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+{
+ "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