Question

3. The bearing capacity of shallow foundations (i.e. footings) depends on the soil shear strength (ie. ф and c). The ultimate bearing capacity (qultnei) of the soil is calculated using the following equation: Dr qult.net bearing capacity failure surface i. Frictional resistance along failure surface due to self-weight of soil below footing level footing level. Ny. No, and Ng are dimensionless bearing capacity factors that depend on the ii. Cohesive resistance along failure surface below footing level. iii. Frictional resistance of soil below footing level due to overburden above internal angle of friction (p): Ny 2(N + 1) tan(p) Nc = (Nq-1)/tan(φ)

Answer 1

Given data: Depth Dr = 2 feet. Square footing B*B FOS = 1.5 (A): For moist gravel ' = 45°, 4 moist = 130 pcf, for gravel c= 0. the ultimate bearing capacity is given by:qu = = ON EyN+yD,(N, -1) No = tan’(45° + 45 )e= tané = 134.87 N, = 2(N, +1) tan o = 2(134.87+1) tan 45 | = 271.74 que =R 130x271.74 +130x2x(134.87 –1) =Bx17663+34806.2

guita [Bx17663+34806.2] now, allowable bearing capacity = . pacity" FOST 1.5 ] since the supported load should be 100kips = 100000 pounds. So, -XB2 Equating vertical load and allowable stress Bx17663+34806.27 100000 = 1.5 150000 =17663B° +34806.2B2 solving the above equation B= 1.57 Take B=2 feet, hence footing sizeshould be 2x 2 feet.

B (B) Ysop = 100 pcf,0 = 28°,c = 400 psf v is given by:1. = BYN,+ xD,(N, -1)+cN. No = tan”(45° + 23 ) *28 = 14.8 2 2(N +1) tan o = 2(14.8+1) tan 28 =16.8 N. - N, -1-26 tan o 2x100x16.8+100×2×(14.8–1)+400x 26 = 840B + 2760+10400 = 840B +13160. 840B +1316 quut Laiv = 25 = 1.5 )

Equate the allowable load with awl stress: [840B +1316022 100000 = 000 = 1.5 XB 150000 = 840B +13160 B2 solving: B = 3 feet Hence adopt 3*3 feet footing dimensions.