Machine Design Report PDF

UNIVERSITI TENAGA NASIONAL
COLLEGE OF ENGINEERING
DEPARTMENT OF MECHANICAL ENGINEERING
MEMB333 - MACHINE DESIGN

SEMESTER 1 18/19
PROJECT TITLE : DOUBLE GEAR REDUCER
GROUP MEMBERS:
1. ID:ME0101875 NAME: ABU DZAR BIN JOHAR
2. ID:ME0100760 NAME: AHMAD AIMAN BIN MUHAMAD ISMAIL
3. ID:ME0101519 NAME: AHMAD AMIRUN AQIL BIN AHMAD PADHLI
4. ID:ME0100043 NAME: AHMAD FADHLI BIN ABDUL MALEK
5. ID:ME0100926 NAME: AHMAD SYAHIR BIN MOHD JAMEL
SECTION: 03B
LECTURER : DR. SAMI SALAMA HUSSEN HAJJAJ

TABLE OF CONTENTS
ABSTRACT................................................................................................................................................ 1
INTRODUCTION ....................................................................................................................................... 2
LITERATURE REVIEW ............................................................................................................................... 4
OBJECTIVE ............................................................................................................................................... 6
DESIGN STEPS I ........................................................................................................................................ 7
GEAR........................................................................................................................................................ 7
SHAFT .................................................................................................................................................... 27
BEARINGS .............................................................................................................................................. 66
DESIGN STEPS II ..................................................................................................................................... 75
CASING BOX .......................................................................................................................................... 75
SPRINGS ................................................................................................................................................ 77
FASTENERS ............................................................................................................................................ 81
DISCUSSION........................................................................................................................................... 85
CONCLUSION......................................................................................................................................... 88
REFERENCES .......................................................................................................................................... 89
APPENDICES .......................................................................................................................................... 90
ABSTRACT
The main components of this course are the study of machine theories and standard
methods of machine designs components. For this project, we were requested to design a speed
reducer by using a double – step reduction gear by choosing the types of gear which are Spur
or Helical gears. Our job was to design a speed reducer which consists of gears, shafts, keys,
casing, bearing, springs and fasteners by considering the factor of safety, the gear material,
quality level of the material that being used, the design life, the total weight of all components
as well as the strength for every component. For this project, Helical gear was picked as our
study case as Helical gears has more advantages compared to spur gears. Some of the
advantages are more work can be transmitted via helical gears, the life expectancy of helical
gears is more as they vibrate less, and less noise is produced when the gears meshes.
The theories that we have learned before in class and lab were used for this project. The
first step was to decide the important parameters for each component, the prior decisions was
important in order to fulfil the required factor of safety. Software such as Microsoft Excel and
SolidWorks were used to do calculations and simulations to verify the mentioned parameters
so it is inclined with the assumptions and calculations that have been done manually from the
prior decisions.
1
INTRODUCTION
Speed reducers are invented to deliver power at slower speed than the power that been
supplied by the motor. The speed reducer transmits the required power by reducing the speed
and increases the torque. The goal of this project is to design a speed reducer to reduce the
given rotation speed from 1200 rpm to 150 rpm. To reduce a much higher rotation speed to a
much lower speed, a double step – reduction is needed. Spur or Helical gears were suggested
to be used for in the system. The input must be aligned to the output shaft.
Figure 1 : Typical speed reducer
Helical gear was chosen for the designed gearbox. The input power transmitted to the
input shaft equals to the ID number of the first person in the group which is 1875 Watt. The
shafts designed are supported by two ball or roller bearings. The desired overall reliability was
set to 95% and the designed life was set to 20 kilohours. The minimum value for the factor of
safety was set to 2.0 and any lower values than that will not be accepted.
For every designed component, design decisions and assumptions, design calculations
and analysis, design findings and design discussion were made in order to fully achieve the
2
objectives of this project. For design decisions and assumptions, the initial value of parameters
was assumed based on the knowledge obtained from class and general knowledge. The decision
was not made based on gut feelings but with good reasoning. The design calculations and
analysis were done manually and by using Microsoft Excel as well in order to prove and verify
the calculations. The analysis was done by checking the factor of safety obtained after several
parameters and materials for every component were chosen. The results were then tabulated in
a table to show the summary of the design. The design findings and design discussion were
made in order to discuss if there were any issues or anomalies. The costing for every designed
components were calculated and were tabulated.
3
LITERATURE REVIEW
Gearbox are generally known as transmission unit that consists gears, shafts, bearing,
spring, fasteners and etc. Gearbox usually used to provide speed and torque by converting high
speed input to lower speed output and vice versa. A gearbox that converts speed input into
single output is known as single stage gearbox and usually has two gears and 2 shafts. For
double stage gearbox, it changes the output speed twice and it has more than two gears and
shafts. For certain cases that the difference between the input speed and the desired output
speed differs greatly, more stages of the gear train are needed. The gear trains are formed to
make sure that the force acting on the gears are not to be above the endurance limit of the
components.
Figure 2: Design for double step reducer gear
Gears that are lubricated will allow easier transmission of mechanical energy from the
driving component to output devices. By reducing the friction between gears, it reduces wasted
energy that is created when the transmission of energy when the gears are rotated. With less
friction, it can also reduce the force acting on the gear surface, hence maximizing the life
expectancy of the gears.
4
In order to maximise the design life of the gearbox, proper installation and maintenance
play huge roles. Each component needed to be carefully selected, and the installation needs to
be accurate. The selection for the material entails much more than simply picking out one from
catalogues.
5
OBJECTIVE
• To design and analyse several machine components such as gears, shafts, keys,
couplings, fasteners, springs, bearings and casing box.
• To determine the most suitable dimension for the components’ designed gearbox.
• To make use of knowledge obtained during lecture by performing analysis and
calculation to design speed reducer by performing complex analysis and calculations.
• To make use of CAD Software such as SolidWorks to model the designed component
and performing complex analysis and calculation.
6
DESIGN STEPS I
GEAR
The reason we choose the helical gear instead of the spur gear is because of the
functionality and the advantages. We decided that the most important point of choosing a gear
type is its functionality, its life expectancy and its frequency of maintenance.
Helical gear has complete face engagement, unlike spur gear that is incomplete, as one
or two teeth are always in contact with another gear. This makes helical gear vibrates less and
less noisy, makes it have higher life expectancy. This results to the saving of the cost in the
long run for the system, as it made changing the gears lesser virtually non-existence.
Other than that, helical gears also have higher torque capacity. By using a complex
design of having angular teeth to give higher load transmission, high efficiency gear mesh can
be achieved, rather than spur gear, which loads being transmitted through fewer teeth. More
teeth transferring the load alleviate the load between each tooth.
For the designation process, there are several things that is needed to be assumed to get
the project going. First, the size of the gear. As we have creating a mini boat rotor in mind, we
decided to make the diameter across for both pinion and gear in both sets to be 17.3205 inches.
For the face width, we made it to be 1 inch. The reason we choose these numbers to be our
model dimensions, is because it fits to what we planned and the size of the model’s gearbox.
Next is the quality factor of the model. We chose grade 7, as we don’t need precise gears for it
to function correctly. Lower grade may affect its reliability, so we maximise between its cost
and reliability, which is grade 7. We assumed the overload factor, Ko, for the driven machine
to have a uniformed power source and moderate shock acting on it
n2 = 1200 rpm
7
n5= 150 rpm
150 𝑟𝑝𝑚 1
e = 1200 𝑟𝑝𝑚 = 8
N2 + N3 = N4 +N5
𝑛 𝑁 𝑛 𝑁
So: 𝑛2 = 4 = 𝑁3 , 𝑛4 = 2 = 𝑁3
3 2 5 4
N3 = 4 N2 and N5 = 2N4
N2 & N4 are pinion
2𝑘
Using equation (13.11), 𝑁𝑝 = (1+2𝑚) 𝑠𝑖𝑛2 ∅
(𝑚 + √𝑚2 + (1 + 2𝑚) sin2 ∅)
2(1)
𝑁2 = (1+2(4)) 𝑠𝑖𝑛2 30°
(4 + √42 + (1 + 2(4)) sin2 30°)
= 16 teeth (minimum)
2(1)
𝑁4 = (1+2(2)) 𝑠𝑖𝑛2 30°
(2 + √22 + (1 + 2(2)) sin2 30°)
𝑁4 = 30 teeth (minimum)
For alignment: N2 + N3 = N4 +N5
8
N2 + 4N2 = N4 +2N4
5N2 = 3N4
5
Np = 3 N2, assume N2 = 18 teeth, minimum 17 teeth
N2 = 18 teeth
5(18) = 3N4
N4 = 30 teeth
N5 = 2(30)
= 60 teeth
N3 = 4(18)
= 72 teeth
N2 = 18 teeth, N3 = 72 teeth, N4 = 30 teeth, N5 = 60 teeth
P = 6 teeth/in (coarse)
9
Gear 2
1875
H = = 2513 hp
746
𝑁
d = 𝑝 cos Ψ
d = diameter of gear
18
d2 = 6 cos 30° = 3.4641 inch
𝜋𝑑𝑛
V2 = 12
𝜋(3.4641)(1200)
V2 = 12
V2 = 1088.28 feet/min
𝐻
WT2 =33000 𝑉
2.513
= 3300 (1088.28 )
= 76.2142 lb f
Wr2 = WT2 tan ∅1
= 76.2142tan 22.8°
= 32.031 lb f
Wa2 = WT2 tan 𝜔
= 76.2142 tan 30°
= 44.0023 lb f
𝑑
T2 = WT2(2 )
3.4641
= 76.2142( )
2
= 132.007 lbf in
10
Gear 3
1875
H = = 2513 hp
746
𝑁
d = 𝑝,
72
d3 = 6cos 30° = 13.8564 inch
𝜋𝑑𝑛
V3 = 12
𝜋(13.8564 )(300)
V3 = 12
V3 = 1088.28 feet/min
𝐻
WT3 =33000 𝑉
2.513
= 3300 (1088.28 )
= 76.2142 lb f
Wr3 = WT3 tan ∅1
= 76.2142 tan 22.8°
= 32.031 lb f
Wa3 = WT3 tan 𝜔
= 76.2142 tan 30°
= 44.0023 lb f
𝑑
T3 = WT2 (2 )
13.8564
= 76.2142 ( )
2
= 528.027 lbf in
11
Gear 4
1875
H = = 2513 hp
746
𝑁
d = 𝑝,
30
d4 = 6 cos 30° = 5.7735 inch
𝜋𝑑𝑛
V4 = 12
𝜋(5.7735 )(300)
V4 = 12
V4 = 453.45 feet/min
𝐻
WT4 =33000 𝑉
2.513
= 33000 (453.45 )
= 182.914 lb f
Wr4 = WT4 tan ∅1
= 182.914 tan 22.8°
= 76.8745lb f
Wa4 = WT4 tan 𝜔
= 182.914 tan 30°
= 105.605lb f
𝑑
T4 = WT4 (2 )
5.7735
= 182.914 ( )
2
= 528.027 lbf in
12
Gear 5
1875
H = = 2513 hp
746
𝑁
d = 𝑝,
60
d5 = 6 cos 30° = 11.547 inch
𝜋𝑑𝑛
V5 = 12
𝜋(11.547 )(150)
V5 = 12
V5 = 196.35 feet/min
𝐻
WT5 =33000 𝑉
2.513
= 33000 (196.35 )
= 182.914 lb f
Wr5 = WT5 tan ∅1
= 182.914 tan 22.8°
= 76.8745 lb f
Wa5 = WT5 tan 𝜔
= 182.914 tan 30°
= 105.605 lb f
𝑑
T5 = WT5 (2 )
11.547
= 182.914 ( )
2
= 1056.05 lbf in
13
addendum, a
1 1
Addendum, a = (𝑃) = (6) = 0.1667
1.25 1.25
Dedendum, b = ( )= ) = 0.2083
𝑃 6
Face width, F = 1 inch
Quality factor = 7
Grade 2, steel
Commercial loading
Reliability = 0.95
For Bending and Wear
Gear 2 & 3
Quality factor, Qv = 7
Reliability, R = 0.95
Hardness ration, Ch = 1
Brinell Hardness, HB = 200
14
Allowable Bending stress, St = 102HB + 16400 psi
= 102(200) + 16400 psi
= 36800 psi
Allowable Contact stress, Sc =349HB + 34300 psi
= 349(200) + 34300 psi
= 104100
Z = [(rP + a)2 - rbP2]1/2 + [(rG + a)2 – rbG2]1/2 –(rP +rG)sin Øn
3.46 3.46cos(22.8) 2 1/2 13.86 13.86cos(22.8)2 1/2 3.46 13.86

= [( + 0.167)2 - ] + [( 2 + + a)2 – ] –( 2 + )sin
2 2 2 2
20
=1.42
𝜋
PN = 𝑃 x cos Øn
𝜋
= 6 x cos 20
= 0.49
𝑃𝑁
Bending stress geometry factor, mN = 0.95𝑍
0.49
= 0.95(1.42)
= 0.36
cosØsinØ mG
Surface geometry factor, I = . 𝑚𝐺+1
2𝑚𝑛
15
cos(22.8)sin(22.8) 4
= . 4+1
2𝑚𝑛
= 0.39
Lewis form factor, YP = 0.309
Lewis form factor, YG = 0.432
𝐹√𝑌 0.0535
Size Factor for pinion, KSP = 1.192( )
𝑃
1√0.309 0.0535
= 1.192( )
6
= 0.88
𝐹√𝑌 0.0535
Size Factor for gear, KSG = 1.192( )
𝑃
1√0.432 0.0535
= 1.192( )
6
= 1.06
Load distribution Factor, KM = 1 + Cmc(CpfCpm + CmaCe)
= 1 +1(0.003867(1) + 0.185(1))
16
= 1.189
Reliability factor, KR = 0.658 – 0.0759ln(1-0.95)
= 0.885
𝐴+ √𝑉 B
Dynamic factor ,KV =( )
𝐴
34.94+ √1088.28 0.731

=( )
34.94
= 1.626
Stress cycle factor pinion, Ynp = 1.3558N2-0.0178
= 1.3558(18)-0.0178
= 0.98
Stress cycle factor pinion, YnG = 1.3558N3-0.0178
= 1.3558(72)-0.0178
=1
17
Pitting resistance stress cycle,Zn = 1.4488N2-0.0323
= 1.4488(18)-0.0323
= 0.88
Pitting resistance stress cycle,Zn = 1.4488N3-0.0323
= 1.4488(72)-0.0323
= 0.95
Geomety factor, J’P = 0.46
Geomety factor, J’G = 0.54
Modifying factor, JP = 0.432
Modifying factor, JG = 0.539
1
Elastic coefficient, Cp =[ 1−𝑣𝑝2 1−𝑣𝐺2
]1/2
𝜋( + )
𝐸𝑝 𝐸𝐺
1
=[ 1−0.3^2 1−0.3^2 ]1/2
𝜋( + )
3(107 ) (2.3(109 )
18
= 3232.7
𝑃𝑑 𝐾𝑚𝐾𝐵
Bending stress for pinion, σ = WtKOKvKS 𝐹 𝐽
6 cos 30° (1.19)(1)

= (76.21)(1.25)(1.63)(0.88) 1 0.43
= 1948.27 psi
𝑃𝑑 𝐾𝑚𝐾𝐵
Bending stress for gear, σ = WtKOKvKS 𝐹 𝐽
6 cos 30° (1.19)(1)

= (76.21)(1.25)(1.63)(1.06) 1 0.54
= 1880.14 psi
𝐶𝑓 𝐾𝑚
Contact stress for pinion, σc = Cp(WtKOKvKS )
𝐼 𝑑𝑝𝐹
1 (1.19)
= 3232.7[(76.21)(1.25)(1.63)(0.88)0.39 1(3.464) ]
= 35376.12 psi
19
𝐶𝑓 𝐾𝑚
Contact stress for gear, σc = Cp(WtKOKvKS 𝐼 )
𝑑𝑝𝐹
1 (1.19)
= 3232.7[(76.21)(1.25)(1.63)(1.06)0.39 1(13.86) ]
= 19398.6 psi
𝑆𝑡𝑌𝑛⁄𝐾𝑇𝐾𝑅
Pinion bending factor of safety, Sf = σ
(36800)(0.98)⁄(1)(0.89)
= 1948.27
= 20.83
𝑆𝑐𝑍𝑛𝐶𝐻⁄𝐾𝑇𝐾𝑅
Pinion wear factor of safety, SH = σc
(104100)(0.87)(1)⁄(1)(0.89)
= 35376.12
= 2.92
As 2.922 > 20.83, G2 threat comes from bending.
20
𝑆𝑡𝑌𝑛⁄𝐾𝑇𝐾𝑅
Gear bending factor of safety, Sf = σ
(36800)(1)⁄(1)(0.89)
= 1880.14
= 22.13
𝑆𝑐𝑍𝑛𝐶𝐻⁄𝐾𝑇𝐾𝑅
Gear wear factor of safety, SH = σc
(104100)(0.94)(1)⁄(1)(0.89)
= 19398.6
= 5.33
As 5.332 > 22.13, G3 threat comes from wear.
21
Gear 2 3 4 5
Mating rate, 4 2
mg
Helix Angle, Ψ 30
(°)
Pressure angle, 20
normal, Øn (°)
Pressure angle, 22.8
tangentianal,
Øt(°)
Contact type, k 1
Number of 18 72 30 60
teeth, N
Speed of gears, 1200 300 300 150
RPM
Diameter, d 3.4641 13.8564 5.7735 11.547
(in)
Velocity, 1088.28 1088.28 453.45 453.45
(ft/min)
Transmitted 76.2142 76.2142 182.914 182.914
load, Wt, (lbf)
Radial load, 32.031 32.031 76.8745 76.8745
Wr,(lbf)
Axial load, 44.0023 44.0023 105.605 105.605
Wa, (lbf)
Resultant load, 93.6525 93.6525 224.766 224.766
W, (lbf)
Torque, T, 132.007 528.027 528.027 1056.05
(lbfin)
Face width, F, 1
(in)
22
Bending and wear
Quality factor, Qv 7
Reliability, R 0.95
Hardness ratio factor, Ch 1
Allowable bending stress, St 36800
Allowable contact stress, Sc 104100
Brinell Hardness, HB 200
Surface strength Geometry factor, I 0.3908
Load distribution factor, Km 1.1888
Cmc, uncrowned teeth, 1
Cpf 0.003867
Cma 0.1849
Ce, for other conditions aside gearing 1
adjusted at assembly, ect.
Cpm, straddled mounted pinion 1
Cma, commercial, enclosed units A = 0.127
B = 0.0158
C = -0.930(10^-4)
Reliability factor, Kr 0.8854
Eg (lbf/in^2) 2300
Ep (psi) 30(10^6)
Poisson ratio 0.3
Overload factor, Ko 1.25
23
Bending and wear for gear 2 & 3
Gear 2 3
Lewis form factor, Y 0.309 0.4324
Size factor, Ks 0.88 1.05
Dynamic factor, Kv 1.63 1.63
Stress cycle factor, Yn 0.98 1.00
Pitting resistance stress 0.88 0.95
cycle, Zn
Z 1.42
Geometry factor J’ 0.460 0.540
Modifying factor J 0.432 0.539
Pn 5.20
Bending stress geometry 0.36558
factor, mn
Contact stress, 35376.1222 19398.6
Bending stress 1948.26682 1880.14
Factor of safety for bending, 20.8385372 22.1331
Sf
Factor of safety for wear, Sh 2.92149933 5.32778
24
Bending and wear for gear 4 & 5
Gear 4 5
Lewis form factor, Y 0.303 0.412
Size factor, Ks 0.88 1.05
Dynamic factor, Kv 1.625 1.625
Stress cycle factor, Yn 0.976 1.00
Pitting resistance stress 0.879 0.9499
cycle, Zn
Z 1.42
Geometry factor J’ 0.460 0.540
Modifying factor J 0.432 0.539
Pn 5.20
Bending stress geometry 0.36558
factor, mn
Contact stress, 38742.4 21236.4
Bending stress 1947.2 1877.7
Factor of safety for bending, 20.85 22.16
Sf
Factor of safety for wear, Sh 2.668 4.867
25
For our design, there are several things that is needed to be assumed to get the project
going. The size of the gear. As we have creating a mini boat rotor in mind, we decided to make
the diameter across for both pinion and gear in both sets to be 17.3205 inches. For the face
width, we made it to be 1 inch. The reason we choose these numbers to be our model
dimensions, is because it fits to what we planned and the size of the model’s gearbox.
The quality factor of our model. When we chose grade 7, as we don’t need precise gears
for it to function correctly. The issues are that the lower grade may affect the reliability, so we
needed maximise between the cost and reliability, which are grade 7. We assumed that the
overload factor, Ko, for the driven machine to have a uniformed power source and moderate
shock acting on it.
Other than that, the anomalies that we discover is there were some of the factor are
same and can be reused on other equation to calculate other gear or pinion but the output are
difference because of the value of m on each gear sets.
26
SHAFT
SHEAR MOMENT CALCULATION
SHAFT A
PLANE Z-X
Calculation for the reaction at the support beam
1. A beam is in equilibrium when it is stationary relative to an inertial reference frame. The
following conditions are satisfied when a beam, acted upon by a system of forces and moments,
is in equilibrium.
ΣFx = 0: HA = 0
ΣMA = 0: The sum of the moments about the pin support at the point A:
P1*1.5 + RB*3 = 0
ΣMB = 0: The sum of the moments about the roller support at the point B:
27
- RA*3 - P1*1.5 = 0
2. Calculate reaction of roller support at the point B:
RB = ( - P1*1.5) / 3 = ( - 76.2142*1.5) / 3 = -38.11 (lbf)
3. Calculate reaction of pin support at the point A:
RA = ( - P1*1.5) / 3 = ( - 76.2142*1.5) / 3 = -38.11 (lbf)
4. Solve this system of equations:
HA = 0 (lbf)
5. The sum of the forces about the Oy axis is zero:
ΣFy = 0: - RA + P1 - RB = - 38.11*1 + 76.2142 - 38.11*1 = 0
Consider first span of the beam 0 ≤ x1 < 1.5
Determine the equations for the shear force (Q):
Q(x1) = - RA
The values of Q at the edges of the span:
Q1(0) = - 38.11 = -38.11 (lbf)
Q1(1.50) = - 38.11 = -38.11 (lbf)
Determine the equations for the bending moment (M):
M(x1) = - RA*(x1)
The values of M at the edges of the span:
M1(0) = - 38.11*(0) = 0 (lbf*in)
M1(1.50) = - 38.11*(1.50) = -57.16 (lbf*in)
Consider second span of the beam 1.5 ≤ x2 < 3
Q(x2) = - RA + P1
Q2(1.50) = - 38.11 + 76.21 = 38.11 (lbf)
28
Q2(3) = - 38.11 + 76.21 = 38.11 (lbf)
M(x2) = - RA*(x2) + P1*(x2 - 1.5)
M2(1.50) = - 38.11*(1.50) + 76.21*(1.50 - 1.5) = -57.16 (lbf*in)
M2(3) = - 38.11*(3) + 76.21*(3 - 1.5) = 0 (lbf*in)
PLANE Y-X
29
is in equilibrium.
ΣFx = 0: HA = 0
P1*1.5 + RB*3 = 0
- RA*3 - P1*1.5 = 0
RB = ( - P1*1.5) / 3 = ( - 32.03105*1.5) / 3 = -16.02 (lbf)
RA = ( - P1*1.5) / 3 = ( - 32.03105*1.5) / 3 = -16.02 (lbf)
HA = 0 (lbf)
ΣFy = 0: - RA + P1 - RB = - 16.02*1 + 32.03105 - 16.02*1 = 0
Q(x1) = - RA
Q1(0) = - 16.02 = -16.02 (lbf)
Q1(1.50) = - 16.02 = -16.02 (lbf)
M(x1) = - RA*(x1)
30
M1(0) = - 16.02*(0) = 0 (lbf*in)
M1(1.50) = - 16.02*(1.50) = -24.02 (lbf*in)
Q(x2) = - RA + P1
Q2(1.50) = - 16.02 + 32.03 = 16.02 (lbf)
Q2(3) = - 16.02 + 32.03 = 16.02 (lbf)
M(x2) = - RA*(x2) + P1*(x2 - 1.5)
M2(1.50) = - 16.02*(1.50) + 32.03*(1.50 - 1.5) = -24.02 (lbf*in)
M2(3) = - 16.02*(3) + 32.03*(3 - 1.5) = 0 (lbf*in)
31
TOTAL REACTION AT THE SUPPORT AND TOTAL SHEAR MOMENT
32
SHAFT B
PLANE Z-X
is in equilibrium.
ΣFx = 0: HA = 0
- P1*1.5 + P2*5.5 + RB*7 = 0
33
- RA*7 + P1*5.5 - P2*1.5 = 0
RB = ( P1*1.5 - P2*5.5) / 7 = ( 76.2142*1.5 - 182.9141*5.5) / 7 = -127.39 (lbf)
RA = ( P1*5.5 - P2*1.5) / 7 = ( 76.2142*5.5 - 182.9141*1.5) / 7 = 20.69 (lbf)
HA = 0 (lbf)
ΣFy = 0: RA - P1 + P2 - RB = 20.69*1 - 76.2142 + 182.9141 - 127.39*1 = 0
DRAW DIAGRAMS FOR THE BEAM
Q(x1) = + RA
Q1(0) = + 20.69 = 20.69 (lbf)
Q1(1.50) = + 20.69 = 20.69 (lbf)
M(x1) = + RA*(x1)
M1(0) = + 20.69*(0) = 0 (lbf*in)
M1(1.50) = + 20.69*(1.50) = 31.03 (lbf*in)
Consider second span of the beam 1.5 ≤ x2 < 5.5
Q(x2) = + RA - P1
34
Q2(1.50) = + 20.69 - 76.21 = -55.53 (lbf)
Q2(5.50) = + 20.69 - 76.21 = -55.53 (lbf)
M(x2) = + RA*(x2) - P1*(x2 - 1.5)
M2(1.50) = + 20.69*(1.50) - 76.21*(1.50 - 1.5) = 31.03 (lbf*in)
M2(5.50) = + 20.69*(5.50) - 76.21*(5.50 - 1.5) = -191.08 (lbf*in)
Consider third span of the beam 5.5 ≤ x3 < 7
Q(x3) = + RA - P1 + P2
Q3(5.50) = + 20.69 - 76.21 + 182.91 = 127.39 (lbf)
Q3(7) = + 20.69 - 76.21 + 182.91 = 127.39 (lbf)
M(x3) = + RA*(x3) - P1*(x3 - 1.5) + P2*(x3 - 5.5)
M3(5.50) = + 20.69*(5.50) - 76.21*(5.50 - 1.5) + 182.91*(5.50 - 5.5) = -191.08 (lbf*in)
M3(7) = + 20.69*(7) - 76.21*(7 - 1.5) + 182.91*(7 - 5.5) = 0 (lbf*in)
35
PLANE Y-X
is in equilibrium.
ΣFx = 0: HA = 0
P1*1.5 + P2*5.5 + RB*7 = 0
- RA*7 - P1*5.5 - P2*1.5 = 0
36
RB = ( - P1*1.5 - P2*5.5) / 7 = ( - 32.03105*1.5 - 76.87451*5.5) / 7 = -67.27 (lbf)
RA = ( - P1*5.5 - P2*1.5) / 7 = ( - 32.03105*5.5 - 76.87451*1.5) / 7 = -41.64 (lbf)
HA = 0 (lbf)
ΣFy = 0: - RA + P1 + P2 - RB = - 41.64*1 + 32.03105 + 76.87451 - 67.27*1 = 0
Q(x1) = - RA
Q1(0) = - 41.64 = -41.64 (lbf)
Q1(1.50) = - 41.64 = -41.64 (lbf)
M(x1) = - RA*(x1)
M1(0) = - 41.64*(0) = 0 (lbf*in)
M1(1.50) = - 41.64*(1.50) = -62.46 (lbf*in)
Consider second span of the beam 1.5 ≤ x2 < 5.5
Q(x2) = - RA + P1
Q2(1.50) = - 41.64 + 32.03 = -9.61 (lbf)
Q2(5.50) = - 41.64 + 32.03 = -9.61 (lbf)
37
M(x2) = - RA*(x2) + P1*(x2 - 1.5)
M2(1.50) = - 41.64*(1.50) + 32.03*(1.50 - 1.5) = -62.46 (lbf*in)
M2(5.50) = - 41.64*(5.50) + 32.03*(5.50 - 1.5) = -100.90 (lbf*in)
Consider third span of the beam 5.5 ≤ x3 < 7
Q(x3) = - RA + P1 + P2
Q3(5.50) = - 41.64 + 32.03 + 76.87 = 67.27 (lbf)
Q3(7) = - 41.64 + 32.03 + 76.87 = 67.27 (lbf)
M(x3) = - RA*(x3) + P1*(x3 - 1.5) + P2*(x3 - 5.5)
M3(5.50) = - 41.64*(5.50) + 32.03*(5.50 - 1.5) + 76.87*(5.50 - 5.5) = -100.90 (lbf*in)
M3(7) = - 41.64*(7) + 32.03*(7 - 1.5) + 76.87*(7 - 5.5) = 0 (lbf*in)
38
39
SHAFT C
PLANE Z-X
is in equilibrium.
ΣFx = 0: HA = 0
- P1*1.5 + RB*3 = 0
- RA*3 + P1*1.5 = 0
40
RB = ( P1*1.5) / 3 = ( 182.9141*1.5) / 3 = 91.46 (lbf)
RA = ( P1*1.5) / 3 = ( 182.9141*1.5) / 3 = 91.46 (lbf)
HA = 0 (lbf)
ΣFy = 0: RA - P1 + RB = 91.46*1 - 182.9141 + 91.46*1 = 0
Q(x1) = + RA
Q1(0) = + 91.46 = 91.46 (lbf)
Q1(1.50) = + 91.46 = 91.46 (lbf)
M(x1) = + RA*(x1)
M1(0) = + 91.46*(0) = 0 (lbf*in)
M1(1.50) = + 91.46*(1.50) = 137.19 (lbf*in)
Q(x2) = + RA - P1
Q2(1.50) = + 91.46 - 182.91 = -91.46 (lbf)
Q2(3) = + 91.46 - 182.91 = -91.46 (lbf)
41
M(x2) = + RA*(x2) - P1*(x2 - 1.5)
M2(1.50) = + 91.46*(1.50) - 182.91*(1.50 - 1.5) = 137.19 (lbf*in)
M2(3) = + 91.46*(3) - 182.91*(3 - 1.5) = 0 (lbf*in)
PLANE Y-X
42
is in equilibrium.
ΣFx = 0: HA = 0
- P1*1.5 + RB*3 = 0
- RA*3 + P1*1.5 = 0
RB = ( P1*1.5) / 3 = ( 76.87451*1.5) / 3 = 38.44 (lbf)
RA = ( P1*1.5) / 3 = ( 76.87451*1.5) / 3 = 38.44 (lbf)
HA = 0 (lbf)
ΣFy = 0: RA - P1 + RB = 38.44*1 - 76.87451 + 38.44*1 = 0
Q(x1) = + RA
Q1(0) = + 38.44 = 38.44 (lbf)
Q1(1.50) = + 38.44 = 38.44 (lbf)
43
M(x1) = + RA*(x1)
M1(0) = + 38.44*(0) = 0 (lbf*in)
M1(1.50) = + 38.44*(1.50) = 57.66 (lbf*in)
Q(x2) = + RA - P1
Q2(1.50) = + 38.44 - 76.87 = -38.44 (lbf)
Q2(3) = + 38.44 - 76.87 = -38.44 (lbf)
M(x2) = + RA*(x2) - P1*(x2 - 1.5)
M2(1.50) = + 38.44*(1.50) - 76.87*(1.50 - 1.5) = 57.66 (lbf*in)
M2(3) = + 38.44*(3) - 76.87*(3 - 1.5) = 0 (lbf*in)
44
45
INPUT SHAFT (SHAFT A)
Point 1
INITIAL GUESS OF CALCULATION PARAMETERS AT POINT 1
From Table 7-1
From Table A-20
MATERIAL AISI 1020 (COLD DRAWN)
Sut = 68kpsi
Shoulder fillet- D/d=1.2
well rounded r/d=0.1
Kt 1.7
Kts 1.5
Kf 1.7
Kfs 1.5
Kt (axial) 1.9
Minimum 2
Factor of
Safety (n)
Moment and Force Exerted at A Value

Ma 20.67 lbf.in
Mm 0
Ta 0
Tm 132.007 lbf.in
Pa 44.002 lbf.in
Pm 0
Calculation:
Se was determined from the initial parameters guessed.
𝑘𝑎 = 𝑎𝑆𝑢𝑡 𝑏
𝑘𝑎 = 2.7(68)−0.265 = 0.8826
46
𝑘𝑏 = 0.9 , since the initial diameter is unknown, so size factor is assumed to be 0.9.
𝑘𝑐 = 1 , the value is set to 1 because of the combined load, bending, torsion and axial. (see
page 298 – 299 Shigley’s Mechanical Engineering Design Textbook)
𝑘𝑑 = 1
𝑘𝑒 = 0.868 , (Table 6-5) page 301
𝑆𝑒 = 𝑘𝑎 𝑘𝑏 𝑘𝑐 𝑘𝑑 𝑘𝑒 𝑆𝑒 ′
𝑆𝑒 = (0.8826)(0.9)(0.868)(0.5)(68) = 23.442𝑘𝑝𝑠𝑖
To estimate the small input diameter, at point 1, we use DE-Goodman criterion equation,
modified with axial component to satisfy the overall load applied.
𝜎𝑎 ′ = (𝜎𝑎 2 + 3𝜏𝑎 2 )1/2
2 2 1/2
′
𝑘𝑓 32𝑀𝑎 𝑘𝑡 4𝑃𝑎 𝑘𝑓𝑠 16𝑇𝑎
𝜎𝑎 = [( 3
+ 2
) + 3( ) ]
𝜋𝑑 𝜋𝑑 𝜋𝑑 3
Since Ta = 0, therefore,
𝑘𝑓 32𝑀𝑎 𝑘𝑡 4𝑃𝑎
𝜎𝑎 ′ = +
𝜋𝑑 3 𝜋𝑑2
𝜎𝑚 ′ = (𝜎𝑚 2 + 3𝜏𝑚 2 )1/2
2 2 1/2
′
𝑘𝑓 32𝑀𝑚 𝑘𝑡 4𝑃𝑚 𝑘𝑓𝑠 16𝑇𝑚
𝜎𝑚 = [( + ) + 3 ( ) ]
𝜋𝑑3 𝜋𝑑2 𝜋𝑑 3
Since Mm = Pm = 0
𝑘𝑓𝑠 16𝑇𝑚
𝜎𝑚 ′ = √3
𝜋𝑑3
1 𝜎𝑎 ′ 𝜎𝑚 ′
= +
𝑛 𝑆𝑒 𝑆𝑢𝑡
Therefore,
𝑘𝑓 32𝑀𝑎 𝑘𝑡 4𝑃𝑎 𝑘𝑓𝑠 16𝑇𝑚

1 3 + 2 √3
= 𝜋𝑑 𝜋𝑑 + 𝜋𝑑 3
𝑛 𝑆𝑒 𝑆𝑢𝑡
47
Since d cannot be simplified and put at one side, due to different exponential of d in axial,
bending and torsion, so d was estimated based on the value of diameter of the bore of bearing
that can support resultant load at the axis.
(1.7)(32)(20.67) (1.9)(4)(44.002) (1.5)(16)(132.007)

1 + √3
𝜋(0.4724)3 𝜋(0.4724)2 𝜋(0.4724)3
= +
𝑛 (23442) (68000)
𝑛 = 2.45
It was found that d1=0.4724inch=12mm, is suitable with bore of 12mm. The bearing of bore
12mm also have C10=6.294 (refer bearing calculation) at 95% reliability. The C10 from
manufacturer catalogue (Table 11-2) is 6.89 which is higher than 6.294. Therefore, the
d1=12mm was selected.
D/d=1.2, so D=(1.2)(0.4724)=0.5669 , then a nominal 0.6 inch D cold-drawn shaft diameter
can be used.
r/d=0.1, so r=(0.1)(0.4724)=0.04724. However, r is pre-determined to r=0.6mm=0.0236inch
so that it can fix with the bearing.
New
D/d=0.6/0.4724=1.27
r/d=0.0236/0.4724=0.05
48
From those value, new constants values were found.
Ka No change 0.882569474
kb (0.4724/0.3)^- 0.952569756
0.107
Kc No change 1
Kd No change 1
Ke No change 0.868
Kt (Figure A-15- 1.96
9)
Kts (Figure A-15- 1.65
8)
Kf 1+0.6892(1.95- 1.66
1)
Kfs 1+0.7477(1.65- 1.486
1)
Kt(axial) (Figure A-15- 2.05
7)
𝑆𝑒 = (0.8826)(0.9526)(0.868)(0.5)(68) = 28.5841𝑘𝑝𝑠𝑖
Notch sensitivity equation.
𝐾𝑓 and 𝐾𝑓𝑠 new,
Bending,
√𝑎 = 0.246 − 3.08(10−3 )(68) + 1.51(10−5 )(682 ) − 2.67(10−8 )(683 ) = 0.09799
Torsion,
√𝑎 = 0.190 − 2.51(10−3 )(68) + 1.35(10−5 )(682 ) − 2.67(10−8 )(683 ) = 0.07334
√𝑟 = √0.0236 = 0.1537
1
𝑞=
√𝑎
1+
√𝑟
𝐾𝑓 = 1 + 𝑞(𝐾𝑡 − 1)
1 1
𝐾𝑓 = 1 + (𝐾𝑡 − 1) = 1 + ( ) (1.96 − 1) = 1.66
√𝑎 0.09799
1+ 1+
0.1537
( √𝑟 )
49
1 1
𝐾𝑓𝑠 = 1 + (𝐾𝑡𝑠 − 1) = 1 + ( ) (1.65 − 1) = 1.486
√𝑎 0.07334
1+ 1+
𝑟 0.1537
( √ )
(1.66)(32)(20.67) (2.05)(4)(44.002) (1.486)(16)(132.007)

1 + √3
𝜋(0.4724)3 𝜋(0.4724)2 𝜋(0.4724)3
= +
𝑛 (28584.1) (68000)
𝑛 = 2.47
After only 1 iteration, the suitable d1 = 0.4724 inch, D1 = 0.6 inch, r = 1mm, for shoulder at
point 1.
Point 2
Next, the big D1 will be the small d2 = 0.6 inch for finding the shoulder at point 2.
D2/d2=1.2, D2=(1.2)(0.6)=0.72 inch, since the keyseat safety factor has not been calculated yet,
the D2 is increased to 0.75 inch to make it safer.
New,
D/d=0.75/0.6=1.25,
r/d=0.1, r=(0.1)(0.6)=0.06inch
Ka No change 0.8826
kb (0.6/0.3)^- 0.9285
0.107
Kc No change 1
Kd No change 1
Ke No change 0.868
Kt (Figure A-15- 1.65
9)
Kts (Figure A-15- 1.4
8)
Kf 1+0.6892(1.95- 1.464
1)
Kfs 1+0.7477(1.65- 1.308
1)
Kt(axial) (Figure A-15- 1.7
7)
50
𝑆𝑒 = (0.8826)(0.9285)(0.868)(0.5)(68) = 27.8623𝑘𝑝𝑠𝑖
Bending,
√𝑎 = 0.246 − 3.08(10−3 )(68) + 1.51(10−5 )(682 ) − 2.67(10−8 )(683 ) = 0.09799
Torsion,
√𝑎 = 0.190 − 2.51(10−3 )(68) + 1.35(10−5 )(682 ) − 2.67(10−8 )(683 ) = 0.07334
√𝑟 = √0.06 = 0.2449
1 1
𝐾𝑓 = 1 + (𝐾𝑡 − 1) = 1 + ( ) (1.65 − 1) = 1.464
√𝑎 0.09799
1+ 1 + 0.2449
( √𝑟 )
1 1
𝐾𝑓𝑠 = 1 + (𝐾𝑡𝑠 − 1) = 1 + ( ) (1.4 − 1) = 1.308
√𝑎 0.07334
1+ 1 + 0.2449
( √𝑟 )
Then, the next step is to measure the safety factor at the beam that hold the gear, together with
the new shoulder at point 2.
Moment and Force Exerted at 2 Value

Ma 41.346 lbf.in, 62.00 lbf.in (highest at the beam)
Mm 0
Ta 0
Tm 132.007 lbf.in
Pa 44.002 lbf.in
Pm 0
51
(1.464)(32)(62) (1.7)(4)(44.002) (1.308)(16)(132.007)
1 + √3
𝜋(0.6)3 𝜋(0.6)2 𝜋(0.6)3
= +
𝑛 (27862.3) (68000)
𝑛 = 3.429
Point 3
At point 3, the Ma exerted is 20.67 lbf.in, so we decided to use same end diameter of d4 =
0.4724 inch as d3. Since the biggest diameter, D2 = 0.75 inch, r=0.0236 to fit with bearing, so,
D/d = (0.75/0.4724) = 1.5875
r/d = (0.0236/0.4724) = 0.05
Ka No change 0.8826
kb (0.4724/0.3)^- 0.9526
0.107
Kc No change 1
Kd No change 1
Ke No change 0.868
Kt (Figure A-15-9) 2.15
Kts (Figure A-15-8) 1.73
Kf 1+0.6892(2.15-1) 1.702
Kfs 1+0.7477(1.73-1) 1.494
Kt(axial) (Figure A-15-7) 2.15
𝑆𝑒 = (0.8826)(0.9526)(0.868)(0.5)(68) = 28.5841𝑘𝑝𝑠𝑖
52
Bending,
√𝑎 = 0.246 − 3.08(10−3 )(68) + 1.51(10−5 )(682 ) − 2.67(10−8 )(683 ) = 0.09799
Torsion,
√𝑎 = 0.190 − 2.51(10−3 )(68) + 1.35(10−5 )(682 ) − 2.67(10−8 )(683 ) = 0.07334
√𝑟 = √0.0236 = 0.1537
1 1
𝐾𝑓 = 1 + (𝐾𝑡 − 1) = 1 + ( ) (2.15 − 1) = 1.464
√𝑎 0.09799
1+ 1+
0.1537
( √𝑟 )
1 1
𝐾𝑓𝑠 = 1 + (𝐾𝑡𝑠 − 1) = 1 + ( ) (1.73 − 1) = 1.308
√𝑎 0.07334
1+ 1+
𝑟 0.1537
( √ )
(1.702)(32)(20.67) (2.15)(4)(44.002) (1.494)(16)(132.007)

1 + √3
𝜋(0.4724)3 𝜋(0.4724)2 𝜋(0.4724)3
= +
𝑛 (28584.1) (68000)
𝑛 = 2.492
Then, the calculation is followed by testing the safety factor at the keyseat 1.
The keyseat used is End-mill keyseat (r/d=0.02).
𝐾𝑡 = 2.14
𝐾𝑡𝑠 = 3.0
𝐾𝑡 (𝑎𝑥𝑖𝑎𝑙) = 0(𝑛𝑜𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑡𝑎𝑏𝑙𝑒 7 − 1)
r = (0.02)(0.6) = 0.012 inch
53
Ka No change 0.8826
kb (0.6/0.3)^-0.107 0.9285
Kc No change 1
Kd No change 1
Ke No change 0.868
Kt (Figure A-15-9) 2.14
Kts (Figure A-15-8) 3.0
Kf 1+0.6892(2.14- 1.814
1)
Kfs 1+0.7477(3.0-1) 2.539
Kt(axial) (Figure A-15-7) 1.7
𝑆𝑒 = (0.8826)(0.9526)(0.868)(0.5)(68) = 27.8623𝑘𝑝𝑠𝑖
Bending,
√𝑎 = 0.246 − 3.08(10−3 )(68) + 1.51(10−5 )(682 ) − 2.67(10−8 )(683 ) = 0.09799
Torsion,
√𝑎 = 0.190 − 2.51(10−3 )(68) + 1.35(10−5 )(682 ) − 2.67(10−8 )(683 ) = 0.07334
√𝑟 = √0.012 = 0.1095
54
1 1
𝐾𝑓 = 1 + (𝐾𝑡 − 1) = 1 + ( ) (2.15 − 1) = 1.814
√𝑎 0.09799
1+ 1+
0.1095
( √𝑟 )
1 1
𝐾𝑓𝑠 = 1 + (𝐾𝑡𝑠 − 1) = 1 + ( ) (1.73 − 1) = 2.539
√𝑎 0.07334
1+ 1 +
0.1095
( √𝑟 )
(1.814)(32)(20.67) (0)(4)(44.002) (2.539)(16)(132.007)

1 3 + 2 √3
𝜋(0.4724) 𝜋(0.4724) 𝜋(0.4724)3
= +
𝑛 (27862.3) (68000)
𝑛 = 2.674
Since the safety factor for keyseat is greater than 2, and the other safety factor at all beam and
shoulder is greater than 2, it can be conclude that the beam with calculated dimension is safe
and acceptable.
D1 = D4 = 0.4724 inch
D2 = 0.6 inch
D3 = 0.75 inch
55
Sy 54Kpsi
Ssy 31.158Kpsi
Torque 132.0068813 lbf.in
Shaft Diameter 0.6 inch
Gear hub length 1 inch
t or w 0.196850394 inch
n 3
F 440.0229377 lbf
L (Sy) 0.248368503 inch
L (Ssy) 0.215224006 inch
56
Key Dimension (based on Shaft
Diameter) Inch mm
t or w 0.196850394 5
h 0.196850394 5
L 0.248368503 6.30855997
Keyway Depth 0.078740157 2
Determination of L of key and keyseat.
The Force at the surface of the shaft is,
𝑇 132.007
𝐹= = = 440.0229 𝑙𝑏𝑓
𝑟 0.6⁄
2
𝑡 = 5𝑚𝑚 = 0.19685 𝑖𝑛𝑐ℎ
𝑡, is determined based on shaft diameter where keyseat at, from table 7-6
By DE theory, the shear strength is
𝑆𝑠𝑦 = 0.577𝑆𝑦 = 0.577(54) = 31.158 𝑘𝑝𝑠𝑖
𝑆𝑠𝑦 𝐹
= 𝜏(𝑠𝑡𝑟𝑒𝑠𝑠) =
𝑛 𝑡𝑙
31.158 𝑥 103 440.0229

=
3 0.19685𝑙
𝑙 = 0.2152 𝑖𝑛𝑐ℎ
To resist crushing, the one-half the face of the key is used.
57
𝑆𝑦 𝐹
=
𝑛 𝑡𝑙⁄
2
54 𝑥 103 440.0229
=
3 (0.19685)𝑙⁄
2
𝑙 = 0.2484 𝑖𝑛𝑐ℎ
The hub length is 1 inch, which is greater than the shaft diameter, 0.6 inch. This is good as it
produced stability. The minimum L for the key and keyseat is 0.2484 inch, as if the length is
increase to the gear hub length, it would have ample strength.
58
SHAFT A
SHAFT B
SHAFT C
59
mm inch
Shaft A D1 12 0.472441
D2 15.24 0.6
D3 19.05 0.75
D4 12 0.472441
Shaft B D5 20 0.787402
D6 24.13 0.95
D7 30.48 1.2
D8 24.13 0.95
D9 20 0.787402
Shaft C D10 25 0.984252
D11 36.83 1.45
D12 30.48 1.2
D13 25 0.984252
Safety
Factor, n
Shaft A P1 2.56
P2 3.43
P3 2.49
60
Keyseat 1 2.67
Shaft B P4 3.83
P5 5.78
P6 3.59
P7 3.02
Keyseat 2 3.64
Keyseat 3 2.42
Shaft C P8 3.75
P9 6.09
P10 2.99
Keyseat 4 3.56
61
Shaft Material AISI 1020 CD
Sut 470 Mpa 68 kpsi
Sy 390 Mpa 57 kpsi
Key Material AISI 1018 CD
Sut 440 Mpa 64 kpsi
Sy 370 MPa 54 kpsi
Shoulder Type Input 3

Shoulder Type Shoulder fillet - well rounded
Constant (initial
guess)
Kt 1.7
Kts 1.5
Kf 1.7
Kfs 1.5
Kt (Axial) 1.9
D/d 1.2
r/d 0.1
Ka 0.8826
Kb 0.9
Kc 1
Kd 1
62
P1 P2 P3 P4 P5 P6 P7 P8 P9 P10
Ma 20.67 62 20.67 23.248 69.7432 216.084 72.0006 49.609 148.833 49.635
Mm 0 0 0 0 0 0 0 0 0 0
Ta 0 0 0 0 0 0 0 0 0 0
Tm 132.007 132.007 132.007 528.0276 528.0276 528.0276 528.0276 1056.0555 1056.056 1056.0555
Pa 44.002 44.002 44.002 149.608 149.608 149.608 149.608 105.6055 105.6055 105.6055
Pm 0 0 0 0 0 0 0 0 0 0
Ka 0.8826 0.8826 0.8826 0.8826 0.8826 0.8826 0.8826 0.8826 0.8826 0.8826
Kb 0.9526 0.9285 0.9526 0.9019 0.884 0.884 0.9019 0.8806 0.8621 0.8806
Kc 1 1 1 1 1 1 1 1 1 1
Kd 1 1 1 1 1 1 1 1 1 1
Ke 0.868 0.868 0.868 0.868 0.868 0.868 0.868 0.868 0.868 0.868
Se 28.5841 27.8623 28.5841 27.0637 26.5255 26.5255 27.0637 26.4251 25.8707 26.4251
D/d 1.27 1.25 1.5875 1.2065 1.2632 1.2632 1.2056 1.4732 1.208 1.2192
r/d 0.05 0.1 0.05 0.05 0.1 0.1 0.05 0.04 0.1 0.04
d 0.4724 inch 0.6 inch 0.4724 inch 0.7874 inch 0.95 inch 0.95 inch 0.7874 inch 0.9843 inch 1.2 inch 0.9843 inch
D 0.6 inch 0.75 inch 0.75 inch 0.95 inch 1.2 inch 1.2 inch 0.95 inch 1.45 inch 1.45 inch 1.2 inch
r 0.0236 inch 0.06 inch 0.0236 inch 0.0394inch 0.095 inch 0.095 inch 0.0394 inch 0.0394 inch 0.12 inch 0.0394 inch
Kt 1.96 1.65 2.15 1.95 1.65 1.65 1.95 2.05 1.62 1.95
Kts 1.65 1.4 1.73 1.59 1.38 1.38 1.59 1.7 1.35 1.65
Kf 1.586 1.464 1.702 1.636 1.49 1.49 1.636 1.703 1.48 1.64
Kfs 1.44 1.308 1.49 1.431 1.307 1.307 1.431 1.511 1.29 1.47
Kt (axial) 2.05 1.7 2.15 1.98 1.7 1.7 1.98 1.88 1.65 2
σ'a 3681.2374 4545.13 3937.9692 1401.6792 1595.834 4191.553 3065.7417 1163.249 1455.197 2878.2765
σ'm 15899.89214 7049.633 16497.847 13649.2766 7099.377 7099.377 13649.277 14761.4124 6947.311 14404.799
nf 2.56 3.43 2.49 3.84 5.8 3.53 3.02 3.75 6.09 2.99
ny 2.91 4.92 2.79 3.79 6.56 5.05 3.41 3.58 6.78 3.3
63
Keyseat 1 Keyseat 2 Keyseat 3 Keyseat 4
Ma 62 69.7432 69.7432 148.833
Mm 0 0 0 0
Ta 0 0 0 0
Tm 132.007 528.0276 528.0276 1056.0555
Pa 44.002 149.608 149.608 105.6055
Pm 0 0 0 0
Ka 0.8826 0.8826 0.8826 0.8826
Kb 0.9285 0.884 0.884 0.8621
Kc 1 1 1 1
Kd 1 1 1 1
Ke 0.868 0.868 0.868 0.868
Se 27.8623 26.5255 26.5255 25.8707
Kt 2.14 2.14 2.14 2.14
Kts 3 3 3 3
Kf 1.6 1.87 1.87 1.89
Kfs 2.2 2.62 2.62 2.65
Kt (axial) 0 0 0 0
σ'a 4682.4663 1545.09 4787.1331 1656.7145
σ'm 11847.49592 14207.64 14207.641 14287.2453
nf 2.67 3.64 2.42 3.56
ny 3.45 3.62 3 3.58
64
ANALYSIS AND DISCUSSION
The shaft in our project experience all three kind of external force that produced
moment, torsion and axial force. In shaft analysis, the moment only form Ma, which is moment
amplitude, as the moment is not reversible, same with the axial force, Pa that was produced
from the axial force of the gear. So, Mm and Pm = 0. Only torque is reversible, producing Tm,
because as the torque rotate, it produced maximum and minimum amplitude, so it produces a
mean value.
The initial assumption of Kt and Kts are from table 7-1, as we used the shoulder fillet-
well rounded as the type of shoulder, and End-mill keyseat as the type of the keyseat.The torque
is relatively small because we were using a large gear diameter and of course, helical
gear.While the moment is also relatively small because we were using short shaft to reduce the
arm length, thus reduce the moment produced.
Moving to key, we chose square key because it shape is simple to create while still able
to transmit desired force without breaking apart.The safety factor for key is also set to 3 to
prevent the key from break.
65
BEARINGS
In this mini project, we also include other components such as bearing, which is used
to enable rotation or linear movement, while reducing friction and handling stresses. Similar
to wheels, bearings enable a device to roll which reduces the friction between the surface of
bearing and the surface of the rolling component, in which in our project is the shaft. Bearing
is vital to the design and is crucial to support the loading and stresses due to the rotation of the
shaft. By analysing the design and requirement from the question, we had assumed and decided
to use ball bearing manufactured by manufacturer 2. There are several reasons we use ball
bearing over roller bearings. One of the reasons is the ball bearing is power saving. With the
truest rolling motion of any type of bearing, ball bearings reduce friction appreciably over roller
bearings, and even more over plain bearing, under ideal or laboratory conditions. In reducing
power consumption, depends upon number of factors in the installation, but with other things
being equal, power saving will be greater with ball bearings. We choose to use deep groove
because it can accept load in axial load, as our gear design uses helical gear. As for the
application factor, we use 1.5 as we are going to use it in machinery with moderate impact. The
bore diameter of the bearing must each matched with the diameter of the shafts which are found
in the previous sections.
CALCULATIONS AND ANALYSIS
We had identified that we are using helical gear which will results into axial loading
present at the bearing. So as there are axial loading, Eqn. (11-12) below is used to calculate
the equivalent radial load.
Fe = XiV Fr + Yi Fa (11–12)
66
where i = 1 when Fa/V Fr ≤ e Fa=axial thrust Fr=radial loads
i = 2 when Fa/V Fr > e
In solving the equations, we refer to Table 11-1. We needed to find Fa/Co which corresponds
to the abscissa, e. The value of Co is obtained by assuming the value of minimum shaft diameter
which corresponds to the bore of the bearing. The value of Fa/Co may not fall exactly to the
value at the table, so interpolation is need to find the value of e. As the number of e is found, it
is compared to the value of Fa/VFr to get the value of Xi and Yi. The value of V is also
determine to be 1.2 as we only use bearing with only the outer ring rotates. Then, with all the
variables are known, the value of Fe is obtained.
The value of Fe obtained is then plugged into Eqn. (11-10) below to get catalog load rating,
C10 as the Reliability is determined to be 95%, R=0.95
C10=af FD[ XD/(X0 + (θ − X0)(1 − RD)1/b)] 1/a R ≥ 0.90 (11–7)
RD = Design Reliability
af =application factor
FD = Fe = Equivalent Radial Load
XD = Dimensionless multiple of rating life
X0 = guaranteed, or “minimum,’’ value of the variate
θ = characteristic parameter corresponding to the 63.2121 percentile
value of the variate
b = shape parameter that controls the skewness
a = type of bearing
67
The value of X0, θ, b is the Weibull Parameters Rating Life which are determined from Table
11-6, by choosing Manufacturer 2, as stated in the assumption above.
Manufacturer =2
X0 = 0.02
θ =4.459
b = 1.483
Lr = 106 ; needed to calculate XD
Then, we calculate the value of XD using the Eqn,
XD =[ (60LDnD )/Lr ]
With LD given in the questions and nD depends on Shaft
The af is determined from Table 11-5 by assuming the Type of Application is Machinery with
moderate impact with value 1.5.
68
SAMPLE OF CALCULATIONS
Bearings at Shaft A
Fa = 97.9 N Fr = 183.963 N From Calculation From Shaft section
Co = 3100 N from table 11-2, by assuming bore is 12mm
Fa/Co = 0.03158 ; obtaining e = 0.22512 by interpolation from Table 11-1
Fa/(VFr) =0.4435 so Fa/(VFr)> e, so i = 2
By interpolation: Xi = 0.56, Yi = 1.95419
Fe1 = XiV Fr + Yi Fa
= (0.56*1.2*183.963) + (1.95419*97.9)
= 314.939 N
RD = 0.95 af =1.5 FD = Fe = 314.939 N from calculation above
LD= 20 000 hour
nD = 1200rpm ; from calculation at Shaft Section
= 1440
X0 = 0.02
θ =4.459
b = 1.483
a=3
C10 =af FD[ XD/(X0 + (θ − X0)(1 − RD)1/b)] 1/a R ≥ 0.90
69
C10 =(1.5)(314.939)[( 1440)/ (0.02+(4.459-0.02)(1-0.95)1/1.483)] 1/3
= 6.2942kN ; take a bigger value available at Table 11-2 ; 6.89
Bore = 12mm
From the calculation done, we assume the value of the Co by using the value of minimum
diameter for the shaft to carry the load, so, we obtain 12mm. So, we use the Table 11-2 to get
the value. Then subs into the formula to find the equivalent radial load. As the calculated C10
with 0.95 reliability is somehow nearest and lower than the C10 of 12mm. So we choose both
of the bearing supporting Shaft A to be having 12mm bore which should be able to support the
load acting on them.
Bearings at Shaft B
Fa = 332.86 N Fr = 641.067 N From Calculation From Shaft section
Co = 6200N from table 11-2, by assuming bore is 20mm
Fa/(VFr) =0.4327 so Fa/(VFr)> e, so i = 2
By interpolation: Xi = 0.56, Yi = 1.73313
= (0.56*1.2*641.067) + (1.73313*332.86)
= 1007.69 N
LD= 20 000 hour
nD = 300 rpm ; from calculation at Shaft Section
70
= 360
X0 = 0.02
θ =4.459
b = 1.483
a=3
C10 =(1.5)( 1007.69)[( 360)/ (0.02+(4.459-0.02)(1-0.95)1/1.483)] 1/3
= 12.6868 kN ; take a bigger value available at Table 11-2 ; 12.7
Bore = 20mm
From the calculation done, we assume the value of the Co by using the value of
minimum diameter for the shaft to carry the load, so, we obtain 20mm. So, we use the Table
11-2 to get the value. As for the value of Fr, we calculate Fr at both ends of the shafts, as we
want to avoid failure, we took the bigger value of Fr. Then subs into the formula to find the
equivalent radial load. As the calculated C10 with 0.95 reliability is somehow nearest and lower
than the C10 of 20mm. So we choose both of the bearing supporting Shaft B to be having 20mm
bore which should be able to support the load acting on them.
71
Bearings at Shaft C
Fa = 42.1415 N Fr = 441.4845N From Calculation From Shaft section
Co = 6200N from table 11-2, by assuming bore is 20mm
Fa/(VFr) = 0.0795 so Fa/(VFr)<= e, so i = 1
By interpolation: Xi = 1, Yi = 0
= (1*1.2*441.4845) + (0*42.1415)
= 529.781 N
LD= 20 000 hour
nD = 150 rpm ; from calculation at Shaft Section
= 180
X0 = 0.02
θ =4.459
b = 1.483
a=3
72
C10 =(1.5)( 529.781)[( 180)/ (0.02+(4.459-0.02)(1-0.95)1/1.483)] 1/3
= 5.294 kN ; take a bigger value available at Table 11-2 ; 6.89 Bore = 12 mm
But we use,
Bore= 25mm
From the calculation done, we assume the value of the Co by using the value of minimum
diameter for the shaft to carry the load, so, we obtain 25mm. So, we use the Table 11-2 to get
the value. Then subs into the formula to find the equivalent radial load. As the calculated C10
with 0.95 reliability is somehow nearest and lower than the C10 of 12mm. But in this case, the
value of the bore used is 25mm because the minimum diameter shaft for shaft C is calculated
to be 25mm. So, we choose both of the bearing supporting Shaft C to be having 25mm bore
which should be able to support the load acting on them and also fit the shaft C.
Decided Design for Both Bearing
Fillet C10 ,
Bearing for Shaft Bore,mm OD,mm Width, mm radius, mm kN
A 12 32 10 0.6 6.89
B 20 47 14 1 12.7
C 25 52 15 1 14
73
DISUSSIONS
From all the bearing designed we could say that, in order to fully designed the bearings,
the calculations are very reliant to the design of shafts. We could obviously see the dependency,
when we are calculating the C10 for shaft A and B. The Fr for bearings at shaft cannot be taken
from the lower value of Fr of the bearings of Shaft B as if we design for the lower one, the end
with the bigger value of Fr could fail with the value we used for C10 . From Shaft C also we
could see that even though we get a lesser bore diameter which is 12mm, we have to consider
the minimum value of shaft C diameter, 25mm. This is because the shaft could not fix the
bearing if it were 12mm in diameter. It is also okay for the bore of the bearing to be bigger as
the bigger the value of the bore the bigger the C10 . This may affect our cost a little bit as the
bigger the size of the bearings, the higher the cost it would take, but it is for the greater good
as we want to prevent failure to our system. All of the assumption made such as the application
factor, type of bearings and manufacturer are picked based on the environment and the usage
of the speed reducer used and also affected by the type of gears used in this machine. As for
the V=1.2, the bearings are held by the connections of the bearings and the wall and the rotating
part is the outer ring which is the connection of the shafts and the bearing.
74
DESIGN STEPS II
CASING BOX
Based on SolidWorks, the total mass, the total volume, and the surface area was
obtained and calculated. Grey cast iron was selected as the material for the gearbox due to its
properties. Grey cast iron can withstand extreme temperature as our desire to design the
gearbox to be used for motorboat. Due to the cost and availability, grey cast iron was chosen
over steel.
Mass, m (g) 161724.73
Volume, V (mm3) 22047300.84
Surface Area, (mm2) 2767430.7
Based on the data obtained, the factor of safety was calculated by using SolidWorks
and force obtained, 1586.52 N was obtained by assuming weight of the total assembly that
include shafts, gears, bearings and keys. The minimum factor of safety obtained can be seen as
1.2.
75
Mass, m (g) 161724.73
Volume, V (mm3) 22047300.84
Surface Area, (mm2) 2767430.7
Factor of Safety 1.2
76
SPRINGS
Another component for our machine is spring. Spring is used to reduce the force of the
load acting on other components. Spring is essential as it would avoid damage. We are using
music wire A288. This is because it can withstand higher stresses under repeated loading than
any other spring material. From table 10-4, A= 201 kpsi.inm , m=0.145, ; from Table 10–5, E
= 28.5 Mpsi, G = 11.75 Mpsi (expecting d > 0.064 in). We also used end squared and ground
spring to make the spring stable on flat surface. As for the safety of the springs, we use design
factor at solid height of (ns)d = 1.2. The safety is low, as there are more than one sprigs used
and also the spring is only used to reduce the force acting on the wall. With Robust linearity: ξ
= 0.15 and as-wound spring , Ssy = 0.45Sut from Table 10–6 as they are recommended design
conditions to follow and also cheaper. The decision variable: d = 0.080 in, music wire gage
#30, Table A–28. From Fig. 10–3 and Table 10–6.
F max=Total force of the box and its content/ number of springs used
=(1586.52N/4.45)/ 12
= 29.71lbf
Design Calculations & Analysis: the actual design procedure

201 000
Ssy = 0.450.080.145
= 130 455 ksi
𝑆𝑠𝑦 130455
α = =
ns 1.2
=108713 ksi
8(1+ 𝜉)𝐹𝑚𝑎𝑥 8(1+0.15)(29.71 )

β = =
πd^2 π(0.082 )
= 13594.47 ksi
77
2(108713)−13594.47 2(108713)−13594.47 3(13594.47 )
C = + [[(( )^2) - 4(13594.47 ))]^(1/2)]
4(13594.47 ) 4(13594.47 )
=7.463
D = Cd = 7.463(0.08)
=0.597 in
4𝐶+2 4∗7.463+2
KB = 4C−3 = 4∗7.463−3
= 1.186
8(1+0.15)(29.71)(0.597)
τs = 1.186( )
π(0.083 )
= 120338.6 ksi
130455
ns = 120338.6
=1.1
OD =0.597+0.08
= 0.677 in
(0.08^4)(11750000)(2)
Na = 8(0.5973 )(29.71)
= 19.03 turns
Nt = 19.03 + 2
= 21.03 total turns
78
Ls = (Nt +1)d = (21.03)(0.08)
= 1.6824 in
Lo = y +Ls = 1.6824 + (1 + 0.15)(2)
= 3.9824 in
0.597
Lcr = 2.63( )
0.5
= 3.14 in
𝜋 2 (0.082 )(21.03)(0.597)
Fom = −2.6 4
= -0.515
79
Fmax 29.71 lbf
Type of Spring Squared and Ground Helical
Compression
Material Of Spring Music wire
Ymax 2 in
Fractional Overrun 0.15
Shear Stress, τs (ksi) 120338.6
Number Of Spring Used 12
Factor Of Safety, ns 1.1
Outer Diameter (in) 0.677
Elastic Modulus, G 11.75
Spring Force, Fs 34.1665
Total Coils Nt 21.03 turns
Solid Length, Ls (in) 1.6824
Free Length, L0 (in) 3.9824
Figure of Merit, fom -0.515
From the calculation done, we found out that the number of springs needed is very high
as our load of the box is high. We also found out the free length and solid length of the spring
to be used in our design. The factor of safety of the spring is also reasonable at 1.1, as many
springs are use and is not critical.
80
FASTENERS
We also include fasteners in our design for the machine. Fasteners are used to join or
fixes two or more objects together. In general, fasteners are used to make non-permanent joints,
that is joints that can be removed or dismantled without damaging the joining components. For
our design, we use fasteners of grade 25 cast-iron pressure vessel with 5/8 in-11 UNC × 2 in
grade 5 finished hex head bolt.

35 2
L = 64 +1.50 + 11 = 2.229 in
𝐴𝑑𝐴𝑡𝐸 0.3068(0.226)(30)
Kb = Ddlt+Adld = 0.3068(0.75)+0.226(0.75)
= 5.21 Mlbf/in
0.5774𝜋𝐸𝑑 0.5774𝜋(14)(0.625)
Km = 0.5774𝑙+0.5𝑑 = 0.5774(1.5)+0.50.625)
2ln(5 ) 2ln(5 )
0.5774l+2.5d 0.5774(1.5)+2.5(0.625)
= 8.95 Mlbf/in
𝑘𝑏 5.21
C = kb+km = 5.21+8.95
= 0.368
Fi = 0.75AtSp = 00755(0.226)(85)
= 14.4 kip
N = C(nL) (Ptotal)/(Sp At – Fi)
= 0.368(1.2)(356.52)/85(0.226) − 14.4
= 32.73
81
85(0.226)−14.4
nL = 356.52
0.368( )
33
= 1.2
𝑆𝑝𝐴𝑡 85(0.226)
nP = 𝑃𝑡 = 356.52
C( )+𝐹𝑖 0.368( )+14.4
𝑁 33
= 0.263
𝐹𝑖 14.4
nO = 𝑃𝑡 = 356.52
( )(1−𝐶) ( )(1−0.368)
𝑁 33
= 2.1
82
Type of Member Cast-iron
Nominal Major Diameter 0.625
(mm)
Number of Members 33
Property Class of Bolt 5
Bolt Grip (mm) 1.5
Total Weight, P Total (lbf) 356.52
Force on each Member 10.8 lbf
Pitch, P (mm)
Tensile-Stress Area, At (in2) 0.226
Minor-Diameter Area, Ar 0.202
(in2)
Minimum Proof Strength, SP 85 kpsi
Minimum Tensile Strength, 120
St (MPa)
Minimum Yield Strength, SY 92
(MPa)
kb (M lbf/in) 5.21
km (M lbf/in) 8.95
C 0.368
1-C 0.632
Yielding Factor of Safety, np 0.263
Load Factor, nL 1.2
83
Factor of Safety against Joint 2.1
Separation, n0
Screw is the smallest component used in our assembly and one of the most important
components that we are required to used. This is because screw, fastener and bolt are needed
to tighten or assemble the housing. Housing is something which hold all the assemble
components. Housing is not only to hold all components but to withstand certain criteria such
as temperature, external and internal forces, pressures and etc. In order to tighten this housing
and make sure it does not open easily fastener is needed.
84
DISCUSSION
After attempting numerous analysis and studies in gear, each data was tabulated and
illustrated. Every equation that was used in calculation was taken from the textbook. Based on
our initial assumption as can be seen in the first step, the number of teeth, the diameter pitch
of each gear, were determined. Since we are doing helical gear, we have an extra force coming
from the gear known as axial force. Extra calculations and assumptions need to be made in
order to determine the factor of safety for gears. As we are designing a mini boat rotor in mind,
we assumed the face width to be 1 inch. For gear and pinion, grade 2 steel was chosen as the
material is tough and has high resistance towards harsh environment. In order to verify our
theory, bending and wear analysis were done. From our analysis, for gear 2 wear gave more
threat while gear 3 also the threat came from wear. The same results were obtained for gear 4
and gear 5 as the threat came from wear.
For shaft analysis, AISI 1020 Cold Drawn Low Carbon was chosen, as clearly
mentioned in textbook, page 353, “Many shafts are made form low carbon, cold drawn or hot
rolled steel, such as AISI 1020- 1050 steel. AZO Material website was chosen as one of our
primary sources, in order to determine the best choice for the material to manufacture our
designed shaft. We begin our analysis for shaft, by finding the shear and moment diagram at
different points, as different amount of moment occurred at different point.
One of the few problems that we uncounted was, during the analysis part of the shaft,
we forgot to include axial force as axial force form helical gears were to huge compared to spur
gears and cannot be ignored in calculation. For every shaft, space allocated for gears and
bearings need to be carefully determined in order to synchronise with the given width form the
manufacturers’ catalogue. This is to prevent vibration and to transmit axial force efficiently.
For the shoulder of the shaft, well rounded was chosen in order to fit the bearing perfectly. For
every part of the shaft the minimum factor of safety was set to 2. At intermediate shaft, point
85
5 and output shaft, point 9 the factor of safety is more than 5. At every gear, there will be keys
in order to transmit power and speed from shaft to gear. The average diameter of our shaft was
smaller in size as we set the length of our shaft was set too short, the effect will come at the
moment. In order to get optimum factor of safety, the diameter need to be smaller in size.
The factor of safety for key was set to 3. It was proven in our calculation, that all the
keys with the calculated dimensions can transmit the power efficiently from the shaft to gear
without breaking apart. Our gears are also considered stable because the hub length is much
bigger than the shaft diameter. (text book page 387, example 7-6).
The values of the Co for bearing were obtained by using the value of minimum
diameter for the shaft to carry the load. Reliability of 0.95 was set for bearing, therefore Table
11-2 was used in order to get the desired value of C10. In order to get the resultant force acting
at each bearing, radial force, tangential force as well as the radial force need to be included as
the axial force that exerted from the helical gears to the shaft.
In order to design the housing for the gears set, the diameter pitch for the biggest gear
need to be considered for the height of the gear box. For the width of the housing, the length
of the longest shaft will be considered in order to gat perfect dimension for the housing, The
material that will be used to produce the housing can withstand extreme temperature because
the gearbox will be used for in motorboat. That is why grey cast iron was chosen as the most
suitable material for the housing. In order to get ample stability, springs were added to the
assembly. Music wire was chosen because of is toughness and the cost as well as the
availability in current market.
Bolt and nut were used as the fastener to tighten the joints of surfaces of the gearbox.
25 cast-iron pressure vessels with 5/8 in-11 UNC × 2 in grade 5 finished hex head bolt was
chose as the fattener for the gearbox as the dimension of the bolt perfectly fitted to connect the
two surfaces of the gearbox. Regular maintenance needs to be taken care as the fatigue for the
86
fastener part will start to show first followed by the housing, shaft and gears. A scheduled
maintenance needs to be done in order to have the gearbox in good condition.
87
CONCLUSION
The objectives of the case study which were to design and analyse several machine
components such as gears, shafts, keys, couplings, fasteners, springs, bearings and casing box.
Software such as SolidWorks and Microsoft Excel were used in order to get right data for the
calculation and analysis part. So far, the theories that learned in class were applied in this
project. The theories and the calculation can has been approved through the results of this
project as the gearbox assembly was successfully simulated without any errors. The output
speed will be reduced as mentioned in the introduction of this report. Even though the
calculations have been approved, one must not ignore that the factor of safety for each
component is important. The factor of safety will determine how long the component will work
before it breaks. The lower the factor of safety, the time for the component to fail will be faster.
The factor of safety for each component also highly affected by the material. The selection of
material for each part is very important and the cost that comes with it cannot be ignored. A
good material will come with a good price which makes the overall assembly more expensive.
88
REFERENCES
[1] Richard. G. B. & J. Keith. N. (2015). Shigley’s Mechanical Engineering Design.(2nd
ed.) New York, NY: McGraw-Hill Education.
[2] Gear. (2018, July,4). In Wikipedia, The Free Encyclopedia. Retrieved 3:43,
July,4,2018, from https://en.wikipedia.org/w/index.php?title=Gear&oldid=796859221
[3] Shaft (mechanical engineering). (2017, July 29). In Wikipedia, The Free Encyclopedia.
Retrieved 1:23, July 30, 2018, from
https://en.wikipedia.org/w/index.php?title=Shaft_(mechanical_engineering)&oldid=7
929421 42
[4] Key (engineering). (2017, August 22). In Wikipedia, The Free Encyclopedia. Retrieved
04:41, August 29, 2017,
[5] from https://en.wikipedia.org/w/index.php?title=Key_(engineering)&oldid=79666800
[6] V.S. Panwar & S.P. Mogal (2015). A Case Study on Various Defects Found in a Gear
System. International Research Journal of Engineering and Technology, 2(3), pp. 425-
429. Retrieved from https://www.irjet.net/archives/V2/i3/Irjet-v2i376.pdf
[7] R. C. Hibbeler, Mechanics of Materials, 8th Edition (2013), Prentice Hall International.
[8] R.C. Hibbeler & K.B. Yap, Mechanic for Engineers: Statics, 13th SI Edition, (2013),
Pearson
[9] Foundations of Materials Science and Engineering, by William F Smith and J. Hashemi,
5th Edition in SI Unit, McGraw Hill, 2011.
89
APPENDICES
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91
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What is the overall purpose of the project?

What is the overall purpose of the project?

What components are designed as part of this project?

What components are designed as part of this project?

UNIVERSITI TENAGA NASIONAL

MEMB333 - MACHINE DESIGN

PROJECT TITLE : DOUBLE GEAR REDUCER

LECTURER : DR. SAMI SALAMA HUSSEN HAJJAJ

Figure 1 : Typical speed reducer

components were calculated and were tabulated.

Figure 2: Design for double step reducer gear

expectancy of the gears.

couplings, fasteners, springs, bearings and casing box.

• To make use of knowledge obtained during lecture by performing analysis and

calculation to design speed reducer by performing complex analysis and calculations.

and performing complex analysis and calculation.

to have a uniformed power source and moderate shock acting on it

N2 & N4 are pinion

For alignment: N2 + N3 = N4 +N5

N2 = 18 teeth, N3 = 72 teeth, N4 = 30 teeth, N5 = 60 teeth

Wr2 = WT2 tan ∅1

Wa2 = WT2 tan 𝜔

= 76.2142 tan 30°

Wr3 = WT3 tan ∅1

= 76.2142 tan 22.8°

Wa3 = WT3 tan 𝜔

= 76.2142 tan 30°

Wr4 = WT4 tan ∅1

= 182.914 tan 22.8°

Wa4 = WT4 tan 𝜔

= 182.914 tan 30°

Wr5 = WT5 tan ∅1

= 182.914 tan 22.8°

Wa5 = WT5 tan 𝜔

= 182.914 tan 30°

Face width, F = 1 inch

For Bending and Wear

Brinell Hardness, HB = 200

= 102(200) + 16400 psi

Allowable Contact stress, Sc =349HB + 34300 psi

= 349(200) + 34300 psi

Z = [(rP + a)2 - rbP2]1/2 + [(rG + a)2 – rbG2]1/2 –(rP +rG)sin Øn

3.46 3.46cos(22.8) 2 1/2 13.86 13.86cos(22.8)2 1/2 3.46 13.86

Lewis form factor, YP = 0.309

Lewis form factor, YG = 0.432

Load distribution Factor, KM = 1 + Cmc(CpfCpm + CmaCe)

Reliability factor, KR = 0.658 – 0.0759ln(1-0.95)

34.94+ √1088.28 0.731

Stress cycle factor pinion, Ynp = 1.3558N2-0.0178

Stress cycle factor pinion, YnG = 1.3558N3-0.0178

Pitting resistance stress cycle,Zn = 1.4488N3-0.0323

Geomety factor, J’P = 0.46

Geomety factor, J’G = 0.54

Modifying factor, JP = 0.432

Modifying factor, JG = 0.539

6 cos 30° (1.19)(1)

6 cos 30° (1.19)(1)

As 2.922 > 20.83, G2 threat comes from bending.

RB = ( - P11.5) / 3 = ( - 76.21421.5) / 3 = -38.11 (lbf)

RA = ( - P11.5) / 3 = ( - 76.21421.5) / 3 = -38.11 (lbf)

ΣFy = 0: - RA + P1 - RB = - 38.111 + 76.2142 - 38.111 = 0

M1(0) = - 38.11(0) = 0 (lbfin)

M1(1.50) = - 38.11(1.50) = -57.16 (lbfin)

M(x2) = - RA(x2) + P1(x2 - 1.5)

M2(1.50) = - 38.11(1.50) + 76.21(1.50 - 1.5) = -57.16 (lbf*in)

M2(3) = - 38.11(3) + 76.21(3 - 1.5) = 0 (lbf*in)