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 = 35^o

c = 0

1.5 m

2 m

2 m

1 m

8.24

We have,

s_all = 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 M_B1 = 0 A = Pa*3 *1.5/2 = 43.6275 kN/m

B1 = 3Pa – A = 14.54 kN/m

S M_B2 = 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 = M_max/s_all = 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,

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

S = M_max/s_all = 25.9/170000 = 1.52 * 10^-4 m³

Clay

g_c = 18.2 kN/m³

q_u = 55 kN/m²

Sand

g_s = 17.5 kN/m³

f_s = 34^o

Hc =5 m

Hs =3 m

H =8 m

8.27

C_av =1/2H (g_sK_sH_s²tanf_s+(H-H_s)n’qu)

K_s = 1

N’ = 0.75

C_av = 19.53 kN/m²

From equation 8.110:

g_a = 1/H (g_sH_s+(H-H_s)g_c

g_a = 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,

s_all = 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 M_B1 = 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 M_B2 = 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 M_B2 = 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 = M_max/s_all = 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 =