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 1 of 3.
AULIA SK SALIM said:
1 year ago
g/cc= 9.81KN/m3.
2gm/cc=2*9.81=19.62KN/m3.
Dry density Yd=19.62/(1+w)=19.62/(1+0.1)=17.836KN/m3,
Yd = G.Yw/ (1+ e) , e=0.4855.
Yb = Yw*(G+e*Sr)/(1+e). Sr=0.5561=55.61%.
2gm/cc=2*9.81=19.62KN/m3.
Dry density Yd=19.62/(1+w)=19.62/(1+0.1)=17.836KN/m3,
Yd = G.Yw/ (1+ e) , e=0.4855.
Yb = Yw*(G+e*Sr)/(1+e). Sr=0.5561=55.61%.
GURPREET SINGH KANSAY said:
2 years ago
This is very simple.
Yd = 2/1 + 0. 1 = 1.81.
Void raito e= (G. Yw/Yd) - 1.
E= (2. 7 X 1/1. 81) -1 = 0.49.
Now we know.
S = WG/e = 0. 1 x 2. 7/0. 49 = 0.55.
55% answer.
Yd = 2/1 + 0. 1 = 1.81.
Void raito e= (G. Yw/Yd) - 1.
E= (2. 7 X 1/1. 81) -1 = 0.49.
Now we know.
S = WG/e = 0. 1 x 2. 7/0. 49 = 0.55.
55% answer.
Eniga said:
2 years ago
Pt = (Gs(1+w) /1+e) pw g/cm^3.
2 = (2.7(1+0.1)/1+e)*1.
e = 0.485.
Se = Gs wc.
S 0.458 = 2.7 0.1
S = 0.589.
2 = (2.7(1+0.1)/1+e)*1.
e = 0.485.
Se = Gs wc.
S 0.458 = 2.7 0.1
S = 0.589.
Mohit Pal said:
2 years ago
Simply apply bulk mass unit wt formula
P= bulk mass unit wt
G= specific gravity of soil solids
e=void ratio
S=saturation
Pw= mass unit wt if water =1g/m3
Now,
P = {(G+Se)}/{1+e}
But,
Se = wG
Put values as a substitute for e, and solving
P = {(G+wG)}/{(1)+((wG)/S)}
Then, S = 0.55670 = 55.67%.
P= bulk mass unit wt
G= specific gravity of soil solids
e=void ratio
S=saturation
Pw= mass unit wt if water =1g/m3
Now,
P = {(G+Se)}/{1+e}
But,
Se = wG
Put values as a substitute for e, and solving
P = {(G+wG)}/{(1)+((wG)/S)}
Then, S = 0.55670 = 55.67%.
Lokesh said:
3 years ago
Dry unit weight =bulk unit weight/1+w.
Dry unit weight =1.81.
Also,
Dry unit weight = G * unit weight of water/1+e.
1.81=2.7*1/1+e,
e=0.49,
Se=wG,
S * 0.49 = 0.10 * 2.7.
S = 55.10%.
Dry unit weight =1.81.
Also,
Dry unit weight = G * unit weight of water/1+e.
1.81=2.7*1/1+e,
e=0.49,
Se=wG,
S * 0.49 = 0.10 * 2.7.
S = 55.10%.
Vim said:
3 years ago
Bulk density =((1+w)*G*Yw)/(1+e).
2=((1+.1)*2.7*1)/(1+e) solve.
e=.485 then,
Sr*e=Gw.
Sr= 2.7*.1/.485 = .5567 = 55.67%.
2=((1+.1)*2.7*1)/(1+e) solve.
e=.485 then,
Sr*e=Gw.
Sr= 2.7*.1/.485 = .5567 = 55.67%.
Asutosh said:
3 years ago
BULK UNIT WEIGHT = 9.81* Density = 9.81*2 = 19.62 kN/m^3.
DRY UNIT WEIGHT = (19.62/1+0.1) = 17.836 kN/m^3.
So, e = (G*Yw/Yd)-1 = (2.7*9.81/17.836) = 0.485.
Therefore, S = w*G/e = (0.1*2.7/0.485) = 55.67%
So, Answer is Option (B).
DRY UNIT WEIGHT = (19.62/1+0.1) = 17.836 kN/m^3.
So, e = (G*Yw/Yd)-1 = (2.7*9.81/17.836) = 0.485.
Therefore, S = w*G/e = (0.1*2.7/0.485) = 55.67%
So, Answer is Option (B).
Ethan Russell M. said:
3 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.
Mostafa said:
3 years ago
Thanks @Gilbert Khakhlari.
Maggi said:
4 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%.
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