Advanced Materials and Materials Selection Resit Assessment

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Advanced Materials and
Materials Selection
Second Resit Assessment
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Module Code: MP 4702
Advanced Materials and Materials Selection
2014-15
Second Resit Assessment
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Question 1:
Given Data:
injection temperature=180 °−192°
injection timeat 180° =25 seconds
injection rate at 170 °=66 seconds
To find:
Time for new temperature=?
Solution:
T 1=170° =443.15 k
T 2=180° =453.15 k
K 1=66 s
K 2=25 s
R=8.314
Ea=
R ln K 2
K 1
( 1
T 1 − 1
T 2 )
Ea= 8.314∗ln 2.64
2.256 x 10−3−2.21 x 10−3 =175457.7373 J
mol =175.46 Kj
Now we will find reaction rate at 192º ;
ln ( K 2 ) −ln ( 25 ) = Ea
R ( 1
T 1 − 1
T 2 )
ln ( K 2 )−ln ( 25 )= 175457.73
8.314 ( 1
443.15 − 1
453.15 )=1.20146
ln ( K 2 )=1.20146+3.22=4.42
K 2=83.22 sec−1

Question 2:
Equation of Creep Deformation Arrhenius Equation
Similarities
Tells about stress and fracture relations of a
material. It will describe the relation between
‘stresses applied on polypropylene and
‘deformation or fracture’ occurred in it.
Tells about strain occurred on the material
(polypropylene) due to external factors.
Describes about the fracture or complete
deformation point of a given material.
Gives fracture point details qualitatively from
the temperature and physical relation.
Differences
Gives details about stress/strain, fracture and
deformation with respect to time.
Includes temperature change, strain size,
chemical activity of a material to describe
materials property.
Physical measurements are unit of force,
applied stress etc.
Measurements are in temp (k), chemical
reactions, etc.
Question 3:
The creep resistance of metal alloy systems increases with the concentration of alloying
elements dissolved into solid solution. The presence of alloying elements in interstitial crystal
sites increases the lattice strain and thereby resists the processes of dislocation slip and climbs
that creep of drive.
Dislocation shrinking or creep is a dislocation movement mechanism. This creeping
process appears to dominate at small pressures and elevated pressures. Dislocations can be
carried through sliding in a slide, a method that requires little thermal activation.
Materials used in high-performance systems, such as jet motors, often exceed 1200 ° C at
severe altitudes, leading to a severe problem. Super alloys depending on Co, Ni and Fe are
designed to be extremely susceptible to creep, thus creating the perfect fabric in high-
temperature settings. For example, non-base alloys modeled according to the Ni-Al system,
called α-β alloys, are even dislocation resistant. The μ is the principal FCC matrix, and the α is
the Ni3 (Al), precipitate phase that increases particle enhancement. Solute components
introduced, such as Ta, W, Mo, Fe, Cr, and Co, add to solid-solution hardening and are
frequently reacted to create carbide droplets that fill grain limits to prevent the slipping of grain
borders.
Question 4:
Given:
crack depth=a=0.02 m
run speed=15 rpm

time=8 hrs
Pmax=+6.4∗104 N
Pmin=−6.4∗104 N
A=0.04 m2
∂ max=+1.6 MPa
σmin=−1.6 MPa
∆ Kth=5 MPa m0.5
Kc=16 MPa m0.5
y=1.12
a)
∆ K =∆ σ . Y . √ pi∗a=3.2∗1.2∗√ pi∗0.02=0.898 MPa m0.5
∆ Kth=5 MPa m0.5
Since∆ K < ∆ Kth,….
So the Crack is not growing. So this machine is okay and safe to use for demonstration purpose.
According to the assumptions statement ∆ K < ∆ Kth, this crack is closed in compressive stress or
load condition. Hence, there is no harm or fear of breakdown etc. safety span or life of rod of
steam engine is unlimited. There is no time specific harm in it that would break it down and
machine would not rotate adequately.
B)
Let crack length be ac,
Kc=16 MPa m0.5
3.2∗1.2∗√ pi∗ac=16
ac=6.34
Pmin=−0.92 MN
Pmax=+ 0.92 MN
σmax = 0.92
0.04 =23 MPa

σmin=−23 MPa
∆ σ=23 MPa
∆ K =∆ σ . Y . √ pi∗a
¿ 46∗1.12∗√ pi∗0.02
∆ K > ∆ Kth, so crack grows.
Now in order to calculate number of cycles Nf;
N= 2
( m−2 ) ( C ) ( ∆ σY ) m pi
m
2 ( 1
a 0
m−2
2
− 1
ac
m −2
2 )
m=4
C=4.3∗10−8 m
∆ σ=σmax=23 MPa
N= 2
( 2 ) ( 4.3∗10−8 m ) ( 23∗1.12 ) 4 pi
4
2 ( 1
0.02
4−2
2
− 1
6.34
m−2
2 )=266.71cycles
Question 5:
Let us combine all the cases in a table below, in order to understand better;

Table 1: cases to be studied
Case 3 and 4 suggests that the material used in making of blades from sample B is
showing brittle behavior. From material sciences book “manufacturing process” by hajra
choudhry, it’s stated that blades and knives are made from soft iron (Choudhry, 1971). This is
because hard iron cannot be sharpened or molded. So sample B contains different combination of
materials than sample A. material used in sample A don’t have an ability to gather sharpness at
all. This means, sample A tolerates the mounding process unlike other stainless steel materials. It
employs that the elastic behavior possessed by compositions is really smart. What leads towards
this kind of behavior is the low amount of carbon present in sample A material.
In order to minimize the elongation properties, one need to cold work the blades of
sample A under the presence of aliphatic gases (hydrocarbons). Nitrogen must be introduced in
sample A during knife-making process; its general property is to strengthen the yield by
hardening of solutions, especially through wide variety of steels.
In order to let sample B experience superior performance, one should change the
additives used during manufacturing process. Here are some mounding reasons with their
prevention methods discussed;
Mounding Reasons Prevention Methods
Corrosion Nitriding, galvanizing, electroplating
Pitting Repeated alkaline treatment
Hydrogen embrittlement Lower hardness of stainless steel material, heat
treatment (Asheesh).

Forging and heat treatment increase the hardness in knives and reduce brittleness. Here
you can see the real life example of hardness made by “AISI 1050” and many other materials
after applying heat treatments (Balkhaya & Suwarno, 2017).
Figure 1: hardness of blades done by AISI-1050 material: cases before and after heat treatment are shown
Figure 2: hardness of blades done by AISI-4340 material: cases before and after heat treatment are shown

Figure 3: hardness of blades done by JIS-SKT-4 material: cases before and after heat treatment are shown
Figure 4: hardness of blades done by Spring Steel material: cases before and after heat treatment are shown
Chemical composition of material used in sample A and sample B for applying hardening
on;

Table 2: chemical composition of observed materials
Question 6:
Air=0.6litre
A 1=125 mm2
ρair =1.23 kg/m3
Solution:
a)
power=force∗velocity
F=PA=1.23∗0.6=7.38
P= 7.38
125 =0.05 watts
b)

as failures generate after flaw occurs. If flaws are not taken above critical flaw size, then
this is called maximum admissible flaw, which will not lead the material or resin towards failure
(Coleman, 2017).
MAF for glass soda lime 0080 = 0.75
MAF for Alumina 99.9 = 4
MAF for silica fused = 11.15
c)

Question 7:
Column Buckling:
An assessment of the beam buckling comprises in determining the column's peak mass
before it falls down.
According to Euler buckling theory,
E 1 y' ' =M =−ρy
E 1 y' ' + Py=0
y= Asin √ P
E1 x+ Bcos √ P
E 1 x
P
X