Engineering Science: Applications of Physical Sciences

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3: Engineering Science
Applications of the physical sciences to
engineering problems

Table of Contents
LO1....................................................................................................................................4
Task 1.0......................................................................................................................................4
Task 2.0......................................................................................................................................7
LO2....................................................................................................................................9
Task 3.0......................................................................................................................................9
3.1................................................................................................................................................................9
3.11..............................................................................................................................................................9
3.12..............................................................................................................................................................9
3.2..............................................................................................................................................................10
3.3..............................................................................................................................................................10
3.4..............................................................................................................................................................12
3.5..............................................................................................................................................................12
LO4..................................................................................................................................13
Task 4.0....................................................................................................................................13
4.1..............................................................................................................................................................13
4.11............................................................................................................................................................13
4.12............................................................................................................................................................13
4.2..............................................................................................................................................................14
4.3..............................................................................................................................................................15
4.31............................................................................................................................................................15
4.32............................................................................................................................................................16
4.4..............................................................................................................................................................17
4.41............................................................................................................................................................17
4.42............................................................................................................................................................18
4.5..............................................................................................................................................................19
4.51............................................................................................................................................................19
4.52............................................................................................................................................................20
References.......................................................................................................................21
Table of Figures
Figure 1: Stress-strain curve......................................................................................................6
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Figure 2 – Dimensional change with thermal energy..............................................................11
Figure 3 – Square waveform....................................................................................................14
Figure 4 – Saw-toothed waveform...........................................................................................15
Figure 5 – Triangled waveform...............................................................................................15
Figure 6 – Laminations type....................................................................................................17
Figure 7 – Transformer core....................................................................................................18
Figure 8 – Phase angle.............................................................................................................20
Figure 9 – Simulation...............................................................................................................20
Table of Tables
Table 1: SI units.........................................................................................................................4
Table 2: Impact table of alloys...................................................................................................7
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LO1
Task 1.0
Organisation named International system of units (SI) states and elucidate seven types of
units for the purpose of measurement.
These units are used for measuring several basic quantities also known as Simple Units and
are shown below.
SI Unit Prefix notation Unit Symbols Derived Unit
Length l m meter
Electrical Current I A ampere
Substance’s Amount Mole mol. mole
Temperature Temp. K Kelvin
Weight Mass kg kilogram
Time T s second
Luminous Intensity - cd candela
Table 1: SI units
These units form the base for measuring the quantity and size of metals being used and even
another form of data.
Graphical representation of the quantitative data gathering, using the appropriate
software
Data that is being measured when depicted in a graphical representation is known as
Quantitative data’s graphical representation. It depicts the methods being taken to show that
data is clear any changes over any point. By means of graphs, minute changes can also be
seen and understood clearly. There are several graphical means to show data such as Pie
chart, bar graph, line graph are the most used and very common ways that are used today.
Some ways being used are described following :
a) Histogram: Histogram are just as same to what the bar graph is, they have similar bar
height to represent the data value but with no gaps present between them so they have
common base and height of the bar represents to frequency for that range.
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b) Pie chart: These charts are shown and represented out from a circular disk representation
and this chart divides the value of the representation in the form of ratio area and then
data is filled in all ratios with their respective information in it.
c) Line graph: Line graph forms the most basic type of forms that are available. Line is
simply made with respect to the values along an axis. This line shows all the changes that
are formed by joining all the value points along the series.
d) Bar graph: In this graph, the data is shown by taking the values and then showing it in
form of rectangular bars where data and is based on the user choices of axis value. Each
bar is placed at equal distance from each other and has the same width.
Reflection on application
Scientific research states the formation of scientists and researchers which work together to
communicate and form methods for scientific researches as shown below.
Title: Here it states the topic of study for research and it tells facts like pressure, state,
volume and much more aspects of the topic.
Abstract: It’s like a glossary to the whole research in the most concise way possible. This is
sure written in stepwise manner so that one could make out the idea of the research at the
beginning before going deep into it.
Introduction: This is the point where the researcher’s idea is visible about the research topic
and finds to find a reference in future in the report.
Materials and Methods: These parts describes the methods and tools that were employed
for the experiment. And data are more referenced with the help of graphs, pictures, tables,
and references.
Results: Here the final result of all the researches ,findings, methods are written. The final
conclusion is shown here.
Discussion: At last final verdicts for the report is concluded. It normally provides weather
research is approved or not or fit for publications or not.
Tables and Graph: According to the name’s meaning it represents indexing of all the tables
and graphs being used in the documentation.
References: It describes the links that are being used for the research projects. References
credit others knowledge which was used for future findings for the report and thesis.
Lastly, safety tips about the practical execution are taken care of.
Analysis of all the graphical data presented
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The graph for the strain-stress curve is shown below
Figure 1: Stress-strain curve
Histogram: It's being used to lay the data symmetry. These type of representation are
generally complex in understanding so such graphs analyses are necessary.
Index plot and time series plot: best used to scope out present errors. It takes only a single
argument at once and scans for the absence of presentation parts.
Pie chart: Here data represented in proportions of a fixed area. Here cross-checking for data
provided is weather divided proportionally or not with the each of individual quantity area
taken.
Selection for graph appropriate for data provided: Graph selection is necessary and is decided
based on the different conditions. Like graph plotted by the single variable needs different
types of representations whereas the graph needed to represent two or more relative quantities
would require different types of graphs. Hence for such purposes, proper analyses about
graph selection are carried out by the experts such that proper graph for data is made and a
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proper graph is made. This graph selection is carried out to depict that the graph being
selected is correct or not.
Below is the data analysis of the steel grades is shown:
Alloy Impact Energy [J (ft-lb)]
8630 low-alloy steel 55(41)
1040 carbon steel 180(133)
Monel 400 (nickel alloy) 298(220)
Ferrous superalloy (410) 34(25)
Aluminium bronze, 9% (copper alloy) 48(35)
L2 tool steel 26(19)
5050 solder (nickel alloy) 21.6(15.9)
Table 2: Impact table of alloys
[Slideplayer.com. (2019)]
Task 2.0
Description of the structural properties linked to their respective material properties
Properties of particular metal like its structure and molecular structural strength describes the
what type metal is. Steel has a very special property because of which it's being used
preferably in the application of making buildings. The property which makes it desirable is
the presence of a magnetic effect in steel. Steel as being one the conductors of electricity is
also preferred in electrical applications and one major reason is its availability and easy
accessibility.
Product design of any metal is based on these properties:
Ductility: It’s the property of a metal which defines the bending, twisting of metal into
several shapes to the limit till its breaking point. It is used mostly when wire designing is
required of metals. Plane designing is the field where highly ductile and strong metals are
preferred because of their ability to absorb shocks during turbulence in the air.
Elasticity: Elasticity could be understood as the phenomenon present in metal which checks
metals ability to reform into its shape when the force applied on it is removed. This property
is most important to be examined when industrial products are designed because no one
wants that when the applied force is removed it permanently destroy the structure. Here we
need to know that there exists a limit called elastic limit of metal beyond which the metal will
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deform or will lose it structure and leads to distortion. Elasticity can also be increased or
decreased for a particular metal by designing them in such ways.
Hardness: Metal’s ability to resist distortion, penetration, cutting, rusting are known as the
attributes to the hardness of the metal. Hardness is just the property and could be increased or
decreased for the particular metals by combining them with several other metals present and
changing their compositional properties. To note we should keep in mind that hardness of
metals is in a close link to the metal’s strength.
Malleability: Malleability of metal is defined as metal’s ability which allows them to be
folded and pressed into thin sheets. This is only a measure till the metal is not cracked and is
most important when sheets out of metals are being produced.
[Integrated Publishing, I. (2019)]
Types of degradation
The stain change with respect to the change in strain is defined as elastic hysteresis within the
presence of force applied. In metals, it’s the amount of strain that is required over the metal
that will bring stress in metal that deforms it is known as the minimum required energy for
that particular metal. To note that in metal stress-strain graph follows a linear path as shown
into the graph which is clear representation that the stress is in ratio with the strain until a
breakdown is achieved. The breakdown is known as the maximum force after which the
stress and strain are not in ratio with each other or we can say that is the force after which the
metal could break or distort.
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LO2
Task 3.0
3.1
3.11
Solving R1 and R2, we get
Rection of forces=all applied forces . distace moved by force
R2 ( 2+ 4 )=4 × 45
R2 ( 6 ) =180
R2= 180
6
R2=30 kN
And for finding R1,
R1 ( 2+ 4 ) =2 × 45
R2 ( 6 )=90
R2= 90
6
R2=15 kN
Hence 30 kN and 15 kN is the reaction on either side of the beam respectively.
[Afsar, J. (2014)]
3.12
For calculation of all the reaction forces
UDL as beam load = f = mg
f =9.81 ×39.00
f =382.59 N m−1
As
f =R1 + R2
displacement × UDL=R1 + R2
382.590 ×6.0=R1 + R2
R1 + R2=2295.54 N
The load being bared by the beam is load point dependent. In this case, it would be the same
and equally distributed. Such that,
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R1 =1147.221 N and R2 =1147.221 N
3.2
To understand the application of buoyancy first we’ll have a look into and understand about
what buoyancy is and what it does.
The force that is exerted upon an immersed object by the fluid in presence of gravitational
force is defined as buoyancy. This force exerted is in an upward direction. Note that the force
exerted over the top surface of the object is lesser when compared. This difference in exerted
force is what makes the object to float as applied fluid pressure increases with depth. This
force is equivalent to the volume of the fluid displaced.
Applications
 Hot air balloon
The atmosphere is filled with ample air exerting a buoyant force on each and every
object. Due to this buoyant force, only that’s responsible for the floating and rise in the
hot air balloon. It becomes stationary only when balloon’s weight is greater than the
weight of the air present around it or more when compared to the buoyancy of the air
around it. Same is applied to stationary the position of the balloon by equaling the
buoyant force and weight of balloon.
The weight is controlled by regulating the air’s temperature inside the balloon.
 Submarine
Submarine have ballistic tanks which hold water into it and out of it. Thus, when
submarines need to float over the surface than they pump out the water from the ballistic
tanks. Pumping out water reduces the weight of submarines and increases its buoyancy,
and when it needs to submerge again, they fill this tank with water which in turn increase
the weight and decrease the buoyant force upon the submarine.
[Simscale.com. (2019)]
3.3
Elastoplastic change: Forming of thermal stress in the thermoplastic component is defined as
the thermal effect in the composition of the steel. As temperature or heat is applied over the
metal the heat is uniformly distributed from the outer surface to the inner surface of the metal
and the heat transferred is radially spread across. Hence over the application of heat, stress is
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tangentially formed due to which in a future strain of the elastoplastic are increased with a
rise over temperature.
[Hamidi, A., Tourchi, S. and Kardooni, F. (2017)]
Dimensional Changes: Magnitude of material like material’s mass, change in temperature and
substance phase are factors that decide the amount of heat which is being transferred.
a) Temperature change is in direct proportion to the heat energy being transferred. To
change the mass twice its size we have to apply double the amount of heat to it.
b) Heat transfer also depends over the mass of the material and is in direct proportion to
its mass. When the mass of an object is doubled it would require double the amount of
heat energy.
c) Substances phase are also key factors which decide the heat transfer amount into the
object. If x degree of temperature is being required in order to bring change in metal
A then y would be the temperature that would be required to bring the same amount
of change in metal B.
Figure 2 – Dimensional change with thermal energy
[Opentextbc.ca. (2019)]
Thermal Stresses: With thermal energy on the metal increases the tensile strength of
particular metal. With the increase of thermal energy, it increases the impurity of metal and
loses its capability to welding. Thermal stresses bring out changes in metal composition and
also in their properties and sometimes even in chemical and physical related properties.
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3.4
Following d’Alemberts Principal, it states that –
F1+ F2 + F3 + F4 +… … … …+ F−ma =0
150.0− ( a ×60 ) =0
A=150/60=2.5 m/ s2
3.5
Based on the provided data, let –
R1 = 150 mm
R2 = 500 mm
T1 = 70 C
T2 = 20 C
k =0.090 W /mK
Hence now thermal conductivity and relation with the above equation is as per
Meta l' s Thermal Conductivity= 2 Π ( T i−T o )
ln ( r2
r1 )
k
Q
L =2 Π ( 70−50 )
ln ( 500
150 )
0.090
Q/ L=3.480 W /m
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