HOMEWORK SOLUTION

HW #4

CE 455

FOUNDATION AND EARTH STRUCTURE

(Question no. 3.9, 3.14, 3.15, & 3.5 ..Braja M. Das textbook)

3.9

1 m

B’ = B – 2e = 1.5 – 2*0.15 = 1.2 m

L = L’ = B = 1.5 m

Using eq. 3.25:

qu’ = q Nq Fqs Fqd Fqi + 0.5 g B’ Ng Fgs Fgd Fgi

From table 3.4:

Nq = 37.75, Ng = 56.31, tan f = 0.73

From table 3.5:

Fqs = 1 + B’/L’ * tan f = 1 + 1.2/1.5 * 0.73 = 1.584

Fgs = 1 – 0.4 * B’/L’ = 1 – 0.4 * 1.2/1.5 = 0.68

Fqd = 1 + 2 tan f (1 – sin f)² * Df/B = 1.165

Fgd = 1

Fqi = Fgi = 1

q_u’ = 17*37.75*1.548*1.165+0.5*17*1.2*56.31*0.68 = 1574.826 kN/m²

Qult = q_u’ * B’ * L’ = 2834.687 kN

Qall = Qult/FS = 708.67 kN

3.14 Strip foundation in a two-layered clay, FS = 3

Since the foundation is stripped L » ¥ B/L » 0

c₂/c₁=0.585

from figure 3.22:

c_a/c₁=0.95 c_a=0.95*1000=950 lb/ft²

q_all = q_u/FS = 1417.183 lb/ft²

3.15

B/L = 0.754, c₂/c₁=0.6

From figure 3.22:

c_a/c₁=0.97 c_a = 0.97*71.9 = 69.743 kN/m²

The ultimate bearing capacity = 441.2 kN/m²

The ultimate load = q_t*B*L = 495.2 kN