Revised list of TBCs

53 files changed, 0 insertions, 15551 deletions

diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA.ipynb
deleted file mode 100755
index 7f8e8cd8..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA.ipynb
+++ /dev/null
@@ -1,320 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8adf459ea4014e5f8ac2990e8d05feaccf30880e758648b173bb1fd18b74b81a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Appendix A: Review of Properties of Plane Area"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.1, Page No:486"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A=2000 #Area of the plane in mm^2\n",
-      "Ix=40*10**6 #Momnet of Inertia in mm^4\n",
-      "d1=90 #Distance in mm\n",
-      "d2=70 #Distance in mm\n",
-      "\n",
-      "#Calculations\n",
-      "Ix_bar=Ix-(A*d1**2) #Moment of Inertia along x_bar axis in mm^4\n",
-      "Iu=Ix_bar+A*d2**2 #Moment of Inertia along U-axis in mm^4\n",
-      "\n",
-      "#Result\n",
-      "print Ix_bar\n",
-      "print \"The moment of inertia along u-axis is\",round(Iu,1),\"mm^4\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "23800000\n",
-        "The moment of inertia along u-axis is 33600000.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.2, Page No:486"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "R=45 #Radius of the circle in mm\n",
-      "r=20 #Radius of the smaller circle in mm\n",
-      "h=100 #Depth of the straight section in mm\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "\n",
-      "#Triangle\n",
-      "b=2*R #Breadth in mm\n",
-      "A_t=b*h*0.5 #Area in mm^2\n",
-      "Ix_bar_t=b*h**3*36**-1 #Moment of inertia in mm^4\n",
-      "y_bar1=2*3**-1*h #centroidal axis in mm\n",
-      "Ix_t=Ix_bar_t+A_t*y_bar1**2 #moment of inertia in mm^4\n",
-      "\n",
-      "#Semi-circle\n",
-      "A_sc=pi*R**2*0.5 #Area of the semi-circle in mm^2\n",
-      "Ix_bar_sc=0.1098*R**4 #Moment of inertia in mm^4\n",
-      "y_bar2=h+(4*R*(3*pi)**-1) #Distance of centroid in mm\n",
-      "Ix_sc=Ix_bar_sc+A_sc*y_bar2**2 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Circle\n",
-      "A_c=pi*r**2 #Area of the circle in mm^2\n",
-      "Ix_bar_c=pi*r**4*4**-1 #Moment of inertia in mm^4\n",
-      "y_bar3=h #Distance of centroid in mm\n",
-      "Ix_c=Ix_bar_c+A_c*y_bar3**2 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Composite Area\n",
-      "A=A_t+A_sc-A_c #Total area in mm^2\n",
-      "Ix=Ix_t+Ix_sc-Ix_c #Moment of inertia in mm^4\n",
-      "\n",
-      "#Part 2\n",
-      "y_bar=(A_t*y_bar1+A_sc*y_bar2-A_c*y_bar3)/(A) #Location of centroid in mm\n",
-      "Ix_bar=Ix-A*y_bar**2 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Result\n",
-      "print \"Moment of inertia about x-axis is\",round(Ix),\"mm^4\"\n",
-      "print \"Moment of inertia about the centroidal axis is\",round(Ix_bar),\"mm^4\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Moment of inertia about x-axis is 55377079.0 mm^4\n",
-        "Moment of inertia about the centroidal axis is 7744899.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.3, Page No:488"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "t=20 #Thickness in mm\n",
-      "h=140 #Depth in mm\n",
-      "w=180 #Width in mm\n",
-      "\n",
-      "#Calculations\n",
-      "Ixy_1=0+(h*t*t*0.5*h*0.5) #product of inertia in mm^4\n",
-      "Ixy_2=0+((w-t)*t*(w+t)*0.5*t*0.5) #Product of inertia in mm^4\n",
-      "Ixy=Ixy_1+Ixy_2 #Product of inertia in mm^4\n",
-      "\n",
-      "#Result\n",
-      "print \"The Product of inertia is\",round(Ixy),\"mm^4\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Product of inertia is 5160000.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.4, Page No:495"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "t=30 #Thickness in mm\n",
-      "h=200 #Depth of the section in mm\n",
-      "w=160 #Width in mm\n",
-      "the=50 #Angle in degrees\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "A1=t*h #Area of the web portion in mm^2\n",
-      "A2=(w-t)*t #Area of the flange portion in mm^2\n",
-      "x_bar=(A1*t*0.5+A2*(t+(w-t)*0.5))/(A1+A2) #Location of x_bar in mm\n",
-      "y_bar=(A1*h*0.5+A2*t*0.5)/(A1+A2) #Location of y_bar in mm\n",
-      "\n",
-      "#Simplfying the computation\n",
-      "a=t*h**3*12**-1\n",
-      "b=A1*(200*0.5-y_bar)**2\n",
-      "c=(w-t)*t**3*12**-1\n",
-      "d=A2*(t*0.5-y_bar)**2\n",
-      "Ix_bar=a+b+c+d #Moment of inertia about x-axis in mm^4\n",
-      "\n",
-      "#Simplifying the computation\n",
-      "p=h*t**3*12**-1\n",
-      "q=A1*(t*0.5-x_bar)**2\n",
-      "r=t*(w-t)**3*12**-1\n",
-      "s=A2*((w-t)*0.5+t-x_bar)**2\n",
-      "Iy_bar=p+q+r+s #Moment of inertia about y-axis in mm^4\n",
-      "\n",
-      "#Simplfying the computation\n",
-      "a1=(t*0.5-x_bar)*(h*0.5-y_bar)\n",
-      "a2=(t*0.5-y_bar)*((w-t)*0.5+t-x_bar)\n",
-      "Ixy_bar=A1*a1+A2*a2 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Part 1\n",
-      "#Simplfying the computation\n",
-      "a3=(Ix_bar+Iy_bar)*0.5\n",
-      "a4=(0.5*(Ix_bar-Iy_bar))**2\n",
-      "a5=Ixy_bar**2\n",
-      "I1=a3+np.sqrt(a4+a5) #Moment of inertia in mm^4\n",
-      "I2=a3-np.sqrt(a4+a5) #Moment of inertia in mm^4\n",
-      "\n",
-      "ThetaRHS=-(2*Ixy_bar)/(Ix_bar-Iy_bar) #RHS of the tan term\n",
-      "theta1=arctan(ThetaRHS)*0.5*180*pi**-1 #Angle in degrees\n",
-      "theta2=theta1+90 #Angle in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "Iu=a3+np.sqrt(a4)*np.cos(2*the*pi*180**-1)-(Ixy_bar)\\\n",
-      "      *np.sin(2*the*pi*180**-1) #Moment of inertia in mm^4\n",
-      "Iv=a3-np.sqrt(a4)*np.cos(2*the*pi*180**-1)+(Ixy_bar)\\\n",
-      "      *np.sin(2*the*pi*180**-1) #Moment of inertia in mm^4\n",
-      "Iuv=np.sqrt(a4)*np.sin(2*the*pi*180**-1)+(Ixy_bar)\\\n",
-      "      *np.cos(2*the*pi*180**-1) #Moment of inertia in mm^4\n",
-      "    \n",
-      "    \n",
-      "#Result\n",
-      "print \"The Principal Moment of inertias are as follows\"\n",
-      "print \"I1=\",round(I1),\"mm^4 and I2=\",round(I2),\"mm^4\"\n",
-      "print \"Princial direction are theta1=\",round(theta1,1), \"degrees\"\\\n",
-      "       \" theta2=\",round(theta2,1),\"degrees\"\n",
-      "print \"The moment of inertia along the uv-axis is\",round(Iuv),\"mm^4\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Principal Moment of inertias are as follows\n",
-        "I1= 47240734.0 mm^4 and I2= 11198811.0 mm^4\n",
-        "Princial direction are theta1= 31.6 degrees theta2= 121.6 degrees\n",
-        "The moment of inertia along the uv-axis is 10817183.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.5, Page No:497"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Ix_bar=37.37*10**6 #Moment of inertia in mm^4\n",
-      "Iy_bar=21.07*10**6 #Moment of inertia in mm^4\n",
-      "Ixy_bar=-16.073*10**6 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Calculations\n",
-      "b=(Ix_bar+Iy_bar)*0.5 #Parameter for the circle in mm^4\n",
-      "R=sqrt(((Ix_bar-Iy_bar)*0.5)**2+Ixy_bar**2) #Radius of the Mohr's Circle in mm^4\n",
-      "\n",
-      "#Part 1\n",
-      "I1=b+R #MI in mm^4\n",
-      "I2=b-R #MI in mm^4\n",
-      "theta1=arcsin(abs(Ixy_bar)/R)*180*pi**-1*0.5 #Angle in degrees\n",
-      "theta2=theta1+90 #Angle in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "alpha=(100-theta1*2)*0.5 #Angle in degrees\n",
-      "Iu=round(b,2)+round(R,3)*round(np.cos(alpha*pi*180**-1),2) #MI in mm^4\n",
-      "Iv=round(b,2)-round(R,3)*round(np.cos(alpha*pi*180**-1),2) #MI in mm^4\n",
-      "Iuv=R*np.sin(2*alpha*pi*180**-1) #MI in mm^4\n",
-      "\n",
-      "#Result\n",
-      "print \"The Principal Moment of inertias are as follows\"\n",
-      "print \"I1=\",round(I1),\"mm^4 and I2=\",round(I2),\"mm^4\"\n",
-      "print \"Princial direction are theta1=\",round(theta1,1), \"degrees\"\\\n",
-      "       \" theta2=\",round(theta2,1),\"degrees\"\n",
-      "print \"The moment of inertia along the uv-axis is\",round(Iuv),\"mm^4\"  \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Principal Moment of inertias are as follows\n",
-        "I1= 47241205.0 mm^4 and I2= 11198795.0 mm^4\n",
-        "Princial direction are theta1= 31.6 degrees theta2= 121.6 degrees\n",
-        "The moment of inertia along the uv-axis is 10817230.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 73
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA_1.ipynb
deleted file mode 100755
index 7f8e8cd8..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA_1.ipynb
+++ /dev/null
@@ -1,320 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8adf459ea4014e5f8ac2990e8d05feaccf30880e758648b173bb1fd18b74b81a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Appendix A: Review of Properties of Plane Area"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.1, Page No:486"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A=2000 #Area of the plane in mm^2\n",
-      "Ix=40*10**6 #Momnet of Inertia in mm^4\n",
-      "d1=90 #Distance in mm\n",
-      "d2=70 #Distance in mm\n",
-      "\n",
-      "#Calculations\n",
-      "Ix_bar=Ix-(A*d1**2) #Moment of Inertia along x_bar axis in mm^4\n",
-      "Iu=Ix_bar+A*d2**2 #Moment of Inertia along U-axis in mm^4\n",
-      "\n",
-      "#Result\n",
-      "print Ix_bar\n",
-      "print \"The moment of inertia along u-axis is\",round(Iu,1),\"mm^4\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "23800000\n",
-        "The moment of inertia along u-axis is 33600000.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.2, Page No:486"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "R=45 #Radius of the circle in mm\n",
-      "r=20 #Radius of the smaller circle in mm\n",
-      "h=100 #Depth of the straight section in mm\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "\n",
-      "#Triangle\n",
-      "b=2*R #Breadth in mm\n",
-      "A_t=b*h*0.5 #Area in mm^2\n",
-      "Ix_bar_t=b*h**3*36**-1 #Moment of inertia in mm^4\n",
-      "y_bar1=2*3**-1*h #centroidal axis in mm\n",
-      "Ix_t=Ix_bar_t+A_t*y_bar1**2 #moment of inertia in mm^4\n",
-      "\n",
-      "#Semi-circle\n",
-      "A_sc=pi*R**2*0.5 #Area of the semi-circle in mm^2\n",
-      "Ix_bar_sc=0.1098*R**4 #Moment of inertia in mm^4\n",
-      "y_bar2=h+(4*R*(3*pi)**-1) #Distance of centroid in mm\n",
-      "Ix_sc=Ix_bar_sc+A_sc*y_bar2**2 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Circle\n",
-      "A_c=pi*r**2 #Area of the circle in mm^2\n",
-      "Ix_bar_c=pi*r**4*4**-1 #Moment of inertia in mm^4\n",
-      "y_bar3=h #Distance of centroid in mm\n",
-      "Ix_c=Ix_bar_c+A_c*y_bar3**2 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Composite Area\n",
-      "A=A_t+A_sc-A_c #Total area in mm^2\n",
-      "Ix=Ix_t+Ix_sc-Ix_c #Moment of inertia in mm^4\n",
-      "\n",
-      "#Part 2\n",
-      "y_bar=(A_t*y_bar1+A_sc*y_bar2-A_c*y_bar3)/(A) #Location of centroid in mm\n",
-      "Ix_bar=Ix-A*y_bar**2 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Result\n",
-      "print \"Moment of inertia about x-axis is\",round(Ix),\"mm^4\"\n",
-      "print \"Moment of inertia about the centroidal axis is\",round(Ix_bar),\"mm^4\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Moment of inertia about x-axis is 55377079.0 mm^4\n",
-        "Moment of inertia about the centroidal axis is 7744899.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.3, Page No:488"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "t=20 #Thickness in mm\n",
-      "h=140 #Depth in mm\n",
-      "w=180 #Width in mm\n",
-      "\n",
-      "#Calculations\n",
-      "Ixy_1=0+(h*t*t*0.5*h*0.5) #product of inertia in mm^4\n",
-      "Ixy_2=0+((w-t)*t*(w+t)*0.5*t*0.5) #Product of inertia in mm^4\n",
-      "Ixy=Ixy_1+Ixy_2 #Product of inertia in mm^4\n",
-      "\n",
-      "#Result\n",
-      "print \"The Product of inertia is\",round(Ixy),\"mm^4\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Product of inertia is 5160000.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.4, Page No:495"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "t=30 #Thickness in mm\n",
-      "h=200 #Depth of the section in mm\n",
-      "w=160 #Width in mm\n",
-      "the=50 #Angle in degrees\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "A1=t*h #Area of the web portion in mm^2\n",
-      "A2=(w-t)*t #Area of the flange portion in mm^2\n",
-      "x_bar=(A1*t*0.5+A2*(t+(w-t)*0.5))/(A1+A2) #Location of x_bar in mm\n",
-      "y_bar=(A1*h*0.5+A2*t*0.5)/(A1+A2) #Location of y_bar in mm\n",
-      "\n",
-      "#Simplfying the computation\n",
-      "a=t*h**3*12**-1\n",
-      "b=A1*(200*0.5-y_bar)**2\n",
-      "c=(w-t)*t**3*12**-1\n",
-      "d=A2*(t*0.5-y_bar)**2\n",
-      "Ix_bar=a+b+c+d #Moment of inertia about x-axis in mm^4\n",
-      "\n",
-      "#Simplifying the computation\n",
-      "p=h*t**3*12**-1\n",
-      "q=A1*(t*0.5-x_bar)**2\n",
-      "r=t*(w-t)**3*12**-1\n",
-      "s=A2*((w-t)*0.5+t-x_bar)**2\n",
-      "Iy_bar=p+q+r+s #Moment of inertia about y-axis in mm^4\n",
-      "\n",
-      "#Simplfying the computation\n",
-      "a1=(t*0.5-x_bar)*(h*0.5-y_bar)\n",
-      "a2=(t*0.5-y_bar)*((w-t)*0.5+t-x_bar)\n",
-      "Ixy_bar=A1*a1+A2*a2 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Part 1\n",
-      "#Simplfying the computation\n",
-      "a3=(Ix_bar+Iy_bar)*0.5\n",
-      "a4=(0.5*(Ix_bar-Iy_bar))**2\n",
-      "a5=Ixy_bar**2\n",
-      "I1=a3+np.sqrt(a4+a5) #Moment of inertia in mm^4\n",
-      "I2=a3-np.sqrt(a4+a5) #Moment of inertia in mm^4\n",
-      "\n",
-      "ThetaRHS=-(2*Ixy_bar)/(Ix_bar-Iy_bar) #RHS of the tan term\n",
-      "theta1=arctan(ThetaRHS)*0.5*180*pi**-1 #Angle in degrees\n",
-      "theta2=theta1+90 #Angle in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "Iu=a3+np.sqrt(a4)*np.cos(2*the*pi*180**-1)-(Ixy_bar)\\\n",
-      "      *np.sin(2*the*pi*180**-1) #Moment of inertia in mm^4\n",
-      "Iv=a3-np.sqrt(a4)*np.cos(2*the*pi*180**-1)+(Ixy_bar)\\\n",
-      "      *np.sin(2*the*pi*180**-1) #Moment of inertia in mm^4\n",
-      "Iuv=np.sqrt(a4)*np.sin(2*the*pi*180**-1)+(Ixy_bar)\\\n",
-      "      *np.cos(2*the*pi*180**-1) #Moment of inertia in mm^4\n",
-      "    \n",
-      "    \n",
-      "#Result\n",
-      "print \"The Principal Moment of inertias are as follows\"\n",
-      "print \"I1=\",round(I1),\"mm^4 and I2=\",round(I2),\"mm^4\"\n",
-      "print \"Princial direction are theta1=\",round(theta1,1), \"degrees\"\\\n",
-      "       \" theta2=\",round(theta2,1),\"degrees\"\n",
-      "print \"The moment of inertia along the uv-axis is\",round(Iuv),\"mm^4\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Principal Moment of inertias are as follows\n",
-        "I1= 47240734.0 mm^4 and I2= 11198811.0 mm^4\n",
-        "Princial direction are theta1= 31.6 degrees theta2= 121.6 degrees\n",
-        "The moment of inertia along the uv-axis is 10817183.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.5, Page No:497"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Ix_bar=37.37*10**6 #Moment of inertia in mm^4\n",
-      "Iy_bar=21.07*10**6 #Moment of inertia in mm^4\n",
-      "Ixy_bar=-16.073*10**6 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Calculations\n",
-      "b=(Ix_bar+Iy_bar)*0.5 #Parameter for the circle in mm^4\n",
-      "R=sqrt(((Ix_bar-Iy_bar)*0.5)**2+Ixy_bar**2) #Radius of the Mohr's Circle in mm^4\n",
-      "\n",
-      "#Part 1\n",
-      "I1=b+R #MI in mm^4\n",
-      "I2=b-R #MI in mm^4\n",
-      "theta1=arcsin(abs(Ixy_bar)/R)*180*pi**-1*0.5 #Angle in degrees\n",
-      "theta2=theta1+90 #Angle in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "alpha=(100-theta1*2)*0.5 #Angle in degrees\n",
-      "Iu=round(b,2)+round(R,3)*round(np.cos(alpha*pi*180**-1),2) #MI in mm^4\n",
-      "Iv=round(b,2)-round(R,3)*round(np.cos(alpha*pi*180**-1),2) #MI in mm^4\n",
-      "Iuv=R*np.sin(2*alpha*pi*180**-1) #MI in mm^4\n",
-      "\n",
-      "#Result\n",
-      "print \"The Principal Moment of inertias are as follows\"\n",
-      "print \"I1=\",round(I1),\"mm^4 and I2=\",round(I2),\"mm^4\"\n",
-      "print \"Princial direction are theta1=\",round(theta1,1), \"degrees\"\\\n",
-      "       \" theta2=\",round(theta2,1),\"degrees\"\n",
-      "print \"The moment of inertia along the uv-axis is\",round(Iuv),\"mm^4\"  \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Principal Moment of inertias are as follows\n",
-        "I1= 47241205.0 mm^4 and I2= 11198795.0 mm^4\n",
-        "Princial direction are theta1= 31.6 degrees theta2= 121.6 degrees\n",
-        "The moment of inertia along the uv-axis is 10817230.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 73
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA_2.ipynb
deleted file mode 100755
index 7f8e8cd8..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA_2.ipynb
+++ /dev/null
@@ -1,320 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8adf459ea4014e5f8ac2990e8d05feaccf30880e758648b173bb1fd18b74b81a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Appendix A: Review of Properties of Plane Area"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.1, Page No:486"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A=2000 #Area of the plane in mm^2\n",
-      "Ix=40*10**6 #Momnet of Inertia in mm^4\n",
-      "d1=90 #Distance in mm\n",
-      "d2=70 #Distance in mm\n",
-      "\n",
-      "#Calculations\n",
-      "Ix_bar=Ix-(A*d1**2) #Moment of Inertia along x_bar axis in mm^4\n",
-      "Iu=Ix_bar+A*d2**2 #Moment of Inertia along U-axis in mm^4\n",
-      "\n",
-      "#Result\n",
-      "print Ix_bar\n",
-      "print \"The moment of inertia along u-axis is\",round(Iu,1),\"mm^4\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "23800000\n",
-        "The moment of inertia along u-axis is 33600000.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.2, Page No:486"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "R=45 #Radius of the circle in mm\n",
-      "r=20 #Radius of the smaller circle in mm\n",
-      "h=100 #Depth of the straight section in mm\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "\n",
-      "#Triangle\n",
-      "b=2*R #Breadth in mm\n",
-      "A_t=b*h*0.5 #Area in mm^2\n",
-      "Ix_bar_t=b*h**3*36**-1 #Moment of inertia in mm^4\n",
-      "y_bar1=2*3**-1*h #centroidal axis in mm\n",
-      "Ix_t=Ix_bar_t+A_t*y_bar1**2 #moment of inertia in mm^4\n",
-      "\n",
-      "#Semi-circle\n",
-      "A_sc=pi*R**2*0.5 #Area of the semi-circle in mm^2\n",
-      "Ix_bar_sc=0.1098*R**4 #Moment of inertia in mm^4\n",
-      "y_bar2=h+(4*R*(3*pi)**-1) #Distance of centroid in mm\n",
-      "Ix_sc=Ix_bar_sc+A_sc*y_bar2**2 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Circle\n",
-      "A_c=pi*r**2 #Area of the circle in mm^2\n",
-      "Ix_bar_c=pi*r**4*4**-1 #Moment of inertia in mm^4\n",
-      "y_bar3=h #Distance of centroid in mm\n",
-      "Ix_c=Ix_bar_c+A_c*y_bar3**2 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Composite Area\n",
-      "A=A_t+A_sc-A_c #Total area in mm^2\n",
-      "Ix=Ix_t+Ix_sc-Ix_c #Moment of inertia in mm^4\n",
-      "\n",
-      "#Part 2\n",
-      "y_bar=(A_t*y_bar1+A_sc*y_bar2-A_c*y_bar3)/(A) #Location of centroid in mm\n",
-      "Ix_bar=Ix-A*y_bar**2 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Result\n",
-      "print \"Moment of inertia about x-axis is\",round(Ix),\"mm^4\"\n",
-      "print \"Moment of inertia about the centroidal axis is\",round(Ix_bar),\"mm^4\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Moment of inertia about x-axis is 55377079.0 mm^4\n",
-        "Moment of inertia about the centroidal axis is 7744899.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.3, Page No:488"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "t=20 #Thickness in mm\n",
-      "h=140 #Depth in mm\n",
-      "w=180 #Width in mm\n",
-      "\n",
-      "#Calculations\n",
-      "Ixy_1=0+(h*t*t*0.5*h*0.5) #product of inertia in mm^4\n",
-      "Ixy_2=0+((w-t)*t*(w+t)*0.5*t*0.5) #Product of inertia in mm^4\n",
-      "Ixy=Ixy_1+Ixy_2 #Product of inertia in mm^4\n",
-      "\n",
-      "#Result\n",
-      "print \"The Product of inertia is\",round(Ixy),\"mm^4\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Product of inertia is 5160000.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.4, Page No:495"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "t=30 #Thickness in mm\n",
-      "h=200 #Depth of the section in mm\n",
-      "w=160 #Width in mm\n",
-      "the=50 #Angle in degrees\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "A1=t*h #Area of the web portion in mm^2\n",
-      "A2=(w-t)*t #Area of the flange portion in mm^2\n",
-      "x_bar=(A1*t*0.5+A2*(t+(w-t)*0.5))/(A1+A2) #Location of x_bar in mm\n",
-      "y_bar=(A1*h*0.5+A2*t*0.5)/(A1+A2) #Location of y_bar in mm\n",
-      "\n",
-      "#Simplfying the computation\n",
-      "a=t*h**3*12**-1\n",
-      "b=A1*(200*0.5-y_bar)**2\n",
-      "c=(w-t)*t**3*12**-1\n",
-      "d=A2*(t*0.5-y_bar)**2\n",
-      "Ix_bar=a+b+c+d #Moment of inertia about x-axis in mm^4\n",
-      "\n",
-      "#Simplifying the computation\n",
-      "p=h*t**3*12**-1\n",
-      "q=A1*(t*0.5-x_bar)**2\n",
-      "r=t*(w-t)**3*12**-1\n",
-      "s=A2*((w-t)*0.5+t-x_bar)**2\n",
-      "Iy_bar=p+q+r+s #Moment of inertia about y-axis in mm^4\n",
-      "\n",
-      "#Simplfying the computation\n",
-      "a1=(t*0.5-x_bar)*(h*0.5-y_bar)\n",
-      "a2=(t*0.5-y_bar)*((w-t)*0.5+t-x_bar)\n",
-      "Ixy_bar=A1*a1+A2*a2 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Part 1\n",
-      "#Simplfying the computation\n",
-      "a3=(Ix_bar+Iy_bar)*0.5\n",
-      "a4=(0.5*(Ix_bar-Iy_bar))**2\n",
-      "a5=Ixy_bar**2\n",
-      "I1=a3+np.sqrt(a4+a5) #Moment of inertia in mm^4\n",
-      "I2=a3-np.sqrt(a4+a5) #Moment of inertia in mm^4\n",
-      "\n",
-      "ThetaRHS=-(2*Ixy_bar)/(Ix_bar-Iy_bar) #RHS of the tan term\n",
-      "theta1=arctan(ThetaRHS)*0.5*180*pi**-1 #Angle in degrees\n",
-      "theta2=theta1+90 #Angle in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "Iu=a3+np.sqrt(a4)*np.cos(2*the*pi*180**-1)-(Ixy_bar)\\\n",
-      "      *np.sin(2*the*pi*180**-1) #Moment of inertia in mm^4\n",
-      "Iv=a3-np.sqrt(a4)*np.cos(2*the*pi*180**-1)+(Ixy_bar)\\\n",
-      "      *np.sin(2*the*pi*180**-1) #Moment of inertia in mm^4\n",
-      "Iuv=np.sqrt(a4)*np.sin(2*the*pi*180**-1)+(Ixy_bar)\\\n",
-      "      *np.cos(2*the*pi*180**-1) #Moment of inertia in mm^4\n",
-      "    \n",
-      "    \n",
-      "#Result\n",
-      "print \"The Principal Moment of inertias are as follows\"\n",
-      "print \"I1=\",round(I1),\"mm^4 and I2=\",round(I2),\"mm^4\"\n",
-      "print \"Princial direction are theta1=\",round(theta1,1), \"degrees\"\\\n",
-      "       \" theta2=\",round(theta2,1),\"degrees\"\n",
-      "print \"The moment of inertia along the uv-axis is\",round(Iuv),\"mm^4\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Principal Moment of inertias are as follows\n",
-        "I1= 47240734.0 mm^4 and I2= 11198811.0 mm^4\n",
-        "Princial direction are theta1= 31.6 degrees theta2= 121.6 degrees\n",
-        "The moment of inertia along the uv-axis is 10817183.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example A.5, Page No:497"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Ix_bar=37.37*10**6 #Moment of inertia in mm^4\n",
-      "Iy_bar=21.07*10**6 #Moment of inertia in mm^4\n",
-      "Ixy_bar=-16.073*10**6 #Moment of inertia in mm^4\n",
-      "\n",
-      "#Calculations\n",
-      "b=(Ix_bar+Iy_bar)*0.5 #Parameter for the circle in mm^4\n",
-      "R=sqrt(((Ix_bar-Iy_bar)*0.5)**2+Ixy_bar**2) #Radius of the Mohr's Circle in mm^4\n",
-      "\n",
-      "#Part 1\n",
-      "I1=b+R #MI in mm^4\n",
-      "I2=b-R #MI in mm^4\n",
-      "theta1=arcsin(abs(Ixy_bar)/R)*180*pi**-1*0.5 #Angle in degrees\n",
-      "theta2=theta1+90 #Angle in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "alpha=(100-theta1*2)*0.5 #Angle in degrees\n",
-      "Iu=round(b,2)+round(R,3)*round(np.cos(alpha*pi*180**-1),2) #MI in mm^4\n",
-      "Iv=round(b,2)-round(R,3)*round(np.cos(alpha*pi*180**-1),2) #MI in mm^4\n",
-      "Iuv=R*np.sin(2*alpha*pi*180**-1) #MI in mm^4\n",
-      "\n",
-      "#Result\n",
-      "print \"The Principal Moment of inertias are as follows\"\n",
-      "print \"I1=\",round(I1),\"mm^4 and I2=\",round(I2),\"mm^4\"\n",
-      "print \"Princial direction are theta1=\",round(theta1,1), \"degrees\"\\\n",
-      "       \" theta2=\",round(theta2,1),\"degrees\"\n",
-      "print \"The moment of inertia along the uv-axis is\",round(Iuv),\"mm^4\"  \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Principal Moment of inertias are as follows\n",
-        "I1= 47241205.0 mm^4 and I2= 11198795.0 mm^4\n",
-        "Princial direction are theta1= 31.6 degrees theta2= 121.6 degrees\n",
-        "The moment of inertia along the uv-axis is 10817230.0 mm^4\n"
-       ]
-      }
-     ],
-     "prompt_number": 73
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01.ipynb
deleted file mode 100755
index 8594491e..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01.ipynb
+++ /dev/null
@@ -1,226 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:055c326b63bc3f150b8a5e433c724771cc1733887ec2b1e223ccf712fcd80848"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 01:Stress"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.1, Page No:9"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#NOTE:The notation has been changed to simplify the coding process\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P_AB=4000 #Axial Force at section 1 in lb\n",
-      "P_BC=5000 #Axial Force at section 2 in lb\n",
-      "P_CD=7000 #Axial Force at section 3 in lb\n",
-      "A_1=1.2 #Area at section 1 in in^2\n",
-      "A_2=1.8 #Area at section 2 in in^2\n",
-      "A_3=1.6 #Area at section 3 in in^2\n",
-      "\n",
-      "#Calculation\n",
-      "#S indicates sigma here\n",
-      "S_AB=P_AB/A_1 #Stress at section 1 in psi (T)\n",
-      "S_BC=P_BC/A_2 #Stress at section 2 in psi (C)\n",
-      "S_CD=P_CD/A_3 #Stress at section 3 in psi (C)\n",
-      "\n",
-      "#Result\n",
-      "print \"The stress at the three sections is given as\"\n",
-      "print \"Stress at section 1=\",round(S_AB),\"section 2=\",round(S_BC),\"section 3=\",S_CD\n",
-      "\n",
-      "#NOTE:The answer for the following example for section 1 and section 2\n",
-      "#are incorrect due to rounding in the textbook\n",
-      "#Computed values are correct"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The stress at the three sections is given as\n",
-        "Stress at section 1= 3333.0 section 2= 2778.0 section 3= 4375.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.2, Page No:10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Ay=40 #Vertical Reaction at A in kN\n",
-      "Hy=60 #Vertical Reaction at H in kN\n",
-      "Hx=0 #Horizontal Reaction at H in kN\n",
-      "y=3 #Height in m\n",
-      "x=5 #Distance in m\n",
-      "p=4 #Panel distance in m\n",
-      "A=900 #Area of the member in mm^2\n",
-      "P_C=30 #Force at point C in kN\n",
-      "\n",
-      "#Calculation\n",
-      "#Part 1\n",
-      "#Applying summation of forces in the x and y direction and equating to zero\n",
-      "P_AB=(-Ay)*(x*y**-1) #Force in member AB in kN\n",
-      "P_AC=-(p*x**-1*P_AB) #Force in member AC in kN\n",
-      "#Using stress=force/area\n",
-      "S_AC=(P_AC/A)*10**3 #Stress in member AC in MPa (T)\n",
-      "\n",
-      "#Part 2\n",
-      "#Sum of moments about point E to zero\n",
-      "P_BD=(Ay*p*2-(P_C*p))*y**-1 #Force in memeber AB in kN (C)\n",
-      "S_BD=(P_BD/A)*10**3 #Stress in member in MPa (C)\n",
-      "\n",
-      "#Result\n",
-      "print \"The Stress in member AC is\",round(S_AC,1),\"MPa (T)\"\n",
-      "print \"The Stress in member BD is\",round(S_BD,1),\"MPa (C)\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Stress in member AC is 59.3 MPa (T)\n",
-        "The Stress in member BD is 74.1 MPa (C)\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.3, Page No:11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as num\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A_AB=800 #Area of member AB in m^2\n",
-      "A_AC=400 #Area of member AC in m^2\n",
-      "W_AB=110 #Safe value of stress in Pa for AB\n",
-      "W_AC=120 #Safe value of stress in Pa for AC\n",
-      "theta1=60*3.14*180**-1 #Angle in radians\n",
-      "theta2=40*3.14*180**-1 #Angle in radians \n",
-      "\n",
-      "#Calculations\n",
-      "#Applying sum of forces \n",
-      "#Solving by matrix method putting W as 1\n",
-      "A=num.array([[-cos(theta1),cos(theta2)],[sin(theta1),sin(theta2)]])\n",
-      "B=num.array([[1],[1]])\n",
-      "C=inv(A)\n",
-      "D=C*B\n",
-      "\n",
-      "#Using newtons third law\n",
-      "#Two values of W hence the change in the notation\n",
-      "W1=(W_AB*A_AB)*(D[1,1])**-1 #Weight W in N\n",
-      "W2=(W_AC*A_AC)*(D[0,1])**-1 #Weight W in N\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum value of W allowable is\",round(W2*1000**-1,1),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum value of W allowable is 61.7 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.4, Page No:19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "d=3*4**-1 #Rivet diameter in inches\n",
-      "t=7*8**-1 #Thickness of the plate in inches\n",
-      "tau=14000 #Shear stress limit in psi\n",
-      "sigma_b=18000 #Normal stress limit in psi\n",
-      "\n",
-      "#Calculations\n",
-      "#Design Shear Stress in Rivets\n",
-      "V=tau*(d**2*(pi/4))*4 #Shear force maximum allowable in lb\n",
-      "#Design for bearing stress in plate\n",
-      "Pb=sigma_b*t*d*4 #lb\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum load that the joint can carry is\",round(V),\"lb\"\n",
-      "\n",
-      "#NOTE:The answer in the textbook is off by 40lb"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum load that the joint can carry is 24740.0 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_1.ipynb
deleted file mode 100755
index e69de29b..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_1.ipynb
+++ /dev/null
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_10.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_10.ipynb
deleted file mode 100755
index e69de29b..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_10.ipynb
+++ /dev/null
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_11.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_11.ipynb
deleted file mode 100755
index e69de29b..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_11.ipynb
+++ /dev/null
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_12.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_12.ipynb
deleted file mode 100755
index 8594491e..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_12.ipynb
+++ /dev/null
@@ -1,226 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:055c326b63bc3f150b8a5e433c724771cc1733887ec2b1e223ccf712fcd80848"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 01:Stress"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.1, Page No:9"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#NOTE:The notation has been changed to simplify the coding process\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P_AB=4000 #Axial Force at section 1 in lb\n",
-      "P_BC=5000 #Axial Force at section 2 in lb\n",
-      "P_CD=7000 #Axial Force at section 3 in lb\n",
-      "A_1=1.2 #Area at section 1 in in^2\n",
-      "A_2=1.8 #Area at section 2 in in^2\n",
-      "A_3=1.6 #Area at section 3 in in^2\n",
-      "\n",
-      "#Calculation\n",
-      "#S indicates sigma here\n",
-      "S_AB=P_AB/A_1 #Stress at section 1 in psi (T)\n",
-      "S_BC=P_BC/A_2 #Stress at section 2 in psi (C)\n",
-      "S_CD=P_CD/A_3 #Stress at section 3 in psi (C)\n",
-      "\n",
-      "#Result\n",
-      "print \"The stress at the three sections is given as\"\n",
-      "print \"Stress at section 1=\",round(S_AB),\"section 2=\",round(S_BC),\"section 3=\",S_CD\n",
-      "\n",
-      "#NOTE:The answer for the following example for section 1 and section 2\n",
-      "#are incorrect due to rounding in the textbook\n",
-      "#Computed values are correct"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The stress at the three sections is given as\n",
-        "Stress at section 1= 3333.0 section 2= 2778.0 section 3= 4375.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.2, Page No:10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Ay=40 #Vertical Reaction at A in kN\n",
-      "Hy=60 #Vertical Reaction at H in kN\n",
-      "Hx=0 #Horizontal Reaction at H in kN\n",
-      "y=3 #Height in m\n",
-      "x=5 #Distance in m\n",
-      "p=4 #Panel distance in m\n",
-      "A=900 #Area of the member in mm^2\n",
-      "P_C=30 #Force at point C in kN\n",
-      "\n",
-      "#Calculation\n",
-      "#Part 1\n",
-      "#Applying summation of forces in the x and y direction and equating to zero\n",
-      "P_AB=(-Ay)*(x*y**-1) #Force in member AB in kN\n",
-      "P_AC=-(p*x**-1*P_AB) #Force in member AC in kN\n",
-      "#Using stress=force/area\n",
-      "S_AC=(P_AC/A)*10**3 #Stress in member AC in MPa (T)\n",
-      "\n",
-      "#Part 2\n",
-      "#Sum of moments about point E to zero\n",
-      "P_BD=(Ay*p*2-(P_C*p))*y**-1 #Force in memeber AB in kN (C)\n",
-      "S_BD=(P_BD/A)*10**3 #Stress in member in MPa (C)\n",
-      "\n",
-      "#Result\n",
-      "print \"The Stress in member AC is\",round(S_AC,1),\"MPa (T)\"\n",
-      "print \"The Stress in member BD is\",round(S_BD,1),\"MPa (C)\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Stress in member AC is 59.3 MPa (T)\n",
-        "The Stress in member BD is 74.1 MPa (C)\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.3, Page No:11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as num\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A_AB=800 #Area of member AB in m^2\n",
-      "A_AC=400 #Area of member AC in m^2\n",
-      "W_AB=110 #Safe value of stress in Pa for AB\n",
-      "W_AC=120 #Safe value of stress in Pa for AC\n",
-      "theta1=60*3.14*180**-1 #Angle in radians\n",
-      "theta2=40*3.14*180**-1 #Angle in radians \n",
-      "\n",
-      "#Calculations\n",
-      "#Applying sum of forces \n",
-      "#Solving by matrix method putting W as 1\n",
-      "A=num.array([[-cos(theta1),cos(theta2)],[sin(theta1),sin(theta2)]])\n",
-      "B=num.array([[1],[1]])\n",
-      "C=inv(A)\n",
-      "D=C*B\n",
-      "\n",
-      "#Using newtons third law\n",
-      "#Two values of W hence the change in the notation\n",
-      "W1=(W_AB*A_AB)*(D[1,1])**-1 #Weight W in N\n",
-      "W2=(W_AC*A_AC)*(D[0,1])**-1 #Weight W in N\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum value of W allowable is\",round(W2*1000**-1,1),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum value of W allowable is 61.7 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.4, Page No:19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "d=3*4**-1 #Rivet diameter in inches\n",
-      "t=7*8**-1 #Thickness of the plate in inches\n",
-      "tau=14000 #Shear stress limit in psi\n",
-      "sigma_b=18000 #Normal stress limit in psi\n",
-      "\n",
-      "#Calculations\n",
-      "#Design Shear Stress in Rivets\n",
-      "V=tau*(d**2*(pi/4))*4 #Shear force maximum allowable in lb\n",
-      "#Design for bearing stress in plate\n",
-      "Pb=sigma_b*t*d*4 #lb\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum load that the joint can carry is\",round(V),\"lb\"\n",
-      "\n",
-      "#NOTE:The answer in the textbook is off by 40lb"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum load that the joint can carry is 24740.0 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_13.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_13.ipynb
deleted file mode 100755
index 8594491e..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_13.ipynb
+++ /dev/null
@@ -1,226 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:055c326b63bc3f150b8a5e433c724771cc1733887ec2b1e223ccf712fcd80848"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 01:Stress"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.1, Page No:9"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#NOTE:The notation has been changed to simplify the coding process\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P_AB=4000 #Axial Force at section 1 in lb\n",
-      "P_BC=5000 #Axial Force at section 2 in lb\n",
-      "P_CD=7000 #Axial Force at section 3 in lb\n",
-      "A_1=1.2 #Area at section 1 in in^2\n",
-      "A_2=1.8 #Area at section 2 in in^2\n",
-      "A_3=1.6 #Area at section 3 in in^2\n",
-      "\n",
-      "#Calculation\n",
-      "#S indicates sigma here\n",
-      "S_AB=P_AB/A_1 #Stress at section 1 in psi (T)\n",
-      "S_BC=P_BC/A_2 #Stress at section 2 in psi (C)\n",
-      "S_CD=P_CD/A_3 #Stress at section 3 in psi (C)\n",
-      "\n",
-      "#Result\n",
-      "print \"The stress at the three sections is given as\"\n",
-      "print \"Stress at section 1=\",round(S_AB),\"section 2=\",round(S_BC),\"section 3=\",S_CD\n",
-      "\n",
-      "#NOTE:The answer for the following example for section 1 and section 2\n",
-      "#are incorrect due to rounding in the textbook\n",
-      "#Computed values are correct"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The stress at the three sections is given as\n",
-        "Stress at section 1= 3333.0 section 2= 2778.0 section 3= 4375.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.2, Page No:10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Ay=40 #Vertical Reaction at A in kN\n",
-      "Hy=60 #Vertical Reaction at H in kN\n",
-      "Hx=0 #Horizontal Reaction at H in kN\n",
-      "y=3 #Height in m\n",
-      "x=5 #Distance in m\n",
-      "p=4 #Panel distance in m\n",
-      "A=900 #Area of the member in mm^2\n",
-      "P_C=30 #Force at point C in kN\n",
-      "\n",
-      "#Calculation\n",
-      "#Part 1\n",
-      "#Applying summation of forces in the x and y direction and equating to zero\n",
-      "P_AB=(-Ay)*(x*y**-1) #Force in member AB in kN\n",
-      "P_AC=-(p*x**-1*P_AB) #Force in member AC in kN\n",
-      "#Using stress=force/area\n",
-      "S_AC=(P_AC/A)*10**3 #Stress in member AC in MPa (T)\n",
-      "\n",
-      "#Part 2\n",
-      "#Sum of moments about point E to zero\n",
-      "P_BD=(Ay*p*2-(P_C*p))*y**-1 #Force in memeber AB in kN (C)\n",
-      "S_BD=(P_BD/A)*10**3 #Stress in member in MPa (C)\n",
-      "\n",
-      "#Result\n",
-      "print \"The Stress in member AC is\",round(S_AC,1),\"MPa (T)\"\n",
-      "print \"The Stress in member BD is\",round(S_BD,1),\"MPa (C)\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Stress in member AC is 59.3 MPa (T)\n",
-        "The Stress in member BD is 74.1 MPa (C)\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.3, Page No:11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as num\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A_AB=800 #Area of member AB in m^2\n",
-      "A_AC=400 #Area of member AC in m^2\n",
-      "W_AB=110 #Safe value of stress in Pa for AB\n",
-      "W_AC=120 #Safe value of stress in Pa for AC\n",
-      "theta1=60*3.14*180**-1 #Angle in radians\n",
-      "theta2=40*3.14*180**-1 #Angle in radians \n",
-      "\n",
-      "#Calculations\n",
-      "#Applying sum of forces \n",
-      "#Solving by matrix method putting W as 1\n",
-      "A=num.array([[-cos(theta1),cos(theta2)],[sin(theta1),sin(theta2)]])\n",
-      "B=num.array([[1],[1]])\n",
-      "C=inv(A)\n",
-      "D=C*B\n",
-      "\n",
-      "#Using newtons third law\n",
-      "#Two values of W hence the change in the notation\n",
-      "W1=(W_AB*A_AB)*(D[1,1])**-1 #Weight W in N\n",
-      "W2=(W_AC*A_AC)*(D[0,1])**-1 #Weight W in N\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum value of W allowable is\",round(W2*1000**-1,1),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum value of W allowable is 61.7 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.4, Page No:19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "d=3*4**-1 #Rivet diameter in inches\n",
-      "t=7*8**-1 #Thickness of the plate in inches\n",
-      "tau=14000 #Shear stress limit in psi\n",
-      "sigma_b=18000 #Normal stress limit in psi\n",
-      "\n",
-      "#Calculations\n",
-      "#Design Shear Stress in Rivets\n",
-      "V=tau*(d**2*(pi/4))*4 #Shear force maximum allowable in lb\n",
-      "#Design for bearing stress in plate\n",
-      "Pb=sigma_b*t*d*4 #lb\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum load that the joint can carry is\",round(V),\"lb\"\n",
-      "\n",
-      "#NOTE:The answer in the textbook is off by 40lb"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum load that the joint can carry is 24740.0 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_14.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_14.ipynb
deleted file mode 100755
index 8594491e..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_14.ipynb
+++ /dev/null
@@ -1,226 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:055c326b63bc3f150b8a5e433c724771cc1733887ec2b1e223ccf712fcd80848"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 01:Stress"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.1, Page No:9"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#NOTE:The notation has been changed to simplify the coding process\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P_AB=4000 #Axial Force at section 1 in lb\n",
-      "P_BC=5000 #Axial Force at section 2 in lb\n",
-      "P_CD=7000 #Axial Force at section 3 in lb\n",
-      "A_1=1.2 #Area at section 1 in in^2\n",
-      "A_2=1.8 #Area at section 2 in in^2\n",
-      "A_3=1.6 #Area at section 3 in in^2\n",
-      "\n",
-      "#Calculation\n",
-      "#S indicates sigma here\n",
-      "S_AB=P_AB/A_1 #Stress at section 1 in psi (T)\n",
-      "S_BC=P_BC/A_2 #Stress at section 2 in psi (C)\n",
-      "S_CD=P_CD/A_3 #Stress at section 3 in psi (C)\n",
-      "\n",
-      "#Result\n",
-      "print \"The stress at the three sections is given as\"\n",
-      "print \"Stress at section 1=\",round(S_AB),\"section 2=\",round(S_BC),\"section 3=\",S_CD\n",
-      "\n",
-      "#NOTE:The answer for the following example for section 1 and section 2\n",
-      "#are incorrect due to rounding in the textbook\n",
-      "#Computed values are correct"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The stress at the three sections is given as\n",
-        "Stress at section 1= 3333.0 section 2= 2778.0 section 3= 4375.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.2, Page No:10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Ay=40 #Vertical Reaction at A in kN\n",
-      "Hy=60 #Vertical Reaction at H in kN\n",
-      "Hx=0 #Horizontal Reaction at H in kN\n",
-      "y=3 #Height in m\n",
-      "x=5 #Distance in m\n",
-      "p=4 #Panel distance in m\n",
-      "A=900 #Area of the member in mm^2\n",
-      "P_C=30 #Force at point C in kN\n",
-      "\n",
-      "#Calculation\n",
-      "#Part 1\n",
-      "#Applying summation of forces in the x and y direction and equating to zero\n",
-      "P_AB=(-Ay)*(x*y**-1) #Force in member AB in kN\n",
-      "P_AC=-(p*x**-1*P_AB) #Force in member AC in kN\n",
-      "#Using stress=force/area\n",
-      "S_AC=(P_AC/A)*10**3 #Stress in member AC in MPa (T)\n",
-      "\n",
-      "#Part 2\n",
-      "#Sum of moments about point E to zero\n",
-      "P_BD=(Ay*p*2-(P_C*p))*y**-1 #Force in memeber AB in kN (C)\n",
-      "S_BD=(P_BD/A)*10**3 #Stress in member in MPa (C)\n",
-      "\n",
-      "#Result\n",
-      "print \"The Stress in member AC is\",round(S_AC,1),\"MPa (T)\"\n",
-      "print \"The Stress in member BD is\",round(S_BD,1),\"MPa (C)\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Stress in member AC is 59.3 MPa (T)\n",
-        "The Stress in member BD is 74.1 MPa (C)\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.3, Page No:11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as num\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A_AB=800 #Area of member AB in m^2\n",
-      "A_AC=400 #Area of member AC in m^2\n",
-      "W_AB=110 #Safe value of stress in Pa for AB\n",
-      "W_AC=120 #Safe value of stress in Pa for AC\n",
-      "theta1=60*3.14*180**-1 #Angle in radians\n",
-      "theta2=40*3.14*180**-1 #Angle in radians \n",
-      "\n",
-      "#Calculations\n",
-      "#Applying sum of forces \n",
-      "#Solving by matrix method putting W as 1\n",
-      "A=num.array([[-cos(theta1),cos(theta2)],[sin(theta1),sin(theta2)]])\n",
-      "B=num.array([[1],[1]])\n",
-      "C=inv(A)\n",
-      "D=C*B\n",
-      "\n",
-      "#Using newtons third law\n",
-      "#Two values of W hence the change in the notation\n",
-      "W1=(W_AB*A_AB)*(D[1,1])**-1 #Weight W in N\n",
-      "W2=(W_AC*A_AC)*(D[0,1])**-1 #Weight W in N\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum value of W allowable is\",round(W2*1000**-1,1),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum value of W allowable is 61.7 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.1.4, Page No:19"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "d=3*4**-1 #Rivet diameter in inches\n",
-      "t=7*8**-1 #Thickness of the plate in inches\n",
-      "tau=14000 #Shear stress limit in psi\n",
-      "sigma_b=18000 #Normal stress limit in psi\n",
-      "\n",
-      "#Calculations\n",
-      "#Design Shear Stress in Rivets\n",
-      "V=tau*(d**2*(pi/4))*4 #Shear force maximum allowable in lb\n",
-      "#Design for bearing stress in plate\n",
-      "Pb=sigma_b*t*d*4 #lb\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum load that the joint can carry is\",round(V),\"lb\"\n",
-      "\n",
-      "#NOTE:The answer in the textbook is off by 40lb"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum load that the joint can carry is 24740.0 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_2.ipynb
deleted file mode 100755
index e69de29b..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_2.ipynb
+++ /dev/null
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_3.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_3.ipynb
deleted file mode 100755
index e69de29b..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_3.ipynb
+++ /dev/null
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_4.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_4.ipynb
deleted file mode 100755
index e69de29b..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_4.ipynb
+++ /dev/null
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_5.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_5.ipynb
deleted file mode 100755
index e69de29b..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_5.ipynb
+++ /dev/null
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_6.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_6.ipynb
deleted file mode 100755
index e69de29b..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_6.ipynb
+++ /dev/null
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_7.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_7.ipynb
deleted file mode 100755
index e69de29b..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_7.ipynb
+++ /dev/null
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_8.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_8.ipynb
deleted file mode 100755
index e69de29b..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_8.ipynb
+++ /dev/null
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02.ipynb
deleted file mode 100755
index 5f5f30b9..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02.ipynb
+++ /dev/null
@@ -1,622 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c5e7c024d1eb3a7426e4241fc719f61d59003d11703e75e2542d5874bd3a6f9f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 02:Strain"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Examples No:2.2.1, Page No:36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "#Axial Forces in lb in member AB, BC and CD\n",
-      "P_AB=2000 \n",
-      "P_BC=2000\n",
-      "P_CD=4000\n",
-      "#Other Variables\n",
-      "E=29*10**6 #Modulus of Elasticity in psi\n",
-      "#Length of each member in inches\n",
-      "L_AB=5*12\n",
-      "L_BC=4*12\n",
-      "L_CD=4*12\n",
-      "#Diameter of each member in inches\n",
-      "D_AB=0.5\n",
-      "D_BC=0.75\n",
-      "D_CD=0.75\n",
-      "\n",
-      "#Calculation\n",
-      "#Area Calculation of each member in square inches\n",
-      "A_AB=(pi*D_AB**2)/4\n",
-      "A_BC=(pi*D_BC**2)/4\n",
-      "A_CD=(pi*D_CD**2)/4\n",
-      "\n",
-      "#Using relation delta=(PL/AE) to compute strain\n",
-      "#As stress in Member CD is compression\n",
-      "delta=(E**-1)*((P_AB*L_AB*A_AB**-1)+(P_BC*L_BC*A_BC**-1)-(P_CD*L_CD*A_CD**-1))\n",
-      "\n",
-      "#Result\n",
-      "print \"The elongation in the total structure is\",round(delta,5),\"in\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The elongation in the total structure is 0.01358 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.2, Page No:36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from scipy.integrate import quad\n",
-      "\n",
-      "#Variable Decleration\n",
-      "E=200*10**9 #Modulus of elasticity in Pa\n",
-      "P=10**5 #Force acting in N\n",
-      "\n",
-      "#Calculations\n",
-      "#Using quad integration\n",
-      "#Area has been defined as a quadratic equation to integrate\n",
-      "def integrand(x, a, b):\n",
-      "    return 1/(a * x + b)\n",
-      "a = 160\n",
-      "b = 800\n",
-      "I = quad(integrand, 0, 10, args=(a,b))\n",
-      "#Using delta=(P/E)*I    where I is the integrand\n",
-      "delta=(P*E**-1)*10**6*I[0]\n",
-      "\n",
-      "#Result\n",
-      "print \"The elongation in the member is\",round(delta*1000,2),\"mm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The elongation in the member is 3.43 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.3, Page No:37"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decelration\n",
-      "A_AC=0.25 #Cross Sectional Area in square inch\n",
-      "Load=2000 #Load at point C in lb\n",
-      "E=29*10**6 #Modulus of elasticity in psi\n",
-      "theta=(pi*40)/180 #Angle in radians\n",
-      "L_BC=8 #Length in ft\n",
-      "\n",
-      "#Calculations\n",
-      "#Using sum of forces \n",
-      "P_AC=Load/sin(theta) #Force in cable AC in lb\n",
-      "L_AC=(L_BC*12)/cos(theta) #Length of cable AC in in\n",
-      "\n",
-      "delta_AC=(P_AC*L_AC)/(E*A_AC) #elongation in inches\n",
-      "\n",
-      "delta_C=delta_AC/sin(theta) #displacement of point C in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The displacement of point C is\",round(delta_C,4),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The displacement of point C is 0.0837 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.4, Page No:46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "d=0.05 #Diameter of the rod in mm\n",
-      "P=8000 #Load on the bar in N\n",
-      "E=40*10**6 #Modulus of elasticity in Pa\n",
-      "v=0.45 #Poisson Ratio\n",
-      "L=300 #Length of the rod in mm\n",
-      "\n",
-      "#Calculation\n",
-      "A=((pi*d**2)/4) #Area of the bar in mm^2\n",
-      "sigma_x=-P/A #Axial Stress in the bar in Pa\n",
-      "#As contact pressure resists the force\n",
-      "p=(v*sigma_x)/(1-v)\n",
-      "#Using Axial Strain formula\n",
-      "e_x=(sigma_x-(v*2*p))/E\n",
-      "#Corresponding change in length\n",
-      "delta=e_x*L #contraction in mm\n",
-      "#Without constrains of the wall\n",
-      "delta_w=(-P*(L*10**-3))/(E*A) #Elongation in m\n",
-      "\n",
-      "#Result\n",
-      "print \"The elongation in the bar is\",-round(delta,2),\"mm contraction\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The elongation in the bar is 8.06 mm contraction\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.5, Page No:47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "E=500 #Modulus of elasticity in psi\n",
-      "v=0.48 #Poisson ratio\n",
-      "V=600 #Force in lb\n",
-      "w=5 #Width of the plate in inches\n",
-      "l=9 #Length of the plate in inches\n",
-      "t=1.75 #Thickness of the rubber layer in inches\n",
-      "\n",
-      "#Calculations\n",
-      "tau=V*(w*l)**-1 #Shear stress in rubber in psi\n",
-      "G=E/(2*(1+v)) #Bulk modulus in psi\n",
-      "gamma=tau/G #Shear Modulus \n",
-      "disp=t*gamma #Diplacement in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The displacement of the rubber layer is\",round(disp,4),\"in\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The displacement of the rubber layer is 0.1381 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.6, Page No:52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=10**6 #Force on the member in N\n",
-      "Es=200 #Modulus of elasticity of steel in GPa\n",
-      "Ec=14 #Modulus of elasticity concrete in GPa\n",
-      "As=900*10**-6 #Area of steel in m^2\n",
-      "Ac=0.3**2 #Area of concrete block in m^2\n",
-      "\n",
-      "#Calculation\n",
-      "#Cross Sectional Areas\n",
-      "Ast=4*As #Cross Sectional Area in m^2 of Steel\n",
-      "Act=Ac-Ast #Cross Sectional Area of Concrete in m^2\n",
-      "\n",
-      "#Applying equilibrium to the structure\n",
-      "#Using the ratio of stress and modulii of elasticity we obtain the following eq\n",
-      "sigma_ct=P/(((Es*Ec**-1)*Ast)+Act) #Stress in Concrete in Pa\n",
-      "sigma_st=sigma_ct*Es*Ec**-1 #Stress in Steel in Pa\n",
-      "\n",
-      "#Result\n",
-      "print \"The stress in steel and concrete is as follows\",round(sigma_st*10**-6,1),\"MPa and\",round(sigma_ct*10**-6,3),\"Mpa respectively\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The stress in steel and concrete is as follows 103.6 MPa and 7.255 Mpa respectively\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.7, Page No:52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "#Say the ratio of stress in steel to concrete is R\n",
-      "R=14.286 \n",
-      "sigma_co=6*10**6 #Stress in concrete in Pa\n",
-      "Ast=3.6*10**-3 #Area of steel in m^2\n",
-      "Aco=86.4*10**-3 #Area of Concrete in m^2\n",
-      "\n",
-      "#Calculation\n",
-      "sigma_st=R*sigma_co #Stress in steel in Pa\n",
-      "#Here stress is below the allowable hence safe\n",
-      "P=sigma_st*Ast+sigma_co*Aco #Allowable force in N\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum allowable force is\",round(P*10**-3),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable force is 827.0 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.8, Page No:53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#NOTE:The NOtation has been changed to ease coding\n",
-      "#Variable Decleration\n",
-      "d=0.005 #difference in length in inch\n",
-      "L=10 #Length in inch\n",
-      "#Area of copper and aluminium in sq.in\n",
-      "Ac=2 #Area of copper \n",
-      "Aa=3 #Area of aluminium \n",
-      "#Modulus of elasticity of copper and aluminium in psi\n",
-      "Ec=17000000 #Copper\n",
-      "Ea=10**7 #Aluminium\n",
-      "#Allowable Stress in psi\n",
-      "Sc=20*10**3 #Copper\n",
-      "Sa=10*10**3 #Aluminium\n",
-      "\n",
-      "#Calculation\n",
-      "#Equilibrium is Pc+Pa=P\n",
-      "#Hookes Law is delta_c=delta_a+0.005\n",
-      "#Simplfying the solution we have constants we can directly compute\n",
-      "A=d*Ec*(L+d)**-1\n",
-      "B=Ec*Ea**-1\n",
-      "C=L*B*(L+d)**-1\n",
-      "sigma_a=(Sc-A)*C**-1\n",
-      "\n",
-      "#Using equilibrium equation\n",
-      "P=Sc*Ac+sigma_a*Aa #Safe load in lb\n",
-      "\n",
-      "#Result\n",
-      "print \"The safe load on the structure is\",round(P),\"lb\"\n",
-      "#NOTE:Answer in the textbook has been rounded off and hence the discrepancy"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The safe load on the structure is 60312.0 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.9, Page No:54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=50*10**3 #Load applied in N\n",
-      "x1=0.6 #Length in m\n",
-      "x2=1.6 #Length in m\n",
-      "L1=1 #Length of steel cable in m\n",
-      "L2=2 #Length of bronze cable in m\n",
-      "L=2.4 #Length in m\n",
-      "#Area in m^2\n",
-      "Ast=600*10**-6 #Steel\n",
-      "Abr=300*10**-6 #Bronze\n",
-      "#Modulus of elasticity in GPa\n",
-      "Est=200 #Steel\n",
-      "Ebr=83 #Bronze\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying the equilibrium and Hookes law we solve by matrix method\n",
-      "a=np.array([[x1,x2],[1,-((x1*Est*Ast*L2)/(x2*Ebr*Abr))]])\n",
-      "b=np.array([L*P,0])\n",
-      "y=np.linalg.solve(a,b)\n",
-      "\n",
-      "#Stresses in Pa\n",
-      "sigma_st=y[0]*Ast**-1 #Stress in steel\n",
-      "sigma_br=y[1]/Abr #Stress in bronze\n",
-      "\n",
-      "#Result\n",
-      "print \"The stresses in steel and bronze are as follows\"\n",
-      "print round(sigma_st*10**-6,1),\"MPa and\",round(sigma_br*10**-6,1),\"MPa respectively\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The stresses in steel and bronze are as follows\n",
-        "191.8 MPa and 106.1 MPa respectively\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.10, Page No:62"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=2.5 #Length in m\n",
-      "A=1200 #Cross sectional Area in mm^2\n",
-      "delta_T=40 #Temperature drop in degree C\n",
-      "delta=0.5*10**-3 #Movement of the walls in mm\n",
-      "alpha=11.7*10**-6 #Coefficient of thermal expansion in /degreeC\n",
-      "E=200*10**9 #Modulus of elasticity in Pa\n",
-      "\n",
-      "#Calculation\n",
-      "#Part(1)\n",
-      "sigma_1=alpha*delta_T*E #Stress in the rod in Pa\n",
-      "\n",
-      "#Part(2)\n",
-      "#Using Hookes Law\n",
-      "sigma_2=E*((alpha*delta_T)-(delta*L**-1)) #Stress in the rod in Pa\n",
-      "\n",
-      "print \"The Stress in part 1 in the rod is\",round(sigma_1*10**-6,1),\"MPa\"\n",
-      "print \"The Stress in part 2 in the rod is\",round(sigma_2*10**-6,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Stress in part 1 in the rod is 93.6 MPa\n",
-        "The Stress in part 2 in the rod is 53.6 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.11, Page No:63"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "delta=100 #Increase in the temperature in degreeF\n",
-      "Load=12000 #Load on the beam in lb\n",
-      "#Length in inch\n",
-      "Ls=2*12 #Steel\n",
-      "Lb=3*12 #Bronze\n",
-      "#Area in sq.in\n",
-      "As=0.75 #Steel\n",
-      "Ab=1.5 #Bronze\n",
-      "#Modulus of elasticity in psi\n",
-      "Es=29*10**6 #Steel\n",
-      "Eb=12*10**6 #Bronze\n",
-      "#Coefficient of thermal expansion in /degree C\n",
-      "alpha_s=6.5*10**-6 #Steel\n",
-      "alpha_b=10**-5 #Bronze\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying the Hookes Law and equilibrium we get two equations\n",
-      "a=np.array([[Ls*(Es*As)**-1,-Lb*(Eb*Ab)**-1],[2,1]])\n",
-      "b=np.array([(alpha_b*delta*Lb-alpha_s*delta*Ls),Load])\n",
-      "y=np.linalg.solve(a,b)\n",
-      "\n",
-      "#Stresses\n",
-      "sigma_st=y[0]*As**-1 #Stress in steel in psi (T)\n",
-      "sigma_br=y[1]*Ab**-1 #Stress in bronze in psi (C)\n",
-      "\n",
-      "#Result\n",
-      "print \"The Stress in steel and bronze are as follows\"\n",
-      "print sigma_st,\"psi (T) and\", -sigma_br,\"psi (C)\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Stress in steel and bronze are as follows\n",
-        "11600.0 psi (T) and 3600.0 psi (C)\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.12, Page No:64"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=6000 #Force in lb\n",
-      "Est=29*10**6 #Modulus of elasticity of steel in psi\n",
-      "L1=24 #Length in inches\n",
-      "L2=36 #Length in inches\n",
-      "alpha_1=6.5*10**-6 #coefficient of thermal expansion in /degree F of steel\n",
-      "alpha_2=10**-5 #coefficient of thermal expansion in /degree F of bronze\n",
-      "As=0.75 #Area os steel in sq.in\n",
-      "\n",
-      "#Calculations\n",
-      "delta_T=((P*L1)/(Est*As))/(alpha_2*L2-alpha_1*L1) #Change in temperature in degree F\n",
-      "\n",
-      "print \"The change in the Temperature is\",round(delta_T,1),\"F\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The change in the Temperature is 32.5 F\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02_1.ipynb
deleted file mode 100755
index 5f5f30b9..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02_1.ipynb
+++ /dev/null
@@ -1,622 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c5e7c024d1eb3a7426e4241fc719f61d59003d11703e75e2542d5874bd3a6f9f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 02:Strain"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Examples No:2.2.1, Page No:36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "#Axial Forces in lb in member AB, BC and CD\n",
-      "P_AB=2000 \n",
-      "P_BC=2000\n",
-      "P_CD=4000\n",
-      "#Other Variables\n",
-      "E=29*10**6 #Modulus of Elasticity in psi\n",
-      "#Length of each member in inches\n",
-      "L_AB=5*12\n",
-      "L_BC=4*12\n",
-      "L_CD=4*12\n",
-      "#Diameter of each member in inches\n",
-      "D_AB=0.5\n",
-      "D_BC=0.75\n",
-      "D_CD=0.75\n",
-      "\n",
-      "#Calculation\n",
-      "#Area Calculation of each member in square inches\n",
-      "A_AB=(pi*D_AB**2)/4\n",
-      "A_BC=(pi*D_BC**2)/4\n",
-      "A_CD=(pi*D_CD**2)/4\n",
-      "\n",
-      "#Using relation delta=(PL/AE) to compute strain\n",
-      "#As stress in Member CD is compression\n",
-      "delta=(E**-1)*((P_AB*L_AB*A_AB**-1)+(P_BC*L_BC*A_BC**-1)-(P_CD*L_CD*A_CD**-1))\n",
-      "\n",
-      "#Result\n",
-      "print \"The elongation in the total structure is\",round(delta,5),\"in\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The elongation in the total structure is 0.01358 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.2, Page No:36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from scipy.integrate import quad\n",
-      "\n",
-      "#Variable Decleration\n",
-      "E=200*10**9 #Modulus of elasticity in Pa\n",
-      "P=10**5 #Force acting in N\n",
-      "\n",
-      "#Calculations\n",
-      "#Using quad integration\n",
-      "#Area has been defined as a quadratic equation to integrate\n",
-      "def integrand(x, a, b):\n",
-      "    return 1/(a * x + b)\n",
-      "a = 160\n",
-      "b = 800\n",
-      "I = quad(integrand, 0, 10, args=(a,b))\n",
-      "#Using delta=(P/E)*I    where I is the integrand\n",
-      "delta=(P*E**-1)*10**6*I[0]\n",
-      "\n",
-      "#Result\n",
-      "print \"The elongation in the member is\",round(delta*1000,2),\"mm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The elongation in the member is 3.43 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.3, Page No:37"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decelration\n",
-      "A_AC=0.25 #Cross Sectional Area in square inch\n",
-      "Load=2000 #Load at point C in lb\n",
-      "E=29*10**6 #Modulus of elasticity in psi\n",
-      "theta=(pi*40)/180 #Angle in radians\n",
-      "L_BC=8 #Length in ft\n",
-      "\n",
-      "#Calculations\n",
-      "#Using sum of forces \n",
-      "P_AC=Load/sin(theta) #Force in cable AC in lb\n",
-      "L_AC=(L_BC*12)/cos(theta) #Length of cable AC in in\n",
-      "\n",
-      "delta_AC=(P_AC*L_AC)/(E*A_AC) #elongation in inches\n",
-      "\n",
-      "delta_C=delta_AC/sin(theta) #displacement of point C in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The displacement of point C is\",round(delta_C,4),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The displacement of point C is 0.0837 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.4, Page No:46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "d=0.05 #Diameter of the rod in mm\n",
-      "P=8000 #Load on the bar in N\n",
-      "E=40*10**6 #Modulus of elasticity in Pa\n",
-      "v=0.45 #Poisson Ratio\n",
-      "L=300 #Length of the rod in mm\n",
-      "\n",
-      "#Calculation\n",
-      "A=((pi*d**2)/4) #Area of the bar in mm^2\n",
-      "sigma_x=-P/A #Axial Stress in the bar in Pa\n",
-      "#As contact pressure resists the force\n",
-      "p=(v*sigma_x)/(1-v)\n",
-      "#Using Axial Strain formula\n",
-      "e_x=(sigma_x-(v*2*p))/E\n",
-      "#Corresponding change in length\n",
-      "delta=e_x*L #contraction in mm\n",
-      "#Without constrains of the wall\n",
-      "delta_w=(-P*(L*10**-3))/(E*A) #Elongation in m\n",
-      "\n",
-      "#Result\n",
-      "print \"The elongation in the bar is\",-round(delta,2),\"mm contraction\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The elongation in the bar is 8.06 mm contraction\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.5, Page No:47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "E=500 #Modulus of elasticity in psi\n",
-      "v=0.48 #Poisson ratio\n",
-      "V=600 #Force in lb\n",
-      "w=5 #Width of the plate in inches\n",
-      "l=9 #Length of the plate in inches\n",
-      "t=1.75 #Thickness of the rubber layer in inches\n",
-      "\n",
-      "#Calculations\n",
-      "tau=V*(w*l)**-1 #Shear stress in rubber in psi\n",
-      "G=E/(2*(1+v)) #Bulk modulus in psi\n",
-      "gamma=tau/G #Shear Modulus \n",
-      "disp=t*gamma #Diplacement in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The displacement of the rubber layer is\",round(disp,4),\"in\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The displacement of the rubber layer is 0.1381 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.6, Page No:52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=10**6 #Force on the member in N\n",
-      "Es=200 #Modulus of elasticity of steel in GPa\n",
-      "Ec=14 #Modulus of elasticity concrete in GPa\n",
-      "As=900*10**-6 #Area of steel in m^2\n",
-      "Ac=0.3**2 #Area of concrete block in m^2\n",
-      "\n",
-      "#Calculation\n",
-      "#Cross Sectional Areas\n",
-      "Ast=4*As #Cross Sectional Area in m^2 of Steel\n",
-      "Act=Ac-Ast #Cross Sectional Area of Concrete in m^2\n",
-      "\n",
-      "#Applying equilibrium to the structure\n",
-      "#Using the ratio of stress and modulii of elasticity we obtain the following eq\n",
-      "sigma_ct=P/(((Es*Ec**-1)*Ast)+Act) #Stress in Concrete in Pa\n",
-      "sigma_st=sigma_ct*Es*Ec**-1 #Stress in Steel in Pa\n",
-      "\n",
-      "#Result\n",
-      "print \"The stress in steel and concrete is as follows\",round(sigma_st*10**-6,1),\"MPa and\",round(sigma_ct*10**-6,3),\"Mpa respectively\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The stress in steel and concrete is as follows 103.6 MPa and 7.255 Mpa respectively\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.7, Page No:52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "#Say the ratio of stress in steel to concrete is R\n",
-      "R=14.286 \n",
-      "sigma_co=6*10**6 #Stress in concrete in Pa\n",
-      "Ast=3.6*10**-3 #Area of steel in m^2\n",
-      "Aco=86.4*10**-3 #Area of Concrete in m^2\n",
-      "\n",
-      "#Calculation\n",
-      "sigma_st=R*sigma_co #Stress in steel in Pa\n",
-      "#Here stress is below the allowable hence safe\n",
-      "P=sigma_st*Ast+sigma_co*Aco #Allowable force in N\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum allowable force is\",round(P*10**-3),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable force is 827.0 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.8, Page No:53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#NOTE:The NOtation has been changed to ease coding\n",
-      "#Variable Decleration\n",
-      "d=0.005 #difference in length in inch\n",
-      "L=10 #Length in inch\n",
-      "#Area of copper and aluminium in sq.in\n",
-      "Ac=2 #Area of copper \n",
-      "Aa=3 #Area of aluminium \n",
-      "#Modulus of elasticity of copper and aluminium in psi\n",
-      "Ec=17000000 #Copper\n",
-      "Ea=10**7 #Aluminium\n",
-      "#Allowable Stress in psi\n",
-      "Sc=20*10**3 #Copper\n",
-      "Sa=10*10**3 #Aluminium\n",
-      "\n",
-      "#Calculation\n",
-      "#Equilibrium is Pc+Pa=P\n",
-      "#Hookes Law is delta_c=delta_a+0.005\n",
-      "#Simplfying the solution we have constants we can directly compute\n",
-      "A=d*Ec*(L+d)**-1\n",
-      "B=Ec*Ea**-1\n",
-      "C=L*B*(L+d)**-1\n",
-      "sigma_a=(Sc-A)*C**-1\n",
-      "\n",
-      "#Using equilibrium equation\n",
-      "P=Sc*Ac+sigma_a*Aa #Safe load in lb\n",
-      "\n",
-      "#Result\n",
-      "print \"The safe load on the structure is\",round(P),\"lb\"\n",
-      "#NOTE:Answer in the textbook has been rounded off and hence the discrepancy"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The safe load on the structure is 60312.0 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.9, Page No:54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=50*10**3 #Load applied in N\n",
-      "x1=0.6 #Length in m\n",
-      "x2=1.6 #Length in m\n",
-      "L1=1 #Length of steel cable in m\n",
-      "L2=2 #Length of bronze cable in m\n",
-      "L=2.4 #Length in m\n",
-      "#Area in m^2\n",
-      "Ast=600*10**-6 #Steel\n",
-      "Abr=300*10**-6 #Bronze\n",
-      "#Modulus of elasticity in GPa\n",
-      "Est=200 #Steel\n",
-      "Ebr=83 #Bronze\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying the equilibrium and Hookes law we solve by matrix method\n",
-      "a=np.array([[x1,x2],[1,-((x1*Est*Ast*L2)/(x2*Ebr*Abr))]])\n",
-      "b=np.array([L*P,0])\n",
-      "y=np.linalg.solve(a,b)\n",
-      "\n",
-      "#Stresses in Pa\n",
-      "sigma_st=y[0]*Ast**-1 #Stress in steel\n",
-      "sigma_br=y[1]/Abr #Stress in bronze\n",
-      "\n",
-      "#Result\n",
-      "print \"The stresses in steel and bronze are as follows\"\n",
-      "print round(sigma_st*10**-6,1),\"MPa and\",round(sigma_br*10**-6,1),\"MPa respectively\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The stresses in steel and bronze are as follows\n",
-        "191.8 MPa and 106.1 MPa respectively\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.10, Page No:62"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=2.5 #Length in m\n",
-      "A=1200 #Cross sectional Area in mm^2\n",
-      "delta_T=40 #Temperature drop in degree C\n",
-      "delta=0.5*10**-3 #Movement of the walls in mm\n",
-      "alpha=11.7*10**-6 #Coefficient of thermal expansion in /degreeC\n",
-      "E=200*10**9 #Modulus of elasticity in Pa\n",
-      "\n",
-      "#Calculation\n",
-      "#Part(1)\n",
-      "sigma_1=alpha*delta_T*E #Stress in the rod in Pa\n",
-      "\n",
-      "#Part(2)\n",
-      "#Using Hookes Law\n",
-      "sigma_2=E*((alpha*delta_T)-(delta*L**-1)) #Stress in the rod in Pa\n",
-      "\n",
-      "print \"The Stress in part 1 in the rod is\",round(sigma_1*10**-6,1),\"MPa\"\n",
-      "print \"The Stress in part 2 in the rod is\",round(sigma_2*10**-6,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Stress in part 1 in the rod is 93.6 MPa\n",
-        "The Stress in part 2 in the rod is 53.6 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.11, Page No:63"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "delta=100 #Increase in the temperature in degreeF\n",
-      "Load=12000 #Load on the beam in lb\n",
-      "#Length in inch\n",
-      "Ls=2*12 #Steel\n",
-      "Lb=3*12 #Bronze\n",
-      "#Area in sq.in\n",
-      "As=0.75 #Steel\n",
-      "Ab=1.5 #Bronze\n",
-      "#Modulus of elasticity in psi\n",
-      "Es=29*10**6 #Steel\n",
-      "Eb=12*10**6 #Bronze\n",
-      "#Coefficient of thermal expansion in /degree C\n",
-      "alpha_s=6.5*10**-6 #Steel\n",
-      "alpha_b=10**-5 #Bronze\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying the Hookes Law and equilibrium we get two equations\n",
-      "a=np.array([[Ls*(Es*As)**-1,-Lb*(Eb*Ab)**-1],[2,1]])\n",
-      "b=np.array([(alpha_b*delta*Lb-alpha_s*delta*Ls),Load])\n",
-      "y=np.linalg.solve(a,b)\n",
-      "\n",
-      "#Stresses\n",
-      "sigma_st=y[0]*As**-1 #Stress in steel in psi (T)\n",
-      "sigma_br=y[1]*Ab**-1 #Stress in bronze in psi (C)\n",
-      "\n",
-      "#Result\n",
-      "print \"The Stress in steel and bronze are as follows\"\n",
-      "print sigma_st,\"psi (T) and\", -sigma_br,\"psi (C)\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Stress in steel and bronze are as follows\n",
-        "11600.0 psi (T) and 3600.0 psi (C)\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.12, Page No:64"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=6000 #Force in lb\n",
-      "Est=29*10**6 #Modulus of elasticity of steel in psi\n",
-      "L1=24 #Length in inches\n",
-      "L2=36 #Length in inches\n",
-      "alpha_1=6.5*10**-6 #coefficient of thermal expansion in /degree F of steel\n",
-      "alpha_2=10**-5 #coefficient of thermal expansion in /degree F of bronze\n",
-      "As=0.75 #Area os steel in sq.in\n",
-      "\n",
-      "#Calculations\n",
-      "delta_T=((P*L1)/(Est*As))/(alpha_2*L2-alpha_1*L1) #Change in temperature in degree F\n",
-      "\n",
-      "print \"The change in the Temperature is\",round(delta_T,1),\"F\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The change in the Temperature is 32.5 F\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02_2.ipynb
deleted file mode 100755
index cd669b4f..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02_2.ipynb
+++ /dev/null
@@ -1,614 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3cb8e8b65ee50988938562d2f6cd882ccb93a3ca89c523d4423804cd1b6898ff"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 02:Strain"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Examples No:2.2.1, Page No:36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "#Axial Forces in lb in member AB, BC and CD\n",
-      "P_AB=2000 \n",
-      "P_BC=2000\n",
-      "P_CD=4000\n",
-      "#Other Variables\n",
-      "E=29*10**6 #Modulus of Elasticity in psi\n",
-      "#Length of each member in inches\n",
-      "L_AB=5*12\n",
-      "L_BC=4*12\n",
-      "L_CD=4*12\n",
-      "#Diameter of each member in inches\n",
-      "D_AB=0.5\n",
-      "D_BC=0.75\n",
-      "D_CD=0.75\n",
-      "\n",
-      "#Calculation\n",
-      "#Area Calculation of each member in square inches\n",
-      "A_AB=(pi*D_AB**2)/4\n",
-      "A_BC=(pi*D_BC**2)/4\n",
-      "A_CD=(pi*D_CD**2)/4\n",
-      "\n",
-      "#Using relation delta=(PL/AE) to compute strain\n",
-      "#As stress in Member CD is compression\n",
-      "delta=(E**-1)*((P_AB*L_AB*A_AB**-1)+(P_BC*L_BC*A_BC**-1)-(P_CD*L_CD*A_CD**-1))\n",
-      "\n",
-      "#Result\n",
-      "print \"The elongation in the total structure is\",round(delta,5),\"in\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The elongation in the total structure is 0.01358 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.2, Page No:36"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from scipy.integrate import quad\n",
-      "\n",
-      "#Variable Decleration\n",
-      "E=200*10**9 #Modulus of elasticity in Pa\n",
-      "P=10**5 #Force acting in N\n",
-      "\n",
-      "#Calculations\n",
-      "#Using quad integration\n",
-      "#Area has been defined as a quadratic equation to integrate\n",
-      "def integrand(x, a, b):\n",
-      "    return 1/(a * x + b)\n",
-      "a = 160\n",
-      "b = 800\n",
-      "I = quad(integrand, 0, 10, args=(a,b))\n",
-      "#Using delta=(P/E)*I    where I is the integrand\n",
-      "delta=(P*E**-1)*10**6*I[0]\n",
-      "\n",
-      "#Result\n",
-      "print \"The elongation in the member is\",round(delta*1000,2),\"mm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The elongation in the member is 3.43 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.3, Page No:37"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decelration\n",
-      "A_AC=0.25 #Cross Sectional Area in square inch\n",
-      "Load=2000 #Load at point C in lb\n",
-      "E=29*10**6 #Modulus of elasticity in psi\n",
-      "theta=(pi*40)/180 #Angle in radians\n",
-      "L_BC=8 #Length in ft\n",
-      "\n",
-      "#Calculations\n",
-      "#Using sum of forces \n",
-      "P_AC=Load/sin(theta) #Force in cable AC in lb\n",
-      "L_AC=(L_BC*12)/cos(theta) #Length of cable AC in in\n",
-      "\n",
-      "delta_AC=(P_AC*L_AC)/(E*A_AC) #elongation in inches\n",
-      "\n",
-      "delta_C=delta_AC/sin(theta) #displacement of point C in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The displacement of point C is\",round(delta_C,4),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The displacement of point C is 0.0837 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.4, Page No:46"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "d=0.05 #Diameter of the rod in mm\n",
-      "P=8000 #Load on the bar in N\n",
-      "E=40*10**6 #Modulus of elasticity in Pa\n",
-      "v=0.45 #Poisson Ratio\n",
-      "L=300 #Length of the rod in mm\n",
-      "\n",
-      "#Calculation\n",
-      "A=((pi*d**2)/4) #Area of the bar in mm^2\n",
-      "sigma_x=-P/A #Axial Stress in the bar in Pa\n",
-      "#As contact pressure resists the force\n",
-      "p=(v*sigma_x)/(1-v)\n",
-      "#Using Axial Strain formula\n",
-      "e_x=(sigma_x-(v*2*p))/E\n",
-      "#Corresponding change in length\n",
-      "delta=e_x*L #contraction in mm\n",
-      "#Without constrains of the wall\n",
-      "delta_w=(-P*(L*10**-3))/(E*A) #Elongation in m\n",
-      "\n",
-      "#Result\n",
-      "print \"The elongation in the bar is\",-round(delta,2),\"mm contraction\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The elongation in the bar is 8.06 mm contraction\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.5, Page No:47"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "E=500 #Modulus of elasticity in psi\n",
-      "v=0.48 #Poisson ratio\n",
-      "V=600 #Force in lb\n",
-      "w=5 #Width of the plate in inches\n",
-      "l=9 #Length of the plate in inches\n",
-      "t=1.75 #Thickness of the rubber layer in inches\n",
-      "\n",
-      "#Calculations\n",
-      "tau=V*(w*l)**-1 #Shear stress in rubber in psi\n",
-      "G=E/(2*(1+v)) #Bulk modulus in psi\n",
-      "gamma=tau/G #Shear Modulus \n",
-      "disp=t*gamma #Diplacement in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The displacement of the rubber layer is\",round(disp,4),\"in\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The displacement of the rubber layer is 0.1381 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.6, Page No:52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=10**6 #Force on the member in N\n",
-      "Es=200 #Modulus of elasticity of steel in GPa\n",
-      "Ec=14 #Modulus of elasticity concrete in GPa\n",
-      "As=900*10**-6 #Area of steel in m^2\n",
-      "Ac=0.3**2 #Area of concrete block in m^2\n",
-      "\n",
-      "#Calculation\n",
-      "#Cross Sectional Areas\n",
-      "Ast=4*As #Cross Sectional Area in m^2 of Steel\n",
-      "Act=Ac-Ast #Cross Sectional Area of Concrete in m^2\n",
-      "\n",
-      "#Applying equilibrium to the structure\n",
-      "#Using the ratio of stress and modulii of elasticity we obtain the following eq\n",
-      "sigma_ct=P/(((Es*Ec**-1)*Ast)+Act) #Stress in Concrete in Pa\n",
-      "sigma_st=sigma_ct*Es*Ec**-1 #Stress in Steel in Pa\n",
-      "\n",
-      "#Result\n",
-      "print \"The stress in steel and concrete is as follows\",round(sigma_st*10**-6,1),\"MPa and\",round(sigma_ct*10**-6,3),\"Mpa respectively\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The stress in steel and concrete is as follows 103.6 MPa and 7.255 Mpa respectively\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.7, Page No:52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "#Say the ratio of stress in steel to concrete is R\n",
-      "R=14.286 \n",
-      "sigma_co=6*10**6 #Stress in concrete in Pa\n",
-      "Ast=3.6*10**-3 #Area of steel in m^2\n",
-      "Aco=86.4*10**-3 #Area of Concrete in m^2\n",
-      "\n",
-      "#Calculation\n",
-      "sigma_st=R*sigma_co #Stress in steel in Pa\n",
-      "#Here stress is below the allowable hence safe\n",
-      "P=sigma_st*Ast+sigma_co*Aco #Allowable force in N\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum allowable force is\",round(P*10**-3),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable force is 827.0 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.8, Page No:53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#NOTE:The NOtation has been changed to ease coding\n",
-      "#Variable Decleration\n",
-      "d=0.005 #difference in length in inch\n",
-      "L=10 #Length in inch\n",
-      "#Area of copper and aluminium in sq.in\n",
-      "Ac=2 #Area of copper \n",
-      "Aa=3 #Area of aluminium \n",
-      "#Modulus of elasticity of copper and aluminium in psi\n",
-      "Ec=17000000 #Copper\n",
-      "Ea=10**7 #Aluminium\n",
-      "#Allowable Stress in psi\n",
-      "Sc=20*10**3 #Copper\n",
-      "Sa=10*10**3 #Aluminium\n",
-      "\n",
-      "#Calculation\n",
-      "#Equilibrium is Pc+Pa=P\n",
-      "#Hookes Law is delta_c=delta_a+0.005\n",
-      "#Simplfying the solution we have constants we can directly compute\n",
-      "A=d*Ec*(L+d)**-1\n",
-      "B=Ec*Ea**-1\n",
-      "C=L*B*(L+d)**-1\n",
-      "sigma_a=(Sc-A)*C**-1\n",
-      "\n",
-      "#Using equilibrium equation\n",
-      "P=Sc*Ac+sigma_a*Aa #Safe load in lb\n",
-      "\n",
-      "#Result\n",
-      "print \"The safe load on the structure is\",round(P),\"lb\"\n",
-      "#NOTE:Answer in the textbook has been rounded off and hence the discrepancy"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The safe load on the structure is 60312.0 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.9, Page No:54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=50*10**3 #Load applied in N\n",
-      "x1=0.6 #Length in m\n",
-      "x2=1.6 #Length in m\n",
-      "L1=1 #Length of steel cable in m\n",
-      "L2=2 #Length of bronze cable in m\n",
-      "L=2.4 #Length in m\n",
-      "#Area in m^2\n",
-      "Ast=600*10**-6 #Steel\n",
-      "Abr=300*10**-6 #Bronze\n",
-      "#Modulus of elasticity in GPa\n",
-      "Est=200 #Steel\n",
-      "Ebr=83 #Bronze\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying the equilibrium and Hookes law we solve by matrix method\n",
-      "a=np.array([[x1,x2],[1,-((x1*Est*Ast*L2)/(x2*Ebr*Abr))]])\n",
-      "b=np.array([L*P,0])\n",
-      "y=np.linalg.solve(a,b)\n",
-      "\n",
-      "#Stresses in Pa\n",
-      "sigma_st=y[0]*Ast**-1 #Stress in steel\n",
-      "sigma_br=y[1]/Abr #Stress in bronze\n",
-      "\n",
-      "#Result\n",
-      "print \"The stresses in steel and bronze are as follows\"\n",
-      "print round(sigma_st*10**-6,1),\"MPa and\",round(sigma_br*10**-6,1),\"MPa respectively\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The stresses in steel and bronze are as follows\n",
-        "191.8 MPa and 106.1 MPa respectively\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.10, Page No:62"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=2.5 #Length in m\n",
-      "A=1200 #Cross sectional Area in mm^2\n",
-      "delta_T=40 #Temperature drop in degree C\n",
-      "delta=0.5*10**-3 #Movement of the walls in mm\n",
-      "alpha=11.7*10**-6 #Coefficient of thermal expansion in /degreeC\n",
-      "E=200*10**9 #Modulus of elasticity in Pa\n",
-      "\n",
-      "#Calculation\n",
-      "#Part(1)\n",
-      "sigma_1=alpha*delta_T*E #Stress in the rod in Pa\n",
-      "\n",
-      "#Part(2)\n",
-      "#Using Hookes Law\n",
-      "sigma_2=E*((alpha*delta_T)-(delta*L**-1)) #Stress in the rod in Pa\n",
-      "\n",
-      "print \"The Stress in part 1 in the rod is\",round(sigma_1*10**-6,1),\"MPa\"\n",
-      "print \"The Stress in part 2 in the rod is\",round(sigma_2*10**-6,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Stress in part 1 in the rod is 93.6 MPa\n",
-        "The Stress in part 2 in the rod is 53.6 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.11, Page No:63"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "delta=100 #Increase in the temperature in degreeF\n",
-      "Load=12000 #Load on the beam in lb\n",
-      "#Length in inch\n",
-      "Ls=2*12 #Steel\n",
-      "Lb=3*12 #Bronze\n",
-      "#Area in sq.in\n",
-      "As=0.75 #Steel\n",
-      "Ab=1.5 #Bronze\n",
-      "#Modulus of elasticity in psi\n",
-      "Es=29*10**6 #Steel\n",
-      "Eb=12*10**6 #Bronze\n",
-      "#Coefficient of thermal expansion in /degree C\n",
-      "alpha_s=6.5*10**-6 #Steel\n",
-      "alpha_b=10**-5 #Bronze\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying the Hookes Law and equilibrium we get two equations\n",
-      "a=np.array([[Ls*(Es*As)**-1,-Lb*(Eb*Ab)**-1],[2,1]])\n",
-      "b=np.array([(alpha_b*delta*Lb-alpha_s*delta*Ls),Load])\n",
-      "y=np.linalg.solve(a,b)\n",
-      "\n",
-      "#Stresses\n",
-      "sigma_st=y[0]*As**-1 #Stress in steel in psi (T)\n",
-      "sigma_br=y[1]*Ab**-1 #Stress in bronze in psi (C)\n",
-      "\n",
-      "#Result\n",
-      "print \"The Stress in steel and bronze are as follows\"\n",
-      "print sigma_st,\"psi (T) and\", -sigma_br,\"psi (C)\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Stress in steel and bronze are as follows\n",
-        "11600.0 psi (T) and 3600.0 psi (C)\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2.12, Page No:64"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=6000 #Force in lb\n",
-      "Est=29*10**6 #Modulus of elasticity of steel in psi\n",
-      "L1=24 #Length in inches\n",
-      "L2=36 #Length in inches\n",
-      "alpha_1=6.5*10**-6 #coefficient of thermal expansion in /degree F of steel\n",
-      "alpha_2=10**-5 #coefficient of thermal expansion in /degree F of bronze\n",
-      "As=0.75 #Area os steel in sq.in\n",
-      "\n",
-      "#Calculations\n",
-      "delta_T=((P*L1)/(Est*As))/(alpha_2*L2-alpha_1*L1) #Change in temperature in degree F\n",
-      "\n",
-      "print \"The change in the Temperature is\",round(delta_T,1),\"F\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The change in the Temperature is 32.5 F\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03.ipynb
deleted file mode 100755
index 20004188..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03.ipynb
+++ /dev/null
@@ -1,338 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3b7b2939d423a571df12ac23bad3e1403d5cf67fdb898826039e21bb6d26d9ff"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Chapter 03:Torsion"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.1, Page No:79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=20*10**3 #Power in W\n",
-      "f=2 #Frequency in Hz\n",
-      "t_max=40*10**6 #Maximum shear stress in Pa\n",
-      "G=83*10**9 #Bulk modulus in Pa\n",
-      "theta=(6*pi)/180 #Angle of twist in radians\n",
-      "L=3 #Length in m\n",
-      "\n",
-      "#Calculations\n",
-      "#Strength condition\n",
-      "T=P/(2*pi*f) #Torque in N.m\n",
-      "d1=((16*T)/(pi*t_max))**0.333 #Max allowable diameter in mm\n",
-      "\n",
-      "#Applying torque-twist relationship\n",
-      "d2=((32*T*L)/(G*theta*pi))**0.25 #Diameter in mm\n",
-      "\n",
-      "d=max(d1,d2)\n",
-      "\n",
-      "print \"To satisfy both strength and rigidity conditions d=\",round(d*1000,1),\"mm\"\n",
-      "\n",
-      "#NOTE:The fractional power leads to the discrepancy in the answer\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To satisfy both strength and rigidity conditions d= 58.9 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.2, Page NO:79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Ga=4*10**6 #Bulk modulus of Aluminium in psi\n",
-      "Gs=12*10**6 #Bulk Modulus of Steel in psi\n",
-      "T=10**4 #Torque in lb.in\n",
-      "L1=3 #Length in ft of the Steel bar\n",
-      "L2=6 #Length in ft of the Aluminium bar\n",
-      "d1=3 #Diameter of the Aluminium bar in inches\n",
-      "d2=2 #Diameter of the Steel bar in inches\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Compatibility and equlibrium conditions\n",
-      "a=np.array([[1,1],[(L1*32)/(Gs*pi*d2**4),-((L2*32)/(Ga*d1**4*pi))]])\n",
-      "b=np.array([T,0])\n",
-      "y=np.linalg.solve(a,b)\n",
-      "\n",
-      "#Stresses\n",
-      "t_max_st=(16*y[0])/(pi*d2**3) #Max shear Stress in Steel in psi\n",
-      "t_max_al=(16*y[1])/(pi*d1**3) #Max shear stress in aluminium in psi\n",
-      "\n",
-      "print \"The maximum values of Shear Stresses are as flollows\"\n",
-      "print round(t_max_st),\"psi in Steel and \",round(t_max_al),\"psi in Aluminium\"\n",
-      "#NOTE:The shear stress for steel in the txtbook is off by 3psi"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum values of Shear Stresses are as flollows\n",
-        "3453.0 psi in Steel and  863.0 psi in Aluminium\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.3, Page No:80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "d=2 #Diameter in ft\n",
-      "G=12*10**6 #Bulk Modulus in psi\n",
-      "#Torque in lb.ft\n",
-      "T1=500 #Torque 1 \n",
-      "T2=900 #Torque 2\n",
-      "T3=1000 #Torque 3\n",
-      "#Length in ft\n",
-      "L1=4 \n",
-      "L2=3\n",
-      "L3=5\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying the sum of torques we get\n",
-      "Tab=T1 #Torque at section AB in lb.ft\n",
-      "Tbc=-T2+T1 #Torque at section BC in lb.ft\n",
-      "Tcd=T3-T2+T1 #Torque at Section CD in lb.ft\n",
-      "\n",
-      "#Summing the angle of twists\n",
-      "theta_r=(((Tab*12*L3*12)+(Tbc*12*L2*12)+(Tcd*12*L1*12))*32)/(pi*2**4*G)\n",
-      "theta=(theta_r*180)/pi #Angle in degrees\n",
-      "\n",
-      "print \"The angle of twist is\",round(theta,3),\"degrees\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The angle of twist is 1.62 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.4, Page No:81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=1.5 #Length of the shaft in m\n",
-      "t_B=200 #Torque per unit length in N.m/m\n",
-      "d=0.025 #Diameter of the shaft in m\n",
-      "G=80*10**9 #Bulk Modulus for steel in Pa\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "#After carrying out the variable integration\n",
-      "T_A=0.5*t_B*L #Torque about point A in N.m\n",
-      "#Using equation of max stress\n",
-      "tau_Max=(16*T_A)*(pi*d**3)**-1 #Maximum stress in the shaft in Pa\n",
-      "\n",
-      "#Part(2)\n",
-      "J=(pi*d**4)*32**-1 #Polar moment of inertia in m^4\n",
-      "#After carrying out the computation for angle of twist we get\n",
-      "theta_r=(t_B*L**2)*(3*G*J)**-1 #Angle of twist in radians\n",
-      "theta=theta_r*(180*pi**-1) #Angle of twist in degrees\n",
-      "\n",
-      "#Result\n",
-      "print \"Result for part (1)\"\n",
-      "print \"Maximum Shear Stress in the shaft is\",round(tau_Max/10**6,1),\"MPa\"\n",
-      "print \"Result for part (2)\"\n",
-      "print \"The angle of twist in the shaft is\",round(theta,2),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Result for part (1)\n",
-        "Maximum Shear Stress in the shaft is 48.9 MPa\n",
-        "Result for part (2)\n",
-        "The angle of twist in the shaft is 2.8 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.5, Page No:91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=6 #Length of the tube in ft\n",
-      "t=3*8**-1 #Constant wall thickness in inches\n",
-      "G=12*10**6 #Bulk modulus of the tube in psi\n",
-      "w1=6 #Width on the top in inches\n",
-      "w2=4 #Width at the bottom in inches\n",
-      "h=5 #Height in inches\n",
-      "theta=0.5 #Angle of twist in radians\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "Ao=(w1+w2)*2**-1*h #Area enclosed by the median line in sq.in\n",
-      "S=w1+w2+2*(sqrt(1**2+h**2)) #Length of the median line in inches\n",
-      "#Using the torsional stifness formula we get\n",
-      "k=4*G*Ao**2*t*(L*12*S)**-1*(pi*180**-1) #tortional Stiffness  in lb.in/rad\n",
-      "\n",
-      "#Part(2)\n",
-      "T=k*theta #Torque required to produce an angle of twist of theta in lb.in\n",
-      "q=T*(2*Ao)**-1 #Shear flow in lb/in\n",
-      "tau=q/t #Shear stress in the wall in psi\n",
-      "\n",
-      "\n",
-      "#Result \n",
-      "print \"Part(1) results\"\n",
-      "print \"Torsional stiffness is\",round(k),\"lb.in/deg\"\n",
-      "print \"Part(2) results\"\n",
-      "print \"The shear stress in the wall is\", round(tau),\"psi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part(1) results\n",
-        "Torsional stiffness is 135017.0 lb.in/deg\n",
-        "Part(2) results\n",
-        "The shear stress in the wall is 3600.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=1.2 #Length of the tube in m\n",
-      "tau=40*10**6 #MAximum shear stress  in MPa\n",
-      "t=0.002 #Thickness in m\n",
-      "r=0.025 #Radius of the semicircle in m\n",
-      "G=28*10**9 #Bulk Modulus in Pa\n",
-      "t1=2 #Thickness in mm\n",
-      "t2=3 #thickness in mm\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "q=tau*t #Shear flow causing the stress in N/m\n",
-      "Ao=pi*r**2*0.5 #Area of the semi-circle in m^2\n",
-      "T=2*Ao*q #Torque causing the shear stress in N.m\n",
-      "\n",
-      "#Part(2)\n",
-      "#After computing the median lines integration we get\n",
-      "S=(pi*25*t1**-1)+(2*25*t2**-1) #Length of median line \n",
-      "theta_r=T*L*S*(4*G*Ao**2)**-1 #Angle of twist in radians\n",
-      "theta=theta_r*(180*pi**-1) #Angle of twist in degrees\n",
-      "\n",
-      "#Result\n",
-      "print \"Result for part(1)\"\n",
-      "print \"The torque causing the stress of 40MPa is\",round(T,2),\"N.m\"\n",
-      "print \"Result for part (2)\"\n",
-      "print \"The angle of twist is\",round(theta,1),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Result for part(1)\n",
-        "The torque causing the stress of 40MPa is 157.08 N.m\n",
-        "Result for part (2)\n",
-        "The angle of twist is 5.6 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_1.ipynb
deleted file mode 100755
index 20004188..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_1.ipynb
+++ /dev/null
@@ -1,338 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3b7b2939d423a571df12ac23bad3e1403d5cf67fdb898826039e21bb6d26d9ff"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Chapter 03:Torsion"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.1, Page No:79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=20*10**3 #Power in W\n",
-      "f=2 #Frequency in Hz\n",
-      "t_max=40*10**6 #Maximum shear stress in Pa\n",
-      "G=83*10**9 #Bulk modulus in Pa\n",
-      "theta=(6*pi)/180 #Angle of twist in radians\n",
-      "L=3 #Length in m\n",
-      "\n",
-      "#Calculations\n",
-      "#Strength condition\n",
-      "T=P/(2*pi*f) #Torque in N.m\n",
-      "d1=((16*T)/(pi*t_max))**0.333 #Max allowable diameter in mm\n",
-      "\n",
-      "#Applying torque-twist relationship\n",
-      "d2=((32*T*L)/(G*theta*pi))**0.25 #Diameter in mm\n",
-      "\n",
-      "d=max(d1,d2)\n",
-      "\n",
-      "print \"To satisfy both strength and rigidity conditions d=\",round(d*1000,1),\"mm\"\n",
-      "\n",
-      "#NOTE:The fractional power leads to the discrepancy in the answer\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To satisfy both strength and rigidity conditions d= 58.9 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.2, Page NO:79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Ga=4*10**6 #Bulk modulus of Aluminium in psi\n",
-      "Gs=12*10**6 #Bulk Modulus of Steel in psi\n",
-      "T=10**4 #Torque in lb.in\n",
-      "L1=3 #Length in ft of the Steel bar\n",
-      "L2=6 #Length in ft of the Aluminium bar\n",
-      "d1=3 #Diameter of the Aluminium bar in inches\n",
-      "d2=2 #Diameter of the Steel bar in inches\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Compatibility and equlibrium conditions\n",
-      "a=np.array([[1,1],[(L1*32)/(Gs*pi*d2**4),-((L2*32)/(Ga*d1**4*pi))]])\n",
-      "b=np.array([T,0])\n",
-      "y=np.linalg.solve(a,b)\n",
-      "\n",
-      "#Stresses\n",
-      "t_max_st=(16*y[0])/(pi*d2**3) #Max shear Stress in Steel in psi\n",
-      "t_max_al=(16*y[1])/(pi*d1**3) #Max shear stress in aluminium in psi\n",
-      "\n",
-      "print \"The maximum values of Shear Stresses are as flollows\"\n",
-      "print round(t_max_st),\"psi in Steel and \",round(t_max_al),\"psi in Aluminium\"\n",
-      "#NOTE:The shear stress for steel in the txtbook is off by 3psi"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum values of Shear Stresses are as flollows\n",
-        "3453.0 psi in Steel and  863.0 psi in Aluminium\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.3, Page No:80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "d=2 #Diameter in ft\n",
-      "G=12*10**6 #Bulk Modulus in psi\n",
-      "#Torque in lb.ft\n",
-      "T1=500 #Torque 1 \n",
-      "T2=900 #Torque 2\n",
-      "T3=1000 #Torque 3\n",
-      "#Length in ft\n",
-      "L1=4 \n",
-      "L2=3\n",
-      "L3=5\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying the sum of torques we get\n",
-      "Tab=T1 #Torque at section AB in lb.ft\n",
-      "Tbc=-T2+T1 #Torque at section BC in lb.ft\n",
-      "Tcd=T3-T2+T1 #Torque at Section CD in lb.ft\n",
-      "\n",
-      "#Summing the angle of twists\n",
-      "theta_r=(((Tab*12*L3*12)+(Tbc*12*L2*12)+(Tcd*12*L1*12))*32)/(pi*2**4*G)\n",
-      "theta=(theta_r*180)/pi #Angle in degrees\n",
-      "\n",
-      "print \"The angle of twist is\",round(theta,3),\"degrees\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The angle of twist is 1.62 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.4, Page No:81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=1.5 #Length of the shaft in m\n",
-      "t_B=200 #Torque per unit length in N.m/m\n",
-      "d=0.025 #Diameter of the shaft in m\n",
-      "G=80*10**9 #Bulk Modulus for steel in Pa\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "#After carrying out the variable integration\n",
-      "T_A=0.5*t_B*L #Torque about point A in N.m\n",
-      "#Using equation of max stress\n",
-      "tau_Max=(16*T_A)*(pi*d**3)**-1 #Maximum stress in the shaft in Pa\n",
-      "\n",
-      "#Part(2)\n",
-      "J=(pi*d**4)*32**-1 #Polar moment of inertia in m^4\n",
-      "#After carrying out the computation for angle of twist we get\n",
-      "theta_r=(t_B*L**2)*(3*G*J)**-1 #Angle of twist in radians\n",
-      "theta=theta_r*(180*pi**-1) #Angle of twist in degrees\n",
-      "\n",
-      "#Result\n",
-      "print \"Result for part (1)\"\n",
-      "print \"Maximum Shear Stress in the shaft is\",round(tau_Max/10**6,1),\"MPa\"\n",
-      "print \"Result for part (2)\"\n",
-      "print \"The angle of twist in the shaft is\",round(theta,2),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Result for part (1)\n",
-        "Maximum Shear Stress in the shaft is 48.9 MPa\n",
-        "Result for part (2)\n",
-        "The angle of twist in the shaft is 2.8 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.5, Page No:91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=6 #Length of the tube in ft\n",
-      "t=3*8**-1 #Constant wall thickness in inches\n",
-      "G=12*10**6 #Bulk modulus of the tube in psi\n",
-      "w1=6 #Width on the top in inches\n",
-      "w2=4 #Width at the bottom in inches\n",
-      "h=5 #Height in inches\n",
-      "theta=0.5 #Angle of twist in radians\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "Ao=(w1+w2)*2**-1*h #Area enclosed by the median line in sq.in\n",
-      "S=w1+w2+2*(sqrt(1**2+h**2)) #Length of the median line in inches\n",
-      "#Using the torsional stifness formula we get\n",
-      "k=4*G*Ao**2*t*(L*12*S)**-1*(pi*180**-1) #tortional Stiffness  in lb.in/rad\n",
-      "\n",
-      "#Part(2)\n",
-      "T=k*theta #Torque required to produce an angle of twist of theta in lb.in\n",
-      "q=T*(2*Ao)**-1 #Shear flow in lb/in\n",
-      "tau=q/t #Shear stress in the wall in psi\n",
-      "\n",
-      "\n",
-      "#Result \n",
-      "print \"Part(1) results\"\n",
-      "print \"Torsional stiffness is\",round(k),\"lb.in/deg\"\n",
-      "print \"Part(2) results\"\n",
-      "print \"The shear stress in the wall is\", round(tau),\"psi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part(1) results\n",
-        "Torsional stiffness is 135017.0 lb.in/deg\n",
-        "Part(2) results\n",
-        "The shear stress in the wall is 3600.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=1.2 #Length of the tube in m\n",
-      "tau=40*10**6 #MAximum shear stress  in MPa\n",
-      "t=0.002 #Thickness in m\n",
-      "r=0.025 #Radius of the semicircle in m\n",
-      "G=28*10**9 #Bulk Modulus in Pa\n",
-      "t1=2 #Thickness in mm\n",
-      "t2=3 #thickness in mm\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "q=tau*t #Shear flow causing the stress in N/m\n",
-      "Ao=pi*r**2*0.5 #Area of the semi-circle in m^2\n",
-      "T=2*Ao*q #Torque causing the shear stress in N.m\n",
-      "\n",
-      "#Part(2)\n",
-      "#After computing the median lines integration we get\n",
-      "S=(pi*25*t1**-1)+(2*25*t2**-1) #Length of median line \n",
-      "theta_r=T*L*S*(4*G*Ao**2)**-1 #Angle of twist in radians\n",
-      "theta=theta_r*(180*pi**-1) #Angle of twist in degrees\n",
-      "\n",
-      "#Result\n",
-      "print \"Result for part(1)\"\n",
-      "print \"The torque causing the stress of 40MPa is\",round(T,2),\"N.m\"\n",
-      "print \"Result for part (2)\"\n",
-      "print \"The angle of twist is\",round(theta,1),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Result for part(1)\n",
-        "The torque causing the stress of 40MPa is 157.08 N.m\n",
-        "Result for part (2)\n",
-        "The angle of twist is 5.6 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_2.ipynb
deleted file mode 100755
index 83b15053..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_2.ipynb
+++ /dev/null
@@ -1,346 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e94e252902947538e327888c054726e4c87fafcbcff4841a5585391ce610b765"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Chapter 03:Torsion"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.1, Page No:79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=20*10**3 #Power in W\n",
-      "f=2 #Frequency in Hz\n",
-      "t_max=40*10**6 #Maximum shear stress in Pa\n",
-      "G=83*10**9 #Bulk modulus in Pa\n",
-      "theta=(6*pi)/180 #Angle of twist in radians\n",
-      "L=3 #Length in m\n",
-      "\n",
-      "#Calculations\n",
-      "#Strength condition\n",
-      "T=P/(2*pi*f) #Torque in N.m\n",
-      "d1=((16*T)/(pi*t_max))**0.333 #Max allowable diameter in mm\n",
-      "\n",
-      "#Applying torque-twist relationship\n",
-      "d2=((32*T*L)/(G*theta*pi))**0.25 #Diameter in mm\n",
-      "\n",
-      "d=max(d1,d2)\n",
-      "\n",
-      "print \"To satisfy both strength and rigidity conditions d=\",round(d*1000,1),\"mm\"\n",
-      "\n",
-      "#NOTE:The fractional power leads to the discrepancy in the answer\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To satisfy both strength and rigidity conditions d= 58.9 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.2, Page No:79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Ga=4*10**6 #Bulk modulus of Aluminium in psi\n",
-      "Gs=12*10**6 #Bulk Modulus of Steel in psi\n",
-      "T=10**4 #Torque in lb.in\n",
-      "L1=3 #Length in ft of the Steel bar\n",
-      "L2=6 #Length in ft of the Aluminium bar\n",
-      "d1=3 #Diameter of the Aluminium bar in inches\n",
-      "d2=2 #Diameter of the Steel bar in inches\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Compatibility and equlibrium conditions\n",
-      "a=np.array([[1,1],[(L1*32)/(Gs*pi*d2**4),-((L2*32)/(Ga*d1**4*pi))]])\n",
-      "b=np.array([T,0])\n",
-      "y=np.linalg.solve(a,b)\n",
-      "\n",
-      "#Stresses\n",
-      "t_max_st=(16*y[0])/(pi*d2**3) #Max shear Stress in Steel in psi\n",
-      "t_max_al=(16*y[1])/(pi*d1**3) #Max shear stress in aluminium in psi\n",
-      "\n",
-      "print \"The maximum values of Shear Stresses are as flollows\"\n",
-      "print round(t_max_st),\"psi in Steel and \",round(t_max_al),\"psi in Aluminium\"\n",
-      "#NOTE:The shear stress for steel in the txtbook is off by 3psi"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum values of Shear Stresses are as flollows\n",
-        "3453.0 psi in Steel and  863.0 psi in Aluminium\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.3, Page No:80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "d=2 #Diameter in ft\n",
-      "G=12*10**6 #Bulk Modulus in psi\n",
-      "#Torque in lb.ft\n",
-      "T1=500 #Torque 1 \n",
-      "T2=900 #Torque 2\n",
-      "T3=1000 #Torque 3\n",
-      "#Length in ft\n",
-      "L1=4 \n",
-      "L2=3\n",
-      "L3=5\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying the sum of torques we get\n",
-      "Tab=T1 #Torque at section AB in lb.ft\n",
-      "Tbc=-T2+T1 #Torque at section BC in lb.ft\n",
-      "Tcd=T3-T2+T1 #Torque at Section CD in lb.ft\n",
-      "\n",
-      "#Summing the angle of twists\n",
-      "theta_r=(((Tab*12*L3*12)+(Tbc*12*L2*12)+(Tcd*12*L1*12))*32)/(pi*2**4*G)\n",
-      "theta=(theta_r*180)/pi #Angle in degrees\n",
-      "\n",
-      "print \"The angle of twist is\",round(theta,3),\"degrees\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The angle of twist is 1.62 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.4, Page No:81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=1.5 #Length of the shaft in m\n",
-      "t_B=200 #Torque per unit length in N.m/m\n",
-      "d=0.025 #Diameter of the shaft in m\n",
-      "G=80*10**9 #Bulk Modulus for steel in Pa\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "#After carrying out the variable integration\n",
-      "T_A=0.5*t_B*L #Torque about point A in N.m\n",
-      "#Using equation of max stress\n",
-      "tau_Max=(16*T_A)*(pi*d**3)**-1 #Maximum stress in the shaft in Pa\n",
-      "\n",
-      "#Part(2)\n",
-      "J=(pi*d**4)*32**-1 #Polar moment of inertia in m^4\n",
-      "#After carrying out the computation for angle of twist we get\n",
-      "theta_r=(t_B*L**2)*(3*G*J)**-1 #Angle of twist in radians\n",
-      "theta=theta_r*(180*pi**-1) #Angle of twist in degrees\n",
-      "\n",
-      "#Result\n",
-      "print \"Result for part (1)\"\n",
-      "print \"Maximum Shear Stress in the shaft is\",round(tau_Max/10**6,1),\"MPa\"\n",
-      "print \"Result for part (2)\"\n",
-      "print \"The angle of twist in the shaft is\",round(theta,2),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Result for part (1)\n",
-        "Maximum Shear Stress in the shaft is 48.9 MPa\n",
-        "Result for part (2)\n",
-        "The angle of twist in the shaft is 2.8 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.5, Page No:91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=6 #Length of the tube in ft\n",
-      "t=3*8**-1 #Constant wall thickness in inches\n",
-      "G=12*10**6 #Bulk modulus of the tube in psi\n",
-      "w1=6 #Width on the top in inches\n",
-      "w2=4 #Width at the bottom in inches\n",
-      "h=5 #Height in inches\n",
-      "theta=0.5 #Angle of twist in radians\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "Ao=(w1+w2)*2**-1*h #Area enclosed by the median line in sq.in\n",
-      "S=w1+w2+2*(sqrt(1**2+h**2)) #Length of the median line in inches\n",
-      "#Using the torsional stifness formula we get\n",
-      "k=4*G*Ao**2*t*(L*12*S)**-1*(pi*180**-1) #tortional Stiffness  in lb.in/rad\n",
-      "\n",
-      "#Part(2)\n",
-      "T=k*theta #Torque required to produce an angle of twist of theta in lb.in\n",
-      "q=T*(2*Ao)**-1 #Shear flow in lb/in\n",
-      "tau=q/t #Shear stress in the wall in psi\n",
-      "\n",
-      "\n",
-      "#Result \n",
-      "print \"Part(1) results\"\n",
-      "print \"Torsional stiffness is\",round(k),\"lb.in/deg\"\n",
-      "print \"Part(2) results\"\n",
-      "print \"The shear stress in the wall is\", round(tau),\"psi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part(1) results\n",
-        "Torsional stiffness is 135017.0 lb.in/deg\n",
-        "Part(2) results\n",
-        "The shear stress in the wall is 3600.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3.6, Page No:92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=1.2 #Length of the tube in m\n",
-      "tau=40*10**6 #MAximum shear stress  in MPa\n",
-      "t=0.002 #Thickness in m\n",
-      "r=0.025 #Radius of the semicircle in m\n",
-      "G=28*10**9 #Bulk Modulus in Pa\n",
-      "t1=2 #Thickness in mm\n",
-      "t2=3 #thickness in mm\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "q=tau*t #Shear flow causing the stress in N/m\n",
-      "Ao=pi*r**2*0.5 #Area of the semi-circle in m^2\n",
-      "T=2*Ao*q #Torque causing the shear stress in N.m\n",
-      "\n",
-      "#Part(2)\n",
-      "#After computing the median lines integration we get\n",
-      "S=(pi*25*t1**-1)+(2*25*t2**-1) #Length of median line \n",
-      "theta_r=T*L*S*(4*G*Ao**2)**-1 #Angle of twist in radians\n",
-      "theta=theta_r*(180*pi**-1) #Angle of twist in degrees\n",
-      "\n",
-      "#Result\n",
-      "print \"Result for part(1)\"\n",
-      "print \"The torque causing the stress of 40MPa is\",round(T,2),\"N.m\"\n",
-      "print \"Result for part (2)\"\n",
-      "print \"The angle of twist is\",round(theta,1),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Result for part(1)\n",
-        "The torque causing the stress of 40MPa is 157.08 N.m\n",
-        "Result for part (2)\n",
-        "The angle of twist is 5.6 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04.ipynb
deleted file mode 100755
index c3a3650e..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04.ipynb
+++ /dev/null
@@ -1,460 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:917bb07406192d85aadde0267447c880e5f0a4926ac5c228682662420223d03b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 04:Shear and Moment in Beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4.1, Page No:103"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "%matplotlib  inline\n",
-      "\n",
-      "#Variable Decleration\n",
-      "F1=14 #Force in kN\n",
-      "F2=28 #Force in kN\n",
-      "l1=2 #Length in m\n",
-      "l2=3 #Length in m\n",
-      "Ra=18 #Reaction at point A in kN\n",
-      "Rb=24 #Reaction at point D in kN\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "#Applying the summation of force in y in segment AB\n",
-      "V_ab=Ra # Shear in part AB in kN\n",
-      "#Moment is in the variable form M=18x kN.m\n",
-      "#Segment BC\n",
-      "V_bc=Ra-F1 #Shear in the segment BC in kN\n",
-      "#Moment in the form M=4x+28 kN.m\n",
-      "#Segment CD\n",
-      "V_cd=Ra-F1-F2 #Shear in the segment CD in kN\n",
-      "#Moment in the form -24x+168 kN.m\n",
-      "\n",
-      "#Importing the plotting libraries and computing the plots\n",
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Result\n",
-      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
-      "#Plotting the SHear Force Diagram\n",
-      "\n",
-      "X1=[0,l1,l1+0.0000000001,l1+l2,l1+l2+0.0000000001,l1+l2+l1]\n",
-      "Y1=[V_ab,V_ab,V_bc,V_bc,V_cd,V_cd]\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",
-      "Y2=[0,36,48,0]\n",
-      "X2=[0,l1,l1+l2,l1+l2+l1]\n",
-      "Z2=[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 0x10bc70310>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
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1SNFJlpXuBIxN1o2BsDELhII8AcDdF5nZLsnxY4ExyfEVZjZ1q7d9\nIvn8KuGXxDeXyyBSVdfc2lvuviD5b1gATE6OzwfaZHAdkYyo8EsxagB8lqwLX5l1FR5vLtxb9+Nv\nXdC/Tj5vpG4/N5Vdc2tfV3i8qcI5m+p4TZFKqatHio67rwLeMrOfQFhu2swOq+G0l4Ezk9fuQlhl\ntSZfAN+r4muFvJmKFDkVfikGTczsnQof/wWcD/ws2YZxPlBx43Kv5PHjhM04FgKjCF06le256hXO\neRr4cXJz95hqXlfVNSt776qeaxldSY2WZRZJmFlTd//SzFoS9ho+2t0/jJ1LJG3qNxTZ4s/JDl7b\nAjeo6EuxUotfRKTEqI9fRKTEqPCLiJQYFX4RkRKjwi8iUmJU+EVESowKv4hIifn/jmoHSu49zj8A\nAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10b5afad0>"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4.3, Page No:108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "import matplotlib.pyplot as plt\n",
-      "%matplotlib inline\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=12 #Length of the beam in ft\n",
-      "F=1000 #Force at the tip of the beam in lb\n",
-      "#w=30x Force per length on the beam in lb/ft\n",
-      "a=15 #Constant\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "#Here all the computation is in variable form\n",
-      "#Applying the sum of forces\n",
-      "#V=1000-15x^2\n",
-      "#Applying moment about C\n",
-      "#M=1000x-5x^3\n",
-      "\n",
-      "#Part(2)\n",
-      "#Max BM when shear force is zero\n",
-      "x=(F*a**-1)**0.5 #Length at which BM is max in ft\n",
-      "M_max=F*x-5*x**3 #Max BM in lb.ft\n",
-      "\n",
-      "#Result\n",
-      "b=np.linspace(0,L,20) #Array\n",
-      "c=np.linspace(0,0,20) #Zero array\n",
-      "V=F-(15*b**2) #Shear Force\n",
-      "M=F*b-5*b**3 #Bending Moment\n",
-      "\n",
-      "#Shear force plot\n",
-      "plt.plot(b,V,b,c)\n",
-      "plt.xlabel(\"Length in ft\")\n",
-      "plt.ylabel(\"Shear Force in lb\")\n",
-      "plt.show()\n",
-      "\n",
-      "#Bending Moment plot\n",
-      "plt.plot(b,M)\n",
-      "plt.xlabel(\"Length in ft\")\n",
-      "plt.ylabel(\"Bending Moment in lb.ft\")\n",
-      "plt.show()\n",
-      "\n",
-      "print \"The maximum BM is\",round(M_max),\"lb.ft\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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WM/BK1djYWHQIZeXnq161/GxQ+8/XVoUklYiYGhHTSr1e0upAz4iYnBWNBvbJ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\nAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10bc81f10>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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tje7uVe5eBXR39yvdfa67vxoli+4xgzHgTqCyJmFEHgNqxv30Bh7JKO9lZi3M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\n6iMAAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10bd236d0>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum BM is 5443.0 lb.ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "\n",
-      "Example 4.4.4, Page no:117\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=30 #Force in kN\n",
-      "M=40 #Moment in kN.m\n",
-      "L1=3 #Length in m\n",
-      "L2=4 #Length in m\n",
-      "Ra=14 #Reaction at A in kN\n",
-      "Re=16 #Reaction at E in kN\n",
-      "\n",
-      "#Calculations\n",
-      "#The plotting will done different from that done in the textbook\n",
-      "#Plot for Shear Force\n",
-      "x_shear=[0,0.000000001,L2,L2+0.000000001,L2+L1,L2+L1*2,L2+L1*2+0.00000001]\n",
-      "y_shear=[0,Ra,Ra,Ra-P,Ra-P,Ra-P,0]\n",
-      "#Plot for Bending Moment\n",
-      "x_bm=[0,L2,L2+L1,L2+L1+0.00000001,L2+L1+L1]\n",
-      "y_bm=[0,Ra*L2,Ra*(L2+L1)-P*L1,Ra*(L2+L1)-P*L1+M,0]\n",
-      "#Zero line\n",
-      "x1=[0,0,0,0,0,0,0]\n",
-      "\n",
-      "#Result\n",
-      "print \"The plots below are the answers\"\n",
-      "\n",
-      "plt.plot(x_shear,y_shear,x_shear,x1)\n",
-      "plt.xlabel(\"Distance from point A in m\")\n",
-      "plt.ylabel(\"Shear Force in kN\")\n",
-      "plt.show()\n",
-      "\n",
-      "plt.plot(x_bm,y_bm)\n",
-      "plt.xlabel(\"Distance from point A in m\")\n",
-      "plt.ylabel(\"Bending Moment in kN.m\")\n",
-      "plt.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The plots below are the answers\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": "iVBORw0KGgoAAAANSUhEUgAAAYgAAAEPCAYAAABY9lNGAAAABHNCSVQICAgIfAhkiAAAAAlwSFlz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\nkwlCktTJBCFJ6vT/AXRCVK96moUqAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10bd21990>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x10bdcae90>"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4.5, Page No:119"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import matplotlib.pyplot as plt\n",
-      "%matplotlib inline\n",
-      "\n",
-      "#Variable Decleration\n",
-      "w1=400 #UDL in lb/ft\n",
-      "P=400 #Point load at C in lb\n",
-      "w2=200 #UDL in lb/ft\n",
-      "L1=2 #Length in ft\n",
-      "L2=1 #Length in ft\n",
-      "L3=4 #Length in ft\n",
-      "V_A=0 #Shear force at A in lb\n",
-      "Rb=-1520 #Reaction at B in lb\n",
-      "Rd=-880 #Reaction at D in lb\n",
-      "d=1.6 #Distance in ft\n",
-      "\n",
-      "#Calculations\n",
-      "#The plotting of the Shear force diagram and the Bending Moment Diagram \n",
-      "#Will be done different from that done in the textbook\n",
-      "\n",
-      "#Calculations for Shear Force\n",
-      "Area1=P*L1 #Area of w diagram from A to B\n",
-      "Area2=0 #Area of w diagram from B to C\n",
-      "Area3=w2*L3 #Area of w diagram from C to D\n",
-      "V_B_left=V_A-Area1 #Shear Force at left of B in lb\n",
-      "V_B_right=V_B_left-Rb #Shear force at right of B in lb\n",
-      "V_C_left=V_B_right-Area2 #Shear Force at left of C in lb\n",
-      "V_C_right=V_C_left-P #Shear Force at right of C in lb\n",
-      "V_D_left=V_C_right-Area3 #Shear Force at left of D in lb\n",
-      "V_D_right=V_D_left-Rd #Shear Force at right of D in lb\n",
-      "V_E=0 #Shear Force at E in lb\n",
-      "\n",
-      "#Calculations for Bending Moments\n",
-      "AreaV1=0.5*L1*V_B_left #Area of V diagram\n",
-      "AreaV2=V_C_left*L2 #Area of V diagram from B to C\n",
-      "AreaV3=V_D_left*(L3-d)*0.5 #Area of V diagram from F to D\n",
-      "M_A=0 #Moment at A in lb.ft\n",
-      "M_B=M_A+AreaV1 #Moment about B in lb.ft\n",
-      "M_C=M_B+AreaV2 #Moment about C in lb.ft\n",
-      "M_F=M_C+V_C_right*0.5*d #Moment about F in lb.ft\n",
-      "M_D=M_F+AreaV3 #Moment about D in lb.ft\n",
-      "M_E=0 #Moment about E in lb.ft\n",
-      "\n",
-      "#Result\n",
-      "print \"The following plots are the results\"\n",
-      "\n",
-      "#Plotting\n",
-      "\n",
-      "#Shear Force\n",
-      "x=[0,L1,L1+0.000001,L1+d,L1+d+0.000001,L1+L3,L1+L3+0.000001,L1+L3+L1]\n",
-      "V=[V_A,V_B_left,V_B_right,V_C_left,V_C_right,V_D_left,V_D_right,V_E]\n",
-      "zero=[0,0,0,0,0,0,0,0]\n",
-      "plt.plot(x,V,x,zero)\n",
-      "plt.xlabel(\"Length in ft\")\n",
-      "plt.ylabel(\"Shear Force in lb\")\n",
-      "plt.title(\"Shear Force Diagram\")\n",
-      "plt.show()\n",
-      "\n",
-      "#Bending Moment\n",
-      "x1=[0,L1,L1+L2,L1+L2+d,L1+L3,L1+L3+L1]\n",
-      "BM=[M_A,M_B,M_C,M_F,M_D,M_E]\n",
-      "zero1=[0,0,0,0,0,0]\n",
-      "plt.plot(x1,BM,x1,zero1)\n",
-      "plt.xlabel(\"Length in ft\")\n",
-      "plt.ylabel(\"Bending Moment in lb.ft\")\n",
-      "plt.title(\"Bending Moment Diagram\")\n",
-      "plt.show()\n",
-      "\n",
-      "#The Bending Moment Diagram differs from that in the textbook because of curve type"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The following plots are the results\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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z9Ge4D8Osc7gPo4W5nmFmzcaHpJqU+zPMrNl4hNHE3J9hZs3ECaPJeT5wM2sW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\nccIwM7NC/j98YLpV4JrJ+QAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10b6bd690>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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GUNVPgL0yGVQuE9nSynCZs3gxrFsHzZqFjsRlQsOG1nr8979DR1IxP/5os6GO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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10bf2d310>"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4.6, Page No:127"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P1=15 #Load in kN\n",
-      "P2=25 #Load in kN\n",
-      "P3=50 #Load in kN\n",
-      "R=90 #Load in kN\n",
-      "L1=3.5 #Length in m\n",
-      "L2=2 #Length in m\n",
-      "L3=3 #Length in m\n",
-      "L=12 #Total span in m\n",
-      "\n",
-      "#Calculation\n",
-      "#Part 1\n",
-      "#Maximum Bending Moment at A\n",
-      "R1=R*L1*L**-1 #Reaction 1 in kN\n",
-      "M_A=R1*L1 #Moment about A in kN.m\n",
-      "#Maximum Bending Moment at B\n",
-      "R1_2=R*(L1+(L3-L2))*L**-1 #reaction 1 in kN\n",
-      "M_B=R1_2*(L1+(L3-L2))-P1*L2 #Moment at B in kN.m\n",
-      "\n",
-      "#Maximum Moment at C\n",
-      "R2=(P2+P3)*(L2+L3)*L**-1 #Reaction 2 in kN\n",
-      "M_C=R2*(L2+L3) #Moment at C in kN.m\n",
-      "\n",
-      "M_max=max(M_A,M_B,M_C) #Maximum Bending Moment in kN.m\n",
-      "\n",
-      "#Part 2\n",
-      "R2_2=R*(L-L3)*L**-1 #Reaction 2 in kN\n",
-      "\n",
-      "V_max=max(R1,R2,R1_2,R2_2) #Maximum Shear Force in kN\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum Shear force is\",V_max,\"kN and the Maximum Bending Moment is\",round(M_max,1),\"kN.m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum Shear force is 67.5 kN and the Maximum Bending Moment is 156.3 kN.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04_1.ipynb
deleted file mode 100755
index c3a3650e..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04_1.ipynb
+++ /dev/null
@@ -1,460 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:917bb07406192d85aadde0267447c880e5f0a4926ac5c228682662420223d03b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 04:Shear and Moment in Beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4.1, Page No:103"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "%matplotlib  inline\n",
-      "\n",
-      "#Variable Decleration\n",
-      "F1=14 #Force in kN\n",
-      "F2=28 #Force in kN\n",
-      "l1=2 #Length in m\n",
-      "l2=3 #Length in m\n",
-      "Ra=18 #Reaction at point A in kN\n",
-      "Rb=24 #Reaction at point D in kN\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "#Applying the summation of force in y in segment AB\n",
-      "V_ab=Ra # Shear in part AB in kN\n",
-      "#Moment is in the variable form M=18x kN.m\n",
-      "#Segment BC\n",
-      "V_bc=Ra-F1 #Shear in the segment BC in kN\n",
-      "#Moment in the form M=4x+28 kN.m\n",
-      "#Segment CD\n",
-      "V_cd=Ra-F1-F2 #Shear in the segment CD in kN\n",
-      "#Moment in the form -24x+168 kN.m\n",
-      "\n",
-      "#Importing the plotting libraries and computing the plots\n",
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Result\n",
-      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
-      "#Plotting the SHear Force Diagram\n",
-      "\n",
-      "X1=[0,l1,l1+0.0000000001,l1+l2,l1+l2+0.0000000001,l1+l2+l1]\n",
-      "Y1=[V_ab,V_ab,V_bc,V_bc,V_cd,V_cd]\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",
-      "Y2=[0,36,48,0]\n",
-      "X2=[0,l1,l1+l2,l1+l2+l1]\n",
-      "Z2=[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 0x10bc70310>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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1SNFJlpXuBIxN1o2BsDELhII8AcDdF5nZLsnxY4ExyfEVZjZ1q7d9\nIvn8KuGXxDeXyyBSVdfc2lvuviD5b1gATE6OzwfaZHAdkYyo8EsxagB8lqwLX5l1FR5vLtxb9+Nv\nXdC/Tj5vpG4/N5Vdc2tfV3i8qcI5m+p4TZFKqatHio67rwLeMrOfQFhu2swOq+G0l4Ezk9fuQlhl\ntSZfAN+r4muFvJmKFDkVfikGTczsnQof/wWcD/ws2YZxPlBx43Kv5PHjhM04FgKjCF06le256hXO\neRr4cXJz95hqXlfVNSt776qeaxldSY2WZRZJmFlTd//SzFoS9ho+2t0/jJ1LJG3qNxTZ4s/JDl7b\nAjeo6EuxUotfRKTEqI9fRKTEqPCLiJQYFX4RkRKjwi8iUmJU+EVESowKv4hIifn/jmoHSu49zj8A\nAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10b5afad0>"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4.3, Page No:108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "import matplotlib.pyplot as plt\n",
-      "%matplotlib inline\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=12 #Length of the beam in ft\n",
-      "F=1000 #Force at the tip of the beam in lb\n",
-      "#w=30x Force per length on the beam in lb/ft\n",
-      "a=15 #Constant\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "#Here all the computation is in variable form\n",
-      "#Applying the sum of forces\n",
-      "#V=1000-15x^2\n",
-      "#Applying moment about C\n",
-      "#M=1000x-5x^3\n",
-      "\n",
-      "#Part(2)\n",
-      "#Max BM when shear force is zero\n",
-      "x=(F*a**-1)**0.5 #Length at which BM is max in ft\n",
-      "M_max=F*x-5*x**3 #Max BM in lb.ft\n",
-      "\n",
-      "#Result\n",
-      "b=np.linspace(0,L,20) #Array\n",
-      "c=np.linspace(0,0,20) #Zero array\n",
-      "V=F-(15*b**2) #Shear Force\n",
-      "M=F*b-5*b**3 #Bending Moment\n",
-      "\n",
-      "#Shear force plot\n",
-      "plt.plot(b,V,b,c)\n",
-      "plt.xlabel(\"Length in ft\")\n",
-      "plt.ylabel(\"Shear Force in lb\")\n",
-      "plt.show()\n",
-      "\n",
-      "#Bending Moment plot\n",
-      "plt.plot(b,M)\n",
-      "plt.xlabel(\"Length in ft\")\n",
-      "plt.ylabel(\"Bending Moment in lb.ft\")\n",
-      "plt.show()\n",
-      "\n",
-      "print \"The maximum BM is\",round(M_max),\"lb.ft\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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WM/BK1djYWHQIZeXnq161/GxQ+8/XVoUklYiYGhHTSr1e0upAz4iYnBWNBvbJ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\nAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10bc81f10>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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tje7uVe5eBXR39yvdfa67vxoli+4xgzHgTqCyJmFEHgNqxv30Bh7JKO9lZi3M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\n6iMAAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10bd236d0>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum BM is 5443.0 lb.ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "\n",
-      "Example 4.4.4, Page no:117\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=30 #Force in kN\n",
-      "M=40 #Moment in kN.m\n",
-      "L1=3 #Length in m\n",
-      "L2=4 #Length in m\n",
-      "Ra=14 #Reaction at A in kN\n",
-      "Re=16 #Reaction at E in kN\n",
-      "\n",
-      "#Calculations\n",
-      "#The plotting will done different from that done in the textbook\n",
-      "#Plot for Shear Force\n",
-      "x_shear=[0,0.000000001,L2,L2+0.000000001,L2+L1,L2+L1*2,L2+L1*2+0.00000001]\n",
-      "y_shear=[0,Ra,Ra,Ra-P,Ra-P,Ra-P,0]\n",
-      "#Plot for Bending Moment\n",
-      "x_bm=[0,L2,L2+L1,L2+L1+0.00000001,L2+L1+L1]\n",
-      "y_bm=[0,Ra*L2,Ra*(L2+L1)-P*L1,Ra*(L2+L1)-P*L1+M,0]\n",
-      "#Zero line\n",
-      "x1=[0,0,0,0,0,0,0]\n",
-      "\n",
-      "#Result\n",
-      "print \"The plots below are the answers\"\n",
-      "\n",
-      "plt.plot(x_shear,y_shear,x_shear,x1)\n",
-      "plt.xlabel(\"Distance from point A in m\")\n",
-      "plt.ylabel(\"Shear Force in kN\")\n",
-      "plt.show()\n",
-      "\n",
-      "plt.plot(x_bm,y_bm)\n",
-      "plt.xlabel(\"Distance from point A in m\")\n",
-      "plt.ylabel(\"Bending Moment in kN.m\")\n",
-      "plt.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The plots below are the answers\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": "iVBORw0KGgoAAAANSUhEUgAAAYgAAAEPCAYAAABY9lNGAAAABHNCSVQICAgIfAhkiAAAAAlwSFlz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\nkwlCktTJBCFJ6vT/AXRCVK96moUqAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10bd21990>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x10bdcae90>"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4.5, Page No:119"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import matplotlib.pyplot as plt\n",
-      "%matplotlib inline\n",
-      "\n",
-      "#Variable Decleration\n",
-      "w1=400 #UDL in lb/ft\n",
-      "P=400 #Point load at C in lb\n",
-      "w2=200 #UDL in lb/ft\n",
-      "L1=2 #Length in ft\n",
-      "L2=1 #Length in ft\n",
-      "L3=4 #Length in ft\n",
-      "V_A=0 #Shear force at A in lb\n",
-      "Rb=-1520 #Reaction at B in lb\n",
-      "Rd=-880 #Reaction at D in lb\n",
-      "d=1.6 #Distance in ft\n",
-      "\n",
-      "#Calculations\n",
-      "#The plotting of the Shear force diagram and the Bending Moment Diagram \n",
-      "#Will be done different from that done in the textbook\n",
-      "\n",
-      "#Calculations for Shear Force\n",
-      "Area1=P*L1 #Area of w diagram from A to B\n",
-      "Area2=0 #Area of w diagram from B to C\n",
-      "Area3=w2*L3 #Area of w diagram from C to D\n",
-      "V_B_left=V_A-Area1 #Shear Force at left of B in lb\n",
-      "V_B_right=V_B_left-Rb #Shear force at right of B in lb\n",
-      "V_C_left=V_B_right-Area2 #Shear Force at left of C in lb\n",
-      "V_C_right=V_C_left-P #Shear Force at right of C in lb\n",
-      "V_D_left=V_C_right-Area3 #Shear Force at left of D in lb\n",
-      "V_D_right=V_D_left-Rd #Shear Force at right of D in lb\n",
-      "V_E=0 #Shear Force at E in lb\n",
-      "\n",
-      "#Calculations for Bending Moments\n",
-      "AreaV1=0.5*L1*V_B_left #Area of V diagram\n",
-      "AreaV2=V_C_left*L2 #Area of V diagram from B to C\n",
-      "AreaV3=V_D_left*(L3-d)*0.5 #Area of V diagram from F to D\n",
-      "M_A=0 #Moment at A in lb.ft\n",
-      "M_B=M_A+AreaV1 #Moment about B in lb.ft\n",
-      "M_C=M_B+AreaV2 #Moment about C in lb.ft\n",
-      "M_F=M_C+V_C_right*0.5*d #Moment about F in lb.ft\n",
-      "M_D=M_F+AreaV3 #Moment about D in lb.ft\n",
-      "M_E=0 #Moment about E in lb.ft\n",
-      "\n",
-      "#Result\n",
-      "print \"The following plots are the results\"\n",
-      "\n",
-      "#Plotting\n",
-      "\n",
-      "#Shear Force\n",
-      "x=[0,L1,L1+0.000001,L1+d,L1+d+0.000001,L1+L3,L1+L3+0.000001,L1+L3+L1]\n",
-      "V=[V_A,V_B_left,V_B_right,V_C_left,V_C_right,V_D_left,V_D_right,V_E]\n",
-      "zero=[0,0,0,0,0,0,0,0]\n",
-      "plt.plot(x,V,x,zero)\n",
-      "plt.xlabel(\"Length in ft\")\n",
-      "plt.ylabel(\"Shear Force in lb\")\n",
-      "plt.title(\"Shear Force Diagram\")\n",
-      "plt.show()\n",
-      "\n",
-      "#Bending Moment\n",
-      "x1=[0,L1,L1+L2,L1+L2+d,L1+L3,L1+L3+L1]\n",
-      "BM=[M_A,M_B,M_C,M_F,M_D,M_E]\n",
-      "zero1=[0,0,0,0,0,0]\n",
-      "plt.plot(x1,BM,x1,zero1)\n",
-      "plt.xlabel(\"Length in ft\")\n",
-      "plt.ylabel(\"Bending Moment in lb.ft\")\n",
-      "plt.title(\"Bending Moment Diagram\")\n",
-      "plt.show()\n",
-      "\n",
-      "#The Bending Moment Diagram differs from that in the textbook because of curve type"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The following plots are the results\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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z9Ge4D8Osc7gPo4W5nmFmzcaHpJqU+zPMrNl4hNHE3J9hZs3ECaPJeT5wM2sW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\nccIwM7NC/j98YLpV4JrJ+QAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10b6bd690>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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GUNVPgL0yGVQuE9nSynCZs3gxrFsHzZqFjsRlQsOG1nr8979DR1IxP/5os6GO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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10bf2d310>"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4.6, Page No:127"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P1=15 #Load in kN\n",
-      "P2=25 #Load in kN\n",
-      "P3=50 #Load in kN\n",
-      "R=90 #Load in kN\n",
-      "L1=3.5 #Length in m\n",
-      "L2=2 #Length in m\n",
-      "L3=3 #Length in m\n",
-      "L=12 #Total span in m\n",
-      "\n",
-      "#Calculation\n",
-      "#Part 1\n",
-      "#Maximum Bending Moment at A\n",
-      "R1=R*L1*L**-1 #Reaction 1 in kN\n",
-      "M_A=R1*L1 #Moment about A in kN.m\n",
-      "#Maximum Bending Moment at B\n",
-      "R1_2=R*(L1+(L3-L2))*L**-1 #reaction 1 in kN\n",
-      "M_B=R1_2*(L1+(L3-L2))-P1*L2 #Moment at B in kN.m\n",
-      "\n",
-      "#Maximum Moment at C\n",
-      "R2=(P2+P3)*(L2+L3)*L**-1 #Reaction 2 in kN\n",
-      "M_C=R2*(L2+L3) #Moment at C in kN.m\n",
-      "\n",
-      "M_max=max(M_A,M_B,M_C) #Maximum Bending Moment in kN.m\n",
-      "\n",
-      "#Part 2\n",
-      "R2_2=R*(L-L3)*L**-1 #Reaction 2 in kN\n",
-      "\n",
-      "V_max=max(R1,R2,R1_2,R2_2) #Maximum Shear Force in kN\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum Shear force is\",V_max,\"kN and the Maximum Bending Moment is\",round(M_max,1),\"kN.m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum Shear force is 67.5 kN and the Maximum Bending Moment is 156.3 kN.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04_2.ipynb
deleted file mode 100755
index c3a3650e..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04_2.ipynb
+++ /dev/null
@@ -1,460 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:917bb07406192d85aadde0267447c880e5f0a4926ac5c228682662420223d03b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 04:Shear and Moment in Beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4.1, Page No:103"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "%matplotlib  inline\n",
-      "\n",
-      "#Variable Decleration\n",
-      "F1=14 #Force in kN\n",
-      "F2=28 #Force in kN\n",
-      "l1=2 #Length in m\n",
-      "l2=3 #Length in m\n",
-      "Ra=18 #Reaction at point A in kN\n",
-      "Rb=24 #Reaction at point D in kN\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "#Applying the summation of force in y in segment AB\n",
-      "V_ab=Ra # Shear in part AB in kN\n",
-      "#Moment is in the variable form M=18x kN.m\n",
-      "#Segment BC\n",
-      "V_bc=Ra-F1 #Shear in the segment BC in kN\n",
-      "#Moment in the form M=4x+28 kN.m\n",
-      "#Segment CD\n",
-      "V_cd=Ra-F1-F2 #Shear in the segment CD in kN\n",
-      "#Moment in the form -24x+168 kN.m\n",
-      "\n",
-      "#Importing the plotting libraries and computing the plots\n",
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Result\n",
-      "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
-      "#Plotting the SHear Force Diagram\n",
-      "\n",
-      "X1=[0,l1,l1+0.0000000001,l1+l2,l1+l2+0.0000000001,l1+l2+l1]\n",
-      "Y1=[V_ab,V_ab,V_bc,V_bc,V_cd,V_cd]\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",
-      "Y2=[0,36,48,0]\n",
-      "X2=[0,l1,l1+l2,l1+l2+l1]\n",
-      "Z2=[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 0x10bc70310>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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1SNFJlpXuBIxN1o2BsDELhII8AcDdF5nZLsnxY4ExyfEVZjZ1q7d9\nIvn8KuGXxDeXyyBSVdfc2lvuviD5b1gATE6OzwfaZHAdkYyo8EsxagB8lqwLX5l1FR5vLtxb9+Nv\nXdC/Tj5vpG4/N5Vdc2tfV3i8qcI5m+p4TZFKqatHio67rwLeMrOfQFhu2swOq+G0l4Ezk9fuQlhl\ntSZfAN+r4muFvJmKFDkVfikGTczsnQof/wWcD/ws2YZxPlBx43Kv5PHjhM04FgKjCF06le256hXO\neRr4cXJz95hqXlfVNSt776qeaxldSY2WZRZJmFlTd//SzFoS9ho+2t0/jJ1LJG3qNxTZ4s/JDl7b\nAjeo6EuxUotfRKTEqI9fRKTEqPCLiJQYFX4RkRKjwi8iUmJU+EVESowKv4hIifn/jmoHSu49zj8A\nAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10b5afad0>"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4.3, Page No:108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "import matplotlib.pyplot as plt\n",
-      "%matplotlib inline\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=12 #Length of the beam in ft\n",
-      "F=1000 #Force at the tip of the beam in lb\n",
-      "#w=30x Force per length on the beam in lb/ft\n",
-      "a=15 #Constant\n",
-      "\n",
-      "#Calculations\n",
-      "#Part(1)\n",
-      "#Here all the computation is in variable form\n",
-      "#Applying the sum of forces\n",
-      "#V=1000-15x^2\n",
-      "#Applying moment about C\n",
-      "#M=1000x-5x^3\n",
-      "\n",
-      "#Part(2)\n",
-      "#Max BM when shear force is zero\n",
-      "x=(F*a**-1)**0.5 #Length at which BM is max in ft\n",
-      "M_max=F*x-5*x**3 #Max BM in lb.ft\n",
-      "\n",
-      "#Result\n",
-      "b=np.linspace(0,L,20) #Array\n",
-      "c=np.linspace(0,0,20) #Zero array\n",
-      "V=F-(15*b**2) #Shear Force\n",
-      "M=F*b-5*b**3 #Bending Moment\n",
-      "\n",
-      "#Shear force plot\n",
-      "plt.plot(b,V,b,c)\n",
-      "plt.xlabel(\"Length in ft\")\n",
-      "plt.ylabel(\"Shear Force in lb\")\n",
-      "plt.show()\n",
-      "\n",
-      "#Bending Moment plot\n",
-      "plt.plot(b,M)\n",
-      "plt.xlabel(\"Length in ft\")\n",
-      "plt.ylabel(\"Bending Moment in lb.ft\")\n",
-      "plt.show()\n",
-      "\n",
-      "print \"The maximum BM is\",round(M_max),\"lb.ft\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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WM/BK1djYWHQIZeXnq161/GxQ+8/XVoUklYiYGhHTSr1e0upAz4iYnBWNBvbJ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\nAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10bc81f10>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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tje7uVe5eBXR39yvdfa67vxoli+4xgzHgTqCyJmFEHgNqxv30Bh7JKO9lZi3M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\n6iMAAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10bd236d0>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum BM is 5443.0 lb.ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "\n",
-      "Example 4.4.4, Page no:117\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=30 #Force in kN\n",
-      "M=40 #Moment in kN.m\n",
-      "L1=3 #Length in m\n",
-      "L2=4 #Length in m\n",
-      "Ra=14 #Reaction at A in kN\n",
-      "Re=16 #Reaction at E in kN\n",
-      "\n",
-      "#Calculations\n",
-      "#The plotting will done different from that done in the textbook\n",
-      "#Plot for Shear Force\n",
-      "x_shear=[0,0.000000001,L2,L2+0.000000001,L2+L1,L2+L1*2,L2+L1*2+0.00000001]\n",
-      "y_shear=[0,Ra,Ra,Ra-P,Ra-P,Ra-P,0]\n",
-      "#Plot for Bending Moment\n",
-      "x_bm=[0,L2,L2+L1,L2+L1+0.00000001,L2+L1+L1]\n",
-      "y_bm=[0,Ra*L2,Ra*(L2+L1)-P*L1,Ra*(L2+L1)-P*L1+M,0]\n",
-      "#Zero line\n",
-      "x1=[0,0,0,0,0,0,0]\n",
-      "\n",
-      "#Result\n",
-      "print \"The plots below are the answers\"\n",
-      "\n",
-      "plt.plot(x_shear,y_shear,x_shear,x1)\n",
-      "plt.xlabel(\"Distance from point A in m\")\n",
-      "plt.ylabel(\"Shear Force in kN\")\n",
-      "plt.show()\n",
-      "\n",
-      "plt.plot(x_bm,y_bm)\n",
-      "plt.xlabel(\"Distance from point A in m\")\n",
-      "plt.ylabel(\"Bending Moment in kN.m\")\n",
-      "plt.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The plots below are the answers\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": "iVBORw0KGgoAAAANSUhEUgAAAYgAAAEPCAYAAABY9lNGAAAABHNCSVQICAgIfAhkiAAAAAlwSFlz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\nkwlCktTJBCFJ6vT/AXRCVK96moUqAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10bd21990>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x10bdcae90>"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4.5, Page No:119"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import matplotlib.pyplot as plt\n",
-      "%matplotlib inline\n",
-      "\n",
-      "#Variable Decleration\n",
-      "w1=400 #UDL in lb/ft\n",
-      "P=400 #Point load at C in lb\n",
-      "w2=200 #UDL in lb/ft\n",
-      "L1=2 #Length in ft\n",
-      "L2=1 #Length in ft\n",
-      "L3=4 #Length in ft\n",
-      "V_A=0 #Shear force at A in lb\n",
-      "Rb=-1520 #Reaction at B in lb\n",
-      "Rd=-880 #Reaction at D in lb\n",
-      "d=1.6 #Distance in ft\n",
-      "\n",
-      "#Calculations\n",
-      "#The plotting of the Shear force diagram and the Bending Moment Diagram \n",
-      "#Will be done different from that done in the textbook\n",
-      "\n",
-      "#Calculations for Shear Force\n",
-      "Area1=P*L1 #Area of w diagram from A to B\n",
-      "Area2=0 #Area of w diagram from B to C\n",
-      "Area3=w2*L3 #Area of w diagram from C to D\n",
-      "V_B_left=V_A-Area1 #Shear Force at left of B in lb\n",
-      "V_B_right=V_B_left-Rb #Shear force at right of B in lb\n",
-      "V_C_left=V_B_right-Area2 #Shear Force at left of C in lb\n",
-      "V_C_right=V_C_left-P #Shear Force at right of C in lb\n",
-      "V_D_left=V_C_right-Area3 #Shear Force at left of D in lb\n",
-      "V_D_right=V_D_left-Rd #Shear Force at right of D in lb\n",
-      "V_E=0 #Shear Force at E in lb\n",
-      "\n",
-      "#Calculations for Bending Moments\n",
-      "AreaV1=0.5*L1*V_B_left #Area of V diagram\n",
-      "AreaV2=V_C_left*L2 #Area of V diagram from B to C\n",
-      "AreaV3=V_D_left*(L3-d)*0.5 #Area of V diagram from F to D\n",
-      "M_A=0 #Moment at A in lb.ft\n",
-      "M_B=M_A+AreaV1 #Moment about B in lb.ft\n",
-      "M_C=M_B+AreaV2 #Moment about C in lb.ft\n",
-      "M_F=M_C+V_C_right*0.5*d #Moment about F in lb.ft\n",
-      "M_D=M_F+AreaV3 #Moment about D in lb.ft\n",
-      "M_E=0 #Moment about E in lb.ft\n",
-      "\n",
-      "#Result\n",
-      "print \"The following plots are the results\"\n",
-      "\n",
-      "#Plotting\n",
-      "\n",
-      "#Shear Force\n",
-      "x=[0,L1,L1+0.000001,L1+d,L1+d+0.000001,L1+L3,L1+L3+0.000001,L1+L3+L1]\n",
-      "V=[V_A,V_B_left,V_B_right,V_C_left,V_C_right,V_D_left,V_D_right,V_E]\n",
-      "zero=[0,0,0,0,0,0,0,0]\n",
-      "plt.plot(x,V,x,zero)\n",
-      "plt.xlabel(\"Length in ft\")\n",
-      "plt.ylabel(\"Shear Force in lb\")\n",
-      "plt.title(\"Shear Force Diagram\")\n",
-      "plt.show()\n",
-      "\n",
-      "#Bending Moment\n",
-      "x1=[0,L1,L1+L2,L1+L2+d,L1+L3,L1+L3+L1]\n",
-      "BM=[M_A,M_B,M_C,M_F,M_D,M_E]\n",
-      "zero1=[0,0,0,0,0,0]\n",
-      "plt.plot(x1,BM,x1,zero1)\n",
-      "plt.xlabel(\"Length in ft\")\n",
-      "plt.ylabel(\"Bending Moment in lb.ft\")\n",
-      "plt.title(\"Bending Moment Diagram\")\n",
-      "plt.show()\n",
-      "\n",
-      "#The Bending Moment Diagram differs from that in the textbook because of curve type"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The following plots are the results\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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z9Ge4D8Osc7gPo4W5nmFmzcaHpJqU+zPMrNl4hNHE3J9hZs3ECaPJeT5wM2sW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\nccIwM7NC/j98YLpV4JrJ+QAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10b6bd690>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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GUNVPgL0yGVQuE9nSynCZs3gxrFsHzZqFjsRlQsOG1nr8979DR1IxP/5os6GO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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10bf2d310>"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 4.4.6, Page No:127"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P1=15 #Load in kN\n",
-      "P2=25 #Load in kN\n",
-      "P3=50 #Load in kN\n",
-      "R=90 #Load in kN\n",
-      "L1=3.5 #Length in m\n",
-      "L2=2 #Length in m\n",
-      "L3=3 #Length in m\n",
-      "L=12 #Total span in m\n",
-      "\n",
-      "#Calculation\n",
-      "#Part 1\n",
-      "#Maximum Bending Moment at A\n",
-      "R1=R*L1*L**-1 #Reaction 1 in kN\n",
-      "M_A=R1*L1 #Moment about A in kN.m\n",
-      "#Maximum Bending Moment at B\n",
-      "R1_2=R*(L1+(L3-L2))*L**-1 #reaction 1 in kN\n",
-      "M_B=R1_2*(L1+(L3-L2))-P1*L2 #Moment at B in kN.m\n",
-      "\n",
-      "#Maximum Moment at C\n",
-      "R2=(P2+P3)*(L2+L3)*L**-1 #Reaction 2 in kN\n",
-      "M_C=R2*(L2+L3) #Moment at C in kN.m\n",
-      "\n",
-      "M_max=max(M_A,M_B,M_C) #Maximum Bending Moment in kN.m\n",
-      "\n",
-      "#Part 2\n",
-      "R2_2=R*(L-L3)*L**-1 #Reaction 2 in kN\n",
-      "\n",
-      "V_max=max(R1,R2,R1_2,R2_2) #Maximum Shear Force in kN\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum Shear force is\",V_max,\"kN and the Maximum Bending Moment is\",round(M_max,1),\"kN.m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum Shear force is 67.5 kN and the Maximum Bending Moment is 156.3 kN.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05.ipynb
deleted file mode 100755
index dd430e74..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05.ipynb
+++ /dev/null
@@ -1,560 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d010557f61cf332fc38fdfe2d2ac6c4c0f491cf23de08424aa98b7c064b49092"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 05:Stresses in Beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.1, Page No:142"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable Decleration\n",
-      "b=0.12 #Breadth of the CS of the beam in m\n",
-      "h=0.2 #Depth of the CS of the beam in m\n",
-      "BM_max=16*10**3 #Maximum Bending Moment in N.m\n",
-      "c=0.1 #Distance of the centroid of the CS from the bottom fibre in m\n",
-      "y1=0.025 #Distance in m\n",
-      "BM=9.28*10**3 #Bending Moment in kN.m\n",
-      "\n",
-      "#Calculations\n",
-      "#Preliminary Calculations\n",
-      "I=b*h**3*12**-1 #Moment of Inertia in m^4\n",
-      "\n",
-      "#Part 1\n",
-      "sigma_max=(BM_max*c)/(I) #Maximum bending stress in the beam in Pa\n",
-      "\n",
-      "#Part 2\n",
-      "#Plot variables\n",
-      "x_plot=[0.00000001,c,c+0.000000011,c+c]\n",
-      "y_plot=[sigma_max,0,0,sigma_max]\n",
-      "\n",
-      "#Part 3\n",
-      "y=h*0.5-y1 #Distance of point at which BM is 9.8kN.m\n",
-      "sigma=(BM*y)/I #Bending Stress in Pa\n",
-      "\n",
-      "#Result\n",
-      "print \"The Bending Stress at maximum Bending Moment in the beam is\",sigma_max*10**-6,\"MPa\"\n",
-      "print \"The Bending Stress in part 3 is\",-sigma*10**-6,\"MPa\"\n",
-      "print \"The plot for stress distribution is given below\"\n",
-      "\n",
-      "plt.plot(y_plot,x_plot)\n",
-      "plt.ylabel(\"Distance from top fibre in m\")\n",
-      "plt.xlabel(\"Stress in MPa\")\n",
-      "plt.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Bending Stress at maximum Bending Moment in the beam is 20.0 MPa\n",
-        "The Bending Stress in part 3 is -8.7 MPa\n",
-        "The plot for stress distribution is given below\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x10b690cd0>"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.2, Page No:143"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "wf=6 #Width of the top flange in inches\n",
-      "df=0.8 #Depth of the top flange in inches\n",
-      "dw=8 #Depth of the web portion in inches\n",
-      "ww=0.8 #Width of the web portion in inches\n",
-      "Ra=1600 #Reation at point A in lb\n",
-      "Rb=3400 #Reaction at point B in lb\n",
-      "w=400 #Load on the beam in lb/ft\n",
-      "M_4=3200 #Moment at x=4 ft in lb.ft\n",
-      "M_10=4000 #Moment at x=10 ft in lb.ft\n",
-      "\n",
-      "#Calculations\n",
-      "#Preliminary Calculations\n",
-      "#Area computation\n",
-      "A1=dw*ww #Area of the web portion in sq.in\n",
-      "A2=wf*df #Area of the top flange in sq.in\n",
-      "y1=dw*0.5 #Centroid from the bottom of the web portion in inches\n",
-      "y2=dw+df*0.5 #Centroid from the bottom of the flange portion in inches\n",
-      "\n",
-      "#y_bar computation\n",
-      "y_bar=(A1*y1+A2*y2)/(A1+A2) #centroid of the section in inches from the bottom\n",
-      "\n",
-      "#Moment of Inertia computation\n",
-      "I=(ww*dw**3*12**-1)+(A1*(y1-y_bar)**2)+(wf*df**3*12**-1)+(A2*(y2-y_bar)**2) #Moment of inertia in in^4\n",
-      "\n",
-      "#Maximum Bending Moment\n",
-      "c_top=dw+df-y_bar #distance of top fibre in inches\n",
-      "c_bot=y_bar #Distance of bottom fibre in inches\n",
-      "\n",
-      "#Stress at x=4 ft\n",
-      "sigma_top=-(12*M_4*c_top)*I**-1 #Stress at top fibre in psi\n",
-      "sigma_bot=12*M_4*c_bot*I**-1 #Stress at bottom fibre in psi\n",
-      "\n",
-      "#Stress at x=10 ft\n",
-      "sigma_top2=M_10*12*c_top*I**-1 #Stress at the top fibre in psi\n",
-      "sigma_bot2=-M_10*12*c_bot*I**-1 #Stress at the bottom fibre in psi\n",
-      "\n",
-      "#Maximum values\n",
-      "sigma_t=max(sigma_bot,sigma_bot2,sigma_top,sigma_top2) #Maximum values for stress in tension\n",
-      "sigma_c=min(sigma_top,sigma_top2,sigma_bot,sigma_bot2) #Maximum values for stress in compression\n",
-      "\n",
-      "#Result\n",
-      "print  \"The maximum values of stress are\"\n",
-      "print \"Maximum Tension=\",round(sigma_t),\"psi at x=4ft\"\n",
-      "print \"Maximum Compression=\",round(-sigma_c),\"psi at x=10ft\"\n",
-      "\n",
-      "#NOTE:Answer is differing becuase of the decimal accuracy\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum values of stress are\n",
-        "Maximum Tension= 2583.0 psi at x=4ft\n",
-        "Maximum Compression= 3229.0 psi at x=10ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.3, Page No:145"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=4 #Length of each section in ft\n",
-      "h_ab=4 #Thickness of the front section in inches\n",
-      "h_bd=6 #Thickness of the back section in inches\n",
-      "P=2000 #Point load acting at point A in lb\n",
-      "M_B=8000 #Moment at 4ft in lb.ft\n",
-      "M_D=16000 #Moment at x=8ft in lb.ft\n",
-      "b=2 #Breadth in inches\n",
-      "\n",
-      "#Calculations\n",
-      "S_ab=b*h_ab**2*6**-1 #Sectional Modulus of section AB in in^3\n",
-      "S_bd=b*h_bd**2*6**-1 #Sectional Modulus of section BD in in^3\n",
-      "sigma_B=12*M_B*S_ab**-1 #Maximum bending stress in psi\n",
-      "sigma_D=12*M_D*S_bd**-1 #Maximum bending stress in psi\n",
-      "\n",
-      "#Maximum stress\n",
-      "sigma_max=max(sigma_B,sigma_D) #Maximum stress in psi\n",
-      "\n",
-      "#Result\n",
-      "print \"Comparing the two results we find that the maximum stress is\"\n",
-      "print \"Sigma_max=\",round(sigma_max),\"psi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Comparing the two results we find that the maximum stress is\n",
-        "Sigma_max= 18000.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.4, Page No:146"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "M=15000 #Maximum bending moment in absolute values in lb.ft\n",
-      "S=42 #Sectional Modulus in in^3\n",
-      "\n",
-      "#Calculations\n",
-      "sigma_max=M*12*S**-1 #Maximum stress in the section in psi\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum Bending Stress in the section is\",round(sigma_max),\"psi\"\n",
-      "\n",
-      "#NOTE:The answer differs due to decimal point accuracy"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum Bending Stress in the section is 4286.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.5, Page No:157"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "M_max=60*10**3 #Maximum Bending Moment in kN.m\n",
-      "sigma_w=120*10**6 #Maximum Bending Stress allowed in Pa\n",
-      "M_max_2=61.52*10**3 #max bending moment computed in N.m\n",
-      "\n",
-      "#Section details\n",
-      "mass=38.7 #Mass in kg/m\n",
-      "g=9.81 #Acceleration due to gravity in m/s^2\n",
-      "S=549*10**3 #Sectional modulus of the section in mm^3\n",
-      "\n",
-      "#Calculations\n",
-      "S_min=M_max*sigma_w**-1*10**9 #Minimum Sectional Modulus required in mm^3\n",
-      "\n",
-      "#We selecet section W310x39\n",
-      "w0=mass*g*10**-3 #Weight of the beam in kN/m\n",
-      "sigma_max=M_max_2*S**-1*10**3 #Maximum stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The section chosen is W310x39 with maximum stress as\",round(sigma_max,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The section chosen is W310x39 with maximum stress as 112.1 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.6, Page No:166"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "V_max=24 #Maximum Shear in kN\n",
-      "b=0.160 #Width of the beam in m\n",
-      "h=0.240 #Depth of the beam in m\n",
-      "\n",
-      "#Calculations\n",
-      "I=b*h**3*12**-1 #Moment of Inertia of the beam in m^4\n",
-      "\n",
-      "#Part 1\n",
-      "Q=b*(h*3**-1)**2 #First moment of Area m^3\n",
-      "tau_max=(V_max*Q)*(I*b)**-1 #Maximum Shear Stress in glue in kPa\n",
-      "\n",
-      "#Part 2\n",
-      "tau_max_2=(3.0/2.0)*(V_max/(b*h)) #Shear Stress in kPa\n",
-      "Q_1=b*h*0.5*h*0.25 #First moment about NA in m^3\n",
-      "tau_maxx=(V_max*Q_1)/(I*b) #Shear stress in kPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The Results agree in both parts\"\n",
-      "print \"The maximum stress is\", round(tau_max_2),\"kPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Results agree in both parts\n",
-        "The maximum stress is 938.0 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.7, Page No:167"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "I=310 #Moment of inertia in in^4\n",
-      "V=160 #Shear Force in kips\n",
-      "#Dimension defination\n",
-      "tf=0.515 #Thickness of flange in inches\n",
-      "de=11.94 #Effective depth in inches\n",
-      "tw=0.295 #Thickness of web in inches\n",
-      "wf=8.005 #Width of lange in inches\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "Q=wf*tf*(de-tf)*0.5 #First moment about NA in inch^3\n",
-      "tau_min=(V*Q*10**2)/(I*tw) #Minimum shear stress in web in psi\n",
-      "\n",
-      "#Part 2\n",
-      "A_2=(de*0.5-tf)*tw #Area in in^3\n",
-      "y_bar_2=0.5*(de*0.5-tf) #Depth in inches\n",
-      "\n",
-      "Q_2=Q+A_2*y_bar_2 #First Moment in inches^3\n",
-      "\n",
-      "tau_max=(V*Q_2*10**2)/(I*tw) #Maximum Shear Stress in psi\n",
-      "\n",
-      "#Part 3\n",
-      "V_web=10.91*tw*(tau_min+((2*3**-1)*(tau_max-tau_min))) #Shear in the web in lb\n",
-      "perV=(V_web/V)*100 #Percentage shear force in web in %\n",
-      "t_max_final=V*10**3/(10.91*tw)\n",
-      "\n",
-      "#result\n",
-      "print \"The final shear stress in the web portion is\",round(t_max_final),\"psi\"\n",
-      "#NOTE:Answer differs due to deciaml point accuracy"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The final shear stress in the web portion is 49713.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.8, Page No:168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "I=547 #Moment of inertia in inches^4\n",
-      "d=8.9 #NA deoth in inches\n",
-      "V=12 #Shear Force in kips\n",
-      "h=7.3 #Depth of NA\n",
-      "b=2 #Width in inches\n",
-      "t=1.2 #Thickness in inches\n",
-      "h2=7.5 #Depth in inches\n",
-      "\n",
-      "#Calculations\n",
-      "#Shear Stress at NA\n",
-      "Q=(b*h)*(h*0.5) #First Moment about NA in in^3\n",
-      "tau=(V*10**3*Q)/(I*b) #Shear stress at NA in psi\n",
-      "\n",
-      "#Shear Stress at a-a\n",
-      "Q_1=(t*h2)*(d-h2*0.5) #First moment about NA in in^3\n",
-      "tau1=(V*Q_1)/(I*t) #Shear Stress in psi\n",
-      "\n",
-      "#Result\n",
-      "print \"Comparing two stresses\"\n",
-      "print \"The maximum stress is\",round(max(tau,tau1)),\"psi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Comparing two stresses\n",
-        "The maximum stress is 585.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.10, Page No:175"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "sigma_w=1000 #Working Stress in Bending in psi\n",
-      "tau_w=100 #Working stress in shear in psi\n",
-      "#Dimensions\n",
-      "b_out=8 #Width in inches\n",
-      "h=10 #Depth in inches\n",
-      "b_in=6 #Width in inches\n",
-      "\n",
-      "#Calculations\n",
-      "I=((b_out*h**3)-(b_in*b_out**3))*12**-1 #Moment of inertia in in^4\n",
-      "#Design for shear\n",
-      "Q=(b_out*h*0.5*0.25*h)-(b_in*b_out*0.5*0.25*b_out) #First Moment about NA in in^3\n",
-      "\n",
-      "#Largest P\n",
-      "P=(tau_w*I*2)/(1.5*Q) #P in shear in lb\n",
-      "\n",
-      "#Design for bending\n",
-      "P1=(sigma_w*I)/(60*5) #P in bending in lb\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum allowable P value is\",round(min(P,P1)),\"lb\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable P value is 1053.0 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.11, Page No:182"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A=2630 #Area in mm^2\n",
-      "y_bar=536.6 #Neutral Axis depth from top in mm\n",
-      "tau_w=100 #Allowable stress in MPa\n",
-      "sigma_b_w=280 #Allowable bending stress in MPa\n",
-      "d=0.019 #Diameter of the rivet in m\n",
-      "t_web=0.01 #Thickness of the web in m\n",
-      "I=4140 #Moment of inertia in m^4\n",
-      "V=450 #Max shear allowable in kN\n",
-      "\n",
-      "#Calculations\n",
-      "Q=A*y_bar #first moment in mm^3\n",
-      "Fw=(pi*d**2)*tau_w*10**6 #Allowable force in N\n",
-      "Fw_2=d*t_web*sigma_b_w*10**6*0.5 #Allowable force in N\n",
-      "e=Fw_2*I*(V*10**3*Q*10**-3)**-1 #Allowable spacing in m\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum spacing allowed is\",round(e*1000,1),\"mm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum spacing allowed is 173.4 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05_1.ipynb
deleted file mode 100755
index dd430e74..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05_1.ipynb
+++ /dev/null
@@ -1,560 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d010557f61cf332fc38fdfe2d2ac6c4c0f491cf23de08424aa98b7c064b49092"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 05:Stresses in Beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.1, Page No:142"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable Decleration\n",
-      "b=0.12 #Breadth of the CS of the beam in m\n",
-      "h=0.2 #Depth of the CS of the beam in m\n",
-      "BM_max=16*10**3 #Maximum Bending Moment in N.m\n",
-      "c=0.1 #Distance of the centroid of the CS from the bottom fibre in m\n",
-      "y1=0.025 #Distance in m\n",
-      "BM=9.28*10**3 #Bending Moment in kN.m\n",
-      "\n",
-      "#Calculations\n",
-      "#Preliminary Calculations\n",
-      "I=b*h**3*12**-1 #Moment of Inertia in m^4\n",
-      "\n",
-      "#Part 1\n",
-      "sigma_max=(BM_max*c)/(I) #Maximum bending stress in the beam in Pa\n",
-      "\n",
-      "#Part 2\n",
-      "#Plot variables\n",
-      "x_plot=[0.00000001,c,c+0.000000011,c+c]\n",
-      "y_plot=[sigma_max,0,0,sigma_max]\n",
-      "\n",
-      "#Part 3\n",
-      "y=h*0.5-y1 #Distance of point at which BM is 9.8kN.m\n",
-      "sigma=(BM*y)/I #Bending Stress in Pa\n",
-      "\n",
-      "#Result\n",
-      "print \"The Bending Stress at maximum Bending Moment in the beam is\",sigma_max*10**-6,\"MPa\"\n",
-      "print \"The Bending Stress in part 3 is\",-sigma*10**-6,\"MPa\"\n",
-      "print \"The plot for stress distribution is given below\"\n",
-      "\n",
-      "plt.plot(y_plot,x_plot)\n",
-      "plt.ylabel(\"Distance from top fibre in m\")\n",
-      "plt.xlabel(\"Stress in MPa\")\n",
-      "plt.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Bending Stress at maximum Bending Moment in the beam is 20.0 MPa\n",
-        "The Bending Stress in part 3 is -8.7 MPa\n",
-        "The plot for stress distribution is given below\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x10b690cd0>"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.2, Page No:143"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "wf=6 #Width of the top flange in inches\n",
-      "df=0.8 #Depth of the top flange in inches\n",
-      "dw=8 #Depth of the web portion in inches\n",
-      "ww=0.8 #Width of the web portion in inches\n",
-      "Ra=1600 #Reation at point A in lb\n",
-      "Rb=3400 #Reaction at point B in lb\n",
-      "w=400 #Load on the beam in lb/ft\n",
-      "M_4=3200 #Moment at x=4 ft in lb.ft\n",
-      "M_10=4000 #Moment at x=10 ft in lb.ft\n",
-      "\n",
-      "#Calculations\n",
-      "#Preliminary Calculations\n",
-      "#Area computation\n",
-      "A1=dw*ww #Area of the web portion in sq.in\n",
-      "A2=wf*df #Area of the top flange in sq.in\n",
-      "y1=dw*0.5 #Centroid from the bottom of the web portion in inches\n",
-      "y2=dw+df*0.5 #Centroid from the bottom of the flange portion in inches\n",
-      "\n",
-      "#y_bar computation\n",
-      "y_bar=(A1*y1+A2*y2)/(A1+A2) #centroid of the section in inches from the bottom\n",
-      "\n",
-      "#Moment of Inertia computation\n",
-      "I=(ww*dw**3*12**-1)+(A1*(y1-y_bar)**2)+(wf*df**3*12**-1)+(A2*(y2-y_bar)**2) #Moment of inertia in in^4\n",
-      "\n",
-      "#Maximum Bending Moment\n",
-      "c_top=dw+df-y_bar #distance of top fibre in inches\n",
-      "c_bot=y_bar #Distance of bottom fibre in inches\n",
-      "\n",
-      "#Stress at x=4 ft\n",
-      "sigma_top=-(12*M_4*c_top)*I**-1 #Stress at top fibre in psi\n",
-      "sigma_bot=12*M_4*c_bot*I**-1 #Stress at bottom fibre in psi\n",
-      "\n",
-      "#Stress at x=10 ft\n",
-      "sigma_top2=M_10*12*c_top*I**-1 #Stress at the top fibre in psi\n",
-      "sigma_bot2=-M_10*12*c_bot*I**-1 #Stress at the bottom fibre in psi\n",
-      "\n",
-      "#Maximum values\n",
-      "sigma_t=max(sigma_bot,sigma_bot2,sigma_top,sigma_top2) #Maximum values for stress in tension\n",
-      "sigma_c=min(sigma_top,sigma_top2,sigma_bot,sigma_bot2) #Maximum values for stress in compression\n",
-      "\n",
-      "#Result\n",
-      "print  \"The maximum values of stress are\"\n",
-      "print \"Maximum Tension=\",round(sigma_t),\"psi at x=4ft\"\n",
-      "print \"Maximum Compression=\",round(-sigma_c),\"psi at x=10ft\"\n",
-      "\n",
-      "#NOTE:Answer is differing becuase of the decimal accuracy\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum values of stress are\n",
-        "Maximum Tension= 2583.0 psi at x=4ft\n",
-        "Maximum Compression= 3229.0 psi at x=10ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.3, Page No:145"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=4 #Length of each section in ft\n",
-      "h_ab=4 #Thickness of the front section in inches\n",
-      "h_bd=6 #Thickness of the back section in inches\n",
-      "P=2000 #Point load acting at point A in lb\n",
-      "M_B=8000 #Moment at 4ft in lb.ft\n",
-      "M_D=16000 #Moment at x=8ft in lb.ft\n",
-      "b=2 #Breadth in inches\n",
-      "\n",
-      "#Calculations\n",
-      "S_ab=b*h_ab**2*6**-1 #Sectional Modulus of section AB in in^3\n",
-      "S_bd=b*h_bd**2*6**-1 #Sectional Modulus of section BD in in^3\n",
-      "sigma_B=12*M_B*S_ab**-1 #Maximum bending stress in psi\n",
-      "sigma_D=12*M_D*S_bd**-1 #Maximum bending stress in psi\n",
-      "\n",
-      "#Maximum stress\n",
-      "sigma_max=max(sigma_B,sigma_D) #Maximum stress in psi\n",
-      "\n",
-      "#Result\n",
-      "print \"Comparing the two results we find that the maximum stress is\"\n",
-      "print \"Sigma_max=\",round(sigma_max),\"psi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Comparing the two results we find that the maximum stress is\n",
-        "Sigma_max= 18000.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.4, Page No:146"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "M=15000 #Maximum bending moment in absolute values in lb.ft\n",
-      "S=42 #Sectional Modulus in in^3\n",
-      "\n",
-      "#Calculations\n",
-      "sigma_max=M*12*S**-1 #Maximum stress in the section in psi\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum Bending Stress in the section is\",round(sigma_max),\"psi\"\n",
-      "\n",
-      "#NOTE:The answer differs due to decimal point accuracy"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum Bending Stress in the section is 4286.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.5, Page No:157"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "M_max=60*10**3 #Maximum Bending Moment in kN.m\n",
-      "sigma_w=120*10**6 #Maximum Bending Stress allowed in Pa\n",
-      "M_max_2=61.52*10**3 #max bending moment computed in N.m\n",
-      "\n",
-      "#Section details\n",
-      "mass=38.7 #Mass in kg/m\n",
-      "g=9.81 #Acceleration due to gravity in m/s^2\n",
-      "S=549*10**3 #Sectional modulus of the section in mm^3\n",
-      "\n",
-      "#Calculations\n",
-      "S_min=M_max*sigma_w**-1*10**9 #Minimum Sectional Modulus required in mm^3\n",
-      "\n",
-      "#We selecet section W310x39\n",
-      "w0=mass*g*10**-3 #Weight of the beam in kN/m\n",
-      "sigma_max=M_max_2*S**-1*10**3 #Maximum stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The section chosen is W310x39 with maximum stress as\",round(sigma_max,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The section chosen is W310x39 with maximum stress as 112.1 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.6, Page No:166"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "V_max=24 #Maximum Shear in kN\n",
-      "b=0.160 #Width of the beam in m\n",
-      "h=0.240 #Depth of the beam in m\n",
-      "\n",
-      "#Calculations\n",
-      "I=b*h**3*12**-1 #Moment of Inertia of the beam in m^4\n",
-      "\n",
-      "#Part 1\n",
-      "Q=b*(h*3**-1)**2 #First moment of Area m^3\n",
-      "tau_max=(V_max*Q)*(I*b)**-1 #Maximum Shear Stress in glue in kPa\n",
-      "\n",
-      "#Part 2\n",
-      "tau_max_2=(3.0/2.0)*(V_max/(b*h)) #Shear Stress in kPa\n",
-      "Q_1=b*h*0.5*h*0.25 #First moment about NA in m^3\n",
-      "tau_maxx=(V_max*Q_1)/(I*b) #Shear stress in kPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The Results agree in both parts\"\n",
-      "print \"The maximum stress is\", round(tau_max_2),\"kPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Results agree in both parts\n",
-        "The maximum stress is 938.0 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.7, Page No:167"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "I=310 #Moment of inertia in in^4\n",
-      "V=160 #Shear Force in kips\n",
-      "#Dimension defination\n",
-      "tf=0.515 #Thickness of flange in inches\n",
-      "de=11.94 #Effective depth in inches\n",
-      "tw=0.295 #Thickness of web in inches\n",
-      "wf=8.005 #Width of lange in inches\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "Q=wf*tf*(de-tf)*0.5 #First moment about NA in inch^3\n",
-      "tau_min=(V*Q*10**2)/(I*tw) #Minimum shear stress in web in psi\n",
-      "\n",
-      "#Part 2\n",
-      "A_2=(de*0.5-tf)*tw #Area in in^3\n",
-      "y_bar_2=0.5*(de*0.5-tf) #Depth in inches\n",
-      "\n",
-      "Q_2=Q+A_2*y_bar_2 #First Moment in inches^3\n",
-      "\n",
-      "tau_max=(V*Q_2*10**2)/(I*tw) #Maximum Shear Stress in psi\n",
-      "\n",
-      "#Part 3\n",
-      "V_web=10.91*tw*(tau_min+((2*3**-1)*(tau_max-tau_min))) #Shear in the web in lb\n",
-      "perV=(V_web/V)*100 #Percentage shear force in web in %\n",
-      "t_max_final=V*10**3/(10.91*tw)\n",
-      "\n",
-      "#result\n",
-      "print \"The final shear stress in the web portion is\",round(t_max_final),\"psi\"\n",
-      "#NOTE:Answer differs due to deciaml point accuracy"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The final shear stress in the web portion is 49713.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.8, Page No:168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "I=547 #Moment of inertia in inches^4\n",
-      "d=8.9 #NA deoth in inches\n",
-      "V=12 #Shear Force in kips\n",
-      "h=7.3 #Depth of NA\n",
-      "b=2 #Width in inches\n",
-      "t=1.2 #Thickness in inches\n",
-      "h2=7.5 #Depth in inches\n",
-      "\n",
-      "#Calculations\n",
-      "#Shear Stress at NA\n",
-      "Q=(b*h)*(h*0.5) #First Moment about NA in in^3\n",
-      "tau=(V*10**3*Q)/(I*b) #Shear stress at NA in psi\n",
-      "\n",
-      "#Shear Stress at a-a\n",
-      "Q_1=(t*h2)*(d-h2*0.5) #First moment about NA in in^3\n",
-      "tau1=(V*Q_1)/(I*t) #Shear Stress in psi\n",
-      "\n",
-      "#Result\n",
-      "print \"Comparing two stresses\"\n",
-      "print \"The maximum stress is\",round(max(tau,tau1)),\"psi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Comparing two stresses\n",
-        "The maximum stress is 585.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.10, Page No:175"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "sigma_w=1000 #Working Stress in Bending in psi\n",
-      "tau_w=100 #Working stress in shear in psi\n",
-      "#Dimensions\n",
-      "b_out=8 #Width in inches\n",
-      "h=10 #Depth in inches\n",
-      "b_in=6 #Width in inches\n",
-      "\n",
-      "#Calculations\n",
-      "I=((b_out*h**3)-(b_in*b_out**3))*12**-1 #Moment of inertia in in^4\n",
-      "#Design for shear\n",
-      "Q=(b_out*h*0.5*0.25*h)-(b_in*b_out*0.5*0.25*b_out) #First Moment about NA in in^3\n",
-      "\n",
-      "#Largest P\n",
-      "P=(tau_w*I*2)/(1.5*Q) #P in shear in lb\n",
-      "\n",
-      "#Design for bending\n",
-      "P1=(sigma_w*I)/(60*5) #P in bending in lb\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum allowable P value is\",round(min(P,P1)),\"lb\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable P value is 1053.0 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.11, Page No:182"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A=2630 #Area in mm^2\n",
-      "y_bar=536.6 #Neutral Axis depth from top in mm\n",
-      "tau_w=100 #Allowable stress in MPa\n",
-      "sigma_b_w=280 #Allowable bending stress in MPa\n",
-      "d=0.019 #Diameter of the rivet in m\n",
-      "t_web=0.01 #Thickness of the web in m\n",
-      "I=4140 #Moment of inertia in m^4\n",
-      "V=450 #Max shear allowable in kN\n",
-      "\n",
-      "#Calculations\n",
-      "Q=A*y_bar #first moment in mm^3\n",
-      "Fw=(pi*d**2)*tau_w*10**6 #Allowable force in N\n",
-      "Fw_2=d*t_web*sigma_b_w*10**6*0.5 #Allowable force in N\n",
-      "e=Fw_2*I*(V*10**3*Q*10**-3)**-1 #Allowable spacing in m\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum spacing allowed is\",round(e*1000,1),\"mm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum spacing allowed is 173.4 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05_2.ipynb
deleted file mode 100755
index dd430e74..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05_2.ipynb
+++ /dev/null
@@ -1,560 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d010557f61cf332fc38fdfe2d2ac6c4c0f491cf23de08424aa98b7c064b49092"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 05:Stresses in Beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.1, Page No:142"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable Decleration\n",
-      "b=0.12 #Breadth of the CS of the beam in m\n",
-      "h=0.2 #Depth of the CS of the beam in m\n",
-      "BM_max=16*10**3 #Maximum Bending Moment in N.m\n",
-      "c=0.1 #Distance of the centroid of the CS from the bottom fibre in m\n",
-      "y1=0.025 #Distance in m\n",
-      "BM=9.28*10**3 #Bending Moment in kN.m\n",
-      "\n",
-      "#Calculations\n",
-      "#Preliminary Calculations\n",
-      "I=b*h**3*12**-1 #Moment of Inertia in m^4\n",
-      "\n",
-      "#Part 1\n",
-      "sigma_max=(BM_max*c)/(I) #Maximum bending stress in the beam in Pa\n",
-      "\n",
-      "#Part 2\n",
-      "#Plot variables\n",
-      "x_plot=[0.00000001,c,c+0.000000011,c+c]\n",
-      "y_plot=[sigma_max,0,0,sigma_max]\n",
-      "\n",
-      "#Part 3\n",
-      "y=h*0.5-y1 #Distance of point at which BM is 9.8kN.m\n",
-      "sigma=(BM*y)/I #Bending Stress in Pa\n",
-      "\n",
-      "#Result\n",
-      "print \"The Bending Stress at maximum Bending Moment in the beam is\",sigma_max*10**-6,\"MPa\"\n",
-      "print \"The Bending Stress in part 3 is\",-sigma*10**-6,\"MPa\"\n",
-      "print \"The plot for stress distribution is given below\"\n",
-      "\n",
-      "plt.plot(y_plot,x_plot)\n",
-      "plt.ylabel(\"Distance from top fibre in m\")\n",
-      "plt.xlabel(\"Stress in MPa\")\n",
-      "plt.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Bending Stress at maximum Bending Moment in the beam is 20.0 MPa\n",
-        "The Bending Stress in part 3 is -8.7 MPa\n",
-        "The plot for stress distribution is given below\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x10b690cd0>"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.2, Page No:143"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "wf=6 #Width of the top flange in inches\n",
-      "df=0.8 #Depth of the top flange in inches\n",
-      "dw=8 #Depth of the web portion in inches\n",
-      "ww=0.8 #Width of the web portion in inches\n",
-      "Ra=1600 #Reation at point A in lb\n",
-      "Rb=3400 #Reaction at point B in lb\n",
-      "w=400 #Load on the beam in lb/ft\n",
-      "M_4=3200 #Moment at x=4 ft in lb.ft\n",
-      "M_10=4000 #Moment at x=10 ft in lb.ft\n",
-      "\n",
-      "#Calculations\n",
-      "#Preliminary Calculations\n",
-      "#Area computation\n",
-      "A1=dw*ww #Area of the web portion in sq.in\n",
-      "A2=wf*df #Area of the top flange in sq.in\n",
-      "y1=dw*0.5 #Centroid from the bottom of the web portion in inches\n",
-      "y2=dw+df*0.5 #Centroid from the bottom of the flange portion in inches\n",
-      "\n",
-      "#y_bar computation\n",
-      "y_bar=(A1*y1+A2*y2)/(A1+A2) #centroid of the section in inches from the bottom\n",
-      "\n",
-      "#Moment of Inertia computation\n",
-      "I=(ww*dw**3*12**-1)+(A1*(y1-y_bar)**2)+(wf*df**3*12**-1)+(A2*(y2-y_bar)**2) #Moment of inertia in in^4\n",
-      "\n",
-      "#Maximum Bending Moment\n",
-      "c_top=dw+df-y_bar #distance of top fibre in inches\n",
-      "c_bot=y_bar #Distance of bottom fibre in inches\n",
-      "\n",
-      "#Stress at x=4 ft\n",
-      "sigma_top=-(12*M_4*c_top)*I**-1 #Stress at top fibre in psi\n",
-      "sigma_bot=12*M_4*c_bot*I**-1 #Stress at bottom fibre in psi\n",
-      "\n",
-      "#Stress at x=10 ft\n",
-      "sigma_top2=M_10*12*c_top*I**-1 #Stress at the top fibre in psi\n",
-      "sigma_bot2=-M_10*12*c_bot*I**-1 #Stress at the bottom fibre in psi\n",
-      "\n",
-      "#Maximum values\n",
-      "sigma_t=max(sigma_bot,sigma_bot2,sigma_top,sigma_top2) #Maximum values for stress in tension\n",
-      "sigma_c=min(sigma_top,sigma_top2,sigma_bot,sigma_bot2) #Maximum values for stress in compression\n",
-      "\n",
-      "#Result\n",
-      "print  \"The maximum values of stress are\"\n",
-      "print \"Maximum Tension=\",round(sigma_t),\"psi at x=4ft\"\n",
-      "print \"Maximum Compression=\",round(-sigma_c),\"psi at x=10ft\"\n",
-      "\n",
-      "#NOTE:Answer is differing becuase of the decimal accuracy\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum values of stress are\n",
-        "Maximum Tension= 2583.0 psi at x=4ft\n",
-        "Maximum Compression= 3229.0 psi at x=10ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.3, Page No:145"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=4 #Length of each section in ft\n",
-      "h_ab=4 #Thickness of the front section in inches\n",
-      "h_bd=6 #Thickness of the back section in inches\n",
-      "P=2000 #Point load acting at point A in lb\n",
-      "M_B=8000 #Moment at 4ft in lb.ft\n",
-      "M_D=16000 #Moment at x=8ft in lb.ft\n",
-      "b=2 #Breadth in inches\n",
-      "\n",
-      "#Calculations\n",
-      "S_ab=b*h_ab**2*6**-1 #Sectional Modulus of section AB in in^3\n",
-      "S_bd=b*h_bd**2*6**-1 #Sectional Modulus of section BD in in^3\n",
-      "sigma_B=12*M_B*S_ab**-1 #Maximum bending stress in psi\n",
-      "sigma_D=12*M_D*S_bd**-1 #Maximum bending stress in psi\n",
-      "\n",
-      "#Maximum stress\n",
-      "sigma_max=max(sigma_B,sigma_D) #Maximum stress in psi\n",
-      "\n",
-      "#Result\n",
-      "print \"Comparing the two results we find that the maximum stress is\"\n",
-      "print \"Sigma_max=\",round(sigma_max),\"psi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Comparing the two results we find that the maximum stress is\n",
-        "Sigma_max= 18000.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.4, Page No:146"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "M=15000 #Maximum bending moment in absolute values in lb.ft\n",
-      "S=42 #Sectional Modulus in in^3\n",
-      "\n",
-      "#Calculations\n",
-      "sigma_max=M*12*S**-1 #Maximum stress in the section in psi\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum Bending Stress in the section is\",round(sigma_max),\"psi\"\n",
-      "\n",
-      "#NOTE:The answer differs due to decimal point accuracy"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum Bending Stress in the section is 4286.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.5, Page No:157"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "M_max=60*10**3 #Maximum Bending Moment in kN.m\n",
-      "sigma_w=120*10**6 #Maximum Bending Stress allowed in Pa\n",
-      "M_max_2=61.52*10**3 #max bending moment computed in N.m\n",
-      "\n",
-      "#Section details\n",
-      "mass=38.7 #Mass in kg/m\n",
-      "g=9.81 #Acceleration due to gravity in m/s^2\n",
-      "S=549*10**3 #Sectional modulus of the section in mm^3\n",
-      "\n",
-      "#Calculations\n",
-      "S_min=M_max*sigma_w**-1*10**9 #Minimum Sectional Modulus required in mm^3\n",
-      "\n",
-      "#We selecet section W310x39\n",
-      "w0=mass*g*10**-3 #Weight of the beam in kN/m\n",
-      "sigma_max=M_max_2*S**-1*10**3 #Maximum stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The section chosen is W310x39 with maximum stress as\",round(sigma_max,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The section chosen is W310x39 with maximum stress as 112.1 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.6, Page No:166"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "V_max=24 #Maximum Shear in kN\n",
-      "b=0.160 #Width of the beam in m\n",
-      "h=0.240 #Depth of the beam in m\n",
-      "\n",
-      "#Calculations\n",
-      "I=b*h**3*12**-1 #Moment of Inertia of the beam in m^4\n",
-      "\n",
-      "#Part 1\n",
-      "Q=b*(h*3**-1)**2 #First moment of Area m^3\n",
-      "tau_max=(V_max*Q)*(I*b)**-1 #Maximum Shear Stress in glue in kPa\n",
-      "\n",
-      "#Part 2\n",
-      "tau_max_2=(3.0/2.0)*(V_max/(b*h)) #Shear Stress in kPa\n",
-      "Q_1=b*h*0.5*h*0.25 #First moment about NA in m^3\n",
-      "tau_maxx=(V_max*Q_1)/(I*b) #Shear stress in kPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The Results agree in both parts\"\n",
-      "print \"The maximum stress is\", round(tau_max_2),\"kPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Results agree in both parts\n",
-        "The maximum stress is 938.0 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.7, Page No:167"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "I=310 #Moment of inertia in in^4\n",
-      "V=160 #Shear Force in kips\n",
-      "#Dimension defination\n",
-      "tf=0.515 #Thickness of flange in inches\n",
-      "de=11.94 #Effective depth in inches\n",
-      "tw=0.295 #Thickness of web in inches\n",
-      "wf=8.005 #Width of lange in inches\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "Q=wf*tf*(de-tf)*0.5 #First moment about NA in inch^3\n",
-      "tau_min=(V*Q*10**2)/(I*tw) #Minimum shear stress in web in psi\n",
-      "\n",
-      "#Part 2\n",
-      "A_2=(de*0.5-tf)*tw #Area in in^3\n",
-      "y_bar_2=0.5*(de*0.5-tf) #Depth in inches\n",
-      "\n",
-      "Q_2=Q+A_2*y_bar_2 #First Moment in inches^3\n",
-      "\n",
-      "tau_max=(V*Q_2*10**2)/(I*tw) #Maximum Shear Stress in psi\n",
-      "\n",
-      "#Part 3\n",
-      "V_web=10.91*tw*(tau_min+((2*3**-1)*(tau_max-tau_min))) #Shear in the web in lb\n",
-      "perV=(V_web/V)*100 #Percentage shear force in web in %\n",
-      "t_max_final=V*10**3/(10.91*tw)\n",
-      "\n",
-      "#result\n",
-      "print \"The final shear stress in the web portion is\",round(t_max_final),\"psi\"\n",
-      "#NOTE:Answer differs due to deciaml point accuracy"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The final shear stress in the web portion is 49713.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.8, Page No:168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "I=547 #Moment of inertia in inches^4\n",
-      "d=8.9 #NA deoth in inches\n",
-      "V=12 #Shear Force in kips\n",
-      "h=7.3 #Depth of NA\n",
-      "b=2 #Width in inches\n",
-      "t=1.2 #Thickness in inches\n",
-      "h2=7.5 #Depth in inches\n",
-      "\n",
-      "#Calculations\n",
-      "#Shear Stress at NA\n",
-      "Q=(b*h)*(h*0.5) #First Moment about NA in in^3\n",
-      "tau=(V*10**3*Q)/(I*b) #Shear stress at NA in psi\n",
-      "\n",
-      "#Shear Stress at a-a\n",
-      "Q_1=(t*h2)*(d-h2*0.5) #First moment about NA in in^3\n",
-      "tau1=(V*Q_1)/(I*t) #Shear Stress in psi\n",
-      "\n",
-      "#Result\n",
-      "print \"Comparing two stresses\"\n",
-      "print \"The maximum stress is\",round(max(tau,tau1)),\"psi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Comparing two stresses\n",
-        "The maximum stress is 585.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.10, Page No:175"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "sigma_w=1000 #Working Stress in Bending in psi\n",
-      "tau_w=100 #Working stress in shear in psi\n",
-      "#Dimensions\n",
-      "b_out=8 #Width in inches\n",
-      "h=10 #Depth in inches\n",
-      "b_in=6 #Width in inches\n",
-      "\n",
-      "#Calculations\n",
-      "I=((b_out*h**3)-(b_in*b_out**3))*12**-1 #Moment of inertia in in^4\n",
-      "#Design for shear\n",
-      "Q=(b_out*h*0.5*0.25*h)-(b_in*b_out*0.5*0.25*b_out) #First Moment about NA in in^3\n",
-      "\n",
-      "#Largest P\n",
-      "P=(tau_w*I*2)/(1.5*Q) #P in shear in lb\n",
-      "\n",
-      "#Design for bending\n",
-      "P1=(sigma_w*I)/(60*5) #P in bending in lb\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum allowable P value is\",round(min(P,P1)),\"lb\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable P value is 1053.0 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.5.11, Page No:182"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A=2630 #Area in mm^2\n",
-      "y_bar=536.6 #Neutral Axis depth from top in mm\n",
-      "tau_w=100 #Allowable stress in MPa\n",
-      "sigma_b_w=280 #Allowable bending stress in MPa\n",
-      "d=0.019 #Diameter of the rivet in m\n",
-      "t_web=0.01 #Thickness of the web in m\n",
-      "I=4140 #Moment of inertia in m^4\n",
-      "V=450 #Max shear allowable in kN\n",
-      "\n",
-      "#Calculations\n",
-      "Q=A*y_bar #first moment in mm^3\n",
-      "Fw=(pi*d**2)*tau_w*10**6 #Allowable force in N\n",
-      "Fw_2=d*t_web*sigma_b_w*10**6*0.5 #Allowable force in N\n",
-      "e=Fw_2*I*(V*10**3*Q*10**-3)**-1 #Allowable spacing in m\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum spacing allowed is\",round(e*1000,1),\"mm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum spacing allowed is 173.4 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06.ipynb
deleted file mode 100755
index d8fd7d88..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06.ipynb
+++ /dev/null
@@ -1,450 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d43d7deae359298d302c004afe7a99ad4d721fd85b5d0a2361afaf05440c9100"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 06:Deflection of Beam"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.1, Page No:196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "wo=400 #loading in lb/ft\n",
-      "E=29*10**6 #Modulus of elasticity in psi\n",
-      "I=285 #Moment of inertia in in^4\n",
-      "S=45.6 #Sectional Modulus in in^3\n",
-      "L=8 #Span in ft\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "#Part1 is theoretical in nature hence not coded\n",
-      "\n",
-      "#Part 2\n",
-      "delta_max=((wo*12**-1)*(L*12)**4)/(8*E*I) #maximum deflection in inches\n",
-      "M_max=(wo*12**-1)*(L*12)**2 #Maximum moment\n",
-      "sigma_max=M_max/(2*S) #Maximum bending stress in psi\n",
-      "\n",
-      "#Result\n",
-      "print M_max\n",
-      "print \"The maximum deflection is\",round(delta_max,4),\"in\"\n",
-      "print \"The maximum Bending Stress is\",round(sigma_max),\"psi\"\n",
-      "\n",
-      "#Answer in the textbook for sigma_max is incorrect"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "307200.0\n",
-        "The maximum deflection is 0.0428 in\n",
-        "The maximum Bending Stress is 3368.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.3, Page No:198"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=300 #Point Load in N\n",
-      "R_a=100 #Reaction at A in N\n",
-      "R_c=200 #Reaction at C in N\n",
-      "E=12 #Youngs Modulus in GPa\n",
-      "L1=2 #Length of the load from A in m\n",
-      "L2=1 #Length of the load from C in m\n",
-      "b=0.04 #Width of the CS of the beam in m\n",
-      "h=0.08 #Depth of the CS of the beam in m\n",
-      "\n",
-      "#Claculations\n",
-      "#Moment of inertia \n",
-      "I=b*h**3*12**-1 #Moment of Inertia in m^4\n",
-      "#Flexural Rigidity\n",
-      "FR=E*10**9*I #FLexural rigidity in N.m^2\n",
-      "\n",
-      "#Moments in terms of x are\n",
-      "#Given\n",
-      "#After the variable Calculations we get\n",
-      "C1=-400/3 #Constant\n",
-      "C3=C1 #Constant\n",
-      "C2=0 #Constant\n",
-      "C4=0 #Constant\n",
-      "\n",
-      "#to get max displacement x we have\n",
-      "x=(6.510/2.441)**0.5 #Length at which displacement is maximum in m\n",
-      "v=(0.8138*x**3-6.510*x) #Displacement in mm\n",
-      "\n",
-      "#Largest slope\n",
-      "theta=(2.441*(L1+L2)**2-(7.324*(L1+L2-L1)**2)-6.150)*10**-3#Angle in radians\n",
-      "\n",
-      "#Result \n",
-      "print \"The maximum displacement is\",round(-v,2),\"mm downwards\"\n",
-      "print \"The maximum angle is\",round(theta*180*pi**-1,3),\"degrees anticlockwise\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum displacement is 7.09 mm downwards\n",
-        "The maximum angle is 0.487 degrees anticlockwise\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.4, Page No:200"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "#The computation is mostly variable based hence values will be directly declared \n",
-      "C1=19.20*10**3 #lb.ft^2\n",
-      "C2=-131.6*10**3 #lb.ft^2\n",
-      "C3=14.7*10**3 #lb.ft^2\n",
-      "C4=-112.7*10**3 #lb.ft^2\n",
-      "EI=10**7 #Flexural Rigidity in psi\n",
-      "\n",
-      "#Calculations\n",
-      "v=-(C2*12**3)/(EI*40) #Displacement in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum displacement is\",round(v,3),\"in downwards\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum displacement is 0.569 in downwards\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.6, Page No:210"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L1=3 #Length in m\n",
-      "L2=1 #Length in m\n",
-      "L3=8 #Length in m\n",
-      "L4=4 #Length in m\n",
-      "L5=6 #Length in m\n",
-      "\n",
-      "#Calculations\n",
-      "#Deflection midway\n",
-      "EIv=250*3**-1*L1**3-(50*3**-1*(L1-L2)**4)-(3925*3**-1*L1) #Deflection in N.m^3\n",
-      "\n",
-      "#Deflection at E\n",
-      "EIv_E=250*3**-1*L3**3-(50*3**-1*(L3-L2)**4)+(50*3**-1*(L3-L4)**4)+(650*3**-1*(L3-L5)**3)-(3925*3**-1*L3) #Deflection in N.m^3\n",
-      "\n",
-      "#Result\n",
-      "print \"The deflection at midspan is\",round(-EIv),\"N.m^3 downwards\"\n",
-      "print \"The deflection at point E is\",round(-EIv_E),\"N.m^3 downwards\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The deflection at midspan is 1942.0 N.m^3 downwards\n",
-        "The deflection at point E is 1817.0 N.m^3 downwards\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.8, Page No:223"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "x1=16*3**-1 #Centroid of the triangle in ft\n",
-      "x2=3 #Centroid of the lower parabola in ft\n",
-      "x3=6 #Centroid of the rectangle in ft\n",
-      "x4=20*3**-1 #Centroid of the triangle in ft\n",
-      "#Moment values\n",
-      "M1=4800 #Moment in lb.ft\n",
-      "M2=14400 #Moment in lb.ft\n",
-      "\n",
-      "#Calcualtions\n",
-      "P=((3**-1*4*M1*x2)+(4*M1*x3)+(0.5*4*M1*2*x4))*(x1*8*8*0.5)**-1 #Force P in lb\n",
-      "\n",
-      "#Result\n",
-      "print \"The magnitude of force P is\",round(P,1),\"lb\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The magnitude of force P is 1537.5 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.9, Page No:225"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=300 #Force in N\n",
-      "L1=1 #Length in m\n",
-      "L2=2 #Length in m\n",
-      "R_a=100 #Reaction at A in N\n",
-      "R_c=200 #Reaction at C in N\n",
-      "EI=20.48*10**3 #Flexural Rigidity in N.m^2\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "tC_A=(0.5*(L1+L2)*P*L1-(0.5*L1*P*(L1+L2)**-1))*EI**-1 #First Moment in m\n",
-      "theta_A=tC_A/(L1+L2) #Angle in radians \n",
-      "\n",
-      "#Part 2\n",
-      "tD_A=0.5*L1*R_a*(L1+L2)**-1*EI**-1 #First Moment in m\n",
-      "delta_D=(theta_A*L1-tD_A) #Displacement in m \n",
-      "\n",
-      "#Result\n",
-      "print \"The angle in part 1 is\",round(theta_A*180*pi**-1,3),\"Degrees\"\n",
-      "print \"The displacement in part 2 is\",round(delta_D*1000,2),\"mm downward\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The angle in part 1 is 0.373 Degrees\n",
-        "The displacement in part 2 is 5.7 mm downward\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.10, Page No:227"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P1=150 #Load in lb\n",
-      "P2=30 #Load in lb\n",
-      "R_A=78 #Reaction at A in lb\n",
-      "R_C=102 #Reaction at C in lb\n",
-      "L1=4 #Length in ft\n",
-      "L2=6 #Length in ft\n",
-      "M1=780 #Moment in lb.ft\n",
-      "M2=900 #Moment in lb.ft\n",
-      "M3=120 #Moment in lb.ft\n",
-      "\n",
-      "#Calculations\n",
-      "EI_AC=0.5*(L1+L2)*M1*(2*3**-1)*(L1+L2)-(0.5*L2*M2*(L1+(2*3**-1)*L2)) #Deflection in lb.ft^3\n",
-      "EI_thetaC=EI_AC/(L1+L2) #Deflection in lb.ft^2\n",
-      "\n",
-      "EI_DC=-0.5*L1*M3*2*3**-1*L1 #Deflection in lb.ft^3\n",
-      "EI_deltaD=EI_thetaC*L1-(-EI_DC) #Deflection in lb.ft^2\n",
-      "\n",
-      "#Result\n",
-      "print \"The deflection is\",round(EI_deltaD),\"lb.ft^2 upwards\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The deflection is 1120.0 lb.ft^2 upwards\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.11, Page No:234"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P1=80 #Load in lb\n",
-      "P2=100 #Load in lb\n",
-      "b1=3 #Distance of load from end in ft\n",
-      "b2=2 #Distance of load from end in ft\n",
-      "L=9 #Span of the beam in ft\n",
-      "\n",
-      "#Calcualtions\n",
-      "EI_delta1=(P1*b1*48**-1)*(3*L**2-4*b1**2) #Deflection in lb.ft^3\n",
-      "EI_delta2=(P2*b2*48**-1)*(3*L**2-4*b2**2) #Deflection in lb.ft^3\n",
-      "EI_delta=EI_delta1+EI_delta2 #Deflection at modspan in lb.ft^3\n",
-      "\n",
-      "#Result\n",
-      "print \"The deflection at midspan is\",round(EI_delta),\" lb.ft^3 downward\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The deflection at midspan is 1981.0  lb.ft^3 downward\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.12, Page no:234"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "wo=600 #Load in N/m\n",
-      "L=6 #Span of the beam in m\n",
-      "b=2 #Distance of the load from end in m\n",
-      "a=1 #Distance of the load from end in m\n",
-      "\n",
-      "#Calulations\n",
-      "EI_delta1=wo*384**-1*(5*L**4-12*L**2*b**2+8*b**4) #Deflection in N.m^3\n",
-      "EI_delta2=wo*96**-1*a**2*(3*L**2-2*a**2) #Deflection in N.m^3\n",
-      "\n",
-      "EI_delta=EI_delta1-EI_delta2 #Total Delfection at midspan in N.m^3\n",
-      "\n",
-      "#Result\n",
-      "print \"The total Deflection at midpsan is\",round(EI_delta),\"N.m^3 downwards\"\n",
-      "\n",
-      "#NOTE:The answer varies due to decimal point accuracy"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total Deflection at midpsan is 6963.0 N.m^3 downwards\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_1.ipynb
deleted file mode 100755
index d8fd7d88..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_1.ipynb
+++ /dev/null
@@ -1,450 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d43d7deae359298d302c004afe7a99ad4d721fd85b5d0a2361afaf05440c9100"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 06:Deflection of Beam"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.1, Page No:196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "wo=400 #loading in lb/ft\n",
-      "E=29*10**6 #Modulus of elasticity in psi\n",
-      "I=285 #Moment of inertia in in^4\n",
-      "S=45.6 #Sectional Modulus in in^3\n",
-      "L=8 #Span in ft\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "#Part1 is theoretical in nature hence not coded\n",
-      "\n",
-      "#Part 2\n",
-      "delta_max=((wo*12**-1)*(L*12)**4)/(8*E*I) #maximum deflection in inches\n",
-      "M_max=(wo*12**-1)*(L*12)**2 #Maximum moment\n",
-      "sigma_max=M_max/(2*S) #Maximum bending stress in psi\n",
-      "\n",
-      "#Result\n",
-      "print M_max\n",
-      "print \"The maximum deflection is\",round(delta_max,4),\"in\"\n",
-      "print \"The maximum Bending Stress is\",round(sigma_max),\"psi\"\n",
-      "\n",
-      "#Answer in the textbook for sigma_max is incorrect"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "307200.0\n",
-        "The maximum deflection is 0.0428 in\n",
-        "The maximum Bending Stress is 3368.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.3, Page No:198"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=300 #Point Load in N\n",
-      "R_a=100 #Reaction at A in N\n",
-      "R_c=200 #Reaction at C in N\n",
-      "E=12 #Youngs Modulus in GPa\n",
-      "L1=2 #Length of the load from A in m\n",
-      "L2=1 #Length of the load from C in m\n",
-      "b=0.04 #Width of the CS of the beam in m\n",
-      "h=0.08 #Depth of the CS of the beam in m\n",
-      "\n",
-      "#Claculations\n",
-      "#Moment of inertia \n",
-      "I=b*h**3*12**-1 #Moment of Inertia in m^4\n",
-      "#Flexural Rigidity\n",
-      "FR=E*10**9*I #FLexural rigidity in N.m^2\n",
-      "\n",
-      "#Moments in terms of x are\n",
-      "#Given\n",
-      "#After the variable Calculations we get\n",
-      "C1=-400/3 #Constant\n",
-      "C3=C1 #Constant\n",
-      "C2=0 #Constant\n",
-      "C4=0 #Constant\n",
-      "\n",
-      "#to get max displacement x we have\n",
-      "x=(6.510/2.441)**0.5 #Length at which displacement is maximum in m\n",
-      "v=(0.8138*x**3-6.510*x) #Displacement in mm\n",
-      "\n",
-      "#Largest slope\n",
-      "theta=(2.441*(L1+L2)**2-(7.324*(L1+L2-L1)**2)-6.150)*10**-3#Angle in radians\n",
-      "\n",
-      "#Result \n",
-      "print \"The maximum displacement is\",round(-v,2),\"mm downwards\"\n",
-      "print \"The maximum angle is\",round(theta*180*pi**-1,3),\"degrees anticlockwise\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum displacement is 7.09 mm downwards\n",
-        "The maximum angle is 0.487 degrees anticlockwise\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.4, Page No:200"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "#The computation is mostly variable based hence values will be directly declared \n",
-      "C1=19.20*10**3 #lb.ft^2\n",
-      "C2=-131.6*10**3 #lb.ft^2\n",
-      "C3=14.7*10**3 #lb.ft^2\n",
-      "C4=-112.7*10**3 #lb.ft^2\n",
-      "EI=10**7 #Flexural Rigidity in psi\n",
-      "\n",
-      "#Calculations\n",
-      "v=-(C2*12**3)/(EI*40) #Displacement in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum displacement is\",round(v,3),\"in downwards\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum displacement is 0.569 in downwards\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.6, Page No:210"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L1=3 #Length in m\n",
-      "L2=1 #Length in m\n",
-      "L3=8 #Length in m\n",
-      "L4=4 #Length in m\n",
-      "L5=6 #Length in m\n",
-      "\n",
-      "#Calculations\n",
-      "#Deflection midway\n",
-      "EIv=250*3**-1*L1**3-(50*3**-1*(L1-L2)**4)-(3925*3**-1*L1) #Deflection in N.m^3\n",
-      "\n",
-      "#Deflection at E\n",
-      "EIv_E=250*3**-1*L3**3-(50*3**-1*(L3-L2)**4)+(50*3**-1*(L3-L4)**4)+(650*3**-1*(L3-L5)**3)-(3925*3**-1*L3) #Deflection in N.m^3\n",
-      "\n",
-      "#Result\n",
-      "print \"The deflection at midspan is\",round(-EIv),\"N.m^3 downwards\"\n",
-      "print \"The deflection at point E is\",round(-EIv_E),\"N.m^3 downwards\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The deflection at midspan is 1942.0 N.m^3 downwards\n",
-        "The deflection at point E is 1817.0 N.m^3 downwards\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.8, Page No:223"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "x1=16*3**-1 #Centroid of the triangle in ft\n",
-      "x2=3 #Centroid of the lower parabola in ft\n",
-      "x3=6 #Centroid of the rectangle in ft\n",
-      "x4=20*3**-1 #Centroid of the triangle in ft\n",
-      "#Moment values\n",
-      "M1=4800 #Moment in lb.ft\n",
-      "M2=14400 #Moment in lb.ft\n",
-      "\n",
-      "#Calcualtions\n",
-      "P=((3**-1*4*M1*x2)+(4*M1*x3)+(0.5*4*M1*2*x4))*(x1*8*8*0.5)**-1 #Force P in lb\n",
-      "\n",
-      "#Result\n",
-      "print \"The magnitude of force P is\",round(P,1),\"lb\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The magnitude of force P is 1537.5 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.9, Page No:225"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=300 #Force in N\n",
-      "L1=1 #Length in m\n",
-      "L2=2 #Length in m\n",
-      "R_a=100 #Reaction at A in N\n",
-      "R_c=200 #Reaction at C in N\n",
-      "EI=20.48*10**3 #Flexural Rigidity in N.m^2\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "tC_A=(0.5*(L1+L2)*P*L1-(0.5*L1*P*(L1+L2)**-1))*EI**-1 #First Moment in m\n",
-      "theta_A=tC_A/(L1+L2) #Angle in radians \n",
-      "\n",
-      "#Part 2\n",
-      "tD_A=0.5*L1*R_a*(L1+L2)**-1*EI**-1 #First Moment in m\n",
-      "delta_D=(theta_A*L1-tD_A) #Displacement in m \n",
-      "\n",
-      "#Result\n",
-      "print \"The angle in part 1 is\",round(theta_A*180*pi**-1,3),\"Degrees\"\n",
-      "print \"The displacement in part 2 is\",round(delta_D*1000,2),\"mm downward\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The angle in part 1 is 0.373 Degrees\n",
-        "The displacement in part 2 is 5.7 mm downward\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.10, Page No:227"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P1=150 #Load in lb\n",
-      "P2=30 #Load in lb\n",
-      "R_A=78 #Reaction at A in lb\n",
-      "R_C=102 #Reaction at C in lb\n",
-      "L1=4 #Length in ft\n",
-      "L2=6 #Length in ft\n",
-      "M1=780 #Moment in lb.ft\n",
-      "M2=900 #Moment in lb.ft\n",
-      "M3=120 #Moment in lb.ft\n",
-      "\n",
-      "#Calculations\n",
-      "EI_AC=0.5*(L1+L2)*M1*(2*3**-1)*(L1+L2)-(0.5*L2*M2*(L1+(2*3**-1)*L2)) #Deflection in lb.ft^3\n",
-      "EI_thetaC=EI_AC/(L1+L2) #Deflection in lb.ft^2\n",
-      "\n",
-      "EI_DC=-0.5*L1*M3*2*3**-1*L1 #Deflection in lb.ft^3\n",
-      "EI_deltaD=EI_thetaC*L1-(-EI_DC) #Deflection in lb.ft^2\n",
-      "\n",
-      "#Result\n",
-      "print \"The deflection is\",round(EI_deltaD),\"lb.ft^2 upwards\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The deflection is 1120.0 lb.ft^2 upwards\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.11, Page No:234"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P1=80 #Load in lb\n",
-      "P2=100 #Load in lb\n",
-      "b1=3 #Distance of load from end in ft\n",
-      "b2=2 #Distance of load from end in ft\n",
-      "L=9 #Span of the beam in ft\n",
-      "\n",
-      "#Calcualtions\n",
-      "EI_delta1=(P1*b1*48**-1)*(3*L**2-4*b1**2) #Deflection in lb.ft^3\n",
-      "EI_delta2=(P2*b2*48**-1)*(3*L**2-4*b2**2) #Deflection in lb.ft^3\n",
-      "EI_delta=EI_delta1+EI_delta2 #Deflection at modspan in lb.ft^3\n",
-      "\n",
-      "#Result\n",
-      "print \"The deflection at midspan is\",round(EI_delta),\" lb.ft^3 downward\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The deflection at midspan is 1981.0  lb.ft^3 downward\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.12, Page no:234"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "wo=600 #Load in N/m\n",
-      "L=6 #Span of the beam in m\n",
-      "b=2 #Distance of the load from end in m\n",
-      "a=1 #Distance of the load from end in m\n",
-      "\n",
-      "#Calulations\n",
-      "EI_delta1=wo*384**-1*(5*L**4-12*L**2*b**2+8*b**4) #Deflection in N.m^3\n",
-      "EI_delta2=wo*96**-1*a**2*(3*L**2-2*a**2) #Deflection in N.m^3\n",
-      "\n",
-      "EI_delta=EI_delta1-EI_delta2 #Total Delfection at midspan in N.m^3\n",
-      "\n",
-      "#Result\n",
-      "print \"The total Deflection at midpsan is\",round(EI_delta),\"N.m^3 downwards\"\n",
-      "\n",
-      "#NOTE:The answer varies due to decimal point accuracy"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total Deflection at midpsan is 6963.0 N.m^3 downwards\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_2.ipynb
deleted file mode 100755
index d8fd7d88..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_2.ipynb
+++ /dev/null
@@ -1,450 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d43d7deae359298d302c004afe7a99ad4d721fd85b5d0a2361afaf05440c9100"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 06:Deflection of Beam"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.1, Page No:196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "wo=400 #loading in lb/ft\n",
-      "E=29*10**6 #Modulus of elasticity in psi\n",
-      "I=285 #Moment of inertia in in^4\n",
-      "S=45.6 #Sectional Modulus in in^3\n",
-      "L=8 #Span in ft\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "#Part1 is theoretical in nature hence not coded\n",
-      "\n",
-      "#Part 2\n",
-      "delta_max=((wo*12**-1)*(L*12)**4)/(8*E*I) #maximum deflection in inches\n",
-      "M_max=(wo*12**-1)*(L*12)**2 #Maximum moment\n",
-      "sigma_max=M_max/(2*S) #Maximum bending stress in psi\n",
-      "\n",
-      "#Result\n",
-      "print M_max\n",
-      "print \"The maximum deflection is\",round(delta_max,4),\"in\"\n",
-      "print \"The maximum Bending Stress is\",round(sigma_max),\"psi\"\n",
-      "\n",
-      "#Answer in the textbook for sigma_max is incorrect"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "307200.0\n",
-        "The maximum deflection is 0.0428 in\n",
-        "The maximum Bending Stress is 3368.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.3, Page No:198"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=300 #Point Load in N\n",
-      "R_a=100 #Reaction at A in N\n",
-      "R_c=200 #Reaction at C in N\n",
-      "E=12 #Youngs Modulus in GPa\n",
-      "L1=2 #Length of the load from A in m\n",
-      "L2=1 #Length of the load from C in m\n",
-      "b=0.04 #Width of the CS of the beam in m\n",
-      "h=0.08 #Depth of the CS of the beam in m\n",
-      "\n",
-      "#Claculations\n",
-      "#Moment of inertia \n",
-      "I=b*h**3*12**-1 #Moment of Inertia in m^4\n",
-      "#Flexural Rigidity\n",
-      "FR=E*10**9*I #FLexural rigidity in N.m^2\n",
-      "\n",
-      "#Moments in terms of x are\n",
-      "#Given\n",
-      "#After the variable Calculations we get\n",
-      "C1=-400/3 #Constant\n",
-      "C3=C1 #Constant\n",
-      "C2=0 #Constant\n",
-      "C4=0 #Constant\n",
-      "\n",
-      "#to get max displacement x we have\n",
-      "x=(6.510/2.441)**0.5 #Length at which displacement is maximum in m\n",
-      "v=(0.8138*x**3-6.510*x) #Displacement in mm\n",
-      "\n",
-      "#Largest slope\n",
-      "theta=(2.441*(L1+L2)**2-(7.324*(L1+L2-L1)**2)-6.150)*10**-3#Angle in radians\n",
-      "\n",
-      "#Result \n",
-      "print \"The maximum displacement is\",round(-v,2),\"mm downwards\"\n",
-      "print \"The maximum angle is\",round(theta*180*pi**-1,3),\"degrees anticlockwise\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum displacement is 7.09 mm downwards\n",
-        "The maximum angle is 0.487 degrees anticlockwise\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.4, Page No:200"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "#The computation is mostly variable based hence values will be directly declared \n",
-      "C1=19.20*10**3 #lb.ft^2\n",
-      "C2=-131.6*10**3 #lb.ft^2\n",
-      "C3=14.7*10**3 #lb.ft^2\n",
-      "C4=-112.7*10**3 #lb.ft^2\n",
-      "EI=10**7 #Flexural Rigidity in psi\n",
-      "\n",
-      "#Calculations\n",
-      "v=-(C2*12**3)/(EI*40) #Displacement in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum displacement is\",round(v,3),\"in downwards\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum displacement is 0.569 in downwards\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.6, Page No:210"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L1=3 #Length in m\n",
-      "L2=1 #Length in m\n",
-      "L3=8 #Length in m\n",
-      "L4=4 #Length in m\n",
-      "L5=6 #Length in m\n",
-      "\n",
-      "#Calculations\n",
-      "#Deflection midway\n",
-      "EIv=250*3**-1*L1**3-(50*3**-1*(L1-L2)**4)-(3925*3**-1*L1) #Deflection in N.m^3\n",
-      "\n",
-      "#Deflection at E\n",
-      "EIv_E=250*3**-1*L3**3-(50*3**-1*(L3-L2)**4)+(50*3**-1*(L3-L4)**4)+(650*3**-1*(L3-L5)**3)-(3925*3**-1*L3) #Deflection in N.m^3\n",
-      "\n",
-      "#Result\n",
-      "print \"The deflection at midspan is\",round(-EIv),\"N.m^3 downwards\"\n",
-      "print \"The deflection at point E is\",round(-EIv_E),\"N.m^3 downwards\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The deflection at midspan is 1942.0 N.m^3 downwards\n",
-        "The deflection at point E is 1817.0 N.m^3 downwards\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.8, Page No:223"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "x1=16*3**-1 #Centroid of the triangle in ft\n",
-      "x2=3 #Centroid of the lower parabola in ft\n",
-      "x3=6 #Centroid of the rectangle in ft\n",
-      "x4=20*3**-1 #Centroid of the triangle in ft\n",
-      "#Moment values\n",
-      "M1=4800 #Moment in lb.ft\n",
-      "M2=14400 #Moment in lb.ft\n",
-      "\n",
-      "#Calcualtions\n",
-      "P=((3**-1*4*M1*x2)+(4*M1*x3)+(0.5*4*M1*2*x4))*(x1*8*8*0.5)**-1 #Force P in lb\n",
-      "\n",
-      "#Result\n",
-      "print \"The magnitude of force P is\",round(P,1),\"lb\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The magnitude of force P is 1537.5 lb\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.9, Page No:225"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=300 #Force in N\n",
-      "L1=1 #Length in m\n",
-      "L2=2 #Length in m\n",
-      "R_a=100 #Reaction at A in N\n",
-      "R_c=200 #Reaction at C in N\n",
-      "EI=20.48*10**3 #Flexural Rigidity in N.m^2\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "tC_A=(0.5*(L1+L2)*P*L1-(0.5*L1*P*(L1+L2)**-1))*EI**-1 #First Moment in m\n",
-      "theta_A=tC_A/(L1+L2) #Angle in radians \n",
-      "\n",
-      "#Part 2\n",
-      "tD_A=0.5*L1*R_a*(L1+L2)**-1*EI**-1 #First Moment in m\n",
-      "delta_D=(theta_A*L1-tD_A) #Displacement in m \n",
-      "\n",
-      "#Result\n",
-      "print \"The angle in part 1 is\",round(theta_A*180*pi**-1,3),\"Degrees\"\n",
-      "print \"The displacement in part 2 is\",round(delta_D*1000,2),\"mm downward\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The angle in part 1 is 0.373 Degrees\n",
-        "The displacement in part 2 is 5.7 mm downward\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.10, Page No:227"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P1=150 #Load in lb\n",
-      "P2=30 #Load in lb\n",
-      "R_A=78 #Reaction at A in lb\n",
-      "R_C=102 #Reaction at C in lb\n",
-      "L1=4 #Length in ft\n",
-      "L2=6 #Length in ft\n",
-      "M1=780 #Moment in lb.ft\n",
-      "M2=900 #Moment in lb.ft\n",
-      "M3=120 #Moment in lb.ft\n",
-      "\n",
-      "#Calculations\n",
-      "EI_AC=0.5*(L1+L2)*M1*(2*3**-1)*(L1+L2)-(0.5*L2*M2*(L1+(2*3**-1)*L2)) #Deflection in lb.ft^3\n",
-      "EI_thetaC=EI_AC/(L1+L2) #Deflection in lb.ft^2\n",
-      "\n",
-      "EI_DC=-0.5*L1*M3*2*3**-1*L1 #Deflection in lb.ft^3\n",
-      "EI_deltaD=EI_thetaC*L1-(-EI_DC) #Deflection in lb.ft^2\n",
-      "\n",
-      "#Result\n",
-      "print \"The deflection is\",round(EI_deltaD),\"lb.ft^2 upwards\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The deflection is 1120.0 lb.ft^2 upwards\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.11, Page No:234"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P1=80 #Load in lb\n",
-      "P2=100 #Load in lb\n",
-      "b1=3 #Distance of load from end in ft\n",
-      "b2=2 #Distance of load from end in ft\n",
-      "L=9 #Span of the beam in ft\n",
-      "\n",
-      "#Calcualtions\n",
-      "EI_delta1=(P1*b1*48**-1)*(3*L**2-4*b1**2) #Deflection in lb.ft^3\n",
-      "EI_delta2=(P2*b2*48**-1)*(3*L**2-4*b2**2) #Deflection in lb.ft^3\n",
-      "EI_delta=EI_delta1+EI_delta2 #Deflection at modspan in lb.ft^3\n",
-      "\n",
-      "#Result\n",
-      "print \"The deflection at midspan is\",round(EI_delta),\" lb.ft^3 downward\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The deflection at midspan is 1981.0  lb.ft^3 downward\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6.12, Page no:234"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "wo=600 #Load in N/m\n",
-      "L=6 #Span of the beam in m\n",
-      "b=2 #Distance of the load from end in m\n",
-      "a=1 #Distance of the load from end in m\n",
-      "\n",
-      "#Calulations\n",
-      "EI_delta1=wo*384**-1*(5*L**4-12*L**2*b**2+8*b**4) #Deflection in N.m^3\n",
-      "EI_delta2=wo*96**-1*a**2*(3*L**2-2*a**2) #Deflection in N.m^3\n",
-      "\n",
-      "EI_delta=EI_delta1-EI_delta2 #Total Delfection at midspan in N.m^3\n",
-      "\n",
-      "#Result\n",
-      "print \"The total Deflection at midpsan is\",round(EI_delta),\"N.m^3 downwards\"\n",
-      "\n",
-      "#NOTE:The answer varies due to decimal point accuracy"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total Deflection at midpsan is 6963.0 N.m^3 downwards\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07.ipynb
deleted file mode 100755
index 9d4c3a36..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07.ipynb
+++ /dev/null
@@ -1,123 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:07968d92ed399b52b001ac620304d00a18188f4412f8dbbd944ed3ed617c5439"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 07:Statically Indeterminate Beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.7.3, Page No:251"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=5000 #Load in N\n",
-      "L=2 #Half span in m\n",
-      "\n",
-      "#Calculations\n",
-      "#After carrying put the variable computation \n",
-      "#We obtain three equations which can be solved simultaneously\n",
-      "A=np.array([[1,1,0],[-L*2,0,1],[32,0,-24]]) #Array of the unknowns\n",
-      "B=np.array([P,-P*L,2000]) #Array of RHS\n",
-      "C=np.linalg.solve(A,B)\n",
-      "\n",
-      "#Result\n",
-      "print \"The values are as follows\"\n",
-      "print \"Ra=\",C[0],\"N  Rb=\",C[1],\"N and Ma=\",C[2],\"N.m\"\n",
-      "\n",
-      "#Answer in the textbook is incorrect"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The values are as follows\n",
-        "Ra= 3718.75 N  Rb= 1281.25 N and Ma= 4875.0 N.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.7.4, Page No:252"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "w=60 #Continous Load in lb/ft\n",
-      "L1=3 #Length in ft\n",
-      "L2=9 #Length in ft\n",
-      "\n",
-      "#Calculations\n",
-      "#After carrying out the variable computations we get\n",
-      "A=np.array([[1,1,0,0],[(L1+L2),0,1,1],[0.5*(L1+L2)**2,0,-(L1+L2),0],[6**-1*(L1+L2)**3,0,-0.5*(L1+L2)**2,0]])\n",
-      "B=np.array([w*L2,w*L2*0.5*L2,L2**3*10,L2**4*2.5])\n",
-      "C=np.linalg.solve(A,B)\n",
-      "\n",
-      "#Result\n",
-      "print \"The values are as follows\"\n",
-      "print \"Ra=\",round(C[0]),\"lb Ma=\",round(C[2]),\"lb.ft Rb=\",round(C[1]),\"lb and Mb=\",round(C[3]),\"lb.ft\"\n",
-      "\n",
-      "#NOTE:The answer for Mb in the textbook is incorrect"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The values are as follows\n",
-        "Ra= 190.0 lb Ma= 532.0 lb.ft Rb= 350.0 lb and Mb= -380.0 lb.ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07_1.ipynb
deleted file mode 100755
index 9d4c3a36..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07_1.ipynb
+++ /dev/null
@@ -1,123 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:07968d92ed399b52b001ac620304d00a18188f4412f8dbbd944ed3ed617c5439"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 07:Statically Indeterminate Beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.7.3, Page No:251"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=5000 #Load in N\n",
-      "L=2 #Half span in m\n",
-      "\n",
-      "#Calculations\n",
-      "#After carrying put the variable computation \n",
-      "#We obtain three equations which can be solved simultaneously\n",
-      "A=np.array([[1,1,0],[-L*2,0,1],[32,0,-24]]) #Array of the unknowns\n",
-      "B=np.array([P,-P*L,2000]) #Array of RHS\n",
-      "C=np.linalg.solve(A,B)\n",
-      "\n",
-      "#Result\n",
-      "print \"The values are as follows\"\n",
-      "print \"Ra=\",C[0],\"N  Rb=\",C[1],\"N and Ma=\",C[2],\"N.m\"\n",
-      "\n",
-      "#Answer in the textbook is incorrect"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The values are as follows\n",
-        "Ra= 3718.75 N  Rb= 1281.25 N and Ma= 4875.0 N.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.7.4, Page No:252"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "w=60 #Continous Load in lb/ft\n",
-      "L1=3 #Length in ft\n",
-      "L2=9 #Length in ft\n",
-      "\n",
-      "#Calculations\n",
-      "#After carrying out the variable computations we get\n",
-      "A=np.array([[1,1,0,0],[(L1+L2),0,1,1],[0.5*(L1+L2)**2,0,-(L1+L2),0],[6**-1*(L1+L2)**3,0,-0.5*(L1+L2)**2,0]])\n",
-      "B=np.array([w*L2,w*L2*0.5*L2,L2**3*10,L2**4*2.5])\n",
-      "C=np.linalg.solve(A,B)\n",
-      "\n",
-      "#Result\n",
-      "print \"The values are as follows\"\n",
-      "print \"Ra=\",round(C[0]),\"lb Ma=\",round(C[2]),\"lb.ft Rb=\",round(C[1]),\"lb and Mb=\",round(C[3]),\"lb.ft\"\n",
-      "\n",
-      "#NOTE:The answer for Mb in the textbook is incorrect"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The values are as follows\n",
-        "Ra= 190.0 lb Ma= 532.0 lb.ft Rb= 350.0 lb and Mb= -380.0 lb.ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07_2.ipynb
deleted file mode 100755
index 1ab7377d..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07_2.ipynb
+++ /dev/null
@@ -1,115 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a060d639455e309e0ebbb56cdce11ccc2075e0745b360f129a4aa2a2ff9becd3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 07:Statically Indeterminate Beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.7.3, Page No:251"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=5000 #Load in N\n",
-      "L=2 #Half span in m\n",
-      "\n",
-      "#Calculations\n",
-      "#After carrying put the variable computation \n",
-      "#We obtain three equations which can be solved simultaneously\n",
-      "A=np.array([[1,1,0],[-L*2,0,1],[32,0,-24]]) #Array of the unknowns\n",
-      "B=np.array([P,-P*L,2000]) #Array of RHS\n",
-      "C=np.linalg.solve(A,B)\n",
-      "\n",
-      "#Result\n",
-      "print \"The values are as follows\"\n",
-      "print \"Ra=\",C[0],\"N  Rb=\",C[1],\"N and Ma=\",C[2],\"N.m\"\n",
-      "\n",
-      "#Answer in the textbook is incorrect"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The values are as follows\n",
-        "Ra= 3718.75 N  Rb= 1281.25 N and Ma= 4875.0 N.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.7.4, Page No:252"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "w=60 #Continous Load in lb/ft\n",
-      "L1=3 #Length in ft\n",
-      "L2=9 #Length in ft\n",
-      "\n",
-      "#Calculations\n",
-      "#After carrying out the variable computations we get\n",
-      "A=np.array([[1,1,0,0],[(L1+L2),0,1,1],[0.5*(L1+L2)**2,0,-(L1+L2),0],[6**-1*(L1+L2)**3,0,-0.5*(L1+L2)**2,0]])\n",
-      "B=np.array([w*L2,w*L2*0.5*L2,L2**3*10,L2**4*2.5])\n",
-      "C=np.linalg.solve(A,B)\n",
-      "\n",
-      "#Result\n",
-      "print \"The values are as follows\"\n",
-      "print \"Ra=\",round(C[0]),\"lb Ma=\",round(C[2]),\"lb.ft Rb=\",round(C[1]),\"lb and Mb=\",round(C[3]),\"lb.ft\"\n",
-      "\n",
-      "#NOTE:The answer for Mb in the textbook is incorrect"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The values are as follows\n",
-        "Ra= 190.0 lb Ma= 532.0 lb.ft Rb= 350.0 lb and Mb= -380.0 lb.ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter08.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter08.ipynb
deleted file mode 100755
index 87dd0fdf..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter08.ipynb
+++ /dev/null
@@ -1,721 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:b494877451d53f8b0ca30d008c3144520923dfdb33c6562fbcdeab1b4ca2b7ce"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter08:Stresses due to Combined Loading"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Examlple 8.8.1, Page No:275"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "p=125 #Pressure in psi\n",
-      "r=24 #Radius of the vessel in inches\n",
-      "t=0.25 #Thickness of the vessel in inches\n",
-      "E=29*10**6 #Modulus of Elasticity in psi\n",
-      "v=0.28 #poisson ratio\n",
-      "\n",
-      "#Calcualtions\n",
-      "#Part 1\n",
-      "sigma_c=p*r*t**-1 #Circumferential Stress in psi\n",
-      "sigma_l=sigma_c/2 #Longitudinat Stress in psi\n",
-      "e_c=E**-1*(sigma_c-(v*sigma_l)) #Circumferential Strain using biaxial Hooke's Law \n",
-      "delta_r=e_c*r #Change in the radius in inches\n",
-      "\n",
-      "#Part 2\n",
-      "sigma=(p*r)*(2*t)**-1 #Stress in psi\n",
-      "e=E**-1*(sigma-(v*sigma)) #Strain using biaxial Hooke's Law\n",
-      "delta_R=e*r #Change inradius of end-cap in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"Part 1 Answers\"\n",
-      "print \"Stresses are sigma_c=\",round(sigma_c),\"psi and sigma_l=\",round(sigma_l),\"psi\"\n",
-      "print \"Change of radius of cylinder=\",round(delta_r,5),\"in\"\n",
-      "print \"Part 2 Answers\"\n",
-      "print \"Stresses are sigma=\",round(sigma),\"psi\"\n",
-      "print \"Change in radius of end cap=\",round(delta_R,5),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part 1 Answers\n",
-        "Stresses are sigma_c= 12000.0 psi and sigma_l= 6000.0 psi\n",
-        "Change of radius of cylinder= 0.00854 in\n",
-        "Part 2 Answers\n",
-        "Stresses are sigma= 6000.0 psi\n",
-        "Change in radius of end cap= 0.00358 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.2, Page No:280"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import matplotlib.pyplot as plt\n",
-      "%matplotlib inline\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=40 #Force in kN\n",
-      "b=0.050 #Width in m\n",
-      "h=0.040 #Depth in m\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "A=b*h #Area in m^2\n",
-      "I=(b*h**3)*12**-1 #Moment of inertia in m^4\n",
-      "c=h*0.5 #m\n",
-      "sigma_max=(P*A**-1)+(P*c**2*I**-1) #Maximum stress in MPa\n",
-      "sigma_min=(P*A**-1)-(P*c**2*I**-1) #Minimum stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The Maximum and Minimum Stress are\"\n",
-      "print \"Max=\",sigma_max/1000,\"MPa and Min=\",sigma_min/1000,\"MPa\"\n",
-      "\n",
-      "#Plotting\n",
-      "x=[20,0,-20]\n",
-      "S=[-sigma_min/1000,0,sigma_max/1000]\n",
-      "plt.plot(S,x)\n",
-      "plt.ylabel(\"Distance from Neutral Axis in mm\")\n",
-      "plt.xlabel(\"Stress in MPa\")\n",
-      "plt.title(\"Stress Distribution Diagram\")\n",
-      "plt.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Maximum and Minimum Stress are\n",
-        "Max= 80.0 MPa and Min= -40.0 MPa\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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c4OvAuba/1PFJpecDC4FvAB+0fVW5fhYw1/bmE+yfSiKiz6TC6L1u90lcDbyy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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10b6d0310>"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.3, Page No:281"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variabel Decleration\n",
-      "b=6 #Width in inches\n",
-      "h=10 #Depth in inches\n",
-      "P1=6000 #Force in lb\n",
-      "P2=3000 #Force in lb\n",
-      "L=4 #Length in ft\n",
-      "P=-13400 #Load in lb\n",
-      "M=6000 #Moment in lb.ft\n",
-      "y=5 #Depth in inches\n",
-      "P2=-9800 #Load in lb\n",
-      "M2=-12000 #Moment in lb.ft\n",
-      "\n",
-      "#Calculations\n",
-      "A=b*h #Area in in^2\n",
-      "I=b*h**3*12**-1 #Moment of inertia in in^4\n",
-      "T=(P1*L+P2*L*3)*(6)**-1 #Tension in the cable in lb\n",
-      "\n",
-      "#Computation of largest stress\n",
-      "sigma_B=(P*A**-1)-(M*y*12*I**-1) #Maximum Compressive Stress caused by +ve BM in psi\n",
-      "sigma_C=(P2*A**-1)-(M2*-y*12*I**-1) #Maximum Compressive Stress caused by -ve BM in psi\n",
-      "\n",
-      "sigma_max=max(-sigma_B,-sigma_C) #Maximum Compressive Stress in psi\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum Stress is\",round(sigma_max),\"psi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum Stress is 1603.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.4, Page No:297"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "theta=(60*pi)/180 #Angle in radians (Twice as declared)\n",
-      "sigma_x=30 # Stress in x in MPa\n",
-      "sigma_y=60 #Stress in y in MPa\n",
-      "tau_xy=40 #Stress in MPa\n",
-      "\n",
-      "#Calcualtions\n",
-      "sigma_xdash=0.5*(sigma_x+sigma_y)+0.5*(sigma_x-sigma_y)*cos(theta)+tau_xy*sin(theta) #Stress at x' axis in MPa\n",
-      "sigma_ydash=0.5*(sigma_x+sigma_y)-0.5*(sigma_x-sigma_y)*cos(theta)-tau_xy*sin(theta) #Stress at y' axis in MPa\n",
-      "tau_x_y=-0.5*(sigma_x-sigma_y)*sin(theta)+tau_xy*cos(theta) #Stress at x'y' in shear in MPa\n",
-      "#Result\n",
-      "print \"The new stresses at new axes are as follows\"\n",
-      "print \"sigma_x'=\",round(sigma_xdash,1),\"MPa sigma_y'=\",round(sigma_ydash,1),\"MPa\"\n",
-      "print \"And tau_x'y'=\",round(tau_x_y),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The new stresses at new axes are as follows\n",
-        "sigma_x'= 72.1 MPa sigma_y'= 17.9 MPa\n",
-        "And tau_x'y'= 33.0 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.5, Page No:297"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "sigma_x=8000 #Stress in x in psi\n",
-      "sigma_y=4000 #Stress in y in psi\n",
-      "tau_xy=3000 #Stress in xy in psi\n",
-      "\n",
-      "#Calculations\n",
-      "R=sqrt(((sigma_x-sigma_y)*0.5)**2+tau_xy**2) #Resultant Stress in psi\n",
-      "\n",
-      "#Principal Stresses\n",
-      "sigma1=(sigma_x+sigma_y)*0.5+R #Principal Stress in psi\n",
-      "sigma2=(sigma_x+sigma_y)*0.5-R #Principal Stress in psi\n",
-      "\n",
-      "#Principal Direction\n",
-      "theta1=arctan(2*tau_xy*(sigma_x-sigma_y)**-1)*0.5*180*pi**-1 #Principal direction in degrees\n",
-      "theta2=theta1+90 #Second pricnipal direction in degrees\n",
-      "\n",
-      "#Normal Stress\n",
-      "sigma_xdash=0.5*(sigma_x+sigma_y)+0.5*(sigma_x-sigma_y)*cos(2*theta1*pi*180**-1)+tau_xy*sin(2*theta1*pi*180**-1)\n",
-      "\n",
-      "#Result\n",
-      "print \"The principal stresses are as follows\"\n",
-      "print \"sigma1=\",round(sigma1),\"psi and sigma2=\",round(sigma2),\"psi\"\n",
-      "print \"The corresponding directions are\"\n",
-      "print \"Theta1=\",round(theta1,1),\"degrees and Theta2=\",round(theta2,1),\"degrees\"\n",
-      "\n",
-      "#NOTE:The answer in the textbook for principal stresses is off by 4 units in each case"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The principal stresses are as follows\n",
-        "sigma1= 9606.0 psi and sigma2= 2394.0 psi\n",
-        "The corresponding directions are\n",
-        "Theta1= 28.2 degrees and Theta2= 118.2 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.6, Page No:298"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "sigma_x=40 #Stress in x in MPa\n",
-      "sigma_y=-100 #Stress in y in MPa\n",
-      "tau_xy=-50 #Shear stress in MPa\n",
-      "\n",
-      "#Calculations\n",
-      "tau_max=sqrt(((sigma_x-sigma_y)*0.5)**2+tau_xy**2) #Maximum in-plane shear in MPa\n",
-      "\n",
-      "#Orientation of Plane\n",
-      "theta1=arctan(-((sigma_x-sigma_y)*(2*tau_xy)**-1))*180*pi**-1*0.5 #Angle in Degrees\n",
-      "theta2=theta1+90 #Angle in degrees\n",
-      "\n",
-      "#Plane of max in-plane shear\n",
-      "tau_x_y=-0.5*(sigma_x-sigma_y)*sin(2*theta1*pi*180**-1)+tau_xy*cos(2*theta1*pi*180**-1) \n",
-      "\n",
-      "#Normal Stress\n",
-      "sigma=(sigma_x+sigma_y)*0.5 #Stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum in-plane Shear is\",round(tau_x_y),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum in-plane Shear is -86.0 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.7, Page No:305"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Vairable Decleration\n",
-      "sigma_x=40 #Stress in x in MPa\n",
-      "sigma_y=20 #Stress in y in MPa\n",
-      "tau_xy=16 #Shear in xy in MPa\n",
-      "\n",
-      "#Calculations\n",
-      "sigma=(sigma_x+sigma_y)*0.5 #Normal Stress in MPa\n",
-      "R=sqrt(((sigma_x-sigma_y)*0.5)**2+tau_xy**2) #Resultant Stress in MPa\n",
-      "\n",
-      "#Part 1\n",
-      "sigma1=sigma+R #Principal Stress in MPa\n",
-      "sigma2=sigma-R #Principal Stress in MPa\n",
-      "theta=arctan(tau_xy*((sigma_x-sigma_y)*0.5)**-1)*180*pi**-1*0.5 #Orientation in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "tau_max=18.87 #From figure in MPa\n",
-      "\n",
-      "#Part 3\n",
-      "sigma_xdash=sigma+tau_max*cos((100-theta*2)*pi*180**-1) #Stress in MPa\n",
-      "sigma_ydash=sigma-tau_max*cos((100-theta*2)*pi*180**-1) #Stress in MPa\n",
-      "tau_x_y=tau_max*sin((100-2*theta)*pi*180**-1) #Shear stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The principal Stresses are\"\n",
-      "print \"Sigma1=\",round(sigma1,1),\"MPa and Sigma2=\",round(sigma2,1),\"MPa\"\n",
-      "print \"The Principal Plane is at\",round(theta),\"degrees\"\n",
-      "print \"The  Maximum Shear Stress is\",tau_max,\"MPa\"\n",
-      "print \"Sigma_x'=\",round(sigma_xdash),\"MPa and Sigma_y'=\",round(sigma_ydash,2),\"MPa\"\n",
-      "print \"Tau_x'y'=\",round(tau_x_y,2),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The principal Stresses are\n",
-        "Sigma1= 48.9 MPa and Sigma2= 11.1 MPa\n",
-        "The Principal Plane is at 29.0 degrees\n",
-        "The  Maximum Shear Stress is 18.87 MPa\n",
-        "Sigma_x'= 44.0 MPa and Sigma_y'= 15.98 MPa\n",
-        "Tau_x'y'= 12.63 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.9, Page No:316"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variabel Decleration\n",
-      "sigma_w=120 #Working Stress in MPa\n",
-      "tau_w=70 #Working Shear in MPa\n",
-      "\n",
-      "#Calcualtions\n",
-      "#Section a-a\n",
-      "M=3750 #Applied moment at section a-a in N.m\n",
-      "T=1500 #Applied Torque at section a-a in N.m\n",
-      "\n",
-      "#After carrying out the variable based computation we compute d\n",
-      "d1=((124.62)/(sigma_w*10**3*pi))**0.3333 #Diameter of the shaft in m\n",
-      "d2=((65.6)/(tau_w*10**3*pi))**0.3333 #Diameter of the shaft in m\n",
-      "d=max(d1,d2) #Diameter of the shaft to be selected in m\n",
-      "\n",
-      "#Result\n",
-      "print \"The diameter of the shaft to be selected is\",round(d*1000,1),\"mm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The diameter of the shaft to be selected is 69.2 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.10, Page No:318"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "t=0.01 #Thickness of the shaft in m\n",
-      "p=2 #Internal Pressure in MPa\n",
-      "r=0.45 #Mean radius of the vessel in m\n",
-      "tw=50 #Working shear stress in MPa\n",
-      "\n",
-      "#Calculation\n",
-      "sigma_x=(p*r)/(2*t) #Stress in MPa\n",
-      "sigma_y=(p*r)/t #Stress in MPa\n",
-      "\n",
-      "R=100-67.5 #From the diagram in MPa\n",
-      "tau_xy=sqrt((R**2-(sigma_y-67.5)**2)) #Stress in MPa\n",
-      "\n",
-      "J=2*pi*r**3*t #Polar Moment of inertia in mm^4\n",
-      "\n",
-      "T=1000*(tau_xy*J)/r #Maximum allowable Torque in kN.m\n",
-      "\n",
-      "#Result\n",
-      "print \"The largest allowable Torque is\",round(T),\"kN.m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The largest allowable Torque is 298.0 kN.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.11, Page No:320"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=15 #Length of the shaft in inches\n",
-      "r=3.0/8.001 #Radius of the shaft in inches\n",
-      "T=540 #Torque applied in lb.in\n",
-      "\n",
-      "#Calculations\n",
-      "V=30 #Transverse Shear Force in lb\n",
-      "M=15*V #Bending Moment in lb.in\n",
-      "I=(pi*r**4)/4.0 #Moment of Inertia in in^4\n",
-      "J=2*I #Polar Moment Of Inertia in in^4\n",
-      "\n",
-      "#Part 1\n",
-      "sigma=(M*r)/I #Bending Stress in psi\n",
-      "tau_t=10**-3*(T*r)/J #Shear Stress in ksi\n",
-      "\n",
-      "sigma_max1=13.92 #From the Mohr Circle in ksi\n",
-      "\n",
-      "#Part 2\n",
-      "Q=(2*r**3)/3.0 #First Moment in in^3\n",
-      "b=2*r # in\n",
-      "\n",
-      "tau_V=10**-3*(V*Q)/(I*b) #Shear Stress in ksi\n",
-      "tau=tau_t+tau_V #Total Shear in ksi\n",
-      "\n",
-      "sigma_max2=tau #Maximum stress in ksi\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum normal stress in case 1 is\",sigma_max1,\"ksi\"\n",
-      "print \"The Maximum normal stress in case 2 is\",round(sigma_max2,2),\"ksi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum normal stress in case 1 is 13.92 ksi\n",
-        "The Maximum normal stress in case 2 is 6.61 ksi\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.12, Page No:330"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "ex=800*10**-6 #Strain in x \n",
-      "ey=200*10**-6 #Strain in y\n",
-      "y_xy=-600*10**-6 #Strain in xy\n",
-      "\n",
-      "#Calculations\n",
-      "e_bar=(ex+ey)*0.5 #Strain \n",
-      "R=sqrt(2*300**2)*10**-6 #Resultant \n",
-      "\n",
-      "#Part 1\n",
-      "e1=e_bar+R #Strain in Principal Axis\n",
-      "e2=e_bar-R #Strain in Principal Axis\n",
-      "\n",
-      "#Part 2\n",
-      "alpha=15*180**-1*pi #From the Mohr Circle in degrees\n",
-      "e_xbar=e_bar-(R*cos(alpha)) #Strain in x \n",
-      "e_ybar=e_bar+(R*cos(alpha)) #Strain in y\n",
-      "\n",
-      "shear_strain=2*R*sin(alpha) #Shear starin \n",
-      "\n",
-      "#Result\n",
-      "print \"The principal Strains are\"\n",
-      "print \"e1=\",round(e1,6),\"e2=\",round(e2,6)\n",
-      "print \"The starin components are\"\n",
-      "print \"ex'=\",round(e_xbar,6),\"ey'=\",round(e_ybar,6),\"y_x'y'=\",round(shear_strain,6)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The principal Strains are\n",
-        "e1= 0.000924 e2= 7.6e-05\n",
-        "The starin components are\n",
-        "ex'= 9e-05 ey'= 0.00091 y_x'y'= 0.00022\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.13, Page No:331"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "e_x=800*10**-6 #Strain in x\n",
-      "e_y=200*10**-6 #Strain in y\n",
-      "y_xy=-600*10**-6 #Strain in xy\n",
-      "v=0.30 #Poissons Ratio\n",
-      "E=200 #Youngs Modulus in GPa\n",
-      "R_e=424.3*10**-6 #Strain\n",
-      "e_bar=500*10**-6 #Strain\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "R_sigma=10**-6*R_e*(E*10**9/(1+v)) #Stress in MPa\n",
-      "sigma_bar=10**-6*e_bar*(E*10**9/(1-v)) #Stress in MPa\n",
-      "\n",
-      "#Part 2\n",
-      "sigma1=sigma_bar+R_sigma #Principal Stress in MPa\n",
-      "sigma2=sigma_bar-R_sigma #Principal Stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The principal Stresses are as follows\"\n",
-      "print \"Sigma1=\",round(sigma1),\"MPa and Sigma2=\",round(sigma2,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The principal Stresses are as follows\n",
-        "Sigma1= 208.0 MPa and Sigma2= 77.6 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.14, Page No:336"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "e_a=100*10**-6 #Strain \n",
-      "e_b=300*10**-6 #Strain\n",
-      "e_c=-200*10**-6 #Strain\n",
-      "E=180 #Youngs Modulus in GPa\n",
-      "v=0.28 #Poissons Ratio \n",
-      "\n",
-      "#Calculations\n",
-      "y_xy=(e_b-(e_a+e_c)*0.5) #Strain in xy\n",
-      "e_bar=(e_a+e_c)*0.5 #Strain \n",
-      "R_e=sqrt(y_xy**2+(150*10**-6)**2) #Resultant Strain\n",
-      "\n",
-      "#Corresponding Parameters from Mohrs Diagram\n",
-      "sigma_bar=(E/(1-v))*e_bar*10**3 #Stress in MPa\n",
-      "R_sigma=(E/(1+v))*R_e*10**3 #Resultant Stress in MPa\n",
-      "\n",
-      "#Principal Stresses\n",
-      "sigma1=sigma_bar+R_sigma #MPa\n",
-      "sigma2=sigma_bar-R_sigma #MPa\n",
-      "\n",
-      "theta=arctan(y_xy/(150*10**-6))*180*pi**-1*0.5 #Degrees\n",
-      "\n",
-      "#Result\n",
-      "print \"The Principal Stresses are as follows\"\n",
-      "print \"Sigma1=\",round(sigma1,1),\"MPa and Sigma2=\",round(sigma2,2),\"MPa\"\n",
-      "print \"The plane orientation is\",round(theta,2),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Principal Stresses are as follows\n",
-        "Sigma1= 41.0 MPa and Sigma2= -66.05 MPa\n",
-        "The plane orientation is 33.4 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter08_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter08_1.ipynb
deleted file mode 100755
index 87dd0fdf..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter08_1.ipynb
+++ /dev/null
@@ -1,721 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:b494877451d53f8b0ca30d008c3144520923dfdb33c6562fbcdeab1b4ca2b7ce"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter08:Stresses due to Combined Loading"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Examlple 8.8.1, Page No:275"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "p=125 #Pressure in psi\n",
-      "r=24 #Radius of the vessel in inches\n",
-      "t=0.25 #Thickness of the vessel in inches\n",
-      "E=29*10**6 #Modulus of Elasticity in psi\n",
-      "v=0.28 #poisson ratio\n",
-      "\n",
-      "#Calcualtions\n",
-      "#Part 1\n",
-      "sigma_c=p*r*t**-1 #Circumferential Stress in psi\n",
-      "sigma_l=sigma_c/2 #Longitudinat Stress in psi\n",
-      "e_c=E**-1*(sigma_c-(v*sigma_l)) #Circumferential Strain using biaxial Hooke's Law \n",
-      "delta_r=e_c*r #Change in the radius in inches\n",
-      "\n",
-      "#Part 2\n",
-      "sigma=(p*r)*(2*t)**-1 #Stress in psi\n",
-      "e=E**-1*(sigma-(v*sigma)) #Strain using biaxial Hooke's Law\n",
-      "delta_R=e*r #Change inradius of end-cap in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"Part 1 Answers\"\n",
-      "print \"Stresses are sigma_c=\",round(sigma_c),\"psi and sigma_l=\",round(sigma_l),\"psi\"\n",
-      "print \"Change of radius of cylinder=\",round(delta_r,5),\"in\"\n",
-      "print \"Part 2 Answers\"\n",
-      "print \"Stresses are sigma=\",round(sigma),\"psi\"\n",
-      "print \"Change in radius of end cap=\",round(delta_R,5),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part 1 Answers\n",
-        "Stresses are sigma_c= 12000.0 psi and sigma_l= 6000.0 psi\n",
-        "Change of radius of cylinder= 0.00854 in\n",
-        "Part 2 Answers\n",
-        "Stresses are sigma= 6000.0 psi\n",
-        "Change in radius of end cap= 0.00358 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.2, Page No:280"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import matplotlib.pyplot as plt\n",
-      "%matplotlib inline\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=40 #Force in kN\n",
-      "b=0.050 #Width in m\n",
-      "h=0.040 #Depth in m\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "A=b*h #Area in m^2\n",
-      "I=(b*h**3)*12**-1 #Moment of inertia in m^4\n",
-      "c=h*0.5 #m\n",
-      "sigma_max=(P*A**-1)+(P*c**2*I**-1) #Maximum stress in MPa\n",
-      "sigma_min=(P*A**-1)-(P*c**2*I**-1) #Minimum stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The Maximum and Minimum Stress are\"\n",
-      "print \"Max=\",sigma_max/1000,\"MPa and Min=\",sigma_min/1000,\"MPa\"\n",
-      "\n",
-      "#Plotting\n",
-      "x=[20,0,-20]\n",
-      "S=[-sigma_min/1000,0,sigma_max/1000]\n",
-      "plt.plot(S,x)\n",
-      "plt.ylabel(\"Distance from Neutral Axis in mm\")\n",
-      "plt.xlabel(\"Stress in MPa\")\n",
-      "plt.title(\"Stress Distribution Diagram\")\n",
-      "plt.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Maximum and Minimum Stress are\n",
-        "Max= 80.0 MPa and Min= -40.0 MPa\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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c4OvAuba/1PFJpecDC4FvAB+0fVW5fhYw1/bmE+yfSiKiz6TC6L1u90lcDbyy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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10b6d0310>"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.3, Page No:281"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variabel Decleration\n",
-      "b=6 #Width in inches\n",
-      "h=10 #Depth in inches\n",
-      "P1=6000 #Force in lb\n",
-      "P2=3000 #Force in lb\n",
-      "L=4 #Length in ft\n",
-      "P=-13400 #Load in lb\n",
-      "M=6000 #Moment in lb.ft\n",
-      "y=5 #Depth in inches\n",
-      "P2=-9800 #Load in lb\n",
-      "M2=-12000 #Moment in lb.ft\n",
-      "\n",
-      "#Calculations\n",
-      "A=b*h #Area in in^2\n",
-      "I=b*h**3*12**-1 #Moment of inertia in in^4\n",
-      "T=(P1*L+P2*L*3)*(6)**-1 #Tension in the cable in lb\n",
-      "\n",
-      "#Computation of largest stress\n",
-      "sigma_B=(P*A**-1)-(M*y*12*I**-1) #Maximum Compressive Stress caused by +ve BM in psi\n",
-      "sigma_C=(P2*A**-1)-(M2*-y*12*I**-1) #Maximum Compressive Stress caused by -ve BM in psi\n",
-      "\n",
-      "sigma_max=max(-sigma_B,-sigma_C) #Maximum Compressive Stress in psi\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum Stress is\",round(sigma_max),\"psi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum Stress is 1603.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.4, Page No:297"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "theta=(60*pi)/180 #Angle in radians (Twice as declared)\n",
-      "sigma_x=30 # Stress in x in MPa\n",
-      "sigma_y=60 #Stress in y in MPa\n",
-      "tau_xy=40 #Stress in MPa\n",
-      "\n",
-      "#Calcualtions\n",
-      "sigma_xdash=0.5*(sigma_x+sigma_y)+0.5*(sigma_x-sigma_y)*cos(theta)+tau_xy*sin(theta) #Stress at x' axis in MPa\n",
-      "sigma_ydash=0.5*(sigma_x+sigma_y)-0.5*(sigma_x-sigma_y)*cos(theta)-tau_xy*sin(theta) #Stress at y' axis in MPa\n",
-      "tau_x_y=-0.5*(sigma_x-sigma_y)*sin(theta)+tau_xy*cos(theta) #Stress at x'y' in shear in MPa\n",
-      "#Result\n",
-      "print \"The new stresses at new axes are as follows\"\n",
-      "print \"sigma_x'=\",round(sigma_xdash,1),\"MPa sigma_y'=\",round(sigma_ydash,1),\"MPa\"\n",
-      "print \"And tau_x'y'=\",round(tau_x_y),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The new stresses at new axes are as follows\n",
-        "sigma_x'= 72.1 MPa sigma_y'= 17.9 MPa\n",
-        "And tau_x'y'= 33.0 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.5, Page No:297"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "sigma_x=8000 #Stress in x in psi\n",
-      "sigma_y=4000 #Stress in y in psi\n",
-      "tau_xy=3000 #Stress in xy in psi\n",
-      "\n",
-      "#Calculations\n",
-      "R=sqrt(((sigma_x-sigma_y)*0.5)**2+tau_xy**2) #Resultant Stress in psi\n",
-      "\n",
-      "#Principal Stresses\n",
-      "sigma1=(sigma_x+sigma_y)*0.5+R #Principal Stress in psi\n",
-      "sigma2=(sigma_x+sigma_y)*0.5-R #Principal Stress in psi\n",
-      "\n",
-      "#Principal Direction\n",
-      "theta1=arctan(2*tau_xy*(sigma_x-sigma_y)**-1)*0.5*180*pi**-1 #Principal direction in degrees\n",
-      "theta2=theta1+90 #Second pricnipal direction in degrees\n",
-      "\n",
-      "#Normal Stress\n",
-      "sigma_xdash=0.5*(sigma_x+sigma_y)+0.5*(sigma_x-sigma_y)*cos(2*theta1*pi*180**-1)+tau_xy*sin(2*theta1*pi*180**-1)\n",
-      "\n",
-      "#Result\n",
-      "print \"The principal stresses are as follows\"\n",
-      "print \"sigma1=\",round(sigma1),\"psi and sigma2=\",round(sigma2),\"psi\"\n",
-      "print \"The corresponding directions are\"\n",
-      "print \"Theta1=\",round(theta1,1),\"degrees and Theta2=\",round(theta2,1),\"degrees\"\n",
-      "\n",
-      "#NOTE:The answer in the textbook for principal stresses is off by 4 units in each case"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The principal stresses are as follows\n",
-        "sigma1= 9606.0 psi and sigma2= 2394.0 psi\n",
-        "The corresponding directions are\n",
-        "Theta1= 28.2 degrees and Theta2= 118.2 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.6, Page No:298"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "sigma_x=40 #Stress in x in MPa\n",
-      "sigma_y=-100 #Stress in y in MPa\n",
-      "tau_xy=-50 #Shear stress in MPa\n",
-      "\n",
-      "#Calculations\n",
-      "tau_max=sqrt(((sigma_x-sigma_y)*0.5)**2+tau_xy**2) #Maximum in-plane shear in MPa\n",
-      "\n",
-      "#Orientation of Plane\n",
-      "theta1=arctan(-((sigma_x-sigma_y)*(2*tau_xy)**-1))*180*pi**-1*0.5 #Angle in Degrees\n",
-      "theta2=theta1+90 #Angle in degrees\n",
-      "\n",
-      "#Plane of max in-plane shear\n",
-      "tau_x_y=-0.5*(sigma_x-sigma_y)*sin(2*theta1*pi*180**-1)+tau_xy*cos(2*theta1*pi*180**-1) \n",
-      "\n",
-      "#Normal Stress\n",
-      "sigma=(sigma_x+sigma_y)*0.5 #Stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum in-plane Shear is\",round(tau_x_y),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum in-plane Shear is -86.0 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.7, Page No:305"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Vairable Decleration\n",
-      "sigma_x=40 #Stress in x in MPa\n",
-      "sigma_y=20 #Stress in y in MPa\n",
-      "tau_xy=16 #Shear in xy in MPa\n",
-      "\n",
-      "#Calculations\n",
-      "sigma=(sigma_x+sigma_y)*0.5 #Normal Stress in MPa\n",
-      "R=sqrt(((sigma_x-sigma_y)*0.5)**2+tau_xy**2) #Resultant Stress in MPa\n",
-      "\n",
-      "#Part 1\n",
-      "sigma1=sigma+R #Principal Stress in MPa\n",
-      "sigma2=sigma-R #Principal Stress in MPa\n",
-      "theta=arctan(tau_xy*((sigma_x-sigma_y)*0.5)**-1)*180*pi**-1*0.5 #Orientation in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "tau_max=18.87 #From figure in MPa\n",
-      "\n",
-      "#Part 3\n",
-      "sigma_xdash=sigma+tau_max*cos((100-theta*2)*pi*180**-1) #Stress in MPa\n",
-      "sigma_ydash=sigma-tau_max*cos((100-theta*2)*pi*180**-1) #Stress in MPa\n",
-      "tau_x_y=tau_max*sin((100-2*theta)*pi*180**-1) #Shear stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The principal Stresses are\"\n",
-      "print \"Sigma1=\",round(sigma1,1),\"MPa and Sigma2=\",round(sigma2,1),\"MPa\"\n",
-      "print \"The Principal Plane is at\",round(theta),\"degrees\"\n",
-      "print \"The  Maximum Shear Stress is\",tau_max,\"MPa\"\n",
-      "print \"Sigma_x'=\",round(sigma_xdash),\"MPa and Sigma_y'=\",round(sigma_ydash,2),\"MPa\"\n",
-      "print \"Tau_x'y'=\",round(tau_x_y,2),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The principal Stresses are\n",
-        "Sigma1= 48.9 MPa and Sigma2= 11.1 MPa\n",
-        "The Principal Plane is at 29.0 degrees\n",
-        "The  Maximum Shear Stress is 18.87 MPa\n",
-        "Sigma_x'= 44.0 MPa and Sigma_y'= 15.98 MPa\n",
-        "Tau_x'y'= 12.63 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.9, Page No:316"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variabel Decleration\n",
-      "sigma_w=120 #Working Stress in MPa\n",
-      "tau_w=70 #Working Shear in MPa\n",
-      "\n",
-      "#Calcualtions\n",
-      "#Section a-a\n",
-      "M=3750 #Applied moment at section a-a in N.m\n",
-      "T=1500 #Applied Torque at section a-a in N.m\n",
-      "\n",
-      "#After carrying out the variable based computation we compute d\n",
-      "d1=((124.62)/(sigma_w*10**3*pi))**0.3333 #Diameter of the shaft in m\n",
-      "d2=((65.6)/(tau_w*10**3*pi))**0.3333 #Diameter of the shaft in m\n",
-      "d=max(d1,d2) #Diameter of the shaft to be selected in m\n",
-      "\n",
-      "#Result\n",
-      "print \"The diameter of the shaft to be selected is\",round(d*1000,1),\"mm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The diameter of the shaft to be selected is 69.2 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.10, Page No:318"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "t=0.01 #Thickness of the shaft in m\n",
-      "p=2 #Internal Pressure in MPa\n",
-      "r=0.45 #Mean radius of the vessel in m\n",
-      "tw=50 #Working shear stress in MPa\n",
-      "\n",
-      "#Calculation\n",
-      "sigma_x=(p*r)/(2*t) #Stress in MPa\n",
-      "sigma_y=(p*r)/t #Stress in MPa\n",
-      "\n",
-      "R=100-67.5 #From the diagram in MPa\n",
-      "tau_xy=sqrt((R**2-(sigma_y-67.5)**2)) #Stress in MPa\n",
-      "\n",
-      "J=2*pi*r**3*t #Polar Moment of inertia in mm^4\n",
-      "\n",
-      "T=1000*(tau_xy*J)/r #Maximum allowable Torque in kN.m\n",
-      "\n",
-      "#Result\n",
-      "print \"The largest allowable Torque is\",round(T),\"kN.m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The largest allowable Torque is 298.0 kN.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.11, Page No:320"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=15 #Length of the shaft in inches\n",
-      "r=3.0/8.001 #Radius of the shaft in inches\n",
-      "T=540 #Torque applied in lb.in\n",
-      "\n",
-      "#Calculations\n",
-      "V=30 #Transverse Shear Force in lb\n",
-      "M=15*V #Bending Moment in lb.in\n",
-      "I=(pi*r**4)/4.0 #Moment of Inertia in in^4\n",
-      "J=2*I #Polar Moment Of Inertia in in^4\n",
-      "\n",
-      "#Part 1\n",
-      "sigma=(M*r)/I #Bending Stress in psi\n",
-      "tau_t=10**-3*(T*r)/J #Shear Stress in ksi\n",
-      "\n",
-      "sigma_max1=13.92 #From the Mohr Circle in ksi\n",
-      "\n",
-      "#Part 2\n",
-      "Q=(2*r**3)/3.0 #First Moment in in^3\n",
-      "b=2*r # in\n",
-      "\n",
-      "tau_V=10**-3*(V*Q)/(I*b) #Shear Stress in ksi\n",
-      "tau=tau_t+tau_V #Total Shear in ksi\n",
-      "\n",
-      "sigma_max2=tau #Maximum stress in ksi\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum normal stress in case 1 is\",sigma_max1,\"ksi\"\n",
-      "print \"The Maximum normal stress in case 2 is\",round(sigma_max2,2),\"ksi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum normal stress in case 1 is 13.92 ksi\n",
-        "The Maximum normal stress in case 2 is 6.61 ksi\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.12, Page No:330"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "ex=800*10**-6 #Strain in x \n",
-      "ey=200*10**-6 #Strain in y\n",
-      "y_xy=-600*10**-6 #Strain in xy\n",
-      "\n",
-      "#Calculations\n",
-      "e_bar=(ex+ey)*0.5 #Strain \n",
-      "R=sqrt(2*300**2)*10**-6 #Resultant \n",
-      "\n",
-      "#Part 1\n",
-      "e1=e_bar+R #Strain in Principal Axis\n",
-      "e2=e_bar-R #Strain in Principal Axis\n",
-      "\n",
-      "#Part 2\n",
-      "alpha=15*180**-1*pi #From the Mohr Circle in degrees\n",
-      "e_xbar=e_bar-(R*cos(alpha)) #Strain in x \n",
-      "e_ybar=e_bar+(R*cos(alpha)) #Strain in y\n",
-      "\n",
-      "shear_strain=2*R*sin(alpha) #Shear starin \n",
-      "\n",
-      "#Result\n",
-      "print \"The principal Strains are\"\n",
-      "print \"e1=\",round(e1,6),\"e2=\",round(e2,6)\n",
-      "print \"The starin components are\"\n",
-      "print \"ex'=\",round(e_xbar,6),\"ey'=\",round(e_ybar,6),\"y_x'y'=\",round(shear_strain,6)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The principal Strains are\n",
-        "e1= 0.000924 e2= 7.6e-05\n",
-        "The starin components are\n",
-        "ex'= 9e-05 ey'= 0.00091 y_x'y'= 0.00022\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.13, Page No:331"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "e_x=800*10**-6 #Strain in x\n",
-      "e_y=200*10**-6 #Strain in y\n",
-      "y_xy=-600*10**-6 #Strain in xy\n",
-      "v=0.30 #Poissons Ratio\n",
-      "E=200 #Youngs Modulus in GPa\n",
-      "R_e=424.3*10**-6 #Strain\n",
-      "e_bar=500*10**-6 #Strain\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "R_sigma=10**-6*R_e*(E*10**9/(1+v)) #Stress in MPa\n",
-      "sigma_bar=10**-6*e_bar*(E*10**9/(1-v)) #Stress in MPa\n",
-      "\n",
-      "#Part 2\n",
-      "sigma1=sigma_bar+R_sigma #Principal Stress in MPa\n",
-      "sigma2=sigma_bar-R_sigma #Principal Stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The principal Stresses are as follows\"\n",
-      "print \"Sigma1=\",round(sigma1),\"MPa and Sigma2=\",round(sigma2,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The principal Stresses are as follows\n",
-        "Sigma1= 208.0 MPa and Sigma2= 77.6 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.14, Page No:336"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "e_a=100*10**-6 #Strain \n",
-      "e_b=300*10**-6 #Strain\n",
-      "e_c=-200*10**-6 #Strain\n",
-      "E=180 #Youngs Modulus in GPa\n",
-      "v=0.28 #Poissons Ratio \n",
-      "\n",
-      "#Calculations\n",
-      "y_xy=(e_b-(e_a+e_c)*0.5) #Strain in xy\n",
-      "e_bar=(e_a+e_c)*0.5 #Strain \n",
-      "R_e=sqrt(y_xy**2+(150*10**-6)**2) #Resultant Strain\n",
-      "\n",
-      "#Corresponding Parameters from Mohrs Diagram\n",
-      "sigma_bar=(E/(1-v))*e_bar*10**3 #Stress in MPa\n",
-      "R_sigma=(E/(1+v))*R_e*10**3 #Resultant Stress in MPa\n",
-      "\n",
-      "#Principal Stresses\n",
-      "sigma1=sigma_bar+R_sigma #MPa\n",
-      "sigma2=sigma_bar-R_sigma #MPa\n",
-      "\n",
-      "theta=arctan(y_xy/(150*10**-6))*180*pi**-1*0.5 #Degrees\n",
-      "\n",
-      "#Result\n",
-      "print \"The Principal Stresses are as follows\"\n",
-      "print \"Sigma1=\",round(sigma1,1),\"MPa and Sigma2=\",round(sigma2,2),\"MPa\"\n",
-      "print \"The plane orientation is\",round(theta,2),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Principal Stresses are as follows\n",
-        "Sigma1= 41.0 MPa and Sigma2= -66.05 MPa\n",
-        "The plane orientation is 33.4 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter08_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter08_2.ipynb
deleted file mode 100755
index 87dd0fdf..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter08_2.ipynb
+++ /dev/null
@@ -1,721 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:b494877451d53f8b0ca30d008c3144520923dfdb33c6562fbcdeab1b4ca2b7ce"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter08:Stresses due to Combined Loading"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Examlple 8.8.1, Page No:275"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "p=125 #Pressure in psi\n",
-      "r=24 #Radius of the vessel in inches\n",
-      "t=0.25 #Thickness of the vessel in inches\n",
-      "E=29*10**6 #Modulus of Elasticity in psi\n",
-      "v=0.28 #poisson ratio\n",
-      "\n",
-      "#Calcualtions\n",
-      "#Part 1\n",
-      "sigma_c=p*r*t**-1 #Circumferential Stress in psi\n",
-      "sigma_l=sigma_c/2 #Longitudinat Stress in psi\n",
-      "e_c=E**-1*(sigma_c-(v*sigma_l)) #Circumferential Strain using biaxial Hooke's Law \n",
-      "delta_r=e_c*r #Change in the radius in inches\n",
-      "\n",
-      "#Part 2\n",
-      "sigma=(p*r)*(2*t)**-1 #Stress in psi\n",
-      "e=E**-1*(sigma-(v*sigma)) #Strain using biaxial Hooke's Law\n",
-      "delta_R=e*r #Change inradius of end-cap in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"Part 1 Answers\"\n",
-      "print \"Stresses are sigma_c=\",round(sigma_c),\"psi and sigma_l=\",round(sigma_l),\"psi\"\n",
-      "print \"Change of radius of cylinder=\",round(delta_r,5),\"in\"\n",
-      "print \"Part 2 Answers\"\n",
-      "print \"Stresses are sigma=\",round(sigma),\"psi\"\n",
-      "print \"Change in radius of end cap=\",round(delta_R,5),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part 1 Answers\n",
-        "Stresses are sigma_c= 12000.0 psi and sigma_l= 6000.0 psi\n",
-        "Change of radius of cylinder= 0.00854 in\n",
-        "Part 2 Answers\n",
-        "Stresses are sigma= 6000.0 psi\n",
-        "Change in radius of end cap= 0.00358 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.2, Page No:280"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import matplotlib.pyplot as plt\n",
-      "%matplotlib inline\n",
-      "\n",
-      "#Variable Decleration\n",
-      "P=40 #Force in kN\n",
-      "b=0.050 #Width in m\n",
-      "h=0.040 #Depth in m\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "A=b*h #Area in m^2\n",
-      "I=(b*h**3)*12**-1 #Moment of inertia in m^4\n",
-      "c=h*0.5 #m\n",
-      "sigma_max=(P*A**-1)+(P*c**2*I**-1) #Maximum stress in MPa\n",
-      "sigma_min=(P*A**-1)-(P*c**2*I**-1) #Minimum stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The Maximum and Minimum Stress are\"\n",
-      "print \"Max=\",sigma_max/1000,\"MPa and Min=\",sigma_min/1000,\"MPa\"\n",
-      "\n",
-      "#Plotting\n",
-      "x=[20,0,-20]\n",
-      "S=[-sigma_min/1000,0,sigma_max/1000]\n",
-      "plt.plot(S,x)\n",
-      "plt.ylabel(\"Distance from Neutral Axis in mm\")\n",
-      "plt.xlabel(\"Stress in MPa\")\n",
-      "plt.title(\"Stress Distribution Diagram\")\n",
-      "plt.show()"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Maximum and Minimum Stress are\n",
-        "Max= 80.0 MPa and Min= -40.0 MPa\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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c4OvAuba/1PFJpecDC4FvAB+0fVW5fhYw1/bmE+yfSiKiz6TC6L1u90lcDbyy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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x10b6d0310>"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.3, Page No:281"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variabel Decleration\n",
-      "b=6 #Width in inches\n",
-      "h=10 #Depth in inches\n",
-      "P1=6000 #Force in lb\n",
-      "P2=3000 #Force in lb\n",
-      "L=4 #Length in ft\n",
-      "P=-13400 #Load in lb\n",
-      "M=6000 #Moment in lb.ft\n",
-      "y=5 #Depth in inches\n",
-      "P2=-9800 #Load in lb\n",
-      "M2=-12000 #Moment in lb.ft\n",
-      "\n",
-      "#Calculations\n",
-      "A=b*h #Area in in^2\n",
-      "I=b*h**3*12**-1 #Moment of inertia in in^4\n",
-      "T=(P1*L+P2*L*3)*(6)**-1 #Tension in the cable in lb\n",
-      "\n",
-      "#Computation of largest stress\n",
-      "sigma_B=(P*A**-1)-(M*y*12*I**-1) #Maximum Compressive Stress caused by +ve BM in psi\n",
-      "sigma_C=(P2*A**-1)-(M2*-y*12*I**-1) #Maximum Compressive Stress caused by -ve BM in psi\n",
-      "\n",
-      "sigma_max=max(-sigma_B,-sigma_C) #Maximum Compressive Stress in psi\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum Stress is\",round(sigma_max),\"psi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum Stress is 1603.0 psi\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.4, Page No:297"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "theta=(60*pi)/180 #Angle in radians (Twice as declared)\n",
-      "sigma_x=30 # Stress in x in MPa\n",
-      "sigma_y=60 #Stress in y in MPa\n",
-      "tau_xy=40 #Stress in MPa\n",
-      "\n",
-      "#Calcualtions\n",
-      "sigma_xdash=0.5*(sigma_x+sigma_y)+0.5*(sigma_x-sigma_y)*cos(theta)+tau_xy*sin(theta) #Stress at x' axis in MPa\n",
-      "sigma_ydash=0.5*(sigma_x+sigma_y)-0.5*(sigma_x-sigma_y)*cos(theta)-tau_xy*sin(theta) #Stress at y' axis in MPa\n",
-      "tau_x_y=-0.5*(sigma_x-sigma_y)*sin(theta)+tau_xy*cos(theta) #Stress at x'y' in shear in MPa\n",
-      "#Result\n",
-      "print \"The new stresses at new axes are as follows\"\n",
-      "print \"sigma_x'=\",round(sigma_xdash,1),\"MPa sigma_y'=\",round(sigma_ydash,1),\"MPa\"\n",
-      "print \"And tau_x'y'=\",round(tau_x_y),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The new stresses at new axes are as follows\n",
-        "sigma_x'= 72.1 MPa sigma_y'= 17.9 MPa\n",
-        "And tau_x'y'= 33.0 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.5, Page No:297"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "sigma_x=8000 #Stress in x in psi\n",
-      "sigma_y=4000 #Stress in y in psi\n",
-      "tau_xy=3000 #Stress in xy in psi\n",
-      "\n",
-      "#Calculations\n",
-      "R=sqrt(((sigma_x-sigma_y)*0.5)**2+tau_xy**2) #Resultant Stress in psi\n",
-      "\n",
-      "#Principal Stresses\n",
-      "sigma1=(sigma_x+sigma_y)*0.5+R #Principal Stress in psi\n",
-      "sigma2=(sigma_x+sigma_y)*0.5-R #Principal Stress in psi\n",
-      "\n",
-      "#Principal Direction\n",
-      "theta1=arctan(2*tau_xy*(sigma_x-sigma_y)**-1)*0.5*180*pi**-1 #Principal direction in degrees\n",
-      "theta2=theta1+90 #Second pricnipal direction in degrees\n",
-      "\n",
-      "#Normal Stress\n",
-      "sigma_xdash=0.5*(sigma_x+sigma_y)+0.5*(sigma_x-sigma_y)*cos(2*theta1*pi*180**-1)+tau_xy*sin(2*theta1*pi*180**-1)\n",
-      "\n",
-      "#Result\n",
-      "print \"The principal stresses are as follows\"\n",
-      "print \"sigma1=\",round(sigma1),\"psi and sigma2=\",round(sigma2),\"psi\"\n",
-      "print \"The corresponding directions are\"\n",
-      "print \"Theta1=\",round(theta1,1),\"degrees and Theta2=\",round(theta2,1),\"degrees\"\n",
-      "\n",
-      "#NOTE:The answer in the textbook for principal stresses is off by 4 units in each case"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The principal stresses are as follows\n",
-        "sigma1= 9606.0 psi and sigma2= 2394.0 psi\n",
-        "The corresponding directions are\n",
-        "Theta1= 28.2 degrees and Theta2= 118.2 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.6, Page No:298"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "sigma_x=40 #Stress in x in MPa\n",
-      "sigma_y=-100 #Stress in y in MPa\n",
-      "tau_xy=-50 #Shear stress in MPa\n",
-      "\n",
-      "#Calculations\n",
-      "tau_max=sqrt(((sigma_x-sigma_y)*0.5)**2+tau_xy**2) #Maximum in-plane shear in MPa\n",
-      "\n",
-      "#Orientation of Plane\n",
-      "theta1=arctan(-((sigma_x-sigma_y)*(2*tau_xy)**-1))*180*pi**-1*0.5 #Angle in Degrees\n",
-      "theta2=theta1+90 #Angle in degrees\n",
-      "\n",
-      "#Plane of max in-plane shear\n",
-      "tau_x_y=-0.5*(sigma_x-sigma_y)*sin(2*theta1*pi*180**-1)+tau_xy*cos(2*theta1*pi*180**-1) \n",
-      "\n",
-      "#Normal Stress\n",
-      "sigma=(sigma_x+sigma_y)*0.5 #Stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum in-plane Shear is\",round(tau_x_y),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum in-plane Shear is -86.0 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.7, Page No:305"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Vairable Decleration\n",
-      "sigma_x=40 #Stress in x in MPa\n",
-      "sigma_y=20 #Stress in y in MPa\n",
-      "tau_xy=16 #Shear in xy in MPa\n",
-      "\n",
-      "#Calculations\n",
-      "sigma=(sigma_x+sigma_y)*0.5 #Normal Stress in MPa\n",
-      "R=sqrt(((sigma_x-sigma_y)*0.5)**2+tau_xy**2) #Resultant Stress in MPa\n",
-      "\n",
-      "#Part 1\n",
-      "sigma1=sigma+R #Principal Stress in MPa\n",
-      "sigma2=sigma-R #Principal Stress in MPa\n",
-      "theta=arctan(tau_xy*((sigma_x-sigma_y)*0.5)**-1)*180*pi**-1*0.5 #Orientation in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "tau_max=18.87 #From figure in MPa\n",
-      "\n",
-      "#Part 3\n",
-      "sigma_xdash=sigma+tau_max*cos((100-theta*2)*pi*180**-1) #Stress in MPa\n",
-      "sigma_ydash=sigma-tau_max*cos((100-theta*2)*pi*180**-1) #Stress in MPa\n",
-      "tau_x_y=tau_max*sin((100-2*theta)*pi*180**-1) #Shear stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The principal Stresses are\"\n",
-      "print \"Sigma1=\",round(sigma1,1),\"MPa and Sigma2=\",round(sigma2,1),\"MPa\"\n",
-      "print \"The Principal Plane is at\",round(theta),\"degrees\"\n",
-      "print \"The  Maximum Shear Stress is\",tau_max,\"MPa\"\n",
-      "print \"Sigma_x'=\",round(sigma_xdash),\"MPa and Sigma_y'=\",round(sigma_ydash,2),\"MPa\"\n",
-      "print \"Tau_x'y'=\",round(tau_x_y,2),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The principal Stresses are\n",
-        "Sigma1= 48.9 MPa and Sigma2= 11.1 MPa\n",
-        "The Principal Plane is at 29.0 degrees\n",
-        "The  Maximum Shear Stress is 18.87 MPa\n",
-        "Sigma_x'= 44.0 MPa and Sigma_y'= 15.98 MPa\n",
-        "Tau_x'y'= 12.63 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.9, Page No:316"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variabel Decleration\n",
-      "sigma_w=120 #Working Stress in MPa\n",
-      "tau_w=70 #Working Shear in MPa\n",
-      "\n",
-      "#Calcualtions\n",
-      "#Section a-a\n",
-      "M=3750 #Applied moment at section a-a in N.m\n",
-      "T=1500 #Applied Torque at section a-a in N.m\n",
-      "\n",
-      "#After carrying out the variable based computation we compute d\n",
-      "d1=((124.62)/(sigma_w*10**3*pi))**0.3333 #Diameter of the shaft in m\n",
-      "d2=((65.6)/(tau_w*10**3*pi))**0.3333 #Diameter of the shaft in m\n",
-      "d=max(d1,d2) #Diameter of the shaft to be selected in m\n",
-      "\n",
-      "#Result\n",
-      "print \"The diameter of the shaft to be selected is\",round(d*1000,1),\"mm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The diameter of the shaft to be selected is 69.2 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.10, Page No:318"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "t=0.01 #Thickness of the shaft in m\n",
-      "p=2 #Internal Pressure in MPa\n",
-      "r=0.45 #Mean radius of the vessel in m\n",
-      "tw=50 #Working shear stress in MPa\n",
-      "\n",
-      "#Calculation\n",
-      "sigma_x=(p*r)/(2*t) #Stress in MPa\n",
-      "sigma_y=(p*r)/t #Stress in MPa\n",
-      "\n",
-      "R=100-67.5 #From the diagram in MPa\n",
-      "tau_xy=sqrt((R**2-(sigma_y-67.5)**2)) #Stress in MPa\n",
-      "\n",
-      "J=2*pi*r**3*t #Polar Moment of inertia in mm^4\n",
-      "\n",
-      "T=1000*(tau_xy*J)/r #Maximum allowable Torque in kN.m\n",
-      "\n",
-      "#Result\n",
-      "print \"The largest allowable Torque is\",round(T),\"kN.m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The largest allowable Torque is 298.0 kN.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.11, Page No:320"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "L=15 #Length of the shaft in inches\n",
-      "r=3.0/8.001 #Radius of the shaft in inches\n",
-      "T=540 #Torque applied in lb.in\n",
-      "\n",
-      "#Calculations\n",
-      "V=30 #Transverse Shear Force in lb\n",
-      "M=15*V #Bending Moment in lb.in\n",
-      "I=(pi*r**4)/4.0 #Moment of Inertia in in^4\n",
-      "J=2*I #Polar Moment Of Inertia in in^4\n",
-      "\n",
-      "#Part 1\n",
-      "sigma=(M*r)/I #Bending Stress in psi\n",
-      "tau_t=10**-3*(T*r)/J #Shear Stress in ksi\n",
-      "\n",
-      "sigma_max1=13.92 #From the Mohr Circle in ksi\n",
-      "\n",
-      "#Part 2\n",
-      "Q=(2*r**3)/3.0 #First Moment in in^3\n",
-      "b=2*r # in\n",
-      "\n",
-      "tau_V=10**-3*(V*Q)/(I*b) #Shear Stress in ksi\n",
-      "tau=tau_t+tau_V #Total Shear in ksi\n",
-      "\n",
-      "sigma_max2=tau #Maximum stress in ksi\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum normal stress in case 1 is\",sigma_max1,\"ksi\"\n",
-      "print \"The Maximum normal stress in case 2 is\",round(sigma_max2,2),\"ksi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum normal stress in case 1 is 13.92 ksi\n",
-        "The Maximum normal stress in case 2 is 6.61 ksi\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.12, Page No:330"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "ex=800*10**-6 #Strain in x \n",
-      "ey=200*10**-6 #Strain in y\n",
-      "y_xy=-600*10**-6 #Strain in xy\n",
-      "\n",
-      "#Calculations\n",
-      "e_bar=(ex+ey)*0.5 #Strain \n",
-      "R=sqrt(2*300**2)*10**-6 #Resultant \n",
-      "\n",
-      "#Part 1\n",
-      "e1=e_bar+R #Strain in Principal Axis\n",
-      "e2=e_bar-R #Strain in Principal Axis\n",
-      "\n",
-      "#Part 2\n",
-      "alpha=15*180**-1*pi #From the Mohr Circle in degrees\n",
-      "e_xbar=e_bar-(R*cos(alpha)) #Strain in x \n",
-      "e_ybar=e_bar+(R*cos(alpha)) #Strain in y\n",
-      "\n",
-      "shear_strain=2*R*sin(alpha) #Shear starin \n",
-      "\n",
-      "#Result\n",
-      "print \"The principal Strains are\"\n",
-      "print \"e1=\",round(e1,6),\"e2=\",round(e2,6)\n",
-      "print \"The starin components are\"\n",
-      "print \"ex'=\",round(e_xbar,6),\"ey'=\",round(e_ybar,6),\"y_x'y'=\",round(shear_strain,6)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The principal Strains are\n",
-        "e1= 0.000924 e2= 7.6e-05\n",
-        "The starin components are\n",
-        "ex'= 9e-05 ey'= 0.00091 y_x'y'= 0.00022\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.13, Page No:331"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "e_x=800*10**-6 #Strain in x\n",
-      "e_y=200*10**-6 #Strain in y\n",
-      "y_xy=-600*10**-6 #Strain in xy\n",
-      "v=0.30 #Poissons Ratio\n",
-      "E=200 #Youngs Modulus in GPa\n",
-      "R_e=424.3*10**-6 #Strain\n",
-      "e_bar=500*10**-6 #Strain\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "R_sigma=10**-6*R_e*(E*10**9/(1+v)) #Stress in MPa\n",
-      "sigma_bar=10**-6*e_bar*(E*10**9/(1-v)) #Stress in MPa\n",
-      "\n",
-      "#Part 2\n",
-      "sigma1=sigma_bar+R_sigma #Principal Stress in MPa\n",
-      "sigma2=sigma_bar-R_sigma #Principal Stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The principal Stresses are as follows\"\n",
-      "print \"Sigma1=\",round(sigma1),\"MPa and Sigma2=\",round(sigma2,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The principal Stresses are as follows\n",
-        "Sigma1= 208.0 MPa and Sigma2= 77.6 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8.14, Page No:336"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "e_a=100*10**-6 #Strain \n",
-      "e_b=300*10**-6 #Strain\n",
-      "e_c=-200*10**-6 #Strain\n",
-      "E=180 #Youngs Modulus in GPa\n",
-      "v=0.28 #Poissons Ratio \n",
-      "\n",
-      "#Calculations\n",
-      "y_xy=(e_b-(e_a+e_c)*0.5) #Strain in xy\n",
-      "e_bar=(e_a+e_c)*0.5 #Strain \n",
-      "R_e=sqrt(y_xy**2+(150*10**-6)**2) #Resultant Strain\n",
-      "\n",
-      "#Corresponding Parameters from Mohrs Diagram\n",
-      "sigma_bar=(E/(1-v))*e_bar*10**3 #Stress in MPa\n",
-      "R_sigma=(E/(1+v))*R_e*10**3 #Resultant Stress in MPa\n",
-      "\n",
-      "#Principal Stresses\n",
-      "sigma1=sigma_bar+R_sigma #MPa\n",
-      "sigma2=sigma_bar-R_sigma #MPa\n",
-      "\n",
-      "theta=arctan(y_xy/(150*10**-6))*180*pi**-1*0.5 #Degrees\n",
-      "\n",
-      "#Result\n",
-      "print \"The Principal Stresses are as follows\"\n",
-      "print \"Sigma1=\",round(sigma1,1),\"MPa and Sigma2=\",round(sigma2,2),\"MPa\"\n",
-      "print \"The plane orientation is\",round(theta,2),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Principal Stresses are as follows\n",
-        "Sigma1= 41.0 MPa and Sigma2= -66.05 MPa\n",
-        "The plane orientation is 33.4 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter09.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter09.ipynb
deleted file mode 100755
index ac59e6fc..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter09.ipynb
+++ /dev/null
@@ -1,285 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:19ebf66d9da8e61964dee3081a6c3e2bb440c25ec82a0b4b2dc44b82d335cf92"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter09:Composite Beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.1, Page No:346"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "n=20 #Modular Ratio\n",
-      "sigma_wd=8*10**6 #Maximum bending stress in wood in Pa\n",
-      "sigma_st=120*10**6 #Maximum bending stress in steel in Pa\n",
-      "\n",
-      "#Cross Sectional Details\n",
-      "Awd=45 #Area of wood in mm^2\n",
-      "y_wd=160 #Neutral Axis of from bottom of the wooden section in mm\n",
-      "Ast=15 #Area of steel in mm^2\n",
-      "y_st=5 #Neutral Axis of the Steel section in mm\n",
-      "#Dimensions\n",
-      "ww=150 #width of wooden section in mm\n",
-      "dw=300 #depth of wooden section in mm\n",
-      "ws=75 #width of steel section in mm\n",
-      "ds=10 #depth of steel section in mm\n",
-      "\n",
-      "#Calculations\n",
-      "y_bar=(Awd*y_wd+Ast*y_st)*(Ast+Awd)**-1 #Location of Neutral axis from the bottom in mm\n",
-      "#Moment of inertia \n",
-      "I=(ww*dw**3*12**-1)+(ww*dw*(y_wd-y_bar)**2)+(n*ws*ds**3*12**-1)+(n*ws*ds*(y_bar-y_st)**2) #mm^4\n",
-      "c_top=dw+ds-y_bar #Distance from NA to top fibre in mm\n",
-      "c_bot=y_bar #Distance from NA to bottom fibre in mm\n",
-      "\n",
-      "#The solution will be in different order \n",
-      "M1=sigma_wd*I*10**-12*c_top**-1 #Maximum Bending Moment in N.m\n",
-      "M2=sigma_st*I*10**-12*c_bot**-1 #Maximum Bending Moment in N.m\n",
-      "M=min(M1,M2) #Maximum allowable moment in N.m\n",
-      "\n",
-      "#Result\n",
-      "print \"The Maximum Allowable moment that the beam can support is\",round(M,1),\"kN.m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Maximum Allowable moment that the beam can support is 25.8 kN.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.2, Page No:351"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "dw=8 #Depth of wooden section in inches\n",
-      "da=0.4 #Depth og aluminium section in inches \n",
-      "w=2 #Width of the section in inches\n",
-      "n=40*3**-1 #Modular Ratio\n",
-      "Ewd=1.5*10**6 #Youngs modulus of wood in psi\n",
-      "Eal=10**7 #Youngs Modulus of aluminium in psi\n",
-      "V_max=4000 #Maximum shear in lb\n",
-      "b=24 #Inches\n",
-      "L=72 #Length in inches\n",
-      "P=6000 #Load on the beam in lb\n",
-      "\n",
-      "#Calculations\n",
-      "I=w*dw**3*12**-1+2*(n*w*da**3*12**-1+n*da*4.2**2) #Moment of Inertia in in^4\n",
-      "\n",
-      "#Part 1\n",
-      "Q=(w*dw*0.5)*2+(n*da)*(dw*0.5+da*0.5) #First Moment in in^3\n",
-      "tau_max=V_max*Q*I**-1*w**-1 #Maximum Shear Stress in psi\n",
-      "\n",
-      "#Part 2\n",
-      "delta_mid=(P*b)*(48*Ewd*I)**-1*(3*L**2-4*b**2)\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum shear stress allowable is\",round(tau_max),\"psi\"\n",
-      "print \"The deflection at the mid-span is\",round(delta_mid,4),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum shear stress allowable is 281.0 psi\n",
-        "The deflection at the mid-span is 0.0968 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.3, Page No:356"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "b=300 #Breadth in mm\n",
-      "d=500 #Depth in mm\n",
-      "Ast=1500 #Area of steel in mm^2\n",
-      "n=8 #Modular Ratio\n",
-      "M=70*10**3 #Bending Moment in N.m\n",
-      "\n",
-      "#Calculations\n",
-      "#Let the LHS be C\n",
-      "C=2*n*Ast*b**-1*d**-1 #The LHS computation\n",
-      "h=np.roots([d**-2,C*d**-1,-C])\n",
-      "#Taking only real root\n",
-      "h=h[1] #mm\n",
-      "\n",
-      "sigma_co_max=(2*M)/(b*h*(d-h*3**-1)) #Maximum Compressive Stress in GPa\n",
-      "sigma_st_max=M/((d-h*3**-1)*Ast) #Maximum Stress in Steel in GPa\n",
-      "#Result\n",
-      "print \"The maximum stress in compression is\",round(sigma_co_max*10**3,2),\"MPa\"\n",
-      "print \"The maximum stress in streel is\",round(sigma_st_max*10**3,1),\"MPa\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum stress in compression is 6.39 MPa\n",
-        "The maximum stress in streel is 104.8 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.4, Page No:356"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "sigma_co_w=12 #Maximum stress in compression in MPa\n",
-      "sigma_st_w=140 #Maximum stress in steel in MPa\n",
-      "M=90 #Moment in kN.m\n",
-      "n=8 #Modular Ratio \n",
-      "\n",
-      "#Calculations\n",
-      "#h=0.4068d\n",
-      "#bd^2=0.04266\n",
-      "b=(0.04266/(1.5**2))**0.3333 #Breadth in m \n",
-      "d=1.5*b #Depth in m\n",
-      "h=0.4068*d #Height in m\n",
-      "\n",
-      "#Area of steel\n",
-      "Ast=((M*10**3)/((d-h*3**-1)*sigma_st_w*10**3))*10**3 #Area of steel in mm^2\n",
-      "\n",
-      "#Result\n",
-      "print \"The dimensions of the beam are\"\n",
-      "print \"b=\",round(b*1000),\"mm and d=\",round(d*1000),\"mm\"\n",
-      "print \"Area of steel=\",round(Ast),\"mm^2\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dimensions of the beam are\n",
-        "b= 267.0 mm and d= 400.0 mm\n",
-        "Area of steel= 1859.0 mm^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.5, Page No:357"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A1=75*10**3 #Area 1 in mm^2\n",
-      "A3=19.20*10**3 #Area 3 in m^2\n",
-      "w=750 #Width in mm\n",
-      "w1=350 #Width in mm\n",
-      "d=444.45 #Depth in mm\n",
-      "sigma_co_w=12*10**6 #Maximum Permissible Bending stress in concrete in Pa\n",
-      "sigma_st_w=140*10**6 #Maximum Permissible Bending stress in steel in Pa\n",
-      "n=8 #Modular Ratio\n",
-      "\n",
-      "#Calculations\n",
-      "#After simplfying the equation we get the following \n",
-      "H=np.roots([200,-200**2+A1+A3,-A1*50+100**2*200-600*A3])\n",
-      "h=max(H) #Depth of NA in mm\n",
-      "#Moment Of Inertia\n",
-      "I=w*h**3*3**-1-(w1*(h-100)**3*3**-1)+A3*d**2 #Moment of inertia in mm^4\n",
-      "\n",
-      "M1=sigma_co_w*I*h**-1*(10**-3)**4*10**3 #Largest Bending Moment in concrete in N.m\n",
-      "M2=sigma_st_w*I*(n*d)**-1*(10**-3)**4*10**3 #Largest Bending Moment in Steel in N.m\n",
-      "M=min(M1,M2) #Largest Bending Moment that can be supported safely in N.m\n",
-      "#Result\n",
-      "print \"The largest Bending Moment that can be supported is\",round(M*10**-3,1),\"kN.m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The largest Bending Moment that can be supported is 185.6 kN.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter09_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter09_1.ipynb
deleted file mode 100755
index ac59e6fc..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter09_1.ipynb
+++ /dev/null
@@ -1,285 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:19ebf66d9da8e61964dee3081a6c3e2bb440c25ec82a0b4b2dc44b82d335cf92"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter09:Composite Beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.1, Page No:346"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "n=20 #Modular Ratio\n",
-      "sigma_wd=8*10**6 #Maximum bending stress in wood in Pa\n",
-      "sigma_st=120*10**6 #Maximum bending stress in steel in Pa\n",
-      "\n",
-      "#Cross Sectional Details\n",
-      "Awd=45 #Area of wood in mm^2\n",
-      "y_wd=160 #Neutral Axis of from bottom of the wooden section in mm\n",
-      "Ast=15 #Area of steel in mm^2\n",
-      "y_st=5 #Neutral Axis of the Steel section in mm\n",
-      "#Dimensions\n",
-      "ww=150 #width of wooden section in mm\n",
-      "dw=300 #depth of wooden section in mm\n",
-      "ws=75 #width of steel section in mm\n",
-      "ds=10 #depth of steel section in mm\n",
-      "\n",
-      "#Calculations\n",
-      "y_bar=(Awd*y_wd+Ast*y_st)*(Ast+Awd)**-1 #Location of Neutral axis from the bottom in mm\n",
-      "#Moment of inertia \n",
-      "I=(ww*dw**3*12**-1)+(ww*dw*(y_wd-y_bar)**2)+(n*ws*ds**3*12**-1)+(n*ws*ds*(y_bar-y_st)**2) #mm^4\n",
-      "c_top=dw+ds-y_bar #Distance from NA to top fibre in mm\n",
-      "c_bot=y_bar #Distance from NA to bottom fibre in mm\n",
-      "\n",
-      "#The solution will be in different order \n",
-      "M1=sigma_wd*I*10**-12*c_top**-1 #Maximum Bending Moment in N.m\n",
-      "M2=sigma_st*I*10**-12*c_bot**-1 #Maximum Bending Moment in N.m\n",
-      "M=min(M1,M2) #Maximum allowable moment in N.m\n",
-      "\n",
-      "#Result\n",
-      "print \"The Maximum Allowable moment that the beam can support is\",round(M,1),\"kN.m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Maximum Allowable moment that the beam can support is 25.8 kN.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.2, Page No:351"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "dw=8 #Depth of wooden section in inches\n",
-      "da=0.4 #Depth og aluminium section in inches \n",
-      "w=2 #Width of the section in inches\n",
-      "n=40*3**-1 #Modular Ratio\n",
-      "Ewd=1.5*10**6 #Youngs modulus of wood in psi\n",
-      "Eal=10**7 #Youngs Modulus of aluminium in psi\n",
-      "V_max=4000 #Maximum shear in lb\n",
-      "b=24 #Inches\n",
-      "L=72 #Length in inches\n",
-      "P=6000 #Load on the beam in lb\n",
-      "\n",
-      "#Calculations\n",
-      "I=w*dw**3*12**-1+2*(n*w*da**3*12**-1+n*da*4.2**2) #Moment of Inertia in in^4\n",
-      "\n",
-      "#Part 1\n",
-      "Q=(w*dw*0.5)*2+(n*da)*(dw*0.5+da*0.5) #First Moment in in^3\n",
-      "tau_max=V_max*Q*I**-1*w**-1 #Maximum Shear Stress in psi\n",
-      "\n",
-      "#Part 2\n",
-      "delta_mid=(P*b)*(48*Ewd*I)**-1*(3*L**2-4*b**2)\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum shear stress allowable is\",round(tau_max),\"psi\"\n",
-      "print \"The deflection at the mid-span is\",round(delta_mid,4),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum shear stress allowable is 281.0 psi\n",
-        "The deflection at the mid-span is 0.0968 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.3, Page No:356"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "b=300 #Breadth in mm\n",
-      "d=500 #Depth in mm\n",
-      "Ast=1500 #Area of steel in mm^2\n",
-      "n=8 #Modular Ratio\n",
-      "M=70*10**3 #Bending Moment in N.m\n",
-      "\n",
-      "#Calculations\n",
-      "#Let the LHS be C\n",
-      "C=2*n*Ast*b**-1*d**-1 #The LHS computation\n",
-      "h=np.roots([d**-2,C*d**-1,-C])\n",
-      "#Taking only real root\n",
-      "h=h[1] #mm\n",
-      "\n",
-      "sigma_co_max=(2*M)/(b*h*(d-h*3**-1)) #Maximum Compressive Stress in GPa\n",
-      "sigma_st_max=M/((d-h*3**-1)*Ast) #Maximum Stress in Steel in GPa\n",
-      "#Result\n",
-      "print \"The maximum stress in compression is\",round(sigma_co_max*10**3,2),\"MPa\"\n",
-      "print \"The maximum stress in streel is\",round(sigma_st_max*10**3,1),\"MPa\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum stress in compression is 6.39 MPa\n",
-        "The maximum stress in streel is 104.8 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.4, Page No:356"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "sigma_co_w=12 #Maximum stress in compression in MPa\n",
-      "sigma_st_w=140 #Maximum stress in steel in MPa\n",
-      "M=90 #Moment in kN.m\n",
-      "n=8 #Modular Ratio \n",
-      "\n",
-      "#Calculations\n",
-      "#h=0.4068d\n",
-      "#bd^2=0.04266\n",
-      "b=(0.04266/(1.5**2))**0.3333 #Breadth in m \n",
-      "d=1.5*b #Depth in m\n",
-      "h=0.4068*d #Height in m\n",
-      "\n",
-      "#Area of steel\n",
-      "Ast=((M*10**3)/((d-h*3**-1)*sigma_st_w*10**3))*10**3 #Area of steel in mm^2\n",
-      "\n",
-      "#Result\n",
-      "print \"The dimensions of the beam are\"\n",
-      "print \"b=\",round(b*1000),\"mm and d=\",round(d*1000),\"mm\"\n",
-      "print \"Area of steel=\",round(Ast),\"mm^2\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dimensions of the beam are\n",
-        "b= 267.0 mm and d= 400.0 mm\n",
-        "Area of steel= 1859.0 mm^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.5, Page No:357"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A1=75*10**3 #Area 1 in mm^2\n",
-      "A3=19.20*10**3 #Area 3 in m^2\n",
-      "w=750 #Width in mm\n",
-      "w1=350 #Width in mm\n",
-      "d=444.45 #Depth in mm\n",
-      "sigma_co_w=12*10**6 #Maximum Permissible Bending stress in concrete in Pa\n",
-      "sigma_st_w=140*10**6 #Maximum Permissible Bending stress in steel in Pa\n",
-      "n=8 #Modular Ratio\n",
-      "\n",
-      "#Calculations\n",
-      "#After simplfying the equation we get the following \n",
-      "H=np.roots([200,-200**2+A1+A3,-A1*50+100**2*200-600*A3])\n",
-      "h=max(H) #Depth of NA in mm\n",
-      "#Moment Of Inertia\n",
-      "I=w*h**3*3**-1-(w1*(h-100)**3*3**-1)+A3*d**2 #Moment of inertia in mm^4\n",
-      "\n",
-      "M1=sigma_co_w*I*h**-1*(10**-3)**4*10**3 #Largest Bending Moment in concrete in N.m\n",
-      "M2=sigma_st_w*I*(n*d)**-1*(10**-3)**4*10**3 #Largest Bending Moment in Steel in N.m\n",
-      "M=min(M1,M2) #Largest Bending Moment that can be supported safely in N.m\n",
-      "#Result\n",
-      "print \"The largest Bending Moment that can be supported is\",round(M*10**-3,1),\"kN.m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The largest Bending Moment that can be supported is 185.6 kN.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter09_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter09_2.ipynb
deleted file mode 100755
index ac59e6fc..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter09_2.ipynb
+++ /dev/null
@@ -1,285 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:19ebf66d9da8e61964dee3081a6c3e2bb440c25ec82a0b4b2dc44b82d335cf92"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter09:Composite Beams"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.1, Page No:346"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "n=20 #Modular Ratio\n",
-      "sigma_wd=8*10**6 #Maximum bending stress in wood in Pa\n",
-      "sigma_st=120*10**6 #Maximum bending stress in steel in Pa\n",
-      "\n",
-      "#Cross Sectional Details\n",
-      "Awd=45 #Area of wood in mm^2\n",
-      "y_wd=160 #Neutral Axis of from bottom of the wooden section in mm\n",
-      "Ast=15 #Area of steel in mm^2\n",
-      "y_st=5 #Neutral Axis of the Steel section in mm\n",
-      "#Dimensions\n",
-      "ww=150 #width of wooden section in mm\n",
-      "dw=300 #depth of wooden section in mm\n",
-      "ws=75 #width of steel section in mm\n",
-      "ds=10 #depth of steel section in mm\n",
-      "\n",
-      "#Calculations\n",
-      "y_bar=(Awd*y_wd+Ast*y_st)*(Ast+Awd)**-1 #Location of Neutral axis from the bottom in mm\n",
-      "#Moment of inertia \n",
-      "I=(ww*dw**3*12**-1)+(ww*dw*(y_wd-y_bar)**2)+(n*ws*ds**3*12**-1)+(n*ws*ds*(y_bar-y_st)**2) #mm^4\n",
-      "c_top=dw+ds-y_bar #Distance from NA to top fibre in mm\n",
-      "c_bot=y_bar #Distance from NA to bottom fibre in mm\n",
-      "\n",
-      "#The solution will be in different order \n",
-      "M1=sigma_wd*I*10**-12*c_top**-1 #Maximum Bending Moment in N.m\n",
-      "M2=sigma_st*I*10**-12*c_bot**-1 #Maximum Bending Moment in N.m\n",
-      "M=min(M1,M2) #Maximum allowable moment in N.m\n",
-      "\n",
-      "#Result\n",
-      "print \"The Maximum Allowable moment that the beam can support is\",round(M,1),\"kN.m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Maximum Allowable moment that the beam can support is 25.8 kN.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.2, Page No:351"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "dw=8 #Depth of wooden section in inches\n",
-      "da=0.4 #Depth og aluminium section in inches \n",
-      "w=2 #Width of the section in inches\n",
-      "n=40*3**-1 #Modular Ratio\n",
-      "Ewd=1.5*10**6 #Youngs modulus of wood in psi\n",
-      "Eal=10**7 #Youngs Modulus of aluminium in psi\n",
-      "V_max=4000 #Maximum shear in lb\n",
-      "b=24 #Inches\n",
-      "L=72 #Length in inches\n",
-      "P=6000 #Load on the beam in lb\n",
-      "\n",
-      "#Calculations\n",
-      "I=w*dw**3*12**-1+2*(n*w*da**3*12**-1+n*da*4.2**2) #Moment of Inertia in in^4\n",
-      "\n",
-      "#Part 1\n",
-      "Q=(w*dw*0.5)*2+(n*da)*(dw*0.5+da*0.5) #First Moment in in^3\n",
-      "tau_max=V_max*Q*I**-1*w**-1 #Maximum Shear Stress in psi\n",
-      "\n",
-      "#Part 2\n",
-      "delta_mid=(P*b)*(48*Ewd*I)**-1*(3*L**2-4*b**2)\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum shear stress allowable is\",round(tau_max),\"psi\"\n",
-      "print \"The deflection at the mid-span is\",round(delta_mid,4),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum shear stress allowable is 281.0 psi\n",
-        "The deflection at the mid-span is 0.0968 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.3, Page No:356"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "b=300 #Breadth in mm\n",
-      "d=500 #Depth in mm\n",
-      "Ast=1500 #Area of steel in mm^2\n",
-      "n=8 #Modular Ratio\n",
-      "M=70*10**3 #Bending Moment in N.m\n",
-      "\n",
-      "#Calculations\n",
-      "#Let the LHS be C\n",
-      "C=2*n*Ast*b**-1*d**-1 #The LHS computation\n",
-      "h=np.roots([d**-2,C*d**-1,-C])\n",
-      "#Taking only real root\n",
-      "h=h[1] #mm\n",
-      "\n",
-      "sigma_co_max=(2*M)/(b*h*(d-h*3**-1)) #Maximum Compressive Stress in GPa\n",
-      "sigma_st_max=M/((d-h*3**-1)*Ast) #Maximum Stress in Steel in GPa\n",
-      "#Result\n",
-      "print \"The maximum stress in compression is\",round(sigma_co_max*10**3,2),\"MPa\"\n",
-      "print \"The maximum stress in streel is\",round(sigma_st_max*10**3,1),\"MPa\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum stress in compression is 6.39 MPa\n",
-        "The maximum stress in streel is 104.8 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.4, Page No:356"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "sigma_co_w=12 #Maximum stress in compression in MPa\n",
-      "sigma_st_w=140 #Maximum stress in steel in MPa\n",
-      "M=90 #Moment in kN.m\n",
-      "n=8 #Modular Ratio \n",
-      "\n",
-      "#Calculations\n",
-      "#h=0.4068d\n",
-      "#bd^2=0.04266\n",
-      "b=(0.04266/(1.5**2))**0.3333 #Breadth in m \n",
-      "d=1.5*b #Depth in m\n",
-      "h=0.4068*d #Height in m\n",
-      "\n",
-      "#Area of steel\n",
-      "Ast=((M*10**3)/((d-h*3**-1)*sigma_st_w*10**3))*10**3 #Area of steel in mm^2\n",
-      "\n",
-      "#Result\n",
-      "print \"The dimensions of the beam are\"\n",
-      "print \"b=\",round(b*1000),\"mm and d=\",round(d*1000),\"mm\"\n",
-      "print \"Area of steel=\",round(Ast),\"mm^2\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dimensions of the beam are\n",
-        "b= 267.0 mm and d= 400.0 mm\n",
-        "Area of steel= 1859.0 mm^2\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.9.5, Page No:357"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A1=75*10**3 #Area 1 in mm^2\n",
-      "A3=19.20*10**3 #Area 3 in m^2\n",
-      "w=750 #Width in mm\n",
-      "w1=350 #Width in mm\n",
-      "d=444.45 #Depth in mm\n",
-      "sigma_co_w=12*10**6 #Maximum Permissible Bending stress in concrete in Pa\n",
-      "sigma_st_w=140*10**6 #Maximum Permissible Bending stress in steel in Pa\n",
-      "n=8 #Modular Ratio\n",
-      "\n",
-      "#Calculations\n",
-      "#After simplfying the equation we get the following \n",
-      "H=np.roots([200,-200**2+A1+A3,-A1*50+100**2*200-600*A3])\n",
-      "h=max(H) #Depth of NA in mm\n",
-      "#Moment Of Inertia\n",
-      "I=w*h**3*3**-1-(w1*(h-100)**3*3**-1)+A3*d**2 #Moment of inertia in mm^4\n",
-      "\n",
-      "M1=sigma_co_w*I*h**-1*(10**-3)**4*10**3 #Largest Bending Moment in concrete in N.m\n",
-      "M2=sigma_st_w*I*(n*d)**-1*(10**-3)**4*10**3 #Largest Bending Moment in Steel in N.m\n",
-      "M=min(M1,M2) #Largest Bending Moment that can be supported safely in N.m\n",
-      "#Result\n",
-      "print \"The largest Bending Moment that can be supported is\",round(M*10**-3,1),\"kN.m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The largest Bending Moment that can be supported is 185.6 kN.m\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter10.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter10.ipynb
deleted file mode 100755
index 913e9f20..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter10.ipynb
+++ /dev/null
@@ -1,258 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:39be4bd5a29fce6d0d8085f52939f9fac01e348dbfffcb5b53194aa4be7d5f98"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter10:Columns"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10.1, Page No:369"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Le=7 #Effective length in m\n",
-      "P=450 #Applied axial Load in kN\n",
-      "FOS=3 #Factor of safety \n",
-      "sigma_pl=200*10**6 #Stress allowable in Pa\n",
-      "E=200*10**9 #Youngs Modulus in Pa\n",
-      "end_cond=0.7 #End Condition factor to be multiplied\n",
-      "\n",
-      "#Calculations\n",
-      "Pcr=P*FOS #Critical Load in kN\n",
-      "A=Pcr*sigma_pl**-1*10**9 #Area in mm^2\n",
-      "\n",
-      "#Part 1\n",
-      "I1=10**15*(Pcr*Le**2)*(pi**2*E)**-1 #Moment of Inertia Required in mm^4\n",
-      "#From table selecting appropriate Section W250x73\n",
-      "\n",
-      "#Part 2\n",
-      "I2=10**15*(Pcr*end_cond**2*Le**2)*(pi**2*E)**-1 #Moment of Inertia Required in mm^4\n",
-      "#From table selecting appropriate Section W200x52\n",
-      "\n",
-      "#Lightest Section that meets these criterion is W250x58 section\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"From the above computation we select W250x58 section\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "From the above computation we select W250x58 section\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10.2, Page No:375"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variabel Decleration\n",
-      "E=200*10**9 #Youngs Modulus in Pa\n",
-      "sigma_yp=380*10**6 #Stress allowable in Pa\n",
-      "Le=10 #Length in m\n",
-      "end_cond=0.5 #Support condition factor to be ,ultiplied to length\n",
-      "A=15.5*10**-3 #Area in m^2\n",
-      "\n",
-      "#Calculations\n",
-      "Cc=sqrt((2*pi**2*E)*sigma_yp**-1) #Slenderness Ratio\n",
-      "\n",
-      "#Part 1\n",
-      "S_R1=142.9 #Slenderness ratio \n",
-      "sigma_w=(12*pi**2*E)/(23*S_R1**2) #Allowable Working Stress in Pa\n",
-      "P=sigma_w*A #Maximum Allowable Load in kN\n",
-      "\n",
-      "#Part 2\n",
-      "S_R2=79.37 #Slenderness ratio \n",
-      "N=5*3**-1+((3*S_R2)/(8*Cc))-(S_R2**3*(8*Cc**3)**-1) #Factor Of Safety\n",
-      "\n",
-      "sigma_w2=(1-(S_R2**2*0.5*Cc**-2))*(sigma_yp*N**-1) #Allowable working Stress in Pa\n",
-      "P2=sigma_w2*A #Allowable Load in kN\n",
-      "\n",
-      "#Part 3\n",
-      "S_R3=55.56 #Slenderness Ratio\n",
-      "N3=5*3**-1+((3*S_R3)/(8*Cc))-(S_R3**3*(8*Cc**3)**-1) #Factor Of Safety\n",
-      "\n",
-      "sigma_w3=(1-(S_R3**2*0.5*Cc**-2))*(sigma_yp*N3**-1) #Allowable working Stress in Pa\n",
-      "P3=sigma_w3*A #Allowable Load in kN\n",
-      "\n",
-      "#Result\n",
-      "print \"The results for Part 1 are\"\n",
-      "print \"Maximum Allowable Load P=\",round(P*10**-3),\"kN\"\n",
-      "print \"Part 2\"\n",
-      "print \"Maximum Allowable Load P=\",round(P2*10**-3),\"kN\"\n",
-      "print \"Part 3\"\n",
-      "print \"Maximum Allowable Load P=\",round(P3*10**-3),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The results for Part 1 are\n",
-        "Maximum Allowable Load P= 782.0 kN\n",
-        "Part 2\n",
-        "Maximum Allowable Load P= 2161.0 kN\n",
-        "Part 3\n",
-        "Maximum Allowable Load P= 2710.0 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10.3, Page No:383"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variabel Decleration\n",
-      "E=29*10**6 #Youngs Modulus in psi\n",
-      "sigma_yp=36*10**3 #Stress in psi\n",
-      "L=25 #Length in ft\n",
-      "A=17.9 #Area in in^2\n",
-      "Iz=640 #Moment of inertia in in^4\n",
-      "Sz=92.2 #Sectional Modulus in in^3\n",
-      "P=150*10**3 #Load in lb\n",
-      "e=4 #Eccentricity in inches\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Part 1\n",
-      "a=1.09836\n",
-      "sigma_max=P*A**-1+(P*e*Sz**-1)*a #Maximum Stress in psi\n",
-      "\n",
-      "#Part 2\n",
-      "#After simplification we get the equation to compute N\n",
-      "N=2.19 #Trial and Error yields\n",
-      "\n",
-      "#Part 3\n",
-      "v_max=e*((np.cos(sqrt((P*L**2*12**2)*(4*E*Iz)**-1)))**-1-1)\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum compressive stress in the Column is\",round(sigma_max,2),\"psi\"\n",
-      "print \"The factor of safety is\",N\n",
-      "print \"The maximum lateral dfelection is\",round(v_max,3),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum compressive stress in the Column is 15527.57 psi\n",
-        "The factor of safety is 2.19\n",
-        "The maximum lateral dfelection is 0.393 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10.4, Page No:384"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Le=7 #Effective Length in m\n",
-      "N=2 #Factor of Safety\n",
-      "h_max=400 #Maximum depth in mm\n",
-      "E=200 #Youngs Modulus in GPa\n",
-      "sigma_yp=250 #Maximum stress in yielding in MPa\n",
-      "P1=400 #Load 1 in kN\n",
-      "P2=900 #Load 2 in kN\n",
-      "x1=75 #Distance in mm\n",
-      "x2=125 #Distance in mm\n",
-      "\n",
-      "#Calculations\n",
-      "e=(P2*x2-P1*x1)*(P1+P2)**-1 #Eccentricity in mm\n",
-      "P=N*(P1+P2) #Applied load after factor of safety is considered in kN\n",
-      "\n",
-      "#Part 1 is not computable\n",
-      "I=415*10**-6 #Moment of inertia from the table in mm^4\n",
-      "\n",
-      "#Part 2\n",
-      "P_cr=pi**2*E*10**9*I*Le**-2 #Critical load for buckling in kN\n",
-      "FOS=P_cr*10**-3/(P1+P2) #Factor of safety against buckling in y-axis\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"The critical load for buckling is\",round(P_cr*10**-3),\"kN\"\n",
-      "print \"The factor of safety is\",round(FOS,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The critical load for buckling is 16718.0 kN\n",
-        "The factor of safety is 12.9\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter10_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter10_1.ipynb
deleted file mode 100755
index 913e9f20..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter10_1.ipynb
+++ /dev/null
@@ -1,258 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:39be4bd5a29fce6d0d8085f52939f9fac01e348dbfffcb5b53194aa4be7d5f98"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter10:Columns"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10.1, Page No:369"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Le=7 #Effective length in m\n",
-      "P=450 #Applied axial Load in kN\n",
-      "FOS=3 #Factor of safety \n",
-      "sigma_pl=200*10**6 #Stress allowable in Pa\n",
-      "E=200*10**9 #Youngs Modulus in Pa\n",
-      "end_cond=0.7 #End Condition factor to be multiplied\n",
-      "\n",
-      "#Calculations\n",
-      "Pcr=P*FOS #Critical Load in kN\n",
-      "A=Pcr*sigma_pl**-1*10**9 #Area in mm^2\n",
-      "\n",
-      "#Part 1\n",
-      "I1=10**15*(Pcr*Le**2)*(pi**2*E)**-1 #Moment of Inertia Required in mm^4\n",
-      "#From table selecting appropriate Section W250x73\n",
-      "\n",
-      "#Part 2\n",
-      "I2=10**15*(Pcr*end_cond**2*Le**2)*(pi**2*E)**-1 #Moment of Inertia Required in mm^4\n",
-      "#From table selecting appropriate Section W200x52\n",
-      "\n",
-      "#Lightest Section that meets these criterion is W250x58 section\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"From the above computation we select W250x58 section\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "From the above computation we select W250x58 section\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10.2, Page No:375"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variabel Decleration\n",
-      "E=200*10**9 #Youngs Modulus in Pa\n",
-      "sigma_yp=380*10**6 #Stress allowable in Pa\n",
-      "Le=10 #Length in m\n",
-      "end_cond=0.5 #Support condition factor to be ,ultiplied to length\n",
-      "A=15.5*10**-3 #Area in m^2\n",
-      "\n",
-      "#Calculations\n",
-      "Cc=sqrt((2*pi**2*E)*sigma_yp**-1) #Slenderness Ratio\n",
-      "\n",
-      "#Part 1\n",
-      "S_R1=142.9 #Slenderness ratio \n",
-      "sigma_w=(12*pi**2*E)/(23*S_R1**2) #Allowable Working Stress in Pa\n",
-      "P=sigma_w*A #Maximum Allowable Load in kN\n",
-      "\n",
-      "#Part 2\n",
-      "S_R2=79.37 #Slenderness ratio \n",
-      "N=5*3**-1+((3*S_R2)/(8*Cc))-(S_R2**3*(8*Cc**3)**-1) #Factor Of Safety\n",
-      "\n",
-      "sigma_w2=(1-(S_R2**2*0.5*Cc**-2))*(sigma_yp*N**-1) #Allowable working Stress in Pa\n",
-      "P2=sigma_w2*A #Allowable Load in kN\n",
-      "\n",
-      "#Part 3\n",
-      "S_R3=55.56 #Slenderness Ratio\n",
-      "N3=5*3**-1+((3*S_R3)/(8*Cc))-(S_R3**3*(8*Cc**3)**-1) #Factor Of Safety\n",
-      "\n",
-      "sigma_w3=(1-(S_R3**2*0.5*Cc**-2))*(sigma_yp*N3**-1) #Allowable working Stress in Pa\n",
-      "P3=sigma_w3*A #Allowable Load in kN\n",
-      "\n",
-      "#Result\n",
-      "print \"The results for Part 1 are\"\n",
-      "print \"Maximum Allowable Load P=\",round(P*10**-3),\"kN\"\n",
-      "print \"Part 2\"\n",
-      "print \"Maximum Allowable Load P=\",round(P2*10**-3),\"kN\"\n",
-      "print \"Part 3\"\n",
-      "print \"Maximum Allowable Load P=\",round(P3*10**-3),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The results for Part 1 are\n",
-        "Maximum Allowable Load P= 782.0 kN\n",
-        "Part 2\n",
-        "Maximum Allowable Load P= 2161.0 kN\n",
-        "Part 3\n",
-        "Maximum Allowable Load P= 2710.0 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10.3, Page No:383"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variabel Decleration\n",
-      "E=29*10**6 #Youngs Modulus in psi\n",
-      "sigma_yp=36*10**3 #Stress in psi\n",
-      "L=25 #Length in ft\n",
-      "A=17.9 #Area in in^2\n",
-      "Iz=640 #Moment of inertia in in^4\n",
-      "Sz=92.2 #Sectional Modulus in in^3\n",
-      "P=150*10**3 #Load in lb\n",
-      "e=4 #Eccentricity in inches\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Part 1\n",
-      "a=1.09836\n",
-      "sigma_max=P*A**-1+(P*e*Sz**-1)*a #Maximum Stress in psi\n",
-      "\n",
-      "#Part 2\n",
-      "#After simplification we get the equation to compute N\n",
-      "N=2.19 #Trial and Error yields\n",
-      "\n",
-      "#Part 3\n",
-      "v_max=e*((np.cos(sqrt((P*L**2*12**2)*(4*E*Iz)**-1)))**-1-1)\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum compressive stress in the Column is\",round(sigma_max,2),\"psi\"\n",
-      "print \"The factor of safety is\",N\n",
-      "print \"The maximum lateral dfelection is\",round(v_max,3),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum compressive stress in the Column is 15527.57 psi\n",
-        "The factor of safety is 2.19\n",
-        "The maximum lateral dfelection is 0.393 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10.4, Page No:384"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Le=7 #Effective Length in m\n",
-      "N=2 #Factor of Safety\n",
-      "h_max=400 #Maximum depth in mm\n",
-      "E=200 #Youngs Modulus in GPa\n",
-      "sigma_yp=250 #Maximum stress in yielding in MPa\n",
-      "P1=400 #Load 1 in kN\n",
-      "P2=900 #Load 2 in kN\n",
-      "x1=75 #Distance in mm\n",
-      "x2=125 #Distance in mm\n",
-      "\n",
-      "#Calculations\n",
-      "e=(P2*x2-P1*x1)*(P1+P2)**-1 #Eccentricity in mm\n",
-      "P=N*(P1+P2) #Applied load after factor of safety is considered in kN\n",
-      "\n",
-      "#Part 1 is not computable\n",
-      "I=415*10**-6 #Moment of inertia from the table in mm^4\n",
-      "\n",
-      "#Part 2\n",
-      "P_cr=pi**2*E*10**9*I*Le**-2 #Critical load for buckling in kN\n",
-      "FOS=P_cr*10**-3/(P1+P2) #Factor of safety against buckling in y-axis\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"The critical load for buckling is\",round(P_cr*10**-3),\"kN\"\n",
-      "print \"The factor of safety is\",round(FOS,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The critical load for buckling is 16718.0 kN\n",
-        "The factor of safety is 12.9\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter10_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter10_2.ipynb
deleted file mode 100755
index 913e9f20..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter10_2.ipynb
+++ /dev/null
@@ -1,258 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:39be4bd5a29fce6d0d8085f52939f9fac01e348dbfffcb5b53194aa4be7d5f98"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter10:Columns"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10.1, Page No:369"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Le=7 #Effective length in m\n",
-      "P=450 #Applied axial Load in kN\n",
-      "FOS=3 #Factor of safety \n",
-      "sigma_pl=200*10**6 #Stress allowable in Pa\n",
-      "E=200*10**9 #Youngs Modulus in Pa\n",
-      "end_cond=0.7 #End Condition factor to be multiplied\n",
-      "\n",
-      "#Calculations\n",
-      "Pcr=P*FOS #Critical Load in kN\n",
-      "A=Pcr*sigma_pl**-1*10**9 #Area in mm^2\n",
-      "\n",
-      "#Part 1\n",
-      "I1=10**15*(Pcr*Le**2)*(pi**2*E)**-1 #Moment of Inertia Required in mm^4\n",
-      "#From table selecting appropriate Section W250x73\n",
-      "\n",
-      "#Part 2\n",
-      "I2=10**15*(Pcr*end_cond**2*Le**2)*(pi**2*E)**-1 #Moment of Inertia Required in mm^4\n",
-      "#From table selecting appropriate Section W200x52\n",
-      "\n",
-      "#Lightest Section that meets these criterion is W250x58 section\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"From the above computation we select W250x58 section\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "From the above computation we select W250x58 section\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10.2, Page No:375"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variabel Decleration\n",
-      "E=200*10**9 #Youngs Modulus in Pa\n",
-      "sigma_yp=380*10**6 #Stress allowable in Pa\n",
-      "Le=10 #Length in m\n",
-      "end_cond=0.5 #Support condition factor to be ,ultiplied to length\n",
-      "A=15.5*10**-3 #Area in m^2\n",
-      "\n",
-      "#Calculations\n",
-      "Cc=sqrt((2*pi**2*E)*sigma_yp**-1) #Slenderness Ratio\n",
-      "\n",
-      "#Part 1\n",
-      "S_R1=142.9 #Slenderness ratio \n",
-      "sigma_w=(12*pi**2*E)/(23*S_R1**2) #Allowable Working Stress in Pa\n",
-      "P=sigma_w*A #Maximum Allowable Load in kN\n",
-      "\n",
-      "#Part 2\n",
-      "S_R2=79.37 #Slenderness ratio \n",
-      "N=5*3**-1+((3*S_R2)/(8*Cc))-(S_R2**3*(8*Cc**3)**-1) #Factor Of Safety\n",
-      "\n",
-      "sigma_w2=(1-(S_R2**2*0.5*Cc**-2))*(sigma_yp*N**-1) #Allowable working Stress in Pa\n",
-      "P2=sigma_w2*A #Allowable Load in kN\n",
-      "\n",
-      "#Part 3\n",
-      "S_R3=55.56 #Slenderness Ratio\n",
-      "N3=5*3**-1+((3*S_R3)/(8*Cc))-(S_R3**3*(8*Cc**3)**-1) #Factor Of Safety\n",
-      "\n",
-      "sigma_w3=(1-(S_R3**2*0.5*Cc**-2))*(sigma_yp*N3**-1) #Allowable working Stress in Pa\n",
-      "P3=sigma_w3*A #Allowable Load in kN\n",
-      "\n",
-      "#Result\n",
-      "print \"The results for Part 1 are\"\n",
-      "print \"Maximum Allowable Load P=\",round(P*10**-3),\"kN\"\n",
-      "print \"Part 2\"\n",
-      "print \"Maximum Allowable Load P=\",round(P2*10**-3),\"kN\"\n",
-      "print \"Part 3\"\n",
-      "print \"Maximum Allowable Load P=\",round(P3*10**-3),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The results for Part 1 are\n",
-        "Maximum Allowable Load P= 782.0 kN\n",
-        "Part 2\n",
-        "Maximum Allowable Load P= 2161.0 kN\n",
-        "Part 3\n",
-        "Maximum Allowable Load P= 2710.0 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10.3, Page No:383"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variabel Decleration\n",
-      "E=29*10**6 #Youngs Modulus in psi\n",
-      "sigma_yp=36*10**3 #Stress in psi\n",
-      "L=25 #Length in ft\n",
-      "A=17.9 #Area in in^2\n",
-      "Iz=640 #Moment of inertia in in^4\n",
-      "Sz=92.2 #Sectional Modulus in in^3\n",
-      "P=150*10**3 #Load in lb\n",
-      "e=4 #Eccentricity in inches\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Part 1\n",
-      "a=1.09836\n",
-      "sigma_max=P*A**-1+(P*e*Sz**-1)*a #Maximum Stress in psi\n",
-      "\n",
-      "#Part 2\n",
-      "#After simplification we get the equation to compute N\n",
-      "N=2.19 #Trial and Error yields\n",
-      "\n",
-      "#Part 3\n",
-      "v_max=e*((np.cos(sqrt((P*L**2*12**2)*(4*E*Iz)**-1)))**-1-1)\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum compressive stress in the Column is\",round(sigma_max,2),\"psi\"\n",
-      "print \"The factor of safety is\",N\n",
-      "print \"The maximum lateral dfelection is\",round(v_max,3),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum compressive stress in the Column is 15527.57 psi\n",
-        "The factor of safety is 2.19\n",
-        "The maximum lateral dfelection is 0.393 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10.4, Page No:384"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "Le=7 #Effective Length in m\n",
-      "N=2 #Factor of Safety\n",
-      "h_max=400 #Maximum depth in mm\n",
-      "E=200 #Youngs Modulus in GPa\n",
-      "sigma_yp=250 #Maximum stress in yielding in MPa\n",
-      "P1=400 #Load 1 in kN\n",
-      "P2=900 #Load 2 in kN\n",
-      "x1=75 #Distance in mm\n",
-      "x2=125 #Distance in mm\n",
-      "\n",
-      "#Calculations\n",
-      "e=(P2*x2-P1*x1)*(P1+P2)**-1 #Eccentricity in mm\n",
-      "P=N*(P1+P2) #Applied load after factor of safety is considered in kN\n",
-      "\n",
-      "#Part 1 is not computable\n",
-      "I=415*10**-6 #Moment of inertia from the table in mm^4\n",
-      "\n",
-      "#Part 2\n",
-      "P_cr=pi**2*E*10**9*I*Le**-2 #Critical load for buckling in kN\n",
-      "FOS=P_cr*10**-3/(P1+P2) #Factor of safety against buckling in y-axis\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"The critical load for buckling is\",round(P_cr*10**-3),\"kN\"\n",
-      "print \"The factor of safety is\",round(FOS,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The critical load for buckling is 16718.0 kN\n",
-        "The factor of safety is 12.9\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter11.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter11.ipynb
deleted file mode 100755
index bf1270ca..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter11.ipynb
+++ /dev/null
@@ -1,293 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:9e6efed33049beac69942b90d39a9e8444a663ad0d711d98275d388c059ec74c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter11:Additional Beam Topics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.1, Page No:394"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "V=1000 #Force acting on he section in lb\n",
-      "t=0.5 #Thickness in inches\n",
-      "#Dimensions\n",
-      "d=8 #Depth of the section in inches\n",
-      "wf=12 #Width of the flange in inches\n",
-      "wbf=8 #Width of the bottom flange in inches\n",
-      "\n",
-      "#Calculations\n",
-      "y_bar=((d*t*0)+wbf*t*4+wf*t*8)/(d*t+wbf*t+wf*t) #Location of Neutral Axis in inches\n",
-      "I=d*t*y_bar**2+t*d**3*12**-1+d*t*(d*t-y_bar)**2+wf*t*(8-y_bar)**2 #Moment of Inertia in in^4\n",
-      "\n",
-      "#Top Flange\n",
-      "q1=V*t*t*wf*(8-y_bar)*I**-1 #Shear flow in lb/in\n",
-      "#Bottom Flange\n",
-      "q2=V*t*t*d*y_bar*I**-1 #Shear Flow in lb/in\n",
-      "#Web\n",
-      "qB=2*q1 #Shear Flow in lb/in\n",
-      "qF=2*q2 #Shear Flow in lb/in\n",
-      "\n",
-      "#Max Shear Flow\n",
-      "qMAX=qB+V*t*(8-y_bar)**2*0.5*I**-1 #Maximum Shear Flow in lb/in\n",
-      "\n",
-      "#Result\n",
-      "print \"The Maximum Shear Flow is\",round(qMAX),\"lb/in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Maximum Shear Flow is 133.0 lb/in\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.2, Page No:395"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "V=1000 #Shear Force in lb\n",
-      "#Rest ALL DATA is similar to previous problem\n",
-      "\n",
-      "#Calcualtions\n",
-      "I=t*wf**3*12**-1+t*d**3*12**-1 #Moment of Inertia\n",
-      "\n",
-      "#Part 1\n",
-      "q1=V*t*t*wf*3*I**-1 #Shear Flow in lb/in\n",
-      "q2=V*t*t*d*2*I**-1 #Shear FLow in lb/in\n",
-      "V1=2*3**-1*q1*wf #Shear force carried in lb\n",
-      "V2=2*3**-1*q2*d #Shear force carried in lb\n",
-      "\n",
-      "#Part 2\n",
-      "e=8*V2*V**-1 #Eccentricity in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The Shear Force carried by Flanges is\"\n",
-      "print \"Top Flange=\",round(V1,1),\"lb Bottom Flange=\",round(V2,1),\"lb\"\n",
-      "print \"The eccentricity is\",round(e,3),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Shear Force carried by Flanges is\n",
-        "Top Flange= 771.4 lb Bottom Flange= 228.6 lb\n",
-        "The eccentricity is 1.829 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.3, Page No:403"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "M=32 #Moment  in kN.m\n",
-      "Iy=4.73*10**6 #Moment of inertia in y-axis in mm^4\n",
-      "Iz=48.9*10**6 #Moment of inertia in z-axis in mm^4\n",
-      "Sy=64.7*10**3 #Sectional Modulus in y-axis in mm^3\n",
-      "Sz=379*10**3 #Sectional Modulus in z-axis in mm^3\n",
-      "theta=16.2 #Angle between moment and z-axis in degrees\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "alpha=np.arctan((Iz*Iy**-1)*tan(theta*pi*180**-1))*180*pi**-1 #Angle between NA and z-axis in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "My=-M*np.sin(theta*pi*180**-1) #Bending Moment in y in kN.m\n",
-      "Mz=-M*np.cos(theta*pi*180**-1) #Bending Moment in z in kN.m\n",
-      "\n",
-      "sigma_max=My*Sy**-1+Mz*Sz**-1 #Largest Bedning Stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The angle between the  Neutral Axis and Z-Axis is\",round(alpha,1),\"degrees\"\n",
-      "print \"The maximum Bending Moment is\",abs(round(sigma_max*10**6)),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The angle between the  Neutral Axis and Z-Axis is 71.6 degrees\n",
-        "The maximum Bending Moment is 219.0 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.4, Page No:403"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A=4.75 #Area in inches^2\n",
-      "Iy_dash=6.27 #Moment of inertia in in^4\n",
-      "Iz_dash=17.4 #Moment of inertia in in^4\n",
-      "ry=0.87 #Radius of Gyration in inches\n",
-      "tan_theta=0.44\n",
-      "P=1 #Force in kips\n",
-      "L=48 #Length in inches\n",
-      "y_dash_B=-4.01 #Y coordinate of point B in inches\n",
-      "z_dash_B=-0.487 #Z coordinate of point B in inches\n",
-      "\n",
-      "#Calcualtions\n",
-      "theta=np.arctan(tan_theta) #Angle in radians\n",
-      "Iy=A*ry**2 #Moment of inertia in y in in^4\n",
-      "Iz=Iy_dash+Iz_dash-Iy #Moment of inertia in y in in^4\n",
-      "\n",
-      "#Part 1\n",
-      "alpha=arctan(Iz*Iy**-1*tan_theta)*180*pi**-1 #Angle in radians\n",
-      "beta=alpha-(theta*180*pi**-1) #Angle in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "M=P*L*4**-1 #Moment in kip.in\n",
-      "My=M*np.sin(theta) #Moment in y in kip.in\n",
-      "Mz=M*np.cos(theta) #Moment in z in kip.in\n",
-      "\n",
-      "y_B=y_dash_B*np.cos(theta)+z_dash_B*np.sin(theta) #Y coordinate  in inches\n",
-      "z_B=z_dash_B*np.cos(theta)-y_dash_B*np.sin(theta) #Z coordinate in inches\n",
-      "\n",
-      "#Maximum Bending Stress\n",
-      "sigma_max=My*z_B*Iy**-1-Mz*y_B*Iz**-1 #Maximum Bedning Stress in ksi\n",
-      "\n",
-      "#Result\n",
-      "print \"The angle of inclination of the Neutral axis to the z-axis is\",round(beta,1),\"degrees\"\n",
-      "print \"The maximum Bedning Stress is\",round(sigma_max,2),\"ksi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The angle of inclination of the Neutral axis to the z-axis is 44.1 degrees\n",
-        "The maximum Bedning Stress is 3.69 ksi\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.5, Page No:412"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A1=4 #Area in in^2\n",
-      "A2=6 #Area in in^2\n",
-      "r1=7.8 #Radius in inches\n",
-      "r2=14.8 #Radius in inches\n",
-      "t=0.5 #Thickness in inches\n",
-      "d=4 #Depth in inches\n",
-      "sigma_w=18 #Maximum allowable stress in kips\n",
-      "\n",
-      "#Calculations\n",
-      "A=A1+A2 #Area in in^2\n",
-      "r_bar=(A1*(r1+t)+A2*(r2+d))*(A1+A2)**-1 #Centroidal Axis in inches\n",
-      "#Simplfying the computation\n",
-      "a=(r1+2*t)/r1\n",
-      "b=r2/(r1+t*2)\n",
-      "integral=d*math.log(a)+2*t*math.log(b) #\n",
-      "R=A/integral #Radius of neutral Surface in inches\n",
-      "\n",
-      "#Maximum Stress\n",
-      "#Answers are in variable terms hence not computable\n",
-      "\n",
-      "P=sigma_w/0.7847 #Maximum Allowable load in kips\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum allowable load is\",round(P,1),\"kips\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable load is 22.9 kips\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter11_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter11_1.ipynb
deleted file mode 100755
index bf1270ca..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter11_1.ipynb
+++ /dev/null
@@ -1,293 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:9e6efed33049beac69942b90d39a9e8444a663ad0d711d98275d388c059ec74c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter11:Additional Beam Topics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.1, Page No:394"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "V=1000 #Force acting on he section in lb\n",
-      "t=0.5 #Thickness in inches\n",
-      "#Dimensions\n",
-      "d=8 #Depth of the section in inches\n",
-      "wf=12 #Width of the flange in inches\n",
-      "wbf=8 #Width of the bottom flange in inches\n",
-      "\n",
-      "#Calculations\n",
-      "y_bar=((d*t*0)+wbf*t*4+wf*t*8)/(d*t+wbf*t+wf*t) #Location of Neutral Axis in inches\n",
-      "I=d*t*y_bar**2+t*d**3*12**-1+d*t*(d*t-y_bar)**2+wf*t*(8-y_bar)**2 #Moment of Inertia in in^4\n",
-      "\n",
-      "#Top Flange\n",
-      "q1=V*t*t*wf*(8-y_bar)*I**-1 #Shear flow in lb/in\n",
-      "#Bottom Flange\n",
-      "q2=V*t*t*d*y_bar*I**-1 #Shear Flow in lb/in\n",
-      "#Web\n",
-      "qB=2*q1 #Shear Flow in lb/in\n",
-      "qF=2*q2 #Shear Flow in lb/in\n",
-      "\n",
-      "#Max Shear Flow\n",
-      "qMAX=qB+V*t*(8-y_bar)**2*0.5*I**-1 #Maximum Shear Flow in lb/in\n",
-      "\n",
-      "#Result\n",
-      "print \"The Maximum Shear Flow is\",round(qMAX),\"lb/in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Maximum Shear Flow is 133.0 lb/in\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.2, Page No:395"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "V=1000 #Shear Force in lb\n",
-      "#Rest ALL DATA is similar to previous problem\n",
-      "\n",
-      "#Calcualtions\n",
-      "I=t*wf**3*12**-1+t*d**3*12**-1 #Moment of Inertia\n",
-      "\n",
-      "#Part 1\n",
-      "q1=V*t*t*wf*3*I**-1 #Shear Flow in lb/in\n",
-      "q2=V*t*t*d*2*I**-1 #Shear FLow in lb/in\n",
-      "V1=2*3**-1*q1*wf #Shear force carried in lb\n",
-      "V2=2*3**-1*q2*d #Shear force carried in lb\n",
-      "\n",
-      "#Part 2\n",
-      "e=8*V2*V**-1 #Eccentricity in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The Shear Force carried by Flanges is\"\n",
-      "print \"Top Flange=\",round(V1,1),\"lb Bottom Flange=\",round(V2,1),\"lb\"\n",
-      "print \"The eccentricity is\",round(e,3),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Shear Force carried by Flanges is\n",
-        "Top Flange= 771.4 lb Bottom Flange= 228.6 lb\n",
-        "The eccentricity is 1.829 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.3, Page No:403"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "M=32 #Moment  in kN.m\n",
-      "Iy=4.73*10**6 #Moment of inertia in y-axis in mm^4\n",
-      "Iz=48.9*10**6 #Moment of inertia in z-axis in mm^4\n",
-      "Sy=64.7*10**3 #Sectional Modulus in y-axis in mm^3\n",
-      "Sz=379*10**3 #Sectional Modulus in z-axis in mm^3\n",
-      "theta=16.2 #Angle between moment and z-axis in degrees\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "alpha=np.arctan((Iz*Iy**-1)*tan(theta*pi*180**-1))*180*pi**-1 #Angle between NA and z-axis in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "My=-M*np.sin(theta*pi*180**-1) #Bending Moment in y in kN.m\n",
-      "Mz=-M*np.cos(theta*pi*180**-1) #Bending Moment in z in kN.m\n",
-      "\n",
-      "sigma_max=My*Sy**-1+Mz*Sz**-1 #Largest Bedning Stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The angle between the  Neutral Axis and Z-Axis is\",round(alpha,1),\"degrees\"\n",
-      "print \"The maximum Bending Moment is\",abs(round(sigma_max*10**6)),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The angle between the  Neutral Axis and Z-Axis is 71.6 degrees\n",
-        "The maximum Bending Moment is 219.0 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.4, Page No:403"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A=4.75 #Area in inches^2\n",
-      "Iy_dash=6.27 #Moment of inertia in in^4\n",
-      "Iz_dash=17.4 #Moment of inertia in in^4\n",
-      "ry=0.87 #Radius of Gyration in inches\n",
-      "tan_theta=0.44\n",
-      "P=1 #Force in kips\n",
-      "L=48 #Length in inches\n",
-      "y_dash_B=-4.01 #Y coordinate of point B in inches\n",
-      "z_dash_B=-0.487 #Z coordinate of point B in inches\n",
-      "\n",
-      "#Calcualtions\n",
-      "theta=np.arctan(tan_theta) #Angle in radians\n",
-      "Iy=A*ry**2 #Moment of inertia in y in in^4\n",
-      "Iz=Iy_dash+Iz_dash-Iy #Moment of inertia in y in in^4\n",
-      "\n",
-      "#Part 1\n",
-      "alpha=arctan(Iz*Iy**-1*tan_theta)*180*pi**-1 #Angle in radians\n",
-      "beta=alpha-(theta*180*pi**-1) #Angle in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "M=P*L*4**-1 #Moment in kip.in\n",
-      "My=M*np.sin(theta) #Moment in y in kip.in\n",
-      "Mz=M*np.cos(theta) #Moment in z in kip.in\n",
-      "\n",
-      "y_B=y_dash_B*np.cos(theta)+z_dash_B*np.sin(theta) #Y coordinate  in inches\n",
-      "z_B=z_dash_B*np.cos(theta)-y_dash_B*np.sin(theta) #Z coordinate in inches\n",
-      "\n",
-      "#Maximum Bending Stress\n",
-      "sigma_max=My*z_B*Iy**-1-Mz*y_B*Iz**-1 #Maximum Bedning Stress in ksi\n",
-      "\n",
-      "#Result\n",
-      "print \"The angle of inclination of the Neutral axis to the z-axis is\",round(beta,1),\"degrees\"\n",
-      "print \"The maximum Bedning Stress is\",round(sigma_max,2),\"ksi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The angle of inclination of the Neutral axis to the z-axis is 44.1 degrees\n",
-        "The maximum Bedning Stress is 3.69 ksi\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.5, Page No:412"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A1=4 #Area in in^2\n",
-      "A2=6 #Area in in^2\n",
-      "r1=7.8 #Radius in inches\n",
-      "r2=14.8 #Radius in inches\n",
-      "t=0.5 #Thickness in inches\n",
-      "d=4 #Depth in inches\n",
-      "sigma_w=18 #Maximum allowable stress in kips\n",
-      "\n",
-      "#Calculations\n",
-      "A=A1+A2 #Area in in^2\n",
-      "r_bar=(A1*(r1+t)+A2*(r2+d))*(A1+A2)**-1 #Centroidal Axis in inches\n",
-      "#Simplfying the computation\n",
-      "a=(r1+2*t)/r1\n",
-      "b=r2/(r1+t*2)\n",
-      "integral=d*math.log(a)+2*t*math.log(b) #\n",
-      "R=A/integral #Radius of neutral Surface in inches\n",
-      "\n",
-      "#Maximum Stress\n",
-      "#Answers are in variable terms hence not computable\n",
-      "\n",
-      "P=sigma_w/0.7847 #Maximum Allowable load in kips\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum allowable load is\",round(P,1),\"kips\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable load is 22.9 kips\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter11_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter11_2.ipynb
deleted file mode 100755
index bf1270ca..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter11_2.ipynb
+++ /dev/null
@@ -1,293 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:9e6efed33049beac69942b90d39a9e8444a663ad0d711d98275d388c059ec74c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter11:Additional Beam Topics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.1, Page No:394"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "V=1000 #Force acting on he section in lb\n",
-      "t=0.5 #Thickness in inches\n",
-      "#Dimensions\n",
-      "d=8 #Depth of the section in inches\n",
-      "wf=12 #Width of the flange in inches\n",
-      "wbf=8 #Width of the bottom flange in inches\n",
-      "\n",
-      "#Calculations\n",
-      "y_bar=((d*t*0)+wbf*t*4+wf*t*8)/(d*t+wbf*t+wf*t) #Location of Neutral Axis in inches\n",
-      "I=d*t*y_bar**2+t*d**3*12**-1+d*t*(d*t-y_bar)**2+wf*t*(8-y_bar)**2 #Moment of Inertia in in^4\n",
-      "\n",
-      "#Top Flange\n",
-      "q1=V*t*t*wf*(8-y_bar)*I**-1 #Shear flow in lb/in\n",
-      "#Bottom Flange\n",
-      "q2=V*t*t*d*y_bar*I**-1 #Shear Flow in lb/in\n",
-      "#Web\n",
-      "qB=2*q1 #Shear Flow in lb/in\n",
-      "qF=2*q2 #Shear Flow in lb/in\n",
-      "\n",
-      "#Max Shear Flow\n",
-      "qMAX=qB+V*t*(8-y_bar)**2*0.5*I**-1 #Maximum Shear Flow in lb/in\n",
-      "\n",
-      "#Result\n",
-      "print \"The Maximum Shear Flow is\",round(qMAX),\"lb/in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Maximum Shear Flow is 133.0 lb/in\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.2, Page No:395"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "V=1000 #Shear Force in lb\n",
-      "#Rest ALL DATA is similar to previous problem\n",
-      "\n",
-      "#Calcualtions\n",
-      "I=t*wf**3*12**-1+t*d**3*12**-1 #Moment of Inertia\n",
-      "\n",
-      "#Part 1\n",
-      "q1=V*t*t*wf*3*I**-1 #Shear Flow in lb/in\n",
-      "q2=V*t*t*d*2*I**-1 #Shear FLow in lb/in\n",
-      "V1=2*3**-1*q1*wf #Shear force carried in lb\n",
-      "V2=2*3**-1*q2*d #Shear force carried in lb\n",
-      "\n",
-      "#Part 2\n",
-      "e=8*V2*V**-1 #Eccentricity in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The Shear Force carried by Flanges is\"\n",
-      "print \"Top Flange=\",round(V1,1),\"lb Bottom Flange=\",round(V2,1),\"lb\"\n",
-      "print \"The eccentricity is\",round(e,3),\"in\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Shear Force carried by Flanges is\n",
-        "Top Flange= 771.4 lb Bottom Flange= 228.6 lb\n",
-        "The eccentricity is 1.829 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.3, Page No:403"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "M=32 #Moment  in kN.m\n",
-      "Iy=4.73*10**6 #Moment of inertia in y-axis in mm^4\n",
-      "Iz=48.9*10**6 #Moment of inertia in z-axis in mm^4\n",
-      "Sy=64.7*10**3 #Sectional Modulus in y-axis in mm^3\n",
-      "Sz=379*10**3 #Sectional Modulus in z-axis in mm^3\n",
-      "theta=16.2 #Angle between moment and z-axis in degrees\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "alpha=np.arctan((Iz*Iy**-1)*tan(theta*pi*180**-1))*180*pi**-1 #Angle between NA and z-axis in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "My=-M*np.sin(theta*pi*180**-1) #Bending Moment in y in kN.m\n",
-      "Mz=-M*np.cos(theta*pi*180**-1) #Bending Moment in z in kN.m\n",
-      "\n",
-      "sigma_max=My*Sy**-1+Mz*Sz**-1 #Largest Bedning Stress in MPa\n",
-      "\n",
-      "#Result\n",
-      "print \"The angle between the  Neutral Axis and Z-Axis is\",round(alpha,1),\"degrees\"\n",
-      "print \"The maximum Bending Moment is\",abs(round(sigma_max*10**6)),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The angle between the  Neutral Axis and Z-Axis is 71.6 degrees\n",
-        "The maximum Bending Moment is 219.0 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.4, Page No:403"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A=4.75 #Area in inches^2\n",
-      "Iy_dash=6.27 #Moment of inertia in in^4\n",
-      "Iz_dash=17.4 #Moment of inertia in in^4\n",
-      "ry=0.87 #Radius of Gyration in inches\n",
-      "tan_theta=0.44\n",
-      "P=1 #Force in kips\n",
-      "L=48 #Length in inches\n",
-      "y_dash_B=-4.01 #Y coordinate of point B in inches\n",
-      "z_dash_B=-0.487 #Z coordinate of point B in inches\n",
-      "\n",
-      "#Calcualtions\n",
-      "theta=np.arctan(tan_theta) #Angle in radians\n",
-      "Iy=A*ry**2 #Moment of inertia in y in in^4\n",
-      "Iz=Iy_dash+Iz_dash-Iy #Moment of inertia in y in in^4\n",
-      "\n",
-      "#Part 1\n",
-      "alpha=arctan(Iz*Iy**-1*tan_theta)*180*pi**-1 #Angle in radians\n",
-      "beta=alpha-(theta*180*pi**-1) #Angle in degrees\n",
-      "\n",
-      "#Part 2\n",
-      "M=P*L*4**-1 #Moment in kip.in\n",
-      "My=M*np.sin(theta) #Moment in y in kip.in\n",
-      "Mz=M*np.cos(theta) #Moment in z in kip.in\n",
-      "\n",
-      "y_B=y_dash_B*np.cos(theta)+z_dash_B*np.sin(theta) #Y coordinate  in inches\n",
-      "z_B=z_dash_B*np.cos(theta)-y_dash_B*np.sin(theta) #Z coordinate in inches\n",
-      "\n",
-      "#Maximum Bending Stress\n",
-      "sigma_max=My*z_B*Iy**-1-Mz*y_B*Iz**-1 #Maximum Bedning Stress in ksi\n",
-      "\n",
-      "#Result\n",
-      "print \"The angle of inclination of the Neutral axis to the z-axis is\",round(beta,1),\"degrees\"\n",
-      "print \"The maximum Bedning Stress is\",round(sigma_max,2),\"ksi\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The angle of inclination of the Neutral axis to the z-axis is 44.1 degrees\n",
-        "The maximum Bedning Stress is 3.69 ksi\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11.5, Page No:412"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "A1=4 #Area in in^2\n",
-      "A2=6 #Area in in^2\n",
-      "r1=7.8 #Radius in inches\n",
-      "r2=14.8 #Radius in inches\n",
-      "t=0.5 #Thickness in inches\n",
-      "d=4 #Depth in inches\n",
-      "sigma_w=18 #Maximum allowable stress in kips\n",
-      "\n",
-      "#Calculations\n",
-      "A=A1+A2 #Area in in^2\n",
-      "r_bar=(A1*(r1+t)+A2*(r2+d))*(A1+A2)**-1 #Centroidal Axis in inches\n",
-      "#Simplfying the computation\n",
-      "a=(r1+2*t)/r1\n",
-      "b=r2/(r1+t*2)\n",
-      "integral=d*math.log(a)+2*t*math.log(b) #\n",
-      "R=A/integral #Radius of neutral Surface in inches\n",
-      "\n",
-      "#Maximum Stress\n",
-      "#Answers are in variable terms hence not computable\n",
-      "\n",
-      "P=sigma_w/0.7847 #Maximum Allowable load in kips\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum allowable load is\",round(P,1),\"kips\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable load is 22.9 kips\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12.ipynb
deleted file mode 100755
index 9163c9cf..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12.ipynb
+++ /dev/null
@@ -1,319 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d8a1a6f05cd4df2c46b4f5147d4f831726de5041386c1f65ed2892779a5fb0fb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12:Special Topics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.1, Page No:422"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "W=24*10**3 #Load in kips\n",
-      "E=29*10**6 #Youngs Modulus in psi\n",
-      "L=72 #length in inches\n",
-      "theta=30 #Angle in degrees\n",
-      "\n",
-      "#Calculations\n",
-      "L_ab=L/np.sin(theta*pi*180**-1) #Length of AB in inches\n",
-      "L_ac=L/np.sin((90-theta)*pi*180**-1) #Length of AC in inches\n",
-      "\n",
-      "#Applying the forces in x and y sum to zero\n",
-      "#Applying the Starin energy formula\n",
-      "#Applying Castiglinos theorem \n",
-      "delta_A=91.16*W*E**-1 #Displacement in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The displacement of point A is\",round(delta_A,4),\"in\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The displacement of point A is 0.0754 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.3, Page No:423"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from scipy.integrate import quad\n",
-      "\n",
-      "\n",
-      "\n",
-      "#We will directly compute the integral as function cannot be declared\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "def integrand(x):\n",
-      "    return (800*x-400*x**2)*(4-0.5*x)\n",
-      "I = quad(integrand, 0, 2)\n",
-      "delta=I[0] #Deflection in horizontal direction in N.m^3\n",
-      "\n",
-      "#Part 2\n",
-      "def inte1(x):\n",
-      "    return x**2\n",
-      "I1=quad(inte1,0,4)\n",
-      "I2=quad(inte1,0,3)\n",
-      "def inte2(x):\n",
-      "    return (4-0.5*x)*(4-0.5*x)\n",
-      "I3=quad(inte2,0,2)\n",
-      "\n",
-      "Q=-delta/(I1[0]+I2[0]+I3[0]) #Horizontal reaction in N\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"The Horizontal deflection is\",round(delta),\"N.m^3\"\n",
-      "print \"The Horizontal reaction is\",round(Q,1),\"N\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Horizontal deflection is 1867.0 N.m^3\n",
-        "The Horizontal reaction is -33.9 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.4, Page No:433"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#NOTE:The figure mentions the unit of length as ft which is incorrect\n",
-      "#Variable Decleration\n",
-      "L=30 #Length in m\n",
-      "m=2000 #Mass in kg\n",
-      "v=2 #Velocity in m/s\n",
-      "E=10**5 #Youngs Modulus in MPa\n",
-      "A=600 #Area in mm^2\n",
-      "g=9.81 #Acceleration due to gravity in m/s^2\n",
-      "\n",
-      "#Calculations\n",
-      "k=E*A*L**-1 #Stifness of the cable in N/m\n",
-      "\n",
-      "#Applying the Work-Energy principle \n",
-      "delta_max=np.sqrt((0.5*m*v**2)*(0.5*k)**-1) #Maximum Displacement in m\n",
-      "\n",
-      "P_max=k*delta_max+m*g #Maximum force in N\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum force is\",round(P_max*10**-3,1),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum force is 146.1 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.5, Page No:434"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "b=0.060 #Breadth of the section in mm\n",
-      "d=0.03 #Depth of the section in mm\n",
-      "L=1.2 #Length in m\n",
-      "m=80 #Mass in kg\n",
-      "g=9.81 #Acceleration due to gravity in m/s^2\n",
-      "E=200*10**9 #Youngs Modulus in Pa\n",
-      "e=0.015 \n",
-      "h=0.01 #height in m\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "I=b*d**3*12**-1 #Moment of Inertia in m^4\n",
-      "delta_st=m*g*L**3/(48*E*I) #Mid-span Displacement  in m\n",
-      "n=1+np.sqrt(1+(2*h/delta_st)) #Impact Factor\n",
-      "\n",
-      "#Part 2\n",
-      "P_max=n*m*g #Maximum dynamic load in N at midspan\n",
-      "M_max=P_max*0.5*L*0.5 #Maximum moment in N.m\n",
-      "sigma_max=M_max*e/I #Maximum dynamic Bending Stress in Pa\n",
-      "\n",
-      "#Result\n",
-      "print \"The impact factor is\",round(n,3)\n",
-      "print \"The maximum dynamic Bending Moment is\",round(sigma_max*10**-6,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The impact factor is 5.485\n",
-        "The maximum dynamic Bending Moment is 143.5 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 65
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.7, Page No:440"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable decleration\n",
-      "M=2.21 #Applied moment in kip.ft\n",
-      "d=3 #Diameter of the bar in inches\n",
-      "sigma_y=40 #Yield strength of the of steel in ksi\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "sigma=32*M*12*(pi*d**3)**-1 #Maximum Bending Stress in ksi\n",
-      "T1=np.sqrt((sigma_y*0.5)**2-5**2)/(12*0.18863) #Maximum Allowable torque in kip.ft\n",
-      "\n",
-      "#Part 2\n",
-      "R=np.sqrt((sigma_y**2-5**2)*3**-1) #Maximum shear stress in ksi\n",
-      "T2=np.sqrt(R**2-5**2)/(12*0.18863) #Maximum safe torque in kpi.ft\n",
-      "\n",
-      "#Result\n",
-      "print \"Using the maximum shear stress theory T=\",round(T1,2),\"kip.ft\"\n",
-      "print \"Using the maximum sitrotion energy theory T=\",round(T2,2),\"kip.ft\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Using the maximum shear stress theory T= 8.56 kip.ft\n",
-        "Using the maximum sitrotion energy theory T= 9.88 kip.ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 79
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.8, Page No:448"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "D=250 #Wideness in mm\n",
-      "b=20 #Thickness of the plate in mm\n",
-      "r=50 #Radius  of the hole in mm\n",
-      "e=50 #Eccentricity in mm\n",
-      "sigma_max=150 #Maximum normal stress at the hole in MPa\n",
-      "kb=2 #Stress Concentraion factor \n",
-      "\n",
-      "#Calculations\n",
-      "A=b*(D-2*r)*10**-6 #Area in m^2\n",
-      "I=10**-12*(b*D**3*12**-1-(b*2**3*r**3*12**-1)) #Moment of inertia in m^4\n",
-      "#Simplfying computation\n",
-      "a=2*r*D**-1\n",
-      "kt=3-3.13*a+3.66*a**2-1.53*a**3 #Stress Concentration factor\n",
-      "#Simplfying computation\n",
-      "b=kt*A**-1\n",
-      "c=kb*r*r*10**-6*I**-1\n",
-      "P=10**3*sigma_max*(b+c)**-1 #Maximum Load in N\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum value of P is\",round(P,1),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum value of P is 157.8 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 106
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12_1.ipynb
deleted file mode 100755
index 9163c9cf..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12_1.ipynb
+++ /dev/null
@@ -1,319 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d8a1a6f05cd4df2c46b4f5147d4f831726de5041386c1f65ed2892779a5fb0fb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12:Special Topics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.1, Page No:422"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "W=24*10**3 #Load in kips\n",
-      "E=29*10**6 #Youngs Modulus in psi\n",
-      "L=72 #length in inches\n",
-      "theta=30 #Angle in degrees\n",
-      "\n",
-      "#Calculations\n",
-      "L_ab=L/np.sin(theta*pi*180**-1) #Length of AB in inches\n",
-      "L_ac=L/np.sin((90-theta)*pi*180**-1) #Length of AC in inches\n",
-      "\n",
-      "#Applying the forces in x and y sum to zero\n",
-      "#Applying the Starin energy formula\n",
-      "#Applying Castiglinos theorem \n",
-      "delta_A=91.16*W*E**-1 #Displacement in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The displacement of point A is\",round(delta_A,4),\"in\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The displacement of point A is 0.0754 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.3, Page No:423"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from scipy.integrate import quad\n",
-      "\n",
-      "\n",
-      "\n",
-      "#We will directly compute the integral as function cannot be declared\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "def integrand(x):\n",
-      "    return (800*x-400*x**2)*(4-0.5*x)\n",
-      "I = quad(integrand, 0, 2)\n",
-      "delta=I[0] #Deflection in horizontal direction in N.m^3\n",
-      "\n",
-      "#Part 2\n",
-      "def inte1(x):\n",
-      "    return x**2\n",
-      "I1=quad(inte1,0,4)\n",
-      "I2=quad(inte1,0,3)\n",
-      "def inte2(x):\n",
-      "    return (4-0.5*x)*(4-0.5*x)\n",
-      "I3=quad(inte2,0,2)\n",
-      "\n",
-      "Q=-delta/(I1[0]+I2[0]+I3[0]) #Horizontal reaction in N\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"The Horizontal deflection is\",round(delta),\"N.m^3\"\n",
-      "print \"The Horizontal reaction is\",round(Q,1),\"N\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Horizontal deflection is 1867.0 N.m^3\n",
-        "The Horizontal reaction is -33.9 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.4, Page No:433"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#NOTE:The figure mentions the unit of length as ft which is incorrect\n",
-      "#Variable Decleration\n",
-      "L=30 #Length in m\n",
-      "m=2000 #Mass in kg\n",
-      "v=2 #Velocity in m/s\n",
-      "E=10**5 #Youngs Modulus in MPa\n",
-      "A=600 #Area in mm^2\n",
-      "g=9.81 #Acceleration due to gravity in m/s^2\n",
-      "\n",
-      "#Calculations\n",
-      "k=E*A*L**-1 #Stifness of the cable in N/m\n",
-      "\n",
-      "#Applying the Work-Energy principle \n",
-      "delta_max=np.sqrt((0.5*m*v**2)*(0.5*k)**-1) #Maximum Displacement in m\n",
-      "\n",
-      "P_max=k*delta_max+m*g #Maximum force in N\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum force is\",round(P_max*10**-3,1),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum force is 146.1 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.5, Page No:434"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "b=0.060 #Breadth of the section in mm\n",
-      "d=0.03 #Depth of the section in mm\n",
-      "L=1.2 #Length in m\n",
-      "m=80 #Mass in kg\n",
-      "g=9.81 #Acceleration due to gravity in m/s^2\n",
-      "E=200*10**9 #Youngs Modulus in Pa\n",
-      "e=0.015 \n",
-      "h=0.01 #height in m\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "I=b*d**3*12**-1 #Moment of Inertia in m^4\n",
-      "delta_st=m*g*L**3/(48*E*I) #Mid-span Displacement  in m\n",
-      "n=1+np.sqrt(1+(2*h/delta_st)) #Impact Factor\n",
-      "\n",
-      "#Part 2\n",
-      "P_max=n*m*g #Maximum dynamic load in N at midspan\n",
-      "M_max=P_max*0.5*L*0.5 #Maximum moment in N.m\n",
-      "sigma_max=M_max*e/I #Maximum dynamic Bending Stress in Pa\n",
-      "\n",
-      "#Result\n",
-      "print \"The impact factor is\",round(n,3)\n",
-      "print \"The maximum dynamic Bending Moment is\",round(sigma_max*10**-6,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The impact factor is 5.485\n",
-        "The maximum dynamic Bending Moment is 143.5 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 65
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.7, Page No:440"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable decleration\n",
-      "M=2.21 #Applied moment in kip.ft\n",
-      "d=3 #Diameter of the bar in inches\n",
-      "sigma_y=40 #Yield strength of the of steel in ksi\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "sigma=32*M*12*(pi*d**3)**-1 #Maximum Bending Stress in ksi\n",
-      "T1=np.sqrt((sigma_y*0.5)**2-5**2)/(12*0.18863) #Maximum Allowable torque in kip.ft\n",
-      "\n",
-      "#Part 2\n",
-      "R=np.sqrt((sigma_y**2-5**2)*3**-1) #Maximum shear stress in ksi\n",
-      "T2=np.sqrt(R**2-5**2)/(12*0.18863) #Maximum safe torque in kpi.ft\n",
-      "\n",
-      "#Result\n",
-      "print \"Using the maximum shear stress theory T=\",round(T1,2),\"kip.ft\"\n",
-      "print \"Using the maximum sitrotion energy theory T=\",round(T2,2),\"kip.ft\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Using the maximum shear stress theory T= 8.56 kip.ft\n",
-        "Using the maximum sitrotion energy theory T= 9.88 kip.ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 79
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.8, Page No:448"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "D=250 #Wideness in mm\n",
-      "b=20 #Thickness of the plate in mm\n",
-      "r=50 #Radius  of the hole in mm\n",
-      "e=50 #Eccentricity in mm\n",
-      "sigma_max=150 #Maximum normal stress at the hole in MPa\n",
-      "kb=2 #Stress Concentraion factor \n",
-      "\n",
-      "#Calculations\n",
-      "A=b*(D-2*r)*10**-6 #Area in m^2\n",
-      "I=10**-12*(b*D**3*12**-1-(b*2**3*r**3*12**-1)) #Moment of inertia in m^4\n",
-      "#Simplfying computation\n",
-      "a=2*r*D**-1\n",
-      "kt=3-3.13*a+3.66*a**2-1.53*a**3 #Stress Concentration factor\n",
-      "#Simplfying computation\n",
-      "b=kt*A**-1\n",
-      "c=kb*r*r*10**-6*I**-1\n",
-      "P=10**3*sigma_max*(b+c)**-1 #Maximum Load in N\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum value of P is\",round(P,1),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum value of P is 157.8 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 106
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12_2.ipynb
deleted file mode 100755
index 9163c9cf..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12_2.ipynb
+++ /dev/null
@@ -1,319 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d8a1a6f05cd4df2c46b4f5147d4f831726de5041386c1f65ed2892779a5fb0fb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 12:Special Topics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.1, Page No:422"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "W=24*10**3 #Load in kips\n",
-      "E=29*10**6 #Youngs Modulus in psi\n",
-      "L=72 #length in inches\n",
-      "theta=30 #Angle in degrees\n",
-      "\n",
-      "#Calculations\n",
-      "L_ab=L/np.sin(theta*pi*180**-1) #Length of AB in inches\n",
-      "L_ac=L/np.sin((90-theta)*pi*180**-1) #Length of AC in inches\n",
-      "\n",
-      "#Applying the forces in x and y sum to zero\n",
-      "#Applying the Starin energy formula\n",
-      "#Applying Castiglinos theorem \n",
-      "delta_A=91.16*W*E**-1 #Displacement in inches\n",
-      "\n",
-      "#Result\n",
-      "print \"The displacement of point A is\",round(delta_A,4),\"in\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The displacement of point A is 0.0754 in\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.3, Page No:423"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "from scipy.integrate import quad\n",
-      "\n",
-      "\n",
-      "\n",
-      "#We will directly compute the integral as function cannot be declared\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "def integrand(x):\n",
-      "    return (800*x-400*x**2)*(4-0.5*x)\n",
-      "I = quad(integrand, 0, 2)\n",
-      "delta=I[0] #Deflection in horizontal direction in N.m^3\n",
-      "\n",
-      "#Part 2\n",
-      "def inte1(x):\n",
-      "    return x**2\n",
-      "I1=quad(inte1,0,4)\n",
-      "I2=quad(inte1,0,3)\n",
-      "def inte2(x):\n",
-      "    return (4-0.5*x)*(4-0.5*x)\n",
-      "I3=quad(inte2,0,2)\n",
-      "\n",
-      "Q=-delta/(I1[0]+I2[0]+I3[0]) #Horizontal reaction in N\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"The Horizontal deflection is\",round(delta),\"N.m^3\"\n",
-      "print \"The Horizontal reaction is\",round(Q,1),\"N\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Horizontal deflection is 1867.0 N.m^3\n",
-        "The Horizontal reaction is -33.9 N\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.4, Page No:433"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#NOTE:The figure mentions the unit of length as ft which is incorrect\n",
-      "#Variable Decleration\n",
-      "L=30 #Length in m\n",
-      "m=2000 #Mass in kg\n",
-      "v=2 #Velocity in m/s\n",
-      "E=10**5 #Youngs Modulus in MPa\n",
-      "A=600 #Area in mm^2\n",
-      "g=9.81 #Acceleration due to gravity in m/s^2\n",
-      "\n",
-      "#Calculations\n",
-      "k=E*A*L**-1 #Stifness of the cable in N/m\n",
-      "\n",
-      "#Applying the Work-Energy principle \n",
-      "delta_max=np.sqrt((0.5*m*v**2)*(0.5*k)**-1) #Maximum Displacement in m\n",
-      "\n",
-      "P_max=k*delta_max+m*g #Maximum force in N\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum force is\",round(P_max*10**-3,1),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum force is 146.1 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.5, Page No:434"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "b=0.060 #Breadth of the section in mm\n",
-      "d=0.03 #Depth of the section in mm\n",
-      "L=1.2 #Length in m\n",
-      "m=80 #Mass in kg\n",
-      "g=9.81 #Acceleration due to gravity in m/s^2\n",
-      "E=200*10**9 #Youngs Modulus in Pa\n",
-      "e=0.015 \n",
-      "h=0.01 #height in m\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "I=b*d**3*12**-1 #Moment of Inertia in m^4\n",
-      "delta_st=m*g*L**3/(48*E*I) #Mid-span Displacement  in m\n",
-      "n=1+np.sqrt(1+(2*h/delta_st)) #Impact Factor\n",
-      "\n",
-      "#Part 2\n",
-      "P_max=n*m*g #Maximum dynamic load in N at midspan\n",
-      "M_max=P_max*0.5*L*0.5 #Maximum moment in N.m\n",
-      "sigma_max=M_max*e/I #Maximum dynamic Bending Stress in Pa\n",
-      "\n",
-      "#Result\n",
-      "print \"The impact factor is\",round(n,3)\n",
-      "print \"The maximum dynamic Bending Moment is\",round(sigma_max*10**-6,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The impact factor is 5.485\n",
-        "The maximum dynamic Bending Moment is 143.5 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 65
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.7, Page No:440"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable decleration\n",
-      "M=2.21 #Applied moment in kip.ft\n",
-      "d=3 #Diameter of the bar in inches\n",
-      "sigma_y=40 #Yield strength of the of steel in ksi\n",
-      "\n",
-      "#Calculations\n",
-      "#Part 1\n",
-      "sigma=32*M*12*(pi*d**3)**-1 #Maximum Bending Stress in ksi\n",
-      "T1=np.sqrt((sigma_y*0.5)**2-5**2)/(12*0.18863) #Maximum Allowable torque in kip.ft\n",
-      "\n",
-      "#Part 2\n",
-      "R=np.sqrt((sigma_y**2-5**2)*3**-1) #Maximum shear stress in ksi\n",
-      "T2=np.sqrt(R**2-5**2)/(12*0.18863) #Maximum safe torque in kpi.ft\n",
-      "\n",
-      "#Result\n",
-      "print \"Using the maximum shear stress theory T=\",round(T1,2),\"kip.ft\"\n",
-      "print \"Using the maximum sitrotion energy theory T=\",round(T2,2),\"kip.ft\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Using the maximum shear stress theory T= 8.56 kip.ft\n",
-        "Using the maximum sitrotion energy theory T= 9.88 kip.ft\n"
-       ]
-      }
-     ],
-     "prompt_number": 79
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12.8, Page No:448"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "D=250 #Wideness in mm\n",
-      "b=20 #Thickness of the plate in mm\n",
-      "r=50 #Radius  of the hole in mm\n",
-      "e=50 #Eccentricity in mm\n",
-      "sigma_max=150 #Maximum normal stress at the hole in MPa\n",
-      "kb=2 #Stress Concentraion factor \n",
-      "\n",
-      "#Calculations\n",
-      "A=b*(D-2*r)*10**-6 #Area in m^2\n",
-      "I=10**-12*(b*D**3*12**-1-(b*2**3*r**3*12**-1)) #Moment of inertia in m^4\n",
-      "#Simplfying computation\n",
-      "a=2*r*D**-1\n",
-      "kt=3-3.13*a+3.66*a**2-1.53*a**3 #Stress Concentration factor\n",
-      "#Simplfying computation\n",
-      "b=kt*A**-1\n",
-      "c=kb*r*r*10**-6*I**-1\n",
-      "P=10**3*sigma_max*(b+c)**-1 #Maximum Load in N\n",
-      "\n",
-      "#Result\n",
-      "print \"The maximum value of P is\",round(P,1),\"kN\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum value of P is 157.8 kN\n"
-       ]
-      }
-     ],
-     "prompt_number": 106
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter13.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter13.ipynb
deleted file mode 100755
index 60e5f3b6..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter13.ipynb
+++ /dev/null
@@ -1,136 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:09673653cb16b83119bc0c347f996805dac297e3d0e1db4167910bd881eb2ed7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 13: Inelastic Action"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.13.1, Page No:461"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "d=150 #Depth of the web in mm\n",
-      "wf=100 #Width of the flange in mm\n",
-      "df=20 #Depth of the flange in mm\n",
-      "t=20 #Thickness of the web in mm\n",
-      "\n",
-      "#Calculations\n",
-      "y_bar=10**-3*(((wf*df*(d+df*0.5))+(d*t*d*0.5))/(wf*df+d*t)) #Distance of Neutral Axis in m\n",
-      "#Simplfying the computation\n",
-      "a=wf*df**3*12**-1\n",
-      "b=wf*df*((d+df*0.5)-y_bar*10**3)**2\n",
-      "c=t*d**3*12**-1\n",
-      "f=t*d*((d*0.5)-y_bar*10**3)**2\n",
-      "I=(a+b+c+f)*10**-12 #Moment of inertia in mm^3\n",
-      "\n",
-      "#Limit Moment\n",
-      "yp=(wf*df+d*t)/(2*t) #Plastic Neutral Axis in mm\n",
-      "Myp=I/y_bar #Yielding will start at moment without the stress term to ease computation\n",
-      "mom=10**-9*((t*yp**2*0.5)+(wf*df*(d-yp+10))+(t*25**2*0.5)) #Sum of 1st moments\n",
-      "Ml_Myp=mom*Myp**-1 #Ratio\n",
-      "\n",
-      "#Result\n",
-      "print \"The ratio ML/Myp=\",round(Ml_Myp,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ratio ML/Myp= 1.765\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.13.2, Page No:467"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "E_st=200 #Youngs Modulus of Steel in GPa\n",
-      "sigma_st_yp=290 #Yielding Stress in MPa\n",
-      "E_al=70 #Youngs Modulus of Aluminium in GPa\n",
-      "sigma_al_yp=330 #Yielding Stresss of Aluminium in MPa\n",
-      "A_st=900 #Area of steel rod in mm^2\n",
-      "A_al=600 #Area of Aluminium rod in mm^2\n",
-      "L_st=350 #Length of the steel rod in mm\n",
-      "L_al=250 #Length of the aluminium rod in mm\n",
-      "\n",
-      "#Calculations\n",
-      "#Limit Load\n",
-      "P_st=sigma_st_yp*A_st*10**-3 #Load in limiting condition in kN\n",
-      "P_al=sigma_al_yp*A_al*10**-3 #Load in limiting condition in kN\n",
-      "P_L=P_st+2*P_al #Total Loading in kN\n",
-      "\n",
-      "#Elastic Unloading\n",
-      "#Solving for Pst and Pal using matri approach\n",
-      "A=np.array([[1,2],[L_st*(E_st*A_st)**-1,-L_al*(E_al*A_al)**-1]])\n",
-      "B=np.array([P_L,0])\n",
-      "C=np.linalg.solve(A,B) #Loading in kN\n",
-      "\n",
-      "#Residual Stresses\n",
-      "P_res_st=C[0]-P_st #Residual Load in kN\n",
-      "P_res_al=C[1]-P_al #Residual Load in kN\n",
-      "sigma_st=P_res_st/A_st #residual Stress in Steel in MPa\n",
-      "sigma_al=P_res_al/A_al #residual Stress in Aluminium in MPa\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"The Residual stresses are as follows\"\n",
-      "print \"Sigma_st=\",round(sigma_st*10**3,1),\"MPa and sigma_al=\",round(sigma_al*10**3,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Residual stresses are as follows\n",
-        "Sigma_st= 151.5 MPa and sigma_al= -113.6 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter13_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter13_1.ipynb
deleted file mode 100755
index 60e5f3b6..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter13_1.ipynb
+++ /dev/null
@@ -1,136 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:09673653cb16b83119bc0c347f996805dac297e3d0e1db4167910bd881eb2ed7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 13: Inelastic Action"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.13.1, Page No:461"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "d=150 #Depth of the web in mm\n",
-      "wf=100 #Width of the flange in mm\n",
-      "df=20 #Depth of the flange in mm\n",
-      "t=20 #Thickness of the web in mm\n",
-      "\n",
-      "#Calculations\n",
-      "y_bar=10**-3*(((wf*df*(d+df*0.5))+(d*t*d*0.5))/(wf*df+d*t)) #Distance of Neutral Axis in m\n",
-      "#Simplfying the computation\n",
-      "a=wf*df**3*12**-1\n",
-      "b=wf*df*((d+df*0.5)-y_bar*10**3)**2\n",
-      "c=t*d**3*12**-1\n",
-      "f=t*d*((d*0.5)-y_bar*10**3)**2\n",
-      "I=(a+b+c+f)*10**-12 #Moment of inertia in mm^3\n",
-      "\n",
-      "#Limit Moment\n",
-      "yp=(wf*df+d*t)/(2*t) #Plastic Neutral Axis in mm\n",
-      "Myp=I/y_bar #Yielding will start at moment without the stress term to ease computation\n",
-      "mom=10**-9*((t*yp**2*0.5)+(wf*df*(d-yp+10))+(t*25**2*0.5)) #Sum of 1st moments\n",
-      "Ml_Myp=mom*Myp**-1 #Ratio\n",
-      "\n",
-      "#Result\n",
-      "print \"The ratio ML/Myp=\",round(Ml_Myp,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ratio ML/Myp= 1.765\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.13.2, Page No:467"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "E_st=200 #Youngs Modulus of Steel in GPa\n",
-      "sigma_st_yp=290 #Yielding Stress in MPa\n",
-      "E_al=70 #Youngs Modulus of Aluminium in GPa\n",
-      "sigma_al_yp=330 #Yielding Stresss of Aluminium in MPa\n",
-      "A_st=900 #Area of steel rod in mm^2\n",
-      "A_al=600 #Area of Aluminium rod in mm^2\n",
-      "L_st=350 #Length of the steel rod in mm\n",
-      "L_al=250 #Length of the aluminium rod in mm\n",
-      "\n",
-      "#Calculations\n",
-      "#Limit Load\n",
-      "P_st=sigma_st_yp*A_st*10**-3 #Load in limiting condition in kN\n",
-      "P_al=sigma_al_yp*A_al*10**-3 #Load in limiting condition in kN\n",
-      "P_L=P_st+2*P_al #Total Loading in kN\n",
-      "\n",
-      "#Elastic Unloading\n",
-      "#Solving for Pst and Pal using matri approach\n",
-      "A=np.array([[1,2],[L_st*(E_st*A_st)**-1,-L_al*(E_al*A_al)**-1]])\n",
-      "B=np.array([P_L,0])\n",
-      "C=np.linalg.solve(A,B) #Loading in kN\n",
-      "\n",
-      "#Residual Stresses\n",
-      "P_res_st=C[0]-P_st #Residual Load in kN\n",
-      "P_res_al=C[1]-P_al #Residual Load in kN\n",
-      "sigma_st=P_res_st/A_st #residual Stress in Steel in MPa\n",
-      "sigma_al=P_res_al/A_al #residual Stress in Aluminium in MPa\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"The Residual stresses are as follows\"\n",
-      "print \"Sigma_st=\",round(sigma_st*10**3,1),\"MPa and sigma_al=\",round(sigma_al*10**3,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Residual stresses are as follows\n",
-        "Sigma_st= 151.5 MPa and sigma_al= -113.6 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter13_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter13_2.ipynb
deleted file mode 100755
index 60e5f3b6..00000000
--- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter13_2.ipynb
+++ /dev/null
@@ -1,136 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:09673653cb16b83119bc0c347f996805dac297e3d0e1db4167910bd881eb2ed7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 13: Inelastic Action"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.13.1, Page No:461"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable Decleration\n",
-      "d=150 #Depth of the web in mm\n",
-      "wf=100 #Width of the flange in mm\n",
-      "df=20 #Depth of the flange in mm\n",
-      "t=20 #Thickness of the web in mm\n",
-      "\n",
-      "#Calculations\n",
-      "y_bar=10**-3*(((wf*df*(d+df*0.5))+(d*t*d*0.5))/(wf*df+d*t)) #Distance of Neutral Axis in m\n",
-      "#Simplfying the computation\n",
-      "a=wf*df**3*12**-1\n",
-      "b=wf*df*((d+df*0.5)-y_bar*10**3)**2\n",
-      "c=t*d**3*12**-1\n",
-      "f=t*d*((d*0.5)-y_bar*10**3)**2\n",
-      "I=(a+b+c+f)*10**-12 #Moment of inertia in mm^3\n",
-      "\n",
-      "#Limit Moment\n",
-      "yp=(wf*df+d*t)/(2*t) #Plastic Neutral Axis in mm\n",
-      "Myp=I/y_bar #Yielding will start at moment without the stress term to ease computation\n",
-      "mom=10**-9*((t*yp**2*0.5)+(wf*df*(d-yp+10))+(t*25**2*0.5)) #Sum of 1st moments\n",
-      "Ml_Myp=mom*Myp**-1 #Ratio\n",
-      "\n",
-      "#Result\n",
-      "print \"The ratio ML/Myp=\",round(Ml_Myp,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ratio ML/Myp= 1.765\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.13.2, Page No:467"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "import numpy as np\n",
-      "\n",
-      "#Variable Decleration\n",
-      "E_st=200 #Youngs Modulus of Steel in GPa\n",
-      "sigma_st_yp=290 #Yielding Stress in MPa\n",
-      "E_al=70 #Youngs Modulus of Aluminium in GPa\n",
-      "sigma_al_yp=330 #Yielding Stresss of Aluminium in MPa\n",
-      "A_st=900 #Area of steel rod in mm^2\n",
-      "A_al=600 #Area of Aluminium rod in mm^2\n",
-      "L_st=350 #Length of the steel rod in mm\n",
-      "L_al=250 #Length of the aluminium rod in mm\n",
-      "\n",
-      "#Calculations\n",
-      "#Limit Load\n",
-      "P_st=sigma_st_yp*A_st*10**-3 #Load in limiting condition in kN\n",
-      "P_al=sigma_al_yp*A_al*10**-3 #Load in limiting condition in kN\n",
-      "P_L=P_st+2*P_al #Total Loading in kN\n",
-      "\n",
-      "#Elastic Unloading\n",
-      "#Solving for Pst and Pal using matri approach\n",
-      "A=np.array([[1,2],[L_st*(E_st*A_st)**-1,-L_al*(E_al*A_al)**-1]])\n",
-      "B=np.array([P_L,0])\n",
-      "C=np.linalg.solve(A,B) #Loading in kN\n",
-      "\n",
-      "#Residual Stresses\n",
-      "P_res_st=C[0]-P_st #Residual Load in kN\n",
-      "P_res_al=C[1]-P_al #Residual Load in kN\n",
-      "sigma_st=P_res_st/A_st #residual Stress in Steel in MPa\n",
-      "sigma_al=P_res_al/A_al #residual Stress in Aluminium in MPa\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print \"The Residual stresses are as follows\"\n",
-      "print \"Sigma_st=\",round(sigma_st*10**3,1),\"MPa and sigma_al=\",round(sigma_al*10**3,1),\"MPa\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Residual stresses are as follows\n",
-        "Sigma_st= 151.5 MPa and sigma_al= -113.6 MPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file