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