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 = 𝑻𝒊 𝒘𝒊 = 𝑻 𝒐 𝒘 𝒐
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 𝒘 𝒐 = 𝟖𝟐 ~ 𝟖𝟓 𝑹𝑷𝑴
This will give
𝑻 𝒊
𝑻 𝒐
=
𝒘 𝒐
𝒘 𝒊
= (
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
𝑻 𝒐
=
𝒘 𝒐
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
=
𝑵 𝟐
𝑵 𝟑
𝑵 𝟒
𝑵 𝟓
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.
𝑵 𝟐
𝑵 𝟑
=
𝑵 𝟒
𝑵 𝟓
= √
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 𝑵 𝒑 given by
𝑵 𝒑 =
𝟐𝒌
(𝟏 + 𝟐𝒎) 𝐬𝐢𝐧 𝟐 ∅
(𝒎 + √ 𝒎 𝟐 + (𝟏 + 𝟐𝒎) 𝐬𝐢𝐧 𝟐 ∅)
where 𝒎 is the ratio of the number of teeth on the pinion, 𝑵 𝒑 to the number of teeth on the gear,
𝑵 𝑮 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,
𝑵 𝒑 =
𝟐(𝟏)
(𝟏 + 𝟐(𝟒)) 𝐬𝐢𝐧 𝟐 𝟐𝟎
(𝟒 + √ 𝟒 𝟐 + (𝟏 + 𝟐(𝟒)) 𝐬𝐢𝐧 𝟐 ∅𝟐𝟎)
𝑵 𝒑 = 𝟏𝟔 𝒕𝒆𝒆𝒕𝒉

7
This is the number of teeth on the pinion without interference. So 𝑵 𝟐 = 𝑵 𝟒 = 𝟏𝟔 𝒕𝒆𝒆𝒕𝒉
𝑵 𝟑 = 𝑵 𝟓 = 𝟒. 𝟓𝟒(𝟏𝟔) = 𝟕𝟐. 𝟔𝟒
Check if output speed, 𝒘 𝒐 = 𝒘 𝟓 is within 82-88 RPM with 𝑵 𝟓 = 𝟕𝟐 𝒕𝒆𝒆𝒕𝒉 and with 𝒘 𝒐 =
𝒘 𝟐 = 𝟏𝟕𝟓𝟎 𝑹𝑷𝑴 as the required input
𝒘 𝟓 =
𝑵 𝟐
𝑵 𝟑
𝑵 𝟒
𝑵 𝟓
(𝒘 𝟐)
𝒘 𝟓 = (
𝟏𝟔
𝟕𝟐
)(
𝟏𝟔
𝟕𝟐
)(𝟏𝟕𝟓𝟎) = 𝟖𝟔. 𝟒𝟐 𝑹𝑷𝑴
This is acceptable!
So,
𝑵 𝟐 = 𝑵 𝟒 = 𝟏𝟔 𝒕𝒆𝒆𝒕𝒉
𝑵 𝟑 = 𝑵 𝟓 = 𝟕𝟐 𝒕𝒆𝒆𝒕𝒉
and then,
𝒘 𝟒 = 𝒘 𝟑 =
𝑵 𝟐
𝑵 𝟑
𝒘 𝟐
𝒘 𝟒 = 𝒘 𝟑 =
𝟏𝟔
𝟕𝟐
(𝟏𝟕𝟓𝟎 𝑹𝑷𝑴)
𝒘 𝟒 = 𝒘 𝟑 = 𝟑𝟖𝟖. 𝟗 𝑹𝑷𝑴
For the torque,
𝑯 = 𝑻 𝟐 𝒘 𝟐 = 𝑻 𝟓 𝒘 𝟓

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, 𝒎 𝑮 is defined as the ratio of number of teeth on gear, 𝑵 𝑮 to the number of teeth
on the pinion, 𝑵 𝒑 and is given by:
𝒎 𝑮 =
𝑵 𝑮
𝑵 𝑷
= 𝟒. 𝟓
where 𝑵 𝑮 = 𝟕𝟐 𝒕𝒆𝒆𝒕𝒉 and 𝑵 𝑷 = 𝟏𝟔 𝒕𝒆𝒆𝒕𝒉
And the compound reverted gear train is a spur gear for which the load-sharing ratio, 𝒎 𝑵 = 𝟏
Now for pressure angle, ∅ 𝒕 = 𝟐𝟎°
, the geometry factor, I for all gears which are external is given
by:

10
𝑰 =
𝒄𝒐𝒔∅ 𝒕 𝒔𝒊𝒏∅ 𝒕
𝟐𝒎 𝑵
𝒎 𝑮
𝒎 𝑮 + 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, 𝑸 𝒗 = 𝟕 is assumed
Then, A and B are given by:
𝑩 = 𝟎. 𝟕𝟑𝟏

11
𝑨 = 𝟔𝟓. 𝟏
and for gear 4, 𝑽 = 𝑽 𝟒𝟓 = 𝟐𝟕𝟏. 𝟓 𝒇𝒕/𝒎𝒊𝒏, then 𝑲 𝒗 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, 𝐊 𝐦 is given by
Equation 3
where

12
𝑪 𝒎𝒇 = 𝒇𝒂𝒄𝒆 𝒍𝒂𝒐𝒅 − 𝒅𝒊𝒔𝒕𝒓𝒊𝒃𝒖𝒕𝒊𝒐𝒏 𝒇𝒂𝒄𝒕𝒐𝒓
𝑪 𝒎𝒄 = 𝒍𝒐𝒂𝒅 𝒄𝒐𝒓𝒓𝒆𝒄𝒕𝒊𝒐𝒏 𝒇𝒂𝒄𝒕𝒐𝒓
𝑪 𝒑𝒇 = 𝒑𝒊𝒏𝒊𝒐𝒏 𝒑𝒓𝒐𝒑𝒐𝒓𝒕𝒊𝒐𝒏 𝒇𝒂𝒄𝒕𝒐𝒓
𝑪 𝒎𝒂 = 𝒎𝒆𝒔𝒉 𝒂𝒍𝒊𝒈𝒏𝒎𝒆𝒏𝒕 𝒇𝒂𝒄𝒕𝒐𝒓
The 𝑪 𝒑𝒇 is given by
Equation 4
𝑪 𝒑𝒇 =
𝑭
𝟏𝟎𝒅
− 𝟎. 𝟎𝟑𝟕𝟓 + 𝟎. 𝟎𝟏𝟐𝟓𝑭
where F = face width and d = gear diameter
With F = 2 in. and d = 2.67 in., 𝑪 𝒑𝒇 is given by:
𝑪 𝒑𝒇 =
𝟐
𝟏𝟎(𝟐. 𝟔𝟕)
− 𝟎. 𝟎𝟑𝟕𝟓 + 𝟎. 𝟎𝟏𝟐𝟓(𝟐)
𝑪 𝒑𝒇 = 𝟎. 𝟎𝟔𝟐𝟒
and
𝑪 𝒆 = 𝟏 (All other conditions)
Then, the load distribution factor, 𝑲 𝒎 is given by
𝑲 𝒎 = 𝟏. 𝟐𝟏

13
The contact stress, 𝝈 𝒄 for gears is given by
Equation 5
The basic material for gear 4 will be steel for which elastic coefficient, 𝑪 𝒑 = 𝟐𝟑𝟎𝟎
There is no detrimental surface finish effect for which 𝑪 𝒇 = 𝟏
No overloading for which 𝑲 𝒐 = 𝟏
No detrimental size effect for which 𝑲 𝒔 = 𝟏
Now, for gear 4 diameter, 𝒅 𝒑 = 𝟐. 𝟔𝟕 𝒊𝒏. , transmitted load, 𝑾 𝟒𝟓
𝒕
= 𝟐𝟒𝟑𝟏 𝒍𝒃𝒇 and geometry
factor, 𝑰 = 𝟎. 𝟏𝟑𝟏𝟓, the contact stress, 𝝈 𝒄 for gear 4 is given by:
The allowable contact stress, 𝜎𝑐,𝑎𝑙𝑙 is given by
Equation 6
The gear strength, 𝑺 𝒄 = 𝒆𝒏𝒅𝒖𝒓𝒂𝒏𝒄𝒆 𝒔𝒕𝒓𝒆𝒏𝒈𝒕𝒉 is based upon a reliability, R of 99% for which
the reliability factor, 𝑲 𝑹 = 𝟏

14
From the design specification, the operating temperature is −10◦ to 120◦F, for the which the
temperature factor, 𝑲 𝑻 = 𝟏
For gear life of 12,000 hours and a speed of 𝒘 𝟒 = 𝟑𝟖𝟖. 𝟗 𝑹𝑷𝑴,
the life in revolutions, L is given by:
𝑳 = 𝟔𝟎 ∗ 𝒉𝒐𝒖𝒓𝒔 ∗ 𝒔𝒑𝒆𝒆𝒅
𝑳 = 𝟔𝟎 ∗ 𝟏𝟐𝟎𝟎𝟎 ∗ 𝟑𝟖𝟖. 𝟗
𝑳 = 𝟐. 𝟖 ∗ 𝟏𝟎 𝟖
𝒓𝒆𝒗
From Figure 7 in appendix, the stress-cycle factor for wear, 𝒁 𝒏 = 𝟎. 𝟗 for 𝟏𝟎 𝟖
𝒄𝒚𝒄𝒍𝒆𝒔
For design factor, 𝒏 𝒅 = 𝟏. 𝟐 against wear
And AGMA factor of safety or stress ratio, 𝑺 𝑯 = 𝒏 𝒅 = 𝟏. 𝟐,
For gear 4,
𝝈 𝒄,𝒂𝒍𝒍 = 𝝈 𝒄
Endurance strength, 𝑺 𝒄 is then given by:
From Table 2 in appendix, this strength is achievable with Grade 2 carburized and hardened
with 𝑺 𝒄 = 𝟐𝟐𝟓𝟎𝟎𝟎 𝒑𝒔𝒊 Answer
Now, factor of safety, 𝒏 𝒄 for wear is given by

15
Answer
Gear 4 Bending
Number of teeth on gear 4, 𝑵 𝟒 = 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 = 𝑾𝒕 𝑲 𝒗
𝑷 𝒅
𝑭
𝑲 𝒎
𝑱
where 𝑾𝒕 = 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒, 𝑲 𝒗 = 𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑓𝑎𝑐𝑡𝑜𝑟, 𝑷 𝒅 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑟𝑎𝑙 𝑝𝑖𝑡𝑐ℎ, 𝑲 𝒎 =
𝑙𝑜𝑎𝑑 − 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟, 𝑭 = 𝑓𝑎𝑐𝑒𝑤𝑖𝑑𝑡ℎ, 𝑱 = 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑦 𝑓𝑎𝑐𝑡𝑜𝑟,
Now, for gear 4 diameter, 𝑷 𝒅 = 𝟔 𝒊𝒏. , transmitted load, 𝑾 𝟒𝟓
𝒕
= 𝟐𝟒𝟑𝟏 𝒍𝒃𝒇 , F = 2 in. , and
𝑲 𝒎 = 𝟏. 𝟐𝟏, 𝑲 𝒗 = 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, 𝒀 𝑵 = 𝟎. 𝟗 for 𝟏𝟎 𝟖

16
Now using Grade 2 carburized and hardened as before, the bending strength, from Table 3, is
given by 𝑺 𝒕 = 𝟔𝟓𝟎𝟎𝟎 𝒑𝒔𝒊 Answer
Assume that bending factor of safety, 𝑆 𝐹 = 1 and that 𝐾 𝑇 and 𝐾 𝑅 = 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, 𝒏 𝒄 = 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, 𝑵 𝟓 = 𝟕𝟐 𝒕𝒆𝒆𝒕𝒉 for which, from Figure 8 in appendix, geometry
factor, J = 0.41
Also the speed of gear 5 is different which is 𝒘 𝟓 = 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 ∗ 𝟏𝟎 𝟕
𝒓𝒆𝒗
From Figure 7 and Figure 9 for 𝟏𝟎 𝟕
𝒀 𝑵 = 𝒁 𝑵 = 𝟏
The contact stress, 𝝈 𝒄 from gear 5 is same as from gear 4 since they are in contact:
𝝈 𝒄 = 𝟏𝟔𝟏𝟕𝟎𝟎 𝒑𝒔𝒊
Now for same design factor, 𝑺 𝑯 = 𝒏 𝒅 = 𝟏. 𝟐,
𝑺 𝒄 =
𝑺 𝑯 𝝈 𝒄
𝒁 𝒏
𝑺 𝒄 =
(𝟏. 𝟐)(𝟏𝟔𝟏𝟕𝟎𝟎)
𝟏
𝑺 𝒄 = 𝟏𝟗𝟒, 𝟎𝟒𝟎 𝒑𝒔𝒊

18
From Table 2 in appendix, this strength is achievable with grade 2 carburized and hardened
with 𝑺 𝒄 = 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 𝑷 𝒅 = 𝟔 𝒊𝒏. , 𝑾 𝟒𝟓
𝒕
= 𝟐𝟒𝟑𝟏 𝒍𝒃𝒇 , F = 2 in. , 𝑲 𝒎 = 𝟏. 𝟐𝟏, and 𝑲 𝒗 = 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, 𝒏 𝒄 = 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, 𝑽 𝟐𝟑 for gear 2 is 1223 ft/min, for which the dynamic factor, 𝑲 𝒗 is given
by Equation 2
𝑲 𝒗 =
𝟔𝟓. 𝟏 + √𝟏𝟐𝟐𝟑
𝟔𝟓. 𝟏
𝑲 𝒗 = 𝟏. 𝟑𝟕
Since the transmitted load of 𝑾 𝟐𝟑
𝒕
= 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, 𝑪 𝒑𝒇 from Equation 4 is now:
𝑪 𝒑𝒇 = 𝟎. 𝟎𝟒𝟑𝟕
Then from Equation 3, the corresponding load distribution factor, 𝑲 𝒎 = 𝟏. 𝟏𝟗
With, 𝑾 𝟐𝟑
𝒕
= 𝟓𝟑𝟗. 𝟕 𝒍𝒃𝒇, 𝑲 𝒗 = 𝟏. 𝟑𝟕, F = 1.5 in. , 𝑲 𝒎 = 𝟏. 𝟏𝟗 and 𝒅 𝒑 = 𝟐. 𝟔𝟕 𝒊𝒏. and
𝑰 = 𝟎. 𝟏𝟑𝟏𝟓
the contact stress, 𝝈 𝒄 for gear 2 from Equation 5 is given by:

20
The life in revolution, L for gear 2 with 𝒘 𝟐 = 𝟏𝟕𝟓𝟎 𝑹𝑷𝑴 is given by
𝑳 = 𝟔𝟎 ∗ 𝟏𝟐𝟎𝟎𝟎 ∗ 𝟏𝟕𝟓𝟎
𝑳 = 𝟏. 𝟐𝟔 ∗ 𝟏𝟎 𝟗
𝒓𝒆𝒗
From Figure 7 in appendix, the stress-cycle factor for wear, 𝒁 𝒏 = 𝟎. 𝟖 for 𝟏𝟎 𝟗
𝑺 𝒄 =
𝑺 𝑯 𝝈 𝒄
𝒁 𝒏
𝑺 𝒄 =
(𝟏. 𝟐)(𝟗𝟒𝟎𝟎𝟎)
𝟎. 𝟗
𝑺 𝒄 = 𝟏𝟐𝟓, 𝟑𝟑𝟑. 𝟑𝟑 𝒑𝒔𝒊
From Table 2 in appendix, this strength is achievable with grade 1 flame hardened
with 𝑺 𝒄 = 𝟏𝟕𝟎, 𝟎𝟎𝟎 𝒑𝒔𝒊 Answer
Factor of safety for wear is now
Answer
Gear 2 bending
Number of teeth on gear 2, 𝑵 𝟐 = 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, 𝑾 𝟐𝟑
𝒕
= 𝟓𝟑𝟗. 𝟕 𝒍𝒃𝒇, 𝑲 𝒗 = 𝟏. 𝟑𝟕, 𝑷 𝒅 = 𝟔 𝒊𝒏. , t, F
= 1.5 in. , and 𝑲 𝒎 = 𝟏. 𝟏𝟗, and geometry factor, 𝑱 = 𝟎. 𝟐𝟕 is given by:
From Figure 9 in appendix, the stress-cycle factor for bending, 𝒀 𝑵 = 𝟎. 𝟖𝟖 for 𝟏𝟎 𝟗
Now using grade 1 flame hardened with as before, the bending strength, from Table 3 is
𝑺 𝒕 = 𝟒𝟓𝟎𝟎𝟎 𝒑𝒔𝒊 Answer
Assume that bending factor of safety, 𝑆 𝐹 = 1 and that 𝐾 𝑇 and 𝐾 𝑅 = 1 as before
Then, allowable bending stress is given by
𝜎 𝑎𝑙𝑙 = 𝑆𝑡 𝑌𝑁
𝜎 𝑎𝑙𝑙 = (45000)(0.88)
𝜎 𝑎𝑙𝑙 = 39600 𝑝𝑠𝑖
Now factor of safety for bending is
Answer
and

22
Wear factor of safety, 𝒏 𝒄 = 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, 𝑵 𝟑 = 𝟕𝟐 𝒕𝒆𝒆𝒕𝒉 for which, from Figure 8 in appendix, geometry
factor, J = 0.41
Also the speed of gear 3 is different which is 𝒘 𝟑 = 𝟑𝟖𝟖. 𝟗 𝑹𝑷𝑴
From the speed, the life in revolutions, L of gear 3 is
𝑳 = 𝟔𝟎 ∗ 𝟏𝟐𝟎𝟎𝟎 ∗ 𝟑𝟖𝟖. 𝟗
𝑳 = 𝟐. 𝟖 ∗ 𝟏𝟎 𝟖
𝒓𝒆𝒗
For 108
𝑐𝑦𝑐𝑙𝑒𝑠 from Figures A and C in appendix
𝒀 𝑵 = 𝒁 𝑵 = 𝟎. 𝟗
The contact stress, 𝝈 𝒄 from gear 3 is same as gear 2 since they are in contact:
𝝈 𝒄 = 𝟗𝟒𝟎𝟎𝟎 𝒑𝒔𝒊
Endurance strength, 𝑺 𝒄 is then given by

23
𝑺 𝒄 =
𝑺 𝑯 𝝈 𝒄
𝒁 𝒏
𝑺 𝒄 =
(𝟏. 𝟐)(𝟗𝟒𝟎𝟎𝟎)
𝟎. 𝟗
𝑺 𝒄 = 𝟏𝟐𝟓, 𝟑𝟑𝟑. 𝟑𝟑 𝒑𝒔𝒊
From Table 2 and Figure 10 in appendix, this strength is achievable with grade 1 through
hardened with 𝑺 𝒄 = 𝟏𝟐𝟔, 𝟎𝟎𝟎 𝒑𝒔𝒊 and 𝑺 𝒕 = 𝟑𝟔𝟎𝟎𝟎 𝒑𝒔𝒊 with hardness of 300 𝑯 𝑩 Answer
Now with, 𝝈 𝒄,𝒂𝒍𝒍 = 𝑺 𝒄 𝒁 𝒏 = (𝟏𝟐𝟔𝟎𝟎𝟎)(𝟎. 𝟗) = 𝟏𝟏𝟑𝟒𝟎𝟎 𝒑𝒔𝒊
The factor of safety for wear is
𝒏 𝒄,𝒂𝒍𝒍 =
𝝈 𝒄,𝒂𝒍𝒍
𝝈 𝒄
𝒏 𝒄,𝒂𝒍𝒍 = 𝟏. 𝟐𝟏 Answer
Now for bending, note that due to J = 0.27 instead of J = 0.41, and from Equation 7 with 𝑲 𝒗 =
𝟏. 𝟑𝟕, facewidth, F=1.5, 𝑲 𝒎 = 𝟏. 𝟏𝟗 and 𝑾 𝟐𝟑
𝒕
= 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, 𝒏 𝒄 = 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
𝑭 𝟐𝟑
𝒓
= 𝟓𝟒𝟎 𝐭𝐚𝐧(𝟐𝟎°) = 𝟏𝟗𝟕 𝒍𝒃𝒇
𝑭 𝟒𝟓
𝒓
= 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), 𝑴 𝒂 = 𝟑𝟔𝟓𝟏 𝒍𝒃𝒇 ∙ 𝒊𝒏
Constant/midrange torque at 𝑻 𝒎 = 𝟑𝟐𝟒𝟎 𝒍𝒃𝒇
Midrange bending moment, 𝑴 𝒎 = 𝟎
Maximum torque, 𝑻 𝒂 = 𝟎
A suitable material selected for the shaft is AISI 1020 CD steel. For this material, the ultimate
tensile strength is 𝑺 𝒖𝒕 = 𝟔𝟖 𝒌𝒑𝒔𝒊
From Table 4 in appendix, the surface factor, 𝒌 𝒂 for cold-drawn (CD) steel is
Since the shaft diameters are not known yet, a value of 0.9 for size factor, 𝒌 𝒃 is assumed
Since bending moment is greater than torque, loading factor, 𝒌 𝒄 = 𝟏
No rotating beam endurance limit is known as room temperature, 𝑆 𝑇 so temperature factor
𝒌 𝒅 = 𝟏
Assume 50% reliability for which reliability factor, 𝒌 𝒆 = 𝟏
For 𝑆 𝑢𝑡 ≤ 200 𝑘𝑝𝑠𝑖, the rotary-beam test specimen modification factor, 𝑺 𝒆
′
is given by
𝑺 𝒆
′
= 𝟎. 𝟓 ∗ 𝑺 𝒖𝒕

30
𝑺 𝒆
′
= 𝟎. 𝟓 ∗ 𝟔𝟖 = 𝟑𝟒 𝐤𝐩𝐬𝐢
Now, the endurance limit, 𝑆 𝑒 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦
Equation 9
𝑺 𝒆 = (𝟎. 𝟖𝟖𝟑)(𝟎. 𝟗)(𝟏)(𝟏)(𝟏)(𝟑𝟒)
𝑺 𝒆 = 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 𝒌 𝒇 = 𝒌𝒕 and 𝒌 𝒇𝒔 = 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
𝑫 𝟒 = 𝟐. 𝟎 𝒊𝒏. and 𝒅 = 𝑫 𝟓 = 𝑫 𝟑 = 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
𝒓
𝒅⁄ = 𝟎. 𝟏
With d = 1.625 in., fillet radius is 𝒓 ≅ 𝟎. 𝟏𝟔 𝒊𝒏.
With 𝑺 𝒖𝒕 = 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, 𝒒 𝒔𝒉𝒆𝒂𝒓 = 𝟎. 𝟖𝟓
With
𝑫
𝒅
= 𝟏. 𝟐𝟑 and 𝒓
𝒅⁄ = 𝟎. 𝟏
From Figure 13 in appendix, 𝑲 𝒕 = 𝟏. 𝟔
From Figure 14 in appendix, 𝑲 𝒕𝒔 = 𝟏. 𝟑𝟓
So now,
The fatigue stress-concentration factor from bending, 𝑲 𝒇 is given by
Equation 11
The fatigue stress-concentration factor from torsion, 𝑲 𝒇𝒔 is given by
Now, let’s evaluate the endurance strength, 𝑺 𝒆
𝒌 𝒂 = 𝟎. 𝟖𝟖𝟑 (Same as before)
Since d = 1.625 in. is between 0.11 in. and 2 in.,
𝒌 𝒃 = 𝟎. 𝟖𝟕𝟗𝒅−𝟎.𝟏𝟎𝟕

33
𝒌 𝒃 = 0.835
Now, from Equation 9
𝑺 𝒆 = (𝟎. 𝟖𝟖𝟑)(𝟎. 𝟖𝟑𝟓)(𝟏)(𝟏)(𝟏)(𝟑𝟒)
𝑺 𝒆 = 𝟐𝟓. 𝟏 𝒌𝒑𝒔𝒊
The effective von Mises stress, 𝝈′
at a given point is given by
For stress amplitude, 𝜎 𝑎
′
Equation 12
With 𝑇𝑎 = 0 at point I, 𝜎 𝑎
′
is given by
For midrange stress, 𝝈 𝒎
′
Equation 13
With 𝑀 𝑚 = 0 at point I, 𝜎 𝑚
′
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, 𝜎 𝑎
𝝈 𝒂 =
(𝟏. 𝟒𝟗)(𝟑𝟐)(𝟑𝟔𝟓𝟏)
𝝅(𝟏. 𝟔𝟐𝟓) 𝟑
𝝈 𝒂 = 𝟏𝟐, 𝟗𝟏𝟑. 𝟑𝟑 𝒑𝒔𝒊
Midrange Torsion, 𝜏 𝑚
𝝉 𝒎 =
(𝟏. 𝟑𝟎)(𝟏𝟔)(𝟑𝟐𝟒𝟎)
𝝅(𝟏. 𝟔𝟐𝟓) 𝟑
𝝉 𝒎 = 𝟒, 𝟗𝟗𝟗. 𝟏𝟖 𝒑𝒔𝒊
and at point I,
𝝈 𝒎 = 𝝉 𝒂 = 𝟎
Combined maximum von Mises stress, 𝜎 𝑚𝑎𝑥
′
is given by

35
Equation 15
𝝈 𝒎𝒂𝒙
′
= [(𝟎 + 𝟏𝟐, 𝟗𝟏𝟑. 𝟑𝟑) 𝟐
+ 𝟑(𝟒, 𝟗𝟗𝟗. 𝟏𝟖 + 𝟎) 𝟐
]
𝟏
𝟐⁄
𝝈 𝒎𝒂𝒙
′
= 𝟏𝟓, 𝟓𝟒𝟕. 𝟔𝟓 𝒑𝒔𝒊
Now check if the sum of 𝝈 𝒂
,
+ 𝝈 𝒎
′
is greater than 𝝈 𝒎𝒂𝒙
′
𝝈 𝒂
,
+ 𝝈 𝒎
′
= 𝟏𝟐𝟗𝟏𝟎 + 𝟖𝟔𝟓𝟗 = 𝟐𝟏, 𝟓𝟔𝟗 𝒑𝒔𝒊 ≥ 𝟏𝟓, 𝟓𝟒𝟕. 𝟓𝟔 𝒑𝒔𝒊 ≥ 𝝈 𝒎𝒂𝒙
′
Hence, there will be no yielding
Also check with yielding factor of safety, 𝑛𝑓
For AISI 1020 CD steel, yield strength, 𝑺 𝒚 = 𝟓𝟕 𝒌𝒑𝒔𝒊
𝒏 𝒇 =
𝑺 𝒚
𝝈 𝒎𝒂𝒙
′
=
𝟓𝟕𝟎𝟎𝟎
𝟐𝟏𝟓𝟔𝟗
= 𝟐. 𝟔𝟒 > 𝟏
This confirms there will be no yielding since 𝒏 𝒇 > 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 𝑫 𝟏 = 𝑫 𝟕 and 𝑫 𝟐 = 𝑫 𝟔
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 𝑴 𝒂 = 𝟑𝟕𝟓𝟎 𝒍𝒃 ∙ 𝒊𝒏 while 𝑻 𝒎 =
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 𝑺 𝒖𝒕 = 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, 𝒒 𝒔𝒉𝒆𝒂𝒓 = 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 𝑻 𝒂 = 𝟎 and 𝑴 𝒂 = 𝟑𝟕𝟓𝟎 𝒍𝒃 ∙ 𝒊𝒏 at point I, 𝝈 𝒂
′
is given by
From Equation 12, with 𝑴 𝒎 = 𝟎 and 𝑻 𝒎 = 𝟑𝟐𝟒𝟎 𝒍𝒃 ∙ 𝒊𝒏 at point I, 𝝈 𝒎
′
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 𝑺 𝒖𝒕 = 𝟏𝟎𝟎 𝒌𝒑𝒔𝒊
Now recalculate everything as before
From Table 4 in appendix, surface factor, 𝒌 𝒂
From Equation 9, endurance strength, 𝑺 𝒆 with 𝒌 𝒃 = 0.835
With 𝑺 𝒖𝒕 = 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, 𝑲 𝒕 = 𝟐. 𝟏𝟒. , 𝑲 𝒇 is given by
Next, with 𝑴 𝒂 = 𝟑𝟕𝟓𝟎 𝒍𝒃 ∙ 𝒊𝒏, 𝝈 𝒂
′
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 𝑲 𝒕 = 𝑲 𝒇 = 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 𝑲 𝒇 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. 𝟎𝟏
𝟎. 𝟎𝟒𝟖⁄ = 𝟎. 𝟐𝟎𝟖 and 𝒂
𝒕⁄ = 𝟎. 𝟎𝟔𝟖
𝟎. 𝟎𝟒𝟖⁄ = 𝟏. 𝟒𝟐
𝑲 𝒕 = 𝟒. 𝟑
With 𝑺 𝒖𝒕 = 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, 𝑲 𝒕 = 𝟐. 𝟕
With d = 1 in. , r = 1 in.
With 𝑺 𝒖𝒕 = 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 (𝑫 𝟏 = 𝑫 𝟕)and I (𝑫 𝟐 = 𝑫 𝟔 = 𝟏. 𝟒 𝒊𝒏. )of
the shaft with trial values for other sections of the shaft at K without taking the deflections into
account
𝑫 𝟏 = 𝑫 𝟕 = 𝟏. 𝟎 𝒊𝒏. and 𝑫 𝟐 = 𝑫 𝟔 = 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 𝒅 𝒐𝒍𝒅 = 𝑫 𝟕 = 𝟏. 𝟎 𝒊𝒏 and design factor, 𝒏 𝒅 = 𝟏

44
The ratio
𝒅 𝒏𝒆𝒘
𝒅 𝒐𝒍𝒅
⁄ is given by
𝒅 𝒏𝒆𝒘
𝒅 𝒐𝒍𝒅
⁄ = 𝟏. 𝟐𝟏𝟔
𝟏⁄ = 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, 𝑫 𝟏 = 𝑫 𝟕 = 𝟏. 𝟎 𝒊𝒏.
With deflection, 𝑫 𝟏 = 𝑫 𝟕 = 𝟏. 𝟐𝟏𝟔 𝒊𝒏.
Fatigue Facotr of Safety, 𝒏 𝒇 = 𝟏. 𝟓𝟔

45
At point to right of I
Without deflection, 𝑫 𝟐 = 𝑫 𝟔 = 𝟏. 𝟒 𝒊𝒏.
With deflection, 𝑫 𝟏 = 𝑫 𝟕 = 𝟏. 𝟕𝟎𝟐 𝒊𝒏.
Fatigue Facotr of Safety, 𝒏 𝒇 = 𝟏. 𝟓𝟒
Nominal Diameter, 𝐷4 = 2.0 𝑖𝑛.
At point K
Without deflection, 𝑫 𝟑 = 𝑫 𝟓 = 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 𝒘 𝟑 = 𝒘 𝟒 = 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 ∗ 𝟏𝟎 𝟖
𝒓𝒆𝒗
The load rating for a bearing is given by
Equation 18
where 𝒙 𝑫 = 𝑙𝑖𝑓𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 , 𝒙 𝒐 = 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑎𝑟𝑖𝑎𝑡𝑒,
𝜽 = 𝒄ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑡𝑜 63.2 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑎𝑟𝑖𝑎𝑡𝑒,
𝑹 𝑫 = 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑟𝑒𝑙𝑖𝑎𝑏𝑖𝑙𝑡𝑖𝑦, 𝒂 𝒇 = 𝑑𝑒𝑠𝑖𝑔𝑛 𝑙𝑜𝑎𝑑, 𝑭 𝑫 = 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑙𝑎𝑜𝑑
Assume a ball bearing for both bearing A and bearing B for which a = 3
For 𝒂 𝒇 = 𝟏, , 𝑭 𝑫 = 𝑹 𝑩 = 𝟏𝟗𝟏𝟖 𝒍𝒃𝒇, 𝒙 𝑫 = 𝑳/𝑳 𝟏𝟎 =
𝟐.𝟖∗𝟏𝟎 𝟖 𝒓𝒆𝒗
𝟏𝟎 𝟔
and Weibull parameters given
by

47
the load rating, 𝑭 𝑹𝑩 = 𝑪 𝟏𝟎 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 𝑭 𝑫 = 𝑹 𝑨 = 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, 𝒏 𝒄 = 𝟏. 𝟐𝟓 and bending factor of safety, 𝒏 = 𝟏. 𝟓𝟐
and
Wear factor of safety, 𝒏 𝒄 = 𝟏. 𝟑𝟗 and bending factor of safety, 𝒏 = 𝟐. 𝟒𝟖
and
Wear factor of safety, 𝒏 𝒄 = 𝟏. 𝟒𝟎 and bending factor of safety, 𝒏 = 𝟑. 𝟎𝟒
Gear 3 specifications

50
and
Wear factor of safety, 𝒏 𝒄 = 𝟏. 𝟐𝟏 and bending factor of safety, 𝒏 = 𝟑. 𝟕𝟕
Shafts
Diameters ad Fatigue Factor of Safety
At point M,
Without deflection, 𝑫 𝟏 = 𝑫 𝟕 = 𝟏. 𝟎 𝒊𝒏.
At point to right of I
Without deflection, 𝑫 𝟐 = 𝑫 𝟔 = 𝟏. 𝟒 𝒊𝒏.
With deflection, 𝑫 𝟏 = 𝑫 𝟕 = 𝟏. 𝟕𝟎𝟐 𝒊𝒏.
Fatigue Facotr of Safety, 𝒏 𝒇 = 𝟏. 𝟓𝟒
Nominal Diameter, 𝐷4 = 2.0 𝑖𝑛.

51
At point K
Without deflection, 𝑫 𝟑 = 𝑫 𝟓 = 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, 𝑍 𝑛 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, 𝐻 𝑛

55
Figure 11: Notch sensitivity, q vs. Notch radius, r
Figure 12: Notch sensitivity, 𝑞 𝑠ℎ𝑒𝑎𝑟vs. 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 107
cycles 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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