Soil Properties

To study the properties of a soil mass, it is good to adopt an idealized form of the it as shown
by the below

The soil mass has a total volume V and a volume of solid particles that sums up to VS then
the volume of voids VV is V-VS

From above the void ratio (e) = Volume of voids/Volume of solids = VV / VS

Porosity (n) = Volume of voids / Total volume = VV / V= VV / VS+ VV

= e/1+e

Degree of saturation

The voids of soil may be filled with air or water or both. if air is present the soil is dry
whereas if only water is present the soil is saturated. When both are present the soil is said to
be partially. The degree of saturation is simply

Sr= volume of water /volume of voids =Vw/VV. (It is usually expressed as a % )

Particle specific gravity (GS)

The specific gravity of a material is the ratio of the weight or mass of a volume of the
material to the weight or mass of an equal volume of water

A soil sample with volume of solids (VS) and weight of solids (WS) would have a

GS = (WS)/( VS*γw)

Where γw is the unit weight of water and if the sample has a mass of solids (MS) then

GS = (MS)/( VS*ρw)

Where ρw is the density of water (100kg/m3 or 1.0 Mg/m3 at 20˚ Celsius

This implies GS = (MS)/(VS*ρw)= (WS)/( VS*γw)

The density of the particle ρs=(MS)/(VS) then the specific gravity of a material could be

rewritten as GS = (ρs)/(ρw)

Density and unit weight

Density (ρ) = mass/volume =m/v

Unit weight (γ) =weight/volume =w/v

The unit weight of a material is its weight per unit volume. In soil works the most important
unit weight are as follows

Bulk unit weight (γ) this is the natural in-situ unit weight of the soil

γ = Total weight /Total volume =W/V =(WS +WV)/(VS+VV)

=[GsVs γw +Vv γwSr]/{VS+VV} = γw(Gs +eSr)/(1+e)

Saturated unit weight (γsat)

(γsat) = Saturated weight /Total volume

When soil is saturated Sr=1

This implies (γsat) = γw (Gs +e)/(1+e)

Dry unit weight (γd)

(γd) = Dry weight /Total volume = γw (Gs)/(1+e) since Sr=0

Buoyant unit weight

A soil below the water table will have part of its weight balanced by the buoyant effect of the
water. This up thrust equals the weight of the volume of the water displaced therefore if the
unit volume is considered then:

Buoyant unit weight =saturated unit weight-unit weight of water

Mathematically written = {γw (Gs +e)/(1+e)}- γw = γw {(Gs-1) / (1+e)}

This is also termed as the effective unit weight or submerged unit weight

Density of soils

Similar expressions can be obtained for densities as below

Bulk density, (ρb) = ρw (Gs +eSr)/(1+e)

saturated density ( ρsat) = ρw (Gs +e)/(1+e)

Dry density ( ρd) = ρw (Gs)/(1+e)

Buoyant density ( ρ՛) = ρw {(Gs-1) / (1+e)}

Relationship between W, (γd) and (γ)

(γ) = (ws +ww)/V--------------------------------------------(a)

(γd) =(ws)/V--------------------------------------------------(b)

w=Ww/Ws-------------------------------------------------------(c)

from equation (c) Ww=w*Ws substitute into equation (a)

(γ) = (Ws +w.Ws)/V =Ws(1+w)/V

Implies (γ) = Ws(1+w)/V

w=Ww/Ws= (Vw* γw)/(Vs* γw*Gs) =Vv/ Vs*Gs= e/Gs

Note Vw = Vv , when the soil is saturated

This means w=e/Gs

And e=wGs/Sr
Table one is a summary of the physical relation in soils
Table one Physical relation in soils
QUESTION

In a bulk density determination, a sample of clay with mass 683g was coated with paraffin
wax. The combined mass of the clay and the wax was 690.6g. The volume of the clay and the
wax was found by immersion in water to be 350ml

The sample was then broken open and moisture content and particle specific gravity test gave
17% and 2.73 respectively.

The specific gravity of the wax was 0.89. Determine the bulk density, unit weight, void ratio
and degree of saturation.

SOLUTION

Mass of soil = 683g

Mass of wax = (690.6-683)g=7.6g

Volume of soil =7.6/0.89=8.55ml

Volume of soil =350-8.6 =341.4ml

Ρb=683/341. 4=2.0g/ml=2.0Mg/m3

γb=2.0 *9.81=19.6KN/m3

Ρd= Ρb/(1+w) =2/(1+0.17)=1.71Mg/m3

Ρd= ΡwGs/(1+e)

1.71(1+e) = ΡwGs

1.71+1.71e=1*2.73

1.71e=2.73-1.71

e= (2.73-1.71)/1.71=0.596

Ρb= Ρw(Gs+eSr)/(1+e)

2.0= Ρw(Gs+eSr)/(1+e)

2.0(1.0+0.596) =2.73+0.596Sr
Sr= {2.0(1.0+0.596)-2.73}/0.596

Sr=77%

Example 2

A soil sample has a void ratio of 0.72, moisture content = 12% and Gs = 2.72 determine the
following: a) Dry unit weight, moist unit weight (KN/m3). b) Weight of water in KN/m3 to
be added for 80% degree of saturation. c) Is it possible to reach a water content of 30%
without change the present void ratio. d) Is it possible to compact the soil sample to a dry unit
weight of 23.5 KN/m3.

Solution Givens: e = 0.72 , %w = 12% , Gs=2.72 a)

∗ γdry =Gs × γw / (1 + e)

=2.72 × 9.81/ (1 + 0.72)

= 15.51 KN/m3.

∗ γbulk = γdry(1 + %w) = 15.51 × (1 + 0.12) = 17.374 KN/m3.

b) The original value of γbulk =17.374 KN/m3

The value of γbulk @80% degree of saturation can be calculated as following:

S.e = Gs.w→ %w80% = (0.8×.0.72)/ 2.72 = 0.2117

γbulk,80% = γdry(1 + %w) = 15.51 × (1 + 0.2117) = 18.8 KN/m3 .

So, the of water added= 18.8 − 17.374 = 1.428 KN/m3

c) e = 0.72 , %w = 30% , Gs=2.72 , S30% =??

We know that the max.value of S=1 so, if the value of S30% > 1 → it’s not possible,

but if S30% ≤ 1 → it’s possible.

S.e = Gs.w→ S30% = [(2.72×.0.3)/ 0.72]× 100% = 1.133 > 1 → Not possible.

d) γdry,new = 23.5 KN/m3 → Can we reach to this value after compaction???, to Know
this, we find the maximum possible value of γdry= γZ.A.V (Zero Air Voids)

γZ.A.V =(Gs × γw)/ 1 + emin

→ emin =(Gs.w)/ Smax

=(2.72 × 0.12)/ 1

= 0.3264

→ γZ.A.V =(Gs × γw)/ (1 + emin)

=(2.72 × 9.81)/ (1 + 0.3264)

= 20.12 < 23.5 → Not possible.

Example three:

If a soil sample has a dry unit weight of 19.5 KN/m3, moisture content of 8% and a specific
gravity of solids particles is 2.67. Calculate the following: a) The void ratio. b) Moisture and
saturated unit weight. c) The mass of water to be added to cubic meter of soil to reach 80%
saturation. d) The volume of solids particles when the mass of water is 25 grams for
saturation.

Solution

Given:

γdry = 19.5KN/m3 , %w = 8% , Gs=2.67

a)

γdry =(Gs × γw)/ 1 + e

19.5 =(2.67 × 9.81)/ 1 + e

e = 0.343

b) γbulk = γdry(1 + %w) = 19.5 × (1 + 0.08) = 21.06 KN/m3

γsat = γdry(1 + %wsat) → %wsat means %w @ S = 100%

S.e = Gs.w→ %wsat = S.e /Gs = [(1×0.343)/ 2.67] × 100% = 12.85%

So, γsat = 19.5(1 + 0.1285) = 22 KN/m3

c) γbulk = 21.06 KN/m3 →

Now we want to find γbulk @ 80% Saturation so, firstly we calculate %w @80% saturation:

%w80% =S.e Gs

=0.8 × (0.343) (2.67)× 100% = 10.27%

γbulk,80% = 19.5(1 + 0.1027) = 21.5 KN/m3

Weight of water to be added = 21.5-21.06 = 0.44 KN/m3

Mass of water to be added = 0.44× 1000/ 9.81 = 44.85 Kg/m3

Example four

a) Show the saturated moisture content is: Wsat = γw [ 1 /γd−1/ γs]

𝐇𝐢𝐧𝐭: γs = solid unit weight

Solution S.e = Gs.w , at saturation → S = 1 → wsat =e /Gs→Eq.(1)

γd =(Gs × γw) /1 + e

→ e =[(Gs × γw)/γd]− 1, substitute in Eq.(1) →→

wsat = {[(Gs × γw)/ γd]− 1}Gs=γw/ γd−1/Gs

but Gs =γs/γw→1/ Gs=γw /γs→→

wsat =γw/ γd−γw/ γs

= γw [1/ γd−1/ γs].

b) A geotechnical laboratory reported these results of five samples taken from a single
boring. Determine which are not correctly reported, if any, show your work. 𝐇𝐢𝐧𝐭: take γw =
9.81kN/m3

Sample #1:w = 30%,γd = 14.9 kN/m3,γs = 27 kN/m3 ,clay

Sample #2:w = 20%,γd = 18 kN/m3,γs = 27 kN/m3 ,silt

Sample #3:w = 10%,γd = 16 kN/m3,γs = 26 kN/m3 ,sand

Sample #4:w = 22%,γd = 17.3 kN/m3,γs = 28 kN/m3 ,silt

Sample #5:w = 22%,γd = 18 kN/m3,γs = 27 kN/m3 ,silt

Solution

For any type of soil, the moisture content (w) must not exceed the saturated moisture content,
so for each soil we calculate the saturated moisture content from the derived equation in part
(a) and compare it with the given water content. Sample #1: (Given water content= 30%)

wsat = 9.81[1 /14.9−1 /27] = 29.5% < 30% → not correctly reported.

Sample #2: (Given water content= 20%)

wsat = 9.81[1/18−1/27] = 18.16% < 20% → not correctly reported.

Sample #3: (Given water content= 10%)

wsat = 9.81[1/16−1/26] = 23.58% > 10% → correctly reported.

Sample #4: (Given water content= 22%)

wsat = 9.81[1/17.3−1/28] = 21.67% < 22% → not correctly reported.

Sample #5: (Given water content= 22%)

wsat = 9.81[1/18−1/27] = 18.16% < 22% → not correctly reported

ASSIGNMENT

What is the difference between moisture and degree of saturation?

A certain saturated soil (s=100%) has a moisture content of 25.1% and a specific gravity of
2.68. It also has a maximum index void ratio of 0.84 and minimum index void ratio of 0.33.
Compute its relative density and classify its consistency.

A sample of soil has a volume of 0.45ft3 and a weight of 53. 3lb.After being dried in oven the
weight came down to 45. 1lb.It has a specific gravity of solids of 2.7.

Compute its moisture content and degree of saturation before it was placed in the oven.