HOMEWORK SOLUTION

HW #6

CE 455

FOUNDATION AND EARTH STRUCTURE

(Question no. 4.13, 4.16, 4.18, 4.20, & 4.22 ..Braja M. Das textbook)

4.13 Rigid foundation = 10 ft x 6.5 ft

q_o = 3200 lb/ft²

m_s = 0.3, E_s = 3200 lb/in², Df = 2.95 ft, H = 32 ft

Se = (B*qo/Es) * (1-m_s²) * a_r

We have,

L/B = 10/6.5 = 1.5

From fig 4.18 a_r = 1.1

Elastic settlement Se = ((6.5*3200)/(3200*144))* (1-0.3²) X 1.1

= 0.045 ft » 0.54 in

4.16

We have, q_o = 3200 lb/ft², B = 6.5 ft

L = 10 ft, Df = 2.95 ft, m_s = 0.3, E_s = 3200 lb/in², and H = 32 ft

g = 110 lb/ft³, 5 year time of creep

By using Iz plot the same as for a square foundation.

We also have,

q = Df * g = 2.95 * 110 = 324.5 lb/ft²

Dz (ft)	Iz	Iz * Dz
3.75	0.3	1.125
10.25	0.25	2.5625

S =3.6875

4.18 Continuous foundation

L/B > 10

g = 115 lb/ft3

c₂ = 10 years

q = Df * g = 5* 115 = 575 lb/ft²

Layer	Dz (ft)	Iz	Es (lb/ft^2)	(Iz/Es)*Dz
1	6	0.3125	126000	1.5*10^-5
2	2	0.4625	250560	3.69*10^-6
3	12	0.375	250560	1.8*10^-5
4	12	0.125	208800	7.18*10^-6

S = 4.387*10^-5

4.20 From problem 4.6 we get DP_av = 1157.9 lb/ft² then,

P_o = 4.5*100+3*(122-62.4)+4*(120-62.4)

= 450+178.8+230.4

= 859.2 lb/ft²

P_o + DP_av = 2017.1 lb/ft²

Also P_o < P_c

4.22 We have, Qo = Am + Pn

From first test:

8070 = 144 m + 48 n …………….(1)

From second test:

25800 = 576 m + 96 n …………..(2)

From eq (1) and (2)

m = 33.54 lb/in2

m = 67.5 lb/in

Qo = 33.54 A + 67.5 P

For Qo = 150000

150000 = 33.54 B2 + 270 B

B2 + 8.05B - 4472.3 = 0

Size of square footing (B) = 63 in