Revised list of TBCs

12 files changed, 0 insertions, 6513 deletions

diff --git a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch2.ipynb b/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch2.ipynb
deleted file mode 100755
index 333b1b19..00000000
--- a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch2.ipynb
+++ /dev/null
@@ -1,454 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Ch-2, Casting Processes"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 1 on page no. 46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import sqrt\n",
-      "# Given that\n",
-      "h=15 # Height of spur in cm\n",
-      "l= 50 # Length of cast in cm\n",
-      "w= 25 # weidth of cast in cm\n",
-      "h1= 15 # Height of cast in cm\n",
-      "g= 981 # Acceleration due to gravity in cm/sec**2\n",
-      "Ag= 5 # Cross sectional area of the grate in cm**2\n",
-      "v3= sqrt(2* g * h)\n",
-      "V = l*w*h1\n",
-      "tf1= V/(Ag*v3)\n",
-      "Am = l*w\n",
-      "tf2 = (Am/Ag)*(1/sqrt(2*g))*2*(sqrt(h) - sqrt(h-h1))\n",
-      "print \"Filling time for first design = %0.2f sec, \\nFilling time for second design = %0.2f sec\"%(tf1, tf2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Filling time for first design = 21.86 sec, \n",
-        "Filling time for second design = 43.72 sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 2 on page no. 53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import sqrt, pi\n",
-      "# Given that\n",
-      "h=15 # Height of spur in cm\n",
-      "l= 50 # Length of cast in cm\n",
-      "w= 25 # weidth of cast in cm\n",
-      "h1= 15 # Height of cast in cm\n",
-      "g= 981 # Acceleration due to gravity in cm/sec**2\n",
-      "Ag= 5 # Cross sectional area of the grate in cm**2\n",
-      "Dm = 7800 # Density of molten Fe in Kg/m**3\n",
-      "Neta = 0.00496 # Kinetic viscosity in Kg/m-sec\n",
-      "theta = 90 # Angle in degree\n",
-      "Eq = 25 # (L/D) Equivalent \n",
-      "\n",
-      "v3= sqrt(2* g * h)*(10**(-2))\n",
-      "d= sqrt((Ag*4)/(pi))*(10**(-2))\n",
-      "Re = Dm*v3*d/Neta\n",
-      "f = 0.0791*(Re)**(-1/4)\n",
-      "L=0.12 # in meter\n",
-      "Cd= (1+0.45+4*f*((L/d)+Eq))**(-1/2)\n",
-      "v3_ = Cd*v3\n",
-      "Re_ = (v3_/v3)*(Re)\n",
-      "f_ = 0.0791 *(Re_)**(-1/4)\n",
-      "Cd_ = (1+0.46+4*f_*(L/d + Eq))**(-1/2)\n",
-      "v3__ = Cd_*v3\n",
-      "V = l*w*h1\n",
-      "tf= (V/(Ag*v3__))*(10**-2)\n",
-      "print \"Filling time for first design = %0.1f sec. \"% tf"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Filling time for first design = 31.7 sec. \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 3 on page no. 55"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import sqrt, pi\n",
-      "# Given that\n",
-      "Hi=1.2 # Initial height in m\n",
-      "H= 0.05 # Height in m\n",
-      "g= 9.81 # Acceleration due to gravity in m/sec**2\n",
-      "Dm = 2700 # Density of molten metal in Kg/m**3\n",
-      "Neta = 0.00273 # Kinetic viscosity in Kg/m-sec \n",
-      "d= 0.075 # Diameter in m\n",
-      "D = 1 # Internal diameter of ladle in m\n",
-      "\n",
-      "v3= sqrt(2* g * Hi)\n",
-      "Re = Dm*v3*d/Neta\n",
-      "ef=0.075\n",
-      "Cd= (1+ef)**(-1/2)\n",
-      "ef_=0.82\n",
-      "Re_ = (2+ef_)**(-1/2)\n",
-      "v3_ = sqrt(2*g*H)\n",
-      "Re_ = Dm*v3_*d/Neta\n",
-      "At = (pi/4)*D**2\n",
-      "An = (pi/4)*d**2\n",
-      "Cd= 0.96\n",
-      "tf= (sqrt(2/g))*(At/An)*(1/Cd)*sqrt(Hi)\n",
-      "m = Dm*An*Cd*sqrt(2*g*Hi)\n",
-      "m_ = Dm*An*Cd*sqrt(2*g*Hi*0.25)\n",
-      "print \"\"\"Time required to empty the ladle = %0.2f sec,\n",
-      "Discharge rate are - \n",
-      "  Initially = %0.2f Kg/sec \n",
-      "  When the ladle is 75 percent empty = %0.2f Kg/sec.\"\"\"%(tf,m,m_)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time required to empty the ladle = 91.60 sec,\n",
-        "Discharge rate are - \n",
-        "  Initially = 55.56 Kg/sec \n",
-        "  When the ladle is 75 percent empty = 27.78 Kg/sec.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 5 on page no. 66"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import sqrt, pi\n",
-      "# Given that\n",
-      "thetaF= 1540 # Temperature of mould face in degree centigrade\n",
-      "Theta0 = 28 # Initial temperature of mould in Degree centigrade\n",
-      "L= 272e3 # Latent heat of liquid metal in J/Kg\n",
-      "Dm = 7850 # Density of liquid metal in Kg/m**3\n",
-      "c = 1.17e+3 #Specific heat of sand in J/Kg-K\n",
-      "k = 0.8655 # Conductivity of sand in W/m-K\n",
-      "D= 1600 # Density of sand in Kg/m**3\n",
-      "h = 0.1 # Height in m\n",
-      "b = 10 # Thickness of slab in cm\n",
-      "r =h/2# V/A in meter\n",
-      "\n",
-      "lamda = (thetaF - Theta0)*(D*c)/(Dm*L)\n",
-      "Beta1 = 2*lamda/sqrt(pi)\n",
-      "Alpha = k /(D*c)\n",
-      "ts1 = r**2 /((Beta1**2)*Alpha)#In sec\n",
-      "ts1_=ts1/3600 # In hour\n",
-      "from sympy import symbols, solve\n",
-      "Beta= symbols(\"Beta\") \n",
-      "p=Beta**2 - lamda*(2/sqrt(pi))*Beta -lamda/3\n",
-      "Beta2 = solve(p, Beta) # taking +ve value\n",
-      "print \"Beta2 = %0.2f\" %Beta2[1]\n",
-      "r1 = r/3\n",
-      "ts2 = (r1**2)/((1.75**2)*Alpha) # in sec\n",
-      "ts2_=ts2/3600#in Hour\n",
-      "print \"Solidification time for slab-shaped casting = %0.3f hr,\\nSolidification time for sphere = %0.3f hr\"% (ts1_,ts2_)\n",
-      "# Answer in the textbook are not accurate."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Beta2 = 1.75\n",
-        "Solidification time for slab-shaped casting = 0.671 hr,\n",
-        "Solidification time for sphere = 0.054 hr\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 6 on page no. 73"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import sqrt, pi, exp\n",
-      "# Given that\n",
-      "thetaF= 1540 # Temperature of mould face in degree centigrate\n",
-      "ThetaO = 28 # Initial temperature of mould in Degree centigrate\n",
-      "L= 272e3 # Latent heat of iron in J/Kg\n",
-      "Dm = 7850 # Density of iron in Kg/m**3\n",
-      "Cs = 0.67e+3 #Specific heat of iron in J/Kg-K\n",
-      "C = 0.376e3 #Specific heat of copper in J/Kg-K\n",
-      "Ks = 83 # Conductivity of iron in W/m-K\n",
-      "K = 398 # Conductivity of copper in W/m-K\n",
-      "D= 8960 # Density of copper in Kg/m**3\n",
-      "h = .1 # Height in m\n",
-      "\n",
-      "zeta1=0.98#By solving eqauation- zeta*exp(zeta**2)*erf(zeta)=((thetaF-thetaO)*Cs)/(sqrt(pi)*L), zeta = 0.98\n",
-      "AlphaS = Ks /(Dm*Cs)\n",
-      "ts1 = h**2 / (16*(zeta1**2) * AlphaS)#In sec\n",
-      "ts1_=ts1/3600 # In hour\n",
-      "Phi = sqrt((Ks*Dm*Cs)/(K*D*C))\n",
-      "zeta2=0.815#By solving eqauation- zeta*exp(zeta**2)*(erf(zeta)+Phi)=((thetaF-thetaO)*Cs)/(sqrt(pi)*L), zeta = 0.815\n",
-      "ts2 = h**2 / (16*(zeta2**2) * AlphaS)#In sec\n",
-      "ts2_=ts2/3600 # In hour\n",
-      "erf = lambda zeta2:(thetaF-Theta0)*Cs/L/sqrt(pi)\n",
-      "thetaS= thetaF+(thetaF-(L*(sqrt(pi))*zeta2*(exp(zeta2**2))*erf(zeta2))/Cs)\n",
-      "print \"\"\"Solidification time for slab-shaped casting when the casting is done in a water cooled copper mould = %0.4f hr,\n",
-      "Solidification time for slab-shaped casting when the casting is done in a very thick copper mould = %0.4f hr,\n",
-      "The surface temperature of the mould = %0.1f\u00b0 C\"\"\"% (ts1_,ts2_,thetaS)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Solidification time for slab-shaped casting when the casting is done in a water cooled copper mould = 0.0115 hr,\n",
-        "Solidification time for slab-shaped casting when the casting is done in a very thick copper mould = 0.0166 hr,\n",
-        "The surface temperature of the mould = 685.7\u00b0 C\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 7 on page no. 75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import sqrt, pi, exp\n",
-      "# Given that\n",
-      "thetaF= 1540 # Temperature of mould face in degree centigrade\n",
-      "thetaO = 28 # Initial temperature of mould in Degree centigrade\n",
-      "L= 272e3 # Latent heat of iron in J/Kg\n",
-      "Dm = 7850 # Density of iron in Kg/m**3\n",
-      "Cs = 0.67e+3 #Specific heat of iron in J/Kg-K\n",
-      "C = 0.376e3 #Specific heat of copper in J/Kg-K\n",
-      "Ks = 83 # Conductivity of iron in W/m-K\n",
-      "K = 398 # Conductivity of copper in W/m-K\n",
-      "D= 8960 # Density of copper in Kg/m**3\n",
-      "h = .1 # Height in m\n",
-      "hF = 1420 # Total heat transfer coefficient across the casting-mould interface in W/m**2-\u00b0C\n",
-      "\n",
-      "AlphaS = K /(D*C)\n",
-      "thetaS = 982 #In \u00b0C as in example 2.6\n",
-      "h1= (1+(sqrt((Ks*Dm*Cs)/(K*D*C))))*hF\n",
-      "a = 1/2 + (sqrt((1/4)+Cs*(thetaF-thetaS)/(3*L)))\n",
-      "delta=h/2\n",
-      "ts = (delta+((h1*delta**2)/(2*Ks)))/((h1*(thetaF-thetaS))/(Dm*L*a)) # in sec\n",
-      "ts_ = ts/3600 # in hours\n",
-      "h2= (1+(sqrt((K*D*C)/(Ks*Dm*Cs))))*hF\n",
-      "gama= ((h2**2)/(K**2))*AlphaS*ts\n",
-      "thetaS_ = thetaO + (thetaS-thetaO)*(1-((exp(gama))*(1-(erf(sqrt(gama))))))\n",
-      "print \"Solidification time = %0.4f hr,\\nThe surface temperature of the mould = %0.2f \u00b0 C\"%(ts_,thetaS_)\n",
-      "# The value of the surface temperature of the mould in the book is given as 658.1\u00b0 C, Which is wrong."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Solidification time = 0.0534 hr,\n",
-        "The surface temperature of the mould = 10389.84 \u00b0 C\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 8 on page no. 77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "A= 60*7.5 # Cross sectional area in cm**2\n",
-      "v=0.05 # Withdrawal rate in m/sec\n",
-      "t = 0.0125 # Thickness in m\n",
-      "thetaF= 1500 # Temperature of mould face in degree centigrate\n",
-      "thetaP = 1550 # \n",
-      "thetaO = 20 # Initial temperature of mould in Degree centigrate\n",
-      "L= 268e3 # Latent heat of molten metal in J/Kg\n",
-      "Dm = 7680 # Density of molten metal in Kg/m**3\n",
-      "Cs = 0.67e+3 #Specific heat of molten metal in J/Kg-K\n",
-      "Cm = 0.755e3 #Specific heat of mould in J/Kg-K\n",
-      "Ks = 76 # Conductivity of molten metal in W/m-K\n",
-      "hF = 1420 # Heat transfer coefficient at the casting-mould interface in W/m**2-\u00b0C\n",
-      "Dtheta = 10 # Maximum temperature of cooling water in \u00b0 C\n",
-      "L_ = L+Cm*(thetaP-thetaF)\n",
-      "x=L_ / (Cs*(thetaF-thetaO))\n",
-      "y= hF*t/Ks\n",
-      "print \"L_/(Cs(thetaF-thetaO))=%0.2f,\\nhF*t/Ks=%0.2f\"%(x,y)\n",
-      "z=0.11 # Where z=hF**2 * lm / (v*Ks*Dm*Cs)\n",
-      "lm= (z*v*Ks*Dm*Cs)/(hF**2)\n",
-      "Z=0.28 # Where Z=Q/(lm*(thetaF-thetaO)*sqrt(lm*v*Dm*Cs*Ks))\n",
-      "Q = Z*lm*(thetaF-thetaO)*sqrt(lm*v*Dm*Cs*Ks)\n",
-      "m = Q / (4.2e3*Dtheta)\n",
-      "print \"The mould length = %0.2f meter,\\nThe cooling water requirement = %0.2f Kg/sec\"%(lm,m)\n",
-      "# Answer for The cooling water requirement in the book is given as 5.05 Kg/sec, Which is wrong."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "L_/(Cs(thetaF-thetaO))=0.31,\n",
-        "hF*t/Ks=0.23\n",
-        "The mould length = 1.07 meter,\n",
-        "The cooling water requirement = 48.07 Kg/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 9 on page no. 81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "# Given that\n",
-      "a = 15 # Side of the aluminium cube in cm\n",
-      "Sh = 0.065 # Volume shrinkage of aluminium during solidification\n",
-      "\n",
-      "Vc = a**3\n",
-      "Vr = 3*Sh*Vc\n",
-      "h = ((4*Vr)/pi)**(1/3)\n",
-      "Rr = 6/h # Where Rr= (A/V)r\n",
-      "Rc = 6/a # Where Rc = (A/V)c\n",
-      "print \"(A/V)r=%f, (A/V)c=%.1f\\nHence Rr is greater than Rc\"%(Rr,Rc)\n",
-      "dmin = 6/Rc\n",
-      "Vr_ = (pi/4)*dmin**3\n",
-      "print \"\"\"With minimum value of d Vr=%d cm**3 .\n",
-      "This valume is much more than the minimum Vr necessary.\n",
-      "Let us now consider the top riser when the optimum cylindrical shape is obtained with h=d/2\n",
-      "and again (A/V)r = 6/d. However, with a large top riser,\n",
-      "the cube loses its top surface for the purpose of heat dissipation.\"\"\"%Vr_\n",
-      "Rc_ = 5/a\n",
-      "dmin_=6/Rc_\n",
-      "print \"d should be greater than or equal to %d cm\"%dmin_\n",
-      "Vr__ = (pi/4)*dmin_**2 *floor(h)\n",
-      "print \"The riser volume with minimum diameter is %d cm**3\"%Vr__"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(A/V)r=0.636422, (A/V)c=0.4\n",
-        "Hence Rr is greater than Rc\n",
-        "With minimum value of d Vr=2650 cm**3 .\n",
-        "This valume is much more than the minimum Vr necessary.\n",
-        "Let us now consider the top riser when the optimum cylindrical shape is obtained with h=d/2\n",
-        "and again (A/V)r = 6/d. However, with a large top riser,\n",
-        "the cube loses its top surface for the purpose of heat dissipation.\n",
-        "d should be greater than or equal to 18 cm\n",
-        "The riser volume with minimum diameter is 2290 cm**3\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
diff --git a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch2_1.ipynb b/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch2_1.ipynb
deleted file mode 100755
index 333b1b19..00000000
--- a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch2_1.ipynb
+++ /dev/null
@@ -1,454 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Ch-2, Casting Processes"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 1 on page no. 46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import sqrt\n",
-      "# Given that\n",
-      "h=15 # Height of spur in cm\n",
-      "l= 50 # Length of cast in cm\n",
-      "w= 25 # weidth of cast in cm\n",
-      "h1= 15 # Height of cast in cm\n",
-      "g= 981 # Acceleration due to gravity in cm/sec**2\n",
-      "Ag= 5 # Cross sectional area of the grate in cm**2\n",
-      "v3= sqrt(2* g * h)\n",
-      "V = l*w*h1\n",
-      "tf1= V/(Ag*v3)\n",
-      "Am = l*w\n",
-      "tf2 = (Am/Ag)*(1/sqrt(2*g))*2*(sqrt(h) - sqrt(h-h1))\n",
-      "print \"Filling time for first design = %0.2f sec, \\nFilling time for second design = %0.2f sec\"%(tf1, tf2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Filling time for first design = 21.86 sec, \n",
-        "Filling time for second design = 43.72 sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 2 on page no. 53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import sqrt, pi\n",
-      "# Given that\n",
-      "h=15 # Height of spur in cm\n",
-      "l= 50 # Length of cast in cm\n",
-      "w= 25 # weidth of cast in cm\n",
-      "h1= 15 # Height of cast in cm\n",
-      "g= 981 # Acceleration due to gravity in cm/sec**2\n",
-      "Ag= 5 # Cross sectional area of the grate in cm**2\n",
-      "Dm = 7800 # Density of molten Fe in Kg/m**3\n",
-      "Neta = 0.00496 # Kinetic viscosity in Kg/m-sec\n",
-      "theta = 90 # Angle in degree\n",
-      "Eq = 25 # (L/D) Equivalent \n",
-      "\n",
-      "v3= sqrt(2* g * h)*(10**(-2))\n",
-      "d= sqrt((Ag*4)/(pi))*(10**(-2))\n",
-      "Re = Dm*v3*d/Neta\n",
-      "f = 0.0791*(Re)**(-1/4)\n",
-      "L=0.12 # in meter\n",
-      "Cd= (1+0.45+4*f*((L/d)+Eq))**(-1/2)\n",
-      "v3_ = Cd*v3\n",
-      "Re_ = (v3_/v3)*(Re)\n",
-      "f_ = 0.0791 *(Re_)**(-1/4)\n",
-      "Cd_ = (1+0.46+4*f_*(L/d + Eq))**(-1/2)\n",
-      "v3__ = Cd_*v3\n",
-      "V = l*w*h1\n",
-      "tf= (V/(Ag*v3__))*(10**-2)\n",
-      "print \"Filling time for first design = %0.1f sec. \"% tf"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Filling time for first design = 31.7 sec. \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 3 on page no. 55"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import sqrt, pi\n",
-      "# Given that\n",
-      "Hi=1.2 # Initial height in m\n",
-      "H= 0.05 # Height in m\n",
-      "g= 9.81 # Acceleration due to gravity in m/sec**2\n",
-      "Dm = 2700 # Density of molten metal in Kg/m**3\n",
-      "Neta = 0.00273 # Kinetic viscosity in Kg/m-sec \n",
-      "d= 0.075 # Diameter in m\n",
-      "D = 1 # Internal diameter of ladle in m\n",
-      "\n",
-      "v3= sqrt(2* g * Hi)\n",
-      "Re = Dm*v3*d/Neta\n",
-      "ef=0.075\n",
-      "Cd= (1+ef)**(-1/2)\n",
-      "ef_=0.82\n",
-      "Re_ = (2+ef_)**(-1/2)\n",
-      "v3_ = sqrt(2*g*H)\n",
-      "Re_ = Dm*v3_*d/Neta\n",
-      "At = (pi/4)*D**2\n",
-      "An = (pi/4)*d**2\n",
-      "Cd= 0.96\n",
-      "tf= (sqrt(2/g))*(At/An)*(1/Cd)*sqrt(Hi)\n",
-      "m = Dm*An*Cd*sqrt(2*g*Hi)\n",
-      "m_ = Dm*An*Cd*sqrt(2*g*Hi*0.25)\n",
-      "print \"\"\"Time required to empty the ladle = %0.2f sec,\n",
-      "Discharge rate are - \n",
-      "  Initially = %0.2f Kg/sec \n",
-      "  When the ladle is 75 percent empty = %0.2f Kg/sec.\"\"\"%(tf,m,m_)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time required to empty the ladle = 91.60 sec,\n",
-        "Discharge rate are - \n",
-        "  Initially = 55.56 Kg/sec \n",
-        "  When the ladle is 75 percent empty = 27.78 Kg/sec.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 5 on page no. 66"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import sqrt, pi\n",
-      "# Given that\n",
-      "thetaF= 1540 # Temperature of mould face in degree centigrade\n",
-      "Theta0 = 28 # Initial temperature of mould in Degree centigrade\n",
-      "L= 272e3 # Latent heat of liquid metal in J/Kg\n",
-      "Dm = 7850 # Density of liquid metal in Kg/m**3\n",
-      "c = 1.17e+3 #Specific heat of sand in J/Kg-K\n",
-      "k = 0.8655 # Conductivity of sand in W/m-K\n",
-      "D= 1600 # Density of sand in Kg/m**3\n",
-      "h = 0.1 # Height in m\n",
-      "b = 10 # Thickness of slab in cm\n",
-      "r =h/2# V/A in meter\n",
-      "\n",
-      "lamda = (thetaF - Theta0)*(D*c)/(Dm*L)\n",
-      "Beta1 = 2*lamda/sqrt(pi)\n",
-      "Alpha = k /(D*c)\n",
-      "ts1 = r**2 /((Beta1**2)*Alpha)#In sec\n",
-      "ts1_=ts1/3600 # In hour\n",
-      "from sympy import symbols, solve\n",
-      "Beta= symbols(\"Beta\") \n",
-      "p=Beta**2 - lamda*(2/sqrt(pi))*Beta -lamda/3\n",
-      "Beta2 = solve(p, Beta) # taking +ve value\n",
-      "print \"Beta2 = %0.2f\" %Beta2[1]\n",
-      "r1 = r/3\n",
-      "ts2 = (r1**2)/((1.75**2)*Alpha) # in sec\n",
-      "ts2_=ts2/3600#in Hour\n",
-      "print \"Solidification time for slab-shaped casting = %0.3f hr,\\nSolidification time for sphere = %0.3f hr\"% (ts1_,ts2_)\n",
-      "# Answer in the textbook are not accurate."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Beta2 = 1.75\n",
-        "Solidification time for slab-shaped casting = 0.671 hr,\n",
-        "Solidification time for sphere = 0.054 hr\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 6 on page no. 73"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import sqrt, pi, exp\n",
-      "# Given that\n",
-      "thetaF= 1540 # Temperature of mould face in degree centigrate\n",
-      "ThetaO = 28 # Initial temperature of mould in Degree centigrate\n",
-      "L= 272e3 # Latent heat of iron in J/Kg\n",
-      "Dm = 7850 # Density of iron in Kg/m**3\n",
-      "Cs = 0.67e+3 #Specific heat of iron in J/Kg-K\n",
-      "C = 0.376e3 #Specific heat of copper in J/Kg-K\n",
-      "Ks = 83 # Conductivity of iron in W/m-K\n",
-      "K = 398 # Conductivity of copper in W/m-K\n",
-      "D= 8960 # Density of copper in Kg/m**3\n",
-      "h = .1 # Height in m\n",
-      "\n",
-      "zeta1=0.98#By solving eqauation- zeta*exp(zeta**2)*erf(zeta)=((thetaF-thetaO)*Cs)/(sqrt(pi)*L), zeta = 0.98\n",
-      "AlphaS = Ks /(Dm*Cs)\n",
-      "ts1 = h**2 / (16*(zeta1**2) * AlphaS)#In sec\n",
-      "ts1_=ts1/3600 # In hour\n",
-      "Phi = sqrt((Ks*Dm*Cs)/(K*D*C))\n",
-      "zeta2=0.815#By solving eqauation- zeta*exp(zeta**2)*(erf(zeta)+Phi)=((thetaF-thetaO)*Cs)/(sqrt(pi)*L), zeta = 0.815\n",
-      "ts2 = h**2 / (16*(zeta2**2) * AlphaS)#In sec\n",
-      "ts2_=ts2/3600 # In hour\n",
-      "erf = lambda zeta2:(thetaF-Theta0)*Cs/L/sqrt(pi)\n",
-      "thetaS= thetaF+(thetaF-(L*(sqrt(pi))*zeta2*(exp(zeta2**2))*erf(zeta2))/Cs)\n",
-      "print \"\"\"Solidification time for slab-shaped casting when the casting is done in a water cooled copper mould = %0.4f hr,\n",
-      "Solidification time for slab-shaped casting when the casting is done in a very thick copper mould = %0.4f hr,\n",
-      "The surface temperature of the mould = %0.1f\u00b0 C\"\"\"% (ts1_,ts2_,thetaS)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Solidification time for slab-shaped casting when the casting is done in a water cooled copper mould = 0.0115 hr,\n",
-        "Solidification time for slab-shaped casting when the casting is done in a very thick copper mould = 0.0166 hr,\n",
-        "The surface temperature of the mould = 685.7\u00b0 C\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 7 on page no. 75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import sqrt, pi, exp\n",
-      "# Given that\n",
-      "thetaF= 1540 # Temperature of mould face in degree centigrade\n",
-      "thetaO = 28 # Initial temperature of mould in Degree centigrade\n",
-      "L= 272e3 # Latent heat of iron in J/Kg\n",
-      "Dm = 7850 # Density of iron in Kg/m**3\n",
-      "Cs = 0.67e+3 #Specific heat of iron in J/Kg-K\n",
-      "C = 0.376e3 #Specific heat of copper in J/Kg-K\n",
-      "Ks = 83 # Conductivity of iron in W/m-K\n",
-      "K = 398 # Conductivity of copper in W/m-K\n",
-      "D= 8960 # Density of copper in Kg/m**3\n",
-      "h = .1 # Height in m\n",
-      "hF = 1420 # Total heat transfer coefficient across the casting-mould interface in W/m**2-\u00b0C\n",
-      "\n",
-      "AlphaS = K /(D*C)\n",
-      "thetaS = 982 #In \u00b0C as in example 2.6\n",
-      "h1= (1+(sqrt((Ks*Dm*Cs)/(K*D*C))))*hF\n",
-      "a = 1/2 + (sqrt((1/4)+Cs*(thetaF-thetaS)/(3*L)))\n",
-      "delta=h/2\n",
-      "ts = (delta+((h1*delta**2)/(2*Ks)))/((h1*(thetaF-thetaS))/(Dm*L*a)) # in sec\n",
-      "ts_ = ts/3600 # in hours\n",
-      "h2= (1+(sqrt((K*D*C)/(Ks*Dm*Cs))))*hF\n",
-      "gama= ((h2**2)/(K**2))*AlphaS*ts\n",
-      "thetaS_ = thetaO + (thetaS-thetaO)*(1-((exp(gama))*(1-(erf(sqrt(gama))))))\n",
-      "print \"Solidification time = %0.4f hr,\\nThe surface temperature of the mould = %0.2f \u00b0 C\"%(ts_,thetaS_)\n",
-      "# The value of the surface temperature of the mould in the book is given as 658.1\u00b0 C, Which is wrong."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Solidification time = 0.0534 hr,\n",
-        "The surface temperature of the mould = 10389.84 \u00b0 C\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 8 on page no. 77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "A= 60*7.5 # Cross sectional area in cm**2\n",
-      "v=0.05 # Withdrawal rate in m/sec\n",
-      "t = 0.0125 # Thickness in m\n",
-      "thetaF= 1500 # Temperature of mould face in degree centigrate\n",
-      "thetaP = 1550 # \n",
-      "thetaO = 20 # Initial temperature of mould in Degree centigrate\n",
-      "L= 268e3 # Latent heat of molten metal in J/Kg\n",
-      "Dm = 7680 # Density of molten metal in Kg/m**3\n",
-      "Cs = 0.67e+3 #Specific heat of molten metal in J/Kg-K\n",
-      "Cm = 0.755e3 #Specific heat of mould in J/Kg-K\n",
-      "Ks = 76 # Conductivity of molten metal in W/m-K\n",
-      "hF = 1420 # Heat transfer coefficient at the casting-mould interface in W/m**2-\u00b0C\n",
-      "Dtheta = 10 # Maximum temperature of cooling water in \u00b0 C\n",
-      "L_ = L+Cm*(thetaP-thetaF)\n",
-      "x=L_ / (Cs*(thetaF-thetaO))\n",
-      "y= hF*t/Ks\n",
-      "print \"L_/(Cs(thetaF-thetaO))=%0.2f,\\nhF*t/Ks=%0.2f\"%(x,y)\n",
-      "z=0.11 # Where z=hF**2 * lm / (v*Ks*Dm*Cs)\n",
-      "lm= (z*v*Ks*Dm*Cs)/(hF**2)\n",
-      "Z=0.28 # Where Z=Q/(lm*(thetaF-thetaO)*sqrt(lm*v*Dm*Cs*Ks))\n",
-      "Q = Z*lm*(thetaF-thetaO)*sqrt(lm*v*Dm*Cs*Ks)\n",
-      "m = Q / (4.2e3*Dtheta)\n",
-      "print \"The mould length = %0.2f meter,\\nThe cooling water requirement = %0.2f Kg/sec\"%(lm,m)\n",
-      "# Answer for The cooling water requirement in the book is given as 5.05 Kg/sec, Which is wrong."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "L_/(Cs(thetaF-thetaO))=0.31,\n",
-        "hF*t/Ks=0.23\n",
-        "The mould length = 1.07 meter,\n",
-        "The cooling water requirement = 48.07 Kg/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 9 on page no. 81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "# Given that\n",
-      "a = 15 # Side of the aluminium cube in cm\n",
-      "Sh = 0.065 # Volume shrinkage of aluminium during solidification\n",
-      "\n",
-      "Vc = a**3\n",
-      "Vr = 3*Sh*Vc\n",
-      "h = ((4*Vr)/pi)**(1/3)\n",
-      "Rr = 6/h # Where Rr= (A/V)r\n",
-      "Rc = 6/a # Where Rc = (A/V)c\n",
-      "print \"(A/V)r=%f, (A/V)c=%.1f\\nHence Rr is greater than Rc\"%(Rr,Rc)\n",
-      "dmin = 6/Rc\n",
-      "Vr_ = (pi/4)*dmin**3\n",
-      "print \"\"\"With minimum value of d Vr=%d cm**3 .\n",
-      "This valume is much more than the minimum Vr necessary.\n",
-      "Let us now consider the top riser when the optimum cylindrical shape is obtained with h=d/2\n",
-      "and again (A/V)r = 6/d. However, with a large top riser,\n",
-      "the cube loses its top surface for the purpose of heat dissipation.\"\"\"%Vr_\n",
-      "Rc_ = 5/a\n",
-      "dmin_=6/Rc_\n",
-      "print \"d should be greater than or equal to %d cm\"%dmin_\n",
-      "Vr__ = (pi/4)*dmin_**2 *floor(h)\n",
-      "print \"The riser volume with minimum diameter is %d cm**3\"%Vr__"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(A/V)r=0.636422, (A/V)c=0.4\n",
-        "Hence Rr is greater than Rc\n",
-        "With minimum value of d Vr=2650 cm**3 .\n",
-        "This valume is much more than the minimum Vr necessary.\n",
-        "Let us now consider the top riser when the optimum cylindrical shape is obtained with h=d/2\n",
-        "and again (A/V)r = 6/d. However, with a large top riser,\n",
-        "the cube loses its top surface for the purpose of heat dissipation.\n",
-        "d should be greater than or equal to 18 cm\n",
-        "The riser volume with minimum diameter is 2290 cm**3\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
diff --git a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch3.ipynb b/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch3.ipynb
deleted file mode 100755
index a0d1dfb1..00000000
--- a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch3.ipynb
+++ /dev/null
@@ -1,648 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter - 03, Forming Processes"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 1 on page no. 112"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import tan, sqrt, atan, log\n",
-      "# Given that\n",
-      "A = 150*6 # Cross-section of strips in mm**2\n",
-      "ti = 6 # Thickness in mm\n",
-      "pA = 0.20 # Reduction in area\n",
-      "d = 400 # Diameter of steel rolls in mm\n",
-      "Ys = 0.35# Shear Yield stress of the material before rolling in KN/mm**2\n",
-      "Ys_ = 0.4# Shear Yield stress of the material after rolling in KN/mm**2\n",
-      "mu = 0.1 # Cofficient of friction\n",
-      "\n",
-      "tf =0.8*ti\n",
-      "Ys_a = (Ys + Ys_)/2\n",
-      "r=d/2\n",
-      "thetaI = sqrt((ti-tf)/r)\n",
-      "lambdaI=2*sqrt(r/tf)*atan(thetaI *sqrt(r/tf))\n",
-      "lambdaN = (1/2)*((1/mu)*(log(tf/ti)) + lambdaI)\n",
-      "thetaN  =(sqrt(tf/r))*(tan((lambdaN/2)*(sqrt(tf/r))))\n",
-      "print \"\"\"The final srip thickness is %0.2f mm,\\nThe avg shear yield stress during the process is %0.3f KN/mm**2,\n",
-      "The angle subtended by the deformation zone at the roll centre is %0.4f rad, \n",
-      "The location of neutral point is %0.3f rad.\"\"\"%(tf,Ys_a,thetaI,thetaN)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The final srip thickness is 4.80 mm,\n",
-        "The avg shear yield stress during the process is 0.375 KN/mm**2,\n",
-        "The angle subtended by the deformation zone at the roll centre is 0.0775 rad, \n",
-        "The location of neutral point is 0.023 rad.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 2 on page no. 113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import exp, ceil\n",
-      "# Given that\n",
-      "A = 150*6 # Cross-section of strips in mm**2\n",
-      "w = 150 # Width of the strip in mm\n",
-      "ti = 6 # Thickness in mm\n",
-      "pA = 0.20 # Reduction in area\n",
-      "d = 400 # Diameter of steel rolls in mm\n",
-      "Ys = 0.35# Shear Yield stress of the material before rolling in KN/mm**2\n",
-      "Ys_ = 0.4# Shear Yield stress of the material after rolling in KN/mm**2\n",
-      "mu = 0.1 # Cofficient of friction\n",
-      "v = 30 # Speed of rolling in m/min\n",
-      "tf =0.8*ti\n",
-      "Ys_a = (Ys + Ys_)/2\n",
-      "r=d/2\n",
-      "thetaI = sqrt((ti-tf)/r)\n",
-      "lambdaI=2*sqrt(r/tf)*atan(thetaI *sqrt(r/tf))\n",
-      "lambdaN = (1/2)*((1/mu)*(log(tf/ti)) + lambdaI)\n",
-      "thetaN  =(sqrt(tf/r))*(tan((lambdaN/2)*(sqrt(tf/r))))\n",
-      "Dtheta_a = thetaN/4\n",
-      "Dtheta_b = (thetaI- thetaN)/8\n",
-      "print \"The values of P_after are:\\n\"\n",
-      "for i in range(0,5,1):\n",
-      "    theta = i*Dtheta_a\n",
-      "    y = (1/2)* (tf+r*theta**2)\n",
-      "    lamda = 2*sqrt(r/tf)*atan(theta*sqrt(r/tf))\n",
-      "    p_a = 2*Ys_a*(2*y/tf)*(exp(mu*lamda))\n",
-      "    print \"%0.3f \\t\\t\"%p_a,\n",
-      "\n",
-      "I1 = (Dtheta_a/3) *(0.75+.925+4*(.788+.876)+2*.830)# By Simpson's rule\n",
-      "print \"\\n\\nThe values of P_before are:\\n\"\n",
-      "for i in range(0,8,1):\n",
-      "    theta1 = i*Dtheta_b + thetaN\n",
-      "    y = (1/2)* (tf+r*theta1**2)\n",
-      "    lamda = 2*sqrt(r/tf)*atan(theta1*sqrt(r/tf))\n",
-      "    p_b = 2*Ys_a*(2*y/ti)*(exp(mu*(lambdaI-lamda)))\n",
-      "    print \" %0.3f \\t\\t\"%p_b,\n",
-      "\n",
-      "I2 = (Dtheta_b/3)*(0.925+.75+4*(.887+.828+.786+.759) + 2*(.855+.804+.772))#By Simpson's rule\n",
-      "F = r*(I1 + I2)\n",
-      "F_ = F*w\n",
-      "T = (r**2)*mu*(I2-I1)\n",
-      "T_ =T*w\n",
-      "W = v*(1000/60)/r\n",
-      "P = 2*T_*W\n",
-      "print \"\\n\\nThe roll separating force = %d kN, The power required in the rolling process = %0.2f kW\"%(ceil(F_),P/1000)\n",
-      "# Answer in the book for the power required in the rolling process is given as 75.6 kW\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The values of P_after are:\n",
-        "\n",
-        "0.750 \t\t0.787 \t\t0.829 \t\t0.874 \t\t0.924 \t\t\n",
-        "\n",
-        "The values of P_before are:\n",
-        "\n",
-        " 0.924 \t\t 0.887 \t\t 0.855 \t\t 0.828 \t\t 0.805 \t\t 0.786 \t\t 0.771 \t\t 0.759 \t\t\n",
-        "\n",
-        "The roll separating force = 1908 kN, The power required in the rolling process = 77.38 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 3 on page no. 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "# Given that\n",
-      "A = 150*6 # Cross-section of strips in mm**2\n",
-      "w = 150 # Width of the strip in mm\n",
-      "ti = 6 # Thickness in mm\n",
-      "pA = 0.20 # Reduction in area\n",
-      "d = 400 # Diameter of steel rolls in mm\n",
-      "Ys = 0.35# Shear Yield stress of the material before rolling in KN/mm**2\n",
-      "Ys_ = 0.4# Shear Yield stress of the material after rolling in KN/mm**2\n",
-      "mu = 0.1 # Cofficient of friction\n",
-      "mu_ = 0.005 # Cofficient of friction in bearing \n",
-      "D = 150 # The diameter of bearing in mm\n",
-      "v = 30 # Speed of rolling in m/min\n",
-      "tf =0.8*ti\n",
-      "Ys_a = (Ys + Ys_)/2\n",
-      "r=d/2\n",
-      "thetaI = sqrt((ti-tf)/r)\n",
-      "lambdaI=2*sqrt(r/tf)*atan(thetaI *sqrt(r/tf))\n",
-      "lambdaN = (1/2)*((1/mu)*(log(tf/ti)) + lambdaI)\n",
-      "thetaN  =(sqrt(tf/r))*(tan((lambdaN/2)*(sqrt(tf/r))))\n",
-      "Dtheta_a = thetaN/4\n",
-      "Dtheta_b = (thetaI- thetaN)/8\n",
-      "for i in range(0,5,1):\n",
-      "    theta = i*Dtheta_a\n",
-      "    y = (1/2)* (tf+r*theta**2)\n",
-      "    lamda = 2*sqrt(r/tf)*atan(theta*(pi/180) *sqrt(r/tf))*180/pi\n",
-      "    p_a = 2*Ys_a*(2*y/tf)*(exp(mu*lamda))\n",
-      "\n",
-      "I1 = (Dtheta_a/3) *(0.75+.925+4*(.788+.876)+2*.830)\n",
-      "for i in range(0,8,1):\n",
-      "    theta1 = i*Dtheta_b + thetaN\n",
-      "    y = (1/2)* (tf+r*theta1**2)\n",
-      "    lamda = 2*sqrt(r/tf)*atan(theta1*(pi/180) *sqrt(r/tf))*180/pi\n",
-      "    p_b = 2*Ys_a*(2*y/ti)*(exp(mu*(lambdaI-lamda)))\n",
-      "I2 = (Dtheta_b/3)*(0.925+.75+4*(.887+.828+.786+.759) + 2*(.855+.804+.772))\n",
-      "F = r*(I1 + I2)\n",
-      "F_ = F*w\n",
-      "T = (r**2)*mu*(I2-I1)\n",
-      "T_ =T*w\n",
-      "W = v*(1000/60)/r\n",
-      "P_ = 2*T_*W\n",
-      "Pl = mu_*F_*D*W\n",
-      "P = Pl+P_\n",
-      "print \"The mill power = %0.2f kW\"%(P/1000)\n",
-      "# Answer in the book is given as 79.18 kW  whcih is wrong.\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The mill power = 80.95 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 4 on page no. 118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "mu = 0.25 # Cofficient of friction between the job and the dies \n",
-      "Y = 7 # Avg yield stress of the lead in N/mm**2\n",
-      "h = 6 # Height of die in mm\n",
-      "L = 150 # Length of the strip in mm\n",
-      "V1 = 24*24*150 # Volume of the strip in mm**3\n",
-      "V2 = 6*96*150 # Volume of the die in mm**3\n",
-      "w= 96 # Weidth of the die in mm\n",
-      "from sympy.mpmath import *\n",
-      "K = Y/sqrt(3)\n",
-      "x_ = (h/(2*mu))*(log(1/(2*mu)))\n",
-      "l = w/2\n",
-      "p1= lambda x:(2*K)*exp((2*mu/h)*x)\n",
-      "I1 = quad(p1,[0,x_])\n",
-      "p2 = lambda y:(2*K)*((1/2*mu)*(log(1/(2*mu))) + (y/h))\n",
-      "I2 = quad(p2,[x_,l])\n",
-      "F = 2*(I1+I2)\n",
-      "F_ = F*L\n",
-      "print \"The maximum forging force = %0.2e N\" %F_ \n",
-      "# Answer in the book is given as 0.54*10**6 N which is wrong.\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum forging force = 4.89e+05 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 5 on page no. 119"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "mu = 0.08# Cofficient of friction between the job and the dies \n",
-      "Y = 7 # Avg yield stress of the lead in N/mm**2\n",
-      "h = 6 # Height of die in mm\n",
-      "L = 150 # Length of the strip in mm\n",
-      "V1 = 24*24*150 # Volume of the strip in mm**3\n",
-      "V2 = 6*96*150 # Volume of the die in mm**3\n",
-      "w= 96 # Weidth of the die in mm\n",
-      "K = Y/sqrt(3)\n",
-      "x_ = (h/(2*mu))*(log(1/(2*mu)))\n",
-      "l = w/2\n",
-      "p1 =lambda x:(2*K)*exp((2*mu/h)*x)\n",
-      "I = quad(p1,[0,l])\n",
-      "F = 2*(I)\n",
-      "F_ = F*L\n",
-      "print \"The maximum forging force = %0.2e N\"%F_ \n",
-      "\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum forging force = 2.36e+05 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 6 on page no. 122"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "r = 150 # Radius of the circular disc of lead in mm\n",
-      "Ti = 50 # Initial thickness of the disc in mm\n",
-      "Tf = 25 # Reduced thickness of the disc in mm\n",
-      "mu = 0.25# Cofficient of friction between the job and the dies \n",
-      "K = 4 # Avg shear yield stress of the lead in N/mm**2\n",
-      "\n",
-      "R = r*sqrt(2)\n",
-      "rs = (R - ((Tf/(2*mu)) * log(1/(mu*sqrt(3)))))\n",
-      "p1 = lambda x:(((sqrt(3))*K)*exp((2*mu/Tf)*(R-x)))*x\n",
-      "I = quad(p1,[rs,R])\n",
-      "p2 = lambda y:((2*K/Tf)*(R-y) + ((K/mu)*(1+log(mu*sqrt(3)))))*y\n",
-      "I_ = quad(p2,[0,rs])\n",
-      "F = 2*pi*(I+I_)\n",
-      "print \"The maximum forging force = %0.3e N\"%F \n",
-      "\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum forging force = 3.649e+06 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 7 on page no. 126"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "Di = 12.7 # Intial diameter in mm\n",
-      "Df = 10.2 # Final diameter in mm\n",
-      "v = 90 # Drawn speed in m/min\n",
-      "alpha=6 # Half angle of dia in degree\n",
-      "mu = 0.1# Cofficient of friction between the job and the dies \n",
-      "Y = 207 # Tensile  yield stress of the steel specimen in N/mm**2\n",
-      "Y_ = 414 # Tensile yield stress of the similar specimen at strain 0.5 in N/mm**2\n",
-      "e = 0.5 # Strain\n",
-      "\n",
-      "e_ =2* log(Di/Df)\n",
-      "Y_e = Y + (Y_ - Y)*e_/e\n",
-      "Y__ = (Y+Y_e)/2\n",
-      "phi = 1 + (mu/tan(alpha*pi/180))\n",
-      "Y_f = Y__ * ((phi/(phi-1)) *(1-((Df/Di)**(2*(phi-1)))))\n",
-      "p = Y_f * (pi/4)*(Df**2)*v/60\n",
-      "Dmax = 1- (1/(phi**(1/(phi-1))))\n",
-      "print \"Drawing power = %0.3f kW, \\nThe maximum passible reduction with same die = %0.2f mm\"%(p/1000,Dmax)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drawing power = 25.530 kW, \n",
-        "The maximum passible reduction with same die = 0.50 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 8 on page no. 130"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "Ri = 30 # Inside radius of cup in mm\n",
-      "t = 3 # Thickness in mm\n",
-      "Rb = 40 # Radius of the blank in mm\n",
-      "K = 210 # Shear yield stress of the material in N/mm**2\n",
-      "Y = 600 # Maximum allowable stress in N/mm**2\n",
-      "Beta = 0.05\n",
-      "mu = 0.1# Cofficient of friction between the job and the dies \n",
-      "\n",
-      "Fh = Beta*pi*(Rb**2)*K\n",
-      "Y_r = (mu*Fh/(pi*Rb*t))+(2*K*log(Rb/Ri))\n",
-      "Y_z = Y_r*exp(mu*pi/2)\n",
-      "F = 2*pi*Ri*t*Y_z\n",
-      "Y_r_ = Y/exp(mu*pi/2)\n",
-      "Rp = (Rb/exp((Y_r_/(2*K)) - ((mu*Fh)/(2*pi*K*Rb*t))))-t\n",
-      "print \"Drawing force = %d N,\\nMinimum passible radius of the cup which can drawn from the given blank without causing a fracture = %0.2f mm\"%(F,Rp)\n",
-      "# Answer in the book given as 62680 N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drawing force = 89210 N,\n",
-        "Minimum passible radius of the cup which can drawn from the given blank without causing a fracture = 9.20 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 9 on page no. 135"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "L_ = 20 # Length of the mild steel product in mm\n",
-      "h = 50 # Height of the mild steel product in mm\n",
-      "L = 50 # Horizontal length of the mild steel product in mm\n",
-      "t = 5 # Thickness in mm\n",
-      "l=25 # Length of the bend in mm\n",
-      "E = 207 # Modulus of elasticity in kN/mm**2\n",
-      "n = 517 # Strain hardening rate in N/mm**2\n",
-      "Y = 345 # Yield stress in N/mm**2\n",
-      "mu = 0.1# Cofficient of friction\n",
-      "e = 0.2 # Fracture strain\n",
-      "theta = 20 # Bend angle in degree\n",
-      "\n",
-      "Rp = ((1 /((exp(e) - 1)))-0.82)*t/1.82\n",
-      "Y_1 = Y+n*e\n",
-      "Y_2 = Y + n*(log(1+(1/(2.22*(Rp/t)+1))))\n",
-      "M = ((0.55*t)**2)*((Y/6)+(Y_1/3)) + ((0.45*t)**2)*((Y/6)+(Y_2/3))\n",
-      "Fmax = (M/l)*(1+(cos((atan(mu))+mu*sin(atan(mu)))))\n",
-      "Fmax_ = L_*Fmax\n",
-      "alpha = 90 /((12*(Rp+0.45*t)*M/(E*(10**3)*(t**3)))+1)\n",
-      "Ls = 2*(((Rp+0.45*t)*pi/4) + 50-(Rp+t))\n",
-      "print \"Maximum bending force = %d N,\\nThe required puch angle = %0.2f\u00b0,\\nThe stock length = %0.2f mm\"%(Fmax_,alpha,Ls)\n",
-      "# Answer in the book for maximum bending force is given as 4144 N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum bending force = 4121 N,\n",
-        "The required puch angle = 88.68\u00b0,\n",
-        "The stock length = 89.18 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 10 on page no. 136"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "L_ = 20 # Length of the mild steel product in mm\n",
-      "h = 50 # Height of the mild steel product in mm\n",
-      "L = 50 # Horizontal length of the mild steel product in mm\n",
-      "t = 5 # Thickness in mm\n",
-      "l=25 # Length of the bend in mm\n",
-      "E = 207 # Modulus of elasticity in kN/mm**2\n",
-      "n = 517 # Strain hardening rate in N/mm**2\n",
-      "Y = 345 # Yield stress in N/mm**2\n",
-      "mu = 0.1# Cofficient of friction\n",
-      "e = 0.2 # Fracture strain\n",
-      "theta = 20 # Bend angle in degree\n",
-      "F = 3000 # Maximum available force in N\n",
-      "\n",
-      "Rp = ((1 /((exp(e) - 1)))-0.82)*t/1.82\n",
-      "Y_1 = Y+n*e\n",
-      "Y_2 = Y + n*(log(1+(1/(2.22*(Rp/t)+1))))\n",
-      "M = ((0.55*t)**2)*((Y/6)+(Y_1/3)) + ((0.45*t)**2)*((Y/6)+(Y_2/3))\n",
-      "Fmax = (M/l)*(1+(cos((atan(mu))+mu*sin(atan(mu)))))\n",
-      "Fmax_ = L_*Fmax\n",
-      "lmin  = Fmax_*l/F\n",
-      "Ls = 2*(((Rp+0.45*t)*pi/4) + 50-(Rp+t))\n",
-      "lmax = Ls / 2\n",
-      "Fmax_min  = Fmax_*l/lmax\n",
-      "print \"Minimum value of die length = %0.2f mm,\\nMinimum required capacity of the machine = %d N\"%(lmin,ceil(Fmax_min))\n",
-      "# Answer in the book is give as 2323 N for Minimum required capacity of the machine\n",
-      "\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minimum value of die length = 34.35 mm,\n",
-        "Minimum required capacity of the machine = 2312 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 11 on page no. 141"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "d = 50 # Diameter of the billet in mm\n",
-      "L =75 # Length of the billet in mm\n",
-      "D = 10 # Final diameter of billet in mm\n",
-      "Y = 170 # Avg tensile yield stress for aluminium in N/mm**2\n",
-      "mu = 0.15 # Cofficient of the friction\n",
-      "\n",
-      "l = L - ((d-D)/2)*tan(45*pi/180)\n",
-      "phi = 1+mu\n",
-      "Y_x = Y*(phi/(phi-1))*(((d/D)**(2*(phi-1)))-1)\n",
-      "F = (pi/4)*(d**2)*Y_x + (pi/sqrt(3))*(d*l*Y)\n",
-      "Pf = pi*Y*(d**2)*((phi/(2*mu))*(((d/D)**(2*mu))-1)-log(d/D)) + (pi/sqrt(3))*Y*d*l\n",
-      "Loss_f = (Pf/F)*100\n",
-      "Y_X = Y*4.31*log(d/D)\n",
-      "F_ = (pi/4)*(d**2)*Y_X + (pi/sqrt(3))*(d*l*Y)\n",
-      "Pf_1 = (pi/sqrt(3))*Y*(d**2)*(log(d/D))\n",
-      "Pf_2 = (pi/sqrt(3))*(d*l*Y)\n",
-      "Pf_ = Pf_1+Pf_2\n",
-      "Loss_f_ = (Pf_/F_)*100\n",
-      "print \"Maximum force required for extruding the cylindrical aluminium billet = %d N\"%F\n",
-      "print \"Percent of the total power input will be lost in friction at the start of the operation = %0.2f %%.\"%(Loss_f_)\n",
-      "# Answer in the book given as 2436444 N for max force required for extruding the cylindrical aluminium billet\n",
-      "\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum force required for extruding the cylindrical aluminium billet = 2436266 N\n",
-        "Percent of the total power input will be lost in friction at the start of the operation = 66.02 %.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 12 on page no. 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "d = 50 # Diameter of the steel sheet in mm\n",
-      "t = 3 # Thickness of the steel sheet in mm\n",
-      "e = 1.75 # True fracture strain\n",
-      "Y = 2.1e3 # True fracture stress for the material in N/mm**2\n",
-      "\n",
-      "C_0 = (t/(1.36*exp(e)))*((2*exp(e))-1)/((2.3*exp(e))-1)\n",
-      "p = t*(1/2.45)*((1.9*exp(e))-1)/((2.56*exp(e))-1)\n",
-      "F = Y*C_0*pi*d\n",
-      "W = (1/2)*(F)*(p)*(10**-3)\n",
-      "print \"The proper clearance between die and punch = %0.2f mm\"%C_0\n",
-      "print \"Maximum punching force = %0.2f N, \\nEnergy required to punch the hole = %0.2f J\"%(F/1000,W)\n",
-      "# Answer in the book given as 45.74 J for energy required to punch the hole\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The proper clearance between die and punch = 0.33 mm\n",
-        "Maximum punching force = 108.61 N, \n",
-        "Energy required to punch the hole = 48.10 J\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
diff --git a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch3_1.ipynb b/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch3_1.ipynb
deleted file mode 100755
index 4b196f65..00000000
--- a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch3_1.ipynb
+++ /dev/null
@@ -1,649 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:b01632c5e1d052fcab2b601597f0c6e72de59cc1f1354d4c2bee4e077a3161c7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter - 03, Forming Processes"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 1 on page no. 112"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import tan, sqrt, atan, log\n",
-      "# Given that\n",
-      "A = 150*6 # Cross-section of strips in mm**2\n",
-      "ti = 6 # Thickness in mm\n",
-      "pA = 0.20 # Reduction in area\n",
-      "d = 400 # Diameter of steel rolls in mm\n",
-      "Ys = 0.35# Shear Yield stress of the material before rolling in KN/mm**2\n",
-      "Ys_ = 0.4# Shear Yield stress of the material after rolling in KN/mm**2\n",
-      "mu = 0.1 # Cofficient of friction\n",
-      "\n",
-      "tf =0.8*ti\n",
-      "Ys_a = (Ys + Ys_)/2\n",
-      "r=d/2\n",
-      "thetaI = sqrt((ti-tf)/r)\n",
-      "lambdaI=2*sqrt(r/tf)*atan(thetaI *sqrt(r/tf))\n",
-      "lambdaN = (1/2)*((1/mu)*(log(tf/ti)) + lambdaI)\n",
-      "thetaN  =(sqrt(tf/r))*(tan((lambdaN/2)*(sqrt(tf/r))))\n",
-      "print \"\"\"The final srip thickness is %0.2f mm,\\nThe avg shear yield stress during the process is %0.3f KN/mm**2,\n",
-      "The angle subtended by the deformation zone at the roll centre is %0.4f rad, \n",
-      "The location of neutral point is %0.3f rad.\"\"\"%(tf,Ys_a,thetaI,thetaN)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The final srip thickness is 4.80 mm,\n",
-        "The avg shear yield stress during the process is 0.375 KN/mm**2,\n",
-        "The angle subtended by the deformation zone at the roll centre is 0.0775 rad, \n",
-        "The location of neutral point is 0.023 rad.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 2 on page no. 113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import exp, ceil\n",
-      "# Given that\n",
-      "A = 150*6 # Cross-section of strips in mm**2\n",
-      "w = 150 # Width of the strip in mm\n",
-      "ti = 6 # Thickness in mm\n",
-      "pA = 0.20 # Reduction in area\n",
-      "d = 400 # Diameter of steel rolls in mm\n",
-      "Ys = 0.35# Shear Yield stress of the material before rolling in KN/mm**2\n",
-      "Ys_ = 0.4# Shear Yield stress of the material after rolling in KN/mm**2\n",
-      "mu = 0.1 # Cofficient of friction\n",
-      "v = 30 # Speed of rolling in m/min\n",
-      "tf =0.8*ti\n",
-      "Ys_a = (Ys + Ys_)/2\n",
-      "r=d/2\n",
-      "thetaI = sqrt((ti-tf)/r)\n",
-      "lambdaI=2*sqrt(r/tf)*atan(thetaI *sqrt(r/tf))\n",
-      "lambdaN = (1/2)*((1/mu)*(log(tf/ti)) + lambdaI)\n",
-      "thetaN  =(sqrt(tf/r))*(tan((lambdaN/2)*(sqrt(tf/r))))\n",
-      "Dtheta_a = thetaN/4\n",
-      "Dtheta_b = (thetaI- thetaN)/8\n",
-      "print \"The values of P_after are:\\n\"\n",
-      "for i in range(0,5,1):\n",
-      "    theta = i*Dtheta_a\n",
-      "    y = (1/2)* (tf+r*theta**2)\n",
-      "    lamda = 2*sqrt(r/tf)*atan(theta*sqrt(r/tf))\n",
-      "    p_a = 2*Ys_a*(2*y/tf)*(exp(mu*lamda))\n",
-      "    print \"%0.3f \\t\\t\"%p_a,\n",
-      "\n",
-      "I1 = (Dtheta_a/3) *(0.75+.925+4*(.788+.876)+2*.830)# By Simpson's rule\n",
-      "print \"\\n\\nThe values of P_before are:\\n\"\n",
-      "for i in range(0,8,1):\n",
-      "    theta1 = i*Dtheta_b + thetaN\n",
-      "    y = (1/2)* (tf+r*theta1**2)\n",
-      "    lamda = 2*sqrt(r/tf)*atan(theta1*sqrt(r/tf))\n",
-      "    p_b = 2*Ys_a*(2*y/ti)*(exp(mu*(lambdaI-lamda)))\n",
-      "    print \" %0.3f \\t\\t\"%p_b,\n",
-      "\n",
-      "I2 = (Dtheta_b/3)*(0.925+.75+4*(.887+.828+.786+.759) + 2*(.855+.804+.772))#By Simpson's rule\n",
-      "F = r*(I1 + I2)\n",
-      "F_ = F*w\n",
-      "T = (r**2)*mu*(I2-I1)\n",
-      "T_ =T*w\n",
-      "W = v*(1000/60)/r\n",
-      "P = 2*T_*W\n",
-      "print \"\\n\\nThe roll separating force = %d kN, The power required in the rolling process = %0.2f kW\"%(ceil(F_),P/1000)\n",
-      "# Answer in the book for the power required in the rolling process is given as 75.6 kW\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The values of P_after are:\n",
-        "\n",
-        "0.750 \t\t0.787 \t\t0.829 \t\t0.874 \t\t0.924 \t\t\n",
-        "\n",
-        "The values of P_before are:\n",
-        "\n",
-        " 0.924 \t\t 0.887 \t\t 0.855 \t\t 0.828 \t\t 0.805 \t\t 0.786 \t\t 0.771 \t\t 0.759 \t\t\n",
-        "\n",
-        "The roll separating force = 1908 kN, The power required in the rolling process = 77.38 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 3 on page no. 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "# Given that\n",
-      "A = 150*6 # Cross-section of strips in mm**2\n",
-      "w = 150 # Width of the strip in mm\n",
-      "ti = 6 # Thickness in mm\n",
-      "pA = 0.20 # Reduction in area\n",
-      "d = 400 # Diameter of steel rolls in mm\n",
-      "Ys = 0.35# Shear Yield stress of the material before rolling in KN/mm**2\n",
-      "Ys_ = 0.4# Shear Yield stress of the material after rolling in KN/mm**2\n",
-      "mu = 0.1 # Cofficient of friction\n",
-      "mu_ = 0.005 # Cofficient of friction in bearing \n",
-      "D = 150 # The diameter of bearing in mm\n",
-      "v = 30 # Speed of rolling in m/min\n",
-      "tf =0.8*ti\n",
-      "Ys_a = (Ys + Ys_)/2\n",
-      "r=d/2\n",
-      "thetaI = sqrt((ti-tf)/r)\n",
-      "lambdaI=2*sqrt(r/tf)*atan(thetaI *sqrt(r/tf))\n",
-      "lambdaN = (1/2)*((1/mu)*(log(tf/ti)) + lambdaI)\n",
-      "thetaN  =(sqrt(tf/r))*(tan((lambdaN/2)*(sqrt(tf/r))))\n",
-      "Dtheta_a = thetaN/4\n",
-      "Dtheta_b = (thetaI- thetaN)/8\n",
-      "for i in range(0,5,1):\n",
-      "    theta = i*Dtheta_a\n",
-      "    y = (1/2)* (tf+r*theta**2)\n",
-      "    lamda = 2*sqrt(r/tf)*atan(theta*(pi/180) *sqrt(r/tf))*180/pi\n",
-      "    p_a = 2*Ys_a*(2*y/tf)*(exp(mu*lamda))\n",
-      "\n",
-      "I1 = (Dtheta_a/3) *(0.75+.925+4*(.788+.876)+2*.830)\n",
-      "for i in range(0,8,1):\n",
-      "    theta1 = i*Dtheta_b + thetaN\n",
-      "    y = (1/2)* (tf+r*theta1**2)\n",
-      "    lamda = 2*sqrt(r/tf)*atan(theta1*(pi/180) *sqrt(r/tf))*180/pi\n",
-      "    p_b = 2*Ys_a*(2*y/ti)*(exp(mu*(lambdaI-lamda)))\n",
-      "I2 = (Dtheta_b/3)*(0.925+.75+4*(.887+.828+.786+.759) + 2*(.855+.804+.772))\n",
-      "F = r*(I1 + I2)\n",
-      "F_ = F*w\n",
-      "T = (r**2)*mu*(I2-I1)\n",
-      "T_ =T*w\n",
-      "W = v*(1000/60)/r\n",
-      "P_ = 2*T_*W\n",
-      "Pl = mu_*F_*D*W\n",
-      "P = Pl+P_\n",
-      "print \"The mill power = %0.2f kW\"%(P/1000)\n",
-      "# Answer in the book is given as 79.18 kW  whcih is wrong.\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The mill power = 80.95 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 4 on page no. 118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "mu = 0.25 # Cofficient of friction between the job and the dies \n",
-      "Y = 7 # Avg yield stress of the lead in N/mm**2\n",
-      "h = 6 # Height of die in mm\n",
-      "L = 150 # Length of the strip in mm\n",
-      "V1 = 24*24*150 # Volume of the strip in mm**3\n",
-      "V2 = 6*96*150 # Volume of the die in mm**3\n",
-      "w= 96 # Weidth of the die in mm\n",
-      "from sympy.mpmath import quad\n",
-      "K = Y/sqrt(3)\n",
-      "x_ = (h/(2*mu))*(log(1/(2*mu)))\n",
-      "l = w/2\n",
-      "p1= lambda x:(2*K)*exp((2*mu/h)*x)\n",
-      "I1 = quad(p1,[0,x_])\n",
-      "p2 = lambda y:(2*K)*((1/2*mu)*(log(1/(2*mu))) + (y/h))\n",
-      "I2 = quad(p2,[x_,l])\n",
-      "F = 2*(I1+I2)\n",
-      "F_ = F*L\n",
-      "print \"The maximum forging force = %0.2e N\" %F_ \n",
-      "# Answer in the book is given as 0.54*10**6 N which is wrong.\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum forging force = 4.89e+05 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 5 on page no. 119"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "mu = 0.08# Cofficient of friction between the job and the dies \n",
-      "Y = 7 # Avg yield stress of the lead in N/mm**2\n",
-      "h = 6 # Height of die in mm\n",
-      "L = 150 # Length of the strip in mm\n",
-      "V1 = 24*24*150 # Volume of the strip in mm**3\n",
-      "V2 = 6*96*150 # Volume of the die in mm**3\n",
-      "w= 96 # Weidth of the die in mm\n",
-      "K = Y/sqrt(3)\n",
-      "x_ = (h/(2*mu))*(log(1/(2*mu)))\n",
-      "l = w/2\n",
-      "p1 =lambda x:(2*K)*exp((2*mu/h)*x)\n",
-      "I = quad(p1,[0,l])\n",
-      "F = 2*(I)\n",
-      "F_ = F*L\n",
-      "print \"The maximum forging force = %0.2e N\"%F_ \n",
-      "\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum forging force = 2.36e+05 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 6 on page no. 122"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "r = 150 # Radius of the circular disc of lead in mm\n",
-      "Ti = 50 # Initial thickness of the disc in mm\n",
-      "Tf = 25 # Reduced thickness of the disc in mm\n",
-      "mu = 0.25# Cofficient of friction between the job and the dies \n",
-      "K = 4 # Avg shear yield stress of the lead in N/mm**2\n",
-      "\n",
-      "R = r*sqrt(2)\n",
-      "rs = (R - ((Tf/(2*mu)) * log(1/(mu*sqrt(3)))))\n",
-      "p1 = lambda x:(((sqrt(3))*K)*exp((2*mu/Tf)*(R-x)))*x\n",
-      "I = quad(p1,[rs,R])\n",
-      "p2 = lambda y:((2*K/Tf)*(R-y) + ((K/mu)*(1+log(mu*sqrt(3)))))*y\n",
-      "I_ = quad(p2,[0,rs])\n",
-      "F = 2*pi*(I+I_)\n",
-      "print \"The maximum forging force = %0.3e N\"%F \n",
-      "\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum forging force = 3.649e+06 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 7 on page no. 126"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "Di = 12.7 # Intial diameter in mm\n",
-      "Df = 10.2 # Final diameter in mm\n",
-      "v = 90 # Drawn speed in m/min\n",
-      "alpha=6 # Half angle of dia in degree\n",
-      "mu = 0.1# Cofficient of friction between the job and the dies \n",
-      "Y = 207 # Tensile  yield stress of the steel specimen in N/mm**2\n",
-      "Y_ = 414 # Tensile yield stress of the similar specimen at strain 0.5 in N/mm**2\n",
-      "e = 0.5 # Strain\n",
-      "\n",
-      "e_ =2* log(Di/Df)\n",
-      "Y_e = Y + (Y_ - Y)*e_/e\n",
-      "Y__ = (Y+Y_e)/2\n",
-      "phi = 1 + (mu/tan(alpha*pi/180))\n",
-      "Y_f = Y__ * ((phi/(phi-1)) *(1-((Df/Di)**(2*(phi-1)))))\n",
-      "p = Y_f * (pi/4)*(Df**2)*v/60\n",
-      "Dmax = 1- (1/(phi**(1/(phi-1))))\n",
-      "print \"Drawing power = %0.3f kW, \\nThe maximum passible reduction with same die = %0.2f mm\"%(p/1000,Dmax)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drawing power = 25.530 kW, \n",
-        "The maximum passible reduction with same die = 0.50 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 8 on page no. 130"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "Ri = 30 # Inside radius of cup in mm\n",
-      "t = 3 # Thickness in mm\n",
-      "Rb = 40 # Radius of the blank in mm\n",
-      "K = 210 # Shear yield stress of the material in N/mm**2\n",
-      "Y = 600 # Maximum allowable stress in N/mm**2\n",
-      "Beta = 0.05\n",
-      "mu = 0.1# Cofficient of friction between the job and the dies \n",
-      "\n",
-      "Fh = Beta*pi*(Rb**2)*K\n",
-      "Y_r = (mu*Fh/(pi*Rb*t))+(2*K*log(Rb/Ri))\n",
-      "Y_z = Y_r*exp(mu*pi/2)\n",
-      "F = 2*pi*Ri*t*Y_z\n",
-      "Y_r_ = Y/exp(mu*pi/2)\n",
-      "Rp = (Rb/exp((Y_r_/(2*K)) - ((mu*Fh)/(2*pi*K*Rb*t))))-t\n",
-      "print \"Drawing force = %d N,\\nMinimum passible radius of the cup which can drawn from the given blank without causing a fracture = %0.2f mm\"%(F,Rp)\n",
-      "# Answer in the book given as 62680 N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drawing force = 89210 N,\n",
-        "Minimum passible radius of the cup which can drawn from the given blank without causing a fracture = 9.20 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 9 on page no. 135"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import cos, sin\n",
-      "# Given that\n",
-      "L_ = 20 # Length of the mild steel product in mm\n",
-      "h = 50 # Height of the mild steel product in mm\n",
-      "L = 50 # Horizontal length of the mild steel product in mm\n",
-      "t = 5 # Thickness in mm\n",
-      "l=25 # Length of the bend in mm\n",
-      "E = 207 # Modulus of elasticity in kN/mm**2\n",
-      "n = 517 # Strain hardening rate in N/mm**2\n",
-      "Y = 345 # Yield stress in N/mm**2\n",
-      "mu = 0.1# Cofficient of friction\n",
-      "e = 0.2 # Fracture strain\n",
-      "theta = 20 # Bend angle in degree\n",
-      "\n",
-      "Rp = ((1 /((exp(e) - 1)))-0.82)*t/1.82\n",
-      "Y_1 = Y+n*e\n",
-      "Y_2 = Y + n*(log(1+(1/(2.22*(Rp/t)+1))))\n",
-      "M = ((0.55*t)**2)*((Y/6)+(Y_1/3)) + ((0.45*t)**2)*((Y/6)+(Y_2/3))\n",
-      "Fmax = (M/l)*(1+(cos((atan(mu))+mu*sin(atan(mu)))))\n",
-      "Fmax_ = L_*Fmax\n",
-      "alpha = 90 /((12*(Rp+0.45*t)*M/(E*(10**3)*(t**3)))+1)\n",
-      "Ls = 2*(((Rp+0.45*t)*pi/4) + 50-(Rp+t))\n",
-      "print \"Maximum bending force = %d N,\\nThe required puch angle = %0.2f\u00b0,\\nThe stock length = %0.2f mm\"%(Fmax_,alpha,Ls)\n",
-      "# Answer in the book for maximum bending force is given as 4144 N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum bending force = 4121 N,\n",
-        "The required puch angle = 88.68\u00b0,\n",
-        "The stock length = 89.18 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 10 on page no. 136"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "L_ = 20 # Length of the mild steel product in mm\n",
-      "h = 50 # Height of the mild steel product in mm\n",
-      "L = 50 # Horizontal length of the mild steel product in mm\n",
-      "t = 5 # Thickness in mm\n",
-      "l=25 # Length of the bend in mm\n",
-      "E = 207 # Modulus of elasticity in kN/mm**2\n",
-      "n = 517 # Strain hardening rate in N/mm**2\n",
-      "Y = 345 # Yield stress in N/mm**2\n",
-      "mu = 0.1# Cofficient of friction\n",
-      "e = 0.2 # Fracture strain\n",
-      "theta = 20 # Bend angle in degree\n",
-      "F = 3000 # Maximum available force in N\n",
-      "\n",
-      "Rp = ((1 /((exp(e) - 1)))-0.82)*t/1.82\n",
-      "Y_1 = Y+n*e\n",
-      "Y_2 = Y + n*(log(1+(1/(2.22*(Rp/t)+1))))\n",
-      "M = ((0.55*t)**2)*((Y/6)+(Y_1/3)) + ((0.45*t)**2)*((Y/6)+(Y_2/3))\n",
-      "Fmax = (M/l)*(1+(cos((atan(mu))+mu*sin(atan(mu)))))\n",
-      "Fmax_ = L_*Fmax\n",
-      "lmin  = Fmax_*l/F\n",
-      "Ls = 2*(((Rp+0.45*t)*pi/4) + 50-(Rp+t))\n",
-      "lmax = Ls / 2\n",
-      "Fmax_min  = Fmax_*l/lmax\n",
-      "print \"Minimum value of die length = %0.2f mm,\\nMinimum required capacity of the machine = %d N\"%(lmin,ceil(Fmax_min))\n",
-      "# Answer in the book is give as 2323 N for Minimum required capacity of the machine\n",
-      "\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minimum value of die length = 34.35 mm,\n",
-        "Minimum required capacity of the machine = 2312 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 11 on page no. 141"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "d = 50 # Diameter of the billet in mm\n",
-      "L =75 # Length of the billet in mm\n",
-      "D = 10 # Final diameter of billet in mm\n",
-      "Y = 170 # Avg tensile yield stress for aluminium in N/mm**2\n",
-      "mu = 0.15 # Cofficient of the friction\n",
-      "\n",
-      "l = L - ((d-D)/2)*tan(45*pi/180)\n",
-      "phi = 1+mu\n",
-      "Y_x = Y*(phi/(phi-1))*(((d/D)**(2*(phi-1)))-1)\n",
-      "F = (pi/4)*(d**2)*Y_x + (pi/sqrt(3))*(d*l*Y)\n",
-      "Pf = pi*Y*(d**2)*((phi/(2*mu))*(((d/D)**(2*mu))-1)-log(d/D)) + (pi/sqrt(3))*Y*d*l\n",
-      "Loss_f = (Pf/F)*100\n",
-      "Y_X = Y*4.31*log(d/D)\n",
-      "F_ = (pi/4)*(d**2)*Y_X + (pi/sqrt(3))*(d*l*Y)\n",
-      "Pf_1 = (pi/sqrt(3))*Y*(d**2)*(log(d/D))\n",
-      "Pf_2 = (pi/sqrt(3))*(d*l*Y)\n",
-      "Pf_ = Pf_1+Pf_2\n",
-      "Loss_f_ = (Pf_/F_)*100\n",
-      "print \"Maximum force required for extruding the cylindrical aluminium billet = %d N\"%F\n",
-      "print \"Percent of the total power input will be lost in friction at the start of the operation = %0.2f %%.\"%(Loss_f_)\n",
-      "# Answer in the book given as 2436444 N for max force required for extruding the cylindrical aluminium billet\n",
-      "\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum force required for extruding the cylindrical aluminium billet = 2436266 N\n",
-        "Percent of the total power input will be lost in friction at the start of the operation = 66.02 %.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 12 on page no. 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "d = 50 # Diameter of the steel sheet in mm\n",
-      "t = 3 # Thickness of the steel sheet in mm\n",
-      "e = 1.75 # True fracture strain\n",
-      "Y = 2.1e3 # True fracture stress for the material in N/mm**2\n",
-      "\n",
-      "C_0 = (t/(1.36*exp(e)))*((2*exp(e))-1)/((2.3*exp(e))-1)\n",
-      "p = t*(1/2.45)*((1.9*exp(e))-1)/((2.56*exp(e))-1)\n",
-      "F = Y*C_0*pi*d\n",
-      "W = (1/2)*(F)*(p)*(10**-3)\n",
-      "print \"The proper clearance between die and punch = %0.2f mm\"%C_0\n",
-      "print \"Maximum punching force = %0.2f N, \\nEnergy required to punch the hole = %0.2f J\"%(F/1000,W)\n",
-      "# Answer in the book given as 45.74 J for energy required to punch the hole\n",
-      " "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The proper clearance between die and punch = 0.33 mm\n",
-        "Maximum punching force = 108.61 N, \n",
-        "Energy required to punch the hole = 48.10 J\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch4.ipynb b/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch4.ipynb
deleted file mode 100755
index 312554ae..00000000
--- a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch4.ipynb
+++ /dev/null
@@ -1,1189 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter4 - Machining Processes"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.1 - page 187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import atan, tan, cos , sin, degrees, pi\n",
-      "# Given that\n",
-      "alpha = 10 # Rake angle in Degree\n",
-      "t = 0.4 # Chip thickness  in mm\n",
-      "T = 0.15 # Uncut chip thickness in mm\n",
-      "\n",
-      "r = T/t\n",
-      "phi = degrees(atan((r*cos(alpha*pi/180))/(1-r*sin(alpha*pi/180))))\n",
-      "gama = 1/tan(phi*pi/180) + tan((phi-alpha)*pi/180)\n",
-      "print \"Shear plane angle = %0.1f\u00b0, \\nMagnitude of the shear strain = %0.2f\"%(phi,gama)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Shear plane angle = 21.6\u00b0, \n",
-        "Magnitude of the shear strain = 2.74\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.2 - page 192"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "t1 = 0.25 # Undercut thickness in mm\n",
-      "t2 = 0.75 # Chip thickness in mm\n",
-      "w = 2.5 # Width in mm\n",
-      "alpha = 0 # Rake angle in Degree\n",
-      "Fc = 950 # Cutting force in N\n",
-      "Ft = 475 # Thrust force in N\n",
-      "\n",
-      "r = t1/t2\n",
-      "mu = ((Fc*sin(alpha*pi/180)) + (Ft*cos(alpha*pi/180)))/((Fc*cos(alpha*pi/180))-(Ft*sin(alpha*pi/180)))\n",
-      "phi = degrees(atan((r*cos(alpha*pi/180))/(1-r*sin(alpha*pi/180))))\n",
-      "As = t1*w/sin(phi*pi/180)\n",
-      "Fs = Fc*cos(phi*pi/180) - Ft*sin(phi*pi/180)\n",
-      "T_s = Fs/As\n",
-      "print \"\"\"Coefficient of the friction between tool and the chip = %0.1f,\n",
-      "The ultimate shear stress of the material = %0.1f N/mm**2\"\"\"%(mu,T_s)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Coefficient of the friction between tool and the chip = 0.5,\n",
-        "The ultimate shear stress of the material = 380.0 N/mm**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.3 - page 193"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "alpha = 10 # Rake angle of tool in Degree\n",
-      "v = 200 # Cutting speed in m/min\n",
-      "t1 = 0.2 # Uncut thickness in mm\n",
-      "w = 2 # Width of cut in mm\n",
-      "mu = 0.5 # Avg value of the cofficient of tbe friction\n",
-      "T_S = 400 # Shear stress of the work material in N/mm**2\n",
-      "\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = (90 + alpha - lamda)/2\n",
-      "Fs = (w*t1*T_S)/(sin(phi*pi/180))\n",
-      "R = Fs/(cos((phi+lamda-alpha)*pi/180))\n",
-      "Fc = R*(cos((lamda-alpha)*pi/180))\n",
-      "Ft = R*(sin((lamda-alpha)*pi/180))\n",
-      "print \"Shear angle = %0.1f\u00b0, \\nCutting force = %0.f N, \\nThrust force = %d N,\"%(phi,Fc,Ft)\n",
-      "# Answer in the book for cutting force is given as 420 N and for thrust force is given as 125 N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Shear angle = 36.7\u00b0, \n",
-        "Cutting force = 429 N, \n",
-        "Thrust force = 127 N,\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.4 - page 194"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "alpha = 10 # Rake angle of tool in Degree\n",
-      "v = 200 # Cutting speed in m/min\n",
-      "t1 = 0.2 # Uncut thickness in mm\n",
-      "w = 2 # Width of cut in mm\n",
-      "mu = 0.5 # Avg value of the cofficient of tbe friction\n",
-      "T_S = 400 # Shear stress of the work material in N/mm**2\n",
-      "Cm = 70 # Machining constant in Degree\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = (Cm + alpha - lamda)/2\n",
-      "Fs = (w*t1*T_S)/(sin(phi*pi/180))\n",
-      "R = Fs/(cos((phi+lamda-alpha)*pi/180))\n",
-      "Fc = R*(cos((lamda-alpha)*pi/180))\n",
-      "Ft = R*(sin((lamda-alpha)*pi/180))\n",
-      "# Using Lee and Shaffer relation \n",
-      "phi_ = 45-lamda+alpha\n",
-      "Fs_ = (w*t1*T_S)/(sin(phi_*pi/180))\n",
-      "R_ = Fs_/(cos((phi_+lamda-alpha)*pi/180))\n",
-      "Fc_ = R_*(cos((lamda-alpha)*pi/180))\n",
-      "Ft_ = R_*(sin((lamda-alpha)*pi/180))\n",
-      "print \"\"\"Shear angle = %0.1f\u00b0,\n",
-      "Cutting force = %0.2f N,\n",
-      "Thrust force = %0.2f N\n",
-      "Using Lee and Shaffer relation- \n",
-      " Shear angle = %0.2f\u00b0,\n",
-      " Cutting force = %0.2f N,\n",
-      " Thrust force = %0.2f N,\"\"\"%(phi,Fc,Ft,phi_,Fc_,Ft_)\n",
-      "# Answer in the book for cutting force is given as 486.9 N and for thrust force is given as 144.9 N , When using Lee and Shaffer relation answer in the book for cutting force is given as 481.9 N and for trust force is given as 160.6 N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Shear angle = 26.7\u00b0,\n",
-        "Cutting force = 468.57 N,\n",
-        "Thrust force = 139.37 N\n",
-        "Using Lee and Shaffer relation- \n",
-        " Shear angle = 28.43\u00b0,\n",
-        " Cutting force = 455.48 N,\n",
-        " Thrust force = 135.48 N,\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.5 - page 196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "t1 = 0.25 # Uncut thickness in mm\n",
-      "w = 2.5 # Width of cut in mm\n",
-      "U_0 = 1.4 # In J/mm**3\n",
-      "alpha = 0 # Rake angle in degree\n",
-      "mu = 0.5 # Cofficient of the friction\n",
-      "T_s = 400 # Shear stress in N/mm**2\n",
-      "\n",
-      "lamda = degrees(atan(mu))\n",
-      "Fc = 1000*(t1*w*U_0)*((t1)**(-.4))\n",
-      "phi = 45 + alpha - degrees(atan(mu))\n",
-      "Fc_ = (w*t1*T_s*cos((lamda-alpha)*pi/180))/((sin(phi*pi/180)) *cos((phi+lamda-alpha)*pi/180))\n",
-      "print \"\"\"The order of magnitude of cutting force = %d N,\n",
-      "Using Lee and Shaffer relation-\n",
-      " The order of magnitude of cutting force = %d N.\"\"\"%(Fc,Fc_)\n",
-      "# Answer in the book for cutting force is given as 1517 N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The order of magnitude of cutting force = 1523 N,\n",
-        "Using Lee and Shaffer relation-\n",
-        " The order of magnitude of cutting force = 999 N.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.6 - page 199"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log, sqrt\n",
-      "# Given that\n",
-      "v = 2 # Cutting speed in m/sec\n",
-      "D = 7200 # Density of mild steel in kg /m**3\n",
-      "k = 43.6 # Thermal conductivity in W/m-\u00b0c\n",
-      "c = 502 # Specific heat of the material in J/kg-\u00b0c\n",
-      "t1 = 0.25 # Uncut thickness in mm\n",
-      "w =2 # Width of cut in mm\n",
-      "theta_0 = 40 # Initial temp of the workpiece in Degree\n",
-      "alpha = 0 # Rake angle in degree\n",
-      "mu = 0.5 # Cofficient of the friction\n",
-      "T_s = 400e6 # Shear stress in N/m**2\n",
-      "\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = 45 + alpha - lamda\n",
-      "Fs = (w*t1*T_s)*(10**-6)/(sin(phi*pi/180))\n",
-      "R = Fs / (cos((phi+lamda-alpha)*pi/180))\n",
-      "Fc = R *(cos((lamda-alpha)*pi/180))\n",
-      "r = sin(phi*pi/180)/(cos((phi-alpha)*pi/180))\n",
-      "Ft= Fc *(tan((lamda - alpha)*pi/180))\n",
-      "F = Fc *(sin(alpha*pi/180))+Ft*(cos(alpha*pi/180))\n",
-      "Ws = F*r*v\n",
-      "Wp = Fc*v-F*r*v\n",
-      "zeta = D*c*v*t1*(10**-3)/k\n",
-      "zeta_ = zeta*tan(phi*pi/180)\n",
-      "nu = 0.15 *(log(27.5/(zeta_)))\n",
-      "theta_P = (1-nu)*Wp/(D*c*v*t1*w*(10**-6))\n",
-      "theta_S = 1.13 *(sqrt(1/(D*c*v*t1*(10**-3)*k*(1+tan((phi-alpha)*pi/180))))*(Ws/w)*(10**3))                 \n",
-      "theta = theta_0+theta_S+ theta_P\n",
-      "print \" \\n Maximum temperature along the rake face of the tool = %0.1f\u00b0C.\"%theta\n",
-      "# Answer in the book  is given as 823\u00b0C"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " Maximum temperature along the rake face of the tool = 835.6\u00b0C.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.7 - Page 206"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "theta_ = 40 #Ambient temperature in\u00b0C\n",
-      "v = 2 # Cutting speed in m/sec\n",
-      "D = 7200 # Density of mild steel in kg /m**3\n",
-      "k = 43.6 # Thermal conductivity in W/m-\u00b0c\n",
-      "c = 502 # Specific heat of the material in J/kg-\u00b0c\n",
-      "t1 = 0.25 # Uncut thickness in mm\n",
-      "w =2 # Width of cut in mm\n",
-      "alpha = 0 # Rake angle in degree\n",
-      "mu = 0.5 # Cofficient of the friction\n",
-      "T_s = 400e6 # Shear stress in N/m**2\n",
-      "H = 350 # Hardness of SAE 1040 steel in HV(Vicker hardness)\n",
-      "\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = 45 + alpha - lamda\n",
-      "Fs = (w*t1*T_s)*(10**-6)/(sin(phi*pi/180))\n",
-      "R = Fs / (cos((phi+lamda-alpha)*pi/180))\n",
-      "Fc = R *(cos((lamda-alpha)*pi/180))\n",
-      "r = sin(phi*pi/180)/(cos((phi-alpha)*pi/180))\n",
-      "Ft= Fc *(tan((lamda - alpha)*pi/180))\n",
-      "F = Fc *(sin(alpha*pi/180))+Ft*(cos(alpha*pi/180))\n",
-      "Ws = F*r*v\n",
-      "Wp = Fc*v-F*r*v\n",
-      "zeta = D*c*v*t1*(10**-3)/k\n",
-      "zeta_ = zeta*tan(phi*pi/180)\n",
-      "nu = 0.15 *(log(27.5/(zeta_)))\n",
-      "Theta_0v = ((1-nu)*Wp + Ws)/ (D*c*v*t1*w*(10**-6))\n",
-      "H_ = 1.5 *(H)\n",
-      "theta_lim = 700*((1-(H_/850))**(1/3.1))\n",
-      "v_lim = (theta_lim/309)**(1/0.5)\n",
-      "print \" \\n Maximum speed at which cutting is passible = %0.2f m/sec.\"%v_lim"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " Maximum speed at which cutting is passible = 2.76 m/sec.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.8 - page 212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "alpha = 0 # Rake angle in degree\n",
-      "gama = 3 # Clearance angle in Degree\n",
-      "w = 1 # Maximum length of flank wear allowed in mm\n",
-      "gama_ = 7 # Increased clearance angle in Degree\n",
-      "I_per = (((tan(gama_*pi/180))-(tan(gama*pi/180)))/tan(gama*pi/180))*100\n",
-      "print \" \\n Percentage increase in tool life = %d percent.\"%I_per"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " Percentage increase in tool life = 134 percent.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 9 on page no. 220"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "d= 4 # Depth of cut in mm\n",
-      "f = 0.25 # Feed in mm/stroke\n",
-      "alpha = 10 # Rake angle in degree\n",
-      "shi = 30 # Principal cutting edge angle in Degree\n",
-      "mu =0.6 # Cofficient of friction between chip and tool\n",
-      "T_s = 340 # Ultimate shear stress of cast iron in N/mm**2\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = 45 +alpha-lamda\n",
-      "Fc = f*d*T_s*(cos((lamda-alpha)*pi/180))/((sin(phi*pi/180))*(cos((phi+lamda-alpha)*pi/180)))\n",
-      "Ft = Fc*(sin((lamda-alpha)*pi/180))/(cos((lamda-alpha)*pi/180))\n",
-      "Ff = Ft*(cos(shi*pi/180))\n",
-      "Fn = Ft*(sin(shi*pi/180))\n",
-      "print \"\"\"The three components of machinig force are as follows-\n",
-      "  Thrust force = %d N,\n",
-      "  Feed force component = %d N,\n",
-      "  Normal thrust force component = %d N.\"\"\"%(Ft,Ff,Fn)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The three components of machinig force are as follows-\n",
-        "  Thrust force = 422 N,\n",
-        "  Feed force component = 365 N,\n",
-        "  Normal thrust force component = 211 N.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 10 on page no. 221"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "d= 4 # Depth of cut in mm\n",
-      "f = 0.25 # Feed in mm/stroke\n",
-      "alpha = 10 # Rake angle in degree\n",
-      "shi = 30 # Principal cutting edge angle in Degree\n",
-      "mu =0.6 # Cofficient of friction between chip and tool\n",
-      "T_s = 340 # Ultimate shear stress of cast iron in N/mm**2\n",
-      "N = 60 # Cutting stroke/min\n",
-      "L = 200 # Length of the job in mm\n",
-      "H = 180 # Hardness of the workpiece in BHN\n",
-      "\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = 45 +alpha-lamda\n",
-      "Fc = f*d*T_s*(cos((lamda-alpha)*pi/180))/((sin(phi*pi/180))*(cos((phi+lamda-alpha)*pi/180)))\n",
-      "Fc_ = Fc*(L/1000)\n",
-      "Wav  =Fc_*N/60\n",
-      "t1 = f*cos(shi*pi/180)\n",
-      "U_0 = 0.81 # By using table 4.4 given in the book, In J/mm**3\n",
-      "Uc = U_0*((t1)**(-.4))\n",
-      "Q = f*d*L*N/60\n",
-      "Wav_ = Uc*Q\n",
-      "print \"Avg power consumption = %d W,\\nSpecific power consumption when hardness of the workpiece is 180 BHN = %d W.\"%(Wav,Wav_)\n",
-      "# Answer in the book for Specific power consumption is given as 294 W"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Avg power consumption = 220 W,\n",
-        "Specific power consumption when hardness of the workpiece is 180 BHN = 298 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 11 on page no. 224"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "alpha_b = 6 # Back rake angle in Degree\n",
-      "alpha_s = 10 # Side rake angle in Degree\n",
-      "gama = 7 # Front clearance angle in Degree\n",
-      "gama_ = 7 # Side clearance angle in Degree\n",
-      "Shi = 10 # End cutting edge angle in Degree\n",
-      "shi = 30 # Side cutting edge angle in Degree\n",
-      "r= 0.5 # Nose radius in mm\n",
-      "\n",
-      "k = tan(alpha_b*pi/180) * cos(shi*pi/180) - tan(alpha_s*pi/180) * sin(shi*pi/180)\n",
-      "print \"The value of k=%0.4f,which is near to 0. Hence the case is close to orthogonal one.\\n\"%k\n",
-      "alpha= degrees(atan(((tan(alpha_b*pi/180)) * sin(shi*pi/180) ) + (tan(alpha_s*pi/180) * (cos(shi*pi/180))))/ (sqrt(1+((tan(alpha_b*pi/180)*cos(shi*pi/180)) - (tan(alpha_s*pi/180)*sin(shi*pi/180)))**(2))))\n",
-      "print \"Normal rake angle = %0.1f\u00b0.\"%(alpha)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of k=0.0029,which is near to 0. Hence the case is close to orthogonal one.\n",
-        "\n",
-        "Normal rake angle = 11.6\u00b0.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 12 on page no. 225"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "# Given that\n",
-      "alpha_b = 6 # Back rake angle in Degree\n",
-      "alpha_s = 10 # Side rake angle in Degree\n",
-      "gama = 5 # Front clearance angle in Degree\n",
-      "gama_ = 7 # Side clearance angle in Degree\n",
-      "Shi = 10 # End cutting edge angle in Degree\n",
-      "shi = 30 # Side cutting edge angle in Degree\n",
-      "r= 0.55 # Nose radius in mm\n",
-      "d = 2.5 # Depth of cut in mm\n",
-      "f = 0.125 # Feed in mm/revolution\n",
-      "N = 300 # Rpm of the job\n",
-      "T_S = 400 # Ultimate shear stress of the workpiece in N/mm**2\n",
-      "mu = .6 # Cofficient of the friction between the tool and the chip\n",
-      "\n",
-      "lamda = degrees(atan(mu))\n",
-      "alpha= degrees(atan(((tan(alpha_b*pi/180)) * sin(shi*pi/180) ) + (tan(alpha_s*pi/180) * (cos(shi*pi/180))))/ (sqrt(1+((tan(alpha_b*pi/180)*cos(shi*pi/180)) - (tan(alpha_s*pi/180)*sin(shi*pi/180)))**(2))))\n",
-      "phi = 45 + alpha - lamda\n",
-      "t1 = f*cos(phi*pi/180)\n",
-      "w = d/cos(phi*pi/180)\n",
-      "Fc = w*t1*T_S*(cos((lamda-alpha)*pi/180))/((sin(phi*pi/180))*(cos((phi+lamda-alpha)*pi/180)))\n",
-      "Ft = Fc*tan((lamda-alpha)*pi/180)\n",
-      "Ff = Ft*cos(shi*pi/180)\n",
-      "Fr = Ft*sin(shi*pi/180)\n",
-      "print \"Component of the machining force are as follows -\\n Feed force component = % d N, \\n Normal thrust force component = % d N.\"%(ceil(Ff),ceil(Fr))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Component of the machining force are as follows -\n",
-        " Feed force component =  118 N, \n",
-        " Normal thrust force component =  68 N.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 14 on page no. 230"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "D = 20 # Nominal diameter of the drill in mm\n",
-      "T_S = 400 # Shear yield stress of work material in N/mm**2\n",
-      "N = 240 # Rpm\n",
-      "f = 0.25 # Feed in mm/revolution\n",
-      "mu = 0.6 # Cofficient of friction\n",
-      "\n",
-      "Beta = 118/2 # From the table 4.12 given in the book\n",
-      "shi = 30 # From the table 4.12 given in the book\n",
-      "alpha = degrees(atan(((2*(D/4)/(D)))*tan(shi*pi/180))/sin(Beta*pi/180))\n",
-      "t1 = (f/2)*sin(Beta*pi/180)\n",
-      "w = (D/2)/sin(Beta*pi/180)\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = 45+alpha-lamda\n",
-      "t1 = f/2\n",
-      "Fc = w*t1*T_S*(cos((lamda-alpha)*pi/180))/((sin(phi*pi/180))*(cos((phi+lamda-alpha)*pi/180)))\n",
-      "Ft = w*t1*T_S*(sin((lamda-alpha)*pi/180))/((sin(phi*pi/180))*(cos((phi+lamda-alpha)*pi/180)))\n",
-      "M = .6*Fc*D/1000\n",
-      "F = 5*Ft*sin(Beta*pi/180)\n",
-      "print \"The drilling torque = %0.2f N-m, \\nThrust force = %d N.\"%(M,F)\n",
-      "# Answer in the book for drilling torque is given as 18.2 N-m, and for thrust force is given as 1500 N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drilling torque = 17.85 N-m, \n",
-        "Thrust force = 1376 N.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 15 on page no. 235"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "# Given that\n",
-      "w = 20 # Width of the mild steel block in mm\n",
-      "Z = 20 # No of teeth in milling cutter\n",
-      "D = 50 # Diameter of the milling cutter in mm\n",
-      "alpha = 10 # Radial rake angle in Degree\n",
-      "f = 15 # Feed velocity of the table in mm/min\n",
-      "N =60 # Rpm of the cutter\n",
-      "t = 1 # Depth of cut in mm\n",
-      "mu = 0.5 # Cofficient of friction\n",
-      "T_s = 400 # Shear yield stress in N/mm**2\n",
-      "t_a = 0.0018 # Avg uncut thickness in mm\n",
-      "\n",
-      "Beta = degrees(asin(2*(t/D)))\n",
-      "theta = 2*pi/Z\n",
-      "t1_max = (2*f/(N*Z))*sqrt(t/D)\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = 45+alpha -lamda\n",
-      "Fc_max = ((w*t1_max*T_s*cos((lamda-alpha)*pi/180)))/((sin(phi*pi/180))*(cos(45*pi/180)))\n",
-      "T_max = Fc_max*D/(2*1000)\n",
-      "M_av = (1/2)*(Beta*T_max)/theta\n",
-      "omega = 2*pi*N/60\n",
-      "U_0 = 1.4 # From the table 4.4 given in the book\n",
-      "Uc_ms = U_0*((t_a)**(-0.4))\n",
-      "R = f*t*w/60\n",
-      "U = Uc_ms * R\n",
-      "print \"Power consumption = %0.1f W.\"%U"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power consumption = 87.7 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 16 on page no. 238"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "w = 20 # Width of the mild steel block in mm\n",
-      "Z = 10 # No of teeth in milling cutter\n",
-      "D = 75 # Diameter of the milling cutter in mm\n",
-      "alpha = 10 # Radial rake angle in Degree\n",
-      "f = 100 # Feed velocity of the table in mm/min\n",
-      "N =60 # Rpm of the cutter\n",
-      "t = 5 # Depth of cut in mm\n",
-      "mu = 0.5 # Cofficient of friction\n",
-      "T_s = 400 # Shear yield stress in N/mm**2\n",
-      "t_a = 0.043 # Avg uncut thickness in mm\n",
-      "\n",
-      "Beta = degrees(asin(2*(t/D)))\n",
-      "theta = 2*pi/Z\n",
-      "t1_max = (2*f/(N*Z))*sqrt(t/D)\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = 45+alpha -lamda\n",
-      "Fc_max = ((w*t1_max*T_s*cos((lamda-alpha)*pi/180)))/((sin(phi*pi/180))*(cos(45*pi/180)))\n",
-      "T_max = Fc_max*D/(2*1000)\n",
-      "M_av = (1/2)*(Beta*T_max)/theta\n",
-      "omega = 2*pi*N/60\n",
-      "U_0 = 1.4 # From the table 4.4 given in the book\n",
-      "Uc_ms = U_0*((t_a)**(-0.4))\n",
-      "R = f*t*w/60\n",
-      "U = Uc_ms * R\n",
-      "print \"Power required = %d W.\"%U\n",
-      "# Answer in the book for Power required is given as 817 W"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power required = 821 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 17 on page no. 240"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "B = 20 # Width of the cut in mm\n",
-      "Z = 10 # No of teeth in milling cutter\n",
-      "D = 75 # Diameter of the milling cutter in mm\n",
-      "alpha = 10 # Radial rake angle in Degree\n",
-      "f = 25 # Feed velocity of the table in mm/min\n",
-      "N =60 # Rpm of the cutter\n",
-      "t = 5 # Depth of cut in mm\n",
-      "mu = 0.5 # Cofficient of friction\n",
-      "T_s = 400 # Shear yield stress in N/mm**2\n",
-      "t_a = 0.043 # Avg uncut thickness in mm\n",
-      "t1_max = 0.01\n",
-      "lamda = 0.28 # From the table 4.13 Given in the book\n",
-      "nu = 1400 # From the table 4.13 Given in the book\n",
-      "t1_av = t1_max/2\n",
-      "P = nu*B*t*f*(10**-4)/(6*((t1_av)**(lamda)))\n",
-      "print \"Power required = %0.2f W.\"%P"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power required = 257.16 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 18 on page no. 240"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "w = 20 # Width of the mild steel block in mm\n",
-      "Z = 10 # No of teeth in milling cutter\n",
-      "D = 75 # Diameter of the milling cutter in mm\n",
-      "alpha = 10 # Radial rake angle in Degree\n",
-      "f = 25 # Feed velocity of the table in mm/min\n",
-      "N =60 # Rpm of the cutter\n",
-      "t = 5 # Depth of cut in mm\n",
-      "mu = 0.5 # Cofficient of friction\n",
-      "T_s = 400 # Shear yield stress in N/mm**2\n",
-      "t_a = 0.043 # Avg uncut thickness in mm\n",
-      "\n",
-      "R = f*t*w/60\n",
-      "Uc = 3.3 # Specific energy in J/mm**3 from the table 4.14 Given in the book\n",
-      "U = Uc * R\n",
-      "print \"Power required = %d W.\"%ceil(U)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power required = 138 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 19 on page no. 241"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "d = 25 # Diameter of circular hole in mm\n",
-      "t = 20 # Thickness of the steel plate in mm\n",
-      "D = 27 # Enlarged diameter of hole in mm\n",
-      "c= 0.08 # Cut per tooth in mm\n",
-      "alpha = 10 # Radial rake angle in Degree\n",
-      "mu = 0.5 # Cofficient of friction\n",
-      "T_s = 400 # Shear yield stress in N/mm**2\n",
-      "\n",
-      "lamda=degrees(atan(mu))\n",
-      "phi = 45-lamda+alpha\n",
-      "w = pi*(d+D)/2\n",
-      "Fc = w*c*T_s*(cos((lamda-alpha)*pi/1800)/((sin(phi*pi/180))*(cos(45*pi/180))))\n",
-      "s = 1.75*sqrt(t)\n",
-      "F = 3*Fc\n",
-      "print \"Peak broaching load = %d N.\"%ceil(F)\n",
-      "# Answer in the book  is given as 22323 N which is wrong."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Peak broaching load = 23280 N.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 20 on page no. 246"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "D = 250 # Diameter of the wheel in mm\n",
-      "N = 2000 # Rpm of the wheel\n",
-      "f =5 # Plung feed rate in mm/min\n",
-      "C = 3 # Surface density of active grain in mm**-2\n",
-      "A = 20*15 # Area of mild steel prismatic bar in mm**2\n",
-      "rg = 15 # In mm**-1\n",
-      "\n",
-      "t1 = sqrt(f/(pi*D*N*C*rg))\n",
-      "U_0 = 1.4 # From the table 4.4 given in the book\n",
-      "Uc= U_0*((t1)**(-.4))\n",
-      "R = A*f/60\n",
-      "P = Uc*R\n",
-      "Fc_ = 60000*(P)/(pi*D*A*C*N)\n",
-      "print \"Power requirement during plunge grinding of the mild steel primatic bar = %d W.\"%ceil(P)\n",
-      "# Answer in the book is given as 94 W which is wrong."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power requirement during plunge grinding of the mild steel primatic bar = 943 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 21 on page no. 248"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "w = 25 # Width of mild steel block in mm\n",
-      "d= 0.05 # Depth of cut in mm\n",
-      "D = 200 # Diameter of the wheel in mm\n",
-      "N = 3000 # Rpm of the wheel\n",
-      "f =100 # Feed velocity of table in mm/min\n",
-      "C = 3 # No of grits in mm**-2\n",
-      "rg = 15 # In mm**-1\n",
-      "\n",
-      "t1_max = sqrt(((6*f)/(pi*D*N*C*rg))*sqrt(d/D))\n",
-      "t1_a = t1_max/2\n",
-      "U_0 = 1.4 # From the table 4.4 given in the book\n",
-      "Uc= U_0*((t1_a)**(-.4))\n",
-      "R = w*d*f/60\n",
-      "P = Uc*R\n",
-      "Fc = 60000*(P)/(pi*D*N)\n",
-      "print \"Grinding force = %d N\"%Fc"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Grinding force = 3 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 22 on page no. 251"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "d= 0.05 # Depth of cut in mm\n",
-      "f =200 # Feed rate in mm/min\n",
-      "theta = 850 # Surface temperature in \u00b0C\n",
-      "Theta = 700 # Maximum surface temperature of workpiece surface required to maintain in \u00b0C\n",
-      "\n",
-      "K = theta * (f**0.2)/(d**0.9)\n",
-      "r = Theta/K\n",
-      "C = d*f\n",
-      "Dm = (r*C**0.2)**(1/1.1)\n",
-      "fm = C/Dm\n",
-      "print \"Required depth of cut = %f mm,\\nRequired feed = %d mm/min\"%(Dm,fm)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required depth of cut = 0.041910 mm,\n",
-        "Required feed = 238 mm/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 24 on page no. 260"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that \n",
-      "shi = 30 # Side cutting edge angle in Degree\n",
-      "lamda = 7 # End cutting edge angle in Degree\n",
-      "r = 0.7 # Nose radius in mm\n",
-      "f = 0.125 # Feed in mm\n",
-      "H_max = f/(tan(shi*pi/180)+1/tan(lamda*pi/180))\n",
-      "H_max_ = (f**2)/(8*r)\n",
-      "print \"Maximum height of uneveness in first tool case = %0.3f mm,\\nIn second tool case = %0.4f mm\"%(H_max,H_max_)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum height of uneveness in first tool case = 0.014 mm,\n",
-        "In second tool case = 0.0028 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 25 on page no. 262"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that \n",
-      "Z = 12 # No of teeth\n",
-      "d = 100 # Diameter of cutter in mm\n",
-      "N = 60 # Rpm of cutter\n",
-      "f = 25 # Table feed in mm/min\n",
-      "\n",
-      "H_max = (f**2)/(4*d*(N**2)*(Z**2))\n",
-      "print \"Maximum height of uneveness = %f mm\"%H_max"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum height of uneveness = 0.000003 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 26 on page no. 268"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that \n",
-      "n = 0.25 # Value of exponent of time in Taylor's tool life equation\n",
-      "C = 75 # Value of constant in Taylor's tool life equation\n",
-      "Lc = .15 # Labour cast in $/min\n",
-      "Tc = 2.50 # Total cast of tool in $\n",
-      "t = 2 # Change time for tool in min\n",
-      "\n",
-      "x = (C)**(1/n) # Where x = k/(f**(1/n))\n",
-      "v_opt = ((n*x*Lc)/((1-n)*((Lc*t+Tc))))**(n)\n",
-      "print \"Cutting speed that will be lead to minimum cast = %0.2f m/min\"%v_opt"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Cutting speed that will be lead to minimum cast = 27.42 m/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 27 on page no. 269"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that \n",
-      "L = 300 # Length of the bar in mm\n",
-      "d=30 # Diameter of the bar in mm\n",
-      "f_max = 0.25 # Maximum allowable feed in mm/revolution\n",
-      "Lc = .25 # Labour and overhead cast in $/min\n",
-      "Tc = 2 # Regrinding cast in $\n",
-      "t = 1 # Change time for tool in min\n",
-      "C_X = 2.50 # Cast of tool of material X per piece in $ \n",
-      "C_Y = 3 # Cast of tool of material Y  per piece in $\n",
-      "n_x = 0.1 # Value of exponent of time in Taylor's tool life equation for material X\n",
-      "n_y = 0.16 # Value of exponent of time in Taylor's tool life equation for material Y\n",
-      "C_x = 30 # Value of constant in Taylor's tool life equation for material X\n",
-      "C_y = 76 # Value of constant in Taylor's tool life equation For material Y\n",
-      "\n",
-      "x_x = (C_x)**(1/n_x) # Where x = k/(f**(1/n))\n",
-      "v_opt_x = ((n_x*x_x*Lc)/((1-n_x)*((Lc*t+Tc))))**(n_x)\n",
-      "Rmin_x = C_X+Lc*t+(Lc*pi*L*d/(1000*f_max*v_opt_x)) + (Lc*t*(pi*L*d/(1000*x_x)))*(v_opt_x**(1/n_y))*(v_opt_x**-1)*(f_max**-1)+(Tc*((pi*L*d/(1000*x_x)))*(v_opt_x**(1/n_x))*(v_opt_x**-1)*(f_max**-1))\n",
-      "x_y = (C_y)**(1/n_y) # Where x = k/(f**(1/n))\n",
-      "v_opt_y = ((n_y*x_y*Lc)/((1-n_y)*((Lc*t+Tc))))**(n_y)\n",
-      "Rmin_y = C_Y+Lc*t+(Lc*pi*L*d/(1000*f_max*v_opt_y)) + (Lc*t*(pi*L*d/(1000*x_y)))*(v_opt_y**(1/n_y))*(v_opt_y**-1)*(f_max**-1)+(Tc*((pi*L*d/(1000*x_y)))*(v_opt_y**(1/n_y))*(v_opt_y**-1)*(f_max**-1))\n",
-      "print \"The minimum cost per piece\\n When material X is used = %0.2f $,\\n When material Y is used = %0.2f $\"%(Rmin_x,Rmin_y)\n",
-      "print \"So material Y will be suitable for tool as it has low cast\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum cost per piece\n",
-        " When material X is used = 4.36 $,\n",
-        " When material Y is used = 4.07 $\n",
-        "So material Y will be suitable for tool as it has low cast\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 28 on page no. 271"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that \n",
-      "n = 0.25 # Value of exponent of time in Taylor's tool life equation\n",
-      "C = 75 # Value of constant in Taylor's tool life equation\n",
-      "Lc = .15 # Labour cast in $/min\n",
-      "Tc = 2.50 # Total cast of tool in $\n",
-      "t = 2 # Change time for tool in min\n",
-      "x = (C)**(1/n) # Where x = k/(f**(1/n))\n",
-      "v_opt = ((n*x)/((1-n)*t))**(n)\n",
-      "print \"Optimum cutting speed for maximum production rate for the job = %0.2f m/min\"%(v_opt)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Optimum cutting speed for maximum production rate for the job = 47.92 m/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
diff --git a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch4_1.ipynb b/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch4_1.ipynb
deleted file mode 100755
index 312554ae..00000000
--- a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch4_1.ipynb
+++ /dev/null
@@ -1,1189 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter4 - Machining Processes"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.1 - page 187"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import atan, tan, cos , sin, degrees, pi\n",
-      "# Given that\n",
-      "alpha = 10 # Rake angle in Degree\n",
-      "t = 0.4 # Chip thickness  in mm\n",
-      "T = 0.15 # Uncut chip thickness in mm\n",
-      "\n",
-      "r = T/t\n",
-      "phi = degrees(atan((r*cos(alpha*pi/180))/(1-r*sin(alpha*pi/180))))\n",
-      "gama = 1/tan(phi*pi/180) + tan((phi-alpha)*pi/180)\n",
-      "print \"Shear plane angle = %0.1f\u00b0, \\nMagnitude of the shear strain = %0.2f\"%(phi,gama)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Shear plane angle = 21.6\u00b0, \n",
-        "Magnitude of the shear strain = 2.74\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.2 - page 192"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "t1 = 0.25 # Undercut thickness in mm\n",
-      "t2 = 0.75 # Chip thickness in mm\n",
-      "w = 2.5 # Width in mm\n",
-      "alpha = 0 # Rake angle in Degree\n",
-      "Fc = 950 # Cutting force in N\n",
-      "Ft = 475 # Thrust force in N\n",
-      "\n",
-      "r = t1/t2\n",
-      "mu = ((Fc*sin(alpha*pi/180)) + (Ft*cos(alpha*pi/180)))/((Fc*cos(alpha*pi/180))-(Ft*sin(alpha*pi/180)))\n",
-      "phi = degrees(atan((r*cos(alpha*pi/180))/(1-r*sin(alpha*pi/180))))\n",
-      "As = t1*w/sin(phi*pi/180)\n",
-      "Fs = Fc*cos(phi*pi/180) - Ft*sin(phi*pi/180)\n",
-      "T_s = Fs/As\n",
-      "print \"\"\"Coefficient of the friction between tool and the chip = %0.1f,\n",
-      "The ultimate shear stress of the material = %0.1f N/mm**2\"\"\"%(mu,T_s)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Coefficient of the friction between tool and the chip = 0.5,\n",
-        "The ultimate shear stress of the material = 380.0 N/mm**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.3 - page 193"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "alpha = 10 # Rake angle of tool in Degree\n",
-      "v = 200 # Cutting speed in m/min\n",
-      "t1 = 0.2 # Uncut thickness in mm\n",
-      "w = 2 # Width of cut in mm\n",
-      "mu = 0.5 # Avg value of the cofficient of tbe friction\n",
-      "T_S = 400 # Shear stress of the work material in N/mm**2\n",
-      "\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = (90 + alpha - lamda)/2\n",
-      "Fs = (w*t1*T_S)/(sin(phi*pi/180))\n",
-      "R = Fs/(cos((phi+lamda-alpha)*pi/180))\n",
-      "Fc = R*(cos((lamda-alpha)*pi/180))\n",
-      "Ft = R*(sin((lamda-alpha)*pi/180))\n",
-      "print \"Shear angle = %0.1f\u00b0, \\nCutting force = %0.f N, \\nThrust force = %d N,\"%(phi,Fc,Ft)\n",
-      "# Answer in the book for cutting force is given as 420 N and for thrust force is given as 125 N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Shear angle = 36.7\u00b0, \n",
-        "Cutting force = 429 N, \n",
-        "Thrust force = 127 N,\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.4 - page 194"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "alpha = 10 # Rake angle of tool in Degree\n",
-      "v = 200 # Cutting speed in m/min\n",
-      "t1 = 0.2 # Uncut thickness in mm\n",
-      "w = 2 # Width of cut in mm\n",
-      "mu = 0.5 # Avg value of the cofficient of tbe friction\n",
-      "T_S = 400 # Shear stress of the work material in N/mm**2\n",
-      "Cm = 70 # Machining constant in Degree\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = (Cm + alpha - lamda)/2\n",
-      "Fs = (w*t1*T_S)/(sin(phi*pi/180))\n",
-      "R = Fs/(cos((phi+lamda-alpha)*pi/180))\n",
-      "Fc = R*(cos((lamda-alpha)*pi/180))\n",
-      "Ft = R*(sin((lamda-alpha)*pi/180))\n",
-      "# Using Lee and Shaffer relation \n",
-      "phi_ = 45-lamda+alpha\n",
-      "Fs_ = (w*t1*T_S)/(sin(phi_*pi/180))\n",
-      "R_ = Fs_/(cos((phi_+lamda-alpha)*pi/180))\n",
-      "Fc_ = R_*(cos((lamda-alpha)*pi/180))\n",
-      "Ft_ = R_*(sin((lamda-alpha)*pi/180))\n",
-      "print \"\"\"Shear angle = %0.1f\u00b0,\n",
-      "Cutting force = %0.2f N,\n",
-      "Thrust force = %0.2f N\n",
-      "Using Lee and Shaffer relation- \n",
-      " Shear angle = %0.2f\u00b0,\n",
-      " Cutting force = %0.2f N,\n",
-      " Thrust force = %0.2f N,\"\"\"%(phi,Fc,Ft,phi_,Fc_,Ft_)\n",
-      "# Answer in the book for cutting force is given as 486.9 N and for thrust force is given as 144.9 N , When using Lee and Shaffer relation answer in the book for cutting force is given as 481.9 N and for trust force is given as 160.6 N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Shear angle = 26.7\u00b0,\n",
-        "Cutting force = 468.57 N,\n",
-        "Thrust force = 139.37 N\n",
-        "Using Lee and Shaffer relation- \n",
-        " Shear angle = 28.43\u00b0,\n",
-        " Cutting force = 455.48 N,\n",
-        " Thrust force = 135.48 N,\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.5 - page 196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "t1 = 0.25 # Uncut thickness in mm\n",
-      "w = 2.5 # Width of cut in mm\n",
-      "U_0 = 1.4 # In J/mm**3\n",
-      "alpha = 0 # Rake angle in degree\n",
-      "mu = 0.5 # Cofficient of the friction\n",
-      "T_s = 400 # Shear stress in N/mm**2\n",
-      "\n",
-      "lamda = degrees(atan(mu))\n",
-      "Fc = 1000*(t1*w*U_0)*((t1)**(-.4))\n",
-      "phi = 45 + alpha - degrees(atan(mu))\n",
-      "Fc_ = (w*t1*T_s*cos((lamda-alpha)*pi/180))/((sin(phi*pi/180)) *cos((phi+lamda-alpha)*pi/180))\n",
-      "print \"\"\"The order of magnitude of cutting force = %d N,\n",
-      "Using Lee and Shaffer relation-\n",
-      " The order of magnitude of cutting force = %d N.\"\"\"%(Fc,Fc_)\n",
-      "# Answer in the book for cutting force is given as 1517 N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The order of magnitude of cutting force = 1523 N,\n",
-        "Using Lee and Shaffer relation-\n",
-        " The order of magnitude of cutting force = 999 N.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.6 - page 199"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log, sqrt\n",
-      "# Given that\n",
-      "v = 2 # Cutting speed in m/sec\n",
-      "D = 7200 # Density of mild steel in kg /m**3\n",
-      "k = 43.6 # Thermal conductivity in W/m-\u00b0c\n",
-      "c = 502 # Specific heat of the material in J/kg-\u00b0c\n",
-      "t1 = 0.25 # Uncut thickness in mm\n",
-      "w =2 # Width of cut in mm\n",
-      "theta_0 = 40 # Initial temp of the workpiece in Degree\n",
-      "alpha = 0 # Rake angle in degree\n",
-      "mu = 0.5 # Cofficient of the friction\n",
-      "T_s = 400e6 # Shear stress in N/m**2\n",
-      "\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = 45 + alpha - lamda\n",
-      "Fs = (w*t1*T_s)*(10**-6)/(sin(phi*pi/180))\n",
-      "R = Fs / (cos((phi+lamda-alpha)*pi/180))\n",
-      "Fc = R *(cos((lamda-alpha)*pi/180))\n",
-      "r = sin(phi*pi/180)/(cos((phi-alpha)*pi/180))\n",
-      "Ft= Fc *(tan((lamda - alpha)*pi/180))\n",
-      "F = Fc *(sin(alpha*pi/180))+Ft*(cos(alpha*pi/180))\n",
-      "Ws = F*r*v\n",
-      "Wp = Fc*v-F*r*v\n",
-      "zeta = D*c*v*t1*(10**-3)/k\n",
-      "zeta_ = zeta*tan(phi*pi/180)\n",
-      "nu = 0.15 *(log(27.5/(zeta_)))\n",
-      "theta_P = (1-nu)*Wp/(D*c*v*t1*w*(10**-6))\n",
-      "theta_S = 1.13 *(sqrt(1/(D*c*v*t1*(10**-3)*k*(1+tan((phi-alpha)*pi/180))))*(Ws/w)*(10**3))                 \n",
-      "theta = theta_0+theta_S+ theta_P\n",
-      "print \" \\n Maximum temperature along the rake face of the tool = %0.1f\u00b0C.\"%theta\n",
-      "# Answer in the book  is given as 823\u00b0C"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " Maximum temperature along the rake face of the tool = 835.6\u00b0C.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.7 - Page 206"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "theta_ = 40 #Ambient temperature in\u00b0C\n",
-      "v = 2 # Cutting speed in m/sec\n",
-      "D = 7200 # Density of mild steel in kg /m**3\n",
-      "k = 43.6 # Thermal conductivity in W/m-\u00b0c\n",
-      "c = 502 # Specific heat of the material in J/kg-\u00b0c\n",
-      "t1 = 0.25 # Uncut thickness in mm\n",
-      "w =2 # Width of cut in mm\n",
-      "alpha = 0 # Rake angle in degree\n",
-      "mu = 0.5 # Cofficient of the friction\n",
-      "T_s = 400e6 # Shear stress in N/m**2\n",
-      "H = 350 # Hardness of SAE 1040 steel in HV(Vicker hardness)\n",
-      "\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = 45 + alpha - lamda\n",
-      "Fs = (w*t1*T_s)*(10**-6)/(sin(phi*pi/180))\n",
-      "R = Fs / (cos((phi+lamda-alpha)*pi/180))\n",
-      "Fc = R *(cos((lamda-alpha)*pi/180))\n",
-      "r = sin(phi*pi/180)/(cos((phi-alpha)*pi/180))\n",
-      "Ft= Fc *(tan((lamda - alpha)*pi/180))\n",
-      "F = Fc *(sin(alpha*pi/180))+Ft*(cos(alpha*pi/180))\n",
-      "Ws = F*r*v\n",
-      "Wp = Fc*v-F*r*v\n",
-      "zeta = D*c*v*t1*(10**-3)/k\n",
-      "zeta_ = zeta*tan(phi*pi/180)\n",
-      "nu = 0.15 *(log(27.5/(zeta_)))\n",
-      "Theta_0v = ((1-nu)*Wp + Ws)/ (D*c*v*t1*w*(10**-6))\n",
-      "H_ = 1.5 *(H)\n",
-      "theta_lim = 700*((1-(H_/850))**(1/3.1))\n",
-      "v_lim = (theta_lim/309)**(1/0.5)\n",
-      "print \" \\n Maximum speed at which cutting is passible = %0.2f m/sec.\"%v_lim"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " Maximum speed at which cutting is passible = 2.76 m/sec.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "PROBLEM 4.8 - page 212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "alpha = 0 # Rake angle in degree\n",
-      "gama = 3 # Clearance angle in Degree\n",
-      "w = 1 # Maximum length of flank wear allowed in mm\n",
-      "gama_ = 7 # Increased clearance angle in Degree\n",
-      "I_per = (((tan(gama_*pi/180))-(tan(gama*pi/180)))/tan(gama*pi/180))*100\n",
-      "print \" \\n Percentage increase in tool life = %d percent.\"%I_per"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " \n",
-        " Percentage increase in tool life = 134 percent.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 9 on page no. 220"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "d= 4 # Depth of cut in mm\n",
-      "f = 0.25 # Feed in mm/stroke\n",
-      "alpha = 10 # Rake angle in degree\n",
-      "shi = 30 # Principal cutting edge angle in Degree\n",
-      "mu =0.6 # Cofficient of friction between chip and tool\n",
-      "T_s = 340 # Ultimate shear stress of cast iron in N/mm**2\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = 45 +alpha-lamda\n",
-      "Fc = f*d*T_s*(cos((lamda-alpha)*pi/180))/((sin(phi*pi/180))*(cos((phi+lamda-alpha)*pi/180)))\n",
-      "Ft = Fc*(sin((lamda-alpha)*pi/180))/(cos((lamda-alpha)*pi/180))\n",
-      "Ff = Ft*(cos(shi*pi/180))\n",
-      "Fn = Ft*(sin(shi*pi/180))\n",
-      "print \"\"\"The three components of machinig force are as follows-\n",
-      "  Thrust force = %d N,\n",
-      "  Feed force component = %d N,\n",
-      "  Normal thrust force component = %d N.\"\"\"%(Ft,Ff,Fn)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The three components of machinig force are as follows-\n",
-        "  Thrust force = 422 N,\n",
-        "  Feed force component = 365 N,\n",
-        "  Normal thrust force component = 211 N.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 10 on page no. 221"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "d= 4 # Depth of cut in mm\n",
-      "f = 0.25 # Feed in mm/stroke\n",
-      "alpha = 10 # Rake angle in degree\n",
-      "shi = 30 # Principal cutting edge angle in Degree\n",
-      "mu =0.6 # Cofficient of friction between chip and tool\n",
-      "T_s = 340 # Ultimate shear stress of cast iron in N/mm**2\n",
-      "N = 60 # Cutting stroke/min\n",
-      "L = 200 # Length of the job in mm\n",
-      "H = 180 # Hardness of the workpiece in BHN\n",
-      "\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = 45 +alpha-lamda\n",
-      "Fc = f*d*T_s*(cos((lamda-alpha)*pi/180))/((sin(phi*pi/180))*(cos((phi+lamda-alpha)*pi/180)))\n",
-      "Fc_ = Fc*(L/1000)\n",
-      "Wav  =Fc_*N/60\n",
-      "t1 = f*cos(shi*pi/180)\n",
-      "U_0 = 0.81 # By using table 4.4 given in the book, In J/mm**3\n",
-      "Uc = U_0*((t1)**(-.4))\n",
-      "Q = f*d*L*N/60\n",
-      "Wav_ = Uc*Q\n",
-      "print \"Avg power consumption = %d W,\\nSpecific power consumption when hardness of the workpiece is 180 BHN = %d W.\"%(Wav,Wav_)\n",
-      "# Answer in the book for Specific power consumption is given as 294 W"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Avg power consumption = 220 W,\n",
-        "Specific power consumption when hardness of the workpiece is 180 BHN = 298 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 11 on page no. 224"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "alpha_b = 6 # Back rake angle in Degree\n",
-      "alpha_s = 10 # Side rake angle in Degree\n",
-      "gama = 7 # Front clearance angle in Degree\n",
-      "gama_ = 7 # Side clearance angle in Degree\n",
-      "Shi = 10 # End cutting edge angle in Degree\n",
-      "shi = 30 # Side cutting edge angle in Degree\n",
-      "r= 0.5 # Nose radius in mm\n",
-      "\n",
-      "k = tan(alpha_b*pi/180) * cos(shi*pi/180) - tan(alpha_s*pi/180) * sin(shi*pi/180)\n",
-      "print \"The value of k=%0.4f,which is near to 0. Hence the case is close to orthogonal one.\\n\"%k\n",
-      "alpha= degrees(atan(((tan(alpha_b*pi/180)) * sin(shi*pi/180) ) + (tan(alpha_s*pi/180) * (cos(shi*pi/180))))/ (sqrt(1+((tan(alpha_b*pi/180)*cos(shi*pi/180)) - (tan(alpha_s*pi/180)*sin(shi*pi/180)))**(2))))\n",
-      "print \"Normal rake angle = %0.1f\u00b0.\"%(alpha)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of k=0.0029,which is near to 0. Hence the case is close to orthogonal one.\n",
-        "\n",
-        "Normal rake angle = 11.6\u00b0.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 12 on page no. 225"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "# Given that\n",
-      "alpha_b = 6 # Back rake angle in Degree\n",
-      "alpha_s = 10 # Side rake angle in Degree\n",
-      "gama = 5 # Front clearance angle in Degree\n",
-      "gama_ = 7 # Side clearance angle in Degree\n",
-      "Shi = 10 # End cutting edge angle in Degree\n",
-      "shi = 30 # Side cutting edge angle in Degree\n",
-      "r= 0.55 # Nose radius in mm\n",
-      "d = 2.5 # Depth of cut in mm\n",
-      "f = 0.125 # Feed in mm/revolution\n",
-      "N = 300 # Rpm of the job\n",
-      "T_S = 400 # Ultimate shear stress of the workpiece in N/mm**2\n",
-      "mu = .6 # Cofficient of the friction between the tool and the chip\n",
-      "\n",
-      "lamda = degrees(atan(mu))\n",
-      "alpha= degrees(atan(((tan(alpha_b*pi/180)) * sin(shi*pi/180) ) + (tan(alpha_s*pi/180) * (cos(shi*pi/180))))/ (sqrt(1+((tan(alpha_b*pi/180)*cos(shi*pi/180)) - (tan(alpha_s*pi/180)*sin(shi*pi/180)))**(2))))\n",
-      "phi = 45 + alpha - lamda\n",
-      "t1 = f*cos(phi*pi/180)\n",
-      "w = d/cos(phi*pi/180)\n",
-      "Fc = w*t1*T_S*(cos((lamda-alpha)*pi/180))/((sin(phi*pi/180))*(cos((phi+lamda-alpha)*pi/180)))\n",
-      "Ft = Fc*tan((lamda-alpha)*pi/180)\n",
-      "Ff = Ft*cos(shi*pi/180)\n",
-      "Fr = Ft*sin(shi*pi/180)\n",
-      "print \"Component of the machining force are as follows -\\n Feed force component = % d N, \\n Normal thrust force component = % d N.\"%(ceil(Ff),ceil(Fr))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Component of the machining force are as follows -\n",
-        " Feed force component =  118 N, \n",
-        " Normal thrust force component =  68 N.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 14 on page no. 230"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "D = 20 # Nominal diameter of the drill in mm\n",
-      "T_S = 400 # Shear yield stress of work material in N/mm**2\n",
-      "N = 240 # Rpm\n",
-      "f = 0.25 # Feed in mm/revolution\n",
-      "mu = 0.6 # Cofficient of friction\n",
-      "\n",
-      "Beta = 118/2 # From the table 4.12 given in the book\n",
-      "shi = 30 # From the table 4.12 given in the book\n",
-      "alpha = degrees(atan(((2*(D/4)/(D)))*tan(shi*pi/180))/sin(Beta*pi/180))\n",
-      "t1 = (f/2)*sin(Beta*pi/180)\n",
-      "w = (D/2)/sin(Beta*pi/180)\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = 45+alpha-lamda\n",
-      "t1 = f/2\n",
-      "Fc = w*t1*T_S*(cos((lamda-alpha)*pi/180))/((sin(phi*pi/180))*(cos((phi+lamda-alpha)*pi/180)))\n",
-      "Ft = w*t1*T_S*(sin((lamda-alpha)*pi/180))/((sin(phi*pi/180))*(cos((phi+lamda-alpha)*pi/180)))\n",
-      "M = .6*Fc*D/1000\n",
-      "F = 5*Ft*sin(Beta*pi/180)\n",
-      "print \"The drilling torque = %0.2f N-m, \\nThrust force = %d N.\"%(M,F)\n",
-      "# Answer in the book for drilling torque is given as 18.2 N-m, and for thrust force is given as 1500 N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drilling torque = 17.85 N-m, \n",
-        "Thrust force = 1376 N.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 15 on page no. 235"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "# Given that\n",
-      "w = 20 # Width of the mild steel block in mm\n",
-      "Z = 20 # No of teeth in milling cutter\n",
-      "D = 50 # Diameter of the milling cutter in mm\n",
-      "alpha = 10 # Radial rake angle in Degree\n",
-      "f = 15 # Feed velocity of the table in mm/min\n",
-      "N =60 # Rpm of the cutter\n",
-      "t = 1 # Depth of cut in mm\n",
-      "mu = 0.5 # Cofficient of friction\n",
-      "T_s = 400 # Shear yield stress in N/mm**2\n",
-      "t_a = 0.0018 # Avg uncut thickness in mm\n",
-      "\n",
-      "Beta = degrees(asin(2*(t/D)))\n",
-      "theta = 2*pi/Z\n",
-      "t1_max = (2*f/(N*Z))*sqrt(t/D)\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = 45+alpha -lamda\n",
-      "Fc_max = ((w*t1_max*T_s*cos((lamda-alpha)*pi/180)))/((sin(phi*pi/180))*(cos(45*pi/180)))\n",
-      "T_max = Fc_max*D/(2*1000)\n",
-      "M_av = (1/2)*(Beta*T_max)/theta\n",
-      "omega = 2*pi*N/60\n",
-      "U_0 = 1.4 # From the table 4.4 given in the book\n",
-      "Uc_ms = U_0*((t_a)**(-0.4))\n",
-      "R = f*t*w/60\n",
-      "U = Uc_ms * R\n",
-      "print \"Power consumption = %0.1f W.\"%U"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power consumption = 87.7 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 16 on page no. 238"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "w = 20 # Width of the mild steel block in mm\n",
-      "Z = 10 # No of teeth in milling cutter\n",
-      "D = 75 # Diameter of the milling cutter in mm\n",
-      "alpha = 10 # Radial rake angle in Degree\n",
-      "f = 100 # Feed velocity of the table in mm/min\n",
-      "N =60 # Rpm of the cutter\n",
-      "t = 5 # Depth of cut in mm\n",
-      "mu = 0.5 # Cofficient of friction\n",
-      "T_s = 400 # Shear yield stress in N/mm**2\n",
-      "t_a = 0.043 # Avg uncut thickness in mm\n",
-      "\n",
-      "Beta = degrees(asin(2*(t/D)))\n",
-      "theta = 2*pi/Z\n",
-      "t1_max = (2*f/(N*Z))*sqrt(t/D)\n",
-      "lamda = degrees(atan(mu))\n",
-      "phi = 45+alpha -lamda\n",
-      "Fc_max = ((w*t1_max*T_s*cos((lamda-alpha)*pi/180)))/((sin(phi*pi/180))*(cos(45*pi/180)))\n",
-      "T_max = Fc_max*D/(2*1000)\n",
-      "M_av = (1/2)*(Beta*T_max)/theta\n",
-      "omega = 2*pi*N/60\n",
-      "U_0 = 1.4 # From the table 4.4 given in the book\n",
-      "Uc_ms = U_0*((t_a)**(-0.4))\n",
-      "R = f*t*w/60\n",
-      "U = Uc_ms * R\n",
-      "print \"Power required = %d W.\"%U\n",
-      "# Answer in the book for Power required is given as 817 W"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power required = 821 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 17 on page no. 240"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "B = 20 # Width of the cut in mm\n",
-      "Z = 10 # No of teeth in milling cutter\n",
-      "D = 75 # Diameter of the milling cutter in mm\n",
-      "alpha = 10 # Radial rake angle in Degree\n",
-      "f = 25 # Feed velocity of the table in mm/min\n",
-      "N =60 # Rpm of the cutter\n",
-      "t = 5 # Depth of cut in mm\n",
-      "mu = 0.5 # Cofficient of friction\n",
-      "T_s = 400 # Shear yield stress in N/mm**2\n",
-      "t_a = 0.043 # Avg uncut thickness in mm\n",
-      "t1_max = 0.01\n",
-      "lamda = 0.28 # From the table 4.13 Given in the book\n",
-      "nu = 1400 # From the table 4.13 Given in the book\n",
-      "t1_av = t1_max/2\n",
-      "P = nu*B*t*f*(10**-4)/(6*((t1_av)**(lamda)))\n",
-      "print \"Power required = %0.2f W.\"%P"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power required = 257.16 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 18 on page no. 240"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "w = 20 # Width of the mild steel block in mm\n",
-      "Z = 10 # No of teeth in milling cutter\n",
-      "D = 75 # Diameter of the milling cutter in mm\n",
-      "alpha = 10 # Radial rake angle in Degree\n",
-      "f = 25 # Feed velocity of the table in mm/min\n",
-      "N =60 # Rpm of the cutter\n",
-      "t = 5 # Depth of cut in mm\n",
-      "mu = 0.5 # Cofficient of friction\n",
-      "T_s = 400 # Shear yield stress in N/mm**2\n",
-      "t_a = 0.043 # Avg uncut thickness in mm\n",
-      "\n",
-      "R = f*t*w/60\n",
-      "Uc = 3.3 # Specific energy in J/mm**3 from the table 4.14 Given in the book\n",
-      "U = Uc * R\n",
-      "print \"Power required = %d W.\"%ceil(U)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power required = 138 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 19 on page no. 241"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "d = 25 # Diameter of circular hole in mm\n",
-      "t = 20 # Thickness of the steel plate in mm\n",
-      "D = 27 # Enlarged diameter of hole in mm\n",
-      "c= 0.08 # Cut per tooth in mm\n",
-      "alpha = 10 # Radial rake angle in Degree\n",
-      "mu = 0.5 # Cofficient of friction\n",
-      "T_s = 400 # Shear yield stress in N/mm**2\n",
-      "\n",
-      "lamda=degrees(atan(mu))\n",
-      "phi = 45-lamda+alpha\n",
-      "w = pi*(d+D)/2\n",
-      "Fc = w*c*T_s*(cos((lamda-alpha)*pi/1800)/((sin(phi*pi/180))*(cos(45*pi/180))))\n",
-      "s = 1.75*sqrt(t)\n",
-      "F = 3*Fc\n",
-      "print \"Peak broaching load = %d N.\"%ceil(F)\n",
-      "# Answer in the book  is given as 22323 N which is wrong."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Peak broaching load = 23280 N.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 20 on page no. 246"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "D = 250 # Diameter of the wheel in mm\n",
-      "N = 2000 # Rpm of the wheel\n",
-      "f =5 # Plung feed rate in mm/min\n",
-      "C = 3 # Surface density of active grain in mm**-2\n",
-      "A = 20*15 # Area of mild steel prismatic bar in mm**2\n",
-      "rg = 15 # In mm**-1\n",
-      "\n",
-      "t1 = sqrt(f/(pi*D*N*C*rg))\n",
-      "U_0 = 1.4 # From the table 4.4 given in the book\n",
-      "Uc= U_0*((t1)**(-.4))\n",
-      "R = A*f/60\n",
-      "P = Uc*R\n",
-      "Fc_ = 60000*(P)/(pi*D*A*C*N)\n",
-      "print \"Power requirement during plunge grinding of the mild steel primatic bar = %d W.\"%ceil(P)\n",
-      "# Answer in the book is given as 94 W which is wrong."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power requirement during plunge grinding of the mild steel primatic bar = 943 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 21 on page no. 248"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "w = 25 # Width of mild steel block in mm\n",
-      "d= 0.05 # Depth of cut in mm\n",
-      "D = 200 # Diameter of the wheel in mm\n",
-      "N = 3000 # Rpm of the wheel\n",
-      "f =100 # Feed velocity of table in mm/min\n",
-      "C = 3 # No of grits in mm**-2\n",
-      "rg = 15 # In mm**-1\n",
-      "\n",
-      "t1_max = sqrt(((6*f)/(pi*D*N*C*rg))*sqrt(d/D))\n",
-      "t1_a = t1_max/2\n",
-      "U_0 = 1.4 # From the table 4.4 given in the book\n",
-      "Uc= U_0*((t1_a)**(-.4))\n",
-      "R = w*d*f/60\n",
-      "P = Uc*R\n",
-      "Fc = 60000*(P)/(pi*D*N)\n",
-      "print \"Grinding force = %d N\"%Fc"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Grinding force = 3 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 22 on page no. 251"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "d= 0.05 # Depth of cut in mm\n",
-      "f =200 # Feed rate in mm/min\n",
-      "theta = 850 # Surface temperature in \u00b0C\n",
-      "Theta = 700 # Maximum surface temperature of workpiece surface required to maintain in \u00b0C\n",
-      "\n",
-      "K = theta * (f**0.2)/(d**0.9)\n",
-      "r = Theta/K\n",
-      "C = d*f\n",
-      "Dm = (r*C**0.2)**(1/1.1)\n",
-      "fm = C/Dm\n",
-      "print \"Required depth of cut = %f mm,\\nRequired feed = %d mm/min\"%(Dm,fm)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required depth of cut = 0.041910 mm,\n",
-        "Required feed = 238 mm/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 24 on page no. 260"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that \n",
-      "shi = 30 # Side cutting edge angle in Degree\n",
-      "lamda = 7 # End cutting edge angle in Degree\n",
-      "r = 0.7 # Nose radius in mm\n",
-      "f = 0.125 # Feed in mm\n",
-      "H_max = f/(tan(shi*pi/180)+1/tan(lamda*pi/180))\n",
-      "H_max_ = (f**2)/(8*r)\n",
-      "print \"Maximum height of uneveness in first tool case = %0.3f mm,\\nIn second tool case = %0.4f mm\"%(H_max,H_max_)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum height of uneveness in first tool case = 0.014 mm,\n",
-        "In second tool case = 0.0028 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 25 on page no. 262"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that \n",
-      "Z = 12 # No of teeth\n",
-      "d = 100 # Diameter of cutter in mm\n",
-      "N = 60 # Rpm of cutter\n",
-      "f = 25 # Table feed in mm/min\n",
-      "\n",
-      "H_max = (f**2)/(4*d*(N**2)*(Z**2))\n",
-      "print \"Maximum height of uneveness = %f mm\"%H_max"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum height of uneveness = 0.000003 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 26 on page no. 268"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that \n",
-      "n = 0.25 # Value of exponent of time in Taylor's tool life equation\n",
-      "C = 75 # Value of constant in Taylor's tool life equation\n",
-      "Lc = .15 # Labour cast in $/min\n",
-      "Tc = 2.50 # Total cast of tool in $\n",
-      "t = 2 # Change time for tool in min\n",
-      "\n",
-      "x = (C)**(1/n) # Where x = k/(f**(1/n))\n",
-      "v_opt = ((n*x*Lc)/((1-n)*((Lc*t+Tc))))**(n)\n",
-      "print \"Cutting speed that will be lead to minimum cast = %0.2f m/min\"%v_opt"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Cutting speed that will be lead to minimum cast = 27.42 m/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 27 on page no. 269"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that \n",
-      "L = 300 # Length of the bar in mm\n",
-      "d=30 # Diameter of the bar in mm\n",
-      "f_max = 0.25 # Maximum allowable feed in mm/revolution\n",
-      "Lc = .25 # Labour and overhead cast in $/min\n",
-      "Tc = 2 # Regrinding cast in $\n",
-      "t = 1 # Change time for tool in min\n",
-      "C_X = 2.50 # Cast of tool of material X per piece in $ \n",
-      "C_Y = 3 # Cast of tool of material Y  per piece in $\n",
-      "n_x = 0.1 # Value of exponent of time in Taylor's tool life equation for material X\n",
-      "n_y = 0.16 # Value of exponent of time in Taylor's tool life equation for material Y\n",
-      "C_x = 30 # Value of constant in Taylor's tool life equation for material X\n",
-      "C_y = 76 # Value of constant in Taylor's tool life equation For material Y\n",
-      "\n",
-      "x_x = (C_x)**(1/n_x) # Where x = k/(f**(1/n))\n",
-      "v_opt_x = ((n_x*x_x*Lc)/((1-n_x)*((Lc*t+Tc))))**(n_x)\n",
-      "Rmin_x = C_X+Lc*t+(Lc*pi*L*d/(1000*f_max*v_opt_x)) + (Lc*t*(pi*L*d/(1000*x_x)))*(v_opt_x**(1/n_y))*(v_opt_x**-1)*(f_max**-1)+(Tc*((pi*L*d/(1000*x_x)))*(v_opt_x**(1/n_x))*(v_opt_x**-1)*(f_max**-1))\n",
-      "x_y = (C_y)**(1/n_y) # Where x = k/(f**(1/n))\n",
-      "v_opt_y = ((n_y*x_y*Lc)/((1-n_y)*((Lc*t+Tc))))**(n_y)\n",
-      "Rmin_y = C_Y+Lc*t+(Lc*pi*L*d/(1000*f_max*v_opt_y)) + (Lc*t*(pi*L*d/(1000*x_y)))*(v_opt_y**(1/n_y))*(v_opt_y**-1)*(f_max**-1)+(Tc*((pi*L*d/(1000*x_y)))*(v_opt_y**(1/n_y))*(v_opt_y**-1)*(f_max**-1))\n",
-      "print \"The minimum cost per piece\\n When material X is used = %0.2f $,\\n When material Y is used = %0.2f $\"%(Rmin_x,Rmin_y)\n",
-      "print \"So material Y will be suitable for tool as it has low cast\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum cost per piece\n",
-        " When material X is used = 4.36 $,\n",
-        " When material Y is used = 4.07 $\n",
-        "So material Y will be suitable for tool as it has low cast\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 28 on page no. 271"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that \n",
-      "n = 0.25 # Value of exponent of time in Taylor's tool life equation\n",
-      "C = 75 # Value of constant in Taylor's tool life equation\n",
-      "Lc = .15 # Labour cast in $/min\n",
-      "Tc = 2.50 # Total cast of tool in $\n",
-      "t = 2 # Change time for tool in min\n",
-      "x = (C)**(1/n) # Where x = k/(f**(1/n))\n",
-      "v_opt = ((n*x)/((1-n)*t))**(n)\n",
-      "print \"Optimum cutting speed for maximum production rate for the job = %0.2f m/min\"%(v_opt)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Optimum cutting speed for maximum production rate for the job = 47.92 m/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
diff --git a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch5.ipynb b/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch5.ipynb
deleted file mode 100755
index dc74bcad..00000000
--- a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch5.ipynb
+++ /dev/null
@@ -1,191 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Ch-5, Joining Processes"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 1 on page no. 285"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sympy import symbols, diff, solve\n",
-      "from __future__ import division\n",
-      "# Given that\n",
-      "A = 20 # Value of A in voltage length characteristic equation\n",
-      "B = 40 # Value of B in voltage length characteristic equation\n",
-      "v= 80 # Open circuit voltage in V\n",
-      "I = 1000 # Short circuit current in amp\n",
-      "\n",
-      "l, I=symbols(\"l I\")\n",
-      "i = ((v-A)-(B* l))*(I/v)\n",
-      "V_given = (A+B*l)# Given in the question\n",
-      "V_cal = v-v/1e3*I # Volts # power source characteristic\n",
-      "# equating both V gives I\n",
-      "I = (v-(A+B*l))/(v/1e3) # A\n",
-      "P = V_given*I\n",
-      "#k = derivat(P)\n",
-      "k = diff(P,l)\n",
-      "#L=roots(k)\n",
-      "L=solve(k,l)[0]\n",
-      "I = (v-(A+B*L))/(v/1e3) # A\n",
-      "#Pmax=((v-A)-(B* L))*(I/v)*(A+B*L)\n",
-      "V_given = (A+B*L)\n",
-      "Pmax = V_given*I\n",
-      "print \"Maximum power of the arc = %0.f kVA\"%(Pmax/1000)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum power of the arc = 20 kVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 2 on page no. 288"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "# Given that\n",
-      "N  =25 # No. of bridges per cm**2\n",
-      "r = 0.1 # Radius of bridge in mm\n",
-      "rho = 2e-5 # Resistivity of the material in ohm-cm\n",
-      "v= 5 # Applied voltage in V\n",
-      "Rc = 0.85*rho/(N*pi*r*0.1)\n",
-      "Q = (v**2)/Rc\n",
-      "print \"Rate of heat generated per unit area = %0.2e W/cm**2\"%(Q)\n",
-      "# Answer in the book is given as 1.136e5 W/cm**2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Rate of heat generated per unit area = 1.15e+06 W/cm**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem  on page no. 292"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sin\n",
-      "# Given that\n",
-      "P = 2.5 # Power in kVA\n",
-      "t = 3 # Thickness of steel plate in mm\n",
-      "T = 85 # Percentage of total time when arc is on\n",
-      "alpha = 1.2e-5 # Thermal diffusivity of steel in m**2/sec\n",
-      "k = 43.6 # Thermal conductivity of steel in W/m-\u00b0C\n",
-      "theta_ = 1530 # Melting point of steel in \u00b0C\n",
-      "theta = 30 # Ambient temperature in \u00b0C\n",
-      "gama = 60 # Angle in degree\n",
-      "\n",
-      "C = T/100\n",
-      "Q = C*P*10**3\n",
-      "w = t/sin(gama*pi/180)\n",
-      "theta_m = theta_ - theta\n",
-      "v_max = (4*alpha/(w*(10**-3)))*((Q/(8*k*theta_m*t*(10**-3)))-0.2)\n",
-      "print \"Maximum passible welding speed = %0.4f m/sec\"%(v_max)\n",
-      "# Answer in the book is given as 0.0146 m/sec"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum passible welding speed = 0.0160 m/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 4 on page no. 303"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "t = 1.2 # Thickness of aluminium sheet in mm\n",
-      "t_ = 0.25 # Adhesive thickness in mm\n",
-      "l = 12 # Overlapped length in mm\n",
-      "E = 703 # Modulus of elastisity in N/mm**2\n",
-      "G = 11.9 # Shear modulus of adhesive in N/mm**2\n",
-      "T_S = 0.6 # Ultimate shear stress in N/mm**2\n",
-      "\n",
-      "K = (((l**2)*G)/(2*E*t*t_))**(1/2)\n",
-      "T = T_S/K\n",
-      "print \"The maximum shear stress the lap joint can withstand = %0.4f N/mm**2\"%(T)\n",
-      "# Answer in the book is given as 0.274 N/mm**2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum shear stress the lap joint can withstand = 0.2977 N/mm**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
diff --git a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch5_1.ipynb b/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch5_1.ipynb
deleted file mode 100755
index dc74bcad..00000000
--- a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch5_1.ipynb
+++ /dev/null
@@ -1,191 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Ch-5, Joining Processes"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 1 on page no. 285"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sympy import symbols, diff, solve\n",
-      "from __future__ import division\n",
-      "# Given that\n",
-      "A = 20 # Value of A in voltage length characteristic equation\n",
-      "B = 40 # Value of B in voltage length characteristic equation\n",
-      "v= 80 # Open circuit voltage in V\n",
-      "I = 1000 # Short circuit current in amp\n",
-      "\n",
-      "l, I=symbols(\"l I\")\n",
-      "i = ((v-A)-(B* l))*(I/v)\n",
-      "V_given = (A+B*l)# Given in the question\n",
-      "V_cal = v-v/1e3*I # Volts # power source characteristic\n",
-      "# equating both V gives I\n",
-      "I = (v-(A+B*l))/(v/1e3) # A\n",
-      "P = V_given*I\n",
-      "#k = derivat(P)\n",
-      "k = diff(P,l)\n",
-      "#L=roots(k)\n",
-      "L=solve(k,l)[0]\n",
-      "I = (v-(A+B*L))/(v/1e3) # A\n",
-      "#Pmax=((v-A)-(B* L))*(I/v)*(A+B*L)\n",
-      "V_given = (A+B*L)\n",
-      "Pmax = V_given*I\n",
-      "print \"Maximum power of the arc = %0.f kVA\"%(Pmax/1000)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum power of the arc = 20 kVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 2 on page no. 288"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "# Given that\n",
-      "N  =25 # No. of bridges per cm**2\n",
-      "r = 0.1 # Radius of bridge in mm\n",
-      "rho = 2e-5 # Resistivity of the material in ohm-cm\n",
-      "v= 5 # Applied voltage in V\n",
-      "Rc = 0.85*rho/(N*pi*r*0.1)\n",
-      "Q = (v**2)/Rc\n",
-      "print \"Rate of heat generated per unit area = %0.2e W/cm**2\"%(Q)\n",
-      "# Answer in the book is given as 1.136e5 W/cm**2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Rate of heat generated per unit area = 1.15e+06 W/cm**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem  on page no. 292"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sin\n",
-      "# Given that\n",
-      "P = 2.5 # Power in kVA\n",
-      "t = 3 # Thickness of steel plate in mm\n",
-      "T = 85 # Percentage of total time when arc is on\n",
-      "alpha = 1.2e-5 # Thermal diffusivity of steel in m**2/sec\n",
-      "k = 43.6 # Thermal conductivity of steel in W/m-\u00b0C\n",
-      "theta_ = 1530 # Melting point of steel in \u00b0C\n",
-      "theta = 30 # Ambient temperature in \u00b0C\n",
-      "gama = 60 # Angle in degree\n",
-      "\n",
-      "C = T/100\n",
-      "Q = C*P*10**3\n",
-      "w = t/sin(gama*pi/180)\n",
-      "theta_m = theta_ - theta\n",
-      "v_max = (4*alpha/(w*(10**-3)))*((Q/(8*k*theta_m*t*(10**-3)))-0.2)\n",
-      "print \"Maximum passible welding speed = %0.4f m/sec\"%(v_max)\n",
-      "# Answer in the book is given as 0.0146 m/sec"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum passible welding speed = 0.0160 m/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 4 on page no. 303"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "t = 1.2 # Thickness of aluminium sheet in mm\n",
-      "t_ = 0.25 # Adhesive thickness in mm\n",
-      "l = 12 # Overlapped length in mm\n",
-      "E = 703 # Modulus of elastisity in N/mm**2\n",
-      "G = 11.9 # Shear modulus of adhesive in N/mm**2\n",
-      "T_S = 0.6 # Ultimate shear stress in N/mm**2\n",
-      "\n",
-      "K = (((l**2)*G)/(2*E*t*t_))**(1/2)\n",
-      "T = T_S/K\n",
-      "print \"The maximum shear stress the lap joint can withstand = %0.4f N/mm**2\"%(T)\n",
-      "# Answer in the book is given as 0.274 N/mm**2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum shear stress the lap joint can withstand = 0.2977 N/mm**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
diff --git a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch6.ipynb b/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch6.ipynb
deleted file mode 100755
index 818587af..00000000
--- a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch6.ipynb
+++ /dev/null
@@ -1,654 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Ch-6, Unconventional Machining Process"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 1 on page no. 332"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi, sqrt\n",
-      "from __future__ import division\n",
-      "# Given that\n",
-      "a = 5 # Side of the square hole in mm\n",
-      "t = 4 # Thickness of tungsten plate in mm\n",
-      "d = 0.01 # Diameter of abraisive grains in mm\n",
-      "F = 3.5 # Force for feeding in N\n",
-      "A =25e-3 # Amplitude of tool oscillation in mm\n",
-      "f = 25e3 # Frequency in Hz\n",
-      "Hw = 6900 # Fracture hardness of WC in N/mm**2\n",
-      "\n",
-      "Z = (1/2)*(4*a**2)/(pi*d**2)\n",
-      "lamda = 5\n",
-      "d1 = (d**2)\n",
-      "h_w = (sqrt((8*F*A)/(pi*Z*d1*Hw*(1+lamda))))\n",
-      "Q = (2/3)*((d1*h_w)**(3/2))*Z*f*pi\n",
-      "t = (a**2)*t/(Q*60)\n",
-      "print \"The approximate time required = %0.2f min\"%(t)\n",
-      "# Answer in the book is given as 13.66 min"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The approximate time required = 14.26 min\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 2 on page no. 334"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "r = 1/3 # Ratio of hardness values of copper and steel\n",
-      "R_Q = (r)**(3/4)\n",
-      "R_t = 1/R_Q\n",
-      "P_R = (1-(1/R_t))*100\n",
-      "print \"Percentage change in cutting time when tool is changed from coppper to steel = %d percent(reduction)\"%(P_R)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage change in cutting time when tool is changed from coppper to steel = 56 percent(reduction)\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 3 on page no. 345"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "m = 5 # Romoval rate in cm**3/min\n",
-      "A = 56 # Atomic gram weight in gm\n",
-      "Z = 2 # Valence at which dissolation takes place\n",
-      "D = 7.8 # Density of iron in gm/cm**3\n",
-      "I = (m/60)*(D*Z*96500)/(A)\n",
-      "print \"Current required = %d amp\"%(I)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current required = 2240 amp\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 4 on page no. 345"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "I = 1000 # Current in amp\n",
-      "p1 = 72.5 # Percentage(by weight) of Ni in Nimonic 75 alloy\n",
-      "p2 = 19.5 # Percentage(by weight) of Cr in Nimonic 75 alloy\n",
-      "p3 = 5 # Percentage(by weight) of Fe in Nimonic 75 alloy\n",
-      "p4 = 0.4 # Percentage(by weight) of Ti in Nimonic 75 alloy\n",
-      "p5 = 1 # Percentage(by weight) of Si in Nimonic 75 alloy\n",
-      "p6 = 1 # Percentage(by weight) of Mn in Nimonic 75 alloy\n",
-      "p7 = 6 # Percentage(by weight) of Cu in Nimonic 75 alloy\n",
-      "# From the table 6.3 given in the book\n",
-      "D1 = 8.9 # Density of Ni in g/cm**3\n",
-      "D2 = 7.19 # Density of Cr in g/cm**3\n",
-      "D3 = 7.86 # Density of Fe in g/cm**3\n",
-      "D4 = 4.51 # Density of Ti in g/cm**3\n",
-      "D5 = 2.33 # Density of Si in g/cm**3\n",
-      "D6 = 7.43 # Density of Mn in g/cm**3\n",
-      "D7 = 8.96 # Density of Cu in g/cm**3\n",
-      "A1 = 58.71 # Gram atomic weight of Ni in gm\n",
-      "A2 = 51.99 # Gram atomic weight of Cr in gm\n",
-      "A3 = 55.85 # Gram atomic weight of Fe in gm\n",
-      "A4 = 47.9 # Gram atomic weight of Ti in gm\n",
-      "A5 = 28.09 # Gram atomic weight of Si in gm\n",
-      "A6 = 54.94 # Gram atomic weight of Mn in gm\n",
-      "A7 = 63.57 # Gram atomic weight of Cu in gm\n",
-      "Z1 = 2 # Valence of dessolation for Ni\n",
-      "Z2 = 2 # Valence of dessolation for Cr\n",
-      "Z3 = 2 # Valence of dessolation for Fe\n",
-      "Z4 = 3 # Valence of dessolation for Ti\n",
-      "Z5 = 4 # Valence of dessolation for Si\n",
-      "Z6 = 2 # Valence of dessolation for Mn\n",
-      "Z7 = 1 # Valence of dessolation for Cu\n",
-      "# Above values are given in table 6.3 in the book\n",
-      "D = 100/((p1/D1)+(p2/D2)+(p3/D3)+(p4/D4)+(p5/D5)+(p6/D6)+(p7/D7))\n",
-      "Q = ((0.1035*(10**-2))/D)*(1/((p1*Z1/A1)+(p2*Z2/A2)+(p3*Z3/A3)+(p4*Z4/A4)+(p5*Z5/A5)+(p6*Z6/A6)+(p7*Z7/A7)))\n",
-      "R = Q*I*60\n",
-      "print \"Removal rate = %0.1f cm**3/min\"%(R)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Removal rate = 2.2 cm**3/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 5 on page no. 352"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "V = 10 # DC supply voltage in Volt\n",
-      "k = 0.2 # Conductivity of electrolyte in ohm**-1-cm**-1\n",
-      "f = 0.1 # Feed rate in m/min\n",
-      "Vo = 1.5 # Total overvoltage in Volt\n",
-      "F = 96500 # Faraday constant in coulombs per mole\n",
-      "\n",
-      "A = 55.85 # Atomic gram weight of iron in gm\n",
-      "Z = 2 # Valency of dissolation of iron\n",
-      "rho = 7.86 # Density of iron in gm/cm**3\n",
-      "Yc = k*A*(V-Vo)/(rho*Z*F*(f/60))\n",
-      "print \"Equilibrium gap = %0.2f cm\"%(Yc)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equilibrium gap = 0.04 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 6 on page no. 353"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "S_I1 = 5 # Surface irregulation in micro meter\n",
-      "S_I2 = 8 # Surface irregulation in micro meter\n",
-      "V = 12 # DC supply voltage in Volt\n",
-      "k = 0.2 # Conductivity of electrolyte in ohm**-1-cm**-1\n",
-      "Vo = 1.5 # Total overvoltage in Volt\n",
-      "F = 96500 # Faraday constant in coulombs per mole\n",
-      "Y_min = (S_I1+S_I2)*(10**(-4))\n",
-      "A = 55.85 # Atomic gram weight of iron in gm\n",
-      "Z = 2 # Valency of dissolation of iron\n",
-      "D = 7.86 # Density of iron in gm/cm**3\n",
-      "f_max = (k*A*(V-Vo)/(Z*D*F*Y_min))*60\n",
-      "print \"Largest passible feed rate = %0.1f mm/min\"%(f_max*10)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Largest passible feed rate = 35.7 mm/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 7 on page no. 355"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "f  = 0.2 # Feed rate in cm/min\n",
-      "l = 2.54 # Length of tool face in cm\n",
-      "w = 2.54 # Width of tool face in cm\n",
-      "T_b = 95 # Boiling temperature of electrolyte in \u00b0C\n",
-      "Nita = 0.876e-3 # Viscosity of electrolyte in kg/m-sec\n",
-      "D_e = 1.088 # Density of electrolyte in g/cm**3\n",
-      "c = .997 # Specific heat of electrolyte\n",
-      "V = 10 # DC supply voltage in Volt\n",
-      "k = 0.2 # Conductivity of electrolyte in ohm**-1-cm**-1\n",
-      "T = 35 # Ambient temperature in \u00b0C\n",
-      "Vo = 1.5 # Total overvoltage in Volt\n",
-      "F = 96500 # Faraday constant in coulombs per mole\n",
-      "A = 55.85 # Atomic gram weight of iron in gm\n",
-      "Z = 2 # Valency of dissolation of iron\n",
-      "D = 7.86 # Density of iron in gm/cm**3\n",
-      "Ye = k*A*(V-Vo)*60/(D*Z*F*f)\n",
-      "J = k*(V-Vo)/(Ye)\n",
-      "D_T = T_b -T\n",
-      "v = (J**2)*(l)/(k*D_T*D_e*c)\n",
-      "Re = ((D_e*v*2*Ye)/Nita)*(0.1)\n",
-      "p = 0.3164*D_e*(v**2)*l/(4*Ye*(Re**0.25))*(10**-4)\n",
-      "A = l*w\n",
-      "F = p*A*(10**-1)*(1/2)\n",
-      "print \"Total force acting on the tool = %d N\"%(F)\n",
-      "# Answer in the book is given as 79 N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total force acting on the tool = 103 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 9 on page no. 378"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "# Given that\n",
-      "a = 10 # Side length of a square hole in mm\n",
-      "t = 5 # Thickness of low carbon steel plate in mm\n",
-      "R = 50 # Resistance in relaxation circuit in ohm\n",
-      "C = 10 # Capacitance in relaxation circuit in micro F\n",
-      "V = 200 # Supply voltage in Volt\n",
-      "V_ = 150 # Minimum required voltage for discharge in Volt\n",
-      "\n",
-      "E = (1/2)*C*(10**-6)*(V_**2)\n",
-      "tc = R*C*(10**-6)*log(V/(V-V_))\n",
-      "W = (E/tc)*(10**-3)\n",
-      "v = t*a**2\n",
-      "Q = 27.4*(W**(1.54))\n",
-      "T = v/Q\n",
-      "print \"The time required to complete the drilling operation = %d min\"%(T)\n",
-      "# Answer in the book is given as 306 min"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The time required to complete the drilling operation = 300 min\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 10 on page no. 382"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "R = 50 # Resistance in relaxation circuit in ohm\n",
-      "C = 10 # Capacitance in relaxation circuit in micro F\n",
-      "V = 200 # Supply voltage in Volt\n",
-      "V_ = 150 # Minimum required voltage for discharge in Volt\n",
-      "E = (1/2)*C*(10**-6)*(V_**2)\n",
-      "tc = R*C*(10**-6)*log(V/(V-V_))\n",
-      "W = (E/tc)*(10**-3)\n",
-      "Q = 27.4*(W**(1.54))\n",
-      "Hrms = 1.11*(Q**0.384)\n",
-      "print \"Surface roughness = %0.3f micro meter\"%(Hrms)\n",
-      "# Answer in the book is given as 5.16 micro meter which is wrong"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Surface roughness = 1.350 micro meter\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 11 on page no. 391"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "w = 150 # Width of slot in micro meter\n",
-      "t = 1 # Thickness of tungsten sheet in mm\n",
-      "P = 5 # Power of electron beam in KW\n",
-      "C = 12 # Specific power consumption for tugsten in W/(mm**3/min) from the table 6.7 given in the book \n",
-      "v = (P*(1000)/C)*(1000/(w*t))*(1/600)\n",
-      "print \"Speed of cutting = %0.1f cm/sec\"%(v)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Speed of cutting = 4.6 cm/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 12 on page no. 392"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "# Given that\n",
-      "V = 150e3 # Acceleration voltage in V\n",
-      "D = 76e-7 # Density of steel in kg/mm**3\n",
-      "Delta = 2.6*(10**-17)*((V**2)/D)\n",
-      "print \"Electron range = %d micro meter\"%(ceil(Delta*(10**3)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Electron range = 77 micro meter\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 13 on page no. 395"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "w = 0.015 # Width of slot in cm\n",
-      "t = 1 # Thickness of tungsten sheet in mm\n",
-      "P = 5e3 # Power of electron beam in W\n",
-      "rho_c = 2.71 # Value of volume specific heat for tugsten in J/cm**3\n",
-      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
-      "T_m = 3400 # Melting temperture in \u00b0C\n",
-      "Z = t/10 # In cm\n",
-      "v = (0.1**2)*(P**2)/((T_m**2)*(Z**2)*(k*w*rho_c))\n",
-      "print \"Speed of cutting = %0.1f cm/sec\"%(v)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Speed of cutting = 24.7 cm/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 14 on page no. 399"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "I = 1e5 # Power intensity of laser beam in W/mm**2\n",
-      "T_m = 3400 # Melting temperture of tungsten in \u00b0C\n",
-      "rho_c = 2.71 # Value of volume specific heat for tugsten in J/cm**3\n",
-      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
-      "p_a = 10 # Percentage of beam absorbed\n",
-      "alpha = k/rho_c\n",
-      "H = (p_a/100)*(I)*(100)\n",
-      "tm = (pi/alpha)*((T_m*k)/(2*H))**(2)\n",
-      "print \"Time required for the surface to reach the melting point = %f sec\"%(tm)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time required for the surface to reach the melting point = 0.000053 sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 15 on page no. 400"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "I = 1e5 # Power intensity of laser beam in W/mm**2\n",
-      "d = 200 # Focused diameter of incident beam in micro meter\n",
-      "T_m = 3400 # Melting temperture of tungsten in \u00b0C\n",
-      "rho_c = 2.71 # Value of volume specific heat for tugsten in J/cm**3\n",
-      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
-      "p_a = 10 # Percentage of beam absorbed\n",
-      "H = (p_a/100)*(I)*(100)\n",
-      "alpha = k/rho_c\n",
-      "zeta = 0.5 # Fr0m the standard table\n",
-      " # By solving the equation T_m = ((2*H)*(sqrt(alpha*tm))/k)*((1/sqrt(pi))-ierfc(d/(4*sqrt(alpha*tm))))\n",
-      "tm = 1/((200**2)*(zeta**2)*(alpha))\n",
-      "print \"Time required for the centre of the circular spot to reach the melting point = %f sec\"%(tm)\n",
-      "# Answer in the book is given as 0.00013 sec"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time required for the centre of the circular spot to reach the melting point = 0.000126 sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 16 on page no. 401"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "d = 200 # Diameter of focussed laser beam in micro meter\n",
-      "T_m = 3400 # Melting temperture of tungsten in \u00b0C\n",
-      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
-      "p_a = 10 # Percentage of beam absorbed\n",
-      "H = 2*k*T_m/(d*10**-4)\n",
-      "I = H/(p_a/100)\n",
-      "print \"Minimum value of beam power intensity to achieve the melting = %0.2e W/cm**2\"%(I)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minimum value of beam power intensity to achieve the melting = 7.31e+06 W/cm**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 17 on page no. 403"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "I = 1e5 # Power intensity of laser beam in W/mm**2\n",
-      "t = 0.5 # Thickness of tungsten sheet in mm\n",
-      "d = 200 # Drill diameter in micro meter\n",
-      "P = 3e4 # Energy required per unit volume to vapourize tungsten in J/cm**3\n",
-      "p_e = 10 # Percentage efficiency\n",
-      "T_m = 3400 # Melting temperture of tungsten in \u00b0C\n",
-      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
-      "H = (p_e/100)*(I)*(100)\n",
-      "v = H/P\n",
-      "T = t*(0.1)/(v)\n",
-      "print \"The time required to drill a through hole = %0.4f sec\"%(T)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The time required to drill a through hole = 0.0015 sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
diff --git a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch6_1.ipynb b/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch6_1.ipynb
deleted file mode 100755
index 818587af..00000000
--- a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch6_1.ipynb
+++ /dev/null
@@ -1,654 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Ch-6, Unconventional Machining Process"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 1 on page no. 332"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi, sqrt\n",
-      "from __future__ import division\n",
-      "# Given that\n",
-      "a = 5 # Side of the square hole in mm\n",
-      "t = 4 # Thickness of tungsten plate in mm\n",
-      "d = 0.01 # Diameter of abraisive grains in mm\n",
-      "F = 3.5 # Force for feeding in N\n",
-      "A =25e-3 # Amplitude of tool oscillation in mm\n",
-      "f = 25e3 # Frequency in Hz\n",
-      "Hw = 6900 # Fracture hardness of WC in N/mm**2\n",
-      "\n",
-      "Z = (1/2)*(4*a**2)/(pi*d**2)\n",
-      "lamda = 5\n",
-      "d1 = (d**2)\n",
-      "h_w = (sqrt((8*F*A)/(pi*Z*d1*Hw*(1+lamda))))\n",
-      "Q = (2/3)*((d1*h_w)**(3/2))*Z*f*pi\n",
-      "t = (a**2)*t/(Q*60)\n",
-      "print \"The approximate time required = %0.2f min\"%(t)\n",
-      "# Answer in the book is given as 13.66 min"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The approximate time required = 14.26 min\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 2 on page no. 334"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "r = 1/3 # Ratio of hardness values of copper and steel\n",
-      "R_Q = (r)**(3/4)\n",
-      "R_t = 1/R_Q\n",
-      "P_R = (1-(1/R_t))*100\n",
-      "print \"Percentage change in cutting time when tool is changed from coppper to steel = %d percent(reduction)\"%(P_R)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage change in cutting time when tool is changed from coppper to steel = 56 percent(reduction)\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 3 on page no. 345"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "m = 5 # Romoval rate in cm**3/min\n",
-      "A = 56 # Atomic gram weight in gm\n",
-      "Z = 2 # Valence at which dissolation takes place\n",
-      "D = 7.8 # Density of iron in gm/cm**3\n",
-      "I = (m/60)*(D*Z*96500)/(A)\n",
-      "print \"Current required = %d amp\"%(I)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current required = 2240 amp\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 4 on page no. 345"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "I = 1000 # Current in amp\n",
-      "p1 = 72.5 # Percentage(by weight) of Ni in Nimonic 75 alloy\n",
-      "p2 = 19.5 # Percentage(by weight) of Cr in Nimonic 75 alloy\n",
-      "p3 = 5 # Percentage(by weight) of Fe in Nimonic 75 alloy\n",
-      "p4 = 0.4 # Percentage(by weight) of Ti in Nimonic 75 alloy\n",
-      "p5 = 1 # Percentage(by weight) of Si in Nimonic 75 alloy\n",
-      "p6 = 1 # Percentage(by weight) of Mn in Nimonic 75 alloy\n",
-      "p7 = 6 # Percentage(by weight) of Cu in Nimonic 75 alloy\n",
-      "# From the table 6.3 given in the book\n",
-      "D1 = 8.9 # Density of Ni in g/cm**3\n",
-      "D2 = 7.19 # Density of Cr in g/cm**3\n",
-      "D3 = 7.86 # Density of Fe in g/cm**3\n",
-      "D4 = 4.51 # Density of Ti in g/cm**3\n",
-      "D5 = 2.33 # Density of Si in g/cm**3\n",
-      "D6 = 7.43 # Density of Mn in g/cm**3\n",
-      "D7 = 8.96 # Density of Cu in g/cm**3\n",
-      "A1 = 58.71 # Gram atomic weight of Ni in gm\n",
-      "A2 = 51.99 # Gram atomic weight of Cr in gm\n",
-      "A3 = 55.85 # Gram atomic weight of Fe in gm\n",
-      "A4 = 47.9 # Gram atomic weight of Ti in gm\n",
-      "A5 = 28.09 # Gram atomic weight of Si in gm\n",
-      "A6 = 54.94 # Gram atomic weight of Mn in gm\n",
-      "A7 = 63.57 # Gram atomic weight of Cu in gm\n",
-      "Z1 = 2 # Valence of dessolation for Ni\n",
-      "Z2 = 2 # Valence of dessolation for Cr\n",
-      "Z3 = 2 # Valence of dessolation for Fe\n",
-      "Z4 = 3 # Valence of dessolation for Ti\n",
-      "Z5 = 4 # Valence of dessolation for Si\n",
-      "Z6 = 2 # Valence of dessolation for Mn\n",
-      "Z7 = 1 # Valence of dessolation for Cu\n",
-      "# Above values are given in table 6.3 in the book\n",
-      "D = 100/((p1/D1)+(p2/D2)+(p3/D3)+(p4/D4)+(p5/D5)+(p6/D6)+(p7/D7))\n",
-      "Q = ((0.1035*(10**-2))/D)*(1/((p1*Z1/A1)+(p2*Z2/A2)+(p3*Z3/A3)+(p4*Z4/A4)+(p5*Z5/A5)+(p6*Z6/A6)+(p7*Z7/A7)))\n",
-      "R = Q*I*60\n",
-      "print \"Removal rate = %0.1f cm**3/min\"%(R)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Removal rate = 2.2 cm**3/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 5 on page no. 352"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "V = 10 # DC supply voltage in Volt\n",
-      "k = 0.2 # Conductivity of electrolyte in ohm**-1-cm**-1\n",
-      "f = 0.1 # Feed rate in m/min\n",
-      "Vo = 1.5 # Total overvoltage in Volt\n",
-      "F = 96500 # Faraday constant in coulombs per mole\n",
-      "\n",
-      "A = 55.85 # Atomic gram weight of iron in gm\n",
-      "Z = 2 # Valency of dissolation of iron\n",
-      "rho = 7.86 # Density of iron in gm/cm**3\n",
-      "Yc = k*A*(V-Vo)/(rho*Z*F*(f/60))\n",
-      "print \"Equilibrium gap = %0.2f cm\"%(Yc)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equilibrium gap = 0.04 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 6 on page no. 353"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "S_I1 = 5 # Surface irregulation in micro meter\n",
-      "S_I2 = 8 # Surface irregulation in micro meter\n",
-      "V = 12 # DC supply voltage in Volt\n",
-      "k = 0.2 # Conductivity of electrolyte in ohm**-1-cm**-1\n",
-      "Vo = 1.5 # Total overvoltage in Volt\n",
-      "F = 96500 # Faraday constant in coulombs per mole\n",
-      "Y_min = (S_I1+S_I2)*(10**(-4))\n",
-      "A = 55.85 # Atomic gram weight of iron in gm\n",
-      "Z = 2 # Valency of dissolation of iron\n",
-      "D = 7.86 # Density of iron in gm/cm**3\n",
-      "f_max = (k*A*(V-Vo)/(Z*D*F*Y_min))*60\n",
-      "print \"Largest passible feed rate = %0.1f mm/min\"%(f_max*10)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Largest passible feed rate = 35.7 mm/min\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 7 on page no. 355"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "f  = 0.2 # Feed rate in cm/min\n",
-      "l = 2.54 # Length of tool face in cm\n",
-      "w = 2.54 # Width of tool face in cm\n",
-      "T_b = 95 # Boiling temperature of electrolyte in \u00b0C\n",
-      "Nita = 0.876e-3 # Viscosity of electrolyte in kg/m-sec\n",
-      "D_e = 1.088 # Density of electrolyte in g/cm**3\n",
-      "c = .997 # Specific heat of electrolyte\n",
-      "V = 10 # DC supply voltage in Volt\n",
-      "k = 0.2 # Conductivity of electrolyte in ohm**-1-cm**-1\n",
-      "T = 35 # Ambient temperature in \u00b0C\n",
-      "Vo = 1.5 # Total overvoltage in Volt\n",
-      "F = 96500 # Faraday constant in coulombs per mole\n",
-      "A = 55.85 # Atomic gram weight of iron in gm\n",
-      "Z = 2 # Valency of dissolation of iron\n",
-      "D = 7.86 # Density of iron in gm/cm**3\n",
-      "Ye = k*A*(V-Vo)*60/(D*Z*F*f)\n",
-      "J = k*(V-Vo)/(Ye)\n",
-      "D_T = T_b -T\n",
-      "v = (J**2)*(l)/(k*D_T*D_e*c)\n",
-      "Re = ((D_e*v*2*Ye)/Nita)*(0.1)\n",
-      "p = 0.3164*D_e*(v**2)*l/(4*Ye*(Re**0.25))*(10**-4)\n",
-      "A = l*w\n",
-      "F = p*A*(10**-1)*(1/2)\n",
-      "print \"Total force acting on the tool = %d N\"%(F)\n",
-      "# Answer in the book is given as 79 N"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total force acting on the tool = 103 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 9 on page no. 378"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "# Given that\n",
-      "a = 10 # Side length of a square hole in mm\n",
-      "t = 5 # Thickness of low carbon steel plate in mm\n",
-      "R = 50 # Resistance in relaxation circuit in ohm\n",
-      "C = 10 # Capacitance in relaxation circuit in micro F\n",
-      "V = 200 # Supply voltage in Volt\n",
-      "V_ = 150 # Minimum required voltage for discharge in Volt\n",
-      "\n",
-      "E = (1/2)*C*(10**-6)*(V_**2)\n",
-      "tc = R*C*(10**-6)*log(V/(V-V_))\n",
-      "W = (E/tc)*(10**-3)\n",
-      "v = t*a**2\n",
-      "Q = 27.4*(W**(1.54))\n",
-      "T = v/Q\n",
-      "print \"The time required to complete the drilling operation = %d min\"%(T)\n",
-      "# Answer in the book is given as 306 min"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The time required to complete the drilling operation = 300 min\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 10 on page no. 382"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "R = 50 # Resistance in relaxation circuit in ohm\n",
-      "C = 10 # Capacitance in relaxation circuit in micro F\n",
-      "V = 200 # Supply voltage in Volt\n",
-      "V_ = 150 # Minimum required voltage for discharge in Volt\n",
-      "E = (1/2)*C*(10**-6)*(V_**2)\n",
-      "tc = R*C*(10**-6)*log(V/(V-V_))\n",
-      "W = (E/tc)*(10**-3)\n",
-      "Q = 27.4*(W**(1.54))\n",
-      "Hrms = 1.11*(Q**0.384)\n",
-      "print \"Surface roughness = %0.3f micro meter\"%(Hrms)\n",
-      "# Answer in the book is given as 5.16 micro meter which is wrong"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Surface roughness = 1.350 micro meter\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 11 on page no. 391"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "w = 150 # Width of slot in micro meter\n",
-      "t = 1 # Thickness of tungsten sheet in mm\n",
-      "P = 5 # Power of electron beam in KW\n",
-      "C = 12 # Specific power consumption for tugsten in W/(mm**3/min) from the table 6.7 given in the book \n",
-      "v = (P*(1000)/C)*(1000/(w*t))*(1/600)\n",
-      "print \"Speed of cutting = %0.1f cm/sec\"%(v)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Speed of cutting = 4.6 cm/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 12 on page no. 392"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "# Given that\n",
-      "V = 150e3 # Acceleration voltage in V\n",
-      "D = 76e-7 # Density of steel in kg/mm**3\n",
-      "Delta = 2.6*(10**-17)*((V**2)/D)\n",
-      "print \"Electron range = %d micro meter\"%(ceil(Delta*(10**3)))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Electron range = 77 micro meter\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 13 on page no. 395"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "w = 0.015 # Width of slot in cm\n",
-      "t = 1 # Thickness of tungsten sheet in mm\n",
-      "P = 5e3 # Power of electron beam in W\n",
-      "rho_c = 2.71 # Value of volume specific heat for tugsten in J/cm**3\n",
-      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
-      "T_m = 3400 # Melting temperture in \u00b0C\n",
-      "Z = t/10 # In cm\n",
-      "v = (0.1**2)*(P**2)/((T_m**2)*(Z**2)*(k*w*rho_c))\n",
-      "print \"Speed of cutting = %0.1f cm/sec\"%(v)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Speed of cutting = 24.7 cm/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 14 on page no. 399"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "I = 1e5 # Power intensity of laser beam in W/mm**2\n",
-      "T_m = 3400 # Melting temperture of tungsten in \u00b0C\n",
-      "rho_c = 2.71 # Value of volume specific heat for tugsten in J/cm**3\n",
-      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
-      "p_a = 10 # Percentage of beam absorbed\n",
-      "alpha = k/rho_c\n",
-      "H = (p_a/100)*(I)*(100)\n",
-      "tm = (pi/alpha)*((T_m*k)/(2*H))**(2)\n",
-      "print \"Time required for the surface to reach the melting point = %f sec\"%(tm)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time required for the surface to reach the melting point = 0.000053 sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 15 on page no. 400"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "I = 1e5 # Power intensity of laser beam in W/mm**2\n",
-      "d = 200 # Focused diameter of incident beam in micro meter\n",
-      "T_m = 3400 # Melting temperture of tungsten in \u00b0C\n",
-      "rho_c = 2.71 # Value of volume specific heat for tugsten in J/cm**3\n",
-      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
-      "p_a = 10 # Percentage of beam absorbed\n",
-      "H = (p_a/100)*(I)*(100)\n",
-      "alpha = k/rho_c\n",
-      "zeta = 0.5 # Fr0m the standard table\n",
-      " # By solving the equation T_m = ((2*H)*(sqrt(alpha*tm))/k)*((1/sqrt(pi))-ierfc(d/(4*sqrt(alpha*tm))))\n",
-      "tm = 1/((200**2)*(zeta**2)*(alpha))\n",
-      "print \"Time required for the centre of the circular spot to reach the melting point = %f sec\"%(tm)\n",
-      "# Answer in the book is given as 0.00013 sec"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time required for the centre of the circular spot to reach the melting point = 0.000126 sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 16 on page no. 401"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "d = 200 # Diameter of focussed laser beam in micro meter\n",
-      "T_m = 3400 # Melting temperture of tungsten in \u00b0C\n",
-      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
-      "p_a = 10 # Percentage of beam absorbed\n",
-      "H = 2*k*T_m/(d*10**-4)\n",
-      "I = H/(p_a/100)\n",
-      "print \"Minimum value of beam power intensity to achieve the melting = %0.2e W/cm**2\"%(I)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minimum value of beam power intensity to achieve the melting = 7.31e+06 W/cm**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 17 on page no. 403"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "I = 1e5 # Power intensity of laser beam in W/mm**2\n",
-      "t = 0.5 # Thickness of tungsten sheet in mm\n",
-      "d = 200 # Drill diameter in micro meter\n",
-      "P = 3e4 # Energy required per unit volume to vapourize tungsten in J/cm**3\n",
-      "p_e = 10 # Percentage efficiency\n",
-      "T_m = 3400 # Melting temperture of tungsten in \u00b0C\n",
-      "k = 2.15 # Thermal conductivity of tungsten in W/cm-\u00b0C\n",
-      "H = (p_e/100)*(I)*(100)\n",
-      "v = H/P\n",
-      "T = t*(0.1)/(v)\n",
-      "print \"The time required to drill a through hole = %0.4f sec\"%(T)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The time required to drill a through hole = 0.0015 sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
diff --git a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch7.ipynb b/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch7.ipynb
deleted file mode 100755
index d03ec326..00000000
--- a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch7.ipynb
+++ /dev/null
@@ -1,120 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Ch-7, Manufacturing in the twenty-first century:micromachining, generatice manufacturing, and self assembly"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 1 on page no. 432"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "F = 4e-6 # Maximum feature dimension in meter\n",
-      "t = 5e-6 # Photorist thickness in meter\n",
-      "g = 25e-6 # Allowable gap between the mask and the resist meter\n",
-      "lamda = (F**2)/(t+g)\n",
-      "print \"Maximum allowable wavelength of the exposing light = %d nm\"%(lamda*(10**9))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum allowable wavelength of the exposing light = 533 nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 2 on page no. 440"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import exp\n",
-      "# Given that\n",
-      "d = 5 # Diameter of hole in micro meter\n",
-      "h = 100 # Depth of hole in micro meter\n",
-      "t = 31.58*(d*(exp(h/(60*d))-1))\n",
-      "print \"Time required to machine the hole = %0.1f min\"%(t)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time required to machine the hole = 62.5 min\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 3 on page no. 448"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "# Given that\n",
-      "J = 2 # The threshold value of dose in kJ/cm**3\n",
-      "h = 300 # Height in micro meter\n",
-      "J_o = J*(exp(0.1*sqrt(h)))\n",
-      "print \"The minimum level of exposure of the PMMA surface = %0.1f kJ/cm**3\"%(J_o)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum level of exposure of the PMMA surface = 11.3 kJ/cm**3\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
diff --git a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch7_1.ipynb b/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch7_1.ipynb
deleted file mode 100755
index d03ec326..00000000
--- a/Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch7_1.ipynb
+++ /dev/null
@@ -1,120 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Ch-7, Manufacturing in the twenty-first century:micromachining, generatice manufacturing, and self assembly"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 1 on page no. 432"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# Given that\n",
-      "F = 4e-6 # Maximum feature dimension in meter\n",
-      "t = 5e-6 # Photorist thickness in meter\n",
-      "g = 25e-6 # Allowable gap between the mask and the resist meter\n",
-      "lamda = (F**2)/(t+g)\n",
-      "print \"Maximum allowable wavelength of the exposing light = %d nm\"%(lamda*(10**9))"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum allowable wavelength of the exposing light = 533 nm\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 2 on page no. 440"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from math import exp\n",
-      "# Given that\n",
-      "d = 5 # Diameter of hole in micro meter\n",
-      "h = 100 # Depth of hole in micro meter\n",
-      "t = 31.58*(d*(exp(h/(60*d))-1))\n",
-      "print \"Time required to machine the hole = %0.1f min\"%(t)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time required to machine the hole = 62.5 min\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Problem 3 on page no. 448"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "# Given that\n",
-      "J = 2 # The threshold value of dose in kJ/cm**3\n",
-      "h = 300 # Height in micro meter\n",
-      "J_o = J*(exp(0.1*sqrt(h)))\n",
-      "print \"The minimum level of exposure of the PMMA surface = %0.1f kJ/cm**3\"%(J_o)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum level of exposure of the PMMA surface = 11.3 kJ/cm**3\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}