Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion

Discussion Forum : Soil Mechanics and Foundation Engineering - Section 3 (Q.No. 25)
25.
A moist soil sample of volume 60 cc. weighs 108 g and its dried weight is 86.4 g. If its absolute density is 2.52, the degree of saturation is
54%
64%
74%
84%
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
15 comments Page 1 of 2.

Rhg said:   1 decade ago
Volume of sand = 86.4/2.52 = 34.28.

Volume of voids = 108-86.4 = 25.72.

Volume of water = 60-34.28 = 25.72.

Raihan said:   1 decade ago
r dry = r/ (1+w).

r dry = (86.4/60) =1

44r = 2.52.

Water content = 25%.

Dry = Grw/(1+e)e = 75%.

r = (G+ESR)row/(1+e).

So degree of saturation = 84%.

Abhay said:   10 years ago
Υd = G.Υw/(1+e).

Υd = Ws/V = 86.4/60 = 1.44.

= 1.44 = 2.52*1/(1+e).

=> e = 0.75.

w = Ww/Ws = (108-86.4)/86.4 = 0.25.

e*Sr =w*G.

= 0.75*Sr = 0.25*2.52.

Sr = 0.84 i.e. 84% answer.

BHASKAR said:   8 years ago
Y=108/60=1.8,
Y(dry)=86.4/60=1.44,
G=2.52.
Now Y=Y(dry)(1+w),,
w=0.25.

Y(dry) = GY(w)/1+e,
e = 0.75.

Se = wG,
S =.25 * 2.52/0.75,
S = 84%.

Kishan datta said:   8 years ago
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%.
(1)

Shashank said:   8 years ago
The Critical hydraulic gradient (i) = G-1/(1+e).

Anomie said:   7 years ago
Thanks for the explanation @Anand Joshi.

Jonty said:   7 years ago
Well said Thanks @Abhay.

Joti said:   6 years ago
Thanks @Bhaskar.

Balvinder said:   6 years ago
How you guys are taking G=2. 52?

Please explain.


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