Soil Mechanic -Ch2

1University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& The physical state of a soil sample

Total Volume = VT = Vs + Vw + Va Total Weight = WT = Ws + Ww Porosity (n): is the ratio of void volume.

n = Vv/VT

Void Ratio (e):is the ratio of void volume to solid volume. e = Vv/Vs

now n = Vv/VT = Vv/ Vv+ Vs = Vv/Vs e= e+1Vv/Vs+1

Note also that: e = n / (1 - n) v = 1 / (1 - n)

Note: • n < 1 and is expressed as % • e may be any value greater or smaller than unity.

Example: A soil has a total volume of 250ml and a void ratio of 0.872. What is the volume of solids of the sample?

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2University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Example: A soil has a porosity of 0.45. What is the value of its void ratio?

Degree of saturation (s):is the ratio of water volume to void volume. Sr = Vw/Vv if S=o dry soil (Vw =o) S=100 saturated soil (Vw =Vv) 0 <S < 1→the soil is partially sat. Water Content (w):is the ratio of water weight in a soil sample to the solids weight. wc = Ww/Ws

Specific gravity (GS): specific gravity of soil solids of a soil is defined as the ratio of the density of a given volume of the solids to the density of any equal volume of water at 4ºC.

ws

s

w

s

s

w

ss V

MVM

Gρρρ

γ===

Soil type G Gravel 2.65-2.68

2

Sand 2.65-2.68

3University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Silt 2.66-2.7 Clay 2.68-2.8 Organic soils may be less than 2.0

Example: A sample of oven dried soil had a mass of 306g. The soil was broken down and placed into a jar of internal volume 1000 ml. Water at 20C was then poured into the jar and a rubber stopper placed to seal the jar. The soil and water were thoroughly mixed using an end over end shaker until all air had been removed. The jar was then topped up with water to the 1000ml mark and the total mass of the jar and its contents was found to be 1440.5g. The mass of the empty jar was 250g.

Determine the particle specific gravity of the soil.

or...

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4University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006

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Air Content (A):is the ratio of air volume to total volume. The air-voids volume, Va , is that part of the void space not occupied by water(is the ratio of air volume to total volume). Av= = Va / V Va = Vv - Vw = e - e.Sr = e.(1 - Sr) Air-voids content, Av Av = (air-voids volume) / (total volume) = Va / V = e.(1 - Sr) / (1+e) = n.(1 - Sr) For a perfectly dry soil: Av = n For a saturated soil: Av = 0

5University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& e in term of V, Ws, Gs

1 1 w

w

Gs VVv V Vs V Ve Vs Vs Vs Ws WsGs

1γ

γ

−= = = − = − = −

Bulk (Total)density (ρt)and Dry density (ρd ):

Bulk (Total)unit weight (γt)and Dry Unit weight (γd ):

Saturated Unit weight ( sγ ):for sat .soil S= 100% =1

1 wse Gs

e γγ ++=

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6University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Tutorial Das 2-4 For a given soil , show that ( )

satw wsat sat

Gs γγ ω γ γ=

− −

Solution ( )sat

w wsat satGs γ

γ ω γ γ=− −

Q wsatbγ γ γ= −

∴ 11( )1

wsatw sat b w wsat

e GseGs Gs

e

γγγ ω γ γ ω γ

++= =− −− +

11 1 .(1 ( )1

w

sat satw sat

e Gse GseGs Gs e Gs

e

γω ωγ ω

+++= =− + − +− +

For satγ → S = 100% Q sSe G ω= → e = satω Gs

∴ 1 1 1sat

sat sat sat

Gs Gse Gs e GsGs e eω

ω ω ω++ += = =+ − + + +

∴ ( 1(1 )

sat

sat

GsGs )ωω

+=+

∴ Gs = Gs o.k. 2-5 For a given soil , show that w

sat wsat

nn

γω γ γ= −

Solution

1

wsat

w w

ne Gs ne

γωγ γ

= + −+

Q 1en e= +

. .1 1

. .1 1 1

sat

e w e we e

e Gs e w w e Gs w e wwe e e

.

γ γ

γ γ γγω + +=

+ +−

+ + +

=

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γ−


..w

satw

eGsγω γ=

At satω → .sat Gsω =

∴ satω eGs=

∴ sat satω ω= o.k. 2-6 For a given soil , the following are given : Gs= 2.67 ; moist. Unit weight 3112 /Ib ftγ = ; moisture content 10.8%ω = . Determine : a-Dry unit weight b-Void ratio c-Porosity d-Degree of saturation Solution

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cwt

w Gseγ γ+= +

∴

0.108 1112 (62.4)(2.67)1184.6 1 0.6482112

0.6482 0.3931 1 0.6482

ee

en e

+= += − =

= = =+ +

3(2.67)(62.4). 101.08 /1 1 0.6482

wd

Gs lb fteγγ = = =+ +

Q sSe G ω=

∴ (0.108)(2.67). 44.48%0.6482c GsS e

ω= = =

2-9 For the soil describe in problem 2.6 , determine the weight of water , in pounds to be added per ft3 of soil for saturation ? Solution 1 ft3 of soil weight 112 lb

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Before saturation ωc =0.108


Q ( )(wwc

s

W WWω = ⇒ = )c sWω = 0.108 x 112 = 12.096 lb

After saturation S = 100% → Q sSe G ω=

∴ se G ω= 0.6482 0.24272.67cforsaturatione

Gsω⇒ = = =

After saturation Ww = ωc x Ws = 0.2427 x 112 = 27.1824 lb ∴ Ww after – Ww before = 27.1824 – 12.096 = 15.08 lb added . 2-12 For a moist soil , given the following : V = 0.25 ft3 ; W = 30.75 lb ; ω = 9.8% ; Gs = 2.66 . determine the following :

a. γ (lb/f3 ) b. γd(lb/f3 ) c. e d. n e. S f. Volume occupied by water

Solution

a. 330.75 123 /0.25tW lb ftVγ = = =

c. Q ( )( )1 0.098 1123 62.4 2.661 1c

wt Gse eωγ γ+ += ⇒ =+ +

∴ e =0.4817

b. ( ) 32.66 62.4 112.02 /1 1 0.4817wdGs lb fteγ γ= = =+ +

d. 1en e= + ∴ 0.4817 0.325 32.5%1 0.4817n = = =+

e. sSe G ω= → S x 0.4817 =2.66 x 0.098 S= 0.5411 =54.11%

f. 30.325 0.081250.25v v

vV Vn V ftV= → = ⇒ = Q 0.5411 0.08125

w wv

V VV= ⇒ =S

∴ Vw = 0.0439ft3

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2-15 For a soil , given ρd = 1712 Kg/m3 ; e = 0.51 determine


a. n b. Gs

Solution

0.51 0.3377 33.77%1 1 0.51en e= = = =+ +

3(9.81) (1.712)(9.81) 16.794 /d d KN mγ ρ= = =

( )(9.8116.7941 1

2.585

wd

GsGse

Gs

)0.51

γγ = ⇒ =+ +=

2-17 A soil has unit weight of 126.8 lb/ft3 . Given Gs = 2.67 and ω= 12.6% determine a.dry unit weight (lb/ft3 ) b. void ratio c. porosity d. The weight of water in (lb/ft3 ) of soil needed for full saturation . Solution

1 0.126 1( )( ) 126.8 (2.6)(62.4)1 1c

wt Gse eωγ γ+ += ⇒ =+ +

0.4407=

e ∴

0.4407 0.3058 30.58%1 1 0.4407

en e= = = =+ +

for S = 100% (1)(0.4407) (2.6)

0.1695 16.95%c c

c

Se Gsω ωω

= ⇒ == =

1 lb/ft3 of dry soil has

1

0.1695 0.16951

wc

s

w w

W Ws lbWW W l

ω = ⇒ =

b∴ = ⇒ =

Q

2-18 The saturated unit weight of soil is 20.12 kN/m3 . Given Gs = 2.74 , determine a. γ dry b. e

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32.6 (62.4) 112.61 /1 1 0.4407wdGs lb fteγ γ= = =+ +


c. n d. ωc Solution

2.7420.12 (9.81)1 1wsate Gs e

e eγ γ+ += ⇒ =+ +

∴ e = 0.657

30.657 2.74(9.81) 16.22 /1 1 0.657wdGs kN meγ γ += = =+ +

0.657 0.3964 39.64%1 1 0.657en e= = = =+ +

cSe Gsω=Q Q S = 100% ∴ (1)(0.6574) (2.74) 0.24 24%c cω ω= ⇒ = = 2-19 For a soil given e = 0.86 , cω = 28% and Gs = 2.72 determine

a. moist unit weight (lb/ft3) b. degree of saturation (%)

solution

31 0.28 1( )( ) (2.72)(9.81) 18.362 /1 1 0.86c

wt Gs kN meωγ γ+ += = =+ +

cSe Gsω=Q S x 0.86 = 0.28 x 2.72 ∴ S = 0.8855 = 88.55% 2-20 For a saturated soil ; given γd = 15.29 kN/m3 ; and cω =21% ; determine

a. γsat. b. e c. Gs d. γ moist when the degree of saturation is 50%.

Solution

( )1 wsate Gs

eγ γ+= +

cSe Gsω=Q for S = 100% 1 x e = 0.21 x Gs

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∴ 0.21seG =


0.21 0.21(9.81) 15.29 (9.81)1 1 1wd d

e eGs

e e eγ γ γ= ⇒ = ⇒ =+ + +

∴ e = 0.4865

0.4865 2.3160.21 0.21seG = = =Q

( )1 wsate Gs

eγ γ+= +

30.4865 2.316(9.81) 18.49 /1 0.4865sat kN mγ += =+

For 50% = S cSe Gsω=Q ∴ 0.5 x 0.4865 = cω x 2.316 ∴ cω = 0.105 = 10.5%

31 0.105 1 (2.316)(9.81) 16.889 /1 1 0.4865c

wt Gs kN meωγ γ+ += = =+ +

Or:

32.316 (0.5)(0.4865) (9.81) 16.889 /1 1 0.4865wtGs se kN meγ γ ++= = =+ +

Additional Tutorial(Reading Carefully)

Ex: . A newly excavated soil sample was found to have a mass of 431g and a volume of 221ml. After oven drying the mass of the soil reduced to 401g. Calculate the:

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(a) Moisture content


(b) bulk density and bulk unit weight

(c) dry density and dry unit weight

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13University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Example: A soil has a bulk density of 1.93 Mg/m3 and a moisture content of 23%. If the void ratio of the soil is 0.652, determine its particle specific gravity and degree of saturation.

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Example: A sample of soil was coated with paraffin wax. Before waxing the soil had a mass of 983g and after waxing the total mass was 992g. The sample was then immersed in water and the total volume was found to be 481cc. On removal from the water the sample was broken open and the soil was found to have a moisture content of 18%. If the particle specific gravity was 2.65 and the specific gravity of the wax was 0.89, then determine:

14University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& (a) the bulk density of the soil

(b) the void ratio of the soil

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(c) the degree of saturation of the soil.


Example: Determine moisture content, void ratio, porosity and degree of saturation of a soil core sample. Also determine the dry unit weight, γdData:

• Weight of soil sample = 1013g • Vol. of soil sample = 585.0cm3 • Specific Gravity, Gs = 2.65 • Dry weight of soil = 904.0g

1 From the previous figure we can find:

Moisture content, w%1.12100

)(904)(109

=×==gg

WWw

s

w

Void ratio, e 715.0

1.3419.243

3

3

===cmcm

VVe

s

v

Porosity, n %7.41100

)(0.585)(9.243

3

3

=×==cmcm

VVn

T

v

15

Degree of saturation, S%7.44100

9.243109

=×==v

w

VVS


Dry unit weight, γd 355.1

585904

cmg

VW

T

sd ===γ

Measurement of Soil Properties

1. The in-situ density of a soil is 1.85 Mg/m3. A moisture content determination test on a sample of the soil gave the following results.

Test No. Mass of tin Tin + wet soil Tin + dry soil (g) (g) (g) 1 20.24 30.61 28.73 2 20.36 32.44 30.28

Determine the moisture content and dry density of the soil.

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17University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 2. The bulk density of a soil sample was found to be 1.90 g/ml and the moisture content 12%.

Determine the dry density, void ratio and degree of saturation if the particle specific gravity was 2.68.

What would the moisture content be if the soil were completely saturated at the same void ratio?

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18University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 3. A sample of saturated clay has a volume of 245ml and, after oven drying, has a mass of 453g.

If the particle specific gravity of the soil is 2.75, determine the dry and saturated unit weights of the soil in its natural state.

4. During a particle specific gravity test on a soil sample the following masses were recorded:

Mass of dry soil sample = 450g

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Mass of pycnometer when full of water = 1875g

19University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Mass of pycnometer + soil sample and full of water = 2160g

Determine the particle specific gravity of the soil.

5. A sand deposit was found to have a bulk density of 1.93 Mg/m3 and a moisture content of 16%. Laboratory tests established that the maximum and minimum void ratio values were 0.75 and 0.48 respectively. If the particle specific gravity was 2.65, determine the void ratio, the degree of saturation and the relative density of the deposit.

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20University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 6. An in-situ test to determine the unit weight of a soil was carried out using the sand replacement method. A total of 4.62kg of soil was extracted from the hole which was then refilled using 3.60kg of loose, dry sand having a dry density of 1.57Mg/m3.

A specific gravity determination of the soil particles using a density bottle yielded the following data:

Mass of bottle and stopper = 25.00g Mass of bottle, stopper and oven dried soil = 36.91g Mass of bottle, stopper, soil and distilled water = 62.59g Mass of bottle, stopper and distilled water = 55.21g

The results of a moisture content test on a sample of the soil were as follows:

Mass of tin and wet soil = 24.10g Mass of tin and dry soil = 22.10g Mass of tin = 12.30g

Calculate the:

(a) specific gravity of the particles (b) moisture content (c) bulk unit weight (d) dry unit weight

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(e) degree of saturation

21University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006

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Tutorial

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27University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Additional Questions Solutions:

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28University of Al Anbar Soil Mechanic Mr. Ahmed Amin Al Hity Collage of Engineering 3rd Stage Lecture no.2 Water Resources& Dams Eng. Dept. 2006-2007 Date / 09 / 2006 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Problem no.2

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Soil Mechanic -Ch2

Documents

ratio of water volume

soil sample total volume

ratio of air volume

ratio of void volume

given soil

airvoids volume total

vvvs e

saturated soil vw