mece 4331 honors

Design and Analysis of a Mechanical Device

Compound Reverted Geartrain

MECE 4331: Honors Credit

Date of submission: 12/7/2015

Shahmeer Baweja

(1180891)

i

Abstract

This document provides design, analysis and evaluation of a compound reverted geartrain with

respect to loading, stress and safety factors to obtain specifications for gears, shafts and bearings

which satisfy the customer requirements for the desired power and torque

ii

Table of Contents Abstract ......................................................................................................................................................... i

List of Figures ............................................................................................................................................... iii

List of Tables ................................................................................................................................................ iii

List of Equations .......................................................................................................................................... iii

Introduction .................................................................................................................................................. 1

Gearbox Design Requirements .................................................................................................................... 2

Gearbox Design Specifications ..................................................................................................................... 2

Design Sequence .......................................................................................................................................... 3

Specifications ................................................................................................................................................ 5

Gear Specifications ................................................................................................................................... 5

Gear Diameter ...................................................................................................................................... 5

Gear Face Width, Strength, Material and Safety Factor ..................................................................... 9

Shaft Specifications ................................................................................................................................ 24

Shaft Layout ........................................................................................................................................ 24

Shaft Diameter and Fatigue Safety Factor ......................................................................................... 27

Bearing Specifications ............................................................................................................................ 45

Summary ..................................................................................................................................................... 48

References .................................................................................................................................................. 52

Appendix ..................................................................................................................................................... 53

iii

List of Figures Figure 1: Compound Reverted gear train ..................................................................................................... 1

Figure 2: Rough Sketch of three shafts layout ............................................................................................ 25

Figure 3: Axial dimensions of Intermediate Shaft ....................................................................................... 26

Figure 4: Free Body Diagram of Intermediate Shaft ................................................................................... 27

Figure 5: Shear Force and Bending Moment Diagram of Intermediate Shaft ............................................ 28

Figure 6: Deflection and Slope Plots of Intermediate Shaft ....................................................................... 42

Figure 7: Stress-cycle factor, 𝑍𝑛 vs. Number of load cycles, N ................................................................... 53

Figure 8: Geometry Factor, J vs. Number of teeth for which geometry factor is desired .......................... 53

Figure 9: Stress-cycle factor, 𝑌𝑛 vs. Number of load cycles, N ................................................................... 54

Figure 10: Allowable contact stress numbers, 𝑆𝑐 vs. Brinell Hardness, 𝐻𝑛 ............................................... 54

Figure 11: Notch sensitivity, q vs. Notch radius, r ...................................................................................... 55

Figure 12: Notch sensitivity, 𝑞𝑠ℎ𝑒𝑎𝑟vs. Notch radius, r ............................................................................. 55

Figure 13: 𝐾𝑡 for round shaft with shoulder fillet in bending .................................................................... 56

Figure 14: 𝐾𝑡𝑠 for round shaft with shoulder fillet in torsion .................................................................... 56

Figure 15: 𝐾𝑡𝑠 for round shaft with flat-bottom groove in torsion ............................................................ 57

List of Tables Table 1: Combined Results of Slope and Deflections of Intermediate Shaft at Points of Interest ............. 43

Table 2: Contact Strength, 𝑆𝑐 at 107cycles and 0.99 Reliability for Steel Gears ....................................... 57

Table 3: Bending Strength, 𝑆𝑐 at 107cycles and 0.99 Reliability for Steel Gears ....................................... 58

Table 4: Parameters for Marin Surface Modification Factor ...................................................................... 58

Table 5: First Iteration Estimates for Stress-Concentration Factors, 𝐾𝑡 and 𝐾𝑡𝑠 ...................................... 59

Table 6: Typical Maximum Ranges for Slopes and Transverse Deflections ................................................ 59

List of Equations Equation 1 ..................................................................................................................................................... 5

Equation 2 ................................................................................................................................................... 10

Equation 3 ................................................................................................................................................... 11

Equation 4 ................................................................................................................................................... 12

Equation 5 ................................................................................................................................................... 13

Equation 6 ................................................................................................................................................... 13

Equation 7 ................................................................................................................................................... 15

Equation 8 ................................................................................................................................................... 16

Equation 9 ................................................................................................................................................... 30

Equation 10 ................................................................................................................................................. 30

Equation 11 ................................................................................................................................................. 32

Equation 12 ................................................................................................................................................. 33

Equation 13 ................................................................................................................................................. 33

iv

Equation 14 ................................................................................................................................................. 34

Equation 15 ................................................................................................................................................. 35

Equation 16 ................................................................................................................................................. 42

Equation 17 ................................................................................................................................................. 43

Equation 18 ................................................................................................................................................. 46

1

Introduction

Many industrial applications require the use of a power source from engines or electric motors to

actuate an output in terms of motion and lead to a desired end-result such a toggling of a flip

switch due to a linear motion of a power screw produced from the rotary motion of the shaft in

phase with the motor. Most of the applications that are efficient incorporates the use of shafts in

addition to gears, bearings and belt pulleys. Moreover, the power source from the motor runs

efficiently at a narrow range of rotational speed. For the case of applications that require the

speed to be slower than the speed supplied by the motor, a speed reducer is introduced. A design

of two-stage gear reduction or a compound reverted gear train shown in Figure 1 will accomplish

the goal of reducing the speed for those applications. This speed reducer should be able to

transmit power from the source to the target application with as little as energy loss as possible

while reducing speed, and consequently increasing the torque. For this product, the design and

analysis of the intermediate shaft and its components: gears, bearings along with other shafts are

presented with specifications to satisfy the customer/design requirements

Figure 1: Compound Reverted gear train

2

Gearbox Design Requirements

The following are the requirements set forth by a potential customer or client for a two-stage

gear reduction

Power to be delivered: 20 hp

Input Speed: 1750 RPM

Output Speed: 85 RPM

Output and Input Shaft in-line

Base mounted with 4 bolts

Continuous operation

6-year life, with 8 hours/day, 5 days/week

Low maintenance

Gearbox Design Specifications

The following specifications provides an appropriate framework within the requirements set

forth by the client or customer previously

Power to be delivered: 20 hp

Power efficiency: >95%

Steady state input speed: 1750 RPM

Maximum input speed: 2400 RPM

Steady-state output speed: 82–88 RPM

Usually low shock levels, occasional moderate shock

Input and output shafts extend 4 in outside gearbox

3

Input and output shaft diameter tolerance: ±0.001in

Input and output shafts in-line: concentricity ±0.005in, alignment ±0.001rad

Maximum allowable loads on input shaft: axial, 50 lbf; transverse, 100 lbf

Maximum allowable loads on output shaft: axial, 50 lbf; transverse, 500 lbf

Maximum gearbox size: 14-in x 14-in base, 22-in height

Base mounted with 4 bolts

Mounting orientation only with base on bottom

100% duty cycle

Maintenance schedule: lubrication check every 2000 hours; change of lubrication every

8000 hours of operation; gears and bearing life >12,000hours;

Infinite shaft life; gears, bearings, and shafts replaceable

Access to check, drain, and refill lubrication without disassembly or opening of gasket

joints.

Manufacturing cost per unit: <$300

Production: 10,000 units per year

Operating temperature range: −10◦ to 120◦F

Sealed against water and dust from typical weather

Noise: <85 dB from 1 meter

Design Sequence

Design is an iterative process but there are steps which can be followed in general to make

designing easier to save time. The following steps are not to be followed strictly in the order they

are listed below

4

- Power and Torque requirements – check all the power requirements in order to

determine the sizing of the parts. Determine the speed/torque ratio from input to output

before determining the gear sizing

- Gear specification: Specify the gears with necessary gear ratios through transmitted

loads

- Shaft layout: Specify the axial locations of gears and bearings on the shaft including that

of intermediate shaft. Decide on how to transmit torque from the gears to the shaft (keys,

spline etc.) as well as how to hold the gears and bearings in place (rings, nuts)

- Force Analysis: once the gear diameters are known as well as axial locations of the gears

and bearing are known, begin analyzing the forces on the gears and bearings

- Shaft material selection: Choose suitable material for shaft since fatigue design depends

on the material

- Shaft stress analysis and specifications: (fatigue and static): Determine the stresses at

critical locations, and estimate the shaft diameter

- Shaft design for deflection – check for critical deflections at bearings and gear locations

on the shaft

- Bearing selection and specifications: Select appropriate bearings from the catalog that

will fit in with shaft diameter

- Ring and Key selection – With the shaft diameter already determine, choose appropriate

keys and rings for keep the gears and bearings in place on the shaft

- Final Analysis: Perform a final analysis of the intermediate shaft by determining the

safety factors

5

Specifications

For a successfully working design of the speed reducer conforming to the requirements set forth

by the customer/client, a set of specifications for gears, shafts and bearings are obtained through

the application of knowledge of the equations for determining the load, stress and failure

Gear Specifications

Gear Diameter

For the two-stage gear reduction, the output power will be 2%-4% less than that of the input

power, and so power is approximately constant throughout the system. Torque, on the other

hand, is not constant. For the compound reverted gear train, the power in and power out (H) are

almost equal and is given by product of torque (T) and rotational speed (w)

Equation 1

𝑯 = 𝑻𝒊𝒘𝒊 = 𝑻𝒐𝒘𝒐

For a constant power, the reduction in speed due to speed reducer will result in increase in torque

which is desired for higher efficiency.

From the design specifications, we need 𝒘𝒊 = 𝟏𝟕𝟓𝟎 𝑹𝑷𝑴 and 𝒘𝒐 = 𝟖𝟐 ~ 𝟖𝟓 𝑹𝑷𝑴

This will give 𝑻𝒊

𝑻𝒐=

𝒘𝒐

𝒘𝒊= (

𝟒𝟏

𝟖𝟕𝟓) 𝒎𝒊𝒏[𝟎. 𝟎𝟒𝟔𝟗] 𝒐𝒓 (

𝟏𝟕

𝟑𝟓𝟎)𝒎𝒂𝒙 [𝟎. 𝟎𝟒𝟖𝟔]

The gear ratio/train value for a two-stage gear reduction can achieve a value of up to 100 to 1

and is given by

𝒆 = 𝑻𝒊

𝑻𝒐

= 𝒘𝒐

𝒘𝒊

For 𝑤𝑖 = 1750 𝑅𝑃𝑀 and 𝑤𝑜 = 85 𝑅𝑃𝑀,

6

𝒆 = 𝟖𝟓

𝟏𝟕𝟓𝟎=

𝟏𝟕

𝟑𝟓𝟎=

𝟏

𝟐𝟎. 𝟓𝟗= 𝟎. 𝟎𝟒𝟖𝟔

and for this compound reverted geartrain,

𝒆 = 𝟏

𝟐𝟎. 𝟓𝟗=

𝑵𝟐

𝑵𝟑 𝑵𝟒

𝑵𝟓

The gearbox needs to be as small as possible for which the two-stage gear reduction will be the

same reduction which will satisfy the requirement of the in-line condition for both the input and

output shaft from the gearbox design specification.

𝑵𝟐

𝑵𝟑=

𝑵𝟒

𝑵𝟓= √

𝟏

𝟐𝟎. 𝟓𝟗=

𝟏

𝟒. 𝟓𝟒

The smallest number of teeth on the pinion which can exist without interference needs to be

determined. This is 𝑵𝒑 given by

𝑵𝒑 = 𝟐𝒌

(𝟏 + 𝟐𝒎) 𝐬𝐢𝐧𝟐 ∅(𝒎 + √𝒎𝟐 + (𝟏 + 𝟐𝒎) 𝐬𝐢𝐧𝟐 ∅)

where 𝒎 is the ratio of the number of teeth on the pinion, 𝑵𝒑 to the number of teeth on the gear,

𝑵𝑮 and ∅ is the pressure angle

Let 𝒎 = 𝟒 such there are 4 teeth on the pinion for every tooth on the gear.

For ∅ = 𝟐𝟎 and 𝒌 = 𝟏 for full-teeth,

𝑵𝒑 = 𝟐(𝟏)

(𝟏 + 𝟐(𝟒)) 𝐬𝐢𝐧𝟐 𝟐𝟎(𝟒 + √𝟒𝟐 + (𝟏 + 𝟐(𝟒)) 𝐬𝐢𝐧𝟐 ∅𝟐𝟎)

𝑵𝒑 = 𝟏𝟔 𝒕𝒆𝒆𝒕𝒉

7

This is the number of teeth on the pinion without interference. So 𝑵𝟐 = 𝑵𝟒 = 𝟏𝟔 𝒕𝒆𝒆𝒕𝒉

𝑵𝟑 = 𝑵𝟓 = 𝟒. 𝟓𝟒(𝟏𝟔) = 𝟕𝟐. 𝟔𝟒

Check if output speed, 𝒘𝒐 = 𝒘𝟓 is within 82-88 RPM with 𝑵𝟓 = 𝟕𝟐 𝒕𝒆𝒆𝒕𝒉 and with 𝒘𝒐 =

𝒘𝟐 = 𝟏𝟕𝟓𝟎 𝑹𝑷𝑴 as the required input

𝒘𝟓 =𝑵𝟐

𝑵𝟑

𝑵𝟒

𝑵𝟓(𝒘𝟐)

𝒘𝟓 = (𝟏𝟔

𝟕𝟐)(

𝟏𝟔

𝟕𝟐)(𝟏𝟕𝟓𝟎) = 𝟖𝟔. 𝟒𝟐 𝑹𝑷𝑴

This is acceptable!

So,

𝑵𝟐 = 𝑵𝟒 = 𝟏𝟔 𝒕𝒆𝒆𝒕𝒉

𝑵𝟑 = 𝑵𝟓 = 𝟕𝟐 𝒕𝒆𝒆𝒕𝒉

and then,

𝒘𝟒 = 𝒘𝟑 =𝑵𝟐

𝑵𝟑𝒘𝟐

𝒘𝟒 = 𝒘𝟑 =𝟏𝟔

𝟕𝟐(𝟏𝟕𝟓𝟎 𝑹𝑷𝑴)

𝒘𝟒 = 𝒘𝟑 = 𝟑𝟖𝟖. 𝟗 𝑹𝑷𝑴

For the torque,

𝑯 = 𝑻𝟐𝒘𝟐 = 𝑻𝟓𝒘𝟓

8

From the gearbox design specification, the maximum size of the gearbox needs to be 22 in., for

which the gear tooth size should be maximum which is also the minimal diametral pitch.

The overall height of the gearbox is given by:

where 2/P is the addendum distances for gears 2 and 5

The pitch diameter, 𝒅 is given by 𝒅 =𝑵

𝑷 where P = diametral pitch and N = number of teeth.

Then substituting 𝑵

𝑷 for 𝒅, the following gearbox height is given by:

Solving for diametral pitch, P:

Allowing 1.5 in. for clearances and wall thicknesses, the minimum diametral pitch, P is given

by:

With P = 6 teeth/in as approximate, the following diameter for gears 2, 3, 4 and 5 are:

9

Answer

Gear Face Width, Strength, Material and Safety Factor

With the gear diameters specified, the pitch-line velocity, V and transmitted load, W between

gears 2 and 3, and gears 4 and 5 are given by:

The speed ratio, 𝒎𝑮 is defined as the ratio of number of teeth on gear, 𝑵𝑮 to the number of teeth

on the pinion, 𝑵𝒑 and is given by:

𝒎𝑮 =𝑵𝑮

𝑵𝑷= 𝟒. 𝟓

where 𝑵𝑮 = 𝟕𝟐 𝒕𝒆𝒆𝒕𝒉 and 𝑵𝑷 = 𝟏𝟔 𝒕𝒆𝒆𝒕𝒉

And the compound reverted gear train is a spur gear for which the load-sharing ratio, 𝒎𝑵 = 𝟏

Now for pressure angle, ∅𝒕 = 𝟐𝟎°, the geometry factor, I for all gears which are external is given

by:

10

𝑰 =𝒄𝒐𝒔∅𝒕𝒔𝒊𝒏∅𝒕

𝟐𝒎𝑵

𝒎𝑮

𝒎𝑮 + 𝟏

With pitch-line velocity and transmitted loads obtained for gears 2, 3, 4 and 5, each of the gears

needs to be analyzed for loads, stresses and failures to obtain specifications for face width,

endurance strength, bending strength, material type and safety factors.

Gear 4

Gear 4 Wear

The dynamic factor, 𝐾𝑣 is given by

Equation 2

𝐾𝑣 =𝐴 + √𝑉

𝐴

where

𝑽 = 𝒑𝒊𝒕𝒄𝒉 − 𝒍𝒊𝒏𝒆 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚

The gears need to be of the highest quality so a value for quality number, 𝑸𝒗 = 𝟕 is assumed

Then, A and B are given by:

𝑩 = 𝟎. 𝟕𝟑𝟏

11

𝑨 = 𝟔𝟓. 𝟏

and for gear 4, 𝑽 = 𝑽𝟒𝟓 = 𝟐𝟕𝟏. 𝟓 𝒇𝒕/𝒎𝒊𝒏, then 𝑲𝒗 is given by

The circular pitch, p is given by ratio of 𝝅 to the diametral pitch, P as

𝒑 =𝝅

𝑷

The face width, F is typically 3-5 times the circular pitch, p.

Trying with 4 times the circular pitch, F is given by

𝑭 = 𝟒 (𝝅

𝑷) = 𝟒 (

𝝅

𝟔) = 𝟐. 𝟎𝟗 𝒊𝒏.

Now verify, if this is a good face width for gear 4 with pitch diameter, 𝒅 = 𝟐. 𝟔𝟕 𝒊𝒏. and

diametral pitch, 𝑷 = 𝟔 𝒕𝒆𝒆𝒕𝒉/𝒊𝒏

Entering the above values on globalspec.com, the face width, F for several spur gears in stock

are found to be 1.5 in. or 2.0 in.

Let F = 2.0 in Answer

The load distribution factor, 𝐊𝐦 is given by

Equation 3

where

12

𝑪𝒎𝒇 = 𝒇𝒂𝒄𝒆 𝒍𝒂𝒐𝒅 − 𝒅𝒊𝒔𝒕𝒓𝒊𝒃𝒖𝒕𝒊𝒐𝒏 𝒇𝒂𝒄𝒕𝒐𝒓

𝑪𝒎𝒄 = 𝒍𝒐𝒂𝒅 𝒄𝒐𝒓𝒓𝒆𝒄𝒕𝒊𝒐𝒏 𝒇𝒂𝒄𝒕𝒐𝒓

𝑪𝒑𝒇 = 𝒑𝒊𝒏𝒊𝒐𝒏 𝒑𝒓𝒐𝒑𝒐𝒓𝒕𝒊𝒐𝒏 𝒇𝒂𝒄𝒕𝒐𝒓

𝑪𝒎𝒂 = 𝒎𝒆𝒔𝒉 𝒂𝒍𝒊𝒈𝒏𝒎𝒆𝒏𝒕 𝒇𝒂𝒄𝒕𝒐𝒓

The 𝑪𝒑𝒇 is given by

Equation 4

𝑪𝒑𝒇 =𝑭

𝟏𝟎𝒅− 𝟎. 𝟎𝟑𝟕𝟓 + 𝟎. 𝟎𝟏𝟐𝟓𝑭

where F = face width and d = gear diameter

With F = 2 in. and d = 2.67 in., 𝑪𝒑𝒇 is given by:

𝑪𝒑𝒇 =𝟐

𝟏𝟎(𝟐. 𝟔𝟕)− 𝟎. 𝟎𝟑𝟕𝟓 + 𝟎. 𝟎𝟏𝟐𝟓(𝟐)

𝑪𝒑𝒇 = 𝟎. 𝟎𝟔𝟐𝟒

and

𝑪𝒆 = 𝟏 (All other conditions)

Then, the load distribution factor, 𝑲𝒎 is given by

𝑲𝒎 = 𝟏. 𝟐𝟏

13

The contact stress, 𝝈𝒄 for gears is given by

Equation 5

The basic material for gear 4 will be steel for which elastic coefficient, 𝑪𝒑 = 𝟐𝟑𝟎𝟎

There is no detrimental surface finish effect for which 𝑪𝒇 = 𝟏

No overloading for which 𝑲𝒐 = 𝟏

No detrimental size effect for which 𝑲𝒔 = 𝟏

Now, for gear 4 diameter, 𝒅𝒑 = 𝟐. 𝟔𝟕 𝒊𝒏. , transmitted load, 𝑾𝟒𝟓𝒕 = 𝟐𝟒𝟑𝟏 𝒍𝒃𝒇 and geometry

factor, 𝑰 = 𝟎. 𝟏𝟑𝟏𝟓, the contact stress, 𝝈𝒄 for gear 4 is given by:

The allowable contact stress, 𝜎𝑐,𝑎𝑙𝑙 is given by

Equation 6

The gear strength, 𝑺𝒄 = 𝒆𝒏𝒅𝒖𝒓𝒂𝒏𝒄𝒆 𝒔𝒕𝒓𝒆𝒏𝒈𝒕𝒉 is based upon a reliability, R of 99% for which

the reliability factor, 𝑲𝑹 = 𝟏

14

From the design specification, the operating temperature is −10◦ to 120◦F, for the which the

temperature factor, 𝑲𝑻 = 𝟏

For gear life of 12,000 hours and a speed of 𝒘𝟒 = 𝟑𝟖𝟖. 𝟗 𝑹𝑷𝑴,

the life in revolutions, L is given by:

𝑳 = 𝟔𝟎 ∗ 𝒉𝒐𝒖𝒓𝒔 ∗ 𝒔𝒑𝒆𝒆𝒅

𝑳 = 𝟔𝟎 ∗ 𝟏𝟐𝟎𝟎𝟎 ∗ 𝟑𝟖𝟖. 𝟗

𝑳 = 𝟐. 𝟖 ∗ 𝟏𝟎𝟖 𝒓𝒆𝒗

From Figure 7 in appendix, the stress-cycle factor for wear, 𝒁𝒏 = 𝟎. 𝟗 for 𝟏𝟎𝟖𝒄𝒚𝒄𝒍𝒆𝒔

For design factor, 𝒏𝒅 = 𝟏. 𝟐 against wear

And AGMA factor of safety or stress ratio, 𝑺𝑯 = 𝒏𝒅 = 𝟏. 𝟐,

For gear 4,

𝝈𝒄,𝒂𝒍𝒍 = 𝝈𝒄

Endurance strength, 𝑺𝒄 is then given by:

From Table 2 in appendix, this strength is achievable with Grade 2 carburized and hardened

with 𝑺𝒄 = 𝟐𝟐𝟓𝟎𝟎𝟎 𝒑𝒔𝒊 Answer

Now, factor of safety, 𝒏𝒄 for wear is given by

15

Answer

Gear 4 Bending

Number of teeth on gear 4, 𝑵𝟒 = 𝟏𝟔 𝒕𝒆𝒆𝒕𝒉 for which, from Figure 8 in appendix, geometry

factor, J = 0.27

Then,

the bending stress, 𝝈 is given by

Equation 7

𝝈 = 𝑾𝒕𝑲𝒗

𝑷𝒅

𝑭

𝑲𝒎

𝑱

where 𝑾𝒕 = 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒, 𝑲𝒗 = 𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑓𝑎𝑐𝑡𝑜𝑟, 𝑷𝒅 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑟𝑎𝑙 𝑝𝑖𝑡𝑐ℎ, 𝑲𝒎 =

𝑙𝑜𝑎𝑑 − 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟, 𝑭 = 𝑓𝑎𝑐𝑒𝑤𝑖𝑑𝑡ℎ, 𝑱 = 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑦 𝑓𝑎𝑐𝑡𝑜𝑟,

Now, for gear 4 diameter, 𝑷𝒅 = 𝟔 𝒊𝒏. , transmitted load, 𝑾𝟒𝟓𝒕 = 𝟐𝟒𝟑𝟏 𝒍𝒃𝒇 , F = 2 in. , and

𝑲𝒎 = 𝟏. 𝟐𝟏, 𝑲𝒗 = 𝟏. 𝟏𝟖 and geometry factor, 𝑱 = 𝟎. 𝟐𝟕,

the bending stress for gear 4 is given by Equation 7 is:

From Figure 9 in appendix, the stress-cycle factor for bending, 𝒀𝑵 = 𝟎. 𝟗 for 𝟏𝟎𝟖𝒄𝒚𝒄𝒍𝒆𝒔

16

Now using Grade 2 carburized and hardened as before, the bending strength, from Table 3, is

given by 𝑺𝒕 = 𝟔𝟓𝟎𝟎𝟎 𝒑𝒔𝒊 Answer

Assume that bending factor of safety, 𝑆𝐹 = 1 and that 𝐾𝑇 and 𝐾𝑅 = 1 as before

Then, allowable bending stress is given by

Equation 8

Now, factor of safety for bending is given by

Answer

Gear 4 specification is

and

Wear factor of safety, 𝒏𝒄 = 𝟏. 𝟐𝟓 and bending factor of safety, 𝒏 = 𝟏. 𝟓𝟐

17

Gear 5

Gear 5 bending and wear

Everything is the same for Gear 5 as Gear 4 except a few things

Number of teeth for gear 5, 𝑵𝟓 = 𝟕𝟐 𝒕𝒆𝒆𝒕𝒉 for which, from Figure 8 in appendix, geometry

factor, J = 0.41

Also the speed of gear 5 is different which is 𝒘𝟓 = 𝟖𝟔. 𝟒 𝑹𝑷𝑴

From the speed, the life in revolutions, L of gear 5 is given by:


𝑳 = 𝟔𝟎 ∗ 𝟏𝟐𝟎𝟎𝟎 ∗ 𝟖𝟔. 𝟒

𝑳 = 𝟔. 𝟐 ∗ 𝟏𝟎𝟕 𝒓𝒆𝒗

From Figure 7 and Figure 9 for 𝟏𝟎𝟕 𝒄𝒚𝒄𝒍𝒆𝒔

𝒀𝑵 = 𝒁𝑵 = 𝟏

The contact stress, 𝝈𝒄 from gear 5 is same as from gear 4 since they are in contact:

𝝈𝒄 = 𝟏𝟔𝟏𝟕𝟎𝟎 𝒑𝒔𝒊

Now for same design factor, 𝑺𝑯 = 𝒏𝒅 = 𝟏. 𝟐,


𝑺𝒄 =𝑺𝑯𝝈𝒄

𝒁𝒏

𝑺𝒄 =(𝟏. 𝟐)(𝟏𝟔𝟏𝟕𝟎𝟎)

𝟏

𝑺𝒄 = 𝟏𝟗𝟒, 𝟎𝟒𝟎 𝒑𝒔𝒊

18

From Table 2 in appendix, this strength is achievable with grade 2 carburized and hardened

with 𝑺𝒄 = 𝟐𝟐𝟓𝟎𝟎𝟎 𝒑𝒔𝒊 Answer

So, factor of safety for wear is

Answer

Now for bending, with J = 0.41 instead of J = 0.27 and with facewidth, F = 2 in. , Answer

and with 𝑷𝒅 = 𝟔 𝒊𝒏. , 𝑾𝟒𝟓𝒕 = 𝟐𝟒𝟑𝟏 𝒍𝒃𝒇 , F = 2 in. , 𝑲𝒎 = 𝟏. 𝟐𝟏, and 𝑲𝒗 = 𝟏. 𝟏𝟖 same as for

gear 4,

the bending stress on gear 5 is now given by

And so the corresponding factor of safety for bending is now given by

Answer

Gear 5 specification

and

Wear factor of safety, 𝒏𝒄 = 𝟏. 𝟑𝟗 and bending factor of safety, 𝒏 = 𝟐. 𝟒𝟖

19

Gear 2

Now like the similarity between gear 4 and gear 5, there is similarity between gear 2 and gear 3

Gear 2 wear

The pitch-line velocity, 𝑽𝟐𝟑 for gear 2 is 1223 ft/min, for which the dynamic factor, 𝑲𝒗 is given

by Equation 2

𝑲𝒗 =𝟔𝟓. 𝟏 + √𝟏𝟐𝟐𝟑

𝟔𝟓. 𝟏

𝑲𝒗 = 𝟏. 𝟑𝟕

Since the transmitted load of 𝑾𝟐𝟑𝒕 = 𝟓𝟒𝟎 𝒍𝒃𝒇 from gear 2 (and gear 3) is less than that from

gears 4 and 5, the facewidth, F needs to be less than 2 in.

Let F = 1.5 in. Answer

With the new facewidth, 𝑪𝒑𝒇 from Equation 4 is now:

𝑪𝒑𝒇 = 𝟎. 𝟎𝟒𝟑𝟕

Then from Equation 3, the corresponding load distribution factor, 𝑲𝒎 = 𝟏. 𝟏𝟗

With, 𝑾𝟐𝟑𝒕 = 𝟓𝟑𝟗. 𝟕 𝒍𝒃𝒇, 𝑲𝒗 = 𝟏. 𝟑𝟕, F = 1.5 in. , 𝑲𝒎 = 𝟏. 𝟏𝟗 and 𝒅𝒑 = 𝟐. 𝟔𝟕 𝒊𝒏. and

𝑰 = 𝟎. 𝟏𝟑𝟏𝟓

the contact stress, 𝝈𝒄 for gear 2 from Equation 5 is given by:

20

The life in revolution, L for gear 2 with 𝒘𝟐 = 𝟏𝟕𝟓𝟎 𝑹𝑷𝑴 is given by


𝑳 = 𝟔𝟎 ∗ 𝟏𝟐𝟎𝟎𝟎 ∗ 𝟏𝟕𝟓𝟎

𝑳 = 𝟏. 𝟐𝟔 ∗ 𝟏𝟎𝟗 𝒓𝒆𝒗

From Figure 7 in appendix, the stress-cycle factor for wear, 𝒁𝒏 = 𝟎. 𝟖 for 𝟏𝟎𝟗𝒄𝒚𝒄𝒍𝒆𝒔




𝒁𝒏

𝑺𝒄 =(𝟏. 𝟐)(𝟗𝟒𝟎𝟎𝟎)

𝟎. 𝟗

𝑺𝒄 = 𝟏𝟐𝟓, 𝟑𝟑𝟑. 𝟑𝟑 𝒑𝒔𝒊

From Table 2 in appendix, this strength is achievable with grade 1 flame hardened

with 𝑺𝒄 = 𝟏𝟕𝟎, 𝟎𝟎𝟎 𝒑𝒔𝒊 Answer

Factor of safety for wear is now

Answer

Gear 2 bending

Number of teeth on gear 2, 𝑵𝟐 = 𝟏𝟔 for which, from Figure 8 in appendix below, geometry

factor, J = 0.27 same as gear 4

21

And so from Equation 7, bending stress with, 𝑾𝟐𝟑𝒕 = 𝟓𝟑𝟗. 𝟕 𝒍𝒃𝒇, 𝑲𝒗 = 𝟏. 𝟑𝟕, 𝑷𝒅 = 𝟔 𝒊𝒏. , t, F

= 1.5 in. , and 𝑲𝒎 = 𝟏. 𝟏𝟗, and geometry factor, 𝑱 = 𝟎. 𝟐𝟕 is given by:

From Figure 9 in appendix, the stress-cycle factor for bending, 𝒀𝑵 = 𝟎. 𝟖𝟖 for 𝟏𝟎𝟗𝒄𝒚𝒄𝒍𝒆𝒔

Now using grade 1 flame hardened with as before, the bending strength, from Table 3 is

𝑺𝒕 = 𝟒𝟓𝟎𝟎𝟎 𝒑𝒔𝒊 Answer

Assume that bending factor of safety, 𝑆𝐹 = 1 and that 𝐾𝑇 and 𝐾𝑅 = 1 as before

Then, allowable bending stress is given by

𝜎𝑎𝑙𝑙 = 𝑆𝑡𝑌𝑁

𝜎𝑎𝑙𝑙 = (45000)(0.88)

𝜎𝑎𝑙𝑙 = 39600 𝑝𝑠𝑖

Now factor of safety for bending is

Answer


and

22

Wear factor of safety, 𝒏𝒄 = 𝟏. 𝟒𝟎 and bending factor of safety, 𝒏 = 𝟑. 𝟎𝟒

Gear 3

Gear 3 bending and wear

Everything is the same for Gear 3 as Gear 2 except a few things

Number of teeth for gear 3, 𝑵𝟑 = 𝟕𝟐 𝒕𝒆𝒆𝒕𝒉 for which, from Figure 8 in appendix, geometry

factor, J = 0.41

Also the speed of gear 3 is different which is 𝒘𝟑 = 𝟑𝟖𝟖. 𝟗 𝑹𝑷𝑴

From the speed, the life in revolutions, L of gear 3 is


𝑳 = 𝟔𝟎 ∗ 𝟏𝟐𝟎𝟎𝟎 ∗ 𝟑𝟖𝟖. 𝟗

𝑳 = 𝟐. 𝟖 ∗ 𝟏𝟎𝟖 𝒓𝒆𝒗

For 108 𝑐𝑦𝑐𝑙𝑒𝑠 from Figures A and C in appendix

𝒀𝑵 = 𝒁𝑵 = 𝟎. 𝟗

The contact stress, 𝝈𝒄 from gear 3 is same as gear 2 since they are in contact:

𝝈𝒄 = 𝟗𝟒𝟎𝟎𝟎 𝒑𝒔𝒊


Endurance strength, 𝑺𝒄 is then given by

23


𝒁𝒏

𝑺𝒄 =(𝟏. 𝟐)(𝟗𝟒𝟎𝟎𝟎)

𝟎. 𝟗

𝑺𝒄 = 𝟏𝟐𝟓, 𝟑𝟑𝟑. 𝟑𝟑 𝒑𝒔𝒊

From Table 2 and Figure 10 in appendix, this strength is achievable with grade 1 through

hardened with 𝑺𝒄 = 𝟏𝟐𝟔, 𝟎𝟎𝟎 𝒑𝒔𝒊 and 𝑺𝒕 = 𝟑𝟔𝟎𝟎𝟎 𝒑𝒔𝒊 with hardness of 300 𝑯𝑩 Answer

Now with, 𝝈𝒄,𝒂𝒍𝒍 = 𝑺𝒄𝒁𝒏 = (𝟏𝟐𝟔𝟎𝟎𝟎)(𝟎. 𝟗) = 𝟏𝟏𝟑𝟒𝟎𝟎 𝒑𝒔𝒊

The factor of safety for wear is

𝒏𝒄,𝒂𝒍𝒍 = 𝝈𝒄,𝒂𝒍𝒍

𝝈𝒄

𝒏𝒄,𝒂𝒍𝒍 = 𝟏. 𝟐𝟏 Answer

Now for bending, note that due to J = 0.27 instead of J = 0.41, and from Equation 7 with 𝑲𝒗 =

𝟏. 𝟑𝟕, facewidth, F=1.5, 𝑲𝒎 = 𝟏. 𝟏𝟗 and 𝑾𝟐𝟑𝒕 = 𝟓𝟒𝟎 𝒍𝒃𝒇, bending stress on gear 3 is now

given by

𝝈 =(𝟓𝟒𝟎 )(𝟏. 𝟑𝟕)(𝟔)(𝟏. 𝟏𝟗)

(𝟏. 𝟓)(𝟎. 𝟐𝟕)

𝝈 = 𝟖𝟓𝟖𝟑 𝒑𝒔𝒊

And so the corresponding factor of safety for bending is now given by

Answer

24

Gear 3 specifications

and

Wear factor of safety, 𝒏𝒄 = 𝟏. 𝟐𝟏 and bending factor of safety, 𝒏 = 𝟑. 𝟕𝟕

Shaft Specifications

We will need layout of shafts, including axial locations of gears and bearings in order to move on

to analyzing the forces on the shaft. The force analysis depends not only on the shaft diameters

but also on the axial distances between gears and bearing. These axial distances should be

sufficiently small so as to reduce the possibly of large bending moments even with a small force

applied. This also applies to ensuring that deflections are kept small since they depend on length

terms raised to the third power.

Shaft Layout

With the diameters of gears found, an estimate of the shafts lengths and the distances between

the gears are estimated on a rough sketch shown in Figure 2 below based on the design

specifications. All three shaft are shown and at this point. At this point, bearing widths are

guessed. The bearings and the gears are placed against the shoulders of the shaft on both sides

with little spacing between them. From the figure, the intermediate shaft length is estimated to be

25

11.5 in. in accordance with the maximum width of the gearbox being 14 in. from the gearbox

design specifications

Figure 2: Rough Sketch of three shafts layout

The intermediate shaft that connect spur gear 3 and 4 is considered below in Figure 3 where the

axial dimensions and the general layout have been proposed

26

Figure 3: Axial dimensions of Intermediate Shaft

The transmitted forces from gears 2 and 3, and from gears 4 and 5 was found previously to be

These forces are in tangential direction and there is a second component in radial directions

which needs to be determined

For pressure angle, ∅𝒕 = 𝟐𝟎°,

the radial forces are given by

𝑭𝟐𝟑𝒓 = 𝟓𝟒𝟎 𝐭𝐚𝐧(𝟐𝟎°) = 𝟏𝟗𝟕 𝒍𝒃𝒇

𝑭𝟒𝟓𝒓 = 𝟐𝟒𝟑𝟏 𝐭𝐚𝐧(𝟐𝟎°) = 𝟖𝟖𝟓 𝒍𝒃𝒇

27

Hence,

the transmitted forces from the gears in radial direction are given by:

With the transmitted forces known, all three shafts need to be analyzes for loads, stresses and

failures to obtain specifications for shaft diameters at different sections as well as fatigue safety

factors

For this, the focus is on the intermediate shaft connecting gears 3 and 4

Shaft Diameter and Fatigue Safety Factor

Figure 6 below shows the free body diagram of the intermediate shaft showing the reaction

forces and transmitted forces (both radial and tangent)

Figure 4: Free Body Diagram of Intermediate Shaft

28

From statics, the sum of the forces in the y and z directions are equal to zero and the sum of

moments about any of the points are equal to zero. Using this knowledge, the following reaction

forces at A and B are obtained as follows:

From statics, using the reactions forces and transmitted forces the following shear force and

bending moments diagrams are plotted in Fig 7. The total bending moment, 𝑀𝑡𝑜𝑡 is shown on the

last plot in this figure.

Figure 5: Shear Force and Bending Moment Diagram of Intermediate Shaft

29

The torque in the shaft between the gears 3 and 4 is calculated as

From Figure 5, at point I, the following bending moments and torque are:

Bending moment amplitude (max), 𝑴𝒂 = 𝟑𝟔𝟓𝟏 𝒍𝒃𝒇 ∙ 𝒊𝒏

Constant/midrange torque at 𝑻𝒎 = 𝟑𝟐𝟒𝟎 𝒍𝒃𝒇

Midrange bending moment, 𝑴𝒎 = 𝟎

Maximum torque, 𝑻𝒂 = 𝟎

A suitable material selected for the shaft is AISI 1020 CD steel. For this material, the ultimate

tensile strength is 𝑺𝒖𝒕 = 𝟔𝟖 𝒌𝒑𝒔𝒊

From Table 4 in appendix, the surface factor, 𝒌𝒂 for cold-drawn (CD) steel is

Since the shaft diameters are not known yet, a value of 0.9 for size factor, 𝒌𝒃 is assumed

Since bending moment is greater than torque, loading factor, 𝒌𝒄 = 𝟏

No rotating beam endurance limit is known as room temperature, 𝑆𝑇 so temperature factor

𝒌𝒅 = 𝟏

Assume 50% reliability for which reliability factor, 𝒌𝒆 = 𝟏

For 𝑆𝑢𝑡 ≤ 200 𝑘𝑝𝑠𝑖, the rotary-beam test specimen modification factor, 𝑺𝒆′ is given by

𝑺𝒆′ = 𝟎. 𝟓 ∗ 𝑺𝒖𝒕

30

𝑺𝒆′ = 𝟎. 𝟓 ∗ 𝟔𝟖 = 𝟑𝟒 𝐤𝐩𝐬𝐢

Now, the endurance limit, 𝑆𝑒 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦

Equation 9

𝑺𝒆 = (𝟎. 𝟖𝟖𝟑)(𝟎. 𝟗)(𝟏)(𝟏)(𝟏)(𝟑𝟒)

𝑺𝒆 = 𝟐𝟕 𝒌𝒑𝒔𝒊

A well-rounded shoulder fillet is assumed to be present at location I in Figure 4

Following this, from Table 5 in appendix, the stress concentration factors are: 𝒌𝒕 = 𝟏. 𝟕

(bending) and 𝒌 = 𝟏. 𝟓 (torsion)

For simplicity for now, assume that the shaft is notch-free such that 𝒌𝒇 = 𝒌𝒕 and 𝒌𝒇𝒔 = 𝒌𝒕𝒔

Now, for the estimation of the shaft diameter, 𝐷4 at point I in Figure 4, the DE Goodman

criterion is used which is good for initial design

Equation 10

With an minimum factor of safety, 𝒏 = 𝟏. 𝟓,

31

This value of d = 1.65 in. is an estimate so now, check with d = 1.625 in.

A typical 𝑫 𝒅⁄ ratio for a support at a shoulder is 𝑫

𝒅= 𝟏. 𝟐

So nominal diameter, 𝑫 = 𝟏. 𝟐(𝟏. 𝟔𝟐𝟓) = 𝟏. 𝟗𝟓 𝒊𝒏.

Nominal diameter, D of 2.0 in. can be used

Hence, without taking shaft deflections into account, the following shaft diameter for sections 3,

4, and 5 were obtained as

𝑫𝟒 = 𝟐. 𝟎 𝒊𝒏. and 𝒅 = 𝑫𝟓 = 𝑫𝟑 = 𝟏. 𝟔𝟐𝟓 𝒊𝒏. Answer

The new 𝐷 𝑑⁄ ratio is now given by

𝑫

𝒅=

𝟐. 𝟎

𝟏. 𝟔𝟐𝟓= 𝟏. 𝟐𝟑

From Table 5 in appendix for this well-rounded shoulder fillet

𝒓𝒅⁄ = 𝟎. 𝟏

With d = 1.625 in., fillet radius is 𝒓 ≅ 𝟎. 𝟏𝟔 𝒊𝒏.

With 𝑺𝒖𝒕 = 𝟔𝟖 𝒌𝒑𝒔𝒊, r = 0.16 in. ,

32

from Figure 11 in appendix, notch sensitivity, q = 0.82 and from Figure 12 in appendix, notch

sensitivity shear, 𝒒𝒔𝒉𝒆𝒂𝒓 = 𝟎. 𝟖𝟓

With 𝑫

𝒅= 𝟏. 𝟐𝟑 and 𝒓 𝒅⁄ = 𝟎. 𝟏

From Figure 13 in appendix, 𝑲𝒕 = 𝟏. 𝟔

From Figure 14 in appendix, 𝑲𝒕𝒔 = 𝟏. 𝟑𝟓

So now,

The fatigue stress-concentration factor from bending, 𝑲𝒇 is given by

Equation 11

The fatigue stress-concentration factor from torsion, 𝑲𝒇𝒔 is given by

Now, let’s evaluate the endurance strength, 𝑺𝒆

𝒌𝒂 = 𝟎. 𝟖𝟖𝟑 (Same as before)

Since d = 1.625 in. is between 0.11 in. and 2 in.,

𝒌𝒃 = 𝟎. 𝟖𝟕𝟗𝒅−𝟎.𝟏𝟎𝟕

33

𝒌𝒃 = 0.835

Now, from Equation 9

𝑺𝒆 = (𝟎. 𝟖𝟖𝟑)(𝟎. 𝟖𝟑𝟓)(𝟏)(𝟏)(𝟏)(𝟑𝟒)

𝑺𝒆 = 𝟐𝟓. 𝟏 𝒌𝒑𝒔𝒊

The effective von Mises stress, 𝝈′ at a given point is given by

For stress amplitude, 𝜎𝑎′

Equation 12

With 𝑇𝑎 = 0 at point I, 𝜎𝑎′ is given by

For midrange stress, 𝝈𝒎′

Equation 13

With 𝑀𝑚 = 0 at point I, 𝜎𝑚′ is given by:

Now the fatigue failure criteria for the modified Goodman line is given by

34

Equation 14

𝟏

𝒏=

𝟏𝟐𝟗𝟏𝟎

𝟐𝟓𝟏𝟎𝟎+

𝟖𝟔𝟓𝟗

𝟔𝟖𝟎𝟎𝟎= 𝟎. 𝟔𝟒𝟐

𝒏 = 𝟏. 𝟓𝟔 (Fatigue safety of factor) Answer

Check for yielding

Stress amplitude, 𝜎𝑎

𝝈𝒂 =(𝟏. 𝟒𝟗)(𝟑𝟐)(𝟑𝟔𝟓𝟏)

𝝅(𝟏. 𝟔𝟐𝟓)𝟑

𝝈𝒂 = 𝟏𝟐, 𝟗𝟏𝟑. 𝟑𝟑 𝒑𝒔𝒊

Midrange Torsion, 𝜏𝑚

𝝉𝒎 =(𝟏. 𝟑𝟎)(𝟏𝟔)(𝟑𝟐𝟒𝟎)

𝝅(𝟏. 𝟔𝟐𝟓)𝟑

𝝉𝒎 = 𝟒, 𝟗𝟗𝟗. 𝟏𝟖 𝒑𝒔𝒊

and at point I,

𝝈𝒎 = 𝝉𝒂 = 𝟎

Combined maximum von Mises stress, 𝜎𝑚𝑎𝑥′ is given by

35

Equation 15

𝝈𝒎𝒂𝒙′ = [(𝟎 + 𝟏𝟐, 𝟗𝟏𝟑. 𝟑𝟑)𝟐 + 𝟑(𝟒, 𝟗𝟗𝟗. 𝟏𝟖 + 𝟎)𝟐]

𝟏𝟐⁄

𝝈𝒎𝒂𝒙′ = 𝟏𝟓, 𝟓𝟒𝟕. 𝟔𝟓 𝒑𝒔𝒊

Now check if the sum of 𝝈𝒂, + 𝝈𝒎

′ is greater than 𝝈𝒎𝒂𝒙′

𝝈𝒂, + 𝝈𝒎

′ = 𝟏𝟐𝟗𝟏𝟎 + 𝟖𝟔𝟓𝟗 = 𝟐𝟏, 𝟓𝟔𝟗 𝒑𝒔𝒊 ≥ 𝟏𝟓, 𝟓𝟒𝟕. 𝟓𝟔 𝒑𝒔𝒊 ≥ 𝝈𝒎𝒂𝒙′

Hence, there will be no yielding

Also check with yielding factor of safety, 𝑛𝑓

For AISI 1020 CD steel, yield strength, 𝑺𝒚 = 𝟓𝟕 𝒌𝒑𝒔𝒊

𝒏𝒇 =𝑺𝒚

𝝈𝒎𝒂𝒙′

=𝟓𝟕𝟎𝟎𝟎

𝟐𝟏𝟓𝟔𝟗= 𝟐. 𝟔𝟒 > 𝟏

This confirms there will be no yielding since 𝒏𝒇 > 𝟏

Now we move on to analysis of the components that are on the intermediate shaft, namely keys

and retaining rings. The keys or keyways help gear transmit the torque from the shaft. The gear

and bearings are held in place by retaining rings and supported by the shoulders of the shaft.

These will help determine the shaft diameters at other sections namely 𝑫𝟏 = 𝑫𝟕 and 𝑫𝟐 = 𝑫𝟔

Focus on the keyway to the right of point I, that is, between the intermediate shaft and gear 4.

Estimate, from the shear force and bending moment diagrams from Figure 5, the bending

moment in the key just to the right of point I in Figure 4 to be 𝑴𝒂 = 𝟑𝟕𝟓𝟎 𝒍𝒃 ∙ 𝒊𝒏 while 𝑻𝒎 =

𝟑𝟐𝟒𝟎 𝒍𝒃 ∙ 𝒊𝒏 as before

36

Assume that at the bottom of the keyway, the radius will be r = 0.02d = 0.02(1.625) = 0.0325 in.

With 𝑫

𝒅= 𝟏. 𝟐𝟑 and 𝒓 𝒅⁄ = 𝟎. 𝟎𝟐,

from Figure 13 and H in appendix

and with 𝑺𝒖𝒕 = 𝟔𝟖 𝒌𝒑𝒔𝒊, r = 0.0325 in. ,

from Figure 11 in appendix, notch sensitivity, q = 0.65 and from Figure 12 in appendix, notch

sensitivity shear, 𝒒𝒔𝒉𝒆𝒂𝒓 = 𝟎. 𝟕𝟏

So now, as before with the shoulder fillet, this time it is the keyway at its bottom just to the right

of I, the fatigue factor of safety is

From Equation 12, with 𝑻𝒂 = 𝟎 and 𝑴𝒂 = 𝟑𝟕𝟓𝟎 𝒍𝒃 ∙ 𝒊𝒏 at point I, 𝝈𝒂′ is given by

From Equation 12, with 𝑴𝒎 = 𝟎 and 𝑻𝒎 = 𝟑𝟐𝟒𝟎 𝒍𝒃 ∙ 𝒊𝒏 at point I, 𝝈𝒎′ is given by

Now the fatigue failure criteria for the modified Goodman line given by Equation 14 is:

37

But this fatigue factor of safety to the right of point I is not high. It is closer to 1 so the keyway

turns out to be more critical compared to the shoulder. The best thing is to increase the diameter

at the end of this keyway or use a material of a higher strength

Let’s try with a higher strength material, AISI 1050 CD steel with 𝑺𝒖𝒕 = 𝟏𝟎𝟎 𝒌𝒑𝒔𝒊

Now recalculate everything as before

From Table 4 in appendix, surface factor, 𝒌𝒂

From Equation 9, endurance strength, 𝑺𝒆 with 𝒌𝒃 = 0.835

With 𝑺𝒖𝒕 = 𝟏𝟎𝟎 𝒌𝒑𝒔𝒊, r = = 0.0325 in.,

from Figure 11 in appendix, notch sensitivity, q = 0.72

With 𝑫

𝒅= 𝟏. 𝟐𝟑 and 𝒓 𝒅⁄ = 𝟎. 𝟎𝟐 ,

and from Figure 13 in appendix, 𝑲𝒕 = 𝟐. 𝟏𝟒. , 𝑲𝒇 is given by

Next, with 𝑴𝒂 = 𝟑𝟕𝟓𝟎 𝒍𝒃 ∙ 𝒊𝒏, 𝝈𝒂′ is given by

38

Now the fatigue failure criteria for the modified Goodman line given by Equation 14 is:

Answer

Now let’s shift focus to the groove at point K in Figure 4,

From shear force and bending moment diagrams in Figure 5, there is no torque at K so 𝑇𝑎 = 0

And at this point K,

To check if this location of K is potentially critical, use 𝑲𝒕 = 𝑲𝒇 = 𝟓. 𝟎 as an estimate

Then,

The fatigue factor at point K on the shaft at the groove is now given by

39

But this fatigue safety factor is still very low i.e. very close to 1. Let’s look for a specific

retaining ring to obtain 𝑲𝒇 more accurately. From globalspec.com, the groove specifications for

a retaining ring selection for a shaft diameter of 1.625 are obtained as follows,

,

Now from Figure 15 with 𝒓 𝒕⁄ = 𝟎. 𝟎𝟏𝟎. 𝟎𝟒𝟖⁄ = 𝟎. 𝟐𝟎𝟖 and 𝒂 𝒕⁄ = 𝟎. 𝟎𝟔𝟖

𝟎. 𝟎𝟒𝟖⁄ = 𝟏. 𝟒𝟐

𝑲𝒕 = 𝟒. 𝟑

With 𝑺𝒖𝒕 = 𝟏𝟎𝟎 𝒌𝒑𝒔𝒊, r = = 0.01 in.,

from Figure 11 in appendix, q = 0.65 in.

Then,

A fatigue factor of safety at point K is now

𝑛𝑓 = 1.86 Answer

40

Now check if point M is a critical point

From moment diagram in Figure 5

At point M,

Point M has a sharp should fillet which is required for the bearing for which r/d = 0.02 d = 1 in.

and from Table 5 in appendix, 𝑲𝒕 = 𝟐. 𝟕

With d = 1 in. , r = 1 in.

With 𝑺𝒖𝒕 = 𝟏𝟎𝟎 𝒌𝒑𝒔𝒊, r = 1 in.,

from Figure 11 in appendix, q = 0.7 in.

then,

𝑛𝑓 = 1.56 Answer

41

Now we have for diameters for critical locations, M (𝑫𝟏 = 𝑫𝟕)and I (𝑫𝟐 = 𝑫𝟔 = 𝟏. 𝟒 𝒊𝒏. )of

the shaft with trial values for other sections of the shaft at K without taking the deflections into

account

𝑫𝟏 = 𝑫𝟕 = 𝟏. 𝟎 𝒊𝒏. and 𝑫𝟐 = 𝑫𝟔 = 𝟏. 𝟒 𝒊𝒏. Answer

These above values do not take into consideration of shaft deflection so next we check for

deflection and obtain new and final values for diameters

Deflection, both angular and linear should be checked at bearings and gears. They depend on the

geometry of the shaft including the diameters. Check if the deflections and slopes at gears and

bearings are within acceptable ranges. If they are not then obtain new shaft diameters to resolve

any problems

A simple planar beam analysis will be used. Model the shaft twice using the x-y and x-z plane.

The material for the shaft is steel with Young’s Modulus, E = 30 Mpsi

With shaft length of 11.5 in. and with using the proposed shaft diameters and the knowledge

from statics, Figure 6 below shows the deflections and the slopes at points of interests along the

shaft

42

Figure 6: Deflection and Slope Plots of Intermediate Shaft

From Figure 6 above, deflections and slopes at points of interests are obtained and combined

using the equation

Equation 16

The combined results are shown below in Table 1

43

Table 1: Combined Results of Slope and Deflections of Intermediate Shaft at Points of Interest

In accordance with Table 6 in appendix, the bearing slopes are well below the limits. For the

right bearing slope, the values are within the acceptable range for cylindrical bearings. For the

gears, the slopes and deflections completely satisfy the limits from Table 6 in appendix

If the deflections values are near the limit, bring down the values by determining new shaft

diameters using equation

Equation 17

The slope at the right bearing in near the limit for the cylindrical bearing so increase the diameter

to bring the value down to 0.0005 rad

For 𝒅𝒐𝒍𝒅 = 𝑫𝟕 = 𝟏. 𝟎 𝒊𝒏 and design factor, 𝒏𝒅 = 𝟏

44

The ratio 𝒅𝒏𝒆𝒘

𝒅𝒐𝒍𝒅⁄ is given by

𝒅𝒏𝒆𝒘𝒅𝒐𝒍𝒅

⁄ = 𝟏. 𝟐𝟏𝟔𝟏⁄ = 𝟏. 𝟐𝟏𝟔

Mutliply all the old diameters with the above ratio to obtain new shaft diameters as:

Answer

Shaft specifications

Diameters ad Fatigue Factor of Safety

At point M,

Without deflection, 𝑫𝟏 = 𝑫𝟕 = 𝟏. 𝟎 𝒊𝒏.

With deflection, 𝑫𝟏 = 𝑫𝟕 = 𝟏. 𝟐𝟏𝟔 𝒊𝒏.

Fatigue Facotr of Safety, 𝒏𝒇 = 𝟏. 𝟓𝟔

45

At point to right of I

Without deflection, 𝑫𝟐 = 𝑫𝟔 = 𝟏. 𝟒 𝒊𝒏.

With deflection, 𝑫𝟏 = 𝑫𝟕 = 𝟏. 𝟕𝟎𝟐 𝒊𝒏.

Fatigue Facotr of Safety, 𝒏𝒇 = 𝟏. 𝟓𝟒

Nominal Diameter, 𝐷4 = 2.0 𝑖𝑛.

At point K

Without deflection, 𝑫𝟑 = 𝑫𝟓 = 𝟏. 𝟔𝟐𝟓 𝒊𝒏.



Bearing Specifications

After the specifications for shafts and gears have been obtained, the bearings need to be specified

in terms of diameters just like for shafts and gears. The appropriate bearings need to be selected

for the intermediate shaft with a reliability of 99 %. They are selected based on the rating

catalog, 𝐶10 or load rating, 𝐹𝑅

From the gearbox design specifications, the design life is 12,000 hours, and the speed of the

intermediate shaft was found out be 𝒘𝟑 = 𝒘𝟒 = 𝟑𝟖𝟗 𝑹𝑷𝑴

46

The estimated bore size and width for the bearings are 1 in.

From free body diagram of the forces on the intermediate from Figure 4, the reaction forces at A

and B were as:

The life in revolution of the bearing life just for gears is given by


𝑳 = 𝟔𝟎 ∗ 𝟏𝟐𝟎𝟎𝟎 ∗ 𝟑𝟖𝟖. 𝟗

𝑳 = 𝟐. 𝟖 ∗ 𝟏𝟎𝟖 𝒓𝒆𝒗

The load rating for a bearing is given by

Equation 18

where 𝒙𝑫 = 𝑙𝑖𝑓𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 , 𝒙𝒐 = 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑎𝑟𝑖𝑎𝑡𝑒,

𝜽 = 𝒄ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑡𝑜 63.2 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑎𝑟𝑖𝑎𝑡𝑒,

𝑹𝑫 = 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑟𝑒𝑙𝑖𝑎𝑏𝑖𝑙𝑡𝑖𝑦, 𝒂𝒇 = 𝑑𝑒𝑠𝑖𝑔𝑛 𝑙𝑜𝑎𝑑, 𝑭𝑫 = 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑙𝑎𝑜𝑑

Assume a ball bearing for both bearing A and bearing B for which a = 3

For 𝒂𝒇 = 𝟏, , 𝑭𝑫 = 𝑹𝑩 = 𝟏𝟗𝟏𝟖 𝒍𝒃𝒇, 𝒙𝑫 = 𝑳/𝑳𝟏𝟎 =𝟐.𝟖∗𝟏𝟎𝟖 𝒓𝒆𝒗

𝟏𝟎𝟔 and Weibull parameters given

by

47

the load rating, 𝑭𝑹𝑩 = 𝑪𝟏𝟎 for bearing B is given by

From globalspec.com for available bearings, this load rating is high for a ball bearing with bore

size of 1 in. Check with a cylindrical roller bearing for which a = 3/10, the load rating for

bearing B, 𝐹𝑅𝐵 is now given by

Cylindrical roller bearings are available from several sources closer to thus load rating. From

SKF, a common supplier of bearings, the specifications for bearing B are

Answer

Where C = catalog rating/load rating, ID = internal diameter, OD = outside diameter and

W=width

For bearing A on the left end of the shaft, the corresponding load rating, 𝐹𝑅𝐴 is given by

where 𝑭𝑫 = 𝑹𝑨 = 𝟑𝟕𝟓 𝒍𝒃𝒇

From SKF, this load rating correspond to a deep groove ball bearing with the following

specifications

48

Answer

Where C = catalog rating/load rating, ID = internal diameter, OD = outside diameter and

W=width

Bearing Specifications

Bearing B

Bearing A

Summary

The following is the summary of specifications obtained for intermediate, shaft and bearing for

two-stage gear reduction or a compound reverted gear train which meet the customer

requirements set forth at the beginning of the document

Gears

Gear 4 specification is

49

and

Wear factor of safety, 𝒏𝒄 = 𝟏. 𝟐𝟓 and bending factor of safety, 𝒏 = 𝟏. 𝟓𝟐


and

Wear factor of safety, 𝒏𝒄 = 𝟏. 𝟑𝟗 and bending factor of safety, 𝒏 = 𝟐. 𝟒𝟖


and

Wear factor of safety, 𝒏𝒄 = 𝟏. 𝟒𝟎 and bending factor of safety, 𝒏 = 𝟑. 𝟎𝟒

Gear 3 specifications

50

and

Wear factor of safety, 𝒏𝒄 = 𝟏. 𝟐𝟏 and bending factor of safety, 𝒏 = 𝟑. 𝟕𝟕

Shafts

Diameters ad Fatigue Factor of Safety

At point M,

Without deflection, 𝑫𝟏 = 𝑫𝟕 = 𝟏. 𝟎 𝒊𝒏.



At point to right of I

Without deflection, 𝑫𝟐 = 𝑫𝟔 = 𝟏. 𝟒 𝒊𝒏.

With deflection, 𝑫𝟏 = 𝑫𝟕 = 𝟏. 𝟕𝟎𝟐 𝒊𝒏.

Fatigue Facotr of Safety, 𝒏𝒇 = 𝟏. 𝟓𝟒

Nominal Diameter, 𝐷4 = 2.0 𝑖𝑛.

51

At point K

Without deflection, 𝑫𝟑 = 𝑫𝟓 = 𝟏. 𝟔𝟐𝟓 𝒊𝒏.



Bearings

Bearing B

Bearing A

52

References

Budynas, Richard G, J K. Nisbett, and Joseph E. Shigley. Shigley's Mechanical Engineering Design. New

York: McGraw-Hill, 2011. Print.

53

Appendix

Figure 7: Stress-cycle factor, 𝑍𝑛 vs. Number of load cycles, N

Figure 8: Geometry Factor, J vs. Number of teeth for which geometry factor is desired

54

Figure 9: Stress-cycle factor, 𝑌𝑛 vs. Number of load cycles, N

Figure 10: Allowable contact stress numbers, 𝑆𝑐 vs. Brinell Hardness, 𝐻𝑛

55

Figure 11: Notch sensitivity, q vs. Notch radius, r

Figure 12: Notch sensitivity, 𝑞𝑠ℎ𝑒𝑎𝑟vs. Notch radius, r

56

Figure 13: 𝐾𝑡 for round shaft with shoulder fillet in bending

Figure 14: 𝐾𝑡𝑠 for round shaft with shoulder fillet in torsion

57

Figure 15: 𝐾𝑡𝑠 for round shaft with flat-bottom groove in torsion

Table 2: Contact Strength, 𝑆𝑐 at 107cycles and 0.99 Reliability for Steel Gears

58

Table 3: Bending Strength, 𝑆𝑐 at 107cycles and 0.99 Reliability for Steel Gears

Table 4: Parameters for Marin Surface Modification Factor

59

Table 5: First Iteration Estimates for Stress-Concentration Factors, 𝐾𝑡 and 𝐾𝑡𝑠

Table 6: Typical Maximum Ranges for Slopes and Transverse Deflections

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