HOMEWORK SOLUTION

HW #7

CE 455

FOUNDATION AND EARTH STRUCTURE

(Question no. 5.1, 5.2, 5.6, 5.9, & 5.13 ..Braja M. Das textbook)

5.1 c_u = 2500 lb/ft2, f = 0^o, B = 20 ft, L = 30 ft, D_f = 6.2 ft

Net ultimate bearing capacity:

q_net(u) = q_u – q

From eq. 5.9:

5.2 Mat foundation 6.5 m x 6.5 m

D_f = 1.5 m, allowable settlement = 50.8 mm, g = 16.5 kN/m³

We have,

For large width and from eq. 5.12:

Depth (m)	Field value NF	sv’ (ton/ft^2)	cN = Ö(1/sg)	Ncor
1.5	9	0.26	1.96	17.64
3.0	12	0.52	1.39	16.68
4.5	11	0.78	1.13	12.43
6.0	7	1.03	0.98	6.93
7.5	13	1.29	0.88	11.44
9.0	11	1.55	0.80	8.8
10.5	13	1.81	0.74	9.62

Average = 11.93

We have F_d = 1.099

5.6 Precosolidation pressure (P_c) = 105 kN/m²

20 x 10 m

			Sand g = 16 kN/m^3

2 m

We have:

P_o = 4.2*16+2*(18-9.81)+2.6*(17.5-9.81) = 103.574 kN/m²

and:

DP_av = 1/6 * (DP_t+4DP_m+DP_b)

For top: DP_t = ((30*10³)/(12*10))* I_c, from table 4.3 I_c = 0.606

DP_t = 151.5 kN/m²

For center: I_c = 0.755, DP_m = 188.75 kN/m²

For bottom: I_c = 0.755, DP_b = 188.75 kN/m²

DP_av = 182.54 kN/m²

Overconsolidated clay:

5.9 We have:

Q = 400+500+450+3000+1200+3000+1200+900+350 = 11,000 kN

Also I_x = 1/12 * 16.5*21.5³ = 13665.3 m⁴

I_y = 1/12 * 16.5³*21.5 = 8048.4 m⁴

SM_y’ = 0

11000 x’ = -3800*8.25+3200*8.25

x’ = 0.45 m

SM_x’ = 0

11000 y’ = 4200*3.5+1350*10.75-4200*3.5-1250*10.75

y’ = 0.098 ~ 0.1 m

M_x = Q_ey = 11000*0.1 = 1100 kNm

M_y = Q_ex = 11000*(-0.45) = -4950 kNm

From eq. 5.25:

Point

A : x = -8.25, y = 10.75 m ® q_A = 36.93 kN/m²

B : x = 0, y = 10.75 m ® q_b = 31.86 kN/m²

C : x = +8.25, y = 10.75 m ® q_C = 26.79 kN/m²

D : x = -8.25, y = -10.75 m ® q_D = 25.07 kN/m²

E : x = 0, y = -10.75 m ® q_E = 30.14 kN/m²

F : x = -8.25, y = -10.75 m ® q_F = 35.21 kN/m²

5.13 Subgrade reaction = 18 kN/m²

Plate 1 m x 0.7 m

For a foundation of 5 m x 3.5 m :

From eq. 5.49:

Subgrade reaction of rectangular plate is a function of B/L ratio

and we have:

B/L (plate) = 0.7/1 = 0.7

and for foundation

B/L = 3.5/5 = 0.7

k_foundation = 18 kN/m²