Problem 2 (a) A square footing is to be constructed in sand as shown the sketch
ID: 2313142 • Letter: P
Question
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 pcfExplanation / Answer
(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
Nq=17.26
Np=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+0.5BPNp
Q=120x3x17.260+0.5x10x(125-62.4)x14.78
Q=10839.4 lbf/ft2
Qnu=10839.4-(120*3)=10479.74 lbf/ft2
Qns=Qnu/F
Qns=10479.74/3=3493.24 lbf/ft2
Qsafe=3493.24+(120x3)=3853.24 lbf/ft2
Safe Load=QsafexArea=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)/Es
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
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