• loads on the machine, • rotating masses, • distance of these masses to the rotation axis, • speeds and accelerations, • resisting torques. Two types of torque are distinguished : Start up slewing torque : Cd=Crv+Crc Acceleration slewing torque : Cg=Crv+Crc+Ca Crv = Friction torque of unloaded bearing Crc = Rotating torque due to loads Ca = Accélération torque Cd = Starting torque All these torques are expressed in kNm Crc : ROTATING TORQUE DUE TO LOADS The starting torque required takes into consideration loads on the bearing and friction of the components. Balls type slewing ring : Crc = [ (13,11 MT / Ø m ) + 3 FA + 11,34 FR ] Ø m . 10 -3 Crossed rollers type slewing ring : Crc = [ ( 15,3 MT / Ø m ) + 3,75 FA + 8,19 FR ] Ø m . 10 -3 MT = Resulting moment in kNm Ø m = Raceway mean Ø in meters FA = Axial load in kN FR = Radial load kN Ca : ACCELERATION TORQUE The torque needed to accelerate the loads from the initial speed up to the final speed, during time (t) is defined by : Ca = [ ( (PI)*n*l) / 30 . t ] . 10 ^{-3 }t = Acceleration time in sec. n = Speed variation in RPM (Final speed - Initial speed) l = Moment of inertia of the machine in Kg . m² l = l1 + l2 + l3 + ..... ln where l1 to ln = moments of inertia of the moving loads with regard to rotation axis expressed in Kg . m² Generally, we have : l1 = G1 * r1² ln = Gn * rn² G1 to Gn = Mass of various rotating components expressed in Kg. r1 to rn = Distances between the loads centre of gravity and the ring rotation axis expressed in meters. The friction torque of standard slewing ring is defined in the following graph. ROLLIX, upon request, can supply slewing ring with lower or higher torque values. Note : the resisting torque depends on the support surface flatness and lubrication. LOAD APPLIED ON THE RING : Axial FA : 68 kN + 5 kN = 73 kN Radial FR : 0,29 kN, negligible Moment MT : 5kN * 1,5 m = 7,5 kNm SLEWING TORQUE : Raceway mean ø = 2 meters Crv : according to the graph : 1 kNm Crc=[( 13,11 * 7,5 / 2 ) + (73*3) + (11,34 * 0)] 2.10-3 Crc = 0,536 kNm Slewing torque at start up : Cd = 1 + 0,536 = 1,536 kNm Platform moment of inertia : MR²/2 = ( 6800 / 2² ) / 2 = 13600 Kg.m² Cube moment of inertia : Mr² = 500 * 1,5² = 1125 kg.m² Total moment of inertia : 13600 + 1125 = 14725 Kg.m² Acceleration torque : n= 6 - 2 = 4 RPM Acceleration time : 20 sec Ca = (14725 * π * 4 ) / (20 * 30) 10-3 = 0,3084 kNm Slewing torque during acceleration Cg = 1 + 0,536 + 0,3084 = 1,845 kNm APPLICATION EXAMPLE Platform diameter : 4 m. Platform mass: 6800 kg Cube mass : 500 kg Ball type slewing ring raceway mean Ø : 2 m. Distance from the cube to the rotation axis : 1,5 m. Initial speed : 2 RPM Final speed : 6 RMP Acceleration time : 20 sec. |

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