Answer 1) a) For continuous footing of B = 4' and Df = 3'
Nc = 25.1, Nq = 12.7 and Ny = 9.7
Thus qnult = cNc + YDf (Nq-1) + 0.5YBNy = 600*25.1 +(11.7*110*3)+(0.5*110*4*9.7) = 21055 psf
qns = 21055/4 = 5263.75 psf thus Gross bearing capacity = qns + YDf = 5263.75 +(110*3) = 5593.75 psf
b) Nc = 37.2 , Nq = 22.5 and Ny = 19.7
Thus qnult = 17*1*21.5+ 0.5*17*2*19.7 = 700.4 KN/m2
Thus gross bearing capacity = 700.4/4 + (17) = 192.1 KN/m2
c) qnult = 16.5*2*21.5 + 0.4*16.5*3*19.7 = 1099.56 KN/m2
Thus gross bearing capacity = 1099.56/4 + 16.5*2 = 307.89 KN/m2
Read sections 3.1 through 3.5 pp. 133-143 and solve the following problems: 1. For the following...