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Diffstat (limited to 'Principles_Of_Geotechnical_Engineering')
18 files changed, 3739 insertions, 0 deletions
diff --git a/Principles_Of_Geotechnical_Engineering/Chapter10.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter10.ipynb new file mode 100755 index 00000000..5d4082c6 --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter10.ipynb @@ -0,0 +1,109 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:a91d8543aecc1fd44cfae5fb69812b5536263b0841045f12cbdb74f6f431640e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter10-Stresses in a Soil Mass"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#principal stress and normal stresses and shear stresses\n",
+ "##initialisation of variables\n",
+ "sx= 2000. ##lb/ft^3\n",
+ "sy= 2500. ##lb/ft^3\n",
+ "T= 800. ##lb/ft^3\n",
+ "t= 0.348##radians\n",
+ "##calculations\n",
+ "s1= (sx+sy)/2.+math.sqrt(((sy-sx)/2.)**2+T**2)\n",
+ "s2= (sx+sy)/2.-math.sqrt(((sy-sx)/2.)**2+T**2)\n",
+ "sn= (sx+sy)/2.+(sy-sx)*math.cos(2.*t)/2.-T*math.sin(2*t)\n",
+ "Tn= (sy-sx)*math.sin(2.*t)/2.+T*math.cos(2*t)\n",
+ "##results\n",
+ "print'%s %.2f %s'% ('principle stress s1 = ',s1,' lb/ft^3 ')\n",
+ "print'%s %.2f %s'% ('principle stress s2 = ',s2,' lb/ft^3 ')\n",
+ "print'%s %.2f %s'% ('normal stress = ',sn,' lb/ft^3 ')\n",
+ "print'%s %.2f %s'% ('shear stress = ',Tn,' lb/ft^3 ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "principle stress s1 = 3088.15 lb/ft^3 \n",
+ "principle stress s2 = 1411.85 lb/ft^3 \n",
+ "normal stress = 1928.93 lb/ft^3 \n",
+ "shear stress = 774.22 lb/ft^3 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg262"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate vertical stress increase\n",
+ "##initialisation of variables\n",
+ "x= 3. ##m\n",
+ "y= 4. ##m\n",
+ "P= 5. ##kN\n",
+ "z= 2. ##m\n",
+ "##calculations\n",
+ "r= math.sqrt(x**2+y**2)\n",
+ "k= r/z\n",
+ "I= 3./(2.*math.pi*((r/z)**2+1)**2.5)\n",
+ "s= P*I/z**2\n",
+ "##results\n",
+ "print'%s %.4f %s'% ('verticle stress increase at 2m = ',s,' kN/m^3 ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "verticle stress increase at 2m = 0.0042 kN/m^3 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_Of_Geotechnical_Engineering/Chapter11.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter11.ipynb new file mode 100755 index 00000000..4b82ba4f --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter11.ipynb @@ -0,0 +1,553 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2a29c7b9c2c5d3914e37e2b72bd62a8b948e23c0720d69e0b40a353ece4dd036"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter11-Compressibility of Soil"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg303"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#evaluvate The elastic settlement at the centre of foundation\n",
+ "Tz=150.\n",
+ "b=1.\n",
+ "l=2.\n",
+ "z=5.*b\n",
+ "Es= (10000*2 + 8000*1 +12000*2)/5\n",
+ "a=4.\n",
+ "H=z\n",
+ "m=l/b\n",
+ "n=2.*H/b\n",
+ "F1=0.641 ##from tables 11.1 and 11.2\n",
+ "F2=0.031\n",
+ "u=0.3\n",
+ "Is= F1 + ((2.-u)/(1.-u))*F2\n",
+ "If=0.71 ##from table 11.3\n",
+ "Sef= Tz *a*b/l *(1-u**2)*Is*If/Es\n",
+ "Ser=0.93*Sef\n",
+ "print'%s %.3f %s'%('The elastic settlement at the centre of foundation =',Ser,'m')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The elastic settlement at the centre of foundation = 0.012 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "## one value of e is done \n",
+ "Gs=2.75\n",
+ "A=30.68\n",
+ "Ms=128.\n",
+ "p=1.\n",
+ "Hs=Ms/(A*Gs*p)\n",
+ "H=2.540\n",
+ "Hv=H-Hs\n",
+ "e=Hv/Hs\n",
+ "print'%s %.3f %s'%('the value of e for give values =',e,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of e for give values = 0.674 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg321"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Compression index\n",
+ "e11=0.9\n",
+ "e21=0.8\n",
+ "T2=4.\n",
+ "T1=2.\n",
+ "Cc= (e11-e21)/math.log10(T2/T1) ## from loading branch\n",
+ "e1=0.67\n",
+ "e2=0.655\n",
+ "Cs=(e1-e2)/math.log10(T2/T1)\n",
+ "k=Cs/Cc\n",
+ "T3=12.\n",
+ "e3=e11-Cc*math.log10(T3/T1)\n",
+ "print'%s %.2f %s'%('Compression index Cc= ',Cc,'')\n",
+ "print'%s %.2f %s'%(' Cs/Cc = ',k,'')\n",
+ "print'%s %.2f %s'%(' e3 = ',e3,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Compression index Cc= 0.33 \n",
+ " Cs/Cc = 0.15 \n",
+ " e3 = 0.64 \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg323"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Primary Consolidation Sc in three parts\n",
+ "Gd=14.\n",
+ "Gss=18.\n",
+ "Gsc=19.\n",
+ "Gw=9.81\n",
+ "To= 2.*Gd+4.*(Gss-Gw)+2*(Gsc-Gw)\n",
+ "LL=40.\n",
+ "Cc=0.009*(LL-10)\n",
+ "H=4.\n",
+ "T=100.\n",
+ "e=0.8\n",
+ "Sc= Cc*H*math.log10((To+T)/To)/(1.+e)\n",
+ "print'%s %.2f %s'%('a)Primary Consolidation Sc = ',Sc,' m')\n",
+ "\n",
+ "\n",
+ "Tc=190\n",
+ "Cs=Cc/6\n",
+ "Sc= Cs*H*math.log10((To+T)/To)/(1+e)\n",
+ "print'%s %.2f %s'%(' b)Primary Consolidation Sc =',Sc,'m')\n",
+ "\n",
+ "\n",
+ "Tc=170\n",
+ "Sc= Cc*H*math.log10((To+T)/Tc)/(1+e)+ Cs*H*math.log10(Tc/To)/(1+e)\n",
+ "print'%s %.3f %s'%(' c)Primary Consolidation Sc =',Sc,' m')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)Primary Consolidation Sc = 0.21 m\n",
+ " b)Primary Consolidation Sc = 0.04 m\n",
+ " c)Primary Consolidation Sc = 0.047 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg325"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate The settlement in the field Sc\n",
+ "Gs=18.\n",
+ "Gw=9.81\n",
+ "H=10.\n",
+ "eo=1.1\n",
+ "To=5.*(Gs-Gw)\n",
+ "T1=48.\n",
+ "T=To+T1\n",
+ "e1=1.045 ## void ratio corresponding to T \n",
+ "e=eo-e1\n",
+ "Sc=H*e/(1.+eo)\n",
+ "print'%s %.2f %s'%('The settlement in the field Sc = ',Sc,' m')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The settlement in the field Sc = 0.26 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg329"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Total consolidation settlement of the clay\n",
+ "T=8.5\n",
+ "eo=0.8\n",
+ "Cc=0.28\n",
+ "To=2650.\n",
+ "T1=970.\n",
+ "C1=0.02\n",
+ "t2=5.\n",
+ "t1=1.5\n",
+ "H=8.5*12\n",
+ "epr=Cc*math.log10((To+T1)/To)\n",
+ "ep=eo-epr\n",
+ "C2=C1/(1.+ep)\n",
+ "Sc=epr*H/(1.+eo)\n",
+ "Ss=C2*H*math.log10(t2/t1)\n",
+ "TS=Sc+Ss\n",
+ "print'%s %.1f %s'%('Total consolidation settlement of the clay =',TS,' in')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total consolidation settlement of the clay = 2.8 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate t field\n",
+ "##T50 = Cvtlab /H^2 lab = Cvtfield?H^2 fiels\n",
+ "tl=140.\n",
+ "Hf=3.\n",
+ "Hd=0.025/2.\n",
+ "tf=tl*Hf**2/Hd**2\n",
+ "k=tf/(3600.*24.)\n",
+ "print'%s %.1f %s'%('t field = ',k,' days')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "t field = 93.3 days\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##Tv is directly proportional to U^2\n",
+ "t1=93.333\n",
+ "U2=30.\n",
+ "U1=50.\n",
+ "t2=t1*U2**2./U1**2.\n",
+ "print'%s %.2f %s'%('t2 =',t2,' days')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "t2 = 33.60 days\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#evaluvate Cv\n",
+ "#intilization variable\n",
+ "t90=75.*24.*60.*60. ## time in sec\n",
+ "T90=0.848\n",
+ "Hd=1.5*100. ##in cm\n",
+ "Cv=T90*Hd**2/t90\n",
+ "print'%s %.3f %s'%('Cv =',Cv,' cm^2/sec')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Cv = 0.003 cm^2/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate K and t60\n",
+ "To=3000. ## lb/ft^2\n",
+ "eo=1.1\n",
+ "e1=0.9\n",
+ "e=eo-e1\n",
+ "ea=(eo+e1)/2.\n",
+ "T1=3000. ## lb/ft^2\n",
+ "T=1. ## in\n",
+ "t = 2. ## min\n",
+ "m=(e/T1)/(1.+ea)\n",
+ "U=50.\n",
+ "Tv=0.197\n",
+ "Gw=62.4 ##lb/ft^3\n",
+ "Cv=Tv*(T/(2.*12.)**2)/t\n",
+ "k=Cv*m*Gw *10**7\n",
+ "print'%s %.3f %s'%('a)k = ',k,' x10^-7 ft/min')\n",
+ "\n",
+ "\n",
+ "U=60\n",
+ "Tv=0.286\n",
+ "H=6\n",
+ "t60=Tv*H**2/(Cv*60*24)\n",
+ "print'%s %.1f %s'%(' b)t60 =',t60,' days')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)k = 3.557 x10^-7 ft/min\n",
+ " b)t60 = 41.8 days\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg344"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Cv\n",
+ "t50=19\n",
+ "Hd=2.24/2\n",
+ "Cv=0.197*Hd**2/t50\n",
+ "print'%s %.3f %s'%('Cv = ',Cv,' cm^2/min')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Cv = 0.013 cm^2/min\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg346"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate SC\n",
+ "LL=40.\n",
+ "Cc=0.009*(LL-10)\n",
+ "H=10.*12.\n",
+ "eo=1.0\n",
+ "Gss=120.\n",
+ "Gsc=110.\n",
+ "Gd=100.\n",
+ "To=10.*Gd +10.*(Gss-62.4)+10.*(Gsc-62.4)/2.\n",
+ "\n",
+ "Tt=0.408\n",
+ "Tm=0.232\n",
+ "Tb=0.019\n",
+ "Tav= (Tt+4.*Tm+Tb)/6.\n",
+ "Sc=Cc*H*math.log10((To+Tav*1000.)/To)/(1.+eo)\n",
+ "print'%s %.3f %s'%('Sc =',Sc,' in')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Sc = 0.826 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg356"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#intilization variable\n",
+ "#Calculate total primary\n",
+ "import math\n",
+ "H = 6.\n",
+ "Cc = 0.28\n",
+ "eo = 0.9\n",
+ "Cv = 0.36\n",
+ "To=210.\n",
+ "Tp=115.\n",
+ "Sc= Cc*H*math.log10((To+Tp)/To)/(1+eo)\n",
+ "t2=9.\n",
+ "Hd=3.\n",
+ "Tv=Cv*t2/Hd**2\n",
+ "U=0.67\n",
+ "Tf=0.677*Tp\n",
+ "print'%s %.1f %s'%('Tf =',Tf,' kN/m^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Tf = 77.9 kN/m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_Of_Geotechnical_Engineering/Chapter12.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter12.ipynb new file mode 100755 index 00000000..da95df97 --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter12.ipynb @@ -0,0 +1,325 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:cc62756cbf06ddef68226804d15a2efed303c30289a23f7a88b85756c1a62af7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter12-Shear Strength of Soil"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg378"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Determine the relationships for peak shear strength(tf) and residual shear strength(tr).\n",
+ "D=50 ## in mm\n",
+ "A= math.pi/4. *(D/1000.)**2\n",
+ "## solving for test 1 \n",
+ "N=150.\n",
+ "Sp=157.5\n",
+ "Sr=44.2\n",
+ "Tf=Sp/A\n",
+ "Tr=Sr/A\n",
+ "## from graph\n",
+ "k=math.tan(27/57.3)\n",
+ "k1=math.tan(14.6/57.3)\n",
+ "\n",
+ "print'%s %.3f %s'%('Peak strength Tf = 40+ t*',k,'')\n",
+ "print'%s %.3f %s'%(' Residual strength Tr = t*',k1,'')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak strength Tf = 40+ t* 0.509 \n",
+ " Residual strength Tr = t* 0.260 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Determine\n",
+ "#a.Angle of friction,f\u0004\n",
+ "#b.Angleuthat the failure plane makes with the major principal plane\n",
+ "T3=16. ## lb/in^2\n",
+ "Tf=25. ## lb/in^2\n",
+ "T1=T3+Tf\n",
+ "a= math.asin((T1-T3)/(T1+T3))*57.3 ## Mohr's circle\n",
+ "print'%s %.2f %s'%('a)Angle of friction,a = ',a,'')\n",
+ "b= 45.+ a/2.\n",
+ "print'%s %.2f %s'%(' b)Angle b that the failure plane makes with the major principal plane = ',b,'')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)Angle of friction,a = 26.02 \n",
+ " b)Angle b that the failure plane makes with the major principal plane = 58.01 \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg386"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Find the normal stress s\u0004and the shear stress tfon the failure plane.\n",
+ "#b.Determine the effective normal stress on the plane of maximum shear stress\n",
+ "T1=41.\n",
+ "T3=16.\n",
+ "a=58.\n",
+ "T=(T1+T3)/2. + (T1-T3)*math.cos(2.*a/57.3)/2.\n",
+ "tf=(T1-T3)*math.sin(2.*a/57.3)/2\n",
+ "print'%s %.2f %s'%('a)the normal stress T = ',T,' lb/in^2')\n",
+ "print'%s %.2f %s'%('(b) and the shear stress tf = ',tf,' lb/in^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)the normal stress T = 23.02 lb/in^2\n",
+ "(b) and the shear stress tf = 11.24 lb/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#The equation of the effective stress failure envelope for normally consolidated clayey soilistf \u0001s\u0004tan 30\u0005. A drained triaxial test was conducted with the same soil at a chamberconfining pressure of 10 lb/in.2Calculate the deviator stress at failure.\n",
+ "##For normally consolidated clay, c' \u0004= 0.\n",
+ "a=30.\n",
+ "T3=10.\n",
+ "T1=T3*(math.tan(60/57.3))**2\n",
+ "Tf=T1-T3\n",
+ "print'%s %.2f %s'%('The deviator stress at failure = ',Tf,' lb/in^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The deviator stress at failure = 19.99 lb/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Determine the shear strength parameters.\n",
+ "T13=70.\n",
+ "T1f=130.\n",
+ "T11=T13+T1f\n",
+ "\n",
+ "T23=160.\n",
+ "T2f=223.5\n",
+ "T21=T23+T2f\n",
+ "\n",
+ "a= 2*(math.atan(((T11-T21)/(T13-T23))**0.5) *57.3-45)\n",
+ "c= (T11-T13*((math.tan((45+a/2.)/57.3))**2)/(2*math.tan(45+a/2.)/57.3))\n",
+ "d=c-267\n",
+ "print('the shear strength parameter d = ',d,' kN/m^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "('the shear strength parameter d = ', 20.686836038348247, ' kN/m^2')\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg394"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#a.Consolidated-undrained angle of shearing resistance,f\n",
+ "#b.Drained friction angle,f\u0004\n",
+ "T3=12.\n",
+ "Tf=9.1\n",
+ "T1=T3+Tf\n",
+ "u=6.8\n",
+ "a=math.asin((T1-T3)/(T1+T3))\n",
+ "\n",
+ "a1= math.asin((T1-T3)/(T1+T3-2*u))\n",
+ "\n",
+ "print'%s %.1f %s'%('a)Consolidated-undrained angle of shearing resistance = ',a*57.3,' degrees')\n",
+ "print'%s %.1f %s'%(' b)Drained friction angle =',a1*57.3,' degrees')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)Consolidated-undrained angle of shearing resistance = 16.0 degrees\n",
+ " b)Drained friction angle = 27.8 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg395"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#What would be the deviatorstress at failure, (\u0010sd)f, if a drained test was conducted with the same chamber allaround pressure (that is, 12 lb/in.2)?\n",
+ "T3=12.\n",
+ "a=27.8\n",
+ "T1=T3*(math.tan(59./57.3))**2\n",
+ "Tf=T1-T3\n",
+ "print'%s %.1f %s'%('the deviator stress at failure = ',Tf,' lb/in^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the deviator stress at failure = 21.2 lb/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg400"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Estimate the average undrained shear strength of the clay [that is,cu(VST)].\n",
+ "PI=28.\n",
+ "OCR=3.2\n",
+ "To=160.\n",
+ "Kn=0.11+0.0037*PI\n",
+ "Ko=OCR**0.8 * Kn\n",
+ "Cu=Ko*To\n",
+ "print'%s %.1f %s'%('the average undrained shear strength of the clay =',Cu,' kN/m^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the average undrained shear strength of the clay = 86.7 kN/m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_Of_Geotechnical_Engineering/Chapter13.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter13.ipynb new file mode 100755 index 00000000..97917965 --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter13.ipynb @@ -0,0 +1,435 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0405d0d7063a41d23ac6999d3522bb5dd0c6d2025222a87c1afa5e51af3425e8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter13-Lateral Earth Pressure: \n",
+ "At-Rest, Rankine, and Coulomb"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-430"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Calculate the lateral force Poper unit length of the wall. Also, determine the location ofthe resultant force. Assume that for sand OCR\u00012\n",
+ "OCR=2.\n",
+ "a=30.\n",
+ "Ko=(1.-math.sin(a/57.3))*(OCR)**math.sin(a/57.3)\n",
+ "##at z=0\n",
+ "To1=0.\n",
+ "Th1=0.\n",
+ "u1=0.\n",
+ "##at z=10\n",
+ "To2=10.*100.\n",
+ "Th2=Ko*To2\n",
+ "u2=0.\n",
+ "##at z=15\n",
+ "To3= 10.*100.+5.*(122.4-62.4)\n",
+ "Th3=Ko*To3\n",
+ "u3=5.*62.4\n",
+ "##Lateral force Po =Area 1 +\u0007 Area 2+\u0007 Area3+\u0007 Area 4\n",
+ "Po =(1./2.)*10.*707.+5.*707.+(1./2.)*5.*212.1+(1/2.)*5.*312.\n",
+ "z=((3535.)*(5.+10./3.)+3535.*(5./2.)+530.3*(5./3.)+780.*(5./3.))/Po\n",
+ "print'%s %.1f %s'%('z = ',z,' ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "z = 4.8 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg449"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#a.Rankine active force per unit length of the wall and the location of theresultant\n",
+ "#b.Rankine passive force per unit length of the wall and the location of the resultant\n",
+ "##c=0\n",
+ "a=36.\n",
+ "G=16.\n",
+ "Ka=(1.-math.sin(a/57.3))/(1.+math.sin(a/57.3))\n",
+ "##at z=0 Tp=0\n",
+ "z=6.\n",
+ "To=G*z\n",
+ "Ta=Ka*To\n",
+ "Pa=z*Ta/2.\n",
+ "\n",
+ "print'%s %.1f %s'%('a)Rankine active force per unit length of the wall = ',Pa,' kN/m')\n",
+ "print(' and the location of the resultant is z = 2m')\n",
+ "\n",
+ "\n",
+ "p=36.\n",
+ "G=16.\n",
+ "Kp=(1+math.sin(a/57.3))/(1-math.sin(a/57.3))\n",
+ "##at z=0 Tp=0\n",
+ "z=6.\n",
+ "To=G*z\n",
+ "Tp=Kp*To\n",
+ "Pp=z*Tp/2.\n",
+ "\n",
+ "print'%s %.1f %s'%(' b)Rankine passive force per unit length of the wall = ',Pp,' kN/m')\n",
+ "print (' and the location of the resultant is z = 2m')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)Rankine active force per unit length of the wall = 74.8 kN/m\n",
+ " and the location of the resultant is z = 2m\n",
+ " b)Rankine passive force per unit length of the wall = 1109.2 kN/m\n",
+ " and the location of the resultant is z = 2m\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg450"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Determine the active force Paperunit length of the wall as well as the location and direction of the resultant.\n",
+ "H=12.\n",
+ "a=20.\n",
+ "b=20.\n",
+ "G=115.\n",
+ "c=30.\n",
+ "Oa= math.asin(math.sin(a/57.3)/math.sin(c/57.3))*57.3-a+2.*b\n",
+ "Ka= (math.cos((a-b)/57.3)*math.sqrt(1.+(math.sin(c/57.3))**2.-2.*math.sin(c/57.3)*math.cos(Oa/57.3)))/((math.cos(b/57.3))**2.*(math.cos(a/57.3)+math.sqrt((math.sin(c/57.3))**2.-(math.sin(a/57.3))**2)))\n",
+ "Pa=G*H**2.*Ka/2.\n",
+ "B= math.atan((math.sin(c/57.3)*math.sin(Oa/57.3))/(1.-(math.sin(c/57.3)*math.cos(Oa/57.3))))*57.3\n",
+ "print'%s %.1f %s'%('The active force Pa per unit length of the wall = ',Pa,' lb/ft')\n",
+ "print'%s %.1f %s'%( ' The resultant will act a distance of 12/3 = 4 ft above the bottom of the wall with B = ',B,' degree')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The active force Pa per unit length of the wall = 6423.5 lb/ft\n",
+ " The resultant will act a distance of 12/3 = 4 ft above the bottom of the wall with B = 30.0 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg451"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#determine the force per unit length of the wall for Rankine\u2019s active state. Also find the location of the resultant.\n",
+ "a=30.\n",
+ "Ka1=(1.-math.sin(a/57.3))/(1.+math.sin(a/57.3))\n",
+ "a=35.\n",
+ "Ka2=(1-math.sin(a/57.3))/(1+math.sin(a/57.3))\n",
+ "##at z=0 so T0=0\n",
+ "##atz=3\n",
+ "To=3.*16.\n",
+ "Ta1=Ka1*To\n",
+ "Ta2=Ka2*To\n",
+ "\n",
+ "## At z=6\n",
+ "To=3.*16.+3.*(18.-9.81)\n",
+ "Ta2=Ka2*To\n",
+ "\n",
+ "Pa =(1/2.)*3.*16.+3.*13.0+ (1/2.)*3.*36.1\n",
+ "z= (24 *(3.+3./3.)+39.0*(3/2.)+54.15*(3/3.))/Pa\n",
+ "print'%s %.1f %s'%('The force per unit length of the wall = ',Pa,' kN/m')\n",
+ "print'%s %.1f %s'% (' The location of the resultant = ',z,'m ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The force per unit length of the wall = 117.2 kN/m\n",
+ " The location of the resultant = 1.8 m \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg453"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#a.Maximum depth of the tensile crack\n",
+ "#b.Pabefore the tensile crack occurs\n",
+ "#c. Pa after the tensile crack occurs\n",
+ "Ka= (math.tan(1./57.3))**2.\n",
+ "G=16.5\n",
+ "cu=10.\n",
+ "H=6.\n",
+ "##at z=0\n",
+ "z=0.\n",
+ "Ta=G*z-2.*cu\n",
+ "##zt z=6\n",
+ "z=6.\n",
+ "Ta=G*z-2.*cu\n",
+ "\n",
+ "zo=2.*cu/G\n",
+ "## Before the tensile crack occurs\n",
+ "Pa= G*H**2./2. - 2.*cu*H\n",
+ "print'%s %.1f %s'%('Pa before the tensile crack occurs = ',Pa,' kN/m')\n",
+ "##After the tensile crack occurs\n",
+ "Pa=(H-zo)*Ta/2.\n",
+ "print'%s %.1f %s'%(' Pa after the tensile crack occurs = ',Pa,' kN/m')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pa before the tensile crack occurs = 177.0 kN/m\n",
+ " Pa after the tensile crack occurs = 189.1 kN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg457"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Determine the Rankine active force Paon the retaining wall after the tensile crack occurs.\n",
+ "H=15.\n",
+ "a=10.\n",
+ "G=118.\n",
+ "b=20.\n",
+ "C=250\n",
+ "Zo=2.*C*math.sqrt((1+math.sin(b/57.3))/(1.-math.sin(b/57.3)))/G\n",
+ "##at z=0 Ta=0\n",
+ "##at z=15 \n",
+ "z=15.\n",
+ "K=0.3\n",
+ "Ta=G*z*K*math.cos(a/57.3)\n",
+ "Pa=(H -Zo)*Ta/2.\n",
+ "print'%s %.1f %s'%('The Rankine active force Pa on the retaining wall after the tensile crack occurs = ',Pa,' lb/ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Rankine active force Pa on the retaining wall after the tensile crack occurs = 2339.8 lb/ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg459"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Estimate the active force,Pa , per unit length of the wall. Also, state the direction and location of the resultant force,Pa.\n",
+ "import math\n",
+ "c=30.\n",
+ "b=15.\n",
+ "a=10.\n",
+ "Ka=0.3872 ## from table 13.8\n",
+ "H=4.\n",
+ "G=15.\n",
+ "Pa=G*H**2.*Ka/2.\n",
+ "print'%s %.1f %s'%('The active force per unit length Pa = ',Pa,' kN/m')\n",
+ "print(' The resultant will act at a vertical distance equal to H/3 = 4/3 = 1.33 m above ' ' the bottom of the wall and will be inclined at an angle of 15\u0005to the back face of the wall.')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The active force per unit length Pa = 46.5 kN/m\n",
+ " The resultant will act at a vertical distance equal to H/3 = 4/3 = 1.33 m above the bottom of the wall and will be inclined at an angle of 15\u0005to the back face of the wall.\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg478"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Determine Pae.Also determine the location of the resultant line of action of Pae\u2014that is, .\n",
+ "kh=0.2\n",
+ "kv=0.\n",
+ "H=4.\n",
+ "a=0.\n",
+ "b=0.\n",
+ "c=15.\n",
+ "d=30.\n",
+ "G=15.5\n",
+ "B= math.atan(kh/(1-kv)/57.3)\n",
+ "b1=b+B\n",
+ "a1=a+B\n",
+ "Ka=0.452\n",
+ "Pa=G*H**2.*Ka/2.\n",
+ "Pae=Pa*(1.-kv)*((math.cos(b1/57.3))**2./((math.cos(b/57.3))**2.*(math.cos(B/57.3))**2.))\n",
+ "Ka=0.3014\n",
+ "Pa=G*H**2*Ka/2.\n",
+ "P1=Pae-Pa\n",
+ "z= ((Pa*H/3)+P1*0.6*H)/Pae\n",
+ "print'%s %.1f %s'%('Pae = ',Pae,' kN/m')\n",
+ "print'%s %.1f %s'%(' Z = ',z,' m')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pae = 56.0 kN/m\n",
+ " Z = 1.7 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Determine the magnitude of the active force,Pae.\n",
+ "H=28.\n",
+ "C=210.\n",
+ "b=10.\n",
+ "G=118.\n",
+ "c=20.\n",
+ "kh=0.1\n",
+ "Ka=math.tan(35./57.3)\n",
+ "zo=2.*C/(G*(Ka))\n",
+ "n=zo/(H-zo)\n",
+ "Nac=1.60\n",
+ "Nav=0.375\n",
+ "L=1.17\n",
+ "Pae= G*(H-zo)**2*(L*Nav)-C*(H-zo)*Nac\n",
+ "print'%s %.1f %s'%('The magnitude of the active force, Pae = ',Pae,' lb/ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnitude of the active force, Pae = 19488.8 lb/ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_Of_Geotechnical_Engineering/Chapter14.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter14.ipynb new file mode 100755 index 00000000..e7939c1f --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter14.ipynb @@ -0,0 +1,123 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2af8058b6581fea92b2edf9bd8732fbea78d002de843d3df7b927ba2074ae2e2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter14-Lateral Earth Pressure: \n",
+ "Curved Failure Surface"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Ex1-497"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#a.Coulomb\u2019s theory\n",
+ "#b.Terzaghi and Peck\u2019s wedge theory\n",
+ "#c. Shields and Tolunay\u2019s solution (method of slices)\n",
+ "#d.Zhu and Qian\u2019s solution (method of triangular slices)\n",
+ "G=15.7\n",
+ "a=0.\n",
+ "b=15.\n",
+ "c=30.\n",
+ "H=3.\n",
+ "Kp=4.977 ## from table 13.9\n",
+ "Pp=Kp*G*H**2./2.\n",
+ "print'%s %.1f %s'%('a)the passive force = ',Pp,' kN/m')\n",
+ "## for part b\n",
+ "Kp=4.53\n",
+ "Pp=Kp*G*H**2./2.\n",
+ "print'%s %.1f %s'%(' b)the passive force = ',Pp,' kN/m')\n",
+ "## for part c\n",
+ "Kp=4.13\n",
+ "Pp=Kp*G*H**2/2.\n",
+ "print'%s %.1f %s'%(' c)the passive force =',Pp,' kN/m')\n",
+ "##for part d\n",
+ "Kp=4.56\n",
+ "Pp=Kp*G*H**2/2.\n",
+ "print'%s %.1f %s'%(' d)the passive force =',Pp,' kN/m')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)the passive force = 351.6 kN/m\n",
+ " b)the passive force = 320.0 kN/m\n",
+ " c)the passive force = 291.8 kN/m\n",
+ " d)the passive force = 322.2 kN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg507"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the design strut loads.\n",
+ "G=16.\n",
+ "H=7.\n",
+ "c=30.\n",
+ "Ta=0.65*G*H*(math.tan(30./57.3))**2\n",
+ "A=Ta*3.*3./4.\n",
+ "B1=Ta*3.-54.61\n",
+ "C=Ta*4.*4./4.\n",
+ "B2=Ta*4.-97.08\n",
+ "s=2.\n",
+ "As=A*s\n",
+ "Bs=(B1+B2)*s\n",
+ "Cs=C*s\n",
+ "print'%s %.1f %s'%( 'The strut loads at level A = ',As,' kN')\n",
+ "print'%s %.1f %s'%( ' The strut loads at level B = ',Bs,' kN')\n",
+ "print'%s %.1f %s'%( ' The strut loads at level C =',Cs,' kN')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The strut loads at level A = 109.2 kN\n",
+ " The strut loads at level B = 36.3 kN\n",
+ " The strut loads at level C = 194.1 kN\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file 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 diff --git a/Principles_Of_Geotechnical_Engineering/Chapter16.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter16.ipynb new file mode 100755 index 00000000..250a7911 --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter16.ipynb @@ -0,0 +1,319 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f9b870b8ab70a6506c4707b625840ce3fe5d061d1cd38773c3e0459e51172aa1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Chapter16-Soil-Bearing Capacity for Shallow\n",
+ "Foundations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg587"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine the gross allowable load per unit area (qall) that the foundation can carry.\n",
+ "import math\n",
+ "c=20.\n",
+ "## from table 16.1\n",
+ "Nc=17.69\n",
+ "Nq=7.44\n",
+ "Ng=3.64\n",
+ "\n",
+ "Df=3.\n",
+ "G=110.\n",
+ "q=G*Df\n",
+ "\n",
+ "C=200.\n",
+ "B=4.\n",
+ "\n",
+ "Qu= C*Nc+q*Nq+G*B*Ng/2.\n",
+ "\n",
+ "Fs=3.\n",
+ "Qall=Qu/Fs\n",
+ "print'%s %.1f %s'%('Qa = ',Qall,' lb/ft**2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Qa = 2264.7 lb/ft**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg588"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#determine the size of the footing\u2014that is, the size of B.\n",
+ "G=18.15\n",
+ "qa=30000.*9.81/1000.\n",
+ "\n",
+ "Nc=57.75\n",
+ "Nq=41.44\n",
+ "Ng=45.41\n",
+ "C=0.\n",
+ "q=G*1.\n",
+ "B=1.\n",
+ "(1.3*C*Nc+q*Nq+0.4*G*B*Ng)*B**2/3. == qa\n",
+ "B= math.sqrt(294.3/(250.7+109.9))\n",
+ "print'%s %.1f %s'%(' B = ',B,' m')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " B = 0.9 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg595"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "# Determine the safe gross load (factor of safety of 3) that the footing can carry\n",
+ "B=1.2\n",
+ "L=1.2\n",
+ "c=32.\n",
+ "C=0.\n",
+ "Df=1.\n",
+ "G=16.\n",
+ "Nq=23.18\n",
+ "Ng=22.02\n",
+ "Nc=1.\n",
+ "Lqs=1.+0.1*B*(math.tan(61./57.3)**2.)/L\n",
+ "Lgs=Lqs\n",
+ "Lqd=1.+0.1*Df*math.tan(61./57.3)/B\n",
+ "Lgd=Lqd\n",
+ "Lcs=1.\n",
+ "Lcd=1.\n",
+ "Gs=19.5\n",
+ "q=0.5*G+0.5*(Gs-9.81)\n",
+ "Qu= C*Lcs*Lcd*Nc+q*Lqs*Lqd*Nq+(Gs-9.81)*Lgs*Lgd*B*Ng/2.\n",
+ "Qa=Qu/3.\n",
+ "Q=Qa*B**2.\n",
+ "print'%s %.1f %s'%('the gross load = ',Q,' kN')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the gross load = 311.6 kN\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg601"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Determine the magnitude of the gross ultimate load applied eccentrically for bearing capacity failure in soil.\n",
+ "e=0.1\n",
+ "B=1.\n",
+ "X=B-2.*e\n",
+ "Y=1.5\n",
+ "B1=0.8\n",
+ "L1=1.5\n",
+ "c=30.\n",
+ "Df=1.\n",
+ "Nq=18.4\n",
+ "Ng=15.668\n",
+ "q=1.*18.\n",
+ "G=18.\n",
+ "Lqs=1.+e*(B1/L1)*math.tan(60./57.3)**2.\n",
+ "Lgs=Lqs\n",
+ "Lqd=1.+e*(Df/B1)*math.tan(60./57.3)\n",
+ "Lgd=Lqd\n",
+ "qu=q*Lqs*Lqd*Nq+Lgs*Lgd*G*B1*Ng/2.\n",
+ "Qu=qu*B1*L1\n",
+ "print'%s %.1f %s'%('The magnitude of the gross ultimate load =',Qu,' kN')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnitude of the gross ultimate load = 751.8 kN\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg601"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine the gross ultimate load per unit length that the foundation can carry.\n",
+ "import math\n",
+ "B=1.5\n",
+ "Df=0.75\n",
+ "e=0.1*B\n",
+ "G=17.5\n",
+ "c=30.\n",
+ "C=0.\n",
+ "q=G*Df\n",
+ "Nq=18.4\n",
+ "Ng=15.668\n",
+ "Lqd=1.+0.1*(Df/B)*math.tan(60./57.3)\n",
+ "Lgd=Lqd\n",
+ "Quc=q*Nq*Lqd+Lgd*B*Ng/2.\n",
+ "k=0.8\n",
+ "a=1.754\n",
+ "Qua=Quc*(1.-a*(e/B)**k)\n",
+ "print'%s %.1f %s'%('The gross ultimate load per unit length = ',Qua,' kN')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The gross ultimate load per unit length = 198.7 kN\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg606"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Estimate the ultimate bearing capacity of a circular footing with a diameter of 1.5 m. The soil is sandy.\n",
+ "Qup=280.\n",
+ "Bp=0.7 ## in m\n",
+ "Bf=1.5\n",
+ "Quf=Qup*Bf/Bp\n",
+ "print'%s %.1f %s'%('The ultimate bearing capacity = ',Quf,' kN/m**2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ultimate bearing capacity = 600.0 kN/m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg606"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Determine the size of a square column foundation that should carry a load of 2500 kN with a maximum settlement of 25 mm.\n",
+ "a=2500.\n",
+ "##doing for the first values only\n",
+ "Bf=4.\n",
+ "Bp=0.305\n",
+ "q=a/Bf**2.\n",
+ "Sep=4.\n",
+ "Sef=Sep*(2.*Bf/(Bf+Bp))**2\n",
+ "print'%s %.1f %s'%('Sef = ',Sef,' mm')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Sef = 13.8 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_Of_Geotechnical_Engineering/Chapter18.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter18.ipynb new file mode 100755 index 00000000..642dccef --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter18.ipynb @@ -0,0 +1,124 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c0db9cffec0e49340dfb3b8ea72b7fe58131405a32fcba0dd38be68beb6719b2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter18-Subsoil Exploration"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg642"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the correctedstandard penetration numbers, (N1)60, at various depths\n",
+ "import math\n",
+ "##solving for z=5 only\n",
+ "To=0.275\n",
+ "Cn=To**(-0.5)\n",
+ "N60=8\n",
+ "N160=Cn*N60\n",
+ "print'%s %.1f %s'%('(N1)60 = ',N160,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(N1)60 = 15.3 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg643"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#estimate the average soil friction angle,f\u0003, from z\u00020 to z\u000225 ft\n",
+ "#import math\n",
+ "z=5.\n",
+ "To=0.275\n",
+ "Cn=2./(1.+To)\n",
+ "N60=8.\n",
+ "N160=Cn*N60\n",
+ "print'%s %.1f %s'%('(N1)60 = ',N160,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(N1)60 = 12.5 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg643"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#estimate the average soil friction angle, f\u0003, from z\u00020 to z\u000225 ft\n",
+ "import math\n",
+ "pa=1. ## 14.7 lb/in**2 = 1ton/ft**2\n",
+ "To=0.275 ## ton/ ft**2\n",
+ "N60=8.\n",
+ "c= math.atan((N60/12.2+20.3*(To/pa))*57.3)**0.34\n",
+ "print'%s %.1f %s'%('The average soil friction angle = ',c,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The average soil friction angle = 1.2 \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_Of_Geotechnical_Engineering/Chapter2.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter2.ipynb new file mode 100755 index 00000000..d6a4aa22 --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter2.ipynb @@ -0,0 +1,62 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:d218f94a8c632b085e71b27c7df143c1bf3edcfb5f6cf33f03429d4e281c4928"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter2-Origin of sail and Grain size"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate uniformity coefficient and coefficient of gradation\n",
+ "##initialisation of variables\n",
+ "##from graph\n",
+ "d= 0.15 ##mm\n",
+ "w= 0.17 ##mm\n",
+ "a= 0.27 ##mm\n",
+ "##calculations\n",
+ "C= a/d\n",
+ "c= w**2/(a*d)\n",
+ "##results\n",
+ "print'%s %.1f %s'%('uniformity coefficient = ',C,\"\")\n",
+ "print'%s %.2f %s'% ('coefficient of gradation = ',c,' ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "uniformity coefficient = 1.8 \n",
+ "coefficient of gradation = 0.71 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_Of_Geotechnical_Engineering/Chapter3.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter3.ipynb new file mode 100755 index 00000000..16fb2a1b --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter3.ipynb @@ -0,0 +1,199 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:dd7ff85f7d52d3589a1d7be4767ada04770d7408bb98377fd79ee304b32478a3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter3-Weight- volume relationships"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg 60"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#evaluvate moist and dry density and void ratio and porosity and degree of saturation and volume of water in soil sample\n",
+ "##initialisation of variables\n",
+ "V= 1.2 ##m**3\n",
+ "M= 2350. ##Kg\n",
+ "w= 0.086\n",
+ "G= 2.71\n",
+ "W= 1000. ##kg/m**3\n",
+ "##calculations\n",
+ "R= M/V\n",
+ "D= M/((1.+w)*V)\n",
+ "e= (G*W/D)-1.\n",
+ "n= e/(e+1.)\n",
+ "S= (w*G/e)*100.\n",
+ "v= (M-(M/(1.+w)))/W\n",
+ "##results\n",
+ "print'%s %.1f %s'% ('moist density = ',R,' kg/m**3 ')\n",
+ "print'%s %.1f %s'% ('dry density = ',D,' kg/m**3 ')\n",
+ "print'%s %.3f %s'% ('void ratio = ',e,' ')\n",
+ "print'%s %.3f %s'% ('porosity = ',n,'')\n",
+ "print'%s %.3f %s'% ('Degree of saturation = ',S,' ')\n",
+ "print'%s %.3f %s'% ('volume of water in soil sample = ',v,' m**3 ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "moist density = 1958.3 kg/m**3 \n",
+ "dry density = 1803.3 kg/m**3 \n",
+ "void ratio = 0.503 \n",
+ "porosity = 0.335 \n",
+ "Degree of saturation = 46.349 \n",
+ "volume of water in soil sample = 0.186 m**3 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg 63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calcualte mass of water to be added for full saturation\n",
+ "##initialisation of variables\n",
+ "n= 0.4\n",
+ "G= 2.68\n",
+ "w= 0.12\n",
+ "R= 1000. ##kg/m**3\n",
+ "V= 10. ##m**3\n",
+ "##calculations\n",
+ "d= G*R*(1.-n)*(1.+w)\n",
+ "s= ((1.-n)*G+n)*R\n",
+ "M= s-d\n",
+ "m= M*V\n",
+ "##results\n",
+ "print'%s %.1f %s'%('mass of water to be added for full saturation = ',m,' kg ')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "mass of water to be added for full saturation = 2070.4 kg \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculatesatuarated unit weight and specific gravity and void ratio \n",
+ "##initialisation of variables\n",
+ "d= 16.19 ##kN/m**3\n",
+ "w= 0.23\n",
+ "W= 9.81 ##kN/m**3\n",
+ "##calculations\n",
+ "R= d*(1.+w)\n",
+ "G= d/(W-d*w)\n",
+ "e= w*G\n",
+ "##results\n",
+ "print'%s %.2f %s'%('satuarated unit weight = ',R,' kN/m**3 ')\n",
+ "print '%s %.2f %s'%('specific gravity = ',G,' ')\n",
+ "print'%s %.2f %s'% ('void ratio = ',e,' ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "satuarated unit weight = 19.91 kN/m**3 \n",
+ "specific gravity = 2.66 \n",
+ "void ratio = 0.61 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate relative density of compaction in percentage\n",
+ "##initialisation of variables\n",
+ "G= 2.68\n",
+ "w= 0.12\n",
+ "d= 1794.4 ##kg/m**3\n",
+ "W= 1000. ##kg/m**3\n",
+ "emax= 0.75\n",
+ "emin= 0.4\n",
+ "##calculation\n",
+ "e= (G*W*(1.+w)/d)-1.\n",
+ "D= ((emax-e)/(emax-emin))*100.\n",
+ "##results\n",
+ "print'%s %.1f %s'% ('relative density of compaction in percentage = ',D,' ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "relative density of compaction in percentage = 22.1 \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_Of_Geotechnical_Engineering/Chapter4.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter4.ipynb new file mode 100755 index 00000000..899bd2de --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter4.ipynb @@ -0,0 +1,61 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ec42a40195fbc1251bf05efd578442adffe962bc7bf4cd047dc1251517110ed5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter4-Plasticity and Structure of soil"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg83"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate shrinkage limit of the soil \n",
+ "##initialisation of variables\n",
+ "V1= 24.6 ##cm^3\n",
+ "V2= 15.9 ##cm^3\n",
+ "M1= 44 ##g\n",
+ "M2= 30.1 ##g\n",
+ "W= 1 ##g/cm^3\n",
+ "##calculations\n",
+ "SL= (((M1-M2)/M2)*100)-(((V1-V2)/M2)*W*100.)\n",
+ "##results\n",
+ "print'%s %.1f %s'%('shrinkage limit of the soil = ',SL,' ')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "shrinkage limit of the soil = 17.3 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_Of_Geotechnical_Engineering/Chapter6.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter6.ipynb new file mode 100755 index 00000000..c085ddbe --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter6.ipynb @@ -0,0 +1,148 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7d0e973e8ccf199512f5a464e18af5d87fc63e5365435eef6c9cd5f40b3801c3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter6-Soil Compaction"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg127"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate maximum dry density and optimum moisture content\n",
+ "##initialisation of variables\n",
+ "G= 2.6\n",
+ "LL= 20.\n",
+ "P= 20.\n",
+ "##calclations\n",
+ "R= (4804574.*G-195.55*(LL)**2+156971*(P)**0.5-9527830)**0.5\n",
+ "n= (1.195e-4)*((LL)**2)-1.964*G-(6.617e-5)*(P)+7.651\n",
+ "w= math.e**n\n",
+ "##results\n",
+ "print'%s %.1f %s'% ('maximum dry density = ',R,' kg/m^3 ')\n",
+ "print'%s %.2f %s'%('optimum moisture content = ',w,' ')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "maximum dry density = 1894.2 kg/m^3 \n",
+ "optimum moisture content = 13.34 \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate dry unit weight of compaction in the field and dry unit weight of compaction in the field\n",
+ "##initialisation of variables\n",
+ "do= 1570. ##kg/m^3\n",
+ "mo= 0.545 ##kg\n",
+ "M1= 7.59 ##kg\n",
+ "M2= 4.78 ##kg\n",
+ "M3= 3.007 ##kg\n",
+ "w= 0.102 ##\n",
+ "dmax= 19. ##KN/m^3\n",
+ "##calculations\n",
+ "Ms= M1-M2\n",
+ "Mc= Ms-mo\n",
+ "Vh= Mc/do\n",
+ "Dc= M3/Vh\n",
+ "Du= Dc*9.81/1000.\n",
+ "f= Du/(1.+w)\n",
+ "Rc= f*100./dmax\n",
+ "##results\n",
+ "print'%s %.2f %s'% ('dry unit weight of compaction in the field = ',f,' kN/m^3 ')\n",
+ "print'%s %.1f %s'% ('relative compaction in the field = ',Rc,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "dry unit weight of compaction in the field = 18.55 kN/m^3 \n",
+ "relative compaction in the field = 97.7 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate sustainabilty number\n",
+ "##initialisation of variables\n",
+ "D1= 0.36 ##mm\n",
+ "D2= 0.52 ##mm\n",
+ "D5= 1.42 ##mm\n",
+ "##calculations\n",
+ "Sn= 1.7*(math.sqrt((3./(D5)**2)+(1./(D2)**2)+(1./(D1)**2)))\n",
+ "##results\n",
+ "print'%s %.1f %s'% ('sustainabilty number = ',Sn,' ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "sustainabilty number = 6.1 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_Of_Geotechnical_Engineering/Chapter7.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter7.ipynb new file mode 100755 index 00000000..75e8c79b --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter7.ipynb @@ -0,0 +1,444 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:db9787af9c3d4874a56e96ce130e20ad75db81e8408117171d5753e7537a242c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter7-Permeability"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the hydraulic conductivity in cm/sec.\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "L= 30. ##cm\n",
+ "A= 177. ##cm^2\n",
+ "h= 50. ##cm\n",
+ "Q= 350. ##cm^3\n",
+ "t= 300. ##sec\n",
+ "##claculations\n",
+ "k=Q*L/(A*h*t)\n",
+ "##results\n",
+ "print'%s %.4f %s'% ('hydraulic conductivity = ',k,' cm/sec ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "hydraulic conductivity = 0.0040 cm/sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg169\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the hydraulic conductivity of the soil in in./sec.\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "L= 203. ##mm\n",
+ "A= 10.3 ##cm^2\n",
+ "a= 0.39 ##cm^2\n",
+ "h0= 508. ##mm\n",
+ "h180= 305. ##mm\n",
+ "t= 180. ##sec\n",
+ "##calculations\n",
+ "k= 2.303*a*L*math.log10(h0/h180)/(A*t)\n",
+ "##results\n",
+ "print'%s %.2f %s'% ('hydraulic conductivity of sand = ',k,' in/sec ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "hydraulic conductivity of sand = 0.02 in/sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#The hydraulic conductivity of a clayey soil is 3 107 cm/sec. The viscosity of water at 25\u00b0C is 0.0911 104 g # sec/cm2 \n",
+ "#Calculate the absolute permeability of the soil.\n",
+ "import math\n",
+ "##initialisation of varilables\n",
+ "k= 3e-7 ##cm/sec\n",
+ "n= 0.0911e-4 ##g*sec/cm^2\n",
+ "dw= 1 ##g/cc\n",
+ "##calculations\n",
+ "K= k*n/dw\n",
+ "##results\n",
+ "print'%s %.4f %s'% ('absolute premeability = ',K,' cm^2 ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "absolute premeability = 0.0000 cm^2 \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg170"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#With k 5.3 105 m/sec for the permeable layer, calculate the rate of seepage through it in m3 /hr/m width if H 3 m and a 8\u00b0.\n",
+ "\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "k= 5.3e-5 ##m/sec\n",
+ "H= 3 ##m\n",
+ "a= 0.139 ##radians\n",
+ "##calculations\n",
+ "A= H*math.cos(a)\n",
+ "i= math.sin(a)\n",
+ "q= k*i*A*3600\n",
+ "##results\n",
+ "print'%s %.4f %s'% ('rate of seepage = ',q,' m^3/hr/m ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "rate of seepage = 0.0785 m^3/hr/m \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate flow rate\n",
+ "##initialisation of variables\n",
+ "L= 50. ##m\n",
+ "k= 0.08e-2##m/sec\n",
+ "h= 4. ##m\n",
+ "H1= 3. ##m\n",
+ "H= 8. ##m\n",
+ "a= 0.139 ##radians\n",
+ "##calculations\n",
+ "i= h*math.cos(a)/L\n",
+ "A= H1*math.cos(a)\n",
+ "q= k*i*A\n",
+ "##results\n",
+ "print'%s %.5f %s'% ('flow rate = ',q,' m^3/sec/m ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "flow rate = 0.00019 m^3/sec/m \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg174"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate hydraulic conductivity at void ratio of 0.65\n",
+ "##initialisation of variables\n",
+ "k1= 0.02 ##cm/sec\n",
+ "e1= 0.5 \n",
+ "e2= 0.65\n",
+ "##calculations\n",
+ "k2= k1*(e2**3/(1.+e2))/(e1**3/(1.+e1))\n",
+ "##results\n",
+ "print'%s %.2f %s'% ('hydraulic conductivity at void ratio of 0.65 =',k2,'cm/sec ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "hydraulic conductivity at void ratio of 0.65 = 0.04 cm/sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg177"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate hydraulic conductivity\n",
+ "##initialisation of variables\n",
+ "e= 0.6\n",
+ "D10= 0.09 ##mm\n",
+ "##calculations\n",
+ "k= 2.4622*(D10**2*(e**3/(1+e)))**0.7825\n",
+ "##results\n",
+ "print'%s %.4f %s'% ('hydraulic conductivity = ',k,' cm/sec ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "hydraulic conductivity = 0.0119 cm/sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg177"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate hydraulic conductivity\n",
+ "##initialisation of variables\n",
+ "e= 0.6\n",
+ "D10= 0.09 ##mm\n",
+ "D60= 0.16 ##mm\n",
+ "##calculations\n",
+ "Cu=D60/D10\n",
+ "k= 35*(e**3/(1+e))*(Cu**0.6)*(D10**2.32)\n",
+ "##results\n",
+ "print'%s %.3f %s'% ('hydraulic conductivity =',k,'cm/sec ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "hydraulic conductivity = 0.025 cm/sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg179"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate hydraulic conductivity\n",
+ "##initialisation of variables\n",
+ "k1= 0.302e-7 ##cm/sec\n",
+ "k2= 0.12e-7 ##cm/sec\n",
+ "e1= 1.1\n",
+ "e2= 0.9\n",
+ "e= 0.75\n",
+ "##calcualtions\n",
+ "n= (math.log10((k1/k2)*((1+e1)/(1+e2))))/math.log10(e1/e2)\n",
+ "C= k1/(e1**n/(1+e1))\n",
+ "k= C*(e**n/(1+e))\n",
+ "##results\n",
+ "print'%s %.e %s'% ('hydraulic conductivity =',k,'cm/sec')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "hydraulic conductivity = 5e-09 cm/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-185"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate ration of equivalent hydraulic conductivity\n",
+ "##initialisation of variables\n",
+ "H1= 2. ##m\n",
+ "H2= 3. ##m\n",
+ "H3= 4. ##m\n",
+ "k1= 1e-4 ##cm/sec\n",
+ "k2= 3.2e-2 ##cm/sec\n",
+ "k3= 4.1e-5 ##cm/sec\n",
+ "##calculations\n",
+ "H= H1+H2+H3\n",
+ "Kh= (1./H)*((k1*H1)+(k2*H2)+(k3*H3))\n",
+ "Kv= H/((H1/k1)+(H2/k2)+(H3/k3))\n",
+ "P= Kh/Kv\n",
+ "##results\n",
+ "print'%s %.2f %s'% ('ration of equivalent hydraulic conductivity =',P,' ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ration of equivalent hydraulic conductivity = 139.97 \n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg186"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate rate of water supply\n",
+ "##initialisation of variables\n",
+ "H= 450. ##mm\n",
+ "h= 150. ##mm\n",
+ "k1= 1e-2 ##cm/sec\n",
+ "k2= 3e-3 ##cm/sec\n",
+ "k3= 4.9e-4 ##cm/sec\n",
+ "h1= 300. ##mm\n",
+ "##calculations\n",
+ "Kv= H/(h*(1./k1+1./k2+1./k3))\n",
+ "i= h1/H\n",
+ "q= Kv*i*100.*3600.\n",
+ "##results\n",
+ "print'%s %.2f %s'% ('rate of water supply =',q,' cm/hr ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "rate of water supply = 291.01 cm/hr \n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_Of_Geotechnical_Engineering/Chapter8.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter8.ipynb new file mode 100755 index 00000000..70736c62 --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter8.ipynb @@ -0,0 +1,241 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0e14a7576f4038d6474523e0faf6c534de06cbb509045bb799ae70b4aaeef049"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter8-See page"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate rate of water flow\n",
+ "##initialisation of variables\n",
+ "H1= 12. ##in\n",
+ "H2= 20. ##in\n",
+ "z= 8. ##in\n",
+ "h1= 24. ##in\n",
+ "h= 20. ##in\n",
+ "k1= 0.026 ##in/sec\n",
+ "D= 3. ##in\n",
+ "##calculations\n",
+ "k2= H2*k1/((z/(1.-h/h1))-H1)\n",
+ "i= h1/(H1+H2)\n",
+ "A= math.pi/4.*D**2\n",
+ "keq= (H1+H2)/((H1/k1)+(H2/k2))\n",
+ "q= keq*A*i*3600.\n",
+ "##results\n",
+ "print'%s %.2f %s'% ('rate of water flow = ',q,' in^3/hr ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "rate of water flow = 330.81 in^3/hr \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg208"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate a)How high (above the ground surface) the water will rise if piezometers are placed at points aandb.\n",
+ "#b.The total rate of seepage through the permeable layer per unit length\n",
+ "#c. The approximate average hydraulic gradient at c.\n",
+ "##initialisation of variables\n",
+ "Nd= 6.\n",
+ "H1= 5.6 ##m\n",
+ "H2= 2.2 ##m\n",
+ "k= 5e-5 ##cm/sec\n",
+ "dL= 4.1 ##m\n",
+ "##calculations\n",
+ "H= (H1-H2)/Nd\n",
+ "h1= 5.61-H\n",
+ "h2= 5.61-5.*H\n",
+ "q= 2.38*(H1-H2)*k/Nd\n",
+ "i= H/dL\n",
+ "##results\n",
+ "print'%s %.3f %s'% ('at point a,water will rise to height of = ',h1,' m ')\n",
+ "print'%s %.3f %s'% ('at point b,water will rise to height of =',h2,' m ')\n",
+ "print'%s %.e %s'% ('total rate of seepage per unit lenghth = ',q,' m^3/sec/m ')\n",
+ "print'%s %.3f %s'% ('average hydraulic gradient at c = ',i,' ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "at point a,water will rise to height of = 5.043 m \n",
+ "at point b,water will rise to height of = 2.777 m \n",
+ "total rate of seepage per unit lenghth = 7e-05 m^3/sec/m \n",
+ "average hydraulic gradient at c = 0.138 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg210"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate average rate of flow\n",
+ "##initialisation of variables\n",
+ "k1= 5.67 ##ft/day\n",
+ "k2= 11.34 ##ft/day\n",
+ "##from graph\n",
+ "Nd= 8\n",
+ "Nf= 2.5\n",
+ "H= 20\n",
+ "##calculations\n",
+ "q= math.sqrt(k1*k2)*H*Nf/Nd\n",
+ "##results\n",
+ "print'%s %.2f %s'% ('average rate of flow = ',q,' ft^3/day/ft ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "average rate of flow = 50.12 ft^3/day/ft \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg 212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate seepage under the dam \n",
+ "##initialisation of variables\n",
+ "B= 6. ##m\n",
+ "L= 120. ##m\n",
+ "s= 3. ##m\n",
+ "T= 6. ##m\n",
+ "x= 2.4 ##m\n",
+ "H= 5. ##m\n",
+ "k= 0.008 ##cm/sec\n",
+ "##calculations\n",
+ "b=B/2.\n",
+ "a1= b/T\n",
+ "a2= s/T\n",
+ "a3= x/b\n",
+ "Q= 0.378*k*H*L*36*24\n",
+ "##results\n",
+ "print'%s %.2f %s'% ('seepage under the dam = ',Q,' m^3/day ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "seepage under the dam = 1567.64 m^3/day \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg217"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate seepage rate\n",
+ "##initialisation of variables\n",
+ "b= math.pi/4. ##degrees\n",
+ "a= math.pi/6.##degrees\n",
+ "B= 10. ##ft\n",
+ "H= 20. ##ft\n",
+ "h= 25. ##ft\n",
+ "k= 2e-4 ##ft/min\n",
+ "##calculations\n",
+ "r= H/math.tan(b)\n",
+ "d= 0.3*r+(h-H)/math.tan(b)+B+h/math.tan(a)\n",
+ "L= d/math.cos(a)-math.sqrt((d/math.cos(a))**2-(H/math.sin(a))**2)\n",
+ "q= k*L*math.tan(a)*math.sin(a)*24.*60\n",
+ "##results\n",
+ "print'%s %.4f %s'% ('seepage rate = ',q,' ft^3/day/ft ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "seepage rate = 0.9724 ft^3/day/ft \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_Of_Geotechnical_Engineering/Chapter9.ipynb b/Principles_Of_Geotechnical_Engineering/Chapter9.ipynb new file mode 100755 index 00000000..08984b22 --- /dev/null +++ b/Principles_Of_Geotechnical_Engineering/Chapter9.ipynb @@ -0,0 +1,205 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b2df79b1bfe1bd5bb53b7bb0f72d193619dc333d1434a7c1a2462664a1d1323a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter9-In situ stresses"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate total pressure and pore water pressure and effective stress at point\n",
+ "##initialisation of variables\n",
+ "Ds= 16.5 ##kN/m**3\n",
+ "S= 19.25 ##kN/m**3\n",
+ "g= 9.8 ##kN/m**3\n",
+ "h1= 6. ##m\n",
+ "h2= 13. ##m\n",
+ "##at point A\n",
+ "Sa= 0.\n",
+ "Ua= 0.\n",
+ "Sa1= 0.\n",
+ "##at point B\n",
+ "Sb= h1*Ds\n",
+ "Ub= 0.\n",
+ "Sb1= Sb-Ub\n",
+ "##at point C\n",
+ "Sc= h1*Ds+h2*S\n",
+ "Uc= h2*g\n",
+ "Sc1= Sc-Uc\n",
+ "##results\n",
+ "print'%s %.2f %s'% ('total pressure at C= ',Sc,' kN/m**3 ')\n",
+ "print'%s %.2f %s'% ('pore water pressure at C = ',Uc,' kN/m**3 ')\n",
+ "print'%s %.2f %s'% ('effective stress at point C=',Sc1,' kN/m**3 ')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "total pressure at C= 349.25 kN/m**3 \n",
+ "pore water pressure at C = 127.40 kN/m**3 \n",
+ "effective stress at point C= 221.85 kN/m**3 \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg233"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate maximu depth that can be made in clay\n",
+ "##initialisation of variables\n",
+ "h= 20. ##ft\n",
+ "g= 120. ##kg/ft**3\n",
+ "h1= 12. ##ft\n",
+ "w= 62.4 ##kg/ft**3\n",
+ "##calculations\n",
+ "H= h-(h1*w/g)\n",
+ "##results\n",
+ "print'%s %.2f %s'% ('maximu depth that can be made in clay = ',H,' ft ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "maximu depth that can be made in clay = 13.76 ft \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate effective stress at point a and b and see page force per unit voume\n",
+ "##initialisation of variables\n",
+ "G= 2.68\n",
+ "e= 0.52\n",
+ "g= 9.81 ##kN/m^3\n",
+ "h1= 0.7 ##m\n",
+ "h2= 1 ##m\n",
+ "h3= 1.5 ##m\n",
+ "h4= 2 ##m\n",
+ "##calculations\n",
+ "##for soil A\n",
+ "sa= (G+e)*g/(1.+e)\n",
+ "##point a\n",
+ "Sa= h1*g+h2*sa\n",
+ "u= (h2+h1+h3/2.)*g\n",
+ "Es= Sa-u\n",
+ "##point b\n",
+ "sb= h1*g+h4*sa\n",
+ "ub= (h4+h1+h3)*g\n",
+ "Eb= sb-ub\n",
+ "i= h3/2.\n",
+ "s= i*g\n",
+ "##results\n",
+ "print'%s %.2f %s'% ('effective stress at point a=',Es,' kN/m^2 ')\n",
+ "print'%s %.2f %s'% ('effective stress at point b= ',Eb,'kN/m^2 ')\n",
+ "print'%s %.2f %s'% ('seepage force per unit voume = ',s,' kN/m^3 ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "effective stress at point a= 3.49 kN/m^2 \n",
+ "effective stress at point b= 6.97 kN/m^2 \n",
+ "seepage force per unit voume = 7.36 kN/m^3 \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg239"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate safety factor\n",
+ "##initialisation of variables\n",
+ "C0= 0.357\n",
+ "H1= 30.5 ##ft\n",
+ "H2= 5. ##ft\n",
+ "w= 62.4 ## lb/ft^3\n",
+ "D= 20.\n",
+ "g= 112. ## lb/ft^3\n",
+ "##calculations\n",
+ "G= g-w\n",
+ "FS= D*G/(C0*w*(H1-H2))\n",
+ "##results\n",
+ "print'%s %.1f %s'% ('safety factor =',FS,' ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "safety factor = 1.7 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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