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Mt 201 b material science new Mt 201 b material science new Presentation Transcript

  • MT-­‐201B  MATERIALS  SCIENCE   1  
  • Why  Study  Materials  Science?  1.  ApplicaBon  oriented  Proper&es      2.  Cost  consideraBon    3.  Processing  route     2  
  • ClassificaBon  of  Materials   1.  Metals   2.  Ceramics   3.  Polymers   4.  Composites   5.  Semiconductors   6.  Biomaterials   7.  Nanomaterials     3  
  • Syllabus  1.  IntroducBon  to  Crystallography  2.  Principle  of  Alloy  FormaBon  3.  Binary  Equilibria  4.  Mechanical  ProperBes  5.  Heat  Treatments  6.  Engineering  Materials  7.  Advanced  Materials     4  
  • Recommended  Books  1.  Callister  W.D.,   Materials  Science  and   Engineering  an  Introduc&on  2.  Askeland  D.R.,   The  Science  and   Engineering  of  Materials  3.  Raghavan  V., Materials  Science  and   Engineering-­‐  A  first  Course,  4.  Avener  S.H,   IntroducBon  to  Physical   Metallurgy,   5  
  • The  Structure  of  Crystalline  Solids   CRYSTALLINE  STATE     •   Most  solids  are  crystalline  with  their  atoms  arranged  in  a                regular  manner.   •   Long-­‐range  order:  the  regularity  can  extend  throughout  the            crystal.   •   Short-­‐range  order:  the  regularity  does  not  persist  over            appreciable  distances.  Ex.  amorphous  materials  such  as  glass              and  wax.   •   Liquids  have  short-­‐range  order,  but  lack  long-­‐range  order.   •   Gases  lack  both  long-­‐range  and  short-­‐range  order.   •   Some  of  the  properBes  of  crystalline  solids  depend  on  the        crystal  structure  of  the  material,  the  manner  in  which  atoms,          ions,  or  molecules  are  arranged.     6  
  • Lace  •   SomeBmes  the  term  lace  is  used  in  the  context  of  crystal            structures;  in  this  sense   lace  means  a  three-­‐          dimensional  array  of  points  coinciding  with  atom  posiBons            (or  sphere  centers).    A  point  la*ce     7  
  • Unit  Cells  •  The unit cell is the basic structural unit or building block of the crystal structure and defines the crystal structure by virtue of its geometry and the atom positions within.•  This size and shape of the unit cell can be described in terms of their lengths (a,b,c) and the angles between then (α,β,γ). These lengths and angles are the lattice constants or lattice parameters of the unit cell.  A  point  la*ce     A  unit  cell   8  
  • Bravais  Lace  Table 1: Crystal systems and Bravais Lattices Crystal systems and Bravais Lattice 9  
  • Types of crystalsThree relatively simple crystal structures are found for mostof the common metals; body-centered cubic, face-centeredcubic, and hexagonal close-packed.1. Body Centered Cubic Structure (BCC)2. Face Centered Cubic Structure (FCC)3. Hexagonal Close Packed (HCP) 10  
  • 1.  Body Centered Cubic Structure (BCC)In these structures, there are 8 atoms at the 8 corners andone atom in the interior, i.e. in the centre of the unit cell withno atoms on the faces. 11  
  • 2. Face Centered Cubic Structure (FCC)In these structures, there are 8 atoms at the 8 corners,6 atoms at the centers of 6 faces and no interior atom. 12  
  • 3. Hexagonal Close Packed (HCP)In these structures, there are 12 corner atoms (6 at the bottomface and 6 at the top face), 2 atoms at the centers of theabove two faces and 3 atoms in the interior of the unit cell. 13  
  • Average Number of Atoms per Unit CellSince the atoms in a unit cell are shared by the neighboringcells it is important to know the average number of atoms perunit cell. In cubic structures, the corner atoms are shared by 8cells (4 from below and 4 from above), face atoms are sharedby adjacent two cells and atoms in the interior are shared byonly that one cell. Therefore, general we can write: Nav = Nc / 8 + Nf / 2 + Ni / 1Where,Nav = average number of atoms per unit cell.Nc = Total number of corner atoms in an unit cell.Nf = Total number of face atoms in an unit cell.Ni = Centre or interior atoms. 14  
  • •  Simple cubic (SC) structures: In these structures there are8 atoms corresponding to 8 corners and there are no atomson the faces or in the interior of the unit cell. Therefore,Nc = 8, Nf = 0 and Ni = 0Using above eqn. we get, Nav = 8/8 + 0/2 + 0/1 = 1 15  
  • 2. Body centered cubic (BCC) structures: In these structures, there are 8 atoms at the 8 corners and one atom in the interior, i.e. in the centre of the unit cell with no atoms on the faces. Therefore Nc = 8, Nf = 0 and Ni = 1 Using above eqn. we get, Nav = 8/8 + 0/2 + 1/1 = 2 16  
  • 3.  Face Centered Cubic Structure (FCC): In these structures, there are 8 atoms at the 8 corners, 6 atoms at the centers of 6 faces and no interior atom Therefore Nc = 8, Nf = 6 and Ni = 0 Using above eqn. we get, Nav = 8/8 + 6/2 + 0/1 = 4 17  
  • 4. Hexagonal Close Packed (HCP) Structures:In these structures, there are 12 corner atoms (6 at the bottom face and 6 atthe top face), 2 atoms at the centers of the above two faces and 3 atoms inthe interior of the unit cell.For hexagonal structures, the corner atoms are shared by 6 cells (3 frombelow and 3 from above), face atoms are shared by adjacent 2 cells andatoms in the interior are shared by only one cell. Therefore, in general thenumber of atoms per unit cell will be as: Nav = Nc / 6 + Nf / 2 + Ni / 1Here Nc = 12, Nf = 2 and Ni = 3Hence, Nav = 12 / 6 + 2 / 2 + 3 / 1 = 6 18  
  • Co-­‐ordina&on  Number  Co-­‐ordinaBon  number  is  the  number  of  nearest  equidistant    neighboring  atoms  surrounding  an  atom  under  consideraBon  1.  Simple  Cubic  Structure:    Simple  cubic  structure  has  a  coordinaBon  number  of  6   19  
  • 2.  Body  Centered  Cubic  Structure:    Body  centered  cubic  structure    has  a  coordinaBon  number  of  8   20  
  • 3.  Face  Centered  Cubic  Structure:    Face  centered  cubic  structure  has  a  coordinaBon  number  of  12   21  
  • 4.  Hexagonal  Close  Packed  Structure:    Hexagonal  close  packed  structure  has  a  coordinaBon  number  of  12   22  
  • Stacking  Sequence  for  SC,  BCC,  FCC  and  HCP  •   Lace  structures  are  described  by  stacking  of  idenBcal  planes          of  atoms  one  over  the  other  in  a  definite  manner    •   Different  crystal  structures  exhibit  different  stacking  sequences   1.  Stacking  Sequence  of  Simple  Cubic  Structure:     Stacking  sequence  of  simple  cubic  structure  is  AAAAA…..since  the   second  as  well  as  the  other  planes  are  stacked  in  a  similar  manner   as  the  first  i.e.  all  planes  are  stacked  in  the  same  manner.           A   A   A   23  
  • 2.    Stacking  Sequence  of  Body  Centered  Cubic  Structure:     •   Stacking  sequence  of  body  centered  cubic  structure  is  ABABAB….         •   The  stacking  sequence  ABABAB  indicates  that  the  second  plane          is  stacked  in  a  different  manner  to  the  first.      •   Any  one  atom  from  the  second  plane  occupies  any  one  intersBBal        site  of  the  first  atom.      •   Third  plane  is  stacked  in  a  manner  idenBcal  to  the  first  and  fourth          plane  is  stacked  in  an  idenBcal          manner  to  the  second  and  so  on.     A        This  results  in  a  bcc  structure.     B   A B   24  
  • 3.    Stacking  Sequence  of  Face  Centered  Cubic  Structure:    •   Stacking  sequence  of  face  centered  cubic  structure  is  ABCABC….        •   The  close  packed  planes  are  inclined  at  an  angle  to  the  cube  faces          and  are  known  as  octahedral  planes    •   The  stacking  sequence  ABCABC…  indicates  that  the  second  plane          is  stacked  in  a  different  manner  to  the  first  and  so  is  the  third  from        the  second  and  the  first.  The  fourth  plane  is  stacked  in  a  similar          fashion  to  the  first   25  
  • 26  
  • 4.    Stacking  Sequence  of  Hexagonal  Close  Packed  Structure:         •   Stacking  sequence  of  HCP  structure  is  ABABAB…..     •   HCP  structure  is  produced  by  stacking  sequence  of  the            type  ABABAB…..in  which  any  one  atom  from  the  second          plane  occupies  any  one  intersBBal  site  of  the  first  plane.     •   Third  plane  is  stacked  similar  to  first  and  fourth  similar  to            second  and  so  on.   27  
  • Atomic Packing Factor (APF)Atomic packing factor is the fraction of volume orspace occupied by atoms in an unit cell. Therefore, APF = Volume of atoms in unit cell Volume of the unit cell Since volume of atoms in a unit cell = Average number of atoms/cell x Volume of an atom  APF = Average number of atoms/cell x Volume of an atom Volume of the unit cell 28  
  • 1.  Simple  Cubic  Structures:    In  simple  cubic  structures,  the  atoms  are  assumed  to  be  placed  in  such  a  way  that  any  two  adjacent  atoms  touch  each  other.  If   a  is  the  lace  parameter  of  the  simple  cubic  structure  and   r  is  the  radius  of  atoms,  it  is  clear  from  the  fig  that:  r  =  a/2     APF = Average number of atoms/cell x Volume of an atom Volume of the unit cell = 1 x 4/3 π r3 = 4/3 π r3 = 0.52 a3 (2r)3 APF  of  simple  cubic  structure  is  0.52  or  52%   29  
  • 2.  Body  Centered  Cubic  (BCC)  Structures:  In  body  centred  cubic  structures,  the  centre  atom  touches  the  corner  atoms  as  shown  in  fig.     If   a  is  the  lace  parameter  of  BCC  structure  and     r  is  the  radius  of  atoms,  we  can  write                                                            (DF)2    =  (DG)2  +  (GF)2   Now                                          (DG)2    =  (DC)2  +  (CG)2  and  DF  =  4r   Therefore,                        (DF)2    =  (DC)2  +  (CG)2  +  (GF)2     30  
  •                                              (4r)2    =  a2    +  a2      +  a2   Therefore,                        r  =  a√3  /  4  APF  =  Average  number  of  atoms/cell  x  Volume  of  an  atom                                                                  Volume  of  the  unit  cell                                            2  x  4/3  π  (a√3  /  4)3            =        0.68                                                                  a3   APF  of  body  centered  cubic  structure  is  0.68  or  68%   31  
  • 3.  Face  Centered  Cubic  (FCC)  Structures:                  In  face  centred  cubic  structures,  the  atoms  at  the  centre  of  faces    touch  the                      corner  atoms  as  shown  in  figure.                    If   a  is  the  lace  parameter  of  FCC  structure  and   r  is  the  atomic  radius                                                          (DB)2      =    (DC)2      +      (CB)2                                                    i.e.  (4r)2      =  a2  +  a2                        Therefore,  r  =  a  /  2√2   APF  =  Average  number  of  atoms/cell  x  Volume  of  an  atom                                                                  Volume  of  the  unit  cell                                            =      4  x  4  /  3  x  π  (a/2√2)3      =        0.74                                                                                      a3   APF  of  face  centered  cubic  structure  is  0.74  or  74%   32  
  • 4.  Hexagonal  Close  Packed  (HCP)  Structures  The  volume  of  the  unit  cell  for  HCP  can  be  found  by  finding  out  the  area  of  the  basal  plane  and  then  mulBplying  this  by  its  height   This  area  is  six  Bmes  the  area  of     equilateral  triangle  ABC   Area  of  triangle  ABC  =  ½  a2  sin  60   Total  area  ABDEFG  =  6  x  ½  a2  sin  60                                                                          =  3  a2  sin  60   Now  volume  of  unit  cell  =  3  a2  sin  60  x  c                For  HCP  structures,  the  corner  atoms                are  touching  the  centre  atoms,  i.e.  atoms                  at  ABDEFG  are  touching  the  C  atom.                Therefore  a  =  2r  or  r  =  a  /  2   33  
  • APF  =  Average  number  of  atoms/cell  x  Volume  of  an  atom                                                                  Volume  of  the  unit  cell                                                                      APF  =  6  x  4π/3  r3                                                                                          3  a2  sin  60  x  c                                                                                                          APF  =        6  x  4π/3  (a/2)3                                                                                                3  a2  sin  60  x  c                                                                    APF  =                  π  a                                                                                                3  c  sin  60    The  c/a  raBo  for  an  ideal  HCP  structure  consisBng  of  uniform  spheres  packed  as    Bghtly  together  as  possible  is  1.633.  Therefore,  subsBtuBng  c/a  =  1.633  and  Sin  60o  =  0.866  in  above  equaBon  we  get:  APF  =      π  /  3  x  1.633  x  0.899    =  0.74   APF  of  face  centered  cubic  structure  is  0.74  or  74%   34  
  • Atomic  Packing  Factor  1.  Simple  cubic  structure:  0.52    2.  Body  centered  cubic  structure:  0.68  3.  Face  centered  cubic  structure:  0.74  4.  Hexagonal  close  packed  structure:  0.74     35  
  • Crystallographic  Points,  Planes  and  DirecBons   1.  Point  Coordinates                      When  dealing  with  crystalline  materials  it  olen  becomes  necessary  to   specify  a  parBcular  point  within  a  unit  cell.                      The  posiBon  of  any  point  located  within  a  unit  cell  may  be  specified  in   terms  of  its  coordinates  as  fracBonal  mulBples  of  the  unit  cell  edge  lengths.   36  
  • 37  
  • 2.  Plane  Coordinates     1.  Find  out  the  intercepts  made  by  the  plane  at  the  three   reference  axis  e.g.  p,q  and  r.   2.  Convert  these  intercepts  to  fracBonal  intercepts  by  dividing   with  their  axial  lengths.  If  the  axial  length  is  a,  b  and  c  the   fracBonal  intercepts  will  be  p/a,  q/b  and  r/c.   3.   Find  the  reciprocals  of  the  fracBonal  intercepts.  In  the  above   case  a/p,  b/q  and  c/r.   4.  Convert  these  reciprocals  to  the  minimum  of  whole  numbers   by  mulBplying  with  their  LCM.   5.  Enclose  these  numbers  in  brackets  (parenthesis)  as  (hkl)                  Note:  If  plane  passes  through  the  selected  origin,  either  another   parallel  plane  must  be  constructed  within  the  unit  cell  by  an   appropriate  translaBon  or  a  new  origin  must  be  established  at  the   corner  of  the  unit  cell.     38  
  • 1.   Intercepts:  p,q  and  r.  2.  FracBonal  intercepts:  p/a,  q/b  and  r/c.  3.   Reciprocals:  a/p,  b/q  and  c/r.  4.  Convert  to  whole  numbers  5.  Enclose  these  numbers  in                  brackets  (parenthesis)  as  (hkl)     39  
  • Step  1  :    IdenBfy  the  intercepts  on  the    x-­‐  ,  y-­‐  and  z-­‐  axes.  In  this  case  the  intercept  on  the    x-­‐axis  is  at  x  =  1  (  at  the  point  (1,0,0)  ),  but  the  surface    is  parallel  to  the  y-­‐  and  z-­‐axes  so  we  consider  the    intercept  to  be  at  infinity  (  ∞  )  for  the  special  case    where  the  plane  is  parallel  to  an  axis.    The  intercepts  on  the  x-­‐  ,  y-­‐  and  z-­‐axes  are  thus    Intercepts  :        1  ,  ∞  ,  ∞    Step  2  :    Specify  the  intercepts  in  fracBonal  co-­‐ordinates    Co-­‐ordinates  are  converted  to  fracBonal  co-­‐ordinates  by  dividing  by  the  respecBve    cell-­‐dimension  -­‐  This  gives    FracBonal  Intercepts  :        1/1  ,  ∞/1,  ∞/1        i.e.        1  ,  ∞  ,  ∞    Step  3  :    Take  the  reciprocals  of  the  fracBonal  intercepts    This  final  manipulaBon  generates  the  Miller  Indices  which  (by  convenBon)  should    then  be  specified  without  being  separated  by  any  commas  or  other  symbols.    The  Miller  Indices  are  also  enclosed  within  standard  brackets  (….).    The  reciprocals  of  1  and  ∞  are  1  and  0  respecBvely,  thus  yielding    Miller  Indices  :      (100)  So  the  surface/plane  illustrated  is  the  (100)  plane  of  the  cubic  crystal.   40  
  • Intercepts  :      1  ,  1  ,  ∞     FracBonal  intercepts  :      1  ,  1  ,  ∞   Reciprocal:  1,1,0   Miller  Indices  :      (110)    Intercepts  :      1  ,  1  ,  1    FracBonal  intercepts  :      1  ,  1  ,  1    Reciprocal:  1,1,1  Miller  Indices  :      (111)     41  
  • Intercepts  :      ½    ,  1  ,  ∞    FracBonal  intercepts  :      ½  ,  1  ,  ∞    Reciprocal:  2,1,0  Miller  Indices  :      (210)     Intercepts  :      1/3    ,  2/3  ,  1     FracBonal  intercepts  :      1/3    ,  2/3  ,   1     Reciprocal:  3,  3/2,  1   Miller  Indices  :      (632)     42  
  • Exercise   43  
  • Exercise   44  
  • Exercise   45  
  • Exercise   46  
  • Exercise   47  
  • 48  
  • If  the  plane  passes  through  the  origin,  the  origin    has  to  be  shiled  for  indexing  the  plane   49  
  • 50  
  • Miller  Indices  of  Planes  for  Hexagonal  Crystals    •   Crystal  Plane  in  HCP  unit  cells  is  commonly  idenBfied  by  using  four  indices        instead  of  three.    • The  HCP  crystal  plane  indices  called  Miller-­‐Bravis  indices  are  denoted  by  the        lepers  h,  k,  i  and  l  are  enclosed  in  parentheses  as  (hkil)    • These   four   digit   hexagonal   indices   are   based   on   a   coordinate   system   with   four  axes.  • The  three  a1,  a2  and  a3  axes  are  all  contained  within  a  single  plane      (called  the  basal  plane),  and  at  1200  angles  to  one  another.  The  z-­‐axis  is      perpendicular  to  the  basal  plane.    • The  unit  of  measurement  along  the  a1,  a2  and  a3  axes  is  the  distance          between  the  atoms  along  these  axes.    • The  unit  of  measurement  along  the  z-­‐  axis  is  the  height  of  the  unit  cell.  •   The  reciprocals  of  the  intercepts  that  a  crystal  plane  makes  with  the        a1,  a2  and  a3  axes  give  the  h,  k  and  I  indices  while  the  reciprocal  of  the        intercept  with  the  z-­‐axis  gives  the  index  l.   51  
  • 52  
  • 53  
  • Miller Indices of Directions for Cubic Crystals•  A  vector  of  convenient  length  is  posiBoned  such  that  it          passes  through  the  origin  of  the  coordinate  system.    •   The  length  of  the  vector  projecBon  on  each  of  the  three  axes            is  determined.  •   These  three  numbers  are  mulBplied  or  divided  by  a  common          factor  to  reduce  them  to  the  smallest  integer  values.    •   The  three  indices,  not  separated  by  commas,        are  enclosed  in  square  brackets  [uvw]    •   If  a  negaBve  sign  is  obtained  represent          the  –ve  sign  with  a              bar  over  the  number   54  
  • 55  
  • For  direcBon  not  originaBng  from  origin  the  origin  has  to  be  shiled     56  
  • Examples  of  direcBons  with  shil  of  origin   57  
  • 58  
  • Family  of  Symmetry  Related  Planes   _   (  1  1  0  )   (1  1  0)   _   (  1  0  1  )   (  1  0  1  )   _   (  0  1  1  )   (  0  1  1  )   {  1  1  0  }  {  1  1  0  }  =  Plane  (  1  1  0  )  and  all  other  planes  related    by  symmetry  to  (  1  1  0  )   59  
  • Family  of  Symmetry  Related  DirecBons   [  0  0  1  ]   IdenBcal  atomic  density   IdenBcal  properBes   _   [  1  0  0  ]   〈  1  0  0  〉   [  0  1  0  ]   _   [  1  0  0  ]  [  0  1  0  ]   〈1  0  0〉=  [  1  0  0  ]  and  all  other   z _   [  0  0  1  ]   direcBons  related  to  [  1  0  0  ]   by  symmetry   y x 60  
  • SUMMARY OF MEANINGS OF PARENTHESES q r s represents a point (hkl) represents a plane {hkl} represents a family of planes [hkl] represents a direction <hkl> represents a family of directions 61  
  • Anisotropy  of  crystals   191.1  GPa  Young s  modulus  of  FCC  Cu   130.3  GPa   66.7  GPa   62  
  • Anisotropy  of  crystals  (contd.)   Different  crystallographic   planes  have  different   atomic  density   And  hence   different   properBes   Si  Wafer  for   computers   63  
  • Linear  and  Planar  DensiBes   Linear  Density  •   Linear  density  (LD)  is  defined  as  the  number  of  atoms  per              unit  length  whose  centers  lie  on  the  direcBon  vector        LD  =  number  of  atoms  centered  on  direcBon  vector                                                        length  of  direcBon  vector       The  [110]  linear  density  for   FCC  is:     LD110  =  2  atoms/4R  =  1/2R     64  
  • Planar  Density    •   Planar  density  (PD)  is  defined  as  the  number  of  atoms  per         unit  area  that  are  centered  on  a  parBcular  crystallographic              plane  •     PD  =  number  of  atoms  centered  on  a  plane                                                                    area  of  plane       Planar  density  on  (110)  plane  in  a  FCC  unit  cell   •   Number  of  atoms  on  (110)  plane  is  2   •   Area  of  (110)  plane  (rectangular  secBon)  is        4R  (length)  x  2√2R  (height)  =  8R2√2        PD  =  2  atoms  /  8R2√2  =                          1  /  4R2√2       65  
  • Planar  density  on  (100)  plane  in  a  Simple  Cubic   Structure:   •   Number  of  atoms  on  (100)  plane  is  1   •   Area  of  (100)  plane  (square  secBon)  is        a  x  a  =  a2        PD  =  1  atom  /  a2  =                    =    1  /  a2  Planar  density  on  (110)  plane  in  a  Simple  Cubic  Structure:  •   Number  of  atoms  on  (110)  plane  is  1  •   Area  of  (110)  plane  (rectangular  secBon)  is  √2a2      PD  =  1  atom  /  √2  a2  =                    =    1  /  √2  a2   66  
  • Planar  density  on  (111)  plane  in  a  Simple  Cubic  Structure:  •   Number  of  atoms  on  (111)  plane  is  1/6  x  3  =  0.5  •   Area  of  (111)  plane  (triangle  DEF)  is        1/2  x  (√2a)  x  (0.866  x  √2a)  =  0.866a2        PD  =  0.5  atom  /  0.866a2  =                    =    0.577  /  a2   Planar  density  on  (100)  plane  in  a     Body  Centred  Cubic  Structure:   •   Number  of  atoms  on  (100)  plane   is  1   •   Area  of  (100)  plane  (square   secBon)  is  a  x  a  =  a2      PD  =  1  atom  /  a2  =  1  /  a2   67  
  • Planar  density  on  (110)  plane  in  a  Body   Centered  Cubic  Structure:   •   Number  of  atoms  on  (110)  plane  is  1/4   x  4  +  1  =  2   •   Area  of  (110)  plane  (rectangle  AFGD)  is   a  x  √2a    =  √2a2          PD  =  2  atoms  /  √2a2  =                    =  √2    /  a2  =  1.414  /  a2  Planar  density  on  (111)  plane  in  a  Body  Centered  Cubic  Structure:  •   Number  of  atoms  on  (111)  plane  is  1/6  x  3  +  1  =  1.5  •   Area  of  (111)  plane  (triangle  DEG)  is  ½  x  √2a        √2a  sin60o    =  0.866  a2      PD  =  1.5  atoms  /  0.866a2  =                    =  1.732  /  a2   68  
  • Voids  in  crystalline  structures    We have already seen that as spheres cannot fill entire space → the atomic packing fraction (APF) < 1 (for all crystals)  This implies there are voids between the atoms. Lower the PF, larger the volume occupied by voids.  These voids have complicated shapes; but we are mostly interested in the largest sphere which can fit into these voids  The size and distribution of voids in materials play a role in determining aspects of material behaviour → e.g. solubility of interstitials and their diffusivity  The position of the voids of a particular type will be consistent with the symmetry of the crystal  In the close packed crystals (FCC, HCP) there are two types of voids → tetrahedral and octahedral voids (identical in both the structures as the voids are formed between two layers of atoms)  The tetrahedral void has a coordination number of 4  The octahedral void has a coordination number 6 69  
  • 70  
  • Inters&&al  sites  /  voids   71  
  • Tetrahedral  sites  in  HCP   Octahedral  sites  in  HCP   72  
  • Voids:  Tetrahedral  and  Octahedral  Sites      •    Tetrahedral   and   octahedral   sites   in   a   close   packed   structure   can  be            occupied  by  other  atoms  or  ions  in  crystal  structures  of  alloys.      •   Thus,  recognizing  their  existence  and  their  geometrical  constrains        help  in  the  study  and  interpretaBon  of  crystal  chemistry.      •   The  packing  of  spheres  and  the  formaBon  of  tetrahedral  and            octahedral  sites  or  holes  are  shown  below.   73  
  • 74  
  • What is the radius of the largest sphere that can be placed in a tetrahedralvoid without pushing the spheres apart?To solve a problem of this type, we need to construct a model for the analysis.Use the diagram shown here as a starting point, and construct a tetrahedralarrangement by placing four spheres of radius R at alternate corners of a cube.•  What is the length of the face diagonal fd of this cube in terms of R? Since the spheres are in contact at the centre of each cube face, fd = 2 R.•  What is the length of the edge for such a cube, in terms of R? Cube edge length a = √2 R•  What is the length of the body diagonal bd of the cube in R? bd = √6 R•  Is the center of the cube also the center of the tetrahedral hole? Yes•  Let the radius of the tetrahedral hole be r, express bd in terms of R and r If you put a small ball there, it will be in contact with all four spheres. bd = 2 (R + r). r = (2.45 R) / 2 - R = 1.225 R - R = 0.225 R•  What is the radius ratio of tetrahedral holes to the spheres? r / R = 0.225 75  
  • Derive the relation between the radius (r) of the octahedral void and the radius (R) of the atom in a close packed structure(Assume largest sphere in an octahedral void without pushing the parent atom) A sphere into the octahedral void is shown in the diagram. A sphere above and a sphere below this small sphere have not been shown in the figure. ABC is a right angled triangle. The centre of void is A.   Applying Pythagoras theorem.   BC2 = AB2 + AC2   (2R)2 + (R + r)2 + (R + r)2 = 2(R + r)2   4R2/2 = (R + r)2               2R2  =  (R  +  r)2     √2R  =  R  +    r     r  =  √2R  –  R  =  (1.414  –1)R r  =  0.414  R   76  
  • Single  Crystal  and  Polycrystalline   Stages  of  solidificaBon  of  a  polycrystalline     material  Single  Crystal   77  
  • silicon single crystalMicrograph of a polycrystallinestainless steel showing grainsand grain boundaries 78  
  • 79  
  • Polymorphism   80  
  • Ceramic  Crystal  Structures  •   Ceramics  are  compounds  between  metallic  &  nonmetallic        elements  e.x.  Al2O3,  FeO,  SiC,  TiN,  NaCl  •   They  are  hard  and  briple  •   Typically  insulaBve  to  the  passage  of  electricity  &  heat    Crystal  Structures  •   Atomic  bonding  is  mostly  ionic  i.e.  the  crystal  structure  is      composed  of  electrically  charged  ions  instead  of  atoms.  •   The  metallic  ions,  or  caBons  are  posiBvely  charged  because        they  have  given  up  their  valence  electrons  to  the        nonmetallic  Ions  or  anions,  which  are  negaBvely  charged     81  
  • Ionic  bonding   82  
  • •   In  a  ceramic  material  two  characterisBcs  of  the          component  ions  influence  the  crystal  structure:           1.  Charge  neutrality     2.  The  relaBve  sizes  of  the  caBons  and  anions   83  
  •          1.  Charge  neutrality:  each  crystal  should  be                            electrically  neutral  e.x.  NaCl  and  CaCl2   84  
  • 2.  The  relaBve  sizes  of  the  caBons  and  anions    •     Because  the  metallic  elements  give  up  electrons  when          Ionized,  caBons  are          smaller  than  anions        •     Hence  rc  /  ra  is  less  than  unity    •     Stable  ceramic  crystal  structures  form  when  those          anions  surrounding  a  caBon  are  all  in  contact  with  the          that  caBon   85  
  • •   CoordinaBon  number  is  related  to  the  caBon-­‐anion  raBo    •   For  a  specific  coordinaBon  number  there  is  a  criBcal        or  minimum  rc  /  ra    raBo         86  
  • 87  
  • 88  
  • PredicBng  Structure  of  FeO   89  
  • 90  
  • AX-­‐TYPE  STRUCTURES   •   Equal  number  of  caBons  and  anions  referred  to  as        AX  compounds     A  denotes  the  caBon  and   X  denotes  the  anion   rNa  =  0.102  nm     rCl  =  0.181  nm     r  Na  /  rCl        =  0.564        CaBons  prefer  octahedral  sites       Rock  Salt  Structure   91  
  • AX-­‐TYPE  STRUCTURES  conBnued…   MgO  also  has  a  NaCl  type  structure  rO  =  0.140  nm    rMg  =  0.072  nm      rMg  /  rO        =  0.514        CaBons  prefer  octahedral  sites           92  
  • AX-­‐TYPE  STRUCTURES  conBnued…   93  
  • AmXp-­‐TYPE  STRUCTURES  •   number  of  caBons  and  anions  are  different,            referred  to  as  AmXp  compounds   Calcium  Fluorite  Structure   94  
  • AmBnXp-­‐TYPE  STRUCTURES  •     Ceramic  compound  with  more  than  two  types                    of  caBons,  referred  to  as  AmBnXp  compounds   95  
  • Crystal  defects  (ImperfecBons  in  Solids)  •  Perfect order does not exist throughout a crystalline material on an atomic scale. All crystalline materials contain large number of various defects or imperfections.•  Defects or imperfections influence properties such as mechanical, electrical, magnetic, etc.•  Classification of crystalline defects is generally made according to geometry or dimensionality of the defect i.e. zero dimensional defects, one dimensional defects and two dimensional defects. 96  
  • Crystal defects / imperfections are broadly classifiedinto three classes: 1. Point  defect  (zero  dimensional  defects)              Vacancy,                                          Impurity  atoms  (  subs&tu&onal    and  inters&&al)              Frankel  and  Scho]ky  defect       2.  Line  defect  (one  dimensional  defects)            Edge  disloca&on            Screw  disloca&on,              Mixed  disloca&on       3.  Surface  defects  or  Planer  defects  (two  dimensional   defects)                Grain  boundaries              Twin  boundary                Stacking  faults   97                
  • 1. Point defectsVacancy   98  
  • Vacancy  •  If an atom is missing from its regular site, the defect produced is called a vacancy•  All crystalline solids contain vacancies and their number increases with temperature•  The equilibrium concentration of vacancies Nv for a given quantity of material depends on & increases with temperature according toWhere:N is the total number of atomic sitesQv is the energy required for the formation of a vacancyT is the absolute temperature &k is the gas or Boltzmann s constant i.e. 1.38 x 10-23 J/atom-K or8.62 X 10-5 eV/atom-K 99  
  • 100  
  • Vacancies aid in the movement (diffusion) of atoms 101  
  •  Impurity  atoms  (  subs&tu&onal    and  inters&&al)   102  
  • •  Impurity point defects are of two types 1. Substitutional 2. Interstitial•  For substitutional, solute or impurity atoms replace or substitute for the host atoms•  For interstitial, solute or impurity atoms fill the void or interstitial space among the host atoms•  Both the substitutional and interstitial impurity atoms distort the crystal lattice affecting the mechanical and electrical / electronic properties 103  
  • •  Impurity atoms generate stress in the lattice by distorting the lattice•  The stress is compressive in case of smaller substitutional atom and tensile in case of larger substitutional atom•  These stresses act as barriers to movement of dislocations and thus improve the strength / hardness of a material•  These stresses also act as barriers to the movement of electrons and lower the electrical conductivity (increases resistivity) of the material 104  
  •  Frankel  and  Scho]ky  defects     105  
  • •  Frenkel and Schottky defects occur in ionic solids like ceramics•  An atom may leave its regular site and may occupy nearbyinterstitial site of the matrix giving rise to two defectssimultaneously i.e. one vacancy and the other self interstitial.These two defects together is called a Frenkel defect. This canoccur only for cations because of their smaller size ascompared to the size of anions.•  When cation vacancy is associated with an anion vacancy, the defect is called Schottky defect. Schottky defects are more common in ionic solids because the lattice has to maintain electrical neutrality 106  
  • 2. Line defectsDislocaBons    •   A  missing  line  or  row  of  atoms  in  a  regular  crystal            lace  is  called  a  dislocaBon  •   DislocaBon  is  a  boundary  between  the  slipped  region          and  the  unslipped  region  and  lies  in  the  slip  plane    •   Movement  of  dislocaBon  is  necessary  for  plasBc            deformaBon  •   There  are  mainly  two  types  of  dislocaBons  (a)  Edge          dislocaBons  and  (b)  Screw  dislocaBons   107  
  • Edge DislocationDislocaBon  line  and  b  are  perpendicular  to  each  other   108  
  • Movement  of  edge  dislocaBon   109  
  • ElasBc  stress  field  responsible  for  electron  scapering  and    increase  in  electrical  resisBvity   lace  strain  around   dislocaBon   110  
  • Screw DislocationDislocaBon  line  and  b  are  parallel  to  each  other   111    
  • Movement of Screw Dislocation 112  
  • When Dislocations Interact 113  
  • Mixed DislocationsBy  resolving,  the  contribuBon    from  both  types  of    dislocaBons  can  be    determined   114  
  • DislocaBons    as  seen  under  Transmission  Electron  Microscope    (TEM)   115  
  • 3. Surface defectsGrain  Boundary  •  Grain boundary is a defect which separates grains of different orientation from each other in a polycrystalline material.•  When this orientation mismatch is slight, on the order of a few degrees (< 15 degrees) then the term small- (or low- ) angle grain boundary is used. When the same is more than 15 degrees its is know as a high angle grain boundary.•  The total interfacial energy is lower in large or coarse-grained materials than in fine-grained ones, since there is less total boundary area in the former.•  Mechanical properties of materials like hardness, strength, ductility etc are influenced by the grain size.•  Grains grow at elevated temperatures to reduce the total boundary energy. 116  
  • 117  
  • Coarse and fine grain structure Grain boundaries acting as barriers to the movement of dislocations Deformation of grains during 118   cold working (cold rolling in this case)
  • Twin  Boundary  Twin  boundary  Atoms  on  one  side  of  the  boundary  are  located  in    Mirror  image  posiBons  of  the  atoms  on  the  other  side   119  
  • A twin boundary is a special type of grain boundary across which there isa specific mirror lattice symmetry; that is, atoms on one side of theboundary are located in mirror-image positions of the atoms on the otherside.The region of material between these boundaries is appropriately termeda twin.Twins result from atomic displacements that are produced from appliedmechanical shear forces (mechanical twins), and also during annealingheat treatments following deformation (annealing twins).Twinning occurs on a definite crystallographic plane and ina specific direction, both of which depend on the crystal structure.Annealing twins are typically found in metals that have the FCC crystalstructure, while mechanical twins are observed in BCC and HCP metals.Twins contribute to plastic deformation in a small way 120  
  • Stacking fault•   Occurs  when  there  is  a  flaw  in  the  stacking  sequence  •   Stacking  fault  results  from  the  stacking  of  one  atomic  plane  out  of          sequence  on  another  and  the  lace  on  either  side  of  the  fault  is          perfect  •   BCC  and  HCP  stacking  sequence:  ABABABAB……      with  stacking  fault:  ABABBABAB……or  ABABAABABAB……..  •   FCC  stacking  sequence:  ABCABCABC….      with  stacking  fault:  ABCABCABABCABC……   Stacking  fault   FCC Stacking 121  
  • PlasBc  DeformaBon   122  
  • Principles  of  Alloy  FormaBon  Solid  Solu&on:    •   A  homogeneous  crystalline  phase  that  contains  two  or          more  chemical  species  •   It  is  an  alloy  in  which  the  atoms  of  solute  are  distributed            in  the  solvent  and  has  the  same  structure  as  that  of  the          solvent      Types  of  Solid  Solu&ons:    1.  IntersBBal  solid  soluBon,  ex.  Fe-­‐C   IntersBBal  Solid  Soln    2.  SubsBtuBonal  solid  soluBon,  ex.  Au-­‐Cu     123   SubsBtuBonal  Solid  Soln  
  • 1.  Inters&&al  Solid  Solu&on  Alloys  •   Parent  metal  atoms  are  bigger  than  atoms  of  alloying  metal.  •   Smaller  atoms  fit  into  spaces,  (IntersBces),  between  larger          atoms.   124  
  • Inters&&al  sites   125  
  • 2.  Subs&tu&onal  Solid  Solu&on  Alloys  •   Atoms  of  both  metals  are  of  almost  similar  size.  •   Direct  subsBtuBon  takes  place.   126  
  • Some Solid Solution AlloysAlloy   Unit  Cell  Structure  Copper  -­‐  Nickel   FCC  Copper  -­‐  Gold   FCC  Gold  -­‐  Silver   FCC  Nickel  -­‐  PlaBnum   FCC  Molybdenum  -­‐  Tungsten   BCC  Iron  -­‐  Chromium   BCC   127  
  • Hume-­‐Rothery s  Rules  of  Solid  Solubility   1.  Atomic  size  factor     2.  Crystal  structure  factor     3.  ElectronegaBvity  factor     4.  RelaBve  valency  factor   128  
  • 1.  Atomic  size  factor:  If  the  atomic  sizes  of  solute  and  solvent   differ  by  less  than  15%,  it  is  said  to  have  a  favourable  size   factor  for  solid  soluBon  formaBon.  If  the  atomic  size   difference  exceeds  15%  solid  solubility  is  limited        2.  Crystal  Structure  factor:  Metals  having  same  crystal  structure   will  have  greater  solubility.  Difference  in  crystal  structure   limits  the  solid  solubility   +   A  (fcc)   B  (fcc)   AB  solid  solu&on  (fcc)   129  
  • 3.  Electronega&vity  factor:  The  solute  and  solvent  should  have  similar  electronegaBvity.  If  the  electronegaBvity  difference  is  too  great,  the  metals  will  tend  to  form  compounds  instead  of  solid  soluBons.    If  electronegaBvity  difference  is  too  great  the  highly  electroposiBve    element  will  lose  electrons,  the  highly  electronegaBve  element  will    acquire  electrons,  and  compound  formaBon  will  take  place.    4.  Rela&ve  Valency  factor:  Complete  solubility  occurs  when  the  solvent  and  solute  have  the  same  valency.  If  there  is  shortage  of  electrons  between  the  atoms,  the  binding    between  them  will  be  upset,  resulBng  in  condiBons  unfavourable  for  solid  solubility       130  
  • 131  
  • Phase  Diagrams  Phase  diagrams:  Phase   or   equilibrium   diagrams   are   diagrams   which   indicate   the  phases   exisBng   in   the   system   at   any   temperature,   pressure   and  composiBon.    Why  study  Phase  Diagrams?  •   Used  to  find  out  the    amount  of  phases  exisBng  in  a  given  alloy            with  their  composiBon  at  any  temperature.    •   From  the  amount  of  phases  it  is  possible  to  esBmate  the            approximate  properBes  of  the  alloy.    •   Useful  in  design  and  control  of  heat  treatment  procedures     132  
  • Terms:  System:   A   system   is   that   part   of   the   universe   which   is   under  consideraBon.    Phase:  A  phase  is  a  physically  separable  part  of  the  system  with  disBnct  physical  and  chemical  properBes.  (In  a  system  consis6ng   of   ice   and   water   in   a   glass   jar,   the   ice   cubes   are  one   phase,   the   water   is   a   second   phase,   and   the   humid   air  over   the   water   is   a   third   phase.   The   glass   of   the   jar   is  another  separate  phase.)    Variable:  A  parBcular  phase  exists  under  various  condiBons  of   temperature,   pressure   and   concentraBon.   These  parameters  are  called  as  the  variables  of  the  phase  Component:     The   elements   present   in   the   system   are   called  as  components.  For  ex.  Ice,  water  or  steam  all  contain  H2O  so  the  number  of  components  is  2,  i.e.  H  and  O.   133  
  • Gibb s  Phase  Rule:  The  Gibb s  phase  rule  states  that  under  equilibrium  condiBons,    the  following  relaBon  must  be  saBsfied:                                                                          P  +    F  =  C  +  2  Where,  P  =  number  of  phases  exisBng  in  a  system  under  consideraBon.  F  =  degree  of  freedom  i.e.  the  number  of  variables  such  as                temperature,  pressure  or  composiBon  (concentraBon)  that  can                be  changed  independently  without  changing  the  number  of                phases  exisBng  in  the  system.  C  =  number  of  components  (i.e.  elements)  in  the  system,  and    2  =  represents  any  two  variables  out  of  the  above  three  i.e.                temperature  pressure  and  composiBon. 134  
  • Most   of   the   studies   are   done   at   constant   pressure   i.e.   one  atmospheric   pressure   and   hence   pressure   is   no   more   a  variable.  For  such  cases,  Gibb s  phase  rule  becomes:                                                                                                                                                              P  +    F  =  C  +  1    In   the   above   rule,   1   represents   any   one   variable   out   of   the  remaining  two  i.e.  temperature  and  concentraBon.      Hence,  Degree  of  Freedom  (F)  is  given  by        F  =  C  –  P  +  1     135  
  • ApplicaBon  of  Gibbs  Phase  Rule   •   C   At  point  A   P  =  1,  C  =  2   F  =  C  –  P  +  1   F  =  2  –  1  +1   F  =  2   The  meaning  of  F  =  2  is  that  both  temperature   and  concentraBon  can  be  varied  independently   without  changing  the  liquid  phase  exisBng  in     the  system    At  point  C   At  point  B  P  =  1,  C  =  2   P  =  2,  C  =  2  F  =  C  –  P  +  1   F  =  C  –  P  +  1  F  =  2  –  1  +1   F  =  2  –  2  +1  F  =  2   F  =  1  The  meaning  of  F  =  2  is  that  both  temperature   The  meaning  of  F  =  1  is  that  any  one  variable  and  concentraBon  can  be  varied  independently   out  of  temperature  and  composiBon  can  be      without  changing  the  liquid  phase  exisBng  in     changed  independently  without  altering  the    the  system   liquid  and  solid  phases  exisBng  in  the  system   136    
  • Types  of  Phase  Diagrams:     •   Unary  phase  diagram   •   Binary  phase  diagram   •   Ternary  phase  diagram   137  
  • 1.  Unary  Phase  diagram  (one  component)    The   simplest   phase   diagrams   are   pressure-­‐temperature  diagrams   of   a   single   simple   substance,   such   as   water.   The  axes  correspond  to  the  pressure  and  temperature.     138  
  • 2.  Binary  Phase  diagram  (two  components)    •   A  phase  diagram  plot  of    temperature  against  the            relaBve  concentraBons  of  two  substances  in  a  binary              mixture  called  a  binary  phase  diagram  •   Types  of  binary  phase  diagrams:    1.  Isomorphous    2.  EutecBc    3.  ParBal  EutecBc     139  
  • 3.  Ternary  Phase  diagram  (three  components)    •   A  ternary  phase  diagram  has  three  components.        •   It  is  three  dimensional  put  ploped  in  two  dimensions  at            constant  temperature  •   Stainless  steel  (Fe-­‐Ni-­‐Cr)  is  a  perfect  example  of  a  metal  alloy          that  is  represented  by  a  ternary  phase  diagram.     140  
  • Binary  phase  diagram  The  binary  phase  diagram  represents  the  concentraBon  (composiBon)  along  the  x-­‐axis  and  the  temperature  along  the  y-­‐axis.  These  are  ploped  at  atmospheric  pressure  hence  pressure  is  constant  i.e.  1  atm.  pressure.  These  are  the  most  widely  used  phase  diagrams.    Types  of  binary  phase  diagrams:  • Binary  isomorphous  system:  Two  metals  having  complete  solubility  in  the  liquid  as  well  as  the  solid  state.  • Binary  eutecBc  system:  Two  metals  having  complete  solubility  in  the  liquid  state  and  complete  insolubility  in  the  solid  state.  • Binary  parBal  eutecBc  system:  Two  metals  having  complete  solubility  in  the  liquid  state  and  parBal  solubility  in  the  solid  state.  • Binary  layer  type  system:  Two  metals  having  complete  insolubility  in  the  liquid  as  well  as  in  the  solid  state.   141  
  • Cooling  curve  for  Pure  Metal  (one  component)   142  
  • Cooling  curve  for  an  alloy  /  solid  soluBon    (two  components)   143  
  • 144  
  • Plong  of  Phase  Diagrams   145  
  • Binary  isomorphous  system:    • These  phase  diagrams  are  of  loop  type  and  are  obtained  for        two  metals  having  complete  solubility  in  the  liquid  as  well  as          solid  state.  •   Ex.:  Cu-­‐Ni,  Au-­‐Ag,  Au-­‐Cu,  Mo-­‐W,  Mo-­‐Ti,  W-­‐V.         146  
  • Lever  rule  Finding  the  amounts  of  phases  in  a  two  phase  region  :  1.  Locate  composiBon  and  temperature  in  phase  diagram  2.  In  two  phase  region  draw  the  Be  line  or  isotherm  3.  FracBon  of  a  phase  is  determined  by  taking  the  length    of  the  Be  line  to  the  phase  boundary  for  the  other  phase,    and  dividing  by  the  total  length  of  Be  line     147  
  • %  of  Solid  =  LO  /  LS  X  100=  (Wo-­‐Wi)  /  (Ws-­‐Wi)  X  100    %  of  Liquid  =  OS  /  LS  X  100=  (Ws-­‐Wi)  /  (Ws-­‐Wi)  X  100  or  simply  %  Liquid  =  100  -­‐  %  of  Solid  or  vice  versa     148  
  • Development  of  Microstructure  during  slow  cooling  in     isomorphous  alloys   149  
  • ProperBes  of  alloys  in  Isomorphous  systems     with  variaBon  in  composiBon     (a)  Phase diagram of the Cu-Ni alloy system. Above the liquidus line only the liquid phase exists. In the L + S region, the liquid (L) and solid (S) phases coexist whereas below the solidus line, only the solid phase (a solid solution) exists. (b) The resistivity of the Cu-Ni alloy as a Function of Ni content (at.%) at room temperature 150  
  • 151  
  • Binary  EutecBc  System:      These  diagrams  are  obtained  for  two  metals  having  complete  solubility  (i.e.  miscibility)  in  the  liquid  state  and  complete  insolubility  in  the  solid  state.  Examples:  Pb-­‐As,  Bi-­‐Cd,  Th-­‐Ti,  and  Au-­‐Si.       152  
  • What  is  a  EutecBc?    •   A  eutec6c  or  eutec6c  mixture  is  a  mixture  of  two  or  more  phases        at  a  composiBon  that  has  the  lowest  melBng  point  •   EutecBc  ReacBon:      Liquid    ↔    Solid  A  +  Solid  B   153  
  • Cooling  Curves  in  EutecBc  System   154  
  • Plong  of  EutecBc  Phase  Diagrams   155  
  •    Binary  Par&al  Eutec&c  System  These  diagrams  are  obtained  for  two  metals  having  complete    solubility  (i.e.  miscibility)  in  the  liquid  state  and  parBal  solubility    in  the  solid  state.  Examples:  Pb-­‐Sn,  Ag-­‐Cu,  Sn-­‐Bi,  Pb-­‐Sb,  Cd-­‐Zn  and  Al-­‐Si.     156  
  • 157  
  • Development  of  microstructure  in  binary  par&al  eutec&c  alloys     during  equilibrium  cooling   1.  SolidificaBon  of  the  eutecBc  composiBon   158  
  • 2.  SolidificaBon  of  the  off  -­‐  eutecBc  composiBon   159  
  • 3.  SolidificaBon  of  composiBons  that  range  between  the  room  temperature  solubility  limit  and  the  maximum  solid  solubility  at  the  eutecBc  temperature   160  
  • Uses  of  Eutec&c  /  Par&al  Eutec&c  Alloys    Alloys  of  eutecBc  composiBons  have  some  specific  properBes    which  make  them  suitable  for  certain  applicaBons:  • Since  they  fuse  at  constant  temperature,  they  are  used  for        electrical  and  thermal  fuses.  • They  are  used  as  solders  due  to  their  lower  melBng  temperature.  • Since  eutecBc  alloys  have  low  melBng  points,  some  of  them  are        used  coaBngs  by  spraying  techniques  • Since  they  melt  at  constant  temperature  they  can  be  used  for          temperature  measurement.  •   Majority  of  the  eutecBc  alloys  are  superplasBc  in  character.  SuperplasBcity  is  the  phenomenon  by  which  an  alloy  exhibits  large  extension  (ducBlity)  when  deformed  with  certain  rate  at  some  temperature.  The  alloy  behaves  like  plasBc  and  can  be  formed  into  many  shapes. 161  
  • The  Iron  –  Carbon  System  Allotrophic  TransformaBons  in  Iron   162  
  • Iron  –  Carbon  Phase  Diagram   163  
  • Phases  in  Iron-­‐Carbon  Phase  Diagram  1.  Ferrite:  Solid  soluBon  of  carbon  in  bcc  iron  2.  Austenite:  Solid  soluBon  of  carbon  in  fcc  iron  3.  δ-­‐iron:  Solid  soluBon  of  carbon  in  bcc  iron  4.  Cemen&te  (Fe3C):  Intermetallic  compound  of  iron    and  carbon  with  a  fixed  carbon  content  of  6.67%  by  wt.  5.  Pearlite:  It  is  a  two  phased  lamellar  (or  layered)  structure  composed  of  alternaBng  layers  of  ferrite  and  cemenBte       164  
  • Austenite  Ferrite  and    δ-­‐iron   Cemen&te   165  
  • 166  
  • The  iron-­‐carbon  system  exhibits  three  important  transformaBons  /  reacBons  as  described  below:   Eutectoid  Reac&on:                                                              Solid1  ↔  Solid2  +  Solid3                                            Austenite  ↔  Ferrite  +  CemenBte     Eutec&c  Reac&on:                                            Liquid  ↔  Solid1  +  Solid2                                            Liquid  ↔  Austenite  +  CemenBte     Peritec&c  Reac&on:                                                                      Solid1  +  Liquid  ↔  Solid2                                                                    δ-­‐iron  +    Liquid  ↔  Austenite       167  
  • What  is  Pearlite?  Pearlite  is  a  two  phased  lamellar  (or  layered)  structure  composed    of  alternaBng  layers  of  ferrite  and  cemenBte  that  occurs  in  some    steels  and  cast  irons    100%  pearlite  is  formed  at  0.8%C  at  727oC  by  the  eutectoid  reacBon  /  PearliBc  transfromaBon   Eutectoid  Reac&on:                        Solid1  ↔  Solid2  +  Solid3      Austenite  ↔  Ferrite  +  CemenBte   168  
  • Development  of  microstructures  in  steel  during     slow  cooling  Eutectoid  Steel   169  
  • Hypoeutectoid  Steel   170  
  • Hypereutectoid  Steel   171  
  • Non-­‐Equilibrium  Cooling  •   Non-­‐equilibrium  cooling  leads  to  shil  in  the  transformaBon      temperatures  that  appear  on  the  phase  diagram  •   Leads  to  development  of  non-­‐equilibrium  phases  that  do  not        appear  on  the  phase  diagram       172  
  • Some  common  binary  phase  diagrams  and  important  alloys  belonging  to  these  systems   Cooper  –  Zinc  (Cu-­‐Zn)   Cooper  –  Tin  (Cu-­‐Sn)   Copper  –  Nickel  (Cu-­‐Ni)   Aluminum  –  Silicon  (Al-­‐Si)   Lead  –  Tin  (Pb-­‐Sn)   173  
  • Copper  and  copper  alloys  Proper&es:  1.  It  has  good  ducBlity  and  malleability  2.  It  has  high  electrical  and  thermal  conducBvity  3.  It  is  non-­‐magneBc  and  has  a  pleasing  reddish  colour  4.  It  has  fairly  good  corrosion  resistance  5.   It  has  good  ability  to  get  alloyed  with  other  elements  Major  copper  alloys  1.  Brass:  Alloys  of  copper  and  zinc  2.  Bronzes:  Alloys  of  copper  containing  elements  other  than  zinc                ex.  Copper-­‐Tin  alloys                            Copper-­‐Nickel  alloys     174  
  • Cooper  –  Zinc  (Cu-­‐Zn)   •   Cu-­‐Zn  alloys  exhibit  good      ducBlity  at  lower  amounts      of  Zn.     •   These  alloys  are  mostly        cast  and  formed     •   Widely  used  for  tubes  in      heat  exchangers,  cartridge    cases,  fixtures,  springs,        utensils,  pump  parts,        propeller  shals,  etc   175  
  • Cooper  –  Tin  (Cu-­‐Sn)   •   Cu-­‐Sn  alloys  exhibit  good          ducBlity  and  malleability        along  with  good  corrosion          resistance.     •   Widely  used  for  pumps,      gears,  marine  fings,        bearings,  coins    etc   176  
  • Copper  –  Nickel  (Cu-­‐Ni)   •   Cu-­‐Ni  Complete  solubility  in          each  other     •   Copper  alloy  containing  about        45%  Nickel  has  very  high          electrical  resisBvity     •   Hence  used  for  resistors  and        thermocouple  wires   177  
  • Aluminium  and  aluminium  alloys  Proper&es:  1.  It  is  ducBle  and  malleable  2.  It  is  light  in  weight  3.  It  has  good  thermal  and  electrical  conducBvity  4.  It  has  excellent  ability  to  get  alloyed  with  other  elements                  like  Cu,  Si,  Mg,  etc.  5.  It  has  excellent  corrosion  and  oxidaBon  resistance  6.  It  is  non-­‐magneBc  and  non-­‐sparking    Major  Aluminium  Alloys:  1.  Aluminium-­‐  silicon  2.  Aluminium  –  copper  3.  Aluminium-­‐  Magnesium         178  
  • Aluminum  –  Silicon  (Al-­‐Si)   •   Al-­‐Si  alloys  widely  used        for  casBngs  due  to  their      excellent  fluidity  and  casBng      characterisBcs.     •   Higher  silicon  content  gives      beper  mechanical  properBes,        beper  corrosion  resistance,      Improved  fludity     •   Widely  used  for  automobile      casBngs  like  engine  block  etc   179  
  • Lead  –  Tin  (Pb-­‐Sn)   •   Pb-­‐Sn  alloys  form  a  eutecBc      at  61.9%  Sn  at  183oC.     •   These  alloys  widely  used        as  solders  because  of  their      low  melBng  point  and  flow      characterisBcs       180