Vertical Screw Conveyor - Design Project
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Oct 12, 2017
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About This Presentation
Vertical Screw Conveyor Design Project for MEX5277 - Machine Design
Step by step guide on how to create a Vertical Screw Conveyor.
Power calculation of a vertical screw conveyor.
Slide Content
MACHINE DESIGN
MEX5277
VERTICAL SCREW
CONVEYOR
NAME : P. A. HEWAPATHIRANA
REGISTER NUMBER : 613344004
DATE OF SUBMISSION : 22/09/2017
CENTER : COLOMBO
MEX5277
VERTICAL SCREW
CONVEYOR
NAME : P. A. HEWAPATHIRANA
REGISTER NUMBER : 613344004
DATE OF SUBMISSION : 22/09/2017
CENTER : COLOMBO
i
Table of contents
List of Illustration ii
Introduction iii
Aim and Objective iv
Specifications iv
1. Design Layout 1
2. Motor Selection 3
3. Belt and Pully Design 6
4. Gear design 13
5. Bevel gear design 19
6. Shaft design 22
7. Keyway design 29
8. Bearing selection 31
9. Clutch 33
References 35
Table of contents
List of Illustration ii
Introduction iii
Aim and Objective iv
Specifications iv
1. Design Layout 1
2. Motor Selection 3
3. Belt and Pully Design 6
4. Gear design 13
5. Bevel gear design 19
6. Shaft design 22
7. Keyway design 29
8. Bearing selection 31
9. Clutch 33
References 35
ii
List of Illustrations
Figure 1.1 – design view of the Vertical screw conveyor 1
Figure 1.2 a – Rendered View of the Gear Box 1
Figure 1.2 b – Top view and side view 1
Figure 1.3 – Rendered View of the Screw conveyor and Section View 2
Figure 2.1 – JHS400 screw specifications 3
Figure 3.1 – Selection of V-belt cross section 6
Figure 3.2 – Service factor for drives 7
Figure 3.3 – Pully arrangement 7
Figure 3.4 – A section V-belts 8
Figure 3.5 – Power correction factors for belt pitch length 8
Figure 3.6 – Power correction factor for Arc contact 9
Figure 3.7 – Pulley belt arrangement 10
Figure 4.1 – Gear arrangement 13
Figure 4.2 – No: of teeth vs. Involute factor Y 14
Figure 4.3 – Values of Deformation Factor C (kN/m) for Dynamic load 16
Figure 4.4 – terms used in a gear 18
Figure 6.1 – Shaft design 22
figure 6.2 – WD components 22
Figure 6.3 – Bending moment diagram 23
Figure 6.4 – Shaft design 24
Figure 6.5 – Forces acting on gear 26
Figure 6.6 – Bending moment diagram 27
Figure 7.1 – terms in key 29
List of Illustrations
Figure 1.1 – design view of the Vertical screw conveyor 1
Figure 1.2 a – Rendered View of the Gear Box 1
Figure 1.2 b – Top view and side view 1
Figure 1.3 – Rendered View of the Screw conveyor and Section View 2
Figure 2.1 – JHS400 screw specifications 3
Figure 3.1 – Selection of V-belt cross section 6
Figure 3.2 – Service factor for drives 7
Figure 3.3 – Pully arrangement 7
Figure 3.4 – A section V-belts 8
Figure 3.5 – Power correction factors for belt pitch length 8
Figure 3.6 – Power correction factor for Arc contact 9
Figure 3.7 – Pulley belt arrangement 10
Figure 4.1 – Gear arrangement 13
Figure 4.2 – No: of teeth vs. Involute factor Y 14
Figure 4.3 – Values of Deformation Factor C (kN/m) for Dynamic load 16
Figure 4.4 – terms used in a gear 18
Figure 6.1 – Shaft design 22
figure 6.2 – WD components 22
Figure 6.3 – Bending moment diagram 23
Figure 6.4 – Shaft design 24
Figure 6.5 – Forces acting on gear 26
Figure 6.6 – Bending moment diagram 27
Figure 7.1 – terms in key 29
iii
INTRODUCTION
Generally, conveying is accomplished by a combination of mechanical, inertial,
pneumatic, and gravity forces. Application of screw conveyors can be seen almost in every industry
nowadays. Conveyors utilizing primarily mechanical forces are screw, belt, and mass conveyors.
Screw conveyors are widely used for transporting and elevating particulates at controlled and
steady rates. They are used in many bulk materials applications in industries ranging from
industrial minerals, agriculture (grains), pharmaceuticals, chemicals, pigments, plastics, cement,
sand, salt and food processing.
The main reason why the screw conveyors are popular because they can be used for uniform
and continuous supply of various materials in manufacturing machinery and transport devices in
many industries as stated above. So it is a very effective way of elevating bulk materials.
Some advantages of Vertical Screw Conveyors are,
• Ideal for handling dry to semi-fluid materials
• Capacities up to 6,000 cubic feet per hour.
• Ability to elevate bulk materials up to 30-feet without use of internal bearings.
• Totally enclosed design for dust and vapor-tight requirements.
Since they operate over a wide range of speeds and angles of elevation up to the vertical. In
this mini project, we have focused on and designed a Vertical screw conveyor which is fully
enclosed.
INTRODUCTION
Generally, conveying is accomplished by a combination of mechanical, inertial,
pneumatic, and gravity forces. Application of screw conveyors can be seen almost in every industry
nowadays. Conveyors utilizing primarily mechanical forces are screw, belt, and mass conveyors.
Screw conveyors are widely used for transporting and elevating particulates at controlled and
steady rates. They are used in many bulk materials applications in industries ranging from
industrial minerals, agriculture (grains), pharmaceuticals, chemicals, pigments, plastics, cement,
sand, salt and food processing.
The main reason why the screw conveyors are popular because they can be used for uniform
and continuous supply of various materials in manufacturing machinery and transport devices in
many industries as stated above. So it is a very effective way of elevating bulk materials.
Some advantages of Vertical Screw Conveyors are,
• Ideal for handling dry to semi-fluid materials
• Capacities up to 6,000 cubic feet per hour.
• Ability to elevate bulk materials up to 30-feet without use of internal bearings.
• Totally enclosed design for dust and vapor-tight requirements.
Since they operate over a wide range of speeds and angles of elevation up to the vertical. In
this mini project, we have focused on and designed a Vertical screw conveyor which is fully
enclosed.
iv
AIM
To design a Vertical screw conveyor to transport material to a higher elevation from ground
level.
OBJECTIVE
• Selecting the necessary material to be transported vertically.
• Designing the vertical screw conveyor.
• Obtaining the power required for the design and selection of an appropriate motor.
• Design the drive mechanism.
• Design the gear box.
• Design the shaft and selection of keys.
• Selection of bearings.
• Design a suitable clutch.
SPECIFICATIONS
Transport material : Cement
For a higher transfer rate of the selected material (Cement),
Selected Screw : JHS400
∴ pitch of the screw : 350 mm
∴ Vertical screw conveyor height : 3.15 m
Speed of the screw : 60 rpm
AIM
To design a Vertical screw conveyor to transport material to a higher elevation from ground
level.
OBJECTIVE
• Selecting the necessary material to be transported vertically.
• Designing the vertical screw conveyor.
• Obtaining the power required for the design and selection of an appropriate motor.
• Design the drive mechanism.
• Design the gear box.
• Design the shaft and selection of keys.
• Selection of bearings.
• Design a suitable clutch.
SPECIFICATIONS
Transport material : Cement
For a higher transfer rate of the selected material (Cement),
Selected Screw : JHS400
∴ pitch of the screw : 350 mm
∴ Vertical screw conveyor height : 3.15 m
Speed of the screw : 60 rpm
1
1. DESIGN LAYOUT
Figure 1.1 – design view of the Vertical screw conveyor
Figure 1.2 a – Rendered View of the Gear Box Figure 1.2 b – Top view and side view
1. DESIGN LAYOUT
Figure 1.1 – design view of the Vertical screw conveyor
Figure 1.2 a – Rendered View of the Gear Box Figure 1.2 b – Top view and side view
2
Figure 1.3 – Rendered View of the Screw conveyor and Section View
Figure 1.3 – Rendered View of the Screw conveyor and Section View
3
2. MOTOR SELECTION
Figure 2.1 – JHS400 screw specifications
Screw blade diameter = 400 mm
Drive shaft diameter = 60 mm
Pitch = 350 mm
Screw blade thickness = 4 mm
Capacity (Q) =
?????? (�−�)
2
× � × � × �� × ?????? × 60
4
Where, D = screw diameter (in dm)
d = drive shaft diameter (in dm)
s = pitch (in dm)
n = revolutions per minute
sg = specific weight of the material
i = degree of through filling
� = Progress resistance coefficient
for Cement, sg = 1600 g
� = 6.0
i = 0.1
Maximum Capacity (Q) =
?????? (4−0.6)
2
× 3.5 ×60 ×1600 ×0.1 ×60
4
= 18,303,672.78 kgh
-1
2.1
2. MOTOR SELECTION
Figure 2.1 – JHS400 screw specifications
Screw blade diameter = 400 mm
Drive shaft diameter = 60 mm
Pitch = 350 mm
Screw blade thickness = 4 mm
Capacity (Q) =
?????? (�−�)
2
× � × � × �� × ?????? × 60
4
Where, D = screw diameter (in dm)
d = drive shaft diameter (in dm)
s = pitch (in dm)
n = revolutions per minute
sg = specific weight of the material
i = degree of through filling
� = Progress resistance coefficient
for Cement, sg = 1600 g
� = 6.0
i = 0.1
Maximum Capacity (Q) =
?????? (4−0.6)
2
× 3.5 ×60 ×1600 ×0.1 ×60
4
= 18,303,672.78 kgh
-1
2.1
4
Im = mass flow rate (ton per hour)
� = Progress resistance coefficient
L = length (m)
H = height of the Vertical screw driver
(L sin90
o
)
The driving power of the loaded screw conveyor is given by,
P = PV + PN + Pst
Where, PV = required power to move the material
PN = required power to operate unloaded screw
Pst = required power for the vertical position of screw conveyor
Material travelling (PV) =
�
?????? ×� × ?????? × �
3600
=
18303.672 ×3.15 × 6.0
3600 ×102
= 0.9421 kW
Unloaded Operated Screw (PN) =
� × �
20
=
400 × 3
1000 × 20
= 0.06 kW
Position of the screw (Pst) =
�
?????? ×� × �
3600
=
18303.672 ×(3.15+0.5)
3600 ×102
= 0.1819 kW
∴ using eq
n
2.2,
Screw Conveyor Power = PV + PN + Pst
= 0.9421 + 0.06 + 0.1819
= 1.1840 kW
Losses within the mechanical elements,
Losses due to Gear train = 2% Losses due to belt = 2%
Losses due to drive pully = 3% Losses due to Bearing = 1%
Losses due to pinion pully = 3% Losses due to chain = 5%
Uncountable losses = 5%
2.2
Im = mass flow rate (ton per hour)
� = Progress resistance coefficient
L = length (m)
H = height of the Vertical screw driver
(L sin90
o
)
The driving power of the loaded screw conveyor is given by,
P = PV + PN + Pst
Where, PV = required power to move the material
PN = required power to operate unloaded screw
Pst = required power for the vertical position of screw conveyor
Material travelling (PV) =
�
?????? ×� × ?????? × �
3600
=
18303.672 ×3.15 × 6.0
3600 ×102
= 0.9421 kW
Unloaded Operated Screw (PN) =
� × �
20
=
400 × 3
1000 × 20
= 0.06 kW
Position of the screw (Pst) =
�
?????? ×� × �
3600
=
18303.672 ×(3.15+0.5)
3600 ×102
= 0.1819 kW
∴ using eq
n
2.2,
Screw Conveyor Power = PV + PN + Pst
= 0.9421 + 0.06 + 0.1819
= 1.1840 kW
Losses within the mechanical elements,
Losses due to Gear train = 2% Losses due to belt = 2%
Losses due to drive pully = 3% Losses due to Bearing = 1%
Losses due to pinion pully = 3% Losses due to chain = 5%
Uncountable losses = 5%
2.2
5
Total uncountable losses of = 1184 ×(
102
100
×
103
100
×
103
100
×
105
100
×
102
100
×
101
100
×
105
100
)−1184
the machine
= 271 W
∴ Total Power required = 1455.2126 W
= 1.4552 kW
Motor selection from the Handbook (page No: 04)
Table 2.1 – “4 POLES”, Synchronous motor at 50Hz,
Output (kW) Efficiency (%)
1.5 75.5 ×
2.2 81.5 √
Motor output power = 2.2 ×
81.5
100
= 1.793 kW
1.793 kW > 1.4552 kW (∴ Motor can supply the necessary power required)
∴ selected motor = D100L at 1425 rpm
Transmission ratio =
����� ���
�ℎ��� ���
=
1425
60
≈ 24
∴ Transmission ratio = 1: 24
Transmission ratio of the Belt drive = 2: 1
Transmission ratio of the Gear drive = 4: 1
Transmission ratio of the Chain drive = 3: 1
2 | 24 _
4 | 12 _
3 | 3 _
1
Total uncountable losses of = 1184 ×(
102
100
×
103
100
×
103
100
×
105
100
×
102
100
×
101
100
×
105
100
)−1184
the machine
= 271 W
∴ Total Power required = 1455.2126 W
= 1.4552 kW
Motor selection from the Handbook (page No: 04)
Table 2.1 – “4 POLES”, Synchronous motor at 50Hz,
Output (kW) Efficiency (%)
1.5 75.5 ×
2.2 81.5 √
Motor output power = 2.2 ×
81.5
100
= 1.793 kW
1.793 kW > 1.4552 kW (∴ Motor can supply the necessary power required)
∴ selected motor = D100L at 1425 rpm
Transmission ratio =
����� ���
�ℎ��� ���
=
1425
60
≈ 24
∴ Transmission ratio = 1: 24
Transmission ratio of the Belt drive = 2: 1
Transmission ratio of the Gear drive = 4: 1
Transmission ratio of the Chain drive = 3: 1
2 | 24 _
4 | 12 _
3 | 3 _
1
6
3. BELT AND PULLY DESIG N
Data:
Transmission ratio of the Belt drive = 2: 1
Power consumption of the Motor = 2.2 kW
Motor rpm = 1425 rpm
From the V-Belt drive handbook,
Selection of V-belt cross section, (page No: 3/79)
Figure 3.1 – Selection of V-belt cross section
According to the design power and rpm of motor,
Type of belt = Type A
Service factor for drives, (page No: 3/80)
Operation hours per day = over 16 hrs.
Type of driven mechanism = Extra heavy duty
Service factor = 1.8
3. BELT AND PULLY DESIG N
Data:
Transmission ratio of the Belt drive = 2: 1
Power consumption of the Motor = 2.2 kW
Motor rpm = 1425 rpm
From the V-Belt drive handbook,
Selection of V-belt cross section, (page No: 3/79)
Figure 3.1 – Selection of V-belt cross section
According to the design power and rpm of motor,
Type of belt = Type A
Service factor for drives, (page No: 3/80)
Operation hours per day = over 16 hrs.
Type of driven mechanism = Extra heavy duty
Service factor = 1.8
7
Figure 3.2 – Service factor for drives
From the belt ratio,
Let Pitch diameter of smaller pully (d) = 125 mm
Pitch diameter of larger pully (D) = 250 mm
Pitch length of the belt (L) calculations, (page No: 3/78)
Pitch length of the belt (L) = 2�+1.57(�+�)+
(�−�)
2
4�
Where, C = center distance of drive
Figure 3.3 – Pully arrangement
3.1
Figure 3.2 – Service factor for drives
From the belt ratio,
Let Pitch diameter of smaller pully (d) = 125 mm
Pitch diameter of larger pully (D) = 250 mm
Pitch length of the belt (L) calculations, (page No: 3/78)
Pitch length of the belt (L) = 2�+1.57(�+�)+
(�−�)
2
4�
Where, C = center distance of drive
Figure 3.3 – Pully arrangement
3.1
8
Recommended range for the center distance,
2(�+�) ≥� ≥0.7(�+�)
2(125+250) ≥� ≥0.7(125+250)
750 mm ≥� ≥262.5 ��
∴ Taking minimum C value for calculations,
C = 262.5 mm
Substituting for the eq
n
3.1,
L = (2 ×262.5)+1.57(125+250)+
(250−125)
2
4�
= 1128.6 mm
According to the standard pitch lengths, From Table3C, (page No: 3/69)
Figure 3.4 – A section V-belts
L = 1250 mm
∴ Power correction factor for belt pitch length (type A) = 0.93 (page No: 3/84)
Figure 3.5 – Power correction factors for belt pitch length
Recommended range for the center distance,
2(�+�) ≥� ≥0.7(�+�)
2(125+250) ≥� ≥0.7(125+250)
750 mm ≥� ≥262.5 ��
∴ Taking minimum C value for calculations,
C = 262.5 mm
Substituting for the eq
n
3.1,
L = (2 ×262.5)+1.57(125+250)+
(250−125)
2
4�
= 1128.6 mm
According to the standard pitch lengths, From Table3C, (page No: 3/69)
Figure 3.4 – A section V-belts
L = 1250 mm
∴ Power correction factor for belt pitch length (type A) = 0.93 (page No: 3/84)
Figure 3.5 – Power correction factors for belt pitch length
9
Center distance calculations, (page No: 3/78)
Center distance (C) = �+ √(�
2
−�)
Where, A =
�
4
−??????
(�+�)
8
B =
(�−�)
2
8
=
1250
4
−??????
(250+125)
8
=
(250−125)
2
8
= 165.2378 mm = 1953.1250 mm
Substituting for the eq
n
3.2,
C = 165.2378 + √(165.2378
2
−1953.1250)
= 324.4400 mm
Number of belts calculation,
No: of belts (X) =
���??????�� �����
�??????��� ���� �����
=
����
���??????�1
Where, Nt = Required Power in watts
Ks – Service Factor for Belt Drive
No – Power Rating
Ke – Power Correction Factor for Arc Contact
Kl – Power Correction Factor for Belt Pitch Length
Nt = 2.2 kW Ks = 1.8
Ke = 0.94 Kl = 0.93 (page No: 3/83, page No: 3/84)
Figure 3.6 – Power correction factor for Arc contact
No - Power rating calculations,
From Table9C, (page No: 3/87)
D (mm) Additional Power Ratio
Speed of faster shaft (rpm) 125 2.00 and over
960 1.61 0.12
1425 X Y
1440 2.24 0.17
3.2
3.3
Center distance calculations, (page No: 3/78)
Center distance (C) = �+ √(�
2
−�)
Where, A =
�
4
−??????
(�+�)
8
B =
(�−�)
2
8
=
1250
4
−??????
(250+125)
8
=
(250−125)
2
8
= 165.2378 mm = 1953.1250 mm
Substituting for the eq
n
3.2,
C = 165.2378 + √(165.2378
2
−1953.1250)
= 324.4400 mm
Number of belts calculation,
No: of belts (X) =
���??????�� �����
�??????��� ���� �����
=
����
���??????�1
Where, Nt = Required Power in watts
Ks – Service Factor for Belt Drive
No – Power Rating
Ke – Power Correction Factor for Arc Contact
Kl – Power Correction Factor for Belt Pitch Length
Nt = 2.2 kW Ks = 1.8
Ke = 0.94 Kl = 0.93 (page No: 3/83, page No: 3/84)
Figure 3.6 – Power correction factor for Arc contact
No - Power rating calculations,
From Table9C, (page No: 3/87)
D (mm) Additional Power Ratio
Speed of faster shaft (rpm) 125 2.00 and over
960 1.61 0.12
1425 X Y
1440 2.24 0.17
3.2
3.3
10
Using interpolating method,
X = 2.2
Y = 0.16
∴ No = X + Y
= 2.2 + 0.16
= 2.38
Substituting for the eq
n
2.3,
X =
2.2 ×1.8
0.94 ×0.93 ×2.36
= 1.8794
≈2 �����
∴ two V-belts are required to transmit the power.
Assessing the required no: of belts using standard equation
Figure 3.7 – Pulley belt arrangement
2.3log|
�1
�2
|= �??????�����(??????) where, T1, T2 = tensions of the belt
?????? = angle of contact
� = coefficient of friction
?????? = Nominal include angle
Assuming � = 0.3
From the V-Belt drive handbook,
Table 2A, (page No: 3/66)
T1 + T2 = 200
Table 1, (page No: 3/65)
2?????? = 40
◦
3.4
Using interpolating method,
X = 2.2
Y = 0.16
∴ No = X + Y
= 2.2 + 0.16
= 2.38
Substituting for the eq
n
2.3,
X =
2.2 ×1.8
0.94 ×0.93 ×2.36
= 1.8794
≈2 �����
∴ two V-belts are required to transmit the power.
Assessing the required no: of belts using standard equation
Figure 3.7 – Pulley belt arrangement
2.3log|
�1
�2
|= �??????�����(??????) where, T1, T2 = tensions of the belt
?????? = angle of contact
� = coefficient of friction
?????? = Nominal include angle
Assuming � = 0.3
From the V-Belt drive handbook,
Table 2A, (page No: 3/66)
T1 + T2 = 200
Table 1, (page No: 3/65)
2?????? = 40
◦
3.4
11
Table 7A,
?????? = 157
◦
where,
�−�
�
=
250−125
324.4400
=0.3853 (page No: 3/82)
∴ 2.3log|
�1
�2
| = �??????�����(??????)
=
0.3×157×??????×�����(20)
180
|
�1
�2
| = 11.0919
From eq
n
s 3.4 and 3.5,
T1 = 183.46 N T2 = 16.5339 N
Power per belt = (T1 - T2) V
= (T1 - T2) rω
= (183.46− 16.5339)(
125×10
−3
2
)(
2??????×1425
60
)
= 1556.8 W
∴ No: of required belts =
���??????�� �����
����� ��� ����
= (
2.2×1.8
1.5568
)
= 2.5437
≈ 3 belts
∴ The required no: of belts = 3 belts
Torque on Smaller pulley = (??????
1− ??????
2)
�
2
= (183.46− 16.5339)
125×10
−3
2
= 10.4044 Nm
Torque on Larger pulley = (??????
1− ??????
2)
�
2
= (183.46− 16.5339)
250×10
−3
2
= 20.8088 Nm
3.5
Table 7A,
?????? = 157
◦
where,
�−�
�
=
250−125
324.4400
=0.3853 (page No: 3/82)
∴ 2.3log|
�1
�2
| = �??????�����(??????)
=
0.3×157×??????×�����(20)
180
|
�1
�2
| = 11.0919
From eq
n
s 3.4 and 3.5,
T1 = 183.46 N T2 = 16.5339 N
Power per belt = (T1 - T2) V
= (T1 - T2) rω
= (183.46− 16.5339)(
125×10
−3
2
)(
2??????×1425
60
)
= 1556.8 W
∴ No: of required belts =
���??????�� �����
����� ��� ����
= (
2.2×1.8
1.5568
)
= 2.5437
≈ 3 belts
∴ The required no: of belts = 3 belts
Torque on Smaller pulley = (??????
1− ??????
2)
�
2
= (183.46− 16.5339)
125×10
−3
2
= 10.4044 Nm
Torque on Larger pulley = (??????
1− ??????
2)
�
2
= (183.46− 16.5339)
250×10
−3
2
= 20.8088 Nm
3.5
12
Material selection,
Pulley material - Cast Iron
Due to reduced weight and their low cost.
V- Belt material - Thermoplastic polyester elastomer, with a shore hardness 92A,
Since this is a heavy-duty mechanism, the V-belt temperature should have a vast range.
Therefore, the selected material has a temperature range of -5° to 70° C
Table 3.1 - Pully characteristics
Characteristics Smaller Pully Larger Pully
Material Cast Iron Cast Iron
Pitch Diameter (mm) 125 250
Outside Diameter (mm) 133 258
Torque (Nm) 10.4044 20.8088
Material selection,
Pulley material - Cast Iron
Due to reduced weight and their low cost.
V- Belt material - Thermoplastic polyester elastomer, with a shore hardness 92A,
Since this is a heavy-duty mechanism, the V-belt temperature should have a vast range.
Therefore, the selected material has a temperature range of -5° to 70° C
Table 3.1 - Pully characteristics
Characteristics Smaller Pully Larger Pully
Material Cast Iron Cast Iron
Pitch Diameter (mm) 125 250
Outside Diameter (mm) 133 258
Torque (Nm) 10.4044 20.8088
13
4. GEAR DESIGN
Data:
Transmission ratio = 1:4
Tn = no: of teeth
Nn = gear speed (rpm)
Dn = diameter of the gear (mm)
�
1
�
2
=
�
1
�
2
4
1
=
�
1
�
2
∴ assuming that, D1 = 100 mm
D2 = 400 mm Figure 4.1 – Gear arrangement
Module (m) =
�
�
From the Gears and Shaft handbook, (page No: 10)
standard module series, m = 4
∴ No: of teeth for each Gear,
T1 =
100
4
T2 =
400
4
= 25 = 100
From the reduction of pulleys,
Speed of Gear 1 = 712.5 rpm
Table 4.1 – Gear characteristics
Gear 1 Gear 2
N (rpm) 712.5 178.125
D (mm) 100 400
T 25 100
For the design, 20
◦
stub involute system (y) (page
y 0.133 0.161
4. GEAR DESIGN
Data:
Transmission ratio = 1:4
Tn = no: of teeth
Nn = gear speed (rpm)
Dn = diameter of the gear (mm)
�
1
�
2
=
�
1
�
2
4
1
=
�
1
�
2
∴ assuming that, D1 = 100 mm
D2 = 400 mm Figure 4.1 – Gear arrangement
Module (m) =
�
�
From the Gears and Shaft handbook, (page No: 10)
standard module series, m = 4
∴ No: of teeth for each Gear,
T1 =
100
4
T2 =
400
4
= 25 = 100
From the reduction of pulleys,
Speed of Gear 1 = 712.5 rpm
Table 4.1 – Gear characteristics
Gear 1 Gear 2
N (rpm) 712.5 178.125
D (mm) 100 400
T 25 100
For the design, 20
◦
stub involute system (y) (page
y 0.133 0.161
14
Figure 4.2 – No: of teeth vs. Involute factor Y
Material selection,
Material selected = Alloy Steel (SNCM439)
[4.1]
Tensile strength (??????ut) = 980 N/mm
2
Allowable Static Stress (??????o) =
σ��
3
=
980
3
= 326.6667 N/mm
2
Strength Factor,
Gear 1, Gear 2,
??????o1 = ??????o x 0.133 ??????o2 = ??????o x 0.161
= 43.4467 N/mm
2
= 52.5933 N/mm
2
Here the Strength factor for G1 is less than that of G2
??????o1 < ??????o2
∴ the rest of the calculations are done to the G1.
Figure 4.2 – No: of teeth vs. Involute factor Y
Material selection,
Material selected = Alloy Steel (SNCM439)
[4.1]
Tensile strength (??????ut) = 980 N/mm
2
Allowable Static Stress (??????o) =
σ��
3
=
980
3
= 326.6667 N/mm
2
Strength Factor,
Gear 1, Gear 2,
??????o1 = ??????o x 0.133 ??????o2 = ??????o x 0.161
= 43.4467 N/mm
2
= 52.5933 N/mm
2
Here the Strength factor for G1 is less than that of G2
??????o1 < ??????o2
∴ the rest of the calculations are done to the G1.
15
Torque in Gear 1 = 20.8088 Nm (∵ torque is not changed throughout the shaft)
Circular pitch (Pc) = π x m
= π x 4
= 12.5663 mm
Tangential Load (WT1) =
2�
�
=
2×20.8088
100×10
−3
= 416.176 N
Normal load (WN1) =
�
??????
cos(??????)
=
416.176
�cos(20)
= 442.8852 N
Let normal pressure between tooth is 45 N/mm
-1
Face Width (b) =
442.8852
45
= 9.8419 mm
Pitch line Velocity of the Gear wheel 1
V = rω
=
100×10
−3
2
×
2??????
60
×712.5
= 3.7306 ms
-1
Value of Deformation Factor C, (page No: 30)
Assuming both pinion and gear material is Steel
Taking the maximum tooth error = 0.08 mm
C = 952 x 10
3
Nm-
1
Torque in Gear 1 = 20.8088 Nm (∵ torque is not changed throughout the shaft)
Circular pitch (Pc) = π x m
= π x 4
= 12.5663 mm
Tangential Load (WT1) =
2�
�
=
2×20.8088
100×10
−3
= 416.176 N
Normal load (WN1) =
�
??????
cos(??????)
=
416.176
�cos(20)
= 442.8852 N
Let normal pressure between tooth is 45 N/mm
-1
Face Width (b) =
442.8852
45
= 9.8419 mm
Pitch line Velocity of the Gear wheel 1
V = rω
=
100×10
−3
2
×
2??????
60
×712.5
= 3.7306 ms
-1
Value of Deformation Factor C, (page No: 30)
Assuming both pinion and gear material is Steel
Taking the maximum tooth error = 0.08 mm
C = 952 x 10
3
Nm-
1
16
Figure 4.3 – Values of Deformation Factor C (kN/m) for Dynamic load
W1 =
21×�×(�.�+ �
??????1)
21×�+ √�.�+ �
??????1
=
21×3.7306×(9.8419×952+ 416.176)
21×�+ √9.8419×952+ 416.176
= 4344.35 N
∴ Dynamic Tooth load (WD) = WT + W1
= 416.176 + 4344.35
= 4760.5247 N
For Alloy steel, Hardness = 300 BHN (using Hardness Table)
[4.2]
∴ Static Tooth load (Ws) = ??????e x b x π x m x y where, ??????e = 17.5 x Hardness
= (17.5 x 300) x 9.8419 x π x 4 x 0.133
= 85989.7891 N
For Safety,
Ws ≥ WD
∴ the condition satisfies for the obtained values
Figure 4.3 – Values of Deformation Factor C (kN/m) for Dynamic load
W1 =
21×�×(�.�+ �
??????1)
21×�+ √�.�+ �
??????1
=
21×3.7306×(9.8419×952+ 416.176)
21×�+ √9.8419×952+ 416.176
= 4344.35 N
∴ Dynamic Tooth load (WD) = WT + W1
= 416.176 + 4344.35
= 4760.5247 N
For Alloy steel, Hardness = 300 BHN (using Hardness Table)
[4.2]
∴ Static Tooth load (Ws) = ??????e x b x π x m x y where, ??????e = 17.5 x Hardness
= (17.5 x 300) x 9.8419 x π x 4 x 0.133
= 85989.7891 N
For Safety,
Ws ≥ WD
∴ the condition satisfies for the obtained values
17
Wear Tooth Load (Ww) = Dp x b x Q x k
Where, DP = Pitch circle diameter of the pinion
b = Face width of the pinion
Q = Ratio factor
K = Load-stress factor
Q =
2 ×(�.�)
(�.�)+1
Where, (V.R) =
�1
�2
=
100
25
= 4
=
2 × 4
4+1
= 1.6
K =
(????????????�)
2
�??????� ??????
1.4
{
1
��
+
1
�??????
} Where, ??????�� = Surface endurance limit
?????? = Pressure angle
EP = Young's modulus for the material of the pinion
EG = Young's modulus for the material of the gear in
=
(28×300−70)
2
�??????�(20)
1.4
{
1
189×10
3
+
1
189×10
3
}
= 179.3832 N/mm
2
Substituting values to eq
n
4.1,
Ww = 100 x 9.8419 x 1.6 x 179.3832
= 282420.9101 N
For Safety,
Ww ≥ WD
∴ the condition satisfies for the obtained values
Table 4.2 - Gear Characteristics
Characteristics Gear 1 Gear 2
Material Alloy Steel (SNCM439) Alloy Steel (SNCM439)
Diameter (mm) 100 400
N (rpm) 712.5 178.125
T 25 100
4.1
Wear Tooth Load (Ww) = Dp x b x Q x k
Where, DP = Pitch circle diameter of the pinion
b = Face width of the pinion
Q = Ratio factor
K = Load-stress factor
Q =
2 ×(�.�)
(�.�)+1
Where, (V.R) =
�1
�2
=
100
25
= 4
=
2 × 4
4+1
= 1.6
K =
(????????????�)
2
�??????� ??????
1.4
{
1
��
+
1
�??????
} Where, ??????�� = Surface endurance limit
?????? = Pressure angle
EP = Young's modulus for the material of the pinion
EG = Young's modulus for the material of the gear in
=
(28×300−70)
2
�??????�(20)
1.4
{
1
189×10
3
+
1
189×10
3
}
= 179.3832 N/mm
2
Substituting values to eq
n
4.1,
Ww = 100 x 9.8419 x 1.6 x 179.3832
= 282420.9101 N
For Safety,
Ww ≥ WD
∴ the condition satisfies for the obtained values
Table 4.2 - Gear Characteristics
Characteristics Gear 1 Gear 2
Material Alloy Steel (SNCM439) Alloy Steel (SNCM439)
Diameter (mm) 100 400
N (rpm) 712.5 178.125
T 25 100
4.1
18
Two gear wheels are designed, which are spur gears and the module for all the gears are same.
Module (m) = 4
Figure 4.4 – terms used in a gear
From the Machine Design – R. S. Khurmi -Table 28.1 (page No: 1032)
Table 4.3 – Particulars for 20degree stub involute system
Table 4.4- Gear Particulars
Particulars Gear 1 Gear 2
Addendum (mm) 3.2 3.2
Dedendum (mm) 4 4
Working Depth (mm) 6.4 6.4
Minimum tool depth (mm) 7.2 7.2
Total Thickness (mm) 6.6283 6.6283
Minimum clearance (mm) 0.8 0.8
Fillet radius at root (mm) 1.6 1.6
Two gear wheels are designed, which are spur gears and the module for all the gears are same.
Module (m) = 4
Figure 4.4 – terms used in a gear
From the Machine Design – R. S. Khurmi -Table 28.1 (page No: 1032)
Table 4.3 – Particulars for 20degree stub involute system
Table 4.4- Gear Particulars
Particulars Gear 1 Gear 2
Addendum (mm) 3.2 3.2
Dedendum (mm) 4 4
Working Depth (mm) 6.4 6.4
Minimum tool depth (mm) 7.2 7.2
Total Thickness (mm) 6.6283 6.6283
Minimum clearance (mm) 0.8 0.8
Fillet radius at root (mm) 1.6 1.6
19
5. BEVEL GEAR DESIGN
Data:
Gear ratio = 3:1
Angle (Σ) = 90
o
Input Speed = 178.125 rpm
Output Torque (TG2) = input torque x gear reduction
TG2 = TG1 x {
����ℎ �2
����ℎ �1
}
= 20.8088 x {
100
25
}
= 83.2352 Nm
Power transmission = Tω
= ??????
2??????�
60
= 1552.60 W
Material Selection
Bevel Gear (G) Bevel Pinion (P)
Material = Cast Steel Material = Steel
??????
�� = 70 Nmm
-2
??????
�� = 100 Nmm
-2
Module and Face width
Pinion Pitch angle (??????
�) = ���
−1
(
1
�.�
)
= ���
−1
(
1
3
)
= 18.43
o
Gear Pitch angle (??????
�) = Σ – ??????
�
= 90
o
– 18.43
o
= 71.57
o
Assume that,
TP = 20 TG = (V.R) Tp
= 3 x 20
= 60
Formative no: of teeth per pinion
TEP = TP Sec(??????
�) TEG = TG Sec(??????
�)
= 20 Sec(18.43) = 60 Sec(71.57)
= 21.08 = 189.8
5. BEVEL GEAR DESIGN
Data:
Gear ratio = 3:1
Angle (Σ) = 90
o
Input Speed = 178.125 rpm
Output Torque (TG2) = input torque x gear reduction
TG2 = TG1 x {
����ℎ �2
����ℎ �1
}
= 20.8088 x {
100
25
}
= 83.2352 Nm
Power transmission = Tω
= ??????
2??????�
60
= 1552.60 W
Material Selection
Bevel Gear (G) Bevel Pinion (P)
Material = Cast Steel Material = Steel
??????
�� = 70 Nmm
-2
??????
�� = 100 Nmm
-2
Module and Face width
Pinion Pitch angle (??????
�) = ���
−1
(
1
�.�
)
= ���
−1
(
1
3
)
= 18.43
o
Gear Pitch angle (??????
�) = Σ – ??????
�
= 90
o
– 18.43
o
= 71.57
o
Assume that,
TP = 20 TG = (V.R) Tp
= 3 x 20
= 60
Formative no: of teeth per pinion
TEP = TP Sec(??????
�) TEG = TG Sec(??????
�)
= 20 Sec(18.43) = 60 Sec(71.57)
= 21.08 = 189.8
20
Assuming tooth form factor, Y = 0.154−
0.912
��
YP = 0.154−
0.912
21.08
YG = 0.154−
0.912
189.8
= 0.111 = 0.149
(??????
��×??????
�) = 100 x 0.111 (??????
��×??????
�) = 70 x 0.149
= 11.1 = 10.43
∴ (??????
��×??????
�)> (??????
��×??????
�)
i.e. the gear is weaker
thus, the design should be based upon the gear.
Torque on gear = 83.2352 Nm (from previous calculations)
Tangential load on gear (WT) =
2�
�??????
=
2�
�.�??????
=
2 ×83.2352 × 10
3
� ×60
=
2.7745 × 10
3
�
N
Pitch line Velocity (V) =
??????�??????�??????
60
=
??????�.�??????×(
??????
??????
3
)
60
=
??????�×60×(
178.125
3
)
60
= 0.1865(m) ms
-1
Velocity factor (Cv) =
3
3+�
=
3
3+0.1865�
Length of pitch cone element (L) =
�??????
2�??????�(????????????)
=
��??????
2�??????�(????????????)
=
�60
2�??????�(71.57)
= 31.62(m) mm
5.1
5.2
Assuming tooth form factor, Y = 0.154−
0.912
��
YP = 0.154−
0.912
21.08
YG = 0.154−
0.912
189.8
= 0.111 = 0.149
(??????
��×??????
�) = 100 x 0.111 (??????
��×??????
�) = 70 x 0.149
= 11.1 = 10.43
∴ (??????
��×??????
�)> (??????
��×??????
�)
i.e. the gear is weaker
thus, the design should be based upon the gear.
Torque on gear = 83.2352 Nm (from previous calculations)
Tangential load on gear (WT) =
2�
�??????
=
2�
�.�??????
=
2 ×83.2352 × 10
3
� ×60
=
2.7745 × 10
3
�
N
Pitch line Velocity (V) =
??????�??????�??????
60
=
??????�.�??????×(
??????
??????
3
)
60
=
??????�×60×(
178.125
3
)
60
= 0.1865(m) ms
-1
Velocity factor (Cv) =
3
3+�
=
3
3+0.1865�
Length of pitch cone element (L) =
�??????
2�??????�(????????????)
=
��??????
2�??????�(????????????)
=
�60
2�??????�(71.57)
= 31.62(m) mm
5.1
5.2
21
Assuming that face width is half of L,
Face Width (b) =
�
2
=
31.62�
2
= 15.81(m) mm
From 5.1, 5.2, and 5.3,
WT = (??????
��×C
�) �??????�??????
�(
�−�
�
)
2.7745 × 10
3
�
= 70×(
3
3+0.1865�
)(15.81�)??????�(0.149)(
31.62−15.81
31.62
)
2.7745 × 10
3
�
=
24570.7931�
2
�31.62(3+0.1865�)
3.5705 =
�
2
(3+0.1865�)
0 = m
3
– 0.6659m-10.7115
m = 2.3049 mm and m = -1/1525±1.8218i mm
∴ m = 3
∴ b = 15.85(m)
= 15.81 x 3
= 47.43 mm
Pitch diameters
DG = mTG DG = mTG
= 3 x 20 = 3 x 60
= 60 mm = 180 mm
5.3
Assuming that face width is half of L,
Face Width (b) =
�
2
=
31.62�
2
= 15.81(m) mm
From 5.1, 5.2, and 5.3,
WT = (??????
��×C
�) �??????�??????
�(
�−�
�
)
2.7745 × 10
3
�
= 70×(
3
3+0.1865�
)(15.81�)??????�(0.149)(
31.62−15.81
31.62
)
2.7745 × 10
3
�
=
24570.7931�
2
�31.62(3+0.1865�)
3.5705 =
�
2
(3+0.1865�)
0 = m
3
– 0.6659m-10.7115
m = 2.3049 mm and m = -1/1525±1.8218i mm
∴ m = 3
∴ b = 15.85(m)
= 15.81 x 3
= 47.43 mm
Pitch diameters
DG = mTG DG = mTG
= 3 x 20 = 3 x 60
= 60 mm = 180 mm
5.3
22
6. SHAFT DESIGN
Figure 6.1 – Shaft design
For Gear 1,
TD = 20.8088 Nm (from previous calculations, Section 3)
FD =
��
��
= 416.176 N
WD = 442.8852 N (from previous calculations, Section 4)
Vertical component of WD = WD Cos (20)
= 416.1759 N
Horizontal component of WD = WD Sin (20) figure 6.2 – WD components
= 151.4757 N
For Pulley,
WP = Tension T1 + Tension T2
= 183.46 + 16.5339
= 199.9999
≈200 �
6. SHAFT DESIGN
Figure 6.1 – Shaft design
For Gear 1,
TD = 20.8088 Nm (from previous calculations, Section 3)
FD =
��
��
= 416.176 N
WD = 442.8852 N (from previous calculations, Section 4)
Vertical component of WD = WD Cos (20)
= 416.1759 N
Horizontal component of WD = WD Sin (20) figure 6.2 – WD components
= 151.4757 N
For Pulley,
WP = Tension T1 + Tension T2
= 183.46 + 16.5339
= 199.9999
≈200 �
23
Using Beam Calculations
[6.1]
RAV = 386.971 N
RBV = 229.029 N
RAH = 39.086 N
RBH = 112.914 N
The Resulting Bending Moment for p
t
A and D,
Resultant Bending Moment at A
= √�
��
2
+�
��
2
= 14000 N.mm
Resultant Bending Moment at D
= √�
��
2
+�
��
2
= 11490.7253 N.mm
Maximum bending moment is at D,
Bending Moment at D = 14000 N.mm
Assuming same torque = 20.8088 Nm
Figure 6.3 – Bending moment diagram
Using Beam Calculations
[6.1]
RAV = 386.971 N
RBV = 229.029 N
RAH = 39.086 N
RBH = 112.914 N
The Resulting Bending Moment for p
t
A and D,
Resultant Bending Moment at A
= √�
��
2
+�
��
2
= 14000 N.mm
Resultant Bending Moment at D
= √�
��
2
+�
��
2
= 11490.7253 N.mm
Maximum bending moment is at D,
Bending Moment at D = 14000 N.mm
Assuming same torque = 20.8088 Nm
Figure 6.3 – Bending moment diagram
24
Material selection
Alloy steel
Km = 2 Shear stress (??????) = 0.3 x (Yield stress)
Kt = 1.5 = 147 MNm
-2
Yield stress = 490 MNm
-2
stress (??????) = 0.6 x (Yield stress)
= 294 MNm
-2
Diameter Calculations
Equivalent Twisting moment (Te) = √(�
��)
2
+(�
�×??????)
2
= √(2×14000×10
−3
)
2
+(1.5×20.8088)
2
= 41.9317 Nm
Te =
??????
16
×??????×�
3
41.9317
=
??????
16
×147×10
6
�
3
d = 11.3257 mm
Equivalent Bending moment (Me) =
1
2
[(�
��)+??????
�]
= 14020.9659
Me =
??????
16
×??????×�
3
14020.9659
=
??????
16
×294×10
6
�
3
d = 7.8608 mm
taking the larger of the two values,
d = 11.32 mm
≈12 mm
Material selection
Alloy steel
Km = 2 Shear stress (??????) = 0.3 x (Yield stress)
Kt = 1.5 = 147 MNm
-2
Yield stress = 490 MNm
-2
stress (??????) = 0.6 x (Yield stress)
= 294 MNm
-2
Diameter Calculations
Equivalent Twisting moment (Te) = √(�
��)
2
+(�
�×??????)
2
= √(2×14000×10
−3
)
2
+(1.5×20.8088)
2
= 41.9317 Nm
Te =
??????
16
×??????×�
3
41.9317
=
??????
16
×147×10
6
�
3
d = 11.3257 mm
Equivalent Bending moment (Me) =
1
2
[(�
��)+??????
�]
= 14020.9659
Me =
??????
16
×??????×�
3
14020.9659
=
??????
16
×294×10
6
�
3
d = 7.8608 mm
taking the larger of the two values,
d = 11.32 mm
≈12 mm
25
Figure 6.4 – Shaft design
For Bevel gear (D),
T = 83.2352 Nm
L = 31.62 m
Mean Radius of Gear (Rm) = (�−
�
2
)
�??????
2�
= (31.62−
15.81
2
)
60 ×10
−3
2×31.62
= 2.25 mm
∴ tangential force acting at mean radius (WT) =
�
�??????
=
83.2352
0.0225
= 3699.3422 N
Axial force acting on the bevel gear shaft (WRH) = (WT)tan(ø)Sin(??????
�)
= 3699.3422 x tan(20) x Sin(18.43)
= 425.6744 N
Figure 6.4 – Shaft design
For Bevel gear (D),
T = 83.2352 Nm
L = 31.62 m
Mean Radius of Gear (Rm) = (�−
�
2
)
�??????
2�
= (31.62−
15.81
2
)
60 ×10
−3
2×31.62
= 2.25 mm
∴ tangential force acting at mean radius (WT) =
�
�??????
=
83.2352
0.0225
= 3699.3422 N
Axial force acting on the bevel gear shaft (WRH) = (WT)tan(ø)Sin(??????
�)
= 3699.3422 x tan(20) x Sin(18.43)
= 425.6744 N
26
Radial force acting on bevel gear shaft (WRV) = (WT)tan(ø)Cos(??????
�)
= 3699.3422 x tan(20) x Cos(18.43)
= 1277.3911 N
For Gear (B),
TB = 83.2352 Nm
WD = 442.8852 N (from previous calculations)
WDV = 416.1759 N
WDH = 151.4157 N
Figure 6.5 – Forces acting on gear
Please Turn Over for the Bending Moment calculations.
Radial force acting on bevel gear shaft (WRV) = (WT)tan(ø)Cos(??????
�)
= 3699.3422 x tan(20) x Cos(18.43)
= 1277.3911 N
For Gear (B),
TB = 83.2352 Nm
WD = 442.8852 N (from previous calculations)
WDV = 416.1759 N
WDH = 151.4157 N
Figure 6.5 – Forces acting on gear
Please Turn Over for the Bending Moment calculations.
27
Using Beam Calculations
[6.1]
RAV = 165.29 N
RCV = 1858.286 N
RAH = 45.314 N
RCH = 623.314 N
The Resulting Bending
Moment for p
t
B and C,
Resultant Bending Moment at
B = √�
��
2
+�
��
2
= 7712.4428 N.mm
Resultant Bending Moment at
C = √�
��
2
+�
��
2
= 87502.7551 N.mm
Figure 6.6 – Bending moment diagram
Using Beam Calculations
[6.1]
RAV = 165.29 N
RCV = 1858.286 N
RAH = 45.314 N
RCH = 623.314 N
The Resulting Bending
Moment for p
t
B and C,
Resultant Bending Moment at
B = √�
��
2
+�
��
2
= 7712.4428 N.mm
Resultant Bending Moment at
C = √�
��
2
+�
��
2
= 87502.7551 N.mm
Figure 6.6 – Bending moment diagram
28
Maximum bending moment is at C,
Bending Moment at C = 87502.7551 N.mm
Torque = 83.2352 Nm
Diameter Calculations
Equivalent Twisting moment (Te) = √(�)
2
+(??????)
2
= √(87502×10
−3
)
2
+(83.2352)
2
= 120.7635 x 10
3
N.mm
Te =
??????
16
×??????×�
3
120.7635 x 10
3 =
??????
16
×147×10
6
�
3
d = 16.1135 mm
≈???????????? ????????????
Maximum bending moment is at C,
Bending Moment at C = 87502.7551 N.mm
Torque = 83.2352 Nm
Diameter Calculations
Equivalent Twisting moment (Te) = √(�)
2
+(??????)
2
= √(87502×10
−3
)
2
+(83.2352)
2
= 120.7635 x 10
3
N.mm
Te =
??????
16
×??????×�
3
120.7635 x 10
3 =
??????
16
×147×10
6
�
3
d = 16.1135 mm
≈???????????? ????????????
29
7. KEYWAY DESIGN
Data:
Shaft material = Alloy Steel
Shear stress = 147 MNm
-2
Crushing stress = 294 MNm
-2
Diameter of the shaft (d) = 18 mm
Figure 7.1 – terms in key
From the Machine Design – R. S. Khurmi -Table 13.1 (page No: 472)
Table 7.1 – Key cross section value for Shaft diameter
For 18 mm shaft,
Width (w) = 8 mm
Thickness (t) = 7 mm
7. KEYWAY DESIGN
Data:
Shaft material = Alloy Steel
Shear stress = 147 MNm
-2
Crushing stress = 294 MNm
-2
Diameter of the shaft (d) = 18 mm
Figure 7.1 – terms in key
From the Machine Design – R. S. Khurmi -Table 13.1 (page No: 472)
Table 7.1 – Key cross section value for Shaft diameter
For 18 mm shaft,
Width (w) = 8 mm
Thickness (t) = 7 mm
30
Considering shearing of the key,
T = � ×??????×?????? (
�
2
)
T = � ×147×8 (
18
2
)
T = 10584 (�)
Torque transmitted from shaft (T) =
??????
16
×??????×�
3
=
??????
16
×147×18
3
= 16331.2468 N.mm
∴ 16331.2468 = 10584 (�)
(�) = 15.9043 mm
Considering crushing of the key,
T = � ×??????×
�
2
(
�
2
)
16331.2468 = � ×294×
7
2
(
18
2
)
16331.2468 = 9261 (�)
(�) = 18.1764 mm
∴ shearing (�) < crushing (�)
Length of ley (�) = 18.1764 mm
Considering shearing of the key,
T = � ×??????×?????? (
�
2
)
T = � ×147×8 (
18
2
)
T = 10584 (�)
Torque transmitted from shaft (T) =
??????
16
×??????×�
3
=
??????
16
×147×18
3
= 16331.2468 N.mm
∴ 16331.2468 = 10584 (�)
(�) = 15.9043 mm
Considering crushing of the key,
T = � ×??????×
�
2
(
�
2
)
16331.2468 = � ×294×
7
2
(
18
2
)
16331.2468 = 9261 (�)
(�) = 18.1764 mm
∴ shearing (�) < crushing (�)
Length of ley (�) = 18.1764 mm
31
8. BEARING SELECTION
Data:
Shaft diameter = 18 mm
Shaft speed = 178.9 rpm
Extra heavy Duty = over 16 hrs
Bearing type = Deep Groove Ball bearing
Maximum load (Fr) = 416 + 1277
= 1693 N
From bearing Handbook, (page No: 31)
Equivalent Fr for 1693 N = 2000 N
Equivalent Shaft rpm for 178.9 rpm = 200 rpm
∴ (C/P) = 2.88
L10h = (hours per day) x (weekdays) x (Weeks per year) x (No: of years)
= 16 x 6 x 25 x 2
= 4800 h
C = 4800 x 2.88
= 13824
From bearing Handbook, (page No: 119)
Equivalent Diameter for 18 mm = 20 mm
Equivalent C for 13824 = 15900
∴ D = 52 mm
B = 15 mm
∴ Shaft diameter is altered to 20 mm
8. BEARING SELECTION
Data:
Shaft diameter = 18 mm
Shaft speed = 178.9 rpm
Extra heavy Duty = over 16 hrs
Bearing type = Deep Groove Ball bearing
Maximum load (Fr) = 416 + 1277
= 1693 N
From bearing Handbook, (page No: 31)
Equivalent Fr for 1693 N = 2000 N
Equivalent Shaft rpm for 178.9 rpm = 200 rpm
∴ (C/P) = 2.88
L10h = (hours per day) x (weekdays) x (Weeks per year) x (No: of years)
= 16 x 6 x 25 x 2
= 4800 h
C = 4800 x 2.88
= 13824
From bearing Handbook, (page No: 119)
Equivalent Diameter for 18 mm = 20 mm
Equivalent C for 13824 = 15900
∴ D = 52 mm
B = 15 mm
∴ Shaft diameter is altered to 20 mm
32
Equivalent bending stress (??????
��) =
16
??????�
3
[(�
��)+√(�
��)
2
+(�
�??????)
2
]
=
16
??????20
3
[(1.5×87502)+√(1.5×87502)
2
+(1×83.235)
2
]
= 167.1165 N.mm
-2
∴ ??????
�.���> ??????
��
Torsional stress (??????
��) =
16
??????�
3
√(�
��)
2
+(�
�??????)
2
=
16
??????20
3
√(1.5×87502)
2
+(1×83.235)
2
= 83.5583 N.mm
-2
∴ ??????
�.���> ??????
��
∴ we can conclude that the material and the diameters are safe.
Equivalent bending stress (??????
��) =
16
??????�
3
[(�
��)+√(�
��)
2
+(�
�??????)
2
]
=
16
??????20
3
[(1.5×87502)+√(1.5×87502)
2
+(1×83.235)
2
]
= 167.1165 N.mm
-2
∴ ??????
�.���> ??????
��
Torsional stress (??????
��) =
16
??????�
3
√(�
��)
2
+(�
�??????)
2
=
16
??????20
3
√(1.5×87502)
2
+(1×83.235)
2
= 83.5583 N.mm
-2
∴ ??????
�.���> ??????
��
∴ we can conclude that the material and the diameters are safe.
33
9. CLUTCH
Data:
Torque = 20.8088 Nm
outer diameter of friction surface = d2
inner diameter of friction surface = d1
For uniform wear conditions,
P x r = c (constant)
At inner radius. Intensity of pressure is at maximum.
Pmax x r = c c = 0.3
�2
2
Nmm
-1
Material selection,
Inner
From Machine Design – R. S. Khurmi – Table 24.1 (page No: 887)
Table 9.1 – 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)
9. CLUTCH
Data:
Torque = 20.8088 Nm
outer diameter of friction surface = d2
inner diameter of friction surface = d1
For uniform wear conditions,
P x r = c (constant)
At inner radius. Intensity of pressure is at maximum.
Pmax x r = c c = 0.3
�2
2
Nmm
-1
Material selection,
Inner
From Machine Design – R. S. Khurmi – Table 24.1 (page No: 887)
Table 9.1 – 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)
34
Normal load acting on the friction surface (W) = 2??????�(�
1−�
2)
= 2??????×0.3�
2(1.25�
2−�
2)
= 0.47�
2
2
Mean radius of the friction surface (R) =
�1+�2
2
=
1.25�2+�2
2
= 1.125 �
2
Torque transmitted ( T) = �×�×??????×??????
20.8088 = 2 x 0.4 x 0.47 �
2
2
x 1.125 �
2
0.423 �
2
3
= 20.8088
�
2 = 3.66 mm
�
1 = 1.25 �
2
= 4.575 mm
Normal load acting on the friction surface (W) = 2??????�(�
1−�
2)
= 2??????×0.3�
2(1.25�
2−�
2)
= 0.47�
2
2
Mean radius of the friction surface (R) =
�1+�2
2
=
1.25�2+�2
2
= 1.125 �
2
Torque transmitted ( T) = �×�×??????×??????
20.8088 = 2 x 0.4 x 0.47 �
2
2
x 1.125 �
2
0.423 �
2
3
= 20.8088
�
2 = 3.66 mm
�
1 = 1.25 �
2
= 4.575 mm
35
REFERENCES
1.1 - Screw selected – JHS400
http://www.jh.nl/english/products/2781_jhs400-screw-conveyor
1.2 – Vertical Screw Conveyors Guide selection (Commercial site)
http://www.kwsmfg.com/products/vertical-screw-conveyors.htm
1.3 – Machine Design – R. S. Khurmi
2.1 – Selection of degree of through filling
http://www.kwsmfg.com/services/screw-conveyor-engineering-guide/screw-conveyor-
capacity.htm
2.2 – CAPACITY CALCULATION
https://www.bechtel-wuppertal.com/media/1859/calculations-screw-conveyor-
bechtel.pdf
4.1 – Steel 4340 properties
http://www.astmsteel.com/product/4340-steel-aisi/
4.2 – Steel Hardness conversion table
http://www.steelexpress.co.uk/steel-hardness-conversion.html
6.1 – Online Beam calculator
https://skyciv.com/free-beam-calculator/
REFERENCES
1.1 - Screw selected – JHS400
http://www.jh.nl/english/products/2781_jhs400-screw-conveyor
1.2 – Vertical Screw Conveyors Guide selection (Commercial site)
http://www.kwsmfg.com/products/vertical-screw-conveyors.htm
1.3 – Machine Design – R. S. Khurmi
2.1 – Selection of degree of through filling
http://www.kwsmfg.com/services/screw-conveyor-engineering-guide/screw-conveyor-
capacity.htm
2.2 – CAPACITY CALCULATION
https://www.bechtel-wuppertal.com/media/1859/calculations-screw-conveyor-
bechtel.pdf
4.1 – Steel 4340 properties
http://www.astmsteel.com/product/4340-steel-aisi/
4.2 – Steel Hardness conversion table
http://www.steelexpress.co.uk/steel-hardness-conversion.html
6.1 – Online Beam calculator
https://skyciv.com/free-beam-calculator/
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