Main spindle
Motor
selection
9/30/2018 Aruna kumar.P 1
9/30/2018 Aruna kumar.P 2
Data:
1. Bearing :
Front – 7018C,OD/ID:140/90mm, B: 24mm.
Rear – 7016C, OD/ID:125/80mm, B: 22mm
2. Power Chuck-165WH:
Weight-13.7Kgf, Moment of inertia -0.063 Kgm2.
3 .Pulleys :
Spindle pulley. OD/ID:180/70mm, Weight-7.4 Kgf.
Motor pulley , OD/ID:90/40mm, Weight-2 Kgf.
4. Reduction ratio 2:1.
5. Rotary Cylinder:
Weight : 12 Kgf, Moment of inertia:0.019 Kgm2
6. Spindle Speed range 290-8000 rpm.
7. Machine specification:
Job max. dia.160 mm, Length-350 mm, Weight-
59 Kg.
8. Motor used: 1PH8103;5.5kW,1500-5200rpm
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load inertia
Total load inertia(jL),kgm2:
Jc+Jb+Js1+Jp+JR
Jc: Inertia of chuck.
Jb: Inertia of all bearings
Js1: spindle Inertia.
Jp: Pulley inertia.
Jr: Rotary cylinder inertia
1. Inertia of chuck(Jc):
Refer Manufacturer’s catalogue.
Jc: 0.063kgm2
9/30/2018 Aruna kumar.P 4
load inertia
2. Bearing inertia (jb)=
3*(Front bearing Inertia)+ 2* (Rear bearing Inertia)
Front bearing Inertia(jbf)= m*[(D1/2)2+(D2/2)2)/2
m=1.34kg:D1:140mm; D2:90mm
jbf = (1.34)[(0.140/2)2 + (0.090/2)2]/2 = 4.73x10-3 kgm2.
Rear bearing inertia(jbr): m*[(D1/2)2+(D2/2)2)/2
m=0.966kg; D1:125mm; D2:80mm
=(0.966)*[(0.125/2)2+(0.080/2)2]/2 = 2.69x10-3
Total bearing inertia(jb): 3*(4.73x10-3)+2(2.69x10-3)
=19.6x10-3kgm2= 0.0196kgm2
9/30/2018 Aruna kumar.P 5
spidle inertia
3. Spindle inertia(Js1): (Ws/g)*[(D1/2)2+(D2/2)2)/2
Ws: 20kgf=200N.
D1:80mm(average dia).
D2: 30mm(bore through spindle).
Js1= (200/9.81)*[(0.08/2)2+(0.030/2)2]/2
=0.0045kgm2
4. Pulley inertia (Jp): (Wp/g)*[(D1/2)2+(D2/2)2)/2
Wp: 7.4kgf=74N; D1:OD= 180mm;D2:ID= 70mm.
Jp= (74/9.81)*[(0.180/2) 2+(0.070/2)]2/2 =0.035kgm2
5. Rotary cylinder inertia(Jr):
Refer Manufacturer's catalogue.
Jr= 0.019kgm2
9/30/2018 Aruna kumar.P 6
inertia referred to Motor
Total load Inertia(JL): Jc+Jb+Js1+Jp+JR
JL= 0.063+0.0196+0.0045+0.035+0.019=0.1411kgm2
Load inertia referred to Motor axis(JLr): JL/i2.
i: Reduction ratio= 2:1.
JL= 0.1411kgm2
(JLr)= 0.1411/22 = 0.0352kgm2
Total load inertia at Motor axis(J): Jlr + Jp2
Jp2: Inertia of Motor pulley,kgm2 :
= (Wp2/g)*[(D1/2)2+(D2/2)2)/2
Wp2: 2kgf=20N; OD(D1)= 90mm;ID(D2)= 38mm.
Jp2=(20/9.81)*[(0.090/2)2+(0.038/2)2]/2 = 0.0025kgm2
J= 0.0352+0.0025=0.0371kgm2
9/30/2018 Aruna kumar.P 7
load inertia
Motor details: 1PH8103-1x F
Motor power: 5.6kW,
Speed: 1500-5200rpm
Inertia:0.0172kgm2;
Rated Torque: 35Nm;
Static torque: 38Nm.
Shaft dia:38mm
9/30/2018 Aruna kumar.P 8
Servo Motor selection
Inertia ratio(jt)= Load inertia/Motor inertia.
Selected Motor inertia: 0.0172kgm2
Jt= 0.0371/0.0172= 2.15
If it is less than 5 it is OK.
Acceleration speed(na): 1500rpm.
Static/Acceleration Torque of motor(ta): 38Nm
Angular acceleration(a):Ta/J=38/0.0371=1024rad/sec2
Acceleration time(ta): 2pn/60*a
=2p*1500/(60*1024) =0.153sec=153msec.
9/30/2018 Aruna kumar.P 9
feed Motor
evaluation
9/30/2018 Aruna kumar.P 10
feed axis configuration
Lathe machine Z-axis : Horizontal axis.
Table(X-axis and turret) weight (Wt):
2400N
Cutting fiorce(Fc): 5000N
Ball screw dia(d)/lead(l):32/12mm
Stroke: 350mm.
Ball screw length: 750mm
Ball screw length(L): 1400mm
Max. linear speed(v):
24000mm/min(24m/min)
Friction of sliding surfaces(m): 0.01
Acceleration/deceleration time(ta/td) =
100ms=0.100s
Direct drive (no reduction).
9/30/2018 Aruna kumar.P 11
load and operating speed
Slide Movement is Horizontal. Cutting power(P),kW: 5kW.
Total slide load(Wt)= Cutting speed(v),m/min: 150.
m*W= 0.01*2400=24N. Tangential force(Ft),N: P/v
Max. linear speed(v) : 24m/min. Ft=5*1000*60/150 = 2000N
Ball screw rotational speed(nmax): (v/l) Ff= say 0.7Ft=0.7*2000=1400N
nMax= 24000/12=2000 rpm. Axial force acting on spindle.
Cutting feed force=Axial force acting FP= say 0.5Ft=0.5*2000=1000N
on ball screw. Radial force acts in Horizontal
Feed force(Ff)= 1400N plane.
Total axial load (Fa): Ff+Wt= 1400+24=1424N.
Preload(Fv): Fmax/3= 1424/3=475N
Ball screw efficiency:0.9(Assumed)
9/30/2018 Aruna kumar.P 12
drive torque
Drive Torque during Machining(Tc): Torque referred to Motor axis(TCr)
For pre loaded screw: [(0.9Tp)+Td+Tb+Tf). Tcr: Tc*(n2/n1)
For Non preloaded screw: (Tp+Tb+Tf). n1/n2: Speed reduction
Tp: Torque due to external load. n1:Motor speed, rpm
Tp=Fa*l/2ph=1424*12/2p*0.9= 3023Nmm. n2: Feed screw speed, rpm
Td: Torque due to Preload. Direct drive:
Td= Fv*l/2p = 475*12/2p =908Nmm So n1/n2=1 for our application
Tb: Friction torque of bearings Tcr = Tc =3788Nmm.
Tb= Fa*mb*d/2 =1424*0.002*(25/2)= 36Nmm
For four bearings(Tb):4*36=144Nmm
d: Bore dia of bearing,mm
Tf :Friction torue of Oil seals,couplings etc. Selected motor torque(Tm) should be
Tf=10Nmm (Assumed) 10 to 30% more than Drive torque(Tc)
Total Load/Drive Torque(Tc):
Preloaded screw and Direct drive.
Tc= (0.9Tp+Td+Tf)
Tc=[0.9*3023+908+144+10]=3788Nmm=3.8Nm
9/30/2018 Aruna kumar.P 13
Acceleration torque
Acceleration torque(Ta): Jt*a, Nm
Jt: Inertia referred to motor axis,kgm2
a: Angular acceleration, rad/sec2= 2pn/60*ta
n=Acceleration speed, rpm
ta= Acceleration time, sec.
Jt: [Jl/(n1/n2)2]+(Jg2+Jm)
Jl : Mass Moment of inertia of feed system,kgm2
Jl= Ji+Jbs+Jg1
Ji: Load inertia,kgm2
Jbs: Ball screw inertia, kgm2 n1/n2= Motor speed/Feed screw speed
Jg1: Inertia of Gear/Pulley on Feed screw, kgm2
( For reduction drives)
Jg2: Inertia of Gear/Pulley on Motor axis, kgm2 (
For reduction drives)
Jm: Motor inertia,kgm2.
9/30/2018 Aruna kumar.P 14
inertia calculations
Ji: Load inertia=(W/g)*(l/2p)2 Jg1,2: Gear/ pulley inertia,kgm2
W: Total static load, 2400N Jg1,2= (mg1,2)*(d22+d12)/8
d1: Bore of gear/pulley,m
l: Pitch of ball screw,12 mm=0.012m d2: OD of gear/pulley,m
Ji= (2400/9.81)*(0.012/2p)2 =0.0009 kgm2 mg1,2 = Mass of gears,kg
Motor is directly coupled to ball screw
Jbs= Ball screw inertia=[pr Hence Jg1=Jg2=0
Ld4/32*g](n1/n2)2
r: Density of steel: 78500N/m3
Jm: Motor inertia.
L=Length of ballscrew,750mm=0.75m to be selected from Motor catalogue.
d: dia of ball screw,32mm=0.032m
Jbs=[p*78500*0.75*(0.032)4/(32*9.81)}
=0.0006kgm2
Jl= 0.0009+0.0006=0.0015kgm2
9/30/2018 Aruna kumar.P 15
inertia and acceleration torque
Selected Motor 1FK7063-XXX-
Tm= 7.3Nm;Jm=0.0015kgm2; nm=3000rpm
High short-term overload capacity (250ms)
Tmax= 3*Tm(100 K).
Jm=0.0015kgm2; Jl=0.0015kgm2
Jl/Jm=0.0015/0.0015=1
So selected Motor inertia is Good
Total inertia(Jt)=Jm+Jl
Jt=0.0015+0.0015=0.0030kgm2
Motor torque (Tm): 7.3Nm;
Load/continuous torque(Tc): 3.8Nm
Tm/Tc=7.3/3.8= 1.9.
So selected Motor is safe.
9/30/2018 Aruna kumar.P 16
Servo Motor selection
Angular acceleration(a):
Acceleration speed(na): 2000rpm.
Acceleration Time,sec(ta): 0.100sec.
a=2pn/60*ta=2p*2000/(60*0.100) = 2094rad/sec2.
Acceleration torque(Ta)=ja = 0.0030*2094
=6.28Nm=6280Nmm.
Drive torque required to move Dead load(Td) =Fad*l/2ph
= 0.01(2400)*12/(2p*0.9)= 51Nmm
Max.torque(Tmax): Ta+Td = 6280+51=6331Nmm=6.33Nm.
Continuous torque(Tc)= 3.8Nm
Power(P)= nT/9550=1000*3.8/9550
=0.39kW =390W
9/30/2018 Aruna kumar.P 17
Servo Motor selection
Selected Motor
Siemens Feed Motor : 1FK7063-XXXX-
(Page2/4).
Rated Torque(Mr): 7.3Nm;
Accelration torque(Mo): 11Nm
Inertia(Jm): 0.0015kgm2;
Nax.Speed(nmax.):3000rpm;
Motor Power(Pm): 2.3kW
9/30/2018 Aruna kumar.P 18
thanks
9/30/2018 Aruna kumar.P 19
