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,          qu,(lab) = 11400 lb/in2, f = 36o,  FS = 3

Qp = Ap qp

qp = qu (Nf +1)

qu = qu(lab)/5 = 2280 lb/in2

Nf = tan2(45+f/2) = 3.85

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

Qall = Qp/FS = 110.58 kips

9.12          L = 50 ft, x-section = 16” * 16”

Embedment in sand

g = 117 lb/ft3, f = 37o

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:

Si = (Qwp+xQws)L/ApEp = 0.104”

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

Embedment in sand

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

From figure 9.36

(L/D)cr = 14.5   Lcr = 5.887 m

Lcr < L

Tun = ½ pgLcr2 Ku tan d + pg Lcr Ku tan d (L-Lcr)

From fig 9.36,

d/f = 1, Ku = 2

Tun = 0.5 *0.406/0.3048*4*118*(5.887/0.3048)2*2*tan37o

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

= 179032.6313 lb

Allowable pullout capacity = Tun/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

Tun = 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

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

Allowable pullout capacity = Tun/FS = 726.51 kN