Solution to first semester soil 2015 16

1. Page 1 of 9
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.
--------------------------------------------------------------------------------------------------------------------
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%
---------------------------------------------------------------------------------------------------------------------
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 5
1.5 10

cm/s and 3
2 10
 cm/s, respectively. Estimate the quantity of seepage flow from the canal into the ditch.
---------------------------------------------------------------------------------------------------------------------
Q4: (35%)
An embankment dam is shown in section in Figure Problem 4, the coefficients of
permeability in the horizontal and vertical directions are 6
7.5 10
 cm/s and 6
2.7 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

2. Page 2 of 8
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

3. Page 3 of 9
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
Mw
wc
Ms

    
3
3
3
40
1 40
0 ,
0 40 40
w
Mw g Vw cm
cmVw Vw
Va Since it is fully Saturated
Vv Va Vw cm
     

    
3
3
160
2.7 59.26
1
s
w w
Ms gm
Gs Vs cm
gmVs Vs
cm

 
    

Thus,
3
3
40 59.26
40 _ 59.26 99.26
Vv and Vs cm
Vt Vv Vs cm
 
   
40 40
0.675 0.40
59.26 99.26
Vv
e and n
Vs
    
PART B
3 3 3
200
2.01 1000 2010
99.26
bulk
Wt gm gm kg
Vt cm cm m
     
3 3 3
160
1.612 1000 1612
99.26
dry
ws gm gm Kg
Vt cm cm m
     
3
2010
1612
251 . 1
100
bulk
dry
KN
wc m

   
 
3 3
160
2.7 2700
59.25
s
ws g Kg
vs cm m
    
PART C
Ws, Vt, Vv and Vs and Gs are unchanged.
0.675
0.40
Vv
e remain unchanged
Vs
Vv
n remain unchanged
Vt
 
 
3
Re , 1612dry
ws Kg
main unchanged
Vt m
  
3
2700s
ws Kg
remain unchanged
vs m
  
.
0.7 0.675 . 2.7
. 0.18
S e wc Gs
wc
wc
  
  

3
1 .
1612 1902
18
1
100
bulk
dry
bulk
bulk
wc
kg
m
only the bulk densitychanges






  


Part D

4. Page 4 of 8
2
3
1
2 45 15 8 15 795
2
200 795 159000
Area m
Volumeof Embankment m
 
       
 
  
( ) 3 3
1
1612 15.81
101.98
dry Pit
kg KN
m m
   
(EMB) (PIT)
(PIT) (EMB)
(EMB)
(PIT) (EMB)
(PIT)
3
(PIT)
......(1)
Rearranging
18
159000 181025
15.81
Before considering swelling
Vd
V
d
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%)

5. Page 5 of 9
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 0.86 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

6. Page 6 of 8
Q3: (20%)

7. Page 7 of 9
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 5
1.5 10

cm/s and 3
2 10
 cm/s, respectively. Estimate the quantity of seepage flow from the canal into the ditch.
---------------------------------------------------------------------------------------------------------------------
Figure Problem 3
Solution:
N.B Two possible solution is valid for this question:
intditch Flow otheditchq 
ditch sandy seam silty seam
q q q 
 
15°

8. Page 8 of 8
For the Sandy Seam
K i A
sandy seam
q 

Area of Flow = Thickness of Sandy-Seam× Length of
the Canal
Area of Flow = 0.9×2000= 1800 m2
3 5
2 10 / sec 2 10
sec
m
K cm 
   
Us DS
Sandy Seam Sandy Seam
Ht Ht h
i
L L
 
 

103 4 107
108
(15) 4
15
108 4 104
107 104 3
Us Canal
DS Ditch
DS Ditch
Us DS
Ht Ht He Hp m
Ht Ht y
y y
Tan y m
x
Ht Ht m
h Ht Ht m
     
  
   
   
     
5
3
3
3
2 10 1800
sec 19.32
6.988 10 25.16
sec
h
K i A K A
sandy seam L
m
sandy seam
m m
sandy seam hr
q
q
q



 

   

  

∴ 2 2
19.3 0.8 19.32SandySeamL m  

9. Page 9 of 9
Similarly For the Silty Seam
K i A
silty seam
q 

Area of Flow = Thick. of Silty-Seam× Length of
the Canal
Area of Flow = 0.5×2000= 1000m2
5 7
1.5 10 / sec 1.5 10
sec
m
K cm 
   
Us DS
Silty Seam Silty Seam
Ht Ht h
i
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 2
22.6 2 22.69Silty SeamL m   
7
3
5
3
1.5 10 1000
sec 22.69
1.983 10 0.07
sec
h
K i A K A
Silty seam L
m
Silty seam
m m
Silty seam hr
q
q
q



 

   

  

2 2
22.6 2 22.69Silty SeamL m   
∴
3
25.16 0.071 25.231
sec
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