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’g_f’H_f²tand/2

K’ = 1-sinf = 0.577

g_f’ = 19.8-9.81 = 9.99 kN/m²

d = 0.5_ffill = 12.5^o

Q_n = 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”, c_u = 860 lb/ft² n₂ = 3

Required allowable load-carrying capacity of the pile group:

From equation 9.130 :

SQ_u = n₁n₂ (qA_pc_u + Sa p c_uDL)

From figure 9.22 a = 0.9

SQ_u = 1040 kips

From equation 9.131 :

SQu = L_gB_gcuN_c*+S2(L_g+B_g)c_u DL

Lg = B9 = 2.5*2+1=6’

From figure 9.58:

For L/B_g = 7.5 and L_g/B_g = 1, Nc* = 9

SQ_u = 2136 kips

Allowable capacity = S Q_{u 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 L_g=B_g = 2.75 m

Q_g = 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, e_o1 = 0.8, c_c1 = 0.8

P_o1 = 3*15.72+3(18.55-9.81)+18(19.18-9.81) = 195.19 kN/m²

Ds₁ = 0.208 m

For layer 2

P_o2 = 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

P_o3 = 297.925 kN/m²

Ds₃ = 0.0024 m

Ds_c = SDs_i = 0.2117 m = 217.7 mm