MOD004047 Innovative Product Design And Manufacture

MOD004047 Innovative Product Design And Manufacture

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MOD004047 Innovative Product Design And Manufacture

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MOD004047 Innovative Product Design And Manufacture

1 Download11 Pages / 2,564 Words

Course Code: MOD004047
University: Anglia Ruskin University

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Country: United Kingdom

Question:

Depending on its number of parts, you will be expected to model some or all of your design using both a CAD/CAM system and rapid prototyping. Your tutor will agree with you how much of your design you will be expected to model. Your final design should aim to optimise the manufacturing and assembly costs by the consideration of design for manufacture and assembly principles. In your final report, you will be expected to identify the areas of your design where it has added value to the end user.
The following represents the minimum that one might expect to see included in/with your report.

Identification of the product and a description of the new functions that add value to the end user.
The main stages of the design process – product design specification, concept generation, concept evaluation and selection, detail design as described below.
Detail and assembly drawings together with a parts list that will identify all parts, including standard bought-in components.
The assembly sequence for the design and the manufacturing processes that would be used to manufacture all non-bought-in parts.
The application of DFMA principles and analysis.
A physical model of your design (remember to check with your tutor regarding how much you need to model).

Identification of product and description of new function that add value
The stages of design process
Detail drawings, assembly drawing, and parts list
Assembly sequence and manufacturing processes

Answer:

Introduction
Da Vinci was not only designer but also an artist and a painter and that also reflect in his work as well .His one of the best work was propelling cart and this cart was impossible of design at that time without necessary resources .This cart is a masterpiece having perfect use of gears .All the parts of cart were recreated using solidworks CAD software . Rendering and detailed drawing of all parts are prepared for manufacturing purpose . Analysis of wheel of cart is done and comparing it for prototyping  purpose and cnc manufacturing .
DESIGN PROCESS
The main focus of design process is to recreate the prototype of 3 wheeler cart effectively considering all design aspects .   
DESIGN CONCEPT/SKETCH
Wheel
Wheel provide the rolling motion to cart , the spokes present on its centre balance center of gravity of wheel .

Under the design concept , a rough sketch  of 3 wheeler cart is obtained. This sketch gives us the basic idea about the product visualization. The above sketch shows that gears assembly is surrounded by frame .This frame not only kept all the parts together but also carry out all the load act upon it .1
3 Wheeler cart should be designed similar to propelling cart and all the main components like spring , gears and frame should be used accordingly .
3 Wheels are used in this cart , front single wheels is used for direction purpose and backward two wheels provide the motion

DESIGN SPECIFICATION
Under design specification all the parts of 3 wheeler car should be specify accordingly . All the component are explained below .

Truss/Frames

Truss/Frames are the structure members generally used to carry loads and coupled multiple parts in single unit for smooth operation of assembly. To carry the weight of multiple gear units , here two types of truss are used .
Truss/Frames used are:-

Bottom Truss/frame
Top Truss/frame
Bottom truss

According to its name suggest ,bottom truss present at the bottom of gear assembly .The load of all mechanical components are directly act on this truss .At the surface of bottom truss , multiple threaded holes are present .These holes are used for mounting and rotating  gear shafts . Other holes are used to fastened the bolts with other frame . Fillets are present at corners of truss for stress reduction.
Manufacturing process –
The shape of frame is non uniform , So all the linear parts are joined together with fasteners , at necessary places holes of required diameter are drilled . Fillets are provided by using grinding operation .
The material used for frame is Wood  .Some properties of wood is given below .
Major properties of Wood is shown below 

Property

Value

Units

Elastic Modulus

3000

N / mm^2

Poisson’s Ratio

0.29

N / A

Shear Modulus

300

N / mm^2

Mass Density

160

kg / m^3

Yield Strength

20

N / mm^2

Thermal Conductivity

0.05

W / (m·K)

Table 1.1 Material Properties 

Top truss/frame

Top truss is present at the top of gear arrangement and lock the linear motion of all gears . The top truss share the load of all components with bottom truss .At its surface, multiple threaded holes are present .These holes are used for mounting and rotating  gear shafts. Fillets are present at corners of truss for stress reduction.4
Manufacturing process –
The shape of top truss is similar to bottom truss as compare in area covered but it is also non uniform , So all the linear parts are joined together with fasteners , at necessary places holes of required diameter are drilled . Fillets are provided by using grinding operation
The material used for frame is Wood  .Some properties of wood is given below .
Major properties of Wood is shown below 

Property

Value

Units

Elastic Modulus

3000

N / mm^2

Poisson’s Ratio

0.29

N / A

Shear Modulus

300

N / mm^2

Mass Density

160

kg / m^3

Yield Strength

20

N / mm^2

Thermal Conductivity

0.05

W / (m·K)

Table 1.2 Material Properties

Wheel

There are 3 wheels in this vehicle , front wheel adjust itself according to direction of cart and two back wheels are assembled with the gear arrangement and provide motion to the cart .
The material used for wheel is Acrylic .Some properties of acrylic is given below .
Major properties of Acrylic is shown below 

Property

Value

Units

Elastic Modulus

3000000000

N / m^2

Poisson’s Ratio

0.35

N A

Shear Modulus

890000000

N / m^2

Mass Density

1200

kg / m^3

Tensile Strength

73000000

N / m^2

Yield Strength

45000000

N / m^2

Thermal Conductivity

0.21

W / (m·K)

Specific Heat

1500

J/(kg·K)

Table 1.3 material properties5
Manufacturing process
All 3 wheels are similar and have uniform cross-section so it is manufactured by machining process .Back wheels have taper tube on its inner face which is joined separately to it .
5Wingerter, R., and Lebossiere, (Feb 2011) P., ME 354, Mechanics of Materials Laboratory: Structures, University of Washington
The material used for Back wheel is Acrylic .Some properties of acrylic is given below .
Major properties of Acrylic is shown below 

Property

Value

Units

Elastic Modulus

3000000000

N / m^2

Poisson’s Ratio

0.35

N A

Shear Modulus

890000000

N / m^2

Mass Density

1200

kg / m^3

Tensile Strength

73000000

N / m^2

Yield Strength

45000000

N / m^2

Thermal Conductivity

0.21

W / (m·K)

Specific Heat

1500

J/(kg·K)

Table 1.4 material properties

Gear arrangement

The gear arrangement shows the combination of different gears , shafts , winding unit , hub connecting in such a way to obtain desired output .
This gear arrangement having following parts explained below –

WINDING UNIT

The winding unit is input link in which a key is present .This unit is connected with the spring .As the key is rotated manually or by external means then this rotational motion stored in spring and then further transfer to gears .
The material used for winding unit is Acrylic .Some properties of acrylic is given below .
Major properties of Acrylic is shown below 

Property

Value

Units

Elastic Modulus

3000000000

N / m^2

Poisson’s Ratio

0.35

N A

Shear Modulus

890000000

N / m^2

Mass Density

1200

kg / m^3

Tensile Strength

73000000

N / m^2

Yield Strength

45000000

N / m^2

Thermal Conductivity

0.21

W / (m·K)

Specific Heat

1500

J/(kg·K)

Table 1.5 material properties

INPUT GEAR A

         The winding unit is directly connected to the input gear a and both are having common shaft .6
The material used for input gear a is Acrylic .Some properties of acrylic is given below .
Major properties of Acrylic is shown below 

Property

Value

Units

Elastic Modulus

3000000000

N / m^2

Poisson’s Ratio

0.35

N A

Shear Modulus

890000000

N / m^2

Mass Density

1200

kg / m^3

Tensile Strength

73000000

N / m^2

Yield Strength

45000000

N / m^2

Thermal Conductivity

0.21

W / (m·K)

Specific Heat

1500

J/(kg·K)

Table 1.6 material properties

STEPPED GEAR SHAFT AND CAM GEAR

The input gear A is connected with stepped gear shaft which is further connect with wheel brake hub , other part of gear A is connected Cam gear which help in completing the circuit and provide motion to other gear.
The material used for gear a is Acrylic .Some properties of acrylic is given below .
Major properties of Acrylic is shown below 

Property

Value

Units

Elastic Modulus

3000000000

N / m^2

Poisson’s Ratio

0.35

N A

Shear Modulus

890000000

N / m^2

Mass Density

1200

kg / m^3

Tensile Strength

73000000

N / m^2

Yield Strength

45000000

N / m^2

Thermal Conductivity

0.21

W / (m·K)

Specific Heat

1500

J/(kg·K)

Table 1.7 material properties

CLUTCH SHAFT

The clutch shaft is directly connected with wheels tube which help the 3 wheeler assembly to stop .
The material used for clutch is Acrylic .Some properties of acrylic is given below .
Major properties of Acrylic is shown below 

Property

Value

Units

Elastic Modulus

3000000000

N / m^2

Poisson’s Ratio

0.35

N A

Shear Modulus

890000000

N / m^2

Mass Density

1200

kg / m^3

Tensile Strength

73000000

N / m^2

Yield Strength

45000000

N / m^2

Thermal Conductivity

0.21

W / (m·K)

Specific Heat

1500

J/(kg·K)

Table 1.8 material properties
DESIGN EVAUATION
As a prototype , consider the cart having speed of 10 m/second .
V =10 m/s
Cart Weight = 1000 Grams
Formula
Kinetic energy = ½ x mass of cart x velocity^2
KE =  ½ * 1000/1000 * 10*10  
K.E= 50 Joules
So this cart need 50J to reach speed of 10 m/s
But spring energy is equal to kinetic energy
S.E = K .E
½ * k *x  = 50
k = spring constant= 800 N/m
Apply values we get
x =125 mm
Dia meter of coil = 12 mm
So its Perimeter = 12 x pi= 12*3.14
                    =37.7 mm
No . of rotations = 125 /37.7
                               = 4
So maximum 4 rotations needed to reach the cart to 10 m /second .
DFMA PRINCIPLES & ANALYSIS
DFMA refers to design for manufacture and assembly .DMFA is technique that focus on efficiency in manufacturing and assembly of a product by simplifying all process include under it .It main motives to manufacture and assembly the product in minimum time and at least cost .
Main principles of DFMA

Simplify the automation process by reducing number of parts in assembly .The 3 wheeler assembly is completely automated .
Encourages to use standard parts that are easily available rather than complex manufacturing that increase cost of product .In 3 wheeler assembly ,standard parts like gear ,spring , shaft hub are used .
Reduce the use of flexible components like gaskets, seal ,rubber . As assembly of these parts are more difficult .No flexible part used in 3 wheeler assembly .
Use adhesive and snap type fitting for assembly rather than fasteners .
To simplify manufacturing , parts design should be simple and unnecessary features on parts should be avoided .

Advantages of DFMA

Cost of assembly and manufacturing becomes low .
Time required for assembly and manufacturing is less .
As numbers of parts reduces then chances of parts failure also reduces therefore reliability increases .
High quality of product is achieved .

8Boothroyd, G., Dewhurst, P. and Knight, W., “Product Design for Manufacture and Assembly, 2nd Edition”, Marcel Dekker, New York, 2012.
DESIGN VALIDATION
Consider the factor of safety is 2 and design the wheel of cart , considering its weight of 7000 grams.
Weight on wheels = 7*9.81 N = 70 N ,To maintain factor of safety is 2 , load applied should be 70 x 2 = 140 N
As Applied load is 140 N on wheels centre and check the maximum stress and deflection produced by the wheels .Diameter of shaft is 3 mm.

Solid part

Properties

Revolve1
 

Mass:0.4 kg
Volume:6.e-007 m^3
Density:1200 kg/m^3
Weight:4 N
 

Material Properties

Model Reference

Properties

 

Name:

Acrylic (Medium-high impact)

Model type:

Isotropic

Default failure criterion:

von Mises Stress

Yield strength:

45 MPA

 

Loading
As discussed below , applied load is 140 N while considering the factor of safety of 2 .Applied load is the total weight of cart acting on the wheels , so this load is applied at centre of wheel .
Mesh information

Mesh type

Solid Mesh

Mesher Used:

Standard mesh

Jacobian points

4 Points

Element Size

2 mm

Tolerance

0.01 mm

Mesh Quality

High

Mesh information – Details

Total Nodes

33998

Total Elements

18454

Maximum Aspect Ratio

14

% of elements with Aspect Ratio < 3 18.1 % of elements with Aspect Ratio > 10

0.001

% of distorted elements(Jacobian)

0

RESULT
As all the input parts are completed now the run the analysis and obtained the output result .
Stress
STRESS
Values of stress is given below
Minimum :-  0 MPa
Maximum :- 31.39 MPa
Deflection
DEFLECTION
Value of deflection obtain is given below
Minimum :-  0.00 mm
Maximum :- 0.166 mm
FACTOR OF SAFETY
Factor of safety is the ratio of yield stress to maximum stress generated while applying load. Yield stress is maximum permissible stress for a particular material after which deformation takes plate. Here Yield stress is 45MPa and maximum stress generated is 31.39 MPa .Although this design is safe but factor of safety value of 2 is not achieved
F.O.S = 45/31.39 =1.43
So changes need to be done to reduce the maximum stress value and achieve the factor of safety to 2
CHANGES IN DESIGN
To maintain the factor of safety of 2 and above , necessary changes need to be done accordingly .Earlier the shaft and hub of wheels having diameter of 3 mm .
According to formula below , stress is inversely proportional to Moment of inertia(I) .
For shaft
I =3.14 *D^4 /32
This shows that to reduce the value of maximum stress ,diameter of shaft need to be increased . Now in new design diameter of shaft is increased from 3 mm to 4 mm due to which stress value will reduce .
The below picture shows that diameter of shaft change to 4mm
Now kept all the procedure same i.e fixtures , loading and run the result with new design.
RESULT
As all the input parts are completed now the run the analysis and new results are obtained shown below
STRESS
Values of stress is given below
Minimum :-  0 MPa
Maximum :- 20.65 MPa
Deflection
DEFLECTION
Value of deflection obtain is given below
Minimum :-  0.00 mm
Maximum :- 0.21 mm
FACTOR OF SAFETY
Here Yield stress is 45MPa and maximum stress generated in new design is 20.65 MPa . Although this design is safe and also factor of safety value of 2 is also achieved .
F.O.S = 45/20.65 =2.17
So changes done are effective to achieve the factor of safety to 2 or more .
Conclusion
In 1st case when the diameter of shaft is 3 mm and applied load is 140 N then although the design is safe but factor of safety of 2 need to be achieved .To increase the value of factor of safety , maximum stress occurs due to applied load need to be reduced . So in new design diameter of shaft is increased from 3 mm to 4 mm so that its moment of inertia get increased and maximum stress reduced . So after increasing the diameter of shaft to 4 mm the maximum stress value reduced to20.65 MPa and factor of safety of 2 is successfully achieved . So overall design is safe .9
References
Lynch, Mike (2010-01-18), “When programmers should know G code”, Modern Machine Shop (online ed.).
Davinci, Leonardo (2011). The Notebooks of Leonardo Da Vinci. Lulu. p. 736. 
Fernando (2010). Technological Concepts and Mathematical Models in the Evolution of Modern Engineering Systems.Birkhäuser. ISBN 978-3-7643-6940-8.
Plesha, Michael E (2013). Engineering Mechanics: Statics (2nd ed.). New York: McGraw-Hill Companies Inc. pp. 364–407. ISBN 0-07-338029
Jacob; Papadopoulos (Oct 2016). Introduction to Solid Mechanics: An Integrated Approach. Springer. ISBN 9783319188782.
Wingerter, R., and Lebossiere, (Feb 2011) P., ME 354, Mechanics of Materials Laboratory: Structures, University of Washington
Record, Samuel, 2010. The Mechanical Properties of Wood. J. Wiley & Sons. p. 165. ASIN B000863N3W.
Sclater, Neil. (2011). “Gears: devices, drives and mechanisms.” Mechanisms and Mechanical Devices Sourcebook. 5th ed. New York: McGraw Hill. pp. 131–174. ISBN 9780071704427. Drawings and designs of various gearings.
Khurmi R S, (2014), ‘A text book of machine design’, Eurasia publishing house(P) ltd., New-Delhi, ISBN 9788121925372
Mahadevan K and Reddy K.Balaveera, (2015), ‘Design data hand book’, CBS publishers and Distributors (P) ltd., New-Delhi, ISBN 9788123923154
 Boothroyd, G., Dewhurst, P. and Knight, W., “Product Design for Manufacture and Assembly, 2nd Edition”, Marcel Dekker, New York, 2012.
Smid, Peter (2010), CNC Control Setup for Milling and Turning, New York: Industrial Press, ISBN 978-0831133504, LCCN 2010007023.
Lynch, Mike (2010-01-18), “When programmers should know G code”, Modern Machine Shop (online ed.).
 Fridtjov Irgens (2010), “Continuum Mechanics”. Springer. ISBN 3-540-74297-2
Ramsay, Angus. “Stress Trajectories”. Ramsay Maunder Associates. Retrieved April 2017.

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