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 = Po = ½*H2*Ko*g

Ko = 1 – sinf = 0.5

At rest pressure = Po = 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 = 5o and  f = 32o active earth pressure coefficient Ko = 0.311

a.

 Z (m) sa = g*z*Ko (kN/m2) Pa = ½*g*z2*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.      Pa = ½*g*H2*Ko , H = 7.5 m

Pa = 159.19 kN/m

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

Angle of orientation 5o C.C.W

6.12

a.       a =10o and d = 20o

from table 6.5         Ka = 0.3857

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

z = 12/3 = 4 ft

b. a = 20o and d = 15o

from table 6.6         Ka = 0.4708

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

 2cÖKp

6.17

b.      Pb = ½*H2*sv*Kp+2*H*c*ÖKp

sv = gH and Kp = tan2 (45+f/2) = 1

Pb = ½*182*120+2*18*800 = 37440 lb/ft

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

 18’’

 10o

7.2

 H’

 18’

We have,

H’ = 2.75+18+6tan10o=21.8’

Pa = ½*g1H’Ka

From table 6.2 Ka = 0.294

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

Pv = Pa sina = 8173.65 sin10o = 1419.34 lb/ft

Pn = Pa cosa = 8173.65 cos10o = 8049.47 lb/ft

Overturning moment :

Mo = Pn *(H/3) = (8049.47*21.8)/3 = 58492.82 lb.ft

Resisting moment:

MR =(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+Pp)/Pacosa

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,

Pp = 1/2Kpg2D2+2c2ÖKpD

Kp = tan2(45+f2/2) = 1.89

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

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

= 2.75>1.5

Factor of safety against bearing capacity failure:

qu = c2NcFcdFci+qNqFqdFqi+1/2g2B’NgFgdFgi

From table 3.4 for f = 18o

Nc = 13.1, Nq = 5.26, Ng = 4.07

We have also,

e = B/2-(SMR-SMo)/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

Fcd=1+0.4(D/B’)=1+0.4(4/10.74)=1.15

Fqd = 1+2tanf2(1-sinf2)2(D/B’)=1.12

Fgd = 1

Fci = Fqi = (1-yo/90o)2

y = tan-1(Pn/v) = 18.86o

Fci = Fqi = 0.62

Fgi = (1-y/f)2 ~0

qu = 9079.36 lb/ft2

we have

= 2681.03 lb/ft2      toe

= 1088.7 lb/ft2        heel

SF(bearing capacity)=qu/qtoe=9079.36/2681.03=3.39>3

 0.3 m

 0o

7.4

 6.5 m

We have,

H’ = 6.5+0.8=7.3’

Pa = ½*g1H’Ka

From table 6.2 Ka = 0.26

Pa = ½*18.08*7.32*0.26 = 125.25 kN/m

Pv = Pa sina = 0

Pn = Pa cosa =125.25 kN/m

Overturning moment :

Mo = Pn *(H/3) = (125.25*7.3)/3 = 304.775 kN.m

Resisting moment:

MR =(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+Pp)/Pacosa

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,

Pp = 1/2Kpg2D2+2c2ÖKpD

Kp = tan2(45+f2/2) = 1.7

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

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

= 2.3>1.5

Factor of safety against bearing capacity failure:

qu = c2NcFcdFci+qNqFqdFqi+1/2g2B’NgFgdFgi

From table 3.4

Nc = 10.98, Nq = 3.94, Ng = 2.65

We have also,

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

B/6 = 3.4/6 = 0.56                   e < B/6…ok

Fcd=1+0.4(D/B’)= 1.25

Fqd = 1+2tanf2(1-sinf2)2(D/B’)=1.4

Fgd = 1

Fci = Fqi = (1-yo/90o)2

y = tan-1(Pn/v) = 18.8o

Fci = Fqi = 0.63

Fgi = (1-y/f)2 ~0.064

qu = 365.9 kN/m2

we have

= 201.4 kN/m2       toe

= 15.2 kN/m2         heel

SF(bearing capacity)=qu/qtoe=365.9/201.4=1.8<3