OF THE ' , . . UNIVERSITY OF CALIFORNIA. . Class ELEMENTARY MANUAL ON APPLIED MECHANICS NINTH EDITION THOROUGHLY REVISED GRIFFIN'S SCIENTIFIC TEXT-BOOKS. By PROFESSOR JAMIESON, M.Inst.C.E., M.Inst.E.E., F.R.S.E., Formerly Professor of Engineering in the Glasgow and West of Scotland Technical College; Consulting Engineer and Electrician, 16 Kosslyn Terrace, Kelvinside, Glasgow. INTRODUCTORY MANUALS. Crown 8vo. With Illustrations and Examination Papers. STEAM AND THE STEAM-ENGINE (Elementary Manual of). For First- Year Students, forming an Introduction to the Author's larger Work. Twelfth Edition, Revised. 3s. 6d. " Should be in the hands of EVEBY engineering apprentice." Practical Engineer. MAGNETISM AND ELECTRICITY (Practical Elementary Manual of). For First-Year Students. With Stud. Inst. C.E. and B. of E. Exam. Questions. Eighth Edition, Revised and Enlarged. 3s. 6d. "A THOROUGHLY TRUSTWORTHY Text-Book. PfiACTicAi, and clear." Nature.. APPLIED MECHANICS (Elementary Manual of). 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FORMERLY PROFESSOR OF ENGINEERING, THE GLASGOW TECHNICAL COLLEGE FELLOW OF THE ROYAL SOCIETY, EDINBURGH; MEMBER OF THE INSTITUTION OF ELECTRICAL ENGINEERS; AUTHOR OF "TEXT-BOOKS ON STEAM AND STEAM ENGINES," "ADVANCED APPLIED MECHANICS," "MAGNETISM AND ELECTRICITY," "ELECTRICAL ENGINEERING RULES AND TABLES." NINTH EDITION, THOROUGHLY REVISED diagrams, arithmetical Samples, examination anb Of THE UNIVERSITY OF |L,ONDON CHARLES GRIFFIN' AND COMPANY, LIMITED, EXETER STREET STRAND. 1910 [All rights reserved.] PREFACE TO NINTH EDITION THIS Elementary Manual on Applied Mechanics has been care- fully revised, whilst several important alterations and additions have been made to it. I have added a drawing and description to the end of Lecture VIII of Butters Brothers and Co.'s new 1909 "Jib Crane Arrester." A new Frontis Plate, as well as a new plate for Lecture XXI, showing the latest 1909 improvements in " Alfred Herbert's Electrically Driven Turret Lathe," together with a description thereof, have been duly inserted at the proper places. Owing to a very clear and useful hint which was given in The Electrician, by the reviewer of the previous Edition, I have trans- ferred the former Lectures numbered XXIV and XXV, upon " Bending and Shearing Stresses in Beams and Metal Structures," ' 20 P SCALE DIAGRAM OF A FORCE. Where represents the point of application, the straight line OP (with the arrow-head), shows the direction in which the force acts, and the length of the divided line OP indicates to scale the magnitude of the force.* (See end of this lecture for further definitions.) Forces in Equilibrium. (i) When any number of forces acting upon a body neutralize each other's effects (i.e., leave the body in the same condition as to rest or motion as before th application of the forces), these forces are said to be in equi- librium. ,(2) Forces which are in equilibrium may be applied to or removed from a rigid body without altering its condition as to rest or motion. (3) Two equal and opposite forces destroy each other's effects ; and, conversely, no two forces can destroy each other's effects un- less they are equal and opposite. (4) A force will have the same effect at whatever point in its own direction it may be supposed to act ; and, conversely, if a force have the same effect whether it act at one or other of two given points, then the straight line joining these points (with the suitably directed arrowhead) will be the direction of the force. Action and Reaction. (i) Whenever a fixed rigid body is * In the case of the above figure the force is represented as equal to 22 Ibs. Students will find it convenient to plot down the repiesentation of forces in their exercise books to a scale of ^ of an inch to a pound, 01 hundredweight, or ton, according to the values of the forces. 4 LECTURE I. acted on by a force, then naturally there is at once set up in that body a secondary force, or a force of reaction, equal and opposite in direction to the primary force. (2) Hence action and reaction are equal and opposite, and neutralize each other's effects. For Example. Suppose a weight is placed on a rigid horizontal table. In the table there is set up an opposing force or upward reaction which exactly counterbalances the downward force of the weight. If this were not the case, then motion would take place, and either the table would give way, or the weight would sink through the table ! Resultant and Components. (i) If any number of forces acting upon a body be replaced by a single force which shall have the same effect, then this force is termed the resultant of these forces, and the forces are called the components of their resultant. (2) The operation of finding the resultant of any number of forces is called the composition of forces; and finding the com- ponents is termed the resolution of forces. Resultant of Forces acting in a Straight Line. (i) The resultant of any number of forces acting in the one direction along one straight line is equal to their sum, and acts in that direction. For Example. Let P 1 P,P 3 P 4 be any four forces acting in one direction along one straight line, then their resultant R = P! + P, + P 3 + P 4 (2) If the forces do not all act in one direction, then the re- sultant is equal to the difference between the resultant of those acting in one direction and the resultant of those which act in the opposite direction, and has the direction of the greater of the two resultants. For Example. Let Pfff 4 be any four forces acting along one straight line to the right hand or in a positive direction ; and QiQaQ ^ e an y three forces acting along the same straight line, but in an opposite or left-hand or negative direction, and let Qi + Q, + Q 3 be less than p i + p + p i + P 4 Then the resultant, R = (P, + P, + P, * P 4 ) - (Q, + Q, + Q 8 ) and acts in the same direction as P t P 2 P 3 P 4 , and along the same straight line. If equilibrium existed between these two sets of oppositely directed forces, then their algebraical sum would be zero, or the resultant would vanish ; t.e., (P, + P, + P, + P 4 ) - (Q, + Q, + Q,) - K - O TERMS, ETC., USED IN MEASURING PRESSURES. C A familiar illustration of the above reasoning is the game of " the tug of war," when, say, a batch of sailors are pitted against a corresponding number of soldiers, each batch pulling their utmost at the opposite ends of a rope, and in opposite directions, with the view of obtaining a resultant. Terms aid Definitions used in Measuring or in Calcu- lating the Values of Pressures and their JJffects. The only terms (except in special cases) which are herein used for the above pur- poses, are : ^ (1) Pressure; (2) Total Pressure ; (3) Intensity of Pressure ; (4) Resultant Pressure; (5) Centre of Pressure. DEFINITIONS. (i) Pressure is a general term for the value or amount of force acting between bodies. In certain cases the words push or pull may be used. Note. The general term Pressure has hitherto been loosely used t* mean either the Total Pressure (P) or to indicate the Intensity of Pressure (p). Hence it is desirable to clearly state in all problems and writings which of these is meant when the term pressure is alone used. (2) Total Pressure (P) means the whole force acting between bodies. It is usual to measure or estimate total pressure in Ibs., thus, P=pA. Ibs. But, when the values are great, it may be given or found in cwts., or in tons. (3) Pressure or, more exactly, Intensity of Pressure (p) is the pressure per unit area ; (e.g.) Ibs. per square inch or Ibs. per square fooc. Ex. Let p = 10 Ibs. per sq. in. ; A=ioo sq. ins. Then P =pA> = lox 100=1000 Ibs. (4) Resultant Pressure (P R ) is the mean pressure per unit area (p) multi- plied by the total area (A) under consi Oration. Here, P R =^A. (5) Centre of Pressure (Pc) is the point at which a single force will balance the total pressure. Or, it may be defined as the point at which the resultant pressure acts. Differences in Nomenclature. Students may find in "Examina- tion Questions," or in papers read before various Institutions, and in books by different authors, other terms than those just defined. For example, the word Thrust has been recently introduced to mean the total pressure (P); and the simple word pressure to indicate the Intensity of Pressure (p), or Pressure in Ibs. per square inch (p), as defined above. Moreover, to show the necessity for a more definite distinction between total and intensity of pressure, if you ask an Engineer or Manufacturer the pressure of steam in his boiler, he will say (e.g.) 100 Ibs. ; when it is actually 100 Ibs. per square inch by the steam gauge, i.e., where^; = ioolbs. Resultant Thrust is sometimes used for the aforementioned Resultant Pressure or Pressure Resultant. The Author believes that these new terms only confuse students. It has hitherto been the custom amongst Mechanical Engineers and Naval Architects to chiefly confine the words Thrust and Resultant Thrust to problems and cases dealing with the screw propellers and the thrust blocks in connection with the screw shafts of steamships. DEFINITIONS SCALAR, VECTOR, AND ROTOR. NOTE FOB PAGE 3. The word Vector was used in defining the conditions of equilibrium in frames, consequently it may be as well to define the following terms here : Scalar. A quantity which has no relation to definite direction in space, or which is considered apart from such direction, is called a "Scalar " or " Scalar-Quantity." Vector. A geometrical quantity which is related to a definite direction in space is called a "Vector" or "Vector-Quantity." Vector-Quantity. This requires for its complete determination (i) the magnitude, (2) the direction, and (3) the sense to be given. A vector-quantity may be geometrically represented by a line, if (1) The length of the line represents to scale the magnitude of the quantity. (2) The line be placed in the proper direction. (3) The proper sense or way be given to the line. The sense is usually indicated by an arrowhead on the line. The line itself with its direction and sense is called a Vector-}- Suppose that a force of known magnitude acts along a line from P to ->-Q ; then, the Vector is written down as PQ7 with a bar- line over the two letters P and Q. Any quantity, whether scalar or vector (considered as occupying a definite position in space), is t-aid to be localised. Thus the mass of a body in a given position is a localised scalar, and a force acting on a body at a definite point is a localised vector. Vector /Sum. The sum of a number of vectors is often called the Kesultant Vector, and in relation to this resultant the other Vectors are called Components. To add a number of vectors, place the first anywhere, the beginning of the second to the end of the first, and so on, then the vector from the beginning of the first to the end of the last is the SUM OF THE GIVEN VECTORS (Henrici and Turner}. Rotor. A localised vector is called a Rotor (Clifford}. EXAMPLE. If A, B, C, D denote four vectors acting at the point to find the sum or result- ant. From any convenient point a, draw ab parallel in direction and containing as many units of length as there are units in the vector A. From the end b, draw be parallel to B, and equal in magnitude to it. Similarly, cd parallel and equal to C, and de equal and parallel to D. The resultant in direction and magnitude of the four vectors is the dotted line joining the initial point a to the final point e. ,- ->a If a vector E, equal in magnitude but in the /.'' \ c contrary direction, acts at O, then the five vectors S \ A, B, C, D, E will have a sum equal to zero, or the five vectors are in equilibrium. The polygon abcdea, when applied to forces is called the polygon of forces. Hence, if any number of vectors act at a point and can be represented by the sides of a closed polygon taken in order, the SUM OP VECTORS ACTING AT A POINT, vectors are in equilibrium. NOTES ON ENGINEERING CALCULATIONS. 5& We shall return to the graphic representation of forces, &c., when we come to deal with the parallelogram and triangle of forces and their application to ascertaining the stresses on simple structures.* Note Regarding Engineering Calculations. Engineering students should clearly understand, that there is no necessity or advantage to be gained in working out their arithmetical results to a greater nicety, than the tool.-', rules, gauges, and instruments placed at their disposal will enable them to measure with accuracy. For example, a skilled mechanic who is furnished with a steel footrule and callipers may express his ideas of length to the Vt r a t best to the rH (' OI ) of an inch, which is equivalent to i part in 1200. It would be ridiculous, therefore, to ask him to calculate such lengths to the third decimal place. A carpenter who uses the well-known 3 ft. rule may be perfectly satisfied if he measures to the ^ of an inch, i.e., to I part in 500. A mason will usually be satisfied if he can measure to within of an inch 1 The captain of a sailing vessel could not be expected to spot the position of his ship at sea, to within a couple of nautical miles ; so it is no use to ask him to place his boat on a particular meridian 1 Note on Questions in Proportion. When dealing with all such questions, it is best for the student to ask himself (1) What is required? Then to put the corresponding given value in the 3rd term, with x, y, or 2 in the 4th for the value to be found. (2) Will the answer be greater or less than the value in the 3rd term T Then to put the greater or the less given value (according to this answer) into the 2nd term, and the remaining known quantity in the ist term. The answer for x, y, or z is then equal to the product of the 2nd and yd terms, divided by the ist. EXAMPLES : ist : 2nd : : $rd : 4th (Term*). (a) If, 10 : 100 :: 1000 : x. Then , (6) If, loo This method is more convenient to the elementary student than dealing directly with fractional ratios. * We have intentionally made this Lecture a short one, and have not appended any question*, "because at the first meeting of a session the Lecturer has to give a series of general instructions to his students, and the class is seldom so completely formed as to make it worth while setting any home work until the second meeting. LEOTTJBE II. CONTENTS. Work Unit of Work Examples I. II. III. IV. Work done against a Variable Kcsistance Example V. Diagrams of Work With Uniform Resistance With a Uniformly Increasing Resistance With a Uniformly Decreasing Resistance With a Combination of Uniform and Variable Loads Example VI. Power or Activity Units of Power The Horse-power Unit To find the Horse-power of any working Agent Example VII. Uses of Squared Paper Clark's Adjustable Curve Example VIII. Questions. Work. If a force acts upon a body and causes that body to move through a distance, then the force is said to have done work. It does not matter how long the operation takes, whether a second, a minute, an hour, or a day, or even a year, the same amount of work is done by the force acting through the distance. Time, therefore, does not come into the question of estimating work done, but we must have a force overcoming a resistance through a definite distance. If the force applied be inadequate to over- come the resistance of the body to motion, then no work is done. The amount of work done therefore depends solely upon the pro- duct of the force applied (or the resistance overcome) and the distance through which it acts in its own direction. Or, Work = Force x Distance. Unit of Work.* The unit of work, is the work done in over- coming unit force through unit distance. Tow, since the British unit of force is th^ pound, and unit distance the foot, the British unit of work is called the foot-pound, and is therefore the work done when a resistance of i Ib. is overcome through a distance of i foot. EXAMPLE I. If a weight of i Ib. be elevated a vertical dis- tance of i ft. against the force of gravity, then i foot-pound of * In the case of heavy work the unit foot-ton is sometimes used in this country. A foot-ton simply means one ton raised one foot high against gravity, or a force of one ton exerted through a distance of one foot, or a resistance of one ton overcome for a distance of one foot. In Electrical Engineering the unit of work is the work done in overcoming a resistance of one dyne through a distance of one centimetre. It is called the Erg. Since the weight of i gramme is = 981 dynes, the work done in raising i gramme through a vertical height of i centimetre against the force of gravity is 981 ergs or (g) ergs. One foot-pound i '356 x io 7 ergs. UNIT OF WORK. work has been performed. If 10 pounds be elevated vertically through a distance of 10 ft., then result is (10 x io)= 100 ft.-lbs. of work. Fig. for Example III. UNIT OF WORK. W = i lb. weight. R = i Ib. reaction. ILLUSTRATING WORK DONE. W = Weight in Ibs. P = Pull in Ibs. L = Length in feet. EXAMPLE II. If a body offers a constant resistance to motion in any direction of P Ibs., and if it be forced along a distance of L ft., in that direction, then tbo work done is P x L ft.-lbs. Or, Work done = Force x Distance i.e. Foot pounds = P Ibs. x L feet. Suppose a cart with its load weighs W Ibs. and offers a con- stant resistance of P Ibs. to traction along a road, and that it is pulled through a distance of L feet ; then, The work done = P x L (ft.-lbs.) EXAMPLE III. In drawing a loaded cart along a level road, a horse has to exert a constant pull of 100 Ibs. ; how much work will be done in 10 minutes supposing the horse to walk at the rate of 6000 yards an hour ? Distance in feet through which the resistance of 100 Ibs. is overcome in 10 minutes. _6ooo (yds.) x 10 (m.) 60 (m.) = 3000 ft. 3 (ft-) Work done in 10 minutes = P x L. = 100 x 3000. = 300,000 ft.-lbs. EXAMPLE IV. A traction engine is employed to draw a loaded waggon along a level road where the resistance to be overcome is 8 LECTURE II. 100 Ibs. per ton. How many foot-pounds of work are expended in drawing 10 tons over 100 yards ? TRACTION ENGINE AND LOAD. 1. Tractive force = 100 Ibs. per ton. 2. Total pull, P, = 100 (Ibs.) x 10 (tons) 3. Distance, L, - 100 (yds.) x 3 (ft.) 4. Work done = P x L. = IOO X IO X 100 X 3. 300,000 ft.-lbs. Work Done against a Variable Kesi stance. If the resist- ance varies whilst the force overcom- ing it acts through a known distance, then the work clone will be measured by the product of the average resist- ance and the distance. If the resist- ance varies uniformly, its average can be found by adding its values at the commencement and end of the motion, and dividing by two. EXAMPLE V. Explain the method of estimating the work done by a force, awid define the unit of work. The sur- face of the water in a well is at a depth of 20 feet, and when 500 gallons have been pumped out, the surface is lowered to 26 feet. Find the number of units of work done in the operation, the weight of a gallon of water being 10 Ibs. (S. and A. Exam. 1887.) For an answer to the first part of this question refer to the previous part of this lecture. WORK VARYING UNIFORMLY. i. Weight of water raised = weight of 500 gallons. = 500 x 10 Ibs. Or, P, =5000 Ibs. DIAGRAMS OF WORK. [Distance through which the 2. Mean height water is lifted = -j centre of gravity, G (of raised [ water), has been elevated. Or, 3. Work done 20 + 26 2 21 ft. ft. 5000 x 23. 115,000 ft.-lbs. " , 10 ft. M Plbs. I \ Bibs. DIAGRAM OF UNIFORM WORK. Diagrams of Work. (i) Against a Uniform Resistance. If the resistance overcome is uniform, then the work done may be graphically represented by the area of a rectangle. To find the work done in over- coming a uniform resistance of 5 Ibs. through a distance of 10 ft. : Plot down a vertical line to any convenient scale to represent P (or 5 Ibs.) and a horizontal line to the same scale to represent L (or 10 ft.) Then complete the rectangle. The area P x L or 5 x 10 = 50 ft.-lbs. of work. In the accompanying figure a scale of -^ inch has been used to represent both i Ib. and i ft., consequently each of the SHIM 11 squares represents to scale one foot-pound of work. (2) With a Uniformly Increasing Resistance, If the resist- ance uniformly increases for ex- ample, in the raising of a length of rope or chain vertically by one end from the ground, then the work done may be graphically represented by the ai ea of a right-angled triangle, where P represents the total weight of chain in Ibs., and L its total length in feet. Plbs. DIAGRAM OF WORK FOR AN INCREASING RESISTANCE. The Total Work done = PxL ft.-lbs. Here the work done per foot of length of chain lifted, uniformly increases from a minimum to a maximum, until the whole rope or chain is off the ground. When any known length, I, has been lifted, then the area enclosed by the triangle whose horizontal side is I, and vertical side p represents the work done. IO LECTURE (3) With a Uniformly Decreasing Resistance. If the resistance uniformly decreases, as in the case of winding a rope or chain upon the barrel of a winch or crane, then the work done will also be represented graphically by the area of a right- angled triangle, where P represents the total weight of rope or chain in pounds being lifted at the start, and L its length in feet. DIAGRAM OF WORK FOR A DECREASING KESISTANCE. Px L .. The Total Work done = - ft.-lbs. Here the work done per foot of length of chain lifted, gradually diminishes from a maximum at the start to a minimum, when the last foot is being lifted. As in Case (2), you can at any time know the work done or still to be done from the scale diagram, if you know the length of chain lifted or to be lifted. For example, if I feet have still to come on to the barrel, then the vertical ordinate p on the scale diagram will represent the pull being exerted at the time, and consequently i represents the work still to be done. Or, generally, with any gradually increasing or decreasing resistance the work done is equal to the mean of the average resistance in Ibs. x the distance through which it acts in feet. (4) With a Combination of Uni- form and Variable Loads. "When one part of a load is uniform and another part variable, as in the case of lifting a weight with a chain, by winding the chain on the barrel of a winch or crane, the diagram of work for the uniform load is naturally a rectangle, and for the chain a triangle if the chain is completely wound on to the barrel, or a trapezoid if there is still some portion of it to be lifted.* * See p. 5 of the Author's " Elementary Manual on Steam and the Steam Engine" for how to find the Area of a Trapezoid. ^ 1 X N. n \1 x x (1 It m N. L -> Q K M DIAGRAM OF WORK FOR A COMBINATION OF UNI- FORM AND VARIABLE LOADS. DIAGRAMS OF WORK. II Let P . the uniform pull required to lift the load *r over- come the uniform resistance. L . = the distance the weight is lifted. Pv Pi = t^e weights of chain hanging at the commence- ment and at the finish of the lift. Work done in lifting the uniform load = P x L Work done in lifting the variable load =*- ^? x L .-. Whole work = P x L+Li x L= 2 2 = Area of the figure, DEFC. The diagram DEFC represents the work done and also the variation of the resistance during the lift. The rectangle ABCD represents the work done in overcoming the uniform load, and the trapezoid ABFE the work done in overcoming the variable load. The resistance at any instant of the lift will be represented by the vertical line drawn from the horizontal base DC to the inclined line EF through the point on DC or AB which represents the position of the load at that instant. The part of this vertical fine intercepted between AB and EF will represent the resistance offered by the variable part of the load at the instant considered. Thus, at the commencement of the lift the total resistance is P +p l and represented by DE, at tho end of the lift the total resistance is P + p, and represented by CF. At , J, and j of the lift the total resistances are repre- sented by the vertical lines GH, KL, and MN respectively, whil i the resistances due to the variable part of the load at these poh-ts are represented by the lengths grH, L, and rriN respectively If the final resistance due bo the variable part of the load was zero (as would be the case if the whole of the chain were wound on to the barrel) then the diagram of work for this part of the load would be the triangle AEB. EXAMPLE VI. Explain fully the mode of measuring the work done by a force. What unit is adopted ? A weight of 2 cwts. is drawn from a mine, 30 fathoms deep, by a chain weighing i lb. per linear foot ; find the number of units of work done. (S. and A. Exam. 1893.) Also find the resistance offered when the weight has been raised through J, ^, and j of the whole depth respectively. ANSWER. (i) The work done by a force is measured by the product of the force into the distance throagh whieh that force moves in its own direction. If P be the force in pounds, and L the distance in feet through which it moves in its own direction, then Work done- Px L f t, Ibi, 12 LECTURE II. (2) The unit of work adopted in this country is the work done when a force of one pound is moved through a distance of i foot, and is called the foot-pound (ft.-lb.). (3) Referring to the previous figure, make AB to represent 30 x 6 = 1 80 ft., the depth of the mine, AD to represent 2x112 = 224 Ibs. the weight of material raised, AE to represent i x 1 80= 1 80 Ibs. the weight of chain at beginning of lift. Then assuming the whole of the chain to be wound up, complete the rectangle ABCD, and join E and B. The area of the figure DEBO then represents the work done. . . Work done = area DEBC - \ (DE + CB) x AB. But DE = DA + AE = 224 + 180 = 404 Ibs. CB = DA =224 Ibs., AB = 1 80 ft. .".Work done = \ (404 + 224) x 180 ft.-lbs. = 56,520 ft.-lbs. (4) The resistance at \ lift, or when the weight has been raised 45 ft., is Gh = Gg + gh = 224 + $x 1 80 = 359 Ibs. At ^ lift the resistance is K =K& + kl = 224 + Jx 180 = 314 Ibs. At f lift the resistance is M?i = Mm4- mn= 224 + | x 180 = 269 Ibs. Power or Activity is the rate of doing work* In estimating or testing the power of any agent the time in which the work is done must be noted and taken into account. Consequently, we speak of the activity or power of a man, of a horse, or of an engine, as capable of doing so many foot-pounds of work per minute. Units of Power, f The unit of power adopted in this country is called the horse-power. It is the rate of doing work at 33,000 ft.-lbs. per min., or 550 ft.-lbs. per sec., or 1,980,000 ft.-lbs per hour. The Horse-power Unit was introduced by James Watt, the great improver of the steam engine, for the purpose of reckoning the power developed by his engines. He had ascertained by experiment that an average cart-horse could develop 22,000 foot- pounds of work per minute, and being anxious to give good value to the purchasers of his engines, he added 50 per cent, to this amount, thus obtaining (22,000+ 11,000) the 33,000 foot-pounds per minute unit, by which the power of steam and other engines has ever since been estimated. * The vrordpoweris very frequently misapplied by writers and students, for they often call the mere pull, pressure, or force exercised on or by an agent the power. Students should strenuously avoid this misuse of the word power, and never employ it in any other sense than as expressing a rate of doing work, or activity. t In Electrical Engineering the Unit of Power is called the Watt, and it equals lo 7 ergs per second, or 746 Watts i horse-power. TO FIND THE HORSE-POWEK OF ANT WOKKTNO AGENT. 1 3 To find the Horse-power of any Working Agent. Divide the number of foot-pounds of work which it does in one minute by 33 000 - Let P = Pull exerted or resistance overcome in pounds. L = Length or distance through which P acts. M = Minutes the agent is at work. H.P. = Ilorse-power. Then, H P - FxL p = II. P. x 33000x11 . L _ H. P. x 33000 x M 33000 x M ' L P EXAMPLE VII In what way is the rate of doing work measured hi horse-power ? If 40 cubic feet of water be raised per minute through 330 feet, what horse-power of engine will be required, supposing that there is no loss of friction or other resistances ? Note. I cubic foot of water weighs 62 J Ibs. (S. and A, Exam. 1892). ANSWER. The rate of doing work, as measured in horse-power, is equivalent to 33,000 foot-pounds of work done per minute. ist. i cubic foot of water weighs 62 J Ibs. .40 cubic feet of water weigh 40x62^ = 2500 Ibs. 2nd. Work done per minute = ^ c 2500 (Ibs.) x 330 ft. M I 3 rd. .-. H.P.= PxL = 2500 x 330 = 825000 = 25> 33000 x M 33000 x i 33000 Note. Students will find it a great advantage, as well as a sav- ing of time not to multiply figures together until the last stage of the answer has been reached, and then to cancel all common factors in numerator and denominator. For example, in the answer to the above question we might proceed thus i st. 40 cubic feet of water = 40 x 62 \ Ibs. 2nd. Work done per minute = 40 x 62! x 330 ft. -Ibs. 4 i 40x62^x^0 3rd. .. Horse-power of engine = - 10 10 10 The process consists in this the factor, 330, can be cancelled from numerator and denominator, leaving 100 as the denominator. The factor, 10, can then be divided out of 40 in the numerator and from the 100 in the denominator, thus leaving 4 x 62 J as the numerator and 10 as the denominator. The remainder of the work is evident. LECTURE H. Uses of Squared Paper. Squared paper is made by drawing a number of equally spaced horizontal lines and crossing these by vertical ones at the same distance apart. The paper is conse- quently covered with a large number of little squares, the sides of which are usually one-tenth of an inch in length. In order to facilitate the measurement of distances, every tenth line, and some- times every fifth, is heavier or of a different colour to the others. By the aid of squared paper we can graphically represent how two varying quantities depend upon each other. For example, take the case of a chain being gradually lifted from the ground, as already considered in connection with diagrams of work and shown by the first of the following figures. The length and weight of chain lifted will alter as the upper end is raised ; but the suspended weight will always depend upon the length which hangs freely from the spring balance. In fact, any change in the length lifted will produce a corresponding change in the load registered by the balance. If we note the pull indicated by the balance for different lengths of hanging chain, we shall be able to obtain a line or curve which will show to the eye how these two quantities depend upon one another. Suppose we obtain the following results : Length of "^ chain lifted V in feet. J I 2 3 3'5 4 S 6 7 8 9 10 Weight on chain lifted V 0-8 i '4 2'0 2 '5 27 3'S 4 '3 .TO 5 '6 6'2 7-0 in Ibs. J Corrected ^ values in > Ibs. J 07 I '4 ,, - 2'8 3'5 4*2 4 '9 , 6-3 7-0 We shall represent the lengths of the hanging chain by horizon- tal distances which are termed abscissa, and the corresponding weights by vertical distances called ordinates. We choose such a scale for these quantities as will enable us to get them all upon the squared paper ; at the same time we keep the scales as large as possible. In this case we have chosen five divisions horizontally to represent one foot, and five divisions vertically for one pound. It is, however, not necessary to adopt equal scales for abscissae and ordinates, but we should select the most convenient scale for each according to circumstances. USES OF SQUARED PAPER. I 5 To find the point corresponding to the fourth column in the table, take B at 3-5 on the horizontal scale, and C at 2-5 on the vertical one. Draw B A vertically and C A horizontally. Then, the intersection of these two lines is the point required. In practice, these lines are not actually drawn, but the point is found by the eye with the assistance of the lines on the paper, and a x or is placed to mark its position. When all the points have been thus plotted from the table, we draw a mean line or curve between them. In this case, it is a straight line passing through Abscissae or Length Femt HANGING CHAIN. SQUARED PAPER. RELATION BETWEEN LENGTH AND WEIGHT OP CHAIN. the origin 0. It will be seen that this line does not pass through all the points, but that some of them are on one side and some on the other. This may be due to errors of observation or to irregularities in the chain. If we know that the chain is uniform, the points ought to lie along the straight line we have drawn, and we can correct for errors of observation. Thus, the point A should have been at D, and the correct weight for that length of chain is represented by BD, which is 2*45 Ibs. We can correct the other values in the same way, and so obtain the numbers shown in the third line of the table. When the points lie approximately in a straight line, the nearest mean straight line is best found with the help of a fine thread which is stretched and moved among the points until it 1 6 LECTURE II. lies most evenly amongst them. The positions of its ends are then marked and a line drawn with a straight-edge through these marks. When the points do not lie near a straight line a smooth curve may be drawn through them, either freehand, or by aid of French curves, a thin strip of wood or steel, or Clark's paiant It ~ CLARK'S PATENT ADJUSTABLE CURVE, adjustable curve. This consists of a flexible strip of celluloid with a brass loop A, for the thumb of the left hand, and another B, for the second or third finger, as shown by the accompanying figures. Now, if these two loops are drawn together the celluloid will be formed into some curve, the shape of which can be ad- justed by moving the sliding rod C with the right hand and fixing this curve by means of a cord (not shown in the figure) joining B with a V-shaped groove in A. It may also be used as USES OF SQUARED PAPK*. a set curve for the purpose of transferring a curve from one drawing to another. When we represent distances along one scale of the squared paper and forces along the other, then the areas such as O D B indicate to scale the work done in raising the length of chain O B. This is true, whether the line D is straight or curved. We can, however, represent any two quantities which depend on one another by a curve on squared paper. EXAMPLE VIIT. A body is being acted upon by a variable lifting force. When the body is lifted x feet the force F Ib. is observed. X 15 25 50 70 100 125 150 180 210 F 530 525 516 490 425 300 210 160 110 90 Plot on squared paper and find the average value of .Ffrom x = o to *=2io. What is the work done by F when the body has been lifted 210 feet? (B. of E. 1904.) ANSWER r~ i 500 SK f : si^ij i:::::i . i- F : - 4- 400 -^ | :: r JI T rTrrHrr 5p *- t: :l::::: ^ *vj 300 . t : ...: Tt-f 1 i w |i ^p-:::::::: ?00 ft : jK:i; : :: "* :|::::: - ---[- L - ^* s . , 100 k 1 ::-:(- t H -^ F:|S::| a ""] ' = B 1 7 5^y?c es { |^S S :::: i : C 20 40 60 80 100 120 140 160 180 - 200 210 DIAGRAM OP WORK DONE BY A VARIABLE FORCE. In order that we may obtain the average value of F, we must nse squared paper since the lifting force is variable. Mark off the distances x in feet (to gcale) along the abscissa C, and at each of the values of , plot ordinates corresponding to the values of F in Ibs. Join all the points by a curve as shown in the above diagram. The area of this diagram below the curved line A B, represents the total work done by the variable force Fibs., in passing through a distance of 210 feet. We can obtain the area of the diagram O A B C either by aid of aplanimeter or by inding the average value of F and multiplying this by the distance 1 LECTURE H. through which Fhas moved, viz., 210 feet. But the ordinary process used by engineers for finding the area of a diagram such as A B C is to divide 9-0 into 10 equal parts and then to measure the value of F at the centre of each part. The sum of these values divided by 10 gives us the average value of F. If this number be now multiplied by the value of C the final result will be the area of the diagram in foot-pounds. Thus, by the latter method we find the average value of F=305 Ibs. Hence the total work = average value of Fx distance it acts. Or, the total work done = 3o$ x 2io-64,050 ft.-lbs. LECTURE H. QUESTIONS. Ip LECTURE II. QUESTIONS. 1. Define tbe unit of work. What name is given to this unit? In drawing a load a horse exerts a constant pull of 120 Ibs. ; how much work will be done in 15 minutes, supposing the horse to walk at the rate of 3 miles an hour 1 Ans. 475,200 ft.-lbs. 2. How is the work done by a force measured ? The resistance to traction on a level road is 150 Ibs. per ton of weight moved ; how many foot-pounds of work are expended in drawing 6 tons through a distance of 150 yards 1 Ant. 405,000 ft.-lbs. 3. Distinguish between force and work done by a force. How is each respectively measured ! A traction engine draws a load of 20 tons along a level road, the tractive force on the load being 150 Ibs. per ton. Find the work done upon the load in drawing it through a distance of 500 yards. Ans. 4,500,000 ft.-lbs. 4. Find the number of units of mechanical work expended in raising 136 cubic feet of water to a h/.ight of 20 yards. The weight of a cubio foot of water is 62^ Ibs. Ans, 510,000 ft.-lbs. 5. A weight of 4 tons is raised from a depth of 222 yards in a period of 45 seconds ; calculate the amount of work done. Ant. 5,967,360 ft. -Ibs. 6. A hole is punched through a plate of wrought-iron one-half inch in thickness, and the pressure actuating the punch is estimated at 36 tons. Assuming that tbe resistance to the punch is uniform, find the number of foot-pounds of work done. Ans. 3360 ft. -Ibs. 7. How is work done by a force measured ? Give some examples. Set out a diagram of the work done in drawing a body weighing 10 Ibs. up a smooth incline 4 feet high, marking dimensions. 8. A train of 12 coal waggons weighing 133 tons is lifted by hydraulic power (two waggons being raised at a time) through 20 feet in 12 minutes. Estimate the work done in foot-tons. Taking the average of work done, how many foot-pounds are done per minute? Ant. 2660 ft. -tons ; 496,533*3 ft.-lbs. per minute. 9. The plunger of a force-pump Is 8f inches in diameter, the length of the stroke is 2 feet 6 inches, and the pressure of tbe water is 50 Ibs. per square inch ; find the number of units of work done in one stroke, nd plot out a diagram of work to scale. Ans. 7517 ft.-lbs. 10. A chain 30 feet long, and weighing 100 pounds per yard, lies coiled on the ground. Find by calculation and by a scale diagram of work how many units of work would be expended in just raising it by the top end from the ground. Ans. 15,000 ft.-lbs. 11. A chain, weighing 30 Ibs. to the fathom, is employed to lift i ton to a height of 30 ft. by winding the chain on a barrel. Find by calculation and by a scale diagram of work, how many units of work will be expended (a) when the outer end of the chain is brought home to the barrel ; (b) when 18 feet of it are still hanging free with the weight at the end of it. Ans. (a) 69,450 ft.-lbs. ; (b) 28,320 ft.-lbs. 12. Define the following mechanical terms : Force, work, unit of work, power, activity, and horse-power. A horse drawing a cart at the rate of 2 miles per hour exerts a traction of 156 Ibs. ; find the number of units of work done in one minute. Ans. 27,456 ft. -Ibs. 13. In what way is the rate of doing work measured in horse-power ? If loo cubic feet of water be raised per minute through 330 feet, what horse-power of engine will be required, supposing that there is no loss by friction or other resistances 1 A j i.e., 8 (oz.) x 24" = 24 (oz..) And P x AB = E (or R) i.e., 1 6 oz. x 24" = 24 (oz.) The fulcrum F, where the equilibrant and resultant act, is termed the centre of the two parallel forces, and it is J of the length of the lever from one end, and f from the other end. Reasoning generally trom this particular case, if you have any two unequal unlike parallel forces, P and Q, acting on a body in the directions AP and BQ respectively, and of which Q CENTRE OP PARALLEL UNLIKE FORCES. is the greater force, then if the line AB be drawn perpendicular to the directions of these forces, and prolonged, a single force E 3 parallel to P and Q and equal to Q P, will balance these two forces at a point C, so thai the moments about C are equal and opposite ; or, P x AC = Q x BC Further, a force R, equal and opposite to E, acting at C, will represent the resultant of P and Q. This point, C, is termed the centre of the parallel forces. The position of the point C, which is determined by the above equation, is not affected by the directions of the forces so long as they act at the same points A and B, and have the same mag- nitudes. You may imagine any number of parallel forces acting in one plane being replaced by a single force. For in the above case you have formed a resultant, R, for the two forces P and Q consequently you could find a resultant, R I} for R and any other parallel force say S ; and so on for any number. 28 LECTURE HI. The fined resultant of the whole of the forces would act at a point which would be the centre of the system of the whole of the parallel forces acting on the body. Centre of Gravity. Since gravity attracts towards the earth each particle of matter of which a body is composed, the weight of a body may be considered as the sum of a system of parallel forces. The centre of these parallel forces is called the centre of gravity of the body, and is the point where the resultant of the weights of all the particles composing the body acts. The following statements in small type, which are generally proved as propositions and corollaries in books on Elementary Theoretical Mechanics, should be remembered by the student : 1. If a body is symmetrical, the centre of gravity (or e.g. of the body) coin- cides with the centre of the mass. 2. If a body be uniform, the c.g. coincides with the centre of volume. 3. In a very thin plate of uniform density the c.g. coincides with the centre of surface. 4. If the c.g. of a body be determined for any one position of the body, the same point is the c.g. for every other position. 5. If a body be supported on its centre of gravity, the body will balance in any position. Or, a body will balance about its c.g. in all positions. 6. If a body balance in all positions about a straight line through it, the c.g. lies in that line. 7. If the c.g. be vertically above or below the point of support, the body will rest in that position. Hence, if you balance or support a body from two different points, the che intersection of its diagonals ; of a triangle at the intersection of straight lines drawn from two of the angles to the middle points of the opposite sides i.e., at a distance from one of the angles along one of these lines equal to | of the line ; of the perimeter of a triangle (i.e., of three uniform rods forming a triangle) at the intersection of the two straight lines which bisect two of the angles of the triangle formed by joining the centres of the three uniform rods ; of a polygon at the point of application of the resultant of the parallel forces represented by the areas of the respective triangles into which the polygon may be formed, and where each of these forces is considered to act at the c.g. of its own triangle ; of a pyramid at f of the line from the vertex to the c.g. of the base ; of a cone at of the axis from the vertex ; of the curved surface of a cone at f of the axis from the vertex ; of a. prism at the middle of the line connecting the c.$r.'s of its ends. THE LEVER. 2Q The Lever when its Weight is taken into Account. In this case we have to add the moment due to the weight of the lever, to the moment of P or of W according as it acts along with the one force or with the other ; i.e., according as the e.g. of the lever is on the same side of the fulcrum as P or W. When the lever is of uniform section and density throughout, then the e.g. of the lever is at its middle point, and consequently the whole weight of the lever may be considered as concentrated and acting at that point. A t WEIGHT OF LEVER CONSIDERED. Let AB be a uniform lever, of weight w, acting at its e.g. or middle point C, let a weight, IF, be attached to the end B, then the force P, which will have to be applied to the other end A, in order to balance the whole about the fulcrum F, will be found by taking moments about F. Thus, P x AF 4- w x CF = W x BF Or* p_WxBF-wxCF EXAMPLE I. A uniform lever, 5 ft. long, of 30 Ibs. weight, is placed on a fulcrum 10 in. from one end, and has a weight of 100 Ibs. attached to the short end. What force must be applied, and in what direction, in order to produce equilibrium ? Also, what is the pressure on the fulcrum, and in what direction does the reaction from the fulcrum act ? 1 . Referring to 'the above figure, we find from the question that AB = 5 ft. = 60 in.; BF= 10 in. .*. AF = 50 in. and CF = 20 in. - W = 100 Ibs. and w = 30 Ibs. 2. By the principle of moments The Opposing Moments about the Fulcrum are equal. Consequently, P x AF + w x CF = W x BF p_WxBF-tpxGF AF Substituting the numerical values P = 100x10-30x20 _ 8 lbg 5 * If the e.g. of the lever was on the opposite side of the fulcrum, i.e., on the side of W, then P x AF=* W x BF + to x CF. 3O LECTURE III. 3. P acts vertically downwards, since the moment due to the weight of the lever is not sufficient to equalise the moment due to the weight W about the point F. 4. The pressure on the fulcrum is evidently equal to the sum of all the forces, since all the forces act in one direction, or vertically downwards. It is therefore equal to W + w + P ^ soo 4- 30 + 8 = 138 Ibs. 5. The reaction from the fulcrum is equal and opposite in direc- tion to this resultant. It therefore acts vertically upwards, and is the equilibrant of the whole of the forces, for a vertical force of 138 Ibs. applied to the lever at F would counterpoise or just lift the whole bar with the attached weights P and W. EXAMPLE II. Suppose everything the same as in the previous example but the weight of the lever, which you may consider as now equal to 60 Ibs. ; what force P would be required, and in what direction would it have to act, in order to produce equili- brium ? Also, what would be the resultant or downward pressure at F. 1. You observe at once that the moment of the weight of the lever is greater than the moment of W about the fulcrum. For, w x OF > W x BF Since, 60 x 20 > 100 x 10 Consequently by the principle of moments P must act against w, or vertically upwards, so as to assist W, in order that the oppos- ing moments about the fulcrum may be equal. 2. The formula therefore becomes !0xCF-PxAF Or, >xCF . u?xCF-WxBF -D " Substituting the numerical values, we have 6ox20-iGoxio_ p _ 1 lbs 5o 3. The resultant pivssure at F is equal to the algebraical sum of the forces, or W + w - P = 100 + 60-4 = 156 lbs. And acts vertically downwards. The equilibrant would there- fore be 156 lbs. acting on the lever at Fand vertically upwards. Position of the Fulcrum. In answering questions which give the magnitude of the forces with which they act, and require only an answer for the position of the fulcrum, the student has THE LEVER. 3 1 simply to employ the general formula for the principle of mo- ments, and then to substitute the known numerical values in order to get the unknown. Or, he may reason out the formula into the following shape, and then interpolate the numerical values. Referring to the last figure, suppose that the distance AF is required : Then, neglecting the weight of the lever, we have by the principle of moments Or, PxAF + Wx AF = AF(P + W) =WxBA WxBA AF = P + W Now, taking the weight of the lever into account , we have by thd principle of moments : PxAF + wxCF = Wx BF. Or, r BA PxAF + wxAF + Wx AF = WxBA + M?x W EXAMPLE III. Where should the fulcrum be placed under a uniform lever in order to produce equilibrium, if the lever is 5 ft. long, weighs 30 Ibs., and has weights of 100 and 8 Ibs. respectively hung at its ends. From the above general equation for equilibrium viz. : 8 + 3Q-I- TOO Which proves the data given in Example I. to be correct, 32 LECTURE III. QUESTIONS. LECTURE III. QUESTIONS. 1. Define what is meant by " the moment of a force,* 1 and give an ex ample with a sketch. 2. State " the principle of moments," and apply it to the case of a simple straight lever. 3. A weight of 10 Ibs. on the end of a lever 100 inches from the centre of motion is found to balance a weight of 100 Ibs. at a distance of 10 inches. Explain the natural law which governs matter and motion, upon which the above mechanical fact depends. (Answer this by giving the definition of the principle of moments.) 4. Describe an experiment to prove the equality of the moments when the pull is between the weight and the fulcrum and acts in the opposite direction to the weight. 5. In the case of a straight lever, how would you ascertain the pressure on and the reaction from the fulcrum ? 6. Three forces, of 12, :o and 2 Ibs., act along parallel lines on a rigid body ; show by a sketch how they may be adjusted so as to be in equili- brium 1 Ans. The force of 12 Ibs. must act as the equilibrant to the forces 2 and 10 Ibs. i.e., in a line with their resultant, but in the opposite direc- tion. 7. Two parallel forces of 10 and 12 Ibs. act in opposite directions on a rigid body, and at 2 fe^t apart. Where is the centre of the two forces, and what is their resultant 1 Ans. 10 feet from the force of 12 Ibs., 2 Ibs. 8. Define the " centre of gravity " of a body, and show how you would find it experimentally in the case of any irregular body. Give an example. 9. State the rule which applies when two unequal forces balance on opposite sides of the fulcrum of a straight lever, the weight of the lever being neglected. A uniform straight lever, 4 feet long, weighs 10 Ibs., the fulcrum is at one end ; find what upward force acting at the other end will keep the lever horizontal when a weight of 10 Ibs. is hung at a dis- tance of i foot from the fulcrum. Find also the pressure on the fulcrum and the direction in which it acts. Ans. 7'5 Ibs. ; 12-5 Ibs. downwards. 10. A uniform bar, 4 feet long and weighing 4 Ibs., can turn about a fulcrum at one end, and a weight of 10 Ibs. is hung upon the bar at a dis- tance of i foot from the fulcrum. Find the upward force at the free end which will keep the bar horizontal. Ans. 4-5 Ibs. 11. A uniform bar of metal 10 inches long weighs 4 Ibs., and a weight of 6 Ibs. is hung from one end. Find the fulcrum or point upon which the bar will balance. Ans. 2 inches from the 6 Ibs. weight. 12. Two parallel forces whose magnitudes are 8 and 12 Ibs. respectively, act in the same direction on a rigid body at points 10 inches apart. Find the magnitude and line of action of the resultant of the two forces. Ans. 20 Ibs. at a point 6 inches from the force of 8 Ibs. 13. A uniform lever is 5 feet long, and weighs 10 Ibs., the fulcrum being at one end. A weight of 30 Ibs. is hung at a distance of 4 feet from the fulcrum ; what upward force acting at the middle point of the lever will keep it in a horizontal position 1 Ans. 58 Ibs. 14. Define " moment of a force." How is it measured ? A bar of metal of uniform section weighs 5 Ibs., and a weight of 10 Ibs. hangs from one end. It is found that the bar balances on a knife edge at 9 inches from the end at which the weight hangs ; what is the length of the bar I 4 ft. 6 in. LECTDBE HI. QUESTIONS. JJ 15. State the principle of the lever, and prove it when F and W act on opposite aides of the fulcrum. A weight of 5 Ibs. is hang at one end of a uniform bar, which is balanced over a knife edge at a point 14 inches from the end at which the weight hangs. The bar weighs 30 Ibs. ; find its length. Ant. 32$ inches. 16. State the principle of the lever. A uniform straight bar, 14 inches long, weighs 4 Ibs. ; it is used as a lever, and an 8 Ib. weight is suspended at one end. Find the position of the fulcrum when there is equilibrium. Ant. 2& inches from the 8 Ib. weight. 34 NOTES AND QUESTIONS. ( 35 ) LECTURE IV. CONTENTS. Practical Applications of the Lever The Steelyard, or Roman Balance Graduation of the Steelyard The Lever Safety Valve Example I. Lever Machine for Testing Tensile Strength of Materials Straight Levers acted on by Inclined Forces Bent Levers The Bell Crank Lever Bent Lever Balance Duplex Bent Lever, or Lum- berer's Tongs Turkus, or Pincers Examples II. and 111. Toggle Joints Questions. IN this Lecture we shall give a number of examples of the application of the lever. STEELYARD, OB ROMAN BALANCE. INDEX TO PARTS. F represents Fulcrum. GA Graduated arm. SW Sliding weight. P Pull due to SW SP Scale pan. W represents Weight in SP. AF Distance of P from F. BF Distance of W fromF. 36 LECTURE IV The Steelyard, or Roman Balance, is a straight lever with unequal arms, having a movable or sliding weight on the longer arm. It is very much used by butchers for weighing the carcasses of cattle and sheep, and in such cases it generally has two fulcra and two scales of division corresponding to them, the one set being, say, for hundredweights and the other for pounds. Graduation of the Steelyard. The practical method of gradu- ating the steelyard is to put unit weight (say i Ib.) into the scale pan, SP (or attach it to the hook on the shorter arm if there should be no such pan), and mark the position where the sliding weight, SW, has to be placed in order to cause equilibrum. Mark this position i on the scale. Then put in two units (say 2 Ibs.) into SP, and adjust SW as before, marking its new position as 2 on the scale ; and so on until SW is at the end of the longer arm. In tnis form of steelyard, if the differences of the weights W, corresponding to successive distances, i to 2, to 3, 3\ 4 5 6 7 ]s 9 10 11 12 13 14 15 I j W P STEELYARD WITH THE CENTRE OF GRAVITY IN THE' LONGER ARM AT G. Let w = weight of beam and scale pan, and suppose P at A and to at G to balance W units at B. Then, for equilibrium, we have P x FA + 10 x FG = W x FB . . (5) As before, increase W by one unit, and let P be shifted to A, in order to restore equilibrium, then we must have Subtracting (5) from (6) we get PxAA 1 = FB, _FB Or, AA t p . . , / \ Now increase W by n units, and let P occupy the position A*, then P x FA,. + w x FG = (W + n) x FB . (8) Subtracting (5) from (8) we get PxAA M = nxFB, FB Or, AA,. = n x -p-, That is, AA,, = n x AA, by equation (7) 38 LECTURE IV. Thus we again see that the graduations are equal for equal increments of the weight W. To find the zero of the scale in this case : In equation (5) put W = 0, then PxFO That is, the zero is in the shorter arm at a distance from F, represented by - J p units of distance, the units in this case being the same as those measuring FG. By making W= i in equation (5) the position of the first figure on the scale can be fixed, and then the whole beam graduated, since all the divisions are of the same size. One important point to be observed in this arrangement is, that when the centre of gravity lies in the longer arm, there is a limit to the smallness of the weight which can be weighed in the scale pan, since the sliding weight moves along the longer arm only. Let P coincide with F, then the weight W which must be placed in the scale pan in order to just balance the weight of the beam and scale pan at G is . Any weight less than this cannot be weighed. This is not an objection to the instrument where the weights to be measured are great, as in the case of the butcher's steelyard used for weighing heavy carcasses. When the centre of gravity lies in the shorter arm the graduations will still be equal. The reasoning is the same as in the last two cases. The student can also prove that the zero of the scale is on the longer arm at the point given by the equation In this case all weights, however small, can be weighed. The Lever Safety Valve.* The lever safety valve is a simple * For a more detailed description of safety valves and their action refer to Lecture XXVII. of the author's Elementary Manual on ' Steam and the Steam Engine." THE LEVER SAFETY VALVE. 39 contrivance fixed on the top of a boiler for the purpose of auto- matically preventing the steam exceeding an agreed-upon working pressure. Referring to the next figure, YC is a cast-iron valve chest, containing a tightly-fitted gun-metal valve seat, VS, on which rests a steam-tight gun-metal valve, V. On the centre of the upper side of this valve rests a conical steel pin attached to a straight lever by an eye and bolt. One end of this lever is free to turn en a fulcrum fixed to the upper flange of the valve chest, and a lock-fast cast-iron weight is placed near the other end, so LOCKFAST LEVEB SAFETY VALVE. INDEX TO PARTS. VC represents Valve chest. VS Valve seat. V represents Valve. P i Locking pin. that the downward moment of the weight about the fulcrum balances the upward moment of the steam pressure on the valve about the same fulcrum. Let L = length of lever in inches from fulcrum to the e.g. of the weight, W. F = Distance in inches from fulcrum to centre line of valve, V. ^ = to e.g. of the lever. W = "Weight in Ibs. of the cast-iron counterpoise block. w / = lever. w * = valve. P =: Pressure of steam in Ibs. per square inch* 4<3 LECTURE IV, d = Diameter of valve in inches. A ='Area of valve in square inches = -d 2 . P x A = Total pressure in Ibs. on the valve. Then, by taking moments about the fulcrum, we find the pressure of steam per square inch which will balance the several forces. for the upward moment = the sum of the downward moments. (P x A- W,)F = (W x L) + (W, x G). Or, (PxA)xF = (WxL) + (W,xG) + (W.xF) AxE If we neglect the weight of the lever and the valve Then, (PxA)xF = WxL -?s EXAMPLE I. A valve, 3 inches in diameter, is held down by a lever and weight, the length of the lever being 10 inches, and the valve spindle being 3 inches from the fulcrum. You are to dis- regard the weight of the lever and to find the pressure per square inch which will lift the valve when the weight hung at the end of the lever is 25 Ibs. -10 W =25 Ibs. Referring to the previous figure as well as to the accompanying one, we see from the question that <* = 3" ' A = ^2 = .7854x3x3 = 7.07 sq. ins. ; BF= 3', AF= 10" and W = 25 Ibs. Taking moments about F, we get (PxA)xBF = Wx AF P x 7.07 x 3 = 25 x 10 P = H'8 Ibs. per square inch. Testing Machine. The following figures illustrate a machine which is used for testing the tensile strength of iron, steel and such like materials. It consists of a combination of levers. After LEVER TESTING MACHINE. >HW LEVEB MACHINE FOR TESTING TENSILE STRENGTH OF MATERIALS. A, L, 7 MW f W n R BW a A 2 DIAGRAM OF THE LEVERS. INDEX TO PARTS. LjL, represent Levers. F,F 2 Fulcra. P Pull, or dead weights. A i Where P acts on L,. Specimen under te.->t. BpA,, represents Where R acts on L,,L 2 . B., Where W acts on S. M\V Movable weight. BW Balance weight. HW Hand-wheel and screw for elevat- ing F lf &c. 2 LECTURE IV. mastering the general arrangement of the machine by comparing the index to parts with the side elevation, the student should refer to the accompanying skeleton diagram (where the same index letters have been used), from which he will readily understand how the stresses are transmitted and magnified. Looking at the second of the above figures, or skeleton diagram of the levers, it will be seen that when equilibrium exists between the stress W on the specimen S, and the pull P, applied at A v P x A^\ = E x B^j, and R x A 2 F 2 = W x B 3 F 2 PxA,F, Wx *i*\ ~ A P x A 1 F 1 x A 2 F 8 -L PxA,F, WxB,F 2 ~ . . _ Consequently, W= Straight Levers Acted on b3 Inclined Forces. In the previous Examples and in Leeeure III. we have considered the forces P and W as acting at right angles to the straight levers. In such cases the forces had the greatest advantage, or their turning moments were a maximum. But the principle of moments is equally applicable to inclined forces acting on straight levers and to bent levers. X s / \ W STRAIGHT LEVEES WITH INCLINED FORCES. For, let AjBj be a straight lever acted on by inclined forces, P and W. "Draw from the fulcrum, F, lines at right angles to the produced directions of the forces as shown by the dotted lines in the above figure. Then, the effective arms for the forces P and W are respectively A 8 F and B,F ; and equilibrium takes place when their momenta about F are equal ; i.e., when P x A,F = W x B F Or, P: W::B,F: A,F. Bent Levers. The BeU Crank Lever. The same principle and action hold good in the case of bent levers. Take an ordi- BENT LEVERS. 43 nary right-angle bell crank lever, as shown by the accompanying figure. Here the effective arms are equal to the actual arms of the lever, because the forces have been shown as acting at right angles to their respective arms, or with maximum turning moments. Therefore, P x AF = W x BF. But, if the lever be turned round through any angle by, say, an extra pull at P, then, in order to ascertain the virtual moments we should have to draw lines at right angles from F on the directions of P and W in order to calculate their effective arms. S.P. w BELL CRANK LEVER BENT LEVER BALANCE. Bent Lever Balance. Examine an ordinary bent lever balance, such as is frequently used for weighing letters and light parcels, where the force P is a constant quantity, and the variable force W is represented by the article to be weighed. As showu by the accompanying figure, the effective arms change with each weight to be ascertained, and consequently the scale S of this balance has to be graduated by trial, or by introducing standard pounds, such as SP, or other units, and marking the values on the scale opposite the position where the end of the pointer on P comes to rest. Or, the graduation might be done by plotting the various positions of the arms and values of the forces to scale, In the illustration we have evidently got equilibrium when P x AF = W x BF. Duplex Bent Lever, or Lumberer's Tongs. The accom- panying illustration shows a very useful and simple application of the bent lever, which is used at the end of a winch or crane chain, 44 LECTURE IV. for affixing to and holding fast stones, logs of wood, blocks of ice, or other heavy articles when they have to be lifted. P, P indicate the directions of the pulling forces on the short chains between the ends of the shorter arms and the common link which is attached to the crane chain. F is the common fulcrum, and W, W show the directions of the forces with which the article is gripped. The student will be able to draw a diagram of the forces and calculate their effective moments for himself for any particular case. The Turkus, or Pincers. The ordinary carpenter's turkus, or pincers, which is frequently used for extracting nails from wood, is another familiar illustration of the duplex bent lever. As shown by the accompanying figure, the forces P^Pj, represent the forces with which the pincers is gripped by the hand after the jaws have been closed on the neck of the nail, and the force B TONGS OE DUPLEX BENT LEVEE. TUKKUS, OE PlNCEES. the pressure which has to be exerted by the arm and body in order to extract the nail from the wood i.e., to overcome the frictional resistance, W, between the wood and the nail. As shown by the separate diagram of forces in dotted lines, straight lines have been drawn, not from the joint of the pincers, but from a position representing the fulcrum F (or point where the nose of the pincers rests on the wood), perpendicular to the direc- tions of the forces P and W, in order to obtain the lengths AF and BF of the effective arms of the bent lever. Here again, P x AF = W x BF. EXAMPLE II. The handle of a claw-hammer is 15 inches long, and the claw is 3 inches long. What resistance of a nail would be overcome by the application of a pressure of 50 Ibs 4 at the end of the handle ? BENT LEVERS. 45 You are required to show, by a diagram, the manner in which you arrive at your result. (S. and A. Exam. 1892.) ANSWER. Here we have a simple case of a bent lever, with fulcrum at F, and effective arms, AF, BF, 1 5 and 3 inches long w P-50 LBS. EXAMPLE OP A BENT LEVER. respectively. Let W represent the resistance in Ibs. offered by the nail at B. Then, by taking moments about F, we get W x BF = P x AF Or, W x 3 = 50 x 15 ... W = 5 x I5 = 250 Ibs. 3 EXAMPLE III. State the mechanical law known as the Principle of the Lever. In a pair of pincers the jaws meet at i J inches from the pin forming the joint. The handles are grasped with a force of 50 Ibs. on each handle at a distance of 8 inches from the pin. Find the compressive force on an object held between the jaws, and also the pressure upon the pin. (S. and A. Exam. 1888.) Let P denote the force of 50 Ibs. with which the handles are grasped at a distance of 8 inches from F, the pin. Let W denote P 7 =50/6s. R=316-6lb*. |P,-50/to. PINCERS OR NIPPERS. the compressive force on the object O, and R the resultant reaction or pressure on the pin or fulcrum F. Although there are two levers here, each having a common fulcrum, F, it is best to con- NoU. It is a mistake to speak of the " Principle of the Lever " ; what is evidently meant is the Principle of Moments as applied to the lever. 4< LECTURE IV. sider the action of one lever only. Suppose the lower handle, H,, to be fixed, and consider the action of the upper handle, H t . It then becomes a simple question on the lever. (1) To find W, take moments round F, then W x ij" = 50 x 8" .. W = 266-6 Ibs. (2) To find R, the pressure on the pin F, take moments round O then R x ij" = 50 x (ij + 8") - 50 x 9 J" ,'. R = 316 8 Ibs. Or, since R must be the resultant of 3? and W, we get R = P + W = $o + 266-6 = 316-6 Ibs. The M Toggle," or " Knuckle Joint," consists of a well- known combination of levers. It is characterised by its capability of exerting an enormous force through a short distance by means of very compact and simple elements. This device has been applied in many well-known cases, such as in the stone-crushing machine, certain brakes, printing and several forms of packing presses, and in the familiar frame by which carriage hoods are held in position. The accompanying diagrams illustrate the principle of the " Toggle," and the plan of its application to a cane mill. In Fig. i, A A' are two links jointed at B, G, and D. The point G is fixed, whilst the point 7? is free to move vertically along the line E C. The centre D is connected with the point F by means of a third link, formed partly of a spring S. The point F is supposed to be fixed in relation to its distance from the vertical line E C, but is free to move in a line parallel to E C. Under these conditions, a force acting in the direction of the arrow along the line G H will resist a much greater force acting along the line E (7, and tending to move the point B upwards. The force along G H acts with a leverage equal to the distance from J to C, whilst the force along B D has only a leverage equal to the short length K C. The nearer the centre D approaches the vertical line, the greater will be the leverage of the force acting on the centre D in the direction G H, and the less will be the leverage of the force acting in the direction E G through the link A'. In practice, a few railway buffer springs, corresponding to S in Fig. i, combined with simple links in a compact arrange- ment, occupying little space, are sufficient to resist forces amount- ing to several hundreds of tons. The line E G corresponds to Wie centre line of one of the top cover bolts of a cane mill, G "THE TOGGLE," OR "KNUCKLE JOIflT." 47 corresponding to the nut at the top of the bolt, and therefore one of the fixed points in the system ; whilst B may be taken as representing a point in the top cover itself, which is supposed to be free to move up and down. Fig. 2 illustrates the operation of the system, and directs attention to a feature of great practical interest and value. The FIG. i. FIG. 2. Iti y 4 ^MOCGr E THE PEINCIPLB OP THE TOGGLE JOINT. parts in this figure correspond with those in Fig. i, but it is supposed that a force acting in the direction E C has raised the point B from its original position, as in Fig. i, to the position B'. The point D has consequently been moved further away from the vertical line EC to the position D'. In this position, the leverage of the force acting along G H is reduced, whilst the leverage of the force acting through the link A' is increased, as compared with the previous conditions in Fig. i ; but it will be seen that Z>. in moving to the position />', has compressed the spring S, and therefore increased its resistance. Hence, by the principle of moments : (Force along B D) x K C = ( Force along G F) x J 0. /. Force along B D = (Force along G F) NOTE. See the Tangentometer at end of Lecture XXVIII. LECTURE IV. T> J But the vertical force E B = (Force aloog B D) x ^-^. J C 1 R T . = (Force along G F) x ^-^ x g-j-j. But if the links B D and D C are equal, then B J = J C. J Q 2 Hence, force along E B = (force along G F) x 7^^ TT^. If the movement along B C is very small compared with the length of the links B D and D C, then J C may be considered as a constant length. Hence, when the spring is so adjusted that its resistance to compression is proportional to K C, the pressure E B upon the rollers will be nearly constant. This arrangement of the " toggle joint," therefore, automati- cally permits of light or heavy feeding of the canes in a sugar mill without bringing undue stresses upon the various parts, and thus diminishes the chance of breakdowns. The following illustration shows the actual construction of the " Toggles " as applied to the top roller of an ordinary 3-roller cane-crushing mill. The left-hand half of the illustration shows the posi- tions of the parts when the mill is empty. In this con- dition the top roller rests in its bearings and gix'es the mini- mum openings between the rollers. The right-hand half of the illustration shows the condition of the system when PATENT TOGGLE APPARATUS. the mill is taking a heavy feed. It will be seen flint as the roller rises it lifts the top caps, which are under the contiol THE TOGGLE," OR "KNUCKLE JOINT.'' 49 of the " Toggles," until the upward pressure is balanced by the resistance of the " Toggles." v. TOGGLE JOINT AS APPLIED TO A SUGAR-CANE MILL. The above is an illustration of a mill made by the Messrs. Mirrlees, Watson & Co., Ltd., of Glasgow, with rollers 32 inches in diameter by 60 inches long. The arrangement of their patent " Toggles " in this case is of a special kind. The vertical bolts of the " Toggles " are formed by continuations of the top cover bolts, but the mill covers or caps are fixed by nuts screwed directly down upon them, and do not lift with the roller. The communi- cation between the " Toggles " and the top roller is made by means of plungers formed on the under side of the bottom plates of the " Toggles," and working through the top caps. The only advantage of this arrangement is, that it- permits the top cap to be used as a brace to bind the upper jaws of the mill cheek together, and thus adds in some measure to the strength of the cheek. 5O LECTUEB IV. QUESTIONS. LECTUBE IV. QUESTIONS. 1. Sketch and describe the steelyard, or Roman balance, and explain fully how the graduations on the scale are equal for equal differences in the weights applied to the shorter arm. 2. Sketch and describe a lockfast lever safety valve. A valve, 3 inches in diameter, is held down by a lever and weight, the length of the lever being 30 inches, and the valve spindle being 4 inches from tho fulcrum. You are to disregard the weight of the lever and to find the pressure per square inch which will lift the valve when the weight hung at the end of the lever is 56 Ibs. Ana. 59*4 Ibs. 3. The diameter of a safety valve is 3", its weight 3^ Ibs. ; length of lever is 30", and its weight 16 Ibs. ; the distance from fulcrum to centre of valve is 3", and to e.g. of lever 12". Find where a weight of 50 Ibs. must be placed on the lever in order that steam may just blow off at 70 Ibs. per square inch by gauge. Ans. 25-65 inches from the fulcrum. 4. The safety valve of a boiler is required to blow off steam at 100 Ibs. per square inch by gauge. The dead weight is 100 Ibs., weight of lever 10 Ibs., and of valve 5 Ibs. ; diameter of valve 3^", distance from centre of valve to fulcrum 4", from e.g. of lever to fulcrum 15". Where should you place the weight on the lever? Ans. 36-9 ins. from fulcrum. 5. Sketch and describe a lever machine for testing the tensile strength of materials. If the advantage, or ratio of resistance R to pull P in the first lever, is 56 to I, and of the second lever 40 to I, what stress will be produced on the test specimen when P= 100 Ibs. 1 Ans. 100 tons. 6. A force of 100 Ibs. acts at one end of a straight lever, but at an angle of 60 to it. What force acting at the other end of the lever, at an angle of 45 to it, will keep the lever in equilibrium if the fulcrum be placed half the distance from the first force that it is from the second ? Draw a dia- gram of the forces and their effective arms. Ans. 61.25 lbs - 7. Sketch a bell crank lever, to convey a small movement from one line to another, cutting each other at 60 ; the distances moved through to be as i to 2. 8. The handle of a claw-hammer is 12 inches long, and the claw is a inches long. What resistance of a nail would be overcome by the appli- cation of a pressure of 40 lbs. at the end of the handle ? Show, by a diagram, the manner in which you arrive at your result. Ans. 240 lbs. 9. In a pair of pincers the jaws meet at i^ inches from the pin forming the joint. The handles are grasped with a force of 30 lbs. on each handle at a distance of 7^ inches from the pin. Find the compressive force on an object held between the jaws, and also the pressure upon the pin. Sketch the apparatus and show the direction and values of all the forces. Ans. 1 80 lbs ; 210 lbs. 10. There is a contrivance for obtaining great pressure through a small distance, commonly termed the toggle or toggle joint. Will you explain it, and show wherein its peculiar action and efficiency consist ? 11. Explain the mechanical advantage of the combination known as a toggle joint. Show its application in printing machinery, or in stone crushing machines, or in any other instances with which you are acquainted. NOTES AND QUESTIONS 12. Sketch any one form of toggle joint with which you are acquainted, and poinc out its object. (C. & G., 1903, O., Sec. A.) 13. The figure shows a bent lever AOB with a frictionless fulcrum 0. AOis 12", BO is 24". The force Q, of 1000 Ib. acts at A, what force P acting at B will produce a balance ? Work the question graphically or in any other way. Neglect the weight of the lever. (B. of E. 1904.) Ant. Force P = 550 Ibs. NOTE Before answering this question and any future questions in- volvirxr angles, or the ratios of the sides of a right angled triangle, students should refer to the construction, action and uses of the Tan- gentometer at the end of Lecture XXVIII. LECTURE V. CONTENTS. The Principle of Work Work put in, Work lost, Useful Work Efficiency of a Machine Principle of Work applied to the Lever Experiments I. II. Wheel and Axle The Principle of Moments applied to the Wheel and Axle The Principle of Work applied to the Wheel and Axle Experiment III. The Winch Barrel Example I. Ship's Capstan The Fusee Questions. The Principle of Work.* The principle of work is applicable to all machines, and may be stated as follows : The work put into a machine is equal to the work absorbed by the machine plus the work given out by the machine. Or, WORK PUT IN = LOST WORK + USEFUL WORK. This is an aociom. But, nevertheless, many deluded would-be inventors have spent much time and money in devising "perpetual motion " appliances, or machines which should turn out as much work as, or even more than, was put into them ! 1. When a machine is employed to perform mechanical work, a certain force must be applied to one part of it in order to move the machine and to perform work at another part. The product of this applied force and the distance through which it acts constitute tlie whole work put into the machine. 2. Some of this work must be expended in merely keeping the different parts in motion, against natural resistances due to fric- tion at the fulcra or journals, and friction between moving parts and the air or water in the case of an hydraulic apparatus. The work so absorbed is termed lost work. The mean value of the frictional resistances, multiplied by the mean distance through which they are overcome, constitute the work lost in the mechanism. One great object to be kept in view, in designing most machines, is to minimise this lost work by minimising the internal resistances to motion in the machine * The Principle of Work is usually stated as follows in books on Mecha- nics, but I find that engineering students much prefer the above definition. " If a system of bodies be at rest under the action of any forces, and be moved a very little, no work will be done." " Conversely : If no work is done during this small movement, the forces are in equilibrium." Prof. Goodeve's " Manual of Mechanics," p. 73. PRINCIPLE OF WORK APPLIED TO THE LEVER. S3 itself ; but you must remember that these can never be entirely dis- posed of, as has only too often been conjectured by " perpetual motion " faddists. 3. The remainder goes to do the useful work for which the machine was designed, and therefore 4. The efficiency of a machine = the work S ot out ' the work put in. To impress these facts on the mind of the student we present them in the following condensed form : 1. Work put in = force applied x the distance it acts. 2. Work lost = force absorbed in overcoming internal resistances x the distance it acts. 3. Useful work = force given out x the distance it acts. 4. Efficiency [ = ratio of work got out to ivork put in. 5. Work put in = lost work + useful work. Principle of Work applied to the Lever. In applying the above " principle of work " to the lever, we will take the liberty of neglecting the lost work. We shall therefore assume that the friction at the fulcrum is so small that it may be neglected for the purpose we have in view. (^ EXPERIMENT I. Let F fr A| AjF be a straight lever 3/| without weight, having yy its fulcrum at Fa force, PmNCIPLE OP WoRK APPLIED T0 A W , acting vertically down- T __ j p AI T~ iMByJoLK. wards from the point B 1? and a force, P, acting vertically upwards at the end A lt keeping W in equilibrium. Now imagine the lever elevated to the posi- tion A,F. The work put in at A : = P x the vertical distance from A t to A^. The work got out at B x = W x the vertical distance from B x to B,. Therefore, since we neglect all frictional resistances The work put in = the work got out Or, P x A>A, = W x B^, ie P B ' B > W AA But by Euclid the triangles A 4 FA, and BjFB a are similar in every respect. 54 LECTURE V. Therefore, Hence, Or, W ' Af = BF But this is the equation we proved in Lecture III. with respect to the lever as complying with the " principle of moments." Hence the "principle of work " and the " principle of moments " are in agreement. In the accompanying figure the force P has been shown as elevated through 12", and the force W as elevated through 6", Therefore, P x 12" = W x 6" 4 - T, ' 7 P being half the magnitude of W, it has to be elevated through double the distance in order that the same amount of work may be done in the same time. EXPERIMENT II. Consider the case of a simple lever, where a weight, W, at B is balanced by another weight, P, at A, around PRINCIPLE OP WORK APPLIED TO JL LEVEB. a fulcrum at F, without friction. Let the lever be turned through 90, or a quarter of a revolution i.e., from a horizontal position, AB, to a vertical position, A'B'. Then by the definition of work The work put in at A = P x A'F, and The work got out at B = W x B'F. THE WHEEL AND AXLE. 55 It does not matter in the slightest degree how circuitous the paths P and "W take in passing from their original to their new positions in this case, since all we require to know is the vertical distances through which P is depressed and "W elevated. Consequently, by the " Principle of Work" P x A'F = W But, A'F = AF, and B'F = BF, .-. Substituting AF for A'F, and BF for B'F, We get, P x AF = W x BF But this is the equation for the "principle of moments" which we have again deduced from the " principle of tuork " by another and simpler form of reasoning. We find that this latter method appeals more directly to the minds of young engineering students than the proofs usually found in books on Mechanics. The Wheel and Axle. The wheel and axle has been used for centuries for drawing water by a bucket from a well. It is used by the navvy for lifting the material which he excavates from the earth, by the mason for raising stones, bricks and mortar, and by many other tradesmen for a variety of purposes ; as well as by the quartermaster as a steering- gear, and the able seaman as a capstan. The accompanying illustration shows the form it takes when used for elevating goods in a store or mill.* It is simply a practical arrange- ment for continuing the action of the lever as long as re- quired. So long as a sufficient WHEEL AND AXLE. pull is applied to the rope, which fits into the grooved wheel, to overcome the resistance of the load attached to the chain hook, the weight will be raised. The wheel and axle is therefore a form of lever by which a weight may be raised through any desired height. The Principle of Moments applied to the Wheel and Axle. In the diagram let the larger circle represent the circum- ference of a wheel of radius, R, to the periphery of which a force, * The above figure represents a wheel and axle as supplied by Messrs. P. & W. MacLellan, of Glasgow. LECTURE V. P, is applied. Let the smaller circle represent the circumference of the axle or barrel of radius, r, to the periphery of which is applied a resistance W. Let the forces P and W act in the same direction and verti- cally downwards. Join the points where the lines of action of the forces are tangents to the wheel and axle by a straight line, AB. Then, AB passes through the common centre of the circles i.e., through their common centre of motion or fulcrum F, and AF is the effective arm for the force P, whilst BF is the effective arm for the force W. In fact, AFB is a straight lever in equilibrium, with the fulcrum at F. Therefore, taking moments about F, we have P x AF = W x BF Or, P x R, = W x r. The Principle of Work applied to the Wheel and Axle. EXPERIMENT III. Take a model of the wheel and axle as illus- trated by the accompanying figure. Let, forces, P and W, act in equilibrium, as in the previous case, at radii R and r respectively. w WHEEL AND AXLE. W MODEL TO TEST THE PEINCIPLE OF WOEK APPLIED TO THE WHEEL AND AXLE. Now mark carefully with a piece of coloured chalk or ink the exact positions where the tape supporting P is a tangent to the wheel, THE WINCH BARREL. and where the cord supporting W is a tangent to the barrel. Pull P until the wheel and barrel have just made one complete revolu- tion. Then, neglecting any force required to overcome friction at the bearings of the spindle The work put in by P = P x 27rB, The work got out in raising W = W x 2irr But the work put in = the work got out . P x 27rR = W x 2rrr Cancelling 2n- from each side of the equation We have P x R = W x r. But this is the same equation as we obtained above by applying the "principle of moments." Therefore, we see that the "prin- ciple of moments" and the "principle of work" harmonise. The Winch Barrel. The wheel may be replaced by a handle H, and the mere axle by a barrel or drum D, of any desired size. W SIDE VIEW. END VIEW. WINCH BARREL AND HANDLE. EXAMPLE I. A man exerts a constant force of 30 Ibs. on a winch handle of 15" radius ; what weight will he be able to lift attached to a rope hanging from a barrel of 5" radius ? By the principles of moments and of work ; and interpolating the numerical values Px R-Wxr 30 x i5=Wx 5 ... w = 3^115 = BO Ib.. Ship's Capstan. A partly sectional, partly outside view of this useful machine is illustrated by the following figure : A capstan is generally fixed upon the forecastle of a ship, or near to the side of a quay or dock, for the purposes of warping and LECTURE V. 7 !BF - SHIP'S CAPSTAN. INDEX TO PAETS. OH represents Capstan head. SS ,, Spokes or arms. E Eadius of S. B Barrel. r Radius of B. PR represents Pall and Ratchet. F Frame. CP Capstan pillar. FS Footstep of CP. berthing the vessels. The above illustration shows a capstan as built into a forecastle, where the round turned footstep, FS, of the vertical cast-iron capstan pillar, CP, bears in a cast-iron or cast-steel shoe fitted upon the steel or wrought beams of the main deck. The frame F, which supports the casing for the pall and ratchet gear, may be the beams of the upper or forecastle deck. A strong rope made fast on shore is passed several times round the capstan barrel B, and the slack end of the rope is coiled on deck. The addition of the rope to the barrel increases the effective arm or radius r, at which the resistance of the ship acts by half the diameter of the rope. Eight or any desired less number of wooden spokes, S, S, having their inner ends squared and tapered, are fixed into hollow square holes in the cast-iron capstan head CH. Then, just as many sailors as may be required to SHIP'S CAPSTAN. 59 overcome the resistance of the ship apply themselves to the outer rounded ends of the spokes, and push away as hard as they can. It will be observed that, calling, p, the force applied by each sailor at radius R ; then, when we have two saiJors acting on diametrically opposite spokes /?, 2R, p forms a couple tending to cause rotation of the capstan in one direction. Consequently from the property of couples (as we showed in Lecture III.) this couple can only be balanced by another couple acting in the opposite direction and having an equal moment. Such another couple exists, when the resistance of the ship, W, acting with an arm, r (equal to the distance from centre of capstan to centre of rope), balances the corresponding reaction at the centre of the capstan barrel. Hence, when the force applied by the two sailors is balanced by the resistance to motion of the ship, we have the one couple just balancing the other one. Or Couple p, 2R, p balancing couple W, r, W i.e., p x 2B = W x r In the same way, with two, three, or four pairs of sailors, each pair being supposed to act on diametrically opposite spokes, we have two, three, or four couples acting in one direction, balanced by one couple, viz., the resistance of the ship into the distance from the centre of the capstan barrel and the reaction from that centre.* In the case of four sailors just being able to move the ship, two couples, p, 2R, p+p, 2R, p, balance one couple, W, r, W; *'.., p x 2R.+J9 x 2R = W xr The Fusee. As an illustration of the lever action and of work put into and got out of a machine, we cannot do better than finish this lecture by a description of the construction and action of the simple yet most ingenious contrivance termed the fusee. In good watches and clocks, where the elastic force of a coiled spring is used to drive the works, the fusee is used for the purpose of compensating the gradually diminishing pull of the uncoiling spring. The driving of the works at a constant rate is the object for which a watch or clock is designed. This naturally entails a constant resistance to be overcome, but since one of the most compact and convenient forms of mechanism into * The student should draw a plan of the capstan barrel, and show radial lines to indicate one, two, or more pairs of diametrically opposite spokes with forces, p, acting at their ends, all tending to turn the barrel in one direction. He will then see that a couple formed by resistance to the stress on the rope, and an equal reaction from the centre of motion, will be required to act in the other direction in order that equilibrium may take place. 6o LECTURE V. which mechanical force can be stored is that of a coiled spring, and since the very nature of the spring is such that its force decreases as it uncoils, we must employ some compensating device between this variable driving force and the constant resistance. The fusee does this in a most accurate and complete THE FUSEE FOE A CLOCK OR WATCH. INDEX TO PARTS. F represents Fusee. B Barrel. BW Ratchet wheel. TW represents Toothed wheel. WS ,, Winding square. manner. Looking at the accompanying figures and index to parts, we see that the barrel B, which contains the watch or clock spring, is of uniform diameter, and that between the outside of this barrel and the fusee, or spirally grooved cone, there passes a cord or chain. When the winding key is applied to the winding square WS, and turned in the proper direction, a tension is applied to the cord, and it is wound upon the spiral cone, thus coiling up the spring inside the barrel B ; for the outer end of this spring is fixed to the periphery of the barrel, and the inner end to its spindle or axle, is in direct gear with the works of the clock. When the spring is fully wound up it has the greatest force, but it acts with the least advantage, since then the cord is on the smallest groove of the cone pulley. When the spring is almost uncoiled it acts with the greatest advantage, for then the eord is on the largest groove of the cone. Consequently the radii THE FUSEE. 6 1 of the grooves of this cone are made to increase in proportion as the force applied to the cord decreases in order that there shall be a constant turning effort on the works of the clock. The work put in when winding up the coiled spring, is given up by it in overcoming the frictional resistances of the different parts of the mechanism. Or the work put in = lost work, for the whole of the work put in is devoted to simply keeping the parts of the machine in motion, thus leaving nothing for other work, unless the clock is used to strike a bell or do some other kind of work. 62 LECTURE Y. QUESTIONS. LECTURE V. QUESTIONS. 1. State the " Principle of Work," and explain the manner in which it is applied in determining the relation of a P to W in the lever. A lever, centred at one end, is 15 feet long, and a weight of W Ibs. hangs from the opposite end. The weight W is supported by an upward pressure of 28,270 Ibs. at 13 feet from the fulcrum. Find W. Ans. 24,5006 Ibs. 2. Define work put in, lost in, and got out of a machine, and prove that the work put in = lost work plus the useful work. How are the "advan- tage " and the efficiency of a machine reckoned ? 3. Sketch and describe the wheel and axle. Apply both the "principle of moments" and the "principle of work" to find the relation between the force applied and the weight raised by aid of this machine. A wheel and axle is required so that the force applied at the circumference of the wheel in moving through a distance of 10 feet shall raise a weight of 4 cwts. through a height of 2 feet. If the diameter of the axle is 10 inches, find the force applied in Ibs., and the radius of the wheel in feet. Ans. 89-6 Ibs. ; 2 feet I inch. 4. The crank or handle which turns a windlass is 14 inches in length j what must be the diameter of the axle when a man exerting a force of 60 Ibs. upon the handle raises a tub of coals weighing 2 cwt. 1 Ans. 7^ inches. 5. In a windlass the barrel is 8 inches diameter, the rope is I J inches diameter, and the crank handle 15^ inches long. What force must b applied at the handle to raise 2 cwt. 1 Also, what weight would be raised by a constant force of 30 Ibs. applied at the handle 1 Ans. 66-8 Ibs. ; 100-5 Ibs. 6. A capstan is worked by four men ; each man exerts a constant force of 30 Ibs. at a distance of 4 feet from the axis. A rope of f -inch diameter is wound round the drum, of 5| inches radius. Find the pull on the rope which balances the pressure on handles. Make a diagram showing the action of the forces, and find the pressure on the central shaft of the cap- stan. Ans. 921 6 Ibs. ; 921-6 Ibs. 7. Describe, with a sketch, the spring-barrel and fusee of a clock or of a watch. Explain its action by reference to the principle of moments. 8. A ship's capstan has a ratchet-wheel with two detents or pawls, arranged so that when one is engaged with a tooth of the wheel the point of the other is midway between two teeth. Sketch the arrangement, and say why in this case two detents or pawls are better than one. 9. Find the average horse-power exerted by a winding engine to lift 3 tons from a pit f mile deep at a uniform speed in two minutes, supposing that 30 per cent, of the total work done is lost in friction. (0. & G., 1905, O., Sec. A.) Ans. 576 H.P. LECTURE VI. CONTENTS. Pulleys Snatcn Block Block and Tackle Theoretical Ad- vantage Velocity Ratio The Principle of Work applied to the Block and Tackle Actual or Working Advantage Work put in Work got oat Efficiency Percentage Efficiency Example I. Questions. Pulleys. Suppose you had to elevate a sack of flour from the ground to an upper storey of a mill or store, you might place it upon your back and carry it up the stairs. In doing so, you would expend so many foot-pounds of work. Let the sack of flour be 100 Ibs., your own weight 150 Ibs., and the height to which it is raised be 30 feet. Then the Work done in elevating the flour = 100 Ibs. x 30' = 3000 ft.- Ibs. yourself =150 x 30' = 4500 Total work done a 250 x 30' = 7500 And your efficiency as a machine would be found thus Mechanical efficiency = sefulwork work got out = 3000 ft-lba. total work work put in 7500 ft.-lbs. Or, your percentage efficiency would be found from the propor- tion 7500 : 3000 :: 100 : x 7500 In other words, 60 per cent, of the total work done is lost work, and only 40 per cent, is useful work. If instead of carrying the sack upstairs, you found ready to hand a long rope (with its two ends close to the ground) that had been passed over a smooth iron hook fixed to the outside wall above an outside landing for the particular storey of the building, and, if you attached one end of this rope to the sack and found that by pulling with all your strength (or say with a force of 150 Ibs., i.e., equal to your weight) on the other end, you could just lift the sack. Then, if by this means you elevated the sack to the landing, you would have expended less work than by the iormer method ; for, 6 4 LECTURE VI. Work done in elevating flour = loolbs. x 30'= 3000 ft.-lbs. against friction, &c. = 50 ,, x 30' = 1500 Total work done =150 x 30 = 4500 .-. Mechanical efficiency = Hi 6 *" 1 or, ut^ 3000^.5 work put in 4500 total work And the percentage efficiency is therefore 66-6. For, 4500 : 3000 :: 100 3000 x 100 45 66-6 % Hence 33.3 per cent., or ^ of the total work put in by you in pulling at one side of the rope, is spent in overcoming the friction between the rope and the hook and bending the rope over the hook, whilst only 6 6' 6 per cent., or |, remain for elevating the sack of flour. If, instead of the iron hook you had found a double-flanged deep V-grooved pulley with a rope over it, as in the ac- companying illustration, and that this pulley revolved so easily on its bearings that you had only to pull with a constant force of no Ibs. in order to lift the sack of flour from the ground up to the 3o-feet level, then Work done in elevating flour = 100 Ibs. x 30' = 3000 ft.-lbs. against friction, &c. = 10 x 30'= 300 PULLEY AND WEIGHTS. Total work done no 30 =3300 .*. Mechanical efficiency = lisefnl work . O r ^J^ out = 3ooo = .^ total work ' ' work put in 3300 And the percentage efficiency is 90-9 For 3300 : 3000 :: 100 : x x = 300Q * IPO sf 9Q . 9 % Hence only 9-1 per cent, of the total work put in is lost work In overcoming friction at the pulley bearing and in bending the rope over the pulley. You see, therefore, what a useful machine a pulley is, not only for enabling you to change the direction of a force, but aho for the saving of labour. SNATCH BLOCK, , 65 A pldley is simply a wfael and axle wherein their radii are one and the same, or a lever with equal arms. Hence the principles of moments and of work may be applied to it in the same way ag we applied them to the lever and to the wheel and axle. Snatch Block. If you should require to put the bend of a rope on a pulley, and at the same time prevent the possibility of the rope coming out of the groove, without having to reeve the end of the rope between its cheeks, you would use what is called a snatch block. One form of snatch block is illus- trated by the accompanying figure, where on the side of one cheek there is a sneck or snatch, which is turned to one side, to enable the bend of the rope to be placed around the U groove of the pulley. The snatch then falls down and closes upon the central pin. Another form has a hinged snatch which can be lifted up at right angles to the face of the cheek, and after the rope has been put on the pulley the snatch is closed down and locked by a pin attached to a short chain fixed to the side of the cheek, just like an ordinary front hinge for closing a chest. The single movable pulley, which is used for sup- porting the load to be lifted by a Chinese wind- lass or by a jib crane, is sometimes called a * u '. J"" snatch block (see the illustration of the wheel and ' compound axle in next Lecture, and of jib cranes in Lectures VIII. and XIII.). In the latter case the chain passes from the barrel of the crane over the pulley at the point of the jib, then vertically down, underneath the snatch-block pulley, and vertically upwards to a point on the under side of the jib where it is fixed by an eye-shackle with a bolt and nut. If the load, including the weight of the snatch-block, be W, then, neglecting friction, the W pull P on the chain will be ; f or W is supported by two ver- tical or parallel parts of the chain, each part carrying half the load, or W = 2?. If the load be elevated any distance L, then the chain will have to be pulled in on the barrel a distance of 2L, for by the principle of work The pull x its distance = the load x its distance. Or, P x 2L = W x L. The theoretical advantage is therefore 2 to i, or a certain force would lift double the weight, neglecting friction. Block and Tackle. Passing over the various arrangements of pulleys for lifting weights which are treated of in theoretical 66 LECTURE VI. mechanics, we come to this well-known and Useful contrivance. As will be seen from the accompanying sketch, it consists of a number of pulleys (or sheaves as they are technically termed) free to run round on a turned central iron or steel spindle, and inserted in a block, having their iron divi- sions between each pulley, and strong iron cheeks fixed to a swivel joint- terminating in an iron hook hung from an eye bolt. Three sheaves are shown in this block, but the number may range from one up- wards, according to the size and work to be done. There is a simi- larly constructed block with two sheaves, to which the weight to be raised, or the body to be pulled, is attached, and this is called the movable block, whereas the upper or home one is termed the fixed block. Around the pulleys of both blocks there is reeved a rope with the inner end made fast to an eye on the movable block, whilst the free end hangs from one of the outside sheaves ; but this arrange- ment is frequently reversed, for the inner end of the rope may be at- tached to an eye on the fixed block, and the free end may spring from the other one (see the figure in con- nection with Example I. of this Lecture). The free end of the rope is then ready to be pulled by the hands or by aid of a winch. Now, neglecting friction, and supposing the rope to be perfectly flexible, a force, P, applied to the free end of the rope would be transmitted throughout it to the other end at the movable block. Hence the effect of this force in overcoming a resistance, W, is multiplied by the number, n > of parts of tfie rope which spring from the movable block. BLOCK AND TACKLE. Or, W And (i) The theoretical advantage W P I UNIVERSITY I X^LIFOR!*^^ THB BLOCK AND TACKLE/ 67 (2) The velocity ratio, or ratio of the distance through which P acts, to that through which W is overcome in the same time. Or, Velocity ratio = P ' 8 ^ tapce = ^ W s distance i In the figure there are shown three pulleys in the upper block and two in the lower, with five parts of rope springing from the latter ; therefore in this case n = 5. Here W = nP = 5 P; or, P * ^ - ^ n 5 since P must pass through five times the distance that W does in the same time. The velocity ratio = ^ d ^ stapce = = * W's distance i 1 So that the theoretical advantage and the velocity ratio have the same algebraical expression and numerical value. - (See note, p. 68.) The Principle of Work applied to the Block and Tackle. Using the very kind of block and tackle represented by the previous figure, attach a light Salter's spring balance by its hook to the rope where the hand is shown. Fix such a weight to the lower block that the weight of rope between the blocks, the movable block, and the load are 60 Ibs. Call this W. Now pull the ring of the spring balance until the load rises slowly and uniformly, and note the reading on the balance ; let it be 18 Ibs., and let the weight of this balance and the hanging free end of the rope, which is assisting the arm, be 2 Ibs. Call this total pull of 20 Ibs. P ; then : (3) 27U actual or Dicing advance = wei 8 ht raised = W = 6olbs. = 8 pull applied P 20 Ibs. 1 Lift W up through one foot exactly, and measure the length of rope which you have pulled out from the upper block, and you will find that it is five feet ; hemce,* (4) The work put in = P x n= 20 Ibs. x 5 ft. = 100 ft.-lbs. (5) The work got out = W x i = 60 Ibs. x i ft. = 60 ft.-lbs. Work got out 60 ft.-lbs. (6) The efficiency =- -J _ =8-3 Work put in I00 ft.-lbs. (7) The percentage efficiency = '6 x 100 = 60 % In the same way the efficiency of uny other block and tackle may be found, and the student should carry out a series of * The above results were obtained by the Author from a block and tackle of the same kind as that shown by the previous figure, at a demon- stration in his Junior Applied Mechanics class. 68 LECTUEE VI. experiments in a laboratory or workshop so as to impress the various measurements and the results on his memory. He will find that if the efficiency is over 50 per cent, a comparatively small load will run down and overhaul the free end of the rope, unless it has some restraining force applied to it, or be fixed to some rigid body. It is for this reason that sailors, who work very much with ordinary block and tackle, always " belay " tne free end of the rope when they have adjusted their sails or have heaved up a body to the required height. EXAMPLE I. A tackle, consisting of an ordinary double and treble block, is em- ployed for lifting a weight of 600 Ibs. attached to the double block. What force is required, - neglecting friction ? If the tackle is reversed, so that the weight is attached to the treble block, the free end of the rope being pulled upwards, what force would now be re- quired to lift the weight? (S. and A. Exam. 1892.) BLOCK AND TACKLE ANSWER. First Case. By an inspec- 2ND CASE, EXAMPLE I. ,. - ,. . J .. . T tion or the previous figure in this Lecture, it' will be apparent that the weight W is supported by Jive parts of the rope, or n = 5. *,- -- n W T = 120 Ibs. 5 /Second Case. Here the system is inverted, so that the block with the three pulleys is lowermost, as shown by the accompany- ing figure. In this case it is evident that there are six parts of the rope supporting W, or n = 6. n Note. If a machine be supposed to work without friction, then the ratio of the resd&tanoe overcome to the effort applied is termed the theoretical or hypothetical mechanical advantage. If, however, friction be taken into account and an effort P be able to overcome a resistance W, then the ratio W is termed the mechanical advantage. LBCTURB VI. QUESTIONS. v 69 LBCTUBB VI. QUESTIONS. 1. Suppose that your weight is 10 stone 10 Ibs., and that you lift a weight of \ cwt. on your shoulder r and walk upstairs with it to a height of 20 ft. ; what work have you expended, and what will be your efficiency as a machine 1 Am. 4120 ft-lbs. ; 27 per cent. 2. Suppose that you had a rope passed round a beam of wood, and that you attached 4 cwt. to one end and pulled with a force of 84 Ibs. on the other end and then elevated it 10 ft. : (a) what work have you put in ? (b) what is the percentage efficiency of the arrangement ? (c) what is the percentage of lost work ? ATM. (a) 840 ft. -Ibs. ; (b) 66-6 ; 33-3. 3. Suppose that a weight of 4 cwt. is attached to one end of a rope passed round a pulley, and that you lift it 10 ft. by pulling on the other end of the rope with a force of 70 Ibs. : what percentage of the work done is lost in overcoming the friction at the pulley ? Ans. 20 per cent. 4. What will be the difference, and why, in the tension on the chain of a crane when a snatch-block is used, and when the weight is lifted directly Sketch a snatch-block, and describe its construction and action. 5. In a rope and pulley lifting block with three sheaves in the upper block, and two sheaves in the lower block, find the theoretical advantage gained. Give the reason for your answer, and sketch the arrangement, showing where the rope is to be attached. Arts. W : P : : 5 : I. 6. Sketch an arrangement of 5 equal pulley sheaves for lifting a weight of i ton. What force is exerted on the rope in your arrangement ? Ex- plain the mo^e of arriving at this numerical result by the principle of work. An. With 3 pulleys in upper block and 2 in lower block, P = 448 Ibs. 7. A tackle is formed of two blocks, each weighing 15 Ibs., the lower one being a single movable pulley, and the upper or fixed block having two sheaves ; the parts of the cord are vertical, and the standing end is fixed to the movable block ; what pull on the cord will support 200 Ibs. hung from the movable block, and what will then be the pressure on the point of support of the upper block ? Give a sketch. Ant. 71*6 Ibs. ; 301-6 Ibs. 8. A weighf of 400 Ibs. is being raised by a pair of pulley blocks, each having two sheaves. The standing part of the rope is fixed to the upper block, and the parts of the rope, whose weight may be disregarded, are considered to be vertical. Each block weighs 10 Ibs.; what is the pres- sure at the point from which the upper block hangs 1 An*. 522-5. 9. A tackle, consisting of an ordinary double and treble block, is em- ployed for lifting a weight of 1000 Ibs. attached to the double block. What force is required, neglecting friction ? If the tackl* is reversed, so that the weight is attached to the treble block, the free nd of the rope being pulled upwards, what force would now be required to lift the weight ? Sketch the two arrangements. Ans. 200 Ibs.; 166-6 Ibs. 10. Apply the " principle of work " to find the relation between the force applied and the weight raised by an ordinary set of block and tackle. State what is meant by the following terms : {i) velocity-ratio ; (2) theo retical mechanical advantage ; (3) actual or working advantage ; (4) work put in ; (5) work got out ; (6) efficiency of an apparatus or machine ; (7) percentage efficiency. 11. With an ordinary block and tackle haying 3 pulleys in upper block and 2 in lower block i.e., 5 ropes attached to lower block it is found that a pull of 50 Ibs. is required to raise a weight of 165 Ibs. Find (i) Theo retical advantage and velocity ratio = 5 : i ; (2) Actual advantage =3- 3 : I ; (3) Efficiency of apparatus =-66 ; (4) Percentage efficiency of apparatus =66. 70 LECTURE VI. QUESTIONS. 12. If the upper block of a set of pulleys and tackle has four equal sheaves, and the lower block three equal sheaves, and if a weight of one ton is hung on the lower block, one end of the rope being fixed to the ground and the other end free, what pull upon the free end will raise the weight, and what distance will the weight rise for every yard of increase of length in the free end? If the rope be fastened to the lower block instead of to the ground, what pull will raise the weight ? Ans. 373 Ibs. ; 6 inches ; and 320 Ibs. 13. A machine is concealed from sight except that there are two vertical ropes ; when one of these is pulled down the other rises. How would you find the efficiency of this lifting machine? What do you mean by velocity ratio, and by mechanical advantage ? (S. E. B. 1901.) 14. In a lifting machine an effort of 26*6 Ibs. just raised a load of 2260 Ibs. ; what is the mechanical advantage 1 If the efficiency is 0755, what is the velocity ratio ? Ans. 85 ; 113. (B. of E., 1902.) 15. Distinguish between force, work and rate of work. Find the pull on the draw bar exerted by a locomotive which develops 600 horse-power when travelling at 60 miles an hour the mechanicaf efficiency of the locomotive being taken as 60 per cent. Ans. 2250 Ibs. (.0. & G., 1903, 0.) 16. Define the terms mechanical advantage, velocity ratio, and efficiency, as applied to lifting tackle. In a lifting machine an effort of 26 Ibs. just raises a load of 2200 Ibs.. and the efficiency is 075. Find the values of the mechanical advantage and velocity ratio. If, with the same machine, a load of 12 Ibs. lifts a load of 600 Ibs., what is the new efficiency? (C. & G., 1905, 0., Sec. A. Ans Mech. Adv. = 84-6 ; Vel. Ratio = 112-5 : 1 ; Efficiency = '44. KOTES AND QUESTIONS. LECTURE VII. CONTENTS. The Wheel and Compound Axle, or Chinese Windlass The" Principle of Moments applied to the Wheel and Compound Axle The Principle of Work applied to the Wheel and Compound Axle Ex- amples I. II. Weston's Differential Pulley Block The Principle of Work applied to Weston's Differential Pulley Block Experiment I. Cause of the Load not overhauling the Chain Questions. The Wheel and Compound Axle, or Chinese Windlass. This ingenious contrivance was first devised by the Chinese for the purpose of lifting weights. The theoretical mechanical advantage is very great, but it possesses the disadvantage of re- quiring a long length of rope to lift the weight a small height. Its construction and action will be easily understood from the accompanying side and end views, which are taken from a model SIDE VIEW. END VIEW (Without End Bearing). THE WHEEL AND COMPOUND AXLE. THE WHEEL AND COMPOUND AXLfc. 73 made in the author's engineering workshop for the purpose of demonstrating its action and efficiency to his students. The Principle of Moments applied to the Wheel and Compound Axle. Taking moments about the axle, we have, when there is equilibrium between P and W, The Principle of Work applied to the Wheel and Com- pound Axle. Neglecting friction, and supposing the rope to be perfectly flexible, cause the wheel to make one complete revolu- tion in the direction shown by the arrow near its circumference on the end view. Then, by the principle of work, The work put in = the work got out. Or, P x its distance = W x its distance ; * i.e., P x circumference = W x J of the difference of the cir- of wheel cumferences of the larger and smaller axles.* Or, P x 27rR = W x (27^ - 2irrJ (Dividing both sides of the equation by 2r) Which is the same result as the one above ; consequently the principle of moments and the principle of work agree. EXAMPLE I. In a compound wheel and axle, where the weight hangs on a single movable pulley, the diameters of the two por- tions of the axle are 3 and 2 inches respectively, and the lever handle which rotates the axle is 12 inches in length. If a force \v * If iff raised the circumference of the larger circle on one side, W then is lowered at the same time on the other side, the circumference of the smaller axle ; consequently W will be elevated a distance equal to fto//the difference of the circumferences of two axles, or= (2Trr l -2irr^. 74 LECTURE vn. of 10 Ibs. be applied to the end of the lever handle, what weight can be raised ? ANSWER. Here P= 10 Ibs. ; R= 12" ; r v = 1.5" and r,= i". By the principles of moments and of work .. \V= 10 x 12 x 4 = 480 Ibs. EXAMPLE II. In a compound wheel and axle, let the diameter of the large axle be 6 inches, and that of the smaller axle 4 inches, and the length of the handle 20 inches ; find the ratio of the velocity of the handle to that of the weight raised. ANSWER. Here R = 20" ; r, = 3" ; r, = 2". By the principles of moments and of work ... p-ifri-v; W R P 1(3-2) W~ 20 .Stttf 6y the principle of work P x its distance = W x its distance i x P's distance = 40 x Ws distance .. The velocity ratio, Q P's distance _ 40 Ws distance 1 Weston's Differential Pulley Block. This practical appli- cation of the Chinese windlass is simply a compound axle without the wheel. Or, where R = r r Hence, P x R = (R _ r) 2 where R is the radius of the larger axle- or pulley, and r the radius of the smaller one. A fter describing Weston's differentia] pulley block, we will deduce this formula from the " principle of work " by the same kind of reasoning as we adopted in the case of the wheel and compound axle. We leave the student, however to apply the "principle of moments," whereby he should get the same results, WESTON'S DIFFERENTIAL PULLEY BLOCK. 75 As will be gathered from an inspection of the accompanying out- side view and the small diagram showing the directions of the forces and their arms, it w r ill be seen that the apparatus consists of three parts (i) an upper block ; (2) an endless chain ; (3) a movable lower block or snatch- block. The upper block has a hook with swivel joint, from which the iron frame is suspended. In the centre of this frame is a turned steel axle on which rotates a couple of pul- leys cast in one piece, and there- fore rigidly con- nected together. The one pulley is slightly larger than the other, and both pulleys have V-grooved peripheries with side ridges or teeth cast on the inner sides of the grooves, so as to fit the pitch of the links * c ,1 , . SKELETON FIGURE OF the chain, WESTON > S DIFFERENTIAL which passes PULLEY BLOCK. over them and thereby prevent it slipping over the surface of the pulleys. The lower or movable pulley is simply an ordinary smooth V-grooved pulley with swivel and hook like that already described under the heading "Snatch Block." The endless chain is an ordinary open- linked chain of uniform pitch and size of link. It passes from the position where the hand or pull, P, is applied, over the larger pulley of the upper block, underneath the lower pulley, over the smaller of the upper block pulleys, and back to the starting-point. (See also the small figure.) When a pull, P, is applied at this part of the chain (if there were no friction), it would be transmitted with undimimshed value throughout its whole length where the tension WESTON'S DIFFERENTIAL PULLEY BLOCK. (BY HOLT & WILLETT.) 76 LECTURE VII. can act ; but, as we shall see afterwards, a large proportion of this force is absorbed in overcoming friction. The stress due to the load W is divided equally between the two vertical parte of the chain connected to the lower block, and if W is moved through any distance, the stress must act through double that distance. The Principle of Work applied to Weston's Differential Pulley Block and Tackle. Theoretically (i.e., leaving friction out of account, the weight of the hanging part of the chain and tlie weight of the lower block), we have by the principle of wwk, in one revolution of the upper pulleys P x its distance = W x its distance. P x circumference of\_W (difference of the circumferences of the larger pulley J ~ 2 1 the larger and smaller pulleys, Px27rR = (27rR-27rr) (Dividing each side of the equation by 2r) PxR 2lv (i) The Theoretical Mechanical Advantage or ratio of W to P is found directly from the above equation by simple transposition. W 2R (2) The Velocity Ratio (or ratio of the distance passed through by P to the distance passed through by W in the same time) is also found in the same way. P'S distal 27rR 2R* " Ws distance i(2n-R - 2irr) R r Or, the velocity ratio has the same numerical value as the theoretical advantage. EXPERIMENT I. With a Weston's differential pulley block, having in the upper block one pulley with an effective radius of 4" (i.e., from the centre of the pulley to the centre of the chain which passes round it), and a smaller pulley with an effective radius of 3^", you can just lift a total load of 100 Ibs. (including the dead weight, the lower block, and the hanging parts of the chain) by a pull of 20 Ibs. on the chain. * Dividing numerator and denominator by r does not alter the fraction WESTON'S PULLEY BLOCK AND TACKLE, 77 In this case the theoretical advantage and the velocity ratio are ach equal to 2R 2 x4* 8 16 Or, the pull on the forward side of the chain must act through 1 6 ft. for every foot the load is raised. (3) The Actual or Working Advantage of the machine is, how- ever, only as W IPO Ibs. 6 P = 20 Ibs. = f (4) The Work put in in lifting W i ft. is Px 16 = 20 Ibs. x i6' = 320 ft.-lbs. (5) The Work got out is = W x i = 100 Ibs. x i' = 100 ft.-lbs. //:\ m w - Work got out 100 ft. Ibs. (6) The Efficiency \* =^ r- 5 ir=~ = --- * n = -3125. Work put in 320 ft. Ibs. > 1 *->' (7) The Percentage Efficiency is = .3125x100 = 31.25%. This is a very low efficiency for a machine, but it accounts for one of the useful properties of the Weston's differential pulley block viz., that you can lift a weight by it, then let go your hold of the chain, and the weight will remain hanging in the exact position you left it, without overhauling the chain in the slightest degree. It is therefore an extremely useful appliance in engineering workshops where, for example, a slide valve and its valve casing port face have to be scraped so as to fit each other. After rubbing the valve on the port face, you can lift the valve by aid of a Weston's block, and leave it hanging, without any fear of its overhauling the chain which supports it, until you have scraped off the high or hard parts from the port face, when you can lower it for another rub. Or, in the case of having to adjust the centres of a heavy job to be turned in a lathe, you can lif t the job from the lathe by a Weston's block, and leave it hanging quite free at the most convenient height to be acted upon, until you are ready to lower it again into position. Of course, with such apparatus, although the theoretical advantage is great, the actual or working advantage is small ; yet this property of not overhauling is of such importance that appliances possessing it are constantly being used in every engineering workshop. Cause of the Load not Overhauling the Chain. In the first place, the chain cannot slip round the pulleys of the upper block, because the links of the chain fit into the notches or 78 LECTURE m between the outstanding teeth or ridges cast in their grooves. In the second place, the friction between the pulleys of the upper block and their axle is so great, that more than 50 % of the " work put in " is expended in overcoming it. To prove this, take the case of the above experiment. When the pull P and the load W are both in action, the downward pressure (due to these two forces alone) between the pulleys and their axle is 120 Ibs. (TOO Ibs. for W 4- 20 Ibs. for P). Now, remove the 20 Ibs. pull, and you only relieve the pressure causing friction by ; for TOO Ibs. (or of 120 Ibs.) is still there. But friction is practically proportional to the pressure in such a case, and therefore, although the pull required to lift the load be removed, of the total friction will remain at the upper block, and the friction at the lower block is unaltered. In lifting the load of TOO Ibs. i ft., there was put into the machine 320 ft.-lbs., or 220 ft.-lbs. was lost work, required solely to overcome friction. Consequently, to lower the load of 100 Ibs. i ft. there would have to be expended at least of 220 ft.-lbs., or not less than 183 ft.-lbs. But the load can of itself only give out 100 ft.-lbs. in descending i ft. ; therefore it must be assisted by at least 83 ft.-lbs. (183 ft.-lbs. 100 ft.-lbs.) put into the chain on the slack side, or where it comes down from the smaller pulley. This principle of the weight not running down (or overhauling, as it is technically termed) is common to all machines wherein more than 50 % of the force applied is spent in merely overcoming frictional resistance.* * The student should be most earnestly warned against such expres- sions as the following, which are only too common in books dealing with Applied Mechanics : " By increasing the number of sheaves in a pair of pulley blocks, the power may be increased." Now, power, or the rate of doing work, can never be increased by any continuously acting mechanical device, so long as the work given out depends directly on the work put in. It is simply the force which can be augmented, whilst the distance through which it acts is diminished. Of course, in the case of a pile- driver where the weight is lifted slowly and let go suddenly, so that the rate of giving out work is greater than the rate of putting it in, it is true that the power is increased. But this is not a continuous acting mechani- cal device in the sense referred to above. The fundamental principle to be observed is, that no more work can be got out than has been put in. The term " mechanical powers " should also be avoided, and the expres- sions " simple machines " or " mechanical elements " used instead. LECTURE VII. QUESTIONS^ 79 LECTUBB VII. QUESTIONS. 1. Sketch and describe the wheel and compound axle, or Chinese wind- lass. Apply the " principle of moments " and the " principle of work " to find the formula for the relation between the force applied and the weight raised by this machine. 2. In a compound wheel and axle the diameter of the two parts of the axle are 5 and 6 inches respectively. The weight raised, viz., W y hangs from a single movable pulley in the usual manner, and is supported by a pressure, P, applied perpendicularly to a lever handle 15 inches in length. Find the ratio of P to W. Sketch and describe the compound wheel and axle, and state its inconveniences. Ans. i : 60. 3. A force of 20 Ibs. draws up W Ibs. by means of a wheel and compound axle. The diameter of the wheel is 5 feet, and the diameters of the parts of the compound axle are 9 and 1 1 inches respectively ; find W. Ans. 1 200 Ibs. 4. In a compound wheel and axle, let the diameter of the larger axle be 8 inches, and the radios of the smaller one 2 inches, while the force applied to the handle passes through 47*12 inches in one revolution. Find the ratio of the velocity of the handle to that of the weight being raised. Ans. 7-5:1. 5. Explain the mechanical principle upon which Weston's pulley block is constructed, and give a skeleton diagram showing the direction of all the forces at work. If the weight which is being raised is left hanging, and the pull removed, why does the weight not descend T 6. Sketch and describe Weston's differential pulley block. If the diame- ters of the pulleys are 4 and 4} inches, what weight can be raised by a force of 20 Ibs. ? If the weight to be raised is half a ton, what force must be applied to the leading side of the chain ? (Neglect friction.) Ans. 360 Ibs. ; 62-2 Ibs. 7. Determine the relation between P and W in Weston's differential pulley block (i) by the "principle of moments" ; (2) by the "principle of work. " 8. If a weight is raised by a Weston's differential pulley block at the rate of 5 ft. per minute, and the diameters of the pulley of the compound sheaves are 7 and 8 inches respectively, at what rate must the chain be hauled ? Work out answer in full from the principle of work. Ans. 80 ft. per minute. 9. By experiment with a Weston's differential pulley block it was found that a pull of 15 Ibs. on the leading side of the chain was required to lift a weight of 60 Ibs. (including the weight of the lower pulley and hook). The dimensions of the apparatus were radius of larger pulley, 2 inches ; radios of smaller pulley, 175 inches. Find (i) the theoretical advan- tage ; (2) the actual or working advantage ; (3) the efficiency or modulus ; (4) the percentage efficiency of the apparatus. Why does the weight remain suspended when there is no pull on the chain T Ans. (i) 16 : i ; (2) 4 : i ; (3) -25 ; (4) 25. 10. In a Weston policy block, the diameters of the two pulleys are 8 ms. and 7 ins. respectively, and it is found that a pull of 25 Ib. is sufficient to raise a weight of 240 Ib. Find the efficiency of the tackle. Ans. 30%. (C. & G., 1903, O., Sec. A-) 11. Describe how you would proceed to determine experimentally (i) the velocity ratio, (2) the mechanical efficiency of a Differential Pulley Block. (B. of E., 1904.) I ( 8o ) LECTURE CONTENTS. Graphic Demonstration of Three Forces in Equilibrium- Parallelogram of Forces Triangle of Forces Three Equal Forces in Equilibrium Two Forces acting at Right Angles Resolution of a force into Two Components at Right Angles Resultant of Two Forces acting at any Angle on a Point Resultant of any number of Forces acting at a Point Example I. Stresses in Jib Cranes Examples II. III. Stresses on a Simple Roof Example IV. Questions. IN Lecture I. we explained and illustrated how a force may be represented by a straight line both in direction and magnitude, and we defined the terms components, equilibrant, resultant, re- solution, and composition of forces. We will now discuss briefly the case of three forces in equilibrium when acting towards or from a point, as well as the parallelogram and the triangle of forces with examples, before taking up the inclined plane and friction. Graphic Demonstration Of Three Forces in Equilibrium. EXPERIMENT I. Take a black board which (for convenience of handling and demonstration before a class) may be of the form shown by the accompanying figure. Select two movable clamps, each fitted with a small V-grooved pulley about 2 inches in dia- meter, with a minimum of friction at their bearings, and fix them to the outside of the board as indicated. Pass a very fine flexible cord over the pulleys, and attach to the ends of this cord S hooks. Hang from these hooks weights of say 24 oz. and 32 oz., and from the cord (anywhere between the pulleys) another cord with an S hook and a weight of, say, 40 oz. After a few up-and-down oscillations these three weights will come tp rest in the definite position shown by the figure, and if you disturb them from this position they will invariably return to it again. Consequently, you conclude that the three forces acting from their common point of attachment are in equilibrium, and that the force 40 oz is the equilibrant of the two forces 24 oz. and 32 oz. * This Lecture may require two meetings of a class when the students have had no previous training in Theoretical Mechanics. In any case, it will be well to spend at least one with a revisal hour before the written examination, which should now take place upon the work gone over since the beginning of the session, prior to the Christmas holidays. FORCES IN EQUILIBRIUM. 81 With a piece of finely pointed white chalk, draw lines (from the point where the three forces act) on the black board parallel to the cords, and plot off from this point to any convenient scale (say by aid of a two-foot rule) distances along them to represent their respective magnitudes. Extend from the same point in an upward vertical direction another line, and mark it off to represent 40 oz. This line evidently corresponds, in point of application, direction, and magnitude, to the resultant of the components (24 oz. and $2 oz.), for it is equal and opposite in direction to their equili- GRAPHIC REPRESENTATION OF FORCES IN EQUILIBRIUM. brant. From the extremity of this resultant draw lines joining tlie outer ends of the components (24 oz. and 32 oz.). Then you a parallelogram whoso adjacent sides from the point of have application, represent, both in direction and magnitude, the com- ponent forces, and whose diagonal represents, also both in direc- tion and magnitude, their resultant. If any other pair of convenient weights be selected and applied in the same way, you can find an equilibrant and resultant for them. From these experiments you conclude that a general prin- ciple, termed the " parallelogram of forces," is true without having recourse fa any special mathematical reasoning. LECTUKE VIII. Parallelogram of Forces. If two forces, acting simultane- ously towards or from a point, be represented in direction and magnitude by the adjacent sides of a parallelogram, then the re- sultant of these forces will be represented in direction and magnitude by the diagonal of the parallelogram which passes through their point of intersection. For example, let any two forces, P and Q, act from the point at any convenient angle, say 60, then, if OA and OB be plotted to scale to represent these forces in direction and magnitude, the diagonal OD of the parallelo- gram OADB will represent in direc- tion and to the same scale their resultant R. But the resultant E, is equal and opposite in direction to a force E, which would exactly balance the effect of P and Q, or to a force represented in direction and in magnitude by the line DO. Further, since the side AD is equal and parallel to the side OB, it may be taken to represent Q in direction and magnitude. Hence we have the three sides of the triangle OAD taken in the order OA, AD, DO, representing in direction and magnitude three forces, P, Q, E, in equilibrium, acting from the point O. Hence we have a general proposition termed the " triangle of forces," or a deduction from the "paral- lelogram of forces." Component PARALLELOGRAM OF FORCES. TRIANGLE OF FORCES. Triangle of Forces. If three forces acting towards or from a point are in equilibrium, and a triangle be drawn with its sides TRIANGLE OF FORCES. 83 respectively parallel to those forces taken in due order, then the forces will be represented to scale by the sides of the triangle. CONVERSELY : If three forces acting towards or from a point are represented in direction and to scale by the sides of a triangle taken in due order, these three forces are in equilibrium. For example, let the three forces P, Q and E act from the point 0, and be in equilibrium. Draw a triangle with its sides, P, Q, E, respectively parallel to these forces ; then the sides of this triangle, taken in that order, represent to the same scale these forces. Or, if the triangle, whose sides are respectively P, Q and E, represent in direction and to scale the three forces P, Q and E, as they act from a point 0, these forces are in equilibrium. We have shown by a dotted line the resultant R, and its direc- tion as opposed to E, by the same side of the triangle. It is quite evident that if the forces P, Q and E acted towards the point 0, instead of from it, the triangle P, Q, E would still represent these forces in magnitude, but the direction of all the arrows would have to be pointed the opposite way. SPECIAL CASES. Three Equal Forces in Equilibrium. It can easily be proved by the apparatus used for Experiment I., or by construe tion, that if you have three equal forces in equilibrium they must act at 120 from each other, and that the triangle representing their directions and magnitudes will be an equilateral triangle, or a triangle whose angles are each equal to 60. Two Forces acting at Hight Angles. In this case it can be E roved by the same apparatus, or by Euclid, Book I. Prop. 47, that any two Drees P and Q, acting at right angles to each other, have a resultant K, or arc balanced by a third force E, of such magnitude that Consequently, if you hare any two forces in the proportion of 3 to 4 acting at right angles to each other, their resultant will proportionate!/ be 5. For, suppose P = 3a, Q=4 w = Seoo; and- = i Precisely the same results will be arrived at if (as shown by the right-hand side of the figure) we considered the vertical side BC of the triangle ABC as representing W, and then draw a line from C parallel to R, and a line BD, parallel to P, to meet the line CD. EXAMPLE II. A force of 100 Ibs. is supported on a smooth inclined plane by a force P acting parallel to the base. If the incline be i in 10, find P. ANSWER. Draw a figure exactly the same as that accom- panying Case 2, and mark W = 100 Ibs., 1= 10, and h = i. Then, by the " triangle of forces " : JP __ CB h i i ri W A 1888, and 1891. ist Law. Friction ia directly proportional to the pressure between two surfaces, if they remain in the same condition. 2nd Law. Friction is independent of the areas in contact. 3rd Law. Friction is independent of velocity. Apparatus for Demonstrating First and Second Laws of Friction. Nevertheless, it will be both interesting and instruc- tive to students to have these three laws demonstrated by the following simple apparatus : G.P APPARATUS FOR DEMONSTRATING THE FIRST AND SECOND LAWS OP FRICTION. INDEX TO PARTS. IP represents Inclined plane. I B represents Box for planes. GP Guide pulley. PP Pinching pin. Q Quadrant. | LL Legs. IO4 LECTURE X. The inclined plane IP is fitted with a joint at its left-hand end, and after slackening the pinching pin PP, it may be raised to any desired angle or fixed in a level position by tightening the pin. The desired position is found by reading off the angle oppo- site the plumb-ball line on the graduated degree scale of the quadrant Q. In the box B may be kept planes of glass, brass, iron, steel, &c., as well as the different kinds of wood to be experi- mented upon. These planes are fixed on IP, in a central position, by means of a catch, and the bodies to be laid upon them should be fitted with a small hook opposite their e.g., to which a fine flexible silk cord can be attached and passed over the guide- pulley GP, which should turn very freely on its bearings.* The pull P is best effected by attaching to this cord a small tin pail into which shot may be dropped one by one until the body moves freely on the plane. The pail and shot may then be unhooked and weighed in a balance. Demonstrations of the First Law of Friction. EXPERI- MENT I. Fix the inclined plane IP in a truly horizontal posi- tion. Take from the box B, say, a long strip of planed yellow pine and a small block of the same kind of wood, and let its weight be W. Adjust the strip along the middle line of IP by means PEOOF OF FIRST LAW OF FRICTION. of the sneck or catch, and place the block therein. At- tach the silk thread to the hook on the forward side of the block, and pass the same over the practically f rictionless pulley. Hang a little tin pail from the free end of the silk thread, and drop small shot one by one into the pail until the block moves freely over the yellow pine strip when aided by a little tapping on the table. Unhook the pail containing the shot, and weigh it as carefully as you weighed the block of yellow pine. Let it equal P units. Then P is the force which just overcomes the directly opposing passive resistance, called friction, between the surface of the yellow * The guide-pulley bracket should be fitted with a stiff joint and with a telescope arm, so that the pulley may be raised or lowered in order to bring the direction of the pull P on the cord parallel to the plane, or parallel to its base, or adjusted to any desired angle with respect to the plane, in order to demonstrate Cases i, 2 and 3 of the inclined plane in Lecture IX. By having, say, a " slot along the middle of the plane, and by lowering the pulley, Case 2, wherein the pull on the body is parallel to the base, may be readily demonstrated; and by pulling out the telescope arm of the bracket, and turning up the bracket, Case 3, wherein the pull makes an angle, 0, with the plane, may be verified.- FRICTION. 105 p pine block and strip ; and the ratio ^ is termed the co-efficient of friction. Now put another block of weight W on the top of the one just tested, so as to double the pressure on the sliding surface, and put in shot until the block moves when aided by a little vibration, so as to overcome the greater resistance to starting the body in motion than to keep it moving.* You find on weighing the pail and shot that it is now 2?. Consequently the co-efficient of 2 p P friction has not altered, for =^- is the same fraction as ^. EXAMPLE I. Suppose you take a very small block of woocf (say i/io" thick, 2" long and i" broad; in fact, so light that its weight is negligible), and place a i-lb. weight on the top of it ; you will find that 575 oz. are required to cause motion of this piece of wood over the surface of the yellow pine strip. You therefore conclude that the co -efficient of friction is P c-7C oz. ft= Wi6oz, = * 353 ' r friction = '353 W. Now, place a 2-lb. weight on the upper piece of wood, and you find that it requires more shot in the pail to move it. Weigh the pail and the shot again just after you have obtained free move- ment of the one bodf over thi other, and you will fi*4 that it amounts to 11.5 oz. P 11.5 Consequently, ^ = - -353 as before. If, however, you put a lo-lb. weight on the upper piece of wood, you will obtain a different result, thus proving the first law and the variation therefrom; because in this latter case the pressure is so great, compared with the first and second experi- ments, that the grains of the upper piece of wood enter those of the lower, and bring into play another condition of affairs viz., the gripping action of the one set of grains on the other set. If you had taken a large plank of yellow pine, weighing, say, 100 Ibs., and had placed it on another similar plank, the co-efficient of friction would have a certain value. If you had even put a loo-lb. weight on the upper plank, the co-efficient of friction might not have varied perceptibly. But if you placed a weight of 1000 Ibs. * Statical friction, or the friction of repose, is that resistance which opposes the commencing of the motion. If a body be allowed to rest on another for some time, it requires more force to move it than if it had only been stationary for a few seconds. 106 LBCTtJKB X. on the upper plank, the co-efficient of friction would be con- siderably altered. Hence you observe that this first law only holds good between narrow limits.* Angle of Repose, or Angle of Friction. EXPERIMENT II. Another way of proving the first law of friction is to disconnect the silk thread and the shot-pail from the upper body, and tilt up the inclined plane to such an angle, a, with the horizontal that, (with the aid of a little tapping) the weighted block of yellow pine just slides slowly down the incline. Here we have simply the force of gravity acting on the body and overcoming friction. At the moment the body just begins to slide we have the weight, W, of the body acting vertically downwards, R the reaction from the plane at right angles to the surface, and F, the passive resistance of friction, acting parallel to the plane in the direction of aP in the first figure in this Lecture. Now, these three forces act from the e.g. of the body, and they are in equilibrium. R is equal to the resolved part of W at right angles to the plane (or R = W Cos a), and it represents the pressure between the surfaces. F is the resolved part of W, parallel to the plane (or F = W Sin a). TT and ^ is the co-efficient of friction. XV F _ W Sin a _ m _ h _ height of plane " R W Cos a b base of plane The angle a, to which the plane must be inclined before the free body will slip over the fixed one, has been termed the " anyle of repose" or " angle of friction" Therefore, ike tangent of the angle of repose is equal to the co- efficient of friction. p But was proved by the previous experiment to be also equal to the co-efficient of friction, P F h * ' W - R - 1 Or, P *= /iW and F * Sir Robert Stawell Ball, when Professor of Mechanism at the Royal College of Science, Ireland, tried a careful experiment in the above way with a smooth horizontal surface of pine 72" x 1 1", and a slide, also of pine, 9"x9" grain crosswise. He loaded and started the slide, and applied a force sufficient to maintain it in uniform motion, and he found that on increasing the load from 14 to 112 Ibs., by increments of 14 Ibs., the co- efficient of friction diminished from -336 to -262. From these experiments he constructed the empirical formula for this case that F = '9 + -266 R, where F is the factional resistance and R the reaction from the surface or net load- FRICTION. 107 where fi is the Greek letter universally adopted to represent co- efficients of friction. The accompanying figure is a diagram of the " angles of repose ' for various common materials, together with the numerical values of or their co-efficients of friction. Jbrlcat e ro rtO~ A.NGLES OP REPOSE. Limiting Angle of Resistance, or Sliding Angle. A third way of proving the first law of friction is to place the bodies so that the sliding surface is perfectly level. Then begin by pressing the upper body through the intervention of a compres- sion spring-balance fitted with a sharp point, so that it will not slip off, and with a clinometer to indicate the angle through which it is tilted away from the perpendicular. Now gradually incline your pressure to the perpendicular, until you arrive at such an angle as will just cause the upper body to slide over the under one. This angle is termed the "sliding angle" or " limiting angle of resistance" because it is the limit, or maximum angle which the reaction from the surface can make with the perpendicular to the surfaces, for the reaction must act in the directly opposite direction to the pressing force.* Again, apply the spring- balance, but with double the registered pressure, and you can just incline this force to the same angle as before. If, however, you pre& f . with ten times the former force, you would probably be able to act at a greater angle than before. It will be seen from this experiment that The Limiting Angle of Res-istance = The Angle of Repose. * Here the weight of the tipper body is supposed to be negligible in Comparison with the inclined pressure upon jt t io8 LECTURE X. PKOOF OF SECOND LAW OF FRICTION. Demonstration of the Second Law of Friction. Ex* PERIMENT III. Take a block of planed yellow pine, and cut it into two equal pieces at right angles to the planed surface. Place one piece on the horizontal strip of yellow pine (used in previously demonstrating the first law), with the planed side next to it, and put the other piece on the top of it, as shown by the second figure in this lecture. Now ascertain the horizontal pull, P, required to overcome friction. Then attach the top piece to the bottom one, as shown by the accompanying figure, so that the area of the surface in contact is doubled, and you will find that the same horizontal force, P, will cause it to move. If you take a long planed block and cut it into ten equal pieces, each of the same size as one of the above pieces, and try the experiment in a similar manner, you will be able to increase the area of contact tenfold, and you will then find that the p ratio ^ is not exactly the same with the surface of one block in contact with the strip, as when the surface of the whole ten came into action at once. The result of increasing the area in contact may also be tried by placing the blocks on the inclined plane, and observing the angle to which the plane is tilted when they begin to slide down the plane. APPARATUS FOR DEMONSTRATING THE FIRST AND THIRD LAWS OF FRICTION. INDEX TO PARTS. L represents Lever. F SR SP TP q Fulcrum. Small roller. Scale-pan. Test-piece. Cylinder. H represents Handle. R S P GS Rack. Spiral spring. Pointer. Graduated scaled Base of apparatus. IKICnON LUBRICATION. 109 Demonstration of the Third Law of Friction. The pre- ceding figure represents the apparatus belonging to the Applied Mechanics Department of the Royal College of Science, South Kensington (as described by Prof. Goodeve in his " Manual of Applied Mechanics "), for demonstrating the first and third laws of friction. If the weights, W, be removed from the scale-pan SP, then there will be but a slight pressure between the lower surface of the test-piece TP, and the roller cylinder C. Consequently, on turning the handle H in the direction of the arrow, there will be a slight pull on the cord, causing the pointer P to move a degree or two over the graduated scale GS. The pointer should there- fore be set back to zero. EXPERIMENT IY. Put a weight, W, of say 5 Ibs., into the scale-pan, and turn the cylinder slowly by the handle as before. The pointer deflects so many degrees. Increase the weight W to 10 Ibs., and the pointer instantly indicates twice the amount of friction ; put in 15 Ibs., and it shows treble the friction ; thus demonstrating the first law. Then turn the handle faster and faster, and the pointer remains fairly stationary, thus proving within certain limits that friction is independent of the velocity.* Lubrication. Lubricants, such as tallow, grease, soft soap, and many kinds of oils, are used to reduce friction. Both skill and knowledge are required to decide upon the best kind of lubri- cant and the proper amount for different cases. Lubrication and lubricants should receive greater attention from the engineer, for the satisfactory working and length of life of most machines de- pend so largely upon effective lubrication. Where very heavy pressures and high speeds are experienced as in some cases of electrical machinery, it pays to use the very best kind of oil, and to distribute it to all the bearings from one common centre under pressure by means of a force-pump. It thereby flows in a con- tinuous stream through the bearings to a filtering tank, from which it is again and again pumped on its soothing mission for months on end, without change or great loss in quantity. This is a very different state of matters from the " travelling oil-can " system, where the amount applied may vary, and the times of appli- cation may be erratic, according to the opinion of the attendant. Anti-Friction Wheels. In the case of delicate machinery, such as in Atwood's machine for ascertaining by experiment the acceleration of gravity, and in Lord Kelvin's mouse-mill for driving the paper rollers of his Syphon Recorder, when receiving * Sea Molesworth'i Pocketbook of Engineering Formulae, and the Trans- actions of the Institution of Mechanical Engineers, for results of friction experiment! with shafts ran at different speeds. I 10 LECTURE X. ANTI-FRICTION WHEELS. telegraphic signals from long submarine cables, anti-friction wheels are used for the purpose of reducing the friction to a minimum. The accompanying figures illustrate one pair of anti-friction wheels. The spindle S, which carries the driving- wheel, instead of resting on two ordinary bearings, is supported by two wheels at each end, so that a rolling contact is pro- duced between it and the wheels. This form of contact implies far less friction to begin with, than a sliding or scraping con- tact. Besides, the small amount of force required to overcome the friction between the spindle and the rims of these wheels, has a great advantage or leverage given to it, in as far as, it acts with an arm equal to the radius of the wheels FW\ and FW 2 . This enables it to turn them with great ease at a slow rate in the very small bearings Bj and B 2 . In merely overcoming friction at a bearing, there is a con- siderable advantage in using large pulleys ; for, the force necessary at the periphery of the pulley to overcome the friction at the bear- ing, is inversely proportional to the radius or diameter of the pulley. (See Lecture XI. fig. i). Ball Bearings. Another example of the effect of rolling contact reducing friction is found in the use of ball bearings, which are now so common in all kinds of cycles and in high-class foot-driven lathes.* When it is necessary to move heavy beams, guns, &c., a common practice is to place them on rollers or on two channel iron girders o with round cannon-shot between them, when a com- paratively small force, properly applied, will have the desired effect. We will have to return to this subject in the Advanced Course when dealing with the friction between shafts and their bearings, and the various means that have been adopted for minimising the same. In the meantime, we will complete this Lecture with an example of work done on an incline when friction is included. Work done on Inclines, including Friction. The method of calculating the work expended in moving a body along a smooth inclined plane was fully dealt with in Lecture IX. ; consequently, the student is prepared, after what has been said about friction in * Refer to Lecture XVI., p. 183. WORK DONE ON INCLINES INCLUDING FRICTION. I I I this Lecture, to consider the case of pulling a body up or down a plane when the co-efficient of friction between the body and the plane is known. The total work expended is evidently divisible into two distinct portions 1 i ) The work done with or against tJie action of gravity, accord- ing as the body is moved down or up the inclined plane = W x h (where h is the height of the plane). (2) The work done against friction = F x I (where I is the length of plane passed over). The work to be done against friction is the same whether the body is urged up or down tho incline; for it is equal to the r^-effiomnt of friction x the reaction of the plane x the distance through which it is moved. Or, FxZ = /ixRxJ But by Lecture IX. B x 1 = W x b; .. F x Z = /z x W x b Or, the work done against friction in moving a body along the inclined distance I, is equal to the work done in moving the same body along a horizontal distance b, equal to the base of the incline. If the work to be done in overcoming friction, is equal to the work capable of being done on the body by gravity, the body will be in equilibrium, and the inclination of the plane is equal to the angle of repose. If the work to be done in overcoming friction is less than the work which gravity can do on the body, the body will slide down the incline, or, in technical language, the machine will overhaul. EXAMPLE II. What is the co-efficient of friction, and how is it ascertained ? There is an inclined plane of i foot vertical to 10 feet horizontal; what work is done in moving 700 Ibs. 5 feet along the plane, the co-efficient of friction being -08 ? (S. and A. Exam. 1892.) w FIGURE FOB EXAMPLE II. ANSWER. The co-efficient of friction for two bodies in contact is the passive resistance (opposing the motion of the one over the ofher) divided by the reaction or normal vressure between the surfaces in contact Friction F i.e. , Co-emcient or fnction = -^ . = ^ = u. Reaction R For methods of ascertaining co-efficients of friction, see the text in this Lecture, -a 112 LECTURE X. Total work done = work done against gravity + work done against frwtion. Referring to the accompanying figure, we see that Si) Work done against gravity = W x DE 2) Work done against friction = F x AD Total work done = W x DE + F x AD We have therefore only to substitute the numerical values corre- sponding to these letters in order to arrive at the result. From the ques- tion W = 7oo Ibs. From the figure we see that DE is parallel to BC ; con- sequently by Euclid the A", ADE, and ABC are similar in every respect ; and therefore DE : BC : : AD : AB ; or, DE= B * AP . Ai> But, also by Euclid, AB= ^ AC 2 + BC 2 = V io 2 +i 2 = 10-05 ft. (nearly) Consequently, DB-?55. =JJL = - 497 ft. And, F * AtB From the question we are told that |i = -08, and we learn from Lecture IX. that B : W : : AC : AB ; or, R= WAC = 7io =696 . s lbs . .'. F=/iR=-o8x 696-5 = 55-72 Ibs. Hence Total Work = W x DE + FxAD = 7oo * -497' + 55-72x5' = 347-9 ft.-lbs. + 278-6 ft.-lbs. = 626-5 ft. -Ibs. NOTE. For the work done against friction quite a simple way would have been to hard taken the formula deduced on the previous page Vix. : FxZ=MXWx&=/mXWxAE= o8X7cx>X4-97 = 278'6 ft.-lbs. * IB-BO A 8 -*p-22*=- w APPROXIMATE ANSWER. Since the inclination of the plane is so very small in this case, we might have assumed that R = W; AB = AC, andDE = JBC Then, (1) Work done against gravity = W x DE = 700 x % = 350 ft.-lbs. (2) Work done against friction = F x AD=-o8 x 700 x 5 = 280 ft.-lbs. % TQtal work = W x DE + F x AD = 350 + 280= 630 ft.-lbs. LECTUEE X. QUESTIONS. - 113 LECTURE X. QUESTIONS. 1. What is friction, and how does it act ? What is developed ^ hen force overcomes friction ? How do you measure the result ? 2. Explain by sketches and concise description how the laws of friction may be tested experimentally. What is meant by the " co-efficient of fric- tion," "angle of repose," "angle of friction," and " sliding angle " or 11 limiting angle of resistance " ? 3. How is the co-efficient of friction between two surfaces ascertained approximately by experiment? When two rough surfaces are pressed together, how much may the line of pressure be inclined to the common perpendicular to the surfaces in contact before motion ensues ? 4- What is the co-efficient of friction when the angle of repose is (a) 5 42' ; (6) 1 1 18' ; (c) 16 4* ; (d) 21 48' ; (e) 26 36' : (/) 30 ; (g) 45 T Draw the angles to scale. Ans. (a) -i ; (b) -2; (c)3; (d) -4; (e) -5 ; (/)'5774: (9) i. 5. An inclined plane is 100 feet long and 20 feet high. A body weighing 100 Ibs. is pulled up from the bottom to the top, and then down again. If the co-efficient of friction between the body and the plane is -5, what work was expended in each case ? What would require to be the co- efficient of friction in order that the body might just slide down of its own accord? Ans. 6,900 ft.-lbs. ; 2,900 ft.-lbs. ; /i= _= ^.=-204. b 12 6. What is the co-efficient of friction, and how is it ascertained ? There is an inclined plane of i foot vertical to 5 feet horizontal ; what work is done in moving 100 Ibs. through 100 feet along the plane, the co- efficient of friction being ! 1 Ans. 2940 ft.-lbs. 7. An incline is 80 feet long, with a rise of 20 feet. A body weighing 100 Ibs. is drawn 40 feet along the incline ; what work is expended if the co-efficient of friction is -6 ? Ans. 3,323 ft. -Ibs. 8. A weight of 5 cwts. resting on a Horizontal plane requires a horizontal force of loo Ibs. to move it against friction. What is the co-efficient of friction ? Ans. -18. 9. A plank of oak lies on a floor with a rope attached to it. When the rope is pulled horizontally with a force of 70 Ibs. it just moves, but when pulled at an angle of 30 to the floor a force of 60 Ibs. moves it. What is the weight of the plank and the co-efficient of friction between it and the floor ? Ans. 116-6 Ibs. ; -6. 10. Suppose a locomotive weighs 30 tons, and that the share of this weight borne by the driving wheel is 10 tons. Then, if the co-efficient of friction between the wheels and the rails be -2, what load will the engine draw on the level if the required co-efficient of traction be 10 Ibs. per ton of train load ? What load will this engine draw at the same rate up an incline of i in 20 ? Ans. 448 tons (including engine) ; 36- 72 tons (in- cluding engine). 11. State the laws of friction, and explain the contrivance known as friction wheels. What is the advantage of ball bearings for bicycles? Sketch in section such a bearing. 12. What are lubricants, and for what purposes are they used in machin- ery ? What kind of lubricant would you use for the moving parts of a very high-speed engine and direct-driven dynamo, and how would you apply it so as to be able to use it over and over again ? 114 LECTURE X. QUESTIONS. 13. What is friction ? What is meant by limiting friction, by sliding friction, and by the co-efficient of friction ? A weight of 5 cwts. resting on a horizontal plane, requires a horizontal force of 108 Ibs. to move it against friction. What in that case is the value of the co-efficient of friction 1 Ans. "192. 14. How would you experimentally determine the nature of the friction between clean, smooth surfaces, say of oak, and what sort of law would you expect to find ? 15. Describe any experiment which you have made or seen for finding the laws of solid friction. What are the laws so found ? Are they quite true ? How do they differ from the laws of fluid friction ? 16. Sketch and describe an apparatus for determining the co-efficient of sliding friction between two planed surfaces of oak. If you have made this or a similar experiment describe the behaviour of the sliding piece, and any troubles you may have had. State how you would conduct the experiment so as to establish the principal facts concerning such friction. (B. of E., 1902.) 17. The tractive resistance of a train weighing 335 tons is 1 1 Ibs. per ton. If the effective horse-pov er of the lo 'emotive is 600, estimate the uniform speed obtainable when ascending an incline of I in 200. Ans. 2662 ft. per min. or 30 miles per hour. (C. & G., 1904,0., Sec. A.) NOTES AND QUESTIONS. LECTURE XL CONTENTS. Difference of Tension in the Leading and Following Parts of a Driving Belt Brake Horse-Power transmitted by Belts Examples I. II. Velocity Ratios in Belt Gearing Examples III. IV. Open and Crossed Belts Fast and Loose Pulleys Belt Gearing Reversing Motions Stepped Speed Cones with Starting and Stopping Gear Driving and Following Pulleys in Different Planes Shape of Pulley Face Questions. WE shall devote this Lecture to the transmission of power by belting and to belt-gearing. Difference of Tension in the Leading and Following Parts of a Driving-Beit. In Lecture VI., when discussing the case of the simple pulley, we assumed that the belt or rope passing over the pulley was perfectly flexible, and that there was no fric- tion at the axle of the pulley. Conse- quently, we found that equal weights would balance each other, or that the tension of the two sides of the belt were equal. A little consideration of the subject will show that when one pulley is driven from another one by an endless belt or rope, the tension on the driving side must be greater than that on the following side. i. Take the case of an ordinary vertical pulley with its axle or shaft resting in two bearings (one on each side of the pulley), with a belt or rope passed over it, and with weights at- tached to the free ends of the same. Here we must have a certain amount of friction between the axle and its bearings, which can only be overcome by a force applied to the circumference of the pulley. T , T? _ ( Force required to overcome friction at the circum- .Lieu . J? ! 1 fsrence of the axle or shaft. DIFFERENCE OF TENSION DUE TO FRICTION. DIFFERENCE OF TENSION. 117 Let . . r l = Radius of the axle. p _ /Force required to overcome the friction of the axle t \ when acting at the circumference of the pulley. . . r 2 = Radius of the pulley to centre of belt. Then, F t xr 1= F 2 xr 2 . [Weight attached to the left-hand side of the belt, Let . . W = j and which therefore produces a tension on the I slack side = T,. (Least weight on the right-hand side of the belt that will produce motion, and which therefore produces a tension on the driving side=Tj. Then taking moments about the centre of the axle, we have VV l x r 2 + F 2 x r 2 = W 2 x r a Or, . . T.xr f + F f xr, = T d x r a Dividing both sides of the equation by r, we get T.+F,=T d .-. F 2 = T d -T, Or, expressed in words, the force F 2 , acting at the circumference of the pulley (which is required to overcome the friction of the axle) is equal to the tension T d on the driving or forward side of the pulley, minus the tension T, on the slack or following side. In order that the periphery of the pulley may move at the same rate as the under face of the belt, we must have sufficient tension on each part of the latter, and the co-efficient of friction between them must not have less than a certain value. Too great adhe- sion between them would result in a loss of work, for in that case an extra force would have to be applied solely for the purpose of pulling the belt from the pulley. 2. Take the case of one vertical pulley of diameter D, driving another vertical pulley of diameter d by means of an endless belt, rope, or chain in the direction of the arrows shown on the accom- panying figure. Whenever the pulley T> is moved, the tension on the driving side T d tends to stretch the belt on that side, and this tension increases until the pulley d begins to move ; whereas BELT DRIVIKG . the tension on the following or slack side, T,, is gradually diminished until the difference of the tensions (T d - T a ) produces a uniform velocity of the belt. Of course the tension on the slack side must be sufficient to pro- ITS LECTURE XI. vent the slipping of the belt on either of the pulleys if the periphery of the driven pulley is to keep pace with the peri' phery of the driving one. In order that there may be a mini- mum chance of the belt slipping, its slack side should always run from the top side of the driving pulley. By so arranging the drive, the sag of the belt on the slack side will cause it to encom- pass a greater length of the circumferences of both pulleys. The motion of the belt will be easier, and the wear and tear of the bearings will be less, because there will be less total stress (T d + T,) tending to draw the pulleys together for the transmis- sion of a certain horse-power, than if the slack side left the under side of the driving-pulley. Referring to the previous figure, if the slack side leaves the top side of the pulley D, it grips the same from position 4, round the back of the pulley to 5, and the pulley d from 6 round to 3 ; whereas, if D were rotated in the opposite direction, we should have the slack side entering on it at i, and only gripping it as far as position 8 ; entering on d at 7, and only gripping it to position 2, thus having far less grip on the pulleys and thereby encouraging the natural tendency of the belt to slip on the pulleys.* Brake Horse-power transmitted by Belts. Let. . V = Velocity of belt in feet per minute. . . P = (T d T 4 ) the net pull causing motion in Ibs. Then, B.H.P.= YP Y ( T ^~ T ) 33> 33>- Let . . D Diameter of driving pulley in feet = 2r. Then . vD *= Circumference of driving pulley in feet = 2wr. Let . n = Number of revolutions of pulley per minute. Then . Y = nDn = 2irrn = velocity of belt (with no slip). And, the B.H.P. = 7rPnP = 27rmP 33><> 33> EXAMPLE I. A pulley 6' in diameter is driven at 100 revolu- tions per minute and transmits motion to another pulley by means of a belt without slip. If the tension on the driving side of the belt is 1 20 Ibs. and on the slack side 20 Ibs., what is the brake horse-power being transmitted ? * The previous figure should have been drawn with the full and dotted .Ines at T,, reversed, but the student will easily follow the explanation. BRAKE HORSE- POWER TRANSMITTED BY BELTS.. 1 1 9 ANSWER. Here r = 3'; n = 100; P = (T d - T.) = (i 20 - 20) = loolbs. B H P = 27rmP = 2 x y x 3 x IPO x IPO _ 5 . 71 33,000 33> EXAMPLE II. What must be the number of revolutions per minute of a driving pulley 6' in diameter, in order that it may transmit 5-71 B.H.P. by a belt to another pulley, if the net pull on the belt Ls 100 Ibs. ? ANSWER. Here we have the same data to go upon as in Example I., except that we are given the B.H.P. instead of the revolutions per minute. Then, transposing every quantity except n (the revolutions per minute) to one side of the above equation, we have n = (B.H.P.) x 33,000 = 5-71x33,000 = 100 r 27irP 2 x Hf x 3 x 100 In precisely the same way, if you were given the power to be trans- mitted, the revolutions per minute, the difference of tension on the two sides of the belt, and you were asked for the diameter of the pulley, the formula would appear thus n _(B.H. P.) x 33,000 7TP If it was the difference of tension in the belt that was asked for, then You would (after arranging the formula in this way, so as to keep the unknown quantity on one side of the equation) simply have to substitute the numerical values corresponding to the different symbols, and then cancel out the figures in numerator and denominator, in order to reduce the long multiplication and division to a minimum, and thereby arrive at the result as quickly as possible. Velocity Ratios in Belt Gearing. Let two or more pulleys be connected by belting in the manner shown by the accompany- ing figure. Then, if there is no slipping of the belts, the circum- VELOCITY RATIOS IN BELT GEARING. f erential speeds of the pulleys will be the same as the velocity of the belts passing round them. 120 LECTURE XI. Let Dj, D, * Diameters of the drivers* F 1? F, = Diameters of the followers. N Di , N" Dj = Number of revolutions per minute of the drivers. N p , N p = Number of revolutions per minute of followers. Then, taking the first pair of pulleys, Dj and F, we have Circumferential speed of driven = Circumferential speed of follower i. (Divide both sides by *) Or, The product of the diameter of -\ the driver and its number of revolutions per minute. i The product of the dia- meter of the follower and its number of re- ( volutions per minute. D "N" Or ...... ^-jg . . . (1) i.e., The ratio of the diameters of the pulleys is in the inverse ratio of their speeds or revolutions per minute. Treating the motion of the second set of pulleys in exactly the same way, we have Circumferential speed, of D f = circumferential speed, of F, (Divide both sides by TT) D,N Dj = F,N,, <* ..... f = ?' ' (2) J a -^ 3 (But the revolutions of F, and of D 2 are the same) .'. N ri = N D Or, . . . D, N, - F,N, .-. N F> - ?i/?' X _F,NF Consequently . . ^ -^ -rr 1 i / (Dividing both numerator) ND = _D!_ = F t x F, I and denominator by N Fi )J ]^ p i D 2 = D X x D, ^a Or, we might have arrived at the same result by multiplying equation* (1) and (2) together. Thus VELOCITY RATIOS IN BELT GEARING. 121 Or, Speed of first driver m Product of diameters of followers Speed of last follower Product of diameters of drivers Or, . N Di x Dj x D, = N Fj x F t x F, t.., Speed of first driver x dia-} _ (Speed of last follower x diame- . meters o/JAe drwers J ~ ( ters of the followers. In the same way we may treat any number of drivers and followers by this general formula viz., Speed or number of revolutions} per minute of the first driver [ x the successive diameters oft the drivers ) Speed of the last follower x the successive diameters of the followers. Precisely the same rule holds good for discs driven by contact friction and for wheel gearing, as you will find from the next lecture ; but in friction gearing and wheel gearing the driver and the follower move in different directions, whereas in belt gearing they move in the same or in the opposite direction, according as the driving belts are " open " or " crossed." EXAMPLE III. Referring to the previous figure, suppose that a driving pulley, D p is connected by a belt to a follower, F p whilst it moves at 100 revolutions per minute. If the diameter of the driver is 6' and of the follower 3', what will be the number of revolutions per minute of the follower 1 By the previous formula for two pulleys, D t x N Di = F t x N ?i XT D, x N D 6' x 100 .. -N Pi -t_ * = 200r.p.m. *! 3 EXAMPLE IT. Referring to the previous figure, suppose that a driving pulley D a (4' diameter), is geared to a follower, F t (2' in diameter), and that a second driver D, (4/ diameter), fixed to the same shaft as F p is geared to a second follower F, (i' dia- meter). If D makes 60 revolutions per minute, what is the speed ofF,? By the previous formula for four pulleys, N Di x D t x D, = N p> x F t x F f ** =480 r.p.m. Open and Crossed Belts. By referring to the next figure, the student will observe that the left-hand end view shows what 122 LECTURE XI. is termed an open belt, OB, and that the right hand end view shows a crossed belt, 013. In the case of open belts, the driver and the follower rotate in the same direction (as may be seen from the second and third figures in this Lecture) ; whereas, with crossed belt driving, the follower revolves in the opposite direction to that of the driver, just as it does when direct friction or wheel- gearing is used. Fast and Loose Pulleys. As will be seen from the two front views in the next set of illustrations, the open and the crossed belts are shown passing from the broad driving-pulleys DP, to the broad loose pulleys LP. Loose pulleys are generally bushed with gun-metal, and then bored out so as to fit their shafts easily. This permits them to rotate without turning the shaft upon which they bear. The pulleys, FP, are keyed hard on to the shafts, so that when the belt is forced over upon them by means of the shifting forks, SF, the machines connected with the same are set agoing. This simple combination of fast and loose pulleys there- fore enables a machine to be stopped or started at pleasure, without interfering with the motion of the driving pulley and the belt. In ordinary cases where there is only one driving belt required, the loose pulley is of the same breadth as the fixed pulley.* Belt- Gearing Reversing Motions. In many kinds of machine tools it is desirable to be able to drive the tool first in one direction and then in the opposite direction, as well as to start or stop it. This is frequently effected by a combination of open and crossed belts with fast and loose pulleys, as illustrated by the accompanying figure. From what has just been said about open and crossed belts, as well as fast and loose pulleys, the student will have no difficulty in understanding this arrangement of reversing gear. If applied to a machine for planing metals, the shaft which is keyed to the fixed pulley FP would be connected either through wheel gearing and a rack, or through a central screw, to the travelling table of machine upon which the job to be acted upon is secured. When- ever the table had been moved backwards to the end of the required stroke by the crossed belt, the shifting fork SF would be pushed forward by an outstanding arm or kicker attached to * See the set of figures after the next, where E l is the driving belt engaging the fixed pulley, FP ; and where LP is the loose pulley, to which the belt may be shifted by means of the shifting-fork, SF, whenever it is desirable to stop the speed cones, SCj, SC^ and the machine to which they are connected. In the first front view of the next set of figures, the driving pulley, DP, and both of the loose pulleys, LP, are drawn too narrow. They should have been represented half as wide again, in order to prevent the belts slipping over the outside edge, when the other belt is shifted on to the fixed pulley situated between them. BELT-GEARING REVERSING MOTIONS. 123 the side of the table at such a position as would cause the crossed belt CB to be shifted from the central fixed pulley to its loose one, and at the same time bring over the open belt from its loose pulley to the central fixed one. Whenever the planing tool had finished its cut on the metal, the shifting fork would be pulled backward by another similar outstanding arm or kicker (also attached to the travelling table of the planing machine, at a position just beyond the end of the required stroke for the particular job under operation), thereby shifting the open belt OB from FP, to its loose pulley, LP, and at the same time pulling over the crossed belt, CB, L^FP FRONT VIEW, Forward and Return at same Speed. FRONT VIEW. Quick Return. BELT GEARING REVERSING MOTIONS. INDEX TO PARTS. DP represents Driving pulleys. FP Fixed pulleys". LP Loose pulleys. OB represents Open belts. CB Crossed belts. SF Shifting forks. from its loose pulley to the central fixed pulley, thus causing the table to make the return stroke. '.The left-hand front view, with its accompanying end views, shov the necessary arrangements when the forward and backward velocities of the table are equal. The right-hand front view illus- trates the case wherein the backward or non-planing motion is intentionally made quicker than the forward or cutting stroke, so as to save time, by having the back motion fixed pulley, FP, and its corresponding loose one, LP, made smaller than the forward set. The end views for this latter case would be similar to the former onp. with the exception that the crossed belt would engage P smaller 124 LECTUKE XI. pulley of the same size as shown by the front view. This latter arrangement can evidently be employed to obtain a fast or a slow motion in the same direction, by simply having both belts open or both crossed. Stepped Speed Cones with Starting and Stopping Gear. In many machines, such as lathes, planers and other machine tools, it is very desirable not only to be able to start and stop them, but also to alter their speed so as to suit different classes of END VIEW. SIDE VIEW. STEPPED SPEED CONES WITH STARTING AND STOPPING GEAB. INDEX TO PARTS. HB t ,HB 2 represent Hangingbrackets for supporting shaft, &c. SCj,SC 2 Speed cones. FP Fast pulley. LP Loose pulley. B,,B 2 represent Belts. H Handle. SB Sliding bar. SF Shifting fork. W Weight to fix SB in positions ^ ^.. work, without affecting the motion of the prime motor or that of the shop driving-shaft. These objects are generally attained by a combination of fast and loose pulleys with what are termed " stepped speed cones." The accompanying side and end views illustrate the arrangement as usually carried out in engineering works. When the starting-handle, H, is turned to the right hand, it pulls over the sliding-bar, SB, with its shifting-fork, SF, which moves the belt, B p from the loose pulley, LP, to the fixed one, FP ; thus setting the speed cones, SG p SC 2 , and thereby the DRIVING AND FOLLOWING PULLETS. 12$ machine in motion. When the handle is turned to the left, it pushes the sliding-bar and shifting-fork also in that direction, thus moving the belt from the fixed to the loose pulley, which allow the cones and machine to come to rest. In each case the weight, W, causes a notch in the sliding-bar to engage with its left-hand supporting bracket, thereby preventing the shifting fork from pushing the driving belt too far, or off either pulley, and at the same time ensuring that it remains in the desired position. Both supporting brackets for the sliding-bar, SB, are merely right-angle extensions from the hanging brackets, HBj, HB,, which carry the upper shaft with its cone and pulleys. The upper and lower speed cones, SCj, SO,, are generally made of the same size and shape, but they are always keyed to their respective shafts in opposite directions. Consequently, if it should be desirable to run the machine fast for light work, the belt, B r is shifted on to the largest pulley of the upper cone and the smallest one of the lower cone. If the machine is required to move slowly for heavy cuts, then the belt is placed on the smallest upper pulley and the largest lower one. Any desired intermediate speed between these extremes is obtained by adjusting the belt on one or other of the remaining sets of pulleys of the upper and lower cones. The student can easily prove to himself (by drawing down the arrangement to scale) that such stepped speed cones, if connected by a crossed belt on one pair of its pulleys, will produce the same tension in the belt with any other pair.* With open belt-driving the tightness of the belt will not be the same when on one pair of the pulleys as when on another ; but the difference is so small that it can generally be disregarded in practice without having recourse to tightening or slackening the same. Driving and Following Pulleys in Different Planes. It is often necessary to drive a follower placed in a different plane from the driver. The accompanying set of illustrations show very clearly how this is effected. The important precaution to be observed is, that the leading or on-going part of the belt must enter upon the follower in a fair or direct line with its plane o/ rotation. If this rule be attended to, then power may be trans- mitted between two non-parallel shafts, as shown by the first tigure, even if their centi-e lines are in planes at right angles to each other i.e., when the belt is working with quarter-twist. When two shafts are in planes at right angles to each other, and * The algebraical proof of this will be considered in our "Advanced Book on Applied Mechanics." The student should refer to the general view and to the detail drawings of the stepped speed cones in the foot- driven screw-eutting lathe illustrated in Lecture XVI. 126 LECTURE XI. Tullis's Thick-sided Leather Chain Belt, Working Quarter- twist, and Transmitting l^ower between two shafts which are not parallel. No Guide Pul- leys are required for this drive. Flat Belt Working Quarter-twist and Transmitting Power between two right- angled shafts, with leading Guide Pulley (GP) to remove the twist from the Belt before it enters upon the Follower, and to give the belt more grip on the pulleys. Flat Belt Transmitting Power over Guide Pulleys between two non -parallel shafts in the same plane. Flat Belt Transmitting Power between two parallel shafts not in the same plane by aid of guide pulleys (GP). SHAPE OF PULLEY FACE. I2/ it is found desirable to remove the twist from the belt before it enters upon the follower, then a guide-pulley, GP, must be used & shown by the second figure. When the shafts are parallel, but not in the same plane, then the power must be transmitted by aid of two guide-pulleys, as seen from an inspection of the third figure. Or, should the shafts not be parallel, but in the same plane, two guide-pulleys are necessary, as in the fourth figure. Guide-pulleys, if supported by spindles running in adjustable bearings or brackets, may be made serviceable as tightening- pulleys for the purpose of taking up the slack of the belt, and thus giving the necessary grip for transmitting more power with a steadier drive than can be obtained without them. Shape of Pulley Face. The student will have observed that the faces or rims of the fast and loose pulleys, as well as those of the stepped cones in the previous set of figures, are slightly curved. This convex curvature, or double coning, is purposely done in order to ensure that the belt may ride easily and fairly in the centre line of the pulley face without inclining to either side. A flat band, if placed on the smaller end of a revolving straight conical pulley, will naturally tend to rise to the larger end of the cone. Consequently, if each half of the face of a pulley is coned (or, which amounts to the same thing, if the rim of the pulley be curved so as to have its largest diameter in the middle of its face), each half of the breadth of the belt will have an equal tendency towards the middle of the pulley's rim. When very fast driving and sudden severe stresses are brought to bear upon a machine, as in the case of circular saws, morticing machines, and emery-wheel grinders, it is found necessary to fit the pulleys with side flanges, in addition to curving their rims, in order to prevent the belts from sliding off the pulley's face to one "ide or to the other. N.B. The student may be referred to the Author's Text-book on "Applied Mechanics and Mechanical Engineering," Chapters XVII. and XVIII., for further information on belt, rope, and chain gearing. 128 LECTUKE XL QUESTIONS LECTURE XI. QUESTIONS. 1. In machinery, where one pulley drives another by means of an end- Jess belt, there is a difference of tension in the two parts of the belt. Why is this ? The pulley on an engine shaft is 5 feet in diameter, and it makes 100 revolutions per minute. The motion is transmitted from this pulley to the main shaft by a belt running on a pulley, and the difference in tension between the tight and slack sides of the belt is 1 15 Ibs. What is the work done per minute in overcoming the resistance to motion of the main shaft ? Ant. 180,642 ft.-lbs. 2. Deduce from the "principle of work" a formula for the brake horse-power transmitted by a belt. The pull on the driving side of a belt is 200 Ibs. and on the following side 100 Ibs., whilst the belt has a velocity of 990 ft. per minute. Find the number of units of work performed in two minutes and the B.H.P. transmitted. Ans. 198,000 ft.-lbs., 3 B.H.P. 3. State and prove the rule for estimating the relative speeds of two pulleys connected by a belt. Also, the velocity ratio between the first driver and the last follower in belt gearing, where there are two or more drivers and a corresponding number of followers. [A main shaft carrying a pulley of 12 inches diameter and running at 60 revolutions per minute, communicates motion by a belt to a pulley of 12 inches diameter, fixed to a countershaft. A second pulley on the countershaft, of 8| inches dia. meter, carries on the motion to a revolving spindle which is keyed to a pulley of 4^ inches diameter. Sketch the arrangement and find the number of revolutions per minute made by this last pulley. Ana. 123-5 4. Two pulleys are connected by a driving belt, and the sum of their diameters is 30 inches ; one pulley makes 2 revolutions while the other makes 3 revolutions ; find their respective diametere, Ans. 18", 12". 5. An engine works normally at 106 revolutions per minute. At that speed it was found that it drove by belting a dynamo at 420 revolutions per minute, but to show off the electric lights at their normal candle power the dynamo had to be run at 460 revolutions per minute. At what speed was the engine being driven 1 Ans. 116 revolutions per minute. 6. A pulley of 3 feet radius rotates at 100 revolutions per minute and transmits motion to another pulley of 36 inches diameter. If there is 10 per cent, slip on the belt what will be the speed of the follower ? What will be the net driving pull on the belt if 5 B.H.P. is transmitted by it ? Ans. 180 revolutions per minute ; 97-2 Ibs. 7. Sketch an arrangement of pulleys and bands for obtaining a reversing motion from a shaft driven at a constant rate in one direction, and describe the action of the combination. 8. Sketch a combination of fast and loose pulleys as used for setting in motion, or stopping machinery. Explain the construction adopted for re- taining a flat belt upon a pulley, pointing out where the fork is to be applied, and why. 9. Sketch and describe a good form of slow forward and quick return for a shaping machine. 10. Sketch and describe an arrangement for driving the table of a plan- ing machine by means of a screw, so that the table may travel 50 per cent, faster in the return than in the forward or cutting stroke. LECTURE XI. QUESTIONS^ I2Q 11. What is the object of using guide-pulleys in machinery ? Mention Instances of their use, and show how the directions of their axes are ascertained. 12. Describe, with a sketch, the mode of reversing the motion of the table in a planing machine, when a screw is employed to drive the table. 13. A rope transmits 20 horse-power to a rope pulley of 8 feet diameter ; draw a section of the rope in its groove. If the pulley makes 100 revolu- tions per minute, what is the speed of the rope in feet per minute ? What is the difference of the tensile forces in the rope on the two sides of the pulley ? As it is the difference between the tensile forces in a belt or rope that is important for power, why is it necessary to have any pull on the slack side ? ^n*. 2513 ft. per min., and 262-5 lbs - 14. What are cone or speed pulleys ? Describe the use of such pulleys in any machine with which you are acquainted. The spindle of a lathe can, by moving the belt on its cone pulleys, be driven at four hundred revolutions per minute when at its greatest and at 100 revolutions per minute when running at its lowest speed. If the revolutions of the driving shaft are kept constant throughout, and the largest diameter of the speed cones is 20", what must be the diameter of the smallest steps on the pulleys ; the speed pulleys on the two shafts being of the same size ? Sketch the pulleys in position. An*. 10 inches. 15. Upon what does the limiting difference of tensions in the tight and slack sides of a moving belt depend? If the working stress in a belt of sectional area a square inches be /pounds per square inch, and the ratio of the tensions in the tight and slack sides be m, find the horse-power that can be safely transmitted when the speed of the belt is v feet per second. (0. & G., 1903, 0., Sec. A.) 550 1 6. A belt transmits 60 H.P. to a pulley 16 inches in diameter running at 263 revolutions per minute. What is the difference of the tensions on the tight and slack sides 1 Ans. T d - T, = 1796 lbs. (B. of E., 1904-) 17. State the condition that has to be satisfied in order that the same belt may work on two or more pairs of pulleys keyed to parallel shafts (i) when the belt is crossed, (2) when open. Explain also why a belli always climbs to the section of the pulley where the diameter is greatest. (C. & G., 1905, O., Sec. A.) 18. Explain any means with which you are acquainted for determining the brake horse-power of an engine provided with a fly-wheel. A rope is wrapped once round a fly-wheel. One end of the rope carries a weight of 500 lbs., and the other end is led upwards, and is attached to a spring balance. When the revolutions are 105 per minute, the pull in the spring balance varied between 10 and 20 lbs. If the diameter of the wheel be 8 ft., and of the rope I in., find the average brake horse-power. Ant. 39-6 B.H.P. (C. & G., 1005, 0., Se^. A.) 130 LECTURE XII CONTENTS. Velocity Ratio of Two Friction Circular Discs Pitch Surfaces and Pitch Circles Pitch of Teeth in Wheel Gearing Rack and Pinion Velocity Ration in Wheel Gearing Example I. Principle of Work applied to Wheel Gearing Examples II. III. Questions. Velocity Ratio of Two Circular Friction Discs. If two truly centred circular discs or cylindrical rollers, having their shafts parallel to each other and free to turn in fixed bearings, be brought into firm contact ; then, if one of them be driven round, and if there be no slipping, the other one will rotate in the opposite direction with the same circumferential speed or surface velocity (see the next figure). Consequently, their velocity ratio will be in the inverse ratio to their diameters. This may be proved in exactly the same way as we found the velocity ratio of two pulleys driven by a belt in Lecture XI. Let D x = Diameter of the driving disc. F x = Diameter of the following disc. N D = Number of revolutions per minute of D r N, = Number of revolutions per minute of F r Then, The peripheral velocity of D l = Peripheral velocity of F t i.e ..... irD^-irFjlSr^ Or, .... D 1 N D1 = F 1 N F1 i.e. The Driver's diameter x its speed = Follower's diam? x its speed. Speed of the Driver = Diameter of Follower Speed of the Follower Diameter of Driver This velocity ratio may also be proved in the following way : Let the two circles centred at A and B represent a cross flection of the two friction discs in contact at C ; and let them move by rolling contact through the angles 6 and respectively in the same time. Since the magnitude of an angle in circular measure is PITCH SURFACES AND PITCH CIRCLES. 13! always = the length of the arc subtended by the angle at the centre of \he circle -r the radius of the circle. VELOCITY RATIO OF Two CIRCULAR Discs. Then, \ a But, the arc DC = the arc EC since there is no slipping. Consequently, DC The angular velocity of circle A _ 6 r, _ r 3 The angular velocity of circle B < EC r l Or,* The angular velocity or speed of driver, A _ Radius of follower B The angular velocity or speed of follower B Radius of driver A Pitch Surfaces and Pitch Circles. In the case of the two discs or rollers just considered, their cylindrical surfaces are termed the pitch surfaces; and the two circles in the previous figure (which is simply a representation of their cross section, or section in the plane of their rotation) are called the pitch circles. * The angular velocity of a rotating disc is the angle described by its radius in unit time. The relation between angular velocity and linear velocity may be shown thus: Let w=the angular velocity; whilst v = the linear velocity of a point at radius r from the centre of motion when the disc makes * revo- lutions in unit time ; Then wxr=v ; or, w=-; but v=2vrn t 132 LECTURE XII. When the resistance to motion of the follower is great, the discs have to be provided with teeth in order to prevent slipping. Consequently, the pitch surfaces and the pitch circles of such toothed rollers, toothed wheels, or spur wheel and pinion, are the surfaces and the circles of their rolling contact.* Pitch of Teeth in Wheel Gearing. The linear or the circular distance from the centre of one tooth to the centre of the next one, or the distance from one edge of a tooth to the corre- sponding edge of its neighbouring one, as measured on the pitch circle, is termed the pitch of the teeth of a wheel. Let D = Diameter of a wheel or pinion at its pitch circle. p = Pitch of the teeth in the wheel or pinion. n = Number of teeth in the wheel or pinion. ThenvrD = p x n For the circumference of the pitch circle must bo equal to the pitch between any two neighbouring teeth x the number of teeth in the wheel or pinion ; since the pitch between each pair of teeth must be the same all round the pitch circle, otherwise the wheel would not gear properly with any other wheel or pinion of the same pitch. Back and Pinion. If a straight bar of iron be furnished with teeth on one side it is called a rack. It may therefore be con- sidered as a wheel of infinite radius. When a rack has a pinion of the same pitch geared with it, the two form the useful combi- nation termed the rack and pinion. It is employed for moving to and fro the tables of planing machines and large saw benches, as well as for elevating and lowering sluices in dams, &c. Pinion END VIEW. SIDE VIEW. BACK AND PINION APPLIED TO A SAW-MILL TABLE. The accompanying illustrations show the second of these appli- cations, where two parallel racks are fitted to the under side of * When a large toothed wheel gears with a small one, the larger is termed a &pur- wheel and the smaller a pinion. It is not possible in the space allotted to this elementary manual to enter into the best forms of the teeth of different kinds of wheel gearing. This subject is taken up in our "Advanced Text Book," Vol. I., Part II. VELOCITY RATIO IN WHEEL GEARING. 133 two movable tables or platforms. Upon the upper side of one of the tables is laid a log of wood adjusted in the desrred position by wedges. The tables are each carried and guided by four rollers turning on fixed spindles. To the projecting end of the pinion shaft there is fitted a lever handle, so that by merely turning this handle in one direction, the racks, tables, and log of wood are pushed forward upon the projecting circular saw which revolves between the platforms, and if turned in the opposite direction they are drawn backwards. The pinions with their shaft and handle, have no linear motion, for the shaft is simply free to rotate in fixed bearings. The rack and pinion with their handle constitute a modification of the wheel and axle, or lever and winch barrel, where the re- sistance offered by the rack and its load is overcome by a force applied to the handle. Every revolution of the handle turns the pinion, and consequently moves the rack through a linear dis- tance equal to the circumference of the pinion's pitch circle. The principles of moments and of work can therefore be applied to this machine in exactly the same way as we applied them to the wheel and axle and the winch. If P = Pull acting on the handles, R = Radius of handle, r = Radius of pinion's pitch circle, W = Weight or resistance overcome ; Then . . P x 2n-R = W x 2wr P x R = W x r Theoretical advantage . . * . P r W s velocity Velocity Ratio in Wheel Gearing. From what has been said about belt gearing, pitch surfaces, pitch circles, and pitch of teeth, it must be at once apparent to the student that the same rule which was worked out in Lecture XL, in connection with belt gearing, will equally apply to the case of wheel gearing, where there are an equal number of drivers and followers. In the accompanying figure, where there are three drivers and three followers, Let D I} D,, D, = Diameters of the drivers. F p F f , F, = Diameters of the followers. N D , N Pt = Number of revolutions in the same time of th first driver and the last follower. Then, following the same reasoning as was expounded in Lecture XL for the velocity ratio of belt gearing, we have 134 LECTURE XII. The speed of the first driver x ' the successive diameters of the drivers x T> 2 x D 3 Or, . :The speed of the last follower x the successive diameters of the followers. N F3 x F L x F 2 x F 3 Fs/ T^ v 1R^ i X -C X J? - x D 2 X D 3 The speed of first driver _ Product of the diameters of the followers. tThe speed of last follower Product of the diameters of the drivers. In the above equation we may substitute the radii, or the cir- H SIDE VIEW. PLAN. WHEEL GEAKING IN A TEIPLE PUKCHASE WINCH cumferences, or the number of teeth in the drivers and in the followers respectively, for their diameters ; consequently, Let r D , r D3 , r D = Radii of the respective drivers. C D[> C v C D ^ = Circumferences ,, Ti 1 , n D , n 3 = Number of teeth in 3 = Radii of the respective roilowers. /, C F 2 , C F 3 = Circumferences S w_, , n p * = Number of teeth in Then, Or, N x r J), , XT* 1 Dl L x C D x C D x CD = N F x C r x C F x 123 312 PRINCIPLE OF WORK APPLIED TO WIIEEL-GEAKING. 135 Or, N D x nv x W D x n^ = N^ x n > a EXAMPLE I. Three drivers of 10, 20, and 30 teeth each, gear respectively with three followers of 40, 80, and 120 teeth each. Ascertain the velocity ratio between the first driver and the tost follower. By the above formula N D x TI DI x n D ^ x n 3 = N F;j X n ^ x n p ^n r ^ N n Fl x n p x n F Substituting the coire-^ spending numerical I N D _ 40 x 80 x 120 4x4x4 64 values for the letters, j ^ ~~ i x 20 x 30 = i "1 we get } Principle of Work applied to Wheel-gearing. Beferring to the previous figure, it is perfectly evident from the former applications to other machines of the " principle of work," that, neglecting friction, the force applied (to the handles of the machine) x the distance through which it acts, will be equal to the weight raised x the distance through which it is elevated. Let P = Push applied to the handles in Ibs. R = Radius or leverage at which P acts. w W = Weight raised by the rope on the barrel B. r = Radius or leverage with which W acts. Dj, D,, D s = Diameters of the driving wheels. F,, F 8 , F,= Diameters of the following wheels. N D = Number of revolutions of the first driver, "D v or of the handles, H. N P = Number of revolutions in the same time of the last follower, F 3 , or of the barrel, B. Then, by the principle of work and neglecting friction P x its distance * = W x its distance. t.., . P x 2?rR x N D = W x 2?rr x Np (Divide both sides of the equation by 2ir) .% PxRxN D =WxrxN P ~ PxR N F P 8 N Ps xr = or = * It is evident that in order to 'obtain the distance through which P acts, we must multiply the circumference of the circle described by the handles by the number of revolutions they make ; and in the same way the circumference of the fearrel must be multiplied by the revolutions which it makes in the same time, in order to get W's distance. Of THE UNIVERSITY 136 LECTURE XII. But by the previous equation for velocity ratios, N F D, x D 2 x P, ND'-^XF.XF, P x R D! x D 2 x P 8 "W^Tr = F 1 xF a xF, Or, P x R x F! x F 2 x F 3 = W x r x D x x D 2 x D, Hence the general rule for work done in wheel-gearing "Pxits leverage x the diameters (or radii, or circumferences, or number of teeth) of all the followers = W x its leverage x the diameters (or radii, or circumferences, or number of teeth) of all the drivers. EXAMPLE II. It four men exert a constant force of 15 Ibs. each on the handles of a compound crab or winch (such as that illustrated by the previous figure), and if the leverage of the handles is 15", whilst the weight to be raised acts on the barrel or drum at a leverage of 5'', what load will they lift if the respective diameters of the drivers are 12", 20", and 20"; and of the followers, 36'', 80'' and 100", neglecting friction? ANSWER. In this case, P = 4x15 = 60 Ibs.; R=i5"; r=5"; 1^=12"; D 8 =2o"; D 3 =2o"; ^ = 36"; F, = So", and F 3 = 100". By the above formula and by substituting the corresponding numerical values \ve have P x R x F t x F 8 x F 3 = W x r x D L x D, x D, 60 x 15'' x 36" x 80" x ioo"= W x 5" x 12" x 20" x 20" 3345 60 x 4ft xg0x $0x|00 $ x U x x ^0 Or, .... W = 6o x 3 x 3 x 4X 5 = 10,800 Ibs. EXAMPLE III. If 40 % of the force applied to the handles be absorbed in overcoming internal friction in the above example of a winch, what weight can then be raised by the four men, each acting, as before, with a constant force of 15 Ibs. ? ANSWER. If 40 % of the applied force be lost in overcoming friction, then only 60 % is left for effective work, or the efficiency or modulus of the machine is said to be 0-6.* Consequently, 100 : 60 : : 10,800 Ibs. : x Ibs. 60 x 10,800 .. x= -? = 6480 Ibs. 100 * The term modulus of a machine is only another expression for the more ppropriate phrase, efficiency of a machine. LECTURE XII. QUESTIONS. 137 LECTURE XII. QUESTIONS. f . When two circular discs with fixed centres are in firm contact and roll uniformly together, state and prove the rule for estimating their relative speeds of rotation. 2. Define the pitch circle of a toothed wheel. When two pitch circles, A and B, of diameters 2 and 3 respectively, roll together, prove that the angular velocity of A is to that of B as 3 to 2. Three spur wheels, A, B, C, with parallel axes, are in gear. A has 8 teeth, B has 32 teeth, and C has 42 teeth. How many turns will A make upon its axis while C goes round 8 times ? Why is B termed an idle wheel J Ant. 42 turns. (Sec Aote to Question 10 re'ldle Wheel.) 3. What is the pitch of a tooth in a spur wheel ? Two parallel shafts, whose axes are to be as nearly as possible 2 feet 6 inches apart, are to be connected by a pair of spur wheels, so that while the driver runs at 100 revolutions per minute, the follower is required to run at only 25 revolu- tions per minute. Sketch the arrangement, and mark on each wheel its diameter and the number of teeth, supposing the pitch of a tooth to be ii inch. Ans. The follower is 48 inches diameter with 120 teeth. The driver is 12 30 4. Define the "pitch surface" and the "pitch circle" of a toothed wheel. Two parallel axes are at a distance of 10 inches, and they are to rotate with velocities as the numbers 2 and 3 respectively. What should be the diameters of the pitch circles of a pair of wheels which would give this motion. Find pitch of teeth on the smaller wheel if the larger has 24 teeth 1 Ans. 12 ins. and 8 ins. ; 1-57 inch. 5. Sketch and describe the "rack and pinion " and give instances from personal observation of its application. A pinion of 3-2" diameter has teeth of i" pitch, and gears with a straight rack applied to a sluice gate. If the weight of the sluice and rack be 100 Ibs. and the lever handle descnbes a circle of 40*2" in each turn, what force must be applied to the handle to lift the gate ? How many feet will the sluice be lifted by six turns of the handle ? Ans. 25 Ibs. ; 5 ft. 6. Sketch the arrangement known as the rack and pinion. Apply the " principle of moments " and the " principle of work" to find the relation between the force applied and the weight raised by aid of this machine. A pinion has sixteen teeth of |-inch pitch in gear with a rack. If the pinion makes 3^ turns, through what distance has the rack been moved ? If the pinion is turned by a "handle 14 inches long, and with a force of 35 Ibs. applied to the handle, find the force with which the rack is urged lorward. A ns. 49 inches ; 223 Ibs. 7. Deduce the formula for the velocity ratio in wheel gearing where there are three drivers and three followers, and state the rule derived therefrom in general terms. Three drivers of 20, 30, and 40 teeth respectively gear with three followers of 40, 60, and 80 teeth. If the first driver makes 160 revolutions, how many revolutions will the last follower make 1 Ans. 20. 8. In the previous question, if the handles attached to the first driver have each a radius of 15% and the drum connected to the last follower be 15" diameter, what force must be applied to the handles in order that they may lift 1120 Ibs. supposing that the efficiency of the machine is 70 per cent. 1 Ans. 100 Ibs. 9. The hour and minute hands of a clock are on the same arbor or axis, and the hour hand takes its motion from the minute hand. Devise some train of wheels for connecting the two hands. 138 LECTURE XII. QUESTIONS. 10. How would you determine the "pitch circles," and the proper " pitch of the teeth " for a pair of spur-wheels ? What would be the diameter of the pitch circle of a spur-wheel having 80 teeth of f-inch pitch 1 Ant. 19 inches. Three spur-wheels A, B, C are on parallel axes, and are in gear. A has so teeth, B has 35 teeth, and C has 55 teeth. How many revolutions upon its axis will be made by A for every 4 revolutions of C ? Why is B called an idle wheel and what is its use ? Ans. 22 revs. Note re "Idle or Intermediate Wheel" When a wheel is carried on a separate axle and is interposed between two other wheels (or is introduced into a train of wheels), merely for the purpose of changing the relative directions of rotation of the first and last wheel, then such intermediate wheel is called an idle wheel, because it does not affect the numerical value of the train, but only its sign. For examples, see Vol. I. of my " Text Book of Applied Mechanics and Mechanical Engineering," Lectun* HOTES AND QUESTIONS* 139 ( 140 ) LECTUKE XIII. CONTENTS. Single-purchase Winch or Crab Example I. Double-pur- chase Winch or Crab Example II. Wheel Gearing in Jib-Cranes Questions. IN this Lecture we will apply the principles and formulae dis- cussed in the previous one to a few practical applications of gearing in machines for lifting weights. Single- pur chase Winch or Crab. The comparatively small working advantage of the simple hand-driven wheel and axle or SINGLE-PURCHASE WINCH OR CRAB. By Messrs. Loudon Bros., Glasgow. . handle and winch barrel (illustrated in Lecture V.) renders it unfit for lifting greater weights than one or two hundredweight. Con- sequently, whenever heavier loads have to he raised by manual SINGLE-PURCHASE WINCH OR CRAB. 141 labour, one of the most useful machines that can be employed is the single-purchase crab. As will be seen from the accompanying perspective view, this machine consists of a pair of lever handles fitted to the squared ends of a round shaft carrying a pinion. This pinion gears with a spur-wheel keyed to a lower shaft, upon which is also fixed a drum or barrel. To a hook or eye on the inside neck of the left-hand flange of this barrel the rope or chain (to be connected to the load) is attached. Therefore, the turning of the handles causes the barrel to rotate and wind the rope upon it, thereby elevating the load. Both shafts turn in bearings bored in the cast-iron end standards or A frames. These frames are bound tightly together and kept at a fixed distance apart by three wrought-iron collared stays, secured on the outside by screw nuts. To the outside right-hand end of the barrel shaft there is keyed a friction pulley acted on by a steel brake-strap, for the purpose of enabling the labourers to lower a load gently or quickly without enduring the stress and danger of pulling back on the handles. In fact, after applying the brake-strap by its outstanding handle, they can lift the claw pawl which is hinged on the top stay (and which keeps the pinion in gear with the spur- wheel when in the position shown on the figure) and by pulling the upper shaft to the right, disengage the pinion from its wheel. Then, by adjusting the pawl into the other groove of this shaft, they are free to lower the load by the brake without having the handles flying round. Between the right hand flange of the barrel and its neighbouring A frame there is a ratchet-wheel (not seen on the figure). This ratchet-wheel is generally cast along with the barrel. Its pawl, which is hinged to the inner side of the standard, can therefore be dropped down so as to engage with a tooth of the stop-wheel, whenever it is necessary to cease heaving up a heavy weight; thereby preventing the machine overhauling, and giving the labourers freedom to leave the handles and attend to other duties. EXAMPLE I. In a single-purchase crab the lever handles are each 1 6" long, the diameter of the barrel is 8" ; the pinion or driver has 12 teeth, and the wheel or follower 60 teeth. If two mei apply a constant force of 20 Ibs. each to the handles, and are just able to raise a weight of 600 Ibs. to a height of 20 feet in two minutes, find (i) the theoretical advantage; (2) the working advantage ; (3) the work put in for every foot the weight is Hf ted ; (4) the work got out for every foot the weight is lifted ; (5) the efficiency; (6) the percentage efficiency of the machine; (7) the H.P. developed by the two men. ANSWEB. Referring to the notation in last Lecture, we have P = 2 x 20 ibs. 40 ltw. ; R=i6"; r = 4"; n D = 12 teeth; n, = 6o 142 LECTUKE teeth ; W T = the theoretical weight that would be raised if there were no friction j W A = 600 Ibs. (the actual weight raised) ; h = 20 feet. W ( i ) Theoretical advantage = -pr By the principle of work (neglecting friction.) P x by its distance* = W T x its distance.* P X 27rR X n y = W T X 27JT X 71 D P x K, x n = WT x r (Substituting the above numerical values we get) i ft x _, W T 800 20 Consequently, . . -^~ = -- = -T j- 40 * W A 600 Ibs. 15 (2) Working advantage =^p~= 40 lbs . = Y (3) Work put in for every foot W A is raised. From equation (i) we see that for every foot W A is raised P must have gone 20 through 20 feet, since the velocity ratio is .'. P x 20 = 40 Ibs. x 20 = 800 ft.-lbs. (4) Work got out for every foot W A is raised = W A x i' = 6oo Ibs. x i' = 600 ft.-lbs. . _ . Work got out 600 ft.-lbs. (<;) The efficiency *= ^TF r 2 . =o - FT-TI = !$ Work put in 800 ft.-lbs. /;> (6) Tlie percentage efficiency =-75 x 100 = 75% / \ rr-L TT r> j 7 7 T J7 . , Work put in per minute (7) The H.P. developed by the two men = 33,000 -u- v 8000 ft.-lbs. 1 i ^.H.r. = - = bare, or of a horse-power per man. * It is evident that P x i turn of handles Number_of teeth in the driver W T x i turn of barrel = NumbeTof teeth irTthelollower. \)r, . . P x 27rK : W DOUBLE-PURCHASE WINCH OR CRAB. 143 Double-Purchase Winch or Crab. It will be observed, from an inspection of the accompanying photographic view of a " Double- purchase Crab," that the chief difference between it and the single- purchase one is, that it has another pinion and wheel, with a view of increasing the actual or the working advantage, and thus enabling the same manual force to lift a greater load, although by taking a longer time. It is also larger, heavier, and stronger. DOUBLE-PURCHASE WINCH OB CRAB. By Messrs. London Bros., Glasgow. As will be seen from the figure, it may be used as a single-purchase winch by simply lifting the claw-pawl hinged on the top stay, and pushing the handle shaft forward until its left-hand pinion gears with the large spur wheel, and then letting the pawl drop on to bearing to the right hand of the two collars on this shaft. By so doing, the right-hand pinion or first driver (when in double-pur- chase gear) is freed from the first follower, and both are inactive during the time it is used in single purchase, but the second 144 LECTURE XIII. driver is still in gear and is turned round by the spur wheel. The brake strap pulley is keyed to the second shaft (carrying the first follower and second driver), and can be used for lowering the load without the handles coming into action (as described in the pre~ vious case) by placing the claw-pawl between the two collars in the first motion shaft. When the pawl is in this position, both of the pinions on this shaft are out of gear. The machine may be locked arid the load left suspended by dropping the ratchet into the ratchet-wheel cast on the right-hand end of the barrel in the same way as with the single-purchase crab. A triple-purchase winch was illustrated in Lecture XII., and the student should again refer to the plan and the side elevation of its gearing. EXAMPLE II. Four men exert a force of 20 Ibs. each, on the handles of a double-purchase crab, which are 15" long. The driving pinions have 12 teeth each, the followers 24 and 48 teeth respectively, and the diameter of the barrel is 10". Find the weight that can be raised if 25 per cent, of the work put in be absorbed in overcoming friction. ANSWER. Here P = 4X2o = 8o Ibs. ; R = 1 5" ; n DL = 1 2 ; n Dt = 12; 7i Fi = 24; n F> = 48; r=s". By the formula deduced in the previous lecture from the principle of work (neglecting friction), P x R x n f x n p = W T x r x n D xn D 324 After cancelling, we get 8ox3X 2 x 4 = W T = 1920 Ibs. If 1920 Ibs. of work be expended by the men and 25 per cent. of this be lost work, there remains 75 per cent, as useful work. Or, . . 100 : 75 :: 1920 Ibs. : W A . Weight actually raised = W A = 1440 Ibs. Wheel Gearing in Jib Cranes. In Lecture VIII. the side view of a jib crane was given for the purpose of exemplifying the stresses on the jib, tie-rods, and central pillar. We now illustrate A swing jib crane on a bogie and rails, to show that the frame- work and lifting gear are simply those of an inverted double- purchase crab with the toothed wheels placed outside the standards instead of inside as in the ordinary winch. The snatch block pulley (previously referred to in Lecture VII.), to the hook of which the load is attached, doubles the theoretical purchase or advantage of the winch gearing, and therefore one, two or more men can lift nearly double the weight by aid of this simple addition to the machine. Large crones of th'S description are WHEEL GEARING IN JIB CRANES. 145 fitted with slewing or horizontal turning gear, to enable the load when lifted to be swung round before depositing it in a truck. hold of a ship, or on a machine tool. This latter gear consists of a horizontal wheel on the top of the vertical central cast-iron 146 LECTURE supporting boss, with which is geared a bevel pinion, actuated by aid of a lever handle. In order to prevent the whole machine being capsized by a heavy load, there is a back balance weight, and further the bogie wheels can be clamped to the rails. The back balance weight also tends to cancel the severe right angle stress on the central pillar which was specially taken notice of in Lecture VIII. We will defer the description of heavy steam power cranes, tripods and shear legs to our Advanced Course. LBCTUEB XIII. QUESTIONS (continued) 10. What do you understand by the efficiency of a machine, and how is it measured? In a single purchase crab, the pinion has 12 teeth and the wheel has 78 teeth, the diameter of the barrel being 7 inches, and the length of the lever handle 14 inches. It is found that the application of a force of 15 Ibs. at the end of the handle suffices to raise a weight of 280 Ibs. Find the efficiency of the machine. Ans. 072 ; or 72 per cent. jo. In a crane an effort of 122 Ibs. just raises a load of 3265 Ibs. What is the mechanical advantage 1 If the efficiency be 60 per cent., what is the velocity ratio 1 Ans. Mech. Adv. 2676 : i ; Vel. Ratio 44-6 : I. (B. of E., 1903.) 11. In a crane, a force of 3 Ibs. applied at the handle is found to raise a weight of 42 Ibs., and a force of 8 Ibs. a load of 120 Ibs. If the relation between the force applied and the weight raised is represented by the straight line law, obtain the equation expressing the relationship between them ; and if the velocity ratio between the force applied and the weight raised is 18, estimate the efficiency of the crane when lifting a load of 200 Ibs. Am. F=5W/78 + 4/i3 ; 84*6%. (0. & G., 1904, 0., Sec. A.) 12. In an electrically driven overhead crane a weight of 5 tons is raised at the rate of 90 feet per minute. What is the horse-power? Convert this into watts. The motor drives through gearing whose efficiency is 70 per cent. How many amperes of current must be supplied to the motor at a voltage of 220 if the efficiency of the motor is 87 per cent. ? Ans. 30-5 H.P. 22,787 watts ; 170 amperes. [Note that i horse power = 746 watts ; and i ampere multiplied by i volt is i watt.J (B. of E. 1905.) LECTUEE Xm. QUESTIONS. LECTUBE XIII. QUESTIONS. 1. Where wheelwork is employed to modify motion, as in a crane, or in the double-geared headstock of a lathe, how is the change of motion calculated ? Write down the formula employed. 2. Sketch a side elevation and end view of a single purchase crab, and describe the same by aid of an "index to parts." Apply the principle of work in solving the following question : The lever handle of a crab is three times the diameter of the drum, and the wheelwork consists of a pinion of 16 teeth driving a wheel of 80 teeth ; what weight will be lifted by a force of 30 Ibs. acting at the end of the lever handle 1 An*. 900 Ibs. 3. Describe, with a freehand sketch, a single purchase lifting crab. The leverage of the handle of the crab is 16 in., and there is a pinion of 20 teeth driving a wheel of 100 teeth, the diameter of the barrel being 8 in. Assign the relative proportions of the working parts, and estimate the theoretical advantage. What weight would be raised by a man exerting a force of 15 Ibs. on the lever handle, neglecting friction ? Ans. 300 Ibs. 4. A weight of 4 cwt. is raised by a rope which passes round a drum 3 feet in diameter, having on its shaft a toothed wheel also 3 feet in diameter. A pinion, 8 inches in diameter, and driven by a winch-handle 1 6 inches long, gears with the wheel. Find the force to be applied to the winch-handle in order to raise the weight. Ans. 112 Ibs. 5. In a lifting crab the lever handle is 14 inches long, the diameter of the drum is 6 inches, and the wheel and pinion have 57 and n teeth respect- ively. Find the weight in pounds which could be raised by a force of 50 Ibs. applied to the lever handle, friction being neglected. Ans. 1209 Ibs. 6. In a crane there is a train of wheelwork, the first pinion being driven by a lever handle ; and the last wheel being on the same axis as the chain barrel of the crane. The wheelwork consists of a pinion of 1 1 gearing with a wheel of 92, and of a pinion of 12 gearing with a wheel of 72, the diameter of the barrel being 18 inches and that of the circle described by the end of the lever handle being 36 inches ; find the ratio of the pull to the weight raised, friction being neglected. Ans. 1 1 : 1 104. 7. In a 3O-ton crane the tension of the chain as it runs on the winding barrel is 7$ tons, the barrel is 2 feet in effective diameter, and the spur wheel connected with it is 4 feet in diameter on the pitch line ; what pressure will come upon the teeth of the spur wheel, supposing such pressure to act on the pitch line (friction is neglected) ? A n.. 375 tons. 8. The crank of an engine is 2' long, and the diameter of the fly-wheel is icf ; also the fly-wheel has teeth on its rim, and drives a pinion 3' in diameter. If the mean pressure on the crank pin be 7^ tons, what is the mean driving pressure on the teeth of the pinion 1 Ans. 3 tons. 9. Draw to scale a side elevation, end view and plan of a double purchase crab, and describe the same by aid of an " index to parts." If four men each exert a constant force of 15 Ibs. on the handles of such a crab ; if the handles have a leverage of 1 6 inches whilst the barrel is 16 inches diameter, and if the drivers have 12 teeth each while the followers have 24 and 60 teeth respectively ; find the weight which they could balance neglecting friction. If 30 per cent, of the work put in, be taken up in overcoming friction, what load can they lift? State (i) theoretical advantage; (2) working advantage ; (3) work put in when lifting the load i foot ; (4) the work got out ; (5) the percentage efficiency ; (6) the height through which they would lift the load in i minute if each man de- veloped i H.P. AM. 1200 Ibs. ; 840 Ibs.; (i) 20 : i ; (2) 14: 1 (3) 1200 ft. -Ibs. ; (4) 840ft.-lbs-, (5) 70 per cent.; (6) 13- 75 ft. ( '48 ) LECTURE XIV. CONTENTS. Screws The Spiral, Helix, or Ideal Line of a Screw Thread The Screw viewed as an Inclined Plane Characteristics and Con- ditions to be Fulfilled by Screw Threads Different Forms of Screw Threads Whitworth's V-Threads Whitworth's Tables of Standard V-Threads, Nuts and Bolt Heads Seller's V-Thread The Square Thread The Hounded Thread The Buttress Thread Right and Left-hand Screws The Screw Coupling for Railway Carriages Single, Double and Treble Threaded Screws Backlash in Wheel and Screw Gearings Questions. Screws. Every one is more or less familiar with the form and uses of the screw nail for securing pieces of wood together, and of the bolt with its nut for fixing metal plates m position ; but every one is not so familiar with the principle upon which screws are generated and act, or with the best shape to be given to a screw under different circumstances. We shall therefore endeavour in this Lecture to explain these points in an elementary manner, instancing a few examples of the practical applications of screws, but reserving for the following Lecture questions on the work done by screws and their efficiency. The Spiral, Helix, or Ideal Line of a Screw Thread. A very good idea of the form of a screw is obtained from the accom- panying figure, which represents one means of elevating or trans- SPIRAL OK SCREW FOR MOVING GRAIN. ferring grain, flour or other powdered substances from one part of a milling works to another. It consists of a steel band twisted around a cylindrical shaft in a continuous and uniformly pitched spiral. This shaft and screw are placed in a trough, tube or pipe. The grain or powdered substance is fed in at one end of the pipe, and by rotating the screw with a wheel or lever fixed THE SCREW. 149 to one end of the shaft, the loose material is gradually pressed forward until it reaches the other end, from which it may be dropped into sacks or put through another process. It is evident from an inspection of the figure that as the screw is turned round by the lever, the particles of matter are forced along the face of the continuous inclined plane formed by the spiral steel band. The principle upon which the screw acts is, therefore, a combination of tfw inclined plane and the lever. To bring this view of the case still more forcibly before the student, take a cylinder and fix along the side thereof parallel to FORMING A SCREW THREAD ON A CYLINDER, its axis (by gum or drawing pins) a rectangle, ACDE, of paper or white cloth, having its sides, AC and DE exactly equal to the circumference of the cylinder. Then, when the envelope is wound round the cylinder by the turning of the handle, H (in the direction shown by the arrow at P), it exactly covers its cylindrical surface. On the outside of this rectangle when unfolded, draw any con- venient number of parallel inclined black lines, AB, &c., equi- distant from each other as shown by the figure, and again wrap it round the cylinder. These lines will be found to form a con- tinuous spiral, helix, or screw-thread Line from one end cf the cylinder to the other. And the side AC of the right-angled triangle ACB forms the circumference, BC the pitch, AB the length of the thread (for one complete turn of the cylinder), and the angle BAC is the inclination or angle of the screw. The Screw Viewed as an Inclined Plane. Take another cylinder having an evenly pitched screw-thread line drawn upon it. Cut a sheet of flexible cardboard into the form of a right- angled triangle with its height BC or h equal to the pitch (or dis- LECTURE XIV. tance between two consecutive threads when measured parallel to the axis of the cylinder) ; AC or b equal to the circumference of the screw and wrap it round the cylinder, taking care to keep BC parallel to the axis. Then the hypothenuse AB or length I of the inclined plane will coincide with the contour of the screw- thread for one complete turn, and BAG or, a, is the angle of the thread to the plane at right angles to the axis of the cylinder. Now conceive this screw-thread instead of being a mere line to be an inclined plane of known breadth, as in the case of the grain elevator.* Let the total weight of material being urged FIGURE TO PROVE THAT A SCREW THREAD is AN INCLINED PLANE. forward or upwards by the turning of the screw be W Ibs., and let the resistance due to this load be uniformly distributed along the screw thread or inclined plane. Then, comparing the first and the third figures, it is evident that any small portion of the load having a weight ~YV 2 Ibs. will have a corresponding reaction R 2 Ibs., and will require a part P 2 Ibs. (of the total force, P, applied to turn the screw at the radius at which this portion is situated) to move it along the screw-plane against the frictional resistance F r Imagine the work done to be transferred to the inclined plane, AB, then any portion of the load having a weight W t Ibs. will have a corresponding reaction R x Ibs., and will require a part P x Ibs. (of the total force, P, applied parallel to the base to pull the whole load up the inclined plane) to move it along the plane against the frictional resistance F r Now, these forces act in identically the same way as the second case of the inclined plane, which was discussed in Lecture IX., consequently W R AC AC b CB CB h AB AB I * Or, that the screw-thread has a certain depth as measured radially from the axis of cylinder. SCREW THREADS. I$I Or P __ C B _ h eight _ h _ pitch of thread W AC base b circumference of screw. We therefore see that a screw may be treated as an inclined plane where the force turning the screw i.e., overcoming the resistance to motion acts parallel to the base of the incline The same reasoning may be applied to any screw turning in a nut or to a nut turning on a screw. Characteristics of and Conditions to be Fulfilled by Screw Threads. The essential characteristics of a screw-thread are its pitch, depth, and form. The principal conditions to be fulfilled by a screw-thread are : (i) efficiency; (2) strength; (3) durability. (1) The efficiency depends on the pitch and the friction, and hence on the pitch and form of thread. (2) The strength depends upon the form or the shearing thick- ness and depth, or area of the cross section parallel to the axis. (3) The durability depends chiefly on the depth that is, upon the extent of bearing surface. Different Forms of Screw Threads. Sir Joseph Whitworth, the famous tool and gun manufacturer, was so impressed with the great inconvenience and loss of money which arose from the use of different pitches and forms of threads for screws and nuts, that he published the following tables giving the dimensions of what has now become known as the Whitworth standard. Prior to 1841, the year in which Whitworth proposed the adoption of standard sizes for screws, and for several years afterwards, differ- ent engineering works in this country not only used different pitches for screws of the same diameter, but it was no uncommon thing to find a want of uniformity in the same shop. Now, every one in Great Britain and her colonies uses the Whitworth standard sizes for V-threaded bolts and nuts of J-inch and upwards, and the British Association standard for smaller screws in electrical and philosophical instruments. Whitworth's V Thread.* The following figures of a Whit- worth thread and nut, together with the tables, will serve to give full information regarding the number of threads per inch for different diameters of screw-bolts, nuts and bolt-heads, th the right- and the left-hand screws into their respective nuts, which consequently draws the hooks to ward each other, and couples the carriages tightly together. EXAMPLE. If the pitch of each screw is J", the length of the lever arm or distance from the axis of the screw to the centre of the ball is 14"; and if the railway porter pulls the ball with a force of 40 Ibs. when the carriages are brought tightly together, what will be the tension on the screw threads ? ANSWER. Here ;?= J"; 6 = 27rll = 2 x x 14" = 88"; P = 4olbs. The formula for the ratio of P to W in the case of a single screw given in this Lecture is But there are two screws, and for every complete turn made by P, the stress W would be moved through twice tho pitch of one screw or through 2 x J" = i". P x b 40 x 88 .-. W = - = - - = 3520 Ibs. NOTE. We may answer this question directly from the " Principle of Work." Students 'should be trained to work out each question from first principles rather than from formulae ; for, by a too free use of formulae thej are apt to lose sight of principles. Let the lever make one complete turn, then eacli nut will advance along its own screw a distance equal to thepitch. Therefore the two nuts, and con- sequently the two carriages, will be brought nearer by a distance equal to twice, the" pitch, or, = 2 x p. By the principle of work, and neglecting friction Work got out = Work put in Or, . . Wx2/> = Px2irR _Px2irR Or, Single, Double, and Treble-threaded Screws. As has been previously stated, both the efficiency and the forward distance traversed in a single turn of a screw are directly as the pitch of 158 LECTURE XIV. the thread, but the strength is proportional to the area of its cross section. Now, if for any purpose requiring a rapid movement of the nut or of a screw, the pitch must be increased ; and if the screw consisted of a single- threaded square one, where the depth, thickness of the thread, and the width of the groove are each equal to half the pitch, the strength of the shaft upon which the screw is cut would be unnecessarily reduced. If the groove be made shallower and narrower, then two threads with two spaces having the same pitch as the single one, can be cut upon it so as to present about the same area of bearing surface to the pressure and at the same time afford quite as great a shearing thickness without interfering with the velocity ratio.* If a very great velocity ratio should be required, then three or more threads with corresponding grooves may be cut in the shaft and nut. Backlash in Wheel and Screw-Gearings. Backlash is the slackness between the teeth of wheels in gear or between a screw and its nut. Suppose that two wheels are in gear, and that you move one of them in a certain direction until it turns the other, and then reverse the motion ; if you can now move the pitch circle through, say, | inch, before the second wheel responds, this distance is the amount of backlash. In the same way, suppose you turn a screw in one direction until its nut moves, and then reverse the motion, the angle or proportion of a turn which you can now make before the nut responds, is the backlash of the screw and its nut. If a great amount of backlash be present in wheel-gearing, it causes vibration and a disagreeable rattling noise ; and where severe stresses and sudden stoppages are common, the teeth are liable to be stripped. It can only be thoroughly prevented by cutting the teeth most accurately of the best rolling contact form by a tooth-cutting machine. All screws and nuts that are much worked are liable to backlash as they become worn, although when new they may have been very free from it, so that the best way of taking up the slack is to form the nut in two parts with flanges connected by screw-bolts, which may be tightened from time to time so as to take up the wear, and thus keep one side of the threads in one half of the nut, bearing hard against one side of the threads of the screw, and those m the other half against the other Bide. * The screw of the fly-press, figured on p. 249, is a double-threaded one. LECTUBE XIV. QUESTION LECTUEB XIV. QUESTIONS. 1. Explain how a screw is a combination of the lever and inclined plane, and illustrate your remarks. Find the theoretical advantage or ratio of W to P in the case of a screw of I inch pitch and 3-2 inches diameter ; if the lever or spanner key be 7 feet long. Ans. 528 : 1. 2. Given a cylinder and a sheet of paper of sufficient size to cover the cylindrical surface, show how you would trace an evenly pitched spiral or screw line on the cylinder. Mark on your sketch the pitch, circumference, \nd angle of the screw-thread. 3. Trace a screw-thread line on a cylinder. Draw a triangle to repre- sent the pitch, circumference and angle of the thread, and show the direction of all the forces on the supposition that there is a total pressure, Wlbs., on the end of the cylinder acting parallel to its axis and balanced by a force, P Ibs., acting at its circumference in a plane at right angles to the axis, with a total friction of F Ibs. on the screw-thread. 4. What are the essential characteristics of a screw-thread ? Upon which of these do (i) the efficiency, (2) the strength, (3) the durability of a screw depend? 5. Sketch and describe all the forms of screw-threads which you have seen in practice. State their representative advantages and disadvantages, and for which kind of work each kind is most suitable. 6. Define the pitch of a screw. In the Whitworth angular screw-thread, what is the angle made by opposite sides of the thread ? To what extent is the thread rounded off at the top and bottom ? Distinguish between a tingle and a double-threaded screw ; in what cases should the latter be used ? Why are holding down bolts made with angular threads ? 7. Distinguish between a right-handed and a left-handed screw. Sketch the screw-coupling which is commonly used to connect two railway carriages, and explain the action of the combined screws. If the pitch of each acrew is f inch and the lever-arm from the axis of the screw to the centre of the ball is 12 inches, with what force will the carriages be pulled together by a force of 50 Ibs. applied to the ball on the end of the arm ? Ans. 5028 Ibs. 8. Draw a single, double, and treble square-threaded screw to a iVth scale, where the outside diameter of the screw-thread is 10 inches and the pitch 6 inches. Explain the advantages of using a double or treble thread instead of a single one for transmitting rapid motion against a considerable resistance. 9. Why is the angular-threaded Whitworth or Seller's screw better adapted than the square, rounded, or buttress thread for the bolts which are used to bind pieces of machines, &c., together ? 10. What is meant by backlash? How may backlash be prevented in a crew, and in wheel gearing 2 LECTUEE XV. CONTENTS. Efficiency, &c., of a Combined Lever, Screw, andPulleyGear Example I. Bottle Screw- Jack Example II. Traversing Screw- Jack Screw Press for Bales Screw Bench Vice Example III. Endless Screw and Worm- Wheel Combined Pulley, Worm, Worm- Wheel and Winch Drum Worm- Wheel Lifting Gear Example IV. Questions. Efficiency, &c., of a Combined Lever, Screw, and Pulley Gear. Construct an apparatus of the following description, having a horizontal Whitworth V-screw of, say, p" pitch, with cylindrical ends and flanges supported by bearings, so that the screw cannot move longitudinally, but with a nut free to travel from one end of the screw to the other, along a slide or guide APPARATUS FOR DEMONSTRATING THE ACTION AND EFFICIENCY OF SCREW GEAR, INDEX TO PARTS. P represents Pull on pulley rope W represents Weight to be lifted. GP Guide Pulley. N Nut. fci Screw. Radius of pulley. Base or support SCBZT, LEV23, ASL PULLEY GEAR. l6l which prevents it from turning round. Apply a force, P, to a rope passed over the V-grooved pulley of radius, R, keyed to the end of the screw shaft, until it moves the nut with the hook, rope, and weight, W, attached thereto, ao shown by the accom- panying side elevation, plan and end view of the apparatus.* EXAMPLE I. If the radius, R, of tho turning-pulley be 12", the pitch, p, of the screw i", and the gross pull, P, required to lift a weight of TOO Ibs. be 4 Ibs. : find (i) the velocity ratio; (2) the theoretical advantage ; (3) the working advantage ; (4) the work put in to lift W i foot ; (5) the work got out ; (6) the percentage efficiency. ANSWER. We have got in this question all the necessary data required to find the various answers except n, the number of turns which the screw will have to make in order to lift W i foot. Since the pitch of the screw is i", each turn thereof will elevate or lower the weight i", according as it is turned the one way or the other ; consequently, if the screw makes 12 turns, the nut and the weight will move through 12", therefore n=i2 turns. * It is evident that, in addition to the friction between the screw and the nut, there is friction at the several bearings, at the nut slide, and in the bending of the ropes. Consequently, if the student were to place in fucceftsion weights at W of, say, 10, 20, 30, 40 Ibs., &c., and ascertain by aid of a Sailer's spring balance (hooked into the rope which passes round the turn- ing-pulley), the corresponding pulls required to lift these several weights, and to plot down the results on squared paper with the weights as abscissa and the pulls as ordinates, and then to draw a line through the inter- sections of the vertical and horizontal lines drawn through the correspond- ing values, he would obtain a characteristic curve for the friction of the machine as a whole. If he took the precaution to balance the initial friction of the machine (when there was no weight attached at W) by hanging such a small weight at P as would just move the nut towards the turning-pulley, he would find upon repeating the above experiments (keeping the small additional weight on all the time) and replotting the results as now recorded by the spring balance, that the second frictional curve would approach much nearer to a straight line than the former one. In fact, its deviation therefrom would simply prove that the friction of the movable bearing surfaces was not directly proportional to the load. To arrive at the characteristic friction curve for the screw alone, he would have to find out by trial the proportion of the several pulls applied, which were spent in overcoming friction at all other points except between the screw and the nut. To those students who have the time and opportunity for carrying out experiments in applied mechanics, the apparatus illustrated above will prove interesting and instructive. The figures are drawn from the machine constructed in the author's engineering workshop for the purpose of enabling his students to make similar tests to those suggested above. A square, or a rounded, or a buttress-thread may be substituted for the V- Whit worth one, and sound information may thus be obtained about different forms of screws, which will make a stronger and more lasting impression on some students than merely studying books. 1 62 LECTURE XY. By the principle of work : / \ m-i rr j * T> * - P s distance in i turn of driving pulley (1) The Velocity Ratio _, ,. . . . ... jfP^- * v ' W s distance in the same tune ~ _ 0* of pulley _ 2?rB. _ 75 '4 pitch of screw p 1 (2) The Theoretical Ad-\ _ Weight lifted if there were no friction vantage . . I Pull applied ^ p 1 (3) The Working Advan- \ _ W = 100 Ibs. = 25 tage . . J P 4 Ibs. 1 (4) The Work Put in to] y X I 2 (5) The Wart 0* va | = w ,_ lbs ,. fl . raising W i /ooi j (6) 31 - _ Effid cieTicy . J J Work got out w " Work put in X IPO ft.-lbs. # 301-56 ft,ibs: = Bottle Screw-Jack. The importance of the screw as a simple machine for exerting great pressures, is very well ex emplified by the screw-jack. This tool is used for replacing locomotives and railway carriages upon their rails, for elevating heavy girders into position, or for overcoming any great resistance through a small space which cannot be effected by a labourer and a lever. As will be seen from the accompanying figure it consists of a strong hollow bottle-shaped casting, with a projecting handle for facilitating the carrying of the tool from one place to another. In the upper end of the casting a square-threaded screw is cut * It is evident that with such a low percentage efficiency the weight when hanging from the rope will not be able to overhsrul the machine. The student can calculate what pitch of screw would be required with the same co-efficient of friction before overhauling could take place. BOTTLE SCREW-JACK, x 163 parallel with the axis, and into this nut there is fitted a steel screw terminating in a spherical head, having two holes bored through it at right angles to each other. Into one or other of these holes an iron lever bar is fixed, so that by pulling or pushing on the outer end of the bar the screw is turned, and thus the head is gradually raised from the base. To avoid the tearing, grinding action that would ensue between the head and the object acted upon, the former is provided with a loose crown fitted on a central pin projecting from the round head. Let L = Length of the lever arm in inches from centre of jack to where the force is applied. p = Pitch of screw in inches. P=^Pull or push applied at radius L. ft W = Weight lifted or resistance over- come. Then, by the Principle of Work, and neglect- ing friction, we have in one turn of lever P x its distance = W x its distance !* P ~"Wxp -C ' BOTTLE SCKEW- JACK. EXAMPLE II. A weight of 10 tons has to be lifted by a screw- jack, in which the pitch of the screw is J". What length of lever will be required if a force of 70 Ibs. be applied at the end of it ? (i) Neglecting friction ; (2) if the modulus or efficiency of the tool is only '4. ANSWER. (i) By the previous formula (neglecting friction) 1120 L X27T /?0X#X22 22 (2) Taking friction into account we see from the question that the efficiency is = -4, therefore the percentage efficiency is 40, or 60 per cent, of the work put in is lost work required to overcome friction between the screw and its nut. But as the length of the lever is directly proportional to the work put in, the theoretical length of the lever found above is only 40 per cent, of the actual or working length required. 164 LECTURE XV. .-. 40: 1001:25-45 :L 3 L 100x25-45 = 63 . 6 ,, 40 Traversing Screw-Jack. It is very often convenient, when using a strong heavy screw-jack, to be able to move the head a short distance to one side or the other, when near the object to which it is to be applied ; or, after having raised a load with one or more jacks, to be able to traverse the jacks forward or back- ward through a short distance until the load is brought into TRAVERSING SCREW-JACK WITH KATCHET-LEVERS. (By P. & W. MacLellan, Glasgow. ) the desired position. These movements may be effected with a jack of the form shown by the accompanying figure. Further, this jack is provided with a side foot-step attached to and pro- jecting from the lower end of the vertical screw. This foot- step can be placed under the flange of a low beam or rail, where it would be inconvenient or perhaps impossible to get the top head under- neath the same. The nut of the horizontal traversing screw is SCREW PRESS FOR BALES. 165 formed in, or fitted to the bottom of the vertical casting, and this screw is turned by a ratchet-lever which may be slipped on to one or other of the squared ends of its shaft. The upward and down- ward movement of the vertical screw is also affected by a ratchet- lever, and in this case without turning the screw, for the ratchet- wheel is fixed to the nut of its screw. The pawl of the ratchet may be locked on one side or the other, so as to enable the ratchet- wheel and the vertical screw-nut to be turned round in either direction for elevating or lowering the load. Screw Press for Bales. When soft goods or hay have to be transported they may be squeezed into small bulk by means of a SCREW PEESS FOB BALES. (By London Bros., Glasgow.) screw press, and bound firmly when under the press, by strips of hoop-iron passed round them and then riveted before the pressure is relieved. The bound bundle is then termed a bale. The operation will be understood by an inspection of the accompany- ing figure. The loose material is placed in the space between the rigid base and the movable plate of the press, the doors are closed and locked, the pressman applies himself to the end of the lever with a force, P, thereby turning the nut of the screw and forcing the movable plate downwards with a pressure, W, until the i66 LECTURE XV. desired compression of the goods has been attained. The doors are then opened and the strips of hoop iron (which were previously or are now placed in the grooves of the base and upper plate) are brought together and riveted. The lever is then turned in the opposite direction to relieve the pressure, and the bale is removed to the store or ship to make room for another quantity of goods being subjected to a similar action. The same formula as we used for the screw-jack and for Example I. in connection with the combined lever, screw and pulley gear, naturally applies to this press, and to any similar appliance, such as a letter-copying press.* Screw Bench Vice. A bench vice is essentially an instru- ment for seizing and holding firmly any small object whilst it is being acted upon by a chisel, file, drill, saw, or emery cloth, &c. Looking at the figure which illustrates the following example, it will be seen that the vice is a combination of two levers, a square- threaded screw, and a nut. The object to be gripped is placed END VIEW. SIDE VIEW. SCEEW BENCH VICE. between the serrated jaws JJ. The lever handle H, on being turned, forces the screw S into its long nut, and thereby presses forward the outer jaw upon the object, by aid of the flange on the screw-head. This jaw is a lever, having a fulcrum at F, and there- fore the pressure on the object is less than that on the screw- collar in the proportion of SF to OF. The bent flat spring between the limbs of the fixed and movable jaws serves to force the movable jaw away from the fixed one when the screw is turned backwards, and thus relieves the object without having to pull * Kefer to index for page where the illastration of the Fly-press occurs.. The statical pressures produced by this machine when used for punching holes, &c., may be treated in the same way. BENCH YICE. * 167 this jaw back by the hand. It will be observed that the fixed jaw should have been continued to the floor level by a vertical supporting leg, in the case of such a big vice intended for rough heavy engineering work. EXAMPLE III. Sketch an ordinary bench vice. Apply the principle of work to find the gripping force obtained when a man exerts a pressure of 20 Ibs. at the end of a lever 18 inches long, the screw having four threads per inch, the length from the hinge to the screw being 18 inches, and the length from the hinge to the jaws being 24 inches. (S. Juft = Diameters or number of teeth in followers. Then, Or, * What was said in Lectures XII. XIII. and XIV. enables the student to pee clearly the velocity ratio between the cut screw and the leading screw. We need scarcely remind the student that the above statements refer to the pitch of a screw as the distance between two consecutive threads, and not to the number of threads per inch. If the number of threads per inch of its length are taken as the pitch, instead of the distance between two threads, the reverse ratio will hold good. Since a pitch of " means 4 threads to the inch, a pitch of $" means 3 threads to the inch, and a pitch of " means 2 threads to the inch. Or, the number of threads per inch is inversely proportional to the distance between two consecutive threads of the screw. LECTURE XVL When the train of wheels is a compound one, as in this case, the two intermediate multiplying or reducing wheels, / and d v are fixed to any outstanding movable arm or quadrant at the left-hand end of the lathe, so as to bring them into gear with d l and /,. (See second view of the previous figure.) If the train of wheels is a simple one, as in the first, third, and fourth views referred to above, where there is only one driver, d, and one follower,/, with, when necessary, one or more idle pulleys, IP, simply for the purpose of connecting d and /and of giving J the desired direction of rotation, then **-7, JV / Should the pitch of a screw be expressed by the number of threads per inch of its length as is usually the case in tables of screws and change wheels then you can either convert this number into the pitch proper, by taking its reciprocal (i.e., by making the number of threads per inch the denominator of a fraction, with i for the numerator) or you may say Let t c = Threads per inch of screw to be cut. ^ = Threads per inch of guiding screw. Then, since the number of threads per inch are inversely proportional to the distance between any two consecutive threads, * PC Or, . . t x ^ x d a = t x /, x f t If the train is a simple one, then i-=/;or,^ x d-i ff x/ la a EXAMPLE* I. The lathe illustration further on, has a guiding screw of J" pitch, or 4 threads to the inch. Calculate the number of teeth in the change wheel to be fixed to the end of the guiding or leading screw in order to cut a screw of 8 threads td the inch when the driver on the lathe-spindle has 40 teeth. Compare the answer with the change-wheel table printed above the general view of the screw-cutting lathe, further on in this Lecture. ANSWER. Here t e = 8 ; t 4 ; d = 40 ; and you are required to find/. NUMBER OF TEETH IN CHANGE -WHEELS. 179 By above formula, ! =/ ; or, * - .*./~ 8 x * = 80 teeth. t g d ' 4 40 4 By using the previous formula, we have p c = J" and p g = J" ... & * s r,|=^... /= i44o = 8_X P<7 / 4 / 8 4 It is at once evident from this example that you avoid having to multiply and divide by sometimes awkward fractions if you consider the number of threads per inch as the measure of the pitch of the screw, instead of the distance between two threads. EXAMPLE II. The guiding screw of a lathe is %' pitch, and you are required to cut screws of $" and ^V' pitch respectively. Determine the number of teeth in the follower, given the use of a driver having 20 teeth. ANSWER. For a screw of ^" pitch, or 10 threads per inch, and using a driver of 20 teeth, we get by the above formula for a simple train, i_ = / ; or IO - IO X I0 - 100 -/ t g d' '22 10 20 d For a screw of " pitch the number of threads per inch will be 20, and using a driver of 20 teeth, we find from the formula for a compound train i. '_x/ t < d, 20 x 40 Here we multiplied numerator and denominator by 20, in order to obtain suitable wheels, of which d v will have 20 teeth. (See in the previous figure the second of the end views showing change wheels.) Movable Headstock for a Common Lathe. Before describing a complete screw-cutting lathe we will explain the use and construction of this part of a common small lathe for ordinary work. As will be seen from the accompanying rough sketch, it consists of a cast-iron poppet-head planed on its under side, so as to engage the breadth of the top of the shears. It may be bolted thereto in any desired position (along the length of the bed) by an underneath iron plate placed across the shears, and a single vertical bolt. The upper portion of the head is cylindrical, and is bored for about seven-eighths of its length to receive a round i8o LECTURE XVI. hollow steel mandril, M, and for the remaining one-eighth to receive the spindle S. The mandril is fitted in front with a tapered centre, 0, and behind with a screw nut, N". The centre is for carrying one end of the job to be operated upon by the turning tool, and the nut is for engaging the screwed part of the spindle S. On the back end of the spindle there is a collar, c (kept in position by a larger collar or guard, G, with small screws), and a hand-wheel, H\V.* Consequently, by turning this wheel in MOVABLE HEADSTOCK FOR A COMMON LATHE. one direction the mandril and its centre are forced forward, and when moved in the opposite direction they are screwed backwards. To prevent the mandril turning round, it is fitted with a longi- tudinal slot on its underside, into which fits the flattened or rounded end of a small screw, s. A fixing stud, FS, with a handle, enables the mandril to be clamped to the head when it has been adjusted by the hand wheel and screwed spindle. Description of a Screw-cutting Lathe. By the favour of Messrs. John Lang & Sons we are enabled to give a general view, with a complete set of reduced working drawings, carefully indexed to every detail, of the very strong and superior 6-inch centre screw-cutting lathe, lately presented to the Author's Elec- trical Engineering Laboratory and Engineering Workshop by Mr. Andrew Stewart, of Messrs. A. & J. Stewart, and Clydesdale. This lathe weighs, with all its chucks and supernumerary parts, over 1 5 cwt. It has a bed 6 feet long, and admits a bar 3 feet * This arrangement of collar and guard is neither good nor strong, although frequently adopted in the case of small foot-lathes. The collar should be inside the bored head, behind the nut N. MOVABLE HEADSTOCK OF A SCBEW-CUTTING LATHE. l8l 2 inches between its centres. The bed is 9^ inches broad and 6 J inches deep. The gap is 9 inches wide and 6 inches deep ; consequently the lathe can swing a job of 20 inches diameter clear of the leading screw, and one of 24 inches diameter when this screw is withdrawn from its bearings. The speed-cone has three pulleys, each 2 J inches broad, the diameter of the largest being 8 inches and that of the smallest 4 inches. The makers have planed and scraped the bed to a true bearing surface, and have so fixed the gap piece that it cannot wear loose or spring the bed. k _EH_ N\\\\\\\Vy\\\\\\\\\\\V ' , , ._!. *-y - ' -j---^ ^ , END VIEW. LONGITUDINAL SECTION. MOVABLE HEADSTOCK OP SCREW-CUTTING LATHE. INDEX TO PARTS. JBP represents Bottom part. TP Top part. S Spindle, or mandril. SC Steel centre. SS Steel screw. HW Hand wheel. BH represents Binding handle. AS Adjusting screw. ES EH B P Eccentric spindle. Eccentric handle. Bolt for clamping. Plate under B. The movable or loose headstock is gripped to the bed by an eccentric motion worked by a handle, so that it may be instantly clamped in position without the trouble of finding a key to fit the usual nut, and then screwing it gradually home. The upper part of this head, which carries the mandril or spindle, has a side adjustment by means of a side screw, whereby the steel centre may be truly aligned with the corresponding centre of the fast headstock, or it may be moved to the one side or to the other in the 'case of taper turning. A small oil-holder is cast on the back side of the head to facilitate the oiling of the steel centre without having to look for an oil-can. 182 LECTURE XVI. FRAMING, ETC., FOE A SCREW -CUTTETO LATHE. 183 184 FAST HEADSTOCK OF SCREW-CUTTING LATHE. H- 8 1 g^l. o 1S&I a j M su cc >- J5 .3 3 ?, | > ^ &mi * ilil:- * 3 I so ipr^i==i!i t^ii M liiidi^^K-|Uil] rrTr^ r^ C3 H &- Ld h-J WWW DESCRIPTION OF A SCREW-CUTTING ..LATHE. I8 S The spindle of the fast headstock is made of hard crucible steel ground accurately cylindrical, where it fits into parallel gun- metal bearings. These bearings are of extra diameter and length. This spindle is bored hollow for 1 2 inches of its length, in order to admit small rods for making terminals and screws in electrical engineering work. The speed-cone is turned inside and outside, and properly balanced. A specially strong and simple reversing gear has been fitted to the back end of this headstock, whereby the machine-cut steel pinions for turning right and left hand screws may be put into or out of gear by simply depressing or elevating a reversing handle. The back-motion gear is actuated by means of a handle and eccentrics on each end of the back- motion shaft ; whilst the front wheel (or last follower, F 2 , as we have symbolled it in the formula) is locked to the cone or thrown out of gear therewith in the usual way viz., by a bolt fitting into a sliding slot in the cone and a projecting nut on the side of the toothed wheel. The saddle has T slots on its upper side for the purpose of bolting work to it that requires boring out, and which necessitates the removal of the slide rest. A quick hand traverse motion is provided for the saddle by means of a rack and pinion motion, quite independent of the sliding motion of the leading screw. The leading screw is turned to the standard pitch of J inch, or four threads to the inch. The engaging nut is made in halves, so that it may grip the leading screw fairly at the top and bottom of the threads.* * In order to make the construction and action of the split nut which engages the leading screw- clearer, we show here an en- larged view with the halves of the nut, N< *N, slightly apart, and the disc handle re- moved, so as to bring into full view the two eccentric slots, ES, which guide the two steel pins, P and P, fixed on N and N. By comparing this view with the others under heading " Saddle and Slide," the student will see how, by merely turning the disc handle DH the disc D is moved round through nearly a quarter of a circle, and the eccentric slots ES cause the pins, P, P, to move closer to or further away ENLAEGED VIEW OF SPLIT NUT FOB from the centre of the disc D, LEADING SCREW, &c. and consequently move the two parts of the nut, N, N, in or out of gear with the leading screw. I 86 SADDLE AND SLIDE FOR A SCREW-CUTTING LATHE. 1". . oo -H ^3 be far, 02 -g be 3 fl H-IJM .'! ^ S#!3A.S| | I iieisi J 5d PH g 3 fr o . 2 ^s,||l||- g ^^-a-g S).s|.S g 0^*1*3^ p EH tt EH H h t-5 ^5 ^ S , a_r CHANGE-WHEEL GEAR. i8 7 *" r-3 02 ^^ ~" A~* rC ^ be ticS t>o fe i flC S3 i 60 .2 i-i ^-** CD zr ^ W E? QQ vU C ^iliii fe ft 1 88 LECTURE XVI. A compound slide rest is fitted to the top of the saddle, having large bearing surfaces with adjustments for taking up the wear, and a swivel arrangement for conical boring. All the toothed wheels, including the change-wheels, have had their teeth cut directly from the solid casting, by the makers' special tool for that purpose, so that back-lash, and consequently noise and vibration arising from fast-speed driving may be mini- mised as far as possible. The driving shaft has anti-friction steel roller-bearings. It is connected to the foot-treadle at each end by a pulley, chain, and crank. The driving-cone is so stepped that the belt has equal tension on any corresponding pair of driving and driven pulleys. It is sufficiently heavy to act as a fly-wheel. It is balanced along with the treadle to secure an easy, steady drive. A power-drive may be applied if desired, but the author believes that, as students should work in pairs or in sets of three in a laboratory, they will take a deeper interest in their experiments if they have turned out everything by their own skill and labour, than if motive power were freely supplied to them. Of heavy chucks there are a very complete set, including a four- jaw expanding chuck, clement driver, drill chucks for both the fast and loose head spindles, , Friction plate. AP Adjusting pin. FC Friction clutch centre. FBS Friction J^ack shaft. C f Collar or ruff on shaf tFBS SS Sliding shaft. ^2, L 3 Levers for working SS. INDEX TO PARTS. PP represents Pulley for working force pump. LS .. Lathe spindle. RCj, RC 2 ,, Friction ring carriers. G lf G 2 Gluts. TO Toggle carrier. T Toggle. S 15 S 2 Studs. FRj, FRo Friction rings. S 3 , S 4 Shoes for friction r and toggle carrier. Lj Lever for actuating R and RC 2 BS Back gear shaft. B!, B 2 Bearings for LS. B 3 , B 4 Bearings for BS. LECTURE XVI. J 2 . (So202WOo26o^HOWcc isli: tn c5 . ^pq * HOwl ^S N , ** **#%*##? * nti \ / if Jj&J <(n, / HEXAGON TUKRET LATHE. LECTURE XVI ^ J c/2 ^ 02 PQ PQ m 02 PH O 02 W OF THE UNIVERSITY OF HEXAGON TURRET LATHE. STAD SL/DE ADJUST/NO SCREW FUNNEL FOR 1MB ft J GATING 3TADy LIQ CLAMP ffACK P/N/ON SMAfT STOP CLAMP \ WORM StiA, CUTTER CLAMP/MG SCPW* CUTTER SL/DE TOP iNCLIME BODV ELEVATING SCREWS *- ItJ- SUPPORT FACE CUTTER SL/DE G/B SCffHS& CUTTER SL/DE CLAMP THE PATENT ROLLER STEADY TURNER. The following improvements have been made in this lathe : 1. The motor is mounted on a hinged base plate, by which it can be elevated and depressed, and it drives the headstock by means of a belt. The swinging movement of the motor serves to adjust the tension of the belt. 2. The geared headstock provides sixteen speeds within itself, thus enabling a constant speed motor to be used, and renders the lathe equally applicable to direct or alternating-current motors. 3. The driving pulley of the lathe contains an epicyclic reverse gear for giving the spindle speeds in every direction. 4. The feed motion now gives nine feeds instead of three as formerly. 5. The turning tool holders are fitted with roller steadies, enabling much higher cutting speeds to be used than formerly. 6. A scale and adjustable pointer are fitted to the bed to enable direct measurements to be made without using a rule. CHUCKS TURRET-HEAD AND SADDLE. 1 93 of 8 B.H.P. It is a semi-enclosed motor with speed variation obtained by inserting resistance in the field coils from a 60- point shunt controller. Consequently any desired speed is easily got between 1150 and 1850 revolutions per minute. The motor M and spindle MS is attached to the rawhide pinion P,. This pinion drives by means of the compound gear train F v P,, F 2 with P 3 , the first shaft FBS. As will be seen from the sec- tional plan the shaft FBS carries two spur pinions P 3 and P 5 , either of which can be put in or out of gear with follower wheels F 3 and F 5 by means of the friction clutches CS^ CS 2 . The levers L S , L 3 with central spindle SS actuate the friction clutches CS 1? CS 2 . Hence, for every speed of the motor this shaft FBS can give two different speeds to the sleeve or followers F 3 and F 4 on the lathe spindle LS which is driven from shaft FBS. These follower wheels correspond to the cone pulley on an ordinary lathe, and drive the spindle LS direct through the friction ring carrier RC 2 , which slides along the key K, fixed in the lathe spindle LS ; or, through double gearing P 4 , F 4 , BS, P 6 , F 6 and friction ring carrier RCj connected to lathe spindle LS by key K 2 , the change from the one to the other is made by means of the lever L r The friction clutches on the shaft FBS are of the expanding ring type operated by means of wedge surfaces P and rollers R 1? R 2 , whilst the clutches RC 1? R0 3 on the spindle LS are also of the expanding ring type, but actuated by toggles T. (See Index for toggle joints.) The lathe shown by the frontispiece is of a later type. It is driven by a constant speed motor mounted on a hinged base plate. This base plate provides for the necessary tightening of an endless belt. Chucks. The lathe may be fitted either with an automatic or universal chuck. The former chuck is recommended where the bars to be worked are practically straight and cylindrical, whilst the latter is used on bars which are badly out of round or not straight. Looking at the front outside view of lathe, it will be seen how that the bar to be turned is gripped by an automatic chuck worked by the longer handle parallel to and beside L r The automatic chuck may be opened and closed whilst the machine is running, and it has the advantage of holding finished work without bruising. Turret-Head and Saddle. It will be seen from the frontis- plates and the accompanying views, Figs. 2 and 3, that the saddle can be moved either by hand or power, as in the ordinary lathe. In the former method of working, the saddle is moved by nvans 194 LECTURE XVI. of the pilot handle PH acting through wheels SPj, SW : , RP to the rack R; whilst, by the second method, motion is taken from the FIG. 5. THE "COVENTRY" SELF-OPENING Dix HEAD WITH ROUGHING AND FINISHING ATTACHMENT. F IGt 6. TURNING TOOLS VIEW SHOWING RIGHT AND LEFT-HAND CUTTEES. feed or traverse shaft FS through the spur wheels SP 8 and worm W, worm wheel WW, and wheels SP p SW 1? RP X to the rack R. CHANGE WHEEL GEAR, ETtt 195 The worm wheel is carried in a cradle CH, hinged round the shaft WS, and held in position by the stop or trip lever TL and spring SS 2 . Whenever the hand traverse is to be used, the traverse worm W is released by the cradle dropping due to its own weight, after releasing the catch or trip lever TL. When the handle TL is pressed to the right, the bar connected to it turns on a fulcrum, and causes the worm cradle handle CH to be dropped, thus stopping any further traverse of the saddle along the bed of the lathe. This traverse may also be stopped automatically by means of the trip lever TL coming into contact with the stop S, which is set at the desired position on the stop rod SR, and actuating the feed trip lever FTL. The hexagon head can be rotated into any position by turning the handle H, plunger pinion PP, and plunger P 2 , so that P, is freed from the cast steel bush B. Bolted to the faces of the hexagon head are the specially shaped tools, while boring bars may be passed through the circular holes shown in the hexagon head and held firmly by bolts. Change Wheel Gear. It will be seen from the illustrations in Fig. 4 that the traverse or clutch shaft CS is rotated from the mandril through the feed pulley FP, and the pairs of wheels SW^ SW 6 . Any pair of wheels may be put into gear by means of the hand lever HL, which acts upon the internal sliding rod SR, and slides a cotter key SO so as to fix SW 4 , SW 5 or SW 6 to the shaft SR. The handle or reversing lever RL puts the claw clutch RC in gear with bevel pinions BP X or BP, to give either a forward or backward rotation or stoppage of the shaft CS. Consequently, three variations of speed for cutting purposes may be obtained by altering the position of the vertical hand lever HL, while the direction of rotation of the feed shaft may be changed by turning the reversing lever RL. Turning Tool Holders as supplied with the lathe are one of its most important features. A general idea of the different tools on the turret or hexagon head is obtained from the frontis- plate. When another tool is required to be fetched up to the work, this is effected by turning the short handle H seen near top of saddle downwards, thus releasing the turret, which can now be turned as desired, afterwards bringing the handle back to its normal, horizontal position, and firmly fixing the hexagon head. Separate views of the turning tool holder, cut-off tool rest and self-opening die head for the hexagon head are also given, see Fig. 6 on page 194 and plate facing page 193. Rapid Turning with High Speed Steel. The student- will be interested to learn something regarding the recent im- 196 LECTURE XVI. provements which have been made in rapidly turning out work with the most modern electric-driven lathes, and the new kinds of high speed steel. Consequently, we have made a short abstract from three papers read recently before Engineering Societies where these matters were specially dealt with and discussed.* FIG. 7. CUT-OFF TOOL REST SHOWING STEPPED CUT-OFF TOOL. Description of Lathe. The lathe on which all the tests were made was supplied by Armstrong, Whitworth & Co., and was on.e of their 15" centre screw-cutting lathes, taking in 9' 6" between the centres, but for these experiments 18" headstocks were fitted. The fast headstock had both double and treble back * See Proc. Inst. Eng. and Ships, in Scotland, vol. xlvii. 1904, for Mr. Charles Day's Paper on " Experiments with Rapid Cutting Steel Tools." Also see Mr. J. M. Gledhill's paper on "High Speed Tool Steel," read before the Coventry Engineering Society in March 1904, and Mr. P. V. Vernon's paper on " Speeds of Machine Tools," read before the Manchester Association of Engineers on November 14, 1903. EESULTS I9/ gears, the ratios being 14.9 to i and 42.5 to i. The headstock was specially fitted with a 3-step cone suitable for a 6" belt. The lathe was driven by a direct current shunt-wound motor of 120 E.H.P., with a large air-cooled rheostat. The speed of the motor could be varied between 150 and 300 revolutions per minute at no load on the lathe, and from 60 to 300 revolutions with heavy cuts by means of the rheostat. The lathe was driven by two intermediate countershafts having 10" belts. Results. The diagrams show the maximum cutting speeds successfully used in each experiment made with rapid cutting steel tools, and the curves show the average speeds during each set of trials. Figs. 8 and 9 show the average results obtained from those tools which finished in such condition as to warrant atten- tion, whilst the dotted lines in the figures show the maximum results obtained with any tool which finished in a satisfactory condition. It was found that no single make of steel proved to be superior to all others in every respect, but it would appear that the average curves are those which may be taken as standards of all round comparison for use in general engineering shops. The following empirical formulae give approximately the cutting speeds which may be adopted for different areas of cut upon different materials, and the curves show the results obtained therefrom : For soft steel S * g6 + 1 2 A x 0.013 medium steel S = *' 23 + 5 Ax 0.015 , hard steel S= A T ' 77 +5 Ax 0.027 soft cast iron S = + 20 A x 0.02 i-5 Ax 3 hard S= Where: S = Cutting speed in ft. per minute. A = Area of section of cut in sq. inches, i.e., traverse in inches multiplied by the depth of cut in inches, 198 LECTUKE XVI. FIG. 8. VARIATION OF CUTTING SPEED WITH AREA OF CUT STEEL. 3 1 FIG. 9. VARIATION OF CUTTING SPEED WITH AREA OF CUT CAST IRON. POWEB BECORDS CUTTING SPEED RESULTS, ETC. 199 Power Records. Very careful records were taken of the power used for the various cuts and spaces. Also, data were obtained of the forces brought to bear on the cutting tools. Cutting Speed Results. Table I. gives the average results obtained from the tools which finished in good condition. The horse-powers given are the gross horse-powers as calculated from the readings of electrical instruments attached to the motor, and include the motor losses with any countershaft friction. The nei horse-powers required to overcome the resistance to cutting are not given in the Table. These are only required for determining the cutting force on the tool point. Endurance Results. Table II. gives the average result* obtained from soft forg< d steel and medium cast iron, which maintained their average cutting edge in fair condition foi 60 minutes or longer. Comparison of Results. For the purposes of comparing results which may be obtained with the new steels against those obtainable with ordinary Mushet steel and ordinary water- hardened steel, tools made of the.se materials were tested, and the average results are also given in Table II. It will be noted that the new steels give decidedly improved results, and that with them the cutting speed can be about twice as fast as with ordinary Mushet steel, and three or four times as fast as with ordinary water-hardened steel. An item of interest which may be mentioned here is that the ordinary Mushet steel can be very greatly improved by treating it in the same manner as the new steels when tempering. This is a point of value, as it enables greatly improved results to be obtained from existing tools. It will be seen from an inspection of Table I. that, when much metal has to be removed, this may be done not only more quickly, but also with a less expenditure of energy per Ib. of material removed, if a heavy cut is taken at a comparatively low speed in preference to a lighter one at a high speed. Cutting Forces and H.P. for Lathes. The figures showing the cutting force on tool points should prove of great service to machine-tool designers. The information regarding the horse-powers is worthy of special attention, for it is this element which perhaps will form the greatest difficulty in the way of using existing lathes efficiently with the new high-speed cutting tools. A lathe on which a cut of f " x J" on soft steel can be taken is by no means 2OO LECTURE XVI. - Ifi- ON O ^1^*1-* ** 0*^00 Tj-QsO t-. c\ 1-1 vo O oo vn u~> o ro ooo *l- ON *o N r^oo ON ro ^ "-i vo >-> co in o\ to >-i rj-O CO C^ * i-< <* N COVO -i CO Tf w cf N O t-^OO O rj- N r^OOOOOVO N 4J -O . u c ls ^C S v- c O N w OOO vO 10 ON 1-1 * O i-t N ro 10 O >-t O 10 10 t^ O O\OO Tf- O t^OO rf -> r^ N- O ONOO O * N vo CO^O ^d" *O CO IO (O d N ^O C^ *O ^"VO LO W N CO W w M C< M JA|' vo >-osO vo >-osO Tj-iomioroM N rriM LOTJ-VO IOM N row M M N M d d d d d c> o* o* d o* d o" d d d c5 o" 6 d o* d d d d B.2 ooqoqooqqqoo-qqooooooooo 6 d d d d d dodddddoddoo d d d d 6 H.S VOvOMMvOvONN CNVQ O) CJ vO 'O N NOVO M NvDvO M qo M qqMooH-,qo M Mqo-'ooM ddddddddddddddddddddddd 's P.-S oo vc >O t^OO OOVO'J -l OON >-iiOWr.__ tovo VO ^O 1OCO 4 s ** COVO ON t^\O O ONVO ddddddddddddddddddddddd f*} 10 wi-^i-iMTj- csoo t^ Ot^vo 10 10 t-^ rj-oo oo vo co >-i oo oo co rooo ONo to w 5 ^ P 5 Ci N ro ro rf ON N ^ O ~ a. u c ON W fO M X ^C w S."" M OJ : 's 7) : : "S - S C 0) c o> o c 0) a W t3 IE 'O 1 oi "S 5 ^ s > M H 3 S 3 1 '5 0> /a 3 - 1 - O y M P 2 ^ ^ a Q. o > K 0> t> >^ r_j Jj OD rt c5 CO oi ec u ^ j & ( ^ a a Q to T3 'O to ^ T3 w B O : : : g 1 00 s s r jj -^ S. o I X s s a 3 s. s s S i S 202 LECTURE XVI. an abnormal one, and this duty can be carried out on most good lathes of, say, 12" centres, but the driving cones, the countershaft, and the belts connected with few such lathes would be suitable for 24 h.p. Further than this, the line shafts in most engineering shops are too light to drive many lathes using 20 h.p. each, or anything approaching that figure. Example. As a small example of what the Herbert Turret Lathe (see previous figs.) has done with this new high-speed steel, we quote the following from the Proceedings of the Coventry Enginering Society, as found in the paper read by Mr.J. M. Gledhill on March 4, 1904 : Sample No. i was a i" bolt 6" long in shank, with i" round head, i^" deep, and with the point screwed for 2". This was finished complete in 5 minutes 28 seconds, using " A. W. " tools. The following are the details : Reducing i-J- in. bar to i in., 6in. long, 96 cuts per inch . . 4 min. o sees. Screwing ..... 40 Cutting- off .... 28 Idle movements, etc. . . . 20 Total . . 5 min. 28 sees. The above time does not include the facing of the back of the head, which would require J minute. This bolt, made in the ordinary way with ordinary tool-steel, with a good operator, requires 15 minutes. Forging and Hardening the Tools. In forging, annealing, and hardening crucible steels it is essential that the most suitable temperatures should be found for all of these processes, and then accurate means be taken to ensure such temperatures being actually obtained as near as practically possible. This can only be effected by the skilful use of pyrometers or other scientific heat-recorders, for to work on the old-fashioned lines of judging by the eye is no criterion of actual temperature, and is no longer advisable. It is now known that every composition of steel has its own definite temperature that is best suited for obtaining from it the most satisfactory results, and the nearer this can be worked to the better, any deviation from the correct temperature, up or down, involving a corresponding difference in the efficiency of the steel.* * Students may refer here to the illustrated descriptions of these Pyro- meters in the Author's Elementary and Advanced Steam Books. FORGING AND HARDENING THE TOOLS 203 Having obtained a bar of, say, the Armstrong- Whitworth ot " A. W." brand of tool steel, it is necessary to cut off the required lengths ; and this must be done at a forging heat. The lengths must not be broken off cold, as this tends to cause cracks in the bars. For forging, the steel should be placed in the fire, and slowly but thoroughly heated, taking care that the heating has penetrated to the centre of the bar, and then forged at a bright red heat. Whilst forging, the bar should not be allowed to get lower than a good red. After the tool is forged it should be laid down in a dry place and allowed to cool slowly. To harden the tool, the nose only should be raised to a white melting heat and then cooled with an air blast. To obtain the maximum efficiency from this " A. W. " steel it is essential that the nose of the tool shall be raised to a white melting heat as described, for if during this heating the point of the nose becomes fused or melted, no harm whatever has been done. The tool is then ready for use after grinding on a wet stone. Another method which may be described of preparing the tools is as follows : Forge the tool as before, and when cold roughly grind to shape on a dry stone or dry emery wheel. The tool then requires heating to a white heat, just short of melting, and cooling in the air blast. This method also lends itself for cooling the tools in oil; before which the temperature has to be lowered from the white heat to a good red heat (about 1600 F.) either by the air blast or in the open, and the tool then quenched in oil. Tools hardened by the latter method are specially good where the retention of a sharp edge is a desideratum, as in finishing tools, capstan and automatic lathe tools, brass workers' tools, iiF in until the level in the pipe is 7 feet ^ ' ' above the base of the tank. Find the pressure on the vertical side and the reduc- tion of pressure when the water in the pipe is allowed to sink ij feet. (The weight of a cubic foot of water = 62.5 Ibs.) (S. and A. Exam. 1890.) ANSWER. In the first case, Height from e.g. of side to free surface = H t = 5.5'. Area of this vertical side in sq. ft. = A = 3' x 4"= 12 sq. ft. Weight of a cubic foot of water = W =62.5 Ibs. By the above formula, The total pressure P^H^AW. .-.?! = 5.5' x 12x62.5 = 4125 Ibs. In the second case, when the free surface is lowered by i J ft., everything remains the same except the H, which is now reduced from Hj to H 2 = 4'. By the formula, P 3 = H 2 AW. .'. P 2 = 4X 12 x 62.5 = 3000 Ibs. Consequently, the reduction in pressure is the difference between these pressures. Or (P t - P 2 ) - 4125 Ibs. - 3000 Ibs. -1125 Ibs. in inches to those in feet, areas in square inches to those in square feet, and weights of cubic inches to those of cubic feet. This is because the usual units of measurement in hydraulics are feet, square feet, and cubic feet. Before attempting the more difficult questions on page 213, he ihould study a few pages of the next Lecture. LECTURE AVJLL. QUESTIONS, 2 1 3 LECTURE XVII. QUESTIONS. 1. Define the terms liquid, hydro-statics, hydro-dynamics, and hydraulics. 2. Give the chief properties of a liquid, stating wherein it differs from a solid and a gas. 3. Describe and illustrate any experiment, other than the one referred to in this Lecture, to prove the law of transmission of pressure by liquids. State Pascal's law. 4. Describe the nature of fluid pressure. A mass of stone when in water appears to be lighter than when it is situated in the open air. Will you explain the cause of this fact, and state the difference of weight per cubic foot of water displaced ? 5. What is meant by " head" in relation to water supplies for developing power t Give an example. 6. Explain how the pressure on the base of a vessel depends solely upon the area of the base and its depth from the free surface. Illustrate your remarks by showing a series of connected vessels of very different shapes, but with each of their bases of the same size and on the same level, and filled with water to the same height. 7. Sketch and describe Sir Wm. Thomson's wire-testing machine, and explain how such a great force is obtained thereby from such a small quantity of water. 8. How is the pressure of water on a given area ascertained ? A tank, in the form of a cubical box, whose sides are vertical, holds 4 tons of water when quite full ; what is the pressure on its base, and what is the pressure on one of its sides ? Ans. 4 tons ; 2 tons. 9. A water tank is 13 feet square and 4 feet 6 inches deep ; find the pres- sure upon one of the sides when the tank is full. Ans. 8226-56 Ibs. 10. State approximately the increase of pressure to which a diver would be exposed when working at a depth of 50 feet below the surface of fresh water. Ans. About 22 Ibs. per square inch. 11. In the vertical plane side of a tank holding water, there is a rectan- gular plate whose depth is I foot and breadth 2 feet, the upper edge being horizontal, and 8 feet below the surface of the water ; find the pressure on the plate. Ans. 1062-5 Ibs. 12. The base of a rectangular tank for holding water is a square, 16 square feet in area. The sides of the tank are vertical, and it holds 250 gallons of water when quite full. Find the depth of the tank and th pressures on each side and on the base when quite filled with water. Am. 2-5 feet ; 781-25 Ibs. ; 2500 Ibs. 13. A rectangular tank for holding water has a vertical side whose dimensions are 4 feet vertical by 5 feet horizontal. An open pipe is in- serted into the cover of the tank, and water is poured in until the level in the pipe is 10 feet above the base of the tank. Find the pressure on the vertical side and the reduction of pressure when the water in the pipe is allowed to sink 2 feet. Ans. 10,000 Ibs. ; 2500 Ibs. 14. A gauge in a water pipe indicates a pressure of water equal to 40 Ibs. on the square inch. What is the depth of the point below the free surface ? Sketch and explain the action of some form of gauge suitable for the above purpose. Ant. 92*16 ft. FRESH LECTURE XVIII. CONTENTS. Useful Data regarding Fresh and Salt Water Examples I. II. III. IV. Centre of Pressure Immersion of Solids Law of Archimedes Floating Bodies Example V. Atmospheric Pressure The Mercurial Barometer Example VI. Low Pressure and Vacuum Water Gauges Example VII. The Siphon Distinction between Solids, Liquids and Gases Definitions of perfect, viscous, and elastic Fluids Cohesion Questions. tTseful Data regarding Fresh, and Salt Water. We will commence this Lecture by giving some useful data regarding the weights, = of water displaced by block, represented by abed. x = amount by which the water rises in the tank when the block is immersed. Then clearly, V 2 = "V\ + v. Or V.-V^v. N ow V a - V x = volume represented by A^BjBjA,, = cross sectional area of tank x x t = 4 2 xa;=i6aj cub. ft. i6x = v or x= ^- ft. 16 But, by the principle of Archimedes we know that The, weight of water displaced by block = The weight of the block. s. = i25 Ibs. v = = 2 cub. ft. x = -^ = - ft. = 1 J inches. Next, we have to find ^e pressure on one of the vertical sides of the tank. Here the depth of the centre of gravity of the area of the side subjected to pressure below the free surface of the water is Total pressure on side = P = HAW Or, ... P = 1220-7 Ibs. Atmospheric Pressure. Surrounding the earth's surface there is a deep belt of air, which gets rarer and lighter the higher we rise from the earth. If we consider the case of a com- plete vertical column of this air, we find that it produces an average pressure on the earth's surface of about 15 Ibs. ; or, in- other words, we say that the atmosphere produces an average pressure of 1 5 Ibs. on the square inch, for we find that it will balance a vertical column of mercury of about 30 inches, or a vertical column of water of 34 feet. We do not experience any inconvenience from this normal pressure of the atmosphere,. because we are so constituted as to be able to resist it. Should we r however, enter the closed compressed air-chamber of the under- ground workings of a railway tunnel (such as those in operation- THE MERCURIAL BAROMETER. 221 at the present time for the construction of the London Tube Railway), or the caissons of a great bridge while they are being sunk (as in the case of the Forth Bridge), or go down into the sea in a diving-dress or diving-bell, then we do feel a most uncom- fortable sensation in our ears, eyes, or nearly -{- of a Ib. per sq. in. The Siphon is simply a bent tube for withdrawing liquids from a higher to a lower level by aid of the atmospheric pressure. It is used in chemical laboratories and works for emptying acids THE SIPHON. , 223 from carboys, in breweries and distilleries for extracting beer from vats and spirits from casks, in the crinal glass tube of Lord Kelvin's recorder for conveying ink from the ink-pot to the telegraph message-paper ; and on a large scale for draining low-lying districts, such as the fens of Lincolnshire. The conditions for the success- ful working of a siphon are, that 1. The liquid shall be carried THE SIPHON. by the outer limb of the tube to a lower level than the surface of the supply. 2. The vertical height from the free surface of the liquid being drained to the top of the bend of the siphon shall not be greater than the height of the water barometer at the time say only 30 feet on account of the necessary deduction of 3 or 4 feet to be made from the full height of 34 feet, due to having to overcome the friction of the pipe. 3. The end of the siphon dipping into the liquid to be drained, shall not become uncovered. To start the siphon, either the tube must be filled with liquid, the ends closed, and the siphon inverted, with the shorter limb under the fluid to be drained, before uncovering the ends ; or, whilst the end of the shorter limb is in the liquid a partial vacuum must be formed in the siphon tube by extracting the air from the end of the longer leg. The principle upon which the siphon acts is as follows : A vacuum having been formed in the tube, the pressure of the atmosphere acting on the free surface of the liquid to be drained, forces it up the shorter limb, and having turned the highest point of the n the liquid descends the longer limb by the action of gravity with a velocity proportional to the J 'difference of levels between the outlet and the free surface of the source of supply. The outflowing liquid is always acting as a water-tight piston at the bend of the f|, and in this way keeping up the vacuum there, until either the inlet and the outlet free surfaces come to a level (when the siphon stops for want of " head "), or, when the difference of level between the free surface of the supply and the top of the bend exceeds the height supportable by the atmosphere, when it stops for want of breath or atmospheric pressure. p 224 LECTURE XVIII. .Distinction between Solids, Liquids, and Gases. At the very commencement of this book we referred to the fact that Matter exists under three conditions. (i) Solids ; (2) Liquids ; (3) Gases. We shall now define and distin- guish concisely between the three states of matter. (1) A Solid is matter in such a condition, that the molecules cannot move freely amongst themselves, and consequently it retains its shape and volume unless acted upon by a force. (2) A Liquid is a collection of inter-mobile particles of matter, which offer great resistance to change of volume, but little to change of shape. (3) A Gas is matter in its most subdivided state, and which readily yields to the slightest force tending to change its shape or its volume. We thus see that the chief characteristic distinctions between these three states of matter are, that (1) A Solid resists both change of shape and of volume. (2) A Liquid only resists change of volume. (3) A Gas resists neither change of shape nor of volume. (4) A Fluid may be either a liquid or a gas. (5) A Viscous Fluid is a liquid which offers more or less resistance to motion amongst its particles, e.g., treacle, tar, and heavy oils, &c. (6) An Elastic Fluid is a gas whose volume will increase indefinitely. (7) Cohesion is a property of matter common to both solids and liquids. It causes more or less resistance to the separation of the molecules of matter. LECTURE XVIII. QUESTIONS. 225 LECTURE XVIII. QUESTIONS. 1. What are the respective specific gravities and the weights per cubic foot and per gallon of fresh and of salt water ? 2. A cylindrical vessel, 120 inches long and 10 inches in diameter, is sunk vertically in water, so that the base, which is horizontal, is at a depth of 100 inches below the surface of the water. Find the upward pressure in pounds on the base of the vessel. Ans. 284-2 Ibs. 3. A cubical box or tank with a closed lid, the length of a side of which is 5 feet, rests with its base horizontal, and an open vertical pipe enters one of its sides by an elbow. The tank is full of fresh water, and the pipe contains water to the height of 10 feet above the top of the tank. What are the pressures of water on the top, bottom, and sides of the tank 1 Ans. 15,625 Ibs. ; 23,437.5 Ibs. ; 19,531-25 Ibs. 4. A water tank 10' long, 10' wide, with an inclined base 10' deep at one end and 5* at the other end, is filled with fresh water. Find the pressure in pounds on each of the four sides and on the base. Ans. 31,250 Ibs. ; 7,812-5 Ibs. ; 18,229-16 Ibs. ; 52,500 Ibs. ; 10,421 *S Ibs.* 5. A lock gate is 12 feet wide, and the water rises to a height of 8 feet from the bottom of the gate. What pressure in pounds does it sustain ? The weight of a cubic foot of water is 62^ Ibs. Ans. 24,000 Ibs. 6. A vertical rectangular sluice gate, measuring 2 feet horizontal by 3 feet vertical, is immersed so that its upper side is 4 feet below the surface of the water pressing on it. Find the pressure on tbe gate : you are required to explain the reasoning on which your calculation is founded., Ant. 2062*5 Ibs. 7. What is meant by the " centre of pressure " in the case of a plane surface immersed in & liquid ? If the plane be a horizontal circle, where does the centre of pressure act ? If it be a vertical rectangle 10 feet wide and 6 feet deep, immersed in water so that the upper edge of the rectangle Is flush with the surface of the water, where does the " centre of pressure " act ? Ans. at the centre of the circle ; 4 feet below surface of water. 8. State the law discovered by Archimedes, and the conditions for a bodj In equilibrium floating in a liquid. A cy Under 10 feet long and 2 feet in diameter floats in fresh water, with 2 feet prelecting from the surface ; find tbe weight of the cylinder. Ans. 1571 Ibs. 9. A rectangular tank, 5 feet square, is filled with water to a height of 7 f feet. A rectangular block of wood, weighing 312-5 Ibs., and having a sec- tional area of 5 square feet, is placed in the tank, and floats with its sH*s vertical and with its section horizontal. How much does the water rise in the tank, and what is now the pressure on one vertical side of the tank I Ans. 2-4 inches ; 9875-4.^)8. 10. The mercurial barometer registers 31"; calculate the height of columns Df fresh and of salt water that will balance the corresponding pressure. Ans. 35-13 ft., 34-24 ft. n. Sketch and describe a mercurial barometer. State how it is made, and how it acts as a register of the pressure of the atmosphere. 12. Describe some simple form of gauge which would enable you to measure the pressure at which gas is supplied, and explain the principle on which it is constructed. * In the answers given, ^12$ is assumed to be 11*3. 226 LECTURE XVIII. QUESTIONS. 13. Sketch and explain the action of the siphon, and give a few practical examples of its use. Also state under what circumstances it fails to work. 14. The bottom of a water-tank measures f in length and 3' 4" in width. When the tank contains 900 gallons of water, what will be the depth of the water, and what would be the pressure on the bottom, on each side and. end of the tank respectively ? One gallon of water weighs 10 Ibs. One cubic foot weighs 62-3 Ibs. Ans. Pressure on bottom of tank = 9000 Ibs. Pressure on each side = 8382 Ibs. Pressure on one end of tank - 3991 Ibs. Depth of centre of pressure is 6*2 feet. 15. Draw the diagram of water-pressure on the side of a tank wi h vertical sides, 12 feet high, and filled with water. Deduce the vertical depth of the centre of pressure below the top edge of the tank. 16. Name the chief physical properties of a liquid, and show in what respect a liquid differs from a gas and from a solid. How is the pressure of water on the vertical sides of a tank calculated ? A water-tank is 10' long, 10' wide, and 10' deep. When it is filled with water, what will be the force with which the water acts on one side of the tank 1 Ans. 31,250 Ibs. 17. Describe how you would carry out an experiment to determine the discharge of water through a sharp edged circular orifice on the sid of a water tank. (B. of B., 1904.) LECTURE XIX. CONTENTS. Hydraulic Machines The^Common Suction Pump Example I. The Plunger, or Single-acting Force Pump Example II. Force Pump with Air Vessel Continuous-delivery Single-acting Force Pump without Air Vessel Combined Plnnger and Bucket Pump Double-acting Force-Pump Example III. Centrifugal Pumps Example IV. Questions. Hydraulic Machines. The Common Suction Pump consists of a bored cast-iron barrel PB, terminating in a suction pipe, SP, fitted with a perforated end or rose R, which dips into the well from which the water is to be drawn. The object of the rose is to prevent leaves or other matter getting into the pump, that might clog and spoil the action of the valves. At the junction between the barrel and suction pipe there is fitted a suction valve SY, of the hinged clack type faced with leather. The piston or bucket B is worked up and down in the barrel of the pump by the force P, applied to the end of the handle H, being commu- nicated to it through the connecting link of the hinged piston- rod PR. In the centre and at the top of the bucket is fixed the clack delivery valve DY, which is also faced with leather in order to make it water-tight. The bucket is sometimes packed with leather ; but, as shown by the figure, a coil of tightly woven flax rope wrapped round the packing groove would be more suitable in the present instance. Action of the Suction Pump. (i) Let the barrel and the suc- tion pipe be filled with air down to the water-line, and let the bucket be at the end of the down stroke. Now raise the bucket to the end of the up-stroke by depressing the pump handle. This tends to create a vacuum below DY ; therefore the air which filled the suction pipe opens SY, expands, and fills the additional volume of the barrel. Consequently, according to Boyle's law, its pres sure must be diminished in the inverse ratio to the enlargement of its volume.* This enables the pressure of the atmosphere * The student may refer to Lecture XII. of the Author's Elementary Manual on " Steam and the Steam Engine," for an explanation and demon- stration of Boyle's law ; where it is shown that if p=the pressure of a gas and t>=its volume, then at a uniform temperature pv=o constant, or p varies as. 228 LECTURE XIX. (which acts constantly on the surface of the water in the well) to force a certain quantity of water up^the suction pipe, until the weight of this column of water and the pressure of the air (between it and the delivery valves) balance the pressure of the outside atmosphere. COMMON SUCTION PUMP. INDEX TO PARTS. H represents Handle. P Push or pull at A. F Fulcrum of H. PR Plunger rod. IB Pump barrel. S Spout. SP represents Suction pipe. R Rose. SV B DV Suction valve. Bucket or piston. Delivery valve. (2) In pressing the bucket to the bottom of the barrel by elevating the handle, the suction valve closes and the delivery valve opens, thereby permitting the compressed air in the barrel to escape through the delivery valve into the atmosphere. (3) Raise and depress the piston several times so as to produce the above actions over again, and thus gradually diminish the volume of the air in the pump to a minimum. Then water will have been forced by the pressure of the atmosphere up the suction pipe and into the pump, if the bucket and the valves are tight, SINGLE ACTING POECE PUMP. 2 29 and if the delivery valve when at the top of its stroke be not more than the height of the hydro-barometric column above the water line of the well.* (4) The bucket now works in water instead of in air. In fact, the machine passes from being an air-pump to be a water one. During the down-stroke water is forced through the delivery valve. During the up-stroke this water is ejected through the spout ; at the same time more water is forced up through suction pipe and valve to supply the place of the vacuum created by the receding piston. The water is therefore discharged only during the up-stroke in the case of the pump illustrated by the figure. Should it, however, be fitted with an air-tight piston-rod and pump cover, and should the pump handle be moved rapidly, more water will be taken into the barrel than can escape from the spout during the up-stroke. Consequently, the compression of the pent- up air between the surface of the water in the barrel and the cover, will cause the water to flow out in a more or less continuous stream during the down-stroke. In other words, the top cover and the portion of the pump above the spout may be converted into an air vessel, the precise action of which will be explained later on. EXAMPLE I. If the cross area of the bucket of a suction pump be 20 sq. in. and if water be raised 24 ft. from its surface in the well, what is the pull on the pump rod ? ANSWER. The pull P on the pump rod is evidently equal to the weight of a column of water of height H=24 ft., and the area of the bucket in sq. ft. = A= 204- 144. Therefore, by the formula employed for the pressure of a liquid on a surface in Lectures XVII. and XVIII. P = 24 'x x 62-5 = 208 Ibs. 144 The Plunger, or Single-acting Force Pump. The upper or outer end of the barrel of this pump is provided with a stuffing- box and gland, through the air-tight packing of which the solid pump plunger works. During the up or outward stroke of the plunger a vacuum is * Theoretically, such a pump should be able to lift water from a depth of 34 feet below the highest part of the stroke of the delivery valve, but practically, owing to the imperfectly air-tight fitting of the piston and the valves, it is not used for withdrawing water from wells more than 20 to 25 feet below this position of the delivery valve. In fact, such a pump fre- quently requires a bucket or two of water to be poured into it above the delivery valve in order to make it work at all, if it should have been left for some time without being worked. 230 LECTURE XIX. created in the pump barrel, and consequently air is expanded into it from the suction pipe. This pipe is attached to the flange of the suction valve-box. During the down or inward stroke the suction valve closes, and the pent-up air in the barrel is forced through the delivery valve. This action goes on precisely in the manner just explained in the case of the suction pump, until the water rises into the barrel. Then the inward stroke of the plunger drives water through the delivery valve to any desired height (or against any reasonable back pressure, as in the case of a feed THE PLUNGER FORCE PUMP. SV represents Suction valve. DV Delivery valve. INDEX TO PARTS. PB PP Ch Checks for valves. SB and G represents Pump barrel. , Pump plunger. , Stuffing box and gland. pump for a steam boiler) consistent with the strength of the pump and the power applied. The eye of the plunger may be attached to a connecting-rod actuated by a hand lever, as in the case of the suction pump, or it may be worked from one eccentric or crank revolved by a steam engine or other motor. By whichever way it is worked, the force applied to the plunger must be sufficient to overcome the friction between the plunger FOECE PUMP WITH AIR VESSEL. 231 and the packing, the resistance due to sucking the water from the source of supply, and of driving the same up to the place where it is delivered. With this pump (as in the case of the suction pump), the water is only delivered during one out of every two strokes of the plunger, and consequently, in an intermittent or pulsating fashion. In order to make the supply continuous we have to use one or other of the devices about to be described. EXAMPLE II. In a single-acting plunger force pump the cross area of the plunger is 10 sq. in., and its distance from the surface of the water in the well, when at the end of its outv* a~d or suction stroke, is 20 ft. During the inward stroke the water is pumped up to a height of 100 ft. above the end of the plunger. What forces are required to move the pump plunger during (i) an " out," and (2) an in-stroke (neglecting the forces to overcome friction). ANSWER. (i) P x = H X AW =20' 10 x x 62*5 = 86-8 Ibs. pull. 144 10 (2) P 8 = H,AW = ioc/ x x 62-5 = 434 Ibs. pressure. Force Pump with Air Vessel. In the following figure of a force pump the only points of difference worth noticing between FOECE PUMP WITH AIR VESSEL. INDEX TO PARTS. SP represents Suction pipe. SV B PB G Suction valve. Barrel of pump. Plunger barrel. Packing gland. Plunger rod. DV represents Delivery valve. S AV AC DP Stop for DV. Air vessel. Air cock. Delivery pipe. 232 LECTURE XIX. it and the previous one are : (i) The plunger, instead of being solid, is a hollow trunk or barrel, with the connecting rod fixed to an eye-bolt at its lower end. (2) The suction and the delivery valves are both at one side, instead of being fixed on opposite sides of the pump. (3) There is an air vessel. Action of the Air Vessel. During the inward or delivery stroke of the plunger, part of the water forced from the barrel goes up the delivery pipe, and the remainder enters the air vessel, and consequently compresses the air in AV. During the out- ward or non-delivery stroke of the plunger the compressed air in the air vessel presses the rest of the water into the delivery pipe. In this simple way a continuous flow of water is main- tained in the delivery pipe, and with far less shock, jar, and noise than in the previous case. Where very smooth working is re- quired, an air vessel is also put on to the suction side of the pump. Should the air in the air vessel become entirely absorbed by the water, the fact will be noticed at once, by the noise and the intermittent delivery. Then the pump should be stopped, the air cock AC opened, and the water run out. When the air vessel is full of air, the air cock should be shut and the pump started again. Continuous-delivery Pump without Air Vessel. A fairly continuous delivery may be obtained by making the plunger of the piston form, and the pump rod exactly half its area, as shown by the accompanying figure. During the down stroke, half the water expelled by the piston from the under side of the pump barrel goes up the delivery pipe, and the other half is lodged above the piston, to be in turn sent up the delivery pipe during the up-stroke. Where very high pressures are required, such as in the filling of an accumulator ram, pumps working on this principle, but of the following form, are frequently used. The *>- ' , Direction of flow. or circulating the cooling water through the condenser tubes of steam-engines, as well as for dredging soft- bottomed rivers. The original type of centrif ugafpump had straight radial blades, but it has been found, that if these are curved in the direction and manner shown by the accompanying figure, there is less shock due to the quick flowing water and greater efficiency. They may be driven by belts, direct coupled steam-engfnes, turbines, or electric motors as preferred. V BON-ACCORD CENTRIFUGAL PUMP. 237 Details. As will be seen from the vertical cross-section and side view, the chamber consists of a snail-like outer pump casing P C, supported upon two flanged feet S F, connected to a suction pipe S P, and a delivery pipe D P. In the centre is fitted the shaft S, which carries an impeller wheel I W, that rotates between the tapered inside faces of the pump casing P C, and a removable side cover S C. Action when circulating cold water through a steam condenser. Should this pump be situated below the level of the supply water, the air is driven out of the pump and its pipes by this head of water. In such a case, the pump can be started straight away by its motor. But, where the pump is situated above the suction supply, then the mere rotation of the impeller wheel I W does not produce a sufficient vacuum to make the water rise into it ; and consequently, the pump casing has either to be filled with water through the nipple hole (beside the lifting eye-bolt) or a steam ejector with a sluice valve are added in certain cases. Supposing that the pump is fairly started, then the mere rota- tion of I W inside the water-tight casing not only gives the kinetic energy and pressure to the water contained therein, to force the same right through the delivery pipe D P and the condenser tubes, butalso tokeepup the necessary vacuum in the purnp,so as to ensure a continuous feed of water through the suction pipe. It will be observed, thf.t the incoming water is divided by the sharp, knife- edged portion of cast iron, at the volute V, and that it flows equally up each side to the centre of the wheel, whereby the same is subjected to balanced side pressures. Should the interior of the pump require to be inspected, the attendant may first open the cleaning door C D, but if he finds that any adjustment is required, then he can take off the side cover S C. When this is removed, he will obtain a clear view of the whole of I W, and he may remove the same from the tapered end of the shaft S, by unscrewing the nut N. 238 LECTURE XIX. EXAMPLE IV. A centrifugal pump is to lift 6'2 cubic feet of water per second to a height of 7 feet ; how much horse-power must be supplied to it if its efficiency is 45 per cent ? It is direct-driven by a continuous current electro-motor whic v works at 200 volts. How many amperes of current must be supplied to the motor, if its efficiency is 85 per cent ? (B. of E., 1904.) Useful work done) TTT i , * by pump in lifting L = Wei g ht of w ^r ln , lbs ' x dlstance m ft - throu S V the water J which it is raised. =6-2 x 62-5 x 7 = 2712-5 ft.-lbs. But, Efficiency of \ _ Useful work done by centrifugal pump in ft.-lbs. Pump, ij f J Total work done in driving the centrifugal pump. 45 ) 2 7 I2 '5 ft.-lbs. 100 J ~~ Total work done. .-.Total work done j 2?I2 . 5X in driving Cen- j- = - ft.-lbs. trifugal Pump J 45 And this work is done per second. .*. H.P. required J 2 7i2' r x 100 to drive the Cen- ]-= =10'96H.P. trifugal Pump J Also i H.P. = 746 watts, and i watt = i volt x i ampere. Hence. Watts) , givenoutby Motor) = Io '9 6 * 746- watts. And, Efficiency of) Watts given out. Motor J ~~ Watts taken in. 85 ^ 10-96 x 746 100 J ^ Watts taken in. .*. Watts taken in) 10-96 x 746 x 100 by Motor /" "~8j~ But, Watts taken) n -n -rr ^ in by Motor ) = Currenfc m Amperes x Pressure in Yolte. .% Current in am-) _ Watts taken in by Motor, peres J ~~ Pressure in Yolts, }_ io - 96 x 746 x 100 85 x 200 4088 -5 =48-1 Amperes. LECTUEE XIX. QUESTION*. 239 LECTURE XIX. QUESTIONS. 1. Explain the manner in which the pressure of the atmosphere Is mada serviceable in the case of the common suction pump. Sketch and explaip by an index the details of this pump. 2. Describe, with a sketch, an ordinary suction or lifting pump, and ex plain its action. If the diameter of the bucket is 4", and the spout is 20? above the free surface of the well, what is the tension on the pump-rod in the tsp-stroke ? Arts. 109 Ibs. 3. Sketch and describe a force pump, drawing a section so as to show the packing of the plunger and the construction of the valves. How is an air-vessel applied to such a pump ? Why is the air-vessel dispensed with when pumping water into an accumulator ? 4. Explain the use of an air-vessel in connection with a force pump. Sketch a section through a double-acting force pump, showing the valves and the connection of the pump with the air-vessel, and explain the action of the pump. Water is forced p to 100 feet above the air-vessel ; what proportion of the volume of the air-vessel is occupied with water, and what is the pressure of the air therein? Ans. 74*6 per cent. ; 43*35 Ibs. per sq. in. above the atmospheric pressure. 5. The leverage to the end of the handle of a common force pump is five times that to the plunger, ana the area of the plunger is 5 square inches ; what pressure at the end of the lever handle will produce a pressure of 45 Ibs. per square inch on the water within the barrel 1 Ans. 45 Ibs. 6. A force pump is used to raise water from a well to a tank. The piston has a diameter of r6", and is 20' above the free surface of the water in the well, and 40' below the mouth of the delivery pipe leading into the tank. Find the force required to work the pump (i) Neglecting friction; (2) when 30% is spent in overcoming friction ; (a) when sucking, (6) when forcing, (c) what is the work put in and got out per double stroke of 6" t Ans. (a) (i) 17-45 Ibs- 5 (2) 24-93 Ibs. ; (&) (i) 34'9 Ibs. ; (2) 49-86 Ibs. ; (c) 37-39 ft.-lbs. ; 26-17 ft. -Ibs. 7. What is the difference between a double-acting and a single-acting pump ? The area of the plunger of a force pump being 3 square inches, find the pressure upon it when water is forced up to a height of 2of. Ans. 26-04 Ibs. 8. Describe, with a sketch, some form of pump which will deliver half the contents of the barrel at each respective up-stroke and down-stroke of the pump-rod. Name the valves. 9. Sketch and describe a "double-acting force pump." If the diameter of the piston be 12", the stroke 3', the distance from pump to well 2d, from pump to position for delivering the water 40', and if the number of strokes per minute be 40, what is (i) the theoretical horse-power required to work the pump, (2) the actual, if 30 per cent, of the power be spent against friction. Ans. (i) 1071 ; (2) 15-3. 10. What is the difference between a single and a double acting pressure pump ? Sketch in section a double-acting force pump for working at high pressure, showing the arrangement of valves, and indicate of what material the several parts should be constructed. 11. Sketch and describe the construction and action of some form of pump by which you could raise water from a well where the level of the Q 240 LECTURE XIX. QUESTIONS. water is 45 feet below the surface of the ground. Explain fully where you would fix the pump, and give reasons for the arrangement which you propose to adopt. 12. Sketch in section and describe the action of the ordinary lifting pump. In such a pump the pump -rod is f inch in diameter, and the pump barrel is 5 inches in diameter, while the spout at which the water is delivered is 20 feet above the surface of the pump bucket when the latter is at its lowest point ; what", would be the maximum tension on the pump rod in the upstroke of the pump, neglecting the weight of the pump rod and the pump bucket, also the weight of water below bucket in suction pipe (the weight of a cubic foot of water is 62*5 Ibs.) ? -4ns. 166*6 Ibs. 13. Describe the construction and action of an ordinary suction-pump for raising water from a well. If 200 gallons of water are raised per hour from a depth of 20 feet, and if the efficiency of the pump is 60 per cenk, wfcat horse- power is being given to the pump ? Ans. -034. LECTURE XX CONTENTS. Bramah's Hydraulic Press Bramah's Leather Collar Pack ing Examples I. II. Large Hydraulic Press for Flanging Boiler Plates The Hydraulic Jack Weem's Compound Screw and Hydraulic Jacfc Example III. The Hydraulic Bear or Portable Punching Machine The Hydraulic Accumulator Example IV. Questions. Bramah's Hydraulic Press. This useful machine was in- vented by Pascal, but he could not make the moving parts water- tight. Bramah, about the year 1 796, discovered a means by which this difficulty was effectually overcome ; and thus the instrument has been handed down to us under his name. As may be seen from the following figure, it consists of a single-acting force pump in connection with a strong cylinder containing a plunger or ram, which is forced outwards from the cylinder through a tight collar by the pressure of the water delivered into the cylinder from the force pump. From what was said in Lecture XIX. about force pumps, we need not particularise about this part of the machine, except to gay that the suction and delivery valve boxes can be disconnected from the pump, and the valve cover-checks removed at any time for the purpose of examining the parts, or of legrinding the valves into their seats. The plunger extends through a stuffing- box and gland filled with hemp packing, and is guided by a cen- trally bored bracket bolted to the top flange of the pump. The lever fits through a slot in this guide-bar, whereby it has an easy free motion, when communicating the force applied through it to the pump plunger. The relief- valve RV has a loaded lever, ad- justed like the lever safety valve in Lecture IV., so as to rise and let the water escape when the pressure exceeds a certain amount. It may also be used for taking the pressure of the object under compression, or for lowering the ram R by simply lifting the little lever and pressing down the table T, when the water flows easily from the cylinder, and out of DP by the relief valve. The delivery pipe DP is made of solid drawn brass, and the ram cylinder is carefully rounded at the bottom end, instead of being flat, in order that it may be naturally of the strongest shape.* * Lo the case of largp cylinders for very great pressures, the lowtr oz LECTUEE XX. 2 4 2 The guide pillars are securely bolted to the base B by nuts and iron washrs, not shown. The cup leather packing CL deserves special attention, because it formed the chief improvement by VERTICAL SECTION OF A BRAMAH HYDRAULIC PRESS, Made in the Engineering Workshop of The Glasgow Technical College. INDEX TO PARTS. L represents Lever. DV represents Delivery valve. p Pressure on L at A. EV , Eelief valve. F Fulcrum of L. DP , Delivery pipe. B L's connection with EC , Earn cylinder. plunger's guide- E , Earn or plunger. rod. CL , Cup leather packing. PP ,, Pump plunger. T , Top, table, or T piece. Q ,, Keaction or stress W , Weight lifted, or on plunger PP. total pressure on E. PB ,, Pump barrel. CO , Cross girder. IP ,, Inlet pipe. GP , Guide pillars. SV ,, Suction valve. BB , Base block. inner end of the cylinder should be carefully rounded off, both inside and outside. For, if left square, or nearly square, the crystals formed in the casting of the metal naturally arrange themselves whilst cooling in such a manner as to leave sn initial stress, and consequent weakness, invit- ing fracture along the lines joining the inside to the outside corners of the cylinder end. The severe shocks and stresses to which this weak line of division is subjected during the working of the press would sooner or later force out the end of the cylinder, in the shape of the frustum of a cone, unless the cylinder had been made unnecessarily thick and strong at the bottom end. BRAMAH'S LEATHER COLLAR PACKING. 243 Bramah on Pascal's press. It consists of a leather collar of fl section, placed into a cavity turned out of the neck of the cylinder, and kept there by the gland of the cylinder cover. The following figure shows an enlarged section of Bramah's packing suitable for a huge press, where the desired shape of the leather collar LC is maintained by an internal brass ring, BR, and an outside metal guard ring GE., resting on a bedding of hemp H. It will be observed at once, from an inspection of this figure, that the water which leaks past the easy fit between the plunger or ram R, and the cylinder C, presses one of the sharp edges of the leather collar against the ram, and the other edge against the side of the bored cavity in the neck of the cylinder, with a force directly propor- tional to the pressure of the water in the cylinder. By this simple automatic action, the greater the pressure in the cylinder the tighter does the leather collar grip the ram and bear on the cylinder's neck. Bramah's Leather Collar Packing. This collar is made from a flat piece of new strong well-tanned leather, thoroughly soaked in water, and forced into a metal mould of the requisite ENLARGED VIEW OP BRAMAH'S LEATHER COLLAR FOB A BIG HYDRAULIC PRESS. INDEX TO PARTS. B represents Ram. C Cylinder. G Gland of C. LC Leather collar. BR represents Brass ring. GR Guard ring. H Hemp bedding. size and shape until it has assumed the form of a \J collar. The central or disc portion of the leather is then cut out, and the cir- cular edges are trimmed up sharp in the bevelled manner shown by the above figure. U5CTTJRE XX. Formula for the Pressure on the Ram of a Bramah Press. Referring again to the first figure in this Lecture, it will be found that by taking moments about the fulcrum at F, we obtain the pressure or reaction Q on the plunger of the force pump Therefore, neglecting weight of lever and friction, we get PxAF-QxBF. .-. Q-^ Further, by Pascal's law for the transmission of pressure by liquids, enunciated in Lecture XVII., we know that the statical pressure Q is transmitted with undiminished force to every corresponding area of the cross section of the ram. Or, , . Q : W : : area of plunger : area of ram. .*. W x area of plunger = Q x area of ram. Wx7rr a = Qx7rR a Where r = radius of plunger, and R = radius of ram, both in the same unit^ Substituting the previous value for Q, and dividing each side of the equa- tion by TT, we get- Since the radius of a circle is directly proportional to its diameter, we may write the formula thus, where D is the diameter of the ram and d the diameter of the plunger, both in the same unit PxAF I? EXAMPLE I. In a small Bramah press, P= 50 Ibs., AF= 20 in., BF= 2 in., area of plunger = i sq. in., whilst area of rani = 14 sq. in. Find W, neglecting friction and weight of lever. ANSWER. By the above formula w _PxAF 7rR a BF X ^r7* Substituting co x 20 IA values, we get W = ^ x = 7000 Ibs. 2 T EXAMPLE II. In Bramah's original press at South Kensington the plunger is 3" in diameter, and it acts at a distance of 6 inches from the fulcrum, which is at one end of a lever 10 feet 3 inches long, carrying a loaded scale-pan at the other end. What should be the pressure of the water in the press in order to lift a weight of 3 cwt. in the scale-pan, neglecting the weight of the lever ? Make a diagram of the arrangement. (S, and A Exam. 1892.) HYDRAULIC PRESS. 245 ANSWER. Here d = 3 in., consequently the area of the plunger / 'x" = s. in. BF = 6"; AF=io' 3" = i2"; P = 3 cwt. = 3x112 = 336 Ibs. ; and we have to find the pressure per sq. in. on the ram that will balance P, acting with the stated advantage, since the area of the ram is not given. By the formula w _PxAF x areaof 1 sq. in._ 33 6x 123" x I sq. i".- 9811bg BF area of plunger 6" 7 sq. in. .'. Pressure per sq. inch on ram of press = 984 Ibs. Large Hydraulic Press for Flanging Boiler Plates, &c. As an example of the practical application of the Bramah press to modern boiler-making, the accompanying illustration shows the form which it takes when worked by a high -pressure water supply derived from a central accumulator, which may at the same time be used to work cranes, punching, riveting, and other similar machine tools, in the same works. The operation of flanging, say the end tube-plates of the cylin- drical barrel of a locomotive boiler, is carried out in the following manner : The ram B, is lowered to near the bottom of the hydraulic cylinder HC, thus leaving room to place the boiler plate (which has been heated all round the outside edge) on the movable table T r High-pressure water is then admitted from the central accumulator to the auxiliary cylinders ACj thus forc- ing the side rams SB, SB,, with their table T r and the plate P, vertically upwards, until the upper surface of the plate bears hard against the bearers B, B, or internal part of the dies. Water from the same source is now admitted into the hydraulic cylinder HC, which forces up the ram B-^ with its table T lf supporting columns SO, SO, and the external part of the dies D, D, until the latter has quietly and smoothly bent the hot edge of the plate round the curved corner of the internal bearer B, B. The ram B, is now lowered, carrying with it the table T x and dies D, by letting out water from HC, and then the table T,, with the flanged plate, are lowered by letting out water from AC. The plate is removed from its table, allowed to cool, placed in position in the barrel of the boiler, marked off' for the rivet holes, drilled and riveted in the usual manner. The student will now under- stand what a useful and powerful servant a hydraulic press is to the engineer in the hands of a skilful workman, for it can be made to do work in the manner indicated above in far less time, and with far greater certainty of uniformity and exactitude, than the boiler-smith could turn out, with any number of hammermen to help him. It is fast replacing, the steam-hammer for crossing. work, and the steam or belt-driven punching and riveting 246 LECTURE XX. LARGE HYDRAULIC PRESS FOR FLANGING BOILER PLATES.* * The above figure is a reduced copy of one from Prof. Henry Kobinson's book on " Hydraulic Machinery," published by Messrs. Charles Griffin & Co., bnt it has been indexed according to the Author's style of symbols, and described in an elementary manner. THE HYDRAULIC JACK. 247 INDEX TO PABTS. HO represents Hydraulic cylinder. R Ram of HC. 0, Columns supporting Y. Y Yoke or cross- head. BB Bearers of the internal die ring. P Plate to be flanged. DD 80 8R AO Dished die or external die ring. Supporting columns for DD. T-pu-ce or movable table for DD. T-piece or movable table for P. Side rams for T e rams or r Auxiliary cylind ers. machines, the steam screw and wheel-gear worked cranes, screw and wheel-gear hoists, as well as the screw press for making up bales of goods mencioned in Lecture XV. For with it, you can bring to bear a force of a few pounds on the square inch or as many tons, by merely turning the handle of a small cock, and with a certainty of action unattainable by any other means. The Hydraulic Jack is a combined hydraulic press and force pump, arranged in such a compact form as to be readily portable, and applied to lifting heavy weights through short distances. It therefore effects the same objects as the screw-jack described in Lecture XV., but with less manual effort or greater mechanical advantage. The base on which the jack rests is continued upwards in the form of a cylindrical plunger, so as to constitute the ram of the hydraulic cylinder HC. Along one side of this ram there is cut a grooved parallel guide slot GS, into which fits a steel set pin, screwed through the centre of a nipple cast on the side of the cylinder (not shown in the drawings) for the purpose of guiding the latter up and down without allowing it to turn round. The top of the ram is then bolted with a water-tight cup leather CL, by means of a large washer and screw-bolt. The action of this cup leather is precisely the same as the leather collar in the cylinder of the Bramah press already de- scribed ; but it has only to be pressed by the water in one direction viz., against the sides of the truly-bored cast-steel cylinder, instead of against both the ram and the cylinder neck, as in the previous case. The head 11 and upper portion of the machine is of square section, and is screwed on to the hydraulic cylinder in the manner shown by the figure. It contains a water reservoir WR, which may be filled or emptied through a small hole by taking out the screw-plug SP.* In the centre line of the head- This screw plug SP is slackened back a little to let the air in or out 248 LECTURE XX. THE HYDRAULIC JACK. THE HYDRAULIC JACK. 249 piece there is placed a small force pump, the lower end of which is screwed into the centre of the upper end of the hydraulic cylin- der. This pump is worked by the up-and-down movement of a handle placed on the squared outstanding end of the turned crank shaft CS. To the centre of the crank shaft there is fixed a crank 0, which gears with a slot in the force-pump plunger P, and thus the motion of the handle is communicated to the pump plunger in a reduced amount, corresponding to the inverse ratio of the lengths of the handle and the crank from the fulcrum or centre of the crank shaft. By comparing the right-hand section of the water reservoir, and the section on the line AB, with the vertical left-hand section of the jack, it will be seen where the inlet and delivery valves IV and DV are situated. On raising the pump plunger P, water is drawn from WE, into the lower end of the pump barrel through IV, and on depressing the plunger this water is forced through the delivery valve DV into the hydraulic cylinder, thus causing a pressure between the upper ends of the cylinder and the ram, and thereby forcing the cylinder, with its grooved head H, and footstep S, upwards, and elevating whatever load may have been placed thereon. Both the inlet and outlet valves>are of the kind known as " mitre valves." They have a chamfer cut on one or more parts of their turned spindles, so as to let the water in and out along these channels. The yalves are assisted in their closing action by small spiral springs SS, bearing in small cups or hollow centres, as shown more clearly in the case of DV by the enlarged section on AB. Weems' Compound Screw and Hydraulic Jack. This is a jack combining some of the advantages of the ordinary screw-jack with those of the hydraulic one. It is often desirable to be able to bring the head or footstep into trial contact with the load before applying the water pressure. This can easily be done by turning the nut at the foot of the screw, cut on the ram of the jack. The arrangement will at once be understood from the figure. It will be observed that the load may also be lowered by turning this nut, or by the screw-tap which permits water to flow from the cylinder back into the cistern, as in the previous case. The bottom nut may be screwed hard up to the foot of the hydraulic cylinder, so as to sustain the whole load, and thus prevent overhauling through leakage of the water. When it is necessary to lower the load or the head of the jack, of the top of the water reservoir when working the jack. There is gene- rally another and separate screw plug opening (as will be seen by the following figure of Weems' patent jack) for filling or emptying the water reservoir, quite independent of the above-mentioned one, which is used in tfci caje lor both purposes. 250 LECTURE XX. the relief valve or lowering screw, is turned so as to permit the water to escape from the hydraulic cylinder back into the water reservoir, as clearly shown by the drawing. This may be done very gently by simply giving this screw a very small part of a complete turn ; in other words, by throttling the passage between the hydraulic cylinder and the water reservoir. Or it may be done quickly by turning it through one or more revolutions. This passage can then be closed by screwing the plug home on its seat. Mr. Croydon Marks, in his book on " Hydraulic Machinery," illustrates and describes another method of lowering the jack-head (first introduced by Mr. Butters, of the Royal Arsenal, Woolwich), where, by a particular arrange- ment, the inlet and delivery valves are acted upon by an extra de- pression of the handle, and conse- quent movement of the pump plunger. He also gives the main dimensions, with a drawing, of the standard 4- ton pattern as used by the British Government, where the ram has a diameter D = 2", the pump plunger a diameter d = i" ; and the ratio of the lever- age of the handle to the crank is 1 6 to i. Therefore from the previous formula we find that, The Theoretical Advantage = WBEMS' COMPOUND SCREW AND HYDRAULIC JACK. WAF _ ~ i X i 2 THE HYDEAUUC BKAB. 251 And he instances two trials by Mr. W. Anderson, the Inspector- general of Ordnance Factories, to determine the efficiency of these jacks, where, with a pressure on the end of the working handle of 76 Ibs., the theoretical load should have been 76 Ibs. x theoretical ad vantage = 76x64 = 4864 Ibs., instead of which it was only 3738 Ibs.; . . 4864 Ibs. : 3738 Ibs. : 100 : x Or, . . x= = 77 % efficiency In a second trial, a load of 1064 ^ DS - required a pressure of 22 Ibs. on the handle, and consequently the efficiency at this lighter load, as might be expected, was less, or only 74 % . EXAMPLE III. With a hydraulic jack of the dimensions given above, and of 77 % efficiency, it is desired to lift a load of 4 tons ; what force must be applied to the lever handle ? ANSWER. By the previous theoretical formula, w _ P x AF D> -IF" '"? . p _ W x BF # ~sr~ "E 5 p_4X 224 x I X ^ 16 2 But the efficiency of the machine is only 77%: consequently 140 Ibs. is 77 per cent, of the force required .. 77 : 100 :: 140 Ibs. ixlba. x= I40XIOO = 181-81 Ibs. 77 The Hydraulic Bear, or Portable Punching Machine. This is another very useful application of the hydraulic press and force pump. It is used in every iron or steel shipbuilding-yard and bridge-building works. By comparing the drawing with the index to parts, and taking into consideration the fact that its construction and action are so very similar to the hydraulic jack already described in full detail, we need say nothing more than direct the student's attention to the action of the raising cam, and to the means by which the apparatus is lifted and suspended. In order to raise the punch for the admittance of a plate between it and the die D, the relief valve B-Y must first be turned back- wards, and the lever L depressed. This causes the corner of the raising cam RC to force the hydraulic ram HB upwards, and the water from the hydraulic cylinder HC back into the water LECTURE XX. reservoir \VR. The relief valve may now be closed and the plate adjusted in position. Then the pump lever can be worked up and down until the punch P is forced through the plate, and the punching drops through the die D and the hole in the metal frame MF, on to the ground, or into a pail placed beneath to receive it. SIDE VIEW AND SECTION. END VIEW AND SECTION. THE HYDEAULIC BEAK, OK PORTABLE PUNCHING MACHINE. INDEX TO PARTS. PL represents Pump lever. CS Crank shaft. C PP WR IV DV BV Crank. Pump plunger. Water reservoir. Inlet valve. Delivery valve. Belief valve. HC represents Hydraulic cylinder. CL Cup leather. HB BC L P D MF Hydraulic ram. Baising cam. Lever for BC. Punch. Die ring. Metal frame. The whole bear is suspended by a chain (worked by a crane or other form of lifting tackle) attached to a shackle, whose bolt passes through a cross hole in the back of the metal frame MF, just above, but a little to the front of the centre of gravity of the machine. This hole and shackle are not shown in the drawing, but the student can easily understand that the hole would be bored a little above where the letters RC appear on the side view, TL8 HYDRAULIC ACCUMULATOR. 253 and that the chain would pass clear of the pump Jever, since it works well to the right-hand side of the bear. The Hydraulic Accumulator. The demand for hydraulic power to work elevators, cranes, swing bridges, dock gates, presses, punching and riveting machines, are also pro- vided. The release of the left-hand weight is gently effected by aid of a pneumatic pipe and ball, as shown at the foot of the figure. This permits the other right-hand weight with its rider O to start falling freely without either jerk, swing or vibration. A correction for the friction of the wheel bearings may be made by putting a little fine wii'e on the descending weight when two equal cylindrical weights are used, until no slowing down is observable after setting these weights in motion. The weight of the cord or thread passing over the pulley may be balanced by a length of the same kind and size, attached* to the bottoms of the right and left weights ; or, by hanging therefrom loose, vertical lengths of such thread. Either of these plans enables the experimenter to raise and lower the weights into position for adjustment or starting an experiment. Method of Usiny the Machine. If the two weights attached to the ends of the cord are equal, then no motion will ensue, since the downward force of gravity acting upon one weight balances its downward force upon the other through their tensions on the connecting thread. The balanced wheel is unaffected by the action of gravity. But, if the right-hand weight be made heavier than the left-hand weight, by placing a very small wire weight to balance the friction of the wheel bearings, plus a rider o upon it, then, it will be set in motion, solely due to the action of gravity upon the rider, until the rider is caught by the ring. The ring, although intercepting the rider, permits the weight with the little wire weight to pass clear through it without touching. The stop-stand may be clamped at any desired position on the vertical scale. Now, referring to the two following figures, let us suppose that only a rider ^ be placed on the right-hand weight whose bottom is level with the * George Atwood w.is born in 1745, educated at Cambiidge, where he became a Fellow and Tutor of Trinity College. He published a few treatises ou Mechanics and Engineering, and died in 1807. The author first experimented with the old Atwood's Machine in 1868-9, made by Professor Copeland in 1796, which belonged to the Natural Philosophy Department of Aberdeen University, and again with an identically similar machine at the College of Science and Arts, Glasgow, from 1880-87. He had it fitted with automatic electrical time starting and recording apparatus. t This mass may be arrived at by trial as explained under the following case VI. IMPROVED AT WOOD'S MACHINE. 263 LINE DIAGRAM TO REPRESENT THE DISTANCES, TIMES, AND THE VELOCITIES OF THE RIGHT- HAND FALLING WEIGHT. WEIGHTS, RIDERS, AND OVERWEIGHTS. IMPROVED ATWOOD'S MACHINE. By E. E. Becker of London. 264 LECTURE XXI. zero of scale and top level with A on the line diagram. Then if th ring intercepts this rider at B one second after the weights have begun moving, the motion during that time has been uniformly accelerated, and the velocity is represented to scale by line BE.* But, when the right-hand weight passes the ring, its motion will thereafter be uniform until its base is arrested by the stop-stand with the velocity which it had acquired when its top just passed through the ring, viz., CD which is equal to BE. We may thus find the velocity gained in the first second, and, by changing the mass of the weights and riders, the positions of the ring and of the stop-stand, we can try the following and other ex- periments, wherein the rider may be replaced by one or other of the circular overweights which pass clear through the ring, and cause uniform acceleration until they come to the stop-stand. Noting the Times in Seconds. These may be done by aid of a simple pendulum beating distinct seconds, or a laboratory clock or a split seconds stop-watch. It is simplest and best to start the right-hand weight with its lower surface at zero upon the beat of the sixtieth second of a minute ; to so set the ring that it catches the rider at an exact number of seconds from the start, and that the base of the right-hand weight is arrested by the stop-stand S, at an exact number of seconds from the time of starting. This saves any confusion and trouble, arising from noting parts of a second, and ensures that the two known distances passed through by the lop of the right-hand weight, viz., from A to B and from A to C, are accomplished in exact whole seconds. The distance CS is simply equal to the height of the right-hand weight. I. To prove, for uniform velocity, that the velocity is equal to the dis- tance passed through divided by the time ; that is, v = h/t ; or, h = vt. II. To prove, that the distances described from rest are proportional to the squares of the times ; that is, h oc &. III. To prove, that the distances from rest during acceleration are half those described in the same timet after the motion has become uniform. IV. To prove, that the acceleration (or increase of velocity per second) is equal to the velocity (at any instant) divided by the time from rest ; that is, a = vjt ; or, v = at. V. To prove, that the distance moved through in a certain time is equal to half the acceleration multiplied by the square of the time ; that is, A = i a 2 ; or, a = 2A/2 ; and v^ = 2ah = 2al. VI. To prove, that when a force produces uniformly accelerated motion in a body, or a system of bodies, then the acceleration is directly propor- tional to the force and inversely proportional to the total mass moved ; ra or, a = -T, ; but F = M a = m g * The scientific meaning of the term acceleration is an increase of velocity per unit of time. And, since velocity is a rate or distance per unity of time, we see that acceleration means an increase of distance passed over by a body per unit of time per unit of time ; or, an increase of distance per second per second. Whereas, when traffic managers of rail- ways or shipping companies speak or write of the speeds of their trains or ships being accelerated, they simply mean that their average velocity has been increased so as to cover their normal distance between certain stations or ports in less time than before. It is only during the times of getting up to their full speed that the train or ship has ( + ) or positive acceleration, and when slowing down that ( - ) or negative acceleration is experienced. EXPERIMENTS WITH ATWOOD'S MACHINE. 265 RESULTS OF EXPERIMENTS WITH AN ATWOOD'S MACHINE. No. of Experiment. Position of Ring B from zero A in ins. Position of stop-stands from zero A in inches. Time in seconds to ring B from rest. Time in seconds to stop-stand 8 from rest. I If 8| I 3 2 7 21 2 4 3 I5l 47i 3 6 J CASE I. When accelerating force is removed then the velocity become* uniform and is equal to the distance passed through the time ; that w, v = ~ ; or, h = v t Exp. (i) Distance from ring to stand = 8f - if = 7 inches. Time taken to move =31=2 seconds. .. j = l = = 3 "5" = velocity acquired in 1 second. Exp. (2) Distance from ricg to stand = 21 7 = 14 inches. Time taken to move =4 2=2 seconds. .. Vt =-1 = = 7" = velocity acquired in 2 seconds. Exp. (3) Distance fron* ring to stand = 47 - isf = 3ii inches* Time taken to move , =6 3=3 seconds. .. v 3 = = s^= 10 "5" = velocity acquired in 3 seconds. t s 6 CASE II. The distances described from rest are proportional to the square* of the times. Exp. (i) Distance in ist second = if inches = = f x i, or : Exp (2) 2nd =7 = 7 = | x 4, : Erp.(3) 3rd =i 5 f =*^= x 9, 32 CASE III. The distances from rest during acceleration are half those described after the motion becomes uniform, in the same time. Exp. (i) Distance passed through from rest in i second = (If x 1) inches. Distance after motion becomes uniform in (31 = 2 seconds) is 8f if =7 . * . Distance in I second after motion becomes uniform = 3^ inches . = (If x 2) Exp. (2) Distance passed through from rest in 2 seconds =(7xl) Distance after motion becomes uniform in (4 2 = 2 seconds) is 21-7 ... . . = 14 Distance in 2nd second after motion becomes uniform =(7x2) t, CASE IV. Uniform acceleration is equal to velocity -r time ; or, a = vjt. From the detailed working out of Case I. we see, that the change of Telocity = (v x - v e ) = (t^-t^) = (t>,- a ) = 3-5" per MO. 266 LECTURE XXI. Hence, since the change of velocity in 3 seconds is 10-5" ; we get, - = I0 5 = 3-5 = a the acceleration. CASE V. Distance moved through = \ acceleration X time 2. Or, h = i a *2. Exp. (3) Distance h moved through in time (t) 3 seconds = 15!". And, from Case IV. the acceleration a = 3-5" per sec. per sec. Hence, by substituting known values in the equation w A = a <2. 5 x 32 = I5 | inches> Now, from Case IV. v = a t, . . v* = a2 2 ; or, f2 = v 2/a2, But from Case V. h = a 2 ................... . 2 2 A/a. Hence, 2A/a = t>2/ a 2. Or, 2 a A = 1>2. Or generally, t>2 = 2 a Z where Z is the distance passed through by a body having an acceleration a. Corollary to Case V. If a body has a certain initial velocity u before it becomes uniformly accelerated, then the final velocity v, will be the sum of the initial and accelerated velocities. From the previous definition of uniform acceleration, viz., that it is the increase of velocity per unit of time from the commencement of the acceleration, we see that u + a . . The average or mean velocity will be half the initial and final velocities, Or i[u + (u + a )] = u + \ a t. And, the distance I moved through or height h through which the body passed will be tlds mean velocity multiplied by the time of its motion from the instant that the noted acceleration a began. That is, I = h = (average velocity) x (time of motion) j Or, h = (u + % a t) x t. .'. h = ut + % a t%. The above refers to positive acceleration or getting up speed : but the same applies to negative acceleration or retardation of a body's initial velocity if we apply the sign instead of the + sign. Here, I = ut - \ a &, Hence generally, / = ut + % a I 2 , And, if M = o, or the body's motion is started from rest, we get as in Case IV. To prove the above by the Atwood's machine : First, place an overweight and then a rider upon the right-hand weight; _jid bring lower surface of the latter opposite the zero mark of the scale. Second, set the ring at say 15!" below zero A, so that the rider is caught at B, the end of the 3rd second and calculate the velocity with which the whole moving mass is going at the instant the rider is caught by the ring ; let this velocity = u inches per second. Third, note the time t in seconds and the acceleration a per second per second which the overweight O now causes the remainder to possess in passing through any convenient height h inches between the upper edge of the ring, and the upper side of the right-hand weight when its bottom is arrested by the stop-stand S. Fourth, check the average of an odd number of trials by substituting the values for h, u, t and a in the formula h = ut + i a t*. EXPERIMENTS WITH ATWOOD^S alA CHINE. 267 E VI. When a constant force produces uniform acceleration of a body or system of bodies, the acceleration is proportional to the force and inversely proportional to the total mass being moved by the force. In the previous Experiments with the Atwood's machine : If m =- Mass of the rider = ^. g ,, w = Weight of the rider = | oz. or '5 oz. F = Force produced by gravity on rider = mg = *5 oz. ,, M = Total mass moved = (right + left weight + thread -4- equivalent mass active at circumference of wheel to overcome its inertia + wire weight to balance friction W -f to of wheel bearings + mass of rider) = - W = Weight due to whole moving mass M (minus mass m of weight %) = 3 Ib. 6| oz. = 5475 oz. Then W + w = Total weight in motion up f o position of ring. And a = Acceleration produced on Mass M by gravity acting on m = 3 '5 inches or (3*5 -f 12 feet) per second per second. = Value of the acceleration produced by the force of gravity upon a freely falling body (to be found from the result of the previous experiments). Hence, from the above statement we see that the following relations should hold good, viz. : F F a= B ;andy=- ir , Or, Ma = F = mg M - A . or g - a ^ a (W + M) 4 g _ a (W + w) 'ma' m w -J- g 10 Inserting the values obtained from the previous experiment*. We get, , = a = 33-*3 . per se, per sec. It has been proved by the most careful experiments carried out at Greenwich Observatory, that the force of gravity there, when reduced to sea level, produces an acceleration of 32*1912 ft. per sec. per sec. But, Aberdeen, where the above experiments were made, is nearer the centre of gravity of the earth, the value may well be nearer 32*2 which is con- sidered the usual average value, although 32 is often taken when only approximate results are desired. Regarding Inertia of the Wheel. The value of the small weight x which forms part of W in the last equation and which balances the inertia of the aluminium wheel, may be found from two experiments producing two different accelerations cij and a% by two different riders TOJ and tn., respectively, and substituting their known values in the previous equation, until it is found that uniform velocity is obtained in each of these trials after the riders m^ and m^ have been caught by the ring. Let, W = W + Xy Where, W = the combined weight of the left and right-hand equal weights + the small wire on the latter which is required to balance the friction of the axle in its bearings And x = the weight required to balance the inertia of the wheel. 268 LECTURE XXI. Then, since gravity has a constant value v e get W+w (W+x + wJ, 9 = a ^ or in Exp. (i) g = a x -- ^ (W+X + Wo) and in Exp. (2)g - a z -- Hence in Exp. (i)-^p 1 = (TT+ x +^) and in Exp. (z)-~ = (W + x + wj Or in Exp. (i) a; = -^- 1 -- (W + wj and in Exp. (2) x = -^p = (W + w a ) When Experiments (i) and (2) give the same value for x under the above conditions it balances the inertia of the wheel in both cases. It will now be interesting to find from the foregoing data what should be the ratio of the mass of a rider to the whole mass moved in order to produce an acceleration of i foot per second per second ; since the dynamical unit of force which is sometimes called the "BritL-h Absolute Unit of Force " called the Poundal, is that force which, acting for i second upon a mass of i lb., imparts to it a velocity of I foot per second. Here M = ( W + ").= lib. 9 a = i ft. per sec. per sec. }i m -^-, and w has to be found T 9 From Case VI. we see that m a i But, since the masses are proportional to the weights < We get, M _ W + w _ 32*2 m w i * And, since (W + w) = i lb. = f-L-\ lb. = i oz. (nearly). Or, a rider of ^ lb., i.e., oz. would be required to be placed upon the right-hand weight of an Atwood's machine where the sum of all the moving weights reckoned as before, was equivalent to i lb. in order to produce an acceleration of i foot per second per second. Hence, speaking generally, the value of a poundal (or so-called absolute dynamical unit of force) is equal to ^ or- lb. in the ordinary engineers gravitation unit of force ; and a force of i pound weight is equal to 32 or g poundals. But, we shall not bother the elementary student in this book with poundals, since, as we said before, the gravitation unit of force or force which will sustain a weight of i lb. of matter is sufficiently absolute for any particular place and for our purpose. It will at once be seen from the following proportion, that no rider, however heavy, if it were placed upon any weight, however small (whjen the latter was balanced against the force of gravity) could produce an acceleration of 32*2 feet per second per second. For, the impressed acceleration a would also have to be equal to g or 32*2, when we get a W 32'2I That is, W could have no weight became u> GALILEO'S AND KATER'S EXPERIMENTS, 269 The application of the foregoing explanations, formula? and reasoning to ordinary applied mechanics questions will be seen, when we come to deal with constant forces acting in any direction upon masses of matter for a known time and producing a certain acceleration ; such as the acceleration produced upon the piston and piston rod, &c., of a steam engine by the pressure of the steam up to the point of cut off. Or, upon the fly wheel due to the mean force upon the crank pin during so many revolutions ; or upon the starting or the stopping of railway trains, centri- fugal machines, dynamos, and many other prime movers, or motors where it is advisable to raise their speed as quickly as possible to a permissible maximum or normal velocity and to stop them quickly. Galileo's Experiments. More than 100 years before Atwood was born, the famous Italian philosopher Galileo (about 1583-1630) experi- mented upon falling bodies of different kinds and weights by letting them fall simultaneously but separately from the top of the leaning tower at Pisa and found that they reached the ground at the same instant, or from the same height in practically the same time. He also let spherical balls roll down a straight, smooth, inclined plane (where the friction between the ball and the plane was a minimum) and found that the distances through which the ball passed in successive seconds varied directly as the square of the times from rest ; and that the velocities acquired by the ball varied directly as the times from rest ; or, Times in seconds varied as I : 2:3 : t Distances passed varied as 12 : 22 : 32 : ^ gt 2 Speeds acquired varied as i :2 :3 :gt. Pendulum Experiments. Galileo (about 1583), Huyghens (about 1650-70), and Captain Kater, an Englishman (about 1813-32), experimented with pen- dulums to determine not only the length of the seconds pendulum but also to ascertain the value of the force of, and acceleration due to gravity at different places. The investigation of -this subject must be deferred until harmonic motion has been explained. It may, however, be stated that for a seconds pendulum The length l = (^~) = 39*13983 inches at Greenwich (reduced to sea level) where g = 32'i9i2feet persecondper second. EATER'S REYEESTBLE 270 LECTURE XXI. Rater's Reversible Pendulum. This pendulum consists of a vertical bar carrying a brass weight near each end, together with two similar boxwood weights, between the brass weights which are so attached as to compensate for the air displacement of the brass ones. The pendulum may be hung from one or other of two movable steel knife edges working on agate planes. We now test to see if the lower weight has been cor- rectlj placed, because, the time taken by the pendulum to perform say 100 swings about the top movable steel knife edge should be equal to the time taken to perform 100 swings about the other knife edge. If the times are not quite equal then one of the brass weights or its movable knife edge must be readjusted until the time of a swing when suspended from either end must be exactly equal. The length of an isochronous, simple equiva- lent pendulum is precisely equal to the final adjusted distance between the knife edges. Therefore, knowing the length of the pendulum arid its time of oscillation, the acceleration due to gravity can be f..und from the gt 2 previously stated formula I = - With proper care Kater's pendulum can be made to give a very accurate result for the value of the? acceleration due to gravity. The velocity attained by a body sliding down any straight inclined plane varies directly as the square root of the height.* Let AB be a smooth inclined plane of length I making an angle 6 with the horizontal, as in the figure, Eesolve the weight of the smooth-faced To PEOVE THAT THE VELOCITY ATTAINED BY A BODY SLIDING DOWN ANY INCLINED PLANE DEPENDS UPON THE HEIGHT OF THE PLANE. body W as represented to scale by the dotted vertical line ab, into two components, one cb parallel to the plane and the other ac perpendicular to it. Now, the latter component force ac has no effect on the motion of the body, since its surface and that of the plane are supposed to be smooth, and therefore the former component cb is the only force which causes the body to move down the plane. The magnitude of this force F = ab sin = W sin 6. * It has been proved in the Author's more Advanced Text-Book on Applied Mechanics and Mechanical Engineering, Vol. II., Lecture XXII. > that although the kinetic energy of a spherical ball, or of a cylinder of a Certain mass, after it falls or slides or rolls down any path through a definite height is of the same value ; yet the velocity and momentum are of different values for the cases of rolling and of sliding. VELOCITY, ETC., IMPARTED TO A FALLING BODY. Also, it has been shown from Case VI. in connection with the Atwood's machine, &c.. that the acting force = mass moved x accelera- W tion attained by body ; that is, F = x a Or, Wsin0x# =Wa Further, it was shown in connection with Case V. that t>2 = 2 a I. Now, substituting the values for a and I in this equation and figure, we get v2 = 2g sin x BA. But, ||= sin 0, or -?5= BA. .^ = 20sin0 x j^- e = 2 i 2g \ energy. / The Velocity, the Momentum, and the Kinetic Energy attained by a body in passing from a given height along any path are each respectively the same, when the friction between the path and the body is negligible. The following figure illustrates an arrangement to prove, that the velocity acquired by a APPARATUS TO PROVE THAT THE VELOCITY, THE MOMENTUM AND THE KINETIC ENERGY IMPARTED TO A FALLING BODY is THE SAME WHICHEVER PATH IT FOLLOWS, IF FRICTION BE NEGLIGIBLE OR NEGLECTED. body falling freely down any curved path is the same whatever be the shape of the path. It shows one straight and two curved paths. Each path is made of a smooth glass plate, or a pair of parallel polished metal rods, along which a truly spherical billiard ball may be allowed to roll down from a platform. The ball when it reaches the lowest position, i> allowed to travel up an adjustable and very smooth, plate glass, inclined plane. It will be noticed that in each instance, the ball ascends to the same height on this plane. Consequently we conclude from the results of the experiment that the velocity, the momentum and the kinetic energy (or energy of motion) of the ball when commencing to ascend this inclined plane must be respectively the same. Further, if the ball experienced no frictional resistance along its pajth, then it would run up the inclined plane until it reached exactly the same level from which it started with no 272 LECTURE XXI. initial velocity on the other side. The resistance of the air and any slight want of smoothness in the paths make this ideal condition unattainable in practice. But, with care, the results are sufficiently accurate to prove, that if A, the height from start to foot of path, W the weight of the ball, and g the acceleration due to gravity remain constant throughout the experiment, and since the ball ascends along the incline always to the same height for a given angle of inclination, the kinetic energy attained at the foot of each path is the same, for nothing else could send it to the same height unless the kinetic energy ^ mvz ; the momentum mv ; and velocity v were the same in each case. Seeing, however, that the body started from rest at a height A above the lowest point of its path, its potential energy before starting = WA, and since no energy is assumed to have been lost we get WA oc Wt> 2 /2. Hence, v2 oc 2 gh, or t;2 oc A since 2 g is a constant, i.e. t v oo ^/h. Consequently under these circum- stances, the velocity, momentum and kinetic energy attained by a body falling down or sliding over any inclined or curved path is solely dependent upon the vertical height of the fall and quite independent of the shape or length of the path taken by the body when friction is eliminated or neglected. Comparison of Dynamical Formulae for Linear and Angular motion when due to any Force producing a constant Acceleration. KINEMATIC FORMULA. Or, formulae dealing with pure motion in the abstract. Linear, Angular. v=at KINETIC FORMULAE Or, formulae dealing with the models of real bodies under the action of actual forces. F _ F _v a -v 1 _T.M. Tr/ - Tir ~ 7~~" . tt j ( Angular ) _ Twisting Moment (Acceleration} "Moment of Inertia _ Twisting moment "Mass x (radius) 3 T.M.XI Momentum = Im- pulse or Force x Angular ^ Momentum time it acts or moment . of momen- tum of the rotating 1M^-F* body. ....ilw 5 Kinetic Energy = Stored work. Or, fMoment^ i of r I Impulse J :T.M..x0 Kinetic Energy = Stored Work. PATH OF A PROJECTED BODY. 273 The Path of a Projected Body which then falls under the Action of Gravity is a Parabola. The accompanying illustra- tion shows a suitable form of apparatus for observing the path, which a body, projected horizontally, describes under the combined actions of the projecting force and of gravity. The whole apparatus should be firmly secured to the wall, and the ball should always be allowed to roll lown from the top of the concave quadrant " race," in order that the path traced on the blackboard may be definitely followed. When a body has a certain velocity imparted to it, such as a stone thrown at an object, or water issuing under pressure from a pipe, or a bullet fired from a rifle in any direction, other than that of a truly vertical one, it has two different motions imparted to it. The one direction and motion is the result of the initial velocity, while the other is due to the earth's attractive force, termed the force of gravity. In the same way, the ball, when it just leaves the lower end of the quadrant "race," has both a horizontal and a vertical motion. The resultant motion is deter- mined from the relative values of these component motions, by the prin- ciple of the parallelogram of velocities. The ball will not move along a straight line, but will describe a parabolic curve, as shown by the dotted curved line in the blackboard of the figure. If the ball had left the quadrant with a velocity in any direction (other than that of a vertical one) it would still have described a parabolic path. Now referring to the two upper left and right figures of a cone, we see that the parabolic curve BVC is the outline of the section of the cone GAL, by a plane passing through VO and parallel to the opposite generating line AL of the cone. Generally speaking, a parabola is a curve traced by a point, which moves in such a manner, that the distance from the point at any instant to the focus of the curve is always equal to the normal distance of the same point from the directrix of the curve. Directrix and Focus. If a point so move that the ratio of its distance from a given fixed point, to its perpendicular distance from a fixed straight line be a constant, it describes a conic section, of which the fixed straight line is termed the directrix and the fixed point the focus. The constant ratio referred to is termed the eccentricity, and its magnitude determines the nature of the conic. Thus, if in the figure DOD be the directrix and F the focus, and if the point P moves so that the ratio of its distance frocr if, is to its distance PN from DOD, be a constant, then P will trace out a conic section which will be a parabola, an ellipse or a hyperbola, according as the ratio in question is equal to, less or greater than unity That is as PF is equal to, less than or greater than PN or as FV = , VO. Proof of Equation to the Parabola. In the right-hand upper figure, let P be a point on the curve BVC, then the distance of P from the focus F is equal to the length of the perpendicular line PN, let fall from the same point P upon the directrix line DOD. Any conic section made by a plane, such as VE, which cuts the two extreme generating lines such as AG and AL of the cone GAL is called an ellipse. An example of an ellipse is shown by the curve BLCGB, of which F! and F 2 are the/oc*. Any conic section made by a plane such as VH, parallel to the axis AC IB called a hyperbola. Referring to the right hand top figure, the equation of a parabola, viz,. y a =4ax may be proved as follow : C; 274 LECTURE XXI. INDEX OF PARTS. A for apex of cone. AQ ,, axis ,, VC section BVC parabola at VC. DOD ,, directrix of para bola. V ,, vertex. F focus. OUTLINE VIEW OF CONE WITH SECTION PLANE. VC FOR PARABOLA ; VH FOR HYPERBOLA ; AND VE FOR ELLIPSE. PROJECTED VIEW OF CONE SHEWING PARABOLIC CURVE BVC, AND ELLIPSE BLCGB. APPARATUS FOR DETERMINING THE PATH OF A BO.JY PBOJECTD HORIZONTALLY ASD FALLING FREELY. CENTRIFUGAL FORCE. 2/5 By Euclid I. -4 7. FP 2 = FM 2 + MP 2 Or, MP 2 = FP 2 - FM 2 That is, MP 2 = P]S T2 - FM 2 Or, MP 2 - (PT + TN) 2 - (VM - VF) 2 Tluit i-, MP 2 = 2 TX x 2 PT (But MP = y: \ Al-o PT = x .-. T, 2 = 4 aa; I And TN = a. Centrifugal Force due to Motion in a Circle. EXPERI- MENT I. When a body such as a stone is attached to a cord and whirled round and round in a circle, the hand experiences a pull in the direction of the string, which is in tension under the action of a force, and the faster the body is moved the greater becomes the stress in the string, just as David of old must have felt it before he let go that pebble from his sung which went so straight for Goliath's brow. The stone is constantly tending to fly off at a tangent, and is only kept moving in the < ircular path by the reaction pulling it towards the centre of motion. The pull from the centre of motion is called the centri- fugal or centre-flying force, and the exactly equal and opposite reaction is termed the centripetal or centre-seeking force. It may be proved by geometry that each of these forces is equal to EXPERIENCING THE EFFECT OP CENTRIFUGAL FORCE. the weight of the body x the square of the velocity 4- the acceleration due to gravity x the radius of the circle described by the body.* * At present the student mast accept the above formula as correct. We shall have occasion to deduce the formula by aid of geometry in the Advanced Course. (See Text-Book on " Applied Mechanics and Mechanical Engineering," Vol. II., lectures xxii. and xxiii.) 276 Or, LECTURE XXI. p - gr Ibs. "Where P = Pull on the cord, or the centrifugal force in Ibs. M W = Weight of the body in Ibs. w v = velocity of the body in feet per second, >* 9 gravity's acceleration = 32' per second in one second, M r = radius from centre of motion to e.g. of body in feet, Centrifugal Force Machine. General Description. This machine enables the student to systematically experiment and prove the laws con- nected with centrifugal force. A polished wooden beam about 2 ft, in length is rotated by means of CENTRIFUGAL FORCE EXPERIMENTAL MACHINE. the pulley at any speed. Attached to the centre of this beam is an accurately machined box with one side flexible. The box is filled with mercury or coloured alcohol. An upright glass tube is fitted into the centre of this box, and exactly in the axis of rotation, so that the height of the mercury can be measured whatever be the speed of the machine. The height of the mercury in the tube can be adjusted by means of a screw at the side of the box. Fastened to the centre of the flexible side of the box is a long brass rod, which in turn is attached to one end of a flat brass rod, whilst the other end of the rod is connected to the end of the beam. This arrangement of supporting the long brass rod allows it to CENTRIFUGAL FORCE MACHINE. 2/7 move backward and forward with less friction than if it were made to elide on a bearing. This long rod is graduated, and a sliding weight can be clamped at any position from the box. Hence, we see that if the centre of the flexible side of the box be pulled or pushed, the mercury will rise or fall in the vertical glass tube ; although, we cannot observe the yielding of the flexible side. With a fixed vertical scale alongside the glass tube it is easy to measure the rise and fall of the mercury in the latter. The weight shown on the left-hand side can be used for balancing the weight on the movable rod, and in this case it should be clamped at the same distance from the axis of the instrument. Prior Adjustments. Before commencing any experiments it is necessary to pull the end of the long movable brass rod with a force of i, 2, or 3 Ibs., &c., by means of a spring balance, to note the respective heights of the column of mercury in the glass tube. Hereby we can afterwards tell the value of our scale measurements. We can also make a number of experiments with the sliding weights removed from the rods, to find the centrifugal force of the remaining parts of the machine when run at different speeds, in order that these readings may be subtracted from subsequent observations to get true results. Working of the Centrifugal Machine. One experimenter turns the handle of a separate pulley, which drives the pulley shown in the figure by means of a rope. At the same time he keeps his eye upon the height of the mercury column in order to keep the speed constant, and he counts the number of revolutions per minute which his hand makes. It is easy with this instrument to test the law which is usually given, that if a body be compelled to move in a curved path it exerts a force directed outwards from the centre of motion, and the amount in pounds is found by multiplying the mass of the body by the square of the velocity in feet per second, and dividing by the radius of the curved path we get Wv^ the centrifugal force =P= . Or, if we multiply the mass by the radius of the circle and by the square of the angular velocity in radians per second, we get ir 2 =P, the centrifugal force urging the weights from the centre ,of motion in a radial direction. 278 LECTURE XXI. EXPERIMENT II. Take a pail and half fill it with water. Attach a rope to the centre of the handle, and swing it round and round your head. The water does not fall out, even if you swing it in a vertical plane, if the velocity be sufficient to cause the centrifugal force to be greater than the force of gravity. EXAMPLE I. A small tin pail, containing i Ib. of water, with a rope attached to its handle, is to be whirled in a vertical circle. If the distance from the hand or centre of motion, to the surface of the water be 2 feet, what is the least number of revolutions per minute that you can give it in order not to spill any of the water ? ANSWER. Here P must be at least equal to i Ib., for W = i Ib. and r = 2 feet, whilst g = 32. By the formula Or, v = 8 x 60 = 480 ft. per minute. Now a circle of 2 feet radius = 12*56 feet circumference. 480 .*. - 7 = 38*2 revolutions per minute. 12*56 Consequently, if you whirl the pail at 40 revolutions per minute, there will be no fear of any water coming out of it even when it is upside down at the highest part of the circle. EXPERIMENT III. Turn a disc of wood with a small barrel on one side of the centre. Fit the wheel and the barrel so truly with a turned axle that when the axle is supported by eye hooks at each end for bearings, a cord wound round the barrel and then pulled sharply, will cause the wheel to revolve freely at a high speed without vibration or oscillation. Now bore a hole through the disc near its circumference, and run in molten lead into this hole. Again spin the wheel rapidly, when it will be found to hobble to such an extent as to shake itself almost out of the bearings. The centrifugal force due to the unbalanced piece of lead asserts itself so thoroughly that when it reaches the highest position of its revolution round the axis, it overcomes gravity, and lifts the whole wheel and barrel clean out of the bearings. It thereby creates such a disturbance as to leave a distinct impres- sion on the mind of the student. Next bore another hole through the disc of the same size as the former one, and at the same distance from the axle, but diametri- cally opposite to the front hole, and run in the same weight of BALANCING HIGH-SPEED MACHINERY. 2/9 lead into it. Again spin the wheel, and it will be found to run smoothly. This experiment conveys to the young engineer a most useful lesson, for it not only shows him the effect of centrifugal force due to want of balance, but it also gives him an idea how to rectify the evil. Balancing High-speed Machinery. All high-speed machi- nery, whether revolving or reciprocating, should as far as possible be most carefully balanced, in order to prevent centrifugal force coming into play and creating that horrid vibration and noise with which it is always more or less accompanied. There is nothing tends so much to the heating of bearings, and to the quick wearing out of brasses and other bearing surfaces as unbalanced moving parts ; besides which, at very high velocities they become actually dangerous, and have frequently been known to cause destruction to life and property.* Centrifugal Stress in the Arms of a Ply-Wheel. If the arms of a fly-wheel or pulley are not properly proportioned to resist the centrifugal force due to the mass of the revolving rim ; or, if tthe casting has been carelessly cooled, so as to set up internal stresses between the arms and the boss or the rim, the wheel may give way. In fact, there is no fly-wheel or pulley made that would not burst, under the very great stress of centrifugal force, if you only ran it fast enough. The student will observe from the formula that the centrifugal force or stress in the arms of a fly- wheel is directly proportional to the square of the velocity, so that by merely doubling the number of revolutions per minute you quadruple the stress in the arms, and if the speed be increased three times, the stress becomes nine times as great. EXAMPLE II. Each segment of a fly-wheel, with its correspond- ing arm to which it is attached, weighs 1000 Ibs., and the mass may be taken as collected at a distance of 4 ft. from the axis of the wheel. If each arm has a breaking stress of 100,000 Ibs., what is the maximum number of revolutions per minute that the fly-wheel could be run at without breaking the arms, neglecting the binding strength of the rim of the wheel ? ANSWER. By the previous formula for centrifugal force 1000 X V 2 100,000 --= -- .. tf =12,800 32 x 4 * See Mr. C. A. Matthey's paper and the discussion on " The Mechanics of the Centrifugal Machine," in the Transactions of the Institution of Engineers and Shipbuilders of Scotland for Session 1898-99. 28O LECTURE XXI. 113 ft. per second, fully. = 113 x 60 = 6780 ft. per minute. .*. v = Or, v Now, the circumference of a circle of 4' radius = 25 ft. 6780 . ,. ,\ f = 271 revolutions per minute. Energy. In applied mechanics energy means the capability of doing work.* Potential Energy is that form of energy which a body possesses in virtue of its position or its condition. For example, when a body of 10 Ibs. is lifted 10 ft. high, it has a potential energy of 100 ft.-lbs. ; for it takes that amount of work to lift the 10 Ibs. through the 10 ft. ; and if then allowed to fall, it would naturally give out the same quantity of work, either in overcoming friction, or, if it fell freely, it could be usefully employed to that amount and no more. Potential energy may also be due to a condition of a body, such as the potential energy in the coiled spring of a watch or clock, which when wound up does work in moving the mechanism. We have also the case of potential energy in a lump of coal which when burned gives out heat, that will raise steam to be used in a steam engine for doing work. Or, in the case of an electric battery, where plates of copper and zinc are respectively placed in solutions of sulphate of copper and zinc, and on being suitably connected by wires to an electric motor, will give out electrical energy, which may be converted into mechanical work by the motor, and thereby effect some useful purpose. Kinetic Energy (E^) is energy due to motion. For example, in the first instance of potential energy the weight of 10 Ibs., in falling freely down through 10 ft., had stored up in it, due to its motion, an amount of accumulated work equivalent to 100 ft.-lbs. Accumulated Work (E,). If a body of weight W Ibs. be raised to a height h feet above the earth The potential energy stored up E^=W& (ft.-lbs.) Now, if the body be allowed to fall freely, under the action of gravity, through h feet, it would have a velocity at the end of time t seconds of v feet per second. Referring back to the formulas for falling bodies previously given in this lecture we see that h - -- .-. Wh = ft.-lbs. *g *g * We have specially avoided using this term hitherto, as students are liable to confuse it with force, work, and power. ACCtTMTTLATED WORK. 281 Therefore the kinetic energy or accumulated work stored up in a moving body is expressed by the formula If a body of weight W Ibs. were impressed forward along a perfectly smooth plane for a distance of I feet, by a force F Ibs., causing an acceleration of, a, feet per second ; then the previous set of formula? for linear velocity would apply when the reaction from the plane cancelled the force of gravity. Here, . . F-/ and I - But the Wvrk Done through distance I = F x I = E / . Therefore in this case the accumulated work stored up in the moving body would be expressed by the formula Accumulated work in a Hotating Body. If a body of W Ibs. be concentrated at a distance of r feet from the centre of motion, and be rotated so that it has a velocity of v feet per second, then The Accumulated Work =- - ft. -Ibs. 2 9 The Energy of a Rotating Ply -wheel is a good example of accumulated work. If the pressure of steam in the cylinder and the point of cut-off be kept constant, and if one or other of the machines which are being driven by the engine be thrown out of circuit or, in other words, if the belt be moved to the loose pulley the load on the engine will be lessened, and the engine will have a tendency to increase in speed. If, however, it be provided with a very heavy fly-wheel, the surplus power of the engine will be stored up in the fly-wheel, so that the increase of speed will not be so great as if it had a light one, or none at all. If a machine should be suddenly brought into circuit again after a short time, then the load on the engine will be as quickly increased ; but the etored-up energy in the fly-wheel will enable it to overcome this sudden demand for power, so that the speed of the engine will not be greatly altered. The fly-wheel, therefore, acts as a regulator of speed, not only for alterations of load, but also for the variable pressures which exist in the cylinder of an engine. This ia 282 LECTURE XXI. particularly noticeable in the case of gas engines, where the almost instantaneous explosion of gas in the cylinder at the beginning of a stroke creates an immense force, which would urge the piston forward at lightning speed, if it were not for the very heavy fly-wheel with which the engine is provided. The fly-wheel stores up some of this sudden force and gives it out again during the intervening strokes when there is no explosion, thus tending to a uniformity of speed which would be conspicuous by its absence if the gas engine had only a light fly-wheel, or none at all. In fact, the motion of gas engines would be so erratic without fly-wheels as to prevent their application to many pur- poses for which they are admirably adapted when aided by very heavy ones. Radius of Gyration.* It will be evident, almost without ex- planation, that in the case of a fly-wheel or a rotating disc, those parts which are furthest from the centre of motion must accu- mulate more energy than those of the same weight which are nearer to that centre, because they move at a greater velocity. There is, however, for every body a mean radius of rotation, termed " radius of gyration" (k) which is at such a distance from the centre of motion, that if the whole mass of the body were concen- trated there, the same kinetic energy or accumulated work would be developed at the same speed or number of revolutions per minute. The length of this mean radius varies with the shape of the rotating body, and requires a knowledge of higher mathe- matics for its computation ; so we will assume that in the case of a fly-wheel it is at the e.g. of the rim, or that the distance is given in any question requiring solution. EXAMPLE III. A fly-wheel weighing 10,000 Ibs. has a mean radius of rotation, k=*r= 5 feet, and turns normally at 100 revolu- tions per minute. Owing to the load being diminished, the speed increases to no revolutions per minute ; what reserve energy is stored up in the fly-wheel, which is fit to overcome any sudden increase of load ? ANSWER. Let v l = the velocity in feet per second, at the nor- mal speed n^ revolutions per minute, And v 2 = the velocity at the increased speed n a revo- lutions per minute ; 2 x 22 x q x 100 Vj = 271-7*^! = ^ = 52*4 ft. per sec. 7 x6o = 2 = 57*6 ft. per sec. 5 The radius of gyration is called the swing radius by some engineers. THE FLY-PRESS: Stored energy at speed n^ = L r> 7 j Reserved stored energy = - 2 _ 2 g W 2 , ' - v = 80,375 ft -Ibs. The Ply press. This machine is used, in the form shown by the figure, either for embossing or stamping pieces of metal with THE FLY-PRESS. INDEX TO PARTS. D represents Disc supporting M. M Metal to be stamped. Punch or die. F Frame of machine. S represents Screw. N Nut for S. L Lever arms. B Balls or weights. some design, or for punching thin metal plates. The piece of met :il M, to be embossed or punched, is laid on a disc D, and the die or punch P is caused to come down on M with a large amount of stored-up energy, due to the operator taking hold of 284 LECTURE XXI. one or other of the heavy balls B, and giving them a very rapid turn round. The result of this movement is to send the quickly pitched square-double-threaded screw rapidly through its nut N, thereby forcing the guided square carrying the punch straight downwards, and causing the latter to overcome the resistance of the hard metal. Neglecting friction at the screw and the guide, and considering the combined weight of the two balls as = W Ibs., and v = their velocity in feet per second at the instant the punch meets the metal M, then The stored energy, or energy of the blow, = - ft.-lbs. If . 1= Length the punch or die goes into the metal in feet, And E, = Ilesiotance overcome (mean) in Ibs., Then ..... 1^ = ft.-lbs. 2g EXAMPLE IV. Distinguish between energy and power. What is the unit of power in this country ? In a fly-press two balls, each weighing 60 Ibs., are moving with a linear velocity of 15 feet per second, what is the measure of the energy existing in the balls (take ^ = 32) ? What is the power required to raise 6600 gallons of water up 150 feet in 30 minutes! A gallon of water weighs 10 Ibs. (S. and A. Exam. 1893.) ANSWER. (i) Energy is the capability of doing work which a body may possess on account of its position, or condition, or motion. Power is the rate of doing work, or the work done in a given time. The unit of power in this country is the horse-power and is the rate of doing work equivalent to 33,000 ft.-lbs. per minute. (2) Here W = combined weight of the two balls =120 Ibs. v = linear velocity of balls = 15 ft. per sec. Then energy existing in balls = 421-87 ft.-lb8. 2x32 (3) Weight of water raised 6600 x 10 per minute ~^ -- = 220 lbs - .". Work done per minute = 2200 x 150' = 330,000 ft.-lbs. ,". Power required 33> 000 _ 10 ENERGY STORED IN A ROTATING BODY. 285 To find Experiment ally the Energy Stored in the Rotating Mass Of a Fly-wheel. Description of the Experimental Machine. The accompanying figure shows a fly-wheel mounted on an axle. This axle is supported on ball bearings, thereby reducing the friction to a minimum. The diameter of the fly-wheel is 18 inches and weighs about 100 Ibs. If the centre of gravity of the fly-wheel is not exactly in the axis, then it is better to place the wheel as in the figure. One end of a cord is looped over a pin on the axle, and after being wound several times round the axle, the other end is led over an aluminium pivoted pulley, and attached to the 7-lb. weight. Object of Experiments. The object of this experiment is to illustrate , EXPERIMENTAL APPARATUS FOE DETERMINING THE ENERGY STORED UP IN A ROTATING FLY-WHEEL. the "Principle of Work" or the law of the "Conservation of Energy." Since energy cannot be generated or destroyed, the quantity given to a machine can be traced in its transmission through the machine, as clearly pointed out in Lecture V. Hence the amount of energy given to the fly- wheel and the attached weight in this experiment is measured by the pull of the earth on the 7-lb. weight multiplied by the distance through which the latter falls. Conservation of Energy. Part of this energy is stored up in the fly-wheel as kinetic energy, part of it is used to turn the wheel against the friction of the bearings, and part is stored as kinetic energy in the falling weight. The last two items of the total energy are converted into heat energy, and 286 LECTURE XXI. gt neral formula. fEnergy given to^ the apparatus ! during tha [ Kinetic } energy E* - = ! stored +- Energy " converted into heat falling of the [ weight. , in the ^fly- wheel. ; by friction. the several sub-divisions of the total energy may be connected under one f Kinetic energy in falling weight at 1 the instant when I the cord is released (^ from the axle. Determination of Speed of Machine by the Fly-wheel. After setting the mark upon the rim of the fly-wheel opposite the fixed mark, and with the 7-lb. weight on the cord to keep it taut, turn the wheel by hand to see as near as possible how many turns the wheel makes in falling a certain number of feet Let this distance through which the weight is to fall in the experiment be h feet, and let n be the number of turns made by the fly-wheel during this fall. Now determine by aid of a stop-watch how long it takes the 7-lb : weight to fall through h feet whilst turning the fly-wheel, and tabulate your results as follows : Weight Height Turns Time taken Velocity of Revolutions on cord in Ibs. of fall in feet of fly- wheel (n) foi weight to fall (h) weight at the instant when per second of 1 fly-wheel at feet cord leaves the the instant axle in feet when cord per second leaves the axle. Final Calculations. As we have found the total number of revolutions made by the wheel until the cord drops off its pin, and also observed the time taken in seconds. Then, as we know that the speed increases uniformly during tMs interval of time, the mean speed is just half the speed at the end of the time interval ; consequently, if we divide t lie number of revolutions by the number of seconds in which they were performed, and multiply this quotient by 2, we will get the number of revolutions per second made by the wheel when the weight just ceases to act. You can test the accuracy of your result by counting the number of revolutions of the fly-wheel from the time that the cord drops off until the wheel comes to rest due to its own friction and dividing by the time which has elapsed, then multiply this quotient by 2 and this will give you the required aged of the fly- wheel. MOTION ON BICYCLE AND RAILWAY CUHVES. 287 Motion on a Curved, Inclined, or "Banked" Track. Take as an example a cycling track, and let us suppose that the bicycle is moving with uniform velocity of v feet per second round a smooth circular course of radius OB equal to r feet. Then it is necessary to find at what angle to the horizontal plane the track should be inclined or " banked " in order that the bicycle may keep in its circular path. We see from the diagram that two forces are acting on the bicycle Of<- B = Radius r feet j Sectional Elevation of Bicycle, \ Course. Oj<- -0 B= -Radius r feel - Plan of above Bicyde Course. SECTIONAL ELEVATION AND PLAN OF A BICYCLE TRACK. (i) Its own weight W ; and (ii) the reaction R, which is perpendicular to the smooth track. These two forces have a resultant horizontal force P = , which is acting towards the centre of the horizontal circle in which the bicycle moves. Hence, if we take any vector line to represent the weight W and from the upper end thereof draw a horizontal line, and from the lower end draw another vector line inclined at an angle 6 to the vertical until it meets the horizontal line in a point. The sides of the triangle will represent in direction and magnitude the forces W, R, and P, and tan = = = f _ Li ^rj v -, from which we can quite easily obtain the required angle of the inclination of the track and the height h for the desired width of the same. Railway Curves. If the lines of a railway curve be laid at the same level, then the centripetal thrust of the rails on the wheel of trains passing round curves act on the flanges of the wheels, and the centri- fugal thrust of the wheel on the track would tend to push it sideways out T 288 -,LECTURE XXI. of its place. Bat, in order to have this action and reaction normal to the track, the outer rail is raised and the track thereby inclined to the hori- zontal. The amount of this super-elevation which is suitable for a par- ticular speed is quite easily calculated. SECTIONAL ELEVATION OF A RAILWAY CUEVE Let G = gauge in inches between the rails. v = velocity in feet per second of the train. r = radius of curve in feet. h = height of super-elevation of one rail above the other in inches. 6 = the angle of " banking " or inclined plane. Now draw the triangle ABC to represent the three forces, W, K, and P as indicated in the previous case. Then AC = AB sin = AB tan 6, since 6 is always very small. Since the retarding force acts for one minute, or sixty seconds . - The average retarding force = change of momentum -r 60 2000 3 x 6080 ** * n 32 x 60 x 60 Or, = 5 ' 2 7 tons. EXAMPLE VIII. A railway train starting from rest along a level line acquires a speed of 30 miles per hour in five minutes. What has been the mean pull between the engine and the train ; the resistances to motion being taken at 10 Ibs. per ton, and weight of train exclusive of engine 150 tons? ANSWER. Constant pull due to resistances =10x150=1 500 Ibs. ico x 2240 Mass of tram = m = j ^--150 x 70 Velocity of train*. = 30 miles per hour = 3 x S 28 feet P or 60 x 60 second. =* v = 44 feet per second. Momentum of train =-m x v**- 150 x 70 x 44 = 462,000. And the time taken to produce this momentum = 5 x 60 seconds. Average pull for change"! __ (mv 9 - mvA _ 462,000 _ n nf t.h A TvmmAnt.iim f ~ *~1 ~ TTTTT ~ J 54 lbs - of the momentum J y $ I 5 x 60 .-. Total mean pull between the) ,, engine and carriages j = 15 + '54 = 34 Ibs. Mass is defined as the quantity of matter in a body. It is measured by the weight of the body in pounds at London divided by the acceleration of gravity at the same place i.e., Mass = t0-r-<7; or, w in Ibs. 4- 32 '2. Inertia is defined as that property of matter whereby it tends to remain in a condition of rest or of uniform motion. Hence, we have Newton's first law of motion as stated in the second page of this Lecture, and which is sometimes termed the " Law of Inertia." NOTE. The force in Ibs. is the distance-rate at which work is done in foot-pounds, and it is also the lime-rate at which momentum is produced or destroyed. Moment of Momentum. Unit moment of momentum or angular momentum is unit momentum at unit perpendicular distance. WORK DONE BY A VARIABLE FORCE. 2 9 I EXAMPLE IX. A body weighing 3220 Ibs. was lifted vertically by rope, there being a damped spring balance to indicate the pulling force of .Fib. on the rope. When the body had been lifted x ft. from its position of rest, the pulling force was automatically recorded as follows : X 18 43 60 74 95 in 130 F 7700 7680 743 713 6770 5960 5160 39/0 Find approximately the work done on the body when it had risen 115 ft. How much of this is stored as potential and how much as kinetic energy ? What is, then, the velocity of the body ? (S. E. B. 1900.) Answer. Use squared paper, and plot the values of x along the 8000 00 10 20 30 40 50 60 70 80 90 100 110 120 C 3000 DIAGRAM OP WORK DONE BY A VARIABLE FORCE. abscissae OC, with the corresponding values of F as ordinates. Then draw a curve AB through the several points. Dmde OC into, say, 15 equal parts, and from the middle point of each division read off upon the vertical scale the corresponding values of F. The sum of these ordinates divided by 15 will give the mean force F in Ibs. This force multiplied by the distance through which it acts, viz., 130 ft. gives the total work done by the force or the area of the diagram OAJBC in ft. -Ibs. LECTURE XXL Total work done by force F=a,rea, of diagram OABO in ft. Ibs. Work done by force F in\_ f f . passing through 115 ft. J~ " " ="6951 xus =799, 360 ft. -Ibs. But, work done in lifting the\_ w , -vwisonft Ih* weight W through h feet j-WA-322ox 115 -370,300 ft.-lbs. Hence, work done by force] ("Work done in^l C K'netic energy, "\ Fibs, in passing through [ = \ lift.ng weight j- -f -j E^, in the body at V 115 ft. J [through 1 15 ft. J I end of the lift. J Or, 799'36o =370,3 00 = 799. 3 6 ~ 370.300=429,060 ft.-lbs. ,_ 429, 060 xzg 429,060x64 W 3220 ' ^=8528. v= ^8528 = 92*3 feet per second. Poten'ial Energy E^ in body at height of 115 ft. = 370, 300 ft.-lbs. Kinetic Energy E^ in body at height of 115 ft. =4=29,060 ft.-lbs. LECTURE XXI. QUESTIONS. 293 LECTURE XXL QUESTION!. 1. A body moves in a circle with a uniform velocity ; show that It must be acted on by a constant force tending towards the centre, and find the magnitude of the force in terms of the radius of the circle, and of the mass and velocity of the body. 2. A body weighing 2\ Ibs., fastened to one end of a thread 4 feet long, is swung round in a circle, of which the thread is the radius ; what will be its velocity when the tension of the thread is a force of 20 Ibs. (^=32) ! Ans. 32 feet per second. 3. When an unbalanced wheel is set in rapid rotation, a considerable amount of shake and vibration is experienced. You are required to explain this result from first principles, and to state the mechanical laws which appear to be at work. How would you calculate the amount of pull that this unbalanced weight exerts ? 4. What primary law in mechanics asserts itself when some revolving piece of machinery moves at a high velocity, and is unbalanced f A weight of i Ib. is placed on the rim of a wheel 2 feet in diameter, which revolves upon its axis and is otherwise balanced. The linear velocity of the rim being 30 feet per second, what is the pull on the axis as caused by the weight of i Ib. 1 Ans. 28*1 Ibs. 5. A segment of a fly-wheel, with the arm to which it is attached, weighs 3500 Ibs., and the mass of the portion may be taken as collected at a dis- tance of 8 feet from the axis of the wheel, which makes 40 revolutions per minute. What is the force tending to pull away the segment and arm from the boss of the wheel f Ant. 15,365 Ibs. 6. Define kinetic energy. How does it differ from potential energy ? If a velocity of 300 ft. per second is impressed on a weight of 10 Ibs., what is the measure of the energy now imparted to the weight. Ans. 14,062-5 ft.-lbs. 7. State the rule for finding the amount of work stored up in a given weight when moving with a given velocity. A weight of 6 cwt. moves with a velocity of 20 feet per second ; how many units of work are stored up in it ? Ans. 4200 ft.-lbs. 8. Write down the formula for the amount of energy stored up in a given weight when moving with a given velocity. Describe, with a sketch, the action of a fly-press. If each ball of the press weighs 50 Ibs., and the work stored up in the balls is 400 ft.-lbs., find the velocity with which they are moving. Take the number 32 to represent g. Ans. 16 feet per second. 9. Account for the storing up of energy in a rotating fly-wheel. If the weight of the rim be doubled while the rate of rotation remains unchanged, how much is the energy increased ? Ans. Twice. 10. State the formula for the energy stored up in a fly-wheel, on the sup- position that the whole of the material is collected in a heavy rim of given mean radius. Apply the formula to show (i) the effect of doubling the number of revolutions per minute ; (2) the effect of doubling the weight ; (3) the effect of increasing the mean radius in the proportion of 3 to 2. 11. A fly-wheel weighs 2j tons, and its mean rim has a velocity of 40 feet per second, If the wheel gives out 10,000 ft.-lbs. of energy, how much is its velocity diminished t An*, x -455 feet per second. 2 9 4 LECTURE XXI. QUESTIONS. 12. Explain the use of the fly-wheel in any machine with which yon are acquainted. To what class of machines is such a wheel usually applied T What is the kinetic energy in a wheel revolving at 150 revolutions per minute, if the wheel loses 5000 ft.-lbs. of energy when its speed is reduced to 147 revolutions per minute ? Ans. 126,263 ft.-lbs. 13. A fly-wheel of a shearing machine has 150,000 foot-pounds of kinetic energy stored in it when its speed is 250 revolutions per minute ; what energy does it part with during a reduction of speed to 200 revolutions per minute ? Ans. 54,000 ft.-lbs. If 82 per cent, of this energy given out is imparted to the shears during a stroke of 2 inches, what is the average force due to this on the blade of the shears ? (S. E. B. 1902.) Ans. 265,680 Ibs. 14. A fly-wheel is required to store 12,000 ft.-lbs. of energy as its speed increases from 98 to 102 revolutions per minute ; what is its kinetic energy at 100 revolutions per minute ? (S. E. B. 1900.) Ans. 150,000 ft.-lbs. 15. A machine is found to have 300,000 foot-pounds stored in it as kinetic energy when its main shaft makes 100 revolutions per minute ; if the speed changes to 98 revolutions per minute, how much kinetic energy has it lost 1 (S. E. B. 1901.) Ans. 11,880 ft.-lbs. 16. What do you understand by work, potential and kinetic energy? A bullet weighing i oz. leaves the muzzle of a rifle with a velocity of 1350 feet per second ; what is the kinetic energy of the bullet in ft.-lbs. ? Ans. 1780 ft.-lbs. 17. If a gun delivers 400 bullets per minute, each weighing 0*5 oz., with 2000 feet per second horizontal velocity ; neglecting the momentum of the gases, what is the average force exerted upon the gun ? (S. E. B. 1900.) Ans. 12*94 Ibs. 1 8. A bullet of O'l lb., with a speed of 2200 feet per second, is fired into the middle of a block of wood of 30 Ibs., which is at rest but free to move ; find the speed of the block and bullet afterwards. What is the loss of kinetic energy in foot-pounds? ' V S. E. B. 1902.) Ans. 7-3 ft. per sec. ; 7537 ft.-lbs. 19. A man and his bicycle weigh 170 Ibs. ; he has a speed indicator (not a mere counter). When going at 10 miles an hour on a level road he suddenly ceases to pedal, and in 15 seconds finds that his speed is 8 miles an hour. What is the force-resisting motion ? (S. E. B. 1901.) Ans. I lb. 20. A car weighing z\ tons and carrying 40 passengers, the average weight of each of them being 145 pounds, is travelling on a level rail at the rate of 6 miles an hour. What is its momentum in engineer's units ? If the propelling force be withdrawn, what average force in pounds must be exerted to bring the car to rest in two seconds ? and supposing the force to be constant, what distance would the car travel before it came to rest? Ans. 3135 Ibs. -ft. -sees. ; 1567-5^8.; 8*8 ft. 21. A car is drawn by a pull of P Ibs., varying in the following way t being seconds from the time of starting : P t 1020 980 882 720 702 650 713 722 805 2 5 8 10 13 16 19 22 The retarding force of friction is constant and equal to 410 lb. Plot P 410, and the time t, and find the time average of this excess force. What does this represent when it is multiplied by 22 seconds ? (S. E. B. 1902.) Ans. 366 Ibs. LECTUEE XXI. QUESTIONS, 295 22. A body weighing 1610 Ibs. was lifted vertically by a rope, there being a damped spring balance to indicate the palling foro F Ib. of the rope. When the body had been lifted x feet from its position of rest, the palling force was automatically recorded as follows : X ii 20 34 45 55 66 76 F 4010 39i5 3763 3532 3366 3208 3100 3007 Find approximately the work done on the body when it has risen 70 feet. How much of this is stored as potential energy, and how much as kinetic energy 1 Ans. 247,000 ft.-lbs. ; 1 12,700 ft. -Ibs. 5134,300 ft.-lbs. (3. E.B. 1901.) 23. A tramcar, weighing 15 tons, suddenly had the electric current cut off. At that instant its velocity was 16 miles per hour. Reckoning time from that instant, the following velocities, V, and times, t, were noted : V. Miles per hour 16 14 12 10 t. Seconds - - o 9*3 21 35 Calculate the average value of the retarding force and find the average value of the velocity from t = o to t = 35. (B. of E. 1903.) Ans. Retarding Force=264 Ibs.; Average Velocity =1274 miles per hour. 24. A projectile has kinetic energy = 1,670,000 foot-pounds at a velocity of 3000 feet per second. Later on its velocity is only 2000 feet per second, how much kinetic energy has it lost ? What is the cause of this loss of energy 1 Ans. 927,778 ft. -pounds. (B. of E. 1903.) 25. A man weighing 160 Ib. is in a lift which starts to descend with an acceleration of 2 feet per second per second. What force is exerted by the man upon the floor of the lift ? What would the force be if the lift were descending at a uniform speed ? Ans. 150 Ibs. ; 160 Ibs. (B. of E. 1903.) 26. In a gun, of which the internal diameter is 6 inches, a projectile weighing 100 Ib. has imparted to it in a distance of 12 feet a velocity of 2500 feet per second. Find the average pressure of the gases on the base of the projectile up to the time it leaves the gun. (Neglect friction and the energy of rotation of the projectile.) Ans. Average Pressure = 7200 Ibs. per sq. inch. (B. of E. 1903.) 27. A weight of 120 Ib. falls to the ground from a height of 18 feet and just rebounds. If the time of contact between weight and ground be the energy of rotation of the projectile.) Ans. Average Pressure = 28, 800 Ibs. per sq. inch. (B. of E. 1903.) 28. A train weighing 250 tons is moving at 40 miles per hour, what is its momentum in engineers' units ? If this momentum is destroyed in ten seconds, what is the average force acting on the train during these ten seconds. Define what is meant by force by people who have to make exact calculations. Ans. Average Force = 456 tons ; Momentum = 14 670 ton- feet seconds ? (B . o f E. 1904.) 29. A fly wheel weighs 24,000 Ib., its mean radius (or rather radius of gyration) is 10 feet, it revolves at 75 revolutions per minute what is its kinetic energy ? If suddenly disconnected from its engine, in how many revolutions will it come to rest, if we know that in each revolution the energy wasted in overcoming friction is 3000 foot-pounds ? Ans. E* = 2,315,051 ft.-lbs. ; and 772 revs. (B. of E. 1904.) 296 LECTURE XXI. QUESTIONS. 30. An ordinary steam engine has a stroke of 18 ins., and the connecting rod is 36 ius. long. The crank shaft makes 400 turns per minute. Find the velocity of the piston, in feet per minute, when it has moved through one-quarter of the stroke, reckoned from the back end. Ans. ^=1760 feet per minute. (C. & G., 1905, 0., Sec. A.) 31. A cricket ball, weighing 0-28 lb. f reaches the batsman when it is travelling horizontally at 96 feet per second ; what is its momentum in engineers' units ? The batsman drives the ball straight back to the bowler with the same speed ; what has been its change of momentum ? If the time of the blow is one-thirtieth of a second, find the average magnitude of the force exerted by the bat upon the ball. (B. of E. 1905.). Ans. Change of momentum = i'68 units ; Average Force = 50*4 Ibs. 32. A casting is bolted to an angle plate on the face plate of a lathe. The casting, angle plate, and bolts are equivalent to 75 Ibs. at a radius of 4^ inches. In what position must a weight of 20 Ibs. be fixed to the face plate to effect a balance 1 Ans. i6| inches from centre of plate. (B. of E. 1905.) 33. The angular position D of a rocking shaft at any time t is measured from a fixed position. Successive positions at intervals of 1/50 second have been determined as follows : Time t, ) seconds j o"oo 0*02 0-04 o'o6 o'o8 O'lO 0*13 0-14 o'i6 0-18 Position D, ) radians J o'io6 o"2o3 o'337 0-487 0*651 0-819 0-978 I'm I '201 1-222 Find the change of angular position during the first interval from t = o-o to t = o - o2 ; calculate the mean angular velocity during this interval in radians per second, and set this up on a time base as an ordinate at the middle of the interval. Eepeat this for the other intervals, tabulating the results, and drawing the curve showing approxi- mately angular velocity and time. Kead off the angular velocity when t = 0-075 second. (B. of E. 1905.) 297 LECTUKE XXII. CONTENTS. Some Properties of Materials employed by Mechanics Essen- tial Properties Extension Impenetrability Contingent Properties Divisibility Porosity Density Cohesion Compressibility and Dilatability Rigidity Tenacity Malleability Ductility Elas- ticity Fusibility Load, Stress, and Strain Total Stress and Inten- sity of Stress Tensile Stress and Stress Example I. Compressive Stress and Strain Example II. Limiting Stress or Ultimate Strength Safe Loads and Elasticity Limit of Elasticity Hooke's Law Factors of Safety Modulus of Elasticity Ratio of Stress to Strain- Examples III.-V. Resilience or Work Done in Extending or Com- pressing a Bar within the Elastic Limit Examples VI.-IX. Single Riveted Lap Joints Example X. Questions. Some Properties of Materials employed by Mechanics. The properties of matter are almost innumerable, but they may be divided into two classes: (i) Essential properties; (2) Con- tingent properties. The essential properties are those without which matter cannot possibly exist. The contingent properties are those which we find matter possessing, but without which we could conceive it to exist. Essential Properties 1. Extension means that property by which every body must occupy a certain bulk or volume. When we say that one body has the same volume as another, we do not mean that it has the same quantity of matter, but only that it occupies the same space.* 2. Impenetrability means that every body occupies space to the exclusion of every other body, or that two bodies cannot exist in the same space at the same time. Contingent Properties. 1. Divisibility means that matter may be divided into a great but not an infinite number of parts. The ultimate particles of matter are termed atoms, derived from a Greek word signifying indivisible. 2. Porosity signifies that every body contains throughout its mass minute spaces or insfcerstices to a greater or less extent. This has been proved to be the case with every known substance. These spaces are supposed to be filled with a highly elastic fluid called ether. For example, when the steel or cast-iron cylinder of a hydraulic * For Simple Rules of Mensuration see the Author's Elementary Manual on "Steam and the Steam Engine," Lectures I., II., III. 9 8 LECTURE xxn. press is subjected to enormous pressure, water will ooze through the metal from the interior to the outside. 3. Density is that property by which one body differs from another in respect of the quantity of matter which it contains. * Let M p M, = Masses of two bodies Let Vj, V, = Volumes of two bodies. Let Dj, D s = Densities of two bodies. If V, - V,, then gi = S ; if D, = D,, then - ,, M. If both vary, then ^ 4. Cohesion is that property by which particles of matter mutually attract each other at insensible or indefinitely small dis- tances. It is therefore different from gravitation, since the latter acts at all distances. It is evident that without this property we could not have a solid, for if a solid body be lifted by one part, the remainder sticks to it, and the whole is kept together by cohesion. 5. Compressibility and Dilatability are properties common to all bodies, by which they are capable of being compressed like a sponge or extended like a piece of india-rubber in a greater or less degree. 6. Rigidity signifies the stiffness to resist change of shape when acted on by external forces. Unpliable materials which possess this property in a large degree are termed hard, whilst u hose which readily yield to pressure, without disconnection, are called soft. Substances which cannot resist a change of shape without breaking are termed brittle, whilst those that do resist and at the same time change their form are said to be tough. 7. Tenacity is the resistance (due to cohesion) which a body offers to being pulled asunder, and is measured by the tensile strength in Ibs. per square inch of the cross section of the body. We will consider this property in the case of metals, &c., when lealinsr with stress and strain. 8. Malleability is that property by which certain solids may be pressed, rolled, or beaten out from one shape to another with- out fracture. It is therefore a property depending upon the softness, toughness, and tenacity of the material. Gold possesses this property in a higher degree than any other metal, and con- * The Density of a substance is either the number of units of mass in a unit of volume, in which case it is equal to the heaviness (i.e., weight of unit volume of substance in unit weight) ; or it is the ratio of the mass of a eriven volume of the substance to the mass of an equal volume of water, m which case it is equal to the specijic gravity. PROPERTIES OF MATERIALS. 299 Bequently sheets of gold are procurable of less than one-thousandth of an inch in thickness. Copper is one of the most useful of the malleable metals, and it may be beaten out into most elaborate shapes from the solid ingot. The Swedish iron of which horse- shoe nails are made is also very malleable, and is therefore highly prized by the blacksmith. Lead, although possessing softness, is not sufficiently tenacious to be considered a very malleable metal, but still it finds one of its most useful applications in the form of rolled lead sheathing for roofs of houses and interiors of water tanks, fec. 9. Ductility* is" that property by which some metals may be drawn down through a die-plate into wire or tubes. This pro- perty depends chiefly on toughness and tenacity. For example, we find that the very fine pianoforte wire used with Lord Kelvin's deep-sea sounding machine is both hard and rigid, but possesses great toughness and tenacity. The copper wire used for electrical conductors becomes harder and harder as it gets drawn down to smaller and smaller sizes, and it has therefore to be annealed in order to comply with the many bendings and unbendings which it has afterwards to undergo in winding and unwinding it upon bobbins whilst twisting it into a stranded conductor or in covering it with a dielectric of cotton, silk, gutta-percha, or india-rubber, 000 lbs - P 61 " S Q- ^h- (2) Original length = L = 10' = 120", and the increase of length = I = i". Let e * strain or extension per unit of length, i.e., per inch in this case, increase of length I i" .-. The Strain, or e = i i Jh = f = " " * 0083 original length L 120 Compressive Stress and Strain. If the line of action of a load be along the axis of a bar, shore, strut, or pillar, so as to tend to compress or shorten the same, the reaction per square inch of cross section is termed the compressive stress, and the diminution per unit of length is called the compressive strain. EXAMPLE II. A vertical support in the form of a hollow pillar, having 2 square inches cross section of metal, is 10 feet long. With a load of 10,000 Ibs. resting on the top, it is found to be compressed ^ of an inch in length, (i) What is the stress? (2) What is the strain ? ANSWER. (i) Here P = 10,000 Ibs., and A =* 2 sq. inches. Let p = stress or compression per sq. in. of cross section in Ibs. P 10,000 .. The stress, or p = -r- = = 5000 per square inch. A. 2 (2) Original length = L = 10' = 120", and the diminution of length = J = A" Let e =- strain or compression per unit of length, t.e., per inch in this case, diminution in length -i" or . - origiDaI lengtl f = ^ - -00083 Limiting Stress or Ultimate Strength. For every kind of material and every way in which a load is applied, there must be a value, which, if exceeded, causes rupture or fracture of tn 302 LECTURE XXII. body. The greatest stress which the material is capable of withstanding is called the limiting stress or ultimate strength per square inch of cross section of the substance, for the particular way in which the load is applied. Factors of Safety. The ratio of the ultimate strength or limiting stress to the safe working load is called the factor of safety. This factor of necessity varies greatly with different materials, and even with the same material, according to circumstances. For materials which are subjected to oxidation or to internal changes of any kind, the factor of safety must of necessity be larger than in those which are always kept dry or are well painted and carefully handled. There is no condition in engineering structures which requires a more careful calculation, or estimate of the necessary factors of safety, than that of railway bridges, which are exposed to all sorts of weathers and to extremely variable live loads. The skill of the engineer is therefore brought out, when he designs structures so as to include all possible circumstances to which they may be subjected, and so proportions the material at his disposal, that there shall be a minimum of internal stress and strain, with a maximum resistance to dead or live loads for a minimum cost of material and workmanship.* TABLE OF ULTIMATE STRENGTH AND WORKING STRESS OP MATE- RIALS WHEN IN TENSION, COMPRESSION, AND SHEARING. i Ultimate Strength. Working Stress. Tons per sq. inch. Tons per sq. inch. Materials. Ten- Com- Shear- Ten- Com- Shear- sion. pression. ing. sion. pression. ing. Cast iron . 7-5 45 14 i-5 9 3 Wrought-iron bars 25 20 20 5 3-5 4 Steel bars . 45 70 3 9 9 5 Copper bolts 15 25 3 5 Brass sheet . . 14 3 Safe Loads and Elasticity. As a rule, however, the object of the engineer is not to put such a stress on his materials of construction as will cause rupture or destruction, but rather to * For other tables relating to the Strength of Materials in Engineering Constructions, Factors of Safety, &c., refer to Rankine's "Rules and Tables," Molesworth's "Pocket Book of Engineering Formulae," D. K. Clarke's "Rules and Tables," "The Practical Engineer's Pocket Book;" and for Electrical Engineering Materials to Munro and Jarnieson's " Pocket Book of Electrical Rules and Tables." LIMIT OF ELASTICITY HOOKE'S LAW. 303 make machines and raise structures that will withstand all rea- sonable forces likely to be brought to bear upon them. Conse- quently, he is quite as much interested in what may be termed safe loads as in ultimate or destructive ones. He therefore requires to know what loads can be safely applied to materials under different circumstances, so as to comply with that most useful property termed elasticity, which we again define as the capability of regaining their original size, shape, and even strength, after the removal oftht forces which caused a change of form in them. Limit of Elasticity Hooke's Law. So long as the stress or reaction per square inch of cross section does not exceed a certain limit, called the limit of elasticity, then the material will return to its original shape, size, and strength, after the removal of the load. This limit has been ascertained for most materials of construction by elaborate experiments, which are to be found tabulated in the Proceedings of the Institutions of the Civil and Mechanical Engineers, and in such books as Rankine's " Rules and Tables," Molesworth's " Pocket Book of Engineering Formula?," and D. K. Clark's " Rules and Tables." For example, with a bar of good wrought iron the elastic limit is only reached after a stress of 24,000 Ibs. per square inch has been brought to bear upon it, and in a similar degree every other material has a corresponding limit, beyond which it is not safe to stress it, for tear that it should be overstrained, and thus lose, to a certain extent, its property of recuperation or restitution, or take a permanent set. Within this limit, Hooked Law holds good for metal bars under the action of forces tending to elongate or compress them. This law states that : (1) The amount of extension or compression for the same bar is in direct proportion to the stress. (2) The extension or compression is directly proportional to the length. (3) The extension or compression is inversely proportional to the cross sectional area ; consequently, if the area be doubled the extension or compression will be halved, or the resistance to the load will be doubled. Let P = Pull, push, or load in Ibs. on the bar. A = Area of cross section of the bar. L = Length of the bar before the load was applied. n I = Length by which the bar is extended or compressed. w p = Stress or load per square inch of cross section = P/A. P Then, so> long as Y- does not exceed the elastic limit, I varies directly U JO4 LECTURE XXIL I P as P for the same bar ; or ^ varies directly as -^> for different bari of the same material and subjected to the same conditions. In other words, so long as the stress does not exceed the elastic limit, the strain will be proportional to the stress. Modulus of Elasticity, or Ratio of Stress to Strain. As we have just indicated, by HOOKE'S LAW, if a metal under test be gradually subjected to a stress, and if the load does not exceed the limits of elasticity of the material, the strain will be in pro- portion to the load. Consequently, the ratio of the stress to the strain is a constant Quantity for eaoh particular substance within the limits of Hooke s ^aw, and is termed the Modulus of Elasticity of the substance. But . . Stress oc Strain Stress = Ex Strain.* Where E represents a constant number or modulus depending on the natural elasticity of each material JP . F Stress ^ PL^ Strain " l_ '' Al L .., ; . . . PL = AZE Or, imagine it is pure imagination that a substance could be elongated to double its length or compressed to zero by sub jecting it to a certain load, we should then have an index value, or constant number, or modulus, by which we could compare it with every other substance which behaved likewise under similar circumstances. This imaginary value is termed the Modulus of Elasticity. For example, take a bar of wrought iron of i square inch cross section, which is found to stretch 2 4 a o^ o o o P ar ^ ^ ^ s length under a stress of i lb., and consequently by Hooke's Law twice that amount under a stress of 2 Ibs., and so on ; then this number (24,000,000) is called the Modulus of Elasticity of the iron bar. For, if the elasticity of the bar were perfect, it is evident that a stress of 24,000,000 Ibs. would produce a strain or elongation equal to the length of the bar, or, fjoo = i. In other words, the length of the bar would be doubled under this stress. Consequently, we have the following definition, * Since stress is reckoned by so many Ibs. per square inch of cross section of a material, and strain is simply an abstract number, it follows that the Modulus of Elasticity (E) must also be reckoned by so many Ibs. per square inch. MODULUS OF ELASTICITY. 30S DEFINITION. The Modulus of Elasticity of any substance is that load which would double its length on the supposition that the elongation was proportional to the stress, and that the cross section t>f the bar was of unit area, or one square inch, and supposing the bar to remain perfect during the operation. From this we again see that Mod ulus of Elasticity = 8tres8 = E = ? / * strain A/ L Or, .... PL = AZE MODULI OP ELASTICITY TO STRETCHING. (See Rankine's Rules and Tables for complete Data.) Material Modulus of Elasticity in Ibs. per sq. in. in round numbers. Material. Modulus of Eia.-. i. R= E X - 2 - mch-lbs. Work done per unit^ _ p 1 of volume j~2E This latter equation gives the strain energy stored in the 6ar, since the material is still elastic. II. When the Load is Suddenly Applied without Initial Velocity. Let I be the elongation or shortening of the bar when the load P is suddenly applied, but without initial velocity. Then, the work done by the external load must be equal to the energy tored up in the bar. Let P max be the maximum stress per square inch which is produced in the bar. p The me*n stress will be max as its initial value was o. Hence, the work done on the bar = - fnax = -. , and the work done by external load = PZ. Therefore, ^J^^^^pf, 2 2L I P or, P,^=c2P; andEj- = 2-^. Consequently, the maximum intensity of stress induced in a bar by the sudden application of a constant load, without initial velocity, is double the intensity of the stress produced by the load itself. III. When the Load is Suddenly Applied and with an Initial Velocity. Let a weight of W Ibs. be dropped from a height of h inches upon a bar, and that the maximum stress produced in the bar was P max Ibs. per square inch, whilst the elongation or compression was I inches. 3O8 LECTURE XXII. Work done by the falling weight = W (A + J) = energy stored up in the bar. P J, EAZ 2 Therefore, -^= - =W (h + l)=Wh if he value of I is small. If v=velocity of load P at moment of impact in ft. per sec., then kinetic energy, E* = g^ = ^ = W ( h + *) ft , lbs . 2y 2 A. 1 2 EXAMPLE VI. Calculate the resilience of a steel tie-bar, i inch in diameter and 4 feet long if the elastic limit is reached under a load ot 20 tons, and modulus of elasticity = 13,000 tons per square inch. Answer. Work done or resilience, 11 = = - inch-tons. EXAMPLE VII. A round bar of steel is 20 feet long and i inch in diameter. Find the tensile load which, if suddenly applied, would cau^e an instantaneous elongation of the bar of 'I inch. Taking = 13,000 tous per square inch. . T7I \ J P= =- where symbol letters represent the values stated 2Ll in the" text. p _ 1 3,000x785 ix 'I 2 X 2O X 1 2 EXAMPLE VIII. Determine the greatest weight that can be dropped from a height of i foot on a bar of steel which is i inch in diameter and 10 feet long. The modulus of elasticity, =13,000 tons per square inch, and the elastic limit 18 tons per square inch. Also, find the alteration in the length of the bar. Answer. p m P=A = -i7 13000x2 RESILIENCE WITH EXAMPLES. 309 EXAMPLE IX. In a tensile test of a piece of flat wrought iron bar the following results were obtained : (i) Original dimensions of cross-section, 2*02 inches by 0-51 inch. (ii) Final dimensions of cross-section at point of fracture. I -49 inches by 0*39 inch. (iii) Gross load at limit of elasticity, 36,000 Ibs. (iv) Gross load at fracture, 59,000 Ibs. (v) Total final extension on a length of 10 inches, 1*63 inches, extension when load was 22,000 Ibs., -0075 inch. Find from the above data, (a) the modulus of elasticity, (6) limit of elasticity, and tenacity in Ibs. per square inch ; also (c) the reduction of area per cent., and (d) the approximate work done in fracturing this specimen. (L.U.B.Sc. Eng. 1903.) Answer. (a) Young's] pL modulus of V E = -T-J elasticity J ** _ 22,OOOX 10 220,000 2 '02 X5 1 X -0075 ~ -007726 E = 28,473,440 Ibs. per square inch. (b) Stress an elastic limit I Load at elastic limit _ 36,000 in tons per j ~~ Original area ~~ i -0302 square inch J =34,944 Ibs. per square inch. (c) Percentage] Originalarea-final area contraction of >- = ,. . . : - - x 100 area J Original area ^(2-02 x -51) -(i '49 x -39) IOQ 2'O2X '51 (d) Work done^ fnch in frac = Work dope in stretching the test-piece taring ^e | Volume of the test-piece specimen J /Mean load between starting-point and point of\ _\ fracture x distance moved through in inches / ""/Area in square inches x length of test-piece in) \ inches / IO _ Final length - original length on" io & inches / ~" Original length * I -^^=16-3 per cent, 310 LECTURE XXII. Single-Riveted Lap Joints. (See the author's text-book on Steam, ., re Riveted Joints.) Let p = pitch of rivets in inches. d = diameter of rivet hole in inches. t = thickness of plate in inches. \ ft = tearing resistance or tensile stress of the plate in Ibs. per square inch. / 8 ^ shearing resistance or shear stress of the rivet in Ibs. pel square inch. n SINGLE-RIVETED" T^AP JOINT. Then for a single-riveted lap joint, the area of plate under tensile stress = (p - d)t, Trd~ and the area of rivet under shear stress = - 4 Hence, for equal strength, the following equation must be true : In practice, d = i '2 y , or = - is usually taken for boilers. ft(p-d) =/ x i'44 = "36 x X/r Or, The values to be taken for/ t and/ s in any given case depends upon the number of rows of rivets ; upon the material which is used (iron or steel), and on whether the holes have been punched or drilled. But, in practice ft ranges from 35,000 to 67,000 Ibs. per square inch, and / from 43,000 to 53>oo Ibs. per square inch. Whilst for iron plates and iron rivets, with drilled holes, the ratio may be taken as equal to '94; which would give for the single-riveted lap joint (p-d)o*94= 1*131. SINGLE-RIVETED LAP JOINT. 311 EXAMPLE X. In a single-riveted lap joint the thickness of the plate ia | inch, and the diameter of the rivet is I inch. If the tearing resistance of the plate is 60,000 Ibs. per square inch, and the shearing resistance cf the rivet is 50,000 Ibs. per square inch, find the proper pitch of the rivets. (C. & G., 1905, O., Sec. B.) Answer. Substituting the numerical values given by the question in the above formula, we get : (p- = '95 + i = 1'95 inches. 312 LECTURE XXII. QUESTIONS. LECTURE XXII. QUESTIONS. 1. State and define the essential and contingent properties of matter, and give the names of those engineering materials with which you happen to be practically acquainted, that best exemplify each property. 2. What is the meaning of the term ductility as applied to wrought iron ? Describe, with sketches, some apparatus for testing a piece of metal as to ductility. If a uniform bar of iron 10 inches long is found to stretch i inches at the time of fracture, what is the measure of the ductility of the material of the bar 1 Ans. 15 per cent. 3. Give the approximate breaking tensile stress for a bar of cast iron of one square inch sectional area, and the same for a bar of wrought iron 1 What is the meaning of the term ductility as applied to wrought iron, and how is the ductility of iron measured ? 4. What must be the diameter in inches of a round rod of wrought iron in order to sustain a load of 50 tons ? It is given that a bar of iron i square inch in section will just support a load of 25 tons. Ans.d=\/ -- = i'6". 5. What is the modulus of elasticity of a substance ? A round bar of iron, 12 feet long and i^ square inches in sectional area, is held at one end and pulled by a force till it stretches inch ; find the force, the modulus of elasticity being 30,000,000, Ans. 39,063 Ibs. 6. A round bar of steel i" in diameter and 10 feet long, is fi.xed at its upper end, and a load is applied to the bottom end and stretches it -05". Find the load if the modulus of elasticity is 30,000,000. Ans. 9817 '5 Ibs. 7. Find the dimensions of a transverse section of a square rod of fir to sustain a suspended load of 10 tons, the rod being held vertically. The breaking load of a rod of fir one square inch in section is 6 tons. Ans. 1-29 inches. 8. Find the extension produced in a bar of wrought iron 4 feet long and 2 square inches in section by a suspended weight of 4! tons, the modulus of elasticity of the material being 29,000,000 pounds per square inch. Ans. -009 inch. 9. What do you understand by the terms stress, strain, and modulus of elasticity ? A tie-rod, 100' long and 2 square inches cross area, is stretched .75" under a tension load of 32,000 Ibs. What is the intensity of the stress, the strain, and the modulus of elasticity under these circumstances? Ans. 16,000 Ibs. per square inch; 0-000625; 25,600,000. 10. Define what is meant by " dead load," " live load," " limiting stress," "limit of elasticity," and " factors of safety." 11. What do you understand by stress and strain respectively? If an iron rod, 50 ft. long, is lengthened by in. under the influence of a stress, what is the strain ? If the rod is 2 sq. in. in section, and the load 1 1,000 Ibs., what is the modulus of elasticity ? Ans. '000417 ; 13,200,000. 12. Find the stress produced in * pump-rod 4" diameter, lifting a bucket 28" diameter if the pressure on the top of the bucket be 6 Ibs. per square inch in addition to the atmosphere, and the vacuum below the bucket be 26" by gauge. Beckon each 2" of vacuum = i Ib. Ans. 931 Ibs. per sq. inch. 13. If the rod in question is 5' long, find its extension if the modulus of elasticity = 9,000,000. Ans. '006 inch. LECTURE XXII. QUESTIONS. 313 14. What do you understand by the terms tensile, compressive and thearing strength respectively of any material? Define "modulus of elasticity." If a wrought-iron bar of I square inch sectional area just breaks under a tensile stress of 60,000 Ibs., what would be the area of the section of a tie-rod which would just support a load of 20 tons? Ans. 75 sq. inch. 15. A wrought-iron tie bar, f inch in diameter, has a modulus of elasticity of 28,000,000 Ibs. per square inch. Its length is 23 inches ; find the load under which the bar will extend '015 of an inch. Find also the stress per square inch. Ans. 8067*6 Ibs. and 18,261 Ibs. per sq. inch. 1 6. How would you find out for yourself the behaviour of steel wire loaded in tension till it breaks ? What occurs in the material ? Use the words "stress" and "strain" in their exact senses. 17. An iron rod, of i inch diameter and 12 feet in length, stretches 3/32-inch under a load of 6 tons suspended at its extremity. Determine the stress, strain, and modulus of elasticity of the bar. Ans. 17, 112*3 Ibs. per sq. in. ; 0.00065 ; E = 26,526,615 Ibs. per sq. in. 1 8. What do you mean by Stress, Strain, and Modulus of Elasticity T A wire 10' long and | sq. inch in sectional area is hung vertically, and a load of 450 Ibs. is attached to its extremity, when the wire stretches 015" in length. What are the stress and strain respectively? And also the modulus of elasticity? Ans. 3600 Ibs. per sq. inch; -000125; E = 28,800,000 Ibs. per sq. inch. 19. An iron wire is loaded with gradually increasing tensile loads till it breaks. We want to know its modulus of elasticity, its elastic limit stress, and its breaking stress. What measurements and calculations do we make? (S. E. B. 1900.) 20. Sketch apparatus and describe a laboratory experiment by which you could find E, Young's modulus of elasticity, for an iron wire 10 feet long and 0*05 inch diameter. How would you secure the upper end of the wire ? How apply the load ? And how measure the elongation ? How would yon plot your results and how deduce the value of E? (S. E. B. 1902.) 21. Describe an experiment by which you could determine E, Young's modulus of elasticity, by stretching an iron wire. (B. of E. 1903.) 22. A bolt z\ inches diameter has a tensile load of 30 tons, what is the stress ? What is the strain if Young's modulus of elasticity is 3 x io 7 pounds per square inch ? What is the elongation of a part which when unloaded was 102 inches long ? (B. of E. 1904.) Ans. Stress = 13,689*8 Ibs. per sq. inch. Strain = '000456. Elongation = '0465 inch. 23. Two bars of equal length, both;of rectangular section but of different materials, are firmly riveted together at their ends and subjected to a pull so that they are compelled to stretch the same amount. If A 1? A 3 represent their sectional areas, and E lf E 2 the values of Young's modulus for the two materials, show that when the pull is P Ibs. the intensities of stress induced in the two bars are PE, PE 2 and A! Ej + A 2 B a A x Ej + A 2 E 2 respectively the limits of elasticity being not exceeded. (C. & G., 1004, 0., Sec. B.) 3*4 NOTES AND QUESTION'S. 24. A bar of mild steel, of rectangular section, is 2 inches wide and J inch thick, and is 10 inches long. If Young's modulus is 12,500 tons per square inch, find the amount the bar stretches when the load on it is 10 tons. How much work is then stored up in the bar ? Ans. I = -008 inch ; work stored up in bar = 89*6 in.-lbs. (C. & G., 1905, 0., Sec. B.) 25. An iron column is 12 inches in external diameter, and the metal is i ^ inch thick. The load on the column is 125 tons. What is the com- pressive stress in the metal ? By what amount will the column be shortened, if its length is 15 feet, and if Young's modulus of elasticity is 12,500,000 pounds per square inch? Ans. f c = 6630 Ibs. per square inch ; I = -095 inch. (B. of E. 1905.) 26. The following results were obtained during a tensile test of a mild gteel bar f inch in diameter : Total load on^| the bar in V 0-88 176 2-64 S-Sa 4-40 5-28 6-16 7-04 tons . . J Elongation on"\ a length of 1 8 inches stated j 0-0012 O'O024 0-0035 0-0047 0-0061 0-0075 0-0088 0-0102 in inches . J (a) Plot a curve on squared paper, going evenly through the points, to show the relation between the load and the elongation. (b) Find the load necessary to cause an elongation of 0*0040 inch. (c) Find the total work done in inch-tons upon this 8" length of the bai during the test. (B. of E. 1905.) Ans. (b) Load = 3-0^ tons. (c) Total work done -0-036 inch-ton. NOTES AND QUESTIONS. LECTURE XXIII. CONTENTS. Stresses on Chains Shearing Stress and Strain Example I. Torque or Twisting Moment Torsion of wires Table giving the strength, moduli of Elasticity and Kigidity of various materials Strength of Solid Kound Shafts Example II. Table giving the Horse-Power which steel shafting will transmit at various speeds Strength of Hollow Kound Shafts Kelation between the Twisting Moment and Horse-Power transmitted by shafting, as well as the diameter necessary to transmit a given Horse-Power Examples, III. IV, Questions. IN this Lecture we will continue the subject of " strength of mate- rials," and finish the course with reasons for the shapes generally given to sections of cast iron, wrought iron, and steel girders. ^ Stresses on Chains. The only stress to which the sides of the links of chains are subjected under ordinary circumstances, is that of tension. This stress tends to bring the sides of the links closer together, and consequently we find that large chain cables for mooring ships (where very sudden and severe stresses are encoun- tered) have a cast-iron stud or wedge fitted between the inner sides of the links. These studs most effectually keep the sides of the links apart, and prevent any link jamming a neighbouring one. They add materially to the strength of the chain, for they are in compression whilst the sides of the links are in tension. Being composed of cast iron, which offers the immense resistance to compression of fully 45 tons per square inch,* there is not much fear of their giving way before the sides of the links. The strength of a stud-link may be taken as equal to double the strength of a rod of wrought iron, of the same diameter and quality of material as that of which the chain is composed, whereas the strength of an open-link chain is only about 70 per cent, of this amount, even with perfect we! ding. f In Molesworth's " Pocket-Book of Engineering Formulae," the student will find at page 54 a formula for the safe load on chains viz. W=7.i^ Where W = Safe load in tons. d = Diameter of iron in inches. * See Table of the Ultimate Strengths and Safe Working Loads given In Lecture XXII. t Some well-known authorities give less than 70 per cent SHEARING STRESS AND STRAIN. 317 Now, Btich a formula is very easy of application, but the student should never rest content until he finds out how the constants have been arrived at, and what relation the various symbols have towards each other. If he refers back to the short table of " Ultimate Strengths and Working Loads " given in the previous Lecture, he will find opposite wrought-iron bars and under tension, the value 5 tons per square inch as the safe working load. Con- sequently, applying what was said above about perfect stud-link chains, he will see that rtwice the load of a rod of the same diameter and quality as that of which the chain is composed. .. W = 2 x 5 x cross area of the chain iron. W=2 x 5 x-d*=2* 5x^x^^ = 7-8^ 4 7 This is near enough to the constant given by the above empirical formula to enable him to see how it has been obtained. Chains which are subjected to many sudden jerks (such as lift- ing chains for cranes and slings) become in time crystalline, or short in the grain, and consequently brittle and unsafe. The best precaution to adopt in order to periodically remove this en- forced internal condition, is to draw them once a year very slowly through a fire, thus allowing them to become heated to a dull red, and then to cool them slowly in a heap of ashes. This method is followed at Woolwich Arsenal and some other Government works. Shearing Stress and Strain. The action which is produced by shearing and punching machines on iron, steel, or copper plates, &c., is to force one portion of the metal across an adjacent portion. The shearing stress is the reaction per square inch opposing the load or pressure applied to the shears or punch, and the shearing strain is the deformation per unit length or volume. Rivets holding boiler plates together, fulcra of levers, the pins of the links of the chain of a suspension bridge, the cotter keys of a pump rod, are all subjected to shearing stresses and strains. The ultimate and the working shearing stresses for a few engineering materials were given in a table in Lecture XXII. In the case of loaded beams (which we will consider shortly in connection with bending moments) the shearing force at any point or any transverse section thereof is equal to the algebraical sum of all the forces on either side of the point or section. EXAMPLE I. A steel punch i" diameter is used in a large shipyard punching machine to make holes in steel plates i" thick. What will be the total shearing stress or least pressure required ? 3r8 LECTUKE XXI1T. ANSWER. Referring to the table in last Lecture, we see that the ultimate shearing strength or shearing stress for steel bars (which we will assume to be the same as for plates) is 30 tons per square inch. Now a hoi 3 i" diameter has a circumference = ird = 3* 14", and since the plate is i" thick, the area of the resisting section must be the circumference of the hole x its depth, or = 3' 14" x i" = 3'i4sq. in., and the total pressure required = 30 tons x 3-14 = 94-2 tons. Torque, or Twisting Moment.* In the case of a shaft having a lever, pulley, or wheel fixed to it with a force P Ibs., applied at radius R feet from the centre of the shaft, then The twisting moment T.M. is = P x R Ibs.-feet. Or, The torque = P x R x 12 Ibs.-inches. * The term torque was devised by the late Professor James Thomson, of Hlasgow University, to signify the twisting or torsional moment. The Ibs. -feet of torque must not be confused with ft. -Ibs. of work ; or with resilience (which is the work done in straining a body, as measured by the elongation or compression in feetxthe mean load causing the strain). It will therefore save confusion, if we take the force applied at the end of the arm in Ibs., and the leverage or arm in inches, and then multiply them, so as to get the torque in Ibs. -inches. APPARATUS FOR SHOWING THAT THE ANGLE OF TORSION IN ROD is PROPORTIONAL TO ITS LENGTH, AND THE VARIATIONS IN THE RIGIDITY OF DIFFERENT MATERIALS. TORSION OF WIRES. Torsion in Bods. It will be seen from the previous figure, that the front puller carries a cord with known weights suspended from it. The spindle of this pulley moves in ball bearings, and carries at its inner end a three-jaw chuck. This chuck is for holding one end of the rod under test, whilst the ther end is for tightly clamping it to the back bracket of the machine. The torsional couple is applied by means of weights hung from the afore-mentioned cord, and the torsion in degrees is read off on the dial. To eliminate errors the weight should be hung first on one side of the pulley, then on the other side, and the mean of the two readings taken. It will be noticed, that both the pointer and the scale can be readily moved to any distance from the fixed end of the rod. It can thus be shown that the angle of torsion in a rod is proportional to its length and that different materials have different rigidities. Torsion of Wires. In the figure, AB represents a wire held firmly at the top end of a supporting rod, which is 8 feet long. A pulley is fixed firmly to the wire at B, and this pulley is acted upon by two cords which tend to turn it without moving its centre sideways, i.e., they act on the pulley with a turning moment only. But, the pulley can only turn by giving a twist to the wire. Hence, if a light pointer be fastened to the wire at 0, the former moves over a dial, and the angle turned through by the wire and the pointer is called the total angle of twist at 0. Similar pointers fixed to the wire at D and E give the angle of torsion at these points. The dials at D and E are supported upon adjustable sliders, so that they may be moved up or down the vertical rod in order that the torsion of any length of rod may be measured. (i) If the length AE be i foot, and the distance between the dials E and D, D and C, be also i foot, then we should find that the angles of twist at E,D and are as i : 2 : 3 respectively. That is, the angle of twist is proportional to the length of wire twisted. (ii) By varying the twisting moment and noting the angle of twist which is produced at each variation, you can prove that the angle of torsion or twist is proportional to the twisting moment. (iii) If you try different sizes of wires of the same material and length, and apply to each of them the same twisting moment, then you will find that the amount of twist produced in them, will be inversely as the fourth power of the diameter of the wires. (iv) If you take wires of the same diameter and length, but of different materials, and apply the same twisting moment to them, then you will find that the amount of twist will be inversely pro- portional to the modulus of rigidity of the material. Note, the Modulus of Rigidity of a Material in Ibs. per square inch may be defined as the ratio of shearing stress to shearing strain. This constant or coefficient of shearing elasticity for each particular material is indicated in the following table by the letter 0. 320 LECTUEE XXIII. EXPEEIMBNTAL APPAEATUS FOB MEASURING TH] TOESION OF WIRES. STEENGTH OF SOLID ROUND SHAFTS. 321 BTBBNGTH, MODULI OF ELASTICITY, AND RIGIDITY OP VARIOUS MATERIALS. Materials. Breaking strength to resist tension, n tons per square inch. Modulusof elasticity, E, in tons per square inch. Modulus of rigidity. C, in tons per square inch. Aluminium - bronze ) (90% copper Und 40 6500 2500 10% aluminium) . J Brass wire . . 20 to 25 5000 to 6500 2OOO tO 23OO Oast-iron .... 5 { o 15 4500 to 7000 1700 to 2700 Charcoal-iron (hard- V "?S to 40 drawn) .... Charcoal-iron I OJ *T \ 12,500 to 13,500 5000 to 5500 (annealed) . . } 30 Copper (cast) . . . 8 to 12 5000 to 6000 1900 to 2300 (rolled) . . 13 to 16 5500 to 7500 2100 tO 2900 wire (annealed) 18 to 20 (hard-drn.) 26 to 30 Delta Metal (forged) 22 tO 24 6350 234O German-silver wire . 30 4800 l8OO Gun-metal (90% cop- per and 10% tin) . Muntz metal (rolled or forged) . . . V 12 tO 17 \ 22 5000 6350 IQOO 2340 Phosphor bronze 16 to 18 ) wire (hard-drawn) J- 45 to 70 > 6000 to 7000 2300 to 2700 Platinum .... 10,500 4000 Steel (ordinary) . . (annealed) . . 7o IOO \ 13,000 to 14,000 5200 to 5700 Strength of Solid Round Shafts. It is evident trom the above, that a shaft subjected to a twisting moment must offer a sufficient resistance thereto, otherwise it would be twisted, or sheared, or ruptured through by the torque. It may be proved that in the case of solid round shafts their resistance to torsion is directly proportional to the cubes cf their diameters when made of the same material and quality.* * This is evident from the fact that the shaft must offer a moment of resiitance, or shearing moment, equal to the twisting moment at the instant of rupture. Now, the area to be sheared, is the cross area of the shaft =!LD a . where D is the diameter of the shaft. The mean arm or leverage at 4 which this resistance acts is equal to half the radius of the shaft, for at the centre the arm is=o, and at the circumference it is=r, the radius of the shaft. The mean arm is therefore= And, if the shearing resistance per tauwre inch of cross section of the material be=/, the product of these three quantities will be the total thearing moment, and must equal the twitting moment viz. =P x R, where P is the force applied at the end of the 322 LECTURE XXIII. Let Dj, !> D, - Diameters of three shafts, i", 2", and 3" dia- meter respectively. Tj, T r T 3 Torques which they will respectively resist when stressed to the same extent. Then, . . T^T.rT.iiD/iD/.-D/ Or, . . . T^T,:!,:: i : 2' : 3 ::i : 8 : 27. In -other words, the strengths of the three solid shafts will be as 1:8:27. A good wrought- iron shaft of i" diameter has been found to withstand a torque of 800 lbs,-ft., or 9600 Ibs.-inches, which means that they will resist 800 Ibs. force at i foot, or 12" leverage, or 400 Ibs. at 2 feet, or 24", and so on. Or, . P x R' = 800 Ibs.-feet of torque i.e., . P x R" = 9600 Ibs. -inch torque. EXAMPLE II. On the above basis, what force acting at the circumference of a pulley 20" diameter will break a wrought- iron shaft 2" diameter? ANSWER. By the above rule we have the proportion : T I: T, -D^iD/ But Tj - P! x R/' = 800 Ibs x i a" AndT,= P 8 x R 8 " - P, x 10" ... P^: P 2 R, :: J>*:J>f i.e., P a R a x D^ = P^ x D, 1 P.R. x D, 8 800 x 12" x 8 <0r > p . - iy x R, -nr^" ' 76801bs lever or circumference of the pulley, and K the length of the arm or radius of the wheel or pulley. Consequently, P x E =-J^( ^-D 2 x - \ =f g "- ' W But S is a constant quantity for any particular material. Also, TT and 16 are constants. .. P x R vary as D 3 . At the instant of rupture the strength of the shaft just balances or is equal to the twisting moment P x R. .*. The strength of shaft varies as D 8 . This is the same as the general statement in the text above. Without Borne such algebraical explanation, students are sorely puzzled how the cube of the diameter crops up ; or still more so when they see the following which appears in some text-books. :Such a statement is, however, quite evident after the above analysis. (We must leave the consideration of hollow shafts, tubes, &c.. to our Advanced Course.) H.P. TRANSMITTED BT STEEL SHAPTa 323 5-1 C? 02 I JOd 'SAag O W Tj-vO OO O CM 'i-vO OO O CM **-$ 00 OvoOvoQVOOvoOvoO VO < MMMMIH O\ H - O lOO ")O OO *O O OO lOO lOO CO OOO O CO U10O O CO lOOO O ci tot^O w lotxo M iotxO oi lOt^OO woo o M r^cogvb ooo ^o M H M CI 0* 01 COCOCOCO-'i-Ti-rl-Tj-iO lOvO \O txCO 00 Os O O w HI M ' Ot 01 COCOCO^-^trfvO' & Scl 3 1 I a \ vp op ON p ci ^ jovp oo p\ p M co t ^ yo co M ON txvp .""t- w p op vp ^ ji VOOO M vooo M Vf Vx O COvb O COvb ON C1 O 00 vo CO HI O\ Cx, vo W O 00 vb M HI C1 01 COCOCO^^-^vovovo vovO tx txOO ONO O HI Ot COrfrfvo ONM ti ' . M f* y^^p

^ ?*& For co^yoyovp fxoo ON ONO M H p - ton-inches. N 2240 x N N But, T.M.=^|&. Hence, ^'=63,03^ If we assume the safe values of the stress f t to be as follows : cast-Iron 3600 Ibs. per square inch ; wrought-iron = 9000 Ibs. per square inch ; and steel * 13,500 Ibs. per square inch, then the diameter in inches for a round shaft in terms of the horse-power to be transmitted is, for Oast-iron Of course, the twisting moment is here assumed to remain constant at its mean value. In practice the twisting moment varies in many cases, and to allow for this it is usual to take the maximum twisting moment from i -3 to 1*5 times the mean twisting moment, thus the values of diameter d as found above are slightly increased. It should also be borne in mind that shafts in practice are subjected to bending as well as twisting, owing to the loads due to the weights of STRENGTH OF SHAFTS. 325 pulleys and the pulls of the belts. Hence, if the usual rule for the dia- \ 3 /TT p meter of a wrought-iron shaft is <*=3'3/y ^7- - when torsion only is con- sidered, then it wiU be d=ex3'3/y S^ 1 when tending is taken into account. Some values of the coefficient c are given in the following table : Kind of Shaft. Value of c. Propeller shafts of steamships, and shafts with similar load ... ria Line shafting in mills, etc i '3 Crank-shafts and shafting subjected to shocks, such as i -42 EXAMPLE III. (a) Find the diameter of a solid steel propeller shaft to transmit 12,000 H.P. at 80 revolutions per minute. (6) If the shaft is to be hollow, find its external diameter, from strength considerations, when its internal diameter is two-thirds of its external diameter. Answer. Let D 1= r diameter of the tolid steel propeller ihaft. D= outer diameter of the hollow steel propeller shaft. a = inner diameter of the hollow steel propeller shaft. N= number of revolutions per minute of shaft. (a) Keferring to the previous article on the relation between the dia- meter of shaft necessary to transmit a given horse-power, we deduced the following formula for a solid round steel shaft, when the stress // was assumed as 13,500 Ibs. per square inch. Substituting the numerical values given by the question, we get . =2-9 //?=2-9 / i5o= 15-4 inches. But, to allow for variations in stress, bending moments, and shocks to which the propeller shaft may be submitted, we find from the table that the value of the coefficient c is 1*13. Hence D l = cd= 1-13 x 15*4= 17*4 inches for the solid round shaft. II P (6) From the above, we see that D 4 -d 4 H.P. 326 LECTURE XXIII. Jfow, substituting the values given by the question, we get Or, iD a *= .. P a 8 = 3x2158-5 =6475-5. . . D a = #6475-5 =18-64 inches. If bending is taken into account, then D 2 = 1-13 X 18-64 =21 inches for the hollow round shaft. EXAMPLE IV. The screw shaft of a marine engine is 10 ins. diameter, and the revolutions 100 per minute. It is replaced by twin screw shafts rotating 500 times a minute. If the total horse-power developed in the two cases be the same, and the working stress is also the same in the twin screw shafts as in the single screw shaft, find the proper diameter of shafts in the second case and compare their weights. (C .&(*., 1905, O.3a,fr, Answer. Let D! = diameter of single screw shaft. n D 2 = diameter of one of the twin screw shafts. ,, N! = number of revs, per minute of single shaft. N 2 = twin screw shafti. H.P. = total horse-power to be developed in each case. ft. = working stress in Ibs. per square inch. Now, Dj = \/ $'i x ^ ; ~ * and, D, = // x " x 33-o J*f. 5-1 x 12 x 33,000 x H.P. .-. IV = ZTTNj/ D a 8 5-1 x 12 x 33,000 x H.P. Or in" i.e., D x = 1-442 D 2 . . . D a = ~ = 7" But, the weights of shafts are proportional to their cross sectional areas if their lengths are equal. | Dj a (1-442 D 2 ) a _ 2-08 1-04 Hence, ^-, = -^r~ ~T ~ ' Therefore, the weights of the shafts in the two cases are approximately the same, but the diameter of th^ single screw shaft is nearly i times the diameter of one of the twin screw shafts. LECTURE XXm. QUESTIONS. 327 LECTURE XXIII. QUESTIONS. 1. An open link chain is constructed of round wrought-iron rod, inch in diameter ; calculate what is the probable breaking load of the chain. Wrought-iron chains are liable to deterioration by constant use ; what change do they undergo, and what precaution is taken to prevent their breaking 1 Ans. io tons. 2. A steel punch inch in diameter is employed to punch a hole in a plate f inch in thickness. What will be the least pressure necessary in order to drive the punch through the plate when the shearing strength of the material is 35 tons per square inch ? An*. 51*56 tons. 3. Define what is meant by ** shearing etress and* strain," "torqifc. n- j~?1tiDg moment." Show by an example that a shaft subjected to torqu* iaars a shearing stress tending to sever it afe right angles to its axis. 4. What is meant by the twisting moment " of a shaft ? If a wrought iron shaft i inch in diameter breaks in torsion by a force of 800 Ibs. at the end of a lever i foot long, what force at the end of a lever 2 feet long will break a shaft of the same material, but 2 inches in diameter 1 Find also the diameter of a wrought-iron shaft to resist a force of 2 tons at a distance of 18 inches from its centre. Ans. 3200 Ibs. 2 inches full. 5. If a shaft, 2 inches in diameter, is found equal to the transmission of 4 horse-power, what amount of power can be transmitted by a shaft 4 inches in diameter, all other questions remaining the same ? Ans. 32 horse-power. 6. If a revolving shaft, which is 2 inches diameter, is found sufficiently strong to transmit 4 horse-power, how much power may be transmitted by a shaft which is 3 inches in diameter, supposing all the other conditions to be the same, and that the iron of both shafts is subjected to the same stress? Ans. 13*5 H.P. 7. If 8co Ibs. at the end of a 12-inch lever be a safe stress to apply to a wrought-iron bar one square inch in section, find the effort which a shaft 2 inches in diameter can transmit at the circumference of a pulley one foot in diameter, and making 300 revolutions per minute. Find also the horse-power transmitted. Ans. 8893 Ibs. ; 254 H.P. 8. If a wrought-iron shaft of i inch diameter is broken by the torsion of a load of 800 Ibs. acting at the end of a 12-inch lever, find the weight which, when applied to the end of the same lever, would break a*shaft of the same material, but 3 inches in diameter. State, in general terms, the reasoning by which you arrive at the result. Ans. 21,600 Ibs. 9. Suppose that a shaft of i inch diameter may be safely subjected to a torque of 2000 Ib. -inches; what torque will a 2j inch shaft safely resist ? Calculate the horse-power which may be safely transmitted by the latter shaft if its speed is 150 revolutions per minute. (B. of E., 1902.) Ans. 22,780 Ib. inches; 54 horse-power. 10. A wire of Siemens' steel O'i inch diameter is to be twisted till it breaks. Sketch the arrangement and show how the angle of twist and the twisting moment are measured, how the results may be plotted on squared paper, and the sort of results that may be expected. In what way may a wire of twice this diameter be expected to behave f (B. of E. 1901.) 328 LECTURE XXIII. QUESTIONS. n. If a shaft 4 inches in diameter will safely withstand a torque of 120,000 lb. -inches, what torque would a g-inch shaft take? What H.P. would the former transmit at 200 revolutions per minute, and what would the latter transmit at 50 revolutions per minute ? (B. of E., 1903). Ans. Torque = 1,367,000 lb. -inches ; H.P a = 3847 and H.P. 2 = uio. 12. Compare the strengths and weights of a solid wrought-iron shaft and a hollow steel shaft of the same external diameter assuming the internal diameter of the hollo w shaft half the external, the working stress of steel ij times that of iron, and the densities of wrought-iron and steel to be the same. (C. & G-., 1903, O. , Sec. B.). ^ ns Strength of solid W.I. shaft_32_ i Strength of hollow steel shaft ~ 45 ~ 1^4* Weight of solid W.I, shaft _ 4 = ij Weight of hollow steel shaft" 3 T~" 13. The propeller shaft of a vessel, whose engines develop 1000 horse- power at 60 revolutions per minute, is 8" dia. Assuming the shaft sub- jected to pure torsion, and that the maximum twisting moment on the haft is I J times the mean, estimate the maximum shear stress induced in the shaft (C.&G., 1904,0., Sec. B.) Ans. Max shear stress = 13, 050 Ibs. per sq. inch. 14. The figure shows the skeleton mechanism of a direct-acting steam- engine. A is the cross -head, B is the connecting-rod, and C is the crank. The connecting rod is 4 cranks long, and in the position shown in the figure, the crank has turned through an angle of 45 from the dead centre in a clock-wise direction. - A force F, due to the steam pressure on the piston of 12,000 pounds, acts upon the cross-head. Find graphically, or in any other way, the thrust in the connecting rod, and the magnitude of the force It between the cross-head and slide bar. All friction to be neglected. (B. of E. 1905.) Ans. F = ^12,200 Ibs. ; B = 12,430 Ibs. ; and B = 2220 Ibs. 15. Describe, with a sketch of the apparatus, how you would experi- mentally determine the law connecting the twisting moment and the angle of twist for a mece of steel wire. (B. of E. 1905.) LECTURE XXIY. CONTENTS. Hooke's Coupling or Universal Joint Double Hooke's Joint Sun and Planet VTheels Cams Heart Wheel or Heart-shaped Cam Cam for Intermittent Motion Quick Return Cam Example Pawl and Ratchet Wheel Reversible Pawl Masked Ratchet Silent Feed Watt's Parallel Motion Parallel Motion Questions. IN this and the following Lecture we shall examine a few of the many devices for transmitting circular motion and for converting it into rectilinear motion, or vice versd, together with other miscellaneous mechanisms. Hooke's Coupling or Universal Joint. This is a contrivance sometimes used for connecting t\vo intersecting shafts. Each of the shafts ends in a iork, F 1? F,, which embraces two arms of the crosspiece, O. The four arms of this cross are of equal length. As Cj rotates, F x and F 2 describe circles in planes perpendicular to their respective axes. Since these planes are inclined to each other the angular velocity of C 2 at any HOOKE'S JOINT. instant is different from that of C 1 , but the mean angular velocities are equal to one another, because at one instant C 2 goes faster than C t , and at another slower. This joint will not M ork when the two shafts are inclined at 90, or any smaller angle, to each other. Double Hooke's Joint. The variable velocity ratio obtained with a Hooke's joint may be obviated by the use of two joints instead of one. The forks are connected by an intermediate link, C 2 , which must be carried on corresponding arms of the two crosses, as shown in the next figure. If the intermediate shaft be equally inclined to the other two shafts, the irregularities caused in the motion by its transmission through the first coupling are exactly neutralised by the equal and opposite ones caused by the second joint. The first and third DOUBLE HOOKE'S JOINT. 330 LECTURE XXIV. shafts, therefore, revolve with the same velocity at every instai t. The double joint works equally well whether the two extrer e axes are inclined as shown in the figure, or are parallel to ea h other but not in line. Both the single and double Hooke's joint are, as a rule, used only for light work, such as for astronomical instruments. Sun and Planet Wheels. This device was invented by Watt to convert the oscillatory motion of the beam in his engines into the circular motion of the flywheel. As will be seen from the SUN AND PLANET WHEELS. first figure, it consists of a wheel D, rigidly fixed to the connect- ing rod D B, and kept in gear with another wheel C, by the link DEC. The wheel C, is keyed to the flywheel shaft. As the beam oscillates up and down, the connecting-rod pulls D up one side of C, and pushes it down the other. It thereby causes C to rotate, and with it the shaft and flywheel.* Cams. Cams are usually of the form of discs or cylinders. They rotate about an axis, and give a reciprocating motion to * See Vol. I., Lecture XIX., of the author's text-book on "Applied Mechanics " for a description of epicyclic trains and the application of the formula to this case. Watt first applied this motion to his " Double Acting Steam Engine" in 1784. See Lecture XVIII. of the author's elementary manual on "Steam and the Steam Engine." HEART WHEEL OR HEART-SHAPED CAM. 331 Borne point in a rod by means of the form of their periphery or surface, or by grooves in their surface. The cam generally revolves uniformly round its axis, whilst the reciprocating motion may be of any nature, depending on the shape of the cam, and may be in a plane inclined at any angle to the axis of rotation. In the following examples, uniformity of rotation is assumed in the case of the cam, and the motion of the reciprocating piece takes place in a plane perpendicular to the axis. Heart Wheel or Heart-shaped Cam. Suppose that it is required to give a uniform reciprocating motion to a bar moving vertically be- tween guides, and in a line passing through C, the centre of motion of the cam plate. Let the sliding bar be at its lowest position, as shown, and when in its highest position let its extremity be at the point 6. The distance thus moved is called the travel and will be passed over during one-half revolution of the cam. The required curved outline may be ob- tained in the following manner : "With centre C, describe circles passing through the extreme posi- tions of the end of the rod. Divide the travel into, say, six equal parts at the points i, 2, 3, (fee. Divide the semi-circumference into the &ame number of equal parts by radial lines C i', C 2', o01 mcn It will be seen from the two figures that micrometer screw gauges MICROMETER SCREW GAUGE. FOR MEASURING ALL SIZES LESS THAN 0-3 INCH BY THOUSANDTHS OF AN INCH. STARRETT MICROMETER GAUGE. are made of various ranges and styles. If a mere portable gauge is required to suit different sizes, then one of the best forms is that made by the Starrett Co., U.S.A. In which the position of the movable end is determined by inserting a hardened steel tapered -pin into hardened steel bushed holes. For fixed measuring machines of great accuracy the Whitworth Millionth Measuring Machine is still considered the standard in this country. 358 LECTURE XXVI. Sir Joseph Whit worth's Early Realisations of Me- chanical Accuracy. "Whitworth's Standard Measuring Machine ' r will be illustrated and described near the end of this Lecture. But, it may be mentioned here, that the whole subject of accurate, scientific, mechanical measurement and its standardisation had remained in great confusion and uncertainty until Whitworth first carefully considered and then made, about the year 1840, mechanically perfect flat-surface plates. Second, he made standard screw threads, screw taps and dies, as well as parent leading screws for lathes, &c. Lastly, a standard or parent Measuring Machine, which was to be an instrument of such extreme precision that it could detect the difference of one one-millionth of an inch in the end measurement of short standard bars. These three early steps in his career were realised in succession. In fact, he could never have made the standard measuring machine if he had not previously made and drilled his men into producing the two former sets of tools. All the present-day accurate machine-shop surface-plates, screw threads, standard bars, gauges and fixed measuring machines, may be said to be the outcome of Whitworth's skill, perseverance and forethought, in systematising the production of standard tools. He also advertised the results and sold accurate copies of his correct parent tools not only amongst British engineers but throughout the civilised world. Sir Joseph Whitworth, before commencing the afore-mentioned difficult tasks, was satisfied that the most practical means of workshop measure- ment was to be founded on the truth of surface and the sense of touch. He maintained to his dying day, that the most delicate sense of the mechanic and mechanician was that of touch. In confirmation thereof, he showed that when a piece of metal had parallel end faces and was so held between the two fair-in line and parallel measuring planes of his machine, that the piece being tested just gravitated slowly downwards, due to its own weight between these two faces. Then, any good mechanic could at once detect the difference in bringing these two plane parts or distance faces nearer together or further apart by the minute difference of one one-millionth of an inch ! Of course, such extreme accuracy is not required in ordinary tool-making and engine -building works, but it is required in some works. Improved Equivalents Micrometer Gauge. This new instru- ment has certain advantages over the ordinary screw Micrometers. By means of the screw on the hub and the two divided discs, readings can be taken up to '300 inch with the various equivalents. Or, measurements may be made in the same way as with the older style of gauge for all measure- ments less than one inch by TTTOTT f an inch between the jaws. On one side of the disc appear decimals of an inch, decimals of mtlli-- metre standard wire gauge and Stubs' round hole wire gauge. On the other side, fractions of an inch by -fa inch, and screwing sizes for B.A. threads. Equivalents to other gauges, such as the "British Standard Wire Gauge," "Stubs' round hole wire gauge," and fractions of an inch on the metric standard, can be read off without the necessity of reference tables ; whilst at the same time the jaws are set to give an exact size in decimals of an inch. It will thus be seen, that by turning the disc until any decimal or other number appears, several equivalents can be read simultaneously, which will be especially convenient for telegraph and electrical engineers. The hub is fitted with a small milled thumb piece, projecting beyond the ordinary hub, by means of which the speed of turning the screw IMPROVED EQUIVALENTS MICROMETER GA-'GE. 359- Tnay be increased. The anvil end of the Micrometer is flush with the screw spindle, thus allowing of the close calipering of projections or ledges. NOTE. I am indebted for the figure and description of this improved gauge to the patentees and makers, Messrs. Grimshaw and Baxter, of 29 Goswell Road, London. 3'60 LECTUEE XXVI. A New Set of English Gauges (Windless Patents). Descrip- tion of the Following Six Figures. Fig. I shows the lower half of a pocket- case^" x 5" x i ") containing a complete set of these Caliper Gauges. It will be observed that both the external and the internal measuring holders are divided in the centre, to allow of the insertion of one or more of the steel blocks (illustrated on the right hand of the case) between either of these two holders. Fig. II indicates the arranging of the External Gauge to measure inch. This can be effected in a very short time by means of the " quick-grip lock-nut device." Fig. Ill gives a photographic proof of the perfectly flat surfaces and the very superior finish of the faces of the seven cast-steel blocks, by their clinging together after having been wrung together. Professor Tyndall was the first scientist to prove that* perfectly flat surfaces adhere together, due to the mere molecular attraction between the great number of bearing points when brought into close contact. He entirely disposed of the previously held theory that the adherence (of, say, two good surface plates) was due to the exclusion of the air between them, and, therefore, to atmospheric pressure. Fig. IV shows the internal measuring holder arranged to measure 3! inches. It is usually finished with flat ends, but it can be supplied with spherical ends having a radius slightly smaller than the smallest cylinder of 2\ inches which the gauge will enter. Fig. V exemplifies how the f ", ", ft* y and & blocks, when wrung together, are used to check the i* block by the external gauge. This process may be reversed by placing the i" piece between the jaws and the five smaller pieces in the holder. This method of self -checking is one of the best points of these gauges. Fig. VI is a view of the new 1910 pattern of a " Combined Limit and Double-ended Caliper Gauge." When the jaws marked (+, +) and { ) are opposite each other, as in the photo, the gauge is set for limit measurements, the (+, -f) gap being slightly larger and the { > ) gap slightly smaller than the exact measurement aimed at. But, when slackened, turned round and then readjusted (so that the (+ > & ) markings come opposite each other), both ends become exact caliper or snap gauges. They now indicate the precise size due to the number and breadths of the round steel standard-sized blocks, which have been inserted in the centre, between the two lock-nuts. It may be mentioned, that all the parts of these several gauges are made of the best British cast-steel, properly hardened and finished. Range of Measurements by the External and In- ternal Holders. In the up-to-date standard sets there are but seven steel blocks, viz., i", ", f", ", ^ %' and T V*. This gives the ex- ternal measuring holder a range from T y to z\\ by T V of an inch, or forty different sizes ; and the internal measuring holder a range from 2%" to 4", also by r Vths of an inch, or twenty-four different sizes. In addition to this, forty end sizes from r V" to 2 J" are obtainable by wringing the blocks together in combinations. Thus, we get a range of over 100 separate standard gauge sizes with one small pocket instrument. Mr. H. M. Budgett, of the Crown Works, Chelmsford, informs the author that he is now (1910) adding ^" and ^" measuring pieces to the above-mentioned seven blocks. This will greatly increase the range of the sizes obtainable with this instrument. In fact, the outside gauge will then give 160 sizes and the internal one 96 sizes, whilst the pieces FOR DESCRIPTION AND DETAILS SEE PROFESSOR JAMIESON'S MANUAL OF APPLIED MECHANICS. FIG. I. SHOWING COMPLETE SET IN CASE, HALF ACTUAL SIZE. These Gauges are made by the Crown Works. Chelmsford, Engla FIG. II. SHOWING EXTERNAL GAUGE BEING ARRANGED TO MEASURE --in. FIG. Ill, SHOWING HOW THE VARIOUS PIECES WILL ADHERE WHEN WRUNG TOGETHER FIG IV. SHOWING THE INTERNAL GAUGE ARRANGED TO MEASURE 34 in. FIG. V. SHOWING HOW THE GAUGES ARE SELF-CHECKING, FOR DESCRIPTION AND DETAILS SEE PROFESSOR JAMIESON'S MANUAL OF APPLIED MECHANICS. FIG. VI. SHOWING COMBINED LIMIT AND DOUBLE ENDED CALIPER GAUGE. WHITWORTH MILLIONTH MEASURING MACHINE. 361 themselves will give 163 sizes ; so that 419 different exact sizes could thereby be measured ! Uses of the Gauges with a Surface-Plate. In every up- to-date tool-making and engineering workshop there should be a set of standard surface-plates. Then, the previously described gauges may be used in many ways with such a surface -plate. For example, an exact height gauge may be formed by wringing one arm of the external measuring holder with any desired number of blocks, and then using them on the surface-plate. The National Physical Laboratory Certificate. We finish the description of these British-made gauges in reproducing from a photograph the Certificate granted in August 1909 by The National Physical Laboratory, with two objects in view : First, to show the extreme accuracy of the gauges, where none of the seven blocks had an error of one one hundred-thousandth of an inch ! Second, to bring to the notice of Engineering Students, that if ever they should be fortunate enough to devise anything new in regard to ther- mometers, pyrometers, barometers, micrometers, cyclometers, galvano- meters, electrometers, or steam, vacuum and mechanical gauges, &c., they may have the accuracy of their invention or improvement tested by an absolutely impartial and reliable judge at a comparatively small cost. Certificate of Examination of Seven End Gauges. By the National Physical Laboratory, Teddington. For : H. M. Budgett, Crown Works, Chelmsford. Form : Cylindrical, i" in diameter with a " hole through the centre. The end faces are perpendicular to the axis of the cylinder. These gauges have been compared with the Laboratory Standards, and the mean lengths at 62 F. have been found to be : Gauge. Length at 62 F. Gauge. Length at 62 F. t 0.062 50 inch. 0.25000 inch. 0.12500 ,, 0.37500 T met9r. This instrument con- ists of a horizontal metre rule, divided into inches and parts of an inch along its upper scale, and into centimetres and millimetres along its lower scale. At the centre of this rule there is attached one end of a second rule, half a metre in length and perforated at various points. A small slot is cut in the second rule to enable the degrees marked on the semicircle to be read. Attached to this second rule by means of a pin, are (i) a third rule, (2) a plumb-line. The third rule is graduated into cm. and mm., and has a fine slot cut down its centre to enable the divisions on the first or horizontal rule to be read. In order to avoid parallax a cursor or fine dividing line connects the vertical with the horizontal rules. The whole instrument is made of boxwood and fixed on a firm central base. 811k thread Plumb-Line. Plumb Weight BERRIDGE'S TANGENTOMETER. (As made by Harris & Co,, Ltd., Birmingham.) Uses of The Tangentometer. This instrument einblea teachers to explain graphically the meanfng of the several trigonometrical ratios to technical students. It also enables a student to measure for himself various angles and to prove his answers to questions by noting the ratio of the actual lengths of the sides of a right-angled triangle. He can then compare his results with the Table of Trigonometrical values for "Functions of Angles " at the end of this book. FOR EXAMPLE, taking the above scale figure of the full-sized instrument, we see, that the centre line of the second rule (which is 50 cms. long) lies at an angle of 48 to the centre line of the first or horizontal rule. And, the third or vertical rule hangs at right angles to the first rule. The length of the plumb-line of the third rule is 37 cms. to where it cuts the centre line of the first rule at right, angles at 33-5 cms. from its centre or zero mark. USES OF THE TAMJENTOMETER. 365 Hero, rhe first or horizontal rule is called the base; the second or 50 cm. rule i- th.j hypothenuse ; and the third or vertical rule is called the per- pend cid r. Then, we have the following ratios for the acuie angle of 48* in the lull-sized instrument.* Sine of the Tangent Cotangent Secant Cosecant Perpendicular HyputbtiuUae _ 5^ _ Hjputuenuse PerriPTidicnlar p , Perpendicular 37 cms. Hvpofhennse ^o cms. Table for 8 * j Perpendicular 37 cms. i Radian = Unit angle in circular measure = = 57'29. .-. 48 -i- 57 -29 = 8378 of a radian. See Table for the radian of 48*. * When resolving a force into its two component forces, the student must pay particular attention to the ( + ) or { ) sign of the value of the angle. See Functions of Angles in Castle's "Practical Mathematics for Beginners," chapter xvi., or other book on Trigonometry. NOTE. I am indebted to Ludw. Loewe and Co. Ltd., 30 Farringdon Road, London, for the use of their Electros of " Limit Gauges." I have to thank Mr. H. M. Budgett, of the Crown Works, Chelmsford r for six excellent views of his company's new English Gauges, and the Louis Gassier Co., Limited, for the liberty to reproduce the Micrometer Gauges and the W'hitworth measuring machine which appeared in Cassier's Magazine, September 1901. The attention of Students who are interested in gauges and accurate work is directed to " The Specification of the Engineering Committee on Standards " ; and to comments upon the same in The Electrician for Aug. 1906. 366 LECTURE XXVI. QUESTIONS. LECTURE XXVI. QUESTIONS. 1. Describe any micrometer screw gauge with which you are acquainted suitable for measuring to the rgV* of an inch. Sketch and describe care- fully the method of graduation and the position of the gauge when set to measure -374 inch. 2. Sketch and describe the construction and use of external and internal workshop gauges, by means of which the size of a spindle (say 2 inches diameter), and that of a hole into which it fits, may be ensured within specified limits of accuracy. State any advantages due to this system of working. (B. of E., 1902.) 3. Sketch a cylindrical one inch external gauge, and describe generally the measuring machine which you would require to employ, and the manner of using the same, in order to construct another gauge of the like kind, but measuring i -005 in diameter. How is the gauge worked down to the right size and finished ? 4. Sketch and describe the construction and action of a *' Whitworth Millionth Measuring Machine." For which purposes have you seen it used? 5. Sketch and describe an " Equivalents Micrometer Gauge " and state the advantages which it possesses over an ordinary gauge. 6. Sketch and describe fully an " External Limit Gauge " for a 2} inch shaft, such as the one illustrated by the second figure in this Lecture. Explain concisely and clearly how you would use such a gauge. Show your calculations in full for the percentage error in 2^ inches diameter, if a turner just makes a shaft as a tight fit for the "go on " end of the gauge. Also for the percentage error if he should so reduce the diameter of a shaft, to let the "not go on " end of the gauge fit the shaft. 7. Describe by aid of sketches Sir Joseph Whitworth's early attempts towards and realisations of mechanical accuracy with standard surface- plates, screws, gauges, and measuring machines. 8. Sketch and describe the new set of English Gauges as made by the Crown Works, Chelmsf ord, according to Windley's Patents. 9. Show how the gauges in the previous question are self-checking. Also, show by calculation, the exact range or number of different measure- ments which can be made with seven blocks and the external holders. If two extra blocks, -fa* and -fa", were added, prove how many more measure- ments could be made. 10. Sketch neatly and describe concisely Sir Joseph Whitworth's Millionth Measuring Machine. State the pitch of screw, number of teeth on worm- wheel, and number of divisions on the graduated wheel to measure with such accuracy. 11. Sketch, index and describe, any fixed good workshop measuring machine and state how you would use it. 12. Find the sine, cosine, tangent, cotangent, secant and cosecant of an angle in a right-angled triangle whose base is 40 cms, and perpendicular 20 cms. Also express 27 in radian measure. ( 367 ) APPENDIX A (pages 368 to 370). (I) General Instructions by the Board of Education for their Examina- tions on Applied Mechanics, Stage 1. (ii) General Instructions by the City and Guilds of London Institute for their Examination on Mechanical Engineering, Ordinary Grade. (iii) Rules and Syllabus of Examinations by the Institution of Civil Engi- neers for Admission of Students. APPENDIX B (pages 371 to 401). Board of Education's Exam. Papers in Applied Mechanics, Stage I. The City and Guilds of London Institute's Ordinary Exam. Papers in Mechanical Engineering. And the Institution of Civil Engineers' Exam. Papers * Elementary Mechanics, arranged in the order of the Lectures. APPENDIX C (pages 403 to 410). The latest Exam. Papars pertaining to Mechanics and set by the govern- ing bodies enumerated under Appendix A. APPENDIX D (pages 411 to 413). (i) Units of Measurement and their Definitions ; Practical Electrical Units and their Symbol Letters. (ii) Examination Tables, Useful Constants, Logarithm?, Antilogarithra and Functions of Angles. 2 A ( 368 ) Appendix A. May Examination on Subject VII. APPLIED MECHANICS.* BY THE BOARD OF EDUCATION, SECONDARY BRANCH, SOUTH KENSINGTON, LONDON Stage 1. GENERAL INSTRUCTIONS. If the regulations are not attended to, your paper will be cancelled. Immediately before the Examination commences, the following REGULATIONS ARE TO BE READ TO THE CANDIDATES. Before commencing your work, you are required to fill up the numbered slip which is attached to the blank examination paper. You may not have with you any books, notes, or paper other than that supplied to you for use at this examination. You are not allowed to write, draw, or calculate on your paper of questions. You must not, under any circumstances whatever, speak to or com- municate with another candidate. Those superintending the examina- tion are not at liberty to give any explanation bearing upon the paper. You must remain seated until your papers have been collected, and then quietly leave the examination room. No candidate will be allowed to leave before the expiration of one hour from the commencement of the examination, and none can be re-admitted after having once left the room. All papers, not previously given up, will be collected at 10 o'clock. If any of you break any of these regulations, or use any unfair means, you will be expelled, and your paper cancelled. Before commencing your work, you must carefully read the following instructions. Put the number of the question before your answer. You are to confine your answers strictly to the questions proposed. Such details of your calculations should be given as will show the methods employed in obtaining arithmetical results. The value attached to each question is shown in brackets after the question. A table of logarithms and functions of angles and useful constants and fc rriiulffi is supplied to each candidate. The examination in this subject lasts for three hours, * See Appendix C for the latest Exam. Papers." APPENDIX A. 369 CITY AND GUILDS OF LONDON INSTITUTE. DEPARTMENT OF TECHNOLOGY. Technological Examinations. MECHANICAL ENGINEERING.* OBDINARY GRADE PART L (FIRST YEAR'S COURSE.) INSTRUCTIONS. No Certificates will be given on the results of this Examination (First Year's Course), but the Candidates' successes will be notified to the Centre where they were examined. To obtain a Certificate, it is essential that Candidates should pass both in Part I. and Part II. ; the Examination in Part II. witt be hdd on Thursday, May 3 Jot 7 p.m. Candidates may take both Parts I. and IL in the same year. The class of Certificate and the order of Prize will be determined by the results of the Examination in Part II. only. The maximum number of marks obtainable is affixed to each question. The number of the question must be placed before the answer in the worked paper. Three how* allowed for thit paper. The Candidate is at liberty to use divided scales, compasses, set squares, calculators, slide rules, and tables of logarithms. A piece of squared paper to be given to each Candidate, if required. The Candidate is not expected to answer more than vine question*, wSaich must be selected from (too Sections only. * See Appendix C for the latest Exam. Papers. t This date is only approximate, and subject to a slight alteration eaoh year. APPENDIX A. The Institution of Civil Engineers* Rules for Admission of Students. SYLLABUS OF THE EXAMINATIONS.* 1. ENGLISH (one Paper, time allowed, 3 hours). A general Paper com- prising questions in Geography, History and Literature. 2. MATHEMATICS (two Papers, time allowed, 3 hour* for each). Papers comprise questions in Arithmetic; Algebra; Geometry (Euclid I. -IV. ) ' and Trigonometry. 3. Two subjects, to be selected by the Candidate from the following ten : a language is not compulsory, but in any case not more than one language may be taken (time allowed, 3 hours for each Paper) : LATIN, GREEK, FRENCH, GERMAN, ITALIAN, SPANISH. Elementary Mechanics of solids and fluids, t ELEMENTARY PHYSICS, including heat, light, Electricity and Magnetism. ELEMENTARY CHEMISTRY. GEOMETRICAL AND FREEHAND DRAWING. * See Appendix C for the latest Exam. Papers. f My Manuals on Applied Mechanics, Magnetism and Electricity, are suitable for Young Engineers desiring to prepare by correspondence or otherwise for these subjects, of which the most recent examples in Mechanics are given in Appendix C, whilst those on Electricity and Magnetism are printed in the Appendix of the Eighth Edition of the latter work. Candi- dates should write direct at once to the Author of this book for his C.E. Prospectus, which gives full details of Tuition for these Examinations. RULES OF THE EXAMINATIONS. The Examinations are held in London in February and October annually, on four days beginning on the second Tuesday in each of those months. The February Studentship Examination may, in the discretion of the Council, be held also in Manchester, Glasgow and Newcastle-on-Tyne. The Council will consider an Application from a person who is duly recommended for Admission as a Student of the Institution, to present himself for the Studentship Examination. , Applications to attend the Associate Membership Examination will be received from Students of the Institution who are not less than 21 years nor more than 26 years of age on the last day for entry. Arrangements may be made for the examination of Candidates In India 1 or in the Colonies, after submitting duly completed proposals for Election. All applications for Rules, Forms and Admission, &c., must be made through the Secretary, the Institution of Civil Engineers, Great George Street, Westminster, London, S.W. ORDINARY QUESTIONS. 371 Appendix B. See Appendix D for Practical Electrical Units LECTURE II. ORDINARY QUESTIONS. 1. In a shale mine in order to drain one of the pits a treble ram pump, driven by an electric motor, is employed. The rams are 9^ inches in diameter by 1 2-inch stroke, they each make 3475 strokes per minute, and the height to which the water is lifted is 393 feet. Fnd: (a) How many gallons of water this pump can lift per minute. (6) How many foot-pounds of useful work are done per minute, (c) The useful horse-power when the pumps are running steadily. (B. of E., 1906.) 2. A windmill is employed to drive a pump which has to lift water from a well and deliver it into an overhead tank. It was found that when the windmill works steadily under the action of a uniform wind for a period of i hour, 5000 gallons of water are raised from the well and delivered into the tank the average height of lif t is 60 feet. What under these conditions is the useful horse-power of the windmill ? (B. of E., 1907.) 3. An electrical hoist is employed in raising coal from the hold of a ship and delivering it into railway cars, the amount of lift being 125 feet. If the coal is raised at the rate of 2400 Ibs. per minute, what is the useful horse-power ? Convert this into watts. If the current is supplied at a voltage of 250, and if the efficiency of the whole arrangement is 50 per cent., how many amperes of current must be supplied to the motor working the hoist ? (B. of E., 1907.) 4. Two closely coiled spiral springs were made out of round steel wire, J-inch diameter. The one spring, A, had a mean diameter of coil of 4 inc^ es and the other, B, had a mean diameter of coil of 5 inches ; both springe had 12 complete coils. These two springs were tested by loads extending them axially, and the results of the tests are shown in the table below : , , Axial load \ in pounds / 2 4 6 8 10 12 H 16 18 20 Extension "| of the spring A. I Inches. J 0-26 0-52 079 1*06 1-32 i'59 1-86 2'12 2'39 2-66 Extension ) of the I spring B. f Inches. ) 0-51 I '02 i'53 2-04 2 '5 5 3-06 3'57 4-09 4'6o 5-12 Plot the results on squared paper. Given that the law connecting the extension of these springs with their mean diameter of coil is of the form Extension of B /Mean diameter of coil of B\n Extension of A ~ \Mean diameter of coil of A) what is the probable value of n ? (B. of E., 1908.) APPENDIX B. 5. Show how to determine the work done by a variable force[~moving"in its own direction. A cage weighing 1200 Ibs, is raised 300 ft, by a windlass having a wire rope weighing i Ibs. per foot run. Show, by a diagram to scale, the work done at any stage, and mark on it the numerical values for a lift of 100, 200, and 300 feet respectively. (C. & G., 1909, O., Sec. A.) 6. An engine cylinder, fitted with a Joy valve gear, has a stroke of 26 inches, and a connecting-rod 74 inches long. In the accompanying LINE DIAGEAMS OF PROBLEM ON JOY'S VALVE GEAR. sketch the link AB is pivoted to the connecting-rod at A, and to the free end B of a swinging link BC centred at C. The motion for operating the valve is taken from a point D on the link AB. Draw the path of thefpoint D for a complete revolution of the crank. Take a scale of lin. = i ft. (C. & G., 1909, 0., Sec. A.) LECTURE IV. ORDINARY QUESTIONS. i.l The right-angled bell crank lever, centred at A, shown[in the[sketch is attached to a spring by one of its arms, and to another^lever, centred^at B, by the other arm. If the spring requires a direct pull of 20 Ibs. hi order to stretch it 2[inches, find what force P, applied as shown, will stretch the spring this amount.C ! (B. of E., 1906.) 24" ~~2-'-* 22 oj 1 J '--4'-* A B ? -> ( nT* BELL CRANK LEVER AND SPIRAL SPRING. ORDINARY QUESTIONS. 373 2. An ordinary bell-pull, shown in the sketch, is in equilibrium. Deter- mine in any way you please the magnitude of the force Q and the magni- tude and direction of the resultant thrust upon the supporting pin A. (B. of E., 1907.) P=10lbs. ORDINAEY BELL-PULL LEVEB. 3. The figure shows the mechanism known as a knuckle joint.^A^force of 50 Ibs. is applied at the point A, its line of action being perpendicular to the line BC. Determine graphically, or in any other way, the vertica thrust delivered by the block D. Neglect friction. How will this thrust vary as the block D descends ? (B. of E., 1908.) TOGGLE OR KNUCKLE JOINT. 3/4 APPENDIX B. LECTURE V. ORDINARY QUESTIONS. i. Describe how you would determine experimentally the coefficient of sliding friction between two pieces of metal of any convenient size when the speed of rubbing is low. (B. of E., 1906.) LECTURE VI. ORDINARY QUESTIONS. f I. Describe any one form of lifting tackle with which you are acquainted and explain with reference to it the terms " velocity ratio," " force ratio," and " efficiency." Explain how you would determine their numerical values for all loads up to the full capacity of the tackle. (n (C. & O., 1908, 0., Seo. A.) LECTURE VII. ORDINARY QUESTIONS. I. The sheave of a differential pulley block consists of two parts which have diameters of 8 and 9 inches respectively. What is the velocity ratio of the mechanism when a load is being raised ? If the mechanical efficiency of the pulley is 32 per cent., what pull must be exerted in order to raise a load of 2 tons ? (B. of E., S. 1, 1909.) LECTURE VIII. ORDINARY QUESTIONS. 1. The weight of a span of telegraph wire is 127 Ibs. At one end the wire makes an angle of 5 and at the other an angle of 7 with the horizontal, what are the pulling forces at these ends ? (B. of E., S. i , 1909. ) 2. A simple Warren girder is as sketched. Loads of 3 and 4 tons are carried at the two joints of the top member. Find, analytically or other- wise, the forces in the different members, (C. & G., 1906, 0., Sec. B.) 1 3 TONS 14 TONS SIMPLE WARREN GIRDER. 3. Explain one method of determining the stresses in the members of a pin-jointed frame, j ORDINARY QUESTIONS. \4 Tons 375 5 Tons ^ ___. LOADED ROOF TRUSS. Determine the stresses in all the members of the roof truss loaded, as shown in the figure. (C. &. G., 1908, O., Sec. B.) 4. A simple crane is of the form shown in the diagram, and carries 5 tons at C. Determine the stress in AB, AC, and BC, and find the magnitude of the balance weight W so that there shall be no bending moment on the post BD. (C. & G., 1908, O., Sec. D.) j 5 Tons LINE DIAGRAM OF A SIMPLE CRANE. 5. "A crane of 'the form shown by the accompanying sketch carries a load of 3 tons. The reaction at the upper bearing ia horizontal. Deter* Horizontal, * Reaction, LINE DIAGRAM OP A WALL CRANE. 376 APPENDIX B. mine the resultant pressure on the footstep bearing, and the stresses in the members of the crane, assuming that all the connections are pin-joints. (C. & G., 1909, 0., Sec. D.) LECTUBE X. ORDINARY QUESTIONS. 1. A machine weighing 8 tons is dragged slowly along a horizontal floor. If the coefficient of friction between the base of the machine and the floor is 0*35, find in pounds the magnitude of the pull, and the normal pressure on the floor when (a) the line of pull is horizontal, (6) the line of pull makes an upward angle of 30 with the horizontal. (B. of E., 1907.) 2. The length of a journal is 9 inches, and diameter 6 inches, and it carries a load of 3 tons. What horse-power is absorbed when making 100 revo- lutions per minute, taking the coefficient of friction as "015, and how many thermal units are radiated away per minute when the temperature of the bearing remains constant ? (C. & G., 1907, 0., Sec. A.) 3. An electric locomotive draws a train of 700 tons up an incline of i in 100 at a steady speed of 10 miles per hour. If the frictional resistances are equal to 15 Ibs. per ton, what is the total pull exerted on the train and what is the horse-power ? Find the current consumption in amperes, if the voltage is 625, and if 60 per cent, of the electric energy supplied to the locomotive is spent in hauling the load.*] (B. of E., S. i, 1909.) LECTURE XI. ORDINARY QUESTIONS. 1. Hemp ropes are employed to transmit power from the engine shaft to the driving pulleys on the different floors in a spinning factory. The maximum tension in a rope is twice the minimum tension, the breaking strength of one rope is 5700 Ibs., and it is desired to have a factor of safety of 30. Find the maximum horse-power which can be safely transmitted by one of these hemp ropes at a speed of 70 feet per second. (B. of E., 1906.) 2. It is required to transfer 3 horse-power to a pulley 16 inches diameter by belting. The revolutions per minute are 100, the tension in the tight side of the belt is i times that in the slack side, the thickness of the belt is -/TJ- inches, and the maximum working stress allowable is 320 Ibs. per square inch. Find the least width of belt. (C. & G., 1906, 0., Sec. A.) 3. In a rope brake on a fly-wheel 8 feet diameter, the ropes being i inch diameter the load is 500 Ibs., and the pull on the spring balance varies from 10 to 20 Ibs. during a test. Find the brake horse-power, the revolu- tions being 105 per minute. (C. & G., 1907, 0., Sec. A.) 4. Two shafts, which are not parallel, and do not intersect, are to be connected by a belt passing over suitably placed pulleys. Explain what are the necessary conditions to be observed in order that the belt shall remain on the pulleys. A horizontal shaft, running along one side of a machine shop, drives another horizontal shaft at right angles to the first shaft and 20 ft. blow. Sketch a suitable arrangement for the belt-drive if both shafts are to revolv at the same speed. 1 (C. & G., 1908, 0., Sec. A.) ORDINARY QUESTIONS. LECTURE XII. ORDINARY QUESTIONS. 377 I. An engine having a stroke of 12 inches, and a connecting-rod 24 inches long, centre to centre, makes 300 revolutions per minute. Find graphically the velocity of the piston at six intermediate positions of the stroke, and draw a curve showing the velocity of the piston at any instant. (C. & G,, 1908, O., Sec. A.) LECTURE XIII. ORDINARY QUESTIONS. I. A crane, tested in the usual way, and in which the velocity ratio is 40, gave the following results : Weight lifted (W) ICO 300 500 700 Force applied (P) 8-5 17-0 25-6 34 '2 Plot a curve showing the relation between P and W on a W base, and, OIL the same base, plot a curve of efficiency. (C. & G., 1906, 0., Sec. A.) LECTURE XV. ORDINARY QUESTIONS. i. Taking the mean diameter of the thread of a i-inch bolt to be 0*92- inches, the number of the threads to the inch being 8, and the coefficient of friction O'I7 ; find the turning couple required to overcome an axial force of 2i tons, and the efficiency under this load. (C. & G., 1906, O., Sec. A.) LECTURE XVI. ORDINARY QUESTIONS. i. To do the cutting work in a small screw cutting lathe it is found: that 0-47 H.-P. is required, and that the frictional losses in the gearing, bearings, &c., absorb another 0*21 H.-P. How many foot-pounds of work per minute is the driving-belt giving to the lathe ? j The countershaft is driven by an electric motor, and the countershaft and belts absorb 0-17 H.-P. How many watts must the motor give off in order to keep the lathe running ? If the voltage is 220, how many amperes will the motor require, assuming; that its own efficiency is 89 per cent. ? i H.-P. = 746 watts, and amperes multiplied by volts = watts. (B. of E., 1906.) 3/8 APPENDIX B. 2. The table of a drilling machine is raised by a hand- wheel, to the spindle of which is attached a single-threaded worm which meshes with a worm- wheel having 40 teeth. Compound with the worm-wheel is a spur-pinion, having 19 teeth of i-inch pitch, which meshes with a rack on the frame of the machine. Sketch the arrangement and find how many turns of the handle are required to raise the table through 2 feet. (C. & G., 1906, 0., Sec. A.) 3. In the feed gear of a drilling machine, in which a rack is used to give the traverse of the spindle, the spindle is rotated by a bevel- wheel of 18 teeth keyed on the driving shaft, gearing with one of 32 teeth on the sleeve surrounding the spindle. The greatest and least diameters of the pulleys on the speed cone for the driving shaft are 7 inches and 4 inches, and this cone drives a similar speed cone on the horizontal speed shaft. On this shaft is a single-threaded worm which gears with a worm-wheel of 45 teeth on the vertical feed shaft. A single-threaded worm on this shaft gears with a wheel of 30 teeth, turning on a horizontal stud, and to which is attached a pinion of 15 teeth gearing with, a rack of -inch pitch, which gives the required feed. Find the least and greatest number of revolutions of the drill spindle per inch of feed. (C. & G., 1906, O., Sec.A.) 4. The traverse shaft of a lathe is driven from the headstock mandrel by belting, the greatest diameter of the speed cone at the extremity of the mandrel being 5 inches. This drives a similar cone on the transverse shaft, and the smallest diameter is 2 inches. A worm on the traverse shaft meshes with a single-threaded worm-wheel, having 40 teeth, turning on a stud carried by the saddle. At the front end of this spindle is. a spur-wheel of 15 teeth, meshing with a wheel of 45 teeth, which turns on a stud carried by the apron ; and compound with this last wheel is a pinion of 12 teeth, which meshes with the rack of -inch pitch, attached to the lathe bed. Sketch the mechanism and find the traverse of the saddle per revolution of the headstock mandrel. (C. & G., 1907, 0., Sec. A.) 5. A lathe is driven by a belt running on the 1 2-inch diameter pulley of its speed cone, which then revolves at 200 revolutions per minute, and the back gear of the lathe reduces this speed in the ratio of 9 to i. Under these working conditions it is found that when a certain cut is being taken off a bar, 6 inches diameter, the horse-power transmitted by the belt is 0*60. What is the pressure on the cutting tool in a direction tangential to the turned surface, if we assume that 75 per cent, of the power trans- mitted through the belt is lost in frictional and other wasteful resistances ? (B. of E., 1908.) 6. Show how screws, differing in pitch from the leading screw, can be cut in a lathe. If the leading screw of a lathe has three threads per inch, and is right-handed, what arrangements of change would you use to cut (i) a right-handed screw of four threads to the inch, (ii) a left-handed screw of eleven threads to the inch ? You may assume that you have a set of change wheels with teeth varying from 20 to 100 by differences of five teeth (C. & G., 1908, 0., Sec. A.) 7. Describe, with the help of neatly-drawn sketches which should be roughly to scale, any form of loose head-stock or poppet-head for a small lathe, with which you have had practical experience. Show how the spindle or poppet is advanced or withdrawn, and how it is clamped. (B. of E., S. i., 1909.) ! 8. A small machine tool is driven direct by an electric motor. How would you determine the horse-power absorbed in the process of cutting *he material ? (B. of E., S. i, 1909.) ORDINARY QUESTIONS. 379 9. The countershaft of a drilling machine makes 240 revolutions per minute, and it carries a stepped pulley, the diameters of which are 12, 9 and 6 inches respectively. This drives an intermediate shaft by a belt and pulley with similar steps. The intermediate shaft drives the drill spindle by a bevel wheel of 30 teeth gearing with one of 40 teeth on the drill spindle. Calculate the possible speeds of the drill spindle, and also determine the diameter of the largest drill you can use if the circumferential cutting speed is limited to 240 inches per minute. (C. & G., 1909, 0., Sec. A.) LECTURE XVIII. ORDINARY QUESTIONS. I. The following results were obtained during an experiment to deter- mine the quantity of water which would be discharged through a small circular orifice in the side of a tank. The diameter of the orifice, which had sharp edges, was I inch. Number of experiment Duration of experiment Actual discharge Head of water above centre of orifice Minutes. Lbs. Inches. I 15 57 6 1-5 2 15 660 2'0 3 15 733 2'5 4 15 827 3'27 5 15 915 4-01 6 15 1,011 5*o 7 10 737 6-0 8 10 788 7'0 jj- Plot on squared paper a curve to show the relation between the dis- charge in Ibs. per minute, and the head of water above the centre of the orifice. From your curve determine the discharge in gallons per hour when the head of water was 5^ inches. (B. of E., 1907.) 2. A straight balk of timber is 20 feet long and 12 inches square in cross- section : its weight per cubic foot is 43*5 Ibs. If a weight of 112 Ibs. is placed on the centre of the balk when it is floating in water, find the depth to which the balk will be immersed. (B. of E., S. i, 1909.) LECTUEE XIX ORDINARY QUESTIONS. I. The rim of a turbine is going at 50 feet per second ; 100 Ibs. of fluid enter the rim each second, with a velocity in the direction of the run's motion of 60 feet per second, leaving it with no velocity in the direction O f the wheel's motion. What is the momentum lost per second by the fluid ? This is force. _^What work is done per second upon the wheel ? (B. of E., 1907.) ' APPENDIX B. 2. A hydraulic press has a ram 6 inches in diameter : water is supplied to the press from a single-acting pump, which has a plunger i inch in diameter with a stroke of i inches. Neglecting frictional and other losses in the pump and press, find the average rate (in foot-pounds per minute) at which the pump works, if it makes 100 working strokes per minute, while the press is exerting a force of 70 tons. (B. of E., 1907.) 3. A centrifugal pump, driven by an electric motor directly coupled to it, is found during a test to deliver 320 gallons of water per minute into ,n overhead tank, the mean height of lift being 65 feet. What useful horse-power is the motor doing ? If during the test the electric motor takes 35 amperes of current at 440 volts, what is the combined efficiency of the whole plant ? (B. of E., 1908.) 4. A turbine, which gives off 50 horse-power to a belt running on a pulley on its shaft, is supplied with water which, as it enters the turbine, is under a head of 125 feet. If 75 per cent, of the total energy of the entering water is thus utilised, what work is done per pound of water, and how many gallons of water pass through the turbine per working day of 10 hours ? (B. of E., S. i, 1909.) LECTURE XX. ORDINARY QUESTIONS. i. Describe, with the help of neatly drawn sketches, which should be roughly to scale, a hydraulic jack, showing clearly all valves. (B. of E., 1908.) :, the wa1 2. In a hydraulic crane, with a ram 8 inches diameter, the water pressure is 800 Ibs. per square inch, and the velocity of lift is increased eight-fold by the use of a four-sheaved pulley block. What load can this crane lift if its mechanical efficiency is 40 per cent. ? How many gallons of power water will be used in lifting the load 50 feet ? (B. of E., 1908.) LECTURE XXI. ORDINARY QUESTIONS. 1. A cycle track is approximately elliptical in shape, the maximum radius of curvature being 1 50 yards and the minimum 50 yards. Find at each of these two places, the ratio which the centrifugal force bears to the weight, if the speed of the racing cyclist is 25 miles per hour. What would be the two inclinations of the track to the horizontal if the track is laid so as to be perpendicular to the resultant force in each case ? (B. of E., 1906.) 2. A railway truck weighing 10 tons starts from rest down an incline mile long of i in 250. If the frictional and other resistances are equiva- lent to 8 Ibs. per ton weight of the truck, with what velocity will the truck be moving when it gets to the end of the incline ? How far would it then run along a level stretch of the line before coming to rest ? (B. of E., 1906.; ORDINARY QUESTIONS. 381 3. Two adjacent positions, G lt G 2 of the centre of mass G of a balance weight were obtained by geometrical construction from a skeleton diagram of the mechanism. These positions, measured in feet from two perpen- dicular axes, were found to be as follows : X y 01 0-167 0*078 o, 0-352 0-146 The displacement O l G 2 took place in 1/50 second. Find the x and y components of the mean velocity of G for this interval. Plot the points G lt G. 2 , on squared paper. (B. of E., 1906.) 4. A traction engine travels at 6 miles per hour ; the road wheels are 6 feet in diameter and are driven through 5 to i gearing. Find the angular velocity in radians per second of the fly- wheel on the engine shaft. (B. of E., 1907.) 5. The rim of a cast-iron pulley has a mean radius of 12 inches; the rim is 6 inches broad, and inch thick, and the pulley revolves at the rate of 1 50 revolutions per minute ; what is the centrifugal force on the pulley rim per inch length of rim ? One cubic inch of cast-iron weighs 0-26 Ib. (B. of E., 1907.) 7. With an automatic vacuum brake a train, weighing 170 tons and going at 60 miles an hour on a down gradiant of I hi 100, was pulled up in a distance of 596 yards. Find the total resistance per ton in pounds, and the time taken to stop the train. (C. & G., 1907, 0., Sec. A.) 8. A horizontal jet of water issues at a velocity of 20 feet per second fiom the 2 -inch diameter nozzle of a hose pipe, and strikes a vertical wall. What is the mass of water, in engineers' units, which strikes the wall per second ? What is the momentum of this quantity of water ? What is the force on the wall ? It is assumed that no water splashes back. (B. of E., 1908.) 9. A man, whose weight is 14 stone, stands on the floor of a lift. What force does he exert on it (i) when the lift is stationary, (ii) when it is des- cending with an acceleration of 10 feet per second per second, and (iii) when it is ascending with the same acceleration ? (B. of E. 1908.) 10. A train is running round a circular curve of 2000 feet radius at a speed of 50 miles per hour. A weight is suspended by a thin cord from the roof of one of the carriages ; at what inclination to the vertical will the cord hang ? (B. of E., 1908.) 11. The fly-wheel of a punching machine weighs i tons, and has a radius of gyration of 3 feet. It is turning at the rate of 130 revolutions per minute when the punching of a hole is started, but at the completion of the operation of punching it is found to be turning at the rate of only 125 revolutions per minute. How many foot-pounds of work have been expended in punching the hole and overcoming the frictional resistances of the machine ? Note. If k is radius of gyration, this means that we may imagine the mass of the whole wheel to be at the distance k from the axis, and this enables us to calculate its kinetic energy. (B. of E., 1908.) 382 APPENDIX B. 12. Explain how to determine the velocity of a moving body by con- sidering the space described in a given time. A motor car, starting from rest, travels a distance of s feet in t seconds, in accordance with the following table : t I 2 3 4 5 6 7 8 8 5 16 33 56 85 1 20 161 208 Draw a curve showing the distance travelled at any time within this period, and from this curve determine the velocity of the car at the ends of the third, fourth, and fifth seconds respectively. (C. & G., 1908, O., Sec. A.) 13. A pulley, 3 feet in diameter, has a peripheral speed of 2000 feet per minute. It is unbalanced to an amount which may be represented by a mass of 0*5 Ibs. at a radius of I foot. Calculate the unbalanced force on the pulley-shaft, and determine the positions at the pulley-rim of two masses of 0-4 Ibs. to give a perfect balance. (C. & G., 1908, O., Sec. A.) ^ 14. An experiment with a small Pelton water-wheel gave results shown Jn the annexed table : Mean Revs, per Min. Cubic Feet of Water passing through Wheel per sec. Speed of Jet. Feet per sec. V. Peripheral Speed of Vane. Feet per Sec. V. Ratio. V T Efficiency of Wheel. Per Cent. 1090 0870 44-0 975 0824 63*8 885 0834 72-4 645 0840 72-8 540 0840 66-8 460 0847 6i-5 385 0834 56-1 - 265 0860 40-1 The cross-sectional area of the nozzle in square feet was 0*001043. Th mean diameter of the bucket was 107 inches. Fill in the third, fourth, and fifth columns of the table. Plot a curve to show the variation of efficiency with variation of ratl y taking efficiency in vertical ordinates and -y in horizontal abscissae. T (B. of E., S. i, 1909.) ORDINARY QUESTIONS. 383 15. Define the term " horse-power." The tension on the draw-bar of a locomotive is 800 Ibs. when the speed is 45 miles per hour. The weight of the train, excluding the locomotive, is 200 tons. If the efficiency of the locomotive is 65 %, find the horse-power of the engine. Also find the accelerating force exerted at the draw- bar to change the speed from 45 to 50 miles per hour in one minute. (C. & G., 1909, O., Sec. A.) 16. Define the terms " velocity " and " acceleration," and show how to determine the acceleration of a body moving in a straight line when its velocity is known at each instant. The accompanying diagram is to be 6 CUEVE REPRESENTING THE VELOCITY OF BODY. drawn so that each unit square is of I inch side. The curve ABC represents the velocity of a body during a given interval of time, it is circular from A to B, with centre at D, and from B to C it is a straight line parallel to the horizontal axis. Draw to scale a curve showing the acceleration of the body at any instant, and mark on it the numerical values of the ac- celeration at the end of each second. (C. & G., 1909, O., Sec. A.) ft 17. Show how the resultant velocity of a body may be obtained when the component velocities impressed upon it are known. A goods engine moving at the rate of 20 miles per hour, has equal coupled wheels 54 inches in diameter, and the crank-pins move in circles 24 inches in diameter. Determine the velocities of the crank-pins relative to the rails for three equally spaced positions reckoned from the lowest positions of the crank- pins. (C. & G., 1909, O., Sec. A.) ^ 1 8. Assuming the expression for the law of centrifugal force, show how to obtain a formula for the stress /in the rim of a fly-wheel in the form where w is the weight of a unit volume of the rim, v is the velocity, and g is the gravitation constant. Determine the limiting speed in revolutions per minute of a cast-iron fly-wheel rim having a mean diameter of 10 feet, when the allowable stress is 2400 Ibs. per square inch. The weight of one cubic inch of cast-iron may be taken as 0*28 Ibs. (C. & G., 1909, O., Sec. A.) 2 B APPENDIX B. 19. A r hammer head weighing 3-22 Ibs. moving at 30 feet per second is stopped in o'ooi second : what is the average force of this blow in pounds ? (B. of E., S. i, 1909.) 20. A locomotive is travelling at 60 miles per hour. The driving-wheels are 6 feet 6 inches in diameter. What is the angular velocity of the driving- wheels in radians per second ? ' If the stroke of the piston is 26 inches, what is the mean speed of the piston relatively to the cylinder in feet per minute, assuming there is no slip of the wheels ? (B. of E., S. i, 1909.) 21. A fly-wheel, which weighs 18 tons, when mounted on its axis and rotated, is found to be out of balance, and, in order to bring it into balance it is found necessary to fix a counter-weight of 420 Ibs. to the wheel, at a distance of 90 inches from the axis of the shaft. What was the distance of the centre of gravity of the unbalanced wheel from the axis of the shaft ? Show by a sketch where you would fix the counter-weight. (B. of E., S. i, 1909.) 22. A fly-wheel weighs 8 tons, its radius of gyration is 5 feet 5 inches, and it is rotating at a speed of 90 revolutions per minute. How many foot-pounds of energy are stored up in it ? (P- ' If this wheel were supported in two bearings, each 12 inches in diameter, and if the coefficient of friction were 0*01, how much energy is wasted hi overcoming friction in one revolution, and how many revolutions would this fly-wheel make before coming to rest after the turning force was cut off ? ' (B. of E., S. i, 1909.) | LECTURE XXII. ORDINARY QUESTIONS. 1. A tie-bar in a roof is made of steel angle bar ; the section of the steel angle bar is 4 inches by 4 inches by inch, and the tie- bar when finished in the workshop is 20 feet in length. When in position in the roof the tie- bar may during a gale have to resist a total pull of 22^ tons ; what is the tensile stress per square inch in the metal of the tie-bar under these con- ditions, and how much would the tie-bar lengthen under this load ? ,-, Young's modulus of elasticity is 12,500 tons per square inch. ^ T (B. of E., 1906.) 2. A knuckle joint is required to withstand a tensile force of 10 tonsi The safe working stress, both in tension and shear, may be taken as 9000 Ibs. per square inch. Find the diameters of the rod and pin, and sketch the joint, roughly, to scale. (C. & G., 1906, 0., Sec. B.) 3. A rectangular test-bar, in tension, gave the following results : Total 1 load > in Ibs. J 8,000 16,000 24,000 32,000 34ooo 40,000 48,000 56,000 60,000 SS.ooo Exten-^ sionin f inches. J '002 0044 0070 '0103 'Oi6 Tgc P470 r 3 fi 2'S 2-9 Sketch a curve showing the relation between force and extension on any suitable scale squared paper may be used and infer the stress at the ORDINAEY QUESTIONS. 385 elastic limit and the maximum stress, the original dimensions of the bar being 1763 inches by '611 inches. If the distance between the gauge points is 10 inches, find the coefficient of elasticity (E) of the bar.^, - ,-- (C. & G., 1906, p., Sec. B.)^? 4. Explain what you mean by the efficiency of a riveted joint, and point out on what it depends. In a marine boiler the diameter is 12 feet, the working pressure is 200 Ibs. per square inch, and the longitudinal joints are butt joints with double straps treble riveted. If the ultimate stress is 62,000 Ibs. per sq. inch, the factor of safety 5, and the efficiency of the joint -8, find the thickness of the plate required, and make a rough sketch of the joint. (C. & G., 1906, O., Sec. B.) 5. In order to connect together the two halves of a long tie-rod, an eye is forged at the end of one half, and a fork (into which the eye enters at the end of the other half, and a pin is passed through the two sides of the fork and through the eye. If the total pull in the tie-rod is 16 tons, and if the shearing stress in the metal of the bolt is not to exceed 8000 Ibs. per square inch, what diameter would you make the pin ? (B. of E., 1907.) 6. A strut is built up out of two pieces of T-steel, each 6 inches by 3 inches by f inch, riveted back to back. If this strut supports a load of 22-3 tons, what is the compressive stress per square inch ? If a total load of 105 tons would destroy this strut, what is the factor of safety ? (B. of E., 1907.) 7. A bar, of rectangular section, 1 75 inches wide and 0*6 1 inches thick, is found under a load of 20,000 Ibs. to have stretched 0*0056 inch. Find the stress induced, and, if the length be 10 inches, find Young's modulus. (C. & G., 1907, O., Sec. B.) 8. Find the thickness of the plates of a cylindrical boiler 50 inches in diameter to sustain a pressure of 50 Ibs. per square inch, the working stress being 4000 Ibs. per sq. inch and the efficiency of the joint being o - 6o. (C. & G., 1907, O., Sec. B.) 9. If in the last question the joint is a lap joint double riveted, and the diameter of the rivets is f inch, find the pitch, the shear stress of rivets, being 4000 Ibs. per square inch. (C. & G., 1907, 0., Sec. B.) 10. A piece of steel is to be tested in tension ; show how you would proceed to make a test, and indicate, by means of a diagram, how the force and extension vary with each other. (C. & G., 1907, O., Sec. B.) 11. A copper trolley wire, which is 0*45 inch in diameter and 60 feet in length, is found to elongate 0-075 m ch under a certain pulL If the mod- ulus of elasticity (Young's modulus) of this quality of copper is known to be 15,000,000 Ibs. per square inch, what is the total pull in the trolley wire ? (B. of E., 1908.) 12. Explain the meanings of the terms " stress," " strain," and " modulus of elasticit y, " by reference to the case of a rod under tensional stress. A piece of bo iler plate, 2 inches by f inch in cross-section, has a load of 12,000 Ibs. applied to it in a testing machine. The modulus of elasticity of the material e xpressed in inches and pounds is 31,000,000. Calculate the values of the stress and strain and determine the increase of length,in a len gth of 12 inches, due to the applied load. (C. & G., 1908, 0., Sec. B.) I 3 . Make a sketch of a knuckle joint connecting an eccentric rod to a val ve spindle, and assuming that the total load on the latter is 4000 Ibs., dete nnine the dimensions of the various parts, and design the joint. Show yo ur calculations clearly, and state what working stesses you have assumed (C. & G., 1908, 0., Sec. B.) 386 APPENDIX B, 14. A tension member, 8 inches by J inch in cross section, has a riveted butt-joint with cover plates on each side. The total load on the member is 72,000 Ibs. Design and draw a joint for this member, and show all your calculations. (C. & G., 1908, O. t Sec. B.) 1 5. A hollow cylinder, 10 inches mean diameter, 10 feet long, and ij inches thick, is to be oast with its axis vertical. Taking the specific gravity of cast iron as 7-5, find the pressure on the bottom of the mould when it is full of metal. One side of a mould for a cast-iron casting is a rectangle, 3 feet deep by 2 feet wide. Find the whole pressure on the side of the mould. (C. & G., 1908, 0., Sec. D.) 1 6. In an experiment with a hollow cast-iron column, 6 feet long,'* 5 inches in external diameter, and 4 inches in internal diameter, it was found that under a compressive load of 30 tons the column shortened by 0*063 inch; what is the value of Young's Modulus (E) in pounds per square inch for this cast-iron ? J^When the load was increased to 192 tons, the column broke; what was the compressive stress in tons per square inch at the instant of fracture. (B. of E., S. i, 1909.) LECTUEE XXIII. ORDINARY QUESTIONS. 1. Describe how you would determine experimentally the modulus of rigidity of either a block of india-rubber or a steel rod. (B. of E., 1906.) 2. In a direct-acting steam-engine mechanism the stroke of the piston is 2 feet and the crank shaft makes 1 50 revolutions per minute. What is the speed of the crank shaft in radians per second ? What is the speed of the crank pin in feet per second ? What is the mean speed of the piston in feet per minute ? (B. of E., 1906.) 3. A solid cylindrical shaft is 5 inches in diameter. Find the external diameter of a hollow shaft of same material, the internal diameter of which is two-thirds the external and which shall have the same strength. Com- pare the weights in the two cases. If the safe working stress be 4 tons per square inch, and the revolutions per minute zoo, find the greatest horse- power which can be safely transmitted. (C. & G., 1906, O., Sec. B.) 4. A motor having a turning moment T is coupled directly to a shaft making N revolutions per minute. Show how to calculate the work trans- mitted by the shaft, and obtain a formula for the horse-power transmitted in^ terms of N and T and a constant. Determine the horse-power trans- mitted by a shaft making 800 revolutions per minute if the turning moment 16,000, measured in pounds and inches. (C. & G., 1908, 0., Sec. B.) 5. A chain is to be used for lifting a load of 5 tons. Assuming a safe working stress on the chain of 4 tons per square inch, find the diameter of the iron of the chain. (C. & G., 1908, 0., Seo. D.) 6. A shaft 3 inches in diameter transmits a twisting moment of 66,ooo-Ib. inches and the flange couplings are bolted together by four bolts spacedi on a circle of 5 inches diameter. Determine the nature and amount of stress on each bolt and determine its diameter if the allowable stress is 12,000 Ibs. per square inch. (C. & G., 1909, O., Sec. B.) 7. Explain what kind of stress is produced in a shaft by a twisting OKDINARY QUESTIONS. 387 moment, and make a diagram showing how the stress varies across the section of a shaft.'* A piece of tubing, 2 inches in external diameter, and inch thick, is'used as a shaft. Assuming that the stress upon it is uniformly distributed, determine the twisting moment it will transmit if the allowable stress is_ 1 2,000 Ibs. per square inch. (C. & G., 1909, 0., Sec. B.) LECTURE XXIV. ORDINARY QUESTIONS. i. Sketch~a"single Hooke's joint, and explain in general terms/ how the angular velocity ratio varies during a revolution. What is the object of a double Hooke's joint ? (C. & G. t 1906, 0., Sec. A.) LECTUBE XXV. OBDINABY QUESTIONS. 1. Sketch, ^andjdescribe the action of, the pin and slot mechanism aa applied to shaping machines. In such a mechanism, the distance between the two centres of rotation is 3"inches, and the time ratio has to be two. If the line of stroke produced pass'through the centre of the variably rotating crank, and is perpendicular to' the line of centres, find the length of the crank radius, and also of the slotted-link, for a stroke of 10 inches. (C. & G., 1906, O., Sec. A.) 2. InTthe four-bar mechanism shown in the sketch, the bar A is"a fixed bar ; the* bars B and D rotate about the fixed centres OAB and OA D, and they are coupled together at their outer ends by the bar C ; the bar B FOUR BAR MECHANISM. revolves~with uniform velocity round its fixed axis" OAB at 50 revolutions per minute. Find in any way you please the position of the bar D when 388 APPENDIX B. the bar B is turned in a clockwise direction through angles of 30, 60, and 90 from the position shown in the sketch. Prepare a table similar to the one shown, and obtain and enter up the results required to complete the table : Angle turned through by the bar B. Angle turned through by the bar D. Mean angular velocity of the bar D in radians per second during each interval . 30. . . 60. 90. (B. of E., S. i, 1907.) 3. Sketch the arrangement in a planing machine in which bevel gears are used, explaining how the motion is reversed, and how a quick-return motion is obtained. Sketch also a " shipper " mechanism in which cams or lugs are used. (C. & G., 1907, 0., Sec. A.) P 4. The piston of a vertical engine has a travel of 18 inches, and the con- necting-rod is 36 inches in length between centres of bearings. The line of action of the piston cuts the horizontal position of the crank 4 inches from the centre of the crank-shaft. Draw the curve of position of the mid-point of the connecting rod for a complete revolution of the crank. Use a scale of th. (C. & G., 1908, O., Sec. A.) W$- Describe, with the help of neatly-drawn sketches which should be roughly to scale, a belt gear for giving a slow cutting speed and a quick return motion suitable for use in a planing machine, the table of which is traversed to and fro by a screw or rack. (B. of E., S. i, 1909.) STUD. INST. C.E. QUESTIONS. 389 LECTUEE II. STUD. I. C.E. EXAM. QUESTIONS. 1. Explain the use of diagrams for determining the results of experiment. Plot the following experimental values of P and W and obtain the relation between them : P pounds ... 8 16 24 40 W o 200 400 800 (Stud. I. C. E., Feb. 1906.) 2. Show how the work done by a variable force can be represented graphically. Assuming that the resistance of a spiral spring", is propor- tional to its extension, and that a load of 24 Ibs. extends the spring 0-25 inch, determine the work done in extending the spring i inch. (Stud. L C. E., Oct. 1906.) 3. Define the terms " force " and " work" A spring is compressed and the relation between the compressive force and the compression is as given below. Find graphically the work done in the process. Force in Ibs. . . I o I 15 I 35 | 65 I no I 170 I 300 Compression in inches . | O | I j 2 | 3 | 4 | 5 | 6 (Stud. I. C. E., Oct. 1907.) LECTURE III. STUD. I. C.E. EXAM. QUESTIONS. 1. A metre rule (weight 50 grams) rests on the edge of a table with 20 centimetres projecting over the edge. On the other end rests a 2O-gram weight. How far from the edge of the table may a 5oo-gram weight be hung before the rod tilts ? (Stud. I. C. E., Feb. 1905.) 2. Prove that the moment about any axis of three forces in equilibrium is zero, and extend the theorem to any number of coplanar forces in equilibrium. Determine the tension of the rope wound on a capstan 2 feet in diameter when 10 men, each weighing 12 stone, are pushing horizontally on the capstan- bars 4 feet from the deck at a radial distance of 8 feet, the vertical through a man's centre of gravity overhanging his toes a distance of I -foot. (Stud. I. C. E., Oct. 1905.) 3. Prove that the centre of gravity of a triangular plate of uniform thickness is on the line joining an apex to the centre of the opposite side, and at a distance from this apex of two-thirds of the length of this median line. Also show how to determine by experiment the centre of gravity of an irregular plate of uniform thickness. (Stud. I. C. E., Feb. 1906.) 4. Show how to determine the resultant of two parallel forces. A horizontal bar, 6 feet long, is supported at each end by rings depending from spring balances. Determine the position of the centre of gravity of the bar if the spring balances indicate loads of 40 Ibs. and 50 Ibs. respectively. (Stud. I. C. E., Oct. 1906.) 5. Explain what is meant by the moment of a force. Show that the sum of the moments of two forces in a plane with respect to a point in that plane is equal to the moment of their resultant. Also show that the moment of a couple, with respect to any axis at right angles to the plane of the couple is invariable. (Stud. I. C. E., Oct. 1906.) 6. Show how to determine experimentally the centre of gravity of an irregular body, and in particular explain how you would proceed to deter- mine the centre of gravity of a metal plate shaped to the section of a tram- way rail (Stud I. C. E., Feb. 1907.) 39 APPENDIX B. 7. What do you understand by the " centre of gravity " of a body ? A balk of timber weighs 800 Ibs. One end rests on the ground,and the other on a " V " support placed on a weigh-bridge. The weight recorded is 320 Ibs. The weigh-bridge is then moved so that the " V " is I foot nearer the end that is resting on the ground and the weigh-bridge registers 360 Ibs. Find how far the centre of gravity of the balk is from that end. (Stud. I. C. E., Oct. 1907.) 8. What is a " couple " ? How is a couple specified, and how can it be represented ? A pair of compasses is opened so that the legs are at 90. Couples are applied to the legs whose moments are respectively 3 Ib.-foot units and 4 Ib.-foot units, and they twist in opposite ways. Find what couple must be applied at the hinge to equilibrate the two, and the axis of that couple. (Stud. I. C. E., Oct. 1907.) 9. Define the term " centre of gravity. " A cylindrical vessel is 5 feet deep and weighs 100 pounds ; when it is empty its centre of gravity is 2 feet above its base. It is gradually filled with water. Plot to scale a curve showing the relation between the depth of water in the vessel and the height of the new centre of gravity, if the vessel when just full can contain 500 pounds of water. (Stud. I. C. E., Feb. 1908.) LECTURE IV. STUD. I.C.E. EXAM. QUESTIONS. 1. Answer, giving reasons, the following questions on the balance : (i) What conditions must be satisfied in order that a balance may be true ? (ii) If the scale-pan knife-edges are above the middle knife-edge show that the sensitiveness of the balance increases with the load, (iii) Why must the scale-pans be suspended freely from the beam ? (Stud. I. C. E., Feb. 1905.) 2. Explain the action of one form of lever weighing-machine. A loo-ton testing machine, using a single lever for weighing the pull on a test-piece, is arranged so that the line of action of the pull is 4 inches distant from the fulcrum, and this pull is balanced by a weight of 5000 Ibs. on the long arm of the lever. Calculate the distance the weight moves from its zero position to balance the full load of 100 tons. (Stud. I. C. E., Feb. 1907. ) LECTURE V. STUD. I.C.E. EXAM. QUESTIONS. i. Explain how work is computed when a force moves its point of appli- cation in any direction. Find the work done per minute by a force pulling a body weighing 400 Ibs. over a rough plane at the rate of 5 miles an hour, if the coefficient of friction is 0-25. (Stud. I. C. E., Feb. 1906.) LECTURE VI. STUD. I.C.E. EXAM. QUESTIONS. I. Write a short essay on the use and principle of a machine, bringing in the meaning of the terms : velocity or displacement ratio, effort, load, advantage, efficiency. Describe some experiment you have made with a machine, and illustrate the meaning of the above terms by numerical examples. (Stud. I. C. E., Feb. 1905.) STUD. INST. C.E. QUESTIONS. 39* 2. Give sketches of three systems of pulleys, and state their mechanical advantage. Explain which system is preferable for a long pull, as in hoisting a weight ; and which is to be preferred for a strong pull, as in setting up a backstay. (Stud. I. a E. f Oct. 1905.) LECTURE VTI. STUD. I.C.E. EXAM. QUESTIONS. I. Explain the principle of action of the " Weston " or differential pulley, and show how to determine the displacement or velocity ratio. In such a machine the velocity ratio was found to be 20, and in order to lift a weight of 420 Ibs. a pull of 40 Ibs. was exerted. Determine the efficiency- of the machine for this load. (Stud. L C. E., Feb. 1906.) LECTUBE VITE. STUD. I.C.E. EXAM. QUESTIONS. 1. A rod AB, the weight of which may be neglected, is hinged at A, andl a weight of 14 Ibs. is suspended from its middle point. A string is fastened to the end B, and when the system is at rest the rod is inclined at 30 to- the horizontal, and the string makes an angle of 90 with the rod. Find the pull along the string. (Stud I. C. E., Oct. 1904.) 2. Two rods AC, BC are freely jointed together at C, and a load of 28 Ibs. is suspended from C. The two ends A and B are connected by a horizontal string. If the system be placed vertically with A and B on a smootk A B Two RODS AC AND BC WITH STRING AB. floor, find by a graphic method the thrusts along AC, BC, and the pull of the string, when AB = 5 feet, AC = 4 feet, BC = 3 feet. * I (Stud. I. C. E., Feb. 1905.) ' 3. Show in a diagram the forces which maintain equilibrium in a book held horizontally between a finger and thumb, and draw a graphical deter- mination of their magnitude. A boat is propelled by two sculls, each 9 feet long and 6 feet from the rowlock to the blade, and the sculler pulls each hand with a force of 20 Ibs. Prove that the thrust on each rowlock is 30 Ibs., but the propulsive torch- on the boat is 20 Ibs., and that the boat moves about double as fast as tee- hands pull. L (Stud. I. C. E., Oct. 1905.) 392 APPENDIX B. ^4.: A flexible oord is carried by two pegs A and D in the same horizontal line and 24 inches apart. A weight of 8 Ibs. hangs at B and an unknown weight, at C,J thereby, causing the cord to assume the form shown in the tflgure. ( A *s 6' *~ tO' * 8" +D FLEXIBLE CORD WITH ATTACHED WEIGHTS. Find by a graphical construction the magnitude of the unknown weight srad the tensions in the parts AB, BG and CD of the cord. (Stud. I. C. E., Feb. 1906.) 5. Show how to determine the resultant of a number of forces meeting at a point. Six forces acting at a point are parallel to the sides of a regular hexagon taken in order, and their magnitudes are 4, 6, 7, 9, 8 and 3 pounds respectively. Find the resultant, assuming that all the forces are directed towards the point. (Stud. I. C. E., Oct. 1906.) 6. Obtain the graphical condition for the equilibrium of a number of tf orces acting at a point. A jointed frame is loaded as shown in the figure. .Determine the stresses in the members of the frame. (Stud. I. C. E., Feb. 1907.) ,4000 3000 Ibs A JOINTED LOADED FRAME. STUD. INST. C.E. QUESTIONS. 393 7. Give the conditions that forces acting on a rigid body may be hi equilibrium. ! I ' A uniform plate, weighing 5 Ibs., is made in the form of a right-angled area whose sides are 3, 4, and 5 feet. It is hung up by means of a string and a peg so that the 5 -foot side is horizontal The peg is at the angle joining the 4- and 5-foot sides; the string is at the other acute angle and makes an angle of 60 with the 5 -foot side. Find graphically the tension in the string. (Stud. I. C. E., Oct. 1907.) 8. What condition must be fulfilled in order that a system of forces acting on a body, which are all in one plane, but not acting at one point, may be hi equilibrium Show how the construction can be used to find the supporting forces required for a structure when loaded. (Stud. I. C. E., Oct. 1907.) 9. If three forces act on a rigid body, what conditions must be fulfilled in order that equilibrium may be maintained ? A uniform beam of timber weighs 200 pounds. One end rests on the ground, the other has a cord attached to it. This cord is pulled till the beam makes 30 with the horizontal, and then the cord makes 60 with the horizontal Find (graphically or otherwise) the tension hi the cord and the force on the ground. Ans. : Tension in cord = 100 Ibs ; reaction = 173 2 Ibs. (Stud. L C. E., Feb. 1908.) LECTURE IX. STUD. I. C.E. EXAM. QUESTIONS. 1. Define horse-power, and prove that a locomotive of H horse-power can draw a train of W tons against a resistance of r Ibs. per ton at a speed of f iff miles an hour. Calculate the horse-power of a locomotive drawing a train of 200 tons up an incline of i in 200 at 50 miles per hour, taking the road and air re- sistance at this speed as 28 Ibs. per ton. (Stud. I. C. E., Oct. 1905.) 2. Define the terms " work " and " power." A car weighing 3 tons is running at 20 miles per hour up a uniform slope of i in 50. The frictional resistances are 50 pounds per ton. Find the H.-P., and the work done in 20 minutes. Ans., H.-P. = 13*8; work done in 20 mins. = 9, 100,800 ft. Ibs. (Stud. I. C. E., Feb. 1908.) LECTURE X. STUD. I.C.E. EXAM. QUESTIONS. i. What is meant by coefficient of friction ? Show how the coefficient of friction between wood and wood can be determined. (Stud. I. C. E., Oct. 1904.) If a machine, such as a screw-jack, does not overhaul, show that the work done against the friction must be more than 50 per cent, of the total work the man does. (Stud. I. C. E., Oct. 1904.) 394 APPENDIX B. 2. Define the coefficient of limiting friction, and prove that it is the tangent of the slope of the incline on which the body is on the point of sliding. Determine geometrically the greatest slope on which a four-wheeled carriage can be held by the brakes, applied to the hind pair of wheels. (Stud. I. C. E., Oct. 1905.) LECTUHE XI. STUD. I.C.E. EXAM. QUESTIONS. 1. A belt-pulley has a diameter of 5 feet, and it is delivering 12 H.-P. to a line of shafting. If the pulley make 120 revolutions per minute, find the force along the tight part of the belt. (Stud. I. C. E., Oct. 1904.) 2. How can the brake horse-power of an engine be determined. If the diameter of the brake-pulley is 3 feet, speed 250 revolutions per minute, load 24 Ibs., and the spring-balance which takes up the rope at the slack end registers 4 Ibs., find the horse-power. (Stud. I. C. E., Feb. 1905.) 3. Explain the meanings of the terms " work " and " energy," and define the practical unit of work used by British engineers. Find the brake horse-power of an engine making 300 revolutions per minute if the tensions on the tight and slack sides of the brake strap are 72 Ibs. and 6 Ibs. respectively, and the brake wheel is 5 feet in diameter. (Stud. I. C. E., Feb. 1907.) 4. Define the term " horse-power." A pulley whose diameter is 4 feet is making 250 revolutions per minute. A belt is put on the pulley and a weight of 50 Ibs. is hung from one end. A spring balance attached to the other end reads 12 Ibs. Find the horse- power delivered to the pulley. If the angle of embrace of the belt is 120 find the extra force on the bearing due to the brake. (Stud. I. 0. E., Oct. 1907.) LECTURE XIII. STUD. I.C.E. EXAM. QUESTIONS. P I. Describe any experiment made by you to determine the efficiency of a machine. In a test of a hand-crane, with gear having a velocity ratio of 1 50 : I , it was found that an effort of 25 Ib. at the handles raised a load of I ton. Determine the efficiency of the machine for this load. (Stud. I. C. E., Feb. 1907.) 2. Explain the terms " velocity ratio," " force ratio," and " efficiency," as applied to a machine. A lifting crab has the following relation between the force on the handle in pounds and the weight lifted in tons. Draw curves connecting the force ratio and the efficiency with the load, and find the efficiency of the crab at loads of 5 tons and 1 5 tons. The velocity ratio is 500. Force in pounds . . . 30 60 90 120 Load in tons . . . 3-3 ^ 7-4 12*5 17*5^. Ans. ^Efficiencies are 0*558 and O'6987 respectively. (Stud. I. C. E., Feb. 1908.) STUD. INST. C.E. QUESTIONS. 395 LECTURE XVII. STUD. I.C.E. EXAM. QUESTIONS. 1. Show that the total pressure upon a flat plate immersed in water is proportional to the depth of its centre of figure below the surface. Calculate the pressure per square inch on a horizontal plate at a depth of 10 feet, assuming that a cubic foot of water weighs 62*4 Ibs. (Stud. I. C. E., Oct. 1906.) 2. Show how to find the total pressure on any submerged plane sur- face. A sluice gate is 4 feet wide and 6 feet high ; the bottom is 20 feet below the. surf ace of the water ; find the total pressure on the gate. [One cubic foot of water weighs 62$ Ibs.] (Stud. I. C. E., Oct. 1907.) LECTUBE XVIII. STUD. I.C.E. EXAM. QUESTIONS. 1. At the bottom of a barometer there is a bubble of air, of which the density is 0-0013 grams per centimetre, and diameter o'3 millimetre. If this rises up the tube its volume increases. Explain why. At what height will its volume be doubled? (Stud. I. C. E., Feb. 1905.) 2. Show how to determine the total pressure and the centre of pressure of a vertical rectangular plate immersed in water with one edge at the surface. Find the total pressure and the depth of the centre of pressure for a plate 12 feet square forming one side of a tank full of water. (Stud. I. C. E., Feb. 1906.) 3.| Define" the term "specific gravity," and show how to determine the specific gravity of (i) a solid, (ii) a liquid. (Stud. I. C. E., Oct. 1906.) 4.' Explain how you would proceed to demonstrate experimentally that a body floating in water displaces a quantity of liquid equal to itself in weight. The displacement of a tug-boat in sea-water was found to be 1560 cubic feet. Calculate the weight of the boat, assuming that a cubic foot of sea- water weighs 64 Ibs. (Stud. I. C. E., Feb. 1907.) 5. Define the term "centre of pressure" for a surface immersed in a fluid. A rectangular sectioned water-channel has a board put vertically across it, which is held up against the water-pressure by two horizontal bars, one at the bottom of the channel, the other 18 inches up the sides. Find the depth of water in the channel which will just overset the board, and the pressure then existing on the board if the breadth of the channel is 5 feet. (Stud. I. C. E., Feb. 1908.) LECTURE XIX STUD. I.C.E. EXAM. QUESTIONSI 1. Describe by aid of a sketch the action of an ordinary suction pump for raising water from a well. Calculate the work required to raise 80 gallons of water through a height of 20 feet, if the pump has an efficiency of 64 per cent. (Stud. I. C. E., Feb. 1907.) 2. Sketch and describe a force pump suitable for raising water from a well. 396 APPENDIX B. It is required to raise water through a total height of 80 feet by a pump of 6 inches stroke with a barrel 3 inches in diameter. If 20 strokes per minute are made, find the gallons pumped per hour, and the horse-power required. (Stud. I C. E., Oct. 1907.) 3. Describe, with sketches, some kind of double-acting force-pump. A pump is delivering water into a boiler in which the pressure is 120 Ibs. per square inch above atmospheric pressure. Find the work done in foot-pounds per pound of water delivered to the boiler. Find also the horse-power of the pump if it delivers 2,000 gallons per hour and its efficiency is 60 per cent. Ans. : Work done per Ib. of water delivered = 276 ft.lbs. ; and H.-P. of pump = 47. (Stud. I. C. E., Feb. 1908.) LECTURE XX. STUD. I.C.E. EXAM. QUESTIONS. i. The pressure of water in a high-pressure main is 700 Ibs. weight per square inch. A load of 2 tons is to be lifted by means of a ram driven from the main. Find the sectional area of the ram, and the cubic feet of water used per H.-P.- hour. (Stud. I. C. E., Oct. 1904.) | LECTURE XXI. STUD. I.C.E. EXAM. QUESTIONS. i. What is the difference between acceleration and velocity ? Plot to scale the following velocities in terms of the time, and write down the times from the start, when the acceleration is zero ; also state the period during which the acceleration is negative. Find by the method of equidistant co-ordinates the average velocity. Time in Velocity in Feet Time in Velocity in Feet Seconds. per Second. Seconds. per Second. 5-0 1 I0'0 2 13-0 3 H-S 4 IS'S 5- iS'8 6 15-8 7 IS'S 8 167 9 18-3 10 20-3 11 22-5 (Stud. I. C. E, Oct. 1904.) 2. A train of 120 tons is found to increase its speed from 20 miles per hour to 40 miles per hour in ten minutes. If the frictional resistance is 2000 Ibs. weight, find the force which must be pulling the tram assuming it to be a constant force. Find average and final horse-power. " (Stud. I. 0. E., Oct. 1904.) 3. Water is projected horizontally from a nozzle. If the point at which it strikes the floor is 6 feet below the nozzle and 5 feet from the vertical line drawn through the nozzle, find the velocity with which the water is projected, (g = 32*2 feet per sec. per sec.) (Stud. I. C. E., Oct. 1904.) 4. A 2-ton fly-wheel drops in speed from 100 revolutions per minute to 90 revolutions per minute. If the mean radius is 5 feet, find the work given up by the fly-wheel. (Stud. I. C. E., Oct. 1904.) STUD. INST. C.E. QUESTIONS. 397 5. A lo-ton truck, moving at the rate of 4 feet per second strikes an 8-ton truck which is standing at rest. If the two move off together after the impact, find the velocity they start with. If the resistance due to- friction, &c., is 200 Ibs. weight, find how far they will run before coming: to rest. (Stud. I. C. E., Oct. 1904.) 6. A ball weighing 10 Ibs. is making 50 revolutions per minute in a horizontal circle of 5 feet radius. Find the force in Ibs. weight acting: upon it towards th*e centre of the circle. (Stud. I. C. E., Oct. 1904.) 7. The distances passed over from rest in I, 2, 3, 4, 5 ... etc., units of time are respectively 0-05, 0-3, 0-58, 0*95, 1-4, 2'O, 27, 3-5, 4-4, 5 35, &S+ 77, 9-0, 10-4, i r6, 13-2, 14-9, 16-6, 18-4, 20-3, 22-3, 24-3, 267, 28-8, 31-2,. 337 units of length. Plot the space-time curve. The unit of time is T fo second, and the unit of length i centimetre. Find the acceleration and show how far it is constant. (Stud. I. C. E., Feb. 1905.) 8. Give examples of how a diagram can be used (i) to correct the observation or measurement of an experiment (See, for example, Question i.) (ii) to find the relation between two quantities, e.g., the effort and load in a machine, (iii) to find the average value of a quantity, e.g., to find the average velocity of a body. (Stud. I. C. E., Feb. 1905.) 9. A shot is projected from a gun. Explain why (i) The momentum of the shot is equal (under certain conditions) to the momentum of the gun. (ii) The energy of the shot when it leaves the gun is greater by far than the energy communicated to the gun. Example : shot 100 Ibs. ; gun i ton ; velocity of shot 1,200 feet per second. Find velocity of recoil of the gun, and the energy of the gun. (Stud. I. C. E., Feb. 1905.) 10. In the car of a balloon a piece of iron is hung from a spring balance. The balance registers 5 Ibs. when the car is at rest. What will it register when the car is rising with an acceleration of 2 feet per second per second ? (Stud. I. C. E., Feb. 1905.) 11. Give some account of the advance made in dynamics by either Galileo or Newton. Show how to find the acceleration towards the centre of a circle of a body moving with uniform velocity in the circle. (Stud. I. C. E., Feb. 1905.) 12. Continuous brakes are now capable of reducing the speed of a train of 3f miles an hour every second, and take 2 seconds to be applied ; show in a tabular form the length of an emergency stop at a speed of 3f , 7^, 1 5, 30, 45, 60 miles an hour. Compare the resistance with gravity ; express the resisting force in Ibs. per ton ; calculate the coefficient of adhesion of the brake-shoe and rail with the wheel, and sketch the mechanical arrangement. (Stud. I. C. E., Oct. 1905.) 13. If W tons is transported from rest to rest a distance feet in t seconds, being accelerated for a distance s 1 and time ^ by a force of P 1 tons up to 398 APPENDIX B. velocity v feet per second, and then brought to rest by P 2 tons acting'f or Jt 2 seconds through * 2 feet, prove the formulas 3?-- oo T? s, (iii) V = 2- Supposing i in m is the steepest incline a train can crawl up, and I in n Is the steepest incline on which the brakes can hold the train, prove that the quickest run up an incline of i in p from one station to stop at the next, a distance of a feet, can be made in V- ( I I\ m + n) V second^ fc- 5) (5- i\ H f ? J Calculate for w = 50, n = 5, p = 100, a = 5280. (Stud. I. C. E., Oct. 1905.) 14. Determine the motion of a circular hoop of radius a feet, whirling in -a vertical plane on a round stick held horizontally, if released when the centre is moving with velocity V feet per second at an angle a with the horizon, and prove that it will make - - revolutions per second in the air. V Prove that the tension in the hoop will be the weight of a length feet of the rim. (Stud. I. C. E., Oct. 1905.) 1 5. Explain how velocities may be compounded. Determine the apparent velocity and direction of rain-drops failing vertically with a velocity of 20 feet per second with reference to a bicyclist moving at the rate of 12 miles an hour. (Stud. I. C. E., Feb. 1906.) 1 6. Show that if a body starting from rest and moving in a straight line is accelerated / feet per second per second it will describe a distance s in t seconds expressed by the formula s = \ft* A train starting from rest receives a uniform acceleration of 0*25 foot per .-a econd per second for one minute. Calculate the distance travelled. (Stud. I. C. E., Feb. 1906.) 17. Define potential and kinetic energy. Find the gam of potential energy of a train weighing 320 tons after mounting an incline 4 miles long of i in 200, and find its kinetic energy -when moving at 30 miles an hour. (Stud. I. C. E., Feb. 1906.) 1 8. Explain how forces are measured, and distinguish between the mass of i Ib. and the weight of i Ib. Determine what force will be necessary to change the velocity of a masa -of 400 Ibs. from 15 to 25 feet per second in 8 seconds. (Stud. I. C. E., Feb. 1906.) STUD. INST. C.E. QUESTIONS. 399 19. Prove that the acceleration a of a body moving with velocity v in a circular path of radius r is expressed by the formula Calculate the force required to constrain a locomotive weighing 50 tons to move in a circle of 400 feet radius when its velocity is 30 miles an hour. (Stud. I. C. E., Feb. 1906.) 20. Explain the terms " moment of inertia," and " radius of gyration," and determine their values for the case of a circular disk of mass m and radius r, when rotating about an axis passing through the centre and perpendicular to the plane of the disk. Give numerical values when the mass is 20 Ibs. and the radius is 2 feet. (Stud. I. C. E. , Feb. 1906.) 21. Define the terms " velocity " and " acceleration " in the case of a body moving in a straight line. A motor-car starting from rest and uniformly accelerated acquires in 2 minutes a velocity of 30 miles an hour. Find the acceleration. (Stud. I. C. E., Oct. 1906.) 22. A heavy ball, attached to a string 30 inches long, is whirled round in a horizontal circle with constant velocity. Make a diagram showing the forces acting on the ball, and calculate the velocity when the string is BO- inclined that the ball moves in a circle of 24 inches radius. (Stud. I. C. E., Oct. 1906.) | 23. Explain what is meant by kinetic energy and deduce an expression for the kinetic energy of a circular disk, rotating about an axis passing through its centre of figure and perpendicular to the plane of the disk. Calculate the kinetic energy of a disk having a radius of 2 feet and weighing; 400 Ibs. when revolving at the rate of 240 revolutions per minute. (Stud. I. C. E., Oct. 1906.) 24. Show how velocities may be compounded. A stone is dropped from a balloon 80 feet above the ground and moving horizontally at the rate of 12 miles an hour. Determine the velocity and direction of the stone when it strikes the ground. (Stud. I. C. E., Feb. 1907.) 25. Define the terms " velocity " and " acceleration." A tram-car start- ing from rest covers s feet in t seconds in accordance with the following; Table : t s ' 2 3 4 5 6 7 8 9 10 4 II 21 34 5Q 69 9i no 144 175 Plot the space-time curve and from it determine the velocity of the body at the end of each second, and show your results to scale upon a time base. Explain how to determine the acceleration from this latter curve,, and determine its value at the end of the fifth second. (Stud. I. C. E., Feb. 1907.) 26. Define " angular velocity " and " angular acceleration " for a body revolving about a fixed axis, and deduce a formula for the angle^turned through by a shaft starting from rest and accelerated uniformly. 2C 4OO APPENDIX B. The spindle of a dynamo is uniformly accelerated, and in 10 seconds from starting it is found to be revolving at the rate of 600 revolutions per minute. Find the acceleration and the number of revolutions it has made. (Stud. I. C. E., Feb. 1907.) 27. Prove that the acceleration (a) of a body moving in a circle of radius y2. r with velocity v is expressed by the formula a = A pulley is found to be out of balance to an amount which may be represented by a mass of 4 oz. at a radius of i foot. Determine the unbalanced force when the shaft is "making 1200 revolutions per minute. ' (Stud. I. C. E., Feb. 1907.) 28. Explain what you understand by " acceleration." A railway carriage is accelerating at 3 feet per sec. per sec. Find the acceleration possessed by a stone dropping from its roof. If the carriage is 8 feet high, find the time taken to fall and the distance the stone travels. (Stud. I. C. E., Oct. 1907.) 29. Define " angular acceleration," and show how it is related to linear acceleration. A hoop whose diameter is 3 feet, is rolling along the ground and comes to rest in 10 seconds, after rolling 240 feet. If it is retarded uniformly, nd the value of the angular retardation. (Stud. I. C. E. Oct. 1907.) 30. A weight is suspended by a string and rotates in a horizontal circle. Find the forces acting on the weight. Such a weight rotates at 20 revolutions per minute when the radius of its circular path is 3 feet. Find the length of the suspending string. (Stud. I. C. E. Oct. 1907.) 31. Define " kinetic energy " and " potential energy." A steamer weighing 2000 tons is proceeding at 20*1 miles per hour. When steam is cut off its speed drops to 19*9 miles per hour after it has moved through 200 feet. Find the mean force retarding it. If it speeds up again to its first speed in half a minute, find approximately the work done in foot-tons, and the horse-power required. 'r (Stud. I. C. E., Oct. 1907.) ' "1 3 2. ^Explain how velocities can be represented and combined. Y Two men, A and B, are 5 miles apart, A being due west of B. They start walking at the same moment : A walks to the south-east at 4 miles per hour, and B walks at 3 miles per hour in such a direction as to meet A on his road. Find graphically the two possible times taken for the two men to meet. Ans. : 53 minutes and i hour 46 minutes. (Stud. I. C. E., Feb. 1908.) 33. Define the term acceleration, and show that if a curve be plotted -connecting the velocity of a body and the time, the area under the curve is the distance traversed, and the slope of the curve measures the accelera- tion. A body starts from rest with a uniform acceleration. After the lapse of a certain time it is found that in successive intervals of 5 seconds and 7 seconds it traverses 62^ feet and 129*- feet respectively : find the accelera- tion at the above time Ans. : 3 feet per sec. per sec. (Stud. I. C. E., Feb. 1908.) 34. Explain what you understand by kinetic and potential energy. A weight of 420 Ibs. is lifted by a force which varies as follows : o I 2 3 4 5 6 7 Force in pounds 700 610 490 390 380 450 650 800 STUD. INST. C.E. QUESTIONS. 40 1 Plot a curve connecting the force and height, and hence find ^"poten- tial and kinetic energy of the body, and the work done by the force when the body is 6 feet from the ground. Ans. : EP = 2730 ft-lb. : EK = 2359 ft.-lb., and work done by the force = 5089 ft.-lb. (Stud. I. C. E., Feb. 1908.) 35. Explain how energy is stored in a fly-wheel, and obtain an expression for this energy. If such a wheel stores 1,000 foot-lb. when rotating at 1 revolution per second, find the work that must be done to change its speed from 10 revolutions per second to 20 revolutions per second. Ans. : EK = *Iw2. Work done in changing the speed = 300,000 foot-lb. (Stud. I. C. E., Feb. 1908.) 36. Explain what you understand by the term " centripetal force." A weight of 20 pounds is hung by a string 10 feet long. It is pulled to one side so as to be 6 feet horizontally away from the vertical If the weight is then let go, find from the energy equation the velocity of the weight at the moment it passes through the lowest point, and deduce the total tension in the string at that moment. Ans.: V = ii'32 feet per sec. ; F = 8 Ib. (Stud. I. C. E., Feb. 1908.) LECTURE XXII. STUD. I. C.E. EXAM. QUESTIONS. 1. Prove that the increase of pressure per foot vertically downwards in a liquid of specific gravity s is 0^433 x s Ibs. per square inch. Mining in ground of uniform density at a depth of h feet, determine the percentage of coal that can be won, leaving sufficient as pillars for the support of the roof, supposing the coal to crush' under its own weight in a column k feet high. (Stud. I. C. E., Oct. 1905.) 2. Investigate the mechanical advantage of the smooth screw, and explain generally how the wind drives a windmill, and a screw propeller propels a steamer. Prove that a platelayer who can apply a force of 28 Ibs. will be apt to break the screw-bolts if provided with a lever more than 3 feet long ; the screw having 8 threads to the inch, and the breaking tension of the bolt being 30 tons per square inch. (Stud. I. C. E., Oct. 1905.) 3. Explain the meanings of the terms " stress," " strain " and " modulus of elasticity " with reference to a bar in tension. A tie-bar, 3*5 square inches in section and 16 feet long, stretches 0*05 inch under a load of 28,000 Ibs. Find the values of the stress, strain, and modulus of elasticity. (Stud. I. C. E., Oct. 1906.) LECTURE XX1IL STUD. I. C.E. EXAM. QUESTIONS. i. A machine is operated by a shaft making N revolutions per minute and transmitting a twisting moment T. Deduce an expression for the horse-power delivered to the machine and calculate its numerical value if the shaft makes 1 10 revolutions per minute, and the twisting moment is 2000, the units being pounds and feet. (Stud. I. C. E., Feb. 1907.) 4O2 NOTES AND QUESTIONS. B. OF E. QUESTIONS, MAY igiO. 403 May, 1910, Examination on Subject VII. APPLIED MECHANICS. STAGE I. GENERAL INSTRUCTIONS. See APPENDIX A. You must not attempt more than EIGHT questions ; EITHER No. i OR No. 22 : must be one of these eight, but not both. The remaining seven questions may be selected from Nos. 2 to 21. The questions in Series A are framed to be more particularly suitable for the Building Trades, and those in Series B /or Mechanical Engineers. SERIES A. 1. Describe, with the help of good sketches, onlyonz of the following, (a), (6),(c),or(d): (a) A mortar-mixing machine. (&) The method of securing the cutting chisels into the cutter blocks of a wood-planing machine. (c) Any form of friction clutch suitable for use with a speed cone or reversing pulleys. (d) Any form of vernier calliper suitable for measuring the dimensions of a test-bar to the nearest thousandth of an inch. (B. of E., 1910.) 2. Describe, with sketches, an apparatus to verify the rule for finding the compressive and tensile forces in the jib and tie of a crane. Do the ex- perimental results exactly agree with the rule, and if not, what is the probable reason ? (B. of E., 1910.) 3. Answer only one of the following, (a), (6), or (c) : (a) Two of the tests specified in order to determine the quality of Portland cement are the determination of the tensile strength of (i) a mortar of neat cement, (ii) a mortar with sand. Describe carefully how the specimens would be made and tested. (&) You are supplied with a length of steel-wire one-eighth of an inch in diameter. You are asked to find (i) Young's Modulus for the material, (ii) the limit of elasticity, (iii) the breaking stress. Explain how you would carry out the test. (See my Adv. Vol. IT.) (c) You wish to know the strength and stiffness of a large timber beam. It is to be built in at the ends ; to be about 20 feet long between supports ; to be, say, 10 inches broad and 12 inches deep, and it is to be loaded uniformly all over. You therefor test a small beam of the same kind of timber ; describe exactly how you would make the test, and how would you use your results ? (See my Adv. Vol. II.) (B. of E., 1910.) 44 APPENDIX C. 4. A wooden beam is built into a wall at one end. Eight feet from the wall there is a hook hi the beam, and from this hook is suspended a weight of i ton. What is the bending moment in the beam (i) at the wall, (ii) at 3 feet from the wall ? Describe the nature of the compressive and tensile stresses throughout any section. (B. of E., 1910.) 5. Let the length of a strut divided by the diameter of its section be called x. W is the maximum load carried. Tests were made on a set of cast-iron struts all of the same section but of different lengths, with the following results : X 10 15 20 25 30 W 64,000 53,500 44,800 33,700 24,IOO Plot a curve showing how the strength depends upon x. What is the maximum load when the length is 18 times the diameter ? (See my Adv. Vol. II.) (B. of E., 1910.) 6. There is a triangular roof-truss ABC ; AC is horizontal and 10 feet long. The angle EC A is 25 and BAG is 55 ; there is a vertical load of 5 tons at B. What are the compressive forces in BA and BC. What are the vertical supporting forces at A and C. Find these answers any way you please. (B. of E., 1910.) 7. A man's hand on the handle of a crane moves 120 feet when the weight is lifted i foot ; 35 per cent, of the total energy given by the man is wasted in friction. A load of 1.5 tons is being lifted. What force is being exerted by the hand ? (B. of E., 1910.) 8. Roughly, what is the weight of a cubic foot of brickwork ? There is a brick building 80 feet long and 50 feet wide. The foundations carry the following weight : First, the volume of brickwork is 24,000 cubic feet, the roof and floors weigh altogether 200 Ib. per horizontal square foot of the area ; the machinery weighs altogether 1 50 tons. What is the total weight to be carried ? What is the breadth of the foundation wall at the footings, if the load there is not to exceed i tons per square foot ? (B. of E., 1910.) 9. It is necessary to keep the " surface level " of water in a shaft at a depth of 30 feet. When left to itself the level rises 4 feet in i minute. The shaft is circular, 6 feet in diameter. What is the weight of water entering per minute ? This water is lifted by a pump whose efficiency is 0.45. What horse-power must be supplied to the pump ? . (B. of E., 1910.) 10. Answer only one of the following questions, (a) or (6) : (a) Describe briefly, sketches are hardly needed, how Portland cement is manufactured. Give a reason for each part of the process. What is your notion of what occurs (i) when cement sets, (ii) when it slowly hardens as it gets older ? (6) Describe briefly, sketches are hardly needed, how any kind of steel used for girders is manufactured. Give a reason for each part of the process. (B. of E., 1910.) B. OF E. QUESTIONS, MAY 19 IO. 405 11. You are given a 4-ton screw-jack. How would you experimentally determine its efficiency under various loads ? What sort of results would you expect to obtain ? (B. of E., 1910.) SERIES B. 12. The depth of water outside the gate of a dry dock is 25 feet. What is the total water pressure on the gate if the width of the gate is 40 feet ? The weight of i cubic foot of salt water is 64 Ib. (B. of E., 1910.) 13. In a hydraulic cylinder, i square foot in cross-section, the piston moves through a distance of i foot. The pressure of the water is 1400 Ib. per square inch. What work is done on the piston ? What is the work done per gallon of water used ? (B. of .,1910.) 14. The pull in the draw-bar between locomotive and train is 13 Ib. per ton when on the level ; the train weighs 200 tons, what is the total force ? If the tram is being pulled up an incline of i in 80 (a vertical rise of i foot in a rail distance of 80 feet), what is the additional pull required ? What is the total pull ? The speed is 2,500 feet per minute. What is the horse- power exercised in drawing the train ? (B. of E., 1910.) 15. As a particle of water flows without friction, its height h feet above- datum level, its pressure p (in Ib. per sq. ft.), and its speed v (in feet per second), may all alter, but the sum 20 W I remains constant. Here g = 32.2 and w = 62.3. The particle flows from a place A where v = O, p = O, and h = 80, to a place B where p = O,. and h = 40 ; what is the value of v at the place B. (B. of E., 1910.) 16. The motion of a body of 3,220 Ib. is opposed by a constant frictional resistance of 2,000 Ib. It starts from rest under the action of a varying force F Ib. whose value is here given at the instants at which the body has passed x feet from rest : F S.HO 2,870 2,630 2,700 , X 5 IO 15 As more work is being done upon the body than what is being wasted in friction, what is the speed of the body when it has moved 1 5 feet from rest ? (B. of E., 1910.) 17. An electric motor is employed for lifting purposes. In lifting 80 tons of grain 100 feet high it is found that 20 Board of Trade units of Energy have to be paid for. A Board of Trade unit is i kilowatt for i hour, and a horse-power is 0.746 kilowatt. The cost is twopence per unit. What is the cost of i horse-power hour usefully done ? What is the ratio of useful work to the electric energy supplied ? (See Appendix D of this book.) t^, (B. of E., 1901.) 406 APPENDIX C. 1 8. The radial speed of the water in the wheel of a centrifugal pump is 6 feet per second. The vanes are directed backwards at an angle of 35 degrees to the rim. What is the real velocity of the water relatively to the vanes ? What is the component of this which is tangential to the rim ? (B. of E., 1910.) ip.'A projectile leaves the muzzle of a gun at 2,000 feet per second, its path being inclined at 20 upwards. What are the horizontal and vertical components of its velocity ? In 3 seconds how far has it travelled hori- zontally ? What is its vertical height above the gun ? Neglect resistance of the atmosphere. (B. of E., 1910.) 2O.*A block of cast-iron, 3 inches by 4 inches by 3 inches, is fastened to the arm of a wheel at the distance of 3 feet from the axis. The wheel makes 2,000 revolutions per minute. What is the force tending to fracture the fastening ? One cubic inch of cast-iron weighs 0.26 Ib. (B. of E., 1910.) .,f an drives air vertically downward through an opening, 8 feet in diameter, with a velocity of 30 feet per second ? The air weighs 0.08 Ib. per cubic foot. What weight of air is driven downward per second ? What is its mo- mentum ? (B. of E., 1910.) 22. Describe, with sketches, only one of the following, (a), (b), (c), or (d) : (a) The shaft bearing of any modern fast running machine such as a water turbine or dynamo machine. (6) An hydraulic appliance in use by hydraulic companies or their customers, such as an accumulator, or a motor, or a force pump. (c) Any form of sensitive drilling machine of the " pillar " type. What are the advantages of this type of machine for small accurate work ? (d) Any form of quick return motion suitable for use on a shaping machine. Show how the stroke of the tool may be varied. (B. of E., 1910.) STUD. INST. C.E, QUESTIONS, OCT. 1909, 407 Appendix C. The Institution of Civil Engineers' Examinations for Admission of Students, October 1909. (vii.) ELEMENTARY MECHANICS. Not more than EIGHT questions to be attempted by any Candidate. 1. Define the terms relative velocity and absolute velocity. A bicycle has 28-inch wheels, and is being ridden at 20 miles an hour. Find the velocity of a point on the rim 14 inches from the ground, (i) relative to the rider, (2) relative to the ground. (Stud. I. C. E., Oct. 1909.) 2. Explain the term acceleration, and show how it is to be measured when non-uniform. The velocity of a body at given times is as given in the schedule below. Draw the curve connecting the two quantities, and find the space the body has moved. (Stud. I. C. E., Oct. 1909.) Time in seconds. . o o i 2 3 4 5 6 7 8 Velocity in feet per'l second . . . .J 0'95 3'8o 5-00 4*60 3-15 1-65 075 o 3. Give the equations for the motion of a body which has a uniform acceleration. An airship is travelling in a horizontal line at 30 miles per hour towards an object on the ground on which it is desired to drop a shell. The ship is 1600 feet above the ground. Find where it must let the shell go in order to hit the object. (Stud. I. C. E., Oct. 1909.) 4. Show how to find the acceleration produced by a force acting on a body. A tramcar whose weight is 14 tons is being pulled horizontally along a track by a force of 2000 Ibs. ; the track friction is 20 Ibs. per ton. Find the acceleration produced in miles per hour per second. (Stud. I. C. E., Oct. 1909.) 5. Enunciate the conditions that must be fulfilled in order that three non-parallel forces may be in equilibrium. A uniform plank AB is pivoted smoothly at the end A, and has a rope attached to the end B. The plank is 10 feet from A to B, and weight 100 Ibs. The rope is 8 feet long and is fixed to a point C, such that C is 408 APPENDIX C. at the same height as A and the angle ABC is a right angle. Determine graphically the tension in the rope and the pressure on the pivot. (Stud. I. C. E., Oct. 1909.) 6. Show that a body which is moving with uniform angular velocity w in a circle of radius r has an acceleration u?2>. towards the centre of the circle. A flywheel has an internal trough turned on it to contain cooling water. Find the least possible number of revolutions per minute that will permit the retention of the water if the diameter of the trough is 8 feet. (Stud. I. C. E.. Oct. 1909.) 7. What is the moment of a force ? A timber balk is 42 feet long : a small cross-bar is placed underneath, and 2 feet from one end ; the force required to lift the other from the ground is 600 Ibs. : the cross-bar is moved to 6 feet from the end, and the force is then 500 Ibs. ; find the weight of the balk, and the distance of its centre of gravity from the end. (Stud. I. C. E., Oct. 1909.) 8. Give the condition that a body may stand in equilibrium on a plane. A triangle ABC is cut out of a thick board : the side AB is 10 inches, and the angle ABC is 30. Find the other sides if the triangle is just stable when standing on the side AB on a horizontal plane. (Stud. I. C. E., Oct. 1909.) 9. Define the terms work, power, horse-power. A car weighs i^ ton and its engine is working at a constant horse-power, It is running on a road for which the total frictional resistance is 50 Ibs per ton. A five-mile run takes 20 minutes, and in the 5 miles the total rise is 400 feet. The car's speed is the same at each end of the run. Find its HP. (Stud. I. C. E., Oct. 1909.) 10. What is the meaning of the efficiency of a machine ? A screwjack has a screw whose pitch is inch, and the force is applied by a lever 15 inches long. It is found that 5 tons is lifted by a force of 56 Ibs. applied at right angles to the lever. Find the efficiency of the jack. (Stud. I. C. E., Oct. 1909.) 1 1 . What is the specific gravity of a body ? A thin metal tube is 32 inches long, and one square inch in sectional area. The bottom inch is filled with a metal. When the tube is put in a vessel of water, it floats with 2 inches out of the water. Find the specific gravity of the metal if the tube weighs 0*6 Ib. (Stud. I. C. E., Oct. 1909.) 12. A plane surface is immersed in a fluid : find an expression for the pressure exerted on the same. A cube of 3 feet side is placed in water 4 feet deep, with one face of the cube horizontal. Find the pressure on each side. (Stud. I. C. E., Oct. 1909.) STUD. INST. C.E. QUESTIONS, FEB. 1910. 409 Appendix C. The Institution of Civil Engineers' Examinations for Admission of Students, February, 1910. ELEMENTARY MECHANICS. Not mare than EIGHT questions to be attempted by any Candidate. (The weight of a cubic foot of water may be taken as 62% Ibs. and g as 32.) 1. Explain how two velocities can be combined. A cyclist is riding along a straight road which runs at 30 to a straight piece of railway line. He sees an engine on that line when he is looking in a direction making a constant angle of 45 with his direction of motion. He is travelling at 12 miles an hour : find the velocity of the engine. 2. Establish the equation for the space traversed by a body moving with constant acceleration. A balloon has a total weight of I ton and is at rest 900 feet above the ground. It suddenly lets fall i cwt. of ballast : neglecting friction, find how high it will have risen when the ballast reaches the ground. 3. A curve is drawn connecting time and velocity of a moving body : show how the space traversed in a given interval can be found from it. Such a curve is drawn with a horizontal scale of I inch = I second and a vertical scale of i inch = i foot per second. It is a semi-circle of 3 inches radius with its centre at the 3 seconds point on the horizontal line. Find the total space traversed, and the acceleration at the end of the first second. 4. Give the theorem known as the triangle of forces. A uniform heavy bar is 5 feet long and weighs 20 Ibs. Two strings 3 and 4 feet long are attached to the ends, the other extremities of the strings being fixed to a peg. Draw the position when the bar hangs- freely ; determine the triangle of forces and the tensions in the strings. 5. If a body he moving in a circular path of radius r with a velocity v,. t'2 show that there must be a constraining force of the amount acting: along the radius. A cyclist and his machine together weigh 12 stone; he is turning a corner with a velocity of 15 miles per hour and the radius is 80 feet. Find the force acting. 6. What is the condition that must be fulfilled in order that a body- acted on by parallel forces shall be at rest ? 410 APPENDIX 0. Two heavy uniform rods are fixed at an angle and pivoted at the junction point. The shorter rod is 6 feet long and weighs 10 Ibs. : the longer rod is 8 feet long and weighs 15 Ibs. When suspended by the pivot the shorter rod is horizontal. Find the angle at which the rods are fixed together. 7. Explain the meaning of the terms worlt and power, and give the usual units in which they are measured. A locomotive is steadily pulling a train weighing 500 tons (total) at 25 milts an hour up a slope of I in 200. The frictional resistance is 10 Ibs. a ton. Find the HP. exerted in traction. 8. Explain the meaning of the term momentum. A vessel of 2000 tons is starting to tow one of 1000 tons. The instant before the rope becomes taut, the vessels are moving on the same line with respective velocities of 7 and 3 knots. Find the common velocity at the moment the rope becomes taut. 9. What do you understand by the velocity ratio and the efficiency of a machine ? A hand crane has a handle radius of i foot. It is found that forty-two turns are required to raise the load one foot, and that a tangential force of 37 pounds will just raise 5 tons. Find the efficiency of the crane. 10. Define the centre of gravity of a body, and show how you would experimentally find the centre of gravity of a lamina, A triangle has sides which are respectively 5, 4 and 3 inches. To each a square is attached externally, and in the plane of the triangle, the squares being all cut from the same sheet. Draw the figure to scale and mark on it the centre of gravity of the area formed by the three squares. 11. Define the term specific gravity and show how you would measure it for a body. A yard of wire weighs 70 grams in air and 61 grams in water. Find the specific gravity and the section of the wire [yard = 91 \ centimetres, i cubic centimetre of water weighs I gram.] 12. How would you find the pressure exerted by a head of water on a dam? A dam is 50 feet high and has a vertical internal face. Find the total pressure per foot run and the position at which it may be considered to act. APPENDIX D. THE CENTIMETRE, GRAMME, SECOND, OR C.G.S. SYSTEM OF UNITS OF MEASUREMENT AND THEIR DEFINITIONS.* I. Fundamental Units. The C.G.S. and the practical electrical units are derived from the following mechanical units. The Centimetre as a unit of length ; the Gramme as a unit of mass ; and the second as a unit of time. The Centimetre (cm) is equal to 0*3937 inch in length, and nominally represents one thousand-millionth part, or I000 1 OOOOQ of a quadrant of the earth. The Gramme (gm) is equal to 15 '432 grains, and represents the mass of a cubic centimetre of water at 4 C. Also, I Ib. of 16 oz. is equal to 45 3 '6 grammes. Mass (M) is the quantity of matter in a body. The second (s) is the time of one swing of a pendulum making 86,i64'09 swings in a sidereal day, or the 1/86,400 part of a mean solar day. II. Derived Mechanical Units. Area (A or cm2). The unit of area is the square centimetre. Volume (V or cms). The unit of volume is the cubic centimetre. Velocity (v or cm/s) is rate of change of position. It involves the idea of direction as well as that of magnitude. Velocity is uniform when equal distances are traversed in equal intervals of time. The unit of velocity is the velocity of a body which moves through unit distance in unit time, or the velocity of one centimetre per second. Momentum (Mv, or gm x cm/s) is the quantity of motion in a body, and is measured by mass x velocity. Accleration (a or cm/s2) is the rate of change of velocity, whether that change takes place in the direction of motion or not. The unit of acceleration is the acceleration of a body which undergoes unit change of velocity in unit time, or an acceleration of one centimetre-per-second per second. The acceleration due to gravity is considerably greater than this, for the change of velocity imparted by gravity to falling bodies in one second is about 981 centimetres per second (or about 32*2 feet per second). The value differs slightly in different latitudes. At Greenwich the value of the acceleration due to gravity is g = 981*17 ; at the Equator / = 978*1, and at the North Pole g = 983*1. * The Author is indebted to his Publishers, Charles Griffin and Co., for liberty to abstract the following pages on this subject from the latest edition of Munro and Jamieson's " Pocket-book of Electrical Rules and Tables for Electricians and Engineers," to which the student is referred for further values and definitions, and values of Practical Electrical Unit* of Measurement and Testing Rules, &c. A.J. 412 APPENDIX D. Force (F or /) is that which tends to alter a body's natural state of rest or of uniform motion in a straight line. Force is measured by the rate of change of momentum which it produces, or mass x acceleration. The Unit of Force, or Dyne, is that force which, acting for one second on a mass of one gramme, gives to it a velocity of one centimetre per second. The force with which the earth attracts any mass is usually called the *' weight " of that mass, and its value obviously differs at different points of the earth's surface. The force with which a body gravitates i.e. its weight (in dynes), is found by multiplying its mass (in grammes) by the value of g at the particular place where the force is exerted. Work is the product of a force and the distance through which its acts. The unit of work is the work done in overcoming unit force through unit distance, i.e. in pushing a body through a distance of one centimetre against a force of one dyne. It is called the Erg. Since the " weight " of i gramme is I x 981, or 981 dynes, the work of raising one gramme through the height of one centimetre against the force of gravity is 981 rgs or g ergs. One kilogramme-metre = 100,000 (g) ergs. One foot-pound as 13,825 (g) ergs = 1-356 x 10? ergs. Energy is that property which, possessed by a body, gives it the capa- bility of doing work. Kinetic energy is the work a body can do in virtue of its motion. Potential energy is the work a body can do in virtue of its position. The unit of energy is the Erg. Power or Activity is the rate of working. The unit is called the Watt(Wp) = io7 ergs per second, or the work done at the rate of one Joule (J) per second. One Horse-power (H.P.) = 33,000 ft.-lbs. per minute = 550 ft.-lbs. per second. But as seen above under Work, i ft. Ib. = i'356 x 10? ergs, and under Power, i Watt 10? ergs per second. Hence, a Horse-power = 550 x 1*356 x 10? ergs per sec. = 746 Watts. If E = volts, = amperes, and R = ohms. ; then, by Ohm's Law C = E/R, Also, EC = C2R = E2 R = Watts. Therefore, H.P. = = = . 746 746 74 6R- UNITS OF MEASUREMENT AND THEIR DEFINITIONS. 413 PRACTICAL ELECTRICAL UNITS. 1. As a Unit of Resistance (R), th International Ohm (ohm or w), based upon the ohm which is 10$ units of resistance in the C.G.S. system of electro -magnetic units, is represented by the resistance offered to an unvarying electric current by a column of mercury at the temperature^ of melting ice, 14-4521 grammes in mass, of a constant cross-sectional area and of the length of 106*3 centimetres. 2. As a Unit of Current (C), the International Ampere (A), which is one-tenth of the unit of current of the C.G.S. system of electro-magnetic units, and which is represented sufficiently well for practical use by the unvarying current which, when passed through a solution of nitrate of silver in water, and in accordance with the International specifications, deposits silver at the rate of o'OOinS grammes per second. 3. As a Unit of Electro-motive Force (E) the International Volt (V), which is the E.M.F. that, steadily applied to a conductor whose resistance is one International Ohm, will produce a current of one Inter- national Ampere, and which is represented sufficiently well for practical use by $$$ of the E.M.F. between the poles or electrodes of the voltaic cell known as Clark's cell, at a temperature of 15 Centigrade, and prepared in the manner described in the International specification. 4. As the Unit of Quantity (Q) the International Coulomb (A xs), which is the quantity of electricity transferred by a current of one Inter- national Ampere in one second. 5. As the Unit of Capacity (K) the International Farad (Fd), which is the capacity of a conductor charged to a potential of one International Volt by one International Coulomb of electricity. 6. As a Unit of Work the Joule (J) (or Watt-second (W x $),) which is ID? units of work in the C.G.S. system, and which is represented sufficiently well for practical use by the energy expended in I second in heating an International Ohm. 7. As the Unit of Power (P w ) the International Watt (W P ), which is equal to 10? units of power in the C.G.S. system, and which is represented sufficiently well for practical use by the work done at the rate of one Joule per second. The Kilowatt (K.W.) = 1000 Watts = i horse-power. 8. As the Unit of Induction (L) the Henry (Hj ), which is the induction in the circuit when the E.M.F. induced in this circuit is one International Volt, while the inducing current varies at the rate of one ampere per second. g. The Board of Trade Commercial Unit of Work or (B.T.U.) is the Kilowatt-hour (K.W.-hr.) = 1000 Watt-hours = i H.P. working for one hour. Or say 10 amperes flowing in a circuit for i hour at a pressure of 100 volts. Note. For further simple explanations with Examples, see 7th Edition of Prof. Jamieson's " Manual of Magnetism and Electricity," pp. 87 to 94, and 222 to 224. Also latest Edition of Munro and Jamieson's Electrical Pocket Book, both published by Charles Griffin & Co. Ltd., London. 414 APPENDIX D. EXAMINATION TABLES. USEFUL CONSTANTS. 1 Inch a 25*4 millimetres. t Gallon = -1606 cubic foot = 10 Ibs. of water at 62* F. I Naut = 6080 feet. I Knot = 6080 feet per hour. Weight of 1 Ib. in London = 445,000 dyne*. One pound avoirdupois = 7000 grains = 453-6 gramme*. 1 Cubic foot of water weighs 62*3 Ibs. at 65* F. 1 Cubic foot of air at 0* 0. and 1 atmosphere, weighs *0807 Ib. 1 Cubic foot of Hydrogen at 0* 0. and 1 atmosphere, weighs -00557 Ib* 1 Foot-pound 1-8562 x 10' ergs. 1 Horse-power-hour = 83000 x 60 foot-pounds. 1 Electrical unit 1000 watt-hours. *V. Kquiv^t to .nit Bgnlf. H, I. {, gj ; } ** 1 Horse-power = 38000 foot-pomnds per minute = 746 watts. Volts x amperes = watts. 1 Atmosphere = 14*7 Ib. per square inch = 2116 Ibs. per square foot a 760 m.m. of mercury = 10* dynes per sq. cm. nearly. A Column of water 2*3 feet high corresponds to a pressure of 1 Ib. per square inch. Absolute temp., t * V 0. + 273*-7. Regnaulfs H = 606'6 + '305 6 0. m 1082 + -805 8* F. fu 1.0*46^479 log M l> = 61007 -f- where log ^B = 3-1812, log W C a fi-0871, p is in pounds per square inch, t is absolute temperature Centigrade^, u is the volume in cubic feet per pound of steam. = 3-1416. One radian = 57 '3 degrees. To convert common into Napierian logarithms, multiplj by 2-3021 The base of the Napierian logarithms is = 2-7183. The value of g at London. 32-182 feet per sec. per sec. TABLE OF LOGARITHMS. 415 10 1 2 3 4 5 6 7 8 9 123 456 789 0000 0043 0086 0123 0170 0212 0253 0294 0334 0374 4 8 12 17 21 25 29 33 37 11 12 13 14 15 16 17 13 19 0414 0792 1139 0453 0328 1173 0492 0864 1206 0531 OS99 1239 0569 0934 1271 0607 0969 1303 0645 1004 1335 0682 1038 1367 0719 1072 1399 0755 1106 1430 4 8 11 3 7 10 3 6 10 15 19 23 26 30 34 14 17 21'24 28 31 13 16 19J23 26 29 1461 1761 2041 1492 1790 2068 1523 1818 2095 1553 1847 2122 1534 1875 2148 1614 1903 2175 1644 1931 2201 1673 1959 2227 1703 1987 2253 1732 2014 2279 369 368 358 12 15 18 11 14 17 11 13 16 21 24 27 20 22 25 18 21 24 2304 2553 2788 2330 2577 2810 2355 2601 2833 2380 2625 2356 2405 2648 2878 2430 2672 2900 2455 2695 2923 2480 2718 2945 2504 2742 2967 2529 2765 2989 257 257 247 10 12 15 9 12 14 9 11 13 17 20 22 16 19 21 16 18 20 20 21 22 23 3010 3032 3054 3075 3096 3118 3139 3160 3131 3201 246 8 11 13J15 17 19 3222 3424 3617 3243 3444 3636 3263 3464 3655 3284 34S3 3674 3304 3502 3692 3324 3522 3711 3345 3541 3729 3365 3560 3747 3385 3579 3766 3404 3393 3784 246 246 246 8 10 1214 16 18 8 10 1214 15 17 7 9 11113 15 17 24 25 26 27 28 29 30 31 32 33 3S02 3979 4150 3820 3997 4166 3833 4014 4183 3356 4031 4200 3S74 4048 4216 3392 4065 4232 3909 40S2 4249 3927 4099 4265 3945 4116 4281 3962 4133 4298 2.45 235 235 7 9 11 12 14 16 7 9 10'l2 14 15 7 8 1011 13 15 4314 4472 4624 4330 4487 4639 4346 4502 4654 4362 4518 4669 4378 4533 4683 4393 4548 4698 4409 4564 4713. 4425 4579 4728 4440 4594 4742 4456 4609 4757 23568 9J11 13 14 23568 9>11 12 14 1 3 4 1 6 7 9 10 12 13 4771 4786 4800 4814 4S29 4843 4S57 4871 4886 4900 134 679 678 578 568 10 11 13 10 11 12 9 11 12 9 10 12 4914 5051 5185 4923 5065 5193 4942 5079 5211 4955 5092 5224 4969 5105 5237 4933 5119 5250 4937 5132 5263 5011 5145 5276 5024 5159 5289 5038 5172 5302 134 134 134 34 35 36 5315 5441 5563 5328 5453 5575 5340 5465 55S7 5353 5478 5599 5366 5490 5611 5373 5502 5623 5391 5514 5635 5403 5527 5647 5416 5539 5658 5428 5551 5670 134 124 124 568 567 567 9 10 11 9 10 11 8 10 11 37 33 39 40 41 42 43 56S2 5798 5911 5694 5809 5922 57 ;. 5821 5933 5717 5832 5944 5729 5843 5955 5740 5855 5966 5752 5866 5977 5763 5877 5988 5775 5888 5999 5786 5899 6010 123 123 123 5 6 7 567 467 8 9 10 8 9 10 s a 10 6021 C031 6042 6053 6064 6075 6035 6096 6107 6212 6314 6415 6117 123 456 8 9 10 6128 6232 6335 6138 6243 6345 6149 6253 6355 6160 6263 6365 6170 6274 6375 6180 6284 6385 6191 6294 6395 6201 6304 6405 6222 6325 6425 123 123 123 456 456 456 789 789 789 44 45 46 6435 6532 6628 6444 6542 6637 6454 6551 6646 6464 6561 6656 6474 6571 6665 6484 6580 6675 6493 6590 6634 6503 6599 6693 6513 66C9 6702 6522 6618 6712 123 123 123 456 456 456 7 8 9 789 778 47 43 49 50 6721 6812 6902 6730 6821 6911 6739 6330 6920 6749 6839 6928 6753 6848 6937 6767 6857 6946 6776 6866 6955 6785 6875 6964 6794 6884 6972 6803 6S93 6981 123 123 123 455 445 445 678 678 678 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 123 345 678 51 Pi -7 53 7076 7lf:0 7243 7324 7034 7168 7251 7093 7177 7259 7101 7185 7267 7348 7110 7193 7275 7118 7202 7284 7126 7210 7292 7135 7218 7300 7143 7226 7308 7152 7235 7316 123 122 122 345 345 345 678 677 667 7332 7340 7356 7364 7372 7380 7388 7306 122 345 667 416 TABLE OP LOGARITHMS. Continued. i r 1 2 3 4 5 6 7 8 9 1 2 3 456 789 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 122 345 567 56 57 58 7482 7559 7634 7490 7566 7642 7497 7574 7649 7505 7582 7657 7513 7589 7664 7520 7597 7672 7528 7604 7679 7536 7612 7686 7543 7619 7694 7551 7627 77a 122 1 2 2 112 345 345 344 567 567 567 59 60 61 7709 7782 7S53 7716 7789 7860 7723 7796 7868 7731 7S03 7875 7738 7810 78S2 77-^5 7818 7889 7752 7825 7896 7760 79.J3 7767 7C39 7910 7774 78-16 79. 7 112 112 112 344 344 344 567 566 566 62 63 64 7924 7993 80G2 7931 8000 8069 7938 8007 8075 7945 8014 80S2 7952 8021 80i9 7959 028 8096 7966 80 5 8102 7973 8041 8109 7980 8048 8116 7917 8055 8122 112 112 112 334 334 334 566 556 556 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 81 9 112 334 556 66 67 68 8195 8261 8325 8202 8267 8331 8209 8274 8338 8215 8280 8344 8222 8287 8351 8228 8293 8357 8235 8299 8363 8241 83C6 8370 8248 8312 8376 SC54 8319 8382 112 112 112 334 334 334 556 556 456 69 70 71 8388 8451 8513 8395 8457 8519 8401 8403 8525 8407 8470 8531 8414 8476 8537 8420 8482 8543 8426 8488 8549 8432 8494 8555 8439 8500 8561 8445 85C6 8567 1 1 2 112 112 234 234 234 456 456 455 72 73 74 8573 8633 8692 8579 8G39 8698 8585 8645 8704 S591 8651 8710 8597 8657 8716 S603 8663 8722 8609 8669 8727 8615 8675 8733 8621 8681 8739 8627 86S6 8745 112 112 112 234 234 234 455 455 455 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 112 233 455 76 77 78 8808 8S65 8921 8814 8871 8927 8820 8876 8932 8S25 8S82 8938 8831 SS87 943 8837 8;93 8949 8842 8899 8954 8848 8904 8960 8854 8910 8965 S8f.9 8915 8971 112 112 112 233 233 233 455 445 445 79 SO 81 8976 9031 9085 C9S2 C036 9090 S9S7 9042 9096 8993 9047 9101 89C8 90C3 9106 9004 90C3 9112 9009 90C3 9117 9015 9069 9122 9020 9074 9128 9025 9079 9133 112 112 112 233 233 233 445 445 445 82 S3 84 9138 9191 9243 9143 9196 9248 9149 9201 9253 9154 92C6 9258 9159 9212 9263 9165 9217 9269 9170 9222 9274 9175 9227 9279 91 SO 9232 9284 9186 9238 9289 112 112 112 233 233 233 445 445 445 85 9294 9299 9304 9309 9315 9320 9325 C330 9335 9340 112 233 445 86 87 88 9345 9395 9445 9350 9400 9450 9355 9405 9455 9360 9410 9460 9365 9415 9465 9370 9420 9469 9375 9425 9474 C3GO 9430 9479 9385 9435 9484 9390 9440 94S9 112 Oil Oil 233 223 223 445 344 344 89 90 91 9494 9542 9590 9499 9547 9595 9504 9552 9600 9509 9557 9605 9513 9562 9609 9518 9566 9614 9523 9571 9619 9528 9576 9624 9533 9581 9628 9538 95S6 9633 Oil Oil Oil 223 223 223 344 344 344 92 93 94 9088 96S5 9731 9643 9689 9736 9647 9694 9741 9652 9699 9745 9657 9703 9750 9661 9708 9754 9666 9713 9759 9671 9717 9763 9675 97-22 9768 9680 9727 9773 Oil Oil Oil 223 223 223 344 344 344 95 9777 9?82 9786 9791 9795 9800 9805 9809 9814 9818 Oil 223 344 96 97 98 9S23 9868 9912 9827 9872 9917 9832 9877 9921 9836 9881 9926 9341 9886 9930 9845 9890 9934 9S50 9894 9939 9854 9899 9943 9859 9903, 9948 9S63 9908 9952 Oil Oil Oil 223 223 223 344 344 344 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 Oil 223 334 TABLE OF ANTILOGARITHMS. 417 1 2 3 4 5 6 7 1 8 9 1 2 3 456 789 00| 1000 1002 1005 1007 1009 1012 1014 1016 1019 1021 001 111 222 01 02 03 1023 1047 1072 1026 1050 1074 1028 1052 1076 1030 1054 1079 1033 1057 1081 1035 1059 1084 1038 1062 1086 1040 1064 1089 1042 1067 1091 1045 1069 1094 001 001 001 111 111 111 222 222 222 041 06 1096 1122 1148 1099 1125 1151 1102 1127 1153 1104 1130 1156 1107 1132 1159 1109 1135 1161 1112 1133 1164 1114 1140 1167 1117 1143 1169 1119 1146 1172 Oil Oil Oil 1 2 ' 1 2 j 1 2| 222 222 222 07 08 09 1175 1202 1230 1178 1205 1233 1180 1-208 1236 1183 1211 1239 1186 1213 1242 1189 1216 1245 1191 1219 1247 1194 1222 1250 1197 1225 1253 1199 1227 1256 Oil Oil 1 1 1 2 1 2 112 222 223 223 10 1259 1262 1265 1268 1271 1274 1276 1279 1282 1285 Oil 112 223 11 12 13 1288 1318 1349 1291 1321 1352 1294 1324 1355 1297 1327 1353 1300 1330 1361 1303 1334 13.5 1306 1337 1363 1309 1340 1371 1312 1343 1374 1315 1346 1377 Oil Oil Oil 122 122 122 223 223 233 14 15 1C 13SO 1413 1445 13S4 1416 1449 13S7 141 1452 1390 1422 1455 1393 1426 1459 1396 14_>9 1462 1400 1432 1466 1403 1435 1469 1406 1439 1472 1409 1442 1476 Oil Oil 1 1 1 2 2 122 1 2 2 233 233 233 17 is 19 1479 1514 1549 1483 1517 1552 14S6 1521 1556 14S9 1524 1560 1493 1523 1563 1496 1531 1567 1500 1535 1570 1503 1538 1574 1507 1542 1578 1510 1545 1581 1 1 Oil 1 1 122 122 122 233 233 333 20 1585 1589 1592 1596 1600 1603 1607 1611 1614 1613 Oil 122 333 21 22 23 1022 1660 1093 1626 1603 1702 1620 1667 1706 1633 1671 1710 1637 1675 1714 1641 1679 1713 1644 16^3 1722 1643 1687 1726 1652 1690 1730 1656 1694 1734 Oil 1 1 Oil 222 222 222 333 333 334 24 25 26 1738 1778 1820 1742 1782 1824 1746 1786 1828 1750 1791 1832 1754 1795 1837 1758 1799 1841 1762, 1803" 1845 1766 1307 1849 1770 1811 1854 1774 1816 1858 Oil Oil Oil 222 222 223 334 334 334 27 28 29 1862 1905 1950 1866 1910 1954 1871 1914 1959 1875 1919 1963 1879 1923 1968 1884 1928 1972 1888 1932 1977 1892 1936 1982 1897 1941 1986 1901 1945 1991 Oil 1 1 Oil 223 223 223 334 344 344 30 1995 2000 2004 2009 2014 2018 2023 2028 2032 2037 Oil 223 344 31 32 33 2042 2089 2138 *2046 2094 2143 2051 2099 2148 2056 2104 2153 2061 2109 2153 2065 2113 2163 2070 2118 2168 2075 2123 2173 2080 2123 2173 20S4 2133 2183 Oil Oil Oil 223 223 223 344 344 344 34 35 36 2188 2239 2291 2193 2244 2296 2198 2249 2301 2203 2254 2307 2203 2259 2312 2213 2265 2317 2218 2270 2323 2223 2275 2328 2228 2280 2333 2234 2286 2339 112 112 112 233 233 233 4 5 4 5 4 5 37 38 39 2344 2399 2455 2350 2404 2460 2355 2410 2466 2360 2415 2472 2366 2421 2477 2371 2427 2483 2377 2432 2489 2382 2438 2495 2333 2443 2500 2393 2449 2506 112 112 112 233 233 233 4 5 4 5 5 5 40 2512 2518 2523 2529 2535 2541 2547 2553 2-559 2564 112 234 455 41 42 43 2570 2630 2692 2576 2636 2698 2582 2642 2704 2588 2649 2710 2594 2655 2716 2600 2661 2723 2606 2667 2729 2612 2673 2735 2618 2679 2742 2624 26S5 2743 112 112 112 2 3 2 3 3 3 455 456 456 44 45 46 2754 2818 28S4 2761 2825 2891 2767 2831 2897 2773 2833 2904 2780 2S44 2911 2786 2851 2917 2793 2853 2924 2799 2864 2931 2805 2371 2938 2312 2S77 2944 112 112 112 3 3 3 3 3 3 456 556 556 47 48 49 2951 3020 :3000 .2953 3027 3097 2965 3034 3105 2972 3041 3112 2979 3048 3119 2985 3055 3126 2992 3062 3133 2999 3069 3141 3006 3076 3148 3013 30S3 3155 112 112 112 3 3 i 3 4 3 4 556 566 566 TABLE OF ANTILOGARITHMS. continued. 1 2 3 * 5 6 7 8 9 1 2 3 456 789 50 3162 3170 3177 3184 3192 3199 3206 3214 3221 3228 112 344 567 51 53 3236 3311 3388 3243 3319 3396 3251 3327 3404 3258 3334 3412 3266 3342 3420 3273 3350 3428 3281 3357 3436 32S9 3365 3443 32% 3373 3451 3304 3381 3459 2 2 2 2 2 2 3 4 5 3 5 3 5 567 567 667 54 55 56 3467 3548 3631 3475 3556 3639 34S3 3565 3648 3491 3573 3656 3499 3581 3664 3508 35S9 3673 3516 3597 3 81 3524 3606 3690 3532 3614 3698 3540 3622 3707 2 2 2 2 2 3 3 5 3 5 3 5 667 677 678 57 58 59 3715 SS02 3S90 3724 3811 3899 3733 3819 3908 3741 3 28 39J7 3750 3837 3926 3758 3846 3936 3767 3855 3945 3776 3864 3954 3784 3.S73 3963 3793 3882 3972 123 123 123 345 446 455 678 678 678 60 3981 3990 3999 4009 4018 4027 4036 4046 4055 4064 123 456 678 61 62 63 4074 4169 4266 4083 4178 4276 4093 4188 4285 4102 4198 4295 4111 4207 4305 4121 4217 4315 4130 4227 4325 4140 4236 4335 4150 4246 4345 4159 42C6 4355 123 123 123 456 456 456 789 789 789 64 65 66 4365 4467 4571 4375 4477 4581 4385 4487 4592 4395 4498 4603 4406 4508 4613 4416 4519 4624 4426 4529 4634 4436 4539 4645 4446 4550 4656 4457 4560 4567 123 123 123 450 456 456 789 789 7 9 10 67 68 69 4677 47S6 4898 46S8 4797 4909 4G99 4OS 49iO 4710 4819 4932 4721 4S31 4943 4732 4842 4955 4742 4853 4966 4753 4864 4977 4764 4S75 4989 4775 4887 5000 123 123 123 457 467 567 8 9 10 8 9 10 8 9 10 70 5012 5023 5G35 5047 50:^ 5070 5082 5093 5105 5117 1 2 4 567 8 9 11 71 "72 73 5129 5248 5370 5140 5260 5383 5152 5272 5395 5164 5284 5408 517J 5297 5420 51S8 5309 5433 5200 6321 5445 5212 63.-J3 5458 5224 5346 5470 5236 5358 5483 1 2 1 2 1 3 567 567 368 8 10 11 9 10 11 9 10 11 74 75 76 5495 5623 5754 5508 5636 5768 5521 5049 5781 5534 5662 5794 5546 5675 5808 5559 56S9 5821 5572 5702 5S34 5585 5715 5S48 5598 5728 5861 5610 5741 5875 1 3 1 3 134 568 578 578 9 10 12 9 10 12 9 11 12 77 78 79 5888 C026 6166 5902 6039 6180 916 6053 6194 5929 CC67 6209 5943 6Gol 6223 5957 6095 6237 6970 6109 6252 5984 (124 6266 5998 6138 C281 G012 6152 6295 134 134 134 578 678 679 10 11 12 10 11 13 10 11 18 80 6310 C324 6339 6353 6368 6383 6397 6412 6427 6442 134 679 10 12 13 81 82 83 6457 6607 6761 6471 6622 6776 G-JS6 6637 6792 6501 6653 6i08 6516 6G68 SS23 6531 66S3 6839 6546 6699 6S55 6561 6714 6S71 6577 6730 6887 6592 6745 6902 235 2 C 5 235 689 089 689 11 12 14 11 12 14 11 13 14 84 35 86 6918 7079 7244 6934 7096 7261 6950 7112 7278 6C66 7129 7295 692 7145 7311 6998 7161 7328 7015 7173 7345 7031 7194 7362 7047 7211 7379 7CC3 7228 7396 235 235 235 6 8 10 7 8 10 7 8 10 11 13 15 12 13 15 ,12 13 15 87 88 89 7413 7586 7762 7430 7603 7780 7447 76:>1 7798 7464 7638 7816 7482 7656 7834 7499 7674 7852 7516 7091 7S70 7534 7709 78t9 7551 7727 7907 75C8 7745 7925 235 2 4 5 245 7 9 10 7 9 11 7 9 11 12 14 16 121418- 13 14 16 90 7943 7962 7980 7998 8017 8035 8054 8072 8891 8110 246 7 9 11 13 15 17 91 92 93 8128 8318 8511 8147 8337 8531 8166 8356 8551 8185 8570 8204 8395 8590 8222 8414 8610 8241 8433 8630 8260 8453 8650 8279 8472 8670 8299 8492 8690 240 240 240 8 9 11 8 10 12 8 10 12 [13 15 17 14 15 17 J14 16 18 94 95 96 8710 8913 9120 8730 8933 9141 8750 8954 9162 8770 8974 9183 8790 8995 9204 8810 9016 9226 8S31 9036 9247 8851 9057 9268 8872 9078 9290 8892 9099 9311 |2 4 6 240 240 8 10 12 8 10 12 8 11 13 14 If 18 15 17 19 15 17 19 97 98 99 9333 9550 9772 9354 9572 9795 9376 9594 9817 9397 9616 9S40 9419 9638 98C3 9441 9661 9886 9462 9683 9908 9484 9705 9931 9506 9727 C954 9528 9750 9977 247 247 257 9 11 13 9 11 13 9 11 14 15 17 20 16 18 20 16 18 20 TABLE OF FUNCTIONS OF IHOLM. 419 1 Angle. ; Chord. fine. Tangent. Co- tangent. Cosine. Degrees Radians. 000 C ao 1 1-414 1-5703 90 1 0175 017 l '0175 0175 57-2900 9998 1-402 1-5533 89 2 034D 035 0349 0349 28-6363 9994 1-3S9 1-5359 83 3 0524 052 0523 0524 19-0311 99S6 1-377 1-51S4 87 4 0698 070 0698 0699 14-3007 9976 1-364 1-5010 86 5 OS73 087 0872 0875 11-4301 9962 1-351 1-4S35 85 6 1047 105 1045 1051 9-5144 9945 1-338 1-4661 84 7 1222 122 1219 1228 8-1443 9925 1325 1-4486 83 8 1396 140 1392 1405 7-1154 9903 1-312 1-4312 82 9 1571 157 1564 1584 6-3133 9S77 1-299 1-4137 81 10 1745 174 1736 1763 5-6713 9848 1-2S6 i-rso 80 11 1929 192 1903 1944 5-1446 9316 1-272 l-37t3 79 12 2094 209 2079 2126 4-7046 97S1 1-259 1-3614 73 13 2269 228 2250 2309 4-3315 9744 1-245 1-3439 i H 2443 244 2419 2493 4-0108 9703 1-231 1-3265 Z ' w a 2613 261 2588 8679 87821 9659 1-213 13090 5 16 2793 278 2756 2S67 3-4S74 9613 1-204 1-2915 4 17 2967 296 2924 3057 32709 9563 1-190 1-2741 3 { 13 3142 313 3090 3249 3-0777 9511 T176 1-2566 2 ' 19 3316 330 3256 3443 2-9042 9455 1-161 1-2302 1 20 3491 347 3420 3640 2-7475 9397 1-147 1-2217 21 3665 364 3584 3S39 2*6051 9336 1-133 "2043 69 22 3840 382 3746 4040 2-4751 9272 1-118 1S63 6* 23 4014 399 3907 4245 2-3559 9205 1-104 1694 67 24 4189 416 4067 4452 2-2460 9135 1-089 1519 66 25 4363 433 4226 4663 2-1445 9063 1-075 1345 65 20 4538 450 4384 4877 2-0503 89S8 1-060 1-1170 64 "7 4712 467 4540 5095 1-9626 8910 1-045 1-0996 63 23 4387 484 4695 5317 1-SS07 8829 1-030 1-0821 62 29 5061 501 4848 5543 1-SC40 8746 1-015 1-0647 61 30 5236 518 5000 5774 1-7321 8660 1-000 1-0472 60 31 5411 534 5150 6009 1-6643 8572 985 1-0297 59 32 5585 551 5299 6249 1-6003 84SO 970 1-0123 58 33 5760 568 5446 6494 1-5399 8387 954 9943 57 34 5934 585 5592 6745 1-4826 8290 939 9774 56 35 6109 601 5736 7002 1-4281 8192 923 9599 55 36 G283 618 5S78 7265 1-3764 8090 908 9425 54 37 6458 635 6018 7536 1-3270 79S6 892 9250 53 33 6632 651 6157 7813 1-2793 7880 877 9076 52 39 6S07 668 6293 8098 1-2349 7771 861 8901 51 40 C9S1 684 6428 8391 1-1918 7660 845 8727 50 41 71. '6 700 6561 8693 1-1504 7547 829 8552 49 42 7330 717 6691 9004 1-1106 7431 813 8373 43 43 7505 733 6820 9325 1-0724 7314 797 8203 47 44 7679 749 6947 9657 1-0355 7193 781 8029 46 45 7S54 765 7071 1-0000 1-0000 7071 765 7854 45 Radians. Degrees. Cosine. Co- tangent. Tangent. Sine. Chord. Angle. INDEX ACCELERATION, definition, 260, 421 ,, due to gravity, 260 ,, unit of, 260 Accumulated work, 280 ,, work in a fly-wheel, 281 ,, work in a rotating body, 281 Accumulator, hydraulic, 253-2^5 Action and reaction, 3 ,, of the common suction- pump, 227-229 Activity, definition of, 12, 412 Actual or working advantage, 67 Admission of Students to the Insti- tution of Civil Engineers, rules and syllabus of Exam?, for, 370 Air vessel, action of an, 232 ,, force pump with, 231 Ampere, 413 Angle of repose, io5 ; resistance, 107 Angular and linear motion, 272 ,, velocity, 260 Anti-friction wheels, 109 Anti-logarithms, Appendix D, 417 Applied Mechanics, definition of, I Archimedes, law of, 217-219 Atmospheric pressure, 220 Atwood's machine, 262-269 Axle, wheel and, 55-57 ,, compound, 72-74 BA.CKLA.SS, in whael and screw- gearings, 158 B ick motion gear in a lathe, 1 74-1 8 5 Balance, Roman, 35 ,, bent lever, 43 Balancing fast spsed machinery,270 Biles, screw for pissing, 165 Ball bearings, no Birometer, the mercurial, 221 Bear, hydraulic, or portable punch- ing machine, 251 Bsaringa, ball, 1 10 ,, roller, 183 B3d or shears of a lathe, 183, 187. B3ll crank lever, 43 Belt-gaaring reversing motions, 122- 124 shape of pulley faces for, 127 ,, ,, velocity ratio of pul- leys in, 119-121 Baits, brake horse-power trans- mitted by, 118 ,, difference of tension in, 116 ,, open and crossed, 121 ,, tendency of, to run on high- est parts of pulleys, 127 Bench vice, screw, 166-167 B^nt lavar balance, 43 ,, levers, 42 ,, ,, duplex, 43 Bavel wheel and clutch reversing gsar, 345 Bioyole and railway curves, motion on, 287 Block and tackle, 65-68 ,, snatch, 65 Block, Weston's differential pullev. 74-77 BDird of Education Exam., instruc- tions for, 368 Baird of Trade unit, 413 Bjiies, formulae for falling, 261 pith of projected badies which fall under the action of gravity, 273 ,, proof by Atwood's michine of formulae for falling, 262-269 BDiler plates, large hydraulic press for flanging, 245-247 . single riveted lap joints, 310 B)lt, holding down, 156 B>n-accord centrifugal pump, 236 421 422 INDEX Bottle screw- jack, 162-164 Brake^ horse-power transmitted by belts, 118 Bramah's hydraulic press, 241-245 leather collar packing, 243 Bucket pump, combined plunger r and, 234 Buttress screw thread, 155 c CALCULATIONS, note regarding en- gineering, 5 Cams, 330-334 Capstan, ships, 57-59 Carpenter's turkus, 44 Centimetre, 411 Centre of gravity, 28 ,, of parallel forces, 26 ,, of pressure, 217 Centrifugal force, 275 ,, force machine, 277 pumps, 236-238 ^ ,, stress in fly-wheels, 279 Centripetal force, 276 Chains, stresses in, 316 Change wheels in a lathe, 176-179, 183, 184, 187 ,, ,, reversing plate or quadrant, 183, 187 Chinese windlass, 72 Circle, definition of pitch, 131 Circular discs, velocity ratio of, 130 City and Guilds exam., instruction ^ for, 369 Civil Engineers, Students' Exam., rules and syllabus of, 370 Clarke's patent adjustable curve, 16, 17 Click, reversible, 335 Co-efficient of friction, 105, in Cohesion of matter, 224, 298 Combined lever, screw, and pulley gear, 160 ,, plunger and bucket pump, 234 Comparison of dynamical formulae ^ for linear and angular mo- tion, 2721 Components and resultant of forces, 4 , 9 of a force at right angles to each other, 84 Composition and resolution of forces, 82, 84 ; velocities, 260 Compound, Weems', screw and hydraulic jack, 249-251 Compound wheel and axle, 72-74 Compressibility of matter, 298 Compressing a bar within the elastic limit, work done in, 307 Compressive stress and strain, 301 Cones, stepped speed, 124 Constants appendix D, 414 Coulomb, 413 Couple, definition of, 25 Coupling joint, Hooke's, 329 ,, screw, for carriages, 156 Crab, double purchase, 143, 144 single purchase, 140-142 Crania, stresses in various membciB and jib arrester, 85, 92 Crank, bell, lever, 43 Curve, focus and directrix of a, 272 Curves, motion on bicycle and rail- way, 287 Cylinder, forming a screw thread on a, 149 D DEAD load, definition of, 500 Density of matter, 298 Differential pulley blocks, Weston's, 74-77 Dilatibility of matter, 298 Directrix of a curve, 272 Discs, velocity ratio of, 130 Distinction, solids, liquids, gases, fluids, &c., 224 Double acting force pump, 234 ,, Hooke's joint, 329 ,, purchase winch or crab, 143, 144 ,, threaded screws, 157 Driving belt, difference of tension in a, 116 Ductility, definition of, 299 Duplex bent levers, 43 Dynamical formulae for linear and angular motion, comparison of, 272 E EFFICIENCY, apparatus for deter- mining, of screw gear, 160 INDEX 423 Efficiency of combined lever, screw, and pulley fear, 160 a machine, defini- tion of, 53 ,, of screws, 151 Elasticity, definition of, 300 ,, limits of, 303 ,, modulus of, 304-306 ,, safe loads and, 302 ,, table of moduli of, 305 Ellipse, 273 Elliptic wheels for quick return, 351-353 Endless screw and worm-wheel, 168 Energy, kinetic, potential, 280,412 Eqglish gauges, 360 Equilibriant of parallel forces, 25 , Jlibriurn, conditions of, In O*M of floating bodies, 218 Equilibrium, forces in, 3 ,, graphic demonstration of three forces in, 80 Erg, 6, 412 Exams. B. of E. C and G ; Inst. C.E., Appendix C, 403 Experimentally determining the energy stored up in a rotating fly-wheel, 285 Extending a bar within the elastic limit, work done in, 307 Extension of matter, 297 FACTORS of safety, 302 Falling bodies, formulae for, 261 ,, proof of formulae for, by Atwood's machine, 262-269 Farad, 413 Feed, silent, 340-342 Floating bodies, conditions of equi- librium, 218 Fluids, solids, gases, 224 Fly- press, the, 283 Fly-wheels, centrifugal stress in, 279 ,, energy stored up in, 281 Fly-wheel, to find experimentally the energy stored up in a rotating, 285 Focus of a curve, 272 Force^definition of, i, 412 .. [elements of a, 2 Force, moment of a, 21 ,, pump, single-acting, 229-231 ,, pump, double-acting, 234 ,, pump with air vessel, 231 ,, resolution of a, into two com- ponents at right angles, 84 ,, unit of, 2, 412 Forces, centrifugal, centripetal, 275 ,, graphic representation of, 3 ,, in equilibrium, 3 parallel, 25-28 ,, parallelogram of, 82 ,, resultant of, 4 ,, straight lever acted on by inclined, 42 ,, three equal, in equilibrium, 83 triangle of, 82 two, at right angles, 83 Forging and hardening luri-lng tools, 202 Friction, angle of, 106 ,, anti-, wheels, 109 ,, circular discs, velocity ratio of, 130 ,, co-efficient of, 105, in ,, definition of, 101 ,, heat developed by, 102 ,, inclined plane with, 110- 112 ,, inclined plane without, 93 .,, laws of, 103-109 ,, cone reversing gear, 344 Fulcrum, position of, in a lever, 30 ,, pressure on, and reaction from, 25 Fuscc, the, 59-61 Fusibility, definition of, 300 G GALILEO'S and Kater's pendulum experiments, 269 Gas gauge, 222 ; gases, 224 Gauge limits, micrometer, &c., 357 ,, Stubs' wire, 358 Gauges, low pressure, &c., 221 Gear, back motion in lathe, 174- 192 ,, change wheels in lathe, 176- 195 ,, efficiency of combined levr, screw and pulley, 160 ,, screw and pulley, 160 4 2 4 INDEX Gear, starting and stopping, 124 ,, worm -wheel lifting, 170 Gearing, backlash in wheel and screw, 158 ,, belt, reversing motions, 122-124 ,, belt, shape of pulley faces ,, for, 127 ,, pitch of teeth in wheel, 132 ,, principle of work applied to wheel, 135 ,, velocity ratio of pulleys in belt, 119-121 ,, velocity ratio in wheel, 123 ,, wheel, in jib-cranes, 144- 146 Grain, screw for moving, 148 Gramme, 41 1 Graphic representation of forces, 3 ,, ,, ,, velocities, 260 Gravity, acceleration due to, 260 ,, centre of, 28 ,, specific, definition of, 214 Gyration, radius of, 282 iH HARDENING the tools for a lathe, 202 Head or pressure of a liquid, 205 Headstock, fast or fixed, of a lathe, 184, 185 ,, movable, for a common lathe, 179, 180 ,, movable, for a screw- cutting lathe, 181 Heart-shaped cam, 3^1 Heat developed by friction, 102 ,, relation between, and work, 104 Helix of a screw thread, 148 Henry, 413 Herbert's hexagon turret lathe, 188- 195 Hollow round shafts, strength of, 324 Homogeneous materials, 299 Hooke's coupling joint, 329 ,, double ,, 329 law, 303 Horse-power brake, transmitted by belts, 118 definition of, 12, 412 . >, of working agent, 13 Huyghen's pendulum experiments, 269 Hydraulic accumulator, 253-255 bear, 251 jacks, 247-251 machines, 227-255 press, Bramah's, 241-245 press, large, 245-247 Hydraulics, 207-258 Hyperbola, 273 IDLE wheel (note on), 138 Immersion of solids in fluids, 217 Impenetrability of matter, 297 Improved Standard Measuring Ma- chine, 363 Inclined forces, straight lever acted on by, 42 ,, plane, the screw as an, 149 ,, planes, principle of work, 93-99, 270-272 applied to, 97, no, 112 Indicator, motion for Richard's, 344 Inertia definition, 290 Intermittent motion for cam, 332 Internal and external limit gauges, 355-357 Isotropic material, 299 Instructions for Board of Education Exams., 368 ,, ,, City and Guilds Et ,, of pulleys in belt gearing, 119-121 ,, of two friction cir- cular discs, 130 ,, t in wheel gearing, 133 Reaction, action and, 3 ,, from fulcrum of a lever, 2 5> 30 Relation between twisting moment, diameter, and horse power transmitted by shafting, 224 Repose, angle of, 106 Resilience, definitions, 307, 318 Resistance, electrical unit, 413 ,, limiting angle of, 107 ,, passive, or friction, 101 ,, work in overcoming a uniform, 6-9 ,, work in overcoming a variable, 8-1 1 Resolution and composition of forces, 82-84 ; of velocities, 260 Resolution of a force into two components at right angles, 84 Resultant and components, 4 ,, of parallel forces, 25-28 ,, two forces at angle and any number at a point, 84 ,, pressure and thrust, 5 Reversible pawl, 335 ,, pendulum, 269 428 INDEX Reversing by friction cones and bevel wheels, 344 ,, gear, bevel wheel and clutch, 345 friction cone, 344 ,, quick return, 346 ,, Whitworth, 345 motions, 343 for belt gearing, 122-124 plate for change wheels, 183-187 Rigidity, definition of, 298 ,, modulus of, 319, 321 Rivets, pitch, 310 ,, shearing stress of, 310 Rods, tension in pump, 187 ,, torsion of, 318 Roller or ball bearings, no, 183 Roman balance, 35 Roof truss, stresses in a, 88, 89 Rotating body, accumulated work in a, 281 Rotor, 50 Rounded screw threads, 155 Rules and syllabus of Exams, -for admission of Students to the Institution of Civil Engineers, 370 SADDLE and slide rest of a lathe, 176, 185 Safe loads and elasticity, 302 Safety, factors of, for materials, 302 ,, * valve, 38-40 Sawing machine, vertical, 337 Scalars, 50 Screw bench vice, 166, 167 ,, combined with lever and pul- ley, 1 60 ,, -coupling for railway car- riages, 156 ,, -cutting lathe, description of a, 180-188 ,, -cutting lathe,self-acting, 182 ,, -cutting mechanism in a lathe, 174-188 ,, endless and worm-wheel, 168 ,, -gauges, micrometer, 357 ,, gear, apparatus for demon- strating efficiency of, 160 ,, gearing, backlash in, 158 ,, -jack, bottle, 162-164 Screw-jack, compound hydraulic and, 249 ,, -jack, traversing, 164 ,, leading, of a lathe, 175, 186; split nut for engaging, 185 ,, -press for compressing bales, 165 ,, or spiral, for grain, 148 ,, pressure or thrust, 5 ,, viewed as an inclined plane, 149 Screws, right and left-hand, 156 ,, single-, double-, and treble- threaded, 157 ,, strength, durability, and efficiency of, 151 Screw thread, 148 ,, buttress, 155 ., ,, rounded, 155 ,, forming a, on a cylin- der, 149 ,, square, 154 Sere \v threads, characteristics of, 151 ,, different forms of, 151-156 ,, Seller's, 154 ,, ,, Whitworth, 151-154 Second of Time, 411 Shafts, strength of hollow, 324 ,, strength of solid round, 322 Shaping machine, 350 Shearing strength of rivets, 310 Shears or bed of a lathe, 183, 186 Ship's capstan, 57-59 Silent feed, 336 Single-riveted lap joints, 310 Siphon, the, 222 Slide rest of a lathe, 176, 185 Sliding angle, 107 Slotting machine, vertical, 341 ,, ,, Whitworth 's, 349 Sluice gate, pressure on, 217 Snatch block, 65 Solid shafts, strength of, 322 Solids, immersion, 217 ; defn., 224 Specific gravity, 214, 298 Speed cones, 124 Split nut for engaging leading screw o a lathe, 185 Squared paper, 14-17 Standard measuring machines, 355- 366 Starting and stopping gear, 124 Starrett micrometer gauge, 357 Steel, high speed, 195 INDEX 429 Steel, specific gravity test, 203 Steelyard, 35-38 Stepped speed cones, 124 Straight levers acted on by inclined forces, 42 Strain, compressive stress and, 301 definition of, 300 ,, shearing stress and, 317 ,, tensile stress and, 301 Strength of materials, machine for testing, 40-42 of materials, ultimate, 301 of solid and hollow round shafts, 322-324 Stress, centrifugal, on fly- wheels, 2 79 ,, definition of, 300 ,, intensity of, 300 ,, limiting,or ultimate strength, 301 ,, shearing and strain, 317 ,, total, 300 Stresses in chains, 316 ,, in jib cranes, 85-89 in simple roof truss, 88, 89 , , tensile and compressive,3O I Stubs' wire gauge, 358 Students' Exams., Institution of Civil Engineers, 389-401, 407 Suction pump, common, 227-229 Sun and planet wheels, 330 Surface, definition of pitch, 131 immersed in liquid, 211 Swing radius, 282 TABLE of melting points of metals, 300 , , of moduli of elasticity, 305 ., of power that steel shafting will transmit at various speeds, 323 ,, of ultimate strengths of materials, 302 Tackle, block and, 65-68 Tangentometer, 364 Tearing strength of plates, 310 Teeth, pitch of, in wheel gearing, 132 Tenacity, definition of, 298 Tensile strength of materials, machine for testing, 40-42 ,, stress and strain, 501 Tension in driving[belts, 116 Tension in pump rods, 229 Test of steel, by specific gravity method, 203 ,, specimen, work done per cubic inch in fracturing a, 309 Testing machine, 40-42 Theoretical advantage, 66-68 Toggle joint, 46-49 Tongs, Lumberer's, 43 Torsion of rods and wires, 318-320 Torque or twisting moment, defini- tion of, 322 Transmission of power by belting, 118; by liquids, 208 Traversing screw-jack, 164 Treadle lathe, self-acting screw- cutting, 182 Triangle of forces, 82 Turning tools, forging and harden- ing, 202 ,, tool holders, 194 ,, with high speed steel, 195 Turkus, carpenter's, 44 Turret-lathe, 188-195 Twisting moment, 322 u ULTIMATE strength of materials 301 Uniform velocity, definition of, 259 Unit of acceleration, 260 force, 2, 412 ,, horse-power, 12, 412 ,, power, 12, 412 ,, velocity, 259, 411 ,, work, 6, 412 Units, C.G.S.,4ii, Practical, 413 Universal joint, Hooke's, 329 Useful constants Appendix D, 414 ,, data regarding fresh and salt water, 214 ,, work in a machine, 52 Uses of squared paper, 14-17 VACUUM water gauges, 221 ^ Variable resistance, work done against a, 8 ,, velocity, definition of ,24 59 Valve, lever safety, 38-40 V-screw-thread, Seller's, 154 ,, ,, "Whitworth's, 151 430 Vector's, $a Velocities, composition and resolu- tion of, 260 ,, graphic representation of, 260 Velocity, and motion, 259 ,, angular, 260 ,, attained by a body sliding down any smooth in- clined plane, 270 ,, definition of, 260,411 ,, linear, with uniform ac- celeration, 260 ,, linear, 259 ,, uniform and variable, 259 ,, unit of, 259, 411 ,, ratio of change wheels in a lathe, 176-179 ,, definition of, 67 , , pulleys in belt-gear- ing, 119-121 two friction circular discs, 130 ,, ,, in wheel-gearing, Vessel, action of an air, 232 ,, ,, force pump with, Viscous fluid, 224 Volt, 41 3 w WATER gauges, low pressure and vacuum, 221 ,, useful data regarding fresh and salt, 214 Watt, 413 Watt's parallel motion, 338 Weems' compound screw and hy- draulic jack, 249-251 Weston's differential pulley block, 74-77 Wheel and axle, 55-57 ,, ,, compound axle, 72-74 Wheel gearing, backlash in, 158 ,, ,, in jib cranes, 1 44- 1 46 ,, ,, pitch of teeth in, 132 ,, ,, principle of work applied to. INDEX Wheel gsiring, velocity ratio in, 133 Wheels, anti-friction, 109 ,, change, in a lathe, 176-179, 183, 184, 187 ; quadrant or reversing plate for, 183, 187 ,, fly-, centrifugal stress in, 279 ,, pawl and ratchet, 334 Wheels, sun and planet, 330 Whitworth's millionth measuring machine, 358 quick return motion, 347 ,, reversing gear, 345 ,, slotting machine, 349 ,, V-screw-threads, 151- 154 Winch barrel, 57 ,, double purchase, 143-144 ,, single purchase, 140-142 Wincli drum combined with pulley, worm, and worm wheel, 168 Windlass, Chinese, 72 Wire gauge, Stubs', 358 ,, -testing machine, Lord Kel- vin's, 210 Wires, torsion of, 319-321 Work, accumulated, 280-282 ,, definition of, 6, 4 2 ,, diagrams of, 9 ,, done against variable resist- ances, 8, 17 ,, done in extending or com- pressing a bar within the elastic limit, 307 ,, done on inclines, 110-112 ,, done per cubic inch in frac- turing a test-bar, 309 ,, principle of, 52 ,, relation between, and heat, 102 ,, transmitted by belts, 118 ,, unit of, 6, 412 Workshop measuring machine, 363 Worm-wheel, screw and worm, 168 ,, lifting ffear, 170 Worssam's silent feed, 338 Co. LIMITED TavistoclTSTrEef, Covent Garden , Loudon THIS BOOK IS DUE ON THE LAST DATE STAMPED'BELOW AN INITIAL PINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN rms BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO 5O CENTS ON THE FOURTH DAY AND TO $1.OO ON THE SEVENTH DAY OVERDUE. FED 4 1944 LD 21-100rn-7,'40 (6936s) VB I 1 077 204114