Zupčanici.pdf

Gears

Power transmission elements in rotational motion • power transmission with form – without slipping • rigid: gear and chain transmissions • flexible: toothed belt transmission

nord.com

• power transmission with friction – slipping • flexible: friction wheel and belt transmissions

asconveyorsystems.co.uk

democrazy.encikbeliau.com/ giveng.com 19.5.2017

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utterpower.com achrnews.com

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Power transmission elements in rotational motion

sew-eurodrive 19.5.2017

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Power transmission elements in rotational motion • why power transmission element is needed? – to decrease/increase the speed – to decrease/increase the torque

• power P = Tω = constant T = torque ω = angular speed

Airila 2003 19.5.2017

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Gear drives

moventas.com

images.google.fi

valtra.com 19.5.2017

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Gear drives

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https://www.youtube.com/watch?v=n4BuwzACiDg&list=PLPhKo_uriS6rTBM GXJ9TAukYWECEVFBXH

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Gear wheel pairs a) b) c) d) e) f) g)

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Spur gear Helical gear Internal gear Pinion and rack Bevel gear Crossed helical gear Worm gear

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Gear ratio • gear ratio i = ω1/ ω2 = n1/n2 = z2/z1 = r2/r1 = T2/T1 ω = angular velocity n = rotational speed z = number of teeth T = torque T = Fur Fu = tangential load r = pitch radius

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Example • Define total gear ratio, output torque and output speed, if – number of teeth are: z1 = 18; z2 = 47; z3 = 26; z4 = 71; z5 = 23; z6 = 50 – torque T1 = 12 Nm – rotational speed n1 = 960 r/min – efficiency η = 0,91

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Example • Solution – total gear ratio: itot = – output torque To = – output speed no =

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Problem 1 Determine the angular velocity ratio for the gear train shown in the figure. If the shaft carrying gear A rotates at 1750 rpm clockwise, determine the speed and direction of the shaft carrying gear E. (A: 10.5 & 167 rpm counter clockwise)

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Involute toothing • involute υ = invα = tanα - α

• α is pressure angle - for spur gears normally α = 20o

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Tooth geometry

p = circular pitch d = pitch circle da = addentum circle hf = dedentum c = clearance 19.5.2017

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db = base circle df = dedentum circle h = whole depth ha = addentum j = backlash 13

Module • module m [mm] – m=d/z d = pitch diameter z = number of teeth

– with the help of the module it is possible to calculate e.g. • centre distance a = m*(z1+z2)/2 • circular pitch p = m*π

– the module defines the manufacturing tool – the module is standardised (ISO 54) 19.5.2017

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Contact ratio • is the average number of teeth in contact as the gears rotate together – in general 1.4 - 1.8, for slow speeds 1.1

• guideline values for the number of teeth – – – – –

for high-speed gears z1min = 16 for medium speed gears z1min = 12 for slow-speed gears z1min = 10 for external pair of gears z1 + z2 ≥ 24 for internal pair of gears z2 - z1 ≥ 10

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Addendum modification • to standardize centre distance and/or improve the strength of tooth dedemdum • also pressure angle changes

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Calculation of the tooth geometry of a spur gear pair

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Problem 2 An external gear drive consists of a gear with 38 teeth and a pinion with 15 teeth. The module is 10 mm and the pressure angle 20o. a) Determine the pitch, the centre distance and the base diameter for the pinion and the gear. (A: 31.4 mm, 265 mm, 141.0 mm and 357.1 mm)

a) In mounting these gears, the centre distance was incorrectly made 6 mm larger. Compute the new values of the pressure angle and the pitch diameters. (A: 22.7o, 152.8 mm and 387.2 mm) 19.5.2017

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Problem 3 Following data of an external spur gear pair is given: module m = 6 mm, pressure angle a = 20o, clearance c = 1.2 mm, number of teeth: z1 = 19 pcs and z2 = 87 pcs and addendum modification coefficients: x1 = + 0.38 and x2 = + 0.35.

Calculate following geometrical values of the gear: a) pitch diameters (A: 114 mm & 522 mm) c) pitch (A: 18.8 mm) e) reference centre distance (A: 318 mm) g) shortening of addendum (A: 0.203 mm) i) addendums (A: 8.08 mm & 7.90 mm) k) tip diameters (A: 130.2 mm & 537.8 mm)

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b) base diameters (A: 107.1 mm & 490.5 mm) d) working pressure angle (A: 21.95o) f) centre distance (A: 322 mm) h) tooth depth (A: 13.3 mm) j) dedentums (A: 5.22 mm & 5.40 mm) l) root diameters (A: 103.6 mm & 511.2 mm).

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Helical gears • with helical gear the contact is smoother => lower noise and less vibration • bigger number of teeth in contact => better strength • longer teeth in helix angle => better bending strength • better power transmission capacity • for downside there is axial force => helix angle β = 8°…15° even to 30° (with double-helical gear even 45°) 19.5.2017

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Calculation of the tooth geometry of a helical gear pair

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Problem 4 Following data of an external helical gear pair is given: normal module mn = 5 mm, centre distance aw = 340 mm, pressure angle an = 20o, facewidth b = 70 mm, number of teeth: z1 = 30 pcs and z2 = 100 pcs, helix angle b = 15o, clearance c = 1.25 mm and addendum modification coefficient x1 = + 0.4.

Calculate following geometrical values of the gear:

a) pitch diameters (A: 155.3 mm & 517.6 mm) b) transverse pressure angle (A: 20.6o) c) base diameters (A: 145.3 mm & 484.4 mm) d) transverse pitch (A: 16.3 mm) e) transverse base pitch (A: 15.2 mm) f) reference centre distance (A: 336.5 mm) g) working pressure angle (A: 22.2o) h) sum of the addendum modification coefficients (A: 0.73) i) addendum modification coefficient x2 (A: 0.33).

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Tooth forces • spur gear – tangential force Ft Ft = FNcosα = T1/r1 = T2/r2

– radial force Fr Fr = FNsinα = Fttanα FN gear tooth force α pressure angle T1 ja T2 gear torques r1 ja r2 pitch radius

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Tooth forces • helical gear – tangential force Ft Ft = T1/r1 = T2/r2

– radial force Fr Fr = Fttanαt = Fttanαn/cosβ

– axial force Fa Fa = Fttanβ αn pressure angle αt transverse pressure angle β helix angle 19.5.2017

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Stresses of the tooth • bending stress on the root of the tooth σF • surface pressure σH

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Dimensioning of gears • strength and power transmission calculations – can be based on following: • • • • •

bending stress on the root of the tooth (fatigue) surface pressure/pitting (fatigue) surface grooves (high pressure, small speed) abrasive wear (foreign particles) scoring (high speeds, inadequate lubrication)

– normally two first are most critical

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Strength calculation of spur wheels • SFS-ISO 6336-1: CALCULATION OF LOAD CAPACITY OF SPUR AND HELICAL GEARS. PART 1: BASIC PRINCIPLES, INTRODUCTION AND GENERAL INFLUENCE FACTORS. • SFS-ISO 6336-2: CALCULATION OF LOAD CAPACITY OF SPUR AND HELICAL GEARS. PART 2: CALCULATION OF SURFACE DURABILITY (PITTING). • SFS-ISO 6336-3: CALCULATION OF LOAD CAPACITY OF SPUR AND HELICAL GEARS. PART 3: CALCULATION OF TOOTH BENDING STRENGTH. • SFS-ISO 6336-5: CALCULATION OF LOAD CAPACITY OF SPUR AND HELICAL GEARS. PART 5: STRENGTH AND QUALITY OF MATERIALS. • SFS-ISO 6336-6: CALCULATION OF LOAD CAPACITY OF SPUR AND HELICAL GEARS. PART 6: CALCULATION OF SERVICE LIFE UNDER VARIABLE LOAD. • SFS-ISO 9085: CALCULATION OF LOAD CAPACITY OF SPUR AND HELICAL GEARS. APPLICATION FOR INDUSTRIAL GEARS.

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Deformations • deformations of teeth and shafts have an effect on tooth contact • contact shape (figure) is controlled with a color • stiffness of the tooth is taken into consideration in exact power transmission calculations • stiffness of the shaft is taken into consideration in calculations with a longitudinal load distribution factor for bending stress

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Efficiency of the gear • efficiency of the gear teeth ηh

– during contact the teeth are rolling but also sliding  friction causes power loss – for one spur gear pair ηh = 0,96…0,99

• efficiency of the bearings ηL

– for rolling bearings ηL = 0,98…0,99 – for slide bearings ηL = 0,96

• total efficiency of the gear ηtot ηtot = (ηhηL)I (ηhηL)II (ηhηL)III… – I, II, III, … number of gear pairs 19.5.2017

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Tolerances of spur gears • Recommended accuracy grades (IS0 1328) – Class 6 • very accurate gear wheel pairs • speed normally > 20 m/s

– Class 8 • normal in mechanical engineering • speed 5…20 m/s

– Class 11 • no special accuracy demands • speed normally < 5 m/s

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Drawing data of a gear wheel

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Noise

noise can be effected by:     

speed tooth contact accuracy class contact ratio number of teeth

 small module and big number of teeth  more teeth in contact

 design of the gearbox

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 shape of the tooth

 relief of tooth addemdum or barrel-shaped gear teeth

   

straight/helical gear teeth shafts bearings materials

Spur and helical gear units • normally helical gears

– bigger power transmission capacity – reduced noise

• materials of the gears – – – –

case hardening steel (D < 400…600 mm) tempered steel spheroidal iron plastics (small diameters)

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Spur and helical gear units

• normally max transmission ratio in one gear pair is 5 – – – –

1 gear pair max i = 5 2 gear pairs max i = 52 = 25 3 gear pairs max i = 53 = 125 4 gear pairs max i = 54 = 625

• lubrication

– oil bath (v < 4 m/s) – splash lubrication (v < 14 m/s) – pressure lubrication (v > 14 m/s)

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Spur and helical gear units • Parallel shaft

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Bevel gear units

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Bevel gear units

nord.com

straight-tooth 19.5.2017

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helical

spiral

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Problem 5 A bicycle is normally driven by the pedals through a roller chain transmission to the back wheel. The power efficiency of such a chain drive is 95 % if it is well lubricated and rather heavily loaded. To avoid getting oil on the trousers from the chain, a design change is considered. The chain drive is to be changed to a shaft with two sealed ball bearings inside a tube in the frame and a bevel gear drive at the pedals and another bevel gear drive at the back wheel. At the speed of 20 km/h the chain-driven bike requires 220 W to the rear wheel, and the same real wheel power is required for the new design. How large does the input power have to be for the new drive to run 20 km/h if the newly designed shaft rotates at 1200 rpm and the bearing and seal friction torque in each bearing is 0.1 Nm. The power efficiency for the bevel gears is 0.98. Also, calculate the total power efficiency for the new drive when the bicycle is driven at 20 km/h. (A: 255 W & 86.4%)

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Worm gear units

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Worm gear units • big gear ratio

– high-speed gears i = 5…15 – slow-speed gears i = 5…70

• power loss – gear teeth loss – bearing loss – idle loss

• use of torus

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Worm gear units • total power loss PL = PLz + PLb + PLi PLz is power loss in toothing PLb is power loss of bearings PLb = kP1 k = 0,005…0,01 (4 roller bearings) k = 0,02…0,03 (4 slide bearings) P1 is input power

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Worm gear units • total power loss

PLi is idle power loss PLi = 10-7a(n1/60)4/3(ν50+90) [kW] a is centre distance [mm] n1 is rotational speed of the worm [r/min] ν50 is viscosity of the oil in 50 oC [mm2/s]

• total efficiency

η = (P1 - PL)/P1 = P2/(P2+PH) P1 is input power

P2 is output power

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Planetary gears • internal centre gear/sun gear S • planet gears P • external centre gear/ring gear/annulus R • arm/planet carrier A

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Planetary gears • advantages – – – –

power can be divided to several driven or drive shafts big gear ratio and power compared to it’s size many gear ratio possibilities symmetrical structure, drive and driven shaft inline

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Planetary gears • limitations and disadvantages – number of teeth has compatibility condition => limits to get a certain ratio with standard gear wheels – circumferential load does not divide equally between planet gears – phenomenon of tooth contact in mesh is more complex compared to spur gear – the axial forces of helical gear are more tricky in planet gear (bearings and centre gears without bearings) – internal gear teeth demands special machinery for manufacturing 19.5.2017

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Gear unit selection • Selection is based on required input power of the application in normal operation P • Input power of the gear (rated power of the motor) P1 P1 = P/η or P1 = (M∙n2)/(9550∙η) [kW] where

η is efficiency of the gear unit M is required torque of the application [Nm] n2 is output speed of the gear [r/min]

• Efficiency e.g. for helical, parallel shaft and helical-bevel gears – – – –

1-stage: 0,985 2-stage: 0,970 3-stage: 0,955 4-stage: 0,940

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Gear unit selection • The gear unit is selected so that nominal power ≥ service factor ∙ required power • Service factor takes into account torque impulses (from driven or driving machine) and operational time (with worm gear units also the ambient temperature and cyclic duration factor ED should be considered) • Normally the operation (of uniformity) is classified to three groups – A: uniform operation – B: moderate shocks, non-uniform operation – C: heavy shocks, extreme non-uniform operation 19.5.2017

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Gear unit selection • Examples of operation classes – Group A (uniform): light screw conveyors, fans, assembly belts, light conveyor belts, small agitators, elevators, cleaning machines, filling machines, testing machines, belt conveyors – Group B (moderate shocks): decoilers, feed drives for wood processing machines, hoists, balancing machines, tapping units, heavy conveyor belts, winches, sliding doors, stall dunging machines, packaging machines, cement mixers, crane travelling mechanisms, mills, bending machines, gear pumps – Group C (heavy shocks): stirrers and mixers, shears, presses, centrifuges, rolling stands, heavy winches and lifts, grinding mills, stone crushers, bucket elevators, punching machines, hammer mills, eccentric presses, folding machines, roller tables, tumbling barrels, choppers, shredders, vibrators

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Gear unit selection • If the gear selection is based on the nominal power of the motor (input power of the gear), following condition must be valid 2 ∙ nominal torque of the gear ≥ starting torque of the motor

• Max power or torque which the gear can take for short times (max twice an hour about 20 seconds) is the double of the nominal power or torque of the gear • The thermally transferable power (thermal limit) should not be exceeded over a longer time period (3 hours) so that the gear unit does not overheat • The permissible overhung and axial forces on the output shaft of the gear unit should not be exceeded.

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Gear unit selection/Example • Select a Nord worm geared motor for a winch based on the following data – – – – – –

Output torque 35 Nm Output speed 20 – 25 r/min Operating time 8 h/day Start/stop frequency 100 cycles/h Max ambient temperature 40 oC Load time 45 min/h

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Gear unit selection/Example • The required minimal operating factor fBmin for an application is calculated as follows: fBmin = fB0 • fB1 • fB2 • The operating factor f takes into account load type A, B or C, the frequency of activation and daily run time. The operating factor f takes into account different ambient temperatures. The operating factor f takes into account intermittent operation. The diagram below is used when determining the operating factors f , f and f . B0

B1

B2

B0

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B1

MatJo Nord: Universal – Worm Gear Units

B2

Gear unit selection/Example • Examples of load types for gear units: A B

C

Light screw conveyors, fans, assembly belts, light conveyor belts, small agitators, elevators, cleaning machines, filling machines, testing machines and belt conveyors. Decoilers, feed drives for wood processing machines, hoists, balancing machines, tapping units, mid-sized stirrers and mixers, winches, sliding doors, stall dunging machines, packaging machines, bending machines and gear pumps. Scissors, presses, punchers, nut bevelling machines, polishing and grinding drums, agitators and choppers.

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MatJo Nord: Universal – Worm Gear Units

Gear unit selection/Example

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MatJo Nord: Universal – Worm Gear Units

Gear unit selection/Example

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MatJo Nord: Universal – Worm Gear Units

Literature • Shigley, Mechanical Engineering Design. McGrawHill 2002. • Hamrock, Fundamentals of Machine Elements. McGraw-Hill 2000. • Juvinall & Marshek, Fundamentals of Machine Component Design. John Wiley & Sons 2003. • Airila etc. Koneenosien suunnittelu. WSOY 2003.

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