{
"metadata": {
"name": "",
"signature": "sha256:f9b870b8ab70a6506c4707b625840ce3fe5d061d1cd38773c3e0459e51172aa1"
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"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": {}
}
]
}