DRIVE CHAINS - Tsubaki

Before Use For Safe UseStandard Roller Chains Lube-Free Roller Chains Heavy Duty Roller Chains Corrosion Resistant Roller Chains Specialty Roller Chains Accessories Selection Handling31T from driven sprocket < 400mmOuter diameter 398 mm PCD d2=376.60 (mm)Number of teeth of drive sprocket =31=21 PCD d=255.63 (mm)1.51PZ'nChain speed =1000 = 38.12136 PZ'n 38.12136Chain speed =1000 1000 = 1000 =28.8m/min50 m/min,so it is possible to select by allowable load.Small sprocket revolution 36r/minRPM Kn=1.03Number of teeth of small sprocket 21T Number of teeth factor Kz=1.1022Chain tension Fw = Conveyor roller rotation torque 1000 Chain tension Fw=Conveyor roller rotation torque 1000 d2d222= 3.31000 =17.5 (kN) = 3371000 =1790 (kgf)376.6 376.6Design chain tension F'w=FwKsKnKz =17.51.31.031.10=25.8 (kN)RS120-1 (Max. allowable load: 30.4kN) can be used.Check the conveyance speed (selection conditions, 30 m/min)(Conveyor roller external diameter +2Belt thickness)Conveyance speed at this point =n2100021 (Conveyor roller external diameter +2Belt ) =n1 31 100021 (380+210) =36 31 1000=30.6 (m/min)(Step 3) Calculate from acceleration/deceleration timeThe small sprocket was decided as RS120 21T from thecalculations in step 2.Thus, calculate using the same pitch and number of teeth.If the acceleration/deceleration time is known, use that valuefor the calculation.The following is calculated assuming it is unknown.Working torque Tm=(Ts+Tb)=(0.116+0.122)=0.119 (kNm)2 2d 255.63Load torque T = Fw =17.5 (21000i) (2100050)=0.045 (kNm)Motor shaft conversion moment of inertia I of load side2Conveyance speedI=M( 2n1 )230.6 =6000( 21800 ) =0.044 (kgm 2 )Moment of inertia of the motor Im=0.088 (kgm 2 )Acceleration time of the motorn1Gts=(Im+I) 4375(TmT) 10001800 G=(0.088+0.044) 4375(0.1190.045) 1000=0.34 (s)Deceleration time of the motorn1 Gtb=(Im+I) 4375(Tm+T) 10001800 G=(0.088+0.044) 4375(0.119+0.045) 1000=0.15 (s)As tb < ts, chain tension during deceleration Fb is larger thanchain tension during acceleration Fs. Thus, use the following.Chain tension during decelerationConveyance speed (Conveyor roller external diameter +2Belt thickness)Fb=M +Fw(tb601000) d230.6 (380+210) =6000 +17.5(0.15601000) 376.6 =39.2 (kN)31T from driven sprocket 400mmOuter diameter 398 mm PCD d2=376.60 (mm)Number of teeth of drive sprocket =31=21 PCD d=255.63 (mm)1.51=28.8m/min50 m/min,so it is possible to select by allowable load.Small sprocket revolution 36r/minRPM Kn=1.03Number of teeth of small sprocket 21T Number of teeth factor Kz=1.10Design chain tension F'w=FwKsKnKz =17901.31.031.10=2640 (kgf)RS120-1 (Max. allowable load: 3100kgf) can be used.Check the conveyance speed (selection condition 30s, m/min)(Conveyor roller external diameter +2Belt thickness)Conveyance speed at this point =n2100021 (Conveyor roller external diameter +2Belt ) =n1 31 100021 (380+210)=36 31 1000=30.6 (m/min)(Step 3) Calculate from acceleration/deceleration timeThe small sprocket was decided as RS120 21T from thecalculations in step 2.Thus, calculate using the same pitch and number of teeth.If the acceleration/deceleration time is known, use that valuefor the calculation.The following is calculated assuming it is unknown.Working torque Tm=(Ts+Tb)=(11.9+12.5)=12.2 (kgfm)2 2d 255.63Load torque T= Fw =1790 (21000i) (2100050)=4.58 (kgfm)Motor shaft conversion GD 2 of the load sideGD =M( 2 2Conveyance speedn1 )230.6 =6000( 1800 ) =0.176 (kgfm 2 )GD 2 of the motor GD 2 m=0.352 (kgfm 2 )Acceleration time of the motorts=(GD 2 m+GD 2 n1) 375(TmT)1800=(0.352+0.176) 375(12.24.58)=0.34 (s)Deceleration time of the motortb=(GD 2 m+GD 2 n1) 375(Tm+T)1800=(0.352+0.176) 375(12.2+4.58)=0.34 (s)As tb < ts, chain tension during deceleration Fb is larger thanchain tension during acceleration Fs. Thus, use the following.Chain tension during decelerationConveyance speed (Conveyor roller external diameter +2Belt thickness)Fb=M +Fw(tb60G) d230.6 (380+210) =6000 +1790(0.1560G) 376.6 =4000 (kgf)143