solution to first semester soil 2015 16

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First Semester Exam (2015-2016) Salahaddin University Date 07/01 /2016

College of Engineering Time: 90 Minutes

Dams & Water Resources Eng. Dept. Subject: Soil Mechanics

Class: 3rd Year Lecturer: Chener S. Qadr

Q1: (25%)

A saturated soil mass was found to be 200 gm. After oven-drying, the mass become 160 gm. The specific gravity

of the soil was found to be 2.7.

a) Determine: e and n and

b) bulk , dry , s in Kg/m3 and pcf

c) If the saturation degree is reduced to 70%. What change in bulk , dry , s e, n will occur ?

d) If a homogenous embankment is to be constructed from the same soil above. The specification requires the

embankment to be compacted to a dry unit weight of 18 KN/m3. Assume the swell index = 1.2.

The dimension of the embankment are as follow: The embankment is 15 meter height and 200 meter wide.

- Side slopes for upstream and downstream: 3H: 1V and the crest width =8 m.

Determine the weight of soil from the borrow pit required to construct the embankment.

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Q2: (20%)

Classify the following soil according to Unified Soil Classification System. Would you consider this soil as good

backfilling material? Justify your answer.

Particle Size(mm) 3’’ 4.75 2.0 0.85 0.425 0.15 0.075 Cc=0.8 LL=30%

% Finer 94 73 71 63 49.8 28.6 11% Cu=12 PL=18%

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Q3: (20%)

A 2 Km long trapezoidal canal is dug parallel to an existing ditch, as shown in Figure Problem 3. Two permeable

Silty and Sandy seam of 0.5 m and 0.9 m. thickness respectively cuts across the otherwise impermeable clay.

Pumping tests revealed an average hydraulic conductivity of the Silty and Sandy soils to be equal to 51.5 10

cm/s and 32 10 cm/s, respectively. Estimate the quantity of seepage flow from the canal into the ditch.

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Q4: (35%)

An embankment dam is shown in section in Figure Problem 4, the coefficients of

permeability in the horizontal and vertical directions are 67.5 10 cm/s and

62.7 10 cm/s, respectively.

- Draw the transformed section (Scale 1:500).

- calculate and then draw the phreatic (Seepage) line

- Determine the pore water pressure at point A and B and

- Estimate the quantity of seepage through the dam using L. Casagrande’s method

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Figure Problem 3

a Natural Section

a Transformed Section

Figure Problem 4: a) Natural Section b) Transformed Section

First Semester Exam (2015-2016) Salahaddin University Date 07/01 /2016

15°

Phreatic Line

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College of Engineering Time: 90 Minutes

Dams & Water Resources Eng. Dept. Subject: Soil Mechanics

Class: 3rd Year Lecturer: Chener S. Qadr

Instructor Solution to 1st Semester Exam 2015-206

Please Note that this Solution may contain some numeric errors, please correct it whenever you came across an error

Q4 is a routine Question and hence, the instructor did not saw any necessity for providing the solution.

Q1: (25%)

HINT: More than one solution is valid for Question 1

Solution:

Part A

Given 200 160 2.7wt ws Gs

200 160. 100 100 25%

160

Mwwc

Ms

33

3

401 40

0 ,

0 40 40

w

Mw gVw cm

cmVw Vw

Va Since it is fully Saturated

Vv Va Vw cm

3

3

1602.7 59.26

1

s

w w

Ms gmGs Vs cm

gmVsVs

cm

Thus,

3

3

40 59.26

40 _ 59.26 99.26

Vv and Vs cm

Vt Vv Vs cm

40 400.675 0.40

59.26 99.26

Vve and n

Vs

PART B

3 3 3

2002.01 1000 2010

99.26bulk

Wt gm gm kg

Vt cm cm m

3 3 3

1601.612 1000 1612

99.26dry

ws gm gm Kg

Vt cm cm m

3

20101612

251 .1

100

bulkdry

KN

w c m

3 3

1602.7 2700

59.25s

ws g Kg

vs cm m

PART C

Ws, Vt, Vv and Vs and Gs are unchanged.

0.675

0.40

Vve remain unchanged

Vs

Vvn remain unchanged

Vt

3Re , 1612dry

ws Kgmain unchanged

Vt m

32700s

ws Kgremain unchanged

vs m

.

0.7 0.675 . 2.7

. 0.18

S e w c Gs

w c

w c

3

1 .

1612 190218

1100

bulkdry

bulkbulk

w c

kg

m

only the bulk density changes

Part D

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2

3

12 45 15 8 15 795

2

200 795 159000

Area m

Volumeof Embankment m

( ) 3 3

11612 15.81

101.98dry Pit

kg KN

m m

(EMB) (PIT)

(PIT) (EMB)

(EMB)

(PIT) (EMB)

(PIT)

3

(PIT)

......(1)

Rearranging

18159000 181025

15.81

Before considering swelling

Vd

Vd

dV V

d

V m

3 3

(PIT)

1.2

181025 1.2 217230

Considering Swelling

V m m

The Swelling Factor

And that is the Solution.

Q2: (20%)

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Classify the following soil according to Unified Soil Classification System. Would you consider this soil as good

backfilling material? Justify your answer.

Particle Size(mm) 3’’ 4.75 2.0 0.85 0.425 0.15 0.075 Cc=0.8 LL=30%

% Finer 94 73 71 63 49.8 28.6 11% Cu=12 PL=18%

Solution:

Step 1) Passing no. 200 Sieve =11% < 50% → Go to Coarse grained Chart

Step 2) GRAVEL = P3inch – P4.75 mm = 94-73 =21%

SAND= P4.75 mm - P0.075 mm = 73-11 =62%

Fines (Silt / Clay) = P0.075 mm = 11%

∴

Sand =62% > Gravel (21%) → Go Sand-Section

Step 3) Fines ( Silt+ Clay) =11% it is between 5-12%

Step 4) 12 0.8 6 1 3u b cu CtC Cc So the soil is poorly Graded

For LL=30 and PL =18 → PI=LL-PL = 30 – 18 = 12%

For LL=30 AND PI =12

From Plasticity Chart,

Fines =CL

Step 5) ∴ Group Symbol : SP-SC

Step 6) 21 15%Gravel

Step 7) ∴Group Name: Poorly Graded SAND with Clay and Gravel

And that is the Solution.

Illustrative Step by step of using the Chart

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Q3: (20%)

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A 2 Km long trapezoidal canal is dug parallel to an existing ditch, as shown in Figure Problem 3. Two permeable

Silty and Sandy seam of 0.5 m and 0.9 m. thickness respectively cuts across the otherwise impermeable clay.

Pumping tests revealed an average hydraulic conductivity of the Silty and Sandy soils to be equal to 51.5 10

cm/s and 32 10 cm/s, respectively. Estimate the quantity of seepage flow from the canal into the ditch.

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Figure Problem 3

Solution:

N.B Two possible solution is valid for this question:

intditch Flow otheditchq

ditch sandy seam silty seamq q q

15°

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For the Sandy Seam

K i Asandy seam

q

Area of Flow = Thickness of Sandy-Seam× Length of

the Canal

Area of Flow = 0.9×2000= 1800 m2

3 52 10 / sec 2 10sec

mK cm

Us DS

Sandy Seam Sandy Seam

Ht Ht hi

L L

103 4 107

108

(15) 415

108 4 104

107 104 3

Us Canal

DS Ditch

DS Ditch

Us DS

Ht Ht He Hp m

Ht Ht y

y yTan y m

x

Ht Ht m

h Ht Ht m

5

33

32 10 1800

sec 19.32

6.988 10 25.16sec

hK i A K A

sandy seam L

m

sandy seam

m m

sandy seam hr

q

q

q

∴ 2 219.3 0.8 19.32Sandy SeamL m

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Similarly For the Silty Seam

K i Asilty seam

q

Area of Flow = Thick. of Silty-Seam× Length of

the Canal

Area of Flow = 0.5×2000= 1000m2

5 71.5 10 / sec 1.5 10sec

mK cm

Us DS

Silty Seam Silty Seam

Ht Ht hi

L L

107 104

107 104 3

Us Canal DS Ditch

Us DS

Ht Ht m and Ht Ht m

h Ht Ht m

2 222.6 2 22.69Silty SeamL m

7

35

31.5 10 1000

sec 22.69

1.983 10 0.07sec

hK i A K A

Silty seam L

m

Silty seam

m m

Silty seam hr

q

q

q

2 222.6 2 22.69Silty SeamL m

∴

3

25.16 0.071 25.231sec

ditch sandy seam silty seam

m

ditch

q q q

q

And that is the Solution. To Question 3

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Question 4

Routine Solution, Student should not face difficulty in addressing such problems

solution to first semester soil 2015 16

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