problems_II_4 - problem_II_4_1.nb Problem II.4.1...

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(* Problem II.4.1 *) Apply[Clear,Names["Global`*"]]; Off[General::spell]; Off[General::spell1]; "no.208 radial ball" B10 = 4000; (* hour *) n = 1200; (* rpm *) L = B10 * 60 * n; (* rev *) "Assumption: the life given is for 90% reliability" LR = 90*10^6; Print["LR = ",LR, " rev"]; d = 40; (* mm - the bore - from Table.II.4.1 *) Print["d = ",d, " mm"]; Cap = 9.40 * 10^3; (* N - from Table II.4.2 *) Print["C = ",Cap, " N"]; Kr = 1; (* from Figure II.4.9 *) Print["Kr = ",Kr]; "Assumption: the loading is steady" Ka = 1; (* from Table II.4.3 *) Print["Ka = ",Ka]; Fr = Cap (LR / L)^(3/10); (* from eq II.4.22 *) Print["Fr = ",Fr, " N"]; no.208 radial ball Assumption: the life given is for 90% reliability LR = 90000000 rev d = 40 mm C = 9400. N Kr = 1 Assumption: the loading is steady Ka = 1 Fr = 6631.06 N problem_II_4_1.nb 1
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(* Problem II.4.3 *) Apply[Clear,Names["Global`*"]]; Off[General::spell]; Off[General::spell1]; "no.207 angular ball bearing" Fr = 250; (* lb *) Fa = 150; (* lb *) n = 1000; (* rpm *) "Assumption: 25° angular ball bearing" r = N[Fa/Fr]; Print["Fa/Fr = ",r]; "for 0 < Fa/Fr < 0.68 => Fe = Fr" Fe = Fr; Fe = Fe * 4.448222; (* convert from lb to N *) Print["Fe = ", Fe, " N"]; "for steady loading Ka = 1 (Table II.4.3)" Ka = 1; "Assumption: the life given is for 90% reliability" LR = 90*10^6; Print["LR = ",LR, " rev"]; Kr = 1; (* from Figure II.4.9 *) Print["Kr = ",Kr]; d = 35; (* mm - the bore - from Table.II.4.1 *) Print["d = ",d, " mm"]; Cap = 8.20*10^3; (* N - from Table II.4.2 *) Print["C = ",Cap, " N"]; L = Kr LR (Cap / (Ka Fe) )^(10/3); Print["L = ",L, " rev"]; B10 = L / (60 n); Print["B10 = ", B10, " hr"]; no.207 angular ball bearing Assumption: 25° angular ball bearing Fa ê Fr = 0.6 for 0 < Fa ê Fr < 0.68 => Fe = Fr Fe = 1112.06 N for steady loading Ka = 1 H Table II.4.3 L Assumption: the life given is for 90% reliability LR = 90000000 rev problem_II_4_3.nb 1
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Kr = 1 d = 35 mm C = 8200. N L = 7.02318 × 10 10 rev B10 = 1.17053 × 10 6 hr problem_II_4_3.nb 2
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Problem II.4.5. From eq. L = K r L R ( C F r ) 3 . 33 , for identical bearings with the same L R , C , and F r we have: L K r L R ( C F r ) 3 . 33 ± ± ± ± ± 90% = L K r L R ( C F r ) 3 . 33 ± ± ± ± ± 60% = L K r L R ( C F r ) 3 . 33 ± ± ± ± ± 98% (1) or L K r ± ± ± ± ± 90% = L K r ± ± ± ± ± 60% = L K r ± ± ± ± ± 98% (2) From Figure II.4.9 for 90% reliability K r = 1 . 0; for 60% reliability K r = 6 . 0; for 98% reliability K r = 0 . 33; L 90% 1 . 0 = L 60% 6 . 0 = L 98% 0 . 33 (3) For L 90% = 12000 hours, L 60% = 6 × 12000 = 72000 hours, and L 98% = 0 . 33 × 12000 = 3960 hours. A higher reliability requirement (fewer bearing failures) means a shorter life. 1
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(* Problem II.4.7 *) Apply[Clear,Names["Global`*"]]; Off[General::spell]; Off[General::spell1]; Fr = 3; (* kN *) Fa = 2; (* kN *) n = 1200; (* rpm *) L = 6000 * 60 * 900; LR = 90*10^6; Print["LR = ",LR, " rev"]; Kr = 0.33; (* for 98% reliability from Figure II.4.9 *) Print["Kr = ",Kr]; r = N[Fa/Fr]; Print["Fa/Fr = ",r]; "a)" "for 0.35 < Fa/Fr < 10 => Fe = Fr (1+1.115 (Fa/Fr - 0.35)) - for radial ball bearings" Fe = Fr (1+1.115 (Fa/Fr - 0.35)); Print["Fe = ", Fe, " kN"]; "for light to moderate impacts Ka = 2 (Table II.4.3) - radial ball bearing"
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problems_II_4 - problem_II_4_1.nb Problem II.4.1...

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