Projetos e Elementos de Maquinas - Juvinall - Eixos - Exercicios Resolvidos

SOLUTION (17.1) Known: A simply supported steel shaft is connected to an electric motor with a flexible coupling.\n\nFind: Determine the value of the critical speed of rotation for the shaft.\n\nSchematic and Given Data:\n\nAssumptions:\n1. Bearing friction is negligible.\n2. The bearings supporting the shafts are accurately aligned.\n3. The shaft remains linearly elastic.\n\nAnalysis:\n1. For the simply supported uniform load case:\n\nw = Ap = πd²ρ/4 where ρ = 0.28 lb/in³ for steel\nw = π(0.25)²/4 (0.28) = 0.0137 lb/in.\n\n2. From Appendix D-2,\nδst = 5wL^4/(384EI) for a uniform load distribution\nwhere E = 30 x 10⁶ psi (Appendix C-1)\nI = πd⁴/64 = π(0.25)⁴/64 = 1.92 x 10⁴ in⁴ (Appendix B-1)\n\n17-1 SOLUTION (17.2) Known: A simply supported aluminum shaft is connected to an electric motor with a flexible coupling.\n\nFind: Determine the value of the critical speed of rotation for the shaft.\n\nSchematic and Given Data:\n\nAssumptions:\n1. Bearing friction is negligible.\n2. The bearings supporting the shafts are accurately aligned.\n3. The shaft remains linearly elastic.\n\nAnalysis:\n1. For the simply supported uniform load case:\n\nw = Ap = πd²ρ/4 where ρ = 0.10 lb/in³ for aluminum\n17-2 SOLUTION (17.3) Known: A simply supported steel shaft is connected to an electric motor with a flexible coupling.\n\nFind: Determine the value of the critical speed of rotation for the shaft.\n\nSchematic and Given Data:\n\nAssumptions:\n1. Bearing friction is negligible.\n2. The bearings supporting the shafts are accurately aligned.\n3. The shaft remains linearly elastic.\n\n17-3 δst = 5wL^4 / 384EI for a uniform load distribution\nwhere E = 30 × 10^6 psi (Appendix C-1)\nI = πd^4 / 64 = π(1.0)^4 / 64 = 0.0491 in^4 (Appendix B-1)\nδst = 5(0.220)(10)(20)^4 / (384(30 × 10^6)(4.91 × 10^-3)) = 0.194 × 10^4 in.\n3. Using Fig. 17.5(c), to find the shaft critical speed\nnc = √(5g / 48δst) = √(5(32.2 ft/s^2)(12 in / ft)\n / (4(0.194 × 10^4 in.))\nnc ≈ 4989 rpm SOLUTION (17.4)\nKnown: A simply supported steel shaft is connected to an electric motor with a flexible coupling.\nFind: Determining the value of the critical speed of rotation for the shaft.\nSchematic and Given Data:\n\nAssumptions:\n1. Bearing friction is negligible.\n2. The bearings supporting the shafts are accurately aligned.\n3. The shaft remains linearly elastic.\nAnalysis:\n1. For the simply supported uniform load case:\n\nw = Ap = πd^2ρ / 4 where ρ = 0.28 lb/in^3 for steel\nw = π(1.0)^2 / 4 (0.28) = 0.220 lb/in.\n2. From Appendix D-2,\nδst = 5wL^4 / 384EI for a uniform load distribution\nwhere E = 30 × 10^6 psi (Appendix C-1)\nI = πd^4 / 64 = π(1.0)^4 / 64 = 0.0491 in^4 (Appendix B-1) SOLUTION (17.5D)\nKnown: A simply supported steel shaft is connected to an electric motor with a flexible coupling.\nFind: Plot nc versus shaft diameter d from 0.10 in to 3.0 in.\nSchematic and Given Data:\n\nAssumptions:\n1. Bearing friction is negligible.\n2. The bearings supporting the shafts are accurately aligned.\n3. The shaft remains linearly elastic.\nAnalysis:\n1. To plot nc versus d we select the following values for d: 0.1, 0.25, 0.5, 1.0, 1.5, 2.0, 2.5, and 3.0. We illustrate the calculation procedure for d = 0.25 for the simply supported load case: n_c (rpm)\n\nDiameter (in.)\n\n0.1\t125\n0.25\t311\n0.5\t623\n1.0\t1246\n1.5\t1869\n2.0\t2492\n2.5\t3115\n3.0\t3738\n\n17-7 n_c (rpm)\n\nCritical speed vs. diameter\n\nComment: By examining δ_st = 5wL^4 / 384EI, we find that δ_st is linearly proportional to 1/d^2.\n\nMoreover, since the shaft critical speed n_c ∝ √(I/δ_st), we can conclude that n_c ∝ √d^2 = d.\n\nThe fact that n_c ∝ d matches what we observed in the plot of n_c versus d.\n\nSOLUTION (17.6D)\n\nKnown: A simply supported steel shaft is connected to an electric motor with a flexible coupling.\n\nFind: Plot n_c versus bearing spacing from 1 in. to 20 in.\n\nSchematic and Given Data:\n\nMotor\n\nFlexible Coupling\n\n0.25 in. dia. shaft\n\nL = bearing space = 1.0 in. to 20 in.\n\n17-8 Assumptions:\n1. Bearing friction is negligible.\n2. The bearings supporting the shafts are accurately aligned.\n3. The shaft remains linearly elastic.\n\nAnalysis:\n1. To plot n_c versus L, we selected the following values for L: 1, 2, 4, 6, 8, 12, 16, 20. We illustrate the calculation procedure for L = 20 in. for the simply supported load case:\n\nwL/2\n\nwL/2\n\nw = Ap = πd^2/4 ρ, where ρ = 0.28 lb/in³ for steel\n\nw = π(0.25)²/4 (0.28) = 0.0137 lb/in.\n\n2. From Appendix D-2,\n\nδ_st = 5wL^4 / 384EI for uniform load\n\nwhere E = 30 x 10^6 psi (Appendix C-1)\n\nI = πd^4 / 64 = π(0.25)⁴ / 64 = 1.92 x 10⁴ in⁴ (Appendix B-1)\n\nδ_st = 5(0.0137)(20)⁴ = 4.98 x 10³ in.\n\n3. Using Fig. 17.5(c), to find the shaft critical speed\n\nn_c = √(5g / 48δ_st) = √(5(32.2 ft/s²)(12 in/ft) / 4(4.98 x 10³ in.))\nn_c = 311 rpm for L = 20 in.\n\n4. Repeat above steps 1 to 4 for each of the other selected values of L.\n\n17-9 5.\nBearing spacing (in.)\tCritical speed (rpm)\n1\t124600\n2\t31150\n4\t7787\n6\t3461\n8\t1947\n12\t865\n16\t487\n20\t311\n\n6.\nCritical speed vs. bearing spacing\n\nComment: By examining δs,t = 5wL4 / 384EI, we find that δs,t is linearly proportional to L4.\nMoreover, since the shaft critical speed nc ∝ √(1/δs,t), we can conclude that nc ∝ 1/√L.\n\nThe fact that nc ∝ 1/√L matches what we observe in the plot of nc versus L.\n\n17-10