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+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:4c2cef1cd7673363b2ef6cb15cb908c88921a9f728253dd7a799bca223f736c1"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter15-Slope Stability"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex1-pg518"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "#a.The factor of safety against sliding along the soil-rock interface.\n",

+      "#b.The height,H, that will give a factor of safety (Fs) of 2 against sliding alongthe soil-rock interface.\n",

+      "Gs=17.8\n",

+      "Gw=9.81\n",

+      "C=10.\n",

+      "c=20.\n",

+      "b=15.\n",

+      "H=6.\n",

+      "G=Gs-Gw\n",

+      "Fs= C/(Gs*H*math.cos(b/57.3)*math.cos(b/57.3)*math.tan(b/57.3))+G*math.tan(c/57.3)/(Gs*math.tan(b/57.3))\n",

+      "print'%s %.2f %s'%('a)The factor of safety = ',Fs,' ')\n",

+      "Fs=2.\n",

+      "H=2.247/(Fs-0.61)\n",

+      "print'%s %.2f %s'%(' b)H= ',H,' m')\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "a)The factor of safety =  0.98  \n",

+        " b)H=  1.62  m\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex3-pg529"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "#a.Determine the maximum depth up to which the excavation can be carried out.\n",

+      "#b.Find the radius,r, of the critical circle when the factor of safety is equal to 1(Part a).\n",

+      "#c. Find the distance  . BC\n",

+      "Cu=40.\n",

+      "G=17.5\n",

+      "b=60.\n",

+      "a=35.\n",

+      "c=72.5\n",

+      "m=0.195\n",

+      "Hc=Cu/(G*m)\n",

+      "r=Hc/(2.*math.sin(a/57.3)*math.sin((c/2)/57.3))\n",

+      "BC=Hc*((1./math.tan(a/57.3))-(1./math.tan(b/57.3)))\n",

+      "print'%s %.1f %s'%('a)The maximum depth Hc = ',Hc,' m')\n",

+      "print'%s %.2f %s'%(' b)The radius, r = ',r,' m')\n",

+      "print'%s %.3f %s'%(' c)The distance BC.= ',BC,' m')\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "a)The maximum depth Hc =  11.7  m\n",

+        " b)The radius, r =  17.28  m\n",

+        " c)The distance BC.=  9.973  m\n"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex4-pg531"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "#a.Determine the undrained cohesion of the clay (Figure 15.13).\n",

+      "#b.What was the nature of the critical circle?\n",

+      "#c. With reference to the toe of the slope, at what distance did the surface of sliding intersect the bottom of the excavation?\n",

+      "Gs=17.29\n",

+      "d=9.15\n",

+      "d1=6.1\n",

+      "D=d/d1\n",

+      "a=40.\n",

+      "m=0.175\n",

+      "b=40.\n",

+      "H=6.1\n",

+      "Cu=H*Gs*m\n",

+      "print'%s %.1f %s'%('a)The undrained cohesion of the clay Cu = ',Cu,' kN/m^2')\n",

+      "print(' b)The nature of the critical circle is midpointcircle')\n",

+      "d=1.5\n",

+      "b=40.\n",

+      "n=0.9\n",

+      "D1=n*H\n",

+      "print'%s %.1f %s'%(' c)Distance = ',D1,' m')\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "a)The undrained cohesion of the clay Cu =  18.5  kN/m^2\n",

+        " b)The nature of the critical circle is midpointcircle\n",

+        " c)Distance =  5.5  m\n"

+       ]

+      }

+     ],

+     "prompt_number": 1

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex5-pg534"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "#a.Determine the maximum depth up to which the cut could be made.\n",

+      "#b.How deep should the cut be made if a factor of safety of 2 against sliding is required\n",

+      "Fs=1.\n",

+      "b=56.\n",

+      "Kh=0.25\n",

+      "M=3.66\n",

+      "Cu=500.\n",

+      "G=100.\n",

+      "Hc=Cu*M/G\n",

+      "print'%s %.1f %s'%('a)The maximum depth =',Hc,' ft')\n",

+      "Fs=2.\n",

+      "H=Cu*M/(G*Fs)\n",

+      "print'%s %.1f %s'%(' b)H= ',H,' ft')\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "a)The maximum depth = 18.3  ft\n",

+        " b)H=  9.2  ft\n"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex6-pg541"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "#a.Find the critical height of the slope.\n",

+      "#b.If the height of the slope is 10 m, determine the factor of safety with respect to strength.\n",

+      "b=45.\n",

+      "c=20.\n",

+      "C=24.\n",

+      "G=18.9\n",

+      "m=0.06\n",

+      "Hc=C/(G*m)\n",

+      "Cd=G*Hc*m\n",

+      "Fc=C/Cd\n",

+      "print'%s %.1f %s'%('a)Critical height of slope = ',Hc,'')\n",

+      "#calculate the factor of safety using spencers solution\n",

+      "import math\n",

+      "%matplotlib inline\n",

+      "import warnings\n",

+      "warnings.filterwarnings('ignore')\n",

+      "import numpy\n",

+      "from math import tan\n",

+      "import matplotlib\n",

+      "from matplotlib import pyplot\n",

+      "phid=numpy.array([20,15,10,5])*math.pi/180.\n",

+      "mx=numpy.array([.06,.083,.105,.136])\n",

+      "cdx=numpy.array([11.34,15.69,19.85,25.7])\n",

+      "m=.06\n",

+      "g=18.9\n",

+      "cd=24.\n",

+      "#calculations\n",

+      "Hcr=cd/g/m\n",

+      "leng=len(phid)\n",

+      "Fcd=numpy.zeros(leng)\n",

+      "Fphi=numpy.zeros(leng)\n",

+      "tanphid=numpy.zeros(leng)\n",

+      "for i in range(0,leng):\n",

+      "\ttanphid[i]=math.tan(phid[i])\n",

+      "\tFphi[i]= tan(phid[0])/tan(phid[i])\n",

+      "\tFcd[i]=cd/cdx[i]\n",

+      "\n",

+      "\n",

+      "#results\n",

+      "print'%s %.2f %s'%('The value of Hcr (in m) = ',Hcr,'')\n",

+      "print 'from graph, Fss=1.4'\n",

+      "pyplot.plot(Fcd,Fphi)\n",

+      "pyplot.xlabel('Fc assumed')\n",

+      "pyplot.ylabel('Fc calculated')\n",

+      "pyplot.title('Graph of Fc assumed vs Fc calculated')\n",

+      "pyplot.show()\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "a)Critical height of slope =  21.2 \n",

+        "The value of Hcr (in m) =  21.16 \n",

+        "from graph, Fss=1.4\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

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dBxddBEcfnW4Xamb5qdZeSQJuBGaWSgqZe4Ejs+33Al4vTApWGzbeGH7yE7j/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\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x2d2ff30>"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex7-pg544"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# using Michalowski\u2019s solution.\n",

+      "import math\n",

+      "FSs=1.\n",

+      "c=20.\n",

+      "G=18.9\n",

+      "C=24.\n",

+      "Hcr=C/(G*math.tan(c/57.3)*0.17)\n",

+      "print'%s %.1f %s'%('a)Critical height Hc = ',Hcr,' m')\n",

+      "H=10.\n",

+      "k=C/(G*H*math.tan(c/57.3))\n",

+      "Fs=4.*math.tan(c/57.3)\n",

+      "print'%s %.1f %s'%(' b)Fs = ',Fs,'')\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "a)Critical height Hc =  20.5  m\n",

+        " b)Fs =  1.5 \n"

+       ]

+      }

+     ],

+     "prompt_number": 7

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex8-pg560"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "# Determine the factor ofsafety,Fs . Use Table 15.3.\n",

+      "W=22.4\n",

+      "C=20.\n",

+      "a=70.\n",

+      "s=math.sin(a/57.3)\n",

+      "c=math.cos(a/57.3)\n",

+      "l=2.924\n",

+      "Wn=W*s\n",

+      "Wn1=W*c\n",

+      "##doing this to all values\n",

+      "F1=30.501\n",

+      "F2=776.75\n",

+      "F3=1638.\n",

+      "Fs=(F1*C+F3*math.tan(C/57.3))/F2\n",

+      "print'%s %.2f %s'%('Fs = ',Fs,'')\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Fs =  1.55 \n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex9-pg560"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "#using Michalowski\u2019s solution\n",

+      "C=20.\n",

+      "G=18.5\n",

+      "r=0.25\n",

+      "H=21.62\n",

+      "C=25.\n",

+      "b= math.atan(0.5)\n",

+      "##from table 15.3 \n",

+      "m=1.624\n",

+      "n=1.338\n",

+      "Fs=m-n*r\n",

+      "print'%s %.1f %s'%(' The value of Fs for D= 1 is',Fs,'')\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        " The value of Fs for D= 1 is 1.3 \n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex10-pg561"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#calculate the factor of safety using spencers solution\n",

+      "import math\n",

+      "%matplotlib inline\n",

+      "import warnings\n",

+      "warnings.filterwarnings('ignore')\n",

+      "import numpy\n",

+      "from math import tan\n",

+      "import matplotlib\n",

+      "from matplotlib import pyplot\n",

+      "beta=numpy.array([26.57,26.57,26.57,26.57])\n",

+      "Fs=numpy.array([1.1,1.2,1.3,1.4])\n",

+      "phid=25*math.pi/180. #degrees\n",

+      "#calculations\n",

+      "print 'From spencers graphs,'\n",

+      "cd=numpy.array([0.0455,0.0417,0.0385,0.0357])\n",

+      "phia=numpy.array([18.,19.,20.,21.])*math.pi/180.\n",

+      "leng=len(phia)\n",

+      "Fss=numpy.zeros(leng)\n",

+      "for i in range(0,leng):\n",

+      "\tFss[i]= tan(phid)/tan(phia[i])\n",

+      "\n",

+      "#results\n",

+      "print 'From graph, a footing of dimensions Fs=1.3'\n",

+      "pyplot.plot(Fs,Fss)\n",

+      "pyplot.xlabel('Fs assumed')\n",

+      "pyplot.ylabel('Fs calculated')\n",

+      "pyplot.title('Graph of Fs assumed vs Fs calculated')\n",

+      "pyplot.show()"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "From spencers graphs,\n",

+        "From graph, a footing of dimensions Fs=1.3\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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ZmRXk/wN5SXo27nRWHQAAAABJRU5ErkJggg==\n",

+       "text": [

+        "<matplotlib.figure.Figure at 0x5567810>"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Ex11-pg561"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "import math\n",

+      "#using Michalowski\u2019s solution\n",

+      "C=20.\n",

+      "G=18.5\n",

+      "H=21.62\n",

+      "c=25.\n",

+      "r=0.25\n",

+      "Fs=3.1*math.tan(c/57.3)\n",

+      "print'%s %.1f %s'%('Fs = ',Fs,'')\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Fs =  1.4 \n"

+       ]

+      }

+     ],

+     "prompt_number": 10

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file