GEAR AND SHAFT DESIGN - Amazon S3 · 2 Gear Design 2.1 Gear A (pinion) & B (gear) Considering the transmitted power, P = 12.0 hp, the shaft 1 motor coupled gear speed, n 2, and the

Faculty of Engineering

GEAR AND SHAFT DESIGN

Project 2ME-321 - Kinematics and Dynamics of Machines

Prepared byPavel SheringID #20523043

Instructor: Eihab Abdel-Rahman3a Mechatronics Engineering

July 5, 2016

Table of Contents

1 Problem Details . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 Gear Design . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1 Gear A (pinion) & B (gear) . . . . . . . . . . . . . . . . . . . 2

2.1.1 Bending Stress . . . . . . . . . . . . . . . . . . . . . . 32.1.2 Contact Stress . . . . . . . . . . . . . . . . . . . . . . 6

2.2 Gear C (pinion) & D (gear) . . . . . . . . . . . . . . . . . . . 82.2.1 Bending Stress . . . . . . . . . . . . . . . . . . . . . . 92.2.2 Contact Stress . . . . . . . . . . . . . . . . . . . . . . 12

3 Shaft Design . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.1 Motor Shaft Force Analysis . . . . . . . . . . . . . . . . . . . 153.2 Reducer Shaft (Shaft 1) Force Analysis . . . . . . . . . . . . . 16

3.2.1 Reducer Shaft 1 Arrangement and Specifications . . . 183.3 Reducer Middle Shaft (Shaft 2) Force Analysis . . . . . . . . 23

3.3.1 Reducer Shaft 2 Arrangement and Specifications . . . 273.4 Output Reducer Shaft (Shaft 3) Force Analysis . . . . . . . . 31

3.4.1 Reducer Shaft 3 Arrangement and Specifications . . . 33

4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

i

List of Figures

1 Full system diagram . . . . . . . . . . . . . . . . . . . . . . . 1

2 Reducer Diagram . . . . . . . . . . . . . . . . . . . . . . . . . 1

3 Input Sheave coupled to the Motor Shaft with a belt drive . . 15

4 Shaft 1 Free Body Diagram . . . . . . . . . . . . . . . . . . . 17

5 Shaft 1 Torque Diagram . . . . . . . . . . . . . . . . . . . . . 17

6 Shaft 1 Shear and Moment Diagram . . . . . . . . . . . . . . . 18

7 Shaft 1 Design . . . . . . . . . . . . . . . . . . . . . . . . . . . 19




11 Shaft 2 Design . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

12 Output Gear D coupled to the Conveyor with a chain drive . . 31




16 Shaft 3 Design . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

ii

List of Tables

1 Gear A & B Properties . . . . . . . . . . . . . . . . . . . . . . 4

2 Bending Stress Paramters . . . . . . . . . . . . . . . . . . . . 5

3 Gear C & D Properties . . . . . . . . . . . . . . . . . . . . . . 10

4 Bending Stress Paramters . . . . . . . . . . . . . . . . . . . . 11

5 SAE 1040 CD Steel Properties . . . . . . . . . . . . . . . . . . 15

6 Combined Shear and Bending Moment Forces . . . . . . . . . 18

7 Shared Parameters for all Shafts . . . . . . . . . . . . . . . . . 19

8 Parameters for D1 . . . . . . . . . . . . . . . . . . . . . . . . 20

9 Parameters for D′1 . . . . . . . . . . . . . . . . . . . . . . . . 20

10 Parameters for D2 . . . . . . . . . . . . . . . . . . . . . . . . 21

11 Parameters for D4 . . . . . . . . . . . . . . . . . . . . . . . . 22

12 Parameters for D′4 . . . . . . . . . . . . . . . . . . . . . . . . 22

13 Parameters for D5 . . . . . . . . . . . . . . . . . . . . . . . . 23

14 Required Diameters for Shaft 1 . . . . . . . . . . . . . . . . . 24


16 Parameters for D1 . . . . . . . . . . . . . . . . . . . . . . . . 27

17 Parameters for D2 . . . . . . . . . . . . . . . . . . . . . . . . 28

18 Parameters for D′2 . . . . . . . . . . . . . . . . . . . . . . . . 28

19 Parameters for D4 . . . . . . . . . . . . . . . . . . . . . . . . 29

20 Parameters for D′4 . . . . . . . . . . . . . . . . . . . . . . . . 29

21 Parameters for D5 . . . . . . . . . . . . . . . . . . . . . . . . 30



24 Parameters for D1 . . . . . . . . . . . . . . . . . . . . . . . . 35

25 Parameters for D2 . . . . . . . . . . . . . . . . . . . . . . . . 35

26 Parameters for D′2 . . . . . . . . . . . . . . . . . . . . . . . . 36

27 Parameters for D4 . . . . . . . . . . . . . . . . . . . . . . . . 37

28 Parameters for D5 . . . . . . . . . . . . . . . . . . . . . . . . 37

29 Parameters for D′5 . . . . . . . . . . . . . . . . . . . . . . . . 38


iii

1 Problem Details

The purpose of this report is to design a system to power the conveyor driven

machine with a 12.0 horse power, 1150 rpm motor under the assumption that

each shaft in the system transmits 12.0 hp. The system is to consist of v-belt

driven input shaft connected to the motor, a 6:1 reducer and an output shaft

connected to the conveyor through a chain drive delivering the power form the

motor.

Figures 1 and 2 illustrate the system problem. The gears of the system are

designed first, as

Figure 1: Full system diagram

Figure 2: Reducer Diagram

1

2 Gear Design

2.1 Gear A (pinion) & B (gear)

Considering the transmitted power, P = 12.0 hp, the shaft 1 motor coupled

gear speed, n2, and the application (Figure 9-24 [1]) to determine a trial value

for the diametral pitch, Pd. The overload factor, KO, is determined from

Table 9-7 [1] to be 1.75, considering both the power source being electric

motor (uniform) and the driven machine is heavy-duty conveyor (heavy).

Calculate the pinion speed and torque based on the gear coupled to the

motor pinion.

n2R2 = nmotorRmotor

n2 = nmotorRmotor

R2

= 1150 rpm5.6 in

8.4 in

= 766.6667 rpm

n2 = nP

(1a)

T1 =P

n2

=63000 · 12.0 hp

766.6667 rpm

= 986.0870 lb · in

T1 = TP

(1b)

Solve for transmitted power in equation 25.

P = 12.0 hp ·KO

= 12.0 hp · 1.75

= 21.0 hp

(2)

From Figure 9-24 [1] match the pinion speed on the x-axis with 766.6667

rpm and on the left y-axis power of 21.0 hp to determine the diametral pitch

PdAB= 8 in or standard module of m = 3.0mm with 20◦ full depth teeth.

To solve for the diameter of pinion A, DP using equation 3, choosing the

number of teeth from Table 8-3 to be greater than 16, thus making the calcu-

lations easy NP = 24 teeth.

DP =NP

PdAB

=24 teeth

8 in

= 3.0 in

(3)

2

By specification, the reducer must be 1/6 reduction in speed and using

two shafts, thereby making the velocity ratio (VR) to be 3 · 2. Using this to

calculate the diameter of Gear B.

NG = V R ·NA

= 3 · 24

= 72 teeth

(4a)

DG =NG

PdAB

=72 teeth

8 in

= 9.0 in

(4b)

Calculate the gear speed and torque.

nG = nPNP

NG

= 766.6667 rpm · 1

3

= 255.5556 rpm

(5a)

TG =P

nG

=63000 · 12.0 hp

255.5556 rpm

= 2958.2607 lb · in

(5b)

Solve for center distance and pitch line speed.

C =NP +NG

2PdAB

=24 + 72 teeth

2 · 8 in

= 6 in

(6a)

vtAB=πDPnP

12

=π · 3.0 · 766.6667

12

= 602.1386 ft/min

(6b)

Calculate the face width for Gears A and B and transmitted load, Wt.

F =12

PdAB

=12

8 in

= 1.5 in

(7a)

WtAB= 33000 · P

vt

= 33000 · 12.0

602.1386

= 657.6559 lb

(7b)

Based on the pitch line speed (in range of 0 - 800 fpm) which determines

the Quality number Av of A10 form Table 9-3 for mining conveyor application.

The gear and pinion properties are summarized in Table 1.

2.1.1 Bending Stress

Bending stress is calculated Equation 8

st =WtPd

FJKOKsKmKBKv (8)

Specifying Ks = 1.0, size factor, due to Pd > 5.

3

Table 1: Gear A & B Properties

Properties Pinion (Gear A) Gear (Gear B)

Pitch Diameter (D) 3.0 in 9.0 inNumber of Teeth (N) 24 72Speed (n) 766.6667 rpm 255.5556 rpmTorque (T) 986.0870 lb · in 2958.2607 lb · inDiametral Pitch (Pd) 8 inFace Width (F) 1.50 inQuality (Av) A10Center Distance (C) 6 inPitch Line Speed (vt) 602.1386 fpmTransmitted Torque (Wt) 657.6559 lb

To calculating load distributing factor Km, loading factor, Cpf and align-

ment, Cma factor must be determined first.

F = 1.50 in

DP = 3.0 in

Cpf =F

10DP

− 0.0375 + 0.0125F when 1.0 < F < 15

Cpf =1.5

10 · 3.0− 0.0375 + 0.0125 · 1.5

= 0.03125

(9)

Solving for the mesh alignment factor Cma for commercially enclosed gear-

box.

Cma = 0.127 + 0.0158F − 1.093 · 10−4F 2

Cma = 0.127 + 0.0158 · 1.5− 1.093 · 10−4 · 1.52

= 0.1504

(10)

Compute the load distribution factor Km.

Km = 1.0 + Cpf + Cma

Km = 1.0 + 0.03125 + 0.1504

= 1.1817

(11)

Commercially made spoked gears are used which are expected to have a

4

well-supported rim, thus the rim thickness factor, KB = 1.0.

Using Av = A10 determine the dynamic loading factor from Table 9-20

Kv = 1.25.

Specify the tooth form, the bending geometry factors for the pinion and

the gear from Figure 9-15 JP = 0.36, JG = 0.42.

The bending stress is determined using the summarized parameters in Ta-

ble 2.

Table 2: Bending Stress Paramters

Parameters Pinion Gear

Transmitted Torque (Wt) 657.6559 lbPitch Diameter (D) 3.0 in 9.0 inDiametral Pitch (Pd) 8.0 inFace Width (F) 1.50 inGeometry factor (J) JP = 0.36 JG = 0.42Overload factor (KO) 1.75Size factor (Ks) 1.0Load distributing factor (Km) 1.1817Rim thickness factor (KB) 1.0Dynamic loading factor (Kv) 1.25

(st)P =WtPd

FJPKOKsKmKBKv

(st)P =(657.6559 lb)(8.0 in)

(1.5 in)(0.36)(1.75)(1.0)(1.1817)(1.0)(1.25)

= 25.1855ksi

(12)

(st)G =WtPd

FJGKOKsKmKBKv

(st)G =(657.6559 lb)(8.0 in)

(1.5 in)(0.42)(1.75)(1.0)(1.1817)(1.0)(1.25)

= 21.5876ksi

(13)

The recommended design life for general industrial systems is 20,000 -

30,000 hours, thus an average of maximum and minimum is used to solve for

the number of cycles. The number number of loading applications for the

system is q = 1.0.

5

(NC)P = 60LnP q

= (60)(25000 hr)(766.6667 rpm)(1.0)

= 1.15 · 109 cycles

(14)

(NC)P = 60LnP q

= (60)(25000 hr)(255.5556 rpm)(1.0)

= 3.8333 · 108 cycles

(15)

Calculate the stress cycle factor for bending, YN .

(YN)P = 1.3558(NC)−0.0178P

= 1.3558(1.15 · 109)−0.0178

= 0.9352

(16a)

(YN)G = 1.3558(NC)−0.0178G

= 1.3558(3.8333 · 108)−0.0178

= 0.9537

(16b)

With no specific uncertainty for the current problem, the service factor,

SF = 1.0, and reliability factor KR = 1.0 for 0.99 reliability, meaning one

failure in 100. These parameters are used to calculate the adjusted bending

stresses.

(sat)P = (st)P(SF )KR

(YN)P

= (25.1855)(1.0)(1.0)

(0.9353)

= 26.9277 ksi

(17a)

(sat)G = (st)G(SF )KR

(YN)G

= (21.5876)(1.0)(1.0)

(0.9537)

= 22.6356 ksi

(17b)

Calculating the minimum required hardness for Grade 1 steel, as Grade 2

is significantly more expensive.

(HB)P =(sat)P − 12.80

0.0773

=26.9277− 12.80

0.0773

= 182.7646

(18a)

(HB)G =(sat)G − 12.80

0.0773

=22.6356− 12.80

0.0773

= 127.2393

(18b)

2.1.2 Contact Stress

Contact stress is calculated using Equation 42.

6

sc = Cp

√Wt

FDpIKOKsKmKv (19)

The gears will be made from steel, thus the elastic coefficient, Cp = 2300.

The pitting geometry factor from Figure 9-21 I = 0.108, for 20◦pressure angle.

Solving for contact stress, using parameters stated above and in Table 2

sc = Cp

√WtAB

FDpIKOKsKmKv

sc = 2300

√657.6559 lb

(1.5 in)(3.0 in)(0.108)(1.75)(1.0)(1.1817)(1.25)

= 136.0306 ksi

(20)

Calculate the stress cycle factor, ZN using the number of cycles from Equa-

tions 14 and 15.

(ZN)P = 1.4488(NC)−0.023P

= 1.4488(1.15 · 109)−0.023

= 0.8966

(21a)

(ZN)G = 1.4488(NC)−0.023G

= 1.4488(3.8333 · 108)−0.023

= 0.9196

(21b)

The same service factor, SF = 1.0 and reliability factor KR = 1.0 are used

to calculated the adjusted contact stress.

(sac)P = sc(SF )KR

(ZN)P

= (136.0306)(1.0)(1.0)

(0.8966)

= 151.7183 ksi

(22a)

(sac)G = sc(SF )KR

(ZN)G

= (136.0306)(1.0)(1.0)

(0.9196)

= 147.9237 ksi

(22b)

Calculating the minimum required hardness for Grade 1 steel.

(HB)P =(sac)P − 29.10

0.322

=151.7183− 29.10

0.322

= 380.8022

(23a)

(HB)G =(sac)G − 29.10

0.322

=147.9237− 29.10

0.322

= 369.0177

(23b)

The HB from contact stress is approximately the same for the gear and

7

the pinion, which will constrain the material choice, along with the maximum

value of the contact stress.

Gear and the pinion will be made form SAE 1340 OQT 700 with HB =

444, Yield Strength = 197 ksi, and Tensile Strength = 221 ksi.

2.2 Gear C (pinion) & D (gear)

The transmitted power is still P = 12.0 hp, the shaft 2 coupled pinion speed,

nB = 255.5556 rpm. The overload factor, KO = 1.75 as it is the same appli-

cation.

Calculate the pinion speed and torque based on the pinion C related to

gear B.

nC = nB

nC = 255.5556 rpm(24a)

TC = TB

TC = 2958.2607lb · in(24b)

Solve for transmitted power in equation 25.

P = 12.0 hp ·KO

= 12.0 hp · 1.75

= 21.0 hp

(25)

From Figure 9-24 [1] match the pinion speed on the x-axis with 255.5556

rpm and on the left y-axis power of 21.0 hp to determine the diametral pitch

PdCD= 5 in or standard module of m = 5.0 mm with 20◦ full depth teeth.

To solve for the diameter of pinion A, DP using equation 26, choosing

the number of teeth from Table 8-3 to be greater than 12, thus making the

calculations easy NP = 20 teeth.

DP =NP

PdCD

=20 teeth

5 in

= 4.0 in

(26)

As mentioned before the reducer must be 1/6 reduction in speed and using

two shafts, thereby making the velocity ratio (VR) to be 3 · 2. Using this to

calculate the diameter of Gear D.

8

NG = V R ·NP

= 2 · 20

= 40 teeth

(27a)

DG =NG

PdCD

=40 teeth

5 in

= 8.0 in

(27b)

Calculate the gear speed and torque.

nG = nPNP

NG

= 255.5556 rpm · 1

2

= 127.7778 rpm

(28a)

TG =P

nG

=63000 · 12.0 hp

127.7778 rpm

= 5916.5207 lb · in

(28b)

Solve for center distance and pitch line speed.

C =NP +NG

2PdCD

=20 + 40 teeth

2 · 5 in

= 6 in

(29a)

vtCD=πDPnP

12

=π · 4.0 · 255.5556

12

= 267.6172 ft/min

(29b)

Calculate the face width for Gears C and D and transmitted load, Wt.

F =12

PdAB

=12

5 in

= 2.4 in

(30a)

WtAB= 33000 · P

vt

= 33000 · 12.0

267.6172

= 1479.7255 lb

(30b)

Based on the pitch line speed (in range of 0 - 800 fpm) which determines

the Quality number Av of A10 form Table 9-3 for mining conveyor application.

The gear and pinion properties are summarized in Table 3.

2.2.1 Bending Stress

Bending stress is calculated Equation 31

st =WtPd

FJKOKsKmKBKv (31)

Specifying Ks = 1.0, size factor, due to Pd = 5.

To calculating load distributing factor Km, loading factor, Cpf and align-

ment, Cma factor must be determined first.

9

Table 3: Gear C & D Properties

Properties Pinion (Gear A) Gear (Gear B)

Pitch Diameter (D) 4.0 in 8.0 inNumber of Teeth (N) 20 40Speed (n) 255.5556 rpm 127.7778 rpmTorque (T) 2958.2607 lb · in 5916.5207 lb · inDiametral Pitch (Pd) 5 inFace Width (F) 2.40 inQuality (Av) A10Center Distance (C) 6 inPitch Line Speed (vt) 267.6172 fpmTransmitted Torque (Wt) 1479.7255 lb

F = 2.40 in

DP = 4.0 in

Cpf =F

10DP

− 0.0375 + 0.0125F when 1.0 < F < 15

Cpf =2.4

10 · 4.0− 0.0375 + 0.0125 · 2.4

= 0.0525

(32)

Solving for the mesh alignment factor Cma for commercially enclosed gear-

box.

Cma = 0.127 + 0.0158F − 1.093 · 10−4F 2

Cma = 0.127 + 0.0158 · 2.4− 1.093 · 10−4 · 2.42

= 0.1643

(33)

Compute the load distribution factor Km.

Km = 1.0 + Cpf + Cma

Km = 1.0 + 0.0525 + 0.1643

= 1.2168

(34)

Commercially made spoked gears are used which are expected to have a

well-supported rim, thus the rim thickness factor, KB = 1.0.

Using Av = A10 determine the dynamic loading factor from Table 9-20

10

Kv = 1.25.

Specify the tooth form, the bending geometry factors for the pinion and

the gear from Figure 9-15 JP = 0.325, JG = 0.385.

The bending stress is determined using the summarized parameters in Ta-

ble 4.

Table 4: Bending Stress Paramters

Parameters Pinion Gear

Transmitted Torque (Wt) 1479.7255 lbPitch Diameter (D) 4.0 in 8.0 inDiametral Pitch (Pd) 5.0 inFace Width (F) 2.40 inGeometry factor (J) JP = 0.325 JG = 0.385Overload factor (KO) 1.75Size factor (Ks) 1.0Load distributing factor (Km) 1.2168Rim thickness factor (KB) 1.0Dynamic loading factor (Kv) 1.25

(st)P =WtPd

FJPKOKsKmKBKv

(st)P =(1479.7255 lb)(5.0 in)

(2.40 in)(0.325)(1.75)(1.0)(1.2168)(1.0)(1.25)

= 25.2478ksi

(35)

(st)G =WtPd

FJGKOKsKmKBKv

(st)G =(1479.7255 lb)(5.0 in)

(2.40 in)(0.385)(1.75)(1.0)(1.2168)(1.0)(1.25)

= 21.3131ksi

(36)

The recommended design life for general industrial systems is 20,000 -

30,000 hours, thus an average of maximum and minimum is used to solve for

the number of cycles. The number number of loading applications for the

system is q = 1.0.

11

(NC)P = 60LnP q

= (60)(25000 hr)(255.5556 rpm)(1.0)

= 3.8333 · 108 cycles

(37)

(NC)P = 60LnP q

= (60)(25000 hr)(127.7778 rpm)(1.0)

= 1.9167 · 108 cycles

(38)

Calculate the stress cycle factor for bending, YN .

(YN)P = 1.3558(NC)−0.0178P

= 1.3558(3.8333 · 108)−0.0178

= 0.9537

(39a)

(YN)G = 1.3558(NC)−0.0178G

= 1.3558(1.9167 · 108)−0.0178

= 0.9655

(39b)

With no specific uncertainty for the current problem, the service factor,

SF = 1.0, and reliability factor KR = 1.0 for 0.99 reliability, meaning one

failure in 100. These parameters are used to calculate the adjusted bending

stresses.

(sat)P = (st)P(SF )KR

(YN)P

= (25.2478)(1.0)(1.0)

(0.9537)

= 26.4735 ksi

(40a)

(sat)G = (st)G(SF )KR

(YN)G

= (21.3131)(1.0)(1.0)

(0.9655)

= 22.0747 ksi

(40b)


(HB)P =(sat)P − 12.80

0.0773

=26.4735− 12.80

0.0773

= 186.5416

(41a)

(HB)G =(sat)G − 12.80

0.0773

=22.0747− 12.80

0.0773

= 119.9832

(41b)

2.2.2 Contact Stress

Contact stress is calculated using Equation 42.

12

sc = Cp

√Wt

FDpIKOKsKmKv (42)

The gears will be made from steel, thus the elastic coefficient, Cp = 2300.

The pitting geometry factor from Figure 9-21 I = 0.094, for 20◦pressure angle.

Solving for contact stress, using parameters stated above and in Table 4

sc = Cp

√WtCD

FDpIKOKsKmKv

sc = 2300

√1479.7255 lb

(2.4 in)(4.0 in)(0.094)(1.75)(1.0)(1.2168)(1.25)

= 151.9506 ksi

(43)

Calculate the stress cycle factor, ZN using the number of cycles from Equa-

tions 14 and 15.

(ZN)P = 1.4488(NC)−0.023P

= 1.4488(3.8333 · 108)−0.023

= 0.9196

(44a)

(ZN)G = 1.4488(NC)−0.023G

= 1.4488(1.9167 · 108)−0.023

= 0.9344

(44b)

The same service factor, SF = 1.0 and reliability factor KR = 1.0 are used

to calculated the adjusted contact stress.

(sac)P = sc(SF )KR

(ZN)P

= (151.9506)(1.0)(1.0)

(0.9196)

= 165.2355 ksi

(45a)

(sac)G = sc(SF )KR

(ZN)G

= (151.9506)(1.0)(1.0)

(0.9344)

= 162.6184 ksi

(45b)


(HB)P =(sac)P − 29.10

0.322

=165.2355− 29.10

0.322

= 422.7810

(46a)

(HB)G =(sac)G − 29.10

0.322

=162.6184− 29.10

0.322

= 414.6534

(46b)

The HB from contact stress is approximately the same for the gear and

13

the pinion, which will constrain the material choice, along with the maximum

value of the contact stress.

Gear and the pinion will be made form Cold Drawn SAE 1340 with HB =

160, Yield Strength = 71 ksi, and Tensile Strength = 80 ksi.

14

3 Shaft Design

All shafts are designed using the previously designed gears, and a safety factor

of N = 3. The shafts will be machined from cold drawn SAE 1040 steel (Table 5

for material properties from Appendix 3 [1]).

Table 5: SAE 1040 CD Steel Properties

sy (ksi) su (ksi) sn (ksi)

71 80 31

Figure 12 illustrated the v-belt drive, from the left view.

Figure 3: Input Sheave coupled to the Motor Shaft with a belt drive

3.1 Motor Shaft Force Analysis

Calculating tangential force, FN , and torque, T.

FN = 126000P

Dn

= (126000)12 hp

(5.6 in)(1150 rpm)

= 234.7826 lb

(47a)

T = 63000P

n

= 6300012 hp

1150 rpm

= 657.3913 lb · in

(47b)

15

Solve for the bending force at the sheave, FB, using the factor of 1.5 for

the v-belt sheave.

FB = 1.5FN

= 1.5(234.7826 lb)

= 352.1739 lb↖

(48)

Solve for forces in x and y directions.

FBx = FB sin 35◦

= 352.1739 sin 35◦

= 201.9987 lb←

(49a)

FBy = FB cos 35◦

= 352.1739 cos 35◦

= 288.4840 lb ↑

(49b)

3.2 Reducer Shaft (Shaft 1) Force Analysis

The torque at Gear A is calculated in equation 1b. TA = 986.0870 lb · in.

The forces for gear A on shaft 1, are equal in magnitude to the input sheave

(motor pinion) but opposite in direction.

FB = 352.1739 lb↘

FBx = 201.9987 lb→

FBy = 288.4840 lb ↓

(50)

Calculate the tangential force, WtA , and radial force, WrA for 20◦ pressure

angle, φ.

WtA =TADA

2

=986.0870

3.02

= 657.3913 lb→

(51a)WrA = WtA tanφ

= 657.3913 lb tan 20◦

= 239.2709 lb ↓

(51b)

Solve for the bearing reaction forces using the free body diagram (Figure 6).

Calculating the reaction forces of the bearings, to create shear force and

moment diagrams.

16

Figure 4: Shaft 1 Free Body Diagram

∑MBx = 0

0 = FA(6 in)− FC(3 in) +Dx(6 in)

Dx =FC(3 in)− FA(6 in)

6 in

=657.3913 lb(3 in)− 201.9987 lb(6 in)

6 in

= 126.9695 lb

(52a)

∑Fx = 0

0 = −FA +Bx − FC +Dx

Bx = FA + FC −Dx

= 201.9987 lb + 657.3913 lb− 126.9695 lb

= 732.6930 lb

(52b)

∑MBy = 0

0 = FA(6 in) + FC(3 in)−Dy(6 in)

Dy =FA(6 in) + FC(3 in)

6 in

=288.4840 lb(6 in) + 239.2709 lb(3 in)

6 in

= 408.1195 lb

(52c)

∑Fy = 0

0 = −FA +By + FC −Dy

By = FA − FC +Dy

= 288.4840 lb− 239.2709 lb + 408.1195 lb

= 457.3326 lb

(52d)

0 2 4 6 8 10 12z (in)

0

200

400

600

800

1000

T (

lb*i

n)

Torque Diagram

Figure 5: Shaft 1 Torque Diagram

Solving for shear and moment forces at points A, B, C, and D using the

formulas below and the final values are summarized in Table 15

17

300200100

0100200300400500600

V (

lb)

Horizontal Plane Vertical Plane

0 2 4 6 8 10 12z (in)

2000

1500

1000

500

0

500

M (

lb.in)

0 2 4 6 8 10 12z (in)

Figure 6: Shaft 1 Shear and Moment Diagram

Vtotal =√V 2x + V 2

y

Mtotal =√M2

x +M2y

(53)

Table 6: Combined Shear and Bending Moment Forces

Point Shear (lb) Moment (lb · in)

A 352.1740 0B 556.9 2113.0437C 427.4141 1284.1891D 427.4141 0

3.2.1 Reducer Shaft 1 Arrangement and Specifications

Figure 7 illustrates the design of the first shaft.

Table 7 summarizes the shared parameters used for all diameter calcu-

18

Figure 7: Shaft 1 Design

lations. A 99.0% reliability factor, CR is used for the design, along with a

standard safety factor of 3.0 for dynamic machines. The size factor for the

shafts is assumed, since it is required for the diameter calculation. By design

CS = 0.81.

Solve for the adjusted endurance limit, s′n.

s′n = snCSCR

= (31.0 ksi)(0.81)(0.81)

= 20.3391 ksi

(54)

Table 7: Shared Parameters for all Shafts

Parameters

Tensile Strength su = 80 ksiYield Strength sy = 71 ksiEndurance Strength sn = 31 ksiAdjusted Endurance Strength s′n = 20.3391 ksiReliability CR = 0.81Safety Factor N = 3.0Size Factor CS = 0.81

To calculate the shaft diameters, Equation 61 is used for points at which

torque or bending moment forces are present. Otherwise the point is at pure

shear, and Equation 58 is used.

D3 =32N

π

√(KtM

s′n

)2

+3

4

(T

sy

)2

(55)

19

D =

√2.94KtV N

s′n(56)

Solving D1

At point A, requires a profile keyseat (Kt = 2.0), and a well-rounded fillets

(Kt = 1.5), and retaining ring groove (Kt = 3.0). At that point, there is no

moment, however, there is torque.

Table 8: Parameters for D1

Parameters

Stress Concentration Factor Kt = 2.0Combined Bending Moment M = 0Torque T = 986.0870 lb · in

D31 =

32N

π

√(KtM

s′n

)2

+3

4

(T

sy

)2

=32(3.0)

π

√((2.0)(0lb · in)

20.3391ksi

)2

+3

4

(986.0870 lb · in

71 ksi

)2

= 0.3675

D1 = 0.7163 in

(57)

Check that the groove diameter is smaller than the shaft diameter. There

is no moment or shear at the groove.

Table 9: Parameters for D′1

Parameters

Stress Concentration Factor Kt = 2.5Combined Shear Force V = 352.1740 lb

20

D′1 =

√2.94KtV N

s′n

=

√2.94(3.0)(352.1740 lb)(3.0)

20.3391ksi

= 0.6769 in

(58)

Set D′1 = D1 = 0.7163. The preferred size for D′1 = 0.75, thus D1 =

1.06 ∗D′1 = 0.7950. The preferred size for D1 = 0.8000 in.

Solving D2

At point B contains a bearing, thus sharp fillet is required, (Kt = 2.5).


Parameters

Stress Concentration Factor Kt = 2.5Combined Bending Moment M = 2113.0437 lb · inTorque T = 986.0870 lb · in

D32 =

32N

π

√(KtM

s′n

)2

+3

4

(T

sy

)2

=32(3.0)

π

√((2.5)(2113.0437lb · in)

20.3391ksi

)2

+3

4

(986.0870 lb · in

71 ksi

)2

= 7.9452

D2 = 1.9954 in

(59)

The next preferred size for D2 is 2.00 in. D′2 is set to 6% than D2 to

minimize bearing outer race interference. Therefore D′2 = 2.2000 in.

Calculate D3 after D4 because it is used as support for gear B.

Solving D4

At point A, requires a profile keyseat (Kt = 2.0), and a well-rounded fillets

(Kt = 1.5), and retaining ring groove (Kt = 3.0).

21


Parameters

Stress Concentration Factor Kt = 2.0Combined Bending Moment M = 1284.1891Torque T = 986.0870 lb · in

D34 =

32N

π

√(KtM

s′n

)2

+3

4

(T

sy

)2

=32(3.0)

π

√((2.0)(1284.1891lb · in)

20.3391ksi

)2

+3

4

(986.0870 lb · in

71 ksi

)2

= 3.8762

D4 = 1.5709 in

(60)

The retaining ring groove has no torsion.


Parameters

Stress Concentration Factor Kt = 2.0Combined Bending Moment M = 1284.1891Torque T = 0 lb · in

D′34 =32N

π

√(KtM

s′n

)2

+3

4

(T

sy

)2

=32(3.0)

π

√((2.0)(1284.1891lb · in)

20.3391ksi

)2

+3

4

(0 lb · in

71 ksi

)2

= 3.8588

D′4 = 1.5685 in

(61)

The preferred size for D′4 = 1.6000 is approximately the same as D4 in,

thus adjust D4 by 6%. D4 = 1.6000 · 1.06 = 1.6960 in. Therefore, preferred

diameter for D4 = 1.7500 in.

22

Solving D3

The D3 has to be bigger than D4 to retain it on one side. Therefore, make D3 =

1.15D4 = 2.0125 in. Thus the preferred size for D3 = 2.2000 in. However,

D′2 = D3 therefore D3 must be increased further. New D3 = 2.5000 in.

Solving D5

At point D contains a bearing, thus sharp fillet is required, (Kt = 2.5).


Parameters


D5 =

√2.94KtV N

s′n

=

√2.94(3.0)(427.4141 lb)(3.0)

20.3391ksi

= 0.7457 in

(62)

The next preferred size is 0.7500 in. D′5 is set to 15% than D5 to minimize

bearing outer race interference. Therefore D′5 = 0.8625 = 1.0000 in.

Summary

Table 30 displays all the diameters for Shaft 1. It is assumed that the bearings

for standard size diameters are available. Based on the choice of bearings may

require change in D’ diameters to prevent interference with the outer rase of

the bearing. None of the diameters exceed 2.5 in which was the assumption

for CS = 0.81.

3.3 Reducer Middle Shaft (Shaft 2) Force Analysis

The torque at Gear B is calculated in equation 5b. TB = 2958.2607 lb · in.

23

Table 14: Required Diameters for Shaft 1

Diameter Calculated (in) Preferred (in)

D1 0.7950 0.8000D′1 0.7163 0.7500D2 1.9954 2.0000D′2 2.1200 2.2000D3 1.8688 2.5000D4 1.5709 1.7500D′4 1.5685 1.6000D5 0.7457 0.7500D′5 0.8625 1.0000

Calculate the tangential force, WtB (Equation 7b), and radial force, WBr

for 20◦ pressure angle, φ.

WtB = 657.6559 lb←

WrB = WtB tanφ

= 657.3913 lb tan 20◦

= 239.2709 lb ↑

(63)

Solve for gear C tangential force, WtC , and radial force, WrC for 20◦ pres-

sure angle, φ.

WtC =TCDC

2

=2958.2607

4.02

= 1479.1304 lb←

(64a)WrC = WtC tanφ

= 1479.1304 lb tan 20◦

= 538.3594 lb ↑

(64b)

Solve for the bearing reaction forces using the free body diagram (Fig-

ure 10).


24


moment diagrams.

∑MAx

= 0

0 = FB(3 in) + FC(8 in)−Dx(11 in)

Dx =FB(3 in) + FC(8 in)

11 in

=657.3913 lb(3 in) + 1479.1304 lb(8 in)

11 in

= 1252.83 lb

(65a)

∑Fx = 0

0 = −Ax + FB + FC −Dx

Ax = FB + FC −Dx

= 657.3913 lb + 1479.1304 lb− 1252.83 lb

= 883.6917 lb

(65b)

∑MAy

= 0

0 = −FB(3 in) + FC(8 in)−Dy(11 in)

Dy =−FB(3 in) + FC(8 in)

11 in

=−239.2709 lb(3 in) + 538.3594 lb(8 in)

11 in

= 326.2784 lb

(65c)

∑Ay = 0

0 = Ay − FB + FC −Dy

Ay = FB − FC +Dy

= 239.2709 lb− 538.3594 lb + 326.2784 lb

= 27.1899 lb

(65d)

0 1 2 3 4 5 6 7 8 9z (in)

0

500

1000

1500

2000

2500

3000

T (

lb*i

n)

Torque Diagram



formulas below and the final values are summarized in Table 15.


y

Mtotal =√M2

x +M2y

(66)

25

1000

500

0

500

1000

1500

V (

lb)


0 1 2 3 4 5 6 7 8 9z (in)

4000350030002500200015001000

5000

500

M (

lb.in)

0 1 2 3 4 5 6 7 8 9z (in)




A 884.1185 0B 310.1582 2652.3556C 1294.5963 3907.1872D 1294.5963 0

26


Figure 11 illustrates the design of the second shaft.


Refer to Table 7 for the shared parameters used for all diameter calcula-

tions.

Solving D1

At point A contains a bearing, thus sharp fillet is required, (Kt = 2.5).


Parameters


D1 =

√2.94KtV N

s′n

=

√2.94(3.0)(884.1185 lb)(3.0)

20.3391ksi

= 1.072 in

(67)

The next preferred size is D1 = 1.2000 in. D′1 is set to 6% than D1 to

minimize bearing outer race interference. Therefore D′1 = 1.2720 = 1.5000 in.

27

Solving D2

At point B, requires a profile keyseat (Kt = 2.0), and a well-rounded fillets



Parameters


D32 =

32N

π

√(KtM

s′n

)2

+3

4

(T

sy

)2

=32(3.0)

π

√((2.0)(2652.3556lb · in)

20.3391ksi

)2

+3

4

(2958.2607 lb · in

71 ksi

)2

= 8.0458

D2 = 2.0038 in

(68)




Parameters


D′32 =32N

π

√(KtM

s′n

)2

+3

4

(T

sy

)2

=32(3.0)

π

√((2.0)(2652.3556lb · in)

20.3391ksi

)2

+3

4

(0 lb · in

71 ksi

)2

= 7.9699

D′2 = 1.9975 in

(69)

28

D′2 is set to D2 because the two diameters are approximately the same.

The preferred size for D′2 = 2.2000. D2 = 1.06D′2 = 2.3320 = 2.4000 in.

Solving D3

The D3 has to be bigger than D2 to retain it on one side. Therefore, make

D3 = 1.15D2 = 2.7600 in. Thus the preferred size for D3 = 2.8000 in.

Solving D4

At point C, requires a profile keyseat (Kt = 2.0), and a well-rounded fillets



Parameters


D34 =

32N

π

√(KtM

s′n

)2

+3

4

(T

sy

)2

=32(3.0)

π

√((2.0)(3907.1872lb · in)

20.3391ksi

)2

+3

4

(2958.2607 lb · in

71 ksi

)2

= 11.7921

D4 = 2.2761 in

(70)




Parameters


29

D′34 =32N

π

√(KtM

s′n

)2

+3

4

(T

sy

)2

=32(3.0)

π

√((2.0)(3907.1872lb · in)

20.3391ksi

)2

+3

4

(0 lb · in

71 ksi

)2

= 11.7404

D′4 = 2.2728 in

(71)



Solving D5

At point D contains a bearing, thus sharp fillet is required, (Kt = 2.5).


Parameters


D5 =

√2.94KtV N

s′n

=

√2.94(3.0)(1294.5963 lb)(3.0)

20.3391ksi

= 1.2978 in

(72)

The next preferred size is D5 = 1.4000 in. D′5 is set to 6% than D5 to

minimize bearing outer race interference. Therefore D′5 = 1.4840 = 1.5000 in.

Change D1 = 1.4000 in to match D5.

Summary

Table 22 displays all the diameters for Shaft 2.

30



D1 1.072 1.4000D′1 1.2720 1.5000D2 2.3320 2.4000D′2 2.0038 2.2000D3 2.7600 2.8000D4 2.5440 2.6000D′4 2.2761 2.4000D5 1.2978 1.4000D′5 1.4840 1.5000

3.4 Output Reducer Shaft (Shaft 3) Force Analysis

The torque at Gear D is calculated in equation 28b. TA = 5916.5207 lb · in.

Figure 12: Output Gear D coupled to the Conveyor with a chain drive

The forces for gear D on shaft 1, are equal in magnitude to Gear C but

opposite in direction.

31

WtD = 1479.1304 lb→

WrD = WtD tanφ

= 1479.1304 lb tan 20◦

= 538.3594 lb ↓

(73)

Forces at the sprocket is calculated.

FN = FB

=TD

Dconveyorsprocket

2

=5916.5207 lb · in

4.2 in2

= 2817.3908 lb↗

FBx = FB sin 20◦

= 2817.3908 sin 20◦

= 963.6044 lb→

FBy = 2817.3908 cos 20◦

= 2647.4813 lb ↑

(74)

Solve for the bearing reaction forces using the free body diagram (Fig-

ure 15).



moment diagrams.

32

∑MAx

= 0

0 = −FB(3 in)− Cx(6 in) + FD(12 in)

Cx =−1479.1304(3 in) + 963.6044(12 in)

6 in

= 1187.6436 lb

(75a)

∑Fx = 0

0 = Ax − FB − Cx + FD

Ax = FB + Cx − FD

= 1479.1304lb + 1187.6436lb− 963.6044lb

= 1703.1696 lb

(75b)

∑MAy = 0

0 = −FB(3 in)− Cy(6 in) + FD(12 in)

Cy =−FB(3 in) + FD(12 in)

6 in

=−538.3594(3 in) + 2647.4813(12 in)

6 in

= 5025.7829 lb

(75c)

∑Fy = 0

0 = Ay − FB − Cy + FD

Ay = FB + Cy − FD

= 538.3594 lb + 5025.7829 lb− 2647.4813 lb

= 2916.661 lb

(75d)

0 2 4 6 8 10 12z (in)

0

1000

2000

3000

4000

5000

6000

T (

lb*i

n)

Torque Diagram



formulas below and the final values are summarized in Table 23


y

Mtotal =√M2

x +M2y

(76)


Figure 16 illustrates the design of the first shaft.

33

3000

2000

1000

0

1000

2000

3000

V (

lb)


0 2 4 6 8 10 12z (in)

0

2000

4000

6000

8000

10000

12000

14000

16000

M (

lb.in)

0 2 4 6 8 10 12z (in)




A 3377.5 0B 2388.8 10132.6C 2817.4 16904.3D 2817.4 0


34

Table 7 summarizes the shared parameters used for all diameter calcula-

tions. However, CS = 0.75 for larger diameters of 5.129 in. Thus the new

value for s′n = snCSCR = 18.8325.

Solving D1

At point A contains a bearing, thus sharp fillet is required, (Kt = 2.5).


Parameters


D1 =

√2.94KtV N

s′n

=

√2.94(2.5)(3377.5 lb)(3.0)

18.8325ksi

= 1.9886 in

(77)

The D′1 = 1.15D1 = 2.2869, the preferred size for D1 = 2.0000. The

preferred size for D′1 = 2.4000 in.

Solving D2

At point B, requires a profile keyseat (Kt = 2.0), and a well-rounded fillets



Parameters


35

D32 =

32N

π

√(KtM

s′n

)2

+3

4

(T

sy

)2

=32(3.0)

π

√((2.0)(10132.6lb · in)

18.8325ksi

)2

+3

4

(5916.5207 lb · in

71 ksi

)2

= 32.9563

D2 = 3.2061 in

(78)




Parameters


D′32 =32N

π

√(KtM

s′n

)2

+3

4

(T

sy

)2

=32(3.0)

π

√((2.0)(10132.6lb · in)

18.8325ksi

)2

+3

4

(0 lb · in

71 ksi

)2

= 32.8825

D′2 = 3.2037 in

(79)



Solving D4

At point C contains a bearing, thus sharp fillet is required, (Kt = 2.5).

36


Parameters


D34 =

32N

π

√(KtM

s′n

)2

+3

4

(T

sy

)2

=32(3.0)

π

√((2.0)(16904.3lb · in)

18.8325ksi

)2

+3

4

(5916.5207 lb · in

71 ksi

)2

= 54.9023

D4 = 3.8000 in

(80)

The preferred size for D′4 = 1.06D4 = 4.0280 = 4.2000 in.

Solving D3

D3 must be greater than D′4, thus D3 = 1.06 ∗D′4 = 4.452 = 4.5000 in.

Solving D5

At point D contains a bearing, thus sharp fillet is required, (Kt = 2.5) to

locate the sprocket. Moment is zero at this point.


Parameters

Stress Concentration Factor Kt = 2.0Combined Bending Moment M = 0Torque T = 5916.5207 lb · in

37

D35 =

32N

π

√(KtM

s′n

)2

+3

4

(T

sy

)2

=32(3.0)

π

√((2.0)(0lb · in)

18.8325ksi

)2

+3

4

(5916.5207 lb · in

71 ksi

)2

= 2.2053

D5 = 1.3016 in

(81)

solve for the groove, Kt = 3.0.


Parameters


D′5 =

√2.94KtV N

s′n

=

√2.94(3.0)(2817.4 lb)(3.0)

18.8325ksi

= 1.9886 in

(82)

D5 must be greater than D′5, the preferred size for D′5 = 2.0000. D5 =

1.06 ∗ D′5 = 2.1200 = 2.2000 in. D′′5 must be greater than D5, thus D′′5 =

1.15 ∗D5 = 2.5300 = 2.6000 in.

Summary

Table 30 displays all the diameters for Shaft 1. It is assumed that the bearings

for standard size diameters are available. Based on the choice of bearings may

require change in D’ diameters to prevent interference with the outer rase of

the bearing. None of the diameters exceed 2.5 in which was the assumption

for CS = 0.81.

38



D1 1.9886 2.0000D′1 2.2869 2.4000D2 3.4450 3.5000D′2 3.2061 3.2500D3 4.4520 4.5000D4 3.8000 3.8000D′4 4.0280 4.2000D5 2.1200 2.2000D′5 1.9886 2.0000D′′5 2.5300 2.6000

39

4 Conclusion

In conclusion, SAE 1040 cold-drawn steel is used for the shafts, SAE 1340

OQT 700 steel for gears A & B, and cold-drawn SAE 1340 steel for gears C &

D. The dimensions of the gears are summarizes at the end of each gear pair

design, and dimensions of the shaft are at the end of each shaft design section.

40

5 References

[1] P. Robert L. Mott, Machine Elements in Mechanical Design. Pearson,2014.

41

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