Projetos e Elementos de Maquinas - Juvinall -exercicios Resolvidos - Rolamentos

SOLUTION (14.1D)\nKnown: A web site address is given as http://www.grainger.com.\nFind: Locate a thrust ball bearing with a 0.50 in. bore, 0.875 in. outside diameter, and a 0.375 in. width. List the manufacturer, description, and price for the bearing.\nAnalysis: The web site provides the following information:\nMfg. Name: Boston Gear\nDescription: Thrust Ball Bearing, 1/2\" Bore 7/8\" OD 3/8\" W\nPrice: $6.12\n\nSOLUTION (14.2D)\nKnown: A web site address is given as http://www.grainger.com.\nFind: Locate a motor grade ball bearing with a double shield and a 0.9842 in. diameter bore. List the manufacturer, description, and price for the bearing.\nAnalysis: The web site provides the following information:\nMfg. Name: NTN\nDescription: MTR Grade Ball Bearing, ABEC 1-DBL Shield.-9842 Bore-NTN #6205ZZ\nPrice: $9.83\n\nSOLUTION (14.3)\nKnown: The site http://patent.womplex.ibm.com provides a database of patent descriptions from the U.S. Patent & Trademark Office, including images for the last 17 years.\nFind:\n(a) Search the patent database using the phrase \"ball bearing\" and record the number of applicable patents found by the search engine.\n(b) Identify the subject of patent 5,340,314.\nAnalysis:\n(a) A search found 267 applicable patents.\n(b) The subject of patent 5,340,314 is dirt ejecting ball transfer unit that ejects contaminants. SOLUTION (14.4)\nKnown: A No. 204 radial ball bearing has a 5000 hr B-10 life at 900 rpm.\nFind: Determine the bearing radial load capacity.\nSchematic and Given Data:\n1800 rpm\nNo. 204 Radial Bearing\nF_r = ?\n90% reliability\nAssume: steady loading\nB-10 life = 5000 hours\n\nAssumptions:\n1. Table 14.2 accurately gives the bearing capacity.\n2. Ball bearing life varies inversely with the 10/3 power of the load (i.e., Eq. (14.5a) is accurate).\n3. The life given is for a 90% reliability.\n4. The loading is steady.\nAnalysis:\n1. From Table 14.2, for a 204 bearing the bore is 20 mm.\n2. From Table 14.2, for L_R = 90 x 10^6 rev and a 200 series bearing, C = 3.35 kN.\n3. From Fig. 14.13, K_3 = 1.0 for a steady load.\n4. From Eq. (14.5a), L = K_r L_R (C/F_e K_a)^{3.33}\n5. Substituting and solving for F_e: F_e = F_r = C(L_R/L)^{0.3}\n6. Substituting values:\nF_r = 3.35 kN\n(90 x 10^6 rev/(5000 hr)(60 min/hr)(900 rev/min)) = 2409 N. SOLUTION (14.5)\nKnown: A No. 208 radial ball bearing carries a radial load of 200 lb and a thrust load of 150 lb at 1800 rpm.\nFind: Determine the bearing B-10 life.\nSchematic and Given Data:\n1200 rpm\nNo. 208 Radial Bearing\nF_r = 200 N; F_t = 150 N\n90% reliability\nsteady loading\nB-10 life = ?\n\nAssumptions:\n1. Table 14.2 accurately gives the bearing capacity.\n2. Ball bearing life varies inversely with the 10/3 power of the load (i.e., Eq. (14.5a) is accurate).\n3. The life given is for a 90% reliability.\n4. The load F_e can be found from Eq. (14.3).\nAnalysis:\n1. From Table 14.1, for a 208 bearing the bore is 40 mm.\n2. From Table 14.2, for L_R = 90 x 10^6 rev and a 200 series bearing, C = 9.40 kN = 2112.3 lb.\n3. From Fig. 14.13, K_r = 1.0.\n4. From Table 14.3, K_a = 1.0 for a steady load.\n5. The ratio F_t/F_r = 150 lb/200 lb = 0.75.\n6. The equivalent load from Eq. (14.3) is F_e = F_r [1 + 1.115((F_t/F_r) - 0.35)] = 200 lb[1 + 1.115(150/200 - 0.35)] = 289.2 lb.\n7. From Eq. (14.5a), L = K_r L_R(C/F_e K_a)^{3.33}.\n8. Substituting values into Eq. (14.5a):\nL = 90 x 10^6 rev[2112.3 lb/289.2 lb]^{3.33} = 6.76 x 10^{10} rev\n(60 min/hr(1200 rev/min))\n= 938,763 hr.\nComment: The life of 938,763 hours corresponds to about 107 years of continuous operation where the bearing runs 24 hours/day and 7 days/week--a long life! SOLUTION (14.6)\nKnown: A radial contact ball bearing has a given radial load.\nFind: Determine the radial load change required to (a) double the life and (b) triple the life.\nSchematic and Given Data:\n\nAssumptions:\n1. Ball bearing life varies inversely with the 10/3 power of the load.\n2. The life given is for a 90% reliability.\n\nAnalysis:\n1. Let L1 and F1 be the original life and load for the bearing. Let L2 and F2 be the new life and load.\n2. Since L1/L2 = (F2/F1)^{10/3}, F2/F1 = (L1/L2)^{3/10}\n3. To double the life, L2 = 2L1, and F2/F1 = (1/2)^{3/10} = 0.812\n4. To triple the life, L2 = 3L1, and F2/F1 = (1/3)^{3/10} = 0.719\n\nComment: To double the bearing life the radial load must be reduced to 0.812 of its original value; to triple the bearing life the radial load must be reduced to 0.719 of its original value. SOLUTION (14.7)\nKnown: Certain bearings are rated for a load capacity based on a life of 10^6 revolutions.\nFind: Determine the value by which the bearing rated capacities should be multiplied so they can be compared with the ratings in Table 14.2 which are based on a 90 x 10^6 revolution life.\nSchematic and Given Data:\n\nAssumptions:\n1. Ball bearing life varies inversely with the 10/3 power of the load.\n2. A 90% reliability is required.\n\nAnalysis: From Eq. (14.1b), Creq = F/(L/LR)^{0.3}. For identical bearings with the same radial load Fr and life L\nCreq/(1/LR^{0.3})^{10^6} = Creq/(1/LR^{0.3})^{90 x 10^6}\nSolving for Creq|90 x 10^6 gives\nCreq|90 x 10^6 = Creq|10^6 (L) * ((1/LR)^{0.3}) / ((1/LR)^{10^6})\nor Creq|90 x 10^6 = Creq | 10^6 (0.259)\n\nSOLUTION (14.8)\nKnown: A No. 204 radial ball bearing has a 5000 hr B-10 life at 1800 rpm.\nFind: Determine the bearing radial load capacity.\nSchematic and Given Data:\n\nAssumptions:\n1. Table 14.2 accurately gives the bearing capacity.\n2. Ball bearing life varies inversely with the 10/3 power of the load (i.e., Eq. (14.5a) is accurate).\n3. The life given is for a 90% reliability.\n4. The loading is steady.\n\nAnalysis:\n1. From Table 14.1, for a 204 bearing the bore is 20 mm.\n2. From Table 14.2, for LR = 90 x 10^6 rev and a 200 series bearing, C = 3.35 kN.\n3. From Fig. 14.13, for 90 percent reliability, Kr = 1.0.\n4. From Table 14.3, Ka = 1.0 for a steady load.\n5. From Eq. (14.5a), L = Kr(LR/CFrKa)^{3/3}.\n6. Substituting and solving for Fc: Fe = Fr = C(LR/L)^{0.3}\n7. Substituting values:\nFr = 3.35 kN (90 x 10^6 rev (5000 hr/60 min/hr (1800 rev/min))^{0.3} = 1957 N SOLUTION (14.9)\nKnown: A No. 204 radial ball bearing carries a radial load of 200 lb and a thrust load of 150 lb at 1800 rpm.\n\nFind: Determine the bearing B-10 life.\n\nSchematic and Given Data:\n1200 rpm\nNo. 204 Radial Bearing\nF_r = 200 N, F_t = 150 N\n90% reliability\nsteady loading\nB-10 life = ?\n\nAssumptions:\n1. Table 14.2 accurately gives the bearing capacity.\n2. Ball bearing life varies inversely with the 10/3 power of the load (i.e., Eq. (14.5a) is accurate).\n3. The life given is for a 90% reliability.\n4. The load F_e can be found from Eq. (14.3).\n\nAnalysis:\n1. From Table 14.1, for a 204 bearing the bore is 20 mm.\n2. From Table 14.2, for LR = 90 x 10^6 rev and a 200 series bearing, C = 3.35 kN = 752.8 lb.\n3. From Fig. 14.13, for 90 percent reliability, K_r = 1.0.\n4. From Table 14.3, K_a = 1.0 for a steady load.\n5. The ratio F_r/F_t = 150 lb/200 lb = 0.75.\n6. The equivalent load from Eq. (14.3) is:\n F_e = F_r[1 + 1.115(F_r/F_t - 0.35)] = 200 lb[1 + 1.115(150/200 - 0.35)] = 289.2 lb.\n7. From Eq. (14.5a), L = K_1L_r(C/F_eK_a)^3.33\n8. Substituting values into Eq. (14.5a):\n L = 90 x 10^6 rev [752.8 lb]^(3.33) / [289.2 lb]\n = [2.18 x 10^9 rev\n (60 min/hr)(1200 rev/min)] = 30,277 hr.\n\nComment: The life of 30,277 hours corresponds to about 3.5 years of continuous operation where the bearing runs 24 hours/day and 7 days/week. SOLUTION (14.10)\nKnown: A No. 204 radial ball bearing has 90% reliability and carries a radial load of 1000 N and a thrust load of 250 N.\n\nFind: Determine the B-10 bearing life.\n\nSchematic and Given Data:\n3500 rpm\nNo. 204 Radial Ball Bearing\nF_r = 1000 N, F_t = 250 N\n90% reliability\nLight-moderate shock loading\nL = ? hr life\n\nAssumptions:\n1. Ball bearing life varies inversely with the 10/3 power of the load (i.e., Eq. (14.5a) is accurate).\n2. The equivalent load can be accurately estimated using Eq. (14.3).\n\nAnalysis:\n1. From Table 14.2, the rated load capacity, C = 3.35 kN.\n2. From Fig. 14.13, for 90 percent reliability, K_r = 1.0.\n3. From Table 14.3, K_a = 1.5 for light-moderate shock loading.\n4. F_r/F_t = 2500/1000 = 0.25 < 0.35.\n5. From Eq. (14.5a), L = K_rL_r(C/F_eK_a)^3.33\n = (1)(90 x 10^6)(3.35)(B^3)^3.33 = 1.307 x 10^9 revs\nL = 1.307 x 10^9 rev\n min hr\n3500 rev 60 min\n = 6224 hours\n\nComment: Inspection of Table 14.4 for representative bearing design lives would suggest that this bearing would be suitable for a gearing application used intermittently, where service interruption is of minor importance. SOLUTION (14.11)\nKnown: A bearing has a life of 5000 hr for 90% reliability.\n\nFind: Estimate the lives for 50% reliability and 99% reliability.\n\nSchematic and Given Data:\n90% reliability\n50% reliability\n99% reliability\nL_90 = 5000 hr\nL_50 = ?\nL_99 = ?\n\nAssumption: Bearing life varies inversely with the 10/3 power of the load (i.e., Eq. (14.2a) is suitable).\n\nAnalysis:\n1. From Eq. (14.2a), L = K_Lr(C/F_r)^3.33\n or \n K_Lr(C/F_r)^3.33 / K_Lr(C/F_r)^3.33 = K_Lr(C/F_r)^3.33 / K_Lr(C/F_r)^3.33 \n2. From Fig. 14.13, for 90% reliability, K_r = 1.0; for 99% reliability, K_r = 0.21; for 50% reliability, K_r = 5.0.\n3. For L_90 = 5000 hours, L_50 = (5)(5000) = 25,000 hours and L_99% = (0.21)(5000) = 1,050 hours\n\nComment: A higher reliability requirement (fewer bearing failures) means a shorter life. SOLUTION (14.12)\nKnown: A No. 211 radial ball bearing has a life of 5000 hr for 90% reliability.\n\nFind: For the same application, estimate the life for 90% reliability for (a) a L11 bearing, (b) a 311 bearing, and (c) a 1211 bearing.\n\nSchematic and Given Data:\n\nNo. 211\nL = 5000 hr\n\nNo. L11\nL = ?\n\nNo. 311\nL = ?\n\nNo. 1211\nL = ?\n\nAssumptions:\n1. Bearing life varies inversely with the 10/3 power of the load (i.e., Eq. (14.1a) is suitable).\n2. The loading conditions are identical for the bearings.\n\nAnalysis:\n1. From Table 14.2, for the 211 bearing, C = 12.0 kN.\n2. Also from Table 14.2, for the\n (a) L11 bearing, C = 8.2 kN\n (b) 311 bearing, C = 18.0 kN\n (c) 1211 bearing, C = 14.9 kN\n3. From Eq. (14.1a), L = LR(C/Fr)3.33\n4. For identical loading conditions (i.e., the same value of Fr) and for bearing rating capacities where LR = 90 x 106 revolutions,\n\n L211 L11 L311 L1211\n 12.0^3.33 8.2^3.33 18.0^3.33 14.9^3.33\n\n5. Since L211 = 5000 hr,\n L11 = 1407 hr,\n L311 = 19,291 hr and\n L1211 = 10,280 hr\n\n14-10