Problem 2 a A square footing is to be constructed in sand as

Problem 2 (a) A square footing is to be constructed in sand as shown the sketch below. Determine the column load, P, which the footing will support when the footing is just safe against bearing capacity failure. Indicate the computations on which your answer is based. (15 pts) (b) Under the applied load, and for the footing width of 10 ft, would you expect the footing settlement to be less than 1 in, about 1 in., or more than 1 in.? Circle one, (10 pts) P = column load only Show Work 1.0 ft 2.0 ft GWT (Water Table) B=10ft Clean Sand 15 Nave20 Ysat = 125 pcf Yabove GWT 120 pcf

Solution

(a)

B=10 feet

D=3 feet

As the soil given is a cohesionless sand, based on the data given above let us select a suitable angle of internal friction for the soil

let us assume that the angle of internal friction as 28

From terzaghis table, we get

N_q=17.26

N_p=14.78

The ultimate bearing capacity of a strip footing is given by

unit weight above the water table=120 pcf

saturated unit weight=125 pcf

unit weight of water=62.5 pcf

Q=CNc+QNq+0.5BPN_p

Q=120x3x17.260+0.5x10x(125-62.4)x14.78

Q=10839.4 lbf/ft²

Q_nu=10839.4-(120*3)=10479.74 lbf/ft²

Q_ns=Q_nu/F

Q_ns=10479.74/3=3493.24 lbf/ft²

Qsafe=3493.24+(120x3)=3853.24 lbf/ft²

Safe Load=Q_safexArea=3853.24 lbf/ft²x3x10

P=115597.2 lbs

P=115.59 kips

(b) Settlement:

The settlement comprises of immediate settlement, primary settlement and secondary settlement:

In case of sand evaluation of primary settlement is considered sufficient because, there will be no consolidation in case of sand,

settlement=qB(1-u²)/E_s

consider u=0.3(poissons ratio of sand)

Es=4 ksi (assume)

B=10 feet

q=115 /(120x30)

q=0.031 ksi

settlement=0.031x120x(1-0.3²)/4

settlement=0.84 in

The assumptions here are based on the generalized standard conditions and may vary a little