Two-stage, compound reverted gear train.

Design Requirements

Power to be delivered: 22 kW

Power efficiency :> 95 %

Steady state Input speed: 45 rev/s

Maximum Input speed: 55 rev/s

Steady state output speed: 3-4 rev/s

Maximum gearbox size: 400 mm X 400 mm base, 700 mm height

Base mounted with 4 bolts

Manufacturing cost per unit: < $ 500

**Part1: Speed, Torque, and Gear Ratios**

Wout= 3+4 / 2 = 3.5 rev/s

W2= Input W = 55 rev/s (given)

e(Gear Ration) = W5/W2 = 3.5/55 =

1/15.71

Number of teeth N2/N3 = N4/N5 =

1/3.96

N2(Pinion) = N4

From Eq. 13.11 p.686 N2 = 15.4 .= 16 teeth

From the gear ratio N3 = 63.36

check whether to take 64 or 63, substituting in the W5 eq. , we find that both work.

So we decided to take

N3 = 63

W3 = W4 = (16/63)*55 =

13.96 rev/s

Part 2

Gear Specs.

Max. Gearbox Height = 700mm = 27.5inch

Pmin = 3.115 .=

4 teeth/inch

from Eq. 18.3

Now, d2 = d4 = N2/P = 16/4 =

4 inch

d3 = d5 = N5/P = 63/4 =

15.75 inch

Shaft Speeds were previously determined.

V23(Pitch line velocity) = 57.56ft/sec =

3453.6ft/min

V45= 14.42ft/sec =

865.2ft/min

from Eq. 13-34

Wt23(Transmitted load) =

281.9 lbf

Wt45 =

1125.17 lbf

from Eq. 13-35

This is needed to calculate the Wear and Bending on the Gears.

**Compound Reverted Gear Design Project**

Finding the Torques

T2 = 63.66 Eq. 18-1

T3 = 250.8

T5 = 63.66

T5 = 987.1

Wear and Bending for Gears

Since

Wt45 > Wt23

, which means that Gear 4 is transmitting the Largest Load, knowing that it is the smallest gear.

I(Geometric Factor)

Pressure angle = 20deg ,

Mn( load-sharing ratio) = 1 for Spur Gears

Mg ( Gear Ratio) = 3.96

I= 0.1283

from Eq. 14.23

Kv (dynamic factor)

Qv(Trans. accuracy lvl number) = 5

Kv = 1.48 from Eq. 14.27

A= 50+56(1-B)

B = 0.25 (12-Qv)^2/3

Face Width is 3 to 5 times the circular Pitch

F = 4*Pi/P =

3.142 inch

Now from the chosen product F= 2.5inch

Pitting resistance (Contact Stress)

σc = 90 997.2 psi from Eq. 14-16

To find Life factor Zn

L4 = 7.5e8 rev

Zn = 0.85 from Fig.14.15

Contact Strength Sc = 150000 from table 14-6

Factor of safety of gear 4 nc = 1.4

*Note: Treatment Grade 1 Nitride 83.5 HR table 14-6

Gear 4 Bending

J = 0.27 from fig 14-6

Kb = 1 [everything is the same as before]

σ = 11 900.9 psi from Eq. 14-15

stress-cycle fac. Yn = 0.85 from fig 14-14

St= 36 840 from figure 14-3

σall = St * Yn = 31 314 psi

n= 2.63

Gear 5 Bending and Wear

Everything is the same as gear 4 except J, Yn, Zn

σc= 90 997.2 psi Eq.14-16

σ = 7837.2 psi

nc = 1.4

n= 3.9

Gear 2 Wear

I = 0.1283 Eq. 14.23

Kv = 1.947 Eq. 14.27

Load in Gear 2 , 3 is less than Gear 4.

F = 2inch

σc= 53451.6 psi Eq.14-16

Sc = 85 5522 psi

n = 1.28

Gear 2 Bending

As in Gear before the constants J, Yn... are found

σ = 4106.3 psi

σall = 21599 psi

n = 5.26

Gear 3

Wear and Bending

σc =53451.6 psi

σ = 4056.2 psi

nc = 2.67

n= 8.6

Products to be used

Spur Gear from Rush Gears inc.

Part #F463

N= 63 DP=4 d=15.75

F = 2 & 2.5

Cost : 55$

Spur Gear from Rush Gears inc.

Part #F416

N= 16 DP=4 d=4

F= 2 & 2.5

Cost: 20$

Products to be Used

Products to be Used

12v Fluid Oil Pump (for Lubrication)

12 Volt / 5 Amp DC Motor

Oil Transfer Flow Rate: 3 Liters Per Minute

Cost 27.5$

The Lubrication method is introduced at the end.

From XtremepowerUS

Products to be used

Red Line 50104 Lightweight Gear Oil

Cost 16.5$

The same procedure as before

**Outline**

**By: Imad Al-Rabdi, Obada Yaghi and Omar Ghanem**

Project overview

Products to be used

Gear Specs, speeds and Ratios

Shaft Stress analysis

Deflection check

Bearing Selection and Key Design

Lubrication

part 4

Bearing Selection

From Stress analysis,

Ray = 715.5 N Rby = 1561 N

Raz = -112.5 N Rbz = -3641 N

R resultant: Ra = 725 N

Rb = 3962 N

LD = 7.5e8 rev

*Starting with bearing B since it has higher loads.

Frb = 59.5 KN from Eq. 11-7 [Ball bearing]

Frb = 4.337 kN [ Roller bearing]

*Kinetic loads

we used loads given, using eq 11,7 p,587 to find the dynamic load that was used to choose the bearing.

Referring to www.SKF.com

Bearing B:

C = 46 KN d = 25mm w = 17mm

Shoulder d = 52mm to 35mm max fillet radius = 1.1mm

Bearing A: ( Deep Groove, ball bearing)

C = 23.4 KN d= 25mm w = 14

shoulder diameter = 55mm to 32 mm

Bearing of Input Shaft

D=20mm

R inpt = 604N

LD = 2.9e7

Ball bearing = 14286 N from Eq. 11.7

Cylindrical bearing = 10418.8 N

Bearing of output Shaft

D=15mm

R out = 3340.3 N

LD = 1.89e7

Ball bearing = 14789 N from Eq. 11.7

Cylindrical bearing = 3882.4 N

Final Bearing values

Bearing for input shaft , cylindrical bearing

D = 25.5 KN d= 20mm w=25mm fillet r= 1.1mm

Bearing for output shaft, cylindrical bearing

D = 47.5KN d= 15mm fillet r = 1.1mm

Products to be used

Cylindrical roller Bearing

www.SKF.com

Cost = 25$

Gear 4 Wear

Part 5

Key Design

T = 254 N.m from stress analysis part

Bore diameter = 38mm

gear hub length L3 = 50mm L4 = 63.5mm

From table 7.6 p.391

choosing square key with side dimensions t = 10mm

Choosing 1010 CD with S= 300 MPa, noting that it should be less than shaft Sy

F= T/r = 13368.4 N

n=2

L = 2.9e-3

W= 0.01782m

Since the load on load on input shaft and output shaft is less than that on the intermediate shaft, the same specs. are used as before.

T = 10mm

L = 18mm

Part2 & Part3

Stress analysis and Deflection check

Lubrication

Lubrication is very essential in such systems to insure the best mech. eff. as well as avoiding failure caused by heat and metal to metal contact.

The energy to run the Pump is coming from the running shaft as a parasitic load.

Thank you for Listening

:)

Products to be used

LSX metalic Bolts and washers

from Metaltek Manufacturing

Cost : 14$

Coupling