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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, ๐๐ 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, ๐๐ vs. Number of load cycles, N ................................................................... 54
Figure 10: Allowable contact stress numbers, ๐๐ vs. Brinell Hardness, ๐ป๐ ............................................... 54
Figure 11: Notch sensitivity, q vs. Notch radius, r ...................................................................................... 55
Figure 12: Notch sensitivity, ๐๐ โ๐๐๐vs. Notch radius, r ............................................................................. 55
Figure 13: ๐พ๐ก for round shaft with shoulder fillet in bending .................................................................... 56
Figure 14: ๐พ๐ก๐ for round shaft with shoulder fillet in torsion .................................................................... 56
Figure 15: ๐พ๐ก๐ 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, ๐๐ at 107cycles and 0.99 Reliability for Steel Gears ....................................... 57
Table 3: Bending Strength, ๐๐ 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, ๐พ๐ก and ๐พ๐ก๐ ...................................... 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: 82โ88 RPM
Usually low shock levels, occasional moderate shock
Input and output shafts extend 4 in outside gearbox
3
Input and output shaft diameter tolerance: ยฑ0.001in
Input and output shafts in-line: concentricity ยฑ0.005in, alignment ยฑ0.001rad
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: โ10โฆ to 120โฆF
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 โ 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 โ 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 โ 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
๐ฏ = ๐ป๐๐๐ = ๐ป๐๐๐
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 ๐๐ = ๐๐๐๐ ๐น๐ท๐ด and ๐๐ = ๐๐ ~ ๐๐ ๐น๐ท๐ด
This will give ๐ป๐
๐ป๐=
๐๐
๐๐= (
๐๐
๐๐๐) ๐๐๐[๐. ๐๐๐๐] ๐๐ (
๐๐
๐๐๐)๐๐๐ [๐. ๐๐๐๐]
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
๐ = ๐ป๐
๐ป๐
= ๐๐
๐๐
For ๐ค๐ = 1750 ๐ ๐๐ and ๐ค๐ = 85 ๐ ๐๐,
6
๐ = ๐๐
๐๐๐๐=
๐๐
๐๐๐=
๐
๐๐. ๐๐= ๐. ๐๐๐๐
and for this compound reverted geartrain,
๐ = ๐
๐๐. ๐๐=
๐ต๐
๐ต๐ ๐ต๐
๐ต๐
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.
๐ต๐
๐ต๐=
๐ต๐
๐ต๐= โ
๐
๐๐. ๐๐=
๐
๐. ๐๐
The smallest number of teeth on the pinion which can exist without interference needs to be
determined. This is ๐ต๐ given by
๐ต๐ = ๐๐
(๐ + ๐๐) ๐ฌ๐ข๐ง๐ โ (๐ + โ๐๐ + (๐ + ๐๐) ๐ฌ๐ข๐ง๐ โ )
where ๐ is the ratio of the number of teeth on the pinion, ๐ต๐ to the number of teeth on the gear,
๐ต๐ฎ and โ is the pressure angle
Let ๐ = ๐ such there are 4 teeth on the pinion for every tooth on the gear.
For โ = ๐๐ and ๐ = ๐ for full-teeth,
๐ต๐ = ๐(๐)
(๐ + ๐(๐)) ๐ฌ๐ข๐ง๐ ๐๐(๐ + โ๐๐ + (๐ + ๐(๐)) ๐ฌ๐ข๐ง๐ โ ๐๐)
๐ต๐ = ๐๐ ๐๐๐๐๐
7
This is the number of teeth on the pinion without interference. So ๐ต๐ = ๐ต๐ = ๐๐ ๐๐๐๐๐
๐ต๐ = ๐ต๐ = ๐. ๐๐(๐๐) = ๐๐. ๐๐
Check if output speed, ๐๐ = ๐๐ is within 82-88 RPM with ๐ต๐ = ๐๐ ๐๐๐๐๐ and with ๐๐ =
๐๐ = ๐๐๐๐ ๐น๐ท๐ด as the required input
๐๐ =๐ต๐
๐ต๐
๐ต๐
๐ต๐(๐๐)
๐๐ = (๐๐
๐๐)(
๐๐
๐๐)(๐๐๐๐) = ๐๐. ๐๐ ๐น๐ท๐ด
This is acceptable!
So,
๐ต๐ = ๐ต๐ = ๐๐ ๐๐๐๐๐
๐ต๐ = ๐ต๐ = ๐๐ ๐๐๐๐๐
and then,
๐๐ = ๐๐ =๐ต๐
๐ต๐๐๐
๐๐ = ๐๐ =๐๐
๐๐(๐๐๐๐ ๐น๐ท๐ด)
๐๐ = ๐๐ = ๐๐๐. ๐ ๐น๐ท๐ด
For the torque,
๐ฏ = ๐ป๐๐๐ = ๐ป๐๐๐
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, ๐ is given by ๐ =๐ต
๐ท where P = diametral pitch and N = number of teeth.
Then substituting ๐ต
๐ท for ๐ , 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, ๐๐ฎ is defined as the ratio of number of teeth on gear, ๐ต๐ฎ to the number of teeth
on the pinion, ๐ต๐ and is given by:
๐๐ฎ =๐ต๐ฎ
๐ต๐ท= ๐. ๐
where ๐ต๐ฎ = ๐๐ ๐๐๐๐๐ and ๐ต๐ท = ๐๐ ๐๐๐๐๐
And the compound reverted gear train is a spur gear for which the load-sharing ratio, ๐๐ต = ๐
Now for pressure angle, โ ๐ = ๐๐ยฐ, the geometry factor, I for all gears which are external is given
by:
10
๐ฐ =๐๐๐โ ๐๐๐๐โ ๐
๐๐๐ต
๐๐ฎ
๐๐ฎ + ๐
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, ๐พ๐ฃ is given by
Equation 2
๐พ๐ฃ =๐ด + โ๐
๐ด
where
๐ฝ = ๐๐๐๐๐ โ ๐๐๐๐ ๐๐๐๐๐๐๐๐
The gears need to be of the highest quality so a value for quality number, ๐ธ๐ = ๐ is assumed
Then, A and B are given by:
๐ฉ = ๐. ๐๐๐
11
๐จ = ๐๐. ๐
and for gear 4, ๐ฝ = ๐ฝ๐๐ = ๐๐๐. ๐ ๐๐/๐๐๐, then ๐ฒ๐ is given by
The circular pitch, p is given by ratio of ๐ to the diametral pitch, P as
๐ =๐
๐ท
The face width, F is typically 3-5 times the circular pitch, p.
Trying with 4 times the circular pitch, F is given by
๐ญ = ๐ (๐
๐ท) = ๐ (
๐
๐) = ๐. ๐๐ ๐๐.
Now verify, if this is a good face width for gear 4 with pitch diameter, ๐ = ๐. ๐๐ ๐๐. and
diametral pitch, ๐ท = ๐ ๐๐๐๐๐/๐๐
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, ๐๐ฆ is given by
Equation 3
where
12
๐ช๐๐ = ๐๐๐๐ ๐๐๐๐ โ ๐ ๐๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐
๐ช๐๐ = ๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐
๐ช๐๐ = ๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐
๐ช๐๐ = ๐๐๐๐ ๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐
The ๐ช๐๐ is given by
Equation 4
๐ช๐๐ =๐ญ
๐๐๐ โ ๐. ๐๐๐๐ + ๐. ๐๐๐๐๐ญ
where F = face width and d = gear diameter
With F = 2 in. and d = 2.67 in., ๐ช๐๐ is given by:
๐ช๐๐ =๐
๐๐(๐. ๐๐)โ ๐. ๐๐๐๐ + ๐. ๐๐๐๐(๐)
๐ช๐๐ = ๐. ๐๐๐๐
and
๐ช๐ = ๐ (All other conditions)
Then, the load distribution factor, ๐ฒ๐ is given by
๐ฒ๐ = ๐. ๐๐
13
The contact stress, ๐๐ for gears is given by
Equation 5
The basic material for gear 4 will be steel for which elastic coefficient, ๐ช๐ = ๐๐๐๐
There is no detrimental surface finish effect for which ๐ช๐ = ๐
No overloading for which ๐ฒ๐ = ๐
No detrimental size effect for which ๐ฒ๐ = ๐
Now, for gear 4 diameter, ๐ ๐ = ๐. ๐๐ ๐๐. , transmitted load, ๐พ๐๐๐ = ๐๐๐๐ ๐๐๐ and geometry
factor, ๐ฐ = ๐. ๐๐๐๐, the contact stress, ๐๐ for gear 4 is given by:
The allowable contact stress, ๐๐,๐๐๐ is given by
Equation 6
The gear strength, ๐บ๐ = ๐๐๐ ๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐ is based upon a reliability, R of 99% for which
the reliability factor, ๐ฒ๐น = ๐
14
From the design specification, the operating temperature is โ10โฆ to 120โฆF, for the which the
temperature factor, ๐ฒ๐ป = ๐
For gear life of 12,000 hours and a speed of ๐๐ = ๐๐๐. ๐ ๐น๐ท๐ด,
the life in revolutions, L is given by:
๐ณ = ๐๐ โ ๐๐๐๐๐ โ ๐๐๐๐๐
๐ณ = ๐๐ โ ๐๐๐๐๐ โ ๐๐๐. ๐
๐ณ = ๐. ๐ โ ๐๐๐ ๐๐๐
From Figure 7 in appendix, the stress-cycle factor for wear, ๐๐ = ๐. ๐ for ๐๐๐๐๐๐๐๐๐
For design factor, ๐๐ = ๐. ๐ against wear
And AGMA factor of safety or stress ratio, ๐บ๐ฏ = ๐๐ = ๐. ๐,
For gear 4,
๐๐,๐๐๐ = ๐๐
Endurance strength, ๐บ๐ is then given by:
From Table 2 in appendix, this strength is achievable with Grade 2 carburized and hardened
with ๐บ๐ = ๐๐๐๐๐๐ ๐๐๐ Answer
Now, factor of safety, ๐๐ for wear is given by
15
Answer
Gear 4 Bending
Number of teeth on gear 4, ๐ต๐ = ๐๐ ๐๐๐๐๐ for which, from Figure 8 in appendix, geometry
factor, J = 0.27
Then,
the bending stress, ๐ is given by
Equation 7
๐ = ๐พ๐๐ฒ๐
๐ท๐
๐ญ
๐ฒ๐
๐ฑ
where ๐พ๐ = ๐ก๐๐๐๐ ๐๐๐ก๐ก๐๐ ๐๐๐๐๐, ๐ฒ๐ = ๐๐ฆ๐๐๐๐๐ ๐๐๐๐ก๐๐, ๐ท๐ = ๐๐๐๐๐๐ก๐๐๐ ๐๐๐ก๐โ, ๐ฒ๐ =
๐๐๐๐ โ ๐๐๐ ๐ก๐๐๐๐ข๐ก๐๐๐ ๐๐๐๐ก๐๐, ๐ญ = ๐๐๐๐๐ค๐๐๐กโ, ๐ฑ = ๐๐๐๐๐๐ก๐๐ฆ ๐๐๐๐ก๐๐,
Now, for gear 4 diameter, ๐ท๐ = ๐ ๐๐. , transmitted load, ๐พ๐๐๐ = ๐๐๐๐ ๐๐๐ , F = 2 in. , and
๐ฒ๐ = ๐. ๐๐, ๐ฒ๐ = ๐. ๐๐ and geometry factor, ๐ฑ = ๐. ๐๐,
the bending stress for gear 4 is given by Equation 7 is:
From Figure 9 in appendix, the stress-cycle factor for bending, ๐๐ต = ๐. ๐ for ๐๐๐๐๐๐๐๐๐
16
Now using Grade 2 carburized and hardened as before, the bending strength, from Table 3, is
given by ๐บ๐ = ๐๐๐๐๐ ๐๐๐ Answer
Assume that bending factor of safety, ๐๐น = 1 and that ๐พ๐ and ๐พ๐ = 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, ๐๐ = ๐. ๐๐ and bending factor of safety, ๐ = ๐. ๐๐
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, ๐ต๐ = ๐๐ ๐๐๐๐๐ for which, from Figure 8 in appendix, geometry
factor, J = 0.41
Also the speed of gear 5 is different which is ๐๐ = ๐๐. ๐ ๐น๐ท๐ด
From the speed, the life in revolutions, L of gear 5 is given by:
๐ณ = ๐๐ โ ๐๐๐๐๐ โ ๐๐๐๐๐
๐ณ = ๐๐ โ ๐๐๐๐๐ โ ๐๐. ๐
๐ณ = ๐. ๐ โ ๐๐๐ ๐๐๐
From Figure 7 and Figure 9 for ๐๐๐ ๐๐๐๐๐๐
๐๐ต = ๐๐ต = ๐
The contact stress, ๐๐ from gear 5 is same as from gear 4 since they are in contact:
๐๐ = ๐๐๐๐๐๐ ๐๐๐
Now for same design factor, ๐บ๐ฏ = ๐๐ = ๐. ๐,
Endurance strength, ๐บ๐ is then given by:
๐บ๐ =๐บ๐ฏ๐๐
๐๐
๐บ๐ =(๐. ๐)(๐๐๐๐๐๐)
๐
๐บ๐ = ๐๐๐, ๐๐๐ ๐๐๐
18
From Table 2 in appendix, this strength is achievable with grade 2 carburized and hardened
with ๐บ๐ = ๐๐๐๐๐๐ ๐๐๐ 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 ๐ท๐ = ๐ ๐๐. , ๐พ๐๐๐ = ๐๐๐๐ ๐๐๐ , F = 2 in. , ๐ฒ๐ = ๐. ๐๐, and ๐ฒ๐ = ๐. ๐๐ 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, ๐๐ = ๐. ๐๐ and bending factor of safety, ๐ = ๐. ๐๐
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, ๐ฝ๐๐ for gear 2 is 1223 ft/min, for which the dynamic factor, ๐ฒ๐ is given
by Equation 2
๐ฒ๐ =๐๐. ๐ + โ๐๐๐๐
๐๐. ๐
๐ฒ๐ = ๐. ๐๐
Since the transmitted load of ๐พ๐๐๐ = ๐๐๐ ๐๐๐ 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, ๐ช๐๐ from Equation 4 is now:
๐ช๐๐ = ๐. ๐๐๐๐
Then from Equation 3, the corresponding load distribution factor, ๐ฒ๐ = ๐. ๐๐
With, ๐พ๐๐๐ = ๐๐๐. ๐ ๐๐๐, ๐ฒ๐ = ๐. ๐๐, F = 1.5 in. , ๐ฒ๐ = ๐. ๐๐ and ๐ ๐ = ๐. ๐๐ ๐๐. and
๐ฐ = ๐. ๐๐๐๐
the contact stress, ๐๐ for gear 2 from Equation 5 is given by:
20
The life in revolution, L for gear 2 with ๐๐ = ๐๐๐๐ ๐น๐ท๐ด is given by
๐ณ = ๐๐ โ ๐๐๐๐๐ โ ๐๐๐๐๐
๐ณ = ๐๐ โ ๐๐๐๐๐ โ ๐๐๐๐
๐ณ = ๐. ๐๐ โ ๐๐๐ ๐๐๐
From Figure 7 in appendix, the stress-cycle factor for wear, ๐๐ = ๐. ๐ for ๐๐๐๐๐๐๐๐๐
Now for same design factor, ๐บ๐ฏ = ๐๐ = ๐. ๐,
Endurance strength, ๐บ๐ is then given by:
๐บ๐ =๐บ๐ฏ๐๐
๐๐
๐บ๐ =(๐. ๐)(๐๐๐๐๐)
๐. ๐
๐บ๐ = ๐๐๐, ๐๐๐. ๐๐ ๐๐๐
From Table 2 in appendix, this strength is achievable with grade 1 flame hardened
with ๐บ๐ = ๐๐๐, ๐๐๐ ๐๐๐ Answer
Factor of safety for wear is now
Answer
Gear 2 bending
Number of teeth on gear 2, ๐ต๐ = ๐๐ 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, ๐พ๐๐๐ = ๐๐๐. ๐ ๐๐๐, ๐ฒ๐ = ๐. ๐๐, ๐ท๐ = ๐ ๐๐. , t, F
= 1.5 in. , and ๐ฒ๐ = ๐. ๐๐, and geometry factor, ๐ฑ = ๐. ๐๐ is given by:
From Figure 9 in appendix, the stress-cycle factor for bending, ๐๐ต = ๐. ๐๐ for ๐๐๐๐๐๐๐๐๐
Now using grade 1 flame hardened with as before, the bending strength, from Table 3 is
๐บ๐ = ๐๐๐๐๐ ๐๐๐ Answer
Assume that bending factor of safety, ๐๐น = 1 and that ๐พ๐ and ๐พ๐ = 1 as before
Then, allowable bending stress is given by
๐๐๐๐ = ๐๐ก๐๐
๐๐๐๐ = (45000)(0.88)
๐๐๐๐ = 39600 ๐๐ ๐
Now factor of safety for bending is
Answer
Gear 2 specification
and
22
Wear factor of safety, ๐๐ = ๐. ๐๐ and bending factor of safety, ๐ = ๐. ๐๐
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, ๐ต๐ = ๐๐ ๐๐๐๐๐ for which, from Figure 8 in appendix, geometry
factor, J = 0.41
Also the speed of gear 3 is different which is ๐๐ = ๐๐๐. ๐ ๐น๐ท๐ด
From the speed, the life in revolutions, L of gear 3 is
๐ณ = ๐๐ โ ๐๐๐๐๐ โ ๐๐๐๐๐
๐ณ = ๐๐ โ ๐๐๐๐๐ โ ๐๐๐. ๐
๐ณ = ๐. ๐ โ ๐๐๐ ๐๐๐
For 108 ๐๐ฆ๐๐๐๐ from Figures A and C in appendix
๐๐ต = ๐๐ต = ๐. ๐
The contact stress, ๐๐ from gear 3 is same as gear 2 since they are in contact:
๐๐ = ๐๐๐๐๐ ๐๐๐
Now for same design factor, ๐บ๐ฏ = ๐๐ = ๐. ๐,
Endurance strength, ๐บ๐ is then given by
23
๐บ๐ =๐บ๐ฏ๐๐
๐๐
๐บ๐ =(๐. ๐)(๐๐๐๐๐)
๐. ๐
๐บ๐ = ๐๐๐, ๐๐๐. ๐๐ ๐๐๐
From Table 2 and Figure 10 in appendix, this strength is achievable with grade 1 through
hardened with ๐บ๐ = ๐๐๐, ๐๐๐ ๐๐๐ and ๐บ๐ = ๐๐๐๐๐ ๐๐๐ with hardness of 300 ๐ฏ๐ฉ Answer
Now with, ๐๐,๐๐๐ = ๐บ๐๐๐ = (๐๐๐๐๐๐)(๐. ๐) = ๐๐๐๐๐๐ ๐๐๐
The factor of safety for wear is
๐๐,๐๐๐ = ๐๐,๐๐๐
๐๐
๐๐,๐๐๐ = ๐. ๐๐ Answer
Now for bending, note that due to J = 0.27 instead of J = 0.41, and from Equation 7 with ๐ฒ๐ =
๐. ๐๐, facewidth, F=1.5, ๐ฒ๐ = ๐. ๐๐ and ๐พ๐๐๐ = ๐๐๐ ๐๐๐, bending stress on gear 3 is now
given by
๐ =(๐๐๐ )(๐. ๐๐)(๐)(๐. ๐๐)
(๐. ๐)(๐. ๐๐)
๐ = ๐๐๐๐ ๐๐๐
And so the corresponding factor of safety for bending is now given by
Answer
24
Gear 3 specifications
and
Wear factor of safety, ๐๐ = ๐. ๐๐ and bending factor of safety, ๐ = ๐. ๐๐
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, โ ๐ = ๐๐ยฐ,
the radial forces are given by
๐ญ๐๐๐ = ๐๐๐ ๐ญ๐๐ง(๐๐ยฐ) = ๐๐๐ ๐๐๐
๐ญ๐๐๐ = ๐๐๐๐ ๐ญ๐๐ง(๐๐ยฐ) = ๐๐๐ ๐๐๐
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, ๐๐ก๐๐ก 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), ๐ด๐ = ๐๐๐๐ ๐๐๐ โ ๐๐
Constant/midrange torque at ๐ป๐ = ๐๐๐๐ ๐๐๐
Midrange bending moment, ๐ด๐ = ๐
Maximum torque, ๐ป๐ = ๐
A suitable material selected for the shaft is AISI 1020 CD steel. For this material, the ultimate
tensile strength is ๐บ๐๐ = ๐๐ ๐๐๐๐
From Table 4 in appendix, the surface factor, ๐๐ for cold-drawn (CD) steel is
Since the shaft diameters are not known yet, a value of 0.9 for size factor, ๐๐ is assumed
Since bending moment is greater than torque, loading factor, ๐๐ = ๐
No rotating beam endurance limit is known as room temperature, ๐๐ so temperature factor
๐๐ = ๐
Assume 50% reliability for which reliability factor, ๐๐ = ๐
For ๐๐ข๐ก โค 200 ๐๐๐ ๐, the rotary-beam test specimen modification factor, ๐บ๐โฒ is given by
๐บ๐โฒ = ๐. ๐ โ ๐บ๐๐
30
๐บ๐โฒ = ๐. ๐ โ ๐๐ = ๐๐ ๐ค๐ฉ๐ฌ๐ข
Now, the endurance limit, ๐๐ ๐๐ ๐๐๐ฃ๐๐ ๐๐ฆ
Equation 9
๐บ๐ = (๐. ๐๐๐)(๐. ๐)(๐)(๐)(๐)(๐๐)
๐บ๐ = ๐๐ ๐๐๐๐
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: ๐๐ = ๐. ๐
(bending) and ๐ = ๐. ๐ (torsion)
For simplicity for now, assume that the shaft is notch-free such that ๐๐ = ๐๐ and ๐๐๐ = ๐๐๐
Now, for the estimation of the shaft diameter, ๐ท4 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, ๐ = ๐. ๐,
31
This value of d = 1.65 in. is an estimate so now, check with d = 1.625 in.
A typical ๐ซ ๐ โ ratio for a support at a shoulder is ๐ซ
๐ = ๐. ๐
So nominal diameter, ๐ซ = ๐. ๐(๐. ๐๐๐) = ๐. ๐๐ ๐๐.
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
๐ซ๐ = ๐. ๐ ๐๐. and ๐ = ๐ซ๐ = ๐ซ๐ = ๐. ๐๐๐ ๐๐. Answer
The new ๐ท ๐โ ratio is now given by
๐ซ
๐ =
๐. ๐
๐. ๐๐๐= ๐. ๐๐
From Table 5 in appendix for this well-rounded shoulder fillet
๐๐ โ = ๐. ๐
With d = 1.625 in., fillet radius is ๐ โ ๐. ๐๐ ๐๐.
With ๐บ๐๐ = ๐๐ ๐๐๐๐, r = 0.16 in. ,
32
from Figure 11 in appendix, notch sensitivity, q = 0.82 and from Figure 12 in appendix, notch
sensitivity shear, ๐๐๐๐๐๐ = ๐. ๐๐
With ๐ซ
๐ = ๐. ๐๐ and ๐ ๐ โ = ๐. ๐
From Figure 13 in appendix, ๐ฒ๐ = ๐. ๐
From Figure 14 in appendix, ๐ฒ๐๐ = ๐. ๐๐
So now,
The fatigue stress-concentration factor from bending, ๐ฒ๐ is given by
Equation 11
The fatigue stress-concentration factor from torsion, ๐ฒ๐๐ is given by
Now, letโs evaluate the endurance strength, ๐บ๐
๐๐ = ๐. ๐๐๐ (Same as before)
Since d = 1.625 in. is between 0.11 in. and 2 in.,
๐๐ = ๐. ๐๐๐๐ โ๐.๐๐๐
33
๐๐ = 0.835
Now, from Equation 9
๐บ๐ = (๐. ๐๐๐)(๐. ๐๐๐)(๐)(๐)(๐)(๐๐)
๐บ๐ = ๐๐. ๐ ๐๐๐๐
The effective von Mises stress, ๐โฒ at a given point is given by
For stress amplitude, ๐๐โฒ
Equation 12
With ๐๐ = 0 at point I, ๐๐โฒ is given by
For midrange stress, ๐๐โฒ
Equation 13
With ๐๐ = 0 at point I, ๐๐โฒ is given by:
Now the fatigue failure criteria for the modified Goodman line is given by
34
Equation 14
๐
๐=
๐๐๐๐๐
๐๐๐๐๐+
๐๐๐๐
๐๐๐๐๐= ๐. ๐๐๐
๐ = ๐. ๐๐ (Fatigue safety of factor) Answer
Check for yielding
Stress amplitude, ๐๐
๐๐ =(๐. ๐๐)(๐๐)(๐๐๐๐)
๐ (๐. ๐๐๐)๐
๐๐ = ๐๐, ๐๐๐. ๐๐ ๐๐๐
Midrange Torsion, ๐๐
๐๐ =(๐. ๐๐)(๐๐)(๐๐๐๐)
๐ (๐. ๐๐๐)๐
๐๐ = ๐, ๐๐๐. ๐๐ ๐๐๐
and at point I,
๐๐ = ๐๐ = ๐
Combined maximum von Mises stress, ๐๐๐๐ฅโฒ is given by
35
Equation 15
๐๐๐๐โฒ = [(๐ + ๐๐, ๐๐๐. ๐๐)๐ + ๐(๐, ๐๐๐. ๐๐ + ๐)๐]
๐๐โ
๐๐๐๐โฒ = ๐๐, ๐๐๐. ๐๐ ๐๐๐
Now check if the sum of ๐๐, + ๐๐
โฒ is greater than ๐๐๐๐โฒ
๐๐, + ๐๐
โฒ = ๐๐๐๐๐ + ๐๐๐๐ = ๐๐, ๐๐๐ ๐๐๐ โฅ ๐๐, ๐๐๐. ๐๐ ๐๐๐ โฅ ๐๐๐๐โฒ
Hence, there will be no yielding
Also check with yielding factor of safety, ๐๐
For AISI 1020 CD steel, yield strength, ๐บ๐ = ๐๐ ๐๐๐๐
๐๐ =๐บ๐
๐๐๐๐โฒ
=๐๐๐๐๐
๐๐๐๐๐= ๐. ๐๐ > ๐
This confirms there will be no yielding since ๐๐ > ๐
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 ๐ซ๐ = ๐ซ๐ and ๐ซ๐ = ๐ซ๐
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 ๐ด๐ = ๐๐๐๐ ๐๐ โ ๐๐ while ๐ป๐ =
๐๐๐๐ ๐๐ โ ๐๐ 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 ๐ซ
๐ = ๐. ๐๐ and ๐ ๐ โ = ๐. ๐๐,
from Figure 13 and H in appendix
and with ๐บ๐๐ = ๐๐ ๐๐๐๐, r = 0.0325 in. ,
from Figure 11 in appendix, notch sensitivity, q = 0.65 and from Figure 12 in appendix, notch
sensitivity shear, ๐๐๐๐๐๐ = ๐. ๐๐
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 ๐ป๐ = ๐ and ๐ด๐ = ๐๐๐๐ ๐๐ โ ๐๐ at point I, ๐๐โฒ is given by
From Equation 12, with ๐ด๐ = ๐ and ๐ป๐ = ๐๐๐๐ ๐๐ โ ๐๐ at point I, ๐๐โฒ 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
Letโs try with a higher strength material, AISI 1050 CD steel with ๐บ๐๐ = ๐๐๐ ๐๐๐๐
Now recalculate everything as before
From Table 4 in appendix, surface factor, ๐๐
From Equation 9, endurance strength, ๐บ๐ with ๐๐ = 0.835
With ๐บ๐๐ = ๐๐๐ ๐๐๐๐, r = = 0.0325 in.,
from Figure 11 in appendix, notch sensitivity, q = 0.72
With ๐ซ
๐ = ๐. ๐๐ and ๐ ๐ โ = ๐. ๐๐ ,
and from Figure 13 in appendix, ๐ฒ๐ = ๐. ๐๐. , ๐ฒ๐ is given by
Next, with ๐ด๐ = ๐๐๐๐ ๐๐ โ ๐๐, ๐๐โฒ is given by
38
Now the fatigue failure criteria for the modified Goodman line given by Equation 14 is:
Answer
Now letโs 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 ๐๐ = 0
And at this point K,
To check if this location of K is potentially critical, use ๐ฒ๐ = ๐ฒ๐ = ๐. ๐ 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. Letโs look for a specific
retaining ring to obtain ๐ฒ๐ 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 ๐ ๐โ = ๐. ๐๐๐. ๐๐๐โ = ๐. ๐๐๐ and ๐ ๐โ = ๐. ๐๐๐
๐. ๐๐๐โ = ๐. ๐๐
๐ฒ๐ = ๐. ๐
With ๐บ๐๐ = ๐๐๐ ๐๐๐๐, r = = 0.01 in.,
from Figure 11 in appendix, q = 0.65 in.
Then,
A fatigue factor of safety at point K is now
๐๐ = 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, ๐ฒ๐ = ๐. ๐
With d = 1 in. , r = 1 in.
With ๐บ๐๐ = ๐๐๐ ๐๐๐๐, r = 1 in.,
from Figure 11 in appendix, q = 0.7 in.
then,
๐๐ = 1.56 Answer
41
Now we have for diameters for critical locations, M (๐ซ๐ = ๐ซ๐)and I (๐ซ๐ = ๐ซ๐ = ๐. ๐ ๐๐. )of
the shaft with trial values for other sections of the shaft at K without taking the deflections into
account
๐ซ๐ = ๐ซ๐ = ๐. ๐ ๐๐. and ๐ซ๐ = ๐ซ๐ = ๐. ๐ ๐๐. 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 Youngโs 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 ๐ ๐๐๐ = ๐ซ๐ = ๐. ๐ ๐๐ and design factor, ๐๐ = ๐
44
The ratio ๐ ๐๐๐
๐ ๐๐๐ โ is given by
๐ ๐๐๐๐ ๐๐๐
โ = ๐. ๐๐๐๐โ = ๐. ๐๐๐
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, ๐ซ๐ = ๐ซ๐ = ๐. ๐ ๐๐.
With deflection, ๐ซ๐ = ๐ซ๐ = ๐. ๐๐๐ ๐๐.
Fatigue Facotr of Safety, ๐๐ = ๐. ๐๐
45
At point to right of I
Without deflection, ๐ซ๐ = ๐ซ๐ = ๐. ๐ ๐๐.
With deflection, ๐ซ๐ = ๐ซ๐ = ๐. ๐๐๐ ๐๐.
Fatigue Facotr of Safety, ๐๐ = ๐. ๐๐
Nominal Diameter, ๐ท4 = 2.0 ๐๐.
At point K
Without deflection, ๐ซ๐ = ๐ซ๐ = ๐. ๐๐๐ ๐๐.
With deflection, ๐ซ๐ = ๐ซ๐ = ๐. ๐๐๐ ๐๐.
Fatigue Facotr of Safety, ๐๐ = ๐. ๐๐
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, ๐ถ10 or load rating, ๐น๐
From the gearbox design specifications, the design life is 12,000 hours, and the speed of the
intermediate shaft was found out be ๐๐ = ๐๐ = ๐๐๐ ๐น๐ท๐ด
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
๐ณ = ๐๐ โ ๐๐๐๐๐ โ ๐๐๐๐๐
๐ณ = ๐๐ โ ๐๐๐๐๐ โ ๐๐๐. ๐
๐ณ = ๐. ๐ โ ๐๐๐ ๐๐๐
The load rating for a bearing is given by
Equation 18
where ๐๐ซ = ๐๐๐๐ ๐๐๐๐ ๐ข๐๐ , ๐๐ = ๐๐๐๐๐๐ข๐ ๐ฃ๐๐๐ข๐ ๐๐ ๐กโ๐ ๐ฃ๐๐๐๐๐ก๐,
๐ฝ = ๐โ๐๐๐๐๐ก๐๐๐๐ ๐ก๐๐ ๐๐๐๐๐๐๐ก๐๐ ๐๐๐๐๐๐ ๐๐๐๐๐๐๐ ๐ก๐ 63.2 ๐๐๐๐๐๐๐ก๐๐๐ ๐๐ ๐กโ๐ ๐ฃ๐๐๐๐๐ก๐,
๐น๐ซ = ๐๐๐ ๐๐๐๐ ๐๐๐๐๐๐๐๐๐ก๐๐ฆ, ๐๐ = ๐๐๐ ๐๐๐ ๐๐๐๐, ๐ญ๐ซ = ๐๐๐ ๐๐๐๐ ๐๐๐๐
Assume a ball bearing for both bearing A and bearing B for which a = 3
For ๐๐ = ๐, , ๐ญ๐ซ = ๐น๐ฉ = ๐๐๐๐ ๐๐๐, ๐๐ซ = ๐ณ/๐ณ๐๐ =๐.๐โ๐๐๐ ๐๐๐
๐๐๐ and Weibull parameters given
by
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the load rating, ๐ญ๐น๐ฉ = ๐ช๐๐ 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, ๐น๐ ๐ต 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, ๐น๐ ๐ด is given by
where ๐ญ๐ซ = ๐น๐จ = ๐๐๐ ๐๐๐
From SKF, this load rating correspond to a deep groove ball bearing with the following
specifications
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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
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and
Wear factor of safety, ๐๐ = ๐. ๐๐ and bending factor of safety, ๐ = ๐. ๐๐
Gear 5 specification
and
Wear factor of safety, ๐๐ = ๐. ๐๐ and bending factor of safety, ๐ = ๐. ๐๐
Gear 2 specification
and
Wear factor of safety, ๐๐ = ๐. ๐๐ and bending factor of safety, ๐ = ๐. ๐๐
Gear 3 specifications
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and
Wear factor of safety, ๐๐ = ๐. ๐๐ and bending factor of safety, ๐ = ๐. ๐๐
Shafts
Diameters ad Fatigue Factor of Safety
At point M,
Without deflection, ๐ซ๐ = ๐ซ๐ = ๐. ๐ ๐๐.
With deflection, ๐ซ๐ = ๐ซ๐ = ๐. ๐๐๐ ๐๐.
Fatigue Facotr of Safety, ๐๐ = ๐. ๐๐
At point to right of I
Without deflection, ๐ซ๐ = ๐ซ๐ = ๐. ๐ ๐๐.
With deflection, ๐ซ๐ = ๐ซ๐ = ๐. ๐๐๐ ๐๐.
Fatigue Facotr of Safety, ๐๐ = ๐. ๐๐
Nominal Diameter, ๐ท4 = 2.0 ๐๐.
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At point K
Without deflection, ๐ซ๐ = ๐ซ๐ = ๐. ๐๐๐ ๐๐.
With deflection, ๐ซ๐ = ๐ซ๐ = ๐. ๐๐๐ ๐๐.
Fatigue Facotr of Safety, ๐๐ = ๐. ๐๐
Bearings
Bearing B
Bearing A
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References
Budynas, Richard G, J K. Nisbett, and Joseph E. Shigley. Shigley's Mechanical Engineering Design. New
York: McGraw-Hill, 2011. Print.
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Appendix
Figure 7: Stress-cycle factor, ๐๐ vs. Number of load cycles, N
Figure 8: Geometry Factor, J vs. Number of teeth for which geometry factor is desired
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Figure 9: Stress-cycle factor, ๐๐ vs. Number of load cycles, N
Figure 10: Allowable contact stress numbers, ๐๐ vs. Brinell Hardness, ๐ป๐
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Figure 11: Notch sensitivity, q vs. Notch radius, r
Figure 12: Notch sensitivity, ๐๐ โ๐๐๐vs. Notch radius, r
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Figure 13: ๐พ๐ก for round shaft with shoulder fillet in bending
Figure 14: ๐พ๐ก๐ for round shaft with shoulder fillet in torsion
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Figure 15: ๐พ๐ก๐ for round shaft with flat-bottom groove in torsion
Table 2: Contact Strength, ๐๐ at 107cycles and 0.99 Reliability for Steel Gears
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Table 3: Bending Strength, ๐๐ at 107cycles and 0.99 Reliability for Steel Gears
Table 4: Parameters for Marin Surface Modification Factor
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Table 5: First Iteration Estimates for Stress-Concentration Factors, ๐พ๐ก and ๐พ๐ก๐
Table 6: Typical Maximum Ranges for Slopes and Transverse Deflections
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