Solved Problems in Soil Mechanics

[Pages:186]Solved Problems in Soil Mechanics

Based on "Principles of Geotechnical Engineering, 8th Edition"

Prepared By: Ahmed S. Al-Agha

February -2015

Chapter (3) & Chapter (6)

Soil Properties &

Soil Compaction

Solved Problems in Soil Mechanics

Soil Properties & Soil Compaction

Useful Formulas:

You should know the following formulas:

Vtotal = Vsolid + Vvoids Vtotal = Vsolid + Vair + Vwater

Wtotal = Wsolid + Wwater (Wair = 0 , Wsolid = Wdry)

dry

=

Gs ? w 1+e

,

dry

=

moist (1 + %w)

,

dry

=

Wdry Vtotal

,

solid

=

Wdry Vsolid

moist

=

Gs

?

w(1 1+

+ e

%w)

,

sat

=

Gs

?

w (1 + 1+e

Ges)

(S

=

1)

Z.A.V

=

Gs ? w 1 + Gsw

(S

=

1

e

=

emin

=

Gsw/1)

S. e = Gs. w

,

S

=

Vwater Vvoids

e , (at saturation S = 1 wsat = Gs)

w = Weight of water = Ww = Wwet-Wdry ? 100%

Weight of solid Ws

Wdry

e

=

Vvoids Vsolid

=

VT - Vs

Vs

e , n=1+e

,

n

=

Vvoids Vtotal

Gs =

solid water

,

solid

=

Wdry Vsolid

,

water

=

Wwater Vwater

water = 9.81KN/m3 = 62.4Ib/ft3 , 1ton = 2000Ib , 1yd3 =27ft3

Air

content

(A)

=

Vair Vtotal

Dr

=

emax - e emax - emin

Relative

Compaction(R.

C)

=

dry,max,field dry,max,proctor

?

100%

Vsolid must be constant if we want to use the borrow pit soil in a construction site

or on earth dam or anywhere else.

Page (1)

Ahmed S. Al-Agha

Solved Problems in Soil Mechanics

Soil Properties & Soil Compaction

1. (Mid 2014):

a) Show

the

saturated

moisture

content

is:

Wsat

=

w

[1

d

-

1]

s

: s = solid unit weight

Solution

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

d

=

Gs ? w 1+e

e

=

Gs

? w d

-

1,

substitute in Eq. (1)

wsat

=

Gs

? w d

Gs

-

1

=

w d

-

1 Gs

but

Gs

=

s w

1 Gs

=

w s

wsat

=

w d

-

w s

=

w

1 [d

-

1 s]

.

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. : take w = 9.81kN/m3 Sample #1: w = 30%, d = 14.9 kN/m3, s = 27 kN/m3 , clay

Sample #2: w = 20%, d = 18 kN/m3, s = 27 kN/m3 , silt

Sample #3: w = 10%, d = 16 kN/m3, s = 26 kN/m3 , sand

Sample #4: w = 22%, d = 17.3 kN/m3, s = 28 kN/m3 , silt

Sample #5: w = 22%, d = 18 kN/m3, s = 27 kN/m3 , silt

Solution

For any type of soil, the mositure content (w) must not exceeds 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%)

11 wsat = 9.81 [14.9 - 27] = 29.5% < 30% not correctly reported. Sample #2: (Given water content= 20%)

Page (2)

Ahmed S. Al-Agha

Solved Problems in Soil Mechanics

Soil Properties & Soil Compaction

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.

Page (3)

Ahmed S. Al-Agha

Solved Problems in Soil Mechanics

Soil Properties & Soil Compaction

2. (Mid 2013):

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

Givens:

dry = 19.5KN/m3 , %w = 8% , Gs=2.67

a)

dry

=

Gs ? w 1+e

19.5

=

2.67 1

? +

9.81 e

e

=

0.343

b)

moist = dry(1 + %w) = 19.5 ? (1 + 0.08) = 21.06 KN/m3

sat = dry(1 + %wsat) %wsat means %w @ S = 100%

S.e

=

Gs.w

%wsat

=

S.e Gs

=

1?0.343 2.67

?

100%

=

12.85%

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

c)

moist = 21.06 KN/m3 Now we want to find moist @ 80% Saturation so, firstly we calculate %w @80%

saturation: S. e 0.8 ? 0.343

%w80% = Gs = 2.67 ? 100% = 10.27% moist,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

Page (4)

Ahmed S. Al-Agha

Solved Problems in Soil Mechanics

Soil Properties & Soil Compaction

Another solution:

VT = 1m3

The water content before adding water (%w1) = 8%

The water content after adding water (%w2) = 10.27% @80%saturation

w

=

Weight of water Weight of solid

=

Ww Ws

Ws :

dry

=

Ws VT

Ws

=

19.5

?

1

=

19.5KN

Ww = Ws ? w Ww,1 = Ws ? w1 , and Ww,2 = Ws ? w2

Then, Ww,1 = 19.5 ? 0.08 = 1.56 KN

Ww,2 = 19.5 ? 0.1027 = 2 KN

Weight of water to be added = 2-1.56= 0.44 KN

Mass

of

water

to

be

added

=

0.44?

1000 9.81

=

44.85

Kg

d)

Mw = 25grams for saturation S = 100% %wsat = 12.85%

Ww

=

(25

?

10-3)Kg

?

9.81 1000

=

24.525

?

10-5KN

Ws

=

Ww w

=

24.525 ? 10-5 0.1285

=

190.85

?

10-5

KN

Now, Gs =

solid

water

solid = 2.67 ? 9.81 = 26.2KN/m3

solid =

Ws

Vs

Vs =

Ws = 190.85?10-5 = 7.284 ? 10-5 m3

solid

26.2

=72.84 cm3

Page (5)

Ahmed S. Al-Agha

Solved Problems in Soil Mechanics

Soil Properties & Soil Compaction

3. (Mid 2013):

An earth dam require one hundred cubic meter of soil compacted with unit weight of 20.5 KN/m3 and moisture content of 8%, choose two from the three borrow pits given in the table below, knowing that the first must be one of the two borrow pits, the specific gravity of solid particles is 2.7. Choose the most economical choice.

Borrow pit No.

1 2 3

Void ratio

0.6 1 0.75

Cost($/m3)

1 1.5 1.7

Available volume (m3)

80 100 100

Some Explanations about the problem:

Borrow pits:

.

Available Volume:

.

, 100 :

,

Vs .

Solution

The first step is to find the value of Vs for earth dam that must be maintained in

borrow pits.

dry

=

moist 1 + %w

=

Gs ? w 1+e

1

20.5 + 0.08

=

2.7 ? 9.81 1+e

e

=

0.395

e

=

VT

- Vs

Vs

0.395

=

100 - Vs

Vs

Vs

=

71.68

m3

The value of Vs = 71.68 m3 must be maintained on each borrow pit.

Page (6)

Ahmed S. Al-Agha

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