# 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/cm

^{3}. If the specific gravity of soil particles and water are 2.7 and 1.0, the degree of saturation of the soil isDiscussion:

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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