Gear Drives Schedule and Reading Assignment Quiz Bolted Joint Qualifying Thursday March 19Th

Home , Friction drive, Gear, Gear train, Involute gear

MIT OpenCourseWare http://ocw.mit.edu

2.72 Elements of Mechanical Design Spring 2009

For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 2.72 Elements of Mechanical Design Lecture 12: Belt, friction, gear drives Schedule and reading assignment Quiz Bolted joint qualifying Thursday March 19th

Topics Belts Friction drives Gear kinematics

Reading assignment • Read: 14.1 – 14.7

• Skim: Rest of Ch. 14

© Martin Culpepper, All rights reserved 3 Belt Drives Why Belts? Torque/speed conversion Cheap, easy to design Easy maintenance Elasticity can provide damping, shock absorption Image by dtwright on Flickr.

Keep in mind Speeds generally 2500-6500 ft/min Performance decreases with age

Images removed due to copyright restrictions. Please see: Image by v6stang on Flickr. http://www.tejasthumpcycles.com/Parts/primaryclutch/3.35-inch-harley-Street-Belt-Drive.jpg http://www.al-jazirah.com.sa/cars/topics/serpentine_belt.jpg

Usually composite structure Rubber/synthetic surface for friction Steel cords for tensile strength

Driven Pulley Slack Side

d2 d 1 ω ω1 2

Driving vbelt Pulley Tight Side

θ2

θ1

dspan

dcenter

θ2 d2

θ1 d1 ω1 ω2

dspan

dcenter

⎛ d −d ⎞ ⎛ d −d ⎞ θ =π −2sin−1 ⎜ 2 1 ⎟ θ =π +2sin−1 ⎜ 2 1 ⎟ 1 ⎜2d ⎟ 2 ⎜ ⎟ ⎝ center⎠ ⎝ 2dcenter⎠

θ2 d2

θ1 d1 ω1 ω2

dspan

dcenter

2 2 ⎛ d2 −d1 ⎞ 2 2 1 dspan = dcenter−⎜ ⎟ L = 4d −(d −d ) + (dθ +d θ ) ⎝ 2 ⎠ belt center 2 1 2 1 1 2 2

θ2 d2

θ1 d1 ω1 ω2

dspan

dcenter

d ω d1 d 2 1 = 2 vb = ω1 = ω2 2 2 d2 ω1

© Martin Culpepper, All rights reserved 10 Elastomechanics Elastomechanics → torque transmission Kinematics → speed transmission Link belt preload to torque transmission Proceeding analysis is for flat/round belt

Driven Pulley Slack Side

d1 ω2 ω1

Driving vbelt Pulley Tight Side

y x dS

μdN F+dF

dN d/2

•Tensile force (F) •Normal force (N) dθ •Friction force (μN) •Centrifugal force (S)

y dS x

Using small angle approx: μdN F+dF dθ dθ ΣFy = 0 = −(F + dF) − F + dN + dS dN d/2 2 2 Fdθ = dN + dS

dθ ΣFx = 0 = −μdN − F + (F + dF) μdN = dF

y dS x

Let m be belt mass/unit length μdN 2 F+dF ⎛ d ⎞ dS = m⎜ ⎟ ω2dθ dN d/2 ⎝ 2⎠ Combining these red eqns: 2 ⎛ d ⎞ dF = μFdθ − μm⎜ ⎟ ω 2dθ dθ ⎝ 2 ⎠ 2 dF ⎛ d ⎞ − μF = −μm⎜ ⎟ ω 2 dθ ⎝ 2 ⎠

Let the difference in tension between the loose side (F2) and the tight side (F1) be related to torque (T) T F F − F = 1 1 2 d 2

Solve the previous integral over contact angle and apply F1 and T F2 as b.c.’s and then do a page of algebra:

μθ F T e contact +1 2 Ftension = d eμθcontact −1 2 μθcontact ⎛ d ⎞ 2 2e F1 = m⎜ ⎟ ω + Ftension μθ ⎝ 2 ⎠ e contact + 1 Used to find stresses 2 in belt!!! ⎛ d ⎞ 2 2 F2 = m⎜ ⎟ ω + Ftension ⎝ 2 ⎠ eμθcontact + 1 © Martin Culpepper, All rights reserved 15 Practical Design Issues Pulley/Sheave profile Which is right?

Manufacturer → lifetime eqs Belt Creep (loss of load capacity) Lifetime in cycles A B C

Idler Pulley Design Catenary eqs → deflection to tension Large systems need more than 1 Idler

ALTIDL P/S Water pump & fan

Idler

Crank

Images by v6stang on Flickr.

© Martin Culpepper, All rights reserved Figure by MIT OpenCourseWare. 16 Practice problem Delta 15-231 Drill Press 1725 RPM Motor (3/4 hp) Images removed due to copyright restrictions. Please see 450 to 4700 RPM operation http://www.rockler.com/rso_images/Delta/15-231-01-500.jpg Assume 0.3 m shaft separation What is max torque at drill bit? What size belt? Roughly what tension?

© Martin Culpepper, All rights reserved 18 Friction Drives Why Friction Drives? Linear ↔ Rotary Motion Images removed due to copyright restrictions. Please see http://www.beachrobot.com/images/bata-football.jpg Low backlash/deadband http://www.borbollametrology.com/PRODUCTOS1/Wenzel/ Can be nm-resolution WENZELHorizontal-ArmCMMRSPlus-RSDPlus_files/rsplus.jpg

Keep in mind Preload → bearing selection Low stiffness and damping Needs to be clean Low drive force

Motor and Transmission/Coupling Drive Roller

Drive Bar Concerned with: •Linear Resolution •Output Force Backup •Max Roller Preload Rollers •Axial Stiffness

d d Δδ = Δθ ⋅ wheel wheel bar 2 Δθ d v = ω wheel Δδ bar wheel 2

Force output found from static analysis Either motor or friction limited

2Twheel Foutput = where Foutput ≤ μFpreload dwheel

−1 −1 ⎛ ⎞ For metals: 2 2 ⎜ 1 1 ⎟ ⎛ 1 − ν wheel 1−ν bar ⎞ 3σ E = ⎜ + ⎟ Re = + y e ⎜ ⎟ ⎜ d ⎟ τ = E E ⎜ wheel rcrown ⎟ max ⎝ wheel bar ⎠ ⎝ 2 ⎠ 2 1 ⎛ 3F R ⎞ 3 Variable Definitions ⎜ preload e ⎟ acontact = ⎜ ⎟ ⎝ 2 Ee ⎠ Shear Stress Equation

acontact Ee ⎛ 1 + 2ν wheel 2 ⎞ τ wheel = ⎜ + ⋅(1 +ν wheel )⋅ 2(1 +ν wheel )⎟ 2π Re ⎝ 2 9 ⎠

3 3 2 16π τ max Re Fpreload , max = 3 2 ⎛1+ 2ν wheel 2 ⎞ 3Ee ⎜ + ⋅(1+ν wheel )⋅ 2(1+ν wheel )⎟ ⎝ 2 9 ⎠

−1 ⎛ ⎞ ⎜ ⎟ 4a E 1 1 1 1 k = e e k = ⎜ + + + ⎟ tangential axial ⎜ k ⎟ (2 −ν )(1+ν ) k shaft torsion ktangential kbar ⎜ 2 ⎟ ⎝ dwheel ⎠ 3πEd 4 k = shaft shaft 4 L3 πGd 4 k = wheel torsion 32 L

EA k = c , bar bar L

Drive performance linked to motor/transmission Torque ripple Angular resolution

Images removed due to copyright restrictions. Please see

http://www.borbollametrology.com/PRODUCTOS1/Wenzel/WENZELHorizontal-ArmCMMRSPlus-RSDPlus_files/rsplus.jpg

Keep in mind Image from robbie1 on Flickr. Requires careful design Attention to tooth loads, profile

Image from jbardinphoto on Flickr.

Images removed due to copyright restrictions. Please see http://elecon.nlihost.com/img/gear-train-backlash-and-contact-pattern-checking.jpg http://www.cydgears.com.cn/products/Planetarygeartrain/planetarygeartrain.jpg

Helical Gears Gradual tooth engagement → low noise Shafts may or may not be parallel Thrust loads from teeth reaction forces Tooth-tooth contact pushes gears apart

Images from Wikimedia Commons, http://commons.wikimedia.org

Worm Gears Low transmission ratios Pinion is typically input (Why?) Teeth sliding → high friction losses

Rack and Pinion

Rotary ↔ Linear motion Images from Wikimedia Commons, http://commons.wikimedia.org Helical or straight rack teeth

Pinion, m2 F (t) + a b k k

Figure by MIT OpenCourseWare. Rack, m1

© Martin Culpepper, All rights reserved Viscous damping, c Rack & Pinion 28 Tooth Profile Impacts Kinematics Want constant speed output Conjugate action = constant angular velocity ratio Key to conjugate action is to use an involute tooth profile

Output speed of gear train “Real” involute/gear

“Ideal” involute/gear

ωout, [rpm]

Non or poor involute

time [sec]

v v v v v ω1 r2 v = ω1 × r1 = ω2 × r2 = ω2 r1

Model gears as two pitch circles Contact at pitch point

Pitch Circles Meet @ Pitch Pt.

ω 2 ω2 r1 r2 r1 r2 ω ω1 1 v

-Ng = # of teeth, Dp = Pitch circle diameter

Ng Diametral pitch, P : PD = D Dp

πDp π Circular pitch, PC: PC = = Ng PD

•Gear is specified by Pitch Point diametral pitch and Φ

pressure angle, Φ Pitch Circle

Base Circle

Images from Wikimedia Commons, http://commons.wikimedia.org

DB/2

DP/2 Image removed due to copyright restrictions. Please see http://upload.wikimedia.org/wikipedia/commons/c/c2/Involute_wheel.gif

= DD PB cosΦ

Pitch Point Pitch Circle DB L3 Ln = n Δθ

L2 2

3 2 1 L1 Base Circle

DB/2

Power out: T y ω Gear train out out Power in: Tin y ωin

ω Transmission ratio for elements in series: TR = (proper sign)⋅ out ωin

N N D N ω From pitch equation: P = 1 = 2 = P 1 = 1 = 2 11 2 1 D D 2 1 2 D2 N2 ω1

For Large Serial Drive Trains: Productof driving TR = (proper sign)⋅ teeth Productof driven teeth

TR = ?

in out Example 2: driven

drive

driven drive TR = ? driven drive in out

Product of driving teeth TR = (proper sign)⋅ Product of driven teeth

Gear - 1 5 N = 9 1 4

Gear - 2

N2 = 38 2 Gear - 3 N = 9 3 1

Gear - 4

N4 = 67

Gear - 5 3 N5 = 33

Planet Planet Ring Planet gear gear gear Arm

Planet ω Arm 2

Planet Sun gear gear

How do we find the transmission Ring Arm ratio? gear

Image removed due to copyright restrictions. Please see http://www.cydgears.com.cn/products/Planetarygeartrain/

planetarygeartrain.jpg Train 1 Planet gear Sun

Ring Arm gear

Train 2 Planet gear Sun

© Martin Culpepper, All rights reserved 38 Planetary Gear Train TR If we make the arm Sun Gear stationary, than this is Planet Gear Ring Gear a serial gear train: ω ω −ω ra = ring arm = TR ω sa ωsun −ωarm N N N TR = − sun ⋅ planet = − sun N planet Nring Nring Arm ω pa ω planet −ωarm = = TR ωsa ωsun −ωarm N TR = − sun N planet

ω 0 −ω ra = arm = TR ω sa ωsun −ωarm N N N TR = − sun ⋅ planet = − sun Arm N planet Nring Nring TR ω = ω = ω output arm TR −1 sun

Given: Shaft TSH (ωSH) find motor TM (ωSH) Geometry dominates relative speed (Relationship due to TR)

2 Unknowns: TM and ωM with 2 Equations: Transmission ratio links input and output speeds Energy balance links speeds and torques

T(ω): ⎛ ω ⎞ T()ω = T ⋅⎜1− ⎟ Motor torque-speed curve S ⎜ ω ⎟ ⎝ NL ⎠ T(ω) P(ω) obtained from P(ω) = T(ω) y ω

( 0 , TS ) Speed at maximum power output:

(ωNL , 0 ) ω ⎛ ω 2 ⎞ P()ω = T ()ω ⋅ω = T ⋅⎜ω − ⎟ S ⎜ ⎟ ⎝ ωNL ⎠ P(ω ) Motor power curve

ω NL ωPMAX = 2 PMAX

(ω , 0 ) ⎛ωNL ⎞ NL PMAX = TS ⋅⎜ ⎟ 4 ⎝ ⎠ ωPMAX ω

T(ω) A = Motor shaft torque-speed curve What is the torque-speed curve for the screw driver? B 1 Train ratio = /81 A

C ω

SCREW DRIVER SHAFT

MOTOR SHAFT TSH, ωSH TM, ωM

Electric GT-1 GT-2 Screw Driver Shaft Motor

System boundary

GEAR train # 1 GEAR train # 2

P(ω) C = Motor shaft power curve D What is the power-speed curve for the screw driver?

1 Train ratio = /81 C

ω SCREW DRIVER SHAFT

MOTOR SHAFT TSH, ωSH TM, ωM

Electric GT-1 GT-2 Screw Driver Shaft Motor

System boundary

GEAR train # 1 GEAR train # 2