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Engineering Design Report

Fish conveyor belt

Auckland University of Technology

Semester 1 2017

2nd June 2017

Client

Salty Dog Ltd.

Team 8

Amila Perera (1312372)

Walter Hsu (14848572)

Yuede Chen (16936911)

Simon Lu (1116527)

Company name:

Inno-tek Ltd.

Mentors

Dr David White

Dr Marcel

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Innotek Engineering “Where innovation meets technology”

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Brief This report details the designs for a fish conveyor compiled for Salt Dog Ltd, it is the development of detail components for an industrial fish conveyor specifically for the fish processing plant located at Castle Point. After several clinical sessions with the client, David White, an extensive list of specifications was produced based on the demands required by client to establish the initial design concepts. The report demonstrates the employment of design methodology approach to the task given by client, showing how the design specifications led to the initial design concept, iterations of the design concepts and further developed to embodiment design and finally the to the detail design. This report also employs the use of multiple design catalogues for the corresponding parts as well as detailed computer aided design CAD software to produce virtual models for the conveyor as well as the components that it is made of.

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Table of contents

1) Task clarification

Brief ..................................................................................................................................... 2 Introduction ........................................................................................................................ 7 Layout .................................................................................................................................. 7 Location ............................................................................................................................... 8 Orientation of pallet ........................................................................................................... 9 Design specifications ........................................................................................................ 10

2) System requirements

Power Transmission ......................................................................................................... 11 Electric Motor ................................................................................................................... 11 Worm Gear Unit (WGU) ................................................................................................... 12 Chain Drive ............................................................................ Error! Bookmark not defined. Summary - Velocity Ratio Transmission Ratios ............................................................... 13 Kinematic Analysis ............................................................................................................ 14 Shaft Estimations .............................................................................................................. 14 Total Inertia and Torque................................................................................................... 17 Power ................................................................................................................................ 19

3) Selecting power transmission components

Electric Motor Selection ................................................................................................... 21 Soft starter selection ........................................................................................................ 24 Coupling selection ............................................................................................................ 27 Worm gear unit (WGU) Selection .................................................................................... 33 Chain selection .................................................................................................................. 36

4) Material selection (slats) Material Selection ............................................................................................................. 41 Slat Material Selection...................................................................................................... 42 Material Index ................................................................................................................... 42 Slat Design ......................................................................................................................... 46 Deflection of slats calculation .......................................................................................... 47 Slats design ........................................................................... Error! Bookmark not defined. Finite Element Analysis..................................................................................................... 50 Cost of Slats ....................................................................................................................... 50

5) Frame design Ergonomics ....................................................................................................................... 53 Ergonomics of the human body ....................................................................................... 53 Conveyor height................................................................................................................ 54 Pallet dimensions .............................................................................................................. 54 Frame Design .................................................................................................................... 56 Loading analysis ................................................................................................................ 60 Stress analysis for frame top ............................................................................................ 62 Structure (Frame) ................................................................. Error! Bookmark not defined. Material Index ................................................................................................................... 68

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6) Shaft design Head shaft Force Analysis ................................................................................................ 76 Loading configuration ....................................................................................................... 76 Force Calculations ............................................................................................................. 77 Overall Shear forces and Bending moments ................................................................... 81 Head Shaft Sizing - Standard AS1403 .............................................................................. 82 Standard AS1403 Head Shaft Diameter Results .............................................................. 84 Head Shaft Sizing – First Principles .................................................................................. 85 Head Shaft Diameter Results Using First Principles ........................................................ 86 Head Shaft Final Profile .................................................................................................... 88 Tail Shaft Sizing - Standard AS1403 ................................................................................. 95 Tail Shaft Sizing – First Principles ..................................................................................... 96 Tail Shaft Final Profile ....................................................................................................... 98 Bearing Selection .............................................................................................................. 99 Procedure of bearing selection ........................................................................................ 99 Load rating and life ............................................................... Error! Bookmark not defined. Basic rating life and basic dynamic load rating ............................................................. 100 Head shaft bearings ........................................................................................................ 101 Radial Load Calculations for Head Shaft Bearing Points ............................................... 101 Bearing selection at A ..................................................................................................... 101 Bearing selection at B ..................................................................................................... 103 Tail shaft bearings ........................................................................................................... 106 Head shaft keyway selection .......................................................................................... 109 Shearing consideration ................................................................................................... 110 Head Shaft Keyway Lengths ........................................................................................... 114 Tail shaft keyway selection............................................................................................. 115

7) Safety

Identify areas that require guarding .............................................................................. 118 Guarding .......................................................................................................................... 120 Workplace safety ........................................................................................................... 113 Failure Mode and Effect Analysis(FMEA) ...................................................................... 124 FMEA analysis of shaft design ........................................................................................ 125 FMEA analysis of frame design ............................................ Error! Bookmark not defined.

8) Maintenance

Maintenance checklist.................................................................................................... 127 Installation ...................................................................................................................... 129 Linking the frame section together ............................................................................... 129 Attachments to the ground............................................................................................ 129

9) Commissioning Commissioning ................................................................................................................ 130 Design phase ................................................................................................................... 130 Construction phase ......................................................................................................... 130 Acceptance phase ........................................................................................................... 131 Integrated system testing .............................................................................................. 131

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10) Lifecycle analysis

Lifecycle analysis ............................................................................................................. 133 Eco Audit ......................................................................................................................... 134 Energy analysis ................................................................................................................ 135 Energy analysis of the total environmental burden. .................................................... 135 Detailed analysis of the material and manufacturing energy cost. ............................. 136 Transportation energy cost ............................................................................................ 136 Motor energy use and energy cost of disposal ............................................................. 137 Disposal and end of life potential .................................................................................. 137 Material and manufacturing CO2 footprint .................................................................. 138 Energy cost for electric motor use ................................................................................ 139 Disposal of frame and shaft slats ................................................................................... 139 End of life potential ........................................................................................................ 139 Final conveyor ................................................................................................................. 140 Design drawings .............................................................................................................. 141

11) Appendices

Work contribution .......................................................................................................... 161

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1) Task Clarification

Introduction The Salty Dog Co is located at Castle Point, NZ, the client is in needs of a conveyor to transport crates filled with fish from the wharf outside the factory to the inside of the factory. Due to fact that Castle Point being relatively segregated from the major cities, the final product would have to be transported and assembled on site. The environment which product will be operated in will be in brine conditions which raises corrosion issues. These pallets will be loaded and unloaded by hand which means the design must comply with legal lifting requirements as well as being ergonomic to the workers to operate.

Fig A, Illustration of the conveyor

Layout The total length of the conveyor is 36m, however different parts of the conveyor will be exposed to different weather conditions as well as different floor conditions. As shown from the Fig B below, design consideration for the feet mounting will be required to support the conveyor on the wood as well and the concrete surface. Also, having to consider that the parts of the conveyor exposed to outdoors must withstand weather conditions such as heavy rain strong winds and tides.

Fig B, Layout of the conveyor

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Location Castle point is a beachside town located north east of Wellington on the Wairarapa coast of Wellington. Important characteristics of the town include having extremely strong winds with speeds up to 15 miles per hour.

Fig, Map showing Castle Point on the north island of NZ

The location is also quite wet throughout the year, with humidity ranging up to 86% during certain times of the year. With a combination of Castle Point’s relatively wet conditions, high winds and coastal location are strong indicatives of the corrosive nature of the place. Furthermore, narrow roads that led up to the Castle Point could post transport challenges if the final assembly of the conveyor were to be transported on site by conventionally trucks.

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Orientation of pallet

For 800mm:

800𝑚𝑚 × 35 𝑝𝑎𝑙𝑙𝑒𝑡 𝑝𝑒𝑟 min = 28 𝑚/𝑚𝑖𝑛 This cannot work as 28m/min is greater than the specified speed which the conveyor is travelling. For 650mm:

65𝑚𝑚 × 35 𝑝𝑎𝑙𝑙𝑒𝑡𝑠 𝑝𝑒𝑟 min = 22.75 𝑚/𝑚𝑖𝑛 22.75m/min is less than the 27m/min so it will work. To meet the demands specified by the design brief, the pallet must travel on its shorter side of 650mm along the conveyor. This also means that the width of the conveyor must also be designed so that the it can fit the longer side of the pallet.

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Design specifications

To effectively design the desirable product by the client, a comprehensive list based on the client demands and design requirements was made to demonstrate the expectations as well as considerations. The list also includes detailed engineering specifications for the corresponding requirements.

Requirement Engineering specification

Target value

System requirement

Minimal maintenance Hours <1 hour

Easy to clean Hours 2 hours

Continuous operation Hours 8h/day

Longevity Service lifetime (years)

15

Load requirement

Meet legal lifting requirement

Human carrying load limit (N)

Structure support pallet weight

Upwards support force (KN)

>45KN

Cost requirement Mid-price Cost(NZD) <$32,000 ex gst

Ease to manufacture and assemble

Number of individual parts

Spatial requirement Minimal footprint Area (m^2) -

Ergonomic height Height (centimetres) 80

Environmental requirement

Noise Not to exceed “harsh” (decibels)

80

Corrosion resistant Corrosion rate of

materials mA cm-2 -

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2) System requirements

Power Transmission This section of the report consists of the determination of power transmission of the entire conveyor system by looking into aspects such as

• Determination of initial elemental requirements

• Kinematic analysis

• Torque and Power of the elements in the system

• Selecting power transmission elements

Determination of initial elemental requirements

I. Electric Motor At 50 Hz frequency supply

• 2 pole = 2900rpm (loaded)

• 4 pole = 1440rpm (loaded)

As the number of poles increase the speed of the motor decreases and the amount of wiring required for it also increases. This causes the cost to increase. Therefore, to get a balance between speed and cost, a 4-pole electric motor has been chosen. Using the ABB Motor Guide catalogue, the basic electric motor specifications has been chosen as shown below. Therefore the overall velocity ratio can be calculated by dividing the rotational speed of the motor with the rotational speed of the conveyor as shown from Equation .. below.

𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑅𝑎𝑡𝑖𝑜𝑂𝑣𝑒𝑟𝑎𝑙𝑙 = 𝑛𝑀𝑜𝑡𝑜𝑟

𝑛𝐶𝑜𝑛𝑣𝑒𝑦𝑜𝑟=

1440

21.6= 66.7 𝑟𝑝𝑚

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II. Worm Gear Unit (WGU) The selection for the basic parameters of the worm gear unit (WGU) is selected from the worm gear catalogue single reduction, PM series of PW type. Using this it is observed as shown in figure C below that limited options for velocity ratio options are available. Using figure C the worm gear unit of 40:1 has been chosen for further calculations.

Figure C

III. Chain Drive For calculating the chain reduction required, the following equation is used.

𝐶ℎ𝑎𝑖𝑛 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 = 40𝑛

1=

66

1

𝑛 = 66

40= 1.65

Therefore, a chain drive of 1.65 reduction should be selected. By considering the chain reduction options available from Roller chain design guide the number of teeth for the driver sprocket and the driven sprocket (that is attached in the head shaft) can be determined

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From this the number of teeth of each sprocket is determined.

Sprocket Number of Teeth

Driver Sprocket (Pinion) 23

Driven Sprocket (In head shaft) 38

Summary - Velocity Transmission Ratios

Power transmission Element Ratio

WGU 40:1

Chain 1.65:1

Check

𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑟𝑎𝑡𝑖𝑜 = 40 × 1.65 ∶ 1 = 66 ∶ 1

𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑅𝑎𝑡𝑖𝑜 = 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑟𝑎𝑡𝑖𝑜𝑂𝑣𝑒𝑟𝑎𝑙𝑙 It can be concluded that the selection for the WGU and the chain drive is appropriate.

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Kinematic Analysis

𝑛𝐶𝑜𝑛𝑣𝑒𝑦𝑜𝑟 = 21.6

𝑆𝑡𝑎𝑟𝑡𝑖𝑛𝑔 𝑡𝑖𝑚𝑒 = 2 𝑠𝑒𝑐𝑜𝑛𝑑𝑠

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑤(0) = 0 𝑟𝑎𝑑𝑠−1

Converting revolutions per second into radians per second

𝑤 = 2𝜋𝑛

60=

2𝜋 × 21.6

60

Using linear equation of motion 𝑣 = 𝑢 + 𝑎𝑡

For angular displacements 𝑤 = 𝑤0+∝ 𝑡

∝ = 𝑤 − 𝑤0

𝑡

𝛼 =𝑤

𝑡=

2𝜋 × 21.6

60= 1.131 𝑟𝑎𝑑𝑠−1

Shaft Estimations The conveyor system consists of a head shaft as well as a tail shaft at each end of the conveyor. Both these shafts consist of 2 sprockets. For the calculation of the start-up torque, continues torque and power of the system, the following assumptions have been initially made in the shaft.

• Shaft length = 1m

• Slat length = 920mm

• Shaft material = mild steel

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Inertia of the shaft

𝐷𝑒𝑛𝑠𝑖𝑡𝑦 = 𝜌𝑚𝑖𝑙𝑑 𝑠𝑡𝑒𝑒𝑙 = 7850 𝑘𝑔𝑚−3

𝑉𝑜𝑙𝑢𝑚𝑒 = 𝑉𝑆ℎ𝑎𝑓𝑡 = 𝜋𝐷2

4𝐿

= 𝜋

4 × (80 × 10−3)2 × 1

𝑽𝑺𝒉𝒂𝒇𝒕 = 𝟓. 𝟎𝟐𝟔𝟓 × 𝟏𝟎−𝟑 𝒎𝟑

𝑚𝑆ℎ𝑎𝑓𝑡 = 𝜌𝑉𝑆ℎ𝑎𝑓𝑡

= 7850 × 5.0265 × 10−3

𝒎𝑺𝒉𝒂𝒇𝒕 = 𝟑𝟗. 𝟒𝟔 𝒌𝒈

To find the moment of inertial of the shaft.

𝐼𝑆ℎ𝑎𝑓𝑡,𝑥 = 1

2 𝑚𝑅2

= 1

2 × 39.46 × (40 × 10−3)

𝑰𝑺𝒉𝒂𝒇𝒕,𝒙 = 𝟎. 𝟎𝟑𝟏𝟓𝟕 𝒌𝒈𝒎𝟐

Length = 1m

X D

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Inertia of the Sprockets

The sprockets are hollow inside

𝑉𝑜𝑙𝑢𝑚𝑒 = 𝑉𝑆𝑝𝑟𝑜𝑐𝑘𝑒𝑡 = 𝜋(𝐷0

2 − 𝐷𝑖2)

4𝐿

𝑉𝑆𝑝𝑟𝑜𝑐𝑘𝑒𝑡 = 𝜋(398.252 − 802)2

4× 10−6 × 25 × 10−3 𝑚3

𝑽𝑺𝒑𝒓𝒐𝒄𝒌𝒆𝒕 = 𝟐. 𝟗𝟖𝟖𝟓 × 𝟏𝟎−𝟑 𝒎𝟑

𝑚𝑆𝑝𝑟𝑜𝑐𝑘𝑒𝑡 = 𝜌𝑉𝑆𝑝𝑟𝑜𝑐𝑘𝑒𝑡

= 7850 × 2.9885 × 10−3

𝒎𝑺𝒑𝒓𝒐𝒄𝒌𝒆𝒕 = 𝟐𝟑. 𝟒𝟓𝟗𝟕 𝒌𝒈 ≈ 𝟐𝟑. 𝟒𝟔 𝒌𝒈

𝐼𝑆𝑝𝑟𝑜𝑐𝑘𝑒𝑡,𝑥 = 1

2 𝑚𝑆𝑝𝑟𝑜𝑐𝑘𝑒𝑡(𝑅𝑖

2 − 𝑅02)

= 1

2 × 23.46 ×

(398.252 − 802)2

4× 10−6

𝑰𝑺𝒉𝒂𝒇𝒕,𝒙 = 𝟎. 𝟓𝟐𝟓𝟏 𝒌𝒈𝒎𝟐

Length =25mm

X

𝐷𝑖 𝐷𝑜

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Total Inertia and Torque

𝐼𝑇𝑜𝑡𝑎𝑙,𝑥 = 𝐼𝑆ℎ𝑎𝑓𝑡,𝑥 + 𝐼𝑆𝑝𝑟𝑜𝑐𝑘𝑒𝑡,𝑥

= 2(0.03157 + 2 × 0.5251)

= 2.16354 𝑘𝑔𝑚2

𝑇1 = 𝐼𝑡𝑜𝑡𝑎𝑙,𝑥 𝛼

= 2.16354 × 1.131

𝑰𝑻𝒐𝒕𝒂𝒍,𝒙 = 𝟐. 𝟒𝟒𝟕 𝑵𝒎

Let: 𝑚𝑐 = 𝑚𝑎𝑠𝑠𝑠𝑡𝑎𝑡𝑠+𝑘3+2𝐶ℎ𝑎𝑖𝑛𝑠 𝑝𝑒𝑟 𝑚𝑒𝑡𝑒𝑟

𝑚𝑝 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎 𝑝𝑎𝑙𝑙𝑒𝑡

Using Newton Equation

𝐹 = 𝑚𝑎

= (36𝑚𝑐 + 50𝑚𝑝)𝑎

= (36 × 22.7 + 50 × 46) × 27

60 × 2

𝑭 = 𝟕𝟎𝟏. 𝟑𝟕 𝑵

Determining torque

𝑇2 = 𝑇𝑆𝑡𝑎𝑟𝑡,𝑙𝑜𝑎𝑑 = 𝐹𝑟 = 𝐹 ×𝑃𝐶𝐷

2

= 701.37 × 398.25 × 10−3

2

𝑻𝟐 = 𝟏𝟑𝟗. 𝟔𝟔 𝑵𝒎

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Therefore the total torque,

𝑇𝑇𝑜𝑡𝑎𝑙 𝑠𝑡𝑎𝑟𝑡 = 𝑇1 + 𝑇2

= 2.45 + 139.66

𝑻𝑻𝒐𝒕𝒂𝒍 𝒔𝒕𝒂𝒓𝒕 = 𝟏𝟒𝟐. 𝟏𝟏 𝑵𝒎 The running torque,

𝑇𝑅𝑢𝑛𝑛𝑖𝑛𝑔 = 𝜌𝑉𝑔�̅� = 𝑚𝑔�̅�

= (36𝑚𝑐 + 50𝑚𝑝)𝑔�̅�

= (36 × 22.7 + 2300) × 9.81 × 40 × 10−3

𝑻𝑹𝒖𝒏𝒏𝒊𝒏𝒈 = 𝟏𝟐𝟐𝟑. 𝟏𝟗 𝑵𝒎

The total running torque,

𝑇𝑇𝑜𝑡𝑎𝑙,𝑅𝑢𝑛𝑛𝑖𝑛𝑔 = 𝑇𝑇𝑜𝑡𝑎𝑙,𝑆𝑡𝑎𝑟𝑡 + 𝑇𝑅𝑢𝑛𝑛𝑖𝑛𝑔

= 142.11 + 1223.19

𝑻𝑻𝒐𝒕𝒂𝒍,𝑹𝒖𝒏𝒏𝒊𝒏𝒈 = 𝟏𝟑𝟔𝟓. 𝟑 𝑵𝒎

Summary

Torque during start-up Torque during continues operation

1365.3 Nm 1223.19 Nm

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Power Starting power

𝑃𝑆𝑡𝑎𝑟𝑡 = 𝑇𝑇𝑜𝑡𝑎𝑙,𝑅𝑢𝑛𝑛𝑖𝑛𝑔𝑤

= 1365.3 × 2𝜋 ×21.6

60

𝑃𝑆𝑡𝑎𝑟𝑡 = 3088.2358 𝑊

Running power

𝑃𝑅𝑢𝑛𝑛𝑖𝑛𝑔 = 𝑇𝑅𝑢𝑛𝑛𝑖𝑛𝑔𝑤

= 1223.19 × 2𝜋 ×21.6

60

𝑃𝑅𝑢𝑛𝑛𝑖𝑛𝑔 = 2766.7906 𝑊

Summary

Power during start-up Power during continues operation

3088.2 W 2766.8 W

Chain of Transmission

Power Transmission

𝑃𝑆ℎ𝑎𝑓𝑡 = 𝑃𝑆𝑡𝑎𝑟𝑡 = 3.1𝑘𝑊

𝑃𝐶ℎ𝑎𝑖𝑛 𝑆𝑦𝑠𝑡𝑒𝑚 = 3.1

0.98= 3.16 𝑘𝑊

𝑃𝑊𝐺𝑈 = 3.16

0.85= 3.72 𝑘𝑊

𝑃𝑀𝑜𝑡𝑜𝑟 = 3.72

0.85= 4.38 𝑘𝑊

Motor WGU Chain

System Shaft

𝜂 = 85% 𝜂 = 85% 𝜂 = 98%

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Speed Transmission

𝑁𝑀𝑜𝑡𝑜𝑟 = 1440 𝑟𝑝𝑚

𝑁𝑊𝐺𝑈 = 1440 𝑟𝑝𝑚

𝑁𝑀𝑜𝑡𝑜𝑟 =1440

𝑉. 𝑅𝑊𝐺𝑈=

1440

40= 36 𝑟𝑝𝑚

𝑁𝑀𝑜𝑡𝑜𝑟 =1440

𝑉. 𝑅𝐶ℎ𝑎𝑖𝑛 𝐷𝑟𝑖𝑣𝑒=

1440

1.65= 21.8 𝑟𝑝𝑚

Torque Transmission

𝑃 = 𝑇𝑤 = 𝑇 ×2𝜋𝑁

60

𝑇 =30𝑃

𝜋𝑁

𝑇𝑀𝑜𝑡𝑜𝑟 = 30 × 𝑃𝑀𝑜𝑡𝑜𝑟

𝜋𝑁𝑀𝑜𝑡𝑜𝑟=

30 × 4.38 × 103

𝜋 × 1440= 29.05 𝑁𝑚

𝑇𝑊𝐺𝑈 = 30 × 𝑃𝑊𝐺𝑈

𝜋𝑁𝑊𝐺𝑈=

30 × 3.72 × 103

𝜋 × 1440= 24.67 𝑁𝑚

𝑇𝐶ℎ𝑎𝑖𝑛 𝐷𝑟𝑖𝑣𝑒 = 30 × 𝑃𝐶ℎ𝑎𝑖𝑛 𝐷𝑟𝑖𝑣𝑒

𝜋𝑁𝐶ℎ𝑎𝑖𝑛 𝐷𝑟𝑖𝑣𝑒=

30 × 3.16 × 103

𝜋 × 36= 838.22 𝑁𝑚

𝑇𝑆ℎ𝑎𝑓𝑡 = 30 × 𝑃𝑆ℎ𝑎𝑓𝑡

𝜋𝑁𝑆ℎ𝑎𝑓𝑡=

30 × 3.1 × 103

𝜋 × 21.8= 1357.93 𝑁𝑚

Summary

Elements N (rpm) T (Nm) P (kW)

Shaft 21.8 1357.93 3.1

Chain Drive 36 838.22 3.16

WGU 1440 24.67 3.72

Motor 1440 29.05 4.38

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3) Selecting power transmission elements Using the determined speed, torque and power from the previous section and using the appropriate catalogues, the power transmission elements are selected as follows.

I. Electric Motor Selection During the selection of the electric motor previous motor selections parameters on it are used as requirements. Therefore, the requirements include

• 50Hz power supply

• 4 pole – 1500rpm

• Power of motor > 4.38kW

Step 1: Electric motor specification

Step 2: Selection criterion

• Ground mounted

• Minimum foot print (Area)

• Light weight

• Cost

• Warranty

• Insulation availability if needed

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The electric motor has been chosen by referring to Brooks Electric Motor Catalogue

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Dimensions and Frame Sizes

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II. Soft starter selection A motor soft starter is a device used with AC electrical motors to temporarily reduce the load

and torque in the power train and the electrical current surge of the motor during start-up. This is a soft stop unit for 1 phase or 3 phase AC motors from 0.37 to 15kW. This soft starter model is Altistart 01, it is extremely compact with combine simplicity (ease of cabling, fast configuration) and efficiency. It also reduces machine wear and minimises the maintenance time.

Other advantages of this includes

• Adjustable starting and deceleration times: 5 to 10 seconds

• Adjustable starting torque

• Integrated bypass contactor

• Simple dialogue by 2 LEDs

• Signalling on completion of starting

• 24 V integrated power supply (ATS01N2...)

• “Boost” function for difficult starting conditions (ATS01N2...

Selection procedure of the soft starter Step 1: Requirements The requires of this include

• Motor full load current < 12A

• Motor voltage between 110 – 480V

• Motor power = 5.5kW

Step 2: selection of soft starter via comparison

25

Thereby comparison soft starter selected is ATSO1N112FT Its main properties include

Complementary Properties

26

Contractual Warranty

Environmental properties

27

III. Coupling selection Coupling is a device used to connect two shafts together at their ends for transmitting power. Selection To select the correct type and size of the coupling, these should be known

• Power to be transmitted

• Characteristics of the drive

• Speed in rpm

• Dimensions of the shaft to be connected

Selection procedure Step 1: Nominal power in kW to be transmitted This can be obtained from table below Therefore, K = 4.38kW

28

Step 2: Selecting appropriate load classification from table 21, denoted either S, M or H. Using Reynolds Coupling Catalogue, Table 1

29

Step 3: Determination of service factors From table 2 and Table 3 of the Reynolds coupling catalogue, service factors can be determined. For service factor 𝑓𝐷 is chosen based on the intensity of the machine characteristics as well as the drive input as shown in Figure ..

For service factor 𝑓𝑠 This is chosen from table 3 based on number of starts per hour of the machine.

The conveyor system is only required to be operational for 8 hours per day and to be turned off once at the end of the session of the day leading to 0-1 start per hour from the above table. But 1-30 starts per hour has been chosen to obtain higher safety of the design.

30

Step 4: Selection power 𝑘𝑠

𝑘𝑠 = 𝑘 × 𝑓𝐷 × 𝑓𝑠

= 4.38 × 1.25 × 1.2

𝑘𝑠 = 6.57 𝑘𝑊 Step 5: Equivalent power at 100rpm 𝑝𝑒𝑞

𝑝𝑒𝑞 = 𝑘𝑠 × 100

𝑟𝑝𝑚=

6.57 × 102

1440

𝑝𝑒𝑞 = 0.45625 𝑘𝑊

Step 6: Coupling selection guide

31

Step 7: Selecting the type of coupling

From the above table, Spider Couplings have been chosen with comparison to the other couplings due to possessing lower kW at 100rpm.

1.12 𝑘𝑊 > 𝑃𝑒𝑞

And so, the maximum torque 107 𝑁𝑚 > 29.05 𝑁𝑚 (𝑇𝑀𝑜𝑡𝑜𝑟)

32

Step 8: Checking for acceptance of required shaft diameters If the shaft dimeter exceeds maximum permissible, then re-select the next larger coupling. For Spider couplings from Coupling catalogue,

From the table above, a spider coupling with a bore dimeter of 30mm has been chosen. This is of Aluminium Bronze possessing higher resistance against corrosion in comparison with coupling made of Cast iron. Also during the selection power per 100rpm and maximum speed has been taken into consideration as shown below

𝑃𝑜𝑤𝑒𝑟 (𝑘𝑊)𝑝𝑒𝑟 100 𝑟𝑝𝑚 = 0.56 > 0.456 (𝑃𝑒𝑞)

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑝𝑒𝑒𝑑 (𝑟𝑝𝑚) = 4800 > 1440 (𝑁𝑀𝑜𝑡𝑜𝑟)

33

IV. Worm gear unit (WGU) Selection Worm gear unit is an arrangement of a worm (gear in the form of a screw) meshed with a worm gear. It reduces the rotational speed or transmit higher torque.

Selection procedure The same initial procedure has been undertaken as for the selection of the coupling. The first 2 steps are summaries in the table below as follows

Steps Description Selection

Step 1 Load Classification Chain – Medium Impulsive

Step 2 Service factors 𝑓𝐷 = 1.25

𝑓𝑆 = 1.2

𝑓𝑇 = 1.0

Step 3: Determination of mechanical service factor (MSF) Mechanical service factor can be determined from the equation below

𝑀𝑆𝐹 = 𝑓𝐷 × 𝑓𝑆 = 1.25 × 1.2 = 1.5 Step 4: Determination of torque

𝑇𝑀𝑎𝑥 = 𝑆𝑡𝑎𝑟𝑡 − 𝑢𝑝 𝑡𝑜𝑟𝑞𝑢𝑒 = 1365.3 𝑁𝑚

𝐶ℎ𝑎𝑖𝑛 𝑑𝑟𝑖𝑣𝑒 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 = 1.65

𝑇𝑜𝑟𝑞𝑢𝑒@𝑊𝐺𝑈 = 1.3653 × 103

1.65= 827.45 𝑁𝑚

Worm gear

Worm wheel

34

With 98% efficiency of the chain drive

𝑇𝑜𝑟𝑞𝑢𝑒@𝑊𝐺𝑈 = 827.45

0.98= 844.34 𝑁𝑚

𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑡𝑜𝑟𝑞𝑢𝑒 = 844.34 × 𝑀𝑆𝐹

= 844.34 × 1.5

= 1266.51 𝑁𝑚

𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑡𝑜𝑟𝑞𝑢𝑒 = 865.8 × 𝐹𝑇 = 865.8 𝑁𝑚 Step 5: Determination of power

𝑃𝑜𝑤𝑒𝑟 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑚𝑎𝑐ℎ𝑖𝑛𝑒 = 𝐴𝑏𝑠𝑜𝑟𝑏𝑒𝑑 𝑡𝑜𝑟𝑞𝑢𝑒 (𝑁𝑚) × 𝑆𝑝𝑒𝑒𝑑 (𝑟𝑝𝑚)

9550

= 1357.93 × 21.58

9550= 3.0685𝑘𝑊

𝑃𝑀𝑜𝑡𝑜𝑟 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑=5.5𝑘𝑊 > 𝑃𝐴𝑏𝑠𝑜𝑟𝑏𝑒𝑑≈3.1𝑘𝑊

35

Step 6: Selecting WGU Basic requirements

• 𝑃𝑀𝑜𝑡𝑜𝑟 = 4.38 𝑘𝑊

• Output Speed = 21.58 rpm

• MSF > �̅�

From the Worm gear catalogue under the 5.5kW power output

In the figure above, it can be see that some sections are highlighted in blue and green colour. Focusing on the blue highlighted areas, this is not taken into consideration due to

• 𝑀𝑆𝐹 = 1.5 > 𝑆𝐹 → 𝑇ℎ𝑒𝑟𝑒𝑏𝑦 𝑑𝑖𝑠𝑜𝑏𝑒𝑦𝑖𝑛𝑔 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡𝑠

• 𝑟𝑝𝑚 < 21.58 𝑟𝑝𝑚 → 𝐻𝑎𝑣𝑖𝑛𝑔 𝑎 𝑙𝑜𝑤𝑒𝑟 𝑟𝑝𝑚 𝑡ℎ𝑎𝑛 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑚𝑖𝑛𝑖𝑚𝑢𝑚

Due to one or both cases on each product highlighted in blue is avoided. Focusing on the two products that are highlighted in green, it can be seen that all the requirements have been made.

36

V. Chain selection After a reduction of 40: 1 by the worm gear unit, it would require further reduction of 1.65 to achieve close to the desired rpm of 21.6 rpm Desired chain reduction ratio = 1.65: 1 From the Renold roller chain catalogue pg 102

𝑍2 = 38 𝑡𝑒𝑒𝑡ℎ for driving sprocket 𝑍1 = 23 𝑡𝑒𝑒𝑡ℎ for driven pinion

Drive ratio

𝑍2

𝑍1=

38

23= 1.65

Application factor 𝑓1 and tooth factor & tooth factor, 𝑓2 Factor 𝑓1 takes account of any dynamic overloads depending on the chain operating conditions. The value of factor 𝑓1 can be chosen directly or by analogy using chart 2.

37

Tooth Factor 𝑓2 The use of a tooth factor further modifies the final power selection. The choice of a smaller diameter sprocket will reduce the maximum power capable of being transmitted since the load in the chain will be higher.

For a sprocket size of 23 teeth, 𝑓2 is found to be 0.83

𝒇𝟏 = 1.5 𝒇𝟐 = 0.83

VI. Chain drive selection

𝑃𝑡𝑟𝑎𝑛𝑠,𝑝𝑖𝑛𝑖𝑜𝑛 = 3.16𝑘𝑊

The 𝑤 angular velocity of the pinion must be determined to choose the pitch required for the chain system. This is can be done by using the velocity ratios.

𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝑠𝑝𝑟𝑜𝑐𝑘𝑒𝑡

𝐷𝑟𝑖𝑣𝑒𝑛 𝑝𝑖𝑛𝑖𝑜𝑛=

𝑁𝑝𝑖𝑛𝑖𝑜𝑛

𝑁𝑤ℎ𝑒𝑒𝑙

1.65 = 𝑁𝑝

36

38

𝑁𝑝 = 59.5𝑟𝑝𝑚 ≅ 60𝑟𝑝𝑚

Calculate selecting power

𝑃𝑠𝑒𝑙𝑒𝑐𝑡 = 𝑃𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑛,𝑝𝑖𝑛𝑖𝑜𝑛 = 𝑓1 × 𝑓2

= 3.16𝑘𝑊 × 1.5 × 0.83

= 3.93𝑘𝑊 ≅ 4𝑘𝑊

Using the selection chart from the Renold Roller Chain Catalogue, using the simplex chain guide.

39

Pitch selection: 11

4 " (31.75mm), pitch simplex chain

Lubrication type: Drip feed type 2

Chain length calculation

From selected pitch selection: 11

4 " (31.75mm)

Renold’s chain catalogue,

Contemplated centre distance: C = 1200mm

Chain length in pitches: To find the chain length in pitches (L) for any contemplated centre distance of a two-point drive, use the formula below:

𝐿 = 𝑍1 + 𝑍2

2+

2𝐶

𝑃+ (

(𝑍2 − 𝑍1

2𝑃 )2

× 𝑃

𝐶)

40

𝐿 = 38 + 23

2+

2 × 1200

31.75+ (

(15

2 × 31.75)

2

× 31.75

1200)

𝐿 = 106.12 𝑝𝑖𝑡𝑐ℎ𝑒𝑠

𝐿 ≅ 106 𝑝𝑖𝑡𝑐ℎ𝑒𝑠

𝑁𝑜𝑚𝑖𝑛𝑎𝑙 𝑐ℎ𝑎𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ, 𝐿𝑚𝑚 = 106 × 31.75𝑚𝑚 = 3365.5𝑚𝑚 Chain elongation must be considered as the chain and sprockets must have enough clearance to avoid interference. The allowable chain elongation is at least 2% of the nominal length of chain. This will account for safe chain wear.

0.02 × 𝑛𝑜𝑚𝑖𝑛𝑎𝑙 𝑐ℎ𝑎𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ = 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑎𝑙𝑙𝑜𝑤𝑒𝑑

𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑎𝑙𝑙𝑜𝑤𝑒𝑑 = 0.02 × 3365.5𝑚𝑚

= 67.31𝑚𝑚

𝑇𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ, 𝐿𝑡𝑜𝑡𝑎𝑙 = 3365.5𝑚𝑚 + 67.31𝑚𝑚

= 3432.81𝑚𝑚

Chain length calculation The contemplated centre distance is usually larger than the exact centre distance. The centre distance is calculated to give installers allowance when during installation. This distance is of horizontal distance of the 2 gears (pinion and wheel).

𝐶𝑒𝑥𝑎𝑐𝑡 = 𝑃

8 [2𝐿 − 𝑍2 − 𝑍1 + √(2𝐿 − 𝑍2 − 𝑍1) − (

𝑃

3.88(𝑍2 − 𝑍1)2)]

𝐶𝑒𝑥𝑎𝑐𝑡 = 31.75

8 [2 × 106 − 38– 23 + √(2 × 106 − 38– 23) − (

𝑃

3.88(38– 23)2)]

𝑪𝒆𝒙𝒂𝒄𝒕 = 𝟏𝟏𝟕𝟑. 𝟖𝟔𝒎𝒎

41

Material Selection Selection of materials during the engineering design process is a critical component of the designs ability to function and survive in the real-world conditions to which it is exposed to (Trivedi, 2014). The aims of material selection are to identify the materials that will have the appropriate properties for the shape and dimensions of the design to carry out its’ purposes intended at the lowest possible cost (Materials and Design, 2000). In this report, the material selections process will be done following the methodologies of Michael Ashby incorporating the Eco-Design Strategy. The design strategy will be based on the minimum carbon foot print and energy consumption design method, focusing on the factors:

• Cost of the material

• Its ability to manufacture and lifecycle

• Environmental considerations

• Chemical properties

• Physical properties

• Mechanical attributes

Following the material selection would be an Eco-audit, this audit will be done on CES and it draws on Grant’s environmental data to study the impact of the material selection on the environment throughout its entire lifecycle.

42

Slat Material Selection

Material Index Using the material index allows the ease of material selection for the materials and their functionality for the given tasks. The slats come under the category of panels (loaded in bending) in the material indices of generic components.

Function:

• The slats are the primary load bearers, supporting 50 crates over the distance of the

conveyor of 36 m.

Primary Objective:

• Reduce the mass of the slats.

Secondary objectives:

• Reduce material costs.

• Limited carbon foot print.

• Limited Embodied energy.

Constraints:

• The slats must support the uniformly distributed load of the crates within a certain

deflection limit.

• Stiffness limited design.

• Fixed length design.

43

Following Ashby’s method for panel mass minimisation Indices for Stiffness & Strength

Using CES level 3 materials selection to show the range of all possible materials:

Once all possible materials are present, limitations are set (based on requirements). This will allow for the selection of materials from a more defined pool. Limiting factors:

• The material must have excellent durability to prolonged periods of UV radiation

exposure from sunlight.

• The material must be able to handle working temperatures between 0℃ − 40℃

• The material must have resistance to prolonged exposure to both fresh and salt water

conditions.

44

• The material must be able to function under small vibrational conditions (vibrations

caused by the moving conveyor)

• Price range up to a maximum of 60𝑁𝑍𝐷

𝑘𝑔

• Young’s modulus up to between 4 𝑡𝑜 500 𝐺𝑃𝑎

• Density up to 5000𝑘𝑔

𝑚3 (Low density reduces weight)

• Non-Magnetic, as to not interfere with other components of the conveyor

After setting the limiting factors, the resulting material selection can be seen below:

From the list of the remaining materials by CES, the first 10 materials were chosen. They were first chosen in the ascending order of cost.

As seen in the table above glass ceramics and Alumino minimizes the mass objective, however they are not a valid realistic choice for the conveyor slats. Glass ceramics do not work will with vibrations that may occur during option and in the event of earthquakes, the damage could be quite substantial.

45

The next two possible options (coloured yellow) lowers the objective cost the most, however the cost secondary cost objective quite high and with a maximum working temperature of 50℃ the material may be affected by high working temperatures and UV radiation heating over the course of its lifetime. It also only has one processing method available (Polymer injection moulding), therefore the remaining two materials are PE-HD (20-30% long glass fiber) and PE-HD(30% glass fiber). The main difference between the two options is the Young’s modulus and cost. PE-HD (30% glass fiber) has a higher main objective but has a lower cost. Both are non-recyclable so, the ruling factor is the deflection calculation which follows.

46

Slat Design After the material selection of the slats the design of slat dimensions is the next step. To ensure the slat is able meet the requirements of supporting the load, it is important to analyse the deflection of the slat in a simulated working environment.

I. Initial calculations

• Crate has a mass of 50 kg

• Crate dimension are 65 x 80 x 30

• Approximately 4 slats per crate including gaps between the slats

(650𝑚𝑚

150𝑚𝑚= 4.3333 ≈ 4 )

• Each slat supports 12.5 kg (50𝑘𝑔

4 𝑠𝑙𝑎𝑡𝑠)

Having similar density properties as the material chosen by the head chief engineer, the design of the slats will be consistent with the calculations previously.

II. Factor of safety Many factors affect the structural and machine components. These factors have been taken into consideration and assigned a value to match. The repeated impact loading of the crates and other design aspects also take in these factors:

• Likely ness of structural or machine abuse (Repeated unexpected impact loading)

• Load magnitude

• Thermal cycling or extreme temperature effects (Including heating due to solar

radiation)

• Seriousness of failure (Singular or compounding effects)

80 cm cm

30 cm

65 cm cm

47

• Dimensional control and effects on assembly (Internal stresses/ induced stress during

assembly)

• Effects of associated assemblies or components (load distribution)

Threat values are based on < 1 being average threat and > 1 being greater threat. The total value will be calculated into the

Definition Threat value

Seriousness of failure

1

Effects of associated assemblies or components

1.2

Dimensional control and effects on assembly

1.4

Thermal cycling or extreme temperature effects

1.7

Likely ness of structural or machine abuse

1.8

Load magnitude

1.7

Factor of safety is the average of the threat levels: 1.52333≈ 1.5 (𝐹𝑂𝑆)

III. Deflection of slats calculation

• 𝑞 = 𝑈𝑛𝑖𝑓𝑜𝑟𝑚𝑙𝑦 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑙𝑜𝑎𝑑 = 147.15 𝑁𝑚

• 𝐿 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑒𝑎𝑚 = 800𝑚𝑚

• 𝐸 = 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑚𝑜𝑑𝑢𝑙𝑢𝑠(𝐸𝑙𝑜𝑛𝑔 = 5.865 𝐺𝑃𝑎, 𝐸𝑠ℎ𝑜𝑟𝑡 = 5.52 𝐺𝑃𝑎)

• 𝐼 = 𝑆𝑒𝑐𝑜𝑛𝑑 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 = 8 × 10−6𝑚4

• 𝜕𝑥 = 𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑥 = 0.4𝑚

48

UDL force per slat:

𝑞 =12.5 × 9.81

0.8 × 1.5 = 147.15 𝑁𝑚

Assume max deflection in the centre:

𝜕𝑥 =𝑞𝑥

24𝐸𝐼(𝐿3 − 2𝐿𝑥2 + 𝑥3)

𝜕𝑥(𝑙𝑜𝑛𝑔) = −1.92 × 10−4𝑚

𝜕𝑥(𝑠ℎ𝑜𝑟𝑡) = −1.77 × 10−4𝑚

From here it can be seen that the PE-HD (30% glass fibre) is the choice for slats.

IV. Slat Manufacturing Manufacturing method plays a major role in the limitation of available materials. The manufacturing would preferably be done in New Zealand. CES has shown several available methods to which the production of the material can be done.

• Polymer Injection moulding – Acceptable

• Polymer Extrusion – Acceptable

• Polymer Thermoforming – Limited

Looking into available New Zealand companies that may provide the services:

• Galantai Plastics Group - Auckland

• Machinetech Ltd - Auckland

• Lane Plastics - Auckland

• Millennium Plastics

Comparing the prices and availability of the processes (these prices are from CES, comparing cost per unit):

Process Cost per unit (NZD) Availability (NZ)

Injection Moulding $64.70 Readily Available

Polymer Extrusion $27.40 Available

Polymer Thermoforming will not be used for this material as it is only limited, meaning the process may cause problems to the slat and reduce functionality. Polymer Extrusion is the cheapest and most effective way. As the slats are not complicated in shape and dimension, the need for machining and dimensional control is reduced. Therefore polymer extrusion will be chosen.

49

V. Slats design

𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 = 2400000𝑚𝑚3

Other possible designs: Cut length wise:

𝑇𝑜𝑡𝑎𝑙 𝑉𝑜𝑙𝑢𝑚𝑒𝑁𝑒𝑤 = 3360000𝑚𝑚3

𝑉𝑜𝑙𝑢𝑚𝑒𝑒𝑚𝑝𝑡𝑦 = 𝑉𝑜𝑙𝑢𝑚𝑒𝑁𝑒𝑤 − 𝑉𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 = 3360000 − 2400000 = 960000

𝑉𝑜𝑙𝑢𝑚𝑒𝑒𝑚𝑝𝑡𝑦 = 96000 = 20 × 𝑑𝑒𝑝𝑡ℎ × 𝑙𝑒𝑛𝑔𝑡ℎ

Assume length of 500mm

96000 = 20 × 𝑑𝑒𝑝𝑡ℎ × 500 𝑑𝑒𝑝𝑡ℎ = 9.6𝑚𝑚

This designed was not followed through due to redundant complexities without much gain.

50

VI. Finite Element Analysis As the price of slats are based a mass per kg, the height of the slats are important in terms of overall cost of the material. Assuming a thickness of 20mm:

𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =𝑀𝑎𝑠𝑠

𝑉𝑜𝑙𝑢𝑚𝑒

∴ 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑎 𝑠𝑙𝑎𝑡 𝑖𝑠 = 1230 × 2.4 × 10−3 = 2.952𝑘𝑔

𝑝𝑒𝑟 𝑠𝑙𝑎𝑡

Applying a UDL force of 147.15 𝑁/𝑚 or an equivalent pressure of 1226.25𝑁

𝑚2

Using solidworks to perform a finite element analysis on the slat (PE-HD(30% glass fiber)), it can be seen that the deflection at the centre is almost negligible at 1.942 × 10−1𝑚𝑚. The slat material is appropriate for the functionality of the conveyor.

VII. Cost of Slats The cost for slats can be separated into two main areas:

• Material costs

• Manufacturing costs

Material costs are the cost directly involved with raw material costs, while manufacturing costs include costs such as cost of overheads and direct and indirect costs. These costs include but not limited to:

• Tooling costs

• Capital costs

• Labour costs

• Overhead rates

51

Using CES part cost estimator with the manufacturing method mentioned above:

Environmental burden will be fully defined in the lifecycle analysis and Eco Audit done using CES.

VIII. Number of slats required Determining the number of slats used by the conveyor is an important part of the cost analysis. The calculated value would be a calculated estimate with a slight variance that errs to side of upward rounding.

• 𝐶𝑜𝑛𝑣𝑒𝑦𝑜𝑟 𝑙𝑒𝑛𝑔𝑡ℎ = 36𝑚

• 𝑆𝑝𝑟𝑜𝑐𝑘𝑒𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟𝑠 = 0.4 𝑚

• 𝐺𝑎𝑝 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑠𝑙𝑎𝑡𝑠 ≈ 5𝑚𝑚

• 𝑆𝑙𝑎𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 0.15 𝑚

𝑇𝑜𝑡𝑎𝑙 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑜𝑛𝑣𝑒𝑦𝑜𝑟 = 36 × 2 + 4 × 𝜋 × 0.4 = 77.0265𝑚

52

Assuming a gap of 5𝑚𝑚 between each slat:

77.0265 = 5(𝑥 − 1) + 150𝑥 𝑥 = 497 ≈ 500 𝑠𝑙𝑎𝑡𝑠

0.4m radius

36m

53

Ergonomics

The client has specifically stated that the final product must comply with legal safety requirements during operation. According to HSWA Health and Safety at Work Act 2015, it is the most recent reformation of the work place safety act to ensure the well-being of workers [1]. The overview of designers, manufacturers, importers, suppliers, installers constructors and commissioners includes: -Duty to ensure plant, substances or structures are without health and safety risks -Duty to test -Duty to provide information From the designer’s perspective to design a conveyor, it is a necessary to take into consideration of the stress on the body of a repetitive action during manual labour. This is important as manual labouring and post several health issues such as serious back injuries, musculoskeletal disorders including occupational overuse syndromes, acute injuries such as sprains and strains of muscles or tendons, injuries sustained through slips, trips and falls [1].

Ergonomics of the human body From Fig D, it is shown that despite the slight variance in handling loads, both males and females both have their strongest manual handling capabilities between the waist/elbow height and the knuckle height. Which means the conveyor height should be designed with the consideration of the typical ground to elbow height of male and females.

Fig D, Guideline Maximum Handling loads

54

Conveyor height According to the employment indicated by manufacturing and construction, most of the workers are found to be males. Therefore, the height of male should be prioritized for deciding the height of the conveyor.

Fig D, Employment in selected industries and sex in NZ [2]

According from “Society at a Glance 2009” [3], The height of average males in NZ was found to be around 177cm and females are found to be 164cm for ages between 20-49. From that, the average corresponding elbow height was found to be 109cm as it is the preferred region for manual labour.

Pallet dimensions

The height of the pallet is found by assuming the density of water as it would give us a the largest expect volume occupied. However, to prevent spillage, the height of the pallet to decide to be around 30 cm. The height of conveyor is found by the different the elbow height found

𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓𝑐𝑜𝑛𝑣𝑒𝑦𝑜𝑟 = 𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑒𝑙𝑏𝑜𝑤 − 𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑟𝑎𝑡𝑒

𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑜𝑛𝑣𝑒𝑦𝑜𝑟 = 109𝑐𝑚 − 30𝑐𝑚 = 79𝑐𝑚 ~ 80𝑐𝑚

55

Fig E, Illustration of the ergonomic height of the conveyor

56

5) Frame Design To investigate the desired frame design, a requirements list is created specifically for the frame based on the constraints derived from ergonomics and load requirements, client overall demands and desirable attributes. Constraints -Must be able to support up to the load of 2000kg -Conveyor frame must be at a height that is accessible by the operator’s elbow, which in this case is to be around 80cm from the ground. -Sections of the structure to constraint to be Environmental requirements -Corrosion resistance (mainly from exposure to salt water, strong winds and sand particles) -Design consideration for seismic activity Desirable Attributes -Low cost -Easy maintenance that can be achieved by having simple frame design -Minimal footprint -Clearance space underneath the structure that will hold the conveyor for possible placement of motor or for the ease of cleaning and maintenance. -Minimise the mass of the components that make up the structure -Lifetime should be up to 20 years Safety requirements -Frame design must consideration possible guarding to be implemented.

57

Frame design concept generation

Type A: Truss frame

The reasoning behind a truss frame is that it allows the load to be distributed to the sided of the frame for a better support. The shape allows for structure more support and therefore allows fewer materials to achieve the same function. Type B: Solid frame

Solid block frames allow for a simpler installation process as the whole beam will be extruded as a whole piece. A solid frame also means that the inside of the structure will be enclosed and act a guarding to minimise hazards exposure. Type C: Simply supported frame

58

Concept Selection The selection matrix for initial concept frame design was made according to the requirements of conveyor. The matrix gave 1 to 3 rating based on minimal design footprint, Hazzard exposure, cost, manufacture, assembly, material, design and transportation. These were then added up for each concept to get the total point and the concepts with the highest one was chosen.

Items/Types Type A Type B Type C

Minimum footprint 3 2 3

Hazzard exposure 2 3 2

Cost 3 1 3

Ease of manufacture 3 1 3

Ease of assembly 1 3 2

Maintenance 2 1 2

Usage of materials 3 1 3

Parts availability 3 1 3

Easy to design 1 3 2

Easy to transport 2 1 3

Total Point 23 17 26

*Note: 3=very good 1=not good Concept C scored the highest from as show from the design matrix, presenting advantages in minimal footprint, east of manufacture, cost, parts availability and ease of transportation.

59

Conveyor shaft mounting sketch

Fig E , Sketch of the mounting of the shaft on to the assembly

Fig F, Sketch of the head shaft and tail shaft

60

Loading analysis

Frame Stress Analysis Calculations The frame is analysed for four columns (legs) and the length of the structure where uniformly distributed load is applied as seen from the figure below.

The bending caused by the applied force and moment is only analysed in the length section in columns as the legs are of a square area. The stress analysis for frame under UDL is analysed for the length direction as it is in the direction of motion of the conveyor.

Stress analysis for legs For the lengthwise section, the neutral axis is this orientation through the beam: The total weight of the structure and supports = 597 N/0.65m. The length of the total frame = 36m Number of sections it has been divided = 6

Therefore, the length of a frame section = 36

6= 6𝑚

Total weight on 1 frame structure = 597

0.65 × 6 = 5510.8 𝑁

As there are 4 legs on 1 frame section,

𝑊1

𝐷1 Neutral Axis

𝐷2 𝑊2

61

Weight occupied by 1 leg (F) = 5510

4= 1377.7 N

Due to dividing the total weight at the centre of the section to vertical concentrated loads to the 4 legs, this results in a moment. This is calculated from the equation below

𝑀 = 𝐹 × 𝑑 Where d = distance between the centre of the frame section to the leg. The moment of inertia is to be calculated for the hollow area. This is calculated as follows

𝐼𝐻𝑜𝑙𝑙𝑜𝑤 = 𝐼𝑂𝑢𝑡𝑒𝑟𝐴𝑟𝑒𝑎 − 𝐼𝐼𝑛𝑛𝑒𝑟𝐴𝑟𝑒𝑎 Since the area is a square, the moment of inertia can be determined for each area as

𝐼 = (𝑏)4

12

The stress is determined by using the following equation where it incorporates the bending moment and inertia as follows

𝜎 = 𝑀𝑦

𝐼

Where y = width of the leg (76mm in design) By using Microsoft Excel spreadsheet this is determined as follows

Vertical beams on the length of the frame

Area Dimensions

Total Weight (N) 5510.8 length (m) 3

Force per leg (N) 1377.7 b1 (m) 0.076

Moment (Nm) 4133.1 d1 (m) 0.076

y (m) 0.076 b2 (m) 0.06

I1 (xE-6 m^4) 2.78 d2 (m) 0.06

I2 (xE-6 m^4) 1.08

I (xE-6) 1.7

Sigma (in MPa) 184.8

Knowing that the maximum yield strength is 250 MPa, this incorporates a safety factor of 1.35. Therefore, it can be concluded that the dimensions of the design for the legs of each sections is appropriate and would produce a safer design structure.

62

Stress analysis for frame top

The top of the frame for each section is under UDL as seen from the above figure. The length of a frame section is designed to be 6m. The length section stress analysed below Total weight on 1 frame structure = 5510.8 𝑁 (𝑎𝑠 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑏𝑒𝑓𝑜𝑟𝑒) As there are 2 length sections on 1 frame section,

Weight occupied by 1 length section (F) = 5510

2= 2755.4N

The inertia is calculated for a rectangular area section as shown from the equation below

𝐼 = 𝑏𝑑

12× (𝑏2 + 𝑑2) 𝐼𝐻𝑜𝑙𝑙𝑜𝑤 = 𝐼𝑂𝑢𝑡𝑒𝑟𝐴𝑟𝑒𝑎 − 𝐼𝐼𝑛𝑛𝑒𝑟𝐴𝑟𝑒𝑎

And the stress is calculated as before,

𝜎 = 𝑀𝑦

𝐼

63

Where y = distance from neutral axis to the outer - Thickness (38mm in design)

Horizontal Beam UDL Area Dimensions Total Weight (N) 5510.8 length (m) 3 Force per side (N) 2755.4 b1 (m) 6 Moment (Nm) 8266.2 d1 (m) 0.82 y (mm) 0.038 b2 (m) 5.848 I1 (xE-6 m^4) 15.04 d2 (m) 0.668 I2 (xE-6 m^4) 11.28 I (xE-6) 3.8

Sigma (in MPa) 83.6

Therefore, the design is well within limitations.

64

Frame FEA Analysis Fixing at the legs

Application of load on to the frame

65

Meshed Design

Stress Analysis

66

Displacement Analysis

It can be concluded that the results of the stress analysis is very evenly spread out with a stress that is of much lower value compared to the yield stress of the material. Hence this will lie on the elastic range. The displacement too is very small of 0.9mm at maximum assuring safer and better design with respect to the material selection as well as the frame design that have been chosen.

67

The resonant frequency of the ground vs building/structure resonance

The period of earthquake is typically at 0.1 to 100 seconds according to earthquake.usgv.gov A magnitude between 2-4 is likely the largest magnitude of earthquake Castle point will experience. Assuming the period of earthquake is around 0.3 seconds Frequency = 1/period = 1/0.3seconds = 3Hz All building/structure have a natural frequency, if the period of the ground motion matches the natural frequency of the building, it will undergo large oscillation. Model the structure of the frame as shown below:

Assumptions: -lumped masses on each section -Rigid beam -Neglect axial deformations in beams and column. To obtain the natural frequency of the system: Find the deflection of the beam when the total load of the slats and pallets is acting on the frame. This will allow obtain the spring constant and natural damping properties of the system. From that, the natural frequency of the system can be obtained. Compare the natural frequency of the conveyor structure with and the natural frequency of the earthquake. If the frequencies are apart, this means that movement by the earthquake will have little effect on the oscillation of the frame structure.

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CES of Frame To find a suitable material for the frame, CES was used to help with selecting the appropriate materials for the working conditions and loads at Castle Point. The frame will be susceptible to harsh weather condition, corrosive environments (salt water), moment loading and the potential threat of an earthquake. This takes in to consideration, the functionality and material requirements that are needed to fulfil the design requirements.

Material Index Using the material index allows the ease of material selection for the materials and their functionality for the given tasks. The frame come under the category of beam (loaded in bending) in the material indices of generic components.

Function:

• The frame must support bending moments and provide a stable working platform.

Primary Objective:

• Reduce the mass of the frame

Secondary objectives:

• Reduce material costs.

• Limited carbon foot print.

• Limited Embodied energy.

Constraints:

• The frame must support the total load of the conveyor components with a certain

deflection limit.

• Stiffness limited design.

• Length of the frame is fixed (36 m)

Following Ashby’s method for panel mass minimisation Indices for Stiffness & Strength

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Limiting factors:

• The material must have excellent durability to prolonged periods of UV radiation

exposure from sunlight.

• The material must be able to handle working temperatures between 0℃ − 40℃

• The material must have resistance to prolonged exposure to both fresh and salt water

conditions.

• The material must be able to function under small vibrational conditions (vibrations

caused by the moving conveyor)

• Non-Magnetic, as to not interfere with other components of the conveyor

Based on the cheapest materials that have the highest Young’s modulus (greater than 200 GPa)

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The top material candidates are:

• Low alloy steel (0.958 − 1.14 𝑁𝑍𝐷/𝑘𝑔)

• Low carbon steel (0.889 − 1.07 𝑁𝑍𝐷/𝑘𝑔)

• Stainless steel (7.79 − 8.47 𝑁𝑍𝐷/𝑘𝑔)

Option are to choose a stainless steel material that has very good corrosion resistance or a material with low corrosion resistance and apply a galvanic coating to help prevent corrosion.

Cost of stainless steel (not including manufacturing and transportation):

• Total volume of one section of frame is 73470220 𝑚𝑚3 = 0.07347022𝑚3.

• Average cost of stainless steel being 8.13 𝑁𝑍𝐷/𝑘𝑔

• Density of stainless steel 7.85 × 103𝑘𝑔/𝑚3

• Number of frame sections is 6

𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =𝑀𝑎𝑠𝑠

𝑉𝑜𝑙𝑢𝑚𝑒

𝑀𝑎𝑠𝑠 = 𝑉𝑜𝑙𝑢𝑚𝑒 × 𝐷𝑒𝑛𝑠𝑖𝑡𝑦

𝑀𝑎𝑠𝑠 = 0.007347022 × 7.85 × 103 = 57.7741 𝑘𝑔

𝑇𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡 𝑜𝑓 𝑓𝑟𝑎𝑚𝑒𝑠 ( 𝑆𝑡𝑎𝑖𝑛𝑙𝑒𝑠𝑠 𝑠𝑡𝑒𝑒𝑙) = 57.741 × 8.13 × 6 = 2813.343 𝑁𝑍𝐷

71

Cost of low carbon steel (not including manufacturing and transportation):

• Total volume of one section of frame is 73470220 𝑚𝑚3 = 0.07347022𝑚3.

• Average cost of stainless steel being 0.9795 𝑁𝑍𝐷/𝑘𝑔

• Density of stainless steel 7.85 × 10^3

• Number of frame sections is 6

𝑀𝑎𝑠𝑠 = 0.007347022 × 7.85 × 103 = 57.7741 𝑘𝑔

𝑇𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡 𝑜𝑓 𝑓𝑟𝑎𝑚𝑒𝑠 ( 𝑙𝑜𝑤 𝑐𝑎𝑟𝑏𝑜𝑛 𝑠𝑡𝑒𝑒𝑙) = 57.741 × 0.9795 × 6 = 339.344 𝑁𝑍𝐷

Cost of galvanising and/or painting Using CES to determine the possible galvanising or paint coating protections:

Possible materials include:

• Aluminium commercially pure, 1-0

• Zinc commercially pure, High grade, min 99.9%

• Zinc-copper alloy

• Molybdenum, 360 grade, stress relieved

• Molybdenum alloy 363, TZM

72

From the selection of possible materials, the choice came down between Aluminium, Zinc (commercially pure) and molybdenum (360 grad, stress relieved).

Zinc has the best galvanic potential and the second best price. Although it may not have the most environmentally friendly, it is able last the entire life of the conveyor of 15 years. Zinc galvanising is an option used for protecting metals in both the maritime and aircraft industry. The cost of zinc galvanising is roughly 4.36 𝑁𝑍𝐷/𝑘𝑔

(Barlow, 2014) With the added cost of galvanising, the total price of the frame for low alloy steels comes to 5.3395 𝑁𝑍𝐷/𝑘𝑔 bringing the total materials cost of the frame to:

𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑠 𝑐𝑜𝑠𝑡 𝑜𝑓 𝑓𝑟𝑎𝑚𝑒𝑠 ( 𝑙𝑜𝑤 𝑐𝑎𝑟𝑏𝑜𝑛 𝑠𝑡𝑒𝑒𝑙) = 57.741 × 4.36 × 6

= 1510.5 𝑁𝑍𝐷 Therefore, the chosen method is to use low alloy steel with zinc galvanising.

73

Foot support These are several ways which the foot support of the leg can be attached to the ground. This can include drilling a hole into the ground and re-cement the feet into the ground as shown in Fig I or by wielding metal parts to together and fastening the base into the ground using concrete screws shown in Fig G.

Fig G, cementing the support pillars into the ground

To prevent the conveyor from swaying during strong winds up to 15 miles per hour, it is important to have a strong base support. The advantage of burying the post into the ground is little to not lateral bracing or base suppose required and keeps the structure from swaying with little work.

Fig H, Different types of ground support based on requirements

In effect the different types of footing at act to spread the load of the deck over a larger surface so the ground can adequately support it []. The benefits of having the footing installed lower into the ground prevents the conveyor from moving as foot shrinks during winter. However, the disadvantage of drilling into the ground is the accumulation of water and other substances inside the cavities of the concrete causing corrosion and other detrimental effects which would require a replacement.

74

Feet support using a metal base and held in place by screws

Figure I, feet support held in place on wood and concrete grounds

This method of foot mounting is a more typical method, the feet are held in place by 4 bolts at the base. Since the bolts will be drilled into different both concrete and wood floors, different types of bolts would need to be employed. The pillar support is wielded to the metal base as shown in Fig I. This method is more convenient over drilling the support pillars into the ground as the supports can be easily maintained or replaced. The support feet will most likely be made of iron, therefore, it must be coated in a layer of paint to prevent corrosion. Initial design sketches The concept shown on Fig J1 consist of an adjsutable thread so that the height of the feet can be manually adjusted for different heights. Fig J2 demonstrates a concept that spread out the load of the conveyor over a larger area therefore it is able to withstand a larger load. Fig J3 concept is the simplest of the the 3 cocnepts with the pillar direct welded to the base plate.

Fig J1 Fig J2 Fig J3

75

Chosen foot mounting configuration

The concept design is extremely simple, which makes it extremely easy to produce and maintain. There are few components that make up the part which makes the part easily replaces in case of failure or corrosion. Due to the simple design characteristics, installation can be done by a regular technician with very simple tools and components.

Figure K, Diagram of feet supports

76

Shaft design Head shaft Force Analysis

Loading configuration Planes YZ – vertical plane XY – Horizontal plane Loadings 𝐶𝑝 = Conveyor chain pull = 6kN

𝑤𝑠 = Weight of slats and pallets = 0.193kN 𝑃𝑐 = Transmission chain pull = 6.86kN

𝑤𝑠

𝑤𝑠

Z - Axis

Y - Axis

X - Axis

𝑃𝑐, 𝑥

𝑃𝑐 , 𝑦

𝐶𝑝

2

𝐶𝑝

2

𝑃𝑐

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Force Calculations Procedure Followed:

1. Calculate Conveyor Chain Pull “C𝑝”

2. Find weight due to conveyor chain sprockets and slats

3. Find Driving Transmission Chain Pull “Pc” and its X and Y components

4. determining the Loading reactions, Shear forces and Bending moments

5. Determining the overall Shear forces and Bending moments

Finding C𝑝

From Conveyor Chain Selection, the chain pull C𝑝 can be found:

C𝑝 = 9.81 × μc[(2.05 × wc × L) + w]

C𝑝 = 9.81 × μc[(2.05 × 22.7 × 36) + (46 × 50)]

C𝑝 = 5.85 kN ≈ 6kN

Weight of Sprockets Part Number 200571## Weight = 19.7kg

Ws = 19.7 × 9.81 = 193.26N

Finding Pc

Pc =𝑇𝑚𝑎𝑥

0.5𝐷1

Tmax = Theadshaft = 1.3653 𝑘𝑁𝑚

D1 = Dwheel = 398.25 𝑚𝑚

Therefore,

Pc =1.3653 × 103𝑁𝑚

0.5(398.25 × 10−3)≈ 6.86 𝑘𝑁

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To find the X and Y components of Pc, the geometry of the transmission sprockets must be considered.

Calculating angle of elevation 𝑟1 = 199.125 𝑚𝑚

𝑟2 = 120.68 𝑚𝑚

𝐿 = 1173.86 𝑚𝑚

𝑥 = 800 − 250 − 𝑟2 = 429.32 𝑚𝑚

𝛼 = tan−1 (𝑥

𝐿) = tan−1 (

429.32

1173.86) = 20.1°

Calculating 𝑃𝐶 components Pc, 𝑥 = 𝑃𝑐 cos(𝛼) = 6.86 cos(20.1) = 6.44𝑘𝑁 Pc, 𝑦 = 𝑃𝑐 sin(𝛼) = 6.86 sin(20.1) = 2.36𝑘𝑁

1. Determining the Shear forces and the Bending moments

To determine the shear forces and the bending moments of the Figure (the first figure in the shaft design), the shaft has been divided into two perpendicular planes in the horizontal and vertical directions. The horizontal plane is in the XY axis while the vertical plane is in the YZ axis respectively of the Figure (Figure 1). The Shear forces and the Bending moments of each plane is determined using basic solid mechanical calculations and graphical representation is shown below.

𝑃𝑐, 𝑥

𝑃𝑐 , 𝑦 𝑃𝑐

79

Head Shaft Horizontal Plane (X-Y) Shear and Bending Diagrams

Shear Force Diagram

Bending Moment Diagram

80

Head Shaft Vertical Plane (Y-Z) Shear and Bending Diagrams

Shear Force Diagram

Bending Moment Diagram

81

Overall Shear forces and Bending moments For determination of the true shear forces and bending moments of each section and point respectively of the shaft, Pythagoras theorem is used of the obtained results of the two planes as shown below. Thereby the overall radial forces, shear forces and the bending moments are determined.

Radial Reaction Forces:

𝑅𝐶 = √6.442 + 2.362 = 6.86𝑘𝑁

𝑅𝐴 = √4.082 + 2.842 = 4.97𝑘𝑁

𝑅𝐷 = √32 + 0.1932 = 3𝑘𝑁

𝑅𝐸 = √32 + 0.1932 = 3𝑘𝑁

𝑅𝐵 = √3.642 + 0.0972 = 3.64𝑘𝑁

Shear Forces:

𝑆𝐹𝐶𝐴 = √6.442 + 2.362 = 6.86𝑘𝑁

𝑆𝐹𝐴𝐷 = √2.362 + 0.4832 = 2.41𝑘𝑁

𝑆𝐹𝐷𝐸 = √0.6372 + 0.292 = 0.7𝑘𝑁

𝑆𝐹𝐸𝐵 = √3.642 + 0.0972 = 3.64𝑘𝑁

Bending Moments:

𝑀𝐶 = 𝑀𝐵 = √02 + 02 = 0𝑘𝑁𝑚

𝑀𝐴 = √772.82 + 283.22 = 823.06𝑘𝑁𝑚𝑚

𝑀𝐷 = √952.392 + 246.52 = 983.77𝑘𝑁𝑚𝑚

𝑀𝐸 = √436.442 + 11.632 = 436.6𝑘𝑁𝑚𝑚

82

Head Shaft Sizing - Standard AS1403 The obtained Bending moments are used to determine the diameter of the shaft concerned. As there are different bending moment values at different sections of the shaft, diameter of each of these sections are to be determined separately. The process for this is explained in Appendix 1. The process contains an iteration analysis where initially a trial diameter value is set. After following the steps mentioned in Appendix 1, a value for the diameter of that section of the shaft is obtained. If this value is smaller than the trail diameter value, this same process is repeated until the obtained value is of lowest. This process has been carried out in Microsoft Excel and the obtained results with each iteration for that section is represented as shown below.

Excel Shaft Sizing Iterations For Section C-A of the Shaft

C-A

Iteration 1 Iteration 2 M_q 823 M_q 823 T_q 1400 T_q 1400 T_e 1685.183 T_e 1685.183

Trail D 35 Trial D 55 L_CA 120 L_CA 120 L/D 3.428571 L/D 2.181818 K_s 1.35 K_s 1.47 K_key 1.5 K_key 1.5 K_bearing 1.6 K_bearing 1.6 K_step 0 K_step 0

K 1.6 K 1.6

D 49.43764 D 50.34168

It is observed from above that for Iteration 1, the section sizing of C-A resulted in a larger diameter than the trial diameter obtained by the equivalent torque formula. Therefore, a higher value for diameter has been put forward as the trial diameter of the next iteration (Iteration 2). Iteration 2 resulted in a smaller diameter than the trial diameter, which means the newly obtained diameter is in between the two boundary conditions (obtained diameter of first iteration and the trail dimeter of the second iteration) resulting an appropriate size for the shaft. The following consists of the iterations for the rest of the shaft sections.

83

For Section A-D of the Shaft

A-D

Iteration 1 Iteration 2 M_q 984 M_q 984 T_q 1400 T_q 1400 T_e 1795.715 T_e 1795.715 Trail D 40 D1 55 Trail D 55 D1 60 L_AD 76 D 40 L_AD 76 D 55 L/D 1.9 D1/D 1.375 L/D 1.381818 D1/D 1.09091

Delta 0.045 Delta 0.1 K_s 1.35 r 1.4 K_s 1.5 r 1.5 r/D 0.035 r/D 0.02727 K_key 1.5 Z 0.08 K_key 1.5 Z 0.12727 K_bearing 1.6 K_bearing 1.6

K_step 1.6 K_step 1.5 K 1.6 K 1.6 D 50.39851 D 51.51698

For Section D-E of the Shaft

D-E

Iteration 1 Iteration 2 Iteration 3 M_q 436 M_q 436 M_q 436 T_q 1400 T_q 1400 T_q 1400 T_e 1481.714 T_e 1481.714 T_e 1481.714 Trail D 30 Trial D 47.18845 Trial D 50

L_DE 810 L_DE 810 L_DE 810 L/D 27 L/D 17.16522 L/D 16.2 K_s 1.3 K_s 1.43 K_s 1.45 K_key 1.5 K_key 1.5 K_key 1.5 K_bearing 1.6 K_bearing 1.6 K_bearing 1.6 K_step 0 K_step 0 K_step 0 K 1.6 K 1.6 K 1.6 D 47.18845 D 48.19913 D 48.35064

84

For Section E-B of the Shaft

E-B

Iteration 1 Iteration 2 M_q 0 M_q 0 T_q 700 T_q 700 T_e 697.1505 T_e 697.1505 Trail D 30 D1 60 Trail D 40 D1 60 L_EB 120 D 30 L_EB 120 D 40

L/D 4

D1/D 2 L/D 3

D1/D 1.5

Delta 0

Delta 0.03

K_s 1.25 r 2.9 K_s 1.35 r 2.9

r/D 0.096667 r/D 0.072

5

K_key 1.5 Z 0.096667 K_key 1.5 Z 0.102

5 K_bearing 1.6 K_bearing 1.6 K_step 1.5 K_step 1.5

K 1.6 K 1.7

D 36.46752 D 37.62567

Standard AS1403 Head Shaft Diameter Results

The obtained results for the diameters for each section of the shaft is then verified using first principles. The results obtained using first principles would be lower as to the results obtained from the Standard AS1403. It is noted that the resulted values obtained are of minimum required values of diameters.

85

Head Shaft Sizing – First Principles The process followed for sizing the shaft diameters by first principles can be found under Appendix 2. These obtained values from the process is also analysed and calculated using Microsoft Excel as shown below to obtain the minimum diameter at each section.

C No Step A No Step D No Step

M 0 M 823 M 983.77 T 1400 T 1400 T 1400 K_fm 1 K_fm 1 K_fm 1

K_ft 1 K_ft 1 K_ft 1 K_m 1.25 K_m 1.25 K_m 1.25 K_t 1 K_t 1 K_t 1 Te 1400 Te 1737.333 Te 1863.382 No FS No FS No FS fs 1.13E+08 fs 1.13E+08 fs 1.13E+08

d 0.039818 d 0.042789 d 0.0438

E No Step

M 436.6 T 700 K_fm 1 K_ft 1 K_m 1.25 K_t 1

Te 887.6052 No FS fs 1.13E+08

d 0.034207

There is no moment or torque at B and so the minimum diameter is dependent on the shear loading.

86

Calculation for minimum shaft diameter at B

𝜏 =𝑉𝑄

𝐼𝑡

For a solid circular shaft

𝜏 =4𝑉

3𝐴

Endurance limit = 225 MPa

Endurance in shear = 225

2= 112.5 𝑀𝑃𝑎

By rearranging Equation above,

𝐴 =4𝑉

3𝜏=

4 × 3.64 × 103

3 × 112.5 × 106= 4.31 × 10−5 𝑚2

𝜋

4𝑑2 = 4.31 × 10−5

𝑑𝐵 = 7.4 × 10−3𝑚 = 7.4 𝑚𝑚

This shows that the shaft diameter will be decided by design and not from failure.

Head Shaft Diameter Results Using First Principles From the results obtained by using first principles, the shaft design representation of the minimum diameters is as follows.

39.8mm 42.79mm 43.8mm 35.14mm 7.4mm

120mm 76mm 120mm 30mm

87

Design Consideration: Head Shaft Profile Modification for Bearing Fits Even though the Standard AS1403 accounts for the stress in the shaft, it does not take into consideration aspects such as manufacture and for of practical use. Therefore this initial design obtained before is to be modified in a way that is of use in the real world applications. Identification of problem In Bearing: In the head shaft AS1403 shaft diameter design, there is a very large step down from section A-B to B-F. This step down is of 20mm. When the bearing is selected the bearing height should be twice more than the step height. Therefore if this is further considered as it is, it would require a bearing height of 40mm. In Gear: The conveyor gear at B is at a step in the shaft. Therefore the conveyor gear hub will not sit flat on the shaft causing rotation around point E that would seriously damage the system. Problem Analysis For bearing, by tapering the shaft from point E to point B in such a way as to get a step down of 2mm would prevent having the shaft sliding out of place longitudinally. Also this would solve the gear problem where the gear then can sit flat on the shaft eliminating the problem of having an uneven surface at the conveyor gear.

88

Comparison between First Principles and Standard 1403 Sizing

Point First Principles Minimum Diameter

Standard AS1403 Minimum Diameter

C 39.82mm 55mm

A 42.79mm 55mm

D 43.80mm 60mm

E 35.14mm 60mm

B 7.40mm 40mm

It should be noted that the diameters of the first principle shaft sizing are smaller than the Standard 1403 shaft sizing. This shows that the Standard shaft sizing is safe to be considered. The minimum diameters obtained for each section of the shaft does not include design considerations such as bearings and keyway fittings.

Head Shaft Final Profile

The final head shaft profile is based on the Standard AS1403 Sizing refined through design considerations of gears and bearings as seen from Figure below.

A similar process as above was followed to obtain the minimum diameter for each section of the tail shaft design.

89

Head Shaft FE Analysis For analysis of the head shaft it was decided to divide the shaft in half during analysis as to obtain accurate results. Therefore the head shaft has been divided in to two parts such that one part consists of the section without power transmission elements applied on it while the other does. Head shaft side without power transmission elements

Stress (von-Mises)

Displacement

90

Head shaft with power transmission section Stress (von-Mises)

Displacement

91

Tail Shaft Force Analysis From the above Figure, it is observed that the tail shaft is at the other end of the conveyor system to the head shaft which behaves as an idler for the chain system to change direction and return to the head shaft underneath the conveyor system. Tail Shaft Loading Configuration Planes Planes YZ – vertical plane XY – Horizontal plane

Tail Shaft Head Shaft

𝑤𝑠

𝑤𝑠

Z - Axis

Y - Axis

X - Axis

𝐶𝑝

2

𝐶𝑝

2

Loadings 𝐶𝑝 = Conveyor chain pull = 6kN

𝑤𝑠 = Weight of slats and pallets = 0.193kN

92

Tail Shaft Horizontal Plane (X-Y) Loading

𝐶𝑝

2= 𝑃1 = 3 𝑘𝑁 𝐴𝑦 = 3 𝑘𝑁

𝐶𝑝

2= 𝑃2 = 3 𝑘𝑁 𝐵𝑦 = 3 𝑘𝑁

Shear Force Diagram

Bending Moment Diagram

93

Tail Shaft Vertical Plane (Y-Z) Loading

𝑤𝑠 = 𝑃1 = 0.193 𝑘𝑁 𝐴𝑦 = 0.193 𝑘𝑁

𝑤𝑠 = 𝑃2 = 0.193 𝑘𝑁 𝐵𝑦 = 0.193 𝑘𝑁

Shear Force Diagram

Bending Moment Diagram

94

Tail Shaft Loading Magnitude Calculations Following the similar process as to the head shaft design, the forces have been split into the vertical and horizontal planes. The resulted values are determined using the Pythagoras theorem to gain the overall results. A symmetricity of the shaft is seen from the bending moment diagrams from the middle vertical line.

Radial Reaction Forces:

𝑅𝐴 = √32 + 0.1932 = 3𝑘𝑁

𝑅𝐷 = √32 + 0.1932 = 3𝑘𝑁

𝑅𝐸 = √32 + 0.1932 = 3𝑘𝑁

𝑅𝐵 = √32 + 0.1932 = 3𝑘𝑁 Shear Forces:

𝑆𝐹𝐴𝐷 = √32 + 0.1932 = 3𝑘𝑁

𝑆𝐹𝐷𝐸 = √02 + 02 = 0𝑘𝑁

𝑆𝐹𝐸𝐵 = √32 + 0.1932 = 3𝑘𝑁 Bending Moments:

𝑀𝐴 = 𝑀𝐵 = √02 + 02 = 0𝑘𝑁𝑚

𝑀𝐷 = 𝑀𝐸 = √2282 + 14.672 = 228.5𝑘𝑁𝑚𝑚

95

Tail Shaft Sizing - Standard AS1403 The tail shaft does not consists of the transmission chain pull, thereby giving a symmetrical and simpler shaft as in comparison to the head shaft design. The shaft diameter iteration analysis process that was followed in the head shaft is too followed here but only for section A-B of the design. For Section A-B of the Shaft

A-B

Iteration 1 Iteration 2

M_q 228.5 M_q 228.5 T_q 0 T_q 0 T_e 262.775 T_e 262.775 Trail D 20 D1 N/A Trail D 30 D1 35 L_AB 962 D N/A L_AB 962 D 30 L/D 48.1 D1/D N/A L/D 32.06667 D1/D 1.166667 Delta N/A Delta 0.07 K_s 1.13 r N/A K_s 1.27 r 1.9 r/D N/A r/D 0.063333

K_key 1.5 Z N/A K_key 1.5 Z 0.133333 K_bearing 1.6 K_bearing 1.6 K_step 1.6 K_step 1.5 K 1.6 K 1.6

D 25.39967 D 25.89896

The middle section diameter of the tail shaft is approximated to a value of 30mm as per Standard AS1403.

96

Tail Shaft Sizing – First Principles In determination of the tail shaft using first principles, similar process to head shaft has been followed. The main difference between the two shafts is the torque loading. That is the torque on the tail shaft is zero as it is an idler shaft that is not been driven by any external forces as the transmission chain pull that is seen on the head shaft. At Conveyor Gears – Points D and E M = 228.5 Nm

𝐾𝑀 = 1.25

𝐾𝑡 = 1 No step so 𝐾𝑓𝑚 = 𝐾𝑓𝑡 = 1

Τ𝑒 = √((𝐾𝑀𝐾𝑓𝑚𝑀)2

+ (𝐾𝑡𝐾𝑓𝑡𝑇)2

)

Τ𝑒 = 285.625 𝑁𝑚

𝑑 = √16𝑇𝑒

𝑓𝑠𝜋

3

= √[16 × 285.625

1.13 × 108 × 𝜋]

3

= 0.0234𝑚 = 23.44𝑚𝑚

The minimum possible diameter for section D-E is then d = 23.44mm However, the diameter will be set by the minimum allowed for the respective sprockets. In this case, the diameter will be of 48 mm. Tail Shaft Ends – Points P and Q At the ends of the shaft, failure depends on shear. Same as the head shaft.

𝜏 =𝑉𝑄

𝐼𝑡=

4𝑉

3𝐴

Endurance in shear = 225

2= 112.5 𝑀𝑃𝑎

A = 4×3000

3×112.5×106= 3.55 × 10−5

𝜋𝑑2

4= 3.55 × 10−5

𝑑 = 6.728 × 10−3 = 𝟔. 𝟕𝒎𝒎

97

Design Consideration: Tail Shaft Minimum Diameter While the standards AS1403 and the first principles gives the minimum permissible diameters for the sections of the shaft, the minimum practical diameter for the components attached to the shaft must be considered. Problem Identification There are the two conveyor sprockets on the tail shaft consists of the same minimum bore diameter of 48mm. These are larger than the minimum diameter of 30mm from the shaft sizing

Problem Analysis The first principles and standards sizes are compared – when the diameter is sufficient, the diameter will be kept constant. When the sprocket requires a larger diameter, the shaft size is increased to the minimum sprocket diameter. As long as the diameters stay the same as or larger than the minimum standards diameters, the shaft will be safe.

Point First Principles Min. Diameter

Standard AS1403 Min. Diameter

Chosen Diameter Reason

A 7.1mm - 30mm -minimise step

C 28.3mm 35mm 48mm -Min Diameter for Sprocket D 28.3mm 35mm 48mm

B 7.1 - 30mm -minimise step

98

Tail Shaft Final Profile This is the final tail shaft design. It is noted that unlike the head shaft this shaft consists of symmetry at the middle and behaves as an idler shaft.

76mm 76mm

30mm 30mm 48mm

99

Bearing Selection

Procedure of bearing selection From table 2.1, the chosen bearing type is the deep groove ball bearing. This is the most suitable based on the conveyor operations. Characteristics such as dual axial direction, the ability to accommodate radial loads are favoured. It is also the most common type of bearing used in variety of fields. Deep groove ball bearings include shield bearings and sealed bearings with grease makes them easier to use [].

fig, Deep groove ball bearing

100

Load rating and life

Basic rating life and basic dynamic load rating A group of seemingly identical bearings when subjected to identical load and operating conditions will exhibit a wide diversity in their durability. This "life" disparity can be accounted for by the difference in the fatigue of the bearing material itself [].

Check for basic load rating is larger than equivalent dynamic load rating:

𝐶𝑟 > 𝑃𝑟

𝐶 : Basic dynamic load rating, g, N {kgf} (𝐶𝑟: radial bearings) 𝑃: Equivalent dynamic load, N {kgf} (𝑃𝑟: radial bearings)

Check for ball bearing life, ball bearing life must be great than 10 years

For ball bearing: 𝐿10 = [𝐶𝑟

𝑃𝑟]

3

> 10 𝑦𝑒𝑎𝑟𝑠

Check that bearing height is greater than the shaft height so that there will be not

overlap of interference between the radial ball bearing and the step in the shaft

D d

Bearing clearance

101

Head shaft bearings

Fig, Illustration of head shaft design

As shown from Fig above, two sets of bearings are required at the two point points of the shaft, point A and point B. Due to the different loading condition of these points (A and B), it is likely that different types of bearings will be required. Bearing selection are determined by the loading shown in bearing dimension tables indicate their load capacity. However, in determining the appropriate bearing type, consideration must also be given to whether the acting load is a radial load only or combined radial and axial load, etc. When ball and roller bearings within the same dimension series are considered, the roller bearings have a larger load capacity and are also capable of withstanding greater vibration and shock loads. []

Radial Load Calculations for Head Shaft Bearing Points From the head shaft loading analysis, the reaction forces are summed to find the radial loads on points A and B. 𝑅𝐴 = 4.97𝑘𝑁 𝑅𝐵 = 3.64𝑘𝑁

Bearing selection at A Ensure that the 𝐶𝑟 > 𝑃𝑟 Fr = 𝑃𝐴= 4.97kN ≈ 5𝑘𝑁 < Cr, OK.

30mm 120mm 76mm 120mm

30mm

Driven gear

Bearings Bearings

Conveyor sprockets

Conveyor sprockets

O C A D E B F

40mm 55mm 60mm

102

From the shaft loading calculation, the diameter of the head shaft at point at was found to be 55mm. From the Renold bearing catalogue, B-12, Deep groove ball bearings

Bearing selection at A is done on excel to calculate the bearing life of particular model of bearings(6811, 6911, and 16011)

103

Bearing at Point A

6811 6911 16011

C_r 8800 C_r 16000 C_r 19400 F_r 5000 F_r 5000 F_r 5000 L_10 5.451776 L_10 32.768 L_10 58.41107 rpm 22 rpm 22 rpm 22 L_10 Hours 4130.133 L_10 Hours 24824.24 L_10 Hours 44250.81

L_10 Years 2.151111 L_10 Years 12.92929 L_10 Years 23.0473

Ensure bearing life > 10 years As shown from the table, both 6911 and 16011 > 10 years, so OK. Ensure there is enough bearing clearance to not interfere with operation.

As seen from the table, 72−55

4= 4.25𝑚𝑚 > 2.5𝑚𝑚(ℎ), OK.

The chosen bearing model for point A is 6911 as it meets of the loading and bearing life requirements.

Bearing selection at B From the shaft loading calculation, the diameter of the head shaft at point at was found to be 40mm. From the Renold bearing catalogue, B-12, Deep groove ball bearings

104

Bearing selection at B is done on excel to calculate the bearing life of particular model of bearings at 40mm diameter (6811, 6911, and 16011)

Bearing at Point B

6808 6908 16008

C_r 6350 C_r 13700 C_r 12600 F_r 3640 F_r 3640 F_r 3640 L_10 5.309053 L_10 53.316 L_10 41.47701

rpm 22 rpm 22 rpm 22 L_10 Hours 4022.01 L_10 Hours 40390.91 L_10 Hours 31421.98 L_10 Years 2.094797 L_10 Years 21.03693 L_10 Years 16.36561

Ensure that the 𝐶𝑟 > 𝑃𝑟 Fr = 𝑝𝐵 = 3.64kN < Cr, so OK. Ensure bearing life > 10 years As shown from the table, both 6908 and 16008 > 10 years, so OK. Ensure there is enough bearing clearance to not interfere with operation. The bearing at point A has a bore diameter of 40mm.

The step height h = 2mm.

As seen from the table, 52−40

4= 3 > ℎ so, OK.

The chosen bearing model for point B is 6908 as it meets of the loading and bearing life requirements.

105

Comparing bearing prices at point A

Comparing bearing prices at point B

106

Tail shaft bearings

Fig, Illustration of tail shaft design

Both of bearings for the tail shaft are of the same type since the diameter of the shaft is the same on the both sides. Radial Load Calculations at Points B and C 𝑅𝐵 = 𝑅𝐶 = 3𝑘𝑁 Therefore, Fr = 𝑃𝐴 = 3kN From the shaft loading calculation, the diameter of the tail shaft at point at was found to be 40mm. From the Renold bearing catalogue, B-12, Deep groove ball bearings Ensure that the 𝐶𝑟 > 𝑃𝑟 All three models have Cr > Fr at 3kN, so OK.

Bearings Bearings

B

A C

D

Sprockets Sprockets

30mm 30mm

107

Bearing Options

Bearing at Point A

6806 6906 16006

C_r 4700 C_r 7250 C_r 11200

F_r 3000 F_r 3000 F_r 3000

L_10 3.845296 L_10 14.114 L_10 52.03437

rpm 22 rpm 22 rpm 22

L_10 Hours 2913.103 L_10 Hours 10692.43 L_10 Hours 39419.98

L_10 Years 1.517241 L_10 Years 5.568973 L_10 Years 20.53124

Ensure bearing life > 10 years As shown from the table, 16006 > 10 years, so OK.

108

The rest of the bearing even though they did the meet the requirements specified the life cycle was low, hence avoided. Bearing Prices

109

Head shaft keyway selection For a rotating shaft, a key is required to connect the rotating element to a shaft. The purpose of the key to prevent relative rotation between the power element and the shaft which can enable torque transmission. For the key to function, a keyway must be present as an insert.

From the head shaft loading analysis, it is known that 𝜙 1=55mm

𝜙 2=60mm

𝜙 3= 40mm

The keyway will be installed on the edge of the rotating element of the shaft

Keyway width

Keyway height

Shaft Diameter

𝜙 2=60mm 𝜙 1=55mm 𝜙 40mm

Keyway length

110

From the keyway chart BS4500 To select the appropriate key, it is important to consider the stresses of the key in shearing and crushing that will cause the key to fail.

Shearing consideration

Forces as the shaft rotates causes shearing stress across the key as shown above The area resisting the shearing:

𝐴𝑠 = 𝑙 × 𝑏 And so, the shear stress on the area:

𝜏𝑠 =𝐹

𝐴𝑠

𝐹 =𝑇

𝑟𝑠ℎ𝑎𝑓𝑡

Using the MAK-A-KEY Steel, the shearing stress of the key is

𝜏𝑠 = 100 𝑀𝑃𝑎 The torque applied on the shaft is calculated to be,

𝑇 = 𝑇𝑆𝑡𝑎𝑟𝑡𝑢𝑝 = 1365.3𝑁𝑚 ≈ 1.4 𝑘𝑁𝑚

111

Shaft at Radius (mm) Force(kN)

C

27.5 50.91

D

30 46.67

Calculating the area of shear:

𝐴𝑠,𝐶 =𝐹

𝜏𝑠=

50910 𝑁

100 × 106𝑃𝑎 = 5.091 × 10−4𝑚2 → 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟 𝑠ℎ𝑎𝑓𝑡 𝑎𝑡 𝐶

𝐴𝑠,𝐷 =𝐹

𝜏𝑠=

46670

100 × 106𝑃𝑎 = 4.667 × 10−4𝑚2 → 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟 𝑠ℎ𝑎𝑓𝑡 𝑎𝑡 𝐷

And the widths of the key for the two shafts are:

𝑏𝐶 = 16 𝑚𝑚

𝑏𝐷 = 18 𝑚𝑚 Solving for length based on shearing failure:

𝑙𝑠,𝐶 =𝐴𝑠,𝐶

𝑏𝐶=

5.091 × 10−4𝑚2

16 × 10−3 𝑚= 31.82 𝑚𝑚 ≈ 32 𝑚𝑚

𝑙𝑠,𝐷 =𝐴𝑠,𝐷

𝑏𝐷=

4.667 × 10−4𝑚2

18 × 10−3 𝑚= 25.93 𝑚𝑚 ≈ 26 𝑚𝑚

Crushing consideration Areas susceptible to crushing stresses: -Hub

Crushing area on key

112

-Key

-Shaft

𝐶𝑟𝑢𝑠ℎ𝑒𝑑 𝑎𝑟𝑒𝑎 = 𝐴𝑐 = 𝑙 × ℎ

2

The stresses on these areas can be found by:

𝜎𝑐 =𝐹

𝐴𝑐

Where,

𝐹 =𝑇

𝑅𝑠ℎ𝑎𝑓𝑡

𝜎𝑐 = 0.6 × 𝜎𝑦 → 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑠𝑡𝑟𝑒𝑠𝑠

Crushing area on key

Crushing area on shaft

113

The table below consist of all the compressive stresses of the different components, this table also takes into the account of the yield stresses of the different materials

Shaft Part Material Yield Stress(MPa)

Compressive Stress (MPa) 𝜎𝑐 = 0.6 𝜎𝑦

G Key MAK-A-KEY 417 250

Hub AISI 304 Stainless Steel

215 129

Shaft AISI 304 Stainless Steel

215 129

C and D Key MAK-A-KEY 417 250

Hub AISI 304 Stainless Steel

215 129

Shaft AISI 304 Stainless Steel

215 129

𝐴 =𝐹

𝜎𝑐

𝑙 =2𝐴

ℎ

Using these formulae, the lengths of the keyway were calculated:

Shaft Part Compressive Stress (MPa) 𝜎𝑐 = 0.6 𝜎𝑦

Area (m2) (10−4)

Length, l (mm)

AB Key 250 2.04

41

Hub 129 3.95

40

Shaft 129 3.95

40

CD Key 250 1.87

34

Hub 129 3.62

66

Shaft 129 3.62

66

114

Choosing the longer lengths to accommodate for the crushing,

Shaft Keyway length due to

crushing (mm)

G 40

C and D 66

Head Shaft Keyway Lengths From analysing both failure modes, it can be deduced that the acceptable keyway lengths are the maximum length calculated.

Head Shaft Keyway Lengths (mm)

C 40

D and E 66

115

Tail shaft keyway selection

Using the MAK-A-KEY Steel, the shearing stress of the key is

𝜏𝑠 = 100 𝑀𝑃𝑎 The torque applied on the shaft is calculated to be,

𝑇 = 0 There is no torque but the chain pull will try and shear the keyway

𝐹 = 3 𝑘𝑁

𝐴𝑠 =𝐹

𝜏= 3 × 10−5 𝑚2

𝑙 =𝐴

𝑏= 2.14 𝑚𝑚

Bearings Bearings

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Like the head shaft, the areas susceptible to crushing stresses will also include hub, key and keyway.

𝐴𝐶 = 𝐶𝑟𝑢𝑠ℎ𝑖𝑛𝑔 𝐴𝑟𝑒𝑎

𝐴𝐶𝐾 = 𝐴𝑐𝑠 = 𝐴𝐶𝐻 = 𝑙 ×ℎ

2

The direct stress on the areas are,

𝜎𝑐 =𝐹

𝐴𝑐

Where,

𝐹 =𝑇

𝑅𝑠ℎ𝑎𝑓𝑡

𝜎𝑐 = 0.6 × 𝜎𝑦 → 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑠𝑡𝑟𝑒𝑠𝑠

Shaft Part Material Yield Stress(MPa)

Compressive Stress (MPa) 𝜎𝑐 = 0.6 𝜎𝑦

At both gears Key MAK-A-KEY 417 250

Hub AISI 304 Stainless Steel

215 129

Shaft AISI 304 Stainless Steel

215 129

𝐴 =𝐹

𝜎𝑐

𝑙 =2𝐴

ℎ

Sigma C F Area(m^2)(E-5) Length (mm)

At both gears Key 250 3000 1.2 2.67 Hub 129 3000 2.3 5.17 Shaft 129 3000 2.3 5.17

Tail Shaft Keyway Length From comparison between values for both shear and crushing failure mode

l = 6 mm

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Safety in Design Safety considerations is important to the fulfilment of technical functions and the protection of humans and the environment []. Designers have recourse to a safety methodology that, following the relevant standard. This includes 3 levels: • direct safety • indirect safety • warnings In case of conveyors, most serious accidents and fatalities involving conveyors result from inadequate guarding between the conveyor belt []. In case of harm resulted from direct and indirect safety hazards, this includes: •Trapped hands – potential cut of crush injuries •Injuries from entanglement •Burn hazards •Bruising and cutting hazard result from friction •Contact or impact injuries To prevent the operator or nearby personnel from experiencing such harms, guarding dangerous parts vital to prevent direct contact. However, indirect methods such as a control system and warnings can also minimize the occurrence of these dangerous scenarios. Some of the ways to control the dangers of scenarios according to work safe NZ includes:

- Fix guards, including interlocked guards. - Install fencing around pulleys to isolate hazards. - Install pop-out rollers to relieve trapping hazards where powered conveyors transfer

goods to conveyors of idle rollers. - Stop/start control switches must be in reach of the operator, including emergency

stops. - When belt conveyors are out of sight of the control start position, audible and visible

warnings must occur before movement. - Never ride on, or cross, conveyors. - Equip crossovers with stairs and handrails. - Never wear ties, loose clothing or gloves. - Do not use conveyors with internal combustion engines Unless adequately ventilated.

Guarding For effective Guarding, it is imperative to recognize and identify the actual or potential hazards, then design the guarding accordingly so they will not affect the functioning of the conveyor.

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Identify areas that require guarding

1. Friction burn hazard from worm gear

2. Electric shock hazard from electric motor

3. Entanglement hazard by chain drive in the area between the transmission sprockets

4. Trapping/crushing hazards between transmission sprockets and chain contact

5. Trapping hazards between the gaps of the slats

6. Trapping hazards between the slats and the conveyor chains

1

2

3

4

5

6

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Guard design/ protective barriers Conveyor side guard

Since the slats will be sitting on top of the chain, there are no trapping hazards involved with the chains as the slats will prevent operator from accessing the chains. Hazard 5 shows that there are trapping hazards between the gaps of the slats. However, effective guarding cannot be implemented there without hindering the function of the conveyor. As a result, operational warnings should be indicated at the side of the frame.

Driven sprocket guard (Head and tail shafts)

fig, guarding for head and tail shafts

The shaft and components such as the sprocket, bearings and gears are exposed at each end of the conveyor. Therefore, a metal guarding as shown on fig will fully cover all the moving components at the end of the conveyor. The guarding also doubles as a table for loading the pallets.

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Chain and gear guard

The hazard typically involved between the transmission chain and gear are impact or rotating shaft-entanglement hazards. Since the gears and chains would not have to be regularly accessible, a fixed guard is therefore suitable in this scenario. The guard is flanged mounted to the worm gear and the conveyor frame. The guarding will prevent access to the moving parts during operation.

Guarding for the Worm gear unit and electric motor

An exposed worm gear can lead to friction burn when the user is in contact with the rotating parts or getting burned when touching the casing of the worm gear. Therefore, the worm gear is covered with a cover with venting holes on the side for to heat to dissipate. The cover

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also protects the worm gear from being overly exposed to the brine and dirt typical of a fish processing plant during operation.

Similarly, to the worm gear unit, the motor post operational hazards such as friction burn and heat generated that can cause harm when in contact with a person during operation. Other than that, it also posts electrical shocking hazards since it is an electric motor. Therefore, it is fully encased in an electric and burn resistant plastic to prevent external contact. In addition, a warming label will also be added to inform the user of such hazards on the cover.

Other safety considerations These are some of the indirect safety protective systems that is typically done to transforms a danger input into an output that alerts the operator. Emergency stop buttons:

The emergency switches will be installed at every 6m of the conveyor at both sides of the conveyor as this is the length of each of the frame section. The purpose of the emergency stop button is to allow any personnel to stop the operation of the conveyor when an emergency is spotted. Warning lights:

The warnings lights will be attach to each side of conveyor to indicate to the operator that the conveyor is in operation or not. This is to account for start-up torque required to start the conveyor so that the operator knows the state of the conveyor. The warning lights also alerts anyone in the proximity of the conveyor when the conveyor is being turned on or off.

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Alarms:

The alarms will work in conjunction with the warning lights to alert anyone in the proximity of the conveyor during start and stop operation. Warning labels/sign: By simply pointing out the dangers and indicate the areas of danger gives the user to revaluate if their course of action can cause direct or indirect harm to themselves or other people. For example, a label mentioning the special features, obstruction or dangerous components can help the user better understand the conveyor even if they have operated a similar conveyor before. Warning labels will be place strategically on areas that is dangerous to the user when they are in contact with during operation, such as on the guarding of chains, gears and other rotating elements as well as the motor to prevent potential shock. Electrical products should be properly grounded to prevent shock. As well as providing electrical interlocks so that high voltage will not be energized.

Fig E, Guards locked by bolts that require a tool to open

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Safety

Workplace safety recommendation list Most accidents causing personal injury can be traced to unsafe work practices by either operating or maintenance personnel. Many accidents occur because the personnel concerned do not realize the danger of improper practices. []

Following are a list of precautions for operating personnel.

1. Do not operate the conveyor without the motor guards in place. 2. Avoid distractions when operating conveyor 3. Never place hands near moving parts. 4. Make sure only trained personnel can operate the conveyor 5. Do not leave the conveyor unattended 6. Make sure all personnel are clear of the conveyor before starting the conveyor 7. Stop the conveyor when someone is in potential danger 8. Stop the conveyor to clear jams or remove foreign objects 9. Do not eat or drink when operating the conveyor 10. Do not wear loose clothing when around the vicinity of the conveyor as it may be

caught in the moving parts 11. Make sure the vicinity is around conveyor doesn’t post any slip or fall hazards always.

A non-slip map can be placed around the conveyor to prevent this.

Following are a list of safety precautions for maintenance personnel.

1. Do not perform maintenance when conveyor is operating. Make sure the motor is disconnected and prevent any personnel from operating the conveyor until maintenance is done.

2. Maintenance personnel should be aware of the hazardous conditions at all time such as broken parts that results in sharp edges.

3. Report all damaged and hazardous parts of the conveyor and irregularities to the proper authority

4. Use proper cleaning equipment for the specific cleaning application. E.g air hoses for chains, long brooms to clean underneath the conveyor.

5. Before restarting the motor, make sure all the guarding is in place. Don’t not operate the conveyor when chain drive is exposed.

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Failure Mode and Effect Analysis(FMEA) FMEA analysis allows the for a detailed analysis for identifying the potential problems within the design. This includes the mode of failure of every component in the system.

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FMEA analysis of shaft design

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The results indicate that the highest RPN is cause by shaft deformation, the reason for this is

due to the high severity of failure as it could lead to severe injury. In combination with a

relatively low level of detection results in a high RPN. In case of an overloading scenario, this

could lead to a hazardous scenario. However, due to presence of the keyway design, the

keyway will typically fail before the shaft as it is made of a softer material when overloading

occurs. This will trigger a repair and parts will be properly re-examined and the state of the

shaft be detected and replaced.