Question

The dimension of the soil sample of a cylindrical shape is provided in Figure 1. In its natural moist state, the soil sample has the total weight of 1350 g before the water content determination. The specific gravity, Gs of the soil is given as 2.72. The values obtained for the determination of water content are: The mass of moist soil sample + mass of empty can = 1.415 kg; the mass of fully dry soil sample + mass of empty can = 1.123 kg; the mass of partially dry soil sample + mass of empty can = 1.227 kg; the mass of empty can = 0.065 kg. h= 15.8 cm d =8.8 cm

i. Natural moist (bulk) density, ρb (3%)

ii. The water content, w

iii. Dry density of the soil, ρd (2%). Kindly note: Use the formula relationship for bulk and dry density

iv. Unit mass of solids, ρs (2%)

v. The degree of Saturation (2%)

vi. Find the volume of air in the soil (2%)

vii. The void ratio, e and porosity, n (4%)

viii. How much water in grams is required to make the soil fully saturated, if it were partially saturated? (3%)

ix. Draw the phase diagram of the soil sample and indicate if the soil is fully saturated, partially saturated or dry? Please note that you must show all the mass-volume relationship with their values provided for (5%

Answer 1

the mass of bulk of soil =1350g =1.350kg

Gs of the soil is given as 2.72

mass of moist soil sample + mass of empty can = 1.415 kg;

the mass of fully dry soil sample + mass of empty can = 1.123 kg;

the mass of partially dry soil sample + mass of empty can = 1.227 kg;

the mass of empty can = 0.065 kg.

h= 15.8 cm d =8.8 cm

volume of cylinder = vol of soil =

mass of soil = 1.123-0.065 =1.058kg

mass of water = 1.415-1.123=0.292kg

i) water content of given soil sample = massof water/ mass of soil = 0.292/1.058 x100=27.59%

ii) density of sample =mass/ volume =1.350/960.975=1.4kg/cm2

iii)density of sample=

iv) mass of soil = 1.123-0.065 =1.058kg

unit wt of soil =

vii)

Void ratio e of soil  

v)

saturation of soil =S =

%

porosity n=

vi)

V=Vv+Va+Vs+Vw

na= % air void

VOLUME OF AIR=Va

na= % air void=Va/Vx100

Va = 1.7001 x 960.975/100 =16.337cm³

e=Vv/Vs

Vs= 16.337/1.477=11.1557cm³

ac= 1-S =1-0.49=0.503

=

Vv= 16.337/0.503 =32.499cm³

viii)

V=Vv+Va+Vs+Vw

Vw= 960.97-(16.337+32.499+11.155)=900.909cm³

mass of partially saturated soil =1.277kg

its water content =%

degree of saturation of soil

Se=Gw

S=(2.72 X 12.7)/1.477=23.39

AS s for fully saturated soil S=100%

the water content should be w=(100x1.477)/2.72=54.30%

mass of water for full saturation of soil =w/Msx100=(54.3x100)/1.058=0.513kg

ix)

V=Vv+Va+Vs+Vw

Vs= 16.337/1.477=11.1557cm³

Va = 1.7001 x 960.975/100 =16.337cm³

Vv= 16.337/0.503 =32.499cm³

Vw= 960.97-(16.337+32.499+11.155)=900.909cm³

Wa = mass of water =0 ws = mass of soil =1.058kg ww=massof water =1.35-1.058=0.292kg

the phase dia of partially saturated soil is given below