Roller Chain - Tsubaki

Drive Chain SelectionExample of Selection for Lifting Transmission Roller ChainDrive ChainSprocket: 14T(PCD : 5.62")Sprocket: 30T(PCD : 12")M = 3000 kg(W = 3000 kgf)M = 6613lbs.Speed Reducer (i = 60)Motor with brakeRoller Chain:SUPER 100Sprocket: 14T(PCD : 6.74”)Roller Chain: SUPER 120(Chain Speed = 6.2 m/min)(Chain Speed = 20.34 ft./min)You are planning to use a lifting transmission machinelike the one on the left, and you are thinking of usingSUPER 120 for the lifting and SUPER 100 for the drivechain. We will now select a chain for drive and for lifting.Motor with brake: 3.7 kW = 5.0 HPMotor shaft rotational speed: n 1 : 1500 r/minSprocket: 14T(PCD : 6.74")Imperial UnitsMetric UnitsProcedure 1: Confirmation of motor characteristicsRated torque:T n = 17.36 ft-lbs.Starting torque:T s = 43.40 ft-lbs.Stalling torque:T b = 52.08 ft-lbs.Motor moment of inertia: I m = 1.42 lbs-ft 2Procedure 2: Calculate chain tension from loadProcedure 1: Confirmation of motor characteristicsRated torque:T n = 2.4 (kgf·m)Starting torque:T s = 6.0 (kgf·m)Stalling torque:T b = 7.2 (kgf·m)Motor moment of inertia: GD 2 m = 0.06 (kgf·m2 )Procedure 2: Calculate chain tension from loadChain tension Fw = M = 6613 lbs. Chain tension Fw = W = 3000(kgf)Chain speed V = 6.2 m/min (20.34 ft/min)........Speed factor: K v = 1.0214-tooth sprocket for lifting...............................Sprocket tooth factor: K c = 1.28Minimal shock...................................................Service factor: K s = 1.3For double strand lifting....................................Imbalance load coefficient K u = 0.6Design chain tensionF’w = Fw × Ks × Kv × Kc × Ku= 6613 × 1.3 × 1.02 × 1.28 × 0.6= 6733 lbs……………………………………… 1Procedure 3: Calculate the chain tension from the motorConverted moment of inertia of the loaded motoroutput shaftI l = M × (V) 2πn1= 6613 × (20.34) 2π × 1500= 0.12318 (lbs.ft 2 )Design chain tensionF’w = Fw × Ks × Kv × Kc × Ku= 3000 × 1.3 × 1.02 × 1.28 × 0.6= 3055 (kgf)……………………………………… 1Procedure 3: Calculate the chain tension from the motorConverted moment of inertia of the loaded motoroutput shaftGD 2 l = W × (V) 2pn1= 3000 × (6.2) 2p × 1500= 0.00519 (kgf·m 2 )A-102Moment of inertia of the motor output shaft (I),Im = 1.42 lbs.ft 2Inertia ratio (R)R = Il= 0.12318Im 1.42= 0.087Moment of inertia of the motor output shaftGD 2 m = 0.06 (kgf·m 2 )As the inertia ratio (R) equals φ.2, the shock factor (K) will be φ.23.Inertia ratio (R) R = GDl2 GD 2 = 0.005190.06= 0.087