HOMEWORK SOLUTION

 

HW #5

 

CE 455

FOUNDATION AND EARTH STRUCTURE

 

(Question no. 4.3, 4.6, 4.7, & 4.10 ..Braja M. Das textbook)

 

 


4.3 B1 = 5 ft, B1 = 10 ft, L1 = 7 ft, L2 = 12 ft

Uniform load on the flexible area = 2500 lb/ft2, depth = 20 ft, by dividing the rectangular area into four small rectangles:

Dp = qo (I1+I2+I3+I3+I4), we have

m1 = B1/Z = 5/20 = 0.25 m3 = B3/Z = 5/20 = 0.25

n1 = L1/Z = 7/20 = 0.35 n3 = L3/Z = 12/20 = 0.6

m2 = B2/Z = 7/20 = 0.35 m4 = B4/Z = 10/20 = 0.5

n2 = L2/Z = 10/20 = 0.5 n4 = L4/Z = 12/20 = 0.6

From table (4.2):

I1 = 0.035855 I2 = 0.06352

I3 = 0.05321 I3 = 0.09473

Increase in pressure Dp = 2500 * 0.247315 = 618.29 lb/ft2

 

4.6 Net foundation load = 900 kN = 202.34 kips

 


 

 

 

 

 

 

We have qo = 202.34/25 = 8.09 kips/ft2

By dividing the footing into 4 equal squares:

2.5 ft

 

2.5 ft

 
 

 

 

 

 

 


Average total stress increase =

H2 = 11 ft, H1 = 3 ft

m = n = 2.5 ft

for Ia (H1) :

m/z = n/z = 2.5/3 = 0.833

from fig. (4.8) : Ia (H1) = 0.213

for Ia (H2) :

m/z = n/z = 2.5/11 = 0.227

from fig. (4.8) : Ia (H2) = 0.1

Total average stress increase =

Effective average stress increase = 1.86 *1000 62.4*11

= 1173.6 lb/ft2 = 56.2 kN/m2

 

4.7 By using the 2:1 method:

Net load of foundation = 900 kN = 203.34 kips

We have, qc = 203.34/25 = 8.09 kips.ft

From eq. (4.43) :

By using 2:1 method:

=

=

=

 

4.10 By using Newmark method:

Dp = IV * N * qu

And IV = 1.\/200 = 0.005

From the chart: N = 49.5

Dp = 0.005 * 49.5 * 2500 = 618.75 lb/ft2