MANUAL OF MACHINE DESIGN CASTLE .>..^^* CORNELL UNIVERSITY LIBRARY Given to the COLLEGE OF ENGINEERING by the Macliine-design dept. TJ 230.C35''""'""'"'*">"-"'™^>' mSmm* machine design, 3 1924 004 578 625 The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924004578625 MACHINE DESIGN MACMILLAN AND CO., Lighted LONDON • BOMBAY • CALCUITA • MADRAS MELBOURNE THE MACMILLAN COMPANY NEW YORK ■ BOSTON ■ CHICAGO DALLAS • SAN FRANCISCO THE MACMILLAI^ CO. OF CANADA, Ltd. TORONTO A MANUAL OP MACHINE DESIGN BY FEANK CASTLE, MJ.Mech.E. LATE OF THE MECHANICAL LABORATORY, ROYAL COLLEQE OF SCIENCE, SOUTH KENSINGTON LECTURER IN PRACTICAL MATHEMATICS, MACHINE CONSTRUCTION AND DHAWING, BUILDING CONSTRUCTION AND ENGINEERING SCIENCE, AT THE MUNICIPAL TECHNICAL INSTITUTE, EASTBOURNE MACMILLAN AND CO., LIMITED ST. MAETIN'S STEEET. LONDON 1919 PREFACE. In a recent Memorandum on the Teaching of Engineering in Evening Technical Schools, the Board of Education, in the sections dealing with Engineering Drawing, directs the attention of teachers and students to the great importance of " Machine Design." During the student's First Year's Course, the memorandum points out, the weights of objects of simple design should be estimated ; also, it is useful, wherever possible, to introduce simple numerical exercises. Simple designs, for example, such as a mild steel eye bolt to carry a snatch block might be set ; or the student might determine the diameter of two long bolts to be used with a 20-ton hydraulic jack to pull a flywheel ofi a shaft ; or determine the diameter of the four bolts supporting the top crosshead of a hydraulic press. In the second year, students might design a cotter connection for a steel tie bar ; estimate the strength of bolts in flanges and cylinder covers ; find the stresses in cotters connecting crossheads to piston rods, and so on. If the student is also studying Practical Mathematics, the sizes and proportions of the machine details he is drawing afford excellent scope for exercises. Thus, with given data he can find the probable stresses in the various parts ; later, he can proceed to carry out the design of some important machine detaU or some machine, such as a winch, or a crane. He can also make working drawings to suitable scales when all the dimensions are obtained. Yi PREFACE ' In the following pages an attempt is made to provide exercises of this kind. As a rule only those details are referred to which admit of numerical calculation. The design of details which depend on empirical data alone are probably most easily obtained from one of the numerous " Pocket books " dealing with the subject. Fully dimensioned drawings may be found in .the author's Machine Construction and Drawing (Macmillan). The scheme of .work outlined in the following pages is that used for several years by the writer, but teachers and others may take the subject in any order they deem best. The exercises are numerous and have been arranged to corre- spond with the usual draughtsman's accuracy ; the dimensions are to the nearest sixteenth of an inch, or to the accuracy obtain- able by four-figure logarithms. The exercises throughout have been checked carefully, and it is hoped that no serious errors wUl be found by those using the book. Without making the book unduly large, it is not possible to deal fully with the important subject of " Compound Stress " ; but students who desire to pursue the subject further should consult Applied Mechanics for Engineers, by J. Duncan (Macmillan). Proofs of some of the more important portions, which have a direct bearing on Machine Design, are included. In other cases, the results alone are given, together with applications of them ; necessary proofs of them may be obtained from any of the numerous books on Strength of Materials. Elementary details — ^for example, the forms and proportions of rivet heads, bolts, screws, and such data — ^are not included ; they may be obtained from the author's Machine Construction and Drawing. Permission' to insert questions from their lespectivfe Examina- tions has been granted by the Board of Education [B.E.], the Lancashire and Cheshire Union of Institutes [L.C.U.], the Union of Educational Institutions [U.E.I. ], the Institution of Civil Engineers [I.C.E.], and the University of London [Lond. B.Sc.]. PREFACE vii The author wishes to express once more, even if only very inadequately, how much he is indebted to Prof. R. A. Gregory and Mr. A. T. Simmons for their Mndly help and valuable assistance during the preparation of the book and its passage through the press. FRANK CASTLE. Hastings. CONTENTS. CHAPTER PAGE I. Mbnsubation. Stress and Stbain 1 II. Riveted Joints. Tie-Babs. Boilers and Pipes. Thick Cylinders 35 III. KNtrcKLE Joints. Suspension Links. Keys. Cot- TERED Joints - . - 65 IV. Beams and Girders. Flitch Beams. Reinforced Concrete Beams. Design of Beams and Girders 88 V. Shafts. Shaft Couplings. Coupling Bolts. Tor- sional Rigidity 129 VI. Belt and Rope Pulleys. Linear and Angular Velocity. Tension and Width of Belts. .Centri- fugal Force 160 VII. Wheels. Wheel Teeth and Arms. Journals. Flywheels 183 VIII. Compound Stress. Struts - 211 IX. Deflection of Beams and Girders 230 X. Chiefly Engine Details 246 XI. Engine Details {Continued) 273 Examination Papers 319 Tables of Logarithms ■ 330 Useful Data- 335 Answers 336 Index .....-- - 345 CHAPTER I. MENSUKATION. STBESS AND STRAIN. The following results in mensuration are of frequent occurrence, and are collected here for convenience of reference. Determination of Akeas. Rectangle. Adjacent sides 6 and h, area = 6/?; length of dia- gonal = Vb^ + h^ ; perimeter = 2(b + h). Parallelogram. Area = 6/1, where 6 denotes the lengthi of one side and h the perpendicidar distance to the opposite side. Square. Area=s^, where s is the length of one side. Triangle. Sides a, b and ; Area = g6A = 560 sin A = 'S/sis -a){s- b){s - e) ; b is the base and h the altitude of the triangle ; 2s = a + b + c. Quadrilateral. The area of a quadrilateral, or any irregular figure, having straight lines for its sides, may be obtained by dividing the figure into triangles ; the sum of the are^s of the triangles is the area required. Trapezium. When two sides of a quadrilateral are parallel lines, the figure is called a trapezium, or trapezoid. If a and 6 are the two parallel sides, and h the perpendicular distance between them, then th.e area is 5(0 + 6)/! ; or, one-half the sum of the pa/raUel sides muUiplied by the perpendicular distance between them. Area of an Irregrular figure. The area of an irregular figure may be obtained approximately by the trapezoidal or mid-ordinate methods. Thus, to find the area of the figure ABCD (Fig. 1), one side of which is a curved line AD, the base AC mav be divided 2 MACHINE DESIGN into an equal number of parts— in tHs case giving 7 ordinates Ui-.y.,. Tte mean ordinate y^ = ^^"'"^^'"^^ and the area of the figure is j/^ x (base BC). This is shown by the area of the rectangle EBCF, where EB =y^. i ;>' 'Z 1 ^ >^ r— , l^. 1^ i ^ FIO. 1. In the mid-ordinate method, intermediate ordinates, midway between those already drawn and shown by dotted lines, are used. The area is the average of the six dotted ordinates multiplied by the length of the base BC. In Fig. 1 the area of the figure ABCD is the sum of the lengths of the ordinates divided by 7, multiplied by 24 (the length of the base). /. area = ^(9-3 +10-8 + 10-2 + 8-6 + 6-8 + 5-8 + 5-8)24 =196-4 sq. units. Using the mid-ordinate method, Area = ^(1 0-5 +10-7 + 8-5 +7-7 + 6-1 +5-6)24 =196-4 sq. units. Simpson's rule. The area of a figure bounded by a curved FIG. 2. line (Fig. 2) may be obtained by Simpson's rule. Divide the base into an even number of parts, thus obtaining an odd number DETERMINATION OF AREAS AND VOLUMES 3 of ordbates. Then, if s denotes the common distance between the ordinates, s Area = — (A + 4B + 2C), where A is the sum of the end ordinates iyi+ !/■,). B is the sum of the even ordinates y^+y^i- y^. C „ „ odd „ or ^3 + ^5. li the ordinates denote meas, then the volume instead of the area will be obtained. Circle. Circumference = 2w =.7t(/ ; area = 7t^ = -of^, where r is the radius and d the diameter. ^ Ellipse. Area = 7ro6, where a and 6 denote the lengths of the semi-axes. Bectangraar prism. Surface = 2(6/1 +6Z+AZ), where 6, h and Z are the three dimensions. Cylinder. Area of curved surface = izclh; where d is the diameter and h the height, or length. Sphere. Area of curved surface =4w2 = 7r(/2, where r is the radius and d the diameter. Cone. Area of curved surface =5(circumference of base) x (slant height). Determination of Volumes. Cylinder or prism. "When the ends are perpendicular to the axis, the volume is (area of one end) x height or length. Cone. Volume = ^jr^/j, where r is the radius of the base and h the height, or altitude. Sphere. Volume =:rni^ = -d^, where *- is the radius and d the diameter. Pyramid. Volume = J(area of base) x altitude. Irregular solids. In some cases where there is no abrupt change in section, the volume may be found by Simpson's rule, the ordi- nates denoting areas. Or we may use the following rule ' in similar cases : Prismoidal formula, a most important case of Simpson's rule, when only three ordinates are given, i.e. the two end ordinates (or areas), and the ordinate midway between them. If the three ordinates are y^, y^, and y^ , Average section = ^(i/j + 1/3 + ^y^. Volume s= ayerage section x length. MACHINE DESIGN Table I. Material. Weight of 1 cub. in. lb. Material. Weight of 1 cub. in. lb. Cast iron 026 Lead 0-41 Wrought iron 0-28 Tin 0-267 Steel - 0-284 Aluminium 0-097 Brass 0-29 Antimony 0-24 Copper 032 Manganese 0289 Zinc - ' 0-26 Water 0036 Expansion by Mat. The coefficient of linear expansion of a material is the expansion per unit length per degree rise in tempera- ture, and is given below for the Fahrenheit scale. Material. Coefficient of expansion. Material. CoefBcient of expansion. Caat iron Wrought iron Steel - Brass 6-2x10-6 6-9x10-6 7-0x10-6 10-4x10-6 Copper Lead Tin Zinc 10-4x10-6 160x10-6 12-0x10-6 16-2x10-6 EXERCISES I. 1. The diameter of a piston is 1 4| in. ; find its area. 2. The area of a cylinder is 626-8 sq. in. ; find its diameter. 3. If the circumference of a shaft is 96-6 in., what is its dia- meter ? 4. The pacMng ring of a piston for a cylinder 9 in. diameter is 9| in. diameter before it is cut ; how much must be cut out of the circumference so that it will just fit the cylinder when sprung into position 1 5. A packing ring for a piston has a piece equal to g in. cut out of its circumference. When sprung into position in a cylinder 1 3 in. diameter, it is found to be ^ in. open ; find the diameter before it was cut. EXERCISES I 5 6. Find the heating surface of a cyliudrical furnace tube 3 ft. diameter, 29 ft. 1 1 1 in. long. 7. If phosphor bronze consists of 90 per cent, copper, 9-5 per cent, zinc and 0'5 per cent, phosphorus, find the weight of each of these materials for a casting weighing 5 cwt. 8. A pump bucket is 5l in. diameter ; if the height of a column of water above it is 28 ft., find the total force on the bucket due to the water pressure. 9. A uniform square bar of wrought iron, side of square 1 ^ in., is 4 ft. 2 in. long ; find its weight. Also find its weight per foot length. 10. Taking the datum required from Table I., show that the weights per foot run of round bar iron of the following diameters are as follows : f in. 1 lb., | in. 2 lb., 1^ in. 4 lb.. If in. 8 lb. Plot these values on squared paper and fiiid the weights of round bars f in., 1 in. and 1 g in. 11. The cross-section of a uniform cast-iron bar 5 ft. 6 in. long is a rectangle 2^ in. by | in. Find (i) its volume and its weight, (ii) its weight per foot run, (iii) side of square bar of same weight per foot length. 12. The internal dimensions of an open water tank are 5 ft. 6 in. by 3 ft. 9 in., depth 2 ft. 3 in. Find the weight of water the tank wiU hold when full. What is the total weight of the tank if it is made of sTieet metal ^ in. thick, allowing 1 6 per cent, for overlapping at the joints ? [1 cub. in. of metal weighs 0-28 lb.] 13. A wrought-iron shaft 2g in. diameter is 20 ft. long ; find its weight. 14. A cast-iron muff or box coupling is 7 in. external diameter, internal diameter 2^ in.,' length of coupling 9 in. ; find its weight. What length of a wrought-iron shaft 2^ in. diameter would have the same weight ? 15. The external and internal diameters of a cast-iron box or mufi coupling are 9 in! and 4 in. respectively. If the length of the coupling is 1 2 in., find its weight. 16. The length of one edge of a cast-iron cube is 1 6^ in. ; find its volume and its weight. 6 MACHINE DESIGN 17. An elliptical cast-iron gjand lias tte dimensions given in Fig. 3 ; find its weight. . -p e > Fia. 3. ■ 18. A brass liner for the gland in Question 17 has the form and dimensions shown in Mg. 4 ; find its weight. Fio. 4. EXERCISES I 19. In a steel joist 1 2 in. depth, the flanges are 6 in. by § in., web 1 in. Find the area of the cross-section and weight of the joist per foot length. 20. The dimensions of a cast-iron flanged coupling are given in Mg. 5 ; find its weight. T 1 >• t 1 1 1 1 ■+ - 1 * <- — -4?i"— 1 . ~<* 1 1 1 1 FIQ. 5. 21. The external diameter of a hollow steel shaft is 21 in., internal diameter 10 in., length 5 yards ; find its weight. Find the diameter of a solid steel shaft of the same length and weight. < - _ m" » <-2--> 4 . ■> <■-/-»• V ""*■ / \:^ / <--3' 1 1 =* t * • 1 1 1 lo 1 1 1 + Q O k— >l 1 iya"> Fig. 6. 22. The dimensions of a cast-iron bracket are given in Pig. 6. There are two bolt holes in the base, diameter 1g in. Find the weight of the bracket. 8 MACHINE DESIGN 23. A boiler contains 500 smoke tubes, each 2^ in. internal diameter ; find the total area for draught through them. What is the total weight of the tubes if the mean thickness of metal is ^ in., length 7 ft. [1 cub. in. of metal weighs "28 lb.] 24. Estimate the weight of a wrought-iron plate girder 40 ft. span, 4 ft. deep overall at centre of span. The flanges consist of three plates each | in. thick, lengths 40 ft., 34 ft. and 23 ft. respectively, thickness of web ^ in., secured to the flanges by angles 6 in. by 6 in. by J in. Composition of Fokces. Parallelogram and triangle of forces. If two forces P and Q act at a point O (Fig. 7), and if the liiles OA and OB represent the two forces in direction and ^P magnitude, then the resultant / ~yj R is obtained in direction and / ,/' magnitude by completing, on / ^y^ / the lines OA and OB as sides, / R/^^ /'' the parallelogram OBDA. Rhas / y^ / the sense from O towards D. / ^y^ / It must be noted carefuUy / ^y^ ' that P and Q must be arranged /y^ , as both pulls, or both pushes, O B Q before constructing the paral- Fio. 7. lelogram. If two forces P and Q act at a point, a force, equal in magnitude but in the reverse direction DO, will give three forces, acting at the point O, which are in equiUbrium. It will be noticed that, to obtain the resultant, it is only neces- sary to draw the triangle OBD, by drawing from B a line "parallel and equal to OA, and joining OD. Such a triangle is called a triangle of forces. The construction indicated is applicable not only to forces, but also to displacements, velocities, accelerations, etc. If three forces P,Q, E act at a point O (Fig. 8), and are in equili- brium, then, if the directions of all three forces and the magnitude of one are known, the magnitudes of the remaining two can be obtained by drawing a line aih parallel to and equal to P (assuming the magnitude of P to be known). Draw he parallel to Q and m parallel to E, thus completing the triangle ahc. The lengths he COMPOSITION OP FORCES 9 and ca, measured on the scale on which ah represents P, will give the magnitudes of Q and E. Fib. 8. Polygon of forces. If a number of forces P, Q,, R, S act at a point O (Fig. 9) the resultant can be obtained by repeated applications of the parallelogram rule. Thus, completiag the parallelogram K __-:-^L 10 MACHINE DESIGN on P and Q as sides, the resultant is given by OL. In a similar manner, combining OL and R, we obtain OM ; this resultant, combined with OS, gives a resultant OK, which is the resultant of the four given forces. A better and neater method is to use the polygon of forces. From any convenient point A (Fig. 9), draw AB parallel to, and containing as many units of length as there' are units of force in, P. From the end B, draw BC parallel and equal in magnitude to Q and in the s'ame sense as Q. Similarly CD is made equal to R, and DE equal to S. The resultant in direction and magnitude is the line joining the initial point A to the final point E. When the final point coincides with the initial point, so that the polygon is a closed figure, the given forces are in equilibrium. The polygon of forces is equally and generally true for all vector quantities— velocities, accelerations, etc. Ex. 1. A couple-close roof truss, span ^6ft., rise J span, is loaded with i ton at the ridge C (Fig. 10). Find the forces in the principal rafters AC and BC and in the tie-beam AB. Fig. 10. Draw a vertical line ab, on any convenient scale equal to ^ ton. Complete the triangle by drawing ao and 6e parallel to AC and BC respectively. Finally, as the reactions bd and da are each J ton, it is only necessary to draw cd, when the lengths of the lines so drawn will give the forces in the various members Thus ac= e6= 1252 lb. The tension in the tie-rod is given by cd and is 1 1 20 lb. COMPOSITION OF FORCES 11 Ex. 2. In a whwrf crane, the post, tie-rod and jib measure' 6, ^ 2 and 1 5 ft. respectively. Find the forces in the jib and tie-rod when a load of 7 tons is suspended by a chain passing over the pulley at the jib-head ; the lifting chain passes from the pulley to the drum (i) pa/rallel to the jib, (ii) pa/rallel to the tie-rod. Draw a triangle ABC having its sides AB = 6, BC=15 and CA=1 2 units respectively. Neglecting the friction of the pulley, the puU in the chain will be 7 tons on each side of the pulley. FIQ. 11. (i) Draw ab vertical and ao parallel to the direction of the chain (Fig. 11) each equal to 7 tons. Complete the four-sided figure abed by drawing lines bd parallel to the jib and erf parallel to the tie-rod. When these lines are measured to the scale on which ab is equal to 7 tons, the forces in the jib and tie-rod are found to be 24-5 tons and 1 3-9 tons respectively. (ii) Draw as before ab vertical and equal to 7 tons ; draw ac parallel to the tie-rod. Lines drawn as before, determine the forces in the jib and tie-rod ; these are found to be 17-5 tons and 7-1 tons respectively. Bow's notation. In determining the forces in the members of a given structure, it is as important to ascert.ain the nature of the stresses (tension or compression) as to determine their 12 MACHINE DESIGN magnitude. For this purpose, what is Iniown as Bow's notation may be used. In this method two letters are used, one on each side of a force or a member of a structure. Ex, S. The outline sTcetch of a crane is shown in Fig. 12. Find the magnitude and nature of the forces in the members of the structure when a load W is carried. The load is indicated by two letters as AB (Fig. 12), the jib by BC and the tie-rod by AC. FlQ. 12. Draw a line ab representing W in magnitude and direction, then lines be and ac intersecting in c and parallel to the jib and tie-rod respectively determine the magnitudes of the forces in the jib and tie-rod. The nature of the forces is found by reading oil the forces at the pin P in ivatch-hand direction. Thus, ab vertical and down- wards ; be, denoting the force in the jib, acts in a direction towards the pin, indicating that the jib is in compression ; ca, in a direction away from the pin, denotes that the rod CA is in tension. Ex. i. In, a direct-acting horizontal steam engine the diameter of the cylinder is 1 5 in. The steam pressure is 80 U>. per sq. in. Radius of crank 1 ft., length of Connecting rod 5 ft. Find the thrust on the guide bar when tlie crank is vertical. ' AO is the crank, AB the connecting rod (Fig. 13). If P is the COMPOSITION OF FORCES 13 total force on the piston rod, Q tte thrust on the guide, and R the reaction, or force, in the connecting rod, then P =^ X 1 52 X 80 -^ 2240 = 6-3 tons. The directions of the three forces acting at B and the magnitude of P are known. Fia. 13. Draw a triangle boa (Kg. 13), the sides being parallel to the three forces P, Q, R ; then, on the scale on which bo represents P, ca represents Q and ba represents R. 60 = ^^52-12 = ^24, 6-3x1 Q = — 7=-=1-288 tons. V24 It will be noticed that Q = -^ P approximately, where r and I are the lengths of the crank and connecting rod respectively, and this approximation is frequently used to determine the magnitude of Q,. EXERCISES II. 1. There is a triangular roof truss ABC ; AC is horizontal and 1 ft. long, angle BCA is 25° and BAG is 55° ; there is a vertical load of 5 tons at B. Find the forces in BA and BC ; also find the vertical supporting forces at A and C. [B.E.] 2. At the pin joint at one end of a roof principal, there is a rafter making an angle of 30° with the horizontal and a tie-bar 14 MACHINE DESIGN making 10° with the horizontal, the angle between the two members being 20° ; the vertical supporting force at the joint is 5 tons. Find the forces in the rafter and the tie-bar. 3. Due to steam pressure on a piston, a force P of 12200 lb. acts upon a crosshead. Find the thrust (R) in the connecting rod and also the magnitude of the force (Q) between the crosshead and slide bar, neglecting friction, when the crank has turned through an angle of 45° from the dead centre in clock- wise direction. The connecting rod is 4 times the length of the crank. 4. The outline diagram of a crane consists of a triangle ABC, the tie-rod AC =16 ft., the jib BC = 21 ft. and crane post AB = 8 ft. (Fig. 11). Find the forces in the jib and tie- rod when a load of 2 tons is lifted, the direction of the chain being, after leaving the top pulley, parallel to the jib. The diameter of the round tie-rod is 1 -3 in. The proportions of a close-link for use with the above crane is shown in Fig. 14. Calcu- late the dimensions and draw two or three links. Scale fidl size. If P is the load in tons and d the diameter of the round iron from which the chain is made, then P = 3-5rf^. 5. A machine weighing 3 tons is supported by two chains ; one of these inclined at 20° jiQ 5^4 to the horizontal is fastened to an eye-bolt in a wall ; the other is inclined at 75° to the horizontal, and is fastened to an overhead beam. Find the pulling forces in the chains ; the weights of the chains may be neglected. [B.E.] 6. A load of 2^ tons hangs from a crane by a chain 1 8 ft. long. It is required to keep the load 3 ft. from the vertical by means of a horizontal rope attached to the bottom of the chain. Find the tension in the rope. 7. A load W of 200 lb. is hung from a pin P, at which two bars AP and BP meet like the tie and jib of a crane. The angles EXERCISES II 15 WPB and WPA are 30° and 60° respectively. Find the forces in AP and BP. i 8. Tke jib, tie-bar and crane post of a crane measure 20 ft., 16 ft. and 6 ft. respectively. Pind the forces in the various members when a load of 30 cwt. is being lifted. The line of direction of the chain is parallel to the tie-bar. [B.E.] 9. In Fig. 15 the outline diagram of a crane is shown. AB = 30, AC = 10, CB = 24, AD=6, DC = DF=12. If the load Wis 5 tons. FIG. 15. determine the forces in the jib, tie-rod and crane post ; also in the stays DC and D F. 2 tons FIO. 16. 10. The open framework cantilever (Fig. 16) carries a load of 2 tons hung from C. Deternoine graphically the forces in the bars AB, BC and AC. [U.E.I.] 16 MACHINE DESIGN 11. A weight of 20 tons is slung from a crane by two chains in the manner indicated in Fig. 17. .What is the pull P in each chain? ^ [U.E.I. ] y(20 tons Fl6. 17. 12. Fig. 18 is fi dimensioned sketch of an overhanging roof frame. Find the forces in the various bars. [U.B.I.] Stress and Strain. Iioad. The external force, or combination of eictemal forces, acting upon a structure is called the load on the structure. The useful load is the load due to the purpose for which the structure is designed. There is usually in addition a load due to tho weight of the structure, or piece of the structure, to be taken into account. In many cases the latter is neglected, where it is of small amount compared to the former. STRESS AND STRAIN 17 A dead load is one which is applied slowly and remains constant. A live load is one which is changing frequently. The effect of a live load is often taken to be twice the corresponding dead load. Stress and strain. The effect of the application of a load is to produce a change of form or dimensions in the body, called strain ; the internal forces resisting this change are denoted by the term stress. The stress on a body is measured by the force per unit area. _ In the case of a uniformly distributed stress, the stress is obtained by dividing the total force by the area over which it acts. j^a^^ .'. stress = area In the case of variable stress, the stress at any given point is obtained by dividing the force acting on a small area (at or near the point) by the area. ' Thus, if a denotes a very '^ — I » small area and p the load on (') it, the stress =p/a. ^ Tensile stress. When the tendency of the load is to ^"^ -pull the portions of a body / ^fmmwmmmmm : apart, the stress produced is 9/w////j/j/wjwumm//. said to be tensile stress, as <"'^ at (I) (Fig. 19). ^^«- 19- Compressive stress. When the portions of a body are being pushed, or forced together, as at (II), the result is compressive stress. Shearing stress. Shearing occurs, as at (III), when the tendency of the load is to cause one portion of a body to slide relatively to another. A convenient notation is to denote the three states of stress by the sjnnbols ft, /<. and /« respectively. The usual units adopted are either pounds, or tons, estimated per square inch or per square foot. Strain. A bar tends to become longer when subjected to tensile stress, and shorter when the stress is compressive. Strain is measured by the ratio of the alteration of length to the original length. _ . _ alteration in length _jf original length Z ' where x denotes alteration in length and I the original length, both expressed in the same imits. 18 MACHINE DESIGN Elastic constants. Under certain conditions of loading, any elastic material is found to recover its original shape and dimen- sions when the load is removed, provided the material is not stressed beyond a certain limit called the elastic limit. The amount by which the material fails to return to its original condition is called permanent set. The materials in general use in Engineering are elastic, and tlie stress below the elastic limit is always a certain mimber of times the strain, the precise number depending upon the Mnd of material. Hence the ratio - — r- = a constant for a given material, strain ° In the case of a pulled or pushed bar, the constant is called Young's Modulus of elasticity ; it is usually denoted by the letter E. It is also known as the Stretch, or Direct Modulus. ^^- fS=^ - H^ w where I, x and / denote length, alteration of length and stress respectively. The modulus of transverse elasticity of the material, usually denoted by the letter G, is also called the shear or rigidity- modulus. The rigidity modulus refers to a body subjected to shearing ■^..,.....,...,.,.^,.^..^,,, M^^ stress and -changing its sUfe by ° (IV) C shearing strain. no. 19. If one face BC of a block of material is fixed, Fig. 19 (IV), and thcopposite face AD is subjected to shearing stress /, then a change of shape will occur, which may be represented by the parallelogram A'BCD'. If Q denote the shearing strain (in radians), then f As Q is usually very small, it is generally sufficiently accurate to write ^y^' ^ = AB- Values of G are given in Table II. Factor of safety. What is called the safe working stress is obtained by dividing the ultimate strength, i.e, the brealdng STRENGTH OF MATERIALS 19 stress, of a material by a number called the factor of safety. The value of the factor depends not only upon the material, but also on the nature of the stress. According to Unwin, factors of safety may be as in the following table (II.) : Table II. Breaking strengths and elastic constants of materials in tons per sq. in. Average values. Material. Tension. Compression. Shear. Young's Modulus E. Rigidity Modulus G. Cast iron 6 to 10 40 to 60 8 to 10 6,000 to 8,000 2,000 to 3,000 Wrought iron 20 to 26 1 8 to 24 14to18 12,500 5,500 Steel 30 — 24 1 3,500 6,000 Copper (cast) p — — 6,000 2,500 Brass - 8 — 8 . 6,000 2,000 Oak 6 4 1 640 ■ — ■ Factors of safety. Live Load. Material. Dead Load. Stress of one kind. Reversed Stresses. Cast iron ' 4 6 10 Wrought iron 3 5 8 Steel 3 5 8 Timber 7 10 N 15 Safe working stress (lbs. per sol. in.). Average values. Young's Rigidity Material. Tension. Compression. Shear. Modulus E. Modulus G. Cast iron 4,000 1 0,000 2,000 1 8,000,000 7,500,000 Wroufiht iron 1 4,000 1 4,000 1 0,000 28,000,000 1 0,500,000 Mild steel 1 5,000 1 5,000 1 0,000 30,000,000 1 2,000,000 Copper 3,500 4,000 3,000 1 5,000,000 5,500,000 Brass 3,000 — — 1 5,000,000 — Oak 1,300 900 — 1 ,500,000 — 20 MACHINE DESIGN Ex. 1. In a tensite test of a % in. round bar, a load of 5g tons 'produced a stretch of 0'00756 in. in a length of 8 in. Find the stress, strain and Young's modulus of elasticity. '„, load 5-5x2240 „^„„„,, Stress = = = 27,880 lb. per sq. in. area -k ,„ _^.- - X (0-75)2 0-00756 Strain = —^ =0-000945. 27880 Modulus of elasticity =-————- = 29-5 x 1 0*. ■' 0-000945 Ex. 2. The external diameter of a cast-iron column is 12 in., mean thickness of metal 1 in. What is the compressive stress in the metal when the load on the column is 1 00 tons ? If the initial length is 15 ft., by what amount will the column he shortened ? „^ 100x2240 „^„„i, ' - Stress = = 6,480 lb. per sq. m. ^(12^-102) If X denotes tte alteration in length, X 6480 15x12 18000000 .•. ;f = 0-0648 in. Ex. 3, A mild steel tie-har is 2 in. diameter and 5 ft. long. What axial pull can it support if the stress is not to exceed 1 0,000 lb. per sq. in. ? How much will this tie-bar elongate, if Young's modulus of elasticity of the material is 30,000,000 lb. per sq. in. ? [Lond. B.Sc] If P denotes the pull and x the elongation of the bar, P = ^ X 22 X 1 0000 = 3142p lb. T, ,1 . 1 0000 X 5 X 1 2 ^'''^^^^' ^ = ^66ooooo-=°-°2in. Ex. 4. Find the limiting extension of a copper bar 1 5 ft. long, if the stress at the elastic limit is 4000 Th. per sq. in. E =1 2 x 1 0® lb. per sq. in. Let X denote the extension ; then from (1) (p. 18) 4000x15x12 '^ 12000000 =°-°^^"- Any extension of the bar beyond 0-06 in. would produce permanent set. COMPOUND BARS 21 Ex. 5. , The cross-sectional dimensions of a concrete pier 1 8 ft. high are 3 ft. 6 in. hy 2 ft. The stress at the base due to its own weight and a given load is 500 lb. per sq. in. Find the load ; what is the amount of the shortening due to the load ? [1 cub. ft. concrete weighs 1 30 a., E = 2 x 1 0^ lb. per sq. in.] Weight of pier = 3 -5 X 2 X 1 8 X 1 30 lb. Stress due to weight = ^'3.5^^^^^^^'^° =1 6-25 lb. per sq. in. due to load =483-75 lb. per sq. in. ; 483-75 X 3-5x2x144 load = - Stress =E X strain = 2240 E^ r 217-7 tons. 483-75x18x12 ^ 2x10« = 0-052 in. Ex. 6. In a mild steel tie-bar the total pull is 43 tons. If the sectimi is an angle bar, mean thickness | in., find the dimensions. Working stress is not to exceed 6i tons per sq. in. If the initial length is 16 /if., find the extension of the ba/r. Draw full size a suitable section, [e = 1 3,500 «ws per sq. in. J Area = —— = 6-5 = 6-616 sq. in. If a denotes the length of one side (Fig. 20), Area =|a + (a - 1)| = 6-61 6. .-. a = 4-78. If X denotes the extension X 6-5 16x12 13500' ;f = 0-092 in. Compound bars. When a bar formed of two or more difEerent materials firmly fastened together is subjected to a pall, or thrust, then, as the alteration in length of the materials must be the same, and as Young's Modulus has a difEerent value' for each of the difEerent materials, the stresses will not be of the same magnitude. For two difierent materials fastened together let A, E and/denote area of cross-section. Young's Modulus and the stress respectively 22 MACHINE DESIGN for one material, and Aj, Ej and /j the corresponding quantities for tlie other material due to a pull or thrust P. Let X and I denote the alteration in length and the original length respectively ; then we obtain x=il = §'l and P=A/+Ai/i. Ex. 7. A flat steel har 3 in. by f in. is rigidly fastened to two flat bars of brass each 3 in. by g in. Find the total pull on each bar when the compound bar is subjected to a pull of 25 tons. Length of all threel}ars the same. [E for steel 12,500 tons per sq. in., E for brass 4,200 tons per sq.in.] [B.E.] If S is the load on the steel har, B the load on each brass bar, Stress on steel bar = - — = = -ir , strain = y, 3x| 9 ' I Stress ^ 4SZ , „ ^„„ ,, . __=E = -=12,500 (1) Similarly -^=4,200 , (2) 4SZ Dividing (1) by (2), 2L 12500 4BZ 4200 Zx .*. -^=^^7^ or 42S = 375B (3) 3B 42 ^ ' Also 2B + S = 25. .-. S = 25 - 2B. Substitute in (3). 42(25 -2B) = 375B. .-. B = 2-29 tons. S = 25 - 2 X 2-29 = 20-4 tons. Resilience. The work done, or energy stored in a bar when extended or compressed, is the product of the mean force and the alteration in length. If P denotes the force, which is not sup- posed to exceed the elastic limit, and x the alteration in length, Work done=|P;f (1) If /, A, I and E denote the stress, area of cross-section, length and Young's Modulus respectively, then , xE fl '=T °^ ^=E- RESILIENCE 23 Also P=/A. Hence, substituting these values in (1), Work done=^^=li?^I??5)'^?JHE? •(2) ////////7mm7','MM>M/u,T The resilience of a material is the energy that may be stored in unit volume by stressing it up to the elastic limit. The magni- tude can be obtained by substituting in (2) the elastic limit for /, the product ZA being unity. Live load. In the preceding paragraph the load P is assumed to be applied gradually, and is called a dead load. If a load W falls freely on to a structure from a height h, the total fall will be h-\-x, where x is the alteration in length. The load may consist of a disc of weight W, threaded loosely on a bolt and allowed to fall freely through a height h before striking the head of the bolt (Pig. 21). The work done on the bolt must equal that done by gravity on the falling weight. .-. W(A+;r)=lA/;r, but ;r=|. fl\ pM T ■^SFj- Fig. 21. -**«. w(*-f)=l'' The solution of this quadratic gives a/w2F+2AEWZA /=- + - ^ A - M The value of / becomes 2W/A when h=0. When a load is gradually applied to a bar the stress is W/A ; but when suddenly applied a momentary stress is produced which is equal to 2W/A. Thus the effect of a suddenly applied live load is to produce twice the stress of the corresponding dead load. Also from x=fl/E. the momentary alteration of length is twice the amount it would be if the load were gradually applied. In most cases the extension x is negligible in comparison with h ; then we obtain ^ .where V is volume of bar. 24 MACHINE DESIGN Ex. 8, A load of 560 Ih. falls freely throicgh a distance of 1 in. on to a collar at the lower end of a vertical bar 1 ft. long, 1 ^ in. diameter. Find (a) instantaneous stress, (b) instantaneous stretch of the ba/r. [E =1 3,000 tons per sq. in.] [B.E.] tit/ 1 \^ Volume of ^^^=zVo) ^ 120=119-2 cub. in., /2x119-2 560x1 = 2 X 1 3000 X 2240' -V- 3000x2240x560 ^„^^„,, =1 6,540 lb. per sq. m., 119-2 fl A 6540 X 1 20 „ „^„ . . x = i^ =. ,„^^^ — tt^ttt: = 0-0681 9 m. E 1 3000 X 2240 If the load' were gradually applied the stress would be ^ load ^„„ ^ ,T_ f = =563-5 lb. per sq. in. area ^ ^ Ex. 9. A tie-bar, made of steel angle 4 in. by 4 in. by g in. in cross-section, is 1 2 ft. 6 in. long. A sudden tensile load of 23 tons is applied to it. Find (a) the maximum instantaneous tensile stress per square inch in tons, (b) The instantaneous elongation of the tie-bar. (c) The work stored up in the stretched, bar at the period of its maximum elongation in inch-tons. {A) What permanent tensile stress and elongation does this load produce, if it continues to be applied to the bar ? [E=1 2,500 tons per square inch.] [B.E.] Area = (4 x l) + (3^ x 5) = 3| sq. in. 23 X 2 (a) Stress = — ^^- =1 2-27 tons per sq. in. X (6) Ex^ = stress =12-27. . 1 i- 12-27x150 „,„„. . . elongation = x= ' — =0-1 472 m. I \ WT T 4. J 23x2x0-1472 (c) Work stored up = = 3-386 inch-tons. {d) Stress =1 x 1 2-27 = 6-1 35 tons per sq. in. Elongation = ^ x 0-1472 = 0-0736 in. REIKFORCED CONCRETE COLUMN 25 Ex. 10. A bar 2i in. diameter stretches j^j^ in. under a steady load of 1 ton. Assuming the ba/r to be initially unstressed, determine the stress induced in it by a loeight of 400 lb. falling 3 in. before commencing to stretch the ba/r. [£ = 30,000,000 lb. per sq. in.] [Loud. B.Sc] From Z=-^, where x=—-- and /= _2212_, / 1500 ^ . 2.^, 30x106x- X2-52 4 2240x1500 =43-82 in. V = - X 2 -52 X 43 -82 and WA= 400x3. 400x3 _/2x_V 2E /= / I200x2x30x10« ^x 2-52 X 43-82 =18,290 lb. per sq. in. Ex. 11. A ferro-concrete column is ^ 2 in. square. The principal reinforcement consists of four longitudinal steel bars, each If in. diameter, placed near th,e angles of the column [Fig. 22). The total load, supported by this column is 60 tons, and Young's modulus of elas- ticity of the steel is 1 5 times that of the concrete. Determine the compressive stress per sq. in. in the steel bars and in ike concrete. Total area of rods = - x (1 1)^ x 4 = 5-94 sq. in. Area of column =1 2^ - 5 -94 =1 38-06 sq. in. Area of rods are equivalent to 1 5 times their, area in concrete. If fc denotes the compressive stress, /o{l 38-06 + (15 X 5-94)} =60 x 2,240. .'. fc = 591 -6 lb. per sq. in. Stress in the steel =15 x591 -6 = 8,874 lb. per sq. in. Fig. 22. 26 MACHINE DESIGN Loads repeatedly applied and removed. The stresses due to live and dead loads may be calculated by the methods already indicated, but in practice, loads are repeatedly applied and removed, also the stresses not only vary in magnitude but also in character. Thus, in the piston rod of a steam engine, the stresses change rapidly from tensile to compressive. The strength of a piece subjected to repeated applications of stress depends not only on the ultimate strength of the material, but also on the range of the fluctuation of stress. Gerber's equation given by Prof. Unwin may be used, namely, where /^ = maximum stress which may be applied an indefinite number of times. Z^= range of stress, /= ultimate strength of the material, n for ductile iron or steel is about 1-5, for harder qualities n = 2. The working stress which may be used is given by the formula of Launhardt and Weyraucb. -.TiT I ■ , f (. minimum load \ Working stress =:f^ 1 ±—, -. — — = — j-. , 1-5\ 2 (maximum load)/ where / denotes the working stress of the material under dead load only. The + sign is taken when the stresses are of the same kind and the - sign when they are opposite kinds. Ex. 12. A mild steel har is subjected to a dead load o/12 tons full and a live load varying from 30 tons full to 5 tons fush. In this case, Minimum load = 12-5, Maximum load = 12 + 30, Working stress=^4(l+^l|^^) = 0-72/. Hence the safe working load is slightly less than | of that for a steady dead load. The value of / may be taken as 6 tons per sq. in. for wrought iron and 9 tons per sq. in. for steel. TEMPERATURE STRESSES 27 Ex. 13. FiTid the working stress for a steel rod under stress alternately in opposite directions, one of which is four times the other. g Working stress =.j-=(1 -2X5) = 65 tons per sq. in. Messrs. Stanton and Bairstow subjected specimens to reversals of direct stress ; the results of the experiments showed that an alteration of the rate of repetition -from 60 to 800 per minute had no appreciable efiect on the results obtained. Also the limiting stress of iron and steel depends on the range of stress, and is practically independent of the value of the maximum stress. It should be noted that in the experiments by Wohler the repetition of the stresses was approximately at the rate of 60 per minute. Temperature stresses. Advantage may be taken of the elonga- tion of a bar when heated, or the contraction during the process of cooling, to exert a force of comparatively large magnitude. Thus, if a denotes the coefficient of linear expansion (p. 4), then a bar of length L heated through a range of t° F. changes its length by an amount atL. Hence, if the bar be compressed to its initial length, Strain = -r— = at. Assuming the ends of the bar to be rigidly fixed so that the bar cannot expand, then the stress / is given by f=atE, where E = Young's Modulus. Expansion Joints. In a com- paratively long line of piping carrying hot water, or steam, some suitable arrangement must be made to allow for the expan- sion and contraction of the pipes. In the case of copper pipes, one of the pipes may be made in the FIO. 23. form of a loop (Fig. 23). A common and effective form for cast- 28 MACHINE DESIGN iron and other steam pipes is shown in Fig. 24. When the surface of a cast-iron pipe is removed bj- boring or turning, the metal -I- PlO. 24. i ^ is liable to corrosion ; hence the stuffing box and gland are either made of brass, or fitted with brass linings. Ex. 14. A hwr of steel is heated thrcmgh 1 40° F. If the ends of the ba/r are fixed, find the stress induced diie to the rise in temperature. a = 0-000007, E =13,000 tons per sq. in., /=7 X 1 0~^ X 1 40 X 1 3000 =1 2-74 tons per sq. in. As this is greater than the safe stress for mild steel, it shows the importance in design to allow for expansion. Ex. 15. A bar of iron 1 in. diameter and 1 ft. long is heated to 1 00° F. above atmosfheric temperature and then firmly fixed at the ends. After cooling to atmosfheric temperature the end fastenings are found to be :j\j in. nearer than tvhen hot. Find the pull exerted by the bar. [E =1 3,000 tons per sq. in. Coefficient of linear expan- sion of iron 0-0000069.] Extension =1 20 x 1 00 x 6-9 x 1 " * = 0-0828 in. Actual extension = 0-0828 - 0-025 = 0-0578. „, . 0-0578 Strain = , . 120 „, 13000x0-0578 „ „„ , btress = r^r^ = 6-26 tons per sq. in. Total pull =■ 120 xl 2x6-26 = 4-917 tons. SHRINKING AND FORCING FITS 29 Ex, 16. A straight length of steam fifing is 50 ft. long. Find the alteration in length when heated from 32° F. to 200° F. Coefficient of linear exfansion of material ■0000062. Alteration in length = 50 x 1 2(200 - 32) x 6-2 x 1 0"* = |in. Sbiinking and forcing fits. Various parts of machine and engine details require to be firmly fixed, and the process known as Bhrinldiig on, or forcing, is used. In the case of a crank, the crank shaft is made slightly larger than the bore of the crank, the usual standard allowance being 0-001 in. per inch of diameter. Thus, a crank web for a 10 in. shaft would be bored out to 10-0-01 = 9-99 in. When the crank web is heated it can be put in position on the shaft, and on cooling the crank, the contraction binds it firmly on the shaft. As an additional precaution a key is used. The crank pin is secured in a similar manner. Another method is to force the parts together, usually by hydraulic pressure, the force in tons varying feom 7 to 1 times the diameter in inches. The force to press a pin 4 in. diameter would range from 28 to 40 tons. The method adopted when parts are being assembled, either by forcing or by shrinking, depends upon circumstances. A pin is easily forced into place by hydraulic pressure ; the advantage due to this method is that the pressure required to force the parts together is indicated by a gauge. Tests have demonstrated that a shrink fit is superior to a force fit, as the parts are held more securely together. The stresses induced in the materials due to shrinkage can be calculated ; the assumption made is that the shaft, or boss does not change in diameter. This is obviously not the case, since the shaft is elastic, and will contract under the stress during cooling, but the assumption made gives the maximum possible stress. Assuming no alteration in the diameter D of the boss, or shaft, then a length irrf before shrinking becomes ttD after. „^ . alteration in length irfD -d) D-d Stram = — — , , ° = . = --^. origmal length Trrf d If /denotes the stress, then/=E x strain. .-. /=^ (1)> or .=^^ (2) 30 ip^CHINE DESIGN Ex. 17. Find the necessary shrinking allowance for a crank fin 1 in. diameter. Also find the stress in the metal if care is taken to expand the hole in the crank only just sufficient to allow the pin to enter. E = 30,000,000 lb. per sq. in. 10-0-001 x1 = 9-99 in. .*. diameter of bore =9-99 in. '^n V 1 n® y n-ni Stress =E X strain = "q = 30,000 lb. per sq. in. If T denotes the necessary rise of temperature, then from Table I., for a rise of 1 ° F. a steel rod elongates 7x10-* in. per inch length. Hence we obtain : 7x10-6LT = 0-001L. .-. T =142-9° or 143°. Hence the temperature must be raised 1 43° F. to allow the pin to enter. RaUway wheels. The wheels for railway carriages, etc., are fitted with separate tyres made of cast steel ; these are shrunk on, and when new are about 3 in. thick. On account of unequal wear they require to be machined in the lathe from time to time. When by these processes the thickness is reduced to about Ig in., the tyres are removed and replaced by new ones. The shrinking on is effected by boring the tyre a small fraction of an inch less in diameter than the wheel. The allowance is usually about as follows : Diameter in inches 38 50 56 66 Allowance in inches 004 0053 006 0070 The tyre is heated either by a reverberatory furnace or by a ring of gas jets. During the process of cooling it contracts and grips the wheel. As a further precaution the tyre is secured to the wheel by means of about a dozen screwed set bolts, or by a recessed ring of steel. RAILWAY WHEELS 31 Ex. 18. A thin steel tiihe is to be shrunk on to a steel ba/rrel, diameter D, and the resultant tensile stress in the tube (assumed to be uniformly distributed) is to bef lb. per sq. in. Find an expression for the diameter d to which the tube must be bored before shrinking on. It is assumed that any alteration in the diameter of the ba/rrel may be neglected. (a) If D = 20 in., /=1 8,000 lb. per sq. in., find d. lb) If D =12 in., rf=l 1-98 in., find/. [E = 30 X 1 0* lb. per sq. in.] (a) Substituting the given values in (2), . , 30 X 1 0^ X 20 , „ „„ . 1 8000 + 30 X 1 0* i-L\ TK^ /ix ^ 30x108(12-11-98) ^„„.,„i, (6) From (1), /= \ = 50,070 lb. per sq. in. 11 '98 In built-up guns, etc., an initial compressive stress in the internal tube is given by wire which is wound on the external surface under great tension ; in addition, a steel tube is shrunk on the outside of the wire. This initial stress is seen to be necessary when it is stated that the pressure in the chamber of a gun at the moment of the discharge of a shell may reach 25 tons per sq. in. EXERCISES III. 1. Define the following terms : " stress," " strain," " modulus of elasticity," " resilience." Show the relationship which exists between them. [I.C.E.] 2. A wire 1 ft. long, J sq. in. sectional area, is hung vertically and stretches 0-01 5 in. when loaded with 450 lb. Find the stress, strain and modulus of elasticity. [B.E.] 3. A round copper trolley wire is 60 ft. long and g in. diameter. Find the total pull in the wire when the elongation is 0-075 in. [E=1 5,000,000 lb. per sq. in.] , [B.E.] 4. In a tensile tesf. of a f in. round bar, a load of 3 tons produced a stretch of 0-00406 in. on a length of 8 in. Find the stress and the value of E. [B.E.] 5. A round wrought-iron tie-bar, f in. diameter and 23 in. long, elongates 0-01 5 in. under a load. Find the load and the stress in the bar, E = 28 x 1 0* lb. per sq. in, [B.E. ] 32 MACHINE DESIGN 6. Steel fcioycle spokes, area of cross-section 0-00332 sq. in. (16 W.G.), length 12 in., are tightened until the spokes stretch rJ^in. Find the pull on each spoke. E = 36 x 1 0* lb. per sq. in. 7. A steel bar, cross-section 2 sq. in., length 8 ft. 4 in., hangs in a vertical position. A load of 1000 lb. falls through a distance of 1 in. and stretches the bar. Find the maximum stress and elongation. E = 30 x 1 0® lb. per sq. in. 8. Water at a pressure of 800 lb. per sq. in. is admitted suddenly upon a piston 6 in. diameter. If the piston rod is 2 in. diameter and 7 ft. long, find the stress in the rod and the elongation. E = 28 X 1 0^ lb. per sq. in. 9. A load of 1 tons at the centre of a beam deflects it through a distance of ^ in. From what height may a load of 1 ton be allowed to fall to produce the same deflection ? 10. A bolt 2^ in. diameter is subjected to a pull of 30 tons. Find (a) the stress, (6) the strain, (c) the length of a part which when unloaded was 1 02 in. E = 30 x 1 0* lb. per sq. in. [B.E.] 11. A tie-bar is made of steel angle bar, mean thickness f in., length 20 ft. Find the dimensions if the safe allowable stress is 5 tons per sq. in. 12. A compressive load of 30 tons is found to shorten "by 0-063 in. a hollow cast-iron pillar 6 ft. long, 5 in. external and 4 in. internal diameter. !Find the value of E. When the load was increased to 1 92 tons the column broke ; find the compressive stress in tons. [B.E.] 13. A steel angle bar 4 in. hj 4 in. by | in. is subjected to a pull of 22-5 tons. What is the stress and the elongation produced ? E =1 2,500 tons per sq. in. [B.E.] 14. A piece of mild steel 8 in. long, f in. diameter, was tested to destruction. The breaking load was 12-2 tons and the extension 1 -96 in. What is the tensile strength and percentage elongation ? [U.E.I. ] 15. A reinforced concrete column, of rectangular cross-section, has to be designed to carry an axial load of 50 tons. The rein- forcement consists of vertical round steel rods,^one at each corner, strengthened by spiral wound wiring. The cross-sectional area of the four vertical reinforcing rods is to be 2 per cent, of the cross-sectional area of the concrete. If the compressive stress is EXERCISES III 33 not to exceed 500 lb. per sq. in., and if E,=15E(;, determine the cross-sectional dimensions of the column and the diameter of the steel bars. The column is to be designed for pure compressive stress. [B.E.] 16. A concrete pier, rectangular in cross-section, 3 ft. x 2 ft., supports one end of a girder. If the maximum compressive stress in the concrete is not to exceed 500 lb. per sq. in., what is the total compressive load which the girder can transmit to the pier ? If the pier is 1 8 ft. high, and if the modulus of elasticity of the concrete is 2,000,000 lb. per sq. in., how much would the pier shorten when the compressive load comes upon it ? 17. Explain how the tensile stress induced by a suddenly applied or falling load may be estimated, if the stress so produced lies within the elastic limit. A bar of 5 in. diameter mild steel, 30 in. long, is stressed by a weight of 50 lb., which falls freely 2 in. before commencing to stretch the bar. Find the maximum stress induced. E = 30 x 1 0* lb. per sq. in. [B.E.] 18. A railway tyre is bored out smaller than the wheel ; assum- ing the allowance for shrinking on to be 0-038 in. for a wheel 4 ft. diameter, 0-049 in. for 5 ft. and 0-058 in. for a wheel 6 ft. diameter, find in each case the probable stress in the material after shrinking on. E = 29 x 1 0* lb. per sq. in. 19. A short ferro-concrete column is a square of 1 4 in. side in cross-section. The reinforcement consists of four longitudinal round steel rods 2 in. diameter. If the load is 60 tons, find the stress (a) in the concrete, (6) in the steel. [E«=1 SE,,.] [B.E.] 20. A short strut is made out of two pieces of T steel each 6 in. by 3 in. by f in. riveted back to back. If the strut supports a load of 22-3 tons, find the compressive stress. If a load of 1 05 tons will destroy this strut, what is the factor of safety ? [B.E.] 31. The breaking strength of a certain kind of mUd steel in tension is 56,430 lb. per sq. in. A tie-bar of angle steel supports an axial puU of 37,700 lb. Find suitable dimensions if the mean thickness of the metal is g in. Factor of safety 6. 22. A tie-bar has to carry a load of 1 tons. What must be the width of the bar ^ in. thick if there is a rivet hole J in. diameter on its centre line ? Working stress 5 tons per sq. in. [U.E.I.] O.M.D. o 34 MACHINE DESIGN 23. A bar | in. diameter and 8 ft. long is rigidly fixed at one end, and is used to stop suddenly a weight W falling in the direction of the length of the bar. If W falls 2 ft. before coming to rest, find its magnitude if the stress in the bar due to the blow is not to be more than 7 tons per sq. in. E = 1 3,000 tons, per sq. in. [Lond. B.Sc] 24. Define "resilience." A weight of 20 lb. falls freely through 1 ft. and is then suddenly checked by the reaction of a bar of steel | in. diameter, 30 ft. long. Find the maximum stress and strain induced in the bar. £ = 30,000,000 lb. per sq. in. [Lond. B.Sc] 25. A tie-rod having a sectional area of 2 sq. in. is 30 ft. long ; it is extended 0-2 in. by a certain force ; find the value of this force, also the stress, strain and resilience of the bar when thus extended. £ = 30,000,000 lb. per sq. in. [I.C.E.] 26. A thin steel tube is to be shrunk on to a steel barrel, diameter D inches, resultant tensile stress in the tube (assumed to be uniform) is to be / lb. per sq. in. Find an expression for the diameter rf, to which the tube must be bored before shrinking on. The barrel is to be assumed so thick that we may neglect any alteration in its diameter produced by the shrinMng on of the tube. If D = 20 in., /=1 8,000 lb. per sq. in., find rf. £ = 30,000,000 lb. per sq. in. [B.E.] 27. A locomotive wheel is 6 ft. diameter. A steel tyre, 3 in. thick, is shrunk on the wheel. Assuming the wheel is not altered in diameter due to the pressure of the tyre, find the internal diameter of the tyre if after shrinkage the hoop stress in the tyre is 6 tons per sq. in. Coefficient of expansion 0-000007 per degree Fah. Find the least temperature to which the tyre must be heated above that of the wheel. [Lond. B.Sc] 28. A mild steel tie-bar in a bridge is subjected to a dead load of 1 6 tons puU and a live load which varies from 36 tons pull to 10 tons push. If the safe stress for a dead load is 9 tons per sq. in., find the safe working stress. CHAPTER II. EIVETED JOINTS. TIE-BARS. BOILERS AND PIPES. THICK CYLINDERS. KivETED Joints. Riveted joints. The cormectioD between two or more plates may be made ' (a) by the process of welding ; (6) by means of bolts ; (c) by riveting, using a lap or butt riveted joint. In a lap joint the edges of the plates are made to overlap, and are fastened by one or more rows of rivets. In a butt joint the plates are placed edge to edge, and cover plates, or butt straps, are fitted to one or both sides of the plates. A riveted Joint forms a permanent connection between two or more plates, and the connection can only be released by destroying the fastenings. To design a riveted joint, it is necessary to ascertain how such a joint will tend to fracture when subjected to forces. As the whole length of a joint simply consists of a number of strips of plate, each of width equal to the pitch, it is only necessary to consider the resistance of one such strip (Fig. 25). The following notation is convenient : p denotes the pitch or distance between the centres of two consecutive rivets, ft the tearing strength of the plates in lb. per sq. in.,/s the shearing strength of the rivets in lb. per sq. ia., and fc or /ft the crushing, or bearing strength in lb. per sq. in. A riveted joint may give way in several forms : by tearing of the plates, as in Kg. 25 ; by sbearlng of the rivets (Fig. 26) ; by 35 36 MACHINE DESIGN cross-breakiiig (Fig. 27) ; and by crushing: of the rivet, or forcing out a strip of plate, as in Fig. 28 (I) and (II). • r K^ K - 1 6 -P-- _/._- ., -«W •^•m\ FIG. 25. FIG. 27. FIG. 28. Resistance to tearing of a strip of plate of width p, thickness t and tensile strength ft is pt/«. Area of section at ab (Fig. 25) is (p - rf)t. Resistance to tearing = (p-af)t/j (1) Resistance to sbeariag. If P denotes the total pull, or tensile force in the plates, then the joint might fracture by shearing, as in Fig. 26. The mean intensity of shear stress is -, where A is the area of the section of the rivet. resistance to shearing = -(/^/s. .(2) RIVETED JOINTS 37 In the design of the joint for a boiler, the two resistances (1) and (2) only usually require to be considered, and the remaining three cases may be neglected (p. 36). Equating (1 ) and (2), we have for a single riveted lap joint, (p-dWt=ld^fs (3) The ratio of fs to ft may be taken equal to 1 for iron plates and iron rivets, arid equal to 0-8 for steel plates and steel rivets. These ratios correspond to the Board of Trade rules. When t, the thickness of the shell plate in inches, is knowp, the diameter d of the rivet may be obtained from d = c-\/t ; the con- stant e varies fromi -2 to 1 -4 ; the former value is adopted usually. Hence d =1 -2-^/*. From this result the value of d (to the nearest sixteenth of an inch) can be obtained, and the value of p then obtained from (3). Besistance to cross-breaJdng. The practical rule that m = d, or the distance from the centre of the rivet to the edge of the plate 1 ^, gives more than sufficient resistance against this form of fracture. The residts obtained have been confirmed by experiment. The bearing area of one rivet is taken to be dt, and therefore the maximum bearing load per rivet is rft/i,. Resistance to crashing. The joint may fail by the crushing of the rivet or plate (Fig. 28). Equating resistance to shearing and resistance to crushing, we obtain tc „ ^d^fs=dtfi (4) where /» denotes the resistance of the plate, or rivet, to crushing per square inch of projected area. Hence i: fs As /t, is about 2fg, we have rf=2-55t. From this result we find that it is not necessary to consider the crushing resistance, when d is less than 2 -55* with rivets in single shear. Also in the case of rivets in double shear (such as in butt joints with double cover straps), we may neglect the crushing resistance when d is less than 1 -27^. The rivets, especially in hand riveting, are heated before insertion in the rivet holes, and to allow for this the diameter of the rivet is usually slightly less than the diameter of the hole. 38 MACHINE DESIGN The rivets may be closed (a) by hand riveting ; (6) by pneumatic hammers ; (c) by a hydraulic riveting machine. In the case of (c), the great pressure exerted ensures that each rivet fills the hole more completely than in the processes (a) and (6). In the following calculations it is assumed that the diameter of the rivet is the same as that of the hole in which it is fitted. DiUlmg and punoMng. The rivet holes are obtained either by drilling or punching. The former has the advantage that the metal is not damaged in the neighbourhood of the hole, but it is necessary to remove the sharp edges of the hole before the rivet is inserted. Ex. 1. A steel punch f in. diameter is employed to punch a hole in a plate \ in. thick. Find the fmce necessary if the shearing strength of the metal is 60,000 lb. per sq. in. Total area of shearing is the curved strface of a cylinder | in. diameter and ^ in. long. Area = 7tx|x|. Force = tt x | x ^ x 60,000 =70,690 lb. Single riveted lap joint. In a single riveted lap joint, there is one line, or row, of rivets. In a single riveted butt joint, there is a line of rivets on each side of the joint. In a double riveted lap joint, there are two lines ; and in a treble riveted joint, three lines of rivets. ^ single riveted lap joint is shown in Pig. 29. The proportions of a snap head are given, and from these the values can be calculated when the diameter of the rivet is known, the numbers are proportional to the diameter. A graphical construction is preferable to calculation. The form and dimensions of the head may be obtained by the construction in Fig. 30. Thus, at one extremity, A, of the diameter AB, draw a line, AF, at 60° to AB; with B as centre, draw an arc of a circle touching the line AF and intersecting the centre line at C. This construction gives the height of the rivet head, and also the radius of the arc of the HjL< FIQ. 29. RIVETED JOINTS 39 circle forming tte head. The head can be drawn by marking off CD = BE, and using D as centre and taking a radius equal to BE. FIO. 30. JIG. 31. A single riveted butt joint with one or two cover plates (or butt straps) is shown in Fig. 31. The thickness of a single butt FlO. 32. riG. 33. strap is 1 1*, where * is the thickness of the plates. When two butt straps are used the thickness of each is either slightly less than, or equal to, t. A double riveted lap joint is shown at Fig. 32, and a double riveted butt joint in Fig. 33. 40 MACHINE DESIGN Ex. 2. Design a single riveted lap joint to he subjected to steam pressure, thickness of plates § in. Let d and p denote the diameter and ^itch of the rivets respec- tively. rf=1-2-v/i = 0-74in. or rf=|in. The breadth of overlap required is 3rf or 2^ in. From (3), (p-d)tft=^d^U Assuming iron plates and rivets, we have ft =/«, .-. p=— +rf=1-87in. To ensure that the joint is steam tight, the pitch is usually slightly less than that obtained, and would be 1 '75 in. If the plates and rivets are steel, and assuming ■^=0-8, we have ft P = 0^ x0-8j + rf=1-69in. =l|^in. least strength of ioint e or efficiency =-7 ,. f ,..', ^ , ' strength of solid plate where the least strength denotes the strength of the plate to tearing, or the rivets to shearing. Hence the efficiency e is the lesser of the two values 1^:^ and i-i. Pift ptft 1-83 -0-74 . . e= — -— — =59-5 per cent., |(0-74)2 '=^837^375 = ^2-7 per cent. RIVETED JOINTS 41 Ex. 3. In a single riveted, lap joint the shdl plates are \ in. thick, rivets | in. diameter, pitch 2\ in. Assuming the tensile strength of the plates to he equal to the shearing strength of the rivet, find whether the joint will fracture hy tearing or hy shearing. Resistance to tearing = (p - . per sq. in. The material is mild steel, tensile strength 30 tons per sq. in. Assume the efficiency of the longitudinal joint to be 70 per cent. Factor of safety 5. Find the tensile stress in the unperforated plate and in the plate at the joint. From pd=2fte we obtain by substitution prf _ 120 X 90 100 ~2^''2x6x2240^ 70 = 0-574 in. or f in. • Stress in unperforated plate /= - — = 9,408 lb. per sq. in. Stress in plate at joint = 9,408 xJ-^ii =13,440 lb. per sq. in. Ex. 9. A boiler 5 ft. diameter, pressure of steam 1 30 lb. per sq. in., has longitudinal double riveted butt joints with two cover platen. Efficiency of joint to be approximately 75 per cent. Design the joint so that it is equally strong for tension and shea/ring ; find thickness of plates and diameter and pitch of rivets. Tensile stress 12,000 lb. per sq. in., shear stress 10,000 lb. per sq. in. [Lond. B.Sc] ^ pd 130x60 100 „,„„ . , . * = 2fe^^4660- "" 75 =°'4^^ ^"- ^"^ ^ ^°-' rf = 1-2'\/0-433 = 0-79 or |f in. Four rivet gections correspond to a section of the plate equal to the pitch {p-d)tft = nd^f, or (P -il)A ^ 12,000 = it X (11)2 X 10,000. .*. p=4-76 in. , Ex. 10. Two lengths of a mild steel tie-rod, 7 in. by f in., are to be fastened together by a butt joint with dovhle cover plates. Find the diameter and the number of rivets required. Draw the devotion MILD STEEL TIE-ROD JOINT 47 and flan of the joint. The stresses allowed are 6 tons per sq. in. tension, 5 tons per sq. in. shea/ring and 1 tons per sq. in. bearing. In such, a joint of greatest economy, the section of the bar is not reduced by more than one rivet hole. Assuming the resistance of a rivet in double shear to be 175 times that in single shear, we have Shearing load on one rivet = bearing load on one rivet. 1-75^/, = cff/6, 7 Tzfs 16 3x10 - ^ - 7 4rex5 :1-09 =1'in. nearly, tensile strength of plate = (7 - 1)6 = 36 tons. ^fi>CT> CT>^n> 4l> "^^ no. 38. ^ -J .ftN T Shearing resistance of one rivet = - x 1^ x V75 x 5 = 6-87 tons. Projected area of one rivet = 1 x f. Bearing resistance of one rivet = | x 10 = 7-5 tons. As the shearing resistance is less than the bearing resistance, the former must be used in finding the number (n) of rivets. Equating the tensile strength of the plate to shearing resistance of rivets, /. nx 6-87 = 36 or n=5-24. Hence 6 rivets may be used, and the joint may be arranged as in Fia. 38. 48 MACHINE DESIGN The section of the plate for tearing strength will be found to be reduced by the projected area of one rivet hole. Thus, although the joint might fracture at any one of the sections AA, BB, orCC, it cannot fracture along CC without shearing the three rivets in double shear, and along BB without shearing one rivet behind it. Strength along AA is (7-1 )tft. .', efficiency=^=— =86 per cent. As this is the weakest section, the efficiency of the joint is 86 per cent. Thus Sectional area of plate at BB = (7 - 2) x 1 =5 sq. in. Tensile resistance = 5 x 6 = 30 tons. Shearing resistance of 1 rivet = 6 -87 tons. Total resistance at section BB = 30 + 6-87 = 36-87 tons. At section CC : Sectional area of plate at CC = (7 - 3) x 1 =4 sq. in. Tensile resistance = 24 tons. Shearing resistance of three rivets = 3 x 6-87 = 20 -61 tons. .'. Total resistance at section CC = 24 + 20-61 =44-6 tons. Ex. 11. Two parts of a flat tie-bar are connected hy a butt joint with double cover plates. Show that in a joint of greatest economy the section of the bar is not reduced by more than one rivet hole. Design such a joint for a tie-bar 1 2 in. wide, § in. thich. Total load 40 tons. [Lond. B.Sc] Taking the tensile, shearing and bearing stresses to be 6 tons per sq. in., 5 tons per sq. in. and 1 tons per sq. in. respectively. As in Ex. 10, we have , letfi 16 5x10 7 7r/g 7 8 X TT X 5 = 0-909=-- in. I D Tensile strength of plate = n2-— J6 = 66-4 tons. Shearing resistance of one rivet = -(—) x1 -75x5 = 6-04 tons. CAST-IRON PIPES 49 15 5 Bearing resistance of one rivet = — x-x10 = 5'86 tons. IT 7 ,. • 66-4 JNumber of rivets = ;— -- =11-3. 5-86 Hence, using 1 2 rivets, the joint may be arranged as in Fig. 39. ,'q> q> q^^ ^^><^ ^T" ^ 1 ^-4^ < .. -db — diT — i" — ^ — ^ ^t? — ^i?" -^17- ^ vj^ ' ].^ Fig. 89. In structural work the pitch of the rivets is usually some multiple of ^ in., i.e. 2^ in., 3 in., etc. Also, the centres of the rivets are not less than 1 ^rf, from the edge of the plates, where d is the diameter of the rivets. Pipes. Eq. (7) (p. 45) is only applicable to those cases of thin cylinders in which the thickness of the shell is comparatively small in relation to the diameter. For a low internal pressure the thickness obtained from (4) is too small to be of any practical value. In such cases the thickness t is obtained from other practical considerations. An empirical formula which may be used for cast-iron pipes is *=^+0-3, (11) where /= 2,000 lb. per sq. in. The thickness is taken to the nearest ^^ in. Ex. 12. If the pressure in a cast-iron pipe 4 in. bore is 50 lb. per sq. in., find the thickness of the pipe. Safe stress 2,000 lb. per sq. in. From (7) (p. 45), *=g=^ = 0-05 ^- Hence the pipe would be less than ^^ in. thick. C.M.D. D 50 MACHINE DESIGN Suph a pipe would be too fragile to be of any practical use. Moreover, such a thickness provides no margin to allow for corrosion, or for any inequalities which may occur during the process of casting.' Using the above empirical formula, we obtain, from (11), . 50x4 „, „,^ 3. t= ^ +0-3 = 0-35 or I in. 4000 ° Cylinders for compressed gas. The ordinary pressure of gas (oxygen, hydrogen or coal-gas) stored under pressure in cylinders is usually about 1 20 atmospheres, or 1 800 lb. per sq. in. The cylinders are made of lap-welded wrought-iron, or steel or seamless steel. The thickness of .such a cyliader is found by an empirical rule. That given by Unwin is where t denotes the thickness of the cylinder in inches, d the internal diameter, /the safe stress and p the safe working pressure. Values of p may be 6 tons per sq. in. for lap- welded wrought-iron. ,, ,, / ,, ,, ,, steel. „ „ 8 „ „ seamless steel. Ex. 13. The internal diameter of a cylinder for compressed gas, material seamless steel, is 5 in. Find the thickness. Assuming the test pressure to be twice the working pressure, then /) = 3,600 lb. per sq. in. =1 -6 tons per sq. in. 1-6x5 * = 2(8TT:6) = °"*2in. = Ain. Thick cylinders. * In what are known as thick cylinders, i.e. hydraulic cylinders, hydraulic pipes, etc., where the thickness is not negligible compared with the diameter, there are three principal stresses involved : (a) hoop stress q ; (&) radial stress p ; (c) longi- tudinal stress, which may be neglected in stating the relation of q and p. These stresses are perpendicular to one another, and are at 90° to the sections upon which they act. The stresses in a thick cylinder depend upon the manner in which the cylinder has been constructed. Thus, when one cylinder is shrunk on to another cylinder, the initial hoop stress in the outer cylinder will be tensile, and compression in the inner. * For a, fuller discussion, see Applied Mechanics /or Engineers. By J. Duncan (Maomillan). THICK CYLINDERS 51 The relation of q and p is obtained by considering an elementary ring (Fig. 40) having a radius /-, thickness 6r, and unit dimension parallel to the axis of the cylinder. In this figure, compressive stresses are taken to be positive, and tensile stresses negative. Also for simplicity in obtaining the results, the pressure on the outer skin of the cylinder is taken to be greater than the internal pressure ; this makes both q and p compressive stresses, and therefore positive. Thus the inner surface of the elementary ring has a radial stress p, and the outer surface has a radial stress (p +dp). The hoop stress on the ring is q. Consideration of the equilibrium of the forces acting on the ring and produced by these stresses, and also of the strains due to these stresses, leads to the equations P=b+%, (1) (2) where a and b have to be determined for the conditions given in any particular problem. The commonest case is that of a cylinder having external and internal radii R and i-^ respectively, subjected to internal pressure p^, and having no external pressure, so that pg is zero. It will be noted that in this case p is compressive (positive), and q is tensile (negative). For the inner surface, from (1) above, we have Pi- = 6+ J, Px>-x --br^ + a. .(3) 52 MACHINE DESIGN For the outer surface, we obtain P2 = t>+^'=0, or 6= -^2 (4) R2-"' "^ "- R2 Substitute in (3) the value of b from (4), giving whence a = -| s. R2 - r^ From (4), 6= Pj^'f- Substituting these values in (2) gives _PiRV___PiRV_ ^~ R2(R2-A-i2) /■2(R2-ri2i) = -^.('-5) •;: <»> "This equation gives the hoop tension at any radius r. The maximum value is obtained when r has its least value, i.e. at the inner surface, when r = r-^. Substituting this- value r=ri^. Then, from (5), PiTl ''l-'R^^V nV = -P R 10 8 SlQ- 63. 12 moments about the right-hand support (as the weight of the beam may be assumed to act at the centre of the span), Rix16 = (10x12) + (3x8) + (12x2). .•. Ri=10i cwt. R2 = 25-10i=14lcwt. 88 CANTILEVERS AND BEAMS 89 It is better to find Rg by taking moments about tbe left-hand support ; tkus : RgXl 6 = (10x4) + (3x8) +(12x1 4). .•. R2=14lcwt. The work may now be checked by taking the sum of the reactions, and ascertaining whether it equals the total sum of the loads on the beam. As one-half the weight of the beam must rest on each support, the reactions due to the twO loads only could be first obtained, and finally adding to each of these reactions 1 ^ cwt. Ex. 2. A beam having a clear bearing of 20 ft. between the supports carries a concentrated load of^ ton at 4- ft. from the left-hand support and a load of 3 tons uniformly distributed over the right- hand half. The beam weighs | ton. Find the reactions. Expressing each load in cwt., R2x20 = (20x4) + (15x10) + (60x15)=1130. 1130 20 Ri = 95 - 561 = 381 cwt. Or, as in Ex. 2, RjX 20 = (60x5) + (15x10) + (20x1 6) .•. Ri = 38lcwt. Cantilevers and beams. A joist or beam fixed at one end and free at the other is called a cantUever ((I) and (II) Fig. 64). When supported, or fixed at both ends, it is usually termed a beam. Joist or girder. Bending moment. The bending moment at any transverse section of a loaded cantUever or beam is the resultant moment of all the external forces on one side of the section. Shearing force. The shearing force at any section is the algebraic sum of all the external forces on one side of the section. Cantilever, load at the free end, When a load W is applied at the free end of a cantilever of length I (Fig. 64 (I)), the bending moment at a section at a distance x from the free end is Wx ; the maximum value is WZ. Hence M = WZ (1) 90 MACHINE DESIGN CantUever, uniformly distributed load. If a cantilever of length I has a uniformly distributed load of w per foot run, the load on a length X would be wx, and acts at a leverage, ;f/2. ;_ M = — = ~ at the fixed end, or M = ^^ where W = «/Z (2) Beam, central load. In a beam supported at each end and carrying a central load W (Fig. 64 (III)), the reaction of each support is W/2. W Bending moment at a section distant x from one end is -^ x x. The maximum value occurs when x=~. . r, WZ ,,, .. IVl = -27 W The bending moment and shearing force diagrams are shown in Fig. 64 (III), and enable the value to be read for any section. Beam, uniformly distributed load. When the uniformly distributed load on a beam is w per foot run, the reaction at each end is ivl/2. The maximum bending moment occurs at the centre of the beam, and is the resultant moment of all the forces on one side of the centre. , /'wl l\ fwl l\ wl^ \Nl ... The bending moment diagram is a parabola and can be drawn by marking oS the maximum ordinate and using the construction shown in Fig. 64. The diagrams of bending moment and shear- ing force are shown in Fig. 64 (IV). Beam, fixed ends. The two important cases when the ends of the beam are fixed are, a central load W and a uniformly distributed load. In the former the greatest bending moments occur at the centre and the ends, and both have the value u/Z/8. The bending moment and shearing force diagrams are shown at Fig. 64 (V). In the case of a uniformly distributed load, the greatest bending moment occurs at the ends, and is WZ/1 2. The diagrams are given M = BENDING MOMENTS AND SHEAKING FORCES 91 I © poonnnnncxrxxDcp Wl t y w ..iOOCYXXXOOQQOOOO mm ,J, ^ V fe-...,. A K ' 1 (V) • ^ \ b\m. /I If ^ \ S.F. Fig. 64. at (VI) (Fig. 64). It should be noted that in beams fixed at the ends, the beam bends convex upwards near the ends, and convex downwards near the centre. 92 MACHINE DESIGN Graphic couatruotion. The variation and magnitude of the bending moment and the shearing force at the various sections along the length of a simply supported beam may be shown by graphical methods as follows : A beam AB is loaded with 3 loads Wj, Wg and W3 at given distances from the points of support A and B (Fig. 65). Fia. 65. The method adopted is to obtain a polygon consisting of hinged links /gr, gh, kv, vq, which, when supporting the given loads appGed at g, k, V, vertically below W^, Wg, Wg, takes the shape shown. These links are under tension, and the ends / and q are kept apart by the closing link fq, which acts as a strut. The arrangement constitutes a link polygon, and is supported at / and q by reactions Rj, R2, equal to those required in the beam. Draw AB to any convenient scale equal to the length I, and mark on it the position of the loads. Draw to scale ab = Wj, 6c = Wj and e(/=W3. Select any pole 0, and join the points a, b, c and d to 0. BENDING MOMENTS AND SHEARING FORCES 93 Commencing at any point / on tte line of the reaction Rj, draw fg parallel to ao to meet the vertical Une of the load Wj^ in g. From g draw gk parallel to 60, kv parallel to co and uq parallel to do. Join fq and draw eo parallel to fq, then de is the magnitude of the reaction Rg and ea = R^, and the polygon fgkvgf is also a bending- moment diagram for the beam. The bending moment at any section is the product of the ordinate of the bending-moment diagram and the perpendicular distance of the pole from the line ad. If p denotes this polar distance, then the bending moment M at the load W2 is given by nk x p. The polar distance is usually made a convenient multiplier, 5, 10, etc. Thus if p=10 units (linear scale), then M=10xnA. At q we have three forces in equilibrium, viz. R^, the pull in uq and the thrust in fq, and frq is the triangle of forces. By similar triangles dep and rfl, de_:fr .'. dexl=frxp. ,'. fr multiplied by p gives the value of Rj x I. Hence qfr is a diagram showing the moments of the reaction Rg about the left-hand support. In a' similar manner vsr is a moment diagram showing the moments of W3 about the same support, kts for Wg and gft for Wj. At any section EE the bending moment is represented by ms ; this is the difference of the moments of the forces Rg and Wg, and the magnitude is given by ^^ ^ ^ The sbearlng force diagram is obtained by drawing horizontal lines through the points a, 6, c, d, e to intersect the lines through the loads and the two reactions ; the base line of the diagram is the line through e. Ex. 3. A beam AB, Sfom 1 5 ft., is haded, with 2-5 tons at 3 ft. from A, 3 tons at 7 ft. from A and 3 -5 tons at 1 1 ft. from A. Determine the reactions. Draw the bending-moment and shearing-force diagrams and obtain the maximum value of the bending moment. To any convenient linear scale make AB =1 5 ft., and mark on AB the positions of the loads Wi = 2-5, W2=3 and W3 = 3-5 (Fig. 65). On the line of loads make a6 = 2-5, 60 = 3 and erf =3 -5. Select 94 MACHINE DESIGN tte pole at a distance of 5 units (on the same linear scale), join the points a, b, c, d to o and draw the diagram of bending moment ; oe parallel to the closing linefq determines the reactions rfe=R2=4-47 tons and ea=Ri=4-b3 tons. By calculation, R2xl5 = (2-5x3) + (3x7) + {3-5x11) = 67. •". R2 = fi = 4-467 tons. Ri = 9-Ra = 4-533 tons, or by calculation, 3-5x4 + 3x8 + 2-5x12^ tons. 1 15 The work may now be checked as follows : Total sum of loads = 2-5 + 3 + 3-5 = 9 tons. „ reactions = 4-467 + 4-533 = 9 tons. The maximum bending moment is seen to occur at the load Wj. nh measures 4-34. Hence M = 4 -34 x 5 = 21 -7 in.-tons. By calculation, M =i (Rg X 8) - (Wg x4)=4-467 x 8 -14 = 21 -73 in.-tons, or M = (Rj X 7) - (4 X 2-5) = 21 -73 in.-tons. Moment of resistance. Under the action of a load, as in FJg. 66, a beam will be bent as shown. The upper part of the Fia. 66. beam will be in compression and the lower part in tension. A layer, such as the dotted line, which is neither compressed nor extended, is called the neutral layer. The intersection of this layer with any transverse section of the beam is called the neutraJ axis of the section. In any transverse section of the beam, the total compressive force on one side of the neutral axis is equal to the toftal tensile force on the other side. It may be shown that the neutral axis of any section passes through the centre of area of MOMENT OP RESISTANCE 95 Pa the section. The moment of these two equal and parallel forces is called the moment of resistance of this section of the beam to bending. lift and/c denote the greatest tensile and compressive stresses in a transverse section, then the moment of resistance may be written in the forms ftz or /^z, where z denotes the modulus of the section. The meaning and value of the modulus for various sections may be ob- tained as follows : If M is the bending moment at a section where I is the moment of inertia of the section about the neutral axis and / the stress at a distance y from the neutral axis, then the following relations may be obtained : ^-f-^ (5) where E = Young's modulus and R is the radius. In these relations — which may be obtained fey several methods, one of which is given below — ^the important assumption is made that sections plane before bending remain plane during bending, provided the elastic limit is not exceeded. When a beam is bent a length LM=OP (Fig. 67) becomes NQ. Strain = alteration in length NQ-OP original length LM but N(a-OP _MP_y LM ~CM~R' Hence the strain = It should be noted that CM, the. radius of curvature, is usually large compared with MP. As the amount of the eixtension or compression of a short fibre, 96 MACHINE DESIGN having its end terminated by a very small element of area a, w proportional to its distance y from the neutral axis, The resistance = tat?, where A is a constant. The moment of resistance = hay^. Total moment of resistance = k'Zay^, where the symbol 2 denotes the summation of all such terms for the whole area. 2ai/2 is called the second moment', or moment of inertia, and is nsTially denoted by the symbol I. Hence — = k. Under the action of a constant bending moment, the bar will be bent into an arc of a circle, if its cross-section is also uniform. Assume the circle completed, then the length of the neutral axis is 2TrR. A line of particles at a distance y from the neutral axis would have a length 2-rz{R + y). ... strain=?^^(^±^b:^=^, 27rR R as has been shown abpve. Hence, as stress =E x strain, ••• '-% " H M The force on a small element of area a is The moment of this force about the neutral axis is , Eai/^ Hence M = \^y^ or !^=|, (7) R I R and from (6), !^ =^ (8) Equations (7) and (8) are of the utmost importance in the design of beams and girders. From (8), f=My/l, or the stress MOMENT OF INERTIA 97 depends upon tte distance y from the neutral axis. The stress is zero at the neutral axis, where i/ = 0, and is a maximum at the edge, or sldn of the material. \jy is called the modulus of a section, and is usually denoted by the letter z. Hence, from (8), IVl=/z (9) The values of z are tabulated for sections in general use (Tables rV. and V. pp. 102, 104). Moment of inertia. If each element of the area of a plane section be multiplied by the square of its distance from a straight line, and the sum of all such products be taken, the. result is called by engineers the moment of inertia, or the second moment of the area with respect to the straight line. It is usually denoted by I. If the word mass be substituted for that of area, the result is the true moment of inertia of a body about the straight line. If k is the distance at which the total area or mass might be supposed to be concentrated and still have the same I as the area or mass possesses, then l=M^. This becomes for a body of total mass M, I = MA2. k is called the radius of gyration. The value of the moment of inertia for any of the various sections in general use may be obtained from tables such as Table V. (p. 104) or may be found by calculation, using the results in Table IV. (p. 102). Rectangular section. The moment of inertia of a rectangular section about an axis XX (Fig. 68) passing through the centre of area may be obtained as foUows : If 6 and d denote the dimensions. The area of a strip of thickness dx, width 6, is bdx, and its moment of inertia = bdx x x^ -f x^dx=^bd\ From this result we may calculate the modulus of a rectangular section. Thus I ;> d z = -, and (/ = -. 6 M = =/-v 98 MACIJINE DESIGN In beams, joists, etc., the moment of inertia is required about tbe neutral axis XX (Fig. 68). In struts, the least value of the moment of inertia is required, and therefore I is taken about YY in Fig. 68. The two values may be denoted by Ix and ly respec- tively. Thus, if 6 = 2 in., (/=8 in., 2x8^ Ix = = 85 -33 mch-units, 12 Iy=^j-2~ = 5 33 „ „ When the cross-section consists of two equal flanges connected by a web, as in steel or wrought-iron girders (Fig. 69), the neutral axis, passing through the centre of area, is the line XX, and Ix is taken. When used as a stanchion or strut, the minimum value of I is required, and this is about the line YY. The value of Ix is found from Ix = B D^-6f/8 12 and z= BD8-6rf« 6D Thus, if the dimensions of the beam are 9 in. by 4 in., thickness of flanges 0-46 in., thickness of web 0-3 in., the numerical value of the moment of inertia is obtained by substitution. Ix^ (4x9^)-(3-7x(8-08)«) 12 = 80-34 in.-units. It will be noticed that the cross-section is assumed to consist of a series of rectangles as at Fig. 69 (I), but the correct profile is MOMENT OF INERTIA 99 as shown at (II). From this figure an accurate value of I can be obtained — when all fillets, rounded corners and taper of flanges are taken into consideration. From such a figure an accurate value of the area can be found by means of a planimeter * and the moment of inertia by a suitable integrating machine. • The result obtained for a rectangular section may also be found as follows : The force acting on narrow strips taken parallel to the neutral axis is zero at the neutral axis XX and increases uniformly to the surface at AB and CD : these forces may be represented by the areas of the two shaded triangles (Fig. 70). . , 1 db bd Area of upper triangle = -—=—• Centre of area G is at a distance fromC of o o ~q' The resultant force passes through G ; hence, by taking moments about the neutral axis, we have A *i---> B w |jr 1 1 ii Ik 1 r 1 C FIG. 70. D bd d 4^3 = 12 The total amount of area for both triangles is therefore bd^j6, and requires to be multiplied by / in order to give the moment of resistance 5^ rf bd^ bd^ I=J^^=2X 6-=T2- Circle. The moment of inertia of a circle about an axis through the centre C, perpendicular to the plane of the circle, may be found as follows : A thin ring of thickness dx at a distance x from the centre (Fig. 71) has an area Zkx dx. Moment of inertia of ring Fia. 71. =2Tzxdxxx^=2Ttx^dx. * See author's Mcmual of Practical Mathematics (Maomillan). 100 MACHINE DESIGN Hence 1= I 2nx^dx, where R is the radius of the circle (Fig. 71). , ttR* 7cD* 1= — or - — . 2 32 This result is called the polar moment of inertia of the circle. In the case of a ring, if R and *■ are the external and internal radii, D and d the corresponding dianieters, I=^(R*-r*), or ~(P*-d% This may also be written in the form I=^(R2-r2){R2 + r2) = |a{R2 + ,^), where A is the area of the ring. This gives the polar moment of inertia of the cross-section of a , _K hollow cylinder. P^— .J|Vi The value of I for bending (i.e. 1^^. j for the neutral axis in the plane of i!j~ ■ ■Tq -jj the figure instead of the perpen- ! / dicular to it, and forming a I [ diameter) is one-half the preceding ~Ty result, and may be found as Fio. 72. follows : If OM and ON (Fig. 72) are two perpendicular axes in the plane of the area, then Ix = 2a(PM)2, lY=2a(PN)2, where a is an elementary area at P. But 0P2 = PM2+PN2. Hence 2a x OP^ = 2a x PM-i- 2a x PN^ or ^ = Ix + Iy. Hence the polar moment of inertia is equal to the sum of the moments of inertia about any two perpendicular axes through the axis O, and in the plane of the area. In the circle Ix = ly. Hence for bending the value of I is one-half the pola/r, thus the preceding equations become - _ RADIUS OF GYRATION 101 As z=~, .. z= — , or — I 1. y' 32' 32V D / For an elliptical section, axes a and b, 32 Radius of gyration. Values of h for the difierent forms of section are required in strut and other problems. If A* denotes the squa,re of radius of gyration, then 2 _ moment of inertia of section _ I area of section ~a' For a rectangle, ft* = ^ = — . For a hollow (Mcle, k^ = ^^^^^— = .„ . TT/^o jn. ID -(D2-rf2) If D-rf is small, then rf=D-A-, where *■ is a small quantity compared with D. „ D2 + rf2 D2 + D2-2D;r + ;f2 D* Hence — 5— — becomes r— =— approx. To 16 8 Values of ft* for various forms of section are given in the following Table IV. (p. 102), and various values for I beams in Table V. (p. 104). It should be noted carefuUy that all the formulae for the strength of beams given above apply to cases within the elastic limit. Ex, 4. FinA the greatest and, least radii of gyration of a steel joist, fianqes 6 in. by 0-8 in., web 0-6 thick, depth of joist 1 5 in. [Lond. B.Sc] Area = 2(6 x 0-8) + (1 3-4 x 0-6) =17-64 sq. in. Ix= A {(6 x1 53) - (5-4 x1 3-4«)} =605 in.-units. / 605 ^ „„ . *^=Vt7:64=^-«^^^- (2x0-8x6») + (13-4x0-6») _348-596 Iy- ^2 "12 = 29-1 in.-units. 102 MACHINE DESIGN f-H S a i q rQ CD O h o a, e3 eg t>C ^ I- d O Sra So at 11^ I CO -"ISO I CO ■*a:* • PROPERTIES OP SECTIONS 103 -l<^ Q|2 (O %\^ CO + Ito HS Ico ffl I 1(0 HS + OQ I Nl* ► ^■^ 104 MACHINE DESIGN ^ ii-i o a> ,5 3 .2 '^ J3 g 3 § ° m o • i-H H ■ggs 127-6 89-59 90-71 lOCMOOWOO CMCMCM ^ 00T7-inir)^O9cDq5-T-r- GMOCOCMT-COCvlCBin't't cbr^obc<4CN6^h.cocbin N CM M M r- 4 CO 00 CM r- T— t^cDWCMT-f^cDOin--^*0500T-inC0CDO ininin'j-coiocMCOT-T-cM 1 3 r-. CD CO XXX 00 CO CO cDCDCDcoinr--^co^*$' xxxxxxxxxxx in*McvicM05050orar-»cD WORK DONE IN BENDING 105 Ex. 5. A piece of plate steel has to be bent round a drum 5 ft. diameter. Determine the maximum thickness the sted may have, > if the stress is not to exceed 14 tons per sq. in. [E =12,500 tons per sq. in.] [Lond. B.Sc] Eroin(6), fA If * denotes the thickness, then y=^t. 2R/ 2x30x14 „„„. • • t = -~-= ^„^^„ - =0-067 m. E 1 2500 Ex. 6. Find the stress in a strip of tempered steel -031 in. thick due to bending it round a pulley 3 ft. diameter. [E = 36x10* lb. per. sq. in.] From (6), /=^. 0031 „^,^^ „ ,o- ^ 36x106x0-0155 j/=—2— =0-0155, R=18in., /= — = 30,990 lb. per sq. in. Or 13-84 tons per sq. in. Work done in bending a bar or plate. When a bending moment, constant for all transverse sections, is applied to a bar of uniform section, the curve obtained is a circular arc. During application the bending moment increases from zero to a maximum M when the radius of curvature is R. The work done is the product of the mean bending moment and the angle of bending in radians. Work done = - x-, .. EI but "^^r"' .-. work done = —2, where L denotes the length of the bar. Since 3 = ^, we have, by substitution, Work done = 2i|j- 106 IIACHINE DESIGN Ex. 7. Find the work done in bending a mild sted "plate 4 in. mde, 0-08 in. thick and A- ft. long to a radius of 6 ft. 8 in. [E = 30 xlO* lb. per sq. in.] Substituting in the formula --^, where 4 X O-OS* E = 30x10«, 1= , L = 48 and R = 80, , , 30x10«x4x0-083x48 . „ ,^ „ .'. work aone = — — - — — - — - — —-^ =1 -6 it.-lb. 2 X 1 2 x 2 X 80^ Ex. 8. A load of 1 250 lb. is flaced at the centre of a rectangular steel bar 8 ft. long between the supports. If the width is 2 in., find the depth. Safe stress due to bending, ^ 0,000 lb. per sq. in. ., 1250x8x12 ^„„„„. ,, M = = 30,000 m.-lb. 4 2xrf2 - . . . in. 6 /SO.OOO X 3 ''=V 10,000 =^ Ex. 9. Find- the greatest stress due to bending in the guide bar of an engine, assuming the greatest thrust on it to occur at the middle of its span of 3 ft. when the crank is at right angles to the line of stroke. The stroke is 2 ft., length of connecting rod 5 ft. Width of guide bar 6 in., depth 3 in. Diameter of cylinder 1 5 in., pressure of steam 80 lb. per sq. in. The thrust on the guide bar, as in Ex. 4, p. 13, is found to be 1 -288 tons. 1 .288 x 36 •*• ^ Z """^ '^^ in.-tons. M /= — , where z= J x 6 x 3^. . ^ 11-59x6 , „„„^ • • /= ~5 — E — =^ '288 tons per sq. in. D X y Ex. 10. In a rolled steel joist the dimensions are flanges 6 in. by 0-S47- in., web 0-55 in. thick, depth of joist 16 in. Find its moment of inertia about an axis passing through the centre of area and parallel to the flanges. The inner dimensions are 16-(2 x0-847)=14'306 in., and 6 -0-55 = 5-45 in. Ix = ^{(6 X 1 6») - (5-45 X 1 4-3063)} =71 g.g in.-units. BEAMS AND GIRDERS 107 The more accurate value, taking into account fillets, rounded corners, etc., is 725-7 (Table V.). (To obtain the value of 14-306* it may be written? -153 x 2, and four-figure logarithms can then be used.) Ex. 11. A floor has to carry a load of 3 cwi. per sq. ft. The floor joists are 1 2 in. deep by 4| in. thick, omd have a spam, of 14 ft. Determine the distance apart from centre to centre at which these joists must he spaced, if the maximum stress is not to exceed 1000 lb. per sq. in. [Lond. B.Sc] If rf denotes the distance apart, the load W on one joist is ^Xl4x3x112lb. ^ C/Xl4x3x112x14x12 _1000^^.^^.,g, o o . 1000x4-5x144x8 ^^ .QQg ft. p, 13-, j^. •■ 6x42x112x14x12 Ex. 12. A rolled sted joist, 10 in. deep, hasflamges 6 in. wide by f in. thich, and the thickness of the web is \ in. If the maximum intensity of stress is not to exceed 7 tons per sq. in., determine the load ver ft. run which this joist can support on a span of^ 5 ft. [Lond. B.Sc] If IV denotes the load per in. run, then M = -5- in.-lb. I=^ig (6 xlO* -5-5x8-5*) = 21 8-5 in.-units. From M =/- by substitution, y H/x (15x12)2 _ _^^ 218-5 5 =7 X 2240 X — - — 8 5 .-. tt/= 169-1 lb. Load per foot run =1 69 -1 x 1 2 = 2029 lb. Ex. 13. Find the dimensions of a timber beam of^ 2 ft. span and depth equal to twice the width, to hear a uniformly distributed load 0/1000 lb. per lineal foot. Maximum sfo-ess 1000 Ih. per sq. in. [I.C.E.] .. WZ 12,000x12x12 M = - = . 108 MACHINE DESIGN If 6 and rf denote the breadth and depth respectively, then, as26 = rf, 1000 M = - ■ X 46*. _ ^ /1 2,000x1 '~V lOOOx 8x6 = 6-87in., 1x4 rf=2x6-87=13-74in. Ex. 14. A cast-iron water -pipe 2 ft. 6 in. bore, metal 1 5 in. thick, is carried across a stream underneath a bridge, span 26 ft. Find the stress due to weight of water and metal. Assume the pipe to he supported at the ends. Weight of pipe =1 (32-52 _ sqZ) x 26 x 1 2 x 0-26 lb. = 4.445 tons. Weight of water = 1^(2 -5)2 x 26 x 62-3 lb. =3-55 tons. Total weight=4-445 + 3-55 =7-995 tons. ' n /32-5*-30*' ^--30*\ F5~A 32 V 32 WZ 7-995x26x12 M=- = 8 923-2. =7-995x39. „. 7-995x39 „ „ . . Stress = — •^r^r^T— — =0-34 ton per sq. m. 923-2 ^ ^ Girder stays. The flat surfaces on the bop of fire boxes, or combustion chambers, in marine and locomotive boilers are usually GIRDER STAYS 109 supported by girder stays. These girders consist of a solid bar, or two bars riveted together, with distance pieces between them. These girders are supported at the ends and the flat surface of the combustion chamber is supported by bolts (Fig. 73). Each bolt is screwed into the plate and fastened by means of a nut and- washer as shown. Ex. 15. The roof of the combustion chamber of a toiler is strengthened hy a number of girder stays 28 in. span, S'paced 8^ in. apart. Three holts 7 in. apart attach the plates to each stay. The section of the stay is rectangular 8 in. deep and 1 f in. wide at the centre of span. Determine the greatest stress in a stay when the pressure in the hoiler is 200 lb. per sq. in. The three bolts together support |P, and the front and back plates each support |P, where P is the total force exerted by the steam on the part of the roof supported by each stay. In Fig. 73 ^ = 1P=?00><28X8:25^^^^^^^ j^ ^ Ri=R2=l«'- Maximum bending moment M occurs at the centre of the span, and is given by m = (|a, x 1 4) - 7ty =1 4«;, M 14x11,550x6 *~1~ 1 -75 X (7-5)2 =9,856 lb. per sq. in. Lloyd's rule for rectangular stays is (L-P)D^xL " ^°^^^^ pressure = p. L = length of girder. P = pitch of stays. D = distance apart of girders. T= thickness of girder at centre. _ d = depth of girder. C =6,000 if there is one stay to each girder. C= 9,000 if there are two or three stays. C =10,200 if there are four stays. All the dimensions are in inches. no MACHINE DESIGN Substituting the given data, Working pressure = 9000 X 82 X 1 -75 (28-7)8-25x28 = 207-7 lb. per sq. in. Hydraulic accumulator. A suitable form of cap to carry the tank (Kg. 45, p. 59) is shown in Fig. 74. FIG. 74. The bending moment across a section such as CD would be 120-4 W?, where W = — - — tons and Z=1 2 in. .•, M =240-8 in.-tons. RD2 Moment of resistance =/z=f . A section of the arm is shown at Fig. 74 (II). Assuming B = 2 in., /=1 8,000 lb. per sq. in. for cast steel, then we obtain approximately, by considering the section to consist of two rect- angles only, /x2D2 ■'— TT— =240-8 X 2240. /. D=^- 240-8x2240x6 2x18,000 = 9-48 in. HYDRAULIC ACCUMULATOR 111 Bolts. As there are six bolts supporting the load (Fig. 74), assuming the safe stress at 4 tons per sq. inT, 120 '4 Bolt area = ^; — r = 5-017 sq. in. 6x4 ^rfi2 = 5-017. .-. rfi=2-5in. rf=2| in. (p. 74). Foot or base plate. Several forms of base plate may be used, one of which is shown in Fig. 75. 1 I ^ e'e- .V FlQ. 75. The pressure per sq. ft. on the concrete on which it rests, due to the load only, will be 120-4 —^ = 2-849 or 2-85 tons per sq. ft. If the safe pressure for the ground on which it rests is 1 g tons per sq. ft., chosen low to take account of the weights of the parts disregarded in the above calculation. Area of concrete = , _ =80-3 sq. ft. 1-5 Pipe flanges. Assuming the packing of a pipe joint to be firm enough to keep the flanges apart, it would be possible by screwing the nuts too tightly to fracture the flange. Any such fracture would probably occur along a line such as AA (Fig. 44, p. 57), where the bending moment is greatest. 112 MACHINE DESIGN The bending moment at tMs section is M=Pl, where P denotes the load on one bolt. ■■ p=-r''- p is the internal pressure in the pipe. Moment of resistance = JfiT^/, where T is the thickness of the flange, 6 the width at the section AA , and / the safe stress. .6X2/ Hence PZ=- Snbstituting the value of P, •■ 8 .-. T = rf, ''~6'' ' SKpl 4bf' The width 6 can be measured from the end view of the flange, and the value of T obtained. Values found from this equation agree approximately with those given by the Engineering Standards Committee. Design of a beam, or girder. In the design of a beam, or girder, the resistance of the web to bending is usually neglected, and a suitable section obtained by an approximate method. Moment of resistance =a{ftrf, or a^^, where at and «« denote the sectional areas of the tension and compression flanges respec- tively, and d is the distance between the lines passing through the centres of area of the flanges ; but as the thickness of the flanges is usually comparatively small, it is customary to use for d the total depth of the beam. In a built-up girder (Fig. 76) the resistance of the angles connecting the web and top and bottom plates must be taken into account. This may be done by adding to the sectional area of the flange plates the sectional area of the horizontal Umbs of the angles. In some cases the net area of the tension flange is Fig. 76. ROLLED STEEL GIRDERS 113 taken, i.e. the area of the rivet holes are deducted. In the com- pression flange, the gross area is taken ; hence the area of this flange may be obtained and the area of the tension flange deduced from it. As already stated, the formula is only a good approxi- mation, and the area of the rivet holes is usually not deducted. When the greatest bending moment is known, the value of z can be found from the relation M =fz ; then, referring to Table V. p. 104, a suitable section can be obtained. The depth of a girder usually varies from j^ to ^ the span — a value -^ is frequently used, and this limits the deflection of the girder to a reasonable value.* Ex. 15a. A rolled steel joist, 20 ft. span, carries a uniformly distributed load of ^0 tons. Find a suitable section, assuming the depth of the joist to be Jjj span, safe stress 7 tons pfr sq. in. . \Nl 10x20x12 „„„ . ^ M = — = = 300 m.-tons. 8 8 , 20 X 1 2 ^ „ . IVI =fad, where d = — -- — = 1 2 m. Assuming the width of flange to be 5 in., 3*57 Thickness of flange =—— -=0-71 in., or, from M =fz, z = -^ = 42 -9. From Table V. a section 12 in. by 5 in. gives the value of z = 43-48. Ex. 16. Design a suitable form of section for a rolled steel girder to carry a brick wall aver a 20 ft. clear span. Total weight of brickwork = 39,600 lb. Safe stress 7 tons per sq. in., depth of girder 1 4 in. WZ^39,600x20x12^3 ^ 3^ .^ .^^ 8 8 39,600 X 30 ^ ^ a= 3-r = 5-4 sq. in. 7 X 2240 X 1 4 ^ *See Applied Mechanics for Engineers, by J. Duncan (Maomillan & Co.). 114 ]VL4CHmB DESIGN 5-4 If width of flange is 6 in., tliickness = — - =0'9 in. M 39,600 X 30 ^^ ,^ ^=7^ 7x2240 =^^'^^- From Table V. the vfilue of z for a beam 1 4 in. by 6 in. is 76'1 2. Ex. 17. A plate web girder of 40 ft. span and A- ft. deep is required to carry a uniformly distributed load of 3 tons per foot of its length. Estimate the weight of the girder and design a suitable central section. Show clearly how to determine the pitch and diameter of the rivets, uniting the web and angles at the ends, and how to find the necessary lengths of the flange plates. [Lend. B.Sc] M=-— -, where W = 3x 40 =120 tons. .. 120x40x12 .,„„„. , .-. M = =7,200 m. tons. o Area of flanges. If a denotes the area, and assuming the safe stress / as 5 tons per sq. in., then, from M =fad, 7200 ^ a =- — — • = 30 sq. in. 5x48 ^ Assuming the dimensions of the angles to be 6 in. by 6 in. by | in. Area of angles resisting bending = 2 x 6 x | = 6 sq. in. Hence Area of flange = 30 - 6 = 24 sq. in. If the width of the flange is 1 8 in., then the thickness required is 11=1 -33 in. Three plates each J in. thick wiU be ample. Thickness of web. If safe shearing stress be 3 tons per sq. in., then the thickness f is given by 60 i=— — - = 0-44m. 3x45 or ^ in. plate may be used. Pitcii of rivets. Total shear stress per foot length at the ends is calculated thus : Shearing force at support = 60 tons. , 60 /, ,, per foot = —=15 tons. PLATE GIRDERS 115 Assuming the diameter of the rivets to be | in* and shear strength 5 tons per sq. in., n the number of rivets per foot run (as the rivets are in double shear) will be : .". n=2-5 nearly. .•. pitch of rivets =12 -7-2-5 = 5 in. nearly. Hence the cross-section of the girder will be as in Fig. 77 (I). FIG. 77. Lengths of plates. As the bending moment varies from a maximum at the centre of the span to a minimum at the ends, it is not necessary to use the three plates shown throughout the full length of the girder. To obtain the lengths of the two outer plates, it is necessary to draw the bending-moment diagram (Fig. 77 (II)) in which AB is made equal to the span and GH to the maximum bending moment — 7,200 in.-tons. The parabola pass- ing through the three points A, H, B can be drawn by the construction shown. Dividing the line GH into three equal parts by the lines EF and CD, gives the required lengths of the two outer plates as 23 ft. and 34 ft. respectively (neglecting the moments of resistance of the angles). * See Author's Machine Ccfnstruction and Drawing. 116 MACHINE DESIGN Weight of girder. The weight is obtained by multiplying the volume in cubic inches by the weight of 1 cub. in., or 0-28 lb. (2x40x12x18x^. 2x34x12x18x1 2x23x12x18x1 .-. vol. of plates =1 2 x 1 8 (40 + 34 + 23) = 20,952 cub. in. Area of angle = (6 x |) + (5l x |) = 5-75 sq. in. Vol. of angles = 40x1 2x5-75x4=11, 040 cub. in. Vol. of web =45x40x12x1=1 0,800 cub. in. Total vol. =42,792 cub. in. nr • t.^ f -J 42,792x0-28 ^ „^ ^ Weight of girder = ^=5-35 tons. In the preceding calculation no allowance has been made for the weight of the rivet heads, or for any stiffeners that may be used to strengthen the web ; making allowance for these the weight is approximately 5^ tons. In addition, the lengths at each end which rest on the supports have been neglected. If the supports consist of brick walls, then, taking the safe compression of stock bricks in cement, or lias mortar, to be 6 tons per sq. ft., and using a sandstone template, safe load 1 5 tons per sq. ft., its area wiU require to be 62-75-^15=4•2 sq. ft. ; the length a on each wall may be found from =4-2. 144 4-2x144 ~ 18 "' 33-6 in. 5-35 The total weight of the girder would be 5-35 H — -— , or 6 tons. ^ 8 Moment of inertia about a parallel axis. If Ix denotes the moment of inertia of a section about an axis passing through the centre of area, and I the moment of inertia about an axis parallel to it and_at a distance y from it, then I=Ix+Ay'', where A denotes the area of the cross-section. CAST-IRON GIRDERS 117 If o is an indefinitely small area at a distance x from the line XX (Fig. 78) passing through the centre of area, then lx.='2oLxK The moment of inertia of the figure about any line AB parallel to XX and distant y from it for the position of a in Fig. 78 is l = l,a{x+yf = 'Za{x^ + 2xy+y^) = 2ax* + 2a X 2xy + "Say^. But 2ajr^ = Ix, 2a[(^=A^2 and 2a;r = 0, , . J , . . FIG. 78. where x is measured from an axis passing through the centre of area. "Eax is the simple moment of area about this axis, and is therefore zero. .-. I=Ix-fAj/2. Ex. 18. In a cast-iron girder the flanges are 4 in. by 1 in. and 6 in. hyA in. Web^ in. thick; depth of girder 10 in. Find {a) the 'position of the centre of area, (6) moment of inertia about an axis XX {Fig. 79) passing through the centre of area. The results may be obtained either graphically or by calculation, but the latter is usually the more accurate. (a) Taking moments about the lower edge AB, if J denotes the distance from AB of the axis XX passing through the centre of area, then jf(6 + 8 +4) = (6x1) +(8x5) + (4x91) = 81. •■• S^=fi=4iiti. (6) Moment of inertia of lower flange about an axis through its centre of area parallel to AB is jL X 6 X 1 3. I of lower flange about the axis XX is (^ x 6)+(6x42) = 96-5 in.-units. I of web = (^Jg X 1 X 8») + (8 X (i)2) = 44-67. I of upper flange = (3^ x 4 x 1 ») + (4 x 5^) =1 00-33. Hence Ix = 96-5 +44-67 +1 00-33 = 241 -5 in.-units. GrapMcally. The centre of area may be found by a graphical method as follows : Draw the cross-section of the girder to any convenient scale, 118 MACHINE DESIGN and through the centres of area of the flanges and web draw vertical lines as in Fig. 79. The areas of the flanges and web will be 6 units, 4 units and 8 units respectively. On a vertical line mark off to any convenient scale ab = 6, bc = 8 and c(/=4. Join these points to a pole O. From any convenient point, m, draw mn and np parallel to ob and oc respec- tively. Finally lines are drawn from m and p parallel to ao and do. A vertical line qq through the point of intersection, q, determines the centre of area required. The moment of inertia of the given section about an axis through the centre of area may be found by graphic construction, but the results are usually not so accurate as those obtained by calculation: ' Flitch Beams. Flitch beams. Logs of timber are liable to have concealed defects in the form of star shakes and cup shakes, etc. A method fre- quently adopted to counteract these defects and to produce a stronger beam is to saw a log through the axis and to fasten the halves together with a steel, or wrought-iron, plate between them by bolts, nuts and washers, as in Fig. 80. The steel plate may be flat as shown, or an I beam may be used ; the depth is made equal to, or less than, that of the timber. Fia. Ex. 19. A flitched timber beam consists of two timber joists each 4 in. wide by'\2 in. Aeef with a sted plate | in. thick and 8 in. deep placed symmetrically betiveen them and firmly fixed. Span of girder 20 ft. ; ends simply supported. Calculate the maximum uniformly distributed load the beam can carry, if the stress in the timber is not to exceed 1 000 lb. per sq. in. What will then be the stress in the steel ? FLITCH BEAMS 119 [E/or steeZ = 30,000,000 Jb. per sq. in:., E for timber ='i ,500,000 lb. per sq. in.] ^ ^^^ ^^^^^ 30,000.000 t^^'^'i- ^■^■'^ . = 1 : — 20 E for timber 1 ,500,000 ,'. the steel plate is equivalent to timber 20 times as wide, or 1 5 in. by 8 in. Equivalent section for wtole beam has a moment of inertia given by 2x4x123 , (i5-|)83 ,,^^ . I = + -i — —^ — = 1 760 in.-umts. 12 12 1760 = 293-3. If h; is the load per inch run, then M= — =/z. ■ '^x(g°x^2)^ioooxizgg. 8 6 /. «/ =40-74 lb. per inch run. When the stress in the timber at the surface is 1000 lb. per sq. in., the stress at 4 in. above the neutral surface will be I x1 000 = 666-7 lb. per sq. in. The steel will carry 20 times the stress in the timber for an equal strain. .^ ^^^.^^ j^ steel = 20 x 666-7=13,334 lb. per sq. in. Eeinforced Concrete Beams. Reinforced concrete beams. The comparatively high value of the compressive strength of concrete, makes the material of great utility in the form of what is termed reinforced concrete. In Fig. 81 a cross-section of such a beam is shown, the reinforcement consisting of two steel rods imbedded in the concrete. Concrete is comparatively weak under ten- sion, hence the usual assumption is that the whole of the tensile stress is taken by the steel" bars, and the whole of the compression by the concrete. The position of the neutral axis XX may be found approximately by equating the 120 MACHINE DESIGN compressive force in tke cement to the tensile force in the metal. The strain at any point in the section is assumed to be propor- tional to the distance from the neutral axis in the following. If X and d are the distances of the neutral axis and the rein- forcement from the edge AB (Fig. 81), then denoting the stress and modulus of elasticity in the concrete by fc and E^ and in the metal by /j and Ej, the strain in the concrete at the edge AB is ■^ ; and ft ■ ■ ' Ec = IS the strain in the metal. Since these strains are propor- tional to the distances from the neutral axis, fc .ft ^ X Ec ■ Ej d-x' . fc X Ec •(1) The total compressive force in the concrete is (Mean stress) x area = -^ x bx, and total tensile force in the rods=/t x (area of section of rods). Equating these values and substituting in (1), the value of x is obtained. Ex. 20. In a reinforced concrete beam (Fig. 81) 1 6 in. deep and 7 in. wide, the two steel reinforcing rods are each 1 in. diameter, centres 2 in. from lower edge of beam. Find the position of the neutral axis and the moment of resistance of the section when the maximum compressive stress is 500 lb. per sq. in. What is then the tensile stress in the rods ? Take the value of E for steel 1 5 times that for concrete. If X is the distance of the neutral axis from the upper edge of the beam, then, from (14), fc_ X ^JL=. " ft 14-;r 15 210-15;f Total thrust in concrete = — xxx7, 2 ' Total pull in rods =ft x 2 x ^ x 1 ^• Hence -^^il^ i ft 7x 210-15;*- REINFORCED CONCRETE BEAMS 121 Solving this equation, ;r = 6-9 in. 500 Total thrust in concrete = — — x 7 x 6 -9 = 1 2,070 lb. 2 This is also the total pull in the rods. Distances of the centres of pressure from the neutral axis are f x6-9 and 7-1 respectively. Moment of resistance =1 2,070(7-1 +| x 6-9) =141,200 in.-lb. Tensile stress in rods = — ^ =7,683 lb. per sq. in. Ex. 21. In Ex. 20 Jind the stress in the steel rods and concrete at a section where the bending moment is 1 00,000 in.-W. The stresses are obtained as follows : Area of rods = 2x- x1^=1 -57 sq. in. /tXl-57x(7-1 +|x 6-9) =100,000. . ^ 100,000 ^^^^ ,, • • ^"^ 1-57x11 -7 ^^ P^"" ^1- "*• /c = ■'^ k 7 X 6-9 X 1 1 -7 =1 00,000. 200,000 7x6-9x11 -7 = 353-9 lb. per sq. in. Ex. 23. The comjyression area ofaferro-concrete heam is T-shaped {Fig. 82), the limiting stresses a/re 500 lb. per sq. in. in the concrete and 12,000 lb. per sq. in. in the' steel. Calculate (a) the depth oftlie neutral axis below the compression face, (&) the area of the three steel reinforcing bars, (c) the momentof resistance of the beam. Take E for steel 1 5 times that for concrete. (a) Let X be the distance of the neutral axis below the compression face /(, _ X 500 or *■- - .21'i. ---*- 1?-' •. ■/ '--:. 7>:-:|t 1 ■.'•■•/. > -'"./- •n i^ - 1 Fie. 82. 12,000 210-15;r .'. A- = 5 -385 in. 122 MACHINE DESIGN (6) It will tterefore be noted that the neutral axis falls below the slab. In order to simplify the calculation, we may neglect the small part of the concrete in compression ia the part of the web lying above the neutral axis, and take account of the area in the slab only. Compressive stress at top of slab = 500 lb. per sq. in. 2*385 „ „ bottom of slab = 500 x ^ ^^^ = 222 lb. per sq. in. nearly. , , 500 + 222 „„ ,, Average compressive stress on slab= =361 lb. per sq. in. .-. total compressive force on slab = 361 x21 x 3 = 22,743 lb. The total tensile force on the steel will also be 22,743 lb. Hence if d denotes the diameter of the steel -bars, Sy.'^cP=^ 2,000 = 22,743. 4 -^^ 22,743 X 4 ' 36,000 X7t = 0-87 = ^ in. (say). (c) The centre of compression of the slab will be a little above the centre of area of the slab section. Since, however, we are neglecting the small area under compression in the web, it will suffice to assume that the centre ot compression lies at the centre of area of the slab, and therefore at a distance of (14-1 -5) =1 2-5 in. from the reinforcement bars. Hence Moment of resistance = 22,743 x 1 2-5 = 284,300 in.-lb. Travelling loads. When a travelling crane, axles a ft. apart, load W on each wheel (Fig. 83), runs on two similar girders, span ri&. 83. I ft., there are two points at which the bending moment is a maximum, and these maxima occur when one wheel is over the TRAVELLING LOADS 123 points. To pbtain the , position of those points, and also the magnitude of the greatest bending moment, we may proceed as follows : If Ri and R„ denote the reactions at the ends A and B, taking moments about the end B, niXl=\N{{l-a-x) + {l-x)}=\N(2l-2x-a). . „ \N(2l-2x-a) .. Ri = -^ . The bending moment at x ft. from the end A is ., „ Wx{2l-2x-a) M = RiJf= ^ — = -' (1) To find the value of x for which (1) is a maximum. We have ■j- = 2l-4x-a=0. ax I a •• ' = 2-4- Hence, the right or left-hand axle must be at a distance a/4 from the centre of the beam. To find the magnitude it is necessary to substitute this value for ;r in (1). •'• M=w(^-|){2Z-l(2Z-a;-a} = ^(2Z-a)^ (2) Ex. 23. A travelling crane is carried on a carriage, aaHes of which are 4 ft. apart. The carriage runs on two similar uniform steel girders. Design one of the girders if the maximum load is 1 tons equally distributed over the four wheels. Span of girder 20 ft., depth 1 2 in., safe stress 5 tons per sq. in. Draw a cross-section and a part of ike elevation of one girder. Scale 6 I'w. =1 ft. The greatest bending moment from (2) is 2-5x2240 M = —- {(2 X 20 X 1 2) - 48}2 = 544,400 in.-lb. 8x20x12^^ ' ' M =fad, where / = 5 tons, d = depth. .-. 544,400 = 5x2240x120. .'. = 4-05 sq. in. 124 MACHINE DESIGN Assuming width of flanges to be 6 in., tten tMckness is 405 „ „^ . , 1 -— =0-675 m. or ^|. Hence tie flanges would be 6 in. x |^ in., and the web | in. thick. Having obtained the dimensions of the cross-section in this manner, it is interesting to find the weight of one of the girders and the moment of inertia, and proceed to find the stress. EXERCISES VII. 1. A small timber beam 1 in. square in section breaks under ,a central load of 350 lb. on a span of 20 in. What central load will a beam 6 in. wide, 1 in. deep, carry if the span is 20 ft., factor of safety 10? S. A small timber beam, width 1 in., depth 1 in., breaks with a central load of 230 lb. on a span of 2 ft. What is the safe distri- buted load for a beam of the same material, width 1 in., depth 14 in., span 20 ft., factor of safety 8 ? 3. A floor joist, 1 in. deiep by 3 in. wide, span 1 5 ft., carries a distributed load of 1 cwt. per foot run. Find the greatest intensity of stress. 4. Find the maximum load which may be placed at the middle of a bar 2 in. wide and 3 in. deep, if the bar is supported on two points 6 ft. apart. Safe stress 1 0,000 lb. per sq. in. 5. Find the moment of inertia at^d moment of resistance of a rolled iron joist, flanges 3 in. x^ in., web \ in. thick, depth of joist 5 in., safe stress 5 tons per sq. in. 6. A rolled steel joist, 1 8 ft. span, 1 4 in. depth, flanges 6 in. by ^ in., web ^ in., carries a brick wall which gives a uniformly distri- buted load (including the weight of the joisb) of 1 680 lb. per ft. run. A cross girder rests on the centre of the joist and transmits a load of 2 tons. Find the maximum bending moment and stress in the flanges. 1. Rolled iron floor joists, 25 ft. span, centres 8 ft. apart, carry a floor weighing with the load 2 cwt. per foot run. If the depth of the joists is ^\y span, safe stress 5 tons per sq. in., find (a) depth of joist, (6) greatest bending moment, (c) area of each flange, [d) draw the cross-section of the joist, scale 4 in. =1 ft. EXERCISES VII 125 8. A hollow cylindrical beam, 16 in. external diameter, 1^ in. thick, rests on two walls 30 ft. apart. Find the safe distributed load if the safe stress is 9,000 lb. per sq. in. 9. The dimensions of a rolled steel section are 3i in by Si in by 1 in. (Fig. 84). Find the distance of the neutral axis XX from the lower edge AB. Also find the moment of resistance of the section, if the allowable stress is 5 tons per sq. in. 10. A principal rafter in a roof truss inclined at 60° to the horizontal and carries a vertical load of 1 ton at the centre of the span of 1 ft. If the breadth is 5 in., find ^^^ g^ the depth. A small beam 1 in. x 1 in. by 1 2 in. span breaks with a central Joad of 450 lb. Factor of safety 8. 11. A wrought-iron joist, 1 6 ft. clear span, 16 in. deep, flanges 6 in. by 0-88 in.^^web | in. If the safe stress is 4 tons per sq. in., find (a) moment of inertia, (6) moment of resistance, (c) safe uniformly distributed load. 12. A beam 8 ft. span carries a central load of 2000 lb. If the breadth is 2 in., depth 3 in., find the stress at distances J in. and 1 in. from the neutral axis. Also find the maximum stress. 13. The flanges of a steel girder are 6 in. by 0-7 in., web 0-4 in., depth of girder 1 in. If the ultimate strength of the material is 40 tons per sq. in., factor of safety 5, find the safe uniform load for a .span of 1 2 ft. 14. A cast-iron hollow cylindrical beam, external diameter 15 in., thickness 1g in., 35 ft. span, is fixed at the ends. If the safe stress is 1 g tons per sq. in., find the safe distributed load. 15. The rolled steel floor joists for a warehouse 22 ft. span are to be 8 ft. apart centre to centre. If the uniformly distributed load is 2 cwt. per ft. run, depth of joists ^ span, find a suitable section for the joists. Safe stress 7 tons per sq. in. Draw the section : scale 6 in. =1 ft. 16. A cast-iron water pipe, 1 ft. 6 in. bore, metal f in. thick, is supported at two points 36 ft. apart. Find the stress in the metal due to weight of water and pipe. The pipe is running full bore. 126 MACHINE DESIGN 17. The bending moment in a timber beam of rectangular section is 2 ft.-tons. If the maximum stress at the section is 550 lb. per sq. in., and if the depth of the beam is three times the breadth, find the dimensions. 18. A steel beam, span 20 ft., supported at the ends, is 1 6 in. depth, flanges 6 -06 in. by -82 in, web -64 in. Calculate moment of inertia, strength modulus, also safe load at the centre in tons. Safe stress 6 tons per sq. in. 19. A rolled steel beam 1 2 in. deep, flanges 6 in. wide, metal everywhere 5 in. thick. Calculate moment of resistance to bending. Safe stress in tension 6 tons per sq. in. In compression 5 tons per sq. in. 20. An iron bar, 2 in. diameter, is bent into an arc of a circle 400 ft. diameter. Find the maximum stress at any transverse section. Show that if the stress is limited to 4 tons, per sq. in., the diameter of the circle must not be less than 540 ft. [E = 29 x 1 0' lb. per sq. in.] 21. Find the greatest stress in the guide bar of an engine. The stroke is 2 ft. and length of connecting rod 5 ft. The bar is 7 in. wide by 3 in. deep, and the greatest thrust on it may be assumed as taking place when the crank is at right angles to the line of stroke and at the middle of its span of 4 ft. The total load on the piston is 4000 lb. [B.E.] 22. If the safe tensile or compressive stress for a mild steel bar is 20,000 lb. per sq. in., find the maximum load which may be placed at the middle of a uniform steel bar 8 ft. long, 3 in. wide and 4 in. deep. [1 cub. in. weighs 0-284 lb.] 23. A rolled steel joist 10 in. deep, flanges 6 in. by | in., web f in. Find the maximum stress produced by a load of 5 tons at the middle of a span of 1 2 ft. [U.E.I.] ' 24. A reinforced concrete beam, 1 8 in. deep and 9 in. wide, has four bars of steel 1 in. diameter, their axes being 2 in. from the lower face of the beam. Find the position of the neutral axis and the moment of resistance of the section when the greatest compressive stress is 200 lb. per sq. in. What is then the tensile stress in the steel ? Take the value of E for steel 1 5 times that for the concrete. EXERCISES VII 127 25. The floor of a room is supported on timber joists spaced 15 in._ apart centres, clear span 12 ft. The weight carried by the joists, including tbeir own weight, is 90 lb. per sq. ft. of floor. If tbe maximum stress is not to exceed 800 lb. per sq. in. and the depth of the joist is 5 times the thickness, calculate the dimensions. [Lond. B.Sc] 26. A beam 20 in. deep, flanges 71 in. by 1 in., web 0-6 in., is supported at the ends and has a span of 25 ft. If the tensile or compressive stress due to bending is not to exceed 6 tons per sq. in., calculate the maximum distributed load the beam can carry. [Lond. B.Sc] 27. The following particulars of standard rolled joists are taken from a maker's catalogue : Size in inciies. Area of Section. Momont of Inertia about XX. 15x5 12x5 9x7 7x4 12-35 11.47 17 06 4-7 428 261 229 39-2 A beam 1 2 ft. span has to carry a central load of 4 tons and a uniformly distributed load of 1400 lb. per ft. run. Choose a suitable joist. Working stress 7 tons per sq. in. [Lond. B.Sc] 28. The dimensions joi a rolled T bar are 6 in. by 4 in. by | in. Find moment of inertia and radius of gyration about a neutral axis parallel to the top table. [Lond. B.Sc] 29. A flitch beam is formed of two timbers 9 in. by 3 in., and a steel plate 9 in. by g in., all assumea to be rigidly bolted together. Find maximum safe moment of resistance. Stress for timber 1000 lb. per sq. in. [E= 1,000,000 lb. per sq. in., stress for steel 1 5,000 lb. per sq. in., E = 30,000,000 lb. per sq. in.] • [I.C.E.] 30. How do you ascertain the " moment of resistance " of a section, also the " radius of gvration " ? Give an example of each. [I.C.B.] 31. A rolled steel joist has to be designed to carry a load of 2^ tons per foot run across a span of 1 8 ft. Thickness of flanges 1 in., web -f- in. Determine suitable dimensions, stress not to £xceed 7 tP.ns per sq. in, [Lond. B.Sc] 128 MACHINE DESIGN 32. The diagram (Fig. 85) shows a flitched beam made up of two wooden beams and a piece of steel plate 8 in. by | in. Find *—^L maximum stress in steel plate when the maximum stress in timber is 1 200 lb. per sq. in. Determine percentage increase of load the flitched beam will carry compared with the two timbers when not reinforced by a steel plate. [E for wood =1,500,000 lb. per sq. in., E for steel = 30,000,000 lb. per sq. in.] [Lond. B.Sc] 33. The roof of the combustion chamber of a boiler is strengthened by a number of girder stays 21 in. span, spaced 8^ in. apart. Two bolts 7 in. apart attach the plates to each stay. The section of the stay is rectangular, 5^ in. deep and 1 g in. wide at centre of span. Determine greatest stress in stay when pressure in boiler is 225 lb. per sq. in. ' [B.E.] 34. Two plates, each | in. thick, are to be connected to a plate FlO. 83. H in. thick by a pin joint (Fig. 86). the pin. Determine the diameter of FIG. 86. (a) Shearing stress shall not exceed 4 tons per sq. in. (h) On the following assumptions (i) pin is loaded at centre C and supported at A and B, (ii) stress due to bending does not exceed 62 tons per sq. in. [Lond. B.Sc] CHAPTER V. SHAFTS. SHAFT COUPLINGS. COUPLING BOLTS. TOESIONAL EIGIDITY. Shafts. Twisting Moment. The work done by a force P acting tangen- tially on a oircular" shaft, of radius r, through an arc AB, is PxAB=Px/^ (Fig. 87), where denotes the angle AGS in radians. A force P acting on a crank, at a radius r, the force between the teeth of a pair of wheels in gear, or the efEective turning force T2-Ti=P on a pulley (Fig. 88), give in each case a twisting moment or torque, Pr, usually denoted by the letter T. In one revolution the angle described is 2it radians, and 27tN for N revolutions. .'. work done = Pr x 27rN = T x 27rN. Kg. 87. C.M.D. 130 MACHINE. DESIGN T is usually expressed in pounds and inches. Hence, if H is the horse-power transmitted, H g"NT 12x33,000 ^ ' Ex. 1. A shaft transmits a torque of 20,000 in.-lb. at 200 revolu- tions per min. Find, the horse-power transmitted. 2tc X 200 X 20,000 ^„ ^ H = =6o"0. 1 2 X 33,000 Moment of resistance to torsion. One method of obtaining this moment is to assume a circular shaft to be made up of a large number of very thin tubes ; it r^ denotes the radius of one such tube, and /^ the shear stress on its section, then we may obtain the relation f, ;•, ^, f=j or /i=^/, (2) where / denotes the stress at the radius r. If a is the area of cross-section, the shearing force at the radius r^ is /^ x a. Moment of this force is /loz-i. „ resistance of the tube =-ar^, from (2). .". total moment = -2ar^. The quantity Sa/-^ is the polar moment of inertia of the section, and may be denoted by I. Hence, if T denotes the twisting moment, then f T=^I (3) r Comparing this result with that obtained for bending (p. 96), we find M / J T / =-=- and v = -- ix y 1 /■ I for a solid shaft is -;-— , and for a hollow shaft of diameters D and rf, ^^ I=^(D*-rf*) (p. 100). Substituting these values in (3), we olftain •^=1-6°'/' W-d-1-6(^> (^) The value of the safe stress / depends on the material and also on the kind of load to which the rotating shaft i^ subjected. It SHAFTS 131 may be taken as 9000 lb. per sq. in., a value suitable for wrought iron and in many cases for mild steel. Tiie formulae (4) and (5) are only applicable to cases of pure torsion. When the torsion is accompanied by bending, or tension, or compression, the • dimensions may be found as in Chapters VIII. and X. (pp. 211, 246). If in such cases either (4) or (.5) are used, it is necessary to employ a comparatively low value for the safe stress. In the calculation of D* - rf*, the numerical work is sometimes lessened by writing it in the form (D^-d^)(D^+d^), and using a table of areas and squares ; or a table giving the fourth power of various numbers may be used (see Table VI. p. 137). Ex. 2. // the safe twisting moment for a shaft, 1 in. diameter, is 2000 in.-lb., what would be the safe twisting moment for a shaft 2-| in. diameter ? Find the horse-power transmitted by the latter at 1 50 rev. per min., assuming that the torque is constant. Safe torque for shaft 2| in. diameter will be 2000 X (2^)3 = 47,540 in.-lb. T, ,,, ,, 271x150x47,540 ^^„„ From (1), H= — — — TT-—^ — =113-2. ' 12x33,000 Ex. 3. A shaft revolving at 250 rev. per min. is subjected to a constant torque of 20,000 in.-lb. Find the horse-power transmitted. Calculate the diameter of the shaft, if the maximum stress is not^ to exceed 9,000 lb. per sq. in. 271x250x20,000 12x33,000 ~ ' ■ If d denotes the diameter of the shaft, then, from fs/<'^=''', bv substitution, \ 3/20,000x16 „ „^^ . „, . d = \ r.r.^r. =2-245 m. or 2^ in. V TTX 9,000 * Ex, 4. Find the value of T, ,the torque transmitted by a circular shaft (i) when the shaft is solid, diameter 2^ in., (ii) when the shaft is hollow, diameters 1 2 in. and 6 in. respectively. Safe stress in each case, 9,000 lb. per sq. in. |i, T.,>.---ii^&ie2).'.„,6,0in..lb. 132 MACHINE DESIGN Ex. 5. Find the diameter of a shaft to transmit 70 horse-power when the shaft is revolving at 75 rev. per min. The maximum torque is 33 per cent, greater than the mean, and the maximum stress is not to exceed 8000 Ih. per sq. in. [Lond. B.Sc] ^ 70x12x33,000 133 ^„ ^ . „ T = — ^ X — - =78,230 m.-lb. 27t X 75 1 00 _, 778,230x16 ^ ^„ . d=\ — ' ^ =3-68 m. V 8,000Tt Ex. 6. A hollow steel shaft is required to replace a solid wrought- iron shaft of the same diameter. TJie stress that the steel can bear is 35 per cent, more than that of the iron. Find the internal diameter and the saving in weight, neglecting the couplings. [B.E.] Denoting by D and d the external and internal diaraeters of the hollow shaft, the length by I, and the weight per cubic inch by «/, lV»D3 =_/,(^--^_ j, where A =1 -35/,. D3 =1-3503(1-^*). Let^=n, .-. D»=1 -3503(1 -n*); n*=1 -:j-^=0-2592. 1 "35 .*. n = 0-71 36, rf = 0-71360. Weight of solid shaft = wlx -D\ hollow „ =wlx^(D^-d^) = wlxgxO-4907D^. .". saving =100 -49-1 =50-9 per cent. Ex. 7. What diameter, of shaft will be required to transmit 80 horse-power at 60 rev. per min., if the maximum twisting force SHAFTS 133 in each revolution exceeds the mean by 30 per cent, and if the stress is not to exceed 8,000 lb. per sq. in., ? [Lond. B.Sc] ff^ /I > Tij ^ 80 X 1 2 X 33,000 . „ From (1), Mean T= — ^ — ^==84,030 in.-lb. 27t X 60 ' Tir ^ 84,030X130 TT ^^^- ^ = -^100 =fg^8,000rf3. . . 784,030x130x16 ••"^'V 8.0007.X100 =^-^^°- Ex. 8. A hollow shaft is 9 in. external and 4 in. internal diameter. Compare the strength of this shaft with that of a soUd shaft of the same weight per foot run, under the following conditions : (a) To resist torsional stresses, (&) to resist bending stresses. [I.C.E.] (a) If rf denotes the diameter of a solid shaft, and w the weight per cubic inch, we have, by considering one inch length of the shaft, /. rf = 8-062 in. . 7t /9* — 4*\ Torque transmitted by hollow shaft =- ( — - — j/=1 37-6/. Torque of solid shaft =:^ (8-062)3/=1 02-9/. Ratio of torques = =1 '33. (6) As the bending resistance is one-half the torsional, the values are 68-8 and 51 -45 respectively ; hence Ratio of bending moments =1 -SS, Shafts under bending moment. It is frequently necessary to determine the diameter of a cylindrical shaft subjected to bending action (such as a carriage axle and other sinailar cases) in which the twisting moment is neglected. , Ex. 9. An axle is supported on two end journals, and carries a load of 5 tons at a ■point 4 ft. from one journal and 3 ft. from the 134 MACHINE DESIGN other {Fig. 89). Determine the diameter of the part A and the of the journals. The stress allowed is 9,000 U>. per sq. in. Tte reaction Rj is -^ — = 2-14 tons. 5x4 „ R, is — -— = 2-86 tons. Bending moment at A = 2 -1 4 x 48 = 1 02 -7 in.-tons. S.Nv-^^\~>\-.';^'^\N«.\\» «, =0 SHAFTS 135 If d denotes the diameter at A, 1 02-7 X 2240 = ^ X 9,000(/». 32 /102-7X 2240x32 9,0007c . or 6| i Bending moment at the journal F = 2-86 x 3. = 6-38 in. or 6| in. ,=^ 2-86 X 3 X 2240 x 32 9000TC = 2-79 in. or 2| in. For practical reasons, the diameters of the two journals would usually be the same, and each would be 2f in. The bending moment at any point along the length of the shaft may be obtained graphically. Draw to any convenient scale, EF (Fig. 89) to represent the shaft, and mark on it the position of the load. Set off ab equal to the load and mark a point o, so that the horizontal distance om =5 units to the scale on which EF=7 units.' Join a and 6 to o. At any point p in the line denoting the reaction R^, draw pq parallel to oa and qr parallel to ob, and the line oc parallel to the line pr. This determines the reactions R^ = bc and Rj =ca. The triangle ^i?/- is the bending-moment diagram for the shaft ; any ordinate such as qn multiplied by the length om, gives the bending moment at that section. qn is found to measure 1 -71 . Bending moment M =1 -71 x 5 = 8-55 ft.-tons. or M =102-6 in.-tons. The shearing-force diagram is obtained by drawing horizontal lines through the points o, 6 and c, to intersect the vertical lines Ri, W and Rg (Fig. 89). Ex. 10. In a railway waggon the maximum load on a pair of wheels ^s^0 tons ; one wheel takes 7 tons, the other 3 tons. The dis- tance between the rails is 4 ft. 9 in.,''betii>een the centres of the axle ioxes 6 ft. 3 in. Determine the diameter of the axle at the wheel. 136 MACHINE DESIGN Safe stress 1 1 ,000 lb. per sq. in. Draw diagrams of bending moment and shearing force. [Lond. B.Sc] If Rj and Rg denote the loads on the axle boxes (Pig. 90), then, as the distance between the reactions is 57 in. and between the r; ■-49-- -eV'--- FIO. 90. loads 75 in., -rt-here R^ is the load at the axle box nearest to the reaction of 7 tons, ' .-. R, x75 = (7x66) + (3x9)=489. 489 Ri=--— = 6-52 tons. 75 M at the wheel = 6-52 x 9 in.-tons. If rf denotes the diameter of the axle at the wheel, |Vl=/z=11,000x^rf«. '=V — 11 9 X 2240 X 32 000 X TT : 4-96 'in. FOURTH POWERS OF NUMBERS 137 Table VI. Fourth powers of numbers. N N4 N N* u 2-44 6| 2076 1l 506 7 2401 If 9-38 7J 2763 2 160 7i 3164 H 25-6 7| 3607 24 > 39-1 8 4096 2| 57-2 Si 4632 3 81 8* 5220 3| 112 8| 5862 150 9 6562 3| 198 94 7321 4 256 9i 8145 4i 326 9| 9037 ^ 410 10 1 0,000 4| 509 10|' 11,029 5 625 12,155 5;: 760 10| 13,355 5 915 11 14,641 5. 1093 11i 1 6,01 8 6 1296 11 17,490 6i 1526 11 1 9,060 el 1785 12 20,736 EXERCISES VIII. 1, Find the mean twisting moment in a shaft which transmits 30 horse-power at a speed of 90 revolutions per minute, 3. Find the speed of a shaft which transmits 20 ho^se-power if the mean torque is 20,000 in.-lh. 3. A shaft transmits 50 horse-power at 1 20 rev. per min. If the safe stress is 9,000 lb. per sq. in., find the mean twisting moment and the diameter of the shaft. 138 MACHINE DESIGN 4. What is the formula which wUl give the torque that can be transmitted by a shaft of diameter D 1 Find the numerical value when D = 3^ in. Safe stress 1 0* ]b. per sq. in. 5. State the formula for the torsional strength of a hollow shaft, diameters D and d. Given D=14 in., flf=10 in., find the diameter of a solid shaft having the same torsional strength, the material of each shaft being the same. 6. A shaft transmits 20 horse-power at 1 00 rev. per min. What would be the horse-power at 250 rev. per min. ? What should be the diameter of a shaft to transmit 40 horse- power at 250 rev. per min. if the safe stress is 9,000 lb. per sq. in. ? 7. The safe torque for a shaft 4 in. diameter is 1 30,000 in.-lb. Find the safe torque for a shaft 9 in. diameter. Find the horse-power transmitted by the former at 200 rev. per min. and the horse-power transmitted by the latter at 50 rev. per min. 8. A shaft revolves at a speed of 60 rev. per min. and transmits 220 horse-power. If the elastic limit of thematerial under shear stress is 9 tons per sq. in., factor of safety 2-5 on the elastic limit, find the diameter. 9. If N denotes the number of revolutions per minute, H the horse-power transmitted and T mean torque in inch-pounds of a cylindrical shaft, write down a formula giving H when N and T are known. Find H (i) when N = 250, T = 20,000, (ii) N=100, T = 27,610. 10. A steel shaft transmits 1 50 horse-power at 1 1 5 rev. per min. The maximum twisting moment in each revolution exceeds the mean by 35 per cent. If the shearing stress is not to exceed 9,000 lb. per sq. in., find the diameter. [B.E.] 11. A hollow steel propeller shaft is made to replace a solid iron shaft, the diameter being unchanged and equal to 1 2 in. Find the size of the bore in the steel shaft so that its strength is not less than that of the iron one. Shearing strength of steel, 30 tons ; of iron, 22 tons. [B.E.] 13. The diameters of a hollow shaft are 20 in. and 8 in. respec- tively, revolutions 100 per minute. Find the mean twisting moment and horse-power transmitted when the maximum shear stress in the shaft is 6,000 lb. per sq. in. EXERCISES VIII 139 13. Compare the strengtlis to resist torsion of solid and hollow shafts of the same length, weight and diameter, the external diameter of the hollow shaft being double its internal diameter. [B.E.] 14. A hollow shaft, the external and internal diameters of which are 20 in. and 1 2 in. respectively, runs at 70 rev. per min. with a constant maximum shear stress of 6,000 lb. per sq. in. Find the twisting moment and horse-power transmitted. 15. Find the diameter of a screw propeller shaft suitable in the case of an engine of 6,000 horse-power, making 1 20 rev. per min. Surface stress not to exceed 5 tons per square inch. [B.E.] 16. A hollow steel shaft has to transmit 500 horse-power at a speed of 65 rev. per min. The maximum twisting moment per revolution exceeds the mean by 32 per cent. Find the external and internal diameter if the internal diameter is ^^5*^^ °^ ^^^ external. Shearing stress not to exceed 5 tons per sq. in. [B.E.] 17. A hollow steel shaft, internal ^^ths the external diameter, transmits 800 horse-power at 75 rev. per min. The maximum twisting moment exceeds the mean by 35 per cent. Find the dimensions if the shear stress is not to exceed 9,000 lb. per sq. in. Find the weight of the shaft per foot length. [B.E.] 18. A shaft transmits a twisting moment of 6,000 ft.-lb. If the shear stress must not exceed 9,000 lb. per sq. in., find its diameter. If the shaft is made hollow, internal diameter one-half the external, find the diameters. v 19. If 200 horse-power is transmitted safely by a shaft 4 in. diameter running at 1 20 rev. per min., what diameter of shaft will be necessary to transmit 300 horse-power at 1 80 rev. per min. ? ■^ < [U.E.I.] 20. What horse-power could be transmitted by a 3 in. shaft running at 1 40 rev. per min. if the maximum stress be 9000 lb. per sq. in. ? [U.E.I.] 21. A pair of wheels of a railway waggon carries a load of 6 tons, four on one wheel and two on the other, centres of axle boxes 6 ft. 2^ in., gauge of rails 4 ft. 8i in. Find diameter of axle at the wheel seat and at the centre of length of the shaft. Safe stress 5 tons per sq. in. [B.E.] 140 MACHINE DESIGN 22. A steel shaft, 1 2 in. diameter, is subjected to a torque which produces a maximum shear stress of 9,000 lb. per sq. in. What will be the stress at distances 3 in. and 4 in. from the centre ? What wiU be the average shear stress over the shaded area ? What is the tor- sional resistance furnished by this ring ? (Fig. 91). [U.E.I.] 23. A hollow steel propeller shaft has to transmit 8,000 horse-power at llOrev. permin. The maximum twist- ing moment exceeds the mean by 20 per cent., and the maximum shearing stress both in shaft and coupling bolts pig. 91. is not to exceed 8,000 lb. per sq. in. If the internal diameter is ^ths the external, determine the dimensions. [Lond. B.Sc] 24. Find the relative weights of circular and hollow circular shafts of the same strength. The internal diameter of the hoUow shaft is f the external. [Lond. B.Sc] 25. The greatest stress in a square shaft, side a in.', when subjected to a twisting- moment T in.-lbs., is T/0-208a*. Find the ratio of this maximum stress to the maximum stress for a circular section of the same area subject to the same twisting moment. • [Lond. B.Sc] 26. A vessel having a single propeller shaft 1 2 in. diameter, and running at 1 60 rev. per min., is re-engined with turbines driving two equal propeller shafts at 750 rev. per min. and developing 60 per cent, more horse-power. If the working stresses of the new shafts are 10 per cent, greater than that of the old shaft, find their diameters. [Lond. B.Sc] Shaft Couplings. Shaft couplings. For convenience in manufacture and in handling, the shafts used in factories and for similar purposes are usually made in lengths of from 20 to 30 ft. long. These separate lengths are fastened together by means of cast-iron SHAFT COUPLINGS 141 couplings, secured to the shafts by means of keys. Exceptions occur in marine shafts, in which the couplings are in one piece with the shafts. Box or muif coupling. A box or muff coupling, shown in Fig. 92, may be used to connect two lengths of shafting. The coupling may be fastened by means of a key, which extends the full length of the coupling. It is much better to use two keys, as in Fig. 92. With this arrangement it is not necessary that FlQ. 92. the depth of the keyway in each shaft shall be the same. In addi- tion, two keys may be secured more tightly than a single long key, and when a space is left between them as shown, one key can be released and used §,8 a driver to release the other. The ends of the shafts may be enlarged to avoid weakening the shafts by cutting the keyways. Ex. 1. Fill in the missing dimensions in the following : Dimensions of Box Couplings. Diameter of shaft (in. ) 2 2| 3 3i 4 4i 5 Diameter of coupling Length of coupling Diameter of enlarged ends of shafts - 142 MACHINE DESIGN Ex. 2. Find the diameter of a shaft to transmit 80 horse-fovjer at 1 50 rev. per min. Safe stress 9,000 lb. per sq. in. Draw a sectional devotion and flan showing a suitable muff coupling. Scale 3" = 1'. D=^ 6 X 33000 X 80 X 1 2 - „ - 2-67 in. or 2ih in. 27c2x 9000x1 50 ^^ The proportions of the coupling are given in page 141. • Flange couplings. A flange or face plate coupling (except in the case of marine shafts) is usually made of cast iron. The halves of the couplings are bored to fit the shafts, the keyways Fia. 93. out and the holes made for the bolts. The two parts are then keyed to the shafts, the head of the key being at the end of the shaft. Rotation of the bolts during the process of screwing home the nuts is prevented by means of a small snug, or pin, inserted close to the head of the bolt. One coupling is made to enter a short distance — ^ in. or | in. — into the other. This plan ensures that the couplings are in line with each other (Pig. 93). The proportional unit is D, the diameter of the shaft. When the torque, or horse-power, transmitted by a shaft is known, the diameter of the shaft can be obtained by calculation (p. 131). The diameter of the coupling bolts can also be found (pp. 143, 144) ; when these are known the dimensions of the remaining parts of the coupling are obtaiaable from the pro- portional dimensions (Fig. 93). If the ends of the shafts are enlarged, the proportional values (Fig. 92) may be used. COUPLING BOLTS Ex. 3. Fill in the dimensions in the following : Dimensions of Flai^e Couplings. 143 Diameter of shaft D (in.) - 2 2i 3 3^ 4 4| 5 Diameter of boss B Thickness of flange t Length of flange and boss F Radius of bolt cii-ele Diameter of flange To avoid any mischance due to projecting bolts, heads and nuts becoming entangled with clothing, the flanges are in some cases recessed, as in Fig. 94. The flanges are slightly thicker than those in Fig. 93 ; the pro- portions are given in Fig. ^94, the proportional unit being d, the diameter of the bolt. 4iD-^'/S- Fig. 94. Coupling Bolts. Coupling bolts. The bolts connecting the halves of a flange coupling are subjected to shearing force. The resistance to shearing of a bolt, diameter rf, when the safe shearing stress is /s, is If r is the radius of the bolt circle (Fig. 95), then the moment of resistance to shearing of n bolts becomes : c^/s X n X /•. 144 JMACHINE DESIGN Equating tliis value with, the torque of a shaft of diameter D, Id^fyn xr==^f sO^ ■(1) It will be noticed that when D and the allowable stresses fh and fs are given, three factors remain to be determined ; of these n and r are usually determined empirically. FIG. 95. Number of Dolts, n the number of bolts may be 3 for shafts up to 1 1 in. diameter ; from 1 1 in. to 4 in. diameter, 4 bolts ; and 4 in. to 6 in. diameter, 6 bolts. r the radius of the bolt circle may be 1 -50. ft denotes the mean shear stress in the bolts. fs denotes the maximum shear stress in the shaft. If the bolts do not fit the holes Ln the flanges tightly, there may be bending in addition to shear stress, and to allow for this, /j may be l/s, when the bolts and shaft are made of the same material. The screwed part of the bolt has only to resist the tension due to screwing up ; hence this part may be made slightly smaller than the unscrewed part. This also avoids injury to the screw threads when inserted in the bolt holes. The diameter of the bolts may be obtained by using the empirical formula 0-42D ,„, d = — 7:^ + 0-3 (2) Ex. 4. Tivo shafts, each 3^ in. diameter, are connected by cast-'iron flange couplings. Find the diameter of the bolts and the remaining dimensions of the couplings. Shearing stresses not to exceed 4 tons per sq. in. for the shaft and 2 tons per sq. in. for the bolts. Draw COUPLINGS FOR MARINE SHAFTS 145 a sectimal elevation, plan and end view, showing the couplings and portions of the shafts. Scale 6 in. ^t ft. Radius of bolt circle = 1 -5 x 3 -5 = 5 -3 in; ; use 4 bolts. -rf^x 4x5-3x2 4 From (2), / 3-58 V8x5-3 0-42 X 3 -5 \/4 = i^(3-5)«x4. 1 -01 in., sayl in. ■ + 0-3=1 -03 in. The remaining dimensions may be found from the proportions on p. 142. Ex. 5. Two shafts each 3 in. diameter are to he connected by a flange coupling, diameter of bolt circle 5 in. Find the diameter and the number of bolts required. Stress in shafts 1 0,000 lb. per sq. in. ; in bolts 9,000 lb. per sq. in. [Lend. B.Sc] If rf = diameter of bolts and 4 bolts are used (p. 144), ^(/^xQOOOx 4x2-5=-^ X 10000x3*. rf = -866 in., say rf = | in. Couplings for marine shafts. In the shafts for marine engines, the flanges are usually forged in one piece with, and form part of, the shaft. The separate couplings are secured by bolts and are usually filleted into one another, as in Fig. 93. The following table gives the dimensions in inches of some couplings taken from actual practice : Couplings for Marine Shafts. Diameter of shaft (D) 6 n 16i 22^ 23 Diameter of flange 12| 19 32 35 38 Thickness of flange If 2-1 4i 6 5 Diameter of bolt circle H 14^ 25 28| 30| Number of bolts 6 6 8 9 9 Diameter of bolts {d) ii 2| H 4i 4i C.M.D. 146 MACHINE DESIGN In marine couplings the number of bolts usually are 6, 8, 9 or 1 2. In the case of large shafts these bolts are taper, the taper Ijeing | in. per foot of length. Also the bolts are sometimes made with, but often without, heads. Fl8. 96. In marine engine practice, there is no fixed rule for the design of a flange coupling ; an examination of the couplings in use also fails to give any satisfactory rule. It is necessary to obtain the diameter of the coupling bolts before the dimensions of the coupling can be found, and for this purpose the radius of the bolt circle may be taken equal to 0-8D, where D is the diameter of the shaft. The following formula will be found to give fairly good average results : A = D + 3(/ + 2in. (Fig. 96.) B = 0-3D. Taper of bolts | in. per foot length (on diameter). Number of bolts = - + 2. As the bolts fit tightly into the bolt holes and the flanges are forged solid with the shafts, a projecting piece on one coupling fitted into a corresponding recess on the other is not necessary. Instead each flange is recessed to an equal depth to receive a steel disc S. The size of the disc varies from 5 to 6 in. diameter and 1 to 1 1 in. thick. UNIVEESAL COUPLING 147 Hooke's joint or univeisal coupling. When the axes of two shafts are inclined to one another, a form of coupling known as Hooke's Joint or coupling may be used. One advantage in the use of this coupling is that the angle between the axes of the shafts may be altered whilst the shafts are in motion. Thus, in the mechanism of a motor-car, universal joints or couplings must be introduced between the gear box and the back axle. In this FIQ. 97. manner the driving shaft can be out of line in any direction with- out interfering in any way with the transmission of the rotation from the gear box to the back axle. The shafts A and B are forked at the ends as shown in Fig. 97. The joint is made by two pins P and P of equal size ; these allow the forked ends to turn freely about their axes. If the horse-power transmitted by the shaft A to B is known, the mean torque can be obtained from (1) (p. 130) and the diameter of A and B from (4) (p. 130). As the pin P is in double shear, the diameter rf can be calculated from 148 MACHINE DESIGN where fp denotes tlie allowable shear stress on the pin, and/s that on the shaft. The remaining dimensions may be found from the proportions given in Fig. 97. Ex. 6. 7/D = 3 in., d='[^ in. {Fig. 97), calculate the remaining sizes of a Hooke's coupling. Draw a sectional elevation and plan. Scale full size. Ex. 7. Find the shearing stress on each of the 7 bolts in a marine shaft coupling, diameter of bolts 2\ in., radius of bolt circle 11 in. The crank on the shaft is 2 ft. long and the thrust of the connecting rod at right angles to it is 60,000 lb. The torque or twisting moment (T) = 60,000 x 24 in. -lb. If fg denotes the shearing stress, |x2-52x7x11/s = 60,000x24. ^ 60,000x24 „„_ „ .*. fs=-r^ =3811 lb. per sq. m. ^'x 2-52x77 4 Ex. 8. Two lengths of propeller shaft are connected together by 9 bolts, each 3^ in. diameter. The bolts are equally^ spaced round a pitch circle, 3 ft. diameter. What horse-power can be transmitted through the flanged coupling at 90 rev. per min., allowing a shearing stress in the bolts of 2 tons per sq. in. ? [B.E.] If T denotes the twisting moment, assumed constant, then T=^ X (3-5)2 X 9 X 1 8 X 2 X 2240 in.-lb. „ 27rNT Horse-power = 1 2 X 33,000' Substituting the value of T, 27rx90x^x (3-5)2x9x18 X4480 .'. Horse-power = = 9977. ^ 12x33,000 Ex. 9. Determine the diameter of the eight bolts of a solid flanged cowpling for a shaft 14 in. diameter. Diameter of bolt circle 23 in. COUPLING BOLTS 149 Also find, the least thickness of the flange so that shea/ring shall not pccur along AB {Fig. 98). Assume fsfor the bolts as %fs for shaft. Diameter of bolt circle 23 in. Draw a sectional elevation, end vieiv and plan, shmving a fair of flange couplings and portions of the shafts. Scale 3 in. = 1 ft. If rf = diameter of bolts, .". rf = 3-15 in., or 3J in. Area of metal resisting shear at AB =7t x 1 4 x f . /, ■n:x14xtx— /s = :j^x14»x/s. 143 • f= =l5in 1 6 X 98 "* ■ Fis. 98. Ex. 10. A marine engine shaft transmits 4,000 horse-power at 90 rev. per min. If the safe allowable stress is 9,000 lb. per sq. in., find the diameter {a) if solid, (6) if hollow, internal diameter one-half the external. Find the saving in weight per foot length in the latter case, (c) Draiv the sectional elevation and end view, shmoing suitable flange couplings for the solid shaft. , . , ^ /1 6 X 33,000 X 4,000 x 1 Z ^ -,- ;„ t? a- 1 jj ia) d = A. K =1 1 'ee m. Equations 1 and 4 ^ ' \ 2Tt2 X 9;000 X 90 ^ — A li — B <6) D ■7- ■ X 33,000 X 4,000 x 1 2 =11-91 in. Equations 1 and 5 (p. 130). 307t2 X 9,000 X 90 rf = 5'9 in. Saving in weignt. Weight per foot length of the solid shaft is - X (11 -66)2 X 0-284 X 1 2 =363-8 lb. Weight of hollow shaft=^(11 -912-5-952) x0-284x12 = 284-9 lb. Hence, saving in weight per foot length of shaft is 79 lb. 150 MACHINE DESIGN EXEBCISES IX. 1. Sketch with dimensions a cast-iron flange coupling suitable for connecting two lengths of 3 in. steel shafting. Describe how the bolts may yield to an excessive twisting moment, and compare their strengths with that of the shaft. 2. Sketch and dimension a flange coupling for a shaft 3i in. diameter. Calculate the number of bolts required. Diameter of pitch circle 1 1 -25 in., diameter of bolts 1 in. Allow a maximum shearing stress in shaft of 4 tons per sq. in. and shearing stress in bolts 2 tons per sq. in. 3. The crank shaft of an engine is coupled to a dynamo by a shaft 1 ft. 6 in. long and provided with cast-iron flange coup- lings at each end. Find the diameters of the shaft and of the 4 bolts in the coupling if 90 horse-power is transmitted at 240 rev. per min. Maximum stresses not to exceed 3 tons per sq. in. in shaft and 2 tons per sq. in. in bolts. Draw a sectional elevation, end view and plan, showing the couplings. Scale 6 in. =1 ft. 4. Find the shearing stress on each of the 7 bolts in a crank shaft coupling ; the crank on the shaft is 2 ft. 2 in. long and the force acting at right angles to it is 50,000 lb. ; diameter of bolts 2i in., and of bolt circle 1 ft. 1 1 in. [B.E.] 5. Some dimensions of a marine flange coupling are given in Fig. 99. The shaft transmits 30,000 horse-power. If the mean diameter of the bolts is 5| in., find the mean stress in each bolt when the stress in the tunnel shaft 2 ft. diameter is 7Q00 lb. per sq. in. EXERCISES IX 151 6. Sketch a flange coupling suitable for two shafts each 6 in. diameter. Calculate the diameter of the bolts so that the shear- ing stress in them is equal to maximum shear stress in the shaft when it is transmitting 363 horse-power at 120 rev. per min. Number of bolts = 6, diameter of bolt circle =10 in. 7. Knd the diameter of shaft for a marine engine of 4,000 horse- power, making 90 rev. per min., (a) solid shaft ; (b) hollow shaft, internal diameter one-half the external ; (c) if in the couplings for the solid shaft 8 bolts are used, find the mean diameter if the stress in the bolts is equal to the maximum stress in the shaft. Calculate all the dimensions of suitable couplings, and draw a sectional elevation, plan and end view. 8. Find the diameter of a shaft to transmit 220 horse-power at 1 20 rev. per min. Elastic strength of the material is 1 8 tons per sq. in., factor of safety 5. Assume the torque to be constant. 9. Two lengths of a propeller shaft are connected together by a flange coupling, and 9 bolts each 4^ in. diameter, diameter of bolt circle 30 in. Find the horse-power transmitted, at a speed of 90 rev. per min., when the shearing stress in the bolts is 2 tons per sq. in. [B.E.] 10. A shaft 3 in. diameter running at 1 00 rev. per min. trans- mits 1 00 horse-power. What is the mean torque on the shaft ? A flange coupling is keyed to the shaft ; if the key is ^ in. wide, find its length. Shear stress is not to exceed' 8,000 lb. per sq. in. [U.E.I. ] 11. Design a flange coupling for a 3 in. shaft, maximum shear stress in shaft not to exceed 9,000 lb. per sq. in., and for coupling bolts 4,000 lb. per sq. in. Shearing stress on key 8,000 lb. per sq. in. Assuming 4 bolts, bolt circle 9 in. diameter, find the diameter of bolts and the length of the key ^ in. wide. [U.B.I.] 12. Two lengths of a hollow shaft, 10 in. external and 5 in. internal diameter, are connected by a flange coupling. There are 9 bolts, bolt circle 7 in. radius. Determine the diameter of the bolts if the shearing stress in the bolts and the maximum shearing stress in the shaft are both equal to 9,000 lb. per sq. in. [U.E.I.] 13. A 10 horse-power motor drives a shaft having a flanged coupling with ^ in. bolts, radius of bolt circle 2i in. If the speed is 800 rev. per min., find the mean turning moment on the shaft and the mean stress in each bolt. 152 MACHINE DESIGN 14. A shaft, 3 in. diameter, is connected to its next length by flange couplings, bolted together with five bolts, bolt circle 7 in. diameter. The shaft transmits 50 horse-power at 300 rev. per min. Find the diameter of the bolts, stress in bolts not to exceed maximum stress in shaft. [I.C.E.] 15. In a Hooke's joint a shaft A transmits a twisting moment of 47,720 in.-Ib. to another shaft B (Kg. 97). Find the diameters of the shafts and the pins, assuming the safe shearing stress to be 9,000 lb. per sq. in. for the shafts and the pins. The shafts are assumed to be subjected to torsion only. Draw a sectional elevation and plan of the coupling. Scale full size. ToEsiONAL Rigidity. Torsional rigidity. When a shaft is transmitting power, it is important to consider not only the torsional strength but also the stiffiiess. The angle through which one end of a shaft will riG. 100. twist relatively to another section at a distance I from it is usually denoted as the angle of torsion. The reciprocal of this angle may be taken as a measure of the stiffness of the shaft. A circular shaft acted upon by a torque T, as in Fig. 100, will cause any line, OA, to assume a new position such as OB. From the centre C of the circular section draw lines CA and CB ; then denoting the angle ACB by 6, the arc AB by x and the shear stress at A by /, then Strain = (sliding between two parallel planes) -r- (distance be- tween them) (p. 18). strain = V TORSIONAL RIGIDITY 153 From the relation , =r. — — = G, Strain we obtain, x =— , G r Gr (where G is the modulus of transverse elasticity, or modtilus of rigidity) (p. 18). Hence, if Z=1 inch and a. denotes the angular twist per inch length, then , •^ = Ga. r Now ^ = j(p. 130)=Ga. If d denotes the diameter of the shaft, „ 2/Z f G d ,,. = (a constant for a given material) x ^. This result may be written in a more convenient form by substituting the value of / from the formula T=:j^/rfMp.l30) ^ 16T Substituting this value of / in (1), we obtain „ 32Tl 10-2TZ -„. ^=^^Gii=-Grf5-' ^^^ 10-2T and «=-GrfS- It is important to notice that the angle 6 is measured in radians. When the angle is measured in degrees, the number of radians is obtained by multiplying the number of degrees by n and dividing by 1 80. 154 MACHINE DESIGN The angle of torsion is usually a fixed proportion of tlie length of the shaft. The working limit is frequently taken to be one degree for a length of twenty diameters. Torsion-meters. Where turbines are adopted as on board ships to drive the propellers, the shaft horse-power which corre- sponds to brake horse-power, ??%%:i^f?i^^%^^ is measured by a torsion-meter. This measures the angle of twist in a test length of the shaft, from this the torque may be obtained from (2) (p. 153), and the shaft horse-power from (1) (p. 130).* Helical springs. Consider a portion of the spring forming approximately a semicircle (Fig. 101 (11)). If W is the load and R the mean radius of the coUs, the torque on the wire will be WR. The length of the wire is TtR. Hence, from (2), 32WR^ where d is the diameter of the wire. The extension of this portion of the spring due to 6 is OR, therefore the extension for n coils will be 64WR3n . Fia. 101. also Substituting, also WR: W = 47i/R^n ■Kfifi Gd*x .(3) from (3). T6R^64R»n' The length I of wire in the spring is given approximately by l = 2i:Rn. The formulae also apply to springs in compression, but must not be used for springs made of wire not circular in cross-section. *See Duncan's Applied Mechanics for Engineers (Maomillan). TORSIONAL RIGIDITY 155 Helical springs are also called cyUndrioal or cylindrical spiral springs. It should be noted carefuUy that none of the formulae given in this chapter for the strength and rigidity of shafts and springs should be employed for cases in which the elastic limit is exceeded. Ex. 1. Find the diameter of a cylindrical solid shaft to transmit 1 00 kor,se-power, revolutions 50 per minute, if the angle of torsion is limited to 1 ° per 1 ft. length. [G =1 0' lb. per sq. in.] [B.E.] 1° ===-^ radians. 57-3 From (5), eJ"'^''^ Grf* ' 100x33000x12 in.-lb. 27rx50 ■ _ 10-2x100x33000x12x10x12 57-3~ 27tx50x10'x(/* ' or C = ^'^ 2x100x33000x120x12x57-3 IOOttxIO^ = 5-45 in. Ex. 3. A solid cylindrical steel shaft, 4 in. diameter, revolutions 200 per minute, transmits 1 50 horse-power. What will he its angle of twist in degrees in a length of 40 ft. ? Modulus of rigidity, 11 -5x10* R>. per sq. in. ^ 150x33000x12 ..,„„.. ,, T = ;; Tzz^ = 47,260 m.-lb. 27C X 200 If d denotes the angle in degrees, then ex7r _ 32x47,260x40 x12 180 " 7cx11-5x10*x4* .-. 0=4-5°. Ex. 3. A solid steel shaft, 3 in. diameter, when transmitting power, is observed to twist through 17^ degrees on a length of 40 ft. The modulus of rigidity of the steel is 5500 tons per square inch. 156 MACHINE DESIGN What is the twisting moment in inch-tons ? What horse-power is the shaft transmitting if it is revolving at 90 rev. per minA [B.E.] 17-5 XTT "- 180 ' and from (5), 3271 Substituting, 17-5x7r 32Tx40x12 180 "t:x5500x3*" 17-5 XTc^x 5500x3* Horse-power 32x40x12x180 = 27-83 in.-tons. 271 X 90 X 27-83 x 2240 ~ 1 2 X 33000 ~ = 89. Ex. 4. A shaft running at 150 rev. per min. has to transmit 1 20 horse-power. The shaft must not he subjected to a greater stress than 5500 lb. per sq. in., and must not twist more than 1 degree in 1 ft. Determine the minimum diameter. Modulus of rigidity =47 00 tons per sq. in. [Lond. B.Sc] ^ J20X33,000X12 .^ 271X150 16 '■ ' .•. rfi = 3-6in. ; and from (1), g = 2^r '^^^^'^^^1^' 5500 X 71 _ 4700 X 2240 x rfg 180 ~ 2x10x12 rf2=7-2in. Therefore tte minimum suitable diameter is the larger of these, viz. 7-2 in. Ex. 5. A shaft, 3-5 in. diameter, running at 150 rev. per min., transmits 50 horse-power between two pulleys 1 2 ft. apart. Deter- mine (a) the maximum stress in the shaft due to twisting ; (6) the TORSIONAL RIGIDITY 157 angle through which one pulley turns relatively to the other. Modulus of rigidity, 5000 tons per sq. in. Prove the formulae you use. [Lond. B.Sc] (a) From H =- ^'^^"^ 1 2 X 33000' ^ 50x12x33000 „ „„ . , ^ = 27^x150x2240 = ^'^^ '^•-^°'^' 9-38 = ~f(3-5)^ where / denotes the stress. . , 9-38x16 , ,,^ ^ • • ^= 71 X (3-5)8 = -114 tons per sq. m. jis ■u'r. /o\ a 32x9-38x12x12x180, (6) From (2), 6 = ,.^5000x(3-5)^ ^'Srees. :. 6=1-0,5°. Ex. 6. A closely coiled cylindrical spring, 6 in. mean diameter, is made out of f in. steel wire. What direct axial pull can this spring maintain if the maximum intensity of stress per sq. in. is not to exceed 20,000 Th. ? [B.E.] R = 3 in. 3\» Hence t= 3W =:j^ x 20,000 x (^ ^^7^x20,000x0-3753^^ 3x16 EXERCISES X. 1. A destroyer has a solid propeller shaft 9 in. diameter ; the angle of twist given by a torsion meter is 0-15° in a length of 20 in. Find the twisting moment and the horse-power trans- mitted if the number of revolutions are 400 per min., and the modulus of rigidity 5,500 tons per sq. in. 2. The turbine shaft of a De Laval steam turbine is 0-26 in. diameter, number of revolutions 30,000 per minute. Find the horse-power transmitted and the angle of twist in a length of 20 in. when the maximum shear stress in the material is 3,000 lb. per sq. in. G = 5,500 tons per sq. in. 158 MACHINE DESIGN 3. In a cylindrical shaft subjected to twisting moment, (a) the angle of torsion is not to exceed 1 ° per length of 1 ft., (6) the stress is not to exceed 9,000 lb. per sq. in. If the shaft transmits 100 horse-power at 50 rev. per min., find, the diameter to satisfy (a)and(&). G=10' lb. per sq. ia. [B.E.] 4. The angle of torsion of a cylindrical wrought-iron shaft is required not to exceed one degree for each 4 ft. of length, and the stress not to be greater than 8,000 lb. per sq. in. Determine the diameter of the shaft above which the second condition, and below which the first condition will fix a limit to the twisting moment which may be applied. [B.E.] 5. The angle of twist of a cylindrical steel shaft must not exceed 1° per yard length, and the stress must not be greater than 1 0,000 lb. per sq. in. Find the diameter to satisfy both these conditions. G =1 2 x 1 0* lb. per sq. in. [B.E.] 6. A steel shaft transmits power to a distance of 1 00 ft. The maximum shearing stress is not to exceed 4 tons per sq. in., and total twist on the whole length 30°. Find the diameter of the shaft ; also the horse-power transmitted at 90 rev. per min. G =1 2 X 1 06 lb. per sq. in. [B.E.] 7. -A steel shaft is 4 in. diameter, revolutions 200 per minute, transmits 1 50 horse-power. Find the twisting moment in inch- tons. Also find the angle of twist in a length of 40 ft. Modulus of rigidity 5,500 tons per sq. in. 8. A solid steel shaft, 3 in. diameter, when transmitting power is observed to twist through 1 7^° on a length of 40 ft. What is the twisting moment in inch-tons ? G = 5,500 tons per sq. in. [B.E.] 9. A steel shaft, 4 in. diameter, twists through an angle of 6° on a length of 45 ft. What is the twisting moment ? What horse- power is transmitted at 1 00 rev. per min. ? Modulus of rigidity 5,500 tons per sq. in. [U.E.I.] 10. A steel shaft, 4 in. diameter, 200 rev. per min., transmits 1 50 horse-power. Find the angle of twist (in degrees) in a length of 46 ft. Modulus of rigidity 11 ,500,000 lb. per sq. in. : 11. A shaft transmits 2,000 horse-power at 80 rev. per min. If the angle of twist is 1 ° in 20 ft., ratio of mean to maximum twisting moment 1 : 1 -5, find the diameter of the shaft. G =1 2,000,000 lb. per sq. in. EXERCISES X 159 12. The steel shaft of a turbine, 4 in. diameter, 800 rev. per min. when transmitting 188 horse-power, is observed to twisi; through an angle of 0-1 ° in 3 ft. length. Find the value of G. IS. Prove the formula for the angle of twist of a solid cylindrical shaft subjected to pure torsion. The horse-power of a marine steam turbine was found by observing the angle of twist of a 20 ft. length of the propeller shaft at 480 rev. per min. to be 1 •15°. Diameter of shaft 7 in. Modulus of rigidity 1 2,000,000 lb. per sq. in. Neglecting eflect of end-thrust, calculate the horse-power. [Lond. B.Sc] 14. A solid steel shaft is to transmit 25 horse-power at 1 20 rev. per min. The angle of twist is not to exceed 1° in a length of 30 diameters of shaft. Find the diameter of the shaft and the greatest shearing stress. G=1 2,000,000 lb. per sq. in. [Lond. B.So.J 15. A shaft is to transmit a certain horse-power at N rev. per min. Deduce formulae to be used in order to design the shaft for strength and stiffness. Find diameter to transmit 5,000 horse-power at 90 rev. per min. Maximum stress not to exceed 8000 lb. per sq. in. Angle of twist not to exceed 1 ° in a length of 20 diameters. Modulus of rigidity =1 2,000,000. [Lond. B.Sc] 16. Prove that the angular displacement between two sections of a circular shaft separated by a distance L is given by T is twisting moment, I is the second moment of area and C is a constant. Calculate 6 for a shaft 1 20 ft. long, 4 in. diameter, if the twisting moment is 1 5,000 in. -lb. and C = 5,200 tons per sq. in. [Lond. B.Sc] CHAPTER VI. BELT AND HOPE PULLEYS. LINEAR AND ANGULAR VELOCITY. TENSION AND WIDTH OF BELTS. CENTRIFUGAL FORCE. Belt and Eope Pulleys. Velocity ratio. If two pulleys A and B (Fig. 102) are connected by a belt so that the rotation of one causes rotation of the other, then, if no slipping of the belt occurs; wND=Tcnrf, N rf ^ or D' (1) where D and d denote the diameters of B and A respectively, and N and n the number of revolutions of each in the same time. Fig. 102. Ex. 1. Two pulleys are connected by a belt. The larger pulley 4 ft. 6 in. diameter, makes 1 00 rev. per mm. (i) Fmd the speed, of *If f is the thickness of the belt, then N_rf+t n D+t 160 FRICTION OF LEATHER BELTING 161 the smaller pulley whose diameter is 2 ft. 3 in. (ii) Find the speed of the belt, neglecting slip of the belt. (i) nx 2-25 =100x4-5. .'. n — 200 rev. per roin. (ii) Speed of belt =71; x 4-5x1 00 =141 3-9 ft. per min. Friction of a belt. If a pulley A (Fig. 102), by means of a belt, drives a puUey B, then to ensure the necessary friction between the belt and the surface of the pulley, it is necessary that the belt should be stretched tightly over the two pulleys. This initial tension, when A begins to rotate, will be increased on one side and diminished on the other ; these are termed the tight and ' slack sides of the belt. Denoting by Tg and Tj the tensions on the tight and slack sides of the belt, (T2-Ti)V = 33,000H, (2) where V denotes the speed of the belt in feet per minute, and H is the horse-power transmitted. The thickness of leather belts varies from -j^ in. to ^ in., but double belts have twice this thickness. The safe working tension per inch width of belt may be taken as/= 300t (laced joints), where / is the safe tension and t the thickness of the belt. The ratio of Tg to Tj is given by '^=e>^o,. .(3) where \i. denotes the coefficient of friction between the belt and the pulley ; B is the angle in radians subtended at the centre by the belt, and e is the ^--» base of the Napierian ( j^ ^ logarithm, or 2-71 8. C^- The ratio (3) for given values of (x and 6 may be obtained by calculation, 01 verified experimentally. For ^-^ — the latter method a fixed ( r, J pulley or cylinder P is \^_y . necessary, together with A pulleys A, B and C (Fig. 103). These rotate with as little friction as possible, and pro- vide for any angle of lapping differing by rt/2 or 90°. rio. 103. 162 MACHINE DESIGN Loads Tj and Tj may be applied until slipping just occurs. In this manner a series of values of Tg and T;^ may be obtained and substituted for verification in (3), which may be written in the form log;=^ = [jieioge. 'i To obtain the ratio by calculation, assume the belt to embrace an arc AC of length s (Fig. 104), and let T be the tension in the belt at B and T+rfT the tension at D. If BD is a small arc of length ds — ^the length rfs being exaggerated for clearness — ^then the angle BOD may be denoted by dd. When slipping is about to take place, the difference between the tensions, or rfT, must be equal to the amount of frictional resistance, [xp ds, developed over the arc ds, where |x denotes the coefficient of friction and p the normal force per unit length of arc (Fig. 104). Hence (/T T Ts^^P^^V smce P = — r (from analogy with the thin-shell problem discussed on p. 45). Also ds=rdd. • ^- 1 •• rdd'^r' or 't'~^ da- This result may also be obtained by considering a small portion BD of the belt subtending an angle 56, and in equilibrium under the action of forces T, T +dt and the normal forces on BD. Then p >^BD = 2T sin — = t6 approximately. Also [xp X BD =dT, when the belt is on the point of slipping. .-. i>.je=8T, rfT or Y^^- FRICTION OF LEATHER BELTING 163 Integrating between the limits Tg and T^, the tensions at C and A respectively, j :. ioge=?=[jie, 'i or =ef», .(3) where e = 2*71 8 and 6 is the angle AOC in radians. D B S.A FIO. 104. The ratio Tg/T^ in ordinary belt gearing lies between 1 '75 and 3. A value frequently used is 2. T2 = 2Ti. Es.. 3. Find the width of belt -^ in. thick to transmit 5 horse- power ; the belt embracing 40 per cent, of the circumference, speed of belt 800 ft. per min., (a = 0-25, safe stress 320 U>. per sq. in. 40 Angle of lap in radian measure =r^ x 2.n = O-Stt. /. [jie = 0-25xO-87t = 0-6284. -^ = 2-718»'«. /. log ^ = ■ 6284 log 2 -71 i '1 164 MACHINE DESIGN .-. ^^=1 875. 'i (Tg - Ti)800 = 5 X 33,000. 5x 33,000x1-875 800 .-. 72 = 441 -9 lb. If bi denotes the width, of belt, ^g6j'x320 = 441-9, . 441-9x16 3x320 = 7-4 in. Ex. 3. Find the width, of belt to transmit 1 5 horse-power to a pulley . 1 3 in. diameter. Pulley makes 1 600 rev. per min. Coefficient of friction betioeen belt and pulley is 0-22 and angle of contact 210°. Maximum tension in belt not to exceed 45 lb. per inch width. Prove the accuracy of any formula used. [Lond. B.Sc] From (3), p. 163, =r =e''«, where \l = 0-22, and 6 = — Ti 1 210X77 80 2 _ „0'8066 or 1^ = 2-24. 1 2-24' From (2), p.l61, {^2-^^'^='^ 5 x 33,000. V = ,„ X 1 600 ft. per min. 12 Tg =164-3 lb. Width of belt = 1 64-3 45 :3-65 in. Arms of pulleys. The arms of cast-iron pulleys are usually segmental in cross-section, as at (I) (Fig. 105) ; or elliptical, as at (II). The former has a lighter appearance and is preferable to the latter. The thickness of the former may be 0-56 and of the latter 0-46, as in Fig. 105. An approximate method of drawing an ARMS OP PULLEYS 165 elliptical cross-section is shown at (III) (Pig. 105). Divide the major axisAB into six equal parts, using 1 and 5 as centres, draw arcs intersecting at c, e ; these give centres for portions of the curve, completed by using the centres 1 and 5. The arms may be made straight as at (II) (Pig. 106), curved as at (III) or S-armed as at (IV). (Ill) (IV) Fig. 106. The curved arm for the same strength is heavier, but has the advantage that there is less chance of fracture occurring during the process of cooling in the mould. With care, by avoiding abrupt changes of section, and keeping the thickness as uniform as possible, a straight-armed pulley can be cast free from, or with only slight initial strain due to cooling, and this form is preferable, being lighter and stronger than the curved form. - One method of drawing the arms in a curved-arm pulley is shown in, Pig. 106 (II). By drawing a 'line making an angle of 30° through the centre, and from the point of intersection of this line 166 MACHINE DESIGN with the rim an angle of 60°, the centre for one curve is obtained. The remaining centre is at a distance equal to the thickness of the rim (tj) from it. The S-shaped, or doubly curved arm, is drawn by using an angle of 45° as shown at (IV). The mid-point of the radius gives the first centre c ; the remaining centre c^ is obtained by drawing a vertical line through e to intersect a horizontal line drawn from the point of intersection of the line at 45° with the rim of the pulley. The taper in the width of the arm is obtained as in the preceding case. Number of arms. The number of arms is quite arbitrary, such as 4 arms for puUeys up to 2 ft. diameter, and 6 arms for pulleys of larger diameter. If N denotes the number of arms, and R the radius of the puUey, then the bending moment M on each arm is given by M=^: (4) P denotes the effective driving force in pounds at the circum- ference of the pulley, or Tj -T^. Also M =/z. The value of z niay be taken as -^^bH and/= 2,000 lb. per sq. in. (p. 101). ^^ PR 7^ „ ••• N-=32'''*-^- (5) Nave of pulley. The nave, sometimes called the eye, or more frequently the Dosb, may have a diameter 2d for comparatively small shafts (where rf denotes the diameter of the shaft) ; or the thickness of the nave may be obtained from O-144'W-l-0-2, (6) where B is the width and D the diameter of the rim in inches.. If bi is the width of the belt, then B is given by B=|(6i + 0-3) (7) The length of the nave may be from |B to B. TMcimess of rim. tj, the thickness at the edge of the rim, may be *i = ^-Q + 0-13 for single belt, or -— + 0-26 for double belt. BELT AND EOPE PULLEYS 167 Ex. 4. A fulley 3 ft. diameter, making 200 rev. per min., trans- mits 10 horse-power. Find the width of a leather belt, if the maximum tension is not to exceed 80 lb. per inch width. The tension on the tight side is twice that on the slack side. Calculate the dimen- sions of the various parts of the pulley, assuming it to have 6 arms. Find the diameter of the shaft, taking the maximum shear stress as 9,000 U). per sq. in. Draw a sectional elevation and end view. Scale 2 in. =1 ft. Velocity of belt = Sre x 200 = 600k ft. pfer min. (T2-1T2)6007C =10x33000. .-. T2 = 350-1 lb. Width of belt =?^ = 4| in. 80 ° Width of rim = B = |(4| + -3) = 5 -26 in. From (2), p. 161, Px OOOtt: =10x33000. .-. P=175lb. From (4), p. 166, ivi=^-^^— =175x3. From (5), p. 166, 175 x 3=^6*a x 2000, also t = 0-56. , 3/175x3x32 „„. •• '' = V7rxO-25x2000 = ^'^'''-' t=1-1 in., • where 6= width and t= thickness of arm (Fig. 105). From (6), thickness of nave = 0-14^5-37 x 36 + 0-2=1 in. If d denotes the diameter of the shaft, then T=175x18 = -^x9000xd3. 3/ 175x18x16 V QOOOtt =1-213 or 11 in. Leugtb of nave may be from f B to B, or 5 in. ThickneBs of rim =5^^% + 0-1 3 = 0-31 or ^ in. Rope gearing. Rope gearing is extensively used for driving in mills and in factories ; it is also employed to drive many kincfi of machines, cranes, etc. 168 iMACHINE DESIGN The ropes are made of cotton, or hemp, the diameters varying from I in. to 2^ in. The sizes in general use do not usually exceed I 4 XL±. uu ^4 1 1 in. diameter. The speed of the rope varies from 2,000 to 6,000 ft. per min. Average speeds from 4,000 to 5,000 ft. per min. are usual. Large rope flywheels are generally made of cast-iron with the rims grooved to suit the ropes. A section of a wheel rim is shown in Fig. 107. The proportional unit is d, thfe diameter of the rope. ..,.Wr---^'-^— •*! ^ + — ^\. Substituting this value for Tj in V(T2 - Tj), .•.V(T.-T,=v{(T,-f)-.-..(T.-"^)} ,., If H denotes the horse-power transmitted and k a constant, then (2) beconues H=.v{(1-e-.«')(T,-"^)} =A|(1-e-.'')(f,V-'^)} (a) The maximum value of (a) is obtained in the usual manner by difierentiating (a) and equating the resulting expression to zero. .-. -=/,|(l-e-.^>(T,-^| = 0. This is zero when T„ = 0. Substituting the given values, vs <«■ T2 = 350x8xl = 1400lb. w = 0-036 X 8 X i X 1 2 =1 -728 lb. ^-4 32-2x1400 „, „^j,^ 3x1-728 =93-26 ft. per sec. (Ta^Ti)Vx60 33000 Tg - Ti = 1 400 - 700 = 700. ,, 700x93-26x60 ,,„, ^= 3300^-="^'^' TENSION DUE TO CENTRIFUGAL FORCE 179 Ex. 4. Find ike linear speed of a leather belt (8 in. by \ in.) when the power transmitted is a maximum. The tensile stress in the leather is not to exceed 500 Ih. per sq. in. T2 = 500x8xl = 2000]b. «; = 0-036x8xix12=1-728lb. VlS^Y 32-2x2000 _, ^ ,, 3x1-728 =111 -4 ft. per sec. Ex. 5. In Ex. 7 (p. 170), taking into account the tension in the rope due to centrifugal force, find the horse-power transmitted by a rope at a speed of ^ 00 ft. per sec. The maximum tension in the rope is not to exceed 200 lb. per sq. in. Find the centrifugal tension in the rim of the pulley. [1 foot length of rope weighs 0-28d^ Ih., where d denotes the diameter of the rope in inches.'] u/ = weight of 1 ft. length of rope = 0-28 x1 -5* = 0-63 lb. Tension in rope due to centrifugal force = 0-63x1002 ,„^„,, = =195-6 lb. 32-2 Maximum tension in rope as in Ex. 7 is 353 -4 lb. Tension on the tight side available for transmitting power is 353-4 -195-6=1 57-8 lb. T,J-|^^=15-88. 1 9-94 (157-8-1 5-88)60 x100 _ 33000 If / denotes the stress m the rim of the puUey, then ^ 12X0-26V2 12x0-26x100^ „^„ ,, /= or ^— — = 969 lb. per sq. m. EXERCISES XII. 1. Find the tensile stress in the rim of a cast-iron flywheel if the velocity of the riili is 90 ft. per sec. 8. The radius of gyration of a cast-iron flywheel is 8 ft. ; find the number of revolutions per min. if the stress does not exceed 2,000 lb. per sq. in. 180 MACHINE DESIGN 3. If the safe tensile stress in the rim of a cast-iron flywheel is assumed to be 1 ,200 lb. per sq. in., find the limiting speed of the rim in feet per second. 4. "What velocity will produce a stress of 3,000 lb. per sq. in. in the rim of a cast-iron flywheel ? 5. A block of cast iron 2 in. x 3 in. x 4 in. is fastened to one arm of a wheel at a distance of 2 ft. from the axis. What is the force tending to fracture the fastening if the wheel makes 1 ,500 rev. per min. ? 6. Show that the stress per square inch on the rim of a flywheel is equal to the momentum of the amount of rim (per sq. in. of section) which passes a fixed point in imit time. Find the limiting speed of periphery, the material being such that a bar of uniform section, cross-section 1 sq. in., 900 ft. long, may be safely supported by tension. [B.E.] 7. A segment of a flywheel with the arm to which it is attached weighs 3,500 lb. ' This weight may be assumed to act at a distance of 8 ft. from the axis. Find the' centrifugal force in tons when the revolutions are 40 per minute. 8. The mean diameter of a cast-iron flywheel is 1 ft. If the tensile stress in the rim is not to exceed 2 tons per sq. in., find the number of revolutions per minute. [B.B.] 9. The thin rim of a wheel 3 ft. diameter is made of steel weighing 0-28 lb. per cubic inch. How many revolution per minute can this wheel make without the stress due to centrifugals force exceeding 10 tons per sq. in.? How much is the diameter of the wheel increased ? [E = 30 x 1 0^ lb. per sq. in ] [B.B.] 10. The rim of a cast-iron pulley has a mean radius of 1 2 in., the rim is 6 in. broad and | in. thick, and the pulley revolves at 1 50 rev. per min. ; what is the centrifugal force on the pulley rim per inch length of rim ? What is the tensile stress in the puUey rim ? Find the limiting speed of rotation if the tenacity of cast iron is 1 2-5 tons per sq. in. [B.B.] 11. Find the maximum velocity of the rim of a cast-iron pulley if the tensile stress due to centrifugal force is not to exceed 1 000 lb. per sq. in. [B.B.] 12. Find the horse-power transmitted by a rope at a velocity of 100 ft. per sec. The maximum tension in the rope is not to EXERCISES XII 181 exceed 200 lb. per sq. in. ; diameter of rope 1 J in., angle of groove in pulley 45° ; coefficient of friction of the rope on a flat surface 0-25. Weight of 1 foot length of rope = 0-28(/2^ where d is the diameter of the rope. 13. A leather belt, 4 in. by J in., has a speed of 800 ft. per man. Find the increase in the -tension of the belt due to centrifugal force. 1 cub. in. of leather weighs 0-036 lb. 14. Find an expression for the hoop stress set up in a fljTjfheel rim, when the mean rim speed is V ft. per second. [U.E.I.] 15. How many cotton ropes 1 5 in. diameter will be required to transmit 600 horse-power over pulleys 30 ft. apart at a speed of 6,000 ft. per min. ?' Assume slack tension. 16. Prove the formula for the stress due to centrifugal force in a thin revolving hoop. Assuming the formula applies to the rim of a flywheel 1 5 ft. diameter and weighing 450 lb. per cub. ft., find the speed in revolutions per minute when the stress due to centri- fugal force is 2 tons per sq. in. [Lond. B.Sc] 17. Show that the hoop stress / in a revolving ring is given approximately by the expression /=pti2/gf, where p is the density of the material and u is the velocity of the ring. Find the safe number of revolutions for a rotor 6 ft. diameter if the stress is not to exceed 1 8,000 lb. per sq. in. Take p = 480 lb. per cub. ft. [Lond. B.Sc] 18. Deduce an expressipn for the tension in a belt due to centrifugal force, and prove the velocity at which maximum power can be transmitted is ■VjgjZw, where T is the max. tension in the belt and w the weight of belt per foot run. What horse-power can be transmitted per sq. in. of cross- section if the tension in the belt is not to exceed 360 lb. per sq. in., and if the ratio of the tension in the tight to the tension in the slack side is 1 -8 ? Weight of 1 cub. in. =0-036 lb. [Lond. B.Sc] 19. Calculate the centrifugal tension in' a belt which runs over two pulleys at a speed of ^,500 ft. per min. The belt is.8 in. wide by 1% in., and weighs 0-036 lb. per cub. in. If e''* = 2, and the maximum tension is 250 lb. per sq. in., find the horse-power that can be transmitted at the above speed. [Lond. B.Sc] 182 MACHINE DESIGN 20. Determine the velocity at which the power transmitted by- a belt is a maximum. Prove that the ratio of" the tension on the tight to the tension on the slack side is 3 ' where (i. denotes coefficient of friction between belt and pulley, and angle of contact between belt and pulley is 6 radians. [Lond. B.Sc.] CHAPTER VII. WHEELS. WHEEL TEETH AND AEMS. JOUENALS. FLYWHEELS. Wheels. Linear speed. If two cylindrical surfaces, A and B (Fig. Ill), are pressed tigktly together, the rotation of A will cause the rotation of B, and the speed of the rim of A must equal that of B. If D and d are the diameters of the larger and smaller cylinders respec- tively, and N and n the revolutions of the two, Surface speed of larger cylinder =7tDN. „ „ smaller cylinder =7rrfn. .'. 7i;DN=7Trf/j. • 1 = ^ (1) Friction. In numerous forms of friction grips, friction pulleys, friction wheels, etc., the friction between two surfaces pressed together is used to transmit force from one shaft to another in cranes, centrifugal dryers, separators, etc. In other cases, such as in journals and sliding surfaces of various kinds, the friction between the surfaces leads to waste of energy, the measurement of which may be made in suitable units, such as foot-pounds, horse-power, heat developed, etc. If W denotes the normal force pressing two surfaces together, (jt the coefficient of friction, then the force F acting parallel to the surfaces in contact, and required to produce motion is given by F = tiW. 183 184 MACHINE DESIGN Ex. 1. If the coefficient of friction for the surfaces of a crank pin and its brasses is 0-08, what work is lost per minute in the friction of a crank pin 6 in. diameter under a mean load of\ tons at 75 rev. per min. ? Frictional force =1 x 2240 x 0-08 =1 792 lb. Distance it moves through = -— -- x 75 ft. per min. Horse-power lost by friction = 1792x37-571 33000 = 6-4. Friction wheels. Friction wheels of a cylindrical, or conical, form are used in cranes, centrifugal d,ryers, etc. One of the two wheels in contact is usually faced with wood, leather, paper, or other material, to increase the friction between the two sur- faces. If W denotes the normal force pressing two such wheels together (Fig. Ill), (i the coefficient of fric- tion, F the driving force at the circumference is given by F = [xW, where \j. may be taken as 0-3 for wood on metal and 0-4 for leather on metal. Ex. 2. A friction wheel faced with leather drives another 2 ft. 6 in. diameter at 200 rev. per min. If the force press-, ing the wheels together is 500 Z&., find the horse-power transmitted. F = [jiW = 0-4 X 500 = 200 lb. V = 2-57tx200. FV 200 X 200 X 2-5Tr no. 111. H = 33000 ■ 33000 - = 9-5. Toothed wheels. When the resistance to motion is sufficiently great, slipping will occur in the case of friction wheels. To prevent this lost motion, the rollers or cylinders are furnished with teeth, WHEELS 185 and are known as spur wlieels. The surfaces of the cylinders are called the pitch surfaces of the two wheels in gear, the circles iu contact (Fig. Ill) the pitch circles. The teeth of the wheels are so shaped that the ratio of the speeds at any instant is constant. Fig. 112. If p denotes the pitch (distance between the centres of two con- secutive teeth measured on the pitch circle) (Fig. 112), m the number of teeth and d the diameter of the pitch circle, pm=v:d (2) From this relation, when any two of the three terms p, m and d are known, the remaining term can be obtained. .In (2) p is called the circular fitch, measured on the pitch circle. If s is the diametral pitch, then the relation is given by 8 xm=d, ^. , , ... diameter of pitch line or Diametral pitch = r z-r — re ^ number of teeth Ex. 3. Two wheels are required to transmit a velocity ratio of 4 : 1 bettveen two shafts ivhose centres are approximately 25§ in. apart. Find (1) number of teeth in the wheels, (6) actual distance between the shaft centres — diametral pitch 1 -25 in. [Lond. B.Sc] (a) Radii of wheels =^ x 25-625 = 5-1 25 in., and fx 25-625 = 20-5 in. Assuming the diameters of the wheels to be 1 0-25 in. and 41 in. respectively, 1 Q.25 Number of teeth on the small wheel=-— — =8-2. We therefore take the numbers of teeth as 8 and 32, and the diameters of wheels are 8 x 1 -25 =1 in. and 32 x 1 -25 =40 in. (&) Distance apart of centres = 25 in. 186 MACHINE DESIGN Ex. 4. Two parallel shafts are to be connected by spur gearing ; one shaft to run at 1 50 rev. per min., the other at 300 rev. per min.; the axes of the two shafts are to be as nearly as possible 1 ft. 6 in. apart. Find the radii of the pitch circles, also the number of teeth in each wheel — ci/rcula/r pitch = ^ in. If R and r denote the radii, /- _ 1 50 _ 1 R~300~2" Hence, as a first approximation, R = |x18=12in., and r=^x18 = 6in. If N denotes the number of teeth in the larger wheel, then Nx| = 7cx24. 7rx24 N = 0-875 = 86-18. Taking the numbers of teeth to be 86 and 43 respectively, then we obtain 86x0-875 R = ;: =11-98in. Hence ,20 teeth 2-K r=5-99 in. .". distance apart of the shafts =17-97 in. Train of wheels. What is called a train of wheels may be used to increase or diminish speed. A famiUar example is furnished by the mechanism of a crab, winch, or crane, as in Fig. 113. A handle, H, drives a pinion. A, and this drives the barrel, or drum, E, through the wheels B, C and D. If H=16 in. long, E = 8 in. diameter ; wheel A of 20 teeth, gearing with B, a wheel of 80 teeth ; wheel C of 20 teeth, gearing with D, a wheel of 1 00 teeth, then Velocity ratio = |g x i£^- x i£- = 80. If the force or efFort applied at H is 20 lb., then W (the weight taised) is 80 X 20=1 600 lb., neglecting friction. EflSciency. The effect of frictional FIG. 113. and other resistances is to make the MORTICE WHEELS 187 useful work obtained from a macMne always less tlian the work put in ; the ratio of the two is called the efficiency, which may be expressed in various forms, one of which is as follows : Efficiency ■■ work obtained work suppUed" Thus, if in the above example the efficiency is 60 per cent., and the eSort is 20 lb., then Actual load raised = W =1 600 x f^ = 960 lb. Wheel Teeth and Arms. Mortice wheels. The noise and vibration caused by two wheels in gear having cast teeth dressed by hand is considerably reduced when one of the wheels is provided with wooden teeth ; such a Steel Pin Wood Wedges rio. 114. wheel is known as a mortice wbeel. It is found that the wear occurs chiefly on the wooden teeth, and these can be replaced when necessary. A mortice wheel, which is usually made of cast iron, has a series of rectangular grooves in the rim ; into each of these grooves hardwood blocks are tightly fitted. After suitable preparation in the lathe, the teeth can be formed and may be secured either by pins passing through the lower part of each tooth, as at (I) (Kg. 114), or wooden wedges, as at (II), may be used. The wooden cog is usually made slightly thicker than the iron tooth working with it {i.e. thickness of tooth may be 0-6p). Propoitions of wheel teeth. The proportions adopted for the teeth of wheels depends upon the manner in which the teeth are formed. When, as is sometimes the case, the teeth are cast with the wheels, then the clearance {i.e. the difEerence between the 188 MACHINE DESIGN thiclmess of a tootli and tie space between two teeti) is greater than in machine cut teeth, in which the teeth are carefully shaped by machinery. The following (Fig. 115) are average proportions : Pitch of teeth = p. Thickness of tooth t = 0-48p. Width of space S = 0-52p. Height above pitch line H =0-3p. Depth below pitch line D =0-4p. Width of tooth 2p to 3p. Thickness of rim Ri = 0-5p. KG. 115. Arms of wheels. The form adopted for the arms of a wheel depends chiefly on the size of the wheel. For small wheels and change wheels for lathes, the cross-section of the arms may be segmental as at (I), or elliptical as at (II) in Fig. 116. For larger wheels, the arms may be cross-shaped as at (III), or double tee-shaped as at (IV). The latter are generally used in machine moulded wheels. In the case of bevel wheels a tee form shown at (V) (Fig. 116) may be used. Number of arms. The number of arms in a wheel is usually some empirical number such as 4, 6, 8, etc., arranged frequently as follows : — -Wheels under 4 ft. diameter, 4 arms ; from 4 to 8 ft., 6 arms ; from 8 to 1 6 ft., 8 arms ; and from 1 6 to 24 ft., 1 arms. WHEEL TEETH AND ARMS 189 Desigrn of arms. If R denotes the radius of the wheel and N the number of arms, then, as in the case of pulleys, it is assumed that each arm is equally loaded and is fixed at the nave, and free at the rim. PR The bending moment (M) on each arm is IVI = -rr. where P is the load on the teeth tangential to the pitch circle. o CIV) riQ. 116. When the sections of the arms are segmental or elliptical, as at (I) and (II) (Fig. 116), then Moment of resistance =--6^t/, where f = 0-56 in (I) and t = 0-46 in (II). , If t = 0-56, then Moment of resistance = •0496'/ nearly. For the sections (III), (IV) and (V), if 6 is the width and * thickness of the arm exclusive of the feathers, then Moment of resistance = ^6^t/, where b denotes the breadth and/ 1 the thickness of the arin (Fig. 116), taken to be 0-48p (p. 188) ; / may be taken to be 3000 lb. per sq. in., pr, — =i62xO-48p/. ' = Va 6PR 48p X N/' .(3) 190 MACHINE DESIGN Nave of wheel. The diameter of the nave of the wheel for shafts from 2 in. to 4 in. diameter is usually made about 2D and its lengtl^ 1 |D (where D is the diameter of the shaft), or the thick- ness of the nave may be obtained from the formula : T = 0-4v'p^-l-0-2 (4) Strength of wheel teeth. A large number of rules have been formulated for the strength of the teeth of wheels, some of which give widely difierent results. If H denotes the horse-power transmitted by a wheel, V the velocity of the pitch circle in feet per minute, andiPthe force in pounds on the teeth in a direction tangential to the pitch circle, then .(5) 33000 The greatest pressure on one tooth is nP, where n lies between J and 1 . The force Q acting at the edge of a tooth may be taken as |P (Fig. 117). FIO. 117. A tooth is assumed to be a rectangle of thickness *, breadth 6 and height h (the curved form of the tooth is negligible so far as this calculation is concerned). The bending moment at the root AB is M =Q/i. 6*2 Moment of resistance -—/. D Hence Clh: 6-' .(6) STRENGTH OF WHEEL TEETH 191 Substituting forQA and *, we obtain, asQ = §P and h = 0-7p, |PxO-7p = -5--^-^/. 12-1P5 P ''f or p IS proportional to -. The value of b may usually be taken between 2p and 3p. Assuming 6 = 2p, and writing Q = nP, ,. „PxO-7p=?^A /=^ (7) For different wheels, when the strength of the teeth is the same, p (pitch) is proportional to -y/P or p = fty'P, where ft is a constant. The value of k for ordinary mill-gearing naay be taken as 0-055. When subjected to excessive vibration and shock, as in some machine tools, the value of k is given by Unwin as 0-06. The value of h may be found as follows : Assuming /= 2000, t = 0-48p, . Q„ r^-, 2p(0-48p)2 X 2000 .-. fPx0-7p = -^^3^ ^ . .-. ■P = 329-1p2, p = 0-055-v/P (8) ■ The average ultimate tensile strength of cast-iron teeth is about 20,000 lb. per sq. in. ; using a factor of safety of 10 the value of / is 2000 lb. per sq. in. This value may also be used for gun-metal ; for phosphor bronze /= 3000, for cast steel 4000 and for mild steel 5000 lb. per sq. in. In the preceding results, the teeth of the wheels are assumed to have a fairly even bearing throughout their length, but this is not necessarily the case. Spur gears made from patterns have a certain taper (or stri-p) to ensure a clean lift from the sand. If care is not taken that the thick part of one tooth gears with the thin part of the other, the whole of the pressure between the teeth may occur across a corner of the tooth. Also, when the axes are rigidly fixed as in cranes, and in gearing subject to shocks, the tooth is liable to frapture across a corner of the tooth. 192 MACHINE DESIGN Greatest force on the comer =nP. Greatest bending moment = nP x mo = nPA sin 6 (Fig. 118). 6 t^ Moment of resistance = -^/, where 6i=AD = Asec 6. :. nPhsmd= — - — /, ^ 3nP . „n . f=— s-sm20. •■J j2 .(9) no. 118. The maximum value of (9) occurs when sin 20=1 , or 0=45°. .■.f=%- (10) The maximum stress, when the load was assumed to act along O .1 nP the whole length of a tooth, was found to be — -^ — (7) ; hence the . . 3 * ratio IS 5"^='^ "^3 : 1 . Ex. 5. A cast-iron spur ivheel has 50 teeth, 1 in. pitch. Find the horse-power that may he transmitted, revolutions 1 50 per minute. From (8), P = 329-1 lb., V = ^OxJ^xISO ^^ ^^^ ^^_ From 12 H- PV 33000' 329-1 X 50x1 50 1 2 X 33000 " SPUR WHEELS 193 Ex. 6. A spur pinion, 30 teeth, diameter of pitch circle 18m., iindth twice the pitch, is rotating at 1 50 rev. per min. Find the horse-power transmitted. Safe stress 3000 lb. per sq. in. Pitch='^|;^=1-886in. 30 From (7) , P = 329 -1 x (1 -886)2 =i 1 69 lb. PV From H = „„„^^ , where V =1-5 XTTX 150. 33000 ., 1169x1-57i;x150 „^ , • • H= „„„„^ =25-1. 33000 Ex. 7. Two wheels on parallel shafts, axes 2 ft. apart, are to transmit 30 horse-power. One is to run at 1 50 rev. per min. and the other at 300 rev. per min. Find the radii of the pitch circles. Also determine the- diameters' of the two shafts. If R and r denote the radii of the larger and smaller wheel respectively, , r 150 1 r'^soo^^' .*. R=§x24=16in., /■=Jx24=8in. If P denotes the load on the teeth in a direction tangential to the pitch circle, then, from. (5), Px 271x16x150 = 30x33000x1 2. .-. P= 787 -7 lb. Torqae. (T) on the larger wheel =787-7 x 1 6 =rs/'/'- '4 737 -7 xM QOOOtt =1 -925 in. or 2 in. The torque in the second shaft is 787-7 x 8. 3/ 787-7x8x16 V 90007: =1 -529 in. or l^in. Ex. 8. Two shafts connected by spur gea/ring a/re to run ail 50 and 300 rev. per min. respectively, and to transmit 20 horse-pawer. C.M.D. N 194 MACHINE DESIGN The axes of the shafts are to he as nearly as possible 1 8 in. apart. Find the diameters of the shafts and the dimensions of the wheels. Draw elevation, plan and end view showing the wheels in positions on the shafts. Assume f= 4000 lb. per sq. in. to allow for bending. Radius of smaller wheel =^x18 = 6in. „ larger „ =fx18=12in. If P is the force on the teeth, 27i:x1 xPx150 = 20x 33000. P=700lb. Torque (T) on the larger wheel =700 x 1 2 = 8400 in.-lb. .-. 8400=j^/(/» ^ 3/8400x16 „„. f d = J ^ „. =2-2 in. \ 40007r Similarly the diameter of the shaft carrying the smaller wheel ' I4 1^- pjtgjj^ of teeth (p) = 0-055\/700=1 -45 in., from (8) 24ir Number of teeth =t— — . = 52. 1-45 Number of teeth in smaller wheel = 26. Assuming 4 arms (p. 188), ., PR 700x12 -,„„,, . M = — = = 21 00 Ib.-m. 4 4 .". ib'^tf =2\00, t = 1 ■45x0-48 = 0-696, c ^6^x0-696x3000 = 2100. ' = Afe 2100x6 2-457 m. or 2i m. •696 X 3000 As the diameter of the shaft is below 4 in., diameter of naye may be 2d or 4-4 in. and thickness = 2-2 -1-1 =1 -1 in. Orfrom(4), T=0-4^/^ + 0-02 = 0-4^(1 -45)2x12 + 0-2 =1 -37 in. Length of nave=1 •8rf = 3-96 in. Or Iength = 3t = 3x1 -37 = 4-11 in. BEVEL WHEELS 195 Bevel wheels. When the axes of two shafts are at right angles to each other, what are known as bevel wheels are used. The proportions for the teeth are the same as for spur wheels, and these are measured along the outside, AB, of the wheel. The Fia. 119. radiating lines meet at the centre O (Fig. 119), which is the point of intersection of the axes of the shafts on which the wheels are placed. Ex. 9. A pair of bevd wheels transmits 50 Jiorse-power. The no. 120. wheel E (Fig. 120) has 60 teeth, 2 in. pitch ; the wheel F has 45 teeth. Find the driving force between the teeth, and deterniine the diameter 196 MACHINE DESIGN of the shaft {'pure torsion only). Shaft E mahes 1 20 rev. per min. Draw sectional elevation and plan. 60 X 2 From PV = 33,000 x 50, where V = — — x 1 20 = 1 200, „ 33000 X 50 ^ „_,^ ,, P= T7i?^ — =1375 lb. 1200 Radius of pitoli circle = — - — =19-1 in, Torque (T) =19-1 x1375. 3/16x19-1 X1375 QOOOtt = 2-458 in. or 2| in. Arms of wheel. As the diameter 'of the wheel does not exceed 4 ft., 4 arms would be used. The section is shown in Fig. 116 (V). PR Bending moment (M) =— . 1375 X19-1 , ,„,, , ,„ „ ^„ ^ .-. 2. =^6^t/=i6'xO-48p../. / 1375x1 9-1 ~'V4x 0-96x3 x6 „., . „— — =3-7 m. 3000 Nave of wheel. Thickness = 0-4 ^/p^ + 0-2 = 0-4v'4x19-1 +0-2=1 -897 iti. or 1 -9 in. Length = 3x1 -897 = 5-691 or 5 -7 in. JOUENALS. Joumals. If the pressure on a bearing is perpendicular to the axis of a shaft, it is called a journal or Journal hearing (Fig. 121) (I). When the direction of the pressure is parallel to the axis of the shaft, the bearing may be a footstep or pivot hearing (II), or a ooUm hearing, Fig. 121 (HI). JOURNALS 197 The proportions of a journal may be as follows : Z=D to I5D, t=lD, Di=ilD to 1|D (Fig. 121 (I) ). riQ. 121. Design of a joumal. In tke design of a journal it is necessary to consider not only the strength, but also the intensity of the pressure on the projected area of the journal, so as to prevent the forcing out of the lubricant between the surfaces. If P denotes the load and p the safe allowable pressure, then we obtain P=prfZ, where d denotes the diameter and I the length of the journal. The pressure on the bearings is frequently obtained by neglecting the weight of the shaft and assuming the value of P to be that due to the power transmitted by the shaft. Ex. 10. A shaft is required to transmit 20 horse-power at 1 50 rev. per min., (a) using a belt pulley 2 ft. 6 in. diameter, (6) spur wheel, pitch circle 3 ft. diameter. Compare the forces on the bearings in the two cases. The wheels are mounted midway between the bea/rings. (a) If Tg and Tj are the tensions in the tight and slack sides respectively, then, as the velocity of the belt is ttx 2-5x1 50 ft. per min., if T^^^.^, (T2 - Ti)7t X 2-5 X 1 50 = 20 X 33,000 ^ _ 20x 33000 "'' =^^~ 2-57TX150" .-. Tjj=1120lb. Ti = lT2 = 560. The force on each bearing is ^ ^ = 840 lb. 198 MACHINE DESIGN (6) If P denotes the force between the teeth, then 7tx3xPx150 = 20x 33,000. .-. P =466-8 lb. Hence force on each bearing is 233-4 lb. Ex. 11. The load on a shaft journal is 1 5 tons and the shaft transmits a torque of 18 ton-inches. Find the size of the journal. The maximum shearing stress must not exceed 3-4 tons fer sq. in. and the maximum load 140 lb. per sq. in. on the projected area of the journal. Assume I =2d. [B.E.] From T=r-/rf^, we have by substitution, = 2-998 in. , 3/l8x16_ Projected area = lxd = 2cl^. :. 2(/2x140=1 -5x2240. -/ 5 X 2240 280 = 3-46 in. or 3iin. The larger size would be used, viz. 3^ in., and the length =7 in. Pedestal or plummer block. The proportions adopted for the various parts of a pedestal or plummer Wock bearing vary in r ffv^ '/^to^^y .,.L..... Unit D+V2 ria. 122. practice. Some fairly average values are given in Fig. 122. In the pedestal, the unit adopted is D +^, where D is the diameter JOURNALS 199 of the bearing. The length I varies from ID to 2D for ordinary speeds, and may be 3D to 6D for high speeds. The proportional unit for the brasses or steps is t=:0-09D +0-1 . The flanges may be of a circular form, as at (I), Fig. 123, or rectangula/r, as at (II). Unit=t=0-09D+0-l" FIO. 123. The inner part fitting the pedestal may be (Ocular, as at (I) and (11), or octagonal, as at (III). The circular form admits of being machined in the lathe ; a stop pin prevents rotation. The octagonal form (III) is usually fitted with chisel and file. Friction. It is an essential condition for the satisfactory working of a journal that the pressure shall not be sufficient to squeeze out the lubricant. In all cases the friction between the surfaces generates heat, which tends to diminish the friction, but makes the lubricant more fluid ; and if the pressure is suffi- cient to squeeze out the lubricant, the friction suddenly increases anS the journal seizes. The ideal condition for a bearing would be that it should run on a film of oil. The position of the steps and brasses depends on the direction of the pressure between the surfaces. Thus, in railway carriage axles, the pressure is in one direction, and no bottom brass is needed ; also the top brass may be cut away at the sides to reduce weight and clinging surfaces. In shaft journals, where the pressure is downwards, the top brass is frequently omitted. In steam turbines, and in high-speed engines by Messrs. Bellis and Co. and other makers, oil is continuously forced into all the bearings. The frictional resistance is given by where (x is the coefficient of friction and P the load forcing the surfaces together. The work absorbed or lost in friction is RxS, where S is the distance moved through per unit time. 200 MACHINE DESIGN Ex. 12. The two journals of a shaft carry a flywhed weighitig 1 tons midway between them. If the coefficient of friction is 0-075 and the speed 70 rev. per min., find the worJc per piin. and the horse- power absorbed in friction. Diameter of journals ^2 in. R=0-075x5x2240lb. Distance moved per min. =7t x1 x70 ft. Work absorbed = 0-075 x 5 x 2240 x tt x 70 =1 84,800 ft.-lb. per min. .". torse-power = 5-6. Ex. 18. A horizontal shaft is subjected to an axial thrust of 6 tons, taken up by a collar. Mean friction diameter is 6 in., width 1 in. measured radially. Coefficient of friction 0-06. Find the work absorbed per min. if the speed is 60 rev. per min. Also find the horse-poioer. , R = 0-06x 6x2240 lb. Distance =n x 0-5 x 60 ft. per min. Work absorbed = 0-06 x 6 x 2240 x 7t x 0-5 x 60 =76,030 ft.-lb. per min. „ 76,032 „ ^ Horse-power = ^-i^—-- =2-3. ^ 33,000 Ex. 14. Find the necessary horse-power to drive a grindstone xoeighing § ton at 1 50 rei). per min., axle 2 in. diameter, coefficient of friction 0-05. TT 0-75x2240x0-05 X 2 XTcx 150 Horse-power = — - — -—---; ■ =0-2. ^ 1 2 X 33,000 The work expended in friction produces a certain amount of heat given by pg Hi=i-4- B.TH.U., J where S is the surface speed in feet per min. and J is Joule's equivalent, or 778 ft.-lb. Thus, in Ex. 13, Hi=-^^5-— =97-7 b.th.u. per min. Increasing the length of a journal diminishes the pressure per unit area, and also its liability to heating ; the heat .effect is not afEected by altering the diameter. FLYWHEELS 201 Flywheels. Flywheels. The turning efiort exerted during a revolution of the crank shaft of an engine is variable ; and the speed of rotation varies at different points of a revolution, assuming the resist- ance to be constant. This variation is prevented by the use of one or more Ajj-wheels. The function of a flywheel is to store the surplus energy at certain parts of a revolution, so that it can be given out again at other parts. In the case of a single cylinder engine, a flywheel prevents stoppage at the dead points {i.e. when the piston rod, connecting rod and crank are all in the same straight line). When, as in certain types of engines, the variation of turning effort is considerable, as in gas engines, two flywheels may be used. (The diameter of a flywheel is usually from 3 to 5 times the stroke of the piston.) The amount of energy stored in a body of mass M lb. moving in a straight line with velocity V ft. per sec. is given by : iyi\/2 K.E. = kinetic energy = — - ft.-lb ,(1) A mass m^ at a distance r-^ from any axis round which it is rotating with linear velocity u^, or angular velocity i - coq). ., 30x2240x121 x27r ., ■'• ^= o o oo o =700x120-2jt. 2x3 x32-2 S X 6 If S denotes the shearing force, — — =700 x 1 20-2t:. „ 700x120-27rx12 ^„„ „„„ ,, .-. S = = 528,800 lb. EXEBCISES XIII. 1. The axes of two shafts are to be as nearly as possible 4 ft. apart and to be connected by spur wheels, pitch of teeth 1^ in. Find numbers of teeth in wheels and exact distance apart of shafts if one shaft makes 1 50 and the other 450 rev. per min. 208 MACHINE DESIGN 2. In Kg. 126 the diagram gives some particulars of the wheel- work of a crane. Radius of handle 1 6 in., diameter of barrel 15 in. Wheel A, 20 teeth, 1 in. pitch; wheel C, 18 teeth, 1^ in. pitch; B, 90 teeth; D, 1 20 teeth. Find the weight raised, neglecting friction, if two men each exert a force of 30 lb. on the handle. Find the tangential force between the teeth of the wheels. 3. A spur wheel, 1 20 teeth, pitch J in., fastened to a shaft making 120 rev. per min., gears with a wheel of 60 teeth. Find the number of revolutions of the smaller wheel, and the distance apart of the axes of the shafts. 4. The tangential force on the teeth of a spur wheel, 50 in. diameter, is 2000 lb. Find the horse-power trans- mitted ; number of revolu- tions 1 20 per minute. Fia. 126. 5. A flywheel on a horizon- tal axis, 1 5 in. diameter, is rotated by the pull of a rope wrapped round the axle. To one end of the rope a weight of 1 lb. is hung, and this falls through a distance of 5 ft. in 1 sec. Find approximately the moment of inertia of the wheel. Diameter of rope ^ in. 6. A flywheel on a horizontal axis is rotated by the pull of a rope wrapped round the axle. A weight of 4 lb. hung from the end of tins rope is just sufficient to overcome friction ; a weight of 1 6 lb. is now added, and falls 1 2 ft. in 2 seconds. Find the moment of inertia. Radius of axle to centre of rope 1 in. 7. Two spur wheels transmit 50 horse-power, velocity ratio 2 to 1 . Speed of smaller wheel 1 00 rev. per min., axes of shafts to be as nearly as possible 3 ft. apart. Find (a) the diameters of the shafts, (6) the pitch and number of teeth in each wheel, EXERCISES XIII 209 (c) distance between the axes of the shafts. Safe shear stress for shafts 9,000 lb. per sq. in. Calculate the dimensions of the remaining parts of the wheels, assuming width of teeth = 2p, depth 0-7p, thickness at pitch line 0-48jo. Safe stress 3000 lb. per sq. in. Draw elevation, end view and plan. Also draw an enlarged detail of teeth, inserting dimensions. 8. A wheel will safely transmit 20 horse-power if the pitch is 2 in. What should be the pitch of a wheel of the same size and materia] running at the same speed to transmit 40 horse-power ? 9. Two bearings, each 6 in. long and 30 in. apart, centre to centre, carry a shaft. Attached to the shaft are a spur wheel with a boss 4 in. wide, and a pinion, boss 5 in. wide. The pinion transmits through the shaft 1 5 horse-power at 60 rev. per min. Calcidate the diameter of the shaft (torsion only). Safe shear stress 1 0,000 lb. per sq. in. Draw elevation, end view and plan. The pitch circles of the wheels to be shown. 10. If the stress on the teeth of a spur wheel is not to exceed 8000 lb. per sq. in., find the safe load when the pitch p is 1 in., thickness of tooth ^p, depth |p, breadth 1|p. [B.E.] 11. What is the kinetic energy in a flywheel revolving at 250 rev. per min., if the wheel loses 5000 ft.-lb. of energy when its speed is reduced to 175 rev. per minute ? • 12. A flywheel is supported on an axle 2| in. diameter and is rotated hj a cord wound round the axle. A weight of 5 lb. on the end of the cord is just sufficient to overcome friction. A load of 25 lb. is attached to 1?he cord, and 3 seconds after starting from rest the weight has descended 5 ft. Find the moment of inertia of the wheel. If the wheel is a circular disc 3 ft. diameter, what is its weight ? 13. A flywheel is required to store 12,000 ft.-lb. of energy, as its speed increases from 98 to 1 02 rev. per min. Find its kinetic energy at 1 02 rev. per min. ; also the moment of inertia of the wheel. 14. A flywheel is required to give out 8000 ft.-lb. of energy, whilst its speed is reduced by 5 per cent, from a mean speed of 1 80 rev. per min. Determine the weight of the wheel if its radius of gyration is 3 ft. 6 in. 210 MACHINE DESIGN 15. A flywheel weighing 5 tons has a radius of gyration of 5 ft. At what speed must it be run if it has to store 475,000 ft.-lb. ? If the speed increases 5 per cent., how much additional energy will be stored ? [U.E.I.] 16. A flywheel is revolving at 1 00 rev. per min. Its store of kinetic energy being 24,000 ft.-lb., what is its moment of inertia ? An accelerating torque of 960 ft.-lb. acts on it for a second. What is now its store of energy ? Neglect friction. [U.E.I.] 17. A fly-wheel increases in speed from 1 68 to 1 70 rev. per min., and absorbs 8000 ft.-lb. of energy, (a) What is the moment of inertia of the flywheel ? (6) How many additional ft.-lb. of energy will it store if its speed increases to 172 rev. per min. ? (c) What is its total store of energy at this speed ? [U.E.I.] 18. In wheels explain how to estimate the pitch of the teeth to transmit a given horse-power with a given speed. Show that under some circumstances the pitch should be proportional to ■y/P, and under other circumstances to P/6, where P is the pressure between two teeth and 6 is breadth of face of wheel. [B.E.] 19. Design a cast-iron spur wheel to satisfy the following conditions : Diameter of pitch circle 40 in., horse-power trans- mitted 50, 100 rev. per min., greatest working stress allowed 3500 lb. per sq. in. Find the tangential pressure on the teeth and pitch of teeth. Full dimensioned working drawings, but only a few teeth to be drawn. All calculations to be shown in full. . [Lond. B.Sc] 20. Two parallel shafts whose axes are apprcjximately 56 in. apart are to be connected by a pair of toothed wheels, the velocity ratio of the shafts to be 6 to 1 . The diametral pitch of the wheels is to be 1^. Determine appropriate numbers for the teeth in each wheel, and exact distance apart of the axes of the shafts. [Lond. B.Sc] 21. The thrust in a shaft is taken by 8 collars, 26 in. external diameter, diameter of shaft between collars being 17 in. • The thrust pressure is 60 lb. per sq. in. Coefficient of friction 0-04, speed of shaft 90 rev. per min. Find horse-power absorbed by friction of thrust bearing. [Lond. B.Sc] CHAPTER VIII. COMPOUND STRESS. STRUTS. Compound Stress. Bending stress combined with a load on beaiing surface. In details such as spindles, pins, gudgeons, etc., the size must be determined not only to allow the piece to be sufficiently strong, but also to permit of a definite load per unit area of bearing surface (pp. 197, 253, 290). Ex. 1. What should he the diameter and length of the centre spindle for a set of four-sheave blocks for a working load of 50 tons ; also, what is the greatest ^pressure allowable per square inch for slow moving surfaces of a similar character, sheaves to have gun-metal bushes ? Safe tensile stress 4 tons per sq. in. p, the allowable pressure on the projected area of the bearing surface for large crane sheaves, is usually 1 500 to 1 800 lb. per sq.in. Hence ^^^^^ . 7 W , , • •• ^=K ^"^ where W denotes the load, p the safe bearing pressure, I the length and d the diameter of the pin. \Nl W^ Bending moment = — - = — -„ from (a). o opo Moment of resistance = ^fd^- 211 212 MACHINE DESIGN Hence ;r— , = 7r^fd^- 8pd 32 -' '4 '32 X W2 Sizpf Substituting the given values, ij 32 >■ X (50 X 2240)2 ^ - =5-87 in. or 5gin. 1 500 X 4 X 2240 Substituting this value of d in (a), , 50 X 2240 , „ ^„ . ••• ^ = T500ir5T87=^2-72xn. Tension or compression combined with bending. When the line of action of a force P is in the axis of a bar, the stress /^ is given by where A denotes the area of the cross-section. When the stress is tensile, the formula is applicable to bars of any length. But when the stress is com- pressive, it only applies to bars of compara- tively short length ; in longer bars failure takes place by bending. In many practical cases, the force P acts in a line parallel to the axis at a distance x from it (Fig. 127). If two equal and opposite forces, each equal no 127 *° ^■' ^® assumed to act along the axis, then, as in Fig. 127, the loading is equivalent to a direct compressive force P in the axis and a couple Px. If M denotes the bending moment, M=P^=/2Z, where /j denotes the stress and z the modulus of the section. Hence the maximum compressive stress is given by ^ ^ ^ P Px ^n x\ ii: COMPOUND STRESS 213 If the section is rectangular, z = ^bifi and A = 6rf, where 6 and d denote the breadth and depth respectively. Hence, by sub- stitution, ^■^ Q^K "- When;f = 0, ^ = . :. e=0. Integrating again. ^. w/Px^ lx« , xi\ , "^-2(2 ^+12)+" y = when x = 0. .'. e' = 0. 232 MACHINE DESIGN Also y is a maximum when x = l. Substituting, ^-sei-seT' ^'^i where W is the total load. Umform beam carrying a central load. Supported at each end, effective span I, central load W (Fig. 134). Bending moment at a distance x from the centre is given by •"U-y ^4->H-® ff ^ ' ' % / -X FIQ. 1S4. Integratmg, ^^ S^ = T (, "2 " 2 j "^ ''' -" = when ;f = 0. .'. c = 0. ax W /Ix^ x\ Integrating, Ely = ^[^- q) + "'■ y = when x = 0. :. c' = 0. The greatest value of y is at x = y,. .'. 0= ; ^ 48EI (4) It will be noted that y is not the actual deflection, but is the height of the centre line of the beam above the deflected centre of the span. The result obtained may also be found from Eq. (2) by con- sidering half the beam, length \l, fixed at the centre of the span, and loaded at the end with load W/2. a 1, r^ ^- • /o\ 5W(1Z)» WZ3 Substituting in (2), y = ^^^ = ^g^j , DEFLECTION OF BEAMS AND GIRDERS 233 Eq. (4) may be written in the form E=-^ (5) Tte deflection due to a central load admits of being easily measured to a fair degree of accuracy, and this enables the value of E to be obtained more easily than by direct tension or compression. Work done during deflection. The work done by a gradually applied weight W in the deflection of a bar through a distance y is Wj//2. The maximum bending moment, if the bar is simply supported at the ends and W applied at the centre, is M=WZ/4, or W=4M/Z. I'rom (4), j, = ^j. Hence we have, for this particular case. Work done =-3^;. Beam supported at each end and loaded uniformly. Length I inches. Uniformly distributed load of w per inch run (Fig. 135). Taking the origin at the centre, and measuring y as before, we have IVI=i ^Sc2fcQQQQQC» ^|<__ , 4-^' Fig. 13^. Integrating, EI— = ^wPx - ^ujx^ + c. — = when x = 0. .'. = 0. dx 234 MACHINE DESIGN Integrating, Ely = -^^uiPx^ - ^uix* + c. y — when x = -; hence, by substitution, we have or if W = Lvl, maximum value of y=-si^'^l^, .: y=^iWP .(6) Reactions of three supports. The deflection of a beam carrying a uniformly distributed load of W lb. is, from (6), gf^WZ^. If a prop be placed at the centre of the beam, the reaction R (Fig. 136) of it depends upon the deflection of the beam ; this reaction is zero when the prop just touches the beam ; it increases as the prop is raised. When the centre and ends of the beam are at the same level, the upward deflection y produced by R is equal to the downward deflection produced by W if R were absent. t~ — r~ — T FIG. 136. 5WZ» rp = 0. (7) •• 384EI 48EI Hence, R = |W Hence the reactions at each end are j^gW. Continuous beams. A beam resting on more than two supports The bending moments and the R, is called a continuous beam. cxicDcrrxrrxxTT x xxx^ FIO. 137. reactions of the supports are determined by means of ciapeyron's theorem of three moments. In Fig. 137, assume the spans Zj, Zj, etc., to carry uniformly distributed loads Wj, Wg, etc. If the PROPPED CANTILEVERS 235 reactions at the supports are Rq, R^, R^, etc., and the bending moments Mq, M,, Mj, etc., then for the first two spans, 4ljMo + 8 (Zi + ^a) Ml + 4I2M2 = VJ^lj^ + Wj.^^ /S) For the second and third spans, and for every two consecutive spans, similar equations are obtained from the theorem. Thus, for a beam resting on three supports, if 1^ = 1^ = 1 and Wi = W2 = W, then from (8), 4ZMo +1 6ZIVI1 + 4ZM2 = 2WZ2. WZ ButMo = IVl2 = 0. .-. 16ZlV!i = 2WZ2 or IVli =— . WZ ButMi=— --RflZ. .-. R„=|W and Ri = 2x|W = |W. Hence, Rj=R2 = |W and Ri = |W. Propped cantdlever. The deflection at the free end of a canti- lever which is carrying a uniformly distributed load W is, by WZ* Eq. 3, given by g^. T- FIO. 138. If a prop be applied at the free end as in Fig. 138, then the reaction P of the prop when the deflection is removed by the prop, i.e. the free end is level with the fixed end, is given by 8EI 3EI ^ Beam with fixed ends, central load. When the ends of a beam are fixed, the bending moments at the ends A and B may be shown to be equal and opposite to that at the centre, and to be WZ/8. The upward reaction at the support is W/2. The vaiue of M at a distance x from the support is given by », WZ W M =— - --;rX. 8 2 cPy _Wl W ^^dx^" 8 2"' 236 MACHINE DESIGN r.,dy Wlx Wx^ Integrating, ^^d}"~8 ^T'^"- -^^=0 when;f = 0. /. e = 0. ax Wlx^ Wx^ Integrating again, BIxy = -j^ — + "'• 1/ = when X -= 0. .'. c' = 0. The deflection has its greatest value when x=-. •• ^=1-92Fl (^) Beam witli fixed ends. Uniformly distributed load. Let iv be the load per inch ran. It may be shown that the bending moments at the walls are wP/l 2 and at the centre u/Z^/24. The upward reaction at the wall is ujIJ2. Bending moment (M) at distance x from one wall is ^p ,^i ^^2s ^y Integratmg, EI-=— ^ {^-—-—j+c. -^ = when ;r = 0. .". c = 0. ax T , . „, tvPx^ Lulx^ uix* , Integratmg, Ely = -^ 12""^^"^"" {/ = when ;f = 0. .". o' = 0. The maximum value of 1/ is at a- = - . Substituting this value for x, wl* wl* wl* 96 96 1 6 X 24 ■• " 384EI 384Er"" : ^^^^ Deflection of slide bars. The slide bars of an engine should be designed for rigidity. The maximum load on the slide bar is obtained as in Ex. 4 (p. 13) ; and when the dimensions of the bar are known, the stress can be calculated as in Ex. 9 (p. 106). The breadth of a slide bar being qonstant, the depth may be uniform or variable, the greatest depth being at the middle of the bar. Slide bars may be treated as rectangular beams, supported at the DEFLECTION OP BEAMS AND GIRDERS 237 ends and loaded at the centre. Although such bars are bolted together rigidly at the ends, it is not safe to consider the ends as fixed. The deflection may be obtained from 1 Q,P ^ 48 EI ■ .(11) where Q = normal thrust on the slide bar, Z= length of span in inches, E = Young's modulus of elasticity=18 xlO^ for cast iron and 28 xlO^ for wrought iron and steel. I = moment of inertia of cross-section. The most important data with regard to cantilevers, beams, etc., so far as design is concerned, are those which give the greatest values of the bending moment, etc. The most important of those in general use are collected here for reference. Table VIII. How Supported and Loaded. Maxiiuum Bending Moment. '^ / -"■ B . \Nl '"^ / ^ wV- wZ -IT or -— 2 2 ^ %r. ^ nnnooooo '"A^ /-- ,A. _@_ ^W- :W .rYYYYYYXi. % WZ 4 -—- or --- 8 8 WZ "8~ WZ 12 WZ 24 at ends at centre Maximum Shearing Force. W vZ or W W 2" wl W -TT or -TT 2 2 W ~2 wl ~2 Maximum Defiection. WZ» 3 EI 8EI WZ3 48EI 5 WZ^ 384 EI 1 WZ^ 192 EI 1 WZ^ 384 EI W= total load, u/=load per inch E = modulus of elasticity, Z=span in run, I = moment of inertia, inches. 238 MACHINE DESIGN Poisson's ratio. When a bar, or rod, is subjected to tension, there is a corresponding lateral contraction and in compression a lateral dilatation ; each, of these has a definite ratio to the longi- tudinal tension or compression. The lateral contraction and dilatation are readily seen by using a band of india-rubber. The ratio of the transverse to the longitudinal, strain is called Folsson's ratio. „ . , ^. transverse strain Potsson s ratto=, -, — =r- — i — ; — ^• longitudinal strain The value of the ratio for most materials varies roughly from 3 to |. More accurate values are as follows : Material. Folsson's Ratio. Material. Poisson's Batio. Cast iron Wrought iron 0-27 0-28 Steel Copper 0-31 0-38 Ex. 1. A bar of mild steel, 1 in. diameter, when subjected to a twisting moment of 2200 in.-lh. is found to twist through an angle of 2-2° in a length of 20 in. The same bar when tested on a span of 20 in. is found to deflect 0-03 in, under a central load of 264 Ih. Find (a) modulus of rigidity (G), (6) modulus of elasticity (E), (c) Poisson's ratio. From the relation G = 10-2TZ where 6 ■■ 2-2 G = exrf*' 57-3 1 0-2 X 2 200x20x57-3 2-2x1* 1-169x10". radian, From (4), p. 232, y=^,, where j/ = 0-03, I = ^. 48EI 264x20^x4x16 <487r X 0-03 = 2-987x10'. Denoting Young's modulus by E and the modulus of rigidity by G, then from the relation Poisson's ratio =- Poisson's ratio : 2G ' (2-987-2-338) X 10' 2-338x10' = 0-2775. DEFLECTION OF BEAMS AND GIRDERS 239 Ex. 2. Find the maximum deflection in the east-iron guide bar of an engine, stroke 2 ft., connecting rod 5 ft. long, width of guide bar 6 in., depth 3 in. Diameter of cylinder 1 5 in., steam pressure 80 lb. per sq. in. , span 'of guide bar 3 ft. Take E = 1 8 x 1 0* R. per sq. in. Tie value of Q, is 1-288 tons (Ex. 4, p. 13), as 1=36 in. " Substituting in (9), we obtain 1 x1 -288x2240 x36»xl 2 '= 48x18x10«x6x33 =°-0115m. Ex. 3. A timber beam, span ^5 ft., cross-section 12 in.y.5 in., supported at each end^ carries a load W at the centre. If the maximum stress in the material is 2000 U). per sq. in., find the magnitude ofVJ ; also the deflection at the centre. E =1 ,500,000 lb. per sq. in. Wx15x12 2000 ^ ,„„ :. = -^r- X 5 X 1 2^. 4 6 .-. W = 5,336 lb. 1 „ „ ^. 5336 X (15x1 2)3x1 2 „^. Deflection = y = —- — , ^ A, ^ ' — -— , = 0-6 in. " 48 X 1 ,500,000 X 5 X 1 2^ Ex. 4. A wooden plank is 1 2 in. wide, 3 in. thick, span 1 8 ft. A man weighing 1 stone stands at the centre carrying a hod of bricks weighing 90 lb. Find (a) the maximum stress due to the load and the weight of the plank, (b) Deflection in the centre. 1 cub. ft. of wood weighs 45 U>. E =1 ,500,000 lb. per sq. in. Weight of plank = ^-^^ x 1 8 x 45 = 202-5 lb. ,, man and hod =1 40 + 90 = 230 lb. 1=3^x12x33 = 27. ,, . 230x18x12 202-5x18x12 Maximum M = ;; 1 = 4 8 = 1 2,420 + 5,467-5 =1 7,887-5 in.-lb. 17,887-5=^x12x32. ^ 17,887-5x6 „„_„!, f= — '—. =993-8 lb. per sq. m. ■' 12x9 ^ ^ . 5 X 202-5 X (18x1 2)3 230 x (18x1 2)3 Deflection- 384x1,500,000x27 "^48x1,500,000x27 = 0-6564 +1 -1 92 in. =1 -848 in. 240 MACHINE DESIGN Ex. 5. A rolled mild steel cantilever projects 6 ft. from a wall and carries a load W at the free end, the deflection being 0-3 in. If the moment of inertia is 31 5 -3 in.-units and E = 30x10^ lb. per sq. in., find the nmgnitude of W. _ W X 72* From (2), p. 231, 0-3 = or W: 3x30x108x315-3' 0-3x3x30x1 0^x31 5-3 72» X 2240 =t0-18 tons. Ex. 6. A pole made of mild steel tvhe, 6 in. external diameter, ^ in. thick, is fi/tmly fixed in the ground, the top being 1 ft. above the ground. A horizontal pull of 2000 lb. is applied at a point 6 ft. from the ground. Find the deflection at the top of the pole. E =1 3,500 tons per sq. in. [Lond. B.Sc] 1= ^ ^^ — ^ =32-94 m.-umts. ' 64 2000x1202(120-^2.) , ooiN y = 7i — .^^^^ — ?;wT^ — s|-?r-, =1 '39 m- (P- 231). '^ 2x13500x2240x32-94 ^^ ' Ex. 7. A rolled steel joist spans an opening of 20 ft. arid carries a load of 1 3,440 lb. at the centre. The joist is 1 4 in. deep overall, flanges 6 in. by 1 in., web ^ in. Neglectitig the weight of the joist, find the deflection at the centre. [E = 30,000,000 lb. per sq. in.] [Lond. B.Sc] I = A(6x1 4*- 5-5 x12») = 579-9 in.-units. 1 3,440 x (20x1 2)3 . u = ^ — =0-22 in " 48 X 30 X 1 08 X 579-9 Ex. 8. Calculate the deflection at the centre of a timber beam 1 2 in. deep and 1 2 in. wide, having a clear span of 20 ft., loaded centrally. Maximum stress 1 000 U>. per sq. in. E =1 ,000,000 lb. per sq. in. [I.C.B.] If W denotes the central load, WZ f^^ . ,., 4x1000x123 ^„„„,, -7-=^6of2. .. W = -- — ———-— =4,800 lb. 4 6 6x20x12 ' I = ^g X 1 2 X 1 2* =1 728 in.-units. 4800 X (20x1 2)3 « = 5 >. — = 0-8 in. " 48x1,000,000x1728 DEFLECTION OF BEAMS AND GIRDERS 241 Ex. 9. A cast-iron water-pipe, 10 in. external diameter, ^ in. thick, rests on supports 40 ft. apart. Calculate the maximum stress (a) when empty, (b)full of water. Find in each case the corresponding deflection. [I.C.E.] Weight of pipe =^(1 0^ - 92) x 40 x 1 2 x 0-26 =1 863 lb „ water=^ x0-752x40x62-3=1101 lb. I = ^(1 0* - 94) =1 68-9 in.-units. n /10*-9*\ - ^ = 32(-f0-)=^3-7«- Maximum M = -—. , , ^ M 1863x40x12 ^"^ ^^^'z = 8x33-78 =3.310 lb. per sq. m. ,,. , 2964x40x12 ^ „„„ ,, (6) /2 = -8->^-33:78~ = 5'266 lb. per sq. m. Assuming E =7500 tons per sq. in. Deflection, when empty = ;-— — ^- ^ ■^ 384 X 7500 X 2240 x 1 68-9 ■ = 0-946 in. r, a J.- til j: x 5 X (40 X 1 2)^ X 2964 Deflection, full of water = — — — — — - — -^-— — -— -— 384 X 7500 X 2240 x 1 68 -9 =1 -5 in. In the design of a joist, beam, or girder, it is usually of the utmost importance to ascertain if the stiffness is sufficient to prevent undue deflection or sag ; especially in the case of joists or girders supporting floors, the sag may result in the cracking of ceilings, etc. According to Tredgold, when a joist has to carry a ceiling on its under surface, the maximum deflection, or sag, must not exceed -^^ inch per foot length of span. Ex. 10. A fir joist, 20 ft. clear span, carries a uniform had, including its own weight, of 2,400 lb. Find suitable dimensions for the joist (a) for strength, (b) for stiffness. Safe worhing stress 242 MACHINE DESIGN 1 ,000 lb. per sq. in. E =1 ,400,000 lb. per sq. in. Defledlion not to exceed g'jj in. per foot of span. ,. ,. 2,400x20x12 ,„„„„■ 11, (a) M = -^ =72,000 m.-lb. O .« f"/^ . L.^ 6x72,000 ^„„ f^--^- .. 6o^= , ' — = 432. 6 1 ,000 432 Assumibg rf = 9 in. , then 6 = — — = 5-33 in. Ol (6) As tie deflection is not to exceed ^ in. per foot, the maxi- mum deflection = 20 xjjj=^ in. Using the value of 6 found above, we have 5 WP ^^SS^^TT' '<^liereI = T'^x5-33of3. ••• " = 7' 5x2,400iK(20x12)»x12x2 384x1,400,000x5-33 =11 -16 in. Hence the dimensions should be 11 -2 in. x 5-33 in. Ex. 11. A rolled' steel joist has to be designed to carry a had of 22 tons per foot run across a span qf^Qft. Selecting a reasonable depth of section, and assuming thickness of flanges 1 in., web f in., determine suitable dimensions for the cross-section if maximum intensity of stress is not to exceed 7 tons per sq. in. IfE=^ 3,400 tons per sq. in., what is the centrai deflection ? [Lond. B.Sc] W =18x2-5 = 45 tons. ,, 45x18x12 ,„,^. ^ M = — =1 21 5 m.-tons. o If a is the area of one flange, then from the approximate method described on p. 112, and taking d to be 24 in., 1 21 5 M=fad. .•. a=- — —=7-233 or 7 J in. 7 x 24 * Hence the flanges will be 7j in. by 1 in., web f in. and depth = 24 in. " Moment of inertia (I) =^13(7-25 x 24^ - 6-5 x 22*) = 2,584 in. -units. r. n .- 5x45xf18x12)3 ^,^. Deflection =y— ——, — — — 7^;^ — = -1 7 in. " 384x13,400x2,584 EXEECISES XVI 243 EXERCISES XVI. 1. A timber beam 6 in. wide, 1 2 in. deep, span 1 2 ft. 6 in., carries a central load of 4,000 lb. and a distributed load of 200 lb. per ft. run. Find the deflection. E=1 -5 xlO^ lb. per sq. in. 2. A beam of uniform section is supported at the ends and loaded in the centre. Calculate the ratio of depth to span in order ,that the deflection may not exceed looot h of the span when the maximum stress is 8,000 lb. per sq. in., E being 28,000,000 lb. per sq. in. 3. In a wrought-iron girder the upper flange is 9 sq. in. area, the lower 8 sq. in., depth of girder 1 ft., span 1 6 ft. Find the mean stress in each flange. The beam carries a uniformly distributed load of 1 ton per foot run. 4. A built-up feteel girder of uniform section carries a uniformly distributed load of 3,000 lb. per foot run over a span of 96 ft. Given that the deflection at the centre is not to exceed g^^ span, find the depth of the girder if the stress is not to exceed 1 0,000 lb. per sq. in. Also find the moment of inertia of the cross-section. 5. A rolled iron joist, 16 in. deep, span 20 ft., is fixed at the ends and has to carry a uniformly distributed load of 1 ton per foot run. The maximum stress is not to exceed 4 tons per sq. in. Find the dimensions and the deflection of the beam when loaded. Draw a cross-section and elevation. Calculate the average shearing stress in the web. E = 29 x 1 0* lb. per sq. in. 6. Find the maximum deflection in a timber beam, 6 in. wide and 1 2 in. deep, supported on two walls 1 2 ft. apart. The loads consist of a central load of 4,000 lb. and a total distributed load of 2,500 lb. E =1 -75 x 1 0* lb. per sq. in. 7. A wooden plank, 1 2 in. wide and 3 in. thick, rests freely on two supports 20 ft. apart. Find (a) maximum stress due to weight of plank, (6) maximum deflection. 1 cub. ft. weighs 46 lb., E =1 ,600,000 lb. per sq. in. [B.E.] 8. A wooden plank, 12 in. wide, 3 in. thick, rests on two supports 20 ft. apart. A man weighing 1 2 stone stands in the middle of this plank carrying a hod of bricks weighing 84 lb. Find (a) maximum stress due to load and weight of plank, (&) the deflection in the centre. 1 cub. ft. of timber weighs 46 lb., E =1 ,600,000 per sq. in. [B.E.] 244 MACHINE DESIGN 9. A beam of uniform rectangular section supported at the ends, carries a uniformly distributed load over a span of 20 ft. Find the depth of the beam so that the deflection may not exceed 0-25 in. under a maximum stress of 8,000 lb. per sq. in. E = 28x1061b. persq. in. [B.E.] 10. A uniform rectangular beam, supported at the ends, carries a uniformly distributed load. If the stress in the wood is 2000 Ibi per sq. in. and E is 1 ,700,000 lb. per sq. in., determine the ratio of depth of beam to span in order that the deflection may not exceed ^^th of span. [B.E.] 11. A rolled joist, flrmly built in at each end, supports a wall over a clear span of 24 ft., the uniformly distributed load including weight of joist is |.ton per foot nin. Moment of inertia of cross- section is 280 inch-units, depth of joist 1 8 in. Find (a) bending moment, (6) stress in flanges, (c) deflection at centre of span. E = 30 x1 QS lb. per sq. in. [B.E.] 12. A rolled steel joist, depth 1 6 in., flanges 6 in. wide, thickness ^ in., web -^ in. thick, is built into two walls 6 ft. apart. At the centre of the joist a steel column rests and transmits a load of 50 tons. Find the maximum stress in the joist and the deflection. E =30 x1 0^ lb. per sq. in. [B.E.] 13. A part of a machine was loaded by the tightening of a Zg in. rod. The distance between two shoulders on the rod was 1 02 in. when unloaded and 0-07 in. greater when loaded. To find the value of E the rod was laid on supports 80 in. apart ; the deflection with a central load of ^ ton was 0-203 in. Find the load on the rod. [B.E.] 14. Find the depth of a plate web girder, span 1 50 ft., ratio of maximum deflection to span ^s&u- Load uniformly distributed, stress in flanges not to exceed 7 tons per sq. in . E = 26 x 1 0* lb. per sq. in. [B.E.] 15. A built-up steel girder of uniform section carries a uniformly distributed load of 3,000 lb. per foot run over a span of 96 ft. If the deflection at the centre of the girder is not to exceed 1 -f 800 of span and stress 1 1 ,000 lb. per sq. in., find the depth and moment of inertia of the cross-section. E = 25 x 1 0^ lb. per sq. in. [Lond. B.Sc] 16. A wooden cross beam 9 in. by 6 in. carries a uniformly distributed load of 0-1 ton per foot run, and rests on two supports EXERCISES XVI 245 18 ft. apart. Find the deflection at the centre. If a post is placed under the centre of the span so that the centre is brought back level with the two ends, find the reaction of the post. E = 600 tons per sq. in. [Lond. B.Sc] 17. Explain what is meant by the stiffness of a beam. A rolled steel joist, depth 20 in., moment of inertia 1 646 inch-units, is supported at the ends, span 30 ft. If the maximum deflection is not to exceed ^^ span, find greatest uniformly distributed load the -joist can safely carry and corresponding intensity of stress due to bending. Take E =1 2,500 tons per sq. in. [Lond. B.Sc] 18. Derive a formula for the deflection of a beam of uniform section due to a central load. Find the deflection of a cast-iron beam 2 in. wide and 3 in. deep, clear span 2 ft. 6 in., central load of 1 ton. E =1 8,000,000 lb. per sq. in. [I.C.E.] CHAPTER X. CHIEFLY ENGINE DETAILS. Cylinder Details. Thickness of cylinder. As already indicated (p. 49), the thickness of metal in castings is in excess of that required for strength. This generalisation applies to cylinders which must be rigid ; and also, in those cases where no liner is used, allows for reboring. The thickness t may be obtained from * = 355o + °-^' (1) where p denotes the steam pressure and d the 'diameter of the cylinder. Ex. 1. Find the thickness of a cylinder if the diameter is 1 5 in., a/nd the pressure of steam 90 lb. per sq. in. 90 X 1 5 From(l), < = -30^ +0-3 = 0-75 in. Cylinder cover. The thickness varies from t + ^ in. for small cylinders to t + | in. for larger cylinders {t denotes the thickness of the cylinder). To diminish the clearance space the inner surface of the cover is made to correspond with the contour of the piston. Cylinder flanges. The thickness of a cylinder flange, which should not be less than the bolt diameter, is usually equal to it, or the thickness may be from 1 •2t to ^ St. The distance from the centre of the bolt to the edge of the flange should not be less than 1 -Srf (where d denotes the diameter of the bolts) ; this gives a width of flange from 3^ to 3^d ; if studs are used, 3^d will be sufficient. 246 CYLINDER DETAILS 247 Bolts or studs in cylinder cover. These should be of such a size and number as to keep the joint steam tight. If p denotes the steam pressure, D the diameter of the cylinder, rf^ the diameter of the bolt at the bottom of thread and n the number of bolts or studs, then ld,^xnf = lD^p. -Vfr ■(2) / is usually 3,000 lb. per sq. in., but should not exceed 6,000 lb. per sq. in. The pitch of the bolts should not exceed 5d (where (/= diameter of bolt). Studs are preferred to bolts, for the absence of bolt heads allows the flanges to be made narrower ; also bolt heads interfere with the lagging round the cylinder. Ex. 2. The pressure of steam in a cylinder 20 in. diameter is 200 lb. per sq. in. The cylinder cover is secured byM bolts. Find the diameter of the bolts. [B.E.] Substituting in (2), <'i = 20y^-y|^^=1 -252 in. From Table III. (p. 74), rf =1 1 in. Ex. S. The cover of a cylinder 20| in. diameter, subjected to a pressure of^ 20 Tb. per sq. in., is secured by | in. bolts. If the stress in the bolts is not to exceed 5000 Ih. per sq. in., how many bolts will be required ? What unll be the pitch of the bolts if the diameter of the bolt circle is 22 in. ? Prom Table III., rfi = 0-622 in. If n denotes the number of bolts, we obtain n X ^(0-622)2 X 5000 =^(20-75)2 x 1 20. /j=26-7 or 27. 22tu Pitch of bolts =^ = 2-56 or 2^gm. In locomotive practice g in. bolts are chiefly used, and in marine engines 1 in. to 1 J in. bolts. The pitch of the bolts for high pressure cylinders is usually 4 to 5 diameters apart and for locomotive cylinders 6 diameters apart. 248 MACHINE DESIGN Crank Details. Proportions of cranks, cast-iron. Cast-iron cranks are usually stifiened by means, either of a single web as at (I) and (II), T FlQ. 139. Fig. 139, or of a double web (III) ; a section of tlie former at AB is shown in (IV) and of the latter at CD is shown in (V). Values of D and d are obtained by calculation, and the remaining pro- portions may be as in Fig. 139. Width of arm 1-6 D and 1-8 rf respectively. Ex. 4. Draw sectional elevation, end view and plan of a cast-iron crank ; diameter of crank shaft 5 in., crank fin 3 in., throw of crank 1 ft. 4 in. Wrought-iron overhung crank. The proportions of an overhung- wrought-iron crank are given in Fig. 140. The proportional units are D, the diameter of the crank shaft, and of, the diameter of the crank pin. As in the cast-iron crank the arm or web is made taper, the width being measured at the centre of the shaft and at the centre of the crank pin respectively. The part of the crank pin inside the arm of the crank may be parallel as at (I), or tapered as at (II). In the former case the crank pin is made slightly larger than the bore of the crank arm, and the crank arm is either shrunk on and riveted, or the pin is forced in by hydraulic pressure. In the latter case the pressure exerted is about 1 to 1 2 tons per inch diameter. CRANK DETAILS 249 When tke part of the pin inside tie crank arm is made taper, it may be fastened by means of a nut screwed on to the end as shown, or by a cotter, as in Fig. 145. These methods have the advantage thalt the pin can be easily removed for renewal or repairs. ■(h2d^ Ex. 5. Calculate the dimensions of the various parts of a wrought- iron, or miU steel, crank. Crank shaft 7 in. diameter, crank pin 4 in. diameter. Draw elevation, end view and plan. Scale ^ in.='\ ft. Locomotive and marine cranks. A cranked axle for a loco- motive with inside cylinders is shown in Fig. 141. Only one-half Fig. 141. is shown, the remaining dimensions are the same as those given. The cranks are at right angles to each other, as shown. in the end view. The dimensions vary in actual practice, but the proportions given (in terms of d, the diameter of the crank pin) will be found 250 MACHINE DESIGN to give fairly average dimensions. This form of cranked shaft is also largely used for marine engines ; but for the engines for comparatively large ships, it is somewhat untrustworthy, and what are known as built-up crank shafts are usually employed. Ex. 6. The diameter of the crank pin in a locomotive cranked axle is 8 in., radius of arank 1 2 in. Find the remaining dimensions, and draw two views as in Fig. 141 ; also d/ram a 'plan. Built-up crank shafts. Built-up crank shafts are used in comparatively large marine engines. The shaft, crank arms and ^•25 FIO. 142. crank pin are made separately. The crank arms are shrunk on to the shaft and the crank pin, precautions being taken that the keyways are kept in line with each other (Fig. 142). As an additional security, the arms are secured to the shaft by sunk keys — ^these may be either circular or rectangular in cross-section. Sometimes the arms (or cheelss) are forged in one piece with the shaft, and the pin shrunk in. The difierence between the dia- meter of the hole in the arm and the diameter of the shaft, so that the stress induced during cooling is a definite amount, can be ascertained as in p. 30. In Fig. 142 the proportional unit is D. Ex. 7. The diameter of the crank pin of a buiU-up crank is 23 in. Calculate the remaining dimensions of the crank and crank shaft. Find the diameter of the 8 bolts in the coupling if the stress CRANK DETAILS 251 in the bolts is the same as in the shaft, 23 in. diameter. Draw two views as in Fig. 142. Scale 1 in. to afoot. The various dimensions can be obtained from the proportional numbers in Fig. 142. Radius of bolt circle =1 -5 x 23 ^ 2 =1 7-25. ^rf2 X 8 X 1 7-25 xf, =^f, X 233- rf = 4-69 or 4|in. *3"->- FlO. 143. u© 23" > ^ Section at A B In some cases the arms of the crank are bolted to flanges formed on the two parts of the shaft, these flanges being sunk into the arms (Fig. 143). Ex. 8 Draw two views of the built-up crank shaft. Calculate the diameter of the eight bolts as in Ex. 7. Scale ^full size. Ex. 9. Draw elevation, end view and flan of the marine engine crank shaft (Fig. 144). Distance between crank arms 1 ft. 2f in. Each arm is 9^ in. thick. Total length of shaft over flanges = 8 ft. 3 in. Each flange has six 3 in. diameter holes on a circle 1 ft. 1 05 in. diameter. The pin acting as key for the crank pin . is 1 5 in. diameter x 8 in. long, and that for the shaft 2^ in. diam. 252 MACHINE DESIGN xSin.hng. Scale 1 in. =^ ft. The crank shaft is one of three for a triple expansion engine, 2,300 horse-po'wer. Show clearly the angular positions of the three crank pins. [L.C.U.] ._(s? ^|Crankpm"C" «- 2'9%"' •■ Shaft "S" ^^^ — j;^ ;^^il__-2'5ri'L_^U-95^, Crank pin. The dimensions of a crank pin may be determined (a) from consideration of its strength to resist bending action, (b) to satisfy a given allowable pressure on the projected area, or the dimensions may be obtained to satisfy both (a) and (6). The allowable bearing pressure varies from 300 lb. per sq. in. for small high- speed engines, to 900 lb. per sq. in. for large low^speed engines. If d denotes the diameter and Z the length of the journal • • (mm - — Bio. 145. (Fig. 145), P the tangential force on the crank pin, then the maximum bending moment M is given by M =iPZ. CRANK PINS 253 When both ends of the crank pin are fixed, as in marine and locomotive engines, the maximum bending moment IVI =PZ/1 2 ; there is also twisting moment to be taken into account. The size of the crank pin is usually estimated from the safe bearing pressure. If p denotes the safe bearing pressure on the proiected area dl (Fig. 142), then pdl=P, or l=— (3) pa • ^ Substituting this value for I in the expression for M, p2 2pd p2 ^ Hence, — = — yy*. 2pd sa-* -7^ 6P2 w (^) Having obtained the value of d, the length of the journal I can be found from (3). / may be taken as 8,000 lb. per sq. in. The value of p lies between 400 and 700 lb. per sq. in. The length I varies from 1 to 1 Jrf ; frequently Z=1 Jrf. If the size of the crank pin were determined by consideration only of its strength, then, from weobtain ^ = ^1^ (5) Ex. 10. A cylinder is 17 in. diameter, stroke 32 in., boiler fressure 1 00 lb. fer sq. in. Assuming the tangential force on the crank fin is equal to the maximum had on the piston, find the diameter of the crank pin. /= 8,000 lb. per sq. in., p = 600 lb. per sq. in. # TT p=-x172x100 = 22,700lb. 1? tA\ , "/lex (22,700)2 From (4), rf = a/ ^ — ^ = 4-8 m. ^ ' AfTtx 8,000x600 ^°"'- 22,700 254 MACHINE DESIGN If -the length of the pin calculated in this manner is found to be too long, then d must be increased. Thus, assuming l=^ ■2d, then, from (3), px 1.2^2 = 22,700. <'=Vt 22,700 = 5-6 in. •2 X 600 Z = 5-6x1 ■2 = 6-7 in. Ex.'ll. On the crank fin of an overhung cranJc, the total force is 1 3,000 lb., and a pressii/re of 600 1h. per sq. in. is to be allowed. The maximum shin stress is not to exceed 5,000 lb. per sq. in. Find the dimensions of the pin. Discuss briefly somi of the factors which determine the choice of a suitable bearing pressure in journals. [Lond. B.Sc] If d denotes the diameter of the crank pjn, then, from (4), where P =1 3,000, /= 5,000 and p = 600, 1 6 X (1 3,000)2 = 4-1 in. ' 71 X 5,000 X 600 Combined bending and twisting. It is only in comparatively exceptional cases that a shaft is subjected to pure torsion ; the stress is usually of a more complicated character. Thus, the tangential force on a crank pin produces combined twisting and bending action in the crank shaft. In a similar manner, the tangential force on the teeth of a wheel produces bending in addition to twisting moment. "When a wheel is situated close to a bearing, the former may be negligible in com- parison with the latter ; but this is not the case when the wheel is at some appreciable distance from a support ; it is then necessary, in calculating the diameter of a shaft, to consider both bending and twist- ing moment. If M denotes the bending moment, and T the twisting moment at any section of a shaft, the stress produced may be obtained as if due to a simple equivalent twisting moment T^, the value of T^ being given by T^^M+^W+V, (1) FIG. 146. COMBINED BENDING AND TORSION 255 When M and T are known the value of Te can be calculated, and d, the diameter of the shaft, from where / is the maximum allowable stress. The value of Te may also be found graphically as follows : Draw two lines EF, EG at right angles, and to any convenient scale, so that EF = T and EG = IV1; then GF = s/m^ + tT With centre G, radius GE, describe an arc intersecting FG produced in H. .-. FH=M+Vm2 + T2 (Fig. 146). Ex. 12. A solid circular shaft is subjected to a twisting moment 0/160 in.-tons and a bending moment 6/100 in.-tons. Find its diameter. Stress in material not to exceed 8,000 Ih. per sq. in. [Lond. B.Sc] Substituting the given values in (6), Te = 100 + v/l002+l602 = 288-7 in.-tons. d = ^' 28 8-7x2,240x16 . 8,000X71 ~ ^'^' Ex. 13. 50 horse-power is transmitted from a shaft A to shaft C through an intermediate shaft B {Fig. 147). The wheel on shaft B, TIG. 147. 50 in. diameter, runs at 45 rev. per min. The distance bettveen the centres of the shaft bearings is 40 in. Determine the maximum bending and twisting moment on shaft B. Find the diameter of shaft B. Safe stress 9,000 Tb. per sq. in. From the relation H -j^^^^ (p. 130), 256 MACHINE DESIGN where H denotes the horse-power transmitted, N the number of revolutions per mimite and T the torque in in. -lb., ^ 50x12x33,000 ^„ „„„ . ,, T = — ^ =70,030 in.-lb. 2-KX45 If P =the tangential force on the teeth of wheel on shaft B, then „ 70,030 „„„„. ,, P = — ^^— = 2,802 m.-lb. M = -^-— X 20 = 28,020 in.-lb. Te = 28,020 + v/28,0202 + 70,030^ = 1 03,450 in .-lb.' If d denotes the diameter of the shaft, -^ 03.450x16 „„„. -,plied Mechanics for Engineers (Macmillan), where detailed information of the principles of Compound Stresses will be found. 264 MACHINE DESIGN Bending moment (M) =1 9,000 x1 5. J2M 32X19,000X15 ^ 1^ ^ ^Jtx7* 7tx7^ Hence, substituting ttese values in (9), the maximum stress is given by g^g^ ^ a/(8463)2 + 4 x (451 3)^ " = 2— • =1 0,41 6 lb. per sq. in. The angle made by the principal stress q m&j be obtained ''"^'^' •••*-^=^^' .■.« = e6=34'. The angle made by the direction of the principal stress q with the axis of the shaft is the complement of 6 or 23° 26'. Ex. 17. A crank shaft is subjected to a bending moment of 1 0,000 in.-lb. and a twisting moment of 20,000 in.-lh. Find the diameter ; the maximum principal stress is not to exceed 9000 lb. per sq. in. From (1), Te = IVH-V'M2 + T2. M=iT. .'. T, = lT+^/l' + T2=1-618T. "/l 6x1 -61 8x20,000 V 9000 X TT = 2-64in. Combined bending, torsion and axial force. When there is in addition to simultaneous bending and twisting an axial force of thrust or pull, the stress due to the axial force must be added algebraically to the bending stress, before proceeding to find the principal stresses. Ex. 18. A propeller shaft transmits a twisting moment o/ 500,000 ft.-lb., a simultaneous bending moment of 90,000 ft.-lb., and an axial thrust of 24 tons. The shaft is 15 in. external and 8 in. internal diameter. Determine the maximum principal stress. Modulus of section for torsion is 7t/154-8«\ „„„„ — ( | = 609'2. leV 15 / CRANK ARMS 265 Maximum torsional shear stress is ' , 500,000x12 „„^^ „ /= gQgTg — =9'851 lb. per sq. m. Maximum bending stress 90.000x12x2 '' " 609-2 ^ ^'^^^ • P®^ ®1- ^^• Compression due to thrust is 24 X 2240 , = 425-1 lb. per sq. m. |(152-82) Hence direct axial compression is 3546 + 425-1 =3,971 lb. per sq. in. Maximum principal stress = I/+ yj^ +f? =1 985 +1 000VJ(1 -985)2 + (9-851)2 =11,885 lb. per sq. in. Crank arm. The proportions given on pp. 248-250 wiU be found to give ^ood average values, or the values may be calculated by equating the resistance of the arm to bending to the resist- ance of the shaft to twisting. Thus, if h denotes the width and 6 the thickness of the arm, then we obtain where d denotes the diameter of the shaft. From this equation, when the value of 6 or A is known or assumed, the value of the remaining term can be obtained. Ex. 19. A crank (Fig. 153) is acted on by a force P 0/30,000 Ih. Determine the dimensions I and h in terms of d, the diameter of the crank shaft, so that the crank arm, key arid shaft may he equally strong, allowing the stress in the material to he the same for each fo/rt. Width of key Id. [B.E.] btfi iPR=-^/*,where6=|A.. 6- 3A^ f X 30,000x1 6 =—/«. 266 MACHINE DESIGN If /e = 9000 lb. per sq. in., h = 8-28 in. JIG. 153. Shearing resistance of key = Z x ~ x -/. 4 Z' f'-fa"/- ?=14-1 in. Eccentrics. Eccentrics. An eccentric may be described as a modification of a crank in which the crank pin is so large that it envelopes the crank shaft (Fig. 154). It is obvious that a crank can only be used at the end of a shaft to allow the connecting rod to swing to and fro in its path ; but an eccentric allows a recipro- cating motion (such as that required for a slide valve) to be obtained at any convenient place on FiQ. 154. the crank shaft. ECCENTRICS 267 An eccentric is sometimes used for small pumps and also in tte " back centre " of a lathe, in whicli case the throw, or radius, is slightly greater than the depth of the teeth. The eccentric for a steam engine consists of two parts, the sheave and the strap. FIQ 155. FlQ. 156. If the eccentric sheave can be passed on to its place over the eijd of the shaft, it is made in one piece ; if not, and this is the usual arrangement, it is made in two parts. The smaller part of the sheave is made solid and may be of wrought iron or steel. In the larger part the rim is usually connected to the boss by an arm, as in Fig. 155 ; in larger sizes, two or more arms are used. The sheave is secured to the shaft by 268 IVIACHINE DESIGN a key and by one or more set-screws ; these latter also enable the eccentric to be fixed temporarily on the shaft until the position of the key-way in the shaft is determined. *- B — * The two parts of ah eccentric may be fastened by means of studs and nuts as in Kg. 155 ; or by bolts and nuts as in Fig. 156. Various forms of the rim have been used ; those in general use are shown at (I) and (II) (Fig. 157). If a is the area of the back of the slide valve and p the pressure of the steam on it, then the width (B) of the sheave is given by B = O^O^Q^/pa. for locomotives, (1) B = 0-01 Vpa for marine engines. (2) The proportions for the remaining parts may be as follows : A = |, E = F = 0-45B, t = 0-12B. One form of eccentric strap is shown in Fig. 156. The best material for the strap is cast iron, and when it works upon a cast-iron sheave, no lining is required. In small eccentricities the strap is frequently made of brass or bronze. Other materials used for the strap are malleable iron, cast steel and wrought iron. The thickness T may be 0-8B for brass and cast iron, and 0-6B for wrought iron or steel. Slide Valve. Slide valve. The slide valve in its simplest form consists of a box-shaped casting as in Fig. 158. The plan of the steam chest with the slide removed is shown at B. This contains three passages or ports, shown in plan and in the sectional elevation ; the two outer marked S lead to the ends of the cylinder and are called steain ports, the inner or exhaust port, marked E, allows the steam to escape either to the condenser or to the atmosphere. The surface AA of the slide valve is made to slide forwards and backwards over the surface BB, and both surfaces are accurately planed. If the slide valve, when in its middle position, only covers the steam ports, the eccentric CB (Fig. 159) must be placed 90° in advance of the crank CA ; it will be found that by this SLIDE VALVE 269 arrangement one end of the cylinder -will be open to steam and the other end open to exhaust during the whole stroke of the piston. 270 MACHINE DESIGN The provision of outside lap (Fig. 158 ) allows the supply of steam to be shut off at any convenient part of the stroke. Usually the valve is open a' small amount when the crank is at a dead point pA ; this opening is called the lead of the valve. The outside lap and lead necessitate a change in the position of the eccentric ; the position is indicated by CB (Fig. 160), and is obtained as follows: Mark ofi CE= lap + lead. Draw EB at right angles to AA', meeting the circle at B ; join B to C, then CB is the position of the eccentric when the crank is at CA. The direction of motion of the crank is assumed to be in the direction of the hands of Fig. 159. a clock. The position of the eccentric for motion in the oppo- site direction, or anti-clockwise motion, is denoted by CB'. The angle BCD or B'CD' is the angular advance of the eccentric. If p denotes the half-travel of the valve, then, if CA = p, . fl lap + lead sm 6 = —^ . P The lead is usually from j^p in slow, to ^p in fast running engines. The four important points in the stroke of a piston are : (1) Admission of steam, (2) steam cut off, (3) release of steam to exhaust, (4) exhaust closes, compressing the steam left in the cylinder. The points in the case of a simple valve are con- veniently determined by the Zeuner diagram. For this purpose the line CB is rotated backwards through an angle 2d. With a SLIDE VALVE 271 Release radius CA equal to one-half the travel of the valve, describe a circle (Fig. 161). Produce BC to B', and on the line BB' describe two circles. Make CM equa] to the outside lap and CN D equal to the inside lap ; then the four points a, b, c, d are determined, (a) the point at. which steam is admitted, (b) steam is cut ojEE, (c) release (exhaust opens), {d) compression (exhaust closes). The effect of inside lap is to close the exhaust port earlier, thus compressing, or cushioning, the en- closed steam to a greater extent. Ex. 20. The travel of a slide vahe is 4 in., lead 5 in. Cut off takes place at three-fourths stroke. Draw Zeuner's diagram and measure the outside lap and angular advance. Draw a circle of 2 in. radius (Fig. 162). Make AQ = |AA'. Draw QB perpendicular to AA' intersecting the circle at B. At A a circle riG. 162. 5 in. radius is drawn and a line tangential to the circle and ter- minating at B is shown by the line PB ; then DCB' is the angular advance and CM the required outside lap. These are found to be 35° and 0'9 in. respectively. 272 MACHINE DESIGN Ex. 21. Design a valve rod, eccentric sheave and strap for operating a slide valve. The net area of the valve face on which the steam pressure acts may be taJcen as a rectangle 1 in. by 8 in. ; steam pressure 140 lb. per sq. in.; coefficient of friction 0-2. Safe stress for rod 4000 lb. per sq. in. Safe bearing pressure 70 Jh. per sq. in. Maximum pull in valve fod =140x80x0-2 = 2240 lb. ; if rfj denotes the diameter at the bottom of thread, ^rfj^x 4000 =2240. 4 '- I 90 An = 0-844 in. .•. (/=1 in. From (1), 8 = 0-018^140x80=1 -9 in. or 1|| in. If d denotes the diameter of the sheave, then, as the safe bearing pressure is 70 lb. per sq. in., 2240 Bxrf = - /U 994.n =16-84 in. 70x1-9 T = 0-88=1-52 orlySgin. The remaining dimensions are obtained from the values given on p. 268. ^ Strap uoits. Assuming a safe stress of 3000 lb. per sq. in., 7t = 4" 2 X -rfj2 X 3000 = 2240. , 2240 di= ^ I- = 0-689 m. ^ X 6000 From Table III. (p. 74), rf = | in. .'. the two bolts in the sheave would be ^ in. CHAPTER XI. ENGINE DETAILS {Continued). Piston Rods. Piston rods. A piston rod is subjected to rapid alternations of stress, tensile and compressive, in addition to other straining actions ; hence, the value of the safe stress / is comparatively low, and is best determined from existing piston rods in stationary engines, warships, and the mercantile marine. An examination of a number of cases gives a value of 4000 lb. per sq. in. for the two former and 3000 lb. per sq. in. for the mercantile marine. The design of the piston rod may be considered in three parts : (ffl) Attachment to piston ; (6) Body of rod ; (c) Attachment of connecting rod. A fairly good result for the diameters of rods of mode- rate lengths is obtained by considering the rod as subjected to tension only, neglecting the efEects of alternate tensile and compressive stresses and using a comparatively large factor of safety. If D denotes the diameter of the cylinder, p the initial pressure of steam, d the diameter of the rod and / the safe stress. '-4 •(1) Ex. 32. Diameter of a stationary cylinder is 18 in., stroke 34 in., initial pressure of steam 1 50 lb. per sq. in. Find the diameter of the piston rod. From(l), rf = D.y/^=18^^ 150 / ■""V4000" or < 273 = 3-48 or 31 in 274 MACHINE DESIGN Long rods. A comparatively large number of rules, or formulae, has been proposed to obtain the diameter of a long piston, or connecting rod {i.e. when the length of the rod exceeds twenty times its diameter). These are usually either Euler's, or some modification of Eankine's formula. The diameter d may be found from one of the following : P='5' (2) P=— ^ : (3) ■© (/ = 0-0158Dyp (4) ■Kd* For mild steel £ = 29x10", I = -^x= /=21 tons, A = area of 64 cross-section, c = 30,^00 ' ^= length and /r = radius of gyration. D is the diameter of cylinder and p the steam pressure. In (2) and (3) P denotes the bucHing load. This must be divided by a factor of safety, usually 8, since the stress is alternately com- pressive and tensile, in a piston rod or connecting rod. Piston rods are usually made of mild steel ; this material is superior in every respect to wrought iron, and is practically always used ia good work. Ex. 23. A steel piston rod is 4 in. diameter and 5 ft. long. ' Diameter of cylinder 26 in., 48 in. stroke, boiler pressure 90 lb. per sq. in. Find the buckling load and factor of safety. If W denotes the budding load, then, from (3), 21 X ^ X 42 ^ =, — 1 60^x16 =r5 = ^ ^^-3 ^^■ ^+7500'' 4 ■Di i. t i ^ 84re X 2240 Factor of safety = = 8-4. 1 -48 X 90 X ^ X 26^ 4 Ex. 2A. Determine the diameter of a piston rod for a cylinder 60 in. diameter, in which the greatest difference of steam pressure on the two sides of the piston may be assumed to be 28 lb. per sq. in. The rod PISTON KODS 275 is made of mild steel secured to the piston by a tapered end and nwt, and to the crosshead by a cotter. Make a drawing of the rod, inserting dimensions. Length of rod 9 ft. Q in. Scale r^-^ full size. E = 29 X 1 0® lb. per sq. in. Factor of safety 8. Load on pistoii = -x 60^x28, Z=n4iii. Substituting in (2). 7t „ „„ 71^ X 29 X 1 0^ X TTrf* . .-. -X 60^x28= — 5— rr:7o— ST- • 4 8 X 1 1 4^ X 64 ix64 */ 60^x7x8xn4^: V 7t2x29x10« I0« =4-92 in. Or, we may use the Kankine formula (3), thus : ^ X 60^ X 28 X 8 P= j^tttt; = 282-9 tons. 2240 ' 282-9= . . , _____., since /f^=r^. 114^x1 6\ 16 / 1 1142 x16\ ^+(7500^— ST-j TC /. 282-9{(7500rf2) + (1142x16)}=21 x~c/2x750D. rf4 - 1 7-1 5(/2 + (8-575)2 = 475. .■. rf = 5-6in. Or, the result given by (4) is rf = 0-01 58 X 60-^/28 = 5 in. Piston Eod Nuts and Collars. Nuts. Large piston nuts (Fig. 163) may be secured by : (a) Keep plate secured by tap-bolts or by square-necked studs. (6) Cotter recessed into grooves in nut. Thickness of keep plate J in. to | in. according to size of nut (shown at k, Fig. 163) ; tap-bolts or studs | in. to f in. diameter. 276 MACHINE DESIGN When a nut is fitted with a pia or a cotter there are usually 1 2 slots ; depth | in. to ^ in., thickness of cotter rfj/l 2. Fia. 163. If d denotes the diameter of the piston rod (Pig. 164), then rf^ is the diameter calculated for the screwed end. Collar. This is made a good fit in the piston, diameter =(/ + ! in., thickness 0-125(/ to 0-1rf. ; radii ■f'^ in. to | in. are worked at the comers. The collar is also fitted outside, clear of piston. The blunt cone shape at the end of the rod renders it easy to disconnect, PISTON ROD NUTS AND COLLARS 277 when this is necessary for renewal or repair. The taper varies from 1 in 20 to 1 in 4. , sd?z^ ^ Fig. 164. The diameter of the screwed end should be such that the stress does not exceed 5,000 lb. per aq. in. for wrought iron, and 6,000 lb. per sq. in. for steel at the bottom of the thread, when a nut is used, or, through the cotter way, if cotters are fitted. Thickness of nut = frfi to rf^ ; the nut is usually made of wrought iron. 278 MACHINE DESIGN Ex 25. The forged steel piston for a marine engvne isM m. diameter, steam pressure 1 25 Ih. per sq. in. (a) Find the diameter of the rod, safe stress 4000 Ih per sq. in. (6) Find the diameter of the screwed end of the rod, and give a dimensioned sJcetch, of the nut (Fig. 164). (a) From (9), VI 25 4000 = ^ ^'^^ (6) If rfj is the diameter of tte screwed end, -rf 2x6000=^x1 72x1 25. 41 •• ''i=^^V6000 = 2-45 or 2^ in Crosshead and Slide Bars. Crosshead and slide, or guide, bars. The crosshead of an engine connects the piston rod to the connecting rod. The term also A R ^ ^ / !ll "n> p ' includes that part which slides on the guide, or slide bars. The effort on the piston is transmitted through the piston rod to the connecting rpd, and through the connecting rod to the crank pin. When the crank is on a dead centre, and piston rod, connecting rod, and crank are in the same straight line, the efiort on all three is the same, neglecting inertia. As the crank rotates, the connecting rod assumes an inclined position, as in Fig. 165. The angle 6 increases until the crank is at right angles to the centre line of the cylinder ; then it diminishes to zero. CROSSHEAD AND SLIDE BARS 279 Assuming the engine to rotate in the direction shown (i.e. in the direction of the hands of a clock), the piston and connecting rods are in compression ; the force in the connecting rod may be resolved into two component forces, one horizontal in the direc- tion of the piston rod, the other vertical and downwards. On the back- or in-stroke of the piston, the piston and connecting rods are in tension, and the force is again downwards. Hence, in engines running only in one direction, it is necessary to provide only one guide. Upper slides, or guide bars, are usually provided to allow for any accidental upward pressure, such as might occur when the valve is set to give considerable compression. The force on the slide for aU practical purposes has its greatest value when the crank is at right angles to line of centres, as in Fig. 165. The triangle of forces for three forces acting at B would have its sides parallel to the. sides of the triangle BCA. Hence, if BC to some convenient scale represents the load P on the piston, then the force R in the connecting rod is represented by BA and the force Q, on the guide bar by AC. •(1) where I denotes the length of the connecting rod and *- the radius of the crank. If Z = 5/-, then P=V24" ••• ^ = ^^ (2) If A denotes the area of the slide block, p the safe allowable pressure, /-> ^=? (^) The value of Q, is approximately P -^ II r, and this is frequently used instead of the more accurate value. The value of p varies from 40 to 500 lb. per sq. in., but values from 40 to 1 00 lb. are those generally used ; pressures greater than these require special provision for efficient lubrication. In locomotive engines the pressure varies from 40 to 60 lb. Ex. 26. Find the area of a guide bar for a horizontal engine, cylinder 12 in. diameter, 24 in. stroke, connecting rod 60 in. 280 MACHINE DESIGN long, boiler ^pressure 90 lb. per sq. in. Safe pressure on guide 50 i per sq. in. P=^x122x90. 4 — r I / if "' j7 ■ I . t I — ■ I 1 A \\ Substituting in (2), q and from (3), FIG. 166. -x1 2^x90 4 V24 7t x1 2^x90 /24 X 50 = 41 -56 sq. in. When two slide blocks are used, tbe combined area must be equal to that obtained, or the area of each would require to be 20-78 sq. in. A simple form of crosshead is shown in Fig. 166. It is of the CROSSHEAD AND SLIDE BARS 281 centre-slipper type, and slides on a planed bed on tte bed-plate of the engine ; there are cover plates provided on each side. These prevent any tendency in the crosshead to lift when the piston is at the ends of the stroke. 4 toB FIQ. 167. A double-slipper crosshead is shown in Fig. 167. The forked end is of wrought iron, the pin of mild steel and the slippers of cast iron, having comparatively large bearing surfaces. This type is much used for large mill engiaes. The proportional unit for the various dimensions is rf, the diameter of the gudgeon pin. Diameter of crosshead pin, or gudgeon. If R denotes the maximum force in the connecting rod, it may be taken as a load uniformly distributed over the crosshead pin. Rl Bending moment = M = — , o where I is the length of the pin. 282 MACHINE DESIGN TC Moment of resistance = —/rf^- •■ S 32-' ' which determines the strength of the pia. Bearing pressure on projected area =prfZ. .'. pdl = R, or I — —,- pd The most important factor in determining the diameter and length of the pin is the allowable pressure p per square inch on the rubbing surfaces. When pins are thus proportioned, they are usually sufficiently strong to resist the bending and shearing stresses. The velocity of the rubbing, surface on a crosshead pin is less than that on the crank pin ; hence a higher pressure can be allowed on the former. The average value of p from a number of existing cases lies between 1 200 and 1 600 lb. per sq. in. The length of the pin varies from 1 ^d to 2^d ; frequently I = 2d. Hi Substituting the value of Z in — , , o ■ ■ 8pd 32'' ■ ■■■ '-m <*) The value of / may be taken as 9,000 for wrought iron and 1 2,000 lb. per sq. in. for steel. It will be seen from Fig. 165 that if P is the Load on the piston, R=^xP, (5) where I and r denote length of connecting rod and radius of crank respectively. Crosshead cotter. In small engines the piston rod is often screwed into the boss of the crosshead and secured by a check nut. The most general method is to use a cotter, the rod usually being made a taper fit in the crosshead, as shown in Fig. 167. If P = load CROSSHEAD AND SLIDE BARS 283 on piston, t and b denote thickness and breadth respectively of cotter. Area of rod through cotter =- rf* -rft, where d is the diameter of the rod. ^ Then (^cP-dt\f=P, where /= 6,000 lb. per sq. in. for steel. If the cotter and rod are of equal strength, as the cotter is in double shear, 2bt=^d^-dt. 4 For a steel cotter t is usually - and b can be obtained. The dimensions of the cotter may be found from 2btf=P (since the cotter is in double shear). The taper of the cotter may be 1 in 30. Ex. 27. In a horizontal engine the diameter of the cylinder is 1 5 in., stroke 24 in., connecting rod 5 ft. long, boiler fressure 90 Ih. per sq. in. Find the dimensions of (o) piston rod, (&) crosshead , pin or gudgeon. Draw sectional elevation and plan of each detail. Scale 6in= 1 ft. {a) From (1), (p. 273) rf=D>^J=1 5^J^ = 2i in. (6) From (4), ''^^l^p' From (5), R= ^ — P = 5— = -g-P. P =^ X 1 5=* X 90 =1 5,91 lb. 4 „ 1 5,91 X -\/26 , ^ „„„ „ R = — ^ -—^ — =1 6,220 lb. 5 .. , . . , */ 4 X (16,220)2 Substitutmg, '^ = V kx9000x1400 = ^-^^ "»• 284 MACHINE DESIGN If L denotes tte lengtli of the pin, then pd 1400x2-27 ^^' If it is necessary to reduce the length of the pin, then the diameter must be increased ; thus, if L=1 -Zd, pdL =Q becomes p x1 ■Zd^=^ 6,220. / 16,220 .'. d-=J ,^' , ^ = 2-985 or 3 m. \ 1 400 X 1 -3 L=1-3(/ = 3-88 or 3^| in. Connecting Eods. , Connecting rods. Empirical rules are frequently used to determine the diameter of a connecting rod. Thus, its diameter may be made proportional to the diameter of the piston, or in the case of a short rod the diameter may be obtained on the assump- tion that it is subjected to tension only, using a low value for the ^afe stress as in the case of a piston rod (p. 273). Let D denote the diameter of the cylinder ; p the initial, or maximum, steam pressure ; Q the total load on the piston. At the beginning of the stroke, when the piston" rod, connecting rod and crank are. in the same straight line, the load on the connecting rod will be Q, = -D?p. As the piston moves, the connecting rod has an inclined position, and, assuming the steam pressure to remain constant, the maximum force in the connecting rod may be obtained either by calculation or graphically by the triangle of forces. If d denotes the least diameter of a connecting rod. O^ d-^^lj: (1) where /=4000 lb. per sq. in. COimECTING RODS 286 The diameter obtained from (1), as in the corresponding case of a piston rod (p. 273), is only suitable for rods of moderate, or short length. Iiong rods. The dimensions of a long connecting rod may be obtained by one of the following, as on p. 274 : P=^. (2) ° ^"^ (3) 'rr^^' 7500 /(2 where P denotes the buckling load, E = 29x10^ lb. per sq. in., I = moment of inertia =jtrf*/64 for a round rod, Z= length of rod in inches, /=21 tons for mild steel, ft = radius of gyration and A = area of cross-section =-0(2. 4 The safe load is obtained by dividing the buckling load by 8. The length of the rod is usually about 2-5 x stroke ; the factor is frequently 3 in high-speed engines, and from 2 to 2-5 in torpedo boats. If W denotes the load on the piston, then, owing to the obliquity of the connecting rod, the working load on it will be 1 -OSW to 1 -OSW, or it may be found as in the following example. Ex. 28. Find the diameter of a steel connecting rod for an engine in which the maximum load on the piston is 70 tons. -Crank of engine 2 ft. radius, connecting rod 1 0ft. length. Factor of safety 8. If I denotes the length of the connecting rod and r the radius of the crank, Maximum force in rod is given by wVP + r^ 1. Z 10 5 ^ , where -=- = -. HT • t -,«'V25+T 70^26 Maximum force =70 — = — = — = — =71 -39 tons. Making use of Buler's formula (2), where E = 29x10e, I^qT, Z=120in., 286 71 -39 X 2240 MACHINE DESIGN 7r2x29x10«X7rrf* 8x1202x64 -=.^5 39 X 2240 X 8 X 1 20'' x 64 TT^ X 29 X 1 0^ = 6 in. Pig. 168. Or, by employing Rankirie's formula (3), we obtain 21 X- 71 •39x8=- 1 + 1 120^x16' 7500' c/2 CONNECTING RODS 287 .-. rf*-34-62rf2 + (17.31)2 =1363-6. (/2 = 54-24. .-. rf=7-4 in. marine coimectmg rod. One form of marine connecting rod nd is shown in Fig. 168. Assuming the diameter of the crank )ia to be known, then the chief remaining dimensions are the ength of the bearing, the diameter of the bolts, thickness t and T if the cap and base respectively, and the diameter of the onnecting rod. Ex. 29. The high-pressure cylinder in a mwrine engine is 22^ in. Hamster, steam pressure 1 60 lb. per sq. in., diameter of crank nn 1 in. Design a suitable connecting rod. Load on piston = Q =^ x 22-52 x 1 60 = 63,620 lb. Crank pin length. From pdl = Q, where p denotes safe allowable )ressure = 500 lb. per sq. in., rf= diameter and Z= length of oumal, 63,620 , „ .,„ , „„ . .'. l = ,.^J. -^ =12-73 or 12f m. 500 X 1 ^ Bolts for the hig end. Assuming rfi= diameter of bolts and safe tress 4,000 lb. per sq. in., .-. 2 x^rfi^x 4000 = 63,620 lb. 4 '- /63,620x4 (/i = a/— ^— — — =3-18 in. 1 a; sOOOjt The two bolts, as also in the crosshead (Fig. 173), are reduced >art of their length to a diameter equal to the diameter at he bottom of the screw thread. This not only reduces the reight of the bolt, but also makes it more elastic and less liable o fracture under any sudden applied load. Rod dimensions. The diameter o^ the connecting rod at the mall end may be obtained from (1) (p. 284). '-°^lr^^■^^f~l 60 3000' rf = 5-196 or 5jin. In the case of a long rod the diameter at the centre is obtained com (2) or (3) (p. 285) ; then assuming the rod to be conical, he diameter fit the large end is known, 288 MACHINE DESIGN It is difficult to state what the stress is in the T end of the rod ; its thickness is usually found from where I is the distance between the bolt centres, d the diameter of the rod and /= 6,000 lb. per sq. in. 9T^ 63, 620 (1 5 - 5 -1 96) ■=V5| 620x9-8x6 ^ _„ . 4-16 m. X 4 X 6000 It is usual to make this part slightly thicker than the cap, as the load, when the rod is in tension, acts at or near the centre. The extra thickness varies from | in. to J in. Cap or keep plate for big end. The cap, or keep plate, may be considered as a beam with a uniform load Q, lb., span I in., equal to the distance between the bolt centres. •■■ -I- Let 6 and t denote the width and thickness respectively ; then 8 ~ 6-'" Substituting the values and taking /= 6,000 lb. per sq. ia., 9t^ 63,620x15 "6"-' ~ 8 ■ -V=l 620x15x6 „ „^ „,, . — - — ^„^„ =3-64 or 3J4m. X 9 X 6000 ^® [An average value of / from a number of existing cases is found to be 6,000 lb. per sq. in.] Small end. One form of small end for a marine connecting rod is shown in Pig. 169. Load on each journal =63,620/2 lb. ; the allowable pressure p lies between 900 and 1 200 lb. per sq. in. ; 1 000 lb. per sq. in. is taken in the following. CONNECTING RODS If Z =length of journal, then Z = rf to 1 -1 rf. .'. pdl=3^,8^o r rf = J f'f° =d-376or 5ii Vl-1 xlOOO ° Z = 5-376x1-1 =5-9in. 289 m. Flo. 169.— Connecting Kod End. Bolts for small end. The safe allowable stress in these bolts should be less than that in the bolts at the large ends, and may be from 2000 to 4000 lb. per sq. in. As there are 4 bolts, rfi^x 3000x4 = 63,620. / 6362 'sOOTt' :2-598in. = 23 in. 290 MACHINE DESIGN Cap or keep plate for smaU end. The dimensions of the cap at the T end are found in a similar manner to those at the big end, using Q/2 instead of Q. The rod is frequently made a straight taper, the diameter at the crank end being 1 -3 to 1 -4 times that at the small end. ri8. 170.— "Box End." Another form of connecting rod, known as " box end," is shown in Fig. 170. This type is also used in stationary and in locomotive engines. Thickness of brasses, ti = 0-12d + 0-2. ' t2 = 0"08rf + 0-2. Gudgeon pin. If Q, as before, is the load on the high-pressure piston, then pdL = Q„ (4) where rf is the diameter and L the length of the pin (Fig. 171). The gudgeon pin is usually designed solely for bearing surface; when this is done the strength is found to be ample. The intensity of the bearing pressure must not be sufficient to squeeze out the lubricant. The value of p lies between 900 and 1250 lb. per sq. in. If necessary any increased surface should be obtained by lengthening the pin instead of increasing its diameter. If in (4) a value for L or of is given or assumed, then, since Q and p are known, the remaining term can be found. CONNECTING RODS 291 Usually L lies between 1 •2or and 1 •?«/, putting L=1-5rf and assuming p =1200. .". l-SaPx 1200 = 63,620. /. (/ = 5-9 in. L=1-5x5-9 = 8-9in. Fio. 171.— Connecting Eod End. Forked end. Another form of small end, in which the jaws of the forked end are solid, is shown in Fig. 171. To obtain the dimensions of the forked end of the connecting rod, let 6 and t denote the width and thickness respectively. 26f/=P or «4, When 6 or f is given, or assumed, the value of the remaining term may be calculated, or the following empirical values may be used: 6 = rf tol •1rf = 6i in., t=0-53d = 2-8 in., T = 0-6d = 3-e in., IVl = 2rf=11-8Ln.. N = 2-47rf=14lin. Marine engine crossheads. Two forms of marine engine crossheads are shown in Figs. 172 and 173. Fig. 173 shows a form of crosshead frequently used in marine engines, also in inverted-cylinder stationary engines. White metal strips suitably dove-tailed are shown, also a plate fitted to allow the slide and 292 MACHINE DESIGN slipper to expand when working. The portion marked n (Fig. 173) is sometimes omitted to reduce tie length I, where I is Fig. 172. — Crosshead. the distance between the centres of the bolts. The nuts are made of wrought iron to prevent seizing. EIQ. 173.— Crosshead. Gun -metal CONNECTING RODS 293 Locomotive connecting rods. These are usually either rect- angular, or I-shaped, in cross-section. The large and small end may be made with a loose strap and cotter arrangement. The strap is secured to the rod by bolts ; this form has given good results, the only objection is by the bolts slackening back slightly. This causes the strap and rod to wear, and if not detected may entail o-e-a- FIQ. 174. — Locomotive Connecting Rod End. considerable work to rectify. This form is shown in Pig. 174, and consists of a strap S, the rod R and the steps or brasses B. The dimensions of the strap and diameter of bolts may be obtained as foUows : Ex. 30. A locomotive cyUnder is 1 8 in. diameter, stroke 34 in., boiler pressure 1 50 lb. per sq. in. Find the dimensions of the strap and the diameter of the bolts. Bolts. Let di denote the diameter at bottom of thread ; taking the safe shearing stress as 5,000 lb. per sq. in., and neglecting the bending stress, then, as each bolt must shear at two sections, |rfi2x4x 5,000 =P. 294 MACHINE DESIGN P is approximately given by - x 1 8^ x 1 50. ^ V 20,000 .-. rf=1|in. (Table III. p. 74.) Strap. Let 6 = width, and *=t]iickness of strap. 2btf=P. The value of / lies between 2000 and 3500 lb. per sq. in. Assume 6=1 -5* and /= 3000 lb. per sq. in. ; then, substituting, 3*2x3000 = 38,170. ■^^ 38,1 70 9000 6 = 3-1 in. = 2 -06 in. Another form of small end is shown in Kg. 175 ; in this case bolts are not used. The rod end is forged solid with the rod, and the adjustment of the brasses is effected by means of the screw and the wedge-shaped block W. Fig. 175. — locomotive Connecting Eod End. Another form of large end is shown in Fig. 176 suitable for use with outside cylinders, but is also used for inside-cyUnder engines ; in this case the loose strap is replaced by a solid forged end, the brasses being inserted and secured by a cap. The two bolts COUPLING RODS 295 passing througli the cap secure the brasses, and allow for their adjustment. The connecting rod may be made parallel through- out its length or have a straight taper, as in Fig. 176. Fia. 176. — Locomotive Connecting Eod End. Locomotive coupling rods. The connecting rods and coupling rods used on locomotives are made of wrought iron or mUd steel ; the cross-section is either rectangular or I-shape (Fig. 177). In these rods there is, in addition to the stress due to the force y^^^ fi)/;,p: ■,f /A'-. 4- <- i >■ «• B -»- Fig. 177. — Cross-sections ot Coupling Bods. transmitted, the bending stress due to centrifugal fbrce and the weight of the rod. Coupling rods are used in locomotives to couple together one or more pairs of wheels to the crank axle, the object being to increase the tractive force due to the adhesion of a larger number of wheels on the rails. 296 MACHINE DESIGN The greatest stress in the rod occurs when in its lowest position. The motion of each point in the rod relative to the engine, is a circle, the radius being the crank radius (/•) (Fig. 178). ^ ®y (S- t I I 4- y t ^ I'lG. 178. If w is the weight per unit length and I the length of the rod, V the velocity of the crank pin centre, then the bending moment (M) is given by M ( , = («/ + gr /8 or if W is the weight of the rod, M=(w + ^n|in.-lb. V gr /8 In this expression, V is in feet per sec, r in. ft. and I in inches. The stress /is obtained from the relation M.=fz. Values of z are : ' Rectangular section. For I section. 6 BPS - b ifi 6D (I) (Fig. 177). (II) (Fig. 177). The maximum stress in the rod is obtained by adding to the stress just obtained the stress in the rod due to the force trans- mitted by it. Ex. 31. The coupUvg rod of 'a locomotive is 8 /;. 6 in. centre to centre, the section is uniform 4i in. deep and 1| in. wide,- average weight 22 Ih. per foot run. The driving wheels are 6 ft. 6 in. diameter, radius of crank 1 in. Find the maximum tensile^ stress COUPLING RODS 297 in the rod when the erigine is running down an incline at 60 miles per hour vnth the steam shut off. W = 22x8-5 =187 lb, Eevolutions of crank =—- -— = 4-308 per sec. 60x60XTCx6-5 ^ V = 27rx^x4-308 = 22-56ft. persec. Tip-, , 1 87 X (22-56^ 1 02 = 47,610 in.-lb. •|x 1-75x4-52 = 47,610. .". /= 8,063 lb. per sq. in. Ex. 33. A locomotive coupling rod, 8 ft. length, betiveen centres, is of uniform rectangular section 3^ in. deep iy'i^ in. wide. Find tJie maximum stress in the rod lohen the crank, 1 ft. radium, makes 200 rev. per min. Diameter of cylinder 1 8 in., steam pressure 140 lb. per sq. in. [1 cub. in. of the rod weighs 0-284 lb.] «/ = weight of rod per inch length = 3-5 x1 -5 x 0-284 =1 -491 lb. , , 27t X 1 X 200 207t ,, V = -— = -;;- it. per sec. 60 3 ^ M/V* 1 -491 /207t\^ Centrifugal force = — ="30.2" ( "3" ) ^ 20-32 lb. per in. length. / wy^\P (1-491+20-32)96* 21 -81 X 96* 8 Ax1-5x3.5-?l:?yi?^^ 6 8 /i = 8,206 lb. per sq. in. 4 = 35,630 lb. Total load on one piston = - x 1 8^ x 1 40 298 ' MACHINE DESIGN Assuming the rod to transmit one-half the load on one piston and applying EanMne's formula when k^ = S-S^l 2, AA /,x1-5x3-5 /gXl -ex 3-5 ^ 2-203 17,815x2-203 ^ ^,. ,, •■• f^= 1-5x3-5 -7,474 lb. per sq. ID. Hence maximum stress = 8,206 +7,474=1 5,680 lb. per sq. in. EXERCISES XVII. [ Ranhine's formula is to be used in the following exercises on shafts, except where otherwise stated.] 1. Find the diameter of a wrought-iron shaft to transmit a twisting moment of 480,000 in. -lb. and a bending moment of 360,000 in. -lb. Safe stress 9000 lb. per sq. in. 2. A wrought-iron shaft is subjected to a bending moment of 8000 in.-lb. and a twisting moment df 1 5,000 in. -lb. If the safe stress is 9000 lb. per sq. in., find the equivalent twisting moment and diameter of shaft. 3. A shaft transmits 60 horse-power at a speed of 1 35 revolu- tions per minute. If the bending moment on the shaft is three- fourths the twisting moment, find the diameter of the shaft. Safe stress 1 0,000 lb. per sq. in. 4. A shaft 3 in. diameter is subjected to equal bending and twisting momenta. If the surface stress in the material is 9000 lb. per sq. in., find the twisting moment. 5. A hollow shaft, external diameter 15 in., internal 10 in., is subjected to an equal bending and twisting moment; the maximum stress in the material is not to exceed 1 0,500 lb. per sq. in. Find the horse-power transmitted when the shaft makes 85 revolutions .per minute. 6. The maximum twisting moment of a screw shaft is 1 ,400,000 in.-lb. The weight of the screw is 5000 lb., distance of its centre EXERCISES XVII 299 from tke stern bush is 1 8 in. Find the diameter of the shaft. Assume the bending action as a live load (to allow for rough weather). Safe stress 9000 lb. per sq. in. 7. The radius of an overhung crank is 18 in. (Fig. 179). Find the diameter of the journal A if the greatest value of the tangential force at the crank pin is 40,000 lb. Stress not to exceed 5000 lb. per sq. in. 8. Find the diameter and length of the crank pin (Fig. 179) if the tangential force on the crank pin is 40,000 lb. Shear stress not to exceed 9000 lb. per sq. in. (a) Length of crank journal =1 1(/, where d is the diameter of the journal. (6) Load on projected area not to exceed 500 lb. per sq. in. 9. A crank (Fig. 180) is acted on by a force applied at C in the central plane of the crank at right angles to OC. It causes a Fig. 179. FiO. 180. ' bending moment on the shaft equal in magnitude to one-half the twisting moment produced. The crank is secured to the shaft by means of a sunk key. Determine the dimensions marked d, a and h, allowing the stress in the material to be the same for 300 MACHINE DESIGN each part, = 5,000 lb. per sq. in. Force at C 40,000 lb., radius of crank 1 8 in. 10. A wrought-iron shaft is subjected to a bending moment equal to the twisting moment ; if 90 horse-power is transmitted at a speed of 1 30 rev. per min., find its diameter. Safe stress 9,000 lb. per sq. in. 11. Calculate the diameter of the crank pin in each of the two cases (I) and (II) (Fig. 181). The tangential force on the crank ( ^I4'l B _l (I) (II) Fia. 181. pin is 1 25,000 lb., length of pin 1 4 in. Safe stress 9,000 lb. per sq. in. 12. Find the safe load on a crank pin 3 in. diameter, 6 in. long, if the stress in the pin is not to exceed 4,000 lb. per sq. in. 13. The crank shaft of an engine makes 75 revi per min. Find the horse-power transmitted if the mean tangential load on the pin is 30,000 lb. Radius of crank 24 in. 14. Some of the dimensions of a crank shaft are given in Fig. 182. A and B are centres of bearings. The maximum thrust P of 1 tons is balanced by, a turning moment in the shaft. A — fs a gi ■ JTA B G^ — >i< 12 — >; Fib. 182. Determine the greatest bending and twisting moments, and find the diameter of the crank shaft. If the length of the crank pin is 1 \d, where d is its diameter, find the dimensions. EXERCISES XVII 301 15. The tangential force on the crank pin of an overhung crank is 50,000 lb. Find the diameter and length of the crank pin, (a) if the strength of .the pin only is considered, (6) if the bearing pressure is not to exceed 800 lb. per sq. in. 16. The greatest difierence of steam pressure on the two sides of a piston is 28 lb. per sq. in. Diameter of piston 5 ft. The piston rod is made of steel secured to the piston by a tapered end and nut, and to the crosshead by a cotter. If the length of the rod is 7 ft. 6 in., calculate its diameter ; factor of safety 8. E = 29x10'' lb. per sq. in. Make a sketch of the rod, inserting dimensions. [B.E.] 17. Determine the dimensions of the crank pin of a single arm crank for a maximum force along the connecting rod of 68 tons. Allow a bearing pressure of 900 lb. per sq. in. of pro- jected area and a maximum bending stress of 3-5 tons per sq. in. Show by sketches how the pin may be fixed in,the arm. [L.C.U.] 18. A twisting moment of 200 ft.-lb. is transmitted to a shaft I5 in. diameter by a wrought-iron hand lever. The maximum force to be applied is 80 lb. Calculate the dimensions of the lever section (rectangular) at the shaft end ; the width of the section is to be ^ the depth. Allow a maximum bending stress of 8,000 lb. per sq. in. Make a dimensioned sketch of the lever. [L.C.U.] 19. A horizontal steam engine is fitted with a wrought-iron forked crosshead and a pin which carries at its ends two cast-iron slide blocks. Maximum pressure along the piston is 25 tons, ratio of connecting rod to crank is 5 : 1 . Estimate the necessary bearing area of the slide blocks, allowing a bearing pressure of 65 lb. per sq. in. Make a dimensioned sketch of one block. [L.C.U.] 20. Three engines of equal power drive one crank shaft, and the twisting moment is applied at one end only of the shaft. The shaft is made in three similar forgings coupled together by bolts through flanges. Each length of the shaft is mounted on two bearings, and each crank pin is midway between the bearings. Show that the mean twisting moment on the crank pin nearest the end to which the twisting moment is appUed is 23-5H-=-N, where H denotes the horse-power and N the number of revolutions per min. • [B.E.] 302 MACHINE DESIGN 21. In a set of vertical triple-expansion marine engines of 3500 horse-power, screw propeller 1 00 rev. per min., length of stroke 3 ft. 6 in., boiler pressure 1 50 lb. per sq. in., diameter of low-pressure cylinder 6 ft. 4 in., maximum steam pressure 13-4 lb. per sq. in., the three separate portions of the crank shaft are fastened by flange couplings and are each mounted on two bearings 4 ft. 9 in. centres, (a) If 9 bolts are used for each coupling, find their diameter ; diameter of bolt circle 20 in. (6) Find the maxi- mum stress in crank shaft and its diameter. Shearing stress in the bolts and maximum shearing stress in the shaft 9000 lb. per sq. in. 22. The high-pressure cylinder of a triple-expansion engine is 1 2 in. diameter, 2 ft. stroke, length of connecting rod 5 ft. 6 in., pressure of steam 1 80 lb. per sq. in. Find the greatest stress in the guide bars, assuming that the stress is greatest when the crank is at 90° to line of stroke. The bar is rectangular, 7 in. wide by 3 in. deep by 3 ft. long. 23. The coupling rod of a locomotive is 8 fib. from centre to centre ; the section is uniform, 1 -5 in. wide and 4j in.- deep. The crank is 1 2 in. radius, driving wheels 7 ft. diameter. Find the 'maximum stress when the speed is 70 miles per hour, assuming no end thrust along the rod. [1 cub. in. weighs 0-28 lb.] [B.E.] 24. The coupling rod of a 6 ft. four-coupled locomotive is of I section, flanges ^ in. x 2| in., web f in. x 3| in., length between centres 9 ft. The rod is operated by a crank 1 3 in. throw. If the speed is 75 miles per hour, find the greatest stress due to the weight of the rod and centrifugal force. [1 cub. in. weighs 0-28 lb.] [B.E.] 25. Design an eccentric sheave and strap when the load on the valve spindle is 1 ton, bearing pressure between sheave and strap 70 lb. per sq. in. on projected area. Find diameter of bolts if tensile stress is not to exceed 1 g tons per sq. in. The sheave is made in two parts, diameter 10§ in. Distance between centre lines of bolts 6| in. Diameter of crank shaft 5 in. [B.E.] 26. Estimate the diameter of a crank shaft where the crank overhangs. Calculate "the force exerted on the crank pin from .the following data : Distance measured along crank shaft from centre of bearing to centre line of cylinder produced is 11 in., diameter of cylinder 20 in., st«am pressure 200 lb. per sq. in., crank radius 1 ft., connecting rod 5 ft. long, maximum stress in, shaft 5 tons per sq. in. [B.E.] EXERCISES XVII 303 27. Kg. 183 shows am incomplete sectional elevation of a crosshead and a portion of the piston rod. Complete the sectional 'A" thick Inclined at 30° to iie.rtictti Gudgeon pin 2'/t"diar length in brasses 'A"dJar Fig. 183.— Crosshead. elevation and draw a pjan. Scale half size. If the piston is 1 5 in. diameter, steam pressure 90 lb. per sq. in., stroke 30 in., find the stresses (a) in piston rod, (6) in gudgeon pin. 28. Investigate formulae for the maximum principal stress and the maximum shearing stress at a section of a circular shaft subject to a bending moment M and a twisting moment T. [Lond. B.Sc] 29. Define principal stresses, and show how to find the stresses at a point produced by the joint action of stresses in two dimen- sions. A solid circular shaft is subjected to a twisting moment of 1 60 in.-tons and a bending moment of 1 00 in.-tons ; find its diameter so that the stress in the material does not exceed 8000 lb. per sq. in. [Lond. B.Sc] 30. Two views of a suggested arrangement for a crane to lift two tons at fifty feet per minute are shown in Fig. 184. Velocity ratio of -the two shafts to be 5. Find (a) pitch of teeth on wheel C, assuming pressure F between teeth is 400 (pitch)^, (6) diameter 304 MACHINE DESIGN of pinion P^, (c) a suitable diameter for pulley P to take a Belt 6 in. wide, {d) twisting moment and bending moment on shaft (II), (e) equivalent twisting moment on shaft (II), (/) diameter of Kg. 184. — Crane. shaft (II) if maximum stress is 4 tons, per sq. in. Centres of shaft bearings 2 ft. 6 in., pull on tight side of belt twice the pull on slack side. Maximum tension in belt 1 00 lb. per inch width. [Lend. B.Sc] 31. A f hollow propeller shaft 12 in. external diameter, 6 in. internal, is subjected to a twisting moment of 100 ft.-tons, a bending iHonient of ^O ft.-tons and a direct axial thrust of 20 tons. What will be the maximum compressive stress ? [Lond. B.Sc.] 32. Find the internal diameter oi a hollow shaft which is 1 7 in. external diameter and is subjected to a torque of 360,000 ft.-lb. and a bending moment of 90,000 ft.-lb. Assume that the maxi- mum shear stress theory is correct for the elastic failure of a shaft. The maximum direct stress allowed is 1 0,000 'b. per sq. in. [bond. B.Sc] DESIGN OP A DOUBLE PURCHASE CRAB 305 33. If a shaft of diameter d is subjected to a twisting moment T and a bending moment B, show that 6 + ^62 + 1^=— rf3/^, where ft is the maximum tensile stress. Show also that Vb^+T^ =— cPfg, 1 6 where /g is the maximum shear stress in the shaft. You may assume the formulae for the moment of resistance of a shaft to a simple torque and to a simple bending moment. [Lond. B.Sc] Design of a double purchase crab to lift 1 ton. The particulars are as follows : Handles 1 6 in. long. Barrel 7 in. diameter, measured to the centre of the rope. » Gearing (dimensions in inches). Wheel. Number of Teeth. Pitch (inches). Pitch Circle (diameter). Breadth. A B C D 12 31 12 76 2i H 3 3 Calculate the necessary force on each handlk (neglecting friction). Calculate the diameters of tlie various shafts c nd other peafs. Mso find the diameter of the rope if the safe stress sk.3000 Ih.^r sg'.pW. (Fig. 185). Make detail drawings of the vai Uni^ ■jf^s^ScaU^. Draw elevation, end view and plan, showing 0}iji4g p^sf^st^^. Scale ^ full size. If d denotes the diameter of the rope, -rf^x 3000 = 2240. 4 Velocity ratio = (/ = 0-975 in. or tikl^ 2x31 x76x1 31 12x12x3-5l I 306 MACHINE DESIGN If P denotes the force required on eack handle, 12x12x3-5x2240 P=- =14-97 lb. or 15 lb. 2x31 x76x16 If the efficiency instead of unity is assumed to be 0-6, then P=15x-ig- Elevation looking: in the direction of the arrow B Plan Fig. 189. — Boiler Pump. 2. The half plan, half elevation and half end view of the forked small end of a connecting rod are shown in Fig. 190. Calculate : (a) The diameter of the minimum section of the bolts, assuming that the stress allowed in them is 1 -4 times the stress in the minimum section of the connecting rod. (6) The intensity of the bearing pressure on the journals when the stress at the maximum section of the rod is 2 tons per sq. in. (e) Neglecting friction, calculate the horse-power which is developed in' the cylinder to which the rod belongs when the speed is 60 rev. per min. Stroke 5 ft., stress as given in (6). S12 MACHINE DESIGN Draw elevation and plan, inserting all omitted bolts and fastenings. Scale J full size. [B.E.] Half End ^ Half Plan Fia. 190. — Connecting Eod End. 3. Fig 191 shows a slotted bar for driving tbe crank shaft of a double-acting steam engine. The crank pin is held in a single crank disc. The upper part of the bar is a continuation of the piston rod ; the lower part of the bar slides in a guide. The power is all transmitted through the crank pin to the crank shaft. Assuming the mean load on the piston is 2000 lb., calculate : (a) The bearing pressure on the crank pin. (6) The diameter of the two bolts at the ends of the slotted bar so that the tensile stress is 1 800 lb. per sq. in. (c) The depth of the cotter, thickness -^g, safe stress 1750 lb. ~ ...... ^ , . Scale 5 per sq. in. full size. Draw sectional elevation and end view. [B.E.] 4. Fig. 192 shows the outline of a crosshead arranged to work on one slide-bar 7 in. by 3 in. in section. The end of the piston rod is shown, and the size of the small end of the connecting journal, together with some dimensions. Calculate : (a) The bearing pressure on the crosshead pin when the stress Forged steel piston rod 2'4"Diai: \ Crank pin f2%"Dia: Cqttered joint centres 'Bolt Guide flbd & Cross B f forged steel in one piece Di\stttmped Fia. 191. — Slotted Bar Mechanism. Couer secured by 6 bolts 1 3 on each side Slii^. bar | _ I Fig. 192.— Crosshead and Slide Bar. 314 MACHINE DESIGN in the piston rod is 2 tons per sq. in. What bearing pressure would you use for the big end journal ? (6) The normal pressure, between the crosshead and bar when the cranio is at right angles to line of stroke and stress on piston rod is 2 tons per sq. in. (c) The diameter of bolts to secure crosshead cover to the body, assuming (&) gives maximum load. Safe stress 1 -8 tons per sq. in. Draw sectional elevation and plan. Scale 5 fuU size. Length of connecting rod 4 ft. 6 in., crank 1 ft. long. [B.B.] 5 tons riG. 193.— Brake Gear. 5. A flywheel 6 ft. diameter, 6 in. broad on the face, is rigged with brake gear as shown in Fig. 193. A pressure of 5 tons is to be produced between each brake block and the flj^rheel rim when air at 60 lb. pressure is admitted to the brake cylinder. The brake levers are jointed to fixed brackets A and B and secured to a massive wall. Calculate : (a) Diameter of brake cylinder. MISCELLANEOUS EXERCISES XVIII 3l6 (6) Diameter of brake rod, safe tensile stress 4,000 lb. per sq. in.- (c) If the pin connecting the brake block to the lever is 1 J in. diameter, determine the depth of the lever. Safe stress 3 tons per sq. in. The lever is made up of two bars each 1 5 in. thick. Draw elevation and end view. Scale J fuU siz^. [B.E.] 6. The crane hook in Fig. 194 is designed to lift 3 tons. It is carried by a single sheave block 11 in. centre to centre of wire -1^ ■%gt^Jolts Fio. 194. — Crane Hook. rope. The wire rope is | in. diameter. The block A in which the hook swivels is 4 in. wide by 6 in. long secured to the side plates by 2 bolts. Calculate : (a) Diameter of pin on which pulley turns, allowing a bearing load of 1 000 lb. per sq. in. 316 MACHINE DESIGN . (6) Tte diameter of tke bolts securing the block A to the side plates. Safe shearing stress 1 -7 tons per sq. in. (c) The maximum stress in the section PQ. Draw sectional elevation and side elevation. [B.E.] 7. Fig. 195 shows a countershaft, on one end of which is a spur wheel B, and on the other end a belt pulley C, 1 8 in', diameter and Scale I full size. Fig. 195. — Countershaft Driving Gear. 6 in. wide. The spur wheel B is in gear with a pinion A, which is keyed to the spindle of a 5 h.p. motor. The motor spindle makes 600 rev. per min. Calculate : (i) The diameter rf of the shaft calculated for pure torsion only. Shearing stress- 6000 lb. per sq. in. (ii) The effort P acting between the teeth at the pitch line tangential to the pitch circles. (iii) The maximum tension in the belt, assuming the ratio of the tensions to be 2 to 1 . (iv) The width of the belt if the safe stress is 1 20 lb. per inch width. 8. The diagrammatic sketch (Fig. 196) indicates the joint between a valve spindle V and a valve rod R. The end of the valve spindle is tapered 1 in 1 and fits into a corresponding tapered hole in the forked crosshead. The two pieces are drawn together and secured by a cotter tapering 1 in 1 6. Complete the design of the joint and calculate the dimensions A, B and C. MISCELLANEOUS EXERCISES XVIII 317 (o) The dimension A such that the net cross-section is equal in area to the valve spindle V. (6) Dimension B to be determined, assuming the strength of the cotter to resist shearing is 80 per cent, of the spindle V to resist tension. i'la. 196. — Valve Spindle Joint. (c) The dimension C, so that the bearing pressure on the pin is .limited to 1200 lb. per sq. in. when the tensjle stress in the valve spindle V is 3 tons per sq. in. Make drawings of the parts assembled together. Scale | full size, (a) A vertical longitudinal section. (6) An outside plan view in direction of arrow P. Double puTcliase crab. In Fig. 197 an incomplete elevation and plan of a double purchase crab is shown to lift 1 3 tons. Some of the dimensions are as follows : Wheel A, 2 ft. 8^ in. diam. Pinions B, fe, D, each 4^ in. diam. Rope barrel, 11 in. diam. Winch handle, 1 5 in. radius. Brake wheel, 1 ft. 2^ in. diam. Calculate the probable pressure on wheel teeth, twisting moments on shafts, allowing for bending ; force applied to winch handle and pressure necessary on brake handle. Deduce diameters of shafts, pitch of teeth of wheels and any other parts to which 318 MACHINE DESIGN calculation is applicable. Draw general elevation, side elevation and plan ; also separate drawings of details where applicable. [B.E.] "Dc" FIQ. 197. — Double Purchase Crab. TYPICAL EXAMINATION PAPERS. LANCASHIRE AND CHESHIRE UNION OF INSTITUTES. Mechanical Engineering. Engineeeing Drawing. Draw either Example (I.) or (II.) ; the drawing must be fully dimensioned, and the calculations involved in the design must be given. (I.) In Pig. 198 the sketch gives a few main dimensions of -the crank pin end of a connecting rod. The bearing is 6j in. diam x 8 in. I 0/7 cup 1 1 1 1 -fj CO 1 Y Fig. 198. — Connecting Bod End. long, and the maximum load along the rod is 20 tons. The bolts must be designed for a stress of 7500 lb. per sq. in., reckoned on the area at the bottom of the thread, 319 320 MACHINE DESIGN Cap. The width B of the cap and the rod end niust be slightly greater than the diameter of the rod. Thickness A of cap and butt end of rod is given by A=1-086(/ia/-. where rfi= diameter of bolt at the bottom of the thread. (a) Calculate the diameter of the bolts, (6) the dimensions of the cap and butt end, and then make complete dimensioned drawings of the connecting rod end, showing the front elevation and the plan. Scale g full size. ^", 1 r-- — •* _ 1 1 *-- 1 V 1 r i 1 1 1 1 1 1 ? .-F-> <--/r-> > i Y I r- r-- ..J 1 1 1 1 • 1 ',*---L—» Fig. 199. — riange Coupling. (II.) Design a cast-iron flange cowpUng for steel shafts, 3| in. diameter from the given particulars and 'empirical proportions (Fig. 199). Bolts (diameter rf). There are six bolts on a circle of 9| in. diam. The diameter of each bolt must be such that, after allowing for the weakening effect due to tightening up and bending in the hole, the mean shearing resistance per sq. inch of bolt area does not exceed 3400 lb. when the twisting moment on the shaft is 50,000 inch-lb. D = 3iin., B =1-80-1- 0-8, L=1 -20 -1-0-8, F = 0-3D -M -3rf-F0-3, E = 0-25D-|-0-4. Determine the above dimensions to the nearest -jh i^., and also fix the dimensions of the keys, and then make the foUowing views, fully dimensioned : (a) Sectional side elevation ; (&) End elevation. Scale J full size. TYPICAL EXAMINATION PAPERS 321 Questions. Three to be answered. The calculations accompanying the sketches or drawings must be given on the drawing paper. 1. Two one-inch diam. shafts are carried by a single bracket and connected by bevel wheels. One shaft is horizontal and the other, vertically above the first, is inclined upwards at 75° to it. The base of the bracket is 6 in. below the horizontal shaft. Make sketches of a suitable bracket, and give probable thicknesses of metal. 2, A steel column is built up of two channel section bars, 1 in. X 31 in. (Fig. 200) stiffened by lattice bars inclined at 45°. Make dimensioned sketches of a suitable base for the column. 3T>« i 1 J 11 ■ Fifl. 200. — steel Column. , Draw accurately, fuU size, three or four teeth of a chain wheel of 1 5 teeth for a roller chain of 1 -5 in. pitch with rollers 0-85 in. diameter. 4, Determine the dimensions of a simple pin joint for two round rods 1 5 in. diameter. Make a dimensioned sketch of the joint. 5. 50 horse-power is transmitted from shaft A to shaft C through an intermediate shaft B by gearing as shown (Fig. 201). The FIG. 201. wheel on shaft B is 50 in. diam. and runs at 45 rev. per min., and the distance between the centres of the shaft bearings is 48 in. The wheel is mounted centrally on the shaft. Determine the 322 MACHINE DESIGN maximum bending moment on the shaft B due to the power transmitted. What is the twisting moment on this shaft ? 6. The sketch (Fig. 202) represents a butt joint for two tie plates. If the safe load on the joint be 70 tons, find the shearing stress per sq. in. on the rivets (^ m. diam.), assuming that a rivet in double shear is 1 f times as strong as one in single shear. A T ^ V 7^ =~ 4 ^^ \ ; l^'^covers Pio. 202. — Butt Joint for two Tie Plates. Find also the tensile stress per sq. iach in the plate at the section AB. THE UNION OF EDUCATIONAL INSTITUTIONS. Machine Construction and Drawing: Advanced. Four questions only are to be attempted, but either No. 1 or No. 2 must be tried. Drawings must be neatly finished in pencil, and be fully dimensioned. All calculations for the determination of sizes, etc., must be shown and set out in such a way that the method can easily be followed. 1. Dimensioned sketches of the forked end of a connecting rod are given (Fig. 203). The load on the rod is 60,000 lb. Deter- mine (a) The diameter of the bolts if the stress allowed in the body of the bolt is not to exceed 2500 lb. per sq. in. (6) The thickness of the keep plate or cap. The stress due to bending must not e2?ceed 6000 lb. per sq. in. TYPICAL EXAMINATION PAPERS 323 Draw to scale 2 in. =1 ft. The side elevation and plan as given, but using the dimensions of the cap plate and bolt you have ascertained by calculation. To the right of the side elevation, and in projection with it, an end elevation. ■6/JC4"W» JFia. 203. — Connecting Rod End. 2. Design a flanged coupling suitable for a 3 in. shaft. The maximum shear stress allowed in the shaft is 9000 lb. per sq. in., and for the coupling bolts -4000 lb. per sq. in. Shekring stress on key not to exceed 8000 lb. per sq. in. The bolt circle is 9 in. diam., and there are six bolts. Make a fully dimensioned working drawing ; arrangements must be made for correct alignment of the shafts, safety of the keys and bolt heads and nuts. Scale ialf full size. The shaft makes! 80 rev. pe r min. What power is it capable of transmitting ? 3. Calculate the stress in the body of the rod just below the fork of Question 1. 4. What is the beajing press'(H'»'i between the brasses of the rod and crosshead pins ? 324 MACHINE DESIGN 5. A slide valve is to cut ofE when the crank angle is 1 20° from inner dead centre. The maximum opening is to be 2^ in. and the lead ^g in. Find the half travel and lap of the valve, and angle of advance of the eccentric. 6. Calculate the thickness of a cast-iron steam pipe, 4 in. diameter, to carry a pressure of 40 lb. per sq. in. 7. Sketch a connecting rod siiitable for a petrol motor of a motor car. Put on appropriate dimensions. BOARD OF EDUCATION. 1, The diagram (Fig. 204) gives some particulars of the big end of a connecting rod. If the mean total pressure on the piston to which the connecting rod is secured is 51 ,000 lb., and if the efiect ^"diafbolt FI9. 204. — Connecting Rod End. of the obUquity of the connecting rod be neglec,ted, calculate and write down on your drawing, showing your calculations in detail : (a) The width of the journal crank pin brasses, allowing about 850 lb. per sq. in. pressure on the projected area of the crank pin bearing surface. (&) The diameter of the bolt EF, so that the shearing stress shall be about 81 00 lb. per sq. in. TYPICAL EXAMINATION PAPERS 325 (c) The dimension X suck that the mean total pressure on the piston shall produce a tensile stress in the section of the fork along the line EF of about 4500 lb. per sq. in. Use in your calcu- lation the diameter of the bolt obtained in (b) and neglect the effect of the slight taper indicated on the drawing. Complete the design of the connecting rod end, drawing to a scale of one-third full size : {d) A sectional elevation, (e) A sectional plan view, taking the section at the centre line of the crank pin. In each of these views the brasses, the bolts EF and GH, complete with nuts and fastenings, the lubricator, and any other omitted parts and details, are to be shown. Fia. 205. — Toothed Wheel Drive. 2. The diagram (Fig. 205) shows a toothed wheel drive arranged to permit a definite change of velocity ratio between the two shafts AB, CD. The shaft AB is connected direct to a motor of 90 H.P. running at 300 rev. per min. The shaft transmits the 326 MACHINE DESIGN power througt section No. 2. The alteration of velocity ratio is done by sliding the sleeve carrying the pinions along the shaft AB. Design the gear for moving the sleeve and for securing it in position, assuming that the gear is only changed when the shafts are at rest ; you are also required to design the shafts for the power conditions stated above. Calculate and write down on your drawing, showing your calculations in detail : (a) The diameter of the shaft AB at section No. 1, calculated for pure torsion so that the maximum shear stress is about 9000 lb. per sq. in. (&) The diameter of the shaft CD at section No. 2, calculated for pure torsion so that the maximum shear stress is about 9000 lb. per sq. in. (c) The speed of the shaft CD when (i) M gears with N ; (ii) P gears with Q. Draw to a scale one-quarter fuU size : (d) A sectional elevation of the sleeve moving gear, taking the ' section through the shaft AB ; also an end elevation and a plan view, adding any bracket or support, or modifying the form of the cast-iron end pieces carrying the shafts, to suit your form of, sleeve, moving gear. You are not required to draw in detail the end castings supporting the shafts ; show only sufficient of these and of the base plate necessary for the purpose of securing your sleeve moving gear. (e) Make dimensioned drawings of the shafts AB and CD. 3. A short-stroke hydraulic press, similar in design to the one shown in the sketch (Fig. 206), is required to exert a pressure of 22 tons on a die placed between the jaws. The pressure in the supply main is 700 lb. per sq. in. The opening at C is connected to the valve controlling the flow of pressure water to, or exhaust water from, the cylinder. The opening at B is connected per- manently to the pressure main. Calculate and write down on your drawing paper, showing all calculations in detail : (a) The diameter D of the upper part of the ram, allowing for 1 per cent, loss in friction. (6) The diameter of the studs securing the top cover to the cylinder body, assuming that there are 1 2 studs and that the stress due to the pressure of the water does not exceed 2| tons per sq. in. TYPICAL EXAMINATION PAPERS 327 (c) The width of the flanges at the section AA, assuming that the maximum stress is 3 tons per sq. in. The effect of the direct stress over the section may be ignored in your calculation. , '" r-\ V. H ^ To scale one-half full size draw : ' ' {d) A sectional elevation of the right-hand half of the cylinder, showing the ram in section and at the top of its stroke. The 328 MACHINE DESIGN means of preventing water leakage past the ram, both in the cylinder and the stuffing-box, must be clearly indicated. To a scale of one-eighth full size draw an outside elevation and a plan view of the press. 4. The diagram (Fig. 207) gives some particulars of a toothed , wheel drive between an electric motor and a reciprocating pump. 5B.H.P. Motor 900 revs. per min. O V±J _. n > =ite ^ o Co 1. ■S ■«: ■5 +• ■£* to I TJ TYPICAL EXAMINATION PAPERS 329 Calculate, showing all calculations in detail : (a) The speed of the pump in revolutions per minute. (6) The load in pounds acting on the teeth of P in a direction tangential to the pitch circle. (c) The diameter of the shaft within the boss of the wheel Q. The safe shearing stress allowable (due to torsion) is about 1 200 lb. per sq. in. In your calculation neglect the efiect of bending. To a scale of one-half full size draw : {d) A sectional view corresponding to the view given, but omitting the bearings and the motor. The teeth of P and Q must be shown in this view ; show also a means of securing the wheels to their respective shafts, and give full particulars of the flexible coupling R. Half-couplings bolted hard together are unsuitable. (e) A sectional end view projected from (d), the section Ijeing taken at WW, looking in the direction of the arrow V. The whole of the pinion and part of the spur wheel is to be shown ; the teeth may be represented by their pitch circles. LOGARITHMS. 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 0000 0414 0792 "39 146 1 1761 2041 2304 2553 2788 3010 3222 3424 3617 3802 3979 4150 4314 4472 4624 4771 4914 5°Si 5185 5315 S44I 5563 5682 5798 5911 602 1 6128 6232 6335 643s 6532 6628 6721 6812 6902 0043 0453 0S28 "73 1492 0086 0492 0864 1206 1523 1790 1818 2068 2330 2577 2810 3032 3243 3444 3636 3820 3997 4166 4330 4487 4639 4786 4928 5065 5198 5328 5453 SS75 S694 5809 5922 6031 6138 6243 634s 6444 6542 6637 6730 6821 6911 2095 2355 2601 2833 3054 3263 3464 3655 3838 4014 4183 4346 4502 4654 4800 4942 5079 5211 5340 5465 5587 5705 5821 5933 6042 6149 6253 6355 6454 6551 6646 6739 6830 6920 0128 0531 0899 1239 1553 1847 2122 2380 2625 2856 3075 3284 3483 3674 3856 4031 4200 4362 4518 4669 4814 4955 5092 5224 5353 5478 559? 5717 5832 5944 6053 6160 626 6365 6464 6561 6656 6749 6839 6928 0170 0569 0934 1271 1584 187s 2148 2405 2648 2878 3096 3304 3502 3692 3874 4048 4216 4378 4533 4683 4829 4969 5105 5237 5366 5490 561 1 5729 5843 59SS 6064 6170 6274 6375 6474 6571 6758 6848 6937 0212 0607 0969 1303 1614 1903 2175 2430 2672 2900 3"8 3324 3522 37" 3892 4065 4232 4393 4548 4698 4843 4983 5"9 5250 5378 5502 5623 574° 5855 5966 6075 6180 6284 6385 6484 6580 6675 6767 6857 6946 6 0253 0645 1004 1335 1644 1931 2201 2455 2695 2923 3139 3345 354J 3729 3909 4082 4249 4409 4564 4713 4857 4997 5132 5263 S39I 5514 5635 S7S2 5866 5977 6085 6191 6294 639s 6493 6590 6684 6776 6866 6955 0294 0682 1038 1367 1673 1959 2227 2480 2718 2945 3160 3365 3560 3747 3927 4099 4265 4425 4579 4728 4871 501 1 5145 5276 5403 5527 5647 5763 5877 5988 6096 6201 6304 6405 6503 6599 6693 6785 6875 6964 8 0334 0719 1072 1399 1703 1987 2253 2504 2742 2967 3181 3385 3579 3766 3945 4116 4281 4440 4594 4742 4886 5024 5159 5289 5416 5539 5658 5775 5999 6107 6212 6314 641S 6513 6609 6702 6794 6884 6972 0374 07S5 1 106 1430 1732 2014 2279 2529 2765 2989 3201 3404 3598 3784 396S 4133 4298 4456 4609 4757 4900 5038 5172 S3°2 5428 SSSi 5670 5786 5899 6010 6ti7 6222 6325 6425 6522 6618 6712 6803 6893 12 3 913 812 812 711 711 7 10 7 10 7 10 6 9 I 2 I 2 I 2 I 2 I 2 4 5 6 17 21 26 16 20 24 151923 151922 14 1821 14 1720 13 1620 13 16 19 12 15 19 12 IS 17 II 14 17 n 14 17 II 14 16 10 13 16 10 13 IS 10 12 IS 9 12 14 9" 14 9 " 13 8 II 13 8 II 13 8 ID 12 8 10 12 7 9" 7 911 7" 7 6 9 10 810 8 9 6 6 6 6 5 S 5 S 5 5 S 4 4 4 4 4 4 456 4 S 6 4 S 5 4 4 5 4 4 5 3034- 28321 27313s 26 30 33 2S 28 32 242731 23 26 30 22 25 29 22 2S 2b 20 23 26 20 23 26 19222s 19 22 24 18 21 23 182023 172022 16 19 21 16 18 21 16 18 20 151719 15 17 19 14 16 18 141517 131517 12 14 16 12 14 IS "i3'S II 13 14 II 12 14 10 12 13 10 II 13 10 II 12 9 II 12 9 10 12 9 roil LOGARITHMS. 1 2 3 4 5 6 7 8 ^ 123 456 789 50 6990 6998 7007 7016 7024 7033 7042 7050, 7059 7067 I 2 3 3 4 5 678 51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 I 2 3 3 4 5 678 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 122 3 4 5 677 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 122 3 4 5 667 54 A324 7332 7340 7348 7356 7364 7372 7380 7388 7396 122 3 4 5 667 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 122 3 4 5 567 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 122 3 4 5 567 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 I 2 2 3 4 5 567 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 112 3 4 4 567 59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 112 3 4 -^ 567 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 112 3 4 4 5 6 6 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 112 3 4 4 566 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 112 3 3 4 5 6 6 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 112 3 3 4 5 56 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 I I 2 3 3 4 556 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 112 3 3 4 5 5 6 66 8195 8202 8209 821S 8222 8228 8235 8241 8248 8254 112 3 3 4 556 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 1 I 2 3 3 4 5 5 6 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 112 3 3 4 456 69 8388 839s 8401 8407 8414 8420 8426 8432 8439 8445 I 1 2 234 456 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 112 234 456 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 I 1 2 234 4 5 5 72. 8573 8579 858s 8591 8597 8603 8609 8615 8621 8627 112 234 4 5 5 73 8633 8639 864s 8651 8657 8663 8669 8675 8681 8686 I I 2 234 4 5 5 74 ,8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 I I 2 234 4 5 5 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 1 I 2 233 4 5 5 76 8808 8814 8820 8S25 8831 8837 8842 8848 8854 8859 I I 2 233 4 5 5 .77 886s 8871 8876 8882 8887 8893 8899 8904 8910 8915 112 233 4 4 5 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 I 1 2 2 3 3 4 4 5 79 8976 8982 8987 8993 8998 9004 9009 901 5 9020 9025 I I 2 233 44 5 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 I I 2 233 4 4 5 81 9090 9096 9101 9106 9112 9117 9122 9128 9133 I 1 2 233 4 4 5 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 112 233 4 4 5 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 I I 2 2 3 3 4 4 5 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 I I 2 233 4 4 5 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 1 I 2 233 4 4 5 86 9345 9350 9355 9360 9365 9370 9375 9380 938s 9390 1 1 2 233 4 4 5 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 I 1 223 3 4 4 88 9445 9450 9455 9460 9465 9469 9474 9479 94«4 9489 1 1 223 3 4 4 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 I I 223 3 4 4 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 1 1 223 3 4 4 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 95^3 Oil 223 3 4 4 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 1 1 223 3 4 4 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 1 I 223 3 4 4 94 9731 9736 974' 9745 9750 9754 9759 9763 9768 9773 1 1 223 3 44 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 1 1 223 3 44 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 Oil 223 3 4 4 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 Oil 223 3 4 4 98 9912 9417 9921 9926 9930 9934 9939 9943 9948 9952 Oil 223 3 4 4 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 Oil 223 3 3 4 ANTILOGARITHMS. 00 1 2 3 4' 5 6 7 8 9 123 456 789 1000 1002 1005 1007 1009 1012 1014 1016 1019 1021 001 I I I 2 2 2\ 01 I023 1026 102S 1030 1033 1035 1038 1040 1042 1045 I 1 I I 2 2 2 •02 1047 1050 1052 1054 1057 1059 1062 1064 1067 1069 I I I I 2 2 2 03 1072 1074 1076 1079 1081 1084 1086 1089 1091 1094 001 I I I 2 2 2 ■04 1096 1099 1102 1104 1 107 1 109 1112 1114 1117 1119 I I I I 2 2 2 2 05 II22 1 125 1 127 1 130 1132 "35 "38 1 140 "43 1 146 I I 112 2 2 2 ■06 1 148' 1151 "53 1156 "59 1161 1 164 1 167 1 169 1172 I I I I 2 2 2 2 •07 II75 1 178 1 180 1 183 1186 1 189 1191 "94 1 197 "99 I I I I 2 2 2 2 ■08 1 202 1205 1208 1211 1213 1216 1219 1222 1225 1227 oil I I 2 223 ■09 1230 1233 1236 1239 1242 1 245 1247 1250 1253 1256 oil I I 2 223 10 I2S9 1262 1265 1268 1271 1274 1276 1279 1282 1285 oil I I 2 223 •11 1288 I29I 1294 1297 1300 1303 1306 1309 1312 1315 oil I 2 2 223 •12 I3I8 I32I 1324 1327 1330 1334 1337 1340 1343 1346 I I I 2 2 223 •13 1349 1352 1355 1358 1361 1365 1368 1371 1374 1377 oil I 2 2 233 •14 1380 ■384 1387 1390 1393 1396 1400 1403 1406 1409 I 1 I 2 2 233 ■15 I4I3 I4I6 1419 1422 1426 1429 1432 1435 1439 1442 oil 12 2 233 •16 1445 1449 1452 1455 1459 1462 1466 1469 1472 1476 oil I 2 2 2 3 3 •17 1479 1483 i486 1489 1493 1496 1500 1503 1507 1510 oil I 2 2 233 •18 1514 I5I7 1521 1524 1528 1531 1535 1538 1542 1545 oil 12 2 233 ■19 1 549 1552 1556 1560 1563 1567 1570 1574 1578 1581 oil 12 2 3 3 3 •20 1585 1589 1592 1596 1600 1603 1607 1611 1614 1618 I I 12 2 3 3 3 •21 1622 1626 1629 1633 1637 1641 1644 1648 1652 1656 I I 2 2 2 3 3 3 •22 1660 1663 1667 167 1 1675 1679 1683 1687 1690 1694 I I 2 2 2 3^3 3 ■23 1698 1702 1706 1710 1714 1718 1722 1726 J 730 1734 I I 2 2 2 3 3 4 •24 1738 1742 1746 1750 1754 1758 1762 1766 1770 1774 I I 2 2 2 3 3 4 •25 1778 1782 1786 1791 1795 1799 1803 1807 tSii 1816 I I 2 2 2 3 3 4 •26 1820 1824 1828 1832 1837 1 841 1845 1849 1854 1858 oil 223 3 3 4 •27 1862 1866 1871 1875 1879 1884 1888 1892 1897 1901 I I 223 3 3 4 •28 190S I9I0 1914 1919 1923 1928 1932 1936 1941 1945 I I 223 3 44 •29 1950 1954 1959 1963 1968 1972 1977 1982 1986 1991 I I 223 3 4 4 •30 ■995 2000 2004 2009 2014 2018 2023 2028 2032 2037 I I 223 3 4 41 •31 2042 2046 2051 2056 2-061 206s 2070 2075 2080 2084 I I 2 2 3 3 4 4 •32 2089 2094 2099 2104 2109 2113 2118 2123 2128 2133 oil 223 3 44 •33 2138 2143 2148 2153 2158 2163 2168 2173 2178 2183 I I 2 2 3 3 44 •34 2188 2193 2198 2203 2208 2213 2218 2223 2228 2234 I I 2 233 4 4 5 ■35 2239 2244 2249 2254 2259 2265 2270 2275 2280 2286 I I 2 233 4 4 5 ■36 2291 2296 2301 2307 2312 2317 2323 2328 2333 2339 I I 2 233 44 5 ■37 2344 235° 2355 2360 2366 2371 2377 2382 2388 2393 I I 2 233 4 4 5 '38 2399 2404 2410 2415 2421 2427 2432 2438 2443 2449 I I 2 233 44 5 ■39 2455 2460 2466 2472 2477 2483 2489 2495 2500 2506 I I 2 233 4 5 5 ■40 2512 2518 2523 2529 2535 2541 2547 2553 2559 2564 I I 2 234 4 5 5 ■41 2570 2576 2582 2588 2594 2600 2606 2612 2618 2624 I I 2 '234 4 5 5 ■43 2630 2636 2642 2649 2655 2661 2667 2673 2679 2685 I I 2 234 4 5 6 ■43 2692 2698 2704 2710 2716 2723 2729 2735 2742 2748 112 3 3 4 456 ■44 2754 2761 2767 2773 2780 2786 2793 2799 2805 2812 I [ 2 3 3 4 456 45 2818 2825 2831 2838 2844 2851 2858 2864 2871 2877 I I 2 3 3 4 556 •46 2884 2891 2897 2904 2911 2917 2924 2931 2938 2944 I I 2 3 3 4 556 •47 2951 2958 2965 2972 2979 2985 2992 2999 3006 3013 I I 2 3 3 4 5S6 ■48 3020 3027 3034 3041 3048 3055 3062 3069 3076 3083 I I 2- 3 4 4 566 ■49 3090 3097 3105 3II2 3"9 3126 3133 3141 3148 31SS 112 3 4 4 566 r 1 2 3 4 5 6 7 8 9 123 4 5 6 7 8 9 50 3162 3170 3177 3184 3192 3199 3206 3214 3221 3228 112 3 4 4 5 6 7 ■51 3236 3243 3251 3258 3266 3273 3281 3289 3296 3304 I 2 2 3 4 5 5 6 7 ■52 33" 3319 3327 3334 3342 3350 3357 3365 3373 3381 122 3 4 5 S 6 7 ■53 3388 3396 3404 3412 3420 3428 3436 3443 345' 3459 12 2 3 4 5 6 6 7 ■54 3467 347S 3483 3491 3499 3508 3516 3524 3532 3540 12 2 3 .4 5 667 55 3548 3SS6 3565 3573 3581 3589 3597 3606 3614 3622 I 2 2 3 •■4 5 677 ■56 3631 3639 3648 3656 3664 3673 3681 3690 3698 3707 I 2 3 3 4 5 678 •57 371S 3724 3733 3741 3750 3758 3767 3776 3784 3793 I 2 3 3 4 5 678 ■58 3802 381 1 3819 3828 3837 3846 3855 3864 3873 3882 I 2 3 A 4 5 678 ■59 3890 3899 3908 3917 3926 3936 3945 3954 3963 3972 I 2 3|4 S 5 678 ■60 3981 3990 3999 4009 401& 4027 4036 4046 4055 4064 I 2 3 4 S 6 678 ■61 4074 4083 4093 4102 4111 4121 4130 4140 4150 4159 I 2 3 4 5 6 789 ■62 4169 4178 4188 4198 420> 4217 4227 4236 4246 4256 I 2 3 4 5 6 789 ■63 4266 4276 428s 429s 4305 431S 4325 4335 4345 4355 I 2 3 4 5 6 789 ■64 4365 437S 4385 4395 4406 4416 4426 4436 4446 4457 I 2 3 4 5 6 7 8 9 65 4467 4477 4487 4498 4508 4519 4529 4539 4550 4560 I 2 3 4 5 6 789 ■66 4571 4581 4592 4603 4613 4624 4634 4645 4656 4667 I 2 3 4 5 6 7 9 10 •67 4677 4688 4699 4710 4721 4732 4742 4753 4764 4775 I 2 3 4 5 7 8 9 10 '68 '^l^^. 4797 4808 4819 4831 4842 4853 4864 487s 4887 I 2 3 4 6 7 8 9 10 ■69 4898 4909 4920 4932 4943 4955 4966 4977 4989 5000 I 2 3 5 6 7 8 9 10 70 5012 S023 5035 5047 5058 5070 5082 5093 5105 5"7 I 2 4 5 6 7 8 9 II ■71 S129 5140 5152 5 164 5176 S188 5200 5212 5224 5236 I 2 4 5 6 7 8 10 II ■72 5248 5260 5272 5284 5297 5309 5321 5333 5346 5358 I 2 4 5 6 7 9 10 II •73 S370 5383 5395 5408 S420 5433 5445 5458 5470 1 3 4 5 6 8 9 10 II ■74 S49S 5508 5521 5534 5546 5559 5572 5585 5598 5610 1-3 4 5 6 8 9 10 12 75 5623 5636 5649 5662 567s 5689 5702 571S 5728 5741 I 3 4 5 7 8 9 10 12 ■76 S7S4 5768 5781 5794 5808 5821 5834 5861 5875 I 3 4 S 7 8 9 11 12 ■77 5888 5902 5916 5929 5943 5957 5970 5984 5998 6012 I 3 4 5 7 8 10 II 12 •78 6026 6039 6053 6067 6081 6095 6109 6124 6138 6152 I 3 4 6 7 8 10 II 13 ■79 6166 6I80 6194 6209 6223 6237 6252 6266 6281 6295 I 3 4 6 7 9 10 II 13 80 6310 6324 6339 6353 6368 6383 6397 6412 6427 6442 I 3 4 6 7 9 10 12 13 •81 6457 6471 6486 6501 6S16 6531 6546 6561 6577 6592 235 6 8 9 II 12 14 •82 6607 6622 6637 6653 6668 6683 6699 6714 ^11° 6745 235 6 8 9 II 12 14 •83 6761 6776 6792 6808 6823 6839 6855 6871 6887 6902 2 3 5 6 8 9 II 13 14 •84 6918 6934 6950 6966 6982 6998 7015 7031 7047 7063 235 6 8 10 " 13 IS 85 7079 7096 7112 7129 7145 7161 7178 7194 721 1 7228 2 3 5 7 8 10 12 13 IS •86 7244 7261 7278 7295 73" 7328 7345 7362 7379 7396 2 3 5 7 8 10 12 13 IS •87 7413 7430 7447 7464 7482 7499 7516 7534 7551 7568 2 3 5 7 9 10 12 14 16 '88 7586 7603 7621 7638 7656 7674 7691 7709 7727 7745 24 5 7 9 n 12 14 16 •89 7762 7780 7798 7816 7834 7852 7870 7889 7907 7925 245 7 9 II 13 14 16 90 7943 7962 7980 7998 8017 803s 8054 8072 8091 8110 246 7 9 11 13 IS 17 •91 8128 8147 8166 8185 8204 8222 8241 8260 8279 8299 246 8 9 II 13 15 17 •92 8318 8337 8356 8375 8395 8414 8433 8453 8472 8492 246 8 10 12 14 IS 17 •93 8511 8531 8551 8570 8590 8610 8630 8650 8670 8690 246 8 10 12 14 16 18 ■94 8710 8730 8750 8770 8790 8810 8831 8851 8872 8892 246 8 10 12 14 16 18 US 8913 8933 8954 8974 8995 9016 9636 9057 9078 9099 246 8 10 12 IS 17 19 •96 9120 9141 9162 9183 9204 9226 9247 9268 9290 93" 246 8 II 13 15 17 19 ■97 9333 9354 9376 9397 9419 9441 9462 9484 9506 9528 247 9 " 13 IS 17 20 ■98 955° 9572 9594 9616 9638 9661 9683 9705 9727 9750 247 9 " 13 16 18 20 99 9772 9795 9817 9840 9863 9886 9908 9931 9954 9977 2 5 7 9 " 14 16 18 20 VAT,UES OF SINE. COSINE, Etc. Ingle. m,nrf1e Sine, Tangent, Cotangent. Cosine. Beg. Radians. UAoms, 0° ai 1 1^414 1^5708 90° 1 •0175 ■017 ■0175 •0175 67^2900 •9998 1^402 1^563S 89 2 •0349 •035 ■0349 •0349 23^6363 ■9994 1^389 1^6359 88 ^3 •0524 •952 •0523 ■0524 19-0811 •9986 1^377 1^5184 87 4 ■0698 •070 •0698 ■0699 14^3006 •9976 1^364 1^5010 86 B •0873 •087 •0872 •0875 11-4301 •9962 1^361 1^4836 86 i •1047 •105 •1045 ■1051 9-5144 ■9945 1^338 1^4661 84 1 •1222 •122 •1219 ■1228 8-1443 •9925 1^325 1-4486 83 8 •1396 •139 •1392 ■1406 r-1154 •9903 1^312 1-4312 82 9 •1571 •157 •1664 •1684 6-3138 •9877 1^299 1^4137 81 10 •1745 •174 •1736 •1763 6-6713 •9848 1^286 1^3963 80 11 •1920 ■192 •1908 - •1944 5-1446 •9816 1-272 1^3788 79 12 •2094 •209 •2079 •2126 4-7046 '9781 1-269 1^3614 78 13 •2269 •226 •2260 •2309 4-3316 •9744 1-24^ 1^S439 77 14 ■2443 •244 •2419 •2493 4-0108 •9703 1-231 1^3265 76 IS •2618 •261 •2588 ■2679 3-7321 •9659 1-217 1-3090 76 16 •2793 •278 •2756 •2867 3-4874 •9613 1-204 1-2916 74 17 •2967 •296 •2924 ■3057 3-2709 •9563 1-190 1^2741 73 18 •3142 •313 •3090 ■3249 S-0777 •9611 1-176 1-2566 72 19 •3316 •330 •3266 ■3443 2-9042 •9455 1-161 1-2392 71 20 •3491 •347 ■3420 ■3640 2-7475 •9397 1-147 1-2217 70 21 •3665 ■364 •3684 ■3839 2-6061 •9336 1^133 1-2048 69 22 •3840 ■382 •3746 ■4040 2-4751 •9272 1^118 1-1868 68 28 •4014 ■399 •3907 •4245 2-3669 •9206 1^104 1-1694 67 24 •4189 •416 •4067 •4452 2-2460 •9136 1^089 1-1519 66 25 •4363 •433 •4226 •4663 2-1445 •9063 1^076 1-1346 65 26 •4538 •450 •4384 •4877 2 ■0603 •8988 1^060 1-1170 64 27 •4712 •467 •4540 ■6095 1^9626 •8910 1^045 1-0996 63 23 •4887 •484 •4695 ■5317 1^8807 •8S29 1-030 1-0821 62 29 •6061 •601 •4848 ■6643 1-8040 •8746 1-016 1-0647 61 30 •6236 •818 •5000 ■5774 ' 1-7321 -8660 1-000 1-0472 60 31 •6411 ■534 •6150 ■6009 1-6643 -8572 ■986 1-0297 59 32 •6586 -•551 ■6299 ■6-249 1-6003 •8480 ■970 1-0123 68 33 •5760 ■568 •5446 ■6494 1-6399 •8387 ■964 -9948 57 84 ■6934 ■585 •6692 ■6745 1-4826 •8290 ■939 •9774 56 35 ■6109 ■601 •5736 •7002 1^4281 •8192 •923 •9599 55 36 •6283 ■618 •5878 •7265 1-3764- •8090 ■908 •9425 54 37 ■6458 ■635 •6018 •7536 1-3270 •7986 ■892 -9260 53 38 •6632 ■651 •6157 •7813 1-2799 •7880 ■877 -9076 52 39 •6807 ■668 ■6293 •8098 1-2349 •7771 ■861 -8901 61 40 •6981 •684 ■6428 •8391 1-1918 •7660 ■846 •8727 60 41 •7166 ■700 ■6561 •8693 1^1504 •7547 •829 •8662 49 42 ■7330 ■717 ■6691 •9004 1-1106 •7431 ■813 •8378 48 43 ■7506 ■733 ■6820 •9325 1-0724 •7314 ■797 ■8203 47 44 •7679 ■749 ■6947 •9667 1-0365 ■7193 •781 ■8029 46 46 •7864 •765 ■7071 l^OOOO 1-0000 ■7071 ■765 ■7864 45 Radians. Beg. Cosine. Cotangent. Tangent. Sine. Chords. Angle. 1 USEFUL DATA. ir=31416 or 3142, i=0-3183, 7r2=9-870. IT 1 in. =2-54 oni.=25-4 mm., 1 metre=39-37 in. =3'281 ft. = 1094 yds. 1yd. =0-9144 metre, 1 mile = 1760 yds. = 5280 ft. = 1 -609 kilom. 6 ft. =1 fathom, 1 kilom. =0-631 mile, 1 knot =6080 ft. per hour. 1 sq. in. =6-451 sq. cm., 1 sq. yd. =08361 sq. m. 1 acre =4840 sq. yds. =0-4047 hectare. 1 cub. in. = 16 -39 c. c. , 1 litre = 1000 o. c. = 1 kg. 1 gallon of water, weighs 10 lb. =0-1604 cub. ft. =4-543 litres. 1 kilog. weighs 2^ lb. =1000 gm. =0-001 tonne. lib. weighs 16 oz. =7000 grains ' =453-6 gm. 1 oub. ft. of water weighs 62-3 lb., 1 radian =57 -3 degrees. 3=32 ft. per sec. per sec. log ir=0-4972 log 2-718 = 0-4343 log 0-7854=1-8951 log 62-3=1-7945 log 1728 = 32375 log 05236 = r-7190 log 0-1604 = 1-2052 To convert common to hyperbolic logarithms multiply by 2-3026 (2-303). (e =2-718.) Volts X amperes = watts. 1 horse-power = 33000 ft. lb. min. = 746 watts. 1 kilowatt =102 kilogram metres per sec. Weight in lb. per cub. in. : Cast iron 0-26, Wrought iron 0-28, Steel 0-284, Brass 029, Copper 0-319, Lead 0412, Tin 0-267, Zinc 026, Aluminium 0-097, Water 0-036, Antimony 0-24, Manganese 0-289. 1 atmosphere=14-7 lb. per sq. in.=2116 lb. per sq. ft. = 10° dynes per sq. cm. A column of water 23 ft. high gives a pressure of 1 lb. per sq. in. Table or Safe Compressive Loads (Average Values). Material. Tona per sq. ft. Material. Tons per sq. ft. Stock bricks in cement ,, ,, lias mortar ,, ,, lime mortar Blue brickwork in lime mortar ,, „ cement mortar 6 6 4 10 12 Limestone Sandstone Hard York stone Portland stone Granite 9 15 15 18 30 335 ANSWERS. Exercises I. (p. 4). 1. 162-4 sq. in. 4. 1-18 in. 7. Copper 504 lb 8. 287-8 lb. 10. 1-48lb. ; 2-64 lb. ; 5-94 lb 11. (i) 123-7 cub. in., 32-18 lb. 12. 2,892 lb. ; 3,259 lb. 14. 74-85 lb., 3-43 ft. 4,492 cub. in., 1,1 68 lb. 16 18, 21 23 SOfin. 2. 28i in. 3. 5.' 1 3-29 in. 6. 282-3 sq. ft. zinc 53-2 lb., phosphorus 2-8 lb. 9. 21 -9 lb., 51 lb. (ii) 5-85 lb. ; (iii) 1 -37 in. 13. 436-4 lb. 15, 159-3 lb. 17. 7-1 lb. 20. 65 lb. 1 -78 lb. 19. 1 5| sq. in., 53-26 lb. 6-1 tons, 18-47 in. 22. 38-9 lb. 13-81 sq. ft., 10,970 lb. 24. 6 tons (seep. 116). Exercises II. (p. 13). Force in Member. Tons. Push. Pull. Reaction at A „ c AB - BC AC 1 3-77 1-23 4-6 2-89 2-6 2. Push in rafter 1 4-4 tons, pull in tie-bar 1 2-7 tons. 336 ANSWERS 337 3. 12^430 lb., 2,372 1b. 4. 7-2 tons, 3-95 tons, 1 -12 in., rf = 0-76 in. 5. 0-78 ton, 2-83 tons. 6. 0-42 ton. 7. 200 lb., 3.46-3 lb. 8. Tie-bar 80 cwt., jib 100 cwt. 9. Member AB BC CD DF 'CA Force in member. Tons 15 12 14-3 14-3 4-1 10. Member. Force in Member. Tons. Push. _ Pull. AC 1-25 CB 2-35 — BD 2-5 — AB — 2-35 12. Force in Member. Member. Tons. Push. PuU. GF — 6,750 FE — 4,500 ED — 2,300 DC 2,000 — CB 2,000 — BA 4,000 . — BE 2,200 — AF 2,9Q0 — 11. 11 -54 tons. Exercises III. (p. 31). 2. 3,600 lb. per sq. in., 0-0001 25 in., 28-8 x 1 0". 3. 303-8 lb. 4. 1 5,220 lb. per sq. ia., 29-97 x 10^. 5. 8,071 lb., 18,270 lb. per sq. in. 6. 99-6 lb. 7. 1 7,320 lb. per sq. in., 0-0577 in. 8. 14,400 lb. per sq. in., 0-0432 in. 9. 0-75 in. 10. (a) 13,690 lb. per sq. in., (6) 0-0004563, (c) 102-9465 in. 11. 3-9 in. X 3-9 in. x | in. 12. 4,851 tons per sq. in., 27-16 tons per sq. in. 13. 4-88 tons per sq. in., 0-0937 in. 14. 27-62 tons per sq. in., 24-5 per cent. 15. 1 72-3 sq. in., 1 -05 in. 16. 1 92-9 tons, 0-054 in. 17. 31 ,920 lb. per sq. in. C.M.D. T 338 MACHINE DESIGN 18. 29,000 lb. per sq. in., 28,520 lb. per sq. in., 29,000 lb. per sq. in. 19. (a) 361 -4 lb. per sq. in. ; (6) 5,421 lb. per sq. in. 20. 7,721 lb. per sq. in., 4-7. ' 21. 4-25in. by4-25in. by0-5in. 22. 4-6 ia. 23. 1 79-1 lb. .24. 30,080 lb. per sq. in., 0-001 . 25. 33,330 lb., 1 6,665 lb. per sq. in., 0-00056, 1 -488 in.-tons. 26. 19-98 in. 27. 71 -96 in., 79-4°. 28. 6-7 tons per sq. in. Exercises IV. (p. 60). 1. 60 per cent. 2. (a) By shearing (&) by tearing. 3. (a) 2-08 in., 57-9 per cent. ; (6) 1 -81 ia., 51 -7 per cent. ; (c) 1 -78 in., 50-8 per cent. 4. (a) 1-57; (&) 0-95 in., 56-6 per cent. 6. 0-598 in., or fin. 1. 70-3 lb. per sq. in., 1 5,280 lb. per sq. in. 8. 1 1 6 lb. per sq. in. 9. 1 ,1 20 lb. per sq. in., 1 ,093 (see p. 52). 10. 0-842 in., or J in. 11. 0-8 in., or ^ in. 13. 5-93 in. 14. (a) 4-94 in. ; (&) 6-18 in. ; (c) 7-42' in. 15. 34-4 tons. 16. 0-804 in., or \% in. 17. 0-849 ia. or ^ in., 2-54 in., 66-6 per cent. ■ 18. 258 cub. ft., 3-07 in., 0-376 in. or | in. (see p. 49). 19. 0-18 in. 20. 5-2 tons per sq. in., 3-2 tons per sq. in. 21. (a) 6 tons per sq. ia. ; (&) 7-6 tons per sq. in. 24. 2-8 tons per sq. in. ■ 25. 1 -83 in., 6 tons per sq. in. Exercises V. (p. 77). 1. rfi = 0-637in., rf=|in. 2. (yi = 1-2ia., rf = 1lin. 3. (/i = 0-597in., (/ = |in. 4. 2-47 in. 5. 2-58 in. 6. 1 -3 in. 7. 3-1 in. 8. 3-79 in. 9. 2,228 lb. per sq. in. 10. 3-16 tons. 11. 31 -5 tons. 12. (/i = 1-89 in. 13. 1 -78 in., 1 -26 in. (p. 72). 14. 0-87 in., or | in. 15. 0-83 in. 17. 77-3 lb. 18. 26. 19. 2-25 in., 2-39 in. 20. 2-67 in. 21. 6-9 in., 0-041 in. 22. 2-3 tons per sq. in. 23. 2-1 in., or 2| in. ANSWEES 339 Exercises VI. (p. 85). 1, D = 1-22(/, 6 = 1-33rf, t = 0-26rf; 3 in., 3-3 in., 0-78 in. 3. 4 in. 4. 72-5 lb. 5. 4-175 in. 6. 6 = 1-13rf, h = ^■77d, f = 0-278rf, 1 -69 in., 2-65 in., 0-42 in. 7. D = l-23£/, 6 = 1-25rf, t = 0-31rf, 1-23in., 1.25 in., O'SI in. 8. D = 3-5 in., 6 = 2-4in., t = 0-56 in. Enlarged part 2-15 in. 9. 1 -8 in., 9-7 tons per sq. in. Exercises VII. (p. 124). 1, 1 ,750 lb. 2. 563-5 lb. per ft. run. 3. 756-1 lb. per sq. in. 4. 1666-7 lb. J 5. 17-92 in.-units, 35-8 in.-tons. 6. 472-6 in.-tons, 6-26 tons per sq. in. 7. (a) 1 5 in. ; (6) 750 in.-tons ; (c) 1 sq. in. 8. 1 ,51 2 lb. per ft. run. , 9. 1 -058 in., 7-37 in.-tons. 10. 8-1 5 in. 11. (a) 724-6 in.-units ; (6) 386-5 in.-tons ; (c) 1 ton per ft. run.. 12. 5;334 lb. per sq. in., 10,668 lb. per sq. in., 16,000 lb. per sq. in. 13. 1 -506 tons per ft. run. 14. 536-5 lb. per ft. run. 15. Flanges 6 in. by 0-987 in., web 0-5 in., depth 14 in. 16. 1 -08 tons per sq. in. 17. 4 in. by 1 2 in. 18. 730-8 in.-units, 91 -35 in.-units, 9-1 tons. 19. 21 1 -7 in.-tons. 20. 1 2,090 lb. per sq. in. 21, 933-3 lb. per sq. in. 22. 6,500 lb. 23. 4 tons per sq. in. 24. 8-72 in. from upper face, 102,700 in.-lb., 2,500 lb. pei^sq. in. 25. 1 -94 in. by 9-7 in. 26. 1 -05 tons per ft. run. 27, Section 1 2 in. by 5 in. 28. 7-5 in.-units, 1 -1 . , 29. 276,700 in.-lb. 31. Depth 1 8 in., flanges 9-6 in. by 1 in., web | in. thick. 32, 1 2,000 lb. per sq. in., 9-9 per cent. 33. 1 1 ,1 00 lb. per sq. in. 34, 1 -13 in., or 14 in. ; 1 -9 in., or 1|| in. C.M.D. y2 340 MACHINE DESIGN Exercises VIII. (p. 137). 1. 21 ,000 in.-lb. 2. 63 rev. per min. 8. 26,260 in.-lb., 2-46 in. 4. ,-^/D», 67,420 in.-lb. 5. 12-66 in. 6. 50, 1 -786 in. 7. 1 ,480,000 in.-lb., 41 2-5, 1 ,1 74. 8. 5-264 in. 9. (i) 79-36; (ii) 43-81. 10. 3-975 in. 11. 8-624 in. 12, 4,101 in.-tons, 14,570. 13. 5^3 : 6. 14. 3,663 in.-tons, 9,1 1 1 . 15. 1 1 -27 in. 16. 7-264 in., 5-1 in. 17. 9-54 in., 7-63 in., 87-8 lb. 18. 3-4 in., 3-5, 1 -75 in. 19. 4 in. 20. 1 06. 21. 4-1 in., 3-8 in. 22. 4,500 in.-lb., 6,000 in.-lb., 5,250 in.-lb., 51,540 in.-lb. 23. 15-91 in., 9-55 in. 24. 0-5636 : 1. 25. 0-7374 : 1 . 26. 6-656 in. Exercises IX. (p. 150). 2. 4. 3. 2-62 in., 0-595 in., or | in. . 4. 4,061 lb. per sq. in. 5. 7,141 lb. per sq. in. 6. 1 -34 in., or If in. 7. (a) 1 1 -66 in. ; (6) 11 -91 in., 5-95 in. ; (c) radius of bolt circle 9-3 in., diameter 2-3 in. 8. 4-1 8 in. 9. 12,930. 10. 63,020 in.-lb., 6 in. 11. 0-91 8 in., or 1 in., 4-5 in. 12. 1 -93 in., or 2 in. 13. 787-7 ia.-lb., 649-4 lb. per sq. in. 14. 0-62 in., or | in. 15. 3 in., 1 -84 in. Exercises X. (p. 157). 1. 1 0-37 X 1 05 in.-lb., 6,582. 2. 4-93, 2-1 5°. 3. (a) 5-45 in. ; (6) 4-15 in. 4. 3-3 in., 4-4 in. 5. 3-44 in. 6. 3-43 in., 29-5. 7. 21-1 in.-tons, 4-2°. 8. 27-79 in.-tons. 9. 26-81 in.-tons, 95-3. 10. 51 -8°, ANSWERS 341 11. 1 2-89 in. 12, 1 2,1 70,000 lb. per sq. in. 13. 2,743. 14. 5.74 in.^ 1 ^1 1 1 lb. persq. in. 15. 13-1 in., 15 in. 16. 4-2°. Exercises XI. (p. 171). 1, (a) 2 ft., 4 ft. ; (6) 2-99 in. 2. 1 ,755 rev. per min. 3. 42 ia., 24 in. 4. 3-89 in. 5. 389 1b. 6. 4-3 in. 7. 18-75. 8. 3-46. 9. 1 0-67 in. 10. 31 6 lb. 11. 642-8 lb. 12. 5,889 lb. 13. 1 ,076 lb. 14. 251 3-6 ft. per min., 262-5 lb. 15. 8-76 in. 16. 1 ,797 lb., 2,994 lb. 17. 40-75. 18. 12-1. 19. 4.-4 in. ' 20. 1 5-8 in. 21. 342 lb. 22. T2/T1 = 2-353, 201 -3 lb. 23. 51 -5 in., 4 in. 24. 499-5 lb., 266-5 lb., 12 in. 25. 29. Exercises XII. (p. 179). 1. 784-8 lb. per sq. in. 2. 171 -5 rev. per min. 3. 1 1 1 -3 ft. per sec. 4. 1 76 ft. per sec. 5. 9,566 lb. 6. 170-2 ft. per sec. 7. 6-8 tons. 8.410-7. 9. 2,948, 0-0269 in. 10. 5-98 lb., 23-9 lb. per sq. in. 1 ,623 rev. per min. 11. 1 01 -6 ft. per sec. 12. 25-1 . 13. 2-385 lb. 14. V = V'/g/1 2x0-26. 15. 27. 16. 273-5. 17. 1,327 rev. per min. 18. 27-5. 19. 281 -8 lb., 52-1 . Exercises XIII. (p. 207). 1. 50, 150, 47-72 in. 2. 3,840 lb., wheels D and C force = 603-3 lb., wheels B and A force = 150-9 lb. 3. 240 rev., per min., 25-06 in. 4.95-2. 342 MACHINE DESIGN 5. 125 in. -units. 6. 0-48 in.'-units. 1. (a) 3-29 in., 2-6 in. ; (6) 2-8 in., 27, 54; (c) 36-1 in. 8. 2-83 in. 9. 2 in. 10. 666-7 lb. 11. 9,802 ft.-lb. 12. 7-82 in.-vinits, 3-48 lb. 13. 31 2,1 00 ft.-lb., 1 76,1 00 ia.-units. 14, 1 ,21 4 lb. 15. 99-8 rev. per min., 48,700 ft.-lb. 16. 14,090 in.-units, 24,300 ft.-lb. 17. (a) 69,490 in.-units ; (6) S, 100 ft.-lb. ; (c) 350,100 ft.-lb. 19. 1,575 lb., 2-18 in. 20. t9, 54, 59 in. 21. 37-5. Exercises XIV. (p. 219). 1. (a) 5 tons per sq. in. ; (6) 12-5 tons per sq. in., 2-5 tons per sq. in. 2. 2-01 3:1, 9,020 lb. per sq. in., 4,480 lb. per sq. in. 3. 4-1 7 tons per sq. in., 7-29 tons per sq. in., 1 -04 tons per sq. in. 4. 6,802 lb. per sq. in. 5. 2-94 tons per sq. in., 25° 42'. 6. 5-81 tons per sq. in., 34° 43'. 1. 5,342 lb. per sq. in., 61° 54' ; 3,432 lb. per sq. in., 17° 24'. 8. 14,161 lb. per sq. in. 9. 1 ,1 44 lb. per sq. in., - 430-9 lb. per sq. in. 10. 57T(//4a, 3-4:1. 11. 1 0-8 tons, 6,544 lb. per sq. in. r 12. 7-1 tons per sq. in. 13. 5-1 tons per sq. in., 50° 39' to axis of rivet. Exercises XV. (p. 227). 1. 5-49 in., 4-39 in. 2. 2-96 tons per sq. in., 0-0663 in., 9'5. 3. 34,260 lb. 4. 40-8 tons. 5. 1 -06 : 1 . 6. 1-1 in. 7. 37-2 tons. 8. 44-9 tons. 9. 1 38-2 tons. 10. 658 lb., 16,000 lb. 11. 32 tons. 12. .14-9 lb. 13. 4-75 in., 0-475 in. 14. 23-9 tons. ANSWERS 343 Exercises XVI. (p. 243). 1. 0-519 in. 2. cl■.l = ^ :21. 3. Tension 4 tons per sq. in. ; compression 3-55 tons per sq. in. 4. 4 ft., 99,540 in.- units. 5. Flanges 6-25 in. by 1 in., web 0-6 in., 1 -1 9 tons per sq. in., 0-067 in. 6. 0-229 in. 7. 383-3 lb. per sq. in., 0-957 in. 8. (a) 1 ,223 lb. per sq. in., (6) 2-64 in. 9. 1 3-7 in. 10. Z:rf = 13-6:1. 11. (a) 144in.-tons; (6) 4-628 tons per sq. in. ; (c) 0-1 99 in. 12. 4-88 tons per sq. ia., 0-0078 in. 13. 46-2 tons. 14. 22-6 ft. 15. 7 ft., 15,920 in.-units. 16. 0-09 in., 2,520 lb. 17. 1 ,51 7 lb. per ft. run, 1 2,450 lb. per sq. in. 18. 0-01 55 in. Exercises XVII. (p. 298). 1. 8-16 in. 2. 25,000 in.-]b., 2-42 in. 3. 3-1 in. 4. 1 1 5,200 in.-lb. 5. 7,532. 6. 6-1 in. 7. 11-8 in. 8. (a) 5-32 ia., 6-65 in. ; (6) 6-5 in., 1 2-3 in. 9. rf = 10-57 in., a = 10-3 ia., A = 11 -5 in. 10. 3-9 in. 11. 9-96 in., 5-48 in. 12. 7,954 lb. 13. 856-9. 14. IVl =54-55 in.-tons, T = 50 in.-tons, T, = 116-5 in.-tons, rf = 4-6 in., crank pin 3-25 in. by 4-1 ia. 15. (a) 5-9 in. by 7-4 in. ; (6) 7 in. by 8-25 in. 16. 4-4 in. 17. 1 1 -37 in. by 14-88 in. 18. 2-6 in. by 0-88 in. 19. 172-3 sq. in. 21. (a) 2-64 in., (fe) 4,481,000 in.-lb., 13-6 in. 22. 1 -44 tons per sq. in. 28. 1 1 ,920 lb. per sq. in. 24. 1 3,1 70 lb. per sq. in. 25. B = 2-39 in., or 2^ in. ; rf = 0-65 in., or |^ in. 26. 9-26 in. 27. (a) 4,000 lb. per sq. in. ; (&) 2j829 lb. per sq in. 29. 8 in. 30. (a) 2-1 in. ; (b) 6 in. ; (c) 3-1 ft. ; (d) M =6,000 in.-lb. ; T = 5,41 5 in.-lb. ; (e) T, = 1 4,081 in.-lb. ; (/) 2 in. 31. 5-48 in.-tons. 32. 14-56 in. 344 MACHINE DESIGN Miscellaneous Exercises XVIII. (p. 310). 1. (a) 0-00546 cub. ft. ; (6) 1 in. ; (c) 1 in. 2. (o) 3-1 2 in. ; (6) 1 ,533 lb. per sq. in. ; (c) 3,480. 3. (a) 179-3 lb. per sq. in. ; (6) rfi = 0-84in.,rf = 1 in. ; (c) 1 -31 in. 4. (a) 2,262 lb. per sq. in. ; (6) 3-223 tons ; (c) rfi = 0-616 in., rf = |in. 5. (a) 10-3 in. ; (6) 1 -26 in. ; (c) 8-47 in. 6. (a) 3-36 in. ; (6) rfi = 0-7494 in., rf = |- in, 7. (i) 1 -39 in. ; (ii) 262-5 lb. ; (iii)700lb. ; (iv) 5-8 in. 8. (a) 1 -95 in. ; (6) .2-5 in. ; (c) 5 in. Lancashire and Cheshire Union of Institutes (p. 319). I. (a)(/i = 1-95in. ; (&) A = 2-4 io., B = 5 in. II. rfi = 0-816in., B = 7-1 in., L = 5in., F = 1-65 in., E = 1-275in. 5. 33,624 in.-lb., 1 1 1 ,325 in.-lb, 6. 11,1 80 lb. per sq. in., 1 1 ,51 lb. per sq. in. The Union of Educational Institutions (p. 322). 1. (a) 2-76 in. ; (6) 2-68 in. 2. rf = 0-7499iQ., or0-75in. ; ,key 0-75 in. by 5-3 in., 136-3. 3. 1 ,955 lb. per sq. in. 4. 634-9 lb. per sq. in. 6. 0-34 in., or | in. (p 49). Board of Education (p. 324). 1. (a) 6 in. ; (h) of = 2 in. ; (o) 2-833, or 2| in. 2. (a) 2-2 in. ; (b) 3-18 in. ; (c) (i) 150 rev. per min., (ii) 100 rev. per min. 3. (a) D = 11 in., (/ = 10lin. ; (b)d=^ in.; (c)7|in. 4. (a) 200 rev. per min. ; (6) 93-3 lb. ; (c) 1 -88 in. INDEX. Acceleration, 10, 206 ; angular, 206 ; parallelogram and triangle oi, 8. Accumulator, hydraulic, 58 ; bolts, 111; cap, 110; foot or base plate, 111. American high-speed wheels, 176. Angle of advance, 268-271. Angle of torsion, 152 ; of twist, 153. Angles, steel, 21, 32. Angular, acceleration, 206, 207 ; momentum, 207 ; velocity, 207. Area, determination of, 1-3 ; centre of, 117, 118 ; of irregular figures, 1-3 ; table of, 102, 103. Arms, of pulleys, 164-166 ; of wheels, 188, 189. Axis, neutral, 94 ; of tee section, 125. Axles, bending moment in, 133-136 ; railway, 135. Beams, 88-108; bending in, 89-94 bending moment diagrams, 91, 92 cast-iron, 108, 117, 118, 241 continuous, 234, 235 ; cylindrical 108, 241 ; deflection in, 230-242 design of, 112-116 ; encastre, 91 235-237; flitch, 118, 119, 128 moment of- inertia in, 99, 100 moment of resistance in, 94101 nature of stresses in, 94, 95 neutral axis in, 94-99, 117, 118 propped, 234 ; reinforced con Crete, 119-122; sections of, 102, 103 ; shearing-force diagrams, 91 92 ; of I section,. 104, 106, 107 stresses due to bending in, 108; 109, 241 ; stiffness of, 230-242 timber, 107, 239-242. Belt pulleys, 160, 161 ; arms of, 164-166 ; naves of, 166 ; velocity ratio of, 160. Belts, 161-167 ; centrifugal tension in, 177-179 ; driving by, 160-167 ; friction of, 161-163 ; speed of, 168 ; horse-power transmitted by, 163-167 ; strength of, 161, 168 ; tensions of, 161, 163, 168 ; thick- ness of, 161 ; width of, 163, 164, 167. Bending and twisting combined, 254-264. Bending combined with axial load, 210-214. Bending moment, by calculation, 88, 89, 94; by link polygons, 92, 94, 134, 257; combined with torsion, 264-264 ; combined with tension or compression, 212-217 ; combined with torsion and axial force, 264, 265 ; in circular shafts, 133- 136 ; in connecting rods, 284-287; in coupling rods, 295- 298 ; diagrams of, 91, 94, 134, 137 ; resistance to, 94-124 ; table of, 91 ; work done in, 105-106. Bearing surface, 37, 47-49 ; crank pin, 252-254; crank shaft, 266- 264; journals, 196-198, 211; rivets, 37, 47-49. Bearings, 196; collar, 197, 200; crank shaft, 266-264 ; footstep, 196, 197; pedestal, 198-200. Bevel wheels, 195, 196 ; friction, 184 ; strength of, 196 ; teeth of, 195. Boiler pump, 310, 311. 345 346 INDEX Boiler stays, 73, 215, 216. Boilers, 44-46 ; combustion cham- ber, 109 ; dished and hemispheri- cal ends, 45 ; efficiency of joints, 40-43 ; riveted joints for, 35-44 ; stresses in shells, 44, 45 ; thick- ness of shell, 44-46. Bolts, 65 ; connecting rod, 287-289 ; coupUng, 143-149 ; cylinder, 68- 70, 247 ; eye, 76 ; flange, 68-70 ; hydrauUo pipe, 69. Bolts and screws, 65-67, 111 ; strength of, 65-68. Bow's notation, 11, 12. Box or muff couplings, 141. Brake gear, 314, 316. Brake pulley, 308-310. Buckling of struts, 220-227. Cantilevers, 89 ; bending and shear- ing in, 89-91 ; deflection in, 230. 231 ; propped, 235. Caps, connecting rod, 288-290 hydraulic, 110. Cast-iron, beams, 108, 117, 118, 241 columns, 20, 214 ; pipes, 49, 50. Centre of area, 117, 118. Centrtfugal force, 174-179 ; in belts, 177-179; in flywheel rim, 175 176. Chains, 14, 16. Circle, area and circumference of, 3 moment of inertia of, 99, 100, 103, Clapeyron's theorem, 234. Clearance of wheel teeth, 187, 188. Coefficient, of expansion, 4, 27-31 of friction, 73. Collars, 81 ; friction of, 200 piston rod, 276. Columns, eccentrically loaded, 212 214 J Euler's formulae, 221-225, 274, 285 ; ferro-concrete, 25, 32, 33 ; Gordon's formula, 223 ; least radius of gyration, 98, 101, 224 Raukine's formulae, 221-227, 275, 285. Combined bending and torque, 254- 264. Composition of forces, 8. Compound, bars, 21, 22 ; stress, 211- 218. Compound stress, 211, 218. Compression, 1 7 ; combined with bending, 212, 215 ; combined with shearing, 217, 218. Compressive stress, 17 ; table of, 19. Concrete, area of. 111 ; columns, 21, 25; reinforced, beams, 119- 122. Connecting rods, 284-295, 312; forms of ends, 286-295, 319, 323, 324; locomotive, 293-295; marine engines, 286-289 ; station- ary engines, 290-292 ; strength of, 284-291 ; thrust in, 13, 282, 287. Constants, elastic, 18 ; of materials, 19. Continuous beams, 234, 235. Copper, strength of, 19, 20. Cottered joints, 81-83, 280-283. Cotters, 79-81 ; crosshead, 282, 283 ; taper of, 81. Couple-close roof truss, 10. Coupling bolts, 143-149. Coupling joints, 70, 71. Coupling rods, 295-298 ; stress in, 296-298. Couplings, for shafts, 140- 149 ; box, 141 ; face plate or flange, 142, 143 ; marine, 145, 146 ; universal, 147, 148. Crab, design of, 305-310, 317, 318 ; wheelwork of, 186, 208. Crane, forces in, 11, 12, 303, 304; hook, 315, 316 ; jib, 213 ; plat- form, 15 ;- tie rod, 70 ; travelling, 122, 123 ; wharf, 11. Crank, arms, 248-252, 265, 266. Crank pins, 252-256 ; friction of, 184 ; safe pressure on, 253, 254. Crank shaft, 248-264, 312 ; bearings, 256-264 ; safe pressure on, 256. Cranks, 248-254 ; built-up, 250-252 ; cast-iron, 248 ; mild-steel and wrought-iron, 248-252 ; loco- motive, 249 ; marine, 249-252 ; webs, 248, 249. Crosshead pin, 281, 284. Crossheads, 278-284 ; marine, en- gine, 291, 292 ; stationary engines, 292, 303. Cylinder flanges, 246. Cylinders, 246, 247 ; compressed gas, 50 ; hydraulic, 55, 58, 59, INDEX 347 69 ; thick, 50-55 ; thickness of, 44-46, 246. Deflection, 230 ; of beams and- cantilevers, 231-240; of guide bars, 236, 239; table of, 237; work done during, 233. Diagrams, of bending and twisting, 254 ; of bending moment and shearing force, 91, 92, 134. Direct modulus, 18, 152-157. Dished and hemispherical ends, 45. Displacements, parallelogram and triangle of, 8. Eccentric, 266-268. Eccentric loading, 212-218. Efficiency,40, 186, 187 ; of machines, 186, 187 ; of riveted joints, 40-44. Elastic limit, 18, 23, 95, 101. Elasticity, 18, 152-157; table of, 19 ; transverse, 18 ; Young's modulus, 18. Ellipse, area of, 3 ; moment of inertia of, 103. Encastre beams, 90, 91, 237. Energy, kinetic, 201, 205 ; poten- tial, 202, 203; of rotation and translation, 205, 206. Equilibrium of forces at a point, 8-10. Euler's formulae for struts, 221-223. Expansion by heat, 4, 28-31 ; ' stresses due to, 28-31 ; table of, 4. Expansion joints, 27, 28. Eye-bolts, 76. Pace-plate couplings, 142, 143. Factors of safety, 18, 19, 222 ; table of, 19. Ferro-ooncrete, beams, 119-122 ; columns, 21, 25. Flange bolts, 56, 58, 143-149. Flange couplings, 142-146. Flange stress. 111, 112. Flanged joints, 55-58. Flanges, 56-58 ; of beams, 101, 113- 116 ; of pipes, 56-58.X Flanges of girders, 98, 106, 107. Flitched beams, 118, 119, 128. Floor joists, 107, 124-128. Flywheel, 201-205 ; arms and boss of, 205 ; kinetic energy of, 201, 202 ; mass of, 202 ; moment of inertia of, 202, 203; of rolling mill, 207 ; stress in rim, 176, 177 ; work stored in, 201. Force, centrifugal, 174-179 ; moment of a, 88. Forces, acting at a point, 8 ; paralle- logram of, 8 ; triangle of, 8-13 ; polygon of, 9. Forcing fits, 29. Foundation bolts, 80, 85, 86. Fourth powers of numbers, 137. Friction, 73 ; coefficient of, 73, 199 ; of belts, 161-164 ; gearing, 183, 184 ; of journals, 184 ; measure- ment of, 73 ; of ropes, 169, 170. Friction gearing, 183, 184. Funicular polygon, 92, 118, 134, 257. Gasket, 56. Gearing, 183-196 ; bevel, 195, 196 ; friction, 183, 184 ; spur, 184. Gerber's equation, 26. Girder stays, 108. Girders, 88-109; built-up, 112; cast-iron, 117 ; continuous, 234, 235 ; depth of, 113 ; modulus of section 'of, 97-104; plate web, 114; rolled steel, 106, 107, 113; weight of, 116. Gland, weight of, 6. Gordon's formula for struts, 223. Grashof's formula, 52. Grindstone, axle friction of, 200. Guard plate, 275-277. Gudgeons, 281, 290. Guide bars, 13, 106 ; deflection in, 239. Heat, 29-31 ; table of expansion, 4. Helical springs, 154, 157. Hemp ropes, 167-171, 177-179; pulley rims for, 168 ; speed of, 168 ; strength of, 168. Holes, drilling and punching, 38. Hooke's law, 18-25. Hoop stress, 50-55. Horizontal engine, 13, 278 ; guide bar, 13, 106, 239. 348 INDEX Horse-power transmitted, 161-171 ; by belt gearing, 161, 163-167 ; by rope gearing, 168, 169 ; by shafting, 130-133 ; bywheelwork, 192-196. Hydraulic, accumulator, 58 ; cap, 110 ; cylinder, 55, 58, 69 ; flange bolts, 69, 111 ; foot or base plate, 111 ; pipe, 63 ; pipe joint, 67 ; piston, 69 ; press, 326, 327 ; ram, 58 ; tank, 69. Inertia, moment of, 201-203, 206; about a parallel axis, 116, 117 ; beam section, 98, 100 ; circle, 99 ; ellipse, 103 ; polar, 100 ; rota- tional, 206, 207. Initial hoop stress, 63. Iron and steel, 4 ; modulus of elas- ticity, 19 ; riveted joints, 37-44 ; strength of, 19 ; weight of, 4. Joints, cottered, 81 ; coupling, 70 ; expansion, 27, 28 ; hydraulic pipe, 57 ; Imuckle, 71, 72 ; riveted, 35-49 ; spigot and faucet, 66 ; suspension link, 74-76 ; tie- bar, 46-49. Joists, floor, 107, 124-128; rolled steel, 98, 106, 107, 113. Joule's equivalent, 200. Journals, 196-200 ; bearing pressure on, 197, 198 ; bending moment in, 133-136; crankshaft, 266-264; friction of, 199, 200 ; horse-power wasted in friction, 184, 199, 200; strength of, 133-136, 266- 264. Junk ring, 67. Keep plates, 275-277, 288, 290. Keys, 83-85, 207. Kinetic energy, 201-205 ; of rotation and translation, 205, 206. Knuckle joint, 71, 72. Lap riveted joints, 35-44. Launhardt-Weyrauch formula, 26. Lead of slide valve, 270, 271. Leather belts, 160-167 ; friction of, 161-164 ; safe tension in, 161 ; thickness of, 161- Leather packing, 69. Levers, 308. Limit of elasticity, 18, IS. Limiting velocity, 175-177. Linear speed, 183. Link, polygon, 92, 118, 134; sus- pension, 75, 76. Live loads, 16, 21. Loads, 16-19; alternating and varying, 26 ; applied gradually, 17-22; applied suddenly, 23-25; dead and live, 17 ; travelling, 122, 124 ; useful, 16 ; working, 19. Locomotive, coupling-rods, 296-298. Logarithms, 330. Lubrication, 197-200 ; forced, 199. Machines, 186, 187 ; mechanical advantage of, 187 ; velocity ratio of, 186, 208. Marine cranks, 249-252. Mass, measurement of, 201. Materials, strength of, 19. Mechanical advantage, 187. Mensuration, 1-4. Middle-third rule, 213. Mill shafting, 140-144. Moduli of elasticity, 19 ; of beam sections, 102, 103, 104. Modulus, of section, 97-99, 102, 103 ; of elasticity, 18 ; shear or rigidity, 18; tables, 102-104. Moment, 88 ; bending, 88-94 ; of a force, 88 ; of inertia, 97-103 ; of inertia about a parallel axis, 116, 117 ; of resistance in beams, 94 ; of resistance to torsion, 130 ; twisting, 129. Moment of inertia, 96-103, 201-203. Momentum, 206 ; angular, 207 ; rate of change of angular, 207. Mortice wheels, 187. Motion in a circle, 174, 176. Moving loads, 16, 23-25, 122-124. MufE couplings, 141. Naves, of wheels, 190 ; of pulleys, 166. Neck journals, 133-136, 196-200. Neutral axis, 94 ; of tee-seotion, 125. Neutral layer, 94. INDEX 349 Numbers, fourth power of, 137. Nuts, proportions of, 67 ; piston rod, 275. Oak, strength of, 19'; strut, 225. Open franiework cantilever, 15. Overlap in riveted joints, 35-44.. Parallelogram, area of, 1 ; of forces, 8, 9. Patterson's formula, 53. Permanent set, 18, 101. Pipe flanges, 55-57, 111, 112. Pipe joints, 55-57 ; hydraulic, 57 ; spigot and faucet, 57. Pipes, flange joints in, 56 ; hydrau- lic, 53-56 ; spigot and faucet joints, 57 ; stresses in, 55, 108 ; thickness of, 45, 49, 53, 54. Piston rings, 277. Piston rods, 273-278, 312, 313. Pistons, hydraulic, 69 ; marine engine, 276, 277. Pitch, circles, 183, 185 ; diametral, 185 ; of rivets, 35-44, 46-49, 114 ; of screws, 66, 67 ; surface, 185 ; tables of screw thread, 74 ; of wheel teeth, 187, 188. Plate girders, 114-116. Plates, length of girder, 115. Poissou's ratio, 238 ; table of, 238. Polygon of forces, 8-13. Ports, slide valve, 268. Power transmission by, belts, 163- 167 ; ropes, 168-171 ; shafting, 131-136; wheelwork, 184, 192- 196. Principal stresses, 217, 218. Prismoidal formulae, 3. Propeller shafts, 132, 138, 150, 157. Pulleys, arms of, 164-166 ; belt, 160-167; horse-power transmitted by, 163, 165, 167-169; naves of, 166, 167 ; rims of, 166, 167 ; rope, 167, 168 ; speed of rim, 175-177 ; velocity ratio of, 160 ; width of rim of, 166, 167. Radius, of curvature, 95 ; of gyra- tion, 101-104, 201, 204, 205. Railway, waggon, 135 ; wheels, 135. Ram, hydrauhc, 59. Rankine's formulae, for struts, 221- 227 ; for combined bending and torsion, 254-264. Reactions, of beam supports, 88, 89, 92-94 ; by hnk polygon, 92, 93 ; of three or more supports, 234, 235. Rectangle, 1 ; modulus of section, 97, 103, 109 ; moment of inertia, 97-99, 102, 103. Reinforced concrete, beams, 119- 122 ; columns, 25. Repetition of stress, 26, 27. ResiUence, 22-25. Resultant, of forces at a point, 8-10. Rigidity modulus, 18, 162-157. Riveted joints, 35 ; combined lap and butt, 43 ; double lap and butt, 39; efficiency of, 40-43; in tie-bars, 46-48 ; modes of fracture of, 36 ; ratio of tenacity to shear, 37 ; single lap and butt, 38, 40. Rivets, diameter of, 37 ; construc- tion for head, 39 ; proportions for head, 38. Rope gearing, 167 ; centrifugal tension, 177 ; friction of, 169 ; horse-power of, 168, 179. Rope pulleys, 168. Rotational inertia, 206. Saddle keys, 83. Screw jack, 78. Screw threads, square, 67 ; table of Whitworth, 74 ; vee, 66. Second moment of area, 97. Sections, table of area and moduli, 102-104. Shaft horse-power, 154. Shafts, 129-136 ; bending moments in, 133-136 ; couplings for, 140- 143; crank, 256-2'65 ; hollow and solid, 130-136; double-pur- chase crab, 305-308 ; horse-power transmitted by, 130 ; marine, 148, 149 ; maximum stress in, 263, 264 ; moment of resistance to torsion, 130-133 ; propeller shafts, 132, 138, 148 ; pure torque in, 129 ; Rankine's for- 350 INDEX mula for, 261, 262 ; stiffness of, 152-154 ; strength of, 130 ; sub- ject to bending moment, 133-136 ; subject to bending and torsion, 254-264 ; subject to bending, torque and axial force, 264, 265 ; subject to tension or compression combined with shearing, 217, 218. Shear modulus, 18. Shear stress, 17, 18, 218. Shearing force in, beams, 91, 93 ; boiler shell, 46 ; cantilevers, 91 ; rivets, 36, 44 ; tie-rod, 47. Shells, cylindrical, 45 ; spherical, 45 ; thickness of, 45, 46. Shrinking and forcing fits, 29-31 ; crank, 30 ; railway wheels, 30 ; steel tubes, 31. Shrouded wheels, 305-307. Simpson's rule, areas, 2, 3 ; volumes, 3. Slide bars, 13, 106, 278 ; deflection in, 239. Slide blocks, 278-281. Slide valve, 268-272 ; diagram for, 270, 271 ; spindle, 73, 272. Slotted bar mechanism, 312, 313. Spanners, 65, 68. Speed, leather belts, 160-167 ; limit- ing, 175-177; linear, 183; of rope belts, 168-171. Spigot and faucet joints, 56. Split pins, 277. Springs, helical or cylindrical, 154, 155 157. Stanchions, 220-227 ; table of, 224. Stays, boiler, 73, 215, 216 ; crane, 15; girder, 108. Steam cylinders, 246, 247. Steam engine, 246-295 ; connecting rods, 284-295; cranks, 248-252, 265, 266; crank pins, 248-258; crank shafts, 256-265 ; eccentrics, 266-268 ; pistons, 276-278 ; piston rods, 273-278 ; orossheads and slide bars, 278-284 ; slide valves, 268-272. Steel, strength of, 19. Steel angles, 21, 32. Strain, 17 ; due to shrinkage, 29. Strength of, bars subjected to torsion, 130-133 ; beams, 94-110 ; bevel wheels, 195, 196 ; bolts, 66-70 ; connecting rods, 284-288 cranks, 248-252, 265, 266 ; cotters, 79-85, 276, 282, 283 ; cylinders, 45, 50-55 ; flanged beams, 106 108 ; leather belts, 161 ; levers. 308, 309 ; long pillars, 220-224"; pipes, 45, 49, 52 ; pulley arms. 164-167; riveted joints, 40-50 ropes, 168; shafts, 129-136. 254-265 ; slide bars, 106, 239 table of ultimate and safe, 19 wheel arms, 189; wheel teeth, 190 192. Stress, bending, 105, 106, 109; in boiler stay, 73, 215, 216 ; in cast- iron pipe, 108, 241 ; due to contraction in cooling, 27-31 ; compound, 211-218 ; in girder stay, 109 ; hoop, 50-55 ; radial, 50-55 ; shearing, 17, 36-50, 91, 92, 115, 218; table of working, 19; tensile, 17 ; thick cyUnders, 54, 55. Stress and strain, 17-22. Struts, Euler's formulae, 221-225 ; eccentric loading, 212-214 ; ferro- concrete, 25, 32, 33 ; Gordon's ■ formula, 223 ; least radius of gyration, 98, 101 ; table of, 224 ; Rarikine's formulae for, 221-227 ; table, 222. Sudden loads, 24, 25. Sunk keys, 83-85. Suspension bar, 76 ; link, 74. Tank, hydraulic, 59. Tee sections, 125, 214, 215, 296; ferro-concrete, 121. Teeth, of wheels, bevel, 195 ; mortice, 187 ; spur, 184-192 ; proportions of, 187, 188 ; strength of, 190-192. Temperature stresses, 27, 30. Tensile stress, 17 ; table of safe, 19. Tension, 17-22; belts, 161-164; combined with bending, 212-217 ; combined with shearing, 217, 218. Thick cylinders, 50-55. Tie-bars, 20, 21, 46-49, 215. Toothed-wheel drive, 316, 325, 326, 328, 329. INDEX 351 Toothed wheels, 184-196; horse- power transmitted by, 192-196 ; mortice, 187 ; strength of, 189- 193 ; teeth of, 187, 188 ; trains of, 186, 208 ; velooitv ratio of, 186. Torque, 129-131; equivalent, 254, 255. Torsion and bending, 254-264 ; and thrust, 264, 265 ; angle of, 152 ; of bolts, 144 ; moment of re- sistance to, 130 ; of shafts, 129- 133. Torsion meters, 154. Trains of wheels, 186, 208. Travelling loads, 122-124. Triangle offerees, 8-13. Twisting moment, 129 ; and bend- ing combined, 254-264 ; diagrams of, 257 ; equivalent, 254-264 ; on shafts, 130-133. Universal couplings, 147. Unwin's formula for varying loads, 26. Useful load, 16. Valve, engine slide, 268-272; spindle, 73, 316, 317 • Zeuner's diagram for, 271. Velocity ratio, 160, 185, 186. Volumes, determination of, 3. Waggon, railway, 135. Wall stability, 216, 217. Webs, crank, 248 ; of plate girders, 112, 114. Weight, of girder, 116; of gland, 6 ; table of, 4. Wheels, arms of, 188, 189; bevel, 195 ; clearance of teeth of, 187, 188 ; friction, 184 ; horse-power transmitted by, 184, 192-196; mortice, 187 ; railway, 30 ; spur, 184-194; teeth of, 187, 188; trains of, 186, 208. Whitworth screw thread, 66 ; table of, 74. 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