HOMEWORK SOLUTION

HW #11

CE 455

FOUNDATION AND EARTH STRUCTURE

(Question no. 8.24, 8.27,8.30,8.33 ..Braja M. Das textbook)

 3.5 m

 C

 B

 A

 Sand g = 17 kN/m3 f = 35o c = 0

 1.5 m

 2 m

 2 m

 1 m

8.24

We have,

sall = 170 MN/m2 and S = 3 m

Ka = tan2 (45-f/2) = 0.27

Pa = 0.65 g H Ka = 19.39 kN/m2

S MB1 = 0   A = Pa*3 *1.5/2 = 43.6275 kN/m

B1 = 3Pa – A = 14.54 kN/m

S MB2 = 0   c = Pa*3.52*0.5/2 = 59.38 kN/m

B2 = 3.5Pa – c = 8.485 kN/m

 24.2375

a)      For unit width:

 14.54

 19.69

 2 m

 1 m

 B1

 B’

 A

x1 = 34.2375/19.39 = 1.25 m

 30.295

 x2

 29.085

 8.485

 C

 B’

 B2

x2 = 30.295/19.39 = 1.56 m

Then the moments are:

At        A = ½ * 19.39 * 1 = 0.695 kN.m

At        C= ½ * 29.085 * 1.5 = 21.81 kN.m

At        B’= ½ * 14.54 * 0.75 = 5.45 kN.m

At        B” = ½ * 8.485* 0.44 = 1.867 kN.m

Mc is maximum, so:

Sx = Mmax/sall = 21.81/170000 = 1.28*10-4 m3

From table C-1 (Appendix C):

Use PSA-23,  S = 12.8*10-5 m3/m

b)      For the water at level B,

Mmax = (B1+B2)S2/8 = (14.54+8.485)*9/8 = 25.9 kN.m

S = Mmax/sall = 25.9/170000 = 1.52 * 10-4 m3

 Clay gc = 18.2 kN/m3 qu = 55 kN/m2

 Sand gs = 17.5 kN/m3 fs = 34o

 Hc =5 m

 Hs =3 m

 H =8 m

8.27

Cav =1/2H (gsKsHs2tanfs+(H-Hs)n’qu)

Ks = 1

N’ = 0.75

Cav = 19.53 kN/m2

From equation 8.110:

ga = 1/H (gsHs+(H-Hs)gc

ga = 17.94 kN/m3

 20 ft

 C

 B

 A

 g = 118 lb/ft2 f = 0 c = 800 lb/ft3

 3 ft

 8 ft

 10 ft

 5 ft

8.30

We have,

sall = 25000 lb/in2 and S = 12 ft

gH/c = 118*26/800 = 3.8 < 4 ………….stiff clay

Pa = 0.3 g H  = 920.4 lb/ft2

From figure 8.52

S MB1 = 0   A = (920.4*8.5280.5+920.4*3.75*10.67)/10=7007.7 lb/ft2

B1 = (920.4*8.5+920.4*3.75)-7007.7=4267.2 lb/ft2

S MB2 = 0   c = (920.4*4.52*0.5+920.4*3.75*6.67)/8=4042.57 lb/ft2

B2 = (920.4*4.5+920.4*3.75)-4042.57=3550.73 lb/ft2

S MB2 = 0   c = (920.4*4.52*0.5+920.4*3.75*6.67)/8=4042.57 lb/ft2

B2 = (920.4*4.5+920.4*3.75)-4042.57=3550.73 lb/ft2

 x1

 5237.7

c)      For unit width:

 4267.2

 1770

 B1

 B’

 A

x1 = 4267.2/920.4 = 4.6 ft

 3405.37

 x2

 637.2

 3550.73

 C

 B”

 B2

x2 = 3550.73/920.4 = 3.86 ft

Then the moments are:

At        A = ½ * 5 * ((920.4/6.5)*5)(5/3)= 2950 lb.ft

At        B’= (4267.2*4.6)-(920.4*4.62/2)=9891.29 lb.ft

At        C= ½ * 3 *((920.4/6.5)*3)*1 = 637.2 lb.ft

At        B” = (3550.73*3.86)-(920.4*3.862/2)=6849.02 lb.ft

Mmax = 9891.29 lb.ft

Section-modulus S = Mmax/sall = 9891.29*12/25000 = 4.748 in3

8.33

h’/B = 2/3 = 0.67

c2/c1 = 45/30 = 1.5

From fig 8.62 (b)

Nc’(strip) = 5.53

Also,

H/B = 8/3 = 2.67

From fig 8.62 (c)

Fd = 1.46

From equation 8.122

Fs = 1 + 0.2B/L = 1 + 0.2 X 3/8 = 1.075

From equation 8.121

Factor of safety against heaving

SF =