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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 Zirconia«««««««««««...3

1.1 Aging process««««««««««««««««««««««««««4 1.2 Modulus Transfer «««««««««««««««««««««««.«5 

1.3 Pre-stressing««««««««««««««««««««««««««...5 

1.4 Crack deflection or impediment ««««««««««««««««««...5 

1.5 Pull-out ««««««««««««««««««««««««««««.6

1.6 Crack shielding ««««««««««««««««««««««««« 6

1.7 Transformation toughening««««««««««««««««««««...7

2.  Fracture toughness of Annealed glass««««««««««««««««««.7

3.  Question 2 (a)«««««««««««««««««««««««««««...8

4.  Question 2 (b)«««««««««««««««««««««««««««...9

5.  Question 3 (a) Gas carburizing«««««««««««««««««««««10

6.  Question 3b(i)«««««««««««««««««««««««««««..13

7.  Question 3b (ii)«««««««««««««««««««««««««««21

8.  Question 4 (a) Stiff ness/ strength performance index«««««««««««.....22

9.  Question 4(b)«««««««««««««««««««««««««««...23

10. Question 4( c )«««««««««««««««««««««««««««.24 

11. Modulus density chart««««««««««««««««««««««««.25 

12. Strength and density chart««««««««««««««««««««««...27

13. Fracture toughness and density chart««««««««««««««««««..28

14. R eference«««««««««««««««««««««««««««««.29

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O ptimized  tetragonal precipitate  of  300 nm can also be  o btained by  sintering zirconia  at 

1770°c for 3hr and  su bsequently  aging at 1400°C  for 140 hrs. This results in a sta bilized 

zirconia with 10% increase in f racture  toughness. However increase in duration of   aging

more than stated value results in the formation of mono-clinic which promotes a decrease in

K Ic value. The f igure given below illustrates the formation of monoclinic grains.

Fig 2: A). Increase in f racture 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  f ibres and whiskers in

ceramic material. This increases  the  f racture  toughness  u p to 15  times. The presence  of  

whisker and f ibresdeflects the growth of crack formation.

Apart f rom this the f racture 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  sta bilized, 

varying the composition allow different pro perties to be enhanced.

The fracture toughness of Zirconia Ceramic can be increased by the following:

y  Modulus Transfer

y  Pre-stressing

y  Crack deflection or impediment 

y  Bridging

y  Pull-out 

y  Crack shielding

y  Energy dissipation

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1.2 Modulus transfer:

Modulus Transfer generally involves high elastic modulus f ibres in a lower modulus matrix.

Where  f ibres  or polymers  having higher elastic modulus  than the modulus  of  Zirconia 

composite is transferred f rom matrix of  the f ibre such that the higher strength f ibres can carry 

the load.

1.3 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 f racture only occurs 

after an enough load is a pplied to exceed the compressive pre-stress and to build u p a tensile 

stress  large  enough  to initiate  a crack at  a critical  flaw. A compressive pre-stress can be 

achieved by many a pproaches; one of  the a pproaches is to place  the surface in compression by quenching and ion exchange 

1.4 Pull-out:

Pull-out consists of  f ibres, particle or grain. That de bound f rom the ad jacent micro-structure 

and pull out as crack o pen, energy that would normally cause a crackpro pagation is partially 

ex pended by de bonding and by f riction as the f ibers, particle or grain slides against ad jacent 

micro-structure  features. This caneffectively increase  f racture  toughness. Pull-out is  often

accompanied by bridging. This a ppears to be an important mechanism for achieving o ptimumtoughening of Zirconia. The nature of  f ibers reinforced ceramic matrixcomposite is critical. If  

the bonding is too strong to allow the pull out. The f racture travels directly to the f iber and no

 pull out or bridging occurs. The material exhibit  low toughness, if  the f iber matrix interface 

 bond is weak to  allow crack deflection or de bonding. The material can exhibit  high 

toughness.

1.5 Crack shielding:

Crack shielding is a stressed induced micro-structural change. That results in a reduction in

stress at the crack tip. The effect occurs around the crack tip and extends along the crack. The 

thickness of zone and the extent of  the wake both affect the degree of  stress shielding at the 

crack tip.

Several ty pe of crack shielding has been identif ied.

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Micro-Cracking:

This result in crack shielding by  locally reducing the  elastic modulus  and  s preading the 

a pplied 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.

1.6 Transformation toughening:

Transformation toughening is  a relatively new a pproach  to achieve  high  toughening in

Zirconia, where  a Zirconium oxide goes  through  a martensitic phase  transformation f rom

tetragonal  to monoclinic crystal. While cooling through  a  temperature  of   a pproximately 

1150

o

C. By control  of  composition, particle  size  and heat  treatment cycle. Zirconia can be densif ied  at  higher temperature  and cooled  such  that  tetragonal phase is maintained  as 

individual grain or as precipitate  to room temperature. The f igure given below ex plains  the 

transformation of Zirconia f rom tetragonal to monoclinic crystal distribution.

Fig 3. Zirconia Transformation f rom Tetragonal to monoclinic crystal.

1.b). Fracture toughness of annealed Glass:

Given 2a = 1.1 m

A = 0.55m

Stress = 125 Mnm-2

Y = 1.0

We know that f racture toughness 

K lc = Y¥a 

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  1.0*125*¥*0.55 

  164.3 Mn-3/2 

Therefore f racture of  toughness of  f reshly annealedglass is found to be 164.3 Mnm-3/2

1(ii).

Given

= 41 MN m-2

We know that 

R ate of  f ractureoccurrence= (Af-Ai)/t 

Where Ai = 1.1m

Af  = (K lc/Y)²*1/

Af  = 5.092m

R ate of  f racture = (Af -Ai)/t 

  (15.092-0.55)/9

  0.50477m/ day 

Therefore rate of  f racture over 9 day period is found to be 0.50477m/day.

Question 2:

2. a) The mechanism for lu brication of piston ring and the consequent wear phenomenon

have been a su bject extensive investigation for some considera ble time. A recent advance 

in technological review indicates  that,during the  lu brication of  piston ring and cylinder

linear,  a region is  encountered. Where  hydrodynamic lu brication fails because  of  

increased  load, high temperature and decreased s peed thus squeezes out of  oil. However

as per recent investigation the f riction increases suddenly whentransition temperature is 

reached. Which is def ined as the temperature at which  lu brication tend to lose its a bility 

to  lu bricate. The piston ring ru bs 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   lu brication. The a bsorbed layer softens whichin turn de pend u pon nature of  

lu brication.

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Under poor running condition the piston ring is always heated at same s pot, whereas the 

same  f riction energy is  dissipated  over a much greater surface  of   linear. The bulk

temperature at ru bbing s pot 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 loading

could be increased safely. Apart  f rom this,  to minimize  the wear the piston ring can be 

made u p of wear resistance material such as cast iron and steel, which can also be coated 

or treated to enhance the wear resistance.

Ty pically  the compression ring or oil ring can be coated with ³chromium or nitride´ 

 possible plasma s prayed or by physical va pour de posit.

This result in enhanced scuff resistance and further improvement in wear resistance.

For example: The modern diesel  engine have  to  p ring coated with modif ied chromium

coating known as  CKS or GDC. A patent coating f rom Goetz, which  has  aluminium

oxides or diamond particles res pectively included in chromium surface. Where the lower

oil ring is designed to leave a lu bricated flux, a few micrometres thick or bore, as piston

descend and this maintains the lu brication of  linear at all time.

Question no.2

2.b). Ty pe of wear = oxidative wear

Activation energy = 268 Kj/ mol= 268000 j/mol 

T1= 250 °c

T2 = 300 °c

Arrhenius equation for wear rate = Aex p(-Q/RT) 

  Where Q = Activation energy 

  R = universal gas constant = 8.314 j/molK 

  T = temperature in K elvin

  A = ex ponential constant = 1016 

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R ate of wear at 250°C = 1016ex p (-268000/8.314*523) 

  1.708148*10-11m

Wear rate at temperature T2 = 300°C 

  1016ex p( -268/8.314*573) 

  3.700*10-9m

The relative increase in wear rate  due  to increase in temperature  f rom 250oC-300°C is 

found to be 

  3.700*10-9 ± 1.708148*10-11 

  3.68 3*10-9

m

Question 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 suff icient 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 gradientin

hardness, producing a strong wear résistance surface  layer on material. High productionrate gas carburizing is commonly carried  out  on highly  stressed component. Such  as 

gears, bearing and  shafts. Ma jority  of  gears  are gas carburized  as it offers  acce pta ble 

control  of  case  hardening,  surface carbon content  and pro perdiffusion 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.

Increasing carbon content  on the  surface increases wear resistance  andresistance  to 

fatigue life.

As  the carbon level increases a bove 0.8%. There is  a  tendency  to retain austenite. This 

 produces carbide network in the u pper case. A recent study showed  that a  level of  15-

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20% retain austenitic is desira ble both for sliding wear resistance ad resistance to pitting

fatigue.

In general case de pth 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 to p 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 ca pacity.

Fig:A Fig: B

Fig: 4 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 

ex plained by the following mathematical ex pression.

Wear rate = KW/H

Where k = constant 

H = hardness 

W = load 

The following diagram shows increase in hardness in conjunction with case de pth for 135 

modif ied steel, 4340 which were carburized for 45 hours.

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Fid 5: R elationship between hardness and wear resistance.

As  the carbon content  of   steel increases it compresses  the  steel  structure. Which 

inevita ble increases  the  fatigue  strength  of   steel  structure. The  diagram given below

illustrate as how the fatigue strength increased by gas carburizing.

Fig6 : Gas carburizing increasing fatigue strength by compressing the structure.

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3. B) Given Co = 0.2 wt%

Cs = 0.8 wt%

(i)  Condition 1

Conditions A1 A2 A3

Temperature 840 840 840

Time 1 5 8

We know that 

(Cx-Co)/ (Cs-Co) = 1-erf (x/2¥Dt) 

Where Do = Do ex p (-Q/RT) 

Diff usion coefficient for the corresponding temperature D1, D2, D3. 

D1 = Doex p (-Q/RT) 

  2.3*10-5ex p(-148000/8.314*1113) 

  26.03*10-3 

D2 = 2.3*10-5ex p (-148000/8.314*1153) 

  4.5352*10-12 

D3=2.3*10-5ex p (-148000/8.314*1203) 

  8.615*10-12 

D1 D2 D3

Diffusion

coeff icient 

value 

2.60*10-12  4.5352*10-12 8.615*10-12 

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Carbon content at X=0.1mm for condition A1 time = 3600 sec

The distance can be taken into meters= 0.1*10-3m

  Cx-0.2 /0.8-0.2 = 1-erf (0.1*10-3/2¥2.60*10-12*3600) 

  Cx-0.2 / 0.6 = 1-erf (0.5168) 

  Cx- 0.2/0.6 = 1-0.5205   Cx-0.2/ 0.6 =0.4795 

  Cx-0.2 = 0.2877

  Cx = 48 wt%

Condition A1 at X= 0.2= 0.2*10-3m

  Cx-0.2/0.6 = 1- erf (0.2*10-3¥26.03*10-3*3600) 

  Cx-0.2/ 0.6 = 1-erf (1.0336) 

  Cx-0.2/0.6 = 1-0.8427  Cx = 29wt%

Condition A1 at X= 0.3= 0.3*10-3 m

Cx-0.2/ 0.6 = 1- erf (0.3*10-3/2¥26.03*10-3*3600) 

  1-erf (1.55044) 

  1- 0.971622

 Cx = 0.21 wt %

Condition A1 at X = 0.5mm = 0.5*10-3m

Cx-0.2/0.6 = 1- erf (0.5*10-3¥26.03*10-3*3600) 

Cx-0.2/ 0.6 = 1-erf (2.5840) 

Cx-0.2/0.6 = 1-0.9993

Cx = 0.204 wt%

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Similarly the carbon content at x = 0.2*10-3m, 0.5*10-3m is given in the ta ble given below.

Condition B: Temperature = 880oc + 273

  1153oF

  Diffusion coeff icient = 4.5352*10-12 

Conditions B1 B2 B3 

Temperature 880 880 880

Time 1 5 8

Condition B1 at X= 0.1*10-3 at a duration of 1 hours = 3600 sec

  Cx-0.2 /0.8-0.2 = 1-erf (0.1*10-3/2¥4.5352*10-12*3600) 

Cx-0.2 / 0.6 = 1-erf (0.3913) 

Cx- 0.2/0.6 = 1-0.428

Cx-0.2/ 0.6 = 0.572

Cx = 54wt%

Similarly the carbon content at X= 0.2*10-3m, 0.5*10-3m for the surface of gear su bjected to a 

carburizing for 1 hour is given in the ta ble below.

Carbon content for condition A3 at 28800 sec duration.

Distance f rom

surface 

Percentage of carbon content for A3 

X = 0.1*10-3m 68 wt %

X = 0.2*10-3m 57wt%

X = 0.5*10-3m 33wt%

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Condition B2 at X= 0.1*10-3 at a duration of 5 hours = 18000 sec

  Cx-0.2 /0.8-0.2 = 1-erf (0.1*10-3/2¥4.5352*10-12*18000) 

Cx-0.2 / 0.6 = 1-erf (0.1749) 

Cx- 0.2/0.6 = 1-0.195468

Cx-0.2/ 0.6 = 00.8045 

Cx = 68wt%

Similarly the carbon content at x = 0.2*10-3m, 0.5*10-3m is given in the ta ble given below.

Condition B3 at X= 0.1*10-3

at a duration of 8 hours = 28800 sec

  Cx-0.2 /0.8-0.2 = 1-erf (0.1*10-3/2¥4.5352*10-12*28800) 

Cx-0.2 / 0.6 = 1-erf (0.1375) 

Cx- 0.2/0.6 = 1-0.1403162

Cx-0.2/ 0.6 = 0.8596

Cx = 71wt%

Carbon content for condition B1 at 3600 sec duration.

Distance f rom

surface Percentage of carbon content for B1 

X = 0.1*10-3m 54wt %

X = 0.2*10-3m 36wt%

X = 0.5*10-3m 20wt%

Carbon content for condition B2 at 18000 sec duration.

Distance f rom

surface Percentage of carbon content for B

X = 0.1*10-3m 68wt %

X = 0.2*10-3m 57wt%

X = 0.5*10-3m 33wt%

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Similarly the carbon content at x = 0.2*10-3m, 0.5*10-3m for the a bove condition is given in

the ta ble given below.

Condition C

Given Temperature = 930oc+273

= 1203

Diffusion coeff icient = 8.615*10-12

Conditions  C1  C2  C3 

Temperature 930 930 930

Time 1 5 8

For the conditionC1 at a subjectedto a duration of 1 hour = 3600

At X = 0.1*10-3m

  Cx-0.2 /0.8-0.2 = 1-erf (0.1*10-3/2¥8.615*10-12*28800) 

Cx-0.2 / 0.6 = 1-erf (0.2839) 

Cx- 0.2/0.6 = 1-0.3026

Cx-0.2/ 0.6 = 0.6974 

Cx = 61wt%

Carbon content for condition B3 at 28800 sec duration.

Distance f rom

surface Percentage of carbon content for B3 

X = 0.1*10-3m 71wt %

X = 0.2*10-3m 61wt%

X = 0.5*10-3m 40wt%

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Similarly the carbon content at x = 0.2*10-3m, 0.5*10-3m for the a bove condition is given in

the ta ble given below.

For the conditionC2 subjectedto a duration of 5 hour = 18000 sec

At X = 0.1*10-3m

  Cx-0.2 /0.8-0.2 = 1-erf (0.1*10-3/2¥8.615*10-12*18000) 

Cx-0.2 / 0.6 = 1-erf (0.1269) 

Cx- 0.2/0.6 = 1-0.1403

Cx-0.2/ 0.6 = 0.8596

Cx = 71wt%

Similarly the carbon content at x = 0.2*10-3m, 0.5*10-3m for the a bove condition is given inthe ta ble given below.

Carbon content for condition C1 at 3600 sec duration.

Distance f rom

surface Percentage of carbon content for C1

X = 0.1*10-3m 71wt %

X = 0.2*10-3m 44wt%

X = 0.5*10-3m 22wt%

Carbon content for condition C2 at 18000 sec duration.

Distance f rom

surface Percentage of carbon content for C2 

X = 0.1*10-3m 71wt %

X = 0.2*10-3m 63wt%

X = 0.5*10-3m 42wt%

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For the conditionC3 subjectedto a duration of 8 hour = 28800 sec

At X = 0.1*10-3m

  Cx-0.2 /0.8-0.2 = 1-erf (0.1*10-3/2¥8.615*10-12*28800) 

Cx-0.2 / 0.6 = 1-erf (0.100) Cx- 0.2/0.6 = 1-0.2227

Cx-0.2/ 0.6 = 0.7773

Cx = 66wt%

Similarly the carbon content at x = 0.2*10-3m, 0.5*10-3m for the a bove condition is given in

the ta ble given below.

Carbon profile of steel gear with respect to different conditions

Z33\

0

10

 

0

30

40

¡ 

0

60

70

¢ 

0

90

100

0 0.£  0.4 0.6 0.¤  1 1. £  1.4 1.6

%

o

C

a

r

b

o

n

Distance from the surface in "x" mm

Carbon profile for 1st condition at 840 degrees

Carbon content for condition C3 at 28800 sec duration.

Distance f rom

surface Percentage of carbon content for C3 

X = 0.1*10-3m 73wt %

X = 0.2*10-3m 66wt%

X = 0.5*10-3m 48wt%

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3b. (ii):

Given Cx = 0.3 wt %

X = 0.5 mm

X = 0.5*10-3mts 

From the carbon prof ile  as plotted  a bove where  a  steel gear is  su bjected  to  different temperature and time combinations 

The most  suita ble condition to reach  a  de pth  of  0.5mm with  a concentration of  0.3wt%

carbon is found to be A3 

0

20

40

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0 0.2 0.4 0. ¥  0.̈ © ©  .2 ©  .4 ©  .¥ 

%

o

C

a

r

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Distance from the surface "X" mm

Carbon profile for the 2nd condition at 880 c

0

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0 0.2 0.   0.6 0.8 1 1.2 1.   1.6

%

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c

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Distance from the surface "x" in mm

Carbon profile for 3rd Condition at 930C

C1 for 1 hr

C2 for   hr

C3 for 8 hr

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Where 

Temperature is found to be = 840oC 

Duration of carburising = 5 hours 

This can better be ex plained f rom the f igure given below.

Question No. 4:

4.a) Stiffness performance Index:

Given deflection = 5FL / 32Ewt³««««««1

Tensile stress () = 3FL/4wt²««««««««.2

Ste p 1: Identif ication of  function, f ree varia bles and o bjective 

Function: to su pport weight 

Free varia ble: ³t´ thickness 

O bjective: minimize weight 

We know that performance equation m = AL 

Where area = w*t 

Therefore the performance equation can be written as 

M = w*t*L*««««««««««««««««««.3

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We know that = 5FL³/ 32Ewt² 

  t³ = 5FL³/ 32E

  t = (5FL³)1/3 / (32Ew)1/3«««««««««««««««««..4 

Inserting equation 4 into equation eq 3 we get 

M = w*L*(5FL³)1/3/(*32wE)1/3 

M = (5F)1/3/ (*32)1/3*(w*L²)/w1/3*(/E1/3) 

Therefore the performance index is found to be 

  P1 = E1/3/

Strength performance index:

We know that tensile stress  = 3FL/ 4wt² 

  t² = 3FL/4w  t = (3FL)1/2

/ (4w)1/2«««««««««««««««««««««««««..5 

Su bstituting 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 stiff ness performance index greater than 6.0 Gpa 

Stiffness performance chart

  Diamond   Sic  SL3 N4 

  Sialoons   Alumina¶s   Silicon

  Laminates GFRP, KFRP

  Magnesium alloy   Beo 

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Strengthperformance chart

  Diamond   Sic  A2LO3  ZrO3

  Mgo   B  Glass   Laminates 

KFRP, CFRP  Silicon

Please refer to the charts attached for design guide lines.

4.c) Selection of material on the basis of Strength and performance index. The followingf ive materials where selected on the basis of  their performance indices values.

Material Stiff ness performance 

index Strength 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 f racture toughness of plate is concern, which has to be a bove 20 mpa 

I select GFRP laminates as it has f racture toughness of 80 mpa.

Stiff ness performance Index= 2.60

Strength performance index = 21.08.

GFRP laminate proves to be the best material as it has low density and high performance index in relation to the f racture toughness which is a bout 80 mpa.

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Modulus and Densitry chart:

The design guide line was o btained by the following calculation:

Point 1:

Since it is given that the strength performance indices should be greater than 6 Gpa 

½/= 6 

½/ 3 = 6 

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Point 1

= 324 M pa 

Point 2:

½/= 6 

Taking density as 1(mg/m3) 

½/ 1 = 6 

Point 2

= 36M pa 

Strength and Density Chart:

For strength and density chart the two points where calculated by the following methods 

Since it is given in the assignment that the stiff ness performance indices should be greaterthan 1.50

E1/3/ = 1.50

Point 1:

Taking the density as 1mg/m3 

E1/3/1 = 1.50

E= 3.375 Gpa 

Point 2:

Taking the density as 3mg/m3 

E1/3/ = 1.50

E1/3/3 = 1.5 

E = 91.1125 Gpa 

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Strength and density Chart

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Fracture toughness and density chart:

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05-2009.

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uses in design´. 3rd edition, CR C press. Taylor and Francis grou p.

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access: 25-05-2010.

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213-220. Availa ble online: www.sciencedirect.com Date of  access: 25-05-2010.

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ceramics for use in fuel coating´. Thesis. University of Wisconsin- Madison.

Availa ble online: www.scribd.com  date of  access: 25-05-2010.

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