Added(A)/Deleted(D) following books

 A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter11_1.ipynb
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter13_1.ipynb
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter14_1.ipynb
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter15_1.ipynb
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter16_1.ipynb
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter17_1.ipynb
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter21_1.ipynb
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter22_1.ipynb
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter23_1.ipynb
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter26_1.ipynb
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter2_1.ipynb
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter4_1.ipynb
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter5_1.ipynb
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter9_1.ipynb
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/screenshots/21AOQCurve_1.png
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/screenshots/21X&RChart_1.png
A  A_Textbook_of_Production_Engineering_by_P._C._Sharma/screenshots/22CuttingVvsCutterDia_1.png
A  Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_1.ipynb
A  Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_10.ipynb
A  Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_2.ipynb
A  Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_3.ipynb
A  Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_4.ipynb
A  Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_5.ipynb
A  Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_6.ipynb
A  Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_7.ipynb
A  Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_8.ipynb
A  Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_9.ipynb
A  Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/screenshots/1.png
A  Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/screenshots/2.png
A  Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/screenshots/3.png
A  Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/README.txt
A  Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_1.ipynb
A  Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.1_1.ipynb
A  Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_1.ipynb
A  Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_1.ipynb
A  Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_1.ipynb
A  Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_1.ipynb
A  Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_1.ipynb
A  Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_1.ipynb
A  Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_1.ipynb
A  Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_1.ipynb
A  Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/B.M.D_1.JPG
A  Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_1.jpg
A  Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_2.jpg
A  "sample_notebooks/DaudIbrahir Saifi/Chapter_07_1.ipynb"

45 files changed, 26076 insertions, 0 deletions

diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter11_1.ipynb b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter11_1.ipynb
new file mode 100644
index 00000000..7e0c843a
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter11_1.ipynb
@@ -0,0 +1,123 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 11 : Surface Finish"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 11.1 : Page 436"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "C.L.A value = 0.00*10**-6 m \n"
+     ]
+    }
+   ],
+   "source": [
+    "v = 15000 # vertical magnification\n",
+    "h = 100 # horizontal magnification\n",
+    "l = 0.8 # sampling length in mm\n",
+    "a1 = 160 # area above datum line in mm**2\n",
+    "a2 = 90 # area above datum line in mm**2\n",
+    "a3 = 180 # area above datum line in mm**2\n",
+    "a4 = 50 # area above datum line in mm**2\n",
+    "a5 = 95 # area below datum line in mm**2\n",
+    "a6 = 65 # area below datum line in mm**2\n",
+    "a7 = 170 # area below datum line in mm**2\n",
+    "a8 = 150 # area below datum line in mm**2\n",
+    "a = (a1+a2+a3+a4+a5+a6+a7+a8)/(v*h)\n",
+    "CLA= a/l\n",
+    "print \"C.L.A value = %0.2f*10**-6 m \" %(CLA*1000)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 11.2 : Page 436"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " The arithmetic average = 0.23*10**-6 m \n",
+      " The r.m.s. value = 0.248*10**-6 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "# from figure 11.23\n",
+    "y1 = 0.15 # mu_m\n",
+    "y2 = 0.25 # mu_m\n",
+    "y3 = 0.35 # mu_m\n",
+    "y4 = 0.25 # mu_m\n",
+    "y5 = 0.30 # mu_m\n",
+    "y6 = 0.15 # mu_m\n",
+    "y7 = 0.10 # mu_m\n",
+    "y8 = 0.30 # mu_m\n",
+    "y9 = 0.35 # mu_m\n",
+    "y10 = 0.10 # mu_m\n",
+    "y1sqr = y1**2 # mu_m\n",
+    "y2sqr = y2**2# mu_m\n",
+    "y3sqr = y3**2 # mu_m\n",
+    "y4sqr = y4**2 # mu_m\n",
+    "y5sqr = y5**2 # mu_m\n",
+    "y6sqr = y6**2 # mu_m\n",
+    "y7sqr = y7**2 # mu_m\n",
+    "y8sqr = y8**2 # mu_m\n",
+    "y9sqr = y9**2 # mu_m\n",
+    "y10sqr = y10**2 # mu_m\n",
+    "n = 10\n",
+    "yn = (y1+y2+y3+y4+y5+y6+y7+y8+y9+y10)/n # arithmetic average in mu_m\n",
+    "rms = sqrt((y1sqr+y2sqr+y3sqr+y4sqr+y5sqr+y6sqr+y7sqr+y8sqr+y9sqr+y10sqr)/n) # r.m.s value in mu_m\n",
+    "print \" The arithmetic average = %0.2f*10**-6 m \\n The r.m.s. value = %0.3f*10**-6 m\"%(yn,rms)"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter13_1.ipynb b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter13_1.ipynb
new file mode 100644
index 00000000..332359b5
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter13_1.ipynb
@@ -0,0 +1,517 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 13 : Analysis of Metal Forming Process"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 13.1 : page 515"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total drawing load = 2136.5 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "from math import sqrt, pi, tan, exp\n",
+    "sigma_0 = 240 # N/mm**2\n",
+    "d1 = 5 # initial wire diameter in mm\n",
+    "d0 = 5.5 # final wire diameter in mm\n",
+    "x = d1/d0 # mm\n",
+    "alpha = 8 # angle of contact\n",
+    "alpha = alpha*pi/180\n",
+    "mu = 0.1 # coefficient of friction\n",
+    "B = mu/tan(alpha) \n",
+    "sigma_d  = (sigma_0*(1+B)*(1-(x)**(2*B)))/B # N/mm**2\n",
+    "l = 3 # die land in mm\n",
+    "mu = 0.1 # coefficient of friction\n",
+    "r1 = d1/2 # mm\n",
+    "sigma_t = sigma_0 - (sigma_0 - sigma_d)/exp((2*mu*l)/r1) # N/mm**2\n",
+    "dl = sigma_t*pi*(r1)**2 # drawing load in N\n",
+    "print \"Total drawing load = %0.1f N\"%(dl)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 13.2 : page 517"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " The pipe drawing force force on plug mandrels = 0.547 \n",
+      " The pipe drawing forcw on mandrels = 0.368\n",
+      " Use of movable mandrel substantially reduces drawing force\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import log10\n",
+    "alpha = 15 # angle of contact\n",
+    "alpha = alpha*pi/180\n",
+    "bita = 0 # degree\n",
+    "mu = 0.1 # coefficient of friction\n",
+    "mu1 = mu\n",
+    "mu2 = mu\n",
+    "h1 = 1.75 # mm\n",
+    "h0 = 2.5 # mm\n",
+    "B = (mu1+mu2)/(tan(alpha)-tan(bita)) \n",
+    "y1 = (1+B)*(1-(h1/h0)**B)/B  #sigma_d/sigma_0 for plug mandrels in N/mm**2\n",
+    "z = 1/((h0/h1)-1)\n",
+    "y2 = log10(z)# sigma_d/sigma_0 for movable mandrels in N/mm**2\n",
+    "print \" The pipe drawing force force on plug mandrels = %0.3f \\n The pipe drawing forcw on mandrels = %0.3f\"%(y1,y2)\n",
+    "print \" Use of movable mandrel substantially reduces drawing force\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 13.3 : page 517"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Neutral section = 20.3 mm\n",
+      "Backward slip = 18.8 percent\n",
+      "Forward slip = 1.5 percent\n",
+      "Maximum pressure = 132.6 N/mm**2\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import acos, atan, log\n",
+    "h0 = 25 # thickness of plate in mm \n",
+    "h1 = 20 # mm \n",
+    "delta_h = h0-h1 # mm\n",
+    "sigma = 100 # maximum pressure in N/mm**2\n",
+    "D = 500 # rolled diameter in mm\n",
+    "r = D/2 # rolled radius in mm\n",
+    "alpha = acos(1-(delta_h/D)) # angle of contact in radians\n",
+    "mu = tan(alpha) # coefficient of friction\n",
+    "Ho = 2*sqrt(r/h1)*atan(sqrt(r/h1)*mu)\n",
+    "Hn = (Ho - (log(h0/h1))/mu)/2\n",
+    "theta = sqrt(h1/r)*tan(sqrt(h1/r)*(Hn/2)) # radian\n",
+    "hn = h1 + r*theta**2 # neutral section in mm\n",
+    "x = hn/h0 \n",
+    "bs = (1-x)*100 # backward slip\n",
+    "y = hn/h1\n",
+    "fs = (y-1)*100 # forward slip\n",
+    "sigma0 = 2*sigma/sqrt(3)\n",
+    "pn = sigma0*hn*exp(mu*Hn)/h1 #N/mm**2\n",
+    "print \"Neutral section = %0.1f mm\"%(hn)\n",
+    "print \"Backward slip = %0.1f percent\\nForward slip = %0.1f percent\"%(bs,fs)\n",
+    "print \"Maximum pressure = %0.1f N/mm**2\"%(pn)\n",
+    " # Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 13.4 : page 519"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Force = 6328 kN\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt, pi\n",
+    "Do = 250 # diameter in mm\n",
+    "ho = 250 # hieght in mm\n",
+    "delta_h = 100 # mm\n",
+    "h = 150 # mm\n",
+    "sigma0 = 55 # N/mm**2\n",
+    "d = Do*sqrt(ho/(ho-delta_h)) # diameter in mm \n",
+    "mu = 0.42 # coefficient of friction\n",
+    "R = 162.5 # mm\n",
+    "pa = sigma0/2*(h/(mu*R))**2*(exp(2*mu*R/h)-2*mu*R/h-1) # N/mm**2\n",
+    "p = pa*pi*(R)**2 # force in kN\n",
+    "print \"Force = %d kN\"%(p/1000)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 13.5 : page 519"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Sticking radius = 48.5 mm \n",
+      " Total forging load = 11.927 MN\n"
+     ]
+    }
+   ],
+   "source": [
+    "from sympy import symbols, diff, solve, exp\n",
+    "from math import log\n",
+    "from sympy.mpmath import quad\n",
+    "d = 150 # diameter in mm\n",
+    "h = 10 # thickness in mm\n",
+    "R = d/2 # radius in mm\n",
+    "mu = 0.2 # coefficient of friction\n",
+    "sigma_0 = 200 # N/mm**2\n",
+    "Rs = R - (h/(2*mu))*log(1/(sqrt(3)*mu)) # sticking radius in mm\n",
+    "Ps = sigma_0*exp(2*mu*(R-Rs)/h) # pressure at sticking radius in N/mm**2\n",
+    "y=lambda r:2*pi*r*sigma_0*exp(2*mu/h*(R-r))\n",
+    "L_sld = quad(y,[48.5,75])\n",
+    "L_sld = L_sld/1000 # load on sliding portion in kN\n",
+    "Pc = Ps + (2*sigma_0*Rs)/(h*sqrt(3)) # pressure at centre in N/mm**2\n",
+    "L_sp = (Pc+Ps)*pi*(Rs)**2/(2*1000) # load on sticking portion in kN\n",
+    "F_l = L_sld + L_sp # total forging load in kN\n",
+    "print \" Sticking radius = %0.1f mm \\n Total forging load = %0.3f MN\"%(Rs ,F_l/1000)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 13.7 : page 522"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Drawing load = 11.21 kN\n",
+      "Power of motor = 26.32 kW\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt, pi\n",
+    "RA = 0.30\n",
+    "d = 12 # diameter in mm\n",
+    "alpha = 6  # angle of contact in degree\n",
+    "alpha = 6*pi/180 # angle of contact in radian\n",
+    "mu = 0.10 # coefficient of friction\n",
+    "sigma_0 = 240  # N/mm**2\n",
+    "B = mu/tan(alpha)\n",
+    "x = 1 - RA\n",
+    "sigma_d  = (sigma_0*(1+B)*(1-(x)**B))/B # N/mm**2\n",
+    "r1  = sqrt(x)*(d/2) # mm\n",
+    "l = sigma_d*pi*(r1)**2 # load in kN\n",
+    "ita = 98 # efficiency\n",
+    "ita = ita/100\n",
+    "s = 2.3 # drawing speed in m/s\n",
+    "P = (l*s)/ita # kW\n",
+    "print \"Drawing load = %0.2f kN\\nPower of motor = %0.2f kW\"%(l/1000 ,P/1000 )\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 13.8 : page 522"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Drawing load = 27.166 kN\n",
+      " Power rating of motor = 17.66 kW\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt, pi, tan\n",
+    "mu1 = 0.15 # coefficient of friction   \n",
+    "mu2 = 0.18 # coefficient of fricton\n",
+    "alpha = 14 # angle of contact in degree\n",
+    "alpha = alpha*pi/180\n",
+    "bita = 10 # semi-cone angle in degree\n",
+    "bita = bita*pi/180\n",
+    "sigma_0 = 1.40 # kN/mm**2\n",
+    "h0 = 1.5 #mm\n",
+    "h1 = 1 # mm\n",
+    "B = (mu1+mu2)/(tan(alpha)+tan(bita))\n",
+    "sigmad = (sigma_0*(1+B)*(1-(h1/h0)**B))/B # drawing stress in kN/mm**2\n",
+    "d1 = 11 # outside diameter in mm\n",
+    "t = 1 # thickness in mm\n",
+    "d2 = d1-2*t # mm\n",
+    "a = (pi*((d1)**2-(d2)**2))/4 # area in mm**2\n",
+    "l = sigmad*a # load in kN\n",
+    "s = 0.65 # drawing speed in m/s\n",
+    "w = l*s # work in kJ/s\n",
+    "p = w # power in kW\n",
+    "print \" Drawing load = %0.3f kN\\n Power rating of motor = %0.2f kW\"%(l , p)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 13.9 : page 523"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Forging load at start of forging = 18.5625 MN\n",
+      " Forging load at completion of forging = 70.875 MN\n"
+     ]
+    }
+   ],
+   "source": [
+    "sigma_0 = 50 # pressure at start in MPa\n",
+    "B = 0.9 # width in m\n",
+    "h1 = 0.2 # thickness in m\n",
+    "b = 0.3 # tool bite in m\n",
+    "# At commencement of forging\n",
+    "FL = sigma_0*B*b*(1+(b/(4*h1))) # forging load in MN\n",
+    "# At completion of forging\n",
+    "h2 = 0.1 # thickness in m\n",
+    "sigma_0c = 150 # pressure at completion in MPa\n",
+    "FL2 = sigma_0c*B*b*(1+(b/(4*h2))) # forging load in MN\n",
+    "print \" Forging load at start of forging = %0.4f MN\\n Forging load at completion of forging = %0.3f MN\"%(FL , FL2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 13.10 : page 524"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Extrusion load = 34 MN\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt, tan, pi\n",
+    "sigma_0 = 250 # N/mm**2\n",
+    "d1 = 5 # initial wire diameter in mm\n",
+    "d0 = 15 # final wire diameter in mm \n",
+    "r0 = d0/2\n",
+    "r1 = d1/2\n",
+    "x = (r0/r1)**2 # mm\n",
+    "alpha = 45 # angle of contact\n",
+    "alpha = alpha*pi/180\n",
+    "mu = 0.1 # coefficient of friction\n",
+    "B = mu/tan(alpha) \n",
+    "sigma_x0  = (sigma_0*(1+B)*(1-(x)**B))/B # N/mm**2\n",
+    "sigma_x0 = -sigma_x0\n",
+    "l = 37.5 # length 0f billet in mm\n",
+    "tau1 = sigma_0/2 # Mpa\n",
+    "Pe = sigma_x0 + (4*tau1*l)/d0 # extrusion pressure in Mpa\n",
+    "el = Pe*pi*(r0)**2 # extrusion load in MN\n",
+    "print \"Extrusion load = %d MN\"%(el/10000)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 13.11 : page 524"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a) Roll pressure at enter and exit = 242.5 N/mm**2\n",
+      "    Roll pressure at neutral plane = 257.81 N/mm**2\n",
+      "(b) Roll pressure at neutral point when co-efficient of friction is 0.40 = 779.97 N/mm**2\n",
+      "(c) Roll pressure when 35 N/mm**2 tension is applied at neutral point = 220.60 N/mm**2\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import cos, acos, sqrt, tan, exp\n",
+    "h0 = 4.05 # thickness of plate in mm \n",
+    "h1 = 3.55 # mm\n",
+    "D = 500 # rolled diameter in mm\n",
+    "r = D/2 # rolled radius in mm\n",
+    "mu = 0.04 # coefficient of friction\n",
+    "sigma = 210 # N/mm**2\n",
+    "delta_h = h0-h1 # mm\n",
+    "p = 2*sigma/sqrt(3) # N/mm**2\n",
+    "alpha = acos(1-(delta_h/D)) # angle of contact\n",
+    "Ho = 2*sqrt(r/h1)*atan(sqrt(r/h1)*alpha)\n",
+    "Hn1 = (Ho - (log(h0/h1))/mu)/2\n",
+    "theta = sqrt(h1/r)*tan(sqrt(h1/r)*(Hn1/2)) # radians\n",
+    "hn = h1 + 2*r*(1-cos(theta)) # mm\n",
+    "pn1 = p*hn*exp(mu*Hn1)/h1 # roll pressure in N/mm**2\n",
+    "# b) roll pressure when coefficient of friction is 0.4\n",
+    "mu2 = 0.4 # coefficient of friction\n",
+    "Hn2 = (Ho - (log(h0/h1))/mu2)/2\n",
+    "theta = sqrt(h1/r)*tan(sqrt(h1/r)*(Hn2/2)) # radians\n",
+    "hn2 = h1 + r*theta**2 # mm\n",
+    "pn2 = (p*hn2*exp(mu2*Hn2))/h1 # roll pressure in N/mm**2\n",
+    "# c) if tension is applied of 35 N/mm**2\n",
+    "sigma_f = 35 # front tension in N/mm**2\n",
+    "pn3 = (p-sigma_f)*hn*exp(mu*Hn1)/h1 # roll ressure in N/mm**2\n",
+    "print \"(a) Roll pressure at enter and exit = %0.1f N/mm**2\\n    Roll pressure at neutral plane = %0.2f N/mm**2\"%(p , pn1)\n",
+    "print \"(b) Roll pressure at neutral point when co-efficient of friction is 0.40 = %0.2f N/mm**2\"%(pn2)\n",
+    "print \"(c) Roll pressure when 35 N/mm**2 tension is applied at neutral point = %0.2f N/mm**2\"%(pn3)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 13.12 : page 526"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Neutral plane = 1.58 degree\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import acos, atan, sqrt\n",
+    "h1 = 6.35 # thickness in  mm\n",
+    "mu = 0.2 # coefficient of friction\n",
+    "r = 50 # rolled radius in cm\n",
+    "r = r*10 # mm\n",
+    "R = 30 # reduction in percent\n",
+    "h0 = h1*100/(100-R) # mm \n",
+    "delta_h = h0-h1 # mm\n",
+    "alpha = acos(1-(delta_h/(2*r))) # angle of contact\n",
+    "Ho = 2*sqrt(r/h1)*atan(sqrt(r/h1)*alpha)\n",
+    "Hn = (Ho - (log(h0/h1))/mu)/2 \n",
+    "theta = sqrt(h1/r)*tan(sqrt(h1/r)*(Hn/2)) # neutral plane in radians\n",
+    "theta = theta*180/pi # neutral plane in degrees\n",
+    "print \"Neutral plane = %0.2f degree\"%(theta)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter14_1.ipynb b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter14_1.ipynb
new file mode 100644
index 00000000..ebc24f18
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter14_1.ipynb
@@ -0,0 +1,823 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 14 : Theory of Metal Cutting"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.1 : page 563"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Tool life = 18 min.\n"
+     ]
+    }
+   ],
+   "source": [
+    "v1 = 18 # cutting speed in m/min\n",
+    "t1 = 3 # tool life in hours\n",
+    "n = 0.125 # exponent\n",
+    "c = v1*(t1*60)**n # constant\n",
+    "v2 = 24 # cutting speed in m/min\n",
+    "t = (c/v2)**(1/0.125) # tool life in min.\n",
+    "print \"Tool life = %d min.\"%(t)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.2 : page 564"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Cutting speed = 56.5 m/min.\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "c_t = 8 # tool change time in min.\n",
+    "r_t = 5 # tool re-grind time in min.\n",
+    "mr_c = 5 # machine running cost per hour\n",
+    "d = 30 # total depreciation per re-grind in paisa\n",
+    "n = 0.25 # exponent\n",
+    "c = 150 # constant\n",
+    "c_c = mr_c*c_t/60 # total change cost in Rs\n",
+    "r_c = mr_c*r_t/60 # regrind cost in Rs\n",
+    "ct = c_c+r_c+d/100 # tooling cost in Rs\n",
+    "cm = mr_c/60 # machining cost in Rs\n",
+    "v = c*((cm*n)/(ct*(1-n)))**n # cutting speed in m/min.\n",
+    "print \"Cutting speed = %0.1f m/min.\"%(v) "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.3 : page 564"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Drawing load = 14.262 kN\n",
+      "Power rating of motor = 9.27 kW\n",
+      "Shear angle = 31.27 degree\n",
+      "Friction angle = 31.82 degree\n",
+      "Shear stress = 236.1 N/mm**2\n",
+      "Cutting power = 1.134 kw \n",
+      "Chip velocity = 1.114 m/s\n",
+      "shear strain = 2.036\n",
+      "Shear strain rate = 86386.645 s**-1\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi, tan, atan, cos, sin\n",
+    "mu1 = 0.15 # coefficient of friction   \n",
+    "mu2 = 0.18 # coefficient of fricton\n",
+    "alpha = 14 # angle of contact in degree\n",
+    "alpha = alpha*pi/180\n",
+    "bita = 10 # semi-cone angle in degree\n",
+    "bita = bita*pi/180\n",
+    "sigma_0 = 1.40 # kN/mm**2\n",
+    "h0 = 1.5 #mm\n",
+    "h1 = 1 # mm\n",
+    "B = (mu1+mu2)/(tan(alpha)+tan(bita))\n",
+    "sigmad = (sigma_0*(1+B)*(1-(h1/h0)**B))/B # drawing stress in kN/mm**2\n",
+    "d1 = 11 # outside diameter in mm\n",
+    "t = 1 # thickness in mm\n",
+    "d2 = d1-t # mm\n",
+    "a = (pi*((d1)**2-(d2)**2))/4 # area in mm**2\n",
+    "l = sigmad*a # load in kN\n",
+    "s = 0.65 # drawing speed in m/s\n",
+    "w = l*s # work in kJ/s\n",
+    "p = w # power in kW\n",
+    "print \"Drawing load = %0.3f kN\\nPower rating of motor = %0.2f kW\"%(l , p)\n",
+    "t = 0.127 # uncut chip thickness in mm\n",
+    "b = 6.35 # width of cut in mm\n",
+    "v = 2 # cutting speed in m/s\n",
+    "alpha = 10 # rake angle in degrees\n",
+    "fc = 567 # cutting force in N\n",
+    "ft = 227 # thrust force in N\n",
+    "tc = 0.228 # chip thickness in mm\n",
+    "r = t/tc # chip thickness ratio\n",
+    "alpha = alpha*pi/180 # rake angle in radians\n",
+    "phi = atan(r*cos(alpha)/(1-(r*sin(alpha)))) # shear angle \n",
+    "phi1 = phi*180/pi # shear angle\n",
+    "print \"Shear angle = %0.2f degree\"%(phi1) \n",
+    "mu =((fc*sin(alpha)+ft*cos(alpha))/(fc*cos(alpha)-ft*sin(alpha))) #coefficient of friction\n",
+    "bita = atan(mu) # friction angle\n",
+    "bita = bita*180/(pi)\n",
+    "print \"Friction angle = %0.2f degree\"%(bita )\n",
+    "fs = fc*cos(phi)-ft*sin(phi) #shear force in N\n",
+    "taus = (fs*sin(phi))/(b*t) # shear stress\n",
+    "print \"Shear stress = %0.1f N/mm**2\"%(taus)\n",
+    "cp = fc*v/1000 # cutting power in kw\n",
+    "print \"Cutting power = %0.3f kw \"%(cp)\n",
+    "vc = v*r # chip velocity in m/s\n",
+    "print \"Chip velocity = %0.3f m/s\"%(vc)\n",
+    "ss = 1/tan(phi) + tan(phi-alpha) # shear strain\n",
+    "print \"shear strain = %0.3f\"%(ss)\n",
+    "spl = t/sin(phi) # shear plane length\n",
+    "vs = v*cos(alpha)/cos(phi-alpha) # shear velocity\n",
+    "S = vs*10/spl # shear strain rate\n",
+    "S = S*10**3 # shear strain rate\n",
+    "print \"Shear strain rate = %.3f s**-1\"%(S)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.4 : page 566"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "constant = 28.37 \n",
+      "Tool life when cutting speed increased by 20 = 14.76 min.\n",
+      "Tool life when feed rate increased by 20 = 20.38 min.\n",
+      "Tool life when depth of cut increased by 20 = 35.71 min.\n",
+      "Tool life when all taken together after increasing by 20 = 2.98 min.\n"
+     ]
+    }
+   ],
+   "source": [
+    "v = 30 # cutting speed in m/min\n",
+    "feed = 0.3 # feed rate in mm/rev.\n",
+    "d = 2.5 # depth of cut in mm\n",
+    "t = 60 # tool life in min.\n",
+    "c = v*t**0.13*feed**0.77*d**0.37 # constant\n",
+    "print \"constant = %0.2f \"%(c)\n",
+    "v2 = v*1.2 # cutting speed in m/min \n",
+    "t2 = (c/(v2*feed**0.77*d**0.37)) # tool life when cutting speed increased by 20% in min.\n",
+    "t2 = t2**(1/0.13)\n",
+    "f2 = feed*1.2 # feed rate in mm/rev.\n",
+    "t3 = (c/(v*d**0.37*f2**0.77)) # tool life when feed rate increased by 20% in min.\n",
+    "t3 = t3**(1/0.13)\n",
+    "d2 = d*1.2 # depth of cut in mm\n",
+    "t4 = (c/(v*feed**0.77*d2**0.37)) # tool life when depth of cut increased by 20% in min.\n",
+    "t4 = t4**(1/0.13)\n",
+    "t5 = (c/(v2*d2**0.37*f2**0.77)) # tool lfe in min.\n",
+    "t5 = t5**(1/0.13)\n",
+    "print \"Tool life when cutting speed increased by 20 = %0.2f min.\"%(t2)\n",
+    "print \"Tool life when feed rate increased by 20 = %0.2f min.\"%(t3)\n",
+    "print \"Tool life when depth of cut increased by 20 = %0.2f min.\"%(t4)\n",
+    "print \"Tool life when all taken together after increasing by 20 = %0.2f min.\"%(t5)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.5 : page 567"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Force of shear at shear plane = 813.80 N\n",
+      "Friction angle = 0.430 degree\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import atan, cos, sin, pi\n",
+    "t = 0.25 # uncut chip thickness in mm\n",
+    "b = 2.5 # width of cut in mm\n",
+    "v = 2.5 # cutting speed in m/s\n",
+    "alpha = 10 # rake angle in degrees\n",
+    "fc = 1130 # cutting force in N\n",
+    "ft = 295 # thrust force in N\n",
+    "tc = 0.45 # chip thickness in mm\n",
+    "r = t/tc # chip thickness ratio\n",
+    "alpha = alpha*pi/180 # rake angle in radians\n",
+    "phi = atan((r*cos(alpha))/(1-r*sin(alpha))) # shear angle \n",
+    "phi2 = phi*180/pi # shear angle\n",
+    "fs = fc*cos(phi) - ft*sin(phi) #shear force in N\n",
+    "print \"Force of shear at shear plane = %0.2f N\"%(fs)\n",
+    "mu = atan((fc*sin(alpha)+ft*cos(alpha))/(fc*cos(alpha)-ft*sin(alpha))) #friction anglele\n",
+    "print \"Friction angle = %0.3f degree\"%(mu )\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.6 : page 568"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Cutting ratio = 0.323\n",
+      "Chip reduction co-efficient = 3.1\n",
+      "Shear angle = 18.78 degree\n",
+      "shear strain = 3.276\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import cos, atan, pi, tan\n",
+    "t = 0.2 # uncut chip thickness in mm\n",
+    "alpha = 15 # rake angle in degrees\n",
+    "tc = 0.62 # chip thickness in mm\n",
+    "r = t/tc # chip thickness ratio\n",
+    "crc = 1/r # chip reduction coefficient\n",
+    "print \"Cutting ratio = %0.3f\\nChip reduction co-efficient = %0.1f\"%(r , crc)\n",
+    "alpha = alpha*pi/180 # rake angle in radians\n",
+    "phi = atan(r*cos(alpha)/(1-r*sin(alpha))) # shear angle \n",
+    "phi = phi*180/pi # shear angle\n",
+    "print \"Shear angle = %0.2f degree\"%(phi) \n",
+    "ss = 1/tan(phi*pi/180) + tan((phi*pi)/180-(alpha*pi)/180) # shear strain\n",
+    "print \"shear strain = %0.3f\"%(ss)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.7 : page 568"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " n = 0.224\n",
+      " C = 68.4\n",
+      " Cutting speed = 27.37 m/min.\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import log\n",
+    "v1 = 25 # cutting speed in m/min\n",
+    "t1 = 90 # tool life in min.\n",
+    "v2 = 35 # cutting speed in m/min\n",
+    "t2 = 20 # tool life in min\n",
+    "n = log(v2/v1)/log(t1/t2) # exponent\n",
+    "C = v1*(t1)**n # constant\n",
+    "t = 60 # tool life in min.\n",
+    "v = C/(t)**n # cutting speed in m/min.\n",
+    "print \" n = %0.3f\\n C = %0.1f\\n Cutting speed = %0.2f m/min.\"%(n , C,v)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.8 : page 569"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Tangential force on tool face = 858 N\n",
+      " normal force on tool face = 1225.0 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import ceil, cos, sin\n",
+    "t = 0.5 # uncut chip thickness in mm\n",
+    "b = 3 # width of cut in mm\n",
+    "alpha = 15 # rake angle in degrees\n",
+    "alpha = alpha*pi/180 # rake angle in radians\n",
+    "r = 0.383 # chip thickness ratio\n",
+    "mu = 0.7 # average coefficient of friction on tool face\n",
+    "bita = atan(mu) # friction angle\n",
+    "tau = 280 # yield stress in N/mm**2\n",
+    "phi = atan((r*cos(alpha))/(1-r*sin(alpha))) # shear angle \n",
+    "fc = (tau*b*t)/(1/cos(bita-alpha)*cos(phi+bita-alpha)*sin(phi)) # cutting force in N\n",
+    "ft = (fc*(mu-tan(alpha)))/(1+mu*tan(alpha)) # thrust force in N\n",
+    "F = fc*sin(alpha)+ft*cos(alpha) # tangential force on tool face in N\n",
+    "F = ceil(F)\n",
+    "N = fc*cos(alpha)-ft*sin(alpha) # normal force on tool face in N\n",
+    "print \" Tangential force on tool face = %d N\\n normal force on tool face = %0.1f N\"%(F,N)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.9 : page 570"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Cutting force= 2034.3 N\n",
+      " Thrust force = 524.3 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt, cos, sin, sqrt\n",
+    "t = 0.25 # uncut chip thickness in mm\n",
+    "b = 0.5 # width of cut in cm\n",
+    "v = 8.2 # cutting speed in m/min.\n",
+    "alpha = 20 # rake angle in degrees\n",
+    "alpha2 = alpha*pi/180 # rake angle in radians\n",
+    "r = 0.351 # cutting ratio\n",
+    "phi = atan(r*cos(alpha2)/(1-r*sin(alpha2))) # shear angle  in radians\n",
+    "phi2 = phi*180/pi # shear angle in degrees\n",
+    "alpha2 = alpha*pi/180 # rake angle in radians\n",
+    "bita = 35+alpha-phi2 # degrees\n",
+    "s = 1/tan(phi) + tan(phi-alpha2) # shear strain\n",
+    "e = s/sqrt(3) # natural strain\n",
+    "sigma = 784*(e)**0.15 # tensile property in N/mm**2\n",
+    "tau = sigma/sqrt(3) # yield shear stress in N/mm**2\n",
+    "As = (b*10*t)/sin(phi) # shear plane area in mm**2\n",
+    "Fs = tau*As # shear gorce in N\n",
+    "R = Fs/cos(phi+(bita*pi/180)-alpha2)\n",
+    "Fc = R*cos((bita*pi/180)-alpha2) # cutting force in N\n",
+    "Ft = R*sin((bita*pi/180)-alpha2) # thrust force in N\n",
+    "print \" Cutting force= %0.1f N\\n Thrust force = %0.1f N\"%(Fc , Ft)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.10  : page 571"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Shear force = 987 N\n",
+      " Normal force = 1519.2 N\n",
+      " Friction force = 1114.9 N\n",
+      " Kinetic coefficient of friction = 0.781\n",
+      " Specific cutting energy = 4000 N/mm**2\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi, atan, cos, sin, tan\n",
+    "f = 0.2 # feed in mm/rev.\n",
+    "t = 0.2 # uncut chip thickness in mm\n",
+    "alpha = 10 # rake angle in degrees\n",
+    "fc = 1600 # cutting force in N\n",
+    "ft = 850 # thrust force in N\n",
+    "tc = 0.39 # chip thickness in mm\n",
+    "r = t/tc # chip thickness ratio\n",
+    "d = 2 # depth of cut in mm\n",
+    "b = 2 # mm\n",
+    "alpha2 = alpha*pi/180 # rake angle in radians\n",
+    "phi = atan(r*cos(alpha2)/(1-r*sin(alpha2))) # shear angle in radians\n",
+    "phi2 = phi*180/pi # shear angle in degree\n",
+    "fs = fc*cos(phi)-ft*sin(phi) #shear force in N\n",
+    "fn = fc*sin(phi)+ft*cos(phi) # normal force in N\n",
+    "f = fc*sin(alpha2)+ft*cos(alpha2) # friction force in N\n",
+    "mu =((fc*tan(alpha2)+ft)/(fc-ft*tan(alpha2))) #kinetic coefficient of friction\n",
+    "s = fc/(b*t) # specific cutting energy in N/mm**2\n",
+    "print \" Shear force = %d N\\n Normal force = %0.1f N\\n Friction force = %0.1f N\\n Kinetic coefficient of friction = %0.3f\"%(fs , fn ,f , mu)\n",
+    "print \" Specific cutting energy = %d N/mm**2\"%(s)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.11 : page 572"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Centre line average roughness = 43.13*10**-6m\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt, pi, sin, cos\n",
+    "cs = 20 # side cutting edge angle in degree\n",
+    "ce = 30 # end cutting edge angle in degree\n",
+    "f = 0.1 # feed in mm/rev.\n",
+    "r = 3 # nose radius in mm\n",
+    "cs2 = cs*pi/180 # side cutting edge angle in radians\n",
+    "ce2 = ce*pi/180 # end cutting edge angle in radians\n",
+    "h = (1-cos(ce2))*r + f*sin(ce2)*cos(ce2) - sqrt((2*f*r*(sin(ce2))**3)-((f**2)*(sin(ce2))**4))\n",
+    "Ra = h/4 # Centre line average roughness in mm\n",
+    "print \"Centre line average roughness = %0.2f*10**-6m\"%(Ra*10**3)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.12 : page 572"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Back rake angle = 2.61 degree\n",
+      " Side rake angle = 9.67 degree\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import atan, cos, tan, pi, sqrt\n",
+    "i = 0 # inclination angle in degree\n",
+    "alpha = 10 # orthogonal rake angle in degree\n",
+    "lemda = 75 # principal cutting edge angle in degree\n",
+    "alpha = alpha*pi/180 # orthogonal rake angle in radian\n",
+    "lemda = lemda*pi/180 # principal cutting edge angle in radian\n",
+    "alpha_b = atan(cos(lemda)*tan(alpha)+sin(lemda)*tan(i)) #back rake angle in radians\n",
+    "alpha_b = alpha_b*180/pi # back rake angle in degree\n",
+    "alpha_s = atan(sin(lemda)*tan(alpha)-cos(lemda)*tan(i)) # side rake angle in radians\n",
+    "alpha_s = alpha_s*180/pi # side rake angle in degree\n",
+    "print \" Back rake angle = %0.2f degree\\n Side rake angle = %0.2f degree\"%(alpha_b,alpha_s)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.13 : page 572"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Othogonal rake angle = 5.93 degree\n",
+      " Inclination angle = 6.7 degree\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt, pi, atan, sin\n",
+    "alphab = 8 # back rake in degree\n",
+    "alphas = 4 # side rake in degree\n",
+    "cs = 15 # side cutting edge angle in degree\n",
+    "lemda = 90 - cs # approach angle in degree\n",
+    "alphab = alphab*pi/180 # back rake in radian\n",
+    "alphas = alphas*pi/180 # side rake in radian\n",
+    "cs = cs*pi/180 # side cutting edge angle in radian\n",
+    "lemda = lemda*pi/180 # approach angle in radian\n",
+    "alpha = atan(tan(alphas)*sin(lemda)+tan(alphab)*cos(lemda)) # orthogonal rake angle in radian\n",
+    "alpha = alpha*180/pi # orthogonal rake angle in degree\n",
+    "i = atan(sin(lemda)*tan(alphab)-cos(lemda)*tan(alphas)) # inclnation angle in radian\n",
+    "i = i*180/pi # inclnation angle in degree\n",
+    "print \" Othogonal rake angle = %0.2f degree\\n Inclination angle = %0.1f degree\"%(alpha , i)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.14 : page 573"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Cutting power = 0.553 kw\n",
+      " Motor power = 0.65 kw\n",
+      " Specific cutting resistance = 1726.90 N/mm**2\n",
+      " Unit power = 1.727 W/(mm**3)*s\n"
+     ]
+    }
+   ],
+   "source": [
+    "cs = 15 # side cutting edge angle in degree\n",
+    "v = 0.2 # cutting speed in m/s\n",
+    "f = 0.5 # feed rate in mm/rev.\n",
+    "d = 3.2 # depth of cut in mm\n",
+    "fc = 1593*(f)**0.85*(d)**0.98 # cutting force in N\n",
+    "pc = fc*v/1000 # cutting power in kw\n",
+    "ita_mt = 0.85 # efficiency of lathe\n",
+    "pm = pc/ita_mt # motor power in kw\n",
+    "a = f*d # area of uncut chio in mm**2\n",
+    "r = fc/a # specific cutting resistance in N/mm**2\n",
+    "p = pc/(a*v)# unit power in W/(mm**3)*s\n",
+    "print \" Cutting power = %0.3f kw\\n Motor power = %0.2f kw\\n Specific cutting resistance = %0.2f N/mm**2\\n Unit power = %0.3f W/(mm**3)*s\"%(pc,pm,r,p)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.15 : page 574"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Percentage increase = 300 percent\n"
+     ]
+    }
+   ],
+   "source": [
+    "C = 400\n",
+    "n=0.5\n",
+    "a=2 # (T1/T2)**n\n",
+    "b=2**(1/n) # T2\n",
+    "i = (b-1)*100 # percentage increase\n",
+    "print \"Percentage increase = %d percent\"%(i)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.16 : page 574"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The percentage of total energy that goes into overcoming friction at tool chip interface = 31.48 percent\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import acos, sin, pi\n",
+    "t = 0.127 # uncut chip thickness in mm\n",
+    "b = 6.35 # width of cut in mm\n",
+    "v = 1.20 # cutting speed in m/min.\n",
+    "alpha = 10 # rake angle in degrees\n",
+    "fc = 556.25 # cutting force in N\n",
+    "ft = 222.50 # thrust force in N\n",
+    "tc = 0.229 # chip thickness in mm\n",
+    "r = t/tc # chip thickness ratio\n",
+    "R = sqrt((fc**2)+(ft**2))\n",
+    "bita = (acos(fc/R)) + alpha*pi/180 #\n",
+    "f = R*sin(bita) # \n",
+    "fe = f*r # friction energy\n",
+    "p = (f*r*100)/fc # percentage of fricton enrgy and total energy \n",
+    "print \"The percentage of total energy that goes into overcoming friction at tool chip interface = %0.2f percent\"%(p)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.17 : page 575"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Power consumption = 1.31 W\n",
+      " Specific cutting energy = 3.08 W*s/mm**3\n",
+      " Energy consumed = 2.141 kWh\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "D = 300 # diameter in mm\n",
+    "r = 45 # rev/min.\n",
+    "d = 2 # depth of cut in mm\n",
+    "f = 0.3 # feed in mm/rev\n",
+    "fc = 1850 # cutting force in N\n",
+    "ff = 450 # feed force in N\n",
+    "V = 2.5*10**6 # metal removed in mm\n",
+    "v = (pi*D*r)/(60*1000) # cutting velocity in m/s\n",
+    "pc = fc*v/1000 # cutting power in kW\n",
+    "fv = f*r/60*1000 # feed velocity in m/s\n",
+    "fp = fv*ff # feed power in W\n",
+    "mrr = d*f*v*60*1000 # mm**3/min.\n",
+    "ce = pc*1000*60/mrr # specific cutting energy in W*s/mm**2\n",
+    "E = ce*V/(3600*1000) # energy consumed\n",
+    "print \" Power consumption = %0.2f W\\n Specific cutting energy = %0.2f W*s/mm**3\\n Energy consumed = %0.3f kWh\"%(pc,ce,E)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 14.18 : page 576"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Cutting force = 247 N\n",
+      " Thrust force = -2.546 N\n",
+      " Feed force = -2.2 N\n",
+      " Radial force = -1.273 N\n",
+      " Cutting power = 388 watts\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import atan, degrees, sin, cos\n",
+    "D = 100 # diameter in mm\n",
+    "cs = 30 # side cutting edge angle in degree\n",
+    "lemda = 90 - cs # approach angle in degree\n",
+    "d = 2.5 # depth of cut in mm\n",
+    "f = 0.125 # feed in mm/rev.\n",
+    "N = 300 # turning speed of job in rev./min.\n",
+    "mu = 0.6 # coefficient of friction\n",
+    "tau = 400 # ultimate shear stress in Mpa\n",
+    "bita = degrees(atan(mu)) # friction angle in radian\n",
+    "alphas = 10 # side rake angle\n",
+    "alphab = 6 # back rake angle\n",
+    "alpha = degrees(atan(tan(alphas))*sin(lemda*pi/180)+tan(alphab*pi/180)*cos(lemda*pi/180)) # orthogonal rake angle in degree\n",
+    "phi = 45 - ( bita - alpha) # shear angle\n",
+    "Fc = tau*d*f/(1/cos((bita-alpha)*pi/180)*cos((phi+bita-alpha)*pi/180)*sin(phi*pi/180)) # cutting force in N\n",
+    "Ft = Fc*tan((bita-alpha)*pi/180) # thrust component in N\n",
+    "Ff = Ft*sin(lemda*pi/180) # feed force along axis of job in N\n",
+    "Rf = Ft*cos(lemda*pi/180) # radial force normal to axis of job in N\n",
+    "v = pi*D*N/(1000*60) # velocity in m/s\n",
+    "p = Fc*v # power in watts\n",
+    "print \" Cutting force = %d N\\n Thrust force = %0.3f N\\n Feed force = %0.1f N\\n Radial force = %0.3f N\\n Cutting power = %d watts\"%(Fc,Ft,Ff,Rf,p)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter15_1.ipynb b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter15_1.ipynb
new file mode 100644
index 00000000..771df52c
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter15_1.ipynb
@@ -0,0 +1,408 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 15 : Design & Manufacture of Cutting Tools"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 15.1 : page 597"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Horsepower at cutter = 20.60\n",
+      " Horsepower at motor = 34.34\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "w = 10 # width of cut in cm\n",
+    "h = 0.32 # depth of cut in cm\n",
+    "n = 8 # number of teeth in cutter\n",
+    "ft = 0.033 # feed rate per tooth\n",
+    "N = 200 # cutter speed in rpm\n",
+    "ita = 60/100 # efficiency\n",
+    "f = n*ft*N # feed rate in cm/min.\n",
+    "mrr = w*h*f # metal removal rate in cm**3/min.\n",
+    "k = 8.2 # machinibility factor from table 15.3\n",
+    "hpc = mrr/k # horsepower at cutter\n",
+    "hpm = hpc/ita # horsepower at motor\n",
+    "print \" Horsepower at cutter = %0.2f\\n Horsepower at motor = %0.2f\"%(hpc , hpm)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 15.2 : page 603"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " MRR = 2666.67 mm**3/s\n",
+      " Cutting power = 16000 watts\n",
+      " Torque = 1527.89 N.m\n",
+      " Cutting time = 37.5s\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi, sqrt\n",
+    "l = 300 # length in mm\n",
+    "w = 100 # width in mm\n",
+    "f = 0.25 # feed in mm/tooth\n",
+    "d = 3.2 # depth of cut in mm\n",
+    "D = 50 # cutter diameter in mm\n",
+    "n = 20 # number of cutter teeth\n",
+    "N = 100 # cutter speed in rev./min.\n",
+    "tf = f*n*N # table feed in mm/min.\n",
+    "mrr = w*d*tf # metal removal rate in mm**3/min.\n",
+    "mrr = mrr/60 # mm**3/s\n",
+    "p = 6*mrr # cutting power from table 14.2 in watts\n",
+    "omega = 2*pi*N/60 # rpm\n",
+    "T = p/omega # torque in N.m\n",
+    "att = sqrt((D*d)-(d**2)) # added table travel in mm\n",
+    "t = (l+att)/tf # cutting time in min.\n",
+    "t = t*60 # s\n",
+    "print \" MRR = %0.2f mm**3/s\\n Cutting power = %d watts\\n Torque = %0.2f N.m\\n Cutting time = %0.1fs\"%(mrr,p,T,t)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 15.3 : page 610"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(i)Broaching power = 0.1355 kW\n",
+      "(ii) Broach Design\n",
+      "(a) Pitch diameter = 10.35mm\n",
+      " (b) width of land = 3.11 mm \n",
+      "    depth of cutting teeth = 4.14 mm\n",
+      "    Tooth fillet radius= 3.11 mm\n",
+      "(c) Length of toothed portion of broach = 124 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi, sqrt\n",
+    "l = 35 # length of bore in mm\n",
+    "v = 0.15 # cutting speed in m/s\n",
+    "t1 = 0.01 # upper limit in mm\n",
+    "t2 = 0.05 # upper limit in mm\n",
+    "D = 32.25 # finished broach in mm\n",
+    "D1 = 32.25+t2 # mm\n",
+    "d = 32.75 # finish diameter in mm\n",
+    "d1 = 32.75 + t1 #finish diameter of hole in mm\n",
+    "s = 0.05 # mm\n",
+    "B = 1.30 # blunt broach factor\n",
+    "c = 45 # specific cutting force in N/mm**2\n",
+    "n = 3 # number of teeth cutting at a time\n",
+    "F = n*pi*d1*s*c*B # force needed for broaching in N\n",
+    "bp = F*v/1000 # Broaching power in kw\n",
+    "# broach design\n",
+    "p = 1.75*sqrt(l) # pitch in mm\n",
+    "theta = 10 # face angle in degree\n",
+    "alha1 = 1.5 # relief angle for roughing in degree\n",
+    "alha2 = 1.0 # relief angle for finishing in degree\n",
+    "w = 0.3*p # width of land in mm\n",
+    "h = 0.4*p # depth of cutting teeth in mm\n",
+    "r = 0.3*p # tooth fillet radius in mm\n",
+    "T = (d1-D1)/2 # mm\n",
+    "n = T/s # number of cutting teeth\n",
+    "n = round(n)\n",
+    "l = (n+7)*p #length of toothed portion of broach in mm\n",
+    "print \"(i)Broaching power = %0.4f kW\"%bp\n",
+    "print \"(ii) Broach Design\"\n",
+    "print \"(a) Pitch diameter = %0.2fmm\\n (b) width of land = %0.2f mm \\n    depth of cutting teeth = %0.2f mm\\n    Tooth fillet radius= %0.2f mm\"%(p,w,h,r)\n",
+    "print \"(c) Length of toothed portion of broach = %d mm\"%(l)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 15.4  : page 616"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Moment = 1024.2 kgf-mm\n",
+      " Power = 0.097 hp\n",
+      " Average force = 284 kgf\n",
+      " Thrust force = 253.4 kgf\n"
+     ]
+    }
+   ],
+   "source": [
+    "Hb = 200 # brinell hardness\n",
+    "d = 12.7 # diameter in mm\n",
+    "f = 0.254 # feed in mm/rev.\n",
+    "N = 100 # rpm\n",
+    "M = (Hb*(d)**2*f)/8 #moment in kgf-mm\n",
+    "k = 1.1 #material factor\n",
+    "p = (1.25*(d)**2*k*N*(0.056+1.5*f))/(10)**5 # power in kW\n",
+    "T1a = (1.7*M)/d # thrust force kgf\n",
+    "T1b = (3.5*M)/d # kgf\n",
+    "T1 = (T1a+T1b)/2 # average\n",
+    "w = 0.14*d # thickness in mm\n",
+    "T2a = (0.1*pi*(w)**2*Hb)/4 # thrust force kgf\n",
+    "T2b = (0.2*pi*(w)**2*Hb)/4 # thrust force kgf\n",
+    "T2 = (T2a+T2b)/2 # average\n",
+    "avg = T1+T2 # kgf\n",
+    "thrust = 1.16*k*d*(100*f)**0.85 # kgf\n",
+    "print \" Moment = %0.1f kgf-mm\\n Power = %0.3f hp\\n Average force = %d kgf\\n Thrust force = %0.1f kgf\"%(M, p ,avg , thrust)\n",
+    "# Error in textbook"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 15.5  : page 617"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1 Diameter of reamer \n",
+      " Maximum diameter of reamer = 55.030 mm \n",
+      " Minimum diameter of reamer = 55.017 mm \n",
+      " 2 Back taper = 0.05 mm \n",
+      " 3 Values of various angles \n",
+      " Rake angle = 5 degree \n",
+      " Plan approach angle = 45 degree \n",
+      " Circular land = 0.25 to 0.50 mm \n",
+      " Secondary clearance angle = 10 degree \n",
+      " 4 Length of reamer \n",
+      " Length of fluted portion = 82.5 mm \n",
+      " Length of reaming allowance = 0.18 mm \n",
+      " Length of cutting section = 2.25 mm \n",
+      " Length of guiding section = 16 mm \n",
+      " 5 Number of teeth = 14\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import ceil, sqrt\n",
+    "d = 55 # diameter in mm\n",
+    "ul = 0.035 # upper limit in mm\n",
+    "ll = 0.000 # lower limit in mm\n",
+    "Dmax = d+ul # maximum diameter of hole in mm\n",
+    "Dmin = d+ll # minimum diameter of hole in mm\n",
+    "IT = 0.035 # hole tolerence in mm\n",
+    "dmax = Dmax-0.15*IT # maximum diameter of reamer in mm\n",
+    "dmin = dmax-0.35*IT # minimum diameter of reamer in mm\n",
+    "l = ((d/4)+(d/3))/2 # length of guiding section in mm\n",
+    "Z = 1.5*sqrt(d)+2 # number of teeth\n",
+    "Z = ceil(Z)\n",
+    "print \"1 Diameter of reamer \\n Maximum diameter of reamer = %0.3f mm \\n Minimum diameter of reamer = %0.3f mm \\n 2 Back taper = 0.05 mm \\n 3 Values of various angles \\n Rake angle = 5 degree \\n Plan approach angle = 45 degree \\n Circular land = 0.25 to 0.50 mm \\n Secondary clearance angle = 10 degree \\n 4 Length of reamer \\n Length of fluted portion = 82.5 mm \\n Length of reaming allowance = 0.18 mm \\n Length of cutting section = 2.25 mm \\n Length of guiding section = %d mm \\n 5 Number of teeth = %d\"%(dmax,dmin,l,Z) \n",
+    "# Answer vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 15.8 : page 629"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " The width of shank = 14.0 mm\n",
+      " Height of shank = 22.45 mm\n",
+      " Shank overhang = 28.1 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi, sqrt\n",
+    "Pm = 10 # power of motor in kw\n",
+    "v = 40 # cutting speed in m/min.\n",
+    "ita = 70 # efficiency\n",
+    "ita = ita/100\n",
+    "Pc = Pm*ita \n",
+    "Fc = (Pc*1000*60)/v # cutting force\n",
+    "sigmab = 250 # stress in Mpa\n",
+    "B = sqrt((Fc*1.25*6)/(sigmab*1.6)) # width of shank in mm\n",
+    "h = 1.6*B # hieght of shank in mm\n",
+    "l = 1.25*h # shank overang in mm\n",
+    "print \" The width of shank = %0.1f mm\\n Height of shank = %0.2f mm\\n Shank overhang = %0.1f mm\"%(B,h,l)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 15.9 : page 630"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Cutting speed = 0.27 m/s\n",
+      " MRR = 2025 mm**3/min.\n",
+      " Time to cut = 0.74 min.\n",
+      " Power = 115.8 watts\n",
+      " Cutting force = 444 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi, sqrt\n",
+    "l = 150 # length in mm\n",
+    "D = 12.70 # diameter in mm\n",
+    "dia = 12.19 # diameter on centre lathe in mm \n",
+    "N = 400 # spindle speed in rev./min\n",
+    "s = 203.20 # axial speed in mm/min.*####\n",
+    "v = (pi*D*N)/(1000*60) # cutting velocity in m/s\n",
+    "d = (D-dia)/2 # depth of cut in mm\n",
+    "f = s/N # feed in mm/rev.\n",
+    "Dave = (D+dia)/2 # average diameter in mm\n",
+    "V = pi*Dave*N\n",
+    "a = d*f # area of cut in mm**2\n",
+    "mrr = a*V # metal removal rate in mm**3/min.\n",
+    "T = l/(f*N) # machine timing in min.\n",
+    "c = 56 # constant from table \n",
+    "p = d*f*v*60*c # power in watts\n",
+    "omega = (2*pi*N)/60 # rpm\n",
+    "t = p/omega # torque in Nm\n",
+    "Fc = (2*t*1000)/Dave # cutting force in N\n",
+    "print \" Cutting speed = %0.2f m/s\\n MRR = %d mm**3/min.\\n Time to cut = %0.2f min.\\n Power = %0.1f watts\\n Cutting force = %d N\"%(v , mrr , T ,p ,Fc)\n",
+    "# Answers are given wrong in book"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 15.10 : page 630"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " MRR = 209.44 mm**3/s\n",
+      " Cutting power = 104.720 watts\n",
+      " Torque = 1.25 N.m\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi, sqrt\n",
+    "f = 0.2 # feed in mm/rev.\n",
+    "N = 800 # spindle speed in rev./min.\n",
+    "d = 10 # doameter of hole in mm\n",
+    "mrr = pi*(d**2)*f*N/4 # metal removal rate in mm**3/min.\n",
+    "mrr = mrr/60 # mm**3/s\n",
+    "p = 0.5*mrr # cutting power from table 14.2 in watts\n",
+    "omega = 2*pi*N/60 # rpm\n",
+    "T = p/omega # torque in N.m\n",
+    "print \" MRR = %0.2f mm**3/s\\n Cutting power = %0.3f watts\\n Torque = %0.2f N.m\"%(mrr,p,T\n",
+    "                                                                                )"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter16_1.ipynb b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter16_1.ipynb
new file mode 100644
index 00000000..dfe026ea
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter16_1.ipynb
@@ -0,0 +1,65 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 16 : Gear Manufacturing"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 16.1 : Page 648"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Tooth thickness = 7.851 mm\n",
+      " Chordal addendum = 5.091 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sin, pi, cos\n",
+    "n = 34 # number of teeths\n",
+    "m = 5 # module in mm\n",
+    "w = m*n*sin (pi/(n*2)) #tooth thickness in mm\n",
+    "h = m*(1+(n*(1 - cos(pi/(n*2)))/2)) # chordal addendum in mm\n",
+    "print \" Tooth thickness = %0.3f mm\\n Chordal addendum = %0.3f mm\"%(w ,h)\n",
+    "# Answers vary due to round off error"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter17_1.ipynb b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter17_1.ipynb
new file mode 100644
index 00000000..ea15fdfe
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter17_1.ipynb
@@ -0,0 +1,93 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 17 : Thread Manufacturing"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 17.1 : Page 670"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Best wire size = 3.464 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "d = 80 # outside diameter in mm\n",
+    "p = 6  # pitch diameter in mm\n",
+    "d = 0.5774*p # best wire size in mm\n",
+    "print \"Best wire size = %0.3f mm\"%d"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 17.2 : Page 670"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Best wire size = 1.444 mm\n",
+      " Distance over wires = 20.541 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "D = 20 # diameter in mm\n",
+    "p = 2.5 # pitch diameter in mm\n",
+    "d = 0.5774*p # mm\n",
+    "W = D+3*d-1.5156*p# best wire size in mm\n",
+    "print \" Best wire size = %0.3f mm\\n Distance over wires = %0.3f mm\"%(d, W)\n",
+    "# Answer vary due to round off error"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter21_1.ipynb b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter21_1.ipynb
new file mode 100644
index 00000000..4440e2e4
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter21_1.ipynb
@@ -0,0 +1,661 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 21 : Statistical Quality Control"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 21.1 : page 740"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "control limits \n",
+      "For X charts \n",
+      " UCL = 11.40 cm \n",
+      " LCL = 11.10 cm\n",
+      " For R charts \n",
+      " UCl = 0.554 \n",
+      " LCL = 0.000\n"
+     ]
+    },
+    {
+     "data": {
+      "image/png": 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cp7Ry0UX2uerfv+BjFiwwLa2ePaPXyrow3IFkGccfb2uurVrZOmpxyNW9yqQb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\nYII=\n",
+      "text/plain": [
+       "<matplotlib.figure.Figure at 0x7f9638ec0e90>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "%matplotlib inline\n",
+    "from matplotlib.pyplot import plot, subplot, title, show, xlabel, ylabel\n",
+    "n = 10 # number of samples\n",
+    "A2 = 0.577\n",
+    "D3 = 0\n",
+    "D4 = 2.115\n",
+    "# number of defectives\n",
+    "x1 = 11.274\n",
+    "x2 = 11.246\n",
+    "x3 = 11.204\n",
+    "x4 = 11.294\n",
+    "x5 = 11.252\n",
+    "x6 = 11.238\n",
+    "x7 = 11.230\n",
+    "x8 = 11.276\n",
+    "x9 = 11.208\n",
+    "x10 = 11.266\n",
+    "r1 = 0.15\n",
+    "r2 = 0.20\n",
+    "r3 = 0.33\n",
+    "r4 = 0.46\n",
+    "r5 = 0.10\n",
+    "r6 = 0.15\n",
+    "r7 = 0.20\n",
+    "r8 = 0.23\n",
+    "r9 = 0.50\n",
+    "r10 = 0.30\n",
+    "x = x1+x2+x3+x4+x5+x6+x7+x8+x9+x10\n",
+    "r = r1+r2+r3+r4+r5+r6+r7+r8+r9+r10\n",
+    "Xavg = x/n\n",
+    "Ravg = r/n\n",
+    "# for X chart\n",
+    "ucl1 = Xavg + A2*Ravg\n",
+    "lcl1 = Xavg - A2*Ravg\n",
+    "# for R chart\n",
+    "ucl2 = D4*Ravg\n",
+    "lcl2 = D3*Ravg\n",
+    "print \"control limits \\nFor X charts \\n UCL = %0.2f cm \\n LCL = %0.2f cm\\n For R charts \\n UCl = %0.3f \\n LCL = %0.3f\"%(ucl1,lcl1,ucl2,lcl2)\n",
+    "# X chart\n",
+    "x=[1,2,3,4,5,6,7,8,9,10] \n",
+    "y=[11.274,11.246,11.204,11.294,11.252,11.238,11.230,11.276,11.208,11.266]\n",
+    "subplot(211)\n",
+    "plot(x,y)\n",
+    "title(\"X chart\")\n",
+    "xlabel(\"Sample No.\")\n",
+    "ylabel(\"X\")\n",
+    "# R chart\n",
+    "z = [0.15,0.20,0.33,0.46,0.10,0.15,0.20,0.23,0.50,0.30]\n",
+    "subplot(212)\n",
+    "plot(x,z)\n",
+    "title(\"R chart\")\n",
+    "xlabel(\"Sample no.\")\n",
+    "ylabel( \"R\")\n",
+    "show()"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 21.2 : page 741"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Control limits \n",
+      " Fraction defectives \n",
+      " UCL = 0.081\n",
+      " LCL = -0.0212 = 0 (-ve fraction defective is meaningless)\n",
+      " Percent defectives \n",
+      " UCL = 8.1 \n",
+      " LCL = -2.1 = 0 (-ve fraction defective is meaningless)\n"
+     ]
+    },
+    {
+     "data": {
+      "image/png": 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oYApwBktvsTtskLQ6sJvtsyDN+9me02GxOsVLpBeqlfJmaivRh100ux3bVwMv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\ngiAIgrqEogiCIAjq8v8BOsGfviq7ZrsAAAAASUVORK5CYII=\n",
+      "text/plain": [
+       "<matplotlib.figure.Figure at 0x7f96388e8250>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    },
+    {
+     "data": {
+      "image/png": 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+      "text/plain": [
+       "<matplotlib.figure.Figure at 0x7f963889ec50>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "from math import sqrt\n",
+    "%matplotlib inline\n",
+    "from matplotlib.pyplot import plot, subplot, xlabel, ylabel, show, title\n",
+    "from numpy import arange\n",
+    "n = 100 # total number of sub groups\n",
+    "s = 10 # number of samples\n",
+    "# number of defectives\n",
+    "d = [3,2,3,5,3,3,2,4,3,2]\n",
+    "p = sum(d) # total number of defectives\n",
+    "pbar = p/(n*s) # average fraction of defectives\n",
+    "sigmapbar = sqrt(pbar*(1-pbar)/n)\n",
+    "ucl1 = pbar + 3*sigmapbar\n",
+    "lcl1 = pbar - 3*sigmapbar\n",
+    "# percent defective (mean)\n",
+    "pbar = pbar*100\n",
+    "sigmap2 = sqrt(pbar*(100-pbar)/n)\n",
+    "ucl2 = pbar + 3*sigmap2\n",
+    "lcl2 = pbar - 3*sigmap2\n",
+    "print \" Control limits \\n Fraction defectives \\n UCL = %0.3f\\n LCL = %0.4f = %d (-ve fraction defective is meaningless)\\n Percent defectives \\n UCL = %0.1f \\n LCL = %0.1f = %d (-ve fraction defective is meaningless)\"%( ucl1,lcl1,lcl1,ucl2,lcl2,0)\n",
+    "\n",
+    "# control chart for fraction defectives\n",
+    "x = arange(0,10.2,1.111)\n",
+    "y = arange(0,0.081,0.009)\n",
+    "subplot(211)\n",
+    "plot(x, y)\n",
+    "title(\"Control chart for fraction defectives\")\n",
+    "xlabel(\"Samples\")\n",
+    "ylabel(\"Fraction defectives\")\n",
+    "show()\n",
+    "# control chart for percent defect\n",
+    "#z = linspace(0,8.1,10)\n",
+    "z = arange(0.8,9,0.9)\n",
+    "\n",
+    "subplot(212)\n",
+    "plot(x,z)\n",
+    "title(\"Control chart for percent defects\")\n",
+    "xlabel(\"Sample No.\")\n",
+    "ylabel(\"Percent defects\")\n",
+    "show()\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 21.4 : page 742"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Proabilities for 0,1,2 and 3 defectives are : 0.819 ,0.1637, 0.0164, 0.00109\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import exp, factorial\n",
+    "n = 1000 # number of units\n",
+    "s = 4 # random sample\n",
+    "d = 50 # defectives\n",
+    "z = d*s/n\n",
+    "pp0 = exp(-0.2)*1 # poisson probabilities for 0 defectives\n",
+    "pp1 = exp(-0.2)*(z) # poisson probabilities for 1 defectives\n",
+    "pp2 = exp(-0.2)*(z**2/factorial(2)) # poisson probabilities for 2 defectives\n",
+    "pp3 = exp(-0.2)*(z**3/factorial(3))# poisson probabilities for 3 defectives\n",
+    "print \"Proabilities for 0,1,2 and 3 defectives are : %0.3f ,%0.4f, %0.4f, %0.5f\"%(pp0,pp1,pp2,pp3)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 21.5 : page 744"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Proabilities for 0,1,2 and 3 defectives are : 0.8145 0.1715 0.0135 0.000475\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "d = 50 # defectives\n",
+    "l = 1000 # lot of pieces\n",
+    "p = d/l # proability of an event happening\n",
+    "q = 1-p # proability of an event not happening\n",
+    "n = 4 # sample size\n",
+    "p0 = q**n #probabilities for 0 defectives\n",
+    "p1 = 4*(q)**3*p # probabilities for 1 defectives\n",
+    "p2 = 6*(q)**2*p**2 # probabilities for 2 defectives\n",
+    "p3 = 4*q*(p)**3 # probabilities for 3 defectives\n",
+    "print \" Proabilities for 0,1,2 and 3 defectives are : %0.4f %0.4f %0.4f %0.6f\"%(p0,p1,p2,p3)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 21.6 : page 747"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Producers risk = 5 percent\n",
+      " Consumers risk = 13.07 percent\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import ceil\n",
+    "# producer's risk\n",
+    "n = 71 # sample size\n",
+    "AQL = 0.005\n",
+    "LTPD = 0.05\n",
+    "l_s = 500 # lot size\n",
+    "z1 = n*AQL # mean number of defects\n",
+    "pp1 = exp(-z1)+z1*exp(-z1) # poisson proability for 1 or less defective\n",
+    "alpha = (1-pp1)*100 # producer's risk\n",
+    "alpha = ceil(alpha)\n",
+    "# consumer's risk\n",
+    "z2 = n*LTPD # mean number of defects\n",
+    "pp2 = exp(-z2)+z2*exp(-z2) # poisson proability for 1 or less defective\n",
+    "bita = pp2*100 # consumer's risk\n",
+    "print \" Producers risk = %d percent\\n Consumers risk = %0.2f percent\"%( alpha,bita)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 21.7 : page 748"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Preliminary control limits \n",
+      " UCL = 0.118 \n",
+      " LCL = 0.013 \n",
+      " Revised control limits \n",
+      " UCL = 0.109 \n",
+      " LCL = 0.009\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "td1= 20 # total number of days\n",
+    "n1 = 200 # sample size \n",
+    "# number of defectives\n",
+    "d1 = 10 \n",
+    "d2 = 15\n",
+    "d3 = 10\n",
+    "d4 = 12\n",
+    "d5 = 11\n",
+    "d6 = 9\n",
+    "d7 = 22\n",
+    "d8 = 4\n",
+    "d9 = 12\n",
+    "d10 = 24\n",
+    "d11 = 21\n",
+    "d12 = 15\n",
+    "d13 = 8\n",
+    "d14 = 14\n",
+    "d15 = 4\n",
+    "d16 = 10\n",
+    "d17 = 11\n",
+    "d18 = 11\n",
+    "d19 = 26 \n",
+    "d20 = 13\n",
+    "d = d1+d2+d3+d4+d5+d6+d7+d8+d9+d10+d11+d12+d13+d14+d15+d16+d17+d18+d19+d20 # total number of defectives\n",
+    "p1 = d/(n1*td1) # average fraction of defectives\n",
+    "sigmap1 = sqrt(p1*(1-p1)/n1)\n",
+    "ucl1 = p1 + 3*sigmap1\n",
+    "lcl1 = p1 - 3*sigmap1\n",
+    "# revised control limits\n",
+    "td2 = 18 # total number of days\n",
+    "D = d - (d10+d19) # number of defects\n",
+    "p2 = D/(n1*td2)\n",
+    "sigmap2 = sqrt(p2*(1-p2)/n1)\n",
+    "ucl2 = p2 + 3*sigmap2\n",
+    "lcl2 = p2 - 3*sigmap2\n",
+    "print \" Preliminary control limits \\n UCL = %0.3f \\n LCL = %0.3f \\n Revised control limits \\n UCL = %0.3f \\n LCL = %0.3f\"%(ucl1,lcl1,ucl2,lcl2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 21.8 : page 750"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Preliminary control limits \n",
+      " UCL = 89.58 \n",
+      " LCL = 41.08 \n",
+      " Revised control limits \n",
+      " UCL = 86.713 \n",
+      " LCL = 39.121\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt, exp\n",
+    "n1 = 15 # total number of sub groups\n",
+    "# number of defectives\n",
+    "d1 = 77\n",
+    "d2 = 64\n",
+    "d3 = 75\n",
+    "d4 = 93\n",
+    "d5 = 45\n",
+    "d6 = 61\n",
+    "d7 = 49\n",
+    "d8 = 65\n",
+    "d9 = 45\n",
+    "d10 = 77\n",
+    "d11 = 59\n",
+    "d12 = 54\n",
+    "d13 = 84\n",
+    "d14 = 40\n",
+    "d15 = 92\n",
+    "d = d1+d2+d3+d4+d5+d6+d7+d8+d9+d10+d11+d12+d13+d14+d15 # total number of defectives\n",
+    "c1 = d/n1\n",
+    "ucl1 = c1 + 3*sqrt(c1)\n",
+    "lcl1 = c1 - 3*sqrt(c1)\n",
+    "# revised control limits\n",
+    "n2 = 12 # total number of sub groups\n",
+    "D = d - (d4+d14+d15) # number of defects\n",
+    "c2 = D/n2\n",
+    "ucl2 = c2 + 3*sqrt(c2)\n",
+    "lcl2 = c2 - 3*sqrt(c2)\n",
+    "print \" Preliminary control limits \\n UCL = %0.2f \\n LCL = %0.2f \\n Revised control limits \\n UCL = %0.3f \\n LCL = %0.3f\"%(ucl1,lcl1,ucl2,lcl2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 21.9 : page 750"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Preliminary control limits \n",
+      " For X charts \n",
+      " UCL = 3754.85 \n",
+      " LCL = 3165.15\n",
+      " For R charts \n",
+      " UCl = 1080.765 \n",
+      " LCL = 0.000 \n",
+      " Revised control limits \n",
+      " For X chart \n",
+      " UCL = 3673.686 \n",
+      " LCL = 3673.686\n",
+      " For R charts \n",
+      " UCl = 1012.974 \n",
+      " LCL = 0.000\n"
+     ]
+    }
+   ],
+   "source": [
+    "n = 20 # number of samples\n",
+    "A = 1.342\n",
+    "A1 = 1.596\n",
+    "A2 = 0.577\n",
+    "d2 = 2.326\n",
+    "d3 = 0.864\n",
+    "D1 = 0\n",
+    "D2 = 4.918\n",
+    "D3 = 0\n",
+    "D4 = 2.115\n",
+    "# number of defectives\n",
+    "x1 = 3290\n",
+    "x2 = 3180\n",
+    "x3 = 3350\n",
+    "x4 = 3470\n",
+    "x5 = 3080\n",
+    "x6 = 3240\n",
+    "x7 = 3260\n",
+    "x8 = 3310\n",
+    "x9 = 3640\n",
+    "x10 = 4110\n",
+    "x11 = 3220\n",
+    "x12 = 3590\n",
+    "x13 = 4270\n",
+    "x14 = 4040\n",
+    "x15 = 3580\n",
+    "x16 = 3500\n",
+    "x17 = 3570\n",
+    "x18 = 3560\n",
+    "x19 = 2740\n",
+    "x20 = 3200\n",
+    "r1 = 560\n",
+    "r2 = 410\n",
+    "r3 = 200\n",
+    "r4 = 300\n",
+    "r5 = 90\n",
+    "r6 = 650\n",
+    "r7 = 890\n",
+    "r8 = 410\n",
+    "r9 = 1120\n",
+    "r10 = 520\n",
+    "r11 = 580\n",
+    "r12 = 670\n",
+    "r13 = 480\n",
+    "r14 = 250\n",
+    "r15 = 170\n",
+    "r16 = 670\n",
+    "r17 = 440\n",
+    "r18 = 660\n",
+    "r19 = 560\n",
+    "r20 = 590\n",
+    "x = x1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11+x12+x13+x14+x15+x16+x17+x18+x19+x20\n",
+    "r = r1+r2+r3+r4+r5+r6+r7+r8+r9+r10+r11+r12+r13+r14+r15+r16+r17+r18+r19+r20\n",
+    "Xavg = x/n\n",
+    "Ravg = r/n\n",
+    "# for X chart\n",
+    "ucl1 = Xavg + A2*Ravg\n",
+    "lcl1 = Xavg - A2*Ravg\n",
+    "# for R chart\n",
+    "ucl2 = D4*Ravg\n",
+    "lcl2 = D3*Ravg\n",
+    "# Revised control limits\n",
+    "n1 = 15\n",
+    "n2 = 19\n",
+    "X = (x - (x5+x10+x13+x14+x19))/n1\n",
+    "R = (r - (r9))/n2\n",
+    "# for X chart\n",
+    "ucl3 = X + A2*R\n",
+    "lcl3 = X - A2*R\n",
+    "# for R chart\n",
+    "ucl4 = D4*R\n",
+    "lcl4 = D3*R\n",
+    "print \" Preliminary control limits \\n For X charts \\n UCL = %0.2f \\n LCL = %0.2f\\n For R charts \\n UCl = %0.3f \\n LCL = %0.3f \\n Revised control limits \\n For X chart \\n UCL = %0.3f \\n LCL = %0.3f\\n For R charts \\n UCl = %0.3f \\n LCL = %0.3f\"%(ucl1,lcl1,ucl2,lcl2,ucl3,ucl3,ucl4,lcl4)\n",
+    "#  'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 21.10 : page 752"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Producers risk = 26.42 percent\n",
+      " Consumers risk = 9.158 percent\n"
+     ]
+    },
+    {
+     "data": {
+      "image/png": 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YB4bTYcqUsEGEEoCIpMvQoaHQXK9eofLottvGHVH65FwSGDUKzj477ihEJN+d\ndx588QX07g1PPQVbbRV3ROmRU91BS5eGncOWLw8r/kRE0sk9LCJ79ll48klo2TLuiJIT+34CqTZm\nTGimKQGISCaYwf/+L+y1Fxx/PHz/fdwRpV7OtATKy52ddw5jAl3qtfWMiEj9bNgQNq1v1Ajuvz/8\nmQvyqiUwd25YzacEICKZ1rgxTJgAX34J556bX5VHcyYJjB6t3cNEJD7NmoWeiEWL4Mor444mdXKm\nO6hNG2fJEth667ijEZFC9vnnoVrBaafBFVfEHU3d8mqdQK9eSgAiEr+ttw4F57p1g7Ztc3/Kes4k\nAXUFiUi22G67kAiOOALatIHBg+OOqP5ypjtowwbPmRF5ESkMixaFXooxY6Bv37ij+bG8mh2kBCAi\n2Wa//cJg8amnwjPPxB1N/eRMEhARyUaHHho2uRo0CBYujDua5CkJiIg0UJ8+cOutYc/zf/877miS\nkzMDwyIi2ezkkzetPLrddnFHlBglARGRFDn77LCquHfvsBdBLkxrz5nZQbkQp4gIhEVkZWWh8ugW\nW8QXRyKzg5QERERSzD20Ct57L2xgH1flYyUBEZGYlJfDkCGhAunkyaEIXabl1ToBEZFc0qhRmDq6\nejX86ldQURF3RDVTEhARSZOmTeHhh+Gdd8IOZdnYoaEkICKSRq1ahXGBWbPgf/4n7mh+TFNERUTS\nbKutQsG5ww8Pr887L+6INlISEBHJgG23Da2B7t3DLolDh8YdUaAkICKSITvvDDNmQM+eoQR1v35x\nR6QxARGRjPrpT+HRR+H002HevLijURIQEcm4gw8OG9efcAK89lq8sSgJiIjEoFcvuP126N8f3n03\nvjg0JiAiEpMTT4SvvgoF5+bPh+23z3wMSgIiIjE666xQebRXrzBG0K5dZu+v2kEiIlng6qvDFNLZ\ns6F169RcUwXkRERyhHtYRPavf8E//5mayqOxF5Azs75m9o6ZLTazK2p4v52ZzTCzBWb2hpmNSGc8\nIiLZygxuuw06doTBg0P10YzcN12fsM2sEfAv4CjgI+AlYKi7v13lnFKgqbtfZWbtovM7uvuGatdS\nS0BECsK6dTBwIHToAKNHQ1EDPqrH3RI4EPi3uy919/XA/cDAauf8B6js/WoNfF49AYiIFJImTeCh\nh2DJEvjNb9JfeTSdSWA7YFmV75dHx6q6C9jHzFYAC4GL0hiPiEhOaNECHnsMnnoKrr8+vfdK5xTR\nRPLX1cDWN3wBAAAJMUlEQVQCdy8xs12AWWb2M3f/pvqJpaWlP7wuKSmhpKQkVXGKiGSdLbeEJ54I\nlUfbtoVf/3rzP1NWVkZZWVlS90nnmMDBQKm7942+vwqocPebqpzzOHCjuz8TfT8buMLdX652LY0J\niEhBWroUunWDP/wBhg9P7mfjHhN4GdjNzDqbWRNgMDC12jnvEAaOMbOOwB7Ae2mMSUQkp3TuHFoE\nl10G06al/vppXSdgZkcDfwUaAfe4+x/M7BwAd78zmhE0GuhESEh/cPcJNVxHLQERKWgvvhjqDE2e\nDIn2hmuxmIhIHpkzB4YMCYvJunbd/PlxdweJiEgK9egB//gHHHNM2Lw+FVRATkQkhxx3HKxaBX36\nhIJznTo17HpKAiIiOWbEiJAIKiuPduhQ/2spCYiI5KCLL4YvvoC+fWHu3LBncX1oYFhEJEe5h0Vk\nixaFaaTNm2/6vmYHiYjkuYoKOPXUsEPZlClQXLzxPc0OEhHJc0VFcO+94fWIESEpJPXzqQ5IREQy\nq7g4LCJbvhwuvDC5yqNKAiIieaB5c5g6FZ59FkaOTPznNDtIRCRPtGkDM2aEgnNt2yb2MxoYFhHJ\nMx9+CEcdBYsXa3aQiEhBWrcOmjbV7CARkYLUpEli5ykJiIgUMCWBDEh2u7dsksuxg+KPm+LPfkoC\nGZDL/5FyOXZQ/HFT/NlPSUBEpIApCYiIFLCcmSIadwwiIrkoL9YJiIhIeqg7SESkgCkJiIgUsKxO\nAmY2ysw+NrPX444lWWa2g5nNNbM3zewNM7sw7piSYWbNzOwFM1tgZm+Z2R/ijqk+zKyRmb1mZtPi\njiVZZrbUzBZF8b8YdzzJMLMtzexBM3s7+v9zcNwxJcrM9oj+zSu/vsrB39+romfP62Y2wcya1npu\nNo8JmFk34FtgrLvvG3c8yTCzbYBt3H2BmbUCXgGOc/e3Yw4tYWbWwt2/M7PGwHzgMnefH3dcyTCz\nS4CuwBbuPiDueJJhZu8DXd39i7hjSZaZjQGecvdR0f+flu7+VdxxJcvMioCPgAPdfVnc8STCzDoD\nc4C93P17M3sAeNzdx9R0fla3BNx9HvBl3HHUh7uvdPcF0etvgbeBn8QbVXLc/bvoZROgEZBTDyMz\n2x7oB9wN1DlDIovlXNxm1gbo5u6jANx9Qy4mgMhRwJJcSQCRr4H1QIsoAbcgJLIaZXUSyBdRZu4C\nvBBvJMkxsyIzWwB8DMx197fijilJNwOXA0luuJc1HHjSzF42s1/FHUwSdgI+NbPRZvaqmd1lZi3i\nDqqehgAT4g4iGVHL8c/Ah8AKYJW7P1nb+UoCaRZ1BT0IXBS1CHKGu1e4+/7A9kB3MyuJOaSEmdkx\nwCfu/ho5+Gk6cpi7dwGOBs6PukdzQWPg58Dt7v5zYDVwZbwhJc/MmgDHApPjjiUZZrYLcDHQmdD7\n0MrMhtV2vpJAGplZMfAQMM7dH4k7nvqKmvLTgQPijiUJhwIDon71iUAPMxsbc0xJcff/RH9+CkwB\nDow3ooQtB5a7+0vR9w8SkkKuORp4Jfr3zyUHAM+6++fuvgF4mPD7UCMlgTQxMwPuAd5y97/GHU+y\nzKydmW0ZvW4O9AJeizeqxLn71e6+g7vvRGjSz3H30+KOK1Fm1sLMtohetwR6AzkxS87dVwLLzGz3\n6NBRwJsxhlRfQwkfIHLNO8DBZtY8eg4dBdTalZvVewyb2UTgCGBrM1sGXOvuo2MOK1GHAcOBRWZW\n+fC8yt1nxBhTMrYFxkSzI4qA+9x9dswxNUT2ToOrWUdgSvgdpjEw3t1nxhtSUn4NjI+6VJYAZ8Qc\nT1KixHsUkEtjMQC4+8Ko1fsyYTzsVeAftZ2f1VNERUQkvdQdJCJSwJQEREQKmJKAiEgBUxIQESlg\nSgIiIgVMSUBEpIApCUhWM7PyqJzv62Y2KVq4lukYjjCzQxI8d6mZbdXQe5jZOWZ2arLXEUmWkoBk\nu+/cvUtUSnwdcG4iPxRVT0yVI6lj2X019V14s8k93P1Od7+vntcSSZiSgOSS+cCuUUmFUdGmN6+a\n2QAAMxthZlPNbDYwy8xaRpUsF5nZQjMbFJ3X28yeNbNXotZFy+j4UjMrjY4vijYX6QycA/wmapEc\nXjUgM9vazGZGGwfdRZVidWY2PIrxNTP7e7T6GjPrG91jgZnNMrMdq98jiuPSKIYXqlyzs5ktil53\nNbOyqMrojGgPC8zswmhDkYXRqnuRWikJSE6IPtn3BRYBvwNmu/tBQA/gT1VKFXcBTnD3I4FrgS/d\nfT93/xkwx8zaAb8Ferp7V8JmP5dEP+vAp9HxOwib6CwF/g78JWqRVN9UZyTwtLv/lFDkrVMU717A\nycChUSXQCmCYmbUnLOEfFFVoPcndP6jhHg64u/8LaBIlI4DBwP3Rv8et0d/1AGA0cGN0zhXA/tHf\n+Zx6/HNLAcnq2kEiQPMqtZeeBkYBzwHHmtll0fGmhIevA7PcfVV0vCfhoQmAu6+KSkzvDTwb1eVp\nAjxb5X4PR3++Cgyqcry2ctTdgOOj6z9uZl9G5/Yk7Gj2cnSfZsBK4CBC0vigMqY67lH5/aTo73ET\nIbGcDOwJ7EPYbwDCpj8rovMXARPM7BEgZ6vXSmYoCUi2WxN9kv5B9NAb5O6Lqx0/iFC7fpPDNVxz\nlrufUsv9vo/+LCfx34/aEsQYd7+6WozHJHjNqh4AJpvZw4TWwRIz2xd4091rGqvoD3Qn1ML/rZnt\n6+7l9bivFAB1B0kuegL4YeNvM6tMEtUfxrOA86uctyXwPHBYtPEG0bjBbpu53zfAFrW89zRwSnSt\no4G2hBbJbODEqPsHM9vKzDpF9+9e2b1TZSZRrfdw9/cISeka4P7o8L+A9hZt4G5mxWa2d1Q6uJO7\nlxE2cmkDtNzM308KmJKAZLuaZttcDxRHg7dvANdVObfq+TcAbaPppQuAEnf/DBgBTDSzhYSuoD1q\nuW/ltaYBx0eDtodVO+86wkP9DUK3UGU3z9uEsYuZ0X1mAttE9z8beDiKaWK1e7xaZfC56t/lAWAY\noWsId18HnAjcFF3nNeAQQrfQfdHg8avA/7n71zX8/UQAlZIWESloagmIiBQwJQERkQKmJCAiUsCU\nBERECpiSgIhIAVMSEBEpYEoCIiIFTElARKSA/X+1yQKhcBgkSwAAAABJRU5ErkJggg==\n",
+      "text/plain": [
+       "<matplotlib.figure.Figure at 0x7f9638ae2950>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "from math import exp\n",
+    "%matplotlib inline\n",
+    "from matplotlib import pyplot as plt\n",
+    "\n",
+    "n = 50 # sample size\n",
+    "rn = 2 # rejection number\n",
+    "AQL = 0.02\n",
+    "LTPD = 0.08\n",
+    "# Producer's risk\n",
+    "z1 = n*AQL # mean number of defectives\n",
+    "pp1 = exp(-z1)+z1*exp(-z1) # poisson proability for 1 or less defective\n",
+    "alpha = (1-pp1)*100 # producer's risk\n",
+    "# consumer's risk\n",
+    "z2 = n*LTPD # mean number of defectives\n",
+    "bita = (exp(-z2)+z2*exp(-z2))*100 # consumer's risk\n",
+    "d1 = 1 # incoming defective in percent\n",
+    "z3 = n*d1/100 # average number of defective\n",
+    "ppa1 = exp(-z3)+z3*exp(-z3) # proability of acceptance\n",
+    "ppa1 = ppa1*100\n",
+    "ppr1 = 100-ppa1 # proability of rejection\n",
+    "AOQ1 = ppr1*0 + ppa1*d1/100\n",
+    "d2 = 2 # incoming defective in percent\n",
+    "z4 = n*d2/100 # average number of defective\n",
+    "ppa2 = exp(-z4)+z4*exp(-z4) # proability of acceptance\n",
+    "ppa2 = ppa2*100\n",
+    "ppr2 = 100-ppa2 # proability of rejection\n",
+    "AOQ2 = ppr2*0 + ppa2*d2/100\n",
+    "d3 = 4 # incoming defective in percent\n",
+    "z5 = n*d3/100 # average number of defective\n",
+    "ppa3 = exp(-z5)+z5*exp(-z5) # proability of acceptance\n",
+    "ppa3 = ppa3*100\n",
+    "ppr3 = 100-ppa3 # proability of rejection\n",
+    "AOQ3 = ppr3*0 + ppa3*d3/100\n",
+    "d4 = 6 # incoming defective in percent\n",
+    "z6 = n*d4/100 # average number of defective\n",
+    "ppa4 = exp(-z6)+z6*exp(-z6) # proability of acceptance\n",
+    "ppa4 = ppa4*100\n",
+    "ppr4 = 100-ppa4 # proability of rejection\n",
+    "AOQ4 = ppr4*0 + ppa4*d4/100\n",
+    "d5 = 8 # incoming defective in percent\n",
+    "z7 = n*d5/100 # average number of defective\n",
+    "ppa5 = exp(-z7)+z7*exp(-z7) # proability of acceptance\n",
+    "ppa5 = ppa5*100\n",
+    "ppr5 = 100-ppa5 # proability of rejection\n",
+    "AOQ5 = ppr5*0 + ppa5*d5/100\n",
+    "print \" Producers risk = %0.2f percent\\n Consumers risk = %0.3f percent\"%( alpha,bita)\n",
+    "x = [1,2,4,6,8]\n",
+    "y = [0.91,1.4716,1.624,1.194,0.733]\n",
+    "plt.plot(x,y)\n",
+    "plt.title(\"AOQ curve\")\n",
+    "plt.xlabel(\"Percent dectives\")\n",
+    "plt.ylabel(\"AOQ of lot\")\n",
+    "plt.show()\n",
+    "\n",
+    "\n",
+    "                    "
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter22_1.ipynb b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter22_1.ipynb
new file mode 100644
index 00000000..0b492269
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter22_1.ipynb
@@ -0,0 +1,365 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 22 : Kinematics of Machine Tools"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 22.2 : page 778"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Spindle Speeds are :\n",
+      "159\t227\t325\t466\t667\t954\t"
+     ]
+    },
+    {
+     "data": {
+      "image/png": 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4oV4Z6zREYNIkCAqCESM8LY1HyczN5N4v7iUtO40NIzdQq6pn15pUNI4cgR49YP9+w/qf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\nAABJRU5ErkJggg==\n",
+      "text/plain": [
+       "<matplotlib.figure.Figure at 0x7f69f0f20e90>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Range of cutting speed for 12 mm diameter = 11.5 m/min.\n",
+      " Range of cutting speed for 36 mm diameter = 10.5 m/min.\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "import numpy as np\n",
+    "from math import pi\n",
+    "%matplotlib inline\n",
+    "from matplotlib.pyplot import plot, show, title, text, xlabel, ylabel\n",
+    "\n",
+    "d1 = 10 # min. dia of cutter in mm\n",
+    "d2 = 60 # max. dia of cutter in mm\n",
+    "v = 30e3 # operating speed in mm/min\n",
+    "z= 6 # required speeds\n",
+    "nmin = v / (pi * d2) # n_min in rpm\n",
+    "nmax = v / (pi * d1) # n_max in rpm\n",
+    "phi = (nmax / nmin)**(1/(z-1))\n",
+    "spindle_speeds = [1]*6 # initializing\n",
+    "print \"Spindle Speeds are :\" \n",
+    "for i in range(0,6):\n",
+    "    spindle_speeds[i] = (phi**i)*nmin\n",
+    "    print '%d\\t'%spindle_speeds[i],\n",
+    "    \n",
+    "\n",
+    "cutter_dia = [1]*6 # initializing\n",
+    "for i in range(0,6):\n",
+    "    cutter_dia[i] = (v/pi)/spindle_speeds[i]\n",
+    "y = [0,v/1e3]\n",
+    "plot([0,cutter_dia[0]], y, [0,cutter_dia[1]], y, [0,cutter_dia[2]], y, [0,cutter_dia[3]], y, [0,cutter_dia[4]], y, [0,cutter_dia[5]], y)\n",
+    "title(\"cutting velocity Vs Cutter diameter\")\n",
+    "xlabel('Cutter Diameter(mm)')\n",
+    "ylabel('Cutting velocity(m/min)')\n",
+    "text(5,28,'%d'%spindle_speeds[5])\n",
+    "text(9,25,'%d'%spindle_speeds[4])\n",
+    "text(12,22,'%d'%spindle_speeds[3])\n",
+    "text(15,18,'%d'%spindle_speeds[2])\n",
+    "text(18,15,'%d'%spindle_speeds[1])\n",
+    "text(20,11,'%d'%spindle_speeds[0])\n",
+    "show()\n",
+    "# from graph\n",
+    "vmax1 = 36 # m/min\n",
+    "vmin1 = 24.5 # m/min\n",
+    "r1 = vmax1 - vmin1 # Range of cutting speed for 12 mm diameter in m/min\n",
+    "vmax2 = 36.5 # m/min.\n",
+    "vmin2 = 26 # m/min.\n",
+    "r2 = vmax2 - vmin2 # Range of cutting speed for 36 mm diameter in m/min\n",
+    "print \" Range of cutting speed for 12 mm diameter = %0.1f m/min.\\n Range of cutting speed for 36 mm diameter = %0.1f m/min.\"%(r1 , r2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 22.2 : page 790"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Number of teeth on driver and driven are :- \n",
+      " t1 = 25 ,T1 = 55\n",
+      " t2 = 29 ,T2 = 51 \n",
+      " t3 = 32 ,T3 = 48 \n",
+      " t4 = 36 ,T4 = 44 \n",
+      "The actual running speed of driven shaft will be : \n",
+      " n1 = 68.18 rev/min\n",
+      " n2 = 85.29 rev/min \n",
+      " n3 = 100.00 rev/min \n",
+      " n4 = 122.73 rev/min\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import floor, ceil\n",
+    "from sympy import solve, symbols\n",
+    "m = 2.5 # module in mm\n",
+    "phi = 1.2 # common ratio\n",
+    "n = 150 # speed in rev/min. of driving shaft\n",
+    "n1 = 70 # speed in rev/min. of driven shaft\n",
+    "n2 = (phi)**1*n1 # speed in rev/min. of driven shaft\n",
+    "n3 = (phi)**2*n1 # speed in rev/min. of driven shaft\n",
+    "n4 = (phi)**3*n1 # speed in rev/min. of driven shaft\n",
+    "T1=symbols('T1')\n",
+    "t1=n1/n*T1\n",
+    "expr = t1+T1-80\n",
+    "T1=solve(expr,T1)[0]\n",
+    "t1=n1/n*T1\n",
+    "T2=symbols('T2')\n",
+    "t2=n2/n*T2\n",
+    "expr = t2+T2-80\n",
+    "T2=solve(expr, T2)[0]\n",
+    "t2=n2/n*T2\n",
+    "\n",
+    "T3=symbols('T3')\n",
+    "t3=n3/n*T3\n",
+    "expr=t3+T3-80\n",
+    "T3=solve(expr, T3)[0]\n",
+    "t3=n3/n*T3\n",
+    "T4=symbols('T4')\n",
+    "t4=n4/n*T4\n",
+    "expr=t4+T4-80\n",
+    "T4=solve(expr, T4)[0]\n",
+    "t4=n4/n*T4\n",
+    "t1 = floor(t1) # number of teeth on driving shaft\n",
+    "T1 = ceil(T1) # number of teeth on driven shaft\n",
+    "t2 = ceil(t2) # number of teeth on driving shaft\n",
+    "T2 = floor(T2) # number of teeth on driven shaft\n",
+    "t3 = floor(t3) # number of teeth on driving shaft\n",
+    "T3 = ceil(T3) # number of teeth on driven shaft\n",
+    "t4 = ceil(t4) # number of teeth on driving shaft\n",
+    "T4 = floor(T4) # number of teeth on driven shaft\n",
+    "# running speeds\n",
+    "n1 = n*t1/T1\n",
+    "n2 = n*t2/T2\n",
+    "n3 = n*t3/T3\n",
+    "n4 = n*t4/T4\n",
+    "print \"Number of teeth on driver and driven are :- \\n t1 = %d ,T1 = %d\\n t2 = %d ,T2 = %d \\n t3 = %d ,T3 = %d \\n t4 = %d ,T4 = %d \"%(t1,T1,t2,T2,t3,T3,t4,T4)\n",
+    "print \"The actual running speed of driven shaft will be : \\n n1 = %0.2f rev/min\\n n2 = %0.2f rev/min \\n n3 = %0.2f rev/min \\n n4 = %0.2f rev/min\"%(n1,n2,n3,n4)\n",
+    "# Answer of n3 is given wrong in book\n",
+    "# Answer vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 22.3 : page 791"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Ta = 20 Tb = 32 \n",
+      " Tc = 22 Td = 30 \n",
+      " Te = 23 tf = 29 \n",
+      " Th = 19 Tj = 28 \n",
+      " Ti = 23 Tg = 23\n"
+     ]
+    }
+   ],
+   "source": [
+    "from sympy import symbols, solve\n",
+    "z = 6 # number of steps\n",
+    "n1 = 180 # rev/min\n",
+    "n2 = 100 # rev/min\n",
+    "Rn = n1/n2 \n",
+    "phi = (Rn)**(1/(z-1)) # common ratio\n",
+    "n3 = phi*n2 # rev/min\n",
+    "n4 = (phi)**2*n2 # rev/min\n",
+    "n5 = (phi)**3*n2 # rev/min\n",
+    "n6 = (phi)**4*n2 # rev/min\n",
+    "n7 = 225 # speed of input shaft in rev/min\n",
+    "Ta=symbols('Ta')\n",
+    "tb=n7/n5*Ta\n",
+    "Ta=solve(tb+Ta-52, Ta)[0]\n",
+    "tb=n7/n5*Ta\n",
+    "tb = ceil(tb)\n",
+    "Tc=symbols('Tc')\n",
+    "td=n7/n6*Tc\n",
+    "Tc=solve(td+Tc-52, Tc)[0]\n",
+    "td=n7/n6*Tc\n",
+    "Tc = ceil(Tc)\n",
+    "Te=symbols('Te') \n",
+    "tf=n7/n1*Te\n",
+    "Te=solve(tf+Te-52, Te)[0]\n",
+    "tf=n7/n1*Te\n",
+    "tf = ceil(tf)\n",
+    "Th=symbols('Th')\n",
+    "tj=n2/n5*Th\n",
+    "Th=solve(tj+Th-46, Th)[0]\n",
+    "Th = ceil(Th)\n",
+    "tj=n2/n5*Th\n",
+    "tj = floor(tj)\n",
+    "Ti=symbols('Ti')\n",
+    "tg=n5/n5*Ti\n",
+    "Ti=solve(tg+Ti-46, Ti)[0]\n",
+    "tg=n5/n5*Ti\n",
+    "print \" Ta = %d Tb = %d \\n Tc = %d Td = %d \\n Te = %d tf = %d \\n Th = %d Tj = %d \\n Ti = %d Tg = %d\"%(Ta,tb,Tc,td,Te,tf,tj,Th,Ti,tg)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 22.4 : page 793"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Common ratio = 1.32 \n",
+      " Spindle speeds are : 334.23 , 441.0 , 581.92 , 767.85 ,1013.18 and 1336.9 rev/min.\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt, pi\n",
+    "v = 21 # cutting speed in rev/min.\n",
+    "z = 6\n",
+    "dmin = 5 # daimeter in mm\n",
+    "dmax = 20 # daimeter in mm\n",
+    "nmax = 1000*v/(pi*dmin) # spindle speed in rev/min.\n",
+    "nmin = 1000*v/(pi*dmax) # spindle speed in rev/min.\n",
+    "phi = (nmax/nmin)**(1/(z-1)) # common ratio\n",
+    "n1 = nmin # rev/min.\n",
+    "n2 = phi*n1 # rev/min.\n",
+    "n3 = (phi)**2*n1 # rev/min.\n",
+    "n4 = (phi)**3*n1 # rev/min.\n",
+    "n5 = (phi)**4*n1 # rev/min.\n",
+    "n6 = (phi)**5*n1 # rev/min.\n",
+    "print \" Common ratio = %0.2f \\n Spindle speeds are : %0.2f , %0.1f , %0.2f , %0.2f ,%0.2f and %0.1f rev/min.\"%(phi,n1,n2,n3,n4,n5,n6)\n",
+    "# 'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 22.5 : page 793"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Three gear ratios between input and intermediate shaft i1 = 1 i2 = 0.79 i3 = 0.630 \n",
+      " The two ratios between intermediate and output shaft i4 = 1 i5 = 0.500 \n",
+      " number of teeth for each pair between input and intermediate shaft t1 = 27/27 ,t2 = 24/30 , t3 = 21/33 \n",
+      " number of teeth for each pair between output and intermediate shaft = t4 = 34/34 ,t5 = 20/48 \n",
+      " Output speeds \n",
+      " n1 = 371  rev/min , n2 = 466 rev/min , n3 = 583 rev/min \n",
+      " n4 = 890 rev/min, n5 = 1120 rev/min , n6 = 1400 rev/min\n"
+     ]
+    }
+   ],
+   "source": [
+    "# from fig. 22.18A\n",
+    "# Three gear ratios between input and intermediate shaft\n",
+    "nmax = 1400 # maximum speed in rev/min.\n",
+    "i1 = 1/1\n",
+    "i2 = 1/1.26\n",
+    "i3 = 1/(1.26)**2\n",
+    "# The two ratios between intermediate and output shaft\n",
+    "i4 = 1/1\n",
+    "i5 = 1/(1.26)**3\n",
+    "# number of teeth for input and intermediate shaft\n",
+    "t1 = 27/27\n",
+    "t2 = 24/30\n",
+    "t3 = 21/33\n",
+    "#  number of teeth for output and intermediate shaft\n",
+    "t4 = 34/34\n",
+    "t5 = 20/48\n",
+    "# output speeds in rev./min\n",
+    "n1 = t3*t5*nmax\n",
+    "n2 = t2*t5*nmax\n",
+    "n3 = t1*t5*nmax\n",
+    "n4 = t3*t4*nmax\n",
+    "n5 = t2*t4*nmax\n",
+    "n6 = t1*t4*nmax\n",
+    "print \" Three gear ratios between input and intermediate shaft i1 = %d i2 = %0.2f i3 = %0.3f \\n The two ratios between intermediate and output shaft i4 = %d i5 = %0.3f \\n number of teeth for each pair between input and intermediate shaft t1 = 27/27 ,t2 = 24/30 , t3 = 21/33 \\n number of teeth for each pair between output and intermediate shaft = t4 = 34/34 ,t5 = 20/48 \\n Output speeds \\n n1 = %d  rev/min , n2 = %d rev/min , n3 = %d rev/min \\n n4 = %d rev/min, n5 = %d rev/min , n6 = %d rev/min\"%(i1 , i2 , i3 , i4 , i5 , n1 , n2, n3,n4,n5,n6)\n",
+    "# Answer vary due to round off error"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter23_1.ipynb b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter23_1.ipynb
new file mode 100644
index 00000000..fd58642f
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter23_1.ipynb
@@ -0,0 +1,396 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 23 : Production Planning & Control"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 23.1 : page 812"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Forecast = 90\n"
+     ]
+    }
+   ],
+   "source": [
+    "d1 = 90 # demand for first quarter\n",
+    "d2 = 100 # demand for second quarter\n",
+    "d3 = 80 # demand for third quarter\n",
+    "sa = (d1+d2+d3)/3 # simple average\n",
+    "print \"Forecast = %d\"%(sa)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 23.2 : page 812"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Forecast = 600\n"
+     ]
+    }
+   ],
+   "source": [
+    "d1 = 300 # demand for july\n",
+    "d2 = 350 # demand for august\n",
+    "d3 = 400 # demand for september\n",
+    "d4 = 500 # demand for october\n",
+    "d5 = 600 # demand for november\n",
+    "d6 = 700 # demand for december\n",
+    "# assuming n = 3 , where n is number of time period\n",
+    "forecast = (d6+d5+d4)/3 # forecast\n",
+    "print \"Forecast = %d\"%(forecast)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 23.3 : page 813"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Forecast by weighted moving average = 625\n"
+     ]
+    }
+   ],
+   "source": [
+    "d1 = 500 # demand for october\n",
+    "d2 = 600 # demand for november\n",
+    "d3 = 700 # demand for december\n",
+    "w1 = 0.25 # relative weight with december\n",
+    "w2 = 0.25 # relative weight with november\n",
+    "w3 = 0.5 # relative weight with october\n",
+    "f = w1*d1 + w2*d2 + w3*d3 # forecast\n",
+    "print \"Forecast by weighted moving average = %d\"%(f)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 23.4 : page 814"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Forecast for january = 281 units\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import ceil\n",
+    "alpha = 0.7 # smoothing coefficient\n",
+    "d1 = 250 # demand for november\n",
+    "d2 = 300 # demand for december\n",
+    "f1 = 200 # forecast for november\n",
+    "f2 = alpha*d1 + (1-alpha)*f1 # forecast for december\n",
+    "f3 = alpha*d2 + (1-alpha)*f2 # forecast for january\n",
+    "f3 = ceil(f3)\n",
+    "print \"Forecast for january = %d units\"%(f3)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 23.5 : page 817"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total cost = Rs 613416.41\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "from math import sqrt\n",
+    "s = 600 # set up cost per lot in Rs\n",
+    "c = 6 # unit cost of item in Rs\n",
+    "a = 100000 # annual demand for item\n",
+    "i = 25 #  annual carrying charges of average inventory\n",
+    "i = 25/100 \n",
+    "k = c*i # carrying cost factor in unit/year\n",
+    "n = sqrt(2*s*a/k) # most economic lot size\n",
+    "tc = a*c + s*a/n + k*n/2 # total cost in Rs\n",
+    "print \"Total cost = Rs %0.2f\"%(tc)\n",
+    "#  'Answers vary due to round off error'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 23.6 : page 835"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Most economical ordering quantity = 365 units\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "a = 8000 # annual requirement of parts\n",
+    "c = 60 # unit cost of part in Rs\n",
+    "r = 150 # ordering cost per lot in Rs\n",
+    "i = 30 # annual carrying charges of average inventory\n",
+    "i = 30/100 \n",
+    "k = i*c # carrying cost per unit per year\n",
+    "n = sqrt(2*r*a/k) # most economical order quantity\n",
+    "print \"Most economical ordering quantity = %d units\"%(n)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 23.7 : page 841"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Most economic lot size = 2000 parts\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "a = 12000 # annual requirement\n",
+    "c = 5 # unit cost of part\n",
+    "s = 60 # set up cost per lot\n",
+    "p = 18750 # production rate per year\n",
+    "i = 20 # inventory carrying cost\n",
+    "i = 20/100 \n",
+    "k = i*c # carrying cost per unit per year\n",
+    "n = sqrt(2*s/(1/a-1/p)*k) # Most economic lot size\n",
+    "print \"Most economic lot size = %d parts\"%(n)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 23.8 : page 841"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Economical order quantity = 1250 units\n",
+      " order cost = Rs 750\n",
+      " carrying cost = Rs 750\n",
+      " Total inventory cost = Rs 1500\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "a = 15625 # annual requirement of parts\n",
+    "c = 12 # unit cost of part in Rs\n",
+    "r = 60 # ordering cost per lot in Rs\n",
+    "k = 1.2 # inventory carrying cost per unit \n",
+    "n = sqrt(2*r*a/k) # economical order quantity\n",
+    "oc = r*a/n # ordering cost in Rs\n",
+    "cc = k*n/2 # carrying cost in Rs\n",
+    "tc = oc + cc # total inventory cost in Rs\n",
+    "print \" Economical order quantity = %d units\\n order cost = Rs %d\\n carrying cost = Rs %d\\n Total inventory cost = Rs %d\"%(n , oc , cc , tc)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 23.9 : page 842"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Increase in inventory cost = Rs 450\n",
+      " Discount offered = Rs500\n",
+      " offer is worth accepting\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "# case a\n",
+    "a = 50 # annual requirement of parts in tonnes\n",
+    "c = 500 # unit cost of part in Rs\n",
+    "r = 100 # ordering cost per order in Rs\n",
+    "i = 20 # inventory carrying cost \n",
+    "i = i/100\n",
+    "d = 2 # discount of purchase cost in percent\n",
+    "k = i*c # inventory carrying cost per unit \n",
+    "n1 = sqrt(2*r*a/k) # economical order quantity\n",
+    "oc1 = r*a/n1 # ordering cost in Rs\n",
+    "cc1 = k*n1/2 # carrying cost in Rs\n",
+    "tc1 = oc1 + cc1 # total inventory cost in Rs\n",
+    "# case b\n",
+    "n2 = 25 # order per lot\n",
+    "oc2 = r*a/n2 # ordering cost in Rs\n",
+    "cc2 = k*n2/2 # carrying cost in Rs\n",
+    "tc2 = oc2 + cc2 # total inventory cost in Rs\n",
+    "i = tc2-tc1 # increase in cost in Rs\n",
+    "d_o = d*c*a/100 # discount offered\n",
+    "print \" Increase in inventory cost = Rs %d\\n Discount offered = Rs%d\"%(i,d_o)\n",
+    "print \" offer is worth accepting\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 23.10 : page 843"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Economic order qunantity = 4000 components\n",
+      " Re-order point = 24000 components\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "a = 1000000 # annual requirement of parts\n",
+    "r = 32 # ordering cost per lot in Rs\n",
+    "k = 4 # inventory carrying cost per unit \n",
+    "d1 = 250 # number of working days\n",
+    "d2 = 2 # days for safety stock\n",
+    "d3 = 4 # lead time in days\n",
+    "eoq = sqrt(2*r*a/k) # economical order quantity\n",
+    "oc = r*a/eoq # ordering cost in Rs\n",
+    "cc = k*eoq/2 # carrying cost in Rs\n",
+    "tc = oc + cc # total inventory cost in Rs\n",
+    "ss = a*d2/d1 # safety stock\n",
+    "ro_p = ss+eoq*d3 # reorder point\n",
+    "print \" Economic order qunantity = %d components\\n Re-order point = %d components\"%(eoq ,ro_p)"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter26_1.ipynb b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter26_1.ipynb
new file mode 100644
index 00000000..b7473fc2
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter26_1.ipynb
@@ -0,0 +1,69 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 26 : Plant Layout"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 26.1 : Page 901"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Number of machines required = 1.62\n",
+      "If machine is to be used exclusively for part considered two machines required\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "N = 100000 # annual output of parts\n",
+    "s = 2 # expected scrap\n",
+    "t = 105 # estimated time per part in s\n",
+    "ita = 80 # production efficiency of machine\n",
+    "a = 2300 # number of working hours\n",
+    "output = (3600*ita)/(t*100) # parts required per hour\n",
+    "pr = N*(100+s)/(a*100) # output from one machine per hour\n",
+    "mr = pr/output # machines required\n",
+    "print \"Number of machines required = %0.2f\"%mr\n",
+    "print \"If machine is to be used exclusively for part considered two machines required\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter2_1.ipynb b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter2_1.ipynb
new file mode 100644
index 00000000..252c5ac6
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter2_1.ipynb
@@ -0,0 +1,743 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 2 - Press Tool Design"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.1 : page 132"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Blanking pressure = 226 kN\n",
+      " Piercing pressure = 108.573 KN\n",
+      " Total pressure required = 334.8 KN\n",
+      " piercing punch diameter = 2.44 cm\n",
+      " blanking punch diametre = 4.96 cm \n",
+      " press capacity = 267.81 KN\n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "from math import pi, sqrt\n",
+    "D = 50 # Diameter of washer in mm\n",
+    "t = 4 # thickness of material in mm\n",
+    "d = 24 # diameter of hole in mm\n",
+    "p = 360 # shear strength of material in N/mm**2\n",
+    "F1 = pi*D*t*p # blanking pressure in N\n",
+    "F2 = pi*d*t*p # piercing pressure in N\n",
+    "F = F1 + F2 # total pressure in N\n",
+    "d1 = d + 0.4 # piercing die diameter in mm\n",
+    "d2 = D - 0.4 # blank punch diameter in mm\n",
+    "c = 0.8*F # press capacity in N\n",
+    "print \" Blanking pressure = %d kN\\n Piercing pressure = %0.3f KN\\n Total pressure required = %0.1f KN\"%(F1/1000,F2/1000,F/1000)\n",
+    "print \" piercing punch diameter = %0.2f cm\\n blanking punch diametre = %0.2f cm \\n press capacity = %0.2f KN\\n\"%(d1/10 , d2/10 , c/1000)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.2 : page 133"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Diameter at first draw  = 8.25 cm\n",
+      " Diameter at second draw  = 6.19 cm\n",
+      " Diameter at third draw  =4.950 cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt, pi\n",
+    "h = 10 # height of cup in cm\n",
+    "d = 5 # diameter of cup in cm\n",
+    "D = sqrt(d**2 + 4*d*h) # blank diameter in cm\n",
+    "# height daimeter ratio is 2 . Therefore from table 2.9 total reductions are 3\n",
+    "r1 = 0.45*D # first reduction is 45% \n",
+    "d1 = D - r1 # diameter at first draw in cm\n",
+    "r2 = d1*0.25 # second reduction is 25% \n",
+    "d2 = d1 - r2 # diameter at second draw in cm\n",
+    "r3 = d2*0.2 # third reduction is 20% \n",
+    "d3 = d2 - r3 # diameter at third draw in cm\n",
+    "print \" Diameter at first draw  = %0.2f cm\\n Diameter at second draw  = %0.2f cm\\n Diameter at third draw  =%0.3f cm\"%( d1 , d2 , d3)\n",
+    " "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.3 : page 133"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "bending force = 44.3 KN\n"
+     ]
+    }
+   ],
+   "source": [
+    "K = 1.20 # die-opening factor\n",
+    "L = 37.5 # Length of strip in cm\n",
+    "T = 2.5 # thickness of strip in mm\n",
+    "sigma_ut = 630 # tensile strength in N/mm**2\n",
+    "W = 16*T # width of die opening in mm\n",
+    "F = (K*L*10*sigma_ut*T**2)/W # bending force in N\n",
+    "print \"bending force = %0.1f KN\"%( F/1000)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.4 : page 134"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " blanking force  = 74.25 KN\n",
+      " work done  = 27.84 Nm\n"
+     ]
+    }
+   ],
+   "source": [
+    "b = 25 # width of blank in mm\n",
+    "l = 30 # length of blank in mm\n",
+    "tau = 450 # ultimate shear stress of material in N/mm**2\n",
+    "t = 1.5 # thickness of metal strip in mm\n",
+    "p = 2*(l + b) # perimeter of blank in mm\n",
+    "f = p*t*tau # blanking force in N\n",
+    "Pt = 0.25*t # punch travel in mm\n",
+    "w = f*Pt # work done in Nmm\n",
+    "print \" blanking force  = %0.2f KN\\n work done  = %0.2f Nm\"%(f/1000,w/1000)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.5 : page 134"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " clearance = 0.075 mm\n",
+      " blank die opening size = 25.35 mm \n",
+      " blanking punch size = 25.20 mm\n",
+      " piercing punch size = 12.75 mm\n",
+      " piercing die size = 12.90 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "t = 1.5 # thickness in mm\n",
+    "c = 0.05*t # clearance in mm\n",
+    "D = 25.4 # outside diameter in mm \n",
+    "D_o = D - 0.05 # blank die opening in mm\n",
+    "B_s = D_o - 2*c # blanking punch size in mm\n",
+    "d = 12.7 # internal diameter in mm \n",
+    "P_s = d + 0.05 # piercing punch size in mm\n",
+    "D_s = P_s + 2*c # piercing die size in mm\n",
+    "print \" clearance = %0.3f mm\\n blank die opening size = %0.2f mm \"%(c ,D_o)\n",
+    "print \" blanking punch size = %0.2f mm\\n piercing punch size = %0.2f mm\\n piercing die size = %0.2f mm\"%(B_s,P_s,D_s )"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.6 : page 135"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "shear force when both punch act at same time and no shear is applied = 50.27 kN\n",
+      "cutting force when punches are staggered = 33.5 kN\n",
+      "cutting force when there is penetration and shear on punch = 23.8 kN\n"
+     ]
+    }
+   ],
+   "source": [
+    "D = 25.4 # outside diameter in mm \n",
+    "d = 12.7 # internal diameter in mm\n",
+    "t = 1.5 # thickness in mm\n",
+    "tau = 280 # ultimate shearing strength in N/mm**2\n",
+    "F = pi*(D + d)*t*tau # total cutting force in N\n",
+    "F_s = pi*D*t*tau # cutting force when punches are staggered in N\n",
+    "k = 0.6 # penetration\n",
+    "i = 1 # shear of punch in mm\n",
+    "F_p = (t*k*F)/(k*t +i)# cutting force when both punches act together in N \n",
+    "print \"shear force when both punch act at same time and no shear is applied = %0.2f kN\"%( F/1000)\n",
+    "print \"cutting force when punches are staggered = %0.1f kN\"%(F_s/1000)\n",
+    "print \"cutting force when there is penetration and shear on punch = %0.1f kN\"%(F_p/1000)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.7 : page 135"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Amount of shear on punch = 1 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "D = 60 # hole diameter in mm\n",
+    "tau = 450 # ultimate shear strength in N/mm**2\n",
+    "t = 2.5 # thickness in mm\n",
+    "F = pi*D*t*tau # maximum punch force in N\n",
+    "w = F*0.4*t # work done in Nm\n",
+    "f = F/2 # punching force in N\n",
+    "k = 0.4 # penetration percentage\n",
+    "i = k*t*(F-f)/f # shear on punch in mm\n",
+    "print \"Amount of shear on punch = %d mm\"%( i)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.8 : page 136"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Value of back scrap = 4.7 mm\n",
+      " Value of scrap bridge = 3.2 mm \n",
+      " Width of strip = 11.00 cm\n",
+      " Length of one piece of stock needed to produce one part = 2.82 cm\n",
+      " Number of parts = 84.0 blanks\n",
+      " Scrap remaining at the end = 2.88 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import ceil\n",
+    "# from fig 2.27\n",
+    "w = 2.5 # cm\n",
+    "t = 3.2 # strip thickness in mm\n",
+    "h = 10 # cm\n",
+    "a = t + 0.015*h*10 # back scrap and front scrap in mm\n",
+    "b = t # scrap bridge in mm\n",
+    "W = h + (2*a)/10 # width of strip in cm\n",
+    "W = ceil(W) # cm\n",
+    "s = w + b/10 # length of one piece of stock in cm\n",
+    "L = 240 # length of strip in cm\n",
+    "N = (L-b)/s # number of parts\n",
+    "y = L - (N*s + b/10)  # scrap remaining at the end in mm\n",
+    "print \" Value of back scrap = %0.1f mm\\n Value of scrap bridge = %0.1f mm \"%( a , b  )\n",
+    "print \" Width of strip = %0.2f cm\\n Length of one piece of stock needed to produce one part = %0.2f cm\"%( W , s)\n",
+    "print \" Number of parts = %0.1f blanks\\n Scrap remaining at the end = %0.2f mm\"%( N , y)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.9 : page 137"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(i) size of blank = 118.20 mm \n",
+      "(ii) Percentage reduction = 57.7 percent \n",
+      "(iii) Number of draws = 2 \n",
+      "(iv) Radius on punch = 4 mm \n",
+      " Die Radius = 6 mm \n",
+      "(v) Die clearance for first draw = 0.87 mm \n",
+      " die clearance for second draw = 0.88 mm\n",
+      " Punch diameter for second draw = 48.4 mm \n",
+      " Die opening diameter for second draw = 50.16 mm \n",
+      " Punch diameter for first draw  = 63.412 mm \n",
+      " Die opening diameter for first draw = 66.752 mm\n",
+      "(vi) Drawing pressure = 91.97 mm \n",
+      "(vii) Blank holding pressure = 30 kN \n",
+      "(viii) Press capacity = 150 kN\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi, sqrt\n",
+    "# from figure 2.73\n",
+    "t = 0.8 # thickness in mm\n",
+    "d = 50 # shell diameter in mm\n",
+    "r = 1.6 # radius of bottom corner in mm\n",
+    "h = 50 # height in mm\n",
+    "D = sqrt(d**2 + 4*d*h) # shell blank size in mm\n",
+    "el = 6.4 # extra length required to add in shell blank size\n",
+    "D = D + el # mm\n",
+    "pr = 100*(1-(d/D)) # percentage reduction\n",
+    "ratio = h/d \n",
+    "n = 2 # number of draws\n",
+    "R1 = 45 # first reduction\n",
+    "D1 = D - R1*D/100 # diameter at first reduction in mm\n",
+    "R2 = 100*(1-(d/D1)) # second reduction \n",
+    "PR = 4*t # punch radius in mm\n",
+    "PR = ceil(PR)\n",
+    "DR = 6 # die radius in mm\n",
+    "DC1 = 0.87 # die clearance for first draw in mm\n",
+    "DC2 = 0.88 # die clearance for second draw in mm\n",
+    "PD2 = d - 2*t # punch diameter for second draw in mm\n",
+    "DD2 = PD2 + 2*DC2 # Die opening diameter for second draw in mm\n",
+    "PD1 = D1 - 2*t # punch diameter for first draw in mm\n",
+    "DD1 = D1 + 2*DC1 # Die opening diameter for first draw in mm\n",
+    "# Drawing pressure\n",
+    "c = 0.65 # constant\n",
+    "sigma = 427 # N/mm**2\n",
+    "F = pi*d*t*sigma*(D/d-c) # Drawing pressure in mm\n",
+    "Bhp = 30.8 # blanking holding pressure in kN\n",
+    "pc = 150 # press capacity in kN\n",
+    "print \"(i) size of blank = %0.2f mm \\n(ii) Percentage reduction = %0.1f percent \\n(iii) Number of draws = %d \\n(iv) Radius on punch = %d mm \\n Die Radius = %d mm \\n(v) Die clearance for first draw = %0.2f mm \\n die clearance for second draw = %0.2f mm\"%( D , pr,n,PR,DR,DC1,DC2)\n",
+    "print \" Punch diameter for second draw = %0.1f mm \\n Die opening diameter for second draw = %0.2f mm \\n Punch diameter for first draw  = %0.3f mm \\n Die opening diameter for first draw = %0.3f mm\\n(vi) Drawing pressure = %0.2f mm \\n(vii) Blank holding pressure = %d kN \\n(viii) Press capacity = %d kN\"%( PD2 , DD2 ,PD1 ,DD1 , F/1000 ,Bhp , pc)  \n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.10 : page 138"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Developed length = 187.22 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "# from figure 2.74\n",
+    "l1 = 76 - ( 2.3 + 0.90) # length1 in mm\n",
+    "l2 = 115 - (2.3 + 0.90) # length2 in mm\n",
+    "t = 2.3 # mm\n",
+    "r = 0.90 # inner radius in mm\n",
+    "k = t/3 # mm\n",
+    "B = 0.5*pi*(r + k) # bending allowance in mm\n",
+    "d = l1 + l2 + B # developed length in mm\n",
+    "print \"Developed length = %0.2f mm\"%( d)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.11 : page 139"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "bending force = 111.25 kN\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import floor\n",
+    "t = 3.2 # thickness of blank in mm\n",
+    "l = 900 # bending length in mm\n",
+    "sigma = 400 # ultimate tensile strength in N/mm**2\n",
+    "k = 0.67 # die opening factor\n",
+    "r1 = 9.5 # punch radius in mm\n",
+    "r2 = 9.5 # die edge radius in mm\n",
+    "c = 3.2 # clearance in mm\n",
+    "w = r1 + r2 + c # width between contact points batween die and punch in mm\n",
+    "F= (k*l*sigma*t**2)/w # bending force in N\n",
+    "F=floor(F/10)*10 # N\n",
+    "print \"bending force = %0.2f kN\"%(F/1000)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.12 : page 139"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "bending force  = 145.24 kN\n"
+     ]
+    }
+   ],
+   "source": [
+    "k = 1.33 # die opening factor\n",
+    "l = 1200 # bend length in mm\n",
+    "sigma = 455 # ultimate tensile strength in N/mm**2\n",
+    "t = 1.6 # blank thickness in mm\n",
+    "w = 8*t # width of die opening in mm\n",
+    "F = k*l*sigma*t**2/w # bending force in N \n",
+    "print \"bending force  = %0.2f kN\"%(F/1000)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.13 : page 140"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Force required when bottoming is used  = 18.2 tonnes\n",
+      "Force required when bottoming is not used  = 1.263 tonnes\n"
+     ]
+    }
+   ],
+   "source": [
+    "c = 1.25 # clearance in mm\n",
+    "r1 = 3 # die radius in mm\n",
+    "r2 = 1.5 # punch radius in mm\n",
+    "sigma = 315 # ultimate tensile strength in MPa\n",
+    "t = 1 # thickness in mm\n",
+    "l = 50 # width at bend in mm\n",
+    "w = r1 + r2 +c # width between contact points on die and punch in mm\n",
+    "F = 0.67*l*sigma*t**2/w # bending force in N\n",
+    "F_p = 0.67*sigma*l*t # pad force in N\n",
+    "sigma_c = 560 # setting pressure in MPa\n",
+    "b1 = 2 # beads on punch\n",
+    "b = b1*r1 # mm\n",
+    "F_b = sigma_c*l*b # bottoming force in N\n",
+    "F_o = F_p + F_b # Force required when bottoming is used in N\n",
+    "F_n = F +F_p # Force required when bottoming is not used in N\n",
+    "print \"Force required when bottoming is used  = %0.1f tonnes\"%(F_o/(9.81*1000))\n",
+    "print \"Force required when bottoming is not used  = %0.3f tonnes\"%( F_n/(9.81*1000))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.14 : page 140"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "cutting force with parallel cutting edges = 3.825 MN\n",
+      "cutting force when shear is considered = 91.48 kN \n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sin, pi\n",
+    "w = 2 # width in mm\n",
+    "t = 5 # thickness in mm\n",
+    "theta=6 # shear in degrees\n",
+    "tau = 382.5 # ultimate shear stress in MPa\n",
+    "F = w*t*tau*1000 # cutting force in N\n",
+    "l = t/sin(theta*pi/180) # length to be cut in mm\n",
+    "F_i = l*t*tau # cutting force in N\n",
+    "print \"cutting force with parallel cutting edges = %0.3f MN\\ncutting force when shear is considered = %0.2f kN \"%(F/10**6 ,F_i/1000)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.15 : page 141"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Blank diameter = 220 mm \n",
+      " Thickness ratio = 0.453 \n",
+      " Diameter for first draw = 121 mm \n",
+      " Depth of first draw = 61.39 mm \n",
+      " Press capacity = 177.92 kN\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi, sqrt\n",
+    "d1 = 105 # inside diameter in mm\n",
+    "h = 90 # depth in mm\n",
+    "t = 1 # thickness in mm\n",
+    "D = sqrt(d1**2+4*d1*h) # blank diameter in mm\n",
+    "tr = t*100/D # thickness ratio\n",
+    "# from table safe drawing ratio is 1.82\n",
+    "r = 1.82 # draw ratio\n",
+    "d2 = D/r # diameter for first draw in mm\n",
+    "d = 130 # Let diameter for first draw in mm\n",
+    "ratio1 = D/d # D/d for first draw \n",
+    "ratio2 = d/d1 # D/d for second draw\n",
+    "h1 =((D)**2-(d)**2)/(4*d) # Depth of first draw in mm\n",
+    "sigma = 415 # N/mm**2\n",
+    "c = 0.65 # constant\n",
+    "pc = pi*d*t*sigma*(D/d - c) # press capacity in kN\n",
+    "print \" Blank diameter = %d mm \\n Thickness ratio = %0.3f \\n Diameter for first draw = %d mm \\n Depth of first draw = %0.2f mm \\n Press capacity = %0.2f kN\"%(D,tr,d2,h1,pc/1000)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.16 : page 142"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Diameter after first draw = 14.7 \n",
+      " Diameter after second draw = 8.82 \n",
+      " Percentage reduction after third draw = 9 percent\n",
+      " Height of cup after first draw = 11.02 cm\n",
+      " Height of cup after second draw = 22.29 cm\n",
+      " Height of cup after third draw = 25.00 cm\n",
+      " Drawing force = 56.817 kN\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi, sqrt\n",
+    "d = 80 # diameter in mm\n",
+    "h = 250 # height in mm\n",
+    "D = sqrt((d**2+4*d*h))/10 # blank diameter in cm\n",
+    "D1 = 0.5*D # diameter after first draw in cm\n",
+    "# let reduction be 40% in second draw\n",
+    "D2 = D1-0.4*D1 # diameter after scond draw in cm\n",
+    "R = (1 - (d/(10*D2)))*100 # percentage reduction for third draw\n",
+    "l1 = ((D)**2-(D1)**2)/(4*D1) # height of cup after first draw in cm\n",
+    "l2 = ((D)**2-(D2)**2)/(4*D2) # height of cup after first draw in cm\n",
+    "l3 = ((D)**2-(d/10)**2)/(4*d/10) # height of cup after first draw in cm\n",
+    "t = 3 # mm\n",
+    "sigma = 250 # N/mm**2\n",
+    "C = 0.66\n",
+    "F = pi*d/10*t*sigma*((D*10/d)-C) # drawing force in kN\n",
+    "print \" Diameter after first draw = %0.1f \\n Diameter after second draw = %0.2f \\n Percentage reduction after third draw = %d percent\"%(D1,D2,R)\n",
+    "print \" Height of cup after first draw = %0.2f cm\\n Height of cup after second draw = %0.2f cm\\n Height of cup after third draw = %0.2f cm\"%(l1,l2,l3)\n",
+    "print \" Drawing force = %0.3f kN\"%(F/1000)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 2.17 : page 143"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total developed length = 1003.39 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "# from figure 2.75 (a)\n",
+    "r1 = 30 # radius in mm\n",
+    "t = 10 # thickness in mm\n",
+    "h1 = 300 # height in mm\n",
+    "ir1 = r1-t # inner radius of bends in mm\n",
+    "L1 = h1-(ir1+t) # mm\n",
+    "alpha1 = 90 # degree\n",
+    "r2 = 2*t # mm\n",
+    "k = 0.33*t # mm\n",
+    "L2 = alpha1*2*pi*(r2+k)/360 # mm\n",
+    "w = 200 # mm\n",
+    "L3 = w-2*(t+ir1)# mm\n",
+    "L4 = L2 #mm\n",
+    "h2 = 100 # mm\n",
+    "L5 = h2 -(t+ir1) # mm\n",
+    "r3 = 150 #mm \n",
+    "ir2 = r3 - t # inner radius in mm\n",
+    "alpha2 = 180 # degree\n",
+    "L6 = alpha2*2*pi*(ir2+k)/360 # mm\n",
+    "dl = L1+L2+L3+L4+L5+L6 # Total developed length in mm\n",
+    "print \"Total developed length = %0.2f mm\"%( dl)\n",
+    "# Answers vary due to round off error"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter4_1.ipynb b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter4_1.ipynb
new file mode 100644
index 00000000..cfbc5330
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter4_1.ipynb
@@ -0,0 +1,863 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 4 : Cost Estimating"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.1 : page 206"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Total cost = Rs 13225\n",
+      " Selling price = Rs 14812\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "d_m = 5500 # cost of direct material in Rs\n",
+    "d_l = 3000 # manufacturing wages in Rs\n",
+    "# factory overhead is 100% 0f manufacturing wages\n",
+    "f_o = (100*d_l)/100 # factory overheads in Rs\n",
+    "FC = d_m + d_l + f_o # factory cost in Rs\n",
+    "nm_o = 15*FC/100 # non-manufacturing overheads in Rs\n",
+    "tc = FC+nm_o # total cost in Rs\n",
+    "p = 12*tc/100 # profit in Rs\n",
+    "sp = tc+p # selling price in Rs\n",
+    "print \" Total cost = Rs %d\\n Selling price = Rs %d\"%(tc,sp)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.2 : page 207"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Selling price = Rs 99000\n"
+     ]
+    }
+   ],
+   "source": [
+    "# given\n",
+    "OS_RM = 20000 # opening stock of raw materials in Rs\n",
+    "CS_RM = 30000 # closing stock of raw materials in Rs\n",
+    "TP_RM = 170000 # total purchase in year in Rs\n",
+    "OS_FG = 10000 # opening stock of finished goods in Rs\n",
+    "CS_FG = 15000 # closing stock of finished goods in Rs\n",
+    "sales = 489500 # sales of finished goods in Rs\n",
+    "D_W = 120000 # direct wages in Rs\n",
+    "F_E1 = 120000 # factory expenses in Rs\n",
+    "NM_E = 50000 #  non-manufacturing expenses in Rs\n",
+    "DMC = OS_RM + TP_RM - CS_RM # direct material cost\n",
+    "FC = DMC + D_W + F_E1 # factory cost\n",
+    "TC = FC + NM_E # total cost\n",
+    "FG_S = OS_FG + TC - CS_FG # cost of finished goods sold in Rs\n",
+    "P = sales - FG_S # profit in Rs\n",
+    "F_E2 = (F_E1)/D_W*100 # factory expenses in percent\n",
+    "NM_C = (NM_E)/FC*100 # non-manufacturing expenses to factory cost\n",
+    "P_C = (P/FG_S)*100 # profit to cost of sales\n",
+    "dm = 20000 # direct material in Rs\n",
+    "dw = 30000 # direct wages in Rs\n",
+    "fe = dw # factory expenses\n",
+    "fc = dm+dw+fe # factory cost in Rs\n",
+    "nme = NM_C*fc/100 # non-manufacturing expenses in Rs\n",
+    "tc = fc+nme # total cost in Rs\n",
+    "p = (P_C*tc)/100 # profit in Rs\n",
+    "sp = tc+p # selling price in Rs\n",
+    "print \"Selling price = Rs %d\"%(sp) "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.3 : page 208"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "factory cost = Rs 8.39 \n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "d = 38 # diameter of bar in mm\n",
+    "l = 25 # length of bar in mm\n",
+    "p = 8.6 # density gm/cm**3\n",
+    "g = 9.81 # acceleration due to gravity in m/s**2\n",
+    "w = (pi*d**2*l*p*g)/(4*10**6) # weight of material in N\n",
+    "mc = w*1.625 # material cost in Rs\n",
+    "lc = (2*90)/60 # labour cost in Rs\n",
+    "fo = 0.5*lc # factory overheads in Rs\n",
+    "fc = mc + lc + fo # factory cost in Rs\n",
+    "print \"factory cost = Rs %0.2f \"%fc\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.4 : page 208"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Time taken = 7.908 Hours\n"
+     ]
+    }
+   ],
+   "source": [
+    "sp = 65 # selling price in Rs\n",
+    "profit = 0.2*sp # profit in Rs\n",
+    "tc = sp - profit # total cost in Rs \n",
+    "P = (sp - profit)/1.4 # production cost in Rs\n",
+    "DM = 15 # cost of direct material in Rs\n",
+    "W = (P - DM)/ 1.4 # direct labour cost in Rs\n",
+    "tt = W/2 # time taken in hours\n",
+    "print \"Time taken = %0.3f Hours\"%(tt )\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.5 : page 209"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "profit =  Rs 600\n"
+     ]
+    }
+   ],
+   "source": [
+    "mp = 6000 # market price of machine in Rs\n",
+    "d = 0.2*mp # discount in Rs\n",
+    "sp = mp - d # selling price of factory in Rs\n",
+    "mc = 400 # material cost in Rs\n",
+    "lc = 1600 # labour cost in Rs\n",
+    "fo = 800 # factory overheads in Rs\n",
+    "F = mc + lc + fo # factory cost in Rs\n",
+    "se = 0.5*F # selling expenses in Rs\n",
+    "profit = sp - (F + se ) # Rs\n",
+    "print \"profit =  Rs %d\"%(profit)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.6 : page 209"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Economic lot size = 1060 pieces\n",
+      "Time for each set up = 21.21 hours\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "a = 1500 # requirements of components\n",
+    "s = 30 # cost of each set up in Rs\n",
+    "k = 0.2 # charge factor\n",
+    "c = 5 # cost of each part in Rs\n",
+    "N = 5*sqrt(a*s)/(k*c) # economic lot size\n",
+    "print \"Economic lot size = %d pieces\"%(N)\n",
+    "S = (N*s)/a # time for each set up in hours\n",
+    "print \"Time for each set up = %0.2f hours\"%(S )\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.7 : page 209"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Unit time to change the cutter = 0.083 min\n"
+     ]
+    }
+   ],
+   "source": [
+    "Tc = 2 # time taken by cutter per cycle in minutes\n",
+    "Tk = 10 # time taken to change cutter in minutes\n",
+    "T = 240 # tool life in minutes\n",
+    "t = (Tc*Tk)/T # time to change the cutter in min.\n",
+    "print \"Unit time to change the cutter = %0.3f min\"%(t )\n",
+    "# Error in textbook"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.8 : page 209"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Unit tool change time per cycle = 1.67 min\n"
+     ]
+    }
+   ],
+   "source": [
+    "Tk = 360 # time taken by tool to cut before sharpening in min.\n",
+    "Tc = 20 # time taken to change the tool in min.\n",
+    "T = 4320 # time taken before it is discarded in min.\n",
+    "t = (Tc*Tk)/T #  tool change time per cycle in min.\n",
+    "print \"Unit tool change time per cycle = %0.2f min\"%(t )"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.9 : page 209"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total time taken to check dimensions = 67 min\n"
+     ]
+    }
+   ],
+   "source": [
+    "Tc = 10 # time taken to check hole in secs\n",
+    "F = 2 # frequency of checking dimension\n",
+    "tc = Tc*F # time taken to check one piece in secs\n",
+    "N = 200 # number of pieces\n",
+    "Tc = tc*(N + 1) # Total time in sec\n",
+    "print \"Total time taken to check dimensions = %d min\"%(Tc/60)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.10 : page 209"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Direct labour cost = Rs 220.5\n"
+     ]
+    }
+   ],
+   "source": [
+    "forgings = 40\n",
+    "setup = 4 \n",
+    "Tc = 12 # machining time in min. per forging\n",
+    "nmt = 21 # non-machining in min. per forging\n",
+    "st = 45 # set up time per set up\n",
+    "ts = 5 # total sharpening in min. per forging\n",
+    "f = 20 # fatigue in percent\n",
+    "f = f/100\n",
+    "pn = 5 # personal needs in percent\n",
+    "pn = pn/100\n",
+    "Tk = 10 # tool chanhe time in min.\n",
+    "T = 8 # tool life in hours\n",
+    "ct = 15 # checking time with 5 checks in 15 secs\n",
+    "R = 1.4 # performance factor\n",
+    "dlc = 5 # direct labour cost in Rs per hour\n",
+    "tt = Tc+nmt # machining and non-machining time in min.\n",
+    "ft = f*tt # fatigue time in min.\n",
+    "pnt = pn*tt # personal needs in min.\n",
+    "t = (Tc*Tk)/(T*60) # total sharpening time in min. per forging\n",
+    "mct = (ts*ct)/60 # measuring and checking time in min.per forging\n",
+    "su = Tc + nmt+ pnt + ft + t + mct # sum of times in min.\n",
+    "tf = su*forgings # time for 40 forgings in min.\n",
+    "tst = st*setup # total set up time in min.\n",
+    "Te = tf+tst # total estimated time in min.\n",
+    "Ta = Te*R # total actual time in min.\n",
+    "lc = (Ta*dlc)/60 # direct labour cost in Rs\n",
+    "print \"Direct labour cost = Rs %0.1f\"%(lc)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.11 : page 216"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total time = 0.22 min.\n"
+     ]
+    }
+   ],
+   "source": [
+    "# from figure 4.4\n",
+    "v = 100 # cutting speed in m/min\n",
+    "D = 50 # mm\n",
+    "l1 = 76 # mm\n",
+    "f = 7.5 # feed in mm/rev.\n",
+    "# Case 1 , time to turn 38 mm diameter by 76 mm length of cut\n",
+    "N1 = (1000*v)/(pi*D)# r.p.m\n",
+    "tm1 = l1*10/(f*N1) # min.\n",
+    "# Case 2 , time to turn 25 mm by 38 mm length\n",
+    "N2 = (1000*v)/(pi*38) # r.p.m\n",
+    "l2 = 38 # mm\n",
+    "tm2 = l2*10/(f*N2) # min\n",
+    "tt = tm1 + tm2 # total time in min\n",
+    "print \"Total time = %0.2f min.\"%(tt)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.12 : page 218"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Time to turn external relief = 0.17 min.\n"
+     ]
+    }
+   ],
+   "source": [
+    "# from figure 4.5\n",
+    "v = 60 #  cutting speed m/min.\n",
+    "f = 0.375 # feed in mm/rev\n",
+    "D = 38 # mm\n",
+    "N = (1000*60)/(pi*D) # rev/min\n",
+    "l = 32 # mm\n",
+    "Tm = l/(f*N) # min\n",
+    "print \"Time to turn external relief = %0.2f min.\"%(Tm )"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.13 : page 220"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The machining time to face on lathe = 1.91 min.\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "# from figure 4.11\n",
+    "l = 7.5 # cm\n",
+    "Dave = (25+ 10)/2 # average diameter in cm\n",
+    "v = 27 # cutting speed in m/min\n",
+    "f = 0.8 # feed in mm/rev\n",
+    "N = (1000*v)/(pi*Dave*10) # r.p.m.\n",
+    "tm = l*10/(f*N) # min.\n",
+    "print \"The machining time to face on lathe = %0.2f min.\"%(tm)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.14 : page 221"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Time taken to drill hole = 0.174 min.\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "D = 12.7 # diameter in mm\n",
+    "d = 50 # depth in mm\n",
+    "v = 75 # cutting speed in m/min.\n",
+    "f = 0.175 # feed in mm/rev\n",
+    "l = d + 2*0.29*D # lemgth of drill travel in mm\n",
+    "N = (1000*v)/(pi*D) # r.p.m.\n",
+    "tm = l/(f*N) # min\n",
+    "print \"Time taken to drill hole = %0.3f min.\"%(tm)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.15 : page 223"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Time required for shaper to complete one cut = 60 min\n"
+     ]
+    }
+   ],
+   "source": [
+    "k = 1/4 # return time to cutting ratio\n",
+    "l = 900 + 2*75 # length of stroke in mm\n",
+    "v = 6 # cutting stroke in m/min\n",
+    "f = 2 # feed mm/stroke\n",
+    "w = 600 # breadth in mm\n",
+    "N = (v*1000)/(l*1.25) # r.p.m\n",
+    "N = round(N)\n",
+    "time =  w/(f*N) # min\n",
+    "print \"Time required for shaper to complete one cut = %d min\"%(time )"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.16 : page 223"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Time taken to broach a four spline brass = 0.0928 min\n"
+     ]
+    }
+   ],
+   "source": [
+    "l = 70 # length of stroke in cm\n",
+    "cs = 11 # cutting speed in m/min\n",
+    "rs = 24 # return speed in m/min\n",
+    "tm = (l/(100*cs)) + (l/(100*rs)) # min\n",
+    "print \"Time taken to broach a four spline brass = %0.4f min\"%(tm)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.17 : page 228"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Table feed = 265 mm/min. \n",
+      " Total cutter travel = 255.9 mm\n",
+      " Time required to machine the slot = 0.965 min.\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "v = 50 # cutting speed in m/min\n",
+    "D = 150 # diameter of face cutter in mm\n",
+    "N = (1000*v)/(pi*D) # r.p.m.\n",
+    "f = 0.25 # feed mm/tooth\n",
+    "n = 10 #number of tooth\n",
+    "tf = N*f*n # table feed in mm/min\n",
+    "l = 200 # length of work piece in mm\n",
+    "d = 25 # depth of slot in mm\n",
+    "tot = sqrt(D*d - d**2) # total overtravel in mm\n",
+    "tct = l + tot # total cutter travel in mm\n",
+    "time = tct/tf # min.\n",
+    "print \" Table feed = %d mm/min. \\n Total cutter travel = %0.1f mm\\n Time required to machine the slot = %0.3f min.\"%(tf , tct ,time )"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.18 : page 228"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Cutting time = 0.163 min.\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "D = 63.5 # diameter of plain milling cutter in mm\n",
+    "w = 30 # width of block in mm\n",
+    "l = 180 # length of block in mm\n",
+    "f = 0.125 # feed in mm/tooth\n",
+    "n = 6 # no. of teeth\n",
+    "N = 1500 # spindle speed in r.p.m\n",
+    "tot = (D - sqrt(D**2 - w**2))/2 # total over travel in mm\n",
+    "tct = l + tot # total cutter travel in mm\n",
+    "Tm = tct/(f*n*N) # cutting time in min\n",
+    "print \" Cutting time = %0.3f min.\"%(Tm)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.19 : page 229"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The milling time = 0.29 min.\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "# from figure 4.17\n",
+    "d = 19 # depth of cut in mm\n",
+    "D1 = 5 # diameter of round bar in cm\n",
+    "v = 50 # cutting speed in m/min\n",
+    "n = 8 # number of teeth\n",
+    "f = 0.2 # feed in mm/tooth\n",
+    "l = 2*sqrt(d*D1*10 - d**2) # length of chord in mm\n",
+    "D2 = 10 # daimeter of cutter in cm\n",
+    "overrun = sqrt(D2*10*d+D1*10*d-d**2) - sqrt(D1*10*d-d**2)# mm\n",
+    "tt = l + overrun # table travel in mm\n",
+    "N = (1000*v)/(pi*D2*10) # r.p.m\n",
+    "tm = tt/(f*n*N) # time in min.\n",
+    "print \"The milling time = %0.2f min.\"%(tm )"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.20 : page 230"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Time required to grind the shaft = 0.64 min.\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "w = 50 # width of grinding wheep in mm\n",
+    "f = w/2 # feed in mm\n",
+    "t = 0.25 # toatal stock in mm\n",
+    "d = 0.025 # depth of cut in mm\n",
+    "n = t/d # number of cuts \n",
+    "v = 15 # cutting speed in m/min\n",
+    "D = 38 # diameter in mm\n",
+    "N = (1000*v)/(pi*D) # r.p.m.\n",
+    "l = 200 # length of part in mm\n",
+    "Tm = (l*10)/(f*N) # min.\n",
+    "print \"Time required to grind the shaft = %0.2f min.\"%(Tm)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.21 : page 231"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Time to cut the threads = 2.05 min\n"
+     ]
+    }
+   ],
+   "source": [
+    "v = 6 # cutting speed in m/min\n",
+    "n = 5 # number of cuts\n",
+    "D = 44 # diameter in mm\n",
+    "N = (1000*v)/(pi*D) # r.p.m\n",
+    "f = 0.5 # feed in cm\n",
+    "l = 8.9 # length of cut in cm\n",
+    "Tm = (l*n)/(f*N) # time in min\n",
+    "print \"Time to cut the threads = %0.2f min\"%(Tm)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.22 : page 232"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total production timr per piece = 3.01 min\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "vt = 40 # cutting speed for turning in m/min\n",
+    "vs = 8 # cutting speed for cutting and knurling in m/min\n",
+    "ft = 0.4 # feed for turning in mm/rev.\n",
+    "ff = 0.2 # feed for forming in mm/rev\n",
+    "d1 = 25 # diameter in mm\n",
+    "l1 = 50 # mm\n",
+    "N1 = 1000*vt/(pi*d1) # spindle speed in rev./min.\n",
+    "time1 = l1/(ft*N1) # min.\n",
+    "tt = 2*time1 # total time in min.\n",
+    "d2 = 15 # mm\n",
+    "N2 = 1000*vt/(pi*d2)# rev/min.\n",
+    "l2 = 30 # mm\n",
+    "time2 = l2/(ft*N2) # min.\n",
+    "eft = 0.15 # end forming time in min.\n",
+    "d3 = 10 # mm\n",
+    "N3 = 1000*vs/(pi*d3) # rev./min.\n",
+    "l3 = 15 # mm\n",
+    "f = 1.5 # feed in min.\n",
+    "time3 = l3/(f*N3) # min.\n",
+    "N4 = 1000*vs/(pi*d1) # rev./min.\n",
+    "l4 = 10 # mm\n",
+    "time4 = l4/(ft*N4) # min.\n",
+    "time5 = 0.15 # time for chamfering in min.\n",
+    "Dave = d1/2 # mm\n",
+    "N5 = 1000*vt/(pi*Dave) # r.p.m.\n",
+    "time6 = Dave/(N5*ff) # min,\n",
+    "tmt = tt+time2+time3+time4+time5+time6+eft # total machining time in min.\n",
+    "t = 0.05 # min.\n",
+    "ht = time5+6*time6+4*t+3*t # handling time in min.\n",
+    "tot = ht+tmt # total handling time in min.\n",
+    "ct = 15*tot/100 # contingency in min.\n",
+    "tct = tot+ct # total cycle time in min.\n",
+    "st = 60 # set up time for turret lathe\n",
+    "p = 100 # total pieces\n",
+    "stp = st/p # set up time per piece in min.\n",
+    "tpt = tct+stp # Total production timr per piece in min.\n",
+    "print \"Total production timr per piece = %0.2f min\"%(tpt)\n",
+    "#Answers vary due to round off error"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter5_1.ipynb b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter5_1.ipynb
new file mode 100644
index 00000000..8c47225e
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter5_1.ipynb
@@ -0,0 +1,1707 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 5 : Economics of Tooling"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.1 : page 238"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Value of machine at the end of 10 years through straight line depreciation method = Rs 137500\n",
+      "Value of machine at the end of 10 years through sinking fund method = Rs 178773\n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "Co = 250000 # original value of machine tool in Rs\n",
+    "Cs = 25000 # salvage value in Rs\n",
+    "n = 20 #useful life in years\n",
+    "d = (Co-Cs)/n # depreciation per year in Rs\n",
+    "v1 = Co - 10*d # value of machine tool at the end of 10 years in Rs\n",
+    "s = Co - Cs # sum at the end of useful life in Rs\n",
+    "i = 8/100 # annual interst rate \n",
+    "D = (s*i)/((1 + i)**n-1) # annual deposit\n",
+    "a  = D*((1+i)**10-1)/i #amount at the end of 10 years in Rs\n",
+    "v2 = Co - a # value at the end of 10 years\n",
+    "print \"Value of machine at the end of 10 years through straight line depreciation method = Rs %d\"%(v1)\n",
+    "print \"Value of machine at the end of 10 years through sinking fund method = Rs %d\"%(v2)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.2 : page 249"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Annual investment using sinking fund factor = Rs 3491 /- per year\n",
+      "Annual investment using capital recovery factor = Rs 23491 /- per year\n"
+     ]
+    }
+   ],
+   "source": [
+    "p = 200000 # present worth in Rs\n",
+    "i = 10 # annual interest rate\n",
+    "i = 10/100\n",
+    "n = 20 # number of years\n",
+    "a1 = (p*i)/((1+i)**n-1) # annual investment using sinking fund factor in Rs\n",
+    "a2 = (p*i*(i+1)**n)/((i+1)**n-1)# annual investment using capital recovery factor in Rs\n",
+    "print \"Annual investment using sinking fund factor = Rs %d /- per year\"%(a1)\n",
+    "print \"Annual investment using capital recovery factor = Rs %d /- per year\"%(a2)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.3 : page 252"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total cash in flows = Rs 85483.68\n",
+      "Total cash out flows = Rs 85489.44\n",
+      "Since cash out flows are more than cash in flows therefore project is not economical\n"
+     ]
+    }
+   ],
+   "source": [
+    "# cash in flows\n",
+    "a = 21240 # annual revenue in Rs\n",
+    "i = 10 # annual interest rate\n",
+    "i = 10/100\n",
+    "n = 5 # perod in years\n",
+    "f1 = 8000 # salvage value in Rs\n",
+    "p1 =  (a*((i+1)**n-1))/(i*(i+1)**5)# annual revenue in Rs\n",
+    "p2 = f1/(i+1)**5 #present worth in Rs\n",
+    "t1 = p1 + p2 # total cash in flows in Rs\n",
+    "# cash out flows\n",
+    "I = 40000 # investment in Rs\n",
+    "f2 = 12000 # annual payment in Rs\n",
+    "p3 = (f2*((1+i)**5-1))/(i*(1+i)**5) # annual payments in Rs\n",
+    "t2 = I + p3 # total cash out flows in Rs\n",
+    "print \"Total cash in flows = Rs %0.2f\\nTotal cash out flows = Rs %0.2f\"%(t1 ,t2) \n",
+    "print \"Since cash out flows are more than cash in flows therefore project is not economical\"\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.4 : page 252"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Net P.W.of Machine A= Rs -115293.66\n",
+      " Net P.W.of Machine B = Rs-129345.49\n",
+      " It is clear that Net P.W of Machine A is less nagative as compared to that of Machine B , therefore Machine A is economcal.\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Machine A\n",
+    "f1 = 2000 # annual benefit from better production quality in Rs\n",
+    "i = 10 # interest rate\n",
+    "i = 10/100\n",
+    "f2 = 12000 # salvage value in Rs\n",
+    "f3 = 8000 # operating and maintenance cost in Rs\n",
+    "I1 = 100000 # initial cost in Rs\n",
+    "n = 5 # years\n",
+    "p1 = (f1*((1+i)**n-1))/(i*(i+1)**n)\n",
+    "p2 = f2/(1+i)**n\n",
+    "c1 = p1 + p2 # cash in flows in Rs\n",
+    "p3 = (f3*((1+i)**n-1))/(i*(i+1)**n)\n",
+    "c2 = I1 + p3 # cash out flows in Rs\n",
+    "Pa = c1 - c2 # net P.W.in Rs\n",
+    "#Machine B\n",
+    "I2 = 60000 # initial cost in Rs\n",
+    "f4 = 16000 # operating and maintenance cost in Rs\n",
+    "f5 = 14000 # reconditioning at the end of third year in Rs\n",
+    "p4 = (16000*((1+i)**5-1))/(i*(1+i)**5)\n",
+    "p5 = f5/(1+i)**5\n",
+    "Pb = -I2 - p4 - p5 # net P.W.in Rs\n",
+    "print \" Net P.W.of Machine A= Rs %0.2f\\n Net P.W.of Machine B = Rs%0.2f\"%(Pa ,Pb)\n",
+    "print \" It is clear that Net P.W of Machine A is less nagative as compared to that of Machine B , therefore Machine A is economcal.\"\n",
+    "# Answers vary due to round off erro"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.5 : page 254"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " P.W. of expenses for A = Rs 89421\n",
+      " P.W. of expenses for B = Rs 79441.68\n",
+      " As the expenses of machine B are less , so this is economical\n"
+     ]
+    }
+   ],
+   "source": [
+    "#machine A\n",
+    "c1 = 20000 # manual cost in Rs\n",
+    "c2 = 40000 # operating cost in Rs\n",
+    "n1 = 2 # machine life in years\n",
+    "i = 10 # interest rate \n",
+    "i = 10/100\n",
+    "crf1 = ((1+i)**n1-1)/(i*(i+1)**n1) # capital recovery factor\n",
+    "pw1 = c1+c2*crf1 # present worth in Rs\n",
+    "# machine B\n",
+    "c3 = 50000 # manual cost in Rs\n",
+    "c4 = 30000 # operating cost in Rs\n",
+    "n2 = 4 # machine life in years\n",
+    "i = 10/100 # interest rate \n",
+    "crf2 = ((1+i)**n2-1)/(i*(i+1)**n2) # capital recovery factor\n",
+    "pw2 = c3+c4*crf2 # present worth in Rs for 4 years\n",
+    "pw3 = (pw2*crf1)/crf2 # present worth in Rs for 2 years\n",
+    "print \" P.W. of expenses for A = Rs %d\\n P.W. of expenses for B = Rs %0.2f\" %(pw1,pw3)\n",
+    "print \" As the expenses of machine B are less , so this is economical\"\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.6 : page 255"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Annual cost of machine A = Rs 16303.12\n",
+      " Annual cost of Machine B = Rs 15841.48\n",
+      " Machine B will be economical\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Machine A\n",
+    "i = 8 # # interest rate\n",
+    "i = i/100 # interest rate\n",
+    "n1 = 10 # economic life in years\n",
+    "CRF1 = i*(1+0.08)**n1/((1+i)**n1-1) # capital recovery factor\n",
+    "p1 = 46000 # first cost in Rs\n",
+    "s1 = 8000 # salvage value in Rs\n",
+    "o1 = 10000 # operating charges in Rs\n",
+    "AC1 = (p1-s1)*CRF1 + s1*i + o1 # annual cost in Rs\n",
+    "#Machine B\n",
+    "n2 = 15 # economic life in years\n",
+    "CRF2 = i*(1+0.08)**n2/((1+i)**n2-1) # capital recovery factor\n",
+    "p2 = 60000 # first cost in Rs\n",
+    "s2 = 10000 # salvage value in Rs\n",
+    "o2 = 9200 # operating charges in Rs\n",
+    "AC2 = (p2-s2)*CRF2 + s2*i + o2 # annual cost in Rs\n",
+    "print \" Annual cost of machine A = Rs %0.2f\\n Annual cost of Machine B = Rs %0.2f\" %(AC1, AC2 )\n",
+    "print \" Machine B will be economical\"\n",
+    "# Error in textbook"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.7 : page 257"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " ERR = 2.5813\n",
+      " Internal rate of return = 20.88 percent\n",
+      " Since Internal rate of return is > M.A.R.R , therefore project is feasible\n"
+     ]
+    }
+   ],
+   "source": [
+    "a = 100000 # Ej(p/f,e%,j) in Rs\n",
+    "n = 5 # life in years\n",
+    "e = 20 # M.A.R.R.\n",
+    "e = e/100 # M.A.R.R.\n",
+    "i = e\n",
+    "A = 32000 # savings in Rs\n",
+    "s = 20000 # salvage value in Rs\n",
+    "b = ((A*(((i+1)**n)-1)/i)+s)/a # (F/p,I,5)\n",
+    "i2 = (b)**(1/n)-1 # internal rate of return\n",
+    "print \" ERR = %0.4f\\n Internal rate of return = %0.2f percent\"%(b , i2*100)\n",
+    "print \" Since Internal rate of return is > M.A.R.R , therefore project is feasible\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.9 : page 259"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "ERRR = 21.25 percent\n",
+      "Since E.R.R.R is > M.A.R.R. therefore project is feasible.\n"
+     ]
+    }
+   ],
+   "source": [
+    "e = 20 # M,A.R.R.\n",
+    "i = e # interest rste\n",
+    "i = i/100 \n",
+    "n = 5 # life in years\n",
+    "s = 32000 # annual net savings in Rs\n",
+    "p = 100000 # present worth in Rs\n",
+    "S = 20000 # salvage value in Rs\n",
+    "a = (p-S)*(i/((1+i)**n-1)) # (p-s)(A/F,e%,n)\n",
+    "E = (s-a)/p # E.R.R.R\n",
+    "print \"ERRR = %0.2f percent\"%(E*100)\n",
+    "print \"Since E.R.R.R is > M.A.R.R. therefore project is feasible.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.10 : page 259"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " ERRR1 = 7.24 percent \n",
+      " ERRR2 = 8.87 percent\n",
+      " Only machine B is accepteble\n"
+     ]
+    }
+   ],
+   "source": [
+    "# machine A\n",
+    "r_e1 = 9600 #cash flow in Rs\n",
+    "p1 = 46000 # intial cost in Rs\n",
+    "s = 0 # salvage value\n",
+    "e = 8 # M.A.R.R\n",
+    "e = e/100\n",
+    "i = 8 # investment rate\n",
+    "i = i/100\n",
+    "n = 6 # life in years\n",
+    "x = i/((1+i)**n-1) \n",
+    "ERRR1 = (r_e1 - (p1-s)*x)/p1\n",
+    "#machine B\n",
+    "r_e2 = 7200 #cash flow in Rs\n",
+    "p2 = 32000 # intial cost in Rs \n",
+    "ERRR2 = (r_e2 - (p2-s)*x)/p2\n",
+    "print \" ERRR1 = %0.2f percent \\n ERRR2 = %0.2f percent\"%(ERRR1*100 ,ERRR2*100)\n",
+    "print \" Only machine B is accepteble\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.11 : page 261"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Investment cost = Rs 21000\n",
+      " Unamortized investment = Rs 15000\n"
+     ]
+    }
+   ],
+   "source": [
+    "pmv = 15000 # present market value in Rs\n",
+    "ss = 6000 # sum needed to make it serviceable in Rs\n",
+    "ic = ss + pmv #  investment cost in Rs\n",
+    "pbv = 30000 # present book value in Rs\n",
+    "sv = 15000 # salvage value in Rs\n",
+    "ui = pbv - sv # unamortized investment in Rs\n",
+    "print \" Investment cost = Rs %d\\n Unamortized investment = Rs %d\"%(ic , ui)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.13 : page 262"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Existing machine = Rs 87931.647 \n",
+      " Proposed machine = Rs 72549.08\n",
+      " Since the equivalent annual cost of proposed machine is less than that of the existing machine , therefore, the replacement is justified.\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Existing machine\n",
+    "pmp = 100000 # present market price in Rs\n",
+    "io = 50000 # immediate overhauling in Rs\n",
+    "asl = 5 # additional service life in years\n",
+    "aoc = 50000 # annual operating cost in Rs\n",
+    "svo = 10000 # salvage value after overhauling in Rs\n",
+    "pc = io + pmp # present cost in Rs\n",
+    "i = 10 # interest rate\n",
+    "i = 10/100\n",
+    "crf1 = (i*(1+i)**asl)/((1+i)**asl - 1) # capital recovery factor\n",
+    "AC1 = (pc - svo)*crf1 + svo*i + aoc # average cost in Rs\n",
+    "# proposed machine\n",
+    "n = 10 # expected economic life in years\n",
+    "ic = 300000 # initial cost in Rs\n",
+    "sv = 100000 # salvage value in Rs\n",
+    "o = 30000 # annual operating cost in Rs\n",
+    "crf2 = (i*(1+i)**10)/((1+i)**10 - 1)\n",
+    "AC2 = (ic - sv)*crf2 + sv*i + o # average cost in Rs\n",
+    "print \" Existing machine = Rs %0.3f \\n Proposed machine = Rs %0.2f\"%(AC1 , AC2)\n",
+    "print \" Since the equivalent annual cost of proposed machine is less than that of the existing machine , therefore, the replacement is justified.\"\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.15 : page 265"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Number of years of economic repair life = 14.53 years\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "c = 20000 # first cost of machine in Rs\n",
+    "s = 1000 # scrap value in machine in Rs\n",
+    "b = 180 # annual increase in cost of repairs in Rs\n",
+    "n = sqrt(2*(c-s)/b) # years\n",
+    "print \"Number of years of economic repair life = %0.2f years\"%(n)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.16 : page 266"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The number of years in which the new machine will pay for itself = 3.945 years\n"
+     ]
+    }
+   ],
+   "source": [
+    "Cn = 72000 # cost of new machine installed and tooled in Rs\n",
+    "Co = 28000 # cost of new machime installed and tooled in Rs\n",
+    "p = 16 # hourly pieces\n",
+    "Nn = 2200*p # estimated annual production on new machine \n",
+    "Ko = 17200 # present book value of old machine in Rs\n",
+    "So = 6400 # scrap value of old machine in Rs\n",
+    "Sn = 8000 # probable scrap value of old machine in at the end of its useful life Rs\n",
+    "oco = 2.5 # opreator cost per hour \n",
+    "mco= 48 # machine cost\n",
+    "ro = 10 # production rate per hour\n",
+    "ocn = 2 # opreator cost per hour \n",
+    "mcn= 62 # machine cost\n",
+    "rn = 16 # production rate per hour\n",
+    "Po = (oco+mco)/ro # labour and machine cost per unit on old machine in Rs\n",
+    "Pn = (ocn+mcn)/rn # labour and machine cost per unit on new machine in Rs\n",
+    "i = 6 # interest on investment\n",
+    "i = i/100\n",
+    "t = 6 # annual taxes\n",
+    "t = t/100\n",
+    "d = 10 # annual allowance for depreciation \n",
+    "d = d/100\n",
+    "m = 3 # annual allowance for maintenance\n",
+    "m = m/100\n",
+    "n = ((Cn-Sn)+(Ko-So))/((Nn*(Po-Pn)) - Cn*(i+t+d+m))\n",
+    "print \"The number of years in which the new machine will pay for itself = %0.3f years\"%(n)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.17 : page 267"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Unit production cost on turret lathe = Rs 9.18 per piece\n",
+      " Unit production cost on engine lathe = Rs 14.55 per piece\n",
+      " Turret lathe will be more economical than two engine lathe\n"
+     ]
+    }
+   ],
+   "source": [
+    "C = 80000 # cost of new machine installed and tooled in Rs\n",
+    "nel = 2 # number of engine lathes\n",
+    "c = 32000*nel # first cost of engine lathe\n",
+    "N = 4000 # annual production of turret lathe\n",
+    "n = 3800 # annual production in engine lathe\n",
+    "nhp1 = 4 # hp motor\n",
+    "L = 2256*nhp1 # annual labour cost of turret lathe\n",
+    "w = 5 # wage in per hour\n",
+    "time = 2300 # hours\n",
+    "l = time*nel*w # labour cost of engine lathe\n",
+    "nhp2 = 2.5 # hp motor\n",
+    "pr = 0.35 # power rate in kwh\n",
+    "p = (nel*nhp2*746*time*pr)/1000 # power cost\n",
+    "P = (nhp1*746*time*pr)/1000 # power cost\n",
+    "F = 480 #saving\n",
+    "I  = 6/100 # interest rate\n",
+    "T = 4/100 # tax rate\n",
+    "D = 10/100 # allowance for depreciation in engine lathe\n",
+    "M = 6/100 # allowance for maintenance in engine lathe\n",
+    "B = 55/100 # labour burden in engine lathe\n",
+    "i  = 6/100 # interest rate\n",
+    "t = 4/100 # tax rate\n",
+    "d = 10/100 # allowance for depreciation in turret lathe\n",
+    "m = 6/100 # allowance for maintenance in turret lathe\n",
+    "X = (L + B*L + P +C*(I+T+D+M) - F)/N\n",
+    "x = (l+l*B + p + c*(i+t+d+m))/n\n",
+    "print \" Unit production cost on turret lathe = Rs %0.2f per piece\\n Unit production cost on engine lathe = Rs %0.2f per piece\"%(X , x)\n",
+    "print \" Turret lathe will be more economical than two engine lathe\"\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.18 : page 268"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Amount invested in turret lathe over the cost of engine lathe = Rs 79562\n"
+     ]
+    }
+   ],
+   "source": [
+    "X = 9.16 # production cost on turret lathe\n",
+    "N = 4000 # annual requirement\n",
+    "c = X*N # cost for 4000 pieces on turret lathe\n",
+    "n = 3800 # production of engine lathe\n",
+    "l = 23000 # labour cost\n",
+    "p = 3002 # power cost\n",
+    "i = 6 # interest rate\n",
+    "i = i/100\n",
+    "t = 4 # tax rate\n",
+    "t = t/100\n",
+    "d = 10 # allowance for depreciation in turret lathe\n",
+    "d = d/100\n",
+    "m = 6 # allowance for maintenance in turret lathe\n",
+    "m = m/100\n",
+    "b = 55/100 #labour burden\n",
+    "a = i+t+d+m \n",
+    "tc = 64000 # first cost of engine lathe\n",
+    "c1 =(N*(l*(1+b)+p))/n+(tc*a) # cost for engine lathe\n",
+    "s = c1-c # savings \n",
+    "amt = s/a # amount invested in turret lathe over the cost of engine lathe\n",
+    "print \"Amount invested in turret lathe over the cost of engine lathe = Rs %d\"%(amt)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.19 : page 268"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Years in which new machine will pay for itself = 1.40 years\n"
+     ]
+    }
+   ],
+   "source": [
+    "Cn = 60000 # cost of new machine\n",
+    "Sn = 5000 # scrap value of new machine\n",
+    "So = 1000 # scrap value of old machine\n",
+    "Nn = 200000 #annual production\n",
+    "I = 10 # interest rate\n",
+    "I = I/100\n",
+    "M = 7 # allowance for maintenane\n",
+    "M = M/100\n",
+    "T = 6 # annual taxes\n",
+    "T = T/100\n",
+    "D = 1/10 # allowance for depreciation\n",
+    "lco = 300 # labour charges for old machine\n",
+    "m = 12 # months\n",
+    "rco = 15000 # running charges for old machine\n",
+    "pro = 50000 # production rate for old machine\n",
+    "lcn = 500 # labour charges for new machine\n",
+    "rcn = 10000 # running charges for old machine\n",
+    "prn = 200000# production rate f\n",
+    "Po = (lco*m + rco)/pro # labour and machine cost on old machine\n",
+    "Pn = (lcn*m + rcn)/prn # labour and machine cost on new machine\n",
+    "n =((Cn-Sn)-So)/((Nn*(Po-Pn))-Cn*(I+T+D+M)) #years\n",
+    "print \"Years in which new machine will pay for itself = %0.2f years\"%(n)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.20 : page 270"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Number of pieces when one run is made and cost is Rs 3000 = 886 pieces\n",
+      "Annual production when 5 runs are made per year = 972 pieces\n",
+      "Annual production when fixture pay for itself = 1531 pieces\n",
+      "Economical investment when 1530 pieces for single run with savings Rs 1.50 per piece = Rs 2997\n",
+      "Annual profit when 950 pieces made per year in 6 runs and saving in labour cost Rs 2 per piece = Rs 635 per year\n"
+     ]
+    }
+   ],
+   "source": [
+    "a = 1.50 #saving in labour\n",
+    "b=55/100 # burden applied on labour\n",
+    "T = 4/100 # allowance for taxes\n",
+    "M = 5/100 # allowance for maintenance\n",
+    "I = 8/100 # interest rate\n",
+    "D = 50/100 # allowance for depreciation\n",
+    "H = 2 # years to amortize the investment\n",
+    "S = 50 # yearly cost for set up\n",
+    "C = 3000 # first cost\n",
+    "N1 = (C*(I+T+M+D)+S)/(a*(1+b)) # annual production when 1 run is made\n",
+    "r = 5 # number of runs\n",
+    "N2 = (C*(I+T+M+D)+S*r)/(a*(1+b)) # annual production when 1 run is made\n",
+    "D1 = 100/100 # allowance for depreciation\n",
+    "N3 = (C*(I+T+M+D1)+S)/(a*(1+b)) # production when D = 100\n",
+    "n1 = 1530 # pieces\n",
+    "C1 = (n1*(a*(1+b))-S)/(I+T+M+D1) # economical investment\n",
+    "n2 = 950 # pieces\n",
+    "a1 = 2 # labour cost\n",
+    "r1 = 6 # number of runs\n",
+    "S1 = r1*S # yearly cost\n",
+    "V = n2*a1*(1+b)-C*(I+T+M+D)-S1 #profit\n",
+    "print \"Number of pieces when one run is made and cost is Rs 3000 = %d pieces\"%(N1)\n",
+    "print \"Annual production when 5 runs are made per year = %d pieces\"%(N2)\n",
+    "print \"Annual production when fixture pay for itself = %d pieces\"%(N3)\n",
+    "print \"Economical investment when 1530 pieces for single run with savings Rs 1.50 per piece = Rs %d\"%(C1)\n",
+    "print \"Annual profit when 950 pieces made per year in 6 runs and saving in labour cost Rs 2 per piece = Rs %d per year\"%(V)\n",
+    "#  'Answers vary due to round off error'\n",
+    " "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.21 : page 271"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total number of pieces produced = 20114\n"
+     ]
+    }
+   ],
+   "source": [
+    "a = 0.125 #saving in labour cost per unit\n",
+    "b = 0.4 # overhead applied on direct labour saved\n",
+    "D = 1/2 # allowance for depreciation\n",
+    "C = 2400 # first cost\n",
+    "I = 6/100 # interest rate\n",
+    "T = 4/100 # allowance for taxes\n",
+    "M = 10/100 # allowance for maintenance\n",
+    "S = 80 # cost of set up\n",
+    "N = (C*(I+T+D+M)+S)/(a*(1+b)) # pieces per year\n",
+    "t = N*2 # total number of pieces\n",
+    "print \"Total number of pieces produced = %d\"%(t)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.22 : page 271"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total number of pieces produced = 24685\n"
+     ]
+    }
+   ],
+   "source": [
+    "a = 0.125 # saving in labour cost per unit\n",
+    "b = 0.4 # overhead applied on direct labour saved\n",
+    "D = 1/2 # allowance for depreciation\n",
+    "C = 2400 # first cost\n",
+    "I = 6/100 # interst rate\n",
+    "T = 4/100 # allowance for taxes\n",
+    "M = 10/100 # allowance for maintenance\n",
+    "n = 6 # number of baches\n",
+    "S = 80 # cost of set up\n",
+    "s1 = S*n # total set up cost\n",
+    "N = (C*(I+T+D+M)+s1)/(a*(1+b)) # pieces\n",
+    "t = N*2 # total number of pieces\n",
+    "print \"Total number of pieces produced = %d\"%(t)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.23 : page 271"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Number of years taken by fixture of Rs 2000 = 5.33 years\n",
+      " profit made when fixture of Rs 1600 = Rs 175\n"
+     ]
+    }
+   ],
+   "source": [
+    "C1 = 2000 # first cost  small tool in Rs\n",
+    "N = 5000 # parts per year\n",
+    "n = 5 # number of batches\n",
+    "S = 50*n # cost of set up\n",
+    "a = 0.15 # saving in labour cost per unit\n",
+    "b = 50/100 # burden applied on direct labour saved\n",
+    "I = 10/100 # interest rate\n",
+    "T = 5/100 # allowance for tax\n",
+    "M = 10/100 # allowance for maintenance\n",
+    "H = C1/((N*a*(1+b))-(C1*(I+T+M))-S) # years\n",
+    "C2 = 1600 # cost of fixture\n",
+    "D = 1/H # allowance for depreciation\n",
+    "V = N*a*(1+b)-C2*(I+T+D+M)-S # profit\n",
+    "print \" Number of years taken by fixture of Rs 2000 = %0.2f years\\n profit made when fixture of Rs 1600 = Rs %d\"%(H ,V)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.24 : page 272"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Minimum number of components to be produced if cost of fixture to be recovered = 500\n"
+     ]
+    }
+   ],
+   "source": [
+    "c1 = 3 #  machine cost per component using existing euipment in Rs\n",
+    "c2 = 1 # machine cost using fixture in Rs\n",
+    "s = c1 - c2 # saving in machine cost per piece\n",
+    "f= 1000 # cost of fixture in Rs\n",
+    "N = f/2 # components\n",
+    "print \"Minimum number of components to be produced if cost of fixture to be recovered = %d\"%(N) "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.25 : page 272"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Number of pieces which must be produced to break even so that fixture may pay for itself = 7174 pieces per year\n"
+     ]
+    }
+   ],
+   "source": [
+    "C = 1000 # cost of fixture\n",
+    "Co = 700 # cost of old fixture\n",
+    "Cs = 250 # scrap value\n",
+    "a = 10 #saving per piece in paisa\n",
+    "a = a/100\n",
+    "b = 30 # overhead applied on direct labour saved\n",
+    "b = b/100\n",
+    "I = 8 # interest rate\n",
+    "I = I/100\n",
+    "M = 3 # allowance for maintenance\n",
+    "M = M/100\n",
+    "T = 12 # allowance for tax\n",
+    "T = T/100\n",
+    "H = 3/2 # amortization\n",
+    "D = 1/H # allowance for depreciation\n",
+    "N = (C*(I+T+D+M)+(Co-Cs)*I)/(a*(1+b)) # pieces per year\n",
+    "print \"Number of pieces which must be produced to break even so that fixture may pay for itself = %d pieces per year\"%(N)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.26 : page 273"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Cost for new fixture = Rs 1264\n"
+     ]
+    }
+   ],
+   "source": [
+    "N = 9000 # number of pieces\n",
+    "Co = 700 # cost of old fixture\n",
+    "Cs = 250 # scrap value\n",
+    "a = 10 #saving per piece in paisa\n",
+    "a = a/100\n",
+    "b = 30 # overhead applied on direct labour saved\n",
+    "b = b/100\n",
+    "I = 8 # interest rate\n",
+    "I = I/100\n",
+    "M = 3 # allowance for maintenance\n",
+    "M = M/100\n",
+    "T = 12 # allowance for tax\n",
+    "T = T/100\n",
+    "H = 3/2 # amortization\n",
+    "D = 1/H # allowance for depreciation\n",
+    "C = (N*a*(1+b)-(Co-Cs)*I)/(I+T+D+M) # cost in Rs\n",
+    "print \"Cost for new fixture = Rs %d\"%(C)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.27 : page 273"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Time taken to amortize the fixture = 2.7 years\n"
+     ]
+    }
+   ],
+   "source": [
+    "n = 6500 # yearly production\n",
+    "c = 1350 # cost of fixture\n",
+    "a = 10 #saving per piece in paisa\n",
+    "a = a/100\n",
+    "b = 30 # overhead applied on direct labour saved\n",
+    "b = b/100\n",
+    "I = 8 # interest rate\n",
+    "I = I/100\n",
+    "M = 3 # allowance for maintenance\n",
+    "M = M/100\n",
+    "T = 12 # allowance for tax\n",
+    "T = T/100\n",
+    "co = 700 # cost of old fixture\n",
+    "cs = 250 # scrap value\n",
+    "H = (c)/((n*a*(1+b))-I*(co-cs)-c*(I+T+M)) # amotization in years\n",
+    "print \"Time taken to amortize the fixture = %0.1f years\"%(H)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.28 : page 273"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 25,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "profit = Rs 237 \n"
+     ]
+    }
+   ],
+   "source": [
+    "n = 9000 # production of pieces per year\n",
+    "c = 1000 # fixture costs\n",
+    "Co = 700 # cost of old fixture\n",
+    "Cs = 250 # scrap value\n",
+    "a = 10 #saving per piece in paisa\n",
+    "a = a/100\n",
+    "b = 30 # overhead applied on direct labour saved\n",
+    "b = b/100\n",
+    "I = 8 # interest rate\n",
+    "I = I/100\n",
+    "M = 3 # allowance for maintenance\n",
+    "M = M/100\n",
+    "T = 12 # allowance for tax\n",
+    "T = T/100\n",
+    "h = 1.5 # amortization\n",
+    "D = 1/h # allowance for depreciation\n",
+    "V = (n*a*(1+b))-(c*(I+T+D+M))-((Co-Cs)*I) # profit\n",
+    "print \"profit = Rs %d \"%(V) \n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.29 : page 274"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Break even point = 1000 pieces \n",
+      " If fixed cost is 125000 and variable cost is Rs 90 per unit then break even point = 1136 pieces\n",
+      " Number of components to get profit of Rs 20000 = 1200 pieces\n"
+     ]
+    }
+   ],
+   "source": [
+    "fc1 = 100000 # fixed cost in Rs\n",
+    "vc1 = 100 # variable cost in Rs per unit\n",
+    "sp = 200 # selling price in Rs per unit\n",
+    "q1 = fc1/(sp-vc1) # quantity of production at break even point\n",
+    "fc2 = 125000 # fixed cost in Rs\n",
+    "vc2 = 90 # variable cost in Rs per unit\n",
+    "q2 = fc2/(sp-vc2) # quantity of production at break even point\n",
+    "p = 20000 # profit in Rs\n",
+    "q3 = (fc1 + p)/(sp-vc1) # quantity of production at profit of Rs 20000\n",
+    "print \" Break even point = %d pieces \\n If fixed cost is 125000 and variable cost is Rs 90 per unit then break even point = %d pieces\\n Number of components to get profit of Rs 20000 = %d pieces\"%(q1 , q2 ,q3)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.30 : page 275"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 27,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Break even point = 7500 pieces\n"
+     ]
+    }
+   ],
+   "source": [
+    "fc1 = 12000 # fixed cost for machine A in Rs\n",
+    "fc2 = 48000 # fixed cost for machine B in Rs\n",
+    "n1 = 6 # unit production cost in Rs per piece for machine A\n",
+    "n2 = 1.2 # unit production cost in Rs per piece for machine B\n",
+    "q = (fc2-fc1)/(n1-n2) # break even point\n",
+    "print \"Break even point = %d pieces\"%(q)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.31 : page 275"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 28,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Break even quantity for a component which can be produced on either the capstan lathe or single spindle automatic = 1373 pieces\n"
+     ]
+    }
+   ],
+   "source": [
+    "# capstan lathe\n",
+    "tc1 = 300 # total cost in Rs\n",
+    "mc1 = 2.5 # material cost per piece in Rs\n",
+    "olc1 = 5 # operation labour cost per hour in Rs\n",
+    "ct1 = 5 # cycle time per piece in min.\n",
+    "slc1 = 20 # setting up labour cost in Rs per hour\n",
+    "st1 = 1 # setting up time in hour\n",
+    "mo1 = 300/100 # machine over heads of operation labour cost\n",
+    "o1 = mo1*olc1 # overheads of capstan lathe in Rs per hour\n",
+    "fc1 = tc1 + slc1*st1 + o1*st1 # fixed cost of capstan lathe in Rs\n",
+    "vc1 = mc1 + (olc1*ct1)/60 + (o1*ct1)/60 # variable cost in Rs\n",
+    "# Automatic (single spindle) \n",
+    "tc2 = 300 # total cost in Rs\n",
+    "cc2 = 1500 # cost of cams in Rs\n",
+    "mc2 = 2.5 # material cost per piece in Rs\n",
+    "olc2 = 2 # operation labour cost per hour in Rs\n",
+    "ct2 = 1 # cycle time per piece in min.\n",
+    "slc2 = 20 # setting up labour cost in Rs per hour\n",
+    "st2 = 8 # setting up time in hour\n",
+    "mo2 = 1000/100 # machine over heads of operation labour cost\n",
+    "o2 = mo2*olc2 # overheads of single spindle in Rs per hour\n",
+    "fc2 = tc2 + cc2 + slc2*st2 + o2*st2 # fixed cost of single spindle in Rs\n",
+    "vc2 = mc2 + (olc2*ct2)/60 + (slc2)/60 # variable cost in Rs\n",
+    "q = (fc2-fc1)/(vc1-vc2) # break even quantity\n",
+    "print \" Break even quantity for a component which can be produced on either the capstan lathe or single spindle automatic = %d pieces\"%(q)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.32 : page 277"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 29,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Break even point = 43 pieces\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Engine lathe\n",
+    "t = 12 # time/piece in min.\n",
+    "l = 7 # overhead cost/hr\n",
+    "o = 4 # direct labour cost/hr\n",
+    "s = 2 # set up time in hour\n",
+    "sr = 8 # set up rate per \n",
+    "# turret lathe\n",
+    "T = 5 # time/piece in min.\n",
+    "L = 5 # overhead cost/hr\n",
+    "O = 8 # direct labour cost/hr\n",
+    "S = 8 # set up time in hour\n",
+    "SR = 8 # set up rate per \n",
+    "q = 60*(S*SR-s*sr)/(t*(l+o)-T*(L+O)) # break even point\n",
+    "q = round(q)\n",
+    "print \"Break even point = %d pieces\"%(q)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.33 : page 277"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 30,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Break even point = 384000 pieces\n"
+     ]
+    }
+   ],
+   "source": [
+    "from sympy import symbols, solve\n",
+    "fc1 = 80000 # fixed cost for turret lathe in Rs\n",
+    "fc2 = 32000 # fixed cost for engine lathe in Rs\n",
+    "n1 = 16 # production of pieces per year in turret lathe\n",
+    "n2 = 10 # production of pieces per year in engine lathe\n",
+    "vc1 = 2 #operators cost in turret lathe\n",
+    "vc2 = 2.5 # operators cost in engine lathe\n",
+    "Q=symbols('Q')\n",
+    "expr = (fc1+1/n1*vc1*Q)-(fc2+2.5*Q/10)\n",
+    "Q=solve(expr,Q)[0]\n",
+    "print \"Break even point = %d pieces\"%(Q)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.34 : page 277"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 31,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The point at which the automatic lathe will be justified = 5.56 \n"
+     ]
+    }
+   ],
+   "source": [
+    "st1 = 15 # set up time for engine lathe in min.\n",
+    "ut1 = 15 # unit time for engine lathe in min.\n",
+    "st2 = 90 # set up time for automatic lathe in min.\n",
+    "ut2 = 1.5 # unit time for engine lathe in min.\n",
+    "q = (st2-st1)/(ut1-ut2) # quantity of production\n",
+    "print \"The point at which the automatic lathe will be justified = %0.2f \"%(q)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.35 : page 278"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 32,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Quantity of pieces at Break even point = 86 pieces\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Automatic lathe\n",
+    "p = 30 # number of pieces produced per hour \n",
+    "l = 4 # labour rate per hour in Rs\n",
+    "d = 4.50 # hourly depreciation rate per machine in hour\n",
+    "s = 4 # set up time in hour\n",
+    "# turret lathe\n",
+    "P = 10 # number of pieces produced per hour \n",
+    "L = 4 # labour rate per hour in Rs\n",
+    "D = 1.50 # hourly depreciation rate per machine in hour\n",
+    "S = 2 # set up time in hour\n",
+    "q = (P*p*(S*L+S*D-s*l-s*d))/(P*(l+d)-p*(L+D)) # quantity of pieces at break even point\n",
+    "print \"Quantity of pieces at Break even point = %d pieces\"%(q)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.36 : page 279"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 33,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Quantity of production at break even point = 760 pieces\n"
+     ]
+    }
+   ],
+   "source": [
+    "Pa = 8.4 # unit tool process cost for method A in Rs\n",
+    "Pb = 14.8 # unit tool process cost for method B in Rs\n",
+    "Ta = 6480 #total tool cost for method A in Rs \n",
+    "Tb = 1616 #total tool  cost for method B in Rs\n",
+    "q = (Ta-Tb)/(Pb-Pa) # break even point\n",
+    "print \"Quantity of production at break even point = %d pieces\"%(q) "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.37 : page 283"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 34,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(i) Cost per unit for machine A = Rs 2.75\n",
+      " Cost per unit machine B = Rs 1.75\n",
+      "machine B will be preferred\n",
+      "(ii) Cost per unit for machine A = Rs 3.88\n",
+      " Cost per unit machine B = Rs 4.00\n",
+      "machine A will be preferred\n",
+      "(iii) Annual production to make cost per piece equal for two machines = 4500 pieces\n"
+     ]
+    }
+   ],
+   "source": [
+    "# machine A\n",
+    "ic1 = 50000 # initial cost\n",
+    "hoc1 = 10 # hourly operating charges\n",
+    "pp1= 5 # pieces produced per hour\n",
+    "i = 15 # interest rate\n",
+    "i = i/100\n",
+    "oh = 2000 # operating hours\n",
+    "fc1 = ic1*i # fixed cost\n",
+    "vc1 = oh*hoc1 # variable cost\n",
+    "tc1 = fc1+vc1 # total charges\n",
+    "ao1 = oh*pp1 # annual output\n",
+    "c1 = tc1/ao1 # cost per unit\n",
+    "# machine B\n",
+    "ic2 = 80000 # initial cost\n",
+    "hoc2 = 8 # hourly operating charges\n",
+    "pp2= 8 # pieces produced per hour\n",
+    "fc2 = ic2*i # fixed cost\n",
+    "vc2 = oh*hoc2 # variable cost\n",
+    "tc2 = fc2+vc2 # total charges\n",
+    "ao2 = oh*pp2 # annual output\n",
+    "c2 = tc2/ao2 # cost per unit\n",
+    "print \"(i) Cost per unit for machine A = Rs %0.2f\\n Cost per unit machine B = Rs %0.2f\"%(c1,c2)\n",
+    "print \"machine B will be preferred\"\n",
+    "# machine A\n",
+    "ao3 = 4000 # annual output\n",
+    "oc3 = ao3*hoc1/pp1 # operating charges \n",
+    "tc3 = oc3+fc1 # total annual charge\n",
+    "c3 = tc3/ao3 # cost/piece \n",
+    "# machine B\n",
+    "ao4 = 4000 # annual output\n",
+    "oc4 = ao4*hoc2/pp2 # operating charges \n",
+    "tc4 = oc4+fc2 # total annual charge\n",
+    "c4 = tc4/ao4# cost/piece\n",
+    "print \"(ii) Cost per unit for machine A = Rs %0.2f\\n Cost per unit machine B = Rs %0.2f\"%(c3,c4)\n",
+    "print \"machine A will be preferred\"\n",
+    "A = hoc1/pp1 # operating cost per piece on machine A\n",
+    "B = hoc2/pp2 # operating cost per piece on machine B\n",
+    "Q = fc2 - fc1 # annual production\n",
+    "print \"(iii) Annual production to make cost per piece equal for two machines = %d pieces\"%(Q )"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.38 : page 284"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 35,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Sales at break even point = 88000 units\n",
+      " Sales at net income of Rs9000 and corporate tax rate being 5.5 = Rs 102645.50\n",
+      " Sales per unit if B.E.P brought down to 10000 units = Rs 8.80 per unit\n"
+     ]
+    }
+   ],
+   "source": [
+    "As = 80000 # annual sales in Rs\n",
+    "vc = 64000 # variable expenses in Rs\n",
+    "c = 16000 # contribution in Rs\n",
+    "fc = 24000 # fixed expenses in Rs\n",
+    "l = 8000 # losses in Rs\n",
+    "p = 9000 # profit in Rs\n",
+    "s1 = fc + vc # sales at B.E.P in Rs\n",
+    "s2 = (fc + vc + p)/0.945 # sales at net income of Rs9000 and corporate tax rate being 5.5%\n",
+    "q = 10000 # quantity of units\n",
+    "sp = (fc+vc)/q # selling price per unit in Rs\n",
+    "print \" Sales at break even point = %d units\"%(s1 )\n",
+    "print \" Sales at net income of Rs9000 and corporate tax rate being 5.5 = Rs %0.2f\\n Sales per unit if B.E.P brought down to 10000 units = Rs %0.2f per unit\" %(s2 , sp)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.39 : page 285"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 36,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Break even point = 1000 pieces\n"
+     ]
+    }
+   ],
+   "source": [
+    "fc = 55000 # fixed cost in Rs\n",
+    "vc = 45 # variable cost per piece in Rs\n",
+    "sp = 100 # selling price per piece in Rs\n",
+    "p = (vc/sp)*100 # percentage of variable cost to \n",
+    "pm = 100 - p # profit margin\n",
+    "bep = ((55000/55)*100)/100 # Break even point\n",
+    "print \"Break even point = %d pieces\"%(bep)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.40 : page 288"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 37,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Total cost per piece for capstan lathe = Rs 5.99\n",
+      " Total cost per piece for turret lathe = Rs 7.47\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "f1 = 335 # fixed cost in Rs for capstan lathe\n",
+    "k = 0.25 # stock carrying factor in paise per piece\n",
+    "k = k/100\n",
+    "N1 = sqrt(f1/k) # pieces for capstan lathe\n",
+    "a1 = 4.16 # variable cost per piece for capstan lathe\n",
+    "tc1 = a1+f1/N1+k*N1 # total cost for capstan lathe\n",
+    "f2 = 2120 # fixed cost in Rs for turret lathe\n",
+    "N2 = sqrt(f2/k) # pieces for turret lathe \n",
+    "a2 = 2.863 # variable cost per piece for turret lathe\n",
+    "tc2 = a2+f2/N2+k*N2 # total cost for turret lathe\n",
+    "print \" Total cost per piece for capstan lathe = Rs %0.2f\\n Total cost per piece for turret lathe = Rs %0.2f\"%(tc1 , tc2)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.41 : page 289"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 38,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Economic order quantity = 2390 or 2400 units\n",
+      " Totl cost = Rs 41020 per year\n",
+      " I = 0.6944\n",
+      " It is clear that inventory cost will get increased very greatly\n"
+     ]
+    }
+   ],
+   "source": [
+    "from sympy import symbols, diff, solve\n",
+    "R=500 # cost of ordering in Rs per order  \n",
+    "A=12000 #annual consumption units\n",
+    "C=3.00 # unit cost of item\n",
+    "K=1.5 # unit storage cost\n",
+    "I1=0.2 # interest rate\n",
+    "N = symbols('N')\n",
+    "G =C*A+I1*C*N/2+K*N/2+A*R/N # total cost per year\n",
+    "expr=G.diff(N)\n",
+    "N=solve(expr, N)[1]\n",
+    "O = A/N # number of orders\n",
+    "N1 = 2400 # units\n",
+    "tc = C*A + I1*C*N1/2 + K*N1/2 + A*R/N1 # total cost in Rs\n",
+    "I2 = (2*R*A)/(C*N1**2) \n",
+    "print \" Economic order quantity = %d or %d units\\n Totl cost = Rs %d per year\\n I = %0.4f\"%(N,N1,tc,I2)\n",
+    "print \" It is clear that inventory cost will get increased very greatly\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.42 : page 291"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 39,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Total cost for lowest cost curve = Rs 300062.50\n",
+      " Total cost for next higher curve = Rs 306123.72\n",
+      " Total cost for highest curve = Rs 10320001.00 \n",
+      " Comparing all total cost lowest is Rs 300,062.50 for an order quantity of 10,000.\n",
+      " N = 10,000 and No. of orders = 4\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "A = 40000 # number of units per year\n",
+    "I = 25 # carrying cost in percent\n",
+    "I = I/100\n",
+    "C1 = 8 # cost for 0 < N < 1000 per unit in Rs\n",
+    "C2 = 7.5 # cost for 1000 < N < 10000 per unit in Rs\n",
+    "C3 = 7.25 # cost for N >= 10000 per unit in Rs\n",
+    "R = 250 # ordering cost per order in Rs\n",
+    "N = 10000 # units\n",
+    "N1 = sqrt(2*R*A/(I*C3)) # optimal quantity for lowest curve\n",
+    "G1= C3*A+(A*R)/N+I*C3*N/2 # total cost in Rs\n",
+    "N2 = sqrt(2*R*A/(I*C2)) # optimal quantity for higher curve\n",
+    "G2= C2*A+(A*R)/N2+I*C2*N2/2 # total cost in Rs\n",
+    "N3 = sqrt(2*R*A/(I*C1)) # optimal quantity for highest curve\n",
+    "G3 = C1*A+(A*R)+1 # total cost in Rs\n",
+    "print \" Total cost for lowest cost curve = Rs %0.2f\\n Total cost for next higher curve = Rs %0.2f\\n Total cost for highest curve = Rs %0.2f \"%(G1,G2,G3)\n",
+    "print \" Comparing all total cost lowest is Rs 300,062.50 for an order quantity of 10,000.\"\n",
+    "print \" N = 10,000 and No. of orders = 4\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 5.43 : page 291"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 40,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The lot size = 8333 components\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import ceil\n",
+    "from sympy import symbols, diff, solve\n",
+    "c = 50000 # components\n",
+    "R=500 # cost of ordering in Rs per order  \n",
+    "A=12000 #annual consumption units\n",
+    "C=3.00 # unit cost of item\n",
+    "K=1.50 # unit storage cost\n",
+    "I=0.2 # interest rate\n",
+    "\n",
+    "N = symbols('N')\n",
+    "G=0.02*N+1500000/N\n",
+    "y=G.diff(N)\n",
+    "N=solve(y, N)[1]\n",
+    "l = c/N # number of lots\n",
+    "l = ceil(l)\n",
+    "ls = c/l # lot size\n",
+    "print \"The lot size = %d components\"%ls"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter9_1.ipynb b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter9_1.ipynb
new file mode 100644
index 00000000..549b20dd
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/Chapter9_1.ipynb
@@ -0,0 +1,476 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 9 : Limits, Tolerances & Fits"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 9.1 : Page 376"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Hole tolerence = 0.02 mm\n",
+      " Shaft tolerence = 0.02 mm\n",
+      " Allowance = 0.03 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "h1 = 37.52 # high limit of hole in mm\n",
+    "h2 = 37.50 # low limit of hole in mm\n",
+    "s1 = 37.47 # high limit of shaft in mm\n",
+    "s2 = 37.45 # low limit of shaft in mm\n",
+    "ht = h1-h2 # hole tolerence in mm\n",
+    "st = s1-s2 # shaft tolerence in mm\n",
+    "a = h2-s1 # allowance in mm\n",
+    "print \" Hole tolerence = %0.2f mm\\n Shaft tolerence = %0.2f mm\\n Allowance = %0.2f mm\"%(ht ,st ,a)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 9.2 : Page 377"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " High limit of hole = 75.075 mm\n",
+      " High limit of shaft = 74.90 mm\n",
+      " Low limit of shaft = 74.825 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "t = 0.075 # tolerence in mm\n",
+    "h2 = 75 # low limit of hole in mm\n",
+    "a = 0.10 # allowance in mm\n",
+    "h1 = h2+t  # high limit of hole in mm\n",
+    "s1 = h2-a # high limit of shaft in mm\n",
+    "s2 = s1-t # low limit of shaft in mm\n",
+    "print \" High limit of hole = %0.3f mm\\n High limit of shaft = %0.2f mm\\n Low limit of shaft = %0.3f mm\"%(h1 ,s1 , s2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 9.3 : Page 377"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " High limit of hole = 75.225 mm\n",
+      " Low limit of shaft = 75.2625 mm\n",
+      " High limit of shaft = 75.4875 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "t = 0.225 # tolerence in mm\n",
+    "h2 = 75 # low limit of hole in mm\n",
+    "a = 0.0375 # interference in mm\n",
+    "h1 = h2+t  # high limit of hole in mm\n",
+    "s2 = h1+a # low limit of shaft in mm\n",
+    "s1 = s2+t # high limit of shaft in mm\n",
+    "print \" High limit of hole = %0.3f mm\\n Low limit of shaft = %0.4f mm\\n High limit of shaft = %0.4f mm\"%(h1 ,s2 , s1)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 9.4 : Page 378"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Tolerence for hole = 0.011 mm\n",
+      " Tolerence for shaft = 0.007 mm\n",
+      " Fundamental deviation for hole = 0.00 mm\n",
+      " Fundamental deviation for shaft = 0.001 mm\n",
+      " Low limit of hole = 60 mm\n",
+      " High limit of hole = 60.011 mm\n",
+      " High limit of shaft = 59.99 mm\n",
+      " Low limit of hole = 59.99 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "s1 = 50 # diameter of step1 in mm\n",
+    "s2 = 80 # diameter of step2 in mm\n",
+    "d = (s1*s2)**(1/2) # mm\n",
+    "i = (0.45*(d)**(1/3)+0.001*d)/10**3 # mm\n",
+    "t1 = 25*i # tolerence for hole in mm\n",
+    "t2 = 16*i # tolerence for shaft in mm\n",
+    "a1 = 0 # fundamental deviation for hole in mm\n",
+    "a2 = 5.5*(d)**0.41 # fundamental deviation for shaft in microns\n",
+    "a2 = a2/10**4 # mm\n",
+    "h1 = 60 # low limit of hole in mm\n",
+    "h2 = h1+t1 # high limit of tolerence in mm\n",
+    "s1 = h1 - t2 # high limit of shaft in mm\n",
+    "s2 = s1-t2 # low limit of shaft in mm\n",
+    "print \" Tolerence for hole = %0.3f mm\\n Tolerence for shaft = %0.3f mm\"%(t1 ,t2)\n",
+    "print \" Fundamental deviation for hole = %0.2f mm\\n Fundamental deviation for shaft = %0.3f mm\"%(a1 , a2 )\n",
+    "print \" Low limit of hole = %d mm\\n High limit of hole = %0.3f mm\\n High limit of shaft = %0.2f mm\\n Low limit of hole = %0.2f mm\"%(h1 ,h2 , s1 , s2)\n",
+    "# Answers vary due to round off error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 9.5 : Page 379"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Basic size = 30 mm\n",
+      " Shaft tolerence = 0.013 mm\n",
+      " Hole tolerence = 0.020 mm\n",
+      " High limit of shaft = 29.995 mm\n",
+      " Low limit of shaft = 29.982 mm\n",
+      " High limit of hole = 30.020 mm \n",
+      " Low limit of hole = 30.000 mm\n",
+      " Maximum clearance = 0.038 mm\n",
+      " Minimum clearance = 0.005 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "b = 30 # basic size in mm\n",
+    "s1 = 0.005 # maximum limit of shaft in mm\n",
+    "s2 = 0.018 # minimum limit of shaft in mm\n",
+    "h1 = 0.020 # maximum limit of hole in mm\n",
+    "h2 = 0.0 # minimum limit of hole in mm\n",
+    "t1 = s2-s1 # shaft tolerence in mm\n",
+    "t2 = h1-h2 # hole tolerence in mm\n",
+    "Sh = b-s1 # high limit of shaft in mm\n",
+    "Sl = b-s2 # low limit of shaft in mm\n",
+    "Hh = b+h1 # high limit of hole in mm\n",
+    "Hl = b+h2 # low limit of hole in mm\n",
+    "c1 = Hh-Sl # maximum clearance in mm\n",
+    "c2 = Hl-Sh # minimum clearance in mm\n",
+    "print \" Basic size = %d mm\\n Shaft tolerence = %0.3f mm\\n Hole tolerence = %0.3f mm\"%(b,t1,t2)\n",
+    "print \" High limit of shaft = %0.3f mm\\n Low limit of shaft = %0.3f mm\\n High limit of hole = %0.3f mm \\n Low limit of hole = %0.3f mm\"%(Sh,Sl,Hh,Hl)\n",
+    "print \" Maximum clearance = %0.3f mm\\n Minimum clearance = %0.3f mm\"%(c1,c2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 9.6 : Page 379"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Hole basis system \n",
+      " Lower limit of hole = 25 mm\n",
+      " Higher limit of hole = 25.006 mm\n",
+      " Higher limit of shaft = 24.990 mm \n",
+      " Lower limit of shaft = 24.986 mm\n",
+      " Shaft basis system \n",
+      " high limit of shaft = 25.000 mm\n",
+      " lower limit of shaft = 24.996 mm\n",
+      " lower limit of hole = 25.010 mm\n",
+      " upper limit of hole = 25.016 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "from sympy import symbols, solve\n",
+    "minc = 0.01 # minimum clearance in mm\n",
+    "bs = 25 # basic size in mm\n",
+    "maxc = 0.02 # maximum clearance in mm\n",
+    "x=symbols('x')\n",
+    "y=1.5*x\n",
+    "expr = y+0.01+x-0.02\n",
+    "x=solve(expr, x)[0]\n",
+    "y=1.5*x\n",
+    "# hole basis system\n",
+    "low_h1 = bs # low limit of hole in mm\n",
+    "high_h1 = bs+y # high limit of hole in mm\n",
+    "u_s = low_h1-minc # upper limit of shaft in mm\n",
+    "low_s1 = u_s-x # lower limit of shaft in mm\n",
+    "# shaft basis system\n",
+    "high_s = bs # high limit of shaft in mm\n",
+    "low_s2 = bs-x # low limit of shaft in mm\n",
+    "low_h2 = bs+minc # low limit of hole in mm\n",
+    "high_h2 = low_h2+y # high limit of hole in mm\n",
+    "print \" Hole basis system \\n Lower limit of hole = %d mm\\n Higher limit of hole = %0.3f mm\\n Higher limit of shaft = %0.3f mm \\n Lower limit of shaft = %0.3f mm\"%(low_h1,high_h1,u_s,low_s1)\n",
+    "print \" Shaft basis system \\n high limit of shaft = %0.3f mm\\n lower limit of shaft = %0.3f mm\\n lower limit of hole = %0.3f mm\\n upper limit of hole = %0.3f mm\"%(high_s,low_s2,low_h2,high_h2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 9.7 : Page 380"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " lower limit of hole = 100.002 mm\n",
+      " higher limit of hole = 100.010 mm\n",
+      " upper limit of shaft = 99.989 mm\n",
+      " lower limit of shaft = 99.978 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import ceil\n",
+    "bs = 100 # basic size in mm\n",
+    "s1 = 120# diameter of step1 in mm\n",
+    "s2 = 80 # diameter of step2 in mm\n",
+    "d = (s1*s2)**(1/2) # mm\n",
+    "d = ceil(d)\n",
+    "i = (0.45*(d)**(1/3)+0.001*d)/10**3 # mm\n",
+    "t1 = 16*i # tolerence for hole in mm\n",
+    "t2 = 25*i # tolerence for shaft in mm\n",
+    "G = (2.5*(d)**0.34)/10**3 # fundamental deviation for hole in mm\n",
+    "e = (11*(d)**0.11)/10**3 # fundamental deviation for shaft in microns\n",
+    "# Hole\n",
+    "LLh = bs+G # lower limit of hole in mm\n",
+    "HLh = LLh+t1 # higher limit of hole in mm\n",
+    "# shaft\n",
+    "ULs = bs-e # upper limit of shaft in mm\n",
+    "LLs = ULs-t2 # lower limit of shaft in mm\n",
+    "print \" lower limit of hole = %0.3f mm\\n higher limit of hole = %0.3f mm\\n upper limit of shaft = %0.3f mm\\n lower limit of shaft = %0.3f mm\"%(LLh,HLh,ULs,LLs)\n",
+    "# Error in textbook"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 9.8 : Page 381"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Hole basis system \n",
+      " Lower limit of journal = 100 mm\n",
+      " Higher limit of bearing = 100.005 mm\n",
+      " Higher limit of journal = 99.998 mm \n",
+      " Lower limit of journal = 99.994 mm\n",
+      "shaft basis system \n",
+      " upper limit of journal = 100.000 mm\n",
+      " lower limit of journal = 99.996 mm\n",
+      " lower limit of bearing = 100.002 mm\n",
+      " upper limit of bearing = 100.007 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "tb = 0.005 # tolerence on bearing in mm\n",
+    "tj = 0.004 # tolerence on journal in mm\n",
+    "a = 0.002 # allowance in mm\n",
+    "#hole-basis system\n",
+    "b = 100 # basic size in mm\n",
+    "Bl = b # lower limit of bearing in mm\n",
+    "Bh = Bl+tb # higher limit of bearing in mm\n",
+    "Jh = Bl-a # higher limit of journal in mm\n",
+    "Jl1 = Jh - tj # lower limit of journal in \n",
+    "# shaft-basis system\n",
+    "Ju = b # upper limit of journal in mm\n",
+    "Jl2 = Ju-tj # lower limit of journal in mm\n",
+    "Bl = Ju+a # lower limit of bearing in mm\n",
+    "Bu = Bl+tb # upper limit of bearing in mm\n",
+    "print \"Hole basis system \\n Lower limit of journal = %d mm\\n Higher limit of bearing = %0.3f mm\\n Higher limit of journal = %0.3f mm \\n Lower limit of journal = %0.3f mm\"%(Bl,Bh,Jh,Jl1)\n",
+    "print \"shaft basis system \\n upper limit of journal = %0.3f mm\\n lower limit of journal = %0.3f mm\\n lower limit of bearing = %0.3f mm\\n upper limit of bearing = %0.3f mm\"%(Ju,Jl2,Bl,Bu)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 9.9 : Page 381"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Hole basis system \n",
+      " upper limit of shaft = 100.120 mm\n",
+      " lower limit of hole = 100.000 mm\n",
+      " higher limit of hole = 100.035 mm\n",
+      " lower limit of shaft = 100.085 mm\n",
+      " Shaft basis system \n",
+      " upper limit of shaft = 100.000 mm\n",
+      " lower limit of shaft = 99.965 mm\n",
+      " lower limit of hole = 99.880 mm\n",
+      " upper limit of hole = 99.915 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Hole-basis system\n",
+    "b = 100 # basic size in mm\n",
+    "i1 = 0.12 # maximum interference in mm\n",
+    "i2 = 0.05 # minimum interfernce in mm\n",
+    "t = (i1-i2)/2 # tolerence in mm\n",
+    "Sh = b+i1 # upper limit of shaft in mm\n",
+    "Hl = b # lower limit of hole in mm\n",
+    "Hh = b+t # higher limit of hole in mm\n",
+    "Sl1 = Sh-t #lower limit of shaft in mm  \n",
+    "# shaft-basis system\n",
+    "Su = b # upper limit of shaft in mm\n",
+    "Sl2 = b-t # lower limit of shaft in mm\n",
+    "Hl1 = b-i1 # lower limit of hole in mm\n",
+    "Hu = Hl1+t # higher limit of hole in mm\n",
+    "print \" Hole basis system \\n upper limit of shaft = %0.3f mm\\n lower limit of hole = %0.3f mm\\n higher limit of hole = %0.3f mm\\n lower limit of shaft = %0.3f mm\"%(Sh,Hl,Hh,Sl1)\n",
+    "print \" Shaft basis system \\n upper limit of shaft = %0.3f mm\\n lower limit of shaft = %0.3f mm\\n lower limit of hole = %0.3f mm\\n upper limit of hole = %0.3f mm\"%(Su,Sl2,Hl1,Hu)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 9.10 : Page 382"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " lower limit of hole = 100 mm\n",
+      " upper limit of hole = 100.016 mm\n",
+      " upper limit of shaft = 99.972 mm\n",
+      " lower limit of shaft = 99.964 mm\n"
+     ]
+    }
+   ],
+   "source": [
+    "aa = 0.04 # average allowance in mm\n",
+    "a = 0.012 # allowance in mm\n",
+    "Max = aa+a # maximum allowance in mm\n",
+    "Min = aa-a # minimum allowance in mm\n",
+    "t = (Max-Min)/3 # tolerence in mm\n",
+    "ts = t # tolerence in shat in mm\n",
+    "th = 2*t # tolerence in hole in mm\n",
+    "b = 100 # basic size in mm\n",
+    "Hl = b # lower limit of hole in mm\n",
+    "Hu = b+th # upper limit of hole in mm\n",
+    "Su = b-0.028 # upper limit of shaft in mm\n",
+    "Sl = Su-ts # lower limit of shaft in mm\n",
+    "print \" lower limit of hole = %d mm\\n upper limit of hole = %0.3f mm\\n upper limit of shaft = %0.3f mm\\n lower limit of shaft = %0.3f mm\"%(Hl,Hu,Su,Sl)"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/screenshots/21AOQCurve_1.png b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/screenshots/21AOQCurve_1.png
new file mode 100644
index 00000000..0716fa4f
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+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/screenshots/21AOQCurve_1.png
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/screenshots/21X&RChart_1.png b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/screenshots/21X&RChart_1.png
new file mode 100644
index 00000000..c09406c5
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+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/screenshots/21X&RChart_1.png
diff --git a/A_Textbook_of_Production_Engineering_by_P._C._Sharma/screenshots/22CuttingVvsCutterDia_1.png b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/screenshots/22CuttingVvsCutterDia_1.png
new file mode 100644
index 00000000..9ecdd1d0
--- /dev/null
+++ b/A_Textbook_of_Production_Engineering_by_P._C._Sharma/screenshots/22CuttingVvsCutterDia_1.png
diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_1.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_1.ipynb
new file mode 100644
index 00000000..63c6e5ec
--- /dev/null
+++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_1.ipynb
@@ -0,0 +1,644 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 01 - Introduction To Heat Transfer"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.1 - Page : 8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# given data\n",
+      "t1=38   # in degree C\n",
+      "t2=21   # in degree C\n",
+      "k=0.19   # unit less\n",
+      "x=4   #in cm\n",
+      "x=x*10**-2   # in meter\n",
+      "# Formula q=k*A*(t1-t2)/x\n",
+      "q_by_A=k*(t1-t2)/x  \n",
+      "print \"The rate of heat transfer is :\",round(q_by_A,3),\"W/m**2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rate of heat transfer is : 80.75 W/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.2 - Page : 8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "t_i=120   # in degree C\n",
+      "t_o=40   # in degree C\n",
+      "K=0.04   # unit less\n",
+      "x=0.06   #in m\n",
+      "Q=50   # in W\n",
+      "print \"Assuming steady state heat transfer in the wall\"\n",
+      "# Rate of heat transfer across the wall = Rate of electrical energy dissipation in the furnance\n",
+      "# Formula Q= K*A*(t_i-t_o)/x \n",
+      "A=Q*x/(K*(t_i-t_o)) \n",
+      "print \"Area of wall = %0.4f square meter\" %A"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Assuming steady state heat transfer in the wall\n",
+        "Area of wall = 0.9375 square meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.3 - Page : 10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "t_f=30   # in degree C\n",
+      "t_s=400   # in degree C\n",
+      "d=0.04   #in m\n",
+      "h=20   # in W/m**2K\n",
+      "l=1   #in meter\n",
+      "A=pi*d*l#\n",
+      "q=h*A*(t_s-t_f)   # in W\n",
+      "print \"Rate of heat loss = %0.3f watt\" %q"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rate of heat loss = 929.911 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.4 - Page : 10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "t_s=100   # in degree C\n",
+      "t_w=80   # in degree C\n",
+      "d=2*10**-3   #in m\n",
+      "h=3000   # in W/m**2 degree C\n",
+      "L=100   #in mm\n",
+      "L=L*10**-3   # in meter\n",
+      "A=pi*d*L#\n",
+      "# Heat loss by convection = Electric power supplied\n",
+      "# Formula h*A*(t_s-t_w) = Q\n",
+      "Q= h*A*(t_s-t_w)#\n",
+      "print \"Electric power supplied = %0.1f watt\" %Q"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Electric power supplied = 37.7 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.5 - Page : 13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "A=0.6*0.9   # in square meter\n",
+      "x=.025   # in meter\n",
+      "t_s=310   # in degree C\n",
+      "t_f=15   # in degree C\n",
+      "h=22   # in W/m**2 degree C\n",
+      "K=45   # in W/m degree C\n",
+      "Q_rad=250   # in W\n",
+      "# Heat transfer through the plate = Convection heat loss + radiation heat loss\n",
+      "# Formula Q_cond = Q_conv + Q_rad\n",
+      "# -K*A*dt/dx = h*A*(t_s-t_f)+ Fg12*sigmaA(Ts**4-Ta64)\n",
+      "t_i=x*(h*A*(t_s-t_f)+Q_rad)/(K*A)+t_s \n",
+      "print \"The inside plate temperature = %0.2f degree C\" %t_i"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The inside plate temperature = 313.86 degree C\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.6 - Page : 13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "T1=50   # in degree C\n",
+      "T1=T1+273   # in K\n",
+      "T2=20   # in degree C\n",
+      "T2=T2+273   # in K\n",
+      "d=5*10**-2   #in m\n",
+      "h=6.5   # in W/m**2K\n",
+      "l=1   #in meter\n",
+      "epsilon=0.8#\n",
+      "sigma=5.67*10**-8#\n",
+      "A=pi*d*l   # in Square meter\n",
+      "q_conv = h*A*(T1-T2)   # in W/m\n",
+      "print \"The heat loss by convection = %0.1f W/m\" %q_conv\n",
+      "# formula q= sigma*A*F_g12*(T1**4-T2**4) = sigma*A*epsilon*(T1**4-T2**4)    (since A1<<A2, so F_g12=epsilon)\n",
+      "q_rad = sigma*A*epsilon*(T1**4-T2**4)   # in W/m\n",
+      "print \"Heat loss by radiation = %0.0f W/m\" %q_rad\n",
+      "q_total= q_conv+q_rad#\n",
+      "print \"Total heat loss = %0.3f W/m\" %q_total"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat loss by convection = 30.6 W/m\n",
+        "Heat loss by radiation = 25 W/m\n",
+        "Total heat loss = 55.672 W/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.7 - Page : 19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.integrate import quad\n",
+      "# given data\n",
+      "T1=1350   # in degree C\n",
+      "T2=50   # in degree C\n",
+      "L=25*10**-2   #in meter\n",
+      "# Formula q= -k*A*dT/dx\n",
+      "# or q/A= -k*dT/dx\n",
+      "# let q/A = q_by_A\n",
+      "def integrand(T):\n",
+      "    return -0.838*(1+0.0007*T)\n",
+      "ans, err = quad(integrand, T1, T2)\n",
+      "def integrand(x):\n",
+      "    return 1\n",
+      "ans2, err2 = quad(integrand, 0, L)\n",
+      "q_by_A=ans/ans2#\n",
+      "print \"Heat transfer rate per square meter through  the cylinder = %0.0f watt\" %q_by_A\n",
+      "\n",
+      "# Note : Answer in the book is incorrect"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate per square meter through  the cylinder = 6493 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.9 - Page : 21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "K_A=0.5   # in W/m degree C\n",
+      "K_B=0.8   # in W/m degree C\n",
+      "Ti_A=600   # inside temp. of slab A in degree C\n",
+      "To_B=100   # outside temp. of slab B in degree C\n",
+      "t_A=4*10**-2   # thickness of slab A\n",
+      "t_B=6*10**-2   # thickness of slab B\n",
+      "# Heat transfer rate per square meter through the slab A\n",
+      "# q/A = +K_A * ( Ti_A - T) / t_A         (1)\n",
+      "# Heat transfer rate through slab B\n",
+      "# q/A = +K_B * ( T - To_B) / t_B        (2)\n",
+      "# Equating Eqns (1) and (2)\n",
+      "# K_A*(Ti_A - T)/t_A  =  K_B*(T - To_B)/t_B\n",
+      "T=t_A*t_B/(K_A*t_B+K_B*t_A)*(K_A*Ti_A/t_A + K_B*To_B/t_B)\n",
+      "print \"T, intermediate temperature of slab A and B is :\",round(T,3),\"degree C\"\n",
+      "#Putting the value of T in Eq(1), we get\n",
+      "q_by_A= K_A*( Ti_A - T) / t_A#\n",
+      "print \"Steady state heat transfer rate per square meter is :\",round(q_by_A,3),\"W/m**2\"\n",
+      "#Note : Answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "T, intermediate temperature of slab A and B is : 341.935 degree C\n",
+        "Steady state heat transfer rate per square meter is : 3225.806 W/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.10 - Page : 22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "La=3*10**-2   # in meter\n",
+      "Aa=1   # in m**2\n",
+      "ka=150   # in W/m-K\n",
+      "\n",
+      "Lb=8*10**-2   # in meter\n",
+      "Ab=0.5   # in m**2\n",
+      "kb=30   # in W/m-K\n",
+      "\n",
+      "Lc=8*10**-2   # in meter\n",
+      "Ac=0.5   # in m**2\n",
+      "kc=65   # in W/m-K\n",
+      "\n",
+      "Ld=5*10**-2   # in meter\n",
+      "Ad=1   # in m**2\n",
+      "kd=50   # in W/m-K\n",
+      "\n",
+      "T1=400   # in degree C\n",
+      "T2=60   # in degree C\n",
+      "\n",
+      "Ra=La/(ka*Aa)#\n",
+      "Rb=Lb/(kb*Ab)#\n",
+      "Rc=Lc/(kc*Ac)#\n",
+      "Rd=Ld/(kd*Ad)#\n",
+      "#The equivalent resistance for Rb and Rc\n",
+      "Re=Rb*Rc/(Rb+Rc)#\n",
+      "#Total Resistance\n",
+      "sigmaR=Ra+Re+Rd#\n",
+      "# heat transfer rate per square meter\n",
+      "q=(T1-T2)/sigmaR#\n",
+      "print \"Heat transfer rate per square meter = %0.2e Watt\" %q"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate per square meter = 1.18e+05 Watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.11 - Page : 23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "k_Al=202   # in W/mK\n",
+      "x_Al=0.005   # in m\n",
+      "del_T=80   # in degree C\n",
+      "R_contact=0.88*10**-4   # in m**2K/W\n",
+      "sigmaR=x_Al/k_Al+R_contact+x_Al/k_Al   # in m**2K/W\n",
+      "q=del_T/sigmaR   # in W/m**2\n",
+      "#Temperature drop across the rough surface\n",
+      "del_T=q*R_contact   #in degree C\n",
+      "print \"Temperature drop across the rough surface = %0.3f degree C\" %del_T"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature drop across the rough surface = 51.198 degree C\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.12 - Page : 24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "T1=100   # in degree C\n",
+      "T2=10   # in degree C\n",
+      "A=3*5   #in square meter\n",
+      "x=40*10**-2   # thickness in m**2\n",
+      "k=1.6   # in W/mk\n",
+      "h=10   # in W/m**2k\n",
+      "# Total resistance in heat flow path\n",
+      "sigmaR=x/(k*A)+1/(h*A)#\n",
+      "# so heat transfer rate\n",
+      "q=(T1-T2)/sigmaR   # in Watt\n",
+      "q=q*10**-3   #in kW\n",
+      "print \"Heat transfer rate = %0.3f kW\" %q\n",
+      "\n",
+      "# Note: Answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate = 3.857 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.13 - Page : 25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.integrate import quad\n",
+      "# given data\n",
+      "k='2.0+0.0005*T'   # in W/m-k\n",
+      "A=3*5   #in square meter\n",
+      "T1=150   # in degree C\n",
+      "T2=50   # in degree C\n",
+      "L=20*10**-2   # thickness in m**2\n",
+      "# Formula q= -k*A*dt/dx\n",
+      "def integrand(T):\n",
+      "    return 2.0+0.0005*T\n",
+      "ans, err = quad(integrand, T1, T2)\n",
+      "def integrand(x):\n",
+      "    return 1\n",
+      "ans2, err2 = quad(integrand, 0, L)\n",
+      "\n",
+      "q=-A*ans/ans2   # in Watt\n",
+      "q=q*10**-3   #in kW\n",
+      "print \"Rate of heat transfer = %0.3f kW\" %q"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rate of heat transfer = 15.375 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.14 - Page : 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "T1=300   #in degree C\n",
+      "T2=50   #in degree C\n",
+      "x2=2*10**-2   # thickness of boiler wall in m\n",
+      "tc2=58   # thermal conductivity of wall in W/mk\n",
+      "x3=0.5*10**-2   # thickness of outer surface of the  wall in m\n",
+      "tc3=116*10**-3   # thermal conductivity of outer surface of the wall in W/mk\n",
+      "R1=2.3*10**-3   # in k/W\n",
+      "R2=x2/tc2#\n",
+      "R3=x3/tc3#\n",
+      "sigmaR=R1+R2+R3   # Total Resistance\n",
+      "q=(T1-T2)/sigmaR#\n",
+      "print \"Heat transfer rate per unit area = %0.3f W/m**2\" %q\n",
+      "# Note: Answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate per unit area = 5464.687 W/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.15 - Page : 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "Tf=80   # in degree C\n",
+      "I=200   # in amp\n",
+      "h=4000   # in W/m**2degree C\n",
+      "rho=70*10**-6#\n",
+      "L=100   # in cm\n",
+      "R=0.1   # in ohm\n",
+      "d=3   # in mm\n",
+      "d=d*10**-3#\n",
+      "As= pi*d#\n",
+      "#Formula I**2*R= h*As*(Tw-Tf)\n",
+      "Tw= I**2*R/(h*As)+Tf#\n",
+      "print \"Central temperature of the wire = %0.f \u00b0C\" %Tw"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Central temperature of the wire = 186 \u00b0C\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.16 - Page : 27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# given data\n",
+      "E=500   #Absorb solar energy in W/m**2\n",
+      "epsilon= 0.9#\n",
+      "T_s= 280   # in K\n",
+      "T_infinite=300   # in K\n",
+      "h_c=20   # in W/m**2degree C\n",
+      "T_sky=280   # in K\n",
+      "sigma=5.67*10**-8#\n",
+      "# Formula E= h_c*(T_p-T_infinite)+epsilon*sigma*(T_P**4-T_s**4)\n",
+      "# On simplication T_P= 340.6-0.255*T-p**4\n",
+      "T_p= 315.5   # in K\n",
+      "print \"Equilibrium Temperature of the plate = %0.1f K\" %T_p"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Equilibrium Temperature of the plate = 315.5 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_10.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_10.ipynb
new file mode 100644
index 00000000..119f2bce
--- /dev/null
+++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_10.ipynb
@@ -0,0 +1,426 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 - Mass Transfer"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  10.1 - Page : 291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# given data\n",
+      "P1=4   # in bar\n",
+      "P2=2   # in bar\n",
+      "T=25   # in degree C\n",
+      "Dhp=9*10**-8   # in m**2/s\n",
+      "S=3*10**-3   # in kg mole/m**3 bar\n",
+      "del_x=0.5*10**-3   # thickness in m\n",
+      "#(a) The molar concentration of a gas in terms of solubility\n",
+      "CH1=S*P1   # in kg mole/m**3\n",
+      "CH2=S*P2   # in kg mole/m**3\n",
+      "#(b) Molar diffusion flux of hydrogen through plastic memberence is given by Fick's law of diffision\n",
+      "#N_H= N_h/A = Dhp*(CH1-CH2)/del_x#\n",
+      "N_H= Dhp*(CH1-CH2)/del_x   # in kg mole/s-m**2\n",
+      "print \"Molar diffusion flux of hydrogen through the membrane = %0.3e kg mole/s-m**2\" %N_H\n",
+      "#Mass_d_Flux= N_H*Molecular_Weight \n",
+      "Molecular_Weight=2#\n",
+      "Mass_d_Flux= N_H*Molecular_Weight \n",
+      "print \"Molar diffusion flux = %0.3e kg/s-m**2\" %Mass_d_Flux"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Molar diffusion flux of hydrogen through the membrane = 1.080e-06 kg mole/s-m**2\n",
+        "Molar diffusion flux = 2.160e-06 kg/s-m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  10.2 - Page : 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "T=25   # in degree C\n",
+      "T=T+273   # in K\n",
+      "P=1#\n",
+      "V1=12   #Molecular volume of H2 in cm**3/gm mole\n",
+      "V2=30   #Molecular volume of Air in cm**3/gm mole\n",
+      "M1=2   # Molecular weight of H2\n",
+      "M2=29   # Molecular weight of Air\n",
+      "#The diffusion coefficient for gases in terms of molecular volumes may be express as\n",
+      "D_AB= .0043*T**(3/2)/(P*(V1**(1/3)+V2**(1/3)))*(1/M1+1/M2)**(1/2)#\n",
+      "print \"The diffusion coefficient for gases in terms of molecular volumes = %0.3f cm**2/sec\" %D_AB"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diffusion coefficient for gases in terms of molecular volumes = 2.997 cm**2/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  10.3 - Page : 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "T=300   # temp of gas mixture in K\n",
+      "D_HN2=18*10**-6   # in m**2/s at 300 K, 1 bar\n",
+      "T1=300   # in K\n",
+      "D_HO2=16*10**-6   # in m**2/s at 273 K, 1 bar\n",
+      "T2=273   # in K\n",
+      "O_2=0.2#\n",
+      "N_2=0.7#\n",
+      "H_2=0.1#\n",
+      "#The diffusivity at the mixture temperature and pressure are calculated as \n",
+      "# D1/D2 = (T1/T2)**(3/2)*(P2/P1)\n",
+      "D_HO2= (T/T2)**(3/2)*1/4*D_HO2#\n",
+      "D_HN2= (T/T1)**(3/2)*1/4*D_HN2#\n",
+      "#The composition of oxygen and nitrogen on a H2 free basis is \n",
+      "x_O= O_2/(1-H_2)#\n",
+      "x_N= N_2/(1-H_2)#\n",
+      "# The effective diffusivity for the gas mixture at given temperature and pressure is\n",
+      "D= 1/(x_O/D_HO2+x_N/D_HN2)   # in m**2/s\n",
+      "print \"Effective diffusivity = %0.3e m**2/s\" %D"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Effective diffusivity = 4.524e-06 m**2/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  10.4 - Page : 293"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "d=3   # in mm\n",
+      "d=d*10**-3   # in meter\n",
+      "T=25   # in \u00b0C\n",
+      "T=T+273   # in K\n",
+      "D= 0.4*10**-4   # in m**2/s\n",
+      "R= 8314#\n",
+      "P_A1=1   # in atm\n",
+      "P_A1=P_A1*10**5   # in w/m**2\n",
+      "P_A2=0#\n",
+      "C_A2=0#\n",
+      "x2= 15   # in meter\n",
+      "x1= 0#\n",
+      "A= pi/4*d**2#\n",
+      "M_A= D*A/(R*T)*(P_A1-P_A2)/(x2-x1)   # in kg mole/sec\n",
+      "N_B= M_A#\n",
+      "M_B= M_A*29   # in kg/sec\n",
+      "print \"Value of N_B = %0.3e kg mole/sec\" %N_B\n",
+      "print \"Value of M_B = %0.3e kg/sec\" %M_B\n",
+      "# Note : The value of M_B in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of N_B = 7.608e-13 kg mole/sec\n",
+        "Value of M_B = 2.206e-11 kg/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  10.5 - Page : 294"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "P=3   # in atm\n",
+      "P=P*10**5   # in N/m**2\n",
+      "r1=10   # in mm\n",
+      "r1=r1*10**-3   # in m\n",
+      "r2=20   # in mm\n",
+      "r2=r2*10**-3   # in m\n",
+      "R=4160   # in J/kg-K\n",
+      "T=303   # in K\n",
+      "D=3*10**-8   # in m**2/s\n",
+      "S=3*0.05# # Solubility of hydrogen at a pressure of 3 atm in m**3/m**3 of rubber tubing\n",
+      "del_x=r2-r1   # in m\n",
+      "L=1   # in m\n",
+      "Am=2*pi*L*del_x/log(r2/r1)#\n",
+      "#Formula P*V= m*R*T\n",
+      "V=S#\n",
+      "m=P*V/(R*T)   # in kg/m**3 of rubber tubing at the inner surface of the pipe\n",
+      "C_A1=m#\n",
+      "C_A2=0#\n",
+      "#Diffusion flux through the cylinder is given\n",
+      "M=D*(C_A1-C_A2)*Am/del_x#\n",
+      "print \"Diffusion flux through the cylinder = %0.1e kg/sm\" %M"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diffusion flux through the cylinder = 9.7e-09 kg/sm\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  10.6 - Page : 295"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "R=4160   # in J/kg-K\n",
+      "M=2#\n",
+      "D_AB=1.944*10**-8   # in m**2/s\n",
+      "R_H2=R/M#\n",
+      "S=2*0.0532# # Solubility of hydrogen at a pressure of 2 atm in cm**3/cm**3 of pipe\n",
+      "P=2   # in atm\n",
+      "P=P*1.03*10**5   # N/m**2\n",
+      "T=25   # in degree C\n",
+      "T=T+273   # in K\n",
+      "r1=2.5   # in mm\n",
+      "r1=r1*10**-3   # in m\n",
+      "r2=5   # in mm\n",
+      "r2=r2*10**-3   # in m\n",
+      "del_x=r2-r1   # in m\n",
+      "L=1   # in m\n",
+      "#Formula P*V= m*R*T\n",
+      "V=S#\n",
+      "m=P*V/(R*T)   # in kg/m**3 of pipe\n",
+      "# So, Concentration of H2 at inner surface of the pipe\n",
+      "C_A1=0.0176   # in kg/m**3\n",
+      "# The resistance of diffusion of H2 away from the outer surface is negligible i.e.\n",
+      "C_A2=0#\n",
+      "Am=2*pi*L*del_x/log(r2/r1)#\n",
+      "# Loss of H2 by diffusion \n",
+      "M_A= D_AB*(C_A1-C_A2)*Am/del_x#\n",
+      "print \"Loss of H2 by diffusion = %0.3e kg/s\" %M_A\n",
+      "\n",
+      "\n",
+      "#Note: In the book , they put wrong value of C_A1 to calculate M_A, so the answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Loss of H2 by diffusion = 3.101e-09 kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  10.7 - Page : 296"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "from math import log\n",
+      "# given data\n",
+      "Px1= 0.14   # in bar\n",
+      "Px2= 0#\n",
+      "P=1.013   # in bar\n",
+      "Py1=P-Px1# # in bar\n",
+      "Py2=P-Px2# # in bar\n",
+      "D=8.5*10**-6   # in m**2/s\n",
+      "d=5   # diameter in meter\n",
+      "L=1   # in mm\n",
+      "L=L*10**-3   #in meter\n",
+      "M=78   # molecular weight\n",
+      "Am_x= 1/4*pi*d**2*M#\n",
+      "R=8314#\n",
+      "del_x=3   # thickness in mm\n",
+      "del_x=del_x*10**-3   # in m\n",
+      "T=20   # in degree C\n",
+      "T=T+273   # in K\n",
+      "P=P*10**5   # in N/m**2\n",
+      "m_x= D*Am_x*P*log(Py2/Py1)/(R*T*del_x)#\n",
+      "# The mass of the benzene to be evaporated\n",
+      "mass= 1/4*pi*d**2*L#\n",
+      "density=880   # in kg/m**3\n",
+      "m_b= mass*density#\n",
+      "toh=m_b/m_x   # in sec\n",
+      "print \"Time taken for the entire organic compound to evaporate = %0.3f seconds\" %toh\n",
+      "\n",
+      "\n",
+      "# Note: Answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time taken for the entire organic compound to evaporate = 643.788 seconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  10.8 - Page : 297"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "A=0.5   # in m**2\n",
+      "Pi=2.2   # in bar\n",
+      "Pi=Pi*10**5   # in N/m**2\n",
+      "Pf=2.18   # in bar\n",
+      "Pf=Pf*10**5   # in N/m**2\n",
+      "T=300   # in K\n",
+      "S=0.072   # in m**3\n",
+      "V=0.028   # in m**3\n",
+      "L=10   # in mm\n",
+      "L=L*10**-3   # in meter\n",
+      "R=287#\n",
+      "# Diffusivity of air in rubber D\n",
+      "# Initial mass of air in the tube\n",
+      "mi= Pi*V/(R*T)   # in kg\n",
+      "#final mass of air in the tube\n",
+      "mf= Pf*V/(R*T)   # in kg\n",
+      "# Mass of air escaped\n",
+      "ma = mi-mf   #in kg\n",
+      "# Formula Na = ma/A = mass of air escaped / Time elapsed * area\n",
+      "A=6*24*3600*0.5#\n",
+      "Na = ma/A   #in kg/sm**2\n",
+      "# Solubility of air should be calculated at mean temperature\n",
+      "S_meanTemperature=(2.2+2.18)/2   # in bar\n",
+      "#Solubility of air at the mean inside Pressure is \n",
+      "S=S*S_meanTemperature   # in m**3/m**3 of rubber\n",
+      "V1=S#\n",
+      "V2=0.072#\n",
+      "T1=T#\n",
+      "T2=T#\n",
+      "P1=2.19*10**5   # in N/m**2\n",
+      "P2=1*10**5   # in N/m**2\n",
+      "# The corresponding mass concentration at the inner and outer surface of the tube, from gas equation are calculated as\n",
+      "Ca1= P1*V1/(R*T1)   # in kg/m**3\n",
+      "Ca2= P2*V2/(R*T2)   # in kg/m**3\n",
+      "# The diffusion flux rate of air through the rubber is given by\n",
+      "# Na = ma/A = D*(Ca1-Ca2)/del_x, here\n",
+      "del_x=L#\n",
+      "D=Na*del_x/(Ca1-Ca2)#\n",
+      "print \"Diffusivity of air in rubber = %0.3e m**2/s\" %D"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diffusivity of air in rubber = 7.905e-11 m**2/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_2.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_2.ipynb
new file mode 100644
index 00000000..b5e164ff
--- /dev/null
+++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_2.ipynb
@@ -0,0 +1,515 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 02 - General Heat Conduction Equation"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  2.5 - Page : 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import log, pi\n",
+      "# given data\n",
+      "r1=5   # in cm\n",
+      "r2=5+4   # in cm\n",
+      "r3= 9+2.5   # in cm\n",
+      "k1=0.0701   # in W/mK\n",
+      "k2=0.1   # in W/mK\n",
+      "L=20   # in m\n",
+      "T1=234.36   # in degree C\n",
+      "T3=24   # in degree C\n",
+      "sigmaR= (log(r2/r1)/(2*pi*k1*L) + log(r3/r2)/(2*pi*k2*L))#\n",
+      "\n",
+      "# Part (i)\n",
+      "q=(T1-T3)/sigmaR   # in watt\n",
+      "print \"Heat transfer rate = %0.3f watt\" %q\n",
+      "\n",
+      "# Part(ii)\n",
+      "# Formula q= (T1-T2)/(log(r2/r1)/(2*pi*k1*L))\n",
+      "T2 =T1- (q*(log(r2/r1)/(2*pi*k1*L)))#\n",
+      "print \"Interface temperature of insulation = %0.3f degree\" %T2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate = 2439.473 watt\n",
+        "Interface temperature of insulation = 71.585 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  2.6 - Page : 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from numpy import pi\n",
+      "# given data\n",
+      "k_brick=0.93   # in W/mK\n",
+      "k_insulation=0.12   # in W/mK\n",
+      "k_wood=0.175   # in W/mK\n",
+      "k_Al=204   # in W/mK\n",
+      "k1=k_brick#\n",
+      "k2=k_insulation#\n",
+      "k3=k_wood#\n",
+      "T1=200   # in degree C\n",
+      "T4=10   # in degree C\n",
+      "x1=10*10**-2   # in m\n",
+      "x2=25*10**-2   # in m\n",
+      "x3=1*10**-2   # in m\n",
+      "A=0.1   # in m**2\n",
+      "sigmaR= x1/(k1*A)+x2/(k2*A)+x3/(k3*A)#\n",
+      "q1=(T1-T4)/sigmaR#\n",
+      "print \"Heat transfer rate without rivet = %0.2f Watt\" %q1\n",
+      "\n",
+      "# Heat transfer rate with rivet\n",
+      "d=3*10**-2   # in meter\n",
+      "x=x1+x2+x3#\n",
+      "k_rivet=k_Al#\n",
+      "A_rivet=pi*d**2/4   # in m**2\n",
+      "R_rivet= x/(k_rivet*A_rivet)#\n",
+      "A_eff=A-A_rivet   # in m**2\n",
+      "sigmaRw= 1/A_eff*(x1/k1+x2/k2+x3/k3)   # in k/W\n",
+      "R_eq= R_rivet*sigmaRw/(R_rivet+sigmaRw)   # in k/W\n",
+      "q2=(T1-T4)/R_eq   # in watt\n",
+      "print \"Heat transfer rate with rivet = %0.3f Watt\" %q2\n",
+      "percentIncrease=(q2-q1)*100/q1   # percent increase in heat flow due to rivet\n",
+      "print \"Percentage increase in heat flow due to rivet = %0.f %%\" %math.ceil(percentIncrease)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate without rivet = 8.45 Watt\n",
+        "Heat transfer rate with rivet = 84.497 Watt\n",
+        "Percentage increase in heat flow due to rivet = 900 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  2.7 - Page : 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "k_cu=384   # in W/mK\n",
+      "k_s=1.75   # in W/mK\n",
+      "k1=k_cu#\n",
+      "k2=k_s#\n",
+      "hi=221   #  in W/m**2K\n",
+      "ho=3605   #  in W/m**2K\n",
+      "Ti=100   # in degree C\n",
+      "To=125   # in degree C\n",
+      "r1=0.2   # in m\n",
+      "r2=0.02+0.006   # in m\n",
+      "r3=0.026+0.003   # in m\n",
+      "ri=0.02   # in m\n",
+      "L=1   # in m\n",
+      "# Part(i)\n",
+      "Ao= 2*pi*r3*L#\n",
+      "Ai= 2*pi*r1*L#\n",
+      "# Formula Uo= 1/Ao*sigmaR\n",
+      "Uo= 1/(r3/(ri*hi) + r3/k1*log(r2/r1) + r3/k2*log(r3/r2) + 1/ho)  # in w/m**2K\n",
+      "print \"Overall heat transfer coefficient based on outer area = %0.3f W/m**2K\" %Uo\n",
+      "\n",
+      "#Part(ii)\n",
+      "del_T= To-Ti#\n",
+      "q=Uo*Ao*del_T#\n",
+      "print \"Water to air heat transfer rate = %0.3f W/m\" %q\n",
+      "\n",
+      "#Part (iii)\n",
+      "# Formula q= T/(log(r3/r2)/(2*pi*k*L)) , where T=T2-T3 and k=k_s\n",
+      "k=k_s#\n",
+      "T= q*log(r3/r2)/(2*pi*k*L)#\n",
+      "print \"Temperature drop across the scale deposited = %.3f degree C\" %T\n",
+      "\n",
+      "# Note: In Part (i), they put wrong value of r2 and r1 in log(r2/r1) to calculate the value of Uo. So there is some difference in answer of coding and book"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Overall heat transfer coefficient based on outer area = 117.730 W/m**2K\n",
+        "Water to air heat transfer rate = 536.298 W/m\n",
+        "Temperature drop across the scale deposited = 5.326 degree C\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  2.8 - Page : 58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "# given data\n",
+      "k=0.175   # in W/mK\n",
+      "h_infinite=9.3   #  in W/m**2K\n",
+      "T_infinite=30   # in degree C\n",
+      "T_s=70   # in degree C\n",
+      "d=10*10**-3   # in m\n",
+      "r=d/2#\n",
+      "L=1   # in m\n",
+      "rc=k/h_infinite   # in m\n",
+      "CriticalThickness = rc-r   # in meter\n",
+      "CriticalThickness=CriticalThickness*10**3#\n",
+      "print \"Critical thickness = %0.1f mm\" %CriticalThickness\n",
+      "\n",
+      "q1=2*pi*r*L*h_infinite*(T_s-T_infinite)   # in W/m\n",
+      "q2= (T_s-T_infinite)/(log(rc/r)/(2*pi*k*L)+1/(2*pi*rc*h_infinite))   # in W/m\n",
+      "PerIncHeatDiss= (q2-q1)*100/q1#\n",
+      "print \"Percentage increase in heat dissipation rate = %0.3f %%\" %PerIncHeatDiss\n",
+      "#Also q1=I1**2*R with bare cable\n",
+      "#     q2=I2**2*R with insulated cable\n",
+      "I2_by_I1 = sqrt(q2/q1)#\n",
+      "# ( I2-I1 ) / I1 = (I2_by_I1 -1) / 1\n",
+      "# Percentage increase in current carrying capacity\n",
+      "PerIncCurrent = (I2_by_I1 -1) / 1 *100#\n",
+      "print \"Increase in current carrying capacity = %0.f %%\" %math.floor(PerIncCurrent)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Critical thickness = 13.8 mm\n",
+        "Percentage increase in heat dissipation rate = 61.845 %\n",
+        "Increase in current carrying capacity = 27 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  2.9 - Page : 59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "k_in=0.3   # in W/mK\n",
+      "k_gw=0.038   # in W/mK\n",
+      "ro=1.5   # in cm\n",
+      "ho=12   #  in W/m**2 degree C\n",
+      "rc=k_in/ho   # in m\n",
+      "rc=rc*10**2   # in cm\n",
+      "print \"Critical radius = %0.1f cm\" %rc\n",
+      "if ro<rc:\n",
+      "    print \"Since radius of insulation (\",round(ro,3),\"cm) is less than critical radius of insulation (\",round(rc,3),\"cm)\"\n",
+      "    print \"so heat transfer rate will increase by adding thsi insulation and hence it is not effective\"\n",
+      "ro=ro*10**-2   # in meter\n",
+      "# For effective insulation\n",
+      "# ro>=rc\n",
+      "# Kin/ho<= ro\n",
+      "roho=ro*ho   # in W/mK\n",
+      "# Kin<= ro*ho\n",
+      "print \"Maximum value of thermal conductivity = %0.2f W/mK\" %roho"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Critical radius = 2.5 cm\n",
+        "Since radius of insulation ( 1.5 cm) is less than critical radius of insulation ( 2.5 cm)\n",
+        "so heat transfer rate will increase by adding thsi insulation and hence it is not effective\n",
+        "Maximum value of thermal conductivity = 0.18 W/mK\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  2.10 - Page : 60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "d=1.2*10**-3   # in m\n",
+      "r=d/2   # in m\n",
+      "rc=1.8*10**-3   # in m\n",
+      "T1=100   # in degree C\n",
+      "T_infinite=30   # in degree C\n",
+      "k=0.3   # in W/mK\n",
+      "h=10   # in W/m**2K\n",
+      "L=1   # in m\n",
+      "ke=5.1*10**7#\n",
+      "q=(T1-T_infinite)/(log(rc/r)/(2*pi*k)+1/(2*pi*rc*h))   # in W/m\n",
+      "# Volume of wire for one meter length\n",
+      "vol= pi*r**2*L   # in m**3\n",
+      "print \"In steady state heat transfer process, the heat produced by the wire is dissipated to surrounding.\"    \n",
+      "# Heat produced per unit volume of the wire\n",
+      "HeatProduced= q/vol   # in w/m**2\n",
+      "# Formula HeatProduced= I**2*R = I**2/ke\n",
+      "I=sqrt(HeatProduced*ke)   # in amp/m**2\n",
+      "# Area of wire\n",
+      "A= pi*r**2#\n",
+      "# so current carrying capacity of the given wire\n",
+      "Current= I*A#\n",
+      "print \"The current carried by the copper wire = %0.3f amphere\" %Current"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In steady state heat transfer process, the heat produced by the wire is dissipated to surrounding.\n",
+        "The current carried by the copper wire = 20.698 amphere\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  2.11 - Page : 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "d_i=.1   # inner dia in m\n",
+      "r_i=d_i/2   # in m\n",
+      "Ti=473   # in K\n",
+      "T_infinite=293   # in K\n",
+      "k=1   # in W/mK\n",
+      "h=8   # in W/m**2K\n",
+      "rc=k/h   # in m\n",
+      "print \"Critical radius = %0.3f meter\" %rc\n",
+      "#when\n",
+      "ro=rc#\n",
+      "q_by_L= (Ti-T_infinite)/(log(rc/r_i)/(2*pi*k)+1/(2*pi*rc*h))   # in W/m\n",
+      "print \"Heat loss per meter length of pipe = %0.3f W/m\" %q_by_L\n",
+      "\n",
+      "# Note: To calculate the value of q_by_L the calculation is wrong in the book so answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Critical radius = 0.125 meter\n",
+        "Heat loss per meter length of pipe = 590.189 W/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  2.12 - Page : 62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "r1=100*10**-3   # in m\n",
+      "r2=200*10**-3   # in m\n",
+      "q1=1.16*10**5   # in W/m**2\n",
+      "t2=30   # in degree C\n",
+      "k=50   # in W/mK\n",
+      "L=1   # in m\n",
+      "# Total heat passing through the cylinder q\n",
+      "#q=q1*2*pi*r1*L                     (1)\n",
+      "# and heat conducted through the cylinder\n",
+      "# q= 2*pi*k*L(t1-t2)/log(r2/r1)     (2)\n",
+      "# From (1) and (2)\n",
+      "t1= t2+ q1*2*pi*r1*L*log(r2/r1)/(2*pi*k*L)   # in degree C\n",
+      "print \"Temperature of inner surface = %0.2f degree C\" %t1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature of inner surface = 190.81 degree C\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  2.13 - Page : 62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "d1=1*10**-3   # in m\n",
+      "d2=3*10**-3   # in m\n",
+      "r1=d1/2#\n",
+      "r2=d2/2#\n",
+      "kp=384   # in W/mK\n",
+      "kw=0.35   # in W/mK\n",
+      "rho=1.96*10**-8   # in Wm\n",
+      "t_s=95   # in degree C\n",
+      "t_infinite=40   # in degree C\n",
+      "h=8.75   # in W/m**2K\n",
+      "q_by_L= (t_s-t_infinite)/(log(r2/r1)/(2*pi*kp)+1/(2*pi*r2*h))#\n",
+      "# Also q_by_L = I**2*R/L = I**2*rho/(pi/4*d**2)\n",
+      "I= sqrt(q_by_L*(pi/4*d1**2)/rho)   # in amp\n",
+      "print \"The maximum steady state current = %0.3f amphere\" %I"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum steady state current = 13.481 amphere\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  2.16 - Page : 66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log, sqrt\n",
+      "from math import pi\n",
+      "from __future__ import division\n",
+      "# given data\n",
+      "d1=10*10**-3   # in mm\n",
+      "r1=d1/2#\n",
+      "K=0.2   # in W/mK\n",
+      "T_max=177   # in degree C\n",
+      "T_infinite=27   # in degree C\n",
+      "ho=10   # in W/m**2K\n",
+      "R=10# # in W/m\n",
+      "rc=K/ho   # in m\n",
+      "x=rc-r1   # in m\n",
+      "q_by_L= (T_max-T_infinite)/(log(rc/r1)/(2*pi*K)+1/(2*pi*ho*rc))#\n",
+      "# Also q_by_L = I**2*R\n",
+      "I= sqrt(q_by_L/R)   # in amp\n",
+      "print \"The maximum possible current = %0.3f amphere\" %I\n",
+      "\n",
+      "# Note: Answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum possible current = 2.811 amphere\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_3.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_3.ipynb
new file mode 100644
index 00000000..870158aa
--- /dev/null
+++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_3.ipynb
@@ -0,0 +1,632 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 - Fins Heat Transfer From Extended Surface"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  3.1 - Page : 84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from numpy import pi\n",
+      "from math import sqrt\n",
+      "# given data\n",
+      "d=20   # in mm\n",
+      "d=d*10**-3   #in m\n",
+      "h=5   # in W/m**2K\n",
+      "T_0=100   # in degree C\n",
+      "T_infinite=20   # in degree C\n",
+      "K=15   # in W/m-K\n",
+      "#(i)Temperature distribution equation\n",
+      "rho=pi*d   # in m\n",
+      "A=pi*d**2/4   #in square meter\n",
+      "m=sqrt(h*rho/(K*A))#\n",
+      "print \"(i) Temperature distribution equation is = theta/theta_0 = (T-T_infinite)/(T_0-T_infinite) = e**- %0.2f*x\" %m\n",
+      "#(ii)Heat loss from the rod\n",
+      "t_0=100   # in degree C\n",
+      "t_infinite=20   # in degree C\n",
+      "q=sqrt(K*A*h*rho)*(t_0-t_infinite)#\n",
+      "print \"(ii) Heat loss from the road is :\",round(q,3),\"watt\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(i) Temperature distribution equation is = theta/theta_0 = (T-T_infinite)/(T_0-T_infinite) = e**- 8.16*x\n",
+        "(ii) Heat loss from the road is : 3.078 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  3.2 - Page : 85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log, pi\n",
+      "from __future__ import division\n",
+      "# given data\n",
+      "d=3   # in cm\n",
+      "d=d*10**-2   #in m\n",
+      "h=20   # in W/m**2K\n",
+      "T1=140   # in degree C\n",
+      "T2=100   # in degree C\n",
+      "L=15*10**-2   # in meter\n",
+      "T_infinite=30   # in degree C\n",
+      "T_0=140   # in degree C\n",
+      "# Let at\n",
+      "x=0#T_0=T1#\n",
+      "x=15   #in cm\n",
+      "x=x*10**-2   # in m\n",
+      "T=100   # in degree C\n",
+      "rho=pi*d#\n",
+      "A=pi*d**2/4#\n",
+      "# Formula (T-T_infinite)/(T_0-T_infinite) = %e**-m*x\n",
+      "m=log((T_0-T_infinite)/(T-T_infinite))/x#\n",
+      "# Formula m=sqrt(h*rho/(k*A))\n",
+      "k=h*rho/(m**2*A)#\n",
+      "print \"Thermal conductivity of the rod material = %0.3f W/m-k\" %k"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thermal conductivity of the rod material = 293.699 W/m-k\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  3.3 - Page : 86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import tanh\n",
+      "# given data\n",
+      "t=1   # in mm\n",
+      "t=t*10**-3   # in meter\n",
+      "L= 10   # in mm\n",
+      "L= L*10**-3   # in meter\n",
+      "k= 380   # W/mK\n",
+      "To= 230   # in \u00b0C\n",
+      "T_inf= 30   # in \u00b0C\n",
+      "h= 40   # in W/m**2K\n",
+      "B= 1   # in meter\n",
+      "Ac= B*t   # in m**2\n",
+      "rho= 2*(B+t)#\n",
+      "m= sqrt(h*rho/(k*Ac))#\n",
+      "# Part(a)\n",
+      "nita= tanh(m*L)/(m*L)*100   # fin efficiency in %\n",
+      "print \"Fin efficiency = %0.3f %%\" %nita\n",
+      "\n",
+      "# Part(b)\n",
+      "N=1000/9+1   # number of fin\n",
+      "Af= N*rho*L   # in square meter\n",
+      "A1= 1   # plate area in m**2\n",
+      "A2= N*1*1*10**-3   # Area where fins are attached in square meter\n",
+      "Au= A1-A2   # in square meter\n",
+      "q_T= N*sqrt(h*rho*k*Ac)*(To-T_inf)*tanh(m*L)+Au*h*(To-T_inf)   # in W/m**2\n",
+      "print \"Total heat transfer per square meter of plane wall surface = %0.3f kW/m**2\" %(q_T*10**-3)\n",
+      "\n",
+      "# Part(c)\n",
+      "A=1*1   # in m**2\n",
+      "q= h*A*(To-T_inf)   # in W/m**2\n",
+      "print \"Heat transfer if there were no fins attached = %0.0f kW/m**2\" %(q*10**-3)\n",
+      "\n",
+      "# Note : Answer of part(b) in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fin efficiency = 99.303 %\n",
+        "Total heat transfer per square meter of plane wall surface = 24.934 kW/m**2\n",
+        "Heat transfer if there were no fins attached = 8 kW/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  3.4 - Page : 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt, tanh\n",
+      "# given data\n",
+      "w=5*10**-2   # in meter\n",
+      "L=1   # in meter\n",
+      "t=2.5*10**-2   # in meter\n",
+      "h=47   # in W/m**2K\n",
+      "k=16.3   # in W/mK (for 18.8 steel)\n",
+      "T_0=100   # in degree C\n",
+      "T_infinite=20   # in degree C\n",
+      "Ac=w*t   # in square meter\n",
+      "rho=2*(w+t)#\n",
+      "m=sqrt(h*rho/(k*Ac))#\n",
+      "q_fin=k*Ac*m*(T_0-T_infinite)*((tanh(m*L)+h/(k*m)  )/(1+h/(m*k)*tanh(m*L)))\n",
+      "print \"The heat lost by the fin of one meter length is :\",round(q_fin,3),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat lost by the fin of one meter length is : 30.32 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  3.5 - Page : 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import floor\n",
+      "# given data\n",
+      "w=1   # in meter\n",
+      "L=2.5*10**-2   # in meter\n",
+      "t=0.8*10**-3   # in meter\n",
+      "l=1   # in meter\n",
+      "T_0=150   # in degree C\n",
+      "T_infinite=40   # in degree C\n",
+      "h=20   # in W/m**2K\n",
+      "k=65   # in W/mK (for 18.8 steel)\n",
+      "Ac=w*t#\n",
+      "d=5*10**-2   # Cylinder dia in meter\n",
+      "rho=2*(w+t)#\n",
+      "rho=floor(rho)#\n",
+      "\n",
+      "m=sqrt(h*rho/(k*Ac))#\n",
+      "mL=m*L#\n",
+      "# heat transfer rate from 12 fins\n",
+      "q_fin=12*k*Ac*m*(T_0-T_infinite)*((tanh(m*L)+h/(k*m)  )/(1+h/(m*k)*tanh(m*L)))#\n",
+      "print \"Heat transfer rate from 12 fins is :\",round(q_fin,3),\"watt\"\n",
+      "Au=pi*d*l-12*w*t#\n",
+      "qu=h*Au*(T_0-T_infinite)#\n",
+      "print \"Now heat transfer from unfinned surface area is :\",round(qu,3),\"watt\"\n",
+      "q=q_fin+qu#\n",
+      "print \"Total head transfer rate from the cylinder is :\",round(q,3),\"watt\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate from 12 fins is : 1155.94 watt\n",
+        "Now heat transfer from unfinned surface area is : 324.455 watt\n",
+        "Total head transfer rate from the cylinder is : 1480.395 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  3.6 - Page : 89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sinh\n",
+      "# given data\n",
+      "T_0=100   # in degree C\n",
+      "T_infinite=30   # in degree C\n",
+      "T_L=100   # in degree C\n",
+      "d=6*10**-3   # copper rod dia in meter\n",
+      "L=50*10**-2   # developed length in meter\n",
+      "Ac=pi*d**2/4   # in square meter\n",
+      "rho=pi*d   # in meter\n",
+      "h=30   # in W/m**2K\n",
+      "k=330   # in W/mK \n",
+      "m=sqrt(h*rho/(k*Ac))#\n",
+      "#(i) Temperature distribution equation for the fin\n",
+      "# (T-T_infinite)/(T_0-T_infinite)=([(T_L-T_infinite)/(T_0-T_infinite)]*sinh(m*x)+sinh(m*(L-x)))/sinh(m*L)\n",
+      "#Temperature at\n",
+      "x=0.25   # in m\n",
+      "T= (((T_L-T_infinite)/(T_0-T_infinite))*sinh(m*x)+sinh(m*(L-x)))/sinh(m*L)*(T_0-T_infinite)+T_infinite#\n",
+      "print \"(i) Temperature at the centre of the rod is :\",round(T,1),\"degree C\"\n",
+      "print \"(ii) Heat transfer rate from the fin- This is equivalent to two fins of length 25 cm long with insulated tip\"\n",
+      "L=25*10**-2   # in meter\n",
+      "q=2*sqrt(h*rho*k*Ac)*(T_0-T_infinite)*tanh(m*L)#\n",
+      "print \"Heat transfer by the rod is :\",round(q,2),\"watt\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(i) Temperature at the centre of the rod is : 49.6 degree C\n",
+        "(ii) Heat transfer rate from the fin- This is equivalent to two fins of length 25 cm long with insulated tip\n",
+        "Heat transfer by the rod is : 9.76 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  3.7 - Page : 91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import cosh\n",
+      "# given data\n",
+      "T_0=100   # in degree C\n",
+      "T_infinite=25   # in degree C\n",
+      "d=5*10**-2   # in meter\n",
+      "L=15*10**-2   # in meter\n",
+      "h=8   # in W/m**2K\n",
+      "k=20   # in W/mK \n",
+      "rho=pi*d   # in meter\n",
+      "Ac=pi*d**2/4   # in square meter\n",
+      "m=sqrt(h*rho/(k*Ac))#\n",
+      "#(ii) Temperature at free end i.e. at \n",
+      "x=L\n",
+      "# Formula  (T_L-T_infinite)/(T_0-T_infinite)= 1/(cosh(m*L)+h/(k*m)*sinh(m*L) )\n",
+      "T_L=(1/(cosh(m*L)+h/(k*m)*sinh(m*L) ))*(T_0-T_infinite)+T_infinite#\n",
+      "print \"(ii) Temperature at free end is :\",round(T_L,3),\"degree C\"\n",
+      "\n",
+      "#(iii) Heat flow out the source means heat transfer from the fin\n",
+      "q_f=sqrt(h*rho*k*Ac)*(T_0-T_infinite)*((h/(k*m)+tanh(m*L))/(1+h*tanh(m*L)/(k*m)))\n",
+      "print \"(iii) Heat flow out the source :\",round(q_f,3),\"watt\"\n",
+      "\n",
+      "# (iv) Heat flow rate at free end\n",
+      "q_L=h*Ac*(T_L-T_infinite)\n",
+      "print \"(iv) Heat flow rate at free end is :\",round(q_L,3),\"watt\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(ii) Temperature at free end is : 76.739 degree C\n",
+        "(iii) Heat flow out the source : 12.089 watt\n",
+        "(iv) Heat flow rate at free end is : 0.813 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  3.8 - Page : 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "import math\n",
+      "from math import sqrt, tanh\n",
+      "# given data\n",
+      "T_0=150   # in degree C\n",
+      "T_infinite=40   # in degree C\n",
+      "w=1   # in m\n",
+      "t=0.75*10**-3   # in m\n",
+      "d=5*10**-2   # in meter\n",
+      "L=25*10**-3   # in meter\n",
+      "k=75   # in W/mK \n",
+      "h=23.3   # in W/m**2K\n",
+      "N=12   # numbers of fins\n",
+      "Ac=w*t   #in square meter\n",
+      "rho=2*(w+t)   # in meter\n",
+      "delta=Ac/rho#\n",
+      "L_c=L+delta#\n",
+      "ML_c=L_c*sqrt(h*rho/(k*Ac))\n",
+      "q_fin= N*sqrt(h*rho*k*Ac)*(T_0-T_infinite)*tanh(ML_c)#\n",
+      "q_fin=math.floor(q_fin)#\n",
+      "A_0=pi*d*w-12*Ac\n",
+      "q_unfin= h*A_0*(T_0-T_infinite)#\n",
+      "q_total=q_fin+q_unfin#\n",
+      "print \"Rate of heat transfer is :\",round(q_total,1),\"watt\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rate of heat transfer is : 1711.5 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  3.9 - Page : 95"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "# given data\n",
+      "L=0.06   # in meter\n",
+      "A=4.64*10**-4   # in m**2\n",
+      "rho=0.12   # in m\n",
+      "h=442   # in W/m**2\n",
+      "T_0=773   # in K\n",
+      "T_infinite=1143   # in K\n",
+      "K=23.2   # in W/mK\n",
+      "m=sqrt(h*rho/(K*A))#\n",
+      "q=sqrt(h*rho*K*A)*(T_0-T_infinite)*tanh(m*L)#\n",
+      "print \"Heat transfer rate is :\",round(q,3),\"watt\"\n",
+      "\n",
+      "# Note: Answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate is : -279.457 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  3.10 - Page : 95"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from math import cosh\n",
+      "# given data\n",
+      "L=0.12   # in meter\n",
+      "t=.15*10**-2   # thickness in m\n",
+      "K=55.5   # in W/mK\n",
+      "h=23.5   # in W/mK\n",
+      "T_L=357   # in K\n",
+      "T_0=313   # in K\n",
+      "\n",
+      "# Formula m=sqrt(h*rho/(K*A)) and rho=pi*d and A=pi*d*t, putting value of rho and A\n",
+      "m=sqrt(h/(K*t))#\n",
+      "mL=m*L#\n",
+      "mL=math.floor(mL)#\n",
+      "# Formula (T_L-T_infinite)/(T_0-T_infinite)= 1/cosh(m*L)\n",
+      "T_infinite=(T_L-T_0/cosh(mL))/(1-1/cosh(mL))#\n",
+      "T_infinite=math.ceil(T_infinite)#\n",
+      "measurement_error=T_infinite-T_L#\n",
+      "print \"Measurement Error is :\",round(measurement_error,0),\"K\"\n",
+      "\n",
+      "# Note: In the book, Unit of answer is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Measurement Error is : 16.0 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  3.11 - Page : 96"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "# given data\n",
+      "k=20   # in W/mK\n",
+      "T_L=150   # in degree C\n",
+      "T_0=70   # in degree C\n",
+      "L=12*10**-2   # in meter\n",
+      "h=80   # in W/m**2K\n",
+      "t=3*10**-3   # in m\n",
+      "# Formula m=sqrt(h*rho/(K*A)) and rho=pi*d and A=pi*d*t, putting value of rho and A\n",
+      "m=sqrt(h/(k*t))#\n",
+      "# Formula (T_L-T_infinite)/(T_0-T_infinite)= 1/cosh(m*L)\n",
+      "T_infinite=(T_L-T_0/cosh(m*L))/(1-1/cosh(m*L))#\n",
+      "PercentageError=(T_infinite-T_L)*100/T_infinite#\n",
+      "print \"Percentage Error is :\",round(PercentageError,2),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percentage Error is : 1.35 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  3.12 - Page : 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import acosh\n",
+      "# given data\n",
+      "k=30   # in W/mK\n",
+      "h=100   # in W/m**2K\n",
+      "T_infinite=300   # in degree C\n",
+      "d=2*10**-2   # in m\n",
+      "t=1*10**-3   # in m\n",
+      "err=1   # in % of applied temperature difference\n",
+      "# Formula m=sqrt(h*rho/(K*A)) and rho=pi*d and A=pi*d*t, putting value of rho and A\n",
+      "m=sqrt(h/(k*t))#\n",
+      "\n",
+      "# From (T_L-T_infinite)/(T_0-T_infinite)= 1/100 = 1/cosh(m*L)\n",
+      "L=acosh(100)/m   # in meter\n",
+      "L=L*10**3   # in mm\n",
+      "print \"Minimum length os pocket is :\",round(L,2),\"mm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum length os pocket is : 91.77 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  3.13 - Page : 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "k=32   # in W/m**2 degree C\n",
+      "h=14.8   # in W/m**2 degree C\n",
+      "t_o=480   # in degree C\n",
+      "t_i=55   # in degree C\n",
+      "t_a=20   # in degree C\n",
+      "d=2.5*10**-2   # in m\n",
+      "rho=pi*d   # in m\n",
+      "Ac=pi*d**2/4   # in m**2\n",
+      "m=sqrt(h*rho/(k*Ac))#\n",
+      "# From (t-t_a)/(t_o-t_a) = cosh(m(L-x))/cosh(m*L)\n",
+      "L=acosh((t_o-t_a)/(t_i-t_a))/m#  # at x=L,t=t_i\n",
+      "print \"Length of shaft specified between the motor and the pump is :\",round(L,4),\"meter\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Length of shaft specified between the motor and the pump is : 0.3798 meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_4.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_4.ipynb
new file mode 100644
index 00000000..9a0797ed
--- /dev/null
+++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_4.ipynb
@@ -0,0 +1,760 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 - Transient (Unsteady State) Heat Conduction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  4.1 - Page :  102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# given data\n",
+      "L=1   # in m\n",
+      "rho=1600   # in kg/m**3\n",
+      "k=40   # in w/mK\n",
+      "Cp=4*10**3   # in J/kgK\n",
+      "a=900   # in  degree C\n",
+      "b=-300   # in  degree C/m\n",
+      "c=-50   # in  degree C/m**2\n",
+      "Qg=1*10**3   # in kW/m**2\n",
+      "A=10   # area in m**2\n",
+      "#t=a+b*x+c*x**2 at any instant, so\n",
+      "# dtBYdx= b+2*c*x\n",
+      "# d2tBYdx2 = 2*c, then\n",
+      "\n",
+      "# Part(a)\n",
+      "#q1= -k*A*dtBYdx , at\n",
+      "x=0#\n",
+      "q1= -k*A*(b+2*c*x)   # in w\n",
+      "#q2= -k*A*dtBYdx , at\n",
+      "x=L#\n",
+      "q2= -k*A*(b+2*c*x)   # in w\n",
+      "E_stored= (q1-q2)+Qg*A*L   # in watt\n",
+      "print \"The rate of change of energy storage = %0.1e watt\" %E_stored\n",
+      "\n",
+      "# Part(b)\n",
+      "alpha= k/(rho*Cp)   # in m**2s\n",
+      "d2tBYdx2 = 2*c#\n",
+      "dtBYdtoh= alpha*(d2tBYdx2+Qg/k )   # in degree C/sec\n",
+      "print \"Rate of change of temperature = %0.3e degree C/sec\" %dtBYdtoh\n",
+      "print \"Since dt by dx is independent of x. Hence time rate of charge of temperature throughout wall will remain same.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rate of change of energy storage = -3.0e+04 watt\n",
+        "Rate of change of temperature = -4.688e-04 degree C/sec\n",
+        "Since dt by dx is independent of x. Hence time rate of charge of temperature throughout wall will remain same.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  4.2 - Page :  109"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import exp\n",
+      "from numpy import log, pi\n",
+      "# given data\n",
+      "k=40   # in W/mK\n",
+      "rho=7800   # in kg/m**3\n",
+      "C=450   # in J/kgK\n",
+      "d=20*10**-3   # in m\n",
+      "r=d/2#\n",
+      "t_i=400   # in degree C\n",
+      "t=85   # in degree C\n",
+      "t_infinite=25   # in degree C\n",
+      "h=80   # in W/m**2K\n",
+      "#l_s=V/A = (4/3*pi*r**3)/(4*pi*r**2) = r/3\n",
+      "l_s=r/3   # in m\n",
+      "Bi= h*l_s/k#\n",
+      "# since Biot number is less than 0.1, hence lumped heat capacity system analysis can be applied\n",
+      "\n",
+      "# Part(a)\n",
+      "# Formula (t-t_infinite)/(t_i-t_infinite)= %e**(-h*A*toh/(rho*V*C)) = %e**(-h*toh/(rho*l_s*C))\n",
+      "toh= -log((t-t_infinite)/(t_i-t_infinite))*(rho*l_s*C)/h   # in sec\n",
+      "print \"The time require to cool the sphere = %0.3f sec\" %toh\n",
+      "\n",
+      "# Part(b)\n",
+      "# dtBYdtoh = h*A*(t_i-t_infinite)/(rho*V*C) = h*(t_i-t_infinite)/(rho*l_s*C)\n",
+      "dtBYdtoh = h*(t_i-t_infinite)/(rho*l_s*C)   # in degree C/sec\n",
+      "print \"Initial rate of cooling = %0.3f degree C/sec\" %dtBYdtoh\n",
+      "\n",
+      "# Part(c)\n",
+      "A=4*pi*r**2#\n",
+      "toh=60#\n",
+      "q_in= h*A*(t_i-t_infinite)*exp(-h*toh/(rho*l_s*C))   # in watt\n",
+      "print \"Instantaneous heat transfer rate = %0.3f watt\" %q_in\n",
+      "\n",
+      "# Part(d) Total energy transferred during first one minute\n",
+      "V=4/3*pi*r**3#\n",
+      "TotalEnergy = rho*C*V*(t_i-t_infinite)*(1-exp(-h*toh/(rho*C*l_s)))#\n",
+      "print \"Total energy transferred during first one minute = %0.3f watt\" %TotalEnergy\n",
+      "\n",
+      "# Note: Answer of first and last part in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The time require to cool the sphere = 268.015 sec\n",
+        "Initial rate of cooling = 2.564 degree C/sec\n",
+        "Instantaneous heat transfer rate = 25.013 watt\n",
+        "Total energy transferred during first one minute = 1855.401 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  4.3 - Page :  111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import exp\n",
+      "# given data\n",
+      "k=40   # in W/mK\n",
+      "rho=8200   # in kg/m**3\n",
+      "C=400   # in J/kgK\n",
+      "D=6*10**-3   # in m\n",
+      "R=D/2#\n",
+      "t_i=30   # in degree C\n",
+      "t_infinite1=400   # for 10 sec in degree C\n",
+      "t_infinite2=20   # for 10 sec in degree C\n",
+      "h=50   # in W/m**2K\n",
+      "\n",
+      "# Part(a)\n",
+      "#l_s= V/A = R/3\n",
+      "l_s= R/3   # in m\n",
+      "#toh= rho*V*C/(h*A) = rho*C*l_s/h\n",
+      "toh= rho*C*l_s/h   # in sec\n",
+      "print \"Time constance = %0.1f sec\" %toh\n",
+      "\n",
+      "# Part (b)\n",
+      "Bi= h*l_s/k#\n",
+      "# since Bi < 0.1 , hence lumped heat capacity analysis is valid. Now , temperature attained by junction in 10 seconds when exposed to hot air at 400 degree C\n",
+      "toh=10   # in sec\n",
+      "# (t-t_infinite1)/(t_i-t_infinite1)= %e**(-h*A*toh/(rho*V*C)) = %e**(-h*toh/(rho*l_s*C))\n",
+      "t= exp(-h*toh/(rho*l_s*C))*(t_i-t_infinite1)+t_infinite1   # in degree C\n",
+      "\n",
+      "print \"The junction is taken out from hot air stream and placed in stream of still air 20 degree C.\"\n",
+      "print \"The initial temperature in this case = \",round(t,3)\n",
+      "t_i=t#\n",
+      "toh=20   # in sec\n",
+      "t= exp(-h*toh/(rho*l_s*C))*(t_i-t_infinite2)+t_infinite2   # in degree C\n",
+      "print \"The temperature attained by junction = %0.3f degree C\" %t\n",
+      "\n",
+      "# Note: In the last, calculation to find the value of t is wrong so Answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time constance = 65.6 sec\n",
+        "The junction is taken out from hot air stream and placed in stream of still air 20 degree C.\n",
+        "The initial temperature in this case =  82.314\n",
+        "The temperature attained by junction = 65.939 degree C\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  4.4 - Page :  112"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "k=8   # in W/mK\n",
+      "alpha=4*10**-6   # in m**2/s\n",
+      "h=50   # in W/m**2K\n",
+      "D=6*10**-3   # in m\n",
+      "R=D/2#\n",
+      "T=0.5   # where T = (t-t_infinite)/(t_i-t_infinite)\n",
+      "#l_s= V/A = R/3\n",
+      "l_s= R/2   # in m\n",
+      "Bi= h*l_s/k#\n",
+      "# since Bi < 0.1 , hence lumped heat capacity analysis can be applied\n",
+      "# toh= rho*V*C/(h*A) = rho*C*l_s/h = k*l_s/(h*alpha)\n",
+      "toh=  k*l_s/(h*alpha)   # in seconds\n",
+      "print \"Time constant = %0.f seconds\" %toh\n",
+      "# It is given that (t-t_infinite)/(t_i-t_infinite) = 0.5 =  %e**(-h*A*c /(rho*V*C)) = %e**(-h*c/(rho*l_s*C)) = %e**(-h*alpha*c/(l_s))\n",
+      "# or (t-t_infinite)/(t_i-t_infinite) = %e**(-h*alpha*c/(l_s)#\n",
+      "c= -log(T)*l_s/(h*alpha)   # in sec\n",
+      "print \"The time required to temperature change to reach half of its initial value = %0.1f seconds\" %c"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time constant = 60 seconds\n",
+        "The time required to temperature change to reach half of its initial value = 5.2 seconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  4.5 - Page :  113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "#t=450-500*x+100*x**2+150*x**3 at any instant, so\n",
+      "# dtBYdx= -500+200*x+450*x**2\n",
+      "L=0.5   # thickness of the wall in meter\n",
+      "k=10   # in W/mK\n",
+      "# Rate of heating entering in the wall per unit area, at\n",
+      "x=0#\n",
+      "#q1= -k*dtBYdx\n",
+      "q1= -k*(-500+200*x+450*x**2)   # in W/m**2\n",
+      "# Rate of heat going out of the wall per unit area , at\n",
+      "x=L#\n",
+      "q2= -k*(-500+200*x+450*x**2)   # in W/m**2\n",
+      "E=q1-q2   # in W/m**2\n",
+      "print \"Heat energy stored per unit area = %0.0f W/m**2\" %E"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat energy stored per unit area = 2125 W/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  4.6 - Page :  114"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# given data\n",
+      "k=385   # in W/mK\n",
+      "h=100   # in W/m**2K\n",
+      "delta =2*10**-3   # thickness of plate in meter\n",
+      "A=25*25   # area of plate in square meter\n",
+      "rho=8800   # kg/m**3\n",
+      "C=400   # J/kg-K\n",
+      "# l_s= V/A= L*B*delta/(2*L*B) = delta/2\n",
+      "l_s= delta/2   # in meter\n",
+      "Bi= h*l_s/k#\n",
+      "# since Bi < 0.1 , hence lumped heat capacity analysis can be applied\n",
+      "\n",
+      "# Part(i)\n",
+      "# toh= rho*V*C/(h*A) = rho*C*l_s/h\n",
+      "toh= rho*C*l_s/h   # in second\n",
+      "print \"Time constant = %0.1f seconds\" %toh\n",
+      "\n",
+      "# Part(ii)\n",
+      "t_i=400   # in degree C\n",
+      "t=40   # in degree C\n",
+      "t_infinite=25   # in degree C\n",
+      "# (t-t_infinite)/(t_i-t_infinite) =  %e**(-h*A*toh /(rho*V*C)) = %e**(-h*toh/(rho*l_s*C)) \n",
+      "toh= -log((t-t_infinite)/(t_i-t_infinite))*rho*C*l_s/h   # in sec\n",
+      "print \"The time required for the plate to reach the temperature of 40 degree C = %0.1f seconds\" %toh"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time constant = 35.2 seconds\n",
+        "The time required for the plate to reach the temperature of 40 degree C = 113.3 seconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  4.7 - Page :  114"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "k=380   # in W/mK\n",
+      "delta =6*10**-2   # thickness of plate in meter\n",
+      "rho=8800   # kg/m**3\n",
+      "C=400   # J/kg-K\n",
+      "# l_s= V/A = delta/2\n",
+      "l_s= delta/2   # in meter\n",
+      "t=80   # in degree C\n",
+      "t_i=200   # in degree C\n",
+      "t_inf=30   # in degree C\n",
+      "hw= 75   # in W/m**2K\n",
+      "ha= 10   # in W/m**2K\n",
+      "\n",
+      "# Part(i)\n",
+      "# ha*A*(t-t_inf_a)+ hw*A*(t-t_inf_w) = -rho*V*C*dtBYdtho, since t_ini_a = t_inf_w = t_inf = 30 degree C\n",
+      "# (ha+hw)*A*(t-t_inf)= -rho*V*C*dtBYdtho\n",
+      "# (ha+hw)/(rho*C*V)*A*dtoh = -dt/(t-t_inf)\n",
+      "# integrate('(ha+hw)/(rho*V*C)*A','toh',0,toh) = integrate('1/(t-t_inf)','t',t_i,t)\n",
+      "toh= -rho*l_s*C/(ha+hw)*log((t-t_inf)/(t_i-t_inf))#\n",
+      "print \"Time required to cool plate to 80 degree C is :\",round(toh,1),\"seconds =\",round(toh/60,2),\"minutes\"\n",
+      "\n",
+      "# Part (ii)\n",
+      "t= -rho*l_s*C/(2*ha)*log((t-t_inf)/(t_i-t_inf))#\n",
+      "print \"Time required to cool plate in only air is :\",round(t,1),\"seconds =\",round(t/60,2),\"minutes\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time required to cool plate to 80 degree C is : 1520.4 seconds = 25.34 minutes\n",
+        "Time required to cool plate in only air is : 6461.5 seconds = 107.69 minutes\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  4.8 - Page :  116"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "k=45   # in W/m degree C\n",
+      "d =0.1   # in meter\n",
+      "l =0.30   # in meter\n",
+      "t=800   # in degree C\n",
+      "t_i=100   # in degree C\n",
+      "t_infinite=1200   # in degree C\n",
+      "h= 120   # in W/m**2 degree C\n",
+      "alpha=1.2*10**-5   # in meter\n",
+      "rhoC= k/alpha#\n",
+      "V=pi/4*d**2*l   # in m**3\n",
+      "A= pi*d*l + 2*pi/4*d**2   # in m**2\n",
+      "# l_s= V/A = (pi/4*d**2*l)/(pi*d*l + 2*pi/4*d**2) = d*l/(4*l+2*d**2)\n",
+      "l_s = d*l/(4*l+2*d**2)#\n",
+      "Bi= h*l_s/k#\n",
+      "# since Bi < 0.1 , hence lumped heat capacity analysis can be applied\n",
+      "# (t-t_infinite)/(t_i-t_infinite)  =  %e**(-h*A*toh /(rho*V*C)) = %e**(-h*toh/(rho*l_s*C)) = %e**(-h*toh/(rhoC*l_s))\n",
+      "toh = -log((t-t_infinite)/(t_i-t_infinite))*rhoC*l_s/h   # in sec\n",
+      "\n",
+      "# So, the velocity of ingot passing through the furnace\n",
+      "FurnaceLength = 8*100   # in cm\n",
+      "time = toh#\n",
+      "Velocity = FurnaceLength/time   # in cm/sec\n",
+      "print \"Maximum speed = %0.4f cm/sec\" %Velocity"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum speed = 1.0291 cm/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  4.9 - Page :  117"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "rho=8500   # in kg/m**3\n",
+      "C=400   # J/kgK\n",
+      "toh=1   # in sec\n",
+      "h= 400   # in W/m**2 degree C\n",
+      "t=198   # in degree C\n",
+      "t_i=25   # in degree C\n",
+      "t_infinite=200   # in degree C\n",
+      "\n",
+      "# Part (1)\n",
+      "# toh =rho*V*C/(h*A) = rho*C*l_s/h\n",
+      "l_s= toh*h/(rho*C)#\n",
+      "# l_s = V/A = r/3 \n",
+      "r=3*l_s   # in m\n",
+      "r=r*10**3   # in mm\n",
+      "d=2*r   # in m\n",
+      "print \"Junction diameter needed for the thermocouple = %0.3f mili miter\" %d\n",
+      "\n",
+      "# Part(ii)\n",
+      "# toh= -rho*V*C/(h*A)*log((t-t_infinite)/(t_i-t_infinite))  \n",
+      "toh = -toh*log((t-t_infinite)/(t_i-t_infinite))#\n",
+      "print \"Time required for the thermocouple junction to reach 198 degree C = %0.3f seconds\" %toh"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Junction diameter needed for the thermocouple = 0.706 mili miter\n",
+        "Time required for the thermocouple junction to reach 198 degree C = 4.472 seconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  4.10 - Page :  118"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "L=40*10**-2   # in m\n",
+      "k=1.5   # in W/mK\n",
+      "A=4   # in square meter\n",
+      "alpha=1.65*10**-3   # in m**2/h\n",
+      "#T = 50-40*x+10*x**2+20*x**3-15*x**4 , so\n",
+      "# dtBYdx= -40+20*x+60*x**2-60*x**3\n",
+      "# d2tBYdx2 = 20+120*x-180*x**2\n",
+      "\n",
+      "# Part (a) Heat entering the slab\n",
+      "#q1= -k*A*dtBYdx , at\n",
+      "x=0#\n",
+      "qi= -k*A*(-40+20*x+60*x**2-60*x**3)   # in w\n",
+      "print \"Heat entering the slab = %0.0f watt\" %qi\n",
+      "# Heat leaving the slab\n",
+      "#ql= -k*A*dtBYdx , at\n",
+      "x=L#\n",
+      "ql= -k*A*(-40+20*x+60*x**2-60*x**3)   # in w\n",
+      "print \"Heat leaving the slab = %0.2f watt\" %ql\n",
+      "\n",
+      "# Part (b) Rate of heat storage\n",
+      "RateOfHeatStorage = qi-ql   # in watt\n",
+      "print \"Rate of heat storage = %0.2f watt\" %RateOfHeatStorage\n",
+      "\n",
+      "# Part (c) Rate of temperature change\n",
+      "# d2tBYdx2 = 1/alpha*dtBYdtoh\n",
+      "# dtBYdtoh= alpha*d2tBYdx2, at\n",
+      "x=0#\n",
+      "dtBYdtoh = alpha*(20+120*x-180*x**2)   # in degree C/h\n",
+      "print \"The rate of temperature change at entering the slab = %0.3f degree C/h\" %dtBYdtoh\n",
+      "# dtBYdtoh= alpha*d2tBYdx2, at\n",
+      "x=L\n",
+      "dtBYdtoh = alpha*(20+120*x-180*x**2)   # in degree C/h\n",
+      "print \"The rate of temperature change at leaving the slab = %0.3f degree C/h\" %dtBYdtoh\n",
+      "\n",
+      "# Part (d) for the rate of heating or cooling to be maximum\n",
+      "# dBYdx of dtBYdtoh = 0\n",
+      "# dBYdx of (alpha*d2tBYdx2) =0\n",
+      "# d3tBYdx3 = 0\n",
+      "x=120/360   # in meter\n",
+      "print \"The point where rate of heating or cooling is maximum = %0.3f meter\" %x"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat entering the slab = 240 watt\n",
+        "Heat leaving the slab = 157.44 watt\n",
+        "Rate of heat storage = 82.56 watt\n",
+        "The rate of temperature change at entering the slab = 0.033 degree C/h\n",
+        "The rate of temperature change at leaving the slab = 0.065 degree C/h\n",
+        "The point where rate of heating or cooling is maximum = 0.333 meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  4.11 - Page :  119"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "k=40   # in W/m degree C\n",
+      "d =12*10**-3   # in meter\n",
+      "t=127   # in degree C\n",
+      "t_i=877   # in degree C\n",
+      "t_infinite=52   # in degree C\n",
+      "h= 20   # in W/m**2 degree C\n",
+      "rho=7800   # in W/m**2K\n",
+      "C=600   # in J/kg K\n",
+      "r=d/2   # in meter\n",
+      "#l_s = V/A = r/3\n",
+      "l_s =  r/3#\n",
+      "Bi= h*l_s/k#\n",
+      "# since Bi < 0.1 , hence lumped heat capacity analysis can be applied\n",
+      "# (t-t_infinite)/(t_i-t_infinite)  =  %e**(-h*A*toh /(rho*V*C)) = %e**(-h*toh/(rho*l_s*C)) = %e**(-h*toh/(rho*C*l_s))\n",
+      "toh = -log((t-t_infinite)/(t_i-t_infinite))*rho*C*l_s/h   # in sec\n",
+      "print \"Time required for cooling process =\",round(toh,3),\"seconds =\",round(toh/60,2),\"minutes\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time required for cooling process = 1122.215 seconds = 18.7 minutes\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  4.12 - Page :  127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "D=10*10**-2   # in m\n",
+      "b=D/2#\n",
+      "h= 100   # in W/m**2 degree C\n",
+      "T_o=418   # in degree C\n",
+      "T_i=30   # in degree C\n",
+      "T_infinite=1000   # in degree C\n",
+      "\n",
+      "print \" (A) For copper cylinder \"\n",
+      "k=350   # in W/mK\n",
+      "alpha=114*10**-7   # in m**2/s\n",
+      "Bi= h*b/k#\n",
+      "theta_0_t = (T_o-T_infinite)/(T_i-T_infinite)#\n",
+      "Fo=18.8#\n",
+      "# Formula Fo= alpha*t/b**2\n",
+      "t=Fo*b**2/alpha#\n",
+      "print \"Time required to reach for the cylinder centreline temperature 418 degree C =\",round(t,3),\"seconds =\",round(t/3600,3),\"hours\"\n",
+      "\n",
+      "# (2) Temperature at the radius of 4 cm\n",
+      "theta_0_t = 0.985#\n",
+      "# Formula theta_0_t = (T-T_infinite)/(T_o-T_infinite)\n",
+      "T= theta_0_t*(T_o-T_infinite)+T_infinite   # in degree C\n",
+      "print \"Temperature at the radius of 4 cm = %0.2f degree C\" %T \n",
+      "print \"It has very less temperature gradients over 4 cm radius\"\n",
+      "\n",
+      "print \" (B) For asbestos cylinder \"\n",
+      "k=0.11   # in W/mK\n",
+      "alpha=0.28*10**-7   # in m**2/s\n",
+      "Bi= h*b/k#\n",
+      "theta_0_t = (T_o-T_infinite)/(T_i-T_infinite)#\n",
+      "Fo=0.21#\n",
+      "# Formula Fo= alpha*t/b**2\n",
+      "t=Fo*b**2/alpha#\n",
+      "print \"Time required to reach for the cylinder centreline temperature 418 degree C =\",round(t,3),\"seconds =\",round(t/3600,1),\"hours\"\n",
+      "\n",
+      "# (2) Temperature at the radius of 4 cm\n",
+      "theta_x_t = 0.286#\n",
+      "# Formula theta_x_t = (T-T_infinite)/(T_o-T_infinite)\n",
+      "T= theta_x_t*(T_o-T_infinite)+T_infinite   # in degree C\n",
+      "print \"Temperature at the radius of 4 cm = %0.3f degree C\" %T \n",
+      "print \"It has large temperature gradients\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (A) For copper cylinder \n",
+        "Time required to reach for the cylinder centreline temperature 418 degree C = 4122.807 seconds = 1.145 hours\n",
+        "Temperature at the radius of 4 cm = 426.73 degree C\n",
+        "It has very less temperature gradients over 4 cm radius\n",
+        " (B) For asbestos cylinder \n",
+        "Time required to reach for the cylinder centreline temperature 418 degree C = 18750.0 seconds = 5.2 hours\n",
+        "Temperature at the radius of 4 cm = 833.548 degree C\n",
+        "It has large temperature gradients\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  4.13 - Page :  133"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "D=5*10**-2   # in m\n",
+      "b=D/2#\n",
+      "h= 500   # in W/m**2 degree C\n",
+      "k=60   # in W/m**2K\n",
+      "rho=7850   # in kg/m**3\n",
+      "C=460   # in J/kg\n",
+      "alpha=1.6*10**-5   # in m**2/s\n",
+      "T_i=225   # in degree C\n",
+      "T_infinite=25   # in degree C\n",
+      "t=2   # in minute\n",
+      "\n",
+      "# Part(i)\n",
+      "Bi= h*b/k#\n",
+      "Fo= alpha*t/b**2#\n",
+      "theta_0_t = 0.18#\n",
+      "# Formula theta_0_t = (T_o-T_infinite)/(T_i-T_infinite)\n",
+      "T_o= theta_0_t*(T_i-T_infinite)+T_infinite   # in degree C\n",
+      "print \"Centreline Temperature of the sphere after 2 minutes of exposure = %0.f degree C \" %T_o\n",
+      "\n",
+      "# Part(2)\n",
+      "depth= 10*10**-3   # in meter\n",
+      "r=b-depth   # in meter\n",
+      "rBYb=r/b#\n",
+      "theta_x_t = 0.95#\n",
+      "# Formula theta_x_t = (T-T_infinite)/(T_o-T_infinite)\n",
+      "T= theta_x_t*(T_o-T_infinite)+T_infinite   # in degree C\n",
+      "print \"The Temperature at the depth of 1 cm from the surface after 2 minutes = %0.1f degree C\" %T\n",
+      "\n",
+      "# Part (3)\n",
+      "BiSquareFo= Bi**2*Fo#\n",
+      "QbyQo= 0.8   # in kJ\n",
+      "A=4/3*pi*b**3#\n",
+      "Qo= rho*A*C*(T_i-T_infinite)   # in J\n",
+      "Qo=Qo*10**-3   # in kJ\n",
+      "# The heat transffered during 2 minute, \n",
+      "Q= Qo*QbyQo   # in kJ\n",
+      "print \"The heat transffered during 2 minutes = %0.3f kJ\" %Q"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Centreline Temperature of the sphere after 2 minutes of exposure = 61 degree C \n",
+        "The Temperature at the depth of 1 cm from the surface after 2 minutes = 59.2 degree C\n",
+        "The heat transffered during 2 minutes = 37.814 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_5.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_5.ipynb
new file mode 100644
index 00000000..d27ae294
--- /dev/null
+++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_5.ipynb
@@ -0,0 +1,802 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 - Forced Convection Heat Transfer"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.1 - Page :  152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import sqrt\n",
+      "# given data\n",
+      "rho=1.14   # in kg/m**3\n",
+      "k=2.73*10**-2   # in W/mK\n",
+      "Cp=1.005   # in kg/kgK\n",
+      "v= 16*10**-6   # in m**2/s\n",
+      "Pr=0.67#\n",
+      "# Other data given in the problem are\n",
+      "V=2   # in m/s\n",
+      "w=20*10**-2   # in m\n",
+      "t_infinite= 10   # in degree C\n",
+      "t_s=65   # in degree C\n",
+      "x=0.25   # in m from leading edge\n",
+      "# Re= rho*Vx/miu = V*x/v\n",
+      "Re= V*x/v#\n",
+      "#Since Re<5*10**5 , hence the flow is a laminar flow\n",
+      "#(a) Boundary layer thickness\n",
+      "delta= 5*x/(sqrt(Re))   # in m\n",
+      "delta=delta*10**2   # in cm\n",
+      "print \"Boundary layer thickness = %0.3f cm\" %delta\n",
+      "\n",
+      "#(b) Thermal boundary layer thickness\n",
+      "delta_t= delta/Pr**(1/3)   # in cm\n",
+      "print \"Thermal boundary layer thickness = %0.3f cm\" %delta_t\n",
+      "\n",
+      "#(c) Local friction coefficient\n",
+      "Cfx= 0.664/sqrt(Re)#\n",
+      "print \"Local friction coefficient = %0.3f\" %Cfx\n",
+      "Cf=2*Cfx#\n",
+      "print \"Average friction coefficient %0.3f\" %Cf\n",
+      "\n",
+      "#(d) Total drag force\n",
+      "A=.25*.2   # in m**2\n",
+      "toh_o=Cf*(rho*V**2/2)#\n",
+      "F=toh_o*A#\n",
+      "print \"Total drag force = %0.3f N\" %F\n",
+      "\n",
+      "#(e) \n",
+      "# Formula Nux= hx*x/k = 0.332*Re**(1/2)*Pr**(1/3)\n",
+      "hx= 0.332*k/x*Re**(1/2)*Pr**(1/3)   # in W/m**2K\n",
+      "print \"Local heat transfer coefficient = %0.3f W/m**2K\" %hx\n",
+      "h=2*hx#\n",
+      "print \"Average heat transfer coefficient = %0.3f W/m**2K\" %h\n",
+      "#(f)\n",
+      "q=h*A*(t_s-t_infinite)#\n",
+      "print \"Rate of heat transfer = %0.3f W/m**2K\" %q\n",
+      "\n",
+      "#Note: In the book, they calculated wrong value of Re so all the answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Boundary layer thickness = 0.707 cm\n",
+        "Thermal boundary layer thickness = 0.808 cm\n",
+        "Local friction coefficient = 0.004\n",
+        "Average friction coefficient 0.008\n",
+        "Total drag force = 0.001 N\n",
+        "Local heat transfer coefficient = 5.608 W/m**2K\n",
+        "Average heat transfer coefficient = 11.216 W/m**2K\n",
+        "Rate of heat transfer = 30.844 W/m**2K\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.2 - Page :  153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# given data\n",
+      "rho=998   # in kg/m**3\n",
+      "k=.648   # in W/mK\n",
+      "v= 0.556*10**-6   # in m**2/s\n",
+      "Pr=3.54#\n",
+      "V=2   # in m/s\n",
+      "t_infinite= 10   # in degree C\n",
+      "t_s=90   # in degree C\n",
+      "Re=5*10**5#\n",
+      "A=1*1   # in m**2\n",
+      "# Re= rho*Vx/miu = V*x/v\n",
+      "x=Re*v/V   # in m\n",
+      "print \"Length of the plate = %0.3f m\" %x\n",
+      "# Nu = h*x/k =Pr**(1/3)*(0.037*Re**0.8-872)\n",
+      "x=1#\n",
+      "Re= V*x/v#\n",
+      "h= Pr**(1/3)*(0.037*Re**0.8-873)*k/x   # in W/m**2\n",
+      "q=h*A*(t_s-t_infinite)#\n",
+      "print \"Heat transfer from entire plate = %0.f kW\" %int(q*10**-3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Length of the plate = 0.139 m\n",
+        "Heat transfer from entire plate = 444 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.3 - Page :  154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "rho=1.06   # in kg/m**3\n",
+      "K=.0289#\n",
+      "v= 18.97*10**-6   # in m**2/s\n",
+      "Pr=0.696#\n",
+      "V=2.2   # in m/s\n",
+      "L=0.9   # in m\n",
+      "B=0.45   # in m\n",
+      "t_infinite= 30   # in degree C\n",
+      "t_s=90   # in degree C\n",
+      "#(a) For first half of the plate\n",
+      "x=L/2   # in m\n",
+      "Re=V*x/v#\n",
+      "# Nu = h*x/K = 0.664*Re**(1/2)*Pr**(1/3)\n",
+      "h= 0.664*Re**(1/2)*Pr**(1/3)*K/x   # in W/m**2 degree C\n",
+      "A=x*B#\n",
+      "Q1=h*A*(t_s-t_infinite)   # in watt\n",
+      "print \"Heat transfer rate from first half of the plate = %0.f watt\" %Q1\n",
+      "\n",
+      "#(b) Heat transfer from entire plate\n",
+      "x=L   # in m\n",
+      "Re=V*x/v#\n",
+      "# Nu = h*x/K = 0.664*Re**(1/2)*Pr**(1/3)\n",
+      "h= 0.664*Re**(1/2)*Pr**(1/3)*K/x   # in W/m**2 degree C\n",
+      "A=L*B#\n",
+      "Q2=h*A*(t_s-t_infinite)   # in watt\n",
+      "print \"Heat transfer rate from entire plate = %0.3f watt\" %Q2\n",
+      "\n",
+      "#(c) From next half of the plate\n",
+      "Q3= Q2-Q1#\n",
+      "print \"Heat transfer rate from next half of the plate %0.3f\" %Q3"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate from first half of the plate = 105 watt\n",
+        "Heat transfer rate from entire plate = 148.342 watt\n",
+        "Heat transfer rate from next half of the plate 43.448\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.4 - Page :  155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "from numpy import pi\n",
+      "# given data\n",
+      "rho=985   # in kg/m**3\n",
+      "k=.654   # in W/mK\n",
+      "Cp=4.18   # in kgJ/kgK\n",
+      "Cp=Cp*10**3   # in J/kgK\n",
+      "v= 0.517*10**-6   # in m**2/s\n",
+      "Pr=3.26#\n",
+      "V=1.2   # in m/s\n",
+      "t_s=85   # in degree C\n",
+      "t_i=40   # in degree C\n",
+      "t_o=70   # in degree C\n",
+      "Ax=15*35   # in mm\n",
+      "P=15+35#\n",
+      "de=4*Ax/(2*P)   # in mm\n",
+      "de=de*10**-3   # in m\n",
+      "Re=V*de/v#\n",
+      "# Formula Nu= h*de/k = 0.023Re**0.8*Pr**0.4\n",
+      "h=0.023*Re**0.8*Pr**0.4*k/de   # in W/m**2K\n",
+      "m=pi*de**2*V*rho/4#\n",
+      "d=de#\n",
+      "L=m*Cp*log((t_s-t_i)/(t_s-t_o))/(pi*d*h)#\n",
+      "print \"The length of tube = %0.3f meter\" %L"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The length of tube = 4.407 meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.5 - Page :  156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "k=.026   # in W/mK\n",
+      "v= 16.8*10**-6   # in m**2/s\n",
+      "miu=2*10**-5   # in kg/ms\n",
+      "Pr=0.708#\n",
+      "V=15   # in m/s\n",
+      "x=2   # in m\n",
+      "A=2*1   # in m**2\n",
+      "Re=V*x/v#\n",
+      "del_t=40-10   # in degree C\n",
+      "# since Re > 3 *10**5, hence turbulent flow at x=2 m length of laminar flow region is x_L then\n",
+      "Re_1=3*10**5#\n",
+      "# Re_1 = 3*10**5 = V*x_L/v\n",
+      "x_L= Re_1*v/V#\n",
+      "\n",
+      "# Part (a)\n",
+      "#Nu= h*x_L/k = 0.664*Re_1**(1/2)*Pr**(1/3)#\n",
+      "h= 0.664*Re_1**(1/2)*Pr**(1/3)*k/x_L   # in W/m**2\n",
+      "print \"The average heat transfer coefficient over the laminar boundary layer = %0.3f W/m**2 \" %h\n",
+      "\n",
+      "# Part(b)\n",
+      "#Nu= h*x/k = (0.037*Re**0.8-872)*Pr**(1/3)#\n",
+      "h= (0.037*Re**0.8-872)*Pr**(1/3)*k/x   # in W/m**2\n",
+      "print \"The average heat transfer coefficient over entire plate = %0.3f W/m**2\" %h\n",
+      "\n",
+      "# Part (c)\n",
+      "q=h*A*del_t#\n",
+      "print \"Total heat transfer rate = %0.3f watt\" %q\n",
+      "\n",
+      "# Note: Calculation of the part(a) in this book is wrong, so answer of the part(a) in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average heat transfer coefficient over the laminar boundary layer = 25.083 W/m**2 \n",
+        "The average heat transfer coefficient over entire plate = 32.910 W/m**2\n",
+        "Total heat transfer rate = 1974.607 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.6 - Page :  158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "rho=997   # in kg/m**3\n",
+      "k=0.608   # in W/mK\n",
+      "Cp= 4180   # in J/kg K\n",
+      "miu=910*10**-6   # in Ns/m**2\n",
+      "d=30*10**-3   # in m\n",
+      "m=0.02   # in kg/s\n",
+      "t_o=30   # in degree C\n",
+      "t_i=20   # in degree C\n",
+      "Re= 4*m/(pi*d*miu)#\n",
+      "q_desh=12*10**3   # in W/m**2\n",
+      "# since Re < 2300, flow is laminar one\n",
+      "\n",
+      "# Part(a)\n",
+      "# Nu = h*d/k = 4.36\n",
+      "h=4.36*k/d#\n",
+      "print \"Heat transfer coefficient = %0.3f W/m**2K\" %h\n",
+      "\n",
+      "# Part (b)\n",
+      "L=m*Cp*(t_o-t_i)/(q_desh*pi*d)#\n",
+      "print \"Length of pipe = %0.3f meter\" %L\n",
+      "\n",
+      "# Part(c)\n",
+      "# q_desh= h*(t_infinite-t_o)\n",
+      "t_infinite = q_desh/h+t_o#\n",
+      "print \"The inner tube surface temperature at the outlet = %0.3f degree C\" %t_infinite\n",
+      "\n",
+      "# Part(d)\n",
+      "f=64/Re#\n",
+      "print \"Friction Factor = %0.3f \" %f\n",
+      "\n",
+      "# Part(e)\n",
+      "V=4*m/(pi*d**2*rho)   # in m/s  ( because m= rho*V*A , m= rho*V*pi*d**2/4 )\n",
+      "del_P= f*L*rho*V**2/(d*2)   # in N/m**2\n",
+      "print \"The pressure drop in the pipe = %0.3f N/m**2\" %del_P\n",
+      "\n",
+      "# Note: In part(b) value of L is miss printed actual value is .739 m"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer coefficient = 88.363 W/m**2K\n",
+        "Length of pipe = 0.739 meter\n",
+        "The inner tube surface temperature at the outlet = 165.804 degree C\n",
+        "Friction Factor = 0.069 \n",
+        "The pressure drop in the pipe = 0.679 N/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.7 - Page :  159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "from math import log\n",
+      "# given data\n",
+      "rho=977.3   # in kg/m**3\n",
+      "kf=0.665   # in W/mK\n",
+      "Cp= 4186   # in J/kg K\n",
+      "miu=4.01*10**-4   # in kg/m-s\n",
+      "Pr=2.524#\n",
+      "d=0.02   # in m\n",
+      "m=0.5   # in kg/s\n",
+      "t_o=70   # in degree C\n",
+      "t_i=20   # in degree C\n",
+      "t_s=100   # in degree C\n",
+      "Re= 4*m/(pi*d*miu)#\n",
+      "\n",
+      "# Since Re > 2300, flow is turbulent flow. Then Nusselt Number\n",
+      "# Nu = h*d/k = 0.023*Re**0.8*Pr**0.4\n",
+      "h=0.023*Re**0.8*Pr**0.4*kf/d   # in W/m**2\n",
+      "print \"Average heat transfer coefficient = %0.3f W/m**2\" %h\n",
+      "L=m*Cp*log((t_s-t_i)/(t_s-t_o))/(pi*d*h)   # in meter\n",
+      "print \"Length of tube = %0.3f meter\" %L\n",
+      "# Note: Calculation of Re is wrong so the answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average heat transfer coefficient = 9207.036 W/m**2\n",
+        "Length of tube = 3.549 meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.8 - Page :  160"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "rho=977   # in kg/m**3\n",
+      "k=0.608   # in W/mK\n",
+      "Cp= 4180   # in J/kg K\n",
+      "miu=910*10**-6   # in poise\n",
+      "d=0.02   # in m\n",
+      "m=0.02   # in kg/s\n",
+      "t_o=40   # in degree C\n",
+      "t_i=10   # in degree C\n",
+      "q_desh= 20*10**3   # in W/m**2\n",
+      "\n",
+      "# Part (a)\n",
+      "Re= 4*m/(pi*d*miu)#\n",
+      "print \"Reynold number = %0.3f\" %Re\n",
+      "\n",
+      "# Part(b)\n",
+      "# Nu = h*d/k = 4.364\n",
+      "h=4.364*k/d#\n",
+      "print \"Heat transfer coefficient = %0.2f W/m**2K\" %h\n",
+      "\n",
+      "# Part (c)\n",
+      "# q= q_desh*A = m*Cp*(t_o-t_i)\n",
+      "# q_desh *( pi*d*l) = m*Cp*(t_o-t_i)\n",
+      "l=m*Cp*(t_o-t_i)/(q_desh*pi*d)#\n",
+      "print \"Length of pipe = %0.3f meter\" %l"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reynold number = 1399.164\n",
+        "Heat transfer coefficient = 132.67 W/m**2K\n",
+        "Length of pipe = 1.996 meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.9 - Page :  161"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "rho=7.7*10**3   # in kg/m**3\n",
+      "k=12   # in W/mK\n",
+      "Cp= 130   # in J/kg degree C\n",
+      "Pr=0.011#\n",
+      "delta=8*10**-8   # in m**2/s\n",
+      "d=0.06   # in m\n",
+      "m=4   # in kg/s\n",
+      "t_i=200   # in degree C\n",
+      "del_t=25   # in degree C\n",
+      "miu=rho*delta#\n",
+      "Re= 4*m/(pi*d*miu)#\n",
+      "# From correlation  Nu =h*d/k = 4.82+0.0185*Pe**0.827\n",
+      "Pe=Re*Pr#\n",
+      "h=(4.82+0.0185*Pe**0.827)*k/d   # in W/m**2K\n",
+      "# Length of tube required by doing every balance\n",
+      "#  m*Cp*del_t = h*A*(t_s-t_b) = h*(pi*d*l)*(t_s-t_b)   # its given (t_s-t_b) = 40 degree C\n",
+      "l= m*Cp*del_t/(h*(pi*d)*40)   # in meter\n",
+      "print \"Length of tube = %0.3f meter\" %l"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Length of tube = 0.678 meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.10 - Page :  162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "d=0.058   # in m\n",
+      "t_infinite=30   # in degree C\n",
+      "t_s=155   # in degree C\n",
+      "V=52   # in m/s\n",
+      "T_f=(t_s+t_infinite)/2   # in degree C\n",
+      "T_f=T_f+273   # in K\n",
+      "# Fluid properties at 92.5 degree C and 1 atm\n",
+      "miu= 2.145*10**-5   # in kg/ms\n",
+      "Pr=0.696#\n",
+      "P=1.0132*10**5#\n",
+      "R=287#\n",
+      "k=0.0312   # in W/mK\n",
+      "rho=P/(R*T_f)   # in kg/m**3\n",
+      "Re=rho*V*d/miu#\n",
+      "C=0.0266#\n",
+      "n=0.805#\n",
+      "# Nu = h*d/k = C*(Re)**n*Pr**(1/3)\n",
+      "h=C*(Re)**n*Pr**(1/3)*k/d   # in W/m**2K\n",
+      "#So, heat transfer rate per unit length from cylinder\n",
+      "q_by_L= h*(pi*d)*(t_s-t_infinite)   # in W/m\n",
+      "print \"Heat transfer rate per unit length from cylinder = %0.3f W/m\" %q_by_L\n",
+      "# Note: Calculation of q_by_L in the book is wrong , so the answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate per unit length from cylinder = 3914.183 W/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.11 - Page :  163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "delta=15.68*10**-6   # in m**2/s\n",
+      "t_infinite=25+273   # in K\n",
+      "t_s=80+273   # in K\n",
+      "t_infinite=25+273   # in K\n",
+      "k=0.02625   # in W/m degree C\n",
+      "Pr=0.708#\n",
+      "miu_infinite=1.846*10**-5   #in kg/ms\n",
+      "miu_s= 2.076*10**-5   # in kg/ms\n",
+      "d=10*10**-3   # in m\n",
+      "V=5   # in m/s\n",
+      "A=4*pi*(d/2)**2#\n",
+      "Re=V*d/delta#\n",
+      "Nu= 2+ (0.4*Re**(1/2)+0.06*Re**(2/3))*Pr**0.4*(miu_infinite/miu_s)**(1/4)#\n",
+      "# Nu = h*d/k\n",
+      "h=Nu*k/d   # in W/m**2K\n",
+      "# heat transfer rate\n",
+      "q=h*A*(t_s-t_infinite)   # in watt\n",
+      "print \"Heat transfer rate = %0.3f watt\" %q"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate = 1.456 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.12 - Page :  164"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# given data\n",
+      "Cp=4179   # in J/kg-K\n",
+      "rho= 997   # in kg/m**3\n",
+      "V=2   # in m/s\n",
+      "miu= 855*10**-6   # in Ns/m**2\n",
+      "Pr=5.83#\n",
+      "k=0.613#\n",
+      "Do=6   #outer dia in cm\n",
+      "Di=4   #inner dia in cm\n",
+      "# de= 4*A/P = 4*pi/4*(Do**2-Di**2)/(pi*(Do+Di))\n",
+      "# or\n",
+      "de= Do-Di   # in cm\n",
+      "de=de*10**-2   # in m\n",
+      "Re= rho*V*de/miu#\n",
+      "# Since Re > 2300, hence flow is turbulent. Hence using Dittus Boelter equation\n",
+      "# Nu= 0.023*Re**0.8*Pr**0.4 =h*de/k\n",
+      "h= 0.023*Re**0.8*Pr**0.4*k/de   # in W/m**2K\n",
+      "print \"Heat transfer coefficient = %0.f W/m**2K\" %math.floor(h)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer coefficient = 7752 W/m**2K\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.13 - Page :  165"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "Cp=0.138   # in KJ/kg-K\n",
+      "m=8.33   # in kg/sec\n",
+      "Pr=0.0238#\n",
+      "k=8.7   # in W/mk\n",
+      "d=1.5*10**-2   # in m\n",
+      "miu=1.5*10**-3   # in kg/ms\n",
+      "Re=4*m/(pi*miu*d)#\n",
+      "Pe=Re*Pr#\n",
+      "# Nu = h*d/k = 7+0.025*Pe**0.8\n",
+      "h= (7+0.025*Pe**0.8)*k/d   # in W/m**2 degree C\n",
+      "print \"Heat transfer coefficient = %0.3f W/m**2 degree C\" %h"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer coefficient = 29255.771 W/m**2 degree C\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.14 - Page :  166"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "rho=887   # in kg/m**3\n",
+      "Pr=0.026#\n",
+      "k=25.6   # in W/mk\n",
+      "d=2.5*10**-2   # in m\n",
+      "miu=0.58*10**-3   # in kg/ms\n",
+      "V=3   # in m/s\n",
+      "Re=rho*V*d/(miu)#\n",
+      "Pe=Re*Pr#\n",
+      "Nu = 4.8+0.015*Pe**0.85*Pr**0.08\n",
+      "h= Nu*k/d   # in W/m**2 degree C\n",
+      "print \"Heat transfer coefficient = %0.3f W/m**2 degree C\" %h\n",
+      "#Note: There is some difference in coding and book answer because they did not take aqurate calculation"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer coefficient = 15217.633 W/m**2 degree C\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  5.15 - Page :  166"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "delta=38.1*10**-6   # in m**2/s\n",
+      "Pr=501#\n",
+      "Prs=98#\n",
+      "K=0.138   # in W/mk\n",
+      "T_infinite=353   # in K\n",
+      "T_s=423   # in K\n",
+      "V=2   # in m/s\n",
+      "d=12.5*2*10**-3   # in m\n",
+      "Re=V*d/delta#\n",
+      "n=0.36# for Pr >= 10\n",
+      "C=0.26   # for Re between 10**3 and 2*10**5\n",
+      "m=0.6   # for Re between 10**3 and 2*10**5\n",
+      "Nu= C*Re**m*Pr**n*(Pr/Prs)**(1/4)#\n",
+      "h= Nu*K/d   # in W/m**2 degree C\n",
+      "A=pi*25*10**-3#\n",
+      "del_t=T_s-T_infinite#\n",
+      "# Formula q=h*A*del_t\n",
+      "q_by_L = h*A*del_t#\n",
+      "print \"Initial rate of heat loss per meter length of cylinder = %0.3f\" %q_by_L\n",
+      "# Note: calculation in the book is wrong  so answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Initial rate of heat loss per meter length of cylinder = 8260.795\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_6.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_6.ipynb
new file mode 100644
index 00000000..e5eca739
--- /dev/null
+++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_6.ipynb
@@ -0,0 +1,543 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 - Free Convection"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  6.1 - Page : 177"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# given data\n",
+      "# (i) when\n",
+      "x=.3   # in m\n",
+      "T_s=100   # in degree C\n",
+      "T_infinite=30   # in degree C\n",
+      "T_f=(T_s+T_infinite)/2   # in degree C\n",
+      "T_f=T_f+273   # in K\n",
+      "Bita=1/T_f#\n",
+      "# Other fluid properties at film temperature\n",
+      "Pr=0.703#\n",
+      "K=0.0301   # in W/mK\n",
+      "T=1.8*10**-5       # in m**2/s\n",
+      "g=9.81#\n",
+      "del_T=T_s-T_infinite#\n",
+      "Gr=(g*Bita*del_T*x**3)/T**2#\n",
+      "Ra=Gr*Pr#\n",
+      "print \"Rayleigh Number is = %0.2e\" %Ra\n",
+      "#Since Ra<10**9, hence flow is laminar, then correlation for vertical plate in laminar flow\n",
+      "# Formula Nu=0.59*Ra**(1/4)=h*x/K\n",
+      "h=0.59*Ra**(1/4)*K/x   # in W/m**2K\n",
+      "A=2*.3*.5#\n",
+      "q1=h*A*(T_s-T_infinite)#\n",
+      "print \"Heat transfer rate from the plate, when the vertical height is 0.3 m :\",round(q1,3),\"W\"\n",
+      "\n",
+      "#(ii) when\n",
+      "x=0.5   # in m\n",
+      "Gr=(g*Bita*del_T*x**3)/T**2#\n",
+      "Ra=Gr*Pr#\n",
+      "# Formula Nu=0.59*Ra**(1/4)=h*x/K\n",
+      "h=0.59*Ra**(1/4)*K/x   # in W/m**2K\n",
+      "q2=h*A*(T_s-T_infinite)#\n",
+      "print \"Heat transfer rate from the plate, when the vertical height is 0.5 m :\",round(q2,3),\"W\"\n",
+      "PercentageDecrease=(q1-q2)/q1*100#\n",
+      "print \"Percentage decreases in heat transfer rate when x=0.5 m as compared to when x=0.3 m is :\",round(PercentageDecrease,3),\"%\"\n",
+      "\n",
+      "#Note : In the book ,In part (b), calculation of getting the value of h is wrong "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rayleigh Number is = 1.19e+08\n",
+        "Heat transfer rate from the plate, when the vertical height is 0.3 m : 129.844 W\n",
+        "Heat transfer rate from the plate, when the vertical height is 0.5 m : 114.277 W\n",
+        "Percentage decreases in heat transfer rate when x=0.5 m as compared to when x=0.3 m is : 11.989 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  6.2 - Page : 178"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "Pr=0.694#\n",
+      "K=0.0296   # in W/mK\n",
+      "rho=1.029   # in kg/m**3\n",
+      "miu=20.6*10**-6   # in poise\n",
+      "x=.2   # in m\n",
+      "T_s=110   # in degree C\n",
+      "T_infinite=30   # in degree C\n",
+      "T_f=(T_s+T_infinite)/2   # in degree C\n",
+      "T_f=T_f+273   # in K\n",
+      "Bita=1/T_f#\n",
+      "g=9.81#\n",
+      "del_T=T_s-T_infinite#\n",
+      "Gr=(rho**2*g*Bita*del_T*x**3)/miu**2#\n",
+      "Ra=Gr*Pr#\n",
+      "#since Rayleigh number is less than 10**10, hence\n",
+      "Nu=0.68*Pr**(1/2)*Gr**(1/4)/((.952+Pr)**(1/4))#\n",
+      "h=Nu*K/x#\n",
+      "A=2*0.2*1#\n",
+      "q=h*A*(T_s-T_infinite)#\n",
+      "print \"Heat transfer rate is :\",round(q,1),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate is : 194.7 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  6.3 - Page : 179"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "d=7.5*10**-2   # in m\n",
+      "x=2   # in m\n",
+      "T_s=70   # in degree C\n",
+      "T_infinite=10   # in degree C\n",
+      "del_T=T_s-T_infinite#\n",
+      "g=9.81#\n",
+      "calculation=4.5*10**10# # value of g*Bita*rho**2*C_p/(miu*k)\n",
+      "K=2.75*10**-2   # in W/mK\n",
+      "# g*Bita*rho**2*C_p/(miu*k) = g*Bita*rho**2/miu**2 * miu*C_p/k = (g*Bita*del_T*x**3/T**2 * miu*C_p/k)/(del_T*x**3)\n",
+      "GrxPr= calculation*del_T*x**3# # value of Gr*Pr\n",
+      "Nu= 0.13*(GrxPr)**(1/3)#\n",
+      "# Formula Nu = h*x/k\n",
+      "h= Nu*K/x   # in W/m**2K\n",
+      "A=2*pi*d#\n",
+      "q=h*A*(del_T)   # in W\n",
+      "q=q*60*60   # in J/h\n",
+      "print \"Heat transfer rate = %0.3e J/h\" %q"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate = 5.067e+06 J/h\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  6.4 - Page : 180"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "m=15   # in kg\n",
+      "C_p=420   # in J/kg K\n",
+      "T_s=200   # in degree C\n",
+      "T_infinite=30   # in degree C\n",
+      "T_f=(T_s+T_infinite)/2   # in degree C\n",
+      "T_f=T_f+273   # in K\n",
+      "Pr=0.688#\n",
+      "K=0.0321   # in W/mK\n",
+      "delta=23.18*10**-6   # in m**2/s\n",
+      "Bita=1/T_f#\n",
+      "g=9.81#\n",
+      "x=0.3   # in m\n",
+      "del_T=T_s-T_infinite#\n",
+      "Gr=(g*Bita*del_T*x**3)/delta**2#\n",
+      "Ra=Gr*Pr#\n",
+      "#Since Ra<10**9, hence it is laminar flow using the relation\n",
+      "# Formula Nu=0.59*Ra**(1/4)=h*x/K\n",
+      "h=0.59*Ra**(1/4)*K/x   # in W/m**2K\n",
+      "print \"(i) Heat transfer coefficient is :\",round(h,3),\"W/m**2K\"\n",
+      "\n",
+      "# (b) Initial rate of cooling\n",
+      "# Formula h*A*(T_s-T_infinite) = m*C_p*dt_by_toh\n",
+      "A=2*0.3*0.5#\n",
+      "dt_by_toh = h*A*(T_s-T_infinite)/(m*C_p)   # in degree C/sec\n",
+      "dt_by_toh=dt_by_toh*60   # in degree C /min\n",
+      "print \"(ii) Initial rate of cooling of the plate is :\",round(dt_by_toh,3),\"degreeC per min\"\n",
+      "\n",
+      "#(c) Time taken by plate to cool from 200 degree C to 50 degree C\n",
+      "T_i=200   # in degree C\n",
+      "T=50   # in degree C\n",
+      "# Formula (T-T_infinite)/(T_i-T_infinite)= %e**(-h*A*toh/(m*C_p))#\n",
+      "toh= -log((T-T_infinite)/(T_i-T_infinite))*m*C_p/(h*A)   # in sec\n",
+      "toh=toh/60   # in min\n",
+      "print \"(iii) Time required to cool plate from 200 degree C to 50 degree C is :\",round(toh,2),\"minutes\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(i) Heat transfer coefficient is : 6.97 W/m**2K\n",
+        "(ii) Initial rate of cooling of the plate is : 3.385 degreeC per min\n",
+        "(iii) Time required to cool plate from 200 degree C to 50 degree C is : 107.46 minutes\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  6.5 - Page : 182"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "rho=0.8   # in kg/m**3#\n",
+      "C_p=1.01   # in KJ/kg K\n",
+      "Pr=0.684#\n",
+      "d=15*10**-2   # diameter in meter\n",
+      "K=0.035   # in W/mK\n",
+      "delta=2.78*10**-5   # in m**2/s\n",
+      "g=9.81#\n",
+      "x=2   # in m\n",
+      "T_s=250   # in degree C\n",
+      "T_infinite=30   # in degree C\n",
+      "T_f=(T_s+T_infinite)/2   # in degree C\n",
+      "T_f=T_f+273   # in K\n",
+      "Bita=1/T_f#\n",
+      "del_T=T_s-T_infinite#\n",
+      "#Heat loss from vertical part by free convection\n",
+      "Gr=(g*Bita*del_T*x**3)/delta**2#\n",
+      "Ra=Gr*Pr#\n",
+      "#Since Ra>10**9, hence turbulent flow\n",
+      "# Formula Nu= h*x/K =0.13*Ra**(1/3)\n",
+      "h=0.13*Ra**(1/3)*K/x   # in W/m**2K\n",
+      "A=2*pi*d#\n",
+      "q1=h*A*del_T   # w\n",
+      "q1=q1*10**-3   # in kW\n",
+      "print \"Heat loss from vertical part is :\",round(q1,3),\"kW\"\n",
+      "#Heat loss for Horizontal part\n",
+      "# here\n",
+      "x=d#\n",
+      "Gr=(g*Bita*del_T*x**3)/delta**2#\n",
+      "Ra=Gr*Pr#\n",
+      "#Since Ra<10**9, hence laminar fluid flow\n",
+      "# Formula Nu= h*x/K =0.53*Ra**(1/4)\n",
+      "h=0.53*Ra**(1/4)*K/x   # in W/m**2K\n",
+      "A=pi*d*8#\n",
+      "q2=h*A*del_T   # w\n",
+      "q2=q2*10**-3   # in kW\n",
+      "print \"Heat loss for horizontal part is :\",round(q2,3),\"kW\"\n",
+      "\n",
+      "#Heat loss by radiation\n",
+      "sigma=5.67*10**-8#\n",
+      "epsilon=0.65   # emissivity of steel\n",
+      "A=pi*d*10#\n",
+      "T_s=T_s+273   # in K\n",
+      "T_infinite=T_infinite+273   # in K\n",
+      "q3=sigma*A*epsilon*(T_s**4-T_infinite**4)   # in w\n",
+      "q3=q3*10**-3   # in kW\n",
+      "print \"Heat loss by radiation is :\",round(q3,3),\"kW\"\n",
+      "#Total heat loss\n",
+      "theta=q1+q2+q3#\n",
+      "print \"Total heat loss is :\",round(theta,3),\"kW\"\n",
+      "\n",
+      "\n",
+      "#Note : value of q3 and theta in the book is wrong so answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat loss from vertical part is : 1.572 kW\n",
+        "Heat loss for horizontal part is : 6.447 kW\n",
+        "Heat loss by radiation is : 11.53 kW\n",
+        "Total heat loss is : 19.549 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  6.6 - Page : 183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "rho=1.205   # in kg/m**3#\n",
+      "C_p=1006   # in J/kg K\n",
+      "Pr=0.71#\n",
+      "K=0.0256   # in W/mK\n",
+      "delta=1.506*10**-5   # in m**2/s\n",
+      "T_s=35   # in degree C\n",
+      "T_infinite=5   # in degree C\n",
+      "T_f=(T_s+T_infinite)/2   # in degree C\n",
+      "T_f=T_f+273   # in K\n",
+      "Bita=1/T_f#\n",
+      "del_T=T_s-T_infinite#\n",
+      "g=9.81#\n",
+      "# Formula 1/x= 1/Lh + 1/Lv\n",
+      "Lh=50   # in cm\n",
+      "Lv=50   # in cm\n",
+      "x=Lh*Lv/(Lh+Lv)   # in cm\n",
+      "x=x*10**-2   # in m\n",
+      "# Formula Gr=(g*Bita*del_T*x**3)/delta**2#\n",
+      "Gr=(g*Bita*del_T*x**3)/delta**2#\n",
+      "Ra=Gr*Pr#\n",
+      "# Formula Nu= h*x/K =0.53*Ra**(1/4)\n",
+      "h=0.53*Ra**(1/4)*K/x   # in W/m**2K\n",
+      "A=2*(0.5+0.5)#\n",
+      "q=h*A*del_T   # w\n",
+      "print \"Heat loss per meter length of pipe is :\",round(q,3),\"watt\"\n",
+      "\n",
+      "# Note: In the book, value of h is wrong due to place miss value of x, so the answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat loss per meter length of pipe is : 272.624 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  6.7 - Page : 184"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.integrate import quad\n",
+      "# given data\n",
+      "L=3   # in m\n",
+      "delta=0#\n",
+      "hx='10*x**(-1/4)'\n",
+      "# (a) Average heat transfer coefficient\n",
+      "def integrand(x):\n",
+      "    return 10*x**(-1/4)\n",
+      "\n",
+      "ans, err = quad(integrand, delta, L)\n",
+      "\n",
+      "h=1/L*ans\n",
+      "print \"(a) Average heat transfer coefficient is :\",round(h,2),\"W/m**2K\"\n",
+      "\n",
+      "# (b) Heat transfer rate\n",
+      "A=3*.3   # in m**2\n",
+      "Tp=170   # plate temp. in degree C\n",
+      "Tg=30   # gas temp. in degree C\n",
+      "del_T=Tp-Tg#\n",
+      "q=h*A*del_T   # in W\n",
+      "print \"(b) Heat transfer rate is :\",round(q,3),\"W\"\n",
+      "\n",
+      "# (c) \n",
+      "x=2   # in m\n",
+      "qx_by_A= 10*x**(-1/4)*(Tp-Tg)#\n",
+      "print \"Local heat flux 2 m from the leading edge is :\",round(qx_by_A,1),\"W/m**2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Average heat transfer coefficient is : 10.13 W/m**2K\n",
+        "(b) Heat transfer rate is : 1276.524 W\n",
+        "Local heat flux 2 m from the leading edge is : 1177.3 W/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  6.8 - Page : 185"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "Pr=0.712#\n",
+      "K=0.026   # in W/mK\n",
+      "delta=1.57*10**-5   # in m**2/s\n",
+      "T_s=320   # in K\n",
+      "T_infinite=280   # in K\n",
+      "del_T=T_s-T_infinite#\n",
+      "T_f=(T_s+T_infinite)/2   # in K\n",
+      "Bita=1/T_f#\n",
+      "d1=20   # in cm\n",
+      "d2=30   # in cm\n",
+      "x=(d2-d1)/2   # in cm\n",
+      "x=x*10**-2   # in m\n",
+      "g=9.81#\n",
+      "Gr=(g*Bita*del_T*x**3)/delta**2#\n",
+      "Ra=Gr*Pr#\n",
+      "# Formula Nu= h*x/K =0.228*Ra**(0.226)\n",
+      "h=0.228*Ra**(0.226)*K/x   # in W/m**2K\n",
+      "A=pi*(d1*10**-2)**2#\n",
+      "q=h*A*del_T   # w\n",
+      "print \"Heat transfer rate is :\",round(q,1),\"watt\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate is : 11.4 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  6.9 - Page : 186"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "K=0.0278   # in W/mK\n",
+      "rho=1.092   # in kg/m**3\n",
+      "miu=19.57*10**-6   # in kg/ms\n",
+      "Cp=1007   # in kg/kg degree C\n",
+      "epsilon=0.9#\n",
+      "sigma=5.67*10**-8#\n",
+      "d=75+2*25   # in mm\n",
+      "d=d*10**-3   # in meter\n",
+      "T_s=80   # in degree C\n",
+      "T_infinite=20   # in degree C\n",
+      "T_f=(T_s+T_infinite)/2   # in degree C\n",
+      "T_f=T_f+273   # in K\n",
+      "Bita=1/T_f#\n",
+      "g=9.81#\n",
+      "del_T=T_s-T_infinite#\n",
+      "Pr=miu*Cp/K#\n",
+      "Gr=(rho**2*g*Bita*del_T*d**3)/miu**2#\n",
+      "# Formula Nu= h*d/K = 0.53*(Gr*Pr)**(1/4)#\n",
+      "h= 0.53*(Gr*Pr)**(1/4)*K/d#\n",
+      "#(a) Heat loss from 6 m length of pipe\n",
+      "A=pi*d*6#\n",
+      "Q_conv=h*A*del_T#\n",
+      "Q_rad=epsilon*sigma*A*((T_s+273)**4-(T_infinite+273)**4)#\n",
+      "#total heat transfer rate\n",
+      "Q=Q_conv+Q_rad#\n",
+      "print \"Total heat transfer rate = %0.f W\" %Q\n",
+      "\n",
+      "# (b) Overall heat transfer coefficient\n",
+      "# Formula Q=U*A*del_T\n",
+      "U=Q/(A*del_T)#\n",
+      "print \"Overall heat transfer coefficient is :\",round(U,3),\"W/m**2 degree C\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total heat transfer rate = 1863 W\n",
+        "Overall heat transfer coefficient is : 13.178 W/m**2 degree C\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_7.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_7.ipynb
new file mode 100644
index 00000000..1a92384e
--- /dev/null
+++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_7.ipynb
@@ -0,0 +1,610 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 - Radiation Heat Transfer"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  7.1 - Page : 212"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import exp\n",
+      "# given data\n",
+      "lamda=2*10**-6   # in m\n",
+      "C1=0.374*10**-15#\n",
+      "T=2000+273   # in K'\n",
+      "C2=1.4388*10**-2#\n",
+      "\n",
+      "#(a)\n",
+      "# Formula Eb_lamda= (C1*lamda**-5)/(exp(C2/(lamda*T))-1)\n",
+      "Eb_lamda= (C1*lamda**-5)/(exp(C2/(lamda*T))-1)#\n",
+      "print \"Monochromatic emissive power at 2 micro wavelength = %0.2e W/m**2\" %Eb_lamda\n",
+      "\n",
+      "#(b)\n",
+      "# Formula lamda_max * T =2898 # in micro m K\n",
+      "lamda_max= 2898/T   # in micro m\n",
+      "print \"Wave-length at which the emission is maximum = %0.3f micro m\" %lamda_max\n",
+      "\n",
+      "#(c)\n",
+      "Elamdab_max=1.285*10**-5*T**5   # in W/m**2-m\n",
+      "print \"Maximum emissive power = %0.4e W/m**2-m\" %Elamdab_max\n",
+      "\n",
+      "#(d)\n",
+      "sigma=5.67*10**-8#\n",
+      "E=sigma*T**4#\n",
+      "print \"Total emissive power = %0.3e W/m**2\" %E\n",
+      "\n",
+      "#Note: Answer of  part (a) in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Monochromatic emissive power at 2 micro wavelength = 5.15e+11 W/m**2\n",
+        "Wave-length at which the emission is maximum = 1.275 micro m\n",
+        "Maximum emissive power = 7.7965e+11 W/m**2-m\n",
+        "Total emissive power = 1.513e+06 W/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  7.2 - Page : 213"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "lamda=2*10**-6   # in m\n",
+      "C1=0.374*10**-15#\n",
+      "T=2000+273   # in K'\n",
+      "C2=1.4388*10**-2#\n",
+      "\n",
+      "epsilon=0.3#\n",
+      "sigma=5.67*10**-8#\n",
+      "T1=300   # in K\n",
+      "T2=200   # in K\n",
+      "del_T=T1-T2#\n",
+      "h=12   # in W/m**2 degree C\n",
+      "d=4*10**-2   # diameter in m\n",
+      "l=1   # in m\n",
+      "A=pi*d*l#\n",
+      "# Heat transfer rate by radiation,\n",
+      "q_r= epsilon*sigma*A*(T1**4-T2**4)   # in W\n",
+      "# Heat transfer rate by convection,\n",
+      "q_c=h*A*del_T   # in W\n",
+      "# Total heat transfer,\n",
+      "q=q_r+q_c#\n",
+      "# Formula q=U*A*del_T\n",
+      "U=q/(A*del_T)   # Overall heat tranfer coefficient\n",
+      "print \"Overall heat tranfer coefficient = %0.3f W/m**2 degree C\" %U\n",
+      "#Note: Value of q_c is wrong in the book, so the answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Overall heat tranfer coefficient = 13.106 W/m**2 degree C\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  7.3 - Page : 214"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# given data\n",
+      "epsilon=0.5#\n",
+      "T1=1200   # in K\n",
+      "T2=300   # in K\n",
+      "#(a) Heat transfer rate between the two plates is \n",
+      "# Formula Fg12=1/((1/epsilon1+(1/epsilon2-1)*A1/A2))\n",
+      "epsilon1=epsilon#\n",
+      "epsilon2=epsilon#\n",
+      "A1byA2=1#\n",
+      "Fg12=1/(1/epsilon1+(1/epsilon2-1)*A1byA2)#\n",
+      "# Formula q12= sigma*A*Fg12*(T1**4-T2**4)  \n",
+      "sigma=5.67*10**-8#\n",
+      "q12byA=sigma*Fg12*(T1**4-T2**4)   # in W/m**2\n",
+      "print \"Heat transfer rate between the two plates = %0.4e W/m**2\" %q12byA\n",
+      "\n",
+      "#(b)\n",
+      "epsilon3=.05#\n",
+      "Fg13=1/(1/epsilon1+(1/epsilon3-1)*A1byA2)#\n",
+      "Fg32=1/(1/epsilon3+(1/epsilon2-1)*A1byA2)#\n",
+      "# q13=q32\n",
+      "# sigma*A*Fg13*(T1**4-T3**4) = sigma*A*Fg32*(T3**4-T2**4)  \n",
+      "T3= ((T1**4+T2**4)/2)**(1/4)#\n",
+      "T3=math.floor(T3)#\n",
+      "q13byA=sigma*Fg13*(T1**4-T3**4)   # in W/m**2\n",
+      "print \"Heat transfer rate if a radiation shield with an emissivity of 0.05 on both sides is placed \"\n",
+      "print \"between the two plates = %0.3e W/m**2 \" %q13byA"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate between the two plates = 3.9038e+04 W/m**2\n",
+        "Heat transfer rate if a radiation shield with an emissivity of 0.05 on both sides is placed \n",
+        "between the two plates = 2.789e+03 W/m**2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  7.4  - Page : 215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "T1=800+273   # in K\n",
+      "A= 5*6   # in square meter\n",
+      "epsilon=0.45#\n",
+      "sigma=5.67*10**-8#\n",
+      "q=epsilon*sigma*A*T1**4   #in watt\n",
+      "print \"Energy emitted by a grey surface = %0.5e watt\" %q"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy emitted by a grey surface = 1.01465e+06 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  7.5 - Page : 215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# given data\n",
+      "A=5   # in m**2\n",
+      "intensity=660   # in W/m**2\n",
+      "alpha=6/11#\n",
+      "rho=6/22#\n",
+      "toh=6/33#\n",
+      "energy_absorbed= intensity*alpha*A   # in watt\n",
+      "print \"Energy absorbed = %0.f watt\" %energy_absorbed\n",
+      "energy_transmitted=intensity*rho*A   #in watt\n",
+      "print \"Energy transmitted = %0.f watt\" %energy_transmitted\n",
+      "energy_emitted= intensity*toh*A   # in watt\n",
+      "print \"Energy emitted = %0.f watt\" %energy_emitted"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy absorbed = 1800 watt\n",
+        "Energy transmitted = 900 watt\n",
+        "Energy emitted = 600 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  7.5 - Page : 216"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "T1=200+273   # in K\n",
+      "T2=100+273   # in K\n",
+      "A= 1*2   # in square meter\n",
+      "sigma=5.67*10**-8#\n",
+      "x_D= 1/4#\n",
+      "y_D= 1/2#\n",
+      "Fg12= 0.033#\n",
+      "q12= Fg12*sigma*A*(T1**4-T2**4)   # in watt\n",
+      "print \"The net heat exchange between two surfaces = %0.2f watt\" %q12"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The net heat exchange between two surfaces = 114.88 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  7.6 - Page : 216"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "d=20*10**-2   #diameter of pipe in m\n",
+      "l=1   # length of pipe in m\n",
+      "s=30*10**-2   # side of duct in m\n",
+      "A1=pi*d*l   # area of pipe in m**2\n",
+      "A2=4*s*s   # area of duct in m**2\n",
+      "epsilon1=0.8#\n",
+      "epsilon2=0.9#\n",
+      "T1=200+273   # in K\n",
+      "T2=20+273   # in K\n",
+      "# Formula Fg12=1/((1/epsilon1+(1/epsilon2-1)*A1/A2))\n",
+      "Fg12=1/((1/epsilon1+(1/epsilon2-1)*A1/A2))#\n",
+      "# Heat transfer rate between pipe and duct\n",
+      "sigma=5.67*10**-8#\n",
+      "q12=sigma*Fg12*A1*(T1**4-T2**4)   # in W\n",
+      "print \"Heat transfer rate between pipe and duct = %0.3f W\" %q12\n",
+      "#Note: Answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate between pipe and duct = 1053.148 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  7.10 - Page : 220"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "D=150*10**-3   # in m\n",
+      "H=400*10**-3   # in m\n",
+      "T1=500   # in K\n",
+      "epsilon=0.7#\n",
+      "# Formula F11=(4*H)/(4*H+D)\n",
+      "F11=(4*H)/(4*H+D)#\n",
+      "sigma=5.67*10**-8#\n",
+      "A1=pi*D*H#\n",
+      "q=sigma*A1*epsilon*T1**4*((1-F11)/(1-F11*(1-epsilon)))#\n",
+      "print \"Heat Heat loss for cavity = %0.3f W\" %q\n",
+      "#Note: Answer in the book is not accutate"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat Heat loss for cavity = 55.227 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  7.11 - Page : 221"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "epsilon1=.04#\n",
+      "epsilon2=epsilon1#\n",
+      "T1=-153+273   # in K\n",
+      "T2=27+273   # in K\n",
+      "h_fg=209   # in kJ/kg\n",
+      "h_fg=h_fg*10**3   # in J/kg\n",
+      "d1=20*10**-2   # in m\n",
+      "d2=30*10**-2   # in m\n",
+      "A1=d1**2   # in square meter\n",
+      "A2=d2**2   # in square meter\n",
+      "A=4*pi*(d2-d1)**2#\n",
+      "Fg12=1/((1/epsilon1+(1/epsilon2-1)*A1/A2))#\n",
+      "sigma=5.67*10**-8#    \n",
+      "q12=sigma*A*Fg12*(T1**4-T2**4)   # in W\n",
+      "print \"Net radiant heat transfer rate = %0.3f watt\" %q12\n",
+      "print \"Negative sign indicates that heat flows into the sphere\"\n",
+      "q12=-q12 #\n",
+      "m=q12*60/h_fg #\n",
+      "print \"Rate of evaporation per minutes = %0.3e kg/min\" %m"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Net radiant heat transfer rate = -1.577 watt\n",
+        "Negative sign indicates that heat flows into the sphere\n",
+        "Rate of evaporation per minutes = 4.526e-04 kg/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  7.12 - Page : 221"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "T1=500   # in K\n",
+      "T2=300   # in K\n",
+      "sigma=5.67*10**-8#\n",
+      "A=2   # surface area of each plate in m**2\n",
+      "#(a) If the plates are perfectly black\n",
+      "F12=1#\n",
+      "q12=sigma*A*F12*(T1**4-T2**4)#\n",
+      "print \"Radiation heat transfer between two black parellel plates = %0.4e watt\" %q12\n",
+      "\n",
+      "#(b) If the plates are gray surface\n",
+      "#in this case\n",
+      "F12=1#\n",
+      "#A1=A2, so\n",
+      "A1byA2=1\n",
+      "epsilon1=.4#\n",
+      "epsilon2=epsilon1#\n",
+      "#Fg12=1/(1/epsilon1+(1/epsilon2-1)*A1byA2)#\n",
+      "Fg12=1/((1-epsilon1)/epsilon1 + 1/F12 + ((1-epsilon2)/epsilon2)*A1byA2)#\n",
+      "q12=sigma*A*Fg12*(T1**4-T2**4)   # in W\n",
+      "print \"Heat transfer rate = %0.3e watt\" %q12"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Radiation heat transfer between two black parellel plates = 6.1690e+03 watt\n",
+        "Heat transfer rate = 1.542e+03 watt\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  7.13 Page No - 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "T1=800   # in K\n",
+      "T3=200   # in K\n",
+      "sigma=5.67*10**-8#\n",
+      "d1=20*10**-2   # in m\n",
+      "d2=30*10**-2   # in m\n",
+      "d3=40*10**-2   # in m\n",
+      "A1=4*pi*(d1/2)**2   # in m**2\n",
+      "A2=4*pi*(d2/2)**2   # in m**2\n",
+      "A3=4*pi*(d3/2)**2   # in m**2\n",
+      "epsilon1=0.2#\n",
+      "epsilon2=epsilon1\n",
+      "epsilon3=epsilon1\n",
+      "Fg12=1/(1/epsilon1+(1/epsilon2-1)*A1/A2)#\n",
+      "Fg23=1/(1/epsilon2+(1/epsilon3-1)*A2/A3)#\n",
+      "# Under steady state condition \n",
+      "# q12 = q23\n",
+      "# A1*Fg12*sigma*(T1**4-T2**4) = A2*Fg23*sigma*(T2**4-T3**4)\n",
+      "T2 = ((A2*Fg23*T3**4/(A1*Fg12)+T1**4)/(A2*Fg23/(A1*Fg12) + 1))**(1/4)\n",
+      "print \"Steady state temperature of the intermediate sphere = %0.f K\" %T2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Steady state temperature of the intermediate sphere = 604 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  7.14  Page No : 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "T1=400   # in K\n",
+      "T2=500   # in K\n",
+      "T3=1200   # in K\n",
+      "alpha1=0.70#\n",
+      "alpha2=0.6#\n",
+      "alpha3=0.4#\n",
+      "# First part\n",
+      "sigma=5.67*10**-8#\n",
+      "qa=alpha3*sigma*T3**4#\n",
+      "print \"The rate of energy absorption = %0.2e W/m**2\" %qa\n",
+      "# Second part\n",
+      "qa=alpha1*sigma*T1**4#\n",
+      "print \"The rate of emission of radiation energy = %0.2f W/m**2\" %qa\n",
+      "# Note : Answer of the first part in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rate of energy absorption = 4.70e+04 W/m**2\n",
+        "The rate of emission of radiation energy = 1016.06 W/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  7.15 Page No : 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "d1=100   # in mm\n",
+      "d1=d1*10**-3   # in m\n",
+      "d2=100+10*2   # in mm\n",
+      "d2=d2*10**-3   # in m\n",
+      "l=1   # in m\n",
+      "A1byA2=d1**2/d2**2#\n",
+      "A1=pi*d1*l   # in m**2\n",
+      "sigma=5.67*10**-8#\n",
+      "T1=120+273   # in K\n",
+      "T2=35+273   # in K\n",
+      "epsilon1=.8#\n",
+      "epsilon2=.1#\n",
+      "Fg12=1/(1/epsilon1+(1/epsilon2-1)*A1byA2)#\n",
+      "# Radiant heat transfer from the tube\n",
+      "q=A1*Fg12*sigma*(T1**4-T2**4)\n",
+      "print \" Radiant heat transfer from the tube = %0.3f W/m\" %q\n",
+      "#Note: Answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Radiant heat transfer from the tube = 35.282 W/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_8.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_8.ipynb
new file mode 100644
index 00000000..dbf7d5a0
--- /dev/null
+++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_8.ipynb
@@ -0,0 +1,903 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 - Heat Exchanger"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.1 - Page : 248"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import log\n",
+      "# given data\n",
+      "t_hi=80   # in degree C\n",
+      "t_ci=30   # in degree C\n",
+      "t_ho=40   # in degree C\n",
+      "Mh=0.278   # in kg/s\n",
+      "Mc=0.278   # in kg/s\n",
+      "Cph=2.09# # in kJ/kg degree C\n",
+      "Cpc=4.18   #  in kJ/kg degree C\n",
+      "U=24   # in W/m**2 degree C\n",
+      "# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)\n",
+      "t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci   # in degree C\n",
+      "del_t1=t_hi-t_co   #in degree C\n",
+      "del_t2=t_ho-t_ci   #in degree C\n",
+      "del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#\n",
+      "Cph=Cph*10**3   # in J/kg degree C\n",
+      "q=Mh*Cph*(t_hi-t_ho)#\n",
+      "#Formula q=U*A*del_tm\n",
+      "A=q/(U*del_tm)   # in m**2\n",
+      "print \"Surface area of heat exchange = %0.1f square meter\" %A"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Surface area of heat exchange = 53.2 square meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.2 - Page : 249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "t_hi=160   # in degree C\n",
+      "t_ci=25   # in degree C\n",
+      "t_ho=60   # in degree C\n",
+      "Mh=2   # in kg/s\n",
+      "Mc=2   # in kg/s\n",
+      "Cph=2.035# # in kJ/kg degree C\n",
+      "Cpc=4.187   #  in kJ/kg degree C\n",
+      "U=250   # in W/m**2 K\n",
+      "d=0.5   # in m\n",
+      "# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)\n",
+      "t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci   # in degree C\n",
+      "del_t1=t_hi-t_co   #in degree C\n",
+      "del_t2=t_ho-t_ci   #in degree C\n",
+      "del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#\n",
+      "Cph=Cph*10**3   # in J/kg degree C\n",
+      "q=Mh*Cph*(t_hi-t_ho)#\n",
+      "#Formula q=U*pi*d*l*del_tm\n",
+      "l=q/(U*pi*d*del_tm)#\n",
+      "print \"Length of the heat exchanger = %0.2f meter\" %l"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Length of the heat exchanger = 18.22 meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.3 - Page : 251"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "from __future__ import division\n",
+      "# given data\n",
+      "t_hi=110   # in degree C\n",
+      "t_ci=35   # in degree C\n",
+      "t_co=75   # in degree C\n",
+      "Mh=2.5   # in kg/s\n",
+      "Mc=1   # in kg/s\n",
+      "Cph=1.9# # in kJ/kg K\n",
+      "Cpc=4.18   #  in kJ/kg K\n",
+      "U=300   # in W/m**2 K\n",
+      "\n",
+      "# Energy balance Mc*Cpc*(t_co-t_ci) = Mh*Cph*(t_hi-t_ho)\n",
+      "t_ho=t_hi- Mc*Cpc*(t_co-t_ci)/(Mh*Cph)   # in degree C\n",
+      "del_t1=t_hi-t_co   #in degree C\n",
+      "del_t2=t_ho-t_ci   #in degree C\n",
+      "del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#\n",
+      "Cph=Cph*10**3   # in J/kg degree C\n",
+      "q=Mh*Cph*(t_hi-t_ho)#\n",
+      "#Formula q=U*A*del_tm\n",
+      "A=q/(U*del_tm)#\n",
+      "print \"Area of the heat exchanger = %0.2f square meter\" %A"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Area of the heat exchanger = 14.92 square meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.4 - Page : 252"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "Fi=0.00014   # in m**2 degree C/W\n",
+      "hi=2000   # in W/m**2degree C\n",
+      "Fo=0.00015   # in m**2 degree C/W\n",
+      "ho=1000   # in W/m**2degree C\n",
+      "di=3*10**-2   # in m\n",
+      "do=4*10**-2   #in m\n",
+      "ro=do/2#\n",
+      "ri=di/2#\n",
+      "k=53   # in W/m degree C\n",
+      "Uo=1/(do/di*1/hi+ do/(2*k)*log(ro/ri) + 1/ho + do*Fi/di + Fo)#\n",
+      "print \"The overall heat transfer coefficient = %0.3f W/m**2 degree C\" %Uo\n",
+      "# Note : Answer in the book is not accurate"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The overall heat transfer coefficient = 473.509 W/m**2 degree C\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.5 - Page : 252"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# given data\n",
+      "V=0.15   # in m/s\n",
+      "di=2.5*10**-2   # in m\n",
+      "\n",
+      "delta=0.364*10**-6   # in m**2/s\n",
+      "k=0.668   # in W/m degree C\n",
+      "Pr=2.22#\n",
+      "\n",
+      "Re=V*di/delta#\n",
+      "# Formula Nu= hi*di/k = 0.023*Re**0.8*Pr**0.3\n",
+      "hi=0.023*Re**0.8*Pr**0.3*k/di   # in W/m**2 degree C\n",
+      "\n",
+      "# Now, Reynold number for flow of air across the tube\n",
+      "delta=18.22*10**-6   # in m**2/s\n",
+      "k=0.0281   # in W/m degree C\n",
+      "Pr=0.703#\n",
+      "d=2.5*10**-2   # in m\n",
+      "u=10   # in m/s\n",
+      "Re=u*d/delta#\n",
+      "Re=math.floor(Re)#\n",
+      "#The Nusselt number for this case\n",
+      "Nu=(0.04*Re**0.5+ 0.006*Re**(2/3))*Pr**0.4\n",
+      "# Formula Nu= ho*do/k\n",
+      "do=di#\n",
+      "ho=Nu*k/do   # in W/m**2 degree C\n",
+      "print \"Heat transfer coefficient = %0.3f W/m**2 degree C\" %ho\n",
+      "U=1/(1/hi+1/ho)#\n",
+      "print \"The overall heat transfer coefficient neglecting the wall resistance = %0.3f W/m**2 degree C\" %U\n",
+      "l=1   # in m\n",
+      "Ti=90   # in degree C\n",
+      "To=10   # in degree C\n",
+      "q=U*pi*d*l*(Ti-To) #\n",
+      "print \"Heat loss per meter length of the tube = %0.3f W/m\" %q\n",
+      "# Note: Answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer coefficient = 7.931 W/m**2 degree C\n",
+        "The overall heat transfer coefficient neglecting the wall resistance = 7.882 W/m**2 degree C\n",
+        "Heat loss per meter length of the tube = 49.523 W/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.6 - Page : 253"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given data\n",
+      "t_hi=83   # in degree C\n",
+      "t_ho=45   # in degree C\n",
+      "t_ci=25   # in degree C\n",
+      "Mh=5   # in kg/min\n",
+      "Mc=9   # in kg/min\n",
+      "Cph=4.18# # in kJ/kg K\n",
+      "Cpc=2.85   #  in kJ/kg K\n",
+      "# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci) \n",
+      "t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci   # in degree C\n",
+      "print \"t_co = %0.2f degree C\" %t_co\n",
+      "if(t_co>t_ho) :\n",
+      "    print \"since t_co > t_ho, hence counter flow arrangment will be suitable\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "t_co = 55.96 degree C\n",
+        "since t_co > t_ho, hence counter flow arrangment will be suitable\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.7 - Page : 254"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "# (a) For parallel flow arrangment\n",
+      "del_t1=60-10   # in degree C\n",
+      "del_t2=40-30   # in degree C\n",
+      "del_tm=(del_t1-del_t2)/log(del_t1/del_t2)   # in degree C\n",
+      "q=100*10**3   # in W\n",
+      "U=75   # in W/m**2 degree C\n",
+      "# Formula q=U*A*del_tm#\n",
+      "A=q/(U*del_tm)#\n",
+      "print \"Area for paraller flow arrangment = %0.1f square meter\" %A\n",
+      "# (b) For counter flow heat exchange\n",
+      "del_t1=60-30   # in degree C\n",
+      "del_t2=40-10   # in degree C\n",
+      "# In this case\n",
+      "del_tm=(del_t1+del_t2)/2   # in degree C\n",
+      "A=q/(U*del_tm)#\n",
+      "print \"Area For counter flow heat exchange = %0.2f square meter\" %A"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Area for paraller flow arrangment = 53.6 square meter\n",
+        "Area For counter flow heat exchange = 44.44 square meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.8 - Page : 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "Cp=4180   # in J/kg degree C\n",
+      "miu=0.86*10**-3   # in kg/m-s\n",
+      "Pr=60#\n",
+      "k=0.60   # in W/m degree C\n",
+      "h_fg=2372400   # in W\n",
+      "ho=6000   # in W/m**2 degree C\n",
+      "di=2*10**-2   # in m\n",
+      "d_o=3*10**-2   # in m\n",
+      "t_co=35   # in degree C\n",
+      "t_ci=15   # in degree C\n",
+      "M=0.9#\n",
+      "Re=4*M/(pi*di*miu)#\n",
+      "# since Re > 2300, hence flow inside tube is a turbulent flow.\n",
+      "# Hence  Nu= hi*di/k = 0.023*Re**0.8*Pr**0.4\n",
+      "hi= 0.023*Re**0.8*Pr**0.4*k/di#\n",
+      "Uo= 1/(1/10213.6*(d_o/di)+1/ho)#\n",
+      "del_t1=50-15   # in degree C\n",
+      "del_t2=50-35   # in degree C\n",
+      "del_tm=(del_t1-del_t2)/log(del_t1/del_t2)   # in degree C\n",
+      "# Formula q= Uo*pi*d_i*L*del_tm = M*Cp*(t_co-t_ci)\n",
+      "L= M*Cp*(t_co-t_ci)/(Uo*pi*d_o*del_tm)#\n",
+      "print \"Length of tube = %0.3f meter\" %L\n",
+      "q=M*Cp*(t_co-t_ci)   # in watt\n",
+      "m=q/h_fg#\n",
+      "print \"Rate of condensation = %0.3f kg/sec\" %m"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Length of tube = 10.604 meter\n",
+        "Rate of condensation = 0.032 kg/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.9 - Page : 257"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "Cph=3850# # in J/kg degree C\n",
+      "t_hi=100   # in degree C\n",
+      "t_ci=20   # in degree C\n",
+      "t_ho=50   # in degree C\n",
+      "Mh=8   # in kg/s\n",
+      "Mc=10   # in kg/s\n",
+      "Cpc=4.18*10**3   #  in J/kg degree C\n",
+      "U=400   # in W/m**2 degree C\n",
+      "# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)\n",
+      "t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci   # in degree C\n",
+      "# Heat load\n",
+      "q=Mh*Cph*(t_hi-t_ho)   # in W\n",
+      "# (a) Parallel flow\n",
+      "del_t1=90   # in degree C\n",
+      "del_t2=3.16   # in degree C\n",
+      "del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#\n",
+      "A=q/(U*del_tm)#\n",
+      "print \"Surface area for parallel flow = %0.3f meter square\" %A\n",
+      "# (b) Counter flow heat exchanger\n",
+      "del_t1=53.16   # in degree C\n",
+      "del_t2=40   # in degree C\n",
+      "del_tm_counter= (del_t1-del_t2)/log(del_t1/del_t2)#\n",
+      "A=q/(U*del_tm_counter)#\n",
+      "print \"Surface area for counter flow heat exchanger = %0.1f meter square\" %A\n",
+      "#(c) One shell pass and two tube pass.\n",
+      "#here\n",
+      "t1=10   # in degree C\n",
+      "t2=46.84   # in degree C\n",
+      "T1=100   # in degree C\n",
+      "T2=50   # in degree C\n",
+      "P=(t2-t1)/(T1-t1)#\n",
+      "R=(T1-T2)/(t2-t1)#\n",
+      "F=0.88#\n",
+      "del_tm=F*del_tm_counter   # in degree C\n",
+      "A=q/(U*del_tm)#\n",
+      "print \"Surface area for one shell pass and two tube pass = %0.2f meter square\" %A\n",
+      "# (d) For cross flow, correction factor \n",
+      "F=0.9#\n",
+      "del_tm=F*del_tm_counter#\n",
+      "A=q/(U*del_tm)#\n",
+      "print \"Surface area for cross flow = %0.3f meter square\" %A"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Surface area for parallel flow = 148.486 meter square\n",
+        "Surface area for counter flow heat exchanger = 83.2 meter square\n",
+        "Surface area for one shell pass and two tube pass = 94.56 meter square\n",
+        "Surface area for cross flow = 92.456 meter square\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.10 Page No : 258"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import exp\n",
+      "# given data\n",
+      "Cpc=4.18*10**3   #  in J/kg degree C\n",
+      "Mc=1   # in kg/s\n",
+      "Mh=2.4   # in kg/s\n",
+      "Cph=2050# # in J/kg degree C\n",
+      "t_hi=100   # in degree C\n",
+      "t_ci=20   # in degree C\n",
+      "C_c=Mc*Cpc   # in W/degree C\n",
+      "C_h=Mh*Cph   # in W/degree C\n",
+      "U=300   # in W/m**2 degree C\n",
+      "A=10   # in m**2\n",
+      "C_min=C_c#\n",
+      "C_max=C_h#\n",
+      "N= A*U/C_min#\n",
+      "C=C_min/C_max#\n",
+      "# Effectiveness for counter flow heat exchanger\n",
+      "epsilon= (1-exp(-N*(1-C)))/(1-C*exp(-N*(1-C)))#\n",
+      "# Total heat transfer\n",
+      "q=epsilon*C_min*(t_hi-t_ci)   # in watt\n",
+      "print \"Total heat transfer = %0.3f kW\" %(q*10**-3) \n",
+      "t_co=t_ci+epsilon*C*(t_hi-t_ci) #\n",
+      "print \"Exit temperature of water = %0.3f degree C\" %t_co"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total heat transfer = 144.170 kW\n",
+        "Exit temperature of water = 49.303 degree C\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.11 Page No : 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "t_hi=135   # in degree C\n",
+      "t_ci=20   # in degree C\n",
+      "t_ho=65   # in degree C\n",
+      "t_co=50   # in degree C\n",
+      "# Energy balance  Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)\n",
+      "# C =C_min/C_max = Mh*Cph/( Mc*Cpc)\n",
+      "C= (t_co-t_ci)/(t_hi-t_ho)#\n",
+      "epsilon=(t_hi-t_ho)/(t_hi-t_ci)#\n",
+      "# Also epsilon = epsilon_parallel = (1-exp(-NTU*(1+C)))/(1+C)\n",
+      "NTU= -log(1-epsilon*(1+C))/(1+C)#\n",
+      "# if the existing heat exchanger is to be used as counter flow mode, its NTU will not change, i.e.\n",
+      "epsilon_c= (1-exp(-NTU*(1-C)))/((1-C*exp(-NTU*(1-C))))#\n",
+      "# Exit temperature\n",
+      "# (i) Hot fluid\n",
+      "t_ho=t_hi-epsilon_c*(t_hi-t_ci)   # in degree C\n",
+      "print \"Exit temperature for hot fluid = %0.3f degree C\" %t_ho\n",
+      "\n",
+      "#(ii) Cold fluid\n",
+      "t_co= t_ci+epsilon_c*C*(t_hi-t_ci)#\n",
+      "print \"Exit temperature for cold fluid = %0.1f degree C\" %t_co\n",
+      "\n",
+      "# (iii) # If the parallel flow heat exchanger is too long, then body fluid will have common outlet temperature (t)\n",
+      "# From MCp_h*(t_hi-t) = MCp_c*(t-t_ci)\n",
+      "\n",
+      "t=(C*t_hi+t_ci)/(1+C)#\n",
+      "print \"The minimum temperature to which the oil may be cooled by increasing the tube length with parallel flow operation,\"\n",
+      "print \"in degree C = %0.1f\" %t"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Exit temperature for hot fluid = 55.909 degree C\n",
+        "Exit temperature for cold fluid = 53.9 degree C\n",
+        "The minimum temperature to which the oil may be cooled by increasing the tube length with parallel flow operation,\n",
+        "in degree C = 54.5\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.12 - Page : 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "t_hi=78   # in degree C\n",
+      "t_ci=23   # in degree C\n",
+      "t_ho=65   # in degree C\n",
+      "t_co=36   # in degree C\n",
+      "# Energy balance  Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)\n",
+      "# C =C_min/C_max = Mh*Cph/( Mc*Cpc)\n",
+      "C= (t_co-t_ci)/(t_hi-t_ho)#\n",
+      "epsilon=(t_hi-t_ho)/(t_hi-t_ci)#\n",
+      "# Formula epsilon = (1-exp(-N*(1+C)))/(1+C)\n",
+      "N= -log(1-epsilon*(1+C))/(1+C)#\n",
+      "# When flow rates of both fluids are doubled , the deat capacity ratio will not change, i.e.\n",
+      "# C=1\n",
+      "#  MCp_new =2* MCp_old\n",
+      "# N=U*A/C_min=N/2\n",
+      "N=N/2#\n",
+      "epsilon=(1-exp(-N*(1+C)))/(1+C)#\n",
+      "# exit temperature\n",
+      "t_ho=t_hi-epsilon*(t_hi-t_ci)   # in degree C\n",
+      "t_co= t_ci+epsilon*(t_hi-t_ci)#\n",
+      "print \"Exit temperature  in degree C is : \",round(t_ho,2),\"and\",round(t_co,2)\n",
+      "# Note: Answer in the book is wrong due to put wrong value of t_ci in second last line"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Exit temperature  in degree C is :  70.47 and 30.53\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.13 - Page : 262"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "t_hi=125   # in degree C\n",
+      "t_ci=22   # in degree C\n",
+      "Mh=21   # in kg/s\n",
+      "Mc=5   # in kg/s\n",
+      "C_ph=2100   # in J/kg K\n",
+      "C_pc=4100   # in J/kg K\n",
+      "Ch=Mh*C_ph   # in Js/kg\n",
+      "Cc=Mc*C_pc   # in Js/kg\n",
+      "C_min=Cc   # in Js/kg\n",
+      "C_max=Ch   # in Js/kg\n",
+      "U=325   # in W/m**2 K\n",
+      "d=2.2*10**-2   # in m\n",
+      "l=5   # in m\n",
+      "total_tube=195   # number of total tubes\n",
+      "A=pi*d*l*total_tube\n",
+      "NTU=U*A/C_min#\n",
+      "C=C_min/C_max#\n",
+      "epsilon = (1-exp(-NTU*(1-C)))/(1-C*exp(-NTU*(1-C)))#\n",
+      "t_co= t_ci+epsilon*(t_hi-t_ci)#\n",
+      "t_ho= t_hi-epsilon*Cc/Ch*(t_hi-t_ci)#\n",
+      "print \"Exit temperature  in degree C :\",round(t_co,1),\"and\",round(t_ho,2)\n",
+      "# Total heat transfer\n",
+      "q=epsilon*C_min*(t_hi-t_ci)#\n",
+      "print \"Total heat transfer = %0.3f kW\" %(q*10**-3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Exit temperature  in degree C : 82.8 and 96.73\n",
+        "Total heat transfer = 1246.589 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.14 - Page : 263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import exp\n",
+      "# given data\n",
+      "t_hi=94   # in degree C\n",
+      "t_ci=15   # in degree C\n",
+      "Mw=0.36   # in kg/s\n",
+      "Mo=0.153   # in kg/s\n",
+      "C_po=2*10**3   # in J/kg K\n",
+      "C_pw=4.186*10**3   # in J/kg K\n",
+      "U=10.75*10**2   # in W/m**2 K\n",
+      "A=1   # in m**2\n",
+      "Ch=Mo*C_po   # in kW/K\n",
+      "Cc=Mw*C_pw   # in kW/K\n",
+      "C_min=Ch   # in W/K\n",
+      "C_max=Cc   # in W/K\n",
+      "C=C_min/C_max#\n",
+      "NTU=U*A/C_min#\n",
+      "# Effectiveness\n",
+      "N=NTU#\n",
+      "epsilon = (1-exp(-N*(1-C)))/(1-C*exp(-N*(1-C)))#\n",
+      "mCp_min=C_min#\n",
+      "q_max= mCp_min*(t_hi-t_ci)   # in W\n",
+      "q_actual= epsilon*q_max   # in W\n",
+      "print \"Total heat transfer = %0.3f watt\" %q_actual\n",
+      "# Outlet temp. of water\n",
+      "t_co= q_actual/Cc+t_ci   # in degree C\n",
+      "print \"Outlet temperature of water = %0.3f degree C\" %t_co\n",
+      "# Outlet temp. of oil\n",
+      "t_ho=t_hi-q_actual/Ch   #in degree C\n",
+      "print \"Outlet temperature of oil = %0.3f degree C\" %t_ho\n",
+      "#Note: Evaluation of Cc and Ch in the book is wrong so the Answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total heat transfer = 22987.488 watt\n",
+        "Outlet temperature of water = 30.254 degree C\n",
+        "Outlet temperature of oil = 18.877 degree C\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.15 - Page : 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "U=1800   # in W/m**2 degree C\n",
+      "h_fg=2200*10**3   # in J/kg\n",
+      "t_ci=20   # in degree C\n",
+      "t_co=90   # in degree C\n",
+      "del_t1=120-20   # in degree C\n",
+      "del_t2=120-90   # in degree C\n",
+      "del_tm=(del_t1-del_t2)/log(del_t1/del_t2)   # in degree C\n",
+      "Mc=1000/3600   # in kg/s\n",
+      "Cc=4180   # in kg/s\n",
+      "# Rate of heat transfer\n",
+      "q=Mc*Cc*(t_co-t_ci)   # in watt\n",
+      "# Formula q=U*A*del_tm\n",
+      "A=q/(U*del_tm)#\n",
+      "print \"Surface area = %0.3f square meter\" %A\n",
+      "#Rate of condensation of steam\n",
+      "ms=q/h_fg   # in kg/sec\n",
+      "print \"Rate of condensation of steam = %0.4f kg/sec\" %ms"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Surface area = 0.777 square meter\n",
+        "Rate of condensation of steam = 0.0369 kg/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.16 - Page : 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# given data\n",
+      "Mh=10000/3600   # in kg/sec\n",
+      "Mc=8000/3600   # in kg/sec\n",
+      "Cph=2095   # in J/kg K\n",
+      "Cpc=4180   # in J/kg K\n",
+      "t_hi=80   # in degree C\n",
+      "t_ci=25   # in degree C\n",
+      "t_ho=50   # in degree C\n",
+      "U=300   # in W/m**2 K\n",
+      "# Energy balance Mh*Cph*(t_hi-t_ho) = Mc*Cpc*(t_co-t_ci)\n",
+      "t_co= Mh*Cph*(t_hi-t_ho)/(Mc*Cpc)+t_ci   # in degree C\n",
+      "del_t1=t_hi-t_co   #in degree C\n",
+      "del_t2=t_ho-t_ci   #in degree C\n",
+      "del_tm= (del_t1-del_t2)/log(del_t1/del_t2)#\n",
+      "q=Mh*Cph*(t_hi-t_ho)#\n",
+      "#Formula q=U*A*del_tm\n",
+      "A=q/(U*del_tm)   # in m**2\n",
+      "print \"Surface area of heat exchange = %0.2f square meter\" %A"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Surface area of heat exchange = 19.23 square meter\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  8.17 - Page : 266"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "from numpy import pi\n",
+      "# given data\n",
+      "ho=5000   # in W/m**2 degree C\n",
+      "rho=988.1   # in kg/m**3\n",
+      "K=0.6474#\n",
+      "D=555*10**-9   # in m**2/s\n",
+      "Pr=3.54#\n",
+      "n=100#\n",
+      "d_i=2.5*10**-2   # in m\n",
+      "ri=d_i/2#\n",
+      "d_o=2.9*10**-2   # in m\n",
+      "ro=d_o/2#\n",
+      "Cp=4174   # in J/kg degree C\n",
+      "Mc=8.333   # in kg/s\n",
+      "Mw=Mc#\n",
+      "t_c1=30   # in degree C\n",
+      "t_c2=70   # in degree C\n",
+      "t_n1=100   # in degree C\n",
+      "t_n2=t_n1   # in degree C\n",
+      "R_fi=0.0002   # in m**2 degree C/W (In the book, there is miss print in this line,they took here R_fi = .002)\n",
+      "# Heat gain by water\n",
+      "Q=Mc*Cp*(t_c2-t_c1)#\n",
+      "# Also Q= U*A*del_tm\n",
+      "del_t1=t_n1-t_c1   #in degree C\n",
+      "del_t2=t_n2-t_c2   #in degree C\n",
+      "del_tm= (del_t1-del_t2)/log(del_t1/del_t2)# \n",
+      "# Mw= 1/4*pi*d_i**2*V*rho*N, here\n",
+      "N=n#\n",
+      "V=4*Mw/(pi*d_i**2*rho*N)#\n",
+      "# Formula Re=V*d_i/v, here\n",
+      "v=D#\n",
+      "Re=V*d_i/v#\n",
+      "# Formula Nu= hi*d_i/K = 0.023*Re**0.8*Pr**0.33\n",
+      "hi= 0.023*Re**0.8*Pr**0.33*K/d_i#\n",
+      "# Formula 1/Vi= 1/hi + R_fi + ri/ro*1/ho\n",
+      "Vi= 1/(1/hi + R_fi + ri/ro*1/ho)   # in W/m**2 degree C\n",
+      "#Formula Q = Vi*(N*pi*d_i*L)*del_tm\n",
+      "L= Q/(Vi*(N*pi*d_i)*del_tm)#\n",
+      "print \"Length of the tube bundle = %0.1f m\" %L"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Length of the tube bundle = 4.6 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_9.ipynb b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_9.ipynb
new file mode 100644
index 00000000..a03c60e2
--- /dev/null
+++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/chapter_9.ipynb
@@ -0,0 +1,280 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 - Condensation And Boiling"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  9.1 Page No : 277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "from __future__ import division\n",
+      "# given data\n",
+      "h_fg=2256*10**3   # in J/kg\n",
+      "rho=970   # in kg/m**3\n",
+      "rho_v=0.596   # in kg/m**3\n",
+      "k=0.66   # in W/mK\n",
+      "miu=3.7*10**-4   # in kg/m-s\n",
+      "T_sat=100   # in degree C\n",
+      "T_s=40   # in degree C\n",
+      "L=1.5   # in m\n",
+      "d=0.09   # in m\n",
+      "g=9.81#\n",
+      "# heat transfer coefficient\n",
+      "#h_bar = 1.13*[ rho*g*(rho-rho_v)*h_fg*k**3/(miu*L*(T_sat-T_s))]**(1/4)   # in W/m**2k\n",
+      "h_bar= 1.13*(rho*g*(rho-rho_v)*h_fg*k**3/(miu*L*(T_sat-T_s)))**(1/4)#\n",
+      "# heat transfer rate\n",
+      "q=h_bar*pi*d*L*(T_sat-T_s)   # in watt\n",
+      "print \"Heat transfer rate = %0.3f kW\" %(q*10**-3)\n",
+      "#rate of condensation\n",
+      "m=q/h_fg   # in kg/s\n",
+      "print \"Rate of condenstion = %0.4f kg/s\" %m"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate = 105.276 kW\n",
+        "Rate of condenstion = 0.0467 kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  9.2 Page No : 278"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "from __future__ import division\n",
+      "# given data\n",
+      "h_fg=2310*10**3   # in J/kg\n",
+      "rho=980   # in kg/m**3\n",
+      "k=0.67   # in W/mK\n",
+      "Cp=4.18#\n",
+      "delta=.41*10**-6   # in m**2/s\n",
+      "miu=rho*delta#\n",
+      "T_sat=70   # in degree C\n",
+      "T_s=55   # in degree C\n",
+      "L=1   # in m\n",
+      "d=0.03   # in m\n",
+      "g=9.81#\n",
+      "N=5#\n",
+      "# (a) for Horizontal tube\n",
+      "h_bar = 0.725*(rho**2*g*h_fg*k**3/(N*miu*d*(T_sat-T_s)))**(1/4)   # in W/m**2k\n",
+      "# heat transfer rate\n",
+      "q=h_bar*pi*d*L*N**2*(T_sat-T_s)   # in watt\n",
+      "print \"Heat transfer rate for horizontal tube = %0.3f kW\" %(q*10**-3)\n",
+      "#rate of condensation\n",
+      "m=q/h_fg   # in kg/s\n",
+      "print \"Rate of condenstion = %0.3f kg/s\" %m\n",
+      "\n",
+      "# (b) For Vertical tube\n",
+      "h_bar = 1.13*(rho**2*g*h_fg*k**3/(miu*L*(T_sat-T_s)))**(1/4)   # in W/m**2k\n",
+      "# heat transfer rate\n",
+      "q=h_bar*pi*d*L*N**2*(T_sat-T_s)   # in watt\n",
+      "print \"Heat transfer rate for vertical tube = %0.3f kW\" %(q*10**-3)\n",
+      "#rate of condensation\n",
+      "m=q/h_fg   # in kg/s\n",
+      "print \"Rate of condenstion = %0.3f kg/s\" %m"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate for horizontal tube = 236.364 kW\n",
+        "Rate of condenstion = 0.102 kg/s\n",
+        "Heat transfer rate for vertical tube = 229.269 kW\n",
+        "Rate of condenstion = 0.099 kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  9.3 Page No : 280"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# given data\n",
+      "h_fg=2392*10**3   # in J/kg\n",
+      "rho=993   # in kg/m**3\n",
+      "k=0.63   # in W/mK\n",
+      "miu=728*10**-6   # in kJ/m-s\n",
+      "N=10#\n",
+      "T_sat=45.7   # in degree C\n",
+      "T_s=25   # in degree C\n",
+      "d=4*10**-3   # in m\n",
+      "g=9.81#\n",
+      "h_bar = 0.725*(rho**2*g*h_fg*k**3/(N*miu*d*(T_sat-T_s)))**(1/4)   # in W/m**2k\n",
+      "m=300/(60*60)#\n",
+      "# Formula m=q/h_fg\n",
+      "q=m*h_fg#\n",
+      "print \"Heat transfer rate = %0.3f kW\" %(q*10**-3)\n",
+      "# Formula q=h_bar*pi*d*L*N**2*(T_sat-T_s)\n",
+      "L=q/(h_bar*pi*d*N**2*(T_sat-T_s))#\n",
+      "print \"Length of tube = %0.3f m\" %L\n",
+      "\n",
+      "# Note: Answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transfer rate = 199.333 kW\n",
+        "Length of tube = 1.068 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  9.4 Page No : 280"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sin\n",
+      "from numpy import pi\n",
+      "# given data\n",
+      "h_fg=2400*10**3   # in J/kg\n",
+      "rho=993   # in kg/m**3\n",
+      "rho_v=0.0563   # in kg/m**3\n",
+      "t_mf=(40+30)/2   # in degree C\n",
+      "k=0.625   # in W/mK\n",
+      "miu=728*10**-6   # in kJ/m-s\n",
+      "x=0.25#\n",
+      "T_sat=40   # in degree C\n",
+      "T_s=30   # in degree C\n",
+      "g=9.81#\n",
+      "\n",
+      "# (a) Thickness of condensate film\n",
+      "delta=(4*k*(T_sat-T_s)*miu*x/(rho*(rho-rho_v)*g*h_fg))**(1/4)   # in meter\n",
+      "print \"Thickness of condensate film = %0.5f mm\" %(delta*10**3)\n",
+      "\n",
+      "#(b) Local value of heat transfer coefficient\n",
+      "hx=k/delta   # in W/m**2\n",
+      "L=0.5   # in m\n",
+      "hm=4/3*(L/x)**(1/4)*hx#\n",
+      "print \"Average heat transfer coefficient = %0.1f W/m**2\" %hm\n",
+      "# The heat transfer rate\n",
+      "A=0.5*0.5   # in m**2\n",
+      "q=hm*A*(T_sat-T_s)   # in watt\n",
+      "print \"The heat transfer rate = %0.2f kW\" %(q*10**-3)\n",
+      "\n",
+      "# (c) \n",
+      "theta=45   # in degree\n",
+      "h_vertical=hm#\n",
+      "h_inclined=h_vertical*(sin(theta*pi/180))**(1/4)#\n",
+      "print \"Average heat transfer coefficient when plate is inclined at 45 degree = %0.1f W/m**2K\" %h_inclined"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of condensate film = 0.11832 mm\n",
+        "Average heat transfer coefficient = 8375.5 W/m**2\n",
+        "The heat transfer rate = 20.94 kW\n",
+        "Average heat transfer coefficient when plate is inclined at 45 degree = 7680.4 W/m**2K\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  9.5 Page No : 282"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given correlataion\n",
+      "#h_A=5.56*(det_T)**3\n",
+      "#h_P=h_A*(rho/rho_a)**0.4\n",
+      "del_T=25   # in degree C\n",
+      "h_A=5.56*(del_T)**3   # in W/m**2K\n",
+      "print \"The heat transfer coefficient = %0.3f kW/m**2K\" %(h_A*10**-3)\n",
+      "# and at 20 bar\n",
+      "rho=20#\n",
+      "rho_a=1#\n",
+      "h_P=h_A*(rho/rho_a)**0.4   # in W/m**2\n",
+      "print \"Value of h_P = %0.1f kW/m**2\" %(h_P*10**-3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat transfer coefficient = 86.875 kW/m**2K\n",
+        "Value of h_P = 287.9 kW/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/screenshots/1.png b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/screenshots/1.png
new file mode 100644
index 00000000..52d1ad2a
--- /dev/null
+++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/screenshots/1.png
diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/screenshots/2.png b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/screenshots/2.png
new file mode 100644
index 00000000..e212cfb1
--- /dev/null
+++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/screenshots/2.png
diff --git a/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/screenshots/3.png b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/screenshots/3.png
new file mode 100644
index 00000000..d1284fe6
--- /dev/null
+++ b/Heat_And_Mass_Transfer_by_Vijay_K._Dwivedi/screenshots/3.png
diff --git a/Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/README.txt b/Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/README.txt
new file mode 100644
index 00000000..832a4ac3
--- /dev/null
+++ b/Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Aviral Yadav
+Course: btech
+College/Institute/Organization: ABES Engineering College
+Department/Designation: Mechanical Engineering
+Book Title: Modern Electronic Instrumentation And Measurement Techniques
+Author: A. D. Helfrick And W. D. Cooper
+Publisher: Dorling Kindersly Pvt. Ltd. India
+Year of publication: 2009
+Isbn: 9788131708880
+Edition: 1
\ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_1.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_1.ipynb
new file mode 100644
index 00000000..c10821ba
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_1.ipynb
@@ -0,0 +1,277 @@
+{

+ "metadata": {

+  "name": "chapter no.10.ipynb"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 10:Theory of Failures"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.10.10.1,Page No.401"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "P_e=300 #N/mm**2 #Elastic Limit in tension\n",

+      "FOS=3  #Factor of safety\n",

+      "mu=0.3 #Poissoin's ratio\n",

+      "P=12*10**3 #N Pull \n",

+      "Q=6*10**3 #N #Shear force\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let d be the diameter of the shaft\n",

+      "\n",

+      "#Direct stress\n",

+      "#P_x=P*(pi*4**-1*d**3)**-1\n",

+      "#After substituting values and further simplifying we get\n",

+      "#P_x=48*10**3\n",

+      "\n",

+      "#Now shear stress at the centre of bolt\n",

+      "#q=4*3**-1*q_av\n",

+      "#After substituting values and further simplifying we get\n",

+      "#q=32*10**3*(pi*d**2)**-1\n",

+      "\n",

+      "#Principal stresses are\n",

+      "#P1=P_x*2**-1+((P_x*2**-1)**2+q**2)**0.5\n",

+      "#After substituting values and further simplifying we get\n",

+      "#p1=20371.833*(d**2)**-1\n",

+      "\n",

+      "#P2=P_x*2**-1-((P_x*2**-1)**2+q**2)**0.5\n",

+      "#After substituting values and further simplifying we get\n",

+      "#P2=-5092.984*(d**2)**-1\n",

+      "\n",

+      "#q_max=((P_x*2**-1)**2+q**2)**0.5\n",

+      "\n",

+      "#From Max Principal stress theory\n",

+      "#Permissible stress in Tension\n",

+      "P1=100 #N/mm**2 \n",

+      "d=(20371.833*P1**-1)**0.5\n",

+      "\n",

+      "#Max strain theory\n",

+      "#e_max=P1*E**-1-mu*P2*E**-1\n",

+      "#After substituting values and further simplifying we get\n",

+      "#e_max=21899.728*(d**2*E)**-1\n",

+      "\n",

+      "#According to this theory,the design condition is\n",

+      "#e_max=P_e*(E*FOS)**-1\n",

+      "#After substituting values and further simplifying we get\n",

+      "d2=(21899.728*3*300**-1)**0.5  #mm\n",

+      "\n",

+      "#Max shear stress theory\n",

+      "#e_max=shear stress at elastic*(FOS)**-1\n",

+      "#After substituting values and further simplifying we get\n",

+      "d3=(12732.421*6*300**-1)**0.5 #mm\n",

+      "\n",

+      "#Result\n",

+      "print\"Diameter of Bolt by:Max Principal stress theory\",round(d,2),\"mm\"\n",

+      "print\"                   :Max strain theory\",round(d2,2),\"mm\"\n",

+      "print\"                   :Max shear stress theory\",round(d3,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Diameter of Bolt by:Max Principal stress theory 14.27 mm\n",

+        "                   :Max strain theory 14.8 mm\n",

+        "                   :Max shear stress theory 15.96 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.10.10.2.Page No.402"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "M=40*10**6 #N-mm #Bending moment\n",

+      "T=10*10**6 #N-mm #TOrque\n",

+      "mu=0.25 #Poissoin's ratio\n",

+      "P_e=200 #N/mm**2 #Stress at Elastic Limit\n",

+      "FOS=2\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let d be the diameter of the shaft\n",

+      "\n",

+      "#Principal stresses are given by\n",

+      "\n",

+      "#P1=16*(pi*d**3)**-1*(M+(M**2+T**2)**0.5)\n",

+      "#After substituting values and further simplifying we get\n",

+      "#P1=4.13706*10**8*(d**3)**-1     ............................(1)\n",

+      "\n",

+      "#P2=16*(pi*d**3)**-1*(M-(M**2+T**2)**0.5)\n",

+      "#After substituting values and further simplifying we get\n",

+      "#P2=-6269718*(pi*d**3)**-1       ..............................(2)\n",

+      "\n",

+      "#q_max=(P1-P2)*2**-1\n",

+      "#After substituting values and further simplifying we get\n",

+      "#q_max=2.09988*10**8*(d**3)**-1\n",

+      "\n",

+      "#Max Principal stress theory\n",

+      "#P1=P_e*(FOS)**-1\n",

+      "#After substituting values and further simplifying we get\n",

+      "d=(4.13706*10**8*2*200**-1)**0.33333 #mm \n",

+      "\n",

+      "#Max shear stress theory\n",

+      "#q_max=shear stress at elastic limit*(FOS)**-1\n",

+      "#After substituting values and further simplifying we get\n",

+      "d2=(2.09988*10**8*4*200**-1)**0.33333\n",

+      "\n",

+      "#Max strain energy theory\n",

+      "#P_3=0\n",

+      "#P1**2+P2**2-2*mu*P1*P2=P_e**2*(FOS)**-1\n",

+      "#After substituting values and further simplifying we get\n",

+      "d3=(8.62444*10**12)**0.166666\n",

+      "\n",

+      "#Result\n",

+      "print\"Diameter of shaft according to:MAx Principal stress theory\",round(d,2),\"mm\"\n",

+      "print\"                              :Max shear stress theory\",round(d2,2),\"mm\"\n",

+      "print\"                              :Max strain energy theory\",round(d3,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Diameter of shaft according to:MAx Principal stress theory 160.52 mm\n",

+        "                              :Max shear stress theory 161.33 mm\n",

+        "                              :Max strain energy theory 143.2 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.10.10.3,Page No.403"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "f_x=40 #N/mm**2 #Internal Fliud Pressure\n",

+      "d1=200 #mm #Internal Diameter\n",

+      "r1=d1*2**-1 #mm #Radius\n",

+      "q=300 #N/mm**2 #Tensile stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#From Lame's Equation we have,\n",

+      "\n",

+      "#Hoop Stress\n",

+      "#f_x=b*(x**2)**-1+a    ..........................(1)\n",

+      "\n",

+      "#Radial Pressure\n",

+      "#p_x=b*(x**2)**-1-a      .........................(2)\n",

+      "\n",

+      "#the boundary conditions are\n",

+      "x=d1*2**-1 #mm \n",

+      "#After sub values in equation 1 and further simplifying we get\n",

+      "#40=b*100**-1-a           ..........................(3)\n",

+      "\n",

+      "#Max Principal stress theory\n",

+      "#q*(FOS)**-1=b*100**2+a           ..................(4)\n",

+      "#After sub values in above equation and further simplifying we get\n",

+      "\n",

+      "#From Equation 3 and 4 we get\n",

+      "a=80*2**-1\n",

+      "#Sub value of a in equation 3 we get\n",

+      "b=(f_x+a)*100**2\n",

+      "\n",

+      "#At outer edge where x=r_0 pressure is zero\n",

+      "r_0=(b*a**-1)**0.5 #mm\n",

+      "\n",

+      "#thickness\n",

+      "t=r_0-r1 #mm\n",

+      "\n",

+      "#Max shear stress theory\n",

+      "P1=b*(100**2)**-1+a #Max hoop stress\n",

+      "P2=-40 #pressure at int radius (since P2 is compressive)\n",

+      "\n",

+      "#Max shear stress\n",

+      "q_max=(P1-P2)*2**-1\n",

+      "\n",

+      "#According max shear theory the design condition is\n",

+      "#q_max=P_e*2**-1*(FOS)**-1\n",

+      "#After sub values in equation we get and further simplifying we get\n",

+      "#80=b*(100**2)**-1+a\n",

+      "#After sub values in equation 1 and 3 and further simplifying we get\n",

+      "b2=120*100**2*2**-1\n",

+      "\n",

+      "#from equation(3)\n",

+      "a2=120*2**-1-a\n",

+      "\n",

+      "#At outer radius r_0,radial pressure=0\n",

+      "r_02=(b2*a2**-1)**0.5\n",

+      "\n",

+      "#thickness\n",

+      "t2=r_02-r1\n",

+      "\n",

+      "#Result\n",

+      "print\"Thickness of metal by:Max Principal stress theory\",round(t,2),\"mm\"\n",

+      "print\"                     :Max shear stress thoery\",round(t2,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Thickness of metal by:Max Principal stress theory 41.42 mm\n",

+        "                     :Max shear stress thoery 73.21 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.1_1.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.1_1.ipynb
new file mode 100644
index 00000000..89760142
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.1_1.ipynb
@@ -0,0 +1,22 @@
+{

+ "metadata": {

+  "name": "chapter no.1.ipynb"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter No.1:Introduction"

+     ]

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_1.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_1.ipynb
new file mode 100644
index 00000000..da0bc7fc
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_1.ipynb
@@ -0,0 +1,2700 @@
+{

+ "metadata": {

+  "name": "chapter no.2.ipynb"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 2:Simple Stresses And Strains"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.1,Page No.14"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "P=45*10**3 #N #Load\n",

+      "E=200*10**3 #N/mm**2 #Modulus of elasticity of rod\n",

+      "L=500 #mm #Length of rod\n",

+      "d=20 #mm #Diameter of rod\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "A=pi*d**2*4**-1 #mm**2 #Area of circular rod\n",

+      "p=P*A**-1 #N/mm**2 #stress\n",

+      "e=p*E**-1 #strain\n",

+      "dell_l=(P*L)*(A*E)**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"The stress in bar due to Load is\",round(p,5),\"N/mm\"\n",

+      "print\"The strain in bar due to Load is\",round(e,5),\"N/mm\"\n",

+      "print\"The Elongation in bar due to Load is\",round(dell_l,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The stress in bar due to Load is 143.23945 N/mm\n",

+        "The strain in bar due to Load is 0.00072 N/mm\n",

+        "The Elongation in bar due to Load is 0.36 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.2,Page No.15"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "A=15*0.75 #mm**2 #area of steel tape\n",

+      "P=100 #N #Force apllied\n",

+      "L=30*10**3 #mm #Length of tape\n",

+      "E=200*10**3 #N/m**2 #Modulus of Elasticity of steel tape\n",

+      "AB=150 #m #Measurement of Line AB \n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "dell_l=P*L*(A*E)**-1 #mm #Elongation\n",

+      "l=L+dell_l*10**-3 #mm #Actual Length \n",

+      "AB1=AB*l*L**-1 #m Actual Length of AB\n",

+      "\n",

+      "#Result\n",

+      "print\"The Actual Length of Line AB is\",round(AB1,2),\"m\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Actual Length of Line AB is 150.0 m\n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.3,Page No.15"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Let y be the yield stress\n",

+      "\n",

+      "y=250 #N/mm**2 #yield stress\n",

+      "FOS=1.75 #Factor of safety\n",

+      "P=140*10**3 #N #compressive Load\n",

+      "D=101.6 #mm #External diameter\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "p=y*(FOS)**-1 #N/mm**2 #Permissible stress\n",

+      "A=P*p**-1 #mm**2 #Area of hollow tube\n",

+      "\n",

+      "#Let d be the internal diameter of tube\n",

+      "d=-((A*4*(pi)**-1)-D**2)\n",

+      "X=d**0.5\n",

+      "t=(D-X)*2**-1 #mm #Thickness of steel tube\n",

+      "\n",

+      "#result\n",

+      "print\"The thickness of steel tube is\",round(t,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The thickness of steel tube is 3.17 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 23

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.4,Page No.16"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=25 #mm #diameter of steel\n",

+      "L=200 #mm #length of stee\n",

+      "P=80*10**3 #KN #Load\n",

+      "P1=160*10**3 #N #Load at Elastic Limit\n",

+      "P2=180*10**3 #N #Max Load\n",

+      "L1=56 #mm #Total Extension\n",

+      "dell_l=0.16 #mm #Extension\n",

+      "\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "A=pi*d**2*4**-1 #Area of steel #mm**2\n",

+      "\n",

+      "p=P1*A**-1 #Stress at Elastic Limit #N/mm**2\n",

+      "E=P*L*(A*dell_l)**-1\n",

+      "\n",

+      "#result\n",

+      "print E"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "203718.327158\n"

+       ]

+      }

+     ],

+     "prompt_number": 24

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.5,Page No.16"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=20 #mm #Diameter of bar\n",

+      "d2=14.7 #mm #Diameter at neck\n",

+      "L=200 #mm #guage Length \n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,10,20,30,40,50,60]\n",

+      "Y1=[0,32,64,95,127,160,190]\n",

+      "Z1=[0,0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Extension in divisions\")\n",

+      "plt.ylabel(\"Load in kN\")\n",

+      "plt.show()\n",

+      "\n",

+      "A=pi*4**-1*d**2 #mm**2 #Area of Bar\n",

+      "A2=pi*4**-1*d2**2\n",

+      "\n",

+      "P=45 #KN #Load obtained from graph\n",

+      "dell=0.143 #mm #Divisions\n",

+      "\n",

+      "#Modulus of Elasticity\n",

+      "E=P*L*(dell*A)**-1 \n",

+      "\n",

+      "BL=93*10**3 #N #Breaking Load\n",

+      "\n",

+      "#Nominal stress at Breaking point\n",

+      "sigma=BL*A**-1 #KN/mm**2 \n",

+      "\n",

+      "#True stress at breaking Point\n",

+      "sigma1=BL*A2**-1\n",

+      "\n",

+      "#Percentage Elongation \n",

+      "dell_l=(A-A2)*A**-1*100\n",

+      "\n",

+      "#Result\n",

+      "print\"The Value of ELongation is\",round(E,2),\"N/mm**2\"\n",

+      "print\"The Nominal stress at the Breaking Point\",round(sigma,2),\"KN/mm**2\"\n",

+      "print\"The True stress at the Breaking Point\",round(sigma1,2),\"KN/mm**2\"\n",

+      "print\"The Percentage Reduction in Area is\",round(dell_l,2)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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4G6puRNyttBR++UvrgLzp06120hV3hYu4RVpaWpWXBa6Hx4/PCAkJ4eTJkzRv3pyCggKa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\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x55dab30>"

+       ]

+      },

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Value of ELongation is 200.33 N/mm**2\n",

+        "The Nominal stress at the Breaking Point 296.03 KN/mm**2\n",

+        "The True stress at the Breaking Point 547.97 KN/mm**2\n",

+        "The Percentage Reduction in Area is 45.98\n"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.6,Page No.19"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "P=40*10**3 #N #Load\n",

+      "L1=160 #mm #Length of Bar1\n",

+      "L2=240 #mm #Length of bar2\n",

+      "L3=160 #mm #Length of bar3\n",

+      "d1=25 #mm #Diameter of Bar1\n",

+      "d2=20 #mm #diameter of bar2\n",

+      "d3=25 #mm #diameter of bar3\n",

+      "dell_l=0.285 #mm #Total Extension of bar\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "E=P*4*(dell_l*pi)**-1*(L1*(d1**2)**-1+L2*(d2**2)**-1+L3*(d3**2)**-1)\n",

+      "\n",

+      "#Result\n",

+      "print\"The Young's Modulus of the material\",round(E,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Young's Modulus of the material 198714.72 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 10

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.7,Page No.19"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "E1=2*10**5 #N/mm**2 #modulus of Elasticity of material1\n",

+      "E2=1*10**5 #N/mm**2 #modulus of Elasticity of material2\n",

+      "P=25*10**3 #N #Load \n",

+      "t=20 #mm #thickness of material\n",

+      "b1=40 #mm #width of material1\n",

+      "b2=30 #mm #width of material2\n",

+      "L1=500 #mm #Length of material1\n",

+      "L2=750 #mm #Length of material2\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "A1=b1*t #mm**2 #Area of materila1\n",

+      "A2=b2*t #mm**2 #Area of material2\n",

+      "\n",

+      "dell_l1=P*L1*(A1*E1)**-1 #Extension of Portion1\n",

+      "dell_l2=P*L2*(A2*E2)**-1 #Extension of portion2\n",

+      "\n",

+      "#Total Extension of Bar is\n",

+      "dell_l=dell_l1+dell_l2\n",

+      "\n",

+      "#Result\n",

+      "print\"The Total Extension of the Bar is\",round(dell_l,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Total Extension of the Bar is 0.39 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 15

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.8,Page No.20"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=1000 #mm #Length of Bar\n",

+      "l=400 #mm #Length upto which bire is drilled \n",

+      "D=30 #mm #diameter of bar\n",

+      "d1=10 #mm #diameter of bore\n",

+      "P=25*10**3 #N #Load\n",

+      "dell_l=0.185 #mm #Extension of bar\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "L1=L-l #Length of bar above the bore\n",

+      "L2=400 #mm #Length of bore\n",

+      "\n",

+      "A1=pi*4**-1*D**2 #Area of bar\n",

+      "A2=pi*4**-1*(D**2-d1**2) #Area of bore\n",

+      "\n",

+      "E=P*dell_l**-1*(L1*A1**-1+L2*A2**-1)\n",

+      "\n",

+      "#Result\n",

+      "print\"The Modulus of ELasticity is\",round(E,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Modulus of ELasticity is 200735.96 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 24

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.11,Page No.23"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "t=10 #mm #Thickness of steel\n",

+      "b1=60 #mm #width of plate1\n",

+      "b2=40 #mm #width of plate2\n",

+      "P=60*10**3 #Load\n",

+      "L=600 #mm #Length of plate\n",

+      "E=2*10**5 #N/mm**2\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Extension of taperong bar of rectangular section\n",

+      "dell_l=P*L*(t*E*(b1-b2))**-1*log(b1*b2**-1)\n",

+      "\n",

+      "A_av=(b1*t+b2*t)*2**-1 #Average Area #mm**2\n",

+      "dell_l2=P*L*(A_av*E)**-1 \n",

+      "\n",

+      "#PErcentage Error\n",

+      "e=(dell_l-dell_l2)*(dell_l)**-1*100\n",

+      "\n",

+      "#Result\n",

+      "print\"The Percentage Error is\",round(e,2)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Percentage Error is 1.35\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.12,Page No.23"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=1.5 #m #Length of steel bar\n",

+      "L1=1000 #m0 #Length of steel bar 1\n",

+      "L2=500 #m #Length of steel bar 2\n",

+      "d1=40 #Diameter of steel bar 1\n",

+      "d2=20 #diameter of steel bar 2\n",

+      "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",

+      "P=160*10**3 #N #Load\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "A1=pi*4**-1*d1**2 #Area of Portion 1\n",

+      "\n",

+      "#Extension of uniform Portion 1\n",

+      "dell_l1=P*L1*(A1*E)**-1 #mm\n",

+      "\n",

+      "#Extension of uniform Portion 2\n",

+      "dell_l2=4*P*L2*(pi*d1*d2*E)**-1 #mm\n",

+      "\n",

+      "#Total Extension of Bar\n",

+      "dell_l=dell_l1+dell_l2\n",

+      "\n",

+      "#Result\n",

+      "print\"The Elongation of the Bar is\",round(dell_l,2),\"mm\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": []

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.14,Page No.25"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Portion AB\n",

+      "L_AB=600 #mm #Length of AB\n",

+      "A_AB=40*40 #mm**2 #Cross-section Area of AB\n",

+      "\n",

+      "#Portion BC\n",

+      "L_BC=800 #mm #Length of BC\n",

+      "A_BC=30*30 #mm #Length of BC\n",

+      "\n",

+      "#Portion CD\n",

+      "L_CD=1000 #mm #Length of CD\n",

+      "A_CD=20*20 #mm #Area of CD\n",

+      "\n",

+      "P1=80*10**3 #N #Load1\n",

+      "P2=60*10**3 #N #Load2\n",

+      "P3=40*10**3 #N #Load3\n",

+      "\n",

+      "E=2*10**5 #Modulus of Elasticity\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "P4=P1-P2+P3 #Load4\n",

+      "\n",

+      "#Now Force in AB\n",

+      "F_AB=P1\n",

+      "\n",

+      "#Force in BC\n",

+      "F_BC=P1-P2\n",

+      "\n",

+      "#Force in CD\n",

+      "F_CD=P4\n",

+      "\n",

+      "#Extension of AB\n",

+      "dell_l_AB=F_AB*L_AB*(A_AB*E)**-1\n",

+      "\n",

+      "#Extension of BC\n",

+      "dell_l_BC=F_BC*L_BC*(A_BC*E)**-1\n",

+      "\n",

+      "#Extension of CD\n",

+      "dell_l_CD=F_CD*L_CD*(A_CD*E)**-1\n",

+      "\n",

+      "#Total Extension\n",

+      "dell_l=dell_l_AB+dell_l_BC+dell_l_CD\n",

+      "\n",

+      "#Result\n",

+      "print\"The Total Extension in Bar is\",round(dell_l,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Total Extension in Bar is 0.99 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.15,Page No.26"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=800 #mm #Length of bar\n",

+      "F1=30*10**3 #N #Force acting on the bar\n",

+      "F2=60*10**3 #N #force acting on the bar\n",

+      "L=800 #mm #Length of bar\n",

+      "d=25 #mm #diameter of bar\n",

+      "L_AC=275 #mm #Length of AC\n",

+      "L_CD=150 #mm #Length of CD\n",

+      "L_DB=375 #mm #Length of DB\n",

+      "E=2*10**5 #Pa #Modulus of elasticity\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let P be the Reaction on tne Bar from support at A\n",

+      "\n",

+      "#Shortening of Portion AC\n",

+      "#dell_l_AC1=P*L_AC*(A*E)**-1\n",

+      "\n",

+      "#Shortening of Portion CD\n",

+      "#dell_l_CD1=(30+P)*L_CD*(A*E)**-1\n",

+      "\n",

+      "#Extension of Portion DB\n",

+      "#dell_l_DB1=(30-P)*L_DB*(A*E)**-1\n",

+      "\n",

+      "#Total Extensions=1*(A*E)**-1*(P*L_AC-(30+P)*L_CD+(30-P)*L_DB)\n",

+      "#As Supports are unyielding,Total Extensions=0\n",

+      "\n",

+      "#After substituting values in above equation and Further simplifying we get\n",

+      "P=(30*375-150*30)*800**-1\n",

+      "\n",

+      "#Reaction of support A\n",

+      "R_A=P\n",

+      "\n",

+      "#Reaction of support B\n",

+      "R_B=30-P\n",

+      "\n",

+      "#Cross-sectional Area\n",

+      "A=pi*4**-1*d**2\n",

+      "\n",

+      "#Stress in Portion AC\n",

+      "sigma1=P*10**3*A**-1 #N/mm**2\n",

+      "\n",

+      "#Stress in Portion CD\n",

+      "sigma2=(30+P)*10**3*A**-1 #N/mm**2\n",

+      "\n",

+      "#Stress in Portion DB\n",

+      "sigma3=(30-P)*10**3*A**-1 #N/mm**2\n",

+      "\n",

+      "#Shortening of Portion AC\n",

+      "dell_l_AC2=P*10**3*L_AC*(A*E)**-1 #mm \n",

+      "\n",

+      "#Shortening of Portion CD\n",

+      "dell_l_CD2=(30+P)*10**3*L_CD*(A*E)**-1 #mm \n",

+      "\n",

+      "#Extension of Portion DB\n",

+      "dell_l_DB2=(30-P)*10**3*L_DB*(A*E)**-1 #mm \n",

+      "\n",

+      "#result\n",

+      "print\"The Reactios at two Ends are:R_A\",round(R_A,2),\"KN\"\n",

+      "print\"                            :R_B\",round(R_B,2),\"KN\"\n",

+      "print\"Stress in Portion AC\",round(sigma1,2),\"N/mm**2\"\n",

+      "print\"Stress in Portion CD\",round(sigma2,2),\"N/mm**2\"\n",

+      "print\"Stress in Portion DB\",round(sigma3,2),\"N/mm**2\"\n",

+      "print\"Shortening of Portion AC\",round(dell_l_AC2,3),\"mm\"\n",

+      "print\"Shortening of Portion CD\",round(dell_l_CD2,3),\"mm\"\n",

+      "print\"Shortening of Portion DB\",round(dell_l_DB2,3),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Reactios at two Ends are:R_A 8.44 KN\n",

+        "                            :R_B 21.56 KN\n",

+        "Stress in Portion AC 17.19 N/mm**2\n",

+        "Stress in Portion CD 78.3 N/mm**2\n",

+        "Stress in Portion DB 43.93 N/mm**2\n",

+        "Shortening of Portion AC 0.024 mm\n",

+        "Shortening of Portion CD 0.059 mm\n",

+        "Shortening of Portion DB 0.082 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.19,Page No.29"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "h=4 #m #height of Pillars\n",

+      "P=20 #KN #Load at M\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let P_A,P_B,P_C,P_D be the forces introduced  in the Pillars\n",

+      "#Sun of All Vertical Forces\n",

+      "#P_A+P_B+P_C+P_D=20  ....................(1)\n",

+      "\n",

+      "#Sum of moment about AB, we get\n",

+      "#P_D+P_C=12   ....................(2)\n",

+      "\n",

+      "#Sum of Moment about AD\n",

+      "#P_C+P_B=8    ....................(3)\n",

+      "\n",

+      "#Let dell_l_A,dell_l_B,dell_l-C,dell_l_D be the deformations of Pillars A,B,C,D respectively\n",

+      "#Diagonals AC and BD will remain straight Lines even after the Load is applied.\n",

+      "#Deflection of central Point is given by (dell_l_A+dell_l_C)*2**-1 & (dell_l_B+dell_l_D)*2**-1\n",

+      "\n",

+      "#dell_l_A+dell_l_C=dell_l_B+ell_l_D\n",

+      "#P_A*L*(A*E)**-1+P_C*L*(A*E)**-1=P_B*L*(A*E)**-1+P_D*L*(A*E)**-1\n",

+      "\n",

+      "#Since Pillars are identical in Length,cross-sectional area,material Property\n",

+      "#P_A+P_C=P_B+P_D      ..............(4)\n",

+      "\n",

+      "#From Equations 1 and 4 we get\n",

+      "#P_B+P_D=10      ....................(5)\n",

+      "  \n",

+      "#Substracting Equation 3 from Equation 2 we get\n",

+      "#P_D-P_B=4     ....................(6)\n",

+      "\n",

+      "#Adding Equation 5 and 6 we get\n",

+      "\n",

+      "P_D=14*2**-1\n",

+      "P_C=12-P_D\n",

+      "P_B=8-P_C\n",

+      "\n",

+      "#Now substituting values of P_B,P_C,P_D in equation1 we get\n",

+      "P_A=20-(P_B+P_C+P_D)\n",

+      "\n",

+      "#Result\n",

+      "print\"The Forces Developed in the Pillars are:P_A\",round(P_A,2),\"KN\"\n",

+      "print\"                                       :P_B\",round(P_B,2),\"KN\"\n",

+      "print\"                                       :P_C\",round(P_C,2),\"KN\"\n",

+      "print\"                                       :P_D\",round(P_D,2),\"KN\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Forces Developed in the Pillars are:P_A 5.0 KN\n",

+        "                                       :P_B 3.0 KN\n",

+        "                                       :P_C 5.0 KN\n",

+        "                                       :P_D 7.0 KN\n"

+       ]

+      }

+     ],

+     "prompt_number": 1

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.20,Page No.31"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "sigma=150 #N/mm**2 #Stress\n",

+      "P=40*10**3 #N #Load\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#LEt P_A.P_B,P_C,P_D be the forces developed in wires A,B,C,D respectively\n",

+      "\n",

+      "#Let sum of all Vertical Forces=0\n",

+      "#P_A+P_B+P_C+P_D=40   ..........................(1)\n",

+      "\n",

+      "#Let x be the distance between each wires\n",

+      "#sum of all moments=0\n",

+      "#P_B*x+P_C*2*x+P_D*3*x=40*2*x\n",

+      "\n",

+      "#After further simplifying we get\n",

+      "#P_B+2*P_C+3*P_D=80     ..........................(2)\n",

+      "\n",

+      "#As the equations of statics ae not enough to find unknowns,Consider compatibilit Equations\n",

+      "\n",

+      "#Let dell_l be the increse in elongation of wire\n",

+      "\n",

+      "#dell_l_B=dell_l_A+dell_l\n",

+      "#dell_l_C=dell_l_A+2*dell_l\n",

+      "#dell_l_D=dell_l_A+3*dell_l\n",

+      "\n",

+      "#Let P1 be the force required for the Elongation of wires,then\n",

+      "#P_B=P_A+P1    ]\n",

+      "#P_C=P_A+2*P1  ]\n",

+      "#P_D=P_A+3*P1  ]     ................................(3)  \n",

+      "\n",

+      "#from Equation (3) and (1) we get\n",

+      "#2*P_A+3*P1=20      ................................(4)\n",

+      "\n",

+      "#from Equation (3) and (2) we get\n",

+      "#6*P_A+14*P1=80 \n",

+      "\n",

+      "#subtracting 3 times equation (4) from (3) we get\n",

+      "P1=20*5**-1\n",

+      "\n",

+      "#from Equation 4 we get\n",

+      "P_A=(80-14*P1)*6**-1\n",

+      "P_B=P_A+P1\n",

+      "P_C=P_A+2*P1\n",

+      "P_D=P_A+3*P1\n",

+      "\n",

+      "#Let d be the diameter required,then\n",

+      "d=(P_D*10**3*4*(pi*150)**-1)**0.5\n",

+      "\n",

+      "#result\n",

+      "print\"The Required Diameter is\",round(d,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Required Diameter is 11.65 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 16

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.21,Page No.32"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "P=20*10**3 #N #Load\n",

+      "d=6 #mm #diameter of wire\n",

+      "E=2*10**5 #N/mm**2 \n",

+      "L_BO=4000 #mm #Length of BO\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let theta be the angle between OA and OB and also between OC and OB\n",

+      "theta=30\n",

+      "\n",

+      "#Let P_OA,P_OB,P_OC be the Forces introduced in wires OA,OB,OC respectively\n",

+      "#Due to symmetry P_OA=P_OC   (same angles)\n",

+      "\n",

+      "#Sum of all Vertical Forces=0\n",

+      "#P_OA*cos(theta)+P_OB+P_OC*cos(theta)=P\n",

+      "\n",

+      "#After further simplifyinf we get\n",

+      "#2*P_OA*cos(theta)+P_OB=20  ...............(1)\n",

+      "\n",

+      "#Let oo1 be the extension of BO\n",

+      "#oo1=L_A1o1*(cos(theta))**-1\n",

+      "\n",

+      "#From relation we get\n",

+      "#P_OB*L_BO=P_OA*L_AO*(cos(theta))**-1\n",

+      "\n",

+      "#But L_AO=L_BO*(cos(theta))**-1\n",

+      "\n",

+      "#After substituting value of L_AO in above equation we get\n",

+      "#P_OB=0.75*P_OA    .......................(2)\n",

+      "\n",

+      "#substituting in Equation 1 we get\n",

+      "#2*P_OA*cos(theta)+0.75*P_OA=20\n",

+      "\n",

+      "P_OA=20*(2*cos(theta*pi*180**-1)+0.75)**-1\n",

+      "\n",

+      "P_OB=0.75*P_OA\n",

+      "\n",

+      "A=pi*4**-1*d**2\n",

+      "\n",

+      "#Vertical displacement of Load\n",

+      "dell_l_BO=P_OB*10**3*L_BO*(A*E)**-1\n",

+      "                    \n",

+      "#Result\n",

+      "print\"Forces in each wire is:P_OA\",round(P_OA,2),\"KN\"\n",

+      "print\"                      :P_OB\",round(P_OB,2),\"KN\"\n",

+      "print\"Vertical displacement of Loadis\",round(dell_l_BO,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Forces in each wire is:P_OA 8.06 KN\n",

+        "                      :P_OB 6.04 KN\n",

+        "Vertical displacement of Loadis 4.27 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 22

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.22,Page No.34"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L_s=L_a=L=500 #mm #Length of bar\n",

+      "A_a=50*20 #mm #Area of aluminium strip\n",

+      "A_s=50*15 #mm #Area of steel strip\n",

+      "P=50*10**3 #N #Load\n",

+      "E_a=1*10**5 #N/mm**2 #Modulus of aluminium \n",

+      "E_s=2*10**5 #N/mm**2 #Modulus of steel\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let P_a and P_s br the Load shared by aluminium and steel strip\n",

+      "#P_a+P_s=P     ..................(1)\n",

+      "\n",

+      "#For compatibility condition,dell_l_a=dell_l_s\n",

+      "#P_a*L_a*(A_a*E_a)**-1=P_s*L_s*(A_s*E_s)**-1    .....(2)\n",

+      "\n",

+      "#As L_a=L_s we get\n",

+      "#P_s=1.5*P_a   .................(3)\n",

+      "\n",

+      "#From Equation 1 and 2 we get\n",

+      "P_a=P*2.5**-1\n",

+      "\n",

+      "#Substituting in equation 1 we get\n",

+      "P_s=P-P_a\n",

+      "\n",

+      "#stress in aluminium strip \n",

+      "sigma_a=P_a*A_a**-1\n",

+      "\n",

+      "#stress in steel strip\n",

+      "sigma_s=P_s*A_s**-1\n",

+      "\n",

+      "#Now from the relation we get\n",

+      "dell_l_a=dell_l_s=P_s*L_s*(A_s*E_s)**-1\n",

+      "\n",

+      "#result\n",

+      "print\"Stress in  Aluminium strip is\",round(sigma_a,2),\"N/mm**2\"\n",

+      "print\"Stress in  steel strip is\",round(sigma_s,2),\"N/mm**2\"\n",

+      "print\"The Extension of the bar is\",round(dell_l_s,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stress in  Aluminium strip is 20.0 N/mm**2\n",

+        "Stress in  steel strip is 40.0 N/mm**2\n",

+        "The Extension of the bar is 0.1 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.23,Page No.35"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "D_s=20 #mm #Diameter of steel\n",

+      "D_Ci=20 #mm #Internal Diameter of Copper\n",

+      "t=5 #mm #THickness of copper bar\n",

+      "P=100*10**3 #N #Load\n",

+      "E_s=2*10**5 #N/mm**2 #modulus of elasticity of steel\n",

+      "E_c=1.2*10**5 #N/mm**2 #Modulus of Elasticity of Copper\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "A_s=pi*4**-1*D_s**2 #mm**2 #Area of steel\n",

+      "D_Ce=D_s+2*t #mm #External Diameterof Copper Tube\n",

+      "\n",

+      "A_c=pi*4**-1*(D_Ce**2-D_Ci**2) #mm**2 #Area of Copper\n",

+      "\n",

+      "#From static Equilibrium condition\n",

+      "#Let P_s and P_c be the Load shared by steel and copper in KN\n",

+      "#P_s+P_c=100 ....................................(1)\n",

+      "\n",

+      "#From compatibility Equation,dell_l_s=dell_l_c\n",

+      "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",

+      "\n",

+      "#Substituting values in above Equation we get\n",

+      "#P_s=1.3333*P_C \n",

+      "\n",

+      "#Now Substituting value of P_s in Equation (1),we get\n",

+      "P_c=100*2.3333**-1 #KN\n",

+      "P_s=100-P_c #KN\n",

+      "\n",

+      "#Stress in steel\n",

+      "sigma_s=P_s*10**3*A_s**-1 #N/mm**2\n",

+      "\n",

+      "#Stress in copper\n",

+      "sigma_c=P_c*10**3*A_c**-1 #N/mm**2\n",

+      "\n",

+      "#Result\n",

+      "print\"Stresses Developed in Two material are:sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",

+      "print\"                                      :sigma_c\",round(sigma_c,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stresses Developed in Two material are:sigma_s 181.89 N/mm**2\n",

+        "                                      :sigma_c 109.14 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.24,Page No.36"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "A_C=230*400 #mm #Area of column\n",

+      "D_s=12 #mm #Diameter of steel Bar\n",

+      "P=600*10**3 #N #Axial compression\n",

+      "#E_s*E_c=18.67\n",

+      "n=8 #number of steel Bars\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "A_s=pi*4**-1*D_s**2*n #Area of steel #mm**2 \n",

+      "A_c=A_C-A_s #mm**2 #Area of concrete\n",

+      "\n",

+      "#From static Equilibrium condition\n",

+      "#P_s+P_c=600   .........(1)\n",

+      "\n",

+      "#Now from compatibility Equation dell_l_s=dell_l_c we get,\n",

+      "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",

+      "\n",

+      "#Substituting values in above Equation we get\n",

+      "#P_s=0.1854*P_c\n",

+      "\n",

+      "#Now Substituting value of P_s in Equation (1),we get\n",

+      "P_c=600*1.1854**-1\n",

+      "P_s=600-P_c\n",

+      "\n",

+      "#Stress in steel\n",

+      "sigma_s=P_s*10**3*A_s**-1 #N/mm**2\n",

+      "\n",

+      "#Stress in copper\n",

+      "sigma_c=P_c*10**3*A_c**-1 #N/mm**2\n",

+      "\n",

+      "#Result\n",

+      "print\"Stresses Developed in Two material are:sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",

+      "print\"                                      :sigma_c\",round(sigma_c,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stresses Developed in Two material are:sigma_s 103.72 N/mm**2\n",

+        "                                      :sigma_c 5.56 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.25,Page No.36"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "P=200*10**3 #N #Load\n",

+      "A_a=1000 #mm**2 #Area of Aluminium\n",

+      "A_s=800 #mm**2 #Area of steel\n",

+      "E_a=1*10**5 #N/mm**2 #Modulus of Elasticity of Aluminium\n",

+      "E_s=2*10**5 #N/mm**2 #Modulus of ELasticity of steel\n",

+      "sigma_a1=65 #N/mm**2 #stress in aluminium\n",

+      "sigma_s1=150 #N/mm**2 #Stress in steel\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let P_a and P_s be the force in aluminium and steel pillar respectively\n",

+      "\n",

+      "#Now,sum of forces in Vertical direction we get\n",

+      "#2*P_a+P_s=200  .........................................(1)\n",

+      "\n",

+      "#By compatibility Equation dell_l_s=dell_l_a we get\n",

+      "#P_s=1.28*P_a  ..........................................(2)\n",

+      "\n",

+      "#Now substituting value of P_s in Equation 1 we get\n",

+      "P_a=200*3.28**-1 #KN\n",

+      "P_s=200-2*P_a   #KN\n",

+      "\n",

+      "#Stress developed in aluminium\n",

+      "sigma_a=P_a*10**3*A_a**-1 #N/mm**2 \n",

+      "\n",

+      "#Stress developed in steel\n",

+      "sigma_s=P_s*10**3*A_s**-1 #N/mm**2 \n",

+      "\n",

+      "#Part-2\n",

+      "\n",

+      "#Let sigma_a1 and sigma_s1 be the stresses in Aluminium and steel due to Additional LOad\n",

+      "\n",

+      "P_a1=sigma_a1*A_a #Load carrying capacity of aluminium\n",

+      "P_s1=1.28*P_a1\n",

+      "\n",

+      "#Total Load carrying capacity \n",

+      "P1=2*P_a1+P_s1 #N \n",

+      "\n",

+      "P_s2=sigma_s1*A_s #Load carrying capacity of steel\n",

+      "P_a2=P_s2*1.28**-1\n",

+      "\n",

+      "#Total Load carrying capacity\n",

+      "P2=2*P_a2+P_s2\n",

+      "\n",

+      "#Additional Load\n",

+      "P3=P1-P\n",

+      "\n",

+      "#Result\n",

+      "print\"Stresses Developed in Each Pillar is:sigma_a\",round(sigma_a,2),\"N/mm**2\"\n",

+      "print\"                                    :sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",

+      "print\"Additional Load taken by pillars is\",round(P3,2),\"N\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "78.0487804878\n",

+        "Stresses Developed in Each Pillar is:sigma_a 60.98 N/mm**2\n",

+        "                                    :sigma_s 97.56 N/mm**2\n",

+        "Additional Load taken by pillars is 13200.0 N\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.26,Page No.37"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=500 #mm #Length of assembly\n",

+      "D=16 #mm #Diameter of steel bolt\n",

+      "Di=20 #mm #internal Diameter of copper tube\n",

+      "Do=30 #mm #External Diameter of copper tube\n",

+      "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel\n",

+      "E_c=1.2*10**5 #N/mm**2 #Modulus of Elasticity of copper\n",

+      "p=2 #mm #Pitch of nut\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let P_s be the Force in bolt and P_c be the FOrce in copper tube\n",

+      "#P_s=-P_s\n",

+      "\n",

+      "dell=1*4**-1*2 #Quarter turn of nut total movement\n",

+      "\n",

+      "#dell=dell_s+dell_c\n",

+      "\n",

+      "#Area of steel\n",

+      "A_s=pi*4**-1*D**2\n",

+      "\n",

+      "#Area of copper\n",

+      "A_c=pi*4**-1*(Do**2-Di**2)\n",

+      "\n",

+      "#dell=P*L*(A_s*E_s)**-1+P*L*(A_c*E_c)**-1\n",

+      "P=dell*(1*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1*L**-1 #LOad\n",

+      "\n",

+      "P_s=P*A_s**-1\n",

+      "P_c=P*A_c**-1\n",

+      "\n",

+      "#result\n",

+      "print\"stress introduced in bolt is\",round(P_s,2),\"N/mm**2\"\n",

+      "print\"stress introduced in tube is\",round(P_c,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "stress introduced in bolt is 107.91 N/mm**2\n",

+        "stress introduced in tube is 55.25 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.27,Page No.39"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "D=20 #mm #Diameter of Bolts\n",

+      "Di=25 #m #internal Diameter\n",

+      "t=10 #mm #Thickness of bolt\n",

+      "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity\n",

+      "p=3 #mm #Pitch\n",

+      "theta=30 #degree\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let P_s be the Force in each bolt and P_c be the FOrce in copper tube\n",

+      "#From Static Equilibrium condition\n",

+      "#P_c=2*P_s\n",

+      "\n",

+      "#As nut moves by 60 degree.If nut moves by 360 degree its Longitudinal movement is by 3 mm\n",

+      "dell=theta*360**-1*p\n",

+      "\n",

+      "#From Compatibility Equaton we get\n",

+      "#dell=dell_c+dell_s\n",

+      "\n",

+      "\n",

+      "A_s=pi*4**-1*"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": []

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.28,Page No.40"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=9 #m #Length of rigid bar\n",

+      "L_b=3000 #Length of bar\n",

+      "A_b=1000 #mm**2 #Area of bar\n",

+      "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brasss bar\n",

+      "L_s=5000 #mm #Length of steel bar\n",

+      "A_s=445 #mm**2 #Area of steel bar\n",

+      "E_s=2*10**5 #N/mm**2 #Modulus of elasticity of steel bar\n",

+      "P=3000 #N #Load\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#From static equilibrium Equation of the rod after appliying Load is\n",

+      "#P_b+P_s=P   ......................(1)\n",

+      "\n",

+      "#P_b=1.8727*P_s ..................(2)\n",

+      "\n",

+      "#NOw substituting equation 2 in equation 1 we get\n",

+      "P_s=P*2.8727**-1\n",

+      "P_b=P-P_s\n",

+      "\n",

+      "d=P_s*L*P**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Distance at which Load applied even after which bar remains horizontal is\",round(d,2),\"m\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Distance at which Load applied even after which bar remains horizontal is 3.13 m\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.29,Page No.41"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "A_b=1000 #MM**2 #Area of brass bar\n",

+      "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brass\n",

+      "A_s=600 #N/mm**2 #Area of steel rod\n",

+      "E_s=2*10**5 #N/mm**2 #Modulus of eLasticity of steel bar\n",

+      "P=10*10**2 #N #Load\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let P_b be the tensile force in brass bar and P_s be the compressive force in steel bar\n",

+      "#Now taking moment about A we get static Equilibrium condition as\n",

+      "#P_b+2*P_s=27500 ......................................(1)\n",

+      "\n",

+      "#Now from deformed shape  we get\n",

+      "#dell_s=2*dell_b\n",

+      "\n",

+      "#P_s*L_s*(A_s*E_s)**-1=P_b*L_b*(A_b*E_b)**-1\n",

+      "#Further simplifying we get\n",

+      "#P_s=1.2*P_b   .........................................(2)\n",

+      "\n",

+      "#Now substituting equation 1 in equation 2 we get\n",

+      "P_b=27500*3.4**-1\n",

+      "P_s=1.2*P_b\n",

+      "\n",

+      "#Tensile stress in brass bar \n",

+      "sigma_b=P_b*A_b**-1\n",

+      "\n",

+      "#compressive stress in steel bar\n",

+      "sigma_s=P_s*A_s**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Compressive Stress in Bar is\",round(sigma_s,2),\"N/mm**2\"\n",

+      "print\"tensile Stress in Bar is\",round(sigma_b,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Compressive Stress in Bar is 16.18 N/mm**2\n",

+        "tensile Stress in Bar is 8.09 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 15

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.30,Page No.44"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=12.6 #m #Length of rail\n",

+      "t1=24 #Degree celsius\n",

+      "t2=44 #degree celsius\n",

+      "alpha=12*10**-6 #Per degree celsius\n",

+      "E=2*10**5 #N/mm**2 #Modulus of ELasticity\n",

+      "gamma=2 #mm #Gap provided for Expansion\n",

+      "sigma=20 #N/mm**2 #Stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "t=t2-t1 #Temperature Difference\n",

+      "\n",

+      "#Free Expansion of the rails\n",

+      "dell=alpha*t*L*1000 #mm \n",

+      "\n",

+      "#When no expansion joint is provided then\n",

+      "p=dell*E*(L*10**3)**-1\n",

+      "\n",

+      "#When a gap of 2 mm is provided,then free expansion prevented is\n",

+      "dell_1=dell-gamma\n",

+      "p2=dell_1*E*(L*10**3)**-1\n",

+      "\n",

+      "#When stress is developed,then gap left is\n",

+      "gamma2=-(sigma*L*10**3*E**-1-dell)\n",

+      "\n",

+      "#Result\n",

+      "print\"The minimum gap between the two rails is\",round(dell,2),\"mm\"\n",

+      "print\"Thermal Developed in the rials if:No expansionn joint is provided:p\",round(p,2),\"N/mm**2\"\n",

+      "print\"                                 :If a gap of is provided then   :p2\",round(p2,2),\"N/mm**2\"\n",

+      "print\"When stress is developed gap left between the rails is\",round(gamma2,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The minimum gap between the two rails is 3.02 mm\n",

+        "Thermal Developed in the rials if:No expansionn joint is provided:p 48.0 N/mm**2\n",

+        "                                 :If a gap of is provided then   :p2 16.25 N/mm**2\n",

+        "When stress is developed gap left between the rails is 1.76 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 23

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.31,Page No.45"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "t=20 #degree celsius\n",

+      "E_a=70*10**9 #N/mm**2 #Modulus of Elasticicty of aluminium\n",

+      "alpha_a=11*10**-6 #per degree celsius #Temperature coeff of aluminium\n",

+      "alpha_s=12*10**-6 #Per degree celsius #Temperature coeff of steel\n",

+      "L_a=1000 #mm #Length of aluminium \n",

+      "L_s=3000 #mm #Length of steel\n",

+      "E_a=7*10**4 #N/mm**2 #Modulus of Elasticity of aluminium\n",

+      "E_s=2*10**5 #N/mm*2 #Modulus of Elasticity of steel\n",

+      "A_a=600 #mm**2 #Area of aluminium\n",

+      "A_s=300 #mm**2 #Area of steel\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Free Expansion \n",

+      "dell=alpha_a*t*L_a+alpha_s*t*L_s\n",

+      "\n",

+      "#support Reaction\n",

+      "P=dell*(L_a*(A_a*E_a)**-1+L_s*(A_s*E_s)**-1)**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Reaction at support is\",round(P,2),\"N\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Reaction at support is 12735.48 N\n"

+       ]

+      }

+     ],

+     "prompt_number": 26

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.33,Page No.48"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "D=25 #mm #Diameter of Brass\n",

+      "De=50 #mm #External Diameter of steel tube\n",

+      "Di=25 #mm #Internal Diameter of steel tube\n",

+      "L=1.5 #m #Length of both bars\n",

+      "t1=30 #degree celsius #Initial Temperature\n",

+      "t2=100 #degree celsius #final Temperature\n",

+      "E_s=2*10**5 #N/mm**2 #Modulus of ELasticity of steel bar\n",

+      "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brass bar\n",

+      "alpha_s=11.6*10**-6 #Temperature Coeff of steel\n",

+      "alpha_b=18.7*10**-6 #Temperature coeff of brass bar\n",

+      "d=20 #mm #diameter of pins\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "t=t2-t1 #Temperature Difference\n",

+      "A_s=pi*4**-1*(De**2-Di**2) #mm**2 #Area of steel\n",

+      "A_b=pi*4**-1*D**2 #mm**2 #Area of brass\n",

+      "\n",

+      "#Let P_b be the tensile force in brass bar and P_s be the compressive force in steel bar\n",

+      "#But from Equilibrium of Forces \n",

+      "#P_b=P_s=P\n",

+      "\n",

+      "#Let dell=dell_s+dell_b\n",

+      "dell=(alpha_b-alpha_s)*t*L*1000\n",

+      "\n",

+      "P=dell*(1*(A_s*E_s)**-1+1*(A_b*E_b)**-1)**-1*(L*1000)**-1\n",

+      "P_b=P_s=P\n",

+      "\n",

+      "#Stress in steel\n",

+      "sigma_s=P_s*A_s**-1\n",

+      "\n",

+      "#Stress in Brass\n",

+      "sigma_b=P_b*A_b**-1\n",

+      "\n",

+      "#Area of Pins\n",

+      "A_p=pi*4**-1*d**2\n",

+      "\n",

+      "#Since,the force is resisted by two cross section of pins\n",

+      "tou=P*(2*A_p)**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Stress in steel bar is\",round(sigma_s,2),\"N/mm**2\"\n",

+      "print\"Stress in Brass bar is\",round(sigma_b,2),\"N/mm**2\"\n",

+      "print\"Shear Stresss induced in pins is\",round(tou,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stress in steel bar is 14.2 N/mm**2\n",

+        "Stress in Brass bar is 42.6 N/mm**2\n",

+        "Shear Stresss induced in pins is 33.28 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 14

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.34,Page No.49"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "b_s=60 #mm #width of steel Bar\n",

+      "t_s=10 #mm #thickness of steel Bar\n",

+      "b_c=40 #mm #width of copper bar\n",

+      "t_c=5  #mm #thickness of copper bar\n",

+      "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel bar\n",

+      "E_c=1*10**5 #N/mm**2 #Modulus of Elasticity of copper bar\n",

+      "alpha_s=12*10**-6 #Per degree celsius #Temperature coeff of steel bar\n",

+      "alpha_c=17*10**-6 #Per degree celsius #Temperature coeff of copper bar\n",

+      "L_s=L_c=L=1000 #mm #Length of bar\n",

+      "t=80 #degree celsius\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "A_s=b_s*t_s #Area of steel bar\n",

+      "A_c=b_c*t_c #Area of copper bar\n",

+      "\n",

+      "#Let P_s be the tensile force in steel bar and P_c be the compressive force in copper bar\n",

+      "#The equilibrium of forces gives \n",

+      "#P_s=2*P_c\n",

+      "\n",

+      "#Let dell=dell_s+dell_b\n",

+      "dell=(alpha_c-alpha_s)*t\n",

+      "\n",

+      "P_c=dell*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1\n",

+      "P_s=2*P_c\n",

+      "\n",

+      "#Stress in copper \n",

+      "sigma_c=P_c*A_c**-1\n",

+      "\n",

+      "#Stress in steel\n",

+      "sigma_s=P_s*A_s**-1\n",

+      "\n",

+      "#Change  in Length of bar\n",

+      "dell_2=alpha_s*t*L+P_s*L_s*(A_s*E_s)**-1\n",

+      "\n",

+      "#result\n",

+      "print\"Stress in copper is\",round(sigma_c,2),\"N/mm**2\"\n",

+      "print\"Stress in steel is\",round(sigma_s,2),\"N/mm**2\"\n",

+      "print\"the change in Length is\",round(dell_2,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stress in copper is 30.0 N/mm**2\n",

+        "Stress in steel is 20.0 N/mm**2\n",

+        "the change in Length is 1.06 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 24

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.35,Page No.50"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "P=2*10**5 #N #Weight\n",

+      "L=1 #m #Length of each rod\n",

+      "A_c=A_s=A=500 #mm**2 #Area of each rod\n",

+      "t=40 #degree celsius #temperature\n",

+      "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel rod\n",

+      "E_c=1*10**5 #N/mm**2 #modulus of Elastictiy  of copper rod\n",

+      "alpha_s=1.2*10**-5 #Per degree Celsius #temp coeff of steel rod\n",

+      "alpha_c=1.8*10**-5 #Per degree Celsius #Temp coeff of copper rod\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let P_s be the force in each one of the copper rods and P_s be the force in steel rod\n",

+      "#2*P_c+P_s=P   .....................(1)\n",

+      "\n",

+      "#Extension of copper bar=Extension of steel bar\n",

+      "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",

+      "#after simplifying above equation we get\n",

+      "#P_s=2*P_c      ........................(2)\n",

+      "\n",

+      "#Now substituting value of P_s in Equation 1 we get\n",

+      "P_c=P*4**-1\n",

+      "P_s=2*P_c\n",

+      "\n",

+      "#Now EXtension due to copper Load\n",

+      "dell_1=P_c*L*1000*(A_c*E_c)**-1\n",

+      "\n",

+      "#Part-2\n",

+      "\n",

+      "#Due to rise of temperature of40 degree celsius\n",

+      "\n",

+      "#As bars are rigidly joined,let P_c1 be the compressive forccesdeveloped in copper bar and P_s1 be the tensile force in steel causing changes\n",

+      "#P_s1=2*P_c1\n",

+      "\n",

+      "#dell_s+dell_c=(alpha_c-alpha_s)*t*L      .......................................(3)\n",

+      "#P_s1*L*(A_s*E_s)**-1+P_c1*L*(A_c*E_c)**-1=(alpha_c-alpha_s)*t*L   ................(4)\n",

+      "#After substituting values in above equation and further simplifying we get,\n",

+      "P_c1=(alpha_c-alpha_s)*t*L*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1  #.................(5)\n",

+      "P_s1=2*P_c1\n",

+      "\n",

+      "#Extension of bar due to temperature rise\n",

+      "dell_2=alpha_s*t*L+P_s1*L*(A_s*E_s)**-1\n",

+      "\n",

+      "#Amount by which bar will descend\n",

+      "dell_3=dell_1+dell_2\n",

+      "\n",

+      "#Load carried by steel bar\n",

+      "P_S=P_s+P_s1\n",

+      "\n",

+      "#Load carried by copper bar\n",

+      "P_C=P_c-P_c1\n",

+      "\n",

+      "#Part-3\n",

+      "\n",

+      "#Let P_c1_1=P_c                 #For convenience\n",

+      "#Rise in temperature if Load is to be carried out by steel rod alone\n",

+      "P_c1_1=P_c\n",

+      "\n",

+      "#From equation 5 \n",

+      "t=P_c1_1*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)*(alpha_c-alpha_s)**-1\n",

+      "\n",

+      "#result\n",

+      "print\"Extension Due top copper Load\",round(dell_1,2),\"mm\"\n",

+      "print\"Load carried by each rod:P_s\",round(P_s,2),\"N\"\n",

+      "print\"                        :P_c\",round(P_c,2),\"N\"\n",

+      "print\"Rise in Temperature of steel rod should be\",round(t,2),\"degree Celsius\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Extension Due top copper Load 1.0 mm\n",

+        "Load carried by each rod:P_s 100000.0 N\n",

+        "                        :P_c 50000.0 N\n",

+        "Rise in Temperature of steel rod should be 333.33 degree Celsius\n"

+       ]

+      }

+     ],

+     "prompt_number": 26

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.36,Page No.53"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "t=40 #degree celsius #temperature\n",

+      "A_s=400 #mm**2 #Area of steel bar\n",

+      "A_c=600 #mm**2 #Area of copper bar\n",

+      "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel bar\n",

+      "E_c=1*10**5 #N/mm**2 #Modulus of Elasticity of copper bar\n",

+      "alpha_s=12*10**-6 #degree celsius #Temperature coeff of steel bar\n",

+      "alpha_c=18*10**-6 #degree celsius #Temperature coeff of copper bar\n",

+      "L_c=800 #mm #Length of copper bar\n",

+      "L_s=600 #mm #Length of steel bar\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let P_s be the tensile force in steel bar and P_c be the compressive force in copper bar\n",

+      "#Static Equilibrium obtained by taking moment about A\n",

+      "#P_c=2*P_s\n",

+      "\n",

+      "#From property of similar triangles we get\n",

+      "#(alpha_c*Lc-dell_c)*1**-1=(alpha_s*L_s-dell_s)*2**-1\n",

+      "#After substituting values in above equations and further simplifying we get\n",

+      "P_s=(2*alpha_c*L_c-alpha_s*L_s)*t*(L_s*(A_s*E_s)**-1+4*L_c*(A_c*E_c)**-1)**-1\n",

+      "P_c=2*P_s\n",

+      "\n",

+      "#Stress in steel rod\n",

+      "sigma_s=P_s*A_s**-1 #N/mm**2 \n",

+      "\n",

+      "#Stress in copper rod\n",

+      "sigma_c=P_c*A_c**-1 #N/mm**2\n",

+      "\n",

+      "#Result\n",

+      "print\"Stress in steel rod is\",round(sigma_s,2),\"N/mm**2\"\n",

+      "print\"STress in copper rod is\",round(sigma_c,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stress in steel rod is 35.51 N/mm**2\n",

+        "STress in copper rod is 47.34 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 4

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.37,Page No.61"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=20 #mm #Diameter of bar\n",

+      "P=37.7*10**3 #N #Load\n",

+      "L=200 #mm #Guage Length \n",

+      "dell=0.12 #mm #Extension\n",

+      "dell_d=0.0036 #mm #contraction in diameter\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Area of bar\n",

+      "A=pi*4**-1*d**2\n",

+      "\n",

+      "#Let s and dell_s be the Linear strain and Lateral strain\n",

+      "s=dell*L**-1\n",

+      "dell_s=dell_d*d**-1\n",

+      "mu=dell_s*s**-1 #Poissoin's ratio \n",

+      "\n",

+      "#dell=P*L*(A*E)**-1\n",

+      "E=P*L*(dell*A)**-1 #N/mm**2 #Modulus of Elasticity of bar\n",

+      "\n",

+      "#Modulus of Rigidity\n",

+      "G=E*(2*(1+mu))**-1 #N/mm**2\n",

+      "\n",

+      "#Bulk Modulus \n",

+      "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",

+      "\n",

+      "#result\n",

+      "print\"Poisson's ratio is\",round(mu,2)\n",

+      "print\"The Elastic constant are:E\",round(E,2)\n",

+      "print\"                        :G\",round(G,2)\n",

+      "print\"                        :K\",round(K,2)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Poisson's ratio is 0.3\n",

+        "The Elastic constant are:E 200004.71\n",

+        "                        :G 76924.89\n",

+        "                        :K 166670.59\n"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.38,Page No.62"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=100 #mm #Diameter of circular rod\n",

+      "P=1*10**6 #N #Tensile Force\n",

+      "mu=0.3 #Poisson's ratio\n",

+      "E=2*10**5 #N/mm**2 #Young's Modulus \n",

+      "L=500 #mm #Length of rod\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Modulus of Rigidity\n",

+      "G=E*(2*(1+mu))**-1 #N/mm**2\n",

+      "\n",

+      "#Bulk Modulus \n",

+      "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",

+      "\n",

+      "A=pi*4**-1*d**2 #mm**2 #Area of Circular rod\n",

+      "#Let sigma be the Longitudinal stress\n",

+      "sigma=P*A**-1 #N/mm**2 \n",

+      "\n",

+      "s=sigma*E**-1 #Linear strain\n",

+      "e_x=s\n",

+      "\n",

+      "#Volumetric strain\n",

+      "e_v=e_x*(1-2*mu)\n",

+      "\n",

+      "v=pi*4**-1*d**2*L\n",

+      "#Change in VOlume\n",

+      "dell_v=e_v*v\n",

+      "\n",

+      "#Result\n",

+      "print\"Bulk Modulus  is\",round(E,2),\"N/mm**2\"\n",

+      "print\"Modulus of Rigidity is\",round(G,2),\"N/mm**2\"\n",

+      "print\"The change in Volume is\",round(dell_v,2),\"mm**3\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Bulk Modulus  is 200000.0 N/mm**2\n",

+        "Modulus of Rigidity is 76923.08 N/mm**2\n",

+        "The change in Volume is 1000.0 mm**3\n"

+       ]

+      }

+     ],

+     "prompt_number": 21

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.39,Page No.62"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=500 #mm #Length of rectangular cross section bar\n",

+      "A=20*40 #mm**2 #Area of rectangular cross section bar\n",

+      "P1=4*10**4 #N #Tensile Force on 20mm*40mm Faces\n",

+      "P2=2*10**5 #N #compressive force on 20mm*500mm Faces\n",

+      "P3=3*10**5 #N #Tensile Force on 40mm*500mm Faces\n",

+      "E=2*10**5 #N/mm**2 #young's Modulus \n",

+      "mu=0.3 #Poisson's Ratio\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let P_x,P_y,P_z be the forces n x,y,z directions\n",

+      "\n",

+      "P_x=P1*A**-1\n",

+      "P_y=P2*A**-1\n",

+      "P_z=P3*A**-1\n",

+      "\n",

+      "#Let e_x,e_y,e_z be the strains in x,y,z directions\n",

+      "e_x=1*E**-1*(50+mu*20-15*mu)\n",

+      "e_y=1*E**-1*(-mu*50-20-mu*15)\n",

+      "e_z=1*E**-1*(-mu*50+mu*20+15)\n",

+      "\n",

+      "#Volumetric strain\n",

+      "e_v=e_x+e_y+e_z\n",

+      "\n",

+      "#Volume\n",

+      "V=20*40*500 #mm**3\n",

+      "#Change in Volume \n",

+      "dell_v=e_v*V #mm**3\n",

+      "\n",

+      "#Result\n",

+      "print\"The change in Volume is\",round(dell_v,2),\"mm**3\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The change in Volume is 36.0 mm**3\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.41,Page No.65"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "E=2.1*10**5 #N/mm**2 #Young's Modulus \n",

+      "G=0.78*10**5 #N/mm**2 #Modulus of Rigidity\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Now using the relation\n",

+      "#E=2*G*(1+mu)\n",

+      "mu=E*(2*G)**-1-1 #Poisson's ratio\n",

+      "\n",

+      "#Bulk Modulus \n",

+      "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",

+      "\n",

+      "\n",

+      "#Result\n",

+      "print\"The Poisson's Ratio is\",round(mu,2)\n",

+      "print\"The modulus of Rigidity\",round(K,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Poisson's Ratio is 0.35\n",

+        "The modulus of Rigidity 227500.0 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.42,Page No.65"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "G=0.4*10**5 #N/mm**2 #Modulus of rigidity\n",

+      "K=0.75*10**5 #N/mm**2 #Bulk Modulus \n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Young's Modulus\n",

+      "E=9*G*K*(3*K+G)**-1\n",

+      "\n",

+      "#Now from the relation\n",

+      "#E=2*G(1+2*mu)\n",

+      "mu=E*(2*G)**-1-1 #POissoin's ratio \n",

+      "\n",

+      "#result\n",

+      "print\"Young's modulus is\",round(E,2),\"N/mm**2\"\n",

+      "print\"Poissoin's ratio is\",round(mu,2)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Young's modulus is 101886.79 N/mm**2\n",

+        "Poissoin's ratio is 0.27\n"

+       ]

+      }

+     ],

+     "prompt_number": 25

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.43,Page No.65"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "b=60 #mm #width of bar\n",

+      "d=30 #mm #depth of bar\n",

+      "L=200 #mm #Length of bar\n",

+      "A=30*60 #mm**2 #Area of bar\n",

+      "A2=30*200 #mm**2 #Area of bar along which expansion is restrained\n",

+      "P=180*10**3 #N #Compressive force\n",

+      "E=2*10**5 #N/mm**2 #Young's Modulus\n",

+      "mu=0.3 #Poissoin's ratio\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#The bar is restrained from expanding in Y direction\n",

+      "P_z=0\n",

+      "P_x=P*A**-1 #stress developed in x direction\n",

+      "\n",

+      "#Now taking compressive strain as positive\n",

+      "#e_x=P_x*E**-1-mu*P_y*E**-1      .......................(1)\n",

+      "#e_y=-mu*P_x*E**-1+P_y*E**-1       ....................(2)\n",

+      "#e_z=-mu*P_x*E**-1-mu*P_y*E**-1  ......................(3)\n",

+      "\n",

+      "#Part-1\n",

+      "#When it is fully restrained\n",

+      "e_y=0\n",

+      "P_y=30 #N/mm**2 \n",

+      "e_x=P_x*E**-1-mu*P_y*E**-1\n",

+      "e_z=-mu*P_x*E**-1-mu*P_y*E**-1\n",

+      "\n",

+      "#Change in Length \n",

+      "dell_l=e_x*L #mm\n",

+      "\n",

+      "#Change in width\n",

+      "dell_b=b*e_y\n",

+      "\n",

+      "#change in Depth\n",

+      "dell_d=d*e_z\n",

+      "\n",

+      "#Volume of bar\n",

+      "V=b*d*L #mm**3\n",

+      "#Change in Volume\n",

+      "e_v=(e_x+e_y+e_z)*V #mm**3\n",

+      "\n",

+      "#Part-2\n",

+      "#When 50% is restrained\n",

+      "\n",

+      "#Free strain in Y direction\n",

+      "e_y1=mu*P_x*E**-1\n",

+      "\n",

+      "#As 50% is restrained,so\n",

+      "e_y2=-50*100**-1*e_y1\n",

+      "\n",

+      "#But form Equation 2 we have e_y=-mu*P_x*E**-1+P_y*E**-1 \n",

+      "#After substituting values in above equation and furthe simplifying we get\n",

+      "P_y=e_y2*E+d\n",

+      "\n",

+      "e_x2=P_x*E**-1-mu*P_y*E**-1 \n",

+      "e_z2=-mu*P_x*E**-1-mu*P_y*E**-1\n",

+      "\n",

+      "#Change in Length \n",

+      "dell_l2=e_x2*L #mm\n",

+      "\n",

+      "#Change in width\n",

+      "dell_b2=b*e_y2\n",

+      "\n",

+      "#change in Depth\n",

+      "dell_d2=d*e_z2\n",

+      "\n",

+      "#Change in Volume\n",

+      "e_v2=(e_x2+e_y2+e_z2)*V #mm**3\n",

+      "\n",

+      "#REsult\n",

+      "print\"Change in Dimension of bar is:dell_l\",round(dell_l,2),\"mm\"\n",

+      "print\"                             :dell_b\",round(dell_b,4),\"mm\"\n",

+      "print\"                             :dell_d\",round(dell_d,2),\"mm\"\n",

+      "print\"Change in Volume is\",round(e_v,2),\"mm**3\"\n",

+      "print\"Changes in material when only 50% of expansion can be reatrained:dell_l2\",round(dell_l2,2),\"mm\"\n",

+      "print\"                                                                :dell_b2\",round(dell_b2,4),\"mm\"\n",

+      "print\"                                                                :dell_d2\",round(dell_d2,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Change in Dimension of bar is:dell_l 0.09 mm\n",

+        "                             :dell_b 0.0 mm\n",

+        "                             :dell_d -0.01 mm\n",

+        "Change in Volume is 93.6 mm**3\n",

+        "Changes in material when only 50% of expansion can be reatrained:dell_l2 0.1 mm\n",

+        "                                                                :dell_b2 -0.0045 mm\n",

+        "                                                                :dell_d2 -0.01 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 23

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.44,Page No.72"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "P=10*10**3 #N #Load\n",

+      "E=2*10**5 #N/mm**2 #Young's Modulus\n",

+      "d2=12 #mm #Diameter of bar1\n",

+      "d1=16 #mm #diameter of bar2\n",

+      "L1=200 #mm #Length of bar1\n",

+      "L2=500 #mm #Length of bar2\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let A1 and A2 be the cross Area of Bar1 & bar2 respectively\n",

+      "A1=pi*4**-1*d1**2 #mm**2\n",

+      "A2=pi*4**-1*d2**2 #mm**2\n",

+      "\n",

+      "#Let p1 and p2 be the stress in Bar1 nad bar2 respectively\n",

+      "p1=P*A1**-1 #N/mm**2\n",

+      "p2=P*A2**-1 #N/mm**2\n",

+      "\n",

+      "#Let V1 nad V2 be the Volume of of Bar1 and Bar2\n",

+      "V1=A1*(L1+L1)\n",

+      "V2=A2*L2\n",

+      "\n",

+      "#Let E be the strain Energy stored in the bar\n",

+      "E=p1**2*(2*E)**-1*V1+p2**2*V2*(2*E)**-1\n",

+      "\n",

+      "#result\n",

+      "print\"The Strain Energy stored in Bar is\",round(E,2),\"N-mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Strain Energy stored in Bar is 1602.6 N-mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 74

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.45,Page No.73"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Bar-A\n",

+      "d1=30 #mm #Diameter of bar1\n",

+      "L=600 #mm #length of bar1\n",

+      "\n",

+      "#Bar-B\n",

+      "d2=30 #mm #Diameter of bar2\n",

+      "d3=20 #mm #Diameter of bar2\n",

+      "L2=600 #mm #length of bar2\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Area of bar-A\n",

+      "A1=pi*4**-1*d1**2\n",

+      "\n",

+      "#Area of bar-B\n",

+      "A2=pi*4**-1*d2**2\n",

+      "A3=pi*4**-1*d3**2\n",

+      "\n",

+      "#let SE be the Strain Energy\n",

+      "#Strain Energy stored in Bar-A\n",

+      "#SE=p**2*(2*E)**-1*V\n",

+      "#After substituting values and simolifying further we get\n",

+      "#SE=P**2*E**-1*0.4244\n",

+      "\n",

+      "#Strain Energy stored in Bar-B\n",

+      "#SE2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1\n",

+      "#After substituting values and simolifying further we get\n",

+      "#SE2=0.6897*P**2*E**-1\n",

+      "\n",

+      "#Let X be the ratio of SE in Bar-B and SE in Bar-A\n",

+      "X=0.6897*0.4244**-1\n",

+      "\n",

+      "#Part-2\n",

+      "\n",

+      "#When Max stress is produced is same:Let p be the max stress produced\n",

+      "\n",

+      "#Stress in bar A is p throughout \n",

+      "#In bar B:stress in 20mm dia.portion=p2=p\n",

+      "\n",

+      "#Stress in 30 mm dia.portion\n",

+      "#p1=P*A2*A3**-1\n",

+      "#After substituting values and simolifying further we get\n",

+      "#p1=4*9**-1*p\n",

+      "\n",

+      "#Strain Energy in bar A\n",

+      "#SE_1=p**2*(2*E)**-1*A1*L1\n",

+      "#After substituting values and simolifying further we get\n",

+      "#SE_1=67500*p**2*pi*E**-1\n",

+      "\n",

+      "#Strain Energy in bar B\n",

+      "#SE_2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1\n",

+      "#After substituting values and simolifying further we get\n",

+      "#SE_2=21666.67*pi*p**2*E**-1\n",

+      "\n",

+      "#Let Y be the Ratio of SE in bar B and SE in bar A\n",

+      "Y=21666.67*67500**-1\n",

+      "\n",

+      "#result\n",

+      "print\"Gradually applied Load is\",round(X,2)\n",

+      "print\"Gradually applied Load is\",round(Y,2)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Gradually applied Load is 1.63\n",

+        "Gradually applied Load is 0.32\n"

+       ]

+      }

+     ],

+     "prompt_number": 4

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.46,Page No.74"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables  \n",

+      "\n",

+      "W=100 #N #Load\n",

+      "E=2*10**5 #N/mm**2 #Young's Modulus \n",

+      "h=60 #mm #Height through Load falls down\n",

+      "L=400 #mm #Length of collar\n",

+      "d=30 #mm #diameter of bar\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "A=pi*4**-1*d**2 #mm**2 #Area of bar\n",

+      "\n",

+      "#Instantaneous stress produced is\n",

+      "p=W*A**-1*(1+(1+(2*A*E*h*(W*L)**-1))**0.5)\n",

+      "\n",

+      "#Now the EXtension of the bar is neglected in calculating work doneby the Load,then\n",

+      "P=(2*E*h*W*(A*L)**-1)**0.5\n",

+      "\n",

+      "#Let percentage error be denoted by E1\n",

+      "#Percentage error in approximating is\n",

+      "E1=(p-P)*p**-1*100\n",

+      "\n",

+      "#Instantaneous Extension produced is\n",

+      "dell_l=round(P,3)*E**-1*L\n",

+      "\n",

+      "#Result\n",

+      "print\"The Instantaneous stress is\",round(p,2),\"N/mm\"\n",

+      "print\"Percentage Error is\",round(E1,2)\n",

+      "print\"The Instantaneous extension is\",round(dell_l,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Instantaneous stress is 92.27 N/mm\n",

+        "Percentage Error is 0.15\n",

+        "The Instantaneous extension is 0.18 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 23

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.47,Page No.75"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=20 #mm #Diameter of steel bar\n",

+      "L=1000 #mm #Length of bar\n",

+      "E=2*10**5 #N/mm**2 #Young's Modulus \n",

+      "p=300 #N/mm**2 #max Permissible stress\n",

+      "h=50 #mm #Height through which weight will fall\n",

+      "w=600 #N #Load\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#ARea of steel bar\n",

+      "A=pi*4**-1*d**2\n",

+      "\n",

+      "#Instantaneous extension is\n",

+      "dell_l=p*L*E**-1 #mm \n",

+      "\n",

+      "#Work done by Load \n",

+      "#W=W1*(h+dell_l)\n",

+      "\n",

+      "#Volume of bar\n",

+      "V=round(A,2)*L\n",

+      "#Let E1 be the strain Energy\n",

+      "E1=p**2*(2*E)**-1*V\n",

+      "\n",

+      "#Answer in Book for Strain Energy is Incorrect \n",

+      "\n",

+      "#Now Equating Workdone by Load to strain Energy \n",

+      "W1=E1*51.5**-1\n",

+      "\n",

+      "#Now when w=600 N\n",

+      "#Let W2 be the Work done by the Load\n",

+      "#W2=w(h2*dell_l)\n",

+      "\n",

+      "h=E1*w**-1-dell_l\n",

+      "\n",

+      "#Result\n",

+      "print\"The Max Lodad which can Fall from a height of 50 mm on the collar is\",round(W1,2),\"N\"\n",

+      "print\"the Max Height from which a 600 N Load can fall on the collar is\",round(h,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Max Lodad which can Fall from a height of 50 mm on the collar is 1372.54 N\n",

+        "the Max Height from which a 600 N Load can fall on the collar is 116.31 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 40

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.48,Page No.76"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "D_s=30 #mm #Diameter of steel rod\n",

+      "d=30 #mm #Internal Diameter of copper tube\n",

+      "D=40#mm #External Diameter of copper tube\n",

+      "E_s=2*10**5 #N/mm**2 #Young's Modulus of Steel rod\n",

+      "E_c=1*10**5#N/mm**2 #Young's Modulus of copper tube\n",

+      "P=100 #N #Load\n",

+      "h=40 #mm #height from which Load falls\n",

+      "L=800 #mm #Length \n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Area of steel rod\n",

+      "A_s=pi*4**-1*D_s**2\n",

+      "\n",

+      "#Area of copper tube\n",

+      "A_c=pi*4**-1*(D**2-d**2)\n",

+      "\n",

+      "#But Dell_s=dell_c=dell\n",

+      "#p_s*E_s**-1*L=p_c*L*E_c\n",

+      "#After simplifying furthe we get\n",

+      "#p_s=2*p_c\n",

+      "\n",

+      "#Now Equating internal Energy to Workdone we get\n",

+      "p_c=(2*P*h*L**-1*(4*A_s*E_s**-1+A_c*E_c**-1))**0.5\n",

+      "p_s=2*p_c\n",

+      "\n",

+      "#Result\n",

+      "print A_s\n",

+      "print\"STress produced in steel is\",round(p_s,2),\"N/mm**2\"\n",

+      "print\"STress produced in copper is\",round(p_c,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "706.858347058\n",

+        "STress produced in steel is 0.89 N/mm**2\n",

+        "STress produced in copper is 0.44 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 2.2.49,Page No.77"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "dell=0.25 #mm #Instantaneous Extension\n",

+      "\n",

+      "#Bar-A\n",

+      "b1=25 #mm #width of bar\n",

+      "D1=500 #mm #Depth of bar\n",

+      "\n",

+      "#Bar-B\n",

+      "b2_1=25 #mm #width of upper bar\n",

+      "b2_2=15 #mm #Width of Lower Bar\n",

+      "L2=200 #mm #Length  of upper bar\n",

+      "L1=300 #mm #Length of Lower bar\n",

+      "\n",

+      "E=2*10**5 #N/mm**2 #Young's Modulus of bar\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Strain\n",

+      "e=dell*D1**-1 \n",

+      "\n",

+      "#Load\n",

+      "p=e*E\n",

+      "\n",

+      "#Area of bar-A\n",

+      "A=pi*4**-1*25**2\n",

+      "\n",

+      "#Volume of bar-A\n",

+      "V=A*D1\n",

+      "\n",

+      "#Let E1 be the Energy of Blow\n",

+      "#Energy of Blow\n",

+      "E1=p**2*(E)**-1*V\n",

+      "\n",

+      "#Let p2 be the Max stress in bar B When this blow is applied.\n",

+      "#the max stress occurs in the 15mm dia. portion,Hence, the stress in 25 mm dia.portion is\n",

+      "#p2*pi*4**-1*b2_2**2*(pi*4**-1*b2_2**2=0.36*p\n",

+      "\n",

+      "#Strain Energy of bar B\n",

+      "#E2=p**2*(2*E)**-1*v1+1*(2*E)**-1*(0.36*p2)**2*v2\n",

+      "#After substituting values and Further substituting values we get\n",

+      "#E2=0.1643445*p2**2\n",

+      "\n",

+      "#Equating it to Energy of applied blow,we get\n",

+      "p2=(12271.846*0.1643445**-1)**0.5\n",

+      "\n",

+      "#Stress in top portion\n",

+      "sigma=0.36*p2\n",

+      "\n",

+      "#Extension in Bar-1\n",

+      "dell_1=p2*E**-1*L1\n",

+      "\n",

+      "#Extension in Bar-2\n",

+      "dell_2=0.36*p2*E**-1*L2\n",

+      "\n",

+      "#Extension of bar\n",

+      "dell_3=dell_1+dell_2\n",

+      "\n",

+      "#Result\n",

+      "print\"Instantaneous Max stress is\",round(sigma,2),\"N/mm**2\"\n",

+      "print\"extension in Bar is\",round(dell_3,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Instantaneous Max stress is 98.37 N/mm**2\n",

+        "extension in Bar is 0.51 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 1

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_1.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_1.ipynb
new file mode 100644
index 00000000..f136c4ad
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_1.ipynb
@@ -0,0 +1,1588 @@
+{

+ "metadata": {

+  "name": "chapter no.3.ipynb"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 3:Shear Force And Bending Moment Diagrams in Statically Determinate Beams"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 3.3.1,Page No.100"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L_AC=L_CD=1 #m #Length of AC & CD\n",

+      "L_DB=1.5 #m #Lengh of DB\n",

+      "L=3.5 #m #Length of Beam\n",

+      "F_B=10 #KN #Force at pt B\n",

+      "F_C=F_D=20 #KN #Force at pt C & D\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "R_A=F_C+F_D+F_B #KN #Force at support A \n",

+      "\n",

+      "#Shear Force Calculations\n",

+      "\n",

+      "#S.F At pt B\n",

+      "V_B1=0  #KN \n",

+      "V_B2=F_B #KN\n",

+      "\n",

+      "#S.F At pt D\n",

+      "V_D1=V_B2 #KN\n",

+      "V_D2=V_D1+F_D #KN\n",

+      "\n",

+      "#S.F At pt C \n",

+      "V_C1=V_D2 #KN\n",

+      "V_C2=V_D2+F_C #KN\n",

+      "\n",

+      "#S.F At Pt A\n",

+      "V_A1=V_C2 #KN\n",

+      "V_A2=V_C2-R_A #KN\n",

+      "\n",

+      "#Bending Moment Calculations\n",

+      "\n",

+      "#B.M At Pt B\n",

+      "M_B=0 #KN.m\n",

+      "\n",

+      "#B.M AT Pt D\n",

+      "M_D=F_B*L_DB #KN.m\n",

+      "\n",

+      "#B.M At pt C\n",

+      "M_C=F_B*(L_DB+L_CD)+F_D*L_CD #KN.m\n",

+      "\n",

+      "#B.M At pt A\n",

+      "M_A=F_B*L+F_D*(L_CD+L_AC)+F_C*L_AC\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,0,L_DB,L_DB,L_CD+L_DB,L_CD+L_DB,L_CD+L_DB+L_AC,L_CD+L_DB+L_AC]\n",

+      "Y1=[V_B1,V_B2,V_D1,V_D2,V_C1,V_C2,V_A1,V_A2]\n",

+      "Z1=[0,0,0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in m\")\n",

+      "plt.ylabel(\"Shear Force in kN\")\n",

+      "plt.show()\n",

+      "\n",

+      "#Plotting the Bendimg Moment Diagram\n",

+      "\n",

+      "Y2=[M_B,M_D,M_C,M_A]\n",

+      "X2=[0,L_DB,L_DB+L_CD,L_AC+L_CD+L_DB]\n",

+      "Z2=[0,0,0,0]\n",

+      "plt.plot(X2,Y2,X2,Z2)\n",

+      "plt.xlabel(\"Length in m\")\n",

+      "plt.ylabel(\"Bending Moment in kN.m\")\n",

+      "plt.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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qqUhkAwBEANcsxmWHvYZgs3B+O0MeWfzxxx/riSeeUFlZmerr6wMftHXr1vBSAgBiSsg1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\nRxEAgOUoAgCw3P8DfIlLuPJXvGsAAAAASUVORK5CYII=\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5dde2d0>"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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pqQavCQgIEADwi1/84he/rPgKCAiw6e+yYlJMDQ0NuPfee/H999+jZ8+eGDhw\nINavX4/g4GC5SyMickmKmWJq3749PvroI4wcORKNjY2YPn06mwMRkYwUM4IgIiJlUUzMtaWsrCwE\nBQXhnnvuwZIlS4y+Zt68ebjnnnsQHh6OwsJCiStsnbn6s7Oz4e3tjcjISERGRuLNN9+UoUrjpk2b\nBrVajbCwMJOvUfK5N1e/ks89AJSWliIuLg4hISEIDQ3Fhx9+aPR1SvwdWFK7ks9/XV0dYmJiEBER\nAY1Gg5deesno65R47gHL6rf6/Nu8qiyShoYGISAgQDh16pRQX18vhIeHC0VFRQav2bZtmzBq1ChB\nEAQhLy9PiImJkaNUoyypf8+ePcLYsWNlqrB1P/74o1BQUCCEhoYafV7J514QzNev5HMvCIJQVlYm\nFBYWCoIgCNXV1UJgYKDD/PtvSe1KP//Xrl0TBEEQbt68KcTExAj/+te/DJ5X6rlvZq5+a8+/4kYQ\nllwwl5GRgeTkZABATEwMqqqqUFFRIUe5d7D0gj9BoTN7Q4YMQZcuXUw+r+RzD5ivH1DuuQcAX19f\nREREAAA8PT0RHByM8+fPG7xGqb8DS2oHlH3+O3bsCACor69HY2MjunbtavC8Us99M3P1A9adf8U1\nCGMXzJ07d87sa86ePStZja2xpH6VSoV9+/YhPDwcCQkJKCoqkrpMmyn53FvCkc59SUkJCgsLERMT\nY/C4I/wOTNWu9PPf1NSEiIgIqNVqxMXFQaPRGDyv9HNvrn5rz79iUkzNLL1g7vYuaOn7xGZJHVFR\nUSgtLUXHjh2xY8cOJCYm4sSJExJU1zaUeu4t4SjnvqamBpMmTcKyZcvg6el5x/NK/h20VrvSz7+b\nmxsOHTqEK1euYOTIkcjOzoZWqzV4jZLPvbn6rT3/ihtB9OrVC6WlpfqfS0tL4efn1+przp49i14K\nudmzJfV7eXnph4KjRo3CzZs3UVlZKWmdtlLyubeEI5z7mzdv4qGHHsLjjz+OxMTEO55X8u/AXO2O\ncP4BwNvbG6NHj8ZPP/1k8LiSz31Lpuq39vwrrkEMGDAAv/32G0pKSlBfX4/09HSMGzfO4DXjxo3D\nl19+CQDIy8uDj48P1Gq1HOXewZL6Kyoq9P8VcvDgQQiCYHSuUImUfO4tofRzLwgCpk+fDo1Gg2ef\nfdboa5T6O7CkdiWf/z/++ANVVVUAgOvXr2P37t2IjIw0eI1Szz1gWf3Wnn/FTTGZumDus88+AwDM\nnj0bCQk5anozAAADy0lEQVQJ2L59O/r164dOnTph9erVMld9iyX1b9q0CZ988gnat2+Pjh07YsOG\nDTJXfcsjjzyCnJwc/PHHH+jduzdef/113Lx5E4Dyzz1gvn4ln3sA2Lt3L9auXYv+/fvr/8/99ttv\n48yZMwCU/TuwpHYln/+ysjIkJyejqakJTU1NmDp1KoYPH+4wf3ssqd/a888L5YiIyCjFTTEREZEy\nsEEQEZFRbBBERGQUGwQRERnFBkFEREaxQRARkVFsEOSUjG1P0ZaWLl2K69evW3W8zMxMk9vXEykR\nr4Mgp+Tl5YXq6mrRPr9v37746aef0K1bN0mORyQHjiDIZRQXF2PUqFEYMGAAhg4diuPHjwMAnnzy\nSTzzzDOIjY1FQEAANm/eDEC3M+acOXMQHByMESNGYPTo0di8eTOWL1+O8+fPIy4uDsOHD9d//quv\nvoqIiAj85S9/wYULF+44/po1a/D000+3esyWSkpKEBQUhKeeegr33nsvHnvsMezatQuxsbEIDAxE\nfn6+GKeJ6BYb70tBpGienp53PHb//fcLv/32myAIupu93H///YIgCEJycrKQlJQkCIIgFBUVCf36\n9RMEQRC+/vprISEhQRAEQSgvLxe6dOkibN68WRAEQfD39xcuXbqk/2yVSiV8++23giAIwvz584U3\n33zzjuOvWbNGmDt3bqvHbOnUqVNC+/bthV9//VVoamoSoqOjhWnTpgmCIAhbt24VEhMTrT0tRFZR\n3F5MRGKoqanB/v37MXnyZP1j9fX1AHTbNTfvPBocHKy/AUxubi6SkpIAQL+/vikeHh4YPXo0ACA6\nOhq7d+9utR5Tx7xd3759ERISAgAICQnBAw88AAAIDQ1FSUlJq8cgshcbBLmEpqYm+Pj4mLyHsIeH\nh/574T/LciqVymDvf6GV5Tp3d3f9925ubmhoaDBbk7Fj3q5Dhw4Gn9v8HkuPQWQPrkGQS+jcuTP6\n9u2LTZs2AdD9QT5y5Eir74mNjcXmzZshCAIqKiqQk5Ojf87LywtXr161qobWGgyRErFBkFOqra1F\n79699V9Lly7FunXrsHLlSkRERCA0NBQZGRn617e8K1jz9w899BD8/Pyg0WgwdepUREVFwdvbGwAw\na9YsPPjgg/pF6tvfb+wuY7c/bur7299j6mcl3cmMnBNjrkStuHbtGjp16oRLly4hJiYG+/btQ48e\nPeQui0gSXIMgasWYMWNQVVWF+vp6LFq0iM2BXApHEEREZBTXIIiIyCg2CCIiMooNgoiIjGKDICIi\no9ggiIjIKDYIIiIy6v8BsxKV3Pt7cnUAAAAASUVORK5CYII=\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5df0ab0>"

+       ]

+      }

+     ],

+     "prompt_number": 15

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 3.3.2,Page No.101"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "w1=10 #KN/m #u.d.L\n",

+      "F_D=20 #KN #Force at pt D\n",

+      "F_C=30 #KN #Force at pt C\n",

+      "L_DB=4 #m #Length of DB\n",

+      "L_CD=L_AC=2 #m #Length of AC & CD\n",

+      "L=8 #m #Length of Beam\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let R_A And R_B be the Reactions at pt A and B \n",

+      "#R_A+R_B=90 \n",

+      "#Now Taking moment at A,M_A we get\n",

+      "R_A=(w1*L_DB*(L_DB*2**-1)+F_D*L_DB+F_C*(L_CD+L_DB))*L**-1\n",

+      "R_B=90-R_A\n",

+      "\n",

+      "#Shear Force Calculations\n",

+      "\n",

+      "#S.F At Pt B\n",

+      "V_B1=0 #KN\n",

+      "V_B2=R_B #KN\n",

+      "\n",

+      "#S.F At pt D\n",

+      "V_D1=R_B-w1*L_DB #KN\n",

+      "V_D2=V_D1-F_D #KN\n",

+      "\n",

+      "#S.F at Pt C\n",

+      "V_C1=V_D2 #KN\n",

+      "V_C2=V_C1-F_C  \n",

+      "\n",

+      "#S.F at PT A\n",

+      "V_A1=V_C2 #KN\n",

+      "V_A2=V_C2+R_A #KN\n",

+      "\n",

+      "#Bending Moment Calculations\n",

+      "\n",

+      "#B.M At Pt B\n",

+      "M_B=0 #KN.m\n",

+      "\n",

+      "#B.M At Pt D\n",

+      "M_D=-R_B*L_DB+w1*L_DB*L_DB*2**-1 #KN.m\n",

+      "\n",

+      "#B.M At PT C\n",

+      "M_C=-R_B*(L_DB+L_CD)+w1*L_DB*(L_DB*2**-1+L_CD)+F_D*L_CD #KN.m\n",

+      "\n",

+      "#B.M At Pt A\n",

+      "M_A=-R_B*L+w1*L_DB*(L_DB*2**-1+L_CD+L_AC)+F_D*(L_CD+L_AC)+F_C*L_AC\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,0,L_DB,L_DB,L_CD+L_DB,L_CD+L_DB,L_CD+L_DB+L_AC,L_CD+L_DB+L_AC]\n",

+      "Y1=[V_B1,V_B2,V_D1,V_D2,V_C1,V_C2,V_A1,V_A2]\n",

+      "Z1=[0,0,0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in m\")\n",

+      "plt.ylabel(\"Shear Force in kN\")\n",

+      "plt.show()\n",

+      "\n",

+      "#Plotting the Bendimg Moment Diagram\n",

+      "\n",

+      "Y2=[M_B,M_D,M_C,M_A]\n",

+      "X2=[0,L_DB,L_DB+L_CD,L_AC+L_CD+L_DB]\n",

+      "Z2=[0,0,0,0]\n",

+      "plt.plot(X2,Y2,X2,Z2)\n",

+      "plt.xlabel(\"Length in m\")\n",

+      "plt.ylabel(\"Bending Moment in kN.m\")\n",

+      "plt.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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Pni0aGhr0vo4HrxERkcSmlo+IiMi8WApERCRhKRARkYSlQEREEpYCERFJWApE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\nICIiCUuBiIgk/w9P4ODmR+1IBgAAAABJRU5ErkJggg==\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5be93d0>"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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Gt27dTE9mRiwY9ovtQ+zHlSuAv780HCkgQO00tkGRgnHr1i2kp6ejpKQEdXV1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\nx6hRo1BfX48XX3wRgYGBasd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+       "text": [

+        "<matplotlib.figure.Figure at 0x5de54f0>"

+       ]

+      }

+     ],

+     "prompt_number": 16

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 3.3.3,Page No.102"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L_DB=L_CD=1.5 #m #Length of DB & CD\n",

+      "L_AC=3 #m #Length of AC\n",

+      "F_D=80 #KN #Force at Pt D\n",

+      "w=40 #KN/m #u.v.l\n",

+      "L=6 #Length of beam\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let R_A and R_B be the Reactions at Pt A & B respectively\n",

+      "#R_A+R_B=140 \n",

+      "#Taking moment at B we get,M_B\n",

+      "R_A=(1*2**-1*L_AC*w*(1*3**-1*L_AC+(L_CD+L_DB))+F_D*L_DB)*L**-1\n",

+      "R_B=140-R_A\n",

+      "\n",

+      "#Shear Force Calculations\n",

+      "\n",

+      "#S.F at B\n",

+      "V_B1=0 #KN\n",

+      "V_B2=R_B #KN\n",

+      "\n",

+      "#S.F At D\n",

+      "V_D1=V_B2 #KN\n",

+      "V_D2=V_D1-F_D #KN\n",

+      "\n",

+      "#S.F at C\n",

+      "V_C=V_D2 #KN\n",

+      "\n",

+      "#S.F At A\n",

+      "V_A1=V_C-1*2**-1*w*L_AC #KN\n",

+      "V_A2=V_A1+R_A #KN\n",

+      "\n",

+      "#Bending Moment Calculations\n",

+      "\n",

+      "#B.M At B\n",

+      "M_B=0 #KN.m\n",

+      "\n",

+      "#B.M At D\n",

+      "M_D=-R_B*L_DB\n",

+      "\n",

+      "#B.M At C\n",

+      "M_C=F_D*L_CD-R_B*(L_DB+L_CD)\n",

+      "\n",

+      "#B.M At A\n",

+      "M_A=F_D*(L_CD+L_AC)-R_B*L+1*2**-1*w*L_AC*(1*3**-1*L_AC)+R_A\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,0,L_DB,L_DB,L_DB+L_CD,L_DB+L_CD+L_AC,L_DB+L_CD+L_AC]\n",

+      "Y1=[V_B1,V_B2,V_D1,V_D2,V_C,V_A1,V_A2]\n",

+      "Z1=[0,0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in m\")\n",

+      "plt.ylabel(\"Shear Force in kN\")\n",

+      "plt.show()\n",

+      "\n",

+      "#Plotting the Bendimg Moment Diagram\n",

+      "\n",

+      "X2=[0,L_DB,L_CD+L_DB,L_AC+L_CD+L_DB]\n",

+      "Y2=[M_B,M_D,M_C,M_A]\n",

+      "Z2=[0,0,0,0]\n",

+      "plt.plot(X2,Y2,X2,Z2)\n",

+      "plt.xlabel(\"Length in m\")\n",

+      "plt.ylabel(\"Bending Moment in kN.m\")\n",

+      "plt.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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UiIhIxlAgIiIZQ4GIiGQMBSIikjEUyKlYe9qMDRs24OrVqxbf3wcffOB0U8GT\nY+J1CuRUBg8eLF+Zag2+vr748ssvceedd9pkf0S2xiMFcnpnz57F/fffjwkTJuC+++7D6dOnAQCP\nPvoonnrqKUybNg1+fn7IysoC0D7baWpqKoKCghAVFYV58+YhKysLmzZtQk1NDSIiIjpdrfziiy8i\nJCQEU6ZMwXfffddl/ytWrMDq1asBAJ988glmzJjRZZ0dO3bgySef7LGuG1VUVCAwMBDJyckYPXo0\nFi1ahAMHDmDatGkICAjAiRMn+v4XR67JFjd3ILKVQYMGdVk2c+ZMUVZWJoQQorCwUMycOVMIIURS\nUpKIj48XQghRUlIi/P39hRBCvPfee2Lu3LlCCCHq6urEHXfcIbKysoQQXW/EIkmS2L9/vxBCiJUr\nV4pXXnmly/6bm5tFcHCwyM/PF6NHjxbnzp3rss6OHTvE8uXLe6zrRuXl5cLd3V18/fXXoq2tTYSF\nhYklS5YIIYTIzs4WsbGxZv+uiLrjUHMfEd2qpqYmfPHFF1i4cKG8rKWlBUD7/D0dM7UGBQWhvr4e\nAFBQUID4+HgAkO+NYEr//v0xb948AEBYWBgOHjzYZZ1f/OIX2Lp1K6ZPn46NGzfC19e3x5pN1XUz\nX19feVKz4OBgzJo1CwAwZswYVFRU9LgPIlMYCuTU2traMHToUBQXF3f7fv/+/eXn4n/tNUmSOt0z\nQPTQdlOpVPJzNzc3tLa2drveyZMnMXz48F7fFKq7um42YMCATvvu2KanOojMYU+BnNqQIUPg6+uL\nffv2AWj/gj158mSP20ybNg1ZWVkQQqC+vh6HDh2S3xs8eDAuXbp0SzWcP38er732mnwDl+7m6e8p\neIhsiaFATqW5uRk+Pj7yY8OGDXjnnXewbds2hISEYMyYMcjJyZHXv3EK6I7nCxYsgFqthlarxSOP\nPILx48fL97NdtmwZ5syZIzeab97+5imlhRBISUnB+vXr4e3tjW3btiElJUUewjK1rannN29j6rWz\nTW1NtsNTUom6ceXKFXh4eODixYsIDw/H559/jhEjRihdFpHVsadA1I0HHngAjY2NaGlpwUsvvcRA\nIJfBIwUiIpKxp0BERDKGAhERyRgKREQkYygQEZGMoUBERDKGAhERyf4fYbq4IcfF0QUAAAAASUVO\nRK5CYII=\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x58f0c10>"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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CBUEIDRWEv/9d7kpIbVz53SnLDGPjxo0YOHAg9u3bh/T0dKSlpQEAjEYjsrKy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\ngg==\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x562f910>"

+       ]

+      }

+     ],

+     "prompt_number": 14

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 3.3.4,Page No.104"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "M_D=120 #KN.m #B.M at Pt D\n",

+      "F_C=40 #KN #Force at Pt C\n",

+      "w1=20 #KN.m\n",

+      "L_DB=1.5 #m #Length of DB\n",

+      "L_CD=1.5 #m #Length of CD\n",

+      "L_AC=3 #m #Length of AC\n",

+      "L=6 #m #Length of Beam\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let R_A And R_B be the Reactions at pt A and B \n",

+      "#R_A+R_B=100\n",

+      "#Now Taking Moment At Pt B We get,M_B\n",

+      "R_A=-(M_D-F_C*(L_CD+L_DB)-w1*L_AC*(L_AC*2**-1+L_CD+L_DB))*L**-1\n",

+      "R_B=100-R_A\n",

+      "\n",

+      "#Shear Force Calculations\n",

+      "\n",

+      "#S.F At Pt B\n",

+      "V_B1=0\n",

+      "V_B2=R_B\n",

+      "\n",

+      "#S.F at Pt D\n",

+      "V_D=V_B2 #KN\n",

+      "\n",

+      "#S.F At Pt C\n",

+      "V_C1=V_D #KN\n",

+      "V_C2=V_C1-F_C\n",

+      "\n",

+      "#S.F At Pt A\n",

+      "V_A1=V_C2-w1*L_AC #KN\n",

+      "V_A2=V_A1+R_A\n",

+      "\n",

+      "#Bending Moment Calculations\n",

+      "\n",

+      "#B.M At Pt B\n",

+      "M_B=0 #KN.m\n",

+      "\n",

+      "#B.M At Pt D\n",

+      "M_D1=M_B-R_B*L_DB #KN.m\n",

+      "M_D2=M_B+M_D-R_B*L_DB\n",

+      "\n",

+      "#B.M At Pt C\n",

+      "M_C=M_D-R_B*(L_CD+L_DB)\n",

+      "\n",

+      "#B.M At Pt A\n",

+      "M_A=M_D-R_B*L+F_C*L_AC+w1*L_AC*L_AC*2**-1\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,0,L_DB,L_DB+L_CD,L_DB+L_CD,L_DB+L_CD+L_AC,L_DB+L_CD+L_AC]\n",

+      "Y1=[V_B1,V_B2,V_D,V_C1,V_C2,V_A1,V_A2]\n",

+      "Z1=[0,0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in m\")\n",

+      "plt.ylabel(\"Shear Force in kN\")\n",

+      "plt.show()\n",

+      "\n",

+      "#Plotting the Bendimg Moment Diagram\n",

+      "\n",

+      "Y2=[M_B,M_D1,M_D2,M_C,M_A]\n",

+      "X2=[0,L_DB,L_DB,L_CD+L_DB,L_AC+L_CD+L_DB]\n",

+      "Z2=[0,0,0,0,0]\n",

+      "plt.plot(X2,Y2,X2,Z2)\n",

+      "plt.xlabel(\"Length in m\")\n",

+      "plt.ylabel(\"Bending Moment in kN.m\")\n",

+      "plt.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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+       "text": [

+        "<matplotlib.figure.Figure at 0x55dcbf0>"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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sNP3OaDQiICAAQUFByMrKssvrp6QAFy4An39ul6cnBdBoxDWr1auBoiK505Ac1HDIzh7M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\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x55ca4b0>"

+       ]

+      }

+     ],

+     "prompt_number": 1

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 3.3.5,Page No.105"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "F_C=20 #KN #Force at Pt C\n",

+      "F_D=40 #KN #Force at pt D\n",

+      "w=20 #KN.m #u.d.l \n",

+      "L_AD=L_DB=2 #m #Length of AD & DB\n",

+      "L_BC=1 #m #Length of BC\n",

+      "L=5 #m #Length of Beam\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#LEt R_A and R_B be the reactions at A & B respectively\n",

+      "#R_A+R_B=100 \n",

+      "#Now Taking Moment at B,M_B we get\n",

+      "R_A=-(F_C*L_BC-F_D*L_DB-w*L_AD*(L_AD*2**-1+L_DB))*(L_AD+L_DB)**-1\n",

+      "R_B=100-R_A\n",

+      "\n",

+      "#Shear Force Calculations\n",

+      "\n",

+      "#S.F At pt C\n",

+      "V_C1=0 #KN\n",

+      "V_C2=-F_C #KN\n",

+      "\n",

+      "#S.F At PT B\n",

+      "V_B1=V_C2 #KN\n",

+      "V_B2=V_C2+R_B #KN\n",

+      "\n",

+      "#S.F At Pt D\n",

+      "V_D1=V_B2 #KN\n",

+      "V_D2=V_D1-F_D #KN\n",

+      "\n",

+      "#S.F At Pt A\n",

+      "V_A1=V_D2-w*L_AD #KN\n",

+      "V_A2=V_A1+R_A #KN\n",

+      "\n",

+      "#Bending Moment Calculations\n",

+      "\n",

+      "#B.M At Pt C\n",

+      "M_C=0 \n",

+      "\n",

+      "#B.M At Pt B\n",

+      "M_B=F_C*L_BC\n",

+      "\n",

+      "#B.M At Pt D\n",

+      "M_D=F_C*(L_BC+L_DB)-R_B*L_DB\n",

+      "\n",

+      "#B.M At Pt A\n",

+      "M_A=F_C*L-R_B*(L_DB+L_AD)+F_D*L_AD+w*L_AD*L_AD*2**-1\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,0,L_BC,L_BC,L_BC+L_DB,L_BC+L_DB,L_BC+L_DB+L_AD,L_BC+L_DB+L_AD]\n",

+      "Y1=[V_C1,V_C2,V_B1,V_B2,V_D1,V_D2,V_A1,V_A2]\n",

+      "Z1=[0,0,0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in m\")\n",

+      "plt.ylabel(\"Shear Force in kN\")\n",

+      "plt.show()\n",

+      "\n",

+      "#Plotting the Bendimg Moment Diagram\n",

+      "\n",

+      "Y2=[M_C,M_B,M_D,M_A]\n",

+      "X2=[0,L_BC,L_BC+L_DB,L_BC+L_DB+L_AD]\n",

+      "Z2=[0,0,0,0]\n",

+      "plt.plot(X2,Y2,X2,Z2)\n",

+      "plt.xlabel(\"Lenght in m\")\n",

+      "plt.ylabel(\"Bending Moment in kN.m\")\n",

+      "plt.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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DoVBVVUVFRQWHDx82t8/9uRxhYWGEhobWeT4zM5OkpCR8fX0JDAwkODiY3NzcyzqWiFxY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\niJj+H1U+eMmY8O6MAAAAAElFTkSuQmCC\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5d9c6f0>"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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hwfgdp6SYHY1vc8uS1I0bNxIfH0/nzp2JiYkhJiam0Qli165dzq8zMzOJi4sD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\nRESc/j+iaB8fTwtkoQAAAABJRU5ErkJggg==\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5c1ead0>"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 3.3.6,Page No.107"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L_BC=L_EB=L_AD=1 #m #Length of spans BC,ED,AD\n",

+      "L_ED=2 #m #Length of ED\n",

+      "w=60 #KNm #u.d.l\n",

+      "F_C=20 #KN Pt Load at C\n",

+      "L=5 #m #Span of beam \n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let R_A & R_B be the reactions at A & B respectively\n",

+      "#R_A+R_B=80 \n",

+      "#Taking Moment At A,we get M_A\n",

+      "R_B=(F_C*L+1*2**-1*L_ED*w*(2*3**-1*L_ED+L_AD))*(L_AD+L_ED+L_EB)**-1\n",

+      "R_A=80-R_B\n",

+      "\n",

+      "#Shear Force Calculations\n",

+      "\n",

+      "#S.F At C\n",

+      "V_C1=0 #KN\n",

+      "V_C2=-F_C #KN\n",

+      "\n",

+      "#S.F At B\n",

+      "V_B1=V_C2 #KN\n",

+      "V_B2=V_C2+R_B #KN \n",

+      "\n",

+      "#S.F aT E\n",

+      "V_E=V_B2 #KN\n",

+      "\n",

+      "#S.F AT D\n",

+      "V_D=V_B2-1*2**-1*L_ED*w #KN\n",

+      "\n",

+      "#S.F At A\n",

+      "V_A1=V_D #KN \n",

+      "V_A2=V_D+R_A\n",

+      "\n",

+      "#Bending Moment Calculations\n",

+      "\n",

+      "#B.M at C\n",

+      "M_C=0 #KN.m\n",

+      "\n",

+      "#B.M at B\n",

+      "M_B=F_C*L_BC #KN.m\n",

+      "\n",

+      "#B.M at E\n",

+      "M_E=F_C*(L_EB+L_BC)-R_B*L_EB #KN.m\n",

+      "\n",

+      "#B.M at D\n",

+      "M_D=F_C*(L_ED+L_EB+L_BC)-R_B*(L_ED+L_EB)+1*2**-1*L_ED*w*1*3**-1*L_ED #KN.m\n",

+      "\n",

+      "#B.M  at A\n",

+      "M_A=1*2**-1*L_ED*w*(2*3**-1*L_ED+L_AD)-R_B*(L_AD+L_ED+L_EB)+F_C*L\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,0,L_BC,L_BC,L_EB+L_BC,L_ED+L_EB+L_BC,L_AD+L_ED+L_EB+L_BC,L_ED+L_EB+L_BC+L_AD]\n",

+      "Y1=[V_C1,V_C2,V_B1,V_B2,V_E,V_D,V_A1,V_A2]\n",

+      "Z1=[0,0,0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in m\")\n",

+      "plt.ylabel(\"Shear Force in kN\")\n",

+      "plt.show()\n",

+      "\n",

+      "#Plotting the Bendimg Moment Diagram\n",

+      "\n",

+      "X2=[0,L_BC,L_BC+L_EB,L_EB+L_BC+L_ED,L_EB+L_BC+L_ED+L_AD]\n",

+      "Y2=[M_C,M_B,M_E,M_D,M_A]\n",

+      "Z2=[0,0,0,0,0]\n",

+      "plt.plot(X2,Y2,X2,Z2)\n",

+      "plt.xlabel(\"Lenght in m\")\n",

+      "plt.ylabel(\"Bending Moment in kN.m\")\n",

+      "plt.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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Cr9fzLAHAgQMHUFpaig8++AAbNmzAvn37Ot1OUUUhJCSkwx1Vq6qqEBoaKjAR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\nAABJRU5ErkJggg==\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x58a6f70>"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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UlD0DZoHZoDk7OxvPPfec/SqwAoNm98HQmcjAloBZINmcwq5du5rcTwEAHnnkEeuqkgCb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\nkeDZfAAAAABJRU5ErkJggg==\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x567bcf0>"

+       ]

+      }

+     ],

+     "prompt_number": 28

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 3.3.7,Page No.109"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L_BC=1 #m #Length of BC\n",

+      "L_DB=2 #m #Length of DB\n",

+      "L_AD=4 #m #Length 0f AD\n",

+      "M_D=30 #KN.m #Moment at D\n",

+      "w=45 #KN/m #u.d.l\n",

+      "L=7 #m #Span of beam\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let R_B & R_A be the Reactions at B & A respectively\n",

+      "#R_B+R_A=180+P  ............(1)\n",

+      "\n",

+      "#Now Taking Moment about A,we get\n",

+      "#R_B=7*P+390   ...............(2)\n",

+      "\n",

+      "#Since R_A & R_B Are Equal\n",

+      "#2*R_B=180+P  ...................(3)\n",

+      "\n",

+      "#From equation 1 and 3 we get\n",

+      "#3*(180+P)=7P+390\n",

+      "#After simplifying Further above equation we get\n",

+      "P=150*4**-1 #KN\n",

+      "R_A=R_B=(180+P)*2**-1\n",

+      "F_C=P\n",

+      "\n",

+      "#Shear Force Calculations\n",

+      "\n",

+      "#S.F At C\n",

+      "V_C1=0 #KN\n",

+      "V_C2=-P #KN\n",

+      "\n",

+      "#S.F At B\n",

+      "V_B1=V_C2 #KN\n",

+      "V_B2=V_C2+R_B #KN \n",

+      "\n",

+      "#S.F At D\n",

+      "V_D=V_B2 #KN\n",

+      "\n",

+      "#S.F At A\n",

+      "V_A1=V_D-w*L_AD #KN\n",

+      "V_A2=V_A1+R_A #KN\n",

+      "\n",

+      "#Bending Moment Calculations\n",

+      "\n",

+      "#B.M at C\n",

+      "M_C=0 #KN.m \n",

+      "\n",

+      "#B.M at B\n",

+      "M_B=F_C*L_BC #KN.m\n",

+      "\n",

+      "#B.M at D\n",

+      "M_D1=F_C*(L_BC+L_DB)-R_B*L_DB #KN.m\n",

+      "M_D2=M_D1+M_D\n",

+      "\n",

+      "#B.M At A\n",

+      "M_A=w*L_AD*L_AD*2**-1+M_D-R_B*(L_AD+L_DB)+P*L\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,0,L_BC,L_BC,L_DB+L_BC,L_DB+L_BC+L_AD,L_DB+L_BC+L_AD]\n",

+      "Y1=[V_C1,V_C2,V_B1,V_B2,V_D,V_A1,V_A2]\n",

+      "Z1=[0,0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in m\")\n",

+      "plt.ylabel(\"Shear Force in kN\")\n",

+      "plt.show()\n",

+      "\n",

+      "#Plotting the Bendimg Moment Diagram\n",

+      "\n",

+      "X2=[0,L_BC,L_DB+L_BC,L_DB+L_BC,L_AD+L_DB+L_BC]\n",

+      "Y2=[M_C,M_B,M_D1,M_D2,M_A]\n",

+      "Z2=[0,0,0,0,0]\n",

+      "plt.plot(X2,Y2,X2,Z2)\n",

+      "plt.xlabel(\"Lenght in m\")\n",

+      "plt.ylabel(\"Bending Moment in kN.m\")\n",

+      "plt.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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+       "text": [

+        "<matplotlib.figure.Figure at 0x5d6f150>"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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+       "text": [

+        "<matplotlib.figure.Figure at 0x5d79470>"

+       ]

+      }

+     ],

+     "prompt_number": 12

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 3.3.8,Page No.110"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=6 #m #Span Of beam\n",

+      "w=30 #KN/m #u.d.l\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Due to Symmetry\n",

+      "#Let R_B and R_C be the reactions at B & C Respectively\n",

+      "R_B=R_C=w*L*2**-1 #KN\n",

+      "\n",

+      "#Let a be the overhang.The Max -ve moment occurs at the support and max +ve moment at middle of the beam\n",

+      "#Now Equating these two equations we get\n",

+      "#30*a*a*2**-1=90*(3-a)-w*L*2**-1*L*4**-1\n",

+      "#After simplifying we get an equation as\n",

+      "#a**2+6*a-9=0\n",

+      "x=1\n",

+      "y=6\n",

+      "z=-9\n",

+      "\n",

+      "p=y**2-4*x*z\n",

+      "\n",

+      "a1=(-y+p**0.5)*2**-1\n",

+      "a2=(-y-p**0.5)*2**-1\n",

+      "\n",

+      "#Now Length cannot be negative,so taking a1 into Consideration\n",

+      "\n",

+      "L_CD=L_AB=a1\n",

+      "L_BC=L-2*a1\n",

+      "\n",

+      "#Shear Force Calculations\n",

+      "\n",

+      "#S.F At D\n",

+      "V_D=0\n",

+      "\n",

+      "#S.F At C\n",

+      "V_C1=V_D-w*L_CD #KN\n",

+      "V_C2=V_C1+R_C #KN\n",

+      "\n",

+      "#S.F At B\n",

+      "V_B1=-w*(L_BC+L_CD)+R_C\n",

+      "V_B2=V_B1+R_B\n",

+      "\n",

+      "#S.F At A\n",

+      "V_A=round(V_B2,2)-round(w*L_AB,2)\n",

+      "\n",

+      "#Bending Moment Calculations\n",

+      "\n",

+      "#B.M At D\n",

+      "M_D=0\n",

+      "\n",

+      "#B.M At C\n",

+      "M_C=w*L_CD*L_CD*2**-1 #KN.m\n",

+      "\n",

+      "#B.M At B\n",

+      "M_B=w*(L_BC+L_CD)*(L_BC+L_CD)*2**-1-R_C*L_BC*L_BC*2**-1\n",

+      "\n",

+      "#B.M At A\n",

+      "X=w*L*L*2**-1\n",

+      "Y=-R_C*(L_AB+L_BC)-R_B*L_AB\n",

+      "M_A=X+Y\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,L_CD,L_CD,L_CD+L_BC,L_CD+L_BC,L_CD+L_BC+L_AB]\n",

+      "Y1=[V_D,V_C1,V_C2,V_B1,V_B2,V_A]\n",

+      "Z1=[0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in m\")\n",

+      "plt.ylabel(\"Shear Force in kN\")\n",

+      "plt.show()\n",

+      "\n",

+      "#Plotting the Bendimg Moment Diagram\n",

+      "\n",

+      "X2=[0,L_CD,L_BC+L_CD,L_AB+L_BC+L_CD]\n",

+      "Y2=[M_D,M_C,M_B,M_A]\n",

+      "Z2=[0,0,0,0]\n",

+      "plt.plot(X2,Y2,X2,Z2)\n",

+      "plt.xlabel(\"Lenght in m\")\n",

+      "plt.ylabel(\"Bending Moment in kN.m\")\n",

+      "plt.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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lXLhwAQBw6dIlHDhwwKleBejt7Q0/Pz+UlpYCAA4ePIiwsLAub6vKm9f6yt3d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\nAElFTkSuQmCC\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5945590>"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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LUVxcrNtAqS48PDxqdH5WVhby8vIQEBAAAJg4cSJ27NjBhEGkEDc3ID5e7BWe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\nICIiWf4/U4J8H+TjvqoAAAAASUVORK5CYII=\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x58e1750>"

+       ]

+      }

+     ],

+     "prompt_number": 25

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 3.3.9,Page No.112"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "F_F=6 #KN #Force at  F\n",

+      "w1=w2=w=3 #KN.m #u.d.l\n",

+      "M_D=24 #KN.m \n",

+      "L_AB=L_CD=L_DE=L_EF=4 #m #Length of AB,CD,DE,EF\n",

+      "L_BC=2 #m #Length of BC\n",

+      "L=18 #m #Span of Beam\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#LEt R_B and R_E be the Reactions at B & E respectively\n",

+      "#R_B+R_E=42\n",

+      "\n",

+      "#Taking Moment At Pt B,M_B\n",

+      "R_E=(F_F*(L_BC+L_CD+L_DE+L_EF)+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC)-w*L_AB*L_AB*2**-1-M_D)*(L_BC+L_CD+L_DE)**-1\n",

+      "R_B=42-R_E #KN\n",

+      "\n",

+      "#Shear Force Calculations\n",

+      "\n",

+      "#S.F aT F\n",

+      "V_F1=0 #KN \n",

+      "V_F2=-F_F #KN\n",

+      "\n",

+      "#S.F at E\n",

+      "V_E1=V_F2 #KN\n",

+      "V_E2=V_E1+R_E #KN\n",

+      "\n",

+      "#S.F aT C\n",

+      "V_C=V_E2-w*(L_CD+L_DE) #KN\n",

+      "\n",

+      "#S.F at B\n",

+      "V_B1=V_C #KN \n",

+      "V_B2=V_C+R_B #KN\n",

+      "\n",

+      "#S.F At A\n",

+      "V_A=V_B2-w*L_AB #KN\n",

+      "\n",

+      "#Bending Moment Calculations\n",

+      "\n",

+      "#B.M At F\n",

+      "M_F=0\n",

+      "\n",

+      "#B.M At E\n",

+      "M_E=F_F*L_EF #KN.m\n",

+      "\n",

+      "#B.M At D\n",

+      "M_D1=F_F*(L_DE+L_EF)-R_E*L_DE+w*L_DE*L_DE*2**-1 #KN.m\n",

+      "M_D2=M_D1-M_D\n",

+      "\n",

+      "#B.M At C\n",

+      "M_C=F_F*(L_CD+L_DE+L_EF)-R_E*(L_CD+L_DE)+w*(L_CD+L_DE)*(L_CD+L_DE)*2**-1-M_D\n",

+      "\n",

+      "#B.M At B\n",

+      "M_B=F_F*(L_BC+L_CD+L_DE+L_EF)-R_E*(L_BC+L_CD+L_DE)-M_D+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC)\n",

+      "\n",

+      "#B.M At A\n",

+      "M_A=w*L_AB*L_AB*2**-1-R_B*L_AB+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC+L_AB)-R_E*(L_AB+L_BC+L_CD+L_DE)+F_F*L-M_D\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,0,L_EF,L_EF,L_EF+L_DE+L_CD,L_EF+L_DE+L_CD+L_BC,L_EF+L_DE+L_CD+L_BC,L_EF+L_DE+L_CD+L_BC+L_AB]\n",

+      "Y1=[V_F1,V_F2,V_E1,V_E2,V_C,V_B1,V_B2,V_A]\n",

+      "Z1=[0,0,0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in m\")\n",

+      "plt.ylabel(\"Shear Force in kN\")\n",

+      "plt.show()\n",

+      "\n",

+      "#Plotting the Bendimg Moment Diagram\n",

+      "\n",

+      "X2=[0,L_EF,L_DE+L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC+L_AB]\n",

+      "Y2=[M_F,M_E,M_D1,M_D2,M_C,M_B,M_A]\n",

+      "Z2=[0,0,0,0,0,0,0]\n",

+      "plt.plot(X2,Y2,X2,Z2)\n",

+      "plt.xlabel(\"Lenght in m\")\n",

+      "plt.ylabel(\"Bending Moment in kN.m\")\n",

+      "plt.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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+       "text": [

+        "<matplotlib.figure.Figure at 0x5d75ab0>"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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VxMfHIzs7G/369cOvv/6KcePG4bHHHnNknEROixVL5KzMziBqa2uRmJiIMWPG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\nfyRzHAAAAABJRU5ErkJggg==\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5c44130>"

+       ]

+      }

+     ],

+     "prompt_number": 17

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 3.3.10,Page No.114"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L_DC=L_BA=2 #m #Length of BA & DC\n",

+      "L_CB=1 #m #Length of CB\n",

+      "F_A=10 #KN #Force at pt A\n",

+      "F_B=20 #KN #Force at pt B\n",

+      "w=4 #KN.m #u.d.l\n",

+      "L=5 #m #Length of beam\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let R_D be the reactions at Pt D\n",

+      "R_D=F_B+F_A+w*L_DC #KN\n",

+      "\n",

+      "#Shear Force Calculations\n",

+      "\n",

+      "#S.F at Pt A\n",

+      "V_A1=0 #KN\n",

+      "V_A2=F_A #KN\n",

+      "\n",

+      "#S.F At Pt B\n",

+      "V_B1=V_A2\n",

+      "V_B2=F_B+F_A\n",

+      "\n",

+      "#S.F at Pt C\n",

+      "V_C=F_B+F_A #KN \n",

+      "\n",

+      "#S.F At Pt D\n",

+      "V_D1=V_B2+w*L_DC\n",

+      "V_D2=F_B+F_A+w*L_DC-R_D\n",

+      "\n",

+      "#B.M At Pt A\n",

+      "M_A=0\n",

+      "\n",

+      "#B.M At Pt B\n",

+      "M_B=F_A*L_BA\n",

+      "\n",

+      "#B.M at Pt C\n",

+      "M_C=F_B*L_CB+F_A*(L_BA+L_CB) #KN\n",

+      "\n",

+      "#B.M At Pt D\n",

+      "M_D1=F_A*L+F_B*(L_CB+L_DC)+w*L_DC*L_DC*2**-1\n",

+      "M_D2=(F_A*L+F_B*(L_CB+L_DC)+w*L_DC*L_DC*2**-1)-M_D1\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,0,L_BA,L_BA,L_BA+L_CB,L_BA+L_CB+L_DC,L_BA+L_CB+L_DC]\n",

+      "Y1=[V_A1,V_A2,V_B1,V_B2,V_C,V_D1,V_D2]\n",

+      "Z1=[0,0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in m\")\n",

+      "plt.ylabel(\"Shear Force in kN\")\n",

+      "plt.show()\n",

+      "\n",

+      "#Plotting the Bendimg Moment Diagram\n",

+      "\n",

+      "Y2=[M_A,M_B,M_C,M_D1,M_D2]\n",

+      "X2=[0,L_BA,L_CB+L_BA,L_CB+L_BA+L_DC,L_CB+L_BA+L_DC]\n",

+      "Z2=[0,0,0,0,0]\n",

+      "plt.plot(X2,Y2,X2,Z2)\n",

+      "plt.xlabel(\"Lenght in m\")\n",

+      "plt.ylabel(\"Bending Moment in kN.m\")\n",

+      "plt.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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+       "text": [

+        "<matplotlib.figure.Figure at 0x5738d10>"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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4EoAKvyJSQ8nAz6nwKyKNoWTgZ1T4FZGmUDLwcSr8iogRlAx8kAq/ImI0JQMf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\nCwAAAABJRU5ErkJggg==\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5788db0>"

+       ]

+      }

+     ],

+     "prompt_number": 12

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 3.3.11,Page No.115"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "w=20 #KN/m #u.v.l\n",

+      "F_C=40 #KN #Force at Pt C\n",

+      "M_D=40 #KN.m #Moment at pt D\n",

+      "L_AB=3 #m #Length of AB\n",

+      "L_BC=1 #m #Length of BC\n",

+      "L_CD=L_DE=2 #m #Length of CD & DE\n",

+      "L=8 #8 #Length of beam\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let R_A & R_E be the Reactions at A & E respectively\n",

+      "#R_A+R_E=70\n",

+      "\n",

+      "#Taking Moments At Pt A we get,M_A\n",

+      "R_E=(F_C*(L_AB+L_BC)+1*2**-1*L_AB*w*2+40)*L**-1\n",

+      "R_A=70-R_E\n",

+      "\n",

+      "#shear Force Calculations\n",

+      "\n",

+      "#S.F At Pt E\n",

+      "V_E1=0\n",

+      "V_E2=R_E #KN\n",

+      "\n",

+      "#S.F aT pt D\n",

+      "V_D=V_E2\n",

+      "\n",

+      "#S.F At PT C\n",

+      "V_C1=V_D\n",

+      "V_C2=V_D-F_C #KN\n",

+      "\n",

+      "#S.F At Pt A\n",

+      "V_A1=V_C2-(1*2**-1*w*L_AB)\n",

+      "V_A2=V_A1+R_A\n",

+      "\n",

+      "#Bending Moment Calculations\n",

+      "\n",

+      "#B.M At Pt E\n",

+      "M_E=0\n",

+      "\n",

+      "#B.M At Pt D\n",

+      "M_D1=M_E-R_E*L_DE\n",

+      "M_D2=M_D1+M_D\n",

+      "\n",

+      "#B.M At Pt C\n",

+      "M_C=-R_E*(L_DE+L_CD)+M_D\n",

+      "\n",

+      "#B.M At Pt B\n",

+      "M_B=-R_E*(L_DE+L_CD+L_BC)+M_D+F_C*L_BC\n",

+      "\n",

+      "#B.M At Pt A\n",

+      "M_A=-R_E*L+M_D+(1*2**-1*L_AB*w*2)+F_C*(L_BC+L_AB)\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,0,L_DE,L_CD+L_DE,L_CD+L_DE,L_CD+L_DE+L_AB,L_CD+L_DE+L_AB]\n",

+      "Y1=[V_E1,V_E2,V_D,V_C1,V_C2,V_A1,V_A2]\n",

+      "Z1=[0,0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in m\")\n",

+      "plt.ylabel(\"Shear Force in kN\")\n",

+      "plt.show()\n",

+      "\n",

+      "#Plotting the Bendimg Moment Diagram\n",

+      "\n",

+      "Y2=[M_E,M_D1,M_D2,M_C,M_B,M_A]\n",

+      "X2=[0,L_DE,L_DE,L_CD+L_DE,L_DE+L_CD+L_BC,L_AB+L_BC+L_CD+L_DE]\n",

+      "Z2=[0,0,0,0,0,0]\n",

+      "plt.plot(X2,Y2)\n",

+      "plt.xlabel(\"Lenght in m\")\n",

+      "plt.ylabel(\"Bending Moment in kN.m\")\n",

+      "plt.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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KRQ4dOoS8vDz4+/sjISEBe/bswX333Sc6Vo/cdNNNAIDhw4fjrrvucqh1zVQq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\niEjGUiAiItn/A+2hg5gYC1MHAAAAAElFTkSuQmCC\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5df5230>"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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PZcuW8dRTT5GZmcnOnTtp1KgRI0aMsPs+cvTnxbp1U4fwLFyoO4nzTCaYMkWt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\nGhPHAAAAAElFTkSuQmCC\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5dff130>"

+       ]

+      }

+     ],

+     "prompt_number": 14

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 3.3.12,Page No.116"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "F_G=10 #KN #Force at Pt G\n",

+      "F_B=F_E=15 #KN #Force at Pt B & E\n",

+      "w=20 #KN/m #U.d.L\n",

+      "L_FG=L_EF=L_DE=L_CD=L_BC=L_AB=1 #m #Lengths of FG,EF,DE,CD,BC,AB respectively\n",

+      "L=6 #m #Length of beam\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#LEt R_F & R_A be the Reactions at E & A respectively\n",

+      "#R_F+R_A=60\n",

+      "\n",

+      "#Taking Moment At Pt A,M_A\n",

+      "R_F=(F_G*L+F_E*(L_AB+L_BC+L_CD+L_DE)+w*L_CD*(L_AB+L_BC+L_CD*2**-1)+F_B*L_AB)*(L_AB+L_BC+L_CD+L_DE+L_EF)**-1\n",

+      "R_A=60-R_F\n",

+      "\n",

+      "#Shear Force Calculations\n",

+      "\n",

+      "#S.F At G\n",

+      "V_G1=0 #KN \n",

+      "V_G2=F_G #KN\n",

+      "\n",

+      "#S.F At F\n",

+      "V_F1=V_G2 #KN\n",

+      "V_F2=V_F1-R_F\n",

+      "\n",

+      "#S.F At E\n",

+      "V_E1=V_F2 #KN\n",

+      "V_E2=V_F2+F_E\n",

+      "\n",

+      "#S.F At D\n",

+      "V_D=V_E2\n",

+      "\n",

+      "#S.F At C\n",

+      "V_C=V_E2+w*L_CD\n",

+      "\n",

+      "#S.F At B\n",

+      "V_B1=V_C\n",

+      "V_B2=V_B1+F_B\n",

+      "\n",

+      "#S.F At A\n",

+      "V_A1=V_B2\n",

+      "V_A2=V_B2-R_A\n",

+      "\n",

+      "#Bending Moment Calculations\n",

+      "\n",

+      "#B.M At Pt G\n",

+      "M_G=0\n",

+      "\n",

+      "#B.M At F\n",

+      "M_F=F_G*L_FG \n",

+      "\n",

+      "#B.M At E\n",

+      "M_E=F_G*(L_FG+L_EF)-R_F*L_EF\n",

+      "\n",

+      "#B.M At D\n",

+      "M_D=F_G*(L_FG+L_EF+L_DE)-R_F*(L_EF+L_DE)+F_E*L_DE\n",

+      "\n",

+      "#B.M At C\n",

+      "M_C=F_G*(L_FG+L_EF+L_DE+L_CD)-R_F*(L_EF+L_DE+L_CD)+F_E*(L_DE+L_CD)+w*L_CD*L_CD*2**-1\n",

+      "\n",

+      "#B.M At B\n",

+      "M_B=F_G*(L_FG+L_EF+L_DE+L_CD+L_BC)-R_F*(L_EF+L_DE+L_CD+L_BC)+F_E*(L_DE+L_CD+L_BC)+w*L_CD*(L_CD*2**-1+L_BC)\n",

+      "\n",

+      "#B.M At A\n",

+      "M_A=F_G*L-R_F*(L_EF+L_DE+L_CD+L_BC+L_AB)+F_E*(L_DE+L_CD+L_BC+L_AB)+F_B*L_AB+w*L_CD*(L_CD*2**-1+L_BC+L_AB)\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,0,L_FG,L_FG,L_FG+L_EF,L_FG+L_EF,L_FG+L_EF+L_DE,L_FG+L_EF+L_DE+L_CD,L_FG+L_EF+L_DE+L_CD+L_BC,L_FG+L_EF+L_DE+L_CD+L_BC,L_FG+L_EF+L_DE+L_CD+L_BC+L_AB,L_FG+L_EF+L_DE+L_CD+L_BC+L_AB]\n",

+      "Y1=[V_G1,V_G2,V_F1,V_F2,V_E1,V_E2,V_D,V_C,V_B1,V_B2,V_A1,V_A2]\n",

+      "Z1=[0,0,0,0,0,0,0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in m\")\n",

+      "plt.ylabel(\"Shear Force in kN\")\n",

+      "plt.show()\n",

+      "\n",

+      "#Plotting the Bendimg Moment Diagram\n",

+      "\n",

+      "X2=[0,L_FG,L_EF+L_FG,L_EF+L_FG+L_DE,L_EF+L_FG+L_DE+L_CD,L_EF+L_FG+L_DE+L_CD+L_BC,L_EF+L_FG+L_DE+L_CD+L_BC+L_AB]\n",

+      "Y2=[M_G,M_F,M_E,M_D,M_C,M_B,M_A]\n",

+      "Z2=[0,0,0,0,0,0,0]\n",

+      "plt.plot(X2,Y2)\n",

+      "plt.xlabel(\"Lenght in m\")\n",

+      "plt.ylabel(\"Bending Moment in kN.m\")\n",

+      "plt.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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+       "text": [

+        "<matplotlib.figure.Figure at 0x57502f0>"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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Xmnfs2FFmtKCkEJzUi0GscPq0cWr5xRehc2ero3GPW0khMTHxghf+9ttvqx1Y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\nkcIBAAAAAElFTkSuQmCC\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x58b0ed0>"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 3.3.13,Page No.117"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "L_AB=L_BC=L_CD=L_DE=L_EF=1 #m #LEngth of AB,BC,CD,DE,EF respectively\n",

+      "M_A=50 #KN/m #Moment at A\n",

+      "w=5 #KN/m #u.v.l\n",

+      "F_D=10 #KN\n",

+      "w2=5 #KN/m #u.d.l\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let R_B & R_E be the Reactions at B and E respectively\n",

+      "#R_B+R_E=20\n",

+      "\n",

+      "#Taking Moment At Pt B,M_B\n",

+      "R_E=(w2*L_EF*(L_EF*2**-1+L_DE+L_CD+L_BC)+w*L_BC*2**-1*2*3**-1+50+F_D*(L_BC+L_CD))*3**-1\n",

+      "R_B=17.5-R_E #KN\n",

+      "\n",

+      "#Shear Force Calculations\n",

+      "\n",

+      "#S.F At F\n",

+      "V_F=0\n",

+      "\n",

+      "#S.F aT E\n",

+      "V_E1=-w2*L_EF #KN\n",

+      "V_E2=V_E1+R_E\n",

+      "\n",

+      "#S.F at D\n",

+      "V_D1=R_E-w2*L_EF #KN\n",

+      "V_D2=V_D1-F_D #KN\n",

+      "\n",

+      "#S.F At C\n",

+      "V_C=V_D2\n",

+      "\n",

+      "#S.F aT B\n",

+      "V_B1=-L_BC*w*2**-1-F_D+R_E-w2*L_EF\n",

+      "V_B2=V_B1+R_B\n",

+      "\n",

+      "#Bending Moment Calculations\n",

+      "\n",

+      "#B.M at F\n",

+      "M_F=0 #KN.m\n",

+      "\n",

+      "#B.M At E\n",

+      "M_E=w2*L_EF*L_EF*2**-1 #KN.m\n",

+      "\n",

+      "#B.M at D\n",

+      "M_D=-R_E*L_DE+w2*L_EF*(L_EF*2**-1+L_DE) #KN.m\n",

+      "\n",

+      "#B.M At C\n",

+      "M_C=F_D*L_CD*R_E*(L_CD+L_DE)+w2*L_EF*(L_EF*2**-1+L_DE+L_CD) #KN.m\n",

+      "\n",

+      "#B.M At B\n",

+      "M_B=F_D*(L_CD+L_BC)-R_E*(L_BC+L_CD+L_DE)+w2*L_EF*(L_EF*2**-1+L_BC+L_CD+L_DE)+1*2**-1*L_BC*w*2*3**-1\n",

+      "\n",

+      "#B.M At A\n",

+      "M_A1=w*L_EF*(L_EF*2**-1+L_AB+L_BC+L_CD+L_DE)-R_E*(L_AB+L_BC+L_CD+L_DE)+F_D*(L_AB+L_BC+L_CD)+1*2**-1*L_BC*w*(2*3**-1*L_BC+L_AB)-R_B*L_AB\n",

+      "M_A2=M_A1+M_A\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,L_EF,L_EF,L_DE+L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC]\n",

+      "Y1=[V_F,V_E1,V_E2,V_D1,V_D2,V_C,V_B1,V_B2]\n",

+      "Z1=[0,0,0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in m\")\n",

+      "plt.ylabel(\"Shear Force in kN\")\n",

+      "plt.show()\n",

+      "\n",

+      "#Plotting the Bendimg Moment Diagram\n",

+      "\n",

+      "X2=[0,L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC+L_AB,L_CD+L_DE+L_EF+L_BC+L_AB]\n",

+      "Y2=[M_F,M_E,M_D,M_C,M_B,M_A1,M_A2]\n",

+      "Z2=[0,0,0,0,0,0,0]\n",

+      "plt.plot(X2,Y2,X2,Z2)\n",

+      "plt.xlabel(\"Length in m\")\n",

+      "plt.ylabel(\"Bending Moment in kN.m\")\n",

+      "plt.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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+       "text": [

+        "<matplotlib.figure.Figure at 0x5c2a7b0>"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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LlixREhMTLY8fMWKEsm/fvjuex4ZfSQi7euUVRXnxRa1TuIdr1xRFr1eUb77R\nOonnseWz0+oYxvfff0/zm/5s8vHxobi4mJYtW3LXXXc1sqapcnNzOXz4MAMGDKC4uBi9Xg+AXq+n\nuLgYUM/kMBqNlmuMRiNms9kury9EQ1VWwurVshWIvXh7qycUylYhzsnqGMb06dMZMGAAEydORFEU\ntm3bRmxsLJcuXSIkJKTRAcrKynj88cdZtmzZHbvh6nS6Oo+Dre2+RYsWWb6PiIiwnBYohL3t2KEO\ndIeGap3EfcyYAZMmqV1TXjItx2EyMjLIqOcOmTZNq92/fz979+5Fp9MxcOBA+vXr19CMt7h27Rpj\nx45l1KhRvHB9xDAoKIiMjAwCAgIoLCxk6NChfPPNN5ZjYRcuXAjAyJEjWbx4MQMGDLj1F5IxDNGE\npk6FiAj4+c+1TuI+FAXuvx/efhsGD9Y6jeewy/bmAJWVlRQVFVFRUWH5q75Lly6NCqcoCvHx8bRv\n356//OUvltvnz59P+/btWbBgAYmJiZSUlNwy6J2ZmWkZ9M7JybmjlSEFQzSVc+fU8y5yc+H6RD5h\nJ3/8I2Rnwz/+oXUSz2GXgvHWW2+xePFi/P39aXbTWZPHjh1rVLg9e/YwePBg7r//fsuHfkJCAv37\n9ycmJoYzZ87cMa12yZIlJCUl4e3tzbJlyxhRw2HJUjBEU3nrLfWQpH/+U+sk7sdsVld+m83g5Btk\nuw27FIxu3bqRmZlZ61GtzkYKhmgqffqofwlHRmqdxD1FRcHs2Wq3n3A8u6z07tKli2V7cyGE6vBh\ntUtq2DCtk7ivuDh4912tU4ibWW1hzJo1i+zsbMaMGWOZXivnYQhP9/zz0K6dOpNHOEZZmbo/17ff\nwvWZ9sKBGnUeRrUuXbrQpUsXysvLKS8vtxygJISn+uknWL8eMjO1TuLefH1h/HjYsAHmzdM6jQDZ\nrVaIetu0Cd55Bz75ROsk7m/HDli4EA4e1DqJ+2tUC2PevHksW7aMcePG1fjEaWlpjU8ohAtKSoKZ\nM7VO4RmGDYOiIvjvf6FXL63TiFpbGAcOHKBfv361rgR01tXT0sIQjpSfry4qy8+Hli21TuMZ5s9X\nV3xfX7srHMRuC/dciRQM4UhLlsCZM2qXlGgaX3+tbnuemws3LQUTdtaoLqnevXvX+cRHjx5teDIh\nXJCiQHKybIzX1EJDoWNHyMiA4cO1TuPZai0Y27ZtA2DFihUAzJgxA0VRWLduXdMkE8LJ7NmjnnfR\nv7/WSTwfqZjLAAASz0lEQVTPjBnqmgwpGNqy2iUVFhbGkSNHbrktPDycw4cPOzRYQ0mXlHCUmTPV\nv3b/7/+0TuJ5ioogOFgdO2rVSus07skuK70VRWHPnj2Wn/fu3SsfyMLjlJbC1q3w5JNaJ/FMAQHw\n8MPq/wZCO1YX7iUlJTFz5kx+/PFHANq2bUtycrLDgwnhTDZtgiFDZMWxluLi1DGk6dO1TuK5bJ4l\nVV0w2rRp49BAjSVdUsIRHn1Und45frzWSTzXlSvQqZO6JqNTJ63TuB+7TKu9evUqmzdvJjc3l4qK\nCssTv/rqq/ZLakdSMIS9ZWerB/nk5YGPj9ZpPNvTT6tjGb/4hdZJ3I9dxjAmTJhAWloaPj4++Pr6\n4uvrSysZdRIeJDlZnaUjxUJ71bOlhDastjBCQ0P5+uuvmypPo0kLQ9hTRQXcd5+6p5EdjrAXjVRV\nBV27QloaPPCA1mnci11aGI888ogs0hMea/t26NxZioWz8PJSZ6qtXat1Es9ktYURHBxMTk4OXbt2\npUWLFupFTrzSW1oYwp4mT4boaHj2Wa2TiGrffANDh6pjSt5W53kKW9ll0Ds3N7fG200mU0NzOZQU\nDGEvP/wAgYFw+jQ4+eRAj9O/P/z2tzBypNZJ3IdduqRMJhN5eXns2rULk8lEq1at7PaBPGvWLPR6\n/S37Vp0/f56oqCh69OhBdHQ0JSUllvsSEhLo3r07QUFBbN++3S4ZhKjNunUwbpwUC2ckx7dqw2rB\nWLRoEX/84x9JSEgAoLy8nCfttNx15syZpKen33JbYmIiUVFRZGdnM3z4cBKv72mclZVFSkoKWVlZ\npKenM2fOHKqqquySQ4jbKQqsWgWzZmmdRNTkiSfggw/UFfii6VgtGFu2bCE1NdUyldZgMFBqp/+V\nBg0aRLt27W65LS0tjfj4eADi4+PZen0vgNTUVKZNm4aPjw8mk4nAwEAy5YxM4SCHDqkfRkOGaJ1E\n1KRDB4iIgM2btU7iWawWjBYtWuDldeNhly5dcmig4uJi9Nf3X9Dr9RQXFwNQUFCA0Wi0PM5oNGI2\nmx2aRXiu5GR1s0Evq/8PEVqZMUNmSzU1q3MMpkyZwnPPPUdJSQl///vfSUpKYvbs2U2RDZ1Oh06n\nq/P+mixatMjyfUREhNOeDiic09WrsGGDnCPt7MaOheeeUw+06tJF6zSuJyMjo9YTVWtjtWD88pe/\nZPv27fj5+ZGdnc3vfvc7oqKiGprRKr1eT1FREQEBARQWFuLv7w+oXWF5eXmWx+Xn52MwGGp8jpsL\nhhD1tXUrhIerC/aE87rrLnXa87p18PLLWqdxPbf/Mb148WKr19jU4I6OjuaNN95gwYIFREZGNjig\nLcaPH8+aNWsAWLNmDRMnTrTcvmHDBsrLyzl16hQnTpygv5xkIxwgOVkGu11F9WwpmUnfNGotGPv2\n7SMiIoLHHnuMw4cPExoaSu/evdHr9Xz00Ud2efFp06bxyCOP8O2339K5c2eSk5NZuHAhO3bsoEeP\nHnz66acsXLgQgJCQEGJiYggJCWHUqFGsWLGizu4qIRrizBk4cACu/50inNwjj8BPP0n3YVOpdeFe\n3759SUhI4Mcff+SZZ54hPT2dhx56iG+++YYnnnjijlP4nIUs3BON8bvfQWEhXD+ZWLiARYvgwgVY\ntkzrJK6tUSu9bz6aNTg4mOPHj1vukyNahTuqqoLu3SElBfr10zqNsNXJk+ppfGaz7CjcGI1a6X1z\nd89dd91lv1RCOKnPP1fPi+7bV+skoj66dVML/ccfa53E/dXawmjWrBktW7YE4MqVK9x9992W+65c\nuWI5TMnZSAtDNFRcnDo76sUXtU4i6utvf4NPPoGNG7VO4rrssvmgq5GCIRri4kV1Lv+JE9Cxo9Zp\nRH1duAAmk7pRZNu2WqdxTXbZfFAIT5CSAsOHS7FwVe3aQVQUbNqkdRL3JgVDCCApSd0KRLguOb7V\n8aRLSni848fV1sWZM3IgjysrLweDATIz1WNcRf1Il5QQNkhOVge8pVi4tubNYepUeO89rZO4L2lh\nCI927Zo62J2RAT17ap1GNFZmJkyfDtnZIBtB1I+0MISwIj0dfvYzKRbu4sEH1S3pv/xS6yTuSQqG\n8GhJSbLRoDvR6eT4VkeSLinhsc6eVVsWZ86An5/WaYS95OaqW7uYzdCihdZpXId0SQlRh/fegwkT\npFi4G5MJQkPhww+1TuJ+pGAIj6Qo0h3lzuT4VseQLinhkfbvh2nT1K1AZDaN+/nxR3X223ffQfv2\nWqdxDdIlJUQtqld2S7FwT23awKhR6pYvwn6khSE8zpUrYDTCf/6j/le4pw8/VA/E2rdP6ySuQVoY\nQtRgyxZ1vr4UC/cWHa12SWVna53EfUjBEB5HBrs9g7c3xMbKViH25HIFIz09naCgILp3787SpUu1\njiNcTG4uHDmiTqcV7q96B9uqKq2TuAeXKhiVlZXMnTuX9PR0srKyWL9+/S1njQthzZo16uwoWdDl\nGcLD1WN39+7VOol7cKn9OTMzMwkMDMRkMgHwxBNPkJqaSnBwsLbBbKAoUFmpflVVOf77qir15DGD\nAe69Vz4gQX1PkpPVMQzhGXS6G2syBg3SOo3rc6mCYTab6dy5s+Vno9HIV199dcfjZs5smg/l+nwP\n6qZozZqpX47+XqdTj60sKICiImjdGjp1Ur8Mhpq/9/dXr3VXu3apRTQ8XOskoilNnw733w/Ll8Pd\nd2udxjm98optj3OpgqGzcdL86lM3Pc4EOMlhKlXXv65p8No/XP86evONxde/DmkQSCuTQLdY6xCi\nyc2Dln/UOoSTOQXk1u8SlyoYBoOBvLw8y895eXkYa5gbqWTIOoyGKC9XWyMFBeqX2Xzn92azeoZE\ndavk5lbKza2VTp2gZUutf6MbSkrUPYZOnpSVv55o7Vr1vO9t27RO4pwCA+Ek1v8gd6mFexUVFfTs\n2ZNPPvmETp060b9/f9avX3/LGIYs3HO8sjIoLKy9oFTfdvfddXeBGQyg14OPj+Mzv/MOfPKJ+qEh\nPE9ZmbruJjtb7XoVtwoMhJMnrX92ulQLw9vbm7/+9a+MGDGCyspKnn76aZcY8HY3vr7Qvbv6VRtF\ngfPn7ywo//0vbN9+47bvv4cOHayPr3To0LhtPJKTYdGihl8vXJuvL4wbBxs2wPPPa53GdblUC8MW\n0sJwLRUV6rkU1lorZWXqbK+6WiudOtW8VfnXX8PIkXD6tHsP6ou67dgBL78MBw5oncT5uGULQ7gf\nb+8bH/p1uXJF7Qa7vZAcOXLjNrNZLQi3F5Jjx9RT2KRYeLZhw9R/Q1lZEBKidRrXJC0M4TYUBS5e\nvLO1UlwMv/iF7B0lYP589Q+HhAStkzgXW1sYUjCEEB7j2DEYPVrtnvRyqX0uHMvWgiFvmRDCY/Tu\nrU6gyMjQOolrkoIhhPAocnxrw0nBEEJ4lNhYSE2FS5e0TuJ6pGAIITxKQAA89BBs3ap1EtcjBUMI\n4XHi4tRzMkT9yCwpIYTHuXxZXaMzcmTjdhBwF2lpcOmSTKsVQogaHTgg531Xa94cpkyRgiGEEMIG\ntnx2yhiGEEIIm0jBEEIIYRMpGEIIIWwiBUMIIYRNpGAIIYSwiRQMIYQQNpGCIYQQwiaaFIxNmzbR\nq1cvmjVrxqFDh265LyEhge7duxMUFMT27dsttx88eJDevXvTvXt35s2b19SRhRDC42lSMHr37s2W\nLVsYPHjwLbdnZWWRkpJCVlYW6enpzJkzx7KQ5Oc//zmrVq3ixIkTnDhxgvT0dC2iu5QM2fTfQt6L\nG+S9uEHei/rRpGAEBQXRo0ePO25PTU1l2rRp+Pj4YDKZCAwM5KuvvqKwsJDS0lL69+8PQFxcHFtl\nq0mr5P8MN8h7cYO8FzfIe1E/TjWGUVBQgPGmg5eNRiNms/mO2w0GA2azWYuIQgjhsbwd9cRRUVEU\nFRXdcfuSJUsYN26co15WCCGEgzisYOzYsaPe1xgMBvLy8iw/5+fnYzQaMRgM5Ofn33K7wWCo8Tm6\ndeuGTvYrtli8eLHWEZyGvBc3yHtxg7wXqm7dull9jMMKhq1u3h1x/PjxxMbG8tJLL2E2mzlx4gT9\n+/dHp9PRunVrvvrqK/r378+7777L888/X+Pz5eTkNFV0IYTwKJqMYWzZsoXOnTvz5ZdfMmbMGEaN\nGgVASEgIMTExhISEMGrUKFasWGFpLaxYsYLZs2fTvXt3AgMDGTlypBbRhRDCY7ndeRhCCCEcw6lm\nSTVGeno6QUFBdO/enaVLl2odR1OzZs1Cr9fTu3dvraNoKi8vj6FDh9KrVy9CQ0NZvny51pE0c/Xq\nVQYMGEBYWBghISG8/PLLWkfSXGVlJeHh4R4/CcdkMnH//fcTHh5uWbpQG7doYVRWVtKzZ0927tyJ\nwWDgwQcfZP369QQHB2sdTRO7d+/G19eXuLg4jh07pnUczRQVFVFUVERYWBhlZWX07duXrVu3euy/\ni8uXL9OyZUsqKip49NFHeeONN3j00Ue1jqWZP//5zxw8eJDS0lLS0tK0jqOZrl27cvDgQe655x6r\nj3WLFkZmZiaBgYGYTCZ8fHx44oknSE1N1TqWZgYNGkS7du20jqG5gIAAwsLCAPD19SU4OJiCggKN\nU2mnZcuWAJSXl1NZWWnTB4S7ys/P58MPP2T27NlypDPY/B64RcEwm8107tzZ8nP1gj8hquXm5nL4\n8GEGDBigdRTNVFVVERYWhl6vZ+jQoYSEhGgdSTMvvvgir7/+Ol5ebvER2Cg6nY7IyEj69evHypUr\n63ysW7xbsu5C1KWsrIzJkyezbNkyfH19tY6jGS8vL44cOUJ+fj6ff/65x26L8cEHH+Dv7094eLi0\nLoC9e/dy+PBhPvroI95++212795d62PdomDcvuAvLy/vlq1EhOe6du0ajz/+OE8++SQTJ07UOo5T\naNOmDWPGjOHAgQNaR9HEF198QVpaGl27dmXatGl8+umnxMXFaR1LM/feey8AHTt2ZNKkSWRmZtb6\nWLcoGP369ePEiRPk5uZSXl5OSkoK48eP1zqW0JiiKDz99NOEhITwwgsvaB1HUz/88AMlJSUAXLly\nhR07dhAeHq5xKm0sWbKEvLw8Tp06xYYNGxg2bBhr167VOpYmLl++TGlpKQCXLl1i+/btdc6udIuC\n4e3tzV//+ldGjBhBSEgIU6dO9diZMADTpk3jkUceITs7m86dO5OcnKx1JE3s3buX9957j127dhEe\nHk54eLjHbotfWFjIsGHDCAsLY8CAAYwbN47hw4drHcspeHKXdnFxMYMGDbL8uxg7dizR0dG1Pt4t\nptUKIYRwPLdoYQghhHA8KRhCCCFsIgVDCCGETaRgCCGEsIkUDCGEEDaRgiGEEMImUjCEx3L0NiFv\nvvkmV65cqdfrbdu2zeO35xfOS9ZhCI/l5+dnWeXqCF27duXAgQO0b9++SV5PCEeTFoYQNzl58iSj\nRo2iX79+DB48mG+//RaAp556innz5jFw4EC6devG5s2bAXUH2Dlz5hAcHEx0dDRjxoxh8+bNvPXW\nWxQUFDB06NBbVlT/+te/JiwsjIcffpizZ8/e8fqrV6/mf//3f+t8zZvl5uYSFBTEzJkz6dmzJ9On\nT2f79u0MHDiQHj16sH//fke8TcJTKUJ4KF9f3ztuGzZsmHLixAlFURTlyy+/VIYNG6YoiqLEx8cr\nMTExiqIoSlZWlhIYGKgoiqJs2rRJGT16tKIoilJUVKS0a9dO2bx5s6IoimIymZRz585Znlun0ykf\nfPCBoiiKMn/+fOX3v//9Ha+/evVqZe7cuXW+5s1OnTqleHt7K19//bVSVVWl9O3bV5k1a5aiKIqS\nmpqqTJw4sb5vixC18ta6YAnhLMrKyti3bx9Tpkyx3FZeXg6o+w1V73YbHBxMcXExAHv27CEmJgbA\ncs5EbZo3b86YMWMA6Nu3Lzt27KgzT22vebuuXbvSq1cvAHr16kVkZCQAoaGh5Obm1vkaQtSHFAwh\nrquqqqJt27YcPny4xvubN29u+V65PvSn0+luOVNBqWNI0MfHx/K9l5cXFRUVVjPV9Jq3a9GixS3P\nW32Nra8hhK1kDEOI61q3bk3Xrl15//33AfUD+ujRo3VeM3DgQDZv3oyiKBQXF/PZZ59Z7vPz8+Pi\nxYv1ylBXwRFCa1IwhMe6fPkynTt3tny9+eabrFu3jlWrVhEWFkZoaChpaWmWx9+8DXb1948//jhG\no5GQkBBmzJhBnz59aNOmDQDPPvssI0eOtAx63359Tdtq3357bd/ffk1tP3vy1t3C/mRarRCNdOnS\nJVq1asW5c+cYMGAAX3zxBf7+/lrHEsLuZAxDiEYaO3YsJSUllJeX8+qrr0qxEG5LWhhCCCFsImMY\nQgghbCIFQwghhE2kYAghhLCJFAwhhBA2kYIhhBDCJlIwhBBC2OT/Afgh7irtHHa4AAAAAElFTkSu\nQmCC\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x1a108f0>"

+       ]

+      }

+     ],

+     "prompt_number": 18

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_1.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_1.ipynb
new file mode 100644
index 00000000..09aec3f9
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_1.ipynb
@@ -0,0 +1,1628 @@
+{

+ "metadata": {

+  "name": "chapter no.4.ipynb"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 4:Stresses in Beams"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.1,Page no.130"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=5000 #mm #Length of Beam\n",

+      "a=2000 #mm #Length of start of beam to Pt Load\n",

+      "b=3000 #mm #Length of Pt load to end of beam\n",

+      "A=150*250 #m**2 #Area of beam\n",

+      "b=150 #mm #Width of beam\n",

+      "d=250 #mm #Depth of beam\n",

+      "sigma=10#N/mm**2 #stress\n",

+      "l=2000 #m #Load applied from one end\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Moment of Inertia\n",

+      "I=1*12**-1*b*d**3 #m**4\n",

+      "\n",

+      "#Distance from N.A to end\n",

+      "y_max=d*2**-1 #m\n",

+      "\n",

+      "#Section Modulus\n",

+      "Z=1*6**-1*b*d**2 #mm**3\n",

+      "\n",

+      "#Moment Carrying Capacity\n",

+      "M=sigma*Z #N-mm\n",

+      "\n",

+      "#Let w be the Intensity of the Load in N/m,then Max moment\n",

+      "#M_max=w*L**2*8**-1 #N-mm\n",

+      "#After substituting values and further simplifying we get\n",

+      "#M_max=w*25*100*8**-1\n",

+      "\n",

+      "#EQuating it to moment carrying capacity,we get max intensity load\n",

+      "w=M*(25*1000)**-1*8*10**-3\n",

+      "\n",

+      "#Part-2\n",

+      "\n",

+      "#Let P be the concentrated load,then max moment occurs under the load and its value\n",

+      "#M1=P*a*b*L**-1 #N-mm\n",

+      "\n",

+      "#Equting it to moment carrying capacity we get\n",

+      "P=M*1200**-1*10**-3 #N\n",

+      "\n",

+      "#Result\n",

+      "print\"Max Intensity of u.d.l it can carry\",round(w,3),\"KN-m\"\n",

+      "print\"MAx concentrated Load P apllied at 2 m from one end is\",round(P,3),\"KN\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max Intensity of u.d.l it can carry 5.0 N-mm\n",

+        "MAx concentrated Load P apllied at 2 m from one end is 13.021 KN\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.2,Page no.131"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "D=70 #mm #External Diameter\n",

+      "t=8 #mm #Thickness of pipe\n",

+      "L=2500 #mm #span \n",

+      "sigma=150 #N/mm**2 #stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Internal Diameter \n",

+      "d=D-2*t #mm\n",

+      "\n",

+      "#M.I Of Pipe\n",

+      "I=pi*64**-1*(D**4-d**4) #mm**4\n",

+      "\n",

+      "y_max=D*2**-1 #mm\n",

+      "Z=I*(y_max)**-1 #mm**3\n",

+      "\n",

+      "#Moment Carrying capacity\n",

+      "M=sigma*Z #N*mm\n",

+      "\n",

+      "#Max moment int the beam occurs at the mid-span and is equal to\n",

+      "#m=P*L*4**-1\n",

+      "\n",

+      "#Equating Max moment to moment carrying capacity we get,\n",

+      "#M=P*2.5*L*4**-1\n",

+      "#After substituting and simplifying we get\n",

+      "P=4*M*(L)**-1*10**-3 #N\n",

+      "\n",

+      "#Result\n",

+      "print\"Max concentrated load that can be applied at the centre of span is\",round(P,3),\"KN\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max concentrated load that can be applied at the centre of span is 5.22 KN\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.3,Page no.132"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Flanges Dimension\n",

+      "b1=180 #mm #Width\n",

+      "d1=10 #mm #Thickness\n",

+      "\n",

+      "D=500 #mm #Overall depth\n",

+      "t=8 #mm #Thickness of web\n",

+      "\n",

+      "#Plate Dimensions\n",

+      "b2=240 #mm #Width\n",

+      "t2=12 #mm #Thickness\n",

+      "\n",

+      "sigma=150 #N/mm**2 #Stress\n",

+      "L=3000 #mm #span\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Distance of centroid from bottom fibre\n",

+      "y_bar=(b2*t2*(D+t2*2**-1)+b1*d1*(D-t1*2**-1)+(D-2*t1)*t*D*2**-1+(b1*t1*t1*2**-1))*(b2*t2+b1*d1+b1*d1+(D-2*d1)*t)**-1\n",

+      "\n",

+      "#M.I of section\n",

+      "I=(1*12**-1*b2*t2**3+b2*t2*(D+t2*2**-1-y_bar)**2+1*12**-1*b1*d1**3+b1*d1*(D-t1*2**-1-y_bar)**2+1*12**-1*b1*t1**3+b1*t1*(t1*2**-1-y_bar)**2+1*12**-1*t*(D-2*t1)**3+t*(D-2*t1)*(D*2**-1-y_bar)**2)\n",

+      "\n",

+      "#Section Modulus\n",

+      "Z=I*(y_bar)**-1 #mm**3\n",

+      "\n",

+      "#Moment or Resistance\n",

+      "M=sigma*Z\n",

+      "\n",

+      "#Let Load on Cantilever be w/m Length \n",

+      "#Max M.I produced\n",

+      "#M_max=w*L**2**-1 \n",

+      "\n",

+      "#Now Equating Moment of resistance to Max moment,we get Max load\n",

+      "#4.5*w=M\n",

+      "#After rearranging and further simplifying we get\n",

+      "w=M*4.5**-1*10**3*10**-9\n",

+      "\n",

+      "#Result\n",

+      "print\"Moment of Resistance is\",round(M,2),\"KN-mm\"\n",

+      "print\"Load the section can carry is\",round(w,3),\"KN/m\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Moment of Resistance is 198770121.83 KN-mm\n",

+        "Load the section can carry is 44.171 KN/m\n"

+       ]

+      }

+     ],

+     "prompt_number": 26

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.4,Page no.134"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Flange (Top)\n",

+      "b1=80 #mm #Width \n",

+      "t1=40 #mm #Thickness\n",

+      "\n",

+      "#Flange (Bottom)\n",

+      "b2=160 #mm #width\n",

+      "t2=40 #mm #Thickness\n",

+      "\n",

+      "#web\n",

+      "d=120 #mm #Depth\n",

+      "t3=20 #mm #Thickness\n",

+      "\n",

+      "D=200 #mm #Overall Depth\n",

+      "sigma1=30 #N/mm**2 #Tensile stress\n",

+      "sigma2=90 #N/mm**2 #Compressive stress\n",

+      "L=6000 #mm #Span\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Distance of centroid from bottom fibre\n",

+      "y_bar=(b1*t1*(D-t1*2**-1)+d*t3*(d*2**-1+t2)+b2*t2*t2*2**-1)*(b1*t1+d*t3+b2*t2)**-1 #mm\n",

+      "\n",

+      "#Moment of Inertia\n",

+      "I=1*12**-1*b1*t1**3+b1*t1*(D-t1*2**-1-round(y_bar,2))**2+1*12**-1*t3*d**3+t3*d*(d*2**-1+t2-round(y_bar,2))**2+1*12**-1*b2*t2**3+b2*t2*(t2*2**-1-round(y_bar,2))**2\n",

+      "\n",

+      "#Extreme fibre distance of top and bottom fibres are y_t and y_c respectively\n",

+      "\n",

+      "y_t=y_bar #mm\n",

+      "y_c=D-y_bar #mm\n",

+      "\n",

+      "#Moment carrying capacity considering Tensile strength \n",

+      "M1=sigma1*I*y_t**-1*10**-6  #KN-m\n",

+      "\n",

+      "#Moment carrying capacity considering compressive strength \n",

+      "M2=sigma2*I*y_c**-1*10**-6  #KN-m\n",

+      "\n",

+      "#Max Bending moment in simply supported beam 6 m due to u.d.l\n",

+      "#M_max=w*L*10**-3*8**-1\n",

+      "#After simplifying further we get\n",

+      "#M_max=4.5*w\n",

+      "\n",

+      "#Now Equating it to Moment carrying capacity, we get load carrying capacity\n",

+      "w=M1*4.5**-1 #KN/m\n",

+      "\n",

+      "#Result\n",

+      "print\"Max Uniformly Distributed Load is\",round(w,3),\"KN/m\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max Uniformly Distributed Load is 5.096 KN/m\n"

+       ]

+      }

+     ],

+     "prompt_number": 16

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.5,Page no.136"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "from scipy.integrate import quad\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Flanges\n",

+      "b=200 #mm #Width\n",

+      "t=25 #mm #Thickness \n",

+      "\n",

+      "D1=500 #mm #Overall Depth\n",

+      "t2=20 #mm #Thickness of web\n",

+      "\n",

+      "d=450 #mm #Depth of web\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Consider,Element of Thickness \"y\" at Distance \"dy\" from N.A \n",

+      "#Let Bending stress \"sigma_max\"\n",

+      "\n",

+      "#Stress on the element \n",

+      "#sigma=y*(D*2**-1)*sigma_max   ..............(1)\n",

+      "\n",

+      "#Area of Element\n",

+      "#A=b*dy       .................................(2)\n",

+      "\n",

+      "#Force on Element \n",

+      "#F=y*250**-1*sigma_max*b*dy\n",

+      "\n",

+      "#Let M be the Moment of resistance\n",

+      "#M=y*250**-1*sigma_max*b*dy*y\n",

+      "\n",

+      "#Moment of Resistance of top flange be M1\n",

+      "def integrand(y, b, D):\n",

+      "     return b*y**2*D**-1\n",

+      "b=200 \n",

+      "D=250\n",

+      "\n",

+      "X = quad(integrand, 225, 250, args=(b,D))\n",

+      "\n",

+      "Y=2*X[0]\n",

+      "\n",

+      "#M1=Y*sigma\n",

+      "\n",

+      "#Now Moment of Inertia I section is\n",

+      "X=b*D1**3\n",

+      "Y=(b-t2)*d**3\n",

+      "I=(X-Y)*12**-1*10**-8\n",

+      "\n",

+      "#Moment acting on the entire section\n",

+      "#since sigmais the value at y=250\n",

+      "y_max=250\n",

+      "Z=I*10**8*y_max**-1\n",

+      "#M=sigma*Z   \n",

+      "#After Simplifying Further we get\n",

+      "#M2=Z*sigma\n",

+      "\n",

+      "#Percentage Moment resisted by Flanges\n",

+      "P1=2258333.3*(2865833.3)**-1*100\n",

+      "\n",

+      "#Percentage Moment resisted by web\n",

+      "P2=100-P1\n",

+      "\n",

+      "#Result\n",

+      "print\"Percentage Moment resisted by Flanges\",round(P1,2),\"%\"\n",

+      "print\"Percentage Moment resisted by web\",round(P2,2),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Percentage Moment resisted by Flanges 78.8 %\n",

+        "Percentage Moment resisted by web 21.2 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 38

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.6,Page no.137"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Flanges\n",

+      "b1=200 #mm #Width\n",

+      "t1=10 #mm #Thickness\n",

+      "\n",

+      "#Web\n",

+      "d=380 #mm #Depth \n",

+      "t2=8 #mm #Thickness\n",

+      "\n",

+      "D=400 #mm #Overall Depth\n",

+      "sigma=150 #N/mm**2\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Area\n",

+      "A=b1*t1+d*t2+b1*t1 #mm**2\n",

+      "\n",

+      "#Moment of Inertia\n",

+      "I=1*12**-1*(b1*D**3-(b1-t2)*d**3)\n",

+      "\n",

+      "#Bending Moment\n",

+      "M=sigma*I*(D*2**-1)**-1\n",

+      "\n",

+      "#Square Section\n",

+      "\n",

+      "#Let 'a' be the side\n",

+      "a=A**0.5\n",

+      "\n",

+      "#Moment of Resistance of this section\n",

+      "M1=1*6**-1*a*a**2*sigma\n",

+      "\n",

+      "X=M*M1**-1\n",

+      "\n",

+      "#Rectangular section\n",

+      "#Let 'a' be the side and depth be 2*a\n",

+      "\n",

+      "a=(A*2**-1)**0.5\n",

+      "\n",

+      "#Moment of Rectangular secction\n",

+      "M2=1*6**-1*a*(2*a)**2*sigma\n",

+      "\n",

+      "X2=M*M2**-1\n",

+      "\n",

+      "#Circular section\n",

+      "#A=pi*d1**2*4**-1\n",

+      "\n",

+      "d1=(A*4*pi**-1)**0.5\n",

+      "\n",

+      "#Moment of circular section\n",

+      "M3=pi*32**-1*d1**3*sigma\n",

+      "\n",

+      "X3=M*M3**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Moment of resistance of beam section\",round(M,2),\"mm\"\n",

+      "print\"Moment of resistance of square section\",round(X,2),\"mm\"\n",

+      "print\"Moment of resistance of rectangular section\",round(X2,2),\"mm\"\n",

+      "print\"Moment of resistance of circular section\",round(X3,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Moment of resistance of beam section 141536000.0 mm\n",

+        "Moment of resistance of square section 9.58 mm\n",

+        "Moment of resistance of rectangular section 6.78 mm\n",

+        "Moment of resistance of circular section 11.33 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 25

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.7,Page no.139"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "F=12 #KN #Force at End of beam\n",

+      "L=2 #m #span\n",

+      "\n",

+      "#Square section \n",

+      "b=d=200 #mm #Width and depth of beam\n",

+      "\n",

+      "#Rectangular section\n",

+      "b1=150 #mm #Width\n",

+      "d1=300 #mm #Depth\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Max bending Moment\n",

+      "M=F*L*10**6 #N-mm\n",

+      "\n",

+      "#M=sigma*b*d**2\n",

+      "sigma=M*6*(b*d**2)**-1 #N/mm**2\n",

+      "\n",

+      "#Let W be the central concentrated Load in simply supported beam of span L1=3 m\n",

+      "#MAx Moment\n",

+      "#M1=W*L1*4**-1\n",

+      "#After Further simplifying we get\n",

+      "#M1=0.75*10**6 #N-mm\n",

+      "\n",

+      "#The section has  a moment of resistance\n",

+      "M1=sigma*1*6**-1*b1*d1**2\n",

+      "\n",

+      "#Equating it to moment of resistance we get max load W\n",

+      "#0.75*10**6*W=M1\n",

+      "#After Further simplifying we get\n",

+      "W=M1*(0.75*10**6)**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Minimum Concentrated Load required to brek the beam\",round(W,2),\"KN\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Minimum Concentrated Load required to brek the beam 54.0 KN\n"

+       ]

+      }

+     ],

+     "prompt_number": 30

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.8,Page no.140"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=3 #m #span\n",

+      "sigma_t=35 #N/mm**2 #Permissible stress in tension\n",

+      "sigma_c=90 #N/mm**2 #Permissible stress in compression\n",

+      "\n",

+      "#Flanges\n",

+      "t=30 #mm #Thickness\n",

+      "d=250 #mm #Depth\n",

+      "\n",

+      "#Web\n",

+      "t2=25 #mm #Thickness\n",

+      "b=600 #mm #Width\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let y_bar be the Distance of N.A from Extreme Fibres\n",

+      "y_bar=(t*d*d*2**-1*2+(b-2*t)*t2*t2*2**-1)*(t*d*2+(b-2*t)*t2)**-1\n",

+      "\n",

+      "#Moment of Inertia\n",

+      "I=(1*12**-1*t*d**3+t*d*(d*2**-1-y_bar)**2)*2+1*12**-1*(b-2*t)*t2**3+(b-2*t)*t2*(t2*2**-1-y_bar)**2\n",

+      "\n",

+      "#Part-1\n",

+      "\n",

+      "#If web is in Tension\n",

+      "y_t=y_bar #mm\n",

+      "y_c=d-y_bar #mm\n",

+      "\n",

+      "#Moment carrying caryying capacity From consideration of tensile stress\n",

+      "M=sigma_t*I*(y_bar)**-1 #N-mm\n",

+      "\n",

+      "#Moment carrying caryying capacity From consideration of compressive stress\n",

+      "M1=sigma_c*I*(y_c)**-1 #N-mm\n",

+      "\n",

+      "#If w KN/m is u.d.l in beam,Max bending moment\n",

+      "#M=wl**2*8**-1\n",

+      "#After further simplifyng we get\n",

+      "#M=1.125*w*10**6 N-mm\n",

+      "w=M*(1.125*10**6)**-1 #KN\n",

+      "\n",

+      "#Part-2\n",

+      "\n",

+      "#If web is in compression\n",

+      "y_t2=178.299 #mm\n",

+      "y_c2=71.71 #mm \n",

+      "\n",

+      "#Moment carrying caryying capacity From consideration of tensile stress\n",

+      "M2=sigma_t*I*(y_t2)**-1 #N-mm\n",

+      "\n",

+      "#Moment carrying caryying capacity From consideration of compressive stress\n",

+      "M3=sigma_c*I*(y_c2)**-1 #N-mm\n",

+      "\n",

+      "#Moment of resistance is M2\n",

+      "\n",

+      "#Equating it to bending moment we get\n",

+      "#M2=1.125*10**6*w2\n",

+      "#After further simplifyng we get\n",

+      "w2=M2*(1.125*10**6)**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Uniformly Distributed Load carrying capacity if:web is in Tension\",round(w,2),\"KN\"\n",

+      "print\"                                               :web is in compression\",round(w2,3),\"KN\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Uniformly Distributed Load carrying capacity if:web is in Tension 73.21 KN\n",

+        "                                               :web is in compression 29.446 KN\n"

+       ]

+      }

+     ],

+     "prompt_number": 26

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.9,Page no.141"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "b1=200 #mm #Width at base\n",

+      "b2=100 #mm #Width at top\n",

+      "\n",

+      "L=8 #m Length\n",

+      "P=500 #N #Load\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Consider a section at y metres from top\n",

+      "\n",

+      "#At this section diameter d is\n",

+      "#d=b2+y*L**-1*(b1-b2)\n",

+      "#After Further simplifying we get\n",

+      "#d=b2+12.5*y #mm\n",

+      "\n",

+      "#Moment of Inertia\n",

+      "#I=pi*64**-1*d**4\n",

+      "\n",

+      "#Section Modulus \n",

+      "#Z=pi*32**-1*(b1+12.5*y)**3\n",

+      "\n",

+      "#Moment \n",

+      "#M=5*10**5*y #N-mm\n",

+      "\n",

+      "#Let sigma be the fibre stress at this section then\n",

+      "#M=sigma*Z\n",

+      "#After sub values in above equation and further simplifying we get\n",

+      "#sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n",

+      "\n",

+      "#For sigma to be Max,d(sigma)*(dy)**-1=0\n",

+      "#16*10**6*pi**-1*((b2+12.5*y)**-3+y*(-3)*(b2+12.5*y)**-4*12.5)\n",

+      "#After Further simplifying we get\n",

+      "#b2+12.5*y=37.5*y\n",

+      "#After Further simplifying we get\n",

+      "y=b2*25**-1 #m\n",

+      "\n",

+      "#Stress at this section\n",

+      "sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Stress at Extreme Fibre is max\",round(y,2),\"m\"\n",

+      "print\"Max stress is\",round(sigma,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stress at Extreme Fibre is max 4.0 m\n",

+        "Max stress is 6.04 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 28

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.10,Page no.143"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "H=10 #mm #Height\n",

+      "A1=160*160 #mm**2 #area of square section at bottom\n",

+      "L1=160 #mm #Length of square section at bottom\n",

+      "b1=160 #mm #width of square section at bottom\n",

+      "A2=80*80 #mm**2 #area of square section at top\n",

+      "L2=80 #mm #Length of square section at top\n",

+      "b2=80 #mm #Width of square section at top\n",

+      "P=100 #N #Pull\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Consider a section at distance y from top.\n",

+      "#Let the side of square bar be 'a'\n",

+      "#a=L2+y*(H)**-1*(b1-b2)\n",

+      "#After further simplifying we get\n",

+      "#a=L2+8*y\n",

+      "\n",

+      "#Moment of Inertia\n",

+      "#I=2*1*12**-1*a*(2)**0.5*(a*((2)**0.5)**-1)**3\n",

+      "#After further simplifying we get\n",

+      "#I=a**4*12**-1\n",

+      "\n",

+      "#Section Modulus \n",

+      "#Z=a**4*(12*a*(2)**0.5)**-1\n",

+      "#After further simplifying we get\n",

+      "#Z=2**0.5*a**3*(12)**-1 #mm**3\n",

+      "\n",

+      "#Bending moment at this section=100*y N-mm\n",

+      "#M=100*10**3*y #N-mm\n",

+      "\n",

+      "#But\n",

+      "#M=sigma*Z\n",

+      "#After sub values in above equation we get\n",

+      "#sigma=M*Z**-1\n",

+      "#After further simplifying we get\n",

+      "#sigma=1200*10**3*(2**0.5)**-1*y*((80+80*y)**3)**-1 .......(1)\n",

+      "\n",

+      "#For Max stress df*(dy)**-1=0\n",

+      "#After taking Derivative of above equation we get\n",

+      "#df*(dy)**-1=1200*10**3*(2**0.5)**-1*((80+8*y)**-3+y(-3)*(80+8*y)**-4*8)\n",

+      "#After further simplifying we get\n",

+      "y=80*16**-1 #m\n",

+      "\n",

+      "#Max stress at this level is\n",

+      "sigma=1200*10**3*(2**0.5)**-1*y*((80+8*y)**3)**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Max Bending stress is Developed at\",round(y,3),\"m\"\n",

+      "print\"Value of Max Bending stress is\",round(sigma,3),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max Bending stress is Developed at 5.0 m\n",

+        "Value of Max Bending stress is 2.455 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 4

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.12,Page no.147"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "b=200 #mm #Width of timber \n",

+      "d=400 #mm #Depth of timber\n",

+      "t=6 #mm #Thickness\n",

+      "b2=200 #mm #width of steel plate\n",

+      "t2=20 #mm #Thickness of steel plate\n",

+      "M=40*10**6 #KN-mm #Moment\n",

+      "#Let E_s*E_t**-1=X\n",

+      "X=20 #Ratio of Modulus of steel to timber\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#let y_bar be the Distance of centroidfrom bottom most fibre\n",

+      "y_bar=(b*d*(b+t)+t2*b2*t*t*2**-1)*(b*d+t2*b2*t)**-1 #mm\n",

+      "\n",

+      "#Moment of Inertia\n",

+      "I=1*12**-1*b*d**3+b*d*(b+t-round(y_bar,3))**2+1*12**-1*t2*b2*t**3+b2*t2*t*(round(y_bar,3)-t*2**-1)**2\n",

+      "\n",

+      "#distance of the top fibre from N-A\n",

+      "y_1=d+t-y_bar #mm\n",

+      "\n",

+      "#Distance of the junction of timber and steel From N-A\n",

+      "y_2=y_bar-t #mm\n",

+      "\n",

+      "#Stress in Timber at the top\n",

+      "Y=M*I**-1*y_1 #N/mm**2\n",

+      "\n",

+      "#Stress in the Timber at the junction point\n",

+      "Z=M*I**-1*y_2\n",

+      "\n",

+      "#Coressponding stress in steel at the junction point\n",

+      "Z2=X*Z #N/mm**2 \n",

+      "\n",

+      "#The stress in Extreme steel fibre \n",

+      "Z3=X*M*I**-1*y_bar\n",

+      "\n",

+      "#Result\n",

+      "print\"Stress in Extreme steel Fibre\",round(Z3,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stress in Extreme steel Fibre 69.67 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 23

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.13,Page no.149"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Timber size\n",

+      "b=150 #mm #Width\n",

+      "b2=120 #mm \n",

+      "d=300 #mm #Depth\n",

+      "\n",

+      "t=6 #mm #Thickness of steel plate\n",

+      "L=6 #m #Span\n",

+      "\n",

+      "#E_s*E_t**-1=20 \n",

+      "#X=E_s*E_t**-1\n",

+      "X=20 \n",

+      "sigma_timber=8 #N/mm**2  #Stress in timber\n",

+      "sigma_steel=150 #N/mm**2 #Stress in steel plate\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "Y\n",

+      "\n",

+      "#Due to  synnetry cenroid,the neutral axis is half the depth\n",

+      "I=1*12**-1*\n",

+      "\n",

+      "\n",

+      "#Result\n",

+      "print Z"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "153.0\n"

+       ]

+      }

+     ],

+     "prompt_number": 34

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.14,Page no.151"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=6000 #mm #Span of beam\n",

+      "W=20*10**3 #N #Load\n",

+      "sigma=8 #N/mm**2 #Stress\n",

+      "b=200 #mm #Width of section\n",

+      "d=300 #mm #Depth of section\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#let x be the distance from left side of beam\n",

+      "\n",

+      "#Bending moment\n",

+      "#M=W*2**-1*x #Nmm   .......(1)\n",

+      "\n",

+      "#But M=sigma*Z   ..........(2)\n",

+      "\n",

+      "#Equating equation 1 and 2 we get\n",

+      "#W*2**-1*x=sigma*Z    ............(3)\n",

+      "\n",

+      "#Section Modulus \n",

+      "#Z=1*6*b*d**2   ...............(4)\n",

+      "\n",

+      "#Equating equation 3 and 4 we get\n",

+      "#b*d**2=3*W*x*sigma**-1 .............(5)\n",

+      "\n",

+      "#Beam of uniform strength of constant depth\n",

+      "#b=3*W*x*(sigma*d**2)    \n",

+      "\n",

+      "#When x=0\n",

+      "b=0\n",

+      "\n",

+      "#When x=L*2**-1\n",

+      "b2=3*W*L*(2*sigma*d**2)**-1 #mm\n",

+      "\n",

+      "#Beam with constant width of 200 mm\n",

+      "\n",

+      "#We have\n",

+      "#d=(3*W*x*(sigma*d)**-1)**0.5\n",

+      "#thus depth varies as (x)**0.5\n",

+      "\n",

+      "#when x=0\n",

+      "d1=0\n",

+      "\n",

+      "#when x=L*2**-1\n",

+      "d2=(2*W*L*(2*sigma*300)**-1)**0.5 #mm\n",

+      "\n",

+      "#Result\n",

+      "print\"Cross section of rectangular beam is:\",round(b2,2),\"mm\"\n",

+      "print\"                                    :\",round(d2,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Cross section of rectangular beam is: 250.0 mm\n",

+        "                                    : 223.61 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 22

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.15,Page no.154"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=800 #mm #Span\n",

+      "n=5 #number of leaves\n",

+      "b=60 #mm #Width\n",

+      "t=10 #mm #thickness\n",

+      "sigma=250 #N/mm**2 #Stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#section Modulus\n",

+      "Z=n*6**-1*b*t**2 #mm**3\n",

+      "\n",

+      "#from the relation\n",

+      "#sigma*Z=M   ...................(1)\n",

+      "#M=P*L*4**-1\n",

+      "#sub values of M in equation 1 we get\n",

+      "P=sigma*Z*4*L**-1*10**-3 #KN #Load\n",

+      "\n",

+      "#Length of Leaves\n",

+      "L1=0.2*L #mm\n",

+      "L2=0.4*L #mm\n",

+      "L3=0.6*L #mm\n",

+      "L4=0.8*L #mm\n",

+      "L5=L #mm\n",

+      "\n",

+      "#Result\n",

+      "print\"Max Load it can take is\",round(P,2),\"KN\"\n",

+      "print\"Length of leaves:L1\",round(L1,2),\"mm\"\n",

+      "print\"                :L2\",round(L2,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max Load it can take is 6.25 KN\n",

+        "Length of leaves:L1 160.0 mm\n",

+        "                :L2 320.0 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.16,Page no.161"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "F=20*10**3 #N #Shear Force\n",

+      "\n",

+      "#Tee section\n",

+      "\n",

+      "#Flange\n",

+      "b=100 #mm #Width\n",

+      "t=12 #mm #Thickness\n",

+      "\n",

+      "#Web\n",

+      "d=88 #mm #Depth\n",

+      "t2=12 #mm #Thicknes\n",

+      "\n",

+      "D=100 #mm #Overall Depth\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Distance of C.G from Top Fibre\n",

+      "y=(b*t*t*2**-1+t2*d*(d*2**-1+t))*(b*t+d*t2)**-1 #mm \n",

+      "\n",

+      "#Moment Of Inertia\n",

+      "I=1*12**-1*b*t**3+b*t*(y-t*2**-1)**2+1*12**-1*t2*d**3+t2*d*(t+d*2**-1-y)**2 #mm**4\n",

+      "\n",

+      "#shear stress at bottom Flange\n",

+      "\n",

+      "#Area above this level\n",

+      "A=b*t #mm**2\n",

+      "\n",

+      "#C.G of this area from N-A\n",

+      "y2=y-t*2**-1\n",

+      "\n",

+      "#Stress at bottom of flange\n",

+      "sigma=F*A*y2*(b*I)**-1 #N/mm**2 \n",

+      "\n",

+      "#sigma2 at same level but in web where width is 12 mm\n",

+      "sigma2=F*A*y2*(t2*I)**-1 #N/mm**2 \n",

+      "\n",

+      "#To find shear stress at N-A\n",

+      "X=t*b*(y-t*2**-1)+t2*(y-t2)*(y-t2)*2**-1 #mm**3\n",

+      "\n",

+      "sigma3=F*X*(t2*I)**-1 #N/mm**2\n",

+      "\n",

+      "#Shear stress at top and bottom fibre is zero\n",

+      "#sigma4 and sigma5 are top and bottom fibre shear stress\n",

+      "sigma4=sigma5=0\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,t,t,y,D]\n",

+      "Y1=[sigma4,sigma,sigma2,sigma3,sigma5]\n",

+      "Z1=[0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in m\")\n",

+      "plt.ylabel(\"Shear Force in kN\")\n",

+      "plt.show()\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force and Bending Moment Diagrams are the results\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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zC9PSYNAgOHXK7ogqnhKCiEgp+fnBypWmkC06Gn76ye6IKpYSgojINfD1hQUL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\niIgIoIQgIiLnKCGIiAighCAiIuf8f5wqyy9KzUKHAAAAAElFTkSuQmCC\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x56a6270>"

+       ]

+      }

+     ],

+     "prompt_number": 15

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.17,Page no.163"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "F=40*10**3 #N #shear Force\n",

+      "\n",

+      "#I-section\n",

+      "\n",

+      "#Flanges\n",

+      "b=80 #mm #Width of flange\n",

+      "t=20 #mm #Thickness\n",

+      "\n",

+      "#Web\n",

+      "d=200 #mm #Depth\n",

+      "t2=20 #mm #Thickness\n",

+      "\n",

+      "#Flange-2\n",

+      "b2=160 #mm #Width\n",

+      "t3=20 #mm #Thickness\n",

+      "\n",

+      "D=240 #mm #Overall Depth\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Distance of N-A from Top Fibre \n",

+      "y=(b*t*t*2**-1+d*t2*(t+d*2**-1)+b2*t3*(t+d+t3*2**-1))*(b*t+d*t2+b2*t3)**-1 #mm\n",

+      "\n",

+      "#Moment of Inertia\n",

+      "I=1*12**-1*b*t**3+b*t*(y-(t*2**-1))**2+1*12**-1*t2*d**3+t2*d*(y-(t+d*2**-1))**2+1*12**-1*b2*t3**3+t3*b2*((d+t+t3*2**-1)-y)**2 #mm**4\n",

+      "\n",

+      "#Shear stress bottom of flange\n",

+      "sigma=F*b*t*(y-t*2**-1)*(b*I)**-1 #N/mm**2\n",

+      "\n",

+      "#At same Level but in web\n",

+      "sigma2=F*b*t*(y-t*2**-1)*(t2*I)**-1 #N/mm**2\n",

+      "\n",

+      "#for shear stress at N.A\n",

+      "X=b*t*(y-t*2**-1)+t2*(y-t)*(y-t)*2**-1 #mm**3\n",

+      "sigma3=F*X*(t2*I)**-1 #N/mm**2\n",

+      "\n",

+      "#Shear stress at bottom of web\n",

+      "\n",

+      "X=b2*t3*((D-y)-t3*2**-1) #mm**3\n",

+      "\n",

+      "#Stress at bottom of web\n",

+      "sigma4=F*X*(t2*I)**-1 #N/mm**2\n",

+      "\n",

+      "#Stress at Lower flange\n",

+      "sigma5=F*X*(b2*I)**-1 #N/mm**2\n",

+      "\n",

+      "#Result\n",

+      "print \"The Shear Force Diagram is the result\"\n",

+      "\n",

+      "#Plotting the Shear Force Diagram\n",

+      "\n",

+      "X1=[0,20,20,140,220,220,240]\n",

+      "Y1=[0,sigma,sigma2,sigma3,sigma4,sigma5,0]\n",

+      "Z1=[0,0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length in mm\")\n",

+      "plt.ylabel(\"Shear Force in N\")\n",

+      "plt.show()\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Shear Force Diagram is the result\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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2vPfee7Ro0cKdZYiIyEXcPmq/YsWKLFy4kD59+pCXl8e4ceMUBCIiNvPIGcgi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\n/w/qZz1xEBCKMwAAAABJRU5ErkJggg==\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5614110>"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.18,Page no.164"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "F=30*10**3 #N #Shear Force\n",

+      "\n",

+      "#Channel Section\n",

+      "d=400 #mm #Depth of web \n",

+      "t=10 #mm #THickness of web\n",

+      "t2=15 #mm #Thickness of flange\n",

+      "b=100 #mm #Width of flange\n",

+      "\n",

+      "#Rectangular Welded section\n",

+      "b2=80 #mm #Width\n",

+      "d2=60 #mm #Depth\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Distance of Centroid From Top Fibre\n",

+      "y=(d*t*t*2**-1+2*t2*(b-t)*((b-t)*2**-1+10)+d2*b2*(d2*2**-1+t))*(d*t+2*t2*(b-t)+d2*b2)**-1 #mm\n",

+      "\n",

+      "#Moment Of Inertia of the section about N-A\n",

+      "I=1*12**-1*d*t**3+d*t*(y-t*2**-1)**2+2*(1*12**-1*t2*(b-t)**3+t2*(b-t)*(((b-t)*2**-1+t)-y)**2)+1*12**-1*d2**3*b2+d2*b2*(d2*2**-1+t-y)**2\n",

+      "\n",

+      "#Shear stress at level of weld\n",

+      "sigma=F*d*t*(y-t*2**-1)*((b2+t2+t2)*I)**-1 #N/mm**2\n",

+      "\n",

+      "#Max Shear Stress occurs at Neutral Axis\n",

+      "X=d*t*(y-t*2**-1)+2*t2*(y-t)*(y-t)*2**-1+b2*(y-t)*(y-t)*2**-1\n",

+      "\n",

+      "sigma_max=F*X*((b+t)*I)**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Shear stress in the weld is\",round(sigma,2),\"N/mm**2\"\n",

+      "print\"Max shear stress is\",round(sigma_max,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Shear stress in the weld is 3.62 N/mm**2\n",

+        "Max shear stress is 4.48 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.19,Page no.165"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Wooden Section\n",

+      "b=300 #mm #Width\n",

+      "d=300 #mm #Depth\n",

+      "\n",

+      "D=100 #mm #Diameter of Bore\n",

+      "F=10*10**3 #N #Shear Force\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Moment Of Inertia Of Section\n",

+      "I=1*12**-1*b*d**3-pi*64**-1*D**4\n",

+      "\n",

+      "#Shear stress at crown of circle\n",

+      "sigma=F*b*D*(d*2**-1-D*2**-1)*(b*I)**-1\n",

+      "\n",

+      "#Let a*y_bar=X\n",

+      "X=b*d*2**-1*d*4**-1-pi*8**-1*D**2*4*D*2**-1*(3*pi)**-1 #mm**3\n",

+      "\n",

+      "#Shear Stress at Neutral Axis\n",

+      "sigma2=F*X*((b-D)*I)**-1 #N/mm**2\n",

+      "\n",

+      "#Result\n",

+      "print\"Shearing Stress at Crown of Bore\",round(sigma,3),\"N/mm**2\"\n",

+      "print\"Shear Stress at Neutral Axis\",round(sigma2,3),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Shearing Stress at Crown of Bore 0.149 N/mm**2\n",

+        "Shear Stress at Neutral Axis 0.246 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 10

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.20,Page no.166"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#flanges\n",

+      "b=200 #mm #width\n",

+      "t1=25 #mm #Thickness\n",

+      "\n",

+      "#web\n",

+      "d=450 #mm #Depth \n",

+      "t2=20 #mm #thickness\n",

+      "\n",

+      "D=500 #mm #Total Depth of section\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Moment Of Inertia of the section about N-A\n",

+      "I=1*12**-1*b*D**3-1*12**-1*(b-t2)*d**3 #mm**4\n",

+      "\n",

+      "#Consider an element in the web at distance y from y from N-A\n",

+      "#Depth of web section=225-y\n",

+      "\n",

+      "#C.G From N-A\n",

+      "#y2=y+(((D*2**-1-t)-y)*2**-1)\n",

+      "\n",

+      "#ay_bar for section at y\n",

+      "#Let ay_bar be X\n",

+      "#X=X1 be of Flange + X2 be of web above y\n",

+      "#X=b*t1*(D*2**-1-t1*2**-1)+t2*(d-t1)*(d-t1+y)*2**-1\n",

+      "#After Sub values and Further simplifying we get\n",

+      "#X=1187500+10*(225**2-y**2)\n",

+      "\n",

+      "#Shear stress at y\n",

+      "#sigma_y=F*(X)*(t2*I)**-1\n",

+      "\n",

+      "#Shear Force resisted by the Element\n",

+      "#F1=F*X*t2*dy*(t2*I)**-1\n",

+      "\n",

+      "#Shear stress resisted by web \n",

+      "#sigma=2*F*I**-1*(X)*dy\n",

+      "\n",

+      "#After Integrating above equation and further simplifying we get\n",

+      "#sigma=0.9578*F\n",

+      "\n",

+      "sigma=0.9578*100\n",

+      "\n",

+      "#Result\n",

+      "print\"Shear Resisted by web\",round(sigma,2),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Shear Resisted by web 95.78 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 1

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.21,Page no.167"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Wooden Beam\n",

+      "\n",

+      "b=150 #mm #width\n",

+      "d=250 #mm #Depth\n",

+      "\n",

+      "L=5000 #mm #span\n",

+      "m=11.2 #N/mm**2 #Max Bending stress\n",

+      "sigma=0.7 #N/mm**2 #Max shear stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let 'a' be the distance from left support\n",

+      "#Max shear force\n",

+      "#F=R_A=W*(L-a)*L**-1 \n",

+      "\n",

+      "#Max Moment\n",

+      "#M=W*(L-a)*a*L**-1\n",

+      "\n",

+      "#But M=sigma*Z\n",

+      "#W*(L-a)*a*L**-1=m*1*6**-1*b*d**2   .....................(1)\n",

+      "\n",

+      "#In Rectangular Section MAx stress is 1.5 times Avg shear stress\n",

+      "F=sigma*b*d*1.5**-1\n",

+      "\n",

+      "#W*(L-a)*L**-1=F                     .....................(2)\n",

+      "\n",

+      "#Dividing Equation 1 nad 2 we get\n",

+      "a=m*6**-1*b*d**2*1.5*(sigma*b*d)**-1\n",

+      "\n",

+      "#Sub above value in equation 2 we get\n",

+      "W=(L-a)**-1*L*F*10**-3 #KN  \n",

+      "\n",

+      "#Result\n",

+      "print\"Load is\",round(W,2),\"KN\"\n",

+      "print\"Distance from Left support is\",round(a,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Load is 21.87 KN\n",

+        "Distance from Left support is 1000.0 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 22

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.22,Page no.168"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=1000 #mm #span\n",

+      "\n",

+      "#Rectangular Section\n",

+      "\n",

+      "b=200 #mm #width\n",

+      "d=400 #mm #depth\n",

+      "\n",

+      "sigma=1.5 #N/mm**2 #Shear stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let AB be the cantilever beam subjected to load W KN at free end\n",

+      "\n",

+      "#MAx shear Force\n",

+      "#F=W*10**3 #KN\n",

+      "\n",

+      "#Since Max shear stress in Rectangular section\n",

+      "#sigma_max=1.5*F*A**-1 \n",

+      "#After sub values and further simplifyng we get\n",

+      "W=1.5*b*d*(1.5*1000)**-1 #KN\n",

+      "\n",

+      "#Moment at fixwed end\n",

+      "M=W*1 #KN-m\n",

+      "y_max=d*2**-1 #mm\n",

+      "\n",

+      "#M.I\n",

+      "I=1*12**-1*b*d**3 #mm**3\n",

+      "\n",

+      "#MAx Stress\n",

+      "sigma_max=M*10**6*I**-1*y_max\n",

+      "\n",

+      "#Result\n",

+      "print\"Concentrated Load is\",round(sigma_max,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Concentrated Load is 15.0 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 14

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 4.4.24,Page no.170"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=4000 #mm #span\n",

+      "\n",

+      "#Rectangular Cross-section\n",

+      "b=100 #mm #Width\n",

+      "d=200 #mm #Thickness\n",

+      "\n",

+      "F_per=10 #N/mm**2 #Max Bending stress\n",

+      "q_max=0.6 #N/mm**2 #Shear stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#If the Load W is in KN/m\n",

+      "\n",

+      "#Max shear Force\n",

+      "#F=w*l*2**-1 #KN\n",

+      "#After substituting values and further simplifying we get\n",

+      "#M=2*w #KN-m\n",

+      "\n",

+      "#Max Load from Consideration of moment\n",

+      "#M=1*6**-1*b*d**2*F_per\n",

+      "#After substituting values and further simplifying we get\n",

+      "w=(1*6**-1*b*d**2*F_per)*(2*10**6)**-1 #KN/m\n",

+      "\n",

+      "#Max Load from Consideration of shear stress\n",

+      "#q_max=1.5*F*(b*d)**-1 #N\n",

+      "#After substituting values and further simplifying we get\n",

+      "F=q_max*(1.5)*b*d #N\n",

+      "\n",

+      "#If w is Max Load in KN/m,then\n",

+      "#2*w*1000=8000\n",

+      "#After Rearranging and Further simplifying we get\n",

+      "w2=8000*(2*1000)**-1 #KN/m\n",

+      "\n",

+      "#Result\n",

+      "print\"Uniformly Distributed Load Beam can carry is\",round(w,2),\"KN/m\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Uniformly Distributed Load Beam can carry is 3.33 KN/m\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_1.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_1.ipynb
new file mode 100644
index 00000000..9684e21b
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_1.ipynb
@@ -0,0 +1,1002 @@
+{

+ "metadata": {

+  "name": "chapter no.5.ipynb"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 5:Deflections Of Beams By Double Integration Methods"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 5.5.2,Page No.192"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=3000 #mm #span of beam\n",

+      "a=2000 #mm\n",

+      "W1=20*10**3 #N #Pt Load Acting on beam\n",

+      "W2=30*10**3 #N #Pt Load Acting on beam\n",

+      "E=2*10**5 #N/mm**2 #Young's Modulus\n",

+      "I=2*10**8 #mm**4 #M.I\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Deflection at free End Due to W2\n",

+      "dell1=W2*L**3*(3*E*I)**-1 #mm\n",

+      "\n",

+      "#Deflection at free end Due to W1\n",

+      "dell2=W1*a**3*(3*E*I)**-1+(L-a)*W1*a**2*(2*E*I)**-1 #mm\n",

+      "\n",

+      "#Total Deflection at free end\n",

+      "dell=dell1+dell2 #mm\n",

+      "\n",

+      "#Result\n",

+      "print\"Deflection at Free End is\",round(dell,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Deflection at Free End is 9.08 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 5.5.4,Page No.193"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "E=2*10**5 #N/mm**2 #Young's Modulus\n",

+      "I=180*10**6 #mm**4 #M.I\n",

+      "W1=20 #N/m #u.d.l\n",

+      "W2=20*10**3 #N #Pt load\n",

+      "L=3000 #m #Span of beam\n",

+      "a=2000 #m #Span of u.d.l\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Displacement of free End due to 20 KN Pt load at free end\n",

+      "dell1=W2*L**3*(3*E*I)**-1 #mm\n",

+      "\n",

+      "#Displacement of free end due to u.d.l\n",

+      "dell2=W1*a**4*(8*E*I)**-1+(L-a)*W1*a**3*(6*E*I)**-1\n",

+      "\n",

+      "#Deflection at free end\n",

+      "dell=dell1+dell2 #mm\n",

+      "\n",

+      "#Result\n",

+      "print\"The Displacement of Free End of cantilever beam is\",round(dell,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Displacement of Free End of cantilever beam is 6.85 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 7

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 5.5.10,Page No.201"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "E=200*10**6 #KN/m**2  #Young's Modulus\n",

+      "I=15*10**-6 #m**4 #M.I\n",

+      "a=4000 #m \n",

+      "L_AB=6 #m #Span of beam\n",

+      "L_CB=2 #m #Length of CB\n",

+      "F_C=18 #KN #force at C\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let V_A & V_B be the Reactions at A & B Respectively\n",

+      "#V_A+V_B=18\n",

+      "#Now taking moment at B,we get M_B\n",

+      "V_A=(F_C*L_CB)*L_AB**-1\n",

+      "V_B=18-V_A\n",

+      "\n",

+      "#Now Taking Moment at distance x\n",

+      "#M_x=6*x-18*(x-4)\n",

+      "#EI*d**2*y*(d*x**2)**-1=6*x-18*(x-4)\n",

+      "\n",

+      "#Now Integrating above equation,we get\n",

+      "#EI*dy*(dx)**-1=C1+3*x**2-9(x-4)**4\n",

+      "\n",

+      "#Again Integrating above equation we get\n",

+      "#EI*y=C2+C1*x+x**3-3*(x-4)**3\n",

+      "\n",

+      "#The Boundary conditions\n",

+      "x=0\n",

+      "y=0  #.....(a)\n",

+      "\n",

+      "x=6\n",

+      "y=0   #....(b)\n",

+      "\n",

+      "#From Boundary Condition(B.C) a we get\n",

+      "C2=0\n",

+      "\n",

+      "#From Boundary Condition(B.C) b we get\n",

+      "#6*C1+216-3*8\n",

+      "#After Further simplifying we get\n",

+      "C1=-(216-24)*6**-1\n",

+      "\n",

+      "#EI*y=-32*x+x**3-3*(x-4)**3\n",

+      "#EI*dy*(dx)**-1=-32+3*x**2-9(x-4)**4\n",

+      "\n",

+      "#For Max Deflection\n",

+      "#Assume it inthe Porion AC i.e x=4=a\n",

+      "#0=-32+3*x**2\n",

+      "x=(32*3**-1)**0.5\n",

+      "\n",

+      "#Value of Max deflection is\n",

+      "ymax=(-32*x+x**3)*(E*I)**-1 #mm\n",

+      "\n",

+      "#slope at mid-span\n",

+      "\n",

+      "#EI*(dy*(dx)**-1)_centre=-32+3*x**2\n",

+      "#at centre ,\n",

+      "x1=3 #m\n",

+      "\n",

+      "#Let (dy*(dx)**-1)_centre=X\n",

+      "X=-(-32+3*x1**2)*(E*I)**-1 #Radian\n",

+      "\n",

+      "#Deflection at Load Point\n",

+      "x2=4 #m\n",

+      "#EI*y_c=-32*x2+x2**3\n",

+      "\n",

+      "y_c=-(-32*x2+x2**3)*(E*I)**-1\n",

+      "\n",

+      "\n",

+      "#Result\n",

+      "print\"Value of Max Deflection\",round(ymax,4),\"mm\"\n",

+      "print\"SLope at mid-span\",round(X,4),\"radian\"\n",

+      "print\"Deflection at the Load Point is\",round(y_c,4),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Value of Max Deflection -0.0232 mm\n",

+        "SLope at mid-span 0.0017 radian\n",

+        "Deflection at the Load Point is 0.0213 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 25

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 5.5.11,Page No.203"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L_CB=2 #m #Length of CB\n",

+      "L_AC=4 #m #Length of AB\n",

+      "M_C=15 #KN.m #Moment At Pt C\n",

+      "F_C=30 #KN\n",

+      "L=6 #m Span of Beam\n",

+      "\n",

+      "#Let X=E*I\n",

+      "X=10000 #KN-m**2\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let V_A and V_B be the reactions at A & B respectively\n",

+      "#V_A+V_B=30\n",

+      "\n",

+      "#Taking Moment a A,we get\n",

+      "V_B=(F_C*L_AC+M_C)*L**-1\n",

+      "V_A=30-V_B\n",

+      "\n",

+      "#Now Taking Moment at distacnce x from A\n",

+      "#M_x=7.5*x-30*(x-4)+15\n",

+      "\n",

+      "#By using Macaulay's Method\n",

+      "#EI*(d**2*x/dx**2)=M_x=7.5*x-30*(x-4)+15\n",

+      "\n",

+      "#Now Integrating above Equation we get\n",

+      "#EI*(dy/dx)=C1+7.5*x**2*2**-1-15*(x-4)**2+15*(x-4)  ............(1)\n",

+      "\n",

+      "#Again Integrating above Equation we get\n",

+      "#EIy=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1..........(2)\n",

+      "\n",

+      "#Boundary Cinditions\n",

+      "x=0\n",

+      "y=0\n",

+      "\n",

+      "#Substituting above equations we get \n",

+      "C2=0\n",

+      "\n",

+      "x=6 #m\n",

+      "y=0\n",

+      "\n",

+      "C1=-(7.5*6**3*6**-1-5*2**3+15*2**2*2**-1)*6**-1\n",

+      "\n",

+      "#EIy_c=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1\n",

+      "#Sub values in Above equation we get\n",

+      "y_c=(93.3333*(X)**-1)\n",

+      "\n",

+      "#Result\n",

+      "print\"The Deflection at C\",round(y_c,4),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Deflection at C 0.0093 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 10

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 5.5.12,Page No.204"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L_AC=L_CD=L_DB=2 #m #Length of AC,CD,DB\n",

+      "F_C=40 #KN #Force at C\n",

+      "w=20 #KN/m #u.d.l\n",

+      "L=6 #m #span of beam\n",

+      "\n",

+      "#Let E*I=X\n",

+      "X=15000 #KN-m**2\n",

+      "\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let V_A & V_B be the reactions at A & B respectively\n",

+      "#V_A+V_B=80\n",

+      "\n",

+      "#Taking Moment B,M_B\n",

+      "V_A=(F_C*(L_CD+L_DB)+w*L_DB*L_DB*2**-1)*L**-1 #KN\n",

+      "V_B=80-V_A #KN\n",

+      "\n",

+      "#Taking Moment at distance x from A\n",

+      "#M_x=33.333*x-40*(x-2)-20*(x-4)**2*2**-1\n",

+      "#EI*(d**2/dx**2)=33.333*x-40*(x-2)-10*(x-4)**2\n",

+      "\n",

+      "#Integrating above equation we get\n",

+      "#EI*(dy/dx)=C1+33.333*x**2*2**-1-20*(x-2)**2-10*3**-1*(x-4)**3\n",

+      "\n",

+      "#Again Integrating above equation we get\n",

+      "#EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4\n",

+      "\n",

+      "#At\n",

+      "x=0\n",

+      "y=0\n",

+      "C2=0\n",

+      "\n",

+      "#At\n",

+      "x=6\n",

+      "y=0\n",

+      "C1=-760*6**-1\n",

+      "\n",

+      "#Assuming Deflection to be max in portion CD and sustituting value of C1 in equation of slope we get\n",

+      "#EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4\n",

+      "#0=-126.667+33.333*x**2**-1-20*(x-2)**2\n",

+      "\n",

+      "#After rearranging and simplifying further we get\n",

+      "\n",

+      "#x**2-24*x+62=0\n",

+      "#From above equations\n",

+      "a=1\n",

+      "b=-24\n",

+      "c=62\n",

+      "\n",

+      "y=(b**2-4*a*c)**0.5\n",

+      "\n",

+      "x1=(-b+y)*(2*a)**-1\n",

+      "x2=(-b-y)*(2*a)**-1\n",

+      "\n",

+      "#Taking x2 into account\n",

+      "x=2.945 #m\n",

+      "C1=-126.667\n",

+      "C2=0\n",

+      "\n",

+      "y_max=(C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3)*X**-1 #mm\n",

+      "\n",

+      "#Max slope occurs at the ends\n",

+      "#At A,\n",

+      "#EI*(dy/dx)_A=-126.667\n",

+      "#At B\n",

+      "#EI*(dy/dx)_B=126.667+33.333*6**2*2**-1-20*4**2-10*2**3\n",

+      "#After simplifying Further we get\n",

+      "#EI*(dy/dx)_B=73.3273\n",

+      "\n",

+      "#Now Max slope is EI(dy/dx)_A=-126.667\n",

+      "#15000*(dy/dx)_=-126.667\n",

+      "\n",

+      "#Let Y=dy/dx\n",

+      "Y=-126.667*X**-1 #Radians\n",

+      "\n",

+      "#Result\n",

+      "print\"Maximum Deflection for Beam is\",round(y_max,4),\"mm\"\n",

+      "print\"Maximum Slope for beam is\",round(Y,4),\"radians\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Maximum Deflection for Beam is -0.0158 mm\n",

+        "Maximum Slope for beam is -0.0084 radians\n"

+       ]

+      }

+     ],

+     "prompt_number": 21

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 5.5.13,Page No.206"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "E=2*10**8 #KN/m**2\n",

+      "I=450*10**-6 #m**4\n",

+      "L_AC=1 #m #Length of AC\n",

+      "L_CD=3 #m #Length of CD\n",

+      "L_DB=2 #m #Length of DB\n",

+      "w=10 #KN/m #u.d.l\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let V_A & V_B be the reactions at A & B respectively\n",

+      "#V_A+V_B=30\n",

+      "\n",

+      "#Taking Moment at distance x from A\n",

+      "#M_x=17.5*x-10*(x-1)**2*2**-1+10*(x-4)**2*2**-1\n",

+      "#EI*(d**2/dx**2)=17.5*x-10*(x-1)**2*2**-1+10*(x-4)**2*2**-1\n",

+      "\n",

+      "#Now Integrating Above equation we get\n",

+      "#EI(dy/dx)=C1+17.5*x**2*2**-1-5*3**-1*(x-1)**2+5*3**-1*(x-4)**3\n",

+      "\n",

+      "#Again  Integrating Above equation we get\n",

+      "#EI*y=C2+C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4+5*12**-1*(x-4)**4\n",

+      "\n",

+      "#At \n",

+      "x=0\n",

+      "y=0\n",

+      "C2=0\n",

+      "\n",

+      "#At \n",

+      "x=6 \n",

+      "y=0\n",

+      "C1=(-17.5*x**3*6**-1+5*12**-1*(x-1)**4-5*12**-1*(x-4)**4)*x**-1\n",

+      "\n",

+      "# 1)Slope at A .i.e at x=0\n",

+      "#EI*(dy/dx)_A=C1=-62.708 #KN-m**2\n",

+      "#let (dy/dx)=X\n",

+      "X=C1*(E*I)**-1 #radiams\n",

+      "\n",

+      "#Deflection at mid-span\n",

+      "x=3 #m\n",

+      "#EI*y_centre=C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**2\n",

+      "y_centre=-(C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4)*(E*I)**-1\n",

+      "\n",

+      "#Maximum Deflection\n",

+      "\n",

+      "#At point of Max deflection (dy/dx)=0\n",

+      "#Assuming it in portion CD\n",

+      "\n",

+      "#0=C1*x+17.5*x**2*2**-1-5*3**-1*(x-1)**3\n",

+      "\n",

+      "#Now Let\n",

+      "#F(x)=C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3\n",

+      "\n",

+      "#Let F(x)=Y\n",

+      "#At \n",

+      "x=2.5\n",

+      "Y1=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",

+      "\n",

+      "#AT\n",

+      "x=3\n",

+      "Y2=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",

+      "\n",

+      "#At\n",

+      "x=2.9 #m\n",

+      "Y3=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",

+      "\n",

+      "#A curve may be plotted for (F(x) and the value for which F(x)=0 may be found\n",

+      "#For F(x)=0 for x=2.92 m\n",

+      "#Therefore y_max occur at x=2.92\n",

+      "\n",

+      "x=2.92 #m\n",

+      "y_max=(C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4)*(E*I)**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Slope at A\",round(X,6),\"mm\"\n",

+      "print\"Deflection at mid-span\",round(y_centre,6),\"mm\"\n",

+      "print\"Maxmimum Deflection is\",round(y_max,5),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Slope at A -0.000697 mm\n",

+        "Deflection at mid-span 0.001289 mm\n",

+        "Maxmimum Deflection is -0.00129 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 32

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 5.5.14,Page No.208"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L_AC=LDE=L_EB=1 #m #Length of AC\n",

+      "L_CD=2 #m #Length of CD\n",

+      "E=200 #KN/mm**2\n",

+      "I=60*10**6 #mm**4 #M.I\n",

+      "F_C=20 #KN #Force at C\n",

+      "F_E=30 #KN #Force at E\n",

+      "w=10 #KN/m #u.d.l\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "X=E*I*10**-6 #KN-m**2\n",

+      "\n",

+      "#Let V_A & V_B be the reactions at A & B respectively\n",

+      "#V_A+V_B=70\n",

+      "\n",

+      "#Taking Moment at distance x from A\n",

+      "#M_x=34*x-20*(x-1)-10*(x-1)**2*2**-1+10*(x-3)**2*2**-1-30*(x-4)\n",

+      "#EI*(d**2y/dx**2)=34*x-20*(x-1)-10*(x-1)**2*2**-1+10*(x-3)**2*2**-1-30*(x-4)\n",

+      "\n",

+      "#Now Integrating Above equation,we get\n",

+      "#EI*(dy/dx)=C1+17*x**2-10*(x-1)**2-5*3**-1*(x-1)**3+5*3**-1*(x-3)**3-15*(x-4)**2\n",

+      "\n",

+      "#Again Integrating Above equation,we get\n",

+      "#EI*y=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4-5*(x-4)**3\n",

+      "\n",

+      "#At\n",

+      "x=0\n",

+      "y=0\n",

+      "C2=0\n",

+      "\n",

+      "#At \n",

+      "x=5 #m\n",

+      "y=0\n",

+      "C1=(-17*3**-1*x**3+10*3**-1*(x-1)**3+5*12**-1*(x-1)**4-5*12**-1*(x-3)**4+5*(x-4)**3)*5**-1\n",

+      "\n",

+      "#EI*y=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4-5*(x-4)**3\n",

+      "C2=0\n",

+      "C1=-78\n",

+      "x=1\n",

+      "y_c=(-78*x+17*3**-1*x)*(X)**-1\n",

+      "\n",

+      "#EI*y_D=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4\n",

+      "x=3\n",

+      "C1-78\n",

+      "C2=0\n",

+      "y_D=(C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4)*(X**-1)\n",

+      "\n",

+      "#EI*y_E=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4\n",

+      "x=4\n",

+      "C1-78\n",

+      "C2=0\n",

+      "y_E=(C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4)*X**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Deflections at C\",round(y_c,5),\"mm\"\n",

+      "print\"Deflections at D\",round(y_D,5),\"mm\"\n",

+      "print\"Deflections at E\",round(y_E,4),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Deflections at C -0.00603 mm\n",

+        "Deflections at D -0.00953 mm\n",

+        "Deflections at E -0.0061 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 45

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 5.5.15,Page No.209"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "E=200 #KN/mm**2 #Modulus of Elasticity\n",

+      "I=300*10**6 #mm\n",

+      "L_AB=L_BC=L_CD=L_DE=1 #m #Length of AB,BC,CD,DE respectively\n",

+      "F_A=20 #KN #Force at A\n",

+      "F_C=10 #KN #Force at C\n",

+      "w=30 #KN/m #u.d.l\n",

+      "\n",

+      "#Let E*I=X\n",

+      "X=E*I*10**-6 #KN-2**2\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let V_E be the reactions at E\n",

+      "V_E=F_A+F_C+w*(L_BC+L_CD) #KN \n",

+      "\n",

+      "#Taking Moment at distance x\n",

+      "#EI*(d**2x/dy**2)=M=-20*x-30*(x-1)**2*2**-1-10*(x-2)+30*(x-3)**2*2**-1\n",

+      "\n",

+      "#Integrating above equation we get\n",

+      "#EI*(dy/dx)=C1-10*x**2-5*(x-1)**3-5*(x-2)**2+5*(x-3)**3\n",

+      "\n",

+      "#Again Integrating above equation\n",

+      "#EI*y=C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1-5*(x-3)**4*4**-1-5*3*(x-2)**3\n",

+      "\n",

+      "#At\n",

+      "#dy/dx=0\n",

+      "x=4 #m\n",

+      "C1=10*x**2+5*(x-1)**3+5*(x-2)**2-5*(x-3)**3\n",

+      "\n",

+      "#AT\n",

+      "x=4\n",

+      "y=0\n",

+      "C2=-C1*4+10*x**3*3**-1+5*(x-1)**4*4**-1-5*(x-3)**4*4**-1+5*3**-1*(x-2)**3\n",

+      "\n",

+      "#Max Deflection and Max slopes occurs at Free end in case of cantilever\n",

+      "y_max=y_A=C2*X**-1\n",

+      "\n",

+      "#EI*(dy/dx)_max=C1\n",

+      "#Let (dy/dx)=Y\n",

+      "Y=C1*X**-1 #radian\n",

+      "\n",

+      "#Now deflection at x=1 #m\n",

+      "C2=-913.333\n",

+      "C1=310\n",

+      "x=1\n",

+      "y_B=(C2+C1*x-10*x**3*3**-1)*X**-1\n",

+      "\n",

+      "#Now Deflection at x=2 #m\n",

+      "C2=-913.333\n",

+      "C1=310\n",

+      "x=2 #m\n",

+      "y_C=(C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1)*X**-1\n",

+      "\n",

+      "#Now Deflection at x=3 #m\n",

+      "C2=-913.333\n",

+      "C1=310\n",

+      "x=3 #m\n",

+      "y_D=(C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1-5*3**-1*(x-2)**3)*X**-1\n",

+      "\n",

+      "y_E=0\n",

+      "\n",

+      "#Result\n",

+      "print\"Max Deflection for Beam\",round(y_A,4),\"mm\"\n",

+      "print\"Max Slope for beam\",round(Y,5),\"radians\"\n",

+      "\n",

+      "#Plotting the ELastic Curve\n",

+      "\n",

+      "Y2=[y_E,y_D,y_C,y_B,y_A]\n",

+      "X2=[L_AB+L_BC+L_CD+L_DE,L_AB+L_BC+L_CD,L_AB+L_BC,L_AB,0]\n",

+      "Z2=[0,0,0,0,0]\n",

+      "plt.plot(X2,Y2,X2,Z2)\n",

+      "plt.xlabel(\"Length in mm\")\n",

+      "plt.ylabel(\"Deflection in mm\")\n",

+      "plt.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max Deflection for Beam -0.0152 mm\n",

+        "Max Slope for beam 0.00517 radians\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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8yjGZmZkkJSWRmZlJSkoKM2bMoLKy8pxilcZbsQJGj4Y//QkWL1aSEWnvaq3R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\nL+mlAAAAAElFTkSuQmCC\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5a29310>"

+       ]

+      }

+     ],

+     "prompt_number": 43

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 5.5.16,Page No.211"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L_BD=L_CB=L_AC=2 #m #Length of BD,CB,AC\n",

+      "F_C=40 #KN #Force at C\n",

+      "F_D=10 #KN Force at D\n",

+      "L=6 #m spna of beam\n",

+      "\n",

+      "#EI is constant in this problem\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let V_A & V_B be the reactions at A & B Respectively\n",

+      "#V_A+V_B=50\n",

+      "\n",

+      "#Taking Moment at Pt A\n",

+      "V_B=(F_D*L+F_C*L_AC)*(L_AC+L_CB)**-1\n",

+      "V_A=50-V_B\n",

+      "\n",

+      "#Now Taking Moment at distance x from A,M_x\n",

+      "#M_x=15*x-40*(x-2)+35*(x-4)\n",

+      "#EI*(d**2*y/dx**2)=15*x-40*(x-2)+35*(x-4)\n",

+      "\n",

+      "#Now Integrating above equation we get\n",

+      "#EI*(dy/dx)=C1+7.5*x**2-20*(x-2)**2+17.5(x-4)**2\n",

+      "\n",

+      "#Again Integrating above equation we get\n",

+      "#EI*y=C2+C1*x+2.5*x**2-20*3**-1*(x-2)**3+17.5*(x-4)**3*3**-1\n",

+      "\n",

+      "#At\n",

+      "x=0\n",

+      "y=0\n",

+      "#we get\n",

+      "C2=0\n",

+      "\n",

+      "#At\n",

+      "x=4 \n",

+      "y=0\n",

+      "#we get\n",

+      "C1=(2.5*4**3-20*3**-1*2**3)*4**-1\n",

+      "\n",

+      "#Now Deflection at C\n",

+      "x=2\n",

+      "C1=-26.667\n",

+      "C2=0\n",

+      "y_C=C2+C1*x+2.5*x**3\n",

+      "\n",

+      "#Now Deflection at D\n",

+      "C1=-21.667\n",

+      "C2=0\n",

+      "y_D=-26.667*6+2.5*6**3-20*3**-1*4**3+17.5*2**3*3**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Deflections Under Loads are:y_D\",round(y_D,4)\n",

+      "print\"                           :y_C\",round(y_C,2)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Deflections Under Loads are:y_D -0.002\n",

+        "                           :y_C -33.33\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 5.5.17,Page No.212"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L_BC=L_EB=2 #m #Length of BC & EB\n",

+      "E=200*10**6 #KN/m**2 #Modulus of eLasticity\n",

+      "I=45*10**-6 #mm**4 #M.I\n",

+      "L_DE=3 #m #Length of DE\n",

+      "L_AD=1 #m #Length of AD\n",

+      "w=20 #KN/m #u.d.l\n",

+      "L=8 #m #span of beam\n",

+      "F_C=30 #KN #Force at C\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let V_A & V_B be the reactions at A & B respectively\n",

+      "#V_A+V_B=90\n",

+      "\n",

+      "#Taking Moment at A,M_A\n",

+      "V_B=(w*L_DE*(L_DE*2**-1+L_AD)+F_C*L)*(L_AD+L_DE+L_EB)**-1\n",

+      "V_A=90-V_B\n",

+      "\n",

+      "#Taking Moment at distance x\n",

+      "#M_x=25*x-20*(x-1)**2*2**-1+20*(x-4)**2*2**-1+65*(x-6)\n",

+      "\n",

+      "#Integrating above equation we get\n",

+      "#EI*(d**2*y/dx**2)=25*x-10*(x-1)**2+10*(x-4)**2+65*(x-6)\n",

+      "\n",

+      "#again Integrating above equation we get\n",

+      "#EI*(dy/dx)=C1+25*x**2*2**-1-10*3**-1*(x-1)**3+10*3**-1*(x-4)**2+65*2**-1*(x-6)**2\n",

+      "\n",

+      "#again Integrating above equation we get\n",

+      "#EI*y=C2+C1*x+25*6**-1*x**3-10*12**-1*(x-1)**4+10*12**-1*(x-4)**4+65*6**-1*(x-6)**3\n",

+      "\n",

+      "x=0\n",

+      "y=0\n",

+      "#Sub these values in above equation,we get\n",

+      "C2=0\n",

+      "\n",

+      "x=6 #m\n",

+      "y=0\n",

+      "C1=-(25*6**-1*6**3-10*12**-1*5**4+10*12**-1*2**4)*6**-1\n",

+      "\n",

+      "#deflection at C is given by\n",

+      "x=8\n",

+      "y_c=(C2+C1*x+25*6**-1*x**3-10*12**-1*(x-1)**4+10*12**-1*(x-4)**4+65*6**-1*(x-6)**3)*(E*I)**-1\n",

+      "\n",

+      "#Assuming y is max in the portion DE,then\n",

+      "#(dy/dx)=0 for that point\n",

+      "\n",

+      "#0=-65.417+25*2**-1*x**2-10*3**-1*x(-1)**3\n",

+      "\n",

+      "#Let F(x)=-65.417+25*2**-1*x**2-10*3**-1*x(-1)**3\n",

+      "#Let z=F(x)\n",

+      "\n",

+      "#AT \n",

+      "x=3\n",

+      "z=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",

+      "\n",

+      "x=2.5\n",

+      "z1=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",

+      "\n",

+      "x=2.4\n",

+      "z2=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",

+      "\n",

+      "#The assumption is max in portion DE\n",

+      "x=2.46\n",

+      "y_max=(-65.417*x+25*6**-1*x**3-10*12**-1*1.46**4)*(E*I)**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Deflection at free end C\",round(y_c,4),\"mm\"\n",

+      "print\"Max Deflection between A and B\",round(y_max,4),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Deflection at free end C -0.0101 mm\n",

+        "Max Deflection between A and B -0.0114 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 1

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 5.5.18,Page No.213"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L_DB=L_AC=L_ED=2 #m #Length of DB & AC\n",

+      "L_CD=4 #m #Length of CD\n",

+      "L_CE=2 #m #Length of CE\n",

+      "F_A=40 #KN #Force at C\n",

+      "F_B=20 #KN #Force at A\n",

+      "E=200*10**6 #KN/mm**2 #Modulus of Elasticity\n",

+      "I=50*10**-6 #m**4 #M.I\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#LEt V_C & V_D be the reactions at C &  D respectively\n",

+      "#V_C+V_D=60\n",

+      "\n",

+      "#Taking Moment At D,M_D\n",

+      "V_C=-(-F_A*(L_AC+L_CE+L_ED)+F_B*L_DB)*L_CD**-1\n",

+      "V_D=60-V_C\n",

+      "\n",

+      "#Now Taking Moment at Distance x from A,\n",

+      "#M_x=-40*x+50*(x-2)+10*(x-6)\n",

+      "\n",

+      "#EI*(d**2*y/dx**2)=-40*x+50*(x-2)+10*(x-6)\n",

+      "\n",

+      "#Now Integrating above Equation we get\n",

+      "#EI*(dy/dx)=C1+20*x**2-25*(x-2)+5*(x-6)**2\n",

+      "\n",

+      "#Again Integrating above Equation we get\n",

+      "#EI*y=C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3\n",

+      "\n",

+      "#At\n",

+      "x=0\n",

+      "y=0\n",

+      "#C2+2*C1=-53.33    ...............(1)\n",

+      "\n",

+      "#At \n",

+      "x=6\n",

+      "y=0\n",

+      "#C2+6*C1=906.667    ...............(2)\n",

+      "\n",

+      "#Subtracting Equation 1 from 2 we get\n",

+      "C1=853.333*4**-1\n",

+      "C2=53.333-2*C1\n",

+      "x=0\n",

+      "y_A=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",

+      "\n",

+      "#Answer For y_A is incorrect in textbook\n",

+      "\n",

+      "#At Mid-span\n",

+      "C1=853.333*4**-1\n",

+      "C2=53.333-2*C1\n",

+      "x=4\n",

+      "y_E=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",

+      "\n",

+      "#Answer For y_E is incorrect in textbook\n",

+      "\n",

+      "#At B\n",

+      "C1=853.333*4**-1\n",

+      "C2=53.333-2*C1\n",

+      "x=8\n",

+      "y_B=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",

+      "\n",

+      "\n",

+      "#Result\n",

+      "print\"Deflection relative to the level of the supports:at End A\",round(y_A,4),\"mm\"\n",

+      "print\"                                                :at End B\",round(y_B,4),\"mm\"\n",

+      "print\"                                                :at Centre of CD\",round(y_E,4),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Deflection relative to the level of the supports:at End A -0.08 mm\n",

+        "                                                :at End B -0.0267 mm\n",

+        "                                                :at Centre of CD 0.0107 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_1.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_1.ipynb
new file mode 100644
index 00000000..4657aa0b
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_1.ipynb
@@ -0,0 +1,1494 @@
+{

+ "metadata": {

+  "name": "chapter no.6.ipynb"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter No.6:Torsion"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.1,Page No.225"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=10000 #mm #Length of solid shaft\n",

+      "d=100 #mm #Diameter of shaft\n",

+      "n=150     #rpm\n",

+      "P=112.5*10**6 #N-mm/sec #Power Transmitted\n",

+      "G=82*10**3 #N/mm**2 #modulus of Rigidity\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "J=pi*d**4*(32)**-1  #mm**3 #Polar Modulus\n",

+      "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",

+      "\n",

+      "r=50 #mm #Radius\n",

+      "\n",

+      "q_s=T*r*J**-1 #N/mm**2 #Max shear stress intensity\n",

+      "Theta=T*L*(G*J)**-1 #angle of twist\n",

+      "\n",

+      "#Result\n",

+      "print\"Max shear stress intensity\",round(q_s,2),\"N/mm**2\"\n",

+      "print\"Angle of Twist\",round(Theta,3),\"radian\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max shear stress intensity 36.48 N/mm**2\n",

+        "Angle of Twist 0.089 radian\n"

+       ]

+      }

+     ],

+     "prompt_number": 4

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.2,Page No.226"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "P=440*10**6 #N-m/sec #Power transmitted\n",

+      "n=280 #rpm\n",

+      "theta=pi*180**-1 #radian #angle of twist\n",

+      "L=1000 #mm #Length of solid shaft\n",

+      "q_s=40 #N/mm**2 #Max torsional shear stress\n",

+      "G=84*10**3 #N/mm**2 #Modulus of rigidity\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#P=2*pi*n*T*(60)**-1  #Equation of Power transmitted\n",

+      "T=P*60*(2*pi*n)**-1 #N-mm #torsional moment\n",

+      "\n",

+      "#From Consideration of shear stress\n",

+      "d1=(T*16*(pi*40)**-1)**0.333333  \n",

+      "\n",

+      "#From Consideration of angle of twist\n",

+      "d2=(T*L*32*180*(pi*84*10**3*pi)**-1)**0.25\n",

+      "\n",

+      "#result\n",

+      "print\"Diameter of solid shaft is\",round(d1,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Diameter of solid shaft is 124.09 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 21

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.3,Page No.227"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "G=80*10**3 #N/mm**2 #Modulus of rigidity\n",

+      "q_s=80 #N/mm**2 #Max sheare stress\n",

+      "P=736*10**6 #N-mm/sec #Power transmitted\n",

+      "n=200\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",

+      "\n",

+      "#Now From consideration of angle of twist\n",

+      "theta=pi*180**-1\n",

+      "#L=15*d\n",

+      "\n",

+      "d=(T*32*180*15*(pi**2*G)**-1)**0.33333\n",

+      "\n",

+      "#Now corresponding stress at the surface is\n",

+      "q_s2=T*32*d*(pi*2*d**4)**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Max diameter required is\",round(d,2),\"mm\"\n",

+      "print\"Corresponding shear stress is\",round(q_s2,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max diameter required is 156.66 mm\n",

+        "Corresponding shear stress is 46.55 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 27

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.4,Page No.228"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=25 #mm #Diameter of steel bar\n",

+      "p=50*10**3 #N #Pull\n",

+      "dell_1=0.095 #mm #Extension of bar\n",

+      "l=200 #mm #Guage Length\n",

+      "T=200*10**3 #N-mm #Torsional moment\n",

+      "theta=0.9*pi*180**-1 #angle of twist\n",

+      "L=250 #mm Length of steel bar\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "A=pi*4**-1*d**2 #Area of steel bar #mm**2\n",

+      "E=p*l*(dell_1*A)**-1 #N/mm**2 #Modulus of elasticity \n",

+      "\n",

+      "J=pi*32**-1*d**4 #mm**4 #Polar modulus\n",

+      "\n",

+      "G=T*L*(theta*J)**-1 #Modulus of rigidity #N/mm**2\n",

+      "\n",

+      "#Now from the relation of Elastic constants\n",

+      "mu=E*(2*G)**-1-1\n",

+      "\n",

+      "#result\n",

+      "print\"The Poissoin's ratio is\",round(mu,3)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Poissoin's ratio is 0.292\n"

+       ]

+      }

+     ],

+     "prompt_number": 30

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.5,Page No.229"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=6000 #mm #Length of circular shaft\n",

+      "d1=100 #mm #Outer Diameter\n",

+      "d2=75 #mm #Inner Diameter\n",

+      "R=100*2**-1 #Radius of shaft\n",

+      "T=10*10**6 #N-mm #Torsional moment\n",

+      "G=80*10**3 #N/mm**2 #Modulus of Rigidity\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n",

+      "\n",

+      "#Max Shear stress produced\n",

+      "q_s=T*R*J**-1 #N/mm**2\n",

+      "\n",

+      "#Angle of twist\n",

+      "theta=T*L*(G*J)**-1 #Radian\n",

+      "\n",

+      "#Result\n",

+      "print\"MAx shear stress produced is\",round(q_s,2),\"N/mm**2\"\n",

+      "print\"Angle of Twist is\",round(theta,2),\"Radian\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "MAx shear stress produced is 74.5 N/mm**2\n",

+        "Angle of Twist is 0.11 Radian\n"

+       ]

+      }

+     ],

+     "prompt_number": 1

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.6,Page No.229"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d1=200 #mm #External Diameter of shaft\n",

+      "t=25 #mm #Thickness of shaft\n",

+      "n=200 #rpm\n",

+      "theta=0.5*pi*180**-1 #Radian #angle of twist\n",

+      "L=2000 #mm #Length of shaft\n",

+      "G=84*10**3 #N/mm**2\n",

+      "d2=d1-2*t #mm #Internal Diameter of shaft\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n",

+      "\n",

+      "#Torsional moment\n",

+      "T=G*J*theta*L**-1 #N/mm**2 \n",

+      "\n",

+      "#Power Transmitted\n",

+      "P=2*pi*n*T*60**-1*10**-6 #N-mm\n",

+      "\n",

+      "#Max shear stress transmitted\n",

+      "q_s=G*theta*(d1*2**-1)*L**-1 #N/mm**2 \n",

+      "\n",

+      "#Result\n",

+      "print\"Power Transmitted is\",round(P,2),\"N-mm\"\n",

+      "print\"Max Shear stress produced is\",round(q_s,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Power Transmitted is 824.28 N-mm\n",

+        "Max Shear stress produced is 36.65 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 14

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.7,Page No.230"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "P=3750*10**6 #N-mm/sec\n",

+      "n=240 #Rpm\n",

+      "q_s=160 #N/mm**2 #Max shear stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#d2=0.8*d2 #mm #Internal Diameter of shaft\n",

+      "\n",

+      "#J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar modulus\n",

+      "#After substituting value in above Equation we get\n",

+      "#J=0.05796*d1**4\n",

+      "\n",

+      "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",

+      "\n",

+      "#Now from Torsion Formula\n",

+      "#T*J**-1=q_s*R**-1    ......................................(1)\n",

+      "\n",

+      "#But R=d1*2**-1 \n",

+      "\n",

+      "#Now substituting value of R and J in Equation (1) we get\n",

+      "d1=(T*(0.05796*q_s*2)**-1)**0.33333\n",

+      "\n",

+      "d2=d1*0.8\n",

+      "\n",

+      "#Result\n",

+      "print\"The size of the Shaft is:d1\",round(d1,3),\"mm\"\n",

+      "print\"                        :d2\",round(d2,3),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The size of the Shaft is:d1 200.362 mm\n",

+        "                        :d2 160.289 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 22

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.8,Page No.231"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "P=245*10**6 #N-mm/sec #Power transmitted\n",

+      "n=240 #rpm\n",

+      "q_s=40 #N/mm**2 #Shear stress\n",

+      "theta=pi*180**-1 #radian #Angle of twist\n",

+      "L=1000 #mm #Length of shaft\n",

+      "G=80*10**3 #N/mm**2\n",

+      "\n",

+      "#Tmax=1.5*T\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "T=P*60*(2*pi*n)**-1 #N-mm #Torsional Moment\n",

+      "Tmax=1.5*T\n",

+      "\n",

+      "#Now For Solid shaft\n",

+      "#J=pi*32*d**4\n",

+      "\n",

+      "#Now from the consideration of shear stress we get\n",

+      "#T*J**-1=q_s*(d*2**-1)**-1\n",

+      "#After substituting value in above Equation we get\n",

+      "#T=pi*16**-1*d**3*q_s\n",

+      "\n",

+      "#Designing For max Torque\n",

+      "d=(Tmax*16*(pi*40)**-1)**0.33333 #mm #Diameter of shaft\n",

+      "\n",

+      "#For max Angle of Twist\n",

+      "#Tmax*J**-1=G*theta*L**-1 \n",

+      "#After substituting value in above Equation we get\n",

+      "d2=(Tmax*32*180*L*(pi**2*G)**-1)**0.25\n",

+      "\n",

+      "#For Hollow Shaft\n",

+      "\n",

+      "#d1_2=Outer Diameter\n",

+      "#d2_2=Inner Diameter\n",

+      "\n",

+      "#d2_2=0.5*d1_2\n",

+      "\n",

+      "# Polar modulus\n",

+      "#J=pi*32**-1*(d1_2**4-d2_2**4)\n",

+      "#After substituting values we get\n",

+      "#J=0.092038*d1_2**4\n",

+      "\n",

+      "#Now from the consideration of stress\n",

+      "#Tmax*J**-1=q_s*(d1_2*2**-1)**-1\n",

+      "#After substituting values and further simplifying we get\n",

+      "d1_2=(Tmax*(0.092038*2*q_s)**-1)**0.33333\n",

+      "\n",

+      "#Now from the consideration of angle of twist\n",

+      "#Tmax*J**-1=G*theta*L**-1\n",

+      "#After substituting values and further simplifying we get\n",

+      "d1_3=(Tmax*180*L*(0.092038*G*pi)**-1)**0.25\n",

+      "\n",

+      "d2_2=0.5*d1_2\n",

+      "\n",

+      "#result\n",

+      "print\"Diameter of shaft is:For solid shaft:d\",round(d,2),\"mm\"\n",

+      "print\"                    :For Hollow shaft:d1_2\",round(d1_2,3),\"mm\"\n",

+      "print\"                    :                :d2_2\",round(d2_2,3),\"mm\"             "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Diameter of shaft is:For solid shaft:d 123.01 mm\n",

+        "                    :For Hollow shaft:d1_2 125.69 mm\n",

+        "                    :                :d2_2 62.845 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 7

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.11,Page No.235"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "P=250*10**6 #N-mm/sec #Power transmitted\n",

+      "n=100 #rpm\n",

+      "q_s=75 #N/mm**2 #Shear stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#From Equation of Power we have\n",

+      "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",

+      "\n",

+      "#Now from torsional moment equation we have\n",

+      "#T=j*q_s*(d/2**-1)**-1\n",

+      "#After substituting values in above equation and further simplifying we get\n",

+      "#T=pi*16**-1**d**3*q_s\n",

+      "d=(T*16*(pi*q_s)**-1)**0.3333 #mm #Diameter of solid shaft\n",

+      "\n",

+      "#PArt-2\n",

+      "\n",

+      "#Let d1 and d2 be the outer and inner diameter of hollow shaft\n",

+      "#d2=0.6*d1\n",

+      "\n",

+      "#Again from torsional moment equation we have\n",

+      "#T=pi*32**-1*(d1**4-d2**4)*q_s*(d1/2)**-1\n",

+      "d1=(T*16*(pi*(1-0.6**4)*q_s)**-1)**0.33333\n",

+      "d2=0.6*d1\n",

+      "\n",

+      "#Cross sectional area of solid shaft\n",

+      "A1=pi*4**-1*d**2 #mm**2\n",

+      "\n",

+      "#cross sectional area of hollow shaft\n",

+      "A2=pi*4**-1*(d1**2-d2**2)\n",

+      "\n",

+      "#Now percentage saving in weight\n",

+      "#Let W be the percentage saving in weight\n",

+      "W=(A1-A2)*100*A1**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Percentage saving in Weight is\",round(W,3),\"%\"\n",

+      "print\"Size of shaft is:solid shaft:d\",round(d,3),\"mm\"\n",

+      "print\"               :Hollow shaft:d1\",round(d1,3),\"mm\"\n",

+      "print\"               :            :d2\",round(d2,3),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Percentage saving in Weight is 29.735 %\n",

+        "Size of shaft is:solid shaft:d 117.418 mm\n",

+        "               :Hollow shaft:d1 123.031 mm\n",

+        "               :            :d2 73.818 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.12,Page No.237"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "d=100 #mm #Diameter of solid shaft\n",

+      "d1=100 #mm #Outer Diameter of hollow shaft\n",

+      "d2=50  #mm #Inner Diameter of hollow shaft\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Torsional moment of solid shaft\n",

+      "#T_s=J*q_s*(d*2**-1)**-1 \n",

+      "#After substituting values in above equation and further simplifying we get\n",

+      "#T_s=pi*16*d**3*q_s ...............(1)\n",

+      "\n",

+      "#torsional moment for hollow shaft is\n",

+      "#T_h=J*q_s*(d1**4-d2**4)**-1*(d1*2**-1)\n",

+      "#After substituting values in above equation and further simplifying we get\n",

+      "#T_h=pi*32**-1*2*d1**-1*(d1**4-d2**4)*q_s  ...........(2)\n",

+      "\n",

+      "#Dividing Equation 2 by 1 we get\n",

+      "#Let the ratio of T_h*T_s**-1 Be X\n",

+      "X=1-0.5**4\n",

+      "\n",

+      "#Loss in strength \n",

+      "#Let s be the loss in strength\n",

+      "#s=T_s*T_h*100*T_s**-1\n",

+      "#After substituting values in above equation and further simplifying we get\n",

+      "s=(1-0.9375)*100\n",

+      "\n",

+      "#Weight Ratio \n",

+      "#Let w be the Weight ratio\n",

+      "#w=W_h*W_s**-1\n",

+      "\n",

+      "A_h=pi*32**-1*(d1**2-d2**2) #mm**2 #Area of Hollow shaft\n",

+      "A_s=pi*32**-1*d**2 #mm**2 #Area of solid shaft\n",

+      "\n",

+      "w=A_h*A_s**-1 \n",

+      "\n",

+      "#Result\n",

+      "print\"Loss in strength is\",round(s,2)\n",

+      "print\"Weight ratio is\",round(w,2)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Loss in strength is 6.25\n",

+        "Weight ratio is 0.75\n"

+       ]

+      }

+     ],

+     "prompt_number": 4

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.13,Page No.239"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "T=8 #KN-m #Torque \n",

+      "d=100 #mm #Diameter of portion AB\n",

+      "d1=100 #mm #External Diameter of Portion BC\n",

+      "d2=75 #mm #Internal Diameter of Portion BC\n",

+      "G=80 #KN/mm**2 #Modulus of Rigidity\n",

+      "L1=1500 #mm #Radial Distance of Portion AB\n",

+      "L2=2500 #mm #Radial Distance ofPortion BC\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "R=d*2**-1 #mm #Radius of shaft\n",

+      "\n",

+      "#For Portion AB,Polar Modulus\n",

+      "J1=pi*32**-1*d**4 #mm**4 \n",

+      "\n",

+      "#For Portion BC,Polar modulus \n",

+      "J2=pi*32**-1*(d1**4-d2**4) #mm**4\n",

+      "\n",

+      "#Now Max stress occurs in portion BC since max radial Distance is sme in both cases\n",

+      "q_max=T*J2**-1*R*10**6 #N/mm**2 \n",

+      "\n",

+      "#Let theta1 be the rotation in Portion AB and theta2 be the rotation in portion BC\n",

+      "theta1=T*L1*(G*J1)**-1 #Radians\n",

+      "theta2=T*L2*(G*J2)**-1 #Radians\n",

+      "\n",

+      "#Total Rotational at end C\n",

+      "theta=(theta1+theta2)*10**3 #Radians\n",

+      "\n",

+      "#Result\n",

+      "print\"Max stress induced is\",round(q_max,2),\"N/mm**2\"\n",

+      "print\"Angle of Twist is\",round(theta,3),\"radians\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max stress induced is 59.6 N/mm**2\n",

+        "Angle of Twist is 0.053 radians\n"

+       ]

+      }

+     ],

+     "prompt_number": 23

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.14,Page No.240"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "q_b=80 #N/mm**2 #Shear stress in Brass\n",

+      "q_s=100 #N/mm**2 #Shear stress in Steel\n",

+      "G_b=40*10**3 #N/mm**2 \n",

+      "G_s=80*10**3 \n",

+      "L_b=1000 #mm #Length of brass shaft\n",

+      "L_s=1200 #mm #Length of steel shaft\n",

+      "d1=80 #mm #Diameter of brass shaft\n",

+      "d2=60 #mm #Diameter of steel shaft\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Polar modulus of brass rod\n",

+      "J_b=pi*32**-1*d1**4 #mm**4 \n",

+      "\n",

+      "#Polar modulus of steel rod\n",

+      "J_s=pi*32**-1*d2**4 #mm**4\n",

+      "\n",

+      "#Considering bras Rod:AB\n",

+      "T1=J_b*q_b*(d1*2**-1)**-1 #N-mm \n",

+      "\n",

+      "#Considering Steel Rod:BC\n",

+      "T2=J_s*q_s*(d2*2**-1)**-1 #N-mm\n",

+      "\n",

+      "#Max Torque that can be applied\n",

+      "T2\n",

+      "\n",

+      "#Let theta_b and theta_s be the rotations in Brass and steel respectively\n",

+      "theta_b=T2*L_b*(G_b*J_b)**-1 #Radians\n",

+      "theta_s=T2*L_s*(G_s*J_s)**-1 #Radians\n",

+      "\n",

+      "theta=theta_b+theta_s #Radians #Rotation of free end\n",

+      "\n",

+      "#Result\n",

+      "print\"Total of free end is\",round(theta,3),\"Radians\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Total of free end is 0.076 Radians\n"

+       ]

+      }

+     ],

+     "prompt_number": 36

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.15,Page No.241"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "G=80*10**3 #N/mm**2 #Modulus of Rigidity\n",

+      "d1=100 #mm #Outer diameter of hollow shft\n",

+      "d2=80  #mm #Inner diameter of hollow shaft\n",

+      "d=80   #mm #diameter of Solid shaft\n",

+      "d3=60  #mm #diameter of Solid shaft having L=0.5m\n",

+      "L1=300 #mm #Length of Hollow shaft\n",

+      "L2=400 #mm #Length of solid shaft\n",

+      "L3=500 #mm #LEngth of solid shaft of diameter 60mm\n",

+      "T1=2*10**6 #N-mm #Torsion in Shaft AB\n",

+      "T2=1*10**6 #N-mm #Torsion in shaft BC\n",

+      "T3=1*10**6 #N-mm #Torsion in shaft CD\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Now Polar modulus of section AB\n",

+      "J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n",

+      "\n",

+      "#Polar modulus of section BC\n",

+      "J2=pi*32**-1*d**4 #mm**4\n",

+      "\n",

+      "#Polar modulus of section CD\n",

+      "J3=pi*32**-1*d3**4 #mm**4\n",

+      "\n",

+      "#Now angle of twist of AB\n",

+      "theta1=T1*L1*(G*J1)**-1 #radians\n",

+      "\n",

+      "#Angle of twist of BC\n",

+      "theta2=T2*L2*(G*J2)**-1 #radians\n",

+      "\n",

+      "#Angle of twist of CD\n",

+      "theta3=T3*L3*(G*J3)**-1 #radians\n",

+      "\n",

+      "#Angle of twist\n",

+      "theta=theta1-theta2+theta3 #Radians\n",

+      "\n",

+      "#Shear stress in AB From Torsion Equation\n",

+      "q_s1=T1*(d1*2**-1)*J1**-1 #N/mm**2 \n",

+      "\n",

+      "#Shear stress in BC\n",

+      "q_s2=T2*(d*2**-1)*J2**-1 #N/mm**2 \n",

+      "\n",

+      "#Shear stress in CD\n",

+      "q_s3=T3*(d3*2**-1)*J3**-1 #N-mm**2\n",

+      "\n",

+      "#As max shear stress occurs in portion CD,so consider CD\n",

+      "\n",

+      "#Result\n",

+      "print\"Angle of twist at free end is\",round(theta,5),\"Radian\"\n",

+      "print\"Max Shear stress\",round(q_s3,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Angle of twist at free end is 0.00496 Radian\n",

+        "Max Shear stress 23.58 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.16,Page No.242"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=1000 #mm #Length of bar\n",

+      "L1=600 #mm #Length of Bar AB\n",

+      "L2=400 #mm #Length of Bar BC\n",

+      "d1=60  #mm #Outer Diameter of bar BC\n",

+      "d2=30  #mm #Inner Diameter of bar BC\n",

+      "d=60   #mm #Diameter of bar AB\n",

+      "T=2*10**6 #N-mm #Total Torque\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Polar Modulus of Portion AB\n",

+      "J1=pi*32**-1*d**4 #mm*4\n",

+      "\n",

+      "#Polar Modulus of Portion BC\n",

+      "J2=pi*32**-1*(d1**4-d2**4) #mm**4\n",

+      "\n",

+      "#Let T1 be the torque resisted by bar AB and T2 be torque resisted by Bar BC\n",

+      "#Let theta1 and theta2 be the rotation of shaft in portion AB & BC\n",

+      "\n",

+      "#theta1=T1*L1*(G*J1)**-1 #radians\n",

+      "#After substituting values and further simplifying we get \n",

+      "#theta1=32*600*T1*(pi*60**4*G)**-1\n",

+      "\n",

+      "#theta2=T2*L*(J2*G)**-1 #Radians\n",

+      "#After substituting values and further simplifying we get \n",

+      "#theta2=32*400*T2*(pi*60**4*(1-0.5**4)*G)**-1 \n",

+      "\n",

+      "#Now For consistency of Deformation,theta1=theta2\n",

+      "#After substituting values and further simplifying we get \n",

+      "#T1=0.7111*T2  ..................................................(1)\n",

+      "\n",

+      "#But T1+T2=T=2*10**6   ...........................................(2)\n",

+      "#Substituting value of T1 in above equation\n",

+      "\n",

+      "T2=T*(0.7111+1)**-1\n",

+      "T1=0.71111*T2\n",

+      "\n",

+      "#Max stress in Portion AB\n",

+      "q_s1=T1*(d*2**-1)*(J1)**-1 #N/mm**2\n",

+      "\n",

+      "#Max stress in Portion BC\n",

+      "q_s2=T2*(d1*2**-1)*J2**-1 #N/mm**2 \n",

+      "\n",

+      "#Result\n",

+      "print\"Stresses Developed in Portion:AB\",round(q_s1,2),\"N/mm**2\"\n",

+      "print\"                             :BC\",round(q_s2,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stresses Developed in Portion:AB 19.6 N/mm**2\n",

+        "                             :BC 29.4 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 1

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.17,Page No.243"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d1=80 #mm #External Diameter of Brass tube\n",

+      "d2=50 #mm #Internal Diameter of Brass tube\n",

+      "d=50  #mm #Diameter of steel Tube\n",

+      "G_b=40*10**3 #N/mm**2 #Modulus of Rigidity of brass tube\n",

+      "G_s=80*10**3 #N/mm**2 #Modulus of rigidity of steel tube\n",

+      "T=6*10**6 #N-mm #Torque\n",

+      "L=2000 #mm #Length of Tube\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Polar Modulus of brass tube\n",

+      "J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n",

+      "\n",

+      "#Polar modulus of steel Tube\n",

+      "J2=pi*32**-1*d**4 #mm**4\n",

+      "\n",

+      "#Let T_s & T_b be the torque resisted by steel and brass respectively\n",

+      "#Then, T_b+T_s=T   ............................................(1)\n",

+      "\n",

+      "#Since the angle of twist will be the same\n",

+      "#Theta1=Theta2\n",

+      "#After substituting values and further simplifying we get \n",

+      "#Ts=0.360*Tb     ...........................................(2)\n",

+      "\n",

+      "#After substituting value of Ts in eqn 1 and further simplifying we get \n",

+      "T_b=T*(0.36+1)**-1 #N-mm\n",

+      "T_s=0.360*T_b\n",

+      "\n",

+      "#Let q_s and q_b be the max stress in steel and brass respectively\n",

+      "q_b=T_b*(d1*2**-1)*J1**-1 #N/mm**2\n",

+      "q_s=T_s*(d2*2**-1)*J2**-1 #N/mm**2\n",

+      "\n",

+      "#Since angle of twist in brass=angle of twist in steel\n",

+      "theta_s=T_s*L*(J2*G_s)**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Stresses Developed in Materials are:Brass\",round(q_b,2),\"N/mm**2\"\n",

+      "print\"                                   :Steel\",round(q_s,2),\"N/mm**2\"\n",

+      "print\"Angle of Twist in 2m Length\",round(theta_s,3),\"Radians\" "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stresses Developed in Materials are:Brass 51.79 N/mm**2\n",

+        "                                   :Steel 64.71 N/mm**2\n",

+        "Angle of Twist in 2m Length 0.065 Radians\n"

+       ]

+      }

+     ],

+     "prompt_number": 12

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.18,Page No.245"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d1=60 #mm #External Diameter of aluminium Tube\n",

+      "d2=40 #mm #Internal Diameter of aluminium Tube\n",

+      "d=40  #mm #Diameter of steel tube\n",

+      "q_a=60 #N/mm**2 #Permissible stress in aluminium\n",

+      "q_s=100 #N/mm**2 #Permissible stress in steel tube\n",

+      "G_a=27*10**3 #N/mm**2 \n",

+      "G_s=80*10**3 #N/mm**2 \n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Polar modulus of aluminium Tube\n",

+      "J_a=pi*32**-1*(d1**4-d2**4) #mm**4\n",

+      "\n",

+      "#Polar Modulus of steel Tube\n",

+      "J_s=pi*32**-1*d**4 #mm**4\n",

+      "\n",

+      "#Now the angle of twist of steel tube = angle of twist of aluminium tube\n",

+      "#T_s*L_s*(J_s*theta_s)**-1=T_a*L_a*(J_a*theta_a)**-1\n",

+      "#After substituting values in above Equation and Further simplifyin we get\n",

+      "#T_s=0.7293*T_a   .....................(1)\n",

+      "\n",

+      "#If steel Governs the resisting capacity\n",

+      "T_s1=q_s*J_s*(d*2**-1)**-1 #N-mm\n",

+      "T_a1=T_s1*0.7293**-1  #N-mm\n",

+      "T1=(T_s1+T_a1)*10**-6 #KN-m #Total Torque in steel Tube\n",

+      "\n",

+      "#If aluminium Governs the resisting capacity \n",

+      "T_a2=q_a*J_a*(d1*2**-1) #N-mm\n",

+      "T_s2=T_a2*0.7293 #N-mm\n",

+      "T2=(T_s2+T_a2)*10**-6 #KN-m #Total Torque in aluminium tube\n",

+      "\n",

+      "#Result\n",

+      "print\"Steel Governs the torque carrying capacity\",round(T1,2),\"KN-m\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Steel Governs the torque carrying capacity 2.98 KN-m\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.19,Page No.247"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "P=225*10**6 #N-mm/sec #Power Trasmitted\n",

+      "q_b=80 #N/mm**2 #Shear stress\n",

+      "n=200 #Rpm\n",

+      "q_k=100 #N/mm**2 #PErmissible stress in Keys\n",

+      "D=300 #mm #Diameter of bolt circle\n",

+      "L=150 #mm #Length of shear key\n",

+      "d=16 #mm #Diameterr of bolt\n",

+      "\n",

+      "#Calculations\n",

+      "T=60*P*(2*pi*n)**-1 #N-mm #Torque\n",

+      "\n",

+      "#Now From Torsion Formula\n",

+      "#T*J**-1=q_s*R**-1\n",

+      "#After substituting values we get\n",

+      "#T=pi*16*d**3*n\n",

+      "#After further simplifying we get\n",

+      "d1=(T*16*(pi*q_s)**-1)**0.33333\n",

+      "\n",

+      "#Let b be the width of shear Key\n",

+      "#T=q_k*L*b*R\n",

+      "#After simplifying further we get\n",

+      "b=T*(q_k*L*(d1*2**-1))**-1 #mm\n",

+      "\n",

+      "#Let n2 be the no. of bolts required at bolt circle of radius\n",

+      "R_b=D*2**-1 #mm \n",

+      "\n",

+      "n2=T*4*(q_b*pi*d**2*R_b)**-1\n",

+      "\n",

+      "#result\n",

+      "print\"Minimum no. of Bolts Required are\",round(n2,2)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Minimum no. of Bolts Required are 4.45\n"

+       ]

+      }

+     ],

+     "prompt_number": 26

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.20,Page No.250"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "T=2*10**6 #N-mm #Torque transmitted\n",

+      "G=80*10**3 #N/mm**2 #Modulus of rigidity\n",

+      "d1=40 #mm \n",

+      "d2=80 #mm\n",

+      "r1=20 #mm\n",

+      "r2=40 #mm\n",

+      "L=2000 #mm #Length of shaft\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Angle of twist \n",

+      "theta=2*T*L*(r1**2+r1*r2+r2**2)*(3*pi*G*r2**3*r1**3)**-1 #radians\n",

+      "\n",

+      "#If the shaft is treated as shaft of average Diameter\n",

+      "d_avg=(d1+d2)*2**-1 #mm\n",

+      "\n",

+      "theta1=T*L*(G*pi*32**-1*d_avg**4)**-1 #Radians\n",

+      "\n",

+      "#Percentage Error\n",

+      "#Let Percentage Error be E\n",

+      "X=theta-theta1\n",

+      "E=(X*theta**-1)*100 \n",

+      "\n",

+      "#Result\n",

+      "print\"Percentage Error is\",round(E,2),\"%\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Percentage Error is 32.28 %\n"

+       ]

+      }

+     ],

+     "prompt_number": 42

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.21,Page No.252"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "G=80*10**3 #N/mm**2 \n",

+      "P=1*10**9 #N-mm/sec #Power\n",

+      "n=300 \n",

+      "d1=150 #mm #Outer Diameter\n",

+      "d2=120 #mm #Inner Diameter\n",

+      "L=2000 #mm #Length of circular shaft\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "T=P*60*(2*pi*n)**-1 #N-mm\n",

+      "\n",

+      "#Polar Modulus \n",

+      "J=pi*32**-1*(d1**4-d2**4) #mm**4\n",

+      "\n",

+      "q_s=T*J**-1*(d1*2**-1) #N/mm**2 \n",

+      "\n",

+      "\n",

+      "#Strain ENergy\n",

+      "U=q_s**2*(4*G)**-1*pi*4**-1*(d1**2-d2**2)*L\n",

+      "\n",

+      "#Result\n",

+      "print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n",

+      "print\"Strain Energy stored in the shaft is\",round(U,2),\"N-mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max shear stress is 81.36 N/mm**2\n",

+        "Strain Energy stored in the shaft is 263181.37 N-mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 51

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.22,Page No.254"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=12 #mm #Diameter of helical spring\n",

+      "D=150 #mm #Mean Diameter\n",

+      "R=D*2**-1 #mm #Radius of helical spring\n",

+      "n=10 #no.of turns\n",

+      "G=80*10**3 #N/mm**2 \n",

+      "W=450 #N #Load\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Max shear stress \n",

+      "q_s=16*W*R*(pi*d**3)**-1 #N/mm**2\n",

+      "\n",

+      "#Strain Energy stored\n",

+      "U=32*W**2*R**3*n*(G*d**4)**-1 #N-mm\n",

+      "\n",

+      "#Deflection Produced\n",

+      "dell=64*W*R**3*n*(G*d**4)**-1 #mm\n",

+      "\n",

+      "#Stiffness Spring\n",

+      "k=W*dell**-1 #N/mm\n",

+      "\n",

+      "#Result\n",

+      "print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n",

+      "print\"Strain Energy stored is\",round(U,2),\"N-mm\"\n",

+      "print\"Deflection Produced is\",round(dell,2),\"mm\"\n",

+      "print\"Stiffness spring is\",round(k,2),\"N/mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max shear stress is 99.47 N/mm**2\n",

+        "Strain Energy stored is 16479.49 N-mm\n",

+        "Deflection Produced is 73.24 mm\n",

+        "Stiffness spring is 6.14 N/mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 53

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.23,Page No.255"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "K=5 #N/mm #Stiffness\n",

+      "L=100 #mm #Solid Length\n",

+      "q_s=60 #N/mm**2 #Max shear stress\n",

+      "W=200 #N #Max Load\n",

+      "G=80*10**3 #N/mm**2\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#K=W*dell**-1\n",

+      "#After substituting values and further simplifying we get\n",

+      "#d=0.004*R**3*n   ........(1) #mm #Diameter of wire\n",

+      "#n=L*d**-1         ........(2)\n",

+      "\n",

+      "#From Shearing stress\n",

+      "#q_s=16*W*R*(pi*d**3)**-1 \n",

+      "#After substituting values and further simplifying we get\n",

+      "#d**4=0.004*R**3*n    .................(4)\n",

+      "\n",

+      "#From Equation 1,2,3\n",

+      "#d**4=0.004*(0.0785*d**3)**3*100*d**-1\n",

+      "#after further simplifying we get\n",

+      "d=5168.101**0.25\n",

+      "n=100*d**-1\n",

+      "R=(d**4*(0.004*n)**-1)**0.3333\n",

+      "\n",

+      "#Result\n",

+      "print\"Diameter of Wire is\",round(d,2),\"mm\"\n",

+      "print\"No.of turns is\",round(n,2)\n",

+      "print\"Mean Radius of spring is\",round(R,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Diameter of Wire is 8.48 mm\n",

+        "No.fo turns is 11.79\n",

+        "Mean Radius of spring is 47.83 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 54

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.24,Page No.255"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "m=5*10**5 #Wagon Weighing\n",

+      "v=18*1000*36000**-1 \n",

+      "d=300 #mm #Diameter of Beffer springs\n",

+      "n=18 #no.of turns\n",

+      "G=80*10**3 #N/mm**2\n",

+      "dell=225\n",

+      "R=100 #mm #Mean Radius\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Energy of Wagon\n",

+      "E=m*v**2*(9.81*2)**-1 #N-mm\n",

+      "\n",

+      "#Load applied\n",

+      "W=dell*G*d**4*(64*R**3*n)**-1 #N \n",

+      "\n",

+      "#Energy each spring can absorb is\n",

+      "E2=W*dell*2**-1 #N-mm\n",

+      "\n",

+      "#No.of springs required to absorb energy of Wagon\n",

+      "n2=E*E2**-1 *10**7\n",

+      "\n",

+      "#Result\n",

+      "print\"No.of springs Required for Buffer is\",round(n2,2)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "No.of springs Required for Buffer is 4.47\n"

+       ]

+      }

+     ],

+     "prompt_number": 66

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.25,Page No.259"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "b=180 #mm #width of flange\n",

+      "d=10 #mm #Depth of flange\n",

+      "t=10 #mm #Thickness of flange\n",

+      "D=400 #mm #Overall Depth \n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "I_xx=1*12**-1*(b*D**3-(b-t)*(D-2*d)**3)\n",

+      "I_yy=1*12**-1*((D-2*d)*t**3+2*t*b**3)\n",

+      "\n",

+      "#If warping is neglected\n",

+      "J=I_xx+I_yy #mm**4\n",

+      "\n",

+      "#Since b/d>1.6,we get\n",

+      "J2=1*3**-1*d**3*b*(1-0.63*d*b**-1)*2+1*3**-1*t**3*(D-2*d)*(1-0.63*t*b**-1)\n",

+      "\n",

+      "#Over Estimation of torsional Rigidity would have been \n",

+      "T=J*J2**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Error in assessing torsional Rigidity if the warping is neglected is\",round(T,2)\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Error in assessing torsional Rigidity if the warping is neglected is 808.28\n"

+       ]

+      }

+     ],

+     "prompt_number": 68

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 6.6.26,Page No.261"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d1=100 #mm #Outer Diameter\n",

+      "d2=95 #mm #Inner Diameter\n",

+      "T=2*10**6 #N-mm #Torque\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus\n",

+      "\n",

+      "#Shear stress\n",

+      "q_max=T*J**-1*d1*2**-1 #N/mm**2 \n",

+      "\n",

+      "#Now theta*L**-1=T*(G*J)**-1\n",

+      "#After substituting values and further simplifying we get\n",

+      "#Let theta*L**-1=X\n",

+      "X=T*J**-1\n",

+      "\n",

+      "#Now Treating it as very thin walled tube\n",

+      "d=(d1+d2)*2**-1 #mm\n",

+      "\n",

+      "r=d*2**-1 \n",

+      "t=(d1-d2)*2**-1\n",

+      "q_max2=T*(2*pi*r**2*t)**-1 #N/mm**2\n",

+      "\n",

+      "X2=T*(2*pi*r**3*t)**-1 \n",

+      "\n",

+      "#Result\n",

+      "print\"When it is treated as hollow shaft:Max shear stress\",round(q_max,2),\"N/mm**2\"\n",

+      "print\"                                  :Angle of Twist per unit Length\",round(X,3)\n",

+      "print\"When it is very thin Walled Tube  :Max shear stress\",round(q_max2,2),\"N/mm**2\"\n",

+      "print\"                                  :Angle of twist per Unit Length\",round(X2,3)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "When it is treated as hollow shaft:Max shear stress 54.91 N/mm**2\n",

+        "                                  :Angle of Twist per unit Length 1.098\n",

+        "When it is very thin Walled Tube  :Max shear stress 53.57 N/mm**2\n",

+        "                                  :Angle of twist per Unit Length 1.099\n"

+       ]

+      }

+     ],

+     "prompt_number": 72

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_1.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_1.ipynb
new file mode 100644
index 00000000..d239a927
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_1.ipynb
@@ -0,0 +1,1309 @@
+{

+ "metadata": {

+  "name": "chapter no.7.ipynb"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter No.7:Compound Stresses And Strains"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 7.1,Page No.269"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "sigma1=30 #N/mm**2 #Stress in tension\n",

+      "d=20 #mm #Diameter \n",

+      "sigma2=90 #N/mm**2 #Max compressive stress\n",

+      "sigma3=25 #N/mm**2\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#In TEnsion\n",

+      "\n",

+      "#Corresponding stress in shear\n",

+      "P=sigma1*2**-1 #N/mm**2\n",

+      "\n",

+      "#Tensile force\n",

+      "F=pi*4**-1*d**2*sigma1\n",

+      "\n",

+      "#In Compression\n",

+      "\n",

+      "#Correspong shear stress\n",

+      "P2=sigma2*2**-1 #N/mm**2\n",

+      "\n",

+      "#Correspong compressive(axial) stress\n",

+      "p=2*sigma3 #N/mm**2 \n",

+      "\n",

+      "#Corresponding Compressive force\n",

+      "P3=p*pi*4**-1*d**2 #N\n",

+      "\n",

+      "#Result\n",

+      "print\"Failure Loads are:\",round(F,2),\"N\"\n",

+      "print\"                 :\",round(P3,2),\"N\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Failure Loads are: 9424.78 N\n",

+        "                 : 15707.96 N\n"

+       ]

+      }

+     ],

+     "prompt_number": 24

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.2,Page No.270"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=25 #mm #Diameter of circular bar\n",

+      "F=20*10**3 #N #Axial Force\n",

+      "theta=30 #Degree #angle \n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Axial stresses\n",

+      "p=F*(pi*4**-1*d**2)**-1 #N/mm**2\n",

+      "\n",

+      "#Normal Stress\n",

+      "p_n=p*(cos(30*pi*180**-1))**2\n",

+      "\n",

+      "#Tangential Stress\n",

+      "p_t=p*2**-1*sin(2*theta*pi*180**-1)\n",

+      "\n",

+      "#Max shear stress occurs on plane where theta2=45 \n",

+      "theta2=45\n",

+      "sigma_max=p*2**-1*sin(2*theta2*pi*180**-1)\n",

+      "\n",

+      "#Result\n",

+      "print\"Stresses developed on a plane making 30 degree is:\",round(p_n,2),\"N/mm**2\"\n",

+      "print\"                                                 :\",round(p_t,2),\"N/mm**2\"\n",

+      "print\"stress on max shear stress is\",round(sigma_max,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stresses developed on a plane making 30 degree is: 30.56 N/mm**2\n",

+        "                                                 : 17.64 N/mm**2\n",

+        "stress on max shear stress is 20.37 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 38

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.3,Page No.272"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "theta=30 #degree\n",

+      "\n",

+      "#Stresses acting on material\n",

+      "p1=120 #N/mm**2\n",

+      "p2=80 #N/mm**2\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Normal Stress\n",

+      "P_n=(p1+p2)*2**-1+(p1-p2)*2**-1*cos(2*theta*pi*180**-1) #N/mm**2\n",

+      "\n",

+      "#Tangential stress\n",

+      "P_t=(p1-p2)*2**-1*sin(2*theta*pi*180**-1)\n",

+      "\n",

+      "#Resultant stress\n",

+      "P=(P_n**2+P_t**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Inclination to the plane\n",

+      "phi=arctan(P_n*P_t**-1)*(180*pi**-1)\n",

+      "\n",

+      "#Angle made by resultant with 120 #N/mm**2 stress\n",

+      "phi2=phi+theta #Degree\n",

+      "\n",

+      "#Result\n",

+      "print\"Normal Stress is\",round(P_n,2),\"N/mm**2\"\n",

+      "print\"Tangential Stress is\",round(P_t,2),\"N/mm**2\"\n",

+      "print\"Angle made by resultant\",round(phi2,2),\"Degree\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Normal Stress is 110.0 N/mm**2\n",

+        "Tangential Stress is 17.32 N/mm**2\n",

+        "Angle made by resultant 111.05 Degree\n"

+       ]

+      }

+     ],

+     "prompt_number": 30

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.4,Page No.272"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Direct Stresses\n",

+      "P1=60 #N/mm**2 \n",

+      "P2=100 #N/mm**2\n",

+      "\n",

+      "Theta=25 #Degree #Angle\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Normal Stress\n",

+      "P_n=(P1-P2)*2**-1+(P1+P2)*2**-1*cos(2*Theta*pi*180**-1) #N/mm**2\n",

+      "\n",

+      "#Tangential Stress\n",

+      "P_t=(P1+P2)*2**-1*sin(Theta*2*pi*180**-1) #N/mm**2\n",

+      "\n",

+      "#Resultant stress\n",

+      "P=(P_n**2+P_t**2)**0.5 #N/mm**2\n",

+      "\n",

+      "theta2=arctan(P_n*P_t**-1)*(180*pi**-1)\n",

+      "\n",

+      "#Result\n",

+      "print\"Stresses on the plane AC is:\",round(P_n,2),\"N/mm**2\"\n",

+      "print\"                            \",round(P_t,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stresses on the plane AC is: 31.42 N/mm**2\n",

+        "                             61.28 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 25

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.6,Page No.278"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Stresses acting on material\n",

+      "p_x=180 #N/mm**2 \n",

+      "p_y=120 #N/mm**2\n",

+      "\n",

+      "q=80 #N/mm**2\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1) #degrees\n",

+      "theta2=theta*2**-1 #Degrees\n",

+      "theta3=theta+180 #Degrees\n",

+      "theta4=theta3*2**-1 #Degrees\n",

+      "\n",

+      "#Stresses\n",

+      "p_1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "p_2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Max shear stress\n",

+      "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Result\n",

+      "print\"Magnitude of Principal stress is:\",round(p_1,2),\"N/mm**2\"\n",

+      "print\"                                 \",round(p_2,2),\"N/mm**2\"\n",

+      "print\"Magnitude of max shear stress is\",round(q_max,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Magnitude of Principal stress is: 235.44 N/mm**2\n",

+        "                                  64.56 N/mm**2\n",

+        "Magnitude of max shear stress is 85.44 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 40

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.7,Page No.279"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#stresses\n",

+      "p_x=60 #N/mm**2\n",

+      "p_y=-40 #N/mm**2\n",

+      "\n",

+      "q=10 #N/mm**2 #shear stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Principal Stresses\n",

+      "p1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Max shear stress\n",

+      "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Inclination of principal stress to plane\n",

+      "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",

+      "theta2=(theta)*2**-1 #degrees\n",

+      "\n",

+      "theta3=(theta+180)*2**-1  #degrees\n",

+      "\n",

+      "#Result\n",

+      "print\"Principal Stresses are:\",round(p1,2),\"N/mm**2\"\n",

+      "print\"                      :\",round(p2,2),\"N/mm**2\"\n",

+      "print\"Max shear stresses\",round(q_max,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Principal Stresses are: 60.99 N/mm**2\n",

+        "                      : -40.99 N/mm**2\n",

+        "Max shear stresses 50.99 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 16

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.8,Page No.280"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#stresses\n",

+      "p_x=-120 #N/mm**2\n",

+      "p_y=-80 #N/mm**2\n",

+      "\n",

+      "q=-60 #N/mm**2 #shear stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Principal Stresses\n",

+      "p1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Max shear stress\n",

+      "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Inclination of principal stress to plane\n",

+      "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",

+      "theta2=(theta)*2**-1 #degrees\n",

+      "\n",

+      "theta3=(theta+180)*2**-1  #degrees\n",

+      "\n",

+      "#Result\n",

+      "print\"Principal Stresses are:\",round(p1,2),\"N/mm**2\"\n",

+      "print\"                      :\",round(p2,2),\"N/mm**2\"\n",

+      "print\"Max shear stresses\",round(q_max,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Principal Stresses are: -36.75 N/mm**2\n",

+        "                      : -163.25 N/mm**2\n",

+        "Max shear stresses 63.25 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 17

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.9,Page No.282"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#stresses\n",

+      "p_x=-40 #N/mm**2\n",

+      "p_y=80 #N/mm**2\n",

+      "\n",

+      "q=48 #N/mm**2 #shear stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Max shear stress\n",

+      "q_max=((((p_x-p_y)*2**-1)**2)+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Inclination of principal stress to plane\n",

+      "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",

+      "theta2=(theta)*2**-1 #degrees\n",

+      "\n",

+      "theta3=(theta+180)*2**-1  #degrees\n",

+      "\n",

+      "#Normal Corresponding stress\n",

+      "p_n=(p_x+p_y)*2**-1+(p_x-p_y)*2**-1*cos(2*(theta2+45)*pi*180**-1)+q*sin(2*(theta2+45)*pi*180**-1) #Degrees\n",

+      "\n",

+      "#Resultant stress\n",

+      "p=((p_n**2+q_max**2)**0.5) #N/mm**2\n",

+      "\n",

+      "phi=arctan(p_n*q_max**-1)*(180*pi**-1) #Degrees\n",

+      "\n",

+      "#Inclination to the plane\n",

+      "alpha=round((theta2+45),2)+round(phi ,2)#Degree\n",

+      "\n",

+      "#Answer in book is incorrect of alpha ie41.25\n",

+      "\n",

+      "#Result\n",

+      "print\"Planes of max shear stress:\",round(p_n,2),\"N/mm**2\"\n",

+      "print\"                           \",round(q_max,2),\"N/mm*2\"\n",

+      "print\"Resultant Stress is\",round(p,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Planes of max shear stress: 20.0 N/mm**2\n",

+        "                            76.84 N/mm*2\n",

+        "Resultant Stress is 79.4 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 40

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.10,Page No.283"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Stresses\n",

+      "p_x=50*cos(35*pi*180**-1)\n",

+      "q=50*sin(35*pi*180**-1)\n",

+      "p_y=0\n",

+      "\n",

+      "theta=40 #Degrees #Plane AB inclined to vertical\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Normal Stress on AB\n",

+      "p_n=(p_x+p_y)*2**-1+(p_x-p_y)*2**-1*cos(2*theta*pi*180**-1)+q*sin(2*theta*pi*180**-1)\n",

+      "\n",

+      "#Tangential Stress on AB\n",

+      "p_t=(p_x-p_y)*2**-1*sin(2*theta*pi*180**-1)-q*cos(2*theta*pi*180**-1) #N/mm**2\n",

+      "\n",

+      "#Resultant stress\n",

+      "p=(p_n**2+p_t**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Angle of resultant\n",

+      "phi=arctan(p_n*p_t**-1)*(180*pi**-1) #degrees\n",

+      "phi2=phi+theta #Degrees\n",

+      "\n",

+      "#Result\n",

+      "print\"Magnitude of resultant stress is\",round(p,2),\"N/mm**2\"\n",

+      "print\"Direction of Resultant stress is\",round(phi2,2),\"Degrees\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Magnitude of resultant stress is 54.44 N/mm**2\n",

+        "Direction of Resultant stress is 113.8 Degrees\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.12,Page No.285"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Direct stresses\n",

+      "p_x=120 #N/mm**2 #Tensile stress\n",

+      "p_y=-100 #N/mm**2 #Compressive stress\n",

+      "p1=160 #N/mm**2 #Major principal stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let q be the shearing stress\n",

+      "\n",

+      "#p1=(p_x+p_y)*2**-1+((((p_x+p_y)*2**-1)**2)+q**2)**0.5\n",

+      "#After further simplifying we get\n",

+      "q=(p1-((p_x+p_y)*2**-1))**2-((p_x-p_y)*2**-1)**2 #N/mm**2\n",

+      "q2=(q)**0.5 #N/mm**2\n",

+      "\n",

+      "#Minimum Principal stress\n",

+      "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Max shearing stress\n",

+      "q_max=(((p_x-p_y)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Result\n",

+      "print\"Shearing stress of material\",round(q,2),\"N/mm**2\"\n",

+      "print\"Min Principal stress\",round(p2,2),\"N/mm**2\"\n",

+      "print\"Max shearing stress\",round(q_max,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Shearing stress of material 10400.0 N/mm**2\n",

+        "Min Principal stress -140.0 N/mm**2\n",

+        "Max shearing stress 150.0 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.14,Page No.291"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "F=40*10**3 #N #Shear Force\n",

+      "M=20*10**6 #Bending Moment\n",

+      "\n",

+      "#Rectangular section\n",

+      "b=100 #mm #Width\n",

+      "d=200 #mm #Depth\n",

+      "\n",

+      "x=20 #mm #Distance from Top surface upto point\n",

+      "y=80 #mm #Distance from point to Bottom\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "I=1*12**-1*b*d**3 #mm**4 #M.I\n",

+      "\n",

+      "#At 20 mm Below top Fibre\n",

+      "f_x=M*I**-1*y #N/mm**2 #Stress\n",

+      "\n",

+      "#Assuming sagging moment ,f_x is compressive p_x=f_x=-24 #N/mm**2\n",

+      "p_x=f_x=-24 #N/mm**2\n",

+      "\n",

+      "#Shearing stress\n",

+      "q=F*(b*I)**-1*(b*x*(b-x*2**-1)) #N/mm**2\n",

+      "\n",

+      "#Direct stresses\n",

+      "\n",

+      "p_y=0 #N/mm**2\n",

+      "\n",

+      "p1=(p_x+p_y)*2**-1+(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "p2=(p_x+p_y)*2**-1-(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Result\n",

+      "print\"Directions of principal stresses at a point below 20mm is:\",round(p1,2),\"N/mm**2\"\n",

+      "print\"                                                          \",round(p2,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Directions of principal stresses at a point below 20mm is: 0.05 N/mm**2\n",

+        "                                                           -24.05 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 36

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.15,Page No.292"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=4000 #mm #Span\n",

+      "W1=W2=W3=2*10**3 #N #Load\n",

+      "\n",

+      "#SEction of beam\n",

+      "b=100 #mm #Width\n",

+      "d=240 #mm #Dept\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let R_A and R_B be the reactions\n",

+      "R_A=R_B=(W1+W2+W3)*2**-1 #KN\n",

+      "\n",

+      "#Now at the section 1.5m from left support A\n",

+      "#Shear Force\n",

+      "F=R_A-W1 #KN\n",

+      "\n",

+      "#B.M\n",

+      "M=R_A*1.5-W1*0.5 #KN-m\n",

+      "\n",

+      "#M.I\n",

+      "I=1*12**-1*b*d**3 #mm**4\n",

+      "\n",

+      "#Bending stress\n",

+      "#f=M*I**-1*y\n",

+      "#After Sub values and further simplifying we get\n",

+      "#f=3.04*10**-2*y\n",

+      "\n",

+      "#As it varies Linearly\n",

+      "\n",

+      "#at distance 0 From NA \n",

+      "f1=0\n",

+      "#at distance 60 mm from NA\n",

+      "f2=1.823 #N/mm**2\n",

+      "#at distance 120 mm from NA\n",

+      "f3=3.646 #N/mm**2\n",

+      "\n",

+      "#Shearing stress\n",

+      "q=F*b*d*2**-1*d*4**-1*(b*I)**-1\n",

+      "\n",

+      "#At 60 mm above NA\n",

+      "q2=F*b*d*4**-1*(d*2**-1-d*8**-1)*(b*I)**-1\n",

+      "\n",

+      "#At 120 mm above NA\n",

+      "q3=0 \n",

+      "\n",

+      "#At NA element is under pure shear\n",

+      "p1=q #N/mm**2\n",

+      "p2=-q #N/mm**2 \n",

+      "\n",

+      "#Inclination of principal plane to vertical\n",

+      "#theta=2*q*0**-1\n",

+      "#Further simplifying we get\n",

+      "#theta=infinity\n",

+      "\n",

+      "#therefore\n",

+      "theta=90*2**-1 #degrees\n",

+      "theta2=270*2**-1 #degrees\n",

+      "\n",

+      "#At 60 mm From NA\n",

+      "p_x=-1.823 #N/mm**2 \n",

+      "p_y=0\n",

+      "q=0.0469 #N/mm**2\n",

+      "\n",

+      "#principal planes\n",

+      "P1=(p_x+p_y)*2**-1+(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "P2=(p_x+p_y)*2**-1-(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Principal planes inclination to hte plane of p_x is given by\n",

+      "theta3=(arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1))\n",

+      "theta4=theta3*2**-1#degrees\n",

+      "\n",

+      "theta5=theta3+180 #Degrees\n",

+      "\n",

+      "#At 120 mm From N-A\n",

+      "p_x2=3.646 #N/mm**2\n",

+      "p_y2=0 #N/mm**2\n",

+      "q2=0 #N/mm**2\n",

+      "\n",

+      "P3=p_x2 #N/mm**2\n",

+      "P4=0 #N/mm**2\n",

+      "\n",

+      "#Answer for P2 at 60 mm  from NA is incorrect\n",

+      "\n",

+      "#Result\n",

+      "print\"Principal Planes at 60 mm from NA:\",round(p_x,2),\"N/mm**2\"\n",

+      "print\"                                  \",round(p_y,2),\"N/mm**2\"\n",

+      "print\"Principal Stresses at 60 mm From NA\",round(P1,4),\"N/mm**2\"\n",

+      "print\"                                   \",round(P2,4),\"N/mm**2\"\n",

+      "print\"Principal Planes at 60 mm from NA:\",round(p_x2,4),\"N/mm**2\"\n",

+      "print\"                                  \",round(p_y2,4),\"N/mm**2\"\n",

+      "print\"Principal Stresses at 60 mm From NA\",round(P3,4),\"N/mm**2\"\n",

+      "print\"                                   \",round(P4,4),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Principal Planes at 60 mm from NA: -1.82 N/mm**2\n",

+        "                                   0.0 N/mm**2\n",

+        "Principal Stresses at 60 mm From NA 0.0012 N/mm**2\n",

+        "                                    -1.8242 N/mm**2\n",

+        "Principal Planes at 60 mm from NA: 3.646 N/mm**2\n",

+        "                                   0.0 N/mm**2\n",

+        "Principal Stresses at 60 mm From NA 3.646 N/mm**2\n",

+        "                                    0.0 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 23

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.16,Page No.295"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=8000 #mm #Span of beam\n",

+      "w=40*10**6 #N/mm #udl\n",

+      "\n",

+      "#I-section\n",

+      "\n",

+      "#Flanges\n",

+      "b=100 #mm #Width\n",

+      "t=10 #mm #Thickness\n",

+      "\n",

+      "D=400 #mm #Overall Depth\n",

+      "t2=10 #mm #thickness of web\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let R_A and R_B be the Reactions at A & B respectively\n",

+      "R_A=w*2**-1*L*10**-9 #KN\n",

+      "\n",

+      "#Shear force at 2m for left support\n",

+      "F=R_A-2*w*10**-6 #KN\n",

+      "\n",

+      "#Bending Moment\n",

+      "M=R_A*2-2*w*10**-6 #KN-m\n",

+      "\n",

+      "#M.I\n",

+      "I=1*12**-1*b*D**3-1*12**-1*(b-t)*(D-2*t2)**3 #mm**4\n",

+      "\n",

+      "#Bending stress at 100 mm above N_A\n",

+      "f=M*10**6*I**-1*b\n",

+      "\n",

+      "#Shear stress \n",

+      "q=F*10**3*(t*I)**-1*(b*t*(D-t)*2**-1 +t2*(b-t2)*145) #N/mm**2\n",

+      "\n",

+      "p_x=-197.06 #N/mm**2 \n",

+      "p_y=0 #N/mm**2\n",

+      "q=21.38 #N/mm**2\n",

+      "\n",

+      "#Principal Stresses\n",

+      "\n",

+      "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Max shear stress\n",

+      "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Result\n",

+      "print\"Principal Stresses are:\",round(P1,2),\"N/mm**2\"\n",

+      "print\"                       \",round(P2,2),\"N/mm**2\"\n",

+      "print\"Max shear stress\",round(q_max,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Principal Stresses are: 2.29 N/mm**2\n",

+        "                        -199.35 N/mm**2\n",

+        "Max shear stress 100.82 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 48

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.18,Page No.298"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=100 #mm #Diameter of shaft\n",

+      "M=3*10**6 #N-mm #B.M\n",

+      "T=6*10**6 #N-mm #Twisting Moment\n",

+      "mu=0.3\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Max principal Stress\n",

+      "\n",

+      "P1=16*(pi*d**3)**-1*(M+(M**2+T**2)**0.5) #N/mm**2 \n",

+      "P2=16*(pi*d**3)**-1*(M-(M**2+T**2)**0.5) #N/mm**2 \n",

+      "\n",

+      "#Direct stress\n",

+      "P=round(P1,2)-mu*round(P2,2) #N/mm**2 \n",

+      "\n",

+      "#Result\n",

+      "print\"Principal stresses are:\",round(P1,2),\"N/mm**2\"\n",

+      "print\"                      :\",round(P2,2),\"N/mm**2\"\n",

+      "print\"Stress Producing the same strain is\",round(P,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Principal stresses are: 49.44 N/mm**2\n",

+        "                      : -18.89 N/mm**2\n",

+        "Stress Producing the same strain is 55.11 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 51

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.19,Page No.299"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=75 #mm #diameter \n",

+      "P=30*10**6 #W #Power transmitted\n",

+      "W=6 #N-mm/sec #Load\n",

+      "L=1000 #mm \n",

+      "N=300 #r.p.m\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#B.M\n",

+      "M=W*L*4**-1 #N-mm\n",

+      "T=P*60*(2*pi*N)**-1 #Torque transmitted\n",

+      "\n",

+      "#M.I\n",

+      "I=pi*64**-1*d**4 #mm**4\n",

+      "\n",

+      "#Bending stress\n",

+      "f_A=M*I**-1*(d*2**-1) #N/mm**2\n",

+      "\n",

+      "#At A\n",

+      "p_x=f_A\n",

+      "p_y=0\n",

+      "\n",

+      "#Polar Modulus\n",

+      "J=pi*32**-1*d**4 #mm**4\n",

+      "\n",

+      "#Shearing stress\n",

+      "q=T*J**-1*(d*2**-1) #N/mm**2\n",

+      "\n",

+      "#Principal Stresses\n",

+      "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Max shear stress\n",

+      "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Bending stress\n",

+      "p_x2=0\n",

+      "p_y2=0\n",

+      "\n",

+      "#Shearing stress\n",

+      "q2=T*J**-1*d*2**-1 #N/mm**2\n",

+      "\n",

+      "#Principal stresses\n",

+      "P3=(p_x2+p_y2)*2**-1+(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",

+      "P4=(p_x2+p_y2)*2**-1-(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Max shear stress\n",

+      "q_max2=(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Answer for Principal Stresses P1,P2 and Max stress i.e q_max is incorrect in Book\n",

+      "\n",

+      "#Result\n",

+      "print\"Principal Stresses at vertical Diameter:P1\",round(P1,2),\"N/mm**2\"\n",

+      "print\"                                       :P2\",round(P2,2),\"N/mm**2\"\n",

+      "print\"Max stress at vertical Diameter        :  \",round(q_max,2),\"N/mm**2\"\n",

+      "print\"Principal Stresses at Horizontal Diameter:P3\",round(P3,2),\"N/mm**2\"\n",

+      "print\"                                       :P4\",round(P4,2),\"N/mm**2\"\n",

+      "print\"Max stress at Horizontal Diameter      :  \",round(q_max2,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Principal Stresses at vertical Diameter:P1 11.55 N/mm**2\n",

+        "                                       :P2 -11.51 N/mm**2\n",

+        "Max stress at vertical Diameter        :   11.53 N/mm**2\n",

+        "Principal Stresses at Horizontal Diameter:P3 11.53 N/mm**2\n",

+        "                                       :P4 -11.53 N/mm**2\n",

+        "Max stress at Horizontal Diameter      :   11.53 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 20

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.20,Page No.302"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d1=100 #mm #External Diameter\n",

+      "d2=50  #mm #Internal Diameter\n",

+      "N=500  #mm #r.p.m\n",

+      "P=60*10**6 #N-mm/sec #Power\n",

+      "p=100 #N/mm**2 #principal stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#M.I\n",

+      "I=pi*(d1**4-d2**4)*64**-1 #mm**4\n",

+      "\n",

+      "#Bending Stress\n",

+      "#f=M*I*d1*2**-1 #N/mm**2\n",

+      "\n",

+      "#Principal Planes\n",

+      "#p_x=32*M*(pi*(d1**4-d2**4))*d1\n",

+      "#p_y=0\n",

+      "\n",

+      "#Shear stress\n",

+      "#q=T*J**-1*(d1*2**-1)\n",

+      "#After sub values and further simplifying we get\n",

+      "#q=16*T*d1*(pi*(d1**4-d2**4))*d1\n",

+      "\n",

+      "#Principal stresses\n",

+      "#P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "#After sub values and further simplifying we get\n",

+      "#P1=16*(pi*(d1**4-d2**4))*d1*(M+(M**2+t**2)**0.5)  ...............(1)\n",

+      "\n",

+      "#P=2*pi*N*T*60**-1\n",

+      "#After sub values and further simplifying we get\n",

+      "T=P*60*(2*pi*N)**-1*10**-6 #N-mm\n",

+      "\n",

+      "#Again Sub values and further simplifying Equation 1 we get\n",

+      "M=(337.533)*(36.84)**-1 #KN-m\n",

+      "\n",

+      "#Min Principal stress\n",

+      "#P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "#Sub values and further simplifying we get\n",

+      "P2=16*(pi*(d1**4-d2**4))*d1*(M-(M**2+T**2)**0.5)*10**-11\n",

+      "\n",

+      "#Result\n",

+      "print\"Bending Moment safely applied to shaft is\",round(M,2),\"KN-m\"\n",

+      "print\"Min Principal Stress is\",round(P2,3),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Bending Moment safely applied to shaft is 9.16 KN-m\n",

+        "Min Principal Stress is -0.336 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 32

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.21,Page No.303"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=150 #mm #Diameter\n",

+      "T=20*10**6 #N #Torque\n",

+      "M=12*10**6 #N-mm #B.M\n",

+      "F=200*10**3 #N #Axial Thrust\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#M.I\n",

+      "I=(pi*64**-1*d**4)\n",

+      "\n",

+      "#Bending stress \n",

+      "f_A=M*I**-1*(d*2**-1) #N/mm**2\n",

+      "f_B=-f_A #N/mm**2\n",

+      "\n",

+      "#Axial thrust due to thrust\n",

+      "sigma=F*(pi*4**-1*d**2)**-1\n",

+      "\n",

+      "#At A\n",

+      "p_x=f_A-sigma #N/mm**2\n",

+      "\n",

+      "#At B\n",

+      "p_x2=f_B-sigma #N/mm**2\n",

+      "\n",

+      "p_y=0 #At A and B\n",

+      "\n",

+      "#Polar Modulus\n",

+      "J=pi*32**-1*d**4 #mm**4\n",

+      "\n",

+      "#Shearing stress at A and B\n",

+      "q=T*J**-1*(d*2**-1) #N/mm**2\n",

+      "\n",

+      "\n",

+      "#Principal Stresses\n",

+      "#At A\n",

+      "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Max shear stress\n",

+      "q_max1=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#At B\n",

+      "P1_2=(p_x2+p_y)*2**-1+(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "P2_2=(p_x2+p_y)*2**-1-(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "#Max shear stress\n",

+      "q_max2=(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",

+      "\n",

+      "\n",

+      "#Result\n",

+      "print\"MAx Principal Stresses:P1\",round(P1,2),\"N/mm**2\"\n",

+      "print\"                      :P2\",round(P2,2),\"N/mm**2\"\n",

+      "print\"Min Principal Stresses:P1_2\",round(P1_2,2),\"N/mm**2\"\n",

+      "print\"                      :P2_2\",round(P2_2,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "MAx Principal Stresses:P1 45.1 N/mm**2\n",

+        "                      :P2 -20.2 N/mm**2\n",

+        "Min Principal Stresses:P1_2 14.65 N/mm**2\n",

+        "                      :P2_2 -62.18 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.22,Page No.311"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#strains\n",

+      "e_A=500 #microns\n",

+      "e_B=250 #microns\n",

+      "e_C=-150 #microns\n",

+      "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",

+      "mu=0.3 #Poissoin's ratio\n",

+      "theta=45 #Degrees\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "e_x=e_A=500\n",

+      "e_45=e_B=250\n",

+      "e_y=e_C=-150 \n",

+      "\n",

+      "#e_45=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(2*theta)+rho_x_y*2**-1*sin(2*theta)\n",

+      "#After sub values and further simplifying we get\n",

+      "rho_x_y=(e_45-(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*cos(2*theta*pi*180**-1))*(sin(2*theta*pi*180**-1))**-1*2\n",

+      "\n",

+      "#Principal strains are given by\n",

+      "e1=(e_x+e_y)*2**-1+(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",

+      "e2=(e_x+e_y)*2**-1-(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",

+      "\n",

+      "#Principal Stresses\n",

+      "sigma1=E*(e1+mu*e2)*(1-mu**2)**-1*10**-6 #N/mm**2\n",

+      "sigma2=E*(e2+mu*e1)*(1-mu**2)**-1*10**-6 #N/mm**2\n",

+      "\n",

+      "#Result\n",

+      "print\"Principal Strains are:e1\",round(e1,2),\"N/mm**2\"\n",

+      "print\"                     :e2\",round(e2,2),\"N/mm**2\"\n",

+      "print\"Principal Stresses are:sigma1\",round(sigma1,2),\"N/mm**2\"\n",

+      "print\"                      :sigma2\",round(sigma2,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Principal Strains are:e1 508.54 N/mm**2\n",

+        "                     :e2 -158.54 N/mm**2\n",

+        "Principal Stresses are:sigma1 101.31 N/mm**2\n",

+        "                      :sigma2 -1.31 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 10

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example No.7.23,Page No.313"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Strains\n",

+      "e_A=600 #microns\n",

+      "e_B=-450 #microns\n",

+      "e_C=100 #micron\n",

+      "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",

+      "mu=0.3 #Poissoin's ratio\n",

+      "theta=240\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "e_x=e_A=600\n",

+      "\n",

+      "#e_A=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(theta)+rho_x_y*2**-1*sin(theta)\n",

+      "#After sub values and further simplifying we get\n",

+      "#-450=(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*(0.5)-0.866*2**-1*rho_x_y   .....................(1)\n",

+      "\n",

+      "#e_C=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(2*theta)+rho_x_y*2**-1*sin(2*theta)\n",

+      "#After sub values and further simplifying we get\n",

+      "#100=(e_x+e_y)*2**-1-0.5*(e_x-e_y)*2**-1*(0.5)-0.866*2**-1*rho_x_y   .....................(2)\n",

+      "\n",

+      "#Adding Equation 1 and 2 we get equations as\n",

+      "#-350=e_x+e_y-(e_x-e_y)*2**-1    ...............(3)\n",

+      "#Further simplifying we get\n",

+      "\n",

+      "e_y=(-700-e_x)*3**-1    #micron              \n",

+      "\n",

+      "rho_x_y=(e_C-(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*cos(2*theta*pi*180**-1))*(sin(2*theta*pi*180**-1))**-1*2 #micron\n",

+      "\n",

+      "#Principal strains\n",

+      "e1=(e_x+e_y)*2**-1-(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",

+      "e2=(e_x+e_y)*2**-1+(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",

+      "\n",

+      "#Principal Stresses\n",

+      "sigma1=E*(e1+mu*e2)*(1-mu**2)**-1*10**-6 #N/mm**2\n",

+      "sigma2=E*(e2+mu*e1)*(1-mu**2)**-1*10**-6 #N/mm**2\n",

+      "\n",

+      "\n",

+      "#Result\n",

+      "print\"Principal Stresses are:sigma1\",round(sigma1,2),\"N/mm**2\"\n",

+      "print\"                      :sigma2\",round(sigma2,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Principal Stresses are:sigma1 -69.49 N/mm**2\n",

+        "                      :sigma2 117.11 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 18

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_1.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_1.ipynb
new file mode 100644
index 00000000..c7043749
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_1.ipynb
@@ -0,0 +1,1512 @@
+{

+ "metadata": {

+  "name": "chapter no.8.ipynb"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter No.8:Thin And Thick Cyclinders And Spheres"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.1,Page No.322"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=3000 #mm #Length\n",

+      "d1=1000 #mm #Internal diameter\n",

+      "t=15 #mm #Thickness\n",

+      "P=1.5 #N/mm**2 #Fluid Pressure\n",

+      "E=2*10**5 #n/mm**2 #Modulus of elasticity\n",

+      "mu=0.3 #Poissoin's ratio\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Hoop stress\n",

+      "f1=P*d1*(2*t)**-1 #N/mm**2\n",

+      "\n",

+      "#Longitudinal Stress\n",

+      "f2=P*d1*(4*t)**-1 #N/mm**2\n",

+      "\n",

+      "#Max shear stress\n",

+      "q_max=(f1-f2)*2**-1 #N/mm**2\n",

+      "\n",

+      "#Diametrical Strain\n",

+      "#Let e1=dell_d*d**-1  .....................(1)\n",

+      "e1=(f1-mu*f2)*E**-1 \n",

+      "\n",

+      "#Sub values in equation 1 and further simplifying we get\n",

+      "dell_d=e1*d1 #mm\n",

+      "\n",

+      "#Longitudinal strain\n",

+      "#e2=dell_L*L**-1      ......................(2)\n",

+      "e2=(f2-mu*f1)*E**-1 \n",

+      "\n",

+      "#Sub values in equation 2 and further simplifying we get\n",

+      "dell_L=e2*L #mm\n",

+      "\n",

+      "#Change in Volume \n",

+      "#Let Z=dell_V*V**-1      ................(3)\n",

+      "Z=2*e1+e2\n",

+      "\n",

+      "#Sub values in equation 3 and further simplifying we get\n",

+      "dell_V=Z*pi*4**-1*d1**2*L\n",

+      "\n",

+      "#Result\n",

+      "print\"Max Intensity of shear stress\",round(q_max,2),\"N/mm**2\"\n",

+      "print\"Change in the Dimensions of the shell is:dell_d\",round(dell_d,2),\"mm\"\n",

+      "print\"                                        :dell_L\",round(dell_L,2),\"mm\"\n",

+      "print\"                                        :dell_V\",round(dell_V,2),\"mm**3\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max Intensity of shear stress 12.5 N/mm**2\n",

+        "Change in the Dimensions of the shell is:dell_d 0.21 mm\n",

+        "                                        :dell_L 0.15 mm\n",

+        "                                        :dell_V 1119192.38 mm**3\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.2,Page No.323"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=2000 #mm #Length\n",

+      "d=200 #mm # diameter\n",

+      "t=10 #mm #Thickness\n",

+      "dell_V=25000 #mm**3 #Additional volume\n",

+      "E=2*10**5 #n/mm**2 #Modulus of elasticity\n",

+      "mu=0.3 #Poissoin's ratio\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let p be the pressure developed\n",

+      "\n",

+      "#Circumferential Stress\n",

+      "\n",

+      "#f1=p*d*(2*t)**-1 #N/mm**2\n",

+      "#After sub values and further simplifying\n",

+      "#f1=10*p\n",

+      "\n",

+      "#f1=p*d*(4*t)**-1 #N/mm**2\n",

+      "#After sub values and further simplifying\n",

+      "#f1=5*p\n",

+      "\n",

+      "#Diameterical strain = Circumferential stress\n",

+      "#Let X=dell_d*d**-1     ................................(1)\n",

+      "#X=e1=(f1-mu*f2)*E**-1 \n",

+      "#After sub values and further simplifying\n",

+      "#e1=8.5*p*E**-1\n",

+      "\n",

+      "#Longitudinal strain\n",

+      "#Let Y=dell_L*L**-1    ......................................(2)\n",

+      "#Y=e2=(f2-mu*f1)*E**-1 \n",

+      "#After sub values and further simplifying\n",

+      "#e2=2*p*E**-1\n",

+      "\n",

+      "#Volumetric strain\n",

+      "#Let X=dell_V*V**-1    \n",

+      "#X=2*e1+e2\n",

+      "#After sub values and further simplifying\n",

+      "#X=19*p*E**-1\n",

+      "#After further simplifying we get\n",

+      "p=dell_V*(pi*4**-1*d**2*L)**-1*E*19**-1 #N/mm**2\n",

+      "\n",

+      "#Hoop Stress\n",

+      "f1=p*d*(2*t)**-1\n",

+      "\n",

+      "X=e1=8.5*p*E**-1\n",

+      "#Sub value of X in equation 1 we get\n",

+      "dell_d=8.5*p*E**-1*d\n",

+      "\n",

+      "Y=e2=2*p*E**-1\n",

+      "#Sub value of Y in equation 2 we get\n",

+      "dell_L=2*p*E**-1*L\n",

+      "\n",

+      "#Result\n",

+      "print\"Pressure Developed is\",round(p,2),\"N/mm**2\"\n",

+      "print\"Hoop stress Developed is\",round(f1,2),\"N/mm**2\"\n",

+      "print\"Change in diameter is\",round(dell_d,2),\"mm\"\n",

+      "print\"Change in Length is\",round(dell_L,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Pressure Developed is 4.19 N/mm**2\n",

+        "Hoop stress Developed is 41.88 N/mm**2\n",

+        "Change in diameter is 0.04 mm\n",

+        "Change in Length is 0.08 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 22

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.3,Page No.324"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=750 #mm #Diameter of water supply pipes\n",

+      "h=50*10**3 #mm #Water head\n",

+      "sigma=20 #N/mm**2 #Permissible stress\n",

+      "rho=9810*10**-9 #N/mm**3\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Pressure of water\n",

+      "P=rho*h #N/mm**2\n",

+      "\n",

+      "#Stress\n",

+      "#sigma=p*d*(2*t)**-1\n",

+      "#After further simplifying\n",

+      "t=P*d*(2*sigma)**-1 #mm \n",

+      "\n",

+      "#Result\n",

+      "print\"Thickness of seamless pipe is\",round(t,3),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Thickness of seamless pipe is 9.197 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 37

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.4,Page No.326"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=2500 #mm #Diameter of riveted boiler\n",

+      "P=1 #N/mm**2 #Pressure\n",

+      "rho1=0.7 #Percent efficiency\n",

+      "rho2=0.4 #Circumferential joints\n",

+      "sigma=150 #N/mm**2 #Permissible stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Equating Bursting force to longitudinal joint strength ,we get\n",

+      "#p*d*L=rho1*2*t*L*sigma\n",

+      "#After rearranging and further simplifying we get\n",

+      "t=P*d*(2*sigma*rho1)**-1 #mm\n",

+      "\n",

+      "#Considering Longitudinal force\n",

+      "#pi*d**2*4**-1*P=rho2*pi*d*t*sigma\n",

+      "#After rearranging and further simplifying we get\n",

+      "t2=P*d*(4*sigma*rho2)**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Thickness of plate required is\",round(t,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Thickness of plate required is 11.9 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 45

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.5,Page No.326"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Boiler Dimensions\n",

+      "t=16 #mm #Thickness\n",

+      "p=2 #N/mm**2 #internal pressure\n",

+      "f=150 #N/mm**2 #Permissible stress\n",

+      "rho1=0.75 #Longitudinal joints\n",

+      "rho2=0.45 #circumferential joints\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Equating Bursting force to longitudinal joint strength ,we get\n",

+      "d1=rho1*2*t*f*p**-1 #mm\n",

+      "\n",

+      "#Considering circumferential strength \n",

+      "d2=4*rho2*t*f*p**-1 #mm\n",

+      "\n",

+      "#Result\n",

+      "print\"Largest diameter of Boiler is\",round(d1,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Largest diameter of Boiler is 1800.0 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 52

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.6,Page No.329"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=250 #mm #Diameter iron pipe\n",

+      "t=10 #mm #Thickness\n",

+      "d2=6 #mm #Diameter of steel\n",

+      "p=80 #N/mm**2 #stress\n",

+      "P=3 #N/mm**2 #Pressure\n",

+      "E_c=1*10**5 #N/mm**2\n",

+      "mu=0.3 #poissoin's ratio\n",

+      "E_s=2*10**5 #N/mm**2\n",

+      "n=1 #No.of wires\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "L=6 #mm #Length of cyclinder\n",

+      "\n",

+      "#Force Exerted by steel wire at diameterical section\n",

+      "F=p*2*pi*d2**2*1*4**-1 #N\n",

+      "\n",

+      "#Initial stress in cyclinder\n",

+      "f_c=F*(2*t*d2)**-1 #N/mm**2\n",

+      "\n",

+      "#LEt due to fluid pressure alone stresses developed in steel wire be F_w and in cyclinder f1 and f2\n",

+      "f2=P*d*(4*t)**-1 #N/mm**2\n",

+      "\n",

+      "#Considering the equilibrium of half the cyclinder, 6mm long we get\n",

+      "#F_w*2*pi*4**-1*d2**2*n+f1*2*t*d2=P*d*d2\n",

+      "#After further simplifying we get\n",

+      "#F_w+2.122*f1=79.58        . ......................................(1)\n",

+      "\n",

+      "#Equating strain in wire to circumferential strain in cyclinder \n",

+      "#F_w=(f1-mu*f2)*E_s*E_c**-1 #N/mm**2\n",

+      "#After further simplifying we get\n",

+      "#F_w=2*f1-11.25    ....................................(2)\n",

+      "\n",

+      "#Sub in equation in1 we get\n",

+      "f1=(79.58+11.25)*(4.122)**-1 #N/mm**2\n",

+      "F_w=2*f1-11.25  #N/mm**2\n",

+      "\n",

+      "#Final stresses\n",

+      "#1) In steel Wire\n",

+      "sigma=F_w+p #N/mm**2\n",

+      "\n",

+      "#2) In Cyclinder\n",

+      "sigma2=f1-f_c\n",

+      "\n",

+      "#Result\n",

+      "print\"Final Stresses developed in:cyclinder is\",round(sigma,2),\"N/mm**2\"\n",

+      "print\"                           :Steel is\",round(sigma2,2),\"N/mm**2\"   "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Final Stresses developed in:cyclinder is 112.82 N/mm**2\n",

+        "                           :Steel is -15.66 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.7,Page No.332"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=750 #mm #Diameter of shell\n",

+      "t=8 #mm #THickness\n",

+      "p=2.5 #N/mm**2\n",

+      "E=2*10**5 #N/mm**2\n",

+      "mu=0.25 #Poissoin's ratio\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Hoop stress\n",

+      "f1=f2=p*d*(4*t)**-1 #N/mm**2\n",

+      "\n",

+      "#Change in Diameter\n",

+      "dell_d=d*p*d*(1-mu)*(4*t*E)**-1 #mm\n",

+      "\n",

+      "#Change in Volume\n",

+      "dell_V=3*p*d*(1-mu)*(4*t*E)**-1*pi*6**-1*d**3\n",

+      "\n",

+      "#Answer for Change in diameter is incorrect in book\n",

+      "\n",

+      "#Result\n",

+      "print\"Stress introduced is\",round(f1,2),\"N/mm**2\"\n",

+      "print\"Change in Diameter is\",round(dell_d,2),\"N/mm**2\"\n",

+      "print\"Change in Volume is\",round(dell_V,2),\"mm**3\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stress introduced is 58.59 N/mm**2\n",

+        "Change in Diameter is 0.16 N/mm**2\n",

+        "Change in Volume is 145608.33 mm**3\n"

+       ]

+      }

+     ],

+     "prompt_number": 56

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.8,Page No.333"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d=600 #mm #Diameter of sherical shell\n",

+      "t=10 #mm #Thickness\n",

+      "f=80 #N/mm**2 #Permissible stress\n",

+      "rho=0.75 #Efficiency joint\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Max Pressure\n",

+      "p=f*4*t*rho*d**-1 #N/mm**2\n",

+      "\n",

+      "#Result\n",

+      "print\"Max Pressure is\",round(p,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max Pressure is 4.0 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 60

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.9,Page No.333"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=1000 #mm #Length of shell\n",

+      "d=200 #mm #Diameter\n",

+      "t=6 #mm #Thickness\n",

+      "p=1.5 #N/mm**2 #Internal Pressure\n",

+      "E=2*10**5 #N/mm**2\n",

+      "mu=0.25 #Poissoin's Ratio\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Change in Volume of sphere\n",

+      "dell_V_s=3*p*d*(1-mu)*(4*t*E)**-1*pi*6**-1*d**3\n",

+      "\n",

+      "#Hoop stress\n",

+      "f1=p*d*(2*t)**-1 #N/mm**2\n",

+      "\n",

+      "#Longitudinal stress\n",

+      "f2=p*d*(4*t)**-1 #N/mm**2\n",

+      "\n",

+      "#Principal strain\n",

+      "e1=(f1-mu*f2)*E**-1\n",

+      "e2=(f2-mu*f1)*E**-1\n",

+      "\n",

+      "V_c=1000 #mm**3\n",

+      "\n",

+      "#Change in Volume of cyclinder\n",

+      "dell_V_c=(2*e1+e2)*pi*4**-1*d**2*L\n",

+      "\n",

+      "#Total Change in Diameter\n",

+      "dell_V=dell_V_s+dell_V_c #mm**3\n",

+      "\n",

+      "#Result\n",

+      "print\"Change in Volume is\",round(dell_V,2),\"mm**3\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Change in Volume is 8443.03 mm**3\n"

+       ]

+      }

+     ],

+     "prompt_number": 66

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.10,Page No.337"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d1=400 #mm #Internal Diameter\n",

+      "t=100 #mm #Thickness\n",

+      "p=80 #N/mm**2 #Fluid pressure\n",

+      "\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Internal Radius\n",

+      "r1=d1*2**-1 #mm\n",

+      "\n",

+      "#Outer Radius\n",

+      "r_o=r1+t #mm\n",

+      "\n",

+      "p1=80 #N/mm**2\n",

+      "p2=0\n",

+      "\n",

+      "#Now From Lame's Euation\n",

+      "#p_x=b*(x**2)**-1-a\n",

+      "#at x=200 #mm \n",

+      "p_x=80 #N/mm**2\n",

+      "#80=b*(200**2)**-1-a     ..........................(1)\n",

+      "\n",

+      "#at x=300 #mm\n",

+      "#p_x2=0\n",

+      "#0=b*(300**2)**-1-a       ...........................(2)\n",

+      "\n",

+      "#Sub equation 2 from 1\n",

+      "#80=b*(200**2)**-1-b*(300**2)**-1\n",

+      "#After Further simplifying we get\n",

+      "b=(50000)**-1*(200**2*300**2*80)\n",

+      "\n",

+      "#From equation 2 we get\n",

+      "a=b*(300**2)**-1\n",

+      "\n",

+      "#Variation of radial pressure p_x;\n",

+      "#p_x=b*(x**2)**-1-a\n",

+      "#After sub values and further simplifying we get\n",

+      "\n",

+      "#Radial pressure Variation\n",

+      "#At \n",

+      "x=200 #mm\n",

+      "p_x=b*(x**2)**-1-a #N/mm**2\n",

+      "\n",

+      "#At\n",

+      "x2=250 #mm\n",

+      "p_x2=b*(x2**2)**-1-a #N/mm**2\n",

+      "\n",

+      "#At \n",

+      "x3=300 #mm\n",

+      "p_x3=b*(x3**2)**-1-a #N/mm**2\n",

+      "\n",

+      "\n",

+      "#Hoop stress Distribution\n",

+      "#Variation of F_x\n",

+      "\n",

+      "#At \n",

+      "x=200 #mm\n",

+      "F_x=b*(x**2)**-1+a #N/mm**2\n",

+      "\n",

+      "#At\n",

+      "x2=250 #mm\n",

+      "F_x2=b*(x2**2)**-1+a #N/mm**2\n",

+      "\n",

+      "#At\n",

+      "x3=300 #mm\n",

+      "F_x3=b*(x3**2)**-1+a #N/mm**2\n",

+      "\n",

+      "#Result\n",

+      "print\"Max Hoop stress is\",round(F_x,2),\"N/mm**2\"\n",

+      "print\"Min Hoop stress is\",round(F_x3,2),\"N/mm**2\"\n",

+      "print\"Plot of Hoop stress\"\n",

+      "\n",

+      "#Plotting Variation of hoop stress\n",

+      "\n",

+      "X1=[x,x2,x3]\n",

+      "Y1=[p_x,p_x2,p_x3]\n",

+      "Y2=[-F_x,-F_x2,-F_x3]\n",

+      "Z1=[0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Y2,X1,Z1)\n",

+      "plt.xlabel(\"Length x in mm\")\n",

+      "plt.ylabel(\"Radial Stress Distribution & Hoop Stress Distribution in N/mm**2\")\n",

+      "plt.show()\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max Hoop stress is 208.0 N/mm**2\n",

+        "Min Hoop stress is 128.0 N/mm**2\n",

+        "Plot of Hoop stress\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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psPpzP4zq6moEBwfj999/N6pCuTGRkD7V1UBaWt0zXZyc6p7p8ucyJqJ2Q9YFiQEBAdJ0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\nbm5uYtOmTUII3b27hBBCoVCIHTt2CCG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+       "text": [

+        "<matplotlib.figure.Figure at 0x50e8310>"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.11,Page No.338"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d_o=300 #mm #Outside diameter \n",

+      "d2=200 #mm #Internal Diameter\n",

+      "p=14 #N/mm**2 #internal Fluid pressure\n",

+      "t=50 #mm #Thickness\n",

+      "r_o=150 #mm #Outside Diameter\n",

+      "r2=100 #mm #Internal Diameter\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#From Lame's Equation\n",

+      "#p_x=b*(x**2)**-1-a #N/mm**2    ...................(1)\n",

+      "#F_x=b*(x**2)**-1+a #N/mm**2   ...................(2)\n",

+      "\n",

+      "#At \n",

+      "x=r2=100 #mm\n",

+      "p_x=14 #N/mm**2\n",

+      "\n",

+      "#Sub value of p_x in equation 1 we get\n",

+      "#14=(100)**-1*b-a    ............................(3)\n",

+      "\n",

+      "#At\n",

+      "x2=r_o=150 #mm\n",

+      "p_x2=0 #N/mm**2\n",

+      "\n",

+      "#Sub value in equation 1 we get\n",

+      "#0=b*(150**2)**-1-a   ......................(4)\n",

+      "\n",

+      "#From Equations 3 and 4 we get\n",

+      "#14=b*(100**2)**-1-b*(100**2)**-1\n",

+      "#After sub values and further simplifying we get\n",

+      "b=14*100**2*150**2*(150**2-100**2)**-1\n",

+      "\n",

+      "#From equation 4 we get\n",

+      "a=b*(150**2)**-1\n",

+      "\n",

+      "#Hoop Stress\n",

+      "#F_x=b*(x**2)**-1+a #N/mm**2\n",

+      "\n",

+      "#At \n",

+      "x=100 #mm\n",

+      "F_x=b*(x**2)**-1+a #N/mm**2\n",

+      "\n",

+      "#At\n",

+      "x2=125 #mm\n",

+      "F_x2=b*(x2**2)**-1+a #N/mm**2\n",

+      "\n",

+      "#At\n",

+      "x3=150 #mm\n",

+      "F_x3=b*(x3**2)**-1+a #N/mm**2\n",

+      "\n",

+      "#If thin Cyclindrical shell theory is used,hoop stress is uniform and is given by\n",

+      "F=p*d2*(2*t)**-1 #N/mm**2\n",

+      "\n",

+      "#Percentage error in estimating max hoop tension\n",

+      "E=(F_x-F)*F_x**-1*100 #%\n",

+      "\n",

+      "#Result\n",

+      "print\"Max Hoop Stress Developed in the cross-section is\",round(F,2),\"N/mm**2\"\n",

+      "print\"Plot of Variation of hoop stress\"\n",

+      "\n",

+      "#Plotting Variation of hoop stress\n",

+      "\n",

+      "X1=[x,x2,x3]\n",

+      "Y1=[F_x,F_x2,F_x3]\n",

+      "Z1=[0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in mm\")\n",

+      "plt.ylabel(\"Radial Stress Distribution & Hoop Stress Distribution in N/mm**2\")\n",

+      "plt.show()\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Max Hoop Stress Developed in the cross-section is 28.0 N/mm**2\n",

+        "Plot of Variation of hoop stress\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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29vaQJAknTpx4bAP379/HiBEjMGzYMMycOdNwT41GA1dXV2RkZGDgwIEcMiIi\nqgaK9hB27dplaARApRoSQmDq1Knw9vY2JAMAeOmll7Bu3TpERERg3bp1GPXwCRFERFQjLNqpnJKS\ngl9++cWwyiggIMCimx88eBDPPfcc/P39DQll4cKF6NGjB8aNG4dLly5BpVIhNjYWLR85H449BCKi\nylN0UjkmJgarV6/GmDFjIITAli1b8F//9V945513rGrQ4sCYEIiIKk3RhODn54ejR4+iWbNmAIDb\nt2+jV69eOHnypFUNWhwYEwIRUaUpXsvo4SMzbXF8JhER2Z7ZSeXJkyejZ8+eRkNGU6ZMsUVsRERk\nQxZNKicmJuLQoUMAgP79+yMoKEj5wDhkRERUaYouOwUAOzs7wyohDhkREdVNZj/dY2JiMHHiRGRn\nZ+P69euYOHGioWopERHVHVxlRERUh3CVERERVRlXGREREYBKrDJ6+IAcrjIiIqqdFNmpnJOTY/S6\n9G2lq41at25tVYMWB8aEQERUaYokBJVKZfjwv3btGjp06GDU4IULF6xq0OLAmBCIiCpN0VpGABAU\nFITk5GSrGrAWEwIRUeUpvsqIiIjqPiYEIiIC8Jhlp0uXLjV0PbKzs7Fs2TKjieXZs2fbLEgiIlKe\nyYRQUFBgmFSeNm0aCgoKbBYUERHZnkWTyjWBk8pERJVXayeVp0yZAhcXF/j5+RmuRUVFwc3NDUFB\nQQgKCsKuXbuUDIGIiCykaEKYPHlyuQ/80vmH5ORkJCcnY+jQoUqGQEREFlI0IfTv3x+tWrUqd51D\nQUREtY/FCWHu3LlITEyEEAIzZ86sUqMrVqxAQEAApk6ditzc3Crdi4iIqofFCaFHjx5YvHgx/P39\nkZeXZ3WD06dPR3p6OlJSUtC+fXvMmTPH6nsREVH1MbnsdOXKlRg+fDg6d+4MABgxYgTWrl0LJycn\ndO3a1eoG27VrZ/h+2rRpCAsLM/neqKgow/dqtRpqtdrqdomI6iKNRgONRlMt9zK57NTX1xe///47\nAODWrVsYMWIEevfujcWLF6Nnz544duyYRQ1otVqEhYUZTljLyMhA+/btAQCfffYZjh07hg0bNpQP\njMtOiYgqrSqfnSZ7CDqdDoWFhbhx4wZGjBiBwYMHY8mSJQCAu3fvWnTzCRMmYP/+/bhx4wY6deqE\nBQsWQKPRICUlBZIkwd3dHatWrbIqcCIiql4mE8KcOXPg4eEBnU4HDw8PODo6QqvVIjY21uIho+++\n+67cNZ5LqnDaAAAWK0lEQVS2RkRUOz12p7JOpzP8929/+xv27NmDoKAgfP7552jbtq2ygXHIiIio\n0hQ/D6EmMCEQEVVerS1dQURETw4mBCIiAsCEQERED5hcZVTq7t272Lx5M7RarWGSWZIk/OMf/1A8\nOCIish2zCWHkyJFo2bIlQkJC0KRJE1vERERENcDsKqOHdyzbElcZERFVnqKrjPr06YMTJ05YdXMi\nInpymO0hdOvWDefPn4e7uzsaN24s/5IkKZ4k2EMgIqo8RTemabVaQyNA2eE2KpXKqgYtDowJgYio\n0hTfqZySkoJffvkFkiShf//+CAgIsKqxSgXGhEBEVGmKziHExMRg4sSJyM7ORlZWFiZOnIjly5db\n1RgREdVeZnsIfn5+OHr0KJo1awYAuH37Nnr16mU430CxwNhDICKqNMVrGTVo0KDC74mIqO4wuzFt\n8uTJ6NmzJ8aMGQMhBLZs2cIzDYiI6iCLJpUTExNx8OBBw6RyUFCQ8oFxyIiIqNIUWWWUn58PJycn\n5OTkAChbblq6/LR169ZWNWhxYEwIRESVpkhCGD58OLZv3w6VSmVIAg9LT0+3qkGLA2NCICKqtFp7\nYtqUKVOwfft2tGvXzrAqKScnB+PHj8fFixehUqkQGxuLli1blg+MCYGIqNIUXWUUGhpq0bWKTJ48\nGbt27TK6Fh0djUGDBuHcuXMIDQ1FdHS0haESEZGSTCaEoqIi3Lx5E9nZ2cjJyTF8abVaXL161aKb\n9+/fH61atTK6FhcXh/DwcABAeHg4tmzZUoXwiYiouphcdrpq1SrExMTg2rVrCAkJMVx3dHTEjBkz\nrG4wKysLLi4uAAAXFxdkZWVZfS8iIqo+JhPCzJkzMXPmTKxYsQJvv/22Io1LklThhDUREdme2Y1p\nTk5O+Pbbb8tdnzRpklUNuri4IDMzE66ursjIyEC7du1MvjcqKsrwvVqthlqttqpNIqK6SqPRQKPR\nVMu9zK4ymjFjhuFf8UVFRfj5558RHByMTZs2WdSAVqtFWFiYYZXRvHnz0KZNG0RERCA6Ohq5ubkV\nTixzlRERUeXZdNlpbm4uxo8fj927d5t974QJE7B//37cuHEDLi4u+J//+R+MHDkS48aNw6VLl7js\nlIiomtk0IRQXF8PX1xfnzp2zqkFLMSEQEVVeVT47zc4hhIWFGb7X6/VITU3FuHHjrGqMiIhqL7M9\nhNLJCkmSYG9vj86dO6NTp07KB8YeAhFRpSm6U1mtVsPT0xO5ubnIyclBw4YNrWqIiIhqN7MJ4auv\nvkLPnj3xww8/YNOmTejZsyfWrFlji9iIiMiGzA4Zde3aFUeOHEGbNm0AADdv3kTv3r05qUxEVAsp\nOmTUtm1bNG/e3PC6efPmaNu2rVWNERFR7WVyldHSpUsBAF26dEHPnj0xatQoAMDWrVvh7+9vm+iI\niMhmTCaEgoICSJIEDw8PPP3004bdyiNHjmT9ISKiOkjRA3KqgnMIRESVp8jGtL/+9a+IiYkx2pj2\ncINxcXFWNUhERLWTyYRQWs303XffLZdtOGRERFT3PHbISKfTYdKkSdiwYYMtYwLAISMiImsotuzU\n3t4ely5dwr1796y6ORERPTnMFrdzd3dHv3798NJLL8HBwQGAnIFmz56teHBERGQ7ZhOCh4cHPDw8\noNfrUVhYaIuYiIioBphNCN7e3uXKXcfGxioWEBER1Qyz+xCCgoKQnJxs9lq1B8ZJZSKiSlNkH8LO\nnTuxY8cOXL16Fe+8846hgYKCApbAJiKqg0wmhA4dOiAkJARbt25FSEiIISE4OTnhs88+s1mARERk\nG2aHjO7fv2/oEeTk5ODKlSvVUtxOpVLByckJdnZ2aNiwIRISEowD45AREVGlKXqm8qBBgxAXFwed\nToeQkBA4Ozujb9++Ve4lSJIEjUaD1q1bV+k+RERUPcyeh5CbmwsnJyf88MMPmDRpEhISEvDTTz9V\nS+PsARAR1R5mE0JJSQkyMjIQGxuL4cOHA6ieWkaSJOGFF15A9+7dsXr16irfj4iIqsbskNE//vEP\nDBkyBH379kWPHj2QlpaGZ555psoNHzp0CO3bt0d2djYGDRoELy8v9O/fv8r3JSIi69SK8xAWLFiA\n5s2bY86cOYZrkiQhMjLS8FqtVkOtVtdAdEREtZdGo4FGozG8XrBggdXD8SYTwqJFixAREYG33367\n3Ky1JElYvny5VQ0CwJ07d1BSUgJHR0fcvn0bgwcPRmRkJAYPHmzURi3IVURETxRFVhl5e3sDAEJC\nQipssCqysrIwevRoAHKJ7T/96U9GyYCIiGyvVgwZVYQ9BCKiylPsPIS1a9ciODgYDg4OcHBwQPfu\n3bFu3TqrGiIiotrN5JDRunXrEBMTg2XLliEoKAhCCCQnJ2Pu3LmQJMlwxCYREdUNJoeMevbsiX//\n+99wd3c3uq7VajF+/Hj8+uuvygbGISMiokpTZMiooKCgXDIA5BpEBQUFVjVGRES1l8mE0KRJE5O/\n9LifERHRk8nkkFHTpk3RpUuXCn8pLS0Nd+7cUTYwDhkREVWaIvsQTp8+bXVARET05OE+BCKiOkSx\nfQhERFR/MCEQERGASiaEnJwcnDhxQqlYiIioBplNCAMGDEB+fj5ycnIQEhKCadOmYdasWbaIjYiI\nbMhsQsjLy1PsCE0iIqo9auwITSIiql3MJoTSIzQ9PDyq9QhNIiKqXbgPgYioDlF0H8K8efOQn5+P\n+/fvIzQ0FG3btsW//vUvqxojIqLay2xC2L17N5ycnLBt2zaoVCqkpaXh008/tUVsRERkQ2YTgk6n\nAwBs27YNr7zyClq0aMFJZSKiOshsQggLC4OXlxcSExMRGhqK69evV0v56127dsHLywvPPPMMFi1a\nVOX7ERFR1Vg0qZyTk4MWLVrAzs4Ot2/fRkFBAVxdXa1utKSkBJ6envjpp5/QsWNHPPvss/juu+/Q\nrVu3ssA4qWyg0WigVqtrOoxagc+iDJ9FGT6LMopOKt++fRv/+7//i7feegsAcO3aNRw/ftyqxkol\nJCSgS5cuUKlUaNiwIV599VVs3bq1SvesyzQaTU2HUGvwWZThsyjDZ1E9zCaEyZMno1GjRjh8+DAA\noEOHDnj//fer1OjVq1fRqVMnw2s3NzdcvXq1SvckIqKqMZsQ0tLSEBERgUaNGgEAmjVrVuVGOSlN\nRFT7mDwxrVTjxo1RVFRkeJ2WlobGjRtXqdGOHTvi8uXLhteXL1+Gm5ub0Xs8PDyYOB6yYMGCmg6h\n1uCzKMNnUYbPQubh4WH175pNCFFRURg6dCiuXLmC1157DYcOHcLatWutbhAAunfvjj/++ANarRYd\nOnTA999/j++++87oPefPn69SG0REVDmPTQh6vR63bt3C5s2bcfToUQBATEwMnJ2dq9aovT3++c9/\nYsiQISgpKcHUqVONVhgREZHtmV12GhISgsTERFvFQ0RENcTspPKgQYOwZMkSXL58GTk5OYavqpgy\nZQpcXFzg5+dnuJaTk4NBgwaha9euGDx4MHJzcw0/W7hwIZ555hl4eXlhz549VWq7tqnoWWzcuBE+\nPj6ws7NDUlKS0fvr27OYO3cuunXrhoCAAIwZMwZ5eXmGn9W3ZzF//nwEBAQgMDAQoaGhRvNw9e1Z\nlFq6dCkaNGhg9JlU355FVFQU3NzcEBQUhKCgIOzcudPws0o/C2HGU089JVQqVbmvqjhw4IBISkoS\nvr6+hmtz584VixYtEkIIER0dLSIiIoQQQpw6dUoEBASI4uJikZ6eLjw8PERJSUmV2q9NKnoWp0+f\nFmfPnhVqtVokJiYartfHZ7Fnzx7D3xgREVGv/7/Iz883fL98+XIxdepUIUT9fBZCCHHp0iUxZMgQ\noVKpxM2bN4UQ9fNZREVFiaVLl5Z7rzXPwmwP4cyZM0hPTzf6On36tHXp7YH+/fujVatWRtfi4uIQ\nHh4OAAgPD8eWLVsAAFu3bsWECRPQsGFDqFQqdOnSBQkJCVVqvzap6Fl4eXmha9eu5d5bH5/FoEGD\n0KCB/L9pz549ceXKFQD181k4Ojoavi8sLETbtm0B1M9nAQCzZ8/G4sWLja7V12chKhj5t+ZZmE0I\nffr0sehaVWVlZcHFxQUA4OLigqysLADyzuiHl6TW501s9f1ZfP3113jxxRcB1N9n8f7776Nz585Y\nu3Yt/va3vwGon89i69atcHNzg7+/v9H1+vgsAGDFihUICAjA1KlTDcPt1jwLkwkhIyMDiYmJuHPn\nDpKSkpCYmIikpCRoNBrcuXOnmv6MikmS9Ng9CNyfUKa+PIuPP/4YjRo1wmuvvWbyPfXhWXz88ce4\ndOkSJk+ejJkzZ5p8X11+Fnfu3MEnn3xitO+gon8hl6rLzwIApk+fjvT0dKSkpKB9+/aYM2eOyfea\nexYml53u3r0ba9euxdWrV40acHR0xCeffGJF2I/n4uKCzMxMuLq6IiMjA+3atQNQfhPblStX0LFj\nx2pv/0lQX5/F2rVrsWPHDuzdu9dwrb4+i1KvvfaaobdU355FWloatFotAgICAMh/b0hICH799dd6\n9ywAGD4rAWDatGkICwsDYOX/F+YmMTZu3FjpiQ9LpKenl5tUjo6OFkIIsXDhwnKTh/fu3RMXLlwQ\nTz/9tNDr9YrEVFMefRal1Gq1OH78uOF1fXwWO3fuFN7e3iI7O9voffXxWZw7d87w/fLly8XEiROF\nEPXzWTysoknl+vQsrl27Zvh+2bJlYsKECUII656FyYSwdetWkZ6ebngdFRUl/Pz8RFhYmLhw4YK1\nf4sQQohXX31VtG/fXjRs2FC4ubmJr7/+Wty8eVOEhoaKZ555RgwaNEjcunXL8P6PP/5YeHh4CE9P\nT7Fr164qtV3bPPos1qxZI3788Ufh5uYmmjRpIlxcXMTQoUMN769vz6JLly6ic+fOIjAwUAQGBorp\n06cb3l/fnsXLL78sfH19RUBAgBgzZozIysoyvL8+PItGjRoZPi8e5u7ubkgIQtSPZ/Hw/xevv/66\n8PPzE/7+/mLkyJEiMzPT8P7KPguTG9P8/Pzw66+/wsHBAdu2bcOsWbPw73//G8nJydi4cSN2795d\nDZ0dIiKqLUxOKjdo0AAODg4AgB9++AFTp05FSEgIpk2bhuvXr9ssQCIisg2TCUEIgYKCAuj1euzd\nuxehoaGGn929e9cmwRERke2YXGU0c+ZMBAUFwdHREd26dcOzzz4LAEhKSkKHDh1sFiAREdnGY4vb\nXblyBdevX0dgYKBht2hGRgbu37+Pzp072yxIIiJSntlqp0REVD+YLV1BRET1AxMC1VrNmzdX9P6f\nf/650fGw1dVefHw8Fi1aVC33IrIlk0NG5s48aN26tSIBEZVydHREQUGBYvd3d3fH8ePH0aZNG5u0\nR1TbmVxlFBwc/NhCSOnp6YoERPQ4aWlpmDFjBrKzs+Hg4IDVq1fD09MTf/7zn9GiRQscP34cmZmZ\nWLx4MV5++WXo9XrMmDED+/btQ6dOndCwYUNMmTIF165dw7Vr1zBw4EA4Ozsb6iR98MEH2LZtG5o2\nbYqtW7ca1YkB5NV3bdq0wfz587F792588skn2L9/v9F71q5di8TERKxYscJkXA/TarUYOnQoevfu\njcOHD6N79+4IDw/HggULkJ2djfXr1+PZZ59FVFSUoQT9pUuXsGzZMhw+fBh79uxBx44dER8fD3t7\ns8ekE5mmwO5qomrRvHnzcteef/558ccffwghhDh69Kh4/vnnhRBChIeHi3HjxgkhhEhNTRVdunQR\nQsi1uF588UUhhBCZmZmiVatWYvPmzUII4xo4QgghSZLYtm2bEEKIefPmiY8++qhc+3fu3BE+Pj7i\n559/Fp6enhWWcVm7dq2YMWPGY+N6WHp6urC3txe///670Ov1IiQkREyZMkUIIZeQGTVqlBBCiMjI\nSNG/f3+h0+nEb7/9Jpo2bWooRzB69GixZcuWxzxNIvMs+ufErVu38McffxhtSHvuuecUS1JEFSks\nLMSRI0cwduxYw7Xi4mIAclnfUaNGAQC6detmOE/j4MGDGDduHAC5ou7AgQNN3r9Ro0YYPnw4APks\n8f/85z/l3tO0aVOsXr0a/fv3R0xMDNzd3R8bs6m4HuXu7g4fHx8AgI+PD1544QUAgK+vL7RareFe\nw4YNg52dHXx9faHX6zFkyBAAcqmZ0vcRWctsQli9ejWWL1+Oy5cvIygoCEePHkXv3r3x888/2yI+\nIgO9Xo+WLVsiOTm5wp83atTI8L14MDUmSZJRrXzxmFXWDRs2NHzfoEED6HS6Ct934sQJODs7W3zw\nSkVxPapx48ZGbZf+zqNxPHzd0niJLGV2lVFMTAwSEhKgUqmwb98+JCcno0WLFraIjciIk5MT3N3d\nsWnTJgDyh+uJEyce+zt9+/bF5s2bIYRAVlaW0Xi/o6Mj8vPzKxXDxYsXsWzZMiQnJ2Pnzp0VHkn4\nuKRTFUrdl6iU2YTQpEkTNG3aFIBcw8jLywtnz55VPDCiO3fuoFOnToavzz//HOvXr8eaNWsQGBgI\nX19fxMXFGd7/8CKI0u9ffvlluLm5wdvbG6+//jqCg4MN/6B54403MHToUEOdrkd//9FFFUIITJs2\nDUuXLoWrqyvWrFmDadOmGYatTP2uqe8f/R1Tr0u/f9x9H3dvIkuZ3ak8evRofP3114iJicHevXvR\nqlUr6HQ67Nixw1YxElXJ7du30axZM9y8eRM9e/bE4cOHy60eIqJKlq7QaDTIz8/H0KFDjcZFiWqz\ngQMHIjc3F8XFxYiIiMCkSZNqOiSiWslkQsjPz4eTk5PJDWrcmEZEVLeYTAjDhw/H9u3boVKpKhyb\n5MY0IqK6hdVOiYgIwGP2ISQlJT32F4ODg6s9GCIiqjkmewhqtRqSJKGoqAiJiYnw9/cHIG/K6d69\nO44cOWLTQImISFkm9yFoNBrs27cPHTp0QFJSEhITE5GYmIjk5GQeoUlEVAeZnUPw9vZGamqq2WtE\nRPRkM1vLyN/fH9OmTcPEiRMhhMCGDRsQEBBgi9iIiMiGzPYQioqKsHLlSvzyyy8A5Cqn06dPR5Mm\nTWwSIBER2QaXnRIREQALhozOnTuHv//970hNTTWcPytJEi5cuKB4cEREZDtmq51OnjwZb731Fuzt\n7bFv3z6Eh4fjT3/6ky1iIyIiGzI7ZBQcHIykpCT4+fnh5MmTRteIiKjuMDtk1KRJE5SUlKBLly74\n5z//iQ4dOuD27du2iI2IiGzIbA8hISEB3bp1Q25uLubPn4/8/HzMmzcPvXr1slWMRERkA5VeZSSE\nQGxsLMaPH69UTEREVANMTioXFhZi6dKl+Mtf/oIvvvgCer0eP/74I3x8fLB+/XpbxkhERDZgsocw\nZswYODk5oXfv3tizZw8uX76MJk2aYPny5QgMDLR1nEREpDCTCcHf3x8nTpwAAJSUlKB9+/a4ePEi\nmjZtatMAiYjINkwOGdnZ2Rl937FjRyYDIqI6zGQPwc7ODg4ODobXRUVFhoQgSRLy8/NtEyEREdkE\naxkREREAC0pXEBFR/cCEQEREAJgQiIjoASYEIiICwIRAREQPMCEQEREA4P8Bc+VeilsXyhwAAAAA\nSUVORK5CYII=\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x58c2e30>"

+       ]

+      }

+     ],

+     "prompt_number": 11

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.12,Page No.339"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d_o=300 #mm #Outside diameter \n",

+      "d2=200 #mm #Internal Diameter\n",

+      "p=12 #N/mm**2 #internal Fluid pressure\n",

+      "F_max=16 #N/mm**2 #Tensile stress\n",

+      "r_o=150 #mm #Outside Diameter\n",

+      "r2=100 #mm #Internal Diameter\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let p_o be the External Pressure applied.\n",

+      "#From LLame's theorem\n",

+      "#p_x=b*(x**2)**-1-a   ..............(1)\n",

+      "#F_x=b*(x**2)**-1+a   ...........................(2)\n",

+      "\n",

+      "#Now At\n",

+      "x=100 #mm\n",

+      "p_x=12 #N/mm**2\n",

+      "#sub in equation 1 we get\n",

+      "#12=b*(100**2)**-1-a    . ..................(3)\n",

+      "\n",

+      "#The Max Hoop stress occurs at least value of x where\n",

+      "x=r1=100 #mm\n",

+      "#16=b*(100**2)**-1+a      .......................(4)\n",

+      "\n",

+      "#From Equations 1 and 2 we get\n",

+      "#28=b*(100**2)**-1+b*(100**2)**-1\n",

+      "#After furhter Simplifying we get\n",

+      "b=28*100**2*2**-1\n",

+      "\n",

+      "#sub in equation 1 we get\n",

+      "a=-(12-(b*(100**2)**-1))\n",

+      "\n",

+      "#Thus At\n",

+      "x2=150 #mm\n",

+      "p_o=b*(x2**2)**-1-a\n",

+      "\n",

+      "#Result\n",

+      "print\"Minimum External applied is\",round(p_o,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Minimum External applied is 4.22 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 18

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.13,Page No.340"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d1=160 #mm #Internal Diameter \n",

+      "r1=80 #mm #External Diameter\n",

+      "p1=40 #N/mm**2 #Internal Diameter\n",

+      "P_max=120 #N/mm**2 #Allowable stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#From Lame's Equation we have\n",

+      "#p_x=b*(x**2)**-1-a       ..........................(1)\n",

+      "#F_x=b*(x**2)**-1+a      ...........................(2)\n",

+      "\n",

+      "#At \n",

+      "x=r1=80 #N/mm**2 \n",

+      "#Sub in equation 1 we get\n",

+      "#120=b*(80**2)**-1+a     ........................(3)\n",

+      "\n",

+      "#The hoop tension at inner edge is max stress\n",

+      "#Hence\n",

+      "#120=b*(80**2)**-1+a    .............................(4)\n",

+      "\n",

+      "#From Equation 3 and 4 we get\n",

+      "b=160*80**2*2**-1 \n",

+      "\n",

+      "#Sub in equation 3 we get\n",

+      "a=-(40-(b*(80**2)**-1))\n",

+      "\n",

+      "#Let External radius be r_o.Since at External Surface is Zero,we get\n",

+      "#0=b*(r_o)**-1-a\n",

+      "#After Further simplifying we get\n",

+      "r_o=(b*a**-1)**0.5\n",

+      "\n",

+      "#Thickness of Cyclinder \n",

+      "t=r_o-r1 #mm\n",

+      "\n",

+      "#Result\n",

+      "print\"Thickness Required is\",round(t,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Thickness Required is 33.14 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 31

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.14,Page No.341"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d_o=300 #mm #Outside diameter \n",

+      "d1=180 #mm #Internal Diameter\n",

+      "p=12 #N/mm**2 #internal Fluid pressure\n",

+      "p_o=6 #N/mm**2 #External Pressure\n",

+      "r_o=150 #mm #Outside Diameter\n",

+      "r=90 #mm #Internal Diameter\n",

+      "\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#From Lame's Equation we have\n",

+      "#p_x=b*(x**2)**-1-a       ..........................(1)\n",

+      "#F_x=b*(x**2)**-1+a      ...........................(2)\n",

+      "\n",

+      "#At \n",

+      "x=r1=90 #N/mm**2 \n",

+      "p=42 #N/mm**2\n",

+      "#Sub in equation 1 we get\n",

+      "#42=b*(90**2)**-1-a       ..............................(3)\n",

+      "\n",

+      "#At \n",

+      "x=r_o=150 #mm\n",

+      "p2=6 #N/mm**2\n",

+      "#sub in equation 1 we get\n",

+      "#6=b*(150**2)**-1-a   ..............................(4)\n",

+      "\n",

+      "#From equations 3 and 4  weget\n",

+      "#36=b*(90**2)**-1-b2(150**2)**-1\n",

+      "#After further simplifying we get\n",

+      "b=36*90**2*150**2*(150**2-90**2)**-1\n",

+      "\n",

+      "#Sub value of b in equation 4 we get\n",

+      "a=b*(150**2)**-1-p_o\n",

+      "\n",

+      "#At \n",

+      "x=r1=90 #mm\n",

+      "F_x=b*(x**2)**-1+a #N/mm**2\n",

+      "\n",

+      "#At \n",

+      "x2=r_o=150 #mm \n",

+      "F_x2=b*(x2**2)**-1+a  #N/mm**2\n",

+      "\n",

+      "#Now if External pressure is doubled i.e p_o2=12 #N/mm**2 We have\n",

+      "p_o2=12 #N/mm**2\n",

+      "#sub in equation 4 we get\n",

+      "#12=b2*(150**2)**-1-a2                    ..........................(5)\n",

+      "\n",

+      "#Max Hoop stress is to be 70.5 #N/mm**2,which occurs at x=r1=90 #mm\n",

+      "#Sub in equation 4 we get\n",

+      "#70.5=b*(90**2)**-1+a2           ................................(6)\n",

+      "\n",

+      "#Adding equation 5 and 6\n",

+      "#82.5=b2*(150**2)**-1+b*(90**2)**-1\n",

+      "#After furhter simplifying we get\n",

+      "b2=82.5*150**2*90**2*(150**2+90**2)**-1\n",

+      "\n",

+      "#Sub in equation 5 we get\n",

+      "a2=b2*(150**2)**-1-12 \n",

+      "\n",

+      "#If p_i is the internal pressure required then from Lame's theorem\n",

+      "p_i=b2*(r1**2)**-1-a2\n",

+      "\n",

+      "#Result\n",

+      "print\"Stresses int the material are:F_x\",round(F_x,2),\"N/mm**2\"\n",

+      "print\"                             :F_x2\",round(F_x2,2),\"N/mm**2\"\n",

+      "print\"Internal Pressure that can be maintained is\",round(p_i,2),\"N/mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Stresses int the material are:F_x 70.5 N/mm**2\n",

+        "                             :F_x2 34.5 N/mm**2\n",

+        "Internal Pressure that can be maintained is 50.82 N/mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 17

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.15,Page No.344"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "r1=200 #mm #Inner Radius\n",

+      "r2=250 #mm #Radius at common surface\n",

+      "r3=300 #mm #Outer radius\n",

+      "p=6 #N/mm**2 #Inital pressure\n",

+      "p2=80 #N/mm**2 #Pressure\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Inner Cyclinder:\n",

+      "\n",

+      "#From Lame's Equation we have\n",

+      "#p_x=b*(x**2)**-1-a       ..........................(1)\n",

+      "#F_x=b*(x**2)**-1+a      ...........................(2)\n",

+      "\n",

+      "#At \n",

+      "x=r1=200 #mm\n",

+      "p_x=0\n",

+      "#0=b1*(250**2)**-1-a1   .................(3)\n",

+      "\n",

+      "#At x=r2=250 #mm\n",

+      "p_x2=6 #N/mm**2\n",

+      "#6=b1*(250**2)-a1    ...................(4)\n",

+      "\n",

+      "#From Equation 3 and 4 we get\n",

+      "b1=6*200**2*250**2*(200**2-250**2)**-1\n",

+      "\n",

+      "#From equation 3 we get\n",

+      "a1=b1*(200**2)**-1\n",

+      "\n",

+      "F_200=b1*(200**2)**-1+a1\n",

+      "F_250=b1*(250**2)**-1+a1\n",

+      "\n",

+      "#For outer cyclinder \n",

+      "#From Lame's Equation we have\n",

+      "#p_x2=b2*(x**2)**-1-a2       ..........................(5)\n",

+      "#F_x2=b2*(x**2)**-1+a2      ...........................(6)\n",

+      "\n",

+      "\n",

+      "#At \n",

+      "x2=r2=250 #mm\n",

+      "p_x2=6 #N/mm**2\n",

+      "#6=b2*(250**2)**-1-a2     ...........................(7)      \n",

+      "\n",

+      "#At\n",

+      "x3=300 #mm\n",

+      "#p_x2=0\n",

+      "#0=b2**2*(300**2)**-1-a2     .................................(8)\n",

+      "\n",

+      "#from equation 7 and 8 we get\n",

+      "b2=6*250**2*300**2*(300**2-250**2)**-1\n",

+      "\n",

+      "#sub in equation 8 we get\n",

+      "a2=b2*(300**2)**-1\n",

+      "\n",

+      "F_250_2=b2*(250**2)**-1+a2\n",

+      "F_300_2=b2*(300**2)**-1+a2\n",

+      "\n",

+      "#When Fluid is admitted\n",

+      "#Let Lame's equation be\n",

+      "#p_x3=b3*(x**2)**-1-a3       ..........................(5)\n",

+      "#F_x3=b3*(x**2)**-1+a3      ...........................(6)\n",

+      "\n",

+      "\n",

+      "#At x=200\n",

+      "p_x3=80 #N/mm**2\n",

+      "#80=b3*(200**2)**-1-a3    ................................(7)\n",

+      "\n",

+      "#At x=300 #mm\n",

+      "#p_x=0\n",

+      "#0=b3*(300**2)**-1-a3     ..............................(8)\n",

+      "\n",

+      "#from Equation 7 and 8 we get\n",

+      "b3=80*200**2*300**2*(300**2-200**2)**-1\n",

+      "\n",

+      "#From Equation 8 we get\n",

+      "a3=b3*(300**2)**-1\n",

+      "\n",

+      "#Hoop stresses \n",

+      "F_200_3=b3*(200**2)**-1+a3 #N/mm**2\n",

+      "F_250_3=b3*(250**2)**-1+a3 #N/mm**2\n",

+      "F_300_3=b3*(300**2)**-1+a3 #N/mm**2\n",

+      "\n",

+      "#Pressure at common surface\n",

+      "p_250=b3*(250**2)**-1-a3 #N/mm**2\n",

+      "\n",

+      "#final stress\n",

+      "f_200=F_200+F_200_3 #N/mm**2\n",

+      "f_250=F_250+F_250_3 #N/mm**2\n",

+      "f_300=F_250_2+F_250_3 #N/mm**2\n",

+      "f_300_2=F_300_2+F_300_3 #N/mm**2\n",

+      "\n",

+      "#Result\n",

+      "print\"final Hoop stress are:f_200\",round(f_200,2),\"N/mm**2\"\n",

+      "print\"                     :f_250\",round(f_250,2),\"N/mm**2\"\n",

+      "print\"                     :f_300\",round(f_300,2),\"N/mm**2\"\n",

+      "print\"                     :f_300_2\",round(f_300_2,2),\"N/mm**2\"\n",

+      "print\"Variation of Hoop stress and Radial stress\"\n",

+      "\n",

+      "#Final stresses\n",

+      "#Variation of hoop stress \n",

+      " \n",

+      "X1=[x,x2,x3,x3]\n",

+      "Y1=[f_200,f_250,f_300,f_300_2]\n",

+      "Z1=[0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in mm\")\n",

+      "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",

+      "plt.show()\n",

+      "\n",

+      "#Due to Fluid\n",

+      "#Variation of hoop stress \n",

+      " \n",

+      "X1=[x,x2,x3]\n",

+      "Y1=[F_200_3,F_250_3,F_300_3]\n",

+      "Z1=[0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in mm\")\n",

+      "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",

+      "plt.show()\n",

+      "\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "final Hoop stress are:f_200 174.67 N/mm**2\n",

+        "                     :f_250 128.83 N/mm**2\n",

+        "                     :f_300 189.43 N/mm**2\n",

+        "                     :f_300_2 155.27 N/mm**2\n",

+        "Variation of Hoop stress and Radial stress\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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MHz4cr7zyCmbPno1///vf1fIBnZeXp/l+8+bNmhVR0dHR+P7771FYWIjMzExcvHgRnTp1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\nEenEREFERDoxURARkU7/H21GIMqBrUIbAAAAAElFTkSuQmCC\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5604510>"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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+       "text": [

+        "<matplotlib.figure.Figure at 0x56e5cb0>"

+       ]

+      }

+     ],

+     "prompt_number": 89

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.16,Page No.348"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "do=200 #mm #Inner Diameter\n",

+      "r_o=100 #mm #Inner radius\n",

+      "d1=300 #mm #outer diameter\n",

+      "r1=150 #mm #Outer radius\n",

+      "d2=250 #mm #Junction Diameter\n",

+      "r2=125 #mm #Junction radius\n",

+      "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",

+      "p=30 #N/mm**2 #radial pressure\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#from Lame's Equation we get\n",

+      "#p_x=b*(x**2)**-1-a       ..........................(1)\n",

+      "#F_x=b*(x**2)**-1+a      ...........................(2)\n",

+      "\n",

+      "#Then from Boundary condition \n",

+      "#p_x=0 at x=100 #mm\n",

+      "#0=b1*(100**2)**-1-a1     .....................(3)\n",

+      "\n",

+      "#p_x2=30 #N/mm**2 at x2=125 #mm\n",

+      "#30=b1*(125**2)**-1-a1    ................................(4)\n",

+      "\n",

+      "#From equation 3 and 4 we get\n",

+      "b1=30*125**2*100**2*(100**2-125**2)**-1\n",

+      "\n",

+      "#From Equation 3 we get\n",

+      "a1=b1*(100**2)**-1\n",

+      "\n",

+      "#therefore Hoop stress in inner cyclinder at junction\n",

+      "F_2_1=b1*(125**2)**-1+a1 #N/mm**2\n",

+      "\n",

+      "#Outer Cyclinder\n",

+      "#p_x=b*(x**2)**-1-a       ..........................(5)\n",

+      "#F_x=b*(x**2)**-1+a      ...........................(6)\n",

+      "\n",

+      "#Now at x=125 #mm\n",

+      "#p_x3=30 #N/mm**2\n",

+      "#30=b2*(125**2)**-1-a2     ..................................(7)\n",

+      "\n",

+      "#At x=150 #mm\n",

+      "#p_x4=0\n",

+      "#0=b2*(150**2)**-1-a2     ...................................(8)\n",

+      "\n",

+      "#From equations 7 and 8\n",

+      "b2=30*150**2*125**2*(150**2-125**2)**-1\n",

+      "\n",

+      "#From eqauation 8 we get\n",

+      "a2=b2*(150**2)**-1\n",

+      "\n",

+      "#Hoop stress at junction \n",

+      "F_2_0=b2*(125**2)**-1+a2 #N/mm**2\n",

+      "\n",

+      "rho_r=(F_2_0-F_2_1)*E**-1*r2\n",

+      "\n",

+      "#Result\n",

+      "print\"Shrinkage Allowance is\",round(rho_r,3),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Shrinkage Allowance is 0.189 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 40

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.17,Page No.350"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d_o=500 #mm #Outer Diameter\n",

+      "r_o=250 #mm #Outer Radius\n",

+      "d1=300 #mm #Inner Diameter\n",

+      "r1=150 #mm #Inner Radius\n",

+      "d2=400 #mm #Junction Diameter\n",

+      "E=2*10**5 #N/mm**2 #Modulus ofElasticity\n",

+      "alpha=12*10**-6 #Per degree celsius\n",

+      "dell_d=0.2 #mm\n",

+      "dell_r=0.1 #mm\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let p be the radial pressure developed at junction\n",

+      "#Let Lame's Equation for internal cyclinder be\n",

+      "#p_x=b*(x**2)**-1-a   ................................(1)\n",

+      "#F_x=b*(x**2)**-1+a     ...............................(2)\n",

+      "\n",

+      "#At \n",

+      "x=150 #mm \n",

+      "p_x=0\n",

+      "#Sub in equation 1 we get\n",

+      "#0=b*(150**2)**-1-a    .........................(3)\n",

+      "\n",

+      "#At \n",

+      "x2=200 #mm\n",

+      "#p_x2=p\n",

+      "#p=b*(200**2)**-1-a    ......................(4)\n",

+      "  \n",

+      "#From Equation 3 and 4\n",

+      "#p=b*(200**2)**-1-b(150**2)**-1\n",

+      "#after further simplifying we get\n",

+      "#b=-51428.571*p\n",

+      "\n",

+      "#sub in equation 3 we get\n",

+      "#a1=-2.2857*p\n",

+      "\n",

+      "#therefore hoop stress at junction is\n",

+      "#F_2_1=-21428.571*p*(200**2)**-1-2.2857*p\n",

+      "#after Further simplifying we geet\n",

+      "#F_2_1=3.5714*p\n",

+      "\n",

+      "#Let Lame's Equation for cyclinder be \n",

+      "#p_x=b*(x**2)**-1-a    .........................5\n",

+      "#F_x=b*(x**2)**-1+a   .............................6\n",

+      "\n",

+      "#At \n",

+      "x=200 #mm\n",

+      "#p_x=p2\n",

+      "#p2=b2*(20**2)**-1-a2    ...................7\n",

+      "\n",

+      "#At\n",

+      "x2=200 #mm\n",

+      "p_x2=0\n",

+      "#0=b2*(250**2)**-1-a2    ....................8\n",

+      "\n",

+      "#from equation 7 and 8 we get\n",

+      "#p2=b2*(200**2)**-1-b2*(250**2)**-1\n",

+      "#After further simplifying we get\n",

+      "#p2=b2*(250**2-200**2)*(200**2*250**2)**-1\n",

+      "#b2=111111.11*p\n",

+      "\n",

+      "#from equation 7\n",

+      "#a2=b2*(250**2)**-1\n",

+      "#further simplifying we get\n",

+      "#a2=1.778*p\n",

+      "\n",

+      "#At the junctionhoop stress in outer cyclinder \n",

+      "#F_2_0=b2*(200**2)**-1+a2\n",

+      "#After further simplifying we get\n",

+      "#F_2_0=4.5556*p\n",

+      "\n",

+      "#Considering circumferential strain,the compatibility condition\n",

+      "#rho_r*r2**-1=1*E**-1*(F_2_1+F_2_0)\n",

+      "#where F_2_1 is compressive and F_2_0 is tensile\n",

+      "#furter simplifying we get\n",

+      "p=0.1*200**-1*2*10**5*(3.5714+4.5556)**-1\n",

+      "\n",

+      "#Let T be the rise in temperature required\n",

+      "#dell_d=d*alpha*T\n",

+      "#After sub values and further simplifying we get\n",

+      "d=250 #mm\n",

+      "T=dell_d*(d*alpha)**-1 #Per degree celsius\n",

+      "\n",

+      "#Result\n",

+      "print\"Radial Pressure Developed at junction\",round(p,2),\"N/mm**2\"\n",

+      "print\"Min Temperatureto outer cyclinder\",round(T,2),\"Per degree Celsius\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Radial Pressure Developed at junction 12.3 N/mm**2\n",

+        "Min Temperatureto outer cyclinder 66.67 Per degree Celsius\n"

+       ]

+      }

+     ],

+     "prompt_number": 13

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 8.8.18,Page No.355"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "%matplotlib  inline\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "d_o=400 #mm #Outer Diameter\n",

+      "r_o=200 #mm #Outer radius\n",

+      "t=50 #mm #Thickness\n",

+      "r1=150 #mm #Internal Radius\n",

+      "p=50 #N/mm**2 #Internal Pressure\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#The Radial Pressure and hoop stress at any radial distance x are given by\n",

+      "#p_x=b*(x**2)**-1-a       ..........................(1)\n",

+      "#F_x=b*(x**2)**-1+a      ...........................(2)\n",

+      "\n",

+      "#Now at\n",

+      "x=150 #N/mm**2\n",

+      "p_x1=50 #N/mm**2\n",

+      "#Sub in equation 1 we get\n",

+      "#50=2*b*(150**3)**-1-a    ...........................(3)\n",

+      "\n",

+      "#At x=200 #mm\n",

+      "p_x2=0\n",

+      "#0=2*b*(200**2)**-1-a     ....................(4)\n",

+      "\n",

+      "#From equation 3 and 4 we get\n",

+      "#50=2*b*(150**3)**-1-2*b*(200**3)**-1\n",

+      "#After further simplifying we get\n",

+      "b=50*150**3*200**3*(200**3-150**3)**-1*2**-1\n",

+      "\n",

+      "#Sub in equation 3 we get\n",

+      "a=b*(200**3)**-1\n",

+      "\n",

+      "#Now At\n",

+      "x=150 #mm\n",

+      "F_x=b*(x**3)**-1+a\n",

+      "\n",

+      "#Now At\n",

+      "x2=160 #mm\n",

+      "F_x2=b*(x2**3)**-1+a\n",

+      "\n",

+      "#Now At\n",

+      "x3=170 #mm\n",

+      "F_x3=b*(x3**3)**-1+a\n",

+      "\n",

+      "#Now At\n",

+      "x4=180 #mm\n",

+      "F_x4=b*(x4**3)**-1+a\n",

+      "\n",

+      "#Now At\n",

+      "x5=190 #mm\n",

+      "F_x5=b*(x5**3)**-1+a\n",

+      "\n",

+      "#Now At\n",

+      "x6=200 #mm\n",

+      "F_x6=b*(x6**3)**-1+a\n",

+      "\n",

+      "#Result\n",

+      "print\"Plot of Variation of hoop stress\"\n",

+      "\n",

+      "#Plotting Variation of hoop stress\n",

+      "\n",

+      "X1=[x,x2,x3,x4,x5,x6]\n",

+      "Y1=[F_x,F_x2,F_x3,F_x4,F_x5,F_x6]\n",

+      "Z1=[0,0,0,0,0,0]\n",

+      "plt.plot(X1,Y1,X1,Z1)\n",

+      "plt.xlabel(\"Length x in mm\")\n",

+      "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",

+      "plt.show()\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Plot of Variation of hoop stress\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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yhIaGAgB8fX01nyMyFgOBnhrl5eVo164dTp8+Xef71ffEFxVTYwqF\nosaT+EQ9U2aVT6gDgBYtWmi9BJOSkoKOHTvq/eCVuup6UsuWLWu0XfmdJ+uoflzfeon0xTkEemq0\nbdsWPXv2xP/+7/8CkH65pqSk1PudoUOHYuvWrRBCICcnR7N9NCBNIj948MCgGjIyMvD111/j9OnT\n2LNnT52PI6wvdBpCrvMSVWIgkMUqLCxEt27dNK9vv/0WGzZswKpVq+Dv7w8fHx/s2rVL8/nqT4Kq\n/HnChAlwdnZGnz59MHXqVPTr1w/29vYAgNmzZ+Oll17STCo/+f0nnywlhMDMmTOxdOlSODk5YdWq\nVZg5c6bmspW272r7+cnvaPtz5c/1nbe+cxPpi8tOqckrKChA69atce/ePQwePBj//ve/azyhj4gk\nnEOgJm/MmDHIz89HcXExPv74Y4YBkRYcIRAREQDOIRARUQUGAhERAWAgEBFRBQYCEREBYCAQEVEF\nBgIREQEA/h+bezx5xlsz+QAAAABJRU5ErkJggg==\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x56b4bf0>"

+       ]

+      }

+     ],

+     "prompt_number": 25

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_1.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_1.ipynb
new file mode 100644
index 00000000..e6d444ca
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_1.ipynb
@@ -0,0 +1,768 @@
+{

+ "metadata": {

+  "name": "chapter no.9.ipynb"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter 9:Columns And Struts"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 9.9.1,Page No.377"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "L=5000 #mm #Length of strut\n",

+      "dell=10 #mm #Deflection\n",

+      "W=10 #N #Load\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Central Deflection of a simply supported beam with central concentrated load is\n",

+      "#dell=W*L**3*(48*E*I)**-1 \n",

+      "\n",

+      "#Let E*I=X\n",

+      "X=W*L**3*(48*dell)**-1 #mm\n",

+      "\n",

+      "#Euler's Load\n",

+      "#Let Euler's Load be P\n",

+      "P=pi**2*X*(L**2)**-1\n",

+      "\n",

+      "#Result\n",

+      "print\"Critical Load of Bar is\",round(P,2),\"N\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Critical Load of Bar is 1028.08 N\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 9.9.2,Page No.377"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=2000 #mm #Length of square column\n",

+      "E=12*10**3 #N/mm**2 #Modulus of Elasticity\n",

+      "sigma=12 #N/mm*2 #stress\n",

+      "W1=95*10**3 #N #Load1\n",

+      "W2=200*10**3 #N #Load2\n",

+      "FOS=3\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#From Euler's Formula\n",

+      "#P=pi**2*E*I*(L**2)**-1  .........(1)\n",

+      "\n",

+      "#Working Load\n",

+      "#W=P*(FOS)**-1\n",

+      "\n",

+      "#Part-1\n",

+      "\n",

+      "#At W1=95*10**3 #N\n",

+      "#W1=P*(3*L**2)**-1\n",

+      "\n",

+      "#Let 'a' be the side of the square\n",

+      "#I=1*12**-1*a**4\n",

+      "\n",

+      "#sub value of I in Equation 1 and further rearranging we get\n",

+      "a=(W1*3*12*L**2*(pi**2*E)**-1)**0.25 #mm\n",

+      "\n",

+      "#From Consideration of direct crushing\n",

+      "#sigma*a**2=W1\n",

+      "#After Reaaranging the above equation we get\n",

+      "a2=(W1*(sigma)**-1)**0.5 #mm\n",

+      "\n",

+      "#required size is 103.67*103.67 i.e a*a\n",

+      "\n",

+      "#Part-2\n",

+      "\n",

+      "#At W2=200*10**3 #N\n",

+      "#W2=P*(3*L**2)**-1\n",

+      "#After substituting values and further Rearranging the above equation we get\n",

+      "a3=(W2*3*12*L**2*(pi**2*E)**-1)**0.25 #mm\n",

+      "\n",

+      "#From consideration of direct compression,size required is\n",

+      "a4=(W2*sigma**-1)**0.5\n",

+      "\n",

+      "#required size is 129.10*129.10 i.e a4*a4\n",

+      "\n",

+      "#Result\n",

+      "print\"For W1 Load Required size is\",round(a*a,2),\"mm**2\"\n",

+      "print\"For W2 Load Required size is\",round(a4*a4,2),\"mm**2\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "For W1 Load Required size is 10747.38 mm**2\n",

+        "For W1 Load Required size is 16666.67 mm**2\n"

+       ]

+      }

+     ],

+     "prompt_number": 16

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 9.9.3,Page No.378"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Flange \n",

+      "b=100 #mm #Width\n",

+      "\n",

+      "D=80 #mm #Overall Depth\n",

+      "t=10 #mm #Thickness of web and flanges\n",

+      "L=3000 #mm #Length of strut\n",

+      "E=200*10**3 #N/mm**2 #Modulus of Elasticity\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Let centroid be at depth y_bar from top fibre\n",

+      "y_bar=(b*t*t*2**-1+(D-t)*t*((D-t)*2**-1+t))*(b*t+(D-t)*t)**-1 #mm \n",

+      "\n",

+      "#M.I at x-x axis\n",

+      "I_x=1*12**-1*b*t**3+b*t*(y_bar-t*2**-1)**2+1*12**-1*t*((D-t))**3+t*((D-t))*((((D-t)*2**-1)+t)-y_bar)**2\n",

+      "\n",

+      "#M.I at y-y axis\n",

+      "I_y=1*12**-1*t*b**3+1*12**-1*(D-t)*t**3 #mm**3\n",

+      "\n",

+      "#Least M.I\n",

+      "I=I_y\n",

+      "\n",

+      "#Since both ends are hinged\n",

+      "#Feective Length=Actual Length\n",

+      "L=l=3000 #mm\n",

+      "\n",

+      "#Buckling Load \n",

+      "P=pi**2*E*I*(l**2)**-1*10**-3 #KN\n",

+      "\n",

+      "#Result\n",

+      "print\"The Buckling Load for strut of tee section\",round(P,2),\"KN\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The Buckling Load for strut of tee section 184.05 KN\n"

+       ]

+      }

+     ],

+     "prompt_number": 10

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 9.9.4,Page No.379"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "D=400 #mm #Overall Depth\n",

+      "\n",

+      "#Flanges\n",

+      "b=300 #mm #Width\n",

+      "t=50 #mm #Thickness\n",

+      "\n",

+      "t2=30 #mm #Web Thickness\n",

+      "\n",

+      "dell=10 #mm #Deflection\n",

+      "w=40 #N/mm #Load\n",

+      "FOS=1.75 #Factor of safety\n",

+      "E=2*10**5 #N/mm**2\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#M.I at x-x axis\n",

+      "I_x=1*12**-1*(b*D**3-(b-t2)*b**3) #mm**4\n",

+      "\n",

+      "#Central Deflection\n",

+      "#dell=5*w*L**4*(384*E*I)**-1\n",

+      "#After sub values in above equation and further simplifying we get\n",

+      "L=(dell*384*E*I_x*(5*w)**-1)**0.25\n",

+      "\n",

+      "#M.I aty-y axis\n",

+      "I=I_y=1*12**-1*t*b**3+1*12**-1*b*t2**3+1*12**-1*t*b**3 #mm**4\n",

+      "\n",

+      "#Both the Ends of column are hinged\n",

+      "\n",

+      "#Crippling Load\n",

+      "P=pi**2*E*I*(L**2)**-1 #N\n",

+      "\n",

+      "#Safe Load\n",

+      "S=P*(FOS)**-1*10**-3 #N\n",

+      "\n",

+      "#Result\n",

+      "print\"Safe Load if I-section is used as column with both Ends hhinged\",round(S,2),\"KN\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Safe Load if I-section is used as column with both Ends hhinged 4123.29 KN\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 9.9.5,Page No.381"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "D=200 #mm #External Diameter\n",

+      "t=20 #mm #hickness\n",

+      "d=200-2*t #mm #Internal Diameter\n",

+      "E=1*10**5 #N/mm**2\n",

+      "a=1*(1600)**-1 #Rankine's Constant\n",

+      "L=4.5 #m #Length\n",

+      "sigma=550 #N/mm**2 #Stress\n",

+      "FOS=2.5\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Moment of Inertia\n",

+      "I=pi*D**4*64**-1-pi*d**4*64**-1\n",

+      "\n",

+      "#Both Ends are fixed\n",

+      "\n",

+      "#Effective Length\n",

+      "l=1*2**-1*L*10**3 #mm\n",

+      "\n",

+      "#Euler's Critical Load\n",

+      "P_E=pi**2*E*I*(l**2)**-1\n",

+      "\n",

+      "A=pi*4**-1*(D**2-d**2) #mm*2\n",

+      "\n",

+      "k=(I*A**-1)**0.5\n",

+      "\n",

+      "#Rankine's Critical Load\n",

+      "P_R=sigma*A*(1+a*(l*k**-1)**2)**-1\n",

+      "\n",

+      "X=P_E*P_R**-1 \n",

+      "\n",

+      "#Safe Load using Rankine's Formula\n",

+      "S=P_R*(FOS)**-1*10**-3 #KN\n",

+      "\n",

+      "#Result\n",

+      "print\"Safe Load by Rankine's Formula is\",round(S,2),\"KN\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Safe Load by Rankine's Formula is 1404.36 KN\n"

+       ]

+      }

+     ],

+     "prompt_number": 39

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 9.9.6,Page No.382"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "L=3000 #mm #Length of column\n",

+      "W=800*10**3 #N #Load\n",

+      "a=1*1600**-1 #Rankine's constant\n",

+      "FOS=4 #Factor of safety\n",

+      "sigma=550 #N/mm**2 #stress\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Effective Length\n",

+      "l=L*2**-1 #mm \n",

+      "\n",

+      "#Let d1=outer diameter & d2=inner diameter\n",

+      "#d1=5*8**-1*d2\n",

+      "\n",

+      "#M.I\n",

+      "#I=pi*64**-1*(d1**4-d2**4) #mm**4\n",

+      "\n",

+      "#Area of section\n",

+      "#A=pi4**-1*(d1**2-d2**2) #mm**2\n",

+      "\n",

+      "#k=(I*A**-1) \n",

+      "#substituting values in above equation \n",

+      "#k=1*16**-1*(d1**2-d2**2)\n",

+      "#after simplifying further we get\n",

+      "#k=0.2948119.d1\n",

+      "\n",

+      "#X=l*k**-1\n",

+      "#substituting values in above equation and after simplifying further we get\n",

+      "#X=5087.9898*d1**-1\n",

+      "\n",

+      "#Crtitcal Load\n",

+      "P=W*FOS #N\n",

+      "\n",

+      "#From Rankine's Load\n",

+      "#P2=sigma*A*(1+a*(X)**2)**-1\n",

+      "#substituting values in above equation and after simplifying further we get\n",

+      "#d1**4-12156618*d1**4-1.96691*10**8=0\n",

+      "#Solving Quadratic Equation we get\n",

+      "#d1**2-12156618*d1-196691000=0\n",

+      "a=1\n",

+      "b=-12156.618\n",

+      "c=-196691000\n",

+      "\n",

+      "Y=b**2-4*a*c\n",

+      "\n",

+      "d1_1=((-b+Y**0.5)*(2*a)**-1)**0.5 #mm\n",

+      "d1_2=((-b-Y**0.5)*(2*a)**-1) #mm\n",

+      "\n",

+      "d2=5*8**-1*d1_1\n",

+      "\n",

+      "#Result\n",

+      "print\"Section of cast iron hollow cylindrical column is:d1_1\",round(d1_1,2),\"mm\"\n",

+      "print\"                                                 :d2  \",round(d2,2),\"mm\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Section of cast iron hollow cylindrical column is:d1_1 146.16 mm\n",

+        "                                                 :d2   91.35 mm\n"

+       ]

+      }

+     ],

+     "prompt_number": 25

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 9.9.7,Page No.383"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "#Let X=(P*A**-1) #Average Stress at Failure \n",

+      "Lamda_1=70 #Slenderness Ratio\n",

+      "Lamda_2=170 #Slenderness Ratio\n",

+      "X1=200 #N/mm**2 \n",

+      "X2=69 #N/mm**2 \n",

+      "\n",

+      "#Rectangular section\n",

+      "b=60 #mm #width\n",

+      "t=20 #mm #Thickness\n",

+      "\n",

+      "L=1250 #mm #Length of strut\n",

+      "FOS=4 #Factor of safety\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Slenderness ratio\n",

+      "#Lamda=L*k**-1\n",

+      "\n",

+      "#The Rankine's Formula for strut\n",

+      "#P=sigma*A*(1+a*(L*k**-1)**-1\n",

+      "\n",

+      "#From test result 1,\n",

+      "#After sub values in above equation we get and further simplifying we get\n",

+      "#sigma_1=200+980000*a   ...................(1)\n",

+      "\n",

+      "#From test result 2,\n",

+      "#After sub values in above equation we get and further simplifying we get\n",

+      "#sigma_2=69+1994100*a   ...................(2)\n",

+      "\n",

+      "#Substituting it in equation (1) we get\n",

+      "a=131*1014100**-1   \n",

+      "\n",

+      "#Substituting a in equation 1\n",

+      "sigma_1=200+980000*a #N/mm**2\n",

+      "\n",

+      "#Effective Length \n",

+      "l=1*2**-1*L #mm\n",

+      "\n",

+      "#Least of M.I\n",

+      "I=1*12**-1*b*t**3 #mm**4\n",

+      "\n",

+      "#Area \n",

+      "A=b*t #mm**2 \n",

+      "\n",

+      "k=(I*A**-1)**0.5\n",

+      "\n",

+      "#Slenderness ratio\n",

+      "Lamda=l*k**-1\n",

+      "\n",

+      "#From Rankine's Ratio\n",

+      "P=sigma_1*A*(1+a*(Lamda)**2)**-1\n",

+      "\n",

+      "#Safe Load\n",

+      "S=P*(FOS)**-1*10**-3 #N\n",

+      "\n",

+      "#Result\n",

+      "print\"Constant in the Formula is:a      \",round(a,6)\n",

+      "print\"                          :sigma_1\",round(sigma_1,2)\n",

+      "print\"Safe Load is\",round(S,2),\"KN\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Constant in the Formula is:a       0.000129\n",

+        "                          :sigma_1 326.6\n",

+        "Safe Load is 38.98 KN\n"

+       ]

+      }

+     ],

+     "prompt_number": 38

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 9.9.8,Page No.385"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "D=200 #mm #Depth\n",

+      "b=140 #mm #width\n",

+      "\n",

+      "#Plate\n",

+      "b2=160 #mm #Width\n",

+      "t2=10 #mm #Thickness\n",

+      "\n",

+      "L=l=4000 #mm #Length\n",

+      "FOS=4 #Factor of safety\n",

+      "sigma=315 #N/mm**2 #stress\n",

+      "a2=1*7500**-1 \n",

+      "I_xx=26.245*10**6 #mm**4 #M.I at x-x\n",

+      "I_yy=3.288*10**6 #mm**4 #M.I at y-y\n",

+      "a=3671 #mm**2 #Area\n",

+      "k_x=84.6#mm\n",

+      "k_y=29.9 #mm\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Total Area\n",

+      "A=a+2*t2*b2 #mm**2\n",

+      "\n",

+      "#M.I\n",

+      "I=I_yy+2*12**-1*t2*b2**3 #mm**4\n",

+      "\n",

+      "k=(I*A**-1)**0.5 #mm\n",

+      "\n",

+      "#Let X=L*k**-1\n",

+      "X=L*k**-1\n",

+      "\n",

+      "#Appliying Rankine's Formula\n",

+      "P=sigma*A*(1+a2*(X)**2)**-1 #N\n",

+      "\n",

+      "#Safe Load\n",

+      "S=P*(FOS)**-1*10**-3 #KN\n",

+      "\n",

+      "#Result\n",

+      "print\"Safe axial Load is\",round(S,2),\"KN\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Safe axial Load is 220.93 KN\n"

+       ]

+      }

+     ],

+     "prompt_number": 48

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 9.9.9,Page No.389"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "E=200*10**3 #N/mm**2 #Modulus of elasticity\n",

+      "sigma=330 #N/mm**2 #Stress\n",

+      "a=1*7500**-1 #Rankine's constant\n",

+      "A=5205 #mm**2 #area of column\n",

+      "I_xx=59.431*10**6 #mm**4 #M.I at x-x axis\n",

+      "I_yy=8.575*10**6 #mm**24#M.I at y-y axis\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Total M.I\n",

+      "I=I_xx+I_yy #mm**4\n",

+      "\n",

+      "#Area of compound Section \n",

+      "A2=2*A #mm**2\n",

+      "\n",

+      "k=(I*A2**-1)**0.5 #mm\n",

+      "\n",

+      "#Equating Euler's Load to Rankine's Load we get\n",

+      "#pi**2*E*I*(L**2)**-1=sigma*A*(1+a*(L*k)**2)**-1\n",

+      "#After Substitt=uting values and further simplifying we get\n",

+      "L=(39076198*(1-0.7975432)**-1)**0.5*10**-3 #m\n",

+      "\n",

+      "#Result\n",

+      "print\"Length of column for which Rankine's formula and Euler's Formula give the same result is\",round(L,2),\"m\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Length of column for which Rankine's formula and Euler's Formula give the same result is 13.89 m\n"

+       ]

+      }

+     ],

+     "prompt_number": 47

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 9.9.10,Page No.387"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "sigma=326 #N/mm**2 #stress\n",

+      "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",

+      "FOS=2 #Factor of safety\n",

+      "a=1*7500**-1 #Rankine's constant\n",

+      "D=350 #mm #Overall Depth \n",

+      "\n",

+      "#Cover plates\n",

+      "b1=500 #mm #width\n",

+      "t1=10  #mm #Thickness\n",

+      "\n",

+      "d=220 #mm #Distance between two channels\n",

+      "\n",

+      "L=6000 #mm #Length of column\n",

+      "\n",

+      "A=5366 #mm**2 #Area of Column section \n",

+      "I_xx=100.08*10**6 #mm**4 #M.I of x-x axis\n",

+      "I_yy=4.306*10**6 #mm**4 #M.I of y-y axis\n",

+      "C_yy=23.6 #mm #Centroid at y-y axis\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "#Symmetric axes are the centroidal axes is\n",

+      "\n",

+      "#M.I of Channel at x-x axis\n",

+      "I_xx_1=2*I_xx+2*(1*12**-1*b1*t1**3+b1*t1*(D*2**-1+t1*2**-1)**2)\n",

+      "\n",

+      "#M.I of Channel at y-y axis\n",

+      "I_yy_1=2*(I_yy+A*(d*2**-1+C_yy)**2)+2*12**-1*t1*b1**3\n",

+      "\n",

+      "#As I_yy<I_xx\n",

+      "#So\n",

+      "I=I_yy_1 #mm**4 \n",

+      "\n",

+      "A2=2*A+2*t1*b1 #Area of channel\n",

+      "\n",

+      "k=(I*A2**-1)**0.5 #mm\n",

+      "\n",

+      "#Critical Load\n",

+      "P=sigma*A2*(1+a*(L*k**-1)**2)**-1 \n",

+      "\n",

+      "#Safe Load\n",

+      "S=P*2**-1*10**-3 #KN\n",

+      "\n",

+      "#Result\n",

+      "print\"Safe Load carrying Capacity is\",round(S,2),\"KN\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Safe Load carrying Capacity is 2717.35 KN\n"

+       ]

+      }

+     ],

+     "prompt_number": 67

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example 9.9.11,Page No.390"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "\n",

+      "#Initilization of Variables\n",

+      "\n",

+      "I=4.085*10**8 #mm**4 #M.I\n",

+      "A=20732.0 #mm**2 #area of column\n",

+      "f_y=250 #N/mm**2 \n",

+      "L=6000 #mm #Length of column\n",

+      "\n",

+      "#Calculations\n",

+      "\n",

+      "k=(I*A**-1)**0.5 #mm\n",

+      "lamda=L*k**-1 #Slenderness ratro\n",

+      "\n",

+      "#From Indian standard table\n",

+      "lamda_1=40 \n",

+      "sigma_a_c_1=139 #N/mm**2\n",

+      "lamda_2=50 \n",

+      "sigma_a_c_2=132 #N/mm**2 \n",

+      "\n",

+      "#Linearly interpolating between these values for lambda=42.744\n",

+      "\n",

+      "sigma_a_c_3=sigma_a_c_1-2.744*10**-1*(sigma_a_c_1-sigma_a_c_2)\n",

+      "\n",

+      "#Safe Load carrying capacity of column\n",

+      "P=sigma_a_c_3*A*10**-3\n",

+      "\n",

+      "#Result\n",

+      "print\"Safe Load carrying capacity is\",round(P,2),\"KN\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Safe Load carrying capacity is 2841.93 KN\n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/B.M.D_1.JPG b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/B.M.D_1.JPG
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diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_2.jpg b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_2.jpg
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diff --git a/sample_notebooks/DaudIbrahir Saifi/Chapter_07_1.ipynb b/sample_notebooks/DaudIbrahir Saifi/Chapter_07_1.ipynb
new file mode 100644
index 00000000..aefd1c71
--- /dev/null
+++ b/sample_notebooks/DaudIbrahir Saifi/Chapter_07_1.ipynb
@@ -0,0 +1,342 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:012ab8557afdcfdae2cdc3da17271647415fc17ab95dd187f4df0903472edf45"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter - 7 : Cathode Ray Oscilloscopes"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  7.1 - Page No : 244"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "l=2.5 # in cm\n",
+      "l=l*10**-2 # in meter\n",
+      "d=.5 # in cm\n",
+      "d=d*10**-2 # in meter\n",
+      "S= 20 # in cm\n",
+      "S= S*10**-2 # in meter\n",
+      "Va= 2500 # in volts\n",
+      "# Formula y = OC*AB/OB = (S*d/2)/(l/2)\n",
+      "y = (S*d/2)/(l/2) # in meter\n",
+      "print \"The value of deflection = %0.f cm\" %(y*10**2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of deflection = 4 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  7.2 - Page No : 244"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " #Given data\n",
+      "R_E1= 5.6 # in kohm\n",
+      "C1= 0.2 # in micro F\n",
+      "V_B1= 6.3 # in volt\n",
+      "V_BE= 0.7 # in volt\n",
+      "TL= 2.5 # trigger level for the Schmitt trigger (UTP,LTP) in volt\n",
+      "del_V1= 2*TL # in volt\n",
+      "I_C1= (V_B1-V_BE)/R_E1 # in mA\n",
+      "print \"Charging current = %0.f mA\" %I_C1 \n",
+      "toh= del_V1*C1/I_C1 # in ms\n",
+      "print \"Time period = %0.f ms\" %toh"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Charging current = 1 mA\n",
+        "Time period = 1 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  7.3 - Page No : 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt \n",
+      "#Given data\n",
+      "L=10 # trace length in cm\n",
+      "DS= 5 # deflection sensitivity in V/cm\n",
+      "V_peakTOpeak= L*DS # in volt\n",
+      "V_peak= V_peakTOpeak/2 # in volt\n",
+      "RMS= V_peak/sqrt(2) # RMS value of unknown as voltage in volt\n",
+      "print \"The value of AC voltage = %0.3f volts\" %RMS "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of AC voltage = 17.678 volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  7.4 - Page No : 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division \n",
+      "#Given data\n",
+      "Y= 2+1/2 # Positive Y-peaks in pattern\n",
+      "X= 1/2+1/2 # Positive X-peaks in pattern\n",
+      "f_h= 3# frequency of horizontal voltage signal in kHz\n",
+      "f_yBYf_x= Y/X \n",
+      "# frequency of vertical voltage signal= f_yBYf_x * f_h\n",
+      "f_v= f_yBYf_x * f_h # frequency of vertical voltage signal in kHz\n",
+      "print \"frequency of vertical voltage signal = %0.1f kHz\" %f_v "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "frequency of vertical voltage signal = 7.5 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  7.5 - Page No : 256"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " #Given data\n",
+      "f_x= 1000 # in Hz\n",
+      "Y= 2 # points of tangency to vertical line\n",
+      "X= 5 # points of tangency to horizontal line\n",
+      "f_y= f_x*X/Y # in Hz\n",
+      "print \"Frequency of vertical input = %0.f Hz\" %f_y"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Frequency of vertical input = 2500 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  7.6 - Page No : 257"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " #Given data\n",
+      "f=2000 # in Hz\n",
+      "T=1/f # in sec\n",
+      "D=0.2 \n",
+      "PulseDuration= D*T # in sec\n",
+      "print \"The value of pulse duration = %0.1f ms\" %(PulseDuration*10**3) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of pulse duration = 0.1 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  7.7 - Page No : 258"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " #Given data\n",
+      "vertical_attenuation= 0.5 # in V/Div\n",
+      "TPD= 2 # time/Div control in micro sec\n",
+      "P= 4*vertical_attenuation # peak-to-peak amplitude of the signal in V \n",
+      "print \"Peak-to-Peak amplitude of the signal = %0.f V\" %P\n",
+      "T= 4*TPD # in micro sec\n",
+      "T=T*10**-6 # in sec\n",
+      "f=1/T # in Hz\n",
+      "print \"The value of frequency = %0.f kHz\" %(f*10**-3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Peak-to-Peak amplitude of the signal = 2 V\n",
+        "The value of frequency = 125 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  7.8 - Page No : 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi \n",
+      "#Given data\n",
+      "C_1N= 36 # in pF\n",
+      "C_2= 150 # in pF\n",
+      "R_1N= 1 # in M ohm\n",
+      "R_1= 10 # in M ohm\n",
+      "R_source= 500 # in ohm\n",
+      "# R_1/(omega*(C_2+C_1N)) = R_1N/(omega*C_1)\n",
+      "C_1= R_1N*(C_2+C_1N)/R_1 # in pF\n",
+      "C_T= 1/(1/C_1+1/(C_2+C_1N)) # in pF\n",
+      "C_T= C_T*10**-12 # in F\n",
+      "f= 1/(2*pi*C_T*R_source) \n",
+      "print \"Signal Frequency = %0.2f MHz\" %(f*10**-6)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Signal Frequency = 18.82 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  7.9 - Page No : 263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " #Given data\n",
+      "f= 20 # in MHz\n",
+      "f=f*10**6 # in Hz\n",
+      "toh= 1/f # in sec\n",
+      "toh=toh*10**9 # in ns\n",
+      "# For one cycle occupying 4 horizontal divisions,\n",
+      "MTD= toh/4 # Minimum time/division in ns/division\n",
+      "# Using the 10 times magnifier to provide MTD\n",
+      "MTD_setting= 10*MTD # minimum time/division setting in ns/division\n",
+      "print \"Minimum time/division setting = %0.f ns/division\" %MTD_setting"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum time/division setting = 125 ns/division\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file