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NORD - Information Calculation Methods and Examples

Getriebebau NORD, Schlicht + Küchenmeister GmbH & Co. Rudolf-Diesel-Str. 1, D- 22941 Bargteheide Telefon: 04532 / 4010, Telefax: 04532 / 401 253

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Physical Formulae

Linear motion

Distance Velocity (speed) Accelleration

Rotating motion s = v ∗ t s t v a = t v =

Angular

ϕ = ω ∗ t

Angular velocity

ω =

Angular accelleration

ϕ = 2 ∗ π ∗ n τ W α = t

Force

F = m ∗ a

Torque

M = J ∗ r = F ∗ r

Power

P = F ∗ v

Power

P = M ∗ ω

Work

W = F ∗ s = P ∗ t

Work

W = M ∗ ϕ = P ∗ t

Kinetic energy

Wkin =

Rotating energy

Wrot =

1 2

∗ m ∗ v2

1 2

∗ J ∗ ω2

Formulae of drive engineering 2 d  FR = m ∗ g ∗  ∗ (µL ∗ + f) + c : µL, f, c, s. ____ tables __1 __,_2 __,_3 __. D 2  

Rolling resistance, -force or

FR = We ∗ m

We for wheel / rail steel s. diagramme 1, 2.

Sliding resitance, -force

FG = m ∗ g ∗ µ

µ

Static friction-force

FH = m ∗ g ∗ µO

µO s. table 4

Windload

F = A ∗ PW

Moment of inertia with refernce to the motor shaft

or

v Jred = 91,2 ∗ m ∗   2  nM   n 2 Jred = J ∗    nM 

Speed

n =

Torque

M =

or

or

2

Rotation

P ∗ 9550 n

F ∗ v 1000 ∗ η M ∗ n PR = 9550 m ∗ a ∗ v 1000 ∗ η J ∗ n2 PB = 91,2 ∗ 1000 ∗ tB ∗ η PB =

Acceleration power

Translation

v ∗ 60 π ∗ D

PR =

Friction power

s. table 4

Translation Rotation

Translation Rotation

NORD - Information

m ∗ g ∗ v 1000 ∗ η

Hubleistung

PHub =

Beschleunigung

aB =

9,55 ∗ v ∗ (MH ± ML) (Jred + JM + JBre + JZ) ∗ n

Beschleunigungszeit

tB =

v aB

Beschleunigungsweg

sB =

v2 2 ∗ aB

Verzögerung

av =

9,55 ∗ v ∗ (MB ± ML) (Jred + JM + JBre + JZ) ∗ n

Verzögerungszeit

tV =

v aV

Verzögerungsweg

sV =

v2 2 ∗ aV

zulässige Schalthäufigkeit

zzul =

Positioniergenauigkeit

Positioniergenauigkeit = ± 0,25 * sv

Bremsarbeit

WB =

Lebensdauer der Bremsbeläge

LN =

Übersetzung

i =

Wirkungsgrad

η =

rücktreibender Wirkungsgrad

η G’ = 2 −

Querkraft

FQ =

Betriebsfaktor

fB =

Massenbeschleunigungsfaktor

maf =

1 − ML ⁄ MH ∗ zo 1 + (Jred + JBre + JZ) ⁄ JM

(Jred + JM + JBre + JZ) ∗ n2 MB ∗ 182,5 MB ± ML Wzul WB ∗ z

n1

n2

=

M2 d2 z2 = = M1 d1 z1

Pab Pzu

η s. Tabelle 5 1 ηG

2 ∗ M2 ∗ fZ ∗ ≤ FQzul D

fZ s. Tabelle 6

M2max M2 Jred JM + JZ + JBre

3

NORD - Information

Formulae symbols and unities a

Acceleration

m/s2

aB

Acceleration (start up)

m/s2

av

Deceleration (braking)

m/s2

A

Area (wind)

m2

c

Additional factor for secondary friction

-

d

Diameter (bearing spigot diameter)

m

dO

Pinion or sprocket diameter

m

d1

Pinion diameter

m

d2

Chain sprocket diameter

m

D

Diameter of the travelling wheel or cable drum or of the sprocket

m

f

Lever arm of rolling friction

m

fB

Service factor

-

fZ

Additional factor for overhung load

-

F

Force, rolling resistance

N

FG

Sliding friction

N

FH

Static friction

N

FQ

Overhung load

N

FQvorh

Existing overhung load

N

FQzul

Permissible overhung load

N

FR

Rolling resistance

N

FW

Wind load

N

g

Gravity (constant: 9,81)

m/s2

i

Reduction

-

iV

Additional reduction (gear, chain, belt ...)

-

J

Moment of inertia

kgm2

JBre

Moment of inertia of the brake

kgm2

JM

Moment of inertia of the motor

kgm2

Jred

Moment of inertia with reference to the motorshaft

kgm2

JZ

Moment of inertia of the z-fan

kgm2

LN

Brake service life until readjustment

h

m

Weight (mass)

kg

maf

Inertia mass acceleration factor

-

mG

Mass of counter weight

kg

mL

Mass with full load

kg

mO

Mass without load

kg

4

NORD - Information

M

Torque

Nm

MB

Braking torque

Nm

MH

Run up torque

Nm

ML

Torque with full load (with reference to the motor shaft)

Nm

MN

Rated torque

Nm

M1

Input torque

Nm

M2

Output torque

Nm

M2max

Maximum permissible output torqe

Nm

n

Speed

1

nM

Motor speed

1

nN

Rated speed

1

n1

Input speed

1

n2

Output speed

1

PW

Wind pressure

N/m2

P

Power

kW

Pab

Required power

kW

Pzu

Supplied power

kW

PB

Acceleration power

kW

PHub

Lifting power

kW

PN

Rated power

kW

PR

Fricition power

kW

r

Radius

m

s

Distance

m

sB

Start up distance

m

sV

Braking distance

m

t

Time

s

tB

Start up time

s

tV

Braking time

s

v

Velocity (speed)

m/s

W

Work

J

We

Standard rolling friction

N/t

Wkin

Kinetic energy

J

Wrot

Rotating energy

J

Wzul

Braking work until readjustment

J

WB

Braking work

J

x

Number of drives

-

/min /min /min /min /min

5

NORD - Information

z

Starting frequency

s/h

zzul

Permissible starting frequency

s/h

zO

Starting frequency with no load

s/h

z1

Number of gear teeth pinion

-

z2

Number of gear teeth gear wheel

-

α

Angular acceleration

1/s2

η

Efficiency

-

Efficiency of gear unit

-

ηG ηG

’

Reverse operating efficiency

-

µ

Coefficient of friction

-

µL

Coefficient of friction for bearings

-

µO

Coefficient of friction (static)

-

ϕ

Angular

°

ω

Angular velocity

1/s

6

NORD - Information

Friction bearing

Sliding bearings

0,005

0,1

µL

table 1: coefficient of friction for bearings µL f steel / steel

0,0005 m

wood / steel

0,0012 m

polymer / steel

0,002 m

hardrubber / steel

0,0077 m

hardrubber / concrete

0,01 - 0,02 m

rubber / concrete

0,015 - 0,035 m

table 2: lever arm of rolling friction f

Friction of anti friction bearings

Friction of sleeve bearings

Friction of guide rollers

0,003

0,005

0,002

c

table 3: rim friction on the wheels c Static friction µO

Sliding friction µ

dry

greased

dry

greased

steel / steel

0,11 - 0,40

0,10

0,10 - 0,30

0,01 - 0,10

steel / last iron

0,18 - 0,25

0,10

0,16 - 0,25

0,05 - 0,10

steel / wood

0,50 - 0,70

0,10

0,20 - 0,50

0,02 - 0,10

steel / polymer

0,20 - 0,50

0,10 - 0,35

steel / rubber

0,40 - 0,50

wood / wood

0,40 - 0,80

0,16

0,20 - 0,50

0,04 - 0,16

table 4: static friction and sliding friction µ η chain

0,90 - 0,96

per complete wrap of the rope around the drum

wire ropes

0,90 - 0,95

per complete wrap

flat polymer belts

0,93 - 0,98

per complete wrap of the rope depending on the material

V-belts

0,85 - 0,95

per complete wrap

rubber belts

0,80 - 0,85

per complete wrap

polymer belts

0,80 - 0,85

per complete wrap

helical inline gear

0,95 - 0,98

oil lubricated depending on the number of the stages

worm gear

0,30 - 0,93

oil lubricated depending on the number of starts of the worm

table 5: efficiency η

fZ helical gears

1,1

z = 17 teeth

chain sprockets

1,4

z = 13 teeth

chain sprockets

1,2

z = 20 teeth

pulleys

1,7

by tensioning influence

pulleys

2,5

by tensioning influence

table 6: additional factor inderming overhung loads fz

7

NORD - Information

Example I.1: Drive arrangement for crane Mass without load of the crane mO Mass without load of the

mk

Load mL

13800 kg 1800 kg 15000 kg

Velocity v

0,17 / 0,66 m/s = 10/40 m/min

Diameter of the travelling wheel D

0,4 m

Number of drives x Additional reduction

2 iv

4,24

Mounting position Switching frequencies Efficiency η

B3 z

60 s/h 0,85

9

NORD - Information

Motor arrangements

Standard rolling friction We We = WO + 30 N/t

W0 = 36 N/t s. diagram

We = 36 N/t + 30 N/t = 66 N/t

30 N/t additional for rim friction

Power P (at maximum velocity)

P =

We ∗ m ∗ v 1000 ∗ η

PO =

66 N ⁄ t ∗ (13,8 t + 1,8 t) ∗ 0,66 m⁄s = 0,80 kW (without load) 1000 ∗ 0,85

PL =

66 N ⁄ t ∗ (13,8 t + 1,8 t + 15,0 t) ∗ 0,66 m⁄s = 1,57 kW (with load) 1000 ∗ 0,85

Pmax =

PL

2

∗

mO + 2 ∗ (mK + mL ) 1,57 kW 13,8 t + 2 ∗ (1,8 t + 15 t) = ∗ 13,8 t + 1,8 t + 15 t mO + mK + mL 2

Pmax = 1,22 kW (one-sided trolley)

Motor data Type Rated output power PN Rated speed nN Rated torque MN Permissible no-load starting frequency zo

0,55 / 2,2 kW 670 / 2740 1/min 7,8 / 7,7 Nm 4000 / 1400 s/h

Motor moment of inertia JM

0,0060 kgm2

Additonal moment of inertia Jz

0,0113 kgm2

Brake moment of inertia JBre

0,0001 kgm2

Braking torque MB (brake 16 adjusted to 8 Nm )

10

100 L/80-20 WU Bre16 Z (2 pieces)

8 Nm

NORD - Information

Gear arrangment

Wheel speed nL nL =

v ∗ 60 π ∗ D

nL =

0,66 m⁄s ∗ 60 = 32 1⁄min π ∗ 0,4 m

Gear output speed n2 n2 = nL ∗ iv n2 = 32 1⁄min ∗ 4,24 = 136 1⁄min

Acceleration factor of mass maf Jred maf = JM + Jz + JBre 0,0810 kgm2 = 4,7 0,0060 kgm2 + 0,0113 kgm2 + 0,0001 kgm2

maf =

Starting freqeuency per hour: 180 (60 times acceleration, switching, deceleration) ⇒ Type of load C, fB = 1,6 Output torque Ma Ma =

PN ∗ 9550 ∗ fB n2

Ma =

2,2 kW ∗ 9550 ∗ 1,6 = 247 Nm 136 1⁄min

For service factor fB = 1,6 the output torque of the gear is 247 Nm. Reduction i i =

nN n2

i =

2740 1⁄min = 20 136 1⁄min

Complete type:

SK 22-100 L/80-20 WU Bre 16 Z PN = 0,55 / 2,2 kW i = 20,03 n2 = 33 / 137 1/min Mounting position B 3 Shaft ø 30 x 60 mm Brake 16 Nm adjusted to 8 Nm Special provision:

special rotor high inertia fan

11

NORD - Information

Example I.2: Drive arrangement for a trolley

Mass without load

mk

Load mL

15000 kg

Velocity v Wheel diameter

1800 kg

0,08 / 0,33 m/s = 5/20 m/min D

0,3 m

Number of drives x Additional reduction

1 iv

Mounting position Switching frequency z Efficiency n

4 B5 60 s/h 0,85

Pairing of material

steel / steel

Guiding

rim friction

Type of bearings (4 wheels)

12

antifriction bearings

NORD - Information

Drive resistance: FR = m ∗ g  

2 d  ∗ ( µ L ∗ + f ) + c  D 2

Fro = 1800 kg ∗ 9,81

m  2 0,06  ∗ (0,005 ∗ + 0,0005 m ) + 0,003m   2 s2  0,3 m 

Fro = 129,5 N ( without load)

FRL = 16800 kg ∗ 9,81

m  2 0,06  ∗  (0,005 ∗ m + 0,0005 m ) + 0,003 m  2 s2  0,3 m 

FRL = 1208,6 N (with load)

Power P (calculation for 2-poles gearmotors)

P =

F∗V 1000 ∗ η

Po =

129,5 N ∗ 0,33 m = 0,05 kW 1000 ∗ 0,85 s

PL =

1208,6 N ∗ 0,33 = 0,47 kW 1000 ∗ 0,85

13

NORD - Information

Motor arrangement Motor data Type

100 L/8-2 WU Bre10 Z

Rated output power PN

0,4 / 1,6 kW

Rated speed nN

670 / 2740 1/min

Rated torque MN

5,7 / 5,6 Nm

Hochlaufmoment MH

9,2 / 8,6 Nm

No-load switching frequency zo

4200 / 1500 s/h

Motor moment of inertia JM

0,0045 kgm2

Moment of high inertia fan Jz

0,0113 kgm2

Brake moment of inertia JBre

0,0001 kgm2

Braking torque MB (brake 10 adjusted on 6 Nm)

6 Nm

Load torque M M =

P ∗ 9550 x ∗ nN

MO =

0,05 KW ∗ 9550 = 0,2 Nm (without load) 2740 1⁄min

ML =

0,47 kW ∗ 9550 = 1,6 Nm (with load) 2740 1⁄min

Reduced moment of inertia Jred Jred =

1 v ∗ 91,2 ∗ m ∗   x  nN 

2

2

 0,33 m⁄s  JredO = 91,2 ∗ 1800 kg ∗  = 0,0024 kgm2 1⁄min 2740   2

 0,33 m⁄s  JredL = 91,2 ∗ 16800 kg ∗  = 0,0222 kgm2 1⁄min 2740   Acceleration aB aB =

14

9,55 ∗ v ∗ (MH − ML)

(Jred ⁄ η + JM + JBre + JZ) ∗ nN

aB =

9,55 ∗ 0,33 m⁄s ∗ (8,6 Nm − 0,2 Nm) = 0,52 m⁄s2 (without load) (0,0024 kgm ⁄ 0,85 + 0,0045 kgm2 + 0,0001 kgm2) ∗ 2740 1⁄min

aB =

9,55 ∗ 0,33 m⁄s ∗ (8,6 Nm − 1,6 Nm) = 0,19 m⁄s2 (with load) (0,0222 kgm ⁄ 0,85 + 0,0045 kgm2 + 0,0001 kgm2) ∗ 2740 1⁄min

2

2

NORD - Information

Decceleration aV av =

9,55 ∗ v ∗ (MB + ML ∗ η2) (Jred ∗ η + JM + JBre + JZ) ∗ nN

aVO =

9,55 ∗ 0,08 m⁄s ∗ (6 Nm + 0,2 Nm ∗ 0,852) = 0,39 m⁄s2 (without load) (0,0024 kgm ∗ 0,85 + 0,0045 kgm2 + 0,0001 kgm2 + 0,0113 kgm2) ∗ 670 1⁄min

aVL =

9,55 ∗ 0,08 m⁄s ∗ (6 Nm + 1,6 Nm ∗ 0,852) = 0,24 m⁄s2 (with load) (0,0222 kgm ∗ 0,85 + 0,0045 kgm2 + 0,0001 kgm2 + 0,0113 kgm2) ∗ 670 1⁄min

2

2

Permissible switching frequency zzul

zzul =

zzul =

1 − ML ⁄ MH ∗ z0 1 + (Jred + JZ + JBre) ⁄ JM

1 − 1,6 Nm ⁄ 8,6 Nm 1 +(0,0222 kgm + 0,0113 kgm2 + 0,0001 kgm2) ⁄ 0,0045 kgm2 2

∗ 1500 s⁄h = 142 s⁄h

The perm. switching frequency is calculated for the acceptance: starting 2-pole with load (every time) is not correct because of delay for switching and running on the 8-pole.

15

NORD - Information

Gear arrangements Wheel speedl nL nL =

v ∗ 60 π ∗ D

nL =

0,33 m⁄s ∗ 60 = 21 1⁄min π ∗ 0,3 m

Gear unit output speed n2 n2 = nL ∗ iv n2 = 21 1⁄min ∗ 4 = 84 1⁄min

Mass acceleration factor maf maf =

Jred JM + Jz + JBre

maf =

0,022 kgm2 = 1,4 0,0045 kgm2 + 0,0113 kgm2 + 0,0001 kgm2

Circuit m per hour: 180 (each 60 accelerations, switching, decelerations) ⇒ type of load B ⇒ fB ≥ 1,3

Output torque Ma

Ma =

PN ∗ 9550 ∗ fB n2

Ma =

1,6 kW ∗ 9550 ∗ 1,3 = 236 Nm 84 1⁄min

For service factor fB = 1,3 the output torque of the gear is 236 Nm.

Reduction i

16

i =

nN n2

i =

2740 1⁄min = 33 84 1⁄min

NORD - Information

Complete type:

SK 22 F - 100 L/8-2 WU Bre 10 Z PN = 0,4 / 0,16 kW i = 34,69 n2 = 19/79 1/min Mounting position B 5 Shaft ø 30 x 60 mm Flange ø 160 mm oder 200 mm Brake 10 Nm adjusted on 6 Nm Special provision:

special rotor (WU-silumin rotor) high inertia fan

17

NORD - Information

Example II.1:

Mass without load m O Load m L

Max. lifting speed v Operation cycle Starting frequency z Efficiency η

18

50 kg 200 kg

midle drum diameter Dm

Positioning

Drive unit for vertical motion

0,208 m 0,24 m/s = 14,4 m/min 8 h/Tag, 40 % ED 360 Hubbewegungen/h 0,8

accuracy

± 1 mm

NORD - Information

Motor arrangement Power P P =

PL =

m ∗ g ∗ v 1000 ∗ η (50 kg + 200 kg) ∗ 9,81

m 2 ⁄s

∗ 0,24

1000 ∗ 0,8

m⁄

s

= 0,74 kW

To get the required accuracy of ± 1 mm we have to choose a polechanging motor.

Motor data Typ

80 L/4-2 Bre8

Rated output power PN

0,60 / 0,75 kW

Rated speed nN

1400 / 2830 1/min

Synchronous speed nsyn

1500 / 3000 1/min

Rated torque MN

4,1 / 2,5 Nm

Run-up torque MH

7,4 / 5,7 Nm 2500 / 1800 s/h

No-load switching frequency zo Motor moment of inertia JM

0,00165 kgm 2

Brake moment of inertia JBre

0,00007 kgm2 7 * 107 J

max. braking work until readjustment Wzul.

0,015 s

Brake reaktion time t2 (DC-connection)

8 Nm

Braking torque MB

Load torque M M =

ML =

P ∗ 9550 nN 0,74 kW ∗ 9550 2830 1 ⁄min

= 2,5 Nm

Switching torque MU MU = 2 * MH4 MU = 2 * 7,4 Nm = 14,8 Nm

reduced moment of inertia Jred  v J red = 91,2 ∗ m ∗   n  N

2

 0,24 m ⁄s  J red = 91,2 ∗ (50 kg ∗ 200 kg) ∗   1  2830 ⁄min 

2

= 0,00016 kgm 2

19

NORD - Information

z0 = 2320 / 1620 s/h = max. perm. switching frequency with no load For this application A 4-2 polemotor (Dahlander-connection) is used. Therefor the half of the zo is criteria.

Permissibleswitchingfrequencyzzul

up motion:

zzul =

zzul =

down motion: zzul =

zzul =

1 − ML ⁄ MH zO ∗ 1 + (Jred + JBre) ⁄ JM 2

1 − 2,5 Nm ⁄ 5,7 Nm 2

2

1 + (0,00016 kgm + 0,00007 kgm ) ⁄ 0,00165 kgm

2

∗

1620 s⁄h = 399 s⁄h (2 poles) 2

2

∗

2320 s⁄h = 846 s⁄h (4 poles) 2

1 − ML ⁄ MU zO ∗ 1 + (Jred + JBre) ⁄ JM 2

1 − 2,5 Nm ⁄ 14,8 Nm 2

2

1 + (0,00016 kgm + 0,00007 kgm ) ⁄ 0,00165 kgm

The mechanicalbraking depends on the positioning speed.The max.braking distance depends on down motion.

Deceleration av av =

av =

9,55 ∗ v ∗ nN4 ⁄ nN2 ∗ (MB − ML ∗ η2) (Jred ∗ η + JM + JBre) ∗ nN4

9,55 ∗ 0,24 m⁄s ∗ 1400 1⁄min ⁄ 2830 1⁄min ∗ (8 Nm − 2,5 Nm ∗ 0,82) (0,00016 kgm2 ∗ 0,8 + 0,00165 kgm2 + 0,00007 kgm2) ∗ 1400 1⁄min

= 2,80 m⁄s2

In case of calculation the deceleration time we have to use the increased speed for the down motion. The cause is the delay for switching and the over-synchronous speed.

Load speed nL nL = nsyn ± ML ⁄ MN ∗ (nsyn − nN) down motion: nL = nsyn + ML ∗ η2 ⁄ MN ∗ (nsyn − nN) nL = 1500 1⁄min +

20

2,5 Nm ∗ 0,8 2 ∗ (1500 1⁄min − 1400 1⁄min) = 1539 1⁄min 4,1 Nm

+: down motion, -: up motion

NORD - Information

Increased speed during braking time ∆n

∆n = ±

9,55 ∗ ML ∗ t2 Jred + JM + JBre

down motion: ∆n =

+: down motion, -: up motion

9,55 ∗ ML ∗ η2 ∗ t2 Jred ∗ η + JM + JBre 9,55 ∗ 2,5 Nm ∗ 0,82 ∗ 0,015 s

∆n =

0,00016 kgm2 ∗ 0,8 + 0,00165 kgm2 + 0,00007 kgm2

= 124 1⁄min

Deceleration time tv (Braking time)

v ∗ (nL + ∆n) ⁄ nN2 a

tv =

0,24 m⁄s ∗ (1539 1⁄min + 124 1⁄min) ⁄ 2830 1⁄min

tv =

m

2,80 ⁄s2

= 0,05 s

Deceleration distance sv (Brakingdistance) 2

 nL + ∆n  v ∗    nN2  sv = 2 ∗ a

2

 1539 1⁄min + 124 1⁄min  0,24 m⁄s ∗   2830 1⁄min   sv = 2 2 ∗ 2,80 m⁄s

Positioning

= 0,004 m

accuracy

The positioningaccuracyis about± 25 % from the deceleration distance sv. Positioning accuracy = ± 25 % * sv = ± 0,25 * 0,004 m = ± 0,001 m

Braking

work WB

WB =

WB =

(Jred ∗ η + JM + JBre) ∗ n2N4

182,5

∗

MB MB ± ML

(0,00016 kgm 2 ∗ 0,8 + 0,00165 kgm2 + 0,00007 kgm2) ∗ (1400 1⁄min)2 8 Nm ∗ = 20 J 182,5 8 Nm ± 0

Because of the same number of up- and down-motion the load torque = 0 Nm.

Brake service life until readjustment LN LN =

Wzul WB ∗ z

LN =

7 ∗ 107J = 9720 h 20 J ∗ 360 1⁄h

21

NORD - Information

Gear arrangements

Gear output speed n2 n2 =

v ∗ 60 π ∗ Dm

n2 =

0,24 m⁄s ∗ 60 = 22 1⁄min π ∗ 0,208 m

^

Mass acceleration factor maf Jred

maf =

JM + JBre 0,00016 kgm2

maf =

0,00165 kgm 2 + 0,00007 kgm2

= 0,09

Switching per hour: 1080 ( each 360 accelerations, change-over, decelerations)

⇒ kind of load A, fB = 1,2

Output

torque Ma

Ma =

PN ∗ 9550 ∗ fB n2

Ma =

0,75 kW ∗ 9550 ∗ 1,2 = 391 Nm 22 1⁄min

Reduction i i =

nN n2

i =

2830 1⁄min = 129 22 1⁄min

Complete

22

type:

SK 2382 A - 80 L/4-2 Bre8 PN = 0,60 / 0,75 kW i = 131,86 n2 = 11 / 21 1/min Mounting position H 1 Hollow shaft ø 35 mm Brake 8 Nm Insolating material class F

NORD - Information

Example III.1: Turntable drive for

processing table

Determine the size of a cd-geared motor for a tuntable with 3 work stations (α = 120°)

Tableweightwithoutload

mO

500 kg

Positioning accuracy

=

± 1 mm

Table diameter

D

2m

Sprocket reduction

iv

3,76

Positions of load

α

120°

Dutyfactor

ED 60 %

Spacedatradius

R

1m

Pulsenumber

360 Takte/h

Ball bearingringdiameter

d

2m

Time of run

16 h/Tag

Cycle time for 120° turn

tges

6s

Efficiency

LoadmL (3 x 750 kg)

mL

2250 kg

Mounting position

η

0,8 V6

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NORD - Information

Distance s ( at a rotation of 120° ) s =

D∗π

3

=

2m∗π

3

= 2,094 m

Acceleration time tB or Deceleration time tV t B = t V = 1 s (acceptance data ) Tablespeed nT s ges ∗ 60 2,094 m ∗ 60 = = 4 1⁄ min π ∗ D ∗ ( t − ( tB + tV ) ⁄ 2 ) π∗2m ∗(6s −(1s +1s ) ⁄ 2)

nT =

Table circumferential velocity v (Ball bearing ring) v =

π ∗ d ∗ nT π ∗2 m ∗ 4 1⁄min = = 0,42 m⁄s 60 60

Momentofinertia J J =

1 1 1 1 ∗ m O ∗ D2 + ∗ m L ∗ d 2 = ∗ 500 kg ∗ (2 m)2 + ∗ 2250 kg ∗ (2 m)2 = 2500 kgm 2 8 8 4 4

Friction power PR (static) PR =

(m O + m L) ∗ g ∗ µ L ∗ v (500 kg + 2250 kg ) ∗ 9,81 m⁄s2 ∗ 0,005 ∗ 0,42 m⁄s = = 0,07 kW 1000 ∗ η 1000 ∗ 0,8

with µL = 0,005 for friction bearing

Acceleration power PB (dynamic) PB =

J ∗ nT2 2500 kgm 2 ∗ (4 1⁄min)2 = = 0,55 kW 91,2 ∗ 1000 ∗ t B ∗ η 91,2 ∗ 1000 ∗ 1 s ∗ 0,8

Power P P = PR + PB (friction + acceleration) P = 0,07 kW + 0,55 kW = 0,62 kW

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NORD - Information

Motor data Type

90 S/8-2 Bre 10

Rated power PN

0,25 / 1,1 kW

Rated speed nN

700 / 2810 1/min

Rated torque MN

3,4 / 3,7 Nm

Hochlaufmoment MH

4,0 / 5,7 Nm

No-load switching frequency z o

9000 / 1500 s/h

Motor moment of inertia JM

0,00235 kgm2

Brake moment of inerta JBre

0,00007 kgm2

Braking torque MB (adjusted at 8 Nm)

8 Nm

LoadtorqueML

M =

PR ∗ 9550 0,07 kW ∗ 9550 = = 0,2 Nm nN 2810 1⁄ min

Reduced moment of inertia Jred  nT   4 1⁄min  2 2 Jred = J ∗   2 = 2500 kgm2 ∗   = 0,00507 kgm 1 n  N  2810 ⁄min

Permissibleswitchingfrequence zzul zzul =

1 − ML ⁄ MH 1 − 0,2 Nm ⁄ 5,7 Nm ∗ zO = ∗ 1500 s⁄h = 453 s⁄h 1 + (Jred + JBre) ⁄ JM 1 + (0,00507 kgm2 + 0,00007 kgm2) ⁄ 0,00235 kgm2

Acceleration aB aB =

9,55 ∗ v ∗ (MH − ML) 9,55 ∗ 0,42 m⁄s ∗ (5,7 Nm − 0,2 Nm) = = 0,90 m⁄s2 2 Jred ⁄ η + JM + JBre) ∗ nN (0,00507 kgm ⁄ 0,8 + 0,00235 kgm 2 + 0,00007 kgm2) ∗ 2810 1⁄min

Acceleration time tB (start up time) tB =

v

aB

=

0,42 m⁄s 0,90 m⁄s2

= 0,47 s

Accelerationdistance sB (start up distance) sB =

v2 0,42 m⁄s2 = = 0,098 m 2 ∗ aB 2 ∗ 0,90 m⁄s2

Change-overtorque MU MU = 2 ∗ MH8 = 2 ∗ 4,0 Nm = 8,0 Nm

At the change over the speed increased and the motor is decelet ad generator-style.

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NORD - Information

Change-over delay aU

aU =

aU =

9,55 ∗ v ∗ (1 − nN8 ⁄ nN2) ∗ (MU + ML ∗ η2) = (Jred ∗ η + JM + JBre) ∗ (nN2 − nN8) 9,55 ∗ 0,42 m⁄s ∗ (1 − 700 1⁄min ⁄ 2810 1⁄min) ∗ (8,0 Nm + 0,2 Nm ∗ 0,8 2)

(0,00507 kgm2 ∗ 0,8 + 0,00235 kgm2 + 0,00007 kgm2) ∗ (2810 1⁄min − 700 1⁄min)

= 1,79 m⁄s2

Change-over time tU v ∗ (1 − nN8 ⁄ nN2)

tU =

aU

=

0,42 m⁄s ∗ (1 − 700 1⁄min ⁄ 2810 1⁄min) 1,79 m⁄s2

= 0,18 s

Change-overdistance sU sU =

(v ∗ (1 − nN8 ⁄ nN2))2 0,42 m⁄s ∗ (1 − 700 1⁄min ⁄ 2810 1⁄min))2 = = 0,028 m 2 ∗ aU 2 ∗ 1,79 m⁄s2

Deceleration a V aV =

9,55 ∗ v ∗ nN8 ⁄ nN2 ∗ (MB + ML ∗ η2) 9,55 ∗ 0,42 m⁄s ∗ 700 1⁄min ⁄ 2810 1⁄min ∗ (8,0 Nm + 0,2 Nm ∗ 0,82) = = 1,80 m⁄s2 (Jred ∗ η + JM + JBre) ∗ nN8 (0,00507 kgm 2 ∗ 0,8 + 0,00235 kgm2 + 0,00007 kgm 2) ∗ 700 1⁄min

Decelerationtime tV v ∗ nN8 ⁄ nN2 0,42 m⁄s ∗ 700 1⁄min ⁄ 2810 1⁄min = = 0,06 s aV 1,80 m⁄s2

tv =

Deceleration distance sV sV =

(v ∗ nN8 ⁄ nN2 )2 (0,42 m⁄s ∗ 700 1⁄min ⁄ 2810 1⁄min)2 = = 0,003 m 2 ∗ aV 2 ∗ 1,80 m⁄s2

Distance s with velocity v s = sges - sB - sU - sV = 2,094 m - 0,098 m - 0,028 m - 0,003 m = 1,965 m

Timetwithvelocity v t =

s 1,965 m = = 4,68 s v 0,42 m⁄s

Pulse duration tges (total cycle time) tges = tB + t + tU + tV = 0,47 s + 4,68 s + 0,18 s + 0,06 s = 5,39 s The required cylce time of 6s is not reached. There is a possibility of driving for a longer period in creep speed.

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NORD - Information

Positioning accuracy The positioningaccuracyisabout± 25% of the deceleration way sV. Positioning accuracy = ± 0,25 * sv = ± 0,25 * 0,003 m = ± 0,00075 m = ± 0,75 mm

Gear arrangements Gear output speed n2 n2 = nT * iV = 4 1/min * 3,76 = 15 1/min

Massacceleration factor maf maf =

Jred 0,00507 kgm2 = = 2,1 JM+JBre 0,00235 kgm2 + 0,00007 kgm2

Switching per hour: 1080 (each 360 accelerations, change-over and decelerations) → kind of load B, fB = 1,5

Outputtorque Ma Ma =

PN ∗ 9550 1,1 kW ∗ 9550 ∗ fB = ∗ 1,5 = 1050 Nm n2 15 1⁄min

Reduction i i =

nN 2810 1⁄min = = 187 n2 15 1⁄min

Complete type:

SK 43 - 90 S/8-2 Bre 8 PN = 0,25 / 1,1 kW i = 169,86 n2 = 4/16 1/min Mounting position V6 Shaft ø 45 x 90 mm Brake 8 Nm

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