From 206d0358703aa05d5d7315900fe1d054c2817ddc Mon Sep 17 00:00:00 2001
From: Jovina Dsouza
Date: Wed, 18 Jun 2014 12:43:07 +0530
Subject: adding book

---
 .../CH10.ipynb                                     | 124 +++++++
 .../CH12.ipynb                                     | 202 +++++++++++
 .../CH14.ipynb                                     | 228 +++++++++++++
 .../CH16.ipynb                                     | 174 ++++++++++
 .../CH17.ipynb                                     | 132 ++++++++
 .../CH18.ipynb                                     | 308 +++++++++++++++++
 .../CH19.ipynb                                     |  61 ++++
 .../CH20.ipynb                                     | 157 +++++++++
 .../CH21.ipynb                                     |  64 ++++
 .../CH3.ipynb                                      | 376 +++++++++++++++++++++
 .../CH4.ipynb                                      | 161 +++++++++
 .../CH5.ipynb                                      | 302 +++++++++++++++++
 .../CH6.ipynb                                      | 342 +++++++++++++++++++
 .../CH7.ipynb                                      | 177 ++++++++++
 .../CH8.ipynb                                      | 111 ++++++
 .../CH9.ipynb                                      | 148 ++++++++
 .../README.txt                                     |  10 +
 .../screenshots/10_2.png                           | Bin 0 -> 12360 bytes
 .../screenshots/8_1.png                            | Bin 0 -> 16879 bytes
 .../screenshots/9_3.png                            | Bin 0 -> 27656 bytes
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 create mode 100644 Materials_science_and_engineering_an_introduction/CH12.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/CH14.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/CH16.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/CH17.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/CH18.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/CH19.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/CH20.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/CH21.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/CH3.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/CH4.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/CH5.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/CH6.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/CH7.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/CH8.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/CH9.ipynb
 create mode 100644 Materials_science_and_engineering_an_introduction/README.txt
 create mode 100644 Materials_science_and_engineering_an_introduction/screenshots/10_2.png
 create mode 100644 Materials_science_and_engineering_an_introduction/screenshots/8_1.png
 create mode 100644 Materials_science_and_engineering_an_introduction/screenshots/9_3.png

(limited to 'Materials_science_and_engineering_an_introduction')

diff --git a/Materials_science_and_engineering_an_introduction/CH10.ipynb b/Materials_science_and_engineering_an_introduction/CH10.ipynb
new file mode 100644
index 00000000..2a7a789c
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH10.ipynb
@@ -0,0 +1,124 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 :Phase Transformation In Metals"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.1 Page no 318"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Computation of Critical Nuclear Radius\n",
+      "\n",
+      "#Given\n",
+      "Hf=-1.16*10**9     # in J/m**3  latent heat of fusion\n",
+      "Y=0.132            # in J/m**2  Surface energy\n",
+      "Tm=1064.0+273.0        # in K  Melting point of gold\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "T=Tm-230.0           # in K  230 is supercooling value\n",
+      "r=-2*Y*Tm/(Hf*(Tm-T))\n",
+      "G=16*math.pi*Y**3*Tm**2/(3*Hf**2*(Tm-T)**2)\n",
+      "\n",
+      "#part (b)\n",
+      "a=0.413*10**-9  # in m  Unit Cell edge length\n",
+      "#unit cells per paticle\n",
+      "u_c=4*math.pi*r**3/(3*a**3)\n",
+      "n=4            #In FCC . no of atoms in per unit cell\n",
+      "U_c=int(u_c)*n\n",
+      "\n",
+      "#Result\n",
+      "print\"(a)Critical Radius is \",round(r/10**-9,2),\"nm\"\n",
+      "print\"Activation free energy is \",round(G,21),\"J\"\n",
+      "print\"(b)Total no. of atoms per critical nucleus are \",U_c\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Critical Radius is  1.32 nm\n",
+        "Activation free energy is  9.68e-19 J\n",
+        "(b)Total no. of atoms per critical nucleus are  548\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 10.2 Page no 335"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determination the rate of recrystallization\n",
+      "\n",
+      "#Given\n",
+      "n=5.0\n",
+      "y=0.3\n",
+      "t=100.0          #in min\n",
+      "\n",
+      "#Calculation\n",
+      "k=-math.log(1-y)/t**n\n",
+      "thalf=(-math.log(1-0.5)/k)**(1/n)\n",
+      "rate=1/thalf\n",
+      "\n",
+      "#Result\n",
+      "print\"Rate is \",rate,\"min**-1\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rate is  0.00875567087531 min**-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH12.ipynb b/Materials_science_and_engineering_an_introduction/CH12.ipynb
new file mode 100644
index 00000000..ee793447
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH12.ipynb
@@ -0,0 +1,202 @@
+{
+ "metadata": {
+  "name": "CH12"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12: Structure and Properties of Ceramics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.1 Page no 418"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Computation of Minimum Caion-to-Anion Radius Ratio forCo-ordination No. of 3\n",
+      "\n",
+      "#For equilateral triangle after joining centres of the atoms Angle = 30\n",
+      "a=30\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "ratio=(1-math.cos(a*math.pi/180.0))/math.cos(a*math.pi/180.0)\n",
+      "\n",
+      "#Result\n",
+      "print\"Cation to anion raio is \",round(ratio,3)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cation to anion raio is  0.155\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.2 Page no 423"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Ceramic Crystal structure prediction\n",
+      "\n",
+      "#Given\n",
+      "r_Fe=0.077  # in nm Radius of iron cation Fe++\n",
+      "r_O=0.140  #in nm  Radius of Oxygen anion O--\n",
+      "\n",
+      "#Calculation\n",
+      "ratio=r_Fe/r_O\n",
+      "\n",
+      "#Result\n",
+      "print\"Ratio is \",ratio\n",
+      "if 0.414<ratio<0.732:  \n",
+      "    print\"Co-ordinaton no. is 6\"\n",
+      "    print\"Structure is Rock Salt type\"\n",
+      "else:\n",
+      "    print \"Coordination no is not 6\"\n",
+      "    \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ratio is  0.55\n",
+        "Co-ordinaton no. is 6\n",
+        "Structure is Rock Salt type\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.3 Page no 424"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Theoretical Density Determination for NaCl\n",
+      "\n",
+      "#Given\n",
+      "A_Na=22.99         # in g/mol\n",
+      "A_Cl=35.45          #in g/mol\n",
+      "r_Na=0.102*10**-7  #in cm Radius of Na+ ion\n",
+      "r_Cl=0.181*10**-7  #in cm Radius of Cl- ion\n",
+      "Na=6.023*10**23  #Avogadro number\n",
+      "\n",
+      "#Calculation\n",
+      "a=2*(r_Na+r_Cl)\n",
+      "V=a**3\n",
+      "n=4                 #For FCC, no. of atoms are 4 per crystal\n",
+      "density=n*(A_Na+A_Cl)/(V*Na)\n",
+      "\n",
+      "#Result\n",
+      "print\"Density is \",round(density,2),\"gm/cm**3\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Density is  2.14 gm/cm**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 12.4 Page no 436"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Computation of the No. of Schottky Defects in KCl\n",
+      "\n",
+      "#Given\n",
+      "Na=6.023*10**23   #Avogadro number\n",
+      "density=1.955     #in g/cm**3\n",
+      "A_K=39.1          #in g/mol\n",
+      "A_Cl=35.45        #in g/mol\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "N=Na*density*10**6/(A_K+A_Cl)\n",
+      "Qs=2.6           # in eV\n",
+      "k=8.62*10**-5    # in eV/K  Boltzmann Constant\n",
+      "T=500.0+273.0        # in K\n",
+      "Ns=N*math.exp(-Qs/(2*k*T))\n",
+      "\n",
+      "#result\n",
+      "print\"No. of Schottky Defects are \",round(Ns,-17),\"/m**3\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "No. of Schottky Defects are  5.31e+19 /m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH14.ipynb b/Materials_science_and_engineering_an_introduction/CH14.ipynb
new file mode 100644
index 00000000..7ade4dbe
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH14.ipynb
@@ -0,0 +1,228 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14: Polymer Structure"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.1 Page No 499"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Computation of average Molecular wt and Degree of Polymerisation\n",
+      "\n",
+      "#Given\n",
+      "#For different molecular wt range ,\n",
+      "#From Table 14.1 (a)\n",
+      "Mi1=7500          #g/mol , Molecular wt\n",
+      "Mi2=12500     \n",
+      "Mi3=17500\n",
+      "Mi4=22500\n",
+      "Mi5=27500\n",
+      "Mi6=32500\n",
+      "Mi7=37500\n",
+      "xi1=0.05           #Number fraction\n",
+      "xi2=0.16\n",
+      "xi3=0.22\n",
+      "xi4=0.27\n",
+      "xi5=0.2\n",
+      "xi6=0.08\n",
+      "xi7=0.02\n",
+      "\n",
+      "#Calculation\n",
+      "xM1=Mi1*xi1\n",
+      "xM2=Mi2*xi2\n",
+      "xM3=Mi3*xi3\n",
+      "xM4=Mi4*xi4\n",
+      "xM5=Mi5*xi5\n",
+      "xM6=Mi6*xi6\n",
+      "xM7=Mi7*xi7\n",
+      "xM=xM1+xM2+xM3+xM4+xM5+xM6+xM7\n",
+      "\n",
+      "#(b)\n",
+      "nC=2          #no of carbon atoms in repeat unit\n",
+      "nH=3          #no of hydrogen atoms in repeat unit\n",
+      "nCl=1         #no of chlorine atoms in repeat unit\n",
+      "MwC=12.01     #Molecular wt of carbon\n",
+      "MwH=1.01       #Molecular wt of Hydogen\n",
+      "MwCl=35.45     #Molecular wt of  chlorine\n",
+      "m=nC*MwC+nH*MwH+nCl*MwCl     #Total wt for PVC\n",
+      "DP=xM/m\n",
+      "\n",
+      "#(c) from fig 14.3 (b)\n",
+      "wi1=0.02           #weight fraction\n",
+      "wi2=0.1\n",
+      "wi3=0.18\n",
+      "wi4=0.29\n",
+      "wi5=0.26\n",
+      "wi6=0.13\n",
+      "wi7=0.02\n",
+      "\n",
+      "#Calculation\n",
+      "wM1=Mi1*wi1\n",
+      "wM2=Mi2*wi2\n",
+      "wM3=Mi3*wi3\n",
+      "wM4=Mi4*wi4\n",
+      "wM5=Mi5*wi5\n",
+      "wM6=Mi6*wi6\n",
+      "wM7=Mi7*wi7\n",
+      "wM=wM1+wM2+wM3+wM4+wM5+wM6+wM7\n",
+      "\n",
+      "#Result\n",
+      "print\"(a)The no Av Molecular wt is\",xM,\"g/mol\"\n",
+      "print\"(b)Degree of Polymerisation is\",round(DP,0)\n",
+      "print\"(c)The weight  Av Molecular wt is\",wM,\"g/mol\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The no Av Molecular wt is 21150.0 g/mol\n",
+        "(b)Degree of Polymerisation is 338.0\n",
+        "(c)The weight  Av Molecular wt is 23200.0 g/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.2 Page No 511"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Computations of the Density\n",
+      "\n",
+      "#Given\n",
+      "Ac=12.01       #in g/mol  Molecular weight of Carbon\n",
+      "Ah=1.008       #in g/mol  molecular weight of hydrogen\n",
+      "a=7.41*10**-8  #in cm\n",
+      "b=4.94*10**-8  #in cm\n",
+      "c=2.55*10**-8  #in cm\n",
+      "Na=6.023*10**23\n",
+      "\n",
+      "#calculation\n",
+      "Vc=a*b*c\n",
+      "n=2\n",
+      "A=(2*Ac)+(4*Ah)\n",
+      "density_c=n*A/(Vc*Na)\n",
+      "\n",
+      "#(b)Percent Crystallinity of Polyethylene\n",
+      "density_a=0.870  # in g/cm**3\n",
+      "density_s=0.925  # in g/cm**3\n",
+      "pc=density_c*(density_s-density_a)*100/(density_s*(density_c-density_a))\n",
+      "\n",
+      "#Result\n",
+      "print\"(a)Density is\",round(density_c,3),\"g/cm**3\"\n",
+      "print\"(b)percentage crystallinity is\",round(pc,1),\"%\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Density is 0.998 g/cm**3\n",
+        "(b)percentage crystallinity is 46.4 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 14.3 Page No 516"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Computations of Diffusion Flux of Carbon dioxide through Plasic Beverage Container\n",
+      "\n",
+      "#function[A]= approx(V,n)\n",
+      "#    A= round(V*10**n)/10**n\n",
+      "#    funcprot(0)\n",
+      "#endfunction\n",
+      "\n",
+      "#print\"\\tExample 14.3\\n\")\n",
+      "\n",
+      "#a\n",
+      "P1=400000.0       # in Pa  Pressure inside the bottle\n",
+      "P2=400.0          # in Pa  Pressure outside the bottle\n",
+      "Pm=0.23*10**-13  #Solubility Coefficient\n",
+      "dx=0.05          # in cm Thickness of wall\n",
+      "\n",
+      "#calculation\n",
+      "J=(-Pm*(P2-P1)/dx)\n",
+      "#(b)Beverage Shell Life\n",
+      "\n",
+      "A=500           #surface area of bottle in cm**2\n",
+      "V_lose=750.0    #cm**3 STP\n",
+      "V=J*A\n",
+      "t=V_lose/round(V,5)\n",
+      "\n",
+      "#Result\n",
+      "print\"Diffusion flux is \",round(J,9),\"cm**3 STP/cm**2-s\"\n",
+      "print\"Time to escape is \",round(t/(3600*24),1),\"days\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diffusion flux is  1.84e-07 cm**3 STP/cm**2-s\n",
+        "Time to escape is  96.5 days\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH16.ipynb b/Materials_science_and_engineering_an_introduction/CH16.ipynb
new file mode 100644
index 00000000..c360c16f
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH16.ipynb
@@ -0,0 +1,174 @@
+{
+ "metadata": {
+  "name": "CH16"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 16: Composites"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 16.1 Page No 589"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Compute the modulus of elasticity\n",
+      "#(b)  find Load carried by each of fiber and matrix phase\n",
+      "#(c) Determine the strain\n",
+      "\n",
+      "#Given\n",
+      "E_gf=69.0     # in GPa  Elasticity of glass fibre\n",
+      "mf_gf=0.4     #Vol % of glass fibre\n",
+      "E_pr=3.4      # in GPa  Elasticity of poyester resin\n",
+      "mf_pr=0.6     #Vol % of polyester resin\n",
+      "\n",
+      "#Calculation\n",
+      "E_cl=(E_pr*mf_pr)+(E_gf*mf_gf)\n",
+      "Ac=250.0         #mm**2\n",
+      "sigma=50.0       #MPa\n",
+      "ratio=(E_gf*mf_gf)/(E_pr*mf_pr)    # ratio=Ff/Fm\n",
+      "Fc=Ac*sigma    #N\n",
+      "Fm=Fc/(ratio+1)\n",
+      "Ff=Fc-Fm\n",
+      "Am=mf_pr*Ac\n",
+      "Af=mf_gf*Ac\n",
+      "sigma_m=Fm/Am\n",
+      "sigma_f=Ff/Af\n",
+      "e_m=sigma_m/(E_pr*10**3)  #Strain for matrix phase\n",
+      "e_f=sigma_f/(E_gf*10**3)  #Strain for fiber phase\n",
+      "\n",
+      "#Result\n",
+      "print\"(a)Modulus of elasticity of composite is \",round(E_cl,0),\"GPa\"\n",
+      "print\"(b)Load carried by each of fiber and matrix phase is \",round(Ff,0),\"N\"\n",
+      "print\"(c)Strain for matrix phase is \",round(e_m,4)\n",
+      "print\"   Strain for fiber phase is \",round(e_f,4)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Modulus of elasticity of composite is  30.0 GPa\n",
+        "(b)Load carried by each of fiber and matrix phase is  11640.0 N\n",
+        "(c)Strain for matrix phase is  0.0017\n",
+        "   Strain for fiber phase is  0.0017\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 16.2 Page No 591"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Elastic Modulus Determination for a Glass Fiber-Reinforced Composite\u2014Transverse Direction\n",
+      "\n",
+      "#Given\n",
+      "E_gf=69      # in GPa  Elasticity of glass fibre\n",
+      "mf_gf=0.4    #Vol % of glass fibre\n",
+      "E_pr=3.4     # in GPa  Elasticity of poyester resin\n",
+      "mf_pr=0.6    #Vol % of polyester resin\n",
+      "\n",
+      "#Calculation\n",
+      "E_ct=E_pr*E_gf/((E_pr*mf_gf)+(E_gf*mf_pr))  #GPa\n",
+      "\n",
+      "#Result\n",
+      "print\"In transverse direction, modulus of elaticity is \",round(E_ct,1),\"GPa\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In transverse direction, modulus of elaticity is  5.5 GPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Design Example 16.1 ,Page No :601"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#A tubular composite shaft is to be designed.\n",
+      "\n",
+      "#Given\n",
+      "Do=70*10**-3        #mm, outside diameter\n",
+      "Di=50*10**-3        #mm, inside diameter\n",
+      "L=1                 #m Length\n",
+      "F=1000              #N load\n",
+      "dy=0.35*10**-3      #mm, deflection\n",
+      "\n",
+      "#Calculation\n",
+      "#Required longitudinal modulus of elasticity\n",
+      "E=(4*F*L**3)/(3*math.pi*dy*(Do**4-Di**4))\n",
+      "Vc=(math.pi*L*(Do**2-Di**2))/4.0\n",
+      "\n",
+      "#Result\n",
+      "print\"(a)longitudinal modulus of elasticity is\",round(E/10**9,1),\"GPa\"\n",
+      "print\"(b)The total tube volume is\",round(Vc*10**6,0),\"cm**3\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)longitudinal modulus of elasticity is 68.3 GPa\n",
+        "(b)The total tube volume is 1885.0 cm**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH17.ipynb b/Materials_science_and_engineering_an_introduction/CH17.ipynb
new file mode 100644
index 00000000..294f5ff9
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH17.ipynb
@@ -0,0 +1,132 @@
+{
+ "metadata": {
+  "name": "CH17"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 17: Corrosion and Degradiation of Materials"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.1 Page No 629"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determination of Electrochemical Cell Characteristics\n",
+      "\n",
+      "#Given\n",
+      "V_Cd=-0.403  #Half Cell Potential of Cd++|Cd\n",
+      "V_Ni=-0.250  #Half Cell Potential of Ni++|Ni\n",
+      "\n",
+      "#calculation\n",
+      "dV=V_Ni-V_Cd\n",
+      "C_Ni=10**-3\n",
+      "C_Cd=0.5\n",
+      "n=2          #Net electron exchange in Redox reaction\n",
+      "V=-dV-(0.0592*math.log10(C_Ni/C_Cd)/n)\n",
+      "\n",
+      "#Result\n",
+      "print\"Standard Cell potential is \",dV,\"V\"\n",
+      "print\"Net EMF is \",round(V,3),\"V\"\n",
+      "if V<0:\n",
+      "    print\"Ni is reduced & Cd is oxidised\"\n",
+      "else:\n",
+      "    print\"Cd is reduced & Ni is oxidised\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Standard Cell potential is  0.153 V\n",
+        "Net EMF is  -0.073 V\n",
+        "Ni is reduced & Cd is oxidised\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 17.2 Page No 637"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Rate of Oxidation Computation\n",
+      "\n",
+      "#Given\n",
+      "#Activation polarisation data for Zn\n",
+      "VZn_Zn2=-0.763\n",
+      "iZn=10**-7\n",
+      "beta_Zn=0.09\n",
+      "#For H2\n",
+      "iH2=10**-10\n",
+      "VH_H2=0\n",
+      "beta_H2=-0.08\n",
+      "\n",
+      "#calculation\n",
+      "#Part i\n",
+      "ic=10**((VH_H2-VZn_Zn2-(beta_H2*math.log10(iH2))+(beta_Zn*math.log10(iZn)))/(beta_Zn-beta_H2))\n",
+      "n=2              #Exchange of 2 electrons\n",
+      "F=96500          #Faradays constant\n",
+      "r=ic/(n*F)\n",
+      "#Part ii\n",
+      "Vc=VH_H2+(beta_H2*log10(ic/iH2))\n",
+      "\n",
+      "#Result\n",
+      "print\"i) Rate of oxiadation is\",round(r,12),\"mol/cm**2-s\"\n",
+      "print\"ii) Corrosion potential is\",round(Vc,3),\"V\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i) Rate of oxiadation is 6.18e-10 mol/cm**2-s\n",
+        "ii) Corrosion potential is -0.486 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH18.ipynb b/Materials_science_and_engineering_an_introduction/CH18.ipynb
new file mode 100644
index 00000000..016bfd26
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH18.ipynb
@@ -0,0 +1,308 @@
+{
+ "metadata": {
+  "name": "CH18"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 18 : Electrical Properties"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 18.1 Page No 682"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Computation of the Room-Temperature Intrinsic Carrier Concentration for Gallium Arsenide\n",
+      "\n",
+      "#Given\n",
+      "sigma=10**-6   # (Ohm-m)**-1  Electrical Conductivity\n",
+      "e=1.6*10**-19  #Coulomb Charge on electron\n",
+      "m_e=0.85       # m**2/V-s  Mobility of electron\n",
+      "m_h=0.04       # m**2/V-s  Mobility of holes\n",
+      "\n",
+      "#Calculation\n",
+      "#ni is Intrinsic carrier concentration\n",
+      "ni=sigma/(e*(m_e+m_h))\n",
+      "\n",
+      "print\"Intrinsic Carrier Concentration is\",round(ni,-11),\"m**-3\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Intrinsic Carrier Concentration is 7e+12 m**-3\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 18.2 Page No  689"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Electrical Conductivity Determination for Intrinsic Silicon at 150\u00b0C\n",
+      "\n",
+      "#Given\n",
+      "e=1.6*10**-19    #Coulomb Charge on electron\n",
+      "ni=4*10**19      #For Si at 423 K   (m**-3)\n",
+      "#Values of m_e and m_h are deduced from graphs at page No.689\n",
+      "m_e=0.06        #m**2/V-s  Mobility of electron\n",
+      "m_h=0.022       #m**2/V-s  Mobility of holes\n",
+      "\n",
+      "#calculation\n",
+      "#sigma is electrical conductivity\n",
+      "sigma=ni*e*(m_e+m_h)\n",
+      "\n",
+      "#result\n",
+      "print\"Electrical Conductivity is \",round(sigma,2),\"(ohm-m)**-1\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Electrical Conductivity is  0.52 (ohm-m)**-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 18.3 Page No 690"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Room-Temperature for Extrinsic Silicon\n",
+      "\n",
+      "#Given (b)\n",
+      "n=10**23 #m**-3  Carrier Concentration\n",
+      "e=1.6*10**-19  #Coulomb  Charge on electron\n",
+      "#From graph 18.18 m_e is calculated corresponding to n=10**23\n",
+      "m_e=0.07  #m**2/V-s  Mobility of electron\n",
+      "\n",
+      "#Calculation\n",
+      "#For extrinsic n-type, the formula used is:\n",
+      "sigma=n*e*m_e\n",
+      "\n",
+      "\n",
+      "#(c)Elevated-Temperature Electrical Conductivity Calculations for Extrinsic Silicon\n",
+      "#From graph 18.19a m_e2 is calculated corresponding to 373 K\n",
+      "m_e2=0.04  #m**2/V-s  Mobility of electron\n",
+      "sigma2=n*e*m_e2\n",
+      "\n",
+      "#Result\n",
+      "print\"Conductivity at n=10**23 is \",sigma,\"(Ohm-m)**-1\"\n",
+      "print\"Conductivity at T=373 K becomes \",sigma2,\"(Ohm-m)**-1\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Conductivity at n=10**23 is  1120.0 (Ohm-m)**-1\n",
+        "Conductivity at T=373 K becomes  640.0 (Ohm-m)**-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Design Example 18.1, Page No:  691"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Specify an impurity acceptor  type\n",
+      "\n",
+      "#Given\n",
+      "c=50           #ohm**-1, room temprature conductivity\n",
+      "Na1=10**22      #m**-3, assumed impurity content value\n",
+      "mu1=0.04        #m**2/Vs, assumed electrical mobility\n",
+      "e=1.6*10**-19  #Electronic charge\n",
+      "NA=6.023*10**23     #Avagadro no\n",
+      "\n",
+      "#Calculation\n",
+      "C=Na1*e*mu1    #Conductivity\n",
+      "#Decreasing  an impurity content\n",
+      "Na2=10**21      #m**-3, \n",
+      "mu2=0.045       #m**2/Vs,\n",
+      "C=Na2*e*mu2\n",
+      "#So we get conductivity = 50 at\n",
+      "Na=8*10**21 \n",
+      "#For Silicon\n",
+      "rho=2.33       # g/cm**3\n",
+      "Asi=28.09      # g/mole\n",
+      "Nsi=(NA*rho*10**6)/(Asi)\n",
+      "Ca=(Na/(Na+Nsi))*100\n",
+      "\n",
+      "#Result\n",
+      "print\"The concentration of acceptor impurities is\",round(Ca,7)\n",
+      "print\"Thus a Silicon material having conductivity 50 ohm**-1  \\nmust contain\",round(Ca,7),\"% boron,aluminium,Gallium or indium .\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The concentration of acceptor impurities is 1.6e-05\n",
+        "Thus a Silicon material having conductivity 50 ohm**-1  \n",
+        "must contain 1.6e-05 % boron,aluminium,Gallium or indium \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 18.4 Page No 693"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Hall Voltage Computation\n",
+      "\n",
+      "#Given\n",
+      "sigma=3.8*10**7  #(Ohm-m)**-1  Electrical Conductivity\n",
+      "m_e=0.0012       #m**2/V-s  Mobility of electron\n",
+      "Rh=-m_e/sigma    #Hall coefficient\n",
+      "Ix=25            #Ampere(A) Current\n",
+      "d=15*10**-3      #m Thickness\n",
+      "Bz=0.6           #Tesla  Magnetic field\n",
+      "\n",
+      "#Calculation\n",
+      "Vh=Rh*Ix*Bz/d\n",
+      "\n",
+      "#Result\n",
+      "print\"Hall coefficient is \",round(Rh,13),\"V-m/A-Tesla\"\n",
+      "print\"Hall Voltage is \",round(Vh,10),\"V\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hall coefficient is  -3.16e-11 V-m/A-Tesla\n",
+        "Hall Voltage is  -3.16e-08 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 18.5 Page No 707"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the Capacitance \n",
+      "\n",
+      "#Given\n",
+      "A=6.45*10**-4    #m**2, area\n",
+      "d=2*10**-3       #m. Plate separation\n",
+      "V=10             #V Potential\n",
+      "Er=6             #Dielectric constant\n",
+      "Eo=8.85*10**-12  #F/m  Constant dielectric constant\n",
+      "#Calculation\n",
+      "E=Er*Eo\n",
+      "C=E*A/d\n",
+      "Q=C*V\n",
+      "D=E*V/d\n",
+      "P=D-Eo*V/d\n",
+      "\n",
+      "#Result\n",
+      "print\"The Capacitance is\",round(C,13),\"F\"\n",
+      "print\"The magnitude of charge stored is \",round(Q,12),\"C\"\n",
+      "print\"The Dielectric displacement is is\",round(D,9),\"C/m**2\"\n",
+      "print\"The Polarization is\",round(P,9),\"C/m**2\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Capacitance is 1.71e-11 F\n",
+        "The magnitude of charge stored is  1.71e-10 C\n",
+        "The Dielectric displacement is is 2.65e-07 C/m**2\n",
+        "The Polarization is 2.21e-07 C/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH19.ipynb b/Materials_science_and_engineering_an_introduction/CH19.ipynb
new file mode 100644
index 00000000..8464a899
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH19.ipynb
@@ -0,0 +1,61 @@
+{
+ "metadata": {
+  "name": "CH19"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 19: Thermal Properties"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 19.1 Page No  733"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation of maximum temperature\n",
+      "\n",
+      "#Given\n",
+      "To=20         # Room Temperature (degree celsius)\n",
+      "sigma=-172    #Mpa  Compressive stress\n",
+      "E=100*10**3   #Mpa Young's modulus\n",
+      "a=20*10**-6   #Celsius**-1  Coefficient of thermal expansion\n",
+      "\n",
+      "#Calculation\n",
+      "Tf=To-(sigma/(E*a))\n",
+      "\n",
+      "print\"Final Temperature is \",Tf,\"C\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Final Temperature is  106.0 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH20.ipynb b/Materials_science_and_engineering_an_introduction/CH20.ipynb
new file mode 100644
index 00000000..c410a6ac
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH20.ipynb
@@ -0,0 +1,157 @@
+{
+ "metadata": {
+  "name": "CH20"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 20 : Magnetic Properties"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 20.1 Page No  84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation of saturation magnetisation and flux density for Nickel\n",
+      "\n",
+      "#Given\n",
+      "b_m=9.27*10**-24   #ampere*m**2  (Bohr Magneton)\n",
+      "Na=6.023*10**23    #atoms/mol  (Avogadro's No.)\n",
+      "d=8.9*10**6        #g/m**3   (density)\n",
+      "uo=4*math.pi*10**-7  #Permitivity of free space\n",
+      "A=58.71             #g/mol  (Atomic weigth of Nickel)\n",
+      "N=d*Na/A            #No. of atoms per cubic meter\n",
+      "\n",
+      "#Calculation\n",
+      "# M is saturation magnetisation\n",
+      "M=0.6*b_m*N  #0.6= Bohr Magneton/atom\n",
+      "#B = Saturation Flux Density\n",
+      "B=uo*M\n",
+      "\n",
+      "#Result\n",
+      "print\"Saturation Magnetisation is \",M,\"A/m\"\n",
+      "print\"Saturation Flux Density is \",round(B,2),\"Tesla\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Saturation Magnetisation is  507834.0 A/m\n",
+        "Saturation Flux Density is  0.64 Tesla\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 20.2 Page No 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation of saturation magnetisation of Fe3O4\n",
+      "\n",
+      "#Given\n",
+      "a=0.839*10**-9   #a is edge length in m\n",
+      "b_m=9.27*10**-24 #ampere*m**2  (Bohr Magneton)\n",
+      "nb=8*4           #8 is no. of Fe++ ions per unit cell\n",
+      "                 #4 is Bohr magnetons per Fe++ ion\n",
+      "#Calculation\n",
+      "M=nb*b_m/a**3  #M is Saturation magnetisation\n",
+      "\n",
+      "print\"Saturation Magnetisation is \",round(M,0),\"A/m\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Saturation Magnetisation is  502278.0 A/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Design Example 20.1 Page No 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design Example 20.1: Designing a cubic mixed-ferrite magnetic material\n",
+      "\n",
+      "#Given\n",
+      "Ms_Fe=5.25*10**5   #Required saturation Magnetisation\n",
+      "b_m=9.27*10**-24   #ampere*m**2  (Bohr Magneton)\n",
+      "a=0.839*10**-9     #a is edge length in m\n",
+      "M=5*10**5          #From previous question result\n",
+      "\n",
+      "#Calculation\n",
+      "nb=Ms_Fe*a**3/b_m\n",
+      "i=8             # No of Divalent ions per unit cell\n",
+      "j=4             #4 is Bohr magnetons per Mn++ ion\n",
+      "n=nb/(i)-j \n",
+      "         \n",
+      "#Result\n",
+      "print\"Replacing percent of Fe++ with Mn++ would produce the required saturation magnetisation\",round(n*100,2)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Replacing percent of Fe++ with Mn++ would produce the required saturation magnetisation 18.1\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH21.ipynb b/Materials_science_and_engineering_an_introduction/CH21.ipynb
new file mode 100644
index 00000000..bb8267f6
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH21.ipynb
@@ -0,0 +1,64 @@
+{
+ "metadata": {
+  "name": "CH21"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 21: Optical Properties"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 21.1 Page No 125"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation of absorption coefficient\n",
+      "\n",
+      "#Given\n",
+      "# x is thickness of glass(mm)\n",
+      "x=200\n",
+      "#It is intensity of non-absorbed radiation\n",
+      "#Io is intensity of non-relected radiation\n",
+      "f=0.98  #f=It/Io\n",
+      "\n",
+      "#Calculation\n",
+      "#b is absorption coefficient\n",
+      "b=-math.log(f)/x\n",
+      "\n",
+      "#result\n",
+      "print\"Absorption coefficient is \",round(b,4),\"mm**-1\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Absorption coefficient is  0.0001 mm**-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH3.ipynb b/Materials_science_and_engineering_an_introduction/CH3.ipynb
new file mode 100644
index 00000000..c521d51b
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH3.ipynb
@@ -0,0 +1,376 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3: Metallic Crystal Structure"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.1 Page No: 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determination of FCC Unit Cell Volume\n",
+      "\n",
+      "#Given\n",
+      "#For FCC a=2*R*math.sqrt(2)\n",
+      "from sympy import Symbol\n",
+      "\n",
+      "#Calculation \n",
+      "R=Symbol('R')   \n",
+      "#Edge Length\n",
+      "a=2*R*round(math.sqrt(2),2)\n",
+      "#Volume determination\n",
+      "V=a**3\n",
+      "\n",
+      "#result\n",
+      "print\"Volume is\",V,\"  m**3\"\n",
+      "print\"which is also equal to 16*sqrt(2)*R**3\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volume is 22.425768*R**3   m**3\n",
+        "which is also equal to 16*sqrt(2)*R**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.2 Page No: 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Computation of the Atomic Packing Factor for FCC\n",
+      "\n",
+      "#Given\n",
+      "#for FCC no. of atoms are 4\n",
+      "n=4\n",
+      "#For FCC a=2*R*math.sqrt(2)\n",
+      "R=1        #say\n",
+      "\n",
+      "#Calculation\n",
+      "#Edge Length\n",
+      "a=2*R*math.sqrt(2)\n",
+      "#Volume determination of cube\n",
+      "Vc=a**3\n",
+      "#Volume of sphere\n",
+      "Vs=n*4*math.pi*R**3/3.0\n",
+      "#Atomic packing Fraction\n",
+      "APF=Vs/Vc\n",
+      "\n",
+      "#Result\n",
+      "print\"Atomic packing fraction is\",round(APF,2)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Atomic packing fraction is 0.74\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.3 Page No: 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Theoretical Density Computation for Copper\n",
+      "\n",
+      "#Given\n",
+      "R=1.28*10**-8        #Atomic radius in cm\n",
+      "A_Cu=63.5            #Atomic wt of copper\n",
+      "n=4                  #For FCC\n",
+      "Na=6.023*10**23       #Avogadro no.\n",
+      "\n",
+      "#Calculation\n",
+      "a=2*R*math.sqrt(2)\n",
+      "Vc=a**3\n",
+      "den=n*A_Cu/(Vc*Na)\n",
+      "\n",
+      "#result\n",
+      "print\"Density is  \",round(den,2),\"g/cm**3\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Density is   8.89 g/cm**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.6 Page No: 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determination of Directional Indices\n",
+      "\n",
+      "#Given\n",
+      "#Projection of given vector\n",
+      "a=1/2.0\n",
+      "b=1\n",
+      "c=0\n",
+      "\n",
+      "x=[2*a,2*b,2*c]\n",
+      "\n",
+      "#Result\n",
+      "print\"The intercept for the given plane is\",x\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The intercept for the given plane is [1.0, 2, 0]\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.8 Page No: 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determination of Directional Indices for a Hexagonal Unit Cell\n",
+      "\n",
+      "#Given\n",
+      "#Projection in terms of unit cell parameter\n",
+      "du=1\n",
+      "dv=1\n",
+      "dw=1\n",
+      "\n",
+      "#Calculation\n",
+      "#For hexagonal system\n",
+      "u=(2*du-dv)/3.0\n",
+      "v=(2*dv-du)/3.0\n",
+      "t=-(u+v)\n",
+      "w=dw\n",
+      "\n",
+      "x=[3*u,3*v,3*t,3*w]\n",
+      "print\"The indices for the given directions are\",x\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The indices for the given directions are [1.0, 1.0, -2.0, 3]\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.9 Page No: 56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determination of Planar (Miller) Indices\n",
+      "\n",
+      "#Given\n",
+      "a=-1\n",
+      "b=1/2.0\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "#Reciprocal\n",
+      "l=0      #Reciprocal of infinity\n",
+      "m=1/a\n",
+      "n=1/b\n",
+      "x=[l,m,n]\n",
+      "\n",
+      "#Result\n",
+      "print\"The intercept for the given plane is\",x\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The intercept for the given plane is [0, -1, 2.0]\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.11 Page No: 59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determination of Miller\u2013Bravais Indices for a Plane Within a Hexagonal Unit Cell\n",
+      "\n",
+      "#Intersection in terms of lattics Parameters\n",
+      "h=1        #Reciprocal of intersection point\n",
+      "k=-1\n",
+      "l=1\n",
+      "i=-(h+k)\n",
+      "\n",
+      "#Calculation\n",
+      "x=[h,k,i,l]\n",
+      "\n",
+      "#Result\n",
+      "print\"The indices of plane are\",x\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The indices of plane are [1, -1, 0, 1]\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 3.12 Page No: 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Interplanar Spacing\n",
+      "\n",
+      "#Given\n",
+      "a=0.2866           #Lattice parameter in nm\n",
+      "h=2\n",
+      "k=2\n",
+      "l=0\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "#(a)\n",
+      "d_hkl=a/(math.sqrt(h**2+k**2+l**2))\n",
+      "\n",
+      "#(b)Diffraction Angle Computations\n",
+      "lam=0.1790     #Wavelength in nm\n",
+      "n=1\n",
+      "theta=math.asin(n*lam/(2*d_hkl))\n",
+      "\n",
+      "#Result\n",
+      "print\"(a)Interplanar spacing is  \",round(d_hkl,4),\"nm\"\n",
+      "print\"(b)Diffraction angle is  \",round(2*theta*(180/math.pi),1),\"degree\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Interplanar spacing is   0.1013 nm\n",
+        "(b)Diffraction angle is   124.1 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH4.ipynb b/Materials_science_and_engineering_an_introduction/CH4.ipynb
new file mode 100644
index 00000000..93bc9a2e
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH4.ipynb
@@ -0,0 +1,161 @@
+{
+ "metadata": {
+  "name": "CH4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4:Imperfections in solids"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.1 Page No: 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Number of Vacancies Computation at a Specified temperature\n",
+      "\n",
+      "#Given\n",
+      "Na=6.023*10**23      #Avogadro No.\n",
+      "den=8.4*10**6        #Density of Copper\n",
+      "A=63.5              #Atomic weight of Copper\n",
+      "\n",
+      "#Calculation\n",
+      "#No. of atomic site per cubic meter\n",
+      "N=Na*den/A\n",
+      "#No. of vacancies at 1000 C\n",
+      "Qv=0.9                 #Activation energy in eV\n",
+      "k=8.62*10**-5          # Boltzmann Constatnt in eV/K\n",
+      "T=1000.0+273.0         #Temperature in K\n",
+      "Nv=N*exp(-Qv/(k*T))\n",
+      "\n",
+      "#Result\n",
+      "print\"No.of vacancies are \",round(Nv,-24),\"/m**3\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "No.of vacancies are  2.2e+25 /m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.3 Page No: 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Composition Conversion- From weight percent to Atom percent\n",
+      "\n",
+      "#Given\n",
+      "#Conversion to Atom percent\n",
+      "C_Al=97.0       #Aluminium wt%\n",
+      "C_Cu=3.0        #Copper wt%\n",
+      "A_Al=26.98      #Atomic wt of Aluminium\n",
+      "A_Cu=63.55      #Atomic wt of Copper\n",
+      "\n",
+      "CAl=(C_Al*A_Cu)/(C_Al*A_Cu+C_Cu*A_Al)\n",
+      "CCu=(C_Cu*A_Al)/(C_Al*A_Cu+C_Cu*A_Al)\n",
+      "\n",
+      "print\"Atomic % of Al is \",round(CAl*100,1),\"%\"\n",
+      "print\"Atomic % of Cu is \",round(CCu*100,1),\"%\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Atomic % of Al is  98.7 %\n",
+        "Atomic % of Cu is  1.3 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 4.4 Page No: 103"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Computations of ASTM Grain Size Number\n",
+      "\n",
+      "#Given\n",
+      "N=45.0           #No. of grains per square inch\n",
+      "\n",
+      "#Calculation\n",
+      "#Dterminin grain size no.  N=2**(n-1)\n",
+      "n=(math.log(N)/math.log(2))+1\n",
+      "\n",
+      "#Number of Grains Per Unit Area\n",
+      "M=85.0\n",
+      "Nm=(100/M)**2*2**(n-1)\n",
+      "\n",
+      "#Result\n",
+      "print\"Grain size no. is \",round(n,1)\n",
+      "print\"No. of grains per inch square are \",round(Nm,1)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Grain size no. is  6.5\n",
+        "No. of grains per inch square are  62.3\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH5.ipynb b/Materials_science_and_engineering_an_introduction/CH5.ipynb
new file mode 100644
index 00000000..5b193d64
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH5.ipynb
@@ -0,0 +1,302 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Diffusion"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.1 Page No. 114"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Diffusion Flux Computation\n",
+      "\n",
+      "#Given\n",
+      "Ca=1.2       #Concentration at A in  kg/m**3\n",
+      "Cb=0.8       #Concentration at B in  kg/m**3\n",
+      "xa=5*10**-3   #Position 1 in m\n",
+      "xb=10*10**-3  #Position 2 in m\n",
+      "\n",
+      "#Calculation\n",
+      "D=3*10**-11   #Diffusion coefficient in m**2/s\n",
+      "J=-D*(Ca-Cb)/(xa-xb)\n",
+      "\n",
+      "#Result\n",
+      "print\"Diffusion flux is \",J,\"kg/m**2-s\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diffusion flux is  2.4e-09 kg/m**2-s\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.2 Page No. 117"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Nonsteady-State Diffusion Time Computation I\n",
+      "from scipy.optimize import fsolve\n",
+      "#Given\n",
+      "Co=0.25        #Initial Conc. in wt%\n",
+      "Cs=1.2         #Surface conc. in wt%\n",
+      "Cx=0.8         #Conc. at any x  in wt%\n",
+      "x=5*10**-4     #Position in m\n",
+      "D=1.6*10**-11  #Diffusion coeff in m**2/s\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "C=1-((Cx-Co)/(Cs-Co))\n",
+      "\n",
+      "def f(z):\n",
+      "    return(0.4210-math.erf(z))\n",
+      "z=fsolve(f,1)\n",
+      "t=x**2/(4.0*D*z**2.0)\n",
+      "\n",
+      "#Result\n",
+      "print\"Time required is \",round(t/3600.0,1),\"h\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time required is  7.0 h\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.3 Page No. 118"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Nonsteady-State Diffusion Time Computation II\n",
+      "\n",
+      "#Given\n",
+      "D500=4.8*10**-14     #Diffusion coefficient at 500 C\n",
+      "D600=5.3*10**-13     #Diffusion coefficient at 600 C\n",
+      "t600=10             #Time in hours to diffuse\n",
+      "\n",
+      "#Calcuation\n",
+      "t500=D600*t600/D500\n",
+      "\n",
+      "#Result\n",
+      "print\"Time to diffuse at 500 C is \",round(t500,1),\"h\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time to diffuse at 500 C is  110.4 h\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.4 Page No. 121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Diffusion Coefficient Determination\n",
+      "\n",
+      "#Given\n",
+      "T=550+273            #in K\n",
+      "D0=1.2*10**-4        #Temperature independent preexponential in m**2/s\n",
+      "Qd=131000            #Activation energy in J/mol-K\n",
+      "R=8.31               #Universal Gas constt\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "D=D0*math.exp(-Qd/(R*T))\n",
+      "\n",
+      "#Result\n",
+      "print\"Diffusion coefficient is \",round(D,14),\"m**2/s\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diffusion coefficient is  5.8e-13 m**2/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 5.5 Page No.121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Diffusion Coefficient Activation Energy and Preexponential Calculations\n",
+      "\n",
+      "#Given\n",
+      "#From graph log D ad 1/T are deducted\n",
+      "inv_T1=0.8*10**-3          #Reciprocal of temp.  in K**-1\n",
+      "inv_T2=1.1*10**-3          #Reciprocal of temp.  in K**-1\n",
+      "logD1=-12.4\n",
+      "logD2=-15.45\n",
+      "R=8.31                   #Gas law Constant in J/mol-K\n",
+      "\n",
+      "#Calculation\n",
+      "Qd=-2.3*R*(logD1-logD2)/(inv_T1-inv_T2)\n",
+      "print\"Activation energy is\",round(Qd/1000,0),\"KJ\"\n",
+      "\n",
+      "#For calculating Peexponential factor\n",
+      "D0=10**(logD2+(Qd*inv_T2/(2.3*R)))\n",
+      "print\"Preexponential factor is\",round(D0,6),\"m**2/s\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Activation energy is 194.0 KJ\n",
+        "Preexponential factor is 5.4e-05 m**2/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Design Example 5.1, Page No.122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Diffusion Temperature\u2013Time Heat Treatment Specification\n",
+      "\n",
+      "#Given\n",
+      "C0=0.2           #Initial concentration in wt%\n",
+      "Cs=1             #Surface conc in wt%\n",
+      "Cx=0.6           #Conc at any position X in wt%\n",
+      "x=7.5*10**-4      #Position in m\n",
+      "D0=2.3*10**-5     #Preexponential factor in m**2/s\n",
+      "R=8.31           #Gas law constant in J/mol-K\n",
+      "Qd=148000        #Activation energy in J/mol\n",
+      "\n",
+      "C=1-((Cx-C0)/(Cs-C0))\n",
+      "z=0.4747\n",
+      "Dt=(x/(2*z))**2\n",
+      "\n",
+      "#Dt=D0*exp(-Qd/RT)*t = value of variable Dt\n",
+      "D=Dt/D0\n",
+      "\n",
+      "#The tempratures are\n",
+      "T1=900.0\n",
+      "T2=950.0\n",
+      "T3=1000.0\n",
+      "T4=1050.0\n",
+      "t1=D/math.exp(-Qd/(R*(T1+273)))/3600.0\n",
+      "t2=D/math.exp(-Qd/(R*(T2+273)))/3600.0\n",
+      "t3=D/math.exp(-Qd/(R*(T3+273)))/3600.0\n",
+      "t4=D/math.exp(-Qd/(R*(T4+273)))/3600.0\n",
+      "\n",
+      "#Result\n",
+      "print\"Temperature in Celcius  are\",T1,T2,T3,T4\n",
+      "print\"Time is respectively \",round(t1,1),\"h,\",round(t2,1),\"h,\",round(t3,1),\"h,\",round(t4,1),\"h\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature in Celcius  are 900.0 950.0 1000.0 1050.0\n",
+        "Time is respectively  29.6 h, 15.9 h, 9.0 h, 5.3 h\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH6.ipynb b/Materials_science_and_engineering_an_introduction/CH6.ipynb
new file mode 100644
index 00000000..e04ca9e8
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH6.ipynb
@@ -0,0 +1,342 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6: Mechanical Properties of Metal"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.1 Page No 140"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Elongation (Elastic) Computation\n",
+      "\n",
+      "E=110*10**3  #Young's modulus of Copper in MPa\n",
+      "sigma=276.0    #Applied stress in MPa\n",
+      "lo=305.0       #Original length in mm\n",
+      "\n",
+      "#Calculation\n",
+      "#Deformation\n",
+      "dl=sigma*lo/E\n",
+      "\n",
+      "#Result\n",
+      "print\"Elongation obtained is  \",round(dl,2),\"mm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Elongation obtained is   0.77 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.2 Page No 142"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Computation of Load to Produce Specified Diameter Change\n",
+      "\n",
+      "#Given\n",
+      "del_d=-2.5*10**-3  #Deformation in dia  in mm\n",
+      "d0=10.0             #Initial dia  in mm\n",
+      "v=0.34           #Poisson ratio for brass\n",
+      "\n",
+      "#Calculation\n",
+      "ex=del_d/d0\n",
+      "ez=-ex/v\n",
+      "E=97*10**3        #Modulus of elasticity in MPa\n",
+      "sigma=ez*E\n",
+      "F=sigma*math.pi*(d0**2)/4.0\n",
+      "\n",
+      "#Result\n",
+      "print\"Applied force is \",round(F,0),\"N\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Applied force is  5602.0 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.3 Page No 146"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#From draph in the question\n",
+      "#stress and strain can be obtained\n",
+      "\n",
+      "#Given\n",
+      "si2=150           # in MPa\n",
+      "si1=0\n",
+      "e2=0.0016\n",
+      "e1=0\n",
+      "d0=12.8*10**-3    #Initial Diameter in m\n",
+      "\n",
+      "#Calculation\n",
+      "#(a)Young's Modulus = stress/strain\n",
+      "E=(si2-si1)/(e2-e1)\n",
+      "\n",
+      "A0=math.pi*d0**2/4.0\n",
+      "sig=450*10**6     #tensile strength in MPa\n",
+      "F=sig*A0\n",
+      "#From stress-strain curve\n",
+      "#Strain corresponding to stress of 345 MPa is 0.06\n",
+      "l0=250     #Initial lengt in mm\n",
+      "e=0.06     #strain\n",
+      "dl=e*l0\n",
+      "\n",
+      "#Result\n",
+      "print\"Modulus of elasticity is \",round(E/10**3,1),\"GPa\"\n",
+      "print\"From the graph the Yield strength is\",l0,\"MPa\"\n",
+      "print\"Maximum load sustained is \",round(F,0),\"N/n\"\n",
+      "print\"Change in length is \",dl,\"mm\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Modulus of elasticity is  93.8 GPa\n",
+        "From the graph the Yield strength is 250 MPa\n",
+        "Maximum load sustained is  57906.0 N/n\n",
+        "Change in length is  15.0 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.4 Page No 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Ductility\n",
+      "\n",
+      "#Given\n",
+      "di=12.8     #Initial dia in mm\n",
+      "df=10.7     #Final dia  in mm\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "#Ductility in terms of Reduction Area \n",
+      "RA = ((di**2-df**2)/di**2)*100\n",
+      "#True-Stress-At-Fracture Computations\n",
+      "Ao=math.pi*di**2*10**-6/4.0\n",
+      "sig=460*10**6    #Tensile strength\n",
+      "\n",
+      "F=sig*Ao\n",
+      "\n",
+      "Af=math.pi*df**2/4.0\n",
+      "sig_t=F/Af\n",
+      "\n",
+      "#Result\n",
+      "print\"percent reduction in area is \",round(RA,0),\"%\"\n",
+      "print\"True stress is \",round(sig_t,1),\"MPa\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "percent reduction in area is  30.0 %\n",
+        "True stress is  658.3 MPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.5 Page No 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculation of Strain-Hardening Exponent\n",
+      "\n",
+      "sig_t=415       #True stress in MPa\n",
+      "et=0.1          #True strain\n",
+      "K=1035.0         # In MPa\n",
+      "\n",
+      "#Calculation\n",
+      "n=(math.log(sig_t)-math.log(K))/math.log(et)\n",
+      "\n",
+      "#Result\n",
+      "print\"Strain - hardening coefficient is \",round(n,2)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Strain - hardening coefficient is  0.4\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 6.6 Page No 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Average Computations\n",
+      "\n",
+      "#Tensile strength at 4 points\n",
+      "n=4.0        #No of points\n",
+      "T1=520\n",
+      "T2=512\n",
+      "T3=515\n",
+      "T4=522\n",
+      "\n",
+      "#Calculation\n",
+      "Tav=(T1+T2+T3+T4)/n\n",
+      "s=(((T1-Tav)**2+(T2-Tav)**2+(T3-Tav)**2+(T4-Tav)**2)/(n-1))**(0.5)\n",
+      "\n",
+      "#Result\n",
+      "print\"The average Tensile strength is\",round(Tav,0),\"MPa\"\n",
+      "print\"The standard deviation  is\",round(s,1),\"MPa\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average Tensile strength is 517.0 MPa\n",
+        "The standard deviation  is 4.6 MPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Design Example 6.1 ,Page No 164"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Specification of Support Post Diameter\n",
+      "\n",
+      "#Given\n",
+      "sig_y=310.0      #Minimum yield strength in MPa\n",
+      "N=5.0            # Conservative factor of safety\n",
+      "\n",
+      "#Calculation\n",
+      "F=220000/2.0    #Two rods must support half of the total force\n",
+      "sig_w=sig_y/N\n",
+      "d=2*math.sqrt(F/(math.pi*sig_w))\n",
+      "\n",
+      "#Result\n",
+      "print\"Diameter of each of the two rods is \",round(d,1),\"mm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter of each of the two rods is  47.5 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH7.ipynb b/Materials_science_and_engineering_an_introduction/CH7.ipynb
new file mode 100644
index 00000000..ffef5715
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH7.ipynb
@@ -0,0 +1,177 @@
+{
+ "metadata": {
+  "name": "CH7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7: Dislocation and Strengthening Mechanisms"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.1 Page no 183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Resolved Shear Stress Computations\n",
+      "\n",
+      "#Direction for given plane\n",
+      "u1=1\n",
+      "v1=1\n",
+      "w1=0\n",
+      "u2=0\n",
+      "v2=1\n",
+      "w2=0\n",
+      "#For lamda\n",
+      "u3=-1\n",
+      "v3=1\n",
+      "w3=1\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "phi=math.acos((u1*u2+v1*v2+w1*w2)/(math.sqrt((u1**2+v1**2+w1**2)*(u2**2+v2**2+w2**2))))\n",
+      "lam=math.acos((u3*u2+v3*v2+w3*w2)/(math.sqrt((u3**2+v3**2+w3**2)*(u2**2+v2**2+w2**2))))\n",
+      "sigma=52       #in MPa, Tensile stress\n",
+      "tr=sigma*math.cos(phi)*math.cos(lam)\n",
+      "trc=30         #in MPa Critical resolved shear stress\n",
+      "sy=trc/(math.cos(phi)*math.cos(lam))\n",
+      "\n",
+      "#Result\n",
+      "print\"(a)The resolved shear stress is \",round(tr,2),\"MPa\"\n",
+      "print\"(b)Yield strength is\",round(sy,1),\"MPa\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The resolved shear stress is  21.23 MPa\n",
+        "(b)Yield strength is 73.5 MPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 7.2 Page no 194"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Tensile Strength and Ductility Determinations for Cold-Worked Copper\n",
+      "\n",
+      "#Given\n",
+      "df=12.2  #Final dia in mm\n",
+      "di=15.2  #Initial dia in mm\n",
+      "\n",
+      "#Calculation\n",
+      "CW = ((di**2-df**2)/di**2)*100\n",
+      "ts=340  #in Mpa  tensile strength, from fig 7.19 (b)\n",
+      "duc=7   #in %  Ductility from fig 7.19 (c)\n",
+      "\n",
+      "#result\n",
+      "print\"Percent Cold Work is \",round(CW,1),\"%\"\n",
+      "print\"Tensile strength is\",ts,\"MPa\"\n",
+      "print\"Ductility is \",duc,\"%\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percent Cold Work is  35.6 %\n",
+        "Tensile strength is 340 MPa\n",
+        "Ductility is  7 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Design Example 7.3 Page no 199"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Description of Diameter Reduction Procedure\n",
+      "\n",
+      "#Given\n",
+      "di=6.4  #Initial dia in mm\n",
+      "df=5.1  #Final dia in mm\n",
+      "\n",
+      "#Calculation\n",
+      "#Cold Work Computation\n",
+      "CW = ((di**2-df**2)/di**2)*100\n",
+      "#From Figures 7.19a and 7.19c, \n",
+      "#A yield strength of 410 MPa \n",
+      "#And a ductility of 8% EL are attained from this deformation\n",
+      "dmid = math.sqrt(df**2/(1-0.215))\n",
+      "\n",
+      "#Result\n",
+      "print\"Cold work is \",round(CW,1),\"%\"\n",
+      "print\"But required ductility and yield strength is not matched at this cold work\"\n",
+      "print\"Hence required Cold work is 21.5 %\"\n",
+      "print\"Hence original diameter for second drawing is \",round(dmid,1),\"mm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cold work is  36.5 %\n",
+        "But required ductility and yield strength is not matched at this cold work\n",
+        "Hence required Cold work is 21.5 %\n",
+        "Hence original diameter for second drawing is  5.8 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH8.ipynb b/Materials_science_and_engineering_an_introduction/CH8.ipynb
new file mode 100644
index 00000000..682a584f
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH8.ipynb
@@ -0,0 +1,111 @@
+{
+ "metadata": {
+  "name": "CH8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8: Failure"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 8.1 Page no 217"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Maximum Flaw Length Computation\n",
+      "\n",
+      "#Given\n",
+      "sigma=40*10**6    # in Pa  Tensile stress\n",
+      "E=69*10**9        #Modulus of elaticity  in pa\n",
+      "Ys=0.3            #Specific surface energy  in N/m**2\n",
+      "\n",
+      "#Calculation\n",
+      "#Maximum length of a surface flaw\n",
+      "a=2*E*Ys/(math.pi*sigma**2)\n",
+      "\n",
+      "#Result\n",
+      "print\"Maximum lemgth of a surface flaw without fracture is \",round(a*10**6,1),\"micro m\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum lemgth of a surface flaw without fracture is  8.2 micro m\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Design Example 8.2 Page no 242"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Rupture Lifetime Prediction\n",
+      "\n",
+      "#Given\n",
+      "T=800.0+273.0      # Temperature in K\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "#From Graph of Fig. 8.32 Larson-Miller Parameter is deduced\n",
+      "L_M=24*10**3\n",
+      "#math.log(t)=((L_M/T)-20)\n",
+      "t=math.pow(10,(L_M/T)-20)\n",
+      "    \n",
+      "#result\n",
+      "print\"Time to rupture is \",round(t,0),\"h\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time to rupture is  233.0 h\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/CH9.ipynb b/Materials_science_and_engineering_an_introduction/CH9.ipynb
new file mode 100644
index 00000000..0953b608
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/CH9.ipynb
@@ -0,0 +1,148 @@
+{
+ "metadata": {
+  "name": "CH9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9: Phase Diagram"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.3 Page No  273"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Determination of Phases Present\n",
+      "#(b)Computation of Phase Compositions\n",
+      "\n",
+      "#Given\n",
+      "C1=40.0    # Overall alloy composition\n",
+      "Cb=98.0\n",
+      "Ca=10.0\n",
+      "\n",
+      "#Calculation\n",
+      "Wa=(Cb-C1)/(Cb-Ca)\n",
+      "Wb=(C1-Ca)/(Cb-Ca)\n",
+      "d_Sn=7.24        # in g/cm**3  density of tin\n",
+      "d_Pb=11.23       # in g/cm**3  density of lead\n",
+      "Ca_Sn=10.0\n",
+      "Ca_Pb=90.0\n",
+      "Cb_Sn=98.0\n",
+      "Cb_Pb=2.0\n",
+      "d_a=100/((Ca_Sn/d_Sn)+(Ca_Pb/d_Pb))\n",
+      "d_b=100/((Cb_Sn/d_Sn)+(Cb_Pb/d_Pb))\n",
+      "Va=Wa/(d_a*((Wa/d_a)+(Wb/d_b)))\n",
+      "Vb=Wb/(d_b*((Wa/d_a)+(Wb/d_b)))\n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print\"Mass fractions for alpha and beta phases are  respectively\",round(Wa,2),\"and\",round(Wb,2)\n",
+      "print\"Density of alpha phase is  \",round(d_a,2),\"g/cm**3\"\n",
+      "print\"Density of beta phase is   \",round(d_b,2),\"g/cm**3\"\n",
+      "print\"Volume fraction of alpha phase \",round(Va,2),\"g/cm**3\"\n",
+      "print\"Volume fraction of beta phase  \",round(Vb,2)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mass fractions for alpha and beta phases are  respectively 0.66 and 0.34\n",
+        "Density of alpha phase is   10.64 g/cm**3\n",
+        "Density of beta phase is    7.29 g/cm**3\n",
+        "Volume fraction of alpha phase  0.57 g/cm**3\n",
+        "Volume fraction of beta phase   0.43\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.4 Page No 299"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Determining ferrite and cementite phase\n",
+      "#(b)Determining proeutectoid ferrite and pearlite\n",
+      "#(c)Determining eutectoid ferrite\n",
+      "\n",
+      "#Given\n",
+      "C0=0.35\n",
+      "Ca=0.022\n",
+      "C_Fe3C=6.7\n",
+      "\n",
+      "#Calculation\n",
+      "Wa=(C_Fe3C-C0)/(C_Fe3C-Ca)\n",
+      "W_Fe3C=(C0-Ca)/(C_Fe3C-Ca)\n",
+      "C_p=0.76\n",
+      "Wp=(C0-Ca)/(C_p-Ca)\n",
+      "W_a=(C_p-C0)/(C_p-Ca)\n",
+      "Wp=(C0-Ca)/(C_p-Ca)\n",
+      "W_a=(C_p-C0)/(C_p-Ca)\n",
+      "Wa=(C_Fe3C-C0)/(C_Fe3C-Ca)\n",
+      "Wae=Wa-W_a\n",
+      "\n",
+      "#result\n",
+      "print\"Mass fraction of total ferritic phase is\",round(Wa,2)\n",
+      "print\"Mass fraction of Fe3C is\",round(W_Fe3C,2)\n",
+      "print\"Mass fraction of Pearlite is\",round(Wp,2)\n",
+      "print\"ass fraction of proeutectoid ferrite is\",round(W_a,2)\n",
+      "print\"Mass fraction of eutectoid ferrite is\",round(Wae,2)\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mass fraction of total ferritic phase is 0.95\n",
+        "Mass fraction of Fe3C is 0.05\n",
+        "Mass fraction of Pearlite is 0.44\n",
+        "ass fraction of proeutectoid ferrite is 0.56\n",
+        "Mass fraction of eutectoid ferrite is 0.4\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Materials_science_and_engineering_an_introduction/README.txt b/Materials_science_and_engineering_an_introduction/README.txt
new file mode 100644
index 00000000..5f1056e2
--- /dev/null
+++ b/Materials_science_and_engineering_an_introduction/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Ankit Garg
+Course: btech
+College/Institute/Organization: DCRUST, Murthal
+Department/Designation: Chemical Engineering
+Book Title: Materials science and engineering an introduction
+Author: William D. Callister
+Publisher: John Wiley & Sons Inc., USA
+Year of publication: 2007
+Isbn: 978-0471736967
+Edition: 7
\ No newline at end of file
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