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

qo = 3200 lb/ft2

ms = 0.3, Es = 3200 lb/in2, Df = 2.95 ft, H = 32 ft

Se = (B*qo/Es) * (1-ms2) * ar

We have,

L/B = 10/6.5 = 1.5

From fig 4.18 ar = 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, qo = 3200 lb/ft2, B = 6.5 ft

L = 10 ft, Df = 2.95 ft, ms = 0.3, Es = 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/ft2)

(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 DPav = 1157.9 lb/ft2 then,

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

= 450+178.8+230.4

= 859.2 lb/ft2

Po + DPav = 2017.1 lb/ft2

Also Po < Pc

 

 

 

 

 

 

 

 

 

 

 

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