HOMEWORK SOLUTION

 

HW #12

 

CE 353

GEOTECHNICAL ENGINEERING I

 

(Question no. 10.2, 10.4, 10.8, ..Al-Khafaji textbook)

 

 


10.2 Square footing B = 1.5 m

D = 1.2 m

Soil sm = 17.6 kN/m3

c = 50 kN/m2

f = 30o

SF = 3

Condition General shear failure

 

Question: Allowable bearing capacity of footing

Solution: for f = 30o we obtain Nc =33 ; Nq = 19.5 ; Ng = 17

 

For square footing

q ult = 1.3 c Nc + q Nq + 0.4 g B Ng

= (1.3x50x33)+(19.5x1.2x22.5)+(0.4x17.6x1.5x17)

= 2145+526.5+179.52

= 2851 kN/m2

 

q all = q ult/SF = 2851/3 = 950.3 kN/m2

 

Q all = q all x B2 = 950.3 x 1.52

= 2138.27 kN


10.4 Strip footing B = 5 ft

D = 4 ft

Soil sm = 100 pcf

c = 750 pcf

f = 25o

SF = 3

Condition General shear failure

 

Question: Allowable bearing capacity of footing

 

Solution:

from fig 10.6 f=17.23 we obtain Nc = 10 ; Nq = 3 ; Ng = 1.1

 

For strip footing:

q ult = 2/3 c Nc + q Nq + 0.5 g B Ng'

= (2/3x750x10)+(100x4x3)+(0.5x100x5x1.1)

= 5000+1200+275

= 6475 psf

 

q all = q ult/SF = 6475/3 = 2158.3 psf


10.8 Square footing Q all = 1200 kN

D = 1.5 m

Soil sm = 18 kN/m3

qu = 120 kN/m2

f = 0o

SF = 2.5

Condition General shear failure

 

Question: Dimension of footing

Solution: for f = 0o we obtain Nc =5.53 ; Nq = 1.0 ; Ng = 0

 

q ult = (Q all x 2.5)/B2 = 3000/B2

 

 

For square footing

q ult = 1.3 c Nc + q Nq + 0.4 g B Ng

(3000/B2) = (1.3x60x5.53)+(18x1.5x1)+(0)

= 458.34 kN/m2

B2 = 3000/458.34 = 6.54

B = = 2.558 ~ 2.6 m