A 600 mm square bearing plate settles by 15 mm in plate load test on a cohesionless soil under an intensity of loading of 0.2 N/ram2. The settlement of a prototype shallow footing 1 m square under the same intensity of loading is
A. 15 mm
B. Between 15 mm and 25 mm
C. 25 mm
D. Greater than 25 mm
Answer: Option B
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A. directly proportional to time and inversely proportional to drainage path
B. directly proportional to time and inversely proportional to square of drainage path
C. directly proportional to drainage path and inversely proportional to time
D. directly proportional to square of drainage path and inversely proportional to time
For Cohesive soil or clay,
(i) Sf/Sp = Bf/Bp
But for Cohesionless soil or sand,
(ii) Sf/Sp = [ Bf (Bp+0.3) / Bp (Bf+0.3) ]^2
Here, it is cohesionless soil, so eq. (ii) will be apply
=》 Sf/0.015 = [ 1 (0.6 +0.3) / 0.6 (1 + 0.3) ]^2
=》 Sf / 0.015 = [ 0.9 / (0.6* 1.3) ]^2
=》 Sf/0.015 = 1.33
=》 Sf = 0.01995 m = 19.95 mm
Which is in between 15mm to 25 mm..
So, Answer B is correct.
Sf=sp(bf(bp+30)÷bp(bf+30)^2 15(100(60+30)÷60(100+30). 9000÷7800 =1.15. 1.15^2 = 1.3225×15=19.83mm. Settlement of plate sp=15mm Width of plate Bf=1m=100cm. Width of plate Bp=600mm=60cm.
Pp=Sf × Wp/Wf
15= Sf × 600/1000
Sf= 25mm....exact
Please someone describe this