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
1
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
1
Book Description Title: Materials Science and Engineering - A First Course Author: V. Raghavan Publisher: Prentice Hall, India Edition: 5 Year: 2007 ISBN: 9788120324558
1
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.
2
Contents Listt of Scilab Codes Lis
4
2 Equ Equili ilibri brium um and and Kine Kinetic ticss
5
3 Cry Crysta stall Geometry Geometry and Structur Structure e Determina Determinatio tion n
7
4 At Atomi omic c Structure Structure and and Chemical Chemical Bondin Bonding g
11
5 St Struc ructu ture re of of Soli Solid d
13
6 Cry Crysta stall Impe Imperfec rfectio tion n
17
7 Ph Phas ase e Di Diag agram ram
21
8 Di Diffu ffusi sion on in in Soli Solids ds
24
9 Pha Phase se Trans Transfor format mation ion
28
10 Elastic Anelastic and Viscoelastic Behaviour
32
11 Plastic Deformation and Creep in Crystalline Materials
34
12 Fracture
38
13 Oxidation and Corrosion
41
14 Conductors and Resistors
42
15 Semiconductors
44
3
16 Magnetic Materials
45
17 Dielectric Materials
48
4
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 . . . . . . . . . . . .
6
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 . . . . .
7
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
8
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
9
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))
11
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
12
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 ” )
13
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
14
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<0.712 then z = 0.81 // fro m t a b l e end end x = z *2* sqrt (D*t*3600) printf ( ” \ n D epth up t o wh ich m ac hi ni ng i s i s n e a r l y %. 2 f mm” ,x*1e3)
26 // n u m er i c al v a l u e o f a ns we r i n book i s
required
0 . 75
Scilab code Exa 8.4 Calculate time required to get required boron con-
centration 1 // C a l cu l a te t i me r e q u i r e d t o g et r e q u i r e d b o ro n
concentration 2 3 4 5 6 7
clc D = 4e -17 / / d i f f u s i o n c o e f f i c i e n t c1 = 0 cs = 3 e26 c_x = 1 e23 / / nu mber o f a to ms x = 2e -6 // d e p t h i n m
29
8 9 10 11 12 13 14 15 16 17 18 19
printf ( ” \ n E xa mp le 8 . 4 ” ) A = cs B = cs - c1 k = ( A- c _x ) / B 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 <0 then printf ( ” \ n S t r e s s t o be a p p l i e d i s z er o ” ) 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 em pe ra tu re %dK e n t i r e l y o ve rc om e by t h er m al f l u c t u a t i o n s ” , T ) 26 27 28 29 30 31 32
end printf ( ” \ n \ n P a rt D : ” ) T = t4 t a u_ a pp = t au _ pn - 4 0* k * T /( v * 1e - 3 0* 1 e 6 ) if tau_app <0 then printf ( ” \ n S t r e s s t o be a p p l i e d i s z er o ” ) 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 em pe ra tu re %dK e n t i r e l y o ve rc om e by t h er m al f l u c t u a t i o n s ” , T ) 33 end
Scilab code Exa 11.3 Calculate the dislocation density in copper
1 2 3 4 5 6
// C a lc u l a te t h e d i s l o c a t i o n d e n s i t y i n c o pp er
clc m u = 44 / / s h e a r m od ul us o f c o pp e r i n GN mˆ −2 b = 2.55 // b u r g er s v e c t o r i n a ng st ro m t au = 35 // s h e ar s t r e s s i n MN mˆ −2 printf ( ” E xa mp le 1 1 . 3 ” )
38
7 l = m u *1 e 9 * b *1 e - 1 0/ ( t au * 1 e6 ) 8 r ho = 1/ l ^2 9 10 printf ( ” \ n D i s l o c a t i o n d e n s i t y i n c o pp e r ˆ−2” ,rho) 11 / / Answer i n b oo k i s 1 e 1 2 mˆ −2
i s %. 1 e
m
Scilab code Exa 11.4 Find the yield stress for a grain size of ASTM 9
1 // Find t h e y i e l d s t r e s s f o r a g r a i n s i z e o f ASTM 9 2 clc 3 s ig ma 1 = 1 20 // i n i t i a l y i e l d s t r e n g t h o f m a t e r i a l
i n MNmˆ −2 4 s ig ma 2 = 2 20 // F in a l
y i el d s tr en gt h o f m at er ia l i n
MN mˆ−2 d 1 = 0. 04 / / i n i t i a l d i a m e t e r i n mm d 2 = 0. 01 // f i n a l d ia me te r i n mm n = 9 / / a st m n um be r printf ( ” E xa mp le 1 1 . 4 ” ) k = ( s i gm a2 - s i g m a1 ) * 1 e 6 / ( 1 / sqrt (d2*1e-3)-1/ sqrt (d1*1 e-3)) 10 s i g ma _ i = s ig m a1 * 1 e 6 - k / sqrt ((d1*1e-3)) 11 d = 1 / sqrt (2^(n-1)*1e4/645) 12 s i g m a _y = s i g ma _ i + k * ( d * 1e - 3 ) ^ ( - 0 . 5) 13 14 printf ( ” \ n Y ie ld s t r e s s f o r a g r a i n s i z e o f ASTM 9 i s %d MN mˆ −2” , ceil (sigma_y/1e6)) 5 6 7 8 9
Scilab code Exa 11.5 Estimate the change in yield strength
39
1 2 3 4 5 6 7
/ / E st im at e t he c ha ng e i n y i e l d s t r en g t h clc n1 = 1 e6 // i n i t i a l number o f p a r t i c l e s n2 = 1 e3 // f i n a l number o f p a r t i c l e printf ( ” \ n E xa mp le 1 1 . 5 ” ) k = ( n 1 / n2 ) ^ ( 1/ 3) printf ( ” \ n Y i e l d s t r en g t h would ha ve d e cr e as e d t o %d%% o f i t s i n i t i a l v a l u e . ” ,100/k)
40
Chapter 12 Fracture
Scilab code Exa 12.1 Estimate fracture strength
1 2 3 4 5 6 7 8 9
/ / E st im at e f r a c t u r e s t r en g t h
clc c = 2 // c ra ck o f h a l f l e n g t h i n m i c r o m e te r Y = 7 0 / / y o u n g s m od ul us i n GN mˆ −2 Gamma = 1 / / s p e c i f i c s u r f a c e e n e r g y printf ( ” \ n E xa mp le 1 2 . 1 ” ) s i g ma _ f = sqrt (2*Gamma*Y*1e9/(%pi*c*1e-6))/1e6 r = Y * 1 e3 / s i gm a _f printf ( ” \ n F r a c t u r e s t r e n g t h o f %d MN mˆ −2 i s 1 /%dth o f y o u n g s m od ul us . ” , ceil ( s i g m a _ f ) , ceil ( r /100)*100) 10 printf ( ” \ n Thus G r i f f i t h s c r i t e r i o n b r i d g e s t h e gap
b e tw e en t h e \ n o bs er ve d a nd i d e a l s t r en g t h s o f b r i t t l e ma te ri al ” )
Scilab code Exa 12.2 Estimate the brittle fracture strength at low tem-
peratures 41
1 // E st im at e t he
b r i t t l e f r a c t u r e s t r e n g t h a t lo w
temperatures 2 3 4 5 6 7 8 9 10
clc G a mm a = 1 .5 / / s p e c i f i c s u r f a c e e n e r g y i n J /mˆ 2 Y = 200 / / Y o u n g s m o du l us i n GN/mˆ 2 c = 2 // h a l f l e n g t h o f c ra ck printf ( ” \ n E xa mp le 1 2 . 2 ” ) s i g ma _ f = sqrt (2*Gamma*Y*1e9/(%pi*c*1e-6)) printf ( ” \ n B r i t t l e
f r a c t u r e s t r e n g t h a t lo w t e m p e r a t u r e s i s %d MNmˆ −2 ” ,sigma_f/1e6) / / a n sw er in book i s 310MNmˆ−2
Scilab code Exa 12.3 Estimate the temperature at which the ductility of
brittle transition occurs at given strain rates 1 / / E st im at e t he t em pe ra tu re a t wh i ch t he
duct ili ty o f b r i t t l e t r a n s i t i o n o c cu r s a t g i ve n s t r a i n rates
2 3 4 5 6 7 8 9 10 11 12
clc Gamma = 2 / / s p e c i f i c s u r f a c e e n e r g y i n J /mˆ 2 Y = 350 / / Y o u n g s m o du l us i n GN/mˆ 2 c = 2 // h a l f l e n g t h o f c ra ck d e_ dt 1 = 1 e -2 // s t r a i n r a t e d e_ dt 2 = 1 e -5 // s t r a i n r a t e printf ( ” \ n E xa mp le 1 2 . 3 ” ) printf ( ” \ n P a rt A : ” ) s i g ma _ f = sqrt (2*Gamma*Y*1e9/(%pi*c*1e-6)) s i gm a _y = s i gm a _f / 1 e 6 T = 1 7 3 60 0 / ( s i gm a _y - 2 0. 6 - 6 1 . 3* log10 (de_dt1)) //
t em p er a tu r e c a l c u l a t i o n 13
42
14 printf ( ” \ n T r a n s i t io n t em pe ra tu re f o r s t r a i n r a t e % . 0 e s ˆ−1 i s %d K” ,de_dt1 ,T) // a ns we r i n book i s
300 K 15 printf ( ” \ n \ n P ar t B : ” ) 16 17 T = 1 7 3 60 0 / ( s i gm a _y - 2 0. 6 - 6 1 . 3* log10 (de_dt2)) //
t em p er a tu r e c a l c u l a t i o n 18 19 printf ( ” \ n T r a n s i t io n t em pe ra tu re f o r s t r a i n r a t e % . 0 e s ˆ−1 i s %d K” ,de_dt2 ,T) // a ns we r i n book i s
230 K 20 // S o l u t io n i n b ook f o r two p a rt s i s d i v i d e d i n t o t h re e p a rt s
43
Chapter 13 Oxidation and Corrosion
Scilab code Exa 13.2 Calculation of required quantity of magnesium
1 2 3 4 5 6 7 8 9 10
// C a l c u l a t i o n o f r e q u i r e d q u a nt i t y o f magnesium
clc j = 1 5 // c u r r e n t d e n s i t y i n mA mˆ −2 m = 0 .0 24 3 / / m o la r m as s o f m ag ne si um F = 96 49 0 // f a ra d c ha rg e n = 2 // c ha r g e on i o n t = 1 0 // t i m e i n y ea r s printf ( ” \ n E xa mp le 1 3 . 2 ” ) a = m * j * 1 e - 3* ( t * 3 6 5 * 2 4* 3 6 0 0) / ( n * F ) printf ( ” \ n A mount o f ma gn es iu m r e q u i r e d mˆ −2” ,a )
44
i s %0 . 1 f kg
Chapter 14 Conductors and Resistors
Scilab code Exa 14.1 Calculate energy difference
1 2 3 4 5 6 7 8 9 10 11 12 13 14
/ / C a lc u l a te e ne rg y d i f f e r e n c e
clc n _x1 = 1 // a to mi c l e v e l n _y1 = 1 // a to mi c l e v e l n _z1 = 1 // a to mi c l e v e l n _x2 = 2 // a to mi c l e v e l L = 1 0 / / l a t t i c e p a r a m e t e r i n mm h = 6 .6 26 e - 34 // p la nk c o n st a n t m _ e = 9 .1 09 e - 31 // mass o f e l e c t r o n i n kg printf ( ” \ n E xa mp le 1 4 . 1 ” ) E 1 = h ^ 2 * ( n _ x1 ^ 2 + n _ y 1 ^ 2 + n _z 1 ^ 2 ) / ( 8* m _ e * ( L * 1 e - 3) ^ 2 ) E 2 = h ^ 2 * ( n _ x2 ^ 2 + n _ y 1 ^ 2 + n _z 1 ^ 2 ) / ( 8* m _ e * ( L * 1 e - 3) ^ 2 ) E = E2 - E1 // e ne rg y d i f f e r e n c e printf ( ” \ n En ergy d i f f e r e n c e i s %. 2 e J ” ,E )
Scilab code Exa 14.2 Calculate conductivity of copper at 300 K
45
1 2 3 4 5 6 7 8 9 10 11
/ / C a lc u l a te c o n d u c t i v i t y o f c op pe r a t 30 0 K clc t au = 2 e -1 4 // c o l l i s i o n t im e o f e l e c t r o n i n s e = 1 .6 02 e - 19 // c ha rg e on e l e c t r o n m _e = 9 .1 e -3 1 // mass o f e l e c t r o n i n kg printf ( ” \ n E xa mp le 1 4 . 2 ” ) n = 6 . 02 3 e 2 3 * 8 9 6 0 / 0. 0 6 3 54 s i gm a = n * e ^ 2* t a u / m _ e printf ( ” \ n C o n d u c t i v i t y o f c op pe r a t 30 0 K i s %. 1 e ohmˆ−1 mˆ −1 ” ,sigma)
Scilab code Exa 14.3 Estimation of resistivity due to impurity scattering
of 1 percent of Nickel in copper lattice 1 // Estimati on of
r e s i s t i v i t y d ue t o i m p u r i t y s c a t t e r i n g o f 1% o f N i c k el i n c op pe r l a t t i c e
2 clc 3 r _c u = 1 .8 e -8 / / r e s i s t i v i t y 4 r _N i_ cu = 7 e -8 / / r e s i s t i v i t y
o f p u r e c o p p e r i n ohm−m o f c o p p e r 4% N i i n ohm−
m 5 6 7 8
p er1 = 4 / / i m p u ri t y i n p e r c e n t p er2 = 1 // i m p ur i t y i n p e rc e nt printf ( ” \ n E xa mp le 1 4 . 3 ” ) r = ( r _ Ni _c u - r _ c u ) * p er 2 / p e r1 // r e s i s t i v i t y
o f
c o p p er 1% Ni i n ohm−m 9 printf ( ” \ n R e s i s t i v i t y due t o i mp ur i ty s c a t t e r i n g o f 1 %% o f N i c k e l i n c o p p er l a t t i c e i s %. 1 e ohm−m” , r)
46
Chapter 15 Semiconductors
Scilab code Exa 15.1 Calculate intrinsic carrier density
1 2 3 4 5 6 7 8 9 10
// Calculate i n t r i n s i c ca rr ie r density
clc r ho = 3 00 0 // r e s i s t i v i t y in ohm m m u_ e = 0 .1 4 m u_ h = 0 .0 5 e = 1 .6 02 e - 19 // c ha rg e on e l e c t r o n printf ( ” \ n E xa mp le 1 5 . 1 ” ) s ig ma = 1/ r ho n = s ig ma / ( ( m u_ e + m u_ h ) *e ) printf ( ” \ n I n t r i n s i c c a r r i e r d e n s i t y )
47
i s % . 3 e mˆ −3” ,n
Chapter 16 Magnetic Materials
Scilab code Exa 16.1 Calculate the net magnetic moment per iron atom
in crystal 1 / / C a l c u l a t e t h e n e t m a gn e ti c moment p e r i r o n atom
i n c r y st a l 2 3 4 5 6 7 8 9 10 11
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 n = 2 / / number o f a to ms p e r u n i t c e l l m = 1750 // S a t u r a t i o n m a g n e t i z at i o n i n kAmˆ −1 m u = 9 .2 73 e - 24 / / b o hr m ag ne to n printf ( ” \ n E xa mp le 1 6 . 1 ” ) m _a t om = m * 1 e3 * ( a *1 e - 10 ) ^ 3 / n m u _b = m _a to m / m u
printf ( ” \ n Net m a g ne t i c m oment p e r i r o n atom i n c r y s t a l i s %. 3 e Amˆ 2 ” ,m_atom) 12 printf ( ” \ n I n u n i t o f mu b , Net m a gn et i c moment p e r i r o n atom i n c r y s t a l i s %. 1 f ” ,mu_b)
48
Scilab code Exa 16.2 Comparison of saturation temperatures
1 2 3 4 5 6 7 8 9 10 11
/ / Com pa ris on o f s a t u r a t i o n t e mp e ra t ur e s
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 m _ n et _G d = 7 // n e t m a g ne t i c m oment o f g a d o l i n i u m m _ n et _ Co = 1 .7 // n et m a gn et i c moment o f c o b a l t t _c _G d = 2 89 // c u r i e t em pe ra tu re f o r Gd printf ( ” \ n E xa mp le 1 6 . 2 ” ) printf ( ” \ n P a rt A : ” ) if m_net_Gd > m_net_Co then printf ( ” \ n At %d K , N et m a g n e t i c moment o f
g a do l in i um i . e . %d i s g r e a t e r t ha n n et m ag ne ti c moment o f c o b a l t i . e . %. 1 f ” ,t1,m_net_Gd ,m_net_Co ) 12 printf ( ” \ n S o , Gd w i l l h ave h i g h e r s a t u r a t i o n magn etiza tion ”) 13 end 14 printf ( ” \ n \ n P ar t B : ” ) 15 if t_c_Gd
w i l l h a ve n e g l i g i b l e m a g n e ti za t i o n \ n a s compared t o Co , whi ch h as h i g h e r c u r i e t e m p e r a t u r e ” ,t2) 18 end
Scilab code Exa 16.4 Calculation of hysteresis loss
1 // C a l c u l a t i o n 2 clc
o f h y s t e r es i s l o s s
49
3 4 5 6 7 8 9 10 11 12
v = 0.01 / / v ol um e i n mˆ 3 x = 1e -4 // a x i s i n t e r c e p t y = 1 e2 // a x i s i n t e r c e p t a = 60 00 0 // H y s t e r e s i s l oo p a re a f = 5 0 // f r eq u en c y i n Hz printf ( ” \ n E xa mp le 1 6 . 4 ” ) e = x * y * a // En erg y l o s s i n o ne l o o p E = e*v // e n e r g y l o s s i n c or e i n on e c y c l e P = E*f // Power l o s s printf ( ” \ n P ower l o s s due t o h y s t e r e s i s i s %d W” ,P)
Scilab code Exa 16.5 Calculation of eddy current loss at normal voltage
and frequency 1 / / C a l c u l a t i o n o f eddy c u r r e n t
l o s s a t n o r ma l
v o l t a g e and f r eq u e nc y 2 clc 3 T o t al 1 = 2 30 0 / / t o t a l i r o n l o s s
i n W a t 4 4 0 V and
5 0 Hz 4 T ot al 2 = 7 50 / / t o t a l
i r o n l o s s i n W a t 2 20 V and 25
Hz 5 printf ( ” \ n E xa mp le 1 6 . 5 ” ) 6 W _e = 1 / 2* ( T o ta l1 - 2* T o t a l 2 ) 7 printf ( ” \ n Eddy c u r re n t l o s s f r e q u e n c y i s %dW” ,4*W_e)
50
a t n o r m a l v o l t a g e and
Chapter 17 Dielectric Materials
Scilab code Exa 17.1 Calculation of relative dielectric constant
1 2 3 4 5 6 7 8 9 10 11
// c a l c u l a t i o n o f r e l a t i v e
d i e l e c t r i c c o n st a n t
clc l = 10 // l e n g t h o f c a p a c i t o r i n mm b = 1 0 // w i d t h o f c a p a c i t o r i n mm d = 2 // d i s t a n c e o f s e p a r a t i o n i n mm c = 1e -9 // c a p ac i t a n c e i n f a ra d e p s i lo n _ 0 = 8 . 85 e - 1 2 // p e r m i t t i v i t y o f f r e e s pa c e printf ( ” \ n E xa mp le 1 7 . 1 ” ) e p s il o n _r = c * d * 1 e - 3/ ( e p s i l on _ 0 * l * 1 e - 3* b * 1 e - 3 ) printf ( ” \ n R e l a t i v e epsilon_r)
d i e l e c t r i c c o n s t a n t i s %d” ,
Scilab code Exa 17.2 Calculate the polarization of a BaTio3 crystal
1 / / c a l c ul a t e the p o l a r iz a t i on 2 clc
51
o f a BaTio3 c r y s t a l
