HOMEWORK SOLUTION

HW #13

CE 455

FOUNDATION AND EARTH STRUCTURE

(Question no. 9.11, 9.12,9.15,9.17 ..Braja M. Das textbook)

9.11 H-section ; HP 14 * 102

Ap = 30”

L = 62 ft, q_u,(lab) = 11400 lb/in², f = 36^o, FS = 3

Qp = Ap q_p

q_p = q_u (Nf +1)

q_u = q_u(lab)/5 = 2280 lb/in²

Nf = tan²(45+f/2) = 3.85

Q_p = 30 * 2280 (3.85+1) = 331740 lb

Q_all = Q_p/FS = 110.58 kips

9.12 L = 50 ft, x-section = 16” * 16”

Embedment in sand

g = 117 lb/ft³, f = 37^o

Allowable working load = 180 kips

Frictional resistance = 110 kips, point load = 70 kips

Ep = 3*10⁶ lb/in², Es = 5*10³ lb/in², us = 0.38 x = 0.57

From equation 9.62:

S_i = (Q_wp+xQ_ws)L/A_pE_p = 0.104”

9.15 x-section = 406 mm * 406 mm, L =10.4 m, FS = 4

Embedment in sand

g_sand = 118 lb/ft³, f = 37^o, Dr = 80%

From figure 9.36

(L/D)_cr = 14.5 L_cr = 5.887 m

L_cr < L

T_un = ½ pgL_cr² K_u tan d + pg L_cr K_u tan d (L-L_cr)

From fig 9.36,

d/f = 1, K_u = 2

T_un = 0.5 *0.406/0.3048*4*118*(5.887/0.3048)²*2*tan37^o

+0.406/0.3048*4*5.887/0.3048*2*tan37 * ((10.4-5.887)/0.3048)

= 179032.6313 lb

Allowable pullout capacity = T_un/FS = 44.76 kips.

9.17 x-section =305 mm * 305 mm

L = 60 m, FS = 3

Top 5 m of clay, cu = 25 kN/m²

Below cu = 55 kN/m² saturated clay

T_un = S Lpa’cu

For cu = 25 kN/m² a’ = 0.9 – 0.00625 cu = 0.74

And cu = 55 kN/m² a’ = 0.9 – 0.00625 cu = 0.56

T_un = 5*0.305*4*0.74*25+55*0.305*4*0.56*55=2179.53 kN

Allowable pullout capacity = T_un/FS = 726.51 kN