Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion

@Balvinder.

Even if the absolute density is given, we get relative density by dividing it by density of water which is in our case 1g/cc and absolute density should have been 2.52 g/cc.

It should have been Relative Density instead of absolute density.

Mass of water=108-86.4 g =21.6 g.
Volume of water=21.6/1 cc=21.6cc.
Now, Mass of dried soil =86.4g and absolute density =2.52.
Therefore Volume of soil solids = 86.4/2.52=34.29cc.
Also, Total volume is given = 60cc.
Hence volume of air =60-(34.29+21.6) cc = 4.11 cc.
Volume of voids = 4.11 + 21.6 = 25.71 cc.
Thus Degree of Saturation =Vol. of water/Vo/. of voids *100=21.6/25.71*100=84%.

@Kishan.

Very nice, Thanks.

A partially saturated sample of soil has a volume of 60 cc and a mass of 92g. The sample is dried in an oven and its dried mass is 73.8g. If the specific gravity of solids be 2.62.

Sind the degree of saturation, water content, void ratio, porosity, bulk unit weight and dry unit weight.

Step1) bulk density(yb)= W/V=108/60 = 1.8.

Step2) dry density (yd)=Wd/V=86.4/60 = 1.44.

Step3) water content determination(w).
Yd=Yb/1+w.

So 1.44= 1.8/1+w,
So 1+w = 1.8/1.44,
So w = (1.8/1.44)-1,
So, w= 1.25-1= o.25,
w = 0.25.

Step4) void ratio determination;
Yd=G.Yw/1+e,
1.44= 2.52/1+e,
e= (2.52/1.44)-1,
e= 0.75.

Step5) degree of saturation determination(SR)
SR= WG/e.
Sr= 0.25 * 2.52/0.75,
Sr= 0.84,
Sr= 84%.

So, the answer is 84%.