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FACULTY OF MECHANICAL AND MANUFACTURING ENGINEERING MECHANICS OF MACHINE GROUP PROJECT TITLE: PROJECT #1: DESIGNING A MECHANISM OF A CONVEYOR SYSTEM

CODE

BDA 20303

COURSE

MECHANICS OF MACHINE (SEMESTER 2 2018/2019)

GROUP MEMBER

1. AARAN HENG JIN HOW (AD170258) 2. CHAI CHUAN YANG (AD170259) 3. CHAN KIAN WUI (AD170220) 4. CHAN YUNG WAI (AD170220) 5. GOH CARMEN (AD170248)

SECTION

7, 8

LECTURER

EN. KHAIRULNIZAM BIN NGADIMON

DATE OF

29 MAY 2019

SUBMISSION

GROUP WORK BREAKDOWN STRUCTURE

1.0 INTRODUCTION OF THE SYSTEM

AARAN HENG JIN HOW

2.0 PROBLEM STATEMENT

CHAI CHUAN YANG

2.1 Outlines and steps taken

CHAN KIAN WUI

2.2 Assumptions

3.0 ANALYSIS AND CALCULATION

GOH CARMEN

3.1 Find the Gripper Force, P and The Distance CHAN YUNG WAI Between 2 Sardine Tins 3.2 Find the power transmission 3.3 Discussion 4.0 CONCLUSION

CHAN YUNG WAI CHAN KIAN WUI

1.0 INTRODUCTION

Nowadays, canned food become more familiar meal that we can see in this modern and competitive world. This is because people nowadays are busy in working or studying and gain some new knowledge. They have spent a lot of time and energy on working and studying. Hence, they will not have a lot of time on cooking their own meal. So, they will try to cook their meal quickly or buy some fast food and outside food. One of the foods that can be cooked quickly is canned food which has already cooked and be pickled inside the can. Hence people can cook the canned food quickly by using microwave or oven. Therefore, canned food become more familiar and common meal that people eat nowadays. Canned food is the food that has been sterilized by heat in a closed, durable container such as tin and aluminium cans, flexible aluminium foil and thermoplastic container including squeeze tubes. Technically, the processes used are highly efficient and used universally. Problems occur rarely but their identification is an important part of food sanitation. Hence, the mechanism of the conveying system is one of the most important part on conveying the canned food. So, that the cans of the canned foods do not break when they are conveyed. Canned food manufacturing factories needed to produce a lot of canned foods so that the quantity and quality products can support the needed in the markets. In order to increase the quality and quantity if the canned food produced, the canned food manufacturing factories needed to have a good machine and mechanism system to run the process of canning the foods. The processing of the canned foods needs a network of conveying systems to efficiency the move of the cans from one position to another position or from one place to another place. For example, the can of the canned food need to be turned upside down so that the expiry date can be printed on the bottom of the can. The machine that can do the process of turning the can upside down has been invented and used in the industry. The machines are hot export stainless steel beverage can rinser, aluminium can rinser and a dumper twist. The machines will be shown at figure below.

Figure 1.1 shows the hot export stainless steel beverage can rinser

Figure 1.2 shows the dumper twist

Most the machines that used to turn or twist the can of canned food are used to rinse, hot and sanitation the cans. While, this technology also can be used to turn or twist the can of canned food upside down in order to print the expiry date on the bottom of the can. So, the dumper twist is used in our design on process of turning or twisting the can. Then, we will analyse the mechanism of the belt conveying system and the twisting process on the design and calculation part.

2.0 PROBLEM STATEMENT

The production line of a sardine manufacturing factory, a conveyor system is used to transfer sardine tins from A to B within speed of 3 cm/s at each conveyor. A mechanism is needed to be used in a production line to turn over sardine tins so that labels can be printed at the tins bottom side. The type of label to be printed is the expiry date of the sardine tin. We have to design a mechanism that can be used to do the job in place. The designing of power transmission system such as gear or belting and its maximum time taken for the sardine tine to be transferred to the next conveyor must be within 15 seconds. Last is the gripping force, given the sardine tin dimensions: 3-inch diameter and 5-inch long. Weight per tin is 250 grams.

Figure 2 shows the design of the mechanism

2.1 Outlines and The Steps Taken To find out the gripping force of the belt, we can find the acceleration through the initial and final velocity of the sardine tin. After applying the Second Newton’s Law which is

F=ma .

We can get the force and find the normal reaction force hence the coefficient of friction. While for the power transmission, we set up the total mass of the belt. Besides, we also set the efficiency of belt roller, and gear head transmission efficiency.

2.2 Assumptions Although the modulus value of conveyor belting is primarily dependent upon the fabric type or fibre used in the belt’s running direction and weaving pattern. In this design, we assume the conveyor belt is inelastic at the no slip condition. Hence, there is no elongation or stretch happened when torsion is occurred. After getting the motor output, we assume a margin of safety in this belt mechanism to get the Torque. Margin of safety we assumed is 2 times. In engineering, the margin of safety is the factor of safety minus one. In design, a factor of safety is normally used for safety related components. With the maximum load, the system allowable torque to carry the load and operate under a normal condition.

3.0 ANALYSIS AND DISCUSSION

3.1 Find the Gripper Force, P and The Distance Between 2 Sardine Tins. Weight of sardine tin = 250g = 0.25kg Initial velocity,

V i = 0 m/s

Final velocity, V f

= 3cm/s = 0.03m/s

Time taken for the sardine to be transferred to the next conveyor, t = 3s P=F s , P=ma F s=μk R N , R N =W = mg

acceleration , a=

change of velocity time taken a=

a=

V f −V i t

0.03−0 3 a=0.01 m/s2

th

2 Newton Law , P= ma

P=0.25 ( 0.01) P=2.5(10−3 )N Normal reaction force , RN =mg RN =0.25 ( 9.81) RN =2.453 N

Since; F s=P=μ k R N −3 2.5 ( 10 ) =μ k ( 2.453 )

μk =1.019 (10−3 ) Hence; F s=μk R N −3 F s=( 1.019 ( 10 ) ) ( 2.453 ) −3

F s=2.5(10 ) N −3

P=F s=2.5 (10 )N ( proven)

Use the open belt drive arrangement, set 2 pulleys having the same diameter 0.03m and the velocity ratio of belt drives, n= 1 and assumptions that the belt used is inelastic, thus the length of the belt passing the driver and driven sheave is the same. Also no slip occurs in the system thus total length of the belt required for the driver and driven sheave is the same.

d 1=diameter of driver sheave

d 2=diameter of driven sheave N 1=Speed of driver sheave

N 2=Speed of driven sheave θ1 ,θ 2=angle of lap 1∧2

x=the distance between2 pullys π d1 N 1=π d 2 N 2

n=

N 2 d1 = =1 N 1 d2

Since the belt is inelastic at the no slip condition that we assumed; The effective speed of driver sheave ,V 1 =The effecive speed of driven sheave , V 2

Given: °

θ1∧θ 2=180

Velocity, V f

= 3cm/s = 0.03m/s

Time taken for the sardine to be transferred to the next conveyor, t = 3s

[

Total Length , T L =2 r 1

[ ()

¿ 2 0.015

( )]

π π +0.27+0.015 2 2

¿ 0.634 m Distance between 2 sardine tins, Distance , s=velocity (time)

¿ 0.03 ( 3 ) ¿ 0.09 m

( π2 ) + x+ r ( π2 ) ] 2

3.2 Find the Power Transmission Total mass of the belt, m b=3 kg Coefficient of the kinetic friction,

−3 μk =1.019 (10 )

Diameter of the pulleys , d 1 ¿ d 2=0.03 m Efficiency of belt roller = 0.95 Gear head transmission efficiency = 0.80 Gear ratio, n = 1 F=μk m

F=( 1.019 ( 10 ) ) ( 3 ) −3

−3 F=3.057 ( 10 )

Load torque , T L =

Fr 2η

3.057 ( 10 ) (0.015) ¿ ¿ T L =¿ −3

−5 T L =2.413 (10 )

Motor output , T M =

T M=

TL n ηG

−5 2.413 ( 10 ) 1 ( 0.80)

T M =3.016 (10−5 ) Assumed margin of safety of 2 times;

( 3.016(10−5 ))(2)=6.033 ( 10−5 ) Torque =T 1 r −5 6.033 ( 10 ) =T 1 ( 0.015) −3 T 1 =4.022( 10 ) Nm

The coefficient friction between the belt and pulley, μk =0.20 T1 T2

=e

T2=

T2=

μθ

T1 e μθ

4.022 (10−3 ) e 0.2(π )

10 (¿¿−3)Nm T 2=2.146 ¿ Max power transmission, P=( T 1−T 2) V −3 −3 P= ( 4.022 ( 10 ) −2.146 ( 10 )) ( 0.03 )

P=5.628(10−5)W 3.3 Discussions In a production line of a sardine manufacturing factory, there have two main components which the mechanism that can be used to transfer sardine tins from the close and seal lid process until the labels printed at the tins bottom side. The design of mechanism is start from after the tin close and seal lid, the tins are moving to the printing process and arranged in upright condition. Then, the sardine tins move along the belt conveyor in a single row to the double helical shaped

railing and the tins are dropped through railing that turned 180 degree upside down, which the helical shaped railing is move from high position to a low position to ensure that the flow of the production line goes smoothly with the help of gravitational force. The incorrect dimension can lead to failure in the mechanism, so dimension of the railing must suit with the dimension of the sardine tins. After that, the belt conveyor transfers the sardine tins to the printing label part with a speed of 3cm/s. There is no distance gap between the two sardine tins. The stamping machine will be activated and prints the Manufacturing date and the Expiry date on the can lids via laser inkjet stamping, when 10 tins arrived at the stamping machine. Then, the grip will be released, and the tins continue to travel in the conveyer for the next processes. From the designed production line, the power transmission system is only applied at the printing systems. In the process of turning over the tins, the double helical shaped railing is used in order to reduce the usage of power which will lead on reducing the total cost of the process. Also, the production system is very convenient as each parts of the system are portable and can be functioned easily. Since this is a short production line, the mechanisms involved are simple and hence the maintenance for the production system will be low. Regarding to the difficulties faced in the designed production line, one of the problems that may occur is low efficiency production. In the double helical shaped railing system, as the tins are dropped one by one, the production line may be slow. The low speed of the conveyor leads to the slow production lines. The second problem is the dimension of the double helical shaped railing. As the dimension of the renter is fixed, the process is only available for certain fixed size of tins. To improve the problems faced in the designed system, the dimension of the double helical shaped railing can be designed to an adjustable dimension in order to suit different sizes of the tins.

4.0 CONCLUSION

In this group project, our task is designing a mechanism that can be used in a production line to turn over sardine tins so that labels can be printed at the tins bottom side. We have successfully design and carried out the analysis. In this project, we found that the dimension of the double helical shaped railing can be designed to an adjustable dimension in order to suit different sizes of the tins. Besides, the speed of the pulley can be increased in order to enhance the efficiency of the production. From this group project, we have learnt about the important role of conveyor system in a production line of sardine manufacturing factory. We also able to design a mechanism that be used in a production line to turn over sardine tins so that the expiry date labels can be printed at the tins bottom size. Besides that, we also have a better understanding on the mechanism of gear and belting system and the calculation of the designing system of mechanism that utilized a gripper. The calculation is theoretically success to stop a sardine tin by a gripper to carry out the printing process. By completing this task, we able to solve different type of problem about mechanics of machine that we have faced with techniques. We will also be able to solve a complex engineering problem in a simpler and easier way as a mechanical engineering students after we have applied the knowledge that we have learned from subject of mechanics of machine. In order to enhance the mechanism efficiency and performance, we can increase the safety margin to 3. Factor of safety is important to avoid the structures and component fail. While the factor of safety cannot be too high, it will be over-design, increase the costing from excessive material usage. We can also change the original material of the conveyor belt to a higher modulus material, this can increase the modulus of the belt hence the total load can be increased without the elongation....