Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion

Given That,

Water Content w = 0.1
Unit Wt of Soil = 2.0 g/cc.
Sp. Gravity of Soil = 2.7.
Unit Wt of water = 1 g/cc.

Dry Unit Wt of Soil = Bulk unit wt/ (1+ w).
= 2/(1+0.1).
=1.818 g/cc.


Now,

Void Ratio (e)=(G * Unit Wt of Water/dry unit wt ) - 1.
= (2.7 * 1 / 1.818) - 1.
= 0.4851 .
Now,

Degree of Saturation (S) = G*w/e.
=2.7*0.1/0.4851.
=0.556.
=55.6% Ans.

But the answer showing is 69.6%.

Read the question properly guys. You all did a common mistake.

Hint: Check the units. It's g/cm^3. It's the unit of density and not unit weight even though the question says so then the question is solved correctly.

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).

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.

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?

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%.

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).

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%.

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%.

We know,

Y = (G+S*e)*Yw/(1+e).

Also, G*w = S*e => e = G*w/S.

Therefore,

Y = (G+G*w)*Yw/(1+G*w/S).

Substitute the values and find S.

You will get answer B as the right option.

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%.