Uft^t §mv^ Mhm^Un ^ mn to 1903 y Cornell University Library v.1-2 Notes on mechanics : lllllllllliil „ 3 1924 031 284 643 olin.anx Cornell University Library The original of tliis bool< is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031284643 NOTES ON MECHANICS. DESIGNED TO BE USED IN CONNECTION WITH RAN- KINE'S APPLIED MECHANICS. BY GAETANO LANZA, S.B., C.E., ASBISTAKT PKOFBBSOR OF MATHEMATICS AHD MECBABICg, MASS. INSTITirTi; OP TECHNOLOGY. PAET I. - STATICS. BOSTON: 1872. Entered, according to Act of Congress, in the year 1872, By GAETANO LANZA, in the OfSce of the Librarian of Congress at Washington. NOTES ON MECHANICS. The fundamental idea of Mechanics is Motion. Force, as far as we have to consider it, may be defined as that which causes, or tends to cause, a body to change its state from rest to motion, or vice verm, or to change its motion tts to direction or speed. Statics is that part of Mechanics which considers the relations of forces as producing pressures, or a tendency to motion. Dynamics considers forces as producing motion, Motion is a'fundamental idea ; it might be defined as change of position. Motion, as far as our senses are capable of recognizing it, is rela-- tive. "We cannot tell whether anything in the universe is abso- lutely at rest or not. All bodies on the earth partake of its motion, and hence even those that seem to be at rest when compared with the position of the earth, are really in motion. A person sleeping on a steamer is at rest with reference to the steamer, but in motion with reference to the earth. Hence we always assume some point as fixed; relatively to which we determine whether a body is in motion or at rest. 4 LAWS OP MOTtON. CONSIDERATION OF THE LAWS OF MOTION. There are three great laws of motion that He at the basis of Mechanics. The proof of these laws lies in a proper conception of motion, and is derived from observation and experiment. FIRST LAW OF MOTION. A hody at rest will remain at rest, and if in motion will continue to move uniformly and in a straight line, unless and until some ex- ternal force acts upon it. This means that matter is inert, that it has no power in itself to change its state from rest to motion, or from motion to rest. If it is at rest it remains at rest until some external cause puts it in mo- tion. If it is already in motion it continues to move uniformly, and in a straight line, that is, with the same speed and in the same direction ; and in order to bring it to rest, or to change its course or its speed, some external force is required. When we see terrestrial matter change its state from rest to mo- tion, or vice versa, we can generally assign the cause. A ball thrown from the hand soon comes to rest; the cause is Gravity and the resistance of the air ; if rolled along the ground Friction causes it to slacken its speed and soon come to rest ; if, however, we diminish the external forces by rolling it on glass or on the ice, it continues its motion longer. Thus we see that the more we diminish the external forces the longer the motion continues, and hence we infer that were we able to remove them entirely the motion would continue forever. In the heavens the uniform motion of the planets shows us that the law is fulfilled there. SECOND LAW OF MOTION. If a hody have two or more motions imparted to it simultaneously it will move so as to preserve them both. SECOND LAW OF MOTION. That is to say, that every force that acts on a body has its full effect entirely independent of any other force or forces that may act on the body. Every relative motion is composed of the motion within a certain space, and also the motion of this space in reference to another space. Thus a person moving on the deck of a ship has two mo- tions in relation to the shore, viz., his own, and that of the ship. Suppose him to move in the direction of motion of the ship at the rate of 10 feet per second, while the ship moves at the rate of 25 feet per second ; he appears to a person on the shore to move at the rate of 2o -|- |0 = 35 feet per second ; if, on the other hand, he walks in the opposite direction at the same rate, he appears to move at the rate of 25 — 10 = 15 feet per second. Suppose a body situated at A (Fig. 1) to have two motions im- parted to it simultaneously, one of which would carry it to B in one second, and the other would carry it to C in one second ; and we wish to determine where it ,will be at the end of one second, and what path it will have pursued. Imagine the body to move in obe- dience to the first alone during one second ; it would thus arrive at B ; then suppose the second motion to be imparted to it instead of the first, and it will arrive at the end of the next second at D, where BD is equal and parallel to AC. Now if the two motions are im- parted simultaneously instead of successively, the same point, D, will be reached in one second, instead of two ; and by dividing up AB and AC into the same (any) number of equal parts we can show that the body will always be situated at some point of the diagonal AD, or that it moves along AD. Hence the diagonal of a parallelogram constructed with two velocities, and with the angle inclosed by them, gives the direction and velocity of the resulting motion. A Force has three characteristics, which, when known, deter- mine it, viz., point of application, direction and magnitude. These can all be represented by a straight line by making the length of the line proportional to the magnitude of the force ; hence a force can be represented in point of application, direction, and magnitude by a straight line. O PAKALLELOGEAM OF FOHCES. Forces are most conveniently referred to our unit of weight as a unit ; they are evidently proportional to the amount of velocity or speed that they communicate to the same body. Thus a force which imparts to a body a velocity of 10 feet per second is twice as great as one that imparts to the same body a velocity of 5 feet per second. Hence from the parallelogram of motions we deduce that of forces, viz. : — If two forces acting at the same point he represented in point of application, direction and magnitude hy two adjacent sides of a par- allelogram, their resultant will he represented hy the diagonal of the parallelogram from their common point of application. Given two forces P and P' acting at the point A and inclined to each other at an angle d^ required to find the magnitude of the re- sultant force and its direction. Let AC (Fig. 1) represent P, AB represent P', and angle BAG = 6, then AD will represent in magnitude and directiop the re- sultant force ; also let DAC = a, then from the triangle. DAC we have AD^ = AC" + CD^* — 2 • AC • CD ■ cos DCA but DCA = 180° — d and .-. cos DCA z= — co& .: R2 =r P^+ P'2 :#F 2PP' cos OT ■ R = V(P^ + P'2 + 2PP' COS 8) This determines the magnitude of R ; and for its direction we have from the same triangle CD : AD = sin o : sin (180° — 0) or P' : R =^ sin a : sin O whence P' . . P sm a = ^ sin ; so also sin (0 — a) = ^ sin EXAMPLE. 1. Find the resultant of two forces equal to 10 and 17 respec- tively, and inclined to each other at an angle of 60°- Solution. R = V 1(10)"+ (17)" + 2-10-17 cos 60°| = V (100 + 289 + 170) = 23.643. sin a = ^^t.f^ sin 60°. TEIANSLE OF POECES. 7 log 10 = 1.0000000 ( R = 23.643 log sin 60° = 9.9375306 Ans. | a = 21° 29' 13" 10.9375306 ( — a = 38° 80' 47" log 23.643 = 1.3737026 log sin a = 9.5638280 a = 21° 29' 13" = angle between the force 17 and resultant. 2. Two forces equal to 47.84 and 75.46 respectively, act at an angle of 73° 14' 21" to each other, find the magnitude and direc- tion of the resultant. TRIANGLE OF FORCES. If we have the two forces OA and OB (Fig. 2) acting through one point O, their resultant is evidently OC ; and in order to bal- ance the two forces we shall have to apoly a force equal in magni- tude and opposite in direction to Qp. Now the sides of the triangle OAC represent in magnitude and in direction the force OA, the force AC and a force equal and opposite to OC, and these three forces, if applied at the same point, would balance each other, hence : — ^ three forces be represented in direction and magnitude hy the three sides of a triangle taken in order, then these forces when applied at one point balance each other. Furthermore we have OC : OA : AC = sin OAC : sin OCA : sin COA, or R : : P : P' = sin » : sin ((J — a) : sin a. If in the value of R, page 6, the angle d becomes 90°, the re- sults become (Fig. 3) P' P R = V (P2 + P"") ; sin a = g-; cos a = g- EXAMPLE. Ex. Let P = 4, P' = 3, and then we have R = V (16 4- 9) = 5, sin a — ^,co&a. — % log 3 = 0.4771213 log 5 = 0.6989700 log sin a = 9.7781513 a = 36° 52' 11" 8 DECOMPOSITION OF FORCES. DECOMPOSITION OF FORCES IN ONE PLANE. If, on the other hand, we have given a force R, makiDg a given angle a with AC (Fig. V) ; we can decompose the force R into two components, one in the direction AC, and the other perpendicular to it ; that is, we can find two forces acting in the directions AC and AB, respectively, that shall be equivalent to the single force R, The two components of R are AC = P, and AB =^ P', and from the figure we derive P = R cos a; P' == R sin a. Suppose now we have given three forces, P, P' and P", making angles «, «' and a'' (Fig. 4) respectively, with the line Ox (which we shall call the x axis), and we wish to determine the resultant of these three forces as to direction and magnitude. We first resolve each one into two components in the directions Oic and Oy, thus the components of P are OA and OA', of P', OB and OB', of P", OC and OC, and moreover, OA = P cos a, OB = P' cos a', OC = P" cos a", OA' = P sin a, OB' = P' sin a', OC = P" sin a". Now these forces are together equivalent to a force P cos a + P' cos a' + P" cos a" in the direction Ox, and a force P sin a + P' sin a' -\- P" sin n" in the direction Ot/. The first may be represented by .Z' P cos a, and the second by .ZP sin a; wheie 2i stands for "sum," either there are three or more forces, and now we can find the resultant of these two, viz., R = y ^ (^- P cos ay + (^ P sin a)% also a„ the angle this resultant makes with Ox, is determined fiom the equation 21 P cos a cos aj5 n • EXAMPLES. 1. Given P = 47.321 « =21° 14' 27" P' = 73.423 a' = 43° 2' 54" T" =43.214 a" = 82° 4' 12" P'" = 23.457 a'" = 112° 41' 45" find the resultant force and its direction. 3. EXAMPLES. tlution. P COS a = 44.1063 P^ sin a = 17.1439 P' cos a' = 53.6559 /P' sin a' = 50.1196 P" cos a" = 5.9619/ P" sin a" = 42.8007 P"'cosa'" = 42^ 404 TP'" sin a'" = 30t9864 i; P cos a = 91.1837 2 P sin ,i = 140.0006 R = ^^91-1837)2 + (140)^^ = v'(8314.4671 + 9600) = ^279 14.4681 == 167.076. logi^Pcosa = 2.1461295 log R = 2.2229141 log cos 6 = 9.9232154 e = 33° 4' 36". Given P = 4.327 a = 77° 19' 47" P' = 45.432 a' == 163° 58' 32" P" = 105.27 „" = 275° 14' 12" Given P = 57.432 a = 33° 4' 56" P' = 43.241 a' = 145° 10' 25" P" = 12.556 «" = ^ 10° 15' 3" P'"= 12.556 a'" = 190° 15' 3" Given P = 123 « = 10° 2' P' = 432 of = Q° 4' P" = 547 a" = 48° 3' P'" =965 " a'" =72° 14' P'^ = 127 „'v = 46° 37' Given P = .0123 a = 37° P' = .0047 a' = 42° 5' 17" P" = 2.3456 a" = 57° 3' P"'= 9.01 Oo a'" = 42° p-v = 7.0003 a" = 60°- P" = 8.001 a^ = 54° 15' Given P = 5 a = COS"^ f P' = 4 a' = P" = 3 a" = 90° 6. 7. Given three forces equal to 17, 15 and 13, respectively ; the angle between the first two is 73°, and between the second two 47°, find the magnitude and direction of the resultant. 8. Given four forces equal to 57, 8, 36 and 25, respectively ; 2 10 COMPOSITION OF FORCES. angle between the 1st and 2d is 60°, between 2d and 3d 30°, be- tween 3d and 4th 45°, find the resultant force and its direction. COMPOSITION OF FORCES IN SPACE. If we have several forces all acting at the same point, of which we wish to find the resultant, we assume first three lines, OX, OY and OZ (Fig. 5), at right angles to each other, whose common origin is the point at which the forces act. Now suppose OE = P to represent one of the forces ; we first resolve it into two forces, OC and OD, one of which acts along OZ, and the other in the plane XOY; we then resolve OD into two components, OA and OB, acting in the directions OX and OY respectively. We thus obtain the three forces OA, 01$ and OC acting in the directions OX, OY and OZ respectively, which are together equiv- alent to the single force OE. These three components are the three edges of a rectangular parallelopiped of which OE = P is the diagonal. Let now a = angle EOX, ^ = EOY, ;' = EOZ and we have from the right angled trianglciiEOA, EOB and EOC respectively the following equations : — OA = OE cos EOX = P cos o. OB — OE cos EOY = P cos |3 OC = OE cos EOZ = P cos r We have also OA^ -^OW = 0D= and OD" +00" = OE" .-. A" -f OB" -h 0C"= OE^ or P" cos" a + P" cos ^ + P" cos" ;- = p" .-. cos" a -\- cos" /? + cos" ?" ^ 1. When two of the angles, a, ^ and y, are given, the third can be determined from the equation above. We proceed in the same way to resolve each of the forces into three components along the axes of x, y and z respectively, and we thus obtain a single force acting along the axis of x equal to ■•"i Pcos a = P cos a + P' cos a -\- P" COS a" -\- &c., SO likewise a force, ^ P COS ^, acting along the axis of ^and - P cos y along COMPOSITION OF FOECES. ir the axis of 2. These three forces produce the same effect as all , the forces with which we started. We next proceed to find a single resultant for these three forces. . Let (Fig.6) OA .= Z P cos a. OB = 2: P cos ^. 00 = ^ P cos y. Compounding OA and OB we find OD to be their resultant, and then by. compounding OD and 00 we find OE to be their re- sultant, and hence the resultant of the three forces OA, OB and OC. We have OE^ =' OD^ + OC^ = OA^ + OB^ + OO" or E^ = (SV cos ay + (rP cos ^y -\- {SY cos 7)^ 4, OA SV cos a , ... Also, cos a, = TyFT = TS — ~" and likewise cos /S, OE 2'Pcos/9 R , cos J', S P cos 7 This gives us the magnitude and direction of the resultant force. Examples. 1. Griven. P = 63.421 a = 53° 15' ^ = 42° 17 P' =49.357 a' = 87° 3' 10" / = 72°3'27" P"= 2.729 ^'=70° 5' 17" /'=45°7'32" P"'=r 4.325 a"'= 90° |3'" = Find the resultant^ force and the angles it makes with the axes respectively. 6S.421 49.357 2.729 4.325 53° 1¥ 87° WW 99° 42° IV 70° 5/17'/ 0° rao3'27// 45° 7' 32'' .05141 .62144 .00000 cos ^ .94997 .34057 1.00000 COS-y ,30767 .00000 Pcos a 37.94605 2.53744 1.60590 0.00000 42.17939 PooS;3 46.92075 46.88766 0.92941 4.; " P cos y 19.51273 15.20541 1.92547 0.00000 36.64361 {S P cos ay = -1779.1009407721 (IT coi^y = 9813.4423063524 (iJP.cos rf = 1342.7541538321 W = 12935.297400956G R = 113.7334. 12 CONDITIONS OP EQUILIBRIUM. log 2 P COS a = 1.6260003 log 2 P cos ;a log E = 2.0558879 log E 9.6701124 = 1.9959100 = 2.0558879 9.9400221 log 2 P cos, V = 1.5639982, los E = 2.0558879 9.5081108 a, = 68° 11' 2" j3, = :: 29' ° 25' 27" ;-, = 71° 12' 16" 2. Given P = 4.324 a =47° 2' § = 65° 7' F =87.465 a' =88° 3' / = 10° h' P" = 6.342 a" = 68° 4' §" — 83° 2' 3. Given P = 5.107 a =90° (3 =90° P' = 7.321 a = 0° P" = 4.325 (3 = 0° P'" = 74.215 a'" =73° 2' 15" f' = ^b°\r 3" 4. Given A P =43.217 a =30° |3 =60° P' =59.017 a =45° r =45° P" = 63 a =72° 3" 14" y =43° 2' 29" P"' = 42 ^ = 83° 7' y =54° 5' 5. Given P = 87 a = 30° j3= 60° P' = 42 a = 45° |3= 45° P" =112 a = 90° |3= 90° P"'= 3 a = 90° ^= 30° P'-= 85 a = 15° ^ = 105° Find in each case the magnitude and direction of the resultant. CONDITIONS OF EQUILIBRIUM. We have seen that in the composition of forces we begin by re- solving each force into three components along three rectangular axes ; and for the final resultant force we have R2 = (J P cos of + (-^P cos (3)2 4- ('S' P cos rY- Now if the forces balance each other, or are in equilibrium, the resultant must be equal to 0, that is, R ^= .•. R^ = .-. {2 P cos ay + (2' P cos /?)2 + (-SP cos jf = 0. But this expression is tJie sum of three squares, none of which RIGID BODIES. 13 can be negative ; hence each one of the quantities in the brackets must be separately equal to 0, hence the conditions for equilibrium* are 2: P cos o = ; 2' P cos ,? = ; and 2' P cos >- = 0. When the forces are all in the same plane, that of (x, y), cos y = 0, since y = 90° in every case, and hence cos^ a + cos^ ;? = 1, or cos /S = sin a, hence the conditions of equilibrium become in this case wT P cos a = ; 2 P sin a = 0, a result which might have been obtained from, the fact that for forces in one plane R^ = (2 P cos «)'' + {ZV sin ay, and hence when R = 0, -T P cos a = and 2Psin a =0. This shows that in order that a set of forces may be in equilibrium the sums of their components along each of the axes must be separately equal toO. STATICS OF RIGID BODIES. Thus far we have always conceived ourTforces'toact uponja sin- gle point, and have not considered the case when we have forces acting at different points of a body. A Rigid Boot is one that does not undergo any alteration of shape, when external forces are applied to it. Strictly speaking no body is absolutely rigid. If a force be applied to a rigid body (Fig. 7),*say at the point A, it comtnunicates motion to this particle ; but since the body does not alter its shape the next particie must move the same amount, and in the same line, and so the next, so that each parti- cle in the line of the force moves just as though the force were ap- plied there. • And moreover if we apply an equal and opposite force at any other point of the line of action of the first, as B, it produces equilibrium, hence in the case of a rigid body the point of appli- cation of a force may be transferred to any other point in the line of action of the force without changing 'the effect. 14 MOMENT OF A FORCE. COMPOSITION OF TWO FORCES ACTING AT DIFFERENT POINTS OF A RIGID BODY. Suppose the force P (Fig. 8) to be applied at the point A, and P'at B. Now, since the point of application of a force may be moved anywhere in its line of action without changing the eflect, we produce the lines representing the directions of the forces till they meet in 0, and suppose P and P' both to act at 0. Now we find by forming the parallelogram, ODFE the resultant of these two forces to be R = OF acting in the direction OF : we may im- agine its point of application to be anywhere in its line of action, say at C. MOMENT OP A FORCE. We have hitherto considered forces so far as they tend to pro- duce motion in a straight line, that is motion of translation. But suppose the point A of 'a rigid body (Fig. 9) to be fixed, and a force to act in the direction, BD ; draw AC perpendicular to BD and transfer the point of application of the force to C, which does not alter the effect ; now the point A being fixed, the only motion that C can have is a motion of rotation around A as a cen- tre. It becomes necessary for us to find a measure for the rotatory effect of a force in regard to any point of a body. Let, as before (Fig. 10), the forces P act at A, and P' at B, and suppose on finding the resultant we find it to be CD. If now we apply at C a force R, equal and opposite to CD, this will, together with P and P', keep the body in equilibrium. Now let us examine the effects of the respective forces P, P' and E in regard to rota- tion around C, since we may regard C as a fixed point, the body being at rest. It is evident that the force P tends to make the body rotate in one direction, and the force P' tends to make it rotate in the oppo- site direction, and since the body has no motion of rotation, and these two are the only forces that have a tendency to give it such but since these areas are equal p OH = jo' OG, or — , == 7=rfj, but STATICAL COUPLES^ 15 a motion, tberefore the rotatory effect of these two forces around C is balanced, and the question arises how to find the condition of such balancing. From C draw the two perpendiculars, p and p', on the directions of the forces, and we have the area of the trian- gle OCG equal to ^ p' • OG ; and that of 0(3h equal to Ip- OH ; £ _ OG P OG P' o I" *„ OH =F •••/= p,orp^=^P, that is, the product of the force P by the perpendicular from C on its direction is equal to the product of P' by the perpendicular from C on its direction, and this is the necessary aiid sufficient con- dition that the two forces, P and P', may produce no rotation around G, bearing in mind, ©f course, that the two tendencies of rotation must be in opposite directions. Definition. The product ■ of a force by its perpendicular dis- tmicefrom a 'point, is called the moment of the force with respect to the point. Since two forces that have the same moment with respect to a poiAt produce the same rotatory effect, therefore we may take the moment of a force as the measure of its rotatory efi'ect. STATICAL COUPLES.' , Suppose (Fig. 11) we have two equal forces acting on a body in parallel but opposite directions ; draw the line AB perpendicu- lar to both, and transfer their points of application to A and B re- spectively. Now it is evident that such a pair of forces can have no single resultant ; for such a resultant would produce motion in a definite direction, and any reason that could be given for assign- ing the motion with respect to one of the forces, would be equally valid with respect to the other, hence the only effect of such a pair of forces is a motion of rotation. Such a pair of forces is called a Statical Couple. Its effect is evidently to produce rotation around an axis perpen- dicular to the plane of the couple. 16 STATICAL COtJPLES. This rotation may be in the direction of the hands of a watch, when it is called right-handed, and is accounted positive, or in the opposite direction, when it is left-handed, and is accounted nega» tive. Tbe rotation produokd bt a couple is the same, no mat- ter at what point we take the axis, provided only this axis be per- pendicular to the plane of the couple ; and the rotatory effect is always measured hy the product of the force at eitJier end multi- plied hy the perpendicular distance between them, which is called the arm of the couple, and the product is called the Moment of the Couple. Let the couple (Fig. 12) be that with arm AB and with force P, and let us take O as the fixed point ; draw from O a perpendic- ular to the forces, and transfer the joint of application of P to D, and of — P to C. Now the rotatory effect of P is P-OD, and that of — P is P-OC ; or the resultant rotatory effect is the difference between these two, since they act in opposite directions, or P-CD = P-AB. . If now the arm of the couple be turned through any angle, it changes neither the direction of rotation nor the rotatory effect ; and the same is the case if the couple be situated in any plane parallel to the plane where it now is. Hence two couples are equivalent whose planes are coincident or parallel, whose moments are equal, and whose sense (direction of rotation) is the same. COMPOSITION OF STATICAL COUPLES. First suppose the couples to be situated in parallel planes, or in the same plane. If their forces are not the same reduce them to the same force, and then place their arms in the same straight line, as in the figure ; the forces P and — P at B evidently balance each other, and may be removed ; and we thus have left the couple whose arm is AC = AB + BC, that is, the sum of the two arms. Multiply through by P, and we have P-AC = P-AB -f P-BC, which shows that MOMENT OF A COUPLE. 17 The moment of the couple, which is the resultant of two or more couples in the same Or parallel plaries, is equal to the sum of the moments of the component couples. EXAMPLES. 1. Convert a couple wliose force is 5, and arm 6, to an equiva- lent couple whose arm is 3. Find the resultant of this, and an- other couple in the same plane, whose force is "1, and arm 8 ; and find the force of the resultant couple corresponding to an arm 5. • Solution. Moment of 1st couple = 5X6 = 30. When arm is 3 force = ^^ = 10. Moment of 2d couple = 7 x 8 = 56. Moment of resultant couple = 30 + 56 = 86. When arm is 5 force = */ ^ 11^. 2. Force. Arm. ■ 12 17 3 8 Given the fol- 5 7 lowing couples in " 6 9 one plane. 12 12 10 9 . 14 6 Convert to couples havings the following. Forte. 5 8 6 Ann. 20 The first and the last three are right handed, the second, third and fourth are left handed ; find the moment of the resultant couple, and also its force when it h&s an arm 11. Arm. 3. Force. Arm. Force 7 9 11 14 12 Given the fol- 15 9 Convert to 6 lowing couples all in the same plane. 12 6 3 1 15 9 2 1 couples having the following. 43 12 - 1 i L 9 30 12 The first four are right handed, last four left handed ; find result- ant couple, and its arm when the force is 25, 3 18 COMPOSITION OF COUPLES. Force. Arm 10 12 4 3 6 9 10 9 5 12 12 3 Force. 8 Arm. The first three are right handed, arid last three left handed ; find resultant conple. REPRESENTATION OF A COUPLE BY A LINE. From the preceding we see that a couple remains always the same as long as its moment does not change ; the direction of its axis (j. e., a line drawn perpendicular to the plane of the couple), ' and the direction' in which it tends to make the body turn remains the same. Hence a couple may be represented by drawing a line in the di- rection of its axis', and laying off a distance representing the mo- ment of the couple on this axis. COMPOSITION OF COUPLES SITUATED IN PLANES IN- CLINED TO EACH OTHER. Suppose we have two couples situated neither in the same plane nor in parallel planes, and we wish to find their single resultant (!Ouple. It is first necessary, if their arms are not equal, to sub- stitute for them equivalent couples having equal arms ; then transfer them in their own planes respectively to a position such that their arms shall lie in the line of intersection of the two planes and make their arms coincide. Suppose after this transformation has been affected 00' (Fig. 14) to be tjie common arm, P and — P the forces of one couple, P' and — P' those of the other. The forces P and P' have for their result- ant E. COMPOSITION OP COUPLES. 19 Now OP and OP' are both perpendicular to 00', hence their plane and consequently the line OR is perpendicular to 00'. — P and — P' give a resultant — R perpendicular also to 00' and par- allel to R. The resultant of the two couples is consequently a couple with the same arm, and with a force R, the diagonal of the parallelogram on OP and OP', so that R = ^ (P2 + p'2 + 2PP' cos d) where d is the angle between the planes of tlie couples. Now if we draw at the line Oa perpendicular to 00' and also to OP, consequently perpendicular to the plane of the first couple ; and Ob perpendicular to the plane of the second ; and make the lengths Oa and 06 proportional to P and P' so that Oa : 06 : : P : P' :: 00' • P : 00' • P' : first we observe that these lines Oa and 06 lie in the same plane with OP and OP', perpendicular to 00' ; and moreover if we con- ■ struct on Oa and 05 a parallelogram the diagonal Oc will be per- pendicular to OR, and Oc:OR :: Oa : OP :: 05 : OF since the angle a06 ^= POP' ; hence if Oa and 06 represent the moments of the couples (of force) P and P' respectively, Oc will represent the moment of the resultant couple whose force Is R, and Oc = y(Oa« 4- 06» + 20a • 06 cos 6) let Oa = L, 06 = M, we have OC — Gr, and, G = y (L^ + M^ + 2LM cos d). We therefore see that we can compound and resolve couples just as we do forces, prQvided we represent a couple by its moment axis. EXAMPLES. 1. Two couples act in planes- inclined to each other at an angle of 65° 5' ; their moments are 43 and 75 respectively, and both are right handed when looking along their axes. Find the moment of the resultant couple and the angle of inclination o£ its axis to that of the first. 20- EXAMPLES. Solution, e = 65° 5', L = 43, M = 75. G = ^{JJ^ + M^ + 2LM cos &) = y 1(43)= + (75)2+ 2-43-75 cos 65° 5'^=^10191.3850 = 100.9523. Now for the angle its axis makes with that of the first, sin a = xttB-S!? sin 65° 5'. log sin 65° 5' '= 9.9575697 log 75 = 1.8750613 1.8326310 log 100.952 = 2.0041149' 9.8285161 a = 42° 21' 33". d — a = 22° 43' 26". If one of the couples were left handed when looking along .its axis as given, we should only look at it from the other end of the axis and the only change made would be = 180° — 65° 5', or ^ = 114° 55' .-. G = ^\{4.Zy + (75)2+2 •43-75 cos 114° 55'^ = y(4756-6150) = 68.9682. log sin 114° 55' = 9.9575697 log 75 = 1.8750613 1.8326310' log 68.9682 = 1.8386488 log sin a = 9.9939822 a = 80° 29' 1" d~a = 34° 25' 59". 2. Given two couples whose axes are inclined to each other at an angle 37° 15' and whose moments are 83 and 112 respectively; hoth right handed. 3. Given L = 57.32, M = 243.12, d = 170° 3', find G and " logik^ = y.8421677 d=ji4:°3'2". This couple and the force 95 at the origin are equivalent to a parallel and equal force at the point llff, 12^*^. 2. F X y3. F x y L ¥ x y 12.3 4 7 14.7 5 6 ' 10 12 13 6.2 — 5 8 11.3 —2—3 6-6 7 12.5 4—10 17.7 5 7 9 14 15 8.2 7—5 18.9 4 3 11 28 53 6.8 4—9 12.3 8—9 49 3 J 5. F X 2^6. F X y 7. F X y 10 7 3 7 51.848 , —8 6 4 3^ 4 3 —5 3 6 '6 3 2 5 7 9—225 — 9 5 '7— 8 12 15 —314 1 i i — 7 17 16 —5 7 9 8. F X y 9. F X y 10. F a; y 9.2 8.4 12.3 127.2 12.4 9.4 10.2 7.3 5.4 7.1 5.6 8.2 835.7 6.37 8.3 12.7 21.4 9 9 12.7 9.8 946.8 4.37 7.2 7 6 6 6.3 4.2 — 7.1 832.8 8.73 9.1 8 9 10 8.3 -^7.1 8.3 946.7 7.65 10.8 11 12 13 9.5 48.1 72.6 832.8 8.51 9.3 7.4 8.2 9 Find the resultant force and th§ coordinate of its point of appli- cation in each of the above. CENTRES OF PARALLEL FORCES. 25 CENTRES OF PARALLEL FORCES. We have seen that in combining any number of parallel forces we obtain, (except when 2: F ^ 0), a single resultant acting in a definite direction whose point of application may be taken any- where along its line of action. If we have two parallel forces (Fig. 19), P and P' applied at the points A and B respectively of a rigid body, from what we have already seen, the resultant ia a force P -|- P' applied at the point C which divides AB inversely as the forces ; now the point of application may be assumed at C, or any other point of CR as long as the forces remain in the same position, but if we suppose them turned around and taking the positions AO and AO' the resultant will still pass through C, but no other point of the line CR, hence C does uot'change as long as the points of application A and B of the forces remain the same, and their relative magni- tude does not change, and they remain parallel, no matter how they are turned. This point C is called the centre of the two parallel forces ; and in general The Centre of a Systf.m ok Parallel Forces is the point of application of the resultant of the' s.y. tern, however the system be turned around, provided only the points of application of the single forces remain the same, and their directions continue parallel to each other. Hence; in finding the centre of a set of parallel forces, we may suppose the forces turned around through any angle whatever, and the centre will remain the same.* examples. Given the following parallel fprces and the coordinates of their points of application, find the centre of each system. 1. F X y 2 2. F X y z ' 75.3 4 9 8 82.6 6 13 48 83.2 _6 5 7 73.1 7 25 65 _9 6 3—8 54.6 —9 3.7 99 12 5—7 3 33.7 12 42 87 4.7 5.3 _8.l 9.7 67.5 —15 31 78 >I now refer the student to Rankiue's App. Mecli., Section 4, for the determination of the coordinates of the centre of a set of pai^lel forces. 2G FOKCES IN A PLANE. F X y z 4. F X y z 63 87 21 1 1 48 93 7 i 1 —1 —1 — 1 74 64 6 2 1 — 1 1 — 1 82 82 3 1 1 1 — 1 91 97 4 1 —1 INCLINED FORCES AT DIFFERENT POINTS ALL ACTING IN THE SAME PLANE. Another mode of demonstrating Rankine's Art. 59. Let CF, (Fig. 20), be one of the forces. We first decompose it into the two forces CD and CE acting in the directions OX and OY respectively, CD = F cos a, CE = F sin a. Now transfer the point of application of CD to B, and of CE to A ; this does not alter their effect. 2d. Apply two forces at O in opposite direc- tions, in the line OY, each equal to CE ^ F sin a ; and two in the direction OX, each equal to F cos a. They do not alter the effect at all. "We have now, instead of the force FC, six forces of which AH and OG form a couple (right handed in the figure), and BL and OK a left handed couple. The moment of the first is AH • AO := F sin a-x, and of the second BL • OB = F cos a-y (a; and y being the coordinates of C) and since these two couples are in the same plane, the moment of their resultant couple will be Fa; sin a — Fy cos a = F (a: sin a — y cos a). We have also a force OM = F cos a, acting along OX and ON = F sin a along OY. We decompose in the same manner each of the forces into two, F cos a along OX, and F sin « along OY, and also a couple F(a; sin a. — y cos «). The resultant force along OX will be ^ F cos a, along OY 2, F sin a, and the moment of the resultant couple M = .i' F (a; sin « — y cos «). The two forces, X F cos a and }l, F sin a, give a resultant R ^ y \{X F cos «)^ + (i^ F sin a)^! EXAMrtES. 27 which we determine as in Art j4' of Rankine. We may then detenmiue the point of application of the single force equivalent to this resultant and the couple by Art. 1^ of Rankine, ui by bh o ^PP'sin a ' ^V^ 2' F cos ,a ' "^ EXAMPLE. 1. SoMioH F a; j; K cas a sin a Fcosa F sin a. Tashiii FycosB 5 3 2 32° .8480S .B2992 4.24025 ■2.64980 7.94880 8.48069 10 1 3 43^ .73135 .68200 7^81350 6.«2000 6.82000 21.94059 7 4 2 54=10' .58543 .81072 4.098DI 5.67594 22.70016 8,19602 —3 7 9 59<' 7' .51329 .85821 -1.53987 -^.57483 —18.02241 —13.85883 2 6 1 35° S' .81866 .67429 1.63730 1.14858 6.89148 1.63730 1 i 2 41° .75471 ,65606 0.75471 0.66806 2.62424 1.50943 16.50390 14.37465 28.96237 ^ 27.80491 27.90491 « 1.0o748 = Moment of nesulL couple. (2" F cos 1 .T = 272.3787152100 (^ F sin <«)* = 206.6305626225 479,0092778325 R=y K^F cos ay + (^F sin ay] = 21.88628 log 16.5039 = 1.2175865 log 11.8862 = 2.3401718 log cos a^. = «1 = F a; y iX 12 27 33 15° 3' 4 32 45 75" 2' 8 —5 —7 63" 5' 9 4 —3 42"" 9' 21 1 2 110° 4' U i —1 163° 2' 9.8774147 41<. 3' 19* F X y a 11 9 8 210° 4' 74 t I 54° 2' 63 * -* 73° 12' 16 i i 87° 43' 25 i ^ 93° 57' 33 ^ 4 82° 9' 28 ■ FORCES IN SPACE, 4. F a; y a 5. F a S' a 4.2 18 27 14° 29' 10 3 2 30° 3.5- 2 4 53" 37' 5 7 — 1 45° 5.7 9 8 82° 51' 3 —2 1 "72° 8.6 10 12 97° 50' 7 1 4 125° 9.3 20 5 81° 14' 3 4 —1 75° 11.2 5 8 93° 17' 27.5 2 7 80° 4' 5 7 9 60° 7. 10 5 17 30° 3 4 10 30° 5 9 82 45° 3 8 11 160° —3 11 91 60° —2 9 2 0° —2 23 83 90° — 1 3 1 60° 1 25 112 0° 3 2 5 75° 9. 7.1 5 3 150° 4 7 2 45° 8.2 9 —2 210° —7 6 3 90° —6.3 3 —4 60° —2 1 <^4 180° 9.4 —2 5 15° 2 4 10° —12,5 1 3 87° Find in each of the above the resultant force, and couple, COMPOSITION or FORCES IN SPACE. Suppose we have a number of forces applied at different points of a body, and in different directions, of which we wish to^ find the resultant ; we refer them all to a set of three rectangular axes. Suppose PR (Fig. 21) to be one of the forces = F. First we draw through P parallels to OX, OY and OZ, and form a rectan- gular parallelopiped ; we thus obtain the three forces PK, PH and PG, which are equivalent to PR. Now let a = RPK, (3 = RPH, and y ::= RPG ; also, let (a; = OA, y = OB, z = OC) be the coordinates of P. We now can transfer the point of application of PK to E with- out altering the effect, and we thus have EL instead of PK. Now introduce at B and at O two forces opposite in direction, each equal to EL, and parallel to EL ; these four forces do not alter FOECE.S IN SPACE. 29 the effect. We now have, instead of the force PK, the five forces EL, BM, BN, OS and OT. The two forces EL and BN form a couple, parallel to the sx plane whose axis lies in the Y axis, the moment of which is EL-EB, but EL = F cos a and EB = z, hence moment = F 2 cos a : the two forces BM and OT form a couple perpendicular to the z axis whose moment is BM ■ OB == F y cos a. Now do the same for the other two . forces PH" and PG, and we shall finally have instead of the force PR three forces, F cos a, F cos f!, F cos y, acting at O in the directions OX, OY, OZ respectively ; together with six couples, two of which are in the zx plane, two in the zy plane, and two in the xy plane. They thus form three couples whose moments are as follows : — Around OX F (y cos y ■ — z cos §) " OY F (2 cos a — x cos 5') " OZ F (a; cos (3 — y cos a) Treat each force in the same way, and we shall have in place of all the forces three forces acting at O ; -T F cos a along OX ; E F cos ^ along OY, and ZY cos 7 along OZ ; also three couples whose moments are as follows : — Around OX Mj ^=- 2 {E {y cos y — z cos §) \ « OY M2 = 2: \F (z cos a — aicos y)\ " OZ M3 = ^^ ^F {x cos |3 — y cos a)l The three forces give a resultant R = V {{ZF cos ay + (ZF cos §y + {2 F cos yYl ^ F cos a „ ZF COB 3 ZF cos 7 cos a, = g ; cos ^, = —■ — g ■ ; cos 7, = g as in Art. 60 of Rankine. Bear in mind also that cos^ a -\- cos^ ^ + cos^ 7=1 The three couples give a resultant M = y(Mi^ + M/ + Ma^) Ml M2 M,, cos A = ij ; cos At = jj ; cos k »= jj- X, n and V being the angles the axis of the resultant couple makes with OX, OY and OZ respectively. 30 FORCES IN space; Example 1, with Solution u: y X a 3 V cos a CO.S p — 14 IS 2 47° 49=' .68200 .65606 1 -1 « 82» 74° .45258 ..84805 9 7 1S° 82° .96593 .21819 2 1 47° 13° .00000 .68200 Ty cos a Fa cos (3 Fa cos a 50.41920 15.74544 16.36800 — 1.79166 33.07395, 17.65062 11.13360 15.27330 67.61510 2.19405 0.00000 0.00000 cosy .32320 .27564 .13917 .73135 F«os I 8.18400 9.65930 0.00000 Foos/3 7.87272 11.02465 2.18190 2.04600 Fcosv 3.87840 3.683.'!2 1.39170 2.19405 23.72684 23.12527 11.04747 Fx cos -y — 54.29760 8.68332 12.52530 4.S Fx COS j8 — 110.21808 11.02465 19.68710 4.C Fy- COS a 106.39200 — 2.94177 77.27440 0.00000 64.09269 101.63372 — 3.S.800B8 135.43460 El = 23.72684 Rj = 23.12527 Rs = 11.04747 — 83.80088 Mi = — 75.46438 180.72463 180.72463 — 256, — 2.13750. 135.43460. 256.18896. R= ^/(Ri^ + R/ + Rs^) = y(1219.7876423594) — 34.92545. M = ^(Mi^ + M/ + Mj^) = v* (83979.89300929) = 289.7928. log E logE = 1.3752398 = 1.5431419 log Rj = 1.1 log E = 1.6431419 log 008 Pi = 9.i " cos «r = 9.8320979 a, = 47° 12' 24" ^ = 48° 32' 15" logEs logE = 1.04S2628 = 1.5431419 log Ml logM log cos (180° ■ 2. P 3 — 1 — 3 7 3. 12 10 5 7 = 0.8299061 = 2.'. log Mj = 2.1317296 log M =2^ logcosy, = 9.5001209 y, ~ 71° 33' 36". logMj logM = 2.4 ■= 2.4620876 - A) = 7.8678185 log cos m = 9.6696420 log cos (180° — v) = 9.9464727 90° 25' 21" iJ. = 62° 8' 18" y = 152° 8' 8". X 2 -3. 5 6 14 3 4 2. V -1 -4 7 3 13 2 3 1 -5 7 6 2 2 1 2 3 a 10° 97° 32° 49° 30° 36° 72° 100° 30° 45° 59° 60° 60° 53° 45° FORCES IN SPACE. 31 4. F X y z a ^ r 3 1 2 3 37° 75° 5 2 — 3 5 87° 23° — 7 — 2 — 1 3 40° 3' 61° 9 — 1 • 4 2 44° 7' 47° — 6 3 3 — 1 90° , 90° 3 2 — 4 — 6 30° b' ■ 79° 55' — 5 — 3 — 1 — 4 25° 37' 80° 8 — 2 3 — 7 61° 5' 30° 5. 7 11 12 10 10° 98° ^W^ 9 13 21 8 77° 14° 16 4 7 9 160° 73° 14 6 3 2 158° 81° 23 9 8 7 '47° 58° 48 1 i 1 32° 1' 81° 9' 17 — 4 3 — 2 56° 55° 65 -i -iV T5V 63° 63° 14 i i iV 72° 72° 4 Suppose after we have reduced one set of forces to a single resultant force acting at O, and a couple ; OP = R, to be the re- sultant force, and OC = M, the axis of the resultant couple ; and let be the angle between these lines. By a principle of projections we know that the projection of the straight line joining two giv6n points on another line is the same as the. projection of any broken line joining the same two points. Now let us project OP = R on the line OC, we obtain OD = R cos Q. Then project in its stead the broken line OABP, and the length of this latter projection must be equal to the former. Now 0A,"-9S. and l>€!, are the coordinates of P, and make with' OC the same angles as the axes OX, OY and OZ, that. is /(, //, and ii, respectively ; hence the length of the projection is equal to OA cos ^ + OB cos //. + OC cos 'j, but OA =^ R cos a, ; OB = R cos (3^ ; OC = R cos j'. 32 CONDITIONS OF KQtJILIBEIUM. .'. R COS S = R COS a, COS A + R cos ^^ cos /* + R cos j'j cos v. .: cos d = cos «, cos -i + cos (3^ cos /j. + cos y^ cos ■^. I. When cos I? = 0, or ^ = 90°, the force lies in the plane of the couple, and we can then reduce them to a single force, acting M at a distance from O eq.ual to 5-, and parallel to R. II. If A = a„ ij. = j3j, and v = ;'„ that is, if the axis of the couple is parallel to the force, then we have a force of translation, and a couple causing rotation around the direction of the force as an axis, and we can simplify it no farther. III. If the axis of the coiiple is oblique to the direction of the, force we can reduce them to a single force, and a couple acting as in the former case. For the couple M may be resolved into two components, one of which acts in a plane perpendicular to the direction of R (M cos 6), and another acting in a plane containing the direction of R ; and the latter on being combined with K gives an equal and parallel force at a distance from the line of action of M'sin d the 1st = — j5 IV. When R = there is only a couple. V. When M = there is a single force. CONDITIONS OF EQUILIBRIUM. The conditions of equilibrium are evidently that there shall be no motion of translation and none of rotation, that is, R = 0, and M = 0. .-. R^ = R,'' + R/ + Rg^ = 0, and M^ = Mi^ + M^^ + Mj^ = 0. .-. Ri^= 0, R2 = 0, R3 = ; Ml = 0, M2 = 0, M3 = ; and these are the six conditions of equilibrium. EXAMPLES. In all the examples of the preceding article reduce the result to a single force and a couple, as in III. PARALLEI, PROJECTIONS. 33 PAKALLEL PROJECTIONS. In order that one figure may be a parallel projection of another, two conditions must be fulfilled. First. Each point of the one must have a corresponding point in the other ; that is, there must be 'some law connecting the two figures, such that as soon as a point is given in the first a point is determined in the second corresponding to it. Second. For each pair of equal and parallel lines in the one there must be a corresponding pair of equal and parallel lines in the other. Thus, suppose we draw in Fig. 23 any two parallel lines, and take AB and CD equal. Now determine the point. A', B', C, D', of (Fig. 24) respectively corresponding to A, B, C, and D, and if the second figure is a parallel projection of the first, A'B' is equal and parallel to CD'. This is the geometric definition,^nd I now proceed to show that ^ / , / if the algebraic definition, — =:«,— ^ o, — = c, is fulfilled, then the geometric is also. I shall limit myself to a plane, and it will be easy for the student to extend it to space. Suppose AB and CD to be a pair of equal and parallel lines in the first figure and that A', B', C, D' are the points corresponding to A, B, C and D respectively, and suppose the algebraic condition, ^ ^= a, y ^= ? to be fulfilled ^Coordinates being as follows : — of A (a;, y), of B (a:', /), of C (x'\ y"). of D {x'", /"), of A' (X, Y) &c.^ , then the tri- angles AEB and JIFD can easily be proved equal to each other; hence AE = CF, but AE =x' — x, and CF = x'" — x" ; again from the conditions we have X x' x" _ v^" _ X ~ X' ~ X" — X'" " " £C fly OC X .^ , . trt If •'• Y^ X "^ « ^ -srw x ^» ^""^ ^viic& X — a; = a; — x therefore X' — X = X'" — X", or A'E' = CF', so also B'E' = D'F', and hence the triangles A'E'B' and CF'D' are equal, and .-. 6 34 PARALLEL PROJECTIONS. A'B' is equal and parallel to CD', which proves the geometric defi- nition which may be otherwise stated thus : — The parallel prelection of- two equal and parallel straight lines is a pair of parallel and equal straight lines. This last definition renders the demonstration of the Proposi- tions enunciated by Eankine very easy. I proceed to give the demonstrations. I. If two equal and parallel straight lines have their parallel projections equal and parallel, it follows that equal portions of the same straight line project into equal portions, and hence that; the parallel projections of two given portions of a straight line are in the same ratio as the portions themselves. A particular ease of parallel projections is the case of project- ing a figure on any plane by parallel projecting lines, as in F;g. 25. CD is a parallel projection of AB, and in this case one-half of CD orone-third of CD would be the parallel projection of one-half of AB or one-third of AB, &c. Hence in this case also the proposition is clear. ' II. The same reasoning will apply to this : — If equal portions, of parallel lines project into equal parallel lines, the parallel projection of a system of parallel lines must be a system of parallel lines whose lengths bear to each other the same ratio as the lines themselves. Thus, if . AB and CD are equal and parallel it is easy to prove that A'B' and CD' are also equal and parallel ; or if AB and CD are parallel, but not equal, that A'B' and Ciy are also parallel and in the same ratio as AB : CD. III. Find the parallel projection 'of each of its vertices, and join them, and we have a closed polygon. IV. The parallel projections of two opposite sides will be equal and parallel, and hence the resulting parallel projection will be a parallelogram. V. The parallel projection of each«of its faces will be a paral. lelogram, and hence the resulting parallel projection will be a par- allelopiped. PARALLEL PROJECTIONS. 35 VI. Divide up the plane surfaces into an indefinite number o£ small rectangles, all equal to each other ; the parallel projections of these rectangles will be parallelograms equal to each other, and the number of rectangles in either of the first surfaces is equal to the number of parallelograms in its parallel projection. VII. Divide the surfaces into small rectangular parallelopipeds and the same reasoning holds. Demonstration of 63 and €4 of Rankine, Suppose we have a pair of forces (Fig. 26), P and Q, applied at the points A and B of a body, and that the lines P' and 22.627 " 8i ^ «b =9 Mb - 24.781 - 27.000 Ma + % 2^t*i (plain) 4 2't#i (dotted) = 35.000 = 133.878 •= 337.324 J. 6)506.202 r' a;*rfa! = 84.367 Ma = 2 tii =2.121 2.236 2.345 2.449 2.549 2.645 2.738 2.8-28 2.915 Mb = 3.000 B^ -f M^ = 5.000 2 2' Ml (plain) = 20.316 42:mi (dotted) = 50.672 6)75.988 x^da = 12.664 f. a; = f i:M J = 6.661 + Now compute the same by the first method of approximation and we have u, 4- Mb •/ L 17.5 i: ui= 151.27 '^x^dx = i (17.5 + 151.27) = 84.385 and 4 12.663 •M^ -(- Mb since in this case s ^= 2.5 and ^ u = 22.826 •••« = f 1:111 = 6.664 + • Note on Art 77 of Rankine. First, to prove that G0G3 (Fig. 32) is parallel to GiG^; sup- pose both triangles removed, and that G is the centre of gravity of the figure ABDFE ; then annex FEC, and the centre of gravity moves in the line GGj, to a point G„ such that GG^: GiG. = Wj . Wj — Wj. Now, mstead of that, suppose the triangle HDF to be PAEALLEL PROJECTIONS. 43 annexed, and the centre of gravity moves instead to a point Gj in the line GGj such that GGj : GgGa = Wi : W„ — Wj. .-. GG„ ; GiG„ - GGs : Q,G, hence the line GjGs is parallel to GjG^, Secondly, from similar triangles we deduce G^Gg : G,G2 = GG, : GGi = Wi : W„ or W SOLUTiaNS OF RANKINE'S EXAMPLES ON CENTRE OF GRAVITY. I and II ane sufficiently clear in the text. m. Take OD for axis of x, and take the moments about AB (Fig. 33) ; the elementary area will be FK (J X sin DOH) nearly, hence we have. /OD FK {dx sin DOH) = ,0D FK (x sip DOH) (dx sin DOH) or dividing out by sin" DOH we have /op «0D FKdx = J FKa!tfa% Now FG : CG = AI : CI, or FG : OD — a; = J (B -:- 5) : OD . _ (OD - X) (B - b) '• ^^ — . 2 0D /: /— OD ^^OD FK dae / FKdx i(B + b)OD ~ B + b - 2 V^ * B + ^./ 44 DlSTEIBtTTED FORCES. The area we know from Geometry is ^ (B + b) DH = i (B + 5) OD sin ODE IV. Areas of similar triangles are to each other as the squares of their homologous sides .*. Wi : Wj = Xi^ : x} as, Wi f a;, — W, f a!, „ W."^ w, — Wj - 5 — w l' 1 _ xf— xj * X^ — £b/ V. Take moments ahout OY, then sin XOY divides out, and we have I xdydx I j ydydx .J. ^^ _t«/_0 , y _. J 8 J 0. r r (fycfe r r " dydx t/Ot/0 J J Let the equation of the parabola be y^ := 4aa;, and the formulae become / xydx f ^ 2a^x^dx f^x'^dx „ 4 r 2/j£B f ^ 2a^x^dx r^^dx fa;^ I y'^cfe 4a f \dx ^-^ 2 1 ydx ia^ I x^dx fajj^ Area = VI. J f ' (?2/«fe sin XOY = P j/cfo; sin XOY = %Xjyi sin XOY .-. W = ^wx^y^ sin XOY. j j xdydx C ' x(yi — y)dx ■"0 /»yi /ixi rtxi ■ — 2/1 r a!&j — r ' xydx 2/iy — f a'lVi Jo Jo CENTEE OF GRAVITY. 45 I I xydx is substituted from its value above.) I ydydx / (yi= — ^^)dx ,,, — ^J » : _ J 0. — / / dydx 2r {y.^~y)dx J J J Jo 2y, f''' yo ^^ ^, Since the centre of gravity is on the axis of symmetry. The area we know from Geometry is J arc X radius = ^(2r d) VIII. ._ fxydx _ !\,^,,-^^-'—')^ _ {-W^—')% f ^(r^ — x')dx Area of segment. J r cos t "'" fydx cos 9 _ ^r° sin' g _ ^r» sin° d 'sin'g sector — triangle r^ — r^ sin cos ^ — sin cos ~T~ 2 2^0 2 fydx 2 -area ,=(,.3in ^cos«) " 2 8 8 fl_l_ '•° cos' g = ^'' ^ ''"^ t^ -i- 3 _ 2(1 -cos e) -cos ^(1 -cos" ff) r%d — sin d cos d) **" (? — sin d cos ^ , 44in^i d — sin'' d cos ^ d — sin ^ «os 46 CENTRE OF GEAVITT. . . , r^ t^ sin cos Area = sector — triangle = -h- — s "• .-. W = ywr%e — sitt d cos e), IX. Area of spandril = rectangle + triaTigle — sector = {r sin 0) {r — r cos 0) + ^r^ sin ^ cos 9 — ^^ = „ 2 sin 5 — sin cos — d "•" ^ 2 /f sin 9 /»r r a;^rsine-V(r''-9;2)| 2 sin S — sin cos d — 6 ir sin 8 /Tsra e nr ^0 Sfdydx area 2 • area r\2 sin fl — sin S cos — 0) 3 sin" (? _ 2 — 3 sin" S cos (? + 3 cos (9 — cos' d = ¥ 2 sin <9 — sin 6 cos ^ — 6 3 sin" (? — 2 sin" (? cos ^ 4 sin" 5 2 sin S — sin cos — X. The centfe of gravity is on tlie middle line. . „ _ A ^ _ ^1«^ - W^2 _ WS"^ — % , Wi w-1 SLENDEK SODS. 47 „ sin ^ sin e r* — r,' . sin «=! = ir ^^. X, = in -T .: X, = f ^^^:^, -^■ Area = {r^ — r^^) „:W = w{f — n^) d. XI. When the elliptic arc is given we know the corresponding circular arc, for the coordinates of its extremity, being x and y, the X of the circle and ellipse are the same, and the y of the. circle = -T-y ot the ellipse, let x', y', he the coordinates of the centre of gravity of the circulair arc, etc. ; then x = x', and y = — y'. XIII. Since the triangular fece is parallel to the x axis, y„ = k. Volume, we know from Geometry, = x^y^Zi .•. W = wxiy-^, J «/ XIV. Find the value of y in the terms of x, thus, Xx—X / x\ y ■■ Vi-- «! — a; : «! .-.y^y^-^-- ■= yj|l__^ „/ a; \ , '( Xm, X* 'i ^1 Volume = J, base X alt. == ^ wzjajj^j. XV. Equation of circle is a:'' -f y'' =/■* .-. y = y (^2 — a;^^.. f cc^ y (ji^'-a:^) dx I - ^ ^''°~'^'> I +~ J^' V (r=^2) ^^ r X y/(r^^af) dx 1 1 (r^ _ a;^) t l ' .a;o XVI. Wi = f wr V, Wj =: f w V; «! = j^ r, iB^ = Jg / is CENTRE OF GRAVITY OF A PLATE. 37r r'» ~ *■' 3^ r* _ /4 ~ 16 *•" 1 7-8-1 "" 16 rs — /' XVII. Area of a section at a distance, x, from the vertex is x^ <" x^ . A -p .•. Eleni,entary volume =^ A. -r^dx sin d A />h . _ _ IlJjl. = — W = i wAA sin A'' Jo XVin. Volumes, of similar cones are to each other as n : n .'.Xq -^ -^ Yj — f h"^ ,. / ^ n-\ . .5 XIX. Conceive the shell divided into a number of shells of small thickness, and let the variable radius be p, then area of zone is 2-p • p(cos p — CQS a), and the distance of the centre of gravity from the centre of the sphere is W i wAA sin e — ^ M A Ta" A' sin e = ^ wAA ( 1 — rr ) ®'° ^ cos p — « + COS ^ _ _ ''Jl ^f'& ^X A<^o'^'^-^o^''')dp I Ydp. 2-1 />^cos ^ — COS a)dp ■ 4 (cos'' |3 — cos'' a) ^ y4 _r'\ cos « + cos ^ . .2 2:rw, s — I (cos ^ — cos a) 2-/<^(cos* ^ — cos\ a)rf^ = ~3^('^ ~ *"' ') (cos (5 ~ cos o) XX. Imagine the shell divided into thin plates by planes perpendicular to OX. We shall thus have a set of additive sec- tors in the outer sphere, and a set of subtractive in the inner. PRACTICAL PROBLEMS.. 49 Area of sectors are 0{r^ — a;*) and 0{r' ^ — x^ respectively, _ and distances of centre of gravity from OX are I V(r^ — x^) '-^ and § V(/ ^ — x^~ f x{r^ — x^)dx — d f x(r' "— 3?)dx ,., ^ — _.L« ./_o • 6 r (r" — x'')dx — of (r' 2 _ x^)dx _ \d{7^ — r'^) __ t^ — r'* — ^olr' — r'^) ~ t^8_^/8* yo = f sin 6 3* r (r^ — x^)^dx—\f' (r'2 — x^)hx ^-(^ — r'^) 3^ r* — r'* sin (9 ^''''^ie(^ _ /3) - 16 ^_/3 e- SOME PRACTICAL PROBLEMS. In the preceding pages have been considered, _y?r«<, the composi- tion and resolution of forces, in whatever way applied to a body ; second, the conditions of equilibrium ; and third, the determination of the centre of gravity of any body whatever. We are thus en- abled to solve an immense number of practical problems relating to tlie stability of structures and machines, and the strength and dimensions of their different parts.. The determination of the suitable dimensions and forms to resist any known state of stress in any one of the parts, will be treated of hereafter, but it is the purpose of this chapter to compute the suitable distribution of the parts and stresses in each of them, re- sulting from any given condition of loading. There are two methods of proceeding in such practical prob- lems, first, by algebraic and numerical calculation, and second, by geometrical construction. Of these two the engineer finds now 50 PRACTICAL PROBLEMS. one and now the other the most convenient. The following prob- lem will exemplify them both : To determine the stresses in the rafters and in the tie beam, also the respective weights sustained by the struts at A and B in the frame ABC (Fig. 34), the weight W being applied at C, and the weight of the' frame itself being disregarded. Solution. Assume the origin of coordinates at C, and the axis of X horizontal; let * represent the stress (of compression) in the rafter CB, and f that in CA, and H the tension in the horizontal tie beam, AB. Now the conditions of equilibrium require tliat ^ P cos a = 0, j; P sin a = and ^^ PjB = The last is not necessary in this case, as the point of application of the force has been taken for origin ; the others give 2 P cos a^ t cos i — t' cos i' ^ -S' P sin a = t sin i + i' sin i' — W = Whence we obtain W (!0S i , W cos t^ t = _■■_ /■■ , ,,, i =^ sin (i + i') sin («' + i') These are the compressions in the rafters. mi .1 L w ■ ., W cos i sin i' . Ine weight at A ^ f sm i = — -. — t-t— j — ^k- ° sm (i + I) W cos i' sin i sin (i-\- i') W cos i cos i' « 11 B = < sin i =^ H = < cos % ^i cos i' sm (i + %) The geometrical solution is given in Fig. 35, where we draw the vertical line, E'D', to any scale to represent the weight "W. We now wish to determine the components of this weight iii the direc- tion of the rafters, or, which is the same thing, the stresses in the rafters. To do this we draw through E and D the lines, EG and DG, respectively parallel to AB and QB, then we know by the principle of the Triangle of forces that EG and GD will repre- sent these required stresses (to the same scale as that to which ED was drawn). We next decompose EG into a horizontal and a ver- tical component, and we have GF for the^stress in the tie bar, AB, PRACTICAL PROBLEMS. 51 and EF for the portion of the weight acting at A ; so likewise FD is the portion acting at B, and by measuring these lines we have the magnitudes of the respective stresses, t — GD, / — EG, H= GF. Wt. at' A — EF ; Wt. at B = FD. If now the algebraic values of these quantities be determined from Fig. 35 we shall obtain the same values as before. Thus the triangle GED gives EG = ED "° ^^^ or « = W '^°'' *' sin EGD "''■—" sin(j + j'') , _,„ W cos i* so « = GD = -; — , . I .,. , etc. sin (%-^ty EXAMPLKS. 1. Given W = 872 lbs., i = 35°, i' = 63°, to find the stresses in all the parts. Solution. _ 872cos63° _^ '~ sin 98° ~ ^^^-^^ . 862 cos 35° = 721.31 sin 98° ^ . 872 cos 35° sin 63° Wt. at A = ^-Qoo = 327.47 sm 9» „ ^ 872 cos 63° sin 35° Wt. at B = ^.^ Qgo = 642.70 „ 872 cos 35° cos 63° H = inrW = 230:35 The student should now make the geometrical construction' and thus compare the results obtained by each method. 2. Given W = 953'lbs., i' = 43°, i == 47° 3. Given W = 545 lbs., i' = 30°, i = 60° 4. Given W = 763 lbs., i' = 27°, » = 59° Perform in each case the algebraic and also the geometrical so- lution, and compare the results. We will now take the case which more commonly occurs in practice, viz., that of a roof uniformly loaded. 52 PAEALI.EL PROJECTIONS. Suppose (Fig. 36) AC and EC to represent a pair of rafters of a roof connected by a tie bar, AB, and we wish to determine the stresses in each. of the parts. In this case we must observe that one-half the weight is borne by each rafter, and hence (W being the totq,! weight) there is one force, JW, acting at the middle of each rafter. Now of this ^W is directly supported at A, and ^W at B, while the remainder, iW + iW = ^W, is the weight at C. In the case before us also i = i', hence the results of the pre- ceding case become ^W cos i _ W sin 2 ^ 4 sin i ^W cos° i W sin 2 I 4 tan i W The weights sustained at A and B respectively are -5-. The ge- ometrical solution is given in Fig. 37. In this case EGD becomes an isosceles triangle. EXAMPLES. 1. Given W = 782 lbs., * = 42° find t, if, and H. 782 .1^ olvtion. * = 4 sin 42° — ^^^"^ 782 _ ^^„,„ ^ - 4 tan 42° " ^^^'^^ 2. Givfen W = 1054 lbs. i = i' = 75° 3. Given W = 1272 lbs. i = i' = 83° 4. Given W = 725 lbs. i = i' = 45° 5. Given W = 324 lbs. i = i' = 30° 6. Given W == 572 lbs. i == i' = 60° Give both the algebraic and the geometrical solution in each case. As another example let us take a Warren Girder (Fig. 38) sup- ported at the points A and B, and loaded uniformly throughout PRACTICAL PROBLEMS. 53 its whole length with a load eqi^al to g lbs. per foot, long measure (including the weight of the girder itself). We wish to determine the tensions in the upper, the lower beam, and in the diagonal braces. The total load of the girder will be gl {I being its length), gl hence the supporting forces at A and B are each equal to -5-, and act upward. Suppose now we wish to find the stress Ri along the upper bar at D4 that (R3) along the lower bar at C, and the stress (Rg) in the diagonal brace DF. We must observe that the forces tending, to turn the cross section, DC, are first the supporting force, ^, acting at' A, and . ffZ glx whose moment with respect to D is „ • AC = -3- ; secondly the force gx uniformly distributed over the portion GD, and which may be assumed to act at E, the middle of GD; hence the mo- jnent of this about D is s~, and hence the total bending mo- glx gx^ ment is -„- — -s" or -f (Z-cc) Now, decomposing the tfensions parallel and perpendicular to GD, and imposing the conditioris of equilibrium, we have . g>- Ri + R3 + Rg cos «■ = and R3 sin i = s" — • gx ' also takiflg moments about D K2« — -J — "2 qx • (^ \ Ri = — Ra — E3 cos «■ = — ^{l — x) — g coil \2~^) These give us the tensions at a joint situated at a distance x 54 ' STRESS. from A. The tensions at the middle would be found by making a: = ;t, and we have ^ — 8A' ' ~ ' 1 8A EXAMPLES. 1. Given g = 700 lbs., I = 60 ft., h = 5 ft., i = 46°, find Ei, Rj and Eg at the middle of the girder, and also at the top of the fourth brace. Solution. 700 X (60)^ At the middle B^ = —R, = — g ^^^ ' = 63000 lbs. Rs = At the top of the fourth brace ?; = 20 ft. 700X20X40 ,,„„„„ .-. E^ = 2^^-g == 56000 lbs. R, = '^? '^,1^ = 7000 V 2 = 9899 lbs. " sm 45 2. Given g = 600 lbs. I ='50 ft. A = 5 ft. * = 45° 3. Given ^ = 930 lbs. Z = 70 ft. A = 7 ft. i = 45° 4. Given ff = 900 lbs. ^ = 80 ft. A = 8 ft. « = 60° 5. Given ^ = 850 lbs. Z = 75 ft. k = 7 ft. » = 70° Find in each case the values of Ri, Rg and R3 at the middle, and also at the top of the fourth brace from the middle. STRESS. Under the head of stress we have to consider forces distributed over surfaces, and not acting at a single point. Thus if we imagine an iron rod suspended from the ceiling with a weight of 1000 pounds attached to it, that weight exerts a stress on the particles nearest to it, those on their neighbors, and so on : and thus the whole rod is in a state of strain. The intensity of a stress is its amount per unit of area. Thus suppose the above rod to be divided into two parts by an ideal plane perpendicular to its axis, and that the area of this nor- STRESS. 55 mal cross section is 12 inches; then the whole stress tending to sepai-ate flie lower from the upper part is 1000 pounds, and hence the intensity of the stress is if |- = 83J pounds per square inch. EXAMPLES. 1. What is the intensity of the stress per square inch on a nor- mal section of a rectangular, rod (sides 10 and 6 inches), to which a weight of 1563 pounds is attached ? 2. "What is the intensity of the stress on a normal section of the rectangular bar of a plough (sides of section 1 .5 and 3 inches) when a dynamometer placed between the horse and the plough in- dicates a pull of 3000 pounds ? 3. What is the total amount of the stress in a rod whose nor- mal cross section is a circle of radius 2 inches, when the intensity of the stress per square inch is 5.7 pounds ? If, on the other hand, the intensity of the stress is variable (dif- ferent at different points of the surface) we must divide the area over which it acts into small rectangles whose sides are A x and J y, parallel to Ox and Oy respectively, and if the intensity of the stress at the point {x, y) be represented by p (p being variable), the amount of stress on this rectangle will he p A x A y nearly; and hence the total amount of stress on the whole area will be P =: // pdxdy ; also the area is S == // dxdy, hence the mean P ffpdxdy intensity IS ^„ = ^= yyS^' UNIFOEM AND CNIFOKMLY VARYING STRESS. The two kinds of stress that we meet with in practice are the Uniform and the Uniformly varying. By a uniform stress we mean one whose intensity is the same at all points of the surface on which it acts. This is the case if we have a rod suspended from the ceiling with a weight attached to it; if we imagine the rod cut through by a plane perpendicular to its axis, the intensity of the stress at all points of this section is the 56 DNirOKMLY VARYING STEESS. same. This iis also the case in tie bars and braces; columns and struts which support a weight, whose length does not su^pass cer- tain limits. In a tic bar the stress is one of extension, and in a strut, of compression. To illustrate Uniformly Varying Stress suppose we have (Fig. 39) a beam firmly imbedded in a wall at one extremity, and loaded at the other with a weight, W ; the effect of this weight is evi- dently to bend the beam, and in this bending the upper fibres are extended and the lower compressed ; , there must be, then, a set of fibres or a lamina somewhere between the two that is neither ex- tended nor compressed. This lamina is called the Neutral Lam- ina. Suppose the beam to be cut by a pair of parallel planes at right angles to its length 'before bending ; it is evident that these planes will diverge after the bending. Let the planes be DE and HG. These include a set of fibres of the same length, but when the weight is applied and the beam is thus bent the cross section HG takes the position LM, such that all the fibres above FK are elon- gated, while those below are compressed (FK lieing a portion of the longitudinal section of the neutral lamina OF). Moreover the amount of tension on any fibre (either of exten- sion or compression) is proportional to its distance from the neu- tral lamina, as can be easily shown. Here, then, the stress on the cross section, HG, tending to sepa- rate the part of the beam beyond this cross section from that be- hind it, varies uniformly from the line in which this cross section cuts the neuti'al lamina or the line. represented in section by K, and this is called the Neutral axis of the cross section HG. When the beam is not bent beyond the limits of elasticity the stress of extension and ■■that of compression are equal, which shows that this neutral axis must pass through the centre of gravity of the cross section. Thus, if the cross section of the beam is a rectangle, AB, CD (Fig. 40), the neutral axis will be EF, if (Fig. 41) it is a circle, the neutral axis will be AB passing through its centre, etc. , GEOMETRICAL REPEESENTATION OF STRESS. 57 As a second exampje let us take the case of the vertical wall of a reservoir. It is an established .fact that the pressure of any. body of water at any point is proportional to the depth of the point below the free upper level. Thus, suppose AB (Fig. 42) to- be a section of such a wall, the pressure at the top (supposing the free upper level to be even with the top of the wall) is nothing ; suppose now that at the bottom to be represented by the line CB ; then that at any point D will, be represented by ED such that ED : CB = AD : AB. Here, then, we have again a case of uniformly varyijig stress varying uniformly from the free upper level of .the water. So likewise in a revetement wall which sustains a bank of earth wc have again a case of uniformly varying stress. This will give the student an idea of some of the cases in which he has to make use of the calculations of both uniform and uniformly varying stress. GEOMETRICAL REPRESENTATION OF STRESS. "VVe have already seen that the simplest way to represent a force acting at a certain point of a body is by means of a straight line drawn from the point of application of the force in the direction of the force, and containing as many units of length as there are units of force in the given force. Now stress is a set of parallel forces distributed over the entire surface on which it acts. These forces, it is true, are balanced by another equal and opposite set of forces, which are brought into' action by the cohesion of the body, but nevertheless since they are forces a straight line suggests itself to our mind as thi? proper sym- bol to represent each one of them. .These lines must evidently be proportional to the amount of stress acting on the unit area, or proportional to the intensity of the stress ; thus at a point where the intensity is twice as great our line must be twice as long. Now since the stress is distributed over the whole surface on \vlMch it acts we shall have an infinite number of such lines, or a volume of a prismatic shape, whose lower base is the surface on- 58 GEOMETRICAL REPRESENTATION OF STRESS. ■which the stress acts, and whose upper base is formed by the ends of all these lines representing the stress at each point. In the case of Uniform Stress the upper base becomes parallel and equal to the lower, as in Fig. 43 ;' if the intensity of the stress at any point D of the circle ABC be represented by UK, the stress at any other point (supposing it to be uniform) will be represented by a line parallel and equal to DE, and we have ilins a cylinder which is a suitable representative of the whole stress on ABC. If the stress be not uniform the upper surface will not be i)aral- lel to the lower, but in the case of uniformly, varying stivss it will be a plane inclined to the plane of the surface on which the stress acts, as in Fig. 44, where the truncated cylinder, ABC DliF, represents a stress acting on the plane of the circle ABC, and va- rying uniformly from the line 00'. Suppose that on the surface, ABCD (Fig. 45), we have a stress varying uniformly from the line GH ; we draw FC to represent its amount at C, and then the stress will be represented by the vol- ume ABCD — EF. Now if we draw through O, the centre of gravity of ABCD, a line perpendicular to the plane, and through P, where it meets the upper- surface, pass a plane parallel to ABCD, we shall have a cylinder- ABCD — IK equal in volume to the former figure for the wedge EIP = KFP. Now this cylinder will represent a uniform stress whose resultant acts of couite through ; and the two wedges in question represent a stress uniformly varying from the line LM, whose resultant is a couple acting in a plane perpendicular to LM. Combining" these two resultants Would merely have the effe'ct of shifting the former resultant to a point situated in the line AC M perpendicular to GH at a distance L = -p- from O. I have taken the stress normal to the surface to illustrate, but if it were oblique the line AC would not necessarily be perj;endic- ular to GH. The line DB parallel to GH is the neutral a.\is, and AC, which passes through O, and the point of application of the resultant stress, is the conjugate axis. CENTRE OF STRESS. 59 CENTRE OF STRESS. "We piast now proceed to find the coordinates of the point of application of the resultant stress, or in other words 'the. centre of stress (the whole is supposed referred to a pair of rectangular axes). Let p denote the intensity of the stress at the point whose coordinates are x and y. Divide up the surface into small rectan- gles, whosfe sides are Ax and /Jy respectively, ai5d we have for the force acting on one of these rectangles pAxJy nearly, and for its moment with reference to the y axis pxJxJy nearly ; hence for the whole surface we have the entire moment in reference to Oy equal to x^ffpdxdy = ffpxdxily, and in reference to Ox y.^ffpdxdy ^= ffpydxdy, {x^,- y^) being the coordinates of the centre of stress ; hence we have _ ffpxdxdy ^ _ ffpydxdy ^ """ - ffpd^dy y» ■ ffpdxdy which must be integrated between the proper limits. The centre of stress is the same as the projection on the plane in question of the centre of gravity of the ideal solid mentioned in Art. 83 of Rankine. EXAMPLES. 1. Where is the centre of stress for a semicircle, the stress varying uniformly from the diameter ? Solution. Here the stress varies as its distance from y, or as the abscissa of the point, hfence p = ax, and the equation of the circle is / + a;* = r\ or y, = ± V(>-« — x^). Xq 4./ 4 2 __ ,_^ 60 TINIirOIlMLT VAETING STRESS. I I axydyax I 1 axdydx J -^Ir' - x')J - 2. Find the centre of stress for a rebtangle, the stress varying uniformly from one side, that is ^ = ax. Solution. ' Xq r r pxdydx j f ax^dydx ah I xHx - I / pdydx I I axdydx ah I xdx _ fj\axydydx _ jj[^ _ i_ j / axdydx ah I fa;c?a; SupposejB ^ m + wa;^ is the law of the stress I I (ma; + nx^)dydx h j (mx + nx^)dx II (m -\- ni^dydx hi (m -\- n£)dx mc^ . no mc + n^^ + T ^#AiA/^ (b«4^«c= c, o.„ _L. ■ „ i and c*'^ SK|^¥74fc[ (2m +i^mcV /b rtC I (m, ■\- nx^)ydydx , 1. / (»* + nx^)dydx MOMENT OF UNIFORMLY VARYING STRESS. We have seen (Fig. 45) that a uniformly varying stress can al- ways be reduced to two, viz., a uniform stress whose intensity is the mean intensity of the entire stress, and 2d, a uniformly varying stress, varying uniformly from a line passing through the centre of gravity of the surface on which the stress acts. This line is called the Neutral Axis of the body for that couditiou of stress. This tJNIFOEMLY VAEYING STRESS. 61 second portion of the stress is. equivalent to a couple, and its tendency is to make the surface on which it acts rotate around the neutral axis ; it is therefore necessary to know the moment of this, couple, so that in structures and machines we may pi-ovide sufficient •strength to resist it. Now if we call the intensity of the stress p, this means that at any point (x, y) the amount of stress (say the number of pounds) on one unit of surface would be p. J3ut we cannot take a unit of surface at that point, for part of it would" be at other points, hence take a small rectangular area Axdy, and we shall be always nearer the truth. Now if we took \ a, unit of surface, the whole amount of stress would he, \p ; so the whole amount of stress on Ax^y is pAxAy, and \i p ■= ax it is /IP = axdx^y. Now this is the force acting on this small rectangle, and we must proceed to get its mo- ment around each of the axes OX, OY and OZ (OX and OY lie in the surface). Now if these .parallel forces which make up the stress are inclined to OX, OY and OZ, respectively, at angles «, |S and J', the moments will be as follows (since z ■:=^ o) (Rankine, Art. 91) : round OX y/P y cos y = axy cos yJxzty. round OY —; A Px cos y = — ax^ cos y Ax Ay. round OZ AV{x cos ^ — y cos «) = (ax" cos ^ ~axy cos (i)AxAy. Now then the total moments round each of these axes will be found by integration, or we have as follows Ml = a cos yffxycMy . = a cos yK. Mj = — a cos yffx^dxdy = — a cos yl. Ms ^= a(cos ^ff3i?dxdy - cos affxydxdy) = «(! cos ^ - K cos a). And we then proceed to compound these couples as Rankiue does in the Text. We thus know the force which has to be re- sisted by the strength of the material, or in a beam in equilibrium we know the resistance it opposes to any definite strain. For the moment of bending and twisting stress, see Text. These formulae all depend on the quantities I and K. The 1st has received the special name of .Moment of Inertia,' for.,reasons to be given hereafter. 62 MOMENT OF INERTIA. The coordinates of the point 6f application of the resultant of the stress are, as we have already found, and as given in Rankine, ffxpdxdn afjx^dxdy ol , ^» = JSfdMy = a/fxdx.4 = P' P^^'P^"- '» "«"'• ^^'^- • ffypdxdy affxdxdy «iK 3'» = ffpdxdy = ^TrjM, = IP ' ^'°°g ^^""''"^ ^^''■ MOMENT OF INERTIA. A TMe quantity'I = ffx^dxdy,' called the moment of inertia of a pla^e surface about the axis of Y, is the sum of the products of t^e elementary areas bv^their distances from the fixed axis. The quantity J = ffypdxdy is the moment of inertia of a body about the axis of x. Thus to get the moment of inertia of the planfe area we divide it into small rectangles by lines drawn' parallel to the axis of x and y, and now if we wish the moment of inertia relatively to Y, we multiply each of these rectangles ^x/Jy by the square of its (iis- tance from Y, and then integrate, and thus obtain ffx^dxdy ; if we wish that about x we have ffypdxdy; if now we wish that about Z we multiply Ax^y by the square of its distance from Z, or by OP* = »•' = x^ + yS and we thus have //(a:" + y')dxdy, but this = // xMxdy + ffypdxdy = I + J. This is what Ran- kine calls the polar Moment of Inertia. EXAMPLES. 1. Find the moment of inertia of a circle about the diameter Y, — Y. Solution. M = r^""'"'' r x'dydx=2f '^if — 3?)^dx = 2 J_Ja;(r' — x")^ + J f V(r« — x^)dx\ = 2 | "^-^^ I = T ~ 64' UNIFORM STRESS. C3 Observe that the last integration need not be carried out, be- cause r ydx or C '^{f — x^)dx is the area of a semicircle, and this we know is equal to -^ " 2. Find the moment of inertia of a rectangle about the axis ' Y, — Y passing through its centre of gravity, and parallel to one side. SoliUion. 3. Find the moment of inertia of the figure (4C i). Solution. ~S 5 Ur - '")- This might also be found by subtracting the moment of inertia of AH' hh' the rectangle JH from that of hh ; thus M = -j-^- — iT ^^ NoTK. liend now Rankine, page 81, from "Jn finding the mompnt of inertia of a surface of complex figure." UNIFORM STRESS. In a rod or bar subject to a stress, either of extension or com- pression, by a load acting in the direction of its length, we have already seen how to estimate the intensity of the stress. The amount of stress which a rod of one square inch cross sec- tion will bear with safety is first determined by experiment. Thig quantity, which we will call K, varies for diiferent materials. For wood it is about 1,000 lbs., for wrought iron 10,000, and for east iron 3,000, when the stress is one of extension. G t BEAMS AND GIRDERS. If then the cross section of any rod be represented by a, the weight it will bear will-be W =^K, for it, is just as though we had a bundle of rods, each of one square inch cross section, and a in number., EXAMPLES. 1. How much weight will a wrought iron tie bar, the , area of whose section is \ square inch, bear with safety? Ans. W = aK = I ■ 10,000 = 5,000 lbs. 2. A wrought iron tie barof circular cross section is to resist a stress of 15,000 lbs, in the direction of its length ; what must be the radius of the cross section ? 3. What must be the* area of the cross section of the tie bar in each of the examples on triangular frames given in Figs. 34, 36 and 38 : 1st, when made of wood, 2d, when wrought iron. APPLICATION OF UNIFORMLY VAP.YING STRESS TO BEAMS AND GIRDERS. A beam is imbedded at one end, and loaded at thej other (Fig. 39) ; it is required to find what load W it will bear vw'th safety. Sohitioii. The force tending to break the beam is the load W, and if the rupture were to occur at the cross section liG, the mo- ment of the breaking couple would be Wa; (x being the perpendic- ular from K on the direction of W). The greatest value of x is evidently Z ( = CP), hence the greatest moment of the coupi.e tending to produce rupture is W/. This is resisted by the stress in the beam, which, as we have already seen, is equivalent to a couple whose moment is al (Riin- kine, Art. 92), and hence we must have W^ = di. to provide against rupture. To determine the value of a, let S be the tension at the iibre situated farthest from the neutral line, and d the distance of this fibre from the neutral line, then • S - S I S = ad, a\-a= -j .: Wl ^ -rL .-. W =: S:7j- a a dl BEAMS AMD &IRDERS. 65 The moment of inertia I, and the quantity d, depend only on the form of the cross section, and this quotient -v is called the moment of resistance of any cross section. Thejvalue of S is dif- ferent for different materials, and has to be ascertained by experi- ment, and it is called the modulus of rupture. It is not, however, safe to load a beam to its utmost, and hence in practice we sub- stitute for S only a portion of its value, which is determined by practice, and this portion is called the coeflScient of safety (see Eankine, Art. 247). If '^ denote this rnnffinlnnt by Kj^fre^hall have the formula "^"^ Wl = Ki#or W = ^l- EXAMPLE. 1. Suppose the cross section of the beam to be a rectangle, where b = '6 inches, A = 6 inches, I = S feet, we shall have / _ K 1 _ K_ is&A' _ KW ^ - I ' d - I ' I I -*■ I ' In this case, W = ^ K %^'^ = \K. If the material is oak, for which the modulus of rupture is 10,000, and for which we take 10-fold security, we have W =:= 500 lbs. If now, instead of having a load at*the end, wo have a uniformly distributed load, we may imagine it acting at the middle of the beam, and then the moment of the breaking couple becomes W^, and we have 11 K I %= K^,orW = 2y.^; or, in words, the beam will bear twice as large a load when it is uniformly distributed over its length, as when applied at the end. Thus in the above case th6 beam would bear ^ = t 35 ' ^ ~ 1,000 lbs. uniformly, distributed over its entire length. ■ 66 BEAMS AND GIRDEBS. Observe that W should include both the weight of the beam and the load. 2. Suppose we have a beam supported at both ends (Fig. 47). and loaded in the middle with a weight W, the load supported at each of the supports is ^W. Rupture will evidently occur at the point where the load is applied, or the middle point, and hence we may regard each half of the beam as a beam firmly imbedded at one extremity and loaded at the other, so that the moment of rup- ture will be in this case, W I WZ Wl ^ KI ,„ 4K] T- 2 = X ••• X = «i = -rf' "'•^ = -Id- If the load is uniformly distributed over the beam we may re- gard ^W in each of the halves as acting at the middle, hence the moment of rupture is ^V_ l_-Wl „ _ 8KI 2 ■ 4 ^ 8 •■■W — ^^^ Thus a beam supported at both ends and loaded uniformly will bear 8 times as much as the same beam imbedded at one end, and loaded at the other. The calculation of the load that can be sup- ported by a beam (Fig. 48) firmly imbedded at both ends involves the discussion of the elastic curve, as the beam is bent more than once, and there are thus three points of ruptuie. 1 do not propose to give now the discussion of this curve, but will give merely the results. For a beam thus imbedded and loaded at the middle we have ^ SKI When the load is uniformly distributed 12KI W ^ -Td- The student can now refer to the table of cross sections among the plates with the corresponding values of I and d calculate^ ; The values of K for different materials are determined experiment- ally, and by consulting Table IV in connection with Art. 247, Bankine. Thus for timber we take ^th of the modulus of rup- EXAMPLES. 67 ture for K, for wrought iron ^th. We will take, in the following exaranles, the following values, for wood K = 1040, for cast iron K = 1000, and for wrought iron K — 9000. EXAMPLES. 1. Given a wooden beam of rectangular cross section, sup- ported at both ends, where b =^ b inches, A = 12 inches, Z = 6 feet. What load will it bear with safety. 2. The following beams of rectangular cross section are of oak, and all supported at both ends. a. 5 = 3 in. b. b =^ 5 in. c. b= Tin. d. J =12 in. A = 8 in. h = 12 in. A = 13 in. A = 12 in. ; = 8 ft. Z = 10 ft. I = 6ft. I =A&. Find what load each one will bear with safety, both when uni- formly distributed over its entire length, and when applied at the middle of it« length. 3. The following cast Iron beams, whose cross sections are hol- low rectangles, are firmly imbedded at both ends. a. / = 10 ft. b.l = 5ft, c. I =U ft. B = 10in. B = 12in. B = 10 in. H=10in. H=14in. H = 15 in. 6 = 8 in. i = 8 in. 5=7 in. k = 8 in. A = 10 in. A = 12 in. 4. Determine the load each one can bear with safety, first when uniformly distributed throughout its whole extent, and secondly, when applied at the middle of its length. 5. The floor of a warehouse which is to sustain, its own weight included, a load of 100 lbs. per square foot, is to be constructed of oaken beams 8 inches broad and 1 2 inches deep, find the distance that can be allowed between the walls, supports, etc.; the beams being placed at a distance of 4 feet from middle to middle of beam. 8KI Solution. The formula' is Wl = —^ = |KiA». 68 STRESS. In this case each beam has to sustain a load of 4 X 100, or 400 lbs. per foot long measure, or ^"- lbs. per inch. ... W = ^l .: ^^-P = I X 1000 X 8 X (12)=' ... p = 4_>ijnoi)_x^8^x^|j|i<-U2 = 46080 sq. in. .-. Z= 214.66 in. = 17.88 ft. 5. Given in the same case 5=10 in., A = 15 in., and the loarl is one of 200 lbs. per square foot. 6. Find the safe load of a cast iron beam, whose cross section is given in Fig. 49, supported at both ends. Given H ^ 5^ in., A = 4.315 in., B = 1.76 in., b = 3 in., I = 6 ft. 7. Do the same for a beam of the section shown | in Fig. 50, Given AB = 2 in., AC = .4 in., DE = 5 in., EF = .5 in,, / = 4 ft. 6 in. 8. Find the safe load of a hollow cylindrical beam of cast iron firmly imbedded at both ends.' D = 5 in., d= i in., Z := 18 ft. 9. Find the diameter of a cylindrical iron rod that it may sus- tain a load of 200 lbs. per foot long measure. Explanation of the work of Eankine's, Art. 95. We ■ must first observe that ffdxdy and ffda^dy' both repre- sent the area in this case of the same surface, hence we can now put dxdy instead of dx'dy' under the // sign. x'^ ^x^ cos^ ^ — 2xy sin ^ cos ^ + 2/^ sin'' ^. ' y''^ = x^ sip^ ^ + 2a;y sin ^ cos ,J + y^ cos ^.^. x'y' = 03^ sin ^ cos § -\- xy (cos'^ IS — siii^ ^) — y^ sin ^,cos j3. Now substitute these values of x'% y' ^ and x'y' in the values (2) of I', J' and K', and change dx'dy' into dxdy, and we have I' = ffx^ cos^ f/ dxdy' — 2ffxy sin ^ cos /3 dxdy -\- ffy^ sin^ ^ doM^y, or I' = I cos^ j3 — 2K sin /S cos ^ + J sin^ j3, and likewise J' = I sin^ ^ + 2K sin ^ cos ^' + J cos= (3, and K' = (I — J) cos fJ sin ^ + K (cos^ |3 — siri^p). Now add the 1st two of (4) together, and we have 1' -)- J' = (I + J) (sin^ ;3 + cos^ ^) = I + J = //(of + y^)dydx. Again multiply them together, and we have STRESS. 69 I'J' = I^sin^ ^ oos^ ^ + IJ sin* /3 — 2IK cos ^ sin' (3 + IJ cos* ^ + 2IK cos» /3 sin (3 + J^ sin^ § cos^ |3 — 2JK cos= |3 sin ^ + 2JK cos (3 sin» (3 — 4K2 cos^ |3 sin^ ^. = (F + -P) sin^ § cos^ (3 + IJ (cos* ^ + sin* ^) + 2IK cos ^ sin ^ (cos*^ — sin""^) — 2JK cos ^ sin § (cos^jS — sin'p) — 4K2 cos^ p sin^ |3. Now square K' and we have K' 2 = (P — 2IJ + J2) sin= (3 cos^ (3 + 2 (IK — JK) cos § sin § (cos^ — sin'p) + K^ (cos* (3 — 2 cos" ^ sin" ^ + sin* ^). ' Now subtract, and we have I'J' _ K'" = IJ (cos* |3 + sin*|3) + 2IJ cos" ^ sin" ^ — 4K" cos" (3 sin" ^ — K" cos* § + 2K" cos" ^ sin" ^ — K" sin* ^ = IJ(cos* ^ + 2 cos" § sin" § + sin* |3) — K" (cos* j3 + 2 cos" (3 sin" j3 + sin* (3) = IJ — K", sincb the paren theses are each equal 1 . In this equation (1' — J')" = (I' + J')' — 4I'J', when I'J' is least the subtrahend in the second member is least, and hence the remainder is greatest, that is (I' — J')" or I' — J' is greatest when I'J' is least, and since I'J' — K'" does not change value. with the change of I',J' and K', let I'J' — K' " = a, now when K = 0, I'J' = a, when K is not then I'J' — K'" = a, or I'J' = a + K'". hence I'J' is least when K' = 0, and hence, when K' = 0, 1' — J' is greatest, and when I' — J' is greatest, I' has its greatest, and J' its least value. Note on Theorem II. "We have already seen that when K' is then I is a maximum and J' a minimum ; now the question arises whether it is always possible to find a pair of axes for which this shall be the case. In order to ascertain, we make K' = in equations 4 of Eankine and now we find the corresponding values of I' and J', and if un- der any posftions of the axes they become imaginary, then for those positions the above conditions can not be fulfilled ; but if, on the other hand, we find (as we shall) that they are always real, .and never imaginary for any position of the axes, then Theorem II fol- 70 STRESS. I lows. Now to test it, make K' = in 4, and from the third equation we obtain (I — J) cos ^ sin ^ + K (cos^ ^ — sin= (3) = 0, or (I — J) , sin 2S — 2K ^^sin2^+Kcps2^ = 0.-.^ = T^Tj' ov — 2K' tan 2^ = j-^^- Make K' ^ in 6 also, and 5 and fi become r + J' r= 1 + J, and I'J' = LI — IT- .-. 1'^ + 2rj^ + J'2= (I + J)2, and 4rj' = 4(IJ — K=) ; subtract and I'2 — 21'J' + J'2 = (I + jy — 4IJ +4K^:=(I — 3y + 4K^ .-. r — J' == V J (I — J)' + 4^"] , and .-. V - I+i. 4- vUi-Jr+4K^L 2 "^ 2 „ _ I + J y| (I — J)'+4K'| '' - ^~ " ^2 Now the quantity under the radical is the sum of two squares, and hence is never negative ; hence follows Theorem II. Equations 9 are obtained directly from 4. by inaking K = 0. iVbie. A pair of rectangular axes which divide any plane figure symmetrically, are principal axes, for K' =ffxydxdy = f^xdx; now if for every value of y there is another, equal with contrary sign, when the proper limits are introduced the value of K' van- ishes, as, for instance, in the circle, EXAMPLES. 1. Knowing the moment of inertia of an ellipse about the major axis to be -j— ,and al)out the mmor -j-, find its moment of inertia about a pair of axes, making an angle § = 30° with these. Solution. r = I cos" ^ + J sin'' |3 = ^ cos= 30° + ~ sin" 30°. CONJUGATE AXK8. 71 J' = I Siu^ ^ + J COS^ ^ = V 8'"' ^^° + ^ <=««' ^'^°- K' = (I — J) COS 30° sin 30° = '-"^—^^ cos 30° sin 30°, or I' = "^(aWi + b% J' = -g- (a^ + jVS), K' = yg* (a2 _ J2)V3. If the ellipse becomes a circle, a = 4 an»l I' = J', and K' = 0. CONJUGATE AXES. We have already seen that the axis conjugate to any neutral axis is the line joining the centre of gravity with the centre of stress for that neutral axis. So that as soon as a neutral axis is given its conjugate is at once determined. We have by Art. 94 (Rankine) the coordinates of the centre of stress, and hence we have the direction of the conjugate axis, for cot = — = -T"' where d is the angle between the neutral and conjugate axis if K = 0, cot = 0, and = 90° .-. when we take one of the principal axes as neutral axis the other becomes its conjugate axis. Theorem IV. We wish to prove that the neutral and conjugate axes are interchangeable. Now to do this we have only to take the conjugate axis as neutral axis, and find the angle between it and its K' conjugate axis ; to do this we have cot d' = -jr, in which we must substitute the values of K' and I' from 4, substituting 6 for /?. K ^ I Now we have cot ^ = 'j •'• tan — ^• tan I •■• ^•" ^ = y(l + tan^«) — V (I' + K")' 1 _ K and cos — ^^^_^ ^^^^ g^ — ^^p _^ j^ay 72 INTERNAL STltKSS. Now substitute in the 1st and last of 4 and we have, IK^ + JI^ — 2IK-^ I(IJ — K^) r K' = I^K — UK + K° — PK _ — K (IJ — K") .-. cot 8' = ^ = —J .-. d' = — .: Q. E. D. Proof of foi-mulaj 13, page 81, Rankiue. — ' — A ) iiquation of tangent is — ^ + -jr ^1, .■. 01= w ^ 1 - ab , ab ab v/'(S'+^")"(^' + T) 'iAi ~ OY' ~ b' ' but from' the equation of ellipse (pole at centre), b' = —^ J .■.n= ^ = (a" cos'' B + h' sin^ p)i- (a^cos^^ + S^sin^jS)^ , II. 't^ = OC^ — OT^ = a'2 — n^, but a'2 + b'^ = a^ + J", . . a'2 = a^ + 52 _ 6'2 .-. <2 = „2 _|_ J2 _ 5/2 _ ;„2 = a-' cos^ p + 0^ sur fi '^ ^ = a^ sin= 3 + b^ cos^ fi' — -3 „ ., . ,., . ,,, ^ ^ a' cos'' fi + b^ snrfi _ (a^ + b*) sin^ fj' cos^ ^ + a^b" (sin^ fj + cos^ ^ — 1) . — - J but sin* ^ + 2 sin^ j3 cos^ |3 + cos* |3 = 1 .-. sin* § + cos* (i; — 1 == _2 sin^"^ cos^ ^ „ (a* — iaW + 5*) sin= 3 cos^ B ■■■ f = ^ '-^ ^- ^ .-. nt = (a^— i^) sin § cos §. . INTERNAL STRESS. Examples on Arts. 96 — 99. 1. A rod of elliptic cross section (semiaxes 1.2 and .7 inches respectively) has suspended to it a weight of 2,000 lbs. What is COMPOUND STRESSES. 73 the intensity of the normal^ and what the tangential stress on a plane inclined at an angle of 60° to the axis of the rod. Solution \st. Area of normal cross section ^ nah =^ jr(1.2)(.7) = 2.638944 sq. inches. Area of oblique section = 2.638944 -=- cos 30° = 3.04717. Total normal component = 2,000 cos 30° = 1732.06 lbs. Total tangential component = 2,000 sin 30° = 1000 lbs. Normal intensity = ^^^^ — 568.41. Tangential intensity = ^MWtt = 328.17. Solution 2d. By formulae p, =p^ cos'' d = ^.-im^^ cos^ 30° = 568.41. Ft =J»x cos (? sin S = ^.^^fe cos 30° sin 30° = 328.17. 2. Find the intensity of the stress on a normal cross section of each of the following struts and ties. Each frame is loaded with a weight of 1,500 lbs. The cross sections of all the above beams are rectangles, whose sides are 2 and 5 respectively. . It is necessary to notice that the whole stress, which is the re- sultant of all the stresses, is a single force acting at a single point ; but the intensity of the stress is the amount per unit of area. COMPOUND STRESSES. We have already seen that stress consists of a distributed force acting on a body, and counteracted by the cohesion, etc., of the body ; thus a body in a state of strain is in equilibrinm, and all our problems of stress, are merely problems of equilibrium, and are solved by imposing the conditions of equilibrium. Moreover, a body may be acted on by stresses in diflerent directions, and in- deed practically always is, though in many eases we need only consider the stress in one direction ; but, for instance, suppose we have a lattice supporting a weight, and that the two cross bars are pinned together, as in the figure, there is a stress acting in the direction of each, lattice on the pins. So suppose a vessel contain- ing water, or any liquid, a stress is exerted in all directions, and 10 74 COMPOUND STRESSES. we must be able, knowing the load, etc., to determine the stress on any given surface. Proof of Eankine's Art., 101. Here we take a prismatic portion of the body, the sides of whose base are parallel to the direction of the 1st stress, and also the plane on which it acts. Now the forces on the plaiiC AB are counteracted of course by a set offerees, and the resultants of each of these sets act through O, and in the same straight line ; hence the forces acting ou the planes AD and BC must be balanced en- tirely, independent of any of the fojces on AB or DC. Kow tup- pose the stress on AD and BC did not act in a direction parallel to AB, we should have their resultants equal, but not directly op- posed ; hence we should have a couple, and there is no opposite couple furnished by the other stress ; hence the body would not then be in equilibrium. .*. Q. E. D. This condition of the stress is e:xhibited in the preceding exam- ple of the lattice. We may have tlyee conjugate stresses acting on a body, as, for instance, a gusset in a corner of an iron T-shaped girder. Note showing the work of Art. 102. Let X, y, zhe the directions of the three stresses. I have not drawn the perpendiculars, not to complicate the figure; we have from spherical trigonometry cos yz = cos sx cos oy -\- sin zx sin xy cos (ZXY). Now the angle ZXY is the angle be- tween the planes ZX and XY, and is the supplement of the angle between two lines drawn perpendicular to these planes ; hence ZXY = 180° — vw .: cos ZXY = — cos vw .: cos yz = cos zx cos xy — sin ex sin xy cos vw cos zx cos xy — cos yz .: cos vw =^ -. -'. ^-. sm zx sin xy From the last we wish to find sin vw. Bin'' WW = 1 — COS'' vw =^\ — I ; —. ^- I \ sin zx Sin xy / cos^ zx cos^ xy — 2 cos zx cos xy cos yz -(- cos^ wa sin^ zx sin^ xy COMPOUND STRESSES. 75 s in' sx sin' xy - cos" z x cos' ccy + 2cos zx cos xy cos ya - cos' yz sin' axe sin' xy (1 -cos'gg)(l -cos'a'y )- cos'^a; cos'cry + 2cos«a; cofa-y cosya - cos'ya sin^ ga; sin' xy 1 - cos' zx - cos' a -y - cos^ ya + 2 cos a-y cos zx cos ys sin^ xy sin' za; Vc .•. sin vw =—. : , and so far itm and icv. sin xy sin zx Pass the plane XOA perpendicular to ZY, and we have a right- angled triangle, XAY, where XA is the measure of the angle he- sin XA tween X and the plane ZY. Now we have sin XYA = ^^ — ; '^ sm a;y but XA = Ma; — 90°, and XYA = tiw. cos 1(X .'. sin uw = -: .•. cos ux := sin xy sin uw. sin xy " Now substitute the value of sin uw just found, and we have Vc cos ux = -; , and so for the other. sin yz Restricted case. I. In this case a;y = 90° .•. cos xy = 0, sin a;y = 1, C = 1 — cos' yz — cos' zx. Make these substitutions in the general formulse. Restricted case. IL Here yz = 90°, zx = 90° .•. cos yz = cos zx = 0, sin ys = sin ea; ^ 1, C = 1 — cos^ xy = sin' xy. Now substitute. EXAMPLES. 1. Given three conjugate stresses, such thata;y^ 30°, sa; = 40°> zy = 60°, find uv, uw, vw, ux, vy, and wz, respectively. Solution. C =: 1 — cos' yz — cos' zx — cos' a;y -|- 2 cos yz cos zx cos a;y = 1 _ cos' 60° — cos' 30° — cos' 40° -f 2 cos 60° cos 80° cos 40°, cos 60° = i = sin 30°, sin 60° = ^^3 = cos 30°, cos 40° := .76604, sin 40° = .64279. C=l — i — f — .586815816 + ^^3 cos 40° = ,076592815816; ^^7=. 27676. COMPOUND STEESSES. _ _ .76604 X .86603 — .5 "=°« '^^ = sin 40° sin 30" ~ .64279 X .5 .5 X .86603 — .76604 cos 40° COS 30° — COS 60° sin 4C )°siE I 30° = .50845. COS 30° COS 60°- — COS 40° sin 30° sin 60° -- — .76906, COS 40° COS 60° — COS 30° .5 X .86603 _ .5 X .76604 — .86603 ''o^ ^'^ "= sin 40° sin 60° " .64279 X .86603 = —.86767. sin vw = .86112 vw = 59° 26' 22" sin uw == .63915 mm = 140° 16' 10" sin uv = .49712 uv = 150° 11' 30" COS ux = .31957 ux = 71° 21' 48" COS V]/ = .43056 vy = 64° 29' 49" COS wz = .56352 wz = 56° 23' 27" 2. Given xy = 25°, yz = 45°, zx == 45°, find as above. Case I. xy = 90°, zx = 50°, xy = 60°. Case II. xy = 30°, zx = 90°, xy = 90°. Now Articles 103 and 104 of Rankine can be read. Note on Art. 103. of Rankine. We must bear in mind that stress is a set of forces balanced by another equal and opposite set of forces. Novp^ a shearing stress is equivalent to a couple, for it tends to draw a part of the fibres one way, and the fibres adjacent to it in the opposite direction, hence being, a couple it can only be balanced by a couple of equal moment and opposite sense. Keeping this in mind, Eankine's demonstration is clear. Note, on Art. 104 of Rankine. The demonstration is clear, only it is necessary to pay attention to the notation. When we have a body referred to three rectangu- lar axes Oaj, Oy and Oz, we call the as plane the plane perpendicu- lar to the axis Oa;, or, which is the same thing, the plane ZOY, etc. then py2, for instance, means the' intensity of the stress on the y plane, in the direction Oz. In Art. 105 of Rankine, we must find the total stresses on OAB, OAC and OBC, respectively, and lay COMPOUND STRESSES. 77 off OF, OE and OD, to represent these, then by con:iponn(!iiig these we obtain the total stress on ABC, then dividing this by the area of ABC we have the intensity of the stress on ABC. The object of Art. 106 is, knowing the direction and intensity of the stress on three given rectangular coordinate planes, to deter- mine the stress on any other given set of rectangular planes. To do this, the resultant stress is first determined on any one plane, ABC. Now resolve this stress into three components parallel to the axes of x, y and z, and here we must notice the notation, viz. : p„j. denotes the intensity of the stress on ABC in the direction Oa;; /)„y that on ABC in the direction Oy ; and p^^ that on ABC iii in the direction Oz ; jo„x' that on ABC in the direction Oa/, etc. Now the component of the stress in the direction Oa;' will be found by projecting the resultant stress on Oa;', this is p^^,, and we obtain the same result by projecting the three components /)„j, jBny, and jB„2, on Oa;' ; we thus obtain p^^, = p„^ cos a;a;' -|- p^^ cos yx^ + p^^ cos zx', and so the values given in the text of pny, and p^^,. We have thus found expressions for the stress on any oblique plane in any given direction, and we may now suppose the oblique plane to be that perpendicular to Ox', and we shall have p^,^, = p^^ cos xx' -\- p^y cos yx' + p,,^ (;os zx', and imposing the same condition on the equations 1 of Art. 105, we obtain Pux — Pxx cos a;a;' + Pxy cos yx' + p^^ cos zx'. p„j = j»iy cos xx' + Jfyy COS yx' -\- p^^ cos zx'. Puz = J»zx COS a;aj' + p^^ cos yx' + p^^ cos zx'. Substitute these values of />„„ p^y and p^^, and we have Px'x' ^ (Pxx cos a:a;' + p^y cos yx' -\- p^^ cos zx') tos xx' + (^xy cos xaf + Pyy COS yx' -\- jByj cos zx') cos 1/x' , ~t~ (Pzx cos a;a;' -|- py^ cos yx' -\- p^^ cos zx') cos zx' = p^,^ Qos^ xx! + jByy cos^ 2/a;' + p^,, cos" zx' + 'i-Pyz cos yxf cos za;' + 2p,iy cos a;aj' cos yx' + '^Pzx. cos zse' cos xx!. By making the same substitution in the other two equations we obtain the values of p^,y, and p^,^, ; then suppose the oblique plane to be, first, that normal to Oy', and second, that normal to Oz', and we obtain the remaining values. 78 COMPOUND STRESSES. We have thus determined the stress on that of the three new coordinate planes, a/Oy', x'Oz' and y'Oz', and now we wish to as- certain whether there can always be found for any state of stress in a body three planes at right angles to each other, on which the stress is wholly nomial, or in other words, three principal axes of stress. To do this we refer back to Art. 105. and impose the condition that the stress shall be wholly normal, or that xr shall equal sen, yr = yn, and zr = zn .•. p„, := p^ cos xn, p„y = jo, cos yn. and p^ ^ Pr cos zn. Substituting these in equations 1, we have equa- tions 2 of this article, from which we must eliminate the cosines. Thus by the method of Undetermined Multipliers, multiply the 2d equation by /, and the third by m, and add all three together, and we have cos xn (p^^ — Pr + Ip^y + mp^) -f cos yn \p^y + I (p^y —p,) + mp^,\ + cos zn\p^^ + Ipy^ -f m{p,^ — p~)\ = ; aud as we can impose two conditions make I'xx — ^r + (Pxy + mp^^ = 0, and p^^ -f l{p^^ — p,) + mpy^ = 0, and hence follows p^^ + ^Pyz + ^{Pzz — jOr)- Now from the first two find I and m, and we have I ^ iPr—P^^)Py.+P.yP.^ ^jjj ^ ^ {Pr~P..){Pr-Pyy)-P.y'' P^Py^ + P^^iPr—PyyY P.yPyz +P.APT—Pyy) Substitute these in the third, and clearing fractions, we obtain Pzx\p.yPy. +Pzx(Pr —Pyy)\ + Pyz\{Pr — P..)Pyz + P.yPz^ + (Pzz —Pr)\(Pr —Px^) (Pt — Pyy) — Px/\ = 0. Whence by performing the operations indicated, we obtain the cubic jOf' — Ap/ -\- BjOj — C = 0. This gives three values of jo„ and hence there are for every condition of stress in a body three planes at right angles to each other, on which the stress is wholly normal. Equations 5 may be obtained by equating the value of cos sn in two of equations 2, and thus obtaining a relation between cos xn and cos yn. STRESS PARALLEL TO ONE PLANE. 7^ To obtain equations 7 we have the whole stress on ABC equal to p • Area ABC. and that on OBC equal to f-^ • Aiea OBC ; pro- ject the 1st on Oa; and we have p • Area ABC • cos rp = /?, ■ Area OBC, or p • Area ABC cos xp =^ pi ■ Area ABC cos xn .: p cos xp =j»i cos xn, and so p cos yp =^ Pi cos yn, and jo cos zp = ps cos zn. To obtain equation 11 project first the whole stress on ABC on the normal On and we obtain p • Area ABC cos np, and then project each of the components of the stress in the same manner, and add their projections and we have p • Area ABC cos np =^ {p ■ Area ABC cos xp) cos xn + (p • Area ABC cos yp) cos yn -\- (p ■ Area ABC cos zp) cos zn. .•. cos np = cos xn cos xp + cos yn cos yp -\- cos zn cos zp, and the rest of the reduction is easy. STRESS PARALLEL TO ONE PLANE. In practical problems we have to do generally with stress par- allel to one plane. ISuch is the case in lattices, in earthwork, and all kinds of structures. The object in the following problems is. Given the condition of stress in a body, that is. Given the stress on any two planes, both as to direction and intensity, to deteimine that on any given plawe whatever. 80 that we may be able to know what resistance the body must oppose in any given direction, and hence what amount of strength is required. Always now it is implied that the stress is parallel to one plane (the plane of the paper in the figures). Note on Problems I and II, Eankine, page 96. We must notice that in finding the resultant stress on the plane AB it would not do to compound merely the intensities of the stress, as we do forces. The force acting on OA is not p^ but Py ■ OA, and that on OB is p^ • OB, so when we have OR for the resultant force this is the force on AB, and the intensity is OR An = ^- 80 EXAMPLES. EXAMPLES. 1. Given two conjugate stresses of intensities, 4 and 5, respec- i\i.\y (4 and 5 lbs. per square inch), jo^ = 4, jOy = 5, their direc- tiuiis inclijned to each other at an angle 60°, what is the intensity of the stress on a plane at right angles to the axis of X. Sdution. In this case OAB becomes a right triangle, right- angled at A, and ABO = 30°. Now take OA = 1 (inch say), then OB = 5775 = 2, and AB = OA tan 60 = V3 .-. cos 60 ' OD = 4, OB = 8, OE = 5, OA = 5, and OR = V(8'' + S'^ + 2-8-5 cos 60°) = V(64 + 25 + 40) = Vl29 11.35 11.35\/3 „_„„„ , := 11.35 .-. pr = — -rr = 5 = 6.oo273 (pounds per square inch say). 2. Given p^ = 1, Py= 9, AOB = 45°, OAB = 30°. 3. Given p^ = b, p^ = —8, AOB = 30°, OAB = 100°. 4. Given p^ = 8, jOy = 10, AOB = 75°, OAB = 90°. 5. Given p^ = —3, py — 5, AOB = 90°, OAB = 45°. 6. Given the stress, Fig. 51, on the plane OA of intensity p^ = 7, and on OB intensity p,j = 5, and angle the 1st makes = 30° with OX, angle xn = 30°. Solution. p^^ = 5 cos 30° = 4.33015. i>.y=Pyx = 5sin30° = 2.5. jjyy = Vl?" — (2.5y\ = V(49 — 6.25) = V(42.75) = 6.5383. ••• Pr = -yiPxJ cos'^xn -\-pyy^ sin^ xn +Pxy^+ ^PxyiPxyrhPyy) ooBXn sin xn\ y}(4.3.301o)2cos-^30°+(42.75)sin230°+6.254-5(10.86845)oos30°sin30°| = VipJ' + pj") = from what follows, V(728.880697) = 26.997. p^ = (6 cos 30°) cos^ 30° + (42.75) sin^ 30° + 5 cos 30° sin 30° == -%-i + 10.68 + 2.16507 = 26.9075. EXAMPLES. 81 ■p, = (2.20815) COS 30° sin 30° + (2.5) (| — I) =.95616 + 1.25 = 2.20616. _^ _ 2.20616 ^° '*'* ~^u ~ 26.9075" logjBt = 0.3436370 logPn= 1.4298733 ' log tan Mr = 8.9137637 nr = 4° 41' 14". 7. Given p^ = 8,p^= 6. Angle p^ makes with OX = 45°, and xn = 40°. 8. py = 7, jBjj = 5. Angle p^ makes with OX = 75°, xn = 45°. Art. 109 of Bankine we have cos xn sill ajw p^ cos^ xn — sin^ xn = p^.—p^; ^"^ ''" ^"^ = ^ ""^ '^'* '^'^ '^''• J o ' 2 -2 sin2aOT 2« and cos 2iBra = cos" xn — sin" xn. .: — ^ — = ^-^ — > cos 2xn p^ — ;>yy „ 2»3i„ or tan 2xn =: ^^—^ — » hence follows what Rankine says about the principal axes. Arts. 110 and 111 are clearly given in Rankine, but I wish to comment a little on their results. From Art. 110 we see that two principal stresses of the same kind and of equal intensity produce on any plane (parallel of course to the stresses) a stress of the same intensity, and normal to the plane of action. This is the case in a perfect fluid, and it is borne out by experiment. If, for instance, we have a vessel of any shape whatever containing a liquid then at any given depth below the free upper level, the pres- sure is the same in all directions, and always normal to the plane of action as we can see by tapping the vessel at any point. If, on the other hand, we have the two stresses of opposite kinds, (one a pull and the other a thrust), then the resultant stress on any plaiie is still of the same intensity, but it is no longer normal to the plane of action, but makes with the axes of principal stress, angles equal to those made by the normal to the plane of action. (Fig. 62.) Thus if OX and OY represent the axes of principal stress, the 11 82 ELLIPSE OP STRESS. stress on OX being a pull, and that on OY a thrust, and if we wish to find the stress on the plane AB (or any plane parallel to it,) we should find by the construction of Art. Ill, OR, where XOR = XOC, and if we wish the stress on DC by the same construction we obtain OP. If, now, COY = BOY = 45°, then the lines OR and OP would coincide with AB and CD respectively, hence follows his conclusion. As an example of this condition of a pull and a thrust, we can take an iron bracket, on which the stress (when loaded) is evidently a pull in a horizontal direction, and a thrust in the verti- cal. ELLIPSE OF STRESS. Before entering on the Problem of the Ellipse of Stress, we must remark that its object is a simplification and also a generaliza-- tion of Problem II, page 97 (Rankine). The practical engineer often finds it muijh more convenient to lay off certain lines to rep- resent the different quantities in his calculations, and then to make a geometrical construction, which enaldes him by simply measuring a certain line to ascertain the result of his calculation, especially if he can make one figure serve him instead of a number of calcu- lations, rather than to make a lengthy numerical calculation every time, in which also he has a great many chances for error. Suppose we have two principal stresses, jo^ and p^, now P^ — 2 2' """Py — 2 — 2 ' hence we may imagine, instead of the two original stresses, p^ and Py, the 4 stresses, viz. : P ~\~ V"^ 1st. A stress of intensity ^-^j^-^-^ acting along OX. 2d. — Py 2 Sd. " " " " ^^ l~/'y jj ,^ QY 4th. 2 ■Py ELLIPSE OF STRESS. 83 1. Now let us find the stress on the plane AB resulting from the 1st and 3d. Since these are of the same kind and of equal intensity, viz, a , they produce a stress of the same intensity, acting in the direction ON perpendicular to AB. Lay off (on any P ~\~ P scale) OM = „ ^ (say an inch to a pound), and this repre- sents the resultant intensity of the stress on AB from the action of the 1st and 3d. Now take the 2d and 4th. These are two stresses of equal in- tensity and opposite kind, hence by Art 112. they produce on AB a stress of the 'same intensity, but whose direction makes with OX an angle equal to that which ON makes with it. If we take P, such that MP = OM, POM is an isosceles triangle, and MPO = MOP, hence PM will be the direction of the resultant stress. Now 7J — P on PM lay off MR = ■ q (on the same scale), and join OE. (We already know that OR will represent the resultant of two forces acting at the same point, represented in direction and mag- nitude by OM and MR respectively, and since it is altogether stress on AB we are considering, we may combine the intensities just as we do the whole stress.) Hence OR represents the resultant in- tensity and direction of the stress on AB. Two things remain to be done. One is to express algebraically the magnitude and direction of O^, and second to see if we cannot find some line straight or curved in which the point R must always fall for every position of the plane AB. Now to find the algebraical value of OR, we have PM = OM = QM, but OM = ^^Y^' '^^°°® P/ = 2 • P-^^ =p.+P„ and RM = ^-^^ ' .-. QR = QM + MR =jo„ PR = PM — MR =py. Now project Rj)^ on OX and PR on OY, and the projections will be p^ cos xn, and pj sin xn, respectively, whence OR^ = 84 ELLIPSE OF STRESS. {p^ COS xny + (py sin xn)", since we shall have a right angled triangle, with OR for hypothenuse, and these projections for sides. .■. pr = V(pJ cos^ xn + p/ sin^ xn). 2. From the trfangle EOM we have sin EOM : sin RMO : : RM : OR : ;- ^^~^^ .- p^. ^= sin RMO • ^''I~^^ , but ROM = nr, and = MQO + MOQ = 2M0Q = 2(90 — A) = 180° — 2xk. .: sin Mr == sin 2xn — k We have thus found the algebraic expressions for the intensity of the resultant stress • and its inclination, and now we proceed to find the locus of the point R.. First the locus of M is a circle, for in whatever position the plane be, OM = „ , and this does not change when AB changes, hence M is always at the same distance from O, and .•. the locus of M is a circle. Now for the locus of R we have seen that RA == p^ and PR = py .-. P^ =/>! + Py Draw perpendiculars From R to the axes of X and Y, so as to obtain the coordinates of R, and we shall have tS"^ QR cos xn =p^ cos xn, y =^ PR sin xn = jBy sin xn, these evidently change for every value of xn, but we have x y -^ = cos xn, — = sin xn. Now square and add, and we have V'^ Pi « x^ «2 a;*^ y^ -^ + r^ = cos" xn + sin^ xn, ot —^ -\- — ^ = 1, where xn has "x Py . /'x Pj disappeared, and we have the equation of an ellipse whose semi- axes are jo^ and p^, respectively ; hence R is always situated on this ellipse. Hence instead of the preceding construction for OR we can construct our ellipse, then draw the perpendicular ON to the plane AB, and then lay off OM =^S^—^, and then draw PM = OM, and where it cuts the ellipse will be R. EXAMPLES. 85 EXAMPLE. It can easily be shown that in a boiler or tank for holding water, the intensity of the pressure in a vertical direction is double that in a horizontal direction. Now suppose we want the intensity of the pressure on any oblique surface. Construct an ellipse with semiaxes 2 and 1 (inches), respectively, draw AB (Fig. 53) repre- senting the plane on which we wish to know the intensity of the stress, draw ON perpendicular to it, lay oif on ON a distance ^-^-^ = |, and draw PM =^ OM, then will OR represent the intensity of the stress on the oblique plane. The intensity and direction being found we can decompose it into a normal and tangential component. The normal component will be the projection of OR ■ on ON, or which is the same, the projection of the coordinates of OR, viz., p^ cos xn and p, sin xn on ON .-. jOn = p^ cos'' xn -\- py sin^ xn. The tangential component will be found by projecting the same on AB .•. jBi = j»x cos xn sin xn — p^ sin xn cos om. Problem 11. The shear is the projection of OR or MR on AB .'. is less than MR, except when MR is parallel to AB, and repre- sents the shear. EXAMPLES. 1. Given two principal stresses of intensities, /i^ ■= 16, Py = 9, respectively. Find the intensity and direction of the stress on a plane where xn = 35°. Also find the normal and tangential com- ponents. Solution. OM = ^""^^^ = ^, MR = ^-^^^ = h OR —p, = VK16)' cos2'35° + (9)2 sin^ 45°| = V(198.4262362884) = 14.08638. sin nr = .-,„ — - sin 2xn = ij^.tV^tf sin 70°. 86 EXAMPLES. jo„ = p^ cos' xn + jOy sin'' xn log 7 = 0.8450980 =10.73610 + 2.96094 = 13.69704 log sin 70° = 9.9729858 0.8180838 log 28.17276 = 1.4498294 p^ ='(jBx — Py) COS xn sin xn log sin nr 9.3682544 _ i^'' — ^y sin 2xn = 3.28891. nr = 13° 30' 8". 2. Given j9, = 23.134. p^ = 17.421. 1st. xre = 41°. 2d. xn = 87° 3d. xn = 5Z°. 3. Given p^ = 13.427. p^ = 14.329. ot = 75°. 4. Givenj9^ = 12.727. py = 5.432. xn = 45°. 5. Given p^ = 32.531. p^ = 17.853. xn = 33°. Find in each of the above, p^, p^,. pt, nr, and also construct the figure in accordance with the data. Also the following : 6. p^ = 10.451. Pj = —8.634. xn = 23°. 7. p^= 9.427. J9y = — 3.231. xra=135°. 8. p^= 8.372. jOy = — 7.438. kw = 127°. 9. p^= 7.241. jBy= 5.327. a» = 475°. 10. p^ = 12.458. Py = 6.453. xn = 37° 5' 22". Problem III. Case 1. RM < OM .-. MOE < 90°, hence MOR is a max itnum when sin MOR is a maximum ; but sin MOR : sin ORM : : MR : OM, or sin MOR = &-=& sin oRM. P^ 'Py Hence sin MOR is greatest when sin ORM is greatest ; but the greatest value of sin ORM is when ORM is 90° ; hence MOR = nr is a max when MR is perpendicular to OR, and then sin maximum i'x— Py . _ij»x — Py r = i -, or max nr = sin ' r^^-^- P^ + Py Px + Py 2. PMN = OMQ = MOP + MPO = 2M0P = 2xn .: 90° -|- max nr xn ;= ^PMN, hence follows xn = g EXAMPLES. 87 3. p, = V{OW — MR2) = V(OM + MR)(OM — MR) = = -yp^Py. In Case 2d p, = V(MR2 - OM^) = V(M R - OM) (MR + OM) EXAMPLE. 1. In tlie 1st of the above examples find the plane for which the tangential stress is greatest. Also find the position of the plane on which the stress is most oblique, and in each case the intensity of the stress. solution. 1st. maxj»t= ^^ ^ ^^ = |, and the plane makes 45° with Ox. log 7 = 0.8450980 2d. max wr = sin-' 3jV = log '25 =1.3979400 16° 15^ 36". log sin nr = 9.4471580 90° -l- 16° 15' 36" Corresponding value of xm = ^ = 58° 7' 48". Intensity = Vjo^Py = 4'3 = 12. 2. Do the same with each of the above examples. Problem IV, This is the converse of the ellipse of stress ; there we had given the two principal stresses p^ and p^'_ and we were required to find the resultant stress on any given plane AB. Suppose, now, we should by Problem I find the stress on AB (construct it), then find it <5n another plane A'B', then suppose the principal axes erased, and that we have given us merely these two planes and the respective stresses on them, and are required to find the axes of principal stress, and also its intensity. This is Prob- lem IV. Now we know that in Problem I, to find R we first lay ofl' on ON a length representing q — ^ then on a line that makes equal angles with x to what ON does, a distance — ^— h ^, and we thus have E. Do the same for the other and we have two triangles, 88 ELLIPSE OF STRESS. where OM = OM' = ^^ "^ ^^ , and MR = M'R' =^^^^^,and the angles are respectively NOR and N'OR', the question then resolves itself into this, viz. — Given OR, NOR, OR', NOR' to construct two triangles, with this side and angle, respectively, whose other two sides shall he respectively equal to each other (one in each triangle). This is done by the construction in the text for IMR' = MR, and OM is common, hence OMR and OM'R' = OMR', are the two required triangles, and OM represents ^^ ^ ^^ , and MR = MR' = ^''~J'^ ' Half their sum will bejB„ and half their difference jOy. Moreover, from Problem I we know that ocn = ^NMR (the tri- angle ORM in either figure is the same), and xn' = JNMR', we thus can construct the axes of principal stress and obtain their intensities by measuring OM and MR. Fig. 54 shows the construction when the stresses are of different kinds, and in this case, by referring to the second figure of Prob- lem I, 2xn. = RMO .-. xn = ^RMO, xn' = iR'MO. The figures now give at once in the 1st case, xn + xn' = ^NMR + ^NMR' = NMS. 2d case, xn — xn' = JRMO — ^R'MO = RMS. So far we have made geometrical constructions, and now we must express the quantities, viz., the intensities of the principal stresses, and their directions. First, to find OM. From the triangle OMR, we have MR= = OM^ + OR^ — 20M-0R cos ROM ; and from OMR', MR'^ = OM^ + OR'^ — 20M-0R cos R'OM ; but OR = p, and ROM = nr. OR' = p', and R'OM = n'r', hence the equations become MR2 = OM^ -\-p^— 2p-0M cos nr, and MR^ = OM^ + p'^ — 2/OM cos n'r' ; subtract and ^^p^ — p'^ — 20M {p cos nr — p' cos n'r'), or OM ?'' -P" - PjL±P3. z{p COS ?ir — p cos nr) 2 EXAMPLES. 89 Now substitute this for OM in the 1st and 2d equations, and we have ^^5-=^ =MR = MR' = \J { (c^^y +P'- (v. + Py)P COS nr ] = \/ { (r^y + P" - (Px + Py) P' «°« "''•' } Now draw a perpendicular from R to OM, and we have OR cos NOR = OM + MR cos NMR (by projection), or p cos nr = ^ „ ^^ + ^ „ ^^ cos 2xn. 2p cos nr — j»x — Py , , . ^. ^mj/ .•. cos 23m = — *-^, and by projecting OR we P^ Py have,^' cos nV =-^^±& + P'~^^ cos 2aw'. „ , 2»' cos mV — »_ — »_ .-. cos 2xn' = — — ^ ^'^• P^—Pj EXAMPLES. 1. Given OR = 17. OR' = 12. wr = 37°. w'/ = 23°. P^+Py _ OM - 289 - 144 145 2 ~ " ~" 2(17 cos 37° — 12 cos 23°) ~ 5.06176 = 28.646. ^'~^^ = V 1820.593316 + 289 — 2(28.646) (17) cos 37°} = V(720.67001152) = 26.845. .-.p^ = 28.646 + 26.845 = 55,491. j9y = 28.646 — 26.845 = 1.801. 2p cos nr — (p„ + «.) 27.15376 — 2(28.646) cos 2xn = = a/ac OAK\ ' P:^—Py 2(26.845) = —56133 .-. 2xn = 124° 8' 52" .-. xn = 62° 4' 26°. „ , 22.092 — 57.292 _,^, ""^ ^^ = 2(26.845) "" —65747. 2xn' = 131° 6' 24" .-. xn' = 65° 33' 12". 12 90 EXAMPLES. Case 3. In this case OR and OR' coincide in direction, and hence we have (Fig. 55) nn' = iNMR + JNMR' = NMS = MOS + OSK = 'ir + 9C°. OR + OR' P+p' Also OS = OM cos MOS = OM cos nr = ^ ^' — 2 i 2 cos nr and OIP = 0S= + MS^. MR= = SW + MS'' ; subtract, and OM^ — MR2 = OS'' — SR2 = (OS + SR) (OS — SE) = OR • OR' = pp' .-. MR2 = OM^ V ( . cos- nr -f^ ) Case 4. In this case the figure becomes Fig. 56, and we have, by projections, MS = p sin nr = p' sin n'r', and, moreover, on resolving OR or OR' into normal and tangential components (viz., along and perpendicular to ON), MS =^ Pti Pa^ P cos nr,p^ = p' CCS n'/, and OM = OG_+OH _ ^^^ ^ P^^!^_ KR = y .KG- + GK ) = i/ J (Pn — ^nO" ^ ^^j I MG » » ' cos.2a;M = cos NMR =^ = ^^ ^. tan 2a!M = tan NMR = "MR p^ — p GK _ 2;,/ MG - p^ -p^- EXAMPLES. 1. Given OR = 17. OR' = 12. nr = 71^ = 37°. Solution. P^+Py _ P-\-p' _ 29 _ 2 2 cos 7tr ~ 2 cos 37° ~ l»-lo5- £i^=l& = V(329.G04025 — 204) = 11.207. ELLIPSE OF STRESS. 91 2. Given 0R = 12.471. 0R'= 1.225. nr = nV= 42°. 3. Given OR = 37.852. OR' =10.01 2. nr = nV = 67°. 4. Given OR = K'ViT. OR' = 12.427. nr = n'/ = 48°. 0. Given p^ = 10.827. jo/ = 8.432. p, =^pl = 3.-75, 6. Given />„ =28.124. p: =20.342. p, =pl =3.1 , 7. Given ;;„ .= 16.327. f„' =12.437. p, =jo/ =6.538. 8. Given jB„, = 18.532. /)„' =15 428. jOt =jo,' =4.836, Problem V. This will be much better understood by taking tl case in which it is applied, viz., earthwork and retaining wall Suppose we have a retaining wall, with a mass of earth piled up behind it. Now the only force which causes a stress is that of gravity, and this acts vertically ; hence, if we imagine a plane AB parallel to the upper surface, the stress (crushing) on this plane is due to the weight of the earth above, and nothing else ; hence it acts in a vertical line ; therefore by the proposition on conjugate stress, the stress on a vertical plane must be in a direction parallel to AB. These are two conjugate stresses. The last is confirmed by the actual state of the case, for if the wall were suddenly re- mov.ed, the earth on top must necessarily begin to move in the direction CD. Now we know that if the obliquity of the stress is too great, that is, if the bank of earth is too steep it will fall, and there is a certain limiting angle of obliquity, which is the natui-al angle of rest different for diffei-ent materials, and which is ascer- tained by experiment. This is called the angle of repose. Now the object of Problem V, as applied to this particular case, would be, Given the slope of the earth, and also the angle of re- pose, to find the ratio between these two conjugate thrusts, that is, the ratio between the vertical stress on the oblique plane (parallel to the upper surface), and the oblique thrust on the vertical plane of the wall. "When this is known- we can easily ascertain the amount of pressure on the wall, for we know the vertical pressure when we know the weight of the earth. Now let

"^= Vli-(F+/r } •■•'" ^^' ~ iv^P'Y 4pp' cos^ nr (p-{-p')'' cos' nr ... 1 _ ain'

= 35°, nr = 17°, find ^• p' c os nr - \/(cos" nr - cos" y )_ cos 17° - V(co8" 17° - cos" 35°), p"~cosM»- + y(cos"»ir-cos"55)~cosl7° +v'(C08"17° -cos"35°) .9 5630- V^(.95630)"- (.81915)" } _ ■95630-.49346 .46284 ~.95630 + y 1 (.95630)"- (.81915)" | "".95630 + .49346~1.44976 t— QiQ9f; ^°S .46284 =T6654309 ^ — .31925. j^g 1.44976 = .1612961 T.5041348 EXAMPLE^. 93 2. Given

= 45°. nr = 0°. 4. Given ^ = 0°. nr = 0°. 5. What is the pressure against the wall at the depth of 10 feet, where = 38°, nr = 16°, and 1 cubic foot of the earth weighs 120 lbs.? 6. Solve the same example when ^ = 72°, nr = 0°. EXAMPLE^ ON ARTICLE 113. Given p = 15, 12, 17, 18, 23. np = 30°, 45°, 73°, 54°, 37° Solution, P np 2np 15 30° 60° 12 45° 90° 17 73° 146° 18 54° 108° 23 37° 74° 85 ■Py — cos 2np .50000 .00000 —.82904 —.30902 .27564 P^^y = -Y- = 42.5. iV{33.8145783424 + 5435.3372880676| = 73.953 .-. jB, = 116.453. jBy = —31.453. sin 2np p cos 2np p sin 2np .86603 7.50000 12.99045 1.00000 0.00000 12.00000 .55919 -14.09368 9.50623 .95106 —5.56236 17.11908 .96126 6.33972 22.10898 —5.81632 73.72474 ^ _ 2p sin 2np 73.72474 tan 2nx = -j^ ~ = ■„,„., 2,pcos 2np 5.8ib32 log 73.72474 = 1.8676132 • log 5.81632 z= 0.7646483 log tan (180 — 2nx) = 1.1029649 180 — 2nx = 85° 29' 21"- 2nx = 94° 30' 39'. nx = 47° 15' 19". 2. Given 3. Given 4. Given 5. Givei p np p np 10 43° 1.125 30° 14 72° 3.427 45° 17 12° 5.642 60° 16 17° 3.721 15° 27 86° * 8.347 75° p np ' 4 10° 12 57° 19 153° 32 149°- 73 190° P 8 7 3 2 1 np 14° 17° 81° 94° 73° 94: EQUILIBRIUM OF FLUIDS. Articles 114, 115, 116 and 117, of Eankine should now be care- fully studied. Note on Art. 116. It is important, in this connection, to keep clearly in mind the meaning of the sj'mbols f^^, etc., hence I repeat them here, viz.) jOxx means the intensity of the stress on a plane perpendicular to Ox in the direction Ox. So jo^y means the intensity of the stress in the direction Oy, on a plane perpendicular to O.r ; and this we have already seen is equal to Py^, or the intensity of the stress in the direction Ox, on a plane perpendicular to Oy. Secondly, -j- represents the rate of variation of the stress along Oz, that is the amount of increase of stress for a unit of dis- dp . tance, and hence -t' ^^ will be the amount of increase for the dis- tance z/a, or the whole intensity will be /) + t"^«- Thirdly. The equations 1 are obtained by imposing the three known conditions of equilibrium, viz., ^ P cos a = 0, ^^ P cos |3 = 0, I' P cos 3' = ; that is, by placing the resultant stress in the direction of each axis equal to 0. ANOTHER VIEW OF ARTICLE 117. Suppose we have a body of liquid confined in a vessel of any shape.; the free upper level will, as we already know, be horizon- tal, and the pressure at any point is due merely to the weight of the water above it. The fact that any molecule, as EFGH, is at rest, shows us that the horizontal pressures acting on the molecule are in equilibrium. Suppose the intensity of the pressure on the plane AB be j», the increase of pressure on the plane CD is equal to the weight of the liquid included between these two planes, AB and CD, thus the EXAMPLES. 95 whole amount of increase is wAJ- surf. CD, where -w is the weight of ,. , , 4i .>^ , . . . . w-^a;surf. CD a unit of volume, and J^ = EG, heuce Us intensity is if~7nS — dp = w A x,OT -^ A X ; hence the total intensity of the stress on the dp plane CD is jo + w A x, or p -\- -i^zl x. If, therefore, jOq is the intensity of the pressure at the free upper /^ dp ■n- dx = i'o "h I 'U'dx, which, when w is the same at all depths, becomes !> = JOo + wa;. PXAMPI.ES. 1. What is the intensity of the pressure of the water contained in a cask, at a depth of 12 feet below the free upper level. Solution, p ^ p^-\- wx. And/>o= 14.7 lbs. w = ^f,^^- lbs. a; = 12 X 12 = 144 inches. .■.p = 14.7 + sa-if^/^iii = 14.7 + 5.2 = 19.9 lbs. This is the pressure at the given depth, but if we were deter- mining the pressure of the water on the sides or bottom of the cask at that depth, we should make a gross error by the solution, for this pressure is counteracted in part by the pressure of the atmosphere outside, and hence the resultant pressure against the sides of the cask is only 6.2 lbs. per square inch, for this case p„ = 0, and .-. p = wx. 2. The free upper level of a reservoir is situated at a height of 80 .ft. above the level of the ground ; what is the intensity of the pressure on a pipe leading from the reservoir at a height of 10ft. above the level of the ground ? 3. What is the intensity of the pressure at the foot of a mill- dam 30 ft. high, the free upper level of the water being 1 ft. above the top of the dam ? 4. What is the total pressure of water on the vertical face of a reservoir wall 10 ft. high, where the free upper level is even with the top of the wall ? 96 FLOATING- BODIES. 5. What is the total normal pressure, and what its mean in- tensity, on a reservoir wall whose back makes with the horizontal an angle of 75° ? Also find its intensity at the foot of the wall. 6. Given the following reservoir walls with an inclined face. 1. height = 10. 2. height = 15. 3. height = 14. a = 60°. a = 58° 3', a = 83°, A SHORTER VIEW OF ARTICLE 120. EQTJILIBKIUM OF A FLOATING BODY. In the case of a floating body, as, for instance, a ship which is partly immersed in water, and thus is in equilibrium, if we im- agine the body removed, its place would be filled with water, and this water is kept in equilibrium by the resultant pressure of the water around it ; but to support it it requires a vertical force equal to the weight of this body of water, and acting vertically upwards, and so likewise to support the floating body we must have a resultant up- ward pressure equal to the weight of the body. Hence A floating body displaces a volume of water whose weight is equal to the weight of the body itself. The centre of gravity of the liquid displaced is called the Centre of Buoyancy of the body. Now the weight of the body acts vertically downward through the centre of gravity ; the upward pressure of the water acts vertically upwards through the centre of buoyancy ; hence for equilibrium we must have these two in the same vertical line. For Article 121 of Rankine a similar reasoning holds. APPARENT WEIGHTS. We have already seen that a body immersed in a liquid is pressed upon by a force whose resultant acts vertically upward and equal in amount to the weight of an equal volume of the liquid ; so also a body in the air is pressed upwards by a force equal to the weight of an equal volume of air. IMMERSED PLANE. 97 Tims if we have two bodies, whose absolute weights are equal, ('. e., which contain the same amount' of matter, and on which, con- sequently, gravity acts equally, suspended one in each pan of a balance of equal arms, if the bodies are not equal in size the air presses unequally on them, and hence their weights are diminishe»;„ Area BC, .-. ;)„ = ^^^^^q =iox^. Art. 125 of Rankine has already been discussed in the problems on earthwork. NOTES ON MECHANICS. PART II.-DYNAMICS. NOTES ON MECHANICS. DESIGNED TO BE USED IN CONNECTION WITH RAN- KINE'S APPLIED MECHANICS. BY GAETANO LANZA, S.B., C.E., ASSISTANT PROPB8SOB 01' MATHEMATICS AKD MECHAHIC8, MASS. INSTITDTK Off TECHNOLOGY. PART II.-DYNAMICS. BOSTON: 1874. Entere(l, according to Act of Congress, in the year 1874, By GAETANO LAKZA, in the Office of the Librarian oE Congi'ess at Washington. lOTES ON MECHANICS. DYNAMICS. Dynamics is that part of Mechanics which considers forces as producing motion. Velocity is the space (number of units of space) traversed by a moving body in one unit of time. Motion is of two kinds, Uniform and Variable. Motion is uniform when the velocity suffers no change, but remains constantly the same at all points of the body's path. Motioii is variable when the velocity is different at different points of the body's path. It is said to be Uniformly Varying when the velocity increases or decreases by equal amounts in equal times. The amount of increase or decrease of a body's velocity in a unit of time is called its Acceleration. UNIFORM MOTION. Let t represent the time ; v the velocity ; and s the space through which a body moves in the time t ; we then have the forro.ul*, s S == pt, or « ^ -■ Thus if a body moves uniformly^with a velocity of 12 feet per second for 5 seconds, the space traversed will be« = 12 X 5 = 60 feet. 4 VARIABLE- AND UNIFORMLY VARYING MOTION. VARIABLE MOTION. When the velocity is variable we have no longer the formula, s * V = — , as the velocity changes during the time t, but if we take a very short time, so that the velocity may be considered nearly uni- form during that interval, calling this short time /Jt, and the space passed through in this time /Is ; the quantity -r: will nearly ex- press the velocity of the body at that part of the path, and passing ds to the limit we shall have the true equation, v = -j-. The veloc- ity at any point of the body's path means the velocity with which the body would move were all forces to stop acting when the body had reached the point in question. UNIFORMLY VARYING MOTION. In this case the velocity receives equal increments in equal times, so that supposing a body to be already moving with a veloc- ity «0) S'^d then a force which imparts to it an increase of velocity, or an acceleration represented byy in each unit of time, it is evi- dent that in t units of time it will have a velocity, ^ =^ Va +ft. ds ds And smce v = -^, we have -^ — Va + Jt; hence, by integra- tion, we obtain s = v^t -{- ^ffi. These are the equations of Uniformly Varying Motion. Differ- entiating the first of these equations, we have dv dt ~ ■'• This is the amount of increase (or decrease) of velocity per unit ds dv d^s . . of time, but «; = ^ .-. ^ = -^ = /; or m words, the second differential coefficient of the space in regard to the time is equal to MEASDEE OF FORCES. 5 the acceleration, or amount of change in the velocity per unit of time. ' dv cPs In the case of Variable Motion, -jt = -r^ still represents the change in velocity per unit of time, and hence the equation, j^:= f,is Still true, with the difference that now / is a variable, while in uniformly varying motion y is a constant. MEASURE OF FORCES. We have seen by the First Law of Motion, that a body at rest will remain at rest, and when in motion will continue to move uni- formly and in a straight line, unless, and until, acted on by some external force. We have also seen by the Second 'Law of Motion that every force which acts on a body produces its full effect, independently of any other force that may be acting on the same body. Consid- ering these two facts together, we find, first, that the motion of a body is not changed as long as no external force acts upon it ; and, secondly, that when a force does act it changes its motion. Now suppose a body to be moving uniformly, and a force to act on the body for a certain length of time, t ; in the direction of its motion the effect of the force is evidently to increase the velocity, and if-y represents the amount of velocity the force would impart in one unit of time, the total increase of velocity imparted at the end of time t, will be fi ; and if the force stop acting the body will again move uniformly, but with a velocity greater by_^. A force which will impart to the same body in the same time a velocity twice as great, is itself twice as great, and hence we can measure forces by the amounts of velocity they impart to the same body in the same time. Next suppose two bodies, equal in every respect, moving sid6 by side with the same velocity, and uniformly. Apply to one of them a force F, in the direction of the body's motion ; the effect of this 6 UNIT OP MASS. force is to increase the velocity with which the body moves, and if we wish to give to the other an equal increase of velocity, so that they shall still move side by side, we must apply an equal force to that. Now let us unite these two bodies into one ; it will still re- quire a force IF to give it the required increment of velocity. Hence we see that if we double the mass to which we wish to impart a certain velocity, we must double the force ; or in other words, employ a force which would impart to the first mass a veloc- ity double the required velocity. Momentum is a term applied to the product obtained by multi- plying the number of units of mass in a body by its velocity. It is merely the amount of velocity per unit of mass, and is that by which we measure forces, since we have already shown that they are proportional to the momenta (or the velocities per unit of mass), they generate in the same time in a body ; hence if F rep- resents the force, m the mass of the body, and v its velocity, we have, F = fnv. Take now the particular case, when the direction of the force poincides with the direction of the body's motion, and we have, THIRD LAW OF MOTION. The Third Law of Motion which was referred to in the Statics, is (as giyen by Sir Isaac Newton) that Action and Reaction are equal. In other wordft that momentum cannot be given to one body without giving an equal and opposite momentum to another. Thus, if a hunter shoots a bird, he receives a harder blow than the bird, only it is distributed over a larger space, and hence the intensity of the blow is not so great.. One man canaot strike another without being struck by him, etc. UNIT OF MASS. We have iu Statics taken weights as the measures of forces, and we have now shown that forces are proportional to the mo- DECOMPOSITION OF FOR0E8. 7 menta they would generate in a unit of time, and now since we have not yet fixedj on a unit of niass we may so choose it that these two measures (their nuni^rical representatives) shall be equal. Kow it is found ,by experiment that the acceleration due to the force of gravity is constant at Ahe same place on the earth's sur- face, and if this acceleration be represented by ^, the momentum pf a falling body of mass jWAirill be mg, which is the dynamica;! measure of the force of gravity acting on the body at that point. W is its statical measure, and we so choose the unit of mass as' to make these equal to each other, or ^ , W W = mg, hence »i = — ;■ or, in other words, the mass of a body is equal to its weight divided by the acceleration due to gravity. If we make' wi = 1, we haver W= g, or in words. The unit of mass is a mass whose Weight is' equal to the acceleration due to gratity, or generally, about 32J times the ttaitf of weights DECOMPOSITION OF FORCES. We have seen that when a force acts on a body in the direction of its motion, the measure of the force is F = m^- If, now, the line of motion of the body (Fig. 1) is AC,- and the force F acts on the body when it has arrived at the point B, the force being inclined to the direction of the body's motion at an angle 0, the effect will be to tifrri the body out of its course, and' if the force continue to act for any length of time the body will describe .a curve. Now tire icffee can be decomposed into' two forces, one actings in the direction of the body's motion, and one perpendicular to its path ; the component in the direction of the body's motion will be F cos & ^ ^Jf^'* "^ com- ponent perpendicular to its direction is F sin 0, and we must find ai 8 DECOMPOSITION OF FORCES. measure for this force in terms of the velocity, and the radius of curvature of the path. To do this we mugt observe, that whatever the path pursued by the body, by drawing the osculatory circle at the point of the path where the body is, we can imagine the body to be moving approximately on the arc of this circle. Let AjAj (Fig. 2) be the portion of this arc described in the time At ,' and let Agi? := v; then (Rankine, Art. 363) chord Are A1A2 = v/lt, chord AjAj = vJt 5 but A2V2 : CA : : vY^ : AjAj .-. A2V2 • AiA2 v^Jt chord vY, = ' CA — r v^ chord ^ ' r arc F sin (5 = wi - Now dVj represents a force which, compounded with Ajf, pro- duces A2V2 ; hence it represents the amount of change in the ve- locity of the body in the time At, approximately, .•. —^ = is approximately the amount of change in velocity per unit of time .•. the true amount of deviation is — , which we obtain by passing to the Kmit ; hence if the mass is m, we have r' Another and more general way to decompose forces is to as- sume, either two rectangular axes in the plabe of the motion, if the path is of single curvature, or, in general, three rectangular ds axes, Ox, Oy, and Oz. We have already seen that i? = -^, and hence vdt = ds. Now this small space, As, can be made the diag- onal of a parallelepiped, whose sides will be Ax, Ay, Az ; hence 1 PI , . .-,■,■, dx dy ^ dz the components 01 the velocity will be -^ > ;^ and -^, respectively, then the components of the acceleration, or the accelerations in UNIFORM AND UNIFORMLY VARYING MOTION. 9 the direction Oa:, Oy and Oz, will be -j^, -j^ and -r^i for these are the limits of the ratios of the increments of velocity to the in- crements of time ; hence if a, ^, and y denote the angles the force makes with Qx, Oy and Oz, we have, s* cos a =^ m -^, s cos p = m—t^, and F cos ;- = t*^- CONSIDERATION OF UNIFORM AND UNIFORMLY VARYING MOTION. We have seen, from considering the laws of motion, that uni- form motion is a state, or condition of equilibrium ; that is, a body which is moving uniformly is either acted on by no force, or by balanced forces. The formulae we have obtained for uniform mo- tion, viz., s ■:= vt, etc., are so simple as to need no farther discussion here. ' . We will now consider uniformly varying motion, and supposing the body to be moving in a straight line under the action of a con- cPs F d^s stant force, F, we shall have, ¥ ^ m -r^ .-. — := —r^ or if F „ ^ cPs rfs „ , , , ^, ^ ^ /. /= W^ •■■ si^fl^ ''"'^'^"^ s = ro« + \ff, as previously developed, where /denotes the actual acceleration of the body per unit of time. As a useful case, we will apply this to falling bodies, and those thrown vertically upward. In either case, the only force acting is that of gravity, and the acceleration due to gravity we represent by g, which is nearly equal to 32^. We shall then have the general formula, h^v^t-\- \gf, and u == gt,^'V~, That is, the vertical height traversed by a falling body in any time t, is v^t + \gt^, where Vo represents the velocity it had at the beginning of the time. If the time be reckoned from when the body is at its highest point, we have v^ = 0, and then h = i^gfi, and i> = 5^*. • 2 10 examples! examples. 1. A Stone is dropped down a precipice, and is heard to strike the bottom in 4 seconds after it started, how deep is the precipice ? Solution. Here h = :^g{\Y = 8(32^) = 257^ feet. 2. How long will a stone, dropped from the top of a precipice, 500 feet deep, take to reach the bottom ? 3. What velocity will the stone, in the first example, have ac- quired when it reaches the bottom of the precipice? 4. What will be that of the stone in the second example ? 5. A body is thrown vertically upward with a velocity of 100 feet per second, to what height will it rise ? Solution, h = Vgt — ^t^ (since gravity acts in a direction opposite to the body's motion) .•. h = lOOt — ^(32^)t% also when it has reached the highest point it must have lost the velocity 100 V = 100 .-. 100 = qt.: t=-^rTr = 3.1088 sec, and A = 100 « — ^(32l)t^ = 155.44 feet. The body will now begin to descend, and, to reach the point from which it started, will take a time equal to 3.1088 sec, and when it reaches that point will have acquired a downward velocity, equal to 100 feet per second again. 6. A body is thrown vertically upwards, and rises to . a height of 50 ft., with what velo /^AC" ■ , /2AB~ /2AC have, AC=,i^«- .-. *' = ^^^■. f.t'^ ^J i^^ = V"^ = i/^ : VACT: .-. < : t'^J^ : VAC = AB : AC, since AC sin d = -yw. Hence the time down the plane, is to the time down the vertical, as the length is to the height. II. From the same formulse, if v denote the velocity at the foot of the plane, i. e., at B, and v' that at C, we have v = ff sin d-t =^ g sin dil — = — z = V2 AB gsinO = ViA.C-g, and V - 9t' = 9^1 (^-f) = V2-AC^ Hence the velocity at the foot of the plane is equal to that at the foot of the vertical. This shows that the velocitiy acquired by a falling body, in fall- ing through a certain height h, is precisely the same whether the path is vertical, or inclined, viz. \/1gh. Hence if the body be obliged to pursue a broken line, or a curved line, in its descent, when it has fallen through a height h, the veloc- ity it has acquired will still be \/1gh. To find the time down a curved line. Suppose a body acted on by gravity to be constrained to move on the curve AC (Fig. 5), in falling from A to P thro ugh a height BJ), the veloc- ity acquired will be \/ig^Yi = \figx, and the space PP' described in the next short interval of time /it, will be v/it, nearly, PP' /1h t= I . This might be at once derived from the formula, CURVILINEAR MOTION. 13 ds v^ ^ ^ v ^ \'2gx, and from this we have a; = g-, as we found before for the height due to the velocity v. CURVILINEAR MOTION. Since a body when acted on by no force, or by balanced forces, will move uniformly, and in a straight line, any deviation, either from a uniform speed, or from a rectilinear direction, must be due to the action of one or more forces. Suppose, first, two uniform motions to be imparted to a body, the resulting motion would be rectilinear and uniform, but differ- ing in direction from both. This would be the case if to a body already moving uniformly, we were to apply suddenly an impulsive force in a different direction from that of the body's motion. A body, on the other hand, with a uniform motion impressed on it, might be acted on by a force in a different direction acting con- stantly. This constantly acting force might be of the same mag- nitude at all points of the body's path, or of varying magnitude ; it might act in parallel lines, or in lines converging to a point, or otherwise. In any of these cases, and also when there is more than one force acting on the body, the resulting motion is curvi- linear, and the velocity generally variable. UNRESISTED PROJECTILE. In the case of an unresisted projectile (i.e., leaving out of ac- count the resistance of the air) we have a body thrown in a cer- tain direction, and thus impressed with a uniform motion in that direction, and acted on constantly by the force of gravity, a force constant in magnitude, and acting in parallel lines. Referring to Fig. 232 of Rankine, the velocity, given to the body in the direction OA, which we shall call ^o, may be con- sidered as made up of the two components, «„ cos d and «„ sin 5, acting in the directions Oa; and O2, respectively. The body 14 UNRKSISTED TROJECTILE. body has thus in a horizontal rlirection a uniform motion ; hence no force acts on it in this direction ; and the equations of motion become d^x d'z d^x d'^z »" S^ = "' ™ dfi = -'"^' °'' d¥ = ^' ^"'^ W^ = -'•'■ Integrating,' and remembering that when i ^ the horizontal and vertical velocities are respectively v^ cos d and v^ sin d, we have dx dz ^ ^ — Vq cos ^ = 0, and -r — z-,, sin = — fft. dx dz , .: -77 = I'o cos p, and j: ^= ''o sm o — gt. Integrating again, and remembering that for i = 0, a; and, z are 0, we have X = v„ cos d'l, and z = f(, sin d-t — -5-. But in order to determine the form of the curvilinear path we must eliminate t from these equations, and we obtain z — X tan a — -^ — „ s~e ; hence the path is a parabola. The velocity at any instant is the resultant of the horizontal and vertical velocities, respectively. .'. v^ = v„ — Iqz :. z = — x :^ -;— ^ 77-. y 2g ig 2(/ Thus the projectile (if thrown upward) at first rises to a height 5-, when it loses all its velocity and begins to descend, and when it has reached the same height as before its velocity is again Vq, as we can see when 2 = 0, v^ — v„^ = 0, or i> = Vg. EXAMPLES. 1. An inclined plane 10 feet long is inclined to the horizon at an angle of 30°, find the time which a body will take to move from the top to the bottom, and the velocity it will have acquired when it arrives there (being acted on by gravity alone). EXAMPLES. 15 Solution. AB = ^^ sin d f .: 20 = g &m 6 t\ (Fig. 6 .) 2 _ 20 _ 20 _ \/20 _ 4.472 _ •■•' ~i(32) — 16 •■• ' ~ "IT ~ 4 ~ .^■■^^^• V = \/2gKE sin — 's/Jgh = v'320 = 4V20 = 17.888. 2. Given in the following right-angled triangle (Fig. 7)^ AB = 10, BAG = 30°. Find the time a body would take whea acted on by gravity alone, to fall through each of the sides respec- tively (AB being vertical). 3. A body acted on by gravity is constrained to move on the' arc of a circle (Fig. 8) from A to C, radius 10 ; find the time of describing the arc (quadrant) and the velocity acquired when it reaches C. 4. An unresisted projectile is thrown upwards at an angle of 45° to the horizon, from a point 10 feet above the surface of the earth, with a velocity of 12 feet per second, find the equation of its path, and the time it will take to reach the earth again. Solution. Assuming the origin at the starting point, equation of path is Z ■=■ X tan — — ' — 32a;2 , x 32»2 x 'M' - 2(12)^^' 'F— V* 144'" -«i/9^ 9" qt"^ We have also z = Wq sin 5 < — -^•, when z^:=- — 10, 3V2 — 10 = 12Vi-f — 16i!= .-. 16<2 — 6V2-if= 10 .-. <2 — -g-< = %^ _ 3V2 /18 160 _ 3V2 j= Vl78 •'• *~ 16 =^ V256 + 256 ~" 16- 5. An unresisted projectile is thrown apwards at aw angle of 37° from the surface of the earth, find the time when it will reach the earth, and the velocity it will acquire when it again reaches the.' earth, the velocity of throwing being 30 feet per second. The student should now read carefully Eankine's Art. 537, and* below I shall give the work, viz., the equation of the path of an' Briresisted projectile is, as we have seen, 16 REVOLVING SIMPLE PENDULUM. qx^ , dz . ^ . ' = tan d-x - j^ji-;;^^, also ^ = ,;„ sm ^ - fft. dx _ . dz _ f„sin d—gt /"^^ V_ ^ r (V'" ^-g'O' V, ," cos" d + rp" sin" g — 2^;„ sin (9f7< + ff''^" 2i;„ sin dgt ■}- gH v^ cos'' S - Vcos'" '^" ' "' = V^^' ^^ ^""''^°'' '^'- ^^*- ^ y ' \dx/ Vg COS S «> L ■ dz ^a; rf's — g ^Sain, ^ — tan e — ^^2 ^^^ g ■: ^2 — ,,^a cos'' '■•^ V "T" rfisM • flfa:" ~ V„ cos (?/ / ' REVOLVING SIMPLE PENDULUM. (Rankine, Art. 539.) The only forces acting on the mass at A (Fig. 9) are the ten- sion on the string P, the centrifugal force Q, and the weight of the mass at W. The first two give a resultant equal and opposite to the weight of the mass at A, as seen in the figure ; moreover, the tri- angle ADE is similar to CAB, and hence we have EC _ AE _ AE A _ W Wv^ AB - ED - "Q" ■■• F - Q ' ^"* Q = ^" h qr , „ .-.-- f^ .-.hv-^gr^. Now when n is the number of turns per second, v := 2/rwr, and hence 4?r^V% = gf^ .: h = . ., ., • EXAMPLES. 1. Given n^= 4, 3, 7, 5, 8, find h in each case. 2. Given A = 4, 7, 30, 29 inches respectively, find n. DEVIATING FORCE. 17 DEVIATING FORCE. We have already seen that a hody acted on by no force, or by balanced forces, moves uniformly and in a straight line, and con- versely ; hence if a body describes a curvilinear path, that body must be acted on by some force, and when the body describes a circular path with a uniform velocity, it is evident that the force acting on it, which causes it to change its direction from that of the tangent, is a force which has no effect in changing its' velocity in the circular path, hence it can have no component in the direc- tion of the tangent, and hence must be wholly normal to the path. It is then called a Deviating Force, as its only effect is to turn the body from its rectilinear course without affecting its velocity. This is equal and opposite to what is callled the centrifugal force, andj from what we have found before, its magnitude is W'b'' WaV WaV „ Q =• = = • Rankme, Art. 540. ^ gr gr g COMPONENTS OF THE DEVIATING FORCE. Since the deviating force Q acts in a direction normal to the tody's path at any point, it may be represented by a line AL in the direction AO (Fig, 10), and this may be decomposed into two' components j one AC, in the direcftiOn AD, and the other AK, in the direction AB, and since the triangle ACL is similar^ to ADO, tve have, AL : AC : CL = AO : AD : OD, or Q : Q, : Qy = r : a; : 2^. ••• Qx = — ) aiid Qy = — ; but Q = — , .-. we have 'rhe negative signs are given because, considering the centrifugal' force as positive, a force directly opposite to it should be accounted' negative. Eankine's seeond. demonstration needs no eomment* 18 STRAIGHT OSCILLATION. STRAIGHT OSCILLATION. Straight Oscillation consists in the oscillation of a body back- wards and forwards in a straight line, Fig. 11. Imagine the body starting at A and moving toward B, acted on by some force which draws it towards the central point C. When it arrives at C the velocity already communicated to it by the action of the force causes it to go on farther, and it would continue to reach farther and farther from C were it not that as soon as it leaves C the force comes again into play, tending to draw it to-wards C, and hence diminishing its velocity,, till when it arrives at B it has lost it all, a^d now the force causes it to return towards C, and thence it goes to A, etc., etc. This force, as a general case, is proportional to its distance from C, that is, at C it is 0, and it has its greatest value at the points A and B, viz., when the body is farthest from C ; or it is equal in every case to x multiplied by a constant, x being the distance from C. Now let us imagine one body moving uniformly in the circumfer- ence of a circle, which is caused, as we have seen, by two deviating forces, — and — ~, at right angles to each other, and then Wa^a; suppose another body to be acted on by the force — alone; imagine both to start together at E, Fig. 12, while the first describes the arc EA, the second would describe EB, for the equation of motion dh: W of the second being m-rj^ = — — a^x, we should have, ^ = — ffl ^ = — «^ cos at. .: -^ = — ar sin ca. .: x = r cos at. Wa^x This force, — - — , has a different value fer- every different posi- tion of the body ; at O it is 0, and it has its greatest value when ar ELLIPTICAL OSCILLATIONS. 19 is greatest, or when x = OE = r ; hence the greatest value of this force IS Q = . 3 For the body revolving in the circle we have, a = 2-n .:— = —; but a = \/^. .: - =2-1/ S-Q ELLIPTICAL OSCILLATIONS. This is given in Art. 543 of Rankine. To derive equations (4), I would refer the student to Straight Oscillation. SIMPLE CIRCULAR PENDULUM. To find the time occupied in a vibration of the simple circular pendulum, we may, as Rankine says, regard the oscillations when the arc is small, as oscillation in a straight line, and then we have, 1 /I , as he deduces, — ^= 2-i/— for the time of a double oscillation, n \ g ' where the quantity n represents the number of turns per second. .'. — is the part of a second, or the number of seconds occupied in one double oscillation, that is, from the time it leaves one extreme point till it returns there. But if the arc of vibration is not very small, it becomes necessary to obtain a more accurate formula. To do this we have merely to refer to the formula developed under the head of Constrained Motion, for the time down any /ds where t represents the time of a single oscillation, viz., the time of passing from A to E, Fig. 13, and where the line DC is taken as axis of X, D being the origin, letting AC = /, BD = h. From the equation of the circle, y'^ = 2lx — x", we have, 20 SIMPLE CIECULAR PENDULUM. dy I — X ds ^ I dx ~~ y " dx ^ y ~~ a/^Ix — x^' /'' Idx /t/ilx — 3? */'2g{h — x) 2/ /.'' dx _ 21 /. " dx V2^J jV/i — X sf-ll — x^/x "" W'^J V^ — ^^ s/-i.l — x) VVJ V/i^^=:^V 2^/ S gJ ^^hx — x\^ W . This cannot be integrated except approximately. Expanding ^1 —^) we have (1 + 4^ + Jf'' ^'''v /J i-/ /^ /•'>/ , a; , 3a;» ., \ dx ' The greatest value of a; is A, and if h is so small that we may x ' omit -VT and all the higher powers, we shall have for an approxi- mate formula, 11 C"^ dx 11 i .,2a;)'' II '=\lg} ,^hx-x^ = V^l — "tIo = "V"^ the same value as was found before for a single oscillation. If, however, the value of h, as compared with I, is too large to render it safe to omit ^, but we can omit the higher powers of X T, we shall have, ,-jLS -. -i2a;- 1 c xdx \^ *-y/-^jversm ^+_J-^^==|^ _ I I S _i2a: , 1/A . -1,2a; .. ^ X* ) ~ V 7 I ^^""^^"^ 'h + Tkj ^^'■'^'^ X -yV/ia:— ar'ij^ )• = n' I / — ( 1 + kt), which is a nearer approximation. Take, however, the formula, t == '^i/~) we can jthus determ- ine the value of t when I is given, or of I when < is given. EXAMPLE. 21 EXAMPLE. 1. Find the length of the simple circular pendulum which is to beat seconds at a place where g = 32J-, Solution t — t: S, — .: f = tt^— .:l = -f .-. 9 32* ^= (3.1416)^ = 3-2591 feet. 2. What is the time of vibration of a simple circular pendu- lum 5 feet long ? 3. What is the time of vibration of a simple circular pendulum 5 inches long ? 4. How long must a simple circular pendulum be which is to beat once in J^ a second? in \ second? in 3 seconds? in 5 seconds? in 10 seconds? SIMPLE CYCLOIDAL PENDULUM. The equation of the cycloid is ^ = a versin ~" — + {iax - x^y' adx _i_ ("^ — x)dx (2a - x)dx jia-x "^ ^iax — x-' ~ V2aa;-a;' ~ V "~S~'^^ V2aa: — x^ i^iax — x^ 3f _ 2a — a; _ ^ ■ ■ rfo;^ X " dx^ X ^_2a-x , d^_2a .^^{^±y,,, /ds -Tr======, becomes Via f dx _ (a\\{ . _i 2x ) »■ /o. The expression is independent of h, so that the time of vibra- tion is the same whether the arc be large or small. A body can be made to vibrate in a cycloidal arc by suspending it by a flexible string between two cycloidal cheeks (Fig. 14) ; this is shown from the fact that the evolute of the cycloid is another 22 SIMPLE CTCLOIDAL PENDULUM. cycloid. To prove this we have, from the equation of the cycloid, . -1 a; , ,„ ,,1 dy /2a — x y = a versin - + (2ax — x')h ^ = y — ^— . t / — ' —^. =: — :; — -,=== •'■ the radius of curvature, V a;' rfa;» xWia — x /■ w>! :M-^ _x§_^ MM _ 2(2«) ^ And we have for the evolute, this relation ds' ^= dp .-. s: == I rf/o, «' being the length of the arc of the evolute, and p the radius of curvature of the original curve. 2a and are the limits of x ; and observing that when x = 2a, /> = 0, we have, s' = p .-. ,' = 2(2a)*V'2a — SB, or if we transform coordinates to B, we have «' = 2(2aa:)^ .-. «'= = 2,ax, which is the equation of another cycloid just equal to the first. This motion along a vertical cycloid may also be obtained by letting a body mov^ along a groove in the form of a cycloid acted on by gravity alpne, and in this case the time of descent of the body to the lowept point O, is precisely the same whether the body is placed either at A, B, C, or any other point of the curve. When the body has reached the lowest point, it has of course acquired suflScignt velocity to cause it to ascend on the other part- of the curve tg a height equal to that from which it started. The converge of the above is true, viz., that a body in pursuing a path such that the time of descent from any position to the low- est point shall be the same, must pursue the arc of a cycloid. The force {icting on the body at any point is the force of gravity acting vertically ; this can be resolved into a tangential and a nor- BESIDUAL FOKCES. 2? mal component, the normal component is resisted by the tension of the string, and the tangential force is g cos 0, where is the angle between the tangent to the curve and the axis of x, dx dx ^ "^^da? + dy'' ^ds ' Now the tangential force can be shown by the Calculus of Varia- tions to vary as the length of the arc from the lowest point, 1"= K«, E being a constant, = ^a;, or s" .: K« = ^r^.-. Ksrfs =^ ffdx .: -^ — gx .: which is the equation of the cycloid. RESIDUAL FOKCES. Rankine, in Art. 546, speaks of Residual Forces ; the result of his reasoning is to show that the weight of a body is not absolutely a measure of the force of gravity that acts on it, for the body is- acted on first by the force of gravity tending to draw it towards- the earth, and secondly, by the centrifugal force which it has in- consequence of its rotation with the earth around the centre of th» earth, which acts contrary to its weight, and hence the weight of » body is really the difference between these two, WOEK. All forces act for a certain amount of time ; thus ill driving si nail with a hammer, there must have been a time when the naiB was half Way in, etc., and hence whenever a force pioduces motion' instead of pressure, it continues to act en the body for a certain time, and through a certain space. If a resistance is overcome through a certain space, We call it so much Work done. Thus if 5 pounds is to be raised through ai height of 3 feet, a certain amount ef force is required, and a cer- tain amount of work is to be done. It is plain that to raise a 6 pound weight through a height of ff feet requires that 3 times as much work should be done as if it- 24: WORK. were to be raised through a height of 1 foot. So, likewise, in 'the same case, 5 times as much work is done as would be in raising a 1 poui;d weight 3 feet. The most common Unit used in measuring Work is that which is performed in raising one pound through a height of one foot, and Is called a fooi pound. In the case of raising a weight, the resistance overcome is grafv- ity, but work may be done against other resistances, as, for in- stance, the work of a locomotive in moving a train along a level track ; here the resistance is friction, and also the resistance of the air. A horse dragging a carriage along the ground, all kinds of tnachinery moved by steam or water, or any other motive power; toills, etc., all require work to be performed to make them fulfill their respective functions. EXAMPLES. 1. The ram of a pile driver weighs 1200 lbs., and has a fall of l5 feet, how much work is performed in raising it? Solution. Work perfortoed = 1200 X 15 = 18,000 foot pounds. 2. How much work is performed by a steam engine in raising 12,000 lbs. of water to a reservoir situated 30 feet above the free upper level? 3. How much work is performed in ra,ising a weight of 650 lbs. through a height of 35 feet ? WOSK Witil KEFEJRENCE TO TIMfi. Thus far -We have considered work independently of the time' Consumed in performing it, but in machinery we must, in most cases, take into account the time ; thus one machine may be capa-" ble of performing 4600 foot pounds of work per minute, and another can perform the same in \ that time. The amount of work generally taken as a unit in machines is what is known as a horse power, and is eq,ual to 33,000 foot pounds per minute ; henee. EXAMPLES. 25 a steam engine capable of raising 16,500 pounds of ore from a mine 40 feet deep in 5 minutes, would in one minute raise 3,300 lbs., and would hence perform 3,300 X 40 = 4(33,000) foot pounds of work per minute, and hence it is an engine of 4 horse power. EXAMPLES. 1. How many horse power would be required to supply a city of 5,000 inhabitants with water, each one using 10 gallons of wa- ter per day, the engihe working 6 hours per day, and the water being raised through a height of 100 feet ? Weight of water = 62 lbs. per cubic footi 2. Required the horse power of an engine which is to give power to 5 tilt hammers, each of which weighs 500 lbs.-, and tnakes 60 lifts per minute, the perpendicular height of each lift being 2 feet? 3. HoW many hours per day Would it be necessary to work an engine of 36 horse power to supply a city of 35,000 inhabitants with waterj each inhabitant using 6 gallons, the water being lifted through a height of 25 feet ? 4. How many gallons of Water can be pumped from ft mine by a 10 horse poWer engine in 11 hours, the depth of the mine being 100 feet? 5. What is the rate in tniles per hour of a train of cars whose Weight is 70 tons, the friction being 7 lbs. to the ton, drawn by an engine of 50 horse power ? We have seen by the first laW of InotioQ- that if a body is in Inotion it Will continue to more in the same direction, and with the same speed, unless and until it is acted on by sofiie external force. We have also seen that, strictly speaking, an impulsite force doest not occur in ' hature, but that every force acts foi- sotne definite time, and hence that When it produces Jnotion it acts through ft certain spacer From these facts we conclude, 1st, that a force when unresisted produces change of velocity, siftee uniform motion is a eonditioti 26 EXAMPLES. of equilibrium of a body. 2d, suppose a force to act on a body- while the body moves through a certain space, it produces a certain velocity which the body retains until it is opposed by some exter- nal force (resistance), and to destroy this velocity an equal amount of worlc must be done to what was done in imparting it to the hod3' ; hence a body which has had a certain amount of work im- pressed upon it is, in its turn, capable of performing that amount of work on another body. Again, suppose the body to overcome a resistance which re- quires a less amount of work than the amount impressed on the body, then the body will continue to move with a less velocity, and will still be capable of performing an amount of work equal to the difference between the amount impressed on it, and the amount of work it has performed. This, in other words, is that the energy exerted (i. e., the amount of energy lost) is equal to the work performed. This is called the principle of Conservation of Energy, and shows us that a machine is not capable of generating, but only of transferring energy, Rankine proves this algebraically in Art. 518. This principle of the Conservation of Energy is very well illus- trated; in the hydraulic press (of which Fig. 17 represents a sec- tion), where the product of the power by the distance through which it moves is equal to the product of the weight by the dis' tance through which it is raised. Energy is, then. Capability for performing Work, Energy is of two kinds. Actual and Potential. To illustrate the 'two, imagine the ram of a pile driver weighing' 1 ton, and having a fall of 15 feet, to have reached a point 5 feet from its point of starting (the top). If it were now to meet the head of the pile it would exert on it 2,000 X 5 = 10,000 foot pounds ; this is its Actual Energy, but it has still 10 feet more ta fall, and if allowed to fall these remaining 10 feet before coming in contact with the head of the pile, it would exert on it 2,000 X 10 = 20,000 foot pounds more. • This latter is its Potential Energy. CONSERVATION OP ENEEGT. 27 Thus Potential Energy is the force multiplied by the distance through which its point of application is capable of being moved. If the force varies at diiferent points of the path, we must mul- tiply each diiferent value of the force by the distance through which its point of application is moved. If it varies at every point we then pass to the limit, and we have /Pcfe. EXAMPLES. 1. Given a force F = ax -\- hx^, find the amount of work done by this force acting through a distance, x =■ a, from the origin. 2. Given P = Vr" — x^, find the amount of work done while its point of application moves through a distance r from 0, on the axis Ox. CONSERVATION OP ENEKGY. - The principle of the Conservation of Energy just enunciated and proved is of the greatest importance in machinery. To fix our thoughts, let us suppose we have a 4 horse power steam en- gine ; the force developed by the engine would be capable of doing 4(33,000) = 132,000 foot pounds of work per minute ; this work causes the first wheel to move, and since this wheel has impressed upon it a force of 4 horse power, it is itself capable of exerting a force of 4 horse power, and this it transfers to its neighbor, that to' the next, and so on ; hence we should expect from the final work- ing piece of the machine an amount of work equal to 4 horse power, and this would be the case were it not for the friction, the weight of the different parts that have to be lifted, and other re- sistances which require that a certain amount of work should be done to overcome them ; hence we never get practically from a machine as much effective work as we impress upon it. What has been said shows that energy is never really lost, but can merely be transferred from one body to another ; hence a body on which work has been performed has so much energy stored, and is in its turn capable of performing that amount of work. 28 VIRTUAL VELOCITIES. VIRTUAL VELOCITIES. The principle of the conservation of energy may be (and is) used to determine the conditions of equilibrium of forces applied to any connected system of points ; it is then known as the prin- ciple of Virtual Velocities. This principle of virtual velocities could be used to demonstrate the parallelogram of forces, and hence we could build on it the whole of Statics. To apply it we imagine the system to move in any way consist- ent with the equilibrium of the system. Thus take, as an exam- ple, a lever of the 1st order, i. e., a bar used to raise a weight. Suppose the lever. Fig. 18, to be at first in the position AE, the power P being applied at A, and the weight at E, when the lever has been caused to take the position BD, the power P has acted through! a height AB ,'. P'AB is the work done in a vertical direc- tion, the weight W has been raised through a height DE ; hence the work done in overcoming the weight is W'DE ; hence P.AP = W.DE...P = Wjjbutjl = ^^ ... P = W^|. PARALLELOGRAM OF FORCES. We will now prove the proposition of the parallelogram of ^rces by means of virtual vplocities. Defimtion. Sjappose the point of application A of the fosce P, Fig. 19, to suffep a small displacement by which it is moved toA4 fhe force P keeping parallel to its former position. If from A' we drop a perpendicular on the first direction of the force, the dis- tence Am is called the virtual velocity of the force P, and hence, virtual velocity of P = Am = AA' cos 6. If q, set of forces are in equilibrium, the total amount of work done by them must be 0, or .2' |??» = 0. Consider three forces, P, Q, and K, Fig. 20, originally applied at A, which point has moved (io A', we have, P-Am -f QA>i -f RAjB = 0. THE SCEEW. 29 Let PAR = ^, QAE = a, PAQ = y ; also A'AP = ; then P-AA' cos d + Q-AA' cos ((? + 7) + R-AA' cos (fj — tf) = : or P cos S + Q cos cos y — Q sin 5 sin 7 + R cos ^ cos d + R sin ^ sin S = ; or (P + Q cos J- + R cos ^)cos (? + (R sin ^ — Q sin|3')sin 5 = 0; but since is arbitrary, by the principle of Indeterminate CoeflS- cient, P + Q cos 7 + R cos ^ := 0, and R sin ^ — Q sin 7 = 0, .-. R cos ^ = — (P + Q cos 7). (1.) R sin j3 = Q sin 7. (2.) . ^ — _Q- = p * ' sin 7 sin ^. sin a and by squaring and adding (1.) and (2.), R2 = P2 + Q2 -f 2PQ cos 7, which proves the proposition of the parallelogram of forces. APPLICATIONS OF THE PRWCIPLE OF VIRTUAL VELOCITIES. THE SCREW. To find the relation between the power and weight in the screw. Let h = the pitch of the screw, Fig. 21, and suppose the power applied at B, the work done by the power in one revolution, is P'2ff>' ; the weight will thus be raised Ihrough a height h ; hence the work done is also = WA .•• P-27rr = WA .-. P = Wa — THE WHEEL AND AXLE. Let a be the radius of the wheel on which the power acts, and h that of the axle on which the weight acts. In one revolution the power moves through 27;a, and the weight through 2n& .-. P-2ffa = W-gffS •• Pa = WJ .-. P = W— • Note on Art. 520, Rankine. The student must take notice that . the force F is a force acting on the body while it has the given motion ; then F cos a denotes the component of the force in the 30 THE WHEEL AND AXLE. direction Ox, and Js cos '/ the component of the motion, so that the product of these two denotes the work done by the force in the direction Oa;. Note on Art. 524. To make it plain, let us take three bodies (Fig. 22) of mass, m, mi and Mg, respectively, at distances x, x^ and X2, from Oy. Sup- pose in one second they move as in the figure, so that x', x{ and x^, are their distances from Qy at the end of the motion, we have then, Xfi being the distance of their centre of gravity from Oy, the formulae, mx -\- THiXi -\- m^Xfi 2mx " m -\- m-i -\- m^ Zm^ , mxf + OT,a!i' + m.^^ i:mx' ° m -{■ m,i -\- m^ 2^m ' ,_^ _ w(!/ — x)-\- OTi(a;/ — a-;) + >«2(a;/ — x^) _ " " m -\- nil -\- 1^2 2fm{^ — X) but a:' — a; = M, «/ — 3:1= Mi, «/ — x^ = u^, x^ — a3„ = m„. 2mu .: Mo == -?: — .". Un2!m = 21 mu. " 2,m " So also Vi^-2m = 2mv, and w^2m = 2mw, Hence if we can ascertain the motion of a' body's centre of gravity, and we know its total mass, we can ascertain its total momentum. In connection with what precedes, Rankine, Chapter II., through Art. 528, should be studied. ACTUAL ENERGY. If a body, whose weight is W, fall through a height h, the force of gravity has impressed on it an energy WA, which it is in its turn capable of exerting on another body ; thus if it were to fall into the pan A, Fig. 23, suspended from a string passing over a wheel, at the other eud of which is another weight W, the impetus COMPONENTS OP ACTUAL ENERGY. 31 of its fall would cause the second weight to be lifted through a height h. Now in falling through a height h, it acquires a velocity V = V 2(/A .-. A ^ K~ ; hence the actual energy of the falling body is WA = -^ = -^ ■ Now it matters not whether the direction of its motion is verti- cal or not, if it have a velocity v imparted to it. The amount of energy impressed on it is -s", and hence its actual energy is -ip = ^ its mass multiplied by the square of the velocity it has ac- quired. COMPONENTS OF ACTUAL ENERGY. Since a body's actual energy is -g— , m whatever direction the body is moving, whether the path be straight or curved when the body's velocity is v, if we decompose the velocity into 3 compo^ nents, in the directions Oaj, Oy and 0«, respectively, these com-- dx dy dz . . , .. ponents Will ^^ '^ f '^ '' 'Jt^ respectively (see my notes, p. 8) ;. dx hence since the velocity in the direction Ox is -^, the body's ac tual energy m that direction is 2\dS) ' ®° ^^ ^^^ direction Oj^it is' 1(1)- »a . 0. =(|y. ..a ...ce (§)■+(!) +(J)'= ^ •-■2{-di) + -2\i) + 2[di) = ^' °' '^^ ""^'^^l ^""^'^ of the body is equal to the sum of the components of the actual energy, in the directions Ox, Oy and Oz. 32 DYNAMICAC MEASURE OP FORCES. RELATION OF ACTUAL AND POTENTIAL ENERGY. EnebgT Stored and Restored. If a body have a velocity V, its actual energy is, as we have seen^ Wv^ . , -g — = Wh, and if it then move to a point where its velocity is v ^ its actual energy becomes • ,j . =s WA' (A and h' being respec-" tively the heights due to the velocities v and *') ; hence the in* w crease of actual energy when m' >■ » is o~('t>'^ ■=— v^), or the loss of W actual energy trhen «' •< « is g-Cf* — «'*)• The 1st W W ^(„'2 _ „2; — W(h' -=-- h), and the 2d, ^(v^ - v"") = W{h - h')\ and this must be equal to the product of the force acting on it by the distance throtigh which its point of £(ppIication has been moved, or in other wwds, to the Potential Energy exe*ted. If, as in the first case. Potential Energy is changed into Actuail Energy, the velocity being increased, the amotmt of actual energy thus gained is said to be stored, and the body is able to do so much more work, but in the Second case, when Actual Enfergy is con* terted back into Potential Energy, it is said to be restoredi IrtNAMCAL MEASURE OE FORCES. It was fof a long time a subject of dispute whether a force waS to be measured by (was proportional to) its momentum (mv), or by the actual energy it had imparted to a body, some contending for the first, saying that to impart to a body a double or treble mo- mentum required a double Or treble force, and the advocates of the latter contending that to impart a double or treble energy re-" quired a double or treble force. The misunderstanding arose from the fact that the disputants did not examine whether they referred their forces to equal times ENERGY DUE TO OBLIQUE FOKCE. 33 or equal spaces, and if this consideration be made we shall find that the two measures are equivalent ; for, to impart to a body a double momentum in the same time requires a double force, and to give it a double actual energy (power of performing work, as, for instance, to lift a double weight) while moving through the same space, requires twice the force, and indeed -y) = '^^r^ for dt as d{-2-) = OT„* ; but ds = vdt .: '^Kt) = m'^f, and ds — -. — dt ds ds d{mv) dv d( — -) d(mv) dt ^ ""dt •■• -AAZ = dt '- ds that is, the increase of actual energy per unit of space is equal to the increase of momentum per unit of time. ENERGY DUE TO OBLIQUE FORCE. To illustrate this take the case of a body moving down an inclined plane (Fig. 24). Suppose a body of weight Vf to start at the top of the plane and slide down, acted on by gravity alone. Here the only force acting is W, and this acts vertically, making with the direction of the body's motion an angle, 1 — d, hence the component of the force in this direction is W cos(„ — O) = "W sin e, and this is the only component that is eifective in imparting energy to the body (the other com- ponent W sin (^j — e 1 = W cos d, being counteracted by the re- action of the plane), hence the actual energy acquired in moving from B to A is W sin S-BA, Take another view of it ; the velocity acquired in falling through 34 EXAMPLES. a height BC is t; = V(2^-BC), and then the actual energy of the body is mv^ Wv^ W20-BC „_^ ^^72 = 72 = ^-^C = ^ "° ''■BA, the same as before. Thus if a force F act on a body, and is inclined at an angle 6 to the direction of the body's motion, the component of this force that is effective in producing motion, or in imparting energy to the body, is F cos 0, and the energy imparted by it in moving through a distance ds is F cos ds .•. the total actual energy imparted by the force will be /F cos d ds. EXAMPLES. What is the, actual energy acquired by a body in falling through' the arc of a circle acted on by gravity alone (Fig. 25) ? 1st iSolution., The force, acting is W, of -syhieh' the' xomponeiit ^, ... ■. /tt a ■■• ■ V .■■'•--■•■* ■ ' effective m generating energy is W cos 1^ — 6), and (:r \ rfy X /tt V « equation of circle being x-'+f= r^ .-. cos (^—9) =1 ••• W cos (|— «) Wy W , , „ „, /' W ' r — V(»'? — x^)ds; butrf»= -^dx .: r —yds = f Wdx — I WcB l^' = Wr. . , 2,d S<)lution. «,== */(2ffh) == V (25'r) .-. Actual energy = Wv^ _ W2gr _ • 2, ;• Wii^t .is,the .actual energy acquired, by a body in falling' through a quadrant of an ellipse (Fig. 26) ? ' .3. What ' is the actual energy acquired, in falling througE- the same arc ■when inverted, as in Fig. 27 ? " -■ • ' ■ >. • 4. What is the "actual ^nei-gy acquired im falling through the inverted arc of a semicycloid ? TOTAL ENERGY. 35 . 5. What is the actual energy acquired by a body undergoing sti'aight oscillation, as described in Art. 542 of Rankine, when it arrives at the centre of the force ? Solution. F=^^ ,. f Fds = f Ydx = f^^^ ff Jo Jo Jog = / xax = — s 9 J ^ff 6. A body moving in the line Ox, and starting from O, is acted on by a force F = ax -|- bx\ find its actual energy when it has reached a point at a distance c from O. TOTAL ENERGY. The actual and potential energy of a body, when added together, give its Total Energy, and this remains unchanged as long as the body is acted on by a reciprocating force. Thus in the case of the ram of the pile driver, described on page^, its actual energy, at a point 5 feet from its point of start- ing is 10,000 foot, pounds, while its potential energy at the same point with 'reference to t;he earth ig 20,000 foot pounds ; these two added together give its total energy, which is the same at all points of its path. Bankine, Ex. Ill, page 504. (Hefer to figure on page 493.) Instead of equation 6, we should have the following, Wv' , f'Qixdx , , Wv" , QtX* w + J »~^r ^ ^^^ '°'^^ ^"^'^^ ^ 'W + ^ ' but from 4 and 5, Art. 541, Eankine, x = r cos at = x^ cos at, dx . . . , and " ^= ;// ^^ — '*'" sin at = — ax^ sin at = — Vf, sm' at; hence by substitution we have, W©" , f^'QiXclx WV . , , QiiBi" cos" a« 2ff ' J Xi 2g . ' 2a!i — ~ — sm" at + —5— COS'' at ; but -w— = —a~ •• Total energy = ~2f + '21^ ^ -l-(sm» at + cos- aO ^ -3- ' Note on Art. 558, Rankine. 3 6 ANGUJtAE MOMENTUM. We have seen by the Third Law of Motion that momentum^ cannot be generated in one body without generating an equal and opposite momentum in another ; hence any momentum caused by an action between two bodies of the same system is neutralized by an equal and opposite momentum, generated in the other body of the system .•. Q. E. D. Art. 559, now follows at once, since by Art, 524 the resultant momentum is = the product of the total mass by the velocity of its centre of gravity. > ANGULAR MOMENTUM. To illustrate angular momentum, let us imagine a wheel revolv- ing on an axis (Fig. 28). Suppose a particle of mass m, at a dis- tance I from the axis of revolution, to be revolving around this axis with a velocity v ; then the force acting on this particle (its "W momentum) is mv, and its moment is mvl = ^-vl. 9 This angular momentum is a couple, and hence can be repre- sented by its moment axis, i. e., by a Hue perpendicular J;o its plane and equal to this angular momentum. Again, since v is the velocity (distance passed over in a unit of time), and I the radius of revolution .-. vl ^= twice the area of the sector described by this particle .•. the Angular Momentum is equal to the product of the mass by the double area swept through by its radius vector. The same proposition is applicable to a body which describes any other curve than a circle, for by taking a sufficiently small portion we may- consider it as part of a circle. ANGULAR IMPULSE. By Angular Impulse is meant the moment of the force which imparts the rotary motion to the body about the axis of revolution and this must be proportional to the amount of angular momentum it imparts, just as we have seen that forces are proportional to the amounts of momentum they impart in the same time .•. Vldt =¥ mldv. ACTUAL ENERGY OP A SYSTEM OP BODIES. ^7 The reasoning used to prove Art. 563 is precisely similar to that used for Art. 558. ACTUAL ENERGY OF A SYSTEM OF BODIES. lu studying Art. 564 of Raukine, we must bear in mind that all our ideas of motion are relative, and that the velocity of a body in motion depends altogether on what we regard as our fixed point from which we estimate its motion. Thus if the body has a veloc- ity V with reference to an external point, and the centre of gravity of the system to which the body belongs has a velocity v' in refer- ence to the same external point, and in the same direction, its velocity, with reference to the centre of gravity of the system, will hev — v'. The quantity ^(u^^ + v^^ + w„^)l'm is the external actual energy, or that due to the motion of its centre of gravity. ^I!m{u'^ + v'^ + «'°),is its internal actual energy, or that due- to the velocity of the body with reference to the centre of gravity of the system. The mutual actions of the bodies can only change „ , „ , . . 2nm the internal energy of, the- system, tor the quantities- t«o = .y. > Q}„ = -= — , v!„ = -s; — , have been shown to remain unaffected " 2, m 2i m by the mutual actions of the bodiegi (Art. 559, Bankine). This relation can also be derived from Art. 524 of Rankine, for when the centre of gravity of the system is the point of reference, ifj = ^j =2 w„ = 0, and .•. .2mu' = 0, 1!mv' = 0, and 2mw' = 0. .COLLISION. Note on Art. 566 of Kankine. The student should read in connection with Art. 566 of Ran- kine, his addendum on page 512. If TWi and mj be the masses of the bodies, and Mi, Mj their veloc- ities before the collision, the two form a system whose centre of gravity hus.a velocity «„ = .;^^ - -^-f";;,,,- " 38 COLLISION. Then the external actual energy is (mi + Jraa) 2 • The velocities in reference to the centre of gravity are % — Mo and Ma — Mo /. the internal actual energy is OTi( M; — UqY m^ju^ — Mq)' 2 + .2 If the energy were consumed in producing internal vibrations and heat, the velocities relative to the centre of gravity would be — - (mi — Mq) = Mj — Ml and — (mj — m„) = m,, — Mj, and hence those relative to the earth would be "■Wi = "o + ("o — Ml) «2 = Mo + (Mo — M2) or «i ^ 2M|, — Ml 1)2 =: 2M(| — Mj, etc., etc. Note on Art. 567, Rankine. If X, y and « are the rectangular coordinates of a body's posi- , ds . . , . ' ], dx dy tion, we have already seen that , is its velocity, and that -tti f.i dz . ' . . . ':■ ■■•■■■ -32, are its components in the directions Oa;, Oy and Oz, respectively, , , - . dx dy _ dz hence the components ot its momentum are m,., m-j, and w it . . . , . d-'x, d^ ^ d^z ' their rates of variation being m ,.2, m^-f , and m ,j- The components of the velocity which tend to turn the body 1 . ,-> ^y ■ , ■. dz about the axis Kjx are i., with au arm 2, and , , with an arm. y acting in the opposite direction ; hence the angular momentum / dii dz\ about « IS rn\zdt—ydt)=''« , . ( dz dx\ about y IS m\x^^ - z^) = v„ , . / dx dy\ about z IS m[y-^ — x-^-) = «,. Their rates of variation in a unit of time are dv^ ( d^y dz dy ^z dydz\ / d'^v d^z\ -dt = "Xdt^ ^ -dtdu-yiu^-duTt)= ""VdP - m), PaiNCIPLE OF d'alembeet. 39 rfyy .. ( dH dxdz d^x di:dx\ / d^z d^x\ -dt = '"i^rfi^ - di dt-'dt^ - dt dt)="'\''df -'d^y dv, ( d^ dydx dhi dxdy\ / d^x dW n = n^6ft» + dtdt-'^dt^ - dtdl) = '^\ydt^-'''-S& Hence follow the equations which Rankine develops. PRINCIPLE OF D'ALEMBERT. If we imagine any train of Mechanism in any way connected to which a set of external forces is applied, these external forces pro- duce certain motions on being combined with the internal forces, which are brought into play by the motions of the parts, these motions are ■ not the direct results of the external applied forces, but if a set of forces which would counteract these motions are applied to the machine, they stop the motion, and hence would counteract the eflFect of the external forces, or be in equilibrium with them. The forces which would directly produce these same motions are called the effective forces. Now it is evident that the internal force between any two of the parts is equal and opposite to the difference between the applied external force aiiidtlie effect- ive force, at that "point. ., We are' thus enabled io determine ' the internal forces acting be- tween the parts of the system. , The components of one of the internal forces will therefore be . : ' .d^x ^ dPy „ d-'z ' _, ■ Rankine, Art. 568. Rankine, Art, 569, should, now be read carefully. ROTATIONS OF KIGID BODIES. Suppose, a rigid body (Fig. 29) to revolve about an axis per- pendicular to the plane of the paper, and passing through O, im- agine a, particle of weight^W situated at a perpendicular .distance, OA = r, from the axis of rotation, and let the angular velocity of the body be a, then, the particle at A will, in a unit of time, de- 40 RADIUS OP GYRATION. scribe an arc' AB = ar ; hence its momentum, which is a measure W of the force that has imparted to it this velocity, is "T"'"') ^^^ ^^ moment of this force, or, in other words, its angular momentum . /W \ W „_ Wr^ with respect to the axis of rotation, is l^—ar)r= "r"'" — " ~a'' The total angular momentum of the body is, consequently, •Wr" a _ „ 9,9 the quantity Z Wr^ is called the moment of inertia of the body with reference to this axis of rotation, and when multiplied by — gives us its angular momentum with reference to the same axis. This illustration will suffice, I think, to make the stnc^ent under- stand the application of the Moment of Inertia of a body with reference to an axis. RADIUS OF GYEATIOBT. Another Definition, The radius of gyration of a body with respect to an axis, is the perpendicular distance from the axis of that point at which, if its .whole mass were concentrated, the angu- lar momentum, and hence the moment of inertia, would remain unaltered. If /) ±=: the Radius of Gyration, the Moment of Inertia would be l^IVf, if the mass were all concentrated at this point ; hence we: must have, i<^l W = ZWr^. which gives the same result as Art. 674 of Eankine. EXAMPLES OF MOMENTS OF INERTIA. The results are on page 518 of Eankine. 1. Find the moment of inertia of a sphere about its diameter. Consider the spTiere as made up of a series of thin plates, all horizontal, i. e-, parallel to the plane xy (Fig, 30). Let one of EXAMPLES OF MOMENT OF INEETIA. 41 these plates be situated at a distance OA above the plane xy, then will AB be its radius. From what we have learned previously, we know that the mo- ment of inertia of the circle (radius AB) is, ?rAB< _ 7r(0B' — OA")" _ 7:{a^—z^Y 2 ~ f ~ 2 ' .-. I = ^wr" = wS^{a^ — z^'^dz = w^ P (a" — z«)"tfe 2. Spheroid of revolution. Let ABFG (Fig. 31) be the ellipse, which by its revolution around OZ generates the spheroid ; let OB = a, OA = r. Equation of ellipse will be a-" z" r' ^ + J = 1 .-. x« = ^(«"-.»). ■ Imagine, as in the previous example, the spheroid to be formed of a series of circular horizontal plates, one of which is situated at a height OC = z, above th^ horizontal plane ; the moment of iner- tia of the circle described by the revolution of CD will be w |(CD)* = «. ^a^ = w^(a'' — s")" ; hence the Moment of Inertia of the whole spheroid is I = ^«,." = Jl^(a' - -Vdz = ^Jl (a" - .")" Va Area of circle = ttCD" = nx\= ^(os* — s") /» Trr" ^ _ I _ 2>^ P — W ~ 5 ■ 42 EXAMPLES. 3. Ellipsoid. Equation of Ellipsoid is, i' _L ^ j_ ?1 - 1 P ^ c'' ^ a'" ^■ Imagine it divided as before, into horizontal plates of elliptical shape, one of which is at a distance OA from O (Fig- 32). Equation of ellipse in plane XZ will be p + 5 - 1 .-. AB= =x^=^ 5(a^ - «^) .-. AB = V («^ - z'). Equation of ellipse in plane YZ will be J' + J = 1 .-. AC'=f= '^,(a^-z^).:AC=^V{a^-z^). Hence (Eankine, Art. 95) Moment of Inertia of Ellipse BCD about axis OZ is ^•(AB" ■ AC + AB-AC) = ^ (AB^ + AC^) = ^ i^-W = -T5-(*' + «')• _^ 7rAB-AC-rf» = TMJ ^ -2(a2 — z'')dz = —^ g i-wabc 2 _ ■'^ _ ^^ + "" = —3— ■■■ P — W ~ 5 4. Spherical shell. W and I will both be obtained by subtract- ing their values for the internal from those for the external sphere. _ 8-w(r° — r'°) _ A-w(r^ — r^») ^ _ _I — 2(>-°-/g) •■•■^~ 15 '*^~ 3 "° ~ "W "~ 5 (/•"-/»)' 5. r' _ /» = (r — r') (r^ + r/ + r'^) = cfr(3r'') nearly, ys _ /5 _ (^ _ ^z) (^4 _|_ ^8/ _^ ^2^/2 _|_ ^/3 _^ ^/4^ _ ^^(5^4^ nearly. These results are obtained by making r = r'. Now substitute in 4, and we have, ^ $-wr*dr ^ . „, „ 2r^ EXAMPLES. 43 6. Circular cylinder. Let (Fig. 33) OA = a, OB = r. Let, as before, the cylinder be divided into a set of thin horizontal circular plates, one of which is situated at a height OC =^ s, above O. Moment of inertia of this plate is w-^dz. W = _ w / Tzr^dz = 2i:wcer^ .: f? ^^ -„• 7. In the Elliptic cylinder the moment of inertia of one of the plates is w^(J2+c2)rf2, •. I = nwbc(b^ -\- c^) r" nwabc 4 /* , Tcwabc „ , ^rf.= -2-(S= + c=). h^ + c^ W = 2w7zbca .: p^ = — j 8. Found by subtraction. 9. Found from 8, like 5. ^ 10. Circular Cylinder (Fig. 34) about the axis OZ ; let OD = r, OE == a. Our horizontal sections will now be rectangles, and the moment of inertia will be %wAC-AB{AW + AC''). Now AC^= OG" — OA? == r^ — 8^ .•. Moment of inertia of plate is ^waV(r* — z^) (a^ + r^ — e"), hence I = f |waV(r« — z^ \ {a^ + r^) — z'ldz = ^wa(a^ + r^) f V(r« — z'')dz — ^waJ_^zW(r^ — z^)dz ■=\wa \\{f - z^)^+(a'+r^)JV(r^-z^)dz-'^JV(r''-^)dz^^ «' + tM = — 4 - I r- , a^ W - 27:war^ ,. f,^ = ^ ^ j + j. 44 EXAMPLES. 11. If the circle becomes an ellipse, we shall have, „2 ,2 wac AC = y, and ^ + |, = 1 .-. AC = |V(6^ - a"). •( c' 12. Found by subtraction. 13. Deduced from 12, like 5. _^ %whc{l^ + e^)dz = — g- (i^ + c'). 15. We must first find the moment of inertia of a rhombus. To do this we have (Fig. 35), y _ * _ *, ^ = - .*. V ^= ~(c — X) c — X c ^ c^ ' . 26 /»° ihc > .-. I = ff^dxdy =1,- / *^(c - x)dx = /^(g^59r ^^,fe>tK'e rt^^BIcl5risa»?v3c:L ^ ^4* ^ W ' 16. OD = 6 (Fig. 36), AC — o, OE = c, .-. BA = y. , Moment of inertia of rectangle is »AB-AC(AB' + AC) = J J (S - e) } p(J _ s). + ^ j. = mi-- ')•+%'('-') _ 2wa&c(c^ + 2a') ~ 3 I c" a' W = iwabc .: p' = ^ = ^ + -g-. CENTRE OF PERCUSSION. 45 Arts. 575 and 576 of Rankine are perfectly cIcJar, and. should be compared with Art. 95. Note on Art. 576, Cor. I. From Equation 2 of Art. 576, I = r,^2:W + I„, but I = p»^W ; I„ = /,'J:W .-. p'i:W = r.^sW + p.^zW ,. ,.^ = r/ + t;\ which shows that p is the hypothenuse of a right triangle, of which j-Q and p^ are the sides. Cor. II. Equation 2 gives I = r^^lW + I,,, ... ^, = I _ roi'JiW. Cor. III. Equation 2 itself shows this. Art. 577 should be read ; and the examples of Art. 578 have already been performed. Art. 579 is clear, if carefully studied, only there are, first, one letter, C, left off in the figure, and an ,. . J'+'c" exponent on, the c in — « EXAMPLE. Find the radius of gyration of a quadrant sector of a circular cylinder about an axis passing through its centre of gravity. CENTRE OP PERCUSSION. The explanation of Centre of Percussion given by Rankine, should be carefully studied, but I shall present it in a different aspect, which I think will furnish a clearer conception of it. First, imagine a body (Fig. 37) revolving about an axis per- pendicular to the plane of the paper, and passing through with an angular velocity a. If, with O as a centre, and a radius OA = r, we describe an arc BC, all particles situated on this arc hare a linear velocity, or, which must have been generated by a force which may be measured by war, where w is the weight of the particle ; the moment of the force is war', and the total angu- lar momentum of the body is Zwar'^ = aSwr"^ = al. The sum of the forces acting on the body '\%)^war^ a2wr, hence the point'of appli- 46 INSTANTANEOUS AXIS. cation of the resultant single force which would produce this rota- tiou IS situated at H, when^H = — y — = -- — , the magnitude of that force being aZwr. This point of application of the resultant single force required to produce the angular velocity, a is the centre of percussion. ANOTHER VIEW. INSTANTANEOUS AXIS. Imagine a body (Fig. 38), and for the sake of simplicity let the particles be conceived to be distributed along a single line AB, and suppose a force F applied at D. We may conceive two equal and opposite forces, each equal to F, applied at C, the centre of gravity of AB, without altering the motion of the body. Then these three forces are equivalent to a single force equal to F, ap- plied at the centre of gravity C, which produces translation of the whole body, and, secondly, a couple whose moment is F*CD, whose effect is to produce rotation around an axis passing through the centre of gravity C. Under this condition of things it is evident that the centre of gravity C will move forward with a velocity due to the force F, the point D will move farther forward, while those on the other side of C will have a less and less velocity, and the particle situated at A (if the force is not too great) will move backward, hence there must be some point, which, for the instant in question, is at rest {i.e., which is changing from moving in one direction to move in another), or about which, for the moment, the body is rotating, and if this point were fixed there would be no strain on the pivot, caused by the force applied at D. An axis passing through this point is called the Instantaneous Axis, or the axis about which the body at that moment tends to rotate. This point is evidently the point of application of the resultant of the forces, which give the body its angular momentum about the axis, and hence the point D is the Centre of Percussion ^ of the body in reference to E. INSTANTANEOUS AXIS. 47 Fig. 39 will make the condition of things clear. Let AB repre- sent the body, C its centre of gravity, D the point of application of the force F, and E the instantaneous axis (D is then the centre of percussion in reference to an axis through E). Let CG repre- sent the distance through which the centre of gravity moves in a F short time, dt, impelled by the force F, then CG = y^dt. Then the point D will, in virtue of the translation, in common with the centre of gravity, move to H, and in virtue of its rota- tion around this centre of gravity, describe the arc HK, such that F-CD , aSwr^ = F'CD .•. « ^ -y — 2, (« being =: the angular velocity) ; F'PT') F'OD^ then the arc HK := aGH-dt = -y — j GB.-dt = -y — ^'^^i but for a short time HL = HK, nearly, hence EC : GH = CG : HL, ^r, nr^— ^ J F-CD'' , EC _ Ilwr^ or EG : CD - ^f. -^^dt .: ^ - cD^-iV; •'• EC-CD = ^ ... CD = ^ 2,w EC--Sw .•.DE = CD + EC = EC + :^f^ = ZrZ = -^> the same expression obtained before. From the value already obtained for the distance from the axis of rotation to the centre of percussion, viz., CB = -^ — > Ran- kine, Fig. 239), we deduce ^ ./2W:^ _ I ^ 1^^^£.= , ._ CB, i.e., the radius of gyration is a mean proportional between the dis- tance EC and the distance r,,, between the axis of rotation and the centre of gravicy. Hence follows the construction given in Ean- kine (Fig. 239, p. 520), viz., at G draw GD perpendicular to CB, and equal to p^, then join CD, then CD^ == p^ + r^ = p% then at D draw CB perpendicular to CD, and B is the centre of percus- sion, for CD" = CG-CB, or p" = nCB .: CB = ^• 48 INCLISTED AXES. If the axis of rotation passes through the centre of gravity of the body, r„ = and CB = oo , hence the centre of percussion moves off to an infinite distance, or in other words there is no cen- tre of percussion (Eankine, Art. 582). EXPLANATIOlf OF THE WOEK OF ABT. 583, EANKINE. Let X, y, z (Fig. 40), be the three original axes, and a;', y', /, the new axes, and let P be the point in consideration, whose coor- dinates with reference to the old axes are OB = x, BA = y, AP = z, and OC = x', is one of its coordinates with reference to the new axes.. Now if CP be projected on Oa;', the length of the projection will be OC = a;'. If instead of that we project the broken line, OBAP, we shall have, Ob cos a:Oa;' + AB cos yOx' + AP cos aOa;' = X cos xx' -\- y cos ya^ -\- z cos ex', hence x' ^ X cos a;a;' + y cos yx' -\- z cos »a/, so also 2/ = a; cos xy' + y cos yj/ + z cos ajf', and s' = a; cos xz! + y cos ya' + z cos ss'. In the same way we should obtain x = af cos x'x -\- 1/ cos i/x + s' cos z'x. y = a/ cos x'y + y' cos y'y -\- s' cos z'y. z^=x' cos x'z + y' cos j/z + z' cos z'z. The following equations have already been proved in the Statics, cos^ xx' -\- cos^ xy' -\- cos^ xzf = 1. cos'' yx' + Cos'' yy' + cos^ yz' = 1. cos'' zx' + cos" zy' + cos'' zz' = 1. cos" a;a:' + cos" yxf -^ cos" zxf = 1. cos" xy^ + cos"y/ + cos" zj/ = 1. cos" xz' + cos" yz' + cos" zz' =^1. The proof of the following equations is precisely similar to that on page 31 and 32 of the 1st part'on Statics. cos yxf cos zx' -\- cos yy' cos «y' + cos yz' cos zz' = 0. cos a;a;' cos za/ + cos xy' cos zy' + cos xz' cos ss' =: Q. cos axe' cos ya;' -J- cos xy' cos yj/ + cos xzf cos ys' = Q. INCLIlirED AXES. 49 Also the following : cos x]/ cos aia' + cos yy' cos yz' + cos zy' cos sa' =^ 0. cos xz' cos za;' + cos ys' cos yai -\- cos ag' cos z^ = 0. cos xx' cos a;y' + cos yx' cos yy' "H cos zx' cos ay' := 0. Let us have, for the sake of brevity, cos xx' = ?i, cos yx' ==■ 4, cos xy' = nil, cos yy' = m^, cos a;z' = Wi, cos ya' = n^. And the last three equations become, W^lMj + ni2n2 + OTaWs := 0. Also Zi^ + 4' + 4'=l. (4.) Eliminating 4 from (2.) and (3.), we have, ll MjWg — : cos zx' = --Is- COS a^ = - Ws. cos aa' = ris. (1-) (2.) (3.) 4 »»iM2 — m.2ni ' and eliminating ^ we have, 4 JWsMi — miWg From (4.) we have, ^ I h^ I , _ j_ (W2W3 — WsWa)'-' + (wgW; — mjUsy _ 1 ^2^ ^2-t-l-;^2 ••■ (miM2 — W2'Jir "^^"4' . ,3 _ (wZiWg — ^2^1)' _ (m^ng — msu^y + (M3M1 — Wing)^ + (miWj — m^rhy ' but the denominator is equal to WW + «3^ + Wini' + «3^) + ms\ni^ + n^) — Im-^m^n-^n^ — 2?Wi»23WiM3 — Im^m^n^g = V(l — n,^) + m,%l — n/) + W(l — n^^ — 2(mi«i>WsW3 + minim^n^ + JWjWg'WsWa). Now squaring (1.) we have + 2(otiWi»22W2 -f~ mTnYm^ng -\- m^n^msn^ = 0. .•. — 2(»Wimi»W2W2 + MiWiOTsWa + m^^^n^ = jWi^Mj^ + wjg^Wa^ + m^n^. .-. The above denominator reduces to 1 _ (miW + mini + mini) + mi^Wj^ + mini + jw^ V = 1 .'. 4 = wiiWj — m^xi and 4 = m^n^ — mrjii, \ = m^ng — msWj. Q. E. D. 7 50 INCLINED AXES. These formulae enable us to obtain the moments of Inertia and moments of Deviation, with reference to any new set of rectangu- lar axes. Note on Art. 584 of Raukine. , The derivation of Equation (4.) from Equation (3.) is precisely aiiulogous to that of Equation (4.) from (3.), Art. 107, Eankine. Every cubic equation has at least one real root, since imaginary roots always enter an equation in pairs, and hence there is always one value ojf Sa;;^, and one position of the axes, for which the mo- ments of deviation are 0. Now by interchanging the axes of X, y and z, we shall find that Sxi", S^i^ Szj^ will be interchanged, and hence these must be the 3 roots of the cubic equation ; hence all the roots are real, and since there are 3 variations in sign, they are positive. To obtain equations (6.) multiply the 1st of (3.) by Sy0, and the 2d by Sariz, and we obtain cos xxi\{^x^ — Bx^)Byz — ^xySizx\ -\- cos yyi\^xySyz — (Sy^ — Six^)^zx\ ^= 0. cos xxi (Sxi^ — 8^^)820; + SxySyz cos yyi (8a;i^ — Sx^)Syz -j- SxySzx' 1 1 cos XX, : cos yy, - ^^^2^^^^ g^^ _j_ q^Szx '' (Sx.^-Sf) Szn^SxySyz, and, in like manner, eliminating cos yy, we obtain, 1 1 cosa;a;i : cos»«i — ^g^^2_g^2)Syj,_(_S^S^3,: (^Sxi'-Sz^)Sxy+ SyzSzx The sum of the roots = A = Sx,^ + 82/1^ + Sz? = 8E^ and so the sum of the products of the roots taken two and two is, B = Sy,'Sz,' + Sz,'Sx,' + S^e.^Sy/, and C = Sx.^Syi'Sz,^ Note on Art. 685. To prove that Sa;^ = (cos^ a)Sx,^ + (oos^ ^Si/t!' + (cos^ 7)8%=, we have, x = x, cos a + y, cos ^ + z^cosy; consequently, a;2 = X,^ cos" a + J/i" gogS ^ _j_ ^2 gog2 J, _|_ 2x^y^ COS a cos ^ + 2a;iZi cos a cos y -\- 2y^ cos j3 cos y. ■•■ fff3?dxdydz = cos" afxi'dxjdy^dz, + cos^ ^ fffyi^dx^^dy^dzj + cos" r fff^i'd'X'idl/idz, + 2 cos a cos ^ffSx^y^dxydy-^dz, + 2 cos a cos yfffx,Zidx,dy,dzi + 2 cos ^ cos r S S SVx'hdx^dy.^dz-^; UNIFORM ROTATIUJN. 51 that is, Sail" = Sx^" cos'^ a + S^i* cos^ ^ + Ssj^ cos'' 7 + 2Sfl3]2/i cos a cos ^ + 2Sa;iZi cos a cos y + 2Syi«i cos (3 cos 7 ; but since the body is referred to its principal axes, Say^i = 0, SajjZ, = 0, and Syj2j = 0. .-. Sa;'' = Sxi" cos^ a + Syi^ cos^ /3 + S^i" cos^ 7. Equations (2.) have been obtained in Art. 583. .-. Sx^ = SR2 — I, Sx;' = SR^ — Ii, Si/i^ = SR2 — I^, Szi^ = SR^ — 13 . . SR'' — I =: SR2(cos^ a + cos^ j3 + cos= 7) — Ij COS^ a — I2 COS^ ^ — I3 COS^ 7 .•. I =^ Ii cos'' a + I2 cos'^ ^ + 13 cos" 7. The remainder of Art. 585 needs no comment. Note on Art. 586, Rankine. Equation (2.) is derived in a manner similar to Equation (1.) of the last article. The equation of an ellipsoid being ^ 4_ ^ _l_ 1' — 1 the equation of a plane tangent to it at the poiuD Xi y^ z^ is, and the length of the perpendicular from the origin on the plane is 1 V a* "^ 6< "*" c* ajj = s cos a, yi := s cos §, »i = s cos 7 .•. 1 2 /cos'' a cos2 j3 cos' y \ i? - * \~^^ + T~ + cW cosl_a c^ cos^' ^ 1 _ ^ UNIFORM ROTATION. (Art. 588, Rankine.) To illustrate the conclusion of this article, we must remember that couples are compounded by the parallel- ogram of forces by means of their moment axes, hence a tendency 52 uniform: rotation. to rotate about two different axes, results (if the body is free) in a rotation about neither of the two, but about one intermediate in position between the two. Now suppose we have a wire, AB (Fig. 41), compelled to rotate around the axis O2, the forces necessary to action the wire to cause this rotation, are a set of forces at all points of the wire (perpendicular to the plane of the paper), and proportional to their distances from Oz ; but these forces have also a moment rela- tively to the axis Ox, equal to the resultant force multiplied by its distance from Oa; ; hence it has also a tendency to rotate around Ox, and if the body wete free it would rotate around neither Oz nor Ox, but around an axis intermediate between the two, and its axis of rotation would continue shifting till the wire came to rotate around a line perpendicular to AB : this is an axis of inertia, and that where the moment of inertia is greatest. Thus, if the axis of rotation, or the axis about which the body is rotating, is not an axis of inertia, the body has a tendency to rotate around other axes also, at right-angles to this (and hence its axis of rotation changes position if the body is free, in which case the axis of rotation is merely an instantaneous axis). In the case of a free body the axis of rotation, or instantaneous axis, mo\es, and comes to change its position so as to approach an axis of inertia, and it only becomes stable when it has reached that axis. If the body is not free the tendency to rotate about other axes causes a stress on the axis which has to be resisted by the strength of the material, or else it will bend, and ultimately break the axis. Hence, in machinery, other things being equal, it is better that all the axes of the wheels, and all the bearings of the rotating pieces should be axes of inertia of the rotating pieces, as this avoids the bending stress on these bearings. To illustrate farther the composition of rotations, we may con- sider the motion of the gyroscope. Suppose the gyroscope (Fig. 42) to have impressed upon if a rotation in the direction of the arrow, it will have, also, another tendency to rotate, caused by the force of gravity ; the first rota» UNIFORM ROTATION. 53 tion about Ox, and the second about Oy ; these two rotations com- pounded by the parallelogram give a rotation about an intermediate axis OP, in the plane xy, hence the whole arm has a retrograde motion in a horizontal plane, or about an 'axis OZ, in addition to its tendency to rotate about an axis parallel to OP ; and the result of compounding these two would be to cause rotation about an in- clined axis, or, in other words, the axis of rotation has risen. Fig. 43 will serve as a iigure for Rankine's work in Art. 588. With the latter part of this article the student should compare Art. 95 of Kankine's statics, as they are precisely analogous. Note on Art. 589, Rankine. We have E =: -„ — , and also E = ^ ; hence the actual en- ergy bears the same relation to linear velocity and weight that it does to angular velocity and moment of Inertia. Again, from (8.) we see that I oc ttrj, hence Note on Art. 590. I. The result of applying any set of forces to a rigid body is to impart to the body (in the most complex case) a translation and a rotation combined, the rotation being in a plane perpendicular to the direction of the translation; leaving out of account the transla- tion which affects the position of the centre of gravity of the body, we have a rotation around a certain line ; hence, considering the centre of gravity as a fixed point, the motion is confined to one plane, and a perpendicular to this plane is the axis of angular momentum. II. If no force acts on the body its angular momentum can not change. in. Refer to Art. 565 of Bankine. Equations (1.) are respectively (4.) of Art. 588, and (2.) of Art. 589, Rankine. 54 UNIFORM ROTATION. Now to derive (2.) we have, Ay j , _ 25;^ Ay _ %E " ^^ F~4-"K"' ^ ** — I ■ ■ P 4- K^ — I .-.9=1 = [ , and from (2.) of Art. 588, p + K^ = Ij2 cos^ a + 1/ cos'' |3 + Is^ cos' 7 ^2 _ r+ K' ^ I,^ cos' a + V cos' ^ + I3' cos' y •'• 2E ~~ I ~ Ii cos' « + I2 cos' fi + I3 cos' y I' + K' . but A is constant, and also E .-. j is constant. The second of equations (1.) gives, 2gE _ V(2ffE) « = 3- .-. « - ^v^^' but the numerator is constant, and from e2[uation (8.) of Art. 588, I oc ry=2 .•. Vl oc jy^, and hetce a oc OR. . \rf. .-. Ij' cos' a + 1/ cos' § + I3' cos' y = "0^2 cos' a + 0^2 cos' ^ + OF' ""^^ '' ' Note on Art. 591 and 592. To illustrate the deviating couple explained in the above articles, let us imagine an ellipse (Fig. 44) constrained to revolve around an axis Oy passing through its centre O, but which is neither the transverse nor the conjugate axis ; the forces which act on the dif- ferent particles to impart to them this rotation are perpendicular to the plane of the paper, and proportional in magnitude to the distances of these particles from Oy ; thus the force acting on a W particle C, at a distance EC =: x from Oy, will be — ax (W be- ing the weight of the particle, and a the angular velocity of the VARIED ROTATION. 55 body), and the linear velocity of the particle will be ux ; but (Art . 537, Rankine) the centrifugal force of this particle, or the force acting on it iu the direction EC, is , equal in this case to WaV _ WA 9 X ~ g ' The moment of this force with reference to an axis passing through 0, perpendicular to the plane of the paper, is Wu^xy _ 9 hence the total moment of the couple tending to turn the body about this latter axis, is 2: ^ = -i:Wxu = . ff 9^9 which proves Rankine's Equation (1.) of Art. 592. The remain- der of the article is very easily deduced. Note on Art. 593. Let Fig. 45 represent a couple, of force F, and arm I, and sup- pose that while it is acting the body moves through an angle a, then the amount of work done will be, F-AA' + F-BB' = YeOA + FSOB = eF(OA + OB) ^ 6FI = Fid. Note on Art. 593, Rankine. In the case in question, 6 = di, hence the energy exerted is Fldi = Mdi ; but di = adt ; hence energy exerted = M.adU When the axis of the couple applied to a body makes an angle

^'^'^ ^™™ -'^^''• F I fa 610, F = -/5 .-. — -^ = / .-. a = ^^. In the case of gyration, by referring to equation fS.), Art. 598, we have for the moment of the couple, IkH lice's M == , or , and this must = FJ = fSl. 9 9 ■' Iflff f =/•••* =V^1 With Art. 611 compare Art. 543, Eankine. If Mj = 2my, then during the time occupied by the body in making one Complete oscillation in the direction Oy, it makes two in the direction Ox ; hence we must have a two lobed curve, and, from symmetry, the point of crossing of the lobes must be on the axis Oa;. Note on Art. 613, Ex. I. In making use of the general equations, as given in this Article, it is important to keep in mind the distinctions between the terms used. (a.) Stresses are the forces that cause the displacements. (6.) Strains are the magnitudes of the displacements them- selves. (c.) The coefficient of elasticity of a substance is the force that would cause in a prism of the substance of a cross section of a MOTIONS OF PLIABLE BODIES. 61 unit of area, an extension equal to its own length (or, in other words, extend it to double its length). Hence if the strains be expressed as fractions of the length of the body, the stress which produces a given strain will be obtained by multiplying the strain by the coefficient of elasticity of the body. In equations (4.) and (5.) of this article, dS expresses the actual magnitude of the strain, while dx expresses the length of the molecule ; hence the strain expressed as a fraction of the length of the particle is d? dS . , ^d-o _ d!^ a =: T- = T- cos 0, and so v = -w- ^^= t- sin a. dX uX (t3C (tx The stress required to produce the strain a will therefore be and to produce the strain v, dS P-y = ^' = ^dx '^° ^■ In this case, equations (2.) of Art. 116, altered as shown in Ex. I. of this Article, become, dp^^ w d^~ dp„ w d^ -T — = — -j^n, and -^j^ = — js ; hence we have, dx g dt^' dx g dr' . d^S . d^d w rf2f Q^ = ^^^ ^^d^-^^^^^-gW ^y = ^dx' = ^^^^•"^=7^^- which equations, as Eankine states, reduce to d^ _ ,^ ^ ^ _ 2^ . de ~ "• dx-"' ^^'^ df ~ " df the first of equations (12.) of Rankine, viz., f =: (f{at -\- x) ■\- ({.{at — x) gives -r: = a-~p'{at + a;) + a. If 2:1 — »(, denotes the mean head, we must have, Q = V(2.9') SAK V(«i — :^) = V(25r) V («i — z^SAK. • v^. , , _ ^V(% - ^)^A /', ^^^ - ^y^ J z' ^ ,-. »i — Za = \ — /z' '^^^ Equation (3.) is derived directly from (2,) by integration, con- sidering y as constant. EXAMPLE. What is the mean head for a rectangular- orifice of efflux 3 ft, wide and 1^ feet high, the lower edge being 2f feet below the level of the water. Note on Art. 628. To fix the ideas, imagine a set of concentric cylinders to be the surfaces of equal head, i. e., that the dynamic head is the same throughout each of these cylinders, while the head is diflFerent in one cylinder from that in another. This is illustrated in Rankine EXAMPLE. 67 Fig. 250. Now the force due to the difference of head in the dif- ferent cylinders cannot have any component in the direction of the cylinder itself; hence if the deviating force is to be furnished en- tirely by the difference of head, the motion must take place in a plane perpendicular to the surface of the cylinder, or as Eankiue expresses it, the radius of curvature must be in a plane perpendic- ular to the surface of equal head. To deduce Equations (2.) we have as follows : . „ . . mN^ 1. The deviatmg force for mass m is • 2. The variation of head per unit of distance normal to the surface of equal head = — -r--, and this into the mass, is propor- tional to the force acting in the direction of the above said normal. The component of this in the direction of the radius of curva- dh ture is — m-^ cos nr .: Y" dh V dh gr dn gr dn Note on Art. 629. To deduce Equation (1.), referring back to Art. 407, we have, V = the mean radial component of the velocity. Now 'i.Tzrb is the portion of the surface of the cylinder through which the current passes, and v2Y/l 1 -i*^ y I =max., or a;y( 1 -a; y )^=xi — x y =max.; now differentiate, and we have 2 i_i / , i\ J_„ . 2 i_i _ 1 + r -a;v 7 -{l+jyy = 0,.^x'y-' IszJ v+1 v^^ 2 / 2 \_v_ .-.»;> = -g- .-.x y = yrfj ••• «= = Vf+TJ^"^ hence follow the remaining equations deduced. To deduce Equation (8.) we have, i;= velocity of sound Xy/(^^::p-l). and J^= {yZfijy^^^ .: — = velocity of sound X f i ^ )^ ^ y — i = velocity of sound X ( i , )2(y — D- MOTION OF LIQUIDS WITH FKICTION. ' EXAMPLES. 1. Suppose the cross section of the channel of an open stream to be a semicircle (radius 20 ft.), what will be the angle of decliv- ity for a velocity of 10 feet per second ? (Use the value/ = a +-- as given in the text.) 70 IMPULSE OF FLUIDS AND SOLIDS. 2. What will be the declivity when the cross section of the above stream is a trapezoid ; breadth at top water = 24 ft. ; breadth at bottom == 12 ft. Depth of water = 4 ft. 3. What is the hydraulic declivity in a uniform pipe of 1 J inch diameter, where u = 8 ft per second ? 4. What in a 2 inch pipe ? IMPULSE OF FLUIDS AND SOLIDS. Note on Art. 648. The jet is at first moving with the velocity v in the direction ah (Fig. 5^), and after leaving the surface it is moving in a direction ow, parallel to ac, with the same velocity v. To find the deviating force we follow the method of Art. 363 of Eankine, and laying off ah and ac, both equal to v, the line he will represent the change of velocity in magnitude and direction, and 5c = 25e = lab sin ^ = 2?? sin -^ The components of this Velocity along and perpendicular to the original direction of the jet, will be represented by Jj^.and do, re- spectively, and from the figure we have hd = ah — ad ==v — v cos ^, or «^ = i> (1 — cos j3), and dc =-v sin ^, or v^ = v sin ^ .■. P : F,, : Fy = 2 sin -|- : 1 — cos § : sin ^ . -P _ F(l— cos^) _ Fsin|3 '■^^ — S ' ^y— W' 2 sin -|- 2 sin | or F. = ^-^(1 _ cos ^) ; F, = ^ ,in ^. Q. E. D. The above applies to a jet iifipinging on a smooth surface which is itself at rest. The case which applies to water wheels, where the surface on which the water impinges is in motion, is treated of in Aft. 649 of Kankine. IMPULSE OP FLUIDS AND SOLIDS. Tl Note on Art. 649. When the surface (as the vane of a water wheel) has a motion of translation in the direction BD, as shown in Fig. 264 of Ean- kine, the direction of motion and the velocity of tie- water rela- tively to surface will be represented by DC, the resultant of the motion of the water and of the surface ; hence the isosceles trian- gle, by which we determine the velocity of deflection will have one side parallel to DC, and the other to the tangent EF, and tke two equal sides will each be equal to DC. To prove Equation (10.) we have ■0^^ = w^ — w' -\- 2wvi cos a ) . „ 5 I 2 1 n t hence by subtractionv jjj^ ;:::; ^2 _|_ ^2 _|_ ^WU COS J' ) *' Vi' — v^^ = 2u(yi cos a — u — w cos j) f^/ .'. ^se,u = Y~ 2w(i'i cos a — M — w cos y), which is the same value as that obtained in Equation (8.). Note on Art. 650. Q being the quantity of water discharged, h the depth of the wheel, and rj the radius of the cylindrical surface of discharge ; the area of this surface is 27irib, and hence the radial velocity of the Q issuing current is m = g » • Npw as the cylindrical surface of discharge is moving with a velocity tfrj, and the velocity of the issuing jet at right angles to its motion (in a radial direction) is u, in order that the guide plates may be inclined in the direction of the resultant motion of the water we must have m tan d = ar^, or tan = — • Note on Art. 656. The changes referred to in this Article are those that occur in the steam contained in the cylinder of a steam engine, and the curves represented in the figures are such as would be obtained on the Indicator card ; their areas represent the energy transferred to the piston from the steam, while the abscissae represent the respec- tive volumes occupied by the steam at different periods in the 72 IMPULSE OF FLUIDS AND SOLIDS. course of the piBton, and the ordinates the corresponding intensi- ties of pressure. To prove Equation (4.) we have from Art. 63.5, Equation 2, p -£''-'* = F^{"-'};;=,-?r(^.'--.-) or r*^ = ^ (Si _ £s) = _2L_^('i_WiV ••• f^fP = ^im \l-{^y~'}. Q. E. D. To deduce Equations (7.) from Equations (6.), we have ^ _ /B ■ 2C\ f- \^+ Trr^ ••• dp /B 2C\ /B , 2C\