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 = q_o (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, q_c = 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 * q_u

And IV = 1.\/200 = 0.005

From the chart: N = 49.5

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