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‡A Calculating driving force F=iƒÊ1A{ƒÊ2B{ƒÊ3C{ƒÊ4D{nFs{nMsj~‚r Find the driving force from the use conditions using the above formula. F = driving force S = safety factor (2 in this example) ƒÊ1, ƒÊ2, ƒÊ3, ƒÊ4 = coefficient of friction of each table ƒÊ1 = 0.15 (erecting) ƒÊ2 = 0.30 (hanging) ƒÊ3 = 0.15 (erecting) ƒÊ4 = 0.15 (erecting) A, B, C and D = load applied to each table A = 25 kgf B = -75 kgf C = 25 kgf D = 125 kgf n = number of tables = 4 Fs = seal friction = 1.2 kgf The value of F is obtained as shown below: F = (0.15 ~ 25 { 0.3 ~ 75 { 0.15 ~ 25 { 0.15 ~ 125 { 4 ~ 1.2) ~ 2 à 107 kgf ‡B Calculating service life N = a~K~i1/2Sj~iWa/Wij~fc/ifw~fv~fe~f.j Find the service life from the use conditions using the above formula. N = service life = required service life (1 million cycles) a = allowable wear amount = 0.1 mm K = coefficient of friction = 5 x 106 S = stroke = 0.25 m Wa = allowable load = 130 kgf (Hanging STF28) Wi = applied load = 75 kgf (Value closest to the allowable load: Select the severest one among the loads found with the load calculating formula.) fc = contact factor = 0.75 (Two single-axis units) fw = load factor = 1.0 (No impact load) fv = velocity factor = 1.0 (0.05 m/s) fe = environmental factor=1.0 (60‹C or less, no foreign matter) fl = lubrication factor = 1.0 (No lubrication) The value of N is found as shown below: N = 0.1 ~ 5 ~ 106 ~ (1/2 ~ 0.25) ~ (130/75) ~ 0.75/ (1.0 ~ 1.0 ~ 1.0 ~ 1.0) = 1,300,000 The required service life of 1 million cycles is satisfied. Type S Type ¦Do not use the STC20 in the lateral or hanging condition. Installation condition Coefficient of friction ƒÊ Erecting 0.15 Lateral 0.17 Hanging 0.30 Technical Information of Slide Shifters 328 Selection Guide Product Information Plastic Bearing Multi-layer Bearing Metallic Bearing Air Bearings Technical Slide Shifter Information Corporate Profi le