Design Final 2.pdf

UNIVERSITY MALAYSIA SARAWAK FACULTY OF ENGINEERING DEPARTMENT OF MECHANICAL AND MANUFACTURING ENGINEERING

KNP 3493: ENGINEERING DESIGN 2 GROUP: KREATECH INC. TITLE: HIGH SPEED ROTATING DRUM DRYER (Hi-RDD) NAME & MATRIC NO

LECTURER

LAI SUK NA

47241

LEW HAO XIAN

47332

MOHD BADRI BIN BAKRI

45685

MUHAMMAD SHUKRI BIN AHMAD TARMEZIE

45745

NUR FIDA’IY ILYASA BINTI MOHD JOHAN

45871

DR ANA SAKURA ZAINAL ABIDIN DR ABANG MOHAMMAD NIZAM ABANG KAMARUDDIN

TABLE OF CONTENTS INTRODUCTION................................................................................................................................. 1 EXECUTIVE SUMMARY .................................................................................................................. 2 DESIGN SPECIFICATION ................................................................................................................ 3 CONCEPT GENERATION ................................................................................................................. 4 TRIZ MATRIX ..................................................................................................................................... 5 CONCEPTUAL SOLUTION ............................................................................................................ 13 CONFIGURATION DESIGN ........................................................................................................... 19 PARAMETRIC DESIGN ................................................................................................................... 20 DETAIL DESIGN ............................................................................................................................... 27 CONCLUSION ................................................................................................................................... 31 REFERENCES ...................................................................................................................................... A PROJECT LOG BOOK AND GRANTT CHART ....................................................................... BBB

INTRODUCTION Drum drying is a method used for drying out liquids from raw materials with drying drum. They were developed in early 1900s and used in drying almost all liquid food materials before spray drying came into use. The common rotary dryer cylinder is tilted slightly compared with horizontal. Materials and hot flue gas goes into the rotary dryer cilinder from the higher end. During the rotation of the cylinder, the material by gravity goes into the lower end. When the wet material is in the process of moving forward in the rotary dryer cylinder body, the lifting plates inside the rotary dryer cylinder makes the material up and down to contact with the hot flue gas completely. Thus, the moisture inside the wet material is evaporated into water vapour, and finally we get the dried materials. In the drum-drying process, pureed raw ingredients are dried at relatively low temperatures over rotating, high-capacity drums that produce sheets of drum-dried product. This product is milled to a finished flake or powder form. Modern drum drying techniques results in dried ingredients which reconstitute immediately and retain much of their original flavor, color and nutritional value. In a drying operation, liquid, slurry, or puree material is applied as a thin layer onto the outer surface of revolving drums that are internally heated by steam. After about three-quarters of a revolution from the point of feeding, the product is dried and removed with a static scraper. The dried product is then ground into flakes or powder. Drum drying is one of the most energy efficient drying methods and is particularly effective for drying high viscous liquid or pureed foods. Some advantages of drum drying include the ability of drum dryers to dry viscous foods which cannot be easily dried with other methods. Drum dryers can be clean and hygienic and easy to operate and maintain. Other products where drum drying can be used are, for example, starches, breakfast cereals, baby food, instant mashed potatoes to make them cold-watersoluble. This project requires the built of a design for a high speed rotating drum dryer. In this report, all of the specific analysis regarding to the design of the product will be shown. By analysing certain features for each of the parts, the most suitable design can be obtained.

1

EXECUTIVE SUMMARY Food drying is a method of food preservation in which food is dried. Dehydration has been used widely for this purpose since ancient times; the earliest known practice is 12,000 B.C. by inhabitants of the modern Middle East and Asia regions. The food drying industry is already a widely known industry throughout the globe. Our product is the best for the simple, light, small and save energy food dryer as we replace heat usage with the high rotation of the drum. The High Speed Rotating Drum Dryer, which fits to dry any food which in granular form about 5mm diameter suitable for any variety of food industry such as the paddy rice, soy beans, salt, dry yeast, barley and wheat processing industry. This product gives simplicity in handling, portable and little power usage. No need to buy the other over expensive product for the same course. Behind the small size, it can handle maximum of 30 kg weight of particles at once with 8 hours per session. Our implementation plan will be carried out in two parts: prepare for manufacturing, and marketing and selling the product. In preparation for manufacturing we plan to hire cheap labor and train them to make a fully assemble of all of each parts, and purchase wholesale components in bulk to cut down on costs. In marketing and selling the product we plan to hire knowledgeable salespeople. To promote our product we will team up with companies selling complimentary products, and involve the company in the food drying or any entrepreneur events. The total estimated cost for the installation and fabrication of High Speed Rotating Drum Dryer is RM1950-RM2000, not including markup. The High Speed Rotating Drum Dryer (Hi-RDD) is one of the food dryer modification in market eager for intentive products and upgrades. We at KreaTech Inc. believe that High Speed Rotating Drum Dryer (Hi-RDD) will create more wave in food drying industry. With our proposal we offer you the chance to join an industry with a strong trend of growth.

2

DESIGN SPECIFICATION Title : Design a High Speed Rotating Drum Dryer (Hi-RDD) Objective : The machine is expected to dry granular sized particles about 5mm diameter and by the given dimension and specification, it be able to accommodate maximum 30 kg weight of particle at once. Engineering Design Specification: High Speed

: 6500 rpm

Long Working Hours

: 8 hours/session

Dimension

: 0.6m (diameter) x 0.6m (length)

Economical

: Low cost minimize the food cost

Efficient

: Minimum loss of power reduce power usage

Safety

: Comply food standard with safer, fresher quality

Type of dryer

: Batch dryer either vertical or horizontal allignment

Operating food Sizing

: Granular-sized food of 5 mm diameter

Drum Capacity

: Accommodation of 30 kg granular food at once

Competition: There are several similar products in the market. The competition range probability is quite big as in the market they used similar or more advanced technologies such as hot air rotary drying food and food dehydrator dryer. The price of the available dryer in the market eventually quite expensive around $500 to $1000 which it gives benefit to our product that can be said as much cheaper with good performances. In the term of technologies that being used in food rotary dryer that available in the market a lot of machine that used spray dryer which the material to be dried is suspended in air, i.e. the liquid is converted into a fog-like mist (atomised), providing a large surface area. The moisture evaporates very quickly so that this process is very useful for materials that are damaged by exposure to heat for any appreciable length of time. Other than that, they also used steam dryer which, is a special drier design that uses superheated steam produced via a heat-exchanger. The drier consists of a pressure vessel in which the water from the product is driven off, turned into steam and then used to dry more products.

Relationship with existing products line: 3

This is a product with new type of technologies that being used which is with the help of resonance. The as the vibration reach the natural frequency of the water molecules, they will vibrate at their maximum amplitude and hence promoted into excited state. Actually this kind of technologies have been explored by certain manufacturer but still not being highlight because it is still new and need to be exposed by years. CONCEPT GENERATION A design concept is an idea that is sufficient developed to evaluate the physical principle that governs its behavior. Throughout a series of discussion, research and meeting, we have map the concept to get a suitable design to fulfill the requirement of project needed. As the result, we have come out with a suitable design of new food rotary drum dryer that enable to solve the problems may occurs.. Below are some of the steps that we had discussed in the concept generation and evaluation process within our group member.

Figure 1: Concept Generation Schematic Diagram

4

Figure 2 : Brainstorming Mind Map. TRIZ MATRIX Worsening Feature 1

9

13

14

17

19

21

Weight

Speed

Stability

Strength

Temperature

Use

of Power

Loss

of

of

of

energy

by consumption

moisture

Rotating

Drum

Drum

the

drum dryer

23

36 of Device complexity

of motor

Drum

5

1

-

Weight of

2, 8

1, 35

28, 27

6, 29

35,12

12, 36,

5, 35

26, 30

15, 38

19, 39

18, 40

14, 38

34, 31

18, 31

3, 31

36, 34

-

28, 33

8, 3

28, 30

8, 15

19, 35

10, 13

10, 28

1, 18

26, 14

36, 2

35, 38

38, 2

28, 38

4, 34

-

17, 9

35, 1

13, 19

32, 35

2, 14

2, 35

15

32

27, 31

30, 40

22, 26

-

30, 10

19, 35

10, 26

35, 28

2, 13

40

10

35, 28

31, 40

25, 28

-

19, 15

2, 14

21, 36

2, 17

3, 17

17, 25

29, 31

16

-

6, 19

35, 24

2, 28

Rotating Drum 9

2, 28

Speed of Drum 13 Stability of

the

13, 38

21, 35

33, 15

2, 39

28, 18

1, 18

8, 13

13, 17

40, 15

26, 14

35

36, 22

2, 28

1, 35

10, 30

6, 38

36, 30

32

22, 40

12, 18

8, 35

19, 13

5, 19

Drum 14 Strength

17 Temper ature 19

19, 24

6

Use

of 28, 31

17, 24

9, 35

3, 14

37, 18

18, 5

27, 28

-

28, 27

20, 19

18, 38

30, 34

-

35, 10

energy by drum

Improving features

dryer 21 Power consum

8, 36

15, 35

35, 32

26, 10

2, 14

16, 6

38, 31

2

15, 3

28

17, 25

19, 37

35, 6

10, 13

2, 14

35, 28

21, 36

35, 18

28, 27

23, 40

28, 38

30, 40

31, 40

39, 31

24, 5

18, 38

26, 30

34, 10

1, 22

2, 13

2, 17

27, 2

20, 19

35, 10

34, 36

28

17, 19

28

13

29, 28

30, 34

28, 29

ption of motor 23 Loss of moisture 36 Device complex

28, 24

-

ity Table 1: TRIZ Contradiction Matrix

7

Statement 1: The weight of the rotating drum should be as light as possible to reduce the inertia of fast spinning speed, but at the same time it should possess high strength in order to hold the 30kg weight of particles to be dehydrated. Step 1: Identify the contradictions Improving feature: Strength Worsening Feature: Weight of rotating drum Step 2: Look at the list of features and identify those important to your contradiction. Strength

14

Weight

1

Step 3 Identify the improving and worsening features from triz matrix above It is found that the features intersect at fourth row and first column. The Altshuller’s Principles involved are 1. Segmentation 8. Anti-Weight 40. Composite Materials 15. Dynamics Step 4: Brainstorming: Principle 1: Segmentation The rotating drum can be designed as an independent part in order to reduce restriction to its motion. Principle 8: Anti-weight The weight of the drum can be compensated by adapting the shape of drum to environment. Moving object always subjected to drag force. In order to reduce the resistance, the shape of the drum can be designed to be streamlined so as to reduce the coefficient of drag. This can be prove by figure below showing the effect of different shape on the drag coefficient

8

Figure 3: Drag Coefficient of Different Shape. (Adapted from The A2A Simulations, 2015) It can be seen that drag coefficient of circular cylinder (Cx= 1.2) and semicircular(Cx = 1.1) are the two best option for design of drum. Principle 40: Composite materials Composite materials not only has the property of light weight, it also possesses relatively high strength. Principle 15: Dynamics According to Triz40 website, under principle of dynamics require the design to change to be optimal or to find an optimal operating condition. From principle of anti-weight, we know that both semicircular and circular cylinder offer lesser drag coefficient. However, circular cylinder has a fix axis of rotation which provide constant centrifugal force. This ensure the drum can rotate at constant rate at magnitude of 𝜔2R. On the other hand, semicircle does not have fix axis of rotation as the R vary, hence it cannot provide constant angular acceleration.( Kalda et al., 2014) Step 5: Conclusion The performance of drum dryer can be improved by designing a circular cylindrical drum with fixed axis of rotation, composed of composite material and independent part supported with frames. Statement 2:

9

The speed of rotating drum should maintain at high angular velocity, that is 6500rpm. However, the temperature of the rotating drum should be maintained low throughout the processing period, which operate for 8 hours per session, in order to prevent quality of food being spoilt. Step 1: Identify the contradictions Improving feature: Speed

Worsening Feature: Temperature

Step 2: Look at the list of features and identify those important to your contradiction. Speed

9

Temperature

17

Step 3 Identify the improving and worsening features from triz matrix above It is found that the features intersect at second row and fifth column. The Altshuller’s Principles involved are 28. Mechanics substitution 30. Flexible shells and thin films 36. Phase transitions 2. Taking out Step 4: Brainstorming: Principle 28. Mechanics substitution According to solution in Triz40, replacement of physical interaction between elements of the drum dryer to electric, magnetic and electromagnetic fields to communicate or give instructions to components of machine is one of the methods to reduce temperature rise, while keeping the efficiency of the machine. Principle 30. Flexible shells and thin films Since humidity will affect the rate of evaporation of moisture from food, it is important to reduced humidity of inlet fresh air. In order to achieve the purpose, the inlet can be fixed with a thin layer of molecular sieve. According to Ackers(1964), molecular sieve is a crystalline metal aluminosilicates having a three dimensional interconnecting network of silica and alumina tetrahedra. Natural water of hydration is removed from this network by heating to produce uniform cavities which selectively adsorb molecules of a specific size. (Molecular Sieves, 2017) It works under the principle of adsorption whereby the cavities between the molecules are able to trap water with bigger molecular size, however, air molecules have smaller size can pass through the film. 10

Principle 36. Phase transitions Phase transition involve absorb and release of heat, which can help in reducing temperature of drum. In order to increase rate of evaporation of processing food, one can incorporates a vacuum within the drum to maintain a low vapor pressure environment. The reduction in pressure also reduces the temperature at which the product moisture evaporates, resulting in improvements in product quality.

Figure 4: Vapor Pressure drying Machine. (adapted from Singh & Heldman, 2009) From Figure 2, the vapor pressure in the drying cabinet is reduced by passing the exhaust air to mechanically refrigerated condenser. In the condenser, the humidity is reduced by condensing vapor to water. (Singh & Heldman, 2009) By doing this, dry air enter the drying cabinet enable moisture in food to evaporate at lower temperature. Principle 2. Taking out In order to prevent heat transfer from rotating motor to the drum dryer, which will indirectly increase the temperature in the drum, it is encouraged that there is a separation between drum and motor. (triz40, n.d.) Step 5: Conclusion Principle 28 cannot be applied because the shaft need connected to the drum to ensure effective energy transfer. The efficiency of the drum dryer can be enhanced by molecular sieve on the inlet, adding a condenser and ensure a distance between motor and drum. 11

Statement 3: The drum dryer should be designed to enable high rate of drying the food while the power consumption of the motor should be, under allowable condition, as low as possible. Step 1: Identify the contradictions Improving feature: Loss of moisture Worsening Feature: Power consumption of motor Step 2: Look at the list of features and identify those important to your contradiction. Loss of moisture Power consumption of motor

23 21

Step 3 Identify the improving and worsening features from triz matrix above It is found that the features intersect at eighth row and eighth column. The Altshuller’s Principles involved are 28. Mechanics substitution 27. Cheap short-living object 18. Mechanical vibration 38. Strong oxidant Step 4: Brainstorming Principle 28. Mechanics substitution This principle is a repetition of Statement 2. Principle 27. Cheap short-living object Replace an inexpensive object with a multiple of inexpensive objects, comprising certain qualities (such as service life, for instance). Use disposable paper objects to avoid the cost of cleaning and storing durable objects. Plastic cups in motels, disposable diapers, many kinds of medical supplies.(Triz40, n.d.) Principle 18. Mechanical vibration Dehydration process can be carried out by using an object’s resonant frequency. Gallego-Juárez et al.(2007) states that application of ultrasonic vibration is effective in dehydration and energy saving by shortening dehydration processing time. Principle 38. Strong oxidant Replace enriched air with pure oxygen. Cut at a higher temperature using an oxy-acetylene torch. 12

Treat wounds in a high pressure oxygen environment to kill anaerobic bacteria and aid healing.(Triz40, n.d.) Step 5: Conclusion From the 4 principles provided by triz, Principle 28, 27 and 38 are not applicable to overcome the contradiction. Principle 28 cannot be applied because the shaft need connected to the drum to ensure effective energy transfer. The loss of substance in this project is the moisture of food, hence, using cheap short living object is unable to solve the problem. Using strong oxidant will require additional element to be fix onto the machine. This increase the cost of production significantly. CONCEPTUAL SOLUTION

Magnify Speed from Motor

Support Rotating Element

Food Drying

Ventilation and Moisture Removal

Figure 5: Function Diagram of Function Magnify

Speed Support

Rotating Food Drying

Ventilation

from Motor

Element

Spur Gear

Support Roller

Single drum dryer

Fan

Helical Gear

Frame

Tumble dryer

Heat pump

Worm Gear

Bearings

Vacuum drum dryer

Condenser

Bevel Helical

and

Moisture Removal

Recirculating batch dryer Table 2: List of possible conceptual solution for the design

Gearbox Selection: Spur Gear The gearbox stated in the table 1 have their own advantages and strength. However, only an gearbox should be selected in our design. Planetary gearbox is very good in endurance 13

and accuracy. Worm gearbox provides a greater speed difference ratio between shafts. Worm gears are used to transmit power at 90 degree. Coaxial helical inclined gearbox are very good in efficiency and quality. Bevel helical gearbox is used to drive non-parallel shafts in rotary motion. The skew bevel helical gearbox is known for their excellent rigidity and strength. Although each of the gear have their advantages, spur gear is selected as it cheaper and efficient power transmission. Support Selection: Rolling Element Bearing (Single Row Deep Groove Ball Bearing) Frame and Roller From the bearing selection, the most common type of the rolling element bearing usually are ball bearing. These bearings can handle both radial and thrust loads but are usually used where the load is relatively small. From this bearing structure, there is not a lot of contact with the balls on the inner and outer races. If the bearing are over-loaded, the ball would deform and ruin the bearing. But, the force to make the ball to deform are much higher. That why roller bearings are able to handle a much heavier, radial load, like conveyor belts and drum dryer. In addition, the efficiency can be improved by using more special rolling element ball bearing that is single or double row deep groove ball bearing. Basically, double row deep groove ball bearing are correspond to single row deep groove ball bearing. Single row deep groove ball bearing the most widely used roller bearing type in the world due to their versatility and overall performance. The advantages of this roller bearing are, excellent in high speed, good radial load capacity and can operate with low noise. As our drum dryer will be operate in high speed, therefore, single row deep groove ball bearing are the suitable choice for bearing selection. But, since the dryer required 30kg load of food of the drying operation, double row deep groove ball bearing are very suitable for bearing arrangement where the load carrying capacity of single row bearing is inadequate. Support roller and frame also needed to support the drum in rotary motion. Selection of Rotary Dryer: Tumble Dryer A single drum dryer is a drum dryer which uses the conduction principle. (Tang, J., Feng, H., & Shen, G., n.d.) The food is compressed into the surface of the shell and the food is then become a thin film. Based on this project, we need to dry 5mm granules of food. This way will alters the shape of the food and indirectly ruin the desired output. 14

Figure 6: Single Drum Dryer (Adapted from:Tang, J., Feng, H., & Shen, G., n.d.) Tumble dryer is used widely in clothes drying industry. It is used as a batch dryer for clothes where the clothes are placed inside the rotating drum. Also there is air ventilation system to allow the moisture content that evaporated from the clothes to be escaped. In this project, tumble dryer is selected as the food granules could be placed inside the rotating drum, the drum is then rotated into high speed. On the other hand, the air vent could be utilised to help in quicker drying by introduce an air flow in high speed rotating drum design. Vacuum dryer is specifically used for the heat sensitive materials. It is also very expensive in terms of the price as the technology used is much more complicated compared to the other types of dryer. The objective of this project is not specifically for the heat sensitive materials. Recirculating batch dryer is a self-contained unit with annular drying chamber. It removes the water from wet grains by forcing heated air through the grain bulk. The system needs high air temperatures with high air flow rates which oppose the idea of this project itself. We cannot mainly use the heat source to dry the materials. Other than that, this machine avoids rapid drying rates as the materials is only expose to the flow of hot air for relatively short time within each cycle. Ventilation and Drying Aid Selection: Fan Ventilation and drying aid mostly used in the rotary drying in various industry. Heat pump and condenser is utilised in the tumble dryer of the washing machine. Heat pump could be a perfect match with the condenser in drying as the heat pump introduce heated air and the moisture will then gather up in the condenser and collected in the circulation. It cost very high. 15

In grain drying application, there is a way that called natural air or low temperature drying. This drying method utilise the fan to introduce the air motion in the bin dryer for grain drying. Hellevang (1994) shows that “A proper sized system may dry the crop more economically than a high temperature dryer” (p.5). Based on this statement, it was believed that if the concept is used in the drum dryer design and the dimension is constructed properly, economical drying using natural air with low temperature is possible because it is cost saving. Furthermore, the high speed rotating drum uses a high speed rotational energy transmitted from the shaft. By introducing the fan blade into the central part of the drum body, wind movement could be introduced. So, we can fully utilise the energy transmitted from the shaft to benefits the ventilation and drying aid. Development of Product Architecture Create Schematic Diagram of The Product The decomposition of rotating drum is based on the function that the component carried out. From figure 1, AC supply is the main power supply which supply energy to the machine in order to operate the desired function of the machine. The motor is responsible to create rotational motion to rotate the drum. On the other hand, drm which rotates at high speed responsible to carry out drying of granular sized material. Furthermore, after each session is done, removal of processed material and restoring of unprocessed material.

Figure 7: Schematic Diagram of a High Speed Rotating Drum Cluster the elements of the schematic

16

Each component involved in the machine is grouped into modules according to their similarity of function. AC supply and motor involve the conversion of energy which act as source of energy to rotate the drum. Rotating drum, together with the delivery of unprocessed granular sized particles will act as the process chamber to ensure the particles are dried at the end of each session. The last module act as material supply. It is the reservoir of raw material which ensure the continuous supply of material to be process without interruption.

Figure 8: Grouping Design Elements into modules

Create a rough geometric layout The geometric layout is arranged such that at the interval between the motor and rotating drum, gears are fixed between them and connected by shaft. This is to allowed effective power transmission. Other application of installing gears is to increase the angular velocity to fulfill the desired speed of the rotating drum. Besides, the increased distance between motor and rotating drum can also act as heat barrier to prevent the transfer of heat to the drum, which will destroy the quality of food. Structural support is needed to be installed on the motor and rotating drum in order to achieve stability. Furthermore, the support also helps to protect the machine from shock and improve appearance of the machine. Dampers will be fixed between rotating drum and structural support in order to absorb vibration. Since the required design of the machine is it able to rotate at 6500 rpm, which is comparatively high. Hence, when 30 kg of load is loaded into the drum, the drum will vibrate

17

thus creating noise and the lifespan of the machine will also be shorten. In a nutshell, mechanical spring and spring dampers will be installed in order to enhance the quality of design.

Figure 9: Geometric layout of energy, material and process modules

Identify the interaction between modules According to Dieter and Schmidt(2009), interactions between neighbouring modules can be classified into four types, namely spatial, energy, information and material. I. Spatial interaction There are several physical interfaces used to connect modules. Firstly, shaft is used to connect motor and rotating drum. Next, mechanical spring is used to connect rotating drum to support. Moreover, gearbox, together with bearing is used to sustain the weight of gears and shaft. II. Energy flow Electrical energy from the AC power supply is converted to rotational energy. Rotational energy supplied to the rotating drum is transferred to the granular size particles. Consequently, the moisture content in the particles achieve resonance frequency with the frequency of rotating drum. Electrons start to delocalized and the particles become unstable, finally, moisture escape from the particles, thus accomplishing drying process(KSHITIJ Education India, 2016). III. Information flow The machine is designed for batch production whereby each session of processing takes 8 hours towards completion. The time is constantly measured by timer installed in the machine 18

and once each session is accomplished, worker will remove the dried particles and replaced with unprocessed particles into the drum. IV. Material flow Material flow involved in the machine is the flow of granular sized particles.

CONFIGURATION DESIGN Force transmission Standard Part i) Gear According to KHK catalog, pinion material is made of AISI 1045 carbon steel. Its surface durability is 85.6Nm and bending stress is 218 Nm. From calculation, the torque transmitted from the drum to the pinion is 32Nm, which is comparatively smaller than the allowable specification, which are 85.6Nm and 218 Nm respectively. For gear, the material is also AISI 1045 carbon steel. The surface durability is 458 Nm whereas bending strength is 630 Nm. As the size of gear increase, its strength is increased by few times compared to pinion strength. The torque transferred to the motor shaft is 72Nm, this show that the gear is capable to sustain the pulling force. As such, it can be said the gear will have very long lifespan and users will not required to replace the part frequently. This helps to saves maintenance cost and labour force. ii) Bearing Bearing is used to support the shaft during the rotary motion. Standard Assembly i) Motor The chosen motor from Alibaba have 37kW and 2950 rpm with an efficiency of 90.5%. The motor is chosen based on the torque required to operate the rotary drum with 30 kg of load. The details would be shown in parametric design. According to www.teco.com.tw, the Special-Purpose Part i) Rotary Drum The rotary drum transmit power from the rotary shaft into the food. The rotational energy is then converted into vibrational energy to excite the water molecules in the food. The materials used should be stainless steel that comply food standard. At the same time, it could

19

withstand 30 kg of load during operation. The drum is designed to be the cylindrical shape in horizontal orientation. ii) Shaft Shaft is very important to transmit power from gear to the drum. The diameter of the shaft depends on the motor selected and the torque required for it to resist. The shaft produced have no uniform diameter. ii)Shaft and Drum Coupling Drum spider is used to connect the shaft and the drum. Bolts and nuts is required to lock the drum spider with the drum. Interfaces and connections There are fixed, non adjustable connection as well as adjustable connection implemented in manufacturing of Hi-RDD. Example of fixed connection is welding at interfaces between the rotating drum and shaft connected to it whereas joints between truss supporting the Hi-RDD also use the same connection. Permanent joint is applied here because both rotating drum and shaft are special purpose part which designed for specific task only. There are rarely replacement of these part during the operating period of the machine. Besides, the interfaces between bearings and supporting frame also use non adjustable connection, that is screw. The non permanent joint is only allowed to be unfastened when there is needs of maintenance. It does not be opened during the operation period. One example of adjustable connection is the bearing and the shaft. The shaft is allowed to rotated along the y-axis. This connection allow one degree of freedom. PARAMETRIC DESIGN Detailed Calculation is Attached to Appendix 5 at page E SUMMARY OF ROTARY DRUM Mass of Rotary Drum: 4.52 kg Dimension of Rotary Drum: (0.6 m long x 0.6 m diameter)

20

Figure 10: Maximum Deflection of Drum at 0.000149 mm. SUMMARY OF MOTOR AND POWER Power=37 kW Speed=2950 rpm

SUMMARY OF SHAFTS

Figure 11: Summary of Output Shaft. 21

Figure 12: Summary of Input Shaft.

Figure 13: Summary of Right Side Shaft Shaft

Maximum Deflection

Output

0.1741mm

Input

0.07132mm

22

Right end

0.01549mm

Table 3: Deflection of Shaft due to Bending

Figure 14: Maximum deflection of Input Shaft at 0.1789 mm due to Torsion.

Figure 15: Maximum deflection of Output Shaft at 0.2934 mm due to Torsion.

23

Figure 16: Maximum deflection of Right Side Shaft at 0.1372 mm due to Torsion.

SUMMARY OF GEAR

Table 4: Summary of Gear Calculation.

24

SUMMARY OF BEARINGS

Figure 17: Bearing Compartment.

Figure 18: Summary of Input Shaft Bearing.

25

Figure 19: Summary of Right Side Shaft Bearing.

Figure 20: Summary of Output Shaft Bearing.

SUMMARY OF SUPPORT Cross section : 0.025m x 0.01m Height

: 0.35m

Width

: 0.6m

Critical Stress : 3.7 MPa SUMMARY OF BOLTS AND NUTS Material

: Stainless steel 304

Size

: M10 (10 mm diameter)

Length

: 50 mm

Critical Stress : 0.13 MPa 26

SUMMARY OF WELDED JOINT Electrode

: E70xx

Shear permissible stress

: 144.6 MPa

Minimum weld length

: 1.73 mm

DETAIL DESIGN After brainstorming about the design we are constructing, we are going to represent the system design in a variety of views where each view uses a different modeling technique. By using a variety of views, different parts of the system can be made clearer by different views. Some views will be at elaborating a systems states whereas other views are better at showing how data flows within the system. Other views are better at showing how different system entities relate to each through class taxonomies for systems that are designed using an objectoriented approach. Make or Buy For the design some parts is special designed and some parts could be bought from the market. It is important to make this decision according to the pricing and which one is cheaper so that the cost to manufacture the product will be lower.

Parts

Modification

Frame

Yes

Drum

Yes

Bolt and Nuts

No

Gears

No

Bearings

No

Shaft

Yes

Coupling

No

Motor

No

Table 5: Special purpose part is needed modification and standard part could just be bought from the market The selection and sizing of the components

27

After analyzing the make/buy decision, we need to complete the selection and sizing of the components. It is necessary to complete these activities before the design can be complete. If the product design is at all complex, it most likely will be necessary to impose a design freeze at some point prior to completion. Which means that beyond certain point, the design do not have to change unless they go through a formal review by a design control board. This is because to reduce the human tendency to continually make slight improvement or changes. With a design freeze, only those last-minute changes that truly affect performance, safety or cost are approved. The selection and sizing of the components also will ease the designing of the product on the dimension. Also, it will ease the work on researching the components and parts that we are going to buy from the store. Lastly, it will act as the standard for all team member in analyzing each part that they are working for. Furthermore, the selection of component for this project is needed for identifying the material of the part that will not affecting to any chemical since we are working with food. Parts

Selection

Sizing

Frame

Stainless Steel 304

0.35m height 0.6mm width 0.8mm long

Drum

Stainless Steel 304

0.6 m long 0.6 m diameter

Bolt and Nuts

Stainless Steel

M10 (50mm length)

Gears

Carbon Steel

Pinion: 25 mm bore 144 mm pitch diameter Gear: 20 mm bore 64 mm pitch diameter

Bearings

SV 30 Stainless Steel

(bore:outer diameter:thickness) (30 mm:42 mm:7 mm) (15 mm:28 mm:7 mm) (25 mm:42 mm:9 mm) (17 mm:26 mm:5 mm)

28

(15 mm:24 mm:5 mm) Shaft

(Stainless Steel)

Detail please refer to parametric

Output AISI 1050 CD

design o f shaft

Input AISI 1006 CD Right Side AISI 1020 CD Coupling

Stainless Steel

250mm diameter

Motor

37 kW/6500 rpm

N/A

Table 6: Selection and sizing of the components Engineering Drawing

Figure 21: Conceptual Model of KreaTech Inc. Hi-RDD

Figure 22: Orthogonal Views of the Rotary Drum. 29

Final cost estimation Parts

Quantity (n)

Final Price (RM)

Frame, Drum

1

137.11

Bolt and Nuts

4

6.90

Gears and Pinion

1, 1

1133.60

Bearings

4

114.50

Shaft

1

42.80

Coupling

1

55.70

Motor

1

428

Roller

1

41.40

Table 7: Pricing of Materials. Total = RM1960.40 Product design specification (PDS) Speed

: 6500 rpm

Dimension

: 0.6m (diameter) x 0.6m (length)

Cost

: RM1960.40

Materials

: Stainless steel 304

Type of dryer

: Horizontal batch dryer

Food sizing

: Granular-sized food of 5 mm diameter

Drum Capacity

: Accommodation of 30 kg granular food at once

30

CONCLUSION As the analysis show, this High Speed Rotating Drum Dryer can hold its own against the competition in food drying industry. Each parts of the product has been analysed specifically and proved to be effectively functional. However, for this kind of product, ease of use and portability

are much more important than performance alone. Granted, not all of the

features turned out to be completely viable in a day-to-day business environment. Nonetheless, this product is an easy recommendation for buyers looking for a much simpler, lighter, and low cost budget food dryer compare to the other products which are for the heavy industry. Additionally, this food dryer is comparatively shock resistant and well protected against damage, thanks to the vibration absorbers. We, the KreaTech Inc. will always find solutions in creating efficient, well working product that provide a good user experience. Through a collaborative effort, much better results can be achieved.

31

REFERENCES Ackers, G. K. (1964). Molecular exclusion and restricted diffusion processes in molecular-sieve chromatography. Biochemistry, 3(5), 723-730.

Gallego-Juárez, J., Rodriguez-Corral, G., Gálvez Moraleda, J., & Yang, T. S. (2007). A New High-Intensity Ultrasonic Technology for Food Dehydration. Retrieved March 12, 2017, from http://digital.csic.es/bitstream/10261/76919/1/435Preprint-food%20dehydration%20(RI35).pdf

Grain storage techniques - Evolution and trends in developing countries... (n.d.). Retrieved March 12, 2017, from http://www.fao.org/docrep/T1838E/ T1838E0w.htm

Hellevang, K. J. (1994, November). Grain Drying. Retrieved March 12, 2017, from https://www.ag.ndsu.edu/graindrying/publications/ae-701-grain-drying

How a Tumble Dryer Works. (2010, September 10). Retrieved March 12, 2017, from http://www.espares.co.uk/advice/0/114/how-a-tumble-dryer-works

Kalda, J., Ainsaar, S., Pungas, T., & Zavjalov, S. (2014, August 2). Problems on Mechanics. Retrieved March 12, 2017, from http://www.ioc.ee/~kalda/ipho/meh_ENG.pdf

Katie Gerard (2014). 6 Most Popular Types of Mechanical Bearings. Retrieved March 12, 2017 from http://info.craftechind.com/blog/bid/371106/6-Most-PopularTypes-of-Mechanical-Bearings

M. G. (2010). Re-circulating batch dryer. Retrieved March 12, 2017, from http://www.knowledgebank.irri.org/training/fact-sheets/postharvest-management/drying-factsheet-category/item/re-circulating-batch-dryer-fact-sheet

Molecular Sieves: Technical Information Bulletin. (2017). Retrieved March 12, 2017, from http://www.sigmaaldrich.com/chemistry/chemical-synthesis/learning-center/ technical-bulletins/al-1430/molecular-sieves.html A

Rotary drum working principle. (2013, March 22). Retrieved March 11, 2017 from http://www.suncomachlinery.com/news/company/rotary-dryer-01.html

Singh, R. P., and Heldman, R. D. Introduction to food engineering. 4th ed. San Diego, CA: Academic Press, 2009. Print.

Single Row Deep Groove, Retrieve March 12, 2017, from http://www.astbearings.com/ single-row-deep-groove.html

Tang, J., Feng, H., & Shen, G. (n.d.). Drum Drying. Retrieved March 12, 2017, from https://labs.wsu.edu/tang/wp-content/uploads/sites/1254/2016/04/book-drumdry-t ang03.pdf

The A2A Simulations Community. (2015, June 27). Retrieved March 12, 2017, from https://a2asimulations.com/forum/viewtopic.php?f=23&t=47896.

Triz40. (n.d.). Retrieved March 12, 2017, from http://www.triz40.com/TRIZ_GB.php

Tumble Dryers Buying Guide. (n.d.). Retrieved March 12, 2017, from http://www.currys.co.uk/gbuk/tumble-dryers-buyers-guide-258-commercial.html

5 Types of Industrial Gearboxes, Their Strengths and Applications. (2016, October 4). Retrieved March 12, 2017, from http://noengwks.org/ 5-types-of-industrial-gearboxes-their-strengths-and-applications/

KSHITIJ Education India. (2016). Retrieved May 26, 2017, from http://www.kshitijiitjee.com/Resonance-energy

Kohara

Gear

Catalog.

(2017).

Retrieved

May

26,

2017,

from

https://www.khkgears.us/catalog/product/

Budynas, R. G., and J. K. Nisbett. Shigley's Mechanical Engineering Design. 10th ed. New York, NY: McGraw Hill, 2015. B

APPENDICES Appendix 1: Selected Gear

Appendix 2: Selected Pinion

C

Appendix 3: Ground Spur Gear Catalog

Appendix 4: Selection of Gear with Precision and Categories

D

Appendix 5: Analysis Evidence Drum Analysis and Motor Selection Mass of rotary drum, m:By referring Shigley’s Mechanical Engineering Design 10th edition text book (pg 1046), the formula to find the mass of hollow cylinder is applicable to estimate the mass of the hollow rotary drum. 𝑚=

𝜋(𝑑12 − 𝑑22 )𝑙𝜌 4

(1)

Where, Outer diameter of the drum, 𝑑1 = 0.6𝑚 Inner diameter of the drum, 𝑑2 = 0.599𝑚 Length of the drum, 𝑙 = 0.6𝑚 Density of stainless steel 304, 𝜌 = 8000𝑘𝑔/𝑚3 Substitute all the values above into equation (1): 𝜋(0.62 − 0.5992 )(0.6)(8000) 𝑚= 4 𝑚 = 4.52𝑘𝑔

Moment of Inertia, I: Moment of inertia of the empty drum (hollow cylinder) is given by the equation as follow: 𝐼=

𝑚 2 (𝑑 + 𝑑22 ) 8 1

(2)

Substitute the values into equation (2) gives: 𝐼=

4.52 (0.62 + 0.5992 ) 8

𝐼 = 0.41𝑘𝑔. 𝑚2 Moment of inertia of the food to be dried, 𝐼𝑓 : Assuming the food as a point mass at the surface of rotary drum, 𝐼𝑓 = 𝑚𝑓 𝑅 2 𝑑2 𝐼𝑓 = 𝑚𝑓 ( )2 2

(3) (4)

Where, E

Mass of food, 𝑚𝑓 = 30𝑘𝑔 Distance of the food from the centre of rotation of the drum, 𝑅 =0.599m Substitute the value above into equation (4) yields: 𝐼𝑓 = 𝑚𝑓 𝑅 2 𝐼𝑓 = 30(

0.599 2 ) 2

𝐼𝑓 = 2.69𝑘𝑔. 𝑚2 Total moment of inertia of the drum with 30kg of load, 𝐼𝑡𝑜𝑡 : 𝐼𝑡𝑜𝑡 = 𝐼 + 𝐼𝑓 𝐼𝑡𝑜𝑡 = 0.41 + 2.69 𝐼𝑡𝑜𝑡 = 3.10𝑘𝑔. 𝑚2

Torque required to move the drum with 30kg of load, 𝑇: 𝑇 = 𝐼𝑡𝑜𝑡 𝛼

(5)

Where, Angular acceleration, 𝛼 =

𝜔 𝑡

Equation (5) finally becomes: 𝑇=

𝐼𝑡𝑜𝑡 𝜔 𝑡

(6)

Angular acceleration, 𝜔 = 6500𝑟𝑝𝑚 = 680.68𝑟𝑎𝑑/𝑠 Time taken for the rotary drum to accelerate to 6500 rpm is given by t. Assuming the desired time, 𝑡 = 15𝑠 Substitute all the value into equation (6): 𝑇=

3.10(680.68) 45

𝑇 = 46.87𝑁. 𝑚

Power required to drive the drum to 6500rpm, P: (7) 𝑃 = 𝑇𝜔 T and 𝜔 is known and be substituted into equation 7: F

𝑃 = 46.87(680.68) 𝑃 = 32𝑘𝑊 From the calculation above it is known that to achieve 6500rpm in 45s, 32kW motor is required to operate the drum dryer. Shear and bending moment analysis (Drum): Fd = 44.34N Wf = 490.5N/m

A

B 0.3m

N1 = 169.32N

0.3m N2 = 169.32N

From the free body diagram above, it is known that: Force on the drum caused by the weight of drum, Fd 𝐹𝑑 = 𝑚 × 9.81 𝐹𝑑 = 4.52 × 9.81 𝐹𝑑 = 44.34𝑁 Force distribution of the food on the drum ignoring the negligible thickness of the drum: 𝑊𝑓 =

𝑚𝑓 × 9.81 𝑙

𝑊𝑓 =

30 × 9.81 0.6

𝑊𝑓 = 490.5𝑁/𝑚 Normal reaction on the drum, N1 and N2: + ∑ 𝑀𝐴 = 0 −0.3(44.34) − 0.3(30 × 9.81) + 0.6𝑁2 = 0 𝑁2 = 169.32𝑁 + ∑ 𝐹𝑦 = 0 G

𝑁1 − 44.34 − 30 × 9.81 + 169.32 = 0 𝑁1 = 169.32N Shear Force and Bending Moment Diagram

H

Shaft Analysis Output Shaft Analysis Shear Force and Bending Moment Analysis Driven Gear

Bearing

Shaft Connect To Drum

0.1m

0.1m

0.1m

For x-y plane:

𝐹𝐺 = 1464.66𝑁

𝐹𝐵1 = 732.33𝑁

𝐹𝐵2 = 732.33𝑁

+ ∑ 𝑀𝐵1 = 0 −0.1(1464.66) + 0.2(𝐹𝐵2 ) = 0 𝐹𝐵2 = 732.33𝑁 + ∑ 𝐹𝑦 = 0 𝐹𝐵1 − 1464.66 + 732.33 = 0 𝑁1 = 732.33N

I

J

For x-z plane: 𝑁𝐺 = 533.33𝑁 𝑁1 = 169.32𝑁

𝑁𝐵1 = 182.00𝑁

𝑁𝐵2 = 520.64𝑁

+ ∑ 𝑀𝐵1 = 0 −0.1(533.33) + 0.2(𝑁𝐵2 ) − 0.3(169.32) = 0 𝑁𝐵2 = 520.64𝑁 + ∑ 𝐹𝑍 = 0 𝐹𝐵1 − 533.33 + 520.64 − 169.32 = 0 𝑁1 = 182.00N

K

To obtain the total moment , the formula should be used is as follow: 2 + 𝑀2 𝑀𝑡𝑜𝑡 = √𝑀𝑥𝑦 𝑥𝑧

For example, the bending moment at the driven gear is: 2 + 𝑀2 𝑀𝑡𝑜𝑡 = √𝑀𝑥𝑦 𝑥𝑧

𝑀𝑡𝑜𝑡 = √73.2332 + 18.20052 𝑀𝑡𝑜𝑡 = 75.46𝑁. 𝑚 After complete the calculation, the bending moment graph is analysed as follow: Reference Position Bearing (A) Fillet (B) Fillet (C) Gear (D) Fillet (E) Bearing (F) Fillet (G) To Drum (H)

Distance, m 0 0.0045 0.0675 0.1 0.1325 0.2 0.2045 0.3

Total Moment, N.m 0 3.4 50.94 75.46 71.17 16.932 16.17 0

L

Total Moment against Distance, N.m 80

Total Moment, N.m

70 60 50 40 30 20 10 0 0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

Distance, m

From the graph above it is known that the highest bending moment is at 0.1m, which is the gear central position. However, the critical point is more likely to be the fillet at 0.1325m (point E) with a moment of 71.17N.m. Selecting Shaft Diameter with a Desired Safety Factor The calculation is only focusing on the point which is subjected to critical bending especially fillet point accounting notch sensitivity. (B, C, E and G) Point E: Fillet at Right Side of the Gear At first, stress concentration factor is to be estimated as the actual dimension are not yet determined. The important references is obtained from Shigley’s Mechanical Design Text 10th edition. The figure are as shown.

M

From the table above, assume well rounded fillet estimated that Kt = Kf = 1.7 and Kts = Kfs = 1.5.

From figure above, the analysis is started with lower strength and relatively inexpensive steel, AISI 1020 CD with 𝑆𝑢𝑡 = 470𝑀𝑃𝑎. Surface Factor, ka: 𝑏 𝑘𝑎 = 𝑎𝑆𝑢𝑡

(8)

N

it is known that factor 𝑎 = 4.51 and exponent 𝑏 = −0.265. Substituting all the value inside the equation 8 yields: 𝑏 𝑘𝑎 = 𝑎𝑆𝑢𝑡

𝑘𝑎 = 4.51(470)−0.265 𝑘𝑎 = 0.883 Size Factor, kb: 𝑑 −0.107 ) = 1.24𝑑−0.107 7.62 𝑘𝑏 = 1.51𝑑−0.157 𝑘𝑏 = (

(9) (10)

From the equation, the diameter of the shaft is needed to compute the surface factor. But the shaft diameter is still an unknown. So, the surface factor is assumed to be 𝑘𝑏 = 0.9, and checked back after the diameter has been determined. Loading Factor, Kc: Bending, 𝑘𝑐 = 1 Axial, 𝑘𝑐 = 0.85 Torsion, 𝑘𝑐 = 0.59 In this shaft analysis, the analysis is combined with both loading factor of torsion and bending. Hence, 𝑘𝑐 = 1 is set to be the parameter for analysis. Temperature Factor, kd/ Reliability Factor, ke/ Miscellaneous-Effects Factor, kf: Since the data for these aspect are not given and limited, 𝑘𝑑 = 𝑘𝑒 = 𝑘𝑓 = 1. The Endurance Limit:

𝑆𝑒 = 𝑘𝑎 𝑘𝑏 𝑘𝑐 𝑘𝑑 𝑘𝑒 𝑘𝑓 𝑆′𝑒

(11) O

Where 𝑆′𝑒 = 0.5𝑆𝑢𝑡 (12) 𝑆𝑒 = 𝑘𝑎 𝑘𝑏 𝑘𝑐 𝑘𝑑 𝑘𝑒 𝑘𝑓 (0.5𝑆𝑢𝑡 ) Substitute all the factors and ultimate stress into the equation (12) yields: 𝑆𝑒 = 𝑘𝑎 𝑘𝑏 𝑘𝑐 𝑘𝑑 𝑘𝑒 𝑘𝑓 (0.5𝑆𝑢𝑡 ) 𝑆𝑒 = 0.883(0.9)(1)(1)(1)(1)(0.5)(470) 𝑆𝑒 = 187𝑀𝑃𝑎 By using the DE-Goodman Criterion, the diameter of the shoulder at point E could be estimated. 1/3 16𝑛 1 1 2 2 1/2 2 2 1/2 𝑑=( { [4(𝐾𝑓 𝑀𝑎 ) + 3(𝐾𝑓𝑠 𝑇𝑎 ) ] + [4(𝐾𝑓 𝑀𝑚 ) + 3(𝐾𝑓𝑠 𝑇𝑚 ) ] }) 𝜋 𝑆𝑒 𝑆𝑢𝑡

(13)

For a rotating shaft with constant bending and torsion, the bending stress is completely reversed and the torsion is steady, 𝑇𝑎 = 𝑀𝑚 = 0. Equation (13) could be reduced to: 𝑑=(

1/3 16𝑛 1 1 1/2 1/2 { [4(𝐾𝑓 𝑀𝑎 )2 ] + [3(𝐾𝑓𝑠 𝑇𝑚 )2 ] }) 𝜋 𝑆𝑒 𝑆𝑢𝑡

(14)

At point E, 𝑀𝑎 = 49.86𝑁. 𝑚, 𝑇𝑚 = 46.87𝑁. 𝑚. From engineeringtoolbox.com, when the materials used is highly reliable, conditions and environment not really severe and loading is important factor, factor of safety is ranged from 1.3-1.5. Safety factor is assumed to be n=1.5. Substituting all the values into equation (14) gives: 1/3 16𝑛 1 1 2 1/2 2 1/2 𝑑=( { [4(𝐾𝑓 𝑀𝑎 ) ] + [3(𝐾𝑓𝑠 𝑇𝑚 ) ] }) 𝜋 𝑆𝑒 𝑆𝑢𝑡 1/3 16(1.5) 1 1 2 1/2 2 1/2 [4[(1.7)(49.86)] ] + [3([1.5(46.87)] )] }) 𝑑=( { 𝜋 187 470

𝑑 = 0.0207𝑚 = 20.7𝑚𝑚 The diameter at shoulder E is selected to be 20mm to fit the gear with bore of 20mm. Using typical ratio of shoulder

𝐷 𝑑

= 1.2, 𝐷 = 1.2(20) 𝐷 = 24𝑚𝑚

So the shaft 24mm diameter shaft could be used. When 𝐾𝑡 = 1.7, assume the r/d=0.1. 𝑑

20

So assume fillet radius, 𝑟 = 10 = 10 = 2𝑚𝑚 𝐷

𝑟

From (Fig. A-15-9), when 𝑑 = 1.2, 𝑑 = 0.1, 𝐾𝑡 = 1.6 P

For Bending: 3 2 − 2.67(10−8 )𝑆𝑢𝑡 √𝑎 = 0.246 − 3.08(10−3 )𝑆𝑢𝑡 + 1.51(10−5 )𝑆𝑢𝑡

√𝑎 = 0.246 − 3.08(10−3 )(68) + 1.51(10−5 )(68)2 − 2.67(10−8 )(68)3 √𝑎 = 0.098√𝑖𝑛 Where, 𝑆𝑢𝑡 in kpsi unit. 𝐾𝑓 = 1 +

𝐾𝑡 − 1 𝑎 1 + √𝑟

Where r is also expressed in inch.

𝐾𝑓 = 1 +

1.6 − 1 0.098 1+ √2/25.4

𝐾𝑓 = 1.15 In torsion, From (Fig A-15-8), 𝐾𝑡𝑠 = 1.35 3 2 − 2.67(10−8 )𝑆𝑢𝑡 √𝑎 = 0.190 − 2.51(10−3 )𝑆𝑢𝑡 + 1.35(10−5 )𝑆𝑢𝑡

√𝑎 = 0.190 − 2.51(10−3 )(68) + 1.35(10−5 )(68)2 − 2.67(10−8 )(68)3 √𝑎 = 0.073√𝑖𝑛 𝐾𝑓𝑠 = 1 +

𝐾𝑓𝑠 = 1 +

𝐾𝑡 − 1 𝑎 1 + √𝑟 1.35 − 1 0.073 1+ √ 2 25.4

𝐾𝑓𝑠 = 1.28 20

−0.107

New 𝑘𝑏 = (7.62) 𝑘𝑏 = 0.902

New 𝑆𝑒 = 0.883(0.902)(0.5)(470) 𝑆𝑒 = 187𝑀𝑃𝑎 𝜎𝑎 , =

32𝐾𝑓 𝑀𝑎 32(1.44)(71.17) = = 104𝑀𝑃𝑎 𝜋𝑑 3 𝜋(0.020)3 Q

1/2

𝜎𝑚 ′

16𝐾𝑓𝑠 𝑇𝑚 2 = [3 ( ) ] 𝜋𝑑 3

=

√3(16)(1.33)(46.87) = 66𝑀𝑃𝑎 𝜋(0.020)3

1 𝜎𝑎 ′ 𝜎𝑚 ′ 104 66 = + = + = 0.697 𝑛𝑓 𝑆𝑒 𝑆𝑢𝑡 187 470 𝑛𝑓 = 1.43 Hence, the safety factor is 1.43 which is close enough to initial assumption of 1.5. The calculation is accepted. Also check for yielding: 𝑛𝑦 =

𝑆𝑦 𝜎𝑚𝑎𝑥 ,

=

𝑆𝑦 390 = = 2.48 𝜎𝑎 ′ + 𝜎𝑚 ′ 91 + 66

The high factor of safety shows the design is safe for yielding.

Keyway Checking At the end of the keyway, the total moment M=50.94N.m. Assuming at the bottom of the keyway r/d=0.02, r=0.02(20)=0.4mm 𝐾𝑡 = 2.14 For Bending: 3 2 − 2.67(10−8 )𝑆𝑢𝑡 √𝑎 = 0.246 − 3.08(10−3 )𝑆𝑢𝑡 + 1.51(10−5 )𝑆𝑢𝑡

√𝑎 = 0.246 − 3.08(10−3 )(68) + 1.51(10−5 )(68)2 − 2.67(10−8 )(68)3 √𝑎 = 0.098√𝑖𝑛 Where, 𝑆𝑢𝑡 in kpsi unit. 𝐾𝑓 = 1 +

𝐾𝑡 − 1 𝑎 1 + √𝑟

Where r is also expressed in inch.

𝐾𝑓 = 1 +

2.14 − 1 0.098 1+ √0.4/25.4

𝐾𝑓 = 1.64 In torsion, From (Fig A-15-8), 𝐾𝑡𝑠 = 3 3 2 − 2.67(10−8 )𝑆𝑢𝑡 √𝑎 = 0.190 − 2.51(10−3 )𝑆𝑢𝑡 + 1.35(10−5 )𝑆𝑢𝑡

R

√𝑎 = 0.190 − 2.51(10−3 )(68) + 1.35(10−5 )(68)2 − 2.67(10−8 )(68)3 √𝑎 = 0.073√𝑖𝑛 𝐾𝑓𝑠 = 1 +

𝐾𝑓𝑠 = 1 +

𝐾𝑡 − 1 𝑎 1 + √𝑟 3−1 0.073 1+ √ 0.4 25.4

𝐾𝑓𝑠 = 2.26 𝜎𝑎 , =

32𝐾𝑓 𝑀𝑎 32(1.64)(50.94) = = 106𝑀𝑃𝑎 𝜋𝑑 3 𝜋(0.020)3 1/2

𝜎𝑚

′

16𝐾𝑓𝑠 𝑇𝑚 2 = [3 ( ) ] 𝜋𝑑 3

=

√3(16)(2.26)(46.87) = 117𝑀𝑃𝑎 𝜋(0.020)3

1 𝜎𝑎 ′ 𝜎𝑚 ′ 106 117 = + = + = 0.815 𝑛𝑓 𝑆𝑒 𝑆𝑢𝑡 187 470 𝑛𝑓 = 1.23 (rejected) The safety factor is too low, it is more critical for the shaft at keyway. Suggested to change the materials of shaft to steel 1050 CD with higher strength but the diameter remains: 𝑏 𝑘𝑎 = 𝑎𝑆𝑢𝑡

𝑘𝑎 = 4.51(690)−0.265 𝑘𝑎 = 0.798 𝑘𝑏 = (

𝑑 −0.107 ) 7.62

𝑘𝑏 = (

20 −0.107 ) 7.62

𝑘𝑏 = 0.902 𝑆𝑒 = 𝑘𝑎 𝑘𝑏 𝑘𝑐 𝑘𝑑 𝑘𝑒 𝑘𝑓 (0.5𝑆𝑢𝑡 ) 𝑆𝑒 = 0.798(0.902)(1)(1)(1)(1)(0.5)(690) 𝑆𝑒 = 248.23𝑀𝑃𝑎

S

𝐾𝑡 = 2.14 For Bending: 3 2 − 2.67(10−8 )𝑆𝑢𝑡 √𝑎 = 0.246 − 3.08(10−3 )𝑆𝑢𝑡 + 1.51(10−5 )𝑆𝑢𝑡

√𝑎 = 0.246 − 3.08(10−3 )(100) + 1.51(10−5 )(100)2 − 2.67(10−8 )(100)3 √𝑎 = 0.0623√𝑖𝑛 Where, 𝑆𝑢𝑡 in kpsi unit. 𝐾𝑡 − 1

𝐾𝑓 = 1 +

𝑎 𝑟

1+√

Where r is also expressed in inch.

𝐾𝑓 = 1 +

2.14 − 1 0.0623 1+ √0.4/25.4

𝐾𝑓 = 1.76 In torsion, From (Fig A-15-8), 𝐾𝑡𝑠 = 3 3 2 − 2.67(10−8 )𝑆𝑢𝑡 √𝑎 = 0.190 − 2.51(10−3 )𝑆𝑢𝑡 + 1.35(10−5 )𝑆𝑢𝑡

√𝑎 = 0.190 − 2.51(10−3 )(100) + 1.35(10−5 )(100)2 − 2.67(10−8 )(100)3 √𝑎 = 0.0473√𝑖𝑛 𝐾𝑓𝑠 = 1 +

𝐾𝑓𝑠 = 1 +

𝐾𝑡 − 1 𝑎 1 + √𝑟 3−1 0.0473 1+ √ 0.4 25.4

𝐾𝑓𝑠 = 2.45 𝜎𝑎 , =

32𝐾𝑓 𝑀𝑎 32(1.76)(50.94) = = 114𝑀𝑃𝑎 𝜋𝑑 3 𝜋(0.020)3 1/2

𝜎𝑚

′

16𝐾𝑓𝑠 𝑇𝑚 2 = [3 ( ) ] 𝜋𝑑 3

=

√3(16)(2.45)(46.87) = 127𝑀𝑃𝑎 𝜋(0.020)3

T

1 𝜎𝑎 ′ 𝜎𝑚 ′ 114 127 = + = + = 0.596 𝑛𝑓 𝑆𝑒 𝑆𝑢𝑡 248 690 𝑛𝑓 = 1.55 (Accepted) The safety factor 1.55 is very close to the assumed safety factor of 1.5 indicating a safe design. Checking groove B for bearing A, from the bending moment diagram: 𝑀𝑎 = 3.4𝑁. 𝑚 𝑀𝑚 = 𝑇𝑚 = 𝑇𝑎 = 0 For bearing shoulder: point B and point G Typical bearing r/d=0.02, r=0.02(30)=0.6 𝐾𝑡 = 2.7 𝐾𝑓 = 1 +

2.7 − 1 0.0623 1+ √0.6/25.4

𝐾𝑓 = 2.21 𝜎𝑎 , =

32𝐾𝑓 𝑀𝑎 32(2.21)(16.17) = = 13.5𝑀𝑃𝑎 𝜋𝑑 3 𝜋(0.030)3

𝑛𝑓 =

𝑆𝑒 248 = = 18.37 , 𝜎𝑎 13.5

Typical bearing r/d=0.02, r=0.02(15)=0.3 𝐾𝑓 = 1 +

2.7 − 1 0.0623 1+ √0.3/25.4

𝐾𝑓 = 2.08 𝜎𝑎 , =

32𝐾𝑓 𝑀𝑎 32(2.08)(3.4) = = 21.3𝑀𝑃𝑎 𝜋𝑑 3 𝜋(0.015)3

𝑛𝑓 =

𝑆𝑒 248 = = 11.64 , 𝜎𝑎 21.3

The safety factor for the bearing shoulder fillet is very safe. From here we can conclude that the design for our output shaft is based on the most critical keyway, with safety factor of 1.55.

U

Summary of Output Shaft Output Shaft (Left Side of the Drum)

Section Diameter(mm) Length(mm)

A 15 10

a 17 63

D 20 65

F 30 72

Fillet Radius(mm)

B 0.3

C 0.4

E 2

G 0.6

H 36 95.5

Material: AISI 1050 CD Steel Safety Factor: 1.55 (point C, most critical) Yielding: 2.29

Input Shaft Bearing

Driver Gear

0.1m

0.1m

V

For x-y plane 𝐹𝐵4 = 732.33𝑁

𝐹𝐵3 = 732.33𝑁

𝐹𝑃 = 1464.66𝑁

+ ∑ 𝑀𝐵1 = 0 −0.1(1464.66) + 0.2(𝐹𝐵2 ) = 0 𝐹𝐵3 = 732.33𝑁 + ∑ 𝐹𝑦 = 0 𝐹𝐵4 − 1464.66 + 732.33 = 0 𝐹𝐵4 = 732.33N Shear Force and Bending Moment Diagram

W

For x-z plane 𝑁𝐵4 = 266.66𝑁

𝑁𝐵3 = 266.66𝑁

𝑁𝑃 = 533.33𝑁

+ ∑ 𝑀𝐵1 = 0 −0.1(533.33) + 0.2(𝐹𝐵2 ) = 0 𝐹𝐵3 = 266.66𝑁 + ∑ 𝐹𝑦 = 0 X

𝐹𝐵4 + 533.33 − 266.66 = 0 𝐹𝐵4 = 266.66N Shear Force and Bending Moment Diagram

Y

After complete the calculation, the bending moment graph is analysed as follow: Distance, m

Reference Position Bearing (I) Fillet (N) Fillet (O) Gear (K) Fillet (E) Fillet (L) Bearing (M)

Total Moment, N.m 0 3.51 52.61 77.94 49.18 3.51 0

0.1 0.1045 0.1675 0.2 0.2325 0.2955 0.3

Total Moment against Distance, N.m 90

Total Moment, N.m

80 70 60 50 40 30 20 10 0 0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

Distance, m

Starting with AISI 1010 CD steel. 𝑆𝑢𝑡 = 370𝑀𝑃𝑎 = 53𝑀𝑝𝑎 𝑏 𝑘𝑎 = 𝑎𝑆𝑢𝑡

𝑘𝑎 = 4.51(370)−0.265 𝑘𝑎 = 0.941 Size Factor, kb: Selected gear bore diameter: 25mm 25 −0.107 ) 7.62 𝑘𝑏 = 0.881 𝑘𝑏 = (

In this shaft analysis, the analysis is combined with both loading factor of torsion and bending. Hence, 𝑘𝑐 = 1 is set to be the parameter for analysis. Since the data for these aspect are not given and limited, 𝑘𝑑 = 𝑘𝑒 = 𝑘𝑓 = 1. The Endurance Limit: Z

𝑆𝑒 = 𝑘𝑎 𝑘𝑏 𝑘𝑐 𝑘𝑑 𝑘𝑒 𝑘𝑓 (0.5𝑆𝑢𝑡 ) 𝑆𝑒 = 0.941(0.881)(1)(1)(1)(1)(0.5)(370) 𝑆𝑒 = 153𝑀𝑃𝑎 Using typical ratio of shoulder

𝐷 𝑑

= 1.2,

𝐷 = 1.2(25) 𝐷 = 30𝑚𝑚 So the shaft 30mm diameter shaft could be used. When 𝐾𝑡 = 1.7, assume the r/d=0.1. 𝑑

25

So assume fillet radius, 𝑟 = 10 = 10 = 2.5𝑚𝑚 𝐷

𝑟

From (Fig. A-15-9), when 𝑑 = 1.2, 𝑑 = 0.1, 𝐾𝑡 = 1.6 For Bending: 3 2 − 2.67(10−8 )𝑆𝑢𝑡 √𝑎 = 0.246 − 3.08(10−3 )𝑆𝑢𝑡 + 1.51(10−5 )𝑆𝑢𝑡

√𝑎 = 0.246 − 3.08(10−3 )(53) + 1.51(10−5 )(53)2 − 2.67(10−8 )(53)3 √𝑎 = 0.121√𝑖𝑛 Where, 𝑆𝑢𝑡 in kpsi unit. 𝐾𝑓 = 1 +

𝐾𝑡 − 1 𝑎 1 + √𝑟

Where r is also expressed in inch.

𝐾𝑓 = 1 +

1.6 − 1 0.121 1+ √2.5/25.4

𝐾𝑓 = 1.43 In torsion, From (Fig A-15-8), 𝐾𝑡𝑠 = 1.35 3 2 − 2.67(10−8 )𝑆𝑢𝑡 √𝑎 = 0.190 − 2.51(10−3 )𝑆𝑢𝑡 + 1.35(10−5 )𝑆𝑢𝑡

√𝑎 = 0.190 − 2.51(10−3 )(53) + 1.35(10−5 )(53)2 − 2.67(10−8 )(53)3 √𝑎 = 0.091√𝑖𝑛

AA

𝐾𝑡 − 1

𝐾𝑓𝑠 = 1 +

𝑎 1 + √𝑟 1.35 − 1 0.091 1+ √ 2.5 25.4

𝐾𝑓𝑠 = 1 +

𝐾𝑓𝑠 = 1.27 𝜎𝑎 , =

32𝐾𝑓 𝑀𝑎 32(1.42)(77.94) = = 73𝑀𝑃𝑎 𝜋𝑑 3 𝜋(0.025)3 1/2

𝜎𝑚

′

16𝐾𝑓𝑠 𝑇𝑚 2 = [3 ( ) ] 𝜋𝑑 3

=

√3(16)(1.27)(71.17) = 76𝑀𝑃𝑎 𝜋(0.025)3

1 𝜎𝑎 ′ 𝜎𝑚 ′ 73 76 = + = + = 0.678 𝑛𝑓 𝑆𝑒 𝑆𝑢𝑡 153 370 𝑛𝑓 = 1.47 Hence, the safety factor is 1.43 which is close enough to initial assumption of 1.5. The calculation is accepted. Also check for yielding: 𝑛𝑦 =

𝑆𝑦 𝜎𝑚𝑎𝑥

,

=

𝑆𝑦 300 = = 2.01 ′ 𝜎𝑎 + 𝜎𝑚 73 + 76 ′

The high factor of safety shows the design is safe for yielding.

Keyway Checking At the end of the keyway, the total moment M=49.18N.m. Assuming at the bottom of the keyway r/d=0.02, r=0.02(25)=0.5mm 𝐾𝑡 = 2.14 For Bending: 3 2 − 2.67(10−8 )𝑆𝑢𝑡 √𝑎 = 0.246 − 3.08(10−3 )𝑆𝑢𝑡 + 1.51(10−5 )𝑆𝑢𝑡

√𝑎 = 0.246 − 3.08(10−3 )(53) + 1.51(10−5 )(53)2 − 2.67(10−8 )(53)3 √𝑎 = 0.121√𝑖𝑛 Where, 𝑆𝑢𝑡 in kpsi unit. 𝐾𝑓 = 1 +

𝐾𝑡 − 1 𝑎 1 + √𝑟

Where r is also expressed in inch. BB

𝐾𝑓 = 1 +

2.14 − 1 0.121 1+ √0.5/25.4

𝐾𝑓 = 1.61 In torsion, From (Fig A-15-8), 𝐾𝑡𝑠 = 3 3 2 − 2.67(10−8 )𝑆𝑢𝑡 √𝑎 = 0.190 − 2.51(10−3 )𝑆𝑢𝑡 + 1.35(10−5 )𝑆𝑢𝑡

√𝑎 = 0.190 − 2.51(10−3 )(53) + 1.35(10−5 )(53)2 − 2.67(10−8 )(53)3 √𝑎 = 0.091√𝑖𝑛 𝐾𝑓𝑠 = 1 +

𝐾𝑓𝑠 = 1 +

𝐾𝑡 − 1 𝑎 1 + √𝑟 3−1 0.091 1+ √ 0.5 25.4

𝐾𝑓𝑠 = 2.21 𝜎𝑎 , =

32𝐾𝑓 𝑀𝑎 32(1.61)(49.18) = = 52𝑀𝑃𝑎 𝜋𝑑 3 𝜋(0.025)3 1/2

𝜎𝑚

′

16𝐾𝑓𝑠 𝑇𝑚 2 = [3 ( ) ] 𝜋𝑑 3

=

√3(16)(2.21)(105.46) = 132𝑀𝑃𝑎 𝜋(0.025)3

1 𝜎𝑎 ′ 𝜎𝑚 ′ 52 132 = + = + = 0.696 𝑛𝑓 𝑆𝑒 𝑆𝑢𝑡 153 370 𝑛𝑓 = 1.44 (Accepted) The safety factor is acceptable. Checking groove N for bearing I, from the bending moment diagram: 𝑀𝑎 = 3.51𝑁. 𝑚 𝑀𝑚 = 𝑇𝑚 = 𝑇𝑎 = 0 For bearing shoulder: point B and point G Typical bearing r/d=0.02, r=0.02(20)=0.4 𝐾𝑡 = 2.7

CC

𝐾𝑓 = 1 +

2.7 − 1 0.121 1+ √0.4/25.4

𝐾𝑓 = 1.87 𝜎𝑎 , =

32𝐾𝑓 𝑀𝑎 32(1.87)(3.51) = = 8.4𝑀𝑃𝑎 𝜋𝑑 3 𝜋(0.020)3

𝑛𝑓 =

𝑆𝑒 153 = = 18.21 𝜎𝑎 , 8.4

Typical bearing r/d=0.02, r=0.02(15)=0.3 𝐾𝑓 = 1 +

2.7 − 1 0.121 1+ √0.3/25.4

𝐾𝑓 = 1.80 𝜎𝑎 , =

32𝐾𝑓 𝑀𝑎 32(1.80)(3.51) = = 19.1𝑀𝑃𝑎 𝜋𝑑 3 𝜋(0.015)3

𝑛𝑓 =

𝑆𝑒 153 = = 8.01 , 𝜎𝑎 19.1

The safety factor for the bearing shoulder fillet is very safe. From here we can conclude that the design for our output shaft is based on the most critical keyway, with safety factor of 1.55.

DD

Summary of Input Shaft

Section Diameter(mm) Length(mm)

I 20 10

J 30 63

K 25 65

L 20 63

Fillet Radius(mm)

N 0.4

O 2.5

P 0.5

Q 0.3

M 15 10

Materials: AISI 1010 CD Steel Safety Factor: 1.44 (most critical point P) Yielding: 2.01 Right Side Shaft 𝑁𝐵6 = 169.32𝑁

𝑁 = 169.32𝑁

𝑁𝐵5 = 338.64𝑁

+ ∑ 𝑀𝐵6 = 0 −0.2(169.32) + 0.1(𝑁𝐵6 ) = 0 𝐹𝐵3 = 338.64𝑁 EE

+ ∑ 𝐹𝑦 = 0 𝑁𝐵6 + 338.64 − 169.32 = 0 𝑁𝐵6 = 169.32N

FF

Reference Position Connection to drum Fillet (U) Bearing (s) Fillet (V) Bearing (T)

Distance, m 0 0.0955 0.1 0.1955 0.2

Total Moment, N.m 0 16.17 16.93 0.762 0

Starting with AISI 1006 CD steel. 𝑆𝑢𝑡 = 330𝑀𝑃𝑎 = 48𝑀𝑝𝑎 𝑏 𝑘𝑎 = 𝑎𝑆𝑢𝑡

𝑘𝑎 = 4.51(330)−0.265 𝑘𝑎 = 0.970 Size Factor, kb: Assuming 𝑘𝑏 = 0.9 In this shaft analysis, the analysis is combined with both loading factor of torsion and bending. Hence, 𝑘𝑐 = 1 is set to be the parameter for analysis. Since the data for these aspect are not given and limited, 𝑘𝑑 = 𝑘𝑒 = 𝑘𝑓 = 1. The Endurance Limit: GG

𝑆𝑒 = 𝑘𝑎 𝑘𝑏 𝑘𝑐 𝑘𝑑 𝑘𝑒 𝑘𝑓 (0.5𝑆𝑢𝑡 ) 𝑆𝑒 = 0.970(0.9)(1)(1)(1)(1)(0.5)(330) 𝑆𝑒 = 144𝑀𝑃𝑎 Checking groove U 𝑑=(

1/3 16𝑛 1 1 1/2 1/2 { [4(𝐾𝑓 𝑀𝑎 )2 ] + [3(𝐾𝑓𝑠 𝑇𝑚 )2 ] }) 𝜋 𝑆𝑒 𝑆𝑢𝑡

1/3 16(1.5) 1 1 2 1/2 2 1/2 [4[(1.7)(16.932)] ] + [3([1.5(0)] )] }) 𝑑=( { 𝜋 144 330

𝑑 = 0.0167𝑚 Referring to the bearing, 17mm bore is chosen. Using typical ratio of shoulder

𝐷 𝑑

= 1.2,

𝐷 = 1.2(17) 𝐷 = 20.4𝑚𝑚 So the shaft 21mm diameter shaft could be used. When 𝐾𝑡 = 2.7, assume the r/d=0.02. So assume fillet radius, 𝑟 = 0.02(17) = 0.34𝑚𝑚 For Bending: 3 2 − 2.67(10−8 )𝑆𝑢𝑡 √𝑎 = 0.246 − 3.08(10−3 )𝑆𝑢𝑡 + 1.51(10−5 )𝑆𝑢𝑡

√𝑎 = 0.246 − 3.08(10−3 )(48) + 1.51(10−5 )(48)2 − 2.67(10−8 )(48)3 √𝑎 = 0.130√𝑖𝑛 Where, 𝑆𝑢𝑡 in kpsi unit. 𝐾𝑓 = 1 +

𝐾𝑡 − 1 𝑎 1 + √𝑟

Where r is also expressed in inch.

𝐾𝑓 = 1 +

2.7 − 1 0.130 1+ √0.34/25.4

𝐾𝑓 = 1.80 Check kb HH

𝑘𝑏 = (

17 −0.107 ) 7.62

𝑘𝑏 = 0.918 New fatigue strength 𝑆𝑒 = 0.970(0.918)(1)(1)(1)(1)(0.5)(470) 𝑆𝑒 = 147 𝜎𝑎 , =

32𝐾𝑓 𝑀𝑎 32(1.92)(16.17) = = 60𝑀𝑃𝑎 𝜋𝑑 3 𝜋(0.017)3

1 𝜎𝑎 ′ 60 = = = 0.411 𝑛𝑓 𝑆𝑒 147 𝑛𝑓 = 2.44 Hence, the safety factor is 2.44 which is high enough to prove the diameter is safe. The calculation is accepted. Also check for yielding: 𝑛𝑦 =

𝑆𝑦 𝜎𝑚𝑎𝑥 ,

=

𝑆𝑦 390 = = 6.50 𝜎𝑎 ′ 60

The high factor of safety shows the design is safe for yielding.

Checking groove V for bearing T, from the bending moment diagram: 𝑀𝑎 = 0.762𝑁. 𝑚 𝑀𝑚 = 𝑇𝑚 = 𝑇𝑎 = 0 For bearing shoulder: point B and point G Typical bearing r/d=0.02, r=0.02(15)=0.3 𝐾𝑡 = 2.7 𝐾𝑓 = 1 +

2.7 − 1 0.121 1+ √0.3/25.4

𝐾𝑓 = 1.80 𝜎𝑎 , =

32𝐾𝑓 𝑀𝑎 32(1.80)(0.762) = = 4.1𝑀𝑃𝑎 𝜋𝑑 3 𝜋(0.015)3

𝑛𝑓 =

𝑆𝑒 147 = = 38.85 𝜎𝑎 , 4.1 II

𝑛𝑦 =

𝑆𝑦 𝜎𝑚𝑎𝑥

,

=

𝑆𝑦 390 = = 95.12 ′ 𝜎𝑎 4.1

Which is very high to prove the shaft diameter is safe. From here we can conclude that the safety factor of the shaft is 2.55 at point U. Yield factor of safety is 6.5. Summary for Right Side Shaft

Section Diameter(mm) Length(mm)

R 21 95.5

S 17 100

Fillet U Radius(mm) 0.34 Materials: AISI 1006 CD Steel

V 0.3

T 15 10

Safety Factor: 2.55 (Most critical point U) Yielding: 6.5

Limits and Fits From Shigley’s Mechanical Design text book, page 389, it has shown that free running fit with symbol of H9/d9 is suitable with high speed application as a clearance fit. Example calculation: Basic size of 25mm: From Table A-11, ∆𝐷 = ∆𝑑 = 0.062 Hole: 𝐷𝑚𝑎𝑥 = 𝐷 + (∆𝐷)ℎ𝑜𝑙𝑒 = 25 + 0.062 = 25.06 𝐷𝑚𝑖𝑛 = 𝐷 = 25𝑚𝑚 From Table A-12, fundamental deviation=-0.02mm Shaft: 𝑑𝑚𝑎𝑥 = 𝑑 + 𝛿𝐹 = 25 − 0.02 = 24.98 JJ

𝑑𝑚𝑖𝑛 = 𝑑 + 𝛿𝐹 − 0.062 = 24.98 − 0.062 = 24.92 The calculation is completed and tabulated in the table below: Fitting(mm) 15 17 20 25 30

Shaft(mm) dmsx 14.95 16.96 19.95 24.92 29.95

dmin 14.91 16.91 19.88 24.98 29.88

Dmin 15 17 20 25 30

Hole(mm) Dmax 15.04 17.04 20.05 25.06 30.05

Deflection Analysis for Bending 𝑃𝐿

Using Maximum deflection=48𝐸𝐼 Where P is critical load, L is length of the shaft, E is elastic modulus of stainless steel 193 x 109 . 𝜋𝑑4 𝐼= 64 Results is tabulated as follow: Shaft Output Input Right end

Maximum deflection 0.1741mm 0.07132mm 0.01549mm

KK

Gear Analysis

LL

MM

NN

OO

PP

QQ

RR

Analysis of Bearing (Selection of Bearing) · Bearing 1 Given load on the Shaft: Fr = 182N, Fa = 732.33 N Shaft diameter = 15mm i. Design Life Requirement, Ld T = 8hr Speed shaft = 6500rpm Ld = 8hr x 60 min/hr x 6500 rev/min = 3.12 x 10^6 ii. Reliability, R Estimation of the reliability with 99% Life factor Adjustment Kr = 0.21

SS

iii.

Application Factor, If Machinery with Light Impact = 1.5

ix. Calculate the dynamic equivalent radial load Pe

Pe1 = (1)182N + (0)732.33N = 182N Pe2 = (0.55)(182N) + (1.45)(732.33) = 1.162 kN

Pe2 > Pe1 thus Pe = 1.162 kN

v. Calculate the basic dynamic load rating requirement TT

vi. With Cd enter a basic load rating table and find the smallest bearing with a load rating of at least Cd Consulting manufacturer’s tables, we need to find a deep-groove bearing with a basic dynamic load rating of at least 4.23kN, a bore diameter of at least 15mm, and an operational speed of at least 6500 rpm. Bearing type : 61902 ZZ Analysis of selecting bearing

UU

VV

WW

Frame Support: Truss Analysis

b

a c

732.33N The calculation will focus at point b which has a highest resultant load at point b.

FBD 1 a

b

Fbc

Ma=0;

570.64N

0.5 𝐹𝑏𝑐 − 0.5(732.33 cos 45) + 0.5(570.64 𝑐𝑜𝑠45) = 0 𝐹𝑏𝑐 = 114.33𝑁 FBD 2 732.33N

570.64N

c

b

Fab XX

Mc=0; 0.5𝐹𝑎𝑏 − 0.5(570.64 𝑐𝑜𝑠45) − 0.5(732.33𝑐𝑜𝑠45) =0 𝐹𝑎𝑏 = 921.3𝑁

𝐹 = 𝑃𝐴 𝑃=

921.3 (0.025 ∗ 0.01)

Use Fab as it is the highest force acted compare to the other trusses.

𝑃 = 3685.2𝑘𝑃𝑎

𝜎𝑦 = 206.807𝑀𝑃𝑎 (steel type 304)

3685.2𝑘𝑃𝑎 < 206.807𝑀𝑃𝑎 Since P<𝜎𝑦 , the frame can withstand the force exerted on the trusses. Bolts and Nuts Material

: Stainless steel 304

Size

: M10 (10mm diameter)

Length

: 50mm

Resultant load on each bolt YY

At A, B, C and D 169.32 4 F   42.33N F 

46.87 4(110) F   0.11N F  

F 2  42.332  0.112  2(42.33)(0.11) cos90  45 F  42.41N

Shear stress on each bolt



42.33 7.85(105 )

  540kPa Moment of inertia

I X  IY 

 (2204 )

64 I X  IY  115(106 )mm 4

  (104 )  I  115(106 )  2  77.782 (44)(10)  64  I  109.68(106 )mm 4

Critical bending stress



46.87(300) (103 ) 6 109.68(10 )

  0.13MPa (smaller than 206.807Mpa) –accepted. The connection can withstand the load on it. Welding From truss calculation, the critical force on the truss is: F  921.3 N

Allowable shear stress E70xx electrode has a tensile stress of 482MPa Using shear welding permissible stress formula: ZZ

 all  0.3(482)  all  144.6 N Weld length

921.3 (144.6  10 )(0.707)(0.005)(2) l  1.73mm l

6

The welding length could be set to 50 mm as 1.73 mm is considered safe but impossible to be welded.

AAA

PROJECT LOG BOOK AND GRANTT CHART LOGBOOK DESIGN PROJECT PROGRESSING Title: Design a High Speed Rotating Drum Dryer (Hi-RDD) DATE: 28 February 2017 ATTENDANCE : 1. Lew Haoxian 2. Mohd Badri 3. Mohd Shukri 4. Lai Sukna 5. Nur Fidaiy ACTIVITY : 1. Discussed the problem statement. Based on the task given, we figured out and highlighted the problem statement that might be involved in this project. 2. Overviewed of what is the rotating drum dryer and its function. We have been searched about rotating dryer drum through internet to get the ideas on how it works. OUTCOME: Each of the members knew what the task is all about and managed to identify the problem statements. Besides, we were being exposed the functions and mechanism of the rotating dryer drum.

BBB

LOGBOOK DESIGN PROJECT PROGRESSING Title: Design a High Speed Rotating Drum Dryer (Hi-RDD) DATE: 2 March 2017 ATTENDANCE : 1. Lew Haoxian 2. Mohd Badri 3. Mohd Shukri 4. Lai Sukna 5. Nur Fidaiy

ACTIVITY : 1. Find the ideas in designing our own dryer drum machine. In this session, we tried to come out the best ideas on how to improve the machine. 2. Identified the constraints that involved in the task and factors that need to be considered. The purpose is to get the ideas and brainstorming the best way to improve the machine beside how we can tackle the problem statement. OUTCOME: Based on the problem statement given, there are the listed ideas on how to improvise the machine: ● Food drying using resonance principle without heating ( Haoxian) ● Rapid drying with using more stable high speed rotating drum (Fida) ● The design of the machine in horizontal way ( Syukri) Based on the constraint and factor that involved in problem statement, there are the listed on how to tackle the problem: 2. Rotating drum should be light to have higher speed ( Sukna) 3. To have the high rate of drying, power transmission is needed to be under allowable condition (Badri)

CCC

LOGBOOK DESIGN PROJECT PROGRESSING Title: Design a High Speed Rotating Drum Dryer (Hi-RDD) DATE: 8 March 2017 ATTENDANCE : 1. Lew Haoxian 2. Mohd Badri 3. Mohd Shukri 4. Lai Sukna 5. Nur Fidaiy ACTIVITY : 1. Divide among the members on assignment one task. Each of the member take part in assignment one that related to design specification of the machine, triz matrix and the conceptual solution.

OUTCOME: 1. In the design specification of the rotating dryer drum, the kinematics and functions of machine part have been explained. 2. In triz method, there are three solution ways to improvise the machine. 3. While in conceptual solution, the selection for rotary dryer, gearbox, and bearing have been made.

DDD

LOGBOOK DESIGN PROJECT PROGRESSING Title: Design a High Speed Rotating Drum Dryer (Hi-RDD) DATE: 12 April 2017 ATTENDANCE: 1. Lew Haoxian 2. Mohd Badri 3. Mohd Shukri 4. Lai Sukna 5. Nur Fidaiy ACTIVITY : 1. In this meeting session, we as a group have been met with Dr Ana Sakura who is one of the coordinators of Design 2 course. 2. We reviewed together the assignment that we have done before this. There were several mistakes have been made in the assignment. 3. In this meeting also, we have discussed about analysis that should we done in the project. 4. Hereby are the listed part for each member: 1. Motor (Hao Xian) 2. Shaft (Shukri) 3. Bearing (Fida’iy) 4. Support (Badri) 5. Gear (Sukna) OUTCOME: 1. The purpose of met up with Dr Ana was reviewed the assignment that we have done. 2. All the mistakes that have been made, we tried to improve in the report later on. 3. Dr Ana also suggested us to divide the analysis of the machine so that we can do the project fairly

EEE

LOGBOOK DESIGN PROJECT PROGRESSING Title: Design a High Speed Rotating Drum Dryer (Hi-RDD) DATE: 8 May 2017 ATTENDANCE: 1. Lew Haoxian 2. Mohd Badri 3. Mohd Shukri 4. Lai Sukna 5. Nur Fidaiy ACTIVITY : 1. In this meeting session, we discussed the progression of our analysis food rotary drum. 2. We also checked from each of our analysis part’s either it correlated with one and another. 3. As the progression still on going, the data from certain part still not be specific thus analysis need to be done step by step 4. We also downloaded the solid work software to do the simulation and each of the member need to do their part in solid work software.

OUTCOME: 1. Each of the members presented their progression on calculation of analysis food rotary drum. 2. The correlation of the analysis is to ensure that each part of the calculation is correct and it will be fits to the machine. 3. The solid work software helps us to do the simulation and take account of the point of maximum deflection of the machine.

FFF

Formulation of problem Engineering Design Specification Triz Method Conceptual Solution Evaluate and Select Concept Decision making Embodiment Design Manual Analysis Solidwork sketch and simulation Analysis using software Detail Design Summarize pricing and bill of materials Complilation of work from group members Project Presentation Final written report

Organisation Chart

Leader: Hao

Secretary: Fida

Suk Na

Shukri

Badri

GGG

29-May

22-May

15-May

8-May

1-May

24-Apr

17-Apr

10-Apr

3-Apr

27-Mar

20-Mar

13-Mar

6-Mar

27-Feb

Grantt Chart