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diff --git a/Principles_Of_Foundation_Engineering/Chapter05.ipynb b/Principles_Of_Foundation_Engineering/Chapter05.ipynb new file mode 100755 index 00000000..ca975aae --- /dev/null +++ b/Principles_Of_Foundation_Engineering/Chapter05.ipynb @@ -0,0 +1,550 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:77a2cf465cec464205dc151afe10d9acafa79fe1b46b30f4a468368f8af3f8ea"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter05:Shallow Foundations: Allowable Bearing Capacity and Settlement"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.1:Pg-212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 5.1\n",
+ "\n",
+ "#first solution\n",
+ "B1=2.5; # in ft\n",
+ "B2=B1;\n",
+ "z=12.5; # in ft\n",
+ "L1=5; # in ft\n",
+ "L2=L1;\n",
+ "m=B1/z;\n",
+ "n=B2/z;\n",
+ "#from table 5.2 of the values using m,n\n",
+ "q=2000; # in lb/ft^2\n",
+ "I=0.0328;\n",
+ "deltasigma=q*4*I; # in lb/ft**2\n",
+ "print round(deltasigma,2),\"change in pressure in lb/ft**2\"\n",
+ "#second solution\n",
+ "Ic=0.131;#from table\n",
+ "deltasigma=q*Ic; # in lb/ft**2\n",
+ "print round(deltasigma,2),\"change in pressure in lb/ft**2\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "262.4 change in pressure in lb/ft**2\n",
+ "262.0 change in pressure in lb/ft**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.2:Pg-215"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 5.2\n",
+ "\n",
+ "qo=100; # in KN/m^2\n",
+ "H1=3; # in m\n",
+ "H2=5; # in m\n",
+ "#from table\n",
+ "IaH2=0.126;\n",
+ "IaH1=0.175;\n",
+ "deltasigma=qo*((H2*IaH2-H1*IaH1)/(H2-H1)); # in kN/m**2\n",
+ "print round(deltasigma,2),\"change in pressure in kN/m**2\"\n",
+ "TS=4*deltasigma; # in kN/m**2\n",
+ "print round(TS,2),\"total change in pressure in kN/m**2\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "5.25 change in pressure in kN/m**2\n",
+ "21.0 total change in pressure in kN/m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.3:Pg-217"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 5.3\n",
+ "H=7;\n",
+ "Gamma=17.5; # in KN/m^3\n",
+ "q0=Gamma*H # in KN/m^2\n",
+ "print q0,\" is pressure change in kN/m**2\"\n",
+ "#part2\n",
+ "#from figure\n",
+ "Ileftside=0.445;\n",
+ "Irightside=0.445;\n",
+ "deltasigma=q0*(Ileftside+Irightside); # in KN/m^2\n",
+ "print round(deltasigma,2),\"is change in stress in kN/m**2\"\n",
+ "#partc\n",
+ "#from figure 5.11\n",
+ "I=0.24;#I'\n",
+ "Dsigma1=43.75*I;#deltasigma1 in KN/m^2\n",
+ "I2=0.495;#I'\n",
+ "Dsigma2=I2*q0;#deltasigma2 in KN/m^2\n",
+ "I3=0.335;#I'\n",
+ "Dsigma3=I3*78.75;#deltasigma3 in KN/m^2\n",
+ "Dsigma=Dsigma1+Dsigma2-Dsigma3; # in KN/m^2\n",
+ "print round(Dsigma,2),\"is total stress increase in A in kN/m**2\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "122.5 is pressure change in kN/m**2\n",
+ "109.03 is change in stress in kN/m**2\n",
+ "44.76 is total stress increase in A in kN/m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.4:Pg-228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 5.4\n",
+ "\n",
+ "zbar=5;\n",
+ "mus=0.3;\n",
+ "F1=0.641;\n",
+ "F2=0.031;\n",
+ "z1=2.0; # in m\n",
+ "z2=1.0; # in m\n",
+ "z3=2.0; # in m\n",
+ "Es1=10000; # in kN/m**2\n",
+ "Es2=8000; # in kN/m**2\n",
+ "Es3=12000;# in kN/m**2\n",
+ "qo=150; # in KN/m^2\n",
+ "#from table 5.4\n",
+ "If=0.709;\n",
+ "Es=(Es1*z1+Es2*z2+Es3*z3)/zbar; # in kN/m**2\n",
+ "print Es,\" is modulus of elasticity in kN/m**2\"\n",
+ "Is=F1+(2-mus)/(1-mus)*F2;\n",
+ "Sc=qo*(1.0/Es-mus**2.0/Es)*Is*If*2;\n",
+ "Scrigid=0.93*Sc; # in m\n",
+ "print round(Scrigid*1000,2),\"is settlement in mm\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "10400.0 is modulus of elasticity in kN/m**2\n",
+ "12.4 is settlement in mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.5:Pg-234"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 5.5\n",
+ "import math\n",
+ "B=5; # in ft\n",
+ "L=10; # in ft\n",
+ "Ef=2.3e6; # in lb/in^2\n",
+ "Eo=1400.0; # in lb/in^2\n",
+ "k=25.0; # in lb/in^2/ft\n",
+ "t=1.0;\n",
+ "mus=0.3;\n",
+ "Df=5.0; # in ft\n",
+ "qo=5000.0; # in lb/ft^2\n",
+ "Ig=0.69;\n",
+ "Be=math.sqrt(4*B*L/math.pi);\n",
+ "If=math.pi/4+1/(4.6+10*(Ef/(Eo+2*Be/2*k))*(2*t/Be)**3);\n",
+ "Ie=1-1/(3.5*math.exp(1.22*mus-0.4)*(Be/Df+1.6));\n",
+ "Se=qo*Be*Ig*If*Ie/Eo*(1-mus**2)/144; # in ft\n",
+ "print round(Se*12,2),\"settlement in inches\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "1.07 settlement in inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.6:238"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 5.6\n",
+ "\n",
+ "import math\n",
+ "import numpy\n",
+ "q=3.06; # in lb/in^2\n",
+ "qbar=25; # in lb/in^2\n",
+ "C1=1-0.5*(q/(qbar-q));\n",
+ "Sum=0;\n",
+ "C2=1+0.2*math.log10(10/0.1);\n",
+ "L=[1, 2, 3, 4, 5];\n",
+ "Dz=[48, 48, 96, 48, 144]; # in inch\n",
+ "Es=[750, 1250, 1250, 1000, 2000]; # in lb/in^2\n",
+ "z=[24, 72, 144, 216, 312]; # in inch\n",
+ "Iz=[0.275, 0.425, 0.417, 0.292, 0.125];\n",
+ "k=numpy.zeros(5)\n",
+ "print \"Layer No.\\t deltaz (in)\\t Es(lb/in**2)\\t z to the middle of the layer (in) Iz at the middle of the layer Iz/delta(z) \\n\"\n",
+ "for i in range(0,5):\n",
+ " k[i]=Iz[i]/Es[i]*Dz[i];\n",
+ " print L[i],\"\\t \\t \",Dz[i],\"\\t\\t \",Es[i],\"\\t\\t \",z[i],\" \\t\\t\\t\\t\\t \",Iz[i],\"\\t\\t \",round(k[i],3)\n",
+ " Sum=Sum+k[i];\n",
+ "\n",
+ "Se=C1*C2*(qbar-q)*Sum; # in inch\n",
+ "print round(Se,2),\"settlement in inches\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Layer No.\t deltaz (in)\t Es(lb/in**2)\t z to the middle of the layer (in) Iz at the middle of the layer Iz/delta(z) \n",
+ "\n",
+ "1 \t \t 48 \t\t 750 \t\t 24 \t\t\t\t\t 0.275 \t\t 0.018\n",
+ "2 \t \t 48 \t\t 1250 \t\t 72 \t\t\t\t\t 0.425 \t\t 0.016\n",
+ "3 \t \t 96 \t\t 1250 \t\t 144 \t\t\t\t\t 0.417 \t\t 0.032\n",
+ "4 \t \t 48 \t\t 1000 \t\t 216 \t\t\t\t\t 0.292 \t\t 0.014\n",
+ "5 \t \t 144 \t\t 2000 \t\t 312 \t\t\t\t\t 0.125 \t\t 0.009\n",
+ "2.54 settlement in inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.7:Pg-244"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 5.7\n",
+ "\n",
+ "Df=1.0; # in m\n",
+ "B=1.75; # in m\n",
+ "L=1.75 # in m\n",
+ "qnet=120.0; # in KN/m^2\n",
+ "N60=10.0;# in m\n",
+ "alpha1=0.14 # for normally consolated sand\n",
+ "alpha2=1.71/(N60)**1.4 # for normally consolated sand\n",
+ "alpha3=1.0 # for normally consolated sand\n",
+ "Se=0.3*alpha1*alpha2*alpha3*(qnet/100)*((B/0.3)**0.7)*((1.25*(L/B)/(0.25+(L/B))))**2\n",
+ "print round(Se*1000,2),\"settlement in mm\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "11.79 settlement in mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.8:Pg-245"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 5.8\n",
+ "\n",
+ "Df=1; # in m\n",
+ "B=1.75; # in m\n",
+ "qnet=120; # in KN/m^2\n",
+ "N60=10; # in m\n",
+ "Fd=1+0.33*Df/B;\n",
+ "Se=2*qnet/N60/Fd*(B/(B+0.3))**2; # in mm\n",
+ "print round(Se,2),\"settlement in mm\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "14.71 settlement in mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.9:Pg-251"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 5.9\n",
+ "\n",
+ "Ny=23.76; \n",
+ "Nq=16.51; \n",
+ "q=3*110.0; # in lb/ft^2\n",
+ "Gamma=110.0; # in lb/ft^3\n",
+ "B=4.0; # in ft\n",
+ "Nqe=0.63*Nq;\n",
+ "Nye=0.4*Ny;\n",
+ "que=q*Nqe+1/2.0*Gamma*B*Nye; # in lb/ft^2\n",
+ "print round(que,2),\" is bearing capacity in lb/ft**2\"\n",
+ "#part 2\n",
+ "V=0.4; # in ft/sec\n",
+ "A=0.32; # given in question\n",
+ "g=9.81; # acceleration constant in m/sec^2\n",
+ "kh=0.26;\n",
+ "k=0.92;#tan(alphae)\n",
+ "Seq=0.174*k*V**2/A/g*kh**-4/A**-4; # in m\n",
+ "print round(Seq,3),\"settelement in m\"\n",
+ "print round(Seq*39.57,2),\"settlement in inches\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "5523.31 is bearing capacity in lb/ft**2\n",
+ "0.019 settelement in m\n",
+ "0.74 settlement in inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.10:Pg-256"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 5.10\n",
+ "\n",
+ "import math\n",
+ "Cc=0.32;\n",
+ "Hc=2.5;\n",
+ "eo=0.8;\n",
+ "sigmao=2.5*16.5+0.5*(17.5-9.81)+1.25*(16-9.81); # in kN/m^2\n",
+ "m1=[2, 2, 2];\n",
+ "z=[2, 3.25, 4.5];\n",
+ "n1=[4, 6.5, 9];\n",
+ "Ic=[0.19, 0.085, 0.045];\n",
+ "Dsigma=[28.5, 12.75, 6.75];#deltasigma\n",
+ "print (\"m1\\t z(m)\\t n1\\t Ic\\t Dsigma \\n\");\n",
+ "for i in range(0,3):\n",
+ " print round(m1[i],2),\"\\t \",round(z[i],2),\"\\t \",round(n1[i],2),\"\\t \",round(Ic[i],2),\"\\t \",round(Dsigma[i],2)\n",
+ "\n",
+ " Dsigmaav=1/6.0*(Dsigma[0]+4*Dsigma[1]+Dsigma[2]);\n",
+ " Sc=Cc*Hc/(1+eo)*math.log10((sigmao+Dsigmaav)/sigmao);\n",
+ "print round(Sc*1000,2),\"settlement in mm\"\n",
+ "#partb\n",
+ "B=1.0; # in m\n",
+ "L=2.0; # in m\n",
+ "z=0.5+1.5; # in m\n",
+ "B=B+z; # in m\n",
+ "L=L+z; # in m\n",
+ "A=0.6; # given in question\n",
+ "#from table\n",
+ "kcr=0.78; # by data\n",
+ "Sep=kcr*Sc;\n",
+ "print round(Sep*1000,2),\"settlement in mm\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "m1\t z(m)\t n1\t Ic\t Dsigma \n",
+ "\n",
+ "2.0 \t 2.0 \t 4.0 \t 0.19 \t 28.5\n",
+ "2.0 \t 3.25 \t 6.5 \t 0.09 \t 12.75\n",
+ "2.0 \t 4.5 \t 9.0 \t 0.04 \t 6.75\n",
+ "46.45 settlement in mm\n",
+ "36.23 settlement in mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.11:Pg-262"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 5.11\n",
+ "\n",
+ "import numpy\n",
+ "N60=(3+7+12+12+16)/5.0;\n",
+ "B=[2, 2.25, 2.3]; # in m\n",
+ "Fd=[1.248, 1.22, 1.215];\n",
+ "Qoac=102000*9.81/1000;#actual Qo\n",
+ "Se=25; # in mm\n",
+ "qnet=numpy.zeros(3)\n",
+ "Qo=numpy.zeros(3) # in kN\n",
+ "print \"B(m)\\t Fd\\t qnet(kN/m**2)\\t \\t Qo \\n\"\n",
+ "for i in range(0,3):\n",
+ " qnet[i]=10/0.08*(B[i]+0.3)**2/(B[i])**2*Fd[i]*Se/25;\n",
+ " Qo[i]=qnet[i]*B[i]**2;\n",
+ " print B[i],\"\\t\",Fd[i],\" \\t \",round(qnet[i],2),\"\\t\\t \",Qo[i],\"\\n\"\n",
+ "print int(Qoac),\"value of Qo in kN\"\n",
+ "print \"since Qo is 1000 kN thus B is equal to 2.3 m from the table\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "B(m)\t Fd\t qnet(kN/m**2)\t \t Qo \n",
+ "\n",
+ "2 \t1.248 \t 206.31 \t\t 825.24 \n",
+ "\n",
+ "2.25 \t1.22 \t 195.88 \t\t 991.63125 \n",
+ "\n",
+ "2.3 \t1.215 \t 194.08 \t\t 1026.675 \n",
+ "\n",
+ "1000 value of Qo in kN\n",
+ "since Qo is 1000 kN thus B is equal to 2.3 m from the table\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |