Skidded helicopter tug design calculation

Final Report
Skidded Helicopter Tug
By
Index No. Name
150127R Dilanjaya H.W.S.
150131A Dilshan K.M.G.L.
150235V Indupama S.V.A.A.

Contents
1. Problem definition
2. Project brief
3. Conceptual design
4. Concept selection
5. Design calculations
a. Scissor jack calculation
i. Axial load calculation
ii. Pitch angle calculation
iii. Screw friction torque calculation
iv. Screw safety check
v. Friction torque calculation
vi. Lifting velocity calculation
vii. Prime mover selection
b. Gear dimension calculation
i. Helical gear module selection
ii. Spur gear module selection
iii. Helical gear strength calculation
iv. Spur gear strength calculation
v. Standard proportions of gear systems
c. Shaft design
i. Placement of the idle gear
ii. Motor shaft diameter calculation
iii. Input shaft diameter calculation
iv. Lay shaft diameter calculation
v. Output shaft diameter calculation
vi. Idle shaft diameter calculation
d. Keys and splines
i. Keys calculation
ii. Splines calculation
e. Bearing selection
i. Bearing calculation
f. Shaft and gear justification
i. Shaft dimension check
ii. Gear dimension check
iii. Fatigue strength calculation
iv. Bending stress concentration calculation
v. Helical gear final dimension
vi. Spur gear final dimension
g. Lubrication and oil seal
i. Lubricant selection
ii. Oil seal selection
h. Gear box casing
i. Gear box casing dimension calculation
i. Energy loss of the gear box
i. Energy loss from gear mesh

ii. Churning power loss
iii. Bearing power loss
iv. Total power loss
6. Final gearbox design
7. Manufacturing process of one major component
8. Engineering drawings

Problem Definition
The project is on designing a Helicopter Tug to be used in airports and aircraft hangers for
taxing small to mid-sized skidded helicopters. A tug with lifting machine is an essential
requirement for a skidded type helicopter. The weights of commonly used helicopters are
varying from 700 kg – 3500 kg with different sizes. The requirement is to reduce the labour
force, ensure labour safety and increase the manoeuvrability.
Given below are the customer requirements:
• The tug should be able to lift a range of helicopters in different weight categories.
• Electric motor driven with battery power is preferred over IC engines.
• Machine should work under different environmental conditions. (ex: light rain)
• The labour requirement should be a single person with minimum technical knowledge.
• Life span should be at least five years.
• Should be able to manoeuvre in tight spaces.
• Both operator and handled equipment safety is highly required.
• Tug should be easily adjustable to different skid sizes.

Project Brief
Background
Sri Lanka Air Force have facilities here to overhaul helicopters in the country. And they work
with helicopters of different sizes and weights on a daily basis. Target of the project is to design
and develop a helicopter tug which can be used by the Air Force in their operations.
Objectives
• Meet the customer requirements
• Versatility
• Simplistic and robust design
• Meet the Machine Design Project Module requirements
Success Criteria
• Meet the deadlines.
• Accurate and successful design.
• A design which can be manufactured in Sri Lanka.
• Using standard components as much possible.
• Project planning
• Safety Aspects
Meeting User Requirements
• A 3 speed gear mechanism was proposed to handle different weight classes of
helicopters.
o 1st
Gear – 700kg – 1500kg
o 2nd
Gear – 1501kg – 2500kg
o 3rd
Gear – 2501kg – 3500kg
• Since almost all the tugs available in the market have used electric motors and in this
situation applying an IC engine would increase the overall size of the machine (Machine
should be able to go under the belly of a helicopter). So single phase AC motor is
suggested based on early power estimations.
• Adjustable skid handles would be able to accommodate helicopters with different skid
sizes.

• Applying a corrosion resistance coating would protect the structure from light rain. And
also it would be required to cover the battery and seal the gearbox.
• The tug would be remote controlled with the gearbox operated only when the machine
is stopped. So a single person would be enough with a basic knowledge of changing a
gear and using a simple remote control to move the tug.
• All the components would be designed with a design consideration of a life span of 5
years. (ex: Bearing life)
• A steering system for the tug is to be implemented.

Conceptual Design
Three members of the group came up with conceptual designs after discussing among each
other. Designs were done with some inspiration from existing commercial models.
Concept 1
Concept 2

Concept Selection
Concept selection was done via a morphological analysis.
Function Solutions
Lifting the load Scissor jack Hydraulic jack Screw jack
Moving the tug
Powered
Tricycle
Remote controlled
Human
pulled with
machine
assistance
Steering the tug
Tricycle setup
with front
wheel turning
Tracked setup
Omni
directional
wheels
Normal
front
steering
setup
Power IC engine AC motor
Handling various
loads
Gearbox
Variable Motor
Power
Helicopter types Wheeled Skidded
• The hydraulic jack method was rejected because it needs a separate hydraulic system
to operate. And the screw jack was not selected because of power train issues and the
scissor jack provides a convenient and horizontal setup.
• Remote controlled setup was proposed because of the safety aspects that comes with
operating the system separately. Human pulled method would ease the design process
but lower the user experience. Powered tricycle is also a feasible method but remote
controlled method was implemented because of its simplicity in using and the ability
to be used in a dangerous environment.
• Omni directional wheels were proposed because of the ease of design and
implementation and great maneuverability. Implementing tracks would be a good
solution because of the increased maneuverability but the cost would be high. A
tricycle setup would increase the stresses in the front wheel and the system won't be
balanced. Normal steering system is also feasible but will require more complex
designing.

• Motor power was selected over IC engine because of the noise and environmental
issues. And the motor powered system is easier to be used indoors. Also an IC engine
would require a flywheel making the tug bigger.
• A gearbox was proposed over variable motor power because a gearbox would provide
the speed reduction needed with increasing torque.
• Design was focused on skidded type helicopters because a simple mechanism would
be enough to pull a wheeled helicopter. We can implement that also to our design as a
modification later.
Sub Function Solutions
Remote controlled
movements
Fully remote
controlled
Remote control
with
manual gearbox
AC motor Single phase Three phase
Gearbox
Constant
mesh
Sliding mesh
• Remote controlled movements with a manual gearbox was proposed because of the
robustness and the reduction of cost over a fully remote controlled system. Since the
gears are changed when the tug is not operational it is safe to go near the tug, consider
the weight of the helicopter and change gear.
• A Single phase AC motor was selected because 3 phase current is not available
frequently.
• Sliding mesh mechanism for the gearbox was selected because as mentioned above,
the gears are changed when the tug is not operating (Fully stopped). So the
mechanism can be simple hence cost effective and small in size.

Description:
Calculate the total axial load on the threaded bar.
Iteration:
Third iteration with the assumption of the maximum total weight that can lift
by using the helicopter tug and thrust bearing number.
Calculation ID:
SHT_AxialLoad_Itr03
Find:
Total axial load on the thread bar screw
Data:
Maximum weight – 3500 kg
Angle of the cross bars – 10.88⁰
Assumption:
Maximum weight – 3500 kg – with the safety factor 1.17
Select the angle of the cross bars is 10.88⁰ using predefined sketches
Formulae:
𝐹 = 𝑊′
cot(𝜃)
Calculation:
𝐹 = 3500 × 9.81 × cot(10.88°)
𝐹 = 178634.919 𝑁
𝐹 = 178.634 𝑘𝑁
178.634 kN
Conclusion:
Total load acting on the threaded bar due to 3500 kg load is 178.634 kN
Reference:
[1] P. M. P. V. P. Anupam Chathurvedi, “An improved scissor Lift working on Lead
Screw Mechanism,” International Journal of Advance Engineering and Research
Development, vol. 4, no. 2, pp. 89-95, 2017.

Description:
Calculate the pitch angle of the lead screw with the initial assumption of the
diameter, pitch and the material.
Iteration:
First iteration with the assumption of the diameter, pitch and the material.
Calculation ID:
SHT_LeadScrew_Itr01
Find:
Pitch angle of the lead screw
Data:
Lead screw outer diameter – 31.75 mm
Lead screw pitch – 6.35 mm
Assumption:
Formulae:
θ=tan-1
( 𝐿𝑒𝑎𝑑
𝜋 × 𝑇ℎ𝑟𝑒𝑎𝑑 𝑏𝑎𝑟 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟⁄ )
Calculation:
θ=tan-1
(6.35
𝜋 × 31.75⁄ )
θ=tan-1(0.06366)
θ=3.643⁰
3.643⁰
Conclusion:
Pitch angle of the lead screw is 3.643⁰
Reference:
[1] “Helix Angle: Definition, Formula & Calculation,” Study.com, 2018. [Online].
Available: https://study.com/academy/lesson/helix-angle-definition-formula-
calculation.html. [Accessed 17 May 2018].

Description:
Calculate the Screw friction torque of the lead screw when the maximum
load (3500 kg) is applied on the tug.
Iteration:
Fifth iteration assuming the material properties and the frictional coefficient
of the material.
Calculation ID:
SHT_ScrewFrictionTorque_Itr05
Find:
Screw friction torque in the thread bar.
Data:
Pitch angle α – 3.64⁰
Axial load – 178.634 kN
Frictional coefficient – 0.11
Assumption:
Formulae:
𝜆 = tan−1
(µ)
𝑇1 = 𝐹′
× d × tan(𝛼 + 𝜆) /2
Calculation:
𝜆 = tan−1
(µ)
𝜆 = tan−1(0.11°)
𝜆 = 6.2773
𝜆 > 𝛼
So that the screw is in self-locking condition.
𝑇1 = 178.634 × 103
× 31.75 × 10−3
× tan(3.64° + 6.2773°)/2
𝑇1 = 495.95 𝑁𝑚
495.95 Nm

Conclusion:
Total load acting on the threaded bar due to 3500 kg load is 495.95 Nm
Reference:
[1] “Friction Factors for Power Screws,” ROYMECHX clone of ROYMECH, 17
January 2013. [Online]. Available:
http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm. [Accessed 12
February 2018].
[2] “Friction and frequency factors,” American Roller Bearing Company, [Online].
Available: https://www.amroll.com/friction-frequency-factors.html. [Accessed 5 2
2018].

Description:
Justify the lead screw pitch is able to withstand the torque and the load when
the maximum load (3500 kg) is applied on the tug.
Iteration:
Fifth iteration assuming the bearing that can withstand to the axial load.
Calculation ID:
SHT_LeadScrewPitch_Itr05
Find:
Screw thread is safe to lift the load.
Data:
Material properties
Tensile strength [σ] – 550 MPa
Shear stress [τ] – 275 MPa
Screw friction torque – T1 – 495.95 Nm
Thread bar outer diameter – D – 31.75 mm
Lead = pitch – p – 6.35 mm
Assumption:
Material is stainless steel
First thread gets 35% of the total load
Formulae:
Method I -
Maximum shear stress according to the Semester 4 Design of machine
element subject
𝜏 𝑚𝑎𝑥 = (𝐹′
× 0.35 × 2)/(𝜋 × 𝑃 × (𝐷 − 𝑃))
Method II -
Direct tensile strength
𝜎 = (𝐹′
× 4)/(𝜋 × 𝐷2
)
𝜏 = (𝑇1 × 16)/(𝜋 × 𝐷3
)
𝜏 𝑚𝑎𝑥 = √ 𝜎2 + 4 × 𝜏2/2
𝜎 𝑚𝑎𝑥 = 𝜎/2 + 𝜏

Calculation:
Method I -
𝜏 𝑚𝑎𝑥 = (𝐹′
× 0.35 × 2)/(𝜋 × 𝑃 × (𝐷 − 𝑃))
𝜏 𝑚𝑎𝑥 = (178.634 × 103
× 0.35 × 2)/(𝜋 × 6.35 × (31.75 −
6.35))
𝜏 𝑚𝑎𝑥 = 246.77 𝑀𝑃𝑎
𝜏 𝑚𝑎𝑥 < [𝜏]
Screw is safe for the first iteration
Method II
𝜎 = (𝐹′
× 4)/(𝜋 × 𝐷2
)
𝜎 = (178.634 × 103
× 4)/(𝜋 × 31.752
)
𝜎 = 225.63 𝑀𝑃𝑎
𝜏 = (𝑇1 × 16)/(𝜋 × 𝐷3
)
𝜏 = (495.95 × 16)/(𝜋 × (31.75 × 10−3
)3
)
𝜏 = 78.92 𝑀𝑃𝑎
𝜏 𝑚𝑎𝑥 =
1
2
√𝜎2 + 4 × 𝜏2
𝜏 𝑚𝑎𝑥 =
11
2
√225.632 + 4 × 78.922
𝜏 𝑚𝑎𝑥 = 137.68 𝑀𝑃𝑎
𝜎 𝑚𝑎𝑥 =
𝜎
2
+ 𝜏
𝜎 𝑚𝑎𝑥 = 225.63/2 + 78.92
𝜎 𝑚𝑎𝑥 = 250.49𝑀𝑃𝑎
𝜏 𝑚𝑎𝑥 < [𝜏] and 𝜎 𝑚𝑎𝑥 > [𝜎]
So that the screw is safe for that axial load.
Conclusion:
The Screw is safe for the given operating region.
Reference:
[1] P. M. P. V. P. Anupam Chathurvedi, “An improved scissor Lift working on Lead
Screw Mechanism,” International Journal of Advance Engineering and Research
Development, vol. 4, no. 2, pp. 89-95, 2017.

Description:
Calculate the torque that need to rotate the screw when the maximum load
(3500 kg) is applied on the tug.
Iteration:
Fifth iteration assuming the bearing that can withstand to the axial load.
Calculation ID:
SHT_ReqTorque_Itr05
Find:
Screw thread is safe to lift the load.
Data:
Axial load – 178.634 kN
Bearing inner diameter – 35 mm
Bearing outer diameter – 80 mm
𝐶 𝑜𝑟– 170 kN
Bearing frictional coefficient – 0.09
Assumption:
Bearing number – 51407
Formulae:
𝑑 𝑚 = (𝑑𝑖 + 𝑑 𝑜)/2
𝑇2 = 𝐹′
× 𝑑 𝑚 × 𝜇 𝑏
Calculation:
𝑑 𝑚 = (𝑑𝑖 + 𝑑 𝑜)/2
𝑑 𝑚 = (35 + 80)/2
𝑑 𝑚 = 57.5 𝑚𝑚
𝑇2 = 𝐹′
× 𝑑 𝑚 × 𝜇 𝑏
𝑇2 = 178.634 × 103
× 57.5 × 10−3
× 0.09
𝑇2 = 924.44 𝑁𝑚
924.44 Nm
Conclusion:
Frictional torque on thrust bearing on the threaded bar due to 3500 kg load is
924.44 Nm

Reference:
[1] “Thrust bearing,” SKF, [Online]. Available: http://www.pkl.hr/download/lezajevi/skf-thrust-
bearings.pdf. [Accessed 10 February 2018].

Description:
Calculate the all rpm values that is sufficient for the helicopter tug.
Iteration:
First iteration assuming the maximum vertical velocity of the lifting scissor
jack
Calculation ID:
SHT_TugRPM_Itr01
Find:
Rpm values of the screw thread bar.
Data:
Maximum vertical velocity – v1– 0.015 m/s
Second maximum vertical velocity – v2 – 0.01 m/s
Third maximum vertical velocity – v3– 0.005 m/s
Scissor jack maximum angle between cross bars – γ – 14.77⁰
Thread bar pitch – p – 6.35 mm
Assumption:
Bearing number – maximum vertical velocity is 0.015 m/s
Formulae:
𝑟𝑝𝑚𝑖 = (60 × 𝑉𝑖 tan(𝛾))/(𝑝)
Calculation:
𝑟𝑝𝑚1 = (60 × 0.015 × tan(14.77°))/(6.35 × 10−3
)
𝑟𝑝𝑚1 = 37.134 , 𝜔1 = 3.889 𝑟𝑎𝑑/𝑠
𝑟𝑝𝑚1
= 37.134
Conclusion:
So that selected maximum rpm value for the screw thread is 37 rpm.
Other two rpm values are selected as 25 rpm and 17 rpm.
So that the new lifting velocity values are,
37rpm – 0.01485 m/s
25 rpm – 0.01004 m/s
17 rpm – 0.00684 m/s
Reference:

Description:
Select the prime mover that can provide torque and power sufficient to the
requirement.
Iteration:
Third iteration
Calculation ID:
SHT_PrimeMover_Itr03
Find:
Power required from the prime mover and the efficient prime mover
Data:
Total torque - Tt – 1420.3857 Nm
Lead screw data
Minimum rpm of the lead screw – 12.378 (1.296 rad/s)
Assumption:
Efficiencies
Gear box efficiency – ηg– 0.80
Scissor jack efficiency – ηj– 0.60
Motor efficiency – ηm– 0.90
Formulae:
𝑃 = 𝑇𝑡 𝜔/(𝜂 𝑔 𝜂 𝑗 𝜂 𝑚)
Calculation:
𝑃 = 1420.385 × 12.378/(0.90 × 0.80 × 0.60)
𝑃 = 4261.157 𝑊
𝑃 = 4.261 𝑘𝑊
𝑃 = 4.261 𝑘𝑊
Conclusion:
Power required for the motor is 4.261 kW. But considering the torque values
and other energy losses, the suitable motor is 5.5 kW motor.
Selected motor specifications
• Motor type – YL132S-4
• Power – 5.5kW
• Voltage – 220V/50Hz

• Pole – 4
• Speed (RPM) – 1400
Reference:
[1] “Electric motor efficiency,” The Engineering ToolBox, [Online]. Available:
https://www.engineeringtoolbox.com/electrical-motor-efficiency-d_655.html.
[Accessed 16 02 2018].
[2] J. Edwards, “Design and Fabrication of Power Scissor Jack.,” Slide player, 2016.
[Online]. Available: http://slideplayer.com/slide/10186142/. [Accessed 16 02 2018].
[3] F. Faulhaber, “A second look at Gearbox efficiencies,” MACHINE DESIGN, 20 June
2002. [Online]. Available: http://www.machinedesign.com/archive/second-look-
gearbox-efficiencies. [Accessed 16 Fecruary 2018].
[4] “Single phase AC motor,” Alibaba.com, [Online]. Available:
https://m.alibaba.com/product/60514541714/50-60HZ-single-phase-ac-
motors.html?s=p&spm=a2706.amp_showroom#show_specifications. [Accessed 17
March 2018].

Description:
Find the suitable modules for the helical gears
Iteration:
Fifth iteration based on the rpm values that calculated above.
Calculation ID:
SHT_HelicalModule_Itr05
Find:
Gear modules for the helical gear pairs
Data:
Motor power – 4.5 kW
Speed of the engine – 1400 rpm
Gear pair name Speed ratio Rpm of the pinion
H01 – H02 6 1400
T01 – T02 5 233.333
Pressure angle - 𝜑 = 20°
Helical gear angle - 𝛼 = 20°
Tooth error action - 𝑒 = 0.0925 𝑚𝑚
(Since the pitch line velocity is less than 12𝑚𝑠−1
and gears are well cut commercial
gears)
Depending factor for deformation factor – 0.111
(Since gears are 20° full depth involute system)
Service factor 𝐶𝑣 = 1.1 (Medium shock intermittent or 3hr per day)
Gear material properties
• Static strength of gear material - 𝜎0 = 510𝑀𝑃𝑎
• Material Brinell hardness number – 444
• Youngs modulus – 15.3 GPa (The average value)
Assumption:
Gear material – Nickel Chromium steel (30 Ni 4 Cr 1)
Speed reduction for the initial stage. Gear pair 1 – 6 and gear pair 2 – 5
Gear module can be changed from 1 to 10 in relevant values those are in the catalogues
Gear face width can be changed from 6 to 20. Since pitch line velocity is less than 12 𝑚𝑠−1
Formulae:
• To avoid interference:
𝑇𝑝𝑖𝑛𝑖𝑜𝑛 ≥
2
√𝐺2+𝑆𝑖𝑛2(𝜑)(1+2𝐺)−𝐺
; 𝐺 > 1
• Normal pitch; 𝑃𝑐 = 𝑚𝜋
𝑃 𝑁 = 𝑃𝐶 cos(𝛼)
• Pressure angle in normal plane
tan(𝜑 𝑁) = tan(𝜑) × cos(𝛼)
• Overlap, minimum face width factor
𝑏 =
1.15𝑃𝑐
tan(𝛼)

• Equivalent number of teeth
𝑇𝐸 =
𝑇
cos3(𝛼)
• 𝐶𝑣 =
6
6+𝑣
• Tangential velocity
𝑣 = 𝜋𝐷 𝑝 𝑁𝑝
• Lewis form factor
𝑦′
= 0.154 −
0.912
𝑇 𝐸
• Tangential tooth load
𝑊𝑇 =
𝑇
𝐷 𝑝 2⁄
= (𝜎0 𝐶𝑣)𝑏𝑚𝜋𝑦′
Calculation:
Calculate the module for the gear pair 1
Minimum number if teeth
𝑇 𝐻1 ≥
2
√62+𝑆𝑖𝑛2(20°)(1+2×6)−6
𝑇 𝐻1 ≥ 15.9470
Selected number of teeth
𝑇 𝐻1 = 20
Torque on the gear
𝑇 =
4.5×60×106
2×1400×𝜋
𝑇 = 30694.1676 𝑁𝑚
Equivalent gear teeth number
𝑇𝐸 =
20
cos3(20°)
𝑇𝐸 = 24.1030
T01
T02
H01
H02
Gear pair 2
Gear pair 1

Selected equivalent gear teeth number
𝑇𝐸 = 25
Lewis form factor
𝑦′
= 0.154 −
0.912
25
𝑦′
= 0.11752
The equation used for calculate the module
2𝑇
𝑚𝑇𝑝
=
6𝜎0
6+
𝜋𝑇𝑝 𝑁 𝑝 𝑚
1000
⁄
𝑥𝑦′
𝜋2
𝑚2
Coefficients
𝑚3
= 974295.7637
𝑚1
= 5400000
𝑚0
= 368330.0112
By using trial and error, selected module is 2.5 mm
Selected face width factor is 14
Face width
𝑏 = 2.5 × 14 = 35𝑚𝑚
Pitch circle diameter
𝐷 𝑝 = 2.5 × 20 = 50𝑚𝑚
𝑇 𝑤 = 6 × 20 = 120
𝐷 𝑤 = 2.5 × 120 = 300𝑚𝑚
Centre distance of the gears = 175mm
Same procedure repeated for the other gears
Gear name Speed rpm Speed ratio
Selected speed
ratio
Minimum number of
teeth pinion
H01 1400
H02 233.33333
T01 233.33333
T02 46.666667
6 6
5 5
15.94703297
15.74045824
Gear name
Selected
number of
teeth - Ti
Torque
transfer - T -
Nmm
Te Te selected Form factor - y'
H01 20 30694.1676 24.10308023 25 0.11752
H02 120 0.1464
T01 16 184165.006 19.28246418 25 0.11752
T02 80 0.1426
6𝑇𝑝 𝜎0 𝑥𝑦′
𝜋2
𝑚3
− 2𝜋𝑇𝑇𝑝 𝑁𝑝 𝑚 − 12𝑇 = 0

Conclusion:
Selected modules for the Gear pair 1 – 2.5mm
Selected module for the Gear pair 2 – 3mm
Gear name m3 m m0 m= Check equation
H01 974295.76 5400000 368330.0112 1 -4794034.247
2 -3373963.901
2.5 1355041.297
3 9737655.61
4 40386598.87
5 94418640.46
H02
T01 779436.61 4320000 2209980.067 1 -5750543.456
2 -4614487.179
2.5 -831283.0202
3 5874808.43
4 30393963.04
5 73619596.31
T02
Gear name
Selected
module -
mm
Face width -
mm
Selected face
width - mm
Pitch circle diameter Ceter distance
H01 2.5 44.8709663 35 50
H02 44.8709663 35 300
T01 3 32.0506902 30 48
T02 32.0506902 30 240
175
144

Reference:
[1] C. c. morse, Standard gears catelogue, 2017.
[2] SDP/SI, Elements of meteric gear technology, 2017.
[3] p. K. &. p. M.M.Mayuran, "Design of gearbox," 2016.
[4] AGMA, "Gear materials," 2014.
[5] K. S. gears, "Gear materuals and heat treatment," Kohara Gear Industry Co. ltd, 2015. [Online].
Available: https://khkgears.net/gear-knowledge/abcs-gears-b/gear-materials-heat-treatments/.
[Accessed 28 March 2018].
[6] ROYMECHX, "Gear efficiency," 20 January 2013. [Online]. Available:
http://www.roymech.co.uk/Useful_Tables/Drive/Gear_Efficiency.html. [Accessed 01 04 2018].
[7] "JIS S35C Mechanical Properties," Mead info, 23 November 2010. [Online]. Available:
http://www.meadinfo.org/2010/11/jis-s35c-mechanical-properties.html. [Accessed 01 04 2018].
[8] B. M. B.V., "Level of accuracy and compliance of the teeth," Tandwiel.info, 2017. [Online].
Available: http://www.tandwiel.info/en/gears/level-of-accuracy-and-compliance-of-the-teeth/.
[Accessed 02 04 2018].
[9] "JIS G4051 Steel for machine structural use," TAI Special Steel, 2017. [Online]. Available:
http://www.astmsteel.com/product/jis-s45c-steel-machine-structural/. [Accessed 02 April 2018].
[10] A. G. Inc, "American standard & Metric custon gear size range," American Gear Inc, 2008.
[Online]. Available: http://www.americangearinc.com/gears.html. [Accessed 02 April 2018].
[11] J. G. R.S. Khurmi, A Text Book of Machine Design, New Delhi: Eurasia Publishing House, 2005.

Description:
Find the suitable modules for the spur gears
Iteration:
Fifth iteration based on the rpm values that calculated above.
Calculation ID:
SHT_SpurModules_Itr05
Find:
Gear modules for the spur gear pairs
Data:
Motor power – 4.5 kW
Speed of the engine – 1400 rpm
Gear pair name Speed ratio Selected speed ratio Rpm of the pinion
T11 – T12 2.745 2.74 46.666
T21 – T22 1.867 2 46.666
T31 – T32 1.261 1.4 46.666
TR1 – TRI 1.25 1.25 46.666
TRI – TR2 1.7 1.7 37.333
Pressure angle - 𝜑 = 20°
Tooth error action - 𝑒 = 0.0925 𝑚𝑚
(Since the pitch line velocity is less than 12𝑚𝑠−1
and gears are well cut commercial
gears)
Assumption:
Formulae:
• To avoid interference:
𝑇𝑝𝑖𝑛𝑖𝑜𝑛 ≥
2
√𝐺2+𝑆𝑖𝑛2(𝜑)(1+2𝐺)−𝐺
; 𝐺 > 1
• Normal pitch; 𝑃𝑐 = 𝑚𝜋
• 𝐶𝑣 =
6
6+𝑣
• Tangential velocity
𝑣 =
𝜋𝐷 𝑝 𝑁 𝑝
60

𝑦 = 0.154 −
0.912
𝑇
𝑊𝑇 =
𝑃𝐶 𝑣
𝑣
= (𝜎0 𝐶𝑣)𝑏𝑚𝜋𝑦
Calculation:
Calculate the module for the gear pair 1
Minimum number if teeth
𝑇11 ≥
2
√2.752+𝑆𝑖𝑛2(20°)(1+2×2.75)−2.75
𝑇11 ≥ 14.821
Selected number of teeth
𝑇11 = 16
Lewis form factor
𝑦 = 0.154 −
0.912
20
𝑦 = 0.097
The equation used for calculate the module
𝑃𝐶 𝑣
𝑣
=
6×104×𝑃𝐶 𝑠
𝜋𝑚𝑇𝑝 𝑁 𝑝
= (𝜎0 𝐶𝑣)𝑥𝑚2
𝜋𝑦𝑦𝑝
3𝜎0 𝑦𝑝 𝑥𝑇𝑝 𝑁𝑝 𝜋2
𝑚3
− 𝜋𝑇𝑝 𝑁𝑝 𝐶𝑠 𝑃𝑚 − 18 × 104
𝑃𝐶𝑠 = 0
Coefficients
𝑚3
= 15011273.5
𝑚1
= 11728612.6
𝑚0
= 900000000
Gear pair 2 1st gear2nd gear 3rd gear Reverse 1.1
T01
T02 T11
T12 T22 T32 TR2
T21 T31
TR1
TRI
Reverse
1.2

By using trial and error, selected module is 4 mm
Selected face width factor is 14
Face width
𝑏 = 4 × 14 = 56𝑚𝑚
𝐷 𝑝 = 2.75 × 16 = 64𝑚𝑚
𝑇 𝑤 = 2.75 × 20 = 44
𝐷 𝑤 = 4 × 44 = 176𝑚𝑚
Centre distance of the gears = 120mm
Same procedure repeated for the other gears
Gear name Speed rpm Speed ratio
Selected speed
ratio
Minimum number of
teeth pinion
Selected number
of teeth - Ti
T11 46.666667 2.74509804 2.75 14.82184459 16
T12 17 44
T21 46.666667 1.86666667 2 14.16075915 16
T22 25 32
T31 46.666667 1.26126126 1.4 13.27580099 20
T32 37 28
TR1 46.666667 1.25 1.25 12.96583978 16
TRI 37.333333 1.7 1.7 13.77566813 20
TR2 21.960784 34
Gear name
Form
factor - y
Facewidth
factor
m3 m m0
T11 0.097 14 15011273.51 11728612.57 900000000
T21 0.097 8 8577870.577 11728612.57 900000000
T31 0.1084 7 10484678.15 14660765.72 900000000
TR1 0.097 14 15011273.51 11728612.57 900000000

Conclusion:
Selected modules for the 1st
gear – 4mm
Selected module for the 2nd
gear – 4mm
Selected module for the 3nd
gear – 5mm
Selected module for the reverse gear – 4mm
Gear name m=
Check
equation
Selected
module - mm
Face width - mm
Pitch circle
diameter
T11 1 -896717339 4 56 64
2 -791638424
3 -506424228
3.5 -268120261
4 48992892.1
5 964680576
6 2330706466
T12 4 56 176
T21 1 -903150742 5 40 80
2 -843105648
3 -680126107
4 -362744896
4.5 -130070156
5 160505210
6 941091432
T22 5 40 160
T31 1 -904176088 5 35 100
2 -830783341
3 -631574456
4 -243641364
4.5 40755530.5
5 395924003
6 1350029714
T32 5 35 140
TR1 1 -896717339 4 56 64
2 -791638424
3 -506424228
3.5 -268120261
4 48992892.1
5 964680576
6 2330706466
TRI 4 56 80
TR2 4 56 136

Reference:
http://www.roymech.co.uk/Useful_Tables/Drive/Gear_Efficiency.html. [Accessed 01 04
2018].
http://www.meadinfo.org/2010/11/jis-s35c-mechanical-properties.html. [Accessed 01 04
2018].
[Accessed 02 04 2018].
http://www.astmsteel.com/product/jis-s45c-steel-machine-structural/. [Accessed 02 April
2018].
[11] J. G. R.S. Khurmi, A Text Book of Machine Design, New Delhi: Eurasia Publishing House,
2005.

Description:
Helical gear strength calculation.
Iteration:
Fifth iteration
Calculation ID:
SHT_HelicalStrength_Itr05
Find:
Find if the helical gears are strength enough to withstand torques and forces
Data:
Helix angle - 20⁰
Pressure angle - 20⁰
Load stress factor – K – 41.5895
• Flexural endurance limit - 𝜎𝑒 = 777 𝑀𝑃𝑎
• Deformation factor – C – 78.5434
Assumption:
Formulae:
• Ratio factor
𝑄 =
2𝑇 𝐺
𝑇 𝑃+𝑇 𝐺
• Minimum face width
𝑏 =
1.15×𝑃𝑐
tan(𝛼)
• Wear tooth load
𝑊𝑤 =
𝐷 𝑝 𝑏𝑄𝐾
cos2(𝛼)
Gear name Module Speed
Number of
teeth
H01 2.5 1400 20
H02 2.5 233.3333 120
T01 3 233.3333 16
T02 3 46.6667 80

𝑣 =
𝜋𝐷 𝑝 𝑁 𝑝
60
𝐶𝑣 =
6
6+𝑣
• Equivalent number of teeth
𝑇𝐸 =
𝑇
cos2(𝛼)
• Form factor
𝑦′
= 0.154 −
0.912
𝑇 𝐸
𝑊𝑇 = 𝜎0 𝐶𝑣 𝑏𝑚𝜋𝑦′
• Static tooth load
𝑊𝑠 = 𝜎𝑒 𝑏𝑃𝑐 𝑦 = 𝜎𝑒 𝑏𝑚𝑦𝜋
• Dynamic tooth load
𝑊𝐷 = 𝑊𝑇 +
21𝑣(𝑏𝐶 cos2(𝛼)+𝑊 𝑇)cos(𝛼)
21𝑣+√𝑏𝐶 cos2(𝛼)+𝑊 𝑇
Calculation:
For H01 – H02 gears
𝐷 𝑝 = 2.5 × 20 = 50𝑚𝑚
Ratio factor
𝑄 =
2×120
120+20
= 1.714286
Minimum face width
𝑏 =
1.15×𝜋×2.5
tan(20⁰)
= 24.8154𝑚𝑚
Selected face width
𝑏 = 35 𝑚𝑚
Wear tooth load
𝑊𝑤 =
50×35×1.714286×41.5895
cos2(𝛼)
= 141297.2397𝑁
Tangential velocity
𝑣 =
𝜋×50×1400
60×1000
= 3.6652 𝑚𝑠−1
𝐶𝑣 =
6
6+3.6652
= 0.62078
Equivalent number of teeth
𝑇𝐸 =
20
cos2(200)
= 24.10
𝑇𝐸(𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑) = 25

Form factor
𝑦′
= 0.154 −
0.912
25
= 0.11752
Tangential tooth load
𝑊𝑇 = 500 × 0.62078 × 35 × 2.5 × 𝜋 × 0.11752 = 10027.219 𝑁
Static tooth load
𝑊𝑠 = 777 × 35 × 2.5 × 0.11752 × 𝜋 = 25100.98 𝑁
Dynamic tooth load
𝑊𝐷 = 10027.219 +
21×3.6652×(35×78.546×cos2(200)+10027.219) cos(200)
21×3.6652+√35×78.546×cos2(200)+10027.219
𝑊𝐷 = 11418.018 𝑁
Since 𝑊𝐷 < 𝑊𝑤, gears are safe.
For gear pair T01 – T02, calculation process is same.
Conclusion:
Helical gears are safe in the operating region
Reference:
Number of
teeth
Pitch circle diameter Rato factor - Q Face width - mm
Wear tooth
load - Ww
H01 2.5 1400 20 50 1.714285714 35 141297.2397
H02 2.5 233.3333333 120 300 35
T01 3 233.3333333 16 48 1.666666667 30 111491.0051
T02 3 46.66666667 80 240 30
Gear name Tangential velocity - v Cv Form factor - y'
Tangential
tooth load - Wt
Static tooth load
- Ws
Dynamic tooth
load
Does helical
gear safe
H01 3.665191429 0.620784394 0.11752 10027.21907 25100.98287 11418.04841 TRUE
H02 3.665191429 0.620784394 0.1464 12491.36208
T01 0.586430629 0.910963819 0.11752 15134.75161 25818.15381 15550.40898 TRUE
T02 0.586430629 0.910963819 0.1426 18364.66626

[Accessed 02 04 2018].

Description:
Spur gear strength calculation and get the desired dimensions.
Iteration:
Fifth iteration.
Calculation ID:
SHT_SpurStrength_Itr05
Find:
Find if the spur gears are strength enough to withstand torques and forces.
Data:
Pressure angle - 20⁰
Load stress factor – K – 41.5895
• Flexural endurance limit - 𝜎𝑒 = 777 𝑀𝑃𝑎
• Deformation factor – C – 78.5434
Assumption:
Formulae:
• Lewis formulae:
𝑊𝑡 = 𝜎0 𝐶𝑣 𝑏𝑚𝑦𝜋
Number of
teeth
Face width
T11 4 46.6667 16 56
T12 4 17 44 56
T21 5 46.667 16 40
T22 5 25 32 40
T31 5 46.667 20 35
T32 5 37 28 35
TR1 4 46.667 16 56
TRI 4 37.333 20 56
TR2 4 16.97 44 56

𝑦 = 0.154 −
0.912
𝑇
• Dynamic load on gear
𝑊𝐷 = 𝐹𝑇 +
21𝑣(𝑏𝐶+𝐹 𝑇)
21𝑣+√𝑏𝐶+𝐹 𝑇
• Wear tooth load
𝑊𝑤 = 𝑑𝑏𝑄𝐾
• Deformation factor
𝐶 =
𝐾.𝑒
1
𝐸 𝑝
+
1
𝐸 𝐺
• Flexural endurance limit
𝜎𝑒 = 1.75 × 𝐵𝐻𝑁
• Surface endurance limit
𝜎𝑒𝑠 = 2.8 × 𝐵𝐻𝑁 − 70
• Load stress factor
𝐾 =
(𝜎 𝑒𝑠)2 𝑆𝑖𝑛(𝜑)
1.4
[
1
𝐸 𝑝
+
1
𝐸 𝐺
]
• Ratio factor
𝑄 =
2𝑇 𝐺
𝑇 𝐺+𝑇 𝑃
Calculation:
Calculate the strength of the gears of T11 – T12
Diameter of the gear
𝑑 = 𝑚𝑇
𝑑 = 4 × 16 = 64𝑚𝑚
Pitch line velocity of the gears
𝑣 =
𝜋𝑑𝑁
60
𝑣 =
𝜋×0.064×46.667
60
𝑣 = 0.1564 𝑚𝑠−1
To avoid failure of the tooth load calculated the design power
𝑃𝑠 = 𝐶𝑠 × 𝑃
𝑃𝑠 = 1 × 4500
𝑃𝑠 = 4.5 𝑘𝑊
Lewis form factor calculation for pinion
𝑦 = 0.154 −
0.912
16
𝑦 = 0.097

𝐶𝑣 =
3
3+𝑣
; gears are full depth involute gears with pressure angle 20⁰ and pitch line velocity
is less than 25𝑚𝑠−1
𝐶𝑣 =
3
3+0.1564
𝐶𝑣 = 0.9505
Face width factor for the gear – 14
Tangential tooth load on the pinion
𝑊𝑡 = 𝜎0 𝐶𝑣 𝑏𝑚𝑦𝜋
𝑊𝑡 = 510 × 0.9505 × 14 × 42
× 0.097 × 𝜋
𝑊𝑡 = 33088.07997 𝑁
𝐹𝑡 =
𝐷𝑒𝑠𝑖𝑔𝑛 𝑝𝑜𝑤𝑒𝑟
𝑃𝑖𝑡𝑐ℎ 𝑙𝑖𝑛𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝐹𝑡 =
4950
0.1564
𝐹𝑡 = 31653.36 𝑁
Since 𝐹𝑡 ≤ 𝑊𝑡; gear pair 1 is safe for the static load condition
Ratio factor
𝑄 =
2𝑇 𝐺
𝑇 𝐺+𝑇 𝑃
𝑄 =
2×44
44+16
𝑄 = 1.4667
Wear tooth load
𝑊𝑤 = 𝑑𝑏𝑄𝐾
𝑊𝑤 = 64 × 56 × 1.4667 × 43.9548
𝑊𝑤 = 231049.8789 𝑁
Dynamic tooth load on gear
𝑊𝐷 = 𝐹𝑇 +
21𝑣(𝑏𝐶+𝐹 𝑇)
21𝑣√𝑏𝐶+𝐹 𝑇
𝑊𝐷 = 31653.36 +
21×0.1564×(56×78.5464+31653.36)
21×0.1564+√56×78.5464+31653.36
𝑊𝐷 = 32266.31 𝑁
Since 𝑊𝑤 ≥ 𝑊𝐷, gears are safe for dynamic load balancing.
Calculations for other gears are same as previous procedure

Gear
number
Speed
RPM
Speed ratio
Selected speed
ratio
Minimum number
of teeth
Selected number
of teeth
Module (mm)
T11 46.666667 2.74509804 2.75 14.82184459 16 4
T12 17 44 4
T21 46.666667 1.86666667 2 14.16075915 16 5
T22 25 32 5
T31 46.666667 1.26126126 1.4 13.27580099 20 5
T32 37 28 5
TR1 46.666667 1.25 1.25 12.96583978 16 4
TRI 37.333333 20 4
T12 16.969697 2.2 2.2 44 4
Gear number
Diameter
(mm)
Center
distance
Calculated
RPM
Pitch line
velocity
Design
power
Lewis form
factor
T11 64 120 46.6666667 0.1563815 5 0.097
T12 176 16.969697 0.1563815 5 0.13327273
T21 80 120 46.6666667 0.19547688 5 0.097
T22 160 23.3333333 0.19547688 5 0.1255
T31 100 120 46.6666667 0.2443461 5 0.1084
T32 140 33.3333333 0.2443461 5 0.12142857
TR1 64 46.6666667 0.1563815 5 0.097
TRI 80 37.3333333 0.1563815 5 0.1084
T12 176 120
Gear
number
Cv
Face width
factor
Face width
(mm)
Wt Ft
Does pinion
safe
T11 0.950455 49470000 14 56 33088.08 31973.09 TRUE
T12 0.950455 67969091 14 56 45461.22 31973.09 TRUE
T21 0.938827 49470000 8 40 29181.48 25578.47 TRUE
T22 0.938827 64005000 8 40 37755.42 25578.47 TRUE
T31 0.924686 55284000 7 35 28104.86 20462.78 TRUE
T32 0.924686 61928571 7 35 31482.78 20462.78 TRUE
TR1 0.950455 49470000 14 56 33088.08 31973.09 TRUE
TRI 0.950455 55284000 14 56 36976.78 31973.09 TRUE
T12
Gear
number
Ratio
factor - Q
Wear tooth
load
Dynamic load -
Wd
Does pinion safe
T11 1.4666667 231049.879 32588.79483 TRUE
T21 1.3333333 187540.486 25747.94373 TRUE
T31 1.1666667 179482.106 20615.13292 TRUE
TR1 1.1111111 175037.787 32163.80488 TRUE

Conclusion:
Spur gears are safe in the operating region
Reference:
[Accessed 02 04 2018].

Skidded helicopter tug design calculation

Description:
Find the standard proportions of gear systems using module of the gear and standard
equations.
Iteration:
First iteration
Calculation ID:
SHT_StdGearProportions_Itr01
Find:
Standard proportions of gear systems
Data:
Assumption:
Formulae:
For spur gears, For 20⁰ full depth involute system, m – modulus
• Pitch circle diameter – 𝑚𝑇
• Addendum – 1𝑚
• Dedendum – 1.25𝑚
• Working depth – 2𝑚
• Minimum total depth – 2.25𝑚
• Tooth thickness – 1.5708𝑚
• Minimum clearance – 0. 25𝑚
• Fillet radius at root – 0. 4𝑚
For helical gears, recommended by American Gear Manufacturer's Association (AGMA).
• Addendum – 0.8𝑚(Maximum)
• Dedendum – 1𝑚(Maximum)
• Minimum total depth – 1.8𝑚
• Tooth thickness – 1.5708𝑚
• Minimum clearance – 0. 2𝑚
Gear number Module
Number of
teeth
Gear number Module
Number of
teeth
H01 2.5 20 H02 2.5 120
T01 3 16 T02 3 80
T11 4 16 T12 4 44
T21 5 16 T22 5 32
T31 5 20 T32 5 28
TR1 4 16 T12 4 44
TRI 4 20

Calculation:
For T11 – modulus = 4
• Pitch circle diameter – 4 × 16 = 64𝑚𝑚
• Addendum – 1 × 4 = 4𝑚𝑚
• Dedendum – 1.25 × 4 = 5𝑚𝑚
• Working depth – 2× 4 = 8𝑚𝑚
• Minimum total depth – 2.25 × 4 = 9𝑚𝑚
• Total thickness –1.5708 × 4 = 6.2832𝑚𝑚
• Minimum clearance – 0. 25 × 4 = 1𝑚𝑚
• Fillet radius at root – 0. 4 × 4 = 1.6𝑚𝑚
For H01 – modulus = 2.5
• Pitch circle diameter – 2.5× 20 = 50𝑚𝑚
• Addendum – 0.8 × 2.5 = 2𝑚𝑚
• Dedendum – 1 × 2.5 = 2.5𝑚𝑚
• Working depth – 2 × 2.5 = 5𝑚𝑚
• Minimum total depth – 1.8 × 2.5 = 4.5𝑚𝑚
• Total thickness –1.5708 × 2.5 = 3.927𝑚𝑚
• Minimum clearance – 0. 25 × 2.5 = 0.0625𝑚𝑚
• Fillet radius at root – 0. 4 × 2.5 = 1𝑚𝑚

Helical gear
Gear name Module Face width
Addendum -
mm
Dedendum - mm
Minimum totol
depth - mm
H01 2.5 35 2 2.5 4.5
H02 2.5 35 2 2.5 4.5
T01 3 30 2.4 3 5.4
T02 3 30 2.4 3 5.4
Gear name
Minimum
clearance -
mm
Thickness of
tooth - mm
Fillet radius at
root
Diameter
H01 0.5 3.927 1 50
H02 0.5 3.927 1 300
T01 0.6 4.7124 1.2 48
T02 0.6 4.7124 1.2 240
Gear name Module Number of teeth
Pitch circle
diameter
Addendum Dedendum
T11 4 16 64 4 5
T12 4 44 176 4 5
Gear name
Working
depth
Minimum total
depth
Total thickness Minimum clearance Fillet radius at root
T11 8 9 6.2832 1 1.6
T12 8 9 6.2832 1 1.6
Pitch circle
diameter
Addendum Dedendum
T21 5 16 80 5 6.25
T22 5 32 160 5 6.25
Gear name
Working
depth
Minimum total
depth
T21 10 11.25 7.854 1.25 2
T22 10 11.25 7.854 1.25 2
Pitch circle
diameter
Addendum Dedendum
T31 5 20 100 5 6.25
T32 5 28 140 5 6.25
Gear name
Working
depth
Minimum total
depth
T31 10 11.25 7.854 1.25 2
T32 10 11.25 7.854 1.25 2

Conclusion:
Proportions of the gears are in the recommended region.
Reference:
[Accessed 02 04 2018].
Pitch circle
diameter
Addendum Dedendum
TR1 4 16 64 4 5
TRI 4 20 80 4 5
TR2 4 34 136 4 5
Gear name
Working
depth
Minimum total
depth
TR1 8 9 6.2832 1 1.6
TRI 8 9 6.2832 1 1.6
TR2 8 9 6.2832 1 1.6

Description:
Find the placement of the idle gear in reverse condition and find the angles.
Iteration:
First iteration
Calculation ID:
SHT_IdleGearPlacement_Itr01
Find:
Angles between vertical plane of TR2 and horizontal plane of TR1
Data:
Assumption:
Formulae:
Cos equation
𝐶𝑜𝑠(𝐴) =
𝑏2+𝑐2−𝑎2
2𝑏𝑐
Calculation:
Centre distance between TR1 and TR2 – 𝑎1 = 120𝑚𝑚
Centre distance between TR1 and TRI – 𝑎2 = 72𝑚𝑚
Centre distance between TRI and TR2 – 𝑎3 = 108𝑚𝑚
𝛼 = cos−1
(
𝑎32+𝑎12−𝑎22
2×𝑎1×𝑎3
)
𝛼 = cos−1
(
1082+1202−722
2×108×120
)
𝛼 = 36.336057°
α
λ
TR
2
TRI
TR
1
a1
a3
a2
Gear name Module Number of teeth Pitch circle diameter
TR1 4 16 64
TRI 4 20 80
TR2 4 34 136

90 − 𝜆 = cos−1
(
𝑎22+𝑎12−𝑎32
2×𝑎1×𝑎2
)
90 − 𝜆 = cos−1
(
722+1202−1082
2×72×120
)
90 − 𝜆 = 62.720387°
𝜆 = 27.2796°
Conclusion:
Idle gear position is fixed using two angles.
𝛼 = 36.336057°
𝜆 = 27.2796°
Reference:

Description:
Find the minimum diameter of the motor input shaft.
Iteration:
Fifth iteration
Calculation ID:
SHT_MotorInputShaft_Itr05
Find:
Minimum diameter of the shafts in the gear box
Data:
Motor power – 4.5kW
Motor RPM – 1400
Helix angle – 20⁰
Pressure angle – 20⁰
Allowable shear stress – 513 MPa
Assumption:
Shaft material – Nickel Chromium steel (Ni/14-46Cr + some combination of Fe Mo Cu
Co Si Ti W Al + others)
Distance from the bearing to neatest gear is 5cm
k = 0.2; assuming that 20% reduction in strength due to keyway
𝐾𝑏 = 1.5 𝐾𝑡 = 1.0 assuming stationary shaft, suddenly applied load and rotating shaft,
gradually applied load
Formulae:
Permissible shear stress (τ) is about
𝜏 = 0.3𝜎𝑒𝑙
or 𝜏 = 0.18𝜎 𝑢, which is less
𝜏 𝑚𝑎𝑥 =
16
𝜋𝑑3 √𝑀2 + 𝑇2
ASME equation
𝑑3
=
16
𝜋(1−𝑘)[𝜏]
√(𝐾𝑏 𝑀)2 + (𝐾𝑡 𝑇)2
Gear number Diameter Gear number Diameter Speed ratio
H01 50 H02 300 6
T01 48 T02 240 5
T11 64 T12 176 2.75
T21 80 T22 160 2
T31 100 T32 140 1.4
TR1 64 T12 80 1.25
TRI 176 2.2

Calculation:
Motor shaft calculation
Gear pair 2
1
st
gear2
nd
gear 3
rd
gear Reverse 1.1
T01
T02 T11
T12
T22 T32
TR2
T21 T31
TR1
TRI
Reverse 1.2
H01
40mm 40mm
VA
VB
VA VB
Fh
40mm 40mm
HA HB
Fv
40mm 40mm
TRI
40mm 40mm
HA
HB

Torque input from H01
𝑇 =
30𝑃
𝜋𝑁
=
30×4500
𝜋×1400
𝑇 = 30.69 𝑁𝑚
Tangential load on H01
𝐹𝑡 =
2𝑇
𝑑
=
2×30.69
50÷1000
𝐹𝑡 = 1227.77𝑁
Axial load on H01
𝐹𝑎 = 𝐹𝑡 tan(𝜑) = 1227.77 × tan(20°)
𝐹𝑎 = 446.87𝑁
Radial load on H01
𝐹𝑟 = 𝐹𝑡 tan(𝛼) = 1227.77 × tan(20°)
𝐹𝑟 = 446.87𝑁
Resultant force on pinion
𝐹 = √ 𝐹𝑡
2
+ 𝐹𝑟
2
= √1227.772 + 446.872
𝐹 = 1306.56𝑁
𝑉𝑏 =
446.87
2
𝑉𝑏 = 223.44 𝑁
𝐻 𝑏 =
1227.77
2
𝐻 𝑏 = 613.88 𝑁
𝑉𝑎 = 446.87 − 223.44
𝑉𝑎 = 223.44 𝑁
𝐻 𝑎 = 1227.77 − 613.88
𝐻 𝑎 = 613.88 𝑁
Vertical bending moment Horizontal bending moment
𝑀1 =
223.44×40
1000
𝑀1 = 8.94 𝑁𝑚
𝑀3 =
613.88×40
1000
𝑀3 = 24.56 𝑁𝑚
Resultant moment
𝑀 = √ 𝑀2
2
+ 𝑀4
2
= √8.942 + 24.562
𝑀 = 26.13 𝑁𝑚
M1
Vertical bending moment diagram
M3
Horizontal bending moment diagram

𝑑 ≥ √
16
𝜋𝜏 𝑚𝑎𝑥
√𝑀2 + 𝑇23
𝑑 ≥ √
16
𝜋×513×103 √26.132 + 20.692
3
𝑑 ≥ 0.0074𝑚 = 0.74𝑐𝑚
Using ASME equation
𝑑 = √
16
𝜋(1−𝑘)[𝜏]
√(𝐾𝑏 𝑀)2 + (𝐾𝑡 𝑇)23
𝑑 = √
16
𝜋(1−0.2)×513×103
√(1.5 × 26.13)2 + (1 × 20.69)23
𝑑 = 0.0085𝑚 = 0.85𝑐𝑚
Conclusion:
ASME equation gives larger diameter than other equation. But in ASME equation
consider several components. So, select the result of ASME equation as the minimum
diameter of the shaft and take forward.
Reference:
[2] PBC, "Round shaft technology," Linear Bearing & shafting, 2017.
[3] saarstahl, "Material specification sheet," Saarsthl C45, 2016.
[4] "Housing and shaft design guide," 2015.
[5] S. E. +. F. s. grades, "NEW ZEALAND Steal," NEW ZEALAND steel plate, 2013.
[6] "C60 (EN) / 1.0601 (EN) High-carbon Steel," Matbase, [Online]. Available:
https://www.matbase.com/material-categories/metals/ferrous-metals/high-grade-steel/material-
properties-of-high-grade-steel-c60.html#properties. [Accessed 10 Apirl 2018].
[7] thread330-241421, "Gearbox shaft material," ENG-Tips.com, 2017. [Online]. Available:
http://www.eng-tips.com/viewthread.cfm?qid=241421. [Accessed 10 April 2018].
[8] "Mechanical properties of metals and metal alloys," engineering-abc, [Online]. Available:
http://www.tribology-abc.com/calculators/uts.htm#shear. [Accessed April 2018].
[9] "S50C Carbon steel JIS G4051 for mechanical structural use," TAI Special steeel, 2016. [Online].
Available: http://www.astmsteel.com/product/s50c-carbon-steel-jis-g4051/. [Accessed 19 April
2018].

[10] "SAE-AISI 1045 (S45C, C45, 1.0503, G10450) Carbon Steel," Makeitfrom.com, 22 April 2018.
[Online]. Available: https://www.makeitfrom.com/material-properties/SAE-AISI-1045-S45C-
C45-1.0503-G10450-Carbon-Steel. [Accessed 01 May 2018].
[11] "Thread: Shaft/Shoulder undercut," PracticalMachinist.com, 1 November 2017. [Online].
Available: http://www.practicalmachinist.com/vb/cnc-machining/shaft-shoulder-undercut-
329633/. [Accessed 20 April 2018].

Description:
Find the minimum diameter of the input shaft.
Iteration:
Fifth iteration
Calculation ID:
SHT_InputShaft_Itr05
Find:
Data:
Motor RPM – 1400
Assumption:
Formulae:
𝜏 𝑚𝑎𝑥 =
16
𝜋𝑑3 √𝑀2 + 𝑇2
ASME equation
𝑑3
=
16
𝜋(1−𝑘)[𝜏]
√(𝐾𝑏 𝑀)2 + (𝐾𝑡 𝑇)2
H01 50 H02 300 6
T01 48 T02 240 5
T11 64 T12 176 2.75
T21 80 T22 160 2
T31 100 T32 140 1.4
TR1 64 T12 80 1.25
TRI 176 2.2

Calculation:
𝑇 =
30𝑃
𝜋𝑁
=
30×4500
𝜋×1400
𝑇 = 30.69 𝑁𝑚
Torque on T01
𝑇 = 30.69 × 6
𝑇 = 184.17 𝑁𝑚
Tangential load on H01
𝐹𝑡 =
2𝑇
𝑑
=
2×30.69
50÷1000
𝐹𝑡 = 1227.77𝑁
Tangential load on T01
𝐹𝑡 =
2𝑇
𝑑
=
2×184.17
48÷1000
𝐹𝑡 = 7673.54𝑁
Axial load on H01
𝐹𝑎 = 𝐹𝑡 tan(𝜑) = 1227.77 × tan(20°)
𝐹𝑎 = 446.87𝑁
Axial load on T01
𝐹𝑎 = 𝐹𝑡 tan(𝜑) = 7673.54 × tan(20°)
𝐹𝑎 = 2792.94𝑁
Radial load on H01
𝐹𝑟 = 𝐹𝑡 tan(𝛼) = 1227.77 × tan(20°)
𝐹𝑟 = 446.87𝑁
Radial load on T01
𝐹𝑟 = 𝐹𝑡 tan(𝛼) = 7673.54 × tan(20°)
𝐹𝑟 = 2792.94𝑁
𝑉𝑏 =
2792.94×(67.5+117.5)+446.87×67.5
67.5+117.5+65
𝑉𝑏 = 2187.43 𝑁
𝐻 𝑏 =
7673.54×(67.5+117.5)+1127.77×67.5
67.5+117.5+65
𝐻 𝑏 = 6009.92 𝑁
𝑉𝑎 = 2792.94 + 446.87 − 2187.43
𝑉𝑎 = 1052.38 𝑁
𝐻 𝑎 = 1227.77 + 7673.54 − 6009.92
𝐻 𝑎 = −435.86 𝑁
VA VB
Fr-H02
Fr-T01
67.5mm 117.5mm 65mm
HA HB
Ft-H02 Ft-T01
67.5mm 117.5mm
65mm
H02
T01
67.5mm 117.5mm 65mm
VA
VB
H02
T01
67.5mm 117.mm 65mm
HA HB

𝑀1 =
1052.38×67.5
1000
𝑀1 = 71.04 𝑁𝑚
𝑀3 =
435.86×67.5
1000
𝑀3 = 29.42 𝑁𝑚
𝑀2 =
2187.43×65
1000
𝑀2 = 142.18 𝑁𝑚
𝑀4 =
6009.92×65
1000
𝑀4 = 390.65 𝑁𝑚
Resultant moment
𝑀 = √ 𝑀2
2
+ 𝑀4
2
= √142.182 + 390.652
𝑀 = 415.72 𝑁𝑚
𝑑 ≥ √
16
√𝑀2 + 𝑇23
𝑑 ≥ √
16
𝜋×513×103 √415.722 + 184.172
3
𝑑 ≥ 0.0165𝑚 = 1.65𝑐𝑚
Using ASME equation
𝑑 = √
16
𝜋(1−𝑘)[𝜏]
√(𝐾𝑏 𝑀)2 + (𝐾𝑡 𝑇)23
𝑑 = √
16
𝜋(1−0.2)×513×103
√(1.5 × 415.72)2 + (1 × 184.17)23
𝑑 = 0.0184𝑚 = 1.84𝑐𝑚
Conclusion:
Reference:
M1
M2
Vertical bending moment
diagram
M M4
Horizontal bending moment
diagram

2018].
329633/. [Accessed 20 April 2018].

Description:
Find the minimum diameter of the Lay shaft.
Iteration:
Fifth iteration
Calculation ID:
SHT_LayShaft_Itr05
Find:
Data:
Motor RPM – 1400
Assumption:
Formulae:
𝜏 𝑚𝑎𝑥 =
16
𝜋𝑑3 √𝑀2 + 𝑇2
ASME equation
𝑑3
=
16
𝜋(1−𝑘)[𝜏]
√(𝐾𝑏 𝑀)2 + (𝐾𝑡 𝑇)2
H01 50 H02 300 6
T01 48 T02 240 5
T11 64 T12 176 2.75
T21 80 T22 160 2
T31 100 T32 140 1.4
TR1 64 T12 80 1.25
TRI 176 2.2

Calculation:
Second gear mesh
Torque on T01
𝑇 = 30.69 × 6
𝑇 = 184.17 𝑁𝑚
Torque on T21
𝑇 = 184.17 × 5
𝑇 = 368.33 𝑁𝑚
𝐹𝑡 =
2𝑇
𝑑
=
2×184.17
48÷1000
𝐹𝑡 = 7673.54𝑁
𝐹𝑡 = 25578.47 𝑁
Radial load on T01
𝐹𝑟 = 𝐹𝑡 tan(𝛼) = 7673.54 × tan(20°)
𝐹𝑟 = 2792.94𝑁
Radial load on T01
𝐹𝑟 = 𝐹𝑡 tan(𝛼) = 25578.47 × tan(20°)
𝐹𝑟 = 9309.80𝑁
Gear pair 2 2
nd
gear
T01
T02
T22
T21
VA
VB
VA VB
Fr-T02 Fr-T21
65mm 120mm 494mm
HA HB
Ft-T02 Ft-T21
65m
m
120mm 494mm

𝑉𝑏 =
9309.80×(65+120)+2792.94×65
65+120+494
𝑉𝑏 = 2803.91 𝑁
𝐻 𝑏 =
25578.47×(65+120)+7673.54×65
65+120+494
𝐻 𝑏 = 7703.68 𝑁
𝑉𝑎 = 2792.94 + 9309.80 − 2803.91
𝑉𝑎 = 9298.83 𝑁
𝐻 𝑎 = 7673.54 + 25578.47 − 7703.68
𝐻 𝑎 = 25548.34 𝑁
𝑀1 =
9298.83×65
1000
𝑀1 = 604.42 𝑁𝑚
𝑀3 =
25548.34×65
1000
𝑀3 = 1660.64 𝑁𝑚
𝑀2 =
2803.91×494
1000
𝑀2 = 1385.13 𝑁𝑚
𝑀4 =
7703.68×494
1000
𝑀4 = 3805.62 𝑁𝑚
Resultant moment
𝑀 = √ 𝑀2
2
+ 𝑀4
2
= √1385.132 + 3805.622
𝑀 = 4049.85 𝑁𝑚
𝑑 ≥ √
16
√𝑀2 + 𝑇23
𝑑 ≥ √
16
𝜋×513×103 √4049.852 + 368.332
3
𝑑 ≥ 0.0345𝑚 = 3.43𝑐𝑚
Using ASME equation
𝑑 = √
16
𝜋(1−𝑘)[𝜏]
√(𝐾𝑏 𝑀)2 + (𝐾𝑡 𝑇)23
𝑑 = √
16
𝜋(1−0.2)×513×103
√(1.5 × 4049.85)2 + (1 × 368.33)23
𝑑 = 0.0392𝑚 = 3.92𝑐𝑚
M1
M2
diagram
M M4
diagram

First gear mesh
VA Gear pair 2 1
st
gear
T01
T02
T11
T12
VB
VA VB
Fr-T02 Fr-T11
65m
m
333mm 281mm
HA HB
Ft-T02 Ft-T11
65m
m
333mm 281mm
T01 T11
Torque (Nm) 184.1650056 506.4537653
Ft 7673.541899 31973.09125
Fr 2792.940843 11637.25351
Resultant force
on pinion
34025.05302
Vb Va Hb Ha
7088.612743 7341.581611 19475.80345 4823.745902
Verical bending
moment
Horizontal bending
moment
M1 477.2028047 M3 313.5434836
M2 1991.900181 M4 5472.700768

Resultant moment
𝑀 = √ 𝑀2
2
+ 𝑀4
2
= √1991.902 + 5472.702
𝑀 = 5823.93 𝑁𝑚
𝑑 ≥ √
16
√𝑀2 + 𝑇23
𝑑 ≥ √
16
𝜋×513×103 √5823.932 + 506.452
3
𝑑 ≥ 0.0387𝑚 = 3.87𝑐𝑚
Using ASME equation
𝑑 = √
16
𝜋(1−𝑘)[𝜏]
√(𝐾𝑏 𝑀)2 + (𝐾𝑡 𝑇)23
𝑑 = √
16
𝜋(1−0.2)×513×103
√(1.5 × 5823.93)2 + (1 × 506.45)23
𝑑 = 0.0443𝑚 = 4.43𝑐𝑚
M1
M2
diagram
M M4
diagram

Reverse gear mesh
Gear pair 2
Reverse
T01
T02
TR1
TRI
VA VB
VA VB
Fr-T02 Fr-TR1
65mm 536mm
78mm
HA HB
Ft-T02 Ft-TR1
65mm 536mm
78mm
T01 TR1
Torque (Nm) 184.1650056 230.206257
Ft 7673.541899 31973.09125
Fr 2792.940843 11637.25351
Resultant force
on pinion
34025.05302
Vb Va Hb Ha
10567.79163 3862.402726 29034.76887 -4735.21952
Verical bending
moment
Horizontal bending
moment
M1 251.0561772 M3 -307789.269
M2 824.2877469 M4 2264.711972

Resultant moment
𝑀 = √ 𝑀2
2
+ 𝑀4
2
= √824.292 + 2264.712
𝑀 = 2410.06 𝑁𝑚
𝑑 ≥ √
16
√𝑀2 + 𝑇23
𝑑 ≥ √
16
𝜋×513×103 √2410.062 + 230.212
3
𝑑 ≥ 0.0289𝑚 = 2.89𝑐𝑚
Using ASME equation
𝑑 = √
16
𝜋(1−𝑘)[𝜏]
√(𝐾𝑏 𝑀)2 + (𝐾𝑡 𝑇)23
𝑑 = √
16
𝜋(1−0.2)×513×103
√(1.5 × 2410.06)2 + (1 × 230.21)23
𝑑 = 0.0330𝑚 = 3.30𝑐𝑚
Conclusion:
Selected layshaft minimum diameter is 4.43cm.
Reference:
M1
M2
diagram
diagram
M3
M4

2018].
329633/. [Accessed 20 April 2018].

Description:
Find the minimum diameter of the output shaft.
Iteration:
Fifth iteration
Calculation ID:
SHT_OutputShaft_Itr05
Find:
Data:
Motor RPM – 1400
Assumption:
Formulae:
𝜏 𝑚𝑎𝑥 =
16
𝜋𝑑3 √𝑀2 + 𝑇2
ASME equation
𝑑3
=
16
𝜋(1−𝑘)[𝜏]
√(𝐾𝑏 𝑀)2 + (𝐾𝑡 𝑇)2
H01 50 H02 300 6
T01 48 T02 240 5
T11 64 T12 176 2.75
T21 80 T22 160 2
T31 100 T32 140 1.4
TR1 64 T12 80 1.25
TRI 176 2.2

Calculation:
Second gear mesh
Torque on T22
𝑇 = 920.83 × 2
𝑇 = 1841.65 𝑁𝑚
𝐹𝑡 = 25578.47 𝑁
Radial load on T22
𝐹𝑟 = 𝐹𝑡 tan(𝛼) = 25578.47 × tan(20°)
𝐹𝑟 = 9309.80𝑁
Gear pair 2 2
nd
gear
T01
T02
T22
T21
VA VB
VA VB
Fr-T21
70mm 450mm
HA HB
Ft-T21
70mm 450mm

𝑉𝑏 =
9309.80×70
70+450
𝑉𝑏 = 1253.24 𝑁
𝐻 𝑏 =
25578.47×70
70+450
𝐻 𝑏 = 3443.26 𝑁
𝑉𝑎 = 9309.80 − 1253.24
𝑉𝑎 = 8056.56 𝑁
𝐻 𝑎 = 25578.47 − 3443.26
𝐻 𝑎 = 22135.22 𝑁
𝑀1 =
8056.56×70
1000
𝑀1 = 563.96 𝑁𝑚
𝑀3 =
22135.22×70
1000
𝑀3 = 1549.47 𝑁𝑚
Resultant moment
𝑀 = √ 𝑀2
2
+ 𝑀4
2
= √563.962 + 1549.472
𝑀 = 1648.91 𝑁𝑚
𝑑 ≥ √
16
√𝑀2 + 𝑇23
𝑑 ≥ √
16
𝜋×513×103 √1648.912 + 1841.652
3
𝑑 ≥ 0.0291𝑚 = 2.91𝑐𝑚
Using ASME equation
𝑑 = √
16
𝜋(1−𝑘)[𝜏]
√(𝐾𝑏 𝑀)2 + (𝐾𝑡 𝑇)23
𝑑 = √
16
𝜋(1−0.2)×513×103
√(1.5 × 1648.91)2 + (1 × 1841.65)23
𝑑 = 0.0313𝑚 = 3.13𝑐𝑚
M1
M3

First gear mesh
T11
Torque (Nm) 2532.268827
Ft 31973.09125
Fr 11637.25351
Resultant force
on pinion
34025.05302
Vb Va Hb Ha
8235.594792 3401.658719 22627.11073 9345.980518
Verical bending
moment
Horizontal bending
moment
M1 1251.810408 M3 3439.320831
Gear pair 2 1
st
gear
T01
T02
T12
T11
VA VB
VA VB
F11
368mm 152mm
HA HB
F11
368mm 152mm

Resultant moment
𝑀 = √ 𝑀2
2
+ 𝑀4
2
= √1251.812 + 3439.322
𝑀 = 3660.05 𝑁𝑚
𝑑 ≥ √
16
√𝑀2 + 𝑇23
𝑑 ≥ √
16
𝜋×513×103 √3660.052 + 2532.272
3
𝑑 ≥ 0.0353𝑚 = 3.53𝑐𝑚
Using ASME equation
𝑑 = √
16
𝜋(1−𝑘)[𝜏]
√(𝐾𝑏 𝑀)2 + (𝐾𝑡 𝑇)23
𝑑 = √
16
𝜋(1−0.2)×513×103
√(1.5 × 3660.05)2 + (1 × 2532.27)23
𝑑 = 0.0392𝑚 = 3.92𝑐𝑚
M1
M3

Third gear mesh
Gear pair 2 3
rd
gear
T01
T02
T32
T31
VA VB
VA VB
Fr-T31
233mm 287mm
HA HB
Ft-T31
233mm 287mm
T31
Torque (Nm) 1289.155039
Ft 20462.7784
Fr 7447.842247
Vb Va Hb Ha
3337.206238 4110.636009 9168.898782 11293.87962
Resultant force
on pinion
21776.03393
Verical bending
moment
Horizontal bending
moment
M1 957.7781902 M3 2631.47395

Resultant moment
𝑀 = √ 𝑀2
2
+ 𝑀4
2
= √957.782 + 2631.472
𝑀 = 2800.36 𝑁𝑚
𝑑 ≥ √
16
√𝑀2 + 𝑇23
𝑑 ≥ √
16
𝜋×513×103 √2800.362 + 1289.162
3
𝑑 ≥ 0.0313𝑚 = 3.13𝑐𝑚
Using ASME equation
𝑑 = √
16
𝜋(1−𝑘)[𝜏]
√(𝐾𝑏 𝑀)2 + (𝐾𝑡 𝑇)23
𝑑 = √
16
𝜋(1−0.2)×513×103
√(1.5 × 2800.36)2 + (1 × 1289.16)23
𝑑 = 0.0352𝑚 = 3.52𝑐𝑚
Conclusion:
Selected layshaft minimum diameter is 3.92cm.
Reference:
M1
M3

Description:
Find the minimum diameter of the idle shaft.
Iteration:
Fifth iteration
Calculation ID:
SHT_IdleShaft_Itr05
Find:
Data:
Motor RPM – 1400
Angle of the gear when meshing
α – 36.34⁰
λ – 27.28⁰
Assumption:
Formulae:
𝜏 𝑚𝑎𝑥 =
16
𝜋𝑑3 √𝑀2 + 𝑇2
ASME equation
𝑑3
=
16
𝜋(1−𝑘)[𝜏]
√(𝐾𝑏 𝑀)2 + (𝐾𝑡 𝑇)2
H01 50 H02 300 6
T01 48 T02 240 5
T11 64 T12 176 2.75
T21 80 T22 160 2
T31 100 T32 140 1.4
TR1 64 T12 80 1.25
TRI 176 2.2

Calculation:
TRI
78mm 78mm
VA
VB
VA VB
Fh
78mm 78mm
HA HB
Fv
78mm 78mm
α
λ
TR2
TRI
TR1
Ft-R2
Fr-R2
Fr-R1
Ft-R1

𝑇 = 30.69 × 30 × 1.25
𝑇 = 11513.03 𝑁𝑚
Tangential load on gear TRI
𝐹𝑡−𝑅1 = 31973.09 𝑁
𝐹𝑡−𝑅2 = 31973.09 𝑁
Radial load on TRI
𝐹𝑟−𝑅1 = 𝐹𝑡−𝑅1 tan(𝛼) = 31973.09 × tan(20°)
𝐹𝑟−𝑅1 = 11637.25𝑁
𝐹𝑟−𝑅2 = 𝐹𝑡−𝑅2 tan(𝛼) = 31973.09 × tan(20°)
𝐹𝑟−𝑅2 = 11637.25𝑁
Horizontal component
𝐹ℎ = 23172.16𝑁
Vertical component
𝐹𝑣 = 5431.64𝑁
Resultant forces on pinion
𝐹 = √ 𝐹𝑣
2
+ 𝐹ℎ
2
= √23172.162 + 5431.642
𝐹 = 23800.25𝑁
𝑉𝑏 =
23172.16
2
𝑉𝑏 = 11586.08 𝑁
𝐻 𝑏 =
5431.64
2
𝐻 𝑏 = 2715.82 𝑁
𝑉𝑎 = 23172.16 − 11586.08
𝑉𝑎 = 11586.08 𝑁
𝐻 𝑎 = 5431.64 − 2715.82
𝐻 𝑎 = 2715.82 𝑁
𝑀1 =
11586.08×78
1000
𝑀1 = 903.71 𝑁𝑚
𝑀3 =
2715.82×78
1000
𝑀3 = 211.83 𝑁𝑚
Resultant moment
𝑀 = √ 𝑀2
2
+ 𝑀4
2
= √211.832 + 903.712
𝑀 = 923.21 𝑁𝑚
𝑑 ≥ √
16
√𝑀2 + 𝑇23
𝑑 ≥ √
16
𝜋×513×103 √928.212 + 1151.032
3
𝑑 ≥ 0.0245𝑚 = 2.45𝑐𝑚

Using ASME equation
𝑑 = √
16
𝜋(1−𝑘)[𝜏]
√(𝐾𝑏 𝑀)2 + (𝐾𝑡 𝑇)23
𝑑 = √
16
𝜋(1−0.2)×513×103
√(1.5 × 928.21)2 + (1 × 1151.03)23
𝑑 = 0.0262𝑚 = 2.62𝑐𝑚
Conclusion:
Minimum idle shaft diameter is taken as 2.62 cm
Reference:
2018].
329633/. [Accessed 20 April 2018].

Shaft
Calculated
Min
m
Diameter
Maxm
Torque
on the shaft
Min
m
Face Width of Gear
Motor (1 Key) 0.0085 30.694 0.035
Input (2 keys) 0.0185 184.165 0.03
Idle (1 key) 0.0262 1151.031 0.056
Lay shaft (1st gear mesh) 0.0443 506.453 0.056
Lay shaft (2nd gear mesh) 0.0443 368.33 0.04
Lay shaft (3rd gear mesh) 0.0443 257.831 0.035
Lay shaft (Reverse gear mesh) 0.0443 230.206 0.056
Description:
Refer the standard catalogues and select the appropriate keys to join the gears to the shafts.
Iteration:
First Iteration
Calculation ID:
SHT_KeyJoints_Itr01
Find:
Suitable key sizes
Data:
Assumption:
Key material – High Carbon Steel
Shear strength – 240MPa
Crushing strength – 250MPa
Formulae:
𝑇𝑜𝑟𝑞𝑢𝑒 = 𝐹𝑜𝑟𝑐𝑒 × 𝑅𝑎𝑑𝑖𝑢𝑠
𝜎 =
𝐹
𝐴
𝐴 = 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑘𝑒𝑦(𝑏) × 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑘𝑒𝑦(𝑙)
Calculation:
Sample calculation for a key in the input shaft:
• Select a nominal diameter from the catalogue considering the minimum shaft
diameter and the height of the key.
• Select the width and the thickness of the key.
𝐹𝑜𝑟𝑐𝑒 =
184.165𝑁𝑚
0.0185𝑚
2
= 19910𝑁
𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑙𝑒𝑛𝑔𝑡ℎ 𝑓𝑜𝑟 𝑠ℎ𝑒𝑎𝑟 =
𝐹
𝜎𝑠 × 𝑏
= 0.014𝑚
𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑙𝑒𝑛𝑔𝑡ℎ 𝑓𝑜𝑟 𝑐𝑟𝑢𝑠ℎ𝑖𝑛𝑔 =
𝐹
𝜎𝑐 × 𝑏
= 0.027𝑚
Compared the minimum face width of the gear (0.03m), the key is safe for operation.

Shaft Nominal Diameter Width (b) Thickness (h) Force
Minm
Length
for Shear
Minm
Length
for Crushing
Motor (1 Key) 0.01 0.003 0.003 7222.118 0.0573 0.0688
Input (2 keys) 0.022 0.006 0.006 19909.730 0.0790 0.0948
Idle (1 key) 0.03 0.008 0.007 87864.962 0.2615 0.3586
Lay shaft (1st gear mesh) 0.05 0.014 0.009 22864.695 0.0389 0.0726
Lay shaft (2nd gear mesh) 0.05 0.014 0.009 16628.894 0.0283 0.0528
Lay shaft (3rd gear mesh) 0.05 0.014 0.009 11640.226 0.0198 0.0370
Lay shaft (Reverse gear mesh) 0.05 0.014 0.009 10393.047 0.0177 0.0330
Conclusion:
Keys for the rest of the joints were calculated following the above procedure.
Reference:
[1] R.A.J.K. GUPTA, A Textbook of Machine Design, EURASIA PUBLISHING HOUSE (PVT.)
LTD. 2005.

Power 4500
RPM (Min) 17
Shaft diameter 39.2
k 0.7
Description:
Calculate the dimensions of the splines for the output shaft of the gearbox using standard
spline tables.
Iteration:
Third Iteration
Calculation ID:
SHT_Splines_Itr03
Find:
Dimensions of the spline joint in output shaft
Data:
Assumption:
Material - Nickel Chromium steel (30 Ni 4 Cr 1)
Crushing strength – 170MPa
Formulae:
𝜎𝑐 =
𝑇
𝑅𝑙𝑘𝑧ℎ
𝑅 =
𝐷 + 𝑑
2
ℎ =
𝐷 − 𝑑
2
− 2𝑓
where,
T – Torsion acting on the joint
R – Mean radius
l – Effective length
k – Factor for none uniformity of load distribution
z – Number of splines
h – Effective contact height
Calculation:
𝑇 =
𝑃𝑜𝑤𝑒𝑟(𝑊)
𝑟𝑝𝑚(𝑟𝑎𝑑𝑠−1)
= 2527.75𝑁𝑚
Select suitable spline dimensions from the table considering the calculated shaft
diameter.
z – 8
d – 42mm
D – 48mm
b (Width) – 8mm
f – 0.4

Calculate the effective length from the above data
𝑅 =
42 + 48
2
ℎ =
48 − 42
2
− 2 ∗ 0.4
𝑙 =
𝑇
𝑅𝜎𝑐 𝑘𝑧ℎ
= 26.82𝑚𝑚
Conclusion:
Since the length of the spline takes the length of the output shaft, the effective length is
safe.
Reference:
[1] ME2080 – Design of Machine Elements module material

Transmitting Power (kW) 5.5
RPM 1400
Allowable Shear Stress (Shaft & Key) (Mpa) 294
Crushing Stress (Shaft & Key) (Mpa) 490
Allowable Shear Stress (Muff Coupling) (Mpa) 15
Shaft Diameter (d) (m) 0.01
Description:
Calculate the dimensions of the coupling between the motor and the motor shaft
Iteration:
First Iteration
Calculation ID:
SHT_Coupling_Itr01
Find:
Dimensions of the muff coupling
Data:
Assumption:
Cast iron was used as the material for muff coupling
Formulae:
T =
π
16
× τc (
D4
− d4
D
)
2l = L = 3.5d
T = l × w × τ ×
d
2
(For Shearing)
T = l ×
t
2
× σc ×
d
2
(For Crushing)

Transmitting Power 5500
RPM 1400
Allowable Shear Stress (Shaft & Key) 294000000
Crushing Stress (Shaft & Key) 490000000
Allowable Shear Stress (Muff Coupling) 15000000
Torque 37.515094
Shaft Diameter (d) 0.01
Hub Outer Diameter (D) 0.033
Hub Length (L) 0.035
Induced Shear Stress in the Sleeve 5361816.4
Is the Sleeve safe for Shear? TRUE
Key Width 0.003
Key Height 0.003
Minimum length for Crushing strength (l) 0.0102082
Total length of the key 0.0204164
Is the Hub length enough for the Key? TRUE
Muff Coupling Sleeve Dimensions
Length 0.035
Inner Diameter 0.01
Outer Diameter 0.033
Key Dimensions
Width 0.003
Length 0.035
Height 0.003
Calculation:
Conclusion:
Reference:
[1] R.A.J.K. GUPTA, A Textbook of Machine Design, EURASIA PUBLISHING
HOUSE (PVT.) LTD. 2005.

Description:
Select the bearings according to the calculated shaft diameters using standard catalogues.
Iteration:
Third Iteration
Calculation ID:
SHT_Bearings_Itr03
Find:
Suitable bearing types and numbers
Data:
T01
T02
H01
H02
Lay Shaft (Only
the 1st mesh is
shown)
Input Shaft
Motor Shaft
A
A
A
B
B
B
1
st
gear2
nd
gear 3
rd
gear
T01
T02 T11
T12
T22 T32
TR2
T21 T31
TR1
TRI
B
B
B
A
A
A
Input Shaft
Lay Shaft
Output Shaft
Idle Shaft

Assumption:
Expected bearing life - 2 hours a day for 365 days a year for 5 years
Formulae:
𝐵𝑒𝑎𝑟𝑖𝑛𝑔 𝐿𝑖𝑓𝑒 = 𝐿10 = (
𝐶 𝑟
𝑃
)3
𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝐵𝑒𝑎𝑟𝑖𝑛𝑔 𝐿𝑜𝑎𝑑 = 𝑃 = 𝑋𝐹𝑟 + 𝑌𝐹𝑎
Calculation:
Sample calculation for Bearing A of Motor Shaft:
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑙𝑖𝑓𝑒 = 2 × 60 × 365 × 5 = 219000 𝑚𝑖𝑛𝑠
𝑆ℎ𝑎𝑓𝑡 𝑅𝑃𝑀 = 1400 𝑟𝑝𝑚
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑣𝑠 𝑖𝑛 𝑚𝑖𝑙𝑙𝑖𝑜𝑛𝑠 = 1400 × 219000 ÷ 106
= 307
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑠ℎ𝑎𝑓𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 8.5𝑚𝑚
𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑠ℎ𝑎𝑓𝑡 𝑑𝑖𝑚𝑎𝑡𝑒𝑟 𝑓𝑟𝑜𝑚 𝑐𝑎𝑡𝑎𝑙𝑜𝑔𝑢𝑒 = 10𝑚𝑚
𝐹𝑎 = 446.87 (𝐹𝑟𝑜𝑚 𝑠ℎ𝑎𝑓𝑡 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛𝑠)
𝐹𝑟 = 653.28 (𝐹𝑟𝑜𝑚 𝑠ℎ𝑎𝑓𝑡 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛𝑠)
𝐹𝑎
𝐹𝑟
= 0.7
Chose a suitable bearing from catalogue according to the calculated diameter
Chosen bearing = 6300
𝐹𝐶𝑜𝑟(𝐶𝑎𝑡𝑎𝑙𝑜𝑔𝑢𝑒) = 3500
𝑓0(𝑓𝑟𝑜𝑚 𝑐𝑎𝑡𝑎𝑙𝑜𝑔𝑢𝑒) = 11.1
𝑓0 × 𝐹𝑎
𝐹𝐶𝑜𝑟
= 1.417
𝑒(𝑓𝑟𝑜𝑚 𝑐𝑎𝑡𝑎𝑙𝑜𝑔𝑢𝑒) = 0.34
𝑋(𝑓𝑟𝑜𝑚 𝑐𝑎𝑡𝑎𝑙𝑜𝑔𝑢𝑒) = 0.56, 𝑌(𝑓𝑟𝑜𝑚 𝑐𝑎𝑡𝑒𝑙𝑜𝑔𝑢𝑒) = 1.31
𝑃 = 0.56 × 653.28 + 1.31 × 446.87
Bearing RPM Shaft Diameter Fa Fr
Motor shaft A 1400 8.5 446.87 653.28
Motor shaft B 1400 8.5 446.87 653.28
Input shaft A 233.33 18.5 2346.07 1139.06
Input shaft B 233.33 18.5 2346.07 6395.62
Lay Shaft A 46.67 44.3 2346.07 27187.97
Lay Shaft B 46.67 44.3 2346.07 30898.16
Output Shaft A 37 39.2 0 23555.8
Output Shaft B 37 39.2 0 24079.26
Idle Shaft A 37.33 26.2 0 11900.12
Idle Shaft B 37.33 26.2 0 11900.12

Bearing RPM
Num of
Revs
in
Millions
Shaft Diameter
Shaft
Diameter
Catelogue
Fa Fr Fa/Fr
Motor shaft A 1400 307 8.5 10 446.87 653.28 0.7
Motor shaft B 1400 307 8.5 10 446.87 653.28 0.7
Input shaft A 233.33 51 18.5 20 2346.07 1139.06 2.1
Input shaft B 233.33 51 18.5 20 2346.07 6395.62 0.4
Lay Shaft A 46.67 10 44.3 45 2346.07 27187.97 0.1
Lay Shaft B 46.67 10 44.3 45 2346.07 30898.16 0.1
Output Shaft A 37 8 39.2 40 0 23555.8 0.0
Output Shaft B 37 8 39.2 40 0 24079.26 0.0
Idle Shaft A 37.33 8 26.2 28 0 11900.12 0.0
Idle Shaft B 37.33 8 26.2 28 0 11900.12 0.0
Bearing Chosen Bearing Fcor f0 f0*Fa/Fcor e X Y P Cr Catelogue Cr Safe?
Motor shaft A 6300 3500 11.1 1.417 0.34 0.56 1.31 951.2365 6414.250102 8200 Safe
Motor shaft B 6300 3500 11.1 1.417 0.34 0.56 1.31 951.2365 6414.250102 8200 Safe
Input shaft A 6304 7900 12.4 3.682 0.42 0.56 1.04 3077.7864 11421.15542 15900 Safe
Input shaft B 6404 13900 11.4 1.924 0.34 0.56 1.31 6654.8989 24695.22728 28500 Safe
Lay Shaft A 6409 45000 12.1 0.631 0.26 1 0 27187.97 59002.54624 77000 Safe
Lay Shaft B 6409 45000 12.1 0.631 0.26 1 0 30898.16 67054.29329 77000 Safe
Output Shaft A 6408 36500 12.3 0.000 0.19 1 0 23555.8 47312.92572 63500 Safe
Output Shaft B 6408 36500 12.3 0.000 0.19 1 0 24079.26 48364.3196 63500 Safe
Idle Shaft A 63/28 14000 12.4 0.000 0.19 1 0 11900.12 23972.79707 26700 Safe
Idle Shaft B 63/28 14000 12.4 0.000 0.19 1 0 11900.12 23972.79707 26700 Safe
𝐶𝑟 = 𝑃 × 𝐿10
1
3 = 6414.25
𝐶𝑟(𝐶𝑎𝑡𝑎𝑙𝑜𝑔𝑢𝑒) = 8200
𝐶𝑟(𝐶𝑎𝑡𝑎𝑙𝑜𝑔𝑢𝑒) > 𝐶𝑟 ∴ 𝑇ℎ𝑒 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑖𝑠 𝑠𝑎𝑓𝑒 𝑡𝑜 𝑜𝑝𝑒𝑟𝑎𝑡𝑒.
Conclusion:
As for the above sample calculation, bearing for all the shafts were selected as below.
Reference:
[1] NTN, Ball and Roller Bearing, NTN, 2009.

Description:
Calculate the deflection of the pinion center and slope at the bearing end and check
whether the shaft is safe.
Iteration:
First iteration
Calculation ID:
SHT_ShaftJustification_Itr01
Find:
Deflection of the shaft
Data:
Youngs modulus of material – 150 GPa
Assumption:
Minimum value of the Youngs module is used in the calculation.
Formulae:
Deflection at pinion center
𝛿 =
𝐹(𝑎+𝑏)3
48𝐸𝐼
• F – force on the shaft
• a + b – shaft length
• E - Youngs modulus
• I – second moment of inertia of the shaft
If 𝛿 < 0.01𝑚, the design is safe. m is the modulus of the gear
Slope at the bearing end
𝛼 =
𝐹𝐿2
16𝐸𝐼
If 𝛼 < 0.0008 rad, the design is safe.
Calculation:
For motor shaft calculation
Shaft final diameter – 10mm
Pinion module – 2.5mm
Defection at pinion center
𝛿 =
1306.56×(40+40)3
48×150×106×(
𝜋×504
64
)
𝛿 = 1.892 × 10−7
Since 𝛿 = 1.892 × 10−7
< 0.01 × 2.5 = 0.025, shaft is safe
𝛼 =
1306.56×(40+40)2
16×150×106×(
𝜋×504
64
)
𝛼 = 2.23 × 10−8
Since 𝛼 = 1.892 × 10−7
< 0.0008, shaft is safe

Similarly, strength of the shaft can be calculated.
Conclusion:
All shafts are safe after selecting the suitable type of bearings and keys
Reference:
Input shaft Units Does safe
Shaft final diameter 20 mm
pinion module 3 mm
Deflection at pinion
centre
2.25636E-06 TRUE
Slope of the bearing
end
8.50626E-08 TRUE
Lay shaft Units Does safe
pinion module 4 mm
center
7.34946E-06 TRUE
end
1.02013E-07 TRUE
Output shaft Units Does safe
Shaft final diameter 35.4 mm
pinion module 40 mm
center
8.61972E-06 TRUE
end
1.56229E-07 TRUE
Idle shaft Units Does safe
pinion module 4 mm
center
4.15931E-07 TRUE
end
2.51285E-08 TRUE

Description:
Calculate the minimum pitch circle diameter of the gears not to fail using standard
formulas.
Iteration:
First iteration
Calculation ID:
SHT_GearJustification_Itr01
Find:
Minimum pitch circle diameters of the gears.
Data:
Assumption:
Formulae:
For pinion
𝑑 𝑚𝑖𝑛 = 2 × 𝑏𝑜𝑟𝑒 + 0.25𝑚
For wheel
𝐷 𝑚𝑖𝑛 = 2 × 𝑏𝑜𝑟𝑒 + 0.1𝑚
Calculation:
For H01 gear – pinion calculation
𝑑 𝑚𝑖𝑛 = 2 × 10 + 0.25 × 2.5
𝑑 𝑚𝑖𝑛 = 20.625𝑚𝑚
𝑝𝑐𝑑 = 50𝑚𝑚 > 𝑑 𝑚𝑖𝑛 = 20.625 ; pinion is safe
For H02 gear – wheel calculation
𝑑 𝑚𝑖𝑛 = 2 × 10 + 0.1 × 2.5
𝑑 𝑚𝑖𝑛 = 40.25𝑚𝑚
𝑝𝑐𝑑 = 300𝑚𝑚 > 𝑑 𝑚𝑖𝑛 = 40.25 ; wheel is safe
Gear number Type
Module -
mm
Bore
diameter
- mm
pcd
H01 pinion 2.5 10 50
H02 gear 2.5 20 300
T01 pinion 3 20 48
T02 gear 3 45 240
T11 pinion 4 45 64
T12 gear 4 5 176
T21 pinion 5 45 80
T22 gear 5 5 160
T31 pinion 5 45 100
T32 gear 5 40 140
TR1 pinion 4 45 64
TRI pinion 4 28 80
TR2 gear 4 40 136

Conclusion:
All the gears and pinions are safe according to the calculations
Reference:
Gear
number
pcd dmin Safe?
H01 50 20.625 Yes
H02 300 40.25 Yes
T01 48 40.75 Yes
T02 240 90.3 Yes
T11 64 57.25 Yes
T12 176 10.4 Yes
T21 80 57.5 Yes
T22 160 10.5 Yes
T31 100 91.25 Yes
T32 140 80.5 Yes
TR1 64 57.25 Yes
TRI 80 57 Yes
TR2 136 80.4 Yes

Description:
Calculate the fatigue endurance limit of the shafts using experimental data and
mathematical model
Iteration:
First iteration
Calculation ID:
SHT_FatiqueEnduarance_Itr01
Find:
Endurance limit of the shaft that undergoes varying stresses
Data:
Shaft material - Ni/14-46Cr + some combination of Fe Mo Cu Co Si Ti W Al + others
Ultimate tensile strength – 330MPa = 47.85ksi
Assumption:
Mathematical model of the fatigue is given only approximate values. Otherwise we have
to model and simulate this using a software.
Operating region of the gearbox is 100⁰F - 200⁰F
Formulae:
𝜎𝑓 =
(0.9𝜎 𝑢𝑡)2
𝜎 𝑒
𝑁
−
1
3
log
0.9×𝜎 𝑢𝑡
𝜎 𝑒
• 𝜎𝑒 = 0.504𝜎 𝑢𝑡 for 𝜎 𝑢𝑡 ≤ 200 𝑘𝑝𝑠𝑖
• 𝜎𝑒 = 100 𝑘𝑝𝑠𝑖 for 𝜎 𝑢𝑡 > 200 𝑘𝑝𝑠𝑖
• 𝜎 𝑢𝑡 − Minimum tensile strength
• 𝜎𝑒 − Rotating beam specimen
Modifying factor
𝜎𝑒
′
= 𝐾𝑎 𝐾𝑏 𝐾𝑐 𝐾𝑑 𝐾𝑒 𝜎𝑒
• 𝜎𝑒
′
− endurance limit of part
• 𝜎𝑒 − endurance limit of specimen
• 𝐾𝑎 − surface factor
𝐾𝑎 = 𝑎𝜎 𝑢𝑡
𝑏
Shaft Name rpm - maximum
Shaft diameter -
mm
Shaft diameter -
inch
Motor input shaft 1400 0.01 0.000393701
Input shaft 233.333 0.02 0.000787402
Lay shaft 46.667 0.045 0.001771655
Output shaft 33.334 0.04 0.001574804
Idle shaft 37.334 0.028 0.001102363

Surface finish Factor “a” kpsi Exponent “b”
Grind 1.34 -0.085
Machined 2.70 -0.265
Cold drawn 2.70 -0.265
Hot rolled 14.4 -0.718
Forged 39.9 -0.995
• 𝐾𝑏 − size factor
Solid rotating shaft
o 𝐾𝑏 = ( 𝑑
0.3⁄ )−0.1133
, 0.11 𝑖𝑛𝑐ℎ ≤ 𝑑 ≤ 2 𝑖𝑛𝑐ℎ
o 𝐾𝑏 = 0.6 𝑡𝑜 0.7 , 𝑑 > 2 𝑖𝑛𝑐ℎ
Hollow shaft or non-rotating solid shaft
o 𝑑 𝑒(𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟) = 0.390𝑑
Rectangular cross section
o 𝑑 𝑒 = 0.808(𝑏ℎ)0.5
For axial load
o 𝐾𝑏 = 1
• 𝐾𝑐 − Load factor
𝐾𝑐 = 0.923 axial load 𝜎 𝑢𝑡 ≤ 220𝑘𝑝𝑠𝑖
𝐾𝑐 = 1 axial load 𝜎 𝑢𝑡 > 220𝑘𝑝𝑠𝑖
𝐾𝑐 = 1 bending
𝐾𝑐 = 0.577 torsion or shear
• 𝐾𝑑 − Temperature factor
Temp ⁰F 𝜎 𝑇
𝜎 𝑅𝑇
⁄ Temp ⁰F 𝜎 𝑇
𝜎 𝑅𝑇
⁄
100 1.008 600 0.963
200 1.020 700 0.927
300 1.024 800 0.872
400 1.018 900 0.797
500 0.995 1000 0.698
Calculation:
For motor input shaft
𝜎𝑒 = 0.504 × 47.85
𝜎𝑒 = 24.1164 𝑘𝑠𝑖
𝜎𝑓 =
(0.9×47.85)2
24.1164
1400−
1
3
log
0.9×47.85
24.1164
𝜎𝑓 = 41.866 𝑘𝑠𝑖
𝐾𝑎 = 1.34 × 47.85−0.085
= 0.965
𝐾𝑏 = 1 ; for axial load

𝐾𝑐 = 0.923
𝐾𝑑 = 1.008
𝜎𝑒
′
= 0.956 × 1 × 0.923 × 1.008 × 24.1164
𝜎𝑒
′
= 21.64 𝑘𝑠𝑖 = 149.21𝑀𝑃𝑎
Calculation steps are same as the previous one.
Modifying factor kpsi Mpa
Motor input shaft 21.64160893 149.218894
Input shaft 21.64160893 149.218894
Lay shaft 21.64160893 149.218894
Output shaft 39.22849255 270.480456
Idle shaft 40.84623313 281.634777

According to the graph, all shafts are not fail under fatigue conditions.
Conclusion:
Shafts are safe from fatigue failures
Reference:
[1] C. Zorowski, "Fatigue Strength and endurance," 2002.
[2] "Motor Vehicle Maintanance & Repair," Stack Exchange, February 2016. [Online]. Available:
https://mechanics.stackexchange.com/. [Accessed 28 04 2018].
[3] "Design fatigue curve for austenitic steels, nickel-chromium-iron alloy, nickel-iron-chromium
alloy, and nickelcopper alloy," [Online]. Available: https://www.researchgate.net/figure/3-Design-
fatigue-curve-for-austenitic-steels-nickel-chromium-iron-alloy_fig31_294693732. [Accessed 02
05 2018].

Description:
Select the suitable fillet radius for the step shafts using stress concentration factors
Iteration:
First iteration
Calculation ID:
SHT_StressConcent_Itr01
Find:
Fillet radius of the shafts
Data:
Maximum shear stress of the material = 513MPa
Assumption:
Formulae:
𝐾𝑡 =
𝜎 𝑚𝑎𝑥
𝜎 𝑛𝑜𝑚
𝜎 𝑛𝑜𝑚 =
32𝑀
𝜋𝑑3
Calculation:
For motor input shaft
𝜎 𝑛𝑜𝑚 =
32×261.31
𝜋×0.0103
𝜎 𝑛𝑜𝑚 = 266.168𝑀𝑃𝑎
𝐾𝑡 =
513
266.168
𝐾𝑡 = 1.927
For the first step, 𝐷 = 𝑑 + 2 𝑚𝑚𝑚
𝐷
𝑑
=
12
10
= 1.2
𝑟
𝑑
= 0.06
𝑟 = 0.6𝑚𝑚
Suitable fillet radius for the end = 0.6mm
Calculation for the other shafts are same.
Shaft Name
Maximum bending
moment - Nm
Diameter -
mm
Motor input shaft 26.131 10
Input shaft 405.667 20
Lay shaft 5823.927 45
Output shaft 3660.049 40
Idle shaft 928.21 28

Conclusion:
Stress concentration factor is calculated and suitable fillet radius is selected using above
chart
Reference:
[1] C. Zorowski, "Fatigue Strength and endurance," 2002.
Shaft Name D - mm D/d
Motor input
shaft
266.1681803 1.927352846 12 1.2
Input shaft 516.5112664 0.993201956 22 1.1
Lay shaft 650.9962502 0.788022972 47 1.04444444
Output shaft 582.5148903 0.880664183 42 1.05
Idle shaft 430.6974045 1.191091459 30 1.07142857

Description:
Finalize the dimensions of the helical gear using final selected shaft diameter.
Iteration:
First iteration
Calculation ID:
SHT_HelicalDim_Itr01
Find:
Dimensions of the helical gears that used shaft diameters
Data:
Bending stress on gear arm – 200MPa
Assumption:
Formulae:
Diameter of the hub
𝑑ℎ = 1.8 × 𝑑 𝑠
Length of the hub
𝑙ℎ = 1.28 × 𝑑 𝑠
Thickness of the rim
• Pinion
𝑡 = 1.8𝑚
• Wheel
𝑡 = 𝑚√
𝑇 𝐺
𝑛
𝑑ℎ − 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑢𝑏
𝑑 𝑠 − 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑠ℎ𝑎𝑓𝑡
𝑙ℎ − 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑢𝑏
𝑇𝐺 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ
𝑚 − 𝑚𝑜𝑑𝑢𝑙𝑒
𝑛 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑟𝑚𝑠
𝑡 − 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑖𝑚
Calculation:
For H01 gear
𝑑ℎ = 1.8 × 10
𝑑ℎ = 18𝑚𝑚
Gear name Type of gear
Number of
teeth
Module
Pitch
circle
diameter
Diameter
of the shaft
Number
of arms
H01 Pinion 20 2.5 50 10 5
H02 Gear 120 2.5 300 20 5
T01 Pinion 16 3 48 20 5
T02 Gear 80 3 240 45 5

𝑙ℎ = 1.28 × 10
𝑙ℎ = 12.8𝑚𝑚
𝑡 = 2.5√
20
5
𝑡 = 4.5𝑚𝑚
Calculation for the other gears are same as previous one.
Conclusion:
If the hub length is smaller than face width, select face width as hub length
Reference:
Gear name
Diameter of
the hub
Length of
the hub
Thicknes
s of the
rim
Thicknes
s of the
rim
H01 18 12.8 4.5 5
H02 36 25.6 12.24745 13
T01 36 25.6 5.4 6
T02 81 57.6 12 13

Description:
Calculate the gear dimensions based on the shaft diameter and calculate the rim, web, hub
dimensions of the spur gears.
Iteration:
First iteration
Calculation ID:
SHT_SpurDim_Itr01
Find:
Dimensions of the spur gears based on shaft diameters
Data:
Material shear strength – 500MPa
Bending strength on arm – 200 MPa
Assumption:
Number of arms for the gears are selected as 4
Formulae:
Pinion is designed as uniform thickness if
𝑃𝐶𝐷 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑛𝑖𝑜𝑛 ≤ (14.75𝑚 + 60)𝑚𝑚
𝑠𝑡𝑎𝑙𝑙𝑖𝑛𝑔 𝑙𝑜𝑎𝑑 =
𝑡𝑎𝑛𝑔𝑒𝑛𝑐𝑖𝑎𝑙 𝑡𝑜𝑜𝑡ℎ 𝑙𝑜𝑎𝑑
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟
𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑎𝑟𝑚 =
𝑠𝑡𝑎𝑙𝑙𝑖𝑛𝑔 𝑙𝑜𝑎𝑑×𝑝𝑐𝑑
2×𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑟𝑚𝑠
Gear name
Tangential
tooth load -
WT
Normal Tooth
load - WN
Number of
teeth
Face width (mm)
T11 31973.091 34025.05302 16 56
T12 31973.091 34025.05302 44 56
T21 25578.473 27220.04242 16 40
T22 25578.473 27220.04242 32 40
T31 20462.778 21776.03393 20 35
T32 20462.778 21776.03393 28 35
TR1 31973.091 34025.05302 16 56
TRI 31973.091 34025.05302 20 56
TR2 31973.091 34025.05302 34 56
Gear name
Module
(mm)
Velocity
factor Cv
Diameter of the
shaft
PCD gear
T11 4 0.95045545 45 64
T12 4 0.95045545 40 176
T21 5 0.93882701 45 80
T22 5 0.93882701 40 160
T31 5 0.924685564 45 100
T32 5 0.924685564 40 140
TR1 4 0.95045545 45 64
TRI 4 0.95045545 28 80
TR2 4 0.95045545 40 136

𝑎1 =
32𝑊𝑠 𝐷 𝐺
𝑛𝜋𝜎 𝑏
𝑏1 =
𝑎1
2
𝑎2 = 𝑎1 −
𝑑
32
𝑏2 =
𝑎2
2
𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑒𝑏 = 1.8𝑚
𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑢𝑏 = 1.8𝑑 𝑠
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑢𝑏 = 1.25𝑑 𝑠
• Pinion
𝑡 = 1.8𝑚
• Wheel
𝑡 = 𝑚√
𝑇 𝐺
𝑛
Calculation:
For T12 gear – not using uniform thickness.
𝑠𝑡𝑎𝑙𝑙𝑖𝑛𝑔 𝑙𝑜𝑎𝑑 =
31973.09
0.95
𝑠𝑡𝑎𝑙𝑙𝑖𝑛𝑔 𝑙𝑜𝑎𝑑 = 33639.76 𝑁
𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑎𝑟𝑚 =
33639.76×40
2×4
= 740.07𝑁𝑚
𝑎1 = √
32×33639.76×176
4×𝜋×200
3
= 42.24𝑚𝑚
𝑏1 =
42.24
2
= 21.12𝑚𝑚
𝑎2 = 42.24 −
176
32
= 36.74𝑚𝑚
𝑏2 =
36.74
2
= 18.37𝑚𝑚

𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑒𝑏 = 1.8 × 4 = 7.2𝑚𝑚
𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑢𝑏 = 1.8 × 40 = 72𝑚𝑚
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑢𝑏 = 1.25 × 40 = 50𝑚𝑚
Selected length of the hub = 56mm
𝑡 = 4√
44
4
= 13.26𝑚𝑚
Other calculations are same as previous one.
Conclusion:
All the pinions are uniform thickness. Some of the wheels must include all dimension
that are calculated, but some of the wheels have some alternations in solid modeling
without significant changes.
Reference:
Gear name
Stalling
load
Bending
moment on
each arm
a1 b1 a2 b2
Thickness of the
web
T11 33639.758 - - - - - 7.2
T12 33639.758 740074.6741 42.24336925 21.12168462 36.74336925 18.37168462 7.2
T21 27245.14 - - - - - 9
T22 27245.14 544902.7933 38.14518392 19.07259196 33.14518392 16.57259196 9
T31 22129.445 - - - - - 9
T32 22129.445 387265.2886 34.04104569 17.02052285 29.66604569 14.83302285 9
TR1 33639.758 - - - - - 7.2
TRI 33639.758 - - - - - 7.2
TR2 33639.758 571875.8845 38.76448003 19.38224002 34.51448003 17.25724002 7.2
Gear name
Diameter
of the hub
Length of the
hub
Does the length
of the hub
higher than the
face width
Selected hub
length
Thickness of
the rim - tR
Thickness of
the rim
selected- tR
T11 81 56.25 TRUE 56.25 7.2 7.2
T12 72 50 FALSE 56 13.26649916 14
T21 81 56.25 TRUE 56.25 9 9
T22 72 50 TRUE 50 14.14213562 15
T31 81 56.25 TRUE 56.25 9 9
T32 72 50 TRUE 50 13.22875656 14
TR1 81 56.25 TRUE 56.25 7.2 7.2
TRI 50.4 35 FALSE 56 7.2 7.2
TR2 72 50 FALSE 56 11.66190379 12

Description:
Find the suitable lubricant oil for the gear box
Iteration:
First iteration
Calculation ID:
SHT_Lubricant_Itr01
Find:
Suitable lubricant type and it’s properties
Data:
Assumption:
1. Calculated gear tooth temperature for maximum RPM of the shaft (1400 rpm)
2. Considered gear tooth temperature using available gear box tooth temperature values.
3. Assumed the normal load is acting on the whole face width
4. Density of the lubricant were not changed while temperature increases
Formulae:
𝐹𝑛 = √𝐹𝑡2 + 𝐹𝑟2
Normal load =
𝐹𝑛
𝐹𝑎𝑐𝑒 𝑤𝑖𝑑𝑡ℎ
Calculation:
For gear T11
Fn =√31973.092 + 11637.252
𝐹𝑛 = 34025.05𝑁
Normal load =
34025.05
56
= 607.59𝑁/𝑚𝑚
Calculation process is same as previous one for other gears
Gear
number
Ft
T11 31973.09
T12 31973.09
T21 25578.47
T22 25578.47
T31 20462.78
T32 20462.78
TR1 31973.09
TRI 31973.09
T12 31973.09

Calculated maximum normal tooth load using above table and used past data of tooth
surface temperature
for calculate operating tooth temperature. To get better values we used a graph.
Normal
load(N/mm)
tooth surface
temperature(C)
110 76
160 79
220 81
270 82
340 84
370 85
450 88
Gear
number
Ft Fr Fn
Face width
(mm)
Fn/b
(N/mm)
T11 31973.09 11637.25 34025.05 56 607.59023
T12 31973.09 11637.25 34025.05 56 607.59023
T21 25578.47 9309.803 27220.04 40 680.50106
T22 25578.47 9309.803 27220.04 40 680.50106
T31 20462.78 7447.842 21776.03 35 622.1724
T32 20462.78 7447.842 21776.03 35 622.1724
TR1 31973.09 11637.25 34025.05 56 607.59023
TRI 31973.09 11637.25 34025.05 56 607.59023
T12 31973.09 11637.25 34025.05 56 607.59023
70
75
80
85
90
95
100
0 50 100 150 200 250 300 350 400 450 500 550 600 650 700
Temperature/C
Normal load N/mm
tooth surface temperature(C)

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