Materials Science and Engineering - A First Course_V. Raghavan

Scilab Textbook Companion for Materials Science and Engineering - A First Course by V. Raghavan1 Created by Vareesh Pratap B.Tech Mechanical Engineering Madan Mohan Malaviya University of Technology College Teacher None Cross-Checked by Aviral Yadav June 2, 2016

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Funded by a grant from the National Mission on Education through ICT, http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilab codes written in it can be downloaded from the ”Textbook Companion Project” section at the website http://scilab.in

Book Description Title:  Materials Science and Engineering - A First Course Author:  V. Raghavan Publisher:   Prentice Hall, India Edition: 5 Year:   2007 ISBN:   9788120324558

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Scilab numbering policy used in this document and the relation to the above book. Exa  Example (Solved example) Eqn  Equation (Particular equation of the above book) AP  Appendix to Example(Scilab Code that is an Appednix to a particular

Example of the above book) For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 means a scilab code whose theory is explained in Section 2.3 of the book.

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Contents Listt of Scilab Codes Lis

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2 Equ Equili ilibri brium um and and Kine Kinetic ticss

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3 Cry Crysta stall Geometry Geometry and Structur Structure e Determina Determinatio tion n

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4 At Atomi omic c Structure Structure and and Chemical Chemical Bondin Bonding g

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5 St Struc ructu ture re of of Soli Solid d

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6 Cry Crysta stall Impe Imperfec rfectio tion n

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7 Ph Phas ase e Di Diag agram ram

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8 Di Diffu ffusi sion on in in Soli Solids ds

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9 Pha Phase se Trans Transfor format mation ion

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10 Elastic Anelastic and Viscoelastic Behaviour

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11 Plastic Deformation and Creep in Crystalline Materials

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12 Fracture

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13 Oxidation and Corrosion

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14 Conductors and Resistors

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15 Semiconductors

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16 Magnetic Materials

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17 Dielectric Materials

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List of Scilab Codes Exa 2.1 Exa 2.2 Exa 3.2 Exa 3.7 Exa 3.8 Exa 4.1 Exa 4.3 Exa 4.4 Exa 5.1 Exa 5.3 Exa 5.4 Exa 5.5 Exa 5.6 Exa 6.1 Exa 6.2 Exa 6.4 Exa 6.5 Exa 6.6 Exa 7.1 Exa 7.2 Exa 7.3

Calculate the entropy increase . . . . . . . . . . . . . Calculation of fraction of atoms with energy equal to or greater than 1eV at temperatures . . . . . . . . . . . . Calculate effective number of lattice point in three cubic space lattice . . . . . . . . . . . . . . . . . . . . . . . Determine interplanar spacing and miller indices . . . Determine structure and lattice parameter of material Calculate frequency and wavelength of radiation   . . . Reconcile the difference of energy . . . . . . . . . . . . Calculation of fraction of hydrogen bonds which breaks during ice melting . . . . . . . . . . . . . . . . . . . . Calculate packing efficiency and density of diamond . Calculate the cation to anion ratio for an ideally close packed HCP crystal . . . . . . . . . . . . . . . . . . . find the size of largest sphere that can fit into a tetrahedral void . . . . . . . . . . . . . . . . . . . . . . . . Find critical radius ratio for triangular coordination . Calculate density of MgO   . . . . . . . . . . . . . . . . Find equilibrium concentration of vacancies in metals at given temperature . . . . . . . . . . . . . . . . . . . . Compute the line energy of dislocation . . . . . . . . . Calculation of down climb of crystal on heating . . . . Calculate surface energy of copper crystal of type 111 Compute the angle at the bottom of groove of a boundary Find degrees of freedom of a system of two components Find minimum number of component in system . . . . Calculate amount of pure water that can be extracted from sea water . . . . . . . . . . . . . . . . . . . . . . 5

5 5 7 8 9 11 12 12 13 14 14 15 15 17 18 18 19 20 21 21 22

Exa 7.5 Exa 8.1 Exa 8.2 Exa 8.3 Exa 8.4 Exa 8.5 Exa 9.1 Exa 9.2 Exa 9.4 Exa 9.6 Exa 10.1 Exa 10.2 Exa 10.3 Exa 11.2 Exa 11.3 Exa 11.4 Exa 11.5 Exa 12.1 Exa 12.2 Exa 12.3 Exa 13.2 Exa 14.1 Exa 14.2 Exa 14.3 Exa 15.1

Calculate proeutectoid ferrite and eutectoid ferrite in 6 tenth percent steel   . . . . . . . . . . . . . . . . . . . . Calculate the rate at which hydrogen escapes through the walls of the steel tank   . . . . . . . . . . . . . . . . Calculate maximum time till which material can be kept at 550 degree Celsius . . . . . . . . . . . . . . . . . . . Calculate minimum depth up to which post machining is to be done . . . . . . . . . . . . . . . . . . . . . . . Calculate time required to get required boron concentration . . . . . . . . . . . . . . . . . . . . . . . . . . Calculate ratio of cross sectional areas   . . . . . . . . . Calculate the critical free energy of nucleation of ice from water and critical radius . . . . . . . . . . . . . . Calculate the change in delta f required to increase nucleation rate . . . . . . . . . . . . . . . . . . . . . . . Calculate delta f of heterogeneous as a fraction of delta f of homogeneous . . . . . . . . . . . . . . . . . . . . . Calculate the free energy change during recrystallization Estimate Youngs modulus of material . . . . . . . . . Calculation of stress in fibers . . . . . . . . . . . . . . Estimate diffusion coefficient . . . . . . . . . . . . . . Calculate the stress required to move the dislocation at given temperature . . . . . . . . . . . . . . . . . . . . Calculate the dislocation density in copper . . . . . . Find the yield stress for a grain size of ASTM 9 . . . . Estimate the change in yield strength . . . . . . . . . Estimate fracture strength . . . . . . . . . . . . . . . . Estimate the brittle fracture strength at low temperatures . . . . . . . . . . . . . . . . . . . . . . . . . . . . Estimate the temperature at which the ductility of brittle transition occurs at given strain rates . . . . . . . . Calculation of required quantity of magnesium   . . . . Calculate energy difference . . . . . . . . . . . . . . . Calculate conductivity of copper at 300 K   . . . . . . . Estimation of resistivity due to impurity scattering of 1 percent of Nickel in copper lattice . . . . . . . . . . . Calculate intrinsic carrier density . . . . . . . . . . . .

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22 24 24 25 26 27 28 29 30 30 32 32 33 34 35 36 36 38 38 39 41 42 42 43 44

Exa 16.1 Exa 16.2 Exa 16.4 Exa 16.5 Exa 17.1 Exa 17.2

Calculate the net magnetic moment per iron atom in crystal . . . . . . . . . . . . . . . . . . . . . . . . . . . Comparison of saturation temperatures . . . . . . . . Calculation of hysteresis loss . . . . . . . . . . . . . . Calculation of eddy current loss at normal voltage and frequency   . . . . . . . . . . . . . . . . . . . . . . . . . Calculation of relative dielectric constant . . . . . . . Calculate the polarization of a BaTio3 crystal . . . . .

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45 46 46 47 48 48

Chapter 2 Equilibrium and Kinetics

Scilab code Exa 2.1  Calculate the entropy increase

1 2 3 4 5 6 7

  // C a l cu l a te t he e nt ro py i n c r e a s e clc d el _h = 6 .0 2   // H ea t a dd ed i n kJ / mol t _m = 2 73 .1 5   / / mean t em p er a tu re i n k e l v i n   printf ( ” \ n E xa mp le 2 . 1 ” ) d el _s = d el _h * 1 e 3 / t_ m   printf ( ” \ n I n c r e a s e i n e nt ro py i s %. 2 f J molˆ −1 Kˆ−1 ” ,del_s)

Scilab code Exa 2.2  Calculation of fraction of atoms with energy equal to

or greater than 1eV at temperatures 1   / / C a lc u la t io n o f f r a c t i o n

o f atoms w i t h e ne rg y e q u a l t o o r g r e a t e r t h a n 1eV a t t e m p e ra t u r e s

2 clc 3 E = 1 // e ne rg y i n e l e c t r o n v o l t 4 e = 1.6 e -19   // c ha rg e on e l e c t r o n

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5 6 7 8 9 10 11

             

k = 1 .3 8 e -2 3   // c o n st a n t t 1 = 300   // t em p er a tu r e i n K t 2 = 15 00   // t em p er a tu re i n K printf ( ” \ n E xa mp le 2 . 2 ” ) printf ( ” \ n \ n P a rt A : ” ) n _N = exp (-(e/(k*t1))) printf ( ” \ n F r a c ti o n o f atoms w i th e ne rg y e qu a l t o o r g r e a t e r th an 1eV a t t em p er a tu re %d K i s %. 2 e ” , t1,n_N)   // n um er ic al v al ue o f a n s w e r i n book i s 2

e −17 12   printf ( ” \ n \ n P ar t B : ” ) 13   n _N = exp (-(e/(k*t2))) 14   printf ( ” \ n F r a c ti o n o f atoms w i th e ne rg y e qu a l t o o r g r e a t e r th an 1eV a t t em p er a tu re %d K i s %. 2 e ” , t2,n_N)   // n u m er i c al v a l u e o f a ns we r i n book i s

4 . 3 e4

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Chapter 3 Crystal Geometry and Structure Determination

Scilab code Exa 3.2   Calculate effective number of lattice point in three

cubic space lattice 1 // C a l c u l a te

e f f e c t i v e number o f l a t t i c e p o i n t i n thr ee cubi c space l a t t i c e

2 clc 3   s c _n = 1 /8 // s h a r i n g o f on e l a t t i c e

p o i nt i n a u n i t

cell 4   s c_N = 8 / / Number o f l a t t i c e p o i n t s i n S i m pl e c u b i c 5 b cc _n _e = 1 /4 // s h a r i n g o f on e e dg e l a t t i c e p o i nt i n

a BCC 6 b cc _N _e = 4 / / Number o f e d g e l a t t i c e p o i n t i n a BCC 7   bcc_n_c = 1 // s h a r i n g o f o ne body c e n t e r l a t t i c e

p o i n t i n a BCC 8 b cc _N _c = 1 / / Number o f bo dy c e n t e r l a t t i c e p o i n t i n a BCC 9 f cc _n _e = 1 /8 // s h a r i n g o f o ne c o r n e r l a t t i c e p o i n t i n a FCC 10 f cc _N _e = 8 / / Number o f c o r n e r l a t t i c e p o i n t i n a FCC 11 f cc _n _f = 1 /2 // s h a r i n g o f o ne f a c e c e n t e r l a t t i c e 10

p o i n t i n a FCC 12 f cc _N _f = 6 / / Number o f f a c e c e n t e r l a t t i c e p o i n t i n a FCC 13   printf ( ” \ n Example 3 . 2 ” ) s c_ n et = s c_ n * s c_ N b c c _n e t = b c c _n _ e * b c c _ N_ e + b c c _ n _c * b c c _N _ c f c c _n e t = f c c _n _ e * f c c _ N_ e + f c c _ n _f * f c c _N _ f   printf ( ” \ n E f f e c t i v e number o f l a t t i c e p o i n t s a r e a s :”) 18   printf ( ” \ n \ n Space l a t t i c e \ t A b b r ev i a t i o n \ t 14 15 16 17

E f f e c t i v e number o f l a t t i c e p o i n t i n u n i t c e l l \ n ”) 19   printf ( ” \ n S im pl e c u b ic \ t \ tS C \ t \ t \ t \ t %d\ n B od y c e n te r c u b i c \ tBCC \ t \ t \ t \ t %d\ n Face c e n t er e d c u b i c \ tFCC \ t \ t \ t \ t % d ” ,sc_net ,bcc_net ,fcc_net )

Scilab code Exa 3.7   Determine interplanar spacing and miller indices

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

  //                        

D e t e r m i n e I n t e r p l a n a r s p a c i n g and m i l l e r i n d i c e s

clc a = 3.16 / / l a t t i c e p a r a m e t e r i n a n g st r o m l 1 = 1   // l i n e number l 2 = 2   // l i n e number l 3 = 3   // l i n e number l 4 = 4   // l i n e number t h e ta 1 = 2 0. 3   // a n g l e f o r l i n e 1 t h e ta 2 = 2 9. 2 // a n g l e f o r l i n e 2 t h e ta 3 = 3 6. 7 // a n g l e f o r l i n e 3 t h e ta 4 = 4 3. 6 // a n g l e f o r l i n e 4 n = 1   // o r d er l a m bd a = 1 .5 4   // w a ve l en g th i n a ng st ro m printf ( ” \ n E xa mp le 3 . 7 ” ) d 1 = l am b da / ( 2* sin (theta1*%pi/180))

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d 2 = l am b da / ( 2* sin (theta2*%pi/180)) d 3 = l am b da / ( 2* sin (theta3*%pi/180)) d 4 = l am b da / ( 2* sin (theta4*%pi/180)) x 1 = a ^ 2/ d 1 ^2 x 2 = a ^ 2/ d 2 ^2 x 3 = a ^ 2/ d 3 ^2 x 4 = a ^ 2/ d 4 ^2   / / where x i s f u n c t i o n o f h , k and l printf ( ” \ n I n t e r p l a n a r s p a c i n g i s %. 3 f a n g s t r o m ” , x1 )   // a ns we r i n book i s 2 . 2 2 0 a ng st ro m 24 if floor ( x1 ) == 2   then 25   printf ( ” \ n \ n F or a ˆ 2 / dˆ 2 = %d \ t R e f l e c t i o n p la ne i s { 1 1 0 } ” ,x1) 26 end 27 28   i f f lo or ( x2 ) == 4   then 29   printf ( ” \ n Fo r a ˆ 2 / dˆ 2 = %d \ t R e f l e c t i o n p la ne i s { 2 0 0 } ” ,x2) 30 end 31 32 i f fl oo r ( x3 ) == 6   then 33   printf ( ” \ n F or a ˆ 2 / dˆ 2 = %d \ t R e f l e c t i o n p la ne i s { 2 1 1 } ” ,x3) 34 end 35 36 i f fl oo r ( x4 ) == 8   then 37   printf ( ” \ n F or a ˆ 2 / dˆ 2 = %d \ t R e f l e c t i o n p la ne i s { 2 2 0 } ” ,x4) 38 end 16 17 18 19 20 21 22 23

               

Scilab code Exa 3.8  Determine structure and lattice parameter of mate-

rial 1 / / D e te rm i ne s t r u c t u r e and l a t t i c e

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p a r am e t er o f  

material 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

                     

               

clc d = 11 4. 6   // d i a m et e r o f p ower c am er a i n a ng st ro m l a m bd a = 1 .5 4   // w a ve l en g th i n a ng st ro m s 1 = 86 s 2 = 100 s 3 = 148 s 4 = 180 s 5 = 188 s 6 = 232 s 7 = 272 printf ( ” \ n E xa mp le 3 . 8 ” ) / / R a di u s R = d /2 if   R==57.3   then k = 1/4   // B ragg a n gl e f a c t o r end a 1 = ( sin (s1*k*%pi/180))^2 a 2 = ( sin (s2*k*%pi/180))^2 a 3 = ( sin (s3*k*%pi/180))^2 a 4 = ( sin (s4*k*%pi/180))^2 a 5 = ( sin (s5*k*%pi/180))^2 a 6 = ( sin (s6*k*%pi/180))^2 a 7 = ( sin (s7*k*%pi/180))^2 c = 2 2   // c o ns t an t t o c o n v er t i n t o i n t e g r a l number

  printf ( ” \ n W it hi n e x pe r i m en t al e r r o r ,

v a l ue s a re a s i n i n t e g r a l r a t i o a r e a s : \ n %d: %d: %d: %d: %d: %d: %d ” , ceil ( c * a 1 ) , ceil ( c * a 2 ) , ceil ( c * a 3 ) , ceil ( c * a 4 ) ,

ceil ( c * a 5 ) , ceil ( c * a 6 ) , ceil ( c * a 7 ) ) 27   printf ( ” \ n S o , t h i s s t r u c t u r e i s FCC and m a t e r i a l i s c o p p er w i th 3 . 6 2 a n gs t ro m l a t t i c e p a r am e t er ” )

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Chapter 4 Atomic Structure and Chemical Bonding

Scilab code Exa 4.1  Calculate frequency and wavelength of radiation

1   / / C a l cu l a te f r eq u en c y and w av el en gt h o f r a d i a t i o n 2 clc 3   E = 1 .6 4 e -1 8   // e ne rg y d i f f e r e n c e b et w ee n two s t a t e s

in J 4 5 6 7 8 9

     

10  

h = 6 .6 26 e - 34   // p l an k s c o n st a n t c = 2. 99 8 e8   // s p e e d o f l i g h t i n m/ s printf ( ” \ n E xa mp le 4 . 1 ” ) n u = E/ h l am bd a = c / nu printf ( ” \ n F re qu en cy o f e mi t te d r a d i a t i o n i s %. 2 e Hz ” ,nu) printf ( ” \ n Wa velen gth o f e m it t ed r a d i a t i o n i s %. 2 e m \ n \ t \ t o r \ t \ t % d a n g s t r o m ” ,lambda ,lambda*1e10) //

a ns we r i n book i s 1 21 0 a ng st ro m

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Scilab code Exa 4.3   Reconcile the difference of energy

1 2 3 4 5

  // R e c o n c i l e t he d i f f e r e n c e o f e ne rg y clc   e _a = 713   // e n th a lp y o f a t o m iz a t io n i n kJ / mol   e _b = 347   // bond e n er g y i n kJ / mol a = 4   // t o t a l number o f atoms i n s i n g l e c r y s t a l

structure 6 b = 2   // number o f a to ms i n a bond 7   printf ( ” \ n E xa mp le 4 . 3 ” ) 8 k = a/b   // e f f e c t i v e number o f bond p er atom 9   e = k *e_b 10   printf ( ” \ n %d kJ s h ou l d be t he e n th a lp y o f   a t o m i z a t i o n o f diamond ” , e) 11   printf ( ” \ n However , %d k J i s v er y c l o s e t o %d k J” ,e , e_a)

Scilab code Exa 4.4 Calculation of fraction of hydrogen bonds which breaks

during ice melting 1   / / C a l cu l a t io n o f f r a c t i o n

o f h yd ro ge n bo nd s wh ich b r e ak s d ur in g i c e m e l t i n g

2 3 4 5 6 7 8

clc d el _h = 6 .0 2   // e nt ha l p y o f f u s i o n i n kJ /mol n = 2   // number o f h y dr o ge n atom i n 1 w a te r atom d el _b = 2 0. 5   / / h yd ro ge n bond e n er g y i n kJ / mol   printf ( ” \ n E xa mp le 4 . 4 ” )   f = d el _h / ( n * d el _b )   printf ( ” \ n F r a c t i o n o f h yd ro ge n b on ds w hi ch b ro ke n i s %. 2 f ” ,f )

15

Chapter 5 Structure of Solid

Scilab code Exa 5.1   Calculate packing efficiency and density of diamond

1   / / C a lc u l a te p ac k in g e f f i c i e n c y and d e n s i t y o f  

diamond 2 3 4 5

clc   n _c = 1/8   / / s h a ri n g o f c o rn e r atom i n a u n it c e l l N_c = 8   / / Number o f c o r n e r a tom s i n u n i t c e l l n_b = 1   // s h a ri n g o f body c e nt e re d atom i n a u n it

cell 6 N_b = 4   // Number o f body c e n t e r e d a to ms i n u n i t

cell 7   n _f = 0.5 // s h a r in g o f f a c e c e n t e r e d atom i n a u n i t

cell 8 9 10 11 12 13

       

N_f = 6 // Number o f f a c e c e n t e r e d a to ms i n u n i t c e l l a = 1 // l et l a t t i c e parameter m = 1 2   // mass o f c ar bo n printf ( ” \ n E xa mp le 5 . 1 ” ) printf ( ” \ n P a rt A : ” ) N = n _c * N _ c + n _ b * N _b + n _ f * N _ f   // e f f e c t i v e number o f  

atoms 14   r = a * sqrt (3)/8 15 p _e = N * 4 /3 * % pi * r ^ 3/ a ^ 3   // p ac ki ng 16

16

effi cien cy

17   printf ( ” \ n P ac k in g e f f i c i e n c y o f diamond i s %. 2 f ” , p_e) 18   printf ( ” \ n \ n P ar t B : ” ) 19   a = 3.57 / / l a t t i c e p a r a m e t e r i n a n g st r o m 20   d = m * 1 . 66 e - 2 7 * N / ( a *1 e - 1 0 ) ^ 3 21   printf ( ” \ n D e n s i t y o f d ia mo nd i s %d Kg/mˆ 3 ” ,d) //

n u me r ic a l a ns we r i n book i s 3 50 0 22   printf ( ” \ n D e n si t y o f diamond i s %. 1 f g /cmˆ3 ” ,d /1000)

Scilab code Exa 5.3  Calculate the cation to anion ratio for an ideally close

packed HCP crystal 1   / / c a l c u l a t e t h e c /a r a t i o

f o r an i d e a l l y c l o s e

p a ck e d HCP c r y s t a l 2 3 4 5 6 7 8 9

clc a = 1   // l e t   PR = a   printf ( ” \ E xa m pl e 5 . 3 ” )   R T = a/ sqrt (3) PT =   sqrt (PR^2-RT^2) c _a = 2* P T/ PR

  / / C a l cu l a t i o n s a re made o n t h e c r y s t a l s t r u c t u r e d ra wn i n b oo k 10   printf ( ” \ n c / a r a t i o f o r an i d e a l l y c l o s e p a c k ed HCP c r y s t a l i s %0 . 3 f ” ,c_a)

Scilab code Exa 5.4  find the size of largest sphere that can fit into a tetra-

hedral void 17

1   / / f i n d t h e s i z e o f l a r g e s t s ph er e t ha t can

f i t i nt o

a t e t r a h e d r a l v oi d 2 3 4 5 6 7 8

       

clc r = 1   // l e t a = 3/4 printf ( ” \ n E xa mp le 5 . 4 ” ) p t = 2* sqrt (2/3)*r s = a *pt - r // s i z e o f s ph er e printf ( ” \ n S i z e o f l a r g e s t s ph er e t h at ca n a t e t r a h e d r a l v oi d i s %. 3 f r ” ,s )

f i t i nt o

Scilab code Exa 5.5 Find critical radius ratio for triangular coordination

1 // f i n d

c r i t i c a l r a d iu s r a t i o f o r t r i a n g u l a r coordination

2 3 4 5

clc t he ta = 60   // a n g l e i n d eg r e e   printf ( ” \ n E xa mp le 5 . 5 ” )   r _ c _a = ( 2/ 3 *2 * sin (theta*%pi/180))-1   // r a t i o

calculation 6   printf ( ” \ n C r i t i c a l r a d i u s r a t i o f o r t r i a n g u l a r c o o r d i n a t i o n i s %0 . 3 f ” ,r_c_a)

Scilab code Exa 5.6   Calculate density of MgO

1   / / C a l c u l a te d e n s i t y o f MgO 2 clc 3 r _m g = 0 .7 8   // r a d i u s o f magnesium c a t i o n i n

angstrom 18

4 5 6 7 8 9 10 11

  r _o = 1 .3 2   // r a d i u s o f o xyg en a ni on i n a ng st ro m n = 4 // e f f e c t i v e number o f u n it c e l l   m _o = 16   // mass o f o xy ge n m _m g = 2 4. 3   / / m as s o f m ag ne si um   printf ( ” \ n E xa mp le 5 . 6 ” ) a = 2 *( r _m g + r_ o ) / / l a t t i c e p a r a m e t e r   d = ( m _ o + m _m g ) * 1 . 66 e - 2 7 * n / ( a * 1 e - 10 ) ^ 3 // d e n s i t y   printf ( ” \ n D e n s i t y o f MgO i s %d Kg /mˆ 3 ” ,d)   / / a n sw er

i s 3 6 10 k g /mˆ 3 12   printf ( ” \ n D e n s i t y o f MgO i s %0 . 2 f g /cm ˆ3 ” ,d/1000)

19

Chapter 6 Crystal Imperfection

Scilab code Exa 6.1 Find equilibrium concentration of vacancies in metals

at given temperature 1   // F ind e q u i l i b r i u m c o n c en t r a ti o n o f v a c a n c i es i n

m et a ls a t g i v en t em p er a tu re 2 3 4 5 6 7

         

clc t 1 = 0   // t em pe ra tu re i n k e l v i n t 2 = 300   // t em pe ra tu re i n k e l v i n t 3 = 900 // t em pe ra tu re i n k e l v i n R = 8. 31 4   // u n i v e r s a l g as c o n s t a n t d e l _h f _a l = 6 8   // E nt ha lp y o f f o r m a ti o n o f a lu mi ni um

c r y s t a l i n KJ 8   d e l _ hf _ ni = 1 68   // E n th a lp y o f f o r m a t io n o f n i c k e l c r y s t a l i n KJ 9   printf ( ” \ n E xa mp le 6 . 1 ” ) 10 11   printf ( ” \ n E q u il i br i um c o n c e n t r a t i o n o f v a c a n c i es o f   a lu mi ni um a t %dK i s 0 ” ,t1) 12   n _N = exp (-del_hf_al*1e3/(R*t2)) 13   printf ( ” \ n E q u il i br i um c o n c e n t r a t i o n o f v a c a n c i es o f   a l um i ni u m a t %dK i s %. 2 e ” ,t2,n_N)   // a ns we r i n

book i s 1 . 4 5 e −12 14   n _N = exp (-del_hf_al*1e3/(R*t3))

20

15   printf ( ” \ n E q u il i br i um c o n c e n t r a t i o n o f v a c a n c i es o f   a l um i ni u m a t %dK i s %. 2 e ” ,t3,n_N)   // a ns we r i n

book i s 1 . 1 2 e −4 16 17   printf ( ” \ n \ n E q u i li b ri u m c o n c en t r a ti o n o f v a c a n c i es o f N i ck e l a t %dK i s 0 ” ,t1) 18   n _N = exp (-del_hf_ni*1e3/(R*t2)) 19   printf ( ” \ n E q u il i br i um c o n c e n t r a t i o n o f v a c a n c i es o f   N i c k e l a t %dK i s %. 2 e ” ,t2,n_N) 20   n _N = exp (-del_hf_ni*1e3/(R*t3)) 21   printf ( ” \ n E q u il i br i um c o n c e n t r a t i o n o f v a c a n c i es o f   N i c k e l a t %dK i s %. 2 e ” ,t3,n_N)   // a ns we r i n bo ok

i s 1 . 78 e −10

Scilab code Exa 6.2  Compute the line energy of dislocation

1 2 3 4 5 6 7 8

  / / Compute t he l i n e e ne rg y o f d i s l o c a t i o n          

clc a = 2.87 / / l a t t i c e p a r a m e t e r i n a n g st r o m b = 2 .4 9   // m ag ni tu de o f b u r g er s v e c t o r i n a ng st ro m G = 80.2   / / s h e a r m od ul us i n GN printf ( ” \ n E xa mp le 6 . 2 ” ) E = G * 1 e9 * ( b *1 e - 10 ) ^ 2 *1 /2 printf ( ” \ n L i ne e ne rg y o f d i s l o c a t i o n i s %. 2 e J mˆ −1 ” ,E)

Scilab code Exa 6.4  Calculation of down climb of crystal on heating

1   // c a l c u l a t i o n

o f down c l im b o f c r y s t a l on h e a t i n g 21

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

         

         

clc a = 1 e10 // t o t a l number o f e dg e d i s l o c a t i o n N = 6 .0 23 e 23   / / A v o g ad r o n um be r R = 8. 31 4   // U n i ve r s a l g as c o ns t an t t 1 = 0 / / i n i t i a l t e m p e r a tu r e i n K t 2 = 10 00   / / F i na l t em pe ra tu re i n K d el _h f = 1 00   // E nt ha lp y o f v ac an cy f o rm a t io n i n KJ d = 2   // l e n g t h o f one s t ep i n a ng st r o m v = 5.5 e -6   // vo lume o f one mole c r y s t a l printf ( ” \ n E xa mp le 6 . 4 ” ) n = N * exp (-(del_hf*1e3)/(R*(t2-t1)))/v k = 1 /( d *1 e - 10 )   // atoms r e q u i r e d f o r 1 m c l im b b = n /( k* a) / / a v e ra g e amount o f c l im b c = b *d *1 e -1 0

  printf ( ” \ n A ver ag e down c li mb o f c r y s t a l )

i s %. 2 em” ,c

Scilab code Exa 6.5  Calculate surface energy of copper crystal of type 111

1   // C a lc u l a te

s u r f a c e e ne rg y o f c o pp er c r y s t a l o f  

type {111} 2 3 4 5 6 7

clc  E =   N_a  n = m = N =

8  9 10   11  

56.4   // bond e n er g y i n KJ = 6 .0 23 e 2 3 // A v o g a d r o s number 1 2   / / number o f b on ds 3   // number o f b r ok e n b on ds 1 .7 7 e19   // number o f atoms i n c op pe r c r y s t a l

type {111}   per mˆ2 printf ( ” \ n E xa mp le 6 . 5 ” ) b _e = 1 /2 * E *1 e 3 * n/ N _a   // bond e n e r g y p e r atom e _ b = b _e * m/ n   // e ne rg y o f b ro ke n bond a t s u r f a c e s _e = e _b * N // s u r f a c e e nt ha l p y o f c op pe r 22

of 

12   printf ( ” \ n S u r f a ce e nt ha l p y o f c op pe r i s %0 . 2 f J m ˆ−2” ,s_e) 13   printf ( ” \ n S u r f a ce e nt ha l p y o f c op pe r i s %d e r g cm ˆ−2” ,s_e*1e3) 14   / / A nswer i n b oo k i s 2 49 0 e r g cmˆ −2

Scilab code Exa 6.6   Compute the angle at the bottom of groove of a

boundary 1   / / Compute t he a n g l e a t t he bottom o f g r oo ve o f a

boundary 2 3 4 5 6 7

       

clc G a m ma _g b = 1   // l e t , e ne rg y o f g r a i n b ou n da ry G a m ma _ s = 3 * G a mm a _g b // e n e r g y o f f r e e s u r f a c e printf ( ” \ n E xa mp le 6 . 6 ” ) t he ta = 2* acos (1/2*Gamma_gb/Gamma_s) printf ( ” \ n A ng le a t t he bottom o f g ro ov e o f a b ou nd ar y i s %d d e g r e e s . ” , ceil (theta*180/%pi))

23

Chapter 7 Phase Diagram

Scilab code Exa 7.1  Find degrees of freedom of a system of two compo-

nents 1   / / F ind d e g re e s o f f re ed om o f a s ys te m o f two

components 2 3 4 5 6 7 8

clc c = 2   / / nu mber o f c o mp o ne n ts   printf ( ” \ n E xa mp le 7 . 1 ” ) for   n = 1 : 4 p = (c -1) +2   // T ot a l v a r i a b l e s f = c -n +2   // d e gr e e o f f re ed om   printf ( ” \ n \ n D e gr ee o f f re ed om f o r two

c o m p o n e n t s wh en \ n n umber o f p ha se s i s %d i s %d” ,n,f) 9 end

Scilab code Exa 7.2  Find minimum number of component in system

24

1 2 3 4 5 6 7 8

/ / F in d minimum n um ber o f c om po ne nt i n s y s t em clc p = 4   // number o f p h a se s o f s ys te m f = 0   // number o f d e g re e o f s ys te m   printf ( ” \ n E xa mp le 7 . 2 ” )   C = f +p -1   / / c o m po n e n ts n um ber   printf ( ” \ n Minimum n umber o f c o mp o ne n ts i n s y s te m i s %d” ,C)

Scilab code Exa 7.3 Calculate amount of pure water that can be extracted

from sea water 1   / / C a l c u l a te amount o f p ur e w at er t h at ca n b e

e x t r a c t ed from s e a w at er 2 3 4 5 6 7 8

       

clc L = 23.3   / / % c o m po s i t io n o f L a = 3.5   // c o n c e n t r a t i o n o f N a c l i n s ea w a te r ice = 0   // % co m p o si t i o n o f i c e printf ( ” \ n E xa mp le 7 . 3 ” ) f _i ce = ( L - a) / (L - i ce ) printf ( ” \ n F r a c t i o n a l amount o f p ur e w at er t h at ca n be e x t r a c t ed fro m s e a w at er i s %0 . 2 f ” ,f_ice)

Scilab code Exa 7.5  Calculate proeutectoid ferrite and eutectoid ferrite in

6 tenth percent steel 1 // Calculate

proeutec toid f e r r i t e in 0.6% st ee l

f e r r i t e and e u t e c t o i d

25

clc a = 0 // l i m i t i n g v al ue   b = 0.8   // l i m i t i n g v al ue   c = 0.6   // p e r ce n t a ge c o mp o s it i on o f c ar bo n   f = 0.88 // f r a c t i o n o f f e r r i t e i n a e u t e c t o i d s t e e l   printf ( ” \ n E xa mp le 7 . 5 ” )   f _ p r o _a l p ha = ( b - c ) /( b - a ) f _p er li te = 1 - f _p ro _a lp ha f _e ut = f * f _ pe r li t e   printf ( ” \ n C om p os i ti o n o f p r o e u t e c t o i d f e r r i t e i s %0 . 2 f ” ,f_pro_alpha) 12   printf ( ” \ n C om p os i ti o n o f e u t e c t o i d f e r r i t e i s %0 . 2 f   ” ,f_eut) 2 3 4 5 6 7 8 9 10 11

26

Chapter 8 Diffusion in Solids

Scilab code Exa 8.1 Calculate the rate at which hydrogen escapes through

the walls of the steel tank 1   / / c a l c u l a t e t he r a t e a t whi ch h y dr og en e s c a p e s

t h r o ug h t h e w a l l s o f t h e s t e e l t a n k 2 3 4 5 6 7 8

         

clc t = 5   // t h i c k n e s s i n mm c = 1 0   // c o n c e n t r a t i o n D = 1e -9 / / d i f f u s i o n c o e f f i c i e n t printf ( ” \ n E xa mp le 8 . 1 ” ) j = D *c /( t*1 e -3) printf ( ” \ n Outward f l u x i s %. 0 e kg mˆ −2 s ˆ−1” ,j)

Scilab code Exa 8.2  Calculate maximum time till which material can be

kept at 550 degree Celsius 1 / / C a l c u l a t e maximum t i me

t i l l w hi ch m a t e r i a l c an b e k ep t a t 5 5 0 d e g r e e C e l s i u s

2 clc

27

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

           

 

 

 

D _0 = 0 .2 4e - 4 / / d i f f u s i o n c o e f f i c i e n t Q = 121 e3 R = 8. 31 4 // U n i ve r s a l g as c o ns t an t T = 550   // t em pe ra tu re i n C e l s i u s k = 0.2   // t h i c k n e s s o f p ur e Al s h e e t i n mm d = 0.1   // p e n e t r a t i o n d ep th i n mm c _x = 0.4   // c o n c e n t r a t i o n i n p e rc e n t ag e A = 2   // C on st an t i n p e r ce n t ag e B = 2 // C on st an t i n p e r ce n t ag e printf ( ” \ n E xa mp le 8 . 2 ” ) x = d -k D _c u_ al = D _0 * exp (-Q/(R*(T+273))) k = ( A- c _x ) / B if   k = =0 .8   then z = 0.9   // fro m t a b l e end t = ( x * 1 e - 3) ^ 2 / ( z ^ 2 * 4* D _ c u_ a l ) // t i m e i n s e c

  printf ( ” \ n M a te r ia l can be k ep t a t %d d e g re e C e l s i u s f o r n e a r l y %d m in ut e ” ,T,t/60) // a ns we r i n b oo k

i s 1 00 min

Scilab code Exa 8.3 Calculate minimum depth up to which post machin-

ing is to be done 1   / / C a l c u l a t e minimum d ep t h u p t o w hi ch p o s t

m ac hi ni ng i s t o be done 2 clc 3   D _ 0 = 0 .7 e -4 / / d i f f u s i o n c o e f f i c i e n t 4   Q = 157   // E ne rg y i n kJ mol ˆ − 1 , c o n s i d e r e d fro m

t a b le 8 .2 5   R = 8. 31 4 // U n i ve r s a l g as c o ns t an t 6   T = 950   // t em pe ra tu re i n C e l s i u s 28

7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

  c 2 = 0.8   // c o n c e n t r a t i o n i n p e rc e nt a g e   c s = 0   // c o n c en t r a ti o n i n p e rc e n t ag e   c _x = 0.6 // c o n c e n t r a t i o n i n p e rc e nt a g e t = 4   // t i me i n h ou rs a = 1 // le t   printf ( ” \ n E xa mp le 8 . 3 ” )   A = cs   B = c2 - cs   D = D_0 *exp (-Q*1e3/(R*(T+273))) k = erf (((A-c_x)/B))*-1 if   k > 0. 7   then if   k 0 .9 9 96 6   then if   k < 0 .9 9 97   then z = 2.55   // fro m t a b l e end end   t = x ^ 2/ ( z ^ 2* 4* D ) // t i m e i n s e c        

  printf ( ” \ n Time r e q u i r e d t o g et r e q u i r e d b or o n c o n c e n t r a t i o n i s %d s e c ” ,t) // a ns we r i n book i s

3 84 5 s e c

Scilab code Exa 8.5  Calculate ratio of cross sectional areas

1 2 3 4 5 6 7

  // C a l c u l a t e r a t i o o f c r o s s s e c t i o n a l a r e a s        

clc r = 1 0   // r a d i u s i n mm t = 4   // t h i c k n e s s i n a n gs tr o m printf ( ” \ n E xa mp le 8 . 5 ” ) r = 2 * % pi * r * 1 e - 3 * t *1 e - 1 0 / ( % pi * ( r * 1 e - 3 ) ^ 2) printf ( ” \ n R a t i o o f c r o s s s e c t i o n a l a r e a s r)

30

i s %. 0 e ” ,

Chapter 9 Phase Transformation

Scilab code Exa 9.1  Calculate the critical free energy of nucleation of ice

from water and critical radius 1 // C a l c u l a te t he

c r i t i c a l f r e e e ne rg y o f n u c l e a t i o n o f i c e f ro m w a te r and c r i t i c a l r a d i u s

2 clc 3 d el _t 1 = 0 // t em pe ra tu re

d i f f e r e n c e i n d e gr e e

Celsius 4   d e l_ t2 = -5 // t em pe ra tu re

d i f f e r e n c e i n d e g r ee

Celsius 5 d el _t 3 = -40 // t em pe ra tu re

d i f f e r e n c e i n d eg re e

Celsius 6 d el _h = 6 .0 2   // e nt ha l p y o f f u s i o n i n kJ /mol 7   T _m = 273   / / mean t e m p e r a t u r e 8 G am ma = 0 .0 76   // e n er g y o f i c e w a t e r i n t e r f a c e 9 10 11 12

in J

/mˆ2   v = 1 9   // m ol ar volume o f i c e   printf ( ” \ n E xa mp le 9 . 1 ” )   printf ( ” \ n P a rt A” )   printf ( ” \ n At d e l t = %d , t h er e i s no s u p e r c o o l i n g . So t h e r e i s no c r i t i c a l r a d i u s ” ,del_t1)   printf ( ” \ n \ n P ar t B” )

13 14 d e l_ f = 1 6 /3 * % p i * ( G a mm a ) ^ 3 * T _m ^ 2 / ( ( d e l _h * 1 e 3 * 1 e 6 / v )

31

15   16   17   18 19 20 21 22 23

       

^2*del_t2^2) r = 2 * T _m * G a m m a / ( - d e l_ h * 1 e 3 * 1 e 6 / v* d e l _ t2 ) printf ( ” \ n C r i t i c a l f r e e e n e r g y o f n u c l e a t i o n i s %. 1 eJ ” ,del_f) printf ( ” \ n C r i t i c a l r a d i u s i s %d a ng s t r o m ” , ceil (r*1 e10)) printf ( ” \ n \ n P ar t C” ) t e m p _r = d e l_ t 3 / d e l _t 2 d e l _ f_ = d e l_ f / t e m p _r ^ 2 r _ = r / te mp _r

  printf ( ” \ n C r i t i c a l f r e e e n e r g y o f n u c l e a t i o n i s % . 1 e J ” ,del_f_) 24   printf ( ” \ n C r i t i c a l r a d i us i s %d a ng st r o m . ” ,   ceil ( r_ *1e10))

Scilab code Exa 9.2   Calculate the change in delta f required to increase

nucleation rate 1   // C a lc u l a te t h e c ha n g e i n d e l f

r e qu i r e d t o

i n c r ea s e n u cl e at i on r a te 2 3 4 5 6 7 8 9 10 11 12 13

     

       

clc T = 3 00   // t em pe ra tu re i n k e l v i n R = 8. 31 4   // u n i v e r s a l g as c o n s t a n t k = 2. 30 3   // c o nv e r s i o n f a c t o r a1 = 1 e42 a2 = 1 e6   // n u c l e a t i o n r a t e a3 = 1 e10   // n u c l e a t i o n r a t e printf ( ” E x a mp l e 9 . 2 ” ) I 1 = ( log10 ( a 1 ) - log10 (a2))*k*R*T   // e xp on en t f a c t o r I 2 = ( log10 ( a 1 ) - log10 (a3))*k*R*T // e xp on en t f a c t o r d el _f = I1 - I 2   // d i f f e r e n c e a = 10^( log10 ( a 3 ) - log10 ( a 2 ) )

32

14 15   printf ( ” \ n

A c ha ng e o f %d KJ m olˆ −1 e n er gy i s r e q u i r e d t o i n c r e a s e n u c l e a t i o n f a c t o r fro m \ n % . 0 e mˆ −3 s ˆ−1 t o %. 0 e mˆ −3 s ˆ −1 ” , ceil (del_f/1

e3),a,a3)

Scilab code Exa 9.4   Calculate delta f of heterogeneous as a fraction of 

delta f of homogeneous 1 2 3 4 5 6 7 8

  / / C a l cu l a te d e l f h e t a s a f r a c t i o n o f d el f h o mo clc G a mm a _a l ph a _d e l = 0 .5   // i n J mˆ −2   G a m m a _ a lp h a _ be t a = 0 .5   / / i n J mˆ −2   G a m m a _ be t a _ de l = 0 . 01   / / i n J mˆ −2

  printf ( ” \ n E xa mp le 9 . 4 ” )   t h et a =   acos ( ( G a m m a _ al p h a _ de l - G a m m a _b e t a _d e l ) / Gamma_alpha_beta) 9   d e l _ f _ ra t i o = 1 / 4* ( 2 - 3 * cos (theta)+( cos (theta))^3) 10 11 12   printf ( ” \ n d e l f h e t i s %0 . 4 f th f r a c t i o n o f   d e l f h o m o . ” ,del_f_ratio)

Scilab code Exa 9.6  Calculate the free energy change during recrystalliza-

tion 1   / / C a l cu l a te t he f r e e

e ne rg y c h a ng e d ur in g

recrystallization 33

2 3 4 5 6 7 8 9

   

   

clc m u = 4 5. 5 e9 b = 2 .5 5 e -1 0 n1 = 1 e9 // i n i t i a l d i s l o c a t i o n d e n s i t y n2 = 1 e13   // f i n a l d i s l o c a t i o n d en s i ty printf ( ” \ n E xa mp le 9 . 6 ” ) E = 1 /2 * m u *b ^ 2* n 2 del_g = E // a s d i f f e r e n c e b et wee n i n i t i a l and f i n a l

d i s l o c a t i o n e ne rg y i s f o u r o r de r m a gn i t u d e 10   printf ( ” \ n F re e e ne rg y c ha ng e d u ri n g r e c r y s t a l l i z a t i o n i s %d J mˆ −3” ,-del_g) 11   // N um er ic al v a lu e o f a ns we r i n book i s 1 48 00

34

Chapter 10 Elastic Anelastic and Viscoelastic Behaviour

Scilab code Exa 10.1   Estimate Youngs modulus of material

1 2 3 4 5 6 7 8 9 10 11 12

// E st im at e y o u n g

s modulus o f m a t e r i a l

clc n = 1 m = 9   A = 7 .6 8 e -2 9   // C on st an t h av in g u n i t J m r _0 = 2 .5 e -1 0   // b on di n g d i s t a n ce i n m   printf ( ” \ n E xa mp le 1 0 . 1 ” )   B = A * r_ 0 ^8 /9   Y = ( 9 0* B / ( r _ 0 ) ^ 1 1 - 2* A / ( r _ 0 ) ^ 3) / r _ 0   printf ( ” \ n Y o u n g ˆ−2” ,Y/1e9)

s modulus o f m a t e r i a l i s %d GN m

35

Scilab code Exa 10.2  Calculation of stress in fibers

1 2 3 4 5 6 7 8 9 10

  // C a l c u l a t i o n o f s t r e s s i n f i b e r s              

clc Y _f = 440 Y _m = 71 s i g m a _ t ot a l = 1 0 0   // t o t a l l oa d printf ( ” \ n E xa mp le 1 0 . 2 ” ) r = Y_f / Y_m s i g m a _ f = r * ( s i g m a _t o t al / 0 .7 ) / ( 1 + r * 3 /7 ) printf ( ” \ n P a rt A : ” ) printf ( ” \ n When l o a d i s a p p l i e d p a r a l l e l t o f i b e r th en , s t r e s s i n f i b e r i s %d MN mˆ −2” ,sigma_f)

11 12   printf ( ” \ n \ n P ar t B : ” ) 13   printf ( ” \ n When l o a d i s

a p p li e d p e r p en d i c u l a r t o f i b e r then , s t r e s s i n f i b e r and m a tr ix i s same i . e . %d MN mˆ −2” ,sigma_total)

Scilab code Exa 10.3   Estimate diffusion coefficient

1 2 3 4 5 6 7 8

// Estimate d i ff u si o n            

coefficient

clc t _r = 100   // r e l a x a t i o n t i m e i n s d = 2.5   // d i s t a n c e i n a ng st ro m printf ( ” \ n E xa mp le 1 0 . 3 ” ) f = 1/ t_r   / / j ump f r e q u e n c y D = ( d *1 e - 10 ) ^ 2* f printf ( ” \ n D i f f u s i o n c o e f f i c i e n t i s %. 2 e mˆ 2 s ˆ −1” ,D )

36

Chapter 11 Plastic Deformation and Creep in Crystalline Materials

Scilab code Exa 11.2 Calculate the stress required to move the dislocation

at given temperature 1   / / C a lc u l a te t h e s t r e s s

r e q u i r e d t o move t h e d i s l o c a t i o n a t g i v en t em pe ra tu re

2 3 4 5 6 7 8 9 10 11 12 13 14 15

                     

clc b = 2   // b u rg e r v e c t o r i n a ng st ro m v = 20* b ^3   // a c t i v a t i o n volume t a u _p n = 1 00 0 // P−N s t r e s s o f c r y s t a l i n MNmˆ −2 k = 1 .3 8 e -2 3   // p h y s i c a l c o ns t an t t 1 = 0   // t em p er at u re i n K t 2 = 100 // t em p er a tu re i n K t 3 = 300 // t em p er a tu re i n K t 4 = 500 // t em p er a tu re i n K printf ( ” \ n E xa mp le 1 1 . 2 ” ) printf ( ” \ n \ n P a rt A : ” ) T = t1 t a u_ a pp = t au _ pn - 4 0* k * T /( v * 1e - 3 0) printf ( ” \ n The s t r e s s r e q u i r e d t o move t he d i s l o c a t i o n a t t e m p e r a t u r e %dK i s %d MNmˆ −2” ,T , tau_app)

37

16 17 18 19

  printf ( ” \ n \ n P ar t B : ” )   T = t2 t a u_ a pp = t au _ pn - 4 0* k * T /( v * 1e - 3 0* 1 e 6 )   printf ( ” \ n The s t r e s s r e q u i r e d t o move t he

d i s l o c a t i o n a t t e m p e r a t u r e %dK i s %d MNmˆ −2” ,T , 20   21   22 23 24 25

tau_app) printf ( ” \ n \ n P a rt C : ” ) T = t3 t a u_ a pp = t au _ pn - 4 0* k * T /( v * 1e - 3 0* 1 e 6 ) if tau_app