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{
"metadata": {
"name": "",
"signature": "sha256:e3a75199f67af72d14bee528a629ae06b2506206625e1ef3a86291ef88f556ed"
},
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"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 03:Shallow Foundations: Ultimate bearing capacity"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.1:Pg-130"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 3.1\n",
"# From Table 3.1\n",
"Nc=17.69;\n",
"Nq=7.44;\n",
"Ny=3.64;\n",
"q=3*115;\n",
"Gamma=115.0; #lb/ft**3\n",
"c=320;\n",
"B=5.0;#ft\n",
"FS=4;#factor of safety\n",
"qu=1.3*c*Nc+q*Nq+0.4*Gamma*B*Ny\n",
"qall=qu/FS; #q allowed\n",
"Q=qall*B**2;\n",
"print Q,\"is allowable gross load in lb\" \n",
"\n",
"# the answer is slightly different in textbook due to approximation but here answer are precise"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"67269.0 is allowable gross load in lb\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.2:Pg-134"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 3.2\n",
"\n",
"from scipy.optimize import fsolve\n",
"import math\n",
"Gamma=105.0;#lb/ft**3\n",
"Gammasat=118.0;#lb/ft**3\n",
"FS=3.0;\n",
"pa=2014.125;#lb/ft**2\n",
"Depth=[5,10,15,20,25]; # in ft\n",
"N60=[4,6,6,10,5]; # in blow/ft\n",
"sigmao=[0,0,0,0,0]; # in lb/ft^2\n",
"phi=[0,0,0,0,0] # in degree\n",
"Gammaw=62.4;\n",
"s=0;\n",
"print \"depth (ft)\\tN60\\t \\tstress(lb/ft**2)\\t phi(degrees)\\n\"\n",
"for i in range(0,5):\n",
" sigmao[i]=2*Gamma+(Depth[i]-2)*(Gammasat-Gammaw);\n",
" phi[i]=math.sqrt(20*N60[i]*math.sqrt(pa/sigmao[i]))+20;\n",
" print \" \",Depth[i],\"\\t \",N60[i],\"\\t\\t \",sigmao[i],\" \\t \\t \\t\",round(phi[i],1),\" \\n\"\n",
" s=phi[i]+s\n",
"\n",
"avgphi=s/(i+1)\n",
"\n",
"print round(avgphi),\"average friction angle in degrees\"\n",
"#using graph get the values of other terms in terms of B and solve for B\n",
"def f(x):\n",
" return-150000/x**2+5263.9+5527.1/x+228.3*x\n",
"x=fsolve(f,4);\n",
"print round(x[0],1),\" is the width in ft\"\n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"depth (ft)\tN60\t \tstress(lb/ft**2)\t phi(degrees)\n",
"\n",
" 5 \t 4 \t\t 376.8 \t \t \t33.6 \n",
"\n",
" 10 \t 6 \t\t 654.8 \t \t \t34.5 \n",
"\n",
" 15 \t 6 \t\t 932.8 \t \t \t33.3 \n",
"\n",
" 20 \t 10 \t\t 1210.8 \t \t \t36.1 \n",
"\n",
" 25 \t 5 \t\t 1488.8 \t \t \t30.8 \n",
"\n",
"34.0 average friction angle in degrees\n",
"4.5 is the width in ft\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.3:Pg-144"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 3.3\n",
"\n",
"import math\n",
"phi=25.0; #degrees\n",
"Es=620.0; #kN/m**2\n",
"Gamma=18.0;#kN/m**2\n",
"Df=0.6;# in m\n",
"B=0.6; # in m\n",
"L=1.2; # in m\n",
"Fqc=0.347;\n",
"Nq=10.66;\n",
"Nc=20.72;\n",
"Ngamma=10.88;\n",
"mu=0.3; # Poisson's ratio\n",
"Fyd=1.0;\n",
"c=48.0;#kN/m**2\n",
"q=Gamma*(Df+B/2);\n",
"Ir=Es/(2*(1+mu)*(c+q*math.tan(phi*math.pi/180.0)));\n",
"print round(Ir,2),\" is value of Ir\"\n",
"Fcc=Fqc-(1-Fqc)/(Nq*math.tan(phi*math.pi/180.0));\n",
"Fcs=1+Nq/Nc*B/L;\n",
"Fqs=1+B/L*math.tan(phi*math.pi/180.0);\n",
"Fys=1-0.4*B/L;\n",
"Fcd=1+0.4*Df/B;\n",
"Fqd=1+2.0*math.tan(phi*math.pi/180.0)*(1-math.sin(phi*math.pi/180.0))**2*Df/B;\n",
"q1=0.6*18;\n",
"Fyc=Fqc;\n",
"qu=c*Nc*Fcs*Fcd*Fcc+q1*Nq*Fqs*Fqd*Fqc+1.0/2*Gamma*Ngamma*Fys*Fyd*Fyc;\n",
"print round(qu,2),\"is ultimate bearing capacity in kN/m**2\"\n",
"\n",
"# the answer is slightly different in textbook due to approximation but here answer are precise"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"4.29 is value of Ir\n",
"469.24 is ultimate bearing capacity in kN/m**2\n"
]
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.4:Pg-156"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 3.4\n",
"import math\n",
"q=110*4.0; #lb/ft**2\n",
"Nq=33.3;\n",
"phi=35.0; # in degree\n",
"Df=4.0; # in ft\n",
"B=6.0; # in ft\n",
"Gamma=110.0;#lb/ft**3\n",
"Ngamma=48.03; #lb/ft**3\n",
"B1=6-2*0.5; # in ft\n",
"Fqi=1;\n",
"Fyi=1;\n",
"Fyd=1;\n",
"Fqs=1;\n",
"Fys=1;\n",
"Fqd=1+2*math.tan(phi*math.pi/180)*(1-math.sin(phi*math.pi/180.0))**2*Df/B;\n",
"qu=q*Nq*Fqs*Fqd*Fqi+1/2.0*B1*Gamma*Ngamma*Fys*Fyd*Fyi;\n",
"Qult=B1*1*qu;\n",
"print round(Qult,2),\" is ultimate bearing capacity in lb/ft\" \n",
"print round(Qult/2000.0,2),\" is ultimate bearing capacity in ton/ft\"\n",
"\n",
"# the answer is slightly different in textbook due to approximation but here answer are precise"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"151738.23 is ultimate bearing capacity in lb/ft\n",
"75.87 is ultimate bearing capacity in ton/ft\n"
]
}
],
"prompt_number": 34
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.5:Pg-158"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 3.5\n",
"\n",
"e=0.5; # in ft\n",
"B=6; # in ft\n",
"k=e/B;\n",
"Gamma=110; # in lb/ft^3 \n",
"q=440;\n",
"print \"get the values of Nqe and Nye from the figure from the value of e/B\"\n",
"Nye=26.8;\n",
"Nqe=33.4;\n",
"Qult=B*1*(q*Nqe+1/2.0*Gamma*B*Nye);\n",
"print round(Qult,2),\" is ultimate bearing capacity in lb/ft\"\n",
"print round(Qult/2000.0,2),\" is ultimate bearing capacity in ton/ft\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"get the values of Nqe and Nye from the figure from the value of e/B\n",
"141240.0 is ultimate bearing capacity in lb/ft\n",
"70.62 is ultimate bearing capacity in ton/ft\n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.6:Pg-159"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 3.6\n",
"\n",
"Df=0.7; # in m\n",
"#from table\n",
"Nq=18.4;\n",
"Ny=22.4;\n",
"q=12.6;\n",
"phi=30; #angle in degree\n",
"L=1.5;# in m\n",
"Fyd=1;\n",
"Gamma=18; # in KN/m^3\n",
"L1=0.85*1.5; # in m\n",
"L2=0.21*1.5; # in m\n",
"B=1.5; # in m\n",
"A=1/2.0*(L1+L2)*B;\n",
"B1=A/L1; #B'\n",
"Fqs=1+B1/L1*math.tan(phi*math.pi/180);\n",
"Fys=1-0.4*B1/L1;\n",
"Fqd=1+2*math.tan(phi*math.pi/180)*(1-math.sin(phi*math.pi/180))**2*Df/B;\n",
"Qult=A*(q*Nq*Fqs*Fqd+1/2.0*Gamma*B1*Ny*Fys*Fyd);\n",
"print round(Qult,2),\" is ultimate load in kN\"\n",
"\n",
"# the answer is slightly different in textbook due to approximation but here answer are precise"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"605.45 is ultimate load in kN\n"
]
}
],
"prompt_number": 41
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.7:Pg-161"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 3.7\n",
"\n",
"e=0.15; # in m\n",
"B=1.5; # in m\n",
"Fqs=1.0;\n",
"L=1.5;# in m\n",
"Gamma=18.0; # in KN/m^3\n",
"q=0.7*18;\n",
"#from table\n",
"Nqe=18.4;\n",
"Nye=11.58;\n",
"Fys=1+(2*e/B-0.68)*(B/L)+(0.43-3/2.0*e/B)*(B/L)**2;\n",
"Qult=B*L*(q*Nqe*Fqs+1/2.0*L*Gamma*Nye*Fys);\n",
"print round(Qult,2),\"is ultimate load in kN\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"803.03 is ultimate load in kN\n"
]
}
],
"prompt_number": 45
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.8:Pg-163"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 3.8\n",
"\n",
"q=16.0;# in kN/m^2\n",
"Nqei=14.2;\n",
"Gamma=16.0 # in kN/m^3\n",
"B=1.5;# in m\n",
"Nyet=20.0;\n",
"Qult=B*(Nqei*q+1/2.0*Gamma*B*Nyet);\n",
"print round(Qult,2),\" is ultimate load in kN/m\"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"700.8 is ultimate load in kN/m\n"
]
}
],
"prompt_number": 48
}
],
"metadata": {}
}
]
}
|
