Questao 2 Prova 2

J U N K Y 8\n$ 0 5 0 3\nNOME: Sam Henrique Lopes - Engenheiro\n1)\nM: 63.83 R B-E A \nM1: 43.X A \nD: 62.93 mm \nT1: 4.74\nM2: 43.X Y \nD2: 62.93 mm \nx: 37.81\n\nFor 29.510, e F0 = 430 W.\n\nDan 384 + Dan 159 = 0.1618 = 97.100\n\nP = 80.109 mm2. P: 5.102 mm.\n\nD: P = D x P.\n\nL.C: K = (L1 x L2) - (T1 x T2)/(F)\n\nY: 378.35 = K: 178.61m\n\nL: L f x L s0 = 3.287 e L = 2.25 m\n\nn: 1nrc = mi 4 = m.i 0.313 kg/m\n\nL\n\nNl: 37 = 4932.410 e Nl: 2463 rpm.\n\nN: D= N 50\n\nWt: 2760 e Wt = 2763.9 e Wt: 289.55 lmd/an\n\nwww.codernil.com.br U C: W T: RT = V3 258.55 x 10.2665 + V2 63.73\n\nPts: F(t) = F. W. (RAB)\n\r\n156/(3.88) = F: 258.55 x Boxio2\nFz: 754.48 BN\n\nV2: C: WT = 258.56 x Boxio3 v2 = 20.61 cn/m\n\nFz: Kumg x \nFz: 4.5x 10.1x 378.6 = Fz: 78.15N\n\nFc: 103.212.064 = Fc: 430.72\nFt: 170.90, 234.488 - 22/15 = Ft: 125.73\n\nA: 10/31 = A: 90 = 4 5.000 mm.\n\nF: Ft = 12523 x 0.5 x 37370 = 0.5 5.5 p.m =\nA 0.00\n\nT: 362.37MPa