Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion
Discussion Forum : Soil Mechanics and Foundation Engineering - Section 1 (Q.No. 19)
19.
A compacted soil sample using 10% moisture content has a weight of 200 g and mass unit weight of 2.0 g/cm3. If the specific gravity of soil particles and water are 2.7 and 1.0, the degree of saturation of the soil is
Discussion:
27 comments Page 2 of 3.
Erifi said:
8 years ago
Yeah, I also got 55.6.
Sujan sarkar said:
8 years ago
Partially saturated soil samples obtained from and Earth fill as a natural moisture contain 22%of Unit Weight of 19.62 kn/m^3. Given a specific gravity 2.7 and unit of water 9.81kn/m^3..determine the degree of saturation, voids ratio, saturated unit weight, dry unit weight of soil.
Can anyone please solve this?
Can anyone please solve this?
Victor n.m said:
7 years ago
G=2.7, YW=9.81 ,
W=22/100=0.22.
Bulk unit wight=19.62
1. Yd=19.62/(1+w)=19.62/(1+0.22)=16.08
2. void ratio (e)=(G*Yw/Yd)-1=(2.7*9.81/16.08)-1=0.647
3. saturatted unit weight (s)=G*W/e=2.7*0.22/0.647=0.918
0.918*100=91.8%.
W=22/100=0.22.
Bulk unit wight=19.62
1. Yd=19.62/(1+w)=19.62/(1+0.22)=16.08
2. void ratio (e)=(G*Yw/Yd)-1=(2.7*9.81/16.08)-1=0.647
3. saturatted unit weight (s)=G*W/e=2.7*0.22/0.647=0.918
0.918*100=91.8%.
(1)
Rahul dangwal said:
7 years ago
Yd = (1+GYw)÷(1+wG÷Sr).
Hasnain malik said:
7 years ago
B is the answer.
UJJVAL said:
7 years ago
W=.1, bulk density = 2g/cc.
Dry density = 2/1 +.1 = 1.8g/cc.
e = [(2.7 * .1)/1.8-1] = .485.
S = (2.7 * 0.1)/.485 = .556.
OR
S = (.556*100)%.
Dry density = 2/1 +.1 = 1.8g/cc.
e = [(2.7 * .1)/1.8-1] = .485.
S = (2.7 * 0.1)/.485 = .556.
OR
S = (.556*100)%.
Anik said:
7 years ago
Very well explained, Thanks @Ujjval.
Maggi said:
6 years ago
A natural soil deposit has a bulk density of 1.90 g/cm^3 and water content of 6 per cent assume G=2.67 assuming the voids ratio to remain constant what will be the degree of saturation at a water content of 16%.
(2)
Mostafa said:
6 years ago
Thanks @Gilbert Khakhlari.
Ethan Russell M. said:
6 years ago
Density (Sat)= Mass(sat)/Vol(sat)
2.0=200/ Vs
Vs=100.
Moisture Content = Wt of Water/Wt dry Soil.
0.1 = (200 - Wdry)/Wdry.
Wdry = 181.81 gm.
Solid Unit Wt = Wdry/Volume (dry).
2.7 = 181.81/V(dry),
V(dry)=67.33.
Void Ratio = Vv / Vs = {V(sat)-V(dry)}/V(dry)
e = (100-67.33) / 67.33
e = 0.485.
se = wGs.
s= {0.1(2.7)/0.485}*100.
s= 55.67.
2.0=200/ Vs
Vs=100.
Moisture Content = Wt of Water/Wt dry Soil.
0.1 = (200 - Wdry)/Wdry.
Wdry = 181.81 gm.
Solid Unit Wt = Wdry/Volume (dry).
2.7 = 181.81/V(dry),
V(dry)=67.33.
Void Ratio = Vv / Vs = {V(sat)-V(dry)}/V(dry)
e = (100-67.33) / 67.33
e = 0.485.
se = wGs.
s= {0.1(2.7)/0.485}*100.
s= 55.67.
(3)
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