From 92cca121f959c6616e3da431c1e2d23c4fa5e886 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Tue, 7 Apr 2015 15:58:05 +0530 Subject: added books --- .../Chapter15.ipynb | 391 +++++++++++++++++++++ 1 file changed, 391 insertions(+) create mode 100755 Principles_Of_Geotechnical_Engineering/Chapter15.ipynb (limited to 'Principles_Of_Geotechnical_Engineering/Chapter15.ipynb') diff --git a/Principles_Of_Geotechnical_Engineering/Chapter15.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter15.ipynb new file mode 100755 index 00000000..d6b731fc --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter15.ipynb @@ -0,0 +1,391 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:22297f0d75def8e7d02cf6683fa1bc18bc061c4f42d017f0fb81a9ce765834b6" + }, + "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": 4 + }, + { + "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" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)Critical height of slope = 21.2 \n" + ] + } + ], + "prompt_number": 6 + }, + { + "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": [ + "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 -- cgit