Mechanical Lifting machine | Design project | Mechanical Engineering Undergraduate

MACHINE DESIGN
MEX5277
MECHANICAL LIFTING
MACHINE
NAME : N.PRASANTH
DUE DATE : 612264658
CENTRE : COLOMBO
DUE DATE : 19/10/2016

Contents
Aim .......................................................................................................................................................5
Objective ..............................................................................................................................................5
Specifications .......................................................................................................................................5
Design Layout.......................................................................................................................................6
Motor Selection .................................................................................................................................10
Belt & Pully Design.............................................................................................................................16
Gear Design........................................................................................................................................27
Chain Driver Design............................................................................................................................43
Bearing Selection ...............................................................................................................................47
Shaft Design .......................................................................................................................................79
Keyway Design ...................................................................................................................................95
Clutch Design ...................................................................................................................................102
Wire Rope Design.............................................................................................................................105
References .......................................................................................................................................109
Appendix ..........................................................................................................................................110

List of Figures
Figure 1: Side view...................................................................................................................................................7
Figure 2: Top view....................................................................................................................................................8
Figure 3: Designing parts in detail ...........................................................................................................................9
Figure 4: Selection of V-belt cross section.............................................................................................................17
Figure 5: Service factor for drives.........................................................................................................................18
Figure 6: Pulley arrangement ................................................................................................................................18
Figure 7: B section V-belts.....................................................................................................................................19
Figure 8: Pulley belt arrangement ........................................................................................................................23
Figure 9: Gear arrangement ..................................................................................................................................28
Figure 10: No: of teeth vs. Involute factor Y..........................................................................................................29
Figure 11: Values of Deformation Factor C (kN/m) for Dynamic load ..................................................................32
Figure 12: Reverse Gear arrangement ..................................................................................................................35
Figure 13: Gear arrangement ................................................................................................................................36
Figure 14: Terms used in a gear.............................................................................................................................42
Figure 15: Terms in key..........................................................................................................................................96
Figure 16: Motor Selectin Chart ..........................................................................................................................110
Figure 17: Power Rating for “B”- Section V-Belt..................................................................................................111
Figure 18: Power correction factors for arc of contact .......................................................................................111
Figure 19: Power correction factors for belt pitch length...................................................................................112
Figure 20: Gear materials ....................................................................................................................................113
Figure 21: Characteristics of roller chains ..........................................................................................................114
Figure 22: Chain Drive selection ..........................................................................................................................115
Figure 23: Factor of safety (n) for bush roller and silent chains.........................................................................116
Figure 24: Values of service factor for bearing....................................................................................................116
Figure 25: Values of X0 and Y0 for bearing...........................................................................................................116
Figure 26: Values of x and y for dynamically loaded bearings ............................................................................117
Figure 27: Bearing Selection (1)...........................................................................................................................118
Figure 31: Values for shaft calculation ................................................................................................................122

List of Tables
Table 01: Pulley characteristics …………………….………………………………………………………………………………………….26
Table 02: Gear Characteristics …………………………………………………….…………………………………………………………….29
Table 03: Gear Characteristics …………………………………………………….…………………………………………………………….37
Table 04: Particulars for 20degree stub involute system ……………….………………………………………………………….42
Table 05: Gear Particulars……..........................................................................................................................42
Table 06: Key cross section value for Shaft diameter…………………………………………………………………………….……97
Table 07: Material selection table……………………………………………………………………………………………………………..103

AIM
To design a Mechanical lifting machine to transport material to a higher elevation from ground
level.
OBJECTIVE
• Selecting the necessary material to be transported upward and downward motion.
• Designing the mechanical lifting machine.
• Obtaining the power required for the design and selection of an appropriate motor.
• Design the drive mechanism.
• Design the gearbox.
• Design the shaft and selection of keys.
• Selection of bearings.
• Design a suitable clutch.
• Design a Wire Rope Design
SPECIFICATIONS
1. Maximum speed of the lifting cabin
Upward motion:- 0.6 m/s , Downward motion:- 0.65 m/s
2. Mass of the cabin = 100 kg
3. Working mass on the cabin = 500 kg
4. Diameter of the cable drum = 0.25 m

Figure 3: Designing parts in detail

The total mass of the cabin is 500 kg which has two direct motions. During these motions, there will be
no accelerations due to the very slow speed needs to be achieved.
 Upward Motion
T1 T1 - Mg = Ma
T1 = Mg (Assuming No Acceleration)
T1 = 600 kg x 9.81 m/s2
T1 = 5886 N
Power = F x V
= 5886 N x 0.6 m/s
(500+100) x g = 3532 W
 Downward Motion
T2
(100 x g) - T2 = Ma
T2 = Mg (Assuming No Acceleration)
T2 = 100×9.81
T2 = 981 N
Power = F * V
= 981 N × 0.65 m/s
100 x g = 638 W

At first, we need to calculate the losses within the mechanical elements.
Losses due to Gear train = 3% Losses due to belt = 4%
Losses due to drive pulley = 3% Losses due to Bearing
couple
= 2%
Losses due to pinion pulley = 3% Losses due to chain = 5%
Uncountable losses = 15%
1) Losses Due to Gear Train
The efficiency of a Gear = 97%
In this design there are 3 different gear mesh can happen. I assume all gear mesh have the same
efficiency of 97%. So losses on the gear 1 & 2, losses on the gear 1 & 3 and losses on the gear 3 & 4
need to be calculated. So the maximum power needed at the cable drum is 3532 W.
Total power loss through Gear = 3532 W – [3532 𝑥 (
97
100
) 𝑥 (
97
100
) 𝑥 (
97
100
)]
= 308 W
2) Losses Due to Bearings
The efficiency of a Bearing couple = 98%
In this design there are 8 different bearing couples are available. I assume all bearings have the same
efficiency of 98%.
Total power loss due To Bearings = 3532 W – [3532 𝑥 (
98
100
)8
]
= 527 W
3) Losses Due to Belts
The efficiency of a Belt = 96%
Total power loss due To Belts = 3532 W – [3532 𝑥 (
96
100
)]
= 141 W

4) Losses Due to Drive pulley
The efficiency of a Drive pulley = 97%
Total power loss due to Drive pulley = 3532 W – [3532 𝑥 (
97
100
)]
= 106 W
5) Losses Due to Pinion pulley
The efficiency of a Pinion pulley = 97%
Total power loss due to Pinion pulley = 3532 W – [3532 𝑥 (
97
100
)]
= 106 W
6) Losses Due to Chain drive
The efficiency of a Chain drive = 95%
95
100
)]
= 177 W
7) Uncountable Loss = 15%
85
100
)]
= 530 W
Total Loss of the Machine = (308+ 527+ 141+ 106+ 106+ 177+ 530) W = 1895 W
Required Motor Power = (3532 + 1895) W = 5427 W = 5.427 kW
The motor need to produce more than 5.5 kW power. So according to the induction motor manual,
I choose 7.5 kW, 8 poles 750 RPM synchronous at 50Hz induction motor. (Refer: Figure 16)

Motor output power = 7.5 𝑘𝑊 𝑥 (
86.8
100
) = 6.51 kW
6.51 kW > 5.427 kW ( Motor can supply the necessary power required)
Selected motor = 1D (1AI) 160L-8 at 719 rpm (Refer: Figure 16)
Transmission Ratios
Upward motion
Cable Drum Diameter (d) = 0.25m
Linear Velocity of the Cable (v) = 0.6 m/S
Angular Velocity of the Drum = 𝜔D
𝜔D =
2𝑉
𝑑
x
60
2𝜋
𝜔D =
2 × 0.6
0.25
×
60
2π
𝜔D = 45.84 r.p.m.
Overall Transmission Ratio =
motor r.p.m
drum r.p.m
=
719
45.84
= 15.68 ≈ 16

Downward motion
Cable Drum Diameter (d) = 0.25m
Linear Velocity of the Cable (v) = 0.65 m/S
Angular Velocity of the Drum = 𝜔D
𝜔D =
2𝑉
𝑑
x
60
2𝜋
𝜔D =
2 × 0.65
0.25
×
60
2π
𝜔D = 49.65 r.p.m.
So, overall Transmission
Ratio =
motor r.p.m
drum r.p.m
=
719
49.65
= 14.48 ≈ 16
Transmission Ratio of the Belt Drive = 2:1
Transmission Ratio of the Gear Drive = 2:1
Transmission Ratio of the Chain Drive = 4:1

Data:
1) Transmission Ratio of the Belt Drive = 2:1
2) Power Consumption of Motor = 7.5 Kw
3) Motor r.p.m. (Speed of Faster Shaft) = 719 r.p.m
From the V-Belt drive handbook,
Selection of V-belt cross section, (page No: 3/79)
According to the design power and rpm of the motor,
Type of belt = Type B
Service factor for drives, (Page No :3/80)
Operation hours per day = over 16 hrs.
Type of driven mechanism = Extra heavy duty
Service factor (Ks) = 1.4
Figure 4: Selection of V-belt cross section

Assumptions:-
1. Pitch Circle Diameter of Smaller Pulley (dp) = 125mm
2. Belt Drive Ratio = 1: 2
∴ Pitch circle diameter of larger pulley (Dp) = 2 × dp = 2 × 125 mm = 250 mm
Center Distance between small and large pulley = C
d
D
C
Figure 5: Service factor for drives
Figure 6: Pulley arrangement

Recommended range for the centre distance is,
∴ taking minimum C value for calculations,
C = 262.5 mm
Recommended/Calculated Pitch Length of the Belt,
L = 2C + 1.57(Dp + dp) + (Dp − dp)2
/4𝐶
L = 2 × 262.5 + 1.57(250 + 125) + (250 − 125)2
/(4 × 262.5)
L = (525) + ( 588.75 ) + (14.88)
L = 1128.6 mm
Where,
L - Pitch Length of Belt in mm
According to the standard pitch lengths, From Table3C, (page No: 3/69)
2(𝑑 + 𝐷) ≥ 𝐶 ≥ 0.7(𝑑 + 𝐷)
2(125 + 250) ≥ 𝐶 ≥ 0.7(125 + 250)
750 mm ≥ 𝐶 ≥ 262.5 𝑚𝑚
Figure 7: B section V-belts

According To the Standard Pitch Lengths,
Nominal Pitch Length of the Belt = 1210mm
Center Distance May Be Calculated From the Following Formula,
C = A + √(A2 − B)
01) A =
L
4
− π
(D + d)
8
A =
1210
4
− π
(250 + 125)
8
A = 155.24mm
02) B =
(D − d)2
8
B =
(250 − 125)2
8
B =1953.12 mm
03) C = 155.24 + √(155.242 − 1953.12)
C = 304.0 mm

Calculation of No of Belts
X =
(NtKs)
(No Ke Kl)
Where,
Nt – Required Power in watts
Ks – Service Factor for Belt Drive (Refer. Figure 5)
No – Power Rating
Ke – Power Correction Factor for Arc Contact
Kl – Power Correction Factor for Belt Pitch Length
X – No of Belts
1. Power Rating for “B”- Section V-Belt, 17mm Wide with 180⁰ Arc of Contact on Smaller Pulley
(Refer. Figure 17)
Speed Of Faster Shaft
(Rev/Min)
Smaller Pulley Pitch
Diameter
Additional Power
Ratio
125mm 2 & Over
720 1.61 kW 0.23kw
Therefore Power Rating (No) = X1 + X2
= 1.61 + 0.23
=1.84 kW

2. Power Correction Factor for Arc of Contact (Refer. Figure 18)
(Dp − dp)
C
=
(250 − 125)
304.0
= 0.41
By using interpolating method,
(0.40 − 0.45)
(0.40 − 0.41)
=
(0.94 − 0.93)
(0.94 − y)
Y = 0.938
∴ Power correction factor for arc contact (Ke) = 0.938
3. Power Correction Factor for Belt Pitch Length. (Refer. Figure 19)
“B” Section
Pitch Length(Mm) Factor
1210 0.87
Power Correction Factor for Belt Pitch Length (Kl) = 0.87
Number of belts:-
X =
𝐷𝑒𝑠𝑖𝑔𝑛 𝑃𝑜𝑤𝑒𝑟
𝐹𝑖𝑛𝑎𝑙 𝐵𝑒𝑙𝑡 𝑃𝑜𝑤𝑒𝑟
=
(NtKs)
(No Ke Kl)
X =
(7.5 × 1.4)
(1.84 × 0.938 × 0.87)
X = 6.99  7.
Therefore 7 belts needed to transmit power.
(Dp-dp)/C Correction Factor , I.E. Proportion Of
180⁰ Rating
0.40
0.41
0.45
0.94
Y
0.93

Assessing the required no: of belts using standard equation
Cross
Section
Symbol
Pitch
Width
(Lp)
Nominal
Top Width
(W)
Nominal
Height (T)
Nominal
Included
Angle (A⁰)
B 14 17 11 40
𝑆𝑖𝑛 ∝ =
𝐷𝑝 − 𝑑𝑝
2𝑐
𝑆𝑖𝑛 ∝=
250 − 125
2 × 304
= 0.205
∝ = sin−1
(0.205)
∝= 11.86°
Angle of lap on the smaller pulley
𝛽 = 180 − 2𝛼
𝛽 = 180° − (2 × 11.86)°
𝛽 = 156.28°
𝛽 = 156.28° ×
𝜋
180
𝛽 = 2.72 𝑟𝑎𝑑.
w
T
A0
Figure 8: Pulley belt arrangement

Pitch line velocity can be obtained from,
V =
𝜋 𝑥 𝑑𝑝 𝑥 𝑁
60
For smaller pulley,
V =
𝜋 𝑥 125 𝑥 719 𝑥 10−3
60
V = 4.7 m/s
Assumptions :- (i) µ = 0.35.
2.3 log (
𝑇1
𝑇2
) = µ. . cosec 𝐴
2⁄
2.3 log (
𝑇1
𝑇2
) = 0.35 x 2.72 x cosec
40
2
= 0.35 x 2.72 x
1
sin20
=
0.35 𝑥 2.72 𝑥 2.724
2.3
log (
𝑇1
𝑇2
) = 1.21
𝑇1
𝑇2
= 101.21
= 16.22
Power (P) = F x V
7.5 x 103
W = (𝑇1 - 𝑇2) x 4.7
7.5 x 103
= (16.22 𝑇2-𝑇2) x 4.7 ( 𝑇1 = 16.22 𝑇2 )
𝑇2 =
7.5 𝑥 103
15.22 𝑥 4.7
𝑇2 = 104.84 N
𝑇1 = 1700.59 N

Power per belt = (T1 − T2) 𝑉
= (T1 − T2) 𝑟𝜔
= (1700.59 – 104.84) 𝑥 (
125 𝑥 10−3
2
𝑥
2𝜋 𝑥 791
60
)
= 8261.33 W
= 8.26 kW
Number of belts =
𝐷𝑒𝑠𝑖𝑔𝑛 𝑃𝑜𝑤𝑒𝑟
𝐹𝑖𝑛𝑎𝑙 𝐵𝑒𝑙𝑡 𝑃𝑜𝑤𝑒𝑟
=
(7.5×1.4)
8.26
= 1.27  1
∴ The required no: of belts = 7 belts
Torque on Small Pulley = (T1 − T2) ×
d
2
= (1700.59 − 104.84) ×
125×10−3
2
= 93.73 Nm
Torque on Large Pulley = (T1 − T2) ×
D
2
= (1700.59 − 104.84) ×
250×10−3
2
= 199.46 Nm

MATERIALS SELECTION
 The pulleys for V-belts are made of cast iron with pressed steel in order to reduce weight.
 V-belts are made of homogeneously rubber or polymer throughout and fibers embedded in the
rubber or polymer for strength and reinforcement.
Table 1 - Pulley characteristics
Characteristics Smaller Pulley Larger Pulley
Material Cast Iron Cast Iron
Pitch Diameter (mm) 125 250
Outside Diameter (mm) 133 258
Torque (Nm) 93.73 199.46

Figure 9: Gear arrangement
T = Number of Teeth
N = Gear speed in r.p.m.
DG = Diameter of the Gear in mm: (RG = Radius of the gear)
UPWARD MOTION
The transmission ratio of Gear drive = 2: 1
𝑁1
𝑁2
=
𝐷 𝐺1
𝐷 𝐺2
2
1
=
𝐷 𝐺1
𝐷 𝐺2
𝐷 𝐺1 = 2 𝐷 𝐺2
Assumptions:-
Diameter of gear 1 (𝐷 𝐺1) = 100 mm
∴ Diameter of gear 2 ( 𝐷 𝐺2) = 200 mm

Module, 𝑚 =
𝐷
𝑇
Selecting Module from That Standard Module, m = 4.
Number of teeth on gear 1 and 2,
𝑇𝐺1 =
100
4
= 25 𝑇𝐺2 =
200
4
= 50
Assumptions:-
1. Speed of the gear 1 = 359.5 r.p.m.
Table 2 - Gear Characteristics
Gear 1 Gear 2
N (rpm) 359.5 179.75
DG (mm) 100 200
T 25 50
For the design, 20◦ stub involute system (y) (Refer. Figure 10)
y 0.133 0.151
Figure 10: No: of teeth vs. Involute factor Y

MATERIALS SELECTION
 For this design 20° Stub involute system gears are used, because it has a strong tooth to take
heavy loads. Using the table of mechanical properties of some gear steels, I chose EN8 Hardened
steel. (Refer. Figure 20)
Ultimate tensile stress (𝜎 𝑢𝑡) = 110 kg/ mm2 = 1078.75 N/ mm2
Allowable static stress ( 𝜎0 ) =
𝜎 𝑢𝑡
3
( 𝜎0 ) =
1078.75 𝑁/𝑚𝑚2
3.
( 𝜎0 ) = 359.58 N/ mm2
Strength factor of gear wheel 1
𝜎01 = 𝜎0 × y
𝜎01 = 359.58 × 0.133
𝜎01 = 47.824 N/ mm2
𝜎02 = 𝜎0 × y
𝜎02 = 359.58 × 0.151
𝜎02 = 54.296 N/ mm2
Here, the strength factor for gear wheel 1 is less than gear wheel 2. ( 𝜎o1 < 𝜎o2)
So, gear 1 is weaker.

From figure 3 we can see the shaft (S2) have a large pulley, clutch and gears 1 & 3 connected to it. I
assume the sped of the shaft is not going to change through the shaft. The large pulley is the first
mechanical component to receive motion and power. The torque on the large pulley is the biggest
on the shaft (S2). I assume the gear 1 and 3 get the same torque.
Torque on gear 1 (TG1) = 199.46 × 103 N-mm.
Circular pitch (Pc) = 𝜋 𝑚 = 𝜋 𝑥 4 𝑚𝑚
(Pc) = 12.566 mm
Tangential load (WT1) =
2𝑇
𝐷
=
2 ×199.46
100 ×10−3
(WT1) = 3989.2 N
Normal load on the tooth (WN1) =
𝑊𝑇
𝑐𝑜𝑠∅
=
3989.2 𝑁
𝑐𝑜𝑠20°
(WN1) = 4245.28 N
Let assume the normal pressure between the teeth is 40 N/mm, therefore necessary face width of
gear 1,
𝑏 =
4245.28
40
= 106.132 ≈ 106 𝑚𝑚
Pitch line velocity of gear wheel 1
𝑣 = 𝑟𝜔
=
100 × 10−3
2
×
2𝜋
60
× 359.5
= 1.882 𝑚/𝑠

Value of Deformation Factor C
According to the table for deformation factor C, the maximum value for tooth error is 0.08 mm.
Therefore our value for deformation factor. Since both gear and pinion both are made of steel,
C= 952 × 103 N/m.
WI1 =
21𝑣 (𝑏.𝐶+𝑊 𝑇1 )
21 𝑣+√ 𝑏.𝐶+𝑊 𝑇1
WI1 =
21 𝑥1.882 𝑥 (106 ×952 + 3989.2)
21 𝑥 1.882 + √(106 ×952)+3989.2
WI1 = 11408.45 N
Dynamic Tooth Load
WD = WT1 + WI1
WD = (3989.2 + 11408.45)
WD = 15397.65 N
Figure 11: Values of Deformation Factor C (kN/m) for Dynamic load

Static Tooth Load (Endurance Strength)
WS1 = 𝜎𝑒 × 𝑏 × 𝜋 × 𝑚 × 𝑦
From steel Hardness conversion table, (Refer. )
40 HRC = 373 B.H.N
𝜎𝑒 = 1.75 × 373 MPa
𝜎𝑒 = 652.75 N/mm2
WS1 = 652.75
𝑁
𝑚𝑚2
× 106 𝑚𝑚 × 𝜋 × 4 𝑚𝑚 × 0.133
WS1 = 36809.88 N
For Safety,
Ws1 ≥ WD
The condition satisfies for the obtained values
Wear Tooth Load
Ww = 𝐷 𝑝 × 𝑏 × 𝑄 × 𝑘
Where,
Ww = Maximum or limiting load for wear in N
DP = Pitch circle diameter of the pinion in mm
b = Face width of the pinion in mm
Q = Ratio factor
K = Load-stress factor (also known as material combination factor) in N/mm2.

1. 𝑄 =
2 ×𝑉.𝑅
𝑉.𝑅+1
V.R =
𝑇 𝐺1
𝑇 𝐺2
=
50
25
= 2
𝑄 =
2 × 2
2 + 1
=
4
3
= 1.333
2. 𝐾 =
(𝜎 𝑒𝑠 )2 ×𝑠𝑖𝑛∅
1.4
{
1
𝐸 𝑝
+
1
𝐸 𝑔
}
Where,
𝜎𝑒𝑠 = Surface endurance limit in MPa or N/mm2
φ = Pressure angle
EP = Young's modulus for the material of the pinion in N/mm2
EG = Young's modulus for the material of the gear in N/mm2.
Both gear and pinion are made of the same material. So, both young modulus values are equal.
Young modulus for EN8 steel = 185 × 103 N/mm2
The surface endurance limit for steel
𝜎𝑒𝑠 = (2.8 × 373 − 70 )𝑁/𝑚𝑚2
σes = 974.4 𝑁/𝑚𝑚2
𝐾 =
(974.4)2 ×𝑠𝑖𝑛20
1.4
{
1
180 ×103
+
1
180 ×103
}
K = 2.577 N/ mm2
W
w = 100 𝑚𝑚 × 106 𝑚𝑚 × 1.333 × 2.577 N/𝑚𝑚2
W
w = 36412.49 𝑁
For Safety,
Ww ≥ WD

Figure 12: Reverse Gear arrangement
DOWNWARD MOTION
Velocity of Cabin in Downwards = 0.65ms-1
Angular velocity of the drum = 49.656 rpm
Angular velocity of the Larger Sprocket = 49.656 rpm
Chain Drive Transmission Ratio = 4: 1
Angular velocity of the Smaller Sprocket = 49.65× 4
= 198.6 r.p.m.
When cabin moves upwards Gear 1 & 2 engage. When the cabin moves downwards Gear 3, 4, & 5
are engaged. So, the Gear 5, gear 2 and small sprocket are all on the shaft 4 (S4). When the lift has
upward motion shaft 4 rotates with the speed of 179.75 rpm but during the downward motion, that
shaft rotates with 198.6 rpm.
Gear 5 rpm = Smaller Sprocket rpm
NG5 = 198.6 r.p.m.

Figure 13: Gear arrangement
Speed Ratio =
𝑁1
𝑁2
=
𝑇2
𝑇1
=
𝜔1
𝜔2
𝑁 𝐺4
𝑁 𝐺5
=
359.5
198.6
= 1.81 ≈ 2
N3
N5
=
DG5
DG3
DG5
DG3
= 2
DG5 = 2 DG3
Assuming Diameter of Gear 3 = 60 mm
∴ D3= 60mm
∴ D5 = 120mm
Gear 3
Gear 4
Gear 5
The Distance between the Two Shafts
D1+D2
2
=
100 + 200
2
= 150 mm

150 𝑚𝑚 =
𝐷 𝐺3
2
+ 𝐷 𝐺4 +
𝐷 𝐺5
2
𝐷 𝐺4 = 150 − (30 + 60)
𝐷 𝐺4 = 60 𝑚𝑚
Table 3 - Gear Characteristics
Module, 𝑚 =
𝐷
𝑇
Selecting Module from That Standard Module, m = 4.
Number of teeth on gear 1 and 2,
𝑇𝐺3 =
60
4
= 15 𝑇𝐺4 =
60
4
=15 𝑇𝐺5 =
120
4
= 30
MATERIALS SELECTION
 For this design 20° Stub involute system gears are used, because it has a strong tooth to take
heavy loads. Using the table of mechanical properties of some gear steels, I chose EN8 Hardened
steel.
Ultimate tensile stress (𝜎 𝑢𝑡) = 110 kg/ mm2 = 1078.75 N/ mm2
Gear 3 Gear 4 Gear 5
N (rpm) 359.5 359.5 198.6
DG (mm) 60 60 120
T 15 15 30
For the design, 20◦ stub involute system (y)
(Refer. Figure 10)
y 0.111 0.111 0.139

Allowable static stress ( 𝜎0 ) =
𝜎 𝑢𝑡
3
( 𝜎0 ) =
1078.75 𝑁/𝑚𝑚2
3.
( 𝜎0 ) = 359.58 N/ mm2
Strength factor of gear wheel 3 & 4
𝜎03 = 𝜎0 × y
𝜎03 = 359.58 × 0.111
𝜎03 = 39.91 N/ mm2
𝜎05 = 𝜎0 × y
𝜎05 = 359.58 × 0.139
𝜎05 = 49.98 N/ mm2
Here, the strength factor for gear wheel 3 is less than gear wheel 5. So, gear wheels 3 & 4 are
weaker. Out of those two gears, number 4 gear is the pinion. I assume the torque received on gear
3 completely transferred onto gear 4.
Circular pitch (Pc) = 𝜋 𝑚
(Pc) = 12.566 mm
Tangential load, WT4 =
2𝑇
𝐷
=
2 ×199.46
60 ×10−3
WT4 = 6648.67 N
Normal load on the tooth, WN1 =
𝑊 𝑇
𝑐𝑜𝑠∅
=
6648.67 𝑁
𝑐𝑜𝑠20°
WN1 = 7075.37 N

Let assume the normal pressure between the tooth is 40 N/mm, therefore necessary face width of
gear 1,
𝑏 =
7075.37
40
= 176.88 𝑚𝑚 ≈ 177 𝑚𝑚
Pitch line velocity of gear wheel 1
𝑣 = 𝑟𝜔
=
60 × 10−3
2
×
2𝜋
60
× 359.5
= 1.129 𝑚/𝑠
Value of Deformation Factor C
According to the table for deformation factor C, the maximum value for tooth error is 0.08 mm.
Therefore our value for deformation factor. Since both gear and pinion both are made of steel.
(Refer. Figure 11)
C= 952 × 103 N/m.
WI4 =
21𝑣 (𝑏.𝐶+𝑊𝑇4 )
21 𝑣+√𝑏.𝐶+𝑊𝑇4
WI4 =
21 ×1.129 𝑥 (177 ×952 + 6648.67)
21 ×1.129 + √(177×952)+6648.67
WI4 = 9390.53 N
Dynamic Tooth Load
WD4 = WT 4+ WI4
WD4 = (6648.67 + 9350.53)
WD4 = 15999.2 N

Static Tooth Load (Endurance Strength)
WS4 = 𝜎𝑒 × 𝑏 × 𝜋 × 𝑚 × 𝑦
From steel Hardness conversion table,
40 HRC = 373 B.H.N
𝜎𝑒 = 1.75 × 373 MPa
𝜎𝑒 = 652.75 N/mm2
WS4 = 652.75
𝑁
𝑚𝑚2 × 177 𝑚𝑚 × 𝜋 × 4 𝑚𝑚 × 0.111
WS4 = 161158.41 N
For Safety,
Ws4 ≥ WD4
Wear Tooth Load
Ww = 𝐷 𝑝 × 𝑏 × 𝑄 × 𝑘
Where,
Ww = Maximum or limiting load for wear in N
DP = Pitch circle diameter of the pinion in mm
b = Face width of the pinion in mm
Q = Ratio factor
K = Load-stress factor (also known as material combination factor) in N/mm2
1. 𝑄 =
2 ×𝑉.𝑅
𝑉.𝑅+1
V.R =
𝑇 𝐺1
𝑇 𝐺2
=
30
15
= 2
𝑄 =
2 × 2
2 + 1
=
4
3
= 1.333

2. 𝐾 =
(𝜎 𝑒𝑠 )2 ×𝑠𝑖𝑛∅
1.4
{
1
𝐸 𝑝
+
1
𝐸 𝑔
}
Where,
𝜎𝑒𝑠 = Surface endurance limit in MPa or N/mm2
φ = Pressure angle
EP = Young's modulus for the material of the pinion in N/mm2
EG = Young's modulus for the material of the gear in N/mm2.
Both gear and pinion are made of the same material. So, both young modulus values are equal.
Young modulus for EN8 steel = 185 × 103 N/mm2
The surface endurance limit for steel
𝜎𝑒𝑠 = (2.8 × 373 − 70 )𝑁/𝑚𝑚2
σes = 974.4 𝑁/𝑚𝑚2
𝐾 =
(974.4)2 ×𝑠𝑖𝑛20
1.4
{
1
180 ×103
+
1
180 ×103
}
K = 2.577 N/ mm2
W
w4 = 60 𝑚𝑚 × 177 𝑚𝑚 × 1.333 × 2.577 N/𝑚𝑚2
W
w4 = 36481.197 𝑁
For Safety,
Ww4 ≥ WD4

Two gear wheels are designed, which are spur gears and the module for all the gears are same.
Module = 4 mm.
From the Machine Design – R. S. Khurmi -Table 28.1 (page No: 1032)
Table 4 – Particulars for 20degree stub involute system
Table 5 – Gear Particulars
Serial No Particulars 20° stub involute system
1) Addendum 3.2 mm
2) Deddendum 4 mm
3) Working Depth 6.4 mm
4) Minimum total depth 7.2 mm
5) Total thickness 6.62832 mm
6) Minimum clearance 0.8 mm
7) Fillet radius at root 1.6 mm
Figure 14: Terms used in a gear

Assumptions:-
Velocity ratio of the driver = 4:1
Hence velocity of small gear Sprocket (Ns) = 198.6 r.p.m
Velocity of large gear Sprocket (NL) = 49.656 r.p.m
According to the table of characteristics of roller chains (Refer. Figure 21)
1. Chain number = 20 B
2. Pitch diameter(P) = 31.75 mm
3. Roller diameter(d1) = 19.05 mm
4. Width between inner plates(b1) = 19.56 mm
5. Simple Breaking load = 64.5 x 103
N
i. Number of teeth on the smaller sprocket, (Ts) = 19 (Refer. Figure 22)
ii. Number of teeth of the large sprocket (TL) = 25 x (
198.6
49.656
) = 75.99 ≈ 76
iii. Design power = Rated power × Service factor.
Service factor (Ks) is the product of various factors k1, k2, and k3.
{From the Machine Design – R. S. Khurmi -Table 28.1 (page No: 1032)}
Load factor (K1) for variable load with heavy shock = 1.5
Lubricant factor (K2) for continuous lubrication = 0.8
Rating factor (K3), for over 16 hours per day = 1.5
Service factor = k1 × k2 × k3
= (1.5) × (0.8) × (1.5)
= 1.8
Design power = 1.8 𝑥 398.92 𝑁𝑚 𝑥 (
198.6 𝑥 2 𝑥 𝜋
60
) w
= 14933.66 W
= 14.94 kW
We find that small gear sprocket speed of 200 r.p.m. the power transmitted for chain No.20B is 16.2
kW strand.

iv. Pitch circle diameter of the smaller sprocket,
ds = P x cosec (
180
𝑇𝑠
)
= 31.75 x cosec (
180
19
)
= 31.75 x 6 mm
= 190.5 mm
Pitch line velocity of smaller sprocket,
Vs =
𝜋 𝑥 𝑑 𝑠 𝑥 𝑁𝑠
60
=
π x 190.5 x 10−3 x 198.6
60
= 1.98 𝑚𝑠−1
Pitch circle diameter of larger sprocket,
dl = P x cosec (
180
𝑇 𝐿
)
= 31.75 x cosec (
180
76
)
= 31.75 x 24 mm
= 762 mm
Load on chain =
8296.47
1.98
= 4190.17 N
∴ Factor of safety =
𝐵𝑟𝑒𝑎𝑘𝑖𝑛𝑔 𝐿𝑜𝑎𝑑
Load on chain
=
64.5 𝑥 103
4190.17
= 15.4
The value is more than the given value on the table, which is equal to 8.55. (Refer. Figure 23)
The minimum center distance between the smaller and larger sprockets should be 30 to 50 times the
pitch. Let us take it like 40 times the pitch.

∴ Center distance between the sprockets,
= 40 x P = 40 × 31.75 = 1270 mm
In order to accommodate initial sag in the chain, the value of center distance is reduced by 2 to 5 mm.
∴ Correct centre distance (X) = 1270 – 4 = 1266 mm
The number of chain links,
𝐾 =
𝑇𝑠 + 𝑇𝐿
2
+
2𝑥
𝑃
+ (
𝑇𝑠 − 𝑇𝐿
2𝜋
)2
×
𝑃
𝑋
𝐾 =
19 + 76
2
+
2 × 1266
31.75
+ {(
76 − 19
2𝜋
)2
×
31.75
1266
}
𝐾 = 129
Length of the chain
L = K.P = 129 x 31.75 = 4095.75 mm

1. Bearing Couple 1
a) Vertical load diagram.
The drive (Small) pulley made out of cast iron. So,
weight of the pulley = 𝑀𝑠𝑝 x g
= 𝑊𝑠𝑝
Density of the pulley = 𝜌 𝑝
= 7850 kg/𝑚3
Total length of the pulley (L) = 60 mm
= 0.06m

Hence,
𝜌 𝑝 =
𝑀𝑠𝑝
𝑉𝑠𝑝
= 𝑀𝑠𝑝 x g
= 𝜌 𝑝 x 𝑉𝑠 𝑝 x g
𝑊𝑠𝑝 = 𝐴 𝑠𝑝 x L x 𝜌 𝑝 x g
=
𝜋(𝑑 𝑝)
2
4
x L x 𝜌 𝑝 x g
=
𝜋∗ (125∗10−3)
2
4
x 0.06 x 7850 x 9.81
= 7.7 N
Applying Moment equation to the vertical plane,
A
(7.7 x 0.07) – (RV1B x 0.14) = 0
RV1B =
7.7 𝑥 0.07
0.14
RV1B = 3.85 N
First of all, considering the vertical loading at C. Let RV1A and RV1B be the reactions of A and
B respectively.
Applying Newton’s second low of vertical plane,
RV1A + RV1B = 7.7 N
RV1A = (7.7 – 3.85)
RV1A = 3.85 N

Forces acting on the bearing A and B,
Bearing 1A Bearing 1B
RV1A = 3.85 N RV1B = 3.85 N
Both bearings A and B have the same reactions forces on the screw in both directions.
Assumptions:
1) Shaft/ Screw rotating speed (N) = 719 rpm
2) Working hours per day = 24 hrs.
3) Average life the bearing = 5 years
The life of the bearing of hours (LH) = 365 x 24 x 5 = 43800 hrs.
L = 60 x N x LH
L = 60 x 719 x 43800 = 1 889 532 000
L = 1889532 𝑥 10 3
rev
The basic dynamic equivalent radial load can be obtained by using the life rating equation,
Where
W = dynamic equivalent radial factor.
V = a rotation factor (1 for all type of bearings when the inner race is rotating).
WR = Radial load = 0 N
WA = Axial load = 7.7 N
For basic static radial load factor (W0) from life rating equation,

X0 and Y0 can be taken as (Refer. Figure 25)
X0 = 0.60
Y0 = 0.50
W0 = ( 0.60 𝑥 0) + (0.50 𝑥 7.7)
W0 = 3.85 N
Basic static load rating (C0),
Where S0 is service factor
I am assuming, my structure to have uniformed steady load. But, sometimes it can get some shocks.
S0 = 2.5 (Refer. Figure 24)
So, the basic static load rating,
C0 = S0 W0
C0 = 2.5 x 3.85 N
C0 = 9.625 N
𝑊 𝐴
𝐶 𝑂
=
7.7
9.625
= 0.8 (Which is greater than e = 0.44, (Refer. Figure 26)
Therefore X = 0.56 and Y = 1
From life rating equation,
W = X. V. WR + Y. WA
W = 0.56 x 1 x 0 + 1 x 7.7
W = 7.7 N

Dynamic Load rating (C),
𝐶 = 𝑊. √(
𝐿
106
)
𝐾
Where, K = 3:
𝐶 = (7.7 𝑁). √(
1889532 𝑥 103
106
)
3
𝐶 = 205𝑁
𝐶 ≈ 0.2 𝑘𝑁
Considering the shaft diameter, principal dimensions, and basic load ratings, I choose deep groove ball
bearing. The bearing number is 61803. (Refer. Figure 27)

2. Bearing Couple 2 & 3
b) Horizontal load diagram

Initial Data:
1) The driven (Large) pulley also made out of cast iron. So,
Weight of the pulley = 𝑀𝐿𝑝 x g
= 𝑊𝐿𝑝
Density of the pulley = 𝜌 𝑝
= 7850 kg/𝑚3
Total length of the pulley (L) = 120 mm
= 0.12 m
Hence,
𝜌 𝑝 =
𝑀 𝐿𝑝
𝑉 𝐿𝑝
= 𝑀𝐿𝑝 x g
= 𝜌 𝑝 x 𝑉𝐿𝑝 x g
𝑊𝐿𝑝 = 𝐴 𝐿𝑝 x L x 𝜌 𝑝 x g
=
𝜋(𝐷 𝑝)
2
4
x L x 𝜌 𝑝 x g
=
𝜋∗ (250∗10−3)
2
4
x 0.12 x 7850 x 9.81
= 453.62 N
2) Mass of the Clutch = 4 kg (Assumption)
Hence weight of the clutch (𝑤𝑐) = 4 x 9.81
= 39.42 N
Width of the Clutch = 100 mm (Assumption)
3) Density of the Steel 𝜌𝑠 = 7700 kg/𝑚3
I assume the gear 1 and 3 get the same torque. (Refer. Page )
(Pc) = 12.566 mm

2𝑇
𝐷
=
2 ×199.46
60 ×10−3
(WT3) = 6648.67 N
𝑊𝑇
𝑐𝑜𝑠∅
=
6648.67 𝑁
𝑐𝑜𝑠20°
(WN3) = 7075.37 N
gear 3,
𝑏 =
7075.37
40
= 176.88 ≈ 177 𝑚𝑚
Combined weight of gear wheels 1 & 3 = 𝑀13 x g
𝑊1 & 3 = (
𝜋
4
x (𝐷 𝐺1)2
x 𝑏 𝐺1 x 𝜌𝑠 x g) + (
𝜋
4
x (𝐷 𝐺3)2
x 𝑏 𝐺3 x 𝜌 𝑝 x g )
=
𝜋
4
𝑥 𝜌𝑠 x g { ⌊(100 𝑥 10−3)2
𝑥 106 𝑥 10−3
] + [ (60 𝑥 10−3)2
𝑥 177 𝑥 10−3⌋ }
=
𝜋
4
x 7700 x 9.81 (1.972 x 10−3
)
= 117 N
The tangential force acting on the gear 1 and 3,
𝐹1 =
𝑀𝑇 𝐺1
𝑅 𝐺1
𝐹3 =
𝑀𝑇 𝐺3
𝑅 𝐺3
𝐹1 =
199.46 × 103
50
𝐹3 =
199.46 × 103
30
𝐹1 = 3989.2 𝑁 𝐹3 = 6648.67 𝑁
Taking larger force acting on the gear 3 and the normal load acting on the tooth of gear 3,
WD3 =
6648.67 𝑁
cos 20°
= 7075.36 𝑁

Horizontal component of WG3, Vertical component of WG3,
𝑊𝐺3𝐻 = 7075.36 × sin 20° 𝑊𝐺3𝑉 = 7075.36 × cos 20°
𝑊𝐺3𝐻 = 2419.92 𝑁 𝑊𝐺3𝑉 = 6648.67 𝑁
First of all, considering the vertical loading at C, E, and D. Let RV3A and RV3B be the reactions of A and
B respectively.
A
[(453.62 x 0.1) + (39.42 x 0.25) + (6765.67 x 0.4815)] – (RV3B x 0.663) = 0
RV3B =
(3312.89)
0.663
RV3B = 4996.82 N
Applying Newton’s second law of vertical plane
RV2A + RV3B = 7258.71
RV2A = 7258.71 – 4996.82
RV2A = 2261.89 N
Now, considering the horizontal loading at D. Let RH2A and RH3B be the reactions of A and C
respectively.
Applying Moment equation to the Horizontal plane,
A
(2419.92 x 0.4815) – (RH3B x 0.663) = 0
RH3B =
1165.19
0.663
RH3B = 1757.45 N

Applying Newton’s second low of horizontal plane
RH2A + RH3B = 2419.92
RH2A = [2419.92] – (1757.45)
RH2A = 662.47 N
RH2A = 662.47 N RH3B = 1757.45 N
RV2A = 2261.89 N RV3B = 4996.82 N
Both bearing 3B have large reactions forces on shaft 2 in both directions.
For Bearing 3B,
Resultant force (fr) = √(4996.82)2 + (1757.45)2
Resultant force (fr) = 5296.87 N
Assumptions:
1) Shaft/ Screw rotating speed (N) = 359.5 rpm
4996.82 N
1757.45 N
F r

L = 60 x N x LH
L = 60 x 359.5 x 43800
L = 944766 𝑥 10 3
rev
Where
WR = radial load = 2419.92 N
WA = axial load = 7258.71 N
X0 and Y0 can be taken as,
X0 = 0.60
Y0 = 0.50
W0 = ( 0.60 𝑥 2419.92) + (0.50 𝑥 7258.71)
W0 = 5081.3 N
S0 = 2.5

C0 = S0 W0
C0 = 2.5 x 5081.3 N
C0 = 12703.25 N
𝑊 𝐴
𝐶 𝑂
=
7258.71
12703.25
= 0.571 (Which is greater than e = 0.44)
W = 0.56 x 1 x 2419.92 + 1 x 7258.71
W = 8613.86 N
𝐶 = 𝑊. √(
𝐿
106
)
𝐾
Where, K = 3:
𝐶 = (8613.86 𝑁). √(
944766 𝑥 103
106
)
3
𝐶 = 84.522 𝑘𝑁
𝐶 ≈ 84.5 𝑘𝑁
Considering the shaft diameter, principal dimensions and basic load ratings, I choose Single row
cylindrical roller bearing. The bearing number is N 308 ECP. (Refer. Figure 28)

3. Bearing Couple 4
Density of the Steel 𝜌𝑠 = 7700 kg/𝑚3
I assume the gear 3 and 4 get the same torque. (Refer. Page )
(Pc) = 12.566 mm

2𝑇
𝐷
=
2 ×199.46
60 ×10−3
(WT4) = 6648.67 N
𝑊𝑇
𝑐𝑜𝑠∅
=
6648.67 𝑁
𝑐𝑜𝑠20°
(WN4) = 7075.37 N
gear 4,
𝑏 =
7075.37
40
= 176.88 ≈ 177 𝑚𝑚
Weight of gear 4 = 𝑀4 x g
𝑊4 = (
𝜋
4
x (𝐷 𝐺4)2
x 𝑏 𝐺4 x 𝜌𝑠 x g)
=
𝜋
4
𝑥 𝜌𝑠 x g ⌊ (60 𝑥 10−3)2
𝑥 177 𝑥 10−3⌋
=
𝜋
4
x 7700 x 9.81 (6.372x 10−4
)
= 37.8 N
𝐹4 =
𝑀𝑇 𝐺4
𝑅 𝐺4
𝐹4 =
199.46 × 103
30
𝐹4 = 6648.67 𝑁
The normal load acting on the tooth of gear 4,
WD3 =
6648.67 𝑁
cos 20°
= 7075.36 𝑁

𝑊𝐺4𝐻 = 7075.36 × sin 20° 𝑊𝐺4𝑉 = 7075.36 × cos 20°
𝑊𝐺4𝐻 = 2419.92 𝑁 𝑊𝐺4𝑉 = 6648.67 𝑁
First of all, considering the vertical loading at C. Let RV4A and RV4B be the reactions of A and
B respectively.
A
(6686.47 x 0.1285) – (RV4B x 0.257) = 0
RV4B =
(859.21)
0.257
RV4B = 3343.23 N
Applying Newton’s second low of vertical plane
RV4A + RV4B = 6686.47 N
RV4A = 6686.47 – 3343.23
RV4A = 3343.24 N
Now, considering the horizontal loading at C. Let RH4A and RH4B be the reactions of A and B
respectively.
A
(2419.92 x 0.1285) – (RH4B x 0.257) = 0
RH4B =
310.96
0.257
RH4B = 1209.96 N

RH4A + RH4B = 2419.92
RH4A = [2419.92] – (1209.96)
RH4A = 1209.95 N
RH4A = 1209.95 N RH4B = 1209.96 N
RV4A = 3343.24 N RV4B = 3343.23 N
Both bearings 4A and 4B have the same reactions forces on the screw in both directions.
For Bearing 4A,
Assumptions:
3343.24 N
1209.95 N
F r

L = 60 x N x LH
L = 60 x 359.5 x 43800
L = 944766 𝑥 10 3
rev
Where
X0 = 0.60
Y0 = 0.50
W0 = ( 0.60 𝑥 2419.92) + (0.50 𝑥 6686.47)
W0 = 4795.19 N
S0 = 2.5

C0 = S0 W0
C0 = 2.5 x 4795.19 N
C0 = 11987.975 N
𝑊 𝐴
𝐶 𝑂
=
6686.47
11987.975
W = 0.56 x 1 x 2419.92 + 1 x 6686.47
W = 8041.625 N
𝐶 = 𝑊. √(
𝐿
106
)
𝐾
Where, K = 3:
𝐶 = (8041.625 𝑁). √(
944766 𝑥 103
106
)
3
𝐶 = 78907.55 𝑁
𝐶 ≈ 79 𝑘𝑁
cylindrical roller bearing. The bearing number is NU 2306 ECP. (Refer. Figure 29)

Assumptions:-
I assume there is no horizontal load acting on the chain drive, small sprocket and large sprocket.
Chain drive small, large sprocket material is AISI type of 304 stainless steel.
Density of 304 stainless steel 𝜌𝑠𝑡 = 8000
𝑘𝑔
𝑚3
Speed Ratio =
𝑁 𝐺4
𝑁 𝐺5
=
𝑇 𝐺5
𝑇 𝐺4
= 2
Torque on gear 5 (TG5) = 2 x 199.46 × 103 N-mm = 398.92 × 103 N-mm
Weight of the gear 5 (WG5) = 𝑀 𝐺5 x g
= (
𝜋
4
x (𝑑 𝐺5)2
=
𝜋
4
× 7850 x g 𝑥 ⌊177 𝑥 10−3
𝑥 (120 𝑥 10−3
)2⌋
= 154.16 N
Weight of the gear 2 (WG2) = 𝑀 𝐺2 x g
= (
𝜋
4
x (𝑑 𝐺2)2
=
𝜋
4
× 7850 x g 𝑥 ⌊106 𝑥 10−3
𝑥 (200 𝑥 10−3
)2⌋
= 241.93 N
Weight of the small gear sprocket = 𝑀𝑠𝑠 x g
𝑊𝑠𝑠 = (
𝜋
4
x (𝑑 𝑠)2
x 𝑏𝑠𝑠 x 𝜌𝑠𝑡 x g)
=
𝜋
4
𝑥 8000 x g ⌊(19.56 × 10−3
× (190.5 × 10−3 )2⌋
= 43.75 N

1) Initial Data: Gear 5
2𝑇
𝐷
=
2 ×398.92
120 ×10−3
(WT5) = 6648.67 N
𝑊 𝑇
𝑐𝑜𝑠∅
=
6648.67 𝑁
𝑐𝑜𝑠20°
(WN4) = 7075.37 N
gear 5,
𝑏 =
7075.37
40
= 176.88 ≈ 177 𝑚𝑚
𝐹5 =
𝑇 𝐺5
𝑅 𝐺5
𝐹5 =
398.92 × 103
60
𝐹5 = 6648.67 𝑁
WD5 =
6648.67 𝑁
cos 20°
= 7075.36 𝑁
𝑊𝐺5𝐻 = 7075.36 × sin 20° 𝑊𝐺5𝑉 = 7075.36 × cos 20°
𝑊𝐺5𝐻 = 2419.92 𝑁 𝑊𝐺5𝑉 = 6648.67 𝑁

2) Initial Data: Gear 2
2𝑇
𝐷
=
2 ×398.92
200 ×10−3
(WT2) = 3989.2 N
𝑊 𝑇
𝑐𝑜𝑠∅
=
3989.2 𝑁
𝑐𝑜𝑠20°
(WN2) = 4245.28 N
gear 2,
𝑏 =
4245.28
40
= 106.132 ≈ 106 𝑚𝑚
𝐹2 =
𝑇 𝐺2
𝑅 𝐺2
𝐹2 =
398.92 × 103
100
𝐹2 = 3989.2 𝑁
WD2 =
3989.2 𝑁
cos20°
= 4245.28 𝑁
𝑊𝐺2𝐻 = 4245.28 × sin 20° 𝑊𝐺2𝑉 = 4245.28 × cos 20°
𝑊𝐺2𝐻 = 1451.97 𝑁 𝑊𝐺2𝑉 = 3989.2 𝑁

First of all, considering the vertical loading at C, E, and D. Let RV6A and RV5B be the reactions of A and
B respectively.
A
[(43.75 x 0.04978) + (4231.13 x 0.15) + (6802.83 x 0.334)] – (RV5B x 0.463) = 0
RV5B =
(2908.99)
0.463
RV5B = 6282.91 N
Applying Newton’s second low of vertical plane
RV6A + RV5B = 11077.71
RV6A = 11077.71 – 6282.91
RV6A = 4794.8 N
Now, considering the horizontal loading at E and D. Let RH6A and RH5B be the reactions of A and B
respectively.
A
[(1451.97 x 0.15) + (2419.92 x 0.334)] – (RH5B x 0.463) = 0
RH5B =
1026
0.463
RH5B = 2215.98 N
RH6A + RH5B = 3871.89
RH6A = [3871.89] – (2215.98)
RH6A = 1655.91 N

RH6A = 1655.91 N RH5B = 2215.98 N
RV6A = 4794.8 N RV5B = 6282.91 N
Both bearing 5B have large reactions forces on shaft 4 in both directions.
For Bearing 5B,
Assumptions:
L = 60 x N x LH
L = 60 x 198.6 x 43800
L = 5219208 𝑥 10 2
rev
6282.91 N
2215.98 N
F r

Where
X0 = 0.60
Y0 = 0.50
W0 = (0.60 𝑥 3871.89) + (0.50 𝑥 11077.71)
W0 = 7861.99 N
S0 = 2.5
C0 = S0 W0
C0 = 2.5 x 7861.99 N
C0 = 19654.97 N

𝑊 𝐴
𝐶 𝑂
=
11077.71
19654.97
W = 0.56 x 1 x 3871.89 + 1 x 11077.71
W = 13245.97 N
𝐶 = 𝑊. √(
𝐿
106
)
𝐾
Where, K = 3:
𝐶 = (13245.97 𝑁). √(
5219208 𝑥 102
106
)
3
𝐶 = 106.648 𝑘𝑁
𝐶 ≈ 107 𝑘𝑁
cylindrical roller bearing. The bearing number is NU 2308 ECP. (Refer. Figure 28)

I assumed there is no horizontal forces/ load in the chain drive. So, shaft 5 does not have any horizontal
forces. Cable drum also made of 304 stainless steel.
Density of 304 stainless steel 𝜌𝑠𝑡 = 8000
𝑘𝑔
𝑚3
1. Weight of the cabin = 100 kg × 9.81 = 981 N
2. Weight of the working load = 500 kg × 9.81 = 4905 N
3. Diameter of the cable drum = 0.25 m =250 mm

Assumptions:-
1. Length of the cable drum =450 mm
2. Rope length = 25m (1 kg/m) = (25 × 9.81) = 245.25
Weight of the cable drum = 𝑀 𝐶𝐷 x g
𝑊𝐶𝐷 = (
𝜋
4
x (𝑑 𝐶𝐷)2
x 𝑏 𝐶𝐷 x 𝜌𝑠𝑡 x g)
=
𝜋
4
𝑥 8000 x g⌊(450 × 10−3
× (250 × 10−3 )2⌋
`= 1733.57 N
Weight of the large gear sprocket = 𝑀𝐿𝑆 x g
𝑊𝐿𝑆 = (
𝜋
4
x (𝑑 𝐿)2
x 𝑏𝑙𝑠 x 𝜌𝑠𝑡 x g)
=
𝜋
4
𝑥 𝜌𝑠𝑡x g ⌊(19.56 × 10−3
× (762 × 10−3 )2⌋
= 700 N
∴ Total load of the cabin and cable drum = (981 + 4905 + 245.25) N = 6191.25 N
Safety factor = 10 %
New total load (with safety factor) = (6191.25 + 619.125) N = 6810.375 N
Torque on Shaft 5,
Speed Ratio =
𝑁 𝑆𝑃
𝑁 𝐿𝑆
=
𝑇 𝐿𝑆
𝑇 𝑆𝑃
= 4
𝑇𝐿𝑆 = 4 𝑥 398.92 × 103
N − mm = 1595.68 × 103
N − mm

First of all, considering the vertical loading at C and D. Let RV7A and RV8B be the reactions of A and
B respectively.
A
{(700 x 0.04978) + (8543.95 x 0.32456)} – (RV8B x 0.59) = 0
RV8B =
3092.27
0.59
RV8B = 5241.14 N
Applying Newton’s second law of vertical plane,
RV7A + RV8B = 9243.95
RV7A = (9243.95 – 5241.14)
RV7A = 4002.81 N
Assumptions:
L = 60 x N x LH
L = 60 x 49.656 x 43800 = 1 889 532 000
L = 130495968 rev

Where
WR = Radial load = 0 N
WA = Axial load = 9243.95 N
X0 and Y0 can be taken as
X0 = 0.60
Y0 = 0.50
W0 = ( 0.60 𝑥 0) + (0.50 𝑥 9243.95)
W0 = 4621.975 N
S0 = 2.5
C0 = S0 W0
C0 = 2.5 x 4621.975 N
C0 = 11554.94 N

𝑊 𝐴
𝐶 𝑂
=
9243.95
11554.94
W = 0.56 x 1 x 0 + 1 x 9243.95
W = 9243.95 N
𝐶 = 𝑊. √(
𝐿
106
)
𝐾
Where, K = 3:
𝐶 = (9243.95). √(
130495968
106
)
3
𝐶 = 46887.45 𝑁
𝐶 ≈ 47 𝑘𝑁
Considering the shaft diameter, principal dimensions, and basic load ratings, I choose Single row deep
groove ball bearings. The bearing number is 6310 M. (Refer. Figure 30)

Calculation of shaft diameters
Power transmission shafts are subjected to the following stresses.
1. Torsion stress due to the torque.
2. Bending stress due to bending moment.
3. Compressive or tensile stresses due to axial force.
4. Combination of all three above type.
A shaft having a diameter d when subjected to combine bending & torsion the bending & torsion
stresses at any point of the shaft are,
σ = 16/ (πd3) × { KbM + √ [(KbM )2 + ( Kt T )2]} (1)
τ = 16/ (πd3) × √ [(KbM) 2 + (Kt T )2] (2)
Where, σ = normal stress
τ = shear stress
The above equations are called ASME code equations & where,
Kt = shock & fatigue factor in torsion.
Kb = shock & fatigue factor for bending.
d = diameter of the shaft max.
M = bending moment of the shaft.
T = torque on the shaft.
Assume the shear stress τ = 0.3 × yield stress
σ = 0.6 × yield stress
For mild steel,
Kb = 1.5
Kt =1.2
Yield stress for mild steel = 490 MN/m2
Refer Figure 31

1) Shaft 1 (S1)
b) Vertical bending moment diagram
c) Resultant bending moment diagram

Since the small pulley mounted on the middle of the shaft, therefore maximum bending moment at
the center of the pulley,
Bending moment at A and B are equal to zero.
MAV1 = MBV1 = 0
Vertical bending moment at C, 𝑀 𝑉1 =
7.7 ×140 × 10−3
4
= 0.2695 𝑁 − 𝑚.
Horizontal bending moment at C, 𝑀 𝐻1 = 0
Resultant bending moment at C,
Mc1 = √(0.2695)2 + (0) = 0.2695 𝑁 − 𝑚.
∴ Maximum bending moment diagram at C, M1 =Mc = 0.2695 N-m.
Let d1 = Diameter of the shaft 1
Twisting moment (Te),
𝑇𝑒 = √(𝐾𝑏 × 𝑀1)2 + (𝐾𝑡 × 𝑇)2
𝑇𝑒 = √(1.5 × 0.2695)2 + (1.2 × 93.73)2 N-m
𝑇𝑒 = 112.47 𝑁 − 𝑚 = 112.47 × 103
𝑁 − 𝑚𝑚
Twisting moment also equals to,
𝑇𝑒 =
𝜋
16
× 𝜏 × 𝑑1
3
𝑑1
3
=
112.47 ×103 ×16
𝜋 ×147
= 3896.635 𝑚𝑚3
𝑑1 = 15.74 mm
Bending moment (Me)
(𝑀𝑒) =
1
2
[(𝑀𝑐1 × 𝐾 𝐵 ) + 𝑇𝑒 ]
𝑀𝑒 =
1
2
[(1.5 × 0.2695) + 112.47] 𝑁 − 𝑚
𝑀𝑒 = 56.44 𝑁 − 𝑚 = 56. .44 × 103
𝑁 − 𝑚𝑚

Bending moments also equals to
𝑀𝑒 =
𝜋
32
× 𝜎 × 𝑑1
3
𝑑1
3
=
56..44×103 ×32
𝜋 ×294
= 1955.42 𝑚𝑚3
𝑑1 = 12.5 mm
∴ Taking larger of the two values, d1 =15.74 mm ≈ 17 mm
Considering both values and availability, I choose the diameter of the shaft to be 17 mm.

b) Horizontal load diagram.
c) Vertical bending moment diagram
d) Horizontal bending moment diagram
e) Resultant bending moment diagram
Considering the Vertical loading of the shaft, bending moment at A and B are equal to zero.
MAV2 = MBV2 = 0
Bending Moment at C, MCV2 = 2261.89 × 100 × 10−3
= 226.1 N-m
Bending Moment at E, MEV2 = ([2261.89 × 0.25] − [453.62 𝑥 0.15] = 497.2 N-m
Bending Moment at D, MDV2 = 4996.82 × 181.5 × 10−3
= 906.93 N-m
Considering the horizontal loading of the shaft, bending moment at A and B are equal to zero.
MAH3 = MBH3 = 0
Bending Moment at C, MCH2 = 662.47 × 100 × 10−3
= 66.247 N-m
Bending Moment at E, MEH2 = 662.47 × 250 𝑥 10−3
= 165.617 N-m
Bending Moment at D, MDH2 = 1757.45 × 181.5 × 10−3
= 318.977 N-m
MC2 = √(226.1)2 + (66.247)2 = 235.6 𝑁 − 𝑚.
Resultant bending moment at E,
ME2 = √(497.2)2 + (165.617)2 = 524 𝑁 − 𝑚.
Resultant bending moment at D,
MD2 = √(906.93)2 + (318.977)2 = 961.388 𝑁 − 𝑚.
∴ Maximum bending moment diagram at D, M2 =MD = 961.388 N-m.

𝑇𝑒 = √(𝐾𝑏 × 𝑀1)2 + (𝐾𝑡 × 𝑇)2
𝑇𝑒 = √(1.5 × 961.388)2 + (1.2 × 199.46)2 N-m
𝑇𝑒 = 1461.81 𝑁 − 𝑚 = 1461.81 × 103
𝑁 − 𝑚𝑚
𝑇𝑒 =
𝜋
16
× 𝜏 × 𝑑2
3
𝑑2
3
=
1461.81 ×103 ×16
𝜋 ×147
= 50645.83 𝑚𝑚3
𝑑2 = 36.99 mm
Bending moment (Me)
(𝑀𝑒) =
1
2
[(𝑀1 × 𝐾 𝐵 ) + 𝑇𝑒 ]
𝑀𝑒 =
1
2
[(1.5 × 961.388) + 199.46] 𝑁 − 𝑚
𝑀𝑒 = 820.771 𝑁 − 𝑚 = 820.771 × 103
𝑁 − 𝑚𝑚
𝑀𝑒 =
𝜋
32
× 𝜎 × 𝑑2
3
𝑑2
3
=
820.771 ×103 ×32
𝜋 ×294
𝑚𝑚3
𝑑2 = 30.53 mm

Since the gear mounted on the middle of the shaft, therefore maximum bending moment at center of
the gear wheel 4,
Vertical bending moment at B, 𝑀 𝑉3 =
6686.47 ×257 × 10−3
4
= 429.6 𝑁 − 𝑚
Horizontal bending moment at B, 𝑀 𝐻3 =
2419.92 ×257 × 10−3
4
= 155.48 𝑁 − 𝑚.
Resultant bending moment at B,
MB = √(429.6)2 + (155.48)2 = 456.87 𝑁 − 𝑚.
∴ Maximum bending moment diagram at B, M3 =MB = 456.87 N-m.
𝑇𝑒 = √(𝐾𝑏 × 𝑀3)2 + (𝐾𝑡 × 𝑇)2
𝑇𝑒 = √(1.5 × 456.87)2 + (1.2 × 199.46)2 N-m
𝑇𝑒 = 725.9 𝑁 − 𝑚 = 725.9 × 103
𝑁 − 𝑚𝑚
𝑇𝑒 =
𝜋
16
× 𝜏 × 𝑑3
3
𝑑3
3
=
725.9×103 ×16
𝜋 ×147
= 25149.51 𝑚𝑚3
𝑑3 = 29.29 mm

Bending moment (Me)
(𝑀𝑒) =
1
2
[(𝑀1 × 𝐾 𝐵 ) + 𝑇𝑒 ]
𝑀𝑒 =
1
2
[(1.5 × 456.87) + 199.46] 𝑁 − 𝑚
𝑀𝑒 = 442.38 𝑁 − 𝑚 = 442.38 × 103
𝑁 − 𝑚𝑚
𝑀𝑒 =
𝜋
32
× 𝜎 × 𝑑33
𝑑3
3
=
442.38×103 ×32
𝜋 ×294
𝑚𝑚3
= 15326.68 𝑚𝑚3
𝑑3 = 24.84 mm
∴ Taking larger of the two values, d3 =29.29 mm  30 mm.

Considering the Vertical loading of the shaft, bending moment at A and B are equal to zero.
MAV4 = MBV4 = 0
Bending Moment at C, MCV4 = 4794.8 × 49.78 × 10−3
= 238.68 N-m
Bending Moment at E, MEV4 = ([4794.8 × 0.15] − [43.75 𝑥 0.1] = 710.47 N-m
Bending Moment at D, MDV4 = 6282.91 × 128.5 × 10−3
= 807.35 N-m
Considering the horizontal loading of the shaft, bending moment at A and B are equal to zero.
MAH4 = MBH4 = 0
Bending Moment at C, MCH4 = 1665.91 × 49.78 × 10−3
= 82.93 N-m
Bending Moment at E, MEH4 = 1665.91 × 152.56 𝑥 10−3
= 254.15 N-m
Bending Moment at D, MDH4 = 2215.98 × 128.5 × 10−3
= 284.75 N-m
MC4 = √(238.68)2 + (82.93)2 = 252.67 𝑁 − 𝑚.
Resultant bending moment at E,
ME2 = √(710.47)2 + (254.15)2 = 754.56 𝑁 − 𝑚.
Resultant bending moment at D,
MD2 = √(807.35)2 + (284.75)2 = 856 𝑁 − 𝑚.
∴ Maximum bending moment diagram at D, Ms =MD = 856 N-m.

𝑇𝑒 = √(𝐾𝑏 × 𝑀1)2 + (𝐾𝑡 × 𝑇)2
𝑇𝑒 = √(1.5 × 856)2 + (1.2 × 398.92)2 N-m
𝑇𝑒 = 1370.34 𝑁 − 𝑚 = 1370.34 × 103
𝑁 − 𝑚𝑚
𝑇𝑒 =
𝜋
16
× 𝜏 × 𝑑4
3
𝑑4
3
=
1370.34 ×103 ×16
𝜋 ×147
= 47476.76 𝑚𝑚3
𝑑4 = 36.2 mm
Bending moment (Me)
(𝑀𝑒) =
1
2
[(𝑀1 × 𝐾 𝐵 ) + 𝑇𝑒 ]
𝑀𝑒 =
1
2
[(1.5 × 856) + 398.92] 𝑁 − 𝑚
𝑀𝑒 = 841.46 𝑁 − 𝑚 = 841.46 × 103
𝑁 − 𝑚𝑚
𝑀𝑒 =
𝜋
32
× 𝜎 × 𝑑4
3
𝑑4
3
=
841.46 ×103 ×32
𝜋 ×294
𝑚𝑚3
𝑑4 = 30.77 mm

5) Shaft 5 (S5)
b) Vertical bending moment diagram

Bending moment at A and D are equal to zero.
MAV5 = MBV5 = 0
Bending Moment at C, MCV5 = 4002.81 × 49.78 × 10−3
= 199.26 N-m
Bending Moment at D, MDV5 = 5241.14 × 265 × 10−3
= 1388.9 N-m
∴ Maximum bending moment at D, M5 =MDV5 = 1388.9 N-m.
𝑇𝑒 = √(𝐾𝑏 × 𝑀5)2 + (𝐾𝑡 × 𝑇)2
𝑇𝑒 = √(1.5 × 1388.9)2 + (1.2 × 1595.68 )2 N-m
𝑇𝑒 = 2689.64 𝑁 − 𝑚 = 2689.64 × 103
𝑁 − 𝑚𝑚
𝑇𝑒 =
𝜋
16
× 𝜏 × 𝑑5
3
𝑑5
3
=
2869.64 ×103 ×16
𝜋 ×147
= 99421.47 𝑚𝑚3
d5 = 46.33 mm
Bending moment (Me)
(𝑀𝑒) =
1
2
[(𝑀5 × 𝐾 𝐵 ) + 𝑇𝑒 ]
𝑀𝑒 =
1
2
[(1.5 × 1388.9) + 1595.68] 𝑁 − 𝑚
𝑀𝑒 = 1839.515 𝑁 − 𝑚 = 1839.515 × 103
𝑁 − 𝑚𝑚
𝑀𝑒 =
𝜋
32
× 𝜎 × 𝑑5
3
𝑑5
3
=
1839.515×103 ×32
𝜋 ×294
𝑚𝑚3
= 63731.78 𝑚𝑚3
𝑑5= 39.94 mm
∴ Taking larger of the two values, d5 =46.33 mm

 There are 05 shafts available in the mechanical lift design, so we have to create 05 keys to to
each shaft.
 The keys needs to be satisfy the following conditions,
1. Allowable shearing stress ≥ Induced shearing stress
2. Allowable crushing stress ≥ Induced crushing stress
The shearing stress can be calculated as follows,
𝑇 = 𝑙 × 𝑤 × 𝜏 ×
𝑑
2
The crushing stress can calculated as follows,
𝑇 = 𝑙 ×
𝑡
2
× 𝜎 ×
𝑑
2
Where,
T: - Torque transmitted to the shaft in N/m
l: - Length of the key in mm
w: - Width of the key in mm
d: - diameter of the key in mm
t :- thickness of the key in mm
Initial data:
Shaft material = Alloy Steel
Shear stress (𝜏 ) = 147 MNm-2
Crushing stress (𝜎 ) = 294 MNm-2
Figure 15: Terms in key

{From the Machine Design – R. S. Khurmi -Table 13.1 (page No: 472)}
Table 6 – Key cross section value for Shaft diameter
1) Keydrive for shaft 1 (S1)
Diameter of the shaft 1 = 17 mm.
For 17 mm shaft,
Width (w) = 6 mm
Thickness (t) = 6 mm
Considering shearing of the key,
T = l × τ × w x (
𝑑
2
)
T = l × 147 × 6 x (
17
2
)
T = 7497 (𝑙1)

Torque transmitted from shaft
T = (
𝜋
16
)× 𝜏 × 𝑑1
3 =
𝜋 𝑥 147 𝑥 173
16
= 141805.79 N.mm
141805.79 = 7497 (𝑙1)
(𝑙1) = 18.91 mm
Considering crushing of the key,
T = 𝑙 ×
𝑡
2
× 𝜎 ×
𝑑
2
141805.79 = 𝑙1 ×
6
2
× 294 ×
17
2
141805.79 = 7497 (𝑙1)
(𝑙1) = 18.91 mm
Shearing (𝑙1) ≤ crushing (𝑙1)
Length of ley (𝑙) = 18.91 mm
2) Keydrive for shaft 2 (S2) & shaft 4 (S4)
For 40 mm shaft,
Width (w) = 14 mm
T = l × τ × w x (
𝑑
2
)
T = l × 147 × 14 x (
40
2
)
T = 41160 (𝑙2)

T = (
𝜋
16
)× 𝜏 × 𝑑2,4
3 =
𝜋 𝑥 147 𝑥 403
16
= 1847256.48 N.mm
1847256.48 = 41160 (𝑙2, 4)
(𝑙2,4) = 44.88 mm
T = 𝑙 ×
𝑡
2
× 𝜎 ×
𝑑
2
1847256.48 = 𝑙2 ×
9
2
× 294 ×
40
2
1847256.48 = 26460 (𝑙2, 4)
(𝑙2, 4) = 69.81 mm
Shearing (𝑙2, 4) ≤ crushing (𝑙2, 4)
For 30 mm shaft,
Width (w) = 10 mm
T = l × τ × w x (
𝑑
2
)
T = l × 147 × 10 x (
30
2
)
T = 22050 (𝑙3)

T = (
𝜋
16
)× 𝜏 × 𝑑3
3 =
𝜋 𝑥 147 𝑥 303
16
= 779311.33 N.mm
779311.33 = 22050 (𝑙3)
(𝑙3) = 35.34 mm
T = 𝑙 ×
𝑡
2
× 𝜎 ×
𝑑
2
779311.33 = 𝑙3 ×
8
2
× 294 ×
30
2
779311.33 = 17640 (𝑙3)
(𝑙3) = 44.17 mm
For 50 mm shaft,
Width (w) = 16 mm
T = l × τ × w x (
𝑑
2
)
T = l × 147 × 16 x (
50
2
)
T = 58800 (𝑙5)

T = (
𝜋
16
)× 𝜏 × 𝑑5
3 =
𝜋 𝑥 147 𝑥 503
16
= 3607922.81 N.mm
3607922.81 = 58800 (𝑙3)
(𝑙3) = 61.36 mm
T = 𝑙 ×
𝑡
2
× 𝜎 ×
𝑑
2
3607922.81 = 𝑙5 ×
10
2
× 294 ×
50
2
3607922.81 = 36750 (𝑙5)
(𝑙3) = 98.17 mm

Initial Data:
1) Torque = 199.46 × 103 N-mm
2) outer diameter of friction surface = d2
3) inner diameter of friction surface = d1
4) Maximum intensity of pressure = 0.3 N/mm2
For uniform wear conditions,
P x r = c (constant)
At inner radius. Intensity of pressure is at maximum.
Pmax x r1 = c
Material selection
{From Machine Design – R. S. Khurmi – Table 24.1 (page No: 887)}
Table 7 – Material selection table
Material = Powder metal on cast iron
Assuming that,
Both sides are effective, n = 2
Ratio of the diameter = 1.25 i.e. (r1/r2 = 1.25)

Normal load acting on the friction surface (W) = 2𝜋𝑐 (𝑟1 − 𝑟2)
= 2𝜋 × 0.3𝑟2 (1.25𝑟2 − 𝑟2)
= 0.47𝑟2
2
Mean radius of the friction surface (R) =
𝑟1+𝑟2
2
=
1.25𝑟2+𝑟2
2
= 1.125 𝑟2
Torque transmitted (T) = 𝑛 × 𝜇 × 𝑊 × 𝑅
199.46 × 103 N-mm = 2 x 0.4 x 0.47 𝑟2
2 x 1.125 𝑟2
0.423 𝑟2
3 = 199.46 × 103
𝑟2 = 77.77 mm
𝑟1 = 1.25 𝑟2
= 97.21 mm

Total load to be hoisted (WT) = 6191.25 N
≈ 6200 N
Down ward motion speed V = 0.65 m/s
1) Rope type: - Steel wire suspension ropes.
6 × 19 rope
2) Factor of safety :- 6 (Jib and pillar cranes)
Since the design load is calculated by taking factor of safety 8 to 2.5 times the factor of safety,
therefore let us take factor of safety is 15.
Hence design load for the wire rope (WD) = 15 x 6200 = 93 kN
3) We find that tensile strength 1250 – 1400 MPa is 425𝑑3
N, where d is the diameter of rope in
mm. equating this tensile strength to the design load, we get
425𝑑2
= 93000
𝑑2
= 218.823
d = 14.79
d ≈ 16 mm
4) Wire diameter (𝑑 𝑤) = 0.063 x d
= 0.063 x 16
= 1.008 mm
Area of rope (A) = 0.38 𝑥 𝑑2
= 0.38 x 162
= 97.28 𝑚𝑚2
5)
a. Weight of the rope W = 0.0383 x (162
)
= 9.8 N/m
Hence W = 9.8 N/m x 450 m
= 4410 N

b. We find that diameter of sheave (D) may be taken as 20 to 30 times of the diameter of the rope
(d). Let us take.
D = 30 d
= 30 x 16
= 480 mm
Hence bending stress,
𝜎𝑏 =
𝐸 𝑟∗ 𝑑 𝑤
𝐷
(𝐸𝑟 = 92 x 103
N/𝑚𝑚2
)
= 42 x 102
x 1.008
= 88.2 N/𝑚𝑚2
And the equivalent bending load on the rope,
𝑊𝑏 = 𝜎𝑏x A
= 88.2 N/𝑚𝑚2
x 97.28 𝑚𝑚2
= 8850 N
There is no additional acceleration on the rope
Hence 𝑊𝑎 = 0
Impact load during starting (When there is no slackness in the rope)
𝑊𝑠𝑡 = 2(W + w)
= 2(6130 + 4410)
= 21084 N
6. We know that the effective load on the rope during normal working (i.e. during uniform lifting or
lowering of the load).
= WT + W + 𝑊𝑏
= 6200 + 4410 + 8850
= 19460 N
Hence actual factor of safety during normal loading, working,
=
93000
19460
= 4.77

Effective load on the rope. During starting
= 𝑊𝑠𝑡 + 𝑊𝑏
= 21084 + 8850
= 29934 N
Hence actual factor of starting during starting,
=
93000
29934
= 3.1
Since the actual factor of safety as calculated above are safe. Therefore a wire rope
of diameter 16mm and 6× 19 type is satisfactory.

References
1. Induction motor manual, Gear and Shaft manual, Keyway manual, SKF bearing manual, and Belt
& pulley manual
2. MEX 5277 Machine design course materials.
3. R. S. Khurmi and J. K. Gupta theory of machines (S. Chand and Company limited 2010).
4. V. B. Bhandari design of machine elements (Tata McGraw-Hill Education 2010).
5. www.machinedesign.com.
6. Shigley machine design 9th edition.
7. www.engineeringbookstore.com.

APPENDIX
Figure 16: Motor Selectin Chart

Figure 17: Power Rating for “B”- Section V-Belt
Figure 18: Power correction factors for arc of contact

Figure 19: Power correction factors for belt pitch length

Figure 21: Characteristics of roller chains

Figure 22: Chain Drive selection

Figure 23: Factor of safety (n) for bush roller and silent chains
Figure 24: Values of service factor for bearing
Figure 25: Values of X0 and Y0 for bearing

Figure 26: Values of x and y for dynamically loaded bearings

Figure 27: Bearing Selection (1)

Figure 31: Values for shaft calculation

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