Bao Last Assignment

8/8/2019 Bao Last Assignment

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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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