This document contains a soil mechanics exam with four questions. Question 1 involves calculating the factor of safety for a cut in stiff clay. Question 2 calculates total stress at a point below two foundations. Question 3 involves drawing shear strength envelopes from triaxial test data. Question 4 determines shear strength parameters from direct shear tests and uses them to calculate initial cell pressure in a triaxial test.

1. Page 1 of 8
Second Semester Exam (2015-2016)
Salahaddin University Date / / 2016
College of Engineering Time: 90 Minutes
Dams & Water Resources Eng. Dept. Subject: Soil Mechanics
Class: 3rd
Year Lecturer: Hawkar Hashim-Chener Salam
Q1: (20%)
A cut is made in a stiff, saturated clay that is underlain by a layer of sand (Figure 1).
What should be the height of the water, h, in the cut so that the factor of safety for stability of the cut
remained above 1.2 ?
Solution:
( )
19.5 9.81
0.9878
9.81
0 5.0 5.0
2
5 (2 ) 3
l=distance between US and DS
3
2
critcal
Heave
exit
sat w
critcal
w
exit
US A
DS B
us DS
exit
i
FOS
i
i
h
i
l
ht ht He hp
ht ht He hp hw
h ht ht hw hw
h hw
i
l
 


 
  


     
    
       
 
 
( )
0.9878
3.0
2
FOS=1.2
3.0
0.9878/1.2
2
1.353
critcal
Heave
exit
i
FOS
hwi
hw
hw m
 




Q2: (30%)

2. Page 2 of 8
a rectangular foundation (3m ×2 m ×0.6) is carrying a 3000 KN load from a column load is adjacent to a
circular footing of 3 m diameter and 0.5 m thickness that carry a stress of 4000 KN, (see Figure 2).
The bases of the both foundations are at 2 meter the ground level and the unit weight of the surrounding
soil is 19.0 KN/m3
. There is no sign of existing water table. Assume concrete unit weight as 24 KN/m3
.
Considering Self-weight, and overburden pressure of the soil, determine the vertical total stress increase
at point A at depth of 4 m below the ground surface.
Figure Q-2
Solution:
Dueto rectangularload( ) ( ) ( )Dueto circularloadA A A     
1) For Circular Loaded Area
( ) Dueto circularload net cA q I  
2 2 2
4000
3 7.07
4 4
load Selft weight
q
Area Area
Selft weight
q
Area Area
Area D m
 
 
 
  
7.07 0.5 24 84.84
Selft weight Area Thickness
concrete
Selft weight KN
  
   
3
4000 84.84
577.77
7.07 7.07
load Selft weight
q
Area Area
KN
q
m
 
  

3. Page 3 of 8
Df= distance from the base of the foundation to the ground level= 2.0 m
577.77 19 2 539.77
net Df
net
q q
q KPa
 
   
Where Ic can be calculated from X/R and Z/R, (See figure 8.11)
For R = 1.5 and X =2.0 m and Z = 2 is distance from the base of the footing to the required point
2 2
1.33 1.33
1.5 1.5
X Z
and
R R
   
For X/R and Z/R , Ic = 0.18 (From figure 8.4)
( )
( ) 539.77 0.18 97.16 KPa
Dueto circularload net c
Dueto circularload
A q I
A


  
   
2) For Rectangular Loaded Area
2
3000
3 2 6
load Selft weight
q
Area Area
Selft weight
q
Area Area
Area m
 
 
  
6 0.6 24 86.4
Selft weight Area Thickness
concrete
Selft weight KN
  
   
3
3000 86.4
514.4
6 6
load Selft weight
q
Area Area
KN
q
m
 
  
Df= distance from the base of the foundation to the ground level= 2.0 m
514.4 19 2 476.4
net Df
net
q q
q KPa
 
   
Re( ) Dueto ctaungular load netA q I  

4. Page 4 of 8
I can be calculated from Table 10.8 or figure 10.21 for given value of m and n
The area is divided into 4 areas as follow:
Area 1 = Area 2 = 1.5 m by 3 m and A3 and A4 = 1 m by 1.5 m
Area 1 and 2 are similar, hence calculation of area is performed and the result of Area 3= Result of Area 4
Area 3 and 4 are similar, hence calculation of area is performed and the result of Area 1= Result of Area2
Areas Z B L m= B/Z n= L/Z I ( Table 10.8 qnet (KPa)
netq I  
1 2 1.5 3 0.75 1.5 0.17 476.4 80.988
2 476.4 80.988
3 2 1 1.5 0.5 0.75 0.107 476.4 -50.97
476.4 -50.97
 60.036
KPa
60.04netq I KPa   
Hence, the vertical total stress increase at point A at depth of 4 m below the ground surface:
Dueto rectangularload( ) ( ) ( )
( ) 97.16 60.04 157.2 .
Dueto circularloadA A A
A KPa
  

   
   
Q3: (30%)

5. Page 5 of 8
For the following results, obtained from a CONSOLIDATED-UNDRAINED Triaxial tests on a normally
consolidated clay soil. Using graphical Method, draw the appropriate shear strength envelope and find
shear strength parameter in term of total and effective stress analysis
Test. No 1 2
Cell Pressure KN/m2
100 200
Deviator stress at failure KN/m2
137 210
Pore pressure at failure KN/m2
28 86
Solution:
Cell Pressure= 3
Deviator stress at failure = d
1 3 d   
1' 1 3' 3
EffectiveStress Total stress porewater pressure
u and u   
 
   
Total Stress Parameters Effective Stress Parameters
3 d 1 U 3' 1'
Test No. 1 100 137 237 28 72 209
Test No. 2 200 210 410 86 114 324
Now a 2 set of Mohr-Circle envelope can be sketched;
1- Total stress envelope
2- Effective stress envelope
From the total stress envelope (next page)
, 24.4
, 15.5o
Cohesion c KPa
Frictionangle 


From the effective stress envelope, next page)
, 3.0
, 27.7o
Cohesion c KPa
Frictionangle 



6. Page 6 of 8
Total Stress Analysis
Effective Stress Analysis
Q4: (25%)

7. Page 7 of 8
A Quick shear box test was carried out on three soil specimens. The cross section of the shear box is
6 cm×6 cm. Results were as follows:
Test 1 Test 2 Test 3
Normal Force, N 201.6 399.6 799.2
Shear Force, N 205.2 270 356.4
A) Determine the shear strength parameters. (10%)
B) An identical soil sample was tested in Triaxial Compression Test. When the sample failed,
the major principal stress was 280 KPa. What initial cell pressure was applied to the sample? (15%)
Test 1 Test 2 Test 3
Normal Force, N 201.6 399.6 799.2
Shear Stress 56 111 222
Shear Force, N 205.2 270 356.4
57 75 99
26 6
0.0036
100 100
Area m
 
   
 
201.6 1
, 56
0.0036 2 1000
Normal force N
Normal Stress KPa
Area m
    
205.2 1
, 57
0.0036 2 1000
Shear force N
Shear stress KPa
Area m
    
For the graph, Cohesion, C = 45 KPa
Friction angle, phi = 14 degree.

8. Page 8 of 8
For Part B.
FOR part b
Form Part one
, 14
14
(45 ) 52
2
90 90 52 38
o
Frictionangle Hence



  
   
Draw a line from POINT A (Major principal stress, 280 KPa, given) with an angle of (90- , i.e. 38)
degree to intersect the failure envelope at point B
- From Point B, Draw BC line that is PERPENDICULAR TO FAILRE ENVELOPE, POINT C
IS THE CENTRE OF THE CIRCLE
- BC is the radius of the circle, hence Minor principal stress can be drawn
The Major Principal Stress, 3 from the plot = 100 KPa.
Deviator Stress = 1 3 280 100 180KPa    
90-
