Question

14 Inclined load on the vertical axes 27 degree. The load is 300 kN/m2 applied to square foot (B= 2 m) over silty sand with capillarity. The water table located at the base of foot. The saturated unit weight above the base is 20 kN/m3, and below the base (G=2.65 and water content =0.471). The friction angle is equal to 29 from CD test. The factor of safety is 3. The factor dg is equal to ut of Select one: estion O a. 1.2 O b. 1.15 O c. 0.885 O d. 1 O e. 0.322 Next page

Answer 1

area of footing = 2*2 = 4m2

total load = 300 KN/m2 * area of footing = 300*4= 1200 KN

due to inclined loading there will be two components Fx and Fy

Fx = 1200 sin (27) = 543.6 KN

Fy = 1200 cos (27) = 1069.2 KN

unit weight of soil abov the base = 20 KN/m3

unit wieght below the base(submerged) = sat. unit wt - wt of water = 20-10 = 10KN/m3

load capacity of soil , FOS = ultimate load capacity of soil / acting load (vertical)

3 = U/1069.2 , therefore U = 3207.6 KN

Dq is the depth factor used in the meyeroffs bearing capacity formula

$D_{q}= 1+2tan\phi (1-sin\phi ^{2})*z/b$

Z = depth of embedment = 1m

Dq = 1.223

therfore opton a is correct