Question

A field density test has been run on a soil containing 18% of its weight in particles retained on the No. 4 sieve. The dry density of the total sample was 128.7 pcf and the water content was 7%. The water content of the gravel fraction was 3%. the specific gravity of the solids was 2.69. Compute the water content and dry density of the soil fraction of the fill that passes the No. 4 sieve.

Answer 1

we know Unit weigth of water  Yw=62.4 pcf

given that the total soil has Yd=128.7 pcf

and water content of soil sample (w)s=7%

Now we know that Bulk unit weight (Yt)= Yd*(1+w) = 128.7*(1+0.07)=137.709 pcf

So, lets assume that volume of total sample is 1 cubic feet

so total weight of sample (W) = 137.709*1=137.709 lb

in which,

weight of soil solids Ws= 128.7 lb

and weight of water Ww= 9.009 lb

Now water content of gravel (w)g= 3%

total wieght of gravel retained = 18% (given )

Therefore weight of gravel (W)g= 0.18*137.709 = 24.78762 lb

soil solid weight of gravel (Ws)g= 24.78762 / (1+0.03) = 24.06565 lb ( we know Ws= W / (1+w)

and weight of water in gravel (Ww)g = 24.78762 - 24.06565 = 0.72197 lb

now total weight of fraction of soil passed (W)p = 82% of W = 0.82 * 137.709 = 112.92138 lb

weight of water in  fraction of soil passed (Ww)p = total weight of water in sample - weight of water in gravel
(Ww)p  = (Ww - (Ww)g) = ( 9.009 - 0.72197) = 8.28703 lb

weight of soil soilids in fraction of soil passed (Ws)p = total weight of soil solids in sample - weight of soil solids in gravel

(Ws)p  = ( (Ws) - (Ws)g ) = ( 128.7 - 24.06565 ) = 104.63435 lb

therefore we know w = (Ww) / (Ws) =

water content of passed fraction (wp) = (Ww)p / (Ws)p = 8.28703 / 104.63435 *100 = 7.91999 %  

(wp) = 7.91999 % Ans

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