HOMEWORK SOLUTION

HW #14

CE 455

FOUNDATION AND EARTH STRUCTURE

(Question no. 9.24, 9.27,9.30,9.31 ..Braja M. Das textbook)

9.24

From equation 9.123:

Qn = pKgfHf2tand/2

K = 1-sinf = 0.577

gf = 19.8-9.81 = 9.99 kN/m2

d = 0.5ffill = 12.5o

Qn = 14.45 kN

9.27

Outside diameter = 460 mm

a)      from equation 9.128:

h = 2(n1+n2-2)d+4D/pn1n2

n1=n2=3

d = 1200 mm

D = 460 mm

p = p * 460

h = 0.88 = 88%

b) using the los angeles group action equation:

9.29     FS = 3, d=30, L = 45   n1 = 3

D = 12, cu = 860 lb/ft2  n2 = 3

Required allowable load-carrying capacity of the pile group:

From equation 9.130 :

SQu = n1n2 (qApcu + Sa p cuDL)

From figure 9.22 a = 0.9

SQu = 1040 kips

From equation 9.131 :

SQu = LgBgcuNc*+S2(Lg+Bg)cu DL

Lg = B9 = 2.5*2+1=6

From figure 9.58:

For L/Bg = 7.5 and Lg/Bg = 1, Nc* = 9

SQu = 2136 kips

Allowable capacity = S Qu lower/FS = 346.67 kips

9.30          L1 = 15 ft, l2 = 20 ft, l3 = 20 ft

Cu1 = 550 lb/ft2, cu2 = 875 lb/ft2, cu3= 1200 lb/ft2

N1 = 4, n2 = 3, d = 40, x-section of pile = 14 * 14

FS = 4

From equation 9.130

SQu = n1n2 (qApcu+SapcuDL)

From figure 9.22:

A1 = 1, a2=1, a3=0.75

SQu = n1n2(qApcu+ap1cuDL1+a2p2cu2DL2+a3p3cu3DL3)=2626 kips

From equation 9.131

SQu = LgBgcuNc* + S2(LgBg)cuDL

Lg = 3*40/12+14/12 = 11.167

Bg = 2*40/12+14/12 = 7.83

From fig 9.58:

Lg/Bg=1.43, L/Bg=7    Nc* = 8.67

SQu = 2800 kips

Alloable capacity = SQulower/SF = 656.5 kips

9.31          Lg=Bg = 2.75 m

Qg = 1335 kN

Dp1 = 1335/(2.75+5)2=22.23 kN/m2

Dp2 = 1335/(2.75+12.5)2=5.74 kN/m2

Dp3 = 1335/(2.75+16.5)2=3.6 kN/m2

Ds1 =

H1 = 10 m, eo1 = 0.8, cc1 = 0.8

Po1 = 3*15.72+3(18.55-9.81)+18(19.18-9.81) = 195.19 kN/m2

Ds1 = 0.208 m

For layer 2

Po2 = 3*15.72+3(18.55-9.81)+18(19.18-9.81)

+2.5(18.08-9.81)=262.715 kN/m2

Ds2 =0.0073 m

For layer 3

Po3 = 297.925 kN/m2

Ds3 = 0.0024 m

Dsc = SDsi = 0.2117 m = 217.7 mm