HOMEWORK SOLUTION

HW #8

CE 455

FOUNDATION AND EARTH STRUCTURE

(Question no. 6.1, 6.10, 6.12, 6.17, 7.2 & 7.4 ..Braja M. Das textbook)

6.1

We have

P₁ = P_o = ½*H²*Ko*g

K_o = 1 – sinf = 0.5

At rest pressure = P_o = ½*12²*0.5*108 = 3888 lb/ft

Its location from the top of the soil = 2/3*12 = 8 ft and z = 4 ft.

6.10

H = 7.5 m

From table 6.2 :

for a = 5^o and f = 32^o active earth pressure coefficient K_o = 0.311

a.

Z (m)	sa = g*z*Ko (kN/m^2)	Pa = ½*g*z^2*Ko (kN/m)
2	11.3204	11.3204
4	22.64	45.28
6	33.96	101.88
7.5	42.45	159.19

b. P_a = ½*g*H²*K_o , H = 7.5 m

P_a = 159.19 kN/m

z = H/3 = 7.5/3 = 2.5 m

Angle of orientation 5^o C.C.W

6.12

a. a =10^o and d = 20^o

from table 6.5 K_a = 0.3857

P_a = ½*Ka*g*H² = ½*0.3857*105*12² = 2915.9 lb/ft

z = 12/3 = 4 ft

b. a = 20^o and d = 15^o

from table 6.6 K_a = 0.4708

P_a = ½*Ka*g*H² = ½*0.4708*105*12² = 3559.25 lb/ft

2cÖK_p

6.17

b. P_b = ½*H²*s_v*K_p+2*H*c*ÖK_p

s_v = gH and K_p = tan² (45+f/2) = 1

P_b = ½*18²*120+2*18*800 = 37440 lb/ft

Z = (1/2*18²*120*6+2*18*500*9)/37440 = 7.44 ft

18’’

10^o

7.2

H’

18’

We have,

H’ = 2.75+18+6tan10^o=21.8’

P_a = ½*g₁H’K_a

From table 6.2 K_a = 0.294

P_a = ½*117*21.8²*0.294 = 8173.65 lb/ft

P_v = P_a sina = 8173.65 sin10^o = 1419.34 lb/ft

P_n = P_a cosa = 8173.65 cos10^o = 8049.47 lb/ft

Overturning moment :

M_o = P_n *(H/3) = (8049.47*21.8)/3 = 58492.82 lb.ft

Resisting moment:

M_R =(18*18/12*5.75+18*1/2*4.67+2.75*12.5*6.25)*150+

(18*6*9.5+1.05*3*8.5)*117 = 184993.235 lb.ft

Factor of safety against overturning:

SF_{(overturning)} = MR/Mo = 3.16 > 2

Factor of safety against sliding:

SF_(sliding) = ((Sv)tan(k₁f₂)+Bk₂f₂+P_p)/P_acosa

Assume k₁ = k₂ = 2/3

Sv = (18*18/12+18*1/2+2.75*12.5)*150+(18.6+1.05*3)*117 = 23560.8 lb/ft

Also,

P_p = 1/2K_pg₂D²+2c₂ÖK_pD

K_p = tan²(45+f₂/2) = 1.89

P_p = ½*1.89*110*4²+2*800Ö1.89*4=10472.04 lb/ft

SF_(sliding) = (23560.8tan(2/3*18^o)+2/3*12.5*800+10472.04)/8049.47

= 2.75>1.5

Factor of safety against bearing capacity failure:

q_u = c₂N_cF_cdF_ci+qN_qF_qdF_qi+1/2g₂B’NgF_gdF_gi

From table 3.4 for f = 18^o

N_c = 13.1, N_q = 5.26, Ng = 4.07

We have also,

e = B/2-(SM_R-SM_o)/Sv) = 12.5/2 – (184993.235-58492.82)/23560.8 = 0.88 ft

B/6 = 12.5/6 = 2.083 e = 0.88’ < B/6

B’= B-2e = 10.74

F_cd=1+0.4(D/B’)=1+0.4(4/10.74)=1.15

F_qd = 1+2tanf₂(1-sinf₂)²(D/B’)=1.12

F_gd = 1

F_ci = F_qi = (1-y^o/90^o)²

y = tan^-1(P_n/v) = 18.86^o

F_ci = F_qi = 0.62

F_gi = (1-y/f)² ~0

q_u = 9079.36 lb/ft²

we have

= 2681.03 lb/ft² toe

= 1088.7 lb/ft² heel

SF_{(bearing capacity)}=q_u/q_toe=9079.36/2681.03=3.39>3

0.3 m

0^o

7.4

6.5 m

We have,

H’ = 6.5+0.8=7.3’

P_a = ½*g₁H’K_a

From table 6.2 K_a = 0.26

P_a = ½*18.08*7.3²*0.26 = 125.25 kN/m

P_v = P_a sina = 0

P_n = P_a cosa =125.25 kN/m

Overturning moment :

M_o = P_n *(H/3) = (125.25*7.3)/3 = 304.775 kN.m

Resisting moment:

M_R =(0.8*3.4*1.7+6.5*0.3*1.25+(6.5*0.3)/2*1)*23.58+

(6.5*2*2.4*18.08) = 753.6 kN.m

Factor of safety against overturning:

SF_{(overturning)} = MR/Mo = 2.47 > 2

Factor of safety against sliding:

SF_(sliding) = ((Sv)tan(k₁f₂)+Bk₂f₂+P_p)/P_acosa

Assume k₁ = k₂ = 2/3

Sv = (0.8*3.4+6.5*0.3+6.5*0.3*0.5)*23.58+(6.5*2*18.8) = 368.15 kN

Also,

P_p = 1/2K_pg₂D²+2c₂ÖK_pD

K_p = tan²(45+f₂/2) = 1.7

P_p = ½*1.7*19.65*1.5²+2*30Ö1.7*1.5=154.93 kN/m

SF_(sliding) = (368.15tan(2/3*15^o)+3.4*30*2/3+154.93)/125.25

= 2.3>1.5

Factor of safety against bearing capacity failure:

q_u = c₂N_cF_cdF_ci+qN_qF_qdF_qi+1/2g₂B’NgF_gdF_gi

From table 3.4

N_c = 10.98, N_q = 3.94, Ng = 2.65

We have also,

e = B/2-(SM_R-SM_o)/Sv) = 3.4/2 – (753.6-304.775)/368.15 = 0.48 m

B/6 = 3.4/6 = 0.56 e < B/6…ok

F_cd=1+0.4(D/B’)= 1.25

F_qd = 1+2tanf₂(1-sinf₂)²(D/B’)=1.4

F_gd = 1

F_ci = F_qi = (1-y^o/90^o)²

y = tan^-1(P_n/v) = 18.8^o

F_ci = F_qi = 0.63

F_gi = (1-y/f)² ~0.064

q_u = 365.9 kN/m²

we have

= 201.4 kN/m² toe

= 15.2 kN/m² heel

SF_{(bearing capacity)}=q_u/q_toe=365.9/201.4=1.8<3