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
Discussion:
15 comments Page 1 of 2.
Umar habib said:
3 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%.
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%.
(9)
Kataki said:
5 years ago
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%.
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%.
(6)
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%.
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)
Geet said:
4 years ago
@Kishan.
Very nice, Thanks.
Very nice, Thanks.
(1)
Romi said:
3 years ago
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.
Sind the degree of saturation, water content, void ratio, porosity, bulk unit weight and dry unit weight.
(1)
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.
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%.
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.
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%.
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%.
Shashank said:
8 years ago
The Critical hydraulic gradient (i) = G-1/(1+e).
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