ENGINEERING 
 LIBRARY 
 
 NOTES 
 
 RANKINE'S 
 APPLIED MECHANICS.
 
 THE LIBRARY 
 
 OF 
 
 THE UNIVERSITY 
 OF CALIFORNIA 
 
 LOS ANGELES 
 
 GIFT OF 
 
 U. of Calif. . 
 Berkeley
 
 '7- 
 
 XT,
 
 NOTES 
 
 RANKINGS APPLIED MECHANICS. 
 
 GEORGE I. ALDEN, B. S., 
 
 PROFESSOR OF THEORETICAL, AND APPLIED MECHANICS 
 
 IN THE WORCESTER FREE INSTITUTE, 
 
 WORCESTER, MASS. 
 
 Third Edition. 
 
 HARTFORD, CONN.: 
 
 PRESS OF THE CASE, LOCKWOOD & BRAINARD COMPANY. 
 1890.
 
 Copyright 
 BY GEORGE I. ALDEN. 
 
 1877.
 
 TA 
 
 35" 
 
 All* 
 
 INTKODTJOTI 
 
 The following pages are the -result of putting in permanent 
 form some of the matter which it ha.s been found expedient or 
 necessary to give by dictation to students in the Worcester 
 Free Institute, who pursue for the first time, the study of Ran- 
 kine's Applied Mechanics. The object in their publication is not 
 to furnish a key, or provide a substitute for diligent study and 
 careful thought on the part of the student, but rather to en- 
 courage him by giving such suggestions, solutions, and refer- 
 ences as experience has shown that the average student requires; 
 thus economizing time in the preparation of the lesson, and also 
 giving the instructor opportunity to devote the time spent in 
 the class room to recitations, and to the application of the prin- 
 ciples and formulae of the lesson, to practical problems. 
 
 To what may be strictly called notes on the "Applied Me- 
 chanics," I have added a brief explanation and illustration of the 
 method of producing the reciprocal diagram of stresses, sub- 
 stantially taken from " Economics of Construction," by R. E. 
 Bow, C. E. Also a separate treatise on strength of beams, and 
 an investigation of a particular problem relating to seven bar 
 parallel motions, known as " Peaucillier's Parallel Motion." 
 
 This work has been prepared from materials drawn from 
 various sources, especially from notes given by Prof. Eustis, of 
 The Lawrence Scientific School, to the class of '68. 
 
 I have also received assistance from George H. White, B. S., 
 a graduate of the Free Institute, and have inserted on several 
 articles of the Applied Mechanics, notes which are entirely his 
 own work. I have endeavored to make proper reference to 
 works from which quotations or extracts have been taken. 
 
 The blank pages at the end are intended to receive such sup- 
 plementary notes as the instructor may find adapted to the 
 needs and capacity of his class. 
 
 GEORGE I. ALDEN. 
 
 WOBCESTEE FREE INSTITUTE, Feb. 1st, 1877. 
 
 NOTE. Some errors found in the earlier editions of "Rankine's 
 Applied Mechanics " have been corrected in later editions, issued since 
 these notes were prepared. References in the notes, to such errors, 
 are allowed to remain for the benefit, of those who may have copies of 
 Uie older editions of the "Applied Mei-hanics." 
 
 713771
 
 INTEGKATIOK 
 
 The following integrals are of frequent occurrence, and are 
 here given for future reference : 
 
 (A) / d x V 2 ** 
 
 In the general formula for integrating by parts, / u d v = 
 
 I v . d u, let u = V' 2 %\ and d x = d v ; then 
 
 x dx 
 
 d u = and x = v 
 
 t/a* x' 
 
 .-. J d x V* x- = x A/"' - x 2 + / - 
 
 v <y a x 
 
 Again, 
 
 /' , fd x (a 2 a; 2 ) / x 2 d x 
 
 .'. fax V a ar == / ; = / , 
 
 J ^/ d 1 x* J */a* x 3 
 
 /' 
 
 4. 
 
 a? d x 
 
 ,- a- a x ft \ a / ~. 
 
 / - - = / a 2 == = a 2 sin- 1 _fL 
 
 t / yV x 2 t/ A/1 a j a 
 
 C. 
 
 fd x A/a 2 * 2 = x V 2 - a:* + 2 sin- 1 + C 
 t/ a 
 
 Let Va 2 + a; 2 =,: Then a 2 -j- x* = 2 2 2^,* 4- rr 2 or
 
 INTEGRATION. 
 
 = z- 2 z x. The differential of this equation is = 2 z 
 
 (z-x)dz 
 
 z 2 z d x 2 x d z .-. dx = 
 
 z 
 
 r dx ( z - *) d z _ r dz _ ]nfr r 
 
 J VV"+^ - (z -x)z ~ J z~ 
 
 z+Va' + xO + C 
 
 (C) fd x vHr+^i 
 
 In the formula for integration by parts, I u d v u v 
 I v d u, let ^d l -J- x 2 = u and d x = d v. Then x ^ v and 
 g ^_g_ _ ^^ 
 
 // x* d x 
 d x Va* + x * = x Va 2 + tf - / y ^ ^ . Again, 
 
 I d x V~~ 
 
 V + a-* 
 
 A4J 
 
 J V a *- 
 
 .: 2 I'd x V a * + x> = x V^ _|_ ^ + a? f * 
 J J V a 2 -j- a; 2 
 
 /> (? x 
 By integral B, we have a 2 / = a 2 log e 
 
 (a + V a * + x 2 ) + C. 
 
 ... fd x v-tf+i?= ^L v^ip^ + ^, i oge (x+ V 
 
 (D) /* 2 d cc Va 2 x 2 
 
 In the formula / udv = uv / v du, let x dx Va' 2 
 = d v, and x = u : then v = - (a 2 x 2 ) and d u = d x 
 - ( 2 - a: 2 )^ + 1 a; (a s - ^ 2 )^
 
 INTEGRATION. 
 
 x> d x 
 
 Transposing the last term of the last member, we have 
 
 - fx* d x V a * x 1 = (a 2 z 2 )^ 4- - fa 2 
 ?,J 3 3 </ 
 
 From integral A we have 
 
 This integral, taken between the limits a and 0, reduces to 
 
 - sin" , which can easily be memorized. 
 o (i 
 
 (E) /"cos 2 d 6. 
 
 Integrals of the powers of sin and cos are found by sub- 
 stituting for these functions of 0, their'values in terms of the 
 multiple angles, as in the following case : 
 
 //>l _u cos 2 sin 2 
 cos 2 d = / d = I- 4- C. 
 t/ 2 24 
 
 ARTICLE 83.* 
 
 In solving the following problems the student should always 
 sketch a figure representing the surface or solid under consid- 
 eration, and one of the elementary parts into which it is conceived 
 to be divided, and determine the limits for the integral by an 
 inspection of this figure. In double or triple integrations, we 
 
 NOTE. The first three integrals are taken from TodhnnterV Integral Calculus. 
 NOTE. Tl.c; abbreviation A. M. will be u^ed for Itaiikino's Applied ilechauics. 
 * The reference in to Article S3 of tlm A^ied Mechanics.
 
 ARTICLE 83. 
 
 must, in general, first integrate with respect to one of the vari- 
 ables, between the proper, limits, and express the result in terms of 
 the other variables. This may be continued until the complete 
 integral is obtained. 
 
 To illustrate this process take the general formulas for center 
 of gravity of a solid, viz: 
 
 I I I x dx'dy dz 
 fff Jx < ly ft 
 
 and similar values for ?/ and z fl . 
 
 The following application of these formulas is found in Tod- 
 hunter's Analytical Statics. 
 
 Problem. Find the center of gravity of the eighth part of an 
 ellipsoid cut off by the three principal planes. 
 
 Let Fig. (1) represent the solid in question, the equation of 
 
 the surface being 1- -y,- 
 
 If we integrate 
 first with respect 
 to z, between the 
 limits z, and zero, 
 we include all the 
 elements (dx dy dz) 
 in the prism P Q. 
 Next integrate 
 with respect to y 
 between the limits 
 y, and zero. AVe 
 thus include all 
 the prisms in the 
 slice between the 
 planes ? L and 
 
 mM. 
 
 F ro m Equa- 
 tion (1), 5 = C 
 
 0) 
 
 / x 2 if 
 1 2 -~- ; from Equati 
 
 ion of ellipse in the plane X Y,
 
 ARTICLE 83. 
 
 / / ' / ' x dx dy dz I j x ax d'y z f 
 f o J^" 1 y^ dx dy dz " y" y* 1 dx dy Zf 
 
 c x dx dy 1/TT^IT' > A x , & dy 
 o a 2 b* t/ o / o b 
 
 /'a /#, / /*a /<//! 1 
 
 f l cfyjy 1 / 1 _^_ ^ / / ' 
 
 . J o . a j j, J oJ o I 
 
 7" /"* x ,fe ^ /'" ,fe -, (" ~ O 
 
 i/o * e/o ^ 
 
 /*" 7 "" 2// /'" 7 ^ 2 / 2 2 
 
 ,/ a'"* 4 t / .^O'" 
 
 The other coordinates of the center of gravity will easily be 
 found by taking moments with reference to the axes OX and 
 OZ, and following the integration as above indicated. 
 
 PROBLEM III. Draw a line through C parallel to E B, form- 
 ing a parallelogram and a triangle. A line joining C and the 
 center of gravity of the triangle will be parallel to O D. 
 
 Let / BOD = 6. Then equating the moment of the trapezoid, 
 about A B, to the moment of this parallelogram and triangle, 
 about the same line, we have 
 
 -. O D. sin . x sin . w = b OD sin . sin . iv -)- 
 
 ~OD 
 
 3 (B+i) 
 
 /3 (B+5) - (B-&)\ TKD / _ E-b \ 
 \ 3 (B + i) / 2 \ 3 (B+i)/' 
 
 PROBLEM V. Taking moments about Y we have for the 
 nrm of the element ydx, the distance x . sin Y X. The dis- 
 tance of the center of gravity from O Y on a line at right angles 
 to Y is x . sin Y O X 
 
 i tx \ f' x \ 
 
 I xydx I xydx 
 
 . -. x sin Y O X = sin Y X J --^ 
 
 /"*! f* x \ 
 
 I y dx y dx 
 
 t/ o t/ o
 
 ARTICLE 83. 
 
 f*x. /:, 
 
 y dx 2 / ydx 
 
 I/O I/O 
 
 The equation of the curve is y 2 = 2 p 1 x. 
 
 PROBLEM VII. In this, as in many problems relating to the 
 circle, it is convenient to use polar coordinates. Let (Fig. 32 
 A. M.) be the pole, p the radius vector, and the variable angle 
 measured from the axis X. The circle is divided into differ- 
 ential areas, by concentric circles dp apart, and by radii making 
 an angle cfy> with each other. 
 
 Then a differential area = p d . d p. 
 
 Its distance from Y = p cos </>. 
 
 Its moment about Y w p 2 d p . cos $ d 0. 
 .-. in this case 
 
 />B />+<*> 
 
 I I p (( p 
 tJ o ' 4> 
 
 +* 
 
 ^ = _2_ R /si 
 ~ 3 " 
 
 sin 0- sin (-0) 
 
 (0) 
 
 /"" xy dx 
 
 PROBLEM VIIT. x = 
 
 / y d cc /"" (y- 2 ic 2 )a rf a; 
 
 t/ r cos i/ r cosB 
 
 r r ydydx r 
 
 J J rcosfl ___ _ t/ r 
 
 r r dy dx 
 
 / O*/ r cos 9 
 
 cos 6 2 
 
 PROBLEM XVI. Take moments about Y. The moment of 
 the annular wedge is equal to the moment of the whole wedge, 
 minus the moment of the interior wedge. (See Art. 76 A. M.) 
 
 Compare with Problem 15. 
 
 PROBLEMS XIX and XX. Conceive a plane passing through 
 O X, (Fig. 39, A. M.) making a variable angle y with the plane 
 X Y. Let p be a radius vector in this plane, making a variable 
 angle with O X. 
 
 Then pd<t>dp is an elementary area in the above described 
 plane. When that plane revolves about O X, through the angle
 
 10 ARTICLE 83. 
 
 dy, this elementary area describes a volume which is measured 
 by the area times the distance through which it moves. This 
 distance is the arc of a circle whose radius is p sin 0, and angle 
 dy, and = p sin <pdy. 
 
 . . Elementary volume = pcfy> . dp . p sin fydy. 
 
 Its distance from the plane Y Z = p cos 0; Z being the 
 third rectangular axis. 
 
 / / / p d . d p . p sin cf> dy . p cos . w 
 
 " 
 
 (j) . d p . p sin d y . w 
 
 j I I p* d p. sin $ cos <j> d <f>. d y 
 
 j I I p 2 d p. sin $ d 0. d y 
 
 Integrating between proper limits, this formula will give the 
 value of x for these, and similar problems. 
 
 ) r ~\ /J ) 2n- 
 
 In problem XIX the limits are p 5- ; y y y 
 
 XX pj/ 0|f; 
 
 In problem XX, and in similar problems, 
 
 / / I pd<f> . d p . p sin (j)dy . p sin ^ cos y . w 
 
 MO 7 ~7* 7 
 
 / / j p d (f> . d p . p sin <p d y . w 
 
 The limits for this integral, in problem XX, are as stated 
 above. 
 
 ARTICLE 91. 
 
 The expressions for moments around X, Y, and O Z, on 
 page 74, are easily obtained by resolving the force P, having 
 its point of application in the plane X Y, into components par- 
 allel to the three axes, and taking the moments of these com- 
 ponents. Equation 4 is reduced by the relation cos 2 a -f- cos 2 ft 
 -j- cos 2 y = 1. See Church's Analytical Geometry, Art. 48, 
 Equa. 4. 
 
 Equation 5 A has for its first member the sum of the rectan- 
 gles of the cosines of the angles which two lines make with 
 three rectangular axes, which is equal to the cosine of the angle 
 between the two lines. See Church's Analytical Geometry, 
 Art. 48, Equa. 5. If this cosine is 0, the angle must be 90. 
 The equation is verified by multiplying the first of equations 5
 
 ARTICLE 95. 11 
 
 by cos a, the second by cos /3, and the third by cos y, and add- 
 ing the products. The second member of this sum must then 
 be reduced by introducing the values of M,, M^ M 3 , from Equa- 
 tion 3. 
 
 ARTICLE 95. 
 
 Equations 3. The point whose coordinates are expressed by 
 this equation, is situated in the angle Y 1 X 1 . 
 
 PROBLEM. 
 
 Find the moment of inertia of a rectangle of length h and 
 breadth 5, about a neutral axis perpendicular to its plane. See 
 Equation 5. 
 
 Equations 13. Complete figure 45, A. M., by drawing a line 
 from C, parallel to T, and call the point where this line inter- 
 sects the line O X n N. From N, draw a line parallel to T C, 
 and call its point of intersection with T, H. The angle 
 T Xj = 90 / X, y = /3 1 . 
 
 Take the most general form of the equation of' a straight 
 
 line, A-|-B?/-|-C=iO, and write it in the form -~- x +7T 
 
 y -\- 1 = 0. Since this equation is true for all values of x and y, 
 it will be true when x or y becomes 0. 
 
 Let the intercept on the axis X = c, and the intercept on 
 Y = d, so that, when y = 0, x = c, and when x =. 0, y = d. 
 
 Make y = 0. Then -^ c + 1 = 0, or -^- = - 
 
 B B 1 
 
 Make x 0. Then -^-d+l = 0, or -^ = ~^ 
 
 These values introduced into the general equation, reduce it 
 
 . (A) 
 
 Now multiply both members of this equation by n, drawn 
 perpendicular to the straight line, and making the angle /3 1 with 
 the axis O X. See line O T, fig. 45, A. M.
 
 12 ARTICLE 95. 
 
 Then x -j- r y = n. But = cos /3 1 and -y- = sin/3 1 
 
 .-. x cos /3 1 + y sin /3 1 = w. (B) 
 
 The equation of a tangent to an ellipse (T C) is a 2 y y 1 -f J* a; 
 
 x 1 = a 2 6 2 (see Olney's General Geometry, Art. 136) 
 
 . -- 1_ jL i. Multiply this equation by n. 
 
 By comparing equations C and B, it will be seen that cos /3 1 , 
 
 x 1 n y 1 n 
 
 -^5-, and am 0' = *_ 
 
 a 2 cos /3 1 & 2 . sin /3 1 
 
 .-. ^ = ^-, and f = - ^- - (D) 
 
 Substitute these values in equation B, for x and y. 
 
 n 
 
 To find the value of n t. 
 The general equation of a normal to an ellipse is y y 1 = 
 
 a* y 1 
 
 JJTJA(X a 1 ). (See Olney's General Geometry, Art. 156). 
 
 In this equation make y = 0, and divide both members by y 1 . 
 - 1 = JL_ ( x _ a i) ; ... _ y x 1 = a 2 x a 2 a 1 ; x 
 
 fr*' ~ o ,^ gl = O N. 
 
 From Equation D, we have a; 1 == 
 
 ... N = ( gl - y )</3 1 
 
 w 
 
 .-.%<= (a 2 & 2 ) cos /3 1 . sin /3 1 . 
 
 PROBLEM IY. 
 
 Use polar coordinates. An elementary area = p d <f> . d p . 
 (See Note on Problem VII of Art. 83.) Its distance from
 
 ARTICLES 102-112. 13 
 
 Y = p cos 0. Its moment of inertia = p d ^ d p (p cos </>) 2 
 
 gral E). 
 
 ARTICLE 102. 
 
 Equations 1 and 2 are given in Ch.au venet's Trigonometry, 
 page 163, formula 53, and page 162, formula 48. 
 
 ARTICLE 105. 
 
 PROBLEM II. According to the preceding notation, the form- 
 ulae near the top of page 91 A. M. should be 
 O D = p xx . area B C -f p yt . area C A -f p zx . area A B. 
 OE=p ff . area B C -j- p . area OCA + p, v . area A B. 
 O F = p xz . area B C + p yz . area C A -f p zz . area A B. 
 
 ARTICLES 110 AND 111. 
 
 The student should carefully study these articles, and mem- 
 orize the theorems, in order to understand Art. 112. 
 
 ARTICLE 112. 
 
 Equation 1. From figure 54, A. M., we have 
 R 2 = OM* + MR* -f 2 CTST. M^TTcos R M N 
 
 Px + A 2 Px P? A 2 A 
 
 = 2 + 2 cos 2 a; w = ^ (1 + cos 2 x n) 
 
 + ^ (1 - cos 2 x n). 
 
 A A A 
 
 But cos 2 a; n = 2 cos 2 icw 1 = 1 2 sin 2 x n. 
 
 A A 
 
 .-. Pr = P? cos 2 x n 4- ^ y 2 sin 2 a; 
 
 4/J A A ) 
 
 ' Pr = ' \ P* cos 2 x n + pf sin 2 x n [ = R. (1) 
 
 Equation 2. From the triangle M R, we have 
 sin N O R : sin R M : : M~R : O~R
 
 14 ARTICLE 112. 
 
 or sin N R : sin (180 2 x n) ; ; Px ~ Py : Pr 
 
 A A p _ p 
 
 .: sin N R = sin n r = sin 2 x n - 
 
 or N R = n r = arc . sin (sin 2 x n Pl ~ Pv \ (2) 
 
 Proof that the locus of the point K is an ellipse whose semi- 
 axes are p x and p f . 
 
 From the construction of figure 54, of the Applied Mechanics, 
 it is readily seen that since OM = MQ = MP, the line 
 
 PR = PM - MR = Pi+Pi-P^ = py . 
 
 Also QR = QM + MR = p * + p * + ^^ = Px . 
 
 A 
 The angle OQM = 90 - xn, 
 
 Therefore for the coordinates of the point R we have 
 
 A A 
 
 x QR cos xn =p x cos xn. 
 
 A A 
 
 y = PR sin xn = p y sin xn, 
 
 Multiplying the first of these equations by p y , the second by p x ; 
 squaring and adding we have 
 
 Plx*+ P lf=plpl 
 which is the equation of an ellipse with semiaxes p x and p v .
 
 ARTICLE 112. 15 
 
 Equations 3 and 4. 
 p n = OM - MR cos (180 - 2 CT) = &LP+ P *~ P * cos 2 zn 
 
 ^ (1 -f cos 2 on) + ^ (1 cos 2 xn) =p x cos 2 z + 
 
 A 
 p y sin 2 . 
 
 j t = MR sin 2 o: = ^ ~ Py. 2 sin xn cos xn = (p x p y ) sin xn 
 
 A 
 cos xn. 
 
 Comparison of figures 54 and 57, A. M. 
 
 In figure 54 suppose a second plane making an angle -with 
 the plane A B, to be drawn through 0. 
 
 Let the same construction be made to represent the compo- 
 nents of the principal stresses on this plane, as is made for AB. 
 
 The distance set off on the normal to this supposed plane, 
 must be equal to OM, and the line corresponding to M'R, must 
 be equal to MR. There will also be a line corresponding to R. 
 Now suppose this second system of lines to be revolved about 0, 
 until the plane coincides with AB. The result will be ORMR'O 
 of figure 57. By the construction of figure 57 it will be seen 
 that MR = MR' and OM is the same for both systems of lines, 
 hence the figure satisfies the condition of the problem-, viz : that 
 
 OM =?^and MR = ^-L . 
 2 2 
 
 Equation 15. From the triangles OMR and OMR r figure 57. 
 MR 2 = (JM 2 + jt; 2 2 ^OM cos nr. 
 MR' 2 = OM 2 + p 1 ' 1 - 2 p' OM cos $/. 
 .-. = p 2 p' 2 2. OM (p. cos nr p 1 cos rcV). 
 
 X - X 2 
 
 ' OM == - (15) 
 
 2 (p. cos nr p' cos n V)
 
 16 ARTICLE 112. 
 
 Equations 16, are found by substituting t^LJLjOL for OM in 
 the above values for MR, and MR'. 
 
 /I/nations 17. For the first of these equations, draw a line 
 from R perpendicular to ON (Fig. 57) and call its intersection 
 with ON, A. 
 
 Then cos 2 xn cos NMR = ^ . MR = ^L-Jjf . 
 
 MR 2 
 
 MA = A OM = OR cos NOR OM = p cos nr J ~^'- 
 
 A 
 _ 2 p cos nr p x p y 
 
 2 
 
 A 
 
 n A 2 . cos nr p p 
 .: cos 2 aj - = * . 
 
 P X ~Py 
 
 A 
 
 The value of cos 2 xr/ is found by drawing a perpendicular to 
 ON, from R', and repeating the above process. 
 
 Equation 27. 
 
 From equation 25, p + p' = V/V^^ . (A) 
 
 COS (ft 
 
 Write equation 25 in the form 
 
 A 
 
 cos 2 (jo 2 -f- 2 _p_?/ -j- ;/ 2 ) = 4 7> ;/ cos 2 r, suotract 4 pp* cos 2 ^> 
 from both members of the equation, and extract the square 
 root. 
 
 p p' = 2 V/' / V cos 2 wr cos 2 (B) 
 
 Then . . __^_ 
 
 COS (p 
 
 From equations A and B, we find 
 
 / A A 
 
 |/ p p' ( cos nr -f- V( cos 2 r cos 2 
 
 A A 
 
 , _ ' (cosnr v(cos 2 nr cos 2 
 
 cos 
 
 A A 
 
 / cos 7/r v(cos 2 nr cos 2 ^>) 
 
 cos ? -j- V( cos 2 nr- cos 2
 
 ARTICLE 124. 17 
 
 ARTICLE 124. 
 Formula for the center of pressure of a fluid, upon a plane surface, 
 
 In figure 62, let BF be the plane and suppose FO to be a line 
 representing the continuation of this plane to O, the surface of 
 the fluid. Let OB be the axis of x, and the axis of y at right 
 angles to it. 
 
 Let the angle that OB makes with the surface of the fluid be 
 represented by Q. 
 
 Let F be the origin of coordinates, and x be measured pos- 
 itively downward. 
 
 dxdy = differential portion of this plane. 
 
 The pressure on a unit's surface of this plane is normal to 
 the surface, and measured by p = w (x -|- OF) sin 0, in which 
 w is the weight of a unit volume of the fluid. (See Art. 110, 
 A. M.) 
 
 Let OF = a. Then p = w (a -f x) sin Q. 
 
 By formula 2 of Art. 89, A. M., 
 
 />-fy /JFB 
 
 / / w x (a -\- x) sin 6 . dxdy 
 
 x = -/+? ' /*'" 
 
 I I w (a -}- a:) sin Q dxdy 
 
 /4y /FB 
 _ y J o (ax + x>}dxdy 
 
 ~ jZf" (a+ ^' 
 (a + x) 
 
 / y /FB 
 
 / / dxdy (a + x) 
 
 t/ -y t/ o 
 
 e surfac 
 uations 
 
 /-FB 
 
 / y (ax -(- a; 2 ) dx 
 
 If the plane surface is symmetrical with respect to the axis x, 
 the above equations become 
 
 f" 
 
 (B) 
 
 If the plane extends to the surface of the fluid, a becomes zero, 
 
 / x 2 ydx 
 and the equations B reduce to x = - ^ , h being the 
 
 depth of the plane surface. (C)
 
 18 ARTICLES 124-159. 
 
 PROBLEM 1. Find the position of the center of pressure on 
 a rectangular plane surface whose depth is h, and breadth I, 
 when its upper edge is parallel to the surface of the fluid, and 
 at a distance a below it. 
 
 From equation B we have * 
 
 yA 
 y (ax -}- x 2 )d x 
 
 x = _ \ But y = - b 
 
 y (a + x) d x 
 
 6 ah 
 
 PROBLEM 2. Find the centers of pressure of the following 
 surfaces, when their bases are in the surface of the fluid. (The 
 depth of each surface = h.) Rectangle, triangle, and parabola. 
 
 214 
 
 Ans. x n = - h, - h, - h, respectively. 
 o 2t i 
 
 PROBLEM 3. The parabola with its vertex in the surface, and 
 base parallel to it. 
 
 Ans. X H = -^ h. 
 
 PROBLEM 4. Find the center of pressure on a triangular sur- 
 face whose depth is A, base parallel to the surface of the fluid, 
 vertex upward and at a distance a below the surface of the fluid. 
 
 4 a It* 4- 3 /i 3 
 
 Ans. x u = -z T distance from vertex. 
 
 6 a h -}- 4 h 2 
 
 ARTICLE 159. 
 
 w 
 
 Equations 4. The load at 5, is resolved into two compo- 
 nents, cos i, which is transmitted through 5 4 and produces 
 a stress on all the pieces of the secondary truss 1534, and 
 sin i, one-half of which is supposed to cause a pull on the 
 
 piece 1 5, and one-half a thrust on the piece 5 3. 
 
 Equations 4 are obtained by computing the stresses due to a 
 
 * NOTE. It will be noticed that ar is measured downward from the top of the plane, 
 and not from the surface of t/te fluid. Also that a is measured on the axis of x.
 
 ARTICLE 159. 19 
 
 w 
 force cos i acting in the direction 54, and combining these 
 
 w 
 stresses with the component sin i above mentioned. 
 
 Draw a line parallel to 54 and equal to cos i. From its 
 
 extremities, draw lines parallel to 14 and 34, and from their 
 intersection draw a line parallel to 13, to meet the first line, to 
 which it is perpendicular. 
 
 W W 
 
 Then from this diagram R 4 , = R 41 = cos i, cosec i 
 
 o o 
 
 cotan i. 
 
 R, 5 = cos i cotan i (from diagram) -\- - w sin i (see above) 
 
 w /cos' 2 i -4- sin' 2 '\ w 
 s - -I r- 1 -- - 1 cosec t. 
 8 \ sin i / 8 
 
 R 51 = cos i cotan i (from diagram) ^ w sin i (see above) 
 
 w /I sin- i sin' 2 i\ w / 1 2 sin 2 i\ w 
 
 IT V sin i / = : ~8~ V sml / = T 
 
 (cosec i 2 sin i). 
 
 The Method of Drawing Reciprocal Diagrams. 
 
 Since the forces acting through any joint of a truss must be 
 in equilibrium, these forces may be represented by a closed 
 polygon whose sides are parallel to their directions. 
 
 A complete diagram for a truss under a given load ought, 
 therefore, to contain as many such polygons as there are joints 
 in the structure. In most cases a variety of diagrams can be 
 drawn, any one of which will correctly represent the stresses in 
 a given truss. The one that fulfills most nearly the condition, 
 that for each joint in the truss, there shall be in the diagram 
 a closed polygon whose sides are parallel to the forces acting 
 through that joint, and taken in the same order as those forces, 
 has been called the reciprocal diagram. 
 
 A method of notation given by R H. Bow, C. E , renders the 
 construction of reciprocal diagrams very easy in most cases. 
 This method consists in placing upon the figure of the truss, a 
 letter in each of the angular spaces formed by the intersection 
 of the forces at each joint (a single letter in an enclosed area 
 answering for all the internal angles formed by its sides), and
 
 20 
 
 RECIPROCAL DIAGRAMS. 
 
 in the diagram the same letter at the junction of two lines rep- 
 resenting concurring forces, that is found in the angle between 
 those forces upon the figure of the truss. 
 
 To illustrate the application of this notation, let us construct 
 the diagram for the frame represented by Fig. 78 A. M., on 
 the supposition that it is loaded with W uniformly distributed 
 over 12 and 13. 
 
 1st. Distribute the load upon the joints, as directed in Art. 
 147 A. M., or for a uniformly distributed load by placing upon 
 each joint half of the load between the adjacent joints. This 
 
 distribution gives - W at the joints 4. 6. 1, 8 and 10, and W 
 
 at the joints 2 and 3. 
 
 2d. Represent the directions of 
 the external forces by lines, as in 
 the accompanying Fig. 1. The 
 forces are all given in direction, 
 the internal forces or stresses be- 
 ing represented in direction by 
 the pieces of the frame. 
 
 3d. Letter the angular spaces 
 formed by concurring forces. 
 
 In Fig. 1, P occupies 
 K the space between 3*5 
 a and the left hand sup- 
 port, 3 5 and 5 7, 5 7 and 
 7 9, 7 9 and 9 2, and 9 2 
 and the right hand sup- 
 port. K occupies the 
 angular space at the left 
 of the left hand support, 
 between it and the load 
 at 3. A occupies the 
 angular space between 
 3 5 and 3 4, 3 4 and 4 5, 
 and 4 5 and 5 3, and so on. 
 4th. Draw a line of 
 loads (vertical in this 
 case) representing the 
 whole load W, and di- 
 vide it into parts repre- 
 senting the external 
 forces acting at t h e 
 Fig. 2. joints. 
 
 Fig. 1.
 
 RECIPROCAL DIAGRAMS. 21 
 
 As the supporting forces here are vertical and equal, they 
 will each be represented by half this line of loads (K P and 
 L P, Fig. 2). 
 
 5th. Letter the line of loads to correspond with the lettering 
 of the truss. In reading from K to , Fig. 1, we cross the line 
 
 W 
 
 representing the load at 3. Therefore K a on the line of 
 
 "W W 
 
 loads, must be For a similar reason a b = - -, and so on. 
 
 Since the loads at 2 and 3 rest directly and vertically upon 
 the supports of the truss (and the supporting forces are vertical), 
 they produce no stress on the frame, and may be left out of 
 account. 
 
 Then a h upon the line of loads will represent the effective 
 load, and a P and h P the supporting forces to sustain that load. 
 The loads at 2 and 3, and the letters K and L could, therefore, 
 be omitted in Figs. 1 and 2. 
 
 6th. Commence at some joint where all but two of the forces 
 are known, draw the polygon of forces for that joint, and by 
 proceeding from one joint to another complete the diagram. 
 
 Commencing at the left hand joint of the truss Fig. 1, and 
 
 "W 
 
 leaving out the load as above directed, v/e have on the line 
 
 1 i 
 
 of loads, Fig. 2, the distance a P, upon which to construct a tri- 
 angle having its sides parallel to 3 4 and 3 5. 
 
 Draw through , Fig. 2, a line parallel to 3 4, Fig. 1, and 
 through P a line parallel to 3 5, and mark their intersection by 
 the letter A. 
 
 Then pass to the joint at 4, draw lines through A and b of 
 Fig. 2, parallel to 4 5 and 4 G of Fig. 1, and mark their inter- 
 section with B. Through B draw a line parallel to 5 G to meet 
 a line through P, parallel to 5 7, and mark the intersection C. In 
 a similar manner complete the diagram represented by Fig. 2. 
 
 The advantage o^f this notation lies in the fact that, if the let- 
 ters in the angular spaces about any joint are taken in order, 
 and the same letters, taken in the same order, are found in the 
 diagram, at the vertices of the polygon representing the forces 
 at that joint, the diagram is correct. Thus in Fig. 1, for the 
 joint 3, we have aAPa, and in Fig. 2, aAPa forms a closed 
 polygon. Reading about joint 1 of the frame we have dDJ&cd, 
 and t/DEeo 7 forms a polygon in the diagram. 
 
 Failing Cases. Whenever we find on arriving at any joint,
 
 22 
 
 RECIPROCAL DIAGRAMS. 
 
 more than two unknown forces, this method apparently fails, 
 since an indefinite number of polygons may then be drawn with 
 corresponding parallel sides. A careful inspection of the truss, 
 in connection with a sketch of the general outline of the diagram, 
 will often reveal some condition fixing the relations of the lines 
 in the diagram. 
 
 This is the case in Fig. 77, A. M. Arriving at joint 5, we find 
 the forces parallel to 5 4, 5 7, and 5 1 unknown ; but by sketching 
 a diagram for this joint, we see that the lines representing the 
 stresses on 4 7 and 1 7 must overlap or be portions of the same 
 
 Fig. 3. 
 
 Diagram of stresses for half the trass represented lij Fig. 77, A. M. 
 Letter that figure in a manner illustrated by Fig. 1. 
 
 line, in order to read correctly for the joints 4 and 7 ; and since 
 5 7 and 7 1 make equal angles with 1 3, the lines on the diagram 
 representing their stresses must intersect midway between lines 
 drawn through e and/, parallel to 1 3. Hence, as CD is the line 
 to be drawn, we may draw the line from C, parallel to 5 4, to a 
 point D, half way between the parallel lines above mentioned. 
 
 There will be no difficulty with the remaining part of the 
 problem. 
 
 Or otherwise : Since the stresses on 5 7 and 5 6 are seen to be 
 equal, and make equal angles with the rafter, we may draw an 
 auxiliary line from B parallel to 5 4, till it meets a line through 
 
 NOTE. For an extended discussion and application of these principles, the student 
 is referred to " Economics of Construction,'' by II. II. I'.ow, C. E.
 
 ARTICLES 160-169. 23 
 
 e parallel to the rafter, which gives E. From E and C draw 
 lines parallel to 5 7 and 5 4, which will intersect in D. It will be 
 seen that, this makes ED = BC, or the stress upon 5 7 equal to the 
 stress upon 5 6. 
 
 PROBLEMS. Draw diagrams for figures 79 and 82, A. M. 
 
 ARTICLE 160. 
 
 EXAMPLE 1. The T 67 of the first, and T 45 of the last equations 
 at the end of this example, should bo omitted. 
 
 ARTICLE 167. 
 
 The funicular polygon is defined in Article 152 instead of 
 Article 150. 
 
 ARTICLE 169. 
 
 Equation 8. Combine the equation a^ -{- x. 2 =. a, with the 
 equation obtained from the proportion y v : 7/ 2 ' * x\ ; x\. 
 
 x 2 x 2 
 
 Equation 9. See equation 5 ; in = -L = ; xf = 4 my v 
 
 4yi 4y 2 
 
 2 __ ,2 
 
 x\ = 4 m y 2 .-. x* -f a:' = 4 m (y, + y t ), .: m = 
 
 4 (y, 4- y 2 ) 
 
 The remaining part of the equation can be obtained by sub- 
 stituting the values of x l and x. 2 in equation 8. 
 
 = fds = C A/ d x* + d if = fdy 
 
 dy> ' _ 
 
 From the equation of the curve. ! = 4 m y; ~=- =A/^L 
 
 dy x ' y ' 
 
 .:*=f<l!> . y 
 Then by Integral C, we have 
 
 ^r lo Se 
 
 = v/y v wi + y + m io ge ( vy + v in + y ) + c. 
 
 When y = 0, s = 0. .-. =z ??i log e v/Hi -(- C or C m \og e Vm,
 
 ARTICLES 169-172. 
 
 Vy (,n + y] -j- m (log, (^y + \/m + y ) log e 
 log e ( V^ + A/^JJ^/) _ 
 
 _*y _Jl _)_ 1 = m tan ' . sec i. See equation 6. 
 
 .-. ,9 = m -< tan i . sec i . -j- log e (tan * -f- sec ') 
 
 (m \^ 
 
 -+- 1 1 .by the binomial 
 
 formula, multiply by dy and integrate two terms of the series. 
 
 ARTICLE 171. 
 
 THE LAST OF EQUATIONS 1. Place the letter G at the intersec- 
 tion of AB and EF prolonged. 
 
 W 
 
 Then DE = | EF = EG !L_Y since EF : EG \\ ef 
 
 (since E G = | B C, for A E = \ A C). 
 
 At the top of page 171. 
 
 Place the letter A in its proper position in Fig. 86. y, is the 
 projection of the line DE upon a line drawn perpendicular to 
 AC. 
 
 ARTICLE 172, 
 
 Equation 11. In Fig. 87, A. M., draw lines through B and 
 x perpendicular to the arrow line through P. It will be seen 
 
 that t . cos i = - . cos / .-. i = arc . cos I cosy) 
 
 \2 1 /
 
 ARTICLES 172-174. 25 
 
 Equation 12. Equation 16 of Art. 169, gives the length of a 
 parabola, from its vertex to a point whose tangent makes the 
 angle i with the axis of x, tangent at the vertex. Equation 1 2 
 gives the length of that portion of the curve included between 
 two points where the inclinations are i and/. This equation is 
 obtained directly from equation 16 Art. 169, by taking that in- 
 tegral between the limits i and /. 
 
 Equation 13. In equation 17, Art. 169, substitute x -f y sin 
 / for x, and y cos/ for y. These values are obtained by an in- 
 spection of Fig. 87, A. M. 
 
 AETICLE 174. 
 
 d 2 u u 2 du d 2 u ludu , /du?\ 
 
 Equation (a). --- - = ; - = - or d I I 
 
 v ' d x 2 a? dx 1 a 1 \dx 2 / 
 
 du 2 u? a d u 
 
 Let v u' 2 -)- a' c z -[- u 
 
 then u 2 -f a 2 c = z 2 -f 2 z u -f- w 2 or a 2 c = z* + 2 z u. (A) 
 Differentiate and divide by 2. 
 
 2 + u 
 
 0= z d z -\- z d u -\- u d z or w =: rt 2 
 
 z 
 
 .-. (? x = a -- .-. x = a log e z ~h a 1&: c i- ^ n which 
 a log e c t is the constant of integration. 
 
 .. = log c Z + log, Cj = log e - 
 
 c, c. 
 
 From equation A above, we have u = 
 
 2 z 
 
 j - - 
 
 e e =Ae Be 
 
 2c, 2 
 
 Equation (c). From Equation 1 we have . = y. Hence, 
 
 differentiate equation (fc), and divide by d x ; the result will be 
 equation (c).
 
 26 
 
 ARTICLE 175. 
 
 Equation 6 is derived from (4), by multiplying by 2 e a , 
 which, reduces it to the form 
 
 AKTICLE 175. 
 Equation 4. By squaring both members of equation 3, and 
 
 solving with respect to 
 
 X <l X 
 
 - , we and -, = 
 d s d ~ 
 
 The 
 
 numerator of the second member is m, and not m 2 . 
 
 Equation 5. Integrate equation 4 by means of integral B. 
 Equation 8. Divide both members of Equation o by m, and 
 
 X g / 72 
 
 write the resulting equation in the form e m -|- \ 1 -(- ~ i - 
 
 Substitute the value of from Equation 6, for the first term 
 
 1 / =L -*\ 
 of the last member, and transpose. The result is - I e * -(- e I 
 
 |/ 1 -|_ -f_ . This value in the first part of the first value of 
 
 y (Equation 8), gives y = y^s 2 + m* m. 
 
 Equations 11. Combine x l -J- x 2 = h, and a;, 2 = ^. 
 
 Equations 12. Putting these values of .r in Equations 6 and 
 8, the following will be found to be the correct results : 
 
 ' 
 
 y,-,^^^ ""- "'"). (c- '"-, ') 
 
 NOTE. The reference in the last line of 
 of 173. as in some editions. 
 
 Article 174, instead
 
 ARTICLE 175. 
 
 27 
 
 Properties of the Catenary. 
 
 I. The radius of curvature may be found by making proper 
 substitutions in the formula 
 
 p = 
 
 d y 
 
 or as follows : 
 
 P = d 
 
 in which ds = (dx* + di 
 d y s 
 
 From Equations 1 and 3, 
 
 = 
 ax 
 
 d 9 y 
 
 = 
 a x 
 
 d s 
 
 dx. ds / ds V* 
 
 d 2 y = - .-. p = d x . d x . d s = m I - - 1 = m . sec 2 ^. 
 m \ d x ' 
 
 II. From the Equation 
 y' 1 = .s 2 -j- m-, (obtained from 
 8 by transferring the origin 
 of coordinates to a distance 
 m below the vertex,)it will 
 be seen that, if a right tri- 
 angle be constructed, having 
 y for its hypotenuse, and m for the side adjacent to the foot of 
 the ordinate, the remaining side will be s, or the length of the 
 curve from the vertex to the point having y as its ordinate. 
 
 d y s 
 
 From Equations 1 and 2. we have tan i = r^- == - 
 
 a x m 
 
 Hence, if the lines are drawn as B D and D A in Fig. 4, tan i 
 
 = tan < A B D = = - ; hence, the line A D is tan- 
 
 JJ -B in 
 
 gent to the curve at A. 
 
 If we draw A C perpendicular to A D, we have from the 
 similar triangles A B D and A B C, in :y\'. y : A C .-. A C 
 
 = - = m I -- 1 = m . sec- A B D = m . sec 2 i = p. See 
 m \ m / 
 
 Equation 17. 
 
 t NOTE. The line A D, prolonged, does not necessarily pass through O.
 
 28 
 
 ARTICLE 83. 
 
 Fig. 6. 
 
 IV. In any 
 position of the 
 rolling p a r - 
 abola, a line 
 drawn from 
 the focus to 
 the point of 
 contact, will 
 be normal to 
 the curve de- 
 scribed by the 
 focus, since it 
 is the instan- 
 taneous radius 
 vector of the 
 describing 
 
 point. (See A. M. at the bottom of page 401.) 
 
 The following shows that this line is equal to m . sec 5 i : 
 Let B, in Fig. 6, be any point in a parabola whose focal dis- 
 tance is w, F its focus, and B E a tangent at the point B, making 
 the angle ft with the axis X. The angle E = < B (from 
 Analytical Geometry), = 90 ft, from figure. 
 
 When the parabola has rolled till B comes into X at B 1 , the 
 line B E will coincide with () X, the line F B will be at F 1 B 1 and 
 make an angle = 90 e ft with X, since a line drawn from B 
 to the focus F will retain its position relative to the tangent, 
 and a line drawn through F 1 _L F 1 B 1 will be tangent to the curve 
 described by the focus, since, as before stated, F 1 B 1 is normal 
 to that curve. 
 Then ft = i. 
 From the polar equation of the parabola, we have p = 
 
 2 m 
 
 when = < B F E. (See Church's Analytical 
 
 Geometry, pp. 128 and 123.) 
 6 = 180 2 (90 ft) = 2 ft 
 = cos 2 i = I '2 sin 2 i 
 2m m 
 
 2 i 
 
 i = F B = F 1 B 1 . 
 
 ~ 2 (1 - sin't) cosH' ~ 
 
 Therefore as F 1 B 1 = m . sec 2 i, and is normal to the curve 
 described, it follows by the first and second geometrical prop- 
 erties that this curve is a catenary.
 
 ARTICLE 181. 29 
 
 ABTICLE 181. 
 
 The Geometrical Construction preceding Equations 5. 
 At this point we need to show : 
 1st. That 2 O^i = A + B. 
 2d. That ap and aq are equal to A and B.. 
 3d. That by construction, the triangle p O'q has a right angle 
 
 atO'. 
 
 4th. That qp is perpendicular to O'a, and tangent to the ellipse. 
 5th. That the angle p O'a is that made by a perpendicular to 
 the tangent qp, with the major axis of the ellipse. 
 
 1st. 
 
 Complete a parallelogram upon O'a and O'B', and call the vt r- 
 tex opposite 0', E. Then by construction, the point E is in tl u 
 prolongation of O'm, which is a diagonal of this parallelogram. 
 
 From Trigonometry, O'E* =. (2 O'm) 2 = 0'1T' 2 -f 0V -f 2 
 O'B' . O'a . cos B'O'a. 
 
 By an examination of Fig. 89, A. M., it will be seen that 
 / B'O'a = / j. Since O'A' and O'B' are conjugate diameters 
 of the ellipse, and O'a = O'A' by construction, let O'B' = A' 
 and O'a = B'. 
 
 Then we have from the preceding equation, (2 O'm) 2 = A' 2 
 -f B' 2 -f 2 A'B' cosj. 
 cosy = cos . (90 A'O'B') = sin of the angle between the 
 
 conjugate diameters A' and B' = - /; and A' 2 -f- B' 2 = A 2 
 
 .A. Jj 
 
 -f- B 2 . See equations 2 and 3 of Art. 157, Church's Analytical 
 Geometry. 
 
 .-. (2 O'm)' = A 2 + B 2 + 2 AB or 2 O'm = A + B . 
 
 2d. 
 
 In a similar manner we obtain B'a = 2 ma = A B, .-. ma 
 A-B
 
 30 ARTICLE 181. 
 
 3d. 
 
 From Geometry 2 m/, 2 -f 2 (Xm 2 = V -f . O 7 If = (2 0' m) 2 
 = (A -|- B) = ~ M \ 
 
 . . as the square of pq is equal to the sum of the squares of Q'q 
 and O'p, tlie triangle has a right angle at 0'. 
 
 4th. 
 
 In the equation on page 187, A. M., c = sec/ = as 
 
 cosj 
 
 will be seen by an inspection of Fig. 89, A. M. By construc- 
 tion, = = cos / = cosB'O''/. This last value 
 
 O' B' c r c 
 
 is obtained from the figure, where it will be seon that/ = WO'a. 
 Then as cos B'O'/t, = -rf-fin *" '^ ws that the angle formed 
 
 by joining B'c/, is 90 at a, also that as the angle at a is 90, the 
 line <] p is parallel to A'O', and therefore tangent to the ellipse 
 at B", the extremity of the diameter conjugate to 0' A'. 
 
 5th. 
 
 Call the angle p 0'<t, W. Draw perpendiculars from a to 
 0'^ and (Yq, and call their intersections with these lines, E and F. 
 
 Then E = ap sin /_p = A sin (f)() B') = A cos B'. 
 
 F = aq sin /_ q = B sin I. 1 /. 
 0V = a'E 2 -|- aF~* = A 2 cos" B' + B' 2 sin' J B'. 
 
 By comparing this equation with equation 13, Art. 95. A M., 
 and the notes upon that equation, it will be seen that E' = /_p 
 0' = the angle that a normal to a tangent makes with the 
 major axis of the ellipse. Hence 0'^ is the direction of the 
 major axis, and O'q at right angles with it, the direction of the 
 minor axis. 
 
 Equations 6 are obtained from the equation of the length of 
 any diameter of an ellipse, in terms of the axes and the angle 
 made by the diameter with one of the axes. 
 
 Let k be the angle made with the major axis, and k' the angle 
 made with the minor axes. 
 
 
 r A* sin 2 k -|- B 2 cos 2 k r A 2 cos 2 k f -f B 2 sin 2 k' 
 (Church's Analytical Geometry, page 1 G9.)
 
 ARTICLE 182. 
 
 31 
 
 The last value in equation 6, which is the value of sin V ob- 
 tained by solving the above equation, should be 
 
 A . /<? r* B 2 
 
 ARTICLE 182. 
 
 Theorem I. Let O s l represent 
 any flexible and inextensible cord, 
 fixed at s l and in the plane X Y, 
 and 'acted upon by forces in that 
 plane. 
 
 Let p be the force per unit of 
 length of the curve, and a and ft 
 the angles that p makes with the 
 axes OX and OY. 
 
 Let T tension at any point, 
 making the angles a and b with 
 OX, Y. Let TO = tension at O, 
 making angles and b with OX, 
 and OY. Let s any length of 
 the curve measured from 0. Let 
 s, = whole length of the curve 
 measured from 0. 
 
 Resolving all the forces parallel 
 
 / . dx 
 
 to the axes I since cos a = - 
 
 and cos b 
 
 - d l \ 
 
 ~ d s/' 
 
 we have 
 
 T (- j p . ds . cos a TO cos G O = 0. 
 
 T ^- 4- / . cfo . cos /3 T cos b = 0. 
 d s / s 
 
 Differentiate equations 1 and 2. 
 
 T . d (\ 4. . d T = - p . ds . cos a. 
 
 \d s* d s 
 
 T = . ds . cos B. 
 
 0) 
 
 (2) 
 (3) 
 
 dy 
 
 Multiply 3 by -. and 4 by , and 
 ds ds 
 
 add.
 
 32 ARTICLES 182-185. 
 
 '^l^-^-p** 
 
 * *(*)+*. (*') = J i (*^+*!)= 0. 
 ds \ds/~ds \dsf 2 \ c/6'- / 
 
 And 
 
 Then rf T = j> (cos a . cos a -{- cos 5 . cos /3) c/s p . di . 
 cos 0. (5) 
 
 In which cos . = the cosine of the angle between the direc- 
 tion of the force p, and a tangent to the curve. For in general, 
 cos = cos (a o) cos a cos a -j- sin a sin a = cos a cos a 
 -f- cos ft . cos b. 
 
 If the force p is normal to the curve, then = 90*, cos = 0, 
 and T will be constant, since its differential is 0. 
 
 In this case, equations 3 and 4 reduce to 
 
 T . d = - p . ds . cos a ; T . d j = - p . ds . cos ft. 
 
 Square and add the above equations. 
 
 T ' 
 
 ==PP- (6) 
 
 In which p is the radius of curvature. (See Church's Cal- 
 culus, Art. 102.) 
 
 Equations 2. See Church's Calculus, Art. 106. 
 
 ARTICLE 183. 
 
 /, x 2 x 2 
 
 x d x = w - % 
 
 /*! a . 2 yfl 
 
 and the equation above 8 should be w I xdx = iv ' 
 
 ARTICLE 185. 
 The equation following 12. When i = 0, the fraction 
 
 i cos . i sin i . 
 ^ r-^-: - is indeterminate. Its value is found bv
 
 ARTICLES 187-197. 33 
 
 taking the third differential co-efficient of the numerator and 
 denominator, and introducing the value i = 0. The result is 
 
 - - Hence for i = 0, p y = w r im ~\ 
 
 Equal ion 13 is found in a form more available for practical 
 use, on page 423 of Rankine's Civil Engineering. 
 
 Equal ion 16. The cos/ of this equation should be sin/. 
 
 AETICLE 187. 
 Equation 4. In Equation 2 we have H = max. value of 
 
 P, . ; max. value of = T * Differentiating both numer- 
 d x d x 
 
 ator and denominator with respect to y to find the value of this 
 
 d. P, 
 
 dy 
 
 vanishing fraction, we have H = - (when y 0). But 
 
 (IP, 
 
 2^ Q . Hence, H = d- x (for y = 0). 
 
 Equation 2. P sin = 
 P 
 
 ARTICLE 191. 
 P 
 
 ARTICLE 197. 
 
 Equations 5 and 6. The last term of the denominator of the 
 last fraction of each equation should be cos 2 0. 
 
 Equation 9. In the equation cos 2 \L, = ^, 
 
 P\ lh 
 substitute the values of i>^ _/> and p. a from Equations 1, 7, and 8, 
 
 cos 2 () 1 
 and reduce to the form cos 2 ^ = : 
 
 sin (b
 
 34 
 
 cos 
 
 \/cos* 
 
 AI 
 COS 2 
 
 JTICLES 
 
 But 
 
 197-199. 
 cos 2 
 
 1 
 
 (1 _ cos 2 0\ 
 
 = - 
 
 sin 
 
 sin 
 
 sm -. 
 sm 
 
 
 sin 
 
 
 sin $ / 
 4/1 ^ 
 
 Vcos' 2 cos' 2 
 
 </< l 
 
 cos' 2 
 
 0)-(l- 
 
 COS' 2 fo) 
 
 .-. cos 
 
 sin 
 2 4, = cos V 1 
 -i sin 
 
 sin' 2 ( 
 
 I 
 
 sin 2 
 
 - sin 
 k 
 
 sin 
 
 sin 
 -i sin i 
 
 sin-0 
 3 
 
 = cos . cos sin -; sin . sm, sm -; 
 
 sm sin 
 
 / . -i sin e\ 1 / . -i sin 0Y 
 
 cos 104- sm -s I .-. ,iL = - I 4- sin -. I 
 
 \ ' sin 07 2 \ ' sm 07 
 
 By changing the construction of Fig. 57, A. M., as directed 
 in Case 2 of Art. 1 1 2, to adapt it to this case, it will be evident 
 
 A 
 that 2 x n' = 2 ^/, as measured in that figure, is > 180 .-. ^/ is 
 
 > 90. Therefore sin ' must be greater than 90, in 
 
 sm 
 
 order that Equation 9 may be true. 
 
 The reference near the bottom of page 216, should be to 
 Problem IV, c. 
 
 ARTICLE 198. 
 
 PROBLEM. Find the pressure against a vertical wall 12 feet 
 high, sustaining a bank of earth which slopes backward from 
 the top of the wall at an angle of 30, the angle of repose of 
 the earth being 45, and its weight 100 Ibs. per cubic foot. 
 
 Ans. 3600 / -\ = 1671 Ibs. nearly. (Length of wall 
 
 = 1 foot.) 
 
 ARTICLE 199. 
 
 PROBLEM. A prismatic column of solid masonry 80 feet high, 
 and weighing 120 pounds per cubic foot, is to be built upon a 
 foundation of earth whose angle of repose is 30. What is 
 the least depth below the surface of the ground at which the 
 foundation course should rest, if the surface of the ground is 
 horizontal, and the earth weighs 100 Ibs. per cubic foot ? 
 
 Ans. ICf ft.-
 
 ARTICLES 199-215. 35 
 
 Equation 5. Draw a trapezoid whose base b is horizontal, and 
 whose parallel sides are vertical and equal to p' and w x, respect- 
 ively. The ordinates of this trapezoid will represent intensity 
 of pressure at any point, and its area, the total pressure upon 
 the surface. Divide the trapezoid into a triangle and a rectan- 
 gle by a line through one extremity of the shorter vertical, par- 
 allel to the base. The distance c b is obtained by equating the 
 moment of the whole trapezoid about the center of b, to the 
 sum of the moments of the above mentioned rectangle and tri- 
 angle. The moment of the rectangle = .-. b . - ~ . b c 
 
 2 ' 6 
 c . , In this equation introduce the value of p' 
 
 as given in Equation 2 of this article, and the resulting equation 
 will be the last value of c given in Equation 5. 
 
 ARTICLE 202. 
 
 Equations 10 and 11. From what precedes these equations, 
 it would seem that they should be obtained by substituting 
 
 and for - in Equations 5 and 7 respectively. 
 The following are the results obtained by such a substitution : 
 ; r-r- w t (x z -\- h x) and w I (x 2 -4- h x}. 
 
 '>/t?lll V ' ' oil V ' / 
 
 ARTICLE 214. 
 
 To obtain Equations 4 and 5 from Equations 1 and 2, divide 
 the numerators and denominators of the second members of 
 those equations by A/~X~, and then make x = infinity in the 
 resulting equations. 
 
 ARTICLE 215. 
 
 Equation 7. Conceive of a circular chimney whose external 
 and internal diameters are t and t 2 B respectively. The 
 outer circumference is TT t, which, multiplied by b, and placed 
 equal to the sectional area of the masonry, gives 
 
 r lt = ^ - I (t - 2 B)' .-. 1 = B (l - 1) .
 
 36 ARTICLES 217-260. 
 
 ARTICLE 217. 
 
 NOTE. The arm of the couple will "be easily obtained by draw- 
 ing from D and F, Fig. 101, A. M., lines perpendicular to H P; 
 from F a line parallel to H P, and a horizontal line through D. 
 
 The first of these lines is - cos 6. The remaining term (y -}- !) 
 t . sin (0 -f- i) is easily obtained. 
 
 ARTICLE 225. 
 
 " Find the center of gravity of the load between the joint of 
 rupture C, and the crown A, and draw through that center of 
 gravity a vertical line." See Fig. 107, A. M. "Then if it be 
 possible, from one point in that vertical line, to draw a pair of 
 lines, one parallel to a tangent to the soffit at the joint of rup- 
 ture, and the other parallel to a tangent to the soffit 'at the 
 crown, so that the former of these lines shall cut the joint of 
 rupture, and the latter the keystone, in a pair of points which 
 are both within the middle third of the arch ring, the stability 
 of the arch will be secure ; and if the first point be the point of 
 rupture, the second will be the center of resistance at the crown 
 of the arch, and the crown of the true line of pressures." 
 Rankine's Civil Engineering, page 442. Let the student make 
 the above calculation and construction for an assumed circular 
 arch. 
 
 ARTICLE 234. 
 
 Example 1. The value of P x is obtained by remembering 
 that the thickness is small compared with the radius of curva- 
 ture, and that the surface of a zone is equal to the circumference 
 of a great circle, multiplied by the altitude of the zone. 
 
 ARTICLE 235. 
 
 Equation 5. This equation will be found in Chauvenet's 
 Trigonometry, Formula 53, Page 163, and Equation 6 in 
 Formula 355, Page 256. 
 
 ARTICLE 249. 
 See reference made in Article 263, to this Art. 
 
 ARTICLE 260. 
 
 After CD in equations 1, read a x. At the end of next to the 
 last line on page 282, read, princiml elementary A-A/Y////.S.
 
 ARTICLE 271. 37 
 
 ARTICLE 271. 
 Equation 1. See Art. 179, Theorem. 
 
 ARTICLE 273. 
 
 Equation 9. In Equation 6, let p = f, divide both numerator 
 and denominator of the last value of p by r 2 , and solve the resul- 
 
 T> 
 
 ting equation with reference to -- 
 
 Figure 119. The general equation of Hyperbolas of Higher 
 Orders, is y m x n = a. 
 
 x y = a is the equation of a hyperbola of the first order, 
 referred to its asymptotes, x y 1 =. a is the equation of a hyperbola 
 of the second order. 
 
 .-. xy 2 a = z' y' 2 . (A) 
 
 Take R as the axis of Y, and a vertical through 0, as the 
 axis of X. 
 
 Then we have 
 
 x : x' ; ; y n : y 1 from equation (A) 
 
 r-A:BB::R_:r- I a* A. M. 
 
 : : o K* : o r f 
 
 By comparing these proportions with equation A, we have 
 rA X r 2 = EB X R 2 = a. (B) 
 
 The area r ABE 
 
 C* x dy = C* dy 
 
 J r V f- * 
 
 (-1\ 
 ), 
 
 By comparing equation C with the last equation in case 2, it 
 will be seen that this area represents case 2, and that a = 
 
 r A x r\ (E) 
 
 From the proportion C A : D B ; ; q : q l , we have C A q l 
 = D B </ . By introducing the values of q l and q , as found in 
 
 the equations at the bottom of page 292, A. M., the equation 
 
 q , 
 . M.,
 
 38 ARTICLE 277. 
 
 becomes - m CA = - _ m DB = (RB - TO) C A 
 
 ( r ~A m) D~B 
 
 R B x C A 7A x D~B = TO (CA DB) m fA-RB) 
 RB (^A rC) (r^A . R~B - H3) = m (rA R~B) 
 r~C (iA - RB) = 7/1 (^A - RB) 
 .-. m = 7G. 
 
 Hence it will be seen that r C represents m of case 1, in the 
 solution of that case. 
 
 Since r C is m, and r A X r* , 
 We have from equation 4, q = m = _ - -- r C. 
 
 But r~A : x' ; ; r' 2 : r 2 .-. ^- = *. 
 
 .: q = x' rC = the segment of the ordinate correspond- 
 ing to y ? y , between C D and the curve A B. 
 Again, p = + m = a/ + r~C = a' + rlS. 
 
 = the entire ordinate from E F to A B. 
 From 6, we have 
 
 Po ++ m = -TA2H1+ r-O - r-A + r~E = AE. 
 
 ARTICLE 277. 
 Differentiate Equation ]. 
 
 ws dx=fds- ^-dx = If. Then^ = log e S + C. 
 
 W 
 
 When a; = 0, S = ; see equation (1) A. M. 
 
 Then = log, ^L + C or C = - log e -^. 
 
 ' 
 
 =J! /S W !Lf! 
 
 e x = V r "7" e '
 
 ARTICLES 295-298. 39 
 
 AETICLE 295. 
 
 Table. Bead m' = -, instead of -. 
 h y 
 
 ARTICLE 298. 
 
 ft] nation 1. Let G, , G. t , G s , and G be the centers of gravity 
 of A, , A 2 , A 3 , and A respectively. 
 
 Take moments about a horizontal line passing midway between 
 the top of the uppar f_a ige, and the bottom of the. lower flange. 
 
 A; = A, (^LVM. _ | : ) _ Ai ^^A+A, _ ^ 
 
 _h_ h_(l h +l>^ A, - (h,+l, 3 -) A 1 -(/^~7 <1 ) A 3 
 
 ' ' Jb ~ 2 2 2 A ' ( ' 
 
 .Equation 2. 
 
 _ A, (S, + *., + 2 /,.) + A. (*, + *,) 
 ~~~~ 
 
 , | 
 
 22 2 
 
 A, (/,, + ^ 2 + 2 7, s ) 4 A 3 (A 2 
 
 _A 2 (/, 2 -f /.,) - 
 ~"~ 
 
 
 The moment of inertia of the section, about its neutral axis 
 will be 
 
 -j -f A, X G~^ + A 2 X G, 
 
 See the end of Art. 95, A. M.
 
 40 ARTICLE 298. 
 
 12 ' 4A 2 L 
 
 4 2 A, A 2 A 3 (A, 4- A 2 4- 2 A 3 ) (A, 4- A 3 ) 
 4 A! A 3 2 (A x 4 A 3 ) 2 , 4 A 2 A : 2 (A, + A 2 4- 2 A 3 ) 2 4- 2 A l A, A,, 
 
 (A, 4 A 2 4- 2 A 3 ) (A 2 4 A 3 ) 
 
 4 A 2 A" (A 2 4- A 3 ) 2 , 4 A 3 A 2 (A 2 4 A 3 ) 2 - 2 A, A 2 A 3 (A 2 4- A :! ) 
 
 (A, -t- A 3 ) 4- A, A 2 (A, 4 A 3 ) 2 J. (D) 
 
 But 2 A l A 2 A 3 (A x 4- A 2 4 2 A 3 ) (A, 4 A 3 ) = 2 A L A 2 A 3 (J tl 4- A :! ) J 
 
 + 2 A, A 2 A 3 (A 2 T A :) ) (/*! + A,)- 
 2 A! A 2 A 3 (A x 4 A 2 4- 2 A 3 ) (A 2 4- A 3 ) 2 = 2 A, A 2 A 3 (A 2 -f A 3 ) 2 
 
 ' 4- 2 A, A 2 A 3 (A 2 4 /< 8 ) (Ai + A 3 ). 
 .-. 2 A t A 2 A 3 [(A! 4- A 2 4 2 A 3 ) ((Aj 4- A 3 ) 4 (A 2 4 A 3 )) (A 2 4- A 3 ) 
 
 = 2 A, A 2 A, [(X -f A,)' 4 (A : 4 A 3 ) (A 2 , A 3 ) 4- (A 2 + /,,)]. 
 = A, A 2 A 3 [(A, 4 A 3 ) 2 4 2 (h, + A 3 ) (A, 4 A 3 ) 4- (A 2 + A,)'-'] 
 
 4- A t A 2 A 3 (A, 4- A 3 ) 2 4- A t A 2 A 3 (A 2 4 A 3 ) 2 
 = Aj A 2 A 3 (A! 4- 7i, -f 2 A 3 ) 2 4 A, A 2 A 3 (A x 4- A,) 2 ^ A 2 A 3 
 
 (A 2 4- /,). 
 This reduces equation D to the form 
 
 A 2 A t 2 4 A, A 2 A s ) 4- (h, 4 A 3 ) 2 (A : A/ 4- A 3 A/ 
 
 4- A, A 2 A,) 4 (A 2 4 /.,)' (A, A 3 2 4- A 3 A 2 2 4- A, 
 
 A 2 A 3 )J. (E) 
 
 A, A 2 2 4 A 2 A 2 4 A, A 2 A 3 = (A, 4- A 2 + A 3 ) A l A. 2 = A A, A r 
 
 A, A, 4 4- A 3 A 2 4 A! A 2 A 3 = A A, A,. 
 
 A 2 A 3 2 + A 3 A 2 2 + A, A 2 A 3 = A A 2 A & . 
 
 These values reduce equation E to the form 
 
 
 4 A, A 3 (A t 4 A 3 ) 2 4- A 2 A 3 (A 2 4- A :i ) 2 ]. (2) 
 Equation 3. See Art. 294, A. M. 
 
 E<] nation 4. This equation can be obtained by considering A,, 
 // and A 3 as being so small that they may be left out of consid- 
 eration. We should then have A = A t 4- A 2 and A' = A 3 = A. 
 These values in equations 1 and 2, give
 
 ARTICLE 300. 
 
 A 1 // A 2 h' A' h l / Aj + A 2 A 2 + A l 
 / %J = TV ~A~ 
 
 [ = _L (A t A 2 (2 //)*) =- h ' ' Al 4l . 
 4 .A. A. 
 
 Introduce these values in equation 3 
 
 /.//'. A, A. 
 
 Then M = mWZ = AI = - ;/ A f -/. V A,. (4) 
 
 ARTICLE 300. 
 
 THE LAST OF EQUATIONS 2. To obtain the last value of 
 --, multiply -T-T- by = , and divide by its equivalent 
 
 d x. 
 
 , 
 
 IX. TVte r/e o/ ". By making the proper substitutions in 
 Equation 5, we have 
 
 n" c 2 = I" f - d x* = c C log e - - 
 
 t/Ot/oC X l/o"c X 
 
 Integrate by parts . / u d v = u v I v d u = the general 
 formula. 
 
 Let loff - = u and d x = d v. 
 &fl c x 
 
 Then c 
 
 r\ c c 
 
 / log. -- = c x log. -- 
 J o " e c x ye c x 
 
 cc 
 a; = z. Then d x = c?z; se = c 2 
 
 r c x dx r d z (c z) 
 
 I / s c log a 2 4- 
 
 J o c _ x / o z 
 
 ge(c-^)4-(c-^)[ o C 
 
 C ^a; = 
 
 = c -j x log c c a; log e (c x) + c log, (c z) - (c x) j- ^
 
 42 ARTICLES 307-312. 
 
 = c I x log e c -f (c x) log e (c - x) - (c x) j- o =c 2 
 .. w" c 2 = e* or n" = 1. 
 
 X. Value of n" c 2 - f f / = c f d x . 
 
 J J V C 2 _ 3.4 </ 
 
 -1 X 
 
 sin 
 c 
 
 Then = sin y ; d x = c cos y d y 
 
 .: c J d x sin - ^ c 2 / T/ cos y d y (by parts) c 2 
 
 | y sin y I sin y d y ( = c 2 } t/ sin y + cos w 1 = c 2 
 ( '-/ ) o ( ) o 
 
 . -' u, -ix - 1 u; i 
 
 sin - sin . sin -- j- cos . sm - r = 
 
 ) 
 
 c o 
 
 ARTICLE 307. 
 
 Remark. Near the bottom of page 334, read, m" is never less 
 than |. 
 
 The last member of Equation 3 should be .; , 
 
 m" m I 
 
 Equation 13 should be M M, = - - M, = 
 
 ,'(#_ a?)' 3 we 2 __ 3 w c 2 /I 4 x 2 
 2 8 ~~8 \3 ~~ ITc 2 " 
 
 ARTICLE 309. 
 Equation 3. Read / y z d y for / y x d y. 
 
 J I/O 
 
 ARTICLE 312. 
 Equation 7. Read for the co-efficient of the second mem-
 
 ARTICLES 314-318. 43 
 
 // "I Q f\ 
 
 ber. Also for (below 7), we should have -^r = 1Q9 
 nearly. 
 
 ARTICLE 314. 
 
 In the first place, suppose the external load W 1 is made up, 
 in part, by the weight of the required beam. Then, having cal- 
 culated the breadth, and found the weight B 1 of the beam, it 
 follows that, if W 1 represents the whole load, W B 1 will 
 
 represent the load exclusive of the beam, or ^ the 
 
 ratio of the gross load to the load exclusive of the beam. 
 
 Equation 1. The student needs to bear in mind that & 1 is the 
 breadth to bear the gross load, minus the weight of the beam, 
 and W is the external load. Since h is a fixed value in this 
 case, we have from M = m W I = nfb A 2 (see page 316), and 
 above ratio, W : W 1 B 1 ; : b : b l 
 
 ' b = w'-'u' < 1} 
 
 Equation '2. The depths being constant, the weights of the 
 beams are proportional to their breadths. 
 
 b 1 W 1 W 
 
 ...B^B::^::^^::!:^^,^ 
 
 B 1 "W 1 
 \\n _ B 1 
 
 Equation 3. The true gross load will be 
 B 1 W W 2 
 
 ARTICLE 318. 
 Equation 5. From (4) 
 d-x P 2 dx . d* x P / (I x
 
 44 ARTICLES 314-318. 
 
 = - I when c 
 c \ 
 
 dx 
 
 a 
 When x = 0, y = .-. c = 0, and the above equation becomes 
 
 y x y y 
 
 or = sin . . a; == a sin 
 a c c
 
 STRENGTH AND DEFLECTION 
 
 OF 
 
 HORIZONTAL BEAMS
 
 STRENGTH OF BEAMS. 
 
 NOTATION. 
 
 I = Length of beam. 
 
 c ^ the length of a beam supported at both ends, or the 
 
 length of a beam fixed at one end, and free at the 
 
 other. 
 
 b = Breadth of beam. 
 A =: Depth of beam. 
 
 y Distance from neutral axis to outer fiber of beam. 
 to = Load per unit of length of beam. 
 W =-- Total load. 
 
 R = Force at right hand support. 
 L = Force at left hand support. 
 M = Moment of external forces, at any cross-section. 
 M = Maximum value of M. 
 F = Shearing force at any cross-section. 
 F = Maximum value of F. 
 
 I = Moment of inertia of cross-section, about its neutral axis. 
 E = Modulus of elasticity. 
 / - Intensity of stress at the outer fiber. 
 y l = Maximum deflection of beam. 
 m = A constant factor depending upon the method of loading 
 
 and supporting. 
 
 n = A constant factor depending upon the form of the beam. 
 k = Coefficient of maximum deflection. 
 O = The Origin of a system of rectangular coordinates, in 
 
 which the axis of the undefiected beam is taken as the 
 
 axis of X. 
 
 The following notes on the strength and deflection of beams, 
 contain the substance of what is found in the A. M., from Arts. 
 288 to 308, inclusive, and may be substituted for those articles. 
 It is believed that, from the simple demonstration of the fun- 
 damental formulae for the restricted case of horizontal beams 
 with parallel vertical loads, the average student will proceed to 
 their application more rapidly and with greater independence 
 and security than by following the Applied Mechanics. These 
 pages are therefore intended to be complete in themselves. 
 
 Rankine's notation has been followed as far as practicable, 
 and whenever introduced the letters stand for the same quanti- 
 ties as in the A. M.
 
 STRENGTH OF BEAMS. 47 
 
 All the notation will be found explained on page 46, or in 
 connection with the subject where it occurs. 
 
 In all problems relating to the strength of materials, two dis- 
 tinct systems of forces must be considered. 
 
 1st. The external forces, consisting of the applied loads, and 
 the supporting forces. 
 
 2d. The stresses, or forces acting amongst the fibers or par- 
 ticles of the material. 
 
 As these two systems of forces stand related to each other as 
 cause and effect, it follows that, so long as equilibrium exists, the 
 two systems must balance each other. 
 
 If we conceive of a cross-section made at any point of a beam, 
 we can imagine the beam broken by the particles sliding upon 
 each other, along that section (shearing), or by the pulling apart 
 or crushing of the fibers in a direction perpendicular to the sec- 
 tion (cross breaking). Since a part of the fibers are extended 
 and part compressed, these stresses, of opposite kinds, form a 
 moment about the neutral axis of the section which tends to pre- 
 vent the beam from bending, and which is equal to the moment 
 of the external forces, or "bending moment," M. 
 
 To find the value of F, take the algebraic sum of all the ex- 
 ternal forces acting upon the portion of the beam included be- 
 tween the section and one end, considering the upward or sup- 
 porting forces as positive, and downward forces or loads upon 
 the beam as negative. 
 
 M is found by taking the moment of all the external forces 
 acting on the beam between the section and one end, about the 
 neutral axis of that section. We have found it convenient in 
 most cases to use right hand rotation as negative, and left hand 
 rotation as positive. 
 
 The maximum values (F () ) of F, and (M (l ) of M, are generally 
 required. The former is evidently some part of the whole load 
 (W), and the latter some portion of the product of the load and 
 length of the beam, and is expressed by the equation M 
 mWl 
 
 Beams of sufficient strength to resist the bending moment, 
 have usually ample strength to resist the shearing force ; hence 
 any further consideration of it here, will be omitted. (See Art. 
 309 A. M., and Table TI in the Appendix.) 
 
 To determine the manner in which the beam resists the bend- 
 ing moment M, notice that when the beam is slightly curved, the 
 layers of fibers upon the convex side are extended, while those 
 upon the concave side are compressed, so that there will be one 
 surface in which the fibers are neither extended or compressed.
 
 48 STRENGTH OF BEAMS. 
 
 This is called the " neutral surface." It is assumed, (which is 
 near enough to the truth for most practical cases,) that this sur- 
 iace passes through the center of gravity of the cross-sec- 
 tion of the beam, and contains the neutral axis of that section. 
 
 Within the limits of proof stress, (see Art. 245 A. M.,) the co- 
 efficients of extension and compression are exactly or nearly 
 equal. From Hooke's law that the strains are proportional to 
 the stresses producing them, we have zero as the stress at the 
 neutral surface, and at any other point it is proportioned to the 
 distance of that point from the neutral surface. 
 
 SECTION 1. Moment of Bending Stress. 
 
 Let p = the intensity of stress at any distance y, fromOZthe 
 neutral axis of the section. 
 
 Let / = intensity of stress at outer fiber. 
 
 d y . d z = an elementary area. 
 p . dy . dz stress on an elementary area. 
 p . y . dy . dz = moment of this stress about the neutral 
 axis. 
 
 ' d . dz . = the total moment. 
 
 / / 
 
 / J ~f 
 
 Since the stress is zero at the neutral surface, and increases to 
 / at the outer edge, and is a uniformly varying stress, we have 
 
 2> .: / : : y : ye w j = y. 
 
 This value introduced in the above integral gives, 
 
 2/0 
 
 which is the moment of bending stress at any section. A few 
 values of I are given in a table at the end of Art. 95, A. M. 
 
 From these values it will be seen that I = a b A 3 , in which a 
 is a constant, depending upon the form of section. 
 
 Also, that y = dxh, where d is a constant, depending upon 
 the position of the neutral axis of the section. 
 
 .-. - - = - =n f b 7t 2 , where n is a constant and = - T - 
 y d h d 
 
 ome 
 
 (B). 
 
 Hence the general formula for the moment of bending stress 
 at any section is,
 
 STRENGTH OP BEAMS. 
 
 TABLE ]. 
 
 49 
 
 VALUES OF 
 
 Rectangle b h, including square, 
 
 Hollow Rectangle, b h on outside, and b h on 
 inside, 
 
 Circle and Ellipse, , 
 
 Hollow Ellipse, outside diameters b and h, in- 
 side Z> y and hj, 
 
 32 
 
 tl - 
 
 V 
 
 Hollow Circle, diameters h and h , 
 
 - 1 - 
 32 
 
 For other forms of cross-section, calculate n from the formula 
 1 
 
 SECTION 2. Transverse Strength of Beams of Uniform Section. 
 
 Calculations upon the transverse strength of beams are based 
 upon the principle of equality of moments. The maximum 
 moment of the external forces must be equated to the moment 
 of bending stress at the cross-section of the beam. 
 
 Thus 
 
 M orWJ ) ( fl 
 
 Mo. of external forces j ^ ~~^~ n J h I /Q\ 
 
 Mo. of internal stress 
 
 PROBLEM I. To find the safe load for a given beam. 
 
 The modulus of rupture of ash being 12,000 pounds per 
 square inch, find the safe load that may be uniformly distrib-
 
 be M = R (c x) w (c x) ( c ~~ x \ = 
 
 50 STRENGTH OF BEAMS. 
 
 uted over a rectangular ash beam whose cross-section b x h is 
 3" x 12", and length I = 20/?., when supported at both ends. 
 
 As the modulus of rupture is usually given in pounds per 
 square inch (see Table IV in the appendix of A. M.), the cross- 
 section and length of the beam should have their dimensions 
 expressed in inches. 
 
 Solution I 20 x 12 = 240" 
 
 The whole load, ~W = wl = 2wc = 20 x 12 X * = 24G<r. 
 The supporting forces R and L will each be we = l'2Qw. 
 
 Let the origin be at the center of the beam, and take a sec- 
 tion at a distance x to the right of 0. Then the moment at that, 
 
 section will 
 
 B(.- )- 
 
 The maximum value of M will be obtained by making 
 
 ^ Or M,, = Re _ -. = Wff _ J^. = *. = 1 W I = 
 
 30 W. 
 
 By Art. 247 A. M., it will be seen that Rankine considered 
 10 as the proper factor of safety for timber, so that the outer 
 fiber should not be under a stress greater than 12000 -4- 10 = 
 1200 Ibs. per square inch. 
 
 .-. / = 1200. I = -^- Substitute in the equation M = 
 and we have 
 
 1200 X 
 
 2 
 
 2880 
 W 2880 Ibs., which is rr^ 144 Ibs. per lineal foot 
 
 of beam. Ans. 
 
 PROBLEM II. Knowing the form of cross-section of <i given beam, 
 and tlit loa<l, to find its dimensions. 
 
 If the modulus of rupture of cast iron beams is 40,000 Ibs., 
 find the cross-section of a rectangular beam 25 feet long, and 
 
 30W = - = 200 X 3 x 144. 
 
 NOTE. The student mii^t continunlly have in mind, that all weights or measures 
 lust be reduced to milts <>r il<e KHUH di-noniinniion 
 NOTE. The subject of lactoi-s <;!' .-alety i* e^yluined in Arts. -245-247, A. M.
 
 STRENGTH OF BEAMS. 51 
 
 b I 
 whose breadth and depth shall have the ratio - = - , for a load 
 
 /I O 
 
 of 200 Ibs. per lineal foot, and a concentrated load P = 2200 
 Ibs. at the center, the beam being supported at both ends. 
 
 ? 25 x 12 = 300" . c = y = 150" . W = (25 x 200) 
 
 -f 2200 = 7200 Ibs. w = 2 . 5 , X ^ = ] -K-'"{ Factor of safety 
 Zo X 1^ * 
 
 = Q}.f= 40000 4- 6 = 3; = g(See Table I) . MO 
 
 ; R = L = 3600. 
 
 Let the origin O, be at the center of the beam, and take a 
 section at a distance x to the right of 0. Then we have 
 x w 
 
 .-. M = Re - Y c ' = 360 X 150 - ^ y x 150 2 = 150 
 (3600 1250) = 352500. 
 
 .. 352500 = iv^v'v 
 
 = = 3.28" nearly. 
 
 PROBLEM III. 
 
 Find the strongest rectangular beam that can be obtained 
 from a given cylindrical beam. 
 
 Let D be the diameter of the circular section of the given 
 beam. We have from the equation of the circle, D 2 = & 2 -\- h? 
 .-. h? = D 2 V. Since for the transverse strength of beams 
 M = n / 5 tf, we have M = nf (D 2 b - Z> 3 ). 
 
 The value of I which makes Mo a maximum is found to be, 
 T 
 
 The corresponding value of h is, 7i D
 
 DEFLKXION OF BEAMS. 
 
 SECTION 3. Equation of the Elastic Curve for Beams. 
 \ 
 
 Resume the 
 equation M = 
 
 Let s equal 
 
 2/0 
 
 the stress on a 
 fiber at a unit's 
 distance from 
 the neutral axis. 
 Then s : / ; ; 
 
 !:,.=. 
 
 Let X be the 
 elongation of a 
 fiber at a unit's 
 distance from 
 the neutral axis. 
 
 Then d X will be the elongation of a fiber whose original length 
 was d x. 3<ufafy&4JL~ 
 
 Let p be the radius of curvature of the neutral ^axis/ Let E 
 be the modulus of elasticity, that is, the force that will elongate 
 a bar whose section is unity, to double its original length, pro- 
 vided the elasticity of the material does not change. 
 
 From Hooke's law, "As the stresses, so the strains," we have 
 
 /I 
 
 j -CJ. 
 d X 
 
 rfX BI 
 
 
 2/0 
 From the similar triai 
 
 Y /7/r 1 . x . 
 
 rfo; 
 
 igles in the figure above, we 
 1 
 
 have 
 
 
 P 
 El 
 
 . (to 
 
 But = 
 
 (dx* + dff = 
 c^a; c? 2 y
 
 DEFLEXION OF BEAMS. 
 
 Since is the tangent of the angle that a tangent line 
 
 to the curve makes with the axis of a-, and in beams is exceed- 
 ingly small when compared with unity, it may be neglected. 
 Then we have 
 
 y - = 
 
 dx* 
 
 dx* 
 
 is the equation sought. 
 
 PROBLEMS. 
 
 1. Find the maximum deflection of a beam loaded uniformly, 
 and supported at the ends. 
 Solution. (Origin at center.) 
 
 w 
 
 = wc(c x) (c- 
 
 __ 
 
 we, . w , . . , w c 3 a; _ 
 
 l y = -g- ( c x )-ii (c ~ ^ +3- +R 
 
 When x = c, y = .-. D = ^ - 
 we, . , w ., w <? x w 
 
 At the point of maximum deflection C - = 0, or the curve 
 
 a x 
 
 is horizontal. In this case it is obviously at the center, where 
 
 NOTE. All the beams in this and the following section are supposed to be of uni- 
 form cross-section, horizontal in position, and with vertically applied loads.
 
 54 DEFLEXION OF BEAMS. 
 
 5 W P 
 
 384~IT ' 
 
 2. Find the maximum deflection of a beam supported at the 
 
 1 P P 
 
 ends, and loaded at the center with P. Ans. 
 
 48 hi I 
 
 3. Find the maximum deflection of a beam supported at the 
 ends, and loaded at the center with P, and uniformly with w. 
 
 4. Find the maximum deflection of a beam fixed at one end, 
 
 I P P 
 and loaded at the free end with P. Ans. . 
 
 5. Find the maximum deflection of a beam fixed at one end, 
 and loaded at the free end with P, and uniformly with w. 
 
 6. Find the values of M and y v for a beam fixed at its ends, 
 and loaded uniformly with w, and at the center, with P. Find 
 also the point of contra-flexure. 
 
 Let w I =. W / 
 
 W 
 
 (from formula E.) where p is the moment required to hold 
 either end of the beam horizontal. 
 
 dy C 
 
 When x = c, y = .-. D = 0. 
 y is a maximum when x = 0. 
 
 NOTE. "Fixed," muaiib Uxud m direction.
 
 DEFLEXION OF BEAMS. 55 
 
 EIv = 4 C - ? C -= _/?_^4- wc *\ 
 12-6 24 8 ~ 6 ~ Y24 24 / 
 
 _ . __ 
 
 192 384 / 384 
 
 To find the value of M Q , substitute the value of p, from equa- 
 tion 2, in 1 
 
 \\ liich gives a maximum value for M, where x = c, 
 /P c w c 2 \ / P I w I-\ /3 P + 2 
 
 M -MT + ir)MT+n>-(Hs- 
 
 To find the point of contra-flexure, make -^ = 0, in equation 
 4, and we have 
 
 = it + 1 4- i/I*iJ PAv : + w /> 
 
 \2WT 2 V 12" W - " * 
 
 = distance from end. 
 
 7. Find the value of R: the point of Max. deflection (a^); the 
 Max. bending moment; and the point of inflection for a beam 
 fixed at L, supported at R, and loaded uniformly. * 
 
 R = f w Z = | W, 
 
 x t = .58 I (nearly.^ from L, 
 
 Max. moment W /, 
 
 Point of inflection at \ 1 from L. 
 
 8. Find the point of contra-flexure, and values of R, y v and 
 M,,, for a beam loaded at the center with P, fixed at L, and sup- 
 ported at R.
 
 56 DEFLEXION OF BEAMS. 
 
 First consider that part of the curve included between the 
 right hand support and the center. 
 
 E 
 
 E I y = (2 c x) 3 -f C a; -f D ....... (3). 
 
 When a; = 2 c, y = .-. D = _ 2 C c. 
 
 V B I y = -f- (2 c - ), C (2 c -^ ...... (4). 
 
 Now consider that part of the beam included between the 
 center and the left hand support, 
 
 EI 
 
 When a/ = 0, -^~ == .-. F = 2 R c 2 --- = -^- (4 R P) 
 
 (4E P) ......................... (6.) 
 
 ^') s -^ ( ')+ Y (411 - P) j 
 
 (P 8 R.) 
 
 ... E Iy> = -?- (2 c-*') 3 _ I- (o_,') 3 
 
 + (P-8R) 
 
 rf y rf y' 
 
 When x = x' = c y = y' and j = -^ . 
 
 c^ a; (/a 
 
 Under these conditions, we have from 2 and 6
 
 DEFLEXION OF BEAMS. 57 
 
 .-. C = (4 R P), and equation 4 becomes 
 
 E I y M |.. (2 _*)_ -^(4R-P)(2 C _*).. (8). 
 We also obtain from the above condition and 7 and 8, 
 
 (P - 8 R) ; = (4 R - P) -f - (P 8 R) = 24 R 6 P _|_ 
 P - 8 R .-. R = j| P .............................. (9). 
 
 To find the point of maximum deflection, substitute the 
 values of R and C in equation 2, and we have 
 
 The maximum deflection is where * = 0, hence to find 
 
 5 c 2 
 
 the point, we have = ^ ( 2 c x )' 2 + -- 
 
 .-. 5 (2 c - a;) 4 = 4 c 2 
 
 (2 c - x) = 2 c jA 
 
 x = 2 c 2 c 4/ _ = >|- /I i/ 1 r= distance from fixed 
 5 \ 5 / 
 
 end. The amornt of maximum deflection will be found by sub- 
 stituting this value of a; in equation 8, which gives ?/, 
 
 - 1 - i/i ))) 
 
 iL 1 i/ 
 
 96 ^125 32 K 5 7 E I 
 
 l. -n PJL 
 
 ;ij 967 E I 
 
 li/rpjl. 
 
 48 r 5 El
 
 58 DEFLEXION OF BEAMS. 
 
 Substitute the value of R in equation 5. 
 Then B I ~^ 2 = A p ( 2 c _ x '} - P (c - x') (11). 
 
 f/2 y' 
 
 To find the point of contra-flexure, make -r-jjjj = 
 Then P (2 c - a/) - P (c - x') = 
 
 a/ = -- - c = Z = point of contra-flexure. 
 To find M from equation 1 1 we have 
 
 M = T ^ (10 c - 5 aS \ 16c + 16.^) = ^ (11 of - 6 c) 
 Making x' = 0, we have 
 
 9. Find the same quantities for a beam fixed at one end, sup- 
 ported at the other, and loaded uniformly, and at the center 
 with P. 
 
 10. Given a beam supported at the ends and center, with a 
 fixed load over its entire length, and an additional load over 
 one span, to find R, and the point of contra-flexure. 
 
 Let w be the intensity of the fixed load, and w { that of the 
 additional load. 
 
 The tendency of this load is to curve the opposite end of the 
 beam upwards, so that instead of requiring a support it would 
 have to be fastened down at that end. 
 
 Let the left hand span be thus loaded, then R will act down- 
 ward. 
 
 Since the fixed load is distributed symmetrically with reference 
 to the center, in calculating for R, it may be considered as all 
 supported at the center. 
 
 Taking the origin of coordinates at L, and the coordinates of 
 the left hand x t and ?/,, and those of the right hand span, x and 
 y, we have the following: 
 For the latter
 
 DEFLEXION OF BEAMS. 
 
 For the former 
 
 -r d- it , w x~ 
 
 d x 2 
 
 These equations contain six constants, A, B, C, D, R, and L. 
 The six equations required for their determination, may be found 
 from the following conditions: 
 
 When x = - , y=0. When x=l, y f 0. 
 
 _ 2 _ u 2 
 
 I d y d y / 
 = X/= 2' Tx~~d^\ 
 d*y d* y/ 
 
 Forming the equations and eliminating we have R = 
 
 16 24 ' 
 
 BP T 7 jw.l\ 7 
 
 5 =: : JU =: - 
 
 y-72 . 
 
 To determine the point of contra-flexure, put ^^==0, remem- 
 bering that M must include the fixed load. 
 
 x = --- ^ = distance from left end.
 
 60 
 
 BENDING MOMENT AND DEFLEXION. 
 
 TABLE 2. 
 
 BEAMS OF UNIFORM SECTION. 
 
 YALTES OF 
 
 M n = mWl. 
 
 Vi 
 
 1. 
 j Loaded at the center with W. 
 1 Supported at the ends. 
 2. 
 j Loaded uniformly with w. 
 ( Supported at the ends. 
 3. 
 j Loaded at free end with W. 
 ( Fixed at one end, free at other. 
 4. 
 j Loaded uniformly with w. 
 \ Fixed as in 3. 
 
 5. 
 ( Loaded with w, and at the 
 ) extreme end with P. 
 ( Fixed as in 3. (w l=W). 
 
 6. 
 j Loaded at center with W. 
 ( Fixed at the ends. 
 
 7. 
 j Loaded uniformly with w. 
 \ Fixed at the ends. 
 
 8. 
 ( Loaded uniformly with w, and 
 -< at the center with P. 
 ( Fixed at the ends. 
 
 9. 
 ( Loaded uniformly with 10, 
 j Fixed at one end and sup- 
 ( ported at the other. 
 
 10. 
 Fixed as in 9, loaded at cen- 
 ter with W. 
 
 >.w, 
 
 I W P 
 
 ; 
 
 5 W I 3 
 
 384 E I 
 
 Wl 
 
 1 W P 
 
 1 
 
 . 1 W P 
 
 2 
 
 8 E I' 
 
 2 
 
 (^+H 
 
 l wi 
 
 1 W P 
 
 8 
 
 192 E I 
 
 1.WI 
 
 12 
 
 1 W / 3 
 384' IT' 
 
 (3P-|_2W\ 
 
 (2P+W,) P 
 384 E I' 
 
 24 / 
 
 ! w? 
 
 2.08 W P 
 
 8 
 
 384 E I 
 nearly. 
 
 3 
 
 i J/TW P 
 
 16 
 
 48^ 5 EI 
 
 NOTE. To calculate the transverse strength of beams, use Ihc fornuil:). m\\ I - 
 /*/*-. Take the value of in\\l from table 2 and the value of /< from pu^e 49.
 
 FORMULA FOR DESIGNING BEAMS. 61 
 
 The preceding table gives the values of Max. bending moment, 
 and Max. deflection for the most common and important cases 
 of loading and supporting. In the column headed j\i y = m W /, 
 the co-efficients of W I which are the values of m for the different 
 cases, are so arranged as to be seen at a glance and readily taken 
 
 W I 3 
 out for use. Also the co-efficients of in the column headed 
 
 W Z ! 
 
 y 1 are the values of ~k in the formula y : = k. , or co-efficients 
 
 Jii 1 
 
 of Max. deflection. (Cases 5 and 8 are not included in the 
 above remarks relating to the co-efficients). 
 
 SECTION 4. Equation for designing a learn whose deflection shall 
 not exceed a given amount. 
 
 In the preceding problems we have seen that the value of y, 
 
 k W P 
 may be expressed by the equation y\ (1). 
 
 -CJ 1 
 
 (2). 
 
 From formula 6 in section 2, we have M or m W I = - 
 mWly 
 
 . . J_ . 
 
 Substitute this value for I in equation (1). 
 
 kWPf_kfl 2 
 ' 
 
 In beams symmetrical with respect to the neutral axis of the 
 
 2k f P 
 
 section, y = h; hence (3) becomes y l = x -4- X -=- 
 
 m xu n 
 
 A 2k f I 
 
 or 7=m x .Tf x T (a) 
 
 In applying this formula, A; and m are to be taken from table 
 (2), or calculated as heretofore explained. E and f are found 
 in the Appendix to the Applied Mechanics, tables 1 and 4 re- 
 spectively ; but the value of / there given, being the ultimate 
 strength of the materials, it must be divided by a proper factor of 
 yifi'tij before being introduced into the formula.
 
 G2 li^AIuri <_>;' UMKOiiil STUENGTH. 
 
 is the ratio of the length of the "beam to the greatest 
 
 deflexion allowable for the particular case in question, and is 
 assumed from experience or from the necessary limitations of 
 the problem. 
 
 PROBLEM. 
 
 Find the dimensions of the rectangular cross-section of a 
 wrought iron beam 20 feet long, which shall deflect but 
 of its length, for a working load of 225 Ibs. per lineal foot. 
 
 5 
 
 See Table 2, 
 
 I \ See Table 1, 
 o ) 
 
 B _ 27000000. 
 
 Working strength of wrought iron, 
 / = 9000. 
 
 ' i = 3565! ? = 12 X 20 = 240"; W = 4500 Ibs; 1 
 = 1000. 
 From formula G, we have 
 
 5 
 
 h __ 2 jfe / I < "384" 
 
 T ~ "^T ' B ' ' ~~T~ * 3000 
 
 50 
 
 720 
 From formula 6, section 2, we have m ~W I = nfbh* 
 
 m Wl I X 45 X 24 81 
 
 : - := - ~ = 
 
 _ x 9000 X 
 
 SECTION 5. Seams of uniform strength. 
 
 CASE 1. Any beam of uniform strength, and uniform depth, 
 not fixed at both ends. 
 
 From the equation of the elastic curve, formula E, section 3. 
 we have
 
 BEAMS OF UNIFORM STRENGTH. 
 
 M 
 
 = *.-. E 
 
 d x> d x* ' I 
 
 M = or -:- = from formula 6, section 2. 
 
 2/0 2/o 
 
 Since the depth is constant, y is constant .-. equation (I) 
 becomes 
 
 cPy M / 
 
 E -~- = -y- = ^ = a constant (2). 
 
 a x- L y 
 
 j -,, f 
 
 = -^ x + G. 
 
 Take the origin so that when x = 0, ^ ^ 0, (generally at 
 the center or end of beam) then C = 
 
 2 _i_ D. When x = 0, y = .-. D = 0. 
 
 .-. E y = -~ x 2 ; y is a maximum when x =. c 
 
 w V ?/ 
 
 yo 
 
 .-. y. = ^= = maximum deflection, c = ? or - I ac- 
 
 2-^2/o 2 
 
 cording to the method of supporting. 
 
 E I 
 
 In section 3, we had the radius of curvature = p = rj 
 
 As E is a constant, and is constant as above shown for 
 
 this case, then p is constant, or the curve is the arc of a circle. 
 
 CASE 2. Beam of uniform strength, uniform breadth, built 
 in at one end. free at the other, and loaded uniformly. 
 
 fP U f 
 
 As before, we have E T-T- = ~ 
 d x* 2/o 
 
 Let y z = maximum half depth of beam, and li z its depth, 
 Then M = - nfb h\ 
 
 (c 
 
 v #y 
 
 & ~~7 : 
 ax- 
 
 # V_ . Vl_ _!_ J_ / *' \ 
 
 xf A* y * y " y z V c x /
 
 64 BEAMS OF UNIFORM STRENGTH. 
 
 E --- = ^ , log e (c x) -f- C 
 
 When x = -~- .-. C = -^ - log e c 
 
 .-. E y = - / log e ( Jdx. Integrate by parts. 
 
 fc( / c \ r x dx 
 
 E v = 1 & 1( >g e I I / ~ 
 
 ?/ z ( \c x / t/ c x 
 
 Make c x = z and integrate. 
 
 When x = 0, y = .-. F = ( - c log e c + c) 
 
 (c - x) - c log e c + 
 
 = -4~ ( lo &< c X ( x ~ c ) + loge ( c - x ) X (c a:) + x j- 
 y is a maximum when x = <= 
 
 ,, = _/^ = _/|_ 
 
 CASE 3. A beam of uniform strength, uniform breadth, 
 built in at one end, and loaded at the other. 
 
 2/1 == TEl/T 
 
 CASE 4. A beam of uniform strength and breadth, loaded 
 at the middle and supported at the ends. 
 
 y *2 
 "' ~ 3 E y, 
 
 CASE 5. A beam of uniform strength and breadth, loaded 
 uniformly, and supported at the ends. 
 
 E~7/ 
 
 NOTE. In case 5 integrate the expression sin~ dx by making ein~ - = z,or ^ 
 -- sin z, and it becomes cz coszdz. Integrate by parts and substitute the value of z.
 
 BEAMS OF UNIFORM STKENGTH. 65 
 
 CASE 6. A beam of uniform strength, depth, and load, fixed 
 at its ends. 
 
 For the plan and elevation of such a beam, see Figs. 144 
 and 143 of A. M. 
 
 Contra-flcxure. 
 
 As the beam is of uniform strength and depth, the radius of 
 curvature will everywhere be the same. (See Case 1.) And as 
 it is fixed at C and A, the point of contra-flexure must be, half 
 way between A and C. (Fig. 143 A. M.) 
 
 ,CB = = . ' (1.) 
 
 Maximum deflection. 
 
 It is obvious that the portion BAB is in the condition of a 
 uniformly loaded beam, supported at its ends, whose length is 
 c L Hence, as the beam is of uniform strength and depth, 
 its deflection will be found by Case 1. 
 
 /(I)' 
 
 /V f I* 
 
 y,' = ------- = -- = -^-= = vertical distance between 
 
 2^y 8E.y 32Ei/ 
 
 A and B. 
 
 By the geometry of the figure, it will be seen that the verti- 
 tical distance between C and A, is twice that between A and B. 
 
 '"' y/ = ~ ' 
 
 Moment of flexure at A. 
 
 Since the portion B A B is similar to a uniformly loaded beam 
 supported at the ends, the supporting forces are , and the mo- 
 
 ment anywhere between B and B is M = I --- x \ -- 
 
 (I-*)' 
 
 The moment at A is given by making x = 0, 
 vc 1 _ We W I
 
 66 BEAMS OF UNIFORM STRENGTH. 
 
 Moment at C. 
 
 From B to C the beam is in the condition of one fixed at one 
 end, loaded at the other with --, and uniformly with w. Hence 
 
 the greatest moment of flexure is at C, where M C = I ~r - 1 
 
 c , / "' ^ \ e _ 3 ?/; c 2 _ 3 W c _ 3 W / 
 2 "'" \ 2 7 4 ~ ~~8" ~T6~~ : : ~32~" 
 
 Comparison of widths at A and C. 
 
 From equations 3 and 4, we see that the moment at C, is three 
 times its value A. Hence, as the depth is constant, the 
 breadth at the end must theoretically be three times the center 
 breadth. 
 
 See theoretical plan of the beam, fig. 144 A. M. 
 
 Moment of flexure anywhere. 
 For the general expression for the moment of flexure, we have 
 
 where p is the moment required to hold the ends horizontal; 
 
 3 AV c 3 W I 
 
 hence it is equal to the moment at U = - = -- 
 
 16 32 
 
 see equation 4. .-. M, = w c(c x) -- (c xf - 
 
 16 
 
 W / /< ., V 3 W I 
 
 ss Ti\"*).r 
 
 = the general equation of the moment of 
 flexure for any point in the beam.
 
 ARTICLE 365. 
 
 Equation 1. The following method of deducing the formula 
 for the centrifugal force of a unit of mass leads to Equation 1, 
 and is a good exercise for the student in the application of some 
 fundamental principles of dynamics : 
 
 Let R the resultant accelerating force acting on a unit of 
 mass. Let the path of this particle be referred to three rec- 
 tangular axes X, Y, and Z. 
 
 Then since the force is measured by the product of mass and 
 
 acceleration, and the mass = 1, we have 
 
 7 - = component of R along X, 
 
 -y-f- = component of R along Y, 
 
 <P z 
 , 2 - = component of R along Z, 
 
 x 
 
 / d 1 i, \ 2 / d 1 z \ 2 / & s \ 2 
 
 + hf) + for) for) 
 
 d? s \ 2 / d? s \ 2 
 
 -j-^-J (subtracting and adding \~^-\ ) 
 
 The square of the radius of curvature of any curve is 
 
 (See 
 
 Church's Calculus, page 146, Equation 2.) 
 
 Substituting the value of the denominator of the second 
 member of this equation for the same quantity in the value of 
 
 ()' 
 
 I , 2 I Since R 2 = the sum of the squares of two quanti- 
 ties, they must be the rectangular components of R; and as 
 one of these, ( p j ^ *^ e force in the direction of the mo-
 
 68 
 
 tion at any instant, the other I I must be the deviating 
 
 \ p 7 
 
 force =. centrifugal force. 
 
 ARTICLE 390. 
 
 Equations 1 and 2. Since the radius vector is normal to the 
 curve, it coincides in direction with the radius of curvature, 
 and both are shifted through the same angle d i in the time d t. 
 The arc described by the point A, Fig. 187, A. M., is measured 
 by p d i ; or since c r is the velocity of A, by c r d t. Hence, 
 Equation (1) p d i = c r d t. 
 
 The distance through which the end T of the radius vector is 
 carried in the time d t, owing to the revolution about C, is b r 2 d t, 
 measured along the arc, whose center is at C. Its projection on 
 a perpendicular to A T is b r 2 d t cos 6, which is the arc subtend- 
 ing the angle through which r has been shifted negatively. 
 This arc divided by r gives the angle 
 
 br, 2 dtcos() . crdt br^dtcosQ / br^cosd\ 
 ; a t == = a I c 
 
 (0 r 2 cos 6\ 
 C f 
 
 ARTICLE 402. 
 
 Read - instead of n, in the paragraph preceding Eq. (6). 
 
 ARTICLE 421. 
 
 d . Q p d . A u p 
 
 Equation 2. Read j-=-*- = ? 
 
 d s ds 
 
 ARTICLE 422. 
 Equation (1). The last member should read ^ 
 
 ARTICLE 423. 
 Equation 2. The second parenthesis of the first member 
 
 /dp dn dp d p \ 
 
 should be I u -=-*- + v -=-! + w . - I. 
 
 \, - -d x dy d z d t / 
 
 ARTICLE 443. 
 The eccentricity of an ellipse is the distance from one focus to 
 
 the center, divided by the transverse axis, or ~ .
 
 ARTICLES 444-462. 69 
 
 ARTICLE 444. 
 
 X Y = V(G H* + X W 2 ). A section of the hyperbole-id 
 through X Y, and perpendicular to A B, Fig. 195, A. M., would 
 be a circle of radius X Y, and the projection of the generating 
 line in the position F E, on the plane of this section would be a 
 vertical line at a distance G H from the center of the section. 
 The point W is where this line cuts the circle, and its distance 
 from the axis of the hyperboloid is evidently 
 
 X Y = (G H 2 -f- X W*)*. Let the student draw the section as 
 above directed. 
 
 ARTICLE 450. 
 
 Equation (1). The velocity of Pj in a plane perpendicular to 
 C, is , C, PI. Since the line of action makes the angle * with 
 the axis c,, it makes (90 ',) with the direction of the motion 
 of Pj. The component of the velocity of P, along the line of 
 action, will, therefore, be a l C, P! cos (90 ',) = a l C, P, sin 
 t\. By the same reasoning the velocity of P 2 along the line of 
 action is a 2 C 2 P 2 sin i v and these are equal by the principle stated 
 in this article. Hence Equation (1). 
 
 
 ARTICLE 458. 
 
 Equation (1). r is the length of cord unwound from the circle 
 whose radius is C, P, (Fig. 198), while a distance q on the circle 
 whose radius is C l I passes the point I. These arcs are propor- 
 tional to the radii of their respective circles, or 
 
 = sin Q, 
 
 ARTICLE 462. 
 
 Equation (3). Substituting values from Equations (1) and (2) 
 of this article in the equation s = 2 I -j- 1 / rdqwe 
 
 have s = 4 r I + I / sin d q. Multiply 
 
 and divide the second member by 2 r,,, and we have * = 8 / *
 
 70 ARTICLES 463-483. 
 
 77+7.) /T ( * ; ' } sin -^ d ~^ = ** VTT+T:, 
 
 ( - cos (^ - o) + 1) = 8 rl (i + ij (l _ S in 0). 
 
 ARTICLE 463. 
 See Willis' Principles of Mechanism, page 137. 
 
 ARTICLE 482. 
 The degree of approximation attained by putting 
 
 Vc* (, r. 2 y c ^L 2 i s seen by squaring both 
 
 members of the inequality. The term ' ' is neglected in 
 the second member. 
 
 ARTICLE 483. 
 
 Equation (1). The first value of y r ' ~~^ 2 is obvi- 
 ous from an inspection of the figure of the conoid. To obtain 
 
 ? _l_ )' (/ <) \2 
 
 the equation r - = --r- 5 ^ i s to ^e considered 
 
 Z A IT C 
 
 that the length of belt for two equal pulleys of radius is L 
 2 c -(- 2 TT >, .which, placed equal to the approximate value of 
 L from Equation (3 A), Article 482, gives 2 c -j- 2 TT r fl = 2 c 
 
 which is also Equation (2) of this article. The manner of using 
 this equation for designing a pair of stepped cones in which 
 two opposite pulleys are of equal radii is explained in the A. M. 
 
 If it is required to complete a pair of cones when the radii of 
 a pair of opposite pulleys are given, the value of r Q may first be 
 found by substituting the given values for r t and r 2 , in equa- 
 tion (2). 
 
 Then replacing r Q by this value, the radius of one of the re- 
 quired pulleys may be assumed and the radius of its opposite 
 pulley calculated by solving the resulting quadratic equation. 
 
 PROBLEM. In a pair of stepped cones let the radii of the end 
 pulleys opposite each other be r, = 24", r., = 6", and let the 
 distance of their centers be c. = 48" required the radii of the
 
 AKTICLES 488-191. 71 
 
 adjacent pair of pulleys on the stepped cones. From (2) we have 
 2r = (r 1 +r 2 )+ (/ ' 1 ~ '^ = 30 -f- ^ l j\ a ,, fr = 3 } nearly) 
 
 32.15 nearly. 
 
 7T C 
 
 Let r, and r. 2 now represent respectively the radii of the 
 greater and smaller of the pulleys to be next calculated. 
 
 We may now fix the value of r, and calculate r.,. It will how- 
 ever generally be easier to find r. t by trial than to solve the 
 quadratic equation. 
 
 Let r, be fixed at 20", and for the first trial let us suppose r, 
 to be equal to 11.8". Then i\ -j- r a = 31.8, i\ r, = 8.2", and 
 
 we have from the formula 31.8 = 32.15 - L-L-Ll 31.7 
 
 nearly. 
 
 Again let r 2 = 11.6" for a second trial, then 31.6 =32.15 
 
 .468 = 31.68. Thus we see that the true value of r, lies be- 
 tween 11.6" and 11.8"; r., = 11.7" very nearly satisfies the 
 equation. 
 
 ARTICLE 488. 
 
 In Fig. 218 let C 2 T, = r, T, T, = /, the angle T, C 2 T, = , 
 and the angle T 2 T l C 2 = /3. Then C 2 A = (/ cos a -f I cos /3) 
 tan (3. 
 
 Placing the above value of C,, A in equation (2), and elimina- 
 ting p by means of the equation, I Gin /> = r sin a , we have r t 
 
 ^ cos a 
 
 ?- 2 sin <| ^ -|- 1 j> from which the velocity of 
 
 j/l-fsin*a 
 I * 
 
 the piston of an engine may be calculated for any given crank 
 angle; for by knowing the number of revolutions, the value of 
 r 2 is easily found. 
 
 ARTICLE 491. 
 
 Equations (2) and (1). Let the axis C 2 be perpendicular to the 
 plane of the paper and let the plane of the axes of the shafts be 
 vertical. The projection on the paper, of the path of F 2 will be 
 the circle F, F 2 B D, and the projection of the path of F l will be 
 the ellipse F, H B E of the accompanying figure.
 
 Suppose F 2 to move to the position F/, Fj will in the same 
 time move to F/. The angle F 2 O F./ is 2 . The angle F T F/ 
 which is also equal to fa, is the projection of the angle de- 
 scribed by F t in the plane of its path. The true magnitude 
 of this angle is F, A, (found by revolving the plane of the 
 ellipse in which the angle lies, about F } B until the ellipse coin- 
 cides with the circle, when F, will fall at A in the vertical line 
 M A.) The angle P., A is fa. (See the A. M.) 
 
 Then M cot 0, = M A. O M tan 2 = M F/. 
 
 tan 2 M F/ O H 
 
 .-. = = = cos i or tan 0, tan 0., = cos i. 
 
 cot fa MA F, 
 
 Differentiating this equation we have 
 d fa tan 0. 2 cos 2 fa tan fa (I -f- tan' 2 0,) 
 d fa tan fa cos' 2 fa tan fa (1 -(- tan' 2 2 ) 
 
 cot ^( 1+ c^-07) 
 
 tan 
 
 COt 0j 
 
 tan 2 = - 
 
 tan 0., -(- cot 2 
 cos i 
 
 cot 0i 
 tan 2 4- cot 2 
 
 Eliminate fa by Equation (1), 
 
 tan 
 
 + 
 
 a, tan 2 0, -f- cos 2 i 
 
 "When F, is in the plane of the axes fa = 0, tan fa = and 
 
 ^1 _ _ which is the first of Equations (1) A. M. In a simi-
 
 ARTICLE 507. 
 
 73 
 
 lar manner by eliminating p,, the second of Equations (1) may 
 
 be obtained. When = p-r = 1 or the 
 
 ! tan' 2 (f> l -{- cos- i 
 
 shafts have equal angular velocities, we find tan' 2 ^ = cos i; <f l 
 tan -1 . Vcos i. (See "Willis' Principles of Mechanism, Article 
 513.) 
 
 ARTICLE 507. 
 
 Equations (3). The square of the perpendicular, drawn from 
 any point in the circumference of a circle to a diameter, is equal 
 to the product of the segments into which the foot of the per- 
 pendicular divides the diameter .-. ( -- \ = T V (2 r T V) 
 = T V . 2 C M. 
 
 An exact parallel motion, accomplished entirely by means of 
 link work (without any sliding piece, as in Watts' Exact Parallel 
 
 Y 
 
 Motion, Article 507, A. M.), has recently been discovered by 
 M. Peaucillier, an outline of which is here given.
 
 74 ARTICLE 507. 
 
 In the accompanying figure, F E G is an equilateral paral- 
 lelogram of bars, jointed at each of its corners. A and C are 
 fixed points; A F and A G are links jointed at their extremi- 
 ties. C E is also a link jointed to the corner (E) of the paralel- 
 ogram, and capable of turning about C. When the point is 
 moved in the plane of the figure, the points F and G must 
 describe arcs of circles about A. E must describe an arc 
 about C. 
 
 When this combination of links is made to oscillate about the 
 fixed points A and C : 
 
 1st. If C is midway between A and E, the point will be 
 guided in a straight line perpendicular to A O. 
 
 2d. If C is in any other position on the lino A E, the point 
 O will move in the arc of a circle. 
 
 To prove these propositions, and also ta find exact relations 
 between the radius of the arc described by ( ). and the position 
 of C on the line A E, is the object of the following analytical 
 investigation : 
 
 Suppose the instrument to be swung aside until the axis A 
 takes the position A B, making an angle with its neutral posi- 
 tion. E will be at E' on this line, and also on the circle whose 
 center is at C. The parallelogram will be partially closed as 
 represented by the dotted lines in the figure. O will have 
 moved to some point B on the new position of the axis, and it 
 is required to find the locus of this point. 
 
 Let O be the origin, X the horizontal, and Y the vertical 
 axis. Then M = x, M B = y. 
 
 Let A F = A G = /; F E = E G = s- A E' = c; A = A ; 
 
 C E = r ; C A = d, and the angle F' E' B = F' B E' = ft. 
 
 Then x = (k y} tan (1) 
 
 y = k (c -f 2 s cos ft) cos (2) 
 
 P = c 2 -f s 2 +.tc cos ft (3) 
 
 A E' c = d cos + V* 1 " rf ' 2 sin* (4) 
 
 Combining 2 and 3 we have 
 
 (/- _ c 2 s-\ P s 2 
 c -| Icos0 = & cos0 (5) 
 
 c f c 
 
 Combining this with (4) 
 
 (? - -S 2 ) COS0 
 
 d COS -4- yV _ (P gin* 
 
 Putting in terms of tan gives
 
 ARTICLE 515. 75 
 
 P fS 2 
 
 l -f tan 2 
 
 y = k -- - -- 
 
 , i/,., _ <(* tan 2 
 -j_ tan 2 1 + tan 2 
 
 P .v 2 
 
 d -f- Vr- 4- (/- 2 (P) tan 2 
 
 Substituting from (1), transposing & and changing the signs 
 of both members 
 
 ( _ y) (P s*) 
 
 (i (/ c - y) + V (& - y)* + ** (''* - c? 2 ) 
 Dropping the factor (& y), clearing of radicals, and placing 
 (P .s 2 ) = H, we have (r 2 rf 2 ) (A - y) 2 -f (r 2 <P) x 2 = n 2 
 2nd (k - y) (7) (r 2 - </-) 7r - 2 /,- (^ d*) y -f (* - a 2 ) 2/ 2 + 
 (r 2 cP) .c* n 1 - 1 n d k -\- 1 n d y, which is apparently 
 the equation of an ellipse (since b- 4 a c < 0. See Church's 
 Analytical Geometry, page 207). 
 
 Since the center of this curve is evidently on the line A 
 produced, which contains one of the principal axes, let us 
 transfer the origin of coordinates to some point on this line at a 
 distance a from O. The formuke for such transformation are 
 x = x'; y = y' -\- a. Then from (7) (r 2 d 2 ) (k a y')* -f 
 ( ; . _ d') d* = n 2 - In d (k a y'). 
 
 Let k a = b, and drop the primes; (r 2 d*) x 2 -f (r 2 d 2 ) 
 / = n* 2 n d b (r 2 - rf*) V + ( 2 n d -f 2 (r 2 d 2 ) b }. y. 
 Jf the origin is at the center of the curve, the term containing y 
 
 must disappear or 2 n d -f- 2 (r 2 d 2 ) b = 0; b = -5 -^ 
 
 7i d 
 
 Replacing b by its value, we have k a = -^ 
 
 nd 
 
 V - d* 
 Since the co-efficients of x* and y 2 are equal, the ellipse has 
 
 equal axis, 'or is a circle of radius a = k + -^- "When d 
 
 = r, a = ex, or the locus is a straight line Q . E . D . 
 
 ARTICLE 515. 
 
 PROBLEM. Find the formula for the work performed upon 
 the piston of a steam engine during one stroke, the length of
 
 76 ARTICLE 531. 
 
 stroke being /, the area of the piston a, the initial pressure of 
 the steam j) l} the steam being cut off at a distance c from the 
 beginning of the stroke. 
 
 Ans. a p l c 1 1 -j- log e I 
 
 ARTICLE 531. 
 
 Equation (1). Let g' = the force of gravity at any point on 
 the earth's surface at the level of the sea. 
 
 Let T = time of one revolution of the earth on its axis. 
 
 Let n = number of revolutions in a unit of time. 
 
 Then n T = 1. 
 
 The centrifugal force of a unit of mass at the level of the sea 
 and latitude X is 
 
 v 2 (2 ,r . R cos X . ) 2 4 7T 2 R 
 
 F __ v _ '_ = _ _ cos X 
 R cos X R cos X T 2 
 
 The component of F acting opposite to gravity is f = F cos X 
 
 4 7T 2 R 2 7T 2 R 
 
 cos 
 
 Gravity being diminished by this amount, if we subtract it 
 from G the real attraction of the earth, we shall have 
 
 The value of ./for latitude 45 has been found by experiment. 
 Denote this value by g^ 
 
 Then from the formula (making X = 45) 
 
 2 7T 2 R 2 7T 2 R 
 
 = -- cos 
 
 For a point at a distance h- above the earth's surface (h being 
 small compared with R) we have from Newton's law of gravity 
 
 (-V 
 
 9 =y''ir L h-* = 9 'w- 
 
 Performing the division to two places 
 
 g ==
 
 ARTICLES 537-557. 77 
 
 ARTICLE 537. 
 
 Equation (2) Let a be the angle which a tangent to the curve 
 makes at any point. 
 
 W W 
 
 Then Q =Wcos a = . 
 
 J~ (dz? ' 
 + 
 
 Since the horizontal component of the velocity of the projec- 
 tile is constant we have 
 
 v cos 
 
 V COS = v COS a. .: COS a = 
 
 v 
 
 v cos 
 
 .-. Q = W. 
 
 ARTICLE 545. 
 
 Equation (1) The component of g urging the pendulum 
 towards its lowest point is g sin 6 in which Q is the inclination 
 of the curve to- the horizon. To show that in the cycloidal pen- 
 dulum this force is proportional to the length of the arc between 
 any position of the pendulum and its lowest position, we con- 
 sider that the length of any curve is s = / d s = I p d Q. 
 From Art. 390 A. M., we have for the cycloid p = 4 r 2 cos 6. 
 .-. s = 4 r 2 / cos d Q = 4 r 2 sin which varies as sin 0. 
 
 */ o 
 
 Hence the vibrations of a cycloidal pendulum are isochronous. 
 
 ARTICLE 552. 
 
 PROBLEM. A body weighing 100 Ibs. starts from a point 48 
 feet above the earth's surface. After falling freely 16 feet it 
 encounters a constant resistance equal to f gravity. Required 
 its potential energy before starting ; its actual energy, when at a 
 distance of 16 feet from the earth ; its potential energy at the 
 same point ; the sum of its actual and potential energies when 
 it reaches the earth ; also the work done in the fall. 
 
 ARTICLE 557. 
 
 Equation 6 should read as follows : 
 W v 2 Ox W vl . Q, x, 0, x, 
 
 sm at 008 at --
 
 78 
 
 ARTICLE 578. 
 
 ARTICLE 578. 
 
 A method of finding the moment of inertia of solids of revo- 
 lution is illustrated in the solution of the following problem. Find 
 the moment of inertia of a cone whose altitude is h and radius 
 of base r, about an axis through 
 its vertex and perpendicular to 
 its geometric axis. Conceive a 
 portion of the cone included be- 
 tween two planes at right angles 
 to its axis, and at a distance tlx 
 apart. Find the moment of inertia 
 of this elementary slice or disc, by 
 applying to it the theorem of Art. 
 576 A. M. 
 
 w*y*dx. x 2 . 
 
 dx - 
 
 X 
 
 dx. This is the formula for problem 
 
 If the moment of inertia about the axis X were required, 
 the axis of inertia would pass through the center of gravity of 
 the disc and be perpendicular to its bases. We should there- 
 
 fore have I x = - / y* 
 2 u o 
 
 I of this Article. 
 
 PROBLEM III. Although the ellipsoid is not a solid of revo- 
 lution, its moment of inertia may be found by the method above 
 explained. 
 
 Let the origin be at the center of the ellipsoid and let X, 
 Y and OZ be in the direction of the semi-axes a, I and c re- 
 spectively. A section perpendicular to O X is an ellipse whose 
 semi-axes are y and z. The moment of inertia of the elemen- 
 tary slice with respect to X is therefore 
 
 w *y* z dx + W * y Z 
 
 d x. 
 
 W_7T / 
 TV -a' 
 
 (See problem in Notes on Art. 95.) 
 T/_> 3 ^' 
 
 The values of z and y are taken from the following equations, 
 j a 2 z 2 4- c 2 x 2 = a 2 c 2 
 \ * if -I- J* x 1 = a 2 V '
 
 ARTICLES 579-590. 79 
 
 4 w TT a b c (b* -f- c 2 ) 
 
 15 
 
 Let the student perform the same problem by integrating the 
 following formula according to the directions in Note on Art. 83, 
 assigning the proper limits. 
 
 I = 10 * dx dy dz (y 2 + z 2 ). 
 
 Theorem. The actual energy of a body revolving about a 
 fixed axis is found by multiplying its moment of inertia by the 
 square of its angular velocity and dividing the product by twice 
 the force of gravity. 
 
 Let p = the radius of gyration of the body, a its angular ve- 
 locity, and W its weight. If the mass of the body were con- 
 centrated at the center of gyration, (that is, at a distance p from 
 the axis of rotation) its velocity would be a p. Its actual energy 
 
 "W" r 2 r/ 2 
 
 would be J| a' P 2 = --- . W p = _ I. 
 
 If the axis of rotation traverses the center of gravity of the 
 body and is a " line of symmetry of the figure of the body," the 
 values of I and p may be taken from the columns headed I and 
 p in the table of Art. 578 A. M. (See Art. 589 A. M.) 
 
 ARTICLE 579. 
 
 V -4- c 2 W -f- c 
 
 Read instead of - under example. 
 6 6 
 
 ARTICLE 590. 
 Equation 5. From equation ('!) of Art. 585, we haves 2 = -- 
 
 From equation (6) of Art. 588 we have cos = - = - = . 
 
 * VF+K" 
 
 _ 2 (I 2 + K 2 ) 
 
 ~T- "P 
 
 Or 1 = (P -f K 2 ) .-. -1 P + K 2 = 11 cos- -f I 2 cos 2 
 /3 + II cos 2 y. Equation (2) 
 
 NOTE. The moment of inertia as here used is defined in Arts. 571, 572, and 573, A. M.
 
 80 ARTICLES 606-648. 
 
 But I = I, cos 2 a + I 2 cos 2 /3 + I 3 cos 2 r . (Eq. (3), Art. 585) 
 
 ,;. IL^_- +k^+lL?r = u ^ + n cos* ,3 + is 
 
 COS 2 y. 
 
 cos 2 y = 0. 
 
 ARTICLE 606. 
 
 First Equation. The moment of the deviating couple in 
 terms of E is M = 2 E tan a. The value of the constant E is 
 the sum of its components with reference to the axes C G and 
 
 C E. and these components are - - - cos 2 a and -r - cos 2 
 
 9 \ 9 
 
 (180 - (90 - )) = - ^-S sin 2 .-. M == (I, - I 2 ) 
 
 w 
 
 COS a sin a = a 2 (p* p 2 2 ) cos a sin a . 
 
 ARTICLE 638. 
 
 Formulae for computing the discharge and diameter of pipes, 
 are given in Article 450 of Rankine's Civil Engineering. 
 
 ARTICLE 648. 
 
 Equations (1), (2), and (3). These equations may be obtained 
 as follows : The velocity of the jet before it meets the surface 
 is v. When the surface has changed its direction through an 
 angle /3, the component of its velocity in the original direction 
 of the jet is v cos /3. The change of velocity is, therefore, v 
 v cos ft = v (1 cos /3). Hence, the change of momentum 
 
 p Q 
 
 which measures the force in this direction, is F, = i 
 
 9 
 
 (1 cos /3). The change of velocity in a direction perpendicu 
 lar the original direction of the jet is v sin j3 .. F y = 
 
 sin /3. The resultant F = v F x 2 -j- F/ = 
 
 <J g r. 
 
 P Q 
 
 COS Ij) = 
 
 P V

 
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