Ball Screw 513-1E 2 kg•m 15-69 –3 -1 10 B × =1 =0.003 (rolling) 0.1 mm m 3,000 min 3,000 Rated rotational speed: None (direct coupling)A=1 f=15 N (without load) AC servo motor Point of Selection of Point J Examples of Selecting a Ball Screw a Ball of Selecting Examples cient of the guide surface fi 2 1 + m m + Reduction gear Positioning accuracy repeatability Positioning accuracy Inertial moment of the motor Frictional coef Guide surface resistance Guide Minimum feed amount Minimum feed service life time Desired s = 0.02mm/pulse 30000h Driving motor Work mass Work Table mass Table Ball screw nut -1 =1m/s =60kg =20kg 1 2 =1000mm max = 0.15s = 0.15s = 0.3 mm/1000 mm S 1 3 m ℓ (Perform positioningfrom (Perform the negative direction) m V n =8min ] Ball screw shaft ] Motor Selection Items Selection Conditions Work Work Mass Stroke length Backlash 0.15mm Backlash Positioning accuracy Deceleration time t [ Screw shaft diameter Lead Nut model No. Accuracy Axial clearance Screw shaft support method Driving motor Number of reciprocations per minute Number Table Mass Mass Table High-speed Transfer Equipment (Horizontal Use) Equipment (Horizontal Transfer High-speed [ Examples of Selecting a Ball Screw a Ball Screw Selecting of Examples Acceleration time t Maximum speed 513-1E
[Selecting Lead Angle Accuracy and Axial Clearance] Selecting Lead Angle Accuracy To achieve positioning accuracy of 0.3 mm/1,000 mm: ±0.3 ±0.09 = 1000 300 The lead angle accuracy must be 0.09 mm/300 mm or higher. Therefore, select the following as the accuracy grade of the Ball Screw (see Table1 on B15- 20 ). C7 (travel distance error: 0.05mm/300mm) Accuracy grade C7 is available for both the Rolled and the Precision Ball Screws. Assume that a Rolled Ball Screw is selected here because it is less costly.
Selecting Axial Clearance To satisfy the backlash of 0.15 mm, it is necessary to select a Ball Screw with an axial clearance of 0.15 mm or less. Therefore, a Rolled Ball Screw model with a screw shaft diameter of 32 mm or less that meets the axial clearance of 0.15 mm or less (see Table13 on B 15-27 ) meets the requirements. Thus, a Rolled Ball Screw model with a screw shaft diameter of 32 mm or less and an accuracy grade of C7 is selected.
[Selecting a Screw Shaft] Assuming the Screw Shaft Length Assume the overall nut length to be 100 mm and the screw shaft end length to be 100 mm. Therefore, the overall length is determined as follows based on the stroke length of 1,000 mm. 1000 + 200 = 1200 mm Thus, the screw shaft length is assumed to be 1,200 mm.
Selecting a Lead With the driving motor’s rated rotational speed being 3,000 min-1 and the maximum speed 1 m/s, the Ball Screw lead is obtained as follows: 1×1000×60 = 20 mm 3000 Therefore, it is necessary to select a type with a lead of 20 mm or longer. In addition, the Ball Screw and the motor can be mounted in direct coupling without using a reduc- tion gear. The minimum resolution per revolution of an AC servomotor is obtained based on the resolution of the encoder (1,000 p/rev; 1,500 p/rev) provided as a standard accessory for the AC servomotor, as indicated below. 1000 p/rev(without multiplication) 1500 p/rev(without multiplication) 2000 p/rev(doubled) 3000 p/rev(doubled) 4000 p/rev(quadrupled) 6000 p/rev(quadrupled)
B15-70 Ball Screw 513-1E 15-71 : a lead of leada: of B 15-70 B Point of Selection of Point [Selecting a Screw Shaft] [Selecting a Screw Shaft] [Selecting Angle Accu- Lead Examples of Selecting a Ball Screw a Ball of Selecting Examples ned in Section fi ficient. Therefore, the Ball Screw should have should Screw Ball the Therefore, ficient. guration for the screw shaft support. guration for the screw shaft support. ) are as follows. 15-35 xed confi B : a rolled Ball Screw with a screw shaft diameter of 32 mm a screw shaft diameter of 32 mm : a rolled Ball Screw with guration is selected as the screw shaft support method. guration is selected as the screw shaft fi xed-fi fi 15-70 B guration requires a complicated structure, needs high accuracy in the xed confi 20mm —— 1000 p/rev 20mm —— xed-supported or xed-supported con fi Shaft diameter Lead 15mm —— 20mm 15mm —— 30mm 20mm —— 20mm 20mm —— 40mm 30mm —— 60mm 30mm —— 1500 p/rev 30mm —— 2000 p/rev 40mm —— 3000 p/rev 60mm —— 4000 p/rev 80mm —— xed-fi , the shaft diameter of 15 mm is insuf is mm 15 of diameter shaft the , 15-70 B Selecting a Screw Shaft Support Method Selecting Selecting a Screw Shaft Diameter Since the assumed type has a long stroke length of 1,000 mm and operates at high speed of 1 m/s, Since the assumed type has a long stroke length select either the fi Since the screw shaft length has to be 1,200 mm as indicated in Section length has to be 1,200 mm as indicated in Since the screw shaft 20, 30, 40, 60 or 80 mm (see Table20 on Table20 (see 20, 30, 40, 60 or 80 mm on a screw shaft diameter of 20 mm or greater. require- the meet that leads and diameters shaft screw of combinations three are there Accordingly, of 20 mm; 20 mm/40 mm; and 30 mm/60 mm. ments: screw shaft diameter of 20 mm/lead that meet the requirements de Those Ball Screw models
To meet the minimum feed amount of 0.02 mm/pulse, which is the selection requirement, the follow- the requirement, selection the is which mm/pulse, 0.02 of amount feed minimum the meet To apply. ing should Lead racy and Axial Clearance] on Axial Clearance] racy and orless; andthe requirement defined inSection on [SelectingScrewa Shaft] However, the fi However, installation. Accordingly, the 513-1E
Studying the Permissible Axial Load Calculating the Maximum Axial Load Guide surface resistance f=15 N (without load)
Table Mass m1 =60 kg
Work Mass m2 =20 kg Frictional coeffi cient of the guide surface = 0.003
Maximum speed Vmax=1 m/s Gravitational acceleration g = 9.807 m/s 2
Acceleration time t1 = 0.15s Accordingly, the required values are obtained as follows. Acceleration:
Vmax α = = 6.67 m/s2 t1 During forward acceleration:
Fa1 = • (m1 + m2 ) g + f + (m1 + m2 ) • = 550 N During forward uniform motion:
Fa2 = • (m1 + m2 ) g + f = 17 N During forward deceleration:
Fa3 = • (m1 + m2 ) g + f – (m1 + m2 ) • = –516 N During backward acceleration:
Fa 4 = –• (m1 + m2 ) g – f – (m1 + m2 ) • = –550 N During uniform backward motion:
Fa5 = –• (m1 + m2 ) g – f = – 17 N During backward deceleration:
Fa 6 = –• (m1 + m2 ) g – f + (m1 + m2 ) • = 516 N
Thus, the maximum axial load applied on the Ball Screw is as follows:
Fa max = Fa1 = 550 N Therefore, if there is no problem with a shaft diameter of 20 mm and a lead of 20 mm (smallest thread minor diameter of 17.5 mm), then the screw shaft diameter of 30 mm should meet the re- quirements. Thus, the following calculations for the buckling load and the permissible compressive and tensile load of the screw shaft are performed while assuming a screw shaft diameter of 20 mm and a lead of 20 mm.
B15-72 Ball Screw 513-1E 15-73 B ) Point of Selection of Point B15-38 Examples of Selecting a Ball Screw a Ball of Selecting Examples =20 (see =17.5 mm =1100 mm (estimate) mm (estimate) =1100 2
1 a ℓ d = 15500 N 4
–1 –1 –1 10 2 4 17.5 1100 = 35500 N = 2 xed: “ =1 m/s =1 m/s =1 m/s 3 3 3 max max max Ph= 60 mm Ph= 20 mm Ph= 40 mm 10 10 10 17.5 xed-fi = 20×× fi × 4 × × × 10 60 60 60 Ph Ph Ph × × × = 116 116 = 2 4 1 2 max max max 1 a ℓ d d V V V × × • 2 = = 1000 min = = = = 1500 min = = = = 3000 min = = η max max max = N N N = 116 116 = 1 2 Factor according to the mounting method the mounting to Factor according Screw-shaft thread minor diameter Screw-shaft thread Since the mounting method for the section between the nut and the bearing, where buckling is nut and the bearing, section between the method for the Since the mounting is “ to be considered, Distance between two mounting surfaces two mounting surfaces Distance between P P Studying the Permissible Rotational Speed Screw shaft diameter: 30mm; lead: 60mm Screw shaft diameter: 20 mm; lead: 20 mm Screw Screw shaft diameter: 20 mm; lead: 40mm Maximum Rotational Speed Permissible Compressive and Tensile Load of the Screw Shaft and Tensile Permissible Compressive Buckling Load on the Screw Shaft Screw Shaft on the Buckling Load Thus, the buckling load and the permissible compressive and the tensile load of the screw shaft are and the permissible compressive and the Thus, the buckling load a Ball Screw that meets these requirements Therefore, axial load. at least equal to the maximum problem. can be used without a ● Maximum speed V Maximum speed V ● Maximum speed V ● Lead Lead Lead 513-1E
Permissible Rotational Speed Determined by the Dangerous Speed of the Screw Shaft Factor according to the mounting method 2 =15.1 (see B 15-40 ) Since the mounting method for the section between the nut and the bearing, where dangerous speed is to be considered, is “fi xed-supported: “ ℓ Distance between two mounting surfaces b =1100 mm (estimate) ● Screw shaft diameter: 20 mm; lead: 20 mm and 40 mm
Screw-shaft thread minor diameter d1=17.5mm
λ d1 7 17.5 7 –1 N1 = 2× 2 10 = 15.1× 2 × 10 = 2180 min ℓb 1100 ● Screw shaft diameter: 30mm; lead: 60mm
Screw-shaft thread minor diameter d1=26.4mm
λ d1 7 26.4 7 –1 N1 = 2× 2 10 = 15.1× 2 × 10 = 3294 min ℓb 1100
Permissible Rotational Speed Determined by the DN Value ● Screw shaft diameter: 20 mm; lead: 20 mm and 40 mm (large lead Ball Screw) Ball center-to-center diameter D=20.75 mm
70000 70000 –1 N2 = = = 3370 min D 20.75 ● Screw shaft diameter: 30 mm; lead: 60 mm (large lead Ball Screw) Ball center-to-center diameter D=31.25 mm
70000 70000 –1 N2 = = = 2240 min D 31.25 Thus, with a Ball Screw having a screw shaft diameter of 20 mm and a lead of 20 mm, the maximum rotational speed exceeds the dangerous speed. In contrast, a combination of a screw shaft diameter of 20 mm and a lead of 40 mm, and another of a screw shaft diameter of 30 mm and a lead of 60 mm, meet the dangerous speed and the DN value. Accordingly, a Ball Screw with a screw shaft diameter of 20 mm and a lead of 40 mm, or with a screw shaft diameter of 30 mm and a lead of 60 mm, is selected.
[Selecting a Nut] Selecting a Nut Model Number Rolled Ball Screw models with a screw shaft diameter of 20 mm and a lead of 40 mm, or with a screw shaft diameter of 30 mm and a lead of 60 mm, are large lead Rolled Ball Screw model WTF variations. WTF2040-2
(Ca=5.4 kN, C 0a=13.6 kN) WTF2040-3
(Ca=6.6 kN, C0 a=17.2 kN) WTF3060-2
(Ca=11.8 kN, C0 a=30.6 kN) WTF3060-3
(Ca=14.5 kN, C 0a=38.9 kN)
B15-74 Ball Screw 513-1E , 5 and Fa 15-75 4 B , Fa 3 ). = 850 mm Point of Selection of Point 3 10 × B15-47 0.15 Examples of Selecting a Ball Screw a Ball of Selecting Examples × 2 a = 13.6 kN). a = 13.6 kN). 0 0.15 + 1 75 75 75 75 (mm) 850 850 × N ℓ 1 ) at 2.5 (see Table1 on Table1 ) at 2.5 (see Travel Travel distance S = 75 mm 3 = 75 mm 3 10 10 × = 1000 – = 1000 – × 3 (N) N 10 =1 m/s 17 550 –17 516 –516 –550 0.15 Fa max 2 = 0.15s = 0.15s = 0.15 × 2 1 3 3 × t × 1 1 Applied axial load • = = max 3 3 10 10 2 + V 1 t × × • 3 1 t t max • • V 2 2 2.5 max 13.6 Motion max – V V S = N = 5.44 kN = 5440 No.1: During No.2: During No.3: During No.4: During No.5: During No.6: During ℓ Deceleration time t Maximum speed V Acceleration time t = = = a forward acceleration forward deceleration S 0 f backward acceleration backward deceleration forward uniform motion 3, 6 2, 5 1, 4 Studying Studying the Permissible Axial Load C uniform backward motion Travel distance during uniform motion distance Travel Travel distance during deceleration Travel distance during acceleration distance during acceleration Travel ℓ ℓ ℓ The subscript (N) indicates a motion number. Calculating the Travel Distance Assuming that this model is used in high-speed transfer equipment and an impact load is applied is load impact an and equipment transfer high-speed used in is model this that Assuming safety factor (f set the static during deceleration, Since the load direction (as expressed in positive or negative sign) is reversed with Fa Since the load direction (as expressed in positive ＊ Based on the conditions above, the relationship between the applied axial load and the travel dis- Based on the conditions above, the relationship tance is shown in the table below. ● ● The obtained permissible axial load is greater than the maximum axial load of 550 N, and therefore, 550 N, and therefore, maximum axial load of is greater than the permissible axial load The obtained with this model. there will be no problem Study the permissible axial load of model WTF2040-2 (C of model WTF2040-2 axial load Study the permissible calculate the average axial load in the two directions. ● 513-1E
Average Axial Load ● Average axial load in the positive direction
Since the load direction varies, calculate the average axial load while assuming Fa 3, 4, 5 = 0N.
3 3 3 3 Fa1 × ℓ1 + Fa2 × ℓ2 + Fa6 × ℓ6 Fam1 = = 225 N ℓ1 + ℓ2 + ℓ3 + ℓ4 + ℓ5 + ℓ6 ● Average axial load in the negative direction
Since the load direction varies, calculate the average axial load while assuming Fa 1, 2, 6 = 0N.
3 3 3 3 Fa3 × ℓ3 + Fa4 × ℓ4 + Fa5 × ℓ5 Fam2 = = 225 N ℓ1 + ℓ2 + ℓ3 + ℓ4 + ℓ5 + ℓ6 Since F am1 = Fam2 , assume the average axial load to be F am = Fam1 = Fam2 = 225 N.
Nominal Life Load factor fW = 1.5 (see Table2 on B 15-48 )
Average load Fm = 225 N
Nominal life L 10m (rev)
3 Ca 6 L10m = α××10 ( F am ) 1 α = fW Assumed model Dynamic load rating Nominal life number Ca(N) L 10m(rev) WTF 2040-2 5400 4.1×109 WTF 2040-3 6600 7.47×109 WTF 3060-2 11800 4.27×1010 WTF 3060-3 14500 7.93×1010
B15-76 Ball Screw 513-1E 15-77 B Point of Selection of Point rev rev Examples of Selecting a Ball Screw a Ball of Selecting Examples rev -1 -1 -1 -1 9 10 10 rev 9 10 10 10 10 × × × -1 × –1 –1 =7.93 =7.47 =4.27 =4.1 =1000 mm 10m 10m 10m 10m S ℓ L L L L Nm = 400 min Nm = 267 min Nm = 267 min Nm = 400 min 9 1000 1000 10 10 9 10 × × 40 60 10 10 10 267 267 400 400 8 8 × × × × × × × 2 2 60 60 4.1 60 60 7.47× 4.27× 7.93× = = 4950000 h = = 2670000 h = = 311000 h = = 311000 = = 171000 h = = 171000 s s ℓ ℓ m m m m × × N N N N n n Ph Ph 10m 10m 10m 10m × × × × L L L L × × 2 2 60 60 60 60 = = = = = 400 min = = min = 267 = = = = m m h h h h L L L L N N Nominal life Nominal life Nominal life Nominal life Average revolutions per minute Average Average revolutions per minute Average revolutions per minute Average Number of reciprocations per minute per minute reciprocations Number of n =8min Average revolutions per minute revolutions per Average Stroke WTF3060-3 WTF3060-2 WTF2040-3 WTF2040-2 Lead: Ph = 60 mm Lead: Ph = 60 mm Lead: Ph = 40 mm Lead: Ph = 40 mm Calculating the Service Life Time on the Basis of the Nominal Life on Life Time Calculating the Service Average Average Revolutions per Minute ● ● ● ● ● ● 513-1E
Calculating the Service Life in Travel Distance on the Basis of the Nominal Life ● WTF2040-2 9 Nominal life L10m=4.1×10 rev Lead Ph= 40 mm -6 L S = L10m × Ph× 10 = 164000 km ● WTF2040-3 9 Nominal life L10m=7.47×10 rev Lead Ph= 40 mm -6 L S = L10m × Ph× 10 = 298800 km ● WTF3060-2 10 Nominal life L10m=4.27×10 rev Lead Ph= 60 mm -6 L S = L10m × Ph× 10 = 2562000 km ● WTF3060-3 10 Nominal life L10m=7.93×10 rev Lead Ph= 60 mm -6 L S = L10m × Ph× 10 = 4758000 km
With all the conditions stated above, the following models satisfying the desired service life time of 30,000 hours are selected. WTF 2040-2 WTF 2040-3 WTF 3060-2 WTF 3060-3
B15-78 Ball Screw 513-1E 15-79 , the most B 15-79 Point of Selection of Point B on Examples of Selecting a Ball Screw a Ball of Selecting Examples Accuracy] . ℃ [Selecting Lead Angle Accuracy and Axial Clear- [Selecting Lead Angle Accuracy Axial and Clearance] 10 seconds because of the structure. The positioning er- 10 seconds because of the structure. ] p) is obtained as follows: 0.05mm/300mm) 1000 ) ± 0.007 + 0.06 = 0.234 mm ´´ × ] 10 5 to Section [Studying the Positioning to Section × 1000 –6 ℓ sin ( × × 10 t × 300 sin 15-70 × . 0.007 mm × × 0.05 B ℓ ± on = p = a = ℓ B15-70 Studying the Orientation Change during Traveling Studying the Thermal Displacement through Heat Generation Displacement through Heat Generation Studying the Thermal Studying the Axial Rigidity Studying the Axial Clearance Studying the Lead Angle Accuracy 150 = = 0.06 mm = WTF2040: axial clearance: 0.1 mm WTF2040: axial clearance: 0.14 mm WTF3060: 12 = C7 (travel distance error: C7 Δ Studying the Positioning Accuracy Positioning Studying the Studying the Rigidity Studying ror due to the pitching is obtained as follows: ror due to the pitching is obtained as follows: ance] compact model WTF2040-2 is selected. Since models WTF2040-2, WTF2040-3, WTF3060-2 and WTF3060-3 meet the selection require- Since models WTF2040-2, WTF2040-3, WTF3060-2 ments throughout the studying process in Section Since the ball screw center is 150 mm away from the point where the highest accuracy is required, it Since the ball screw center is 150 mm away during traveling. is necessary to study the orientation change Assume that pitching can be done within Thus, the positioning accuracy ( Thus, Assume the temperature rise during operation to be 5 Assume Since the load direction does not change, it is unnecessary to study the positioning accuracy on the on accuracy positioning the study to unnecessary is it change, not does direction load the Since basis of the axial rigidity. on clearance is not included in the positioning axial in a given direction only, Since positioning is performed study the axial clearance. As a result, there is no need to accuracy. based on the temperature rise is obtained as follows: The positioning accuracy Accuracy grade C7 was selected in Section C7 was selected Accuracy grade [ [ not particularly neces- and this element is do not include rigidity for selection Since the conditions described here. it is not sary, 513-1E
[Studying the Rotational Torque] Friction Torque Due to an External Load The friction toruque is obtained as follows: Fa•Ph 17×40 T1 = •A = × 1 = 120 N•mm 2π•η 2×π×0.9
Torque Due to a Preload on the Ball Screw The Ball Screw is not provided with a preload.
Torque Required for Acceleration Inertial Moment Since the inertial moment per unit length of the screw shaft is 1.23 × 10-3 kg•cm2 /mm (see the spec- ifi cation table), the inertial moment of the screw shaft with an overall length of 1200 mm is obtained as follows. –3 2 Js = 1.23 × 10 × 1200 = 1.48 kg • cm = 1.48 × 10 ‒4 kg • m2
2 2 Ph 2 –6 2 40 2 –6 –4 2 J = (m1+m2) •A ×10 +Js•A = (60+20) ×1 ×10 +1.48×10 ×1 ( 2 × π ) ( 2 × π ) = 3.39×10–3 kg•m2 Angular acceleration: π π 2 •Nm 2 ×1500 2 ω′ = = = 1050 rad/s 60•t1 60×0.15 Based on the above, the torque required for acceleration is obtained as follows. –3 –3 T2 = (J + Jm ) ×´ = (3.39 × 10 + 1 × 10 ) × 1050 = 4.61N • m = 4.61 × 103 N • mm Therefore, the required torque is specifi ed as follows. During acceleration 3 Tk = T1 + T2 = 120 + 4.61×10 = 4730 N • mm During uniform motion
Tt = T1 = 120 N • mm During deceleration 3 Tg = T1 – T2 = 120 – 4.61×10 = – 4490 N • mm
B15-80 Ball Screw 513-1E 00.15 15- B 15-81 B 2 2.6 4490 Point of Selection of Point 0.85 2 0.85 0.15 Examples of Selecting a Ball Screw a Ball of Selecting Examples 0.15 0.15 120 2 4730 –1 –1 4 t 2 s T 3 4 t t 2 g 3 t T ] 2 2 t t t 1 t T 1 t Doubled: 2000 p/rev Doubled: N mm 2 2 k T can be expressed as follows. can be expressed as follows. 1305 = 4730 N • mm = = 0 = 4490 N • mm = 4730 N • mm = 120 N • mm S g t k max = 2.6 s = 0.15 s = 0.15 s = 0.15 s = 0.15 s = 0.85 s 4 3 1 2 Rated rotational speed of the motor: 3000 min speed of the motor: Rated rotational T T Encoder resolution: 1000 p/rev. resolution: 1000 p/rev. Encoder Maximum working rotational speed : 1500 min rotational speed Maximum working t t T T T t t is the required maximum torque. is Effective Torque Value Motor Torque Minimum Feed Amount Rotational Rotational Speed rms Studying the Driving Motor the Driving Studying T Therefore, the instantaneous maximum torque of the AC servomotor needs to be at least 4,730 AC servomotor needs to be at least 4,730 Therefore, the instantaneous maximum torque of the N-mm. 80 The torque during acceleration calculated in Section [Studying the Rotational Torque] on Torque] [Studying the Rotational calculated in Section The torque during acceleration The selection requirements and the torque calculated in Section [Studying the Rotational Torque] on Torque] [Studying the Rotational in Section The selection requirements and the torque calculated B15-80 During acceleration: As with the rotational speed, the Ball Screw lead is selected based on the encoder normally used for the Ball Screw lead is selected based As with the rotational speed, to study this factor. Therefore, it is unnecessary AC servomotor. an Since the Ball Screw lead is selected based on the rated rotational speed of the motor, it is unneces- is it motor, the of speed rotational rated the on based selected is lead Screw Ball the Since of the motor. the rotational speed sary to study [ The effective torque is obtained as follows, and the rated torque of the motor must be 1305 N•mm or torque is obtained The effective greater. When stationary: During uniform motion: During deceleration: 513-1E
Inertial Moment The inertial moment applied to the motor equals to the inertial moment calculated in Section [Studying the Rotational Torque] on B 15-80 . J = 3.39 × 10–3 kg • m2 Normally, the motor needs to have an inertial moment at least one tenth of the inertial moment ap- plied to the motor, although the specifi c value varies depending on the motor manufacturer. Therefore, the inertial moment of the AC servomotor must be 3.39 × 10–4 kg-m2 or greater.
The selection has been completed.
B15-82 Ball Screw 513-1E 15-83 600 B Point of Selection of Point 1 m 2 Examples of Selecting a Ball Screw a Ball of Selecting Examples m -1 2 kg•m –5 -1 10 × =0.3m/s =10kg =40kg =5 2 1 max = 600mm =0.003 (rolling) = 0.2s = 0.2s = 0.7mm/600mm 0.05mm m s 1 3 f=20 N (without load) m ℓ m Rated rotational speed: 3,000 min Rated rotational speed: 3,000 min n =5min ] ] cient of the guide surface fi Selection Items Selection Conditions Acceleration time t Screw shaft diameter Lead Nut model No. Accuracy Axial clearance Screw shaft support method Driving motor [ Reduction gear Frictional coef None (direct coupling) Inertial moment of the motor Inertial J Guide surface resistance [ Mass Table Vertical Conveyance System System Conveyance Vertical Stroke length Work Work Mass Backlash 0.1mm Backlash Positioning accuracy Positioning accuracy repeatability Positioning feed amount Minimum s = 0.01mm/pulse life time Service motor Driving 20000h AC servo motor Maximum speed V Deceleration time of reciprocations per minute Number t 513-1E
[Selecting Lead Angle Accuracy and Axial Clearance] Selecting the Lead Angle Accuracy To achieve positioning accuracy of 0.7mm/600mm: ±0.7 ±0.35 = 600 300 The lead angle accuracy must be 0.35mm/300 mm or higher. Therefore, the accuracy grade of the Ball Screw (see Table1 on B15-20) needs to be C10 (travel distance error: 0.21 mm/300 mm). Accuracy grade C10 is available for low priced, Rolled Ball Screws. Assume that a Rolled Ball Screw is selected.
Selecting the Axial Clearance The required backlashes is 0.1 mm or less. However, since an axial load is constantly applied in a single direction with vertical mount, the axial load does not serve as a backlash no matter how large it is. Therefore, a low price, rolled Ball Screw is selected since there will not be a problem in axial clear- ance.
[Selecting a Screw Shaft] Assuming the Screw Shaft Length Assume the overall nut length to be 100 mm and the screw shaft end length to be 100 mm. Therefore, the overall length is determined as follows based on the stroke length of 600mm. 600 + 200 = 800 mm Thus, the screw shaft length is assumed to be 800 mm.
Selecting the Lead With the driving motor’s rated rotational speed being 3,000 min –1 and the maximum speed 0.3 m/s, the Ball Screw lead is obtained as follows: 0.3×60×1000 = 6 mm 3000 Therefore, it is necessary to select a type with a lead of 6mm or longer. In addition, the Ball Screw and the motor can be mounted in direct coupling without using a reduc- tion gear. The minimum resolution per revolution of an AC servomotor is obtained based on the resolution of the encoder (1,000 p/rev; 1,500 p/rev) provided as a standard accessory for the AC servomotor, as indicated below. 1000 p/rev(without multiplication) 1500 p/rev(without multiplication) 2000 p/rev(doubled) 3000 p/rev(doubled) 4000 p/rev(quadrupled) 6000 p/rev(quadrupled)
B15-84 Ball Screw 513-1E 15-85 B guration for the fi Point of Selection of Point Examples of Selecting a Ball Screw a Ball of Selecting Examples xed-supported con and Section [Selecting a Screw Shaft] on on Shaft] [Selecting a Screw and Section ), select the fi -1 15-84 B on ) are as follows. ) are B15-35 6mm —— 3000 p/rev 6mm —— 3000 8mm —— 4000 p/rev 8mm —— 4000
15mm —— 10mm 20mm —— 10mm 25mm —— 10mm 10mm —— 1000 p/rev 10mm —— 2000 p/rev 20mm —— 2000 p/rev 40mm —— Table20 on Table20 (see Selecting the Screw Shaft Support Method Selecting Selecting the Screw Shaft Diameter Accordingly, the combination of a screw shaft diameter of 15 mm and a lead 10 mm is selected. the combination Accordingly, Since the assumed Ball Screw has a stroke length of 600 mm and operates at a maximum speed Since the assumed Ball Screw has a stroke min of 0.3 m/s (Ball Screw rotational speed: 1,800 Those Ball Screw models that meet the lead being 10 mm as described in Section [Selecting Lead [Selecting Lead in Section that meet the lead being 10 mm as described Those Ball Screw models Angle Accuracy and Axial Clearance] B15-84 To meet the minimum feed amount of 0.010mm/pulse, which is the selection requirement, the follow- the requirement, selection the is which 0.010mm/pulse, of amount feed minimum the meet To apply. ing should Lead Shaft diameter Lead screw shaft support.
of pulse starting the and mm/pulse, 0.002 is distance feed the mm, 8 or mm 6 being lead the with However, least 150 kpps, and the cost of the commands to the motor driver needs to be at the controller that issues controller may be higher. and greater, the torque required for the motor is also the Ball Screw is greater, In addition, if the lead of thus the cost will be higher. select 10 mm for the Ball Screw lead. Therefore, 513-1E
Studying the Permissible Axial Load Calculating the Maximum Axial Load Guide surface resistance f=20 N (without load)
Table Mass m1 =40 kg
Work Mass m2 =10 kg
Maximum speed Vmax=0.3 m/s
Acceleration time t 1 = 0.2s Accordingly, the required values are obtained as follows. Acceleration
Vmax α = = 1.5 m/s2 t1 During upward acceleration:
Fa1 = (m1 + m2 ) •g + f + (m1 + m2 ) • = 585 N During upward uniform motion:
Fa2 = (m1 + m2 ) •g + f = 510 N During upward deceleration:
Fa3 = (m1 + m2 ) •g + f – (m1 + m2 ) • = 435 N During downward acceleration:
Fa4 = (m1 + m2 ) •g – f – (m1 + m2 ) • = 395 N During downward uniform motion:
Fa 5 = (m1 + m2 ) •g – f = 470 N During downward deceleration:
Fa6 = (m1 + m2 ) •g – f + (m1 + m2 ) • = 545 N Thus, the maximum axial load applied on the Ball Screw is as follows:
Famax = Fa1 = 585 N
Buckling Load of the Screw Shaft Factor according to the mounting method 2 =20 (see B15-38 ) Since the mounting method for the section between the nut and the bearing, where buckling is to be considered, is “fi xed-fi xed: ” ℓ Distance between two mounting surfaces a =700 mm (estimate)
Screw-shaft thread minor diameter d1=12.5 mm
4 4 d1 4 12.5 4 P1 = η2 • ℓ 2 ×10 = 20××2 10 = 9960 N a 700
Permissible Compressive and Tensile Load of the Screw Shaft 2 2 P2 = 116d 1 = 116 × 12.5 = 18100 N Thus, the buckling load and the permissible compressive and tensile load of the screw shaft are at least equal to the maximum axial load. Therefore, a Ball Screw that meets these requirements can be used without a problem.
B15-86 Ball Screw 513-1E 15-87 B ) Point of Selection of Point B15-40 Examples of Selecting a Ball Screw a Ball of Selecting Examples –1 =0.3 m/s =12.5 mm =15.1 (see =15.1 =700 mm (estimate) =700 max 1 2
b ℓ D=15.75 mm V d Ph= 10 mm Ph= 10 mm = 3852 min 7 10 –1 –1 2 700 12.5 xed-supported: ” fi ×× 3 10 = 15.1 × 7 15.75 70000 60 Ph 2 × 1 b ℓ d max 10 D V × 2 70000 = = 1800 min = = = = = 4444 min = λ 2 1 max N N N Ball center-to-center diameter Maximum speed Screw-shaft thread minor diameter Lead Factor according to the mounting method Factor according to the mounting Since the mounting method for the section between the nut and the bearing, where dangerous for the section between the nut and Since the mounting method is “ speed is to be considered, mounting surfaces Distance between two Studying Studying the Permissible Rotational Speed Screw shaft diameter: 15mm; lead: 10mm (large lead Ball Screw) shaft diameter: 15mm; lead: 10mm (large lead Screw Screw shaft diameter: 15mm; lead: 10mm 15mm; lead: 10mm Screw shaft diameter: Screw shaft diameter: 15mm; lead: 10mm Screw shaft diameter: 15mm; Permissible Rotational Speed Determined by the DN Value Speed Determined by the DN Value Permissible Rotational Permissible Rotational Speed Determined by the Dangerous Speed of the Screw Shaft Speed Determined by the Dangerous Permissible Rotational Maximum Rotational Speed Thus, the dangerous speed and the DN value of the screw shaft are met. Thus, ● ● ● 513-1E
[Selecting a Nut] Selecting a Nut Model Number The Rolled Ball Screw with a screw shaft diameter of 15 mm and a lead of 10 mm is the following large-lead Rolled Ball Screw model. BLK1510-5.6
(Ca=9.8 kN, C 0a=25.2 kN)
Studying the Permissible Axial Load Assuming that an impact load is applied during an acceleration and a deceleration, set the static safety factor (fS ) at 2 (see Table1 on B 15-47 ).
C0a 25.2 Famax = = = 12.6 kN = 12600 N fS 2 The obtained permissible axial load is greater than the maximum axial load of 585 N, and therefore, there will be no problem with this model.
Studying the Service Life Calculating the Travel Distance Maximum speed V max=0.3 m/s
Acceleration time t 1 = 0.2s
Deceleration time t 3 = 0.2s ● Travel distance during acceleration
Vmax • t1 3 0.3×0.2 3 ℓ1, 4 = ×10 = ×10 = 30 mm 2 2 ● Travel distance during uniform motion
Vmax • t1 + Vmax • t3 3 0.3×0.2 + 0.3×0.2 3 ℓ2, 5 = ℓS – ×10 = 600 – ×10 = 540 mm 2 2 ● Travel distance during deceleration
Vmax • t3 3 0.3×0.2 3 ℓ3, 6 = ×10 = ×10 = 30 mm 2 2 Based on the conditions above, the relationship between the applied axial load and the travel dis- tance is shown in the table below.
Applied axial load Travel distance Motion ℓ Fa N(N) N(mm) No1: During upward acceleration 585 30 No2: During upward uniform motion 510 540 No3: During upward deceleration 435 30 No4: During downward acceleration 395 30 No5: During downward uniform motion 470 540 No6: During downward deceleration 545 30 ＊ The subscript (N) indicates a motion number.
B15-88 Ball Screw 513-1E 15-89 B ) ) = 492 N = ) 6 ℓ Point of Selection of Point • 3 6 B15-48 rev 9 + Fa 5 ℓ 10 • 3 × Examples of Selecting a Ball Screw a Ball of Selecting Examples rev rev 5 9 9 -1 es the desired service life time of es the desired service life time of 10 10 -1 × × + Fa 4 ℓ = 2.34 6 • 3 –1 =2.34 =2.34 10 4 = 492 N = 492 = 600 min (rev) = 1.5 (see Table2 on Table2 = 1.5 (see =600 mm m am 10m 10m 10 S W × ℓ Ph= 10 mm Ph= 10 mm L L f L F N Ca= 9800 N Ca= 9800 N 3 + Fa 3 ℓ • 492 3 3 × 9800 9 600 1.5 + Fa 2 ℓ 10 ( ) ( 10 600 5× • × 3 = 2 6 × × 2 10 60 2.34 = 23400 km + Fa -6 1 3 ℓ = = 65000 h s • 10 ℓ a 3 am 1 m × C × F N n Ph 10m Ph ×× • L S × α × ℓ ( ) 2 60 1 10m W × 1 f = 2 = = = = = 600 min = m = L 3 = h 10m S N L α L Average Average load Stroke Lead Average revolutions per minute Average Lead L Nominal life Nominal life Number of reciprocations per minute Number of reciprocations n = 5 min Load factor Dynamic load rating Dynamic load Nominal life = (Fa = m a Calculating the Service Life in Travel Distance on the Basis of the Nominal Life Calculating the Service Life in Travel Calculating the Service Life Time on the Basis of the Nominal Life on the Basis of the Nominal Calculating the Service Life Time Average Average Revolutions per Minute Nominal Life Average Axial Axial Load Average F 20,000 hours. With all the conditions stated above, model BLK1510-5.6 satisfi With all the conditions stated above, model 513-1E
[Studying the Rigidity] Since the conditions for selection do not include rigidity and this element is not particularly neces- sary, it is not described here.
[Studying the Positioning Accuracy] Studying the Lead Angle Accuracy Accuracy grade C10 was selected in Section [Selecting Lead Angle Accuracy and Axial Clearance] on B15-84 . C10 (travel distance error: 0.21mm/300mm)
Studying the Axial Clearance Since the axial load is constantly present in a given direction only because of vertical mount, there is no need to study the axial clearance.
Studying the Axial Rigidity Since the lead angle accuracy is achieved beyond the required positioning accuracy, there is no need to study the positioning accuracy determined by axial rigidity.
Studying the Thermal Displacement through Heat Generation Since the lead angle accuracy is achieved beyond the required positioning accuracy, there is no need to study the positioning accuracy determined by the heat generation.
Studying the Orientation Change during Traveling Since the lead angle accuracy is achieved at a much higher degree than the required positioning ac- curacy, there is no need to study the positioning accuracy.
[Studying the Rotational Torque] Frictional Torque Due to an External Load During upward uniform motion:
Fa2•Ph 510×10 T1 = = = 900 N•mm 2×π×η 2×π×0.9
During downward uniform motion:
Fa5•Ph 470×10 T2 = = = 830 N•mm 2×π×η 2×π×0.9
Torque Due to a Preload on the Ball Screw The Ball Screw is not provided with a preload.
B15-90 Ball Screw 513-1E 2 1 × 15-91 –4 B 10 × /mm (see the speci- the (see /mm Point of Selection of Point 2 +0.31 –6 kg•cm kg•cm 10 -4 × 2 10 10 1 Examples of Selecting a Ball Screw a Ball of Selecting Examples × × 2 π × 10 2 ( ) 2 942 = 0.2 N•m = 200 N•mm 942 = 0.2 N•m = 200 × ) –5 = (40+10) 2 10 A × ed as follows. ed as follows. • fi s 2 +J + 5 + –4 –6 10 10 0.2 1800 × × × 2 × A 2 • 2π 2 2 m • π = (1.58 ´ 800 = 0.31 kg • cm 800 = 0.31 × Ph kg = 900 + 200 = 1100 N•mm N•mm = 900 + 200 = 1100 kg • m 2 –4 ) • = 900 – 200 = 700 N•mm = 900 – 200 = 700 N•mm 3 4 × ‒ 3 m ( ) t 60 t –4 ) 10 • 2 10 – T Nmax + T = 900 N•mm 1 1 10 1 × • × 60 × +m 1 = 1030 N•mm = T = 630 N•mm = T = 830 N•mm = T 2π = (J + J = t2 g2 k2 t1 g1 3 k1 T T T T T T T Torque Torque Required for Acceleration = 1.58 = 3.9 = 0.31 S cation table), the inertial moment of the screw shaft with an overall length of 800mm is obtained as of 800mm is obtained with an overall length of the screw shaft table), the inertial moment cation = = = 942 rad/s = ω′ = = During upward uniform motion: During Therefore, the required torque is speci Therefore, the required torque upward acceleration: During fi follows. J Based on the above, the torque required for acceleration is obtained as follows. required for acceleration is obtained as Based on the above, the torque Angular acceleration: Inertial Moment: 3.9 is shaft screw the of length unit per moment inertial the Since During upward deceleration: During downward acceleration: During downward uniform motion: During J = (m During downward deceleration: 513-1E
[Studying the Driving Motor] Rotational Speed Since the Ball Screw lead is selected based on the rated rotational speed of the motor, it is unneces- sary to study the rotational speed of the motor. Maximum working rotational speed : 1800 min –1 Rated rotational speed of the motor: 3000 min –1
Minimum Feed Amount As with the rotational speed, the Ball Screw lead is selected based on the encoder normally used for an AC servomotor. Therefore, it is unnecessary to study this factor. Encoder resolution: 1000 p/rev.
Motor Torque The torque during acceleration calculated in Section [Studying the Rotational Torque] on B15- 90 is the required maximum torque.
Tmax = Tk1 = 1100 N•mm Therefore, the maximum peak torque of the AC servomotor needs to be at least 1100 N-mm.
Effective Torque Value The selection requirements and the torque calculated in Section [Studying the Rotational Torque] on B15-90 can be expressed as follows. During upward acceleration:
Tk1 = 1100 N•mm
t1 = 0.2 s During upward uniform motion:
Tt1 = 900 N•mm
t2 = 1.8 s During upward deceleration:
Tg1 = 700 N•mm
t3 = 0.2 s During downward acceleration:
Tk2 = 630 N•mm
t1 = 0.2 s During downward uniform motion:
Tt2 = 830 N•mm
t2 = 1.8 s During downward deceleration:
Tg2 = 1030 N•mm
t3 = 0.2 s
When stationary(m 2=0):
TS = 658 N•mm
t4 = 7.6 s
B15-92 Ball Screw 513-1E 7.6 × 2 15-93 658 B ＋ 0.2 Point of Selection of Point × 2 or greater. greater. or 2 4 t 1030 kg-m • –5 2 ＋ s 7.6 T 10 1.8 ＋ Examples of Selecting a Ball Screw a Ball of Selecting Examples ＋ × × 3 2 t • 0.2 2 830 g2 ＋ T ＋ ＋ 1.8 4 2 0.2 t t ＋ × • 2 ＋ 2 3 t2 t T 0.2 630 ＋ 2 ＋ ＋ t ＋ 1 t ＋ • 0.2 1 0.2 2 t k2 × 2 ＋ T ＋ 3 t ＋ c value varies depending on the motor manufacturer. depending on the motor manufacturer. c value varies 700 3 fi 1.8 t ＋ 2 ＋ . • t ＋ 2 g1 ＋ 1.8 1 T 0.2 t × 2 ＋ 2 t 2 B15-90 900 • 2 t1 ＋ T kg • m kg 0.2 ＋ –4 1 t × mm 2 • 10 • 2 k1 × 1100 T = 743 N = = J = 1.58 Inertial Moment rms T Normally, the motor needs to have an inertial moment at least one tenth of the inertial moment ap- moment inertial the of tenth one least at moment inertial an have to needs motor the Normally, although the speci plied to the motor, Therefore, the inertial moment of the AC servomotor must be 1.58 AC the inertial moment of the Therefore, The selection has been completed. The The inertial moment applied to the motor equals to the inertial moment calculated in Section [Studying Section in calculated moment inertial the to equals motor the to applied moment inertial The on Torque] the Rotational The effective torque is obtained as follows, and the rated torque of the motor must be 743 N•mm or or N•mm 743 be must motor the of torque rated the and follows, as obtained is torque effective The greater. 513-1E
B15-94