. Differential and integral calculus. Let w = load per unit of length of beam; then at anysection S M=w(l-x) —- = ™ {l-xf. Mechanical Applications 389 dV UL) Hence ^ = Yei^ ~ 2 lx + ^; (px -lx>+ -\. dySince —- = o when x = o ; .*. C = o. Integrating again we have a/ //2*2 /x3 x4\ Since jr = o when x = o; .*. C = o. Equa. («) gives theshape the beam assumes under the action of the load. Repre-senting the maximum deflection by 8 which obviously occurswhen x = I we have Cor. Let W — wl = load on beam ; then y = Wl* 8J5Z Comparing 8 with 8 of § 257 we find 8 = f 8 = 3 8, nearly. That is, the de
. Differential and integral calculus. Let w = load per unit of length of beam; then at anysection S M=w(l-x) —- = ™ {l-xf. Mechanical Applications 389 dV UL) Hence ^ = Yei^ ~ 2 lx + ^; (px -lx>+ -\. dySince —- = o when x = o ; .*. C = o. Integrating again we have a/ //2*2 /x3 x4\ Since jr = o when x = o; .*. C = o. Equa. («) gives theshape the beam assumes under the action of the load. Repre-senting the maximum deflection by 8 which obviously occurswhen x = I we have Cor. Let W — wl = load on beam ; then y = Wl* 8J5Z Comparing 8 with 8 of § 257 we find 8 = f 8 = 3 8, nearly. That is, the deflection is nearly three times as great when theload is concentrated at the end as it would be if uniformly dis-tributed over the beam. 259. Shape and deflection of a beam supported at both ends a?idloaded in ce?iter. WIn this case M — — x\ 2 dy _ jv_ • dx* ~ 2 EI*dx 4 £/ 390 Integral Calculus If x = -, —- = o, since at the middle of the beam the tan-2 ax gent is || to X; ... Hence, dx \EI\ 4/ . V . W /xz Px\ Integrating agam, y = —^ ( - - — )• Since # = o, y == o.; ,*, C = o. When x = - we have 8 = 0 _ .2 £f 260. Shape and de/iection of a beam supported at both ends anduniformly loaded.
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