Post on 04-Apr-2018
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MACHINE COMPONENT DESIGN
BMCD3523
STIRLING ENGINE KITCHEN SMOKE REMOVER
FAKHRURAZI BIN AZIZ
B040910051
HALIMATON BASMATU BINTI ALI
B040910081
MOHD AZARUL SYAZARI BIN CHE AZIZ
B040910167
MOHD IDHAM KHALID BIN ABU HASAN
B040910094
NURUL NADIA BINTI ABDULLAH
B040910227
SYED ZULKARNAIN BIN SYED SAIDINB040910037
DUE DATE
1 JUNE 2012
LECTURE
SHAFIZAL BIN MAT
Bachelor of Mechanical Engineering (Thermal-Fluid), Faculty of Mechanical Engineering,
Universiti Teknikal Malaysia Melaka, Durian Tunggal, Melaka.
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List of Contents
List of Tables.....3
List of Figures............4
Abstract......5
Acknowledgement.6
Introduction.......7
Chapter 1 : Objective and Problem Statement 7-8
1.1.Objective.......71.2.Problem Statement...........8
Chapter 2 : Theory..8-12
Chapter 3 : Methodology..12-14
Chapter 4 : Conceptual design.15-19
4.1.Design 1...............................154.2.Design 2...............................164.3.Design 3..........................16-174.4.Weight Decision Matrix.......17-184.5.Pugh method.......................19
Chapter 5 : Detail design..19-26
Chapter 6 : Discussions and Design Analysis .....27-47
6.1.Analysis due to welding27-306.2.Analysis of shaft31-326.3.Bolt Analysis..32-356.4.Bearing analysis ...36-376.5.Bevel gear analysis38-406.6.Non-Permanent Joints. 40-456.7.Discussion..46-47
Chapter 7 : Conclusion...........................................47
References48
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List of Figures
Figure 1 Screw.........................................................................................................................10
Figure 2 strength of bolt.........................................................................................................11
Figure 3 Gear...........................................................................................................................12
Figure 4: Stirling Engine........................................................................................................12
Figure 5: Stirling Engine Parts..............................................................................................13
Figure 6: Parts involved in power generation......................................................................14
Figure 7: Electricity powered................................................................................................15
Figure 8 : Solar powered........................................................................................................16
Figure 9 : Stirling Engine powered.......................................................................................17
Figure 10 : Blade....................................................................................................................19
Figure 11 : Suction Tunnel.....................................................................................................20
Figure 12 : Gears and Shaft..................................................................................................20
Figure 13 : Bearing................................................................................................................21
Figure 14 : Stirling Engine.....................................................................................................21
Figure 15 : Stove including the stove cabinet.......................................................................22
Figure 16 : Final Rendering..................................................................................................22
Figure 17 : Engine Assembly.................................................................................................23
Figure 18 : Bill of Material....................................................................................................24
Figure 19 : 2D Drawing..........................................................................................................25
Figure 20 : 2D Drawing of Stove Assembly.........................................................................26
Figure 21 : welding part 1.....................................................................................................27
Figure 22 : Welding part 2....................................................................................................29
Figure 23 : shaft part..............................................................................................................31
Figure 24 : bolt part................................................................................................................32
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List of Tables
Table 1 : Table 11-2................................................................................................................36
Table 2 : Table 8-15................................................................................................................40
Table 3 : Table 8-11................................................................................................................41
Table 4 : table 8-1..................................................................................................................41
Table 5 : Table A-30...............................................................................................................42
Table 6 : Table 8-8..................................................................................................................44
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Abstract
In this project, stirling engine concept is being used for green technology aspect to run
the blade. Why the Stirling Engine being choose? It is because the sustainable type of energy
that has a lot more benefits than the conventional energy that uses electricity as the source.
But the main objective in this project is to analysis each parts of mechanical component that
being installed in this engine. The mechanism will be working by allowing compression and
expansion of air or other gas at different temperature levels such that there is a net conversion
of heat energy to mechanical work. Then the process will be transferred into mechanical part.
All of the parameter will be calculated according to the aspects that have been given. Whether
the dimension of the component is suitable for the system or not, is the factor safety is not
below then one or not? And many more aspect have been consider to build this sisten
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Acknowledgement
We owe a great many thanks to a great many people who helped and supported us
during the build this project, many thank to Sir. Shafizal Bin Mat who has given us a chance
to prove that we can do things on our own. He gave a positive perspective in life and sharing
his valuable time with giving us an information to finish this project. Although, the project is
quit challenging, but we can finish it on time.
We would have not finished this project without the support of our team mate whose
has encouragement each other to keep going and struggle to ensure this project was success.
And special gratitude to our classmate whose helped us in researching on different fields
concerning this project and giving a fresh idea that help us a lot to complete this project.
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Introduction
Nowadays, green technology is something that no longer alien to our daily life. We
can see how the scientist manipulated each type of energy to become new useful machinery
that can save our environment. From the light energy that can convert into electrical energy
by using solar panel, from the mechanical energy or also known as vibration energy also can
convert into electrical energy by using Piezoelectric, also from the wind, see waves, and many
more. But here we will focused on stirling engine that can convert onto mechanical energy.
Stirling Engine is one of the reversible thermodynamic cycle application, they work
based on compression and expansion cycling of air or other gas. They are also traditionally
know as an external combustion engine. They are widely can found in industry application
because of the high thermal efficiency, form a quiet and safe operation, ease of operation and
able to working with any form of thermal energy. The most efficient type of sterling engine is
the type that can handle the low temperature difference. Such as temperature between the
hand palm and the ambient temperature.
Advantage of applying this technology are, can operate with any type of heat source, it
is last longer than reciprocating engine type and also having the simple engine mechanisms. it
is also using the single-phase of working fluid that will reduce the explosion rate. lastly, they
also a flexible technology. For the disadvantage, it need huge size of engine to supply large
amount of power, and it will cost of a lot of money. Type of gas being used also need to be
considered, the gas should have a low heat capacity to produced high pressure.
Chapter 1 : Objective and Problem statement
1.1.Objective1.1.1. To design a kitchen smoke remover that uses sustainable energy as source of
energy
1.1.2. To analysis each part of mechanical component1.1.3. To apply the principles of mechanical component design in this project
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1.2.Problem statement
Based on the conventional kitchen smoke remover that being use nowadays, it
is using electricity as the source of energy to run the suction blade. It is not very
environmental friendly because electricity is not one of the sustainable energy.
Furthermore, consumers will be burdened with excessive electricity bills every
month. Now, from what we can see the earth will not allowing us to stick to one
source of energy, the available sources like coal, oil and other will diminish. And
now we need to design a new technology that will use sustainable energy that easy to
get in our daily life. It is our objective to change this type of energy to a more
sustainable energy as the alternative to a more environmental friendly energy.
Chapter 2: Theory
An Exhaust Smoke system was used to remove smoke that produces by kitchen. This
system was work based on stirling engine that placed on the kitchen. There are several part on
the system had been analyze. A small machine components are joined together to form a
larger machine part. The design of joint are important because it was critical part that ensure
that the system were able act with the force react on the system. A week of joint may spoil a
utility of the machine part.
Bolt was one of nonpermanent joint where it was temporary of joining the materials.
To ensure the joint that causing by bolt was safety, there was several analysis was done to
know the factor safety, preload recommendation for the type of bolt, preload stress and stress
the can be supported by the bolt. All the value of preload was used to calculate the load factor,
yielding factor that can cause by the bolt, and also the load factor guarding against joinseparation. The analysis was depending on the type of material of bolt and also the type of
bolt. A different type and material of bolt that used were giving a different value of safety
factor. A mass of part that was used the nonpermanent joint was mostly effected the safety
factor of the bolt were it was concern as force load that act on the joint. The total force that
acts on the plate was divided by the amount of the bolt that used on the plate. The calculations
of all the part of analysis on bolt are based on these equations.
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Torque necessary to achieve preload;
T = K Fi d (1)
Load factor;
nL = ( SP AtFi)/ C(Ptotal/N) (2)
Yielding factor safety;
nP = ( SP At)/ [C(Ptotal/N) + Fi] (3)
Load factor guarding against join separation;
no = Fi/[ (Ptotal/N)(1-C)] (4)
Welding was type of permanent joint where it cannot be dissemble without damaging
the components. The part that was weld is used molecular force to joint the component. The
permanent joints are preferred on the part that face with a high load and its was durability.
Advantage of this type of joint are high load bearing capacity, better distribution of stresses
through the material and also has higher yield strength of the combined structure. The two
plate of metal was melt at the bottom by using high temperature where the aid of filler
material a weld pool was created. The joint made permanent by reducing the work piece
temperature to the room temperature. There were several type of weld such as Gas welding
and also several type of joint such as butt joint, corner joint, edge joint, lap joint and also tee
joint. To ensure the type of weld and joint are safe for the structure, there were analysis that
used to calculate the factor safety of the weld. The depths of weld were the critical part in the
welding design. The type of force that acts on the welding are based on bending or torsional
force. Different types of weld were giving a different analysis. An equation that used onanalyze the welding to find the factor safety are:
Primary shear
= V/A (5)
Secondary shear bending
= Mc/I (6)
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Total shear,
= (2+ 2)1/2 (7)
Factor safety
Fs = Ssy/max (8)
Shaft is rotating member where it used to transmit power or motion. There was several
type of material that used for shaft. The shaft rotates due to the transmitting power from the
stirling engine that produce by the kitchen. The forces that involve are bending and torsional.
The shaft was rotate due to torsional and bending. To find the maximum shearing stress that
act on the shaft, equations below are used;
max = (9)
Screw was also one of the types of non permanent joint. The helical-thread screw was
undoubtably an extremely important mechanical invention. It was the basis of power screw,
which change angular motion to linear motion to transmit power or develop large forces and
threaded fastener. All the threads are made according to right-hand rule unless noted.
The thread geometry of metric M and MJ profiles is shown in figure below;
These assemblies may structural as a load bearing of component that subjected both shear and
tensile stresses.
Figure 4 Screw
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Torque for raising the load;
TR = ( ) (10)
Torque required lowering the load;
TR = ( ) (11)
One of machine element is bearing that constrain the relative motion between two or
more parts to only desired type of motion. It was manufactured to take pure radical loads,
pure thrust load, or a combination of the kinds of loads. It used to allow and also promote a
free linear movement or free rotation around a fixed axis.
In selecting a bearing for a given application, it is necessary to relate the desired load and life
requirements to the published catalog load rating corresponding to the catalog rating life.
Where;
FR(LR nR 60)
1/a
= FD(LD nD 60)
1/a
(12)Where;
FR = catalog rating
LR = rating life
nR = rating speed
FD = desired radial load
LD = desired life
nD = desired life
Figure 5 strength of bolt
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Gear is a rotating machine where it has cut teeth which mesh with another toothed part
in order to transmit torque. The gear that had working in tandem is known as transmission and
it can provide a mechanical advantage through the gear ratio.
Chapter 3 : Methodology
Energy Concept
A Stirling engine is a type of external combustion engine that uses the expansion and
compression of air in order to create a mechanical energy.
Figure 4: Stirling Engine
Figure 6 Gear
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Typical of heat engines, the general cycle consists of compressing cool gas, heating
the gas, expanding the hot gas, and finally cooling the gas before repeating the cycle. The
efficiency of the process has an upper limit set by the efficiency of the equivalent Carnot
cycle operating between the same hot and cold temperatures.
Advantages
-The Stirling engine is noted for its high efficiency compared to steam engines, quiet
operation, and the ease with which it can use almost any heat source. This compatibility with
alternative and renewable energy sources has become increasingly significant as the price of
conventional fuels rises, and also in light of concerns such as peak oil and climate change.
-This engine is currently exciting interest as the core component of micro combined heat and
power (CHP) units, in which it is more efficient and safer than a comparable steam engine.
Figure 5: Stirling Engine Parts
Cooling Compartment
Cool Piston
Heating
Compartment
Hot Piston
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Mechanism of Power Generation
Figure 6: Parts involved in power generation
1) When the stove is ignited, heat will be transmitted by the heat conduction coil to the
Stirling engine in order to generate heat.
2) When enough heat had been accumulated, the Stirling engine will start to operate and starts
to transmit rotational motion to the shaft.
3) The shaft will transmit the power towards the gear compartment.
4) Bearing that is being placed at the shaft will smooth the rotary motion.
4) In gear compartment, the gear will increase the rotational power of the shaft; hence we will
get more power.
5) From gear compartment, a small shaft will transmit the power towards the blade, and the
blade will spins.
6) When the blade spins, smoke from the stove will be sucked to the blade and out of the
kitchen.
Coil
Heat Conductor
Shaft
Gear
Blade
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Chapter 4 : Conceptual design
A concept design could be regarded as the initial proposed design for a particular
product. Usually several design drawings will be made in order to make the best choice from
the options that is given. It will make it easier for a design team to make assessment of the
potential design and make the right choice out of it.
For our design, we had come out with several concept designs that is designed based
on its type of energy used. Out of the three designs, we had chosen the best design that met
our design requirements.
4.1 Design 1
For this design, we used electricity as the source of energy. Electricity could also be
regarded as renewable energy as long as it is based on hydroelectric power. Hence it is
suitable with our energy requirement that requires us to use renewable energy. However,
electricity is not quite suitable due to the fact that it requires monthly bill cost and also some
of it is generated by coal power plant that is not a renewable energy. So, we decided to scrap
this idea.
Figure 7: Electricity powered
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Figure 9 : Stirling Engine powered
4.4.Weight Decision MatrixObjective Tree For The Design Of A Kitchen Smoke Remover
KITCHEN SMOKE REMOVER
COST QUALITY IN SERVICE
ROI Mfg. Cost Reparability Durability Reliability maintenance Safety
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O1 = 1.0
O11 = 0.5 O12 = 0.5
O111 = 0.3 O112 = 0.4 O113 = 0.3 O121 = 0.4 O122 = 0.3 O123 = 0.3 O124 = 0.3
Weighted Decision Matrix
Decision
Criterion
Weight
Factor
Unit
s Electric Stirling engine Solar
Score Rating Score Rating Score Rating
maintenance 0.15 RM 8 1.20 6 0.90 4 0.60
Manufacturingcost
0.20 RM 7 1.40 7 1.40 5 1.00
Reparability 0.15 EXP 5 0.75 9 1.35 8 1.20
Durability 0.20 EXP 6 1.20 8 1.60 8 1.60
Reliability 0.15 EXP 5 0.75 9 1.35 8 1.20
ROI 0.15 HR 5 0.75 7 1.05 4 0.6
Safety 0.20 EXP 8 1.60 5 1.00 7 1.40
7.65 8.65 7.60
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4.5.Pugh methodNumber criteria solar datum Stirling
engine
1 maintenance - +
2 Manufacturing cost - +
3 Reparability + -
4 Durability - +
5 Reliability + -
6 ROI - -
7 Safety - -
plus 2 3
minus 5 4
Chapter 5 : Detail design
Figure 10 : Blade
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Figure 11 : Suction Tunnel
Figure 12 : Gears and Shaft
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Figure 13 : Bearing
Figure 14 : Stirling Engine
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Figure 15 : Stove including the stove cabinet
Final Rendering
Figure 16 : Final Rendering
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Figure 17 : Engine Assembly
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Figure 18 : Bill of Material
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Figure 19 : 2D Drawing
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Figure 20 : 2D Drawing of Stove Assembly
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Chapter 6: Design analysis and Discussions
ANALYSIS OF EACH PART
6.1.Analysis due to welding
part 1
Due to bending force;
By assuming Sy = 300MPa,
Force =250kNh = 8
Primary shear, = V/A, where V = shear force
Secondary shear bending, = Mc/I, where I = 0,707hIu
Total shear, = (2+ 2)1/2
From table 9-2 (pg 488; shigleys)
N0 6; throat area, A = 1.414h (b + d)
Figure 21 : welding part 1
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Location of G,= d/2
X = b/2
Unit second moment of area, Iu =
Calculation;
A = 1.414h (b + d)
= 1.414 (8) (780 + 379.814) = 13119.82 mm2
X = b/2 = 780/2 = 390mm
= d/2 = 379.814/2 = 189.91mm
Iu = = 3 = V/A= (250x 103)/ 13119.82 = 19.06N/mm2
I = 0.707hIu= 0.707(8) (65392793.8) = 369.8616x106
mm4
= Mc/I = 250 x103(550)(189.91)/ 369.8616 x106 = 70.60N/mm2
max= (2+ 2)1/2 = (70.602 + 19.062)1/2 = 73.13N/mm2
Ssy = 0.577Sy = 0.577(300) = 173.1 MPa
Factor safety = Ssy/max = 173.1 / 73.13 = 2.37
Explanation;
Based on the calculation that had done based on the dimension from the solid work
drawing, the value of factor safety are 2.37.Its was due to the dimension of the welding
process where the depth of welding between the plate to the and the exhaust smoke casing are
8mm. Amount of area that are weld are 13119.82mm2 . The forces that attract on the welding
are bending force. By assuming the value of distortion-energy criterion, the stresses are
300MPa where the stresses allowable are 173.1 MPa. Due to the calculation that had done the
value of maximum stress are 73.13MPa. The factor safety is based on the allowable stress
over maximum stress where it was 2.37. The depths of weld in this system are safe. The value
of factor safety shown that the weld between plate and the exhaust smoke casing are safe.
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Part 2
Due to bending force;
By assuming Sy = 300MPa,
Force =250kN
h = 12
Primary shear, = V/A, where V = shear force
Secondary shear bending, = Mc/I, where I = 0,707hIu
Total shear, = (2+ 2)1/2
From table 9-2 (pg 488; shigleys)
N0 6; throat area, A = 1.414h (b + d)
Location of G,= d/2
X = b/2
Unit second moment of area, Iu =
Calculation;
A = 1.414h (b + d)
= 1.414 (12) (780 + 50) = 14083.44 mm2
Figure 22 : Welding part 2
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X = b/2 = 780/2 = 390mm
= d/2 = 50/2 = 25mm
Iu =
= 3 = V/A= (250x 103)/ 14083.44= 17.75N/mm2
I = 0.707hIu= 0.707(12) (2987500) = 25.34595x106
mm4
= Mc/I = 250 x103(252.447)(25)/ 25.34595x106 = 62.25N/mm2
max= (2+ 2)1/2 = (17.752 + 62.252)1/2 = 64.73N/mm2
Ssy = 0.577Sy = 0.577(300) = 173.1 MPa
Factor safety = Ssy/max = 173.1 / 64.73 = 2.67
Explanation;
Based on the calculation that had done based on the dimension from the solid work
drawing, the value of factor safety is 2.67. The depth weld between the plate of the exhaust
smoke casing and plate of shaft are 12mm. The depth welding of the part are greater compare
the depth welding between the plate and the exhaust smoke casing because it was critical partcompare with that. Its because, when the system was running based on the stirling engine
process, the shaft was rotating and that may causing a vibration on the part. The vibrations
were giving a more force on the weld. So, the weld should be greater to circumvent the
system from broken. The amounts of area that are weld are about 14083.44mm2. The forces
that attract on the welding were due to bending force. By assuming the value of distortion-
energy criterion, the stresses are 300MPa where the stresses allowable are 173.1 MPa. The
values of maximum stress that are act at the weld are 64.73MPa. The factor safety is based on
the allowable stress over maximum stress where it was 2.67. The depths of weld in this
system are safe. Its also greater than the value of safety factor of plate and the exhaust smoke
casing are safe.
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6.2.Analysis of shaft
Assuming N=500rpm
Power, P = 20kW
Sy= 300MPa
Mass gear = 0.5 kg
Given( from drawing)
L = 1210mm
d = 30mm
Calculation;
Ssy = 0.577sy = 0.577(300) = 173.1MPa
Bending moment, M = P x L
Wgear = 0.5(9.81) = 4.905 N
P= Wgear= 4.905 N
M= P x L = 4.905(1210) = 5935.05 Nmm
From power,P = T
T = P/ , where = 2N/60
= 2(500)/ 60 = 52.36 rad/s
so, T = 20x103 / 52.36 = 381.97097Nm = 381970.97Nmm
max =
max =
= 72.06 MPa
thus; factor safety = Ssy/max = 173.1/ 72.06 = 2.40
Figure 23 : shaft part
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Explanation;
Shaft is rotating member where it used to transmit power or motion. The material of shaft that
used is steel. The rotating of the shaft are about 500rpm that transmit power about 20 kW. The
mass of the gear that was placed at the shaft are 0.5kg and the value of distortion-energy
criterion, the stresses are 300MPa where the stresses allowable are 173.1 MPa. The length of
the shaft are 1210mm. the length of shaft was greater because it was placed from the stirling
engine at the kitchen and transmit the power to the gear and the rotation of the gear were
move the fan blade on the exhaust smoke. From the analysis that had done, the force that act
on the shaft are bending moment and torque. The diameters of shaft are about 30mm. from
the amount of power that transmit by the shaft, the value of a torque were find that are about
381970.97 Nmm. The bending moment those acts on the shaft are about 5935.05 Nmm. The
maximum shear stress that had calculated based on the equation is 72.06 MPa. Due to the
factor safety calculation, the amount of safety factor of this system is 2.40. This was shown
the type and diameter of shaft was suitable to used in these system that are not exceed to 1.
6.3.Bolt Analysis
Type of material : steel
Type of bolt : Hex Head (upset)
Mass : 20kg
Area : 10 x 40 = 400mm2
Angle = 900C = 1.57 rad
Times, t = 1sec
K = 0.18 (lubricated)N = 3( total bolt)
Figure 24 : bolt part
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From table 8.8 (pg 130:shigleys)
Poisson Ratio : 0.291
Elastic Modulus : 207 Gpa
A : 0.78715
B : 0.62873
From table 8.9 (pg 433)
*Psi = lbf/in2
*Sp ( minimum Proof Strength)
SAE Grade No: 5 (material: Medium carbon, Q & T)
Minimum Proof Strength : 85 kpsi = 586.05 MPa
Minimun Tensile Strength : 120 kpsi = 827.37 MPa
Minimum Yield Strength : 92 kpsi = 634.32 MPa
By assumming Length of screw, L : 30mm
d : 9.53mm
From table 8.7( pg 426)
Based on drawing l : 10mm
LT = 2d + 6mm, L 125mm, d 48mm
Length of unthreaded portion in grip : ld = L -LT
Length of threaded portion in grip: lt = l - ld
Area of unthreaded portion : Ad= d/ 4
Area of thereaded portion : At from table 8.1
Fastener stiffness : kb=
From table 8.1(pg 412)
Nominal Major Diameter, D : 14mm
Pitch, p : 2mm
Tensile-stress Area, At : 125 mm2
Minor-Diameter Area, Ar : 116mm2
Equations that was used: (8.23) (pg 429)C =
(pg 436)
Fp = At Sp
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Fi = 0.90 Fp (permenent connection)
i = Fi/ At
b = Fb/ At = (CP+ Fi) / At
T = K Fi d
nL = ( SP AtFi)/ C(Ptotal/N)
nP = ( SP At)/ [C(Ptotal/N) + Fi]
no = Fi/[ (Ptotal/N)(1-C)]
Calculations :
LT = 2(9.53) + 6mm = 25.06 mm
ld = 3025 = 5mm
lt = 105 = 5mm
Ad = = 71.33 mm2Bolt stiffness;
kb = = 1.8802 x 1012 Mpa
Members stiffness;
km/ (207x 106)(9.53) = (0.78715) exp [(0.62873)(9.53)/10 )]
km = 1.7092 x 109
Mpa
Stiffness constant;
C = (1.8802 x 1012) / (1.8802 x 1012 + 1.7092 x 109 ) = 0.9991
Proof load;
Fp = 125 (856.05 Nmm2) =107.01 kN
Preload recommandation;
Fi = 0.90 (107.01 kN) = 96.31 kN
Preload stress;
i = Fi/ At = 96.31 x103 / 125 = 770.45 N/mm2
Stress under the service ;
b = Fb/ At = (CP+ Fi) / At
Fp = ma, a = x/t2
X= r = 4.77 (1.57) = 7.4889mm
a = 7.4889 / 12 = 7.4889 mm/s
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Fp = 20 (7.4889) = 149.778 N
P = Fp/A = 149.778 / 400 = 0.37 Pa
b = [(0.9991)(0.37) + 96.31)/ 125] = 0.7734N/mm2
Torque necessary to achieve preload;
T = K Fi d = 0.18 (96.31x 103) (9.53) = 162.21 kNmm
Load factor;
nL = ( SP AtFi)/ C(Ptotal/N)
nL = ( 856.05(125)96.31x103) / 0.9991(0.37/3) = 86.804x103
Yielding factor safety;
nP = ( SP At)/ [C(Ptotal/N) + Fi]
nP = ( 856.05(125))/ [0.9991(0.37/3) + 96.31x103] = 1.11
Load factor guarding against join separation;
no = Fi/[ (Ptotal/N)(1-C)]
no = 96.31x103 / [(0.37/3)(1-0.9991)] = 867.66x106
Explanation;
Bolt are the types of nonpermanent joints.On this system the material that used are
steel and the bolt are Hex Head (upset) type. The mass of this system is about 20 kg. The plate
that used is not much heavy where it was suitable for the system. The areas that are involved
with bolt are 400mm2. The major of the bolt are 14mm and the fastener diameter of the bolt is
9.53mm. From all the specification on the system, there are several tables had been refer to
find the accurate value of torque, load factor, yielding factor safety and also the load factor
guarding against join separation. Due to the bolt that had used on the system, the value of
preload and service load stresses are respectively ten percent less than the proof strength.Therefore the types of bolt are suitable applied at this system where the torque that was
applied at the bolt is only about 162.21 kNmm. Based on the theoretical calculation, the
yielding factor of safety that guarding against the static stress exceeding the proof strength
that is about 1.11. Based on the factor safety that is not exceeding 1, the bolt was suitable for
this system. The fillets of the bolt are the point of stress concentration. This type of bolt can
stand with the vibration that produces by the shafts that are including the different frequency.
The bolt was also durable for this system.
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6.4.Bearing analysis
Given:
02-series deep groove ball bearing :-
Desired load = 1.2kN
Desired life LD = 30 000h
Desired speed nD = 400 rpm
Ball bearing; a = 3
Application factor, af = 1.2
The rating life is revolutions.( based on the SKF manufactured )
= 10755.371 N= 10.75 kN
So from the Table 11-2, we obtain;
Table 7 : Table 11-2
The new C10 =12.7 kN
Bore = 20 mmOD = 47 mm
*
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Based on the SKF manufactured :-
X0 = 0.02
= 4.459
b = 1.483
so from the data above
( - X0 ) = 4.439
XD= = = 720
From the new C10,Based on the equation Reliability, R
( * )R < 0.90
For the ball bearing decision making we choose 02-series, deep groove ball bearing on our
product. First we assume the desired load for the ball bearing is about 1.2kN as a load that
will be apply to the ball bearing. Then we assume the desired life for that material is 30 000h.
For this ball bearing we put it as a shaft for the fan and we assume it as 400rpm . it is how fast
if the fan rotate as 400rpm and give the bearing force to rotate. Every bearing that we created
is based from 2 manufactured that is Timken manufactured and SKF manufactured. For the
ball bearing we use is we decide from the SKF manufactured. For SKF manufactured we need
to refer its own rating life for the every purchase of manufactured. For the ball bearing the
value for a is 3.
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6.5.Bevel gear analysis
All the units calculation are in mm units.
Given:
Bevel Gear
N = 400 rpm
Transmit power, H = 40 kW
Solution
The pitch angles are
= 35o =32o
The pitch-line velocity corresponding to the average pitch radius is
V = 1.6755
Therefore the transmitted load is
Which act in the positive z-direction
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Where Wr is in x-direction and Wa is in y-direction
In preparing to take sum of the moments about bearing D, define the position vector from D
to G.
We shall also required a vector from D to C;
Then, summing moments about D gives;\
-------------------1When we replace the details in equation 1 become equation 2.
After the two cross products are taken, the equation become ------
2
From which
T = 5037.30 N.m ---------------3
Now sum the forces to zero. Thus
FD + FC + W = 0
when the details are inserted, Equation 4 becomes
--------4
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Find we see that = 460459 N, and so
+
+
k
Then, from Equation
FD = ( 668i - 9605.57k ) N
In determining shaft and bearing loads for bevel-gear applications, the usual to use the
tangential or transmitted load that would occur if all the forces were concentrated at the
midpoint of the tooth. Here we prepared the picture the bevel gears dimension and the force
distribution for every part of critical point of the bevel gears. The force transmitted to the
bevel gears and it calculated by vector methods. In the bevel gears we have tangential force
Wt, radial force W r, and axial force W a. All the three forces Wt ,Wr, and Wa are at right
angles to each other and can be used to determine the bearing loads by using the methods of
statics.
6.6. Non-Permanent Joints.
1. Tightening torque, T = KFid
Table 8 : Table 8-15
Torque factor,K= 0.20
Preload, Fi = 0.75 Fp
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Proof load, Fp = At Sp
From Table 8-11;
Table 9 : Table 8-11
For M12, Sp = 310 MPa
From table 8-1;
Table 10 : table 8-1
For M12 and Course-Pitch series, At = 84.3 mm2
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Proof load, Fp = At Sp
Fp = (84.3)(310)
Fp = 26133 MPa.mm2
Preload, Fi = 0.75 Fp
Fi = 0.75(26133)
Fi = 19600 MPa.mm2
Therefore;
Tightening torque, T = KFid
T = (0.2)(19600)(8)
T = 31360 N.m
2. Factor safety guarding against yielding.Bolt length, L = l + H
given;
l = 20mm
From table A-30
Table 11 : Table A-30
Height, H = 7.76 mm
Therefore;
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Bolt length, L = 20 + 7.76
= 27.76 mm 28 mm ( From table A-17)
Treaded length, LT = 2d + 6
LT = 2(8) + 6
LT = 22 mm
Length of unthreaded portion in grip, ld= L - LT
ld= 28 - 22
ld= 6 mm
Length of threaded portion in grip, lt= l - ld
lt= 20 - 6lt= 14 mm
Area unthreaded portion, Ad =
Ad =
Ad = 16 mm2
Fastener stiffness: kb = , E = 207 GPa ( Table 8-8 for steel)kb =
kb = 7.252x10
11
Member stiffness, km = AEd exp (Bd /l )
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From table 8-8;
Table 12 : Table 8-8
A= 0.78715
B= 0.62873
E= 207 GPa
Member stiffness, km = AEd exp (Bd /l )
km = (0.78715)( 207 G)( 0.62873)( 0.62873x8/ 20)
km = 1.6763x1012
Stiffness constant of the joint, C =
C =
C = 0.3020
Resultant bolt load, FB = Pb + Fi = CP + Fi , P = 50KPa / 5 = 10KPa
= (0.3020)(10x103) + 19600 MPa
= 22620 N
Resultant load on connected member, Fm = Pm + Fi = (1-C)P - Fi
= (1-0.3020)(10x103) - 19600
= -12620 N
Factor Safety for tension in members, n =
, Ssy = 340 MPa
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=
= 149.70 MPa
Therefore, n =
= 2.27
Factor Safety for bolt, n =
=
= 200 MPa
n =
n = 1.7
In order to find the value of resultant bolt load, FB and resultant load on the connected
members, Fm, first step is identify several parameters required. The purpose of finding
the resultant load is to determine the factor safety. First is need to find the tightening
torque,T where the value that were obtained from the calculation is 31360 N.m.
Before the value factor safety can be determine, the value of member stiffness, Km and
Kb must be calculate. The value of Km obtained is 1.6763x1012 and the value of Kb is
7.252x1011. By substituting into the equation, the value of stiffness constant of the
joint, C can be determine and only then the value of the resultant bolt and the resultant
load on the connected members can be determine. The value that were obtained for FB
is 22620 N and the value for Fm is -12620 N. After that we can determine the stressconcentration and then the factor safety. The value for the factor safety for tension in
members is 2.27 which is guaranteed safe because it is in range. Meanwhile, the factor
safety for bolt is 1.7. Even though it is out of safety range a little bit, but some
observation has been made by surfing the internet where it is found that there are
several companies that considered this kind of values is still in safety zone. Therefore,
it can be conclude that this fastener is suitable for this application.
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6.7.Discussion
After one semester of teaching and learning of Mechanical Component Design, this
report is mainly about designing a product for Machine component Design subject. For this
project, the application of stirling engine is used in order to complete the project.
In this project, there are several factors that need to be considered. Factors that need to
be considered are based on the parts of the design. Most parts must be calculated in order to
ensure that the product meet the safety requirements. Each part of the design must be
calculated to ensure that the product was safe to use. By considering the load of force that act
on the system, the shear stress and factor safety can be calculated. The factor safety of the part
must be greater than one. It was been calculate by referring the dimension of drawing for each
component n solid work. If the factor safety that get was less than one, there are several part
will be considered. That is change the dimension, change the material and also change the
position.
Due to welding part, there was bending and torsional force that act on the bending.
The basic of torsional and bending force are based on type of welding. The areas of welding
also give some effect on the force that act on the plate. In this project, there were two plates
that were involved with welding. Meanwhile, bolt analysis was based on the type of material
that was used and the type of head that used on the machine screw. The stiffnesss and the
bolt strength had been calculated based on the material and type of bolt that used in this
system.
Shaft was the rotating cylinder that used to rotate the gear and provide the torque to
rotate the blade to remove the hot air from the building to the environment. Shaft was the
most important part where it was the main part of transmitted the power from the stirling
engine through the gear to moving the blade. The other part that used in this system is gear.The type of gear that used in this system is bevel gear. The force analysis was used to
determining shaft and bearing loads for the bevel gear application that was to used tangential
and also transmitted load. The pitch of bevel gears is measured at the large end of tooth, and
both the circular pitch and the pitch diameter are calculated.
Screw was non-permanent joint that used in this system. The total amount that used
was 10 pieces. Screw is among the important component as it holds bigger components
together. Screw is used because it is cheap. Besides that, the calculation of the resultanceforce that acting on the fastener or bolt and the resulting force that acting on the members also
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been calculated. From the force that were obtained, the factor safety of the part can be
calculated and the result that were obtained shown that the design is in safety range. On the
other hand, bearing is used to smoothen any rotations in the system. The type of bearing used
is ball bearing. Only one bearing is used in the design.
However, there were also some problems that need to overcome in order to ensure that
this design is successfully done. One of the problems are the design is of the stirling engine
itself. A precise consideration needs to be done in order to make sure the stirling engine is
applicable to the application. First is how to make it functioning. To be functioning the blade
must be connected to the stirling engine and to overcome it, shaft was introduced to the
system where it connect blade to the engine. Since there is a different in angle between the
position of the engine and the blade, therefore, the bevel gear was applied to the shaft of the
blade and the stirling engine.
Chapter 7 : Conclusion
Throughout this project, there is a lot of input were gained, especially for a better
understanding in what have been learn in previous class. The knowledge and theory that
Mr.Shafizal taught in class is obviously were strengthen much more after completing this
project. And for this project, it can be conclude that, based on the calculation that been made
the product which is the Smoke Kitchen Remover functioning according to plan and most
importantly it is safe to be use in daily life. To be precise, the best design has been selected by
using the method of Weighting Matrix and Pugh Method. Besides that, instead of using an
electric current to run the stirling engine, the usage of heat from the stove to run the engine.
So that, the daily electricity consume can be reduced and of course it is an important element
that must be considered nowadays. Asides that, this project required a presentation by videomontage, which is a good added value, since it is the first time student of 3 BMCT been
exposed to that approach of presentation. On top of that, this project is extremely good for
better understanding in Mechanical Engineering Design subject and of course, credit should
be given to the lecturer, Mr.Shafizal Bin Mat for giving out this assignment at the first place.
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