Problem 21(i) A 600 g sample of moist soil with a moisture content of 12% was pl

Question

Problem 21(i) A 600 g sample of moist soil with a moisture content of 12% was placed on a stack of #200 sieve and then "washed" forcing minus #200 sized particles to pass through. The soil retained on #200 was then oven dried and its mass was found to be 400 g. The soil had no gravels. Atterberg's limit tests were performed separately and following results were obtained: LL -20, PL 15. Using USCS classification, classify the soil. Mention the symbol and description. Show your work (% Gravel, % Sand, % Fines and flowchart steps) to get full credit. (5 points)

Explanation / Answer

Q.1) eight of moist soil sample = 600g
water content(w)=12%
so, weight of dry soil =600/(1+w) =600/1.12 =535.7 g
soil, retain on #200 no sieve =400 g
% of soil retain = 400/535.7 =74.67% > 50%
so, it is coarse grain soil.
As the soil had no gravel, so, it is sand.
It contain 100-74.67=25.33% of fine material.
as liquid limit is 20 ,less than 50 , it is low plastic.
From equation of A line Ip=0.73(WL-20)
=0.73(20-20) =0
Here plasticity index = liquid limit -plastic limit = 20-15 =5
so the point is above A line( as 5>0)
so it contain Clay.
so, the soil is Sand contain low plastic Clay.
Group symbol is SC

Q.2)thickness of soil layer is 3ft
void ratio(e) =1.3
New void ratio(e')=0.9
assuming a unit area soil, so total volume = area*thickness =1*3 =3
now ,void ratio = Vv / Vs = 1.3
so, Vv=1.3 Vs
again Vv+Vs = 3 (total volume)
so , 1.3Vs+Vs = 3
Vs =3/2.3=1.304
Now, again Vv' / Vs=0.9
so, Vv' =0.9Vs
=0.9*1.304( by compression,only void is compressed, solid volume remain constant)
=1.174
Now ,total volume =Vv'+Vs =1.174+1.304 = 2.478
thickness = volume / area =2.478/1 =2.478 ft
new thickness = 2.478 ft

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