Design project of transmission, Assignments of Mechanical Engineering

Download Design project of transmission and more Assignments Mechanical Engineering in PDF only on Docsity! HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY OFFICE OF INTERNATIONAL STUDY PROGRAM MACHINE ELEMENTS PROJECT Student: Lê Nhật Tiến Student code: 2052745 Instructor: Prof.Dr. Nguyễn Hữu Lộc Lecturer’s comment: I. Choosing the type of belt 1/ P = 5.5 (kW); n = 1450 (rpm) base on the table 4.22 choose type B Parameters Shaft Motor I II Working Power P, kW 5.3 4.932 4.638 4.5 Ration u 2.302 3.15 1 Rotational speed rpm 1450 629.887 200 200 Torque T, Nmm 34.907 74.776 221.465 214.875 Belt parameters: 𝑏𝑝 = 14 (𝑚𝑚) ; 𝑏0 = 17 (𝑚𝑚); ℎ = 10.5 (𝑚𝑚); 𝑦0 = 4 (𝑚𝑚); 𝐴 = 138(𝑚𝑚2) 2/ The small pulley diameter 𝑑1 = 1.2 𝑑𝑚𝑖𝑛 = 1.2 × 125 = 150 (𝑚𝑚) Choose type 𝑑1 = 160 (𝑚𝑚) 3/ Belt velocity: 𝑣 = 𝜋𝑛1𝑑1 60000 = 𝜋×1450×160 60000 = 58 15 𝜋 ≈ 12.147 ( 𝑚 𝑠 ) < 25 (m/s) 4/ Speed ratio: 𝑢 = 𝑛1 𝑛2 = 1450 629.887 = 2.302 5/ Determine the diameter of driven pulley when the relative creep factor 𝜉 = 0.01 𝑑2 = 𝑢𝑑1(1 − 𝜉) = 2.302 × 160 × 0.99 = 364.637 (𝑚𝑚) According to the standard we choose the value: 𝑑2 = 400 (𝑚𝑚) Accurated calculation the ratio speed of V-belt drive: 𝑢 = 𝑑2 𝑑1 = 400 160 = 2.5 6/ Preliminary determination of center distance a: 2(𝑑2 + 𝑑1) ≥ 𝑎 ≥ 0.55(𝑑2 + 𝑑1) + ℎ 2(400 + 160) ≥ 𝑎 ≥ 0.55(400 + 160) + 13.5 1120 𝑚𝑚 ≥ 𝑎 ≥ 321.5 𝑚𝑚 𝑎 = 𝑑2 = 400 𝑚𝑚 7/ Belt length L: 𝐿 = 2𝑎 + 𝜋 𝑑2 + 𝑑1 2 + (𝑑2 − 𝑑1)2 4𝑎 = 2 × 400 + 𝜋(400 + 160) 2 + (400 − 160)2 4 × 400 = 1715.645 𝑚𝑚 According to the standard, we choose L= 1800 mm. Then calculate the exact center distance a: 𝑘 = 𝐿 − 𝜋 𝑑2+𝑑1 2 = 1800 − 𝜋 400+160 2 = 920.354 ∆ = 𝑑2−𝑑1 2 = 400−160 2 = 120 From here we infer a = 𝑘+√𝑘2−8∆2 4 = 920.354+√920.3542−8×2402 4 = 385.461 𝑚𝑚  Approved  Using the normal V-belt 𝑖 = 𝑣 𝐿 = 12.147 1.8 = 6.748 < 10 ( 𝑠𝑎𝑡𝑖𝑓𝑖𝑒𝑑) 8/ Contact angle of small pulley 𝛼1: 𝛼1 = 180 − 57 𝑑2−𝑑1 𝑎 = 180 − 57 400−160 385.461 = 144.51° = 2.522 (𝑟𝑎𝑑). 9/ Select the number of belts according to the formula: 𝑧 ≥ 𝑃1 [𝑃0]𝐶𝛼𝐶𝑢𝐶𝑙𝐶𝑧𝐶𝑣𝐶𝑟 Where 𝑃1 = 5.5 (𝑘𝑊); 𝑣 = 12.147 ( 𝑚 𝑠 ) ; 𝑑1 = 160 (𝑚𝑚) - Allowable useful power [𝑃0] = 3.57 (𝑘𝑊) - Type B: 𝐿0 = 2240 (𝑚𝑚). 𝜎0𝐻𝑙𝑖𝑚2 = 2𝐻𝐵2 + 70 = 2 × 210 + 70 = 490 (𝑀𝑃𝑎) 𝑠𝐻 - factor of safety, for normalized and structural improvement steel 𝑠𝐻= 1,1. - Determination of allowable contact stress: [𝜎𝐻]1 = 0.9𝜎0𝐻𝑙𝑖𝑚1 𝑠𝐻 𝐾𝐻𝐿1 = 0.9 × 530 1.1 × 1 = 434 (𝑀𝑃𝑎) [𝜎𝐻]2 = 0.9𝜎0𝐻𝑙𝑖𝑚2 𝑠𝐻 𝐾𝐻𝐿2 = 0.9 × 490 1.1 × 1 = 401 (𝑀𝑃𝑎) For straight bevel gears, the allowable contact stress when calculating is selected according to the smallest value from two value [𝜎𝐻]1 và [𝜎𝐻]2, therefore [𝜎𝐻] =[𝜎𝐻]2= 401 MPa. 4) Allowable bending stress: [𝜎𝐹] = 𝜎𝐹𝑙𝑖𝑚 . 𝐾𝐹𝐿. 𝑌𝑅 . 𝑌𝑥 . 𝑌𝛿 . 𝐾𝐹𝐶 𝑠𝐹 Since the coefficients 𝑌𝑅 , 𝑌𝑥 , 𝑌𝛿 in the preliminary design stage have not been determined, the above formula can be written in the form: [𝜎𝐹] = 𝜎𝐹𝑙𝑖𝑚.𝐾𝐹𝐿.𝐾𝐹𝐶 𝑠𝐹 (1) Fatigue bending limits 𝜎𝐹𝑙𝑖𝑚 for normalized and structural improvement steel: σFlim1= 1.75HB1 = 1.75×230 = 402.5 MPa σFlim2= 1.75HB2 = 1.75×210 = 367.5 Mpa Safety factor for bengding stress 𝑠𝐹 = 1.75 (6.13t table) Factor taking into account the effect of reverse operation 𝐾𝐹𝐶 = 1 (one direction) Number of equivalent cycles to the driving gear (pinion), the constant input source: 𝑁𝐹𝐸1 = 60𝑐𝑛1𝐿ℎ = 60 × 1 × 629.887 × 10000 ≈ 3.77 × 108 (𝑐𝑦𝑐𝑙𝑒𝑠) Number of equivalent cycles to the driven gear (wheel): 𝑁𝐹𝐸2 = 60𝑐𝑛2𝐿ℎ = 60 × 1 × 200 × 1000 = 1.2 × 108 (𝑐𝑦𝑐𝑙𝑒𝑠) Because 𝑁𝐹𝐸1 , 𝑁𝐹𝐸2 >𝑁𝐹0 = 5 × 106, therefore 𝐾𝐹𝐿1 = 𝐾𝐹𝐿2 = 1 - Replace into equation (1): [𝜎𝐹]1 = 𝜎𝐹𝑙𝑖𝑚1. 𝐾𝐹𝐿. 𝐾𝐹𝐶 𝑠𝐹 = 402.5 × 1 × 1 1.75 = 230(𝑀𝑃𝑎) [𝜎𝐹]2 = 𝜎𝐹𝑙𝑖𝑚2. 𝐾𝐹𝐿. 𝐾𝐹𝐶 𝑠𝐹 = 367.5 × 1 × 1 1.75 = 210(𝑀𝑃𝑎) 5) Speed ratio: u = 3.15 6) We have: 𝜓𝑏𝑒 = 0.285 → 𝜓𝑏𝑒 . 𝑢 2 − 𝜓𝑏𝑒 = 0.285 × 3.15 2 − 0.285 = 0.523 According to the table 6.19, we choose 𝐾𝐻𝛽 = 1.55 The outside pitch diameter for driving gear: 𝑑𝑒1 = 1688√ 𝑇1𝐾𝐻𝛽 𝑢[𝜎𝐻]2 3 =1688√ 74.776×1.23 3.15×4012 3 = 103.245(𝑚𝑚) 7) From table 6.20 we choose 𝑧1𝑝 = 19 teeth, hardness HB1 and HB2 less than 350 HB, so 𝑧1 = 1,6𝑧1𝑝 = 1,6×19 = 30.4 teeth so choose 𝑧1 = 31 teeth 8) The number of teeth of driven gear 𝑧2 = 𝑢. 𝑧1 = 31 × 3.15 = 97.65 so choose 𝑧2 = 98 𝑡𝑒𝑒𝑡ℎ. ∆𝑢 = 𝑧2 𝑧1 − 𝑢 = 98 31 − 3.15 = 0.0112 < 2%𝑢 (𝑠𝑎𝑡𝑖𝑓𝑖𝑒𝑑) 9) Module of outside pitch diameter: 𝑚𝑒 = 𝑑𝑒1 𝑧1 = 103.245 31 = 3.33, from the standard series we choose 𝑚𝑒 = 4 (𝑚𝑚)  𝑑𝑒1 = 𝑧1𝑚𝑒 = 31 × 4 = 124(𝑚𝑚) 𝑚𝑚 = 𝑚𝑒(1 − 0.5𝜓𝑏𝑒) = 4 × (1 − 0.5 × 0.285) = 3.43 10) The outside pitch of driven gear: 𝑑𝑒2 = 𝑧2𝑚𝑒 = 98 × 4 = 392 (𝑚𝑚) 11) Cone distance: 𝑅𝑒 = 0.5𝑚𝑒√𝑧1 2 + 𝑧2 2 = 0.5 × 4 × √312 + 982 = 205.572(𝑚𝑚) 12) Face width: b = 𝜓𝑏𝑒 × 𝑅𝑒 = 0.285 × 205.572 = 58.588(𝑚𝑚) 13) Pitch angle: 𝛿1 = 𝑎𝑟𝑐𝑡𝑎𝑛 1 𝑢 = 𝑎𝑟𝑐𝑡𝑎𝑛 1 3.15 = 17.613° 𝛿2 = 90° − 17.613° = 72.387° 14) The average pitch diameter: 𝑑𝑚1 = 𝑑𝑒1 (1 − 0.5𝜓𝑏𝑒) = 124(1 − 0.5 × 0.285) = 106.33(𝑚𝑚) 𝑑𝑚2 = 𝑑𝑒2 (1 − 0.5𝜓𝑏𝑒) = 392(1 − 0.5 × 0.285) = 336.14(𝑚𝑚) 15) Peripheral velocity: 𝑣 = 𝜋𝑑𝑚1𝑛1 60000 = 𝜋 × 106.33 × 629.887 60000 = 3.507 (𝑚/𝑠) 16) Contact stress: 𝜎𝐻 = 𝑍𝐻𝑍𝑀𝑍 √ 2𝑇1𝐾𝐻√𝑢2 + 1 0.85𝑑𝑚1 2 𝑏𝑢 = 1.76 × 196 × 0.96 × √2 × 74.776 × 103 × 1.62 × √3.152 + 1 0.85 × 106.332 × 58.588 × 3.15 = 222.51 < [𝜎𝐻] = 401 (𝑀𝑃𝑎) (sastified) Where: 𝑍𝑀 = 196 (𝑀𝑃𝑎 1 2) when the gear made of steel 𝑍𝐻 = √ 2 𝑠𝑖𝑛2𝛼 = √ 2 sin (2 × 20𝑜) = 1.76 𝑍 = √ 4 − 휀𝛼 3 = √ 4 − 1.2 3 = 0.96, 𝑐ℎ𝑜𝑜𝑠𝑒 휀𝛼 = 1.2 𝐾𝐻 = 𝐾𝐻𝛽 × 𝐾𝐻𝑉 = 1.55 × 1.045 = 1.62 (according to the table 6.18 and 6.19) 1) We select the material for making the shaft is steel C45 with the following data (base on the table 10.1) and the value of [𝜎], d and [𝜏] ( according to the table 10.2): 2) Design layout of shafts for strength: Determination of the minimum diameter of the shaft (torque stress): Shaft I: 𝑑1 ≥ 10√ 16𝑇𝐼 𝜋[𝜏] 3 = 10√ 16 × 74.776 𝜋 × 20 3 = 26.703 (𝑚𝑚) According to the standard value series, we choose 𝑑1 = 26 (𝑚𝑚) Shaft II: 𝑑2 ≥ 10√ 16𝑇𝐼𝐼 𝜋[𝜏] 3 = 10√ 16 × 221.465 𝜋 × 20 3 = 38.349 (𝑚𝑚) According to the standard value series, we choose 𝑑2 = 40 (𝑚𝑚) 3) Designing the shaft construction (desgin layout): From the table 10.1 we have these property: 𝛔𝒃 MPa 𝛔𝒄𝒉 MPa 𝝉𝒄𝒉 MPa 𝛔−𝟏 MPa 𝝉−𝟏 MPa [𝝈] MPa 𝒅 𝑚𝑚 [𝝉] MPa Input/output 833 638 383 432 255 70 30 20 80 80 80 c fl w x d 2 d d We use the tapper bearing in this case Shaft I: According to the table 10.3: 𝑇1 = 74.776 (𝑁𝑚) torque in the range 60-80 Nm, so that we choose: 𝑒 = 𝑢 = 𝑓 = 80 (𝑚𝑚) 𝑥 = 10 (𝑚𝑚) 𝑤 = 50 (𝑚𝑚) Shaft II: According to the table 10.3: 𝑇2 = 221.465 (𝑁𝑚) torque in the range 200-400 Nm, so that we choose: 𝑓 = 70 (𝑚𝑚) 𝑥 = 10 (𝑚𝑚) 𝑤 = 50 (𝑚𝑚) 𝑙2 = (1 ÷ 1.5)𝑑2 = 1.5 × 𝑑2 = 1.5 × 38 = 57 (𝑚𝑚) From the figure we infer the formula of distance between driven bevel gear and tapper bearing: 𝑐 ≈ 𝑙2 2 + 𝑥 + 𝑤 2 = 57 2 + 10 + 50 2 = 63.5 (𝑚𝑚) Distance between 2 bearing for the 1 level bevel gear tranmission: 𝑙 ≈ 2 (𝑙2 + 2𝑥 + 𝑤 2 ) = 2 (57 + 2 × 10 + 50 2 ) = 204 (𝑚𝑚) 4) Force acting on the shaft: Analyze forces on the Shaft I: 80 80 80 z y x Determine the position with the maximum equivalent moment: 𝑀 = √𝑀𝑦 2 + 𝑀𝑥 2 + 0.75𝑇2 - At point B: 𝑀𝐵 = √𝑀𝐵𝑦 2 + 𝑀𝐵𝑥 2 + 0.75𝑇2 = √39.0342 + 112.5192 + 0.75 × 74.7762 = 135.564 (𝑁𝑚) - At point C: 𝑀𝐶 = √𝑀𝐶𝑦 2 + 𝑀𝐶𝑥 2 + 0.75𝑇2 = √141.9492 + 0 + 0.75 × 74.7762 = 156.023 (𝑁𝑚) The most dangerous cross section is at point C because the moment at C is maximum Determination of diameter at dangerous cross-section: σ𝐹 = 𝑀 × 103 𝑊 = 32 × 𝑀𝐵 × 103 𝜋 × 𝑑3 ≤ [σ𝐹] 𝑑 ≥ 10 √ 32 𝑀 𝜋[σ𝐹] 3 Base on the table 10.2 page 403 we choose [σ𝐹] = 70 𝑀𝑃𝑎 𝑑𝐴 ≥ 10 √ 32 × 8.235 𝜋 × 70 3 = 10.622 𝑚𝑚 𝑑𝐵 ≥ 10 √ 32 × 135.564 𝜋 × 70 3 = 27.02 𝑚𝑚 𝑑𝐶 ≥ 10 √ 32 × 156.023 𝜋 × 70 3 = 28.316 𝑚𝑚 According to the standard value we choose 𝑑𝐵 = 𝑑𝐶 = 30 𝑚𝑚 𝑑𝐴 = 𝑑𝐷 = 28 𝑚𝑚 All diameter are less than 50 m, so [𝜎] = 70 𝑀𝑃𝑎 satisfied. The elastic section modulus : According to the table 13.1 Exercise book we have the index of key at point A and D: Diameter b h 𝑡1 𝑡2 𝑑𝐴 = 28𝑚𝑚 10 8 5 3.3 𝑑𝐷 = 28𝑚𝑚 10 8 5 3.3 Shaft II has 2 keys so we use: 𝑊 = 𝜋𝑑3 32 − 𝑏𝑡(𝑑 − 𝑡)2 𝑑 Bending stress: 𝜎𝑎 = 𝑀 × 103 𝑊 𝑤ℎ𝑒𝑟𝑒 𝑀 = √𝑀𝑥 2 + 𝑀𝑦 2 𝑀𝐵 = √𝑀𝐵𝑥 2 + 𝑀𝐵𝑦 2 = √112.5192 + 39.0342 = 119.097 𝑁𝑚 𝑀𝐶 = √𝑀𝐶𝑥 2 + 𝑀𝐶𝑦 2 = √0 + 141.9492 = 141.949 𝑁𝑚 Polar modulus of section 𝑾𝟎 is determined by the formula: 𝑊𝑜 = 𝜋𝑑3 16 − 𝑏𝑡(𝑑 − 𝑡)2 𝑑 Torsional stress: 𝜏𝑎 = 𝑇 × 103 𝑊𝑜 The stress concentration factors 𝐾𝜎 and 𝐾𝜏 with keyway and σ𝑏 < 900 𝑀𝑃𝑎(σ𝑏 = 883 𝑀𝑃𝑎) so 𝐾𝜎 = 2.2 and 𝐾𝜏 = 2 table 10.9 Diameter factor 휀𝜎 = 0.88 & 휀𝜏 = 0.81(carbon steel, base on the table 10.4 page 411) For medium carbon steel 𝜓𝜎 = 0.1 𝑎𝑛𝑑 𝜓𝜏 = 0.05 The factor of increase in surface strength when grinding is calculated 𝛽 = 1.8 base on the table 10.5 Safety factor according to bending stress: 𝑠𝜎 = σ−1 𝐾𝜎 𝜎𝑎 휀𝜎𝛽 + 𝜓𝜎𝜎𝑚 Safety factor according to torsion stress: 𝑠𝜏 = τ−1 𝐾𝜏 𝜏𝑎 휀𝜏𝛽 + 𝜓𝜏𝜏𝑚 Because the bending stress varies with symmetrical alternating cycles, so 𝜎𝑚 = 0 and the torsional stress varies with a zero plus cycles so 𝜏𝑚 = 𝜏𝑎( rotate in 1 direction) Safety factor: 𝑠 = 𝑠𝜎𝑠𝜏 √𝑠𝜎 2 + 𝑠𝜏 2 Base on these formula we have the table for these index: Dangerous cross-section 𝑊 (𝑚𝑚3) 𝜎𝑎 𝑊𝑜 (𝑚𝑚3) 𝜏𝑎 휀𝜎 휀𝜏 𝑠𝜎 𝑠𝜏 𝑠 𝑑𝐵 = 30𝑚𝑚 1609.052 74.017 4259.771 17.554 0.88 0.81 4.202 10.217 3.886 𝑑𝐶 = 30 𝑚𝑚 1609.052 74.017 4259.771 17.554 0.88 0.81 4.202 10.217 3.886 Conclusion: all safety factor are greater than [𝑠] = (2.5 ÷ 3) , hencewe do not need to reinforce the structure anymore. Checking the static condition: To prevent the shaft from suffering deformable or being broken when suddenly overloading, we need to check this condition: 𝜎𝑡𝑑 = √𝜎2 + 3𝜏2 ≤ [𝜎𝑞𝑡] Whereas: 𝜎 = 32𝑀𝑚𝑎𝑥 𝜋𝑑3 , 𝜏 = 16𝑇𝑚𝑎𝑥 𝜋𝑑3 𝑎𝑛𝑑 [𝜎𝑞𝑡] = 0.8𝜎𝑐ℎ The most dangerous cross-section is at point C, we infer: 𝜎 = 32𝑀𝑚𝑎𝑥 𝜋𝑑3 = 32×141.949 𝜋×(30×10−3)3 = 53.55 (𝑀𝑃𝑎) where 𝑀𝑚𝑎𝑥 = 𝑀𝐶 70 63.5 204 O z yx Shaft II A B C D O z y O z yx 32.299 Nm 49.701 Nm O y x 39.722 Nm 70 Nm T=221.465 Nm Determine the position with the maximum equivalent moment: 𝑀 = √𝑀𝑧 2 + 𝑀𝑥 2 + 0.75𝑇2 - At point A: 𝑀𝐴 = √𝑀𝐴𝑧 2 + 𝑀𝐴𝑥 2 + 0.75𝑇2 = 0 - At point B: 𝑀𝐵 = √𝑀𝐵𝑧 2 + 𝑀𝐵𝑥 2 + 0.75𝑇2 = √49.7012 + 39.7222 + 0.75 × 221.4652 = 202.072 (𝑁𝑚) - At point C: 𝑀𝐶 = √𝑀𝐶𝑧 2 + 𝑀𝐶𝑥 2 + 0.75𝑇2 = √0 + 702 + 0.75 × 221.4652 = 204.169 (𝑁𝑚) - At point D: 𝑀𝐷 = √𝑀𝐷𝑧 2 + 𝑀𝐷𝑥 2 + 0.75𝑇2 = √0.75 × 221.4652 = 191.794 (𝑁𝑚) The most dangerous cross section is at point C because the moment at C is maximum Determination of diameter at dangerous cross-section: σ𝐹 = 𝑀 × 103 𝑊 = 32 × 𝑀𝐵 × 103 𝜋 × 𝑑3 ≤ [σ𝐹] 𝑑 ≥ 10 √ 32 𝑀 𝜋[σ𝐹] 3 Base on the table 10.2 page 403 we choose [σ𝐹] = 70 𝑀𝑃𝑎 𝑑𝐴 ≥ 10 √ 32 × 0 𝜋 × 70 3 = 0 𝑚𝑚 𝑑𝐵 ≥ 10 √ 32 × 202.072 𝜋 × 70 3 = 30.865 𝑚𝑚 𝑑𝐶 ≥ 10 √ 32 × 204.169 𝜋 × 70 3 = 30.971 𝑚𝑚 𝑑𝐷 ≥ 10 √ 32 × 191.794 𝜋 × 70 3 = 30.333 𝑚𝑚 Because there is a keyway on the shaft, we increase the diameter 5….10%: 𝑑𝐵 = (1.05 ÷ 1.1) × 30.865 = 32.408 ÷ 33.952 mm 𝑑𝐷 = (1.05 ÷ 1.1) × 30.333 = 31.849 ÷ 33.366 𝑚𝑚 According to the standard value we choose 𝑑𝐵 = 𝑑2 = 40 𝑚𝑚 𝑑𝐴 = 𝑑𝐶 = 34 𝑚𝑚 𝑑𝐷 = 32 𝑚𝑚 All diameter are less than 50 m, so [𝜎] = 70 𝑀𝑃𝑎 satisfied. The elastic section modulus : According to the table 13.1 Exercise book we have the index of key at point B and D: Diameter b h 𝑡1 𝑡2 𝑑𝐵 = 40𝑚𝑚 12 8 5 3.3 𝑑𝐷 = 34𝑚𝑚 10 8 5 3.3 Shaft II has 2 keys so we use: 𝑊 = 𝜋𝑑3 32 − 𝑏𝑡(𝑑 − 𝑡)2 𝑑 Bending stress: 𝜎𝑎 = 𝑀 × 103 𝑊 𝑤ℎ𝑒𝑟𝑒 𝑀 = √𝑀𝑥 2 + 𝑀𝑧 2 𝑀𝐵 = √𝑀𝐵𝑥 2 + 𝑀𝐵𝑧 2 = √49.7012 + 39.7222 = 63.624 𝑁𝑚 𝑀𝐶 = √𝑀𝐶𝑥 2 + 𝑀𝐶𝑧 2 = √702 + 0 = 70 𝑁𝑚 Shaft I f e u d m 1 𝜔1 = 629.887 (𝑟𝑝𝑚), choose 𝛼 = 16𝑜; 𝐿ℎ = 10000 (ℎ); d = 28 mm. Static load. 𝐹𝑟𝐵 = √𝑅𝐵𝑥 2 + 𝑅𝐵𝑦 2 = √2812.9782 + 901.4572 = 2953.891 (𝑁) 𝐹𝑟𝐶 = √𝑅𝐶𝑥 2 + 𝑅𝐶𝑦 2 = √1406.4892 + 3163.7422 = 3462.293 (𝑁) 1) According to the table 11.3 , axial load factor: 𝑒 = 1.5𝑡𝑎𝑛𝛼 = 1.5 × 𝑡𝑎𝑛16𝑜 = 0.43 2) The component of auxiliary axial force generated by the radial force: 𝑆𝑖 = 𝑒′𝐹𝑟𝑖 = 0.83𝑒𝐹𝑟𝑖 𝑆2 = 𝑒′𝐹𝑟𝐶 = 0.83𝑒𝐹𝑟𝐶 = 0.83 × 0.43 × 3462.293 = 1235.692 (𝑁) 𝑆1 = 𝑒′𝐹𝑟𝐵 = 0.83𝑒𝐹𝑟𝐵 = 0.83 × 0.43 × 2953.891 = 1054.243 (𝑁) Base on the table 11.1 page 444: 𝑆2 > 𝑆1 and 𝐹𝑎 < 𝑆2 − 𝑆1 , so base on the table 11.1 we have: The calculated axial load: 𝐹𝑎1 = 𝑆2 − 𝐹𝑎 = 1235.692 − 154.9 = 1080.792(𝑁) 𝐹𝑎2 = 𝑆2 = 1235.692 (𝑁) 3) Ratio: 𝐹𝑎2 𝐹𝑟𝐵 = 1235.692 2953.891 = 0.41 < 𝑒 = 0.43 We use the taper roller bearing Therefore according to the table 11.3 , we can consider: 𝑋 = 1 𝑎𝑛𝑑 𝑌 = 0 4) Therefore factor 𝐾𝜎 = 1 with the static load, 𝐾𝑡 = 1 and V=1(due to the rotation of inner ring). 5) Calculate the dynamic equivalent bearing load Q: 𝑄 = (𝑋𝑉𝐹𝑟 + 𝑌𝐹𝑎)𝐾𝜎 𝐾𝑡 = (1 × 1 × 2953.891 + 0 × 1235.692) × 1 × 1 =2953.891 N 6) The life in million revolution L: 𝐿 = 60𝐿ℎ𝑛 106 = 60 × 10000 × 629.887 106 = 377.93 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛 7) Calculate the basic dynamic load rating of bearing: 𝐶 = 𝑄 √𝐿 𝑚 = 2953.891 √377.93 10 3 = 17523.38 (𝑁) 8) According to Appendix (9.4) [55], we choose a light-sized with symbol 7206 with dynamic load capacity C = 31000N and the number of critical revolution when lubricated with grease 𝑛𝑚 = 8500 𝑟𝑝𝑚 9) The life of the bearing ( in million revolutions) determined by the formula: 𝐿 = ( 𝐶 𝑄 ) 𝑚 = ( 31000 2953.891 ) 10 3 = 2530.603 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 The life of the bearing in hours 𝐿ℎ = 106𝐿 60𝑛 = 106 × 2530.603 60 × 629.887 = 66959.179 ℎ𝑜𝑢𝑟𝑠 Shaft II c fl w x 𝜔1 = 200 (𝑟𝑝𝑚), choose 𝛼 = 16𝑜; 𝐿ℎ = 10000 (ℎ); d = 34 mm. Static load. 𝐹𝑟𝐴 = √𝑅𝐴𝑧 2 + 𝑅𝐴𝑥 2 = √508.6442 + 625.4582 = 806.174 (𝑁) 𝐹𝑟𝐶 = √𝑅𝐶𝑥 2 + 𝑅𝐶𝑧 2 = √1740.9412 + 353.7442 = 1776.516 (𝑁) 1) According to the table 11.3 , axial load factor: 𝑒 = 1.5𝑡𝑎𝑛𝛼 = 1.5 × 𝑡𝑎𝑛16𝑜 = 0.43 2) The component of auxiliary axial force generated by the radial force: 𝑆𝑖 = 𝑒′𝐹𝑟𝑖 = 0.83𝑒𝐹𝑟𝑖 𝑆1 = 𝑒′𝐹𝑟𝐶 = 0.83𝑒𝐹𝑟𝐶 = 0.83 × 0.43 × 1776.516 = 634.039 (𝑁) 𝑆2 = 𝑒′𝐹𝑟𝐴 = 0.83𝑒𝐹𝑟𝐴 = 0.83 × 0.43 × 806.174 = 346.655 (𝑁) Base on the table 11.1 page 444: 𝑆1 > 𝑆2 and 𝐹𝑎 = 497.922 > 0, so base on the table 11.1 we have: