diff options
Diffstat (limited to 'Materials_science_and_engineering_an_introduction')
20 files changed, 3077 insertions, 0 deletions
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
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