Chapter 4: Design of

Mechanical Element 1: Spur Gear

DR. AMIR PUTRA BIN MD SAAD

C24-322

amirputra@utm.my | amirputra@mail.fkm.utm.my

mech.utm.my/amirputra

USAGE:

• To transmit torque/ power from one shaft to another shaft.

• To avoid slippage and discontinuity during power transmission, the best

solution is gear drive.

• To get desired rotational speed.

• To change direction of rotation.

• Where distance between drive shaft and driven shaft is very small.

4.1 INTRODUCTION

How gears transmit power and motion?

• Consider two circular wheels A & B mounted on two parallel shafts and the

wheels are pressed against each other.

DriverDriven

Friction Wheel Concept Spoke Concept

(a) (b)

DriverDriven

4.1 INTRODUCTION

(a) (b)

How to prevent this slippage ?

Simply by increasing the number of spokes between the two surfaces. The projection

of small spokes on the periphery of driver and driven wheels (called teeth) increase the

friction resistance.

Driver

Driven

A friction wheel with teeth cut on it is called gear.

4.1 INTRODUCTION

• Spur gears have teeth parallel to the axis of rotation and are used to transmit motion from one shaft to another, parallel, shaft.

• Helical gears have teeth inclined to the axis of rotation. Helical gears are not as noisy, because of the more gradual engagement of the teeth during meshing.

• Bevel gears have teeth formed on conical surfaces and are used mostly for transmitting motion between intersecting shafts.

• Worms and worm gears, The worm resembles a screw. The direction of rotation of the worm gear, also called the worm wheel, depends upon the direction of rotation of the worm and upon whether the worm teeth are cut right-hand or left-hand.

4.1 INTRODUCTION

4.1 INTRODUCTION

4.2 TWO GEARS IN OPERATION

4.3 NOMENCLATURE OF GEAR TEETH

– The pitch circle is a theoretical circle upon which all calculations are usually based; its diameter is the pitch diameter.

– A pinion is the smaller of two mating gears. The larger is often called the gear.

– The circular pitch 𝑝 is the distance, measured on the pitch circle, from a point on one tooth to a corresponding point on an adjacent tooth.

– The module 𝑚 is the ratio of the pitch diameter to the number of teeth.

– The diametral pitch 𝑃 is the ratio of the number of teeth on the gear to the pitch diameter.

– The addendum 𝑎 is the radial distance between the top land and the pitch circle.

– The dedendum 𝑏 is the radial distance from the bottom land to the pitch circle. The whole depth ℎt is the sum of the addendum and the dedendum.

– The clearance circle is a circle that is tangent to the addendum circle of the mating gear.

– The clearance 𝑐 is the amount by which the dedendum in a given gear exceeds the addendum of its mating gear.

– The backlash is the amount by which the width of a tooth space exceeds the thickness of the engaging tooth measured on the pitch circles.

4.3 NOMENCLATURE OF GEAR TEETH

4.3 NOMENCLATURE OF GEAR TEETH

𝑚 =𝑑

𝑁

• Module, m (SI units) :

• Diametral pitch, 𝑃 (English units) :

𝑃 =𝑁

𝑑

𝑝 =𝜋𝑑

𝑁= 𝜋𝑚

• Circular pitch, 𝑝 :

4.4 GEOMETRY OF GEAR

𝑃 = mm per tooth𝑑 = Pitch Diameter, in

𝑃 = teeth per inch𝑑 = Pitch Diameter, in

𝑚 =25.4

𝑃

• Conversion unit from module to diametral pitch:

Actual sizes of gear teeth ofvarious diametral pitches.

Note:In general, fine-pitch gears have P≥ 20; coarse pitch gears have P <20.

4.4 GEOMETRY OF GEAR

Actual sizes of gear teeth of various module size (SI units).

0.2 to 1.0 by increments of 0.1

1.0 to 4.0 by increments of 0.25

4.0 to 5.0 by increments of 0.5

With English units the word “pitch,” without a qualifying adjective, meansdiametral pitch (a “12-pitch gear” refers to a gear with 12 teeth per inch of pitchdiameter) whereas with SI units “pitch” means circular pitch (a “gear of pitch =3.14 mm” refers to a gear having a circular pitch of 3.14 mm).

4.4 GEOMETRY OF GEAR

• The term conjugate refers to the angularvelocities of two profiles in contact andconjugate action occurs when thesevelocities remain constant throughout thecontacting cycle.

• To transmit motion at a constant angular-velocity ratio, the pitch point must remainfixed; that is, all the lines of action for everyinstantaneous point of contact must passthrough the same point P.

𝜔𝑑𝑟𝑖𝑣𝑒𝑟

𝜔𝑑𝑟𝑖𝑣𝑒𝑛=

𝑟𝑑𝑟𝑖𝑣𝑒𝑛𝑟𝑑𝑟𝑖𝑣𝑒𝑟

= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

driving

driven

𝜔𝑑𝑟𝑖𝑣𝑒𝑟

𝜔𝑑𝑟𝑖𝑣𝑒𝑛

𝑟𝑑𝑟𝑖𝑣𝑒𝑛

𝑟𝑑𝑟𝑖𝑣𝑒𝑟

4.5 CONJUGATE ACTION

The involute of a circle is the spiraling curve traced by the end of an imaginary taut string unwinding itself from that stationary circle.

4.6 INVOLUTE OF A CIRCLE

4.7 BASE CIRCLE

𝑟𝑏 = 𝑟 cos𝜙

4.8 TOOTH ACTION

Driver teeth𝑚𝑐 =

𝑞𝑡𝑝=

𝐿𝑎𝑏𝑝𝐴cos𝜙

• A number that indicates the average number of pairs of teeth incontact.

• Gears should not generally be designed having contact ratiosless than about 1.20, because inaccuracies in mounting mightreduce the contact ratio even more, increasing the possibility ofimpact between the teeth as well as an increase in the noiselevel.

4.9 CONTACT RATIO

• Contact portions of tooth profiles that are not conjugate is calledinterference.

• The actual effect of interference is that the involute tip or face of the drivengear tends to dig out the non-involute flank of the driver.

• A portion of its dedendum falls inside the base circle.• If the teeth were produced by generating process (rather than stamping),

then the generating process removes the interfering portion; known asundercutting.

4.10 INTEFERENCE AND UNDERCUTTING

Driver (ccw)

Driven (cw)

4.10 INTEFERENCE AND UNDERCUTTING

4.11 GEOMETRY OF A PAIR OF GEAR

𝜔𝑝

𝜔𝑔= −

𝑑𝑔

𝑑𝑝

𝑐 =𝑑𝑝 + 𝑑𝑔

2= 𝑟𝑝 + 𝑟𝑔

* The minus sign indicates that the twocylinders (gears) rotate in opposite directions

* 𝑟 and 𝑑 is the pitch circle radius and pitch circle diameter.

Speed Ratio (or Gear Ratio):Driving

Driven

Driver(cw)

Driven(ccw)

Pitch Circle

Dedendum Circle coincides with Base Circle

Action Line

Addendum Circle

Start Finish

4.12 PRESSURE LINE

4.13 GEAR TRAIN

* A pair gear → Gear Ratio or Speed Ratio** Gear train → Train Value or Overall Speed Ratio

The speed ratio (or “gear ratio”) of a single pair of external spur gears is expressed by the simple equation

𝜔𝑝

𝜔𝑔=𝑛𝑝𝑛𝑔

= −𝑑𝑔

𝑑𝑝= −

𝑁𝑔

𝑁𝑝

• Gear 3 is an idler gear and affects only the direction of rotation of Gear 6.

• Gears 2, 3, and 5 are drivers, while 3, 4, and 6 are driven gears. We define the train value (or overall speed ratio), e as

• For spur gear, e is positive if the last gear rotates in the same sense as the first, and negative if the last rotates in the opposite sense.

𝑒 =𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑡𝑜𝑜𝑡ℎ 𝑛𝑢𝑚𝑏𝑒𝑟𝑠

𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑒𝑛 𝑡𝑜𝑜𝑡ℎ 𝑛𝑢𝑚𝑏𝑒𝑟𝑠

𝑛𝑙𝑎𝑠𝑡 = 𝑒 𝑛𝑓𝑖𝑟𝑠𝑡

4.13 GEAR TRAIN

(a) (b) (c)

Pinion

4.14 GEAR FORCE ANALYSIS

Driver(cw)

Driven(ccw)

2

3

Driver

Driven

ϕ

b

a

n3

n2

PressureAngle

4.14 GEAR FORCE ANALYSIS

4.14 GEAR FORCE ANALYSIS

𝐹𝑟 = 𝐹𝑡 tan𝜙

where,𝐹𝑡 = Tangential Force, Ib𝐹𝑟 = Radial Force, Ib𝜙 = Pressure Angle, degree

Radial Force, 𝐹𝑟:

4.14 GEAR FORCE ANALYSIS

where,ሶ𝑊 = Transmitted Power, hp

Transmitted Power, ሶ𝑊:

ሶ𝑊 = 𝑇𝜔 =𝐹𝑡𝑉

33000

where,𝐹𝑡 = Tangential Force, N

In SI Units. Transmitted Power, ሶ𝑊:

ሶ𝑊 = 𝑇𝜔 = 𝐹𝑡𝑉

4.14 GEAR FORCE ANALYSIS

𝑉 =𝜋𝑑𝑛

60,000

In SI units. Pitch-line velocity, 𝑉:

where,𝑉 = pitch-line velocity, m/s𝑑 = pitch diameter, mm𝑛 = gear speed, m/s

𝑉 =𝜋𝑑𝑛

12

Pitch-line velocity, 𝑉:

where,𝑉 = pitch-line velocity, feet/min𝑑 = pitch diameter, inches𝑛 = gear speed, rev/min

4.15 SAMPLE PROBLEM

Figure above shows three gears of 𝑃 = 3, 𝜙 = 20°. Gear 𝑎 is the driving, orinput, pinion. It rotates counterclockwise at 600 rpm and transmits 25 hp toidler gear 𝑏. Output gear 𝑐 is attached to a shaft that drives a machine. Nothingis attached to the idler shaft, and friction losses in the bearings and gears canbe neglected. Determine the resultant load applied by the idler to its shaft.

4.15 SAMPLE PROBLEM

Solution

𝑎

𝑏 𝑐

Idler Gear

Input

Output

1

2 3

4.15 SAMPLE PROBLEM

Solution:

𝑎

𝑏

𝑐

Input

Output

1

23

𝐹21

𝐹32

𝐹21𝑡

𝐹21𝑟

𝐹21

𝐹21𝑡

𝐹21𝑟

𝐹32𝑡

𝐹32𝑟

𝐹32 𝐹32𝑡

𝐹32𝑟

𝑅𝑐𝑦

𝑅𝑐𝑥

𝑅𝑏𝑦

𝑅𝑎𝑦

𝑅𝑏𝑥

𝑅𝑎𝑥

𝑇𝑎

𝑇𝑐𝑇𝑏= 0

4.15 SAMPLE PROBLEM

𝑉𝑡1 =𝜋𝑑1𝑛112

=𝜋(4)(600)

12= 628.28 𝑓𝑡/𝑚𝑖𝑛

ሶ𝑊1 = 𝑇1𝜔1 =𝐹21𝑡 𝑉𝑡1

33000↔ 𝐹21

𝑡 =33,000 ሶ𝑊1

𝑉𝑡1= 1313 𝑙𝑏

𝑃 =𝑁

𝑑↔ 𝑑1 =

𝑁1𝑃

=12

3= 4 𝑖𝑛.

𝐹21𝑟 = 𝐹21

𝑡 tan 20𝑜 = 1313 tan 20𝑜 = 478 𝑙𝑏

𝑇𝑎 = 𝐹21𝑡 𝑟1 = 1313

4

2= 2626 𝑙𝑏 ∙ 𝑖𝑛

𝐹21 = 𝐹21𝑡 2

+ 𝐹21𝑟 2

= 13132 + 4782 = 1397 𝑙𝑏

Thank You