ai*, =Aa«a- From Ca set off a line cj)^, whose length represents on some given scale the area Aj, and from Ci a line c^b^^ parallel to it, whose length represents on the same scale the area Aj. Join bi, b^ Then the intersec- tion of bj>i and QCa is the common centre of gravity c. The two triangles are simikr, therefore — ~r = ~Z^) '^^ "-i^i ~ ■'Vs^a < X, Ai *i ^ X, >c, ' N.B. — The lines C,*, arid C^b, \^ /J' are set off on opposite sides of ^ / ^ C„ Cj, and at opposite ends to their respective areas, at any convenient angle ; but it is undesirable to have a very acute angle at c, otherwise the point will not be well defined. When one of the areas, say Aj, is negative, i.e. is the area of a hole or a part cut out of a lamina, then the lines f ,i, and c^b, must be set off on the same side of the line, thus — Then— 'Az Fig. 76a. ai" ■^ 'S <—a>,- — ■, A, f i k'^^ -r-' = ^, or AiJCi = A^j Fig. yti. 64 Mechanics applied to Engineering. Position of Centre of Gravity, or Centroid. Graphical method. Lamina with hole. j^ ^^ ^^ ^^^ ^_ ^f g_ „f ^3^^, . ^2 >. ). SJ^- . Join ^1, ^2, and produce; set off C2K2 and fjKi parallel to one another and equal to A and a respectively; through the end points K5, Kj draw a line to meet the line through ^1, (Tj in "^1.21 which is the c. of Fig. 70. ■■■^'■-s g. of the whole figure. Note.— The lines c^Ks, f,K, need not be at right angles to the line ^,fj, but the line KjK, should not cut it at a very acute angle. Portion of a regular polygon or an arc of a circle, considered as a thin wire. Let A = length of the sides of the poly- gon, or the length of the arc in the case of a circle ; R = radius of a circle inscribed in the polygon, or the radius of the circle itself; C„ = chord of the arc of the polygon or circle ; Y = distance of the c. of g. from the centre of the circle. Then A : R R Y or Y = RC. N.B. — The same expression holds for an arc greater than a semicircle. Moments. 65 See p. 63. Fig. 8a>>. Regard each side of the polygon as a piece of wire of length I; the c. of g. of each side will be at the middle point, and distant _j'i,j'2,_)'3, etc., from the diameter of the inscribed circle; and let the projected f>y length of each side on the ^j/'' diameter be c-i, c,, c,, etc. // The triangles def and Oba Ui;^ . are similar ; .' /{ ,^ , fd Oa / R M-- therefore-V = -^, or - = — ^■ Je ab C-, y^ and J'l = — likewise y^ = -j, 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 ^ wyi + wy.i+, etc. wn where n = number of sides. The w Cancels top and bottom. Substituting the values oiyi,yi, etc., found above, we have— Y = -k, + c,+, etc.) nl (Proof concluded on p. i>i^ 66 Mechanics applied to Engineering. Position of Centre of Gravity or Centroid. Semicircular arc or wire. 9\ \ ^ i I -VLL cs^i^ ' Fig. 8z. Circular sector considered as a thin sheet. ..A ! ^^^ ^\c.(fq.of/^ V — 'RC„ Fig. 82. Semicircular lamina or sheet "-■ofQ- ofs^etr « 4R 2D _^ \ 3ir 3T ^--CgZR > Fig. 83. Parabolic segment. ri \i V = fH where Y = distance of c. of g. from apex. '\c.fy. }A The figure being symmetrical, the c. of •,1 i \ S- li^s °^ ^^^ 3.xis. Fig. 84. Moments. 67 but Ci + ^2 +1 etc. = the whole chord subtended by the sides of the polygon = C, and nl = A. RC, A When n becomes infinitely great, the polygon becomes a circle. The axis of symmetry on which the c. of g. lies is a line drawn from the centre of the circle at right angles to the chord. Y = In the case of the sector of a polygon or a circle, we have to find the c. of g. of a series of triangles, instead of their bases, which, as we have shown before, is situated at a distance equal to two-thirds of their height from the apex ; hence the c. of g. of a sector is situated at a distance = |Y from the centre of the inscribed circle. From the properties of the parabola, we hav h_ H ' = ¥* B>% Area of strip = b .dk = — -j dh rl {Continued on p. 69.) 68 Mechanics applied to Engineering. Position of Centre of Gravity or Centroid, Parabolic segment. Y = f H (see above) where Y, = distance of c. of g. from axis. Fig. 8j. Moments. moment of strip about apex = 69 h.b .dh = — idh moment of whole figure about apex = — f^^fi — — r X -«- H* J o H' 7 = fBH2 the area of the figure = |BH (see p. 30) the dist. of the c. of g. from the apex : |BIP fBH = |H From the properties of the parabola, we have- h p H B" orM H b^ B^ B"(H -K) = ^''H B^y^o = B2H- tm h. B»^ -b-^) Apea area of strip = h^db moment of strip about axis = b . h^b = ^^ (S'b — P)db = 5 J (B'6-i')db H rBV _ ^n B B^L 2 Jo = H/'B* _ B^\ ^ HI B\ 2 4 / 4 moment of whole area) about axis ) the area of the figure = #BH a-" HB^ the distance of the c. of g. )_ 4 _ 3 „ from the axis ) 2r>-H ' ^o Mechanics applied to Engineering. ^Apeoc Position of Centre of Gravity or Centroid. Ex-parabolic segment. < B Y = AH where Y, = distance of c. of g. from axis; Y = distance of c. of g. from apex. Fig. 86. Irregular figure. Y = Mhy +3^+5 >^8+7^4+, etc.) 2('4i+'^2+'4,+.«i+, etc.) where Y = distance of c. of g. from line AB ; 111 = width of strips ; A = mean height of the strips. Y. = '^{ ^'i+3^„+5^'b+7A\+, etc. x 2 V li\+k\+A',+/i',+, etc. J where Y„ = distance of c. of g. from Une CD ; lel = width of strips ; A' = mean height of the strips. Moments. 71 By the principle of moments, we have — Distance of c. of g. of figure from axis f area of rect. X (^ist of its c. of g.| _|area ofpara.| >< fdist. of its c. of g. \ _\ I. from axis J I segment ) \ from axis / Yi = Likewise — area of figure BH X ? - |BH X fB H fB BH X - - |BH X f H Y = ^ = -5-H iBH This is a simple case of moments, in which we have — Distance of c. of g.^ _ moment of each strip about AB area of whole figure from line AB Moment of first strip= w^, x — 2 „ second „ ^wh^y.^ 2 ,, third „ =a/^x^- 2 : wh-i, + whi + w^s +, etc. The area of the first strip = wht, „ „ second „ =whi „ „ third „ =wht and so on. Area of whole figure Distance of c. j w/^i X - + wA, X ^ + wA, X 5^ +,etc. of g. frninl_Y_ ^ ' , 2 line AB J whx + wh^ + wh^ +, etc. one w cancels out top and bottom, and we have — Y ^-w / K + 3^2 + 5^'3 + ih +, etc. 2 V h-i-{-K-\- hi-\- hi-V, etc. and similarly with Yj. The division of the figure may be done thus : Draw a Une, xy, at any angle, and set oif equal parts as shown; project the first, third, fifth, etc., on to xz drawn normal to AB. Fig. 87a. Mechanics applied to Engineering. Position of Centre of Gravity OR Centroid. Wedge. On a plane midway between the ends, and at a height — from base. 3 For frustum of wedge, see Trapezium. Pyramid or cone. On a line drawn from the middle. point of the base to the apex, and at a distance f H from the apex. Fig. Bg. Frusitim of pyramid or cone. t .f,4P«« On a line drawn from the middle point of the base to the apex, and at a height /I — f&\ H |H( ; 1 from the apex, where « = _- 1 * V I - ;;V H or ^ 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 at a height — from .the base. Volume of layer = b'^ . dh moment of layer about apex = b'^ . k . dh But ^-=1 h B. b = h. B H B^/JV moment of layer about apex = -^^dh Yia. igtt. moment of the whole pyramid! _ B^ I ,3 ., B'H^ B^H^ about apex J" iP J o 4H» " 4 volume of pyramid B'H 3 B^H^ distance of c. of g. from apex = „^„ - 3 ' 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) _B^ f ^ .3 about the (imaginary) apex S H'' „ J Hi _ F /- ff HiM - ffV 4 ~ 4 ) • volume of frustum = Bi 4 4 B^H Bi'Hi 3 3 Hi (i.) (ii.) substituting the value -^=^ = n then the distance of the c. of g-\ ilj ^ ag /' '"^'^ ~\ from the apex / (ii.) * Vi-«^> Fio. 91, 74 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 a;,byerecting a perpendicular to cut' the centre line. W, the weight of the engine, is found by weighing it in the ordinary way. Wa 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 hi 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 have- But - = I- x^ G whence ~%^ = Wjj- 75 «i =■ W,G W y = — - ^1 2 G 2 By similar triangles — y A K W,G W ^ ^ ^ /4, = h (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, 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. 76 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 lo = 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 ; Ro = the perpendicular distance between the two axes. Then I„ = I + MR„2, or 1 + ARo" Let xy be the axis passing through the c. of g. Let «;yi be the axis of revolution, parallel to xy and in the plane of the surface or lamina. Let the elemental areas, <hi (h., thi etc., be situated at distances r„ r,, r„ etc., from xy. Then we have — I„ = ai(R„ + ri)2 + «,(R, + ;j)-''+,etc. = c^(R,' + r,' + 2R,rO +, etc. = ajTi" + a.f.^ +, etc. + R„'(a,+a,+,etc.) + 2R|,(airi + tVa + , etc.) But as xy passes through the c. of^. of the section, we have (hr■^ + cv^ +i etc. = o (see p. 58), for some r's are positive and some negative ; hence the latter term vanishes. The second term, ffj + <?, +, etc. = the whole area (A) ; whence it becomes Ro'^A. In the first term, we have simply the second moment, ot moment of inertia, about the axis passing through the c. of g. = I ; hence we get — I„ = 1 + R.2A Moments. 77 We may, of course, substitute Wj, 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 re/olution, 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 \^ Ip = I. + Iv Consider any ele- mental area a, distant r from the pole. I, about ox = ay'' I, » oy = ax^ Tp „ pole = ar- But r'=«2+y and ar'^=ax'^+ay hence I,= I,-f I, 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 the axis xy (k) first moment of surface about x j ) area of surface _ first moment of body about xy volume of body lar 78 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 gyrationV second moment of surface about xy about the axis xy (k?) J = area of surface second moment of body about xy volume of body 2^ I A or = ^ = ' - = "X. or I = Ak' Second Moment, or Moment of Inertia (I). Parcdklogram treated as a thin lamina about its extreme end. O I.= BIP K M o o If the figure be a square — B = H = S we have I,, = — Fio. g6. Radius of gyration H Moments, 79 For other cases of moments of inertia or second moments, see Chapter IX., Beams. Area of elemental strip = "& . dh second moment of strip = 'R.h'' .dh second moment of whole surface !-// Ifl.dh = BH^ -/I. 5! 3 area of whole | _ -dtt surface ) ~ square of radius 1 _ BH^ of gyration ) ~ 3BH ' radius of gyration = —p^ It will be seen that the above reasoning holds, however the parallelogram may be distorted sideways, as shown. <^ -M Fig. 96a, 8o Mechanics applied to Engineering. Second Moment, or Moment of Inertia (I). o < -//■■:> O \ Parallelogram treated as a jj thin lamina about its ; central axis. J BH» S* I = or — 12 12 Radius of gyration H \/l2 < // -> \c.cfp.B Parallelogram treated as a thin lamina about an axis distant 'R^from its c. ofg. Ri;-i H ■^oi Io = Bh(^' + R„') o Fig. gS. Moments. &i This is simply a case of two parallelograms such as the XT above put together axis to axis, each of length — ; \2 ) BH^ Then the second moment of each = — 3 8X3 then the second moment "i _ /BH°\ BH'' of the two together / \ 24 / 12 area of whole surface = BH radius of gyration = k/ - BH' H 2BH- V77 From the theorem given above (p. 76), we have- I, = [ + Ro'A I = ^ ^^+R/BH A = BH 12 " = BH(!i + R/j when Ro = o, lo = I L L- 82 Mechanics applied to Engineering. Secoxd Moment, or Moment of Inertia (I). Fig. gg. Hollow parallelogi-am. Let le = second moment of external figure ; \i = second moment of internal figure ; lo = second moment of hollow figure ; lo = I. - I.. Radius of gyration Triangle about an axis parallel to the base passing through the apex. O Io = BH3 H Triangle about an axis parallel to the base passing through the c. of g. I = BH3 36 H Vil Moments. 83 The lo for the hollow parallelogram is simply the difference between the I, for the external, and the Ij for the internal parallelogram. Area of strip = b.dh; but * = ^ „ =\h.dh second moment of strip = — /z^ . dh rl ^ 3-^ <:^'"l ^ A i? B f^ „ triangle = g h^-dh J <- ^ -> < M- > r V BH* BH^ " - 4H 4 area of triangle = Fig. Tooa. /BH» radius of gyration = / —2— = \y — = -^ / BH 2 V 2 From the theorem on p. 76, we'have — I„ = I + R.^'A I^ _ BH^" I-I„_R„.A R„._('Hy=4H^ J _ BH3 _ 4H2 ^ BH ;^ _ BH ' 492 2 _ BH3 36 /bh» radius of gyration = / _2£«=x/— s= -^= / BH ^ 18 ^,8 V 84 Mechanics applied to Engineering. Second Moment, or Moment of Inertia (I). O Triangle about an axis at the base. _BH' 12 FtG. T02. Radius of gyration H ^6 Trapezium about an axis coinciding with its short base. O 4 <--- H Io= (3B + BQH' Let Bi= «B. V 6(« + 3_ 6(« + i) ^ Fig. 103. Trapezium about an axis coinciding with its long base, -H I.,= (3B, + B)H ' 12 or-77(3«+ i) Fig. ro4. H / 3 "+ I V 6(« 4- i) Moments. 85 From the theorem quoted above, we have- I„ = 1 + Ro'A BIT W 36 + 9 I -BH' R„^ = H" I« = BH 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. BiH' • . For parallelogram, lo = — — for triangle, I,, = (B - BQH' B.ff (B-B.)H ' for trapezmm, lo — • — - — + 7 I„ (3E + BQff Fig. 103a. When the axis coincides with the long base, the I for the /■D "D \TT3 triangle = ^^ — ; then, adding the I for the parallelogram as above, we get the result as given. When « = r, the figures become parallelograms, and I = , as found above, ^ , r BH' When « = o, the figures become triangles, and the I = — — for the first case, as found for the triangle about its apex ; and I = for the second case, as found for the triangle about 12 its base. 86 Mechanics applied to Engineering. Second Moment, or Moment of Inertia (I). Radius ol gyration ma parallel wtm cne oase. O , _ (B.' + 4B.B + B°)H3 36(B. + B) ^^BIP/«Mm«jLi\ 36 \ « + I ) Bi For a close approximation, see next figure. 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 C^ the same as the I for trapezium. For dotted rectangle — J _ (B + B.)H3 orif B, = «B Fig. 106. I = (« + i) 24 Moments. 87 From the theorem on p. 76, we have- I = I„ - R,2A l^ ^ (3B1 + B)ff ^^^^^^ i^^g ^'^ base) 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. ■D'XJ3 The I for a rectangle is (see p. 80). Putting in the value 5-+^ = B', we get- I = _ B + B. ^ H3 _ (B + B,)H3 24 The following table shows the error involved in the above assumption ; it will be seen that the error becomes serious when « < o"S : — Value of «. Approx. method, the correct value being i. °"9 I 001 0'8 i'oo5 07 l-OII 06 I -021 o-S I -039 0-4 I '065 0-3 no? 0-2 ■ •174 The approximate method always gives too high results. 88 Mechanics applied to Engineering. Second Moment, or Moment of Inertia (I). Square about its diagonal. t2 Radius of gyration s ^12 Circle about a diameter. I =: 64 Hollow circle about a dia- meter. D 4 VD.^+D.^ Moments. 89 This may' be taken as two triangles about their bases fsee p. 84). In this case, B = v' 2S --i, /\ -^ / -JIS •■i|. 12 vS */:' 12 area of figure = S^ Fig. ztyjct. / S'' S radius of gyration — a/ j^S^ "~ J~ From the theorem on p. 77, we have I, = 1, + I,; in the circle, Ij = L. Then L= 2L irD* rD* and I, = f = 73^, = -^ (see Fig. 118). The I for the hollow circle is simply the difference between the I for the outer and inner circles. 90 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 y same as in the case given above for the hollow circle. Radius of gyration Ellipse about minor axis. o 64 Moments. 91 The axis 00 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 — --QHb + «) = -V).^b 4 4 whence b — D,^ - W also--D,2(a-^) = -D,2a 4 4 whence a = BJ'x From the theorem on p. 76, we have- I', = I, + A/^ for the outer circle about the c. of g. of figure 64 4 also F, = I, + A,fl^ for the inner circle about the c. of g. of figure 64 4 and I = r. — I', for the whole figure. Substituting the values given above, and reducing, we get the expression given on the opposite page. The second moment, or moment of inertia, of a figure varies directly as its breadth taken parallel to the axis of revolution ; hence the I for an ellipse about its minor axis is simply the I for a circle of diameter Da reduced in the ratio =:-i D2 or-D^xg' 64 D, 64 92 Meclianics applied to Engineering. Second Moment, or Moment of Inertia (I). ^ ^ Ellipse about major axis. Radius of gyration f c- tA I - '^D/D' i^ V 1 / ^^ 4 V Fig. 112. Parabola about its axis. V^ y^ ApeXf A 1 \ 1 = ,^HB' B ^5 < -aS- > Fio. nj. 1 Moments. 93 And for an ellipse about its major axis, the I is that for a circle of diameter Di increased in the ratio — - or^*X°? = 3!^^ 64 D, 64 h = ^{\ - -^ (see p. 69). area of strip =: h .db second moment of strip = b"^ .h. db second moment of whole figure .3 SB^j o "F _ B'l .3 sj = H 2HB 15 second moment for double"! figure shown on opposite > = AHB' page 94 Mechanics applied to Engineering. Second Moment, or Moment of Inertia (I). Radius of gyration Parabola about its base. ■jApejc I = Ji^BH^ Fig. Z14. V^H Irregular figures. 4 49*1 +. etc.) Fig. T15. (See opposite page.) b = Moments. B .h* 95 H* area of strip = b . dh second moment J ^^jj_ ^^,3 _^^ of strip {Yi.-hfBh\dh YC second moment of whole ■! figure / Fig. ii4ff. B fH = ^ {Wh* + h^ - 2h^B)dh B_ \ 2Wk^ H* }_ 3 2K = B(|ff + f H» - 4H') 18 T)XJ3 for the double figure shown^ _ _32_t)tt8 on opposite page ) ' "* Divide the figure up as shown in Fig. 87a. Let the areas of the strips be a^, a^, a^, a^, etc., respectively ; and their mean distances from the axis be r^, r^, r,, r^, etc., respectively. Then Iq = a-^r^ + a^i + a^^ +, etc. But fli = wb^, and a^ = wb^, and so on and n =— , ^2 = -^, ^3 = ^— , and so on 2 2 2 hencel. = «.{^0 + <fj + .3(?j+,et.c.} 4 {bi + 9*a + 25*3 + 49^4 +, etc.) Also k' — (^ + 9*2 + 25^3 + 49^4 +, etc.) ii>{bi + b^ + bs + bi+, etc. ^ ^ / ^1 + 9/^2 + 25<^3 +, etcT 2 V ^j 4. ^^ + ^3 +^ etc. This expression should be compared with that obtained for finding the position of the c. of g. on p. 71. Tke comparison helps one to realize the relation between the first and second moments. 96 Mechanics applied to Engineering. Second Moment, or Moment of Inertia (I). ^ 1 Graphic method. Radius 0^ gyration c <- Af -- \ ^ <- Y^ 9 Fig. ii( 1 5J i Let A = shaded area ; N^..,-.-J»/ Y = distance of c. imv ' ofg. of shaded \ ; S areafromOO; 1 |i|^ H = extreme di- ; 1 j ® mension of 1 J figure mea- ilii '^ sured normal \^\ to 00 ; ° \ ' Then I„ = AYH Second Paraildog c. ofg. Polar Moments of Surfaces or T rram about a pole passing through its I. = ??(H^ + B^) hin Laminae. ,/«•+- / n / / 1 M square of side S — S ■< -H- Fig 117. i=si ' 6 V^ Circle about a pole passing through the c. of g. ^ y I, = — . or D a/8" or ^ Fig. ii8. Moments, 97 Divi.le the figure up into a number of strips, as shown in Fig. ii6j project each on to the base-line, e.g. ab projected to fli^ij join «i and b^ to c, some convenient point on 00, cutting ab in a^^, and so on with the other lines, which when joined up give the boundary of the shaded figure. Find the c. of g. of shaded figure (by cutting out in cardboard and balancing). The principle of this construction is fully explained in Chap. IX., p. 360. See also Barker's "Graphical Calculus," p. 184, and Line- ham's " Text-book of Mechanical Engineering," Appendix. From the theorem on p. 77, we have — I, = I, + I» = ^+4f BH" , HB' 12 12 \y 12 Fig. Tiya. I 4 Thickness of ring = dr area of ring = zirr . dr second moment of ring = 2irr .r^ . dr fR circle = 2ir \ r^ . dr = 2ir I f^ . («R-e) 2«-R^_ 4 2 X 16 2 98 Mechanics applied to Engineering. Second Polar Moment, or Polar Moment of Inertia. ff Hollow circle about a pole passing through the c. of g. a7td normal to the plane. I, = ^(D.* - D<^) Radius of gyration V 8~" or / R.' + R.' Second (Polar) Moments of Solids. Gravitational Units. Bar of rectangular section about a pole passing through its c. of g. , ^1~~ O )f IG. I20. B"! For a circular bar of radius R — 12 ^ L'' + B^ V'^ si- L' + 3R' 12 Cylinder about its axis, o ■Di- <- -R--* : T ttD^HW irR^HW H I»= — > or ■ 32 ^ 2 g Fig. 191. Moments, 99 The Ij, for the hollow circle is simply the difference between the L for the outer and the L 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 /\? + B^ gyration of each being K = a/ — — — (see Fig. 1 1 7), which is the radius of gyration of the bar. The second moment of the LBH W bar will then be K^ (weight of bar), or -^fl{U + ^l- Where W is the weight of 1 cubic inch or foot of the material, according to the units chosen. The cylinder may be regarded as being made up of a great number of thin circular lamins revolving about a pole passing through their centre, the radius of gyration of each being The second moment of cylinder = K^ (weight of cylinder) D^ irD^HW ttD^HW = - X = 8 4 ^ 32 ^ lOO Mechanics applied to Engineering. <■■ R. Second Polar Moment, or Polar Moment of Inertia. g— V-/Pf--> ^ Hollow cylinder about Radius of gyration M I. I a pole passing ihi-ottgh I its axis. H or —(R/ - R/)- or Disc flywheel. i Treat each part sepa- rately as hollow cylinders, and add the results. 1 1 o Fio. IB3. Moments. lOl The Ij, for the hollow cylinder is simply the difference between the I, 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 P 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 [^ Jffe >- the I for the shaded figure Fig. 123a. treated as a plane surface. The construction simply reduces the width of the solid section at each point proportional to its distance from OO ; it is, in fact, the " modulus figure " (see Chap. IX.) of the section. Let y^ = the distance of the c. of g. of the second modulus figure from the axis 00 shown black ; A = the area of the black figure. Ai = the area of the modulus figure then Ii = AiK^ = hyj (see page 96) Ax K^ = : The moment of inertia of the wheel V- Weight of the wheel X k-ycy 102 Mechanics applied to Engineering. Second Polar Moment, or Moment of Flyw heel with arms. Treat the rim and boss separately as hollow cylin- ders, and each arm thus (assumed parallel) — For each arm (see Fig. 120) — L (sectional area),, a p ga 12 F'°-"4. + I2R„") where R, = the radius of the c. of g. of the arm. For most practical purposes the rim only is considered, and the arms and boss neglected. Inertia. Radius of gyration Sphere about its diameter. W or i,rD» w RVf D or-^= V 10 Moments. 103 The arms are assumed to be of rectangular section; if they are not, the error involved will be exceedingly small. The sphere may be regarded as being made up of a great number of thin circular layers of radius rj, and radius of gyration -T=(see p. 96). +2RJ' = 2Rj-/ volume of thin layer = irr^dy second moment of \ „ r^ layer about 00 / = '^''1 '0' X T = -Ai^y-ffdy = j(4Ry+y-4R/Kj»' second moment of hemi- 1 sphere, i.e. of all layers | on one side of the I diameter dd J (4Ry+/-4R/K^ _^r4Ry /_4R/1-y = ^ . 2L 3 ^5 4 \y=o IT r4R° R^ 4Rn second moment of sphere = iV^R" L-3 4 J I04 Mechanics applied to Engineering. Second Polar Moment, or Polar Moment of Inertia. Radius of gyration Sphere about an external axis. w When the axis becomes a tangent, Ro = R ; I = rR=— Fig. 126. Cone about its axis. O I,=^R^H^. 10 ^ or ^D^H^ 160 g I i.e. I of the I for the cir- ■ cumscribing cylinder. ^ 10 Dv'X ^ 40 Moments. 105 From the theorem on p. 7 6, we have- K = = (^R» + I^R'Ro") V = |,rR» The cone may also be regarded as being a great number of thin layers. Volume of thin layer = irr^dk second moment of \ «,, ^ r' '^t*Ji. layer about axis OOj 7rR*/J< second moment of cone 2H' dh v%\y'''i rR*H CHAPTER IV. RESOLUTION OF FORCES. We have already explained how two forces acting on a point may be replaced by one which will have precisely the same efifect on the point as the two. We must now see how to apply the principle involved to more complex systems of forces. Polygon of Forces. — If we require to find the resultant of more than two forces which act on a point, we can do so by finding the resultant of any two by means of the parallelogram of forces, and then take the resultant of this resultant and the , next force, and so on, as shown in the diagram. The resultant of I and 2 is marked Ri.2., !;.,R,i and so on. Then we finally get the resultant Ri. 2.3.4. for the whole system. Such a method is, however, J clumsy. The following will be found much more direct and convenient : Start from any point O, and draw the line i parallel and equal on a given scale to the force 1 ; from the extremity of 1 draw the line 2 equal and parallel to the force 2 ; then, by the triangle of forces, it will be seen that the line R1.2. is the resultant of the forces 1 and 2. From the extremity of 2 draw 3 in a similar manner, and so on with all the forces; then it will be seen that the line Ri,2.3.4. represents the resultant of the forces. In using this con- struction, there is no need to put in the lines R1.2., etc. ■ in the figure they have been inserted in order to make it Fig. 128. Resolution of Forces. 107 clear. Hence, if any number of forces act upon a point in such a manner that Unas drawn parallel and equal on some given scale to them form a closed polygon, the point is in equilibrium under the action of those forces. This is known as the theorem of the polygon of forces. Method of lettering Force Diagrams. — In order to keep force diagrams clear, it is essential that the forces be lettered in each diagram to prevent con- fusion. Instead of lettering the force itself, it is very much better to letter the spaces, and to designate the force by the letters corresponding to the spaces on each side, thus : The force separating a from b is termed the force ab ; likewise the force separating d from b, db. fig- "9. This method of notation is usually attributed to Bow; several writers, however, claim to have been the first to use it. Funicular or Link Polygons. — When. forces in equi- librium act at the corners of a series of links jointed together at their extremities, the force acting along each link can be readily found by a special application of the triangle of forces. Consider the links ag and bg. There are three forces in equili- brium, viz. ab^ ag, bg, acting at the joint. The magnitude of ab is known, therefore the magnitude of the other two acting on the links may be obtained from the triangle of forces shown on the right-hand side, viz. abg. Similarly consider all the other joints. It will be found that each triangle of forces contains a line equal in every respect to a line in the preceding triangle, hence all the triangles may be brought together to form one diagram, as shown to the extreme right hand. It should . be noticed that the external forces form a closed polygon, and the forces in the bars are represented by radial lines meeting in the point or pole g. It will be evident that the form taken up by the polygon depends on the magnitude of the forces acting at each joint. Fig. 130. io8 Mechanics applied to Engineering. Saspecsiou Bridge. — ^Another special application of the triangle of forces in a funicular polygon is that of finding the forces in the chain of a suspension bridge. The platform on which the roadway is carried is supported from the chain by means of vertical ties. We will assume that the weight sup- ported by each tie is known. The force acting on each portion of the chain can be found by constructing a triangle of forces at each joint of a vertical tie to the chain, as shown in the figure above the chain. But bo occurs in both triangles; hence the two triangles may be fitted together, bo being com- mon to each. Likewise all the triangles of forces for all the joints may be fitted together. Such a figure is shown at the side, and is known as a ray or vector polygon. Instead, Fig. 131. however, of constructing each triangle separately and fitting them together, we simply set oiF all the vertical loads ab, be, etc., on a straight line, and from them draw lines parallel to each link of the suspension chain ; if correctly drawn, all the rays will meet in a point. The force then acting on each segment of the chain is measured off the vector polygon, to the same scale as the vertical loads. In the figure the vertical loads are drawn to a scale of 1" = 10 tons ; hence, for example, the tension in the segment ao is 9*8 tons. The downward pressure on the piers and the tension in the outer portion of the chain is given by the triangle aol. If a chain (or rope) hangs freely without any platform suspended below, the vertical load will be simply that due to the weight of the chain itself. If the weight per foot of hori- zontal span were constant, it is easy to show that the curve taken up by the chain is a parabola (see p. 493). In the same chapter, the link and vector polygon construction is employed Resolution of Forces, 109 to deteimine the bending moment due to an evenly distributed load. The bending moment M^ at x is there shown to be the depth D, multiplied by the polar distance OH (Fig. 132) ; ix. Fig. X33. the dip of the chain at x multiplied by the horizontal com- ponent of the forces acting on the links, viz. the force acting on the link at x, or M^ = D,, X OH. In the same chapter, it is also shown that with an evenly distributed load M^^ = -5-, o where w is the load per foot run, and / is the span in feet Hence 8 = D, . OH = D,, or .^ = 8D« where h is the tension at the bottom of the chain, viz. at x. It will, however, be seen that the load on a freely hanging chain is not evenly distributed per foot of horizontal run, because the inclination of the chain varies from point to point. Therefore the curve is not parabolic; it is, in reality, a catenary curve. For nearly all practical purposes, however, when the dip is not great compared with the span, it is suffi- ciently accurate to take the curve as being parabolic. Then, assuming the curve to be parabolic, the tension at any other point, y, is given by the length of the corresponding line on the vector polygon, which is readily seen to be — T, = ^h^ + w'l^ The true value of the tension obtained from the catenary is — T, = >4 + wD, (see Unwin's " Machine Design," p. 421), which will be found to agree closely with the approximate value given above. no Mechanics applied to Engineering. Data for Force Polygons. — Sometimes it is impossible to construct a polygon of forces on accoimt of the incomplete ness of the data. In the case of the triangle and polygon of forces, the follow- ing data must be given in order that the triangle or polygon can be constructed. If there are « conditions in the completed polygon, « — 2 conditions must be given ; thus, in the triangle of forces there are six conditions, three magnitudes and three directions: then at least four must be supplied before the triangle can be constructed, such as — 3 magnitude(s) and i direction(s) Likewise in a five-sided polygon, there are ten conditions, eight of which must be known before the polygon can be constructed. When the two imknown conditions refer to the same or adjacent sides, the construction is perfectly simple, but when the unknown conditions refer to non-adjacent sides, a special construction is necessary. Thus, for example, suppose when dealing with five forces, the forces i, 2, and 4 are completely known, but only the directions, not the magnitudes, of 3 and 5 are known. We proceed thus : Draw lines r and 2 in the polygon of forces. Fig. 133, in the usual way. From the extremity of 2 draw a line of indefinite Fig. 133. Fig. 134. length parallel to the force 3 ; its length cannot yet be fixed, because we do not know its value. From the origin of i draw a line of indefinite length parallel to 5 ; its leng& is also not yet known. From the extremity of 4 in the diagram of forces, Fig. 134, drop a perpendicular ah on to 3, and in the polygon of forces, Fig. 133, draw a line parallel to 3, at a distance ab Resolution of Forces. Ill from it. The point where this line cuts the line S is the extremity of 5. From this point draw a line parallel to 4 ; then by construction it will be seen that its extremity falls on the line 3, giving us the length of 3. The order in which the forces are taken is of no import- ance. Forces in the Members of a Jib Crane. Case I. The weight W simply suspended from the end of the jib. — There is no need to construct a separate dia- gram of forces. Set off be = W, or BC on some convenient scale,^ and draw ca parallel to the tie CA ; then the triangle bac is the triangle of forces acting on the point b. On measuring the force dia- gram, we find there is a compressive force of 15 "2 tons along AB, and a tension force of 9-8 tons along AC. The pressure on the bottom pivot is W (neglecting the weight of the crane itself). The horizontal pull at the top of the crane-post is ad, or 7-9 tons ; and the force (tension) acting on the post between the junction of the jib and the tie is cd, or 6 tons. The bending moment at y will be ad X h, or W X /. For determining the bending stress at y, see Chap. IX. Taking moments about the pivot bearing, we have — rM/> 1 D M-^ f 1 [ ...^£..-1....- c W^//MM i. ''"W m d, Fig. 135. p^x=p^ = '^l W/ or A =/, = — The sections ot the various parts of the structure must be determined by methods to be described later on. The weight of the structure itself should be taken into account, which can only be arrived at by a process of approxi- mation ; the dimensions and weight may be roughly arrived at by neglecting the weight of the structure in the first instance. Then, as the centre of gravity of each portion will be approximately at the middle of each length, the load W must be increased to ' In this case the scale is o'l inch = 2 tons, and W = 7 tons. 112 Mechanics applied to Engineering. W + ^(weight of jib and tie). The downward pressure on the pivot will be W + weight of structure. The dimensions of the structure must then be increased accordingly. In a large structure the forces should be again determined, to allow for the increased dimensions. The bending moment on the crane-post at y may be very much reduced by placing a balance weight Wj on the crane, as shown. The forces acting on the balance-weight members are found in a similar manner to that described above, and, neglect- ing the weight of the structure, are found to be 8" i tons on the tie, and 4*4 tons on the horizontal strut. The balance weight produces a compression in the upper part of the post of 6*8 tons ; but, due to the tie ac, we had a tension of 6"o tons, therefore there is a compression of o'8 ton Fig. 136. Fig. 137. in the upper part of the crane-post The pressure on the lower part of the crane-post and pivot is W -1- Wj -f weight of structure. Then, neglecting the weight of the structure, the bending moment on the post at y will be — W/ - Wi/, W/ The moment Wi*^ should be made equal to — , then the 2 post will never be subjected to a bending moment of more than one-half that due to the lifted load, and the pressure/. and /a will be correspondingly reduced. Case II. The weight W suspended from a chain passing to a barrel on the crane-post. — As both poitions of the chain are Resolution of Forces "3 subjected to a pull W, the resultant R is readily determined. From c a line ac is drawn parallel to the tie ; then the force acting down the jib is ab = i6'4 tons j down the tie ac = 4*4 tons. The bending moments on the post, etc., are determined in precisely the same manner as in Case I. When pulley blocks are used for lifting the load, the pull in the chain between the jib pulley and the barrel will be less than W in the proportion of the velocity ratio. The general effect of the pull on the chain is to increase the thrust on the jib, and to reduce the tension in the tie. In designing a crane, the members should be made strong enough to resist the greater of the two, as it is quite possible that a link of the chain may catch in the jib pulley, and the conditions of Case I. be realized. Forces in the Members of Sheer Legs.— In the type of crane known as sheer legs the crane-post is dispensed with, and lateral stability is given by using two jibs or sheer legs spread out at the foot ; the tie is usually brought down to the level of the ground, and is attached to a nut working in guides. By means of a horizontal screw, the sheer legs can be tilted or " derricked " at will : the end thrust on the screw is taken by a thrust block ; the up- ward pull on the nut and guides is taken by bolts passing down to massive foundations below. The forces are readily determined by the triangle of forces. The line ^^ris drawn parallel to the tie, and represents the force acting on it; then ac represents the force acting down the middle line of the two sheer legs. This is shown more clearly on the projected view of the sheer legs, cd is then ^'°- '38- drawn parallel to the sheer leg ae; then dc represents the force acting down the sheer leg ae ; likewise ad down the leg of, and I 114 Mechanics applied to Engineering. dg the force acting at the bottom of the sheer legs tending to make them spread; ch represents the thrust of the screw on the thrust block and the force on the screw, and bh the upward pull which has to be resisted by the nut guides and the founda- tion bolts. The members of this type of structure are necessarily very Fig. i33«. heavy and long, consequently the bending stress due to their own weight is very considerable, and has to be carefully con- sidered in the design. The problem of combined bending and compression is dealt with in Chapter XII. Forces in a Tripod or Three Legs.— Let the lengths Resolution of Forces. "5 of the legs be measured from a horizontal plane. The vertical height of the apex O from the plane, also the horizontal distances, AB, EC, CA, must be known. In the plan set out the triangle ABC from the known lengths of the sides; from A as centre describe an arc of radius equal to the length of the pole A, likewise from B describe an arc of radius equal to the length of the pole B. They cut in the point o-^. From o^ drop a perpendicular on AB and produce ; similarly, by describing arcs from B and C of their respective radii, find the point o-a, and from it drop a perpendicular on BC, and produce to meet the perpendicular from Ox in O, which is the apex of the tripod. The plan of the three legs can now be filled in, viz. AO, BO, CO. Produce AO to meet CB in D. From O set off the height of the apex above the plane, viz. O^uj, at right angles to AO ; join Af^m. This should be measured to see that it checks with the length of the pole A. Join DiPm. From o-a\ set off a length to a con- venient scale to represent W, complete the parallelogram of forces, then Oxa^ gives the force acting down the leg A, and o-a.\d the force acting down the imaginary leg D, shown in broken line, which lies in the plane of the triangle OBC ; resolve this force down OC and OB by setting off o-ad along OuD equal to o-a.\d, found by the preceding parallelogram, then the force acting down the leg B is Oyj), and that down C is o-aC The horizontal force tending to spread the legs AOm and DOui is given by fd. This is set off at A/,'and is resolved along AC and AB. The force acting on an imaginary tie AC is Ke, and on AB is A^, similarly with the remaining tie. When the three legs are of equal length and are symmetric- ally placed, the forces can be obtained thus — l^J-V - ^ Three equal fe^s Fig. 138*. where / is the force acting down each leg. ii6 Mechanics applied to Engineering. Forces in the Members of a Roof Truss.— Let the roof truss be loaded with equal weights at the joints, as shown ; the reactions at each support will be each equal to half the total load on the structure. We shall for the present neglect the weight of the structure itself. .t The forces acting on each member can be readily found by a special application of the polygon of forces. Consider the joint at the left-hand support BJA or Rj. We have three forces meeting at a point ; the magnitude of one, viz. Rj or ba, and the direction of all are known ; hence we can determine the other two magnitudes by the triangle of forces. This we have done in the triangle ajh. Resolution of Forces 1 1 7 Consider the joint BJIC Here we have four forces meeting at a point ; the magnitude of one is given, viz. be, and the direction of all the others ; but this is not sufficient — we must have at least six conditions known (see p. no). On referring back to the triangle of forces just constructed, we iind that the force bj is known ; hence we can proceed to draw our . polygon of forces cbji by taking the length of bj from the tri- angle previously constructed. By proceeding in a similar manner with every joint, we can determine all the forces acting on the structure. On examination, we find that each polygon contains one side which has occurred in the previous polygon ; hence, if these similar and equal sides be brought together, each polygon can be tacked on to the last, and so made to form one figure con- taining all the sides. Such a figure is shown below the structure, and is known as a " reciprocal diagram." When determining the forces acting on the various members of a structure, we invariably use the reciprocal diagram without going through the construction of the separate polygons. We have only done so in this case in order to show that the reciprocal diagram is nothing more nor less than the polygon of forces. We must now determine the nature of the forces, whether tensile or compressive, acting on the various members. In order to do this, we shall put arrows on the bars to indicate the direction in which the bars resist the external forces. The illustration represents a man's arm stretched out, resisting certain forces. The arrows indi- cate the direction in which he is exerting himself, from which it ^^^ will be seen that when the arrows ' ^^' on his arms point outwards his arms are in compression, and when in the reverse direction, as in the chains, they are in tension ; hence, when we ascertain the directions in which a bar is resisting the external forces acting on it, we can at once say whether the bar is in tension or compression, or, in other words, whether it is a tie or a strut. We know, from the triangle and polygon of forces, that the arrows indicating the directions in which the forces act follow round in the same rotary direction ; hence, knowing the direc- tion of one of the forces in the polygon, we can immediately find the direction of the others. Thus at the joint BJA we ii8 Mechanics applied to Engineering. know that the arrow points upwards from b Xa a; then, con- tinuing round the triangle, we get the arrow-heads as sliown. Transfer these arrows to the bars themselves at the joint in question ; then, if an arrow points outwards at one end of a bar, the arrow at the other end must also point outwards ; hence we can at once put in the arrow at the other end of the bar, and determine whether it is a strut or tie. When the arrows point outwards the bar is a strut, and when inwards a tie. Each separate polygon has been thus treated, and the arrow- heads transferred to the structure. But arrow-heads must not be put on the reciprocal diagram ; if they are they will cause hopeless confusion. With a very little practice, however, one can run round the various sections of the reciprocal diagram by eye, and put the arrow-heads on the structure without making a single mark on the diagram. If a mistake has been made anywhere, it is certain to be detected before all the bars have been marked. If the beginner experiences any difficulty, he should make separate rough sketches for each polygon of forces, and mark the arrow-heads ori each side. At some joints, where there are no external forces, the direction of the arrows will not be evident at first; they must not be taken from other polygons, but from the arrow-heads on the structure itself at the joint in question. For example, the arrows at the joints ABJ and BJIC are perfectly readily obtained, the direc- tion being started by the forces AB and BC, but at the joint JIA the direction of the arrow on the bars JI and JA are known at the joint ; either of these gives the direction for starting round the polygon ahij. The following bars are struts : BJ, IC, GD, FE, JI, GF. The following bars are ties : JA, IH, HA, HG, FA. Some more examples of reciprocal diagrams will be given in the chapter on " Framework St.Tictures." CHAPTER V. MECHANISMS. Professor Kennedy '^ defines a machine as " a combination of resistant bodies, whose relative motions are completely constrained, and by means of which the natural energies at our disposal may be transformed into any special form of work." Whereas a mechanism consists of a combination of simple links, arranged so as to give the same relative motions as the machine, but not necessarily possessing the resistant qualities of the machine parts ; thus a mechanism may be regarded as a skeleton form of a machine. Constrained and Free Motion. — Motion may be either constrained or free. A body which is free to move in any direction relatively to another body is said to have free motion, but a body which is constrained to move in a definite path is said to have constrained motion. Of course, in both cases the body moves in the direction of the resultant of all the forces acting upon it ; but in the latter case, if any of the forces do not act in the direction of the desired path, they automatically bring into play constraining forces in the shape of stresses in the machine parts. Thus, in the figure, let a^ be a crank which revolves about a, and let the force be in the direc- tion of the connecting-rod act on the pin at b. Then, if b were free, it would move off in the direction of the dotted line, but as b must move in a circular path, a force must act along the crank in order to prevent it following the dotted line. This force acting along the crank is readily found by resolving be in ' " Mechanics of Machinery,'' p. 2. 120 Mec/ianics applied to Engineering. a direction normal to the crank, viz. bd, i.e. in the direction in which b is moving, and along the crank, viz. dc, which in this instance is a compression. Hence the path of 6 is determined by the force acting along the connecting-rod and the force acting along the crank. The constraining forces always have to be supplied by the pdrts of the machine itself. Machine design consists in arranging suitable materials in suitable form to supply these constraining forces. The various forms of constrained motion we shall now consider. Plane Motion. — When a body moves in such a manner that any point of it continues to move in one plane, such as in revolving shafts, wheels, connecting-rods, cross-heads, links, etc., such motion is known as plane motion. In plane motion a body may have either a motion of translation in any direction in a given plane or a motion of rotation about an axis. Screw Motion. — ^When a body has both a motion of rotation and a translation perpendicular to the plane of rota- tion, a point on its surface is said to have a screw motion, and when the velocity of the rotation and translation are kept constant, the point is said to describe a helix, and the amount of translation corresponding to one complete rotation is termed t\i& pitch of the helix or screw. Spheric Motion. — When a body moves in such a manner that every point in it remains at a constant distance from a fixed point, such as when a body slides about on the surface of a sphere, the motion is said to be spheric. When the sphere becomes infinitely great, spheric motion becomes plane motion. Relative Motion. — When we speak of a body being in motion, we mean that it is shifting its position relatively to some other body. This, indeed, is the only conception we can have of motion. Generally we speak of bodies as being in motion relatively to the earth, and, although the earth is going through a very complex series of movements, it in nowise affects our using it as a standard to which to refer the motions of bodies ; it is evident that the relative motion of two bodies is not affected by any motions which they may have in common. Thus, when two bodies have a common motion, and at the same time are moving relatively to one another, we may treat the one as being stationary, and the other as moving relatively to it : that is to say, we may subtract their common Mecltan isms. 121 motion from each, and then regard the one as being at rest. Similarly, we may add a common motion to two moving bodies without affecting their relative motion. We shall find that such a treatment will be a great convenience in solving many problems in which we have two bodies, both of which are moving relatively to one another and to a third. As an example of this, suppose we are studying the action of a valve gear on a marine engine; it is a perfectly simple matter to construct a diagram showing the relative positions of the valve and piston. Precisely the same relations will hold, as regards the valve and piston, whether the ship be moving forwards or backwards, or rolling. In this case we, in effect, add or subtract the motion of the ship to the motion of both the valve and the piston. Velocity. — Our remarks in the above paragraph, as regards relative motion, hold equally well for relative velocity. Many problems in mechanisms resolve themselves into finding the velocity of one part of a mechanism relatively to that of another. The method to be adopted will depend upon the very simple principle that the linear velocity of any point in a rotating * /^ body varies directly as the distance of that 'a^^"<^'^ point from the axis or centre of rotation. ^■°- '*^" Thus, when the link OA rotates about O, we have — velocity of A V„ r^ velocity of B ~ Vj ~ rj If the link be rotating with an angular velocity w radians per second (see p. 4), then the linear velocity of a, viz. V. = uir„ and of b, Vj = eo^j, 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 ab be part of a mechanism having a complex motion. The paths of the two end points, a and b, 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 b, which give us the directions in which each is moving at the instant. From the points a, b draw normals aa' and bl/ to the tangents ; then the centre about which a is moving at the instant lies somewhere on the Une ad, likewise with bb' ; 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. \ 23 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 n, 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. Oentrode and Axode. — As the link m moves in such a manner that its end joints a and i 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 tiie 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 r<?»n. «?■-• Fio. 146. roll on one another, because A, B and C, D are the foci of the two ellipses. 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 Odd 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 Obc. Again, let d be I fixed, and consider the motion of b relatively to d. The point- path of one end of b, viz. Oab, describes the arc of a circle about Oad, therefore the virtual centre lies on a produced j for a similar reason, the virtual centre lies on c produced hence it must be at Obd, the meet of the two lines. 0^0 Oaxs Fig. 14S. Odo In a similar manner, consider the link e 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. Ocd. oOcd Oad, Obc Fig. 150. The mechanism shown in Fig. 149 is kinematically similar to the mechanism in Fig. 150. Instead of c sliding to and fro in guides, a link of any length may be substituted, and the fixed link d may be carried round in order to provide a centre from which c shall swing. Then it is evident that the joint Obc moves in the arc of a circle, and if c be infinitely long 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 Oac. Consider the link c as fixed, then the bar d swings about the centre Ocd; hence every point in it moves in a path at right angles to a line drawn from that point to Ocd. Hence the virtual centre lies on the line Ocd, Oad; also, for reasons given below, it lies on the prolongation of the bar b, viz. Oac. 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 Ocui, Oac, Ocd are on one line ; also Obc, Obd, 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, an J let the virtual centre of its motion relative to a be O^^ ; likewise let Oac be the virtual centre of c's motion relative to a. If we Mechanisms. X'Z'J want to find the velocity dof a point in b relatively to a point ii €, we must find the virtual centre, Obc. Let it be at O : then, considering it as a point of b, it will move in the arc i.i struck from the centre Oab; but considering it as a point in c, it will move in the arc 2.2 struck from the centre Oac. But the tangents of these arcs intersect at 0, therefore the point O has a motion in two directions at the same time, which is im- fig. 151. possible. In the same manner, it may be shown that the virtual centre Obc cannot lie any- where but on the line joining Oab, Oac, for at that point only will the tangents to the point-paths at O coincide ; therefore the three virtual centres must lie on one straight line. Relative Linear Velocities of Points in Mechan- isms. — Once having found the virtual centre of any two bars of a mechanism, the finding of the velocity of any point in one bar relatively to that of any other point is a very simple matter, for their velocities vary directly as their ^ virtual radii. In the mechanism shown, let the bar d be fixed; to find the relative velocities of the points i and 2, we have — velocity i _ Obd i _ ^1 velocity 2 Obd 2 r^ Similarly — velocity i _ r-^ velocity 3 r, ^^^ vdocity 3 ^ /3 velocity 4 rt, The relative velocities are not affected in the slightest degree by the skape of the bars. When finding the velocity of a point on one bar relatively to the velocity of a point on another bar, it must be remembered 128 Mechanics applied to Engineering, that at any instant the two bars move' as though they had one point in common, viz. their virtual centre. As an instance of points on non-adjacent bars, we will pro- ceed to find the velocity of the point 8 (Fig. 153) relatively to that of point 9. By the method already explained, the virtual centre Oac is found, which may be regarded as a point in the bar c pivoted at Ocd; likewise as a point in the bar a pivoted Fig. 153- at Oad. As an aid in getting a clear conception of the action, imagine the line Ocd . Oac, also the bar c, to be arms of a toothed wheel of radius /04, and the line Oad . Oac, also the bar a, to be arms of an annular toothed wheel of radius pa, the two wheels are supposed to be in gear, and to have the common point Oac, therefore their peripheral velocities are the same. Denoting the angular velocity of a as o)„ and the linear velocity of the point 8 as Vg, etc., we have — <^ap3 = <^cpiy and <o„ = -^ also Vg = oi^ps Vg = 01^ Substituting the value of (i)„, we have — Vg _ MqPjPs _ piPs ^ i'29 X 076 _ ^ .^^ Vg _ OM.S Y,~ Odd.g~ 2-i6 or paPs 2'20 2'29 X 0'42 = I'02 Mechanisms. 1 29 Similarly, if we require the velocity of the point 6 relatively to that of point 5 — Vg _ Pb V - ^8P6 iT » 6 — Vs Pa Pi ^j — P} V = X5P? V9 P9 P9 whence — = — ' .^5^= P^'P'^" = M.' Vs Vj pePn PsPsPePi PsPe Voac Ps Vfl^ p4 Vg p4pe 1-29 X 0-3 = 0-36 V5 psPs 2-29 X 0-47 Likewise — velocity 7 _ R? _ 2'3i _ velocity 8 Rs 2-20 ~ velocity 6 p^ x velocity 8 X R hence velocity 7 Ps X K.7 X velocity 8 = 0-38 Ps X R7 076 X 2-31 Fig. 154. This can be arrived at much more readily by a graphical process; thus (Fig. 154): With Oad as centre, and p, as K 130 Mechanics applied to Engineering. radius, set off Oad.h = p, along the line joining Oad to 8 ; set off a line Ai to a convenient scale in any direction to represent the velocity of 6. From Oad draw a line through i, and from 8 draw a line Be parallel to /it ; this line will then represent the velocity of the point 8 to the same scale as 6, for the two triangles Oad.8.e and Oad.h.i are similar ; therefore — 8(? Oad • 8 _ Pa _ velocity 8 _ o'8o _ hi Oad.h p^ velocity 6 0-32 From the centre Odd and radius Rj, set off Odd./ = Rj ; 6ia.vf/.g parallel to <?. 8, and from Odd draw a line through e to meet this line in g", then,/^ = V,, for the two triangles Obd.f.g and Obd.i.e are similar; therefore — fg _ Obd.f _ R7 _ velocity 7 _ 2-3 8^ ~ Obd.B ~ Rs ~ velocity 8 ~ 2^ ^ ^'°^ , velocity 6 ih o'^z and —. — H^ = T = —5- = o'38 velocity 7 ^ o'84 "^ 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 m; then the linear velocity V„ of a point A situated at a radius r„ is — V V„ = u)r„ and m = -^ In order to find the relative angular velocities 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 Oad be r^. When the point A is regarded as a point in the bar b, we shall term it B, and its radius about Obd, r^. Let the linear velocity of A be Vju and that of B be Vb, and the angular velocity of A be 0)4, and of B be Mg. Then V^ = Wjj-^, and Vi = m^rj^. But Va = Vb as A and B are the same point ; hence — <"ii ''a {Oad)A Mechanisms. 131 This may be very easily obtained graphically thus : Set off a line he in any direction from A, whose length on some given scale is equal to (Oij join e.Obd; from Oad draw 0«(^/ parallel b Obd Fig. iss. Fig. 156. X.0 e.Obd. Then A/= 0)3, because the two triangles k.f.Oad and A.e.Obd are similar. Hence — he ^ (Obd)B _ (0^ A/ (Oad)A (ub In Fig. 156, the distance A^ has been made equal to h.Oad, and gf is drawn parallel to e.Obd. The proof is the same as in the last case. When a is parallel to e, 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 e, we proceed thus : For con- venience we have numbered the various points instead of using the more cumbersome virtual centre nomenclature (Figs. 157 and 158). The radius 1.6 we shall term r^^, and so on. Then, considering points i and 2 as points of the bar b, we have — V ■ ''2.6 132 Mechanics applied to Engineering. Then, regarding point i as a point in bar c, and regarding point 2 as a point in bar a — V, = w„ X n.. Vo = (o.ra.; J:: _ s ■?-•"" Fig. 158. Then, substituting the values of Vj and Vj in the equation above, we have — Mechanisms. 133 Draw 4.7 parallel to 2.3 ; then, by the similar triangles 1.2. 6 and 1.7.4 — 4-7 ri.4 and ^a., X i-i., = ?-i,, X 4.7 Substituting this value above, we have — ^c 1.6 X ^as '1.8 = -M, or = iiP by similar triangles X 4-7 4-7 5-4 Thus, if the length ^2.3 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 off; 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 Oac 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. the ordinary steam- engine mechanism. ^^^^ y Having given the ' fig. 139. 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 Obd.V velocity X ~ OW.X ■XM 1 34 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 QI2e is similar to the triangle P.OW.X ; hence — OP Obd.V velocity of pin "p7 ~ Obd.^ ~ velocity of cross-head But the velocity of the crank-pin may be taken to be constant. Let it be represented by the radius of the crank-circle OP; then to the same scale Te represents the velocity of the cross- head. Set up fg = P<? at several positions of the crank-pin, and draw a curve through them; then the ordinates of this curve represent the velocity of the cross- head at every point in the stroke, where the radius of the crank - circle repre- sents the velocity of the crank-pin. When the connecting-rod is of infinite length, or in the case of such a mechanism as that shown in Fig, i6o, the line gc (Fig. 159) is always parallel to the axis, and consequently the crosshead- velocity diagram becomes a semicircle. An analytical treatment of these problems will be found in the early part of the next chapter. Another problem of considerable interest in connection with the steam-engine is that of finding the journal velocity, or the velocity with which the various journals or pins riib on their brasses. The object of making such an investigation will be more apparent after reading the chapter on friction. Let it be required to find the velocity of rubbing of (i) the crank-shaft in its main bearings ; (2) the crank-pin in the big end brasses of the connecting-rod ; (3) the gudgeon-pin in the small end brasses. Let the radius of the crank-shaft journal be r„ that of the crank-pin be r^, and the gudgeon-pin r,. Let the number of revolutions per minute (N) be 160. Let the radius of the crank be i'25 feet. Let the radii of all the journals be 0*25 foot. We have taken them all to be of the same size for the sake of comparing Mechanisms. 135 the velocities, although the gudgeon-pin would usually be considerably smaller. (i) V, = 2irr„N or u)^r„ =250 feet per minute in round numbers (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 Oab may be regarded as a part of the bar a pivoted at Oad^ also as a part of the bar b rotating for the instant about the virtual centre Obd; then, by the gearing conception already explained, we have — il' = ]^», or <o. = '^ = Yi. «. Ri' R. R» When the crank-arm and the connecting-rod are rotating in the opposite sense, the rubbing velocity — This has its maximum value when ^ is greatest, i.e. when R» 136 Mechanics applied to Engineering. Rj is least and is equal to the length of the connecting-rod, i.e. at the extreme " in " end of the stroke. Let the connecting-rod be n cranks long ; then this expression becomes — v, = .,4+i) which gives for the example taken — = 314 feet per minute taking « = 4. But when the crank-arm and the connecting-rod are rotating in the same sense, the rubbing velocity becomes — V =.,«.„(.- 1) The polar diagram shows how the rubbing velocity varies at the several parts of the stroke. Conn£ctui'9 rod b? O \ •' tnol/iolder Gegrwilh '"^'•e Off- ^■e'd Fig. i6xa. (3) Since the gudgeon-pin itself does not rotate, the rubbing velocity is simply due to the angular velocity of the connecting- rod. Mechanisms. 117 which has its maximum value when R, is least, viz. at the extreme end of the " in " stroke, and is then 63 feet per minute in the example we have taken. By taking the same mechanism, and by fixing the link b instead of d, we get another familiar form, viz. the oscillating cylinder engine mechanism. On rotating the crank the link d becomes the connecting-rod, in reality the piston rod in this case, and the link c oscillates about its centre, which was the gudgeon-pin in the ordinary steam-engine, but in this case it is the cylinder trunnion, and the link c now becomes the cylinder of the engine. Another slight modification of the same inversion of the mechanism is one form of a quick-return motion used on shaping machines. In Fig. i6ia we show the two side by side, and in the case of the engine we give a polar diagram to show the angular velocity of the link c at all parts of the stroke when a rotates uniformly. We shall again make use of the gearing conception in the solution of this problem, whence we have — o)„R„ = oj^Ra, and <*<« = t> R. 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- sponding 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 of the gear," which is readily determined. The link c is in its extreme position when the link « is at right angles to it ; the cutting angle is 360 — B, and the return angle B (Fig. 161J). Fig. i6ij. •38 Mechanics applied to Engineering. The ratio of the gear R = 360 -g or 0(R + i) = 360° and a = b cos — 2 Let R = 2 ; B = 1 20° J b = 2a. ^ = 3 J ^ — 9°° > ^ — 'i'42a. 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 Q Fig. i6k. a — d COS — : therefore d must be made 2 equal to 2a for a ratio of 2, and i'42a 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«:). This special form was adopted on account of its lending itself readily to a variation of the stroke of the piston as may be required for various powers. This varia- tion is accomplished by shift- ing the point Oad to or from the centre of the fl5rwheel Oab. A very curious develop- ment of the steam-engine mechanism is found in Stan- nah's pendulum pump (Fig. 161/^. The link C is fixed, and the link d simply rocks to and fro ; the link a is the flywheel, and the pin Oab is attached to the rim and works in brasses fitted in an eye in the piston-rod. The velocity of the point Fig. i6irf. Oab is the same as that of any other point in the link b. When C is fixed, the velocity of b relatively to C is the same as the velocity of C relatively to b when b is fixed, whence from p. 133 we have — Mechanisms. 139 ^ Oad 'Oab R. «tfjRj R. ",R, = 0)^ The Principle of Virtual Velocities applied to Mechanisms. — If a force acts on any point of a mechanism and overcomes a resistance at any other point, the work done at the two points must be equal if friction be neglected. In Fig. 162, let the force P act on the link a at the point 2. Find the magnitude of the force R at the point 4 required to keep the mechanism in equilibrium. If the bar a be given a small shift, the path of the point 2 will be normal to the link a, and the path of the point 4 will be normal to the radius 5.4. Resolve P along Fr parallel to a, which component, of course, has no turning effort on the bar ; also along the normal P« in the direction of motion of the point 2. Likewise resolve R along the radius Rr, and normal to the radius R«. Now we must find the relative velocity of the points 2 and 4 by methods previously explained, and shown by the construc- tion on the diagram. Then, as no work is wasted in friction, we have — ^ ' The small simultaneous displacements are proportional to the velocities. 140 Mechanics applied to Engineering. ^H/- e-S J?^ 4 -~L--^ '^3 ,^ ■^^%: e'-^<r\ «-r X "I ^ \ \a d N kg \ "^ ^ / A /'~. V >io -- o ■a 13 V eed°^°'>^_ 1 Seconds V) o / 3 3 4 5 6 7 e a lo II IS 13 /^ /s /e /il Seconds Seconds \ \ \ • Seconds Tio. i6j. Mechanisms^ 141 P„V3 = R„V4 and Rn = ^ ' 4 Velocity and Acceleration Curves. — Let the link a revolve with a constant angular velocity. Curves are con- structed to show the velocity of the point / relatively to the constant velocity of the point e. Divide the circle that e describes into any convenient number of equal parts (in this case 16). The point / will move in the arc of a circle struck from the centre g; then, by means of a pair of compasses opened an amount equal to ef, from each position of e set off the corresponding position of/ on the arc struck from the centre g. By joining up he, ef,fg, we can thus get every position of the mechanism ; but only one position is shown in the figure for clearness. Then, in order to find the relative velocity of e and /, we produce the links a and c to obtain the virtual centre Obd. This will often come off the paper. We can, however, very easily get the relative velocities by drawing a line ^'parallel to c. Then the triangles hej and Obd.e.f are similar, therefore — he _ Obd.e _ velocity of e hj ~ Obd.f velocity of/ The velocity of e is constant ; let the constant length of the link a, viz. he, represent it; then, from the relation above, hj will represent on the same scale the velocity of/. Set off on a straight line the distances on the e curve o. i, 1.2, 2.3, etc., as the base of the speed curve. At each point set up ordinates equal to 4/' for each position of the mechanism. On drawing a curve through the tops of these ordinates, we get a complete speed curve for the point / when the crank a revolves uniformly. The speed curve for the point if is a straight line parallel to the base. In constructing the change of speed curve, each of the divisions o.i, 1.2, 2.3, etc., represents an interval of one second, and if horizontals be drawn from the speed curve as shown, the height X represents the increase in the speed during the interval o.i, i.e. in this case one second; then on the change of speed diagram the height X is set up in the middle of each space to show the mean change in speed that the point / has undergone during the interval 0.1, and so on for each space. A curve drawn through the points so obtained is the rate of change of speed curve for the point /. 142 Mechanics applied to Engineering. The velocity curve is obtained in the same way as the speed curve, but it indicates the direction of the motion as well as its speed, and similarly in the case of the acceleration curve. Let the curve (Fig. 164) 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 , . ,, • , . . dv . change of velocity the acceleration during the interval is — , t.e.—. — - — ; — — -: — -, dt interval of time or the change of velocity in the given interval. By similar triangles, we have — • = — . ■' ° ' 'f* xy dt When dt= 1 second, dv = mean acceleration. Hence, make xy^ = i on the time scale, then z^y^, the subnormal, gives us the acceleration 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 -r. ' ■ Calculus " and Duncan's / / "Practical Curve Tracing" Fig. 164, /T \ ° (Longmans). 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.') 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^ as shown by the arrow ; the angular velocity (It of the body must also be known. If oi be zero, then every • " Graphics," by R. H. Smith, Bk. I. chap. ix. ; or " Kinematics of Machines," by R. J. Durley (J. Wiley & Sons), Mechanisms. H3 point in the body will move with the same velocity as V^. From A draw a line at right angles to the direction of motion as indicated by Va, then the body is moving about a centre situated somewhere on this line, but since we know V^ and 0), we can find the virtual centre P, since coRi = Va. 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 — Vb = «).PB, and Vo = u-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 Va ; (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 ab 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/^ 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 Va 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 Va, and ab its motion at right angles to AB, whence /5 represents the velocity of B in magnitude and direction. Similarly, the point C par- takes of the general motion /a in the direction Va, and, due to the spin of ttie 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 /a 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 Prr^,, R-9 Fig. i66. 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^ _ AF . EQ Vb ED . QF Mechanisms. 145 Tlie points C and B may be regarded as points on the bar AD, whence — PC PB RG Vc = Vi>p^, and Vb = V^p^, and V = V^^ Let the radius of the crank be 16', and the revolutions per minute be 80 ; then — V,= _ 3'i4 X 2 X 16 X 80 = 670 feet per minute Then we' get — Vg = 250 feet per minute Vc = 790 Vd = 515 Vg = 246 „ „ Vh = 792 .. .. Cnmkpbi •, Valve rod • Fig. 167. 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 /, 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 146 Mechanics applied to Engineering. the velocity of A : likewise draw pb 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 j. 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//,/ s, which give respectively the velocities 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 z', : then pvx 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 v^ draw a line parallel to a tangent at V, and from / a line parallel to the valve-rod, meeting in v ; then pv is the required velocity of the valve, and vv-^ is the velocity of " slip " of the die in the link. Cams. — When designing automatic and other machinery it often happens that the desired motion of a certain portion of the machine cannot readily be secured by the use of ordinary mechanisms such as cranks, links, wheels, etc.; cams must then be resorted to, but unless they are carefully constructed they often give trouble. A rotating cam usually consists of a non-circular disc formed in such a manner that it imparts the desired recipro- cating motion, in its own plane of rotation, to a body or follower which is kept in contact with the periphery of the disc. Another form of cam consists of a cylindrical surface which rotates about its own axis, and has one or both edges of its curved surface specially formed to give a predetermined motion to a body which is kept in contact with them thus, causing it to slide to and fro in a direction parallel, or nearly so, to the axis of the shaft. Generally speaking, it is a simple matter to design a cam to give the desired motion, but it is a mistake to assume that any conceivable motion whatever can be obtained by means of a cam. Many cams which work quite satisfactorily at low speeds entirely fail at high speeds on account of the inertia of the follower and its attachments. The hammering action often experienced on the valve stems of internal combustion engines is a familiar example of the trouble which sometimes arises from this cause. Design of Cams. — (i) Constant Velocity Cams. — In dealing with the design of cams it will be convenient to take definite numerical examples. In all cases we shall assume that they rotate at a constant angular velocity. Let it be required to Mechanisms. 147 design a cam to impart a reciprocating motion of 2 inches stroke to a follower moving at uniform speed {a) in a radial path, (5) in a segment of a circle. In Fig. 168 three cams of different dimensions are shown, each of which fulfils condition {a), but it will shortly be shown that the largest, will give more satisfactory results than the smaller cams. The method of construction is as follows : — Take the base circle abed, say 3 inches diameter. Make ce = 2 inches, that is, the stroke of the follower. Draw the semicircle efgsxA divide it into a convenient number of equal parts — say Scale : f ths full size. Fig. 168. 6 — and draw the radii. Divide ag, the path of the follower, into the same number of equal parts, and from each division draw circles to meet the respective radii as shown, then draw the profile of the cam through these points. It will be obvious from the construction that the follower rises and falls equal amounts for equal angles passed through by the cam, and since the latter rotates at a constant angular speed, the follower therefore rises and falls at a uniform speed, the total lift being ce or ag. The two inner cams are constructed in a similar manner, but with base circles of t inch and 5 inch respectively. The form of the cam is the well-known Archimedian spiral. 148 Mechanics applied to Engineering. The dotted profile shows the shape of the cam when the follower is fitted with a roller. The diameter of the roller must never be altered for any given cam or the timing will be upset. In Fig. 169 a constant velocity cam is shown which raises its follower through 2 inches in g of a revolution a to e, keeps Jrd full size. Fig. 169. it there for 5 of a revolution e Xaf, and then lowers it at a con- stant velocity in g of a revolution /to g where it rests for the remaining \, g to a. The construction will be readily followed from the figure. When the follower is attached to the end of a radius bar the point in contact with the cam moves in a circular arc ag. Fig. 170. In constructing the cam the curved path is reproduced at each interval and the points of intersection of these paths and the circles give points on the cam. The profile shown in broken lines is the form of the cam when the radius bar is provided with a roller. The cams already dealt with raise and lower the follower at a constant velocity. At two points, a and if, Figs. 168 and 170, and at four points a, e,f, g, Fig. 169, the velocity of the follower, if it could be kept in contact with the cam, would be instantaneously changed and would thereby require an infinitely great force, which is obviously impossible ; hence such cams Mechanisms. 149 cannot be used in practice unless modified by "easing off" at the above-mentioned points, but even then the acceleration of the follower may be so great at high speeds that the cam face soon wears irregularly and causes the follower to run in an unsatisfactory manner. By still further easing the cam may be made to work well, but by the time all this easing has been Fig. 170. accomplished the cam practically becomes a simple harmonic cam. (ii) Constant acceleration or gravity cam. — This cam, as its name suggests, accelerates the follower in exactly the same manner as a body falling freely under gravity, hence there is a constant pressure on the face of the cam when lifting the follower, but none when falling. The space through which a falling body moves is given by the well-known relation — S = \gfi The total spaces fallen through in the given times are propor- tional to the values of S given in the table, that is, proportional to the squares of the angular displacements of the radius vectors. Time, / . . . . I 2 4 S 6 Space, S . . . . I 4 9 16 -' 2S 36 Velocity, v . . . I 2 3 4 s 6 ISO Mechanics applied to Engineering. Acceleration diagram for the follower. Fig. 17X. ' ' ' 'y/ Fig. 172. Mechanisms. 151 In Fig. 171 the length of the radius vectors which fall outside the circle abed are set off proportional to the values of S given in the table. For the sake of comparison a " constant velocity " cam profile is shown in broken line. A cam of this design must also be " eased " off at e or the follower will leave the cam face at this point. As a matter of fact a true gravity cam is useless in practice; for this reason designers will do well to leave it severely alone. In Fig. 172 the cam possesses the same properties, but the follower is idle during one half the time — dab— and is then accelerated at a constant rate during be for a quarter of a revolution and finally is allowed to fall with a constant retardation during ed, the last quarter of a revolution. (iii) Simple harmonic cam. — In a simple harmonic cam the motion of the follower is precisely the same as that of a crosshead which is moved by a crank and infinitely long con- necting rod, or its equivalent — the slotted crosshead, see Fig. 160. There would be no reason for using such a cam rather than a crank or eccentric if the motion were required to take place in one complete revolution of the cam, but in A Fig. 173. the majority of cases the s. h. m. is required to take place during a portion only of the revolution and the follower is stationary during the remainder of the time. 152 Mechanics applied to Engineering. In the cam abed shown in Fig. 173 the follower is at rest for one half the time, during dab, and has s. h. m. for the re- mainder of the time, bed. The circle at the top of the figure represents the path of an imaginary crank pin, the diameter of the circle being equal to the stroke of the follower. To con- struct such a cam the semi-circumference of the equivalent crank circle is divided up into a number of equal parts, in this case six, to represent the positions of the imaginary crank pin at equal intervals of time. These points are projected on to the line ag, the path of the follower, and give its corresponding positions. From the latter points circles are drawn to cut the corre- sponding radii of the cam circle. If a cam be required to give s. h. m. to the follower without any period of rest the same construction may be used, the cam itself (only one half of which, afe, is shown) then be- coming approximately a circle with its centre at h. The distance ho being equal to the radius of the imaginary as!' crank viz. — . If the cam were made 2 truly circular, as in Fig. 174, it is evident that oh, the radius of the crank, and ih, the equivalent connect- ing rod, would be of constant Jength, hence the rotation of the cam imparts the same motion to the roller as a Fig 174. crank and connecting rod of the same proportions. Size of Cams. — The radius of a cam does not in any way affect the form of motion it imparts to the follower, but in many instances it greatly affects the sweetness of running. A cam of large diameter will, as a rule, run much more smoothly than one of smaller diameter which is designed to give pre- cisely the same motion to the follower. The first case to be considered is that of a follower moving in a radial path in which guide-bar friction is neglected. In Fig. 168 suppose the cam to be turned round until the follower is in contact at the point i (for convenience the follower has been tilted while the cam remains stationary). Draw a tangent to the cam profile at i and let the angle between the tangent and the radius be 6. The follower is, for the instant, acted upon by the equivalent of an inclined plane on the cam face whose angle of inclination is a = 90 - 6. Mechanisms. 153 The force acting normally to the radius, / = W tan (a + ^) (seep. 291), where ^ is the friction angle for the cam face and follower, and W is the radial pressure exerted by the follower. When a + <^ = 90° the follower will jamb in the guides and consequently the cam will no longer be able to lift it, and as a matter of fact in practice unless a + S^" is much less than 90° the cam will not work smoothly. Let the cam rotate through a small angle 8^. Then a point on the cam at a radius p will move through a small arc of llength 8S = p80. During this interval the follower will move through a radial distance hh. Hence, tan a = — ^ Thus for any given movement of the follower, the angle a, which varies nearly as the tangent for small angles, varies inversely as the radius of the cam p at the point of contact. By referring to Fig. 168 it will be seen that the angle a is much greater for the two smaller cams than for the largest. It has already been shown that the tendency to jamb increases as a increases, whence it follows that a cam of small radius has a greater tendency to jamb its follower than has a cam of larger radius. The friction angle ^ is usually reduced by fitting a roller to the follower. When the friction between the guides and follower is considered, we have, approximately, / = (W +/ tan <^i) (tan {a. + <^\) where ^1 is the friction angle between the guide and follower. The best practical way of reducing the guide friction of the follower is to attach it to a radius arm. When the follower moves in a radial path the cam, if symmetrical in profile, will run equally well, or badly, in both directions of rotation, but it will work better in one direction and worse in the other when the follower is attached to a radius arm. In Fig. 170 let the cam rotate until the follower is in contact at the point j, draw a tangent to the profile at this point, also a line making an angle <^, the friction angle, with it. The resultant force acting on the follower is ib where id is the pressure acting normally to the radius of the follower arm iy and ic is drawn normal to the tangent at the point i. The line di is drawn normal to id. The force ib has a clockwise moment ib X r about the pin y tending to lift the follower, if the slope of the cam is such that bi passes through y the cam 1 54 Mechanics applied to Engineering. will jamb, likewise if in this particular case the moment of the force about y is contra-clockwise, no amount of turning effort on the cam will lift the follower. Velocity ratio of Cams and Followers. — The cam A shown 'in Fig. 175 rotates about the centre Oab. Since the follower c moves in a straight guide, the virtual centre Obc is on a line drawn at right angles to the direction of ihotion of c and at infinity. Draw a tangent to the cam profile at the point of contact y which gives the direction in which sliding is taking place at the instant. The virtual centre of A and C therefore lies on a line passing through the point of contact y and normal .to the tangent, it also lies on the line Oab Obc, Fig. 175. Fig. 176. therefore it is on the intersection, viz. Oac. The virtual radius of the cam is Oab . x, i.e. the perpendicular distance of Oab from the liormal at the point of contact. Let the angular velocity of the cam be ua, then 0)^ x Oab . x = V where V is the com- ponent of C's velocity in the direction yx. In the triangle of velocities, V„ represents the velocity of c in the guides, and V, the velocity of sliding. This triangle is similar to the triangle Oab ■ Oac , x . hence V Oab . Oac Oab .X Mechanisms. 1 5 S Substituting the value of V we have V„ _ Oah . Oac (i)j^ . Oab . X Oab . x V and — = Oab . Oac ("a In the case in which the follower is attached to an arm as shown in the accompanying figure, Fig. 176, the virtual centres are obtained in a similar manner, and toSibc . Oac = wjdab . Oac u„ _ Oab . Oac 0)^ Obc . Oac we also have the velocity V^ of the point of the follower in contact with the cam, Y, = ,^j:)bc.y ft), Oab . Oac V, 0), Obc. Oac Obey X &)« V„ Oab. Oac. Obey ^ , ^ — = - — „, „ = Oab . Oac o)« Obc . Oac f Obey \ \0bc.0ac) When the follower moves in a radial path the virtual centre is at infinity and the quantity in the bracket becomes unity; the expression then becomes that already found for this particular case. In Figs. 171, 172, 173, polar velocity diagrams are given for the follower ; the velocity has been found by the method given above, a tangent has been drawn to the cam profile at y, the normal to which cuts a radial line at right angles to the radius Oy in the point z, the length Oz is then trans- ferred to the polar velocity diagram, viz. OV, in some cases enlarged for the sake of clearness. In Figs. 168, 170, the velocity of the follower is zero at a and e, and immediately after passing these points the velocity is finite ; hence if the follower actually moved as required by such cams, the change of velocity and the accele- ration would be infinite at a and e, which is impossible ; hence if such cams be used they must be eased off at the above- mentioned points. In Fig. 169 there are four such points; the polar diagram 156 Mechanics applied to Engineering. indicates the manner in which the velocity changes ; the dotted lines show the effect of easing oif the cam at the said points. The cams shown in Figs. 171, 172, are constructed to give a constant acceleration to the. follower, or to increase the velocity of the follower by an equal amount in each succeed- ing interval of time ; hence the polar velocity diagram for such cams is of the same form as a constant velocity cam, viz. an Archimedian spiral. Having constructed the polar velocity diagram, it is a simple matter to obtain the acceleration diagram from it, since the acceleration is the change of velocity per second. Assume, in the first place, that the time interval between any two adjacent radius vectors on the polar velocity diagram is one second, then the acceleration of the follower — to a scale shortly to be determined — is given by the difference in length between adjacent radius vectors as shown by thickened lines in Figs. 171, r72, 173. The construction of the acceleration diagram for the simple harmonic cam is given in Fig. 173. The radius vector differences on the polar diagram give the mean accele- rations during each interval ; they are therefore set off at the middle of each space on the base line ij, and at right angles to it. It should be noted that these spaces are of equal length and, moreover, that the same interval of time is taken for the radius vector differences in both cams. In the case of the cam which keeps the follower at rest for one half the time, the lifting also has to be accomplished in one half the time, or the velocity is doubled ; hence since the radial acceleration varies as the square of the velocity of the imaginary crank pin, the acceleration for the half-time cam is four times as great as that for the full time cam. Similar acceleration diagrams are obtained by a different process of reasoning in Chapter VI., p. 180. Analytical methods are also given for arriving at the acceleration in such a case as that now under consideration. The particular case shown in Fig. 173 is chosen because the results obtained by the graphic process can be readily checked by a simple algebraic expression shortly to be given. In the above-mentioned chapter it is shown that when a crank makes N revolutions per minute, and the stroke of the cross- head, which is equivalent to the travel of the follower in the case of a simple harmonic cam, is 2R (measured in feet), the maximum force in pounds weight, acting along the centre line Mechanisms. 157 of the mechanism, required to accelerate a follower of W pounds is o'ooo34WRN'^ for the case in which the follower is lifted to its full extent in half a revolution of the cam, such as afe (only one half of the cam is shown). But when there is an idle period, such as occurs with the cam abe. Fig. 173, the velocity with which the follower is lifted is greater than before in the ratio where a. is the angle passed through by the a cam while lifting the follower. Since the radial acceleration varies as the square of the velocity, we have for such a case— The maximum force 'j in pounds weight I _ o'ooo34WRN2 x i8o2 _ ii'osWRN^ required to accele- j ~ ^2 ~ ^,2 rate the follower ) The pressure existing between the follower and the face of a simple harmonic cam, when the follower falls by its own weight is — ,-. , II-03WRN2 maximum pressure, W -\ ^^ ■ a." . . „, iro3WRN2 mmimum pressure, W =^=-r a'' When the follower is just about to leave the cam face — iro3WRN2 w ^ a" and the speed at which this occurs is — Vr Hence a simple harmonic cam must never run at a higher speed than is given by this expression, unless some special provision is made to prevent separation. A spring attached either directly to the follower or by means of a lever is often used for this purpose. When directly attached to the follower, the spring should be so arranged that it exerts its maximum effort when the follower is just about to leave the cam face, II-03WRN2 which should therefore be not less than 5 W, when the follower simply rests on the cam, and n-o-!WRN2 ^^—5 + W, when the follower is below the cam. 158 Mechanics applied to Engineering. The methods for finding the dimensions of such springs are given in Chapter XIV. Instead of a spring, a grooved cam (see Fig. 177), which is capable of general application, or a cam with double rollers Fig. 177. Fig. 178. on the follower (see Fig. 178), can be used to prevent sepa- ration, but when wear takes place the mechanism is apt to be noisy, and the latter is only applicable to cases where the requisite movement can be obtained in 180° movement of the cam. When the follower is attached to an arm, the weight of the arm and its attachments Wa must be reduced to the equivalent weight acting on the follower. Let K = the radius of gyration of the arm about the pivot. r„ = the radius of the c. of g. of the arm about the pivot. r = the radius of the arm itself from the follower to the pivot. We = the equivalent weight acting on the follower when considering inertia effects. W„ = Ditto, when considering the dead weight of the arm. W K W r Then W, = -^-^ and W„ = ^^^^ Then II-03WRN2 iro3W,KRN2 and W becomes a." W„/-. in the expressions above for the case when the follower is attached to a radius arm. Mechanisms, 159 General Case of Cam Acceleration. — In Fig. 179, velocity and acceleration diagrams have been worked out for such a cam as that shown in Fig. 175. The polar velocity diagram has been plotted abonj a zero velocity circle, in which the radius vectors outside the circle represent velocities of the follower when it is rising, and those inside the circle when it is falling ; they have afterwards been transferred to a straight base, and by means of the method given on p. 141, the acceleration diagram has also been constructed. It will be seen that even with such a simple-looking cam the accele- i 6 z ration of the follower varies considerably in amount and rapidly, and the sign not unfrequently changes. Scale of Acceleration Diagrams.— Let the drawing of the cam be -th full size,- and let the length of the radius n vectors on the polar velocity diagram be m times the corre- sponding length obtained from the cam drawing. In Fig. 171, /« = 3. In Fig. 172, m = X. In Fig. 173, m = 2. 27rN Let the cam make N revolutions per mmute, or -g— = radians per second. 9"55 The velocity of the follower at any point where the length of the radius vector is OV (see Fig. 173) measured in feet, is V, = feet per second. i6o Mechanics applied to Engineering. One foot length on the polar diagram represents inch 9"5S»2| feet per second Let the angle between the successive radius vectors be 6 degrees, and let the change of velocity during one of the intervals be SV. Then each portion of the circle subtending 9 represents an interval of time 8i = ^ ^^ = ,-tt- seconds. ^ 360N 6N The acceleration of the follower is — 8V _ 8/ X N2 X « X 6 8/ ^ 'X N2x« feet per second per second. 9"55 Xfnxd 1-59 XmX6) Where 8/ is the difference in length of two adjacent radius vectors, measured in feet. The numeral in the denominator becomes ig-i if 8/ is measured in inches. The force due to the acceleration of the follower is in all W cases found by multiplying the above expressions by — or •§" the equivalent when a rocking arm is used. Useful information on the designing and cutting of cams will be found in pamphlet No. 9 of " Machinery's " Reference Series, published by " Machinery," 27, Chancery Lane ; "A Method of Designing Cams," by Frederick Grover, A.M.LC.E. ; Proceedings I.C.E., Vol. cxcii. p. 258. The author wishes to acknowledge his indebtedness to Mr. Frederick Grover, of Leeds, for many valuable sugges- tions on cams. the figure let A and B be the A rotates at a given speed ; it is required to drive B at some predetermined speed of rota- tion. If the shafts be pro- vided with circular discs a and b of suitable diameters, and whose peripheries are kept in contact, the shaft A will drive B as desired so long as there is sufficient friction at the line of contact, but the latter condition is the twisting moment is small, and no In order to prevent slipping each wheel Toothed Gearing. — In end elevations of two shafts. Fig. 180. only realized when slipping takes place. Mechanisms. 1 6 1 must be provided with teeth which gear with one another, and the form of which is such that the relative speeds of the two shafts are maintained at every instant. The circles representing the two discs in Fig. i8o are known as the pitch circles in toothed gearing. The names given to the various portions of the teeth are given in Fig. i8i. Clearance — ' /f\ —-.^^^^ fioot circle \^ Fig. i8i. The pitch P is measured on the pitch circle. The usual proportions are A = o'sP, B = 0*47 to o'48P, C = 0-53 to 0-52P. The clearance at the bottom of the teeth is o'lP, thus the total depth of the tooth is 07 P. The width is chosen to suit the load which comes on each tooth — for light wheels it is often as small as o'sP and for heavy gearing it gets up to 5 P. Most books on machine design assume that the whole load is concentrated on the corner of the tooth, and the breadth B calculated accordingly, but gearing calculated on this basis is far heavier than necessary. Velocity Ratio. — Referring to Fig. 180, let there be sufificient friction at the line of contact to make the one wheel revolve without slipping when the other is rotated, if this b.e so the linear velocity of each rim will be the same. Let the radius of a be r„, of bhe. r^; the angular velocity of a be a)„, of ^ be wj ; N^ = the number of revolutions of « in a unit of time; N} = the number of revolutions of ^ in a unit of time. Then m^ = — ^— ? = 27rN„, and wj = 27rNj 'a „, '■»_"«_ ^irN^ _ N„ ^^— — — sr ^irr 7\ u)j 2TrNi, Nj M 1 62 Mechanics applied to Engineering. or the revolutions of the wheels are inversely proportional to their respective radii. The virtual centres of a and c, and of b and c, are evidently at their permanent centres, and as the three virtual centres must lie in one line (see p. 126), the virtual centre Oab 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 Oab, the point of contact of the two discs. To insure that the velocity ratio at every instant shall be constant, the virtual centre Oab must always retain its present position. We have shown 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 projection or a tooth, say on A (Fig. 182), 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 Oab. 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 ^ (aW\ surfaces at the point of con- ^5~\^I^Wj \. tact must pass through the /^\~7/w.^/\v virtual centre Oah. If such /A \/ '^^^^^^\p\ ^ normal do not pass through _ ^v.^_^ccE^ _\.J^- Oa*, the velocity ratio will ^ — r ^^ be altered, and if Oab shifts J\ about as the one wheel ^ moves relatively to the Fig. 182. Other, the motion will be jerky. Hence, in designing the teeth of wheels, we must so form them that they fulfil the condition that the normal to their profiles 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. 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 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 Mechanisms. 163 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. 183) ; when on the outside of a circle, the curve traced out is Fig. 184. termed an epicycloid (Fig. 184) ; when on the inside of a circle, the curve traced out is termed a hypocycloid (Fig. 185). If a straight line / (Fig. 186) 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 /, or normal to the virtual 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 ^ ; as the string is unwound d moves in an involute. When setting out cycloidal teeth, only small portions of the cycloids are actually used. The cycloidal portions can d .7^ be obtained by construc- tion or by rolling a cir- cular disc on the pitch circle. By reference to Fig. 187, which represents a model used to demonstrate the theory of cycloidal teeth, the reason why such teeth gear together smoothly will be evident. Fig. 186. 164 Mechanics applied to Engineering. A and B 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 Fig. 187. 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, Mechavisms. i65 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. 184 ; at the same time the pencil-point 2 traces a hypocycloid on the side of the disc B, as in Fig. 185. 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 when friction is neglected 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°. The effect of friction is to increase this angle during the arc of approach by an amount equal to the friction angle between the surfaces of the teeth in contact, and to diminish it by the same amount during the arc of recess. The effect of friction in reducing the efficiency is consequently more marked during approach than during recess, for this reason the teeth of the wheels used in watches and clocks are usually made of such a shape that they do not rub during the arc of approach. In order to keep this angle small, a large rolling circle must be used. In many instances the size of the rolling circle has to be carefully considered. A large rolling circle increases the path 1 66 Mechanics applied to Engineering. of contact and tends to make the gear run smoothly with a small amount of outward thrust, but as the diameter of the rolling circle is increased the thickness of the tooth at the flank is decreased and consequently weakened. If the diameter of the rolling circle be one half that of the pitch circle the flanks become radial, and if larger than that the flanks are undercut. Hence, in cases where the strength of the teeth is the ruling factor, the diameter of the rolling circle is never made less than one half the diameter of the smallest wheel in the train. Generally speaking the same rolling circle is used for all the wheels required to gear together, but in special cases, where such a practice might lead to undercut flanks in the Fig. i88. small wheels of a train, one rolling circle may be used for generating the faces of the teeth on a wheel A, and the flanks of the teeth they gear with on a wheel B, and another size of rolling circle may be used for the flanks of A and the faces of B. In some instances the teeth are made shorter than the standard proportions in order to increase their strength. In setting out the teeth of wheels in practice, it is usual to make use of wooden templates. The template A (Fig. i88) is made with its inner and outer edges of the same radius as the. pitch circle, and the edge of the template B is of the same radius as the rolling circle. A piece of lead pencil is attached by means of a clip to the edge of the template, and having its point exactly on the circumference of the circle. The template Mechanisms. 167 A is kept in place on the drawing paper weight or screws, a pencil run round the convex edge gives the pitch circle. The tem- plate B is placed with the pencil point just touching the pitch circle; it is then rolled, without slipping, on the edge of the template A and the pencil traces out the epicycloid required for the face of a tooth. The template A is then shifted until its concave edge coincides with the pitch circle. The tem- plate B is then placed with the pencil point on the pitch line, and co- inciding with the first point of the epicycloid, then when rolled upon the inside edge of the template A, the pencil traces out the hypocy- cloid which gives the profile of the flank of the tooth. A metal tem- plate is then carefully made to exactly fit the profile of one side of the tooth thus ob- tained, and the rest of the teeth are set out by means of it. It is well known that cycloidal teeth do not work well in practice in cases where it is difficult to ensure ideal conditions as regards constancy of shaft centres, perfection by means of a heavy 1 68 Mechanics applied to Engineering. of workmanship, freedom from grit, etc. For this reason involute teeth are almost universally used for engineering purposes. The model for illustrating the principle of involute teeth is shown in Fig. 189. Here again A and B are parts of two circular discs connected 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. 164. Then, from what has been said on cycloidal teeth, 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, the angle DEF seldom exceeds is|° ; this gives a base circle |f of the pitch circle. In special cases where pinions are required with a small number of teeth, this angle is sometimes increased to 20°. Involute teeth can be set off by templates similar to those shown in Fig. 188, but instead of the template A being made to fit the pitch circle, it is curved to fit the base circle, and the template B is simply a straight-edge with a pencil attached and having its point on the edge itself. If for any reason the distance between the centres of two involute gear wheels be altered by a small amount, the teeth will still work perfectly, provided the path of contact is not less than the pitch. By reference to Fig. 189 it will be evident that altering the wheel centres only alters the diameters of the pitch circles, but does not affect the diameters of the base circles upon which the velocity ratio entirely depends. This is a very valuable property of involute teeth, and enables them to be used in many places where the wheel centres cannot for many reasons be kept constant. If the same angle DEF, Fig. 189, be used in setting out the teeth, all involute wheels of the same pitch will gear with one another. The portions of the flanks inside the base circles are made radial. Mechanisms. 169 Readers interested in mechanical devices for drawing the teeth of wheels and the pistons for rotary pumps and blowers should refer to a paper by Dr. Hele-Shaw, F.R.S., in the British Association Report for 1898, an extract of which will be found in Dunkerley's " Mechanism " (Longmans). The general question of the design of toothed gearing will also be found in standard books on machine design. Readers should also refer to Anthony's " Essentials of Gearing " (D. C. Heath & Co., Boston, U.S.A.), and the series of pamphlets published by " Machinery," 27, Chancery Lane. No. i,iWorra Gearing; No. 15, Spur Gearing; No. 20, Spiral Gearing. Velocity Ratio of Wheel Trains. — In most cases the problem of finding the velocity ratio of wheel trains is Fig. 190. Fig. 191. 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 Hob. form of the wheels used; thus if a in Fig. 190 have a clockwise or + rotation, h will have an anti-clockwise or — rotation; but in Fig. 191 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 = ?:» = I-' = ^ai-^bc ^ N, ' R, T„ Qab.Oac N^ where T,, is the number of teeth in a, and T5 in b, and N„ is the number of revolutions per minute of a and N5 of b. In the case of the three simple wheels in Fig. 192, we have the same peripheral velocity for all of them ; hence — o)„R„ = — wjRj = co„R„ <«5 ^ R„ ^ Tj ^ N^ °' a,, R,. T„ N„ 17 o Mechanics applied to Engineering. 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, since — or - = - — <«s R. Fig. 192. 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. 191, 192, 193, 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„ of the last wheel falls on 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 com- pound, 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 rotation is not altered. If it is desired to get the same velocity ratio with an idle wheel in the train the wheel c must be altered as with Fig. 193. a compound wheel, b in the proportion r„ where b is the driven and V the driver. The velocity ratio V, of this train is obtained thus — and - (OiRy = <o„R„ Substituting the value of — wj, or eoj, we get — <«.R„R» <iR» = R»' andV, = '!^« = |4^: 0)„ RaRj- T,T, N2 Mechanisms. 171 Thus, taking a as the driving wheel, we have for the velocity ratio — Revolutions of driving wheel Revolutions of the last wheel in train _ product of the radii or number of teeth in driven wheels product of the radii or number of teeth in driving wheels The same relation may be proved for any number of wheels in a train. If C were an annular wheel, the virtual centres would be as shown. R„ is on the opposite side of d to R„ ; therefore the wheel c rotates in the opposite sense to a (Fig. 194). In some instances the wheel C rotates on the same axle as a ; such a case as this is often met with in the feed arrange- ments of a drilling-machine (Fig. 195). The wheel A fits loosely on the outside of the threads of the screwed spindle S, and is driven by means of a feather which slides in a sunk keyway, the wheels B and B' are both keyed to the same shaft ; C, however, is a nut which works freely on the screw S. Now, if A and C make the same number of revolutions per minute, the screw will not advance, but if C runs faster than A, the screw will advance ; the number of teeth in the several wheels are so arranged that C shall do so. For example — Let A have 30 teeth, B, 20, B', 21, C, 29, and the screw have four threads per inch : find the linear ad- vance of the screw per revo- lution of S. For one revolu- tion of A the wheel C makes 30 X 21 „„o = 5f2 = i'o86 revo- 20 X 29 ^*° lution. Thus, C makes o-o86 revolution relatively to A per revolution of the spindle, or it advances the screw H:^ Fig. 194. Jig. ips. o"o2i inch per revolution of S. 1- Change Speed Gears. — The old-fashioned back gear 1/2 Mechanics applied to Engineering. so common on machine tools is often replaced by more con- venient methods of changing the speed. The advent of motor-cars has also been responsible for many very ingenious devices for rapidly changing speed gears. The sliding key arrangement of Lang is largely used in many change-speed gears ; in this arrangement all the wheels are kept continuously in mesh, and the two which are required to transmit the power are thrown into gear by means of a sliding cotter or key. In Fig. 196 the wheels on the shaft A are keyed, whereas the wheels on the hollow shaft B are loose, the latter wheels are each provided with six keyways, a sliding key or cotter which passes through slots in the shaft engages with two of these key- ways in one of the desired wheels. The wheel to be driven is Fig. 196. determined by the position of the key, which is shifted to and fro by means of a rod which slides freely in the hollow shaft. The bosses of the wheels are counterbored to such an extent that when the key is shifted from the one wheel to the next both keyways are clear of the key, and consequently both wheels are free. The sliding rod is held in position by a suitable lever and locking gear, which holds it in any desired position. Epicyclic Trains. — In all the cases that we have con- sidered up to the present, the axle frame on which the wheels are mounted is stationary, but when the frame itself moves, its own rotation has to be added to that of the wheels. In the mechanism of Fig. 197, if the bar the fixed, and the wheel a be rotated in clockwise fashion, the point x would approach c, and the wheel b would rotate in contra-clockwise fashion. Mechanisms. 173 If a be fixed, the bar c must be moved in contra-clockwise fashion to cause c and x to approach, but b will still continue to move in contra-clockwise fashion. Let c be rotated through one complete revolution in contra-clockwise fashion; then h will make — N revolutions due to the teeth, where N = 7^, and at the lb same time it will make — 1 revolu- tion due to its bodily rotation round a, or the total revolutions of h will be — N — I or — (N -f i) revolu- tions relative to a, the fixed wheel. '^'°' '*'" The — sign is used because both the arm and the wheel rotate in a contra-clockwise sense. But if b had been an annular wheel, as shown by a broken line, its rotation would have been of the opposite sense to that of c; consequently, in that case, b would make N — i revolutions to one of c. If either idle' or com- pound wheels be intro- duced, as in Fig. 198, we get the revolutions of each wheel as shown for each revolution of the T arm e, where v^j, = ~, and v.^ = : when there is an idle wheel between, or z/,e = V '^ ° 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 all 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, -|- i for wheels that rotate in the same sense as the arm, and - i for wheels that rotate in the opposite sense to the arm. 174 Mechanics applied to Engineering. In the case of simple, i.e. not bevil trains, it should be remembered that wheels on the «th 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." Hence we have — « Sense of rotation of »th wheel. Even Same as arm. Odd Same as arm if V, is less than i. Opposite to arm if V, is greater than I. Epicyclic Bevil Trains. — When dealing with bevil trains the sense of rotation of each wheel must be carefully Fig. igg. considered, and apparently no simple rule can be framed to cover every case; in Fig. 199 the several wheels are marked S for same and O for opposite senses of rotation — arrows on the wheels are of considerable assistance in ascertaining the sense of rotation. The larger arrows indicate the direction in which the observer is looking. Mechanisms. I7S JEEM 11!.""^ . Jl Qlueruer U m/cUui tllli wcu/ Fig. The bevil train shown in Fig. 200 is readily dealt with, thus let T„, T„, etc., represent the number of teeth in a and c respectively — b is an idle wheel, and consequently Tj does not affect the velocity ratio. Fix D, and rotate a contra-clockwise through one revolution, T then c makes ~ revolu- tions in a clockwise direction. Fix a and rotate the arm D through one clockwise revolution. Then since the tooth of b, which meshes with the stationary tooth of a, may be regarded as the fulcrum of a lever ; hence the tooth of b, which meshes with c, moves in the same direction as D, and at twice the speed. T Hence the number of revolutions of cis 7=r + i for one revolution of D. The problem may also be dealt with in the same manner as the simple epicyclic train. 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, + I for wheels that rotate in the same sense as the arm, and — I for wheels that rotate in the opposite sense to the arm. For elementary bevil trains, such as that shown in Fig. 200, wheels on the «th axle rotate in the opposite sense to the arm when n is an even number, and in the same sense when n is an odd number. Hence we have for elementary bevil trains — n Sense of rotation of «th wheel. Odd Same as arm. Even Same as arm if V, is less than i. Opposite to arm if V, is greater than i . 1/6 Mechanics applied to Engineering. In all cases the actual number of revolutions per minute of the several wheels calculated for one revolution of the arm must be multiplied by the number of revolutions per minute of the arm N^, also, if the wheel a rotates, its revolutions per minute must be added, with due regard to the sign, to the revolutions per minute of each wheel calculated for a stationary wheel. The following table may be of assistance in this connection. T„ T. N» N* N„ when ~- = i. N„ when -^ = 1-5. V„+. = 2. V„+. = =-5. lOO -100 -ISO lOO 5 - 100 + 2 X 5=- 90 -I50+2-SX 5 = -i37"S — lOO 5 100-J-2X s= no 150+2-5 X 5= i62-s 100 -s -100— 2X 5=-IIO - 150-2-5 X 5= -162-5 lOO 20 - 100+2 X 20=— 60 - 150+2-5 X 20= — 100 lOO 50 - 100+2 X 50= - 150+2-5 X 50=- 25 100 60 - 100+2 X 60= 20 - 150+2-5 X 60= lOO 70 - 100+2 X 70= 40 - 150+2-5 X 70= 25 lOO 100 -100+2X100= 100 -150+2-5x100= 100 — 100 100 IOO+2XIOO= 300 150+2-5x100= 400 100 — 100 -100— 2X100= -300 — 150 — 2-5 X 100= —400 40 30 - 40+2 X 30= 20 - 60+2-5 X 30= 15 40 -40 -60 S 2X5 = 10 2-5x5=12-5 -100 2(- 100) =-200 2-5(-I00)=-2SO 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, Bi = 16, 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 Bi are fixed together. Let d make one complete clock- wise revolution ; then the other wheels will make — Revs, of B on own axle = — -," z= — |£ 1 h = -I'lS C absolute = -? 4. 1 = || + i =: ^.gj Mechanisms. 177 Revs. E = revs, of B x t^-' + i = - i-i'S X M + i = -0-541 + I = 0-459 Whence for one revolution of E, C makes 4-83 °'459 = io"53 revolutions. The -f- sign in the expressions for the speed of C and E is Fig. 20I. 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 of this type 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 + for odd axes, and — for even axes, always counting the first as " one." 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.l product of teeth in drivers of E j product of teeth in driven wheels Revs.l Ta T,XT,^ 't:xt: -0-541 ofc}=T:=3-«3 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 _ Ne _ — 0-541 + I _ 1 Revs. C Ne 3-83 + I 10-53 178 Mechanics applied to Engineeritig, Particular attention must be paid to the sense of rotation. Bevil 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. This mechanism can also be used as a simple reduction gear — in which case the shaft Fj is not continuous with Fj. The wheel C is then keyed to Fj, or drives it through the set „ „. „ . , J , T. , Revs. E Revs. Fj screw S. Since E is keyed to F, we have =; ;; = =; -. Revs. C Revs. F2 Thus, if F2 is coupled to a motor running at 1053 revs, per min., the low speed shaft Fi will run at 100 revs, per min. for the set of wheels mentioned above. CHAPTER VI. lOYNAMICS OF THE STEAM-ENGINE. Reciprocating Farts. — Oh 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. 160), 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. 202. Let R represent V, the linear velocity of the crank-pin, assumed constant ; then the ordinates Yi, Ya represent to the same scale the velocity of the piston V Y when it is at the positions Ai, Aj respectively, and -^ = ~ i8o Mechanics applied to Engineering. Let the total weight of the reciprocating parts = W. Then— The kinetic energy of the V _ WVi'' reciprocating parts at Aj \ ~ 2g Likewise at A. = 2^-R^ The increase of kinetic energy^ _ WV^ , ^ during the interval A1A2 • j ~ 2°^^' ~^^) 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) _ -pi _ \ moving parts during the interval 3 ~ '^^ ■*!' But V + Y,^ = xi + Ya^ = R hence Y^ — Yi = x^ — x^ and — Increase of kinetic energy of the] -nnn reciprocating parts during the > = (x^ — x^) interval J ^S^ then P(*, - X,) = ^,(«/ - *.») WV where * is the mean distance ' '*^ of the piston from the Dynamics of the Steam-Engine. iSi middle of the stroke ; and when ;«: = R at the beginning and end of the stroke, we have — P = ^ -^ We shall term P the " total 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 00 in the ratio — . K. 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 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 '■- - ' "~- •'' piston, and the whole Fig. 203. rotated together as one solid body, then each point in the body would describe the arc of a circle, and would be subjected to the centrifugal force C = — 5-, 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, 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 first half above, and the second half below the base-line. We show such a diagram in Fig. 2 04. The height of any point in the sloping line ab above the base-line Fig. 204. represents the pressure 182 Mechanics applied to Engineering. 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 -7- =/> where A = the area of the piston. We W will also put w = — , where w is the weight of the reciprocatmg 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. 27rRN ■ and/ '■ 60 6o=^R = o'ooo34ze'RN'' N.B. — -The radius of the crank R is measured in feet. 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. A more accurate method of finding the portion of the connect- ing-rod to be included in the reciprocating parts is to place the rod in a horizontal position with the small end resting on the plat- form of a weighing machine or suspended from a spring balance, the reading gives the amount to be included in the reciprocating parts. 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 j thus — Fig. 205. weight of \ piston -)- piston and ail-rods -f- both cross-heads -f small area of end of con. rod -(- ' ' 1- air- 2 piston ir-pumpplungerf ^j \^.i^^l\ \ Dynamics of the Steam- Engine. 183 The kinetic energy of the parts varies as the square of the velocity; hence the \j\ Values of w in pounds per square inch of piston : — Steam engines with no air pump or other attachments 2 to 4 „ ' „ attachments 3 to 6 In compound and triple expansion engines the reciproca- ting parts are frequently made of approximately the same total weight in each cylinder for balancing purposes, in such cases w is often as high as 6 lbs. for the H.P. cylinder and as low as I lb. for the L.P. Influence of Short Connecting-rods. — In Fig. 159 a velocity diagram is given for a short connecting-rod ; repro- ducing a part of the figure in Fig. 206, we have — Short rod — cross-head velocity _ OX crank-pin velocity OCj Infinite rod- — , cross-head velocity _ OXj crank-pin velocity OCi velocity of cross-head with short rod _ OX velocity of cross-head with long rod OXj But at the " in " end of the stroke .-2J = ^ = li+^ and at the " out " end of the stroke = i — — Thus if the connecting-rod is n cranks long, the pressure at the "in" end is - greater, and at the "out" end - less, than n n ' if the rod were infinitely long. The value of/ at each end of the stroke then becomes — p = o'ooo34wRNV I -|- - 1 for the " in " end p = o'ooo34wRN-( 1 — - ) for the " out " end 184 Mechanics applied to Engineering. Si \ 0) \ a) Si ^° • 1 ^B \ XL % ^^p^ ^ % ^ ^ 1 1- T3 \ \ \ ^8 ^p ^ 1 f' rS \ \ \ 1 nc/v* ^^/ — cc W iL,,^ N<^o '\\ /^ ~-> ---5 3 \ Dynamics of the Steam-Engine. 185 The line ab is found by the method described on p. 133. Set off aa„ = —, also bl>„ = — , and cc. = — , ^ is the position n n n of the cross-head when the crank is at right angles to the centre line, i.e. vertical in this case. The acceleration is zero where 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/. 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; 'the crosshead is then at a distance R(V'«^-f- i — n), "P or very nearly — , from the middle of the stroke. The point/ 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 page 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 draw perpendiculars to cut the circle at the points M and i. The corresponding positions of the cross-head are shown at h and i respectively. In these positions the acceleration curves cross one another, viz. at h„ and 4. It will shortly be shown that when the crank has passed through 45° and 135° the acceleration pressure with the short rod is equal to that with the infinitely long rod. From ]i drop a perpendicular to k„ set off kji„ = L, also k'k = L, and from the point where the perpendicular from k'„ cuts ab draw a horizontal to meet the perpendicular from k, where they cut is another point on the acceleration curve for the short rod ; proceed similarly with /. ~ We now have eight points on the short-rod acceleration curve through which a smooth curve may be drawn, but for ordinary purposes three are sufBcient, say a„, c„ b„. The acceleration pressure at each instant may also be arrived at thus — Let 6 = 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. ' Engineering, }\x\y 15, 1892, p. S3 ; also June 2, 1899. 1 86 Mechanics applied to Engineering. Then— v_ si n (g + g ) V sin (90- a) ^^^^^'^•^°'''' ^,/sin 6 cos a + cos B sin a\ » = V( ) \ cos a / In all cases in practice the angle a is small, consequently cos a is very nearly equal to unity; even with a very short rod the average error in the final result is well within one per cent. Hence— z* = V (sin Q + cos sin a) nearly We also have — L R « = sin 6 sin u. sin a = sin 6 n 1 Substituting this value- — V = v(sin e + cos 6 sin n ti or z) = V^sin 6 + sin 26\ in ) dB~ v(cos 1^ + cos 20\ n J The acceleration of the cross-head /„ = — = —„• — ■" dt de dt Let B = the angle in radians turned through in the time t. Then e = ■ ^'■*' radius N.t ^dB V , . dv Y whence/. = -^._ Substituting the value of — found above, we have — , Vy . , cos2(9\ Dynamics of the Steam- Engine. 187 and the acceleration pressure, when the crank has passed through the angle Q from the " in " end of the stroke, is— wVY cos 2ff\ or/ = o-ooo34ze/RN=('cos B + E2if_^^ When 6 = 45° and 135°, cos 26 = and the expression becomes the same as that for a rod of infinite length. When 6 = and 180° the quantity in brackets becomes 1 + i and I 1 . Correction of Indicator Diagram for Acceleration Pressure.— An indicator diagram only shows the pressure of the working fluid in the cylin- der; 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 " ~— ™™™'-' rr^:. — > 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 does not show the impelling pressure on a piston at every portion of the stroke. The impelling pressure is really the difference be- tween the two pressures on both sides of the piston at any one instant, hence the impelling pressure must be measured between the top line of one diagram and the bottom line of the other, as shown in full lines in Fig. 207. The diagram for the return stroke is obtained in the same manner. The two diagrams are set out to a straight base in Fig. 208, the one above the line and the other below. On the same base line the acceleration diagram is also given to the same scale. The real pressure transmitted along the centre line of the engine is given by the vertical height of the shaded figures. In the case of a vertical engine the accelera- tion line is shifted to increase the pressure on the downward stroke and decrease it on the upward stroke by an amount zc, i88 Mechanics applied to Engineering. see Fig. 209. The area of these figures is not altered in any way by the transformations they have passed through, but it should be checked with the area of the indicator diagrams in order to see that no error has crept in. When dealing with engines 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. ■^'- III! I l|ll I, , Fig. Z08. Fig. 2og, Let the original indicator diagrams be taken with springs of the following scales, H.P. -, I.P. -, L.P.-. Let the areas of oc z the pistons (allowing for rods) be H.P. X, LP. 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 — Corrected height of LP. diagram) ^ f iiSltll x f if pistons were of same size 1 | diagram i i = actual height X — X Dynamics of the Steam-Engine. 189 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 j j-actual height of ^ ^ corrected for scales of springs \ — \ intermediate V X ^ X - and for areas of pistons J ( diagram j « X -J'v^ diagram j = actual height X ^— Similarly for the L.P. diagram — Height of L.P. diagram corrected . ^^^^^j ^^^.^^ ^^. ^^ It is probably best to make this correction for scale and area after having reduced the diagrams to the form given in Fig. 208. Pressure on the Crank-pin. — The diagram given in /^ '~y<^ "^•',^ / -^ '' N^^^^ ~~^ ^ / / /■ 1;*-^— . ■~'\ J— \— -v / 1^ 4 \\ . w Fig. 210. -/ Fig. 208 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. 210. The curve 3 shows that there is too much compression resulting in a negative pressure — / at the end of the stroke ; at the point x the pressure on the pin would be reversed, and, if 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 — / to +p. 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 1 90 Mechanics applied to Engineering. 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, as applied to steam engines, 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 ; and to Haeder and Huskisson's " Handbook on the Gas Engine," Crosby Lockwood, for the application of them to gas and oil engines. 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 j A„ = area of the annular top side in square inches ; W = total weight of the reciprocating parts in lbs. ; c = clearance in feet at top of stroke. At the top, i.e. at the " in end," of the stroke we have — Pi = o-ooo34WRN2(^i -f i^ - W -f i4-7A Assuming isothermal compression of the air, and taking the pressure to be atmospheric at the bottom of the stroke, we have — I47A„(2R -t- f) = P,f whence c = ^ Pi - i4-7Aa Dynamics of the Steam-Engine. 191 Or for adiabatic compression — I4-7A„(2R + <r)'"" = Pi<;^'" 2R Vi47Aa/ 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 difiSculties than in the steam-engine, partly because the ordinary cushioning method cannot be adopted, and further because the eifective 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 lbs. and the springs 800 lbs. Polar Twisting-Moment Diagrams. ^ From the diagrams of real pressures transmitted to the crank-pin that Fig. 211. we have just constructed, we can readily determine the twisting moment on the crank-shaft at each part of the revolution. In Fig. 2ir,let/ be the horizontal pressure taken from such a diagram as. Fig. 209. Then /i is the pressure transmitted 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 /„R ; R, however, is constant, therefore the twisting moment is proportional to/„. By setting off values of p„ radially from the crank-circle we get a diagram showing the 192 Mechanics applied to Engineering. twisting moment at each part of the revolution. /„ is measured on the same scale, say -, as the indicator diagram ; then, if A be the area of the piston in square inches, the twisting moment in pounds feet =/„a:AR, where /„ is measured in inches, and the radius of the crank R is expressed in feet. When the curve falls inside the circle it simply indi- FiG. 2ia. Gates 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. 212 indicator diagrams are given, which have been set down in the manner shown in Fig. 208, and after correct- ing for inertia pressure they have been utilized for construct- ing the twisting moment diagram shown in Fig. 213. The diagrams were taken from a vertical triple-expansion engine made by Messrs. McLaren of Leeds, and by whose courtesy the author now gives them. The dimensions of the engine were as follows : — Dynamics of the Steam-Engine. 193 Diameter of cylinders — High pressure Intermediate Low pressure Stroke 9*01 inches, I4"2S .1 22-47 >. 2 feet Fig. 182*. 194 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 Crankshaft. — In some instances it is more convenient to calculate the twisting moment on the crank- shaft when the crank has passed through the angle 6 from the inner 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 ; Pi = the component of the eflfort acting along the connecting-rod ; , . sin 9 P, = Pi cos o, and sm a = ^- from which a can be obtained, since 6 and n are given. The tangential component — T = Pi cos </) and </) = 90 — (^ ^- a) whence T= ^^cos {9o-(e + a)] = ^' ^'" ^^ + "^ 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 angplar 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. 195 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. 215. 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. Fig. 215. 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 outside edge of the rim : V„ = minimum velocity at C (Fig. 215); V^ = maximum „ E ; W = weight of the flywheel in pounds, usually taken for practical purposes as the weight of the rim ; R„ = radius of gyration in feet of the flywheel rim, usually taken as the external radius for practical purposes. For most practical purposes it is sheer waste of time cal- culating the moments of inertia and radii of gyration for all the rotating -parts, since the problem is not one that permits of great accuracy of treatment, the form of the indicator diagrams does not remain constant if any of the conditions are altered even to a small extent, then again the coefficient -of fluctuation k is not a definitely known quantity, since different authorities give values varying to the extent of two or three hundred per cent. The error involved in using the above approximations is not often greater than five per cent, which is negligible as compared with the other variations, and by adopting them a large amount of time is thereby saved. ' Figures 207, 208, 210, 215 are all constructed from the same indicator diagram. 196 Mechanics applied to Engineering. Then the energy stored in the flywheel at C = - E = ^g W The increase of energy during CE = — (V/— V„^) (i.) This increase of energy is 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„ (Fig. 215)- Let E„ = OT 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 — p, _ m X indicated horse-power of engine x 33000 ... . 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/^a 3. _ OT X I.H.P. X 33000 , Tg- '~ °' 2N ■ • ^ ' But '"' ° = V (approximately) 2 or V. + V. = 2V V — V also — i^r= — 5 = K, " the coefficient of speed fluctuation " and V. - V. = KV v." - V/ = 2KV Substituting this value in (iii.) — -(2KV=) = *^ ^ ^•^■^- ^ 33000 ^ E 2/ 2N " „„A w - 48,5o°.o°o/« X I.H.P. KWRJ ' • ' ^^^-^ 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 : — Dynamics of the Steani-Engine. 197 K = o'oi to o'oa for ordinary electric-lighting engines, but for public lighting and traction stations it often gets as low as o"ooi6 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 o'o4 for factory engines; = o'o6 to o'i6 for rough engines. 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„ 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, since 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 constructing such diagrams as that given in Fig. 215 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 « = 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. Tiiree cylinders. Cranks at 120° O'l 0'35 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 0-27 0-068 0-030 ' 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. 198 Mechanics applied to Engineering. m FOK Gas - AND OiL-ENGINES. Otto cycle. Double acting. Number of cylinders. X 2 4 I 3 Exploding at every cycle .... Single 37 to 4-5 — — 2-3 to 2-8 — Twin or tandem — i-S to I -8 0-3 to 0-4 — 0-3 to 0-4 When missing every alternate charge. Single 8-5 to 9-8 2-5 to 3-0 — — — Gas and oil engines, single acting (Otto) w = i"5 + >Jd „ „ double „ 1X1= 2'5 + 'I'd'Jd d 2 High speed petrol engines, a' = 5 + 3 o d Tandem engines, per line, from i'8 to I'g times the above values. Where d is the diameter of the cylinder in inches. Relation betvireen 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. WV* The energy stored m the wheel = Then from equation (iv.), we also have — The energy stored in the^ _ E„ wheel / ~ 2K and the average work done'! _ E„ per stroke / ^ _ I.H.P. X 33000 ^ ffor a double- 2N )\ acting engine the number of average* _ 2K _ m . strokes stored in flywheel/ ~ E„ ~ 2K ^'' Dynamics of the Steam- Engine. 199 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., i.e. \ 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 = o'o4, be permitted, the numbers given below must be divided by 4. Number of Strokes stored in a Flywheel for Double-acting Steam-Engines.' Cut-off. Single cylinder. Two cylinders. Cranks at right angles. Three cylinders. Cranks at 120°. o-i 18 4'4 2'0 0-2 17 4" I I "9 0-4 16 3'9 I "8 06 IS 3-6 '7 0-8 14 3-S 1-6 End of stroke «3 3-4 i-S Gas-Engines (Mean Strokes). Otto cycle. Double acting. Number of cylinders. I 2 4 z 2 When exploding at Single 185 to 225 — ■ — 115 to 140 — every cycle. Twin or tandem — 75 to 90 15 to 20 — 15 to 20 When missing every alternate charge . Single 425 to 490 125 to ISO — — — 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. ' See note at foot of p. 197. 200 Mechanics applied to Engineering. Gas-Engine Flywheels. — The value of m for a gas- engine can be roughly arrived at by the following method. Fig. 2i6. 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 = 2,. Similarly, when every alternate explosion is missed, m = j. 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. 216. It has been constructed in precisely the same manner as Figs. 207, 208, and 215. When oil- or 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. 201) on account of their Dynamics of the S team-Engine. 201 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. There is no perceptible flicker in the lights when about forty impulse strokes, or i6o 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„ is found thus : Fig. 217. 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. See The Engineer, January 9, 1885. Stress in Flywheel Rims. — If we neglect the effects of 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. 421), in which we have the relation — or P,R„ =/, when t = -i inch The P, in this instance is the pressure on each unit length of rim due to centifugal force. We Fig. 2i3. 202 Mechanics applied to Engineering. shall find it convenient to take the unit of length as i foot, because we take the velocity of the rim in feet per second. Then— WV ^ WV " where W, = the weight of i foot length of rim, i square inch in section = 3 "I lbs. 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, in the above equations, we have — 32"2 /= 0-096 V„' V " / or/= -^ (very nearly) 10 In English practice V„ 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 lbs. per square inch. The above expression gives the tensile stress set up in a thin plain rotating ring, due to centrifugal force ; but it is not the only or even the most important stress which occurs in many flywheel and pulley rims. The direct stress in the material causes the rim to stretch and to increase in diameter, but owing to its attachment to the arms, it is unable to do so beyond the amount permitted by the stretch of the arms, with the result that the rim sections bend outwards between the arms, and behave as beams which are constrained in direction at the ends. If the arms stretched sufficiently to allow the rim to remain circular when under centrifugal stress there would be no bending action, and, on the other hand, if the arms were quite rigid the bending stress in the rim sections between the arms could be calculated by treating them as beams built in at each end and supporting an evenly distributed load equal in intensity to the centrifugal force acting on the several portions of the rim ; neither of these conditions actually hold and the real state of the beam Dynamics of the Steam-Engine. 203 is intermediate between that due to the above-mentioned assumptions, the exact amount of bending depending largely upon the stretch of the arms. A rigid solution of the problem is almost impossible, the results obtained by different authorities are not in agree- ment, owing to arbitrary assumptions being made. In all cases it is assumed that there are no cooling stresses exist- ing in the arms of the wheel, which every practical man knows is not always correct. However, the results obtained by the more complete reasoning are unquestionably nearer the truth than those obtained by the elementary treatment given above. For a more complete treatment the reader is referred to Unwin's "Elements of Machine Design," Part II. (Longmans). The following approximate treatment may be of service to those who have not the opportunity of following the more complete theory. The maximum bending moment in pound-inches on an initially straight beam built in at both ends is — (see p. 529), where w is the evenly distributed load per inch run, and / is the length between supports in inches. In the case of the flywheel, w is the centrifugal force acting on the various portions of the rim, and is ° ^ " — , where 0-26 is the weight of a cubic inch of cast iron, V„ is the rim velocity in feet per second, A the area of the section of the rim in square inches, g the acceleration of gravity, R„ the radius of the rim in feet, Z the tension modulus of the section (see Chapter IX.). Then the bending stress in the rim due to centrifugal force o-26V„^A/' '^ ■'" 12 X 32-2 X R„ X Z The rim section, however, is not initially straight, hence the ordinary beam formula does not rigidly hold. / is taken as the chord of the arc between the arms. As already ex- plained, the stress due to bending is really less than the above expression gives, but by introducing a constant obtained by comparing this treatment with one more complete, we can bring the results into approximate agreement : this constant is about 2'2. Hence o-26V„^A/' ^ V^'A/' ■^^ ~ 2-2 X 12 X 32'2R„Z ~ 327oR„Z 204 Mechanics applied to Engineering. and the resulting tensile stress in the rim is — All who have had any experience in the foundry will be familiar with the serious nature of the internal cooling stresses in flywheel and pulley arms. The foundry novice not unfre- quently finds that one or more of the arms of his pulleys are broken when taken out of the sand, due to unequal cooling ; by the exercise of due care the moulder can prevent such catastrophes, but it is a matter of common knowledge that, in spite of the most skilful treatment it is almost impossible to ensure that a wheel is free from cooling stresses. Hence only low-working stresses should be permitted. When a wheel gets overheated through the use of a friction brake, the risk of bursting is still greater; there are many cases on record in which wheels have burst, in some instances with fatal results, through such overheating. In a case known to the Author of a steam-engine fitted with two flywheels, 5 feet in diameter, and running at 160 revolutions per minute, one of the wheels broke during an engine test. The other wheel was removed with the object of cutting through the boss in order to relieve the cooling stresses; but as soon as the cut was started the wheel broke into several pieces, with a loud report like a cannon, thus proving that it was previously sub- jected to very serious cooling stresses. In order to reduce the risk of the bursting of large flywheels, when made with solid rims, they should always be provided with split bosses. In some cases the boss is made in several sections, each being attached to a single arm, which effectually prevents the arms from being subjected to initial tension due to cooling. Built-up rims are in general much weaker than solid rims ; but when they cannot be avoided, their design should be most carefully considered. The question is discussed in Lanza's " Dynamics of Machinery " (Chapman & Hall). Large wheels are not infrequently built up entirely of wrought iron or steel sections and plates ; in some instances a channel rim has been used into which hard-drawn steel wire is wound under tension. Such wheels are of necessity more costly than cast-iron wheels of the same weight, but since the material is safe under much higher stresses than cast iron, the permissible peripheral speed may be much higher, and conse- quently the same amount of energy may be stored in a much lighter wheel, with the result that for a given speed fluctuation Dynamics of the Steam-Engine. 205 the cost of the built-up wheel may be even less than that of a cast-iron wheel. In the place of arms thin plate webs are often used with great success ; such webs support the rim far better than arms, and moreover they have the additional advantage that they materially reduce the air resistance, which is much more im- portant than many are inclined to believe. 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. TO Thickness of rim. Remarks, lbs. sq. in. inch. 430 18,500 068 Solid rim, 6 arms, 15 ins. diam. 388 15,000 0-56 . ) » » » » 192 3.700 Jointed rim, ,, „ 38" 14,500 0-65 Solid rim, 3 arms, „ 363 13,200 0-38 .. .» ». 38s 14,800 IS „ 6 arms, 24 ins. diam. Two internal igo 3.610 075 flanged joints, ,, „ 305 9.300 Linked joints „ ,, 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 ; — Sharpe on " Flywheels," Manchester Association of Engineers. Haeder and Huskisson's " Hand- book on the Gas-Engine." " Flywheels," " Machinery " Reference books. 2o6 Mechanics applied to Engineering. Arms of Flywheels. — In addition to the unknown tensile stresses in wheels with solid bosses and rims, the arms are under tension due to the centrifugal force acting (i) on the arm itself, (ii) on a portion of the rim, amounting to approxi- mately one-fourth of the length of rim between the arms. In addition to these stresses the arms are subjected to bending due (iii) to a change in the speed of the wheel, (iv) to the power transmitted through the wheel when it is used for driving purposes. The tension due to (i) is arrived at thus — Let Aa = Sectional area of the arm in square inches as- sumed to be the same throughout its length. r = Radius from centre of wheel to any element of the arm in feet. (1) = Angular velocity of the arm. /"i = Radius of inside of the rim of the wheel in feet, fa = Radius of boss of the wheel in feet. w = Weight of a cubic inch of the material. The centrifugal force acting on the element of the arm is g and on the whole arm g Jvi g \ 2 / 6' The tension in the arm due to the centrifugal force acting on one-fourth of the rim between the arms is _ ■w\NJ' Hence the tensile stress in the arm at the boss due to both is 7r Ir --T— + T>PP'°'- Let the 'acceleration of the rim, arising from a change of speed of the shaft, be 8V feet per sec. per sec. Let W be the weight of the rim in pounds, then, since the arms are built in at both ends, the bending moment on them (see p. 504) is 6WR„.8V . ^ J 1 ,. ^ .. mch-pounds, approximately ; the bendmg stress is Dynamics of the Steam-Engine. 207 6WR .8V '^ — , where n is the number of arms and Z the modulus gnT. of the section of the arms in bending. If the flywheel be also used for transmitting power, and P be the effective force acting on the rim of the wheel, the bending stress in the arms is —?, on the assumption that the stress in the most strained arm is twice the mean, which experiments show is a reasonable assumption. Hence the bending stress in the arms due to both causes The strength of flywheel and pulley arms should always be checked as regards bending. Bending Stresses in Locomotive Coupling-rods. — Each point in the rod describes a circle (relatively to the Fig. 219. engine) as the wheels revolve ; hence each particle of the rod is subjected to centrifugal force, which bends the rod upwards when it is at the top of its path and downwards when at the bottom. Since the stress in the rod is given in pounds per square inch, the bending moment on the rod must be ex- pressed in pound-inches ; and the length of the rod / in inches. In the expression for centrifugal force we have foot units, hence the radius of the coupling crank must be in feet. The centrifugal force acting ) ^ ^ o-ooo34«/R„N^ on the rod per men run ^ jt where w is the weight of the rod in pounds per inch run, ox w = o'28A pounds, where A is the sectional area of the rod. The centrifugal force is an evenly distributed load all along the rod if it be parallel. 2o8 Mechanics applied to Engineering. The maximum bending moment ) _ C/^ _ /At' in the middle of the rod \~ i> ~ y (see Chapters IX. and X.) where k^ = 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 — 000034 X o'zS X A X R, X N° X /° X J' ~ 8 X A X K^ ' o-oooor2R„Ny;/ _ R^N^ ^~ k" ' ""^ ~ 84,0001^ The value of k' can be obtained from Chapter III. For a rectangular section, k' = — ; and for an I section, k' = BH° - bh? I2(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 reduces the bending stress. 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 nearly uniform intensity acting over the section of the rod, sometimes in tension and sometimes in compression. This stress is due to the driving effort trans- mitted through the rod from the driving to the coupled wheel, but it is impossible to say what this eiFort may amount to. It is usual to assume that it amounts to one-half of the total pressure on the piston, but a safer method is to calculate it from the maximum adhesion of the coupled wheels. The coefficient of friction between the wheels and rails may be taken at o'3. Dynamics of the Steam-Engine. 209 On account of the bending moment on these rods a certain amount of deflection occurs, which reaches its maximum value when the rod is at the top and bottom of its traverse. When the rod is transmitting a compressive stress it becomes a strut loaded out of the centre, and the direct stress is no longer distributed uniformly. The deflection due to the bending moment already considered is 8 = 384EI 8o7,oooEI if T be the thrust on the rod in pounds. The bending stress due to the eccentric loading is TS_ TR.NV* Z ~ Soy.oooEK'^Z and the maximum stress due to all causes is RoN'/'j)' ( . T/^ 84,oook^ 1 + 9-6iEI - A The + sign refers to the maximum compressive stress and the — sign to the tensile stress. The second term in the brackets is usually very small and negligible. A more exact treatment will be found in Morley's " Strength of Materials," page 263. In all cases coupling rods should be checked to see that they are safe against buckling sideways as struts ; many break- downs occur through weakness in this direction. Bending Stress in Connecting-rods. — In the case of a coupling-rod of uniform section, in which 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 sec- tion, varies from a maximum at the crank-pin to zero at the gudgeon-pin. The centrifugal force acting on any element distant x from the gudgeon-pin is —, where C is the centrifugal ^"^ "°- force acting on it if 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 p org. C • x. — ^^^^ I — 2IO Mechanics applied to Engineering. 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 — •n , C/ / , „ C/ RJ = — X - whence R„ = -r "^2 3 "6 The shear at a section distant x')_^_ fCx x\ from the gudgeon-pin ) 6 V / 2/ C/ C:x^ I The shear changes sign when -^ = — ^ or when x = — ;=: o 2/ V3 But the bending moment is a maximum at the section where the shear changes sign (see p. 482). The bending moment at a section y distant x from R^ — M^ = R^ 5- X - X - = R^a; - -^ "^ / 2 3 " 6/ The position of the maximum bending moment may also be obtained thus — for a maximum value dx ~ 0.1 6 6/ c/ 6 3C^ 6/ and / ^=V-3 By substitution of the values of x and R, and by reduction, we have — _ _c^ _ c/" ■M-inai. — /- — -■ 9V3 150 ^ ^ ^ A- . X RNV^ and the bendmg stress/ = —z ^ ° "^ i64000K^ which is about one-half as great as the stress in a coupling-rod working under the same conditions. Dynamics of the Steam- Engine. 211 The rod is also subjected to a direct stress and to a very small bending stress due to the deflection of the rod, which can be treated by the method given for coupling rods. 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 Balancer — 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 ; Wj = „ „ balance weight ; R = the radius of the c. of g. of the unbalanced portion ; Ri = 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 — 0-00034WRN2 = o-ooo34WiRiN* or WR = WiRi or WR - WiR, = 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." 212 Mechanics applied to Engineering. When a shaft has standing balance, it will also be perfectly balanced at all speeds, //-mafei/ 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 /(iC 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 I will tend to turn about an axis I 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 Fig. 221. 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 c-^ form a couple, tending to turn the shaft about some point A between them. In order to Fig, ! 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 Dynamics of the Steam-Engine. 213 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, 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 Wa and W3 be attached to the shaft in the same plane as Wi and W, i.e. diametrically opposite to them, also having "standing laalance." Then, in order that the axis may be permanent, the following condition must be fulfilled : — <y + c^y-i = c^y^i + hy o-ooo34N^(WRy+W,R,ji'i) = o'ooo34N'(W,R2j/,+W3R3jFs) or WRj/ + WjRi^i - W,R,j/s, - WsR^^a = 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 - WiR, - W^Ra + WaRa = 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, oy a " running balance," are — SWR = o and SWR,)- = o where 2WR 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. 214 Mechanics applied to Engineering. Graphic Treatment of Balance Weights. — Such a problem as the one just dealt vpith 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 'J3 Fig. 223. 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 W4, and the arm yi, 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 C,, Cj, C„ Ci, must be zero, i.e. they must form a closed polygon. Since C is pro- portional to WR, set- oflf W,R„ WR, WjRj, W,R„ to some suitable scale and in their respective directions ; then the closing line of the force polygon gives us W^R, in direction, magnitude, and sense. The radius R, IS given, whence W, is found by dividing by R,. Runfiing balance. There must be no tendency for the axis to wobble ; hence the vector sum of the moments C,_j'„ etc., about a. given plane must be zero, i.e. they, like the forces, must form a closed polygon. We adopt Pro- fessor D^by's method of taking the plane of one of the rotating masses, viz. W, for our plane of reference ; then the force C, has no moment about the plane. Constructing the triangle of mo- ments, we get the value of W,R,^4 from the closing line of the triangle. Then dividing by WjR,, we get the value of V4. W,R,», Dynamics of the S team-Engine. 215 /to A B Tcrr Fig. 224. 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/, and/j about the point O. When the piston A is at the bottom of' its stroke, there is a contra-clockwise mome.n\.,p^y„ due to the acceleration pressure p^ tend- ing to turn the whole engine round in a- contra-clockwise direction about the point O. The force /j 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, ^oJ«i 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 may give rise to very serious vibra- tions unless the foundations be very massive. It must be clearly understood that the cushioning of the steam men- tioned on p. 190 in no way tends to reduce this effect; balance weights on the cranks will partially remedy the evil, but it is quite possible 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 cranks are immediately op- posite and of equal throw. 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 con- necting-rods about the gudgeon-pins are the same, the 2i6 Mechanics applied td Engineering. disturbing effect of the obliquity of the rods will be entirely eliminated. A single cylinder engine can be balanced by a similar device. Let B represent the piston, cross-head, and connecting rod of a single-cylinder engine, and let a " bob-weight " be substituted for the piston and cross-head of the cylinder A. Then, provided the " bob-weight " slides to and fro in a similar manner and fulfils the conditions mentioned above, a perfect balance will be established. In certain cases it may be more convenient to use two " bob-weights," Aj and A^, each attached to a separate connecting rod. Let the distances of the planes of the connecting rods from a plane taken through B be Oil and x^, and the "bob-weights" be Wj and Wa, and the radii of the cranks r^ and r.^, respectively. Let the weight of the reciprocating parts of B be'W and the radius of the crank r. Then using connecting rods of equal moment of inertia about the gudgeon pin for Aj and Aj, and whose com- bined moments of inertia are equal to that of B, we must have — Wi/-i(a:i -f x^ = WrjCj, also '^ir4^Xi + x^= Vfrx^. The same arrangement can be used for three-cylinder engines,^the "bob-weights" then become the weights of the reciprocating parts of the two cylinders Aj and Aj. Any three-cylinder engine can be balanced in a similar man- ner by the addition of two extra cranks or eccentrics to drive suitable "bob-weights," the calculations for arriving at the neces- sary weights, radii of cranks and their positions along the shaft can be readily made by the methods shortly to be discussed. If slotted cross-heads are used in order to secure simple harmonic motion for the reciprocating parts the matter is simplified in that no connecting rods are used and consequently there are no moments of inertia to be considered. A three-cylinder vertical engine having cranks at r2o°, 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 inserting the angles 6, 6 + 120°, and B + 240° in the equation on p, 186 ; the sum of the inertia forces will be found to be zero. The proof was first given by M. Normand of Havre, and will be found in " Ripper's Steam Engine Theory and Practice." There will, however, be small unbalanced forces acting at right angles to the centre line, tending to make the engine rock about an axis parallel to the centre line of the crank-shaft. Dynamics of the Steam- Engine. 217 A four-cylinder engine, apart from a small error due to the obliquity of the rods, can be perfectly balanced j thus — Fig. 226. JtBi, "msf Let the reciprocating masses be Wi, Wa, etc. ; the radii of the cranks be Rj, Rj, etc.; the distance from the plane of refer- ence taken through the first crank bejj'a.ji'a, etc. Then the acceleration pressure, neglecting the obliquity of the rods at each end of the stroke, will be o"ooo34WRN^, with the corresponding suffixes for each cylinder. Since the 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 — t5' .Weight of reciprocating parts. lbs. Radius of crank. Proportional acceleration force. Distance of centre line from plane of reference. Proportional acceleration force moment. I 2 3 4 W,= 7.?o W,= iooo W5=I20O W.=i230 R,= I2" R,= i4" R3 = i4" R, = I2" ■W]R,= 9,000 ■Wi,Rj= 14,000 W3R3= 16,800 W<Ri = 14,800 y^= 40" y,= 80 ;/<=li2" "^^0^= 560,000 W3R;jy3= 1,344,000 W<R4j/4= 1,653,000 2i8 Mechanics applied to Engineering. 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 j 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. The following symbols will be used in the paragraphs on locomotive balancing : — W„ for rotating weights (pounds) to be balanced. Wp, for reciprocating weights (pounds) to be balanced. Wb, for balance weights ; if with a suffix p, as Wsp, it will indicate the balance weight for the reciprocating parts, and so on with other suffixes. R, for radius of crank (feet). R„ „ „ coupling-crank. Rb, „ „ 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. 2 1 1, which gives us — Dynamics of the Steam-Engine, 219 W,R = W,,R3 and W,, = ^ ■Kb The weights included in the W, vary in different types of engines ; we shall consider each case as it arises. Reciprocating Parts of Locomotive. — We have already shown (p. 181) that the acceleration pressure at the Fig. 227. 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-ooo34Wbp . Rb . N^ = o-ooo34Wp .R.N' Wbp . Rb = Wp . R andWBp=^%^ (i.) •Kb 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 Wsp - greater to allow for the 220 Mechanics applied to Engineering. 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 Itistitute of Civil Engineers , vol. civ. ; or Barker's " Graphic Methods of Engine Design ; " also Dalby's " Balancing of Engines." The portion of the connecting-rod which may be regarded as rotating with the crank-pin and the portion as reciprocating with the cross-head may be most readily obtained by find- ing the centre of gravity of the whole rod — let its distance from the crank-pin centre be x, the length of the rod centres /, then the portion to be included in the reciprocating parts is —j-, where W is the weight of the whole rod. The remainder (i— j-Jistobe included in the rotating parts. If the rod be placed horizontally with the small end on a weighing machine or be suspended from a spring balance the reading will give the weight to be included in the reciprocating parts. For most purposes it is sufficiently near to take the weight of the small end together with one-half the plain part as reciprocating, and the big end with one-half the plain part as rotating. Inside-cylinder Engine (uncoupled).- -In this case we have — Wp = weight of (piston -|- piston-rod -f cross-head -|- small end of connecting-rod -|- \ plain part of rod) ; W, = weight of (crank-pin -1- crank- webs ' -f- big end of connecting-rod -P \ plain part of rod). If we arrange balance weights so that their c. of g. rotates in the same plane as the crank-pins, their combined weight would be Wg, -|- Wup, placed at the radius Rb, and if we only counterbalance two-thirds of the reciprocating parts, we get — W R(|Wp + W,) See p. 229, _ J-V^3»vp T T',y ... , »'B0 — t5 V"'J Dynamics of the Steam-Engine. 221 Balance weights are not usually placed opposite the crank- webs as shown, but are 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 the balance weight on one w^heel, and Wj the other, distant y' and yl from the plane of the crank, then — or Wy = W,;/i' 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 WboX y i°off'K/heel opposite 'off "crank Cofe.ifCr^/rmi^/llS. WboZ y On "near" wheel opposite "near^Qrank Flc. 328. rotation permanent. Then, considering the vertical crank shown in Fig. 228, by taking moments, we get the equivalent weights at the wheel centres as given in the figure. We have, from the figure — X = ■ z=y-^i J 2 _y ■\-c 222 Mechanics applied to Engineering. Substituting these values, we get — ^»/ y — c) = Wbi, as the proportion of the balance weight on the " off" wheel opposite the far crank and — "(v + ^) = Wb2, 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^a at N (Fig. 229), opposite the near crank, and one small one Wei at F, opposite the far crank. Such an arrangement \f 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 Wb at M. . ThenWB= VWbi' + Wb On substituting the values given above for Wei and Wj, we have, when simplified — W„ V2Wb 2y V/ + ^ In English practice^ = 2'^c (approximately) On substitution, we get — Wb = o-76Wb, Substituting from ii., we have — o-76R(|W, + W,) Rb Let the angle between the final balance weight and the near crank be a, and the far crank Q + 90. Then a = 180 - 6 andtane = S^=^^ Wbs ;» + <■ Dynamics of the Steam- Engine. 223 Substituting the value oi y for English practice, we get — tan B = — —= o"42Q 3-5 ^ ^ Now, 6 = ^ very nearly ; hence, for English practice, if the quadrant opposite the crank quadrant be divided into W„ = Wb ■ nearly Fig. 230. four equal parts, the balance weight must be placed on the first of these, counting from the line opposite the near crank. Outside-eylinder Engine (uncoupled). — Wp and W, 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 — Rb and the balance weight is placed diametrically opposite the crank. When the planes do not approximately coincide — Let J/ = the distance between the wheel centres ; cylinder centres ; cylinder centre line and the " near " wheel ; cylinder centre line and the " off" wheel. 2 c = X = * = c-y The balance weight required! on the "off" wheel opposite? the " far " crank ' The balance weight required. Wbo^: 2y on the "near" wheel oppo-|= —'^- = ^^(c+y) = W^ site the "near" crdnk ) ^ '-^ 224 Mechanics applied to Engineering. Then W, = ^2W„ 2y 'Jf + c" which is precisely the same expression as we obtained for inside-cylinder engines, but in this casejc = o'8f to o'<)c. On substitution, we get Wb = i'I3Wb„ to i-osW^o, and fl = 6° to 3°. The same reasoning applies to the coupling-rod balance weights Wbo in the next paragraphs. Inside-cylinder Engine (coupled). — In this case we have Wp the same as in the previous cases. Wo = the weight of coupling crank-web and pin ^ -f coupling rod from a to b, or c to d, otbto c (Fig. 195), as the case may be ; Wbo = the weight of the balance weight required to counter- balance the coupling attachments ; Ro = the radius of the coupling crank. Fio. 231. In the case of the driving-wheel of the four-wheel coupled engine, we have Wb arrived at in precisely the same manner as in the case of the inside-cylinder uncoupled engine, and ^^ Rb • The portion of the coupling rod included in the Wo is, in this case, one-half the whole rod. The balance weight Wbo is placed diametrically opposite the coupling crank-pin. After finding Wb and Wbo. tliey are combined in one weight Wbf by the parallelogram of forces, as already described. With this type of engine the balance weight is usually small. Sometimes the weights of the rods are so adjusted that a balance weight may be dispensed with on the driving- wheel. ' See p. 229. Dynamics of the Sieam- Engine. 225 It frequently happens, however, that W^ is larger than Wpo ; in that case Wgp is placed much nearer Wj than is shown in the figure. On the coupled wheel the balance weight Wjo is of the same value as that given above, and is placed diametrically opposite the coupling crank-pin. Fig. 232. In the six-wheel coupled engine the method of treatment is precisely the same, but one or two points require notice. RcWc W'b„ = . Rn The portion of the coupling rod included in the Wo is from b\.o e; whereas in the Wbo the portion is from a to ^ or ^ to d. Coupling cranks * have been placed with the crank-pins ; the balance weights then become very much greater. They are treated in precisely the same way. Some locomotive-builders evenly distribute the balance weights on coupled engines over all the wheels : most authorities strongly condemn this practice. Space will not allow of this point being discussed here. Outside-cylinder Engine (coupled). W is the same as before ; W, is the weight of crank-web ' and pin -|- coupling rod from a\.ob ■\- big end of connecting-rod + half plain part of rod ; Wc is the same as in the last paragraph; Ro = R VV3 = \Vn„ = ' See Proc. Inst. C.£., vol. Uxxi. p. m w, + w,) Rb 122. ' See p. 229 Q 226 Mechanics applied to Engineering. The six-wheel coupled engine is treated in a similar way ; the remarks in the last paragraph also apply here. The above treatment only holds when the planes of the crank-pins and wheels nearly coincide, as already explained when dealing with the uncoupled outside-cylinder engine. On some narrow-gauge railways, in which the wheels are placed inside the frames, the crank and coupling pins are often at a considerable distance from the plane of the wheels. Let the coupling rods be on the outer pins. It will be convenient, | .,. . when dealing with this case, to l^'i jt ~ ^jp^^^^la^i^^;^^ ' iJij', find the distance between the 1^ planes containing the centres of I '«--t}-i — ^— y ' ' gravity of the coupling and r* g connecting rods, viz. C,. „ _ W. Q + (W, -I- fW,)C c I Fig. 234. W, -f W, + fWp The W, must include the weight of the crank-webs and pins all reduced to the radius of the pin and to the distance C. For all practical purposes, C, may be taken as the distance between the insides of the collars on the crank-pin. Then, by precisely similar reasoning to that given above — where W.^'^^tyVp + W.-fW.) In some cases _)> is only osC, ; then — Wb = i-s8Wbo, and 0= 18° Dynamics of the Steam- Engine. 227 Hammer Blow. — If the rotating parts only of a loco- motive are fully balanced there is no variation of load on the rail due to their centrifugal force, but when, in addition, the reciprocating parts are partially or fully Ijalanced the vertical component of the centrifugal force of the excess balance weight over and above that required to balance the rotating parts causes a considerable variation of the load on the rail, tending to lift the wheel off the rail when the balance weight is on top and causing a very rapid increase of rail load — almost amounting to a blow — when the balance weight is at the bottom. The hammer blow is rather severe on the cross girders of bridges and on the permanent way generally. At very high speeds the upward force will actually lift the wheel off the rail if it exceeds the dead weight on the wheel. On some American railroads the proportion of the recipro- cating parts to be balanced is settled by the maximum speed the engine is likely to reach. In arriving at this speed, the balance weight required to completely balance the rotating parts is first determined, then the balance weight for both rotating and reciprocating parts is found, the difference between the two is the unbalanced portion which is responsible for the hammer blow and the tendency to lift the wheel off the rail. Let the difference be W^' then the centrifugal force tending to lift the wheel is o"ooo34VV^RbN* and when this exceeds the dead weight W„ on the rail the wheel will lift, hence the speed of lifting is -sj W N o-ooo34W^Rb When calculating the speed of the train from N it must not be forgotten that the radius of the wheel is greater than Rb. Centre of Gravity of Balance Weights and Crank- webs. — The usual methods adopted for finding the position and weight of balance weights are long and tedious ; the follow- ing method will be found more convenient. The effective balance weight is the whole weight minus the weight of the spokes embedded. Let Figs. 235, 236, 237 represent sections through a part of the balance weight and a spoke ; then, instead of dealing first with the balance weight as a whole, and afterwards deducting 228 Mechanics applied to Engineering. the spokes, we shall deduct the spokes first. Draw the centre lines of the spokes x, x, and from them set off a width w on Fig. ass- each side as shown in Fig. 236, where wi = half the area of the spoke section ; in the case of the elliptical spoke, wf = o'785Di^ o'392Di/ of the rectangular spoke, wi = — 2 o'sD^ By doing this we have not altered either the weight or the position of the centre of gravity of the section of the balance weight, but we have reduced it to a much simpler form to deal with. If a centre line yy (Fig. 235) be drawn through the balance weight, it is only necessary to dealt with the segments on one side of it. Measure the area of the segments when thus treated. Dynamics of the Steam-Engine. 229 Let them be Aj, A^, A3 ; then the weight of the whole balance weight is the sum of these segments — Wb = 2twJ,k^ + A, + A3) where zc„ = the weight per cubic inch of the metal. For a cast-iron weight — Wb = o-52/(Ai + Aj, + A3) For a wrought-iron or cast-steel weight — Wb = o-56/(Ai + A, + A3) all dimensions being in inches. The centre of gravity of each section can be calculated, but it is far less trouble to cut out pieces of cardboard to the shape of each segment, and then find the position of the centre of gravity by balancing, as described on p. 75. Measure the distance of each centre of gravity from the line AB drawn through the centre of the wheel. Let them be ri, r^, r^ respectively ; then the radius of the centre of gravity of the whole weight (see Fig. 235) — Rb- A, + A,-t-A. (^^«P-S8) md WgRj = \ or (0-56; i?(Airi + AaZ-a + Aa^s) ?30 Mechanics applied to Engineering. If there were more segments than those shown, we should get further similar terms in the brackets. Tig. 237- Fig. 238. AVhen dealing with cranks, precisely the same method may be adopted for finding their weight and the position of the centre of gravity. In the figures, the weight of the crank = zto™ X shaded areas. The position of the centre of gravity is found as before, but no material error will be introduced by assuming it to be at the crank-pin. Governors. — The function of a flywheel is to keep the speed of an engine approximately constant during one revolu- tion or one cycle of its operations, but the function of a governor is to regulate the number of revolutions or cycles that the engine makes per minute. In order to regulate the speed, the supply of energy must be varied proportionately to the resistance overcome ; this is usually achieved automatically by a governor consisting essentially of a rotating weight suspended in such a manner that its position relatively to the axis of rotation varies as the centrifugal force acting upon it, and therefore as the speed. As the position of the weight varies, it either directly or indirectly opens and closes the valve through which the energy is supplied, closing it when the speed rises, opening it when it falls. The governor weight shifts its position on account of a change in speed ; hence some variation of speed must always take place when the resistance is varied, but the change in Dynamics of the Steam-Engine. 231 speed can be reduced to a very small amount by suitably arranging the governor. q Simple Watt Governor. — Let the ball shown in the figure be suspended by an arm pivoted at O, and let it rotate round the axis OOi at a constant rate. The ball is kept in equilibrium by the three forces W, the weight of the ball acting vertically downwards (we shall for the present neglect the weight of the arm and its attachments, also friction on the joints) ; C, the centri- fugal force acting horizontally; T, the tension in the supporting arm. f IG. 239. Let H = height of the governor in feet ; h = „ „ „ inches; R = radius of the ball path in feet ; N, = number of revolutions made by the governor per second ; N = number of revolutions made by the governor per minute ; V = velocity (linear) in feet per second of the balls. By taking moments about the pin 0, we have— WV^H CH = WR, hence H = i;R -=WR g^^ o'8i6 47r^R^N/ N 1 Expressing the height in inches, and the speed in revolu- tions per minute, we get — 3513? h = Thus we see that the height at which a simple Watt governor will run is entirely dependent upon the number of revolutions per minute at which it runs. The size of the balls and length of arms make no difference whatever as regards the height when the balls are " floating." The following table gives the height of a simple Watt <• governor for various speeds : — 232 Mechanics applied to Engineering. Change of height Revolutions per minute («). Height of governor in inciics (A). corresponding to a change of speed of 10 revolutions per minute. Inches. 50 14-09 — (54-2) (I2-00) — 60 979 4-30 70 7-19 260 80 S'Si 1-68 90 4-3S i-i6 100 3-52 083 no 2-91 o'6i 120 2 -45 0-46 These figures show very clearly that the change of height corresponding to a given change of speed falls off very rapidly as the height of the governor decreases or as the apex angle Q increases ; but as the governing is done entirely by a cliange in the height of the governor in opening or closing a throttle or other valve, it will be seen that the regulating of the motor is much more rapid when the height of the governor is great than when it is small ; hence, if we desire to keep the speed within narrow limits, we must keep the height of the governor as great as possible, or the apex angle 6 as small as possible, within reasonable limits. Suppose, for instance, that a change of height of 2 inches were required to fully open or close the throttle or other valve ; then, if the governor were running at 60 revolutions per minute, the 2-inch movement would correspond to about 7 per cent, change of speed; at 80, 15 per cent.; at 100, 24 per cent; at 120, 36 per cent. The greater the change of height corresponding to a given change of speed, the greater is said to be the sensitive?KSs of the governor. A simple Watt governor can be made as sensitive as we please by running it with a very small apex angle, but it then becomes very cumbersome, and, moreover, it then possesses very little " power " to overcome external resistances. Loaded Governor. — In order to illustrate the principle of the loaded governor, suppose a simple Watt governor to be loaded as shown. The broken lines show the position of the governor when unloaded. When the load W„ is placed on the balls, the " equivalent Dynamics of the Steam- Engine. 233 height of the simple Watt governor " is increased from H to H^. Then, constructing the triangle of forces, or, by taking W moments about the pin, and remembering that — - acts ver- 2 tically downwards through the centre of the ball, we have — Fig. 240. \V R. r — Then, by precisely the same reasoning as in the case given above, we have — H.= W ^ W 4-—!° o-8i6/ ^ 2 N/ W or h, ■ W + — 3523°! L N" \ W If W„ be m times the weight of one ball, we have — the value of m usually varies from 10 to 50. This expression must, however, be used with caution. Consider the case of a simple Watt governor . both when unloaded and when loaded as shown in Figs. 239 and 240. If the same governor be taken in both instances, it is evident that its maximum height, i^. when it just begins to lift, also its 234 Mechanics applied to Engineering. minimuni apex angle, will be the same whether loaded or unloaded, and cannot in any case be greater than the length of the suspension arm measured to the centre of the ball. The speed of the loaded governor corresponding to any given height will, however, be greater than that of the unloaded governor in the ratio » /i + — to i, and if the engine runs at the same speed in both cases, the governor must be geared up in this ratio, but the alteration in height for any given alteration in the speed of the engine will be the same in both cases, or, in other words, the proportional sensitiveness will be the same whether loaded or unloaded. We shall later on show, however, that the loaded governor is better on account of its greater power. In the author's opinion most writers on this subject are in error ; they compare the sensitiveness of a loaded governor at heights which are physically impossible (because greater even than the length of the suspension arms), with the much smaller, but possible, heights of an unloaded governor. If the reader wishes to appeal to experiment he can easily do so, and will find that the sensitiveness actually is the same in both cases. The following table may help to make this point clear. m has been chosen as i6 : then / m On comparing the last column of this table with that for the unloaded governor, it will be seen that they are identical, or the sensitiveness is the same in the two cases. Change of height corresponding to a change of speed Revolutions per Height of loaded of 30 revolutions minute of governor in per minute of governor. inches. govemorp or 10 revolutions per minute of engine. Inches. 150 14-09 180 979 4 "3° 210 7-19 2 -60 240 551 1-68 270 4'35 116 300 3"5z 0-83 330 2'9I 0'6i 360 2-45 0-46 Dynamics of the Steam- Engine. 235 If, by any system of leverage, the weight W„ moves up and down .t: times as fast as the balls, theabove expression becomes — Porter and other Loaded Governors. — The method of loading shown in Fig. 240 is not convenient, and is rarely adopted in practice. The usual method is that shown in Fig. 241, viz. the Porter governor, in which the links are usually of equal length, thus making x = 2; but this proportion is not always adhered to. Then — - u _ 35230/W + W„N ^' = ^(^ + *") Occasionally, governors of this type are loaded by means of a spring, as shown in Fig. 243, instead of a central weight. The arrangement is, however, bad, since the central load increases as the governor rises, and consequently makes it far too sluggish in its action. Let the length of the spring be such that it is free from load when the balls are right in, «'.*. when their centres coincide with the axis of rotation, or when the apex angle is zero. The reason for making this stipulation will be apparent when we have dealt with other forms of spring governors. Let the pressure on the spring = Pa:, lbs. when the spring is compressed x„ feet ; / = the length of each link in feet ; H, = the equivalent height of the governor in feet ; ^V = the weight of each ball in lbs. ; X. = 2(1 - H.) ; N = revolutions of governor per minute ; Then, as in the Porter- governor, or, by taking moments about the apex pin, and remembering that the force Px^ acts parallel to the spindle, we have — H. _ W + Pa:. _ W + gP/ - zPH. R, ~ C ~ o-ooo34WR,N!' ,HXo-ooo34WN2 + 2P) = W + 2P/ „ W + 2PI . , "• = 0-00034WN' + 2P ^°' ** ^"''*=^^ SovernoT zP/ H, = 'iTrNT2~i — T> >i horizontal „ 236 Mechanics applied to Engineering. In the type of governor shown in Fig. 244, which is frequently met with, springs are often used instead of a dead weight. The value of * is usually a small fraction, consequently a huge weight would be required to give the same results as a Porter or similar type of governor. But it has other inherent defects which will shortly be apparent. Isochronous Governors. — A perfectly isochronous governor will go through its whole range with the slightest Fig. 241. Fig. 242. variation in speed ; but such a governor is practically useless for governing an engine, for reasons shortly to be ^cussed. Fig. 243. ' But when designing a governor which is required to be very sensitive, we sail as near the wind as we dare, and make it very nearly isochronous. In the governors we have considered. Dynamics of the Steam-Engine. 237 the height of the governor has to be altered in order to alter the throttle or other valve opening. If this could be accom- plished without altering the height of the governor, it could also be accomplished without altering the speed, and we should have an isochronous governor. Such a governor can be con- structed by causing the balls to move in the arc of a parabola, the axis being the axis of rotation. Then, from the pro- perties of the parabola, we know that the height of the' governor, i.e. the subnormal to the path of the balls, is constant for all positions of the balls ; therefore the sleeve which actuates the governing valve moves through its entire range for the smallest increase in speed. We shall only consider an approximate form which is very commonly used, viz. the crossed-arm governor. The curve abc is a parabolic arc ; the axis of the parabola is O^; then, if normals be drawn to the curve at the highest and lowest positions of the ball, they intersect at some point d on the other side of the axis. Then, if the balls be suspended from this point, they will move in an approxi- mately parabolic arc, and the governor will therefore be approximately isochronous — ■ and probably useless because too sensitive. If it be de- < sired to make the governor more stable, the points d, d are brought in nearer the axis. The virtual centre of the arms is at their inter- section ; hence the height 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. It can also be arrived at thus. By taking moments about the pin d, W is the weight of the ball, / the length of the arm ad. Fig. 24s. 238 Mechanics applied to Engineering. VV(R ■\-x) = CH„ = 0-00034WRNV/'' - (R + xf ^=s/- ^^+^ 0-00034 kV/'' - (R + xf X H„ - H , „ RH„ o-8i6 x 60^ or thus: - = -^^ and H =^^-^ = - ^, — --^- 2937(jc + R) KH„ when the dimensions of the governor are taken in feet and the speed in revolutions per minute. In some instances Watt governors are made with the arms suspended at some distance, say x, from the axis of rotation, as shown in Fig. 249, but without the central weight. Then the above expression becomes — N- / R-* o-ooo34RV/'-(R-a:)^ and when the expression for H is used (p. 231), the height H is measured from the level of the ball centres to the point where the two arms cross (see Fig. 249). Astronomical Clock Governor. — A beautiful applica- tion of the crossed-arm principle as applied to isochronous governors is found in the governors used on the astronomical clocks made by Messrs. Warner and Swazey of Cleveland, Ohio. Such a clock is used for turning an equatorial telescope 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 Dynamics of the Steam- Engine. 239 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, e, covered with some soft material, is attached to a lug on the link c, and, as the governor rotates, presses on the fixed drum f. The point of suspension, d, is so chosen that the governor is , practically isochronous. The ; weights rest in their cradle until '^V the speed of the governor is sufficiently high to cause them to lift ; when in their lowest position, ^'°' ''*^- the centre line of the weight arm passes through b, and con- sequently the pull along the arm has no moment about this point, but, as soon as the speed rises sufficiently to lift the weights, the centre line of the weight arm no longer coincides with b, and the pull acting along the weight arm now has a moment about b, and thus sets up a pressure between the rotating brake shoe e and the fixed drum /. The friction between the two acts as a brake, and thus checks the speed of the clock. It will be seen that this is an extremely sensitive arrange- ment, since the moment of the force acting along the ball link, and with it the pressure on the brake shoe, varies rapidly as the ball rises ; but since the governor is practically isochronous, only an extremely small variation in speed is possible. It should be noted that the driving effort should be slightly in excess of that required to drive the clock and telescope, apart from the friction on the governor drum, in order to ensure that there is always some pressure between the brake shoe and the drum. Wilson Hartnell Governor. — Another well-known and highly successful isochronous governor is the " Wilson Hartnell " governor. In the diagram, c is the centrifugal force acting on the ball, and / the pressure due to the spring, i.e. one-half the total pressure. As the balls fly out the spring is compressed, and since the pressure increases directly as the compression, the pressure p increases directly (or very nearly so) as the radius r of the balls ; hence we nn.ay write / = Kr, where K is a constant depending on the stiffness of the spring. 240 Mechanics applied to Engineering. Let r^ = nr, frequently n =. \. Then Wj — pr and o'ooo34W>-''«N^ = YJ-' and N'' = K o"ooo34Wn For any given governor the weight W of the ball is con- stant; hence the denominator of the fraction is constant, whence N^, and therefore N, is constant ; i.e. there is only one speed at which the governor will float, and any increase or decrease in the speed will cause the balls to fly right out or in, or, in other words, will close or fully open the governing valve ; therefore the governor is isochronous. There are one or two small points that slightly affect the isochronous cha- FlG. 247. 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. 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. Dealing first with the case of the arm, and afterwards with the ball- Let the vertical section of the arm be a square inches, and all other dimensions be in inches. The centrifugal moment acting on the element is — wadr.rurh _ wau^r^dr 1 2p 1 2"- tan Q Dynamics of the Steam-Engine. and on the whole arm 241 J r^dr 1 2g tan 0. 12X3^ tan 9 which may be written „ R' «-' waR X — X : — a 3 J2g tan The quantity wa^ is the weight of the whole arm VV„, R2 and — IS the square of the .3 radius of gyration of the arm about the axis of rotation, and the product is the moment of Fio. 248. inertia of the arm in pounds weight and inch ^ units. lo)' W„RV The centrifugal moment) acting on the arm ) i2g tan 36^ tan 6 In the case of the ball we have the centrifugal moment for the elemental slice. ■waidf\r' i(i?ki 12g and for the whole ball^ woyr Via?/ (77i)«"^=MVn The quantity outside the brackets is the sum of the moments of the weight of each vertical slice of the ball about the axis of rotation, which is the product of the weight of the ball and the distance of its centre of gravity from the axis. The centrifugal moment of the ball is — • We X Rb X <o' X K This is the value usually taken for the centrifugal moment R 242 Mechanics applied to Engineering. of the governor, but it of course neglects the arms. A common way of taking the arms into account is to assume that the centrifugal force acts at the centre of gravity of the arm; but it is incorrect. By this assumption we get for the centrifugal moment — wd^ R H „ M.aRa)^R= X X — X «)' = — ^ g 2X12 2 48^ tan B W„RV 48^ tan Q Hence the ratio of the true centrifugal moment of the arm to the approximation commonly used is f, thus the error involved in the assumption is 33 per cent, of the centrifugal moment of the arm. In cases in which the weight of the arm is small compared with the weight of the ball the error is not serious, but in the case of some governors in which thick 'stumpy arms are used, commonly found where the balls and arms are of cast or malleable iron, the error may amount to as much as 20 per cent., which represents about 1 1 per cent, error in the speed. The centre of gravity assumption is a convenient one, and may be adhered to without error by assuming that the weight of the arm is f of its real weight. When the arm is pivoted at a point which is not on the axis of rotation, the corresponding moment of inertia of the arm should be taken. The error involved in assuming it to be on the axis is quite small in nearly all cases. In all cases the centripetal moment in a gravity governor is found by taking the moment of the arm and ball about the point of suspension. The weight of the arm and ball is considered as concen- trated at the centre of gravity ; but in the case of the lower links, the bottom joint rises approximately twice as fast as the upper joint, hence its centre of gravity rises 1-5 times as fast as the top joint, and its weight must be taken as I 'S times its real weight; this re- mark also applies to the centrifugal moment, in that case the lower link is taken as twice its real weight. The height of this governor is H„ not h^ ; i.e. the height is measured from the virtual centre at the apex. Dynamics of the Steam-Engine. 243 H = n - '-a 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. 250 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 friction. The speed of these governors is usually calculated by assuming that the mass of each arm is concentrated at its centre of gravity, or that it is a governor having weightless arms carrying equivalent balls, as shown in broken lines in the figure. It can be readily shown by such reasoning as that given above that such an assumption is correct provided the arms are parallel with the spindle, and that the error is small provided that the arms only move through a small angle. These governors work exceedingly well, and keep the speed within very narrow Hmits. The figure is not drawn to scale. Fig. 250. 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 Hartnell- McLaren governor, chosen because it contains many good points, and, moreover, has a great reputation for govern- ing within extremely fine limits (Figs. 251 and 252). The eccentric E is attached to a plate pivoted at A, and suspended by spherical-ended rods at B and C. A curved cam. 244 Mechanics applied to Engineering. 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 alter the speed when running, a leather-covered lever is pu.shed 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 qpeed of the engine increased, and if on the other edge the reverse. 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 extremely sensitive, and, in order to prevent hunting, a dashpot K is attached to the weight arm. In the actual governor two weights 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. A later form of the same governor is shown in Fig. 252, an inertia weight I is attached to the eccentric. The speeding Dynamics of the Steam-Engine. 24S up is accomplished by the differential bevil gear shown. The outer pulley A is attached to the inner bevil wheel A, and the inner pulley B to the adjoining wheel; by applying a brake to one pulley the bevil wheels turn in one direction, and when the brake is applied to the other pulley the wheels turn in the opposite direction, and so tighten or slacken the springs SS. Inertia Effects on Governors. — Many governors rely 246 Mechanics applied to Engineering. 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 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 foi^ce, 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, tiiey 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, since 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 j«/« of the two, and a prompt action is secured ; but if, on the other hand, the inertia moment aqts 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. 253. Dynamics of the Steam- Engine. 247 In every inertia force. case c is the centrifrugal force, and T the dv W dv T = M— 7, or = — ■ — , where W is the weight dv _dt ^ g^'" df of the ball, and ^ the acceleration in feet per second per second. For example, let the centre of gravity of a crank- shaft governor arm and weight be at a radius of 4 inches when the governor is running at 300 revolutions per minute, and let it be at a radius of 7 inches when the governor is running at 312 revolutions per minute, and let the change take place in o'2 second. The weight of the arm and weight is 25 pounds. The change of velocity is — 2 X ^'14 — (7 X 312 — 4 X 300) = 8-59 feet per sec^ond. X 60 SV8-SQ and the acceleration -kt —7-^ = 42*95 feet per sec. per sec, and the force T = n_='-' ^ — ? = 23-4 pounds. 22'2 Sense of rotation. Position of ball. Centrifugal moment. A B Inertia moment for an increase of speed. A B Effect of inertia on governor. + + + I 2 3 I 2 ' 3 -c^3 CxXi Ketards its action No effect Assists its action »i II No effect 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 248 Mechanics applied to Engineering. Scale ■ 0-2? feet 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. 254 we give such a diagram for a simple Watt governor, neglecting the weight of the arms. The axis OOi 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 W„R, which are equal for all positions of the ball, unless the ball is being accelerated or retarded. The centri- fugal moment tends to carry the ball out- wards, the centripetal to bring it back. The four numbered curves show the relation between the moment tendingto make the balls fly out (ordi- nates) and the position of the balls. The centri- petal moment line shows the relation between the moment tending to bring the balls back and the position of the balls, which is independent of the speed. We have — CH = o-ooo34WRN2H = o-ooo34WN''(RH) = KRH The quantity O'ooo34 WN^ is constant for any given ball running at any given speed. Values of KRH have been calcu- lated for various posi- tions and speeds, and the curves plotted, directly as the radius ; The Fig. 254. value of W.R Dynamics of the Steam-Engine. 249 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 vary to a greater extent. For the centrifugal moment, W = weight of (ball + 1 arm + | link). The resultant acts at the centre of gravity of W. For the centripetal moment, W„ = weight of (ball + sleeve + arm + i link). The resultant acts at the centre of gravity of W„ , the weight of the sleeve being re- garded as concentrated at the top joint of the link. It is here assumed that the sleeve rises twice as fast as the top pin of the link. If the centrifugal and centripetal curves coincided, the governor would be isochronous. If the slope of the centrifugal curve be less than that of the centripetal, the governor is too stable ; but if, on the other hand, the slope of the centrifugal curve be greater than that of 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 balls in — consequently the balls are accelerated, and fly out to their extreme position, com- pletely 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. When designing a governor which is intended to regulate the speed within narrow limits, it is important to so arrange it that any given change in the speed of the engine shall be constant for any given change in the height throughout its range. Thus if a rise of 1 inch in the sleeve corresponds to a difference of 5 revolutions per minute in the speed, then each |th of an inch rise should produce a difference of i revolution per minute of the engine in whatever position the governor 250 Mechanics applied to Engineering. may be. This condition can be much more readily realized in automatic expansion governors than in throttling governors. 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. 255 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 8q 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 on the friction of governors and their attachments ; ' 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. 250). The effect of friction is to increase the height of the governor when it is rising, and to reduce it when falling. The exact difference in height can be calculated if the frictional ' See Paper by Ransome, Proc. Inst. C.£., vol. cxiii. j the question was investigated some years ago by one of the author's students, Mr. Eurich, wJio found that when oiled the Watt governor tested lagged behind to the extent of 7'S per cent., and when unoiled I7"5 per cent. Fig. 255. Dynamics of the Steam-Engine. 251 resistance referred to the sleeve is known ; it is equivalent to increasing the weight on the sleeve when rising and re- ducing it when falling. In the well-known Pickering governor, the friction of the governoritself is reduced to a minimum by mount- ing the ballson 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 de- tail several types of go- vernor by the method just described. Porter Governor Diagram. — In this case the centripetal force is greatly increased while the centrifugal is un- affected by the central weight W^, which rises twice as fast as the balls (Fig. 256) when the links are of equal length. Resolve W„ in the directions of the two arms as shown : it is evident that the com- ponent ab, acting along the upper arm, has no moment about O, but l)d = Wo has a centri- petal moment, WoR,, ; then we have — CH = WR -f W„R„ Values of each have been calculated and plotted as in Fig. 254. In the central spring governor W„ varies as the balls liltj in other respects the construction is the same. It should be noticed that the centripetal and centrifugal Fig. 256. 252 Mechanics applied to Engineering. moment curves coincide much more closely as the height of the governor increases j 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 comcident, hence the governor is isochronous. Wilson Hartnell Governor Diagram (Fig. 257). — When constructing the curves a, b, c, d, e, the moment of the weight of the ball on either side of the suspension pin, also the other disturbing causes, have been neglected. We have shown that cr„ = pr, also that c and p vary as R, hence the centrifugal moment lines (shown in full) and the centripetal moment line Oa both pass through the origin, under these conditions the governor is isochronous. A com- mon method of varying the speed of such governors is to alter the load on the spring by the lock nuts at the top ; this has the effect of bodily shifting the centripetal moment line up or down, but it does not alter the slope, such as db, ec, both of which are parallel to Oa. But such an alteration also affects the sensitiveness ; if the centripetal line was db, the governor would hunt, and if ec, it would be too stable. These defects can, however, be remedied by altering the stiffness of the spring, by throwing more or less coils out of action by the corkscrew nut shown in section, by means of which db can be altered to dc and ec to eb. For fine governing both of these adjustments are necessary. When the moment of the weight of the ball and other disturbing causes are taken into account, the curves / and g are obtained. Instead of altering the spring for adjusting the speed, some makers leave a hollow space in the balls for the insertion of lead until the exact weight and speed are obtained. 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 Dynamics of the Steam- Engine. Oi 253 Fro. 257. 254 Mechanics applied to Engineering. rapidly as the centrifugal moment of the balls increases. The governor, which is shown in Fig. 258, 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. Id 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 efiect of the conical 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 t"his case is rather too steep for accurate governing. The centrifugal moment at any instant is — 4 X o-ooo34WRN=H where W is the weight of one ball. And the centripetal moment is — Load on spring x R. See Fig. 258 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'S X n^ X 450^* X I'sS = 260 pound-inches and for the centripetal moment — The load on the spring =111 lbs. ; and R, = 278 centripetal moment = in x 278 = 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 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 256 Mechanics applied to Engineering. 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. Fig. 359. 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 Dynamics of the Steam- Engine. 257 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 poweiTj 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 + 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 + min. load on sp ring \ X the stretch or compression of spring where n = the number of springs employed j express weights in poimdSj 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 jn a governor : — Foot-pounds of energy Type of governor. stored per inch diametei of cylinder. For trip gears and where small resistances have to be overcome ... ... ... .-• ... o'5-o*7S For fairly well balanced throttle-valves 075-1 In the earlier editions of this book values were given for automatic expansion gears, which were bfsed 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 s 258 Mechanics applied to Engineering. times that amount of energy stored is found to be insufficient. If tiie 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 (j.e. the whole area of tha back x the steam pressure) in the case of unbalanced valves. The effort a governor is capable of exerting can also be arrived at approximately by finding the energy stored in the springs, and dividing it by the distance the slide-valve moves while the springs move through their extreme range. Generally speaking, it is better to so design the governor that the valve-gear cannot react upon it, then no amount of pressure on the valve-gear will after the height of the governor ; that is to say, the reversed efficiency of the mechanism which alters the cut-off must be negative, or the efficiency of the mechanism must be less than 50 per cent. On referring to the McLaren governor, it will be seen that no amount of pressure on the eccentric will cause the main weight W to move in or out. Readers who wish to go more fully into the question of governors will find detailed information in " Governors and Governing Mechanism," by H. R. Hall, The Technical Publishing Co., Manchester ; " Dynamics of Machinery," by Lauza, Chapman and Hall ; " Shaft Governors," Trinks and Housum, Van Nostrand & Co., New York. CHAPTER VII. VIBRA TION. Simple Harmonic Motion (S.H.M.).— When a crank rotates at a uniform velocity, a slotted cross-head,, such as is shown in Fig. i6o, moves to and fro with simple harmonic motion. In Chapter, VI. it is shown that the force required to accelerate the cross-head is T, WV^ * /■^ ^^ = 7r^r •. ■ ■ • • «• where W = weight of the cross-head in pounds. V = velocity of the crank-pin in feet per sec. R = radius of the crank in feet. X = displacement of cross-head in feet from the central position. g = acceleration of gravity. K = radius of gyration in feet. Then the acceleration of the crosshead in this position '\'^x _ Y! V i ''* displacement from the middle ■ "rs" ~ r2 ^ 1 of the stroke, V _ /acceleration _ /^jS_ ,..> °' R - V displacement " V W^ • • • . (n^ These expressions show that the acceleration of the cross-head is proportional to the displacement x, and since the force tending to make it slide is zero at the middle of the stroke, and varies directly as x, it is clear that the direction of the acceleration is always towards the centre O. The time t taken by the cross-head in making one complete journey to and fro is the same as that taken by the crank in making one complete revolution. W " 26o Mechanics applied to Engineering Hence t ■■ 27rR = 2TrsJ' displacement acceleration (iii) On referring to the argument leading up to expression (i) in Chapter VI. it will be seen that Pj is the force acting along the centre line of the cross-head when it is displaced an amount x from its middle or zero-force position. When dealing with the vibration of elastic bodies W is the weight of the vibrating body in pounds, and Pj is the force in pounds weight required to strain the body through a distance x feet. When dealing with angular oscillations we substitute thus — Angular. Linear. The moment of inertia of the body I = in pound-fdot units. Mass of the body ( — ). where W is in pounds. The couple acting on the body C = lA in pound-foot units. W Force acting on the body P, = —/ in pounds. The angular displacement (9) in radians. Linear displacement {x) in feet. The angular velocity (oj) in radians per second. The linear velocity (v) in feet per second. The angular acceleration (A) in radians per second per second. . The linear acceleration (/) in feet per second per second. Kinetic energy J Ia>' Kinetic energy - ~v' Hence, when a body is making angular oscillations under the influence of a couple which varies as the angular displacement, the time of a complete oscillation is — "•V le c (iv) These expressions enable a large number of vibration problems to be readily solved. Vibration. 261 Simple Pendulum. — In the case of a simple pendulum the weight of the suspension wire is regarded as negligible when compared with the weight of the bob, and the displacement x is small compared to /. The tension in the wire is normal to the path, hence the only accelerating force is the component of W, the weight of the bob, tangential to the path, hence Pj = W sin 6 = — r- /Wxl / g (v) where / is the length of the pendulum in feet rm. ^go. measured from the point of suspension to the c. of g. of the bob; t is the time in seconds taken by the pendulum in making one complete oscillation through a small arc. The above expression may be obtained direct from (iv) by substituting — for I, and "Wx or W/5 for C. g Then VfBg Compound Pendulum. — When the weight of the arm is not a negligible quantity, the pendulum is termed compound. Let W be the weight of the bob and arm, and /, b^ the distance of their c. of g. from the point of suspension. Then if ;«; be a small horizontal dis- placement of the c. of g. C = W^c = W49 (nearly), and W/' (vi) CofG 4? Fig. 261. where Ko is the radius of gyration of the body about the point of suspension. If K be the radius of gyration of the body about an axis through the centre of gravity, then (see p. 76) 262 and Mechanics applied to Engineering. J ^ VVK vv^ ^ w , S g g If / be the length of a simple pendulum which has the same period of oscillation, then g 1 = K= + // and/„(/-4) = K^ or OG.GOi = K°, and since the position, of G will be the same whether the point of suspension be O or Oj the time of oscil- lation will be the same for each. Oscillation of Springs. — Let a weight W be suspended from a helical spring as shown, and let it stretch an amount 8 (inches) when supporting the load W, where 8 = -^^ (see page 587). D is the mean diameter of the coils in inches d „ diameter of the wire in inches n ,, number of free coils G „ modulus of rigidity in pounds per square inch W „ weight in pounds, Then, neglecting the weight of the spring itself, we have, for the time of one complete oscillation of the spring (from iii) /W^ / W8 / D'Ww t = 2-rsJ s— = 2'r\/ rr— = 2-K\J -— ^ ^ig ^ i2W^ V v^gGd" t = o-904>/ -^^ (vm) t = 2TT\/ ■- , where A is the deflection in feet. For a simple pendulum t = 2ir\J - , hence the time of one complete oscillation of a spring is the same as that of a simple pendulum ' Vibration. 263 whose length is equal to the static deflection of the spring due to the weight W. The weight of the spring itself does not usually affect the problem to any material extent, but it can be taken into account thus : the coil at the free end of the spring oscillates at the same velocity as the weight, but the remainder of the coils move at a velocity proportional to their distance from the free end. Let the weight per cubic inch of the spring = w, the area of the section = a. Consider a short length of spring dl, distant /from the fixed end of the spring, L being the total length of the wire in the spring. Weight of short length = wa . dl V/ L V/ Velocity of element = -j- w , ci , V^ Kinetic energy of element = ' Ml 2g\I waVV'=^„ „ waV\} Kmetic energy of sprmg = ^ I I dl = 3 X 2g\j 3 X 2^ ~ 3\ 2g ) where W, is the weight of the spring. Thus the kinetic energy stored in the spring at any instant is equal to that stored in an oscillating body of one-third the weight of the spring. Thus the W in the expression for the time of vibration should include one-third the weight of the spring in addition to that of the weight itself. Expression (viii.) now becomes — Example. — D = 2 inches d = 0-2 inch W = 80 pounds « = 8 G = 12,000,000 pounds per sq. inch. Find the lime of one complete oscillation, (a) neglecting the weight of the spring, (d) taking it into consideration. 264 Mechanics applied to Engineering. / 2= X 80 X 8 / — 7- /=o'Q04x/ - = o'Q04v 0-267 V 12,000,000 X o'ooib = 0-4668 sec, say 0*47 second. In some instances the deflection x is given in terms of the force Pi ; if the deflection S be woiked out by the expression given on page 587, it will be found to be 1-33" or o'li foot for every 50 lbs. Then- /. '■ = 2W/y/ 80 X o-ii = 0-47 sec. 50 X 32-2 The weight of the spring = o-69«^D = 0-69 X 8 X 0-04 X 2 Wj = 0-44 lb. Then allowing for the weight of the spring — / 8 X 80-15 X 8 ^ , t = o-oo4x/ 2 = o"4074 second 'V 12,000,000x0-0010 say 0-47 sefcond. Thus it is only when extreme accuracy is required that the weight of the spring need be taken into account. In the case of a weight W pounds on the end of a cantilever L inches long, the time of one complete oscillation is found by inserting the value for 8 (see p. 587) in equation (iii), which gives — The value of 8 can be inserted in the general equation for any cases that may arise. Oscillation of Spring-controlled Governor Arms. — In the design of governors it is often of importance to know the period of oscil- lation of the governor arms. This is a case of angular oscillation, the time of which has already been given in equation (iv), viz. ■= 27r^ 10 C Vibration. 265 The I is the moment of inertia of the weight and arm about the pivot in foot and pound units, and is obtained thus (see page 76) g S^ 12 4) where K is the radius of gyration of the weight about an axis parallel to the pivot and passing through the centre of gravity. For a cylindrical weight K'' = — , and for a spherical weight o K^ = — . The weight of the arm, which is assumed to be of 10 uniform section is w. More complex forms of arms and weights must be dealt with by the graphic methods given on page loi. Example. — The cylindrical weight W is 48 pounds, and is 5 inches diameter. The weight of the arm =12 pounds, Zj = 18 inches, 4=12 inches, /j = 8 inches, h = 2 inches. The spring stretches o'3 inch per 100 pounds. If the dimensions be kept in inch units, the value for the moment of inertia must be divided by 144 to bring it to foot units. T = 48 /25 \ 12 y/3 24 + \ \ 3^1 32-2x144X8 ^V"^32-2 X i44^V 12 / 43 = 1-8 pounds-feet units. 0-3 100 X 8 e = —^ when C = hence t= 2 X 2,'^^\/ a J ,Z. ^ a ~ °'^ ^^"^• X o"3 X 12 8 X 100 X 8 Vertical Tension Rod. — For a rod of length / supported at top with a weight W at the lovyer end^ Inserting the values // - / w/ X =-<g and Pi = A/ we get t = ztt^ g^ Torsional Oscillations of Shafts. — If an elastic shaft be held at one end, and a body be attached at the other as shown, it will oscillate with simple harmonic motion if it be turned through a small arc, and then released. The time of one complete oscillation has already been shown (iv) to be /Te = ^W c 266 Mechanics applied to Engineering. on page 579 it is shown that = — ^ GI„ 6^ Fig 264. radians, where M, is the twisting moment in pounds inches, and I^ is the polar moment of inertia of the shaft section in inch' units. G is the modulus of rigidity in pounds per square inch, / is the length in inches. Substituting this value, we get — I2lM^ M.GI„ /12I/ 2,r^ GT Note. — C is in pounds-feet, and M, in pounds-inches units. Writing L for the length in feet, we get — /i44lL / J,L where Ij is the moment of inertia of the oscillating body in pound and • inch units. If all the quantities be expressed in inch units, and the constants be reduced, we get -Wc^. If greater accuracy is required, one-third of the polar moment of inertia of the rod should be added to I^, but in general it is quite a negligible quantity. If there are two rotating bodies or wheels on a shaft, there will be a node somewhere between them, and the time of a complete oscillation will be and It not infrequently happens in practice that engine crank ^3 — 1. , 1 I, M '2 H Vibration. 267 shafts fracture although the torsional stresses calculated from static conditions are well within safe limits. On looking more closely into Ihe matter, it may often be found that the frequency of the crank effort fluctuations agrees very nearly with the frequency of torsional vibration of the shaft, and when this' is the case the strain energy of the system may be so great as to cause fracture. An investigation of this character shows why some engine crank shafts are much more liable to fracture when running at certain speeds if fitted with two fly- wheels, one at each end of the shaft, than if fitted with one wheel of approximately twice the weight and moment of inertia. f,^. ^cs. *^ In both cases the effective length / is roughly one half the distance between the two wheels, or it may be the distance from the centre of the crank pin to the flywheel ; but the moment of inertia I in the one case is only one half as great as in the other, and consequently the natural period of torsional vibration with the two flywheels is only —7= = 07 of the period with the one flywheel. If the lower speed ha'pperis to synchronise, or nearly so, with the crank effort fluctuations, fracture is liable to occur, which would not happen with the higher period of torsional vibration. Vibration and Whirling of Shafts. — If the experi- ment be made of gradually increasing the speed of rotation of a long, thin horizontal shaft, freely supported at each end, it will run true up to a certain speed, apart from a little wobbling at first due to the shaft being not quite true, and will then quite suddenly start to vibrate violently, and will whirl into a single bow with a node at each bearing. As the speed increases the shaft straightens and becomes quite rigid till a much higher speed is reached, when it suddenly whirls into a double bow with a node in the middle. It afterwards straightens and whirls into three bows, and so on. High speed shafts in practice are liable to behave in this way, and may cause serious disasters, hence it is of great importance to avoid running at or near the speed at which whirling is likely to occur. An exact solution of many of the cases' which occur in practice is a very complex matter (see a paper by Dunker- ley, FMl. Trans., Vol. 185), but the following approximate 268 Mechanics applied to Engineering. treatment for certain simple cases gives results sufficiently accurate for practical purposes. The energy stored at any instant in a vibrating body consists partly of kinetic energy and partly of potential or strain energy, the total energy at any instant being constant, but the relation between the two kinds of energy changes at every instant. In the case of a simple pendulum the whole energy stored in the bob is potential when it is at its extreme position, but it is wholly kinetic in its central position, and in inter- mediate positions it is partly potential and partly kinetic. If a periodic force of the same frequency act on the pendulum bob it will increase the energy at each application, and unless there are other disturbing causes the energy will continue to increase until some disaster occurs. In an elastic vibrating structure the energy increases imtil the elastic limit is passed, or possibly even until rupture takes place. When a horizontal shaft rests on its hearings the weight of the shaft sets up bending stresses, consequently some of the fibres are subjected to tension and some to compression, depending upon whether they are above or below the neutral axis, hence when the shaft rotates each fibre is alternately brought into tension and compression at every revolution. The same distribution and intensity of stress can be produced in a stationary shaft by causing it to vibrate laterally. Hence if a shaft revolve at such a speed that the period of rotation of the shaft agrees with the natural period of vibration, the energy will be increased at each revolution, and if this particular speed be persistent the bending of the shaft will continue to increase until the elastic limit is reached, or the shaft collapses. The speed at which this occurs is known as the whirling speed of the shaft. The problem of calculating this speed thus resolves itself into finding the natural period of lateral vibration of the shaft, if this be expressed in vibrations per minute, the same number gives the revolutions per minute at which whirling occurs. The fundamental equation is the one already arrived For the present purpose it is more convenient to use revolutions per minute N rather than the time in seconds of one vibration, also to express the deflection of the shaft 8 in inches. t 27rV vva at. viz.- /^ Vibration. 269 In this case the displacing force Pi is the load W which tends to bend the shaft, and x is the deflection. Reducing the constants we get — In the expression for the deflection of a beam it is usual to take the length (/) in inches, but in Professor Dunkerley's classical papers on the whirling of shafts he takes the length L in feet, and for the sake of comparing our approximations with his exact values, the length in the following expressions is taken in feet. The diameter of the shaft {d) is in inches. The weight per cubic inch = o'zS lb. E is taken at 30,000,000 lbs. . ird* per square inch. The value of I is -7— . For the values of S see Chapter XIII. Parallel Shaft Supported freely at each end. — - _ ^wl^ _ 270WL* _ 384EI EI 25000^' O-ziiird^ , where w = pounds 4 i8y 29S90i/ y First critical speed ■^ "~ ' — ^- iJ \ ■ — ^single bow 25000;^'' munkerley gets ^^ / ) At the second critical speed the virtual length is — , hence N,= ^^=il^ double bow . (9 ,„d N3=^ = ?%^^reblebow. The following experimental results by the author will show to what extent the calculations may be trusted. Steel shaft — short bearings, d = 0-372 inch, L = 5-44 feet. 270 Mechanics applied to Engineering. Condition. Whirling speed R. p.m. Calculated. Experiment. Single bow .... 370 350 to 400 Double bow 1480 1460 to 1550 Treble bow. . . . 3340 3200 to 3500 It is not always easy to determine the exact speed at which whirHng occurs, the experimental results show to what degree of accuracy the speed can be determined with an ordinary speed indicator. Parallel Shaft, supported freely at each end, central load W. — W/^ _ 36WL° 48EI ~ EI ""^-^^ (-»^-«"-^) Experimental results — </= o'-gg^ inch, L = ys-i feet Whirling speed R. p.m. Calculated. Experiment. Unloaded .... 2940 3100 W = 33 pounds . . 1080 1 100 The weight of the shaft itself may be taken into account thus — the deflection coefficient for the weight of the shaft is jfj = say ^, and for the added load, ^, hence 4f of the weight of the shaft should be added to W, in general the weight of the shaft is a negligible quantity. Tapered Shaft, supported freely at each end, cen- tral load. — If the taper varies in such a manner that — is constant except close to the ends where it must be enlarged on account of the shear, but such a departure from the Vibration. 27 1 assumed conditions does not appreciably aifect the deflection, ■py then since p = —- = constant, the beam bends to an arc of a circle, and — s _ ^ = 54WL° 32EI~ EI ^^ 31050^ Note. — The d is the diameter at the largest section If tlie taper varies £ stant, and we have — If tlie taper varies so that the stress is constant, then — is con- W/s _6sWL' N = 26-5EI EI 281001^ Vwi? Shafts usually conform roughly to one of the above con- ditions but it is improbable that the taper will be such as to give a lower whirling speed than the latter value. For calcu- lating a shaft of any profile the following method is convenient. In Chapter XIII., we show that I I S = :^ (2 m .X .dxioi z. parallel shaft. In the case of a tapered shaft this becomes S JmxBx _ Moment of the area of half the be nding moment diagram ~ Moment of the area of half the b.ni. diagram after altering its depth at each section in the ratio of the moment of inertia in the middle to the moment of inertia at that section. W/= area e/gA x h ,„. .. \ ^' = i8EI^ • 1 <^'S.^66.) ^ area ehg X - 3 2/2 Mechanics applied to Engineering. where—, = =^ = ^r a similar correction being made at every ctb 1-^ a" section. The I in the above expressions is for the largest section, . ttD VIZ. —. — . 64 Fig. 266. For some purposes the following method is more convenient. Let the bending moment diagram egh be divided into a num- ber of vertical strips of equal width, it will be convenient to take 10 strips, each — wide. Start numbering from the support (see page 508). Mean bending moment at the middle of the The distance of each strip from the support. W / 1st Strip = — X — "^ 2 40 l_ 40 W 3/ 2nd „ = — X ^- 2 40 3/ 40 .3-"=?x| Si 40 and so on. and so on. Vibration. 273 The respective moments of inertia are Ij, I2, I3, etc., that at the middle of the shaft being I. Then EIVSoIi 20 40 80I2 20 40 + V-r X — X 3- + etc. I 80I3 20 40 / S = W/^ ^i^9_)_^5^49 ,81 ^121 ^169 ^ 225 64oooE\Ii I2 I3 I4 Ib le ~ J7 Is 289 , 36i \ 8==J^a + J-^ + ii + l? + etc. + f,^) 3i42E\a'f 4* (/s* fl'i (/lo / „ WL' / I , , 36i\ + I + I Experimental Results. — d = 0-995 inch, parallel in middle, tapering to 0*5 inch diameter at ends. L = 3"i7 feet. Whirling speed R. p.m. Calculated. Experiment. Parallel for 4 ins. in middle, unloaded 2606 2750 Ditto for I in. . . . 2320 2425 to 2500 Ditto for 4 ins. Loaded, W = 33 lbs. 900 900 to 910 Ditto for 1 in. Loaded, W = 33 lbs. 800 750 to 770 When the middle of a shaft is rigidly gripped by a wheel boss, or its equivalent, of length 4, the virtual length of shaft for deilection and whirling, purposes is/— 4 instead of/. If the framing, on which the- shaft- bearings are- mounted is^ T 274 Mechanics applied to Engineering. not stifF, the natural period of vibration of the frame must be considered. An insufficiently rigid frame may cause the shaft to whirl at speeds far below those calculated. Vertical shafts, if perfectly balanced, do not whirl, but if thrown even a small amount out of balance they will whirl at the same speeds as horizontal shafts of similar dimensions. The following table gives a brief summary of the results obtained by Dunkerley. See "The Whirling and Vibration of Shafts" read before the Liverpool Engineering Society, Session 1894-5, also by the same author in " The Trans- actions of the Royal Society," Vol. 185A, p. 299. N = the number of revolutions per minute of the shaft when whirling takes place d = the diameter of the shaft in inches L = the length of the shaft in feet Description of shaft. Remarks on /. N. L. U nloaded : overhanging from a long bearing which fixes its direction Length of over- hang in feet n645j;. -v^ Unloaded : resting in short or swivelled bearings at each end Distance between centres of bear- ings in feet 3^364^, ■VI * Unloaded : supported as in last case, but one end overhanging c feet Ditto c d For values ofa see Table I. 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 5'34°i "Vl Unloaded : shaft supported in three short or swivelled bearings, one at each end -I L, = shorter span Lj = longer span in feet d For see values of <j Table II. Vibration. 275 Description of shaft. Remarks on /. N. L. Unloaded : shaft supported in long bearings which fix its direction at each end Clear span be- tween inner edges of bearings d 7497 'l. VI Unloaded : long continuous shaft supported on short or swivelled bearings equi- distant Distance between centres of bear- ings in feet 3286^ -Vi Loaded : shaft supported in short or swivelled bear- ings, single pulley of weight fV pounds cen- trally placed between the bearings Distance between the centres of the bearings in feet N - 37,35o^^j i Loaded ; long bearings which fix its direction at each end Clear span be- tween inner edgesof bearings ^ ^'''^"VwL' 276 Mechanics applied to Engineering TABLE I. Value of »■ = ?_. ... va« I -00 7.554 87 075 12,044 no 0-50 20,931 HS 0-33 28.095 168 0-25 to O'lO 31,000 176 Very small 31.590 178 Zero 32,864 iSi TABLE IL Value of r = i h 0> V5. i-oo too7S 32,864 18: 0-5 to 07s 36.884 192 o'33 41,026 203 0-25 43,289 208 0'20 to 0-14 44.312 211 0"I25 to O'lO 47.I2S 217 Very small 50.654 225 For hollow shafts in which — <^, = the external diameter in inches d^ = „ internal „ „ substitute for d in the expressions given above the value — fjd^ + di for unloaded shafts, and substitute for d^ the value — Vi^i* - d^ for loaded shafts. CHAPTER VIII. GYROSCOPIC ACTIO X. When a wheel or other body is rotating at a high speed a considerable resistance is experienced if an attempt be made to rapidly change the direction of its plane of rotation. This statement can be verified by a simple experiment on the front wheel of a bicycle. Take the front wheel and axle out of the fork, suspend the axle from one end only by means of a piece of string. Hold the axle horizontal and spin the wheel rapidly in a vertical plane, on releasing the free end of the axle it will be found that the wheel retains its vertical position so long as it con- tinues to spin at a high speed. The external couple required to keep the wheel and axle in this position is known as the gyroscopic couple. The gyroscopic top is also a famiUar and striking instance of gyro- scopic action, and moreover affords an excellent illustration for demonstrating the principle involved, which is simply the rotational analogue of Newton's second law of motion, and may be stated thus — When a < til ^^^^ ^"^ ^ body, or self-contained system the change of J^^g^^j^^ momentum^ is proportional to the ii^pressed j , |. Hence, if an external couple act for a given time on the wheel of a gyroscope the angular momentum about the axis of the couple will be increased in proportion to the couple. Fig. 267. From Worthington's "Dyna- mics of Rotation." momentum ) generated in a given time 278 Mechanics applied to Engineering. If a second external couple be impressed on the system for the same length of time about the same or another axis the angular momentum will be proportionately increased about that axis. Since angular momenta are vector quantities they can be combined by the method used in the parallelogram of forces. Precession of Gyroscope. — A general view of a gyro- scopic top is shown in Fig. 267, a plan and part sectional elevation in Figs. 268 and 269. Fig. 269. If the wheel were not rotating and a weight W were suspended from the pivot x as shown, the wheel and gimbal ring would naturally rotate about the axis yy., but when the wheel is rotating at a high speed the action of the weight at first sight appears to be entirely different. Let the axis 00 be vertical, and let the gimbal ring be placed horizontal to start with. Looking at the top of the wheel (Fig. 268), let the angular momentum of a particle on the rim be represented by Qm in a horizontal plane, and let 0« I'epresent the angular momentum in a direction at right angles to Om generated in the same time by the moment of W about yy. The resultant angular momentum will be represented by Or, that is to say, the plane of rotation of the wheel will alter or precess into the direction shown by Or; thus, to the observer, it.precesses in a contra-clockwise sense at a rate shortly to be determined. The precession will be reversed in sense if the wheel rotates in the opposite direction or if the external couple tends to lift the right-hand pivot x. If any other particle on the rim be considered, similar Gyroscopic Action. 279 results, as regards the precession of the system will be obtained. Consider next the case of a horizontal force P applied to the right-hand pivot. Let ym (Fig. 270) represent the angular momentum of a particle on the rim of the wheel, which at the instant is moving in a vertical plane. Let yn represent the angular mo- mentum generated by the external couple of P about the axis 00. The resultant angular momentum is repre- sented by yr, thus indicating that the plane of rotation which was vertical before the application of the external moment has now tilted over or precessed to the inclined position indicated by yr. Thus an observer at y-^ look- ing horizontally at the edge of the wheel sees it precess- ing about y-^^y in a contra- clockwise sense. If the sense of rotation of the wheel be reversed, or if P act in the opposite direction, the sense of the precession will also be reversed. We thus get the very curious result with a spinning gyro- scope that when it is acted on by an external couple or twist- ing moment the resulting rotation does not occur about the axis of the couple but about an axis at right angles to it. An easily remembered method of determining the sense in which the gyroscope tends to precess is to consider the motion of a point on the rim of the wheel just as it is penetrating the plane containing the axis of the wheel and the external couple. In Fig. 268 this point is on the axis 00 on either the top or bottom of the rim. In Fig. 270 it is on the axis yy. The diagram of velocity' is constructed on a plane tangential IV Angular momentum = In? ; K Since I and K are constant, the angular momentum generated is pro- portional to V, hence velocity diagrams may be used instead of angular momentum diagrams. 28o Mechanics applied to Engineering. to the rim of the wheel at the point in question and parallel to the axis of rotation, by drawing lines to represent (i) the velocity of the point owing to the rotation of the wheel ; (ii) the velocity of the point due to the extemal coupje ; (iii) the resultant velocity, the direction of which indicates the plane towards which the wheel tends to precess. In the case of rotating bodies in which the precession is forced the sense and direction of lines (i) and (iii) are known ; hence the sense and direction of (ii) are readily obtained, thus supplying all the data for finding the sense of the external couple. When only the sense of the precession is required, it is not necessary to regard the magnitude of the velocities or the corresponding lengths of the lines. Rate of Precession. — When the wheel of a gyroscope is rotating at a given speed, the rate at which precession occurs Ae p* Fig. 271. is entirely dependent upon the applied twisting moment, the magnitude of which can be readily found by a process of reasoning similar to that used for finding centrifugal force. See the paragraph on the Hodograph on p. 17. The close connection between the two phenomena is shown by the following statements : — (Gyroscopic couple) : —3 ((wheel)] "((spinning)) at a constant < /,\ > speed and be constrained in such a .1. ^-A moves ) , . Ca centre O in the plane of manner that it J(precesses)r^°"* l(an axis 00 at right angles motion I to the axis of rotation) j* Let OP, represent the [(^'^^^,^^^^] velocity of the \i^^^l^^ Gyroscopic Action. 281 when in position r and let a[(/^J^f^^| act on the ^ "J, J in such a manner as to bring it into position 2. Then OP2 represents its corresponding \ , '"^ j"^ ^ > velocity, and PiP„ represents, to the same scale, the change of r velocity j I (angular velocity)/- '^he{(eouple)|^^1"'-d '° ^^^^ ^^^ change ofjjj^^tr velocity)] '^g'^^^^y W _ WVS2 \ i/'^^^ WVKS2 WK^^ ^^ \| (^CK = = ii(o = IQo) j or Centrifugal force = mass of body x velocity x angular velocity. Gyroscopic couple {moment of inertia of wheel about the axis of rotation angular ) velocity X S^T'^'' ^^^°"'y of wheel) I of precession. In the above expressions — W = Weight of body or the wheel in pounds. V = Velocity in feet per second of the c. of g. of the body or of a point on the wheel at a radius equal to the radius of gyration K (in feet). i2 = The angular velocity of the body or the wheel in radians per second. to = The rate of precession in radians per second. N = Revolutions per minute of the wheel. n = Revolutions per minute of the precession. For engineering purposes it is generally more convenient to express the speeds in revolutions per minute. Inserting the value of ^ and reducing we get — WK^N« Gyroscopic couple = = o'ooo34WK^N« in pounds- feet. 282 Mechanics applied to Engineering. The following examples of gyroscopic effects may serve to show the magnitude of the forces to be dealt with. The wheels of a locomotive weigh 2000 pounds each, the diameter on the tread is 6 feet 6 inches, and the radius of gyration is z'8 feet. Find the gyroscopic couple acting on the wheels when running round a curve at 50 miles per hour. Radius of curve 400 feet. Find the vertical load on the outer and inner rails when there is no super-elevation, and when the dead load on each wheel is 10,000 pounds. Rail centres 5 feet. N = 215-5 n = r75 Gyroscopic couple = 2000 X 2'8^ X 2is'5 X f75 2937 = 2015 pounds-feet. Arm of couple, 5 feet. Force, 403 pounds. The figure will assist in determining the sense in which the external couple acts. Since the wheels precess about a vertical axis passing through the centre of the curve, we know that the external forces act in a vertical direction. As the wheels traverse the curve their direction is changed from oni to or, hence the external couple tends to move the top of the wheel in the direction on. The force, which is the reaction of the outer rail, therefore acts upwards, which is equivalent to saying that the vertical pressure on the rail is greater by the Gyroscopic Action. 283 amount of the gyroscopic force than the dead weight on the wheel. Since the total pressure due to both wheels is equal to the dead weight upon them, it follows that the vertical pressure on the inner rail is less by the amount of the gyro- scopic force than the dead weight on the wheel. The vertical pressure on the outer rail is 10,009 + 403 = iOi403 pounds, a«id on the inner rail 10,000 — 403 = 9597 pounds. (No super-elevation.) Thus it will be seen that the gyroscopic effect intensifies the centrifugal effect as far as the overturning moment is con- cerned. If the figure represented a pair of spinning gyroscopic wheels which are not resting upon rails, but with the frame supported at the pivot O, the gyroscope would precess in the direction of the arrow if an upward force were applied to the left-hand pivot x, or if a weight were suspended from the right-hand pivot x. In the case of a motor-car, in which the plane of the fly- wheel is parallel to that of the driving wheels, the angular momentum of the flywheel must be added to that of the road wheels, when the sense of rotation of the flywheel is the same as that of the road wheels, and subtracted when rotating in the opposite sense. When the flywheel rotates in a plane at right angles to that of the road wheels, and its sense of rotation is clockwise when viewed from the front of the car, the weight on the steering wheels will be increased and that on the driving wheels will be diminished when the car steers to the chauffeur's left hand, and vice versa when steering to the right ; the actual amount, however, is very small. When a car turns very suddenly, as when it skids in turning a corner on a greasy road, the gyroscopic effect may be so serious as to bend the crank shaft. Lanchester strengthens the crank web and the neck of the crank-shaft next the flywheel, in order to provide against such an accident. CHAPTER IX. FRICTION. W=iV When one body, whether solid, liquid, or gaseous, is caused to slide over the surface of another, a resistance to sliding is experienced, which is termed the " friction " between the two bodies. Many theories have been advanced to account for the friction between sliding bodies, but none are quite satisfactory. To attribute it merely to the roughness between the surfaces is but a very crude and incomplete statement j the theory that the surfaces somewhat resemble a short-bristled brush or velvet pile much more nearly accounts for known phenomena, but still is far from being satisfactory. However, our province is not to account for what happens, but simply to observe and classify, and, if possible, to sum up our whole experience in a brief statement or formula. Frictional Resistance. (F). — If a block of weight W be placed on a horizontal plane, as shown, and the force F applied parallel to the surface be required to make it slide, the force F is termed the frictional resistance of the block. The normal pressure between the surfaces is N. Coefficient of Friction (/i). — Referring to the figure F F . above, the ratio^ or ^ = /*, and is termed the coefficient of friction. It is, in more popular terms, the ratio the friction bears to the normal pressure between the surfaces. It may be found by dragging a block along a plane surface and measuring F and N, or it may be found by tilting the surface as in Fig. 274. The plane is tilted till the block just begins to slide. The vertical Fig. 273. Friction, 285 weight W may be resolved normal N and parallel to the plane F. The friction is due to the normal pressure N, and the Fig. 274. Fig. 27s. force required to make the body slide is F ; then the coefficient F F of friction ^ — ^3& before. But ^ = tan <^, 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, a force Fo (Fig. 275) will -r-^/r.. Fig. 276. Fig. 277. then 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 Fj (Fig. 276). 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." 286 Mechanics applied to Erigineering. 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 abradec!. 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 ' 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 cmves in Fig. 278 show approximately the diflFerence 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 =/x,N it is more nearly given by F = f<,N"''', or F = /h^'I'v'N The variation is really in /a and not in N, but the ex- pression, which is empirical, assumes its simplest form as given above. For dry surfaces ft has the following values :■ — ' The friction of rest has been very aptly termed the " sticktion." Friction. 287 Wood on wood Metal „ Melal on metal o'2S to 0-50 020 to o'6o 0'I5 to o'30 Leather on wood ,, metal Stone on stone 0-25 to o'5o o'3o to o'6o 040 to o"6s These coefficients must always be taken with caution ; they vary very greatly indeed with the state of the sur- faces in contact. The third law given above is far from representing facts ,: in the limit the fourth law becomes a special case of the third. '^ If the surfaces were perfectly '^ clean, and there ij; were no film of air between, this law would pro- bably be strictly accurate, but all experiments show that the friction decreases with velocity of rubbing. IntensUy of jn-essuro Fig. 278. + Morin-. O Ri'iLtiCe. • Westinghouse &■ Ga/ton 30 40 SO 60 Speed in feet per second. Fig. 279. 288 Mechanics applied to Engineering. The following empirical formula fairly well agrees with experiments : — Let ft = coefficient of friction ; K = a constant to be determined by experiment ; V = the velocity of sliding. Ihen u = 7= 2-4VV The results obtained with dry surfaces by various experi- menters are shown in Fig. 279. 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. Wood on wood »» »» Metal on metal Stone on stone Leather on iron 0-S4 0-69 034 074 0-S9 0'46 0-43 0-26 063 0-52 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 f ^ ^ about to consider, we shall for sake of simplicity, assume that Coulomb's laws hold good. Oblique Force re- squired to make a Body slide on a Horizontal Plane. — If an oblique force P act upon a block of weight W, making an angle Q 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. n, together with W. From a draw a line making an angle <^ (the friction angle) with W, cutting P in Friction. 289 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 « 4- W; hence F = /^(« + W). When 6 = o, P„ = fa? = /tW. When 6 is negative, it simply indicates that Pj 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 = 4>; then — P,„in. = Po cos <j> = fxW cos (t> = tan "^ W cos <l> = W sin 4> It will be seen from the figure that P is infinitely great when ab is parallel to be — that is, when P is just on the edge of the friction cone, or when go — 6 = <j>. When 6 = — 90°, P acts vertically upwards and is equal to W. A general expression for P is found thus — ;? = P sin e F = P cos e = /x(W + n) F = //.(W + P sin ff) and P(cos 6 + i>. s\n 6) = fiW /X.W tan </. W hence P = P = cos ^ + /x sin 6 cos 6 + /u, sin sin <f> W cos <^ cos + sin <^ sin 6 _ W sin <^ ~ cos (^ + 6) When P acts upwards away from the plane, the — sign is to be used in the denominator ; and for the minimum value of P, <f> = —6; then the denominator is unity, and P = W sin <j>, 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 tc represent the weight W, and from e draw a line ed, making V 290 Mechanics applied to Engineering. Fig. aSi. an angle a with it ; then, if from b a. line be drawn parallel to the direction of pull Pi, the line M^ 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 id^c is simply the triangle of forces acting on the block ; the line cdi is, of course, normal to the plane. When friction is taken into account, draw the line ce, making an angle <f> (the friction angle) with cd; then ie^ gives the pull Pj required to drag the block up the plane including friction. For it will be seen that the normal pressure on the plane is cdo, and that the friction parallel to the direction of sliding, viz. normal to cd, is — fx, . cdQ ^= tan (^ . cd^ = d^^ Then resolving d^^ in the direction of the pull, we get the pull lequired to overcome the friction dye-^ ; hence the total pull required to both raise the block and overcome the friction is be^. Least Pull.— The least pull required to pull the block up the plane is when be has its least value, i.e. when be is normal to ce ; the direction of pull then makes an angle <^ with the plane, ox B = ^, for cd is normal to the plane, and F,o. 28s. <■* makes an angle <^ with cd. Then ?„,„ = W sin (<A + a) (i-) Friction. 291 Horizontal Pull.— When the body is raised by a hori- zontal pull, we have (Fig. 283) — Pj = W tan (^ + a) (ii.) Fig. 283. Fig. 284. Thus, in all cases, the effect of friction is equivalent to making the slope steeper by , -» an amount equal to the friction I - angle. Parallel Pull. — When the '^ body is raised by a pull parallel ^ /?--- to the plane, we have (Fig. 284) — fig^s. V^^^ + db But ed = dc tan ^ = /idc and dc = W , cos a therefore^ = /t . W . cos a and d3 = W sin o hence P, = W(fj. . cos a -f sin a) , . This may be expressed thus (see Fig. 285) — (Hi) p, = w(.g-fg) or P,L = W//,B + WH or, Work done in dragging a body of weight W up a . plane, by a force / acting parallel to the plane / 'work done in dragging^ the body through the same distance on a horizontal plane against friction. /work done in lifting I the body 292 Mechanics applied to Engineering. General Case. — When the body is raised by a pull making an angle Q with the plane — P = P = of (iv.) COS (e - ^) Substituting the value P„,„. from equation (i.) — W sin («^ + a ) COS (6 - <^) When the line of action of P is towards the plane, as in Po, the B becomes minus, and we get — W sin (<^ + a) cos ( — ^ — <^) F,o. 286. All the above expressions may be obtained from this. When the direction of pull, Po, is parallel to ec, 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 6 < (90 — <j>) 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 <f>, the body must be dragged down, and if a be Po = Note. — The friction now assists the lowering, hence « is set off to the right olcti. Friction. 293 greater, a force must be applied to prevent it from being accelerated. Least pull when body is lowered, <^ < a (Fig. 287). Pmin. =. be = W sin'(a - <^) and 6 = <^ . . . (v.) When <^>a, be^ is the least force required to make the body slide down the plane. P,».„. = ■^i = W sin (<^ - a) . . . . (vi.) when <^ = a, P„|„, is of course zero. The remaining cases are arrived at in a similar manner ; we will therefore simply state them. *<«. 0>O. Least pull Parallel pull Horizontal pull (General case W sin (b — <))) W (sin — |U cos 0) W tan (a - ^) W sin (0 — <f) cos (9 + <p) \V sin (f — a) W (jit cos — sin a) W tan (((> - a) W sin (<^ - 0) cos (fl - <p) Note.— If the line of pull comes below the plane, the angle 9 takes the — sign. In the case of the parallel pull, it is worth noting that when t^ < 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. EflSciency of Inclined Planes. — If an inclined plane be used as a machine for raising or lowering weights, we have — ■ccc ■ useful work done (t.e. without friction) EtSciency = ; ; — ^-^^ — , . , ^ . . — r — i- actual work done (with friction) Inclined Plane when raising a Load. — The maximum efficiency occurs when the pull is least, i.e. when 6 = tji. The useful work done without friction is when = o ; then — 294 Mechanics applied to Engineering. The work done without friction = -^ — from (iv.) cos 6 „ „ with „ = LW sin (<^ + a) from (i.) where L = the distance through 'which the bpdy is dragged ; a = the inclination of the plane to the horizon ; B = the inclination of the force to the plane ; ^ = the friction angle. LW sin a (vii.) Then maximuml _ cos 6 _ sin a efficiency ) LW sin (<^ + a) cos B sin (^ + a) When the pull is horizontal, ^ = a, and — „^ . sin o tan a. / ■■■ \ Efficiency = — rr-r — ^ = - — jj—. — •■ (viii.) cos a . tan (<;* + a) tan (^ + a) when the pull is parallel, 6 = o, and cos 6 = i ; ■,71V- • sin a . cos <i /• \ Efficiency = ^ — -. — -— f .... (ix.) sin (a + (ji) General case, when the line of pull makes an angle 6 with the direction of sliding^ sin a . cos (0 — ^) / > o — ■„ / 1 I — \ • • • (^•) Efficiency = -^ — -■ — 7-7— i — ( ' cos 6 . sin (0 + a) 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 (<^ + a) for a single inclined plane; but here we have two inclined planes, each at an angle F,Q 'j5g, a, hence W moves twice as far for any given movement of P as in the single inclined plane ; hence — P = zW tan (<j> + 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. 293 we have the pull required to withdraw the wedge — - P = 2W tan (a - ,^) Friction. 295 The efficiency of the wedge is the same as that of the inclined plane, viz. — Efficiency = : — , ° , ^ when overcoming a resistance (xi.) reversed | ^ tan (a - ^) ^^^^ withdrawing from a resistance efficiency J tan a (^^^ p 335) 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. 290. Let/ = the pitch of the screw ; ^0 = the mean diameter of the threads ; W = the weight lifted. The useful work done per revolution! _ -^p _ -^^^ ^^^ on the nut without friction I " The force applied at the mean radius ofl ^ fP = Wtan(a+^) the thread required to raise nut I I (see equation ii.) The work done in turning the nut) _ p^^^ through one complete revolution f = 'Wirda tan (a + 4>) Wx^ o tan g Efficiency, when raising the weight, = YV^^^laiToSTT^) tan a sin a cos (a + 0) tan (a + </)) cos a sin (a + 4>) _ sinJ2a_+ « ^) - sin ^ ^ ^ 2 sin (^ '" sin"(2a"+ <^) + sin <^ sin(2a + ^) + sin <^ This has its maximum value when the fractional part is least, or when sin (2a + ^) = i. 296 Mechanics applied to Engineering. Since the sine of an angle cannot be greater than i, then 2a + c^ = 90 and u = 45 . Inserting this value, maximum efficiency = | ^ I = i — 2/i (nearly). 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 1 _ tan a . and thrust collar ] " tan (a + 2^) WP"""^-) In the case of a nut the radius at which the friction acts will be about i^ times that of the threads ; we may then say — efficiency of screw thread and nut ) _ tan o bedding on a flat surface j ~ tan (a + 2*5 A) 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 — reversed efficiency = ^ — (see p. 335) 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 Wn I W screw, in the ratio oi ~ = -, and Wq = — , where 6 is W p (7 cos - cos — 2 2 the angle of the thread. In the Whitworth thread the angle 6 is 55°, hence Wo = 1-13 W, In the Sellers thread 6 = 60° and W„ = i'i5 W J then, taking a mean value of Wo = i'i4 W, we have — efficiency = *^" " tan (a -|- I "141^) Friction. 297 In tlie case of an ordinary bolt and nut, the radius at which the friction acts between the nut and the washer is about i^ times that of the thread, and, taking the same coefficient of friction for both, we have — efficiency \ of a bolt = tan a and nut I tan (a + 2-6+.^) (approx.) The following table may be useful in showing roughly the efficiency of screws. In several cases they have been checked Fig. 291. 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. ElHciency per cent, when Ef&cieocy^er cent, allow- Angle of no friction between nut ing for friction between thread a. and washer or a thrust nut and washer or a collar. thrust collar. Sq. thread. V-thread. Sq. thread. V-thread. 2» 19 17 II 8 ->o 26 23 14 12 4° 32 28 17 16 5° 36 33 21 20 10° ss 52 36 29 20° 67 63- 48 42 45-1 79 75 52 44 In the above table <p has been taken as 8*5°, 01 fi = o'i5. For the efficiency of a worm and wheel see page 344. 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 — 298 Mechanics applied to Engineering. P . BA = W . AC W.AC orP = LetAC = K; Then P = BA W.K J B ///// 1 w But h is usually small compared with R ; hence we may write — ^'°- "^^ P = ^ (nearly) P and W, also K and R, must be measured in the same units, or the value of K corrected accordingly. The above treatment is very rough, but the relation holds fairly well in practice. There is much need for further research in this branch of friction. Values of K. Iron or steel wheels on iron or steel rails ... I. » 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 OOS-0'20 3-S 0"02-0'022 0'04-o-o6 0-04 o-og-o'ii 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 reverse direction to the mo- tion of the wheel, and so hinders its progress. Likewise, as the wheel leaves the heap behind it, the rubber returns to its original Fig. 293. Friction. 299 . 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 length of path rolled over is greater than the hori- zontal distance travelled by the wheel, hence it does not travel its geo- metrical distance ; the amount it falls short of it or the " slip " depends upon the hardness of the surfaces in contact. Even Fig. 294. with the balls in ball bear- ings the " slip " is quite appreciable. Antifriction Wheels. — In order to reduce the friction on an axle it is sometimes mounted on antifriction wheels, as shown. A is the axle in question, B and C are the anti- friction wheels. If W be the load on the axle, the load on each antifriction wheel bearing will be — W„=- W W -1, and the load on both ^ ^' cos d 2 cos Let Ra = the radius of the main axle ; R= „ „ antifriction wheel ; r — „ ,, axle of the antifriction wheel. The rolling resistance on the surface j _ WK of the wheels J "~ R^ cos d The frictional resistance referred toj ^ the surface of the antifriction wheels, V = ~- or the surface of the main axle ) ^ • ^°^ " W /K , u,A The total resistance = ^^1 ^ + -5- ^ cos ^ \R„ R / If the main axle were running in plain bearings, the resistance would be /aW ; hence — /<.R cos B frictio n with plain bearings friction with antifriction wheels k| + ^. 300 Mechanics applied to Engitieering. In some instances a single antifriction wheel i? used, the axle A being kept vertically over the axle of the wheel by means of guides. The main trouble with all such devices is that the axle travels in an endwise direction unless prevented by some form of thrust bearing. One British Railway Com- pany has had large numbers of waggons fitted with a single antifriction disc on each axle bearing ; the roUing resistance is materially less than when fitted with ordinary bearings, but the discs are liable to get " cross cornered," and to give trouble in other ways. Roller Bearings. — There are many forms of roller bear- ings in common use, but unfortunately few of them give really satisfactory results. The friction of even the worst of them is considerably lower than that of bearings provided with ordinary lubrication. If the rollers are not perfectly parallel in them- selves (in a cylindrical bearing), and are not kept absolutely parallel with the shaft, they tend to roll in a helical path, but since the cage and casing prevent them from doing so, they press the cage against the flange of the casing and set up what Lcfmibuduwl Sectiorv. If aW Cross Section, onljine A-B Fig. 295. I/aJf Cross Section- an C0rUreLvi£ is known as " end-thrust," which thereby gives rise to a large amount of friction between the end of the cage and the flange of the bearing, and in other ways disturbs the smoothness of running. Few, if any, roller bearings are entirely free from this defect ; it is moreover liable to vary greatly from time to time both as regards its amount and direction. In general, it is less at high speeds than at low, and it increases with the load on the bearing ; it does not appear to be greatly affected by lubrication. Bearings in which the end-thrust is high nearly always show a high coefficient of friction, and vice versa. High friction is always accompanied with a large amount of wear and vibration. In order tO- reduce wear and to ensure smooth Friction. 301 running, the rollers, sleeve, and liner should be of the hardest steel, very accurately ground and finished. In many of the cheaper forms of roller bearing no sleeve is used, hence the rollers are in direct contact with the shaft. The outer casing is usually split to allow of a bearing on a long line of shafting being replaced when necessary without removing the pulleys and couplings, or without taking the shafting down. This is an undoubted advantage which is not possessed by ball bearings. The reason why ball bearings will not run with split or jointed races will be obvious after reading the paragraphs devoted to ball bearings. The commonest form of roller bearing is that shown in Fig. 295. There is no sleeve on the shaft and no liner in the casing ; the steel rollers are kept in position by means of a gun-metal cage, which is split to allow of the rollers being readily removed. Another cheap form of roller bearing which is extensively used is the Hyatt, in which the rollers take the form of helical springs ; they are more flexible than solid rollers, and conse- quently accommodate them- selves to imperfections of align- ment arid workmanship. — In the Hoffmann short roller bearing. Fig. 296, the length of the rollers is equal to their diameter; the roller paths and rollers are of the hardest steel ground to a great degree of accuracy. The end-thrust is almost negligible and the co- efficient of friction is low ; the bearings will run under a con- siderably higher load than a ball bearing of similar dimensions. The following table gives fair average results for a friction test of an ordinary roller bearing : — Centre of Shaft 1 Fig. 296. Total load 40 revolutions per minute. 400 revolutions per minute. in lbs. V- End thrust in lbs. V- End thrust in lb:i. 2000 4000 6000 8000 10,000 o'oi3i o'oo94 0-0082 00076 0'0072 82 147 212 276 0-0053 0-0035 0-0029 0-0026 0-0024 51 89 128 166 205 303 Mechanics applied to Engineering. A test of a Hoffmann short roller bearing gave the following results : — Total load in lbs. C End thrust in lbs. Temperature air at 62° F. 2000 0-OOI2 None 4000 O'OOIO ■ — 6000 00008 7 76° F. 8000 O'OOIO 84 89° F. 10,000 o-ooii 2X0 96° F. 12,000 O'OOIO 270 94° F. 14,000 O'OOIO 20 95° F. 14,500 O'OOIO 95° F. Diameter of sleeve ... .. 4' 7 5 inches Diameter of rollers . . . .. I'ooinch Length of rollers ... .. I'ooinch Number of rollers . . . .. 14 Revolutions per minute .. 400 Hardened steel highly finished. For details of other types of roller bearings and results of tests, the reader should refer to a paper by the author on " Roller and Ball Bearings," Froceedings of the Institution of Civil Engineers, vol. clxxxix. 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, however, requires careful consideration. Various types of ball races are shown in diagrammatic form in Fig. 297. A is known as a four-point radial bearing, the outer cones screw into the casing, and thereby permit of adjust- ment as the bearing wears. B is a three-point bearing capable of similar adjustment. C is a three-point bearing ; the inner coned rings are screwed on to the shaft, and can be tightened up as desired. None of these forms of adjustment are satis- factory for heavy loads. A four-point thrust bearing is shown at F; the races are ground to an angle of 45° with the axle, but since the circumferential speed of the race at a is greater than at b, the circle aa on the ball tends to rotate at a higher speed than the circle bb, but since this cannot occur, grinding and scratching of the ball take place. In order to avoid this defect, races were made as shown at G; the circle aa was greater than bb in the ratio ~. By this means it was expected that a true rolling motion would occur, but the bearing was Friction. 303 not a success. The three-point bearing H was designed on similar lines, but a considerable amount of grinding of the balls took place. To return to the radial bearings. A two-point bearing is shown at D; the balls rolled between tfro plain cylindrical surfaces. A cage was usually provided for holding the balls in their relative position. In E the balls ran in grooved races. In these bearings a true rolling motion is secured, the balls do not grind or scratch, and the friction is considerably less than in A, B, C. Thrust bearings designed on similar lines are shown at I and J. A more detailed view of such bearings is shown in Fig. 298. The lower race is made with a spherical seat to allow it to swivel in case the shaft gets out of line with its housing, a very wise precaution which greatly increases the life of the bearing. When designing bearings for very heavy loads, the difficulty is often experienced of placing in one row a sufficient number 304 Mechanics applied to Engineering. of balls of the required diameter. In that case two or more rows or rings may be arranged concentrically, but it is almost impossible to get the workmanship sufificiently accurate, and to reduce the elastic strain on the housings to such an ex- tent as to evenly dis- tribute the load on both rings. The one set should therefore be backed with a sheet of linoleum or other soft material which will yield to a sufficient extent to equalize approximately the load on each ring. Such a bearing is shown in Fig. 299. The sheet-iron casing dips into an oil channel for the purpose of excluding dust. The lower half of the housing is made with a spherical seat. Fig. 258. Fig. 299. Modern cylindrical or radial bearings are almost always made of the two-point type; for special purposes plain cylindrical races may be used, but balls running in grooved races will safely carry much higher loads than when they run on plain cylinders. With plain raceS there is no difficulty in inserting the full number of balls in the bearing, but when grooved races are used, only about one-half the number can be inserted unless some special device be resorted to. But Friction. 305 since the load-carrying capacity of a bearing depends upon the number of balls it contains, it is evidently important to get the bearing as nearly filled as possible. After packing in and spacing as many balls as possible, the remaining balls are inserted through a transverse slot in one side of the race ; the depth of this slot is slightly less than that of the groove in which the balls run, hence it does not in any way affect the smoothness of running. See Fig. 300. When two or more rows of balls are used in a cylindrical bearing, each row must be provided with a separate ring. The inner ring or sleeve must be rigidly attached to the shaft, and the outer ring should be backed with linoleum in order Fio. 300. to evenly distribute the load on each row of balls. The housing itself should be provided' with a spherical seating to allow for any want of alignment. A design for such a bearing is shown in the Author's paper referred to above. A special form of bearing has been designed by the Hoffmann Co. with the same object. It is of great importance to attach the sleeve, or inner ring, of the bearing rigidly to the shaft. It is sometimes accom- plished by shrinking ; in that case the shrinkage must not be more , diameter of shaft ^r .1 ■ ^- ■ j j ■ than . If this proportion is exceeded the 2000 ring is liable to crack on cooling, or to expand to such an extent as to jamb the balls. Where shrinking is not resorted 3o6 Mechanics applied to Engineering. to the ring is sometimes made taper in the bore, and is tightened on to the shaft by means of a nut, or by a clamping sleeve, as shown in Fig. 300. In the Skefco ball bearing shown in Fig. 301 the outer ball race is ground to a spherical surface, and the sleeve is provided with grooved races to receive two rows of balls. By this arrangement the full number of balls can be packed in by tilting the inner race, but the great feature of the bearing is the spherical ball race which enables the bearing to be used on a shaft which does not run true, or on a machine in which the frame springs considerably relative to the shaft. Centr e of Sha ft Fig. 301. Fig. 302. Approach of the Balls Race when under Load. — When an elastic ball is placed between two elastic surfaces which are pressed together, the ball yields under the pressure and the surfaces become hollow ; the theory of the subject was first enunciated by Hertz, and afterwards by Heerwagen. The results obtained by the two theories are not identical, but there is no material diiference between them. Experimental re- searches on the strain which steel balls undergo when loaded show that the theories are trustworthy within narrow limits. The following expression is due to Hertz. Let 8 = the amount, in inches, the plates approach one another when loaded. P = the load on the ball in pounds. d = the diameter of the ball in inches. Friction. 307 Then ;V /pa 32,000 ■*■ d Distribution of the Load on the Balls of a Radial Bearing. — The load on the respective balls in a radial bearing may be arrived at by Stribeck's method. Thus — When the bearing is loaded the inner race approaches the outer race by an amount S„. The load on the ball a immediately in the line of loading is p^. The load on the adjacent balls b is less, because they are compressed a smaller amount than a, namely, 8j = 8„ cos aj. Similarly, 8„ = S^ cos a^. Let m be the number of balls in the bearing, then — 360° , 2 X ■?6o° a„ = •s and a„ = -^ m m 3 /pa Then 8„ = - 32,000 p 2 p 2 p s hence ^ = jT = Tl ~ ^^'^•' when d does not vary. P* = PA^ P„8jcosi(^°) 8«^ ' = P„ cos'^ \ m / and P„ = P, cos!' 2 The total load on the bearing W is — W = P„ + 2P, + 2P, + etc. = P.[. + 4c«.i(f) + c.,i<3^) + e,c.}] Exapiples — m 10 IS 20 360 36° 24° 18° W Pa 2'2S 3'44 4-58 4-38 4-37 4' 36 Thus P„ = 4-37W 3oS Mechanics applied to Engineering. This expression is only true when there is no initial " shake " or " bind " in the bearing and no distortion of the ball races. To allow for such deficiencies Stribeck proposes — P„ = 5^and W = ^ m 5 Necessity for Accuracy of Workmanship. — The expression for the approach of the plates shows how very small is the amount for ordinary working conditions. In a one-inch ball the approach is about j^ of an inch ; hence any combined errors, such as hills on the races or balls to the extent of j^ of an inch, will increase the load on the ball by 50 per cent. Extreme accuracy in finishing the races and balls is therefore absolutely essential for success. Friction of Ball Bearings. — The results of experiments tend to show that the friction of a ball bearing — (i) Varies 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 following results, which may be regarded as typical, lend support to the statements (i), (2), (3) : — Load, in lbs. Friction moment, inch-lbs. 1000 S-o 2000 6-0 3000 8-4 4000 12-8 Sooo lyo 5ooo 20'4 7000 25-2 8030 30-4 9000 36-0 10,000 40'o Speed, in revs, per min. (approx.)... Friction moment, inch-lbs. S 20'0 5° 19-8 100 19-7 500 19-8 800 20'I 1000 20'0 Temperature, Fahrenheit .. . Friction moment, inch-lbs. 58° 37-6 65° 38-1 77° 39-0 39"o 98° 38-7 The curves given in Fig. 303 were obtained by an auto- graphic recorder in the author's laboratory. They show clearly Friction. 309 lo'Ol L ^earing 001 1. Same loads in all cases. Mail Searings / 0. 1 .234 SevobtUotu ofSlia/t Fig. 303. 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. Although the friction of a ball bearing is not reduced by lubri- cation, yet a small amount of lubrication is necessary in order to prevent rust and corrosion of the balls and races. Thick grease resembling vaseline, which has been freed from all traces of cor- rosive agents, is used by many makers ; others find that the best lard oil is preferable, but in any case great care must be taken to get a lubricant which will not set up corrosion of the balls and races. Cost of Ball Bearings.— The cost of a ball bearing or a first-class roller bearing is considerably greater than that of an ordinary bearing, but owing to the fact that they are more compact, and that the mechanical efficiency of a machine fitted with ball bearings is much higher than when fitted with ordinary bearings, a considerable saving in metal may be efiected by their use ; with the result that the first cost of some machines, such as electric motors, is actually less when fitted with ball bearings than with ordinary bearings. The quantity of lubricant required by a ball bearing is practically nil, and they moreover require practically no attention. Provided ball bearings are suitably proportioned for the load and speed, and are intelli- gently fitted and used, they possess great advantages over other types of bearings. Safe Loads and Speeds for Ball Bearings. — The following expressions are based on the results of a large number of experiments by the author : — W = The maximum load which may be placed on a bear- ing in pounds. m = The number of balls in the bearing. d = The diameter of the balls in inches. N = The number of revolutions per minute made by the shaft. D = The diameter of the ball race, measured to the middle of the balls, in inches. 3IO Mechanics applied to Engineering. ND + C^ where K and C have the following values :— Type of bearing K C Cylindrical — no grooves 1 ,000,000 2000 Cylindrical — grooved races 2,000,000 to 2,500,000 2000 Thrust*— no grooves 500,000 200 Thrust — grooved races 1 ,000,000 1,250,000 200 Information on ball bearings can also be found in the fol- lowing publications : — Engineering, April 12, 1901 ; December 26, 1902; February 20, 1903. Proceedings of the Institution of Civil Engineers, vols. Ixxxix. and clxxxix. " Machinery " Handbooks—" Ball Bearings." 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, I. The frictional resistance is nearly proportional to the normal pressure between the two surfaces. 2. The frictional resistance is nearly independent of the speed for low pressures. For high pressures it tends to decrease as the speed increases. Lubricated Surf cues. 1. The frictional resistance is almost independent of the pressure with bath lubrication so long as the oil film is not ruptured, and approaches the behaviour of dry surfaces as the lubrication becomes meagre. 2. The frictional resistance of a flooded bearing, when the tempera- ture is artificially controlled, in- creases (except at very low speeds) nearly as the speed, but when the temperature is not controlled the friction does not appear to follow any definite law. It is high at low speeds of rubbing, decreases as the speed increases, reaches a minimum at a speed dependent upon the tem- perature and the intensity of pres- sure ; at higher speeds it appears to increase as the square root of the speed ; and finally, at speeds of over 3000 feet per minute, some believe that it remains constant. Friction. 311 3. The frictional 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. 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. 3. The frictional resistance de- pends largely upon the temperature of the bearing, 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 bear- ing 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. The following instances will serve to show the nature of the experimental evidence upon which the above laws are based. I. The frictional resistance is independent of the pressure with bath lubrication. 312 Mechanics applied to Engineering. Tower's Experiment [jubricant. Pressure in pounds per square inch. 153 205 310 415 S20 625 Frictional resistance in pounds. Olive oil ... Lard oil Sperm oil ... Mineral grease 0-89 0-90 0*64 0-87 0-82 0-84 _ 0-87 o-8o ! 0-86 090 0-87 0-87 0-57 o-SS 0-S9 — — 1-27 I -.35 1-24 I-I2 I-I4 1-25 The results shown in Fig. 304 were, obtained from the author's friction testing machine. In the case of the '"oil 4600 490P fZOO 1000 8O0 OOP fOV ZOO O LOfI DS IN POUNDS SO INCH Fig. 304, bath" the film was ruptured at a pressure of about 400 lbs. per square inch, after which the friction varied in the same manner as a poorly lubricated bearing. It is of interest to note that the friction of a dry bearing is actually less than that of a flooded bearing when the intensity of pressure is low. Friction. 313 2. The manner in which the friction of a flooded bearing varies with the velocity of rubbing is shown in Fig. 305. Curves A and B were obtained from a solid bush bearing such as a' lathe neck by Heiman {Zeitschrift des Vereines Deut- scher Ingenieure, Bd. 49, p. n6i). Curves C and D were 1 Ui/mf %'l A Hc,»*/,n *3 20 a HeiKAua 43 SO c Stmidcck 57 ss D STniaecK 213 25 E GOOOIHAH 75 40 , F GaaoMAH ISO 40 Uubbing Speeet: (Feet per TrUnuteJ. Fig. 305. obtained by Stribeck (Z des V. D. Ing., September 6, 7902) with double ring lubrication. Curves E and F were obtained by the Author with bath lubrication {Proceedings I. C. E. vol. clxxxix.). The erratic fashion in which the friction varies is due to many complex actions, which have not as yet been reduced to rigid mathematical treatment, although Osborne Reynolds, Sommerfeld, Petrofif, and others have done much excellent work in this direc- tion. An examination of the problem on the assumption that the friction is due to the shearing of a viscous film of oil of uniform thickness is of interest, although it does not give results entirely in accord with experiments. Fig. 306. 314 Mechanics applied to Engineering. In the theory of the shear of an elastic body we have the relation (see page 376) — / G AG where/, is the intensity of shear stress. F, is the total resistance to shear. A is the cross-sectional area of the element sub- jected to shear. G is the modulus of rigidity. But in the case of viscous fluids in which the resistance to flow varies directly as the speed S, we have — S_/. _ F. AKS 7=K-AK ^"'^ ^'^—r where K is the coefficient of viscosity. When a journal runs in a solid cylindrical bush of diameter d and length L with a film of oil of uniform thickness inter- posed, the friction of the journal is — If W be the load on the journal, and /x be the coefficient of friction, then _F, _x^LKS '* ~ W ~ W/ lip be the nominal intensity of pressure— W hence M=— = - From this relation we should expect that the coefficient of friction in an oil-borne brass would vary directly as the speed, as the coefficient of viscosity, also inversely as the intensity of pressure and as the thickness of the film. But owing to dis- turbing factors this relation is not found to hold in actual bearings. Osborne Reynolds and Sommerfeld have pointed out that the thickness of the oil film on the " on " side of a brass is greater than on the "off" side. The author has Friction. 3 1 g experimentally proved that this is the case by direct measure- ment, and indirectly by showing that the wear on the " off" side is greater than on the " on " side. The above-mentioned writers have also shown that the difference in the thickness on the two sides depends on the speed of rotation, the eccentricity being greatest at low speeds. Reynolds has shown that the friction increases with the eccentricity ; hence at low velocities the effect of the eccentricity is predominant, but as the speed increases it diminishes with a corresponding reduction in the friction until the minimum value is reached (see Fig. 305). After the minimum is passed the effect of the eccentricity becomes less important, and if the temperature of the oil film remained constant the friction would vary very nearly as the speed, but owing to the fact that more heat is generated in shearing the oil film at high speeds than at low the tempera- ture of the film increases, and thereby reduces the viscosity of ■the oil, also the friction with the result that it increases less rapidly than a direct ratio of the speed. Experiments show that when the temperature is not controlled the friction varies more nearly as the square root of the speed. This square root relation appears to hold between the minimum point and about 500 feet per minute ; above that speed it increases less rapidly than the square root, and when it exceeds about 3000 feet per minute the friction appears to remain constant at all speeds. For a flooded bearing in which the temperature is not controlled the friction appears to follow the law — between the above-mentioned limits. The fact must not be overlooked that the deviation from the straight line law of friction and speed after the minimum is passed is largely due to the fact that the temperature of the film does not remain constant. In some tests made by the author in 1885 {Proceedings Inst. C. Engineers, vol. Ixxxix., page 449) the friction was found to vary directly as the speed when the temperature was controlled by circulating cooling water through the shaft. The speeds varied from 4 to 200 feet per minute with both bath and pad lubrication, and with brasses embracing arcs from 180° to 30°. Tests of a similar character made on white metals on a large testing machine also showed the straight line law to hold between about 15 and 1000 feet per minute with both bath and pad lubrication. 3i6 Mechanics applied to Engineering. Where the temperature of the bearing is controlled, the friction-speed relation appears to closely agree with that de- duced above from viscosity considerations, viz. — <rS 3. The curves given in Fig. 307 show the relation between 8» 5 50 (3 u i" 30 • Specimen A. Temp&rautLu-a conjtroU£d.a£ 120°F. C Specimen A. Tempsralure ctSaiuettto vafi/. + Specimen fl. Temperatur-e controUeoLat t20'*F. 4 Speoimen B. TemperaUi-re aJiojuedt to vary . SCfFt -L. 2.000 4.000 6,000 a,ooo fO,00O /z,ooo /C,OOCf Lo(z<3^ 07V Meexj^iTxg : Pounds ■ Fig. 307. the friction and the temperature. When other conditions are kept constant the relation between the coefficient of friction Friction. 317 ju. and the temperature t may be represented approximately by the empirical expression — constant where /a, is the coefficient of friction at the temperature t F. Thus, if the coefficient at 60° F. is o'oi76 the constant is 0-332, and the coefficient at 120° F. would be 0-005 r. Tower showed that the relation constant ^' = — T~ closely held for many of his tests. When making comparative friction tests of bearing metals it is of great importance to control the temperature of the bearing at a predetermined point for all the metals under test. In all friction testing machines provision should be made for circulating water through the shaft or the bearing for coii • trolling the temperature. The curves in Fig. 307- shows the effect of controlling the TemperaUtre in. Degrees Fakrenheit ft) Fig. 3070:. temperature when testing white antifriction metals, also of allowing it to vary as the test proceeds. 4. Mr. Tower and others have shown that in the case of a 3i8 Mechanics applied to Engineering. 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 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 lbs. sq. inch Coefficient of (A friction tB 150 0-0143 0-0083 250 0-OII2 0-0062 3S0 0-0091 0-0054 450 0-0082 o;ooso 750 0-0075 0-0045 950 0-0083 0-0047 5. The following tests by Thurston will show how much greater is the friction of rest than of motion : — Load in lbs. sq. inch Coefficient of j A^°inst^t"'\ fr"="°" I of starting/ SO 0-013 100 o-oo8 250 0-005 500 0-004 750 0-0043 0-07 0-I3S 0-14 0-15 0-185 1000 0-009 0-18 Oil used, sperm. The ratio between the starting and the running coefficients depends largely upon the viscosity of the oil, as shown by the following tests by the author. See Proceedings Inst. C. E., vol. Ixxxix. p. 433. Coefficient of friction. Running. At starting. Machinery oil Thick valve oil |g Grease 0-0084 0-0329 0-0252 0-0350 0-192 O-171 0-147 0-090 22-9 S-2 S-8 2-6 Friction. 319 6. Experiments by Tower and others show that a steel shaft in a gun-metal bearing seizes at about 600 lbs. square inch under steady running, whereas when dry the same materials ft 9 No load. § 1^ IdOCt<i' SOO lh& so inch' s ■> ■/' Time , 6 M, Second « / inAih, Fin. 308. seize at about 80 lbs.' square inch. The author finds that the seizing pressure increases as the viscosity of the oil increases. 7. Fig. 308 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 fig. 309. 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 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; when the lubrication 320 Mechanics applied to Engineering. is very meagre or with a dry bearing it is a maximum at the crown, and is least at the two sides. When the bearing is flooded with oil the distribution of pressure can be calculated from hydro- dynamical principles, an account of which will be found in Dr. Nicolson's paper, " Friction and Lubrication," read before the Manchester Associa- tion of Engineers, November, 1907. Fig. 310. For the purpose of comparing roughly the intensity of pressure on two bearings, the pressure is assumed to be evenly distri- buted 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, and the nominal pressure W per square inch is — „ where W is the total load on the bearing. Beauchamp Tower's Experiments. — These experi- ments were carried out for a research committee of the Centre Fig. 311. 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. 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 Friction. 321 varied as shown in Fig. 3CI, the pressure being greatest on the "off" side. In this connection Mr. Tower shows that it is useless — worse than useless — to drill an oil-hole on the resultant line of pressure of a bearing, for not only is it irnpossible 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. 312), 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 rio pressure (Fig. 313). The position of the point of minimum pressure depends somewhat on the speed of rotation. Fig. 312. Fig. 313. 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. o Load at which seizing occur- red, in lbs. sq. inch. Coefficient of friction 150 370 0'Oo6 55° 0'Oo6 600 90 0-035 Other of Mr. Tower's experiments are referred to in preceding and succeeding paragraphs. Professor Osborne Reynolds' Investigations, — A theoretical treatment of the friction of a flooded bearing has been investigated by Professor Osborne Reynolds, a full Y 322 Mechanics applied to Engineering. account of which will be found in the Philosophical Transac- tions, Part I, 1886; see also his published papers, Vol. II. p. 228. 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 and Sommerfeld have also done very similar work. A convenient summary of the work done by the above-mentioned writers- will be found in Archbutt and Deeley's " Lubrication and Lubricants." The theory of the distribution of oil pressure in a flooded collar bearing has been recently investigated by Michell, who has successfully applied it to the more difficult problem of producing an oil-borne thrust bearing. See a paper by Newbigin in the Proceedings of the Institution of Civil Engineers, Session 1913-1914; also Zeitschrift fiir Mathematik und Physik, Vol. 52, 1905. 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 oflf, 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 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 Fig. 314. Friction. 323 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 always begins to fuse on the " off " side first. 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° (with white metal bearings this angle is nearer 60°) the wear is on the off side ; if less than 90°, on the "on" side. This wear was measured thus : The four screws, a, a, a, a, were fitted to an overhanging lip - J?iarrL o^ Fig. 315. O-ZS O'J 0-7J t-O Cfwrcts irv cemtact Fig. 316. on the bearing as shown. They were composed of soft brass. Before commencing a run, they were all tightened up to just touch the shaft; on removing the bearing after some weeks' running, it was seen at once which screws had been bearing and which were free. Another set of experiments were made in 1885, to ascertam the effect of cutting away the sides of a bearing. The bearings experimented upon were semicircular to begin with, and the sides were afterwards cut away step by step till the width of the bearing was-only i d. The effect of removing the sides is shown in Fig. 316. 324 Mechanics applied to Engineering. The relation may be expressed by the following empirical formula : — Let R = frictional resistance ; _ width of chord in contact diameter of journal K and N are constants for any given bearing. Then — R = K + NC Methods of Lubricating. — In some instances a small force-pump is used to force the oil into the bearing ; it then becomes equivalent to bath lubrication. Many high-speed Fig. 318.— Collar bearing. Fig. 319.— Pivot or footstep. engines and turbines are now lubricated in this way. The oil is forced into every bearing, and the surplus runs back into Friction. 325 a receiver, where it is filtered and cooled. When forced lubrication is adopted with solid bushes, or in bearings in which the load constantly changes in direction, as in connect- ing rods, the clearance must be considerably greater than when meagre lubrication is supplied. The clearances usually adopted are about one thousandth of the diameter of the shaft for ordinary lubrication, and rather more for flooded bearings. Relaiive Friction of Different Systems of Luhrication. Mode of lubrication. Tower. Goodman. Bath Saturated pad Ordinary pad Syphon I '00 6'48 7-06 I '00 1-32 2-21 4'20 Seizing of Bearings. — It is well known that when a bearing is excessively loaded, the lubricant is squeezed out, and the friction takes place between metal and metal ; the two surfaces then appear to weld themselves together, and, if the bearing be forced round, small pieces are torn out of both surfaces. The load at which this occurs depends much upon the initial smoothness of the surfaces and upon the nature of the material, but chiefly upon the viscosity of the oil. If only the Fig. 320. viscosity can be kept up by artificially keeping the bearing cool, by water-circulation or otherwise, the surfaces will not seize until the pressure becomes enormous. The author has had a bearing running for weeks under a load ol two tons per square inch at a surface velocity of 230 feet per minute with pad lubrication, temperature being artificially kept at 110° Fahr. by circulating water through the axle.' Seizing not unfrequently occurs through unintentional high pressures on the edges of bearings. A very small amount of ' In another instance nearly four tons per square inch for several hours. 326 Mechanics applied to Engineering. spring will cause a shaft to bear on practically the edge of the bearing (Fig. 320), and thereby to set up a very intense pressure. This can be readily avoided by using spherical-seated bearings. For several examples of such bearings the reader is referred to books on " Machine Design." Seizing is very rare indeed with soft white metal bearings ; this is probably due to the metal flowing and adjusting itself when any un- even pressure conies upon it. This flowing action is seen clearly in Fig. 321, which is from photo- graphs. The lower portion shows the bearing before it was tested in a friction-testing machine, and the upper portion after it was tested. The metal began to flow at a temperature of 370° Fahr., under a pressure of 2000 lbs. per square inch : surface speed, 2094 feet per minute. The conditions under which the oil film ruptures in a flooded bearing will be discussed shortly. Bearing Metals. — If a bearing can be kept completely oil-borne, as in Tower's oil-bath experiments, the quality of the bearing metal is of very little importance, because the shaft is not in contact with the bear- ing; but unfortunately, such ideal conditions can rarely be ensured, hence the nature of the bearing metal is one of great importance, and the designer must very carefully consider the conditions under which the bearing will work before deciding upon what metal he will use in any given case. Before going further, it will be well to point out that in the case of bearings running at a moderate speed under moderate loads, practically any material will answer perfectly ; but these are not the cases that cause anxiety and give trouble to all concerned : the really troublesome bearings are those that have to run under extremely heavy loads or at very high speeds, and perhaps both. The first question to be considered is whether the bearing Fig. 32X. Friction. 327 will be subjected to blows or not; if so, a hard tough metal must be used, but if not, a soft white metal will give far better frictional and wearing results than a harder metal. It would be extremely foolish to put such a metal into the connecting-rod ends of a gas or oil engine, unless it was thoroughly encased to prevent spreading, but for a steadily running journal nothing could be better. The following is believed to be a fair statement of the relative advantages and disadvantages of soft white metal for bearings : — • Soft White Metals for Bearings. Advantages. Disadvantages. The friction is much lower than Will not stand the hammering with hard bronzes, cast-iron, etc., action that some shafts are sub- hence it is less liable to heat. jected to. The wear is very small indeed The wear is very rapid at first after the bearing has once got well if the shaft is at all rough ; the bedded (see disadvantages). action resembles that of a new file on lead. At first the file cuts rapidly, but it soon clogs, and then ceases to act as a file. It rarely scores the shaft, even if It is liable to melt out if the the bearing heats. bearing runs hot. It absorbs any grit that may get If made of unsuitable material into the bearing, instead of allowing it is liable to corrode, it to churn round and round, and so cause damage. As far as the author's tests go, amounting to over one hundred different metals on a 6-inch axle up to loads of 10 tons, and speeds up to 1500 revolutions per minute, he finds that ordinary commercial lead gives excellent results under moderate pressure : the friction is lower than that of any other metal he has tested, and, provided the pressure does not greatly exceed 300 lbs. per square inch, the wear is not excessive. A series of tests made by the author for the purpose of ascertaining the effect of adding to antifriction alloys small' quantities of metals whose atomic volume differed from that of the bulk, yielded very interesting results. The bulk metal under test consisted of lead, 80; antimony, 15 ; tin, 5 ; and the added metal, 0*25. With the exception of one or two metals, which for other reasons gave anomalous results, it was found that the addition of a metal whose atomic volume was greater than that of the bulk caused a diminution in the friction, whereas the addition of a metal whose atomic volume was less than that of the bulk caused an increase in the friction, and 328 Mechanics applied to Engineering. metals of the same atomic volume had apparently no effect on the friction. All white metals are improved if thoroughly cleaned by stirrjng in sal ammoniac and plumbago when in a molten state. Area of Bearing Surfaces. — From our remarks on seizing it will be evident that the safe working pressure for revolving bearings largely depends upon their temperature and the lubricant that is used. If the temperature rise abnormally, the viscosity of the oil is so reduced that it gets squeezed out. The temperature that a bearing attains to depends (i) on the heat generated; (2) on the means for conducting away the heat. Let S = surface speed in feet per minute ; W = load on the bearing in pounds ; /„ = number of thermal units conducted away per square inch of bearing per minute in order to keep the temperature down to the desired limit. u,WS The thermal units generated per minute = - — 773 The nominal area of bearing surfacel _ A^WS in square inches, viz. dh J "~ 7734 As a first approximation the following values of /a, and 4 may be assumed : — VaLDES of /i AND tu. Method of lubrication. Value of i^ Bath o'004 Pad 0'0i2 Syphon 0020 Values of tu. Conditions of running. Crank and Continuous Crank and Continuous other pins. running bearing. other pins. running bearing. Maximum temperature of bearing. 140° F. 140° F. 100° F. 100° F. Exposed to currents of cold air or other means of cooling, as in locomotive or car axles 4-7 «-is 2-3 -S 0-5-0-75 In tolerably cool places, as in marine and stationary en- gines 0-75-I 0-3-0-5 0-4-O-S 0-15-0-25 In hot places and where heat is not readily conducted away 0-4-o-S 0-I-0-3 0-2-0-25 ~ Friction. 329 After arriving at the area by the method given above, it should be checked to see that the pressure is not excessive. Bearing. Crank-pins. — Locomotive ... Marine and stationary Shearing machines Gudgeon pins. — Locomotive Marine and stationary Railway car axles Ordinary pedestals, — Gun-metal Good white metal Collar and thrust bearings, — Gun-metal Good white metal Lignum vilse ... Slide blocks. — Cast-iron or gun-metal Good white metal Chain and rope pulleys for cranes, — Gun-metal bush aximum permis- sible pressure in 'bs. per sq. inch. 1500 600 3000 ZOOO 800 200 500 80 200 SO 80 250 1000 Work absorbed in Revolving Bearings. Let W = total load on bearing in pounds ; D = diameter of bearing in inches ; N = number of revolutions per minute ; L = length of journal in inches. For Cylindrical Bearings. — Work done per minute"! _ /uWttDN in foot-pounds j ~ horse-power absorbed = 12 WttDN/* _ /itWDN 12 X 33)000 126,000 A convenient rough-and-ready estimate of the work absorbed by a bearing can be made by assuming that the frictional resistance F on the surface of a bearing is 3 lbs. per square inch for ordinary lubrication, 2 lbs. for pad, r lb. for bath, the surface being reckoned on the nominal area. Work done in overcoming the friction \ _ ttD^LFN per minute in foot-pounds / "~ i^ Flat Pivot. — If the thrust be evenly distributed over the whole surface, the intensity of pressure is — ' 7rR» 330 Mechanics applied to Engineering. pressure on an elementary ring = 2irrp . dr moment of friction on an elementary ring = iirr^iip . dr moment of friction on whole surface = 2irfjipfr' . dr 3 Substituting the value of/ from above — M, =|/x,WR work done per minute in foot-pounds = 5 73 ftWDN 189,000 horse-power absorbed: This result might have been arrived at thus : Assuming the load evenly distributed, the triangle (Fig. 323) 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 f 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 Fig. 322. value for M,; 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. 325).— By similar reasoning to that given above, we get — Fig. 323. Moment of friction 1 _ on collar J ~ "^^^ M _ 2/^W(R,^ - R33) 3(Ri" - R2=) jr=R, dr Conical Pivot. — ^The intensity of pressure p all over the surface is the same, whatever may be the angle a. Let Po be the pressure acting on one half of the cone — VV a sm a Friction. 331 The area of half the surface of the cone is — ■rRL_ ttR' 2 2 sin a A = : . _ Po _ W . 2 sin a _ W _ weight A 2sina.irR^ jtR^ projected area Fig. 324. Fig. : Total normal pressure on any elementary ring = zirrp . dl moment of friction on elementary ring = zttz-V/ ■ dl /, ,, dr \ iirr^apdr \ sin a) sm a Sirixp i moment of friction on whole surface = • fr^ . dr sm a M,= 27r/t/R° 3 sin a Substituting the value of/, we have — 2/iWR M,= J sin o The angle a becomes 90°, and sin » = i when the pivot becomes flat. By similar reasoning, we get for a truncated conical pivot (Fig. 326)— 2/.W(R,^ - R,°) *^^'- 3 sin a(R,2 - R,') 332 Mechanics applied to Engineering. Schiele's Pivot and Onion Bearing (Figs. 327, 328). — 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 «i, 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, and ad to the full-lined curve at a ; then, if h be taken as very small, ba-^ will be practically parallel to ad, and the two triangles aba-^ and acd will be practically similar, and — Fig. 326. ad aa-, , ac Y. aa, — =— 2, 01 ad = -' ac ba ba or ad = ba But ba is the wear of the footstep nornial 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 00 FV 00 f>.p . 27rrN or ba = Vi-iJ-pr where K is a constant for any given speed and rate of wear ; hence — ad : K/u//- K/t/ But h is constant by hypothesis, and /* is assumed to be constant all over the pivot; / we_have already proved to be constant (last paragraph) ; hence ad, 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 = the constant, the wear will be (nearly) even all over the pivot. Although our Friction. 333 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. 327. Fig. 328. The normal pressure at every point is — weight _ W projected area 7r(Ri'' — R^) By similar reasoning to that given for the conical pivot, we have — Moment of friction on an elementary^ _ 2-irr'ii.pdr ring of radius r ) iuTo" (but -J^ = /) = 2TTtu.prdr \ sm a / '^ and moment of friction for whole pivot = 2Ttt)x.p r .dr Mf = 2-jr ffip ^1' - R2' Substituting the value of/, M, = Wju/" 334 Mecltanics 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 liie 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 tiie 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/ -I — -j- when in good condition F = 0-08/ + ^ „ 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„ through a height /«„ ; the work done in overcoming friction be equivalent to raising a weight W, through a height A^. Then, if there were no friction — Supply of energy = useful work done W/4 = W„/5„ or mechanical advantage = velocity ratio When there is friction, we have — Supply of energy = usefiil work done -f- work wasted in friction W/t = WA + W/, Friction. 335 and — , L ■ 1 a: ■ useful work done the mechanical efticiency = — — , 7—= total work done the work sot out or = — i ;-e — the work put in Let t] = the mechanical eflficiency ; then — „ _ WA ^^ WA ' -wh ' WA + w/, Tj 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 — _ A cc • useful work done in lifting W through h Reversed efficiency = — = =-^i r-- . °„, , °, , total work done in lowering W„ through h„ W/^ _ WA-W/^ ''' W„/5„ WA When W acts in the same direction as W„, i.e. when the machine has to be assisted to lower its load, i;, takes the negative sign. In an experiment with a two-sheaved pulley block, the pull on the rope was 170 lbs. when lifting a weight J. of 500 lbs.; the velocity ratio in this case R = -^ = ^. TV,»„ - ^"'''« - 5°° X I _„.-,. 1 hen 17 = ,^r- = =07^5 ' W/4 T70 X 4 •^ The friction work in this case y^/h, was 170 x 4 — 500 X 1 = 180 foot-lbs. Hence the reversed efficiency w, = 500 = o'64, and in order to lower the 500 lbs. weight gently, the backward pull on the rope must be — -—- X 0*64 = 80 lbs. 4 If the 80 lbs. had been found by experiments, the reversed efficiency would have been found thus — 80 X 4 ./- m, = 3l = 0-64 500 X I 336 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/i, = W„^„ ; then, substituting this value in the efficiency expression for forward motion, we have — Thus, in order that a machine which is not fitted with a non-return mechanism may be self-supporting its etficiency cannot be over 50 per cent. This statement is not strictly accurate, because the frictional resistance varies somewhat with the forces transmitted, and consequently is smaller when lowering than when raising the load ; the error is, however, rarely taken into account in practical considerations of efficiency. 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 jj„ j/j, %, etc., the efficiency of the whole machine will be — >SJ^ 17 = iji X r;a X %, etc. If the power be transmitted through n mechan- isms of the same kind, each having an efficiency iji, the efficiency of the whole series will be approxi- mately — ■ p 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. Fra. 329. 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. W Friction. 337 size, and newness of rope ; the second with the lubrication. The following table gives a fairly good idea of the total eflficiency at or near full load of single pulleys j it includes both resistances i and 2 : — Diameter of rope ,, . f Clean and well oiled Maximum U; efficiency! Clean and well oiled, 1 per cent, y ^j^j^ gjiij- ne,,. jopg ] i in. 96 94 Jin. 93 9> 91 91 89 i\ in.r 88 86 chain. 95-97 93-96 These figures are fair averages of a large number of experiments. The diameter of the pulley varied from 8 to 1 2 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 ,^\\\\^^\^^^^^^^^\\^^Kx\\^^^^^^^ efficiency of the whole maybe calculatedapproxi- mately from the known efficiency of the single pulley. The efficiency of a single pulley does not vary greatly with the load uqless 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 1J1, we have the efficiency of the whole — The following table will serve to show how the efficiency varies in different pulley blocks. 338 Mechanics applied to Engineering. Single pulley. Two-sheaved. Three-sheaved. pounds. Old l-in. New J-in. Old i-in. New i-in. Old }-in. New J-in. rope. rope. rope. rope. rope. rope. «4 94 90 _ _ 28 94-S 90-5 80 75 30 24 56 95 91 84 78-S 5° 35 112 96 92 86 91-5 60 41 i6g 87-S 93 65 44 224 — — 89 93 69 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 — 2 2 w -(R - ^) = PR 2 and the velocity ratio- W P v, = — = 2R R- 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. 339 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„ ~r = o'97 X o'97 x — R ox r = o'94R 2R and the velocity ratio = R - 0-94R = 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. Morris High-efficiency Self-sustaining Pulley Block. — In pulley blocks of the Weston type the efficiency rarely exceeds 45 per cent., but in geared self-sustaining blocks it may reach nearly 90 per cent. The self-sustaining mechanism is shown in Fig. 332. When hoisting the load the sprocket wheel A together with Ratchet \ Back *f4s^e/-l__J Brake | Driver in hoisting Drivers irfien hoisting. Fig. 33J. By kind permission of Messrs. Herbert Morris, Ltd.; Lougliborough. the nut N are rotated by means of an endless hand chain running in a clockwise sense of rotation. The nut traverses the quick running thread until the leather brake ring presses on the face of the ratchet wheel, the friction between these surfaces^ also between the back of the ratchet wheel and the back washer, becomes sufficiently great to lock them altogether. The back washer is keyed to the pinion shaft, the pinion gears 340 Mechanics applied to Engineering. into a toothed wheel provided with a pocketed groove for the lifting chain. Let P,, = the pull on the hand-chain. D = the diameter of the hand-chain sprocket wheel A. d„ = the mean diameter of the screw thread (see page 295). _ P = the circumferential force acting at the mean diameter of the screw thread when lifting. W = the axial pressure exerted by the screw when lifting. e = the mechanical efficiency of the gear from the lifting hook to the brake, a = the angle of the screw thread. <^ = the friction angle for the threads and hut which is always less than a. /ij = the coefficient of friction between the brake ring and the ratchet wheel. /i,» = ditto, back washer and ratchet wheel. Di = mean diameter of brake ring. D,„ = mean diameter of back washer bearing surface. Then P = ?i^ = W tan (a + <^) . . . . (i.) (see page 295) ^^= W^. tan (a -J- <^) (ii.) The axial pressure must be at least sufficient to produce enough friction on the brake ring and on the back washer to prevent the load on the hook from running down when the hand chain is released. Hence P D W^iisDj -f j«.,„D,„) must be greater than -^^— (iii.) In order to provide a margin of safety against the load running back, the friction on the back washer may be neg- lected ; then from (ii.) and (iii.) W^i.D, = Wrf„ tan (a -f <^) Dj tan (a -f- <^) ,. ^ and —r — * (iv.) Friction. 341 When these conditions are fulfilled the brake automatically locks on releasing the hand chain. The overall mechanical efficiency of the pulley block can be calculated from the mechanical efficiency of the toothed gearing and the friction of the chain in the pocketed grooves. When lowering the load the hand-chain wheel revolves in the opposite direction, thus tending to relieve the pressure of the brake. At the same time the sleeve on which the hand wheel is mounted bears against the washer and nut at the end of the pinion shaft, so that a drive in the lowering direction can be obtained through the gears. 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 + Wa; The value of K depends upon the type of machine under consideration, and the value of si upon the velocity ratio v^ of the machine. From Fig. 333 it will be seen how largely the efficiency is dependent upon the value of K. The broken and 342 Mechanics applied to Engineering. the full-line efficiency curves are for the same machine, with a large and a small initial resistance. The mechanical efficiency i; = VV Yvr (K + Wx)v, Thus we see that the efficiency increases as the load W K. increases. Under very heavy loads ^^ may become negligible ; hence the efficiency may approach, but can never exceed — Vmar The following experiments : — values give results agreeing well with X K n Rope pulley blocks Chain blocks of the\ Weston type / Self-sustaining geared\ blocks J I + 0-052/, ■Vr 1 + O-Wr V, I + oo09», Vr 2v,dVas. 3 lbs. i-S lbs. W W{i+o-osz',.) + Kz', W W(l+0-IZ/r) + K»r W W(i + ooogz/r) + Kz-, d = diam. of rope in inches. 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 100 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. 165), also the fi, 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, 343 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 — t\ = o'g6 — for rough unfinished teeth — 1/ = o'go — 2-5N 2-5N 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, = 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 ...J Efficiency 10 0*940 5° o"972 100 o'gSo 150 o'9S4 200 0-986 Screw and Worm Gearing.— We have already shown Fig. 334- 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 : — 344 Mechanics applied to Engineering. Velocity of pitch line in feet per) minute .... ) 10 50 100 ISO 200 Efficiency per cent. Angle of thread o, 45° 87 94 95 96 97 30° 82 90 93 94 95 20° 7S 86 90 92 92 ■5° 70 82 87 89 90 10° 62 76 82 «S 86 7° S3 69 76 80 81 s° 4S 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 books on 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 lbs. 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 j L = length of the shafting in feet ; F = the friction horse-power of the shafting. Then F = NLD« 80,000 The horse-power that can be transmitted by a shaft is— H.P. _ to — - (see p. 580) according to the working stress. Friction. 345 )f line shafting on which the« _ horse-power transmitted - friction horse-power Hence the efficiency of line shafting on which there are numerous pulleys is — horse-power transmitted 1\ = 64 ' LND=» 80,000 ND^ = I — 64 L r for a 1250 and I — L 2000 r — L 2960 for a working stress of 5000 lbs. sq. inch ,, 8000 i> i> 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. 349), 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 lbs. The load on the countershaft bearings may be taken at about the same amount or, say, a load on the bearings of 1000 lbs. in all. Let the diameter of the shafting be 3 inches J the S feet length will weigh about 120 lbs., 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. 346 Mechanics applied to Engineering. Consider first the case of a shaft of the same diameter 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 ; /A = the coefficient of friction ; ■q = the efficiency of transmission ; / = the torsional, skin stress on the shaft per square inch; Weight of the) „. r,2T ik * shaft f=W,rR=Llbs. moment of the 1 ., t,3t • u iu friction |=/^W,rR3Lmch-lbs. the maximum] twisting mo- , ^3 ment at the )=-'-i inch-lbs. (see p. 576) motoi end I ^ of the shaft I ^^^ f ??"^"*^^ ) the effective twisting moment at the far end of the trans- \ = — -r — r-r—. 2 — mission n I *^ twisting moment at the motor end _ maximum twisting moment — friction moment maximum twisting moment _ _ friction moment maximum twisting moment /aWttR^L _ 2j^WL " ' ~ /3rR3 * - /. 2 For a hollow shaft in which the inner radius is - of the outer, this becomes — 2«WL/ «" \ Now consider •^he 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, ; Friction. 347 Let the useful twisting moment at the far end of the shaft = Tj. Then the increase of twisting moment dt due to the friction on an elemental length dl = fjiW-n-R^d/ = dt. For the twisting moment / we may substitute — '=-^(seep. 576) or 7rR3 = ?;? J* by substitution, we get — dt= ^Jt^^^ and^^=?^^ i /. Integrating — where e =■ 372, the base of the system of natural logarithms. The efficiency ,' = J" =.^ _ 2>iWL and for a hollow shaft, such as a series of drawn tubes, which are reduced in size at convenient intervals — _ 2) jWLg' The following table shows the distance L to which power may be transmitted with an efficiency of 80 per cent. For or(Unary 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 348 Mechanics applied to Engineering. take a high value, viz. o'ooz. Let the skin stress y^ on the shaft be 8000 lbs. sq. inch, and let « = i"25. Form of shafting Parallel. Taper. Kind of bearings Ordinary. Ball. Oidinary, Ball Solid shaft mth belts ' ,, „ without belts Hollo* „ Feet. 400 6000 9840 Feet. 120,000 197,000 Feet. 76,600 J2S.SSO 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 Driver B n=/ Fig. 335. 71=3 -fV=^ 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 ' A part from the loss in bending rh? belts to and fro as they pass over the pulleys. Friction. 349 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 rj, the efficiency of the whole will be, approximately — % = ij" 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 zi foot-lbs. per square foot of belt passed over the pulleys. Mechanical EfSciency 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 3S~47 P^f cent. Piston and rod 21-33 Ciank-pin 5-7 Cross-head and gudgeon-pin 4-5 Valve and rod ... 2°5 balanced, 22 unbalanced Eccentric strap 4-S Link and eccentric 9 The following instances may be of interest in illustrating the approximate constancy of the friction at all powers : — Experimental Engine, Univeestty College, Lot Syphon Lubrication. London. LH.P B.H.P Friction H.P. ... 2-75 O'O 27s 9 "25 5-63 3-62 1023 7-50 273 11-14. 7-66 3-48 12-34 ^•09 3-25 13-95 II '09 2-86 14-29 11-25 3-04 Experimental Engine, The University, Leeds. Syphon and Pad Lubrication. LH.P. B.H.P. Friction H.P. 2-48 00 248 S-i6 2-35 2-81 6-83 3-94 2-89 8-30 5-61 2-69 11-50 8-70 2-8o 13-84 1082 302 17-02 13-89 3-«3 22-30 19-09 3-21 3S0 Mechanics applied to Engineering. Belliss Engine, Bath (Forced) Lubrication. (See Proc. J.M.E., 1897.) I.H.P. B.H.P. Friction H.P. 49-8 102 '7 147 I 193-6 44'S 97-0 140*6 i860 5-3 S-7 6'S 7-6 217-5 209-5 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. Under the best conditions it may be as low as i lb. per square inch {%e& Engineer, May 30, 1913, p. 574). In ordinary steam engines in good condition the friction pressure amounts to 2J to 3^ lbs. square inch, but in certain bad cases it may amount to 5 lbs. square inch. It has about the same value in gas and oil engines per stroke, or say from 10 to 14 lbs. 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 lbs. square inch, and the friction pressure 3 lbs. square inch, the mechanical efficiency of the engine would be = 94 per cent, if double-acting, and - — '^— 5° S° = 88 per cent, if single-acting. The mean effective pressure in a gas-engine cylinder seldom exceeds 75 lbs. 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 (J ton) „ „ „ (larger sizes) Epicycloidal pulley block ... Morris One-ton steam hoists or windlasses Hydraulic windlass „ jack Cranes (steam) Travelling overhead cranes 30-40 40-47 40-45 75-85 50-70 60-80 80-90 60-70 30-50 Friction. 35 1 T .. draw bar H.P. ,^ „^ Locomotives o5~75 1. ri.r. Two-ton testing-machine, worm and wheel, screw and nut, slide, two collars ... ... ... 2-3 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, 18 sq. inches, two screws and nuts 2-3 (About 1000 H.P. engines, spur-gearing, and engine friction 74 Rope drives 70 Belt , 71 Direct (400-H.P. engines) ... 76 Belts. Coil Friction. — Let the pulley in Fig. 336 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 Tg must be greater than T, 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, - T, = 2/./ Let the angle a embraced by the belt be divided into a a great number, say «, 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 circulav measure, we have — / = T? • n The friction at any point is (neglecting the stiffness of the belt)— ft* = mT - = Tj' - T,' But we may write - as 8a : also Tj - T,' as ST. Then— fiT . 8a = 8T 352 Mechanics applied to Engineering. which in the limit becomes — /iT . da = </r — = li.da We now require the sum of all these small tensions ex- pressed in terms of the angle embraced by the belt : — log, Tj - log. Ti = ixa n log. '1\ ixa = /^° or -©= ±o'4343/^" V l— /J^ where e =■ 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 o'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 /n 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. 336. Friction. 353 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 transmittec = = ^— ^ i^— 33.000 33,000 33.000 t/i - ^ where the friction F is expressed in pounds, and V = velocity in feet per minute. Substituting the value of <f, and putting /A = o'4 and o = 3"i4 (180°), we have the tension on the tight side 3 '5 times that on the slack side. H.P. = ""y^"^'^ 33,000 For single-ply belting Tj may be taken as about 80 lbs. per inch of width, allowing for the laced joints, etc. Let w = width of belt. Then T2=8ow J tr r. 0-7 2 X 8o7</V and H.P.= — ^ = 33.000 600 for single-ply belting; andH.P-«'V .300 for double-ply belting. The number of square feet of belt passing over the pulleys per minute is — • ^ 12 Hence the number of square feet of belt required per minute per horse-power is — vN H.z= 50 square feet per minute for single-ply, and wV 25 square feet per mtaute for double-ply 600 2 A 354 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. 202 we showed that the stress due to centrifugal force was — g where W, is the weight of i foot of belting i square inch in section. W, = 0*43 lb., and V„ = the velocity in feet per second : V = velocity in feet per minute ; hence — o-43V„'' ^ v." ^ V^ 32"2 75 270,000 and the effective tension for the transmission of power is — V= T — 270,000 The usual thickness of single-ply belting is about 0-22 inch, and taking the maximum tension as 80 lbs. per inch of width, this gives -; — = 364 lbs. per square inch of belt, and the power transmitted per square inch of belt section is — P = TaV- 270,000 d\ T,- 3^= 270,000 For maximum power T.= 270,000 and V = = 5700 feet : per minute. Friction. 355 The tension in the belt when transmitting the maximum power is therefore — Ts — r— = 364 — 121 = 243 lbs. per square inch. 270,000 and the maximum horse-power transmitted per square inch of belt section — 072 X 243 X 5700 H.Pma«. = —— = 3° nearly. 33,000 For ropes we have taken the weight per foot run as o'35 lb. 2000 3000 1000 5000 6000 7000 Velocity in Feeifer Minult, Fig. 337. per square inch of section, and the maximum permissible stress as 200 lbs. 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 = i7*i. The curves in Fig. 337 show how the horse-power trans- mitted varies with the speed. The accompanying figure (Fig. 338), 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 356 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 Leeds University. Creeping of Belts. — ^The material on the tight side of 9 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 ; Li = stretched length on the Tj side ; A ^^ )) )» » ■'■1 »i N, = revolutions per minute of driven pulley ; N2= , driving „ di = diameter of driven pulley) measured to the middle (/a = „ driving „ J of the belt ; X = stretch of belt in feet ; E = Yoimg's modulus ; /, Atid/j = stresses corresponding to T, and Tg in lbs. square inch. Then x -. E i,=/+x = = '(■ +® = 7r</,N, A- -(■ -l) = W,N, N, _ (E ^A)d^ (E +/,)-/, If there were no creeping, we should have — E = from 8,000 to io,ooo lbs. per square inch. Taking Friction. 357 Ta = 80 lbs. per inch of width, and the thickness as o'2 2 inch, we have when a — 3-14 — f-i = = 364 lbs. per square inch 0*22 Q _ Q _ and Ti = Trr^-~T. = — = 23 lbs. per inch width 2"12 3'5 /i = — 2_ = 104 lbs. per square inch 0'22 Hence %±fy = ^°'°°° + ^°4 = ^.975 E +/a 10,000 + 364 ^'^ Fig. 338. or the belt under these conditions creeps or slips 2's per cent. When a belt transmits power, however small, there must be some slip or creep. 358 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 + /, 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 o"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 — — — rr-^ = 4168 revolutions per minute 13 X 8 But the true speed would be much more nearly — 140 X 72-22 X 43'22 X o'Q75 X o"o7s „ — i 5j? — ^15 11^ = 3822 revs, per mmute 13'22 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. 339. Rope Driving. — When a rope does not bottom in a grooved pulley, it wedges itself in, and the normal pressure is thereby increased to — p -JL sm - 2 The angle 6 is usually about 45°; hence P, = 2 -6?. The most convenient way of dealing with this increased pressure is to use a false coefficient 2"6 times its true value. Taking /i = 0*3 for a rope on cast iron, the false /i for a grooved pulley becomes 2-6 x 0-3 = o'78. Friction. 359 The value of ei^' now becomes lo'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 loc^ to i6c'^, where c\s the circumference in inches. Then, by similar reasoning to that given for belts, we get for the horse-power that may be transmitted per rope for the former value — „ „ c^V </-V H.P. = , or 3740 374 where d = diameter of rope in inches. The reader should refer to a paper on rope driving by Mr. Coombe, Insf. Mech. Engrs. Proceedings, 1889. Coefficients of Friction. The following coefficients obtained on large bearings will give a fair idea of their friction : — Ball bearings with plain| ^ . cyhndrical ball races, i ^ ^' _, r Flat ball races „ „ o-ooo8 to 0-0012 Ihrust ■jOneflat,one vrace, 3 ,. „ mean 00018 "^^""Sslxwoyraces, 4 » .. .. o'o°S5 Gun-metal bearings r Plain cylindrical journals^ tested by Mr. with bath lubrication / Beauchamp Tower Plain cyhndrical journals V for the Institution' with ordinary lubrication/ of Mechanical Thrust or collar bearingj ^.^ Engineers ^ well lubricated / ^ Good white metal (author) with very meagrej ^.^^ lubrication > Poor white metal under same conditions o*oo3 o'ooi o'oi Reference-books on Friction. " Lubrication and Lubricants," Archbutt and Deeley. "Friction and Lubrication," Dr. J. T. Nicolson, Man- chester Association of Engineers, 1907— 1908. " Cantor Lectures on Friction," by Dr. Hele-Shaw, F.R.S. Published by the Society of Arts. CHAPTER X. STSESS, 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 relatively 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 J ~ 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 XV. 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. 361 s.s. O >ig H <3 V 41 3 4-> 5 «.3 -H -^4 II .a bfl n s 1 M :& Hi-- hI--> .a c .!3 hK >2 ,_^ (3 ,*<, c 4» M O ii 0) u o o J3 s i e ° ai V •-J u e53 c " o tuo "S a •S u »j 2 S-S §■£■3 S ai G 5 ""■ .Sou ^ a ^g g a a •? 01s ,a a I 1 ,r ■I© fVi 362 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. Fenuanent 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 wjiich 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. Ductile 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. 363 point will be reached when the strain will (in good wrought iron and mild steel) suddenly increase to 10 or 20 times its previous amount. This point is termed \ih& 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 xrjo of the length of the bar ; but when the yield point is reached, the stretch would amount to y^, 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. 340 was designed by the author Fig. 340. some years ago, and gives entirely satisfactory results ; it reads to ^q^(,J of an inch ; it is 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 ; 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. 341 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. 364 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. OOlB fs ^4 6 8 10 Stress in. Ions per Sf Inxfi, Fig. 341. 14 Just before the maximum stress is reached, the material appears to be nearly perfectly plastic. It keeps on stretching without any increase in the load. Up to this point the strain on the bar has been evenly distributed (approximately) along its whole length; but very shortly after the plastic state has been reached the bar extends locally, and "stricture" com- mences, «.<f. a local reduction in the diameter occurs, which is followed almost immediately by the fracture of the bar. The extension before stricture occurs is termed the " proportional " extension, and that after fracture the " final " extension, which Stress, Strain, and Elasticity. 365 is known simply as the "extension" in commercial testing. We shall return to this point later on. The stress-strain diagram given in Fig. 342 will illustrate clearly the points mentioned above. Brittle Materials. — Brittle materials at first behave in a similar manner to ductile materials, but have no marked elastic limit or yield point. They break oflf short, and have no ductile or plastic stage. Extension of Ductile Materials. — We pointed out above that the final extension of a ductile bar consisted of two parts — (i) An extension evenly distributed along the whole length of the bar, the total amount of which is consequently Stress orMiltl Steel- Stricture tJu-esT Fis. proportional to the length of the bar ; (2) A local extension at fracture, which is very much greater per unit length than the distributed or proportional extension, and is independent (nearly so) of the length of the bar. Hence, on a short bar the local extension is a very much greater proportion of the whole than on a long bar. Consequently, if two bars of the same material but of different lengths be taken, the percentage of extension on the short bar will be much greater than on the long bar. 366 Mechanics applied to Evgineering. The following results were obtained from a bar of Lowmoor Flo. 343- The local extension in this bar was 54 per cent, on 2 inches. The final extensions reckoned on various lengths, each including the fracture, were as follows : — Length Percentage of extension 10 22 24-5 6" 34 4 41 2 54 (See papers by Mr. Wicksteed \-a Industries, Sept. 26, 1890, and by Professor Unwin, I.C.E., vol. civ.) Hence it will be seen that the length on which the percentage of extension is measured must always be stated. The simplest way of obtaining comparative results for specimens of various lengths is to always mark them out in inches throughout their whole length, and state the percentage of extension on the 2 inches at fracture as well as on the total length T)f the bar. A better method would be to make all test specimens of similar form. Stress Via. 344. t.e. the diameter a fixed proportion of the length ; but any one acquainted with commercial testing knows how impracticable such a suggestion is. Load-strain diagrams taken from bars of similar material, but of different lengths, are somewhat as shown in Fig. 344. Stress, Strain, and Elasticity. 367 If L = original length of a test bar between the datum points ; Li = stretched length of a test bar between the datum points ; Then Lj — L = x, the extension The percentage of extension is — L, — L loox -^— — X roo = — — In Fig. 345 we show some typical fractures of materials tested in tension. Gun metal. Hard steel. Soft Delta steel. Copper. metal. Fig. 345. If specimens are marked out in inches prior to testing, and after fracture they are measured up to give the extension on various lengths, always including the fracture, they will, on plotting, be found to give approximately a law of the form — Total extension = K + «L where K is a constant depending upon the material and the diameter of the bar, and n is some function of the length. Several plottings for different materials are given in Fig. 346. Beduction in Area of Ductile Materials. — The volume of a test bar remains constant within exceedingly small limits, however much it may be strained ; hence, as it exterids 368 Mechanics applied to Engineering. the sectional area of the bar is necessarily reduced. The reduction in area is considered by some authorities to be the best measure of the ductility of the material. Annealed Coppe* Mild Steel Wroughi Iron 4- e a Length. Fig. 346. to. tnches Let A = the original sectional area of the bar ; Ai = the final area at the fracture. Then the percentage of reduction in area is — —^ x 100 A If a bar remained parallel right up to the breaking point, as some materials approximately do, the reduction in area caii be calculated from the extension, thus : Stress, Strain, and Elasticity. 369 The volume of the bar remains constant ; hence — LA = LjAi, or Aj LA and the reduction in area is — A- A, A Then, substituting the value of Aj , we have-^ LiA - LA L, - L * L,A Li L. Thus the reduction in area in the case of a test bar which remains parallel is equal to the extension on the bar calculated on the stretched length. This method should never be used for calculating the reduction in area, but it is often a useful check. The published account of some tests of steel bars gave the following results : — Length of bar, 2 inches ; extension, 6'o per cent. ; reduc- tion in area, 4*9 per cent. ; 6x2 Then x in this case was '— = o" 1 2 inch 100 and Li = 2" 12 inches „ , . . o'i2 X 100 Reduction ui area = \ = S'66 per cent. Thus there is probably an error in measurement in getting the 4'9 per cent., for the reduction in area could not have been less than 5 '66 per cent, unless there had been a hard place in the metal, which is improbable in the present instance. Real and Nominal Stress in Tension. — It is usual to calculate the tensile stress on a test bar by dividing the maximum load by the area of the original section. This method, though convenient and always adopted for commercial purposes, is not strictly accurate, on account of the reduction of the area as the bar extends. Using the same notation as before for the lengths and areas — Let W = the load on the bar at any instant ; W S = the nominal stress on the bar, viz. -r ; . W S, = the real stress on the bar, viz. -r-- 2 B 370 Mechanics applied to Engineering. Then, as the volume of the bar remains constant — L A LA = LiA„ and ^ = . - W S W Ai L ~K SLn OT the real stress Sj = -^r- The diagram of real stress may be conveniently constructed as in Fig. 347 from the ordinary stress-strain diagram. The construction for one point only is given. The length Tl^-'" A Fig. 347. i^ e strain % L of the specimen is set oflf along the strain axis, and the stress ordinate de is projected on to the stress axis, viz. ao. The line ba is then drawn to meet ed produced in c, which gives us one point on the curve of real stress. For by similar triangles we have — S, L. SLi which we have shown above to be the real stress. The last part of the diagram, however, cannot be obtained Stress, Strain, and Elasticity. 371 thus, as the above relation only holds as long as the bar remains parallel; but points on the real stress diagram between ^-Cu. Mid /^teel fIfvurM \ \ (^^ \ Xoad Fig. 348. g and / can be obtained by stopping the test at intervals, noting the load and the corresponding diameter of the bar in the stricture : the load divided by the corresponding stricture area gives the real stress at the instant. Fig. 349. — Steel containing several percentages of carbon. Typical Stress. Strain Curves for Various Materials in Tension. — The curves shown in Figs. 348, 372 Mechanics applied to Engineering. 349, were drawn by the author's autographic recorder (see Engineering, December 19, 1902), from bars of the same length and diameter. Some of the curves in Fig. 350 are curiously serrated, i.e. the metal does not stretch regu- larly (these serrations are not due to errors in the recording apparatus, such as are obtained by recorders which record the faults of the operator as well as the characteristics of the ma- terial). The author finds that all alloys containing iron give a serrated diagram when cold and a smooth diagram when hot, whereas steel does the reverse. This peculiar effect, which is disputed by some, has been independently noticed by Mons. («) Le Chatelier. Artificial Raising of the Fig. 350. — (a) Rolled aluminium . roHed copper i^c) rolled "bull "metal, temp. 400° Fahr. ; (a!) ditto 60° Kahr. N.B.— Bull metal and delta metal ElastiC Limit. The form of behave in practically the same way in „t „ 4. ■_ j 1 the testing-machine. a strcss-strain curve depends much upon the physical state of the metal, and whether the elastic limit has been artificially raised or not. It has been known for many years that if a piece of metal be loaded beyond the elastic limit, and the load be then released, the next time the material is loaded, the elastic limit will approxi- mately coincide with the pre- vious load. In the diagram in Fig- 35 1, the metal was loaded up to the point c, and then re- leased ; on reloading, the elastic limit occurred at the stress cd, whereas the original elastic limit was at the stress ab. Now, if in manufacture, by cold rolling, drawing, or otherwise, the limit had been thus artificially raised, the stress-strain diagram would have been dee. Young's Modulus of Elasticity (E). — We have already Stress, Strain, and Elasticity. 373 stated that experiments show that the strain of an elastic body is proportional to the stress. In some elastic materials the 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, asstiming 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. 4 .[] J _^....'Vj; ~~yiPess~ Fig. 352. In the diagram in Fig. 352 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. 365. 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 — I E 374 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 ori^nal 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 : E=Z X / Then, if we reckon the strain per unit length as on p. 361, ha' thus :- we have - = unit strain, and we may write the above relation Young's modulus of elasticity = — ^ unit stram Thus Young's modulus is often defined as the ratio of the unit stress to the unit 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- 353- 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 — / E 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. 427. Stress, Strain, and Elasticity. 375 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 b|:":;a":: ••^ stress Fig. 353. borne by each? Take the case of a compound tension member. Let E, = Young's modulus for material i ; ^2 ^^ )» J» )) 2 j Ai = the sectional area of i ; ■"2 = )i _ II 2 ; /i = the tensile stress in i ; /i — >i I. 2 ; W = the total load on the bar. Then /2 Eja'/i Eg since the components of the member are attached together at both ends, and therefore the proportional strain is the same in both; 376 Mechanics applied to Engineering. and A2/2 A.E^ Wj which gives us the proportion of the load borne by each of the component members ; W and Wi = 1 + A,E, W,= A1A2 w 1 + AjEj A2E2 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 Fig. 354. element subjected to shear is measured by the slide, x (see p. 361). 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. Stress, Strain, and Elasticity. 177 From similar triangles, we have I G _ / stress G = - = - X strain Relation between the Moduli of Direct and Trans- verse Elasticity. — Let abed be a square element in a perfectly elastic material which is to be subjected to — (i) Tensile stress equal to the modulus stress; then the length / of the line ab will be stretched to 2/, viz. abi, and the 2/— / strain reckoned on unit length will be ■ — - — = i. (2) Shearing stress also equal to the modulus stress ; then the length / of the line ab will be stretched to is/P + P = ^2l= I "41/, and the strain reckoned on unit length 1-41/-/ will be = 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 Fig. 3SS. in tension ; but the stress is proportional to the strain, therefore the modulus qf transverse elasticity is o'4i, or | 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. Strength of Wire. — Surprise is often expressed that the 378 Mechanics applied to Engineering. strength of wire is so much greater than that of the material from which it was made; the great difference between the two is, however, largely due to the fact that the nominal tensile strength of a piece of material is very much less than the real strength reckoned on the final area. The process of drawing wire is equivalent to producing an elongated stricture in the material ; hence we should expect the strength of the wire to approximate to that of the real strength of the material from which it was made (Fig. 356). That it does so will be clear from the following diagrams. In addition to this the skin of the wire is under very severe tensile stress, due to the punishing action of the draw-plate, which causes a compression _ g of the core, with the result that the density of the wire is slightly increased with a correspond- ing increase in strength. Evidence will shortly be given to show that the skin is in tension and the core in compression. The process of wire-drawing very materially raises the elastic limit, and if several passes be made without annealing the wire, the elastic limit may be raised right up to the break- ing point ; the permanent stretch of the wire is then extremely small. If a given material will stretch, say, 50 per cent, in the stricture before fracture, and a portion of the material be stretched, say, 48 per cent., by continual passes through the draw-plate without being annealed, that wire will only stretch roughly the remainder, viz. 2 per cent., before fracture. We qualify this remark by saying roughly, because there are other disturbing factors ; the statement is, however, tolerably accu- rate. If a piece of wire be annealed, the strength will be reduced to practically that of the original material, and the proportional extension before fracture will also approximate to that of the undrawn material. If a number of wires of various sizes, all made from the same material, be taken, it will be found that the real stress on the final area is very nearly the same throughout, although the nominal strength of a small, hard, i.e. unannealed, wire is considerably greater than that of a large wire. The above remarks with regard to the properties of wire also apply to the case of cold drawn tubes and extruded metal bars. The curves given in Fig. 357 clearly show the general effects of wire-drawing on steel ; it is possible under certain conditions to get several " passes " without annealing. The range of elastic extension of wires is far greater than Stress, Strain, and Elasticity. 379 that of the material in its untreated state; in the latter case he elastic extension is rarely more than ^Ao of the length of the bar, but in wires it may reach i^Vo "■ " The elastic ex tension curve for a sample of hard steel wire is given in Fig. 358. In some in- vestigations by the author it was found that Young's Modulus for wires was con- siderably lower than that for the undrawn material. On considering the matter, he concluded that the highly stressed skin of the wire acted as an elastic tube tightly stretched over a core of material, which thereby com- pressed it transversely, and caused it to elongate longi- tudinally. If this theory be correct, annealing ought to increase Young's Modulus for the wire. On appealing to ex- periment it was found that such was the case. A further series I 2 NumAer or Passes Fig 357. 30 40 50 60 70 stress in. tons j)er Sq. IntJi. Tig. 3s8. 90 of experiments was made on wires of different sizes, all made 38o Mechanics applied to Engineering, from the same billet of steel; the results corroborated the former experiments. In every case the hard-drawn steel wires had a lower modulus than the same wires after annealing. The results were — o*i6o 0*160 0*210 o'aio 0-174 0*174 0*146 0*146 0*115 0*115 Elastic limit. Maxi- mum stress. Pounds per sq. inch. 181,700 59,600 78,000 63,800 125,000 71,600 z6o,ooo 7i»3«> 316,000 102,400 126,700 164,800 124,800 189,700 xxi,ooo Extension per cent on 10 ins, 9-6 3'S on 3 ins. 19*0 19*5 5 n 34'S 54'o 46-8 5a'3 31*5 51 'o 30*4 S7'i E. Pounds sq. inch. 25,430,000 28,500,000 37,520,000 37,800,000 25,4x0,000 a7,5oo»ooo 35,300,000 27,200,000 Remarks- Hard drawn Annealed Rolled rod Ditto annealed Rod after one " pass " Ditto annealed Rod after two " Ditto annealed 188,000 72,000 318,900 1x3,800 30*3 35.330,000 37,000,000 31,400,000 Rod after three " paa Ditto annealed Ditto after breaking Wire Bopes. — The form in which wire is generally used for structural purposes is that of wire ropes. The wires are suitably twisted into strands, and the strands into ropes, either in the same or in the opposite sense as the wires according to the purpose for which the rope is required j for details, special treatises on wire ropes must be consulted. The hauling capacity of a wire rope entirely depends upon the strength at its weakest spot, which is usually at the attach- ment of the hook or shackle. The terminal attachment, or the " capping," as it is generally tended, can be accomplished in many ways, but, unfortunately, very few of the methods are at all satisfactory. In the table below the average results of a large number of tests by the author are given. The method adopted for testing purposes in the Leeds University Machine is shown in Fig. 359. After binding the rope with wire, and tightly " serving " with thick tar band in order to keep the strands in position, the ends are frayed out, Stress, Strain, and Elasticity. 381 cleaned thoroughly with emery cloth, and finally a hard white metal 1 end is cast on. With ordinary ropes high efficiencies are obtained, but with very hard steel wires, which only stretch a very small amount before breaking, the wires have not the same chance of adjusting themselves to the variable tension in each (due to imperfect manufacture and capping), and consequently tend to break piecemeal at a much lower load than they would if each bore its full share of the load. On first loading a new wire rope the strain is usually large, due to the tightening up of the strands and wires on one another (see Fig. 359)., but the rope shortly settles down to an elastic condition, then passes an ill- defined elastic limit, and ulti- mately fractures. From its be- haviour during the elastic stage, a value for Young's modulus can be obtained which is always very much lower than that of the wires of which it is com- posed. This low value is largely due to the tightening of the strands, which continues more or less even up to the breaking load. In old ropes which have taken a permanent " set " the tightening effect is reduced to a minimum, and consequently Young's modulus is greater than for the same rope when new. The value of E for old ropes varies from 8000 to 10,000 tons per square inch. The strength is often seriously reduced by wear, corrosion, and occasionally by kinks. Fig. 35g. — Method of capping wire ropes ' A mixture of lead 90 per cent., antimony 10 per cent. 382 Mechanics applied to Engineering. Breaking load. Number Diameter of Section of E. Tons Tensile strengtli of of rope. wires. Rope. Sum of wires. Ratio. ins. sq. ios. tons sq. in. 20 0-0895 0-126 6000 14-0 14-6 096 Ii7\ 24 0-085 0-136 6870 18-0 18-5 0-97 85 30 0-091 0-195 S«40 23-5 24-6 0-96 127 V 30 0-085 0-187 SSSo 8-25 9-1 0-91 49 3" 0-136 0-522 7200 47-2 47-5 0-99 91 Steel 4? 0081 0-220 5400 20-0 20-4 0-9S 90 wire ■^s 0-023 0-232 6330 309 31-75 0-98 13^ ropes 108 0-065 0-358 6020 41-0 42-3 0-97 "5 222 0-044 0-337 5560 35 '4 46-3 0-76 105 222 0039 0-264 6530 24-9 30-0 0-83 95' 19 0-231 0-796 ,3770 6-95 8-27 0S4 8-7 ^l"- 7 0-231 0-293 35yo 225 3-03 0-74 7'7 cable Work done in fracturing a Bar. — Along one axis a load-strain diagram shows the resistance a bar offers to being pulled apart, and along the other the distance tliough which this resistance is over- come; hence the product of the two, viz. the area of the dia- gram, represents the amount of work done in fracturing the bar. Let a = the area of the dia- gram in square inches ; / = the length of the bar in inches (between datum points) \ A = the sectional area of the bar. If the diagram were drawn i inch = i ton, and the strain were full size, then a would equal the work done in fracturing the bar ; but, correcting for scales, we have — N M . - ^ ^ i ! Iioeui TTu tons ■■ Fig. 360. N = work done in inch-tons in fracturing the bar and TT-vj = work done in inch-tons per cubic inch in fracturing ''^'^ the bar Stress, Strain, and Elasticity. 383 Sir Alexander Kennedy has pointed out that the curve during the ductile and plastic stages is a very close approxima- tion to a parabola. Assuming it to be so, the work done can be calculated without the aid of a diagram, thus : Let L = the elastic limit in tons per square inch ; M = the maximum stress in tons per square inch ; X = the extension in inches. Then the work done in inch-tons per) _ ^ . 2 ,^ _ , . square inch of section of bar )~ "t" 3^ / = |(L + 2M) work done in inch-tons per cubic inch = — 7(L + 2M) ^ X , . . But Y X 100 = tf, the percentage of extension hence the work done in inch-tons per) £_,y , ^> cubic inch ) ~ 3oo'' ' The work done in inch-tons per cubic inch is certainly by far the best method of measuring the capacity of a given material for standing shocks and blows. Strictly speaking, in order to get comparative results from bars of various lengths, that part of the diagram where stricture occurs should be omitted, but with our present system of recording tests such a procedure would be inconvenient. The value of the expression for the '' work done " in fractur- ing a bar is evident when one considers the question of bolts which are subjected to jars and vibration. It was pointed out many years ago that ordinary bolts are liable to break off short in the thread when subjected to a severe blow or to long- continued vibration, and further, that their life may be greatly increased by reducing the sectional area of the shank down to that of the area at the bottom of the thread. The reason for this is apparent when one calculates the work done in fracturing the bolt in the two cases ; it is necessarily very small if the section of the shank be much greater than that at the bottom of the thread, because the bolt breaks before the shank has even passed the elastic limit, consequently all the extension is localized in the short length at the bottom of the thread, but when the area of the shank is reduced the extension is evenly distributed along the bolt. The following tests will serve to emphasize this point : — 384 Mechanics applied to Engineering. Diameter of bolt. Length. Work done in fracturing the bolt. Remarks. I in. I „ I3'2 ins. 132 » I0'4 inch-tons 39'9 .. .. Ordinary state Turned shank i:; 14 ins. 14 » 2' 1 5 inch-tons lb-8 „ „ Ordinary state Turned shanlc Jin. f.. 14 ins. 14 .1 175 inch-tons 74 » .. 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 ^ .1 J I 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 Soft Cast iron. Fia. 361 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. 38s Fig. 362. prevent buckling. Such short specimens, however, are incon- venient, for measuring accurately the relations between the stress and the strain. (Jp 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 splittbg 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. 362. If the material reached a perfectly ^, , , . load at any instant (W) , plastic stage, the real stress, t.e. — -. ' ' / , ' '^ ° sectional area at that mstant (Aj) 2 c Fig? 363. 386 Mechanics applied to Engineeiing. would be constant, however much the material was compressed ; then, using the same notation as before — A, - _ W and from above, — = constant Ai Substituting the value of Ai from above, we have — -— ' = constant ; /A. But /A, the volume of the bar, is constant ; hence W/j = 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. 363). 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. 361 (cast iron). The shearing fracture is quite what one might expect from purely theoretical reasoning. In Fig. 365 let the sectional area = A ; Fig. 364.— Asphalte, then the stress on the cross-section S = W Stress, Strain, and Elasticity. 3^7 and the stress on an oblique section aa, making an angle a '»5J5m5^&?;$^J%%M^ A w / tt. N\j /^ a Fig. 366. — Portland cement. with the cross-section, may be found thus : resolve W into components normal N = W cos o, and tangential T = W sin a. 388 Mechanics applied to Engineering. The area of the oblique section aa = Ao= ^ cos a .1. 1 ^ N W cos a W cos* a _ „ , the normal stress = — = — - — = ^ a . cos" a cos a tangential or si ar-) _ T _ W sin a _ W cos a sin a f Ao ing stress f , Aj 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 tiie tangential, and it makes an angle of 90 — o 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) sm o normal stress = S sin* 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 0-71 = 0-5 S. From this reasoning, we should expect compression speci- mens to fail by shearing along planes at 45° to one another, and a cylindrical specimen to form two cones top and bottom, and a cube to break away at the sides and become six pyramids. That this does occur is shown by the illustrations in Figs. 364-366. Real and Nominal Stress in Compression (Fig. 367). — 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 volume remains constant and that the compression 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. 347) — S L c SL] Stress, Strain, and Elasticity. 389 Behavia ar of Materials subjected to Shear. Nature of Shear Stress. — If an originally square plate or block be Sffess Fig. 367. acted upon by forces P parallel to two of its opposite sides, the square will be distorted into a rhombus, as shown in Fig. 368, . *^ v a- b d y§ d C ' ' t— ^ — Fig. 368. ? Fig. 369. p and the shearing stress will be / = -,, taking it to be of unit thickness. This block, however, will spin round due to the couple P . arf or P . 6c, unless an equal and opposite couple be applied to the block. In order to make the following remarks 390 Mechanics applied to Engineering, perfectly general, we will take a rectangular plate as shown in Fig- 369- The plate is acted upon by a clockwise couple, P . ad, or f,.ab . ad, and a contra-clockwise couple, P, . a^ ox f, .ad . ab, but these must be equal if the plate be in equilibrium ; theny^ .ab.ad =fi .ad.ab or/. =/; t.A 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 Pi may be resolved into forces R and E.i acting along the diagonals as shown in Fig. 370. The effect of these forces will be to distort the square into a rhombus exactly as before. (N.B. — The rhombus Fig. 370. Fig. 371. in Fig. 368 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. 371 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/o = the stress normal to the diagonal. Then/oa<r =/,,»/ lab, Oif^Jibc = R But V 2P, or ij 2f,. ab, or ij 2/,. be ='&. hence/, =/, =/,' Stress, Strain, and Elasticity. 391 Thus the intensity of shear stress is equal on all the four edges and the tension and compression 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 f of their tensile strength (see p. 400). Ductile Materials in Shear. — The following results, ob- tained 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 : — ' Material. Nominal tensile strength. Shearing strength. Shearing strength. Tensile strength. Work done per sq. in. of metal sheared through. Cast iron — hard, close grained ,, ordinary . . . „ soft, open grained Best wrought iron .... Mild steel Hard steel Gun-metal Copper Aluminium I4'6 io'9 7-9 22'0 26-6 48-0 13-5 150 8-8 13-5 I2"9 107 i8-i 20'9 34'o 15-2 II-O S7 0-92 i-i8 1-36 •0-82 079 071 I-I3 073 0-65 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 — ■'• n\rt load on punch circumference of hole X thickness of plate 392 Meclianics applied to Engineering. ft Ratio. Wrought iron . . 19-8 24-8 o-8o Mild steel . . . 22-2 28-4 078 Copper. . . . 10-4 147 071 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 i"i 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 of unit thickness. The intensity of the shear stress on the two edges will be equal (see p. 390). Hence — The total shear stress on the face gi = f,.gi = Pj „ „ „ „ hi =f, . M= F „ direct „ „ gi=/,.gi = T Let the resultant direct stress on the face gA, which we are about to find in terms of the other stresses, be /„. Then the total direct stress acting normal to the face^^ =/u-ih = Pj. Stress, Strain, and Elasticity. 393 Now consider the two horizontal forces acting on the «,« C't— ^ Fig. 372. element, viz. T and P, and resolve them normally to the face gh as shown, we get T. and P,. 394 Mechanics applied to Engineering. cos d COS Q alsoP„ + T„ = P„=/«i;4 hence, substituting the above values, we have — ff'B +/.. hi =fggh cos e =f„.'gl and/.+-^=/« Next consider the vertical force acting on the element, viz. Pj. Pi = P„ sin =fagh_s\n 6 =fjd or^.^7=/>'~and ^-^^ = M -. = ^ = tan e Substituting this value of hi in the equation above, we have — or/«/. ^ff=f^ Jtt ~ Juft —J, Fig. 373. Solving, we get /. = J±V^ 7? +^ The maximum tensile stress on the) ^i _ft , /ft , ^ {a.cegh 5 -'■'■- 2 "^ V ^+-/' and the compressive stress on the face) at right angles to gh, viz.j'h i /»=f-vf+^" Some materials are more liable to fail by shear than by tension, hence it is necessary to find the maximum shear stress. Drawy^ at right angles to gh, and// making an angle a with it, the value of which will be determined. The compressive stress onj'h is/^, and the total stress represented by mn isj'h ./„. The maximum shear stress^ occurs on the i&CQJl, and the total shear stress on // is /^ . jl. The tensile stress on tiie face kl IS fa and the total stress is -4/ .^ which is represented by fq. If the tensile stress be + the compressive stress will be — . Since Stress, Strain, and Elasticity. 395 we want to find the shear stress we must resolve these stresses in the direction of the shear. Resolve pg, also mn^ parallel and normal to jl. pr = pq sin a = kl.fa sin a on = tun cos d = —jk.f^ cos a f.Ji = pr — on = kl.fa sin a - jk ./„ cos o fm. = Jjf^ sm a -J^f^ cos a =-f^ cos a sin u, — _/^ cos a sin a = (/„ —f^ cos a sin a = sm 2a, 2 This is a minimum when 2a = 90° and sin Ra = i. Then /^ = yirt JK /» , = \/// + f^ This is the maximum inten- sity of shear stress which occurs when a piece of elastic material is subjected to a direct stress ft and a shear stress /,. The direction of the most stressed section is inclined at 45° to that on which the maximum Intensity of tensile stress occurs. Bi^^*— Compound Stresses. — Let the block ABCD, of uni- form thickness be subjected to tensile stresses f^ and/, acting normal to the mutually perpen- dicular faces AD and DC or BC and AB. The force P^=/^. AD or/«.BC P,=/,.ABor/,.DC P„=/„.AC V,=ft- AC. (/j is the tangential or shear stress). Fig. 374. It is required to find (i) the intensity and direction of the 396 Mechanics applied to Engineering' normal stress /„ acting on the face AC, the normal to which makes an angle Q with the direction of P,. (ii) the in- tensity of the shear or tangential stress y^ acting on the face AC. Resolving perpendicular to AC we have — P„ = P„, + P.« = Py cos + P. sin /;AC =/,AB cos d +/«BC sin Q , ,AB ., ,BC . a /» =/»Xc *^°^ ^ "^-^'AC ^'° /„ =/, cos" e +/, sin» e also P, = P„ - P,. P, = P, sin 6 -V, cos 6 /,A.C =/»AB sin -/,BC cos 6 . .AB . . ,BC . •^' "•^'AC ^'" ~-^'AC *^°^ ^ =^ cos sin B — /, sin 9 cos ft = (/» -/.) cos d sin e =^^^^ sin 2^ This is a maximum when B = 45°. The tangential or shear stress is then = — — —. Thus the maximum intensity of shear stress is equal to one half the difference between the two direct stresses. The same result was obtained on page 395 by a different process. FoiBSOn'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 fraction - is generally known as Foisson's ratio, although in reality Foisson's ratio is but a special value of the fraction, viz. \. Strains resulting from Three Direct Stresses acting at Bight Angles. — In the following paragraph it will Stress, Strain, and Elasticity. 397 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,, and the ■^ strain per unit length -~ will be termed S„ and the stress on the face f^ ; then /x S,='g(seep. 374). Every applied force which produces a stretch or a + strain in its own direction will betermed +, and vice versd. The strains produced Thuiknj&ss Fig. 375. by forces acting in the various directions are shown in tabulated form below. Fio. 376. Force acting oa face of cube. Strain in' direction Xn Strain in directioilj'. Strain id direction *, Si. p« E E« _/x E« p» E» E E« p. -A 'S.n E» 4 398 Mechanics applied to Engineering. These equations give us the strains in any direction due to the stresses/,,^,/, 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. 390) 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^ andyi, which are of opposite sign and act at right angles to one another. The strains are — 4 + 4; = T^f 1 + -^ in the direction (i) » .. (3) E «E -4+4=0 «E ■ «E Thus the strain in two directions has been increased by - due to the superposition of the two stresses, and has been reduced to zero in the third direction. LetS=4(r+i). If a square abed had been drawn on the side of the element, it would have become the rhombus db'dd' after the strain, the AT— l*ac- ..a. .;:, A":>\d' Fig. 378. long diagonal of the rhombus being to the diagonal of the square as i + S to i. The two superposed are shown in Fig. 378. Then we have — ^ + (/+*)»=(i+S)» or 2/» + 2/a: + «» = I + 2S + S» Stress, Strain, and Elasticity. 399 But as the diagonal of the square = i, we have— 2/»= I X And let 7 = So ; x = IS^; then by substitution we have — a/» + 2/% + /%" =1 + 23 + 3" and I + So + — = 1 + 2S + S' 2 Both S and S, are exceedingly small fractions, never more than about YoVo" ^i^'i their squares will be still smaller, and therefore negligible. Hence we may write the above — r I + S„ = I + 2S orSo=2S = ^(x+i) / /E E/ I \ E« ^"'""^ ^=:777rr^( TTi )^ ^(«+^) hence n = 5 y^, and E = — ^ When « = s, G = -j^E = o-42E. « = 4, G = |E = o-4oE. « = 3. G = f E = 0-38E. « = 2, G = |E = 0-33E. Some values of n will be given shortly. We have shown above that the maximum strain in an element subject to shear is — =K.-0 but the maximum strain in an element subject to a direct stress in tension is — »<- E hence — = 3 iG+3 3,- / E = (.^0^ orS = s,(. + I) 400 Mechanics applied to Engineering. S Safe shear stress s; safe tensile stress When « = S il 1 „ « = 4 t i » » = 3 « \ » « = 2 i \ J^ 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 j or, in other words, the shearing strength of a material is only f 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 Longitudiiial Stress, but -with Lateral Strain prevented. in one Direc- tion. — Let a bar be subjected to longitudinal stress, _/^, in the direction x, and be free in the direction y, but be held in the direction z. The strain in the direction 2 due to the stresses _;^ and/j. is f f . . ■~f — ^~= = o, because the strain is prevented m this direction. r Flo. 379. Hence ^ = The strain in the direction x due to these stresses is — Jx Jz Jx Jx _^ Jxi l_\ 15-^ E «E E «^E E Thus the longitudinal strain of a bar held in this manner is only y| as great as when the bar is free in both lateral directions. Bar under Longitudinal Stress with Lateral Strain prevented in both Directions. «E _A_^ _^ _ q\ because the E «E «E I strain is pre- . —fi _f^ _f»_ ^Q [vented in these E «E «E / directions. Strain in direction :«; = ^ — =^ ■ E «E Stress, Strain, and Elasticity. 401 Then -=^ 4- — E «E^«E and /^ = «/, -f^ =fy{n - 1) since /„ =/, The strain in direction x is — E «E(« - 1) E\ «(« - 1)/ "^ E Or the longitudinal strain of a bar held in this manner is only f as great as when the bar is free. \^J [/| Anticlastio Curvature. — When a yecllljt' 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.; due to the extension of the fibres, the tension side of the beam section contracts laterally and the compression side extends. Then, Fig. 380. corresponding to the strain at EE on the beam profile, we have - as much at E'E' on the section ; also, the proportional strain — EE - LL _y LL p E'E' - L'L' _ y L'L' p. ButZ = i.2 Pi » P hence pi = np = 4p and 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 ; 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 stifTer in the ratio of «3 «!"- = yI than if the section were narrow. 2 D 402 Mechanics applied to Engineering. Boiler Shell. — On p. 421 we show that P. = aP,; /. = 2/. ; /. = /. ii I Fig. 381. Fig. 382. Strains. Let « = 4. ' E E« EV2 J ^E ' E« E« B.\2n J "E " ~ E« ^ E ~ E*. i^/ ~ » E 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 -j as great, or a cylindrical boiler shell will stand f= iT4or 14 per cent, more pressure before the elastic limit is reached than is given by the ordinary ring theory. Thin Sphere subjected to Internal Fluid Pressure. — In the case of the sphere, we have P. = P. ; /i = ^. Strains. — c _ /. _ 21 =-6/'i - i^ = s/' ^' E E« EV «/ *E c A - A = _/'/'i 4- ^^ - _ i/' °'~ E« E« E\n^ nj »E *•" E^^E EV n)~*E Stress, Strain, and Elasticity. 403 But^ in this dase ='^in the case given above; .-. S. in this case = | X ^ = |^ Maximum strain in sphere _ -f _ 3. maximum strain in boiler shell \ '' 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. Cylinder exposed to Longitudinal Stress when under Internal Pressure. — When testing pipes under pressure it is a common practice to close the ends by flat plates held in position by one long bolt passing through the pipe and covers, or several long bolts outside the pipe. The method may be convenient, but it causes the pipes to burst at lower pressures than if flanged covers were used. In the case of pipes with no flanges a long rod fitted with two pistons and cup leathers can be inserted in the pipe, and the water pressure admitted between them through one of the pistons, which produces a pure ring stress, with the result that the pipes burst at a much higher pressure than when tested as described above. (i) Strain with simple ring stress = ^ zff Pressure required to burst a thin cylinder/ = — a (ii) Strain with flanged covers = 8„ 2ft Pressure required to burst a thin cylinder = p = ^-~ (iii) When the covers are held in position by a longitudinal bolt, the load on whith is Pj, the longitudinal compression in P the walls of the pipe is-j =7^. The circumferential ring strain due to internal pressure and longitudinal compression — 404 Mechanics applied to Engineering. where m =-^ Jx n 2ft and/ = — j — X -^ n + m a Example. — Let d= 8 ins. / = 05 in. f^ = ro.ooo lbs. sq. in. Simple ring stress / = 1250 lbs. sq. in. With flanged covers/ = 1428 lbs. sq. in. With longitudmal bolt and let «? = 2. "j This is a high value but is sometimes \ p = 830 lbs. sq. in. experienced ' Thus if a pipe be tested with longitudinal bolts under the extreme condition assumed, the bursting pressure will be only 58 per cent, of that obtained with flanged covers. 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 ; Sv = change of volume due to change of pressure ; Bp „ pressure. Sp change of pressure ~ Sv "" change ofvolume per cubic unit of the body ~V V K is termed the coefficient of elasticity of volume. The change of volume is the algebraic sum of all the strains produced. Then, putting p =f^ =fy =/; for a fluid pressure, we have, from the table on p.. 397, the resulting strains — E £«"*"£ E«'^E E« E E« E E« _/ _ pEn _ E« - _ EG 3/« — bp 3« — 6 9G — 3E ButE = ^-^(^(p.399) hence K = '^^^ = i(-^)e and n = 'g + ^^ 3« - 6 ^\« - 2/ 3K - 2G Stress, Strain, and Elasticity. 405 The following table gives values of K in tons per sc^uare inch also of n : — Material. K. n. Water 140 Cast iron 6,000 3'0 to 47 Wrought iron 8,800 3-6 Steel 11,000 3-6 to 4"6 Brass 6,400 31 to 3-3 Copper 10,500 29 to 30 Flint glass ... 2,400 3'9 Indiarubber ... 2-0 The « given above has not been calculated by the above formula, but is the mean of the most reliable published experiments. Strain Energy Stored in a Plate. — Let the plate be subjected to stresses^ and/,,. Strain in direction x = ^ — '^ y = Energy stored inch per cub '1= fy_L E «E strain x stress \E «E/2'^\E «E/2 ~2EV' ^■'' n ) Strength of Plat Plates. — An attempt treatment of the strength of flat plates subjected to fluid pressure is long and tedious. See Mor- ley's "Strength of Materials." The following approximate treat- ment yields results sufficiently accurate for practical purposes, Kectangular Plate of thickness /. Consider a diagonal section d. The moment of the water pressure about the diagonal acting on the triangle efg exact pab X 4o6 Mechanics- applied to Engineering. The moment of the reactions of the edges oi\ ^^^ ^ the plate when freely supported, the resultant is = — X - assumed to act at the middle of each side j 2 2 The resulting moment is equal to the moment of resistance of the plate to bending across the diagonal, hence — ab c ab c , dfi ,^ _,, ^_. p—X~-p—X- = f-r (See Chapter XI) 22-^23'^6 "^ ' pabc _ , ^"■^ , ab But d = y/a^ + 1^ and c = -7=?==^ 2{cP + ^)/" •^^'^^^ „/„8 I »\/a ~ f when freely supported. and / 2 1 psM ~'f w^^'' ^^ edges are rigidly held. For a square plate of side a this becomes ■^ = / when freely supported. V and -Ya = /when the edges are securely bolted. From experiments on such plates, Mr. T. H. Bryson, of Troy, U.S.A., arrived at the expression pj^_ When the length of a is very great as compared with b, the support from the ends is negligible. When the edges are rigidly held, the plate simply becomes a built-in beam of breadth a, depth t, span b — p.ab.~=f-r 12 Pl-r For a circular plate of diameter d and radius r The moment of the water pressure about a^ trr^p 4^ diameter j ~ ~2~ ^ rir The moment of the reactions of the edge of) tn^p 2r the plate when freely supported \ ^ ~^ ^ '^ Siress, Strain, and Elasticity. The. !^(-)=/£j! ^=/ When the edges are rigidly held ^ = / 407 w W Fig. 384. 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)«* W =ft{'W — d)t, where _;^ is the tensile strength of the metal, and t 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 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 metal not so highly stressed, hence the reduction in 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. Strength of an Elementary Fin Joint. — If a bar, perforated at both ends as shown, were pulled apart through the medium of pins, 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. Then the diameter of the holes =d + c no. 38s. LetyS = 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— -^2 4o8 Mechanics applied to Engineering. 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. Let the thickness of this damaged ring be — ; then 2 the equivalent diameter of the hole will be <^-f-f +K, and for equal strength of plate and pins — 4 A riveted joint differs from the pin joint in one important respect : the rivets, when closed, completely (or ought to) fill the holes, hence the diameter of the rivet \^d-\-c when closed. In speaking of the diameter of a rivet, we shall, however, always mean the original diameter before closing. Then, allowing for the increase in the diameter of the rivet, the above expressions become, when the plates are not damaged as in drilling — Fig. 386. {■w-{d^c)W, T{d + cy / When the plates are punched — [w-{d + c+TL)]ff,= r(d+cY -/. The value of c, the clearance of a rivet-hole, may be taken at about ^V of the original diameter of the rivet, or c = o'o^d. The diameter of the rivet is rarely less than | inch or greater than i^ inch for boiler work ; hence, when convenient, we may write c = o'os inch. The equivalent thickness of the ring damaged by punch- ing may be taken at ^ of an inch, or K = 0-2 inch. This value has been obtained by very carefully examining all the most recent published accounts of tests of riveted joints. Stress, Strain, and Elasticity. 409 The relation between the diameter of the rivet and the thickness of the plate is largely a matter of practical conve- nience. In the best modern practice somewhat smaller rivets are used than was the custom a few years ago; taking d = i"i ■// and {d + cf = r^T,t appears to agree with present practice. Instead of using 7^, we may write f^ (see 400), where /, is the tensile strength of the rivet material. The values oifi,f„ B.nAf, may be taken as follows, but if in any special case they differ materially, the actual values should be inserted. Material. Iron ... Steel ... A' fr ... Jo 24 21 K . , W 28 28 \ ^"^ P" square inch. Ways in which a Riveted Joint may fail. — Riveted joints are designed to be equally strong in tearing the plate o o Tearing plate through rivet- holes. Shearing of rivet. -S- J^ Bursting through edge of plate; Fig. 387. -Q- Shearing through edge of plate. Crushing of rivet. and in shearing the rivet ; the design is then checked, to see that the plates and rivets are safe against crushing. Failure through bursting or shearing of the edge of the plate is easily avoided by allowing sufficient margin between the edge of the rivet-hole and the edge of the plate ; usually this is not less than the diameter of the rivet See paragraph on Bearing Pressure. 4IO Mechanics applied to Engineering. Lap and Single Cover-plate Riveted Joints. — When such joints are pulled, the plates bend, as shown, till the two Fig. 388. Fig. 389. lines of pull coincide. This bending action very considerably increases the stress in the material, and consequently weakens the joint. It is not usual to take this bending action into account, although it is as great or greater than the direct stress, and is the cause of the dangerous grooving so often found in lap-riveted boilers.' ' The bending stress can be approximately arrived at thus : The . P/ maximum bendmg moment on the plates is — (see p. 505), where P is the total load on a strip of width w. This bending moment decreases as the plates bend. Then, /( being the stress in the metal between the rivet-holes, the stress in the metal where there are no holes 'i& M — ^— |> \ w ) hence— and the bending moment = -^^ - The plate bends along lines where there are no holes ; hence Z = ^ (see Chap. XI.) 6 and the skin stress due to bending — Stress, Strain, and Elasticity. 411 Single Row of Rivets. Punched iron — {w ~ d — c ■ 4 5 Fig. 390. Fig. 391. Then, substituting 0-05 inch for e on the left-hand side of the equation, and putting in the numerical value of K as given above, also putting {d + cf = i"33/', we have on reduction — / w — d = o-83'^ + o'25 inch Jt w — d= 1*20 inches Thus the space between the rivet-holes (w — d) of all punched iron plates with single lap or cover joints is i"20 inch, or say I5 inch for all thicknesses of plate. By a similar process, we get for — Punched steel — w — d = I '09 inches In the case of drilled plates the constant K disappears, hence w — diso'2 inch less than in the case of punched plates ; then we have for — Drilled iron — Drilled steel — w — d = I'o inch m — d = o"89 inch Double Row of Rivets. — In this case two rivets have to Fig. 393. Fig. 393. 412 Mechanics applied to Engineering. be sheared through per unit width of plate w; hence we have — Punched iron— {iv — d — -Yi)ff.=VL^d^c)^^ 4 5 / IV — d = i-gi"^ + 0-2S inch Jt Punched steel — Drilled iron — Drilled steel — w — d = zt6 inches w — d = i-g2 inches w — d = vg6 inches w — d = V]2 inches Double Cover-plate Joints. — In this type of joint there is no bending action on the plates. Each rivet is in double shear; therefore, with a single row of rivets the space between the rivet-holes is the same as in the lap joint with two rows of rivets. The joint shown in Fig. 394 has a single row of rivets («.*. in each plate). Double Row of Rivets. — In this case there are two rivets in double shear, which is equivalent to four rivets in single shear for each unit width of plate w. Punched iron — (w-d-c-Y:)tf, = ^(d + cj^ 4 5 Fig. 394. ■w-d= 3-347 + 0-25 It r \ i w — d = 4*07 inches k- "*"-■% \ Punched steel — 000 w — d= 3*59 inches K) Drilled iron — 1 w — d= 3-87 inches • Drilled steel— Tig. 395. w — d=i 3'39 inches Stress, Strain, and Elasticity. 413 Diamond Riveting. — In this case there are five rivets in double shear, which is equivalent to ten rivets in single shear, Fig. 396. in each unit width of plate w ; whence we have for drilled steel ' plates — {JV - d - c)if, = ^^{d ^,cf^ 4 5 w — d= 8*4 inches Combined Lap and Cover-plate Joint. — This joint may fail by — (i) Tearing through the outer row of rivet-holes. ' (2) Tearing through the | \ ^m^ inner row of rivet-holes and shearing the outer rivet (single shear). (3) Shearing three rivets in single shear (one on outer row, and two on inner). Making ( i ) = (3), we have for drilled steel — o ' o 1 o o ■---I — j oo6o<i>ooool l^ Fio. 397. (^-u>-d-c)tf, = ^{d + cf^^' 4 5 w — d = 2'7i inches If we make (3) = (2), we get {w-2d- 2c)tf, + -(^ + cf"^ = ^-{d -f cf-^ 4 5 4 5 On reduction, this becomes — w — 2d= 2*07 inches * This joint is rarely used for other than drilled steel plates. 414 Mechanics applied to Engineering. If the value of d be supplied^ it will be seen that the joint will not fail by (2), hence such joints may be designed by making (i) = (3). Fitch of Rivets. — The pitch of the rivets, i.e. the distance from centre to centre, is simply w ; in certain cases, which are IS) very readily seen, the pitch is — . The pitch for a number of joints is given in the table below. The diameters of the rivets to the nearest ^V of an inch are given in brackets. Iron plates and rivets. Steel plates and rivets. Thick- ness of Diameter of rivet. Type of joint.' plate, t. d=i-W7 Punched. Drilled. Punched. Drilled. 1'20 + d. i+d. 109 + rf. 0-89 + d. in. in. \ °-67 GS) 1-87 1-67 176 1-56 A 1 5 078 (i) 1-98 178 I '87 1-67 ' % 0-87 (?) 2-07 1-87 1-96 1-76 \ 0-9S HI) 2-IS 1-95 2-04 1-84 2-i6 + d. 1-96 + d. 1.92 + d. lyi + d. 1 i 0-67 (JU (a) 2-63 283 (a) Z-43 2-63 (a) 2-21 2-S9 (a) 2-01 ■2-39 ji ■ \ o78(i) 2-94 2-74 (a) 2-51 270 (a) 2-31 2-50 B 1 \ i 0-87 (?) 3-03 2-S3 («) 2-76 279 (a) 2-56 2-59 ; 0-9S «i) 3-II 2-91 2-87 2-67 fttf 1-03(1) 319 2-99 295 27s I I -10(11) 3-26 3-06 302 2-82 407 + d. 3'87 + </. 3-59 + d. 3-39 + d. 1 \ 078 (1) 4-68 4^ 4-65 3-99 3-79 1 t 0-87 (?) 4-94 4-74 4-40 4-20 C 1 \ 0-95 (ti) 5-02 4-82 4-54 4-34 II i I -03 (I) S-io 4-90 4-62 4-42 I 1 -10(11) S-I7 4-97 4-69 4-49 14 Ji7(ift) 5-24 5-04 476 4-56 Stress, Strain, and Elasticity. 415 Thick- ness of plate, t. Diameter of rivet. Steel plates and rivets (drilled holes). Inner row. Outer row. 8-4 +rf. 1 1 in, 3 1 I I I I in. 0-9S (}■§) 103 (I) i-io(ij) 1-17(1^) 1-23 (ii) 1-30 (ift) 4-67 4-71 4;7S 4-81 4-85 9-3S 9"43 9-50 D i .■:•.•) 1 ........... • 9"57 9-63 1 , 1 970 271 + d. ^ f I 0-78 (3) 0-87 (?) 0-9S (H) 1-03(1) I'IO(lJ) i-i7(ij|) I -75 1-79 i;83- 1-87 i-gi 1-94 3-49 3-88 It is found that if the pitch of the rivets along the caulked edge of a plate exceeds six times the diameter of the rivet, the plates are liable to pucker up when caulked ; hence in the above table all the pitches that exceed this are crossed out with a horizontal line, and the greatest permissible pitch inserted. Bearing Pressure. — The ultimate, bearing pressure on a boiler rivet must on no account exceed 50 tons per square inch, or the rivet and plate will crush. It is better to keep it below 45 tons per square inch. The bearing area of a rivet is dt, or (^d -{■ c)t. Let_^ = the bearing pressure. The total load on a"] group of Y^ets r ^^^ ^^ m a strip of plate / •'"^ ^ ' of width w j theloadonastripof. ^ (^ _ ^ _ .yt, or {w - d - c - Y.)f,t plate of width ?e// ^ vt> \ ui then (w-d-c)f^ = nf^{d + c)t and/s _ \w-d- c)f. ]]^So tons per sq. inch n{d -f- c) The bearing pressure has been worked out for the joints 4i6 Mechanics applied to Engineering. 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. — ^''offSU-^-^tl?" joint J strength of plate effective width of metal between rivet-holes w — d — c or pitch of rivets w — d — c —K. 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 Q Q strong with and across the grain, then a:,, \ ,''' "■■, the zigzag distance between the two holes, "0' Q X -X-^j ^ should be -. The plate is, however. Fig. 398. weaker along than across the grain, con- 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 Xi = — . 3 Riveted Tie-bar Joints. — When riveting a tie bar, a very high efficiency can be obtained by properly arranging the -._^ rivets. The arrangement shown in Fig. 399 is radically wrong for tension joints. The strength of such a joint is, neglecting the — » clearance in the holes, and damage done by punching — (w — 4d)/^ at aa Fig. 399. -ted abed Fic. 400. Stress, Strain, and Elasticity. Efficiency of Riveted Joints. 417 Type of joint. It:: Thick- ness of plate. i^ Iron plates and rivets. Punched. Drilled 51 48 46 44 [a) 65 67 65 63 67 61 66 60 64 59 62 78 77 76 75 74 73 57 53 51 49 (a) 70 73 70 82 81 79 78 77 76 Steel plates and rivets. Punched. Drilled. 48 43 4' {«) 58. 64 (a) 59 62 («)59 60 58 57 55 74 74 74 72 71 70 54 50 48 46 (a) 64 70 (0)64 67 W64 64 62 6r 59 78 78 77 76 74 73 87 87 86 72 70 69 67 66 65 41 8 Mechanics applied to Engineering. whereas with the arrangement shown in Fig. 400 the strength is — (w — i)fit at aa (tearing only) (k' - ■iisff + -d^f, at bb (tearing and shearing one rivet) 4 {w - 2,<^f + — -^X at cc ( „ „ three rivets) 4 (w - i4)f,t 4- -^d^f. at dd ( „ „ six „ ) 4 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. 399. The joint in Fig. 400 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)ff In the above expressions the constants c and K have been omitted j 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. Tie bar joints are frequently made with double cover plates ; the , bearing stress on the rivets in such joints often requires more careful consideration than the shearing stress. When ties are built up of several thicknesses of plate they should be riveted at intervals in order to keep the plates close together, and so prevent rain and moisture from entering. For this reason the pitch of the rivets in outside work should never exceed 12/, where t is the thickness of the outside plates of the joint. Attention to this point is also important in the case of all compression members, whether for outside or inside work, because the plates tend to open between the rivets. GrcSups of Rivets. — In the Chapter on combined bend- ing 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 Stress, Strain, and Elasticity. 419 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. 401. 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. 402. on the intersection of the lines drawn through the centres of gravity of the cross-sections of the two bars. 420 Mechanics applied to Engineering. In other words, the rivets must be arranged symmetrically about the two centre lines (Fig. 4°i)- Punching E£fects. — 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 ^ 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. 402 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 p lbs. 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 /, is the tensile stress in the material. ^'"- ■'°^' Then/D = if,t or/R =f,t When the cylinder is riveted with a joint having an efficiency ■H, we have — /R =/A In addition to the cylinder tending to burst by tearing the plates longitudinally, there is also a tendency to burst Stress, Strain, and Elasticity. 421 circumferentially. The total bursting pressure in this direction Is/ttR'', and the resistance of the metal Is zttR^,, where /j is the tensile stress in the material. Then/7rR2 = 27rR^ or/R = 2tff and when riveted — /R = 2tfy Thus the stress on a circumferential section is one-half as great as on a longitudinal section. On p. 402 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 /; be strained to r, + «jr, = ^^(1 + «<) r, ■„ „ r, + V. = ''.(i + «.) where « is a small fraction indicating the elastic strain of the metal, which never exceeds yoVo fo'' ^^^^ working stresses (see p. 364). The sectional area of the cylinder will be the same (to all intents and purpose:^) before and after the a^iication of the pressure ; hence — Fig. 404. T(r.=' - r?) = ^{r.^{i + „,y - r,\i + «,)'} which on reduction becomes — ri%n? + 2n^) = r,\n,^ + 2«,) « being a very small fraction, its square is still smaller and may be neglected, and the expression may be written — 422 Mechanics applied to Engineering. The material being elastic, the stress/ will be proportional to the strain n ; hence we may write — ^'' =4 or/,n^ =/,;-." re Ji 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) _ f j elementary ring ) =LflLdr r\ = fir?.r-Hr The total stress on the whole section) _ , i\'-2j of one side of the cylinder ) ~ •'''''' I ^ _ /^«v. '-/<n — I (Substituting the value of /(;»■,' from above) — I = /'■«-//'. This total stress on the section of the cylinder is due to the total pressure//-,; hence — pn=/,r,-/^. Substituting the value of/„ we have — pr,= M- Mr. r^ Dividing by r and reducing, we have — Pr. Pr. -U, Stress, Strain, and Elasticity. 423 For a thin cylinder, we have — pr, =ft Thus a thick cylinder may be dealt with by the same form of expression as a thin cylinder, taking the presstire to act on the external instead of on the internal radius. The diagram (Fig. 404, 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 isfr^ = constant. Lames 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 ;-, of thickness hr. Let the radial pressure on the inner surface of the ring be /, and on the outer surface {p — Sp), 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 !"'=■ 4°5- the figure. The force tending to burst this thin ring = / x 2;- . . (i.) „ prevent bursting ={ ^V/t^S^S These two must be equal — 2pr = 2pr + 2/8r — 2r%p -|- 2/Sr (iii.) We can find another relation between/ and p, which will enable us to solve this equation. The radial stress/ tends to squeeze the element into a thinner slice, and thereby to cause it to spread transversely; the stress / tends to stretch the element in its own direction, and causes it to contract in 424 Mechanics applied to Engineering. thickness and normal to the plane of the paper. Consider the P strain of the element normal to the paper ; due to / it is — - / (see p. 397), and due to/ it is — ~, and the total longitudinal P f strain normal to the paper is ^ p. Both /and/, however, diminish towards the outer skin, and the one stress depends upon the other. Now, as regards the strain in the direction/ both pressures /and/ act together, and on the assumption just made — / — / = a constant = 2a The 2 is inserted for convenience of integration. From (iii.) we have — Ip 2p 2a dp . . . -^ = — — ^ = -f^ m the hmit or r r dr Integrating we get — . b , / = 7. + « , b where a and b are constants. Let the internal pressure above the atmosphere be /<, and the external pressure/, = o, we have the stress on the inner skin — Reducing Barlow's formula to the same form of expression, we get— fi=Pi^^_,.^ Kxperlmental 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 Stress, Strain, and Elasticity. 425 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 100-ton testing machine, a delicate extenso- meter, capable of reading to ^ ,'0 „ of an inch, was fitted on /rttenor of Cylinder per ^ iMeril pressure 'US 2-607!m% Sqmck imefnal.^ jfressare-^ 033T0I SqrinA. oderiml p» press ure By experimenJti. Lame's theory . Barlow^ theory _ W \ ^ x\ -Cyhnder WaU- FlG. 406. the outside j 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 426 Mechanics applied to Engineering. 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, Lamp'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 ah, ag being compression, and bh tension. Then, when subjected to pressure, the curve of stress would have been dc as before, but when combined with ab 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 modem guns by either shrinking rings on one another in such a manner that the internal rings are initially in compression and the externa] 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 eflfect is regulated by drawing the wire through friction dies mounted on a pivoted arm — in effect, a friction brake. Fig. 407. Stress, Strain, and Elasticity. 427 Strength and Coefficients of Elasticity of Materials in Tons square inch. Elastic limit. Breaking strength. E. Tension G. Shear. i Material. ■i J is or com- s 1 s i 1 c 1 Is pression. ■a ■« 3 6 V u i Wrought - iron) iars 5 12-15 10-12 21-24 17-19 ( 11,000- 1 13,000 5,000) 6,000 j 10-30 15-40 Plates with grain 13-15 — — 20-Z2 — — — 5-10 7-13 ,, across ,, H-13 — — 18-20 — .— — ~- 2-6 3-7 Best Yorkshire, ) with grain ... } Best Yorkshire,! across grain i 13-14 lz-14 — 20-23 — 17-19 12,000 — 15-30 40-50 13-14 19-20 IO-3Z 13-ao Steel, o-i % C. ... 13-14 lO-II 21-22 — 16-17 ( 13.000 (14,000 5,0001 6,000 J 27-30 45-50 „ o-=%C.... 17-18 16-17 13-14 30-32 — 34-26 „ 20-23 27-32 „ o-s%C.... 20-21 — 16-17 34-35 — 28-29 „ „ 14-17 17-20 „ l-o%C.... 28-29 22-24 21-23 5^55 ■^ 42-47 14,000 „ 4-5 7-8 Rivet steel 15-17 — 12-14 26-28 — 21-22 ■ — .— 30-35 30-50 Steel castings ... TO-II (not neal an- ed) 30-25 • - (■12,000 to } ~( S-I3 6-13 „ ,, IS-I7 (ann ealed) 30-40 / — ^- ^12,500 } ~- \ ro-zo 15-35 Tool steel 35-45 40-50 — 40-70 (unh en ard- ed) / 14,000 to ^ ~~ f 1-5 1-5 ^ 6g-8o — — 60-80 Chard ened) USjOoo / - \ ni! nil Nickel steel Ni( 4% ) 25-30 - — 35-40 - - - - 30-35 50-5S 45-65 — __ 90-95 — — — — 10-12 24-27 Manganese steel) Mn I % ...; 15 - - 30-35 - - - - 28-33 - ., 2 % IS — — 50-5S — 5-7 — Chrome steel ( Cr 5 % - i 30-40 - - 60-75 - - - - 10-15 — Chrome Vana- } 65-85 wat cr 70-go (hard ened) — — 10-15 45-55 ft „ ••• 60-75 oil oil 65-80 (hard ened) — — 8-18 45-55 Chrome nickel ... 45-50 — ~" , 5S-6o (ann ealed) — — 13-15 50-55 .. ......{ 100- i°5 }air air| 105- xzo (hard ened) — — 6-8 30-35 ti »i ••• \ 115- 130 ^oil oil{ 120- 130 (hard ened) - - 5-8 8-10 Cast iron ■jno mar lim ked) it S 7-X1 3S-6o 8-13 ' 6,000 V 10,000 2.500 \ to 4,000 J pra cally cti- nil Copper 2-4 10-13 anne aled 12-15 20-25 11-12 1 7.000 to — 35-40 50-60 10-12 bard drawn 16-20 — . 7.500 — 3-5 40-55 Gun-metal 3-4 S-6 2 '5-5 g-i6 30-50 8-12 i S.ooo- l 5.5c" 2,000 ( 2,500) 8-15 lO-iS 428 Mechanics applied to Engineering. Elastic limit. Breaking strength. s E. g Material. J 1 Tension or com- G. Shear. 0.0 1 = d B o S g pression. ■K C3 '% 1* 1 *c a i % H t3 W H tS <g M £ 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 — ( 5.50° \ '° 1 — ^ - 8-16 10-22 Rolled IS-2S l6-22 27-34 45-60 — ( 6,000 17-34 27-50 Phosphor bronze 7-9 24-26 — 6,000 2,500 Muntz metal 20-25 (roll- ed) 8-10 — =5-30 — — — — 10-20 30-40 Aluminmm 2-7 - 7-10 - 5-6 J 3,000 ( 5,000 } 1,700 4-iS 30-70 Duralumin 4-S — — 22-24 ( with ( 4,200 i 4,500 \ }- 10-12 30-38 Oak 4-6 ( grain 0*2 0*07 500-700 Soft woods 1-3 x-3{ to 0-4 450-500 CHAPTER XI. BEAMS. The beam illustrated in Fig. 408 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 compressioii, whereas between the two a strip remained unbroken, thus indicating no longi- FlG. 40S. 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. 409 is hinged at tjie 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. 410. It 430 Mechanics applied 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 lower is plain, because it is in --^- FlG. 4og, Fig. 410. compression. Let the sectional area of each block be a ; then, as before, we have "^l = px. But/ =fa, where /= the stress in either block in either tension or compression ; Or- hence W/ = fax or = 2fay The moment of] fthe moment of the internal forces, or the the external I = ! internal moment of resistance of the forces I I 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 The bending moment^ _ J stress on the) (modulus of the at any section ) \ material f ( section The connection between the T lever, the beam model 01 Fig. 410, and an actual beam may not be apparent to some readers, so in Fig. 411 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 Fio. , Beams. 431 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 sag as ^ (yy) shown by the dotted ^ |/o " 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 — or ^ : i.e. the strain is directly proportional to the distance «" ^1 from the point of the pivot. The enlarged figure shows this more distinctly, where e, e^ show the extensions, and c, Cj show the compressions at the distances y, y^ from the pivot. From the similar Fl'G, 41a. triangles, we have also - = ^. But we = y. ; also i = y. ex y-L h yi have previously seen (p. 375) that when a piece of material is strained {i.e. stretched or compressed), the stress varies directly as the ^'asxa, provided the elastic limit has not been passed. Hence, since the strain varies directly as the distance from the pivot, the stress must also vary in the same manner. Let/= stress in outer blocks, and/j = stress in inner blocks ; then — Z = Zor/,=^ /i yx y Then, taking moments about the pivot as before, we have™ W/ = if ay + z/ioyi Substituting the value of/i, we have — W/ = 2fay + z/ayi' If V, = ^. W/ = 2fay + ^* 2 Ay W/ = 2fay(x + \) 432 Mechanics applied to Engineering. Thus the addition of two inner blocks at one-half the distance of the outer blocks from the pivot has only increased the strength of the beam by \, or, in other words, the four-block model will only support x\ times the load (W) that the two- block model will support. If we had a model with a very large number of blocks, or a beam section supposed to be made up of a large number of layers of area a, a^, a^, a,, etc., and situated at distances y, y-^, yi, yz, etc., respectively from the pivot, which we shall now term the neutral axis, we should have, as above — W/ = 2fay + 2A.yi' 4. zAj-a" + zAjI'a' ^_ gtj._ Fig. 414. = ■^2((zy= -j- a^yy^ + a-ij/i + a^y^^ -\-, etc.) The quantity in brackets, viz. each little area (a) multiplied by the square of its distance {y'') 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/ = /I or The bending moment at any section the stress on thei (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 j 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) I second moment, or moment of on the section ) 1 inertia of the section distance of the outermost layer from the neutral axis Beams, 433 But we have also seen that W/ = /Z, and here we have 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— moment' aU = i*^ maximum or skin) (the modulus of any section) ' stress on the section) •^ \ the section 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. 412, in that the strain does increase directly as the distance from the neutral axis. Very delicate experiments clearly 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. 364 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 2 F 434 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 thjui the compression on the lower blocks, with the result that the beam would tend to turn about some point other than the pivot, Fig. 415; 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 «t& perfectly Fig. 415. 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 definite experi- mental proof will be given of the accuracy of the beam theory. Graphical Method of finding the Modulns 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 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 +fiai -^fa^ +, 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. 43 S 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 «i, the reduced area of the layer will be — <« yy- say «! whence we havey^Oj =fa^ Then the sum of the forces acting over the half-section, viz. — fa +/,<?! 4-/2«a +, etc. becomes /a +fal ■\-fai +, etc. or/(a + «i' + flJa' +, etc or /(area of the figure on one side of the neutral axis) or (the whole force acting on one side of the neutral axis) Then, since the intensity of stress all over i!a& figure is the same, the position of the resultant will be at the centroid or centre of gravity of the figure. Let Ai = the area of the figure below the neutral axis ; Aa = „ „ above „ „ d-^ = the distance of the centre of gravity of the lower figure from the neutral axis ; di = the distance of the centre of gravity of the upper figure from the neutral axis. Then the moment of all the forces acting! _ , . on one side of the neutral axis | " -^ ' ^ ' Then the moment of all the forces acting j _ fi\ j \ \ j\ on both sides of the neutral axis f ~ ■^^^•'^' + ^^'^''> = /Z(seep. 354) or Z = A,rfi + Aj^/a We shall shortly show that Ai = Aj = A (see next para- graph). Then Z = A(^i + 4) Z = AD where D is the distance between the two centres of gravity. In a section which is symmetrical about the neutral axis d = di — d, and (^ + (^a = 2;/- 436 Mechanics applied to Engineering. Then Z = 2M or Z = AD The units in which Z is expressed are as follows — Z = U (length units)* ^ ^ j^ ^^i^^j3 y length or Z = AD = area X distance = (length units)'' X length units = (length units)' Hence, if a modulus figure be drawn, say, - full size, the result obtained must be multiplied by «* to get the true value. For example, if a beam section were drawn to a scale of 3 inches = i foot, i.e. \ full size, the Z obtained on that scale must be multiplied by 4' = 64. We showed above that in order to construct this figure, which we will term a "modulus figure," the width of each strip of the section had to be reduced in the ratio ^, which we have previously seen is equal to — . This reduction is easily done thus: Let Fig. 416 represent a section through the indiarubber blocks of the beam model. Join ao, bo, cutting ^ ,. ,# the inner block in c and d. Then by similar triangles — t: zszzi Fig. 416. r ab or w{ ' = <y) the strip in c and d. In the case of a section in which the strips are not all of the same width, the same construction holds. Project the strip ab on to the base-line as shown, viz. db'. Join «'<?, l/o, cutting By similar triangles we have — ^^■^ ^1=7=7. °^-'^ = «^(7) Beams. 437 Several fully worked-out sections will be given later on. By way of illustration, we will work out the strength of the four-block beam model by this method, and see how it agrees with the expression found above on p. 432. The area A of the modulus figure onl _ ^ 1 ^ ' one side of the neutral axis / ' The distance d of the c. of g. of the modulusl _^_+J^ of the c. of g. of the modulus! _ ay + Oiy-. ; neutral axis (see p. 58) / a + Oi' figure from the ButW/=/Z =/X 2Ad or W/ = 2/(a + a/) X ~^rff = ^/^^ + 2>i>" But a,' = a^, .-. W/ = 2fay + ^^^ which is the same expression as we had on p. 431. 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. 438 Mechanics applied to Engineering. liBt 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 fli, (h, ^a> etc. ; and below the neutral axis be a/, ^a', a,', etc. ; and their respective distances above the neutral axis yi, y^, y^, etc. ; ditto below j/, y^, yl, etc : and the stresses in the several layers above the neutral axis be /i, ft, /s, etc. ; ditto below the neutral axis be //, fi, fi- Then, as the stress in each layer varies directly as its distance from y',^i Fig. 4i3. the neutral axis, we have — ^=^. and/.=^^ U y2 yi also ^' = ^ 7i yi The total stress in all the layers] =/i«i + fnHi +,etc. on one side of the neutral I /,, , , .. \ I = -(aiyi + «a;'2 +, etc.) axis 1 j/j The total stress in all the layers) on the Other side of the neutral! =^(fli,'j'i' + a^y^' +, etc.) j y. axis 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 — ay + a^yi + (jys +. etc. = dy' + a/j/ + aiyl +, 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 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 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 Beams. 439 compression. The position of the neutral axis has nothing whatever to do with the relative strengths of the material in tension and compression. In a reinforced concrete beam the modulus of elasticity of the tension rods differs from that of the concrete, which is in compression. Such beams are dealt with later on. 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 s.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. 433, is true). We have shown above on p. 432 that M =/-, and that I . •'' Z = -, where VIS the distance of the skin from the neutral axis. y . . In a symmetrical section y 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. If the maximum tensile stress ft is required, the 7, must be taken as the distance of the tension skin from the neutral axis ; and likewise when the maximum compressive stress /<, is required, the y^ must be measured from the compression skin. Thus we have either — ' We shall show later on that such a great difference as 5 ot 6 is undesirable for practical reasons. 440 Mechanics applied to Engineering. and as - = — , we get precisely the same value for the bending yt y<i moment whichever we take. We also have two values of ^ ■ I ^ ,1 r. Z, VIZ. - = Z, and - = Z. ; andM =y^Z, or/^. We shall invariably take /jZ, 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. G^mpressioTV 6as& Zma, yA / Comji ressiorh h<xse'tmj& Xensio7v 'base Zirve Fig. 419. Fig. 420. Construction for Z^ (Fig. 419). — 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. _j'^; 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 Zo = shaded area above or below the neutral axis X D. Construction for Z, (Fig. 420). — 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. 441 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 /„ as yt is to y^ 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 mo Julus 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^ has been found, and the Z, is required or vice versA, there is no need to construct the two figures, for — Z, = Z. X ^-°, or Z„ = Z, X -^ .;'. yc 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 be made with such a great difference, the tension flange would 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. 421. 442 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. 421. "Massing up" Beam Sections.— Thin hollow beam CT^f /o [■■- i- /■s a Fig. 421. Fig. 422. 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. 422. 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 j 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. < S » Fig. 423. B = H = S (the side of the square) Beams. 443 Those who have frequently to solve problems involving the strength and other properties of rolled sections will do well to get the book of sections issued by The British Standards Committee. Modulus of the section Z. Remarks. BH' 6 BH' The moment of iuettia (sec p. 8o) = H y=2 BH» ., 12 BH« ^ H 6 2 Also by graphic method — (Square) S' 6 The area A = — 4 ^ 2 H H -'=3X2=1 BH H BH» Z = ZAJ =2X -—X T=-fi- 430 444 Meclianics applied to Engineering. Section. Hollow rect- angles and girder sections. One corruga- tion of a trough flooring. Examples of Modulus Figures. " Corrtjor^c.sston' NA- ^ Te.n^torv k Beams, 445 Modulus of the section Z. BH' - bh' 6H Approximate- ly, when the web is thin, as in a rolled joist — Bm where t is the mean thickness of the flange. BH* Moment of inertia for outer rectangle = ;' = inner „ _bh' 12 hollow „ H 2 BH» - bh' BH' - bh* 12 12 BH» - W H 2 6H z = This might have been obtained direct from the Z for the solid section, thus — Z for outer section = —t— ^ . bhf h bh* Z., mner „ =-x- = — „ , ,, BH' bh* BH' - bh* Z „ hollow ,, = = 6 6H 6H The Z for the inner section was multiplied by the ratio -^, 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 nr-«er the truth than B^Hg, where Ho 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 7$ per cent, of tiiat 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 « rivets of diameter d in any one cross-section, the Z will be— (B - nd)^ (approx.) 446 Mechanics applied to Engineering. Section. Examples of Modulus Figures. Square on edge. Fig. 426. Tees, angles, and LI sections. Fig. 427. ■ t I This figure , , H becomes a T or ffAj^. \ .\ u when massed i up about a ver- tical line. Fig. 428. Beams. 447 Modulus of the section Z. 0-II8S' The moment of inertia (see p. 88) = — S §1 12 VzS' = oii8S» BiH,'+B.,IV-3A' 3H, H, = 07H approx. H, = 0-3H „ The moment of inertia for \ _ BiH|* the part above the N.A./ ~ 3 Ditto below = -^— ^ 3 Ditto for whole section = ^'^■' + ^'^' ~ ^'^' 3 ~, , ,, B,H,' + BjHj' - M' Z for stress at top = — — * — ■ — ^-2 3H, 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"3H from the bottom. Rolled sections, of course, have not square comers as shown in the above sketches, but the error involved is not material if a mean thickness be taken. 448 Mechanics applied to Engineering. Section. T section on edge and cruciform sec- tions. Examples of Modulus Figures. 2 ff B > Fig. 439. Unequal flanged sec- tions. ci,,,::::. ...i f f'<' ,.■ c--:.- --ja .if' '"^9 h :■■-::::,:„ 'y Fig. 430. Compressitm base'lim Tension hose line Modulus of the section Z, iW + B/4» 6H AppTOT. Z 6 Beams. 449 ITTJ Moment of inertia for vertical part = ^ 12 „ „ horizontal „ = -^ iH'+Bi' „ ,, whole section = — , iH'+BA' Zfor „ - gjj Or this result may be obtained direct from the moduli of the two parts of the section. Thus — Z for vertical part = -g- „ horizontal „ = _ ^ g (see p. 445) " ''l>°l'='=«'=f°°= 6 +6H- 6H 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. B,H,'+BjH,'-*,V-Vi' 3H, Moment of inertia) B,H,' i,Ai' for upper part ) ~ 3 3 Ditto lower part = ^'"'' ^'^'' Moment ^ of inertia B,H,»+B,n,»-i,.S,>-*A' for whole 3 section Z = Tj- for the stress on the tension flange . . 4SO Section, Mechanics applied to Engineering. Examples of Modulus Figures. ...JBt.-, M/1. h FlO. 432. This _ figure be- ! I comes the' i J, same as '^{T the last i I when -f— massed Hi about a i centre line. Triangle. ///« Trapezium. AfA Modulus of the section Z. Beams. 45 1 The construction given in Fig. 4.30 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 — 12 For stress at apex — BH' 24 The moment of inertia for al BH' , ^ ^. triangle about its c. of g. / 36 ^ ^' H 2H y for stress at base = — , at apex = — J A BH» Z for stress at base = i^ = T BH» apex= ^^ For stress at wide side — BH'/'«'+4« + i'\ 12 V, 2« + I J For stress at narrow side — BIP/'«'+4» + l '\ 12 \ K + 2 ) Approximate value forZ— Moment of ^ BH» ^ »' + 4« + i ^ , ^^ gg. inertia / 36 ^ « + 1 j^ ^' ' y for stress at wide side = H, = — T ^^^y J BH' / «' + 4» + I N Z for stress at wide side = 3^ \ « + 1 7 H / 2« + I N 3 *^ » + i y BH' Z' «' + 4« + I ^ ~ 12 ^ 2« + I y y for stress at narrow side = Hj = — f " j 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. 45.2 Mechanics applied to Engineering. SectloD. Circle. Hollow circle and corrugated section. Examples of Modulus Figuzvt. Fig. 435. Beams. 4S3 Modulus of the section Z. 10*2 The moment of inertia of a circle \ _ "'D* about a diameter (see p. 88) / ~ 64 D ' D - 32 2 ir(D' - D.«) 32D The moment of inertia of a hollow circle \ _ t(D* — Dj*) about a diameter (see p. 88) / ~ 64 D ^=2 Z = t(D' - D(') 64 D 2 ., _ ir(D' - D««) 32D This may be obtained direct from the Z thus — Z for outer circle = irD' 32 '^^''xg' (see p. 445) hollow 32 ^ t(D' - D/) 32D For corrugated sections in which tbe 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. 438. F'°' 438. The strength, then, is simply that of a rectangular section of height H = thickness of plate. 454 Mechanics applied to Engineering. Irregular sections. Bull-headed rail. Examples of Modulus Figures. Fig. 439. Flat-bottomed rail. Fig. 440. Tram rail (distorted). Fig. 441, Beams. 455 Irregular sections. Bulb section. Hobson's patent floor- ing. Fireproof flooring. Examples of Modulus Figures. Fic. 442' One section. Four sections massed up. Fig. 443. Fig. 444- 45 6 Mechanics applied to Engineering. Irregular sections. Fireproof flooring. Examples oC Modulus Figure 4- FlG. 445. Table of hydraulic press. FlQ. 446. Beams. 457 Fig. 447. Shear on Beam Sections. — In the Fig. 447 the rect- angular element abed on the unstrained beam becomes aW(^ when the beam is bent, and the element has undergone a shear. The total shear force on any vertical section = W, and, assum- ing for the present that the shear stress is evenly distributed over the whole section, the mean shear W stress = x, where A = the area of the section ; or we may write it W T. TT . But we have shown (p. 390) 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 x • The shear on vertical planes tends to make the various parts of the beam slide downwards as shown in Fig. 448, a, but the shear on the horizontal planes tends to make the parts of the beam slide as in Fig. 448, b. This action may be illustrated by bending some thin strips of wood, when it will be foimd that they slide over one another in the manner shown, often fail in this manner when tested. In the paragraph above 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. 479). We "*' will now work out one or ra-449- two cases to show the distribution of the shear on a beam FlQ. 448. Solid timber beams By z. * ^^. \ ; Y y^:--\ y\ ^ a; NA A /\ A y 4S8 Mechanics applied to Engineering. section by a graphical method, and afterwards find an analytical expression for the same. In Fig. 449 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 = ft^ on a given scale. 3 is proportional to the area of the strip between I and 3 = 4^ on a given scale. 4 is proportional to the area of the strip between I and 4 = ^A on a given scale. 5 is proportional to the area of the strip between 1 and 5 = y on a given scale. 6 is proportional to the area of the istrip between I and 6 = ^/ on a given scale. Let the width of the modulus figure at any point distant y from the neutral axis = b; then — the shear at v = 2 2 But^ = ?,and* = l? the shear atjF = ?X_^ = ^/_ \{f) 2 2Y 2Y in the figure U =^ ff, = \(j^) 2 2 1 Thus the shear curve is a parabola, as the ordinates //,, etc., vary as y^ ; hence the maximum ordinate kl = \ (mean ordinate) (see p. 30), or the maximum shear on the section is f of the mean shear. In Figs. 450, 45 1 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. 45 1 shows the total shear and the inner figure the intensity of shear at the different layers. The W shear at any section (Fig. 452) ab = — , and the intensity of Beams. 459 W shear on the above assumption = —^, where A„ = the sectional 2 A«, But the area of the web, or the intensity of shear = intensity of shear stress on aa^ = the intensity of shear stress 2ht heights Fig. 4SO. Fig. 451. on ab, hence the intensity of the shear stress between the web W and flange is also = — ;-. We shall make use of this when 2ht 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 JVmJtraZ-pl^zna 2. . r «■■-?-■. d Z M i Fig. 452. there is equal to the total direct tension or compression acting above or below it. 460 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 j S = shear force on section cd. The area of the strip distant j*) _ i j from the neutral axis ) ~ " " the total force acting on the strip =f.b.dy f V Fy But|=^, or/= y- Substituting the value of/ in the above, we have — ■^b.y.dy „ „ FI F M , But M = Y> ""^ y = Y (see p. 432) F Substituting the value of y ii the above, we have — the total force actmg on the stnp = -jO .y .ay But M = S/ (see p. 482) Substituting in the equation above, we have — -^b.y.dy Dividing by the area of the plane, viz. b.I,v{e get — S/fY The intensity of shearing stress. on that plane = fl> I b.y.dy J y JY ^D.y.dy Beams. 461 g the mean intensity of shear stress = -r nrhcre A = the area of the section. S fY IB ^-y-^y The ratio of the maximum"! _ J o intensity to the mean J ~ s A IBJo K = — 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. ri The value of I b.y.dyi^ 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 J = Y, which is the moment of the ^'°' ^^'" shaded area Aj about the N.A., viz. AiY„ where Y, 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 shear in ieam sections increases the deflection over and above that due to the bending moment. The shear effect is negligible in solid beanie of ordinary proportions, but in the case of beams having natrow webs, especially when the length of span is small com- pared to the depth of the section, the shear deflection may be 3 o or 30 per cent, of the total. Consider a short cantilever of length /, loaded at the free end. The deflection due to shear is x. For the present the deflection due to the bending moment is neglected. Work done by W in deflecting the beam by shear = — 2 Let the force required to deflect the strip of area hdh through the distance x be d^, let the shear stress in the strip be/„ then — 462 Mechanics applied to Engineering. X f, dW G Gbdh V (see page 376) (fW =ffidh Beams. 463 Work done in deflecting the strip by) _ xd'W _fflbdh shear 5 - ~^ - 2G Total work done in deflecting the) / V^i -fih^j, _ ^■^ beam section by shear j ~ ^j _^ -'• '"^"' ~ ~ Let a be the area of the modulus figure between h and Hi. Let /i be the skin stress due to bending. The total longi- tudinal force acting on the portion of the beam section between h and Hi is a/i, and the shear area over which this force is distributed is bl, hence /, = -^, since the shear is constant throughout the length. WG/i_H, 6 ~WG/ /■Hi ay/i where Ao = / —7—; which is the area of the figure mno^ between the limits Hj and — H2, Substituting the value of /i in terms of the bending moment M and the modulus of the section Zj AqM^ _ sAqM^ *~WGZl^~2WEZlV SAoW/ X = " -3 for a cantilever, and the total deflection at the free end of a cantilever with an end load, due to bending and shear, is — ^=.S + x = ^ + ^E2? ^^^^ P^^® 5^°^ and for a beam of length L = 2/ supporting a central load Wi= 2W ^ W,U 5A„WiL 48EI''" SEZi^ WiL/U; sAA \6I "^ Z,= / and E = 8A \6I ' Zi= In the case of beams of such sections as rectangles and circles the deflection due to shear is very small and is usually 464 Mechanics applied to Engineering. negligible, especially when the ratio of length to depth is great. In the case of plate web sections, rolled joists, braced girders, etc., the deflection due to the shear is by no means negligible. In textbooks 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 Section. Span L. Depth of section H. From 5. From A. Rolled joist 28" 6" 7,120 12,300 j» »» 3^;; 6" 8,760 12,300 ,) j» 42" 6" 9,290 12,400 »» »» 56" 6" 10,750 12,700 >i a 60" 6" 11,200 12,800 Riveted girder 60' S' 10,200 12,200 »» j» 60' 6' 9,000 12,200 )» )» 75' 4' ir,ooo 12,600 tf 1} 160' 12' 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. The application of the theory given above often presents difficulties, therefore, in order to make it quite clear, the full working out of a hard steel tramway rail is given below. The section of the rail was drawn full size, but the horizontal width of the diagram, i.e^ -7- was drawn 5 of full size, hence the actual Beams. 46s area of mnop was afterwards multiplied by 4. The original drawing has been reduced to 0-405 of full size to suit the size of page. a in square inches. «2 I, in inches. a' b Between Tolal. m and 2 0-38 0-38 0-14 2*20 0-07 2 » 3 o'6o 0-98 0-96 2-35 0-41 3 >. 4 o'6o 1-58 2' 50 3 "03 0-82 4 ., 5 o"64 2-22 4-93 2-25 2'19 5 „ 6 0-3S 2-57 6-6o 0-72 9-17 6 „ 7 0'12 269 7-24 0-42 17-2 O'lO 2-79 7-78 0-42 I8-S 8 „ 9 003 2-82 7-95 0-42 18-9 9 „ 10 — 0-09 2 '73 7-45 0-42 17-7 10 „ II —0-42 2-31 5-34 0-42 12-7 II „ 12 -0-23 2-o8 4-33 i-oo 4*33 12 „ 13 -0-82 1-26 1-59 5-30 0-30 13 .. P — I 26 Depth of section 6-5 ins. Modulus of section (Z). In this case the) g two moduli are the same 5 Moment of inertia of section (I) 48T Ao (area mnof) . . ' 73'9 sq. ins. Span (^ 60 ins. ... 28 ins. Load at which the deflection) g ^^^^ _ _ _ ^^ ^^^^ IS measured 3 Mean deflection A for the) (^ j^^ _ _ ^ -^^^ above loads ) E from -;r^^ tons sq. m. 12,200 . . . 8790 488! Efrom^^(Y4-^°) „ ,, 13.9°° • • • 14,200 E from a tension specimen) ^^^^ i„ cut from the head of rail j ^' J . In the case of a rectangular section of breadth B and depth H the value of Ao is , by substitution in the expres- sion for A for a centrally loaded beam of rectangular section, the deflection due to shear x = g^ and the ratio 2 U 466 Mechanics applied to Engineering. Deflection due to shear 32! Deflection due to bending 1^, the shear deflection is about 2 per cent, of Taking — as the bending deflection. 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 discrepancies may be dealt with under three heads — (r) Discrepancies below the elastic limit (2) » at (3) .. after (1) The discrepancies below the elastic limit are partly due to the fact that the modulus of elasticity (E^) in compression is not always the same as in Compression- tension (E,). For example, suppose a piece of material to be tested by pure tension for E„ and the same piece of material to be afterwards tested as a beam for Ej (modulus of elasticity from a bending test) ; then, by the usual beam theory, the two results should be identical, but in the case ^"'- ■ts'- 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„ is greater than the E„ the compression area A„ of the modulus figure will be smaller than the tension area A,, for the tensions and compressions form a couple, and AcE„ = A,E,. The modulus of the section will now be A. X D, or IM x y^r^'/TT 4 V -Ci, + V Jit fifA TjuhejtE^' Be X D ; thus the Z is 1 /F increased in the ratio . °.„ . If the E, be 10 per cent. ^E. + ^E,' Beams. 46; greater than E„ the Ej 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 Ej greater than E,. There is also another dis- crepancy which has a similar tendency, viz. that some materials do r^ot. perfectly obey Hooke's law; the strain increases slightly more rapidly than the stress (see p. 364). 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 E5 greater than E,. , On the other hand, the deflection due to the shear (see p. 463) is usually neglected in calculating the value Ej, which consequently tends to make the deflection greater than calculated, and reduces the value of Ej. And, again, experimenters often mea- sure the deflection between the bottom of the beam and the supports as shown (Fig. 458). 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 Ej. 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 Fic. , Fig. 458. 468 Mechanics applied to Engineering. 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. . 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. 459, 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" shouldnotbe 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 Fig. 459. Fig. 460. Beams. 469 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.^ This in 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. 461. Fig. 462. 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 afkr the elastic limit has been passed. Reinforced Concrete Beams. — The tensile strength of concrete is from 80 to 250 pounds per square inch, but the compressive strength is from 1500 to 5000 pounds per square inch. Hence a concrete beam of symmetrical section will always fail in tension long before the compressive stress reaches its ' 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. 470 Mechanics applied to Engineering. ultimate value. In order to strengthen the tension side of the beam, iron or steel rods are embedded in the concrete, it is then known as a reinforced beam. The position of the neutral axis of the section can be obtained thus — Let E = the modulus of elasticity of the rods. E„ = the „ „ „ concrete. E , r = :^ = from lo to 12. / = the tensile stress in the rods. /, = the maximum compressive stress in the concrete. a = the combined sectional area of the rods in square inches. The tension in the concrete may be neglected owing to the fact that it generally cracks at quite a low stress. It is as- sunied that originally plane sections of the beam remain plane after loading, hence the strain varies directly as the distance from the neutral axis, or i^ = - and ■'- = The total compressive force acting on the concrete is equal to the total tensile force acting on the rods, then for a section of breadth b. he" E. bi? c E or bcx - = a=r X e 2 E, which may be written — =(</■ obtained. c)ra, from which c can be The quantity bcX - is the moment of the concrete area about the neutral axis, and a— x « is the moment of the rod E. Beams. 47 1 area, increased in the ratio r of the two moduli of elasticity, also about the neutral axis. Hence if the sectional area of the rods be increased in the ratio of the modulus of elasticity of the rods to that of the concrete the neutral axis passes through the centre of gravity, or the centroid, of the section when thus corrected as shown in Fig. 464. The moment of resistance of the section (/Z) is, when considering the stress in the concrete, /o(7)(^+^) or /„©(§. -f.) and when considering the stress in the rods— fa{g-\-e) or fa{^c-\-e) The working stress in the concrete /, is usually taken from 400 lbs. to 600 lbs. per square inch, and / from 10,000 lbs. to 16,000 lbs. per square inch. For the greatest economy in the reinforcements, the moment of resistance of the concrete should be equal to that of the rods, but if they are not equal in any given case, the lower value should be taken. The modulus of the section for the concrete side is— and the corresponding moment of inertia — Similarly, in the case of the reinforced side of the section — I = aer^^c + e\ Example. — Total depth of section 18 inches; breadth 6 inches, four rods | inch diameter, distance of centres from bottom edge i| inch, the compressive stress in the concrete 400 lbs. per square inch, r ~ 12. Find the moment of resistance of the section, the stress in the rods, and the moment of inertia of the section. In this case d= 16 '5 inches, a = 0-785 square inch. — ^=(i6-5 -^)i2 X 0-785 c = 5-8 inches. e = 10-7 inches. 472 Mechanics applied to Engineering. Moment of resistance for the concrete — For the stress in the rods — ^2 X 5-8 /Xo-78s(' + 107 j= loi, 400 inch-lbs. /= 8860 lbs. per square inch The moment of inertia — /6 X 5'8'V2 X 5-8 , \ . , 4 ., ( -^ — 11 2 Y 107 I = 1470 mch -units or (0785 X 107 1470 inch*-units Fig. 465. In reinforced concrete floors, the upper portion is a part of the floor; at intervals reinforced beams are arranged as shown. Example. — ^Total depth = 24 inches, thickness of floor it) = 4 inches, breadth {b) = 22 inches, breadth of web (J>') = 7 inches. Six |" rods, the centres of which are I's" from the tension skin. Compressive stress in concrete (/„) 400 lbs. per sq. inch. /■= 12. Find the moment of resistance, the stress in the rods, and the moment of inertia. Here d = 22*5 inches a = 2'6s square inches. Position of neutral axis — 22 X 4{c - 2) -f 7^^ ^ = 2-65 X 12(22-5 - c) c=r^s' ^ = 3-15" <?= i5'35 Beams 473 Moment of resistance of concrete — \ ^(! X 7'i5 + 15-34) - (^^-^J3-i5 X ^(1 X 3'iS + iS'35)|4oo(i4oo)4oo = 560,000 in.-lbs. Stress on rods — 560,000 =/ X 2-65 X 20'I /= 10,500 lbs. square inch. Moment of inertia — 1400 X 7" 1 5 = 10,000 inch*-units or 2-65 X 15-35 X 12 X 20-I = 9810 „ „ In order to get the two values to agree exactly it would be necessary to express c to three places of decimals. For further details of the design of reinforced concrete beams, books specially devoted to the subject should be consulted. CHAPTER XII. BENDING MOMENTS AND SHEAR FORCES. Bending Moments. — When two ^ equal and opposite couples are applied at opposite ends 'y'/'/y ■'■■:■ /i of a bar in suck 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. 466, the bar ab is subjected to the two equal and opposite couples R . ac and W . be, 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. i R*W Fig. 466. W-R,+lf2 Fig. 467. Loa/i. Svpj ^ort \lioaei hence the term "bending moment." Likewise in Fig. 467 the couples are RiOf and Ra^iT, which have the same effect as the couples in Fig. 466. The bar in Fig. 466 is termed a " canti- lever." The couple R . af is due to the resistance of the wall into which it is built. The bar in Fig. 467 is termed a " beam." When a cantilever or beam is subjected to a bending moment which tends to bend it If there be more than two couples, they can always be reduced to two Stj^ tport Bending Moments and Shear Forces. 475 Fig. 469. concave upwards, as in Fig. 468 (a), the bending moment will be termed positive (+), and when it tends to bend it the reverse way, as in Fig. 468 {b), it will be termed negative (— ). Bending-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. 469 is W . ^tf ; set down from c the ordinate ct^ = 'W .be on some given scale. The bend- ing moment at d=Vf .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. 470 is obtained. The ordinate ddi represents on a given scale the bending mo- ment Ri«(f, likewise cci the bending moment Rioc or Ra^c, also ee^ the bending moment Rj>e. The reactions Ri and Rj are easily found by the principles of moments thus. Taking moments about the point b, we have — Fig. 470. R^ab = Wbc Ri = W.bc ab R, = W-R, In the cantilever in Fig. 469, let W = 800 lbs., be = 675 feet, bd = 4-5 feet. The bending moment ai c = W .be = 800 (lbs.) X 6-75 (feet) = 5400 (Ibs.-feet) Let I inch on the bending-moment diagram = 12,000 (lbs.- , /ii. r ,.\ ■ u 12000 Ibs.-feet feet), or a scale of 1 2,000 (Ibs.-feet) per inch, or j-. — j-v — . 476 Mechanics applied to Engineering. ^, , ■!• S400 (Ibs.-feet) ,. , . Then the ordinate ec, = ,,, , \ = 0*45 (inch) ' 12000 (Ibs.-feet) ^^ ^ ' I (inch) Measuring the ordinate dd^, we find it to be 0*3 inch. Then 0-3 (inch) X "°°° O^s-feet) ^ ^3600 (Ibs.-feet) bending •^ ^ ' I (inch) ( moment at a In this instance the bending moment could hav£ 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. 470), let W = 1200 lbs., ac= $ feet, dc= 5 feet, ad= 2 feet. yv .be 1200 (lbs.) X 3 (feet) R>=^r= 8 (feet) =450 lbs. the bending moment at ^ = 450 (lbs.) X 5 (feet) = 2250 (lbs.-feet) Let 1 inch on the bending-moment diagram = 4000 Ibs.- 4000 (lbs.-feet) feet), or a scale of 4000 (lbs. -feet) per mch, or )■ . > — • ™, , ■,- 2250 (lbs.-feet) , ,. , . Then the ordinate cc, = ,,, ^ J. = o'go (mch) ' 4000 (lbs.-feet ) •' ^ ' I (inch) Measuring the ordinate ddi, we find it to be 0*225 (inch). „, 0-225 (inch) X 4000 (lbs.-feet) ^ 1 900 (lbs.-feet) bending I (inch) ( moment at (/ General Case of Bending IVEoments. — Tke bending moment at any section of a beam is t}u 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. 471 is, taking moments to the left of/— or, taking moments to the right of/— Bendinsr Moments and Shear Forces. All That the same result is obtained in both cases is easily verified by taking a numerical example. Let Wj = 30 lbs., W2 = 50 lbs., W3 = 40 lbs.; ac =■ 2 feet, cd = 2*5 feet, df = i'8 feet,/^ = 2-2 feet, eb = ^ feet. W, W2 w. or f R. Fio. 471. We must first calculate the values of Ri and Ra. Taking moments about b, we have — . R,«^ = V^^cb + y^^db + ^^b „ W,<r^ + ^^b + Ws^-J ^' ^ _ 3o(lbs.) X 9-5(feet)+5o(lbs.) X 7(feet)+4o(lbs.) X 3(feet ) 11-5 (feet) \ =755 0bs-feet)^ 11-5 (feet) ^ ^ .R,=Wi + Wa + W3-R, R, =3o(lbs.)+So(lbs.)+4o(lbs.)-6s-6s (lbs.) = 54-3S (lbs.) The bending moment at/, taking moments to the left of/, = Ri«/-Wie^-W3^ = 65-65 (lbs.) X 6-3 (feet) - 30 (lbs.) X 4"3 (feet) - 50 (lbs.) X I -8 (feet) = 194-6 (lbs.-feet) The bending moment at/, taking moments to the right oif, = R,bf- Wsef = 54'3S (lbs.) X 5-2 (feet) — 40 (lbs.) X 22 (feet) = 194-6 (lbs.-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. 478 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. 472 be loaded until it fails. It 0, ., 1 i i, y':-::^ 11 1 1 II w r/ Fia. 472. Fig. 473. 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 dV, is of the same value, and equal to W. In the case of the beam in Fig. 473, 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 h and c is of the same value, and equal to Rj, 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. FioC'&d ' 1 1 FlG. 474. Bending Moments and Shear Forces. 479 When the sheared part slides in a I clockwise \ '^ t contra-clockwise J direction relatively to the fixed part, we term it a (positive {A-) shear) (negative (— ) shear) 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. 472, 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. 47 3, the shear at all sections between a and c is constant and equal to Rj. It is also positive (-[-), 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 Rj, 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 Rj — R^. 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. 475 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 „ rf=-W, „ W, .. ,. .. c „ d= -t-Wj „ W, 48o Mechanics applied to Engineering. Construct the diagrams separately for each shear as shown, then combine by superposing the — diagram on the + diagram. The unshaded portion shows where the — shear neutralizes the «S W, W m> m iili »i III III 1 w H'+MJ-Mf" iiiiiiiiiiiiwiMiiiiiiir yry_ Fig. 475. IV + shear ; then bringing the + portions above the base-line and the — below, we get the final figure. Bending Moments and Shear Forces. 481 Resultant shear at any section — Between To the right. To the left. a and b W -W, + W^ - (W + W, - W,) = -W b and c w-w, W, - (W + W, - W,) = -(w - W.) (T and a Wj - w, + w orW+W,-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 Wj 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 Bending-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 Wj, W^, W,, — W4, — Wb at distances l^, 4 4i h, h respectively from any given section a, as shown in Fig. 476. The bending moment at a is = W/a + W3/3 — W4/4 or — Wj^ But Wa/j is the area of the shear diagram due to W, between W, and the section a, likewise Wg/j is the area between Wj and a, also — W4/4 is the area between — W4 and a. The positive areas are partly neutralized by the negative areas. The parts not neutralized are shown shaded. The shaded area = Wa4 + Ws4 — W4/4, 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 2 I 482 Mechanics applied to Engineering. - W/i + Wj/j, i.e. to the bending moment at a. Hence we get this relation — The bending moment at any section of a freely supported beam is equal to the area of the shear diagram up to that point. The bending moment is therefore a maximum where the shear changes sign. w. Wz IV3 ' ^s- j:1\ ± ' - h h Fig. 476. Due attention must, of course, be paid to positive and negative areas in the shear diagram. To make this quite clear, we will work out a numerical example. In the figure, let W, = 50 lbs. A = i foot Wj = 80 lbs. 4=2 feet W, = 70 lbs. /, = 4 feet By moments we f W4 = ■s.%2-7, lbs. /« = 5 feet find (W. = 67-8 lbs. 4 = 4 feet The figure is drawn to the following scales — Length i inch = 4 feet load I inch =160 lbs. hence i square inch on the shear diagram = 4 (feet) x 160 (lbs.) = 640 (lbs.-feet) Bending Moments and Shear Forces. 483 The area of the negative part of the shear diagram below the base-line is — o'4oi sq. inch, and the positive part above the base-line is 0*056 sq. inch; thus the area of the shear diagram up to the section a is — o'4oi + o"o56 = 0*345 sq. inch. But I sq. inch on the shear diagram = 640 (Ibs.-feet) bending moment, thus the bending moment at the section a = o*345 X 640 = 221 (Ibs.-feet). The area of the shear diagram to the left of a = o'345 sq. inch, i.e. the same as the area to the right of the section. As a check on the above, we will calculate the bending moment at a by the direct method, thus — The bending moment at a = W^^ + W/j — W/4 = 80 (lbs.) X 2 (feet) + 70 (lbs.) X 4 (feet) - 132-2 (lbs.) X 5 (feet) = 221 (Ibs.-feet) which is the same result as we obtained above from the shear diagram. This interesting connection between the two diagrams can be shown to hold in all cases from load to slope diagrams. If a beam supports a distributed load, it can be represented at every part of the beam by means of a diagram whose height is proportional to the load at each point ; then the amount of load between the abutment and any given point is proportional to the area of the load diagram over that portion of the beam. But we have shown that the shear at any section is the algebraic sum of all the forces acting either to the right or to the left of that section, whence the shear at that section is equal to the reaction minus the area of the load diagram between the section in question and the said reaction. We have already shown the connection between the shear and bending moment diagrams, and we shall shortly show that the slope between a tangent to the bent beam at any point and any other point is proportional to the area of the bending-moment diagram enclosed by normals to the bent beam drawn through those points. 484 Mechanics applied to Engineering. Cantilever with single load at free end. Fig. 477. Cantilever with two loads. W*w, Fig. 478. Bending Moments and SJiear Forces. 485 Bending moment M in ll».-incheft = W(lbs.)X/(in.) M, = W/, M = (/. mn at any section where d = depth of beriding- moment diagram in inches Depth of bend- in^-moment dlagnun In mches. Scale of W, m lbs. = 1 indi. Scale of/, -full size. W/ mn mn Remarks. The only moment acting to the right of X is W/, which is therefore the bending moment at x. Likewise at^. The complete statement of the units for the depth of the bending-moment diagram is as follows : — xn(lbs.) = i inch on diagram, or -^. — ^ ^ ' ^ ' l{mch) «(in.)=l „ „ W (lbs.) /,(inches) W/ m (lbs.) I inch n (inches) mti (inches) M = 1/ . mn mn This is a simple case of combining two such bending-moment diagrams as we had above. The lower one is tilted up from the diagram shown in dotted lines. 486 Mechanics applied to Engineering. Cantilever with an evenly distri- buted load of w lbs. per inch run. Cfg. of loads Hendiuruf TTvoTnentff apfix- Bending Moments and S/iear Forces. 487 Bending moment M in lbs.-inche8. Mx = iv? w(\hi, ) ^(inches)' inches 2 (constant) - ^.p (Ibs.-inches) constant Let W = o)/ Depth 3f bend- inc-moment diagram in inches. Scale of r/, m lbs.=si inch. Scale of/, - full size. 2 M = 1/ , mn W/ 2mn Remarks. Tn statics any system of forces may always be replaced by their resultant, which in this case is siluatedat the centre of gravity of the loads ; and as the dis- tribution of the loading is uniform, the resultant acts'^t a distance - from x. 2 The total load on the beam is wl, or W ; hence the bending moment at * / wl' = wl X — = . At any other sec- 22 tion, V = wl. X - = — =-, ■^ '2 2 Thus the bending moment at any section varies as the square of the distance of the section from the free end of the beam, therefore the bending moment dia- p;ram is a parabola. As the beam is fully covered irith Ioad.s, the sum of the forces to the right of any section varies directly as the length of the beam to the right of the section ; therefore the shearing force at any section varies directly as the distance of that section from the free end of the beam, and the depth of the shearing-force diagram varies in like manner, and is therefore triangular, with the apex at the free end as shown, and the depth at any point distant /, from the free end is wl,, i.e. the sum of the loads to the right of /,, aird the area of the shear diagram up to that point is ^?i2iZ« = !<, ,-.,. •^ 2 2 the bending moment at that point. 488 Mechanics applied to Engineering. Cantilever irregu- larly loaded. Fig. 480. Beam supported at both ends, with a central load. Fig. 481. Bending Moments and Shear Forces. 489 Bending moment Mm lbs.-inches. M«=W/+W,/, "*" 2 M = </ . mn Depth of bending' moment diagram in inches. Scale of W, m lbs. ^ X inch. Scale of/, - full siM. to/." Remarks. This is simply a case of the combina- tion of the diagrams in Figs. 478 and 479. 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. w/ W/ M«_ ^ ^mn M = </ . ««« Each support or abutment takes one- W half the weight = — • The only moment to the right or left . W / W/ of the section a: is — x - = — • 224 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. 490 Mechanics applied to Engineering. Beam supported at both ends, with one load not in the middle of the span. Fig. 482. Beam supported at both ends, with two symmetrically placed loads. R'W Fio. 483, Bending Moments and Shear Forces. 491 Bending moment Min Ibs.-inchcBi W M.= y(AA) M = </ . mn Depth of bending- moment diagram in inches. Scale of W, It lbs. = I inch. Scale of /p - full size. Imn Remarks. Taking moments about one support, we have R,/ = W/,, or R, = H4. The bending moment at x — W. M. = R,/, = -^(/,4) M, = W4 M„ = W/. M = </, mn W4 mn The beam being symmetrically loaded, each abutment takes one weight = W = R. The only moment to the right orthe right-hand section ;ir is W . 4 ; likewise with the left-hand section. At any other section y between the loads, and distant /v from one of them, we have, taking moments to the left of y, R{4 + /,) - W . /, = R . 4 + R . /, - R . 4 = R . 4 01 W . 4, ».«. the bending moment is constant between the two loads. The sum of the forces to the right or left aiy_ = W — R = o, and to the right of the right-hand section the sum of the forces = R ,= W at every section. 492 Mechanics applied to Engineering. Beam supported at both ends, load evenly distributed, w lbs. per inch run. Fig. 484. Bending Moments and Shear Forces. 493 Bending moment M in lbs.-incbes. Let W = a// Mx = 8 N.B.— Be very caieful to reduce the distributed load to pounds per inth tun if the dimen- sions of the beam are in inelus. Depth of bending- moment diagram in inches. Scale ofW, Mlbs. s 1 inch. Scale of/, — full WBt. Remarks. As in the case of the uniformly loaded cantilever, we must replace the system of forces by their resultant. The load being symmetrically placed, the abut- ments each take one-half the load = — • 2 Then, taking moments to the left of *, we have — wl wl I 2 ^a 2^4- wP 8 wl The — shown midway between x and the abut- ment is the resultant of the loads on half the beam, acting at the centre of gravity of the load, viz. - 4 from X, or the abutment. , The bending moment at any other section y, distant I, from the abutment, is : taking moments to the right oty — wl , , ly '"'ly,. =f<4««.?(^) where I,' = I — /,. Thus the bending moment at any section is proportional to the product of the segments into which the section divides the beam. Hence the bending-moment diagram is a parabola, with its axis vertical and under the middle of the beam as shown. The forces acting to the right of the section x = =^0 ; i.e. the shear at the middle section 2 2 is zero. Atthe section>=i»/,- ^ = w (/,--). Hence the shear varies inversely as the distance from the abutment, and at the abutment, where 4 = o, it is wl 494 Mechanics applied to Engineering, Beam supported at two points equi- distant from the ends, and a load of w lbs. per inch run evenly distri- buted. PCTTTTYTXTY^ I ^ y f ] *■■■ ij >i ^/ -11-. ■-^--> J\re^ative B. M. thteto dvorhanffTr^ lo ads Positive B. If^eh^e^^centred- ^euv ComhmeA B. M. SJvear Fio. 485. Bending Mommts mid Shear Forces. 495 Bending moment Depth of bend- inff-moment dUsnun Ibs.-inches. in inches. Scale of W fff Ibfc. =11 tncn. Scale of /, i full siie. n Remarks. The bending-moment diagram for the loads on the overhanging ends is a com- bination of Figs. 479 and 483, and the dia- gram for the load on the central span is simply Fig. 484. Here we see the im- portance of signs for bending moments. The beam will be subject to the smallest M.= «"'• bending moment when M, = M, ; or when Mx 2 ivl^ _ a//," _ w4» mn mn 282 /, = 283/, But /, + 2/, = / substituting the value\ _ -, , ,, _ , of/, above 1 - 2 83/, + 2/, - / or/ =4-83/, or say /, = \f for the conditions of maxi- mum strength of the beam. The shear diagram will be seen to be a combination of Figs. 479 and 484. 496 Mechanics applied to Engineering. Beam supported at each end and irregularly loaded. FlQ. 486. Bending Moments and Shear Forces. 497 Bending Min Ibs.-inches. Depth of bend' ing-moment diagram in inches. Scale of W. M lbs.=i inch. Scale of/, — full size. Msid , mn mn or mn etc Remarks, The method shown in the upper figure is simply that of drawing in the triangular bending- moment diagram for each load treated separately, as in Fig. 481, then adding the ordinates of each to form the final diagram by stepping off with a pair of dividers. In the lower diagram, the heights ag,gh, etc., are set off on the vertical drawn through the abut- ment = W,/„ Wj/j, etc., as shown. The sum of these, of course, = R,/. From the starting- point a draw a sloping line ai, cutting the vertical through W, m the point b. Join gb and produce to e, join he and produce to d, and so on, till the point / is readied ; join fa, which completes the bending-moment dis^am abcdcf, the depth of which in inches multiplied by mn gives the bending moment. The proof of the construction is as follows : The bending moment at any point * is R,4 — W^^,. On the bending-moment diagram -^= j-; or R./X4 K/ = ^ = = =R,/. 0/ if X /. W,/, X r. = W/. and the depth ofl the bending-mo- > = KO = K/ - 0/ = R/, - W^r, ment diagram ) It will be observed that this construction does not involve the calculation of R, and R,. For the shear diagram R, can be obtained thus : Measure off of in inches ; then -^—z = R^, where / is the actual length of the beam in inches. 49^ Mechanics applied to Engineering. Beam supported at the ends and irregularly loaded. Bending Moments and Shear Forces. 499 Bendtne moment Min IbS'-incbes. M = )n.n(Dx Ok) Remakks. Make the height of the load lines on the beam propor- tional to the loads, Tu. — i. W, etc., inches. Drop perpendiculars through each as shown. On a vertical fb \V W set ofif y!r = — , ed = — ?, etc. Choose any convenient point O distant Oh from the vertical. Ok is termed the "polar distance." JoinyO, <0, etc. From any pointy on the line passing through R, draw a line/m parallel to _/0 ; from m draw mK parallel to eO, and so on, till the line through R, is reached in g. Join gf, and draw Oa on the vector polygon parallel to this last line ; then the reaction R, =_/a, and Rj = ia. 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 jr^wj and Oaf axe similar, for Jm is parallel to/D, axA jp to aO, and/w io fa ; also/? is drawn at right angles to the base mp. Hence — height of A ipm _ base of A iiim height of ^ Oaf base of ^ Oaf or^ = OA mp af . h -Ok ••D.-R-, For 7^ = /i and af^ R, ; and let mp = D^, i.e. the depth of the bending-moment diagram at the section x, or R,/, = D,OA = M, = the bending moment at x. By similar reasoning, we have — R,4 = rf)t.Oh also \V,(4 - /,) = rlC X Oh the bending moment at^ = M,, = Rj/j — W,(/j — /,) = Oh{,rt - rK) = 0/5(K/) = OA.D, where D, = the depth of the bending-moment diagram at the section J/. 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 lbs, I „ „ length „ = » inches. Hence the measurements taken from the diagram in inches must be multiplied by mn. The bending moment expressed \ „ ,t^ ..,,, in lbs. .incites at any section ) = M = r« . « . (D . 0-5) Soo Mechanics applied to Engineering. Beam sup- ported at two points with overhanging ends and irre- gularly loaded. Bending Moments and Shear- Forces. 501 Bending moment Min lbs. -inches. where D is the depth of the diagram in inches at that section, and Oh is the polar distance in inches. In Chap. IV. we showed that the resultant of such a system of parallel forces as we have on the beam passes through the meet of the first and last links of the link polygon, viz. through u, where /»« cuts gh. Then, as the whole system of loads may be replaced by the resultant, we have Rj/j = Rj/,. But we have shown above that the triangles juw and Oaf are similar ; hence ^ = — ^. \jh of But jv — /j, therefore af x l, = Oh x uw =: Rj/j, or af— R,. Similarly it may be shown that ab = R^. M = »j. «(Dx 0/0 The loads are set down to the proper scale on the vector polygon as in the last case, A pole O is chosen as before. The vertical load lines are dotted in order to keep them distinct from the reaction lines which are shown in full. Starting from the point/ on the reaction line Ri, a line jm is drawn parallel to oO on the vector polygon, from m a line mk (in the space i) is drawn parallel to bO, and so on till the point u is reached, from « a line is drawn parallel to fO to meet the reaction line R2 in the point g. Join Jg. From the pole O draw a line Oj parallel lojg. Then ai gives the reaction R„ and if the reaction Rj. The bending moment diagram is shown shaded. The points where the bending moment is zero are known as the points of contrary flexure. The construction of the shear diagram will be evident when it is remembered that the shear at any section is the algebraic sum of the forces to the right or to the left of the section. The bending moment at any section of a beam loaded in this manner can be readily calculated. The reactions must first be found by taking moments about one of the points of support. The bending moment at any section X distant 4 from the load ^is Mx = Tdli + 7el, - R,/, + rf/x and the distance ig of the point of contrary flexure from the load ef is obtained thus /« = d^h - R.A " "" — ^ . -r- ef+de- Rj 502 MecJianics applied to Engineering. Beam supported at each end and loaded with an evenly distributed load of TV lbs. per inch run over a part of its length. Beam supported at each end with a distributed load which varies directly as the dis- tance from one end. Bending Moments and Shear Forces. 503 BendinK moment M in Ibs.-incbes. MnM.= — P Remarks. Remembering that the bending moment at any section is equal to the area of the shear diagram up to that section, the maximum bending moment will occur at the section where the sliear chapges sign. R.= 2a/4' / R. 2/,R, _ 2/.R. _ 2/,/, Kj + R, 2wL I V.^x _ 2w4V,« 2 p ^' ' max. — — M M,. \\P_ 9V3 Let ia, be the intensity of loading at any point distant /, from the apex of the load diagram. p. w H' The shear at this point = I^|— I ■U'^ll = Ri — ,- ( /»<// Jo 'Jo 6 W/.' 2/ Therefore the shear diagram is parabolic The shear is zero when — 6 ■2.1 or when /, I The maximum bending moment occurs at the section where the .shear changes sign, and is equal to the area of the shear curve ; hence — / ^P M„ «/3 9^/3 For another method of arriving at this result, see p. 185 504 MecJianics applied to Engineering. Beam built in at both ends and centrally loaded. Ditto with evenly distributed load Cantilever propped at the outer end with evenly distributed load. Beam built in at both ends, the load applied on one of the ends, which slides paral- lel to the fixed end. (jgS -I \ Fig. 491. ^ i*^ o^ ^p Fig. 492. l^fp. Fig. 493. ■ rv : | i W'/f Fig. 494. Bending Moments and Shear Forces. 505 Bending moment Mm Ibs.-!nches. M,= W/ M.= 24 Mx ~ 128 tap M„ = M,= — M, = The determination of these bending moments depends on the elastic properties of the beams, which are fully discussed in Chap. XIII. In all these cases the beam is shown built in at both ends. The beams are assumed to be free endwise, and guided so that the ends shall remain horizontal as the beam is bent. If they were rigidly held at both ends, the pioblem would be much more complex. CHAPTER XIII. DEFLECTION OF BEAMS. Beam bent to the Arc of a Circle. — Let an elastic beam be bent to the arc of a circle, the radius of the neutral axis being p. The length of the neutral axis will not alter by the bending. The distance of the skin from the neutral axis = y. The original length of ) _ the outer skin != 27rp Fig. 495. the length of the outer i ^ ^, . skin after bending ) ^ ■" the strain of the skin ) _ , / i \ due to bending C ^^^ ^' ° ' — 2irp = 27ry But we have (see p. 373) the following relation : — strain _ stress original length modulus of elasticity 2'^' _ > _ / or ■ 27rp But we also have — / = M Substituting this value in the above equation — p EZ whence M = — ^ ; or M = — P P Central Deflection of a Beam bent to the Arc of a Circle. — From the figure we have — Deflection of Beams. ,-^ = (p-8)^+(L^y S07 whence 2p8 — 8' = — 4 The elastic deflection (S) of a beam is rarely more than -p^ of the simn (L) ; hence the 8* ■will not exceed —^ , which is quite 360,000 ^ negligible ; T a hence 2pS = — 4 But p=^l 8 = 8p /if V hence 8 = ML* 8EI \Vc shall shortly give another method for arriving at this result. General Statement regarding Deflection. — In speaking of the deflection of a cantilever or beam, we always mean the deflection measured from a line drawn tangential to that part of the bent cantilever or beam which remains parallel to its unstrained position. The deflec- tion 8 will be seen by referring to the figures shown. The point / at which the tan- gent touches the beam we shall term the " tangent point." When dealing with beams, we shall find it convenient to speak of the deflec- tion at the support, «.& the height of the support above the tangent. Deflection of a Cantilever. — Let the upper diagram (Fig. 498) represent the distribution of bending moment acting on the cantilever, the dark line the bent cantilever, and the straight dotted line the unstrained position of the cantilever. Consider any very small portion ^j/, distant /, from the free end of the cantilever. We will suppose the length jy so small that the radius of curvature p, is the same at both points, y,y. Let the angle subtending yy be 6, (circular measure) ; then the angle So8 Miclianics applied to Engineering. between the two tangents ya, yb will also be 6,. Then the deflection at the extremity of these tangents due to the bending between ^/.j* is— o» = ^- ^ -yy Bute, =-^-^ Pi and from p. 424, we have — '? EI where M, is the mean bending moment between the points y^y- , . . Then by substitution, we have — ^'~ EI where Q, is the "slope" be- tween the two tangents to the bent beam at^_j'; But Mjj'j' = area (shown shaded) of the bending-moment diagram between y, y hence ^, = .gr? Fig. 498. and 8, : A/, EI That is, the deflection at the free end of the cantilever due to the bending between the points y, y is numerically equal to the moment of the portion of the bending-moment diagram over yy about the free end of the cantilever divided by EI. The total deflection at the free end is — 8 = S(8, + 8. -1-, etc.) 8 = ^5A^, + A.4 -f , etc. where the suffix x refers to any other very small portion of the cantilever xx. Thus the total deflection at the free end of the cantilever is Deflection of Beams. 509 numerically equal to the sum of the moments of each little element of area of the bending-moment diagram about the free end of the cantilever divided by EI. But, instead of dealing with the moment of each little element of area, we may take the moment of the whole bending-moment diagram about the free end, i.e. the area of the diagram X the distance of its centre of gravity from the free end ; or 8 = EI where A = the area of the bending-moment diagram ; Lc = the distance of the centre of gravity of the bending- moment diagram from the free end. To readers familiar with the integral calculus, it will be seen that the length that we have termed yy above, is in calculus nomenclature dl in the limit, and the deflection at the free end due to the bending over the elementary length dl is — M„. /„.<// _ 8,= £1 and the total deflection between points distant L and o from the free end is— « = ^I Ml.dl (ii.) where M = the bending moment at the point distant / from the free end. Another calculus method commonly used is as follows. The slope of the beam between P and Q (the distance PQ is supposed to be infinitely small) dy is denoted by — . This ratio dx is constant if the beam is straight, but in bent beams the slope varies from point to point, and the change of slope in a given length dx is — Fig. 499- \dx' dx dx^' Sio Mechanics applied to Engineering. When Q is very small dx becomes equal to the arc subtending the angle Q, and p? = Pq = p, then -j- = — -^, in the limit PQ = dx, and the change in slope in the length dx is <?) . dx M '' ds^ (iii.) This expression will be utilized shortly for finding the deflection in certain cases of loaded beams. Case I. — Cantilever with load W on free end. Length L. Method (i). Fig. 50a A = WL X - = 2 2 S = 2 WL' 3EI EI Method (ii.). — By integration M = W/ W hence 8 = — -, EI '^ = ^' WL' /=o 3EI Method (iii.). — Consider a section of the beam distant x from the abutment. d}v The bending moment M = W(L - a;) = El-^ L — a; = Integrating EI^ "iN d£- L.-^Vc = |f 2 Yl dx where C is the integration constant. When x = o the slope dy -r is also zero, hence C = o. ax Deflection of Beams. S" Integrating again — L^= x"^ , ^^ EI 2 6 W When * = o, the deflection y is zero, hence K = o and _ WL^ _ ^\ •''" E[\ 2 6/ which gives the deflection of the beam at any point distant x from the abutment. An exactly similar expression can be obtained by method (i.). The deflection S at the free end of the beam where :r = L is — '=iEI In this particular case the result could be obtained much more readily by methods (i.) and (ii.). Case II. — Cantilever with load W evenly distributed or w per -unit length. Length L. Method (i.)— XfLx^ 8 = wL* 8EI 8EI or by integration — Method (lu) M = hence S = /=L l\, wJJ 512 Mechanics applied to Engineering. Method (iii.) — Consider a section distant x from the abutment. rr^, , ,• ,, K'fL — xf „,d^y The bending moment M = = El-j-5 w doc r, . x' zL^c" , ^ 2EI dy Integratms Ux -i f- C = r 32 w dx For the reason given in Case I, C = o . LV , «* L^ , ^ 2EI Integrating again f- K = y ° 2123 w as explained above K = o, ^ = 2lEi^^^'^ + *'-4^^^ •which gives the deflection of the beam at any point distant x from the abutment. This is an instance in which the value of J* is found more readily by method (iii.) than by (i.) or (ii.). The deflection 8 at the end of the beam where x='L is — _ w\} _ WL° 8E1 ~ 8EI Case III. — Cantilever with load W not at the free end. A = WL. X ^ = ^- 3 3 L, = L - ^ 3 2EI\^ 3 Fig. 502. •' N.B. — The portion db is straight. 8 = S(l-L') Deflection of Beams. 513 Case IV. — Cantilever with two loads Wi, Wa, neither oj them at the free end. (-^) 2EI \- 3 , WaL,' Wj Fig. S03. Case V. — Cantilever with load unevenly distributed. Length L. Let the bending-moment diagram shown above the canti- lever be obtained by the method shown on p. 487. Then if i inch = m lbs. on the load scale ; I inch = n inches on the length scale ; D = depth of the bending-moment diagram measured in inches ; OH = the polar distance in inches ; M =fthe bending-moment in inch-lbs.; M = w.w.D.OH; hence i inch depth on the bending-moment diagram represents M . =r = m .n . OH mch-lbs. j and i square inch of the bending- moment diagram represents tn .rfi . OH inch-inch-lbs. \ hence — f area of bending-moment N j ^;^ ^ - V diagram in squa re inches y ^ • " ' " ^ ^ ^« 8= El 2 L 514 Mechanics applied to Engineering. The deflection 8 found thus will be somewhere between the deflection for a single end load and for an evenly distributed load ; generally by inspection it can be seen whether it will approach the one or the other condition. Such a calculation is useful in preventing great errors from creeping in. In irregularly loaded beams and cantilevers, the deflection cannot conveniently be arrived at by an integration. Deflection of a Beam freely supported. — Let the lower diagram represent the distribution of bending moment on the beam. The dark line represents the bent beam, and the straight dotted line the unstrained position of the beam. By the same process of reasoning as in the case of the cantilever, it is readily shown that the deflec- tion of the free end or the support is the sum of the moments of each little area of the bend- ing-moment diagram between the tangent point and the free end about the free end ; or, as before, instead of dealing with the mo- ment of each little area, we may take the moment of the whole area of the bending- moment diagram between the free end and the tangent point, about the free end, i.e. the area of the bending-moment diagram between the tangent point and the free end X distance of the centre of gravity of this area from the free end. Then, as before — J^eeJBrul Fig. 505. AL. EI where A and L, have the slightly modified meanings mentioned above. Deflection of Beams. 515 Case VI. — Beam loaded with central load W. Length L. A=^xt L,=- - ...1- \ .■? EI 8 = 4 WL3 Fig. 506. 48EI Or by integration, at any point distant / from the support — M = — i 2 J ~ EI I 2 EI V6 z' <^ n When / = — , we have — 2 WLs 8 = . WL» 2" X 6EI 48EI Case VII. — Beam loaded with an evenly distributed load w per unit length. Length L. 8 = ^'xSjxf^ 16 EI S= 5^L^ _ SWL' 384EI 384EI Fig. 507. Or by integration, at any point distant / from the support (see p. 512) the bending moment is — M=^(/L-/^)=^-^ 2^ 82 where x is the distance measured from the middle of the beam and y the vertical height of the point above the tangent at the middle of the beam. Then by method (iii.) — d'y wU wx 2 5i6 Mechanics applied to Engineering. ^^^ = "8 6" + ^ 16 •w (C = o) (K = o) . so/L* sWL' ^ L o = - „, = o „T when * = - 384EI 384EI 2 Case VIII. — Beam loaded with two equal weights symmetri- cally placed. — By taking the *f W moments ofthe triangular area ._^_ i^ _:^ abc and the rectangle bced, the --Z/— ► — i.^—^y—-i---, deflection becomes — and when L, = L2 = — , this 3 c e Fig. 508. expression becomes — „ 23WL» WL» - , V ^ = Wl = iSEI <"^"'y> or if Wo be the total load — Case IX. — Beam loaded with one eccentric concentrated load. . L W L, Fig. 509. It should be noted that the point of maximum deflection does not coincide with that of the maximum bending moment. We have shown that the point of maximum bending moment in a beam is the point where the shear changes sign, and we Deflection of Beams. 517 shall also show that the bent form of a beam is obtained by constructing a second bending-moment diagram obtained by taking the original bending-moment diagram as a load diagram. Hence, if we construct a second shear diagram, still treating the original bending-moment diagram as a load diagram, we shall find the point at which the second shear changes sign, or where the second bending moment, i.e. the deflection, is a maximum. This is how we propose to find the point of maximum deflection in the present instance. Referring to p. 483, we have — The shear at a section | _ t> _ ^^ distant / from Ri j ^ 2L1 =^.=^(^+^7) + ML = The shear changes sign and the deflection is a maximum when — ML/ LiN ML/ M/2 orwhen/=^^\L.+\)+% = L^ 3^ 3L ^3 where Lj = «L and La = L — Lj, i.e. the shorter of the two segments ; and the deflection at this point is — 3EI 3EIL The deflection 8 under the load itself can be found thus — Let the tangent to the beam at this point o be »«; then we have — But these are the deflections measured from the tangent vx. Let S = slope of the tangent vx; then uv = SLj, and xy = SL2, and the actual deflection under the load, or the vertical distance of o from the original position of the beam, is — 8 = 81 — «?/, or 8 = 8a -f xy whence W_sL^ = 5^V SI. 5i8 Mechanics applied to Engineering. 3EI V Lj + La / and 8 = 5^' + J-ig / ixiJji — K-2i-A2 3EI ' 3EU L, + L, ) which reduces g _ WLi'La' _ WW 3EI(La + L,) 3EIL Case X. — Beam hinged at one end, free at the other, propped in the middle. Load at the free end. The load on the hinge is also equal to W, since the prop is central, and the load on the prop is 2W; hence we may treat it as a beam sup- ported at each end and ^^ centrally loaded with a load zW. Hence the 2WL' central deflection would be ■ if the two ends were kept level ; but the deflection at the free end is twice this amount ; or — 8 = 4WL' ^ WL' 48EI 12EI Case XI. — General case of a learn whose section varies from point to point. (See also p. 270.) (i.) Let the depth of the section be constant, and let the breadth of the section vary directly as the bending moment ; then the stress will be constant. We have — ,, /I EI / 1 M = ■'^ = — , or ^ = - y 9 ^y p But, since/, y, and E are constant in any given case, p is also constant, whence the beam bends to the arc of a circle. (ii.) Let both the depth of the section and the stress vary ; then <- = - if V varies directly as /, -t will be constant, and y 9 y the beam will again bend to the arc of a circle. Deflection of Beams. 519 From p. 507, we have- 8EI g - 3WL^ 8EB^ 32EI Let the plate be cut up into strips, and bring the two long edges of each together, making a plate with pointed ends of the same form as plate i on the plan ; pack all the strips as shown into a symmetrical heap. Looked at sideways, we see a plate railway spring. t: 1 > Fig. 511 Let there be n plates, in this case 5, each of breadth b\ then B = nb. Substitute in the expression above — s _ 3WL» 8E«,5/» If a railway-plate spring be tested for deflection, it may not, probably will not, quite agree with the calculated value on account of the friction between the plates or leaves. The result of a test is shown in Fig. 512. When a spring is very rusty it deflects less, and when unloading more than the formula gives, but when clean and well oiled it much more closely agrees with 520 Medianics applied to Engineering. the formula, as shown by the dotted lines. If friction could be entirely eliminated, probably experiment and theory would agree. In calculating the deflection of such springs, E should be taken at about 26,000,000, which is rather below the value for the steel plates liiemselves. Probably the deflection due to shear is partly responsible for the low modulus of elasticity, and 'V 6 8 10 Locul on Spring 12 from the fact that the small central plate (No. 5) is always omitted in springs. Case XII. Beam unevenly loaded. — Let the beam be loaded as shown. Construct the b ending-moment diagram shown below the beam by the method given on p. 498. Then the bending moment at any section is M = »«.«. D . OH inch-lbs., using the same notation as on p. 499. Then i inch on the vertical M scale of the bending-mpment diagram = — = »«.«. OH inch-lbs., and i inch on the horizontal scale = n inches. Hence one square inch on the diagram = m . «'0H inch-lbs. Then A = a.m. «^0H, where a = the shaded area measured in square inches. The centre of gravity must be found by one of the methods Deflection of Beams. 521 described in Chap. III. Then L„ = « . 4 where /„ is measured in inches, and the deflection — AL, EI a.m. «^0H . /, EI It is evident that the height of the supports above the tangent is the same at both ends. Hence the moment of the areas about the supports on either side of the tangent point must be the same. The point of maximum deflection must be found in this way by a series of trials and errors, which is very clumsy. The deflection may be more conveniently found by a somewhat diflferent process, as in Fig. 514. We showed above that the deflection is numerically equal to the moment of each little element of area of the bending- moment diagram about the free end -i- EL The moment of Fig. 513- Jtefledian, Curve Fig. 514. each portion of the bending-moment diagram may be found readily by a link-and-vector polygon, similar to that employed for the bending-moment diagram itself. Treat the bending-moment diagram as a load diagram ; split it up into narrow strips of width x, as shown by the dotted lines ; draw the middle ordinate of each, as shown in full lines : then any given ordinate x by « is the area of the strip. Set down these ordinates on a vertical line as shown; choose a 5^2 Mechanics applied to Engineering. pole O', and complete both polygons as in previous examples. The link polygon thus constructed gives the form of the bent beam ; this is then reproduced to a horizontal base-line, and gives the bent beam shown in dark lines above. The only point remaining to be determined is the scale of the deflection curve. We have i inch on the load scale of the] first bending-moment diagram also I inch on the length of the bending-1 _ moment diagram j ~ "^ mches and the bending moment at any point M = m .n.Ti . OH ' I = OT lbs. where D is the dept h of the bending-moment diagram at the point in inches, and OH is the polar distance, also expressed in inches. Hence i inch depth of the bending-moment diagram represents Y=r-= wz« . OH inch-lbs., and i square inch of the bending-moment diagram represents ot«*OH inch-inch-lbs. Hence the area xQ represents arDww^OH inch-inch-lbs. ; but as this area is represented on the second vector polygon by D, its scale is xmn'OH. ; hence — s «w«='OHmDiOiHi 8 = Ei EI If it be found convenient to reduce the vertical ordinates of the bending-moment diagram when constructing the deflection vector polygon by say -, then the above expression must be multiplied by r. The following table of deflection constants k will be found very useful for calculating the deflection at any section, if the load W is expressed in tons, E must be expressed in tons per square inch. The length L and the moment of inertia I are both to be expressed in inch units. Example. — A beam 20 feet long supports a load of 3 tons at a point 4 feet from one support : find the deflection at 12 feet from the same support. I = 138 inch^-units. E = 12,000 tons per square inch. The position of the load is ^ = o'2. Deflection of Beams. 523 H < a o U o z Q a Q H PS O pEOj JO nopisoj b b b b ON b i. ^ O r^ u »> =" S 01 < o w 9 N O o Q H W H O s <; o g M t^ N u^ iri w ':^ r^ NO ON b M CO ■ei- 10 NO NO NO ■<d- CNl 8 8 8 8 8 8 8 8 b b b b b b b b b « fo r^ t^ CO VI t^ 00 ro NO 00 M M -^ 8 kH •H M p b b b b b b b b >f2 r-* ■* NTl 10 f^ t^ S" 00 w m NO NO ■^ NO b 13 b b b b b b b b b S 3 a \r\ r^ NO M I'l Tj- f LO li^ 00 On On VO w ND HH M m b p p ^0 b b b b b b b b b '■C u •« „ 00 i/i NO 00 NO U-) 00 N* VI ■s NO ON ON VO NO M b p p ^ •1 b b b b b b b b b *J > ■g 2 vo N NO t^ NO ^ « NO On ON 00 iri 10 Ti* M tS b P p p P P p: b b b b b b b b +j (2 r^ ITN NO ■^ r^ NO vo *:(- NO NO NO w GO ti- CO 8 hH >-4 b p 3 p p b b b b b b b b b r^ u^ ■ 00 t^ r^ fO w 00 M 00 NO m C4 8 p b b b b b b b b b b NO t^ ^ <o LO w 1^ « NO NO NO NJ~) ■^ ro i-i 8 b b b b b b b b b b puoj JO UOlJtSOJ b b b On b 524 Mechanics applied to Engineering. The position at which the deflection is measured = ^ = 0-6. Referring to the table, ^ = 0-0107. Then 8 = 0-0107 X 3 X 240 = 0-24 inch. 12,000 X 138 When a beam supports a number of loads, the deflection due to each must be calculated and the results added. When the loads are not on the even spaces given in the table, the constant can be obtained approximately by interpolation or by plotting. A very convenient diagram for calculating the deflection of beams has been constructed by Mr. Livingstone; it is pub- lished by, "The Electrician" Printing and Publishing Co., Fleet Street, London. Another type of diagram for the same purpose was published in Engineering, January 13, 1913, p. 143. Example. — A beam 20 feet long, freely supported at each end, was loaded as follows : — Load in tons (W). Distance rrom end of beam. Position of load. K . KW 3 S 2 4 3' 6" 7' 6" 11' 8" IS' a" 0-175 0-37S 059 0-76 0-0095 0-0177 0-0194 0-0139 0-0285 0-0885 0-03S8 0-0556 0-2114 Find the deflection under the 2 tons load, tons per sq. inch. I = 630. E = 12,000 8 = 0-2114 X 240' I2-000 X 630 = 0-39 inch. By a graphical process 8 = 0-40 inch. Deflection of Built-in Beams.— When a beam is built in at one end only, it bends down with a convex curvature Fig. 515. Fig. stS. Upwards (Fig. 515); but when it is supported at both ends, it bends with a convex curvature downwards (Fig. 516); Deflection of Beams. 525 and when a beam is built in at both ends (Fig. 517), we get a combined curvature, thus — Fig. 519. Then considering the one kind of curvature as positive and the other kind as negative, the curvature will be zero at the points XX (Fig. 518), at which it changes sign; such are termed "points of contrary flexure." As the beam undergoes no bending at these points, the bending moment is zero. Thus the beam may be regarded as a short central beam with free ends resting on short cantilevers, as shown in Fig. 519. Hence, in order to determine the strength and deflection of built-in beams, we must calculate first the positions of the points *, X. It is evident that they occur at the points at which the upward slope of the beam is equal to the downward slope of the cantilever. We showed above that the slope of a beam or cantilever at any point is given by the expression — A Slope = EI Case XIII. Beam built in at both ends, with central load. ■0-2S—* Fig. A for cantilever 4 2 WT A for beam = Iltl^ X i^ = 2 2 526 Mechanics applied to Engineering. Hence, as the slope is the same at the point where the beam joins the cantilever, we have — WL,= _ WL, '-, or L, = L2 = — 4 4 4 Maximum bending moment in middle of central span^ WLs ^ WL 2 8 Maximum bending moment on cantilever spans — WLi_ WL 8 Deflection of central span- W(2L, 48Er Deflection of cantilever — ^L» 2 ' \ = - — 3EI ..w@". WL« 48EI 384EI 3E1 384E1 WL* Total deflection in middle of central span — WL» 8 = 8^ + 8 = [92EI This problem may be treated by another method, which, in some instances, is simpler to apply than the one just given. Wlien a beam is built in at both ends, the ends are necessarily level, or their slope is zero ; hence the summation of the slope taken over the whole beam is zero, if downward slopes be Deflection of Beams. 527 given the opposite sign to that of upward slopes. Since the slope between any two sections of a beam is proportional to the area of the bending-moment diagram between those sections, the net area of the bending-moment diagram for a built-in beam must also be zero. A built-in beam may be regarded as a free-ended beam having overhanging ends, da, Vb, which are loaded in such a manner that the negative or pier moments are just sufficient to bring those portions of the beam which are over the supports to a level position. Then, since the net area must be zero, we have the areas — feg-adf-gcb^ o But in order that this condition may be satisfied, the area of the pier-moment diagram adcb must be equal to the area of the bending-moment diagram aeb for a freely supported beam, or — , he ad = — Whence the bending moment at the middle and ends is -5-, and the distance between the points of contrary flexure fg = - ; all the other quantities are the same as those found 2 by the previous method. It will be seen that dc is simply the mean height-line of the bending-moment diagram for the free-ended beam. Thus when the ends are built in, the maximum bending moment is reduced to one-half, and the deflection to one- quarter, of what it would have been with free ends. Case XIV. — Beam built in at both ends, with a uniformly distributed load. A. for cantilever- A for beam — f'wLjL, , a/L, + > 2 ^ 3 These must be equal, as explained above— .^26 528 Mechanics applied to Engineering. Let Iq = wLa. Then ^ = ^^^ + ^5^ 326 2 = 3«^ + «' which on solving gives us « = 0732, We also have — L, + L, = - 2 or 1732L2 = - 2 La = o'289L and Li = 0732 X o'289L = o'2iiL ^L. Fig. 5sa. Maximum bending moment in middle of central span — a/La" ^ a; X o-289'L '' ^ wL» 2 2 24 Maximum bending moment on cantilever spans — w\^ + ^' = «' X 0-289L X o-2iiL+"'^°'"''^' 2 2 _ «/L' 13 Deflection of central span — g ^ 5K/(o-578L)* ^ w\} ' 384EI 689EI Deflection of cantilevers due to distributed load — « _ w(o'2iiLy _ wh* ^ 8El 4038EI Deflection of Beams. Deflection due to half-load on central part — 5 _ zfLa X Li' _ w X o'289L x o-2ii'L' 3EI ~~ ^Ei 1105EI Total deflection in middle of central span — K/L* 529 = ^ + 8, + 8„ = 384EI This problem may also be treated in a similar manner to the last case. The area of the parabolic bending- ;v<vj moment diagram axbxc \bf . ac, and the mean ^ height ae = ^bf; whence ae, the bending moment at the ends, is — ^~8 TT and bg, the bending moment in the middle, is— and for the distance xx, we have — -(L,L,) = ^ 2 12 L,(L - L,) = 1? L] = 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 -j-"^ = -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 distribtited load. Since the ends of the beam are guided horizontally the slope of the ends is zero, hence the net area of the bending S30 Mechanics applied to Engineering. moment diagram is also zero. The area of the bending moment diagram A for a freely supported beam is therefore equal to that of the pier moment diagram. and Mo = ^- - M. Mp = Mo x = -\ -Z-. — TTF^ I (See p. 60.) 3VM<j + Mp/ '^ ^ ' Substituting the value of Mq'and reducing — Mp = ^(2/- 3a:) 2A also Mq = -^(3^ - I) where x = The distance of the centre of gravity of the bending moment diagram for a freely supported beam from the nearest abutment (the centre of gravity of the pier moment diagram is at the same distance from the abutment). c = The distance of the centre of gravity of the bending moment diagram from the middle of the beam. A = The area of the bending moment diagram for a freely supported beam. If the beam be regarded as a cantilever fixed at one end, say Q, and free at the other. The moment of the external system of loading between P and Q causes it to bend down- wards, but the pier moment causes it to bend upwards, and Deflection of Beams. 531 since the deflection at P is zero under Ihe two systems of loading it is evident that the moment of the banding-moment diagram due to the external loads between P and Q is equal to the moment of the pier bending-moment diagram, and since the areas of the two diagrams are equal the distance of the centre of gravity of each is at the same distaijce from Q. When the load is symmetrically disposed f = 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. — Wlien 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) _ 3 _ S^L^ when not propped j 384EI Let Wi be the load on the central prop. W,L' Then the upward deflection due to W, = Si = 4SEI 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 — 5WL3 _ WiL'' 384EI ~ 48EI whence Wi = |W 532 Mechanics applied to Engineering. Thus the central support or prop takes | of the whole load ; and as the load is evenly distributed, each of the end supports takes one-half of the remainder, viz. ^ of the load. Fig. 525. The bending moment at any point x distant /j from the end support is — M, = igwL/i — wliX - = z./(^L-9 = ^\3L-84) The points of contrary flexure occur at the points where the bending moment is zero, i.e. when — ^'(3L - 8/,) = or when 3L = B/j or /, = fL Thus the length of the middle span is — . It is readily shown, 4 by the methods used in previous paragraphs, that the maximum wP bending moment occurs over the middle prop, and is there — , 32 or 5 as great as when not propped. When the three supports are not level. Let the load on the prop be — mvL = nW Then the upward deflection due to the prop is — 48EI and the difference of level between the central support and the end supports is — 384 EI ~ 48EI ~ 384EI^^ "^ Deflection of Beams. 533 When the result is negative it indicates that the central support is higher than the end supports ; if « = i the whole WL' load is taken by the prop, and its height is — -^=-z above the end supports. When the load is evenly distributed over the three supports « = I the prop is then below the end supports by 7WL= WL« -^^ = —r-v^ nearly. 11S2EI i6sEI ■' Where there are two props symmetrically placed at a distance x from the middle of the beam, the downward de- flection at these points when freely supported at each end is (see page 516) — ^ ^~384EI 384EI^'4L^ ibx ) If the spans are equal x = -^ and _ 4-s46wU _ WL' ^~ 384EI ~ 88-4EI The upward deflection due to the two props is — o'o309PL' El where P is the load on each prop ; the constant is taken from the table on page 523. When all the supports are level — WL" _ o-o309PL» 88-4EI ~ EI W P = — - = 0-37W 273 And the load on each end support is 0T3W. Case XVII. Beam with the load unevenly distributed, with an uncentral prop. — Construct the bending moment and deflec- tion curves for the beam when supported at the ends only (Fig. 526). Then, retammg the same scales, construct similar curves for the beam when supported by the prop only (Fig. 527). If, due to the uneven distribution of the load, the beam does not balance on its prop, we must find what force must be applied 534 Mechanics applied to Engineering. at one end of the beam in order to balance it. The unbalanced moment is shown by xy (Fig. 527). In order to find the force required at z to balance this, join xz and yz, and from the pole of the vector polygon draw lines parallel to them ; then the intercept x-^y-^, = Wj on the vertical load line gives the required force acting upwards (in this case). ■ A f, y Fig. 529. In Fig. 528 set off 8 and 80 on a vertical. If too small to be conveniently dealt with, increase by the method shown ioj/,ej, and construct the rectangle efgh. If the prop be lowered so that the beam only just touches it,- the whole load will come on the end supports ; the proportion on each is obtained from Ri and R3 in Fig. 526. Divide ^/4 in i in this proportion. As the prop is pushed up, the two ends keep on the end Deflection of Beams. 535 supports until the deflection becomes 8 + So ; at that instant the reaction Rj becomes zero just as the beam end is about to lift ofl" the support, but the other reaction Rj supports the un- balanced force W]. This is shown in the diagram by ee-^ = W, to same scale as Ri and Rj. Join ie and ge-^ ; then, if the three supports be level, the prop will be at the height/. Draw a horizontal from/ to meet ge-^ in gf,; erect a perpendicular. Then the proportion of the load taken by the prop is ^^, by the support Rj is |^, by the /««« /o«o support Rais^. Likewise, if the prop be raised to a height corresponding to /i, the proportions will be as above, with the altered suffixes to the letters. In Fig. 528, we have the final bending-moment diagram for the propped beam when all the supports are level ; comparing it with Fig. 526, it will be seen how greatly a prop assists in reducing the bending moment. It should be noted that in the above constructions there is no need to trouble about the scale of the deflections when the supports are level, but it is necessary when the prop is raised or lowered above or below the end supports. This method, which the author believes to be new, is equally applicable to continuous beams of any number of spans, but space will not allow of any further cases being given. Stiffness of Beams.— The ratio '^^^^^^^°" is termed the span ' stiffness " of a beam. This ratio varies from about the best English practice for bridge work ; it is often as great as 3^ for small girders and rolled joists. By comparing the formulas given above for the deflection, it will be seen that it may be expressed thus — «EI where M is the bending moment- and « is a constant depend- ing on the method of loading. In the above equation we may substitute /Z for M and Zy for I ; then — - g^/ZL^^/L^ nYlLy nEy 536 Mechanics applied to Engineering. Hence for a stiffness of 2^5, we have — i = _i_ = fh L 2000 wEy or 2000/L = wEy Let/= 15,000 lbs. square inch ; E = 30,000,000 „ „ Then «y = L But V = - 2 where d = depth of section (for symmetrical sections) ; then— nd = 2L , d 2 and =r- = - L » Values of «. Beam. Cantilever. (a) Central load ... 12 — End load ... ... — 3 {i) Evenly distributed load g-6 4 {e) Two equal symmetrically placed loads dividing } beam into three equal parts ... ... ...\ ^ ^ {d) Irregular loading (approx.) 11 3*5 Values of ~ , Stifihess. iimro vm ^ .firaffi, central load 6 12 24 Cantilever, end loa.d 1-5 3 6 .ffMOT, evenly distributed load 4-8 9-6 19-2 Cantilever, „ ,, 2 4 8 Beam, two symmetrically placed loads, as in} ., Fig. 423 S"^^ 9-3 l8-6 .ffMOT, irregular loading (approx.) 5-5 II 22 Cantilever, „ „ 175 3-5 7 This table shows tlie 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 — nE.d Deflection of Beams. 537 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 lo per cent., and usually will not amount to more than 5 per cent. CHAPTER XIV. COMBINED BENDING AND DIRECT STRESSES. In the figure, let a weight W be supported by two bars, i and 2, whose sectional areas are respectively Aj and Aj, and the corresponding loads on the bars Rj and R2; then, in order that the stress may be the same in each, W must be so placed that Rj and R2 are pro- portional to the sectional areas of the Ri_Ai m '/////. M f?f y i >? II ,1 bars, or But Ri« = RjZ, M^.■ ■-« - Fig. 530. or Ai« = Aja ; hence W passes through the centre of gravity of the two bars when tlie 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 musi 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 — ^{u + z) = W(« -f x) and when W is at the centre of gravity — R2(« -f z) = W« Thus when W is not placed at the centre of gravity of the section, the section is subjected to a bending moment Wa: in addition to the direct force W. Thus— If an external force W acts on a section at a distance xfrom its centre of gravity, it will be subjected to a dire J force W acting ^, /f. •■^..U-'ii-X- > Fio. 531. Combined Bending and Direct Stresses. S39 uniformly all over the section and a bending moment War. Th(^ 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 i — '■ ^^^^ ^ — ,leMf ied direct stress acting uni- ' ^^^m \juia- formly all over the section may be represented by the diagram aicd, where a6 or cd is the intensity of the tensile stress (+) ; then if the intensity of tensile stress due to bending be represented by 6e (+), and the compressive stress ( — ) hy fc, we shall have — The total tensile stress on the outer skin = ab ■\- be = ae „ „ „ inner „ = dc -fc=df If the bending moment had been stili greater, as shown in side- side Xtnlocuz0<t- sid& Fig. 533, the stress (^ would be — , i.e. one side of the bar would have been in compression. Stresses on Bars loaded out of the Centre. — Let W = the load on the bar producing either direct tensile or compressive stresses ; A = the sectional area of the bar ; Z = the modulus of the section in bending ; *• = the eccentricity of the load, i.e. the distance of the point of application of the load from the centre of gravity- of the section ; /', = the direct tensile stress acting evenly over the section ; J\ —■ the direct compressive stress acting evenly over the section ; 540 Mechanics applied to Engineering, f, = the tensile stress due to bending ; fc = the compressive stress due to bending ; M = the bending moment on the section. W Then j-=/c orf, or/' (direct stress) M W* "2 = "2" ^-^ ^"^-/^ or/ (bending stress) Then the maximum stress on the skinj _ ^ i ^ _ W , Wa: of the section on the loaded side /""•' ■'""a "Z^ Then the maximum stress on the skin^ _ f _ f _ xv/^ i x\ of the section on the unloaded side/ "•' -' ~ \A ~ z) In order that the stress on the unloaded side may not be of opposite sign to the direct stress, the quantity - must be greater ' than -. When they are equal, the stress will be zero on the unloaded side, and of twice the intensity of the direct stress on J X Z the loaded side ; then we have t = v> or — = «. Hence, in Jx cj A. order that the stress may no": change sign or that there may be no reversal of stress in a section, the line of action of the 7 external force must not be situated at a greater distance than — A from the neutral axis. Z For convenience of reference, we give various values of - A in the following table : — Combined Bending and Direct Stresses. 541 i /^^ /^/^^^ i^ -«imnil iimw v5 .S|| 1 t3 (((®)i)S m Mil Will 6 fa C*) i lit 11 '11111 P g,s.s \^-</ • *U u J9 •3^° 11 rt ., •Ss III 3 s o .5 <A ^ V u '% ^ ^ 'd ts ** i^ s s ' ■§ ^ «M 3? :§ 5; a 5- ■3 II W|« Q|«> 1 1 PQ + + + Q 00 "S t K ES. a s. 2i 1 M S ►*a a >o VO H . •< ST- 1 :§ a (5 S K Q * 1 + 1 ■* En n N s Q 8 N -5 _,^ , s- .~i tM ST'^ Q 1 Q T, -5 1 M a + a all 'mm Q H CO P 6 — ,D u M ll |, J tS 1^ 1 S a K *G A =i° w 542 Mechanics applied to Engineering. Af\A. i General Case of Eccentric Loading. — In the above instances we have only dealt with sections symmetrical about the neutral axis, and we showed that the skin stress was much greater on the one side than the other. In order to equalize the skin stress, we frequently use unsymmetrical sections. Let the skin stress at a due to bending and direct stress ,...^..„_^. Fig. 538. = /■'„ ; likewise that at i = f\. W The direct stress all over the section = f = —- A For bending we have — according to the side we are considering; or Wa; = — or -r hence /a = W;cy. andA=/+/=T + — ^■- also/»=/'-/= W A W " A ' V^x y<. I When y^ = y^ the expression becomes the same as we had above. Cranked Tie-bar.— Occasionally tie bars and rods have Fig. 539. 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. 543 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 t. Then, using the same notation as before — B b X = - + u — - 2 2 the direct stress on the\ W ^ wide part of the bar I Bt ~ ' the bending stress on the"! _ Wx \ 2 ^ 2 / wide part of the bar / z" ^ BV the maximum skm stress) _ W \ 2 2 / due to both > Bt '^ Wt 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 + 2u- b) it ~Bt'^ 2BV W Then dividing both sides of the equation by y, and solvmg, we get — B = a/66u + 3^ + 26 Both 6 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 oif 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 imequal, 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. 544 Mechanics applied to Engineering. Bending of Curved Bars. — Let the curved bar in its original condition be represented by the full lines, and after bending by the broken lines. C.A. Fig. S40. Let the distance of any layer from the central axis which passes through the centroid of the section be 4-^ when measured towards the extrados, and —y when measured towards the intrados. Let the area of the cross section be A = S^Sy = 28a The original circumferential length of a) _ ^tj , ^r, _ , layer distant j/ from the central axis 3 ~ ' +^/''i ~ ^i The final length =(R2+j')62 = 4 The strain on the layer = (R2 + y)^^ — (Rj + y)Q-^^ = x a:=RA-RA+J^(^2-<9i) . . . • (i.) a: = o when y = — 6a — Oi = h (ii.) The only layer on which the strain is zero is that at the neutral axis, hence h is the distance of the neutral axis from the central axis of the section. Rj and ^2 are unknown at this stage in the reasoning, hence the expression must be put in another form before h can be calculated. Substituting the value of h in (i.), x= -h{6^-6,)+y{e^-6,) x = (y-A){e!,-eO = {y-A)^6 . . (iii.) Combined Bending and Direct Stresses. 545 For elastic materials we have — 1 = 1=^ and /=^ = E.. . (iv.) '-—~h — ^^-x Substituting the value of x from (iii.) J - j^ = Etf . . . . (vi.) Since the total tension on the one side of the neutral axis is equal to the total compression on the other, the net force on the whole section 2(/Sa) = o or/ifli +/2aa + etc. =//«i' +/2V + etc. o, e4('J^*>. + elc.} - EA^K-'-i+i),. + ett.} ^) <^^) ' 4-a->i ^ ^ R^+y Ri(A - A') . A = -^ = -^, '- (see p. 546) . . . (vii.) Ri+Jf We also have M =/Z =/i^i«i +^^2*2 + etc. or M = 27(/. J . S«) = E A6li;|- ^^-^ ~ '^^ 8g| Ae= , ^ Substituting this value in (vi.) 2 N 271 546 Mechanics applied to Engineering. Substituting the value of /i f=Z-^ M But Hence Ula = o, and Rj -^ — ;— = o / = _ M Ri+y' y — h ~ hK Ri +y The stress at the intrados — ,_ M{y,-h) ^' hA{R,-y,) The stress at the extrados — y:= My. + h) hh(K,+y,) where yi and ^, are the dis- tances of the intrados and the extrados respectively from the central axis. The value of ■^a-.) A' or R: can be found graphically thus : The section of the curved bar is shown in full lines ; the centre of curvature is at O. The points a and b are joined to O ; they cut the central axis in d and b'. At these points erect perpen- diculars to cut the line ab, as shown in «„ and b^ which are points on the new area A', because ab ~Ri+;' Combined Bending and Direct Stresses. 547 Similarly cd "~ Ri — _y The two areas, A and A', must be accurately measured by a planimeter. The author wishes to acknowledge his indebtedness to Morley's " Strength of Materials," from which the above paragraph is largely drawn. Hooks. — In the commonly accepted, but erroneous, theory of hooks, a hook is regarded as a special case of a cranked tie bar, and the stresses are calcu- lated by the expressions given in the paragraph on the " general case of eccentric loading,'' but such a treatment gives too low a value for the tensile stress on the intrados, and too high a value for the compressive stress on the extrados. In spite, however, of its inaccuracy, it will probably continue in use on account of its simplicity; and provided the permissible tensile stress be taken somewhat lower to allow for the error, the method gives quite good results in practice. The most elaborate and com- plete treatment of hooks is that by Pearson & Andrews, "On a '°' "^' Theory of the Stresses in Crane and Coupling Hooks," published by Dulau & Co., 37, Soho Square, W. The application of their theory is, however, by no means simple when applied to such hook sections as are commonly used for cranes. In the graphical treatment the sections must be drawn to a very large scale, since very small errors in drawing produce large errors in the final result. For a com- parison of their theory with tests on large crane hooks, see a paper by the author. Proceedings I.C.E., clxvii. For a comparison of the ordinary theory with tests on- drop forged steel hooks, see a paper by the author. Engineering, October 18, 1901. The hook section shown in Fig. 541 gave the following results — A = 14-53 sq. ins. Aj = 15-74 sq. ins. Ri = 5 ins. h = 0-384 in. M = 85 in.-tons. yi = 2'46 ins, y, = 2-81 ins. Then/, = 12-5 tons sq. in. 548 Mechanics applied to Engineering. The Andrews- Pearson theory gave i3"9 tons sq. in., the common theory 8-8 tons sq. in., and by experiment i3'2 tons sq. in. By this theory/, = 6*2 tons sq. in., the Andrews-Pearson theory gave 4"6, and the common theory 7-6 tons sq. in. 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. 543, resolve W into two com- ponents, W, acting normal to the beam, and P acting parallel with the beam ; then the bending moment at the section x = Wj/i. But Wi = W sin a F.G. 543. ^"'^ ^ = ilHT hence M. = W sin ui X -■ * sm a M, = W/=/Z W/ ^~ Z fr.1 ^ . -11 i.u ..• P W cos O The tension actmg all over the section = x = a — hence y max. = and/min. = W cos a ~A W cos a W/ „, / cos a /\ N.B. — The Z is for the section x taken normal to the beam, ftof a vertical section. Machine Frames sub- jected to Bending and Direct Stresses. — Many machine frames which have a gap, such as punch- ing and shearing machines, riveters, etc., are subject 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 Fig. 544. Combined Bending and Direct Stresses. 549 present unknown, unless a section 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 m p tension. Let the ratio be, say, 3 to i; then, W^^-i\ neglecting the strength of the web, our section will be somewhat as follows : — Make A. = ^A, then X = G, + - approx. '^'* 4 *''G- S45- Z = - = ^4 ^ ^ l±i. (approx.) 4 Z :» sAjH (for tension) But WX =/Z (/being the tensile stress) W (' '(g. + 5) = 3A3/ wCg.4-5) WrG,4-- Ac = Tj ■■ or -— p 3H/ «H/ where n is the ratio of the compressive to the tensile stress, and A, = «A, Having thus approximately obtained the sectional areas of the flanges, complete the section as shown in Fig. 546 ; and as a check on the work, calculate the stresses by finding the centre of gravity, also the Z or the I of the complete section by the method given on page 544, or better by the " curved bar" method. CHAPtER XV. STRUTS. General Statement. — The manner in which short com- pression pieces fail is shown in Chapter X. ; but when their length is great in proportion to their diameter, they bend laterally, tmless 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 XIV. 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 B Fig. 547. remain so when loaded j either or both of the following causes may set up bending : — (i) The one side of the strut may be harder and stiffer than the other ; and consequently the soft side will yield most, and the strut will bend as shown in A, Fig. 547. Struts. 55 1 (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. 548 is evidently W8. Then WS =/Z, using the same notation as in the preceding chapters. If we increase the deflection we shall correspondingly increase the bending moment, and consequently the stress. From above we have — ^ =iz or's'Z, and so on O Oj But as /varies with 8,"s-= a constant, say K; Fig. 548. then W = KZ But Z for any given strut does not vary whqn 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. 549), be applied to it until the deflection is Sj, and its stress /i ; Then W81 + M =/,Z But W81 =/iZ 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 SS2 Mechanics applied to Engineering. 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 Wj 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 Sj and the stress /i ; then we have, as before — WiSi + M =/iZ = W8i = WA + wl^ hence M = w\ Thus, in order to keep the strut in its bent position with a deflection Sj, we must subject it to a + bend- ing moment M, i.e. one which tends to bend the strut in the same direction as WiSi ; hence, if we remove the bending moment M, the deflection will become zero, i.e. the strut will straighten itself. Now, let a load Wa greater than W be placed on the strut, say W s= Wj — a/, and let it again be bent until its deflection = Sj, and the stress f^ by an external bending moment M ; then we have as before — WA + M =/,Z = WA - wS, hence M = —w\ Thus, in order to keep the strut in its bent position with a deflection \, we must subject it to a — bending moment M, i.e. one which tends to bend the strut in the opposite direction to W281 ; 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 — Fig. 549- Condition. When slightly bent by an ex- ternal bending moment M, on being released, the strut will- When supporting a load — Remain bent Straighten itself Bend still further and ultimately buckle W. less than W. greater than W. Struts. SS3 _ 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. As a first approximation, mainly for the sake of getting the form of expression for the buckling load of a slender strut, assume that the strut bends to an arc of a circle. Let / = the eifective length of the strut (see Fig. S5o); 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). Then for a strut loaded thus — * = 8Rt(^^«P- 425) =■• SEP 8EI 8EI or W = —7^ (first approximation) As the strut is very long and the deflection small, the length / remains practically constant, -and the other quantities 8, E, I are also constant for j-,<;. j^^^ 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 — <» ^ ^ ■., S I ^r M/^ W8/2 8=-X-Mx§X--^EI = -7^7 = -7^ 3 2 a„dW = 9^ ' 9-6EI 9-6EI The value obtained by Euler was — _ g^EI _ 9-87EI _ loEI (nearly) 554 Mechanics applied to Engineering. This expression is obtained thus — The bending moment M at any point distant x from the middle of the strut is M = -WS = El^ (see page 510) Multiply each side by — ^.^^ _WS ^ dx ds^ EI dx Integrating (|J=-S(8^ + c) When ^ = o, 8 = A, hence C = - A^ dx Integrating again — ^ = A/^sin-'- + K V W A When a: = o, 8 = 0, therefore K = o Hence 8 = A sin \x^ — j When a; = - 8 = A z 'CVI,)= and sin ( ■"■ EI> The only angles whose sines are = 1 are -, — , etc. We 2 2 require the least value of W, hence — aV EI ~ 2 and W = EI ;ei Struts. 555 It must not be forgotten that this expression is only an approximation, since the direct stress on the strut is neglected. When the strut is very long and slender the direct compressive stress is very small and therefore negligible, but in short struts the direct stress is not negligible consequently for such cases the above expression gives results very far from the truth. Effect of End holding on the Buckling Load. — In the case we have j-ust considered the strut was supposed to be free or pivoted at the ends, but if the ends are not free the stmt behaves in a different manner, as shown in the accompany- ing diagram. Diagram showing Struts of Equal Strength. One end free, the other hxed. /=3L p = Both ends pivoted or rounded. /=L Fig. W = P = loEI loEp' One end rounded or pivoted, the other end built in or fixed. P = 20Ep' Both ends fixed or built in. 1=^ 2 Each strut is supposed to be of the same section, and loaded with the same weight W. 556 Mechanics applied to Engineering. W A W Let P = the buckling stress of the strut, i.e. —, where W = the buckling load of the strut ; A = the sectional area of the strut. We also have r- = p'' (see p. 78), where p is the radius of gyration of the section. Substituting these values in the above equation, we have — ' /=> The " eflfective " or " virtual "length /, shown in the diagram, is found by the methods given in Chapter XIII. 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 shifts 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 mches long; we get P = — z 9000000 1 _ jgj ^^^ jj^^^ 1000 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. 557 for long struts, whose length / is not less than 30 dia- meters for wrought iron and steel, or 12 for cast iron and wood. Notwithstanding the unsatisfactory basis on which it rests, many high authorities 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, contains empirical constants obtained from experiments by Hodgkinson and others; but by making certain assumptions constants can be obtained rationally which agree remarkably well with those 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,' 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 the calculated values of the constant (a), shortly to be considered, would not agree so well with the experimental values. ' Mons. Considire 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. 558 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 — C = -and/= — then S = C +/ S = -- + -=- (the least Z of the section) We have shown above that, on Euler's hypothesis, the maximum deflection of a strut is — g^ M/' _ fZfl loEI loEZy 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 = -, 3 where d is the least diameter of the strut section. By substitution, we have — 8 = 4^. , ^ W , W/72 ,. V ^ SEdZj w/ _ W/ A/d P\ ~ A^' + pZ "^ W If W be the buckling load, we may replace — by P, Then S = p(i + a^^) T, S S Struts. 559 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 — L JJ □! Fig. 532. 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 true for the two extreme cases, viz. for a very short strut, when W = AS, and for a very long strut, in which S =/; then — ,„ 5E</Z loEI W = -j^ 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 the rational value for a agrees so well with the experimental value is strong evidence that the formula is trustworthy. Values of S,/, and E are given in the table below j they S6o Mechanics applied to Engineering. must be taken as fair average values, to be used in the absence of more precise data. Pounds per square in ch. S f E Soft wrought iron Hard „ Mild steel Hard Cast iron „ (hard close - grained\ metal) / Pitchpine and oak 40,000 48,000 67,000 110,000 f 80,000 (no \marked limit) 130,000 8,000 28,000 32,000 45,000 75,000 80,000 130,000 8,000 25,000,000 29,000,000 30,000,000 32,000,000 13,000,000 22,000,000 900,000 Material. Form of section. sEZ a. by experiment. Wrought iron ... OB Tloto,^ TJjtOTOJ • ifetos^ ^ tSntOsi, Tsm '0 555 ML+-LU ?5iito,J, 410553 Mild steel ■i ^ jfc to m • ^ ■ ^ ^ A, HL + J.U ^ 5J11 to 5J, (author) Hard steel ■■ ssj to ^ • BjtOjfe ^ sS,' HL + _LU ^ — ' The discrepancies in these cases may be due to the section being thicker or thinner than the one assumed in calculating the value of a. In the case of hollow sections, angles, tees, etc., the value of o should be worked out. Struts. 561 Material. Formof section. sEZ a by experiment Cast iron ... ■■ raitOTj, tI, • iJlltOT}, ^ iJntOTJs ,1, (Rankine) jjj HL4--LU A to A A' Fitchpine and oak ■i i. <^ m A ^ (author) ,', N.B. — The values of a given in the last column are four times as great as those usually given, due to the / used in our formula being taken equal to L for rounded ends, whereas some other writers take it for square or fixed ends. The values of the constant a have been worked out for the various materials, and are given in tabulated form above ; also values found by experiment as given in Rankine's " Applied Mechanics," and by Bovey, " Theory of Structures and Strength of Materials " (Wiley, New York). Rankine's Strut Formula. — In the above tables it will be noticed that the value of a as found by calculation quite closely agrees with that found by experiment for the solid sections, but the agreement is not so good in the case of hollow or rolled sections, largely due to the fact that a varies with each form of section and with the thickness of the metal. If the value of a be calculated for each section there is no objection to the use of the Gordon formula, but if one value of a be taken to cover all the cases shown in the tables above it is possible that considerable errors may creep in. For such cases it is better to use Rankine's modification of Gordon's formula. Instead of writing in the Gordon formula — S = ^(x+^ we may write ) AV "^ loEjAKV ~ A\ I + loE KV W = AS i + ^R 2 o 562 Mechanics applied to Engineering. where h = f and „ uvw^^ -^ ~ A' '■'■ *^^ equivalent length divided by the /^aj/ radius of gyration of the. section about a line passing through the centroid- of the section. Values of k will be found in Chapters III. and XI. The values of b are as follows : — Value oib. Material. loE By experiment. Wrought iron 5MII Sim Mild steel 5755 7SM Hard steel ?OTn SMI Cast iron TB35 t° -Am ims The discrepancies are due to the assumed value of/ not being suitable for the material experimented upon. The terms " mild " and " hard " steel are very vague. If the properties of the material in the tested struts were known, the dis- crepancies would probably be smaller. It must, however, be borne in mind that the strength of struts cannot be calculated with the same degree of accuracy as beams, shafts, etc. In the case of long cast-iron struts the failure is usually due to tension on the convex side, and not to compression on the concave side. The expression then becomes — T P=: ar' — \ where T = the tensile strength of the material, or rather the tension modulus of rupture, i.e. the tensile stress as found from a bending experiment. The values of/ and T then vary from 30,000 to 45,000 lbs. per square inch, and E (at the breaking point) varies from i t, 000,000 to 16,000,000 lbs. per square inch. The value of a then becomes jg^ for a rectangular section. On calculating some values for P, it will be seen that for long struts where the fracture might occur through the excessive Struts. 563 stress on the tension skin, the value given by this formula agrees fairly well with the values calculated from the original formula ; hence we see that such struts are about as likely to fail by tension on the convex side as by compression on the concave side. The following tables have been worked out by the formula given above to the nearest 100 lbs. per square inch. For those who are constantly designing struts, it will be found convenient to plot them to a large scale, in the same manner as shown in Fig. 553. In order better to compare the results obtained by Euler's and by Gordon's formula, curves representing both are \ ■ , 1 s. s V \ \ \ \ \ \ «? \ ^ \^ \ \^ • Q s; ■* ^ ^ \ j^ ^ ^ s Rl '^^ ■~- ;r -— . ^ — — ^^ ^^ — — Fio. 553. ■(4) Note. — The two curves in many cases practically coincide after 40 diameters. In the figure the Gordon curve has been shifted bodily up, to better show the relation. given, from which it will be seen that they agree fairly well for very long struts, but that Euler's is quite out of it for short struts. The table on the opposite page gives the ultimate or the buckling loads ; they must be divided by a suitable factor of safety to get the safe working load. 564 Mechanics applied to Engineering. Buckling Load of Struts in Pounds per Square Inch of Section. HL HL r or ■■ • + ■■ • + I d XU XU Wrought iron. Mild steel. S 42,600 42,000 43,000 41,700 64,100 63,100 64,700 62,000 10 38,800 37,200 39,900 36,000 56,600 56,300 58,300 Sijooo 20 28,800 25,600 30,900 23,200 38.500 33.500 41,900 29,800 30 20,000 16,800 22,200 14,700 25,000 20,600 28,600 17,500 40 14,000 11,400 16,300 9,600 17,000 13,400 19,700 11,100 50 10,400 8,000 12,400 6,700 11,900 9,200 14,100 7,800 60 7,600 5,900 9,100 4,900 8,700 6,700 10,500 5.400 70 5,800 4,500 7,100 3.700 6,700 5,000 8,100 4,100 80 4,600 3.500 S,6oo 2,900 5,200 3.900 6,300 3,200 90 3.700 2,800 4.500 2,100 4,200 3,100 5,100 2,500 100 3,100 2,300 3.800 1,900 3.400 2,500 4,200 2,100 Hard steel. Cast iron (soft). S 102,000 100,300 104,000 98,600 67,100 64,000 69,200 58,900 10 85,500 79,400 89,600 74.500 45,200 40,000 49,200 32.900 20 51.500 43.300 57.500 37.900 19,600 16,000 22,900 11,900 30 30,800 24,600 36,100 20,400 10,100 8,000 12,100 5,800 40 19,700 15,400 23,700 12,700 6,000 4.700 7.300 3.400 1° 13.500 10,300 16,400 8,500 3.900 3,100 4,800 2,200 60 9.750 7,400 11,900 6,100 2,800 2,200 3.400 1,500 70 7.300 5,500 9,100 4.500 2,100 1,600 2,500 1,100 80 S.70O 4.300 7,100 3.500 1,600 1,200 2,000 870 90 4,600 3.400 5,700 2,800 1,300 980 1,600 690 100 3.700 2,800 4,600 2,200 1,000 790 1,300 560 r or / d Cast iron (hard close-grained). Pitchpine and oak. S 109,000 104,000 112,000 95.700 6,300 5,900 10 73.500 65,000 80,000 53.500 3.800 3.300 20 31,800 26,000 37,200 19,300 1,500 1,200 30 16,400 13,000 19,700 9.400 730 580 40 9.700 7,600 11,900 5.500 430 340 50 6,300 5,000 7,800 3,600 280 220 So 4,600 3,600 5. 500 2,400 200 150 70 3.400 2,600 4,100 1,800 140 1 10 80 2,600 1,900 3.300 1,400 110 90 90 2,100 1,600 2,600 1,100 90 70 100 1,600 1,300 2,100 900 70 60 Struts. 56S Factor of Safety for Struts. Wrought iron and steel Cast iron Timber 350 Dead loads. Live loads. ... 4 8 ... 6 12 ... 5 10 300 250 200 -s 'SISO xlOO 50 b/ /<> / / / / v / / / • / // xy / y * //. / y y y ^ ^ ^ 25 SO 75 100 125 150 175 Weight in, Pounds per ft. Fig. SS4. 200 225 In choosing a section for a column, economy in material is not the only and often not the most important matter to be considered ; every case must be dealt with on its merits. Even as regards the cost the lightest column is not always the cheapest. In Figs. 554 and 555 we show by means of curves how the weight and cost of different sections vary with the load to be supported. Judging from the weight only, the hollow circle would appear to be the cheapest section, but the cost per ton of drawn tubes is far greater than that of rolled sections ; hence on taking this into account, we find the hollow circle the most expensive form of section. The values given in the figures must not be taken as being rigidly accurate ; they vary largely with the state of the market. Designers, however, will find it extremely useful to plot such curves for themselves, not only for struts, but for floorings, cross-girders, roof-coverings, roof-trusses, and many other details which a designer constantly has to deal with. Straight-line Strut Formula. — The more experience S66 Mechanics applied to Engineering. 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 caitfully 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 com- mercial struts, which are not aX^jiSiys perfectly straight, and are 350r 300 250 1 200 1l50 ^100 50 A // / ^ fKJ y<- ^ y y 1/ V ^ / ^ £5 ^ £10 £15 Cost of a 20 foot column. Fig. S5S- £20 not always perfectly centrally loaded, the results are frequently lo 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. Struts. 567 Material. Form of section. M. N. i not to d exceed Wrought iron ... IB 47,000 82s 40 ' • 47,000 900 40 47,000 775 40 Ht-ff-J-U 47,000 1070 30 Mild steel ■■ 71,000 1570 30 # 71,000 1700 30 73,000 1430 30 HL + XU 71,000 1870 30 Hard steel 1^ -114,000 3200 30 • 114,000 3130 30 114,000 2700 30 HL + J.U 114,000 3500 30 Soft cast iron . . . ^ 90,000 4100 • 90,000 4700 90,000 3900 HL+ J-U 90,000 5000 Hard cast iron ... ■1 140,000 6600 • 140,000 7000 > 140,000 6100 HL. + -LU 140,000 8000 Pitchpine and oak Hi 8000 470 10 • 8000 Soo. 10 S68 Mechanics applied to Engineering. XW Columns loaded on Side Brackets. — The barbarous practice of loading columns on side brackets is \Wi unfortunately far too common. As usually carried "n..!... out, the I practice reduces the strength of the cohimn to one-tenth ' of its strength when cen- trally loaded. In Fig. 557 the height of the shaded figure on the bracket of the column shows the relative loads that may be safely placed at the various distances from the axis of the column. It will be perceived how very rapidly the value of the safe load falls as the eccentricity is increased. If a designer will take the trouble to go carefully into the matter, he will find that it is positively cheaper to use two separate centrally loaded columns Fig. 556. instead of putting a side bracket on the much larger column that is required for equal strength. Lety^ = the maximum compressive stress on the material due to both direct and bending stresses ; ff = the maximum tensile stress on the material due to both direct and bending stresses ; / = the skin stress due to bending ; C = the compressive stress acting all over the section due to the weight W ; A = the sectional area of the column. WX W Then/.=/-FC = ^4-^ wx_ w Z A If the column also carries a central and /, = load Wi, the above become — /, = WX W-fWi Z ■*" A /,= WX W + Wi Z A Columns loaded thus almost invariably fail in tension, therefore the strength must be calculated on the /, basis. We have neglected the deflection due to loading • The ten is not used with any special significance here ; may be one-tenth or »ven oue-twenlieth. Fig. 557- Struts. 569 (Fig. 558), which makes matters still worse; the tensile stress then becomes — /.= W(X + 8) W_+ W, The deflection of a column loaded in this way may be obtained in the following manner : — The bending moment = WX area of bending-moment diagram = WXL „ WXL' (approximately) After the column has bent, the bending moment of course is greater than WX, and approximates to W(X + 8), but S is "1^ if = 0-2 ^ S^ff,!^ Fig. 558. Fig. 559. usually small compared with X, therefore no serious error arises from taking this approximation. A column in a public building was loaded as shown in Fig. 559; the deflections given were taken when the gallery was empty. The deflections were so serious that when the gallery was full, an experienced eye immediately detected them on entering the building. The building in question has been condemned, the galleries have been removed, and larger columns without brackets have 570 Mechanics applied to Engineering. been substituted. Thecolumn, as shown, was tested to destruc- tion by the author, with the following results — External diameter 4 '95 inches Internal diameter ... ... ... 3'7o ,, Sectional area ... fi'49sq. inches Modulus of the section 8" 18 Distance of load from centre of column ... 6 inches Height of bracket above base ... 8' 6" Deflection measured, above base 4' 3" Win tons... I 2 I 4 I 6 1 8 I 10 I 12 I 14 I 16 I 18 5 in inches... | o'035 1 o'o69 1 o'loo 1 0'148 1 o'igs 1 0'245 1 0^300 1 0-355 ' 0'42o The column broke at iS'iy tons; the modulus of rupture was i3'3 tons per sq. inch. Judging from the deflection when the weight of the gallery rested on the bracket, it will be seen that the column was in a perilously dangerous state. Another test of a column by the author will serve to emphasize the folly of loading columns on side brackets. Estimated Buckling Load if centrally loaded, about 1000 tons. Length 10 feet, end flat, not fixed. Sectional area of metal at fracture ... 34*3 sq. inches Modulus of section at fracture 75 'O Distance of point of application of load from centre of column, neglecting slight amount of deflection when loaded ... 17 inches Breaking load applied at edge of bracket 65'5 tOnS Bending moment on section when fracture occurred 1114 tons-inches Compressive stress all over section when fracture occurred I '91 tons per sq. inch Skin stress on the material due to bending, assuming the bending formula to hold up to the breaking point ... ... I4"85 ,, ,, 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 l6'76 ,, ,, Tensile strength of material as ascertained from subsequent tests 8'45 ,, ,, Compressive strength of material as ascer- tained from subsequent tests ... ... 30'4 i> i> Thus we see that the column failed by tension in the material on the off side, i.e. the side remote from the load. ' 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 XVI. TORSION. GENERAL THEORY. Let Fig. 560 represent two pieces of shafting provided with y Fig. s6o. 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/ = Sy Let the area of the pin = a, and the shear stress 'on the pin he/,. Then we may write the ab o ve equation — W/ = f^y Now consider the case in which there are two pins, then — Fig. 561. w/ = Sy + Si^/i =fAy +Aaxyi The dotted holes in the figure are supposed to represent the pin-holes in the other disc coupUng. 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 572 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 y =-^'(a/ + «^>'.') Then, if a = Oj, and say y^ _y 4 Thus the inner pin, as in the beam (see p. 432), has only increased the strength by j. 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^, a^, etc., distant y, y^, y^, etc., respectively from the axis of revolution. Then, as before, we have — W/=-^(ay^ + a^j^ + 0^,=' +, 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 I,. Then — The W/ is termed the twisting moment, M,. /, 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. 573 -^ = the modulus of the section = Z^. To prevent confusion, we shall use the suffix / to indicate that it is the polar modulus of the section, and not the modulus for bending. Thus we have M, =/,Zp 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 evidently will be in shear, and will be distorted P Fig. 562. 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 X. we showed that when an element was distorted by shear, as shown in Fig. 563 (a), it was Fig. 363. equivalent to the element being pulled out at two opposite corners and pushed in at the others, as shown in Fig. 563, (b) and {c), hence all along the diagonal section AB there 574 Mechanics applied to Engineering. 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 J 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 p,Q jg^. twisted in the direction of the arrow A, there will be compression normal to the slot, shown by a full Une, 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. XL), 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 th« disc couplings above. The stress on the inner block = fa, and on the outer block = /, ; 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 ai, where — =-^ =-^, or <j!,|/; = aja- <h J, y Fig. 565. Torsion. General Theory. S7S Then the magnitude of the resultant force acting on the two blocks = af, + «,/,! = of, + <f. =/.(« + «i') =f, (sHaded area or area of modulus figure) And the position of the resultant is distant y,, from the centre, where — ay + gi>i i.e. at the centre of gravity of the blocks. Then/.Z, =f,{a + «,') X ^^^^ =f.{ay + ch'y,) =^{ay^ + a^y^) 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_ for a -no Fig. s66. 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 = irD ; 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 ciicumference of circles of radii Oi, Oa, etc. Thus the triangle Oab 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'l', etc., represents the stress in the metal at that layer in precisely the same manner as in the beam modulus figures. Then — The polar modulus of ) S^'^% °^ T^'f^ ^^T ^ f ''f "''^ the section Z, = i "^ ^; °^ S- of modulus figtfrefrom ' ' {_ centre of circle or Z, = Ay. 5/6 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, Fig. 567. 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 assumed to be so. 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 ?2 = Z,. In Chapter III., we showed that I. = — , where D y 32 is the diameter, and y in this case = — ; hence — 2 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. For a hollow shaft — Torsion. General Theory. 577 I„ = — ^^ '-, -where D< = the internal diameter 32 ' hence Z, = .ep = oT96(^^^-j If - of the metal be removed from the centre of the shaft, n ' we have — the external area The internal area = • n — ^ = or Dj^ = -a 4 4» W' Z, = o-i96D{i-i) The strength of a shaft with a sunk keyway has never been arrived at by a mathematical process. Experiments show that if the key be made to the usual proportions, viz. the width of the key = \ diameter of shaft, and the depth of the keyway = \ width of the key, the shaft is thereby weakened about 19 per cent. See Engineering, March 3rd 191 1, page 287. Another empirical rule which closely agrees with experi- ments is : The strength of a shaft having a sunk keyway is equal to that of a shaft whose diameter is less than that of the actual shaft by one-half the depth of the keyway j thus, the strength of a 2-inch shaft having a sunk keyway 0-25 inch deep is equal to a shaft {2 ~j = r87S inches diameter. This rule gives practically the same result as that quoted above. 2 p 578 Mechanics applied to Engineering. Strength of Shafts of Various Sections. „ . ,/D* - DA Fig. 569. Fin. 570. 5 -- Fig. 571. Z, = -^,or Z, = 0-I96D3 z = 0-19603(1- m^J z. irTid-^ Z, = o-r96D<^ Z„ = 0-208S3 g ^ 584M^ GD* . _ 584M/ G(D* - D<*) ^^ 292M/(rf» + D") GDV g _ 4ioM^ S*G Z„ = B<52 where m = 3 + I'Sw 6 2o5M,/(^ + B') ^B»G S Fig. S72. B Any section not containing re-en- trant angles (due to St. Venant). Z.= A* i^ (aPProx.) where A = area of sec I, = polar moment of in y = distance of furthest cdg 22901^/ ^- A*G tion; ertia of section ; e from centre of section, Torsion. General Theory. 579 Twist of Shafts. — In Chapter X., we showed that when an element was sheared, the amount of slide x bore the follow- ing relation : — I G (i.) where f, 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. If Qr be expressed in radians, then — Fig. 573- e,D and Qr — 2fl GD If 6 be expressed in degrees — irD9 X = —7- 360 Substituting the value of x in equation (i.), we have— 360/ G irGD But M, =/.Z, =f:^, and/. = ^ hence 6 - 3^0 X 16 X M/ _ 584M./ for solid circular shafts. Substituting the value of Z, for a hollow shaft in the above, we get — e-. 584M/ G(D^ - D,*) for hollow circular shafts. N.B. — The stiffness of a hollow shaft is the difference of the stifihess 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 580 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 1° in 20 diameters; the stress corresponding to this is arrived at thus — We have above 6 = ^-dfr But when = 1°, / = 20D a-GD G_ then/. - ^g^ ^ ^^p _ ^^^^ For steel, G = 13,000,000; /, = 5670 lbs. per sq, inch Wrought iron, G = 11,000,000;^ = 4800 „ „ Cast iron, G = 6,000,000;/, = 2620 „ „ In the case of short shafts, in which the spring is of no importance, the following stresses may be allowed : — Steel, ^ = 10,000 lbs. per sq. inch Wrought iron,/, = 8000 „ „ Cast iron, yi = 3000 „ „ Horse-power transmitted by Shafts. — Let a mean force of P lbs. act at a distance r inches from the centre of a shaft ; then — The twisting mo- 1 ment on the shaft }■ = P (lbs.) X r (inches) in Ibs.-inches J The work done per ) t, /ii, \ ,, /■ u \ . revolution in foot- = P Obs.) X r (mches) X 2^ lbs. ) " The work done perl _ P (lbs.) x r (inches) x 27rN (revs.) minute in foot-lbs. | ^ where N = number of revolutions per minute. 27rPrN The horse-power transmitted = = H P '^ 12 X 33000 •"' then — _ 12 X 33000 X H.P. /D» ^'■~ 2irN ~ 5-1 _3 _ 12 X 33000 X H.P. X 5-1 _ 321400 H .P. 2tN/. - N^: 64-3 H.P . N taking/, at Sooo lbs. per square inch. D = Torsion. General Theory. ^^i 4-^/ — :^ (nearly) for 5000 lbs, per sq. inch VHPT ~ 3"S\/ ~N~ ^'^^ 7500 lbs. per sq. inch = 3\/ "1^' ^°^ i2)Ooo lbs. per sq. inch In the case of crank shafts the maximum effort is often much greater than the mean, hence in arriving at the diameter ctf the shaft the maximum twisting moment should be taken rather than the mean, and where there is bending as well as twisting, it must be allowed for as shown in the next paragraph. Combined Torsion and Bending. — In Fig. 574 a shaft is shown subjected to torsion only. We have previously seen Fig 574. (Chapter X.) that in such a case there is a tension acting ■^y^ Tension ont^ Fig. S7S- 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. 575 a shaft is shown subjected to tension only. In this case the tension acts normally to a face at 90° with the axis. In Fig. 576 a shaft is shown subjected to both torsion and 582 Mechanics applied to Engineering. tension ; the face on which the normal tensile stress is a maximum will therefore lie between the two faces mentioned above, and the intensity of the stress on this face will be Fig. 576. greater than that on either of the other faces, when subjected to torsion or tension only. We have shown in Chapter X. that the stress /„ normal to the face gh due to combined tension and shear is — /. If the tension be produced by bending, we have — 7,-2 If the shear be produced by twisting— ^' Z, 2Z Substituting these values in the above equation — /«Z = M, = M+ <JW+Mi alsoA X 2Z = /.,Z, = M„ = M + Vm» + M," Torsion,. General Theory. 583 The M„ is termed the equivalent bending moment, the Mjj the equivalent twisting moment, that would produce the same intensity of stress in the material as the com- bined bending and twisting. The construction shown in Fig. 578 is a convenient graphical method of finding In Chapter X. we also showed that — Fig. 578. f,gi--f^hw[\Q and L^ = ^ cos = sin;e /»=^Hll=tan0,or /« cos d —^ = tan M« From this expression we can find the angle of greatest normal tensile stress 6, and therefore the angle at which fracture will probably occur, in the case of materials which are weaker in tension than in shear, such as cast iron and other brittle materials. In Fig. 579 we show the fractures of two cast-iron torsion test-pieces, the one broken by pure torsion, the other by 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 6 ; the agreement is remarkably close. The following results of tests made in the author's laboratory 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 agree well with the calculated values. 584 Mechanics applied to Engineering. Twisting Bending Equivalent Modulus Angle of fracture. moment moment M twisting moment of rupture /if tons per Mt Pounds- inches. Md sq. inch Actual. Calculated. Zero 2300 4600 25-5 0° 0° 777 1925 4000 267 12° 11° nyo 2240 475° 27-1 14° 14° 1228 22SS 4820 231 17° 15" 1308 2128 4628 24-0 19° i6» 2606 137s 4320 20-8 ,,0 31° 2644 766 3520 i6-2 38° 37° 3084 Zero 3084 i6'o 43° 45° Mean of a Pure shear ... 13-0 0° 0° large number „ tension... ... 11-5 0° 0° of tests. L Pure torsion. Fig. 579. ComHued toision and bending. In the case of materials which are distinctly weaker in shear than in tension it is more important to determine the maximum shear stress than the maximum tensile stress, because a shaft subjected to combined bending and torsion will fail in shear rather than in tension. Torsion. General Theory. 5^5 In Chapter X. it is shown that the maximum shear stress — /. max. — \/ — + f^ 4 Hence the equivalent twisting moment which would produce the same intensity of shear stress as the combined bending and twisting is — and the angle at which fracture occurs is at 45° to the face^^ Fig. 577. ' In the case of ductile materials, in their normal state, the angle of fracture, as found by experiment, undoubtedly does approximately agree with this theory, but in the case of crank shafts broken by repeated stress the fracture more often is in accordance with the maximum tension theory. The maximum tension theory is generally known as the " Rankine Theory," and the maximum shear theory as the "Guest Theory," named after the respective originators of the two hypotheses. Example. — A crank shaft is subjected to a maximum bend- ing moment of 300 inch-tons, and a maximum twisting moment of 450 inch-tons. The safe intensity of tensile stress for the material is 5 tons per sq. inch, and for shear 3 tons per sq. inch. Find the diameter of the shaft by the Rankine and the Guest methods. The equivalent bending moment (Rankine) — 2 2 ' = 420 inch-tons. D' 420 _ . , = ~ — D = Q'l; mches. 10-2 5 ^ = The equivalent twisting moment (Guest) — = Vsoo^ -f 450^ = 540 inch-tons. — = ^^-^ D = 07 mches. 5-13 Helical Springs. — The wire in a helical spring is, to all intents and purposes, subject 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. 586 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 — 'xS: Fig. s8o. 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 twistmg 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 / = irDn nearly, or more accurately / = t/{TrT>nY + L*, Fig. 581, 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 moves through a distance 8, where -,and8 = ^. X f But we have -= ^ hence S = -pJ^ "I (i) Torsion. General Theory. 587 AlsoM=/.z,=/i!l^ 2 16 _ 16WD _ 2-55WD ,. . then 8 = ^SP'^ (from i. and ii.) Substituting the value of / = mrD — g^ 2-55D'W^,rD _■ SD'Wn Gd* Gd* D'W« G for steel = 12,000,000 8 = 1,500,000^* DHV« G for hard brass = 5,000,000 8 = Safe Load. — From equation (ii.), W 625,0001^ 2-S5D Experiments by Mr. Wilson Hartnell show that for steel wire the following stresses are permissible : — Diameter of wire. Safe stress {/si- \ inch . . , 70,000 lbs. per square inch I „ ... 60,000 lbs. „ „ J „ ... 50,000 lbs. „ „ When a spring is required to stretch M times its initial length L, let the initial pitch of the coils be/, then L =«/ From (i.) 8 = ^ 8 = L(M - i) = «/(M - 1) = ^~ Gd pd{M. — 1) _ t/, _ 3-14 X 70,000 _ 1^ D^ G 11,000,000 50 and so(M - 1) = — For a close coiled spring, p= d, then — Vso(M-i)=? 588 Mechanics applied to Engineering. Work stored in Springs. The work done in stretching 1 _ W8 or compressing a spring \ ~ 2 f^ X V)lf. ,, . ..... = 2 X 2-55D X Qd ^^'""^ "• ^*i ">•) Jl^nQ (substituting the value of /) = , ^ (inch-lbs.) 19,500,000 putting G = 12,000,000 Weight of Spring. — Taking the weight of i cub. inch of steel = o"28 lb., then — The weight of the spring w = o"j8^d^I X o'28 = o'22d^/ Substituting the value of /, we have — w = Q'Sgnd'D Height a Steel Spring will lift itself (A). work stored in spring ~ weight of spring , /.'^«D /^. , '* ~ r62G X o-6gnd^D ' ri2G '"*^^^ = -^ — = — feet i3'4G 161,000,000 The value of // is given in the following table corresponding to various values of/": — f, (lbs. per square inch) ... 30,000 60,000 90,000 120,000 150,000 A (feet) 5-56 22-4 50-3 89-5 139-8 h also gives the number of foot-pounds of energy stored per pound of spring. 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. 589 ,i35,°°°y ^''""^ 'P""S') o = 2 890.0008^ (brass springs) n4s ) square section' W = ^9°^°°^-S brass Safe Load for Springs -of Square Section. W = ^' — stress 70,000 lbs. per square inch 24,9608' = — ^-^ — » 60,000 lbs. „ „ 20,8ooS' ,, = — i^ „ 50,000 lbs. „ „ Taking a mean value, we have — W = J. — (square section) work stored = ' „ (inch-lbs.) steel 24,680,000 ^ ' weight = o-88«S''D (steel) Height a square-section spring) , fl_ ,, . will lift itself (steel) ] "■ - 260,600,000 ^ > /^ (lbs. per square inch) ... 30,000 60,000 90,000 120,000 150,000 -iCfeet) 3-45 138 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. 5 go Mechanics applied to Engineering, Let /3j and p^ = the mean radii of the spring in inches before and after twisting respectively ; «, and «2 = the number of free coils before and after twisting respectively ; ^1 and Qi = the angles subtended by the wire in inches before and after twisting respectively ; = 36o«i and 360/2, respectively ; 6 = ^1 — ^3, or the angle twisted through by the free end of the spring in degrees ; M, = 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 frefe wire in the spring in inches ; E = M,L _ 36oM,L 2irl(«i — «j) 2irl8 E = 734opi«iM, j.^j. ^jj.^ ^f circular section trO E = 4320Pi«iM, f ^^jj.g pf 5 g section d*e If 6^ be the angle of twist expressed in radians, we have — 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, = WR cos a, which twists the wire about an axis ad. Think of ai 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 al> represent the side view of half an annular disc, suitably attached to the wire at a, and which rotates about an axis ai. Then, by Torsion. General Theory. 591 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 Fig. 582. that which produces the bending moment in the plane ab, i.e. normal to the respective sides of the triangle of moments obi. In our expression for the twist of a shaft on p. 579, we gave the angle of twist 6c in degrees; but if we take it in circular measure, we get * = rd„ where r is the radius of the vyire, and — I -G ^_2l_ M, M, I- rG- rGZ„ " GI„ and 6. = AVR cos g GL Likewise due to bending we have (see p. 590) — a, /M /WR sin g ° ~ EI ~ EI We must now find how these straining actions affect the axial deflection of the spring. 592 Mechanics applied to Engineering. The twisting moment about ab produces a strain be = RS^, which may be resolved into two components, viz. one, ef = R^, sin a, which alters the radius of curvature of the coils, and which we are not at present concerned with ; and the other, bf = R5„ cos a, which alters the axial extension of the spring /WR^cos^a by an amount pj The bending moment about cd in the plane of the imaginary disc ab produces a strain bh = RdJ, of which Ag alters the radius of curvature in a horizontal plane, normal to the axis of the spring ; and bg- = R6,' sin a alters the axial extension of the spring in the same direction as that due to twisting, by an /WR'' sin2 o amount - EI whence the total axial ") = /WR2 extension 8 j < w jx. ^ qj^ On substitution and reduction, we get — cos^ a sm EI '-) 8 = 8«D'W Gd* cos a cos' o + I"2S and for the case of springs in which the bending action is neglected, we get — s wpnv ~ Gd^ cos a Angle. Ratio of deflection allowing for bending and torsion deflection allowing for torsion only K 0-998 IO° 0-992 IS" 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 XVII. 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 1 cubic foot of air (say o'o8o7 lb.) ; P = pressure of wind per square foot of surface exposed (poimds). 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 — Mz'= P or = P g But W = wo hence — = P g or expressed in miles per hour by substituting v - i'466V, and putting in the value of w, we have — p _ o-o8o7 X 1-466'' X V . ^,, r = ! = o'ooi;4V* 32-2 ^^ If, however, the section of the air stream is much greatei 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, tut 2 Q 594 Mechanics applied to Engineering. later experimenters have shown that such a value is probably too high. Martin gives 0*004, Kernot o'oo33, Dines 0*0029, and the most recent experiments by Stanton (see Proceedings I.C.E., vol. clvi.) give 0*0027 for t^^ 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*1 to 1 in the case of circular plates, and a mean of I "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". 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* 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„ = P . sin fl the horizontal pressure P, = P„ sin = P sin^ 6 the vertical pressure P, = P„ cos 6 = P . cos d . sin 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 Fig. 583. will enable a comparison to be made. The experimental values have been reduced to a horizontal pressure of 40 lbs. per Structures. 595 square foot of vertical section of air stream acting on a flat vertical surface. Normal pressure. Vertical pressure. Horizontal pressure. roof. Experi- ment.' P sin 9. Experi- ment. P COS e sin e. Experi- ment. P sin" 9. IO° 20° 3°: 50° 60° 70° 97 i8-i 26'4 33-3 38-1 40-0 41-0 7-0 137 20'0 257 30-6 34-6 37-6 9-6 ii7-o 22-8 25-5 24-5 20'0 i4"o 6-9 12-9 i7'3 197 197 17-3 129 17 6-2 13-2 21-4 29-2 34 -o 38-S 1-7 47 100 16-5 23-5 30-0 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 : — Flat plate Parachute (concave surface), depth diameter Sphere ... Elongated projectile Cylinder Wedge (base to wind) ,, (edge to wind), vertex angle 90° Cone (base to wind) ,, (apex to wind), vertex angle 90° » . ,1 „ 60° Lattice girders o'36-0'4i 0-5 0-S4-0-S7 0'8-o'97 0'6-07 0-9S 054 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 .. s 1 IS 25 52 4 6 65 7-5 7 t? 18 21 23 «3 23 25 30 32 19 28 31 35 40 Z6, 32 34 37 43 ' Deduced from Hutton's experiments by Unwin (see " Iron Roofs and Bridges "). S96 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 as measured by small gauges is always higher than that found from 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. Forth Bridge Experiments. Date. Small revolving gauge. Small fixed gauge. Large fixed gauge, 15' X zo'. Mean. Centre. Corner. Mar. 31, 1886 Jan. 25, 1890 26 27 31 24 >9 18 28J 23* 22 22 In designing structures, it is usual to allow for a pressure of 40 lbs. per square foot. In very exposed positions this may not be excessive, but for inland structures, unless exceptionally exposed, 40 lbs. is unquestionably far too high an estimate. In sheltered positions in towns and cities the wind pressure rarely exceeds 10 lbs. per square foot. 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 " (Broadbent and Co., Huddersfield), Chatley's " The Force of the Wind " (Griffin), Husband and Harby's "Structural Engineering" (Longmans). Weight of Roof Coverings. — For preliminary estimates, the weights of various coverings may be taken as — Covering. Slates Tiles (flat) Corrugated iron Asphalted felt Lead ... Copper Snow Weight per sq. foot in pounds. 8-9 12-20 ii-3i 2-4 S-8 i-ii S 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. 597 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 — w=e^3 8 and for arched roofs — W = ^- + 3 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. wrniD Fig. 584. 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 598 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 / I. jc^. i acting along the bar/^. ^^--;;/V<J^ . ... .^ i Take a section through X^^)/ / X'^J ^ \'"^i-': the Structure a^ ; then ^-(^\ / p \/\N? three forces,/a, qe, pq, [ ^"'^^\/ — aZ — L v X/^^j y. must be applied to the I / \ -y^i *^"'- ^^"^ ^° ^^^P '^'^ J, \' Structure in equili- FiG. 585. brium. Take moments about the point O. The forces pa and qe pass through O, and therefore have no moment about it ; but pq has a moment pq X y about O. pqXy = Wi*i + Waaij + Wa«3 - W,x, + W^, + W^, pq = 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. 586), 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. 586). 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. 599 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. 587. .The funicular polygon has been fully described in Chapters IV. and XII, hence no further description is necessary. The direction of the reaction at the fixed support Fig. 586. 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 6oo Mechanics applied to Engineering. diagram is also constructed ; it presents no difficulties beyond that mentioned in thfe 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. Fig. 587. 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 Boof. — 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 \yjpa on the vector polygon in magnitude and direction (Fig. 588). Its position on the structure must then be deter- mined. This has been done by constructing a funicular polygon Structures. 60 1 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. 588. 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. 589). The bending moment on the post is obtained by the product j>a 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 compressions. But in the present instance, where there is bending in one of the members, we must intiroduce 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 — Fig. 589. 6o2 Mechanics applied to Engineering. F X >-Y = ^ X 3 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. sgo. 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. 590). 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. 603 this diagram as shown in the lower figure ; then, \i d = depth wP of this parabola to proper scale, we have -— = d, where a is the equivalent evenly distributed load due to the train. (The small diagram is not to scale in this case.) Let W, = total rolling load in tons distributed on each pair of rails ; S = span in feet. Then for English railways W, = i-6S + 20 Maximum Shear due to a Rolling Load, Concentrated Rolling Load. — The shear at any section is equal to the algebraic sum of the forces acting to the right or left of any section, hence the shear between the load and either abutment is ^"'- s?'- equal to the reaction at the abutment. We have above — Ri/= Wa;, R, = W^ likewise Rj W^ When the load reaches the abutment, the shear becomes W, W and when in the middle of the span the shear is — . The shear diagram is shown shaded vertically. Uniform Rolling Load, whose Length exceeds the Span. — Let (VrrrrrCrC^ w = the uniform live load per foot run, and zv^ = the uniform dead load per foot run. Let M = ^ 6o4 Mechanics applied to Engineering. The total load on the structure = ■wn = Vf then Ri/= — = — 2 2 Ri = — «2 = Kw" 2/ where K is a constant for any given case. But as the train passes from the right to the left abutment, the shear between the head of the train and the left abutment is equal to the left reaction Rj, but this varies as the square of the covered length of the bridge, hence the curve of shear is a parabola. When the train reaches the abutment, I = n; then Ri = — = — 2 2 The curves of shear for both dead and live loads are shown in Fig. 593. When a train passes from right to left over the point /, the shear is re- versed in sign, because the one shear is positive, and the other negative. The distance x between the two points /, / is known as the " focal distance " of the bridge. The focal distance x can be calculated thiis : The point / occurs where the shear due to the live load is equal to that due to the dead load. Fig. 593- wn The shear due to the live load = — j dead ,, = -^ = w^ 2 m^ (I \ Solving, we get — n = l^U + M^ - /M and X = I — 2n = /(i - 2 VmTIP + 2M) Structures. 605 Determination of the Forces acting on the Members of a Girder.— In the case of the girder given \W'P^''''''"4'^ Changes Fig. 594. as an example, the dead load w^ = 075 ton per foot, or ?-^^ ^^ = 5 tons per joint. The live load w = 175 ton per foot run, or 175 X 13*5 = 23-6 tons on each bottom joint, thus giving 5 tons on each top joint, and 28'6 tons on each bottom joint. The forces acting on the booms can be obtained either by 6o6 Mechanics applied to Engineering. constructing a reciprocal diagram for the structure when fully loaded as shown, or by constructing the bending-moment diagram for the same conditions. The depth of this diagram at any section measured on the moment scale, divided by the depth of the structure, gives the force acting on the boom at that section. The results should agree if the diagrams are carefully constructed. The forces in the bracing bars, however, cannot be obtained by these methods, unless separate reciprocal diagrams are con- structed for several (in this instance six) positions of the train, since the force in each bar varies with each position of the train. The nature of the stress on the bracing bars within the focal length changes as the train passes ; hence, instead of designing the focal members to withstand the reversal of stress, it is usual, for economic reasons, to counterbrace these panels with two tie-bars, and to assume that at any given instant only one of the bars is subjected to stress, viz. that bar which at the instant is in tension, since the other tie-bar is not of a suitable section to resist compression. All questions relating to moving loads on structures are much more readily solved by " Influence Lines " than by the tedious method of constructing reciprocal diagrams for each position of the moving load. By this means the greatest stresses which occur in the various members of the structure can be determined in a fraction of the time required by the older methods. Deflection of Braced Structures. — The deflection produced by any system of loading can be calculated either algebraically by equating the work done by the external forces to the internal elastic work done on the various members of the structure, or graphically by means of the Williott diagram. Readers should refer to Warren's " Engineering Construc- tion in Iron, Steel, and Timber," Fidler's " Practical Treatise on Bridge Construction," Husband and Harby's " Structural Engineering," Burr and Falk's " Influence Lines for Bridges and Roofs." Girder with a Double System of Triangulation. — Most girders with double triangulation are statically indeter- minate, and have to be treated by special methods. They can, however, generally be treated by reciprocal diagrams without any material error (Fig. 595). We will take one simple case to illustrate two methods of treatment. In the first each system is treated separately, and where the members overlap, the forces must be added : in the second the whole diagram will Structures. 607 be constructed in one operation. The same result will, of course, be obtained by both methods. Fig. 595. / V^ 6o8 Mechanics applied to Engineering. In dealing with the second method, the forces mn, hg 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. 596. 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 Structures. 609 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 r ■ — > 3_ T^ the thickness by reducing , the number of plates. The \ i i bending-moment diagram N^ lends itself very readily to the stepping of the plates. Thus suppose it were found Fig. 597. 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.^ The shear in the web may be conveniently obtained from the shear diagram (see Chapter XII.). 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^ = depth of web in inches, / = 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 -^ span. The thickness of web is then readily obtained. If on calculation the thickness comes out less than | inch, and it has been decided not to use a thinner web, the depth in some cases is decreased accordingly within reasonable limits. ' It is usual to allow from 6 inches to 12 inches overlap of the plates beyond the points thus, obtained, 2 R 6io Mechanics applied to Engineering. The web is attached to the flanges or booms by means of angle irons arranged thus : Fig. 599. Fig. 598. The pitch / of the rivets must be such that the bearing and shearing stresses are within the prescribed Hmits. On p. 389 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, which has to be taken by the rivets a, a. Fig. 598. The shep-r per (gross) inch run of web plate along ef The shearing resistance of each rivet is (in double shear) 4 Whence, to satisfy shearing conditions — This pitch is, however, often teo small to be convenient ; then two (zigzag) rows of rivets are used, and/ = twice the above value. The bearing resistance of each rivet is — dtf,=Up Structures. 6ll The bearing pressure /, 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, 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, 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 cff 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 -^^ V^"^ r^/ r^ \ \ '^ ) // r w ¥/ ] d b / ( ^ 1^ ^ 1 1 i~\ -^=H h , \\ ^ - , ; \ KJ v_/ y^ ■ A rs Fig. 6qo. 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. 600. That 6i2 Meclianics 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 on 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. 613 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 6 14 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 stiffeners /,, there should certainly be one at every cross girder, or other con- centrated load, and for the prevention of twisting, well-fitted plate stiffeners 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 = - rouehly 500 ^ ■' 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 Avere 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. 615 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. 541). 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 liie 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 structureis 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 throtigh 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 — pxag-.) and P = = T 2T <;-) 6i6 Mechanics applied to Engineering. T The mean stress on the section = — d hence, when — d ^ . . r * = T, P = 2 X mean intensity of stress d „ . . 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 — 2'67T — -J- = or < 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 j R, = the radius of the arch in feet. — a/S a/S Then d = ^— for brick to ^^ for masonry 2-2 3 or, according to Trautwine — d = where K = i for the best work ; I •12 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 th3 joint at an angle greater than the friction angle. Let // be tie line of thrust, then sliding will occur in 9ie manner Structures. 617 shown by the dotted lines, if the joints be arranged as shown at M— 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 ; 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. 601. Fig. 602. 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 Rj, and it was required to bring it to Rj. If the original polar distance were OoH, the new polar distance required to bring the rise to Rj would be dH = OqH X ^• 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 6i8 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 refei 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. 7v y / B' d OU Fig. 603. 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 itiay touch the inner and one the outer middle half lines. As a matter of fact, an Structures. 6ig 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. 604. 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 CMOwn, Ihe position of the line of thrust may be found thus : ' Fio. 606. ' See also a paper by Bell, I.C.E., vol. xxxiii. p. 68. 620 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 nin'. Both the slope and the vertical position of the element are altered by the straining of the rib. First consider the efifect 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 — AD V . „ AAi . ,. , 6 being expressed in circular measure. Likewise — AjD X , ^ AAi ... AA; = AB A,D=^.^ = e.* . . (11.) 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 'XQy — 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 niti, the rib moves through an angle B„ and the point A receives a vertical displacement S . B^ 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 — %6x^'&B, = o (iv.) By similar reasoning — Mx-, + Si9a = o or S(S - a;)0 + S^A = o (v.) If the arch be rigidly fixed at C — and "SSx = o (vL) Structures. 621 If it be fixed at A — Sa^o and S(S — x)6 = o from v. If it be fixed at both ends, we have by addition — S(S9 -xe + xff) = o 2Se = o Then, since S is constant — 26 = o (vii.) Whence for the three conditions of arches we have — Arch hinged at both ends, %y6 = o from iii. „ fixed „ „ "Zyd, and 39 = o from iii. and vii. „ „ „ one end only, ^yO, and 1,x6 = o from i. and vi. We must now find an expression for 6. 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 po) and after straining p^. Then, using the symbols of Chapter XIII., we have — M„ = 5^, and M^ = ^ Po Pi Then the bending moment on the rib due to the change of curvature when strained IS — ^ Fig. 607. -l-i M = Mi- M„= Ya(---\ \pi Pa' (viii.) But as / remains practically constant before and after the strain, we have — 61 = -, and 5o = Pi Po and 0. - 5o = (9 = /- - -) \Pi. po' or /= . I Pi I P» (ix.) 622 Mechanics applied to Engineering. EI Then we have from viii. and ix. — M/ = Eie ande=- If the arched rib be of constant cross-section, constant ; but if it be not so, then the length / must be taken , I ■ 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 c; 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 d , also adb, cda. are similar ; hence — Fig. 608. F cd \ ab ac 7d ab ad n ac cd F or — = _ = orY .ab = Yi.ac=y[. but H is constant for any given case. Let ac = z. Hence the expression S5_y = o / . may be written Si\Ty = o, since ^y is constant or l^z .y = Then, substituting in a similar manner in the equations above, we have — Arch hinged at both ends, "S^z .y = o „ fixed „ „ Ss . J" = o, and S3 = o „ „ at one end only, Ss . 7 = o, and S.XZ = o Thus, after the median line of the arch has been drawn, a line of thrust for uneven loading is constructed; and the Structures. 623 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 Y. y must be found, some z's being negative, and Fig. 609. 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 versa 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 a's must also Fig. 610. 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 — Jc ^ 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 — Or/= — ; M T . nb (see Fig. 608) 624 Mechanics applied to Engineering. and the maximum stress in the material due to both — T , H.2 °^ = A + ^ 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 - t\ N^ = R2 + ?1 4 4 N" - Ni^' = R= - R,» Fio. 611. Substituting the value of Ni and reducing, we get — _£L+?£1=(R + R,)(R_R,) n^ n But R - Rj = 8 and R + Ri = 2R (nearly) The fraction - is very small, viz. <^^ and is rarely more than n tj jJ^j ; hence the quantity involving its square, viz. is negligible. l"hen we get — N» 9000000' Structures. 62s *-«R . , . (X.) But (^y^N'^-Ra also(^y=p^-(p-R)» N^ - R2 = p-" - (p - R)» from which we get — N^ = 2pR Substituting in x., we have — 8 n and P - 2R also Pi -2R, The bending moment on the rib due to the change of curvature is — M = EI| — ) from viiL Vp Pi/ and the corresponding skin stress is— M My M^ ^~ Z- I - 2I where d is the depth of the rib in inches if R and N are taken in inches. Then, substituting the value of M, we have — i'Eld / R RA Ni' may without sensible error (about i in 2000) be written N»; then— /=||(R-RO E^ /- jja 2 s 626 Meclianics applied to Engineering. The stress at the crown due to the change of curvature on account of the compression of the rib then becomes — -• _ 2E^p and 1=/' (see p. 374) where _/j is the compressive stress all over the section of the rib at the crown ; hence — J ^12 vj-S taking /„ as a preliminary estimate at about 4 tons per square inch ; then - =' — — (about). In the case of the change of curvature due to change of temperature, it is usual to take the expansion and contraction on either side of the mean tempera- ture of 60° Fahr. as \ inch per 100 feet for temperate climates such as England, and twice this amount for tropical climates. Hence for England — n 1200 4800 putting E = 12,000 tons per square inch; •' va Thus in England the stress due to temperature change of curvature is about five-eighths as great as that due to the com- pression change of curvature. The value of p varies from I'zsN to 2'sN, and d from N N — to — . 30 20 Hence, due to the compression change of curvature — /= o'6 to o"8 ton per square inch and due to temperature in England^ / = 0-4 to o"5 ton per square inch In the case of ribs fixed rigidly at each end, it can be Structures. 627 shown by a process similar to that given in Chapter XIII., of the beam built in at both ends, that the change of curvature stresses at the abutments of a rigidly held arch is nearly twice as greatj and the stress at the crown about 50 per cent, greater than the stress in the hinged arch. An arched rib must, then, be designed to withstand the direct compression, the bending stresses due to the line of thrust passing outside the section, the bending stresses due to the change of curvature, and finally it must be checked to see that it is safe when regarded as a long strut ; to prevent side buckling, all the arched ribs in a structure are braced together. Effect of Sudden Loads on Structures. — If a bar be subjected to a gradually increasing stress, the strain increases in proportion, provided the elastic limit is not passed. The work done in producing the strain is given by the shaded area abc in Fig. 612. Let, X = the elastic strain produced ; / = the unstrained length of the bar ; / = the stress over any cross-section ; E = Young's Modulus of Elasticity ; A = sectional area of bar. Then .* = ir ill The work dorie in straining! /Aa: _/°A/ a bar of section A / ~ 2 ~ 2E , A/ P = T^E = \ vol. of bar X modulus of elastic resilience The work done in stretching a bar beyond the elastic Umit has been treated in Chapter X. If the stress be produced by a hanging weight (Fig. 613), and the whole load be suddenly applied instead of being gradually increased, then, taking A as i square inch to save the constant repetition of the symbol, we have — Fig. 6i2. The work stored in the weight w^ due to falling through a height jcf = fx But when the weight reaches c (Fig. 613), only a portion of the energy developed during the fall has been expended in stretching the bar, and the remainder is still stored in the 628 Mechanics applied to Engineering. falling weight ; the bar, therefore, continues to stretch until the kinetic energy of the weight is absorbed by the bar. * • fx The work done in stretching the bar by an amount « is — ; hence the kinetic energy of the weight, when it reaches c, is the difference between the total work done by the weight and the work done in stretch- fx fx ing the bar, ox fx = — . The bar will therefore con- tinue to stretch until the work taken up by it is equal to the total work done by gravity on the falling weight, or until the area ahd'xz equal to the area ageh. Then, of course, the area bed is equal to the area ach. When the weight reaches h, the strain, and therefore the stress, is doubled ; thus, when a load is suddenly applied to a bar or a structure, the stress produced is twice as great as if the load were gradually applied. When the weight reaches ^, the tension on the bar is greater than the weight; hence the bar contracts, and in doing so raises the weight back to nearly its original position at a ; it then drops again, oscillating up and down until it finally "comes to rest at c. See page 265. If the bar in question supported a dead weight W, and a further weight w^ were suddenly applied, the momentary load on the bar would, by the same process of reasoning, be W + 2ze/„. If the suddenly applied load acted upwards, tending to compress the bar, the momentary load would be W — 2Wo. Whether or not the bar ever came into compression would entirely depend upon the relative values of W and Wo. The special case when Wo = 2W is of interest ; the momen- tary load is then — W- 2(2 W) = -3W the negative sign simply indicating that the stress has been StrucHires. 629 (ii) (ill) Fig. 614. reversed. This is the case of a revolving loaded axle. Con- sider it as stationary for the moment, and overhanging as shown — The upper skin of the axle in case (i) is in tension, due to W. In order to relieve it of this stress, i.e. to straighten the axle, a force — W must be applied, as in (ii) ; and,- further, to bring the upper skin into compression of the same intensity, as in (i), a force — 2W must be applied, as shown in (iii). Now, when an axle revolves, every portion of the skin is alternately brought into tension and compres- sion ; 'hence we may regard a revolving axle as being under the same system of loading as in (iii), or that the momentary load is three times the steady load. The following is a convenient method of arriving at the momentary force produced by a suddenly applied load. The dead load W is shown by heavy lines with arrows pointing up for tension and down for compression, the suddenly -n ■ZIV ^ < ^ l?^~ i CM ^1 » I \'V I Fig. 615. 630 Mechanics applied to Engineering. applied load Wi is shown by full light lines, the dynamic increment by broken lines. The equivalent momentary load is denoted W„. In some instances in which the suddenly applied load is of opposite sign to that of the dead load, the dead load may be greater than the equivalent momentary load ; for example, cases 5, 6, 7, 8. In case 7 the momentary load is compressive ; when designing a member which is subjected to such a system of loading it should be carefully considered from the standpoint of a strut to withstand a load W^, and a tie to withstand a load W. In case 8 the member must be designed as a strut to support a load — W, and be checked as a tie for a load W„. In case 3, W may represent the dead load due to the weight of a bridge, and Wj the weight of a train when standing on the bridge. W^ is the momentary load when a train crosses the bridge. There are many other methods in use by designers for allowing for the effect of live loads, some of which give higher and some lower results than the above method. Falling Weight. — When the weight W„ falls through a height h before striking the collar on the bottom of the bar, Fig. 613, the momentary effect produced is considerably greater than 2W„. Let/, = the static stress due to the load z£/„ _ w„ ~ A / = the equivalent momentary stress produced I 2 2E ^=E Work done in stretching the bar — ' or the resilience of the bar per square inch of section Work done by the falling weight, \ ^ W , s _ .,, , s per sq. inch of section j ^ ^^ + •»; -W + x) /../(. V. + f) If a helical buffer spring be placed on the collar, the momentary stress in the bar will be greatly reduced. Let Structures. 63 1 the spring compress i inch for a load of P lbs., and let the maximum compression under the falling weight be S inches, and h the height the weight falls before it strikes the spring. The momentary load on the spring = P8 The momentary load on the bar —f,h. PS =/.A The total work done in stretching the ) _ kf^l P82 bar and compressing the spring j 2E 2 The work done by the falling weight = (W„(/^ + S + a:) from which the value of/^ can be obtained. Experiments on the Repetition of Stress.— In 1870 Wohler published the results of some extremely interesting experiments on the effect of repeated stresses on materials, often termed the fatigue of materials ; since then many others have done similar work, with the result that a large amount of experimental data is now available, which enables us to con- struct empirical formulas to approximately represent the general results obtained. In these experiments many materials have been subjected to tensile, compressive, torsional, and bending stresses which were wholly or partially removed, and in some cases reversed as regards the nature of the stress imposed. In all cases the experiments conclusively showed that when the intensity of the stresses imposed approached that of the static breaking strength of the material, the number of repe- titions before fracture occurred was small, but as the intensity was reduced the number of repetitions required to produce fracture increased, and finally it was found that if the stress imposed was kept below a certain limit, the bar might be loaded an infinite number of times without producing fracture. This limiting stress appears to depend (i) upon the ultimate strength of the material {not upon the elastic limit), and (ii) upon the amount of fluctuation of the stress, often termed the " range of stress " to which the material is subjected. In a general way the results obtained by the different experimenters are fairly concordant, but small irregularities in the material and in the apparatus used appear to produce marked effects, consequently it is necessary to take the average of large numbers of tests in order to arrive at reliable data. The following figures, selected mainly from Wohler's tests, will serve to show the sort of results that may be expected from repetition of stress experiments : — 632 Mechanics applied to Engineering. Material. Krupp's Axle Steel. Tensile strength, varying from 42 to 49 tons per sq. inch. Tensile tress applied Nominal bending stress in tons from per square inch to repetitions before fracture. in tons per square inch from to repetitions before fracture. 38-20 18,741 26-25 1,762,000 33'40 46,286 25-07 1,031,200 28-65 170,170 24-83 1,477,400 26-14 123,770 23-87 5,234,200 23-87 473,766 23-87 40,600,000 22-92 13,600,000 (unbrolcen) (unbroken) Nominal ben revoU from ding stress in a ing azle to Number of repetitions before fracture. 20-1 — 20-1 5S.IOO 17-2 -17-2 1Z7.77S 16-3 -16-3 797,525 15-3 -'5-3 642,675 11 M 1.665,580 I) l» 3,114,160 14-3 -I4'3 4.163.37s ,( r* 45,050,640 Material. Krupp's Spring Steel. Tangle strength, 57-5 tons per sq. inch. Tensile stress applied in tons p in from sr square ch to repetitions before fracture. in tons p in from er square ch to repetitions before iractuiv. 4775 7-92 62,000 38-20 4-77 99.700 >* 15-92 149,800 »J 9-55 176,300 w 23-87 400,050 »» 14-33 619,600 )i 27-83 376,700 »» 2.135.670 I) 31-52 19,673,000 (unbroken) 11 19-10 35,800,000 (unbroken) 42-95 9-SS 81,200 33-41 4-77 286,100 If 14-33 1,562,000 1. 9-55 701,800 19-10 225,300 II 11-94 36,600,000 II 23-87 1,238,900 (unbroken) »» 300,900 If 28-65 33,600,000 (unbroken) Strttcturis, Material. Phceiiix Iron for Axles. Tensile strength, 21 '3 tons per sq. inch. 633 Nominal in a re> bending stress •olving axle. Rounded shoulder. Square shoulder. Conical shoulder. From To Number of repetitions before fracture. iS'3 -15-3 56,430 134 • -13-4 183,145 II-5 -ri-S 909,84.0 %^ - 9-6 4,917,992 8-6 - 8-6 19,186,791 2,063,760 535,302 7-6 - 7-6 132,250,000 (not brolien) 14,695,000 1,386,072 The above figures show very clearly the importance of having well rounded shoulders in revolving axles. Material (Tests by Author). Mild steel cut from an over-annealed crank shaft. Elastic limit, 8'28 tons sq. inch ; tensile strength, 27^98 tons sq. inch. Nominal bending stress in a revolving axle. Number of repetitions before fracture. From To lo-o 95 90 8-5 8-0 -lO'O - 9-S — 9'0 = 8:^ 418,100 842,228 729,221 5,000,000 not broken 5,000,000 not broken The figures given in the above table were obtained from a series of tests made on bars cut from a broken crank-shaft ; the material was in a very abnormal condition owing to pro- longed annealing, which seriously lowered the elastic limit. Other specimens were tested after being subjected to heat by a steel specialist, which raised the elastic limit nearly 50 per cent., but it did not materially affect the safe limit of stress under repeated loading, thus supporting the view, held by many, that the capacity of a given material to withstand repeated loading depends more upon its ultimate or breaking stress than upon its elastic limit. 634 Mechanics applied to Engineering. Various theories have been advanced to account for the results obtained by repeated loading tests, but all are more or less unsatisfactory ; hence in designing structures we are obliged to make use of empirical formulas, which only ap- proximately fit experimental data. Many of the empirical formulas in use are needlessly com- plicated, and are not always easy of application ; by far the simplest is the " dynamic theory " equation, in which it is assumed that the varying loads applied to test bars by Wohler and others produce the same effects as suddenly applied loads. In this theory it is assumed that a bar will not break under repeated loadings if the "momentary stress" (see Fig. 614) does not exceed the stress which would produce failure if statically applied. Whether the assumptions are justified or not is quite an open question, and the only excuse for adopt- ing such a theory is that it gives results fairly in accord with experimental values, and moreover it is easily remembered and applied. The diagram, Fig. 616, is a convenient method of showing Static braaking stress i Fig. 616. to what extent repetition of stress experiments give results in accordance with the " dynamic theory." In this diagram all Structures. 63 5 stresses are expressed as fractions of the ultimate static stress, which will cause fracture of the material. The minimum stress due to the dead load on the material is plotted on the line aoh, and the corresponding maximum stress, which may be applied over four million (and therefore presumably an infinite number of) times is shown by the spots along the maximum stress line. If the " dynamic theory " held perfectly for repeated loading, or fatigue tests, the spots would all lie on the line AB^, since the stress due to the dead load, i.e. the vertical height between the zero stress line and the minimum stress line plus twice the live load stress, represented by the vertical distance between the minimum and maximum stress lines, together are equal to the static breaking stress. 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 o'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 o'S 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. For an application of this diagram to the most recent work on the effect of repeated loading, readers should refer to the Proceedings of the Institution of Mechanical Engineers., November, 1911, p. gro. Assuming that the dynamic theory is applicable to mem- bers of structures when subjected to repeated loads, we proceed thus — let the dead or steady load be termed W,„i„ , and the live or fluctuating load (W,„„ — W„[„), then the equivalent static load is — W = W ■ Z 2('W - W • ) * * c ^ rain, -r ^ V max. * ^ inin./ or using the nomenclature of Fig. 614 W„ = W + 2(W + K'l - W) W^ = W + 2Wi The plus sign is used when both the dead and the live loads act together, i.e. when both are tension or both com- pression, and the minus when the one is tension and the other compression. 636 Mechanics applied to Engineering. For a fuller discussion of this question, readers are referred to the following sources : — Wohler's original tests, see En- gineering, vol. xi., 1871; British Association Report, 1887, p. 424; Unwin's " Testing of Materials " ; Fidler's "Practical Treatise on Bridge Construction " ; Morley's " Theory of Structures " ; Stanton and Bairstow, " On the Resistance of Iron and Steel to Reversals of Direct Stress," Froc. Inst. Civil Engineers, 1906, vol. clxvi. ; Eden, Rose, and Cunningham, "The Endurance of Metals," Froc. Inst. Mech. Efigineers, November, 1911. CHAPTER XVIII. HYDRAULICS. In Chapter X. 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 bddy returns to its original form after the removal of the stress, the body is said to be an elastic solid {t:.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 mider 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 imder 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 lbs. per cubic foot, or o'o36 lb. 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 ( T^" + 2 so 000 the volume at 39' 2° Fahr. or 500° ■! = ^i, absolute as i \ PresBure due to a Given Head. — If a cube of water of 638 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 62'^ will weigh — 5= 0-434 lb. Hence a column of water i foot 144 high produces a pressure of o'434 lb. per square inch. Temp. Fahr. 3»°. ig-J>. 50°. 100°. T50°. «o°. Ice. Water. Weight per 1 ' cubic foot in } lbs ) 57'2 62-417 62-425 62-409 62-00 61 20 60-14 59-84 Volume of a \ given weight, taking water at 39'2° Fahr. as I ) I '091 I •0001 i-oooo 1-0002 1-007 I 020 1-038 1-043 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 ; p = 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 lb. Then p = 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 lb. per square inch, Taking the pressure due to the atmosphere as 14-7 lbs. per square inch, we have the head of water corresponding to the pressure of the atmosphere — 14-7 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'. 617 — Hydraulics. 639 The pressure over any unit area of surface at « = /„ = o-i,j,i,h„ b = P, = o-434'4. 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 in 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. 374- Let V = the dimmution of volume under any given pressure p in pounds per square inch (corresponding to x on p. 374); V = the original volume (corresponding to / on p. 37 4) j K = the modulus of elasticity of volume of water ; p =■ the pressure in pounds per square inch. Fig. 617. Then \ = ^ _/V orK=^ V K = from 320,000 to 300,000 lbs. per square inch. Thus water is reduced in bulk or increased in density by I per cent, when under a pressure of 3000 lbs. 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. 640 Mechanics applied to Engineering. Let Fig. 618 represent an immersed surface. Let it be divided up into a large number of horizontal strips of length ^//j, etc., and of width* each at =-=^-= ^f=: ^^";3 p^-=--- a depth h-^, th, etc., respectively . t^. — .^^^^^ from the surface. Then the / '' ] total pressure on each strip is I^^3^^' PA^t PA^, etc., where pi, p^ etc., are the pressures corre- FiG. 618. spending to A,, ^, etc. But/ = wh, and UJ> = Oi, Ij) = ^a, etc. The total pressure on each strip = wcL^h^, wciji^, etc. Total pressure on whole surface = P„ = w(^Ai + aji^ + etc.) But the sum of all the areas a^, a^, etc., make up the whole area of the surface A, and by the principle of the centre of gravity (p. 58) we have — ajii + aji^ +, etc., = AHo where Ho is the depth of the centre of gravity of the immersed surface from the surface of the water, or — P„ = wAH, — '"n.'-H 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„ = the head of water above the centre of pressure ; Ho = the head of water above the centre of gravity of the surface ; Q = the angle the immersed surface makes with the surface of the water ; lo = the second moment, or moment of inertia of the surface about a line lying on the surface of the water and passing through o ; I = 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. 641 Ro = the perpendicular distance between the two axes; K* = the square of the radius of gyration of the surface about a horizontal axis passing through the c. of g. of the surface ; «„ Oj, etc. = small areas at depths h-^, h^, etc., respectively below the surface and at distances x^^x^, etc., from O. A = the area of the surface. Taking moments about O, we have — p^OiXx ■^■pia^i +, etc. = P*, XT wh^a^Xi + wh^a^^ +, etc. = wAHo^ ° "sin B a/ sin Q{a^oc^ + a^ +, etc.) = z«/AH„-t ° 'sin 61 sin^ Q{a^x^ + a^^ +, etc.) = AH„H, Fig. 619. On p. 76 we have shown that the quantity in brackets on the left-hand side of the equation is the second moment, or moment of inertia, of the surface about an axis on the surface of the water passing through O. Then we have — sin^ e lo = sin^ e(I + ARo") = AH„H„ or sin^ B{k.K^ + AR,") = AH„H, Substituting the value of R,, we get — „ sin^ e K* + Ho" ^' = H^ The centre of pressure also lies in a vertical plane which passes through the c. of g. of the surface, and which is normal to the surface. The depth of the centre of pressure from the surface of the water is given for a few cases in the following table : — z T 642 Mechanics applied to Engineering. Vertical surface. «'. H„. Ht. Rectangle of depth d with upper edge at surface 1 of water / 12 2 i^ Circle of diameter d with circumference touching \ surface of water / 16 2 1^ Triangle of height d with apex at surface ofl water and base horizontal / 18 2</ 3 3rf 4 The methods of finding k* and H,, have been fiilly described in Chapter III. Graphical Method for finding the Centre of Pressure. — In some cases of irregularly shaped surfaces the algebraic method given above is not ^^^^^^^ g , —^ easy of application, but the following ^^^-^^^^^^"^3= graphic method is extremely simple. / \ \ In the figure shown draw a series of lines across, not necessarily equi- distant; project them on to a base- line drawn parallel to the surface of hb the water. In the figure shown only one line, CM, is projected on to the base- line in bb. Join bb to a point on the surface of the water and vertically over the centre of gravity of the immersed surface, which cuts off a line dd ; then we have, by similar triangles — aa bb h^ dd ~ dd ~ h^ or the width of the shaded figure dd at any depth h^ below the surface is proportional to the total pressure on a very narrow strip aa of the surface ; hence the shaded figure may be regarded as an equivalent surface on which the pressure is uniform ; hence the c. of g. of the shaded figure is the centre of pressure of the original figure. It will be seen that precisely the same idea is involved here as in the modulus figures of beams given in Chapter XI. Fig. 620. Hydraulics. 643 Fig. 621. The total normal pressure on the surface is the shaded area A, multiplied by the pressure due to the head at the base-line, or — Total normal pressure = wA^j Practical Application of the Centre of Pressure.— A good illustration of a practical applica- tion of the use of the centre of pressure is shown in Fig. 6a i, which represents a self- acting movable flood dam. The dam AB, -usually of timber, is pivoted to a back stay, CD, the point C being placed at a distance = f AB from the top j hence, when the level of the water is below A the centre of pressure falls below C, and the dam is stable; if, however, the water flows over A, the centre of pressure rises above C, and the dam tips over. Thus as soon as a flood occurs the dam automatically tips over and prevents the water rising much above its normal. Each section, of course, has to be replaced by those in attendance when the flood has abated. Velocity of Flow due to a Given Head. — Let the tank shown in the figure be provided with an orifice in the bottom as shown, through which water flows with a velocity V feet per second. Let the water in the tank be kept level by a supply-pipe as shown, and suppose the tank to be very large compared with the quantity passing the orifice per second, and that the water is sensibly at rest yw. 6sa. and free from eddies. Let A = area of orifice in square feet ; Aj = the contracted area of the jet ; Q = quantity of water passing through the orifice in cubic feet per second ; V = velocity of flow in feet per second ; W = weight of water passing in pounds per second ; h = head in feet above the orifice. Then Q = A„V Work done per second by W lbs. of \ _ ^tt , r m, water falling through h feet / ~ loot-iDs. 644 Mechanics applied to Engineering. ) = But these two quantities must be equal, or — Kinetic energy of ■ the water on \ _ WV* leaving the orifice ) ~ 2g WA = , and h = — and V = V '2'gh 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. 623. 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^, which is found experimentally. Hydraulic Coefficients. — The coefficient of discharge Y^ may be split up into the coefficient of velocity K„ viz. — actual velocity of fl ow and the coefficient of contraction K„ viz. — actual area of the stream area of the orifice Then the coefficient of discharge — K^ = K„K. Hydraulics. 645 The coefficient of resistance — jr _ 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 A, feet vertically while travelling h.^ feet horizontally, where ^1 = \gfi and/ = A /?^ ^ g also h^ = v^ where v^ = 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. — V = ij 2gh where h is the head of water in the tank over the centre of the orifice. Then », = '^^ andK. = -»=- ^ 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 Vi = ij 2gh^, 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,. 646 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 diflScult 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. 624); if not, the water dribbles down the sides and makes the coefficient variable. In this case K, = about 0*97, and K, = about 0-64, giving K^ = o'62. Experiments show that Ki decreases with an increase in the head and the diameter of the orifice, also the sharper the edges the smaller is the coefficient, but it rarely gets below o'sga, and sometimes reaches o'64. As a mean value K^ = o'62. Q = o'62A.\ 2gh Fig. 624. Fig. 625. Rounded Orifice. — If the orifice be rounded to the same form as a contracted jet, the contraction can be entirely avoided, hence K, = i ; but the friction is rather greater than in the plain orifice, K, = o"96 to cgS, 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 where the water leaves the orifice ; as a mean value we may take — Q= o-qTK'J 2gh Hydraulics. 647 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 /k \' same as in the plain orifice. When \^)nhi 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 ^ = Vj ; and the head = h^ the velocity at o = V, ; „ „ = h^ Then the loss of head = ^— = -'- (see p. 673) V 2 ^V. — V )^ and A,= — + '^ -^ The velocities at the sections a and 6 will be inversely as the respective areas. If K„ be the coefficient of contraction at 6, we have Vj = j^; inserting this value in the expression given above, we get — Fig. 626. v„ = V'Hi-') /^gK The fractional part of this expression is the coefficient of velocity K, for this particular form of orifice. The coefficient of discharge Kj, 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.AV2PI The pressure at a is atmospheric, but at b it is less (see 648 Mechanics applied to Engineering. p. 666). If the stream ran clear of the sides of the pipe into the atmosphere, the discharge would be — but in this case it is — Qi = K„Av'2^^. Let h^ — nhf Then Qi = YLjykiJ 2gnhi or the discharge is —^ — times as great as before ; hence we may write — Q, = =^X lL^y.K^2gnh^ -^)h when running a full stream and only h^ when running clear. The difference is due to a partial vacuum at b amounting to hAn{=~-\ — i \. If the head h^ be kept constant and the length of the pipe be increased it will be found that the quantity passing diminishes. The maximum quantity passes when D = 4//4j where D is the diameter of the pipe in feet, and / is the friction coefficient, see p. 679. Readers familiar with the Calculus will have no difficulty in obtaining this result, the relation can also be proved by calculating the quantity of water passing for various values of the length ab. When the pipe is horizontal n = 1, and the vacuum head is Hydraulics. 649 The following results were obtained by experiment : — The diameter of the pipe = 0*945 inches Length h^ = 2 "96 ,, Ka = o"6i2 Head ^j (inches) ... l6-i 131 lO'I 8-1 6-1 0-786 4-1 0-780 3'« Kd 0799 0797 0797 0-792 o'773 Vacuum head by ex- periment in inches 167s I4'37 11-95 10-50 9-00 7-45 6-35 Vacuum head by cal- culation 1 6 '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„ due to the flow of water over the surface of the tank all round the orifice. Then we have — P„ = w/iA + P, Fig. 627. 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, is zero. Let the section of the jet be a, and the area of the orifice A. 650 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 = 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 1 "^ »™P"lse per second, or pressure waSf^ _ K/AV hence a = o'sA or K„ = o'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 A 1 A i 2 Kd 0'6i 0-56 o-SS 0-S4 0-S3 052 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. 6si 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„, the value of K„ 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 — Tr\/' 2-6i8 - r6i8 A^ where A is the area of the orifice, and A„ is the area of the approach channel. The author has, however, obtained a rational value for this coefficient (see Engineering, March ii, 1904), but the article is too long for reproduction here. The value obtained IS — ■K _ o'5 ( ,«* - 2«2 -I- I \ where « = the ratio of the radius of the approach channel tc the radius of the orifice. The results obtained by the two formulas are — ff. Kc Rational. Kc Rankine. 2 0-679 0-672 3 0-645 0-640 4 0-634 0-631 S 0-631 0-626 6 0-629 0-624 8 0-628 0-622 10 0-627 0620 100 0-625 o-6i8 1000 0-625 o-6i8 Diverging Mouthpiece. — This form of mouthpiece is of great interest, in that the discharge of a pipe can be greatly 652 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. 666); 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 pltis 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. 666) — Fig. 623. 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. /„ is also atmospheric ; hence £-=^. w w VV The velocity V is zero, hence — is zero. ^g Then, assuming no loss by friction, we have — H-4 or h orV. ^g = ^2gA and the discharge — Q = = K,V.A, = = KAVa^A ' This reasoning will not hold if the mouthpiece discharges into vacuum. Hydraulics. 6S3 In the case above, the mouthpifece is horizontal, but if it be placed vertically with b below, the proof given above still holds j the h must then be measured from b, 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 below the boiUng point corre- sponding to the temperature, the stream will not be continuous. From the above, we have — w ' 2g "^ W If — ^ becomes zero, the stream breaks up, or when — 2g = ^=34 feet Buty^ = ^ = «, or V„ = «Vj hence ^-X» li_^.l^(«2 - i) = 34 2^ 2.?- or A(n^ — i) = 34 In order that the stream may be continuous, ^n' — 1) 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* fig. 629. 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 t. The close agreement between the experimental and the ' A ' ^=-iJ _B__ ^^r\ -^ 6S4 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. Hj. Vacuum at throat in feet of water. H« Velocity of flow at throat. Feet per second. Ha. By experiment. VwCHs+H,). 69-24 69-24 69-24 12-50 12-50 12-50 8-00 2-00 0-25 58-85 50-78 None 8-50 5-00 1-50 None None None None 33-S 33-S 11-3 33-S 33-S 33-S 8-2 0-52 6s -97 80-97 81-43 37-90 S3-98 54-60 51-67 24-74 6-66 66-78 81-34 ?3 S4-43 54-43 51-70 25-63 7-04 Jet Fniup or Hydraulic Injector. — If the height of the column of water in the vacuum gauge at / (Fig. 629) 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 viS 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. 630. The main trouble that occurs with such a form of pump is that the watei ^"'■^^°- chums 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 a Hydraulics. 655 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. dh » ■B- f — 'i Fig. 631. 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 = '&.dh quantity of water passing strip perl v t? /^a second, neglecting contraction J ~ » • ^ • »" where V = velocity of flow in feet per second, \ \ hhdh = ijzgh, or Hence the quantity of water passing stripl _ . — -ojiji. per second, neglecting contraction f ~ ^ *^ the whole quantity of water Q passing"^ over the notch in cubic feet per>=V*^B second, neglecting contraction J - h' Q = -/^BfWn Q = PH^2^H introducing a coefficient to) ^ T?-2T>tT / — s allow for contraction fQ= ^^^^ ^ *^" where B and H are both measured in feet; where K has values varying from 0*59 to o'64 depending largely on the 656 Mechanics applied to Engineering. proportions of the section of the stream, i.e. 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 = o'62. The following empirical formula by Braschmann gives values of |K for various heads H allowing for the velocity of approach. Let B^ = the breadth of the approach channel in feet. |K = (0-3838 + o-o386l+°-:?gli) 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. Experi- Fin. 633. ments show that K for such a notch is very nearly constant. Hence, in the absence 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 : Area of elementary strip = b . dh ^ b B.-h , ^ B(H - >4) area of elementary strip = ' ~ — '- ■ dh rl velocity of water passing strip = V = ij 2gh = \'^ h^ quantity of water passing] -g/jj _ >■, strip per second, neglect- [ = -i— = — -^2gh^ . dh ing contraction I "■ whole quantity of water Q j f^ = H \dh lole quantity of water Q ] T^ = H passing over the notch in I B (jj - h)h' cubic feet per second, ~"H'^"^Ja = o neglecting contraction Hydraulics. 657 B _P = ^ J h = a ^ = \j^g 3 3 h = \\ Q = |V^(fH? - |Hi) = |ViiAH^ Introducing a coefficient for the contraction of the steam and putting B = 2H for a right-angled notch. where C has the following values. See Engineering, April 8th and 15th, igio. H (feet) ... 0-05 o-io o-is 0-20 0-25 0-30 040 C o'289 0-304 0-306 0-306 0-305 0-304 0-303 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 Fio. 633. Fig. 634. depth, the discharge is obtained by precisely the same reasoning as in the two last cases ; it is — Q = KfBV'2i(H.i - H,?) 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 2 u 658 Mechanics applied to Engineering. both sides of an orifice, the discharge is not free ; the calculation of the flow is, however, a very simple matter. The head producing flow at any section xy (Fig. 634) is Hj — Hj = H j likewise, if any other section be taken, the head producing flow is also H. Hence the velocity of flow V = V 2^H, and the quantity discharged — Q= KAVz^H K varies somewhat, but is usually taken o"62 as a mean value. Flov7 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 a. F^E5 H-h Fig. 635. 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 — Q = KAv'2^^H - h) 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„, 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 /4, i.e. the head over the orifice ; V = velocity of water issuing from the orifice. Hydraulics. 659 Then V„ = ^V, and V = V, + p V = —V + V A V = . Ka Broad-Crested Weir. — The water flows in a parallel stream over the crest of the weir if the sill is of sufficient Fig. 636. breadth to allow the stream lines to take a horizontal direction. Neglecting the velocity of approach, the velocity of the stream passing over the crest is, V = V zgh where h is the depth of the surface below that of the water in the approach channel. Let B = the breadth (in feet) of the weir at right angles to the direction of flow. Then the quantity passing over the weir in cubic feet per second is — Q = B(H->%)V2p The value of h, however, is unknown. If h be large in proportion to H, the section of the stream will be small, and the velocity large; on the other hand, if h be small in pro- portion to H, the section of the stream will be large and the velocity small, hence there must be some value of h which gives a maximum flow. Let /4 = «H; _ Q = BV'2^(H - nYi)>JnK Q = BVii X hV - n") dQ , — 3,, _i „ I, 66o Mechanics applied to Engineering; This is a maximum when — 5«~- = ^ffi or when n = \ Inserting the value of n in the equation for Q, we have — Q = o-sSsBHV'i^ The actual flow in small smooth-topped weirs agrees well with this expression, but in rough masonry weirs the flow is less according to the degree of roughness. 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.) 2'ank of uniform cross-section. Let the area of the surface of the water be A.; „ „ „ stream through the orifice be K^A ; „ greater head of water above the orifice be Hi ; „ lesser ,, „ ^ ^ „ „ xlg • „ head of water at any given instant be h. The quantity of water passing through) _-k- a / — r j, the orifice in the time dt ^ - ii-.* A. v 2^A . dt 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 KJth, which is equal to' that which has passed through the orifice in the interval, or — Y^i^,jlgh.dt= kjh dt = ^ f° /i -^dh p = H, '^=K:f7p' ^~*^* 2A,(VHi-_VH ,) K.A^2^ The time required to empty the tank is — Hydraulics. 661 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 'Q.l.dh in the given interval of 1 ^ time dt. H? | But/=y ^ ill hence '&.l.dh = -^r^h dh «i Fig. 637. dt = BL HiK^AVz^ Ji'dh Integrating, we get- 2BL(Hii - H,J ) 3HiK,AV2i- (iii.) Hemispherical tank. In this case — 1^= 2-SJi-h'' The quantity in the tank is reduced by ir(2Ry4 — h'^)dh in the ^ interval dt. dt = K,AV2ir T = (2R/4 - hyrUh Fig. 638. K,AV2^ (2R/*^ - h^) dh J H, i rate. T = _^ 5 _3 5 / KiAV2^ Hv.) C«J<? ;■« which the surface of the water falls at a uniform 662 Mechanics applied to Engineering. In this case — is constant ; at hence Kik^~2gh = A„ X a constant But K^Av 2^ is constant in any given case, hence the area of the tank A„ at any height h above the orifice varies as ^/h . or the vertical section of the tank must be paraboUc as shown. (v.) Time required to change the level when water is flowing into a tank at constant rate and leaving by an orifice. Let Aa = area of the surface of the water in square feet, when the depth of water above the outlet is h feet. Fic. 638 A. In a tank of geometrical form we may write A„ = C/?" where C and n are constants. Q, = the quantity of water in cubic feet per second running into the tank. A = the area of the orifice in square feet. T, = the time required to raise the level of the water from Hi to H„ feet. T( = the time required to lower the level of the water from H„ to Hi feet. The quantity of water flowing into the tank) _/-,,. in the time dt V The portion of the water which is retained in j _ * the tank in the time dt \~ " dh The quantity of water flowing out of the tank) _ t^ a / — i jj through the outlet in the time dt \~ ^"^'^ "^^"^ " When the water is entering the tank at a constant rate and leaving more slowly at a rate dependent upon the head, we have — Y^iK'Jlgh dt=. Qi dt - A„ dh dt{Q_t - K^aV^) = A,dh = Ch" dh /•Hm in J I ■^ Jh, Q,-K,A^/2^-4 This expression can be integrated, but the final expression is Hydraulics. 663 very long, and moreover in practice the form of the tank or reservoir does not always conform to a geometrical law, hence we use an approximate solution which can be made as accurate as we please by taking a large number of layers between measured contour areas. The above expression then becomes 'H A 1 Qi-K^aV 2^0-^1 A, M A2 Ih A,U + 'Z ,, ,==- + — ZTT^"^ • ' • 2(q,-nVho q,-nVh2 q,-nVHs 2(Q,-NVHji The first and last terms are divided by 2 because we start and end at the middle sections of the upper and lower layers, hence the thickness of these layers is only one half as great as that of the intermediate layers. If we require to find the time necessary to lower the surface, i.e. when the water leaves more rapidly than it enters, we have — ,4 Ai SA , A2 8/% , As U + ^^ ,— — + ^^ , r— — + 2{NVHi-Qe) NVH^-Q, NVHs-Q, 2(NVH„-Q,)3 Where Aj is the area of the surface at a height Hj above the middle of the culvert j and A2 is the area at a height Ha = Hi + 8;^, and A3 at a height H3 = Hg + 8/4, and so on, and N = K^a^/ 2g. Example. — Let Q, = 48 cubic feet per second Hi = 12 ft., hh = 0-5 ft, K^AVzg = 7 Ai = 140 sq. ft., Ag = 151, A3 s= 162, A4=i7o, Ag = 190 Then the time required to raise the level from 12 to 14 feet would be — T ^ 140 X o-5_ 151 X 0-5 r62 X o-5_ 2(48—7^/12) 48 — 7Vi2-5 48 — 7V13 170 X 0-5 , 190 X o'5 H —-^— -\ -p= = 14-3 seconds. 48-7Vi3-s 2(48- 7V 14) (vi.) Time of discharge through a submerged orifice. — In Fig. 639 we have — 664 Mechanics applied to Engineering, and 8Ha + mj^ 8A 8Hi = Ih A^U !+■ Aa X Ab U ht = Fig. 639. T = T = Aa X Ab (Aa+Ab)k^aV2^/* (Aa + Ab)K^aV2^- zAa X Ab i _ . (Aa + Ab)k,aV2/ '' p h -L- Fig. 640. where H and Hi are the initial arid final differences of heads. lime of discharge when the two tanks are connected by a pipe of length L. — The loss of head h, due to LV friction in the pipe is =:=- (see page 681), and the head h, dissipated in eddies at the V outlet is — 2P- hence V = y h + h, _ / h /h_ KD^2^ L KD "*" ig Hence from similar reasoning to that given in the last paragraph we have — T = AaAbVc V H, \dh (Aa + Ab^AJ Hi 2AaAb Vc ■ , (Aa + Ab)a(H -Hx') Where A is the area of the pipe which is taken to be bell- mouthed at entry, if it be otherwise the loss at entry (see p. 673) must be added to the friction loss. Hydraulics. 665 Time required to lower the Water in a Tank when it flpws over a Weir or Notch.^ — Rectangular Notch. — By the methods already given for orifices we have — jdi= ^^%- i \-^dh zK^Bv 2^J Ha Right-angled Vee Notch. — In this case we get by similar reasoning — T = ^SA^ r .-= dh = _25A„^( H,-^-- - H.-^-^ j 8K^//2^-iHa 8K,V2^A -1-5 / -T ^ 5A. / I i_\ Flow-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 lbs. 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^ foot-lbs., or h foot-lbs. per pound of water. By lowering the water to the datum, WA 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 = WA 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^ second, where V = »/ 2gh, or h = — . Then, smce no energy 2g WV is destroyed in the fall, we have VJh = foot-lbs. of energy stored in the falling water when it reaches the datum, or — foot-lbs. per pound of water. This energy, which is due to its velocity, is termed its kinetic energy, or energy of motion ; or — WV The energy of motion = 666 Mechanics applied to Engineering. 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 / at the foot of the pipe will be wh lbs. per square inch. This pressure is capable of overcoming a resistance through a distance / feet, and thereby doing pi foot- lbs, of work ; then, as no energy is destroyed in passing along the pipe, we have// = W^ = -^ foot-lbs. of work done by the water under pressure, or ^ foot-lbs. per pound of water. This w 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 perl _ (energy of)' (energy ofl , (pressure-) pound of water) ~ \ position \ \ motion ft energy ) ig w This may, perhaps, be more clearly seen by referring to the figure. Fig. 64Z 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 — ;i, + Yk + A =^ 4. Yl + A ^ constant ig W 2g W Hydraultcs. 667 The quantity of water passing any given section of the pipe in a given time is the same, or— or AjVi = AjV, Yi = ^ V, A, or the velocity of the water varies inversely as the sectional area — Fig. 642. 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 — H, = ^, and Hj = -^ Wl w and the total heights of the water-columns above datum — w w and the differences of the heights — ^-^ + A w •■■-h = •W 2g 2g V ' — V H, - H, = -1? 11 2.? from the equation given above. Thus we see that, when water is steadily running through 668 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. Ventnri 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 difierence of the heads corresponding to the two pressures be Ho in feet of water (Fig. 643)— V^ — V ' „ ^ = H„, or Vi - V,^ = 2^Ho Let Aj = nKi, ; then Vj = — n hence V/ - (^J = 2^Ho and V^ = / J^Si. • '/Ho'=cVh; Q = A,V, = Aj V / 2^ / where C = A, / 2. The difference of head is usually measured by a mercury gauge (shown in broken lines in Fig. 643), and the tubes above the surface of the mercury (sp. gravity 13-6) should be kept full of water, the mercury head H„ is most conveniently measured in inches. Then we get, Q = C^il3ljlJ& ^ cVr^^Ei;: There is a small loss of head in the short cone due to friction which can be allowed for by the use of a coefficient Hydraulics. 669 of velocity K, which is very nearly constant over a very wide range, its value is from 0-97 to o'gS, or say, o'gys, then — Q = K^cVroS V'H^ = CVh^ very nearly. At very low velocities of flow, where errors are usually of no importance, the value of K„ varies in a very erratic fashion, the reason for which is unknown at present; but for such velocities of flow as are likely to be used in practice the meter gives extremely accurate results. When used for waterworks purposes the meter is always fitted with a recorder and integrator, particulars of which can be obtained of Mr. Kent, of High Holborn, London. Fig. 643. The loss of head on the whole meter often amounts to about — - For experimental data on the losses in divergent pipes, readers should refer to a paper by Gibson, "The Resistance to Flow of Water through Pipes or Passages having Divergent Boundaries," Transactions of the Royal Society of Edinburgh, vol. xlviii. Radiating Currents and Free Vortex Motion. — 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 Oi, «2j then, since the flow is constant, we have — 670 Mechanics applied to Engineering. V2 <h n , ''1 Wiffi = Villi, or — = — = — hence w^ = v-r- »i a.i r^ r^ or the velocity varies inversely as the radius. The plates being horizontal, the energy of position remains constant ; therefore — 2g W 2g W Substituting the value of v^ found above, we have — ■ !!l J.-6 _ '"'''"'' M 2g W 2g. Ti Then, substituting — + - = H 2g w .P2 w from above, and putting r ; ^1"^ = ^, we have — '2 H-/4,= A Then, starting with a value for hy, the h^ for other posi- tions is readily calcu- lated and set down from the line above. If a large number of radial segments were taken, they would form a com- plete cylinder of water, in which the water enters at the ^""" ^^ centre and escapes radially outwards. The distribution 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. 671 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. (ii.) The centrifugal force act- ing horizontally, where V is its velocity in feet per second, and r its radius in feet. (iii.) The fluid pressure, which is equal to the resultant of i. and ii. From the figure, we have — w ■ ae be Fig. 645. which may be written — 'Wgr be 2 TT But — IS constant, say C ; g Then Cfyt^ = "1 be But ac = r therefore C«^ = — be And for any given number of revolutions per second «* 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 0, 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 Vj feet per second, when it reaches B the direction of flow is suddenly changed from AB to BC, and the 672 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 V3 with which the water flows after passing the corner is given by the diagram of velocities ABD, from which we see that the component Vj, normal to BC, is wasted in eddying, and the energy wasted per pound of water IS _1- =— i 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. 646). If the surface be a smooth curve (Fig. 647) in which there is no abrupt change of direction, there will be no loss due to ^ \ Fig. 646. Fig. 647. 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 — V. - X.' = V, n and the loss of energy per pound of water will be— 2£ 2g- 2g\ n' Hydraulics. 673 When « = I, no striking lakes place, and consequently no loss of energy J when «= 00 , i.e. when the surface is stationary, the loss is -i, 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 Y direction, but with a velocity — j hence I ci9^ — ~~ the loss of energy per pound of water * ^nzj^ Cr-^A is precisely the same as in the last para- ^^s>;r:rr (y -ViV ^-^^^ — I ' > ~) Fig. 648 (see also No. 3 graph, viz. ^ ^ . The energy lost ^c'lg p- 67+). 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 « times, is — v.'(--;)' Or if we refer to the velocity in the large section as Vj, we have the velocity in the small section «Vi, 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.' 649 Cs« also No. 4 facing 0*62 to o"66; hence, «* = from i'6i p-674). to i'5 ; then, the loss of head = 2 X 6/4 Mechanics applied to Engineering. Total Loss of Energy due to a Sudden Enlargement and Contraction. — Let the section before the enlargement be termed i, the enlarged section 2, and the section after the enlargement 3, with corresponding suffixes for velocities and pressures. Then for a horizontal pipe we have — W- 2g W 2g 2g\ nJ W Ig; Gibson finds that the loss of energy is slightly greater than this expression gives. The author finds that the actual loss in some cases is nearly twice as great as the calculated. 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 fn^nner ,ip which th,e ^ow takes place in channels of various fofms. He' takes two sheets of glass and' fits them into a suitable frame, whichholds them in position at about yj^ inch apart. , Through this narrow space liquid is caused to flow under ■pressare, 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. 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 sitiuous 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, ITofacep. 674. FLOW OF WATER DIAGRAMS. Kindly supplied by Processor Helt-Shaw, F.R.S. Hydraulics. 675 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. 650. 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. Friction in Pipes. — Contimeotis Flow. — When the flow in a pipe is continuous, i.e. not of an eddying nature, the resistance to flow is entirely due to the viscosity of the fluid. On p. 314 we showed that the resistance to shearing a viscous fluid is — where A is the wetted surface in square feet, K is the 6/6 Mechanics applied to Engineering. coefficient of viscosity, S the speed of shearing (usually denoted by V in hydraulics) in feet per second, / is the thickness of the sheared element in feet, in the case of a pipe of radius R, / = R. Then, without going fully into the question, for which treatises on Hydraulics should be consulted — Let Pi = the initial pressure in pounds per sq. foot. Pa = the final pressure in pounds per sq. foot. L = the length of the pipe in feet. Aj, = the area of the pipe in sq. feet. F, = (Pi - Pe)A, = W„(Hi - H,)A^ F. = W„/iA^ where h„ is the loss of head in feet due to the resistance on a length of pipe L. Then W„/4.A, = ^^^ R (27rRL)KV _ 2LKV ^ 8LKV W„.(xR^)R ~ W„R' ~ W^D'' LV CD" This expression only holds for stream line, or con- tinuous flow, the critical velocity V„ at which the flow changes from continuous to sinuous is always much higher than the velocity V„i at which the flow changes from sinuous to con- tinuous. The critical velocity also largely depends upon the temperature of the water owing to a change in the viscosity. Let Vc = the critical velocity, i.e. the velocity in feet per second at which the flow changes from con- tinuous to sinuous. V„i = ditto at which the flow changes from sinuous to continuous. n and «i = coefficients which depend upon the temperature of the water. D = the diameter of the pipe in feet, then V. = - and V,i = g Hydraulics. 677 Temperature Fahrenheit. «. "i- c. 32 0-25 0-040 52, SCO 40 021 0-034 61,000 60 0-15 0-025 83,500 80 0'12 0-019 io7,coo 100 O'lO 0-016 134,000 120 o'o8 0-013 164,000 140 0-065 o-oii 204,000 160 o'oss 0-009 240,000 180 0-047 0-008 278,000 200 0-040 0-006 328,000 212 0-037 o"oo6 350,000 Mr. E. C. Thrupp has, however, shown that the values ^ +5 r i B 1 1 1 1 1 J A > "••■ ^ «' ^'' -I +1 +3 +5 +7 +9 Logarithms of hydraulic gradient. Fig. 651. given in the above table only hold for very small pipes; in the case of large pipes, channels, and rivers the velocity at 678 Mechanics applied to Engineering. which the water breaks up is very much greater than this expression gives. See a paper on " Hydraulics of the Re- sistance of Ships," read at the Engineering Congress in Glasgow, 1901 ; also Engineering, December 20, 1901, from which the curves in Fig. 651 have been taken. The Osborne Reynolds' law is represented by AC and BD, whereas Thrupp shows that experimental values lie somewhere between AA and BB. The change points from continuous to sinuous flow are shown in Fig. 652. At low velocities of flow the loss of / i y 2 / u s •a •0 / 1 .a r^ / J 1 1 E 1/ B •c / ^ / / Logarithm of velocity. Fig. 6sa. head varies simply as the velocity, therefore the slope of the line AB is i to i. At B the flow suddenly changes to sinuous flow, and at higher velocities the loss of head varies approxi- mately as the square of the velocity, hence the slope of the line CD is 2 to i. When the velocity is decreased the loss of head continues to vary as the square until E is reached, and below that it returns to the state in which it varies simply as the velocity. The point B corresponds to V„, and the point E to V„i. For further details of Reynolds' investigations, the. reader is referred to the original papers in the Philosophical Trans- actions for 1884 and 1893; also to Gibson's "Hydraulics and Hydraulics. 679 its Applications," and Turner and Brightmore's " Waterworks Engineering." Sinuous Flow.— 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 tp the surface in feet per second ; R = frictional resistance in pounds ; Then, R = S/V^ 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 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. 653 and 654, 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 68o Mechanics applied to Engineering. 1-3 B \ 0-3^ 0-2 - ^!??5 r-Af -i. '^^ Qa<£^ COAff.JE SAfID VW£^ ^4A^ *0 2S 30 DISTANCE nan CVTIMren Fig. 653. " ^ „ K ^l ^ -^ — r. m F 01 * ^ ti i .... 1 'n i • A< fr- f i V , \\ .._ *> r ^ ::::: — - -«_ _ / /A '^r S m\ " a 5 Cl >/» pj J£ 1 •A NL ■> ™ si •' ^ ' y ^ « ZO 2S 30 3S DISTANCE FROM curwATen Fig. 6s4- H Hydraulics. 68 1 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 ... o'oi32 lb. „ „ fine „ „ „ varnish tinfoil ... 0-0096 „ ... 0-0043 .. ... 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 Pa 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 = (Pi - Pa)A Then, putting Pi = h^„ and Pj = -^jW^, we have — R = W„A(^i - hi) = W„AA where h is the loss of head due to friction on any length of pipe L; then — W„A/4 = S/V or -^ = LttD/V* hence/J = -.— =-.— /_ LV^ _ LV^ ° W„' R ~4KR where R = hydraulic mean depth (see p. 683). The coefficient -^ has to be obtained by experiment ; according to D'Arcy — K 3200V 12D/ where D is the diameter of the pipe in feet. 682 Mechanics applied to Engineering. 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) :— — = ( I + -r~ ) for clean cast-iron pipes K 5ooo\ 2D/ — = ( I H — ;:r I for incrusted pipes K 25oo\ 2D/ 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 lb. of water moving with a velocity V feet per second is — ; hence the whole energy of motion of the water is dissipated in friction when — V^ LV 2g~ KD 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 in) ^ _ a v _ '^^"^ cubic feet per second j^ - ti — AV — — — From the same formula, we have — ''2400DA vV^ Hydraulics. 683 Substituting this value, we have — Q=38-SD^\/J=38-SD^>/ L Thrupp's Formula for the Plow 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 = — ; Q = quantity of water flowing in cubic feet per second. Then V = C^S where x, C, n are coeflScients depending on the nature of the surface of the pipe or channel. For small values of R, more accurate results will be ob- tained by substituting for the index x the value x + y^ 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. 684 Mechanics applied to Engineering. Siufacc. n. c. X. y- - Wrought-iron pipes l-8o 0-004787 0-65 0-018 0*07 Riveted sheet-iion pipes 1-825 0-005674 0-677 — New cast-iron pipes /I-85 \2-00 0-005347 0-006752 0-67 0-63 — Lead pipes I-7S 0-005224 0-62 — Pure cement rendering /1 74 1 1 '95 0-004000 0-006429 0-67 o-6i — — Brickwork (smooth) z-oo 0-007746 o'6i „ (rough) 2 -co 0-008845 0-625 0-01224 0-50 Unplaned plank 2-00 0-008451 0-615 °03349 0-50 Small gravel in cement 2'00 0-OII8I 0-66 0-03938 0-60 Large „ „ 2-00 0-OI4I5 0-705 0-07590 I -00 Hammer-dressed masonry ... 2-0O 0-OIII7 0-66 0-07825 I -00 Earth (no vegetation) 2 -CO 0-01536 0-72 Rough stony earth 2-00 0-02144 0-78 — If we take x as 0-62, and « = 2, C = 0-0067, wc get — Q = which reduces to — 3-01 C VS Similarly, for new cast-iron pipes — h = . 320oD''»* taking « = 1-85, and x = 0*67. These expressions should be compared with the rougher ones given on pp. 681, 682. 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 the 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 the difference in height between any two points will represent the loss of head due to friction on tiiat 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 Hydraulics. 685 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 nil, 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 i-t 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. 655. 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 ^ to ^ 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 ^ to 1/ is the same as from b to c, and at d there is a drop due to a sudden enlargement, then from dXo e the line is parallel to the upper line. The amount of the drop at each resistance can be calculated by the methods already explained. The pressure at every point in the main is proportional to the height of the virtual slope line above the main ; hence, if the main at any point rises above the virtual slope line, the pressure will be negative, i.e. less than atmospheric, or there will be a partial vacuum at such a point. If the main rises more than 34 feet above the virtual slope line, the water will 686 Mechanics applied to Engineering. break up, and may cause very serious trouble. In waterworks mains great pains are taken to keep them below the virtual slope line, but if it is impracticable to do so, air-cocks are placed at such summits to prevent the pressure falling below that of the atmosphere ; the flow is then due to the virtual slope between the upper reservoir and this point. In certain cases it is better to put an artificial resistance in the shape of a pierced diaphragm or a valve on the outlet end of the pipe, in order to raise the virtual slope line sufficient to bring it above every point of the main, or the same result may be accom- plished by using smaller pipes for the lower reaches. Flow of Water down an Open Channel on a Steep Slope. — Let u = the initial velocity of the water in feet per second. V = the velocity of the water after running along a portion of the channel of length /(feet) measured on the slope. 6 = the angle of the slope to the horizontal. H = the vertical fall of the channel in the length / then H = / sin e. h = the loss of head in feet due to friction while the water is flowing along the length /. R = the hydraulic mean depth of the channel. K = four times the constant in D'Arcy's formula for pipes (multiplied by 4 to make it applicable to channels, and using the hydraulic mean depth instead of the diameter). The velocity of a particle of water running down a slope is the same as that of a particle falling freely through the same vertical height, if there is no friction. When there is friction we have— V^ = u^ + 2^(H - A) The loss of head due to friction is — / (?/' + 2^'-H)KR ^ / {t{' + 2glsine)K R V KR-l-2^/ V KR + 2^/ Hydraulics. 687 This expression is used by calculating in the first place the value for V, taking for H some small amount, say 10 feet. Then all the quantities under the root are constant, except u, hence we may write — v. = -v/!^ + n sin 0)in m + n for the first 10 feet, then the u for the second 10 feet becomes -^ and for the third 10 feet- V. (Vi^ + n sin _ / (V/ + n sin e)m and so on for each succeeding 10 feet. The following table shows a comparison between the results obtained by Mr. Hill's formula ' and that given above — « = 15 feet per second, R=i-S3- K = 15130 (deduced from Mr. Hill's constant). e= 12° 40'. H = 10 feet. / = 45 "7 feet. Values of V. Fall reckoned from slope in feet. Hill. Author. 15-00 15-0 10 28-34 27-9 20 36-22 35-6 30 41-95 413 40 46-42 45-9 5° 50-03 49-S 60 53'oi S2-6 70 55-50 55-2 80 57-60 57:4 90 59-39 59-3 100 60-93 60-9 * Proceedings Institution of Civil Engineers, vol. clxi., p. 345. 688 Mechanics applied to Engineering. If the slope varies from point to point the angle 5 must be taken to suit : similarly, if the hydraulic mean depth varies the proper values of R must be inserted in the expression. An interesting application of the above theory to the formation of ponds at the approach end of culverts will be found in a paper by the author, " The Flooding of the Approach End of a Culvert," Proc. Inst. Civil Engineers, vol. clxxxvi. Flow through a Pipe with a Restricted Outlet. — When a pipe is fitted with a valve or nozzle at the outlet end, the kinetic energy of the escaping water is usually quite a large fraction of the potential energy of the water in the upper reservoir ; but in the absence of such a restriction, the kinetic energy of the escaping water is quite a negligible quantity in the case of long pipes. The velocity of the water can be found thus : Let H = head of water above the valve in feet ; L = length of main in feet ; V = velocity of flow in the main in feet per second ; K = a constant depending on the roughness of the pipe (see p. 68 1) ; D = diameter of the pipe in feet j Vi = velocity of flow tlu'ough the valve opening ; n = the ratio of the valve opening to the area of the V pipe, or « = =^ . »i The total energy per pound of water is H foot-lbs. This is expended (i.) in overcoming friction, (ii.) in imparting kinetic energy to the water issuing from the valve ; or — LV» V,^ V Substitutmg the value of Vj = — , we have — H KD "*■ zgn^ On p. 717 we give some diagrams to show how the velocity of the water in the main varies as the valve is closed ; in all cases we have neglected the frictional resistance of the valve itself, which will vary with the type employed. - In the case of a long pipe it will be noticed that the velocity of flow Hydraulics. 689 in the pipe, and consequently the quantity of water flowing, is but very slightly affected by a considerable closing of the valve, e.g. by closing a fully opened valve on a pipe 1000 feet long to o"3 of its full opening, the quantity of water has only been reduced to 0*9 of its full flow. But in the case of very short pipes the quantity passing varies very nearly in the same proportion as the opening of the valve. Resistance of Knees, Bends, etc. — ^We have already shown that if the direction of a stream of water be abruptly changed through a right angle, the whole of its energy of motion is destroyed ; a similar action occurs in a right-angled knee or elbow in a pipe, hence its resistance is at least equivalent to the friction in a length of pipe about 37 diameters long. In addition to this ^^^^^ loss, the water overshoots the corner, as shown Fig. 656. in Fig- 656, and causes a sudden contraction and enlargement of section with a further loss of head. The losses in sockets, sudden enlargements, etc., can be readily calculated ; others have been obtained by experi- ment, and their values are given in the following table. When calculating the friction of systems of piping, the equivalent lengths as given should be added, and the friction calculated as though it were a length of straight pipe. 2 Y 690 Mechanics applied to Engineering. Nature of resistance. Equivalent length of straight pipe expressed in diameters, on the basis of L =: 36D, 15" 30° 45° ij inch check valve ij inch ball check valve Sluice and slide valves «= Unwin Right-angled knee or elbow (experiments) Right-angled bends, exclusive of resist- ance of socketsi at ends, radius of bend, = 4 diameters Ditto including sockets (experiments) Sockets (screvred) calculated from the! sudden enlargement and contraction ( (average sizes) Ditto by experiment Sudden enlargement to a square-ended) , large area pipe, where n = — =7^ ' "^ small area Sudden contraction Mushroom valves {handle turned throue;h port area area of opening „. J J. V area of pipe Pierced diaphragm » = .f \ area of hole Water entering a re-entrant pipe, such as\ a Borda's mouthpiece ... ... ...J Water entering a square-ended pipe flush) with the side of the tank / /30-40 in plain pipe \ 50-90 with screwed elbow 3-1 S 22-30 24 16-20 H-ff 12 approx. 120-400 27 200 HOC 700-1500 2000-3000 ioo(« — l)' 36(1 -Sb - 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 ' Up to 50 „ Pumping mains _ 3 to S «t Waterworks mains 2 to 3 „ ' Such velocities are unfortunately common, but they should be avoided if possible. CHAPTER XIX. 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. r. 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 utihzed 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 lbs. of water have descended through a height H feet, the work done = WH foot-lbs. Only a part, however, of this will be utiUzed by the motor, for reasons which we will now consider. Fig. 637. Fig. 658. The illustrations. Figs. 657, 658, show various methods of 692 Mechanics applied to Engineering. utilizing the weight-energy of water. Those shown in Fig. 657 are very rarely used, but they serve well to illustrate the principle involved. The ordinary overshot wheel shown in Fig. 658 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 of 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 1 in 8, we get the peripheral velocity of the wheel V„ = zVRj 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 12 feet. Then the velocity of the rim should not exceed 2 v' 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 h = — , or, introducing a coefficient to allow for the friction in the sluice, we may write It ^= = 1-6 foot. One-half 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. 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 FiQ. 659. Hydraulic Motors and Machines. 693' shown. Then by measurement from the diagram, or by a simple calculation, we see that this loss is o'isD, A clearance of about o's 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 = i2'5 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,) _ ..„ o-is X 12-5 feet ]-^9 » Clearance at bottom = o's „ 4' 2 feet 15— 4"2 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. The loss due to the water leaving too early can be largely avoided by arranging the wheel as shown in Fig. 660. Fig. 660. Fig. 661. Pressure Machines. — In these machines the water at the 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 694 Mechanics applied to Engineering. piston in precisely the same manner as in a steam-engine. The work done is the same as before, viz. WH foot-lbs. for f ft .0 ^_____^ n — — - V n , \ ^ Fig. 662, the pressure at the lower level is W„H lbs. per square foot j and the weight of water used per square foot of piston = W„L = W, I Fig. 663. where L is the distance moved through by the piston in feet. Then the work done by the pressure water = W„LH = WH foot-lbs. Several examples of pressure machines are shown in Figs. 661, 662, 663, a and b. Fig. 661 is an oscillating cylinder pressure motor used largely on the continent. Fig. 662 is an ordinary hydraulic pressure riveter. Fig. 663 (a) is a passenger lift, with a wire-rope multiplying arrangement. Fig. 663 {b) is an ordinary ram lift. For details the reader is referred Hydraulic Motors and Machines. 69s to special books on hydraulic machines, such as Blaine ' or Robinson.'' 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 \ 22 feet / Jchain'l I only J 427 633 745 857 969 1081 "93 Water used in gallons at \ 80 lbs. per square incH / 7-5 10 14 16 17 20 21 22 Efficiency per cent, (actual) — ,Si S4 SO 60 S» 61 feS Efficiency per cent, if stroke \ were of fixed length ... / — 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 diflferent sizes, or one cylinder and a differential piston. When lightly loaded, the smaller cylinder is used, and the larger one ' " Hydraulic Machinery " (Spon). ' " Hydraulic Power and Machinery " (Griffin), 696 Mechanics applied to Engineeritig. 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 smallei cylinder. In order to place it beyond his control, the ex- tremely ingenious device shown in Fig. 664 is sometimes used. FULL P/fESSUKE c 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. 'Wlien 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 vid 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 Hydraulic Motors and Machines. 697 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 sufSciently 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. 665) is used. It consists essentially of a vertical cylinder, provided with a long- stroke plunger, which is weighted to give the required pressure, Fig. 66s.* usually from 700 to 1000 lbs. per square inch. With such a means of storing energy, a very large amount of power — ^far in excess 698 Mechanics applied to Engineering. 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. — \l s = the stroke of the accumulator in feet j d = the diameter of the ram in inches ; fi = the pressure in pounds per square inch. Then the work stored in foot-lbs. = o-']2>^cPps Work stored per cubic foot of water in 1 , , . foot-lbs. [ = '44/ Work stored per gallon of water = -^r; — = 23'o4^ Number of gallons required per minute \ _ 33»°°p _ 143' at the pressure / per horse-power | ~ 23'o4/ "" p Number of cubic feet required per minute 1 _ 33>°°° _ 229-2 at the pressure/ l ~ 144/ ~ / 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 1 square inch in section, whose length L in feet is that of the main along which the water is flowing to the motor = o"434L ; area of plunger or piston m = the ratio p -; ? — : -. — area of section of water main Hydraulic Motors and Machines. 699 / = 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. 182, 187 — p = o"ooo34»2(o'434L)RN^ ( i ± - ] p = o"ooor5«LRN^( 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 i 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 lbs. per square inch (see p. 405) ; 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 ; / = 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 lb. ; L = the length of the column of flowing water in feet: 700 Mechanics applied to Engineering. d = the diameter of the pipe in inches j T = the thickness of the pipe in inches ; /, = the tensile stress in the pipe (considered thin) due to the increased internal pressure/; E = Young's modulus of elasticity for the pipe material. Then/, = ^ 2 1 The increase in diameter dae to thel _ fd^ increased pressure I 2TE tip ltd The increase in cross-section = =^=rp X — 2TE 2 The increase in volume of the pipe perl _ \jfd square inch of cross-section ) TE Let a portion of the pipe in question be represented by Fig. 666. Consider a plane section of the pipe, ab, distant L from the valve at the instant the valve is suddenly closed. On account of the yielding of the pipe and the compres- sion of the water, the plane ab still continues to move forward until the spring of the water and the pipe is a maximum, i.e. when the position dV is reached, let the distance between them bearj then, due to the elastic compression of the water, the plane ab moves forward by an amount x^= ~ (see p. 374), and a further amount due to the distention of the pipe of a;, = -,==-, hence— X ill a, a' VtOve ^ ■X' S^ L Fig. 666. —X ' *=-^(^+te) and/ = X Ki+A) But since x is proportional to L in an elastic medium, the pressure/ is therefore independent of the length of the pipe. At the instant of closing the valve the pressure in the immediate neighbourhood rises above the static pressure by an amount / and a wave of pressure starting at the valve is transmitted along the pipe until it reaches the open end, the velocity of which V„ is constant. The time 4 taken by the Hydraulic Motors and Machines. 701 wave in traversing a distance - is -^r; and the distance x^ n «V„ which the plane ab traverses in this time is - , but n *=Xe+^]4X=^^- Xn X niL _ \, Hence — = - = ■—- = mS[, Thus the velocity with which the plane ab travels is constant. Let the velocity of the water at the instant of closing the valve be V, and since the velocity of ab is constant the water con- tinues to enter the pipe at the velocity V for a period of t seconds after closing the valve, i.e. until the pressure wave reaches the open end of the pipe, hence V = - The change of mo-) _ mass of the change of velocity in mentum in the time t )~ water ^ the time t ft _ °'434L X Substituting the value of x, we have — L_ / \/«'(^+T^) and when the elasticity of the pipe is neglected— ^V^ The quantity — is the velocity with which the compression wave traverses the pipe, pr the velocity of pulsation. Inserting numerical values for the symbols under the root, we get the velocity of pulsation 4720 feet per second, i.e. the velocity of sound in water when the elasticity of the pipe is neglected. The kinetic energy of the column of| _ o-434LV^ water per sq. inch of section 5 '^ The work done in compressing the water) _f{x„ + x^ and in the distention of the pipe j ~ " 2 2 Vk ^ TE/ 702 Mechanics applied to Engineering. 2g 2 VK TE/ and/= ■ ^■^'V K ■ d When the elasticity of the pipe is neglected — /i = 63-5V A comparison between calculated and experimental results are given below. The experimental values are taken from Gibson's "Hydraulics and its Applications," p. 217. Sudden Closing of Valve. Velocity in feet per second 0-6 20 3'o 7-S Observed pressure lb. sq. inch 43 "3 173 426 Calculated/ = 63-41; .... 38 127 190 476 Calculated allowing for elasticityl of pipe / 35 116 17s 436 When the valve is closed uniformly in a given time, the manner in which the pressure varies at each instant can be readily obtained by constructing (i.) a velocity-time curve; (ii.) a retardation or pressure curve, as explained on p. 140. But the pressure set up cannot exceed that due to a suddenly closed valve, although it may closely approach it. When the pressure wave reaches the open end of the pipe, the whole column of water is under compression to its full extent, it then expands, and when it reaches its unstrained volume the water at the open end is travelling outwards with a velocity V (very nearly, there is a small reduction due to molecular friction) and overshoots the mark, thus producing a negative pressure, i.e. below the static pressure in the pipe, to be followed by a pressure wave and so on. The time during which the initial pressure is maintained is therefore the time taken by a compression wave in traversing the pipe and 2L returning, viz. — seconds. Hence, if the time occupied in Hydraulic Motors and Machines. 703 closing the valve is not greater than this, the pressure set up at the instant of closing will be approximately that given by the above expression for f, but the length of time during which the pressure is maintained at its full amount will be correspondingly reduced. If the period of closing be greater than — seconds, the pressure set up at the instant of closing will be less than/, but a rigid solution of the problem then becomes somewhat complex. Maximum Power transmitted by a Water-Main. — We showed on p. 580 that the quantity of water that can be passed through a pipe with a given loss of head is — Q = 38-50^ \/^ Each cubic foot of water falling per second through a height of one foot gives — • 62-5 X 60 = 0*1135 horse-power 33,ooo_ •'^ ^ henceH.P. = o-ii3sQH where H is the fall in feet, and Q the quantity of water in cubic feet per second. Then, if h be the loss of head due to friction, the horse- power delivered at the far end of the main L feet away is — H.P. = o-ii35Q(H--4) Substituting the value of Q from above, we have — ap; = 0II3S X 38-50^ (H - h)\/ J Let h = «H. Then, by substitution and reduction, we get the power delivered at the far end — , , /«H»I H.P. = 4-37 (i - «) V -17 ig-inWB^i - «)' H.P.'' These equations give us the horse-power that can be 704 Mechanics applied to Engineering. transmitted with any given fraction of the head lost in friction j also the permissible length of main for any given loss when transmitting a certain amount of power. The power that can be transmitted through a pipe depends on (i.) the quantity of water that can be passed ; (ii.) the effective head, i.e. the total head minus the friction head. Power transmitted P = Q(H — ^) X a constant or P = AV( H - j^ I X a constant (ALV \ AHV ^ 1 X a constant 2400D/ 2401 Then— ^=(^H-f^X a constant When the power is a maximum this becomes zero ; then — LV H 240oD ~ 3 or « = — 3 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^, until a maximum is found. Whence the maximum horse-power that can be transmitted through any given pipe is — H.P. {max) = 1-67 isj 5!£' obtained by inserting n =\va. the equation above. N.B. — H, D, and L are all expressed in feet Velocity MacMnes.— 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. 705 110 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 — V= ./JgR where H is the head of water above the outlet. When friction is taken into account — V = V2^(H - h) where h 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. 667 be mounted on wheels, or otherwise arranged so that it can be moved along horizontally Ftr. 667. 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 un1< V. Velocity rel. to ground Vo. ». Cos*. A B C V V V V \ -V 180° 90° e 1 — I D ^V« + w' + 2vV cos 9 cos 2 Z 7o6 Mechanics applied to Engineering. The velocity Vo will be clear from the diagram. The expression for D is arrived at thus : y = ab cos 6 = v . cos 6 and X = V .sm 6 Vo = ^(V+^f + *= Substituting the values of x and y, and remembering that cos' + sin° 6 = 1, we get the expression given above. If the value of cos 6 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 due t-o a force P for a space of time /, the velocity will be increased to Vj, 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 on") . [distance through which the body I = "»P"'«^ X \ it is exerted = increase in kinetic energy The kinetic energy ~l MV before the impulse j ~ ^ The kinetic energy 1 MVi" after the impulse J ~ ~2 Increase in kinetic 1 M energy J ~ T '^ ~ * ) The distance through which the impulse is exerted is — V, + V , 2 M V, 4- V hence ^(V,» - V) = P/ ' ^ or P^ = M(V, - V) or impulse in time / = change of momentum in time / Hydraulic Motors and Machines. 707 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 — Yt = MV orP=-V M But -T is the mass of water delivered per fig. 668. second. Let W = weight of water delivered per second. r^, W M Then — = -r S i , -^ WV and P = g For another method of arriving at the same result, see P- 593- It should be noticed that the pressure due to an impingmg 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 = where w = the weight of a unit column of water. We have W = o/V. Substituting this value of wV — / = WV 2^ The impinging jet corresponds to a dynamic load, and a column of water to a steady load (see p. 627). 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. 7o8 Mechanics applied to Engineering. Pressure on a Moving Surface due to an Imping- ing Jet. — Let the plate shown in the Fig. 669 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 lbs. as before ; then, if the plates succeed one another very rapidly as in many types of water-wheels, the quantitj- impinging on the plates will also be sensi- bly equal to W. The impinging velocity Fig. 669. is V — , or V f I — - J ; hence the pressure in pounds' weight on the plates is — WV P = And the work done per second on the plates in foot-lbs. — wv<i-J) n (i.) and the energy of the jet is — WV' (ii.) i. 2( \\ hence the efficiency of the jet= Tr = ^l^i ~«> The value of « for maximum efficiency can be obtained by plotting or by differentiation.* 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 ' Efficiency = t) = — ( i j 22 _, _ 1) = — = 2n ' — 2»^ d-n = — 2tt~' + 4« ' = o, when t) is a maximum dn or 2H~' = 4«~' -i = 4, whence « = 2 Hydraulic Motors and Machines. 709 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 f i — - ) > then, sub- stituting this value for W in the equation above, it will be seen 2 ( I \ 2 that the efficiency of the jet = - I i I , and the maximum efficiency occurs when « = 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 Q to the plate, and splits up into two streams. The velocity V may be resolved into Vj normal and V„ parallel to the plate. After im- • ■ ..u » T. 1 V Fig. 670. pingmg, the water has no velocity normal to the plate, therefore the normal pressure — WVi WV sin Q ~ g ~ g Pressure on a Smooth Curved Surface due to an Fig. 671. Impinging Jet. — We will first consider the case in which tjie surface is stationary and the water slides on it without shock ; •JIO Mechanics applied to Engineering. how to secure this latter condition we will consider shortly. We show three forms of surface (Fig. 671), 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 Vj ; hence the resultant pressure on the surface in the direction of cb is — P = WVi g Then, reproducing the diagram of velocities above, we have — _j; = V sin e a; = V cos B y,^=(V -xf+f Then, substituting the values of x and _v and reducing, we have — Fig. 672. V, = Vv'2(i -cose The component parallel to the jet is V - .»: = V(i - cos 0). Thus in all the three cases given above we have the pressure parallel to the jet — _ WV(i - cos e) ^0 — ' The effect of friction is to reduce the length ac (Fig. 671), hence when 8 is less than 90°, the pressure is greater, and when fl is greater than 90°, it is less than Po. The following results were obtained in the author's laboratory. The calculated value being taken as unity. Po by experiment Cone (45°). Hollowed cone (55°). Flat. 1 Approx. hemisphere I '4 I -2 i-o 07 t Pel ton bucket (162°). 08 Hydraulic Motors and Machines. 711 1 ■a £ i •Ss ^'-' ^ <b w " c > iVl ■K ^ > 1 ^ 0--.S O >l« > 8 1> o > « -i ilh Hydratilic Motors and Machines. 713 Pelton or Tangent Wheel Vanes. — The double vane shown in section is usually known as the Pelton Wheel Vane; but whether Pelton should have the credit of the in- vention or not is a disputed point. In this type of vane the angle B approaches 180°, then i — cos 6 = 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 view of such a wheel is shown in Fig. 673. Fig. 673. 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; (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. In 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 ' Reproduced by the kind permission of Messrs. Gilbert Gilkes and Co., Kendal. 714 Mechanics applied to Engineering. very faulty, since the water after striking the outer lip is abruptly changed in direction at the corners a and b, where much of its energy is dissipated in eddying ; then, further, on Fig 674. leaving the vane it strikes the back of the approaching vane, and thereby produces a back pressure on the wheel with a Fig. 675.— Doblo " Tangent Wheel " buckets. consequent loss in efficiency. This action is shown irf Fig. 674. The outer lip is not only unnecessary, but is distinctly wrong in theory and practice. In the Doble vane (Fig, 675) Hydraulic Motors and Machines. 715 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 B cannot practically be made so great as 180°, 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 dQes not occur 1-0 2-0 3-0 Matio of Jet to ivheel velocity Fig. 676. 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. 676 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 lbs. 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 7i6 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 ' 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. 2 3 4 390 480 '55 130 los go ' A similar device is used by Messrs. Gilbert Gilkes & Co., Kendal. Hydraulic Motms and Machines 717 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 phenojnena, 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, 01 0-2 0-3 O* 0-5 0-6 0-7 0-8 0-9 10 c/osed. Satio of Valv» opening to Aretv ofPipe=N. Fu/t op en. Fig. 677. remaining the same, and via versd, 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 tht; conditions are only slightly altered, such as ar? alteration in the length or diameter of the supply pipe. 7i8 Mechanics applied to Engineering. On p. 688 we showed that the velocity (V) of flow at any instant in a pipe is given by the expression — V /IT L . I KD + ign'' In Fig. 677 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, = — 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 lbs. per second. The kinetic energy of the stream issuing from the nozzle is — 62-4AV 2gn^ Inserting the value of V, and reducing, we get — ^, , • • r , O76DV H \3 The kinetic energy of the stream = 2 — / j I2 VkD "*" 2P« V Which may be written — K. = ^i — :— r where B = g ^ n C 1 ~ («5)i/ ^ A3 - (B«J -j- n-\)\ = C(B«J 4- n-l) -\ 5" = C { - f(B«^ + n-\)-\^nl - %n-\)\ Hydraulic Motors and Machines. which may be written — C{-f«S(B«^ + i)-^ X |«-S(2B«^ - i)} and C{-(B«''+i)-^X(2B«^-i)} =o when it has its maximum value, but (B«^ + i) is greater unity, hence 2B«^ — i = o , 1 KD