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Advance manufacturing Management
Assignment 2
Submitted by: Amer Khan
Submitted to: Dr.Bao/Dr Gawne
ID : 2827066
Date of submission : 28-05-2010
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Index
1. Methods of increasing Fracture toughness of Zirconi.32.
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Question 1:
Methods of Improving fracture toughness of Zirconia(Ceramic):
1.a) Zirconia is a composite ceramic noted for its strength, wear and thermal properties, the
fracture toughness Klc of a zirconia can be increased by the following methods. The
engineering strength of ceramic can be improved by engineering microstructure, so that
higher energy is needed for the crack formation.
Zirconia can either have a cubic lattice structure or tetragonallattice structure. The fracture
toughness of cubic zirconia can be increased by adding a small amount of magnesia and by
use of special thermal treatment, the fracture toughness can be increased up to five times.
The figure given below show a partially stabilized zirconia reinforced by magnesia particles
at a higher magnification.
Fig 1 a) Magnesia particle used in zirconia b). Zirconia magnification at 1 m
Aging process: the fracture toughness of zirconia can also be improved by aging process.
According to the journal of fracture toughness and mechanical strength of ceramics. Which
states that transformation of fine tetragonal precipitates immersed within a cubic coarsematrix can enhanced the properties of zirconia. The size of tetragonal precipitate can be
optimized via two steps heat treatment in which material is firstplaced in cubic phase field at
1800 degree Celsius followed by aging in tetragonal cubic phase. At a temperature between
1400-1500 C. such treatment induces decomposition of cubic grains forming tetragonal
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precipitates that grow up to an optimum size resisting the formation of crack and increases
fracturetoughness by 10%.
Optimized tetragonal precipitate of 300 nm can also be obtained by sintering zirconia at
1770c for 3 hr and subsequently aging at 1400C for 140 hrs. This results in a stabilized
zirconia with 10% increase in fracture toughness. However increase in duration of aging
more than stated value results in the formation of mono-clinic which promotes a decrease in
Klc value. The figure given below illustrates the formation of monoclinic grains.
Fig 2: A). Increase in fracture toughness with increase in temperature
B). Formation of mono-clinic grains in Zirconia ceramic.
Fracture toughness of Zirconia can also be improved by the addition of fibres and whiskers in
ceramic material. This increases the fracture toughness up to 15 times. The presence of
whisker and fibresdeflects the growth of crack formation.
Apart from this the fracture toughness can also be increased by the addition of small amount
of oxides such as Yttria or magnesia. And the material is said to be partially stabilized,
varying the composition allow different properties to be enhanced.
The fracture toughness of Zirconia Ceramic can be increased by the following:
y Modulus Transfery Pre-stressingy Crack deflection or impedimenty Bridging
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y Pull-outy Crack shieldingy Energy dissipation
Modulus transfer:
Modulus Transfer generally involve high elastic modulus fibres in a lower modulus matrix.
Where fibres or polymers having higher elastic modulus than the modulus of Zirconia
composite is transferred from matrix to the fiber such that the higher strength fibres can carry
the load.
Pre-stressing:
This involves placing a portion of ceramic under residual compressive stress. A crack cannot
start or extent as long as ceramic is pre-stressed in compression. Tensile fracture only occurs
after an enough load is applied to exceed the compressive pre-stress and to build up a tensile
stress large enough to initiate a crack at a critical flaw. A compressive pre-stress can be
achieved by many approaches; one of the approaches is to place the surface in compression
by quenching and ion exchange
Pull-out:
Pull-out consists of fibres, particle or grain. That debound from the adjacent micro-structure
and pull out as crack open, energy that would normally cause a crack propogation is partially
expended by debonding and by friction as the fibers, particle or grain slides against adjacent
micro-structure features. This caneffectuively increase fracture toughness. pull-out is often
accompanied by bridging. Pullout appears to be an important mechanism of
achievingoptimum toughening of Zirconia. The nature of fibers reinforced ceramic
matrixcomposite is critical. If the bonding is too strong to allow the pull out. The fracture
travels directly to the fiber and no pull out or bridging occurs. The material exhibit low
toughness, if the fiber matrix interface bond is weak to allow crack deflection or debonding.The material can exhibit high toughness.
Crack shielding:
Crack shielding is a stressed induced micro-structural change. That results in a reduction in
stress at the crack tip. The effect occur around the crack tip and extend along the crack. The
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thickness of zone and the extent of the wake both affect the degree of stress shielding at the
crack tip.
Several type of crack shielding have been identified.
Micro-Cracking:
This result in crack shielding by locally reducing the elastic modulus and spreading the
applied stress over many cracks rather than one primary crack.
Ductile Zone:
A ductile zone results in crack shielding by allowing plastic deformation around crack tip.
Transformation toughening:
Transformation toughening is a relatively new approach to achive high toughening in
Zirconia, where a Zirconium oxide goes through a martenstic phase transformation from
tetragonal to monoclinic crystal. While cooling through a temperature of approximately
1150oC.by control of composition, particle size and heat treatment cycle. Zirconia can be
densified at higher temperature and cooled such that tetragonal phase is maintained as
individual grain or as precipitate tp room tempertature. The figure given below explain the
transformation of Zirconia from teragona to monoclinic cysrtall distribution.
1.b). Fracture toughness of annealed Glass:
Given 2a = 1.1 m
A = 0.55
Stress = 125 Mnm-2
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Y = 1.0
We know that fracture toughness
Klc = Ya
1.0*125**0.55 164.3Mn-3/2Therefore fracture of toughness of freshly annealed glass is found to be 232.3 Mnm-3/2
1(ii).
Given
= 41 MN m-2
we know that
Rate of fractureoccurrence = (Af-Ai)/t
Where Ai = 1.1m
Af= (Klc/Y)*1/
Af= 5.092m
Rate of fracture = (Af-Ai)/t
(15.092-0.55)/9 0.50477mTherefore rate of fracture over 9 day period is found to be 0.50477m.
Question 2:
2. a) The mechanism for lubrication of piston ring and the consequent wear phenomenonhave been a subject extensive investigation for some considerable time. A recent
advances in technological review indicates that the during the lubrication of piston ring
and cylinder linear, a region is encountered. Where hydrodynamic lubrication fails
because of increased load, high temperature and decreased speed thus squeezes out of oil.
As per recent investigation the friction increases suddenly when transition temperature is
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reached. Which is defined as the temperature at which lubrication tend to lose its ability
to lubricate. The piston ring rubs against the wall of linear and this result an increase in
wear. However a recent investigation revealed that when metal surfaces are heated in the
presence of lubrication. The absorbed layer softens which in turn depend upon nature of
lubrication.
Under poor running condition the piston ring is always heated at same spot, whereasthe
same friction energy is dissipated over a much greater surface of linear. The bulk
temperature at rubbing spot is higher on piston ring than on linear.
Methods to reduce wear and increase performance:
A common technician found that by matching the molecular chain length of polar
additives to the chain length of the base fluid, wear resistance and maximum loadingcould be increased safely. Apart from this, to minimize the wear the piston ring can be
made up of wear resistance material such as cast iron and steel, which can also be coated
or treated to enhance the wear resistance.
Typically the compression ring or oil ring can be coated with chromium or nitride
possible plasma sprayed or by physical vapour deposit.
This result in enhanced scuff resistance and further improvement in wear resistance.
For example: The modern diesel engine have top ring coated with modified chromium
coating known as CKS or GDC. A patent coating from Goetz, which has aluminium
oxides or diamond particles respectively included in chromium surface. Where the lower
oil ring is designed to leave a lubricated flux, a few micrometres thick or bore, as piston
descend and this maintains the lubrication of linear at all time.
Question no.2
2.b). Type of wear = oxidative wear
Activation energy = 268 Kj/ mol= 268000 j/mol
T1= 250 c
T2 = 300 c
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Arrhenius equation for wear rate = Aexp(-Q/RT)
Where Q = Activation energy R = universal gas constant = 8.314 j/mol.K T = temperature in Kelvin A = exponential constant = 1016
Rate of wear at 250C = 1016exp (-268000/8.314*523)
1.708148*10-11mWear rate at temperature T2 = 300C
1016exp( -268/8.314*573) 3.700*10-9mThe relative increase in wear rate due to increase in temperature from 250-300C is found
to be
3.700*10-9 1.708148*10-11 3.68 3*10-9mQuestion no.3:
Gas carburizing:
Gas carburizing is a case hardening process in which carbon is dissolved on to the surface
layer of steel part at a temperature sufficient to render the steel austenitic characteristics.
Followed by quenching and tempering to form a martensitic micro-structure. The
resulting gradient in carbon content below the surface of part causes a gradient in
hardness, producing a strong wear rsistance surface layer on material. High production
rate gas carburizing is commonly carried out on highly stressed component. Such as
gears, bearing and shafts. Majority of gears are gas carburized as it offers acceptable
control of case hardening, surface carbon content and proper diffusion of carbon into
steel.
Effect of carburizing on gear materials:
A carburized gear tooth may be regarded as a composite structure consisting of lower
carbon in centre and high carbon content on the surface.
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Increasing carbon content on the surface increases wear resistance and resistance to
fatigue life.
As the carbon level increases above 0.8%. There is a tendency to retain austenite. This
produces carbide network in the upper case. A recent study showed that a level of 15-
20% retain austenitic is desirable both for sliding wear resistance ad resistance to pitting
fatigue.
In general case depth of a carburized tooth is a function of diametric pitch. The bigger the
tooth the more case is needed to carry the load that will be imposed on the tooth. Too
much case hardening makes the tooth brittle with a tendency to shatter off the top of the
tooth. However too thin case hardening reduces tooth resistance to pitting. The diagram
given below illustrates the diffusion of carbon on to the surface of gear tooth increasing
the wear capacity.
Fig:A Fig: B
Fig: 3 a)A gas carburized Gear b)Diffusion of carbon on to the surface of tooth.
The wear resistance increases by increase in hardness. The wear rate can better be
explained by the following mathematical expression.
Wear rate = KW/H
Where k = constant
H = hardness
W = load
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The following diagram shows increase in hardness in conjunction with case depth for 135
modified steel, 4340 which were carburized for 45 hours.
Fid 4 : Relationship between hardness and wear resistance.
As the carbon content of steel increases it compresses the steel structure. Which
inevitable increases the fatigue strength of steel structure. The diagram given below
illustrate as how the fatigue strength can be increased by gas carburizing.
Fig5 : Gas carburizing increasing fatigue strength by compressing the structure.
3.b) Given Co = 0.2 wt%
Cs = 0.8 wt%
(i) Condition 1Conditions A1 A2 A3
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Temperature 840 840 840
Time 1 5 8
We know that
Cx-Co/ Cs-Co = 1-erf(x/2Dt)
Where Do = Do exp (-Q/RT)
Diffusion coefficient for the first condition at a temperature of 840C.
D = Doexp(-Q/RT)
2.3*10-5exp(-148000/8.314*1113) 26.03*10-3
Carbon content at X=0.1.
Cx-0.2 /0.8-0.2 = 1-erf(0.1/226*10-3*1)
Cx-0.2 / 0.6 = 1-erf(0.309)
Cx- 0.2/0.6 = 1-0.4284
Cx-0.2/ 0.6 = 0.5716
Cx-0.2 = 0.34296
Cx = 0.54wt%
Condition A2 at X= 0.5
Cx-0.2/0.6 = 1- erf (0.5/226.03*10-3*5)
Cx-0.2/ 0.6 = 1-erf(0.692)
Cx-0.2/0.6 = 1-0.6778
Cx = 0.39 wt%
Condition A3 at X= 1.0
Cx-0.2/ 0.6 = 1- erf(1/226.03*10-3*8)
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1-erf( 1.0956) 1- 0.8427 Cx = 0.29438 wt %
Condition 2:
Condition B1 B2 B3
Temperature 880 880 880
Time 1 5 8
Diffusion coefficientfor 2nd Condition
D =Do exp (-Q/RT)
2.3*10-5exp(-148000/8.314*1153) 4.53*10-12
At time 1 hr and at a distance of 0.1
Cx-0.2/0.6 = 1- erf(0.1/24.53*10-12*1)
1-erf(234902.049)This is the place where Iam getting error.
Question No. 4:
4.a) Stiffness performance Index:
Given deflection = 5FL / 32Ewt 1Tensile stress () = 3FL/4wt.2
Step 1: Identification of function, free variables and objective
Function: to support weight
Free variable: t thickness
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Objective: minimize weight
We know that performance equation m = AL
Where area = w*t
Therefore the performance equation can be written asM = w*t*L*.3
We know that = 5FL / 32Ewt
t = 5FL / 32E t = (5FL )1/3 / (32Ew)1/3 ..4
Inserting equation 4 into equation eq 3 we get
M
= w*L*(5FL )
1/3
/(*32wE
)
1/3
M = (5F)1/3/ (*32)1/3*(w*L)/w1/3*(/E1/3)
Therefore the performance index is found to be
P1 = E1/3/Strength performance index:
We know that tensile stress
= 3FL/ 4wt
t = 3FL/4w
t = (3FL)1/2/ (4w)1/2..5
Substituting equation 5 in equation 3 we get
m = (3F/4) 1/2*(w*L3/2) / w1/2*[/1/2]
Therefore the performance index is found to be P2= ()1/2/ .
4.b) from the charts provided following materials were selected with strength performance
index greater than 1.50 mpa and stiffness performance index greater than 6.0.Stiffness performance chart
Diamond Sic SL3N4 Sialoon Aluminas
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Silicon Laminates
GFRP, KFRP Magnesium alloy Beo
Strengthperformance chart Diamond Sic A2LO3 ZrO3 Mgo B Glass Laminates
KFRP, CFRP SiliconPlease refer to the charts attached for design guide lines.
4.c) Selection of material on the basis of Strength and performance index. The followingfive materials where selected on the basis of their performance indices values.
MaterialStiffness performance
indexStrength performance index
Be (450)1/3/2.1 = 3.6 Sqrt(5000)/2.1= 33.67
Diamond (1000)1/3/ 3.22 = 3.22 Sqrt(10000)/3.1 = 32.25
Sic (600)1/3/3 = 2.81 Sqrt(9000)/3 = 31.6
GFRP(laminates) (60)1/3/1.5 = 2.60 Sqrt(1000)/ 1.5 = 21.08
Sailoon (500)1/3/3 = 2.64 Sqrt(7000)/3.2 = 26.14
As fracture toughness of plate is concern, which has to be above 20 mpa
I select GFRP laminates as it has fracture toughness of 80 mpa.
Stiffness performance Index = 2.60
Strength performance index = 21.08.
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GFRP laminate proves to be the best material as it has low density and high performanceindex in relation to the fracture toughness which is about 80 Mpa.
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