Chapter16-Soil-Bearing Capacity for Shallow Foundations

Ex1-pg587

In [6]:
#determine the gross allowable load per unit area (qall) that the foundation can carry.
import math
c=20.
## from table 16.1
Nc=17.69
Nq=7.44
Ng=3.64

Df=3.
G=110.
q=G*Df

C=200.
B=4.

Qu= C*Nc+q*Nq+G*B*Ng/2.

Fs=3.
Qall=Qu/Fs
print'%s %.1f %s'%('Qa = ',Qall,' lb/ft^2')

Qa =  2264.7  lb/ft^2

Ex2-pg588

In [2]:
import math
#determine the size of the footing—that is, the size of B.
G=18.15
qa=30000.*9.81/1000.

Nc=57.75
Nq=41.44
Ng=45.41
C=0.
q=G*1.
B=1.
(1.3*C*Nc+q*Nq+0.4*G*B*Ng)*B**2/3. == qa
B= math.sqrt(294.3/(250.7+109.9))
print'%s %.1f %s'%(' B = ',B,' m')

 B =  0.9  m

Ex3-pg595

In [3]:
import math
# Determine the safe gross load (factor of safety of 3) that the footing can carry
B=1.2
L=1.2
c=32.
C=0.
Df=1.
G=16.
Nq=23.18
Ng=22.02
Nc=1.
Lqs=1.+0.1*B*(math.tan(61./57.3)**2.)/L
Lgs=Lqs
Lqd=1.+0.1*Df*math.tan(61./57.3)/B
Lgd=Lqd
Lcs=1.
Lcd=1.
Gs=19.5
q=0.5*G+0.5*(Gs-9.81)
Qu= C*Lcs*Lcd*Nc+q*Lqs*Lqd*Nq+(Gs-9.81)*Lgs*Lgd*B*Ng/2.
Qa=Qu/3.
Q=Qa*B**2.
print'%s %.1f %s'%('the gross load = ',Q,' kN')

the gross load =  311.6  kN

Ex4-pg601

In [4]:
import math
#Determine the magnitude of the gross ultimate load applied eccentrically for bearing capacity failure in soil.
e=0.1
B=1.
X=B-2.*e
Y=1.5
B1=0.8
L1=1.5
c=30.
Df=1.
Nq=18.4
Ng=15.668
q=1.*18.
G=18.
Lqs=1.+e*(B1/L1)*math.tan(60./57.3)**2.
Lgs=Lqs
Lqd=1.+e*(Df/B1)*math.tan(60./57.3)
Lgd=Lqd
qu=q*Lqs*Lqd*Nq+Lgs*Lgd*G*B1*Ng/2.
Qu=qu*B1*L1
print'%s %.1f %s'%('The magnitude of the gross ultimate load =',Qu,' kN')

The magnitude of the gross ultimate load = 751.8  kN

Ex5-pg601

In [5]:
#determine the gross ultimate load per unit length that the foundation can carry.
import math
B=1.5
Df=0.75
e=0.1*B
G=17.5
c=30.
C=0.
q=G*Df
Nq=18.4
Ng=15.668
Lqd=1.+0.1*(Df/B)*math.tan(60./57.3)
Lgd=Lqd
Quc=q*Nq*Lqd+Lgd*B*Ng/2.
k=0.8
a=1.754
Qua=Quc*(1.-a*(e/B)**k)
print'%s %.1f %s'%('The gross ultimate load per unit length = ',Qua,' kN')

The gross ultimate load per unit length =  198.7  kN

Ex6-pg606

In [7]:
import math
#Estimate the ultimate bearing capacity of a circular footing with a diameter of 1.5 m. The soil is sandy.
Qup=280.
Bp=0.7 ## in m
Bf=1.5
Quf=Qup*Bf/Bp
print'%s %.1f %s'%('The ultimate bearing capacity = ',Quf,' kN/m^2')

The ultimate bearing capacity =  600.0  kN/m^2

Ex7-pg606

In [2]:
#calculate the value of Cv
import math
#Determine the size of a square column foundation that should carry a load of 2500 kN with a maximum settlement of 25 mm.
a=2500.
##doing for the first values only
Bf=4.
Bp=0.305
q=a/Bf**2.
Sep=4.
Sef=Sep*(2.*Bf/(Bf+Bp))**2
print'%s %.1f %s'%('Sef = ',Sef,' mm')
import math
%matplotlib inline
import warnings
warnings.filterwarnings('ignore')
import numpy
from math import tan
import matplotlib
from matplotlib import pyplot
#given
t=numpy.array([.02,.1,.25,.5,1,2.,4.,8.,16.,30.,60.,120.,240.,480.,960.,1440.])
gauge=numpy.array([3975.,4082.,4102.,4128.,4166.,4224.,4298.,4420.,4572.,4737.,4923.,5080.,5207.,5283.,5334.,5364.])
Hdr=2.24
t50=19.
#calculations
Cv=.197*(Hdr/2)**2 /t50/60.
leng=len(t)
logt=numpy.zeros(leng)
for i in range(0,leng):
	logt[i]=math.log(t[i])

#results
print'%s %.4f %s'%('The value of Cv (cm^2/sec) = ',Cv,'')
pyplot.plot(logt,gauge)
pyplot.xlabel('Time(min) - log scale')
pyplot.ylabel('Dial reading (cm)')
pyplot.title('Graph of dial reading vs time')
pyplot.show()

Sef =  13.8  mm
The value of Cv (cm^2/sec) =  0.0002