NOTES ON THE Graphics of Machine Forces BY ROBERT C. H. HECK, ME. PROFESSOR OF MECHANICAL ENGINEERING IN RUTGERS COLLEGE WITH 39 ILLUSTRATIONS NEW YORK D. VAN NOSTRAND COMPANY 23 MURRAY AND 27 WARREN STREETS 1910 B ASCMCNT OTO n AQ I CORNELL UNIVERSITY LIBRARY Given to the COLLEGE OF ENGINEERING by the Machine-design dept. T I <-.c u Cornell University Library I J i75.n44 Notes on the graphics of machine forces, 3 1924 004 644 757 NOTES ON THE Graphics of Machine Forces BY ROBERT C. H. HECK, M.E. PROPESSOR OF MECHANICAL ENGINEERING IN RUTGERS COLLEGE WITH 39 ILLUSTRATIONS NEW YORK D. VAN NOSTRAND COMPANY 23 MURRAY AND 27 WARREN STREETS 1910 Copyright, ipio, BY D. VAN NOSTRAND COMPANY THE SCIENTIFIC PRESS ROBERT DRUMMOND AND COMPANY BROOKLVN, N. V. PREFACE These notes are intended to serve as text for a graphical course which has hitherto been based upon Herrmann's " Graph- ical Statics of Mechanisms." To the latter text there are the objections that it does not set forth the fundamental mechanical principles clearly enough for students with the usual degree of effective preparation, and that it wastes entirely too much space on detailed and repeated explanations of examples. The prob- lems in this course are in the shape of good-sized drawings of machines — on sheets about 20 in. by 27 in. — ^which are repro- duced from tracings as positive prints, with dark lines on a light ground. The student is thus saved the labor of mere drawing, and at once takes up the force determination. On the drawing there is room for such special notes and suggestions as may be called for; but the emphasized purpose is to have the student think for himself, with needed help and suggestion from the instructor, and not follow a ready worked example. In section J are added some special force constructions which are useful chiefly in the problems of graphical dynamics, but which are needed to round out the presentation of graphical methods for determining impressed forces in machines. It is thought that the title here used, is more appropriate and descriptive than " graphical statics of mechanisms." R. C. H. Heck. New Brunswick, N. J., May, 1 910. The original of tiiis 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/cu31924004644757 CONTENTS A. General Conditions of Problems B. Force Diagram Constructions C. The Action of Friction .... D. Journal Friction ... . . E. The Efficiency of Machines .... F. Resistance to Rolling .... G. Toothed-gear, Chain, and Rope Resistances H. Belt Transmission . ... I. General Procedure . J. Speoal Force Constructions K. Friction and Lubrication PARAGRAPHS PAGE I to 2 I 3 to II 2 12 to l8 9 19 to 26 13 27 to 32 19 33 to 35 23 36 to 42 25 43 to 48 29 49 '33 50 to 60 34 61 to 64 40 GRAPHICS OF MACHINE FORCES A. General Conditions of Problems 1. In order to determine, by the simpler methods of graphical analysis, the principal forces acting in machines, we make the following assumptions: (a) The weight of the machine members, or the force of gravity acting upon them, may be disregarded. (b) The force of inertia need not be taken into account: this idea involves either a slow running of the machine or a practi- cally uniform motion of the parts. A great many problems may be solved with quite sufficient correctness under these assumptions. If the need of greater accuracy or their increase in relative magnitude requires that these forces (which are functions of mass or of mass and motion) be included, the solution becomes much more complicated. (See Section J.) 2. For present purposes, the machine is considered as made up of rigid bodies, or of members which are fixed and definite in form so far as the forces entering into the problem are con- cerned. These are all ipipressed forces, imposed through the contact of one machine part with another. The forces which act upon one piece get at each other and come into equilibrium through the medium of internal forces or stresses within the body. We pass over all questions as to the magnitude or the manner of action of these stresses, and use simply the resultant relations among the imposed forces; which relations are the same as if the forces acted upon and met in a single material particle. GRAPHICS OF MACHINE FORCES B. Force Diagram Constructions 3. In every case, the forces on any member of the machine are to be considered as in equilibrium. The requirements for this condition are as follows: If there are but two forces, they must be equal and opposite on a common line of action. If there are three forces, their action-lines must meet in a common point, and the forces must form a closed triangle. If there are four or more forces, they need not all pass through one point, but they must be reducible to the three-force case. With parallel forces, we must apply separately the require- ments that the algebraic sum of the forces and the similar sum of their moments about any center shall each be equal to zero. Ion3 Fig. I. — Three Forces on Bell-crank. 4. In the three-force example shown in Fig. i, the forces on the bell-crank 3 are to be determined. Rod-pull (2 on 3) is completely known, in line of action, direction, and intensity; rod-pull (4 on 3) is known in line of action, and gives the inter- section P. This point and the center of the bearing 13 determine the line of the third force, the bearing-pressure (i on 3). Having used the drawing of the machine to find the force directions (that is, the angular positions or inclinations of the lines of action), it is often well to draw the force diagram separately, as in Fig. xh. Here the line AB is parallel and equal to (2 on 3), BC is parallel to (4 on 3), and AC is parallel to (i on 3). The inter- FORCE DIAGRAM CONSTRUCTIONS 3 section at C fixes the length or value of the latter two forces. Remember that for equilibrium the direction-arrows point in circular order around the triangle. 5. The two typical cases of determinate conditions under the action of four forces will now be taken up — and these will be considered as sufficiently representing all cases of the equilibrium of more than three forces on one piece. In Fig. 2, the forces on the elevator body 2 are, the load Q, the upward pull P, and the -W. IonZ fa / . A \ c — G, IonZ Fig. 2. — Wall or Bracket Elevator. two guide-bar' pressures Gi and G2;* all the action -lines and the intensity of one force Q being known. Combine the forces in pairs, Q with Gi, P with G2. For equilibrium, the resultants i?i and i?2 must balance, hence must have a common line of action; and the intersections Oi and O2 determine this line. The diagram at b shows how Ri is found from Q and Gi, then reversed for R2 and resolved into P and G2. * Here friction is disregarded, consequently these forces are perpendicular to the surfaces in contact. GRAPHICS OF MACHINE FORCES 6. In Fig. 3 the piece 4, which is made up of the wheels and axle of a locomotive, is subjected to four forces (besides its weight load, not here included) ; these are, the pressures of the two connecting-rods 2 and 3 upon their crank-pins, the pressure of the bearings (parts of frame i) upon the axle, and the tan- gential resistance of the track T. Three lines of action are known, but for the location of the fourth force there is only one determin- ing point. The problem is soluble when two of the forces, as (2 on 4) and (3 on 4), are completely known; whereupon their resultant R can be found, and we drop to the three-force deter- mination, with the intersection P as the second point required to AohT Ton 4 Fig. 3. — Driving Action in Locomotive. fix the unknown force-line. Note that in this figure the force triangles are made right on the drawing of the machine; and the force R, a result in the first triangle, is moved to another place on its line of action in order to enter as a primary quantity into the second triangle. 7. The equilibrium of three parallel forces is illustrated in Fig. 4. Always, the two outer forces P and Q act in the same direction, while the equilibrant R, equal to their sum, lies some- where between them. The only moment equations that need be (I) (2) (3) considered are: With origin on R, Pa=Qb; With origin on P, Ra = Qc; With origin on Q, Rb=Pc. FORCE DIAGRAM CONSTRUCTIONS 5 All of these are expressed graphically in the triangles obtained by drawing the parallel cross-lines AC, DF, and GK, and the diagonal GF. Considering the first relation, for instance, we have from the similar triangles ABG, CBF: or whence CF:CB::AG:AB, P:h::Q:a, Fig. 4. — Parallel Force Relations. For equal moments, the forces must be inversely as their own moment-arms, or directly as the opposite arms. In effect, in Fig. 4, each force is transferred to the line of the other force, P to FC, Q to AG, and there is then a direct proportion to the arms BC> and AB. This idea of interchanging the forces for purposes of graphical construction is used in every case; thus for the second relation, with origin of moments on line of P, we put R at KF and Q at BH and have the forces in the ratio of the distances GK and GH. The cross-lines AB, DF, and GK need not be perpendicular to the force-lines; their parallelism is the essential thing. 8. The typical problems in parallel forces are represented in Fig. 5. In each diagram, full lines show originally known GRAPHICS OF MACHINE FORCES quantities, dotted lines the results got by construction. The figures are lettered in such a manner that the alphabetical order indicates the order in drawing the lines. . The four cases may be briefly set forth as follows : Case I.— The three lines of action and an outer force P are known: draw AB and CD, to carry P over to line of Q; diagonal DEF determines Q in AF and i? in CF, and line FG transfers these forces to their proper lines. Case II. — The three lines of action and the inner force R are known: draw AB-AC and DE-DF, to carry R to line of Q; Fig. s. — Problems in Parallel Forces. diagonal CGE divides R into AG or Q and GD or P, and line GH-GK transfers the forces to their own lines. Case III. — Two forces in same direction are completely known, and tliird (middle) force (known in intensity, as the sum of the other two) is to be located: draw AB, CD, and EF, to carry forces to opposite lines, P to DB, Q to AF; then diagonal FGD locates at G an origin about which P and Q have equal moments, which is therefore a point on line of R. Case IV. — Two forces in opposite directions are completely known, and the third (the outer force beyond the larger of the FORCE DIAGRAM CONSTRUCTIONS two, and known in intensity as their difference) is to be located: draw AB, CD, EF, and diagonal FDG to meet AB produced at G; the latter point locates the line of Q, and DH and EK carry the force over to this line. Fig. 6. — Construction for Distant Intersection. 9.][^Wlien two forces are nearly parallel, so that their inter- section falls far off the drawing board, it is necessary to find the direction of the third force without actually drawing it to the intersection. In Fig. 6, AB and CD are the known force- lines, and the third force is to pass through the point E. Draw a cross-line AC through E and, at a convenient distance, a parallel cross-line BD. The third force-line EF must divide BD in the same ratio as AC. From A draw AD' (at any con- venieiit angle) equal to BD, join CD', and draw EF' parallel to CD'; then lay off point F' on BD, at F, and draw EF. Fig. 7. — Geometrical Constructions. 10. Another scheme for dividing the second cross-line in the same ratio as the first, which involves no measurement or transfer of lengths, is shown in Fig. 7. For the first case, with the pomt E between the known lines, draw the diagonal CB; then EG, parallel to AB, carries the ratio (AE:EC) to (BG:GC); and drawing 8 GRAPHICS OF MACHINE FORCES GF paraUel to CD makes (BF:FD) equal (BG:GC). To apply this exact scheme when E lies outside the known lines, we should have to draw EG' parallel to AB, meeting CB produced, then draw G'F parallel to CD. It is more convenient, and tends to better accuracy, again to make the construction on the middle line, now the known line CD: draw CG parallel to AB, join GD, and draw EF parallel to GD. The essential thing is to have a line paraUel to the base of a triangle, so as to divide the sides in the same ratio. II. When three intersecting forces are not far from parallelism, the force triangle becomes very flat, as at diagram b in Fig. 8, and the intersection U of the sides representing the imknown H -P M L R J c ^/^ < . ■c 3 6 pt= Q Fig. 8.— The Flat Force Triangle. forces is rather indefinite. For accurate determination of these forces, the moment relation must supplement the force triangle. One method is to take an origin on the line of one unknown force, as at G, draw and measure the perpendicular moment- arms GH and GJ, measure the force P, and multiply and divide in the relation, Rb = Pc, (4) Or, as indicated at c, a regular parallel-force construction may be made for one force, as R. In either case, after a second force has been found we go back to the force triangle and use it to get the third force. THE ACTION OF FRICTION C. The Action of Friction 12. In Fig. 9 is represented a block 2 which is pressed down upon the plane i by a force N, normal to the surfaces in con- tact, and is supported by the equal and opposite upward re- action of the plane i. If the pressure between the surfaces is uniformly distributed over the contact-area, as indicated by the diagram at the right, the equivalent concentrated force N will act at the geometrical center of this area. If the block is made to move, as in the direction of the motion-arrow, there will be developed between the surfaces a frictional resistance F, opposing the movement. This frictional force F bears to the normal pres- ■^r- _^ 2 ^ , r- 3 *HHHHHHHr* Fig. 9. — Resistance to Sliding. sure N a ratio which is known as the coefficient of friction. As in every cSse, there is here the inevitable condition of an equal and opposite action and reaction: the friction force F^^ ^n 2 opposes the movement of 2 on i, while the reaction F2 on 1 sim- ilarly opposes the movement of i relative to 2. 13. In Fig. 9 only the conditions right at the contact sur- faces are considered, and nothing is shown as to the detail of the external forces on the block 2. The complete representation of a simple case is given in Fig. 10. Block 2 is pulled downward by its own gravity force W, which necessarily passes through the center of gravity G; and the force P, which moves the block, acts along the line CD, parallel to the plane. To bring the block just to the eve of motion, or to maintain a uniform motion, P must equal F; and the resultant of W and P is the pressure 10 GRAPHICS OF MACHINE FORCES of 2 on I, balanced by that of i on 2. The latter force has F^ ^^ , as one component, while the other is N, equal and opposite to ir. Since P and F are not along the same line, they form a turning couple; but tlieir moment Pa is balanced by the equal and opposite moment Wb of the couple W and A'". Further, since the total pressure (2 on i) or (i on 2) is not central at tlie contact surface, the actual distributed force will be non-imiform, as shown by the pressure diagram at the right. Fig. 10. — Block Sliding on Plane. 14. The arrangement outlined in Fig. 11 is frequently used to illustrate the action of sliding friction, and serves especially to give fuller meaning to the expression " angle of frictior.." The gravity force W is resolved into components, N normal and P parallel to the plane; and the inclination of the plane is just enough to make the active component P equal to the frictional resistance F — ^by " active " component is meant the one in the direction in which the body is capable of moving under its con- ditions of constraint of motion. The angle a, which tlie plane AB makes with the horizontal base BC, or which tlie total pres- sure W makes with tlie normal force -V, is called the angle of friction; and the coefficient of friction is tlie tangent of tliis angle, or F fi * = - \Y =tan a. (5) * (."iioek HtH, same as m. THE ACTION OF FRICTION 11 15. Suppose that under the conditions of Fig. 9 or Fig. 10 a force P is applied which is less than the total friction /xN; then only so much of the frictional resistance will come into play as is needed to balance P. As P increases, the frictional force be- FlG. II. — Sliding on Inclined Plane. 2 ON I Fig. 12. — Limits of Frictional Resistance. tween the surfaces also increases, up to the limit jjN; when this limit is reached, motion is ready to begin. If P becomes greater than F, the line of resultant pressure of i on 2 does not swing beyond the angle a, as shown at AB or AC in Fig. 12, but the excess of P over i^ is a free or unbalanced force, acting to accelerate the moving body. In a reversal of motion, as by an engine crosshead at the end of the stroke, the line of pressure swings through the angle 2a, say from AB to AC. 16. The first question to be answered in any case of sliding friction in a machine is, ^onI To which side of the normal is the pressure- ^^^ i,.— inclination of line inclined? The general principle of all the Pressure-line, friction action is that friction opposes rela tive movement; consequentiy, the pressure acting upon any piece at the contact surface must have a component against the 12 GRAPHICS OF MACHINE FORCES movement of that piece. The simple " rule of thumb " is illustrated in Fig. 13. At about the place where the force-line wUl cross the contact-surface, Crosshatch a little block on each piece to emphasize the contact, and draw arrows indicating the respective directions of relative movement of the two pieces; then draw an inclined line which joins, or tends to join, the taUs of the arrows. This line shows the direction in which the pressure-line slants away from the normal to the surfaces. 17. The machine outlined in Fig. 2 and reproduced in Fig. 14 serves as a very good example of the effect of sliding friction. Fig. 14. — Wall Elevator, with Friction. The first step is to sketch in at A and D the determination of Fig. 13, purposely exaggerating the angle of inclination; then the action-lines of the guiding pressures Gi and G2 are definitely drawn by measuring off along the normals convenient lengths AB and DE and erecting perpendiculars, BC=/iXAB and EF = /iXDE. The intersections Oi and O2 now fix the re- sultant line, and the force diagram takes the form shown in full lines at b. With this is drawn in dotted lines the diagram for downward motion of the elevator, and in dot-and-dash lines the diagram for the ideal case of no friction, reproduced from Fig. 2. In downward movement the weight Q becomes the JOURNAL FRICTION 13 driving force, while P, now the hold-back or brake force, serves as resistance. 1 8. It is of interest to set forth a reason for the fact that, in Fig. 14, the guiding forces d and G2, and consequently the friction on the guide-bar, are greater in downward than in up- ward motion — that is, a reason with a more fundamental basis than the mere appearance of this result in the force-diagram determination. In Fig. 15 are given diagrams of all the vertical forces on the elevator body, the third force F being the resultant of the two frictions and equal to the difference between P and Q. The resultant turning inoment of the three forces must be balanced by the couple made up of the normal components of Gi and G2. a pA 1 r-n-f^ Fig. 15. — The Moment of Friction in Fig. 14. Taking the center of moments on the line of force P, we see that in case a the moment Fn acts against Qm, while in case b these two moments act together. This example shows the advantage of using other lines of reasoning to supplement, and sometimes to check, the direct graphical determination. D. Journal Friction 19. The turning of a journal in its bearing is nothing but a sliding of curved surfaces upon each other, but the conditions as to pressure-distribution are much less simple than with flat surfaces. In Fig. 16a the journal is supposed to fit the bearing with a snug running fit, close but not forced; then under the action of the force N the pressure would be distributed about as shown by the diagram. An actual bearing is somewhat loose 14 GRAPHICS OF MACHINE FORCES on the journal; wear in service tends toward a uniform distribu- tion of pressure; and between the surfaces there is an elastic film of lubricant. For these reasons the curve of distribution will probably take the form at ft in Fig. i6; it is more nearly uniform over a considerable arc at the bottom, but does not cover the entire half-circumference. The essential fact, in any case, is that the total pressure between the surfaces is greater than the resultant N' — this because the oblique pressure away from the resultant-line has only a component for or against N. With the same load and the same coefficient there would be, therefore, more frictional resistance in a bearing than under a slide-block. 'lo»2'. Fig. i6. — Pressure between Journal and Bearing. Fig. 17. — Journal Friction. 20. Actually, the coefficient of friction is, for the same mate- rials and lubricant, quite a good deal less in the bearing than on the slide, chiefly because the lubrication is so much more effective as explained in paragraph 63, following. Further, the coeffi- cient as experimentally determined includes the influence of obliquity of surfaces, since the experiments are made with bearings of actual form. Practically, we treat the problem as if the pres- sure and its resulting friction were concentrated upon a narrow JOURNAL FRICTION 15 flat surface which is central on the main line of thrust. This condition is depicted in Fig. 17, where the fundamental rela- tions are closely analogous to those in Fig. 9. It is assumed that all special details in the action of pressure and friction have exerted their influence in helping to determine the friction-coefficient = F/N= tan a. To oppose rotation about the journal-axis O, the fric- tional force F exerts the resisting moment, M=FXR=4>NXR (6) In order that the force P, which is equal to N/cos a, shall have this moment, it must satisfy the relation, N r=^NR, (7) cos a and pass the center of the journal at the distance, r=R. (10) With the line of total pressure tangent to the friction circle, in- stead of passing through the center, the pressure diagram will be distorted from the symmetrical form in Fig. 16: it will be heavier on the side where the journal surface is descending, or where, to a slight degree, the journal tends to climb in the bearing. 16 GRAPHICS OF MACHINE FORCES 22. Examples illustrating the use of the friction circle are given in Fig. i8, with the simple link-work fully outlined at a as basis. The single force-line along the connecting link 3 is to be located, and the question is, To which side of the friction circle will this line be tangent at each joint? Without friction, the line will go through the centers of the two journals. In case a, the rod is in a state of tension, or is pulling down on the arm 2, hence the " contact " will be on the top of the upper pin — and on the bottom of the lower pin. Noting that the angle between links 2 and 3 is becoming less or " closing," we see that 3 wUl have left-hand or anti-clockwise rotation with reference to 2, or that 2 will turn toward the right on 3 ; and the direction arrows are drawn accordingly. A line joining the tails of these arrows, as in Fig. 13, shows to which side of the friction circle the final force-line will pass. The same determination is made at the lower joint, where the relative movement is the reverse of that at the top; and the force-line is drawn tangent to the two fric- tion circles on the sides indicated. 23. For the preliminary determination of the side to which the true line of force is deflected from the normal, it is proper to center the contact on the force-line for the case of no friction (it is understood that the force-action without friction will have been worked out before the case with friction is taken up). Remember, however, that this shows only the direction of the deflection a, and does not fix the radial line from which this angle a will finally be measured. The friction circle expresses geometrically the condition to be met by the force-line at this one journal-bearing; if the line is tangent to the circle, it wUl make the proper angle a with a radius drawn to the point where it crosses the circumference. To fix the line, another determinant, outside of the single bearing, is required. 24. In parts b and c of Fig. 18 are shown two out of a number of possible variations from case a. At b there is an interchange of pin and eye between the links at each joint, as compared with a; but the final result, in the location of the force-line, is un- changed. At c, however, link 4 is made driver and the direction JOURNAL FRICTION 17 of motion is kept the same, thus changing the stress in rod 3 from tension to compression; and now the force-line is reversed in tangency at both joints. The following general rules govern these and other changes: To interchange journal and bearing or pin and eye at a joint (or to be uncertain, in a problem, as to which part belongs to which piece) will not affect the force-line at this joint, provided that the contacts and motions are correctly represented for the existing arrangement. To reverse either direction of force or direction of motion at a joint will swing the force-line to the other side of the fric- 2on3 4on3 Fig. 18. — Use of the Friction Circle. tion circle, but to reverse both together will leave it unchanged. The two changes act like minus signs in an algebraic multipli- cation, one effecting a reversal, two neutralizing each other. 25. A very useful check upon the detailed determination of the side of tangency to the friction circle (that is, upon the method of Fig. 17) is got by applying the general principle that friction must hinder the motion. Thus in Fig. i8a, " driver " 2 is made to turn about the center 12 by some driving force not shown on the figure, and the force (3 on 2) acts as a resistance to motion; this resistance is given a greater effect by moving it outward from 12, or increasing its moment-arm from that center. 18 GRAPHICS OF MACHINE FORCES Ax the lower joint, however, force (3 on 4) is a driving force, and the hindering effect of friction is shown by the decrease in its moment-arm (from 14) due to tangency on the inner side of the friction circle. This check is not always so obvious or so easy to apply as in this example, but it should constantly be kept in mind. 26. For the purpose of finding the relative motion in a turning joint, the scheme illustrated in Fig. 19 is very effective. Con- sider joint B : the lines AB and CB (of constant length) form two sides of a triangle, with AC as base. Whether AC will increase or decrease as the mechanism moves in the direction of the arrow. Determination of Direction of Motion. determines whether the angle ABC is opening or closing. Point C is moving in a direction perpendicular to the arm CD; and since this direction makes an acute angle with AC, we see that the latter is decreasing. For the joint C, with DB as the variable base, the corresponding angle is obtuse, hence DB is increasing and the angle opening. If C were, for a particular position of the mechanism, moving in a direction perpendicular to AC, it would indicate that there was, for the instant, no turning at all in the joint at B. THE EFFICIENCY OF MACHINES 19 E. The Efficiency of Machines 27. The efficiency of a machine is the ratio of the useful work delivered to the total work put in. Let P stand (as hereto- fore) for the driving force and Q for the useful resistance; and while P moves a distance p* let Q move a distance q*, the ratio oi p to q being determined wholly by the proportions and con- figuration of the mechanism. Further, let the total friction of the machine be combined in a single force P, which is overcome through the distance/. Then the general relation is, Pp = Qq + Ff, (II) or, work put in equals useful work plus work wasted against friction. The efficiency is, -I (") 28. The graphical methods with which we are now concerned determine forces, but not distances or velocities. It is therefore desirable to be able to express the efficiency e as a ratio of forces rather than of work quantities. This end is attained with the help of the ideal case without friction, for which we have the relation, Pp=Qq (13) Suppose that we start P as the known force, and work through to Q. The value without friction, which we call Qo, will be greater than the actual value Q, and the efficiency tvUI be, '—p^Q-oroo ^''^ * These are effective distances, measured in the directions of the forces. If the force is oblique to the path of the point of application, we may use the com- ponent of motion along the force-line for p or q, or we may use the total motion and with it the force-component along the path as P or Q. 20 GRAPHICS OF MACHINE FORCES If, on the other hand, we have Q known and work back to P, tlie actual P (with friction) will be greater tlian tlie ideal Pq, and the efficiency will be, ^=p^=p^=p ^^5) 29. Some machines, notably hoisting machines (in which gravity acts as the load-force Q), can run backward with the same set of forces as in forward running, but with Q now acting as driving force and P as resistance. To distinguish the forces acting under this condition, we bracket them thus, (P), (Q): the work equation takes the form, iQ)q=iP)P+iF)f, (16) and the expressions corresponding to equations (14) and (15) are, The forward efficiency e is always less tlian unity, l}mg some- where between zero and one; the backward efficiency (e) may pass through zero and become negative, and on this negative side may become greater than one. This peculiar state of af- fairs can best be understood with tlie help of an example. 30. The mechanism in Fig. 20, outlined for each case at a, has for its moving parts the wedge 2 and the vertical slide-block 3, with the load Q. In forward motion, the wedge is pushed toward the right and the load lifted; in backward motion, the wedge is withdrawn and the load descends. The two cases are, I, wedge steep or blunt, backward efficiency positive; II, wedge flat or sharp, backward efficiency negative. On diagrams a the various force-directions are indicated, but tlie lines are draw-n so as to keep clear of each other, and no attempt is made to get the proper intersections. As in Fig. 14, dot-and-dash lines are used for forces without friction,* full lines for forward-motion * \\'ithout friction, the forces are the same in either for\\'ard or backward motion. THE EFFICIENCY OF MACHINES 21 forces, dotted lines for the case of backward motion. Diagrams b, with all the forces marked, should require little explanation: a force triangle with Q as the known side is first drawn for piece 3; then force (2 on 3), reversed to (3 on 2), is the base for the construction of the triangle for piece 2. 31. The essential difference between the two cases in Fig. 20 lies in the opposite directions of the force (P). In case I it Fig. 20.' — Reversal of Backward Efficiency. points in the same direction as Pq or P; and Q, driving the machine backward, does a " useful " work in overcoming the resistance (P). In case II the friction effect is so great, relative to the forces along the wedge, that Q alone cannot produce back- ward motion. Force (P), instead of holding the wedge in place and acting as a brake-force, must turn around and help Q. The full effect exerted by Q, if expressed as a horizontal force on the 22 GRAPHICS OF MACHINE FORCES wec'ge 2, is equal to Pq. If the friction in the machine is just enough to balance Q or to neutralize Pq, the backward efficiency (e) is zero. If, as here, the friction exceeds this amount, or (P) has to help Q, this efficiency becomes negative. In a word, the work Qq is supplied at what is now the input end of the machine; but to make it move at all the further work {P)p must be sup- plied at the output end. In the sense of a normal output, this work {P)p is negative, hence the minus sign for (e). 32. A machine which has so much friction that it will not run backward under the action of its load is said to be self-locking. With this property goes a low efficiency in forward running. Ccnsider the case where the friction-work {F)f is just equal to Pop or Qq, and assume that in forward running the lost work Ff is the same as (F)/. Then from equation (11) we have, Pp = Qq+Ff=^2Qq, .... (18) and the value of e is one-half or 50 per cent. If {F)f is greater than Qq, e will fall below this limit. The difference between {F)f and Ff will be small, with a general probability that Ff will be the larger quantity, because the forces in the machine are likely to be a little greater in forward than in backward running. The self-locking machine has the advantages of a high velocity- ratio of P to Q- — for it is because Pq is relatively small that fric- tion can overbalance it' — and is usually a simple device for lifting a big load with a small driving force; also, it makes a safe hoist, because no brake is needed to hold up the load. The fact that at least half, and probably more, of the work put in will be ex- pended in overcoming friction is, however, a decided drawback. The most common examples of this type of machine are the screw and the worm and worm-wheel, both derivatives from the wedge mechanism in Fig. 20. RESISTANCE TO ROLLING 23 F. Resistance to Rolling 33. When a heavy or heavily loaded roller rests on a plane surface, there is always some elastic compression of both roller and plane at the contact. This contact does not really exist along a line, but is spread over a narrow surface; the character of the pressure-distribution is indicated in Fig. 21, but with the width of contact tremendously exaggerated. If the roller advances, there is a continual compression of the surfaces on the side of advance and a release or expansion on the opposite side; the net result is a resistance to rolling, which can be most simply represented by shifting the resulting supporting force N Fig. 21. — Contact in Rest. Fig. 22. — Resistance to Rolling. Fig. 23. — Wheel on Road. through a small distance d toward the side of advance, as in Fig. 22. This force now passes the center of the roller at the distance d, and exerts the moment Nd or Qd, against the turning. If instead of a roller on a firm, smooth, clean track we have a wheel ■on a road covered with dust or mud, sand or loose stone, this effect is very much greater, as indicated by Fig. 23. 34. A long-accepted approximation for the value of d is that it may be taken as a constant at 0.02 inch, being independent of the diameter of the roller, and also of the material, provided that the loading is done with due regard to the strength of the latter. The low resistance obtained with small diameters in ball and roller bearings indicates much smaller values of d for these arrangements. We shall not attempt any general dis- •cussion of this question here; but for the problems in the present 24 GRAPHICS OF MACHINE FORCES course shall assume the constant value d=o.o2 inch. Note that if the machine is drawn to a reduced scale, the distance d must be divided accordingly. 35. The typical conditions under which rolling resistance will enter iato a graphical determination are represented in Figs. 24 to 26. In Fig. 24 the driving force P passes through the center of the roller and is parallel to the track; then at the track surface there is developed an equal and opposite holding friction T, which is a resistance to slipping at the contact; and the moment PR is equal to Qd. The diagonal OC is the common line of the re- sultants of Q and P and of N and T, and serves to determine the value of P in force diagram b. Fig. 24. — Moving Force at Center. T T Fig. 25. — Wheel with Journal. Fig. 26.^Case of Double Rolling. Fig. 25 shows the actual wheel, with the load Q impressed, through a bearing and journal. The force P must now over- come both rolling resistance and journal friction, and the re-- sultant line OC has the added inclination due to the radius of the friction circle, measured horizontally toward the side away from d. In Fig. 26 is represented the case of double rolling, the roller lying between two flat surfaces; this is typical of ball and roller bearings. The displacement d is now measured off at both contacts, m. opposite directions, but the resultant line has the same inclination, and the force P the same value relative to Q, as in Fig. 24. This force is the same, with double resistance, because its moment-arm is twice what it was in Fig. 24; the equation of moments is now PX2R=QX2d. From another TOO THED-GEAR, CHA IN, AND ROPE RESISTANCES 25 point of view, note that in one revolution of the roller P moves 2kR in Fig. 24 and /^nR in Fig. 26, doing twice as much work in the second case, and thus overcoming the two rolling resist- ances, each over the angular distance 2k. G. Toothed-gear, Chain, and Rope Resistances 36. The fundamental conditions as to the transmission~of pressure by gear teeth are set forth in Fig. 27. With involute Driver. Fig. 27. — Line of Tooth Thrust. Fig. 28. — Inch'nation of Thrust Line. profiles, case a, the locus of the point of contact is a straight line AB, at an angle of about 75 deg. * with the line of centers, and this is also the line of tooth thrust. With cycloidal pro- files, case b, the corresponding locus is a double curve ACB, made up of arcs of the two describing circles; and the line of thrust is continually oscillating from CD at 90 deg. to CE at perhaps 60 deg. with the line of centers. For graphical pur- poses we assume either that gears have involute teeth, or that the action of cycloidal teeth may be well enough represented by an average line of thrust, constant at 75 deg. * In some systems the angle is not exactly 75 deg., but here we shall use this angle. 26 GRAPHICS OF MACHINE FORCES 37. Fig. 28 illustrates a simple rule for fixing the direction in which the line of thrust is deflected from the tangent to the pitch circles. If from the pitch point C we run out, in the di- rection of motion, along the tangent CD, the thrust-line AB — or, to be absolutely exact, the part CB — will be swung away from the driver and toward the driven gear. A few trials of possible cases, with sketched-in profiles of the sides of teeth in action, will show that this always holds. 38. The teeth of a pair of gears slide together as they ap- proach the pitch-point, slide apart as they recede from it. The friction action which accompanies, this sliding is illustrated in OfliVEN. Fig. 29. — Friction on Gear Teeth. Fig. 29, which is supposed to represent general, average condi- tions. The dot-and-dash line AB, passing through the pitch- point, is the line of thrust for the case of no friction. At the contacts A and B are made the regular determinations for the deflection due to sliding friction, resulting in the pressure-lines AF and BG. These lines meet at D; and the pressures being assumed equal and laid off in DF and DG, have a resultant DH which is parallel to AB. 39. The distance DE, through which the line DH is shifted from AB, is found as follows: Angle DAB or DBA is the friction-angle a; distance AB is essentially the same as the circular pitch of the teeth, which we TOOTHED-GEAR, CHAIN, AND ROPE RESISTANCES 27 shall call i; and letting fi=tzna be the coefi&cient of friction as heretofore, we have, DE = ^ SEVER, Prof. G. F. Electrical Engineering Experiments and Tests. on Direct-Current Machinery. With diagrams and figures. Second edition, thoroughly revised and enlarged. 8vo, pamphlet. Illus- trated net, $1.00' and TOWNSEND, F. Laboratory and Factory Tests in Elec- trical Engineering. Second Edition, thoroughly revised and enlarged.. 8vo, cloth. Illustrated. 236 pages net, $2.5a. SHELDON, S., Prof., and MASON, HOBART, B.S. Dynamo Elec- tric Machinery; its Construction, Design, and Operation. 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