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# Compound Reverted Gear Design Project

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on 20 January 2015

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#### Transcript of Compound Reverted Gear Design Project

Speed Reducer
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

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
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]
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
Full transcript