HOMEWORK SOLUTION

HW #12

CE 455

FOUNDATION AND EARTH STRUCTURE

(Question no. 9.1, 9.3,9.8,9.10 ..Braja M. Das textbook)

9.1 L = 20 m, x-section = 460 mm * 460 mm

Sand g = 18.6 kN/m³, f = 30^o

a) Meyerhof’s method:

From equation 9.13:

Qp = Ap*Qp = Ap q’ Nq*

From fig. 9.14 Nq* = 55

Limit value of Qp:

Qp : Ap*Qc ; qL=50 Nq* tan f (equation 9.15)

Qp = 0.46*0.46*18.68*20*55 = 4329.34 kN.

b) Vesic method:

From equation 9.20

Qp = Ap. sc’ Ns*

sc’ = ((1+2Ko)/3)q’ ; Ko = 1 – sin f = 0.5

From appendix D.4 = Ns* = 44.8

Qp = 0.46*0.46*44.8*2/3*18.6*20 = 2350.96 kN

c) Janbu

From equation 9.30

Qp = Ap q’ Nq’

From fig. 9.15 Nq*=20

Qp=0.46*0.46*18.6*20*20 = 1574.3 kN

9.3 from equation 9.33:

Qp = q’ Nq* Ap

From fig.16 for L/D = 20/0.46 = 43 Nq* = 23

Qp = 18.6*20*23*0.46*0.46=1810.45 kN

Qs = p DL f

f = K sv’ tan d ; d = 0.8f

And assume Bored or jetted pile K = Ko = 1-sin f = 0.5

Qs = 0.46*4*20*0.5*18.6*20*tan(0.8*30) = 3047.5 kN

Qu = Qp + Qs = 4857.95 kN

Qall = Qu/F.S = 1214.5 kN

9.8 L = 20 m, x-section = 381 mm * 381 mm

Embedment in a saturated clay layer

Clay = g_sat = 18.5 kN/m³, f=0, cu = 70 kN/m3

Water table lies below the tip of the pile

Factor of safety = 3

l method =

f_av = l (sv’ + 2 cu ) equation 9.45

sc’ = (g_sat – g_w)*L = 173.8 kN/m²

From fig 9.20 l = 0.17

fav = 0.17(173.8+2*70) = 53.346

Qs = pLf_av = 1626 kN

Qs_allowable = Qs/FS = 542 kN

Silty clay

gsat = 118 lb/ft³

cu = 700 lb/ft²

9.10

a) a method:

Qs = S a cu p DL

= 1*700*16/12*4*20+0.63*1500*16/12*4*40

a obtained from fig 9.22

Qs = 276.27 kips

b) l method:

fav = l (sv’+2 Cu)

at level 20 ft sv’ = (118-62.4)*20=1112 lb/ft²

at level 60 ft sv = 1112 + (122.4-62.4)*40 = 3512 lb/ft²

sv’ =

cu = (700*20+1500*40)/60 = 1233.33 lb/ft²

from figure 9.20 l = 0.12

f_av = 503.2

Qs = p L f_av = 16/12 *4*60*503.2 = 161.024 kips

c) b method:

Qs = S fp DL

f = B sv’

B = ktanf_R for normally consolidated clay and f_R = 20^o

= (1-sinf_R)tanf_R = 0.24

Qs = 0.24*16/12*4*20*1112+0.24*16/12*4*40*3512

Qs = 208.3 kips