Class X/LSxi Book " .. GopyiightN^. ..#^ M ■*'■ COPYRIGHT DEPOSrr HANCOCK'S APPLIED MECHANICS FOR ENGINEERS •The THE MACMILLAN COMPANY NEW YORK • BOSTON • CHICAGO • DALLAS ATLANTA • SAN FRANCISCO MACMILLAN & CO., Limited LONDON • BOMBAY • CALCUTTA MELBOURNE THE MACMILLAN CO. OF CANADA, Ltd. TORONTO c v./ bridge trusses, in cranes, and in other jointed structures all forces act- <-^-€: (a) ih) Fig. 8 ing on each member are often considered as applied at two points near the ends of that member, and the member may thus be regarded as acted upon by only two forces. Where such is the case the two forces acting upon the member must be equal and opposite and their lines of action must coincide with the line joining the points of application of the forces. If the two forces act towards each other, the member is said to be in compression. If the forces act away from each other, the member is said to be in tension. Thus in Fig. 8 the member is in compression in (a) and in tension in (6), the amount of compression or tension being P. B Fig. 9 Illustration. In Fig. 9 a horizontal pin carrying the weight IT passes through the members AB and BC which CONCURRENT FORCES 15 are inclined respectively at angles 60° and 30° to the horizontal and held at A and by horizontal pins perpen- dicular to the plane ABC. If the weights of AB and BO are small compared to TT, AB and BO may each be re- garded as acted upon by only two forces. The values of these forces may be found as follows: Since AB and BO are in equilibrium under the action of only two forces, those acting at the pins, the forces acting on each member must be along the lines joining the pins. The members themselves must then exert on the pins forces equal and opposite to those that the pins exert on the members. Hence acting on the pin at B^ holding it in equilibrium, are the forces TF", P, and §, acting respectively downward, along AB^ and along OB (Fig. 10). Applying the conditions for equilibrium, 2X =0, 2 Z = 0, there results P cos 60° - Q cos 30° = 0, F sin 60° + Q sin 30° - W= 0. Solving these equations, we find P=.866 W, 16 APPLIED MECHANICS FOE ENGINEERS A graphical solution is obtained by laying out the force triangle for the forces in equilibrium at the point B and measuring the sides representing P and Q (Fig. 11). Problem 5. A weight of 10 ions is supported as shown in Fig. 12. Find the force acting in the tie A and the member B. Problem 6. Two ropes of length 5 ft. and 8 ft. are attached at one end to a weight of 500 lb. The other ends of the rope are at- tached to two points 9 ft. apart on a horizontal line. When the weight is suspended by the ropes, find the tension in each rope. 19. Body in Equilibrium under the Action of Three Forces. — ^ three forces are in equilibrium^ any one of them must be the equilibrant of the other two. Its line of action inust therefore pass through the intersection of the lines of action of the other two. Also the vector of the third force must be equal and opposite to the diagonal of the parallelogram formed on the vectors of the other two forces as sides. Imposing these conditions on any body in equilibrium under the action of three forces will usuall}^ suggest an easy method of finding two of the forces acting on the body when the third one is known. CONCURRENT FORCES 17 Illustration. A gate 9 ft. wide bj 5 ft. high, weigh- ing 80 lb., is hung by two hinges distant respec- tively 6 in. from the top and the bottom of the gate. If the lower hinge carries all of the weight (i.e. the upper hinge exerts no upward force on the gate), find the reactions of the hinges. The weight of the gate may be assumed to act at the center. n Solution. The gate is acted ^^^- ^^ upon by three forces : the weight, 80 lb., acting down- ward at the center, a horizontal force exerted by the upper hinge, and a force unknown in magnitude and direction exerted by the lower hinge. These three forces must pass through a common point, the point A., Fig. 14, and form a closed triangle of forces. Hence, if AB = 80 lb., the forces P and Q in Fig. 14 represent on the same scale the reactions of the lower and upper hinges re- spectively. Tlie values may be found by measurement, or by solving the triangle ABC. Comparing the similar triangles ABO and AED, one a triangle of forces, the other a triangle of distances, we have Fio. 14 Q ^ 4.5 80 4 .-. § = 90 lb. 80 4 .-. P = 120.4 lb. 18 APPLIED MECHANICS FOR ENGINEERS Problems of this nature are more easily solved of the theory of moments developed later. Problem 7, Neglecting the weight of the members AD in Fig. 15, find analytically and graphically the stress in BC reaction at ^. Suggestion. J5C is acted upon by two forces, one at B at C. AD is acted upon by three forces. For the auab^t first find the values of ^0 and BO and compare the - . with the triangle A OB. by use and BC and the and one 2000 LBS Fig. 15 Fig. 16 Problem 8. A post 20 ft. high is hinged at the foot and stands vertically. Two ropes ^C and ED on opposite sides of the post and in a plane perpendicular to the axis of the hinge make angles of 50° and 40° respectively with the horizontal and are attached to the post at distances of 10 ft. and 20 ft. respectively from the bottom. What tension in ED will cause a tension of 500 lb. in BC, and what will then be the reaction at the foot of the post? Solve graphically. Problem 9. Neglecting the weight of the truss in Fig. 16, find analytically and graphically the reaction at A and the tension in BC. Problem 10. Find analytically and graphically the value of the horizontal force P that will just raise the corner A of the block in Fig. 17 from the floor. Find also, graphically, the value of P when A is raised 6 in. from the floor. When A is raised 1 ft. CONCURRENT FORCES 19 Problem 11. The column AB, Fig. 18, is one foot square, 15 ft. long, and weighs 1200 lb. Find graphically the tension in the rope and the reaction at A when the ang-Ie which the column makes with f 41 m' A\ _ Fig. 17 Fig. 18 the horizontal is 0°, 30°, 60°. The pulley at C is small. Consider the weight of the column as acting at its center. Problem 12. A wheel is about to roll over an obstruction. The diameter of the wheel (Fig. 19) is 3' and its weight 800 lb. Find the horizontal force P through the center necessary to start the wheel over the obstruction. Fig. 19 Problem 13, An angle iron, whose weight is 20 lb. and angle a right angle, rests upon a circular shaft, radius 2 in. Find the normal pressure at A and B (Fig. 20). Fig. 20 20. Concurrent Forces in Space. — Let P^, Pg^ ^3^ ®^^' ^® a set of concurrent forces not in one plane, and let their 20 APPLIED MECHANICS FOR ENGINEERS direction angles be respectively Wj, /3j, 7^ ; a^, ^^, y^ ; ccg, /Sg, 73 ; etc. (The direction angles of a force are the Fig. 21 angles at the point of application of the force measured from the positive direction of the coordinate axes to the vector representing the force.) The force P^ may be re- solved into components parallel to the coordinate axes, X^ = Pj cos ocj, Y^ = P^ cos y8j, Z^ = Pj cos 7^. The resultant force may be found in magnitude and direction by an analysis similar to that used in Art. 17. The sum of the components of all the forces parallel to the 2:-axis is 2X= Pj cos cjj -\- Pg cos 0^2 + P3 cos ctg + etc., the sura of the components parallel to the ?/-axis, Sy = Pj cos /Sj + Pg cos 13^ + Pg cos ySg + etc., and the sum of the components parallel to the si-axis, 2Z= Pj cos 7j 4- P2 cos 72 + Pg cos 73 + etc. The original system of forces may now be replaced by a system of three rectangular forces 2X, 2 F, and 2Z CONCURRENT FORCES 21 (Fig. 22). Finally, this system may be replaced by a resultant which is the diagonal of a parallelopiped con- structed with 2X, 2P, ^Z as edges. In magnitude this resultant may be expressed i^= V(2X)2+ (2F)2+ (SZ)2 (see Fig. 22) and its direction given by ^ig. 22 the angles ot, /3, and 7. These angles are given by the equations cos a = -— — , cos p = -— — , cos 7 = For equilibrium R must be ; that is, (sx)2 + (sr)2+(sz)2 = o, and therefore. This gives three equations of condition from which three unknown quantities may be determined. In the preced- ing case of Art. 17 there were only two equations of condition 2X= and 2 1^= 0; consequently, only two un- known quantities could be determined. Problem 14. Prove that if ct, p, y are the direction angles of any straight line, then cos^ a + cos^ /? + cos^ y = 1. Problem 15. Three men (Fig. 23) are each pulling with a force P at the points a, &, and c, respectively. What weight Q can they raise with uniform motion if each man pulls 100 lb. ? Each force makes an angle of 60° with the horizontal and the projections of the forces on a horizontal plane make angles of 120^ with each other. Solve analytically and also graphically by projecting each force on the vertical and horizontal planes. 22 APPLIED MECHANICS FOR ENGINEEliS Problem 16. Three concurring forces act upon a rigid body. Find the resultant in mag- nitude and direction. The forces are defined as follows : 72=-^ Pi = 75 lb. ; a, = 63° 27' ; /3, = 48° 36' ; y^ P, = 80 lb. ; a^ = 153° 44' ; /S^ = 67° 13' Ps = 95 lb. ; «3 = 76° 14' ; ^3 = 147° 2' ; y., = ? Hint, y^, y^, and yg may be found from the relation : cos^ a + cos^ /3 + cos- y = 1. Problem 17. Each leg of a pair of shears (Fig. 24) is 50 ft. long. They are spread 20 Fig. 23 ft. at the foot. The back stay is 75 ft. long. Find the forces acting on each member when lifting a load of 20 tons at a distance of 20 ft. from the foot of the shear legs, neglecting the weight of the structure. 20 TONS Fig. 24 21. Moment of a Force. — The moment of a force with respect to any point is defined as the product of the force and a perpendicular from the point to the line of action CONCURRENT FORCES 23 0/ the force. Let P (Fig. 25) be the force and the point and a the perpendicular distance of the force from the point ; then Pa is the moment of the force with respect to the point 0. This moment is measured in terms of the units of both force and lengthy viz. pound- feet or pound-inches, and is read pound-feet or pound- inches to distinguish it from foot-pounds of work or inch-pounds of work. For convenience the algebraic sign of the moment is said to be positive when the moment tends to turn the body in a (direction counter-clockwise^ and negative when it tends to turn the body in the clockwise direction. The moment may be repre- sented geometrically as follows : let EF represent the magnitude of P, drawn to the desired scale, and draw EO and FO. The area of the triangle OFF = ^ EFa, or EFa = 2 A OFF ; that is, the moment of the force with respect to a point is geometrically represented hy twice the area of the triangle., whose base is the line representing the magnitude of the force and whose vertex is the given point. If the moment of the force is negative, then the moment is represented by minus twice the area of the triangle. 22. Varignon's Theorem of Moments. — The moment of the resultant of two concurring forces with respect to any point in their plane is equal to the algebraic sum of the moments of the two forces with respect to the same point. 24 APPLIED MECHANICS FOR ENGINEERS Let P and Q be any two concurrent forces and a point in tlieir plane. Through draw a line parallel to the line of action of one of the forces, as P. Let the seg- ment AC cut off on the line of Q by this line be taken to D CO D -1 (a) B A ^^j ^B Fig. 26 represent the force Q, and let AB represent the force P to the same scale. Then, using the upper sign for case (a) and the lower for case (i). Fig. 26, mom P + mom Q = 2 (area of OAB ± area of OAC). But area of OAB = area of ABB = area oi AOB. .'. 2 (area of OAB ± area of OAC) = 2 area of OAD = mom B. .*. mom P + mom Q = mom B. 23. Moment of a Force with Respect to a Line. — Let P be any force and AB any line (Fig. 27). The moment of P with respect to AB is defined as follows : Resolve P into two components one of which is parallel to, and the other perpendicular to, AB. The product of the component per- pendicular to AB and the per- pendicular distance of thiscom- YiG, 27 ponent from AB is called the CONCURRENT FORCES 25 moment of P with respect to AB. (In Fig. 27 the mo- ment of P with respect to AB is aPy} More briefly the definition is : The moment of a force with respect to a line is the moment of the projection of the force upon a plane perpendicular to the line, with respect to the inter- section of the line and the plane. The moment is con- sidered positive or negative according as the tendency to rotatiori. by the force is counter-clockwise or clockwise. The sign of the moment will then change if the observer changes from one side of the plane on which the projection is made to the other, and in comparing the moments of several forces with respect to a line the forces should all be projected upon the same plane and their projections viewed from the same side of that plane. 24. Moment of the Resultant of Two Concurrent Forces with Respect to a Line. — The sum of the moments of two con- current forces with respect to any line is equal to the moment of their resultant with respect to that line. Proof : Let P and Q be any two concurrent forces, B their resultant, and AB j/r any line (Fig. 28). Through the intersection of P and Q pass a plane MN perpendicular to AB^ cutting ^^in 0. Project P, ft and B on MJSf, the projections being respec- tively Pj, Qy, and By Fig. 28 From the definition of projection it follows at once thnt Pj is the resultant of P^ and Qy The moments of P, Q, 26 APPLIED MECHANICS FOB ENGINEERS and M with respect to AB are resf)ectively equal to the moments of Pj, Q^^ and R^ with respect to 0. (Definition.) By Art. 22, with respect to 0, mom Pj + mom §j = mom My Therefore, with respect to AB^ mom P + mom Q = mom i?. If there are three or more forces, the application of the above theorem to the resultant of two of the forces and a third force, etc., proves that the sum of the m,oments of any number of concurre7it forces with respect to any line is equal to the moment of their resultayit with respect to that line. CoKOLLARY. From the definition of the moment of a force with respect to a line it follows that the sum of the moments of two forces of equal numerical value but oppo- site in direction and acting in the same line is zero. Use will be made of these principles in a later chapter. CHAPTER III PARALLEL FORCES 25. The Resultant of Two Parallel Forces. — In consid- ering two parallel forces three cases arise : (a) when the forces are in the same direction ; (5) when they are unequal and in opposite directions ; (c*) when they are equal and in opposite directions, but have different lines of action. In case (c) the two forces form a couple. It will be shown later that there is no single force that will replace them. In cases (a) and (5) the two forces have a resultant. Its value and line of action are found as follows : Let the two forces be Pj and Pgi acting at the points A^ and A^. At A^ and A^ in the line A^A^ put in two equal and Fig. 29 opposite forces T^ and T^. These two forces will have no effect as far as the state of rest or motion of the body on which the forces act is concerned. Combine these 27 28 APPLIED MECHANICS FOR ENGINEERS forces with Pj and P^^ respectively, obtaining the result- ants i^j and M^' i'hese resultants may be moved back to the point of intersection of their lines of action B, and resolved into components parallel to their original components. The two forces T^ and T^ then annul each other, and there are left the two forces Pj and P^ acting at B parallel to their original positions. The resultant then in case (a) is P = P^ + P^^ and in case (5) R== P^ — Py Let the line of action of the resultant cut the line A^A^ in (7, and let the distances from to A^ and A^ be re- spectively c?j and c?2* Then from similar triangles, in either case, BCP^ .BC_P^ a^ ij ^2 1 2 By division, remembering that Tj = T2, d^ p; Hence, the resultant of two parallel forces acting in the same direction is equal to their sum^ is parallel to the forces^ and divides the line Joining their points of application in the inverse ratio of the forces. The resultant of two unequal parallel forces acting in opposite directions is equal to their difference^ acts hi the direction of the larger force^ and divides the lifie joining their points of application externally in the inverse ratio of the forces. 26. The Moment of the Resultant of Two Parallel Forces. — Applying the theorem of Art. 24 for the moment of PARALLEL FORCES 29 the resultant of concurrent forces to the forces of the preceding article, mom H = mom M^ + mom i^g = mom Pj + mom T^ + mom P^ + mom T^. But mom T^ + mom T^ = (Art. 24, Cor.). Therefore mom M = mom P^ + mom P^. Or, ^^Ag moment of the resultant of two parallel forces with respect to any line is equal to the algebraic sum of the mo- ments of the two forces ivith respect to that line. 27. The Resultant of Any Number of Parallel Forces in Space. — By combining the resultant of two parallel forces with a third, and that resultant with another, and so on, the theorems of the two preceding articles may be ex- tended at once to any number of parallel forces : (1) The resultant of any number of parallel forces in space is equal to their algebraic sum and acts parallel to the forces. (2) The algebraic sum of the moments of any number of parallel forces in space with respect to any line is equal to the moment of their resultant with respect to that line. It folloAVS at once that if a set of parallel forces is in equilibrium^ the sum of the moments of all the forces with respect to any line is zero., since the resultant of the forces is zero, and hence the moment of the resultant is zero. If, conversely, the sum of the moments of a set of par- allel forces with respect to a line not parallel to the forces is zero, either the resultant of the forces must be zero, or else the line of action of the resultant must intersect the given line. If, however, the sum of the moments 30 APPLIED MECHANICS FOR ENGINEERS of the forces about each of two lines not parallel to the forces which cannot intersect the same line parallel to the lines of action of the forces is zero, the resultant must be zero and the forces are in equilibrium. If the lines of action of the forces all lie in one plane, the most convenient line about which to take moments is perpendicular to the plane of the forces. The condition just stated for equilibrium in this case becomes : If the sum of the moments of a set of parallel forces in one plane tvith respect to each of three points of Fig. 30 the plane not in the same straight line is zero, the forces are in equilibrium. Illustration. Forces of 7 lb., 5 lb., and 10 lb. act per- pendicular to the plane of the lines OA and OB, as shown in Fig. 30. The distances of the lines of action of the forces from OA and OB are respectively 1, 3, 4 units and 2, 1, 5 units. To hnd where the line of action of the resultant cuts the plane OAB : Let a and b be the distances of the line of action of the resultant from OA and OB respectively. The result- ^^'^^^ i2 = 7 + 5 + 10 = 22 1b., acting in the direction of the forces. PARALLEL FORCES 31 Taking moments about the line OA^ a.i2 = 1.7 + 3.5 + 4.10, or a = 2.82 units of space. Taking moments about OB, 5. ^ = 2. 7 + 1-0+5. 10, or 6 = 3.14: units of space. Hence the resultant is equal to 22 lb., acts in the direc- tion of the forces, and at distances of 2.82 and 3.14 units from OA and OB respectively. Fig. 31 Problem 18. Two parallel forces, one of 20 lb. and one of 100 lb., have lines of action 24 in. apart. Find the resultant in magnitude, direction, and line of action : (1) When they are in the same direction. (2) When they are in opposite directions. Problem 19. A horizontal beam of length I is supported at its ends by two piers and loaded with a single load P at a distance of - from one end. Find the reactions of the piers against the beam. Problem 20. The locomotive shown in Fig. 31 is run upon a turntable whose length is 100 ft. Find the position of the engine so that the table will balance. 32 APPLIED MECHANICS FOR ENGINEERS Problem 21. Weights of 100, 50, and 120 lb. are placed at the ver- tices A , B, and C respectively of an equilateral triangle 4 ft. on a side. Find the distances bi the center of the forces from the lines AB and BC. Problem 22. On a square table weights of 70, 80, and 100 lb. are placed as shown in Fig. 32. Find the position of a fourth weight of 90 lb. which will balance the given weights with respect to the two lines AB and CD. Ans. f ft. from AB, 3f ft. from CD in quadrant BOD. Fig. 32 28. Center of Parallel Forces. — In Art. 25 it was shown that the resultant of two parallel forces P^ and P^, acting at the points A^ and A^ respectively, divides the line A-^A^ in the inverse ratio of the forces. If, then, without changing the points of application of the parallel forces Ar C A, 1 t 1 1 1 1 1 Pi 1 1 1 1 1 it: 1 1 1 1 Po 1 1 " 1 1 1 1 1 1 / i__ • 1 1 1 R t . ' -Pl+^2 Fig. 33 Fig. 34 or their magnitudes their direction be changed, the point on ^1^2 through which the resultant passes is not changed, since it must still divide A^A^ in the ratio P^ : Py In the same way, if three parallel forces of fixed value Pj, -Pgi -^3 ^ct at three fixed points A^, A^, A^, and if C PARALLEL FORCES 33 is the point on ^1^2 through which the resultant of Pj and Pg passes, then the point C on O'A^ which divides C'Aq in the ratio P^:(^P^-\- P^ is a point through which the resultant of the three forces always passes no matter what their direction. The same reasoning applies to any number of parallel forces acting at definite points. For any set of parallel forces acting at definite points there is therefore a point through which the resultant always passes no matter what direction the parallel forces may take. This point is called the center of the parallel forces. 29. Graphical Construction for the Center of Parallel Forces. — Let the parallel forces P^ and Pg act at A^ and A^ (Fig. 35). From A^ lay off A^B^ equal to Pg but in the opposite direction to Pg. From A^ lay off A^B^^ equal to P^ and in the same direction as Pj. The inter- section of B^B^ and A-^A^ is then the center of the two forces. For, letting A^C = d^ and OA^ = d^^ we have, by similar triangles A^B^C and A^Bfi, ^ = 4A or ^ = ^ A* d. AB2 d. "2 ^-^2^2 '^2 Hence Ois the point found in Art. 25 to be the center of the forces. 34 APPLIED MECHANICS FOR ENGINEERS If there are more than two parallel forces, the continued application of the construction will locate the center of the forces. Problem 23. Weights of 20, 45, and 70 lb. are suspended from points on a straight line at distances of 2, 5, and 7 ft. respectively from a point O. Find by graphical construction the center of the forces exerted by the weights. Check by the theory of moments. Problem 24. Locate graphically the center of the three forces in the illustration of Art. 27. (N.B. The forces may be assumed to act in the plane GAB without changing the position of the center.) 30. In Regard to Signs. — Given a set of parallel forces, Pj, J*2^ -^3' ^tc-' acting at definite points. Let 01^ be a line in a plane perpendicular to the lines of action of the forces. For convenience assume the lines of action of the forces to be vertical. Let it be agreed that forces acting upward shall be positive and those acting downward negative. Let x-^^^ X2, % ••• be the perpendiculars from 01^ to the lines of the forces P^ P2' ^3^ •*•' X being counted positive if measured from the axis OY to- ward the right and negative if measured from OY toward the left. Then, when observed in the direction YO^ Px is the Fig. 36 p t^ • 1 moment 01 F with respect to OY^ not only in numerical value but in sign. For if both P and X are positive, as P^ and x^^ or both negative, as Pg and a^g, the product is positive. Inspection of the figure shows that the tendency to rotation in both cases is counter- clockwise, which has been defined as positive. PARALLEL FORCES 35 If X is positive and P is negative, as x^ and P2' ^^ ^^ ^ is negative and P is positive, as x^ and P^, the product is negative, and an inspection of the figure shows that the tendency to rotation is clockwise, which has been defined as negative. The sum of the moments of the forces with respect to 1^ is then PiX^ + ^2^2 + -^3^3 + -^4^4 + • ' ' • This sum is denoted by ^Px. 31. Coordinates of the Center of a System of Parallel Forces. — Let Pp Pg, Pg, ••• be a set of parallel forces whose points of applica- tion are (x-^^ y^, 2^), (x^^ Vv ^2)' C^3' Vz^ %)' ••• ii^ rectangular coordinates. Denote the center of the forces by (2:, y, z). As was shown in Art. 28 the position of the center of the forces is independent of the direction in which they act. Assume then ^^^- ^^ that they act parallel to OZ, Taking moments about OY^ we have the moment of the resultant equal to the sum of the moments of the forces, or X{P^ + P2 + P3 + •••) = ^1^1 + ^^2^2 + ^3^3 +••• or x^P — ^Px. Taking moments about OX^ y^P=^Py, 36 APPLIED MECHANICS FOR ENGINEERS Assuming the forces to act parallel to OX and taking moments about OY, we obtain Therefore - _ 2 J>i/ 2P ' y r_21>« ^-2P^ Problem 25. Forces of 40, 65, and 70 lb. act in the same direction from the points (4, 3, 1), (5, - 2, 3), and (2, 3, 6), respectively. Find the resultant and the center of the forces. Problem 26. Three equal parallel forces act at the vertices of a triangle in the same direction ; prove that their resultant acts at the intersection of the medians of the triangle. Solve by applying the theorem of moments, and check by graphical construction. Problem 27. Four equal forces act in the same direction at the vertices of a regular tetrahedron. Find the center of the forces by taking moments. Also locate the center graphically. 32. Arrang-ement of the Work for Computation. — If sev- eral parallel forces and their points of application are given, it is sometimes worth while to tabulate the work in some form such as the following : Forces Coordinates Moments X y Z Px Py Pz ^1 X, Vi ^l P,x, PiVx p - 1 y.^ P. X.2 ?/2 ^2 -^2'^2 ^2.V2 Pf-2 Ps Xo ?/3 ^3 ^3^3 P^y-i P,h P. X, " //4 -^4 P,x, P^Va P,z, Resultant X Jj ~ ^Px ^Py ^Pz PARALLEL FORCES 37 The values of quantities in the columns one, two, three, and four are given except in the last row. The values of the quantities in columns five, six, and seven are then computed and their sums and the sum of column one placed at the foot. The values of x, y^ and z are then easily computed. Problem 28. Using the above arrangement, find the coordinates of the center of the following parallel forces : P^ = 50 lb., P.^ = 100 lb., P3 = 300 lb., P4 = 10 lb., and P.. = - 400 lb., the points of applica- tion being respectively (2, 1, -5), (-1, -2, 4), (2, 1, -2), (-2, 1, 1), (1,1,1). Ans. (3, -4, -14). Problem 29. Parallel forces P^, Pg, P3, and P^ act at the corners of a rectangle 3 ft. by 2 ft. and perpendicular to its plane. Find the point of application of the result- ant, if Pj = 10 lb., P2 = 50 lb.. P., = 100 lb., P4 = 200 lb., Pi and P2 be- ing 2 ft. apart, and Pg on same side as Pg. Problem 30. Eight parallel forces act at the corners of a one-inch cube. Find the point of application of the y- resultant force, if P-^ = 30 lb., Pg = 50 lb., P3 = 10 lb., P4 = 20 lb., P5 = 100 lb., P(i = o lb., P7 = 10 lb., Ps = 40 lb. The subscripts of the forces acting at the various vertices are shown in the figure (Fig. 38). Ans. (.283, .302, .415). Fig. 38 CHAPTER IV CENTER OF GRAVITY 33. Definition of the Center of Gravity. — The center of gravity of a body may be defined as the point of appli- cation of the resultant attraction of the earth for that body, and the center of gravity of several bodies consid- ered together, as the point of application of the resultant attraction of the earth for the bodies. The attention of the student is called to the fact that the forces acting upon the particles of a body, due to the attraction of the earth, are not parallel, but meet in the center of the earth. For all practical purposes, however, they are considered parallel. The center of gravity of many simple bodies can be found by inspection. For example, for a sphere of uni- form material it is evident that the line of action of the earth's attraction always passes through the center. For a rectangular block of uniform material the same is true. If a body be divided up into a number of parts whose centers of gravity are known, the whole weight of the body is the resultant of the known weights of the parts acting at known points, and hence the theorem of moments may be applied to determine the position of the center of gravity of the body; i.e. the equations -_2x^ ,7_2yP 5_2^ 38 CENTER OF GRAVITY 39 may be used where P^, P^^ etc., re^jresent the weights of the known parts, a;^, x^^ etc., the abscissas of the centers of gravity of these parts, etc. As an illustration consider the prob- lem of finding the center of gravity of the solid shown in Fig. 39. The figure represents a Z-iron of the same cross section throughout, and Pj, P^^ and Pg are the weights of the indi- vidual parts (considering the Z-iron as divided into three parts — two legs and the connecting vertical portion). If the w^eight of a cubic inch of iron = .26 lb., Pi = .78 lb., ^2 = 2.08 lb., P3 = 1.04 lb., and therefore i?=3.9 lb. The points of application of P^, P^^ and Pg are (— J^ — !» ^2)' Q, — ^, 5), and (2, — |-, |^), respectively, so that - _-78(-.i)+ -2.08(1) + 1.04(2) _ X = ^ = 3.9 = .70 in.. .78(-i)4-2.Q8(-i:)+l.Q4(-i) ^ _ ^^ .^^^ 3.9 .78(V-)-f2.Q8(5) + l.Q4a) ^.^^ 3.9 in. This point x^ «/, z is, in this case, the center of gravity of tlie Z-iron. 34. Centers of Gravity of Uniform Bodies and of Areas. — If Pj, P2, P3, etc., are the weights of the parts of a body or of several bodies, whose unit weights are, respectively, 7^, 72' 73' ®tc., and volumes F^, V^^ Fg, etc., we may 40 APPLIED MECHANTCH FOR ENGINEERS write JiV^ for P^, ry^V^ for P^, J^^s foi" -^3^ etc. The formulce for rr, «/, z then become ~ ^ 7i ^1^1 + 72 ^^2^^2 + 73 ^3^3 + etc. ^ 27 Fa; 7i ^1 + 72 ^^2 + 73 ^"3 + etc. l,yV ' - 27 FV - S7F2 ^ 27F 27F And if the bodies are all of the same material and so have the same heaviness, 7 is constant and may be taken outside the summation sign, Avhere it cancels out. This gives values for x^ «/, and z^ formulae exactly similar to those of Art. 33, where the P's are replaced by F's. If the bodies are thin plates of the same material, of constant thickness J, we may write for F^, F^, F^, etc. 5Pj, hF^-, hF^^ etc., where the P's represent the areas of the faces of the plates. Making this substitution for the F's, x^ y^ z may be written - _ hF^x^ + ^^2^2 + ^-^3^3 + etc. _ ^Fx ^~ 5Pi + bF^+bF^'+ etc. Yf' - _^Fy -^_^Fz y~ si' ^~ 2P' the 5, being a constant factor, cancels out. These formulge are taken as defining the '' center of gravity of an area " and are much used by engineers for finding the center of gravity of sections of angles, channels, T-sections, Z-sec- tions, etc. Usually, the a^y-plane is taken in the plane of the area and only the values of x and y are needed. CENTER OF GRAVITY 41 0.4 Problem 31. Find the center of gravity of the channel section shown in Fig. 40. Problem 32. Find the center of gravity of the T-section shown in Fig. 41. -;^ -10'^- Fig. 40 ~X^ Fig. 41 Fig. 42 Problem 33. Find the center of gravity of the U -section shovrn in Fig. 42. Given the fact that the center of gravity of a semicircular 4 r area is - — from the diameter. (See Prob. 41.) 3 TT Problem 34. Find the position of the center of gravity of a trapezoidal area, the lengths of whose parallel sides are a^ and a^, re- rj p spectively, and the distance between them h. (See Fig. 43.) Hint. Draw the diago- nal AB and call the tri- angle ACB, Fp and the triangle ABD, F^. Fig. 43 Given, the center of gravity of a triangle is i the distance from the base to the vertex. (See Prob. 39.) Select .1 D as the a;-axis, then 42 APPLIED MECHANICS FOR ENGINEERS y _ ^ii/i±Z^ where y-^ = | h and y^= \ h. The center of gravity is seen to lie on a line joining the middle points of the parallel sides. Problem 35. A cylindrical piece of cast iron, whose height is 6 in. and the radius of whose base is 2 in., has a cylindrical hole of 1 in. radius drilled in one end, the axis of which coincides with the axis of the cylinder. The hole was originally 3 in. deep, but has been filled with lead until it is only 1 in. deep. Find the center of gravity of the body, the unit weight of lead being 710 and of cast iron 450 (Fig. 44). Problem 36. Find the center of gravity of a portion of a reinforced concrete beam. (See Fig. 45.) The beam is reinforced with three half -inch steel rods, centers 1 in. from the bottom of the beam and 1 Z Fig. M ^ 0^:..C)1. -,S ■X Fig. 45 CENTER OF GRAVITY 43 m. H in. from the sides. The center of the middle rod is 4 in. from the sides. (y for steel = 490 lb. per cubic foot ; y for concrete = 125 lb. per cubic foot.) Note. It is seen that the thickness cancels out of the expression for the center of gravity, and might, therefore, have been neglected. 35. Center of Gravity of a Body with a Portion Removed. — It is sometimes convenient in finding the center of gravity of a body to regard it as a larger body from which a por- tion has been removed. The weight of the given body can then be regarded as the resultant of the weight of the larger body acting at its center of gravity and a force equal to the weight of the portion removed acting at ' its center of gravity in the opposite direction. To illustrate this consider the cylinder of Fig. 46 of height jETand radius R from which a cylindrical portion of height h and radius r and having the same axis is removed from one end. If W is the weight of the whole cylinder, and w the weight of the portion removed, then the center of gravity of the remaining body is the center of the two forces W and w acting as shown in the figure. If 7 is the heaviness, W= yirE'^R, w = yirr^h. The distance y from the open end to the center of gravity of the body is therefore -R- -w Y \iM Fig. 46 44 APPLIED MECHANICS FOR ENGINEERS - - 2^2 _ 1 722^2 _ ^2^2 •^ ~ jttEW- yirr^h 2 E'^IT-r^h' In general the formula is, where TFg is the weight removed from the weight W^ and a:j and x^ are the abscissas of the centers of gravity of the weights W^ and TF2, respectively, before the removal of the weight W^- Problem 37. Through a circular disk 1 ft. in diameter a hole 4" in diameter is bored with its axis distant 3" from the axis of the disk. Find the position of the center of gravity of the remainder. Problem 38. From two adjacent corners of a cube of edge 1 ft. cubes of 2-inch and 3-inchi edges are removed. Find the center of gravity of the remaining portion. 36. Center of Gravity Determined by Integration. — In many cases of areas and solids the position of the center of gravity may be determined by integration. The method as applied to areas is as follows : Let F be any area in the 2:j/-plane. In order to deal with actual forces think of F as the face of a thin sheet of uniform thickness and uniform material and let 7 be the weight of the sheet per unit area in the surface F. Divide the area up into elements of area AF. (This may be done in a variety of ways, Fig. 47.) The weight of such an element is then yAF. The weight of the sheet may then be replaced by a set of parallel forces of magni- tude yAF acting at the centers of these elementary areas. CENTER OF GRAVITY 45 The centers of the elementary areas and the elementary areas themselves may be determined approximatel}', and if x' is the abscissa of the approximate center * of AF, then an approximate value for the abscissa of the center of the parallel forces is ^'=%If- CArt.38) The limiting value of x' as the elementary areas are in- definitely decreased is defined as the abscissa, x^ of the center of gravity of the area F. Hence, - _ Lim ^x'yAF Lim l^yAF as AF approaches the limit zero. By a theorem of calculus this may be written Cx'ydF \dF ' x = /^ *The approximate center must of course be such that the approximate center and the true center of gravity of the element have the same limiting position as the elementary area is indefinitely decreased. If both dimensions of AF approach the limit zero as in {h) and (c), Fig. 47, the approximate center may be taken as any point of AF. 46 APPLIED MECHANICS FOR ENGINEERS or, since 7 is constant, Jx'dF fdF^ the integration being taken to inclnde the whole area F. Similarly Jy'dF y^ f dF In deriving the latter formula the forces would be thought of as acting parallel to the a;-axis. In the use of these formulae it must be kept in mind that x' and 7/' are the coordinates of the approximate center of the elementary area AF. Illustration, To find the center of gravity of the area bounded by the a;-axis, the curve i^^y}^ y = sin x^ and the line X ——' 2 Divide the area up into strips ^ parallel to the ?/-axis of width Ax. Then con- sidering any strip whose middle ordinate intersects the curve in the point (2:, ?/), the value of AF is yAx ap- proximately, the value of x^ is a:, and the value of y^ is I (Fig- 48). Fig. 48 CENTER OF GRAVITY 47 Then - _ Jxydx I X sin xdx ^ _ _o _»A J I ydx j 2 sin a:cZ2; Jo ( — a; cos a: + sin x') cos a; 1 '• J^ ydx - I sin^ a;c?a; 2"^ _ 2-^0 77 77 J I ydx I sin a;6?a; *>'0 (i ^ ~ i sin 2 a;) 2 TT — COS a; 8__7r T~" 8" Fig. 49 Problem 39. Find the center of gravity of a triangle whose alti- tude is h and whose base is a. Take the origin at the vertex and draw the x-axis perpendicular to the base. (See Fig. 49.) 48 APPLIED MECHANICS FOR ENGINEERS ix'clF X = -J^ Here dF = aV/.r, and from similar triangles ^4F a' — ~ X, so that dF = - xdx. h A ' and sc h a rh Ji 3Jo \ x\lx > Cxdx~'^-X Jo 2 Jo I- The center of gravity is | the distance from the vertex to the base, and since the median is a line of symmetry, it is a point on the median. It is, in fact, the point where the medians of the triangle intersect. B Fig. 50 Problem 40. Find the center of gravity of a parabolic area shown in Fig. 50, the equation of the parabola being if = 2 px. Here dF = ydx, so that i xydx ■\/2p i '^x^dx -5 x^ \ g ("a ra 1 -3 Ho 5 \ ydx v-/>\ x'^dx I ^^ L It is left as a problem for the student to show that y = \h. Problem 41. Find the center of gravity of a sector of a flat ring outside radius R^ and inside radius R^. (See Fig. 51.) Let the CENTER OF GRAVITY 49 angle of the sector be 2 6. Take the origin at the center and let the a:-axis bisect the angle 2 0. Here dF = pdpda, and x = p cos a, so that I i cos ada • p^dp X = Fig. 51 Integrating the numerator first, \ cos ada r J,/p = ^1 .^^^ cos «cZ« - ^^ , ^'^ (2 sin (9). Integrating the denominator, _ 2 JJi3 - JBs^ sill Therefore, ^ 3 «,. _ jj,. e If i?2 = 0, the sector becomes the sector of a circle, and x becomes _ 2^ sin^ If the sector is a semicircle, that is, if 2 6 = tt, then, since = -, 3 IT 50 APPLIED MECHANICS FOR ENGINEERS Problem 42. Find the center of gravity of a semi-ellipse (Fig. 52) whose equation is .r^ . // therefore (IF = 2 ydr = 2 - V'a^ - x^ - dx, a Fig. 52 hra 2 ~\^ ^\/(jfi — 'xP- dx -lia^-x'y]l X ■= ]) fa If X]" 2- I Va'^ — x'^dx H xVa^ — a:^ + rt2sin-i~ I 3 _4a O^ TT 3 IT 2 *2 CENTER OF GRAVITY 51 Problem 43. Find the center of gravity of the area between the parabola, the ^/-axis, and the line AB in Problem 40. Problem 44. A quadrant of a circle is taken from a square whose sides equal the radius of the cir- cle. (See Fig. 53.) Find the center of gravity of the remaining area. Problem 45. Suppose that the corners A and C of the angle iron in Fig. 54 are cut to the arc of a circle of j\ in. radius and the angle at B is filled to the arc of a circle f in. radius; what would be the change in Ic and ^? Y Fig. 53 J3 C T X Fig. 54 37. Center of Gravity of a Solid. — By dividing a solid into in- finitesimal parts and taking the moments of the weights of these parts with respect to the three coordinate axes, the coordinates of the center of gravity of the solid may be found. As an illustration, suppose it is desired to obtain the center of gravity of a right circular cone of altitude h and radius of base r. Take the a>axis as the axis of the cone with the vertex at the origin. (See Fig. 55.) It is evident, that ^ = and 2 = 0, so that it is only necessary to find X. The Fig. 55 52 APPLIED MECHANICS FOR ENGINEERS volume, dv, cut from the cone by two parallel planes, per- pendicular to OX and separated by a distance dx, is ir^^dx, and the weight of this dv is ^iry^dx — dP. Therefore X = I x' dP i x^iry^d:* J dP I ^iry^dx But from similar triangles y : x::r :h or y = -x. This gives y2 /»A rr*' A2 >2 /'A /v^" 3, -= -rh. h 4 The expressions for ^, ^, and 2, involving c?P, ma}^ be changed to similar ones involving dv^ and these become for homogeneous bodies, since dP = ^dv^ ix'dv _ iy'dv _ iz'dv ^ -_ ^1 , y = ^ , z = ^ j 6??; i ds I dv The center of gravity of thin homogeneous wires of con- stant cross section may be found by replacing the dv in the above formulae by ads^ where a is the constant area of cross section and ds is a distance along the curve. The formulae then become j x'ds I y' ds { z' ds a?=^ — , y = fds' fds f ds Problem 46. Find the center of gravity of a hemisphere, the radius of the sphere being r. Let the equation of the generating circle of the surface be x^ -\- y^ = r^. Divide the hemisphere into CENTER OF GRAVITY 53 slices of thickness dx by planes parallel to the plane face of the hemisphere. Then (x'dP X — —^ , where dP = y-nyHx = y7r(r'^ — x'^)dx. J4P Show that '=h Fig. 56 Problem 47. Find the center of gravity of a portion of circular wire (Fig. 56) of length L and whose chord = 2 b. Take the center of the circular arc as oricrin and let the x'-axis bisect L. Then - J xds X ^ds But .r^ + .y2 = RK .*. 2 xdx + .'. ds = 0. \x/ = y/dx^ + dy'^ = ■dy = —dy. X .*. T R\dy _ radius x chord L arc 54 APPLIED MECHANICS FOR ENGINEERS For a semicircular wire X — Diameter TT Problem 48. Find the center of gravity of a paraboloid of revo- lution. If the equation of the generating curve is y^ = 2px, and the greatest value of x is a, show that 3 Suggestion. Use the same method as that used for the right circular cone. Problem 49. (a) Show that the center of gravity of the circular sector A OB (Fig. 57) of angle 2 a and chord 2 d is given by _ _ 2 d _2 radius x chord 3 a 3 arc Fig. 57 (b) From this result and the known position of the center of gravity of the triangle, prove that the center of gravity of the segment ADB is given by -,^2d^ ^ ~SF where F is the area of the segment. CENTER OF GRAVITY 55 Suggestion. In (a) use polar coordinates. The element of area is pdOdp and ("Cp'' cos ec/Odp X = i^i^'^'f Problem 50. Show that the center of gravity of any pyramid is in a plane parallel to the base which cuts the altitude at | the dis- tance from vertex to base. Problem 51. Show that the center of gravity of a hemispherical surface bisects the radius perpendicular to the plane of the base of the surface. Problem 52. Show that the center of gravity of a spherical zone is midway between the planes of the bases of the zone. Problem 53. Show that the center of gravity of the surface of a right circular cone is at | the distance from the vertex to the base. Problem 54. Show that the center of gravity of the frustum of a right circular cone, the radii of the bases being R and r and the alti- tude H, is distant H(R^ + 2Rr+S r^) 4:(R'^ + Rr + r2) from the base of radius R. Find also the distance of the center of gravity from the base of radius r. Problem 55. A casting is in the form of a hollow cylinder with one end closed. The thickness of the end is | in., the length of the casting is 12 in., and the radii of the inner and outer surfaces of the cylinder are 5^ and 6 in. Find the position of the center of grav- ity of the casting. Problem 56. If the above casting is filled with material | as heavy as the material of the casting, find the position of the center of gravity. Problem 57. A hemispherical shell of inner and outer radii r and R rests upon a hollow cylinder of the same material of height H BQ APPLIED MECHANICS FOR ENGINEERS and inner and outer radii ?• and R. Find the distance of the center of gravity from the base of the cylinder. Problem 58. From a riglit circular cone of height 20 in. and radius of base 10 in. a cylinder of height 10 in. and radius of base 3 in. is removed, the base and axis of the cylinder being in the base and axis of the cone. Find the distance from the base of the cone to the center of gravity of the part remaining. Problem 59. If the cylindrical portion of the preceding problem is filled with material n times as heavy as the material of the cone, find the distance from the base to the center of gravity. 38. Center of Gravity of Counterbalance of Locomotive Drive Wheel. — In Fig. 58 the drive wheel is indicated by the circle and the counterbalance by the portion inclosed by the heavy lines, the point is the center of the wheel, and a is the angle subtended by the counterbalance. The Fig. 58 point 0' is the center of the circle forming the inner boundary of the counterbalance, and /3 is the angle sub- tended by the counterbalance at this point. Let F^ repre- sent the area of the segment of radius r and F^ the area of the segment of radius r^ Also let x-^^ represent the distance of the center of gravity of F^ from 0, and x^' the CENTER OF GRAVITY 57 distance of the center of gravity of F^ from 0' . Then, from Problem 49, X. = 77— =r and .r^' — — — . But a^gi the distance of the center of gravity of Fc^^ from (9, = Xo' — 00 = —-—- — r. cos ^ — r cos - . ^ 3 i<^2 V 2 27 Then the distance of the center of gravity of the counter- balance from is X = ^1-^2 _^2 /3 a r-, cos ^ — r cos - Q where F. = ar cos -, ^ 2 2 and #„ = ^— J — ar-t cos — . Problem 60. If r = 3 ft, r^ = 10 ft., and a =120^, find the posi- tion of the center of gravity of the counterbalance of Art. 38. 39. Graphical Method of Finding the Center of a Set of Parallel Forces in One Plane. — Let P^, P^, Pg, P^ be a set of parallel forces acting at the points A^, A^^ ^g, A^^ respectively, in one plane. Assume the forces to be acting in the direction shown in Fig. 59. Let one of the forces, as Pj, be moved to any point in its line of action, as JB^ and resolved into two components in any two directions, as S^^ J and S^^- Extend the line of one of these compo- 58 APPLIED MECHANICS FOR ENGINEERS nents, aS'^^ g' ^^ ^^^ ^^^^ li^^ *^f ^ second force, Pg' ^^^^ resolve the second force into two components, one of which is equal and opposite to the component, S^^ ^, of the first force on that line. Extend the line of the other component, S^^ 3, of the second force, to cut the third Fig. 59 force and resolve the third force into components as in the case of the second force. Proceeding in this way until the last force is reached there remain then only two unbalanced components of the original forces, one of the components, S^^ -^^ of the first force, and one, S^^ 5, of the last force. The resultant of the original forces is there- fore the resultant of these two unbalanced components. By Art. 27 the resultant of the parallel forces is parallel to the forces. Hence the resultant, P^^ acts in a line, parallel to the original forces, through the intersection, ^g, of the unbalanced components, aS^^^ j and S^^ g. CENTER OF GRAVITY 59 The actual work of locating the line of action of the resultant may be shortened by the following considera- tion : the triangles of forces at the vertices B^^ B^^ B^^ B^ may be placed with equal sides coinciding, as in Fig. 59 (5). The sides representing the forces P^, P^^ Pg, P^ then fall in succession on a straight line, the beginning of any force falling at the end of the preceding one. The lines representing the components S-^^^^ S^^ g, etc., all meet at a common point 0. The rays from are then parallel respectively to the sides of the polygon B^B^B^B^B^. This enables one to draw the polygon B^B<^B^B^B^ without constructing the separate triangles B^B^Up etc. Since S-^^ ^ ^^^ ^5, i ^^^ arbitrary directions given them, the point may be chosen arbitrarily, and the polygon B^B^B^B^B^ constructed by drawing its sides parallel to the corresponding rays from 0. To sum up, the method is as follows : Assume a direction in which the forces act. Lay off in succession the forces on a line parallel to their lines of action. Join their points of intersection with an arbitrary point 0, Fig. 60 (h). Construct a polygon with vertices on the lines of action of the forces and with sides parallel to the corresponding rays from 0. The intersection of the sides of this polygon, B^ in Fig. 60 (a), drawn par- allel to the rays from to the beginning and end of the line of forces, is a point through which the resultant passes. This gives the line of the resultant for the as- sumed direction of action of the forces. The center of the forces is therefore on this line. 60 APPLIED MECHANICS FOR ENGINEERS In the same way another line of action for another as- sumed direction may be determined. The intersection of these two lines is the center of the given forces (Art. 28). Fig. 60 In doing this work the forces may be laid off in suc- cession in any order. It is only necessary to make the sides of the polygon B^B^B^ ... correspond to the rays from to the line of forces; i.e. the side of the polygon connecting any two forces must be parallel to the ray from to the intersection of the same two forces. The polygon B^B^B^ ... is called the equilibrium polygon. A more general discussion will be given in the chapter on non-concurrent forces in one plane. Problem 61. Parallel forces of 8, 12, 16, and 20 lb. act at points (0, 0), (1, 3), (3, I), and (5, 4) respectively. Find graphically the line parallel to the ?/-axis in which the center of the forces lies. Make the construction twice, laying off the forces in different orders. Check by the theorem of moments. Problem 62. Find graphically the distance from the ar-axis to the center of the forces of the preceding problem. Problem 63. AVeights of 1200, 1600, 3000, and 2000 lb. act on a beam at distances of 0, 5, 8, and 12 ft. respeetively from one end. CENTER OF GRAVITY 61 Find graphically where their resultant cuts the beam. Check by moments. Problem 64. Locate graphically the center of gravity of the area of a quadrant of a circle. Suggestion. Divide the area up into a number, say eight, of strips of equal width (Fig. 61). The weights of the strips are then approximately proportional to the ordinates at their middle points and act in the lines of these ordinates. Hence use these ordinates, or any proportional parts of them, as forces and pro- ceed as in finding the center of a set of parallel forces. Fig. (51 Problem 65. Locate graphically the center of gravity of the arc of a quadrant of a circle. Problem 66. Show how to find graphically, approximately, the center of gravity of any area in one plane. Apply the method to finding the center of gravity of a given area. Check by cutting the area out of cardboard and balancing. Problem 67. Find graphically, approximately, the position of the center of gravity of a hemisphere. Suggestion. The hemisphere may be divided up into a number of parts by parallel planes equally spaced. The w^eights of these parts will then be approximately proportional to the squares of the ordinates at the middle of the generating strips of area. Lines pro- portional to the squares of these ordinates may be constructed as shown in Fig. 62 (b) . These lines may then be used as forces and the graphical solution obtained as in the preceding problems. Problem 68, Find graphically the position of the center of gravity of a solid of revolution obtained by revolving a parabola about its axis. Problem 69. By the method of Problem 67 find the position of the center of gravity of the frustum of a cone. Compare the result with that obtained from the answer to Problem 54 by giving particular values to it, r, and H. 62 APPLIED MECHANICS FOR ENGINEERS (a) Fig. 62 40. Simpson's Rule. — When the algebraic equation of a curve is known, it is expressed as y =f(^x)^ and the area between the curve and either axis is usually determined by integration. In Fig. 63 the area ABCD is expressed by the integral ydx = 1 f(x)dx, a *^ a r Fig. 63 CENTER OF GRAVITY 63 when the curve represented by y =f(x) is continuous between A and B. In many engineering problems the curve is such that its equation is not known, so that approximate methods of /-. /I 1 — ] n G 1^ 1 " A n ^ ^ Vo y^ y. 2/3 E 2/4 y, F y. 2/7 y. y. 2/io 7. ' u Fig. 64 F G H Fig. 65 D obtaining the areas under the curve must be resorted to. One of these methods of approximation is known as Simpson s Rule. Suppose the curve in question is the curve AB (Fig. 64) and it is desired to find the area between the portion AB and tlie a;-axis. Divide the length h — a into an even number, /^, of equal parts (here 7i=10). Consider the portion CDEF and imagine it magni- fied as shown in Fig. 65. Pass a parabolic arc through the points C^ G^ i), the axis of the parabola being parallel to the ?/-axis ; then the area CDEF is approximated by the area of the parabolic segment OGDI plus the area of the trapezoid CDEF. F 64 APPLIED MECHANICS FOR ENGINEERS .-. area CaDi:F = \iy^ + y.^EF + | [2/3 - \ {y^ +3/4)]^'^^, since the area of the parabolic segment is | the area of the circumscribing parallelogram. Since EH = = Aa;, this area may be written In a similar way the next two strips to the right will have an area, -^(^4 + 4 ^/^-f-^/g), and the next two strips, an o \nr. area, -^(^g + 4 ?/y + ?/g), and so on. Adding all these so o as to get the total area under the portion of the curve AB^ we get total area = ^^ [y^ + 4(?/i + 2/3 + ^/^ + y^ + y^) + 2(2/2 + ^4 + ^6 + J/s) + ^10]' or in general for n divisions, total area = -^ [?/o + 4(^i + ^^ + ^ 4- ... ?/„_i) + 2(^/2 + ^4 + «/6 + ••• 2/»-2) + ^n]. and this is Simpson's formula for determining approxi- mately the area under a curve. It is easy to see that the smaller Aa:, the closer the approximation will be. 41. Application of Simpson's Rule. — Simpson's Rule may be made use of in determining approximately not only areas, but volumes and moments. For if a volume be divided up by planes perpendicular to the a:-axis, distant Aa; apart, into elements of volume A?;, then dv = Adx, where A is the area of the cross section at any value of x^ CENTER OF GEAVITT Qb and the volume from x = a to x=h may be expressed as JAdx. Hence the volume may be evaluated by Simp- a son's formula, the y of the formula being replaced by A. The sum of the moments of these elements of volume with respect to the ?/-axis is j xdv or I xAdx and may be evaluated by Simpson's formula, replacing y by xA. On account of its use in adding moments Simpson's formula ^0 ^•li A„ ^u .4 ^ X\ y\ y^\ y 1 1 1 1 1 1 1 1 1 1 1 -;^ 1 J- -" "^ ^ y^ 1 ^'^'' y^ -^^^^ • "*- . -~— 4— - ^ /j^^^**""^ ^ y ^ ^^^^^^^^ Fig. 66 may be employed in finding the center of gravity of areas or volumes bounded by lines or surfaces whose equations are not known. Suppose, for example, it is desired to know the volume and position of the center of gravity of a coal bunker of a ship as shown in Fig. ^^. The bunker is 80 ft. long and the areas in square feet are as follows: A^ = 400, A^ = 700, A^ = 650, A^ = 600, A^ = 400. The distance between the successive areas is 20 ft. Applying Simpson's formula for volume, 80 volume = (8) (4) lA, + 4CA, + A,-) + 2A, + A,]. P 6Q APPLIED MECHANICS FOR ENGINEERS Summing the moments, we obtain 80 ^vx (3) (4) [AqXq + 4c(A^x^ + A^x^-) + 2^122^2 + A^J' where Xq = 0, 2-^ = 20, x^ = 40, x^ = 60, x^ = 80. The po- sition of tlie center of gravity from the fore end can now be obtained from the relation 'liVX x = An approximate value of x may also be obtained from the equation ^1 + ^2+^3 + ^4 where v^ = 1(Aq + -4^)20, v^ = ^(A^ + A^}20, etc., and x\ = 10, x'^ = 30, etc. Problem 70. A reservoir with five- foot contour lines is shown in Fig. 67. Find the volume of water and the distance of the center of gravity from the surface of the water, if the areas of the contour lines are as follows: ^o = 0, ^^ = 100 sq. ft., ^2 = 200 sq. ft., ^3 = 500 sq. ft., A^ = 600 sq. ft., A^ = 1000 sq. ft., Aq = 1500 sq. ft., ^- = 2000 sq. ft., ^8 = 2500 sq. ft. Making substitutions in the Simpson's formula, it becomes, for the volume, volume = [^0+4(^1 + A^ + A,+ A,)+2(A,+ A^-\-Aq)+A^-]. (3) (8) Summing the moments by Simpson's formula, we have 40 '^vx = (3) (8) + 2(^422:2 + ^4^4 + ^c^<;) + ^s^s]^ CENTER OF GRAVITY 67 where a;o = ft., x^ = 5 ft., x^ = 10 ft., etc. Then - _ '^vx V Both numerator and denominator are computed by Simpson's formula. Compute X by means of the equa- tion, - _ v-jX'^ + Vqx'^ + v^x'^ + ••• ^1 + ^'2 + *'3 + ••• and compare with the previous re- sult. Problem 71. Compute x for the parabolic area of Fig. 50, by using Simpson's Kule, and compare the result with that obtained by integra- tion. Problem 72. By Simpson's ^^^- ^^ Rule find the area and center of gravity of the rail section shown in Fig. 68. Beginning at the top, the horizontal distances are as follows : n,, = 4" wg = 1.24" u^= 1.0" Wii = 4.08" u^ = 1.18" u^ = 1.24 Wjo = 4.24" u, = 1.0" u, = 2.23 u^ = 2.5" u, = 1.0" u^ = 5.5" n, = 6" Problem 73. Find the center of gravity of the deck beam section shown in Fig. 69. Use the equation - _ F^x\ + F^x'^ + F.,x'^ + etc. i^i -}- ^2 + i^3 + etc. ' dividing the area of the section into convenient areas. Check the result thus obtained with that obtained by balancing a stiff paper model over a knife edge. 68 APPLIED MECHANICS FOR ENGINEERS -1.91- 42. Durand's Rule. — A method of find- / N ing the area of irregular areas was pub- ^ '^ lished by Professor Duraiid in the Un- *' gineering News^ Jan. 18, 189-4. The rule states that the total area of an irregular curve equals -^ (0.4 3/q + 1.1 z/^ + ^2 + ^3 + 3/4 + •• • Jr 1.1 y^_,+OAy,) where a, 5, n, and the ^'s have the same meaning as in Simpson's Rule. The number of divisions may be even or odd. The student is advised to make use of this rule as well as Simpson's Rule and compare the Fig. 69 results. 43. Theorems of Pappus and Guldinus. — Let -F be a plane area and OX a line in its plane not cutting the area F (Fig. 70). Let the plane containing the area F be revolved through the angle a about the line OX as an axis. Let dF be an element of the area at a distance y from OX. Then the _ _ element of volume ^^^- "^^ generated by dF is approximately dv = yadF, and the y^ ^^ Y^ dF \ r ,/^ /^ y^ \ \ \^\^ \ \ \ . \ ^ \ 1 y ^ \ \ '- — ^ CENTER OF GRAVITY 69 total volume generated by the area F is V=Ja7/dF = ai ydF. But the distance of the center of gravity of the area F CydF from OJTis given by y =^^—— — • F ■f> ydF = Fy, and V=Fay. This may be stated as a general principle as follows : The volume of any solid of revolution is equal to the area of the generating figure times the distance its center of gravity moves. Problem 74. Find the volume of a sphere, radius r, by the above method, assuming it to be generated by a semicircular area revolving about a diameter. Problem 75. Assuming the volume of the sphere known, find the center of gravity of the generating semicircular area. Problem 76. Find the volume of a right circular cone, assuming that the generating triangle has a base r and altitude h. Problem 77. Assuming the volume of the cone known, find the center of gravity of the generating triangle. Problem 78. The parabolic area of Problem 40 revolves about the ic-axis ; find the volume of the resulting solid. Problem 79. Find the volume of an anchor ring, if the radius of the generating figure is a and the distance of its center from the axis of revolution is r. 70 APPLIED MECHANICS FOR ENGINEERS Fig. 71 Let the curve AB (Fig. 71), of length Z, be the gen- erating curve of a surface of revolution. The area of the surface generated by ds when revolved through the angle a is dF= ayds^ and the area of the whole surface is i<^= a j yds. The center of gravity of this curve AB is given by the expression I yds .'. I yds = ly^ and F = lay. This may be stated as follows : The area of any surface of revolution is equal to the length of the generating curve times the distance its center of gravity moves. Problem 80. Find the surface of a sphere, radius r, assuming the generating line to be a semicircular arc. Problem 81. Find the center of gravity of a quadrant of a circu- lar wire, radius of the circle r. Use results obtained above. Problem 82. Find the surface of the paraboloid in Problem 78. \jyd,. CHAPTER V COUPLES Fig. 72 44. Definitions. — Two numerically equal forces acting in parallel lines in opposite directions form a couple. The distance between the forces is called the arm of the couple. The product of one of the forces and the arm, with the proper sign prefixed, is called the moment of the couple. The sign is plus if the forces tend to produce counter-clockwise rotation and negative if they tend to produce clockwise rotation. If P is one of the forces, d the distance between them, and M the moment of the couple, then M= ± Pd. The unit moment is that due to unit forces at unit distance apart. If P = 1 lb. and c? = 1 ft., the moment is called 1 lb. -ft. 45. Moment of a Couple about ^^^- "^'^ Any Point in its Plane. — Let be any point in the plane of the couple of Fig. 73 and let a be the distance from to one of the forces, P', as shown. 71 72 APPLIED MECHANICS FOR ENGINEERS (The forces are marked P and P' to distinguish one from the other; P = P'.) Then for the point the sum of the moments of the forces P and P^ with respect to the point is P(a + (^)-P'a, or P^; for the position 0-^ the sum of the moments is P(d-a)-\-F'a, ov Fd ; for the position 0^ the sum of the moments is P'a - P(a - d), or Pd. Hence, the sum of the moments of the forces forming a couple with respect to any point in the plane of the couple is equal to the moment of the couple. 46. Combination of Couples in the Same Plane. — Let two couples of moments M^ and M^ act in the same plane. The forces forming the ^ couples may always be combined in pairs, one force from each couple, into two numerically equal, parallel forces opposite in direction, R and R' (Fig. 74). ^^^- ^^ These resultant forces therefore form a couple. (If they fall in the same line, the moment of the couple is zero.) Let MhQ the moment of this resultant couple. If any point in the plane be chosen, then with respect to this point, M= mom R + mom R' . COUPLES 73 But mom R = mom P^ + mom P^ and mom W = mom P-^ H- mom P^ . (Art. 22.) Therefore, adding, itf = (mom Pj + mom P^') + (mom P^ + mom PgO Therefore, ^^o couples in the same plane are equivalent to a single couple in their plane whose moment is equal to the algebraic sum of the moments of the two given couples. It follows at once that any number of couples in the same plane are equivalent to a single couple in that plane whose moment is equal to the algebraic sum of the moments of the given couples. 47. Equivalent Couples, (a) In the same plane. — Let the forces P^ and P^' form a couple whose moment is M^ = P^dy Let any two parallel lines AB and CD dis- tant d^ apart cut the lines of action of P^ and Pj' in A and C respectively. Move the forces P^ a nd P/ to A and and at these points put in two equal and oppo- site forces, S and S', in the line AC, of such value that « the resultant of Pj and S falls in the line AB. The resultants, P^ and P^', of P^ and S and of Pj' and S' re- spectively, are then numerically equal, parallel, and oppo- site in direction. They therefore form a couple. Fig. 75 74 APPLIED MECHANICS FOR ENGINEERS By similar triangles, Fig. 75, Therefore -f*2^2 — P^d^. Hence, a couple may he replaced hy any other couple in its plane provided only that the moment of the second couple is equal to the moment of the first. (5) In parallel planes. — Let P and P' be the forces of a couple in the plane MN. From two points A and A' in their lines of action draw parallel lines to cut the plane Fig. 76 M'JSJ'', parallel to MJST, in the points B and B' . At B and B' put in pairs of equal and opposite forces P and P' parallel to the forces P and P' in the plane MN. The two forces P and P at ^ and B^ may be combined into a single force 2P acting at the middle of the line AB'. Likewise P' and P' acting at A' and B COUPLES 75 may be replaced by the force 2P' acting at the middle of A'B. But ABA' B' is a parallelogram, and the diagonals bisect each other. Therefore the forces 2 P and 2 P' act at the same point, and, since they are equal and opposite, annul each other. There are left then the forces P at ^ and P' at B\ forming a couple in the plane M'N' ^ equal in moment to the original couple. Hence a couple may be moved to any plane parallel to the plane in which it acts. To sum up (joi) and (K) : A couple acting on any body may he replaced hy any other couple of the same moment acting anywhere in the plane of the couple or in any par- allel plane. 48. Moment of a Couple with Respect to Any Line. Defi- nition. — The moment of a couple with respect to any line is the sum of the moments of the forces forming that couple with respect to that line. Let P and P' form a couple in the plane HN whose moment is M= Pd^ and let AB be any line mak- ing the angle a with the normal to UN. Pass a plane RT perpendicular to AB to intersect the plane UN in HV. The couple may be moved in its plane until the forces are parallel to the line B. V Fig. 77 76 APPLIED MECHANICS FOR ENGINEERS (Art, 47) without changing the moment of the couple with respect to AB. For (Fig. 75) with respect to any line, mom P^ = mom P^ -f- mom S and mom P^' — '^^om P^ + mom S' . (Art. 24.) Adding, mom P^ + mom P^ = mom P^ + mom P^'^ since mom S + mom aS'' = 0. The projections of the forces P and P' on the plane ITT are equal to the forces themselves, and the projection of the distance, d, between them is d cos a. The moment of the couple with respect to AB is therefore Pd cos a, or Mcos a (Arts. 23 and 45). 49. Combinations of Couples in Intersecting Planes. — Let the couples of moments Jf^ and M^ (Fig. 78) be replaced in their planes by equivalent couples having a common arm, a, coinciding with the line of intersection of the planes (Art. 47). The forces forming the couples may then be combined into a pair of numerically equal, parallel, and opposite forces B, and B' which form a couple in a plane containing the line of intersection of the two given planes. i^E7 Fig. 78 COUPLES 77 If a is the angle between the planes, measured between Pj and Pg (Fig- '^^)' ^^^ ^is the moment of the resultant couple, then M= aR = a VPi2 + P^s + 2 F^F^ cos « = V(aPi)2 + (aP2)2+ 2 aP^aPg cos a ' = -y/M^ ■\-M^ + 2 M^M^ cos a. The angle, ^, which the plane of the resultant couple makes with the plane of the couple of moment M-^^ meas- ured from P-^ to P, is given by or tan 6 = tan 6 = Pg^sin a Pj + Pg cos a M^ sin a M^ + M^^ cos a Conversely, the couple of moment itf may be regarded as re- solved into the component couples of moments M-^ and MPd. (Art. 46.) Therefore the system of non-concurrent forces in a plane may be reduced to a single force, H = V (^Xy + (2 1^)^, acting at an arbitrarily selected origin, and a single couple in the plane of the forces whose moment is the sum of the moments of the forces with respect to that origin. For equilibrium, lt = and ^Pd = 0, or SX = 0, SI" = 0, and SjPci = ; that is, for equilibrium, the sum of the components of the forces along each of the two axes is zero and the sum of the moments with respect to any point in the plane is zero. Conversely, if IX = 0, ^Y=0, and ^Pd = 0, for a point in the plane, the resultant force M and the resultant couple both vanish and the forces are in equilibrium. 55. Other Conditions for Equilibrium of Forces in One Plane. — The forces may be combined in succession until there is obtained either a final resultant, including a zero resultant, or a couple (Arts. 13 and 25). In either case the sum of the moments of the forces with respect to 84 APPLIED MECHANICS FOR ENGINEERS any point in the plane is equal to the moment of their resultant force or couple (Arts. 22 and 46). If the sum of the moments of the forces with respect to any selected point is zero, the forces cannot be equivalent to a couple, for the moment of a couple is the same for all points of the plane. The forces may, however, have a resultant force. But if in addition the sum of the moments of the forces about two other points of the plane not in the same straight line with the first point is zero, the resultant must be zero ; for either the resultant or the arm of the resultant must be zero, and the arm can- not be zero for three different points not in the same straight line. Hence, if the sum of the moments of a set of forces in one plane with respect to three points not in the same straight line is zero, the forces are in equilibrium. It is left as an exercise for the student to prove that if the sum of the rectangular components in one direction of a set of forces in one plane is zero and the sum of the moments of the forces with respect to each of two points in the plane not in a perpendicular to the given direction is zero., the forces are in equilibrium. Problem 90. Prove the statement of the last paragi'aph. Problem 91. The following forces in one plane act upon a rigid body: a force of 100 lb. whose line of action makes an angle of 45° with the horizontal, and whose distance from an arbitrarily selected origin is 2 ft. ; also a force of 50 lb. whose line of action makes an angle of 120"^ with the horizontal, and whose distance from the origin is 3 ft. ; and a force of 500 lb. whose line of action makes an angle of 300° with the horizontal and whose distance from the origin is 6 ft. Find the resultant force and the resultant couple. NON-CONCURRENT FORCES 85 1 TON Fig. 85 X Problem 92. It is required to find the stress in the members AB, BC, CD, and CE of the bridge truss shown in Fig. 85. Note. The member AB is the member be- tween A and B, the member CD is the mem- ber between C and D, etc. This is a type of Warren bridge truss. All pieces (members) are pin-connected so that only two forces act on each member. The members are, therefore, under simple tension or compression; that is, in each member the forces act along the piece. Solution of Problem. The reactions of the supports are found by considering all the external forces acting on the truss. Taking moments about the left support, we get the reaction at the right support, equal to 4500 lb. Summing the vertical forces or taking moments about the right- hand support, the reaction at the left-hand support is found to be 3500 lb. Cutting the truss along ah and putting in the forces exerted by the left-hand portion, consider the right-hand portion (see Fig. 86). The forces C and T act along the pieces, forming a system of concurring forces. For equilibrium, '^X = d and 2F = 0, giving two equations, sufficient to determine the un- knowns C and T. The forces in the members CD and CE may now be considered known. Cutting the truss along the line cd and putting in the forces exerted by the remaining portion of the truss, we have the portion represented in Fig. 87. This gives a \l 4500 LB Fig. 86 t 2 TONS Fig. 87 86 APPLIED MECHANICS FOR ENGINEERS system of concurring forces of which C and 2 tons are known, so that from the equations '^X =0 and 2F = the remaining forces C^ and T^ may be found. Problem 93. In the crane shown in Fig. 88 (a) find the forces acting on the pins and the tension in the tie AC. The method of cutting cannot be used in this case since the vertical and horizontal members are in flexure. Taking the horizontal member and consid- ering all of the forces acting upon it, we have the .system of non-con- curring forces shown in Fig. 88(6). Three unknowns are involved, Pg, 1\, and Pg, and these may be determined by three equations "^X = 0, ^y = 0, and 2P^ J^v '^"^^ ^^2,3- Extend the line of aS2,3 to cut the line of Pg and proceed as before. When the last force, P^, is reached, there remain unbalanced only the component, iS'g ^ of the first force Pj, and )S^^ 5 of the last force P^. The force Pg that will bal- ance these components will balance the original forces. Hence Pr is de- termined in magnitude, direction, and line of action as the force that will balance these two compo- nents (Fig. 98). The force Pg is called the equilihrant of the forces P, Fig. 98 A- ^5, with vertices on the lines of The polygon A-^A^ the forces, is known as an equilibrium polygon. If the sides of this polygon were replaced by Aveightless rods, or links, hinged at the vertices, the forces P^ ••• P^ would be just balanced by the tensions (or compressions) in the rods, and the framework would retain its position. An indefinite number of equilibrium polygons may be con- structed for a set of balanced forces. A shorter method of constructing the equilibrium poly- gon and determining the force, Pg, to balance the given forces is the following: One triangle from each vertex NON-CONCURRENT FORCES 95 Fig. 99 ^j, '" Ar^ may be taken and their common sides put together, without changing their directions, to form a closed polygon whose sides are the vectors of the forces laid off in succession (Fig. 99). This polygon is called the force polygon. From the force polygon it is evident that the vector of the force Pg, the e(][uilibrant of the given forces P^ ••• P^, is the closing side of the polygon formed by laying off in succession the vectors of the given forces, the begin- ning of each vector being placed at the end of the preceding one. Having laid off the force polygon and thus de- termined the magnitude and direction of the equilibrant, its line of action is determined by drawing the equilibrium polygon ; but this may now be done without constructing the triangles at the vertices -4 J, A^ •". For the lines (called rays) from the point in the force polygon are parallel to the corresponding sides of the equilibrium polygon, and since the directions of /S'j 2 ^11^^ ^5,1 were arbitrary, the point may be any point in the plane. Join to the vertices of the force polygon and then construct the equilibrium polygon by drawing its sides parallel to the corresponding rays of the force polygon ; i.e. a side of the equilibrium polygon terminating on two lines of force must be parallel to the ray of the force polygon drawn to the intersection of those same two forces. To sum up: To determine graphically the equilibrant of a set of forces in one plane, construct a polygon of the 96 APPLIED MECHANICS FOR ENGINEERS vectors of the given forces. The closing side is the vector of the equilibrant. Draw rays from any point in the plane to the vertices of the force polygon. Construct the equilibrium polygon by drawing lines parallel to the corresponding rays of the force polygon, intersecting on the lines of action of the given forces. The intersection of the two sides of the equilibrium polygon parallel to the rays to the closing side of the force polygon is a point on the line of action of the equilibrant. The equi- librant is thus completely determined. In lettering the rays of the force polygon and the sides of the equilibrium polygon it is sufficient to mark them only with the subscripts of the forces that they connect. Figure 100 illustrates this. Fig. 100 When the forces are in equilibrium, as P^, Pg^ ••* ^5 i^ Fig. 100, both the force polygon and the equilibrium polygon are closed. It is possible to have the force poly- gon close without having the equilibrium polygon close ; as. NON-CONCUBBENT FOBCES 97 for example, P^^ ••• ^4 and a force P^ parallel and equal to the closing side of the force polygon but not passing Fig. 101 through the remaining vertex of the equilibrium polygon (Fig. 101). The forces in this case form a couple, for the forces Pp •••P4 are equivalent to a force equal and oppo- site to Pg passing through the vertex A^ of the equilibrium polygon. Problem 108. Determine graphically the equilibrant of the four forces of Fig. 102. Where does it cut the line J.5? Problem 109. Find graphically the reactions due to the loads in Fig. 103. Suggestion. The force polygon here becomes a straight line. Call the re- 1000 LBS. 500 LBS, jAl A h^'- 2000 LBS. 1500 LBS. ^' IJ Fig. 102 actions P^ and P., and agree that P4 shall follow Ps in the force polygon. P. must then close the polygon. The missing ray to the intersection of P^ and P^ is found by drawing from a ray par- 98 APPLIED MECHANICS FOB ENGINEERS allel to the closing side (dotted) of the equilibrium polygon. Hence 7^4 and P- are determined. Check by moments. Fig. 103 When a truss is acted upon by wind loads, the directions of the supporting forces depend upon the manner in which the truss is attached to the supports. Some trusses are pinned at one end and rest on rollers at the other. The reaction at the roller end may then be assumed to be vertical, the pins taking all of the horizontal load. If the truss is fixed at both ends, the reactions may be assumed to be in parallel lines, or else that the supports take equal parts of the horizontal load. 500 LB. Fig. lOi AB in Fig. 104 carries loads as shown. Problem 110. The beam Find graphically the reactions at A and B, given that the reaction at B is vertical. NON-CONCURBENT FORCES 99 Suggestion. Since the point A is the only known point on the reaction through ^, begin the equilibrium polygon at A as a vertex. Problem 111. Find graphically the reactions of the truss of Fig. 105, assuming that the reactions are in I ^ ''■°'^^' parallel lines. Problem 112. Find graphically the reac- tions of the truss of Fig. 105, assuming that the left-hand reaction is vertical. 2 TONS Fig. 105 58. Stresses in Frames. — Stresses in roof and bridge trusses are usually computed on the supposition that the members are two force pieces; i.e. are pin-connected at the ends and have loads applied only at the joints. The stresses in the truss of Problem 92 were computed on that supposition. The graphical method of determining the stresses is often more easily and quickly applied than the analytical method, except for very simple cases. As an illustration of the graphical method, the stresses in the truss of Fig. 106 are determined. It is convenient here to use Bow's notation. Represent a force acting on the truss, or a member of the truss, by a pair of letters, one on each side of the line of the force or member. Thus the left-hand load on the truss is read AB^ the right-hand reaction is CD, the upper horizontal member of the truss is BF, etc. Determine the reactions, either graphically or by moments. They are CD = -^^s DA = ^-. The loads and reactions form a system in equilibrium. These forces are laid off on a vertical line, 100 APPLIED MECHANICS FOR ENGINEERS known as the load line, in succession in the order in which they are met in going around the outside of the truss, as AB, BO, CD, DA, The force AB is represented on the load line by the same letters in small type, ab, etc. At any joint of the truss three or more forces act in the 5 TONS. Fig. 106 directions of the forces and members that meet at that joint. These forces are in equilibrium and must therefore form a closed polygon when laid off in succession (Art. 16). In constructing the polygons the forces already laid off on the load line are made use of. Also a polygon of forces, all of which are known in direction, cannot be constructed if more than two forces are unknown in magnitude. Hence the force polygon is first drawn for a joint where only two forces are unknown. The determi- nation of the forces acting at this point gives additional known forces at other points, for the force in any member acts equally and in opposite directions on the pins at its ends. NON-CONCURRENT FORCES 101 Choosing the joint of the truss at the left support, the forces DA, AE, and ED must form a closed triangle. Hence using the force da on the load line, complete the force triangle dae by drawing through d and a lines parallel respectively to DE and AE to meet in e. The lines ae and ed represent the forces in AE and EB. The directions of the forces acting at the joint are those given by passing around the triangle dae in the order d-a-e, since the force da acts from d towards a. Next go to the joint ABFE and construct the force polygon for that joint, using the known forces ea and ah, drawing lines through e and h parallel respectively to EF and BF to intersect in /. The polygon ahfea is then the force polygon for the joint ABFE. The directions of the forces acting at the joint are given by passing around the polygon in the order a-h-f-e-a, since the force ah acts from a toward h. The direction in which the forces act at the joint are indicated by putting arrows on the members near the joint. Thus the member AE pushes against the pins at its ends, while EB pulls on the pins at its ends. AE is in compression, EB in tension. The whole stress diagram (Fig. 106 (5)) is thus con- structed upon the load line. The stress in any member is represented by the line in the force diagram terminating in the same letters as those of the member, as^ represents the force in FGr. A check on the correctness of the work is that the last line of the force diagram must be parallel to the cor- responding member of the truss. More accurate results are obtained by drawing the figure to large scales. 102 APPLIED MECHANICS FOR ENGINEERS Problem 113. In the truss of Fig. 107 the apex is distant ^ the length of the span above the lower chord. The panels carry equal loads, half the load of each panel being transferred to the truss at the Fig. 107 pin at the end of that panel. Calling the total load unity, find graphi- cally the stresses in the members of the truss. Write the values of the stresses on the figure of the truss and indicate whether tension or compression. Problem 114. A wind load acts on the truss of Fig. 107, as shown in Fig. 108. Assuming the reactions to be in parallel lines, find the reactions and stresses in the members graphically, and write down the stresses as in the preceding problem. 'A Fig. 108 Suggestion. In finding the reactions the work will be simplified by combining the loads into one load. NON-CONCUEEENT FOECES 103 Problem 115. Assuming the total wind load to be equal to I of the total vertical load in the preceding problems find the stresses when both loads act simultaneously. Problem 116. Find the stresses in the truss of Fig. 109, all members of the truss being of the same length. 2 TONS. 2 TONS. Problem 117. Find the stresses in the members of the cantilever truss of Fig. 110. Problem 118. Find the stresses in the truss of Fig. 111. 59. Method of Substi- tution. — It sometimes happens that in con- FiG. 110 structiog the force diagram a point is reached where it 1 TON 1 TON. 3 TONS. 3 TONS 3 TONS. Fig. Ill 104 APPLIED MECHANICS FOR ENGINEERS is not possible to proceed to another joint at which there are only two unknown forces acting. The temporary removal of two members and the substitution of another member to hold the truss from collapsing will sometimes en- able one to continue the construction of the force diagram. This is illustrated in the case of the Fink truss shown in Fig. 112. Starting at the left support, the force polygons fag, alhg, and fghk can be con- structed. The poly- gons for the remain- ing joints in the left half of the truss then have three sides miss- ing and cannot be completed in the usual way. If the members ML and MN be removed, and a member M N (Fig. 112(5)) be substituted, the truss would stand. Moreover, the stresses in Z>iV, NE^ and EF would not be changed ; for if sections were made across these members in the two cases, the parts of the truss to the left of the cuts would be acted upon by the same forces, and hence would require the same forces in BN, NE^ and NON-CONCUBBENT FOBCES 105 UF to balance them. (The same is true for sections across BIT, HK^ and KF.) With the changed form of the truss the force polygons hhhcmJ and m' cdn can be con- structed. Having located w, we may return to the origi- nal truss and construct the polygon nmcd^ then hcmlkh^ etc. Problem 119. The panels of a Fink truss are equal and carry equal loads. The upper chord is inclined 30° to the horizontal, and the lower chord is horizontal. Calling the total load unity, construct the stress diagram and write the stresses in the members on the figure of the truss, indicating tension and compression. CHAPTER VII MOMENT OF INERTIA 60. Definition of Moment of Inertia. — The study of many problems in mechanics brings to our attention the value of the integral of the form | y'^dF^ where dF repre- sents an infinitesimal area and y is the distance of that element from an axis of reference. The value of this integral taken over a given area is called the moment of inertia of that area with respect to the given axis of ref- erence. In like manner j r^dV and j r^dM taken to in- clude all elements of a given volume or a given mass are called respectively the moment of inertia of the volume, and mass, with respect to the given axis. In these inte- grals dV and dM represent infinitesimals of volume and mass respectively, and r the distance of the infinitesimal element from the axis of reference. We shall designate moment of inertia by the letter /. Thus we write : I^jyHF, I=fr^dV, for area, mass, and volume, respectively. 106 MOMENT OF INERTIA 107 Many problems that confront the engineer involve in their solution the consideration of the moment of inertia. This is the case when the energy of a rotating flywheel, for example, is being determined. The energy of a rotat- ing body (Art. 137) is expressed as follows : Kinetic energy = -^, LA where / is the moment of inertia with respect to the axis of rotation and to is the angular velocity. (See Art. 118.) It is seen that the energy of rotating bodies, having the same angular velocity, or the same speed, is directly pro- portional to their moments of inertia. The quantity, therefore, plays a very important part in the considera- tion of rotating bodies. Also the strength of a beam or column depends upon the moment of inertia of the area of the cross section of the beam or column with respect to a line of the section through the center of gravity of the section. In a later chapter a reason will be seen for the name " moment of inertia." For the present it may be regarded as a name arbitrarily applied to a quantity frequently used in the applications of mechanics. (See Art. 137.) 61. Radius of Gyration. — The moment of inertia of an area involves area times the square of a distance. We may write 1= \ y'^dF = Fk"^^ where F is the area and k is a distance, at which, if the area were all situated, the moment of inertia would be unchanged. This distance k is called the radius of gyration. In a similar way 108 APPLIED MECHANICS FOR ENGINEERS for a mass we write : 1= j yMM== Mk^, and for volume 1= Cy2dV=^Vk\ 62. Units of Moment of Inertia. — The moment of inertia of an area with respect to any axis may be expressed as Fk'^. The area involves square inches, and k"^ is a dis- tance squared. The product is expressed as inches to the fourth power. The moment of inertia of a volume Vk^ requires inches to the fifth power. The moment of in- ertia of a mass requires in addition to Vk"^ the factor ^, g so that pounds and feet per second per second are involved. This is somewhat more complicated since it involves units of weight, distance, and time. The presence of ^ (= 32.2) in the expression requires that all distances be in feet. It is customary to express the moment of inertia of a mass without designating the units used, it being understood that feet, pounds, and seconds were used. 63. Representation of Moment of Inertia. — From the defi- nition of moment of inertia it is evident that an area has a different moment of inertia for every line in its plane. We shall designate the moment of inertia with respect to a line through the center of gravity by Ig with a subscript to indicate the particular gravity line intended. For example, Ig^ indicates the moment of inertia with respect to a gravity axis parallel to x, and Igy indicates tlie moment of inertia with respect to a gravity axis parallel to y. The moment of inertia with respect to a line other than a gravity line will be designated by /, the proper MOMENT OF INERTIA 109 subscript indicating the particular line. Similar sub- scripts will be used to designate the corresponding radii of gyration. It should be noted that moment of inertia is not a quantity involving direction. It has to do only with magnitude and is essentially positive. 64. Moment of Inertia. Parallel Axes. (a) For Areas. In Fig. 113 (a), let OX and 0' X' be two parallel axes in the plane of the area F. If y is the ordinate of an element of area dF referred to 0' X and h the distance between the axes, then I^ = ^(y + },yclF=j^{y'^ ■Y'lhy^ W)dF = CyHF + 2 Ji^ydF + h^^dF = r,-{-2.hyF-^h'^F, where y is the ordinate of the center of gravity of the area Preferred to the axis 0' X' (Art. 36). If 0' X passes through the center of gravity of #, then y is zero and the equation may be written 110 APPLIED MECHANICS FOR ENGINEERS where Ig is the moment of inertia of F with respect to a gravity axis parallel to OX. This relation may be expressed in words as, The moment of inertia of an area with respect to any line in its plane is equal to its moment of inertia with respect to a parallel gravity axis plus the area times the square of the distance between the two axes. (b) For Solids. In Fig. 113 (5), OX and O'X' are parallel axes distant h apart. dM is the mass of an ele- ment of the body distant r from OX and r' from O'X' . If the coordinates of dMsive (^, s), as shown in the figure, then = h^-\-2hy-h r''\ Therefore I, = Cr^dM=f(h^ + 2 % + r'^)dM = h^CdM-{- 2 hCydM+fr'^dM = h^M+2hyM+r,. If 0^ X' passes through the center of gravity of the solid, y = and the equation reduces to From the above formulae it is evident that the moment of inertia of an area or a solid about a gravity axis is less than that for any parallel axis. 65. Moment of Inertia ; Inclined Axis. — It is often desir- able, when I^ and ly are known, to find the moment of inertia with respect to an axis w making an angle a with x. (See MOMENT OF INERTIA 111 Fig. 114.) Here, 7, = Cv^dF and I, = Cw^dF. lu terms of x^ y, and a, Iy,= j (7/ cos a — X sin a)'^dF = I y2 cos^ a c?7' —^\xy cos a sin a c?^^ + \ ^ sin^ a c?JP = cos^ a I ?/2<^jP — 2 sin a cos a I xydF + sin^ « | aj^c?^ = I^ cos^ a — sin 2 « j xydF + i^ sin^ a. In a similar way ly = /j. sin^ a + 2 sin a cos a I xydF + /^ cos^ a, where OF is perpendicular to OW. These are the required formulse for obtaining the mo- ment of inertia with respect to inclined axes. It follows that ^w ~(" -'-V — J-x + -*»/• 112 APPLIED MECHANICS FOR ENGINEERS That is, the sum of the moments of inertia of an area with respect to two rectangular axes in its plane is the same as the sum of the moments of inertia with respect to any other two rectangular axes in the same plane and passing through the same point. This states that the sum of the moments of inertia for any two rectangular axes through a point is constant. It will be seen in Art. 68 that this constant is the polar moment of inertia. 66. Product of Inertia, — The integral i xydF is called a product of inertia^ for want of a better name. In case the area has an axis of symmetry, either the x- or ?/-axis may be taken along such an axis. The product of inertia then becomes zero, since if x is the axis of symmetry, for every + y there is a corresponding — y. A similar rea- soning shows that the product of inertia is zero when y is the axis of symmetry. In such cases and I^ = I^ sin- ^ + ly cos^ a. When I xydF is not equal to zero, it is necessary to select the proper limits of integration and sum the in- tegral over the area in question. This is illustrated in Art. 76. 67. Axes of Greatest and Least Moment of Inertia. — It is often important to know for what axis through the center of gravity the moment of inertia is least or greatest ; that is, what value of a makes i^ or I^ a maximum or a mini- mum. For any area J^, ly, and j xydF are constant after MOMENT OF INERTIA 113 the X- and ?/-axes have been selected. Using the method of the calculus for finding maxima and minima, we have, putting / flT xydF= ^, ^^^ = (7^ - 7^) sin 2 a - 2 z cos 2 a. da Equating the right-hand side to zero, the value of a that gives either a maximum or minimum is seen to be given by the equation tan 2 a = — — , J-y -/.J. or, what is the same thing, sin2a= and cos 2 a= ^ It is seen upon substituting these values of sin 2 a and cos 2 a in do? = 2(Iy — 7^)cos 2 a + 4 z sin 2 a that the positive sign before the radical indicates a mini- mum and the negative sign a maximum value for i^. The equation for tan 2 a is satisfied by two values of 2 a differing by 180°. These values of 2 a correspond to the two signs in the values for sin 2 a and cos 2 a. The values of a corresponding to the two signs therefore differ by 90°. Hence, through any point of the plane, there are two axes at right angles to each other about one of which the moment of inertia is a maximum and about the other a minimum. These axes are known as the principal axes of the area through that point. The axes through the center of 114 APPLIED MECHANICS FOR ENGINEERS gravity of the area for which the moments of inertia was greatest and least are known as the Principal Axes of the Area. This subject will be further discussed in Art. 76. It is seen from the above that when j xydF= 0, the values of a which give maximum or minimum values for I^ and ly are 0° and 90°. This means that the x- and ?/-axes, themselves, are the principal axes. Conversely, if the x- and ?/-axes are principal axes, then j xydF = 0. If either the X- or ^-axis is an axis of symmetry, 1 xydF = and hence the x- and ?/-axes are principal axes. Problem 120. Derive the formula for 7„ written in Art. 65. Problem 121. Prove that when /^ is a maximum, /„ is a minimum ; and that when 7^ is a minimum, /^ is a maximum, by using 7^ 4- /y = constant. 68. Polar Moment of In- ertia. — The moment of in- ertia of an area with re- spect to aline perpendicular to its plane is called the polar moment of inertia of the area. Consider the area represented in Fig. 115 and let the axis be perpendicular to the area at any point 0. Let dF represent an infinitesimal area and let r be its distance from the axis. Representing the polar moment of inertia Fig. 115 MOMENT OF INERTIA 115 by ip, we have I^^^rUF; but r^^x^ + ^2^ so that I^ = jxHF+^y^dF, or p ~ ^y "^ ^oc' That is, the polar momerit of inertia of an area is equal to the sum of the moments of inertia of any two rectangular axes through the same point. It has already been shown that 1^+ Iy = constant (see Art. 65) for any point of an area. 69. Moment of Inertia of Rectangle. — Let it be required to find the moment of inertia of the rectangle shown in Fig. 116 (a), with respect to the axis x. We may write I. = fg^dF. Since dF = bdg, this becomes , 1 * 4, ■rn^/z d'l , 1 1 1 (a) -X Fig. 116 116 APPLIED MECHANICS FOR ENGINEERS To find the moment of inertia witli respect to a grav- ity axis parallel to x we may make use of the formula Ig^ = I^ — Fd'^^ from which we have Igx = ^bh^ and fc2^^ = — The same result may of course be obtained by taking the axes through the center and integrating between the limits 7 7 and - • From comparison we may write the moment of inertia with respect to a gravity line perpendicular to x^ and the polar moment of inertia for the center of gravity 70. Moment of Inertia of a Triangle. — It is required to find the moment of inertia of the triangle shown in Fig. 116 (^) with respect to the axis x^ coinciding with the base of the triangle. We have /^ = I if'dF^ where dF = xdy^ But a; = - (/i — z/), from similar triangles, giving h h h MOMENT OF INERTIA 117 The moment of inertia with respect to the horizontal gravity axis may now be determined. Ig^ = I^- Fd? = gg-, and k% = ^^ • The same results may of course be obtained by direct integration. Problem 122. Find the moment of inertia of tlie area of a triangle with respect to an axis through the vertex parallel to the base. Problem 123. Find the polar moment of inertia of the area of a right triangle for the center of gravity. 71. Moment of Inertia of a Circular Area. — The moment of inertia of a circular area Avith respect to a horizontal gravity axis x^ as shown in Fig. 117, may be found as fol- Fia. 117 lows : /y^ = I if-dF. Here it is simpler to use polar coordi- nates. Changing to polar coordinates, remembering that 118 APPLIED MECHANICS FOR ENGINEERS y = p sin 6, and dF = dp{pdO)^ the integral becomes /_= r^ Cp'^sin'^epdedp Jo Jq ^gx r4 r*2^ = -j 1(1 -COS 2 e)dO r'rO 1 . ^.T' The corresponding radius of gyration is k^^ = ox 2 On account of the symmetry of the figure this is the moment of inertia for any line in the plane through the center of gravity. It follows that and that I. = 7rr* 9 K = 9 * 72. Moment of Inertia of Elliptical Area. — Let it be required to find Ig^. and Igy of the elliptical area shown in Fig. 118. The equation of the bounding curve is and Igy :=fx^dF =Cx^ 2 7/dx. From the equation of the bounding curve a so that /„„ = — I x^ Va^ — x"^ dx a ^0 MOMENT OF INERTIA 119 4^ 6a% ■-(2 x^ — a^)Va^ — x^.-\ sin ^ - 8 8 a , and therefore, kgy = ^ Fig. 118 In a similar way I ox = f,fdF = j f 2 xd7/ ^0 ^^^, and therefore kgj.= a Since Ip = Ig^. -\- Igy, the polar moment of inertia with respect to the center is ab IT (a2 + h'^)^ and k^ = l^a^ + b\ It is seen that when a = b = r^ the equations obtained for the elliptical area are the same as those obtained for the circular area, just as they should be. 120 APPLIED MECHANICS FOR ENGINEERS 73. Moment of Inertia of Angle Section. — When an area may be divided up into a number of triangles, or rec- tangles, or other simple divisions, the moment of inertia of the whole area with respect to any axis is often most easily found by taking the sum of the moments of inertia of the individual parts. Tliis method is often made use of in determininsr the moment of inertia of such areas as the o section of the angle iron, shown in Fig. 119. ffy II y' < — 1 1 y - \ ■ \ J< > y 1 1 \ y 1 y 1 ' ^ =" i ^^ *^ y- ^ 1 1 1 1 .9^ 9^ Fig. 119 We shall now determine the moment of inertia of this section with respect to the horizontal and vertical gravity axes, Igj. and Igy^ and also with respect to an axis v (see Art. 76), making an angle a with the axis x. Consider MOMENT OF INERTIA 121 the section divided into two rectangles, one 5" xf ", which we may call F^, and the other 3|" x |", which we may call jFg. The moment of inertia of the section, with respect to X, is equal to the moment of inertia of F^ with respect to x plus the moment of inertia of F^ with respect to a;, so that ^,. = tV(5)(I)' + 5(f) (-808)2 + _,_(.2^I)3 I + .2^.(5) (1.19)2 = 7.14 in. to the 4th power. Similarly I.. = tV(¥)(I)' + ¥(f)(l-31)^ + tV(|)(5)' + 6(|)(.88)2 = 12.61 in. to the 4th power. Note. The problem of finding the moment of inertia of angle sections, channel sections, Z-bar sections, and the built-up sections shown in Figs. 121-125, is of special interest and importance to engineers, occurring as it does in the computation of the strength of all beams and columns made up of these shapes. Problem 124. Find the moment of inertia of the Z-bar section shown in Fig. 120 for the grav- ity axes gj. and gy. Hint. Divide the area into three rectangles. Problem 125. Com- pute the moment of inertia for the channel section, shown in Fig. 40, 1 Fig. 120 Problem 31, for the horizontal and vertical gravity axes. Problem 126. Required the moment of inertia of the T-section (Fig. 41, Problem 32), also the moment of inertia of the U-section (Fig. 42, Problem 33) with respect to both horizontal and vertical gravity axes. Problem 127. The section shown in Fig. 121 consists of a web section and 1 angles, as shown. Find the moment of inertia of the -.gx 122 APPLIED MECHANICS FOR ENGINEERS whole section with respect to the horizontal gravity axis. Given, the moment of inertia of an angle section with respect to its own gravity axis, g', is 28.15 in. to the 4th power. Problem 128. Consider the section given in Problem 127 to be so taken that it includes two |-inch rivet holes, as indicated by the ^ m m v" 6"V- F='— 9' ^ ^ F^ J.- 1.78 y-^. h6 d '—g' -14 Fig. 122 Fig. 121 position of the rivets in Fig. 121. Compute the moment of inertia of the M^hole section, when the moment of inertia of the rivet holes is deducted. The distance from the center of the rivet hole to the out- side of the angle sec- '^ * T 1.82 ± t 3 ^-^^ — . I I tion may be taken as ~-:~tTi M ^' -^ i"- Compare the result with that obtained in the succeeding problem. ■, 1.78" ^x :rzTr__., gx -1-28- 1 Fig. 125 124 APPLIED MECHANICS FOR ENGINEERS The column is built up of one central plate, two outside plates, and four Z-l)ars, The legs of the Z-bars are equal, and have a length of 3^ in. The moments of inertia of each Z-bar section with respect to its own horizontal and vertical gravity axis are 42.12 and 15.44 in. to the 4th power, respectively. Problem 132. Find the moment of inertia of the section shown in Fig. 12.5, with respect to the horizontal and vertical gravity axes gj. and g,j. This section is made up of plates and angles. The moment of inertia of each angle section with respect to both its own horizontal and vertical gravity axes is 28.15 in. to the 4th power. 74. Moment of Inertia by Graphical Method. — It will often be necessary to find the moment of inertia of a plane section whose bounding curve is of a complicated form, as Fig. 126 in the case when it is necessary to compute the strength of rails or deck beams. The graphical method given below may be used for such cases. Let F be the area bounded by the outer curve of Fig. MOMENT OF INERTIA 125 126 and let OX he a line with respect to which it is desired to find the moment of inertia of F. Draw a line MJV parallel to OX at a convenient dis- tance, a, from OX. Let AB be any chord of the bound- ary curve of F parallel to OX. Project AB on MN, obtaining CD. From C and D draw lines to any point P on OX, cutting AB in A^ and B^ Project A^B^ on MJV, obtaining C^Dj. Join (7j and Z>j to P by lines cutting AB in A^ and B^. If this be done for all positions of AB, a new curve is obtained, as is shown in Fig. 126. Let F" be the area bounded by this curve. Letting AB = x, A^B^ = x', and ^2^2 = ^'^ ^^ ^^ seen from similar triangles that X a ^ x' a - = - and — =-, X y x y P 1*1 X Ct d irom which — - = — or x = — x x" y^ y Then, if J= moment of inertia of F with respect to OX, I=.^yixdy =j-a:^x"dy = a?^dF'\ or I=a;^F". The area F" may be measured by a planimeter or by one of the methods previously explained (Arts. 40-42). If it is desired to measure the area of F" by Simpson's Rule, the area F may first be divided into an even number of strips of the same width by horizontal lines and on the lines thus drawn the points of the auxiliary curve deter- mined by the method described above. The area F" will then be ready for the application of Simpson's Rule. 126 APPLIED MECHANICS FOR ENGINEERS If, in laying off the area F^ 1 inch parallel to (9X repre- sents m inches and 1 inch perpendicular to OX represents n inches, then, measuring a in inches and F" in square inches, the value Jin inches^ is given by It is left for the student to show that the distance, ^, of the center of gravity of F from OX is given by _ aF' where F^ is the area of the curve traced out by A^ and B^ in Fig. 126. Problem 133. By the above method find the moment of inertia of the area of a circle about a diameter. Determine the area of F" by the use of Simpson's Rule. Problem 134. By the above method find the moment of inertia of a rectangle about one side. Take the point P at one of the vertices of the rectangle and show that the boundary curve of F" is a parabola. Problem 135. Find the moment of inertia of the area of the rail section of Fig. 68 with respect to the base line. Using the values of the area and height of the center of gravity found for the figure in Problem 72, find the moment of inertia of the area with respect to a horizontal gravity axis. 75. Moment of Inertia of an Area, (a) by Direct Addition, and (b) by Use of Simpson's Rule. (a) Divide the area into n narrow strips of equal width Ax parallel to tlie line OY with respect to which the moment of inertia is desired (Fig. 127). If the middle lengths of these strips are respectively ^1,^2' '** ^n» MOMENT OF INERTIA and their distances from OY respectively Xy x^, the moment of inertia of the y area is given approximately by the formula, 1= x\A^ + xlA^ + ••• + xlAr,, h — a 127 or, since A.r = , and. n Fig. 127 approximately, A^ = y\^^^ A^ = y^/\x^ •" A^ = y^t^x, the approximate formula, be- comes h — a J = n [o-fi/j+o:-^?/, + ••• +a^;,y,J- Y A more accurate result is usually obtained by the use of Simpson's Rule. (5) Divide the area into an even number of strips of width Aa: parallel to the line with respect to which the moment of inertia is V,, sought (Fig. 128). Since / = I x^ydx^ where y is the variable length of >| the strips, it may be eval- uated by Simpson's Rule as in Art. 41, the quantity .'/o .'/i Ik % -Zi Fig. 128 X hj taking the place of «/ in I ydx. Hence, approximately. jr = h — a 3n 128 APPLIED MECHANICS FOR ENGINEERS Problem 136. By the direct addition formula compute the mo- ment of inertia of a rectangle 4 in. by 2 in. with respect to a 2 in. side, using (1) 5 divisions, (2) 10 divisions. Compare with the exact value. Problem 137. Solve the preceding problem by the use of Simp- son's Rule, using (1) 6 divisions, (2) 10 divisions. Problem 138. Find the moment of inertia of the area of the sec- tion of Fig. 68 with respect to the base by the method of direct addition. Problem 139. Solve the preceding problem by the use of Simp- son's Rule. 76. Least Moment of Inertia of Area. — In considering the strength of columns and struts it is necessary to know the axis about which the moment of inertia of a cross sec- tion is a minimum, since bending will take place about this axis. It was shown in Art. 65 that, if the moment of inertia of the area with respect to two rectangular axes in its plane fs known, the moment of inertia with respect to any other axis, making an angle a with one of these, could be found. It was further developed (Art. 67) that the value of a that would render the moment of inertia a min- imum was given by the equation tan 2 a = 2 CxydF I. - 1. ' In case either of the axes x or ?/ is an axis of symmetry, the value of a given by this criterion is zero, so that, for areas having an axis of symmetry, the axis of least moment of inertia is the axis of symmetry or the one perpendicular to it. As an illustration of the problem in general let it be required to find the least moment of inertia of the angle MOMENT OF INERTIA 129 section shown in Fig. 119 with respect to any axis in the plane of the area through the center of gravity. Let v be the gravity axis making an angle a with the .r-axis. The problem then is to find such a value of a that Ig^ will be a minimum. From Art. 65 we have Z gv = ly^ cos^ a — sin 2 a I xydxdy + Igy sin^ a. In Art. 73 it was found that I^^ ■= 7.14 and 7^^= 12.61. We proceed now to find the value of j xydxdy for the angle section. For this purpose, suppose the section com- posed of two rectangles, F^ (5 in. x f in.), and F^ (^^ in. X f in.), and then find the value of the integral for the two rectangles separately. Considering first the area F^^ and using the double integration, we get I I xydxdy = I xdx\ ^ — -^^ ^ Jl.62 J_.4'/5 ^ c/i.62 L 2 2 _ L 2 2 J In a similar way for F2, we have T' r-% 7 r-!5 r(2.88)2 (-.495)2- I I xydxdy = I xdxl ^ ^^ ^ .495 »^.995 L 2 2 = 4 .025[ai6 62)2 (,995) ^ = 3.29. Therefore, | xydxdy for the whole area of the angle sec- tion is 5.51 in. to the 4th power. From this we find tan 2a = 11^ =2.02. 5.47 130 APPLIED MECHANICS FOR ENGINEERS Therefore 2 a = 63° 40', a=31°50'. The expression for Ig^ now becomes Ig, = 7.14 cos2 (31° 50')- 5.51 sin (63° 40') + 12.61 sin2 (31° 50')= 3.72 in. to the 4th power. This gives the least radius of gyration, kg^ = .84 in. Problem 140. Find the least moment of inei-tia /^,, and least radius of gyration kg.,, of the Z-section shown in Fig. 120. In this case Ig^ = 15.44 in. to the 4th power and Igy = 42.12 in. to tlie 4th power. Anfi. Least /g-. = .5.61 in.- to 4th power and least kg^ = .80 in. Problem 141. An angle iron has equal legs. The section, similar to that in Fig. 119, is 8 in. x 8 in. with a thickness of I in. Find ■igxj ■'■gyi isasi -ioD? anci least iCg^. Ans. Igx = Igy = 48.58 in. to the 4th power, Ig„ = 19.55 in. to the 4th power. kgy = 1.59 in. Problem 142. Find the moment of inertia of the column section, shown in Fig, 124, with respect to an axis v making an angle of 30° with gx. What value of a makes /^y mininmin in this case? 77. The Ellipse of Inertia. — It is interesting to note, at this point, the relations between the moments of inertia with respect to all the lines, in the plane of the area, pass- ing through a point. We have seen that for every point in an area there is always a pair of rectangular axes for one of which the moment of inertia is a maximum and for the other a minimum; that is, there is alwa3^s a pair of prin- cipal axes. The criterion for such axes was found to be 2^ tan 2 a = — zr, -£.,, "^ -'■•r MOMENT OF INERTIA 131 which means, since the tangent of an angle may have any value from to infinity, positive and negative, that for every point there is always a pair of axes such that { = 0, i.e. j xydF= 0. This means that the expression for Z„ may always be reduced to the form I^ = I J. cos^ a-{- ly sin^ a (Art. Q^) by properly selecting the axes of reference, where now Ij. and ly represent the principal moments of inertia. If we divide through by i^, the equation becomes ^2 _ ^2 cos^ ct -\-kl sin^ «. k^ k^ ky ,1 1 cos^a , sin^a or -J _ p2 cos^ at. .p^ sin^ a which is the equation of an ellipse referred to the princi- pal axes of inertia as axes, the coordinates of any point on the ellipse being x = p cos a, y = /o sin a. Hence if OX and OY are respectively the axes of maximum and minimum moments of inertia of an area for a given point 0, there exists an ellipse with minor and major axes on (9X and OF respectively such that the dis- tance from to the ellipse along any line is the reciprocal of the radius of gyration of the area with respect to that 132 APPLIED MECHANICS FOR ENGINEERS Fig. 129 line. This ellipse is called the ellipse of inertia of the area for the given point (Fig. 129). The ellipse of inertia furnishes a graphical method of finding the moment of inertia of an area for any axis through a point when the principal moments of inertia of the area for that point are known. Problem 143. Sketch the ellipse of inertia of the area of a rectangle for (a) the intersection of the diagonals, (6) the middle point of one side of the rectangle. Problem 144. Determine the princi- pal moments of inertia of the area of a rectangle 6 inches by 10 inches for axes passing through a vertex of the rectangle. Sketch the ellipse of inertia. Problem 145. Determine the principal moments of inertia of the area of a circle for a point on its circumference, and sketch the ellipse of inertia of the area for that point. 78. Moment of Inertia of a Thin Plate, (a) with Respect to an Axis in the Plate Parallel to its Faces, (b) with Respect to an Axis Perpendicular to its Faces. — (a) Suppose the plate to be of constant small thickness t and unit weight 7 and let x be the distance of an element of mass c^iltf from the axis, where dM=^ ^tdF (Fig. 130). Fig. 130 MOMENT OF INERTIA 133 Then approximately I=CxHM= ^-tSxHF, But j Tp'dF is the moment of inertia of the area of the face of the plate about the given axis. Therefore, ap- proximately, the moment of inertia of a thin plate with reference to an axis of the plate parallel to its faces equals — times the mometit of inertia of the area of its face with 9 reference to the same axis. (5) Let r be the distance of dM hom. the axis perpen- dicular to the face of the plate (Fig. 130). Then the moment of inertia of the plate with respect to the axis is I=CrHM= ^CrHF. Or, approximately, the moment of inertia of a thin plate with reference to an axis perpendicular to the faces of the ertia of the face of the plate with refer- ence to that axis. 79. Moment of Inertia of Solid of Rev- olution. — Consider the moment of in- ertia of a solid of revolution with respect to its axis of revolution. Imagine the solid cut into thin slices, all of the same thickness, by parallel planes perpen- dicular to the axis of revolution (Fig. 131). Each slice is a circular disk of thickness dy and radius x., and its Fig. 131 134 APPLIED MECHANICS FOR ENGINEERS moment of inertia with respect to the axis of revolution is -dyiraP' - — (Art. 71). The moment of inertia of the 9 2 solid of revolution is the sum of the moments of the small slices, so that the limits of integration and the relation between x and y depending upon the particular solid considered. Problem 146. Prove that the moment of inertia of a right cir- cular cylinder of radius r and altitude h with respect to its axis is Problem 147. Prove that the moment of inertia of a right cir- cular cone with respect to its axis is Ig^ = -^^ Mr^. Problem 148. Prove that the moment of inertia of a sphere with respect to a diameter is | A/r^. Problem 149. The surface of a spheroid is generated by re- volving the ellipse — + ^ = 1 about the x-axis. Prove that the mo- ment of inertia of the solid inclosed with respect to the axis of revolution is /^^ = | Mb"^. Problem 150. Prove that the moment of inertia of a rectangular parallelopiped with edges a, b, and c, with respect to a gravity line parallel to the edge c is Ig^ = ^— (a^ + b^). Problem 151, Prove that the moment of inertia of a slender rod of length a with respect to an axis perpendicular to the rod and passing (a) through the center, (h) through the end, is (a) 12 iVa2 MOMENT OF INERTIA 135 Problem 152. An anchor ring is generated by revolving a circle of radius a about a line in its plane distant b from the center of the circle. Show that the moment of inertia of the mass of the anchor ring with respect to the axis of revolution is M{b- + | a-). 80. Moment of Inertia of a Body by Parallel Sections. — By dividing a body up into thin plates by parallel planes, parallel to the axis with respect to which the moment of inertia of the body is sought, the mo- ment of inertia is made to de- pend upon the moments of inertia of the areas of the sec- tions and their distances from the given axis. As an illustration consider the problem of finding the mo- ment of inertia of a right circu- lar cone with respect to an axis through the vertex perpen- dicular to the axis of the cone (Fig. 132). The moment of inertia of the thin plate with respect to the diameter of its lower face is approximately 71). The moment of inertia of this plate with respect to OX is therefore , . « ^^-^-:- + - dy .y\ a 4: a y 9 or — f^ + ^yV?/, and the X for the cone is then 9 fir Jo V 4 -^x^yAdy. 136 APPLIED MECHANICS FOR ENGINEERS From the figure x may be evaluated in terms of 1/ and the integral evaluated. Problem 153. Prove that the moment of inertia of a right cir- cular cone with respect to an axis through the vertex perpendicular to the axis of the cone is ^V M(r'^ + 4 A^). Problem 154. Using the result of Problem 153, find the moment of inertia of the cone about a gravity axis parallel to the base, and then about a diameter of the base, /o, = — Af ( r^ _) — ] . Problem 155. Prove that the moment of inertia of a right circu- lar cylinder with respect to a diameter of the base is Mi — f- — ) • Problem 156. Prove that the moment of inertia of an elliptical right cylinder of height h and semi-axes of base a and b with respect to the diameter 2 a of the base is All H — ), and with respect to a M gravity line parallel to the diameter 2 a is — (3 6^ + h^). Problem 157, Show that the moment of inertia of a right circu- lar cylinder, altitude h, and radius of base r, with respect to a gravity axis parallel to the base is Igx = ^^^(-7 H j, and find the moment of inertia with respect to an axis parallel to this and at a distance d from the base. Problem 158. It is required to find the moment of inertia of the cast-iron disk flywheel shown in Fig. 133 with respect to its geomet- rical axis. <^ ^ # ^ ^ ^ {^^^^^^^ Fia. 133 MOMENT OF INERTIA 137 Hint. The wheel may be regarded as made up of three hollow cylinders, the moment of inertia of the whole wheel being equal to the sum of the moments of inertia of the three parts. The dimensions are as follows: diameter of wheel 2 ft., width of rim and hub 4 in., thickness of rim and web 2 in., thickness of hub 1^ in., and diam- eter of shaft 2 in. All distances must be in feet. Problem 159. Find the moment of inertia of the castriron fly- wheel shown in Fig. 134 with respect to its axis of rotation. There are six elliptical spokes, and these may be regarded as of the same cross section throughout their entire length. 1 ■'::■■///:■ -.^--10 • > 15 Fig. 134 Problem 160. Find the mass and moment of inertia, with respect to the axis, of a right circular cylinder whose density varies directly as the distance from the axis and is y\ at the outside of the cylinder. '^9 5 81. Moment of Inertia of a Mass ; Inclined Axis. — We shall now study tlie problem of tiiidiiig the moment of inertia of a solid with respect to an axis inclined to the 138 APPLIED MECHANICS FOR ENGINEERS coordinate axes. Suppose the moments of inertia of the body with respect to the three coordinate axes known from the expressions : and let it be required to find the moment of inertia of the body with respect to any other axis V making angles z dM Fig. 135 a, /3, 7 with the coordinate axes. (See Fig. 135.) Let c?i^f equal the mass of an infinitesimal portion of the body and d its distance from the axis V. Since r^ = x^ + y"^ + z\ OA = x cos « + ?/ cos jB -{- z cos 7 and d!^=zr^— OA^ = (aP' + y'^ -\- z^) — {x cos « + ^ cos ^ -\-z cos 7)^ MOMENT OF INERTIA 139 we may write I^ = CdhlM = fli-^''^ + / + 2^) - i-^ cos rt 4- // COS yS + 2 COS yy]d3L This reduces, since cos^ « -f cos^ fi H- cos^ 7 = 1, to /„ = C(f -f- ^2) cos2 adM+fCz'^ + x2 ) cos2 /3(^il[f 4- ) (a:2 + ?/2) cos2 ydM— 2 cos a cos ^ I xydM — 2 cos yS cos 7 I yzdM — 2 cos 7 cos a j xzdM^ or /„ = Jj. cos2 a-\- ly cos2 /3 4- ^2 cos2 7 — 2 cos a cos yS j xydM — 2 cos /S cos 7 j yzdM — 2 cos 7 cos « | xzdM^ which gives the moment of inertia of the body with respect to an inclined axis in terms of the moments of inertia with respect to the coordinate axes and the prod- ucts of inertia j xydM, | yzdM, and | xzdM. 82. Principal Axes. — If the three products of inertia I xydM, I yzdM, and j xzdM are each equal to zero, the expression for I^ reduces to the form ly = Ij. cos2 a 4- ly cos2 /3 -{- I^ cos2 7. In this case the coordinate axes x, y, and z are called the principal axes for the point and the moments /,, I^, and I^, the principal moments of inertia for that point. 140 APPLIED MECHANICS FOR ENGINEERS If the point is the center of gravity of the body, the principal axes are called the principal axes of the body. It can be shown that it is always possible to select the coordinate axes :r, ?/, and z so that the products of inertia given in the expression for /„ will each be zero. It fol- lows that for every point of a body there exists a set of rectangular axes that are principal axes. If the a:?/-plane is a plane of symmetry, the products of inertia j zydM^i\(\. | zxdM-dVQ both zero, for to any term in ^zydM^ as z-^y-^dM^ there corresponds a term, — z^y^dM^ equal in numerical value but opposite in sign. Hence if two of the coordinate planes are planes of symmetry of a body, the three products of inertia with respect to these planes are zero, and the coordinate axes are the principal axes of the body through the origin. Problem 161. What are the principal axes of a sphere through a point on its surface ? What are the values of the principal mo- ments of inertia for that point ? | Mr^, I Mr% | Mr^. Problem 162. Find the moment of inertia of the ellipsoid whose surface is given by the equation -V.2 9/2 ^2 a^ b'2 c^ with respect to the axes a, b, and c, and with respect to an inclined axis OV making angles a, (3, y with a, b, and c, respectively. The volume of an ellipsoid is 3 5 o o and /„ = Ta cos- « + h cos^ jS -\- Ic cos*^ y. 83. Ellipsoid of Inertia. — It is always possible to re- duce the expression for I^ to the form MOMENT OF INERTIA 141 1^ = 1^ cos^ a. + ly cos^ P -{• Iz cos^ 7, by selecting the axes x^ y^ and z so that the products of inertia are zero. Dividing this equation througli by M^ we have kl = kl cos^ a -\- k'^y cos^ ^ + k^ cos^ 7. Let p = — , a = — , b = —^ and c = — . Then the equa- kv fCj. Ky kz tion becomes, on multiplying by p\ p^ cos^ a p^cos^ /3 /9^ cos^ 'y _ 1 G? IP' C^ which is the equation of an ellipsoid of semi-axes a, 5, and c, on the principal axes, the coordinates of any point on the surface being a;=/3cosa, y = p cos /3, z — pco^f^. Hence for any body there exists for each point an ellipsoid such that the distance from the given point to the surface along any line is the reciprocal of the radius of gyration of the body with respect to that line, the axes of the ellipsoid coinciding witli the principal axes for that point. Since one of the semi-axes, a, 6, c, of the ellipsoid is a maximum value and another a minimum value of the dis- tance from the center to points on the surface, it follows that two of the principal moments of inertia of a body for any point are respectively minimum and maximum moments of inertia of the body with respect to lines through the point. 142 APPLIED MECHANICS FOR ENGINEERS Problem 163. Write the equation and construct the inertia ellij)- isoid for the contcr of gravity of a right circular cylinder, allitiidc h and radius r. Problem 164. Construct the inertia ellipsoid for the center of a solid .s]>here of radius r. Problem 165. Show that tlie moment of inertia of the seg- ment of the circle F\ (Fig. 136) with respect to the axis ()Z is the moment of inertia of the sector OBSA , minus \ ah^, the moment of inertia of the tri- angle OAB, or Fig. 136 7^^ := ^^^2 « - sin 2 a], and the moment of inertia of F■^^ with respect to OS is , ( q — ,-. sin a j, the moment of inertia of the sector, minus I ha^, the moment of inertia ^4/4 I \ of the triangle A OB, or 7^^.. = -(«-., sin a + - sin 2 « . b^ O O / Problem 166. Show that the moment of inertia of the counter- balance (Fig. 58), with respect to a line through 0, perpendicular to 05, and in the plane of the wheel, is approximately /,^=[^(2«-sin2«).-^;(2^-si„2^)+iiLip_F,(00')-^]^', = ^; - „r. w^here F^ = t^ — ai\ cos ^. 00' — r, cos ^ — r cos ?, and t is the thick- "9 ^ O ' o 2 ness, as explained in Art. 38. Problem 167. Find the moment of inertia of the counterbalance (Fig. 58), with respect to a line through perpendicular to the plane MOMENT OF INERTIA U3 of the wheel. It may be written Io= [|'(2 a - .^sin 2 a - ^sin «) - '^(2 /S - Urn 2 fi - |siii ^) 7;(00')-]' + ^aWO' 3 84. Moment of Inertia of Locomotive Drive Wheel. — The drive wheel may be represented as in Fig. 137, and may be considered as made up of a tire, rim, twenty elliptical spokes, counterbalance, and equivalent weight on opposite side of center, and hub. Fig. 137 The dimensions of the wheel are as follows : Tire, outside radius 40'^ inside radius 36", width 5". Rim, outside radius 36", inside radius 34", width 4J". Hub, outside radius 10^', inside radius 4|", thickness 8". 144 APPLIED MECHANICS FOR ENGINEERS Counterbalance, outside radius 34", inside radius 7'11.5'^ thickness 7|'^ 20 spokes, 24" long, elliptical 3^'^ by 21". Angle at center subtended by counterbalance, «= 94° 40'. Radius of crank-pin circle 18". 7 = 490 lb. per cu. ft. From the above data the following additional values are computed (Art. 38): Angle subtended by counterbalance at center of inner boundary circle, /3 = 30° 20'. Mass of counterbalance 17.1. Distance of center of gravity of counterbalance from center of wheel 28.8". 28 8 Mass of weights carried by crank-pin = ^^-^- x 17,1 = 27.4. 18 (Since moment of counterbalance = moment of crank- pin weights.) The moment of inertia of the wheel with respect to its axis of rotation is first found. The tire, rim, and hub are hollow cylinders, and their moments of inertia are com- puted from the formula For the counterbalance the formula of Problem 167 is used. The spokes are regarded as elliptical cylinders with the short axis of the ellipse in the plane of the wheel. Tlie moment of inertia of one spoke is computed from the formula TEhf^'U + ^^dx, (Arts. 77, 72, 64.) MOMENT OF INERTIA 145 and 10 per cent is deducted for the parts of the spokes in- cluded in the counterbalance and boss. The moment of inertia of the weights at the crank-pin is computed on the assumption that the whole weight is concentrated at the center of the crank-pin. The results obtained are the following : Tire 7^ = 423 Rim 136 Hub 7 Spokes 81 Counterbalance 120 Weights at crank-pin 62 Total lo =S29 With respect to a gravity axis OZ (Fig. 136) in the plane of the wheel, when the counterbalance is in a posi- tion where the line joining its center of gravity to the center is perpendicular to OZ, we get for the moments of inertia of the various parts : Tire I^^ = 212 Rim 68 Hub 4 Spokes 41 Counterbalance 103 Weights at crank-pin .... 62 Total /^^ = 490 Since the widths of tire and rim are small compared to the diameter of the wheel, their moments of inertia are closely approximate to the values they would have if the L 146 APPLIED MECHANICS FOR ENGINEERS material of tire and rim were in one plane, which would be one half of the corresponding Iq. The same is true, less accurately, of spokes and hub. The values of Iq have been divided by 2 for the corresponding values of Iq^ for the four parts mentioned. The weights at the crank-pin center are at the same distance from OZ as from 0, and the moment of inertia is therefore tlie same as for 0. The moment of inertia of the counterbalance was computed from the formula of Problem 166. Problem 168. Compute the moment of inertia of a pair of drivers and their axle with respect to their axis of rotation. Use the data given above and assume the axle as cylindrical, the diameter being 9}," and the length 68". Ans. 1661. Problem 169. Compute the moment of inertia of the pair of drivers and their axle, given in the preceding problem, with respect to an axis midway between the wheels and perpendicular to the axle. Consider the counterbalance of both wheels in such a position as to give a maximum moment of inertia and the distance between the centers of the wheels 60". Problem 170. Find the moment of inertia of two cast-iron car wheels and their connecting steel axle with respect to («) their axis of rotation, (h) an axis midway between the wheels and perpendicular to the axle. Consider the car wheels as composed of an outside tread, a circular web, and a hub ; each part may be considered a hollow cylinder with the following dimensions: tread, outside radius 16", inside radius 14", width ol" ; web, outside radius 14", inside radius .5^", thickness 1.5" ; hub, outside radius o^", inside radius 2\", width 8" ; axle (considered cylindrical), 5" diameter and 7' 8" long. Dis- tance between centers of wheels 60". According to the assumption made above, the flange has been neglected, the web is considered a hollow disk, and the axle of uniform diameter throughout its length. MOMENT OF INERTIA 147 Problem 171. The value 490 is the greatest vahie for the moment of inertia of a drive wheel with respect to a gravity axis in its plane. The least value will be with respe'ct to an axis at right angles to this through the centers of gravity of the counterbalance and wheel. The student should compute this least moment of inertia. Problem 172. In Problem 169 the drivers have been considered as having their cranks in the same plane. In practice they are 90^ apart. Find the moment of inertia with respect to the axis stated when the wheels are so placed. CHAPTER VIII FLEXIBLE CORDS. THE ARCH. BENDING MOMENTS 85. Introduction. — A cord under tension due to any load may be considered as a rigid body. In the analysis of problems in which such cords are consid- ered, the method of cutting or section may be used. Since the cord is flexible (requiring no force to bend it), it is easy to see that, no matter what forces are acting upon it, it must T' have at any point the direction of the result- ant force at that point, and so must be under simple tension. If the cord is curved, as is the case where it is wrapped around a pulley, the resultant force is in the direction of the tangent. Consider, as the simplest case, a weight W suspended by a cord, as shown in Fig. 138 (a). The forces acting on TFare shown in (J) of the same figure. The cord has been considered cut and the force T^ acting vertically 148 FLEXIBLE CORDS 149 upward, has been used to represent the tension. Summa- tion of vertical forces = gives T = W. In Fig. 138 (c) the weight W^ is supported b}^ two cords. The system of forces acting on the point is shown in (c?), where T' and T' represent the tensions in the cords A and B respectively. ^X = and ^Y = give T' cos a = T" cos /3 and r sin a + T" sin /3 = W^. These two equations are sufficient to determine the unknown tensions T' and T^'. Problem 173. A weight of 500 lb. is attached to the ends of two cords of length 8 ft. and 6 ft., the other ends of the cords being attached to the points A and B respectively, where .1 is lower than B, and is distant 9 ft. horizontally and 4 ft. vertically from B. Find the tensions in the cords. Solve analytically and graphically. SuGGESTiox. Make use of the horizontal and vertical projections of the cords to determine the angles which the cords make with the horizontal. Problem 174. A weight of 275 lb. is knotted at a point C to a rope which passes over two smooth pulleys at A and B distant 50 ft. apart and in the same hori- zontal line, carrying weights of 350 lb. and 300 lb. respectively, as in Fig. 139. Find for what position of C the weights will be in equilibrium. Solve ana- lytically and graphically. Fig. 139 Problem 175. If, in the pre- ceding problem, B is 10 ft. lower than A and distant 50 ft. horizon- tally from A, find the position of C for equilibrium. Solve analyti- cally and graphically. 150 APPLIED MECHANICS FOR ENGIXEERS mmmm. 1) 86. Several Suspended Weights. Analytical Method of Solution. — If two weights W^ and W^ are attached to the cord, as shown in the case of the cord ABQD (Fig. 140), each portion is un- der tension. Consider the cord cut at A and D and represent the tensions by T^ and Tg respectively. From2X=0 and 21^= we have T-^ cos 7 = ^2 cos a and 2\ sin ^ ^ T^ sin « = W^ + ^Y^, A consideration of tlie forces acting at B^ if we call the tension in the portion BC^ T^^ gives, when tlie summation of the X and y forces are each put equal to zero, T^ cos 7 = 7^3 cos fi and 7\ sin 7 - T^^ sin /3 = ^¥^. In a similar way, consider the forces acting on the point (7, and we have T^ cos yS = 2^2 COS a and Tg sin /3 -}- 2^2 ^^^^ " = ^^v Of tlie six equations given above only four are independ- ent ; consequently, of the six quantities 7\, 2^2' ^3' "' A and 7, two must be known in order to determine the other four, or else additional conditions must be given for deter- mining two more independent equations. Such condi- tions may, for example, be the lengths of the cords and FLEXIBLE CORDS I'jl the positions of the two points of support, A and Z>. of iMg. 140. In general, if there are n knots such as B and C of Fig. 140, with the weights TFj, W^, TFg, TF^, etc., attached, it will be possible to get n + 2 independent equations. Tliese will be sufficient to determine the tension in eai-h portion of the cord and its direction, provided the tension at A^ say, and its direction are known. If the weights are close together, tlie curve takes more nearly the form of a smooth curve. Two special cases of this kind are discussed in this chapter in Art. 97 and Art. 100. Problem 176. The points .1 and D of Fig. 140 are distant 20 ft., apart, and are in the same horizontal hne. Eacli of the cords is 10 ft. long and the weights are each 50 lb. Find the tensions in the cords. Problem 177. In Fig. UO W^ = 50 lb., W., = 120 lb., AB = 10 ft., y =: (jO*^. J) and A are on the same level, 20 ft. apart. Find the tensions in CD and BC and tlie lengths of tliese cords. 87. Graphical Method of Solution. — In Fig. 141 (z) is shown a cord carrying the weights IFj, TF^, TFg, and in (z7) the corresponding stresses in the cord and the re- action at the supports. The cord in this position may be thought of as a rigid body in equilibrium under the action of the weights and the reactions of the supports. The weights and the reactions of tlie supports therefore form a system of forces in equilibrium and the relation found in Art. 57 must exist here between the rays to the force polygon and the sides of the equilibrium polygon. In stating the problem for solution enough conditions must be given to determine the construction. If the 152 APPLIED MECIIAMCS FOR ENGINEERS points of support are given and assigned weights are to act in assigned vertical lines, then there exist an indefinite number of solutions. For the tensions in tlie strings can be increased or decreased by shortening or lengthening them. If the string were cut at any point and held in position, the part of the string to one side of the cut would be in ii) Fig. 141 equilibrium under the action of the forces acting upon that part. The horizontal component of the tension at the point w^here the cut is made must then be equal to the horizontal component of the reaction at either support. The horizontal component of the tension of the string is therefore the same at all points and is equal to the hori- zontal component of either reaction. If, now, in addition to having given the points of sup- port, the weights and their lines of action, there is also given the horizontal component of the tension in the string, the graphical construction for the shape of the FLEXIBLE CORDS 153 string and the tensions in the segments can be carried out. The force polygon agf"-ca is laid off, the point h not being determined. Any point is chosen and the rays oa^ og^ ••• oc are drawn. Horizontal and vertical lines are drawn through the points of support as action lines of the horizontal and vertical components of the reactions at these points. Beginning at any convenient point, as on AB^ the equilibrium polygon is drawn with sides respectively parallel to the rays of the force polygon. Parallel to the closing side (dotted) of the equilibrium polygon, a ray (dotted) is drawn from to ca determin- ing the point h. Therefore ch and ha are respectively the vertical components of the reactions of the right and left supports with the given value of the horizontal compo- nent of the tension. The triangle hag is then the triangle of forces acting at the left support, and hg therefore repre- sents the tension in the segment of the cord attached there. The segment of the cord BG may then be drawn parallel to hg from the point of support to the vertical line in which W^ acts. In the same way the triangles hgf^ hfe^ and hed are the force triangles for the points where the weights are attached and hf^ he, and hd represent the tensions in the segments BF, BE, and BD respectively. They give therefore the direction of these segments. The shape of the cord can then be constructed. A check on the accuracy will be that the last segment must pass through the right support. The segments of the cord form an equilibrium polygon, ov string polygon, for which the corresponding ray poly- gon has h as a pole. 15-t APPLIED MECHANICS FOR ENGINEERS Problem 178. Weights of 25, 40, and oO lb, are to he suspended by a cord in lines distant 4, 8, and 11 ft. respectively from the left support. The right support is to be distant 15 ft. horizontally and 3 ft. higher than the left. With a horizontal comi)onent of tension in the cord equal to 30 lb., find by graphical methods the shape of tiie cord, the lengths of the segments, the tensions in the segments, and the reactions at the support. Problem 179. Ten equal weights are to be hung at equal hori- zontal distances on a cord supported at two points on the same level. The horizontal tension in the cord is to equal five times the value of one weight. Draw the shape of the string and determine the maximum tension. Draw on the same figure tlie shape of the cord when the horizontal tension is three times the value of one weight. 88. Locus of the Pole of the String Polygon. — Let the left- and right-hand vertical components of the reactions at the supports be Y^ and Y2 respectively, II the horizontal component of the tension, a^, a^-, a^ the horizontal distances of the weights W^^ TF^, W^ from the left support, I the horizontal distance between the supports, and h the verti- cal height of the right support above the left. Taking moments about the left support, Let y/^ ^^1^1 + ^2^^2 + ^3^3 ^ I Then F2 = Yl + -R. The value of P^ ^^ the value that Pg ^^'ould have if either h ov H were zero, i.e. if the supports were on the same level, or if the string were replaced b}" a rigid body simply- resting on the supports without horizontal pressure. FLEXIBLE CORDS 155 The value Y'^ may be computed analytically and laid off vertically from d on the load line gd (Fig. 142), or it may be located by using the equilibrium polygon for vertical forces only, assuming H to be zero. Y^ is then found by increasing Y'^ by the quantity - H^ which is proportional to H. If through the end, 6', of Y'^ a line is drawn parallel to the line joining the points of support, the pole, ^, of the string polygon is found on this line at a horizon- tal distance H from the load line gd. Conversely, any [)oint on this line may be taken as a pole for a string polygon, and the corresponding value of H is the horizon- tal distance from the point chosen to the load line gd. The locus of the poles of the string polygons is therefore the straight line parallel to the line joining the points of support and passing through the points which would divide the load line into the reactions if the supports were on the same level. 156 APPLIED MECHANICS FOR ENGINEERS By varying the position of the pole the form of the string polj^gon may be changed. 89. Pole Distance and Depth of String Polygon. — The depth of the string polygon^ measured from the line joining the 'points of support^ varies inversely as the pole distance from the load line. For, if d is the depth of the string polygon at a certain point and D the distance of the pole b from the load line (Fig. 143), we have from similar triangles, (d + zy.x:: F/ :i), and z: (^x— a-^) : : W^i D. Eliminating 2, Since aj, Y\^ and Wi are constants, and for the given position X is constant, this equation shows that d varies inversely as D. The proof can be given in like manner for any point. The depth of the string polygon at any point can then be made to have any desired value by a proper choice of FLEXIBLE CORDS 157 the pole. If with a pole distance I) the depth of the string polygon at a certain point is d, and a depth d^ is desired, choose a new pole distant D^ from the load line so that i)j = — , and the string polygon constructed by the use of this pole will be the one sought. Problem 180. Weights of 100, 300, 200 lb. are to be suspended from a cord in lines 3, 5, and 8 ft. respectively from the left support. The right support is distant 10 ft. horizontally and 2 ft. vertically above the left support. Using a scale of 1 in. = 2 ft., construct a string polygon whose depth shall be 4 ft. at the second load. From the diagram scale off the tensions in each part of the cord and the horizontal component of the stress in the cord. Use a force scale of 1 in. = 100 lb. Problem 181. A cord is suspended from tvs^o points on the same level 9 ft. apart. Eight weights of 50 lb. each are to be suspended at equal intervals between the supports. The maximum tension in the cord is to be 300 lb. Draw the shape of the cord. Find the total length of the cord and the horizontal component of the tension. Problem 182. Show that when equal weights are distributed at equal horizontal intervals along a cord suspended between two supports on the same level, the maximum tension in the cord is greater than half the total load on the cord. Problem 183. A cord supported at the ends on the same level carries a load uniformly distributed along the horizontal. Draw approximately the shape of the cord, given that the tension of the cord at the lowest point is one half the total load. Suggestion. Divide the horizontal distance up into small equal intervals and assume that the weight in each interval acts at the center of that interval. With the vertex at the lowest point of the string polygon, construct a parabola to pass through the points of support and compare it witli the string polygon. 158 APPLIED MECHANICS FOR ENGINEERS Problem 184. Make the constructions of the above problem when the points of support are not on the same level. Problem 185. Show how to obtain the ratios of a set of weights to cause a cord to liang in a given string polygon. Hint. Work from the string polygon to the force polygon. 90. The Linked Arch. — Analogous to the problem of finding the form taken by a cord carrying loads at inter- vals is that of finding the form of a set of connected weightless links to sustain a given set of loads in given vertical lines, the links to be in compression instead of tension. Here again if the horizontal component of the Fig. 144 thrust in any link is given, the form of the linkage can be determined. The construction and proof are analogous to those for the cord carrying weights. (See Fig. 144.) The links themselves lie in an equilibrium polygon the pole of which is b. It will be noted that the pole of the link polygon lies on the opposite side of the load line from the pole of the string polygon, and that all poles of link polygons lie on the same straight line as that on which the poles of the string polygons lie. FLEXIBLE CORDS 159 91. The Masonry Voussoir Arch. — In an arch made of links, constructed to carry a given set of loads, any varia- tion of the loads, unless they were all changed simulta- neously in the same ratio, would be apt to cause the collapse of the arch. In an arch built of dressed stone, called voussoirs, owing to the size of the bearing surfaces and the friction between the surfaces, an arch can be built to carry a given set of loads and allow for a given varia- -\ — t r 7 — \ \ \ ' ■^' \ > c / J^ \ ] \ I \ ^ Vl — t? Le-i -^ ^ < 7- \ ' ' / \ I \ / \ ^ fl \ 5 tiJ R Vi W V-t, 1/ ^ 1 (a) Fig. 145 (c) tion in the loads without endangering the structure. The voussoirs can be made of such a depth as to keep the line of the resultant pressure of two adjacent voussoirs on each other in a certain portion of the joint. The broken liiie joining the points of the resultant pressures of the adjacent arch stones on each otlier is called the line of resistance. It is usually considered advisable, though it is not necessary for the stability of the arch, to keep the line of resistance within the middle third of the arch. The intensity of the pressure between two arch stones is assumed to vary uniformly as indicated in Fig. 145. In (a) the resultant pressure falls within the middle third, in (5) at the edge of the middle third, and in (c) outside the middle third. In the latter case there is a tendency 160 APPLIED MECHANICS FOR ENGINEERS for the joint to open. With the line of resistance below the middle third there would be a tendency to open the joint on the upper part of the arch, and also to unduly in- crease the pressure on the lower part of the joint. vVhere the line of resistance falls above the middle third the joint would tend to open on the lower side. It is desired here to present only the elementary prin- ciple of the arch. An adequate treatment of arches may be found in I. O. Baker's " Treatise on Masonry" (John Wiley & Sons). Problem 186. Find the shape of a linkage that would carry the weights of Problem 178 with the same horizontal stress, the links being in compression. Is the link polygon the same as the inverted string polygon of that problem ? Problem 187. Show that when, and only when, the supports are on the same level, the link polygon for a given set of loads and horizontal stress is the inverted string polygon for that set of loads. Show also that in any case the link polygon can be obtained from the string polygon turned through ISO"" in its plane if in constructing the string polygon the loads and their horizontal distances are reversed in order from right to left. Problem 188. An arch is to be constructed with its center line in the form of an arc of a circle, the span being 20 ft., and rising 6 ft. in the middle. Considering the .load on the arch to be vertical only and at any point proportional to the distance from the center line of the arch to a straight horizontal line 6 ft. above the highest point of the center line, sketch approximately the line of resistance. About how deep would the arch stones need to be to keep the line of resist- ance within the middle third ? Suggestion. Assume the loads concentrated at equal horizontal intervals and construct the link polygon with the required height of 6 ft. It may then be seen how the line of resistance may be changed FLEXIBLE COEDS 161 so as to more nearly coincide with the given circle by a slight move- ment of the pole. Problem 189. Find approximately by graphical methods the form of an arch to carry a load uniformly distributed along the horizontal. 92. Bending Moments. — As an application of the equi- librium polygon for parallel forces a brief discussion is here given of bending moments in a horizontal beam sub- jected to vertical forces only. Let Fig. 146 represent a horizontal beam acted on by vertical forces. At any p^ point C pass a vertical plane perpendicular to the axis of the beam. The sum of the moments about a horizontal line in this plane of all the forces acting on the beam to the left of the plane is called the bending moment at that section. Clockwise direction will be counted as positive. Repre- senting the bending moment by M^ the value of iltf at 6' (Fig. 146) is -aj-^ W LBS /ft. I. Fig. 146 M= R^x -{x + a^)P^ - W(x - ay Since the sum of the moments of all the forces acting on the beam is zero, the sum of the moments of the forces to the right of the section is the negative of the sum of the moments to the left of the section. Hence, in comput- ing -bending moments, if it is more convenient, we may take as the bending moment at the section the sum of the moments of all forces acting: on the beam to the rig-ht of the section and consider counter-clockwise as positive. 162 APPLIED MECHANICS FOR ENGINEERS Problem 190. A beam 10 ft. long is supported at the ends and carries loads of 500 lb., 600 lb., 800 lb. at distances of 3 ft., 5 ft., and 8 ft. respectively from the left end. Compute the bending moment at intervals of 1 ft. along the beam, neglecting the weight of the beam. Problem 191. A beam 12 ft. long, supported at the ends, carries a load of 500 lb. at a point 4 ft. from the left end and a uniformly distributed load of 200 lb. per linear foot over the right half of the beam. The beam weighs -30 lb. per foot of length. Compute the bending moment at intervals of 2 ft. along the beam. Using M as ordinate and distance along the beam as abscissa, sketch a curve repre- senting the bending moment at all sections. 93. Bending Moment Diagram. — Using the bending mo- ment as ordinate and distance along the beam as abscissa, a curve may be plotted whicli shows the bending moment Fig. 147 at each section. This curve is called the bending moment diagram. It will now be shown that for concentrated loads the equilibrium polygon represents to a certain scale, the bending moment diagram. Consider a beam supported at the ends carrying loads Pj, P^, Pg lb. at distances of Xp x^, j^^ ft. from the left end FLEXIBLE CORDS 163 respectively. Construct the equilibrium polygon for the forces and reactions, using the following scale : V =m ft. along the beam, 1" = n lb. on the line of loads. Let D be the pole distance in inches, and d the depth in inches of the equilibrium polygon at a point distant — in. m from the left support between the loads Pj and P^. By similar triangles m n and z : — ^I^i = — i : i). m 71 Eliminating z, d = — - — ^ — ^^^^^ — - • ° mnl) By definition the bending moment, M, measured in pound- feet at the given point is M=xR^-{x-x^)F^. ,'. d = -, mnD or M—mnD'd. This formula can as easily be shown to hold for any other point. Hence the depth of the equilibrium polygon in inches multiplied by the number of ft. per inch along the beam^ the number of lb. j^er inch on the load line, and the pole distance in inches gives the value of the bending moment in lb. ft. Tlie depth of tlie equilibrium polygon then represents the bending moment to the scale 1" = mnl) \h.-it. 164 APPLIED MECHANICS FOR ENGINEERS If it is desired to use the pound-inch as the unit of moment, then let 1" =m in. along the beam. To obtain the bending moment diagram to a given scale, 1" = k lb. -ft., it is only necessary to choose D so that mnD = k. For a distributed load the bending moment curve may be obtained approximately by dividing the beam length into small intervals and considering the load on each in- terval to act at the center of that interval. Problem 192. Construct graphically the bending moment for the beam of Problem 190. Let 1 in. = 2 ft. along the beam, 1 in. = 500 lb. on the load line, and choose D so that 1 in. on the moment diagram shall equal 1500 Ib.-ft. of moment. Problem 193. Construct approximately the bending moment diagTam for a beam supported at the ends and carrying a uniformly distributed load. 94. Bending Moment for Beams not Supported at the Ends. — (a) Cantilever Beam. A cantilever beam is shown in Fig. 148. The construction for the bending moment is indicated, the scale factors being the same as in Art. 93. P. 1 4 1 iT^ 2\^ ^ 3\ J. 148 FLEXIBLE COBDS 165 (J) Overhanging Beam. In Fig. 140 the construction is indicated for the bending moment of an overhanging beam. The diagram (i) is the equilibrium polygon. The diagram (ii) is obtained from (i) by replacing the closing line of the polygon and the two lines connected Fig. 149 with it by a horizontal line and drawing a polj^gon with a depth on each load line and reaction line, the same as that of the equilibrium polygon {i). Problem 194. Prove that the metliod indicated of drawing the bending" moment diagram of a cantilever beam is correct. Problem 195. Prove that the method indicated of constructing the bending moment diagram of the overhanging beam is correct. Problem 196. Construct the bending moment diagram of a cantilever beam 12 ft. long carrying loads of 400, 600, and 1000 lb. at distances of 0, 3, and 7 ft. respectively from the free end. Scale the bending moment at 3 ft. and at ft. from the free end and com- pare with computed values. State the scales used. 166 APPLIED MECHANICS FOR ENGINEERS Problem 197. A beam 16 ft. long is supported at two pothts dis- tant respectively 3 ft. from the left end and 4 ft. from the right end. Loads of 300, 500, 1000, and 600 lb. are applied at distances of 0, 6, 10, and 16 ft. respectively from the left end. Construct graphically the bending moment diagram to a convenient scale. Check by com- puting analytically the bending moment at several points. ^^ A /\ R. Fig. 150 95. Tensile and Compressive Stresses in Beams. — Let Fig. 150 represent the portion of the beam of Fig. 146 to the left of the plane section at C. This portion of the beam is in equilibrium under the action of the external forces acting on this portion and the internal forces exerted on it by the portion of the beam to the right of the section. These internal forces can be re- solved into horizontal and vertical forces. The vertical component of the internal forces must be equal and opposite to the sum of the external forces to the left of the section. This latter sum is called the shear at the section. Since the external forces are all vertical, the sum of the horizontal components of the internal forces must be zero. The horizontal forces therefore form a couple. This couple is called the internal stress couple. To produce equilibrium the moment of the internal stress couple must be equal and opposite in sign to the moment about any horizontal line in the section of all the external forces to the left of the section. That is, the internal stress couple at any section of the beam is equal to the bending moment at that section changed in sign. FLEXIBLE COIiDS 107 It can be shown by experiment that when a rod is stretched or compressed within tlie ehistic limit, the amount of elongation or compression is directly proportional to the applied force. Defining unit stress in a rod uniformly stretched or compressed, i.e. without bending, as the total force divided by the cross-sectional area, and unit strain as the amount of elongation or compression divided by the original length, this experimental fact may be expressed by the equation. unit stress = 11 unit strain where -E'is constant for a given material. ^ is called the modulus of elasticity. Experiment shows that it is the same for compression as for tension, for such materials as wood and structural steel. It is also shown by experiment that when a beam is bent, within the elastic limit, a plane section of the beam before bending remains a plane section after bending. When the beam is bent, the upper fibers of the beam are then compressed (or elongated) and the lower fibers are elongated (or compressed) as in Fig. 151. There will be A' B' \ ^ — 1 V- ^ 1 7 t * «• / 1 y ''1 ^ 4.:....-L — i^ B Fig. 151 Iti8 APPLIED MECHANICS FOR ENGINEERS somewhere between top and bottom a surface, called the neutral surface^ where there is neither tension nor com- pression, and the amount of tensile or compressive stress at any point in the section will vary directly as the distance of this point from the neutral surface, since the amount of the elongation or compression of the longitudinal fiber of the beam through the point varies as the distance from the neutral surface. If s is the unit compressive stress at the upper outside fiber of the beam, distant y from the neutral surface, s' the unit stress at a distance y' from the neutral surface, then y s' will be positive or negative according to the sign of y' . Considering the portion of the beam as in equilibrium under the action of the external and internal forces, if dA is an element of area of the cross section distant y' from the neutral surface, we have from 2X= 0, Cs'dA = 0, or "-Cy'dA^O. y^ But i y'dA is equal to the product of the area of the cross section and the ordinate of the center of gravity measured from the origin of ordinates, i,e. from the neutral surface. If y is this ordinate, then ^yA=^. y Therefore y =^i FLEXIBLE CORDS 1G9 and the neutral surface passes through the center of gravity of the section.* Equating moments of external and internal forces about a horizontal gravity axis of the section, Mom of external forces = — mom of internal forces, or M= Cy's'dA = -Cy'HA = -I y^ y where J is the moment of inertia of the area of the cross section with respect to the horizontal gravity axis of the section. This formula gives the stress s at a distance y from the neutral axis of the section. The stress at any other point in the section is obtained by the direct proportion s y' For a beam of constant cross section y and J are constant and hence 8 varies directly as M. The maximum tensile or compressive stress will therefore be found where the bending moment has maximum numerical values. In computing stresses care must be used to have the same units throughout the formula. If stress is reckoned in pounds per square inch, the dimensions of the cross * In the above it is assumed tliat the intersection of the vertical section and the neutral surface is a liorizontal straight line. If the loads lie in one plane which is a plane of symmetry of the beam, this will be the case. Under other conditions the neutral surface may not be liorizontal. 170 APPLIED MECHANICS FOR ENGINEERS section must be in inches and the bending moment in pounds-inches. Problem 198. A beam 4 in. broad by 6 in. deep, 12 ft. long, is supported at the ends and carries loads of 400 lb. and 800 lb. at dis- tances of 5 ft. and 9 ft. respectively from the left end. Sketch the bending moment diagram and find the maximum fiber stress. Aris. 11.50 Ib.-sq. in. under the 800 lb. load. Problem 199. A beam 4 in. broad by in. deep, 12 ft. long, is supported at the ends and carries two equal loads at distances of 4 ft. from the ends. What maximum value may the loads have if the maximum allowable fiber stress is 1000 lb. per sq. in. ? Problem 200. Find the ratio of the loads that could be put on a beam 4" by 6" when placed with the 4" face horizontal and when placed with the 6" face horizontal, the maximum fiber stress to be the same in the two cases. Problem 201. Compare the loads, one concentrated at the middle, the other uniformly distributed over the length of the beam, to cause the same maximum fiber stress in the beam. Problem 202. An I-beam of depth 10 in. and weight 25 lb. per linear foot is 16 ft. long and is supported at the ends. What con- centrated load can it carry in the middle so as to cause a maximum fiber stress of 16,000 lb. per sq. in.? The moment of inertia of the cross section about a horizontal gravity axis is 122 in.^. Take the weight of the beam into consideration. 96. Cords and Pulleys. — When a cord passes over a pulley, without friction, the tension is transmitted along its length undiminished. A weight W attached to a cord which passes vertically over a pulley is raised by a direct downward pull P on the other end of tlie rope. If there is no friction, P is equal to TT, for uniform motion. In the case of a system of pulleys, as shown in Fig. 152, the FLEXIBLE COEDS 171 cord may be considered as under the same tension through- out and parallel to itself in passing from one sheave to tlie other. It is then possible to cut across the cords, just as was done in the case of the bridge truss, Problem 92, where the stress was along the member in each case. Cutting all the cords at C and considering all the forces acting on the sheave B^ we get, calling the tension in the cord P, or the tension in the cord is W/Q. A con- sideration of the upper sheaves gives r= 7 P = 7/6 ( TF). The various cases of cords and pulleys that come up in engineering work may be taken up in a similar way, but in any case of cutting cords, it must be remembered that all cords attaching one part to another must be cut and the tension acting along the cords inserted be- fore the principles of equilibrium can be applied. The consideration of the friction between cords and pulleys will be taken up in Chapter XIII. Problem 203. Suppose the hook of the lower sheave of Fig. 152 attached to a weight of i]50 lb., and a man of weight 150 lb. stands on the weight and lifts hinjself and the weight by pulling on the rope. What force must he exert? If he stands on the ground, what force must he exert to raise a weight of 500 lb. ? 97. Equilibrium of a Flexible Cord Carrying- a Load Uniformly Distributed along the Horizontal. — Let the cord supported at Fig. 152 172 APPLIED MECHANICS FOR ENGINEERS A and B (Fig. 153 (^ Ix ~ X ^ ^ Since c and iv are constants, a value of x can be found for a point on the curve, not necessarily on the cord -between A and J5, for which -^ = 0. If the origin be moved to this dx point on the curve by a translation of axes, equation (4) will take the form ^ = — CS^ dx X This means that the origin is taken at the lowest point of the curve as in Fig. 153 (a) and (6). In (a) the origin is on the cord, and in (J) it is not. Integrating equation (5), y^^^' + o'. But y = when x = 0. Therefore c' = and the equation of the curve becomes ^ 2X Hence the cord hangs in a portion of a parabola with ver- tical axis. 98. The Horizontal Component of the Tension in Terms of the Length of Span and the Deflection. 174 APPLIED MECHANICS FOR ENGINEERS The horizontal distance between the supports is called the span^ I. The vertical distance of the vertex of the parabola below the lower point of support is called the deflection, d. In Fig. 153 (a) and (J) the coordinates of A are ( — Zj, c?) and (Zj, d) respectively, and the coordinates of B are (Zg, d ■]- h'). The coordinates of A and B satisfy the equation of the curve. Hence in case (a) cl=:J^l2 and d-\-h = -^Il Therefore, ?, = J^^, Z. = J^^(Z±5. Therefore Z = /^ + /, = V".~ ("^^ + ^'^ + ^)' and X= w If h = 0, the supports are on the same level and In case (6), 1 = 1^ — \ = ^" (c? + A) — -y^*" -, and -X' = 2(Vd-\-h-vdy 99. Length of Cord. — The length of the curve from the origin to any point (a:, ?/) on tlie curve is given by /- -XV+fST-^- FLEXIBLE CORDS 175 Letting — = a, the equation of the parabola becomes zv or • ft 11 ^ y = - — , from which -f- = -, and hence 2 a ax a ^ 2a \^a^ + l'i-\-ano^. I a ?2 V d^ ^l\-\- a^ log, h + Va2 + l'\ a h + Va2 + li a The total length of the curve is in case (a) S = 8-^-\- %^^ and in case (6) *S'= 8^ — s^. Another formula for the length of the cord may be l+-^by the binomial theorem and integrating the series term by term when a is less than each of the quantities l^ and l^' Thus, »^=£{\ + '^dx -i V^2a2 8a^'^T6«6 128a«'^" V ^ 6a2 40^4 112 «6 1152 a^"^ In terms of the deflection, c?, this becomes, since a = -J-, 4>^ a 8 = 1 +?^_?^_L.^^_15^ . ^ ^ 3/i 5/-J 7 /-J 9 /} 176 APPLIED MECHANICS FOR ENGINEERS 111 like maimer, if c?j = c? + h. If the supports are on the same level, l^ = l^ = - and the expressions for the total length S become respectively 1 a - VP + 4 a^ + a^ log, Z 4- VZ2 + 4 a2 "Za S=l^l '' 1 /s 1 z- Z9 3 23 . a^ 20 25 . «4 56 2' . «« 576 2^ • «» TO ; , 23 . 6^2 25 . t/^ , 2 . 2' . r/6 5 . 29 . ^ , SI 5/3 7/5 9/' For cords for which the deflection is small compared to the length of span the series converge rapidly and three or four terms are sufficient for computing the length. For large values of d the series should not be used. Problem 204. A suspension bridge as shown in Fig. 154 has a span of 1200 ft. and the cable a niaxiniiuu deflection at the center d = 120 ft. The weight of the floor is 2 tons per linear foot. Find Fia. 154 FLEXIBLE COEDS 177 the equation of the cable and the tension at and at B. If the safe strength of cable is 75,000 lb. per sq. in., find the area of wire section of cable necessary to support the floor. Problem 205. Find the length of the cable in the preceding problem. Problem 206. A cable is to be suspended from two points distant 100 ft. ai^art horizontally and 20 ft. vertically. It is to carry a load of 500 lb. per horizontal foot, and the lowest point of the curve is to be 25 ft. below the lower point of support. Find the position of the lowest point of the cord, the value of the horizontal component of the tension, and the maximum tension. Problem 207. Find the length of the cable of the preceding problem. Problem 208. A cable 105 ft. long, carrying a uniform horizontal load, is stretched between two points on the same level distant 100 ft. apart. Find the deflection. Suggestion. Use the first three terms of the series which expresses S in terms of / and d. See how large the fourth term is with the computed value of d. Problem 209. A uniform wire weighing ^ lb. per foot of length is supported between two points 200 ft. apart on the same level and the maximum deflection is 2i ft. Find the horizontal tension, the maxinmm tension, and the length of the wire. Suggestion. Since the deflection is here relatively small, the load is very nearly uniformly distributed along the horizontal. The curve may then be regarded as approximately a parabola. 100. Equilibrium of a Flexible Cord Carrying a Load Uni- formly Distributed along the Cord. — Using a notation like that of Art. 97, the same method leads to an equation like (5) of that article, with x replaced by s; namely, (It/ _ w tx ~ X^' N 178 APPLIED MECHANICS FOR ENGINEERS in which the origin is at a point of the curve where the slope is zero and s is the length of the curve from the origin to the point (a;, ?/). Y -lo- -I ^-1 As. (a) (b) Fig. 155 (c) X Replacing dy by ^ ds^ — dx^ and writing — = a, this w becomes ds^ — dx^ 8^ dx^ or ds ds = - Va^ + s^ • dx. a dx Integrating, Va2 + s2 « loge (s + V s-^ + a^) = _ -^ ^. a s = when 2: = ; .•.(?= log a, log. --^- = - • \ a J a s H- Va^+ g'^ — ^~. Invert, a « + Va2 + §2 e ". FLEXIBLE CORDS 179 Rationalizing the denominator, 8 — Va^ + s^ _ _ g-f^ Add to — ■ = ea. a Then (1) It was found above that -^ = -. dx a dy = I (e" — ^ a)c?2; a X from which V =■ ~ (^" + ^ «) H- c'. 9 But a: = when ?/ = ; . •. c' = — «. .•.2/ + a = ^(e« + e"«)> (2) which is therefore the equation of the curve, the origin being at the lowest point of the curve (Fig. 155 (a) and This curve is called the catenary. The length of the curve follows at once from equation (1) ; namely, OA = s^ = ~(ea — e a) and OB=s.^ = ^(iei-e~a), (Fig. 155) and S = S2± s^ according as the cord hangs as in (a) or (^) of Fig. 155. 180 APPLIED MECHANICS FOR ENGINEERS If the points of support are on the same level, 1^ = 1^ = -, and the total length of cord is The values of y and s may be expressed in a series by making use of the expansion /y*A /yHJ ^^ /y*0 e. = l + . + _ + _ + „ + _+.... It is left for the student to show that c, — / 4- 1 ^1 1 1 ^1 1 101. Representation by Means of Hyperbolic Functions.— From the definitions sinh x = \ (^^ — e"""), cosh x = \ (e^ + e~^), it follows at once that the values of y and «, the arc from to the point (ic, ?/), are given by the formulae y -\- a = a cosh-, a 8 = a sinh-, where a = — • a w A table of hyperbolic sines and cosines is given in the ap- pendix and should be used in the solution of the following problems. Problem 210. A flexible wire weighing \ lb. per foot is supported by two posts 200 ft. apart. The horizontal pull on the wire is 500 lb. Find the deflection at the center and the length of the wire. FLEXIBLE CORDS 181 Problem 211. What pull will be necessary in Problem 210 so that the greatest deflection will not be greater than 6 in.? What is the length of the wire for this case? Problem 212. Find the tension in the wire of Problem 210 at the supports. Problem 213. A wire weighing \ lb. per foot is suspended be- tween two points A and B where B is 20 ft. higher than A and distant 120 ft. horizontally. The horizontal component of the tension of the wire is 50 lb. Find the position of the lowest point of the curve, the deflection, the length of the wire, and the maximum tension. Sketch the curve in which the wire hangs. Suggestion. Assuming that the wire hangs as in (a), Fig. 155, we may write d + 20 = 200 cosh ^, since a = - = 200, 200 w rf = 200 cosh i. Subtracting, 20 = 200 (cosh -|- - cosh ^\ = 400sinh'-^sinh'-aj=^', since cosh x — cosh y = 2 sinh — ^^—^ sinh ~ -^ • 2 2 But ^ + /a = l-'^- Therefore 20 = 400 sinh .3 • sinh ^-^:^, 400 from which the value of l^ — l^ may be found from the table of hyper- bolic sines. Combining this value with ^2 + ^1= 120, the values of /, and /g are determined. The values of d, S, and the maximum tension may then be determined. Problem 214. Solve the preceding problem if B is 40 ft. higher than A<, all other conditions remaining unchanged. 182 APPLIED MECHANICS FOR ENGINEERS Problem 215. A wire is suspended between two points on the same level 240 ft. apart. The deflection at the center is 60 ft. Find the value of «, or — , on the supposition that the load is (1) uniformly IV distributed along the horizontal, (2) uniformly distributed along the wire. Compute the values of // for x = iO and for a; = 80 ft. for both cases and sketch the parabola and the catenary. Suggestion. In case (1) a is obtained by substituting the coordi- nates of the point of support in the equation of the parabola. In case (2) the value of a found for the parabola may be used as a trial value and the correct value of a found by modifying the trial value until a value is found to satisfy the equation y -\- a = a cosh - , where // = 60 and x = 120. a That is, a must satisfy h 1 = cosh -^^ a a Problem 216, Solve the preceding problem if the deflection in the middle is only 6 ft. CHAPTER IX MOTION IN A STRAIGHT LINE (RECTILINEAR MOTION) 102. Velocity. — When a particle moves along a straight line passing over equal spaces in equal times, it is said to have uniform motion. In this case the ratio of the space passed over to the time taken to pass over that space is called the velocity of the particle. If the motion is along a straight line but is not uniform, the ratio of the S2:)ace passed over in any time to the time is called the average velocit}" of the particle for that time, or space. Thus, if the particle moves from P to Pj, a As distance As, in the time A^, then — = average velocity of the particle between P and Py As The limiting value of — as A^ approaches the limit zero is defined to be the velocity of the particle at P. Thus, .= §5 tliat is, tJie velocity is the first derivative of the distance tvith respect to time. The unit of velocity is the unit of space per unit of time, as /it. per sec, mi. per hr., etc. Speed is sometimes used instead of velocity, especially in speaking of the motion of machines or parts of machines. Speed, however, involves only the rate of motion without reference to its direction, while velocity involves both rate of motion and the direc- tion in which the motion takes place. 183 184 APPLIED MECHANICS FOR ENGINEERH Problem 217. Tlie space passed over by a particle moving in a straight line is given by the formula .s = 16 /-, where s is the distance in feet passed over in the time t in seconds. Find the velocity wlien t = Q, when / = 2, and at the end of / sec. What is the velocity of the particle when it has moved 40 ft. V Problem 218. A particle moves back and forth along a straight line, the abscissa of the particle being given by x — k cos (ct). Find the location of the particle and its velocity when / = 0, — , -, '-^ . '2 c c c pA-ect ordinates to represent the velocity for several positions of the moving particle and represent the velocity by a curve. 103. Acceleration for Straight-line Motion. — When the velocity of a particle moving in a straight line increases by equal amounts in equal times, the motion is said to be uniformly/ accelerated^ and the gain in velocity per unit of time is called the acceleration of the particle. If, for straight-line motion, the velocity changes from v at the time t to v -\- Av at the time t -f- A^, then — is called the average acceleration of the particle for the time A^.' /\ 9) (111) The limitinof value of — , i.e. — , is defined to be the At dt acceleration of the particle at the time t. Thus, if a represents acceleration, ~ dt~ dP Another form for a is frequently used. Since a = — and V = — , there results dt vdv = ads by eliminating dt. MOTION IN A STRAIGHT LINE 185 The unit of acceleration is the unit of velocity per unit of time, as ft. per sec. per sec.^ mi. p>er hr. per hr.^ yd. per min. per sec, etc. Problem 219. Find the acceleration of the particle of Problem 217 at tlu' times specified. Derive an equation which expresses the velocity in terms of the space, and from this equation obtain a formula for the acceleration in terms of the space. Problem 220. Using t for abscissa and s, v, a, respectively, as or- dinates, plot the curves which represent the space, velocity, and ac- celeration respectively in terms of the time in Problem 217. Show how the ordinate to the velocity curve is related to the slope, or gradient, of the space curve at any value of t. Likewise for the acceleration and velocity curves. Problem 221. Find the acceleration of the moving particle in Problem 218. Show that the acceleration is proportional to x. Plot curves showing x, v, and a in terms of t. Also a curve showing a in terms of x. Problem 222. Show that if a curve is plotted, using values of time as abscissas and values of velocity as ordinates, the area under the curve between two values of the time is equal to the space passed over by the particle in that interval of time. 104. Constant Acceleration. — When the acceleration is constant, we have the relation dv = a^c?^, a,, representing the constant value of a, and thtrefore Jdv = Uc I dt., and since V = act + VQ'y ds v = — , dt I ds = a^ j tdt -{■ VqI dt. or 8 = ]a^Vi + Vfjt. 186 APPLIED MECHANICS FOR ENGINEERS In ti similar way the relation vdv = a^ds gives /vdv = aA "ds ; V 1'^ therefore -y = ^c^, or s = 2ac These equations of motion give the velocity in terms of time, the distance in terms of time, and the distance in terms of velocity. 105. Freely Falling^ Bodies. — Bodies falling toward the earth near its surface have a constant acceleration. It is usually represented by ^ and equals approximately 32.2 ft. per second per second. The value of g varies slightly with the height above the sea level and the latitude, but for the purposes of engineering it may usually be taken as 32.2. The equations of motion for such bodies are, then, v = gt-\- Vq, s= J^^^ + V' ^_v^-vl^ 2^ If the body falls from rest, Vq =0, and the equations of motion become v = ot, v^ ^=2,- This latter is often written v^ = '2gh, where h = 8. MOTION IN A STRAIGHT LINE 187 106. Body Projected Vertically Upward. — When a body is projected verticall}^ upward from the earth, the accelera- tion is constant and equals — g. If the velocity of pro- jection is Vq, the equations of motion are V = —gt + v^t, Problem 223. A body is projected vertically downward with an initial velocity of 30 ft. per second from a height of 100 ft. Find the time of descent and the velocity with which it strikes the ground. Problem 224. 'A body falls from rest and reaches the ground in 6 sec. From what height does it fall, and with what velocity does it strike the ground ? Problem 225. A body is projected vertically upward and rises to the height of 200 ft. Find the velocity of projection y-^, and the time of ascent. Also find the time of descent and the velocity with which the body strikes the ground. Problem 226. A stone is dropped into a well, and after 2 sec. the sound of the splash is heard. Find the distance to the surface of the water, the velocity of sound being 1127 ft. per second. Problem 227. A man descending in an elevator whose velocity is 10 ft. per second drops a ball from a height above the elevator floor of 6 ft. How far will the elevator descend before the ball strikes the floor of the elevator? Problem 228. In the preceding problem, suppose the elevator going up with the same velocity, find the distance the elevator goes before the ball strikes the floor of the elevator. 107. Newton's Laws of Motion. — Three fundamental laws may be laid down whicli embody all tlie principles in accordance with which motion takes place. These are 188 APPLIED MECHANICS FOR ENGINEERS the result of observation and experiment and are known as Neivtons Laws of Motion. First Law. Every body remains in a state of rest or of uniform motion in a straight line unless acted upon by some unbalanced force. Second Law. When a body is acted upon by an unbal- anced force, motion takes place along the line of action of the force, and the acceleration is proportional to the force applied and inversely proportional to the mass of the body. Third Law. To every action of a force there is always an equal and opposite reaction. The second law states that in case the system of forces acting on the body is unbalanced, the motion is accelerated. Motion takes place in the direction of the resultant force with an acceleration proportional to the force. It also implies that each force of the system produces or tends to produce an acceleration in its own direction propor- tional to the force. That is to say, each force produces its own effect, regardless of the action of the other forces. As a result of this latter fact, if a body is acted upon by a force P and the earth's attraction (r^, we have P: a=^a:g, where G is the weight of the body, ^ the acceleration of grav- it}^ and a the acceleration due to the force P. From this it follows that Y> ^ ^ Tir ^ ri The quantity — , in which G- is the weight of the body in pounds, and g is in /f. /sec. 2, is called the mass of the body in " Engineer's Units.'' MOTION IN A STRAIGHT LINE 189 Fig. 15(i If ^=32.2, the Engineer's I'nit of mass is equivalent to 32.2 pounds. 108. Motion on an Inclined Plane. — A body (See Fig. 156), of weight (r, moves down an inclined plane, without friction, under the action of a force G- sin 6. The acceleration down tlie plane equals the accelerating force divided by the mass (Art. 107) = — -j^ — =g sin 6. The acceleration is y constant. The equations of motion for such a case, then, are (Art. 104) 1? = (Jk ■ s = Vq. This means that the body comes to rest when s lias reached a certain value, v^ viz. ,y' From the original assumption, a — — ks, it is MOTION IN A STRAIGHT LINE 193 seen that the acceleration is greatest when s is greatest, that is, when s = —= ; and is least when s is least, that is, when 8=0. To get the relation between distance and time, the equation v = Vv^ — ks^ may be put in the form ds . == =dt^ V ^5 — k^^ from which — =- sin~i '— = f, or IT sin ^kt = s. This relation between the distance and time shows that as t increases s changes in value from — p to ~7=^ , as- wk \k suming all values between these limits, but never exceed- ing them, since sin ^kt can never be greater than + 1 or less than —1. The motion is, therefore, vibratory or periodic, and is known as harmonic motion. The complete period in this case is '^^^-^' The relation between velocity and time may be found for this case by differentiating the last equation with respect to time. Then, V = Vq cos ^kt This shows that v^ is the greatest value of v. This motion is usually illustrated by imagining a ball attached to two pins by means of two rubber bands or 194 APPLIED MECHANICS FOR ENGINEERS springs, since the force exerted by either of these is pro- portional to the elongation. (See Fig. 161.) Assuming ^ ^ that there is no friction and L ^ that tlie ball is displaced to ±10 a position B by stretching ^'^' ^^^ one of the rubber bands, when released it continues to move backward and forward with harmonic motion. Problem 241. Suppose the ball in Fig. 161, held by two helical springs, to have a weight of 10 lb. and that it is displaced 1 in. from 0. The two springs are free from load when the body is at O. The springs are just alike, and each requires a force of 10 lb. to compress or elongate it 1 in. Find the time of vibration of the body and its velocity and position after ^- sec. from the time when it is released. It has been found by experiment that the force necessary to com- press or elongate a helical spring is proportional to the compression or elongation, 111. Motion with Repulsive Force Acting. — Suppose the force to be one of repulsion and to vary as the distance ; then a = ks^ and vdv — ksds^ so that These equations show that as t increases s also increases and the body moves farther and farther away from the center of force. The motion is not oscillatory. 112. Motion where Resistance Varies as Distance. — I f a body whose weight is 644 lb. falls freely from rest through 60 ft. and strikes a resisting medium (a shaft where fric- MOTION IN A STRAIGHT LINE 195 tion on the sides equals 2 ^= 10 times the distance ; see Fig. 162), since accelerating force equals mass times acceleration, a = a-2 F M a-\os s 9 It is required to find (a) the distance tlie body goes down the shaft before coming to rest ; (l>^ the distance at which the velocity is a maximum ; (c) the total time of fall ; (c?) the velocity at a distance of 10 ft. down the shaft. After striking the shaft the relation be- tween velocity and distance is as follows : £vdv^j^(g-fj,s. Fig. 162 The remainder of the problem is left as an exercise for the student. Problem 242. A ball wliose weight is 32.2 lb. falls freely from rest through a distance of 10 ft. and strikes a 400-lb. spring (Fig. 16;i). Find the compression in the spring. It is to be miderstood that a 400-lb. spring is such a spring that 400 lb. resting upon it compresses it one inch, and 4800 lb. resting on it compresses it one foot, if such compression is possible. After the ball strikes the spring it is acted upon by the attraction of the earth and the resistance of the spring. The acceleration n is then ^-4800.^ M , where s is meas- ured in feet. The relation between velocity and distance is then ob- tained from the relation, Jv„ Jo 196 APPLIED MECHANICS FOR ENGINEERS Problem 243. A 20-ton freicrht car (Fig. 104), moving with a velocity of 4 mi. per hour, strikes a bumping post. The 6U,000-lb. spring of the draft rigging of the car is compressed. Find the com- pression s. Assume that the bumping post absorbs none of the shock. Problem 244. Suppose the car in the preceding problem to be moving with a velocity of 4 mi. per hour, what should be the strength of the spring in the draft rigging so that the com- pression cannot exceed 2 in. ? Problem 245. After the spring in Problem 242 has been compressed so that the ball comes to rest, it begins to regain its Fig. 163 original form. Find the time required to do this and the velocity with which the ball is thrown from the spring. 4 Ml. PER HR. < izzWaaamaad Fig. 164 MOTION IN A STRAIGHT LINE 197 113. Motion when Attractive Force Varies Inversely as the Square of the Distance. — This is the I case of motion (Fig. 165) vvheu two Q bodies in space are considered, since in such cases the attractive force varies directly as the product of their masses and inversely as the square of the dis- tance between their centers of gravity. The same attraction holds between two opposite poles of magnets or between two bodies charged oppositely with Fig. 165 electricity. — k Suppose the acceleration = — — and that the initial velocity is zero. Then, j vdv = — 1 ■'"^ C?8, SO that and cjs__ \'lk_ Vg(^.s dt ~ ^' Sr S2 '0 Sr ^2k ■VsqS 52 — Q vers -1 The time required to reach the center of attraction from the position of rest is obtained by putting s = 0. This gives t = — I -^ It is seen that when s = 0, the velocity is infinite, and therefore the body approaches the center of attraction with increasing velocity and passes through the center, to be retarded on the other side until it reaches a distance — 8q. The motion will be oscillatory. 198 APPLIED MECHANICS FOR ENGINEERS If one of tlie bodies is the earth, of ladius r, and the other is a body of weight G- falling toward it, the equa- tions just derived hold true. In this case it is possible to determine k. The attraction on the body at the sur- face of the earth is 6r, and at a distance s is F^ so that F = Grl —]. The acceleration is therefore -— — = — g{ — \s^J M ^ W. This gives k^ then, equal to r^g. Substituting these values in the above equation, we find v = yJEE 2 VSqS — s^ When s = r at the earth's surface. V = yj'^ V(So - r) = V2 gr^fl r Sq \ 8( If 8q= cc^ v = V2 gr. But this is a value of v that cannot be obtained, since 8q cannot be infinite. So that the velocity is always less than V2^r. It is interesting to notice here that if a body were projected from the earth with a velocity greater than V2^r, it would never return, provided there were no atmospheric resistance. Substituting ^ = 32.2 and r = 3963 mi., V = 6.95 mi. per sec. This is the greatest velocity that a body could possibly acquire in falling to the eartli, and ii body projected upward with a greater velocity would never return (neglecting resistance). MOTION IN A STRAIGHT LINE 199 If the body falls to the earth from a height A, the veloc- ity acquired may be obtained from the foregoing by put- ting s = r and s^^ = li -\- r \ then I'lgrh If h is small compared to r, this may be written, without serious error, which is the formula derived for a freely falling body in Art. 105. Problem 246. A body of mass 10 lb. has an initial velocity of 50 f /s and moves in a medium which resists with a force proportional to the velocity and equal to 1 lb. when the velocity is 4 f /s. How far will the body move before the velocity is reduced to 10 f /s? Would the velocity ever become zero with that law of resistance ? What would be the limiting value of the space passed over? Problem 247. A body of weight W lb. is projected vertically upward with a velocity I^o- If the resistance of the air is equal to kv'^, prove that the height to which the body will rise is ,.JElog.(l+^o). 2 kg ^ V W/ Problem 248. The body of the preceding problem falls again to the earth. Show that the velocity with which it reaches the earth is Problem 249. A man jumps from a balloon and acquires a velocity of 80 f /s before his parachute is fully opened. The man and parachute weigh 150 lb. The resistance of the air varies as the square of the velocity (approximately) and is 1 lb. per square foot of opposing surface \vhen the velocity is 20 f /s. If the diameter of the parachute is 12 ft., what velocity will tlie man have after descending a further distance of 200 ft. ? after 1000 ft. ? 200 APPLIED MECHANIC ti FOli ENGINEERS Problem 250. On a toboggan slide of constant slope assume the friction to be constant and the resistance of the air to be proportional to the square of the velocity. Derive the formula for the velocity in terms of the distance, starting from rest. 114. Relative Velocity. — When we speak of the velocity of a body, it is understood that we mean the velocity of the body relative to the earth, more particularly the point on the earth from which the motion is observed. Since the earth is in motion, it is evident that velocity as gen- erally spoken of is not absolute velocity, and since there is nothing in the universe that is at rest, all velocities must be relative. In everyday life, however, we think of velocities referred to any point on the surface of the earth as being absolute. Suppose two particles, A and B^ to have the velocities Va and Fft, as illustrated in Fig. 166. 77ie motion of B relative to A is obtained by regarding A as at rest and B as having a velocity com- posed of the actual velocity of B and the reversed velocity of A at each instant. Thus, the vector F", which is the result- ant of Vb ^i^d ^ reversed, is the velocity of B relative to A at the instant at which the velocities are Va ''^^^^ ^"6- This is made more evident in the case of constant velocities by the following consideration : if F^ and Vj, are constant, then at the end of a unit of time A would be in the position A^^ and B in B^, and their distance apart and direction from each other would be represented by the line Aj^B^^ while if A re- FiG. 16G MOTION IN A STRAIGHT LINE 201 mained at rest, and B moved with a velocity composed of Vj, and Va reversed, the line AB' would represent the dis- tance and direction of B from A. Since A^B^ and AB' are evidently equal and parallel, the relative positions of the two particles w^ould be the same in one case as in the other. As an illustration, consider the velocity of the wind relative to the sail of an ice boat. Let F^ and V^^ be the velocities of the boat and wind re- spectively (Fig. 167) and SL the direction of the sail, wd:iich is as- sumed to be a smooth plane sur- face. Combining F^ with Vb reversed, the velocity V is obtained, which is the velocity of the wind relative to the boat. This velocity can be resolved into two components, OA and 0^, along and perpendicular to the sail respectively. The component OA will have no effect on the boat on the assumption that the sail is a smooth surface. The component OU may be resolved into two components, OB and 00, respectively, along and per- pendicular to the path of the boat. The vector OB then represents the component of the velocity of the wind which urges the boat forward. Fig. 167 202 APPLIED MECHANICS FOR ENGINEERS From the figure it is clear that there may be a forward component of the wind's velocity on the boat even when the velocity of the boat is greater than the velocity of the wind. It is only necessary for the velocity of the wind relative to the boat to fall in front of the sail. Then as long as the forward pressure of the wind is greater than the resistance of the ice the velocity of the boat will increase. In the case of ice boats, where the resistance is small, the velocity of the boat may greatly exceed the velocity of the wind for high wind velocities. Problem 251. A train is moving with a speed of 60 mi. per hour, and another train on a parallel track is going in the opposite direction with a speed of 40 mi. per hour. What is the velocity of the second train as observed by a passenger on the first? Problem 252. A man in an automobile going at a speed of 40 mi. per hour is struck by a stone thrown by a boy. The stone has a velocity of 30 ft. per second and moves in a direction perpendicular to the direction of motion of the automobile. With what velocity does the stone strike the man? Problem 253. A man attempts to swim across a river, \ mi. wide, which is flowing at the rate of 3 mi. per hour. If he can swim at the rate of 4 mi. per hour, what direction must he take in swim- ming in order to reach a point directly across on the opposite shore. What distance will he swim relative to the water ? How long will it take for him to cross ? Problem 254. An ice boat is moving due north at a speed of 60 mi. per hour, and the wind blows from the southwest with a velocity of 40 mi. per hour. What is the apparent velocity and direction of the wind as observed by a man on the boat? Problem 255. A man walks in the rain with a velocity of 4 mi. per hour. The raindrops have a velocity of 20 ft. per second in a direction making 60° with the liorizontal. How much must the man incline his umbrella from the vertical in order to keep off the rain : («) MOTION IN A STRAIGHT LINE 203 when going against the rain, (b) when going away from the rain ? If he doubles his speed, what change is necessary in the inclination of his umbrella in (a) and (b) ? Problem 256. The light from a star enters a telescope inclined at an angle of 45° with the surface of the earth. The velocity of light is 186,000 mi. per second and the earth (radius 4000 mi.) makes one revolution in 24 hr. What is the actual direction of the star with respect to the earth ? This displacement of light due to the velocity of the earth and the velocity of light is known as aberration of light. Problem 257. A locomotive is moving with a velocity of 40 mi. per hour. Its drive wheels are 80 in. in diameter. What is the tangential velocity of the upper point of the wheels with respect to the frame of the locomotive? What is the tangential velocity of the lowest point? Problem 258. Show that if an ice boat were moving north and the wind had a velocity of 20 mi. per hour from the southwest, the maximum velocity that the boat could attain if there were no friction would be 38.6 mi. per hour if the sail w'ere set at 30° with the direction of the boat's motion. What effect would increasing the angle between the sail and the boat's direction have in this case? Problem 259. With the sail set at 30° with the direction of the boat's motion, in what direction could the boat go fastest if there were no friction? Ans. AVhen the direction of the boat makes an angle of 60° with the direction of the wind. Problem 260. Given that the angle which the sail makes with tlie direction of motion of the ice boat is SO"^, the angle which the direction of the wind makes with the direction of the boat's motion is 45°, the area of the sail is 50 sq. ft., the resistance of the ice to the boat's motion is 16 lb., and given that the pressure of the wind normal to the surface of the sail varies as the square of the velocity (component of relative velocity normal to sail), and is 1 lb. per square foot of opposing surface when the velocity is 15 mi. per hour, find the maximum velocity that the boat can attain when the velocity of the wind is (a) 20 mi. per hour, (b) 30 mi. per hour. Ans. (a) 14.6 mi. per hour. (/;) 31.0 mi. per hour. CHAPTER X CURVILINEAR MOTION 115. Velocity. — Suppose a particle to be moving along the curve of Fig. 168, from A toward B. The average velocity of the particle be- tween A and B is defined as the velocity the particle would have if it moved uni- formly from A to B, i.e. along the chord AB with con- stant speed. The average velocity of the particle between A and B is therefore, 1 ., chord AB average velocity = -— - • time spent between A and B If the chord is represented by Ac and the time by At, then Fig. 168 average velocity = — , Ac and is represented by a vector A (7, of length — , laid off from A on AB. When the time At is made to approach Ac the limit zero, the limiting" value of the ratio — is defined ^ At to be the velocity of the particle at the point A. 204 CURVILINEAR MOTION 205 The limiting direction of the chord is the tangent at A. Hence, calling v the velocity of the particle at the point A^ dc V = — dt dc and is represented by a vector of length — laid off from A along the tangent in the direction of motion. It is shown in calculus that if a single tangent to the curve at A exists, then the limit of the ratio of the arc to its chord, as the arc approaches the limit zero, is unity. Hence, if tlie arc AB is As, then c?s _ 1 dc 1 dc ds ds and v = — — = —-' at dc dt The velocity of a particle moving along a curved path is therefore represented at any point of the path by a vector equal to the value of — laid off on the tangent to the path dt at the given point, and in the direction of the motion. 116. Acceleration in a Curved Path. — Let the velocities at A and B (Fig. 169 (a)) be v and v -\- Av respectively. Lay off the vectors representing these velocities from the same point A (Fig. 169 (^)). The vector CD, from the end of v to the end of v -\- Av, represents the change in velocity in the time A^. Tlie ratio of this vector to the time At is the average accelera- tion for the time A^. Thus the vector CJE represents the average acceleration of the particle between A and B. 206 APPLIED MECHANICS FOR ENGINEERS The limiting value of this average acceleration as A^ approaches zero as a limit is defined to be the acceleration of the particle at A. Letting v, and v, be the projections Fig. 169 of V upon rectangular axes, then the projections of CD on these axes are Av^ and Avy and the projections of CU are ^^ and ^^ Hence A^ A^ and calling a the acceleration of the particle at A, \dtj \dtj If a and a^ are the components of a along the axes, then dv. dv. dt a,. = iiiw. dt Now v^ = v cos and Vy = v sin 6 CURVILINEAR MOTION 207 where is tlie angle which the tangent to the curve at A makes with the .i^-axis. Therefore a^ = cos D V sm u — , dt dt • ndv , ^ a^ = sm u —- + V cos 6 dt d6 dt Since V = ds dt dt dt^ dO A value of ^ may be found as follows : Draw a normal dt to the curve at A (Fig. 170). On this normal there is a point 0', the center of a circle tangent to the curve at A and passing through B. Let r' be the radius of this circle and A^' the angle subtended at 0' by 0' the arc AB. As B i^^Nv j / is made to approach ^ as a limit the tan- \\aAO \s^ ¥ gents to the curve and the circle at B \ "^^/a^,^^^ take the same limit- \\ ^ \ ing position and \\^^y : Fig. 170 Ai9' approaches 1 as a limit. The ratio of the arcs, As' of the circle, and As of the curve, has the limit 1, since they have the same chord and the limit of the ratio of each arc to its cliord is 1. As the point B is made to coincide with A the point 0' takes some limiting position, 0, wliich is called the center of curvature of the curve at A. If r is the value of Ovl, then limit r' = r. 208 APPLIED MECHANICS FOR ENGINEERS Now r'AO'^As', or^ = l. As' r' rrtu £ dO' 1 . d6 ^ 1 6?S -, Ineretore — — = -, or since — - = 1 and — - = 1, ds r du' as' ^ = 1 ^dl^dSd^^lch^v^ ds r dt ds dt r dt r The quantity r is the radius of curvature of the curve at A. Substituting the values found for — and ^, the values 01 a^ and ay become ax = cos e ^— 9 ^ - oi,. A ^^"* ^ ^^ cos 9 <*« = sin y — — - H • 117. Tangential and Normal Components of the Accelera- tion. — The above values of a^ and ay hold for any rectangu- lar axes. Let the a^-axis be taken along the tangent at A and the «/-axis along tlie normal. Then cos^=l, and sin ^ = at the given point, and the values of a^ and ay be- come respectively The normal force and tangential force may now be written, Normal force = ^^^ ; r Tangential force = 3r ^. For all curves except the circle r, the radius of curvature varies from point to point. In the circle, however, it is the radius, and is therefore constant. Tn this case the normal force is usually called tlie centripetal force. CURVILINEAR MOTION 209 Problem 261. A ball weighing 2 lb. is attached to one end of a string 5 ft. long, the other end of which is attached to a fixed point. The ball is pulled aside and released. When the string makes an angle of 30'^ with the vertical, the velocity of the ball is 10 f/s. Find the normal and tangential components of the acceleration and the accel- eration in magnitude and direction at that instant. Find also the ten- sion in the string. (Notice that the force acting tow^ard the center on the body is the difference between the tension in the string and the component of the weight along the direction of the string.) 118. Uniform Motion in a Circle. — A body moving with constant speed, v^ in the circumference of a circle is acted upon by one force, the normal or centripetal force, and this equals That this is true is evident when it is re- membered that the tangential velocity is constant, thus making the tangential accelera- tion zero. An illustration of uniform motion in a circle is seen in the case of the simple governor shown in Fig. 171. When the speed is constant, then a, A, and r are constant. Let T be the tension in the rod supporting the ball, then, since there is no vertical motion 2 y = 0, so that T cos « = a. Fig. 171 Jii;2 Considering the normal force, we have T sin a = , so r thattana=— . From these equations 7^ may be found for any values of a and r. 210 APPLIED MECHANICS FOE ENGINEERS Problem 262. The weighted governor shown ill Fig. 172 is rotated at such a speed that a =30°. Find the forces acting on the longer rods and the stress in the shorter rods. The connections are all pin connections. Problem 263. A type of swing is shown in Fig. 17o. A revolving central post supported by wires ^l and B carries six cars G, each suspended from cross arms D by means of cables 50 ft. long. When the swing is at rest, the cars hang vertically and a = 0; as the speed of rotation increases, a becomes larger. Suppose the car and its load of four passengers to weigh 1000 lb., and the speed to be such that «;= 30°; find the tension in the cables supporting the cars. Assume that a single car is carried by one cable. Problem 264. The same principle that has been seen to hold for motion in a circle enables us to solve a problem that comes up in railroad work. When a train goes around a curve, it is desirable to have the outer rail raised suffi- ciently so that the Fig. 172 CURVILINEAR MOTION 211 wheel pressure will be normal to the rails. It is really the same problem as Problem 263, where the sustaining cable is replaced by a track. (See Fig. 174.) Let r be the radius of curvature, v the veloc- ity of the car, of weight G. Show that the superelevation of the outer rail is given by tan a = — , and so, approximately, h= — , gr where cl is the distance between the rails in feet, r the velocity in feet per second, g is 32.2, ;• is the radius of curvature in feet, and h is the superelevation of the outer rail in feet. Problem 265. Show that, using d = 4.9 ft., this height may be expressed, approximately, as follows : Fig. 174 h = 3r where h and r are in feet and i\ is the velocity in miles per hour. Problem 266. Find the superelevation of the outer rail on a curve of radius 2000 ft. for speeds of 20 mi. per hour, 40 mi. per hour, 60 mi. per hour. Problem 267. What would the superelevation need to be for the above speeds on a curve of radius 500 ft. ? 119. Motion without Friction along" Any Curve in a Vertical Plane. — Let a particle slide down the smooth curved track (Fig. 175), start- ing from rest at the point A dis- FiQ. 175 tant 1/q above a 212 APPLIED MECHANICS FOR ENGINEERS horizontal line of reference. When in the position P, the forces acting on the particle are the weight (r, acting ver- tically downward, and the force exerted by the track, which is normal to the track since there is no friction. The tangential force is then Gr sin 6, and hence for de- scending motion, ---= a sin (9, g dt^ where s is the length of the curve AP measured from A toward P. XT • /} dy -y -i (Ps dv Now sm u = — f- and also = v --. ds dt" ds mi £ dv dy Ihereiore v— - = — g^. ds ds Integrating, -=: -gg-\- C. When y =iy^^ v=0 \ therefore C = gy^. Therefore Hence, the velocity acquired by a body in sliding down a smooth track is the same at any point as if the body had fallen freely through the same vertical distance. The time of descent depends upon the form of the curve. Problem 268. Show that the equation vdv — — gdy holds when the body is ascending as well as when descending, and then prove that the body would rise on the track to the level from which it started. CURVILINEAR MOTION 213 Problem 269, A body of weight G starts from rest at the top of a smooth circular track of radius R in a vertical plane. Find the velocity, acceleration, and force exerted by the track on the body when it has traveled through 60°, 90°, 150°, 180°, 210° of arc. Problem 270. AVith what velocity would a body have to be pro- jected from the lowest point of a smooth circular vertical track in order that its velocity would just keep it from falling from the track at the top? Problem 271. If the body of the preceding problem were pre- vented from leaving the track, what velocity at the lowest point would just carry it around the track ? Problem 272. In the centrifugal railway (Fig. 176), friction neglected, what would have to be the ratio of h to A' so that the car would not leave the track at .4, starting from rest at the height h ? Fig. 17(i Problem 273. If the car in the preceding problem were prevented from leaving the track, what would be the ratio of h to h' to just carry the car past A ? Problem 274. A particle slides from rest from any point on a smooth sphere. Show that it will leave the sphere when it has de- scended vertically through one third of its original vertical distance above the center. 214 APPLIED MECHANICS FOR ENGINEERS Problem 275. A locomotive weighing 175 tons moves in an 800- ft. curve with a velocity of 40 mi. per hour. Find the horizontal pres- sure on the rails, if they are on the same horizontal. Problem 276. If the velocity of the earth was 18 times what it actually is, show that the force of gravity would not be sufficient to keep bodies on the earth near the equator. Take the radius of the earth as 4000 mi., and assuming the above conditions, find at what latitude the body would just remain on the earth. Problem 277. A pail containing 5 lb. of water is caused to swing in a vertical circle at the end of a string 3 ft. long. Find the velocity of the pail at the highest point so that the water will remain in the pail. Find also the velocity of the pail at the lowest point. 120. Simple Circular Pendulum. — The simple circular pendulum consists of a weight Gr suspended by a string without weight, of length Z, in such a way that it is free to move in a circle in a vertical plane due to the action of gravity. (See Fig. 177.) Let B be such a position of the pendu- B lum that its height above the horizontal is h, and (7 any other po- sition designated by the coordinates x and 7/. Let the weight be G and the tension in the string T. These are the only forces acting on the body. The only forces that can produce motion in the circle are those that are tangent to the circle G Fk;. 177 CURVILINEAR MOTION 215 OB. The force T is normal to the circle and so has no tangential component. The force G has a tangential com- ponent — Gr sin 0. The equation of motion is, therefore, vdv = ads = — g sin Ods, arc (J \ 9 where ^ = — =-, where s denotes distance along the curve. Since s = 16, ds — Idd, and hence vdv = — Ig sin 6d6. (1) Approximate Solution. The integration of this equa- tion leads to a complicated relation between the angle and the time. An approximate solution can be obtained for small angles of oscillation by replacing sin 6 by 6. Equation (1) then becomes vdv = - IgOdd. Integrating, v^ _ _ /^^2 _|_ q^ Let v= when 0=9^; then C=^lge\. .-. v= ^Ig-VOl-d^. For the body descending s decreases as t increases and therefore — is neo^ative, and dt dt dt ^1 dO or V6>2 - e^ ^-yR-dt. Integrating, sin~^— = — \-t+ Cj. 216 APPLIED MECHANICS FOR ENGINEERS Let ^ = when 6=0^; then sin~^l = C^ From the angles whose sines are 1 choose — = sin~U= Oy Then ^.« = |_sin->^. (2) As the pendulum swings from the highest point on the right to the point on the left on the same level (Art. 119), i.e. to where = — 6^, the value of t continually increases. 9 Equation (2) shows that sin"^— must continually decrease }_ during this time. Therefore sin ^^ must decrease in this rrr interval from — to sin~i ( — 1), which therefore can only be — — . Hence if t is the time of a single oscillation, 4 ■l-l-c-i) € or t = IT- ''9 This value of t is called the period of vibration. General Solution. For large vibrations the above ap- proximate solution is not sufficiently accurate. Integrating equation (1), vdv = — gl sin OdS., we obtain ~^ — 9^ ^^^ -\- 0. When i; = 0, 6> = B^. .'. C = — gl cos 6^ and ?;2= 2^Z(cos ^ — cos ^j). CURVILINEAR MOTION 217 .♦. l^= -V2^^ (cos (9 -COS (9,) at or -7====== - \/^ 6?L ^^ VcOS ^ — COS 0^ ^ I Q Since cos 6 = 1 — 2 sin^ -, this equation may be written dd 4 = -2yjS.dt. 0. • 2 ^ ^ ^ — sm^ - 2 2 If ^j is the time that it takes for the pendulum to descend from e = d^to 0= 0, then \/sin^-^— sin^ - >/ 2 2 ^2 2 The time of vibration, t^ is double this, or ^^J7 r''^ c^<9 -vr ^Sin2^-sin2- Change the variable by the substitution .6 • d. . . Sin - = sin—i sm V 2 218 APPLIED MECHANICS FOR ENGINEERS This integral is an ellipticititegral. The expression can be integrated only by expanding into a series and integrating term by term. Thus, letting k = sin -i, TT = 2\f^ r ( 1 + -A;2 sin2 ch + ^^ -k^ sin^ 6 ^gJo' 2 ^1.222 ^ 1.2. 3 23 From the reduction formula sin"^ a;c?x = -\ I sin"*"^ xdx n m ^ there follows IT TT P- m J m-in. ^_2 -, _ (m-l)(7^-3)....S.l 7r I sm"* xdx = I sin"' ^ a^aa; — ; ^ ] 7. ^ •> and hence where ^ = sin -^* 2 Problem 278. Using the series, compute correct to three figures the time of vibration of a simple pendulum of length 3 ft. when the jiendulum swings through 20°. Compare with the time obtained from t = TT-v'— CURVILINEAR MOTION 219 Problem 279. Compute the time of vibration of a simple pendu- lum of length 4 ft. when swinging through an angle of 80^. Problem 280. A pendulum vibrates seconds at a certain place and at another place it makes 60 more vibrations in 12 hours. Compare the values of g at the two places. Problem 281. Show that in the simple pendulum for any angle of swing the value of the tension in tlie string is t=g[' + -'1-'''), where h and ij have the meaning given them in Art. 120. Hence show that the tension is equal to the weight w^hen the weight has descended through one third the original vertical distance. Problem 282. The weight of a chandelier is 300 lb., and the dis- tance of its center of gravity from the ceiling is 16 ft. Neglecting the weight of the supporting chain, find how much the tension in the chain will be increased if the chandelier is set swinging through an angle of 8^ measured at the ceiling. 121. Cycloidal Pendulum. — It has been found that a pendukim may be obtained whose period of vibration is constant by allowing the string to wraj:) itself around a cycloid as shown in Fig. 178. The pendulum hangs from tlie point A. AB and AC are cych)idal guides around which the string wraps as the penduhim swings. This causes the length of the pendulum to continually cliange and the pendulum '"bob " to move in another cycloidal curve COB. The equation of this curve referred to the axes X and y is a: = ^ vers i-^jA ~^\^^~ .^^• In Art. 119 it was seen that v^=2g {h — y) represented the velocity of a body moving in a vertical curve when 220 APPLIED MECHANICS FOR ENGINEERS only the force of gravity and a force normal to the path of the curve acted. We may make use of the equation in r Fig. 178 this case, since the same conditions exist. We may write 7. y/dx^ + dy^ . ds at = — '' '— , since v = — V2g{h-i/) dt From the equation of the curve, we find dx = ^ ~2^ • d^^ so that 2y taking the negative sign, since ^ is a decreasing function of y. Therefore ^=\/ — CURVILINEAR MOTION 221 r 9 -1 2 y vers ^ — ^ h The whole time of vibration is twice this value, so that the time of vibration This expression is independent of A, so that all vibrations are made in the same time. The motion is therefore isochronal. Problem 283. Using the calculus formula for radius of curva- ture, 3 r = (h/Yl^ show that the radius of curvature of the cycloid of the above article at any point is r=V/(/-2?/), and hence show that the tension in the string for any position is where G is the weight of the pendulum bob. Problem 284. Assuming a value of h in terms of Z, say h = - , 4 plot the curve representing the tension in terms of //. For what position of the pendulum is the tension a maximum ? Show that for any swing the maximum tension is T=o['-±p) Problem 285. A particle slides from rest down an inverted cycloid. Prove that its vertical velocity is greatest when it has de- 222 APPLIED MECHANICS FOR ENGINEERS scended through one half the original vertical distance above the lowest point of the curve. What is the vertical velocity at that point? 122. Motion of Projectile in Vacuo. — A method, slightly different from the preceding, of dealing with a problem of curvilinear motion, is il- lustrated in the present ar- ticle. It is de- sired to find the path taken ^lo. 179 by a body pro- jected with a velocity Vq at an angle of elevation a, when the resistance of the air is neglected. (See Fig. 179.) Let the point of projection be taken as origin and the 2;-axis horizontal. Then, since there is no horizontal force act- ing on the body, a^ = 0, so that d X /-v doc and — = constant = v^ cos a, therefore x = VqCos a{t). In a similar way we know that the vertical acceleration ay = — g^ since the only force acting is G. Then, g = -, and -^ — — qt-\- constant. dt CURVILINEAR MOTION 223 This equation may be rewritten Vy= — gt -\- constant. To determine this constant of integration, we put f = 0, ajid Vy = Vq sin a ; therefore -^ = — gt -\- v^ sin a dt and y = — ^gt^ -{- Vq sin a(^). Eliminating t between the equations in x and ?/, we get y^xtana- f ^" ^ 2 t^o cos- a as the equation of the path of the projectile. This is evi- dently a parabola, with its axis vertical. Range. To find the range or horizontal distance d we put g = ; then x — and x = d^ .2. J.1 i , VnSm2a so that d = — . a From this it is clear that the greatest range is given when a = 45°, since then c? = — . The Greatest If eight. The greatest height to which the v sm 2 ot projectile will rise is found by putting x = -^^ in the equation of the curve and solving for g. This gives , vl sin^ a and the angle that gives the greatest heiglit is a = 90°. For this case h = -—. Tliis is the case tliat has already 2^ 224 APPLIED MECHANICS FOR ENGINEERS been considered under the head of a body projected verti- cally upward. Problem 286. A fire hose delivers water with a nozzle velocity vq, at an angle of elevation a. How high up on a vertical wall, situated at a distance - — 1 ^ = ^-T7 e r 4-1 = ^> 1- e r +1 This formula shows that v remains always less than the constant quantity b. On account of the velocity not exceeding a fixed value this form of chute is useful in sending packages from one floor of a building to another. Problem 298. Given a helical chute of radius 3 ft. and pitch 2 ft., find the velocity of a small body descending from rest after 5 sec, after 10 sec, if the value of c is .05. CHAPTER XI ROTARY MOTION 125. Angular Velocity and Angular Acceleration of a Par- ticle. — If a particle moves in any plane curve, the rate at which a line joining the particle to a fixed point in the plane is turning is called the angular velocity of the particle with respect to that point. Thus, if 6 is the angle which the line ioininef the particle makes Fig. 184 j &> r with a fixed line through (Fig. 184), the angular velocity, o), of the particle with respect to is dt The rate of change of the angular velocitj^ is called the a^igular acceleration of the particle with respect to the point. Thus, if a is the angular acceleration of the particle with respect to (9, a = — - or a = , dt' dt' The angular acceleration may also be written in the form d(o dO dta CC = — , or a = 0) — . dO dt d9' 232 ROTARY MOTION 233 The angular velocity and angular acceleration of a particle moving in a given path with a given speed de- pend upon the point to which they are referred. Problem 299. Prove that if a particle moves around a circle, the angular velt>city of the particle with respect to a point on the circum- ference is equal at any instant to one half the angular velocity of the particle with respect to the center of the circle, and hence that the angular velocity of the particle at any instant is the same with re- spect to all points on the circumference. Show also that a like statement holds for angular acceleration. Problem 300. Show that if a particle have constant angular acceleration with respect to any point, the following formulae hold : t where 6 is the angle turned through in the time <, Wq the initial an- gular velocity, and w the angular velocity at the instant t. Compare these formulae with those of Art. 104. Problem 301. A flywheel making 100 revolutions per minute is brought to rest in 2 min. Find the angular acceleration a and the angle 6 turned through before coming to rest. Problem 302. A flywheel is at rest, and it is desired to bring it to a velocity of 300 radians per minute in \ min. Find the angular acceleration « necessary and the number of revolutions required. What is the angular velocity 2 : Wj, the velocity of P is zero. The locus of all such points is clearly a straight line through 0. Hence there is a straight line, OP, in the plane of the two intersecting axes of rotation wliich is at rest at the given instant, and about which the body is rotating at that instant. To find the angular velocity, co, of the body about this instantaneous axis OP, consider the motion of the point ROTARY MOTION 1^41 on OA. Draw CU perpendicular to OB and OF per- pendicular to OP. The point C has angular velocit}^ 0)2 about OB, and hence its velocity is UCco^. Its angular velocity about OP is o) and hence its velocity is FCco. .-. FCco=EC(o^. Now jFe= 0(7 sin a and ^0= 0(7 sin (9. (o sin « = ftjg sin ^. a) Again, ^(7 = rj cos a, and EC =r^-\-r^ cos ^. Therefore r^co cos a = ^20^2 + t^w.^ cos 6, But and hence ^2®2 = ^\^v CO cos a = ft)j + 0)2 COS 6. Squaring and adding equations (1) and (2), «■- = 0)2 + «2 -I- 2 «i«2 cos 6. Dividing (1) by (2), tan a = (1)2 sin ^ Wj + cOg cos ^ (2) (3) (4) Equations (3) and (4) are exactly the equations that would be obtained by re- garding Q)j and 0)2 as vec- tors and finding their re- sultant as in Fig. 193. Hence we may represent angular velocity of a body about a line by a vector laid off on that line, equal in length to the numerical value of the angular ve- 242 APPLIED MECHANICS FOR ENGINEERS locity, and may combine simultaneous angular velocities about two intersecting lines by combining their vectors in the usual way. The resultant vector represents the value of the resultant angular velocity and coincides with the instantaneous axis of rotation of the body. The vector representing an angular velocity is made to point opposite to the direction from which the rota- tion appears counter-clockwise to an observer looking along the axis of rotation. It is evident at once, then, that the angular velocity of a body about a line may be regarded as made up of component angular velocities about two or more axes intersecting that line, obtained by using the parallelo- gram law of vectors. In particular a rotation about an axis may be re- garded as made up of the simulta- neous rotations about three axes at / ^^ I right angles to each other and inter- secting the given line (Fig. 194). If the angular velocity of the body about the line is «, the angles which this line makes with the three rectangular axes, a, /S, 7, and the angular velocities of the body about these axes w^, (Oy, &>, respectively, then co^ = 0) cos a, (Oy= CO cos yS, o)^= co cos 7, and ft)2 = ft);. + ft)2 + ft)2. Problem 317. Prove that the vector representation holds for simultaneous rotation about two parallel axes, i.e. that the resultant angular velocity of Wj and o)„ is Fig. 194 ROTARY MOTION 243 and that the instantaueous axis of rotation divides the line joining the axes of Wj and ojo in the inverse ratio of o)^ and w.,. (See Fig. 195.) Problem 318. A wheel is making 120 r. p. m. about a shaft and the shaft at the same time is making 90 r. p. m. about a line perpen- dicular to the axis of the shaft. Find the instantaneous axis of the shaft at any instant. ■r^-^ -fl — ^^^ — >- ■0: i? C VcJo D ■ff^ Fig. 195 Fio. lOfi Problem 319. A body has simultaneous rotations (o^= 1 tt rad./sec, W2 = 3 TT rad./sec, about the parallel lines of Fig. 196 in the direc- tions indicated. Locate the instantaneous axis of rotation and find the angular velocity of the body about that axis. 133. Rotation of the Earth. Foucault's Pendulum. — T.et (o be the angular velocity of the earth about its axis of rotation CF (Fig. 197). This angular velocity can be resolved into two components about rectangular axes CA and CB. If X is tlie latitude of the position of A^ the component of o) about CA is (o sin X. If a plane containing CA could be fixed so as not to turn about CA, tlie rotation of the earth about CA would be indicated by the apparent rotation of the plane in the Fkj. liiT 244 APPLIED MECHANICS FOR ENGINEERS opposite direction about (7A, since our directions are measured with reference to the earth's surface. Such a phine is the plane of vibration of a heavy weight suspended by a long wire. Foucault was the first to demonstrate the rotation of the earth by this method. Problem 320. Show that the rotation of the earth about a di- ameter through a point on the equator is zero. Problem 321. Show that the time for the plane of the Foucault'.s pendulum to turn through 360^ is — — hours, where A is the latitude sin A of the place. Problem 322. The plane of a Foucault's pendulum is observed to turn through 26^ in 2 hr. 20 min. Find the latitude of the place. CHAPTER XII WORK AND ENERGY 134. Definitions. — If a force, constant in niacfnitude and direction, acts at a fixed point of a body, the force is said to do work on the body wlien the point of application of the force has a displacement with a component in the direction in which the force acts. The product of the force and the component of the dis- placement in the direction of the force is defined to be the work done on the body by the force in that displace- ment. If there is a displacement opposite to the direction of the force, work is said to be done against the force. For example, in Fig. 198, let the forces P, 72, (r, iVact upon the block as the block moves from the position A to b'm. i\)S the position B. If 5(7 is perpendicular to the line of P, the displacement in tlie direction of P is AC and tlie work done by P on the block is P - AC. If AB = s, then AC=8 cos 6, and llu? work done by P is Ps cos 0. Tliis 245 246 APPLIED MECHANICS FOR ENGINEERS may be written P cos 6 - s. But P cos 6 is the component of P in the direction of the displacement of the point of application, and therefore the work done by P is equal to the product of the displacement of the point of appli- cation and the component of the force in the direction of the displacement. Since iV and Gr are at right angles to the displacement, no work is done by iVor Gr. Since the displacement is opposite to the direction of i?, work is done against i2, the amount being Rs. If the body had moved from P to ^ during the action of the given forces, the work done by M would be Ps, and the work done against P would be P cos -s . If the point of application changes in the body, the work done on the body is the product of the force and the displacement, the displacement meaning the actual displacement of the force in space plus the relative dis- placement of the moving point of application past the initial point of the force vector in the direction opposite to the direction in which the force acts. If this relative dis- placement is in the direction in which the force acts, it must be subtracted from the displacement of the force in space. For example, in jj Fig. 199 suppose a force P, applied to the block P, drags P along" A and at the '^ & - ^ > Fig. 199 aion^ same time moves A. Let P be the forward force exerted by P on A. If when A moves forward a distance «, B moves relative to ^ a distance s', then the relative dis- WORK AND ENERGY 247 placement of the point of application past the initial point of the force is s' in the direction in wliich the force acts, and hence the work done by F on A is F(^8-\-s' — s') or Fs. On the other hand, a force F acts backward on B and the block B does work against this resistance to an amount equal to ^ (s + s'). Hence the total work done against the forces of resistance between the bodies is Fs\ or the product of the force and the distance one body moves rel- ative to the other. 135. Friction Forces. — Wherever the surfaces of two bodies come in contact and there is motion of one body along the other, or any force tending to produce such motion, each body exerts upon the other a force along their common surface, a force tending to prevent the rela- tive motion of one surface along the other. This resist- ing force is known as friction. The laws of friction are discussed in a later chapter. In general the law of fric- tion of unlubricated surfaces may be expressed by saying that the force of friction for two given surfaces is pro- portional to the normal pressure between the surfaces. The ratio of the force of friction to the normal force is called the coefficient of friction. Consider the work done ao'ainst fric- tion on an axle rotating in a fixed bear- ing (Fig. 200). Here tlie point of ap- ^iq. -200 plication of the friction force, F, changes in the body while the initial point of the force vector re- mains fixed in space. The displacement is then the dis- 248 APPLIED MECHANICS FOR ENGINEERS placement of the outer points of the axle past the fixed point at which F acts. The work done on the axle by friction, in resisting its motion, is therefore the product of jP and the distance through which a point on the circum- ference moves; or in turning through an angle d the work done is Frd. If the bearing is also in motion, as in Fig. 201, then if the bearing moves forward s ft. while the axle rubs a distance s' ft. past Fig'201 ^^^® bearing, the work of F on the bearing is Fs, and the work done against F on the axle is F(s-\-s'), and hence the total work done against friction is Fs' . 136. Units of Work. — The unit of work involves the unit of space and the unit of force. If the distance is measured in feet and the force in pounds, the unit of work is the work done by a force of 1 lb. when the displacement in the direction in which the force acts is 1 ft. This unit of work is called the foot-pound, designated hy ft.-lb. In the c. g. s. system the ei-g is defined as the work done by a force of 1 dyne when the displacement in the direction in which the force acts is 1 cm. The gram- centimeter is the work done by a force of 1 gram working through a distance of 1 cm., etc. 137. Work of Components of a Force. — Let the constant force P (Fig- 202) acting on a body have the displacement WORK AND ENERGY 249 Fig. 202 ABj or 8. Resolve P into any two rectangular com- ponents, X and Y. In the given movement the displace- ment of -X" is AC, or x, and of Y is AD, or y. The work done by P in the given dis- placement is P cos 6 ' 8. But P cos 6 is the projec- tion of P on AB and is there- fore equal to the sum of the projections of X and Y on AB; i.e. P cos 6 = X cos a + !F sin a. .-. P cos 6 ' s = Xs cos « -|- Ts sin a Therefore, ^Ag ?^;orA: do7ie hy a constant force in any dis- placement is equivalent to the work done by any rectangular components of the force in the same displacemeyit. Problem 323. A weight of 20 11). is dragged 50 ft. up a plane inclined 30^ to the horizon by a constant force, P = 25 lb., acting at an angle of 15"^ to the plane. There is a retarding force, R = 5 lb., along the plane. Compute the work done by, or against, each force acting on the body. Find the work done by the horizontal and vertical components of P in the given displacement, and also the work done by the components of W perpendicular to and parallel to the given plane. If the body started from rest, what velocity does it have at the end of the 50 ft. ? Problem 324. A block, yl, weighing .50 lb., is pulleile going at the speed of 30 mi. per lionr comes to the foot of a lull. The power is then shut ofl and the machine allowed to "coast" up hill. If the slope of the hill is 1 ft. in 50, how far up the hill \vill it go, if friction acting down the plane is .06 G', where G is the weight of the machine? 144. Pile Driver. — A pile driver consists essentially of {I hammer of weight G so mounted that it may have a free fall from rest upon the pile (Fig. 211). The safe load to be placed upon a pile after it has been driven is the problem that interests the engineer. This is usually determined b}^ driving the pile until it sinks only a certain fraction of an inch under each blow, then the safe load is a fraction of the resistance offered by the earth to these last blows. The resistance is usually small when the pile begins to penetrate the earth, but increases as penetration proceeds, until finally, due to the last blows, it is nearly constant. If we regard the hammer (r as a freely falling body, and consider the hammer and pile as rigid bodies, and further assume, as is usually done, that R for the last few blows is constant, we may write the work-energy equation. ]■ \] t. Fig. 211 since the final kinetic energy is zero and the weight of the hammer as a working force is negligible. Tlie distance 8^ is the amount of penetration of the pile for the blow in 260 APPLIED MECHANICS FOR ENGINEERS question. But v^ = 2^A, so that i Mvi = ah. We have then as the value for the supporting power of a pile, 11 = ^. A safe value, M' from ^ to |^ of i2, is taken as the safe sup- porting power of piles. The factor of safety and the value of S;^ for the last blow are usually matters of specification in any particular work. This is the formula given by Weisbach and Molesworth. Other authorities give formulse as follows : Trautwine, R = 60 (r VA, if s^ is small, 5 a dt and the angular acceleration as dot a = 9 dt which may also be written Fig. 213 °- " ^p' 1 du They are the forces which acting on c?iff separated from the body would give it the same motion that it has as part of the body. If a^ is the tangential acceleration of dM and a is the angular acceleration of the body, then dT=dMat = radM. The forces dN and dT 'dve the components of the result- ant of all forces acting on d3I. The forces are made up of the action of forces exerted by adjacent particles on dM and any external forces that act on dM. The forces ex- erted by the particles on each other are assumed to be in the lines joining the particles, and to be equal and opposite, according to Newton's laws. Hence in any summation of all the forces dT and f?iV, acting on all the particles, the reactions of the particles on each other would annul and leave only the summation of the impressed forces. There- 268 APPLIED MECHANICS FOR ENGINEERS fore the sum of the moments of the effective forces about the line AB is equal to the sum of the moments of the im- pressed forces about that line; i.e. /• rdT = 2(mom of impressed forces) = SPa. But dT= radM. .'. ^Pa = Cr'^adM= aCr'^dM, or 'ZPa = « J, where / is the moment of inertia of the body about the axis of rotation. The name moment of inertia is suggested for I r^dM since it is seen to be equivalent to the moment of a force which would produce unit angular acceleration of the body about the given axis, against the inertia of the body. The work done by any force P during the rota- tion through the angle dd is Pcos(;)c?s(Fig. 215). But ds = rdd^ and cos (^ = - . Fig. 215 and hence the expression for the work may be written j PadQ., and the total work done by all the impressed forces for any rotation becomes ^(^Pa)de. WORK AND ENERGY 269 It was shown above that ^Pa = a/, and hence the work done may be written d CO 1 1 adO, or since a= co ^ du (Odo) = ^ X(w| — 0)2), where Wq and w^ are the initial and final angular velocities of the body for the given displacement. It follows that the work that the rotating body, with angular velocity coq, could do against resisting forces before coming to rest is I i^o- That is, the kinetic energy of a body rotating about a fixed axis with angular velocity a> is ^ i&)2, where /is the moment of inertia of the body about the axis of rotation. As an illustration, let us consider the case of two weights (See Fig. 216), a^=20 lb. and a^ = 10 lb., suspended from drums rigidly at- tached to each other and of radii 3 ft. and 2 ft. respectively. Let the weight of the two drums and shaft be 644 lb., and the radius of gyration 2 ft. The radius of the axle is one incli and the axle friction 30 lb. The friction acts tangentially to the axle. Assuming that the initial angular velocity &)q is one radian per second, and the final anguLar velocity 18 radians per second, how many revolutions will the drums make? Fig. 216 270 APPLIED MECHANICS FOR EJNGINEERS The external forces actiiiuf on the drains are the ten- sions, 7\ and T^^ in the cords attached to the weights G^ and G^ respectively, the reaction of tlie axle, and the force of friction. Acting on the weight G^ are the forces (tj and 2\ and on G2 the forces (rg and T^- The work-energy equations for these three bodies are respectively, 2 7rr^7iT^ - 2 irr^^nT.^ - 2 Trr^n • 30 = ^ 80(18^ - 12), lirr^HiG^- 2\) 2 7rr2<^2-^2) = ^^[a8r2)2-i], - 9 where r^ is the radius of the large drum, r^ that of the small drum, and r^ that of the axle. Eliminating T^ and T^ by adding the three equations, 2 irr-^nG^ — 2 Trr.^^i (^2 — 2 7^r37^ • 30 = YsO +--1^ +-^Vl82 - 12). 2V ^ ff J Substituting the known values, we obtain 7i = 59.5. Problem 353. In the above illustration, what are the velocities of G^ and G2 when w has its initial and final values ? In what time do the drums make the 59.5 revolutions? Problem 354. The drum in Fig. 217 is solid and has a radius 7* and a thickness h. Initially, it is rotating, mak- ing wo radians per second, but it is brought to rest by the action of a brake. The brake is applied from below by a force P acting at the end of the beam. WOBK AND ENERGY 271 pi The force of friction between the drum and brake is — , where P' is 4 the normal pressure exerted by the beam on the drum. The radius of the axle is r^, and the axle friction (.05) P", where P" is the pres- sure of the axle on the bearing. Required the work -energy equation. Since the drum comes to rest, the final kinetic energy is zero, so that _ \l^^2^ + UL'2 Trrn + (.05) P"2 Tr/y = 0. There are no working forces, so we find the equation reducing to the form : the initial kinetic energy equals the work of resistance. The normal pressure exerted by the beam on the drum maybe found by taking moments about the hinge of the beam. Then p, ^ a +h p b The number of revolutions turned through in coming to rest is designated by n. The equation then becomes 1 r 2 irrn (a -}- b) P , , n-\ nun -loii = ^^ — — h (-Oo) P"2 Trr,n- 2 2 ^ ' V / 1 Problem 355. Suppose the drum in the preceding problem to be 3 ft. in diameter, \\ in. thick, and made of cast iron. It is nuik- ing 4 revolutions per second when the force P = 100 lb. is applied to the beam. The length of the drum is 6 ft., and the rim weighs twice as much as the spokes and hub. If ^ = 1.25 ft., a = b = i ft., and Tj = 1 in., find the number of revolutions that the drum will make before coming to rest. Assume the friction of the brake on the drum to be I the normal pressure, and the friction of the axle (.05) P". Problem 356. The drnm in the preceding problem is making 3 revolutions per second. What force will be required to bring it to rest in 100 revolutions? Problem 357. If the brake in Problem 355 is above instead of below the drum, how will the results in Problems 355 and 356 be changed? 272 APPLIED MECHANICS FOR ENGINEERS Problem 358. so as to rotate due to the weight G. ^ (J= too LBS. FiCx. 218 A square prism as shown in Fig. 218 is mounted The elastic cord runs over the pulley B and meets the square at P', but is connected to the square at P. The mechanism is such that motion begins when P is in the position tjhown, and ceases when tlie prism lias made a quarter turn ; that is, when P reaches P'. The diameter of the journal is 2 in., and the weight on the same is 600 lb. The force of friction on the journal is 60 lb., and on the pulley at B equivalent to 10 lb. acting at the rim of the pulley. Find the tension in the cord w^hen P reaches P'. The cord is elastic, and is made of such material that it elongates, due to a pull of 100 lb., .02 in. in each inch of length. What is the elongation per inch due to the fall of G as stated ? 148. Brake-shoe Testing Machine. — The brake-shoe testing machine owned by the Master Car Builders' Association has been established at Purdue University. It consists of a heavy flywheel attached to the same axle as the car wheel. These are connected with the engine, and may be given any desired rotation. When this has been obtained, they may be disconnected and allowed to rotate. The dimensions and weight of the parts are known so that the kinetic energy of the flywheel and rotating parts may be computed by noting the angular velocity. When the desired velocity has been attained, the brake shoe is brought down on the car wheel. The required normal pressure on the shoe at A (See Fig. 219) is obtained by applying suitable weiglits at B. The system of levers is such that one pound at B gives a nor- mal pressure of 24 lb. on the brake shoe. The weight of WORK AND ENERGY 273 the levers themselves gives a normal pressure of 1233 lb. Provision is also made for measuring the tangential pull of the brake friction ; this, however, is not shown in the fio-ure. f) Fig. 219 The weight of the flywheel, car wheel, and shaft, and all rotating parts is 12,600 lb., and the radius of gyration is V2.16. The weight of 12,600 lb. is supposed to be the greatest weight that any bearing in passenger or freight service will be called upon to carry. The diameter of the flywheel is 48 in., its thickness 30 in., diameter of shaft 7 in., and the diameter of the car wheel is 33 in. The brake-shoe friction is J the normal pressure of the brake shoe on the wheel, and the journal friction may be assumed as (.002) of the pressure of the axle on the bearing. The work-energy equation for the rotating parts after being disconnected from the engine becomes 1 12,600 33 2'ikr*--^^'^'^'-"^'^ + (^-'^^-^-^^^i-^ii 71 + (1233 + 12,600 H- 24 a)(.002)2 tt - 7i = 0. 24 274 APPLIED MECHANICS FOR ENGINEERS Problem 359. The speed is such as to correspond to a speed of traiu of a mile a minute when brakes are applied. What must be the weight G so that a stop may be made in a thousand feet? What is the corresponding normal pressure on the brake shoe? Problem 360. If the speed corresponds to the speed of a train of 100 mi. per hour, what weight G would be necessary to reduce the speed to 60 mi. per hour in one mile? What is the normal pressure on the brake shoe necessary? Problem 361. If the velocity corresponds to a train velocity of 60 mi. per hour, and the apparatus is brought to rest in 220 revolu- tions, the weight G is 100 lb. Find the tangential force of friction acting on the face of the w^heel. What relation does this bear to the normal brake-shoe pressure ? Note. In the preceding problems, the ratio (the coefRcieut of fric- tion, see Art. 135) has been taken as ^. One of the important uses of this testing machine is to determine the coefficient of friction for different types of brake shoes. Experiment shows that it varies generally from I to i, sometimes going as high as y\. 149. Kinetic Energy of a Body having Plane Motion. — Sup- pose a body to have plane motion. Let the angular veloc- ity of the bod}' at a given instant be co and the ve- locity of the center of gravity of the body be Vy Choose coordinate axes so that the origin coin- cides at the given instant with the center of gravity and the a:-axis coincides with the direction of motion of tlie center of gravity, the x- and z-axes being in the plane of motion of the body (Fig. 220). Fig. 220 WORK AND ENERGY 275 Let dM be any element of mass of the body, distant r from the ^-axis. The velocity of dM is then composed of the velocity of any point on tlie ^-axis and the velocity of c?ili" relative to the ^-axis. Hence, v, the velocity of dM, is given by v^ = v'^-{- r^tiP' -|- 2 v-^ro) cos <^, where ^ is the angle whicli r makes with a line parallel to tlie ;3-axis. But r cos = 2, so that v^ — v\ 4- r^ft)^ + 2 v^(tiz. The kinetic energy of the whole body is the sum of the kinetic energy of its particles, or K. E. of Body = ^\{v\ + r'^o? -h 2 v^uiz)dM = I v\CdM+ I ay'-jrUM^ 2 v^coCzdM. Since the a;?/-plane passes through the center of gravity, \dM= 0. (Art. 33.) /' Therefore K. E. of Body = \ Mvi + 1 Iu>^ where M is the mass of the body and / its moment of inertia about a gravity axis perpendicular to the plane of motion. This formula may be expressed in words as follows : The kinetic energy of a body having plane motion is equal to the Icinetie energy the tvhole mass would have if concen- trated at the center of gravity, ivith the velocity of the center of gravity, plus the kinetic energy of rotation that the body 276 APPLIED MECHANICS FOR ENGINEERS would have if the gravity axis were at rest and the body rotating about it ivith the same angular velocity. 150. Work and Kinetic Energy in Plane Motion. — If im- pressed forces act on a body liaviiig plane motion, the work done by tliese forces in any displacement equals the change in kinetic energy of the body in that dis- placement. Proof: The total gain in the kinetic energy of the body is the sun:i of the increments in kinetic energy of all the particles of the body, which, by Art. 143, is the total work done on the particles by all the forces acting on them. In finding the total work done on the particles the work done by the forces that the particles exert on each other adds up to zero and there is left only the work done by the impressed forces. For two particles of the body, m-^ and TTig, remain the same distance apart and exert on each other forces that are equal and opposite and in the line joining the particles. During any dis- placement m^ has a motion made up of the motion of m^ and a motion about m-^ as a center (Fig. 221). For that component of its motion parallel to the motion of m^ the forces exerted by the particles on each other are equal and opposite and have the same displace- ment, and hence the work done by one force is just equal to the work done against the other. For tlie motion of rotation of m^ about m^ the force acting on m^ is perpen- FiG. 221 WORE AND ENERGY 211 dicular to the direction of displacement, and hence no work is done. Hence the forces exerted by the particles on each other do no work, and we may write for any plane motion. Work doue by impressed forces = change in kinetic energy of body. As an illustration, consider a body of circular section, as a hoop, cylinder, or sphere, with center of gravity at the center of the circular section, rolling without slipping down an inclined plane (Fig. 222). The impressed forces acting on the body are its weight, 6r, the normal reaction of the plane, iV, and a retarding friction force, i^, along the plane. The point of application of iV has no displacement in tlie direction in which JV acts, and hence no work is done by N. The point of application of F^ continually changing in the body, has at each instant a motion at right angles to F; for if there is no slipping, the point of the body in contact with the plane at any instant leaves the plane at right angles to the plane. Hence no work is done by F. The total work done in the descent is therefore Gh. If (Oq and Vq are the angular velocity of the body and the linear velocity of its center of gravity respective!}^ at the top of the plane and co and v the corresponding values at the foot, then the work-energy equation is I 70)2 + .] Mv^ - 1 IcD-^ -IMvl^ ah. Fig. 222 If 8 is the distance passed through by the center and 6 278 APPLIED MECHANICS FOR ENGINEERS tlie angle turned through in the same time, s = rO. Therefore or Substituting the work-energy equation becomes ds dt d6 "dt' V = ro). ft) _ ^ — ~ -> (Oq = \i= ^k\ M= a r r y .9 2gh or ly^ — v^ = . Problem 362. Prove that all .solid spheres will roll down the inclined plane at the same rate. Find the velocity at the foot of the plane. Problem 363. A uniform sphere, a uniform disk, and a hoop, starting at the top of an inclined plane, roll from rest to the foot. Find the velocity of each on reaching the foot. In what order do they arrive? Problem 364. Which will roll faster down an inclined plane, a hollow sphere with diaineter of the hollow one half that of the sphere, or a solid uniform disk? 151. Kinetic Energy of Rolling Bodies. — It is convenient to express the kinetic energy or combined rotation and translation of such bodies as rolling wheels in a different form from tliat given in the preceding article. There is some mass M^ that will have the same kinetic energy when translated with a velocity v^ as tlie kinetic energy WOUK AXIJ EX ERG Y 279 of translation plus the kinetic energy of rotation of tlic body of mass 31; that is, 2 2 2' For a wheel rolling on a straight track cor = t^j, where r is the radius. Then 3L=M-h{. This has been called tlie equivalent mass. For example, for a rolling disk, since 1= ^ Mr^, 31^ = | 31, Problem 365. A sphere rolls without slipping down an inclined plane. Show that its kinetic energy is the same at any instant as that of a sphere whose mass is f larger translated with a velocity equal to the velocity of the center of gravity of the rolling sphere. Problem 366. The sphere in the preceding prohlem is made of steel, 12 in. in diameter, and the inclination of the plane is 30°. If Vq = 10 ft. per second, what will be the velocity 10 ft. down the plane? 152. "Work-energy Relation for Any Motion. — The rela- tion between work and energy for the motions considered in this chapter holds for more complicated motions and for motions in general. The limits of the present work will not admit the proof of the general theorem. It may be said, however, that for any motion the work done by the working forces equals the work done by the resisting forces plus the change in kinetic energy. In the case of the motion of a complicated machine, the work done by the working forces equals the work done against the resist- ances plus the gain in kinetic energy of the various parts of the machine. 280 APPLIED MECHANICS FOE ENGINEERS 153. Work Done when Motion is Uniform. — When the motion is uniform, the change in kinetic energy is zero, and tlie work-energy equation reduces to the form : work done equals the work done against the resistance overcome. As an illustration, let us consider the case of a loco- motive moving at uniform speed and represented in Fig. 223. Suppose P the mean effective steam pressure (See Art. 139), F the friction of the piston, F' the friction of the crosshead, F" the journal friction, F'" the crank-pin nizn Fig. 223 friction, T the friction on the rail, M the draw-bar resist- ance, M' the horizontal component of pressure of the drivers' axle on the frame, r, r-^, and r^ the radii of the crank-pin circle, the driver axle, and the crank-pin re- spectively, 6r the weight of the locomotive, and N' and N the normal reactions of the rails on the wheels. Consider both sides cf the locomotive and write the work-energy equation for a distance s, equal to a half turn of the driver (from dead center A to dead center J5), that the locomo- tive travels. Considering the work done on the frame, counting both cylinders, we have E'lra -f- 2(i^-f F')iTa -\- 2 F"7ra = (2 P + R)7ra. (1) WORK AND ENERGY 281 Considering the work done on the rotating and oscillat- ing parts, 4- 2F"(nTa+7Tr^}^2F"''Trr^. (2) Adding these equations there results 4jPr = R^ra +{F+F')^r + 2F"TTri + 2 /^'"-rrr^. (3) If we neglect friction, this equation becomes 4 Pr = iraR^ or r = ^^ R. Here P is the mean effective pressure in one cylinder. This is the formula usually given for the tractive power of a locomotive having single expansion engines. This may be expressed in terms of the dimensions of the cylin- ders and the unit steam pressure. Let p be the unit steam pressure in pounds per square inch, I the length of the cylinder in inches, d the diameter of the cylinder in inches, and d-^ the diameters of the drivers in inches ; then R = d?pl For uniform motion the train resistance cannot exceed the friction force or force of adhesion between the drivers and the rails, since these are the external horizontal forces acting on the engine at any time. This force of adhesion in American practice is usually taken as \ or -J of the weight on the drivers. Problem 367. Derive equation (3) of this article by considering the work done on the whole engine. 282 APPLIED MECHANICS FOR ENGINEERS Problem 368. What resistance R may be overcome by a locomo- tive mo\ iiig at uniform speed, diameter of drivers 62 in., cylinders 16 X 24 in., and a steam pressure on the piston of 160 lb. per square inch ? What should be the weight of the locomotive on the drivers? Problem 369. If the diameter of the drivers of a locomotive is 68 in., and the size of the cylinder is 20 x 24 in., what train resistam^e may be overcome by a steam pressure of 160 lb. per square inch? Problem 370. A locomotive has a weight of 155 tons on the drivers. If the adhesion is taken as |, this allows 31 tons for the drawbar pull. The train resistance per ton of 2000 lb., for a speed of 60 mi. per hour, is 20 lb. Find the weight of the train that can be pulled by the locomotive at the speed of 60 mi. per hour. Problem 371. An 80-car freight train is to be pulled by a single expansion locomotive at the rate of 30 mi. per hour. The weight of each car is 60,000 lb., and the resistance for this speed is 10 lb. per ton. "What must be the weight on the drivers, if the adhesion is ^? CHAPTER XIII FRICTION 154. Friction. — When, one body is made to slide over another, there is considerable resistance offered because of the roughness of the two bodies. A book drawn across the top of a table is resisted by the roughness of the two bodies. The rough pai'ts of the book sink into the rough parts of the table so that when one of the bodies tends to move over the other, the projections interfere and tend to stop the motion. The bearings of machines are made very smooth, and usually we do not think of such surfaces as having projections. Nevertheless they are not perfectly smooth, and when one surface is rubbed over the other, re- sistance must be overcome. This resisting force to the motion of one body over another is known as friction. When the bodies are at rest relative to each other, the friction is known as the friction of rest, or static friction. When they are in motion with respect to each other, the friction is known as t\\Q friction of motion, ov kinetic friction . 155. Coefficient of Friction. — If the body represented in Fig. 224 be pulled along the horizontal plane by the force P, the following forces will be acting on it : the downward force Gr and the reaction R inclined back of the vertical through the angle 6. Th) just prevent it from sliding down the plane (Fig. 226), is sin(a + ^) sin(oc- ^ ) (d) r = 7-r- /^ (r, (n) F = 7-7 /. (jr, ^ ^ COS(cf> - 6) ^ ^ COS ((f>+ 6) where 6 is the antiie of friction. Problem 374. Show that the least values of P in the preceding problem are when <^ = ^ in (a), and when ^ = — ^ in (//) ; i.e. when (a) P = G' sin (a + 0), (h) P = G sin {a - 0). Problem 375. In Fig. 227 the weight G is raised l)y the horizontal force P. If the only friction is between the surfaces of the wedge and the weight, prove that the value of P just sufficifut to raise the weight is Fio. 227 P =G tiiu((i + 6). 288 APPLIED MECHANICS FOB ENGINEERS Problem 376. Defining the efficiency, E, of the wedge as the ratio of the useful work accomplished in raising the weight to the total work done by P (Fig. 227) show that P _ /tan a tan (a -\- 0) Problem 377. For a given value of 0, show that E isa maximum when a = 450 _ ^ . 2 (Since a square-threaded screw may be regarded as an inclined plane, this formula also holds for such a screw.) Problem 378. Find the value of the horizontal force P that will just raise the weight of 500 lb. in Fig. 228, given that the coefficients of friction at J., J5, and C are re- spectively .20, .25, and .30. ■//////^//////yy//. Fig. 228 Suggestion. Consider the wedge and block separately, and the forces that hold them in equilib- rium. 157. Friction of Lubricated Surfaces. — The laws of fric- tion, as given by Morin and stated in the preceding article, hold approximately for rubbing surfaces, when the sur- faces are dry or nearly so ; that is, for poorly lubricated surfaces. If, however, the surfaces are well lubricated so tliat the projections of one do not fit into the other, but are kept apart by a film or layer of the lubricant, the laws of Morin are not even approximately true. The study of the friction of lubricated surfaces, then, may be divided into two parts : (1) tlie study of poorly lubricated bear- ings, and (2) the study of well lubricated bearings, the FRICTION 289 friction of which varies from J to ^ that of dry or poorly lubricated bearings. Since the friction of poorly lubricated bearings is about the same as that of dry surfaces, we shall consider that the laws of Morin hold, and shall confine our attention to the friction of well lubricated bearings. If the lubricant is an oil, the friction of the bearing is no longer due to one surface rubbing over the other, but to the friction between the bearing and the oil, and to the internal fric- tion of the oil. That is, the oil adheres to the two sur- faces, and its own particles attract each other, and the motion of one of the surfaces with respect to the other clianges the positions of the oil particles. It is to be expected, then, that the friction of an oiled bearing will depend upon the viscosity of the oil^ upon the thickness of the layer interposed between the surfaces^ and upon the velocity and form of the bearing. The coefficient of friction is no longer constant, but varies with the temperature, velocity, and pressure. The variation of the coefficient of friction of a paraffine oil with temperature is shown in Fig. 229 when the pressure on the bearing is 33 lb. per square inch and a velocity of rub- ])ing of 296 ft. per minute. It is seen that the coefficient of friction decreases with increase of temperature until a temperature of 80° F. is reached, when it increases rapidly. This means that above this temperature the oil is so thin that it is squeezed out of the bearing, and the conditions of dry bearing are approached. Tlie temperature at which oils show an increasing coefficient of friction is dif- ferent for different oils, even at the same pressure and 290 APPLIED MECHANICS FOR ENGINEERS velocity. The curve in Fig. 229, however, may be re- garded as typical of all oils when the pressure and velocity are constant. w &0 60; 50 40 30 — : -: — r — ~ - ' — — — — ■- ■ — I^ Ip I^ ^ — -t^- ^1 _:_- „_. __„. , _: . 1 if^ ^— T- — , — — — — ^— \ — " — - — " i - -re — — — — - — — — " — " !___ — — :. 1 ■4 i 1 — — r -- 1 1 — . 1 -;- ""■ ;3t zzv. -^ " : 1 \ w ^~ ■ r ~" ._ - ~ 1 — h— TTT' _ ;\^ _ - , V f:3 1 \ ^i i?? cc p V DEF RES ELO FIC F A 3UR CIT ! )N H E \: ::± par; E 33 LB r 296 F£ - . 1 . ui- \FF S. F :et 1- INf ER PEF RIC"" -. SQ. Ml 1 K IL INC SUT ■~~ — ■ — s K I — E^ |:r:^ 1^ ' — — — — ^ — Six; \^ \^ t£. riilrifr \^ .01 .02 .03 coefficient of friction Fig. 229 .01 The following table, due to Professor Thurston, shows the relation between the coefficient of friction and tem- perature for a sperm oil in steel bearings when the veloc- ity of rubbing is 30 ft. per minute: FRICTION 291 PRESStTRE, Lb. Temperature, Coefficient Pressure, Lb. Temperature, Coefficient I'ER Sq. In. Degrees F. OF Friction PER Sq. In. Degrees F. OF Friction 200 150 .0500 100 110 .0025 •200 140 .0250 50 110 .0035 200 130 .0160 4 110 .0500 200 120 .0110 200 90 .0040 200 110 .0100 150 90 .0025 200 100 .0075 100 90 .0025 200 95 .0060 50 90 .0035 200 90 .0056 4 90 .0400 150 110 .0035 It is seen that for a pressure of 200 lb. per square inch as the temperature increases from 90°' F. the coefficient in- creases, indicating that the temperature of 90°, for the given pressure and velocity, was above the temperature at which the oil became so thin as to be squeezed out and the bearing to approach the condition of a dry bearing. For a constant temperature 110° F. and 90° F. the coeffi- cient is seen to decrease with increase of pressure up to a certain point and then to increase. This is a typical be- havior of oils when tlie temperature is constant and the pressure varies. At speeds exceeding 100 ft. per minute, the same author- ity found '' that the heating of the bearings within the above range of temperatures decreases the resistance due to friction, rapidly at first and then slowly, and gradually a temperature is reached at which increase takes place and progresses at a rapidly accelerating rate." The relation between the coefficients of rest and of motion as determined by Professor Thurston for three oils is given 292 APPLIED MECHANICS FOR ENGINEERS below. The journals were cast iron, in steel boxes ; velocity of rubbing 150 ft. per minute and a temperature 115° F. Sperm Oil West Virgini A Oil Lard Pressire, At 150 At At At 150 At At At 150 At At Lb. per ft. per start- stop- ft. per start- stop- ft. per start- stop- Sy. In. min. ing ping min. ing ping mm. ing ping 50 .013 .07 .03 .0213 .11 .025 .02 .07 .01 100 .008 .135 .025 .015 .135 .025 .0137 .11 .0225 250 .005 .14 .04 .009 .14 .026 .0085 .11 .010 500 .004 .15 .03 .00515 .15 .018 .00525 .10 .016 750 .0043 .185 .03 .005 .185 .0147 .0066 .12 .020 1000 .009 .18 .03 .010 .18 .017 .0125 .12 .019 Steel Journals and Brass Boxes 500 1000 .0025 .008 .004 .009 It is seen that the coefficient of friction at starting is much greater than at stopping, and that these are both much greater than the value at a speed of 150 ft. per minute. For an intermittent feed such as is given by one oil hole, without a cup, oiled occasionally. Professor Thurston found for steel shaft in bronze bearings, with a speed of rubbing of 720 ft. per minute, the following coefficients of friction : Oil Pkessure, Lb. per Sy. In. 8 16 32 48 Sperm and lard .... Olive and cotton seed Mineral oils .ir)9-.25 .l(J0-.283 .154-.261 .138-. 192 .107-.245 .145-.233 .086-.141 .101-.168 .086-.178 .077-.144 .079-.131 .094-.222 FRICTION 293 The results show that the coefficient decreases with the pressure within the range reported, but that the results are considerably higher than those for well lubricated bearings. He also found in connection with the same tests that with continuous lubrication sperm oil gave the following coefficients : Pressure, Coefficient Lb. per Sq. In. of Friction 50 .0034 200 .0051 300 .0057 The results of tests of the friction of well-lubricated bearings are summarized by Goodman (^Engineering Neivs^ April 7 and 14, 1888) as follows : (a) The coefficient of friction of well lubricated surfaces is from ^ to Jq- that of dry or poorly lubricated surfaces. (5) The coefficient of friction for moderate pressures and speeds varies approximately inversely as the normal pressure ; the frictional resistance varies as the area in contact^ the nor- mal pressure remaining the same. (c) For low speeds the coefficient of friction is abnormally high., but as the speed of rubbing increases from about 10 to li)0 ft. per minute, the coefficient of friction diminishes, and again rises when that speed is exceeded, varying approxi- mately as the square root of the speed. (f?) The coefficient of friction varies approximately in- versely as the temperature, within certai?i limits; namely, just before abrasion takes place. 158. Method of Testing Lubricants. — To make the matter of the tests of the friction of lubricants clear, it will be 294 APPLIED MECHANICS FOR ENGINEERS convenient to make use of the description of a testing machine used by Dean W. F. M. Goss at Purdue Uni- versity on graphite, and a mixture of graphite and sperm Fig. 230 oil. In making the tests the apparatus shown in Figs. 230 and 231 was used. (See " A Study in Graphite,'' Joseph Dixon Crucible Co.) This apparatus represents, in principle, the machines generally used for testing lubricants. It is therefore shown in some detail. The weight Gr is hung from the shaft upon which it is suspended by the form of box to be tested. The desired speed of rubbing is obtained by means of the cone of pulleys, and the pressure on the bear- ing is adjusted by the spring. The temperature of the bearing is read from the thermometer inserted in the bear- ing. When rotation takes place, the weight Gr is rotated a certain distance dependent upon the friction. This dis- tance is measured on tlie scale A. The forces acting upon FRICTION 295 R-\-0 the peiiduluin G- are shown in Fig. 2ol, where R represents the resistance of the spring, F tlie force of friction, I the distance of the center of gravity of G from the axis of rotation, (^ the angle through which G is deflected, r the radius of the shaft, and / the coeffi- cient of friction. Taking moments about the center of the shaft, we have, when G is lield in the position shown, due to the friction, or r(F^ F^)=Gl^m(i>, rf{R + G + B)=Gl sin (/>. / = Gl sin ^ 7^(2 R^G) Fig. 231 It is customary to take G small compared with R, so that the pressure on both sides of the bearing may be considered equal to R^ tlie resistance of the spring. The formula then becomes f _ Gls'm The spring is easily calibrated so that R may be made any- thing desired by compressing the spring through the ap- propriate distance, as indicated on the scale J^(Fig. 230). The quantities G, I, r, and R are known, and can be re-ad so that /can be calculated. The results of tests made upon a mixture of graphite and oil as a lubricant are given in the pamphlet. The tests were run under 200 lb. per square inch pressure, at a speed of rubbing of 145 ft. per minute. Oil was dropped 296 APPLIED MECHANICS FOR Ei\G INFERS into the bearing at the rate of about 12 drops per minute, showing a coefficient of friction of |. Problem 379. If the weight of the pendulum is 360 lb., the diameter of the shaft -i^ in., distance of the center of gravity of G from the center of shaft 2 ft., the angle (f> 5 degrees, and the average resistance of the spring 1000 lb., find the coefficient of friction. The weight G should not be neglected in this case. 159. Rolling Friction. — The resistance offered to tlie rolling of one body over another is known as rolling fric- tion. It is entirely different from sliding friction, and its laws are not so well understood. When a wheel or cylinder (Fig. 232) rolls over a track, the track is depressed and the wheel dis- torted. The force P necessary to overcome this depression and distortion is known as rolling friction. The forces acting on the wheel are seen from Fig. 232 to be: P the working force, TTthe weight on the wheel, and B the reaction of the track or roadway. This upward pressure E is not quite vertical, but has its point of ap- plication a short distance K' from the vertical. Its line of action passes through the center of the wheel. The distance K' depends chiefly upon the roadway ; it is called the coefficient of rolling friction. It is measured in inches and is not a coefficient of friction in the strict sense that/ is the coefficient of sliding friction. Taking moments about the point of application of i2, Fig. 232 FRICTION 297 we have, approximately, so that K' = WK' = Pr, Pr ^,. „ K'W W or r r When the track or roadway is ehistic or nearly so, we have a condition something like that represented in Fig. 233. The wheel sinks into the ma- terial and pushes it ahead, at the same time it comes up behind the wheel. For a portion of the wheel on each side of the point the roadway is simply compressed ; over the remainder of the surface in contact, however, slipping occurs, as indi- cated by the arrows. The resultant resistance, however, is in front of the vertical through the center, and we have, as in the case of imperfectly elastic roadways, p_K'W r It lias been found by Reynolds (See Phil. Trans. Royal Soc, Vol. 166, Part 1) that when a cast-iron roller rolls on a rubber track, the slippage, due to the elasticity of the track, may amount to as much as .84 in. in 34 in. An elastic roller rolling on a hard track will roll less than the geometrical distance traveled by a point on the circumfer- ence. When the roller and track are of the same material, the roller rolls through less than its geometrical distance. 298 APPLIED MECIIAMCS FOR ENGINEERS 160. Antifriction Wheels. — The axle A, of radius r, carrying a weight W, rests upon two wheels of radius r^^ turning on axles of radius r^ (Fig. 234). The force on each of the bear- ings of wheels B and is W 2cosy8' and if F is the friction at each of the bearings of B and C, and / Fig. 234 the coefficient of sliding friction, 2cOSy8' and the work lost in friction at the bearings of B and (7 in one revolution of the axle A is /■ W z irr cos/3 ^ r^ Since the axle A rolls on B and (7, there is no sliding fric- tion there and the rolling friction is small enough to be neglected. If A were in an ordinary bearing, the work lost per revolution would be flirrW. Therefore the ratio of work lost with antifriction wheels to the work lost with plain bearing is r^ cos yS This ratio decreases as the ratio r^r^ increases and as /8 decreases. FRICTION 299 Problem 380. If W — 4 tons, the radius of the shaft is 2 iu., and the coefficient of friction is .07, what work is lost per revohition? If the shaft makes 3 revolutious per second, w^hat horse power is lost in friction ? Given also (3 = 45°, )\ = f in., and r^ = 4 in. Problem 381. In the case of the shaft mentioned in the preced- ing problem, how much more horse power would it take if the hear- ing were plain? What value of /3 would give the same loss due to friction in both the plain bearing and the one provided with friction wheels ? 161. Resistance of Ordinary Roads. — Resistance to trac- tion consists of axle friction, rolling friction, and grade resistance. Axle friction varies from .012 to .02 of the load, for good lubrication, according to Baker. The tractive power necessary to overcome axle friction for ordinary American carriages has been found to be from 3 lb. to 3| lb. per ton, and for wagons with medium-sized wheels and axles from 3| lb. to 4^ lb. per ton. The total tractive force per ton of load, for wheels 50 in., 30 in., and 26 in., in diameter, respectively, is, according to Baker (^Engineering News^ March 6, 1902) : Tractive Force IN Pounds On macadam roads On timothy and blue grass sod, dry, grass cut . On timothy and V)lue grass sod, wet and springy On plowed ground, not harrowed, dry and cloddy 57 132 173 252 61 145 203 303 70 179 288 374 Rolling resistance is influenced by tlie width of the tire. According to Baker, poor macadam, i)oor gravel, compres- sible earth roads, and, on agricultural lands, narrow tires, 300 APPLIED MECHANICS FOR ENGINEERS usually require less traction. On earth roads composed of dry loam with 2 to 3 in. of loose dust, traction with li-in. tires was 90 lb. per ton, and with 6-in. tires 106 lb. per ton. On the same road when it was hard and dry, with no dust, that is, when it was compressible, the traction was found to be 149 lb. per ton with l|-in. tires and 109 lb. per ton with 6-in. tires. On broken stone roads, hard and smooth, with no dust or loose stones, the traction per ton was 121 lb. witli IJ-in. tires, and 98 lb. with 6-in. tires. Moisture on the surface or mud increases the traction. Morin found that with 44-in. front and 54-in. rear wheels on hard dry roads the traction per ton was 114 lb. with either l|-in. or 3-in. tires. On wood-block pave- ments the traction per ton was 28 lb. with l2-in. tires, and 38 lb. with 6-in. tires. On asphalt, bricks, granite, macadam, and steel-road surfaces, investigated by Baker, the traction per ton varied from 17 lb. to 70 lb., the average being 38 lb. Morin gives the coefficient of rolling friction for wagons on soft soil as .065 in., and on hard roads .02 in. Accord- ing to Kent (" Pocket-Book "), tests made upon a loaded omnibus gave the following results : Pavemknt Granite Asphalt Wood Macadam, graveled Macadam, granite, new Speed, Miles PER Hour 2.87 3.56 a.34 3.45 3.51 Coefficient, Inches .007 .0121 .0185 .0199 .0451 Resistance, pkb Ton, in Lb. 17.41 27.14 41.60 44.48 101.09 FRICTION 301 Problem 382. Compare the resistance oifered to a load of two tons pulled over asphalt, macadam, good earth roads, or wood-block pavement. ^Vidth of tires, fi in. 162. Roller Bearings. — In the roller bearings the shaft rolls on hardened steel rollers as shown in cross section in Fig. 235. The roll- ers are kept in place in some way similar to that shown in the journal of Fig. 236. Such bearings are used where heavy loads are to be car- ried. Tests of roller bearings have been made by Dean C. H. Benjamin (^Machinery^ October, 1905), who determined the follow- ing values for the coefficient of friction. Speed 480 revolutions per minute. DlAMF.TER OK JouRXAi-, IN Inches Roller Bearing Plain Cast-iron Bearing Max. Mill. Average Max. Min. A.verafre IB .036 .010 .026 .160 .099 .117 2A .052 .034 .040 .129 .071 .094 ^^ .041 .025 .030 .143 .076 .104 2H .053 .049 .051 .138 .091 .104 It was found that the coefficient of friction of roller bearings is from ^^ to J that of plain bearings at moderate speeds and loads. As the load on the bearing increased, 802 APPLIED MECHANICS FOR ENGINEEliS the coefficient of friction decreased. Tightening the bear- ing was found to increase the friction considerably. Tests of the friction of steel rollers 1, 2, 3, and 4 in. in diameter are reported in the Trans. Am. Soc. C. E., August, 1894. The rollers were tested between plates 1| in. thick and 5 in. wide, ar- ranged as shown in Fig. 237. Tests were made with the plates and rollers of cast iron, wrought iron, and steel. The friction P' for .0063 Fig. 236 unit load P was found to be V^ 0120 for cast-iron rollers and plates,^ — ^-for wrought iron, and Vr .0073 ► 7" for steel, where r represents the radius of the roller Vr in inches. The rollers were turned and the plates planed, ^ but neither was polished. 163. Ball Bearing^s. — For high speeds and light or moder- ate loads the friction is much W reduced by the use of hardened steel balls instead of the steel rollers. These bearings are now used on all classes of machinery, giving a much greater efficiency except for heavy loads. The principal objection to the ball bearing seems to be due to the fact that there is FRICTION 303 '^^^ O-^ .j^ so little area of contact between the balls and bearing plates. This gives rise to very high stresses over these areas, and consequently a considerable deformation of the balls. When the ball has been changed from its spherical form, it is no longer free to roll, and the friction increases rapidly. Some authorities con- sider a load of from 50 to 150 lb. sufficient for balls varying in size from ^ to ^ inch in diameter. Fig- ure 238 illustrates a type of bear- Fig. 238 ing used for shafts, and Fig. 239 a type used for thrust blocks. The conclusions reached by Goodman from a series of tests on bicycle bearings (Proc. Inst. C. E., Vol. 89) are as follows : (1) The coefficie7it of friction of hall hearings is constant for varying loads^ hence thefrictional resistance varies directly as the load. (2) The friction is U7iaffected by a change of temperature. The bearings were oiled be- fore starting tlie tests. The coefficient of friction for ball bearings was found to be rather higher than for plain bear- ings with bath lubrication, but lower than for ordinary lubrication. Ball bearings will also run easily with a less supply of oil. The followiuLi^ tal)le gives the results of 804 APPLIED MECHANICS FOR ENGINEERS tests of ball bearings. The bearings were oiled before start- ing, and the tests were run at a temperature of 68° F. 19 167 360 Load on Bkahinu Kkvolutions per Min. ItEVOLrTIOXS PER MiN. PtEVOUTTIONS FEB MiN. Coeff. friction Friction, lb. Coetf. friction Friction, lb. Coeif. friction Friction, lb. 10 .0060 .06 .0105 .10 .0105 .10 20 .0045 .09 .0067 .13 .0120 .24 30 .0050 .15 .0050 ,15 .0110 .33 40 .0052 .21 .0052 .21 .0097 .39 50 .0054 .27 .0054 .27 .0090 .45 60 .0050 .30 .0055 .33 .0075 .45 70 .0049 .34 .0054 .38 .0068 .47 80 .0048 .38 .0062 .49 .0060 .48 90 .0050 .45 .0068 .61 .0060 .54 100 .0058 .58 .0069 .69 .0057 ..57 110 .0054 .59 .0065 .71 .0060 .66 120 .0055 .66 .0075 .90 .0057 .68 130 .0058 .75 .0078 1.01 .0062 .81 140 .0056 .78 .0077 1.08 .0060 .84 150 .0060 .90 .0083 1.24 .0062 .93 160 .0075 1.20 .0081 1.29 .0058 .93 170 .0079 1.34 .0078 1.33 .0055 .93 180 .0079 1.42 .0078 1.40 .0053 .95 190 .0087 1.65 .0076 1.44 .0054 1.03 200 .0090 1.80 .0081 1.62 .0060 1.20 Another series of tests, run with a constant load on the bearing of 200 lb. and a temperature of 86° F., shows the variation of the coei^icient of friction with the speed. It is seen that as the speed increased the coefficient and the friction decreased. The preceding table, however, shows, for loads below 175 lb., an increase in the coefficient with increase in speed. In particular, this table shows that for FRICTION 305 loads below 80 lb. the coefficient increased with increase of speed ; for loads between 90 and 175 lb. it increased when the speed was 150 r.p.m. and decreased when it was 350 r.p.m. Beyond 175 lb. the coefficient increased. Kevolutions per Minute Coefficient Friction Friction Pounds 15 .00735 1.47 93 .00465 .93 175 .00375 .75 204 .00345 .69 280 .00300 .60 It seems from the data given that the first conclusion of Goodman's should be changed to read : the coefficient of friction of hall hearings is constant for varying loads^ up to a certain limits heyond which it increases with increase of load. This limit is about 150 lb. in the tests reported. Tests on ball bearings designed for machinery subjected to heavy pressures have been made in Germany. (See Zeit- schrift des Vereins deutsche Ingenieure, 1901, p. 73.) It was found that at speeds varying from 65 to 780 revolu- tions per minute, where the bearing was under pressures varying from 2200 lb. to 6600 lb., the coefficient of friction varied little and averaged .0015. Tests of ball bearings made by Stribeck and reported by Hess (Trans. Am. Soc. M. E., Vol. 28, 1907) give rise to the following conclusions : (a) tlie load that may be put upon a bearing is given by the formula cdhi r = 11.02' 306 APPLIED MECHANICS FOR ENGINEERS where P is the load in pounds on a bearing, consisting of one row of balls, c is a constant dependent upon the mate- rial of the balls and supporting surfaces and determined experimentally, d the diameter of the balls, the unit being I of an inch, and 7^ the number of balls. For modern materials c varies from 5 to 7.5. (5) The coefficient of friction varied from .0011 to .0095. It was independent of speed, "within wide limits," and approximated .0015; this was increased to .003 when the load was about one tenth the maximum. The following values for the coefficient of friction for heavy loads are reported, from observation, with the state- ment that the real values are probably somewhat less : Revolutions per minute 65 100 190 380 580 780 1150 Coefficient of friction for load 840 lb. .0095 .0095 .0093 .0088 .0085 .0074 Coefficient of friction for load 2400 lb. .0065 .0062 .0058 .0053 .0050 .0049 .0047 Coefficient of friction for load 4000 to 92.30 lb. .0055 .0054 .0050 .0050 .0041 .0041 .0040 It should be remembered that the friction of a ball bearing is due to both sliding and rolling friction, the sliding friction being due to the elasticity of the balls and the bearing. (See Art. 159.) Rolling friction is most FRICTION 307 nearly approached wlien the balls are hard and not easily changed from their spherical shape. All materials, how- ever, are deformed under pressure so that perfect rolling friction is impossible. On account of the sliding friction present in roller and ball bearings, it is necessary to use a lubricant to prevent wear. Problem 383. IIow many f-in. balls will be necessary in a ball bearing designed to carry -iOOO lb., if c = 7.5 ? If /= .0015, what work is lost per revolution, the distance from the axis of rotation to the center of balls being one inch ? 164. Friction Gears. — In the friction gears the driver is usually the smaller Avheel, and when there is any differ- ence in the materials of which the wheels are made, the driver is made of the softer material. This latter arrange- ment is resorted to, to prevent flat places being worn on either wheel in case of slipping. These gears have been used for transmitting light roads at high speeds, where toothed gears would be very noisy, or in cases where it is necessary to change the speed or direction of the motion quickly. The use of paper drivers has made possible the trans- mission of much heavier loads by means of such gears. A series of tests, made by W. F. M. Goss, and reported in Trans. Am. Soc. M. E., Vol. 18, on the friction be- tween paper drivers and cast-iron followers, is of interest in this connection. The apparatus used is shoAvn in Fig. 240. The pressure between the wheels was obtained by a mechanism that forced the two wheels together with a pressure P. A brake wheel shown in the figure absorbed the power transmitted. 308 APPLIED MECHANICS FOR ENGINEERS The coefficient of friction was regarded as the ratio of JP to P, as in sliding friction. While tliis is customary, it is not entirely true, since we have the rolling of one IRON FOLLOWER Fig. 240 body over the other. We shall, however, assume that we may call the coefficient of friction /= —• It was found that the coefficient of friction varied with the slippage, but was fairly constant for all pressures up to some point between 150 to 200 lb. per inch of width of wheel face. '^ Variations in the peripheral speed hetiveen 400 and 2800 ft. per minute do not affect the coefficient of friction.^' If the allowable coefficient of friction be taken as .20, the horse power transmitted per inch of width of face of the wheel, for a pressure of 150 lb., is H.P. = ^^^ ^ '^ x-^,7rdxw xJSr ^ .000238 dwN, 33,000 FRICTION 309 where d is the diameter of the friction wheel in inches, w the width of its face in inches, and iV the number of revolutions per minute. Using this formula, the following table is given in the article in question: Horse Power which may he transmitted by Means of Paper Friction Wheel of One Inch Face, when run rxPER a Pressure of 150 Lb. Diameter 1 lEvoi.iTiuxs PEU Minute OF PlLI.EY IN Inches 35 50 75 100 150 200 600 1000 8 .047G .0952 .1428 .1904 .2856 .3808 1.1424 1.904 10 .0595 .1190 .1785 .2380 .3.570 .4760 1.4280 2.380 14 .0833 .1666 .2499 .3332 .4998 .6664 1.9992 3.332 16 .0952 .1904 .2856 .3808 .5712 .7616 2.2848 3.808 18 .1071 .2142 .3213 .4281 .6426 .8.568 2.5704 4.288 24 .1428 .2856 .4284 .5712 .8568 1.1424 3.4272 5.712 30 .1785 .3570 .5355 .7140 1.0710 1.4280 4.2840 7.140 36 .2142 .4284 .6426 .8568 1.2852 1.7136 5.1408 8.560 42 .2499 .4998 .7497 .9996 1.4994 1.9992 5.9976 9.996 48 .2856 .5712 .8568 1.1424 1.7136 2.2848 6.8544 11.420 The value of the coefficient of friction for friction gears (Kent, " Pocket- Book ") may be taken from .15 to .20 for metal on metal ; .25 to .30 for wood on metal ; .20 for wood on com- pressed paper. Problem 384. If the friction wheels are grooverl as shown in Fig. 241, both Fia. 241 310 APPLIED MECHANICS FOR ENGINEERS of cast iron, and the small driver fits into the groove of the larger follower, prove that the force transmitted is F=2fN sm a Problem 385. The speed of the rim of two grooved friction wheels is 400 ft. per minnte. If u = 45°, /= .18, what must be the pressure P to transmit 100 horse power? Problem 386. What horse power may be tiansmittedby the gear- ing in the preceding problem, if P = 0000 lb. and the peripheral velocity is 12 ft. per second ? 165. Friction of Belts. — When a belt or cord passes over a pulley and is acted upon by tensions 2\ and T^, the tensions are unequal, due to the friction of the pulley on the belt. We shall determine the relation between T^ and T^' Let the pulley be represented in Fig. 242. The belt '\ dfj Fig. 242 covers an arc of the pulley whose angle is a. Consider the forces acting upon the belt and suppose T^ and T^ to be tlie tensions in tlie belt on the tiofht and slack sides, FRICTION 311 respectively, and 2^ the tension in the belt at any point of the arc of contact. Consider the forces acting on a por- tion of the belt of length As = rA/3 (Fig. 242). These forces are T, 2^4- A 2^, and Ai^, tangent to the arc, and AP, the normal pressure of the pulley on the belt, which may be regarded as acting at the center of the arc. Let m be the mass of a unit length of the belt. The length of the portion considered is then mrA/S. Suppose the belt to have uniform speed, v. The forces along the tangent will then balance, i.e. T+AF= T-hAT. .-. dF=dT. The acceleration toward the center is — , and the force toward the center is A/3 (T+A7+ T) sin^-AP. 2 .-. (2 T-^AT)sm A^ 2 AP = nn'^A/3. Dividing by A/3 and passing to the limit, remembering A/3] = 1, we have, that lim sin A^ T- dP d^ mv- If /is the coefficient of friction, and the belt is on the point of slipping, - fdp. or dF dP = \dF=\dT. f f ol2 APPLIED MECHAXTCS FOR ENGTNEERS Substituting this value of dP^ we have fdl3 dT T — mv^ Integrating, =/'//3. logg(7^— mv^^ Ti =//3 or loge 7^ =/« ^2 — ^?'2 ' Ti— niv- For low velocities the term mv^ is small compared to T^ and 2^2' ^^^ ^1^6 formula may be written It should be noted that m in the above formula is the mass of a portion of the belt 1 foot long, and that v must be reckoned in feet per second if y is taken as 32.2. Problem 387. A rope makes two complete turns around a post 6 in. in diameter. What maximum tension could be balanced by a force of 100 lb. if/= .3? Problem 388. A weight of 500 lb. is to be lowered by a rope wound round a horizontal drum. If the arc of contact is 450° and the coefficient of friction is .25, what force is necessary to lower the weight uniformly ? Problem 389. The velocity of a belt is 3000 ft./min., the tension in the tight side is 150 lb. per inch of width, the coefficient of friction is .25, and the weight of a portion of the belt 1 ft. long and 1 in. wide is .15 lb. If the belt is on the point of slipping and the arc of contact is 150°, what is the tension in the slack side? FRICTION 313 Problem 390. A rope is wrapped four times around a post and a man exerts a pull of 50 lb. on one end. If the coeHicient of friction is .3, what force can be exerted upon a boat attached to the other end of the rope ? 166. Power Transmitted by a Belt. — From the relation dF=-dT (Art. 165.) there follows dT= I dF, or T^— T^ = jP, the total friction. The work done per second against F is Fv or (7\- T^)v. Hence the horse power transmitted by the belt is HP - (Ti-T,)v 650 ' where T^ and jPg are in pounds and v is in feet per second. Problem 391. Show that the formula for H.P. transmitted by the belt may be reduced to the form ' ' 550 Problem 392. Given a maximum allowable tension, T'j, show that the power transmitted is a maximum when Problem 393. A pulley 4 ft. in diameter making 200 r. p. m. drives a belt that absorbs 20 H.P. The belt is \ in. thick and weighs 56 lb. per cubic foot. If a = tt and/ = .27, how wide must the belt be that the tension may not exceed 75 lb. per inch of width ? 314 APPLIED MECHANICS FOR ENGINEERS Problem 394. What II. P. may be transmitted by a belt 6 in. wide, \ in. thick, weighing 56 lb. per cubic foot, when traveling at 1500 ft. per minute, if (t = 108°,/= .25, and the maximum tension is 300 lb. per square inch ? Problem 395. Find the maximum H. P. that can be transmitted by the belt in the preceding j^roblem, and the corresponding velocity. 167. Transmission Dynamometer. — It has been shown, in Art. 165, that the tension of a belt on the tight side is greater than the tension on the slack side. The transmission dynamometer (the Fronde dynamometer), illus- trated in Fig. 243, is designed to measure tlie difference in these tensions. Let the pulley D be the T^ driver and the pulley U the fol- lower, so that T^ represents the tight side of the belt and T^ the slack side. The pulleys B^ B run loose on the T-shaped frame CBB. This frame is pivoted at A. If we neglect the friction due to the loose pulleys, we have the following forces acting on the T-frame, two forces T^ at the center of the right- liand pulley B^ two forces T^ at the center of the left-hand pulley B^ a measurable reaction P at (7, Taking moments about Fig. 243 and tlie reaction of the pin at A. the pin, we have FRICTION 315 P( CA) = 2 T^iBA) - 2 T^iBA) = 2BA(T,-T,-), so that T^- To = ^%^' The distances CA and ^^ are known, and P may be measured ; the difference, then, T^^ — T^^ may always be obtained. The value T-^ — T^ is then known and the horse power determined by the relation ^'^'~ 33;000 ' where 7i is the number of revolutions per minute, and r is the radius of the machine pulley in feet. 168. Creeping or Slip of Belts. — A belt that transmits power between two pulleys is tighter on the driving side than it is on the following side. On account of this differ- ence in tension and the elasticity of the material, the tight side is stretched more than the slack side. To compen- sate for this greater stretch on one side than on the other, the belt creeps or slips over the pulleys. This slip has been found for ordinary conditions to vary from 3 to 12 ft. per minute. The coefficient of friction when tlie slip is considered is about .27 (Lanza). It has also been found that the loss in horse power in well-designed belt drives, due to slip, does not exceed 3 or 4 per cent of the gross power transmitted, and that ropes are practically as efficient as belts in this respect. For an account of the experimental investigations on this subject the student is referred to Inst. Mecli. Eng., 1895, Vols. 3-4, p. 599, and Trans. Am. Soc. M. E., Vol. 26, 1905, p. 584. 316 APPLIED MECHANICS FOR ENGINEERS 169. Coefficient of Friction of Belting. — The value of the coefficient of friction of belting depends not only on the slip bat also upon the condition and material of the rubbing surfaces. Morin found for leather belts on iron pulleys the coefficient of friction /= .56 wlien dry, .36 when wet, .23 when greasy, and .15 when oily (Kent, "Pocket-Book"). Most investigators, however, including Morin, took no ac- count of slip, so that the best value of /, everything con- sidered, is that given in the preceding article (.27). 170. Friction of a Worn Bearing. — The friction of a bearing that fits perfectly is the friction of one surface sliding over another and is given by the equation F=fN, where F is the force of friction, / the coefficient of friction, and iVis the total normal pressure on the bearing. When, however, the bearing is worn, as is shown much exaggerated in Fig. 244, the friction may be somewhat B Fig. 2M FRICTION 317 different. When motion begins, the shaft will roll up on the bearing until it reaches a point A where slipping be- gins. If motion continues, slipping will continue along a line of contact through A. Let P be a force that causes the rotation, R a force tending to resist the rotation, and J?j the reaction of the bearing on the shaft. There are only three forces acting on the shaft, so that P, i2, and R^ must meet in the point B. The direction of R^ is accord- ingly determined. The normal pressure is N= R^ cos ^, and the force of friction is It is seen that 6 is the angle of friction. The moment of the friction with respect to the center of the axle is Fr = JRxV sin 9. If the axle is well lubricated, so that 6 is small and sin 6 may be replaced by tan 6 =f^ the friction is F = flt„ and the moment Fr = fRxr, The circle tangent to AB of radius r sin 6 is called the friction circle. Since r and 6 are known generally, this circle may be made use of in locating the point A, The shaft will continue to rotate in the bearing so long as the reaction R^ falls within the friction circle, and slipping will begin as soon as the direction of Pj becomes tangent to the friction circle. Problem 396. If the radius of the shaft is 1 in.. 6 = 4°, P = 500 lb., Oj = 3 ft., ag = 2 ft., angle between ai and 02 is 100'^, and 318 APPLIED MECHANICS FOR ENGINEERS P aud R are right angles to n^ and a.^, what resistance R may be overcome by P when slipping occurs ? Problem 397. The radius of a shaft is 1 in., A' = 20 lb., P = 20 lb., r/j = 3 ft., and «., = 2 ft. What force of friction will be acting at the point A, when the angles between P and ai and R and a 2 are right angles? What must be the value of the coefficient of friction? 171. Friction of Pivots. — The friction of pivots presents a case of sliding friction, so that tlie force of friction F equals the coefficient of friction times the normal pressure. That is, r = fN. (a) Flat- end Pivot. Assuming the pivot to press uniformly on the bearing, the friction on an element of area r'dOdr' is -<- ->- f—r'dedr'. (Fig. 245.) The total work done agrainst fric- FiG. 245 tion per revolution is rr /»2,r p Work per rev. = 1)2 irr'f — - r'dr'du %Jo «/ ITT = ^ r Cr"'dr'dd, 7-2 Jo Jo . or 4 irrfP Work per revolution = — e* — o (ft) Collar Bearing or Hollow Pivot. FRICTION 319 Here the work per revolution becomes (Fig. 246) Work per rerolntioii = f *^' T"^ 2 -nr' --4- — ^ r'dr'dQ _ 4_Tr ?•:] — ^'i ~ X r1 - ?i /i* 1^ ^ Fig. 246 (c) Conical Pivot. The conical pivots, illustrated in Fig. 247, do not usually fit into the step the entire depth of the cone. Let the radius of the cone at the top of the step be rj, a half the angle of the cone, and let dP^ be the normal pressure of the bearing on an elementary area whose horizontal projection is r'dr'dO. 820 APTLIED MECHANICS FOR ENGINEERS The vertical com- ponent of c?Pj is equal to the vertical load on the liori- zontal area r'dr'dO which is dP = -^r'dr'de. TTVi Trrf sin a and the force of fric- tion acting on this element is F.G. 247 '^'■1 «'" « The total work per revolution done against friction is therefore Work per revolution = C^ C — f^^ — 2 irr'^dr'dQ Jo Jo irr'i Sin a • _ 4 Trn/P 3 sin a TT If a = ^i this value for work lost reduces to the work lost per revolution, in the case of the flat-end solid pivot. It is easily seen, since sin a is less than unity, that if r^ is nearly equal to r, the friction of the conical bearing is greater than tlie friction of the flat-end bearing. Tliis might have been expected from the wedgelike action of the pivot on the step. It is also easily seen that r^ may FRICTION 321 be taken small enough so that the friction will be less than the friction of the flat pivot. The work lost due to friction in the case of the conical pivot will be equal to, greater, or less than, the work lost, due to friction in the case of the flat-end pivot, according as r, > r sin a. < (c?) Spherical Pivot. Suppose the end of the pivot is a spherical sur- face, as shown in Fig. 248. Let r be the radius of the shaft and r^ the radius of the spherical surface ; then the load per unit of area of P_ horizontal surface is Fig. 248 Trr" The horizontal projection of any elementary ring of the bearing, of radius x^ is 2 irxdx. Tlie load on this area is and the corresponding normal pressure is jn 2 Pxdx r, dP. = — sec p, and the total work done asfainst friction in one revolution IS Jo TrfPx^dx 'o r^ cos yS 322 APPLIED MECHANICS FOR ENGINEERS Here /S is a variable such that 2: = rj sin yS. .*. dx = r^ cos fid/3, and the expression for the work becomes Work per rev. = — ^- — 1 I sui'' pap 4 7rfP)ifa 1 . = — *^ ^1 77 — ^ sm a cos a = 2 ir/Pri r^-5: COtal, Lsm-a J since r = r^ sm a. IT If the bearing is hemispherical, a = — , and the work lost per revolution becomes The friction of flat pivots is often made much less by forcing oil into the bearing, so that the shaft runs on a film of oil. In the case of the turbine shafts of the Niag- ara Falls Power Company (see Art. 143) the downward pressure is counteracted by an upward water pressure. In some cases the end of a flat pivot has been floated on a mercury bath. This reduces the friction to a minimum. (See Engineering, July 4, 1893.) The Schiele pivot is a pivot designed to wear uniformly all over its surface. The surface is a tractrix of revolu- tion ; that is, the surface formed by revolving a tractrix about its asymptote. Its value as a thrust bearing is not as great as was first anticipated. (See American Ma- chinist, April 19, 1894.) FRICTION 323 The coefficient of friction for well-lubricated bearings of flat-end pivots has been found to vary from .0044 to .0221. (See Proc. Inst. M. E., 1891.) For poorly lubri- cated bearings the coefficient may be as high as .10 or .25 for dry bearings. Problem 398. Show that the work lost per revolution for the hemispherical pivot is 2.35 times the work lost per revolution for the flat pivot. Problem 399. The entire weight of the shaft and rotating parts of the turbines of the Niagara Falls power plant is 152,000 lb., the diameter of the shaft 11 in. If the coefficient of friction is con- sidered as .02 and the bearing a flat-end pivot, what work would be lost per revolution due to friction ? Problem 400. A vertical shaft carrying 20 tons revolves at a speed of 50 revolutions per minute. The shaft is 8 in. in diameter and the coefficient of friction, considering medium lubrication, is .08. What work is lost per revolution if the pivot is flat? What horse power is lost ? What horse power if the pivot is hemispherical? Problemi 401. What horse power would be lost if the shaft in the preceding problem was provided with a collar bearing 18 in. out- side diameter instead of a flat-end block? Compare results. Problem 402. A vertical shaft making 200 revolutions per minute carries a load of 20 tons. The shaft is 6 in. in diameter and is provided with a flat-end bearing, well lubricated. If the coefficient of friction is .004, what horse power is lost due to friction? 172. Absorption Dynamometer. — The friction brake shown in Fig. 240 is used to absorb the energy of the mechanism. It may be used as a means of measuring the energy, and when so used it may be called an absorption dynamometer. The weight TF, attached to one end of the friction band, corresponds to the tension in the tight side of an ordinary 324 APPLIED MECHANICS FOR ENGINEERS belt (see Art. 165), while the force measured by the spring S corresponds to the tension on slack side of a belt. Let W= T^ and S= T^\ then T^ — T^e^°-, just as was found in the case of belt tension. The work absorbed per revolu- tion is work = (I'j — 2^2)2 Trr-^, where r^ is the radius of the brake wheel. The horse power absorbed is 33,000 where n is the number of revolutions per minute. In many cases the friction band is a hemp rope, and in such cases it is possible to wrap the rope one or more times around the pulley, making it possible to make T^ — T^ large while T^ is small. The surface of the brake wheel may be kept cool by allow- ing water to flow over the inside surface of the rim, which should be provided with inside flanges for that purpose. 173. Friction Brake. — The friction brake shown in Fig. 249 consists of the lever EC^ the friction band, and the friction wlieel. Such brakes are used on many types of hoisting drums, automobiles, etc. Let the band ten- sions be T^ and T^^ and let W be the force causing the motion, that is, the working force, and P the force applied at the end of the lever EQ in such a way as to retard the rotation of the drum. We have here as before T^ = Tc^e^°- and the work per revolution {T^— T,^^iTr^-\- F^irr^,- By taking moments about A we have for uniform motion T^ — T^= 1^^^ 3^ where r^ is the radius of the shaft and F is the force of friction acting on the shaft. Taking FRICTION 325 moments about (7, we have P{EC) = T^d^ sin S + T^d^ sin /9. Fig. 249 Problem 403. A weight of one ton is being lowered into a mine by means of a friction brake. The radius of the drum is 1^ ft., radius of the friction wheel 2 ft., coefficient of brake friction .30, 8 = 4.5^ /? = 15S r/, = r/^ = 1 ft., EC = ^ ft., radius of shaft 1 in., coefficient of axle friction -Oi, and the weight of the drum and brake wheel is 600 lb. Find T^, T^, and P in order that the weight W may be lowered with uniform velocity. Problem 404. Suppose the weight in the above problem is being lowered with a velocity of 10 ft. per second when it is discovered that the velocity must be reduced one half while it is being lowered the next 10 ft., what pressure P will it be necessary to apply to the lever at £J to make the change? What will be the tension in the friction bands and the tension in the rope that supports PV? Problem 405. If the weight in the above problem has a velocity of 10 ft. per second, and it is required that the mechanism be so con- structed that it could be stopped in a distance of 6 ft . what pressure P on the lever and tensions I\ and T"., would it require? What 326 APPLIED MECHANICS FOR ENGINEERS would be the tension in the rope caused by the sudden stop? Cora- pare this tension with W, the tension when the motion is uniform. 174. Prony Friction Brake. — The Pron}- friction brake may be used as an absorption dynamometer as shown in principle in Fig. 250. Let W be a working force acting on the wheel of radius r and suppose the brake wheel to Fig. 250 be of radius r^. The brake consists of a series of blocks of wood attached to the inner side of a metal band in such a way that it may be tightened around the brake wheel as desired by a screw at B. This band is kept from turning by a lever OAD, held in the position shown by an upward pressure P, at A. Considering the forces acting on the brake and taking moments about the center, we have the couple due to friction, Prj, equal to the moment PiOA),ov ,Fr,^p(^OA~). Considering the forces acting on the wheel, and neglecting axle friction, we get for uniform motion of TF, Fr^^Wr. FRICTION 827 The energy absorbed is used in heating the brake wheel. The wheel is kept cool by water on the inside of the rim. The work absorbed per minute is 2 irr^Fn = 2 7rP( OA)n^ where n is the number of revolutions per minute. The force P may be measured by allowing a projection of the arm at A to press upon a platform scales. The horse power absorbed is ^'^'- 33,000 ' where OA is expressed in feet, and n is the number of revolutions per minute. 33 If OA be taken as — -^, a convenient length, the formula 2 TT reduces to H.P. = 1000 A dynamometer slightly different from the Prony dy- namometer is shown in Fig. 251. It differs only in the In this case the force P is meas- means of measuring P. Fui. 251 328 APPLIED MECHANICS FOR ENGINEERS ured by the angular displacement of a heavy pendulum W\. Taking moments about the axis of W^ and calling /•g the distance from that axis to its center of gravity and fi the angular displacement, we have Fr^ = TTi^g sin jS, so that the horse power absorbed may be written ^•^•" ^4 33,000 ■' where OA^ r^, and r^ are expressed in feet. If OA be taken as , this becomes 27r' Wiv^ sin pn H.P. = n 1000 The student should understand that the rotation of the mechanism at is not in every case due to a weight W being acted upon by gravity. In fact, in most cases, the motion will be due to the action of some kind of engine. This, however, will not change the expressions for horse power. QQ Problem 406. If W^ = 100 lb., OA =^--,r^ = 2 ft., and r^ = Q in., what horse power is absorbed by the brake if y8 is 30"^, and n is 300 revolutions per minute ? 175. Friction of Brake Shoes. — The application of the brake shoe to the wheel of an ordinary railway car is shown in Fig. 252, where W is the axle friction, F the brake-shoe friction, iV the normal pressure of the brake shoe, Gr the weight on the axle, and F^ and N^ the reaction FRICTION 329 Fig. 252 of the rail on the wheeL The brakes on a railway car when applied should be capable of absorbing all the en- ergy of the car in a very short time. The high speeds of modern trains require a system of perfectly working brakes, capable of stop- ping the car when running at its maxi- mum speed in a very short distance. The coefficient of friction between the shoes and wheel for cast-iron wheels at a speed of 40 mi. per hour is about ^, while at a point 15 ft. from stopping the coefficient of friction is increased 7 per cent, or it is about .27. The coefficient for steel- tired wheels at a speed of 6b mi. per hour is .15, and at a point 15 ft. from stopping it is .10. (See Proc. M. C. B. Assoc, Vol. 39, 1905, p. 431.) The brake shoes act most efficiently when the force of friction F is as large as it can be made without causing a slipping of the wheel on the rail (skidding). The normal pressure iV, corresponding to the values of the coefficient of friction given above, varies in brake-shoe tests from 2800 lb. to 6800 lb., sometimes being as high as 10,000 lb. Problem 407. A 20-ton car moving on a level track with a velocity of a mile a minute is subjected to a normal brake-shoe pres- sure of 6000 lb. on each of the 8 wheels. If the coefficient of brake friction is .15^ how far will the car move before coming to rest? 330 APPLIED MECHANICS FOR ENGINEERS Problem 408. In the above problem the kinetic energy of rota- tion of the wheels, the axle friction, and the rolling friction have been neglected. The coefficient of friction for the journals is .002, that for rolling friction is .02. Each pair of v^heels and axle has a mass of 45 and a moment of inertia with respect to the axis of rotation of 37. The diameter of the wheels is 32 in. and the radius of the axles is 2| in. Compute the distance the car in the preceding problem will go before coming to rest. Compare the results. Problem 409. A 30-ton car is running at the rate of 70 mi. per hour on a level track when the power is turned off and brakes ap- plied so that the wheels are just about to slip on the rails. If the coefficient of friction of rest between wheels and rails is .20, how far will the car go before coming to rest ? Problem 410. A 75-ton locomotive going at the rate of 50 mi. per hour is to be stopped by brake friction within 2000 ft. If the coefficient of friction is .25, what must be the normal brake-shoe pressure ? Problem 411. A 75-ton locomotive has its entire weight carried by five pairs of drivers (radius 3 ft.). The mass of one pair of drivers is 271 and the moment of inertia is 1830. If, when moving with a velocity of 50 mi. per hour, brakes are applied so that slipping on the rails is impending, how far will it go before being stopped ? The coefficient of friction between the wheels and rails is .20. 176. Train Resistance. — The resistance offered by a train depends upon a number of conditions, sucli as velocity, acceleration, the condition of track, number of cars, curves, resistance of the air, and grades. No law of resistance can be worked out from a theoretical consideration, be- cause of the uncertainty of the influence of the various factors involved. Formulse have been developed from the results of tests ; the most important of these are given below. FRICTION 331 Let H represent the resistance in pounds and v the velocity in miles per hour. W. F. M. Goss has found that the resistance may be expressed as B= .0003(Z + 347)?;2, where L is the length of the train in feet. (See Engineer- ing Record^ May 25, 1907.) The Baldwin Locomotive Works have derived the formula i? = 3 + - 6 as the relation between the resistance and velocity. When all factors are considered, this becomes i2 = 3 + ^ + .3788(0 + .5682(c?) + .1265(a), 6 where t = grade in feet per mile, c the degree of curvature of the track, and a the rate of increase of speed in miles per hour in a run of one mile. To get the total resistance it is necessary to include, in addition to the above factors, the friction of the locomo- tive and tender. This is given by Holmes (see Kent's "Pocket-Book") as R^= [12 +.3(v- 10)17], where TTis the weight of the engine and tender in pounds and R^ the resistance in pounds due to friction. Other formulse derived as the result of experiments are shown graphically in Fig. 253. 332 APPLIED MECHANICS FOR ENGINEERS RESISTANCE VELOCITY CURVE^ FOR RAILROAD TRAINS I 10 20 30 40 50 60 70 bO 90 100 VELOCITY IN IVIILE8 PER HOUR Fig. 253 FRICTION 333 The formulse themselves are as follows (see Engineering, July 26, 1907) : CuKVB Number Formula Authority 1 ^ = ^^^1:4 Clark 2 ^-^-m Clark 3 ^=^-^^^;62 Wellington •4 290 Deeley 5 i? = .2497 V Laboriette 6 R = 3.36 + .1867 v Baldwin Company 7 R = 4.48 + 284 v Lmidie 8 i^ = 2 + .24 y Sinclair 9 i? = 2.5+ ^^ 65.82 Aspinal It is evident that these formulae do not agree as closely as one would wish. The difference must be due chiefly to the different conditions under which the tests were made. These conditions should be taken into account in any application of the formulae to special cases. CHAPTER XIV DYNAMICS OF RIGID BODIES 177. Statement of D'Alembert's Principle. — A body may be considered as made up of a collection of individual particles held together by forces acting between them. The motion of a body concerns the motion of its individ- ual particles. We have seen that in dealing with such problems as the motion of a pendulum it was necessary to consider the body as concentrated at its center of gravity ; that is, to consider it as a material point. The principle due to D'Alembert makes the consideration of the motion of bodies an easy matter. Consider a body in motion due to the application of certain external forces or impressed forces. Instead of thinking of the motion as being pro- duced by such impressed forces, imagine the body divided into its individual particles and imagine each of the parti- cles acted upon by such a force as would give it the same motion it has due to the impressed forces. These forces acting upon the individual particles are called the effective forces. D'Alembert's Principle, then, states that the im- pressed forces will he in equilibrium with the reversed effec- tive forces. This amounts to saying that the system of effective forces is equivalent to the system of impressed forces. The truth of the principle is evident if it be granted that 334 DYNAMICS OF RIGID BODIES 335 the forces acting between any two particles of the body are equal and opposite and act in the line joining the particles. In any summation of all the forces acting on all of the particles of the body these forces acting between the particles of the body would cancel out and leave the summation of the impressed forces. 178. Simple Translation of a Rigid Body. — A body has pure translation when all points of the bod}^ have at each instant the same velocity. Consider a body (Fig. 254) under the action of the impressed forces Pj, Pg' Pg,-', and let a be the ac- celeration of all particles of the body. Choose axes of reference, taking the rc-axis parallel to the direction of the acceleration. Think of the body as divided up into particles and each particle under the action of the effective forces. Let dM be the mass of any particle of the body. The effective forces acting on dM can be combined into a single force, dT, 2 Fi = - jsin ^dN-\- fcos c^xJiT, 2 mom^vp = I 2 sin (j)dN— | z cos <^dT^ 2 vciomiy — — i z cos (f>dN— J z sin (\idT^ ]Smom,-2 = I rdT. Replacing dN and dT by their values from Art. 179, and replacing r cos (j> and r sin respectively by x and y^ these equations reduce to the forms, DYNAMICS OF RIGID BODIES 339 2Xi = - co^xM - ayM, (1) 2 r;. = - ay^yM + axM, (2) tZ,= Q; (3) 2 mom,,, = ©2 I yzdM — « j xzdM^ (4) 2 mom^y = — 0)2 I xzdM— a J yzdM^ (5) 2 mom^-2 = ai^. (6) These equations hold true at any instant of the motion of the body. Since x and y are the coordinates of the center of gravity, if the axis of rotation passes through the center of gravity, the right-hand sides of equations (1) and (2) reduce to zero. Also if the a;^-plane is a plane of symmetry of the body, I xzdM and I yzdM both reduce to zero, since for every term xQ-\- z)dM there is a corresponding term xQ— z~)dM, and for every term y(^-\-z~)dM there is a corresponding term ?/(— z')dM. Therefore, when the axis of rotation passes through the center of gravity, and is taken as the 2-axis, and the rr^-plane is a plane of symmetry of the body, the six equations for the forces acting on the body become : 2^.= 0, (10 SF, = 0, (20 2Z,= 0; (3') 5:mom,,= 0, (4') S mom.y = 0, (50 2 mom^, = al^. (6') 340 APPLIED MECHANICS FOR ENGINEERS This case is the one that usually comes up in engineer- ing problems, and so these simplified equations are more often used than the six more general equations. It will be noticed that these equations are exactly the same as the conditions for equilibrium as determined in Art. 56, except that 2 mom^-^ is not zero. 181. Reaction of Supports of a Rotating Body. — When a body is rotating about a fixed axis, the reactions of the supports on the body are impressed forces and the appli- cations of equations (1) to (6) of the preceding article will, when the other impressed forces and the angular veloc- ity are given, enable one to find the reactions of the sup- ports. As an illustration suppose the body (Fig. 257) to be a cast-iron sphere, radius 2 in., connected to a vertical axis by a weightless arm whose length is 4 in., the ^ axis being supported at two points as shown. The reac- tions of the lower support are P^, Fy, P^, and of the upper Pi, P;, 0. Let the body be rotated by a cord running over a pulley of radius 1 in. situated 2 in. below P^. Let the constant tension, P, in the cord be 10 lb. and suppose it acts parallel to the ?/-axis. Consider Fig. 257 DYNAMICS OF RIGID BODIES 341 the motion when the sphere is in the xz-^Vdiie. Take the icy-plane through the center of the sphere perpendicular to 2J, then j xzdM and j yzdM are both zero. Using the foot-pound-second system of units, and the weight of cast iron as 450 lb. per cubic foot, we have ^ = |, ^ = 0, a = 8.727 lb., M= .271, and 7, = .0708. Equations (1) to (6) (Art. 180) then become, P, + Pi = -. 1365 0.2, (1) P^ + P;-10= .1855 a, (2) P,- 8.727 = 0, (3) ^•^-p;+ii', = o. (4) P' - -ii'. + K8-727) = 0, (5) if=.0708«. (6) From (6) we obtain a= 11.77 rad./sec.2. Suppose the body to begin to rotate from rest and at the time under consideration it has been rotating 2 seconds. Then « = 2 a = 23.54 rad./sec. Solving the remaining equations, we obtain, P, = - 47.15 lb. Pi= - 27.94 lb. P, = 2.17 1b. PJ=9.42 1b. P, = 8.73 1b. A negative sign indicates that the force acts opposite to the direction assumed in the figure. 342 APPLIED MECHANICS FOR ENGINEERS Problem 412. A flywheel 3 ft. in diameter. The cross section of the rim is 3 in. x 3 in. and is made of cast iron. Neglect the weight of the spokes. This wheel is placed on the axis of Fig. 257, with its center on the axis, instead of the sphere. If the other conditions are the same, find the reactions of the supports. Problem 413. The sphere in Fig. 257 is replaced by a right circular cast-iron cone of height one foot and diameter of base one foot. The vertex is placed at the point of attachment of the sphere, and the base is outward. If the other conditions are the same, find the reactions of the sup- ports. Problem 414. In the problem of the sphere (Art. 181) find the angular veloc- ity at the end of 30 sec. and the reactions oi the supports for such speed. Problem 415. The two spheres of Fig. 258 each weigh 220 lb. and are 12 in. in diameter. The rod connecting them is inclined 30° to the horizontal and is rigidly connected to the vertical axle AB. Find the reactions at A and B when the system is rotating uniformly at 120 r. p. m., neglecting the weight of the rod. Problem 416. The spheres C and D of Fig. 259 weigh respectively 100 lb. and 50 lb. and their centers are 2 ft. and 1^ ft. respectively from the axis I of the shaft. Their planes are 90° from each other. Find the reactions at the supports A and B when they are revolving at the rate of 150 r. p. m., the reactions being com- FiG. 258 DYNAMICS OF RIGID BODIES 343 puted when C is (a) vertically above the shaft, (&) vertically below the shaft. Problem 417. A uniform sphere of radius 1 ft. and weight 2000 lb. is mounted on a horizontal axle of diameter 4 in., round which a small cord is wound carrying a weight of 50 lb. Neglecting fric- tion, find the angular acceleration of the sphere, the linear accelera- tion of the weight, the tension in the cord, and the velocity of the weight at the end of 5 sec. Suggestion. Consider the motion of the sphere and the weight separately, under the action of the forces acting on each. Problem 418. A rectangular slab 6 ft. by 8 in. by 2 in. is sup- ported by an axis through the slab perpendicular to the face 6 ft. by 8 in. at a distance of 1 ft. from one end and 4 in. from the side. The slab is pulled aside to a horizontal position and released. Find the reaction on the sup- port when the slab has ^^ ^' turned through 60° ; when through 90°, the weight of the slab being 320 lb. SuGGESTioisi". Choose z- axis as axis of support and a:-axis along the axis of the slab at the instants in question. Problem 419. Two drums whose radii are r^ = 16 in. and r^ = 12 in. are mounted as shown in Fig. 260. Their combined weight is 200 lb. and k = 14". The forces Gi and G2 act upon the drum, as well as journal friction amounting to 16 lb. The radius of the shaft is 1 in. Find the velocities of G-^ and G2 and the drums when the point of attach- FiG. 260 344 APPLIED MECHANICS FOR ENGINEERS ment of the cord on the small drum has traveled from rest at yl to ^' through 90°. Neglect the friction at B. 182. The Compound Pendulum. — A body rotating under the action of gravity about a horizontal axis is called a com- pound pendulum. Let Fig. 261 represent a com- pound pendulum rotating about an axis through perpendicular to the plane of the paper. Let the distance from the axis of ro- tation to the center of gravity of the body be d. Applying D'Alembert's princi- ple, we have Fig. 261 — Grd sin 6 = aL 0' or, since Y ^J' 7,9 -^0 — -— «^o' a= —^ sin 6. It was shown in Art. 120 that the tangential acceleration of a simple pendulum of length I is at = -g sin 6, and hence its angular acceleration is y, or a = — ^ sin 6. C Hence the angular motion of a compound pendulum is exactly the same as that of a simple pendulum whose DYNAMICS OF RIGID BODIES 345 lengtli, Z, is such that or '"d 7,2 This value, —?, is called the length of the equivalent simple pendulum. The time of a small vibration of a compound pendulum is therefore The point, 0, of the pendulum, about which rotation takes place, is called the center of suspension. The point O on OC such that 00' =^ is called the center of oscillation. It is that point at which the whole mass might be concentrated without changing the time of vibration, the center of suspension remaining the same. Problem 420. A board 4 ft. by 1 ft. by 1 in. vibrates about an axis perpendicular to the 4 ft. by 1 ft. face through a point of the board 18 in. from the center. Find the time of vibration for small oscillations, and the length of the equivalent simple pendulum. Problem 421. Show that the time of vibration is the same (for equal angles of vibration) for all parallel axes of suspension that are at equal distances from the center of gravity of the body. Problem 422. A cast-iron sphere w^hose radius is 6 in. vibrates as a pendulum about a tangent line as an axis. Find the period of vibration and the length of a simple pendulum having the same period. Locate the center of oscillation. 346 APPLIED MECHANICS FOR ENGINEERS 183. Centers of Oscillation and Suspension Interchangeable. — If k is the radius of gyration of the body about the axis through the center of gravity parallel to the axis of sus- pension, then rr\-\ n -i rC -p CL I hereiore i = ; — , a or d^l -d)= k\ In this formula c?and I — d enter in exactly the same way. It follows therefore that if 0' were taken as a center of suspension, would be the center of oscillation. That is, in a compound pendulum the centers of oscillation and suspension are interchangeable. This is easily verified ex- perimentally. Problem 423. In a plane through the center of gravity, (7, of a body circles of radii r and — are drawn with centers at C, where k is r the radius of gyration about a gravity axis perpendicular to the given plane. Show that the time of vibration is the same for all axes of suspension perpendicular to the given plane and passing through a point on the circumference of either circle. 184. Experimental Determination of Moment of Inertia. — The computation of the moment of inertia of many bodies is a difficult matter. It is often convenient, therefore, to use an experimental method in dealing with such bodies. The compound pendulum furnishes a means whereby such determinations may be made. From Art. 182, we find that the time of vibration of a compound pendulum is ^gd DYNAMICS OF RIGID BODIES 347 This may be written IT Multiplying both sides by M^ the mass of the body, we have It thus appears that if d^ the distance from to the center of gravity, is known (the center of gravity may be located by balancing over a knife edge) and also the weight (7, and the body be allowed to swing as a pendulum about as an axis, t may be determined, giving Iq. If Ig be desired, it may be determined from the formula (see Art. 64), Problem 424. The connecting rod of a high-speed engine tapers regularly from the cross-head end to the crank-pin end. Its length is 10 ft., its cross section at the large end 5.59" x 12.58" and at the cross-head end 5.59" x 8.39". Neglecting the holes at the ends, the center of gravity is 64 in. from the cross-head end. The rod is made of steel and vibrates as a pendulum about the cross-head end in 1.3 sec. Compute its moment of inertia about the gravity axis. The student should take such a connecting rod as the one in the preceding problem, or other body, and by swinging it as a pendulum find its period of vibration. Compute the moment of inertia about the axis of suspension and about the gravity axis. 185. Determination of g. — From the preceding article we see that M«r2 77^2 ^ df MtH 348 APPLIED MECHANICS FOR ENGINEERS This relation enables us to determine g, as soon as we know Iq, M, and d, by determining the time of vibration about the point 0, It is evident that -^ — Md is a constant for the body, when the axis is through 0, and that when once determined accurately, the pendulum might be used to determine g for any locality. L TT This constant, -^-—s is known as the pendulum constant. Md Problem 425. A round rod of steel 6 ft. long is made to swing as a pendulum about an axis tangent to one end andperpendicular to its length. The rod is 1 in. in diameter. Determine the pendulum constant. Problem 426. The center of gravity of a connecting rod 5 ft. long is 3 ft. from the cross-head end. The rod is vibrated as a pen- dulum about the cross-head end. It is found that 50 vibrations are made in a minute. Find the radius of gyration with respect to the cross-head end. 186. The Torsion Balance. — A torsion balance consists of a body suspended by a slender rod or wire attached rigidly to the body and at the point of support, the center of gravity of the body lying in the line of the wire (Fig. 262). Let OA be the neutral position of a line in the surface of the disk used as a torsion pendulum (Fig. 262). Let the disk be turned through an angle 6^ and released. The wire is Fig. 262 DYNAMICS OF RIGID BODIES 349 then twisted and exerts a twisting moment on the body, tending to restore it to the neutral position. Experiment shows that this twisting moment exerted by the wire is proportional to the angle of twist. Hence, since 2 mom = /a, we have for the motion of the disk, starting from the posi- tion where 6 = ^q, -oo = i<^f,, (1) the minus sign being used since co -— is the acceleration du counter-clockwise and CO represents the clockwise twist- ing moment. .'. Icod(o=- OOde. Integrating, 2 2 ^ When t=0,e=eQ,co = 0. .-. (7i=C^. .-. Ja)2= a(6'2-6'2). " (2) This equation shows that the angular velocity of the body is the same for negative values of 6 as for the nu- merically equal positive values. The motion is therefore periodic, the body vibrating through equal angles on both sides of the neutral position. Considering the motion from ^ = ^^ to ^= 0, equation (2) may be written de W V^2-6I2 ^I 350 APPLIED MECHANICS FOR ENGINEERS Integrating, 0- \-j-^-^' sin-i- = - x/- • ^ + a. When t=0, = 0^. r. C^ = sin-i 1. Choosing — as the value ofsin~il, we have This equation shows that as t increases sin"i— - must de- crease. Hence, since we chose sin~^"- = — when 6 = 6^^^ 6q 2 the angle — must decrease from — and will therefore reach zero when ^ = 0. /y Hence t = -^\yy when ^ = 0. Therefore the time for the body to make a complete swing, one way, is If m-^ is the twisting moment exerted by the wire when twisted through an angle 6^ m^ = C6^, and the expression for the period becomes The time of vibration is independent of the initial angu- lar displacement. DYNAMICS OF RIGID BODIES 351 187. Determination of the Moment of Inertia of a Body by Means of the Torsion Balance. — The moment of inertia of a body may be found by suspending it by a wire and ob- serving the time of vibration when used as a torsion m balance. The constant (7= -^ is a constant of the wire or rod and depends upon the material and diameter. Know- ing this constant, it would only be necessary to determine the period of vibration in order to find /. For practical purposes, however, it is desirable to elimi- nate from consideration the value m^ ^ For this purpose suppose the disk provided with a sus- pended platform rigidly attached as shown in cross section in Fig. 263. Let t be its time of vibration and / its mo- ment of inertia about the axis of sus- pension. Now place on the disk two equal cylinders H in such a way that their center of gravity is the axis of suspension. Let ^j be the period of vibration of the cylinders and support and ij their moment of inertia. Then — =— . The moment of inertia of the two cyl- inders with respect to the axis of rotation is known ; Fig. 263 call it ^. Then Ii = I+Iv so that I-I '" 352 APPLIED MECHANICS FOR ENGINEERS This gives the moment of inertia of the torsion balance, which, of course, is a constant. The moment of inertia of any body L may now be determined by placing the body on the suspended plat- form with its center of gravity in the axis of rotation and noting the time of vibration. Calling the time of vibra- tion of the body L and the balance ^3 and their moment of inertia ig, we have t = L Let the moment of inertia of L itself be JT^, so that Then h = r-^- This method may be used in finding the moment of inertia of non-homogeneous bodies, provided the center of gravity be placed in the axis of rotation. Problem 427. The moment of inertia of a torsion balance is 6300, where the units of mass, space, and time are respectively the gram, centimeter, and second, and its time of vibration 20 sec. The body L consists of a homogeneous cast-iron disk 3 in. in diameter and 1 in. thick. Find the time of vibration of the balance when L is in place. Compute the moment of inertia of the disk. Problem 428. The same balance as that used in the preceding problem is loaded with a body Z, and the time of vibration is found to be 30 sec. Determine the moment of inertia of L. 188. Rotating Body under the Action of no Forces. — If a body is rotating about a fixed axis, the axis of 2, and is acted upon by no forces, the equations of Art. 180 become DYNAMICS OF RIGID BODIES 353 = -(o^xM-a'^M, (1) = - (o'^M+ axM, (2) = 0, (3) = 0)2 CyzdM - a CxzdM, (4) () = -(o^C xzd M- a CyzdM, (5) 0=aJ,. (6) From these equations it follows that a = 0, CxzdM=0, CyzdM =^0, x = 0, y = 0. Hence, if a body is rotating about a fixed axis and no forces are acting on the body, (1) the angular velocity is constant, (2) the axis of rotation passes throiigh the center of gravity, (3) if the axis of rotation is chosen as the axis of z, then CxzdM= 0, and CyzdM= 0. (When ixzdM=0 and I yzdM=0, the 2-axis is a prin- cipal axis of the body. It can be shown that there are three lines at right angles to each other through any point of the body which are principal axes, and that the ellipsoid of inertia of the body for tllat point has its axes lying on these three lines. Compare Art. 77.) 189. Rotation of a Body about a Fixed Axis with Constant Angular Velocity. — If the angular velocity is constant, a = 0, and the equations of Art. 180 take simpler forms. 2a 354 APPLIED MECHANICS FOR ENGINEERS In addition, at a» given instant, let the a^-axis be chosen through the center of gravity of the body. Then ^ = 0, and equations (1) to (6) of Art. 180 become 2X, = -G,2^i!f, (1) 2 F, = 0, (2) 2Z, = 0, (3) S inoni^^ = ft)2 I yzdM^ (4) 2 momjj/ = — ft)2 J xzdM^ (5) 2 mom,., = 0. (6) The first equation shows that unless x is zero there is a resultant of all the x-components of the impressed forces and that this resultant is R^=- uP'xM. Equations (2) and (3) show that the ^-components of the impressed forces either have a resultant zero or else form a couple, and the same of the ^-components. A particular case is that where i^xzdM=^ 0, and (yzdM= 0, as, for example, where the a^^-plane is a plane of symmetry. Suppose in this case the only impressed forces are the weights and the reactions of the supports, as indicated in Fig. 264. Equations (1) to (6) then become P^4-P; = -a)2^i^f, 0') Py^P[=^^. (2') P.-Tf=0, (30 DYNAMICS OF RIGID BODIES 355 hPy - aP[ = 0, aP',^xW-hP, = 0, 0=0. Fig. 264 From (20 and (4'), Py =0, P^ = 0. From (3'), P, = TT. (4') (5') (6') From (!') and (5'), Px = a: a "" -: W ^—^xaP'M. p; = _ If the weight W were counterbalanced by an equal upward force acting through (7, the only forces acting on the axis would be P, = ^ xay'^M, PI = ^—- xco^M. a -{- a + 356 APPLIED MECHANICS FOB ENGINEERS The resultant of these forces is a single force, and if its distance above the a;^-plane is z\ z'R. = aPL-hP^ a -\-b a -\-h = 0. Therefore, when the a??/-plane is a plane of symmetry, or when j xzdM= and j yzdM = 0, the effect of the rota- tion of the body about the axis of z is to cause an outward pull on the axis of rotation in a line through the center of gravity perpendicular to the axis of rotation and of the same value, MxuP'^ that would be caused by concentrating the whole mass at the center of gravity. Problem 429. A wheel of weight 100 lb. is mounted on an axle and is off center \ inch, the plane of the wheel being perpendicular to the axis. Find the force tending to bend the shaft when the wheelis making 200 r. p m. Problem 430. A thin rod, 2 ft. long, weighing 5 lb. is attached to an axle at an angle of 60°. Find the outward force in magnitude and position in the plane of the rod and axle due to the rotation when mak- ing 150 r. p. m. If the rod is joined to the axle 2 ft. and 1 ft. from the supports (Fig. 265), find the reaction at the supports due to the rotation. Problem 431. A steel disk 3 ft. in diameter and 1 in. thick is not perpendicular to the axis of rotation, but is out of true by -^ of its radius. Find the twisting couple introduced tending to make the shaft wobble. DYNAMICS OF RIGID BODIES 357 Problem 432. A grindstone weighing 200 lb., of radius 2 ft., is making 100 r. p. m. Find the force with which one half of the stone pulls on the other half. Problem 433. Neglecting the weight and the tension in the spokes of a rotating fly- wheel, prove that the tension in the rim is P = iip-r^ yF 9 Fig. 266 where w = angular velocity, y = the heaviness of the material, F = the area of the cross section of the rim, and r = mean radius of the rim (Fig. 266). Problem 434. In the preceding problem, suppose r = 6 ft., F=10 in. by 4 in., and the wheel is made of cast iron. If the tensile strength of the material is 25,000 lb. per square inch, what speed would be attained before the wheel bursts ? Problem 435. Show that if P is the pressure of the side rod of a r> locomotive on the crank pin, r the ra- dius of the crank-pin circle, r' the radius of the driver, and v the velocity of the train, Grv^ dN □cLY G Fig. 267 2P-G^ /2 gr when the side rod is in the lowest position (Fig. 267). Problem 436. Suppose the velocity of a locomotive to be 90 mi. per hour, the radius of the crank-pin circle 20 in., the radius of the drive wheel 40 in., and the weight of the side rod 400 lb. Find the pressure on the crank pins due to the rotation alone. 358 APPLIED MECHANICS FOR ENGINEERS 190. Rotation of a Locomotive Drive Wheel. — The drive wheel of a locomotive (Fig. 268) may be considered for the present as rotating about a fixed axis. We shall con- sider the effect of the weight of the counterbalance on the tire due to rotation only, on the assumption that the tire carries all the weight of the counterbalance. Note. It is to be understood that the wheel center carries part of the weight of the counterbalance, but a complete solu- tion of the problem of the drive wheel is beyond the scope of this book. The above assumption is therefore made. Let M be the mass of J of tire and p the distance of its center of iG. 268 gj^.^^^^^ from the center of wheel. Let iHfj be the mass of the counterbalance, and p^ the dis- tance of its center of gravity from the center of wheel. Then 2 P = 0)2 (Mp + M^p^^. In particular, suppose the diameter of the tread of the tire to be 80 in., distance of the center of gravity of ^ of tire from center 27 in., and mass of ^ of tire 21. The mass of the counterbalance is 20, and the distance of its center of gravity from the center of the wheel 29 in. Substitut- ing these values, we get 2P = a)2 [21 (f I) + 20 (II)] =95.6 6)2. If now we know the speed of rotation of the wheel so that CO is known, we may determine P. Let us take co corresponding to a speed of train of 60 mi. per hour. DYNAMICS OF BIGID BODIES 359 This gives o) = 26.4 radians per second and P = 33,300 lb. Problem 437. If the area of the cross section of the tire is 20 sq. in., find the stress on the metal due to rotation about the axis under the above assumption. If the allowable stress on the metal is 20.000 lb. per square inch, find the speed of train to cause this stress. 191. Standing and Running Balance of a Shaft. — Let weights, TFj^ ^^ W^, W^, etc., be attached to a shaft, the distances of their centers from the center of the shaft being r^, r^, ^3, r^, etc. (Fig. 269), it being assumed that Fig. 269 each weight is symmetric with respect to a plane through its center and perpendicular to the axis of the shaft, so that the pull of the weight on the shaft due to rotation is through the center of gravity of the weight (Art. 189). If the shaft is balanced when at rest in any horizontal position, the sum of the moments of the weights about the axis must be zero for any position. The moment of 360 APPLIED MECHANICS FOR ENGINEERS Wi is Wi times the horizontal component of r^, or it is the horizontal component of W^r^ laid off along r^ The weights will then be in standing balance when, and only when, the sum of the horizontal components of the vec- tors r^TFj, r^W^, ^^3^1 etc., is zero for any horizontal posi- tion of the shaft. Hence the vector polygon must close. For running balance the reactions at each support, due to the centrifugal pull of the weights, must annul or the shaft will tend to wobble. These forces due to rotation are along the radii (Art. 189), and if x-^, x^, 2^3, ... are the distances along the shaft of the weights from one bearing, the reactions at the other bearing are -^r^ccP'M-^^ -^r^ctP'M^^ • .., where I is the distance between the supports, and M^, iHigi ^31 etc., are the masses of the weights. These reactions are therefore proportional to x^^W-^^ ^2^2 ^2' ^^^"> ^^^^ ^^6 in the directions of r-^, rg, etc., re- spectively. In order to balance, the polygon of vectors ^i^i^i + 'V2^2+--* must therefore close. In order that the reactions due to rotation balance at the other bearing, the vector polygon (I - x^y^ TFi -f (Z - x^y^ TF2 + . . . must close. But since the vectors may be taken in any order, this polygon may be plotted in the order DYNAMICS OF RIGID BODIES 361 which clearly will close when the two polygons already considered close. There will then be standing and running balance when two closed polygons can be formed of vectors r^ TTj, r^ "B^, r^W^, etc., and x-^r^W-^, x^r^W^, x^r^W^^ etc., respectively, the vectors having the directions of the perpendiculars from the center line of the axle to the centers of gravity of the corresponding weights. Problem 438. In Fig. 269 assume W^ = 6, W^ = 5, TFg = 4 ; Tj = 4.3, rg = 3, Tg = 6, r^ = 2 ; angle between r^ and r^ = 135°, angle between r^ and r^ = 90° ; x^ = 1.5, 0:2 = 1. Find x^ and the value and position of W^ for running balance. Problem 439. Show that two weights, given in position and magnitude in the same plane perpendicular to the axis of rotation, can be balanced for running by a third given weight in that plane. Show how to determine the position of this third weight, and show that the condition for standing balance is the same as for running balance in the case of weights all in the same plane perpendicular to the axis of rotation. Problem 440. If two weights are attached to the axis in the same plane containing the axis, show how to find the position of a third given weight that will balance the two given weights for running and standing balance. If balanced for standing, will they be balanced for running ? Problem 441. Show that three weights cannot be balanced for running unless they either lie all in one plane perpendicular to the axis of rotation or else all in a plane containing the axis. Problem 442. Show that four weights and the arms of three of them being given and two of the weights definitely located on the axis, the other weights may be placed so as to form standing and running balance, and show how the location of the two weights would be de- termined. Could this be done in more than one way? 362 APPLIED MECHANICS FOR ENGINEERS 192. Rotation of Flywheel of Steam Engine. — In Fig. 270 let a belt run horizontally over a flywheel, the tensions being F^ and P^ where P^ > P^ The effective steam pressure is P, J^' the pressure of the guides on the cross- head. It is normal if friction is neglected. .— TTf^^EEE , -, H ^ni-^, r^ -2a- FiG. 270 The pressure on the crank pin is resolved into tan- gential and radial components, T and i^j. From the relation, 2 moms = aZ, we have for the motion of the wheel Ta - (P2 - ^1)^ = «^- (1) It will be assumed that the resistance of the machinery, as shown by P^ — Pj, is constant. When T=(^P^ — Pi)-, « = 0, and the angular velocity, CO, is either a maximum or a minimum. When ^=0 (Fig. 270), T=0, a is negative, and co is DYNAMICS OF RIGID BODIES 363 therefore decreasing. Hence co has a minimum value in the first quadrant at B^ where T= (Pg — ^i)~' From B^ the angular velocity increases as long as T remains greater than {P^ — P^-, In the second quad- rant as T decreases toward zero at the dead point A, there is a point A^ where jr= (Pg — P\)- ^^ which therefore g) has a maximum value. In the same way the points A^ and B^ in the third and fourth quadrants correspond to minimum and maximum values of o). Equation (1) may be written 7a)c?a) = TadS - {P^ - P^rdO, (!') If s and s' are the arcs passed over in the crank-pin circle and the flywheel circle respectively, then adO = d% and rdd = c?s', and equation (!') becomes on integrating from an initial value ft)Q to any value «, \ iQ^ - 0,2) = frds - (P, - Pj) (s' - «;). Since the work done on one end of the connecting rod equals the work done by the other, neglecting its own change in kinetic energy, CTds = Cpdx, and hence 364 APPLIED MECHANICS FOR ENGINEERS Pdx may be found by read- ing from the indicator card the values of P for successive values of x between the limits x and x^. A different treatment of the above equation may be obtained by considering that the pressure of steam in the cylinder is constant and equal to P' up to the point of cut-off and that beyond this point the pressure varies inversely as the volume. If we assume P constant and equal to P' to the cut-off, then the limits of integration will be regarded accordingly, and we may write J /(<»f - 0,2) = P'ix^ -x,-)-(F,^ Pi) (.s-i - «;>, where x-^ is the value of x at the cut-off. Beyond the cut-off P varies inversely as the volume of steam in the cylinder, or P = ^P'. X Then from the point of cut-off to any value of x, ^f'l X = x,P' log, (^) - (P, - PO (s' - s[). If P[ is the mean effective pressure, we have, considering the work done on the flywheel for the motion from B to A^ neglecting friction and the change in kinetic energy of the connecting rod, 1 7(0,2 - 0,1) = 2 aP[ - (P, - P,-)7rr. Problem 443. Suppose the mean effective steam pressure is 16,000 lb., the radius of the crank -pin circle 18 in., and the radius of the fly- DYNAMICS OF RIGID BODIES 365 wheel 3 ft. If P^- P^= 500 lb., /"^ = 2000, and w^ = 2 tt radians per second, find co^. Problem 444. The flywheel in the above problem has a velocity (0^ = 6 TT radians per second. What constant resistance (Pg — -^i) "^ill change this to 2?? radians per second in 100 revolutions? Problem 445. Find the values of for which w is maximum and minimum in the above problem. Problem 446." In Problem 443 find the value of w when the wheel has turned through 90° from B. Problem 447. If the total pressure in the above problem is 20,000 lb. up to the cut-off at } the stroke, find a = —- = v.[ cos u — - — sin u tan 9 h cos u sec^(p —^ dt \ dt dt dt. = V "cos (^ + COS^ (j) or a = -^^ [«i cos (e + <|>) cos The value of co from equation (4) may be substituted in equation (6) and thus the value of a determined. Problem 448. The connecting rod given in Problem 424, Art. 184, is in use on an engine whose crank has a constant angular velocity of 26 radians per second. The length of the crank is 2 ft., the effective steani pressure on the piston is 16,000 lb. Use the values of M, I, X, and y from Problem 424. Find the values of N', N^, and T when e = 30°. Problem 449. Show that w has its greatest numerical value when o ^ = and 9 = TT, and its least numerical value when 6 =— and 6 = — . 2 2 What are these greatest and least values? Problem 450. Find what values of 6 will make a a maximum or minimum. Locate the crosshead for these values. Problem 451. Find values for T, N', N-^, and the resultant pressure on the crank pin when — tt and when ^ = 0. Use the above data. Problem 452. Assume a force of friction F acting on the cross- head, such that F = .06 N'. In the above case when 6 = 30°, what is the value of F, N', N^, and T? DYNAMICS OF BIGID BODIES 369 Fig. 273 Problem 453. Suppose the steam pressure zero, find T, N', N^, and the resultant crank-pin pressure, if w^ is the same. 195. Angular Momentum or Moment of Momentum. — If the velocity of a particle is resolved into two components, one of which is parallel to a given line and the other in a plane perpen- dicular to the line, the product of the latter component, the perpendicular distance of this component from the given line, and the mass of the particle is called the moment of momentum or angular momentum of the particle with respect to tlie given line. Thus, if v^ is the component of the particle in the plane perpendicular to the given line, d the distance of this component from the given line, and m the mass of the particle, then its angular momentum = mv-^d. (Fig. 273) If the given line be taken as the axis of x^ the component Vj may be resolved into components Vy and v^ parallel to the y- and 2-axes, respectively. Then, since the moment of any vector is equal to the sum of the moments of its components, w^e have angular momentum of the particle with respect to 2:-axis = (yv^ — zvy)m. If dm is the element of mass of any body, then the angu- lar momentum of the body about any line is defined to be the sum of the angular momenta of its particles about that line. Thus, representing the angular momenta of a body 2 B 370 APPLIED MECHANICS FOR ENGINEERS about the x-^ 3/-, and 2-axes by hj,. hy, h^^ respectively, we have 196. Torque and Angular Momentum. — Let dm be the mass of an element of a body, and a^., a^, a^, respectively, the components of its acceleration parallel to the axes. Using D'Alembert's principle, we may equate the sum of the moments of the effective forces to the sum of the moments of the impressed forces. The effective forces acting on dm are a^dm^ aydm^ a^dm^ respectively parallel to the x-^ y-^ and 2-axes. Hence call- ing the torques due to the impressed forces about the axes respectively T^., Ty^ T^, we have T^ = J {ya^ - zay^dm, Ty= I {za^ — xa^dm^ T^= \{xay — ya^dm. ^-r dx dy dz Now ., = -, ^ = j, .. = -, _ ^'^x _ dVy _ dv^ ^ dt ^ dt " dt Also -TXy'^z — ^'^y)='^y'"z + y^z—'^zVy — zay=ya^ — zay. 'd '■• T^=jj^iy^z-zvy)dm. Since the derivative of a sum of terms is the sum of the derivatives of the separate terms, and the integral is the DYNAMICS OF RIGID BODIES 371 limit of a summation, we may write this latter equation Likewise Ty = -^^, or rp^^dh, at d\ dt and T, = ^'' dt That is : The moment of impressed forces about any axis is equal to the rate of change of the angular momentum about that axis. A corollary of this theorem is that when the moment of impressed forces acting on a body about an axis is con- stantly equal to zero, the angular momentum about that axis is constant. 197. Moment of Momentum of a Body with One Point Fixed in Terms of Angular Velocity. — If a body in motion has one point fixed, its motion at any instant is one of instantaneous rotation about an axis passing through the fixed point. Take the fixed point, 0, of the body as origin of coordinates. The angular velocity about the instantaneous axis can be resolved into component angular velocities, oj^., a)^, co^, about the axes (Art. 132). The velocity v^ or — is then due to rotation about OY ^ dt and OZ. Figure 274 shows the component parallel to OX of the velocity due to rotation about OZ to be — r-^(Oz cos /3 or — yw^. o72 APPLIED MECHANICS FOE ENGINEERS In the same way the rc-component of the velocity due to rotation about OY is zcoy. Wz z dx di = ^^y - y^z Similarly and dy -&- = XQ), — 2a)_, dt dz dt y^X - ^03y. Substituting these values in the expressions for A^, we obtain — I (j/^ + z'^)o)^dm — 1 xywydm — i xzco^dm^ or Aj. = 1^03^ — Wyi xydm — (oA xzdm. Similarly, hy = IyOi>y — (nA yzdm — (o^l yxdrn^ and hz = I^(o^ — (^x\ ^xdm — (Oyi zydm. DYNAMICS OF RIGID BODIES 373 The only body whose motion we shall study in this book is a uniform body in the form of a solid of revolution. If one of the axes be taken as the axis of the body, then all of the products of inertia vanish and the expressions for the angular momenta become 198. Vector Representation of Angular Momentum. — Sim- ilar to the representation of angular velocity by a vector, angular momentum about any line may be represented by a vector along the line. Since in the special case we shall study the angular momentum about any axis used is pro- portional to the angular velocity about that axis, it is clear that the angular momentum may properly be treated as a vector in the way defined. The signs of the angular momenta will be the same as those of the corresponding angular velocities, and the vectors will be laid off in the same way as for angular velocities. 199. Kinetic Energy of a Body with One Fixed Point. — The kinetic energ}^ of a body at any instant is the sum of the kinetic energy of its particles, or K. E. = ^ Cv^dm = 1 C^v^ +vl-{- vl^dm. In Art. 197 it was proved that The substitution of these values in the expression for kinetic energy gives K. E. = ljl(f + /)a,J + (a;2 ^ ^2^)^2 ^_ ^^2 _^ x'^)(o';]dm 374 APPLIED MECHANICS FOR ENGINEERS J [zi/Q)yCo^ H- xzoo^o)^ + xi/co^Qyy^dm, or, K. E. = I J o>2 + 1 J 0)2 + ^ J a>2 — Wy«g j zydm — (>>^o>z i xzdtn, — "jpO^ j ocydm. \ For the special case of the body of revolution with one of the coordinate axes coinciding with the axis of the body, or for any body when referred to principal axes, K. E. = - J «2 + i J «2 + - J 0)2 ^* ^* 2 ^x X ^ 2 y y ^ 2 ^z z- 200. Vector Rate of Change Due to Rotation. — Let h be a vector with point of application at 0. Suppose the vector is changing with the time in magnitude and direction but keeps the point of application the ^^' same. Let the vector change in length from li to h-\- Ah in time A^ and in that time turn through the angle A^. The rate of change in the original direction of the vector is then Limit (^ "^ A^)cos AO — h ^t^ Alt and the rate of change perpendicular to the original direc- tion of the vector is A/=0 A^ The first of these quantities may be written Limit -A(l-cosAg) M^^^^^- At At DYNAMICS OF RIGID BODIES 375 = Lim = Lim h '2 ' sin^ -h A0- At + dh ~di . Ad sm — 2 . AS AO sm — • — AS 2 A^ + dh dt _ dh ~ dt' dh dt The second is A^^o^ ^ A^ A^ = A • 1 • — = Aft), where g) is the angular velocity of rotation of the vector. Hence the rate of change in the direction of the vector is dh di' and the rate of change at right angles to the vector is hco. 201. Rate of Change of Angular Momentum of a Body Due to Rotating Axes. — Let the components of the angular mo- mentum of a body with fixed point with respect to axes OX, OY^ OZ be respectively Aj, Ag^ ^31 ^^^ ^^^ ^^^^ frame of axes coincide at a given instant with a fixed frame OXp 37() APPLIED MECHANICS FOR ENGINEERS 0T\, OZ^^ but be rotating about the fixed frame with angular velocities tWj, co^^ co^. Lay off the vectors 7ij, Ag^ ^h ^^^ ^^^ moving axes (Fig. 276). Since the moving axes at the given instant coincide with the fixed axes, the com- ponents of the momentum are the same for the fixed and moving axes at that instant. The rate of change of angular momentum along the fixed ^ axes is then given by the method of the preceding article. Thus along OX^ the z^ h. X, Y rate of change is — l, due to Fig. 276 the change in h^ This is in- creased by Ag&)2 by the rotation of Ag about the i/-axis and by — ^2^3 by the rotation of h^ about the 2-axis. Thus the rate of change of angular momentum about the fixed x- axis is dt - h^(Oo + AoQ)o. 2rr Similarly about the fixed ?/-axis the rate of change is dK dt ^ — AgCOj + 7^lCt> 3' and about the fixed s-axis the rate of change is dt ^ ^ ^ ^ DYNAMICS OF RIGID BODIES 377 Since the moment of impressed forces about any axis is equal to tiie rate of change of the angular momentum about that axis, we may write dhc .2«^3 .3^,2, ^.= -^- Vl+^«1«3' T, dJi^ dt h^a)^ + /igCOj. dm. ax 202. Motion of the Center of Gravity of Any Body. — Let dm be the mass of any element of a body and (x^ y, 2) the coordinates of dm (Fig. 277). The effective forces acting on dm are re- spectively a^dm, aydm^ and a/Lm, Equating the components of the im- pressed and effective forces acting on the body, parallel to the a:;-axis, we have 2Xi = ^ajim. (1) Now if X is the abscissa of the center of gravity of the bod}^ Mx — ^xdm. Differentiating, j^dx ^ ^dx^ dt dt Fig. 277 and (i"X dj X M — - = S — dm = ^ajim. dt^ dt^ 378 APPLIED MECHANICS FOR ENGINEERS Therefore, if ^^ is the a;-component of the acceleration of the center of gravity of the body, equation (1) may be written 2X, = Ma,. In like manner, ■ ^Y, = Ma,, 2Z,. = Ma,. Hence, for any motion of a body the acceleration of the center of gravity is the same as if the whole mass were con- centrated there and all the impressed forces were applied there parallel to their original directions. 203. Gyroscope. Simplest Case — Consider a body in the form of a solid of revolution turning with uniform angular velocity, o), about its geometric axis, which is horizontal, while at the same time ^ this axis, passing always through a fixed point, turns with uniform an- gular velocity, II, about the vertical (Fig. 278). The problem is to de- termine the forces neces- sary to maintain this motion. Let 00 he the axis of the body, moving with the body. Choose fixed axes OX, OY, OZ, so that the a;-axis coincides with 0(7 at the given instant. Let OA be the horizontal axis perpendicular to 00. At the given instant OA coincides with OY. Fig. 278 DYNAMICS OF RIGID BODIES 379 Denote the moment of inertia of the body about OChj (7, and that about OA or OZ by A. Then from Art. 197 the angular momenta about 00^ OA^ and OZ slvq respectively h^ = (7ft), ^2 =0, Ag = AD^. These values of A^, h^^ and Ag, while they are the values of the angular momenta about the fixed axes OX^ OY, OZ, at the given instant, are not values which hold in general, and cannot therefore be differentiated to obtain the torques about these axes. The method of Art. 201 applies here where 0)^ =0, (^2= 0, ft>g = 12. We may write then ^x= ^^^ Agftjg 4-^30)2 = 0, (1) Ty = — -^ — AgWj H- Ajft>g = (7ft)ll, (2) y, = ^ - ^i<»2 + ^2»i = 0- (3) For the motion of the center of gravity of the body, since the center of gravity is moving uniformly in a horizontal circle and is at the given instant on the ic-axis, the acceler- ation is directed toward the point of support along the rc-axis. Hence a^. = — bD,\ ay = 0, a^= 0, where h is the distance from the point of support to the center of gravity. .-. 2X, = -itfm2, (4) 2:F, = 0, (5) 2Zi=i). (Art. 202) (6) 380 APPLIED MECHANICS FOR ENGINEERS The impressed forces necessary to satisfy equations (1) to (6) may, under certain conditions, be only the weight of the body and the reaction of the support. If these are the only impressed forces, equation (2) becomes Wb = (7a>n, where W is the weight of the rotating body. The re- maining equations are then satisfied, so that the necessary and sufficient condition that the body have the motion described is C(aQ = Wb. The motion of the body around the vertical axis is colled, precession. When the angular velocity around the vertical axis is constant, the precession is said to be steady. Problem 454. If the rotating body is a thin disk 2 ft. in diam- eter fastened to a rod of negligible weight which passes through the center of the disk perpendicular to its faces, the center of the disk being 6 in. from the point of support, what will be the rate of pre- cession in a horizontal plane when the angular velocity of rotation of the disk about its axis is 300 r. p. m. ? Ans. 9.79 r. p. m. Problem 455. How would the rate of precession vary with the radius of gyration of the disk if the angular velocity co be kept un- changed ? 204. The Gyroscopic Couple. — Equation (2), Art. 203 shows that when a body of revolution is rotating uniformly about its axis and this axis is rotating uniforml}^ about an axis at right angles to it, there is acting on tlie body at right angles to the other two axes a couple whose value is the product of the moment of inertia of the body about its axis of revolution and the two angular velocities. This DYNAMICS OF RIGID BODIES 381 couple is called the gyroscopic couple. In the case of the body discussed in Art. 203, the gyroscopic couple is (7a)Il. Since action and reaction are equal and opposite, the rotating body will resist the turning of its plane of rota- tion with a couple equal and opposite to the gyroscopic couple. This finds application in the effect of rapidly rotating wheels of cars, automobiles, etc., when turning around curves. In Fig. 278 if the disk represents a wheel of a car rolling in the direction indicated by co and at the same time turning about OZ., the rotation about OZ would be in the opposite direction; i.e. H would be negative in the formula Ty = CcoH. The gyroscopic couple acting on the wheel would then be in the opposite direction, or about 01^ in the direction from OX to OZ. The couple that the wheel would exert, tending to upset the car, would then be about OYiw the direction from OZ to OX. Problem 456. A ship carries a cast-iron flywheel whose rim is 6 ft. outside diameter, 4 in. thick, and 18 in. wide. When it is making 3 revolutions per second, its axis is turned about an axis through the plane of the wheel with unit angular velocity. Find the moment of the couple that tends to turn the wheel about an axis perpendicular to these two axes. Problem 457. A solid cast-iron disk 3 ft. in diameter and 3 in. thick revolves about its axis, making 3000 revolutions per minute. At the same time it is made to turn about an axis in its plane at the rate of 2 revolutions per minute. Find the magnitude of the couple tending to rotate the disk about an axis perpendicular to these two axes. Problem 458. A locomotive is going at the rate of 40 mi. per hour around a curve of 600 ft. radius. The diameter of the drivers is 80 in., and a pair of drivers and axle have a moment of inertia 382 APPLIED MECHANICS FOR ENGINEERS about an axis midway between the wheels and perpendicular to the axle of 3000. What is the magnitude of the couple introduced by the precession al motion of this pair of wlieels ? Give the direction in which it acts. Does it tend to make the locomotive tip inward or outward ? Problem 459. A car pulled by the locomotive in the preceding problem has four pairs of wheels. The moment of inertia of each pair of wheels and their connecting axle, with respect to an axis mid- way between the wheels and perpendicular to the axle, is 320. What is the magnitude of the precessional couple acting upon tlie whole car? Problem 460. The flywheel of an engine on board a ship makes 300 revolutions per minute. The rim has the following dimensions : outside radius 4 ft., inside radius 3| ft., width 12 in. The ship rolls with an angular velocity of ^ a radian per second ; find the toi'que acting on the ship due to the gyrostatic action of the flywheel. 205. Car on Single Rail. — An interesting application of the gyroscope has been made recently in England. A car is run upon a single rail, and is held upright by means of rapidly rotating flywheels. Each car contains two of these wheels rotating in opposite directions, at the rate of 8000 revolutions per minute. Any tendency of the car to tip over, either when running or standing at the station, is righted by the gyroscopic action of the flywheels. The experimental car was so successful in operation that it maintained itself in an upright position even when loaded eccentrically. The action of the flywheels is such as to place the center of gravity of the car and load directly over the rails. See Engineering^ June 7, 1907. Another practical application of the gyroscopic couple is to be found in Schlick's ''stabilizer" for steadying ships. DYNAMICS OF RIGID BODIES 383 206. Gyroscope. Inclined Axis. — Let the angular veloc- ity of the body relative to its own axis, 0(7, be « at any instant, the angular velocity with which the plane con- taining the axis turns about the vertical be 11, and the angle which the axis of the body makes with the vertical be 6. The component about 00 oi the total angular velocity of the body is then co + the component of 11 about 00 or co 4- II cos 0. The angular velocity co is called the velocity of spin. Let dt = ft)^ Then (o' is the angular velocity of the axis of the body in the vertical plane containing the axis. Choose fixed axes so that at the instant in question the a:3-plane contains the axis of the body, the x- axis being horizontal (Fig. 279). Let OA, OB, 00 he a set of moving axes, of which 00 is the axis of the rotating body, OB is hori- zontal, and OA is at right angles to 0(7 and OB. Let 0^1, OB^, 00^ be fixed axes which at the given instant coincide with OA, OB, 00 respectively. The frame of moving axes OABO then has angular velocities about the fixed axes 0^4^, OB^, 00-^ respectively ft)j = — fl sin 6, 0)^ = co', cOq = fl cos 0. The angular velocity 12 may be resolved into components 384 APPLIED MECHANICS FOR ENGINEERS about OA^ OB, (9 C' respectively — XI ain ^, 0, O cos 0. The angular velocities of the body about the axes OA, OB, O' respectively are therefore — fl sin 0, w\ and co + O cos 6. If the moments of inertia of the body about the axes OA, OB, 00 are respectively A., B, C, or, because of sym- metry, A, A, 0, the angular momenta of the body about these axes are respectively A-^ = — A^ sin 6, h^ = Aco' , h^ = 0(^co + n cos 0). The discussion will be divided into two cases : (a) steady precession, (5) unsteady precession. (a) Steady Precession. Suppose the angle and the angular velocities co and XI are all constant. For this case ft)' = and \ = — AD. sin 6, h^ = 0, Ag =: (7(ft> + XI cos ^), J = — XI sin 6, 0)2 = 0, cog = XI cos 6. CO Also, equating the torques and the rates of change of the angular momenta about the axes OA^, OB^, OC-^, we have Ta = ^ - VS + h'->2 = 0' (1) at (O ^^ ^ lit ~ ^^"^^ ^ ^^"^ = (7X1 sin 6'(ft> + XI cos 6) - AD'^ sm d cos 6, (2) DYNAMICS OF RIGID BODIES 385 It is possible for this motion to take place under the action of the weight and the reaction of the support as the only impressed forces for certain related values of ft), fl, and the constants of the body. For if only the weight and the support act on the body, then, since the center of gravity moves as if all the forces were concen- trated there, the only additional conditions to be satisfied are X, = -Mb sin (902, (4) r; = o, (5) Zr = TF, (6) where ^^, T^, Zj. are the components of the reaction of the support along the fixed axes at the given instant, and b is the distance from the point of support to the center of gravity of the body. With only the weight and the reactions of the support as impressed forces, the torques T^ and T^ are both zero and Tf, = Wb sin 6, All of the conditions of motion would then be satisfied if, from equation (2), Wb sin 6 = on sin 6(co + H cos 6) - An"^ sin cos (9, or {A - 0) cos en'^ - Ccoa -^ wb = o. The values of H that satisfy this equation are ^^ Cft)±VaW-4 Wb^A- (7)cos6> 2{A- (7) cos l9 There are therefore two solutions, one solution, or no solution according as C^co^^^WbCA- (7)cos6>. 2c 386 APPLIED MECHANICS FOR ENGINEERS Problem 461. Find the value, or values, of O for steady pre- cession with axis inclined 75° from the vertical of the disk of Prob- lem 454, the thickness of the disk being negligible. What is the least value of o) that would satisfy the conditions for the motion of this disk when = 75°? Problem 462. A conical top is made of wood and is spinning about its axis with a velocity of 20 revolutions per second. The cone has a base of 2 in. and a height of 2 in., and spins on the apex. While spinning steadily with its axis vertical (sleeping), it is dis- turbed by a blow so that its axis is inclined at an angle of 30' with the vertical. Find the velocity of precession and the torque that tends to keep the top from falling. (J) Unsteady Precession. Assume 0, co^ and O to be variables. The equations then become a)j = — 12 sin 0, co^ = co', cOg = II cos 0, h^ = ~ An sin ^, ^2 = Act)', h^ = C(^(o + U, cos ^), T^ = - Afn COS 0'(o' -\- sin e—\-Aco'n cos 6 ^ ^^ ^ -^Cco' (co +n cos 0), (1') Tj, = A^-\- en sin 0(co + n COS 6) - An^ sin cos 0, (2^ r, = (7^(0) + O cos 6>). (3') etc Again suppose the weight and the reaction of the support to be the only impressed forces. Then T^ — 0, Ti, = Wb sin 6, T^ = 0. Equation (3^ then becomes 4-(«4-Xlcos^)=0. at Therefore (o -\~ n cos ^ = a constant. DYNAMICS OF RIGID BODIES 387 For the sake of simplicity suppose the body to have been placed with its geometric axis at rest making an angle 6^ with the vertical, the body then given an angular velocity coq about its axis and released. Then the value of ft) + fl cos 6 at the start was g)q. Therefore throughout the motion ft) + H cos 6 = (Oq. Substituting this value in equation (1^) where ^^ = 0, and replacing co' by — , it becomes (a/V 2 An cos 6dd + A sin edfl - Cco^dd = 0. If this equation be multiplied through by sin 6^ it can then be integrated, for it becomes ^(2 X2 sin e cos Odd + sin2 Od^^ - 0(o^ sin SdO^ 0, in which the quantity inclosed in the parenthesis is ^(O sin2 (9). Integrating, An sin^ 6 + Ocdq cos 6 = constant. At the beginning of motion = 0, and 6 = 6^. Therefore An sin^ 6 = (7a)Q (cos 6^ — cos ^). (5') This equation expresses H in terms of 6. Instead of using equation (2') a simpler equation is ob- tained by considerations of work and energy. The work done from the beginning of motion to the instant under consideration is that done by gravity, i.e. TTKcos^o- cos l9). 388 APPLIED MECHANICS FOR ENGINEERS The gain in kinetic energy in this time is \iA(- n sin (9)2 + ^A(o'^ 4- 1- CcoH - i Ccol (Art. 199) Therefore lA(n^ sin2 e + a)'2) = Tf 6 (cos 0^ ~ cos (9), 2Wb or ft)'2= (cos ^0 — cos ^) — 112 sin2 ^^ A Replacing cos 6^ — cos by — -— from equation (5'), Ccoq co'^ = n(^ -n) sin2 6. (6') \ CcOq J The factors 12 and — H must both be of the same sign since 0)^2 u^ust be positive. Assuming &)q to be positive, it follows that 12 is positive, for if XI were negative, the sec- ond factor would be positive and their product would be negative. Also the values of 12 that make to' zero are and From equation (5') O — ^^ cos 6^ — cos Q ~ A ' sin2 6> from which it can be shown that C(o^ 'A— dd A sin3 d [(cos6'o-cosl9)2 + sin2(9o]. Since sin 6 is positive, —zr is positive; i.e. 12 increases as du 6 increases and decreases as decreases. It follows then that 6 and 12 both increase until 12 = • (o' is then 0, 6 DYNAMICS OF BIGID BODIES 389 changes from increasing to decreasing and continues to decrease until 11 = and = 6q (Equations (6') and (5'))- The axis of the body therefore alternately falls and rises between the values 0=6^ and 6 = 6^, where ^^ is the value obtained by putting Q. = —- — in equation (5'). As the axis of the body falls and rises, the value of H increases from when ^ = ^q to — — when 6 = ^j, decreasing again Ccoq to when 6 = 6^. The condition for this motion is that the value of 6 ob- tained from equations (5') shall be real. Problem 463. Using the body of Problem 454, making wq = 600 r. p. m. and Oq = 30°, find the range of values for and O. Problem 464. For what least value of o)q would the body of the preceding problem have the motion described in this article, the other conditions being the same ? CHAPTER XV IMPACT 207. Definitions. — When two bodies collide, they are said to be subject to impact. When the velocity of the striking surfaces is in the direc- tion of the normal to those surfaces, the impact is said to be direct. Otherwise it is oblique. When the normal to the surfaces at the point of contact (or center of forces exerted by one body on the other) passes through both centers of gravity of the bodies, the impact is said to be central. The phenomena of impact may be best studied by con- sidering the two bodies somewhat elastic. Suppose for simplicity that they are two spheres, M-^ and M^ (Eig. 280), and that they are moving with velocities v^ and v^ and that the impact is central. In Fig. 280 (a) they are shown at the instant when contact first takes place, and in Fig. 280 (5) they are shown some time after first contact when each has been deformed somewhat by the pressure of the other. The dotted lines indicate the original spherical form and the full lines the assumed form of the deformed spheres. When the spheres first touch, the pressure P between them is zero, but as each one compresses the other, the pressure P increases until it becomes a maximum. The compression of the spheres is indicated in the figure by d-^ and d^. We 390 IMPACT 391 (a) shall designate the time during which the bodies are being compressed as the period of deformation. After the compression has reached its maximum value the bodies, if they be partially elastic, begin to separate and to regain their original form. The common ^ pressure P decreases and becomes zero, if the bodies are sufficiently elastic so that they finally separate. We shall designate this pe- riod of separation as the period of restitution^ and the velocities after sep- aration as vl and v^. If the bodies are en- tirely inelastic, there will be no restitution. They will, in that case, remain in contact just as they are when the pressure be- tween them is amaximum and will move on with a common velocity V. In what follows velocities and accelerations toward tlie right will be called positive and those toward the left negative. 208. Direct Central Impact. — When the bodies meet in direct central impact, separation will take place along the (6) Fig. 280 392 APPLIED MECHANICS FOR ENGINEERS line joining the centers of gravity. Let T be the time from the first contact up to the time of maximum pressure, that is, the time of deformation^ and T^ the time from first contact up to the time of separation. Then T^— T represents the time of restitution. We have, from Art. 103, dv = a ' dt. Considering the motion of M^ during the period of deformation, we have dv = adt and \ dv = — —— I Pdt. or M^iV- v^) = - Cpdt, where Vi^ the common velocity of the centers of gravity of the bodies at the end of the period of deformation. In a similar way, ^2(^-^2)= r^dt, Pdt on the right-hand side of the preceding equations cannot be determined since we do not know in general how the pressure P varies with the time ; we do know, however, that they are equal term for term, so that we may eliminate them. We have, then, or (ilTi + iHf 2) V = MxVi + MiVi. If the impact is not too severe, elastic or partially elastic bodies tend to regain their original shape after the defor- mation has reached a maximum and finally separate if they IMPACT 393 possess sufficient elasticity. Using the notation of Art. 207, we have, for the period of restitution, if It is the force of restitution, M^j 'dv=:j Rdt, so that ^i(^i- ^) = - r ^Rdi^ and ^2(4 - F) = r ^Rdt, from which {M^ + M^ V = M^v\ + M^v^^. The value of the integral I Pdt during deformation will not in general be the same as its value during restitution. Call the ratio of I Hdt to I Pdt^ e. This value, which is called the coefficient of restitution^ is constant for given materials. It is unity for perfectly elastic substances, zero for non-elastic substances, and some intermediate value for the imperfectly elastic materials with which the en- gineer is usually concerned. The following values of e have been determined : for steel on steel, e = .bb \ for cast iron on cast iron, g = 1, nearly ; for wood, e = 0, nearly. From the above definition of e, it is seen at once that we may write -V 1 -^1 and e ■■ V- V n which it follows that v[ = :r(i + 0- -ev^. 394 APPLIED MECHANICS FOR ENGINEERS where r=^;t±^^. From the above equations it is seen that 209. Momentum and Kinetic Energy in Impact. — When a constant force acts upon a body for a certain time, the change in the velocity of the body is directly proportional to the force and to the time and inversely proportional to the mass of the body. The product of the force, F^ and the time, ^, is called the impulse of the force^ and the product of the mass and velocity is called the momentum of the body. (The body is supposed to have no rotation.) From Newton's law, F = Ma, it follows that Ft = M(v^ - ^;o), or, when a body is acted upon hy a constant force, the im- pulse of the force is equal to the change in momentum of the body. If the force is not constant, and is represented at any time by P, then | Pdt is the impulse of the force for the time from to ^, and from the equations of the preced- ing article. .r Pdt=-M^(iV-v^^, rPdt = M^{^V-v^\ IMPACT 395 it follows that the impulse of the force is equal to the change of the momentum of the body on which it acts, no other force being assumed to act on the body. From the equations of the preceding article, (i^i + M^^ V= M^v^ + M^v^, and (ilffj + ilfg) ^= ^A + ^^2' it follows that there is no change in the sum of the mo- menta of the bodies in impact, either during the period of deformation or the period of restitution. This is otherwise evident from the fact that the forces acting on the bodies are equal and opposite, and hence what one body gains in momentum the other loses. With the kinetic energy it is different. During the period of deformation the loss in kinetic energy is _ 1 ~~ 2 11-^ 22 M^ + M^ = ^^i^^2 ^, _ , ^2 2(ilfi + 7^2) During the period of restitution there is a gain in kinetic _ e^M^M^ ^ _ .2 ~2(Jf, + i/2) ' ""' since v[ — v'^= — e(^v^ — v^). The total loss in kinetic energy during the impact is there- 396 APPLIED MECHANICS FOR ENGINEERS If the bodies are inelastic, e = 0, and the bodies go on with the common velocity V. The loss in kinetic energy in this case is If the bodies are perfectly elastic, g = 1, and the loss in kinetic energy is zero. Problem 465. A lead sphere whose radius is 2 in. strikes a large mass of cast iron after falling freely from rest through a distance of 100 ft. What is its final velocity ? What is the loss of kinetic energy ? Assume e = 0. Problem 466. A 10-lb. lead sphere is at rest when it is acted upon by another lead sphere, whose radius is 3 in., in direct central impact. The velocity of the latter sphere is 20 ft. per second. What is the common velocity of the two spheres and what is the loss of kinetic energy due to impact ? Problem 467. Prove that if two inelastic bodies, moving in opposite directions with speeds inversely proportional to their masses, collide, both will be reduced to rest. Problem 468. Prove that if two perfectly elastic bodies with equal masses collide, each mass will have after impact the velocity that the other had before. If one of the bodies is at rest and the other strikes it, what will happen ? Problem 469. Figure 281 represents a set of perfectly elastic *- balls of equal mass and size in contact ^ ■ ^ sA A A A A /) jjj a straight line. Another ball, of the same mass and material, moving in the direction of the line of balls, strikes one end of the line. Prove — ^-^^^ N A A A AA ) that the moving ball will be brought ^^^- ^^^ to rest where it strikes and the last ball at the other end of the line will take its velocity, all other balls remaining at rest. Prove also that if two balls, moving together. IMPACT 397 strike the line, the two balls at the other end of the line will take the velocity of the striking balls and the latter will be brought to rest. Problem 470. Prove that if a ball of mass Mi fall from a height H upon a very large mass with a horizontal surface and rebounds to a height h, the coefficient of restitution of the materials used is =V H (Regard M2 as infinite and V = 0.) Problem 471. A bullet weighing 1 oz. is fired horizontally into a box of sand (inelastic) weighing 20 lb., and remains imbedded in the sand. The box is suspended by a string attached to a fixed point 4 ft. from the center of the box of sand. The impact of the bullet causes the box to swing aside through an angle of 42". Find (a) the velocity of imj^act of the bullet, {h) the kinetic energy lost in impact, (c) the greatest tension in the string. Problem 472. Assuming inelastic impact, show that if a pile driver weighing G-^ lb. falls h ft. and drives a pile weighing G^ lb. s ft. into the ground against a constant resistance R, R = _ G\ Gi+ G2S (Hint. Equate work done against resistance plus kinetic energy lost in impact to the work done by gravity on the falling weight.) Problem 473. A wooden pile weighing 1500 lb. is driven by a steel hammer, of weight 2000 lb., falling 20 ft. The penetration at the last blow is observed to be \ in. What is the resistance offered by the earth to the pile ? With a safety factor 6, what load would the pile carry ? 210. Elasticity of Material. — All materials of engineer- ing are imperfectly elastic. Some, however, show almost perfect elasticity for stresses that are rather low. This has been expressed by saying that all materials have a 398 APPLIED MECHANICS FOR ENGINEERS limit (elastic limit) beyond which if the stress be increased the material will be imperfectly elastic. Within the limit of elasticity^ stress is proportional to the deforma- tion produced. Let I^ be the total stress in tension or compression, / the stress, in pounds per square inch of cross section, d be the deformation caused by P and X the deformation per inch of length. Within the limit of elasticity of the material the ratio - is a constant, and P d ^ since / = — -, and \ = -, when F is the area of cross sec- tion and I is the length of the material, it may be written PI . . — — . This constant is called the modulus of elasticity of the material; it is usually represented by U, so that X Fd for tension or compression. For steel U has been found to be about 30,000,000 lb. per square inch. 211. Impact Tension and Impact Compression. — Figure 282 (a) represents a mass iltfg subjected to impact from the mass M^ falling from rest through a height h. The mass M^ is compressed by the impact. Figure 282 (6) represents the body M2 as subjected to impact in tension, the mass in this case being a rod having M-^ attached to one end and the other end attached to a crosshead A. The rod, crosshead, and weight fall freely together through the distance h until A strikes the stops at B^ when one end of the rod suddenly comes to rest and the mass M^ causes tension in the rod due to impact. IMPACT 399 Suppose v^ represents tlie velocity of M^ when impact occurs and l^ the length of M^^, whether it be a tension or compression piece. M, B 2Io J/„ (a) (ft) Fig. 282 In case (5) when the weights strike the stops at B the kinetic energy of the falling weights is equal to ((^i4-^2)^ft.-lb., where (r^ and G-^ are the weights of M^ and i^ in pounds and h is in feet. This kinetic energy is used up in doing work in compressing the stops and in stretching M-^ and M^. If M^ is a rod of small cross section, its length is in- creased a large amount compared to the change in length of M^ and the stops against which the crosshead strikes, 400 APPLIED MECHANICS FOB ENGINEERS and we may without serious error assume that all the ki- netic energy is used up in stretching M^. We may assume too that the stress in M^ is the same throughout its length. Let P^ be the maximum stress induced in the rod. The p average stress is then — ^, and if c?^ is the total increase in the length of the rod, the work done on the rod is 2 P (J m^-^m 2 =(a^ + a^)h. But p^^F^^%, (Art. 210) • • CC'-TM — ' ~: — — — • Here E must be expressed in pounds per square foot and all dimensions in feet if G^ and G-^ are in pounds and h in feet. In case (a) the student can easily show that U"m — "V 2 Gihl F2E2 Problem 474. Derive the formula just given for the case (a). Problem 475. A weight of 500 lb. falls through a distance of 2 ft. in such a way as to put a 1-in. round steel rod in tension. If the rod is 18 ft. long, what will be the elongation due to the impact? Use E = 30,000,000 lb. per square inch for steel. What maximum unit stress is induced in the rod ? IMPACT 401 Problem 476. A cylindrical piece of steel 1 in. high and 1 in. in diameter is subjected to compression by a weight of 20 lb. falling through a distance of 1 in. How much will it be compressed? Problem 477. In the preceding problem what stress (pounds per square inch) was caused in the cylindrical block by the fall of the 20-lb. weight? Problem 478. The safe stress in structural steel for moving loads is usually taken as 12,500 lb. per square inch. Through what height might a 300-lb. weight fall so as to produce tension in a 1-in. steel round bar, 10 ft. long, without exceeding the safe stress ? Problem 479. Two steel tension rods in a bridge, each 2 sq. in. in cross section and 20 ft. long, carry the effect of the impact of a loaded wagon as one wheel rolls over a stone 1 in. high. The weight on the wheel is 2000 lb. What stress is introduced in the tension rods ? I^OTE. For the strength of metals under impact the student is referred to the work of W. K. Hatt, Am. Soc, " Testing Materials," Vol. IV, p. 282. 212. Direct Eccentric Impact. — Let M^ and M^ have impact as shown in Fig. 283, in which the centers of gravity of the bodies are moving along parallel lines and the surfaces of contact are perpendicular to the line of motion. ^ This is known as direct eccentric impact. ^ Let the velocity of M^ before impact be v^, the velocity of the center of gravity of M^ be Vj, and its angular velocity be ft) 3/, 1- (a) The problem will first be solved on the assumption that M^ is acted upon by no forces except that of impact. First period of impact. Let P be the force at any in- 2d 402 APPLIED MECHANICS FOR ENGINEERS stant of the first period of impact, T the time of the first period, F'the velocity of iUfgi (^ the angular velocity of M^^ and V^ the velocity of the center of gravity of M^ at the end of the first period. During the first period, which lasts for a very short time, the position of the body M^ may be regarded as unchanged. We then have for the motion of i^^, _ rPdt = M^ Cdv = M^( V- v^), (1) and for the motion of M^ CPdt = M^ C'dv = Jfi( Fj - Vi), (2) and hP = J^ dt or hrFdt = MJf:^C''d(o = M^k\(o-(o{), (3) where h is the radius of gyration of M^ about a gravity axis perpendicular to the plane of motion. Also, since at the end of the first period the velocity of M^ is the same as the velocity of the point of impact on M^^ ■V=V^ + bco. . (4) The unknowns in these four equations are The equations are linear in these unknowns and the un- knowns can be easily determined from them. Second period ofimp,act. If R is the force of impact at any time during the second period of impact, T the value of ^, ^2 the velocity of M^^ v[ the velocity of the center of IMPACT 403 gravity of M^, and coj the angular velocity of M^ at the end of the second period, then -J^' Edt = M^J"' dv = M^{v^^ - F), (5) y Rdt = M^^^'^dv = M^{v[ - r^), (6) h^''^ Rdt = M^k^j^'' d(o = M^k\(o[ - CO) . (7) Rdt = el Pdt. (8) 7^ «/o Also Here again are four linear equations in the four un- J<»Tf Rdt, vl^ v!), ojj, T which are therefore sufficient to determine the unknowns. (6) Impact on a body with fixed axis. Suppose the body rotating about the fixed axis through (Fig. 284), with angular acceleration a^ and angular velocity ©j at the beginning of im- pact. Choose the origin of coordinates at and the rr-axis coinciding with the direction of motion of the cen- ters of gravity of the bodies. Let X and Y be the reactions of the supporting axis through at any time during the impact. First period of impact. For the first period we have the following equations : Fig. 2»1 404 APPLIED MECHANICS FOB ENGINEERS j^\P-hX)dt = M,(V,-v,\ (2') fJ[Pb - X(h - b)]dt = M^k\cD - a)i). (30 Also, since is fixed, Fi = (A-5)a,, (4') V=h(o. (50 Here are five equations containing five unknowns: f^Pdt, £'xdt, Fi, r, CO, from which the unknowns may be determined. Second period of compact. For the second period the following equations hold : - £' Bdt = M^iv', - V), (6') £\r + X)dt = M,(v[ - Fj), (7') C'lBb - X(h - *)]* = M-J<:\(o[ - o)), (8') v'i = (h-h)co[, (9') XT'/ /»r Edt = ej Pdt. (100 Here again are five equations containing five unknowns : J Rdt, I Xdt, v[, v'^, (o[. These equations are sufficient to determine the unknowns, and thus the motion of the bodies is given after impact. In the above equations k is the radius of gyration of M^ about a gravity axis perpendicular to the plane of motion. IMPACT 405 Problem 480. A uniform steel bar, 4 ft. long, weighing 16 lb., lies at rest on a smooth horizontal plane. It is struck by a steel ball weighing 2 lb., with a velocity of 30 f/s, at right angles to the rod at a point 6 in. from one end. Find the velocity of the ball, the velocity of the center of gravity of the bar, and its angular velocity at the end of the first period, and at the end of the second period of impact, given e = .55. Problem 481. In the preceding problem find the point about which the bar is turning at the end of the first period. Is this point the center of rotation at the end of the second period ? Problem 482. If Mi is at rest when M^ strikes with direct eccen- tric impact, write the equations which determine the motion at the end of the first, and at the end of the second period of impact. Problem 483. If e = between M^ and M^ of Art. 212 (a), find the kinetic energy lost in impact. Problem 484. Suppose Mx to be a rod of steel | in. in diameter and 2 ft. long, and suppose M2 to be a hammer weighing 2 lb. and that its velocity at the time of impact is 20 ft. per second, find V and at time of greatest pressure. Problem 489. A man strikes a blow with a steel rod 1| in. in diameter and 4 ft. long, by holding the rod in the hand and striking the farther end against a stone in such a way as to cause the rod to be under flexure. Where should he grasp the rod in order that he may receive no shock ? Problem 490. Prove that after the blow has ceased to act on the body (Fig. 285) the body would then move in such a way that a circle of the body with radius CA and center C would roll along a straight line. Hence show that every point of the body in this circle would generate a cycloid. It is assumed that no other forces act on the body. 214. Oblique Impact of Body against Smooth Plane. — Let M(Fig. 286) be a sphere moving toward the plane indicated with a velocity at impact of v, the direc- tion of motion making an angle a with the vertical to the plane. After im- IMPACT 409 pact the body M rebounds with a velocity v^ in the direc- tion making an angle (3 with the vertical AB. Since the plane is considered smooth, the effect of the impact will be all in the direction of AB^ and hence the component of the velocity parallel to the plane will not change. . •. Vj sin l3 = V sin a. (1) Assuming that the plane does not move, F= 0, where V is the component of the velocity of M perpendicular to the plane at the time of greatest pressure. Then i ^Pdt = M[0 - (y cos a)], ri and r Rdt = Mi- v^co^P -()'). Therefore J-*Tt Rdt vcosa rpdt or v-^ cos (3 = ev cos a. (2) Dividing (1) by (2), tan /8 = - tan a. e Squaring and adding (1) and (2), i»2 = ^2 (^sii;^2 a-{- e^ cos^ a) . Hence if a, v, and e are known, v-^ and ^ may be deter- mined. If the bodies are perfectly elastic, e = 1, /S = a, and v^ = V. If the body is inelastic, g = 0, /S = — , v^ = v sin a. 410 APPLIED MECHANICS FOR ENGINEERS The body then moves along the plane with a velocity V sin a. Problem 491. A ball is projected with velocity 50 f/s against a smooth plane surface at an angle with the normal of 40°. It leaves at an angle of 50° with the normal. Find e and the velocity at which it leaves. 215. Impact of Rotating Bodies. — Suppose two bodies M^ and M2 revolve about two parallel axes Oj and O2 (Fig. 287) in such a way that impact occurs at a point along the line DE. Let the line along which impact occurs be distant r^ and ^2 respectively from 0-^ and Og. Let P be the force of im- pact at any instant during the first period and M the force at any instant during the second period, ft) J and ft)2 the angular velocities at the beginning, co[ and CO2' the angular velocities at the end of impact, I^ and ig ^^^ moments of inertia of the bodies respectively about Oj and Og, and F'the rectangular component of the velocity of the point of impact in the direction of DJS at the time of greatest pressure. The angular velocities of the bodies at this instant are then — and — Fig. 287 Then, as in Art. 212, (1) IMPACT 411 '•iX^'^* = /iX"^<» = I,(a>[ - r^, (8) Rdt = e^^ Pdt. (5) From (1) and (2), 4(F-ria,i)+:^^(r-r2a,2) = 0, or r= ^1^2(-^1^2^1+Vl^2) , From (1), (3), and (5), F fV or w[ = —(1 + g) — eft)!- Similarly, cog = —(1 + ^) — ^ft)2* (It should be noted that in these formulse the angular velocities of the bodies are reckoned in opposite direc- tions.) Problem 492. If I^= 3000, /g = 15,000, Wj = 1 radian per second, 0)2 = 0, Tj = 2 ft., rg = 3 ft., and e = 0, find V, w{, Wg, and the kinetic energy lost in impact. 412 APPLIED MECHANICS FOR ENGINEERS Problem 493. A well drill is shown in principle in Fig. 288. The drill is supported by a cable that passes over a pulley C and is attached to a friction drum A. When A is held, the drill is raised by the operation of M^ and Afg. Suppose that / is 300 and cd^ = 3 Fig. 288 radians per second; /g = 200 and tOg^O; r^ = 2 ft. and r2 = 6 ft. Assume e = |. Find w[ and Wg. What kinetic energy is lost due to each impact ? IMPACT 413 Problem 494. The moment of inertia of the trip hammer M^, illustrated in principle in Fig. 289, is 100,000 ; that of ilfj is 60,000. Fig. 289 If rj = 3 ft., r-i = 10 ft., (Oj = 2 radians per second, 0J2 = 0, and e = -|, find oj[ and Wo. What is the kinetic energy lost due to each impact? What is the kinetic energy of the hammer? APPENDIX I HYPEEBOLIC FUNCTIONS cosh X = — - — p^' p~ -^ sinh X = 2 , 1 sinh X p/' — e~^ tanh X = — = — cosh X e-^' -\- e ^ HYPERBOLIC FUNCTIONS 1 41' X Coshx Sinli X X Coslix Sinh X 0.01 1.0000500 0.0100002 0.51 1.1328934 0.5323978 .02 .0002000 .0200013 .52 .1382741 .5437536 .03 .0004500 .0300045 .53 .1437686 .5551637 .04 .0008000 .0400107 .54 .1493776 .5666292 .05 .0012503 .0500208 .55 .1551014 .5781516 .06 .0018006 .0600360 .56 .1609408 .5897317 .07 .0024510 .0700572 .57 .1668962 .6013708 .08 .0032017 .0800854 .58 .1729685 .6130701 .09 .0040527 .0901215 .59 .1791579 .6248305 .10 .0050042 .1001668 .60 .1854652 .6366536 .11 .0060561 .1102220 .61 .1918912 .6485402 .12 .0072086 .1202882 .62 .1984363 .6604917 .13 .0084618 .1303664 .63 .2051013 .6725093 .14 .0098161 .1404578 .64 .2118867 .6845942 .15 .0112711 .1505631 .65 .2187933 .6967475 .16 .0128274 .1606835 .66 .2258219 .7089704 .17 .0144849 .1708200 .67 .2329730 .7212643 .18 .0162438 .1809735 .68 .2402474 .7336303 .19 .0181044 .1911452 .69 .2476458 .7460697 .20 .0200668 .2013360 .70 .2551690 .7585837 .21 .0221311 .2115469 .71 .2628178 .7711735 .22 .0242977 .2217790 .72 .2705927 .7838405 .23 .0265668 .2320333 .73 .2784948 .7965858 .24 .0289384 .2423107 .74 .2865248 .8094107 .25 .0314132 .2526122 .75 .2946833 .8223167 .26 .0339908 .2629393 .76 .3029713 .8353049 .27 .03(56720 .2732925 .77 .3113896 .8483766 .28 .0394568 .2836731 .78 .3199392 .8615330 .29 .0423456 .2940819 .79 .3286206 .8747758 .30 .0453385 .3045203 .80 .3374349 .8881060 .31 .0484361 .3149891 .81 .3463831 .9015249 .32 .0516384 .3254894 .82 .3554658 .9150342 .33 .0549460 .3360222 .83 .3646840 .9286; 547 .34 .0583590 .3465886 .84 .3740388 .9423282 .35 .0618778 .3571898 .85 .3835.309 .9561 KiO .36 .0655029 .3678265 .86 ,3931614 .969S)ii!)3 .37 .0692345 .3785001 .87 .4029312 .9839796 .38 .0730730 .3892116 .88 .4128413 0.99.S0.-,S4 .39 .0770189 .3999619 .89 .4228927 1.0122369 .40 .0810724 .4107523 .90 .4330864 .0265167 .41 .0852341 .4215838 .91 .4434234 .0408^)91 .42 .0895042 .4324574 .92 .4539048 .0553S56 .43 .0938888 .4433742 .93 .4645315 .0699777 .44 .0983718 .4543354 .94 .4753046 .084(i7(i8 .45 .1029702 .465.3420 .95 .4862254 .0994843 .46 .1076788 .4763952 .96 .4972947 .1144018 .47 .1124983 .4874<.I59 .97 .5085137 .1294307 .48 .1174289 .4<)S6455 .98 .5198837 .1445726 .49 .1224712 .5098450 .99 .5314057 .159SL>,S8 0.50 1.12762()0 0.5210953 1.00 1.5430806 1.1752012 418 HYPEBBOLTC FUNCTIONS 2 X Oosh X Sinh X X Cosh X Sinh X 1.01 1.5549100 l.lfX)6910 1.51 2.3738201 2.1529104 1:02 .5668948 .2062999 1.52 .3954676 .1767566 1.03 .5790365 .2220294 1.53 .4173563 .2008206 1.04 .5913358 .2378812 1.54 .4394857 .2251046 1.05 .6037945 .2538567 1.55 .4618591 .2496111 1.06 .6164134 .2699576 1.56 .4844787 .2743426 1.07 .6291940 .2861855 1.57 .5073467 .2993014 1.08 .6421375 .3025420 1.58 .5304654 .3244903 1.09 .6552453 .3190288 1.59 ,5538373 .3499117 1.10 .6685186 .3356474 1.60 .5774645 .3755679 1.11 .6819587 .3523997 1.61 .6013494 .4014618 1.12 .7005670 .3642872 1.62 .6254945 .4275958 1.13 .7093449 .3863116 1.63 .6499021 .4539726 1.14 .7232938 .4034746 1.64 .6745748 .4805947 1.15 .7374148 .4207781 1.65 .6995149 .5074650 1.16 .7517098 .4382235 1.66 .7247249 .5345859 1.17 .7661798 .4558128 1.67 .7502074 .5619603 1.18 .78082<)5 .4735477 1.68 .7759650 .5895910 1.19 .7956513 .4914299 1.69 .8020001 .6174806 1.20 .8106556 .5094613 1.70 .8283154 .6456319 1.21 .8258410 .5276436 1.71 .8549136 .6740479 1.22 .8412089 .5459788 1.72 .8817974 .7027311 1.23 .8567610 .5644685 1.73 .9089692 .7316847 1.24 .8724988 .5831146 1.74 .9364319 .7609115 1.25 .8884239 .6019191 1.75 .9641884 .79421 3.0049163 1.33 .0227603 .7582830 1.83 .1971501 .03673 MEASURES METRIC TO CUSTOMARY WEIGHTS Milligrams firams Grams Kilograms Tonnes to Tonnes to No. to to to Avoirdupois to Avoirdupois Net Tons of Gross Tons of tiraiiis Troy Ounces Ounces Pounds 2000 Pounds 2240 Pounds 1 .01543 .03215 .03527 2.20462 1.10231 .98421 2 .03086 •06430 .07055 4.40924 2.20462 1.96841 3 .04630 .09645 .10582 6.61387 3.30693 2.95262 4 .06173 .12860 .14110 8.81849 4.40924 3.93682 5 .07716 .16075 .17637 11.02311 5.51156 4.92103 6 .09259 .19290 .21164 13.22773 6.61387 5.90524 7 .10803 .22506 .24692 15.4.3236 7.71618 6.88944 8 .12346 .25721 .28219 17.63698 8.81849 7.87365 9 .13889 .28936 .31747 19.84160 9.92080 8.85785 1 Kilogram = 1543 2.35639 Grains LINEAR Ml EASURE Millimeters Centimeters Meters Meters Kilomeiters Kilometers No. to 64tlis of an to to to to to Inch Iiulies Feet Yards Statute Miles Nautical Miles 1 2.51968 .39370 3.280833 1.093611 .62137 .53959 2 5.03936 .78740 6.561667 2.187222 1.24274 1.07919 3 7.55904 1.18110 9.842500 3.280833 1.86411 1.61878 4 10.07872 1,57480 13.12.3333 4.374444 2.48548 2.15837 5 12.59840 1.968.50 16.404167 5.468056 3.10685 2.69796 6 15.11808 2.36220 19.685000 6.561667 3.72822 3.23756 7 17.63776 2.75590 22.965833 7.655278 4.34959 3.77715 8 20.15744 3.14960 26.246667 8.748889 4.97096 4.31674 9 22.67712 3.54330 29.527.500 9.842.500 5.59233 4.85633 438 CONVERSION TABLES TABLES FOR CONVERTING UNITED STATES ^WEIGHTS AND MEASURES CUSTOMARY TO METRIC WEIGHTS Giauis . Troy Ounces Avoirdupois Avoirdupois Net Tons of Gross Tons of l\0. to to Ounces Pounds to 2000 Pounds 2240 Pounds Milligrams Grams to Grams Kiloi^rams to Tonnes to Ton:: OS 1 64.79892 31.10348 28.34953 .45359 .90718 1 .01605 2 129.59784 62.20696 56.69905 mi\% 1.81437 2.03209 3 194.39675 93.31044 85.04858 1.36078 2.72155 3.04814 4 259.19567 124.41392 113.39811 1.81437 3.(i2874 4.06419 5 323.99459 155.51740 141.74763 2.26796 4.5.3592 5.08024 () , 388.79351 186.62088 170.09716 2.72155 5.44311 6.09628 7 453.59243 217.72437 198.44669 3.17515 6.35029 7.11233 8 518.39135 248.82785 226.79621 3.62874 7.2.5748 8.12838 9 583.19026 279.93133 255.14574 4.08233 8.16466 9.14442 1 Avoirdup L ois Pound = .INEAR Ml 453.5924277 Grams EASURE 64tlis of an Inches Feet Yards Statute Miles Nautical Miles No. Inch to to to to to to Millimeters Centimeters Meters Meters Kilometers Kilometers 1 .39688 2.54001 .304801 .914402 1.60935 1.85.325 2 .79375 5.08001 .609601 1.828804 3.21869 3.70650 3 1.19063 7.62002 .914402 2.743205 4.82804 5.55975 4 1.58750 10.16002 1.219202 3.657607 6.43739 7.41300 5 1.98438 12.70003 1.524003 4.572009 8.04674 9.26625 6 2.38125 15.24003 1.828804 5.486411 9.65608 11.11950 7 2.77813 17.78004 2.133604 6.400813 11.26543 12.97275 8 3.17501 20.32004 2.4.38405 7.315215 12.87478 14.82600 9 3.57188 22.86005 2.743205 8.229616 14.48412 16.67925 1 Nautit ;al Mile = 1853.25 Meters 1 Gunte r's Chaiu = 20.1168 Meters 1 Fathoi 11 = 1 .829 Meters INDEX The numbers refer to the pages. Absorption dynamometer, 323. Acceleration, 184. angular, 232. components of, 208. constant, 185. in a curved path, 205. of points of a body having plane motion, 237. variable, 191. Angular momentum, 369. vector representation of, 373. Angular velocity, 232, 235. vector representation of, 241. Antifriction wheels, 298. Arch, linked, 158. masonry, 159. Balance, standing and numing of a shaft, 359. Ball bearings, 302. Bending moments, 161. diagram of, 162. graphical construction of, 162. Bow's notation, 99. Brake shoe testing machine, 272. Car on single rail, 382. Catenary, 177. Center of gravity, 38, 39, 43. by graphical methods, 57. by integration, 44. by Simpson's rule, 65. motion of, 377. of counterbalance, 56. Center of instantaneous rotation, 405. Center of oscillation, 345. Center of percussion, 405. Center of suspension, 345. Coefficient of friction, 247, 283. of belting, 316. Coefficient of restitution, 393. CoejSicient of rolling friction, 296. Compound pendulum, 344. Compression, 14. Concurrent forces, 10, 19. Connecting rod, 366. Conservation of energy, 254. Cords and pulleys, 170. Couples, 27, 71. combination of, 72, 76. components of, 78. equivalent, 73. gyroscopic, 380. moment of, 71, 75. substitution of force and couple for a force, 80. vector representation of, 77. Creeping of belts, 315. D'Alembert's principle, 334. Determination of g, 347. Displacement, 3. Durand's rule, 68. Dynamometer, absorption, 323. transmission, 314. Effective forces, 267, 334. Efficiency of wedge, 288. Elasticity of material, 397. modulus of, 398. Ellipse of inertia, 130. Ellipsoid of inertia, 140. Energy, 254. Equilibrant, 94. graphical method of finding, 94. Equilibrium, 1. conditions of, 12, 20, 29, 83, 92. of three forces, 16. of two forces, 14. Falling bodies, 186. Flexible cords, 148. as a catenary, 177. as a parabola, 171. 439 440 INDEX Force, 1. transmissibility of, 6. unit of, 2. vector representation of, 5. Forces, concurrent, 6, 10, 19. in space, 90. non-concurrent, 82. polygon of, 8. resolution of, 7. triangle of, 7. Foucault's pendulum, 243. Friction brake, 324. Prony, 326. Friction circle, 317. Friction, coefficient of, 247, 283. laws of, for dry surfaces, 285. of belts, 310. of brake shoe, 328. of lubricated surfaces, 288. of pivots, 318. of worn bearing, 316. rolling, 296. Friction gears, 307. Guldinus, theorem of, 68. Gyroscope, 379. inclined axis, 383. Gyroscopic couple, 380. Harmonic motion, 192. Helical chute, 229. Horse power, 253. Hyperbolic functions, 180. tables of, appendix. Impact, direct, oblique, central, 390. direct central, 391. direct eccentric, 401. oblique against smooth plane, 408. of rotating bodies, 411. on a body with fixed axis, 403. Impact tension and compression, 398. Impressed forces, 266, 334. Impulse of a force, 394. Inertia, 1. ellipse of, 130. ellipsoid of, 140. moment of, 106. product of, 112. Instantaneous axis of rotation, 238,407. center of rotation, 239, 407. Internal stress couple, 166. Kinetic energy, 254. lost in impact, 395. of a body having plane motion, 274. of a body with one fixed point, 373. of rolling bodies, 279. Line of resistance of arch, 159. Lubricants, testing of, 293. Mass, 1. unit of, 2, 188. engineer's unit of, 188. Method of substitution, 103. Modulus of elasticity, 167, 398. Moment of a force, 22, 24. of the resultant, 23, 25, 28. Moment of inertia, 106, 268. axes of greatest and least, 112, 128. by direct addition, 126. by experiment, 346, 351. by graphical method, 124. by parallel sections, 135. by Simpson's rule, 126. polar, 115. principal moments of inertia, 139. theorem of inclined axes for. 111. theorem of parallel axes for, 109. units of, 108. Moment of momentum, 369. of a body with one fixed point, 371. rate of change of, due to rotating axes, 375. Momentum of a body, 394. Motion, along a curve in a vertical plane, 211. curvilinear, 204. in a circle, 233. in a twisted curve, 227. of center of gravity of body, 377. on inclined plane, 189. rotary, 232. uniform, in a circle, 209. where attraction varies directly as distance squared, 197. where resistance varies as distance, 194. with repulsive force, 194. INDEX 441 Neutral surface, 168. Newton's laws of motion, 187. Pappus, theorem of, 68. Parabola, 171. Parallel forces, center of, 32, 35. resultant of, 27, 29. Pendulum, constant of, 398. compound, 344. conical, 209. cycloidal, 219. simple, 214. Period, of compound pendulum, 345. of deformation, 391. of restitution, 391. of simple pendulum, 216. Piledriver, 259. Pivots, flat end, 318. collar bearing, 318. conical, 319. Schiele's, 322. spherical, 321. Polar moment of inertia, 115. Pole of string polygon, 154. Power, transmitted by a belt, 313. transmitted by friction wheel, 309. Precession, 380. steady, 384. unsteady, 386. Principal axes, 113, 139. Principal moments of inertia, 139. Product of inertia, 112. Projectile in vacuo, 222. in resisting medium, 226. Radius of gyration, 107. Reaction of supports of rotating body, 340. Relative velocity, 200. Resistance of roads, 299. Rigid body, 4. Roller bearings, 301. Rolling friction, 296. Rotation about a fixed axis, 266, 336. with constant angular velocity, 353. Rotation and translation, 365. Rotation, of flywheel of steam en- gine, 362. of locomotive drive wheel, 358. of the earth, 243. under no forces, 352. Shear, 166. Simpson's rule, 62. applications of, 64. Simultaneous rotation about inter- secting axes, 240. Specific gravity, table of, 3. Speed, 183. Steam hammer, 262. Stresses in beams, 166. Stresses in frames, 99. String polygon, 153. depth of, 156. locus of pole of, 154. Substitution, method of, 103. Tension, 14. Torque and angular momentum, 370. Torsion balance, 348. Train resistance, 330. Translation of rigid body, 335. Transmission dynamometer, 314. Unit strain, 167. Unit stress, 167. Unit weights, table of, 3. Varignon's theorem, 23. Vector rate of change due to rota- tion, 374. Vector representation of angular momentum, 373. Vectors, 3. Velocity, 183. of spin, 383. relative, 200. Weight, 3. unit of, 3. Weights suspended by cords, 150. graphical solution of, 151. Work and energy, 245, 255, 276, 279. units of, 248. Work of a variable force, 250. of components of a force, 249. graphical representation of, 251. in uniform motion, 280. 'npHE following pages contain advertisements of a few of the Macmillan books on kindred subjects A Treatise on Hydraulics By hector J. HUGHES, A.B., S.B., M.Am. Soc. C.E. Assistant Professor of Civil Engineering, Harvard University AND ARTHUR T. SAFFORD, A.M., M. Am. Soc. C.E. 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THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New York Alternating Currents and Alternating Current Machinery NEW EDITION, REWRITTEN AND ENLARGED By DUGALD C. JACKSON, C.E. Professor of Electrical Engineering, Massachusetts Institute of Technology; Past-President and Fellow of the American Institute of Electrical Engineers; Past-President of the Society for the Promotion of Engineering Education, etc. AND JOHN PRICE JACKSON, M.E., Sc.D. Commissioner of Labor and Industry, Commonwealth of Pennsylvania; Dean of the School of Engineering, Pennsylvania State College; Fellow of the American Institute of Electrical Engineers; Member of the American Society of Mechanical Engineers, etc. Cloth^ 8vo, g68 pp., ^5.50 net: postage extra This is a new edition of the author's book on Alternating Currents and Alternating Current Machinery which was first published in 1896, and which has now been rewritten and greatly extended. The former editions met so favorable a response that the authors cannot help but believe that it served an important place in connection with the de- velopment of the methods of teaching the subject of alternating currents and their applications ; and they earnestly hope that the new edition will be equally useful. In this edition are maintained the well-known features of the earlier book in which were worked out the characteristics of electric circuits, their self-induction, electrostatic capacity, reactance and impedance, and the solutions of alternating current flow in electric circuits in series and parallel. More attention is paid to the transient state in electric circuits than was th'e case in the original edition. A considerable amount of related matter has been introduced in respect to vectors, complex quan- tities, and P^ourier's series which the authors believe will be useful to students and engineers. The treatment of power and power factor has been given great attention, and a full chapter is now allotted to the hysteresis and eddy current losses which are developed in the iron cores of electrical machinery. More space and more complete treatment have been assigned to synchronous machines and to asynchronous motors and generators. The treatment of the self-inductance and mutual inductance of line circuits and skin effect in electric conductors which was found in the old book has been extended, and it has been supplemented by a treatment of the electrostatic capacity of lines and the influences of dis- tributed resistance, inductance, and capacity. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New York A Textbook of Elementary Foundry Practice By WILLIAM ALLYN RICHARDS, B.S. in M.E. Junior Member A.S.M.E. ; jMember American Gas Institute, A.M.; Chemical Society Instructor in Forge and Foundry Practice in the University High School and the University of Chicago Cloth, 8vo, illusU'ated, 121 pp., %i'2S net This book is intended for use both as a classroom text and as a laboratory guide. It contains some forty exercises, which have been chosen so as to bring out the largest num- ber of principles used in the molder's art. Each operation is so plainly and practically discussed that, even without the direction of an instructor, the student can put the pat- terns into the sand and achieve good results. Descriptions and illustrations of the tools used in the different kinds of molding and discussions of the principles involved are in- cluded. The patterns chosen may be easily obtained, and, as each brings out one or more distinct principles, the stu- dent who has completed this course should be able without further instruction to produce satisfactory results with any reasonable pattern. The book is thoroughly illustrated and contains a glossary and an appendix which will be invaluable foi* reference purposes. CONTENTS Chapter Chapter I. Materials VIII. Molding Exercises with Dry Cores II. Tools IX. Miscellaneous Exercises III. Principles of Molding X. Open Molding, Sweep and Strike rV. Molding Exercises without Work Core XI. Dry-sand Molding V. Matches XII. Cupola Practice or Melting VI. Molding Exercises with Green- XIII. Cleaning Castings sand Cores XIV. Chilled and Malleable Castings VII. Dry Cores XV. Brass Molding THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New York The Elements of Electrical Transmission A Text-book for Colleges and Technical Schools By OLIN JEROME FERGUSON Associate Professor of Electrical Engineering in Union College; Mem. A.I.E.E. ; Mem. N.E.L.A. ; Mem. Soc. Prom. Eng. Ed. Cloth, 8vo, vii-]r4S7 pages, illustrated, index, diagrams, $3.50 net In preparing the material for this book, it has been the author's aim to present those things which should be grouped and articulated in order to afford an elemental study of the very broad subject of the generation, transmission, distribution, and utilization of power by electrical processes. The title of the work only partially covers this field, but was chosen with a view to brevity and convenience. In fact, our language needs some modest word for so large a subject. Necessarily, the material available far exceeds the capacity of any single volume, but the attempt is made to provide a working text-book which may serve as a more or less complete course, depending upon the amount of time available for the subject. It is hoped that it may serve as an outline for even an extended course, supplemented always by that most important source, the teacher. Nevertheless, the book will probably stand or fall as an elementary text-book, which is all that it desires to be. Immediately after publication this book was adopted as a re- quired text for use with a large class of students at the Massa- chusetts Institute of Technology. This adoption combined with the fact that Professor Ferguson was fortunate enough to secure the valuable criticism and advice of Dr. Charles P. Steinmetz while preparing the manuscript, augurs well not only for the accu- racy of the book but also for its availability for use as a class text. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New York miiiiiiiH LIBRARY OF CONGRESS 028 20 1 53