Design of Transmission System with Parametric CAD Implementation

This project presents a comprehensive redesign of an existing power transmission system, encompassing detailed calculations and parametric CAD implementation. The assignment begins with identifying the system's components, including the shaft, pulleys, bearings, and keyways, followed by calculating the power transmitted by the existing system and determining the required power for the redesigned system. The core of the project involves selecting a grooved pulley to replace the flat pulley and designing a V-belt system to replace the flat belt pulley. This includes determining the design power, selecting the appropriate belt type and cross-section, and choosing pulley diameters. The design further extends to the shaft, involving stress, fatigue, and deflection analysis, along with the selection of keys, keyways, and bearings. Section B focuses on the CAD implementation, providing 3D models of the complete assembly and individual parts, along with 2D drawings. The project concludes with a discussion of the design choices and recommendations, along with a conclusion and references.

Design of a Transmission System with Parametric CAD Implementation
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[DATE]
[Company name]
[Company address]

Contents
Section A: Complete System Design Calculations..........................................................................3
A. Identification of Parts:.............................................................................................................3
B. Calculation of power:..............................................................................................................6
C. Selection of a Grooved pulley to replace existing Flat pulley:.............................................11
D. Selection of V belt system to replace the Existing Flat belt pulley AB:...............................16
E. Design of Shaft:.....................................................................................................................22
F. Fatigue Analysis of shaft.......................................................................................................32
G. Design of shaft deflection:....................................................................................................38
h. Keys and Keyways selection:................................................................................................41
I. Selection of bearing:...........................................................................................................43
Section B: CAD implementation:..................................................................................................46
1. 3D CAD models:....................................................................................................................46
1.1 Complete Assembly:........................................................................................................46
Parts............................................................................................................................................47
2. 2D Drawings..........................................................................................................................51
Section C:.......................................................................................................................................57
Discussion and Recommendations:...............................................................................................57
Conclusion:....................................................................................................................................57
References......................................................................................................................................58
Appendix A:...................................................................................................................................59
Appendix B:...................................................................................................................................61

Section A: Complete System Design Calculations
A. Identification of Parts:
In this topic the various parts associated with the system which was examined will be explained
briefly with their operations. The schematic layout of the given system is given at the figure1.
The various parts of the system will be initially identified and then explained briefly.
Figure1. Image illustrates the schematic of the given system.
The various parts present in the system are:
1. A Shaft of 850mm Length.
2. Pulley AB of 250 mm diameter.
3. Pulley CD of 300 mm Diameter.
4. Assembly of two Bearings with housing at the both ends.
5. Keyways
6. Screws.
Part1: Shaft:
The Major part of the system is the shaft. A shaft is a rotating machine element which is used to
transmit motion, torque and power. The shafts are of circular cross section in general. The shaft
plays a key role in power transmission such as gearbox, propeller shaft, wheel axle shafts, etc.
The shafts should be designed for reliability, strength and life. It should have good load bearing
capacity and also the vibrational characteristics must be considered. The shaft designed should

be targeted for low mass and high strength. Material for the shafts are generally low carbon steel,
hot rolled steel, cold drawn steel, etc. (Whitman, M.P. 2015)
Part2: Pulleys
The system consists of two pulleys one of diameter 250 mm and another pulley of 300mm
diameter which are used to receive and transmit power from the source to the final system. One
of the pulley among them is driving pulley another one is driven. A pulley is a mechanical
structure which consists of a wheel attached to a shaft, it is rotated by a belt assembly which is
mounted at the top of the pulley wheel. There are many types of pulleys such as Flat belt pulley,
V-groove pulley, double V-groove pulley, etc. The pulleys are to be selected on the basis of the
power transmission requirements, torque loads acting, type of applications and the reliability
requirements of the system. (Antchak, J.R., et.al 2018)
Part3: Bearings with housing:
A bearing is a machine part which act as a support to the parts in motion and also helps to
confine the motion of the part. The bearing is utilized in almost all machine parts such as
automobiles, pumps, generators, turbines, etc. The bearings are selected on the basis of the
design of the shaft and the service factor requirements. The bearings are of different types. They
can be classified as follows:
1. Radial bearings
2. Thrust bearings
1. Radial bearings:
Figure2. Image illustrates the various parts of a radial bearing assembly.

The radial bearings are also called as sliding contact bearings, they can be used in the application
where the loads are acting along the circumferential direction along the bearings. The figure2
illustrates a typical radial bearing.
They are further classified into four types as:
a) Zero film bearing.
b) Thin film.
c) Thick film.
d) Externally pressurized.
2. Thrust bearings:
Thrust bearings are used in places where axial thrust force acts on the bearing such as office
chairs and other machine components.
Figure3. Illustrates the thrust bearing exploded view.
Bearings are mounted on a supporting house which is used to confine the bearing and the shaft
attached to the bearing. The whole assembly is bolted or attached to a foundation.

Figure4. Image illustrates a typical bearing confined on a housing.
B. Calculation of power:
Power to be transmitted:
The layout diagram of the existing system si provided the data available from the given system
are the geometric properties of the shaft, pulleys, the distance between two pulleys, etc. The
power transmitted is not provided, so the power transmitted by the exsiting system will be
calculated. Then the power flow required by the system to be redesigned will be calculated by
finding the value of the power that is 15 times greater than that of the calculated power of the
existing system. As the power requirement is given as 15 times the proposed system. (Gonen, T.,
2015)
Power transmitted by the existing system:
The power is given by the equation:
P = 2 πNT/ 60 in Watts
Where,
N is the speed in RPM

T is the Toque
Now,
The toque of the existing system can be calculated as:
Torque, T = π
16 ƫ d3
Where,
ƫ is the shear stress in N/m2
d is the diameter of the shaft in m.
T is the toque in N.m
The power of the existing system can be calculated by finding the power transmitted by the
pulley on the existing system as the pulley is mounted on the shaft only.
The power transmitted by the Pulley CD can be calculated as:
P = (T𝛚)
Where,
P is the power transmitted.
T is the Torque transmitted.
𝛚 is the Angular acceleration.
The torque T can be calculated using the relation between the forces acting
on the tight and slack side of the pulley. The forces are acting on the side F1
and F2 for the existing system is given as:
T = (F1-F2) r
Where,
F1 is the Force acting on the tight side.
F2 is the force acting on the slack side.
r is the radius of the pulley.

From the diagram provided in the Appendix A the force transmitted F1 and F2 are taken as:
Force on tight side, F1: 270 N
Force on slack side, F2: 50 N
Radius: r: 0.15 m
Now,
T = (270-50)(0.15)
T = 33 N.m = 24.339 Ft-Ib
The total amount of power transmitted by the pulley CD is:
From the technical specifications the angular velocity of the Shaft: 1500 Rpm: 157 Rad/sec
P = 33(157) = 5181 W = 5.181 KW
P = 7.0441996 PS
The power transmitted by the pulley CD is found as 7.0441996 PS.
The forces acting on the P1 and P2 sides of the Pulley AB when considering the frictional
losses of 10 percent. Can be calculated as follows:
Considering 7.0441996 PS of Power is transmitted from the Pulley CD of diameter 300 mm.
Power loss due to friction: 10 percent
Shaft rotation: 157 Rad/sec
Angle of the pulley: 2A = 450
Diameter of pulley CD: 300 mm
Diameter of the pulley AB: 250 mm
Radial difference between the diameters of the pulleys
Power transmitted by the Pulley AB = (7.0441996) (0.9) [Considering 10 percentage
power loss]
Power, P = 6.34 Ps

Now,
P = (T𝛚)
Where,
P is the power transmitted.
T is the Torque transmitted.
𝛚 is the Angular acceleration.
4.66 = T (157)
Torque, T = 4662/157 = 29.7 N.m = 21.90 Ft-Ibs
The torque transmitted by the pulley AB is found as 21.90 Ft-Ibs.
Now,
T = (P1−¿P2) r
(P1−¿P2) = T/r
(P1−¿P2) = 29.7/ 0.125 = 237.6 N.m
Now,
P1 / P2 = [e^(μθ)] / Sin (A)
Where,
P1 and P2 are the forces acting on tight and slack sides.
θ - Angle of wrap on P1 is calculated as:
θ= 180 - 2A = 180- 45 = 1350 = 2.356 Rad/s
μ Coefficient of friction: 0.4 (Assumed)
We have,
(P1−¿P2) = 237.6 N.m
P1 = 237.6 +¿ P2 N.m
Now,

P 1
P 2 =eμθ ¿ ¿
sin A =6.705
P2 = (P1) / 6.705
P2 = (237.6 +¿P2) / 6.705
6.705 P2 = 237.6 + P2
6.705 P2 –P2 = 237.6
5.705 P2 = 237.6
P2 = 41.6476 N
Substituting the value of P2 in equation (P1−¿P2) = 237.6 N.m
We get,
P1 = 279.24 N (or) 28.47 Kgf
The force P1 is calculated as 279.24 N (or) 28.47 KgF
The force P2 is calculated as 41.676 N (or) 4.249 KgF
Flow chart1. Illustrates the power flow diagram.
Motor Pulley
CD Pulley
AB Output
10% Power
loss

C. Selection of a Grooved pulley to replace existing Flat pulley:
Now the grooved pulley which will replace the existing 300 mm diameter flat belt pulley will be
selected.
The pulley will be selected based on the power transmission criteria of the shaft. The new system
design should be designed to carry 15 times more power than that of the existing system. The
Power transmission requirements of the pulley are calculated as (Tsukamoto, M et.al 2015):
Power required to transmitted by the pulley:
P=15 X ( Power transmitted by pulley CD)
PCD = 7.044 PS
P = 15 X (7.044)
P = 105.66 PS
The flat belt pulley CD is of 300 mm diameter. The power transmitted by the pulley is found as
105.66 PS
The Selection procedure to be followed for choosing a V belt drive system is as follows
(Dalsania, B.,et.al 2015):
Step1: Determine the Design power
a) Calculate service factor
b) Find design power
Step2: Select type of belt
Step3: Select belt cross section
Step4: Select pulley diameter
Step5: Selection of center distance
Step6: Belt length determination
Step1: The design power. The system need to transmit 15 times of the power transmitted by the
existing system. The design power is calculated by multiplying the rated power with service
factor. The service factor table is given below:

The system is about to run 16 hours per day. The service factor is taken as 1.2 and the system is
designed to withstand 15X higher power than the existing system (Pawar, A.R., et.al 2016).
Design power: (105.66) X 1.2 = 126.7 PS (or) 94 KW
Step2: From the table the Type of belt required is selected based on the power and the Rpm.
The minimum belt diameter required for the V groove belt can be found with the help of the
table given below:
Power: 94KW
Rpm: 1500