 Open Access
 Total Downloads : 4448
 Authors : A. Mallikarjuna Rao, G S S V Suresh 2, Priyadarshini D
 Paper ID : IJERTV1IS7487
 Volume & Issue : Volume 01, Issue 07 (September 2012)
 Published (First Online): 25092012
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Alternate Design And Optimization Of Conveyor Pulley Using Finite Element Analysis
A. Mallikarjuna Rao1, G S S V Suresh 2, Priyadarshini D3

Sr. Asst. Professor, Department of Mechanical Engineering, V R Siddhartha Engineering College, Vijayawada520 007, A P, India,

Asst. General manager, Department of Mechanical Engineering, National Mineral Development Corporation (NMDC), Khanij Bhavan, Masab Tank, Hyderabad500 173, A P, India,

M.Tech.(CAD/CAM) Scholar, Department of Mechanical Engineering, V.R.Siddhartha Engineering College, A P, India,
.
Abstract
In this project, the main aim is to alter the design of conveyor pulley by changing the dimensions components with respect to the individual components of the design considered.
The main components of conveyor pulley are shaft, drum or shell, diaphragm plates, locking elements, hub, lagging and bearing assemblies. Designing units of this kind requires precise calculations of all belt tensions and loads in static conditions. The solid works analysis is performed on the total assembly of the conveyor pulley considered as a reference for the existing design and even for the altered design which is the main task of this project. The design and modeling is done in parametric software Pro/Engineer and finite element analysis is analyzed in solid works.
Key words: conveyor pulley, design, calculating diameter, altering the design, analysis.

Introduction
Pulley: Conveyor Pulley is used to transmit the motion power to belt and also Pulleys are necessary to change the direction of belt in any direction, and to form endless loop for continuous operation, and it is also used for the material handling system in various
industries to transfer raw material from one place to another.
Components of Pulley:

Drum or shell

Diaphragm plates

Shaft

Locking elements

Hub

Lagging

Bearing assemblies

Drum or Shell: The drum is the portion of the pulley in direct contact with the belt. The shell is fabricated from either a rolled sheet of steel or from hollow steel tubing .shown in fig.1
Fig No :1
Diaphragm Plates: The diaphragm or end disc of a pulley is circular discs which are fabricated from thick steel plate and which are welded into the shell at each end, to strengthen the drum. shown in fig.2
Fig No:2
Shaft: The shaft is designed to accommodate all the applied forces from the belt and / or the drive unit, with minimum deflection. The shaft is located and locked to the hubs of the end discs by means of locking elements. The shaft is supported on both ends by bearings. Shown in fig.3
Fig No:3
Locking Elements: These are highprecision manufactured items which are fitted over the shaft and into the pulley hubs. The locking Elements attach the pulley firmly to the shaft via the end plates. Shown in fig.4
Fig No: 4
Hubs: The hubs are fabricated and machined housings which are welded into the end plates. The hubs are sized according to the size of the pulley. Shown in fig.5
Fig No: 5
Lagging: It is sometimes necessary or desirable to improve the friction between the conveyor belt and the pulley in order to improve the torque that can be transmitted through a drive pulley. Shown in fig.6
Fig No: 6
Bearing assemblies: Bearings support the rotating shaft and hence the pulley. Which
enable the mass of the pulley assembly plus the belt tension forces to be transmitted for the supporting pulley structure. Shown in fig.7
Fig No:7
PULLEY DESIGN:
Design Considerations for Pulleys:
The procedure for selecting pulleys for a conveyor for any given application involves the evaluation of a number of factors pertinent to the installation Consideration should be given to the following:

Application / Environment

Conveyor design

Angle of Wrap:

Belt selection:

Conveyor duty:

Belt Tension:

Belt width:


Standardisation

Specifications

Layout

Pulley design

Dimensions

Accessories

Drive friction

Materials

For the pulley design the main design of the pulley structure is dependent on the shaft design. Hence the shaft design is the initial consideration of the pulley design.
1) Shaft design :
In designing shafts on the basis of strength, the following cases may be considered:

Shafts subjected to twisting moment or torque only.

Shafts subjected to Bending moment only.

Shafts subjected to Combined twisting and bending moments, and

Shafts subjected to axial loads in addition to combined torsional and bending loads.
Material used for the shafts:
Generally at NMDC, according to the Indian Standards the shaft is designed and carbon steel of grade 40C8 is used.
For the design of the shafts the materials chosen for this thesis purpose are chosen according to the Indian standards are:
1. 37 Mn 2
2. 40 Cr 1 Mo 28
3. 40 Ni 2 Cr 1 Mo 28 4. 35 Ni 1Cr 60
5. 40 C 10 S 18
6. 40 C 15 S 12
Calculating the diameter of the shaft for 40Ni 2Cr1Mo28 material
Conveyor Input Values:
Horizontal length of conveyor belt = 69.919m Lift = 5.5 m
Belt width = 1.2 m Capacity = 2400 TPH
Bulk density of material = 2.4 tonnes/m3 Belt speed = 2m/s
Carrier idler spacing = 1.2 m Return idler spacing = 3m
Wt of one set of carrier idler = 27 kg Wt of one set of return idler = 24 kg Weight of belt per m =27 kg Coefficient of friction, m or f = 0.023 Number of scrapers =3
Length correction factor =1.3
Diameter of pulley
Speed at pulley = 60 rpm
We know that, D = 0.636 m Therefore, Diameter of the pulley, D= 0.7 (or) 700 mm
Radius of pulley, R =0.35m
Power calculation:
To find out the power required to drive the pulley; P=P1+ P2+ P3+PT
Where, P is the total power required
P1 is the power required to move empty belt P2 is the power required to move material horizontally
P3 is the power required to lift the material PT is the power required for accessories Step 1
Calculate the power required to move empty belt: Where, C is
the correction factor
f is the coefficient of friction
l is the horizontal length in m
MC is approximate wt of one set of rotating part of carrying idler in kgf
MR is approximate wt of one set of rotating part of return idler in kgf
S1 is carrying idler spacing S2 is return idler spacing
Mb is approximate wt of belt per m in Kg/m WP is wt of non drive pulley per m in Kg/m V is belt speed in m/s
Now substituting the values to find P1
Weight of non drive pulley per m = 30.25 kg
Angle of wrap = 2100 Coefficient of = 0.4
Weight of drive pulley = 1000 kg
P1=
Step 2
= 4.7KW
Face width of pulley = 1.4 m Length of skirt board = 1.5 m Speed of motor = 1500rpm Gear ratio = 1:25
Shaft material
Allowable bending stress = 1200 kg/cm2 Allowable shear stress = 700 kg/cm2 Service factor (Bending) = 2
Service factor (Torsion) =1.5
Calculate the power required to move the material horizontally. P2=
Where, Q is the design capacity in TPH Now, substituting the values to find P2 =13.671KW
Step 3
Calculate the power required to lift the material. Where; H is lift in m
Now, substituting the values, to find P3.
P3= = 35.967KW
Step 4
Calculate power required for accessories.
PT = Pg + Ps, Where,
Pg = [LSK – (BW + v + 0.7)] v * 0.5
Ps = BW * v * 0.3 * NSP
Where, LSK is length of skirt board in m BW is the belt width
NSP is number of scrapers
Now, substituting the values to find Ps, Pg and finally PT.
Ps = 1.2 * 2 *0.3 * 3 = 2.16
Pg = [15 + (1.2 + 2 + 0.7)] 2 * 0.05 = 1.41 PT = Pg + Ps = 2.16 + 1.41 = 3.27 KW
Total power required,
P = P1 + P2 + P3 + PT
=4.7 + 35.967 + 13.671 + 3.27
P = 57.6 KW
Torque Calculations
Power, P = Speed, N = N = 54.55 rpm.
Now, Power P Torque Tr
Tr = 10083.239 Nm
To, find equivalent tension (Te), Te * R = Tr , Te
Te = 28809.25 N
We know that,
T1 + T2 = 28809.25 . . . . . . . (1)
Also,
Where, is coefficient of friction = 0.4
is angle of wrap = 210 0 = rad.
, 4.265 . .. . (2)
From (1) and (2), we get
(4.265 T2) – T2 = 28809.25, T2 =
T2 = 8823.66 N
T1 = (8823.66)4.265, T1 = 37832.91 N.
Pulley Diagram:
Fig No:8
Free Body Diagram:
Fig No :9 Resolved Forces: Bending moment:
Fig No:9
Determining the Diameter of the Shaft: Materials used for shaft is EN9/BS: 970 (or) equivalent forged quality.
Case 1: Considering only Torque Kt * Tr = * * d3
1.5* 10083.239 = * 700 * 9.81* 104 * d3
d = 0.112 m (or) 112.6 mm
Case 2: Considering only bending moment Km * M = * b * d3
2 * 6540.82 = 104 * d3
d= 0.114m (or) 114.7mm
Case 3: Considering both torque and bending moment: (i) Equivalent Torque, re
=
Tre = 22426.57 Nm, Now,
Tre = * d3
Fig No: 10
Tan = , Tan ) = y = 0.615 tan (0.1) = 1.07 mm.
ymax (or) yact = [3L2 4a2]
Where, P is force acting on shaft = 23360 N a is distance btwn bearing and hub =0.28 m L is the length of the shaft between two bearings= 1.79 m
E is youngs modulus = 2.05 * 1011 N/m2
I is area moment of inertia of the shaft =
= 1.88 * 10 5 m4
22426.57 = * 550 * 9.81* 104
d = 0.95 m (or) 95mm
(ii) Equivalent Bending moment, Me = [(Km * M) + Tre ]
* d3
yact =
= 7.07 * 10 5 (9.2987), = 0.8296 mm
Deflection of the designed shaft is less than the maximum limit. Hence, the design is safe. Tensional Deflection:
Given torsional deflection = 0.26 deg/m
= [(2 * 5440.82) + 22426.58]
Me = 16118.9 Nm
Now, Me = * b * d3
d= ,
d = 0.134 m (or) 134 mm
At equivalent torque, diameter is maximum. Hence, diameter of shaft is d = 124 mm.
Therefore, d = 140 mm. (with reference to preferred numbers.
Case 4: Deflection based diameter Angular deflection:
Given angular deflection = 6 minutes at hub.
Torsional deflection = 0.26 *L
= 0.26 * 1.79 = 0.465 deg.
We know that, =
Where, is the torque = 10083.29 Nm L = 1.79 m
C is rigidity modulus = 0.84 * 1011 N/m2
J is moment of inertia = = 3.7 * 10 5 m 4
= =
= 5.8 * 10 3 rad = 0.33 deg Deflection of the designed shaft is less than
the maximum limit. Hence, the design is safe.
Determining the diameter of hub:
Material used for hub is St.41 W, IS: 2062 1992
We know that, Hub diameter (DH) is equal to
1.6 times of shaft diameter.
DH = 1.6 * d = 1.6 = 0.224 m
To check the design of hub:
Consider hub is made as a hallow material as that of shaft.
Lets us consider hub as a hallow shaft .we know that,
Maximum torque, =
10083.23 =
= 5.391 * 10 6 N/m2.
Allowable shear stress of the material = 500 kg/cm2 = 4.9 * 10 6 N/m2.
Shear stress of the designed hub is less than the allowable shear stress of the material. Hence, the design is safe.
Determining the Thick Ness of End Disc (or) Diaphragm:
Consider diaphragm as a column of length, I and square cross section on which resultant force is acting. Material used for diaphragm is MS as per IS: 226
The procedure to calculate the thickness of the shell is very complicated. Research is being carried out to find out an easier and accurate way to calculating the thickness. The thickness found out from mathematical calculation is very less, which cannot be fabricated. Hence, the thickness is assumed based on the practical application (12 mm in our case).Material used for the shell is St .42, IS: 20621992
Mass of the shell:
Density of the shell = 0.078 *106 N/m3 Area of shell = 2* *R *face width.
Diam eter of the Shaft 
Material consider ed for the shaft 
Stress (N/mm2) or(MPa) 
Strain 
Displacement (mm) 

Min 
Ma x 
M in 
Max 
M in 
Max 

40Ni2Cr 
0 
352. 807 N/m m2 (MP a) 
0.0009 24929 
0 m m 

1Mo28 
N/m 
1.24316 

140 
m2 
0 
e+008 

Diam 
40 Cr 1 
(MP 
mm 

eter 
Mo 28 
a) 

150 Diam eter 
35 Ni 1Cr 60 37 Mn 2 
0 N/m m2 (MP 
455. 759 N/m m2 (MP 
0 
0.0011 5619 
0 m m 
1.32087 e+008 mm 
Volume = area = thickness = 0.073 m3
T1 =37832.91 N, T2 = 8823.66 N
W= mg =581.02 * 9.8 = 5694 N
R= R = 45550.95 N
Now, B.S =
Where, B.S = bending stress = 300 MN/m2
b is disc constant = 1.22 P is resultant load = R/2 = 22775.47 N
= 2324.02 kg
B is distance between two bearings = 179 cm L is distance between two discs =123 cm
A is disc constant = 4.9 t is thickness of disc
D is outer diameter of the disc =67.6 cm On substituting,
t = = 0.36 cm = 3.6 mm Hence, thickness of end disc plate is 4mm.
Similarly the diameter of the shafts for: 1. 37 Mn 2
2. 40 Cr 1 Mo 28
3. 35 Ni 1Cr 60
4. 40 C 10 S 18
5. 40 C 15 S 12 determined as 150, 140,150,160,160 respectively.
Analysis values of the conveyor pulley stress,strain and displacement values: :
Mass = density * volume = 0.078 * 106 *
0.073 =581.03 kg
Forces Acting on Diaphragm:
To find resultant acting on diaphragm:
a) 
a) 

40 C 10 
0 
459. 38 N/m m2 (MP a) 
0.0012 1743 
0 m m 

S 18 
N/m 
1.28931 

160 
m2 
e+008 

Diam 
40 C 15 
(MP 
0 
mm 

eter 
S 12 
a) 
Table No:1
Failures analysis of the conveyor puleey with different diameters :
Dia mete r 
Damage percentag e 
Load factor 
Life cycle(cycl s) 

M in 
Max 
Min 
Max 
Min 
Ma x 

140 
0. 1 
0.1 
4.44 262 
8.40989 e+028 
1e+ 006 
1e+ 006 
150 
0. 1 
0.34 8904 
3.45 074 
8.83039 e+0.28 
286 612 
1e+ 006 
160 
0. 1 
0.37 6359 
3.41 691 
8.83039 e+028 
265 704 
1e+ 006 
Table No:2
160 diameter stress vs strain graphs :
160 Diameter Stress Vs Displacement graphs :
160 failure analysis
150 diameter Stress Vs Strain Graphs
150 diameter Stress Vs Displacement graphs
150 failure analysis
140 diameter Stress Vs Strain Graphs
140 Stress Vs Displacement Graphs
140 failure analysis
Analysing the analysis of the conveyor pulleys with different diameter a new alter design of the conveyor pulley of 160 diameter is designed for the better improvement in the design standards and techniques:
The altered design results for the 160 diameter is the maximum stress that are in 160 diameter is 523.151 N/mm2 or (MPa) and the strain obtained in the conveyor pulley is 0.00132252 and displacement that can be obtained in the total assembly is 1.7271e+008 mm.
160 diameters Stress Vs Strain
160 diamter Stress Vs dispalcement
And in the failure analysis for the altered design of the conveyor pulley the damage percentage is max 0.689413 and the load factor that can be applied for the pulley is max of 8.83039e+028 and the life cycle of the pulley is 1e+006 cycles.
Failure analysis for alter design :
Conclusion:
According to the Indian Standards the materials chosen for design of conveyor pulley followed by the total assembly of the conveyor pulley is safe and can be used for the general conveyor purposes. The alter design of the conveyor pulley without shaft is also proved as a safe design and can be used for the conveyor purpose. Hence, chosen materials are safe for the design and working purposes.
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International Journal of Engineering Research & Technology (IJERT)
ISSN: 22780181
Vol. 1 Issue 7, September – 2012