Expt No.: DESIGN AND DRAFTING CONTROL COMPONENTS GEAR

Date:

AIM:

INTRODUCTION TO GEARS:

Gears are toothed members which transmit power / motion between two shafts by meshing without any slip. Hence, gear drives are also called positive drives. In any pair of gears, the smaller one is called pinion and the larger one is called gear immaterial of which is driving the other.

When pinion is the driver, it results in step down drive in which the output speed decreases and the torque increases. On the other hand, when the gear is the driver, it results in step up drive in which the output speed increases and the torque decreases.

CLASSIFICATION OF GEARS:

Gears are classified according to the shape of the tooth pair and disposition into spur, helical, double helical, straight bevel, spiral bevel and hypoid bevel, worm and spiral gears and this is shown in Fig. 1

Fig. 1.6 (a) Spur gear, (b) helical gear, (c) Double helical gear or herringbone gear, (d) Internal gear , (e) Rack and pinion, (f) Straight bevel gear, (g) Spiral bevel gear, (h)

Hypoid bevel gear , (i) worm gear and (j) Spiral gear

NOMENCLATURE OF SPUR GEAR

Design Procedure:

Given:

n1 = Speed of Pinion n2= Speed of Gear

z1=Number of teeth on Pinionz2=Number of Teeth on GearP = Powerα=Pressure Angleσo1=Allowable Static Stress for Pinionσo2=Allowable Static Stress for Gear

Step: 1To Identify Weaker member Details σ Y Capacity of y RemarksPinion σ1 y1 σ1y1

Gear σ2 y2 σ2y2

The member with smaller σoy is weaker member. The design should be based on weaker member.

For 14 12⁰ involute pressure angle, the lewis factor(y) is

y=0.124−0.684z

For 20 involute pressure angle, the lewis factor(y) is⁰

y=0.154−0.192z

Step:2 Design Based on strength of weaker member

(a)Tangential tooth load due to power transmission

F t=9.55×106 PC s

nr

Where

Cs = Serive factor = 1.5 (Based up on no. of hours /day service)

r = radius of the pinion and gear

(b)Lewis equation for tangential tooth load

F t=σobyp Kv

Where

b = face width = 9.5m to 12.5m

p = pitch

Kv=Velocity factor that depends on velocity

Equate (2) and (1) for strength and find module ‘m’ by trial and error procedure

Equation (1) will be F t=(….)m

and equation (2)F t=(… ..)m2K v

After equating (2) and (1) we get m3 Kv= (.....)

Since m and Kv are interdependent, trial and procedure is to be adopted.

To start first trail, approximately take KV ≈0.5 ,

m3×0.5=RHS

m≈ 3√2×RHS

Select the neatest standard module value

STEP: 3 Dimensions:

(a)Pitch Diameters, for Pinion d1=mz1

for Gear d2=mz2

(b)Centre distance

a=d1+d2

2

(c)Addendum and dedendum for 14.5⁰ involute pressure angle

Addendum (ha) = 1m

Dedendum (hf) = 1.157m

(c)Addendum Diameters

* For pinion da1=d1+2ha

* For gears da2=d2+2ha

(d)Dedendum Diameters

* For pinion d f 1=d1+2hf

* For gears d f 2=d2+2hf