ABSTRACT
Solution Method I-By making use of reduction factors RW1 and RW2 and using Eqs. (12.20) and (12.23), we may write
1 qnu = cNc + yDf{Nq -1)Rw1 +2,yBNy Rw2 Given: Nq =41.4, Ny =42.4 and Nc =57.8
By substituting the known values in the equation for qnu' we have
1 q =30 x 57.8+ 18.5 x 2x 40.4 x 0.813+-x 18.5 x 3x 42.4 x 0.5 =3538 kN/m2
q = 3538 = 1179 kN/m2 na 3
q =30x57.8 + 18.5 x2x 40.4 x 1+.!..x 18.5x 3x 42.4 xO.71 =4064 kN/m2 nu 2
q = 4064 =1355 kN/m2 na 3
Method 2-Using the equivalent effective unit weight method. Submerged unit weight Yb =18.5 - 9.81 =8.69 kN/m3. Per Eq. (12.25) The net ultimate bearing capacity is
_ 1.25 _ 3 Yel - 8.69 + 2 (18.5 - 8.69) - 14.82 kN/m
Ye2 = Yb = 8.69 kN/m 3
1 q =30 x 57.8 + 14.82 x 2 x 40.4 + - x 8.69 x 3 x 42.4 =3484 kN/m2
~-+- 3 m -t--~
Figure Ex. 12.4 Effect of WT on bearing capacity
q = 3484 =1161 kN/m2 na 3
_ 1.25 _ 3 Ye2 - 8.69+ 3 (18.5 - 8.69) -12.78 kN/m
1 qnu = 30x 57.8 +18.5 x 2x 40.4 +ix 12.78 x 3x 42.4 = 4042 kN/m2
q = 4042 =1347 kN/m2 na 3
Example 12.5 A square footing fails by general shear in a cohesionless soil under an ultimate load of QUit = 1687.5 kips. The footing is placed at a depth of 6.5 ft below ground level. Given l/J = 35°, and r= 110 Ib/ft3, determine the size of the footing if the water table is at a great depth (Fig. Ex. 12.5).
