. Applied calculus; principles and applications . (5) by the binomial theorem and integratinggives a Jo aJo L 2a Sa^ 16a^ J _ir . x^ x^ x 1 ~ar^*6a 40a3^ 112^6 * J , x^ x^ , x^ , 1 w; = ^+6^40¥^+ll2^ ^^^ a = g- .*. Total length, /S = 2 si ^24^2 640JY^^7168^6 * * * * ^^ The total length can be found by (8) or (9) to any desireddegree of accuracy and it can be gotten exactly by (7); whend is quite small compared with I, then the third and succeed-ing terms in (8) and (9) are so small that they may be neg-lected giving: wH^ 8 (P Total length, S = l-{-kt-tt^ = ^ + ^Tjapproximately. (10) 244 INTEGR
. Applied calculus; principles and applications . (5) by the binomial theorem and integratinggives a Jo aJo L 2a Sa^ 16a^ J _ir . x^ x^ x 1 ~ar^*6a 40a3^ 112^6 * J , x^ x^ , x^ , 1 w; = ^+6^40¥^+ll2^ ^^^ a = g- .*. Total length, /S = 2 si ^24^2 640JY^^7168^6 * * * * ^^ The total length can be found by (8) or (9) to any desireddegree of accuracy and it can be gotten exactly by (7); whend is quite small compared with I, then the third and succeed-ing terms in (8) and (9) are so small that they may be neg-lected giving: wH^ 8 (P Total length, S = l-{-kt-tt^ = ^ + ^Tjapproximately. (10) 244 INTEGRAL CALCULUS When >S and H are given to find I, a cubic equation results,the solving of which may be avoided by putting S^ for P,since they are nearly equal when c? is small; then, 2^2^ approximately. (11) l = S 145. The Suspension Bridge. — The cables of a suspen-sion bridge are loaded approximately uniformly horizontally,since the roadway is horizontal, or nearly so; and the extraweight of the cables and the hangers near the supports is a. small part of the total load carried by the cables. As shownin Art. 144, under the conditions stated, the curve of the cables is the parabola whose equation \^ y = frjj, where H is the horizontal tension. Example 1. — The Brooklyn Bridge is a suspension bridgewhich has stays and stiffening trusses to prevent span between main towers is 1595 feet. If the sag OD is 128 feet and the weight per foot supportedby each cable is 1200 lbs., without considering the stays orstiffening trusses, to find the terminal and horizontal ten-sions, substitute the numerical values in (3) of Art. 144. Forterminal tension, Ti = H sec <^i orFor horizontal tension, Ti = /7sec0i = (f + HK 2,981,279 lbsvol Sd VZ2 +16^2 = 3,128,929 lbs. CURVE OF A FLEXIBLE CORD — CATENARY 245 Example 2. — A cord loaded uniformly horizontally with1 lb. per foot is suspended from two supports at same eleva-tion and 200 feet apart with a sag at middle o
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