1. Course : S0705 – Soil Mechanic
Year : 2008
TOPIC 3 SOIL STRESS

2. CONTENT TOTAL STRESS EFFECTIVE STRESS STRESS DISTRIBUTION
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3. TOTAL NORMAL STRESS
Generated by the mass in the soil body, calculated by sum up the unit weight of all the material (soil solids + water) multiflied by soil thickness or depth.
Denoted as , v, Po
The unit weight of soil is in natural condition and the water influence is ignored.
z = The depth of point
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4. ·A ·B ·C ·D EXAMPLE A = t,1 x 1 m t,1 = 17 kN/m3 = 17 kN/m2
d,1 = 13 kN/m3
3 m
4 m
B = t,1 x 3 m
= 51 kN/m2
t,2 = 18 kN/m3
d,2 = 14 kN/m3
C = t,1 x 3 m + t,2 x 4 m
= 123 kN/m2
2 m
t,3 = 18 kN/m3
d,3 = 15 kN/m3
D = t,1 x 3 m + t,2 x 4 m + t,3 x 2 m
= 159 kN/m2
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5. EFFECTIVE STRESS
Defined as soil stress which influenced by water pressure in soil body.
Published first time by Terzaghi at 1923 base on the experimental result
Applied to saturated soil and has a relationship with two type of stress i.e.:
Total Normal Stress ()
Pore Water Pressure (u)
Effective stress formula
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6. EFFECTIVE STRESS
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7. EXAMPLE Sand h1 = 2 m t = 18.0 kN/m3 d = 13.1 kN/m3 h2 = 2.5 m
MAT
Clay
t = kN/m3 x
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8. EXAMPLE Total Stress  = d,1 . h1 + t,1 . h2 + t,2 . h3
 =
= kN/m2
Pore Water Pressure
u = w . (h2+h3)
u =
= 70 kN/m2
Effective Stress
’ =  - u = 90.3 kN/m2
’ = d,1 . h1 + (t,2 - w) . h2 + (t,2 - w) . h3
’ = (18-10) (19,8-10) . 4.5
= 90.3 kN/m2
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9. Profile of Vertical Stress
EXAMPLE
Total Stress ()
Pore Water Pressure (u)
Effective Stress (’)
26.2 kPa
26.2 kPa
-2.0
71.2 kPa
25 kPa
46.2 kPa
-4.5
160.3 kPa
70 kPa
90.3 kPa
-9.0
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Profile of Vertical Stress

10. SOIL STRESS CAUSED BY EXTERNAL LOAD
External Load Types
Point Load
Line Load
Uniform Load
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11. LOAD DISTRIBUTION PATTERN
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12. STRESS CONTOUR
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13. STRESS DISTRIBUTION
Point Load P z 2 1 z
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14. STRESS DISTRIBUTION
Uniform Load L z B
L+z
B+z
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15. BOUSSINESQ METHOD
Point Load z P r z
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16. BOUSSINESQ METHOD ]
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17. BOUSSINESQ METHOD
Line Load z q r z x
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18. BOUSSINESQ METHOD Uniform Load Square/Rectangular Circular Trapezoidal
Triangle
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19. BOUSSINESQ METHOD
Rectangular z x y
m = x/z
n = y/z qo
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20. BOUSSINESQ METHOD
Rectangular
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21. BOUSSINESQ METHOD Circular At the center of circle (X = 0)z x z 2r
At the center of circle (X = 0)
For other positions (X  0),
Use chart for finding the influence factor
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22. BOUSSINESQ METHOD
Circular
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23. BOUSSINESQ METHOD
Trapezoidal
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24. BOUSSINESQ METHOD
Triangle
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25. EXAMPLE The 5 m x 10 m area uniformly loaded with 100 kPa Y 5 m
Question :
Find the at a depth of 5 m under point Y
Repeat question no.1 if the right half of the 5 x 10 m area were loaded with an additional 100 kPa Y A B C D E F G H I J
5 m
5 m m m
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26. EXAMPLE Question 1 z total = 23.8 – 20.9 – 20.6 + 18 = 0.3 kPa Item
Area
YABC
-YAFD
-YEGC
YEHD x 15 10 5 y z
m = x/z 3 2 1
n = y/z I
0.238
0.209
0.206
0.18 z
23.8
- 20.9
-20.6
18.0
z total = 23.8 – 20.9 – = 0.3 kPa
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27. EXAMPLE Question 2 z total = 47.6 – 41.9 – 43.8 + 38.6 = 0.5 kPa Item
Area
YABC
-YAFD
-YEGC
YEHD x 15 10 5 y z
m = x/z 3 2 1
n = y/z I
0.238
0.209
0.206
0.18 z
47.6
- 41.9
-43.8
38.6
z total = 47.6 – 41.9 – = 0.5 kPa
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28. NEWMARK METHOD Where : qo = Uniform Load I = Influence factor
N = No. of blocks
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29. NEWMARK METHOD Diagram Drawing
Take z/qo between 0 and 1, with increment 0.1 or other, then find r/z value
Determine the scale of depth and length
Example : 2.5 cm for 6 m
3. Calculate the radius of each circle by r/z value multiplied with depth (z)
4. Draw the circles with radius at step 3 by considering the scale at step 2
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30. NEWMARK METHOD Example, the depth of point (z) = 6 m
z/qo
r/z
Radius (z=6 m)
Radius at drawing
Operation
0.1
0.27
1.62 m
0.675 cm
1.62/6 x 2.5 cm
0.2
0.40
2.40 m
1 cm
2.4/6 x 2.5 cm
0.3
0.52
3.12 m
1.3 cm
3.12/6 x 2.5 cm
0.4
0.64
3.84 m
1.6 cm
3.84/6 x 2.5 cm
And so on, generally up to z/qo  1 because if z/qo = 1 we get r/z = 
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31. NEWMARK METHOD
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32. EXAMPLE
A uniform load of 250 kPa is applied to the loaded area shown in next figure :
Find the stress at a depth of 80 m below the ground surface due to the loaded area under point O’
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33. EXAMPLE v = 250 . 0.02 . 8 = 40 kPa Solution :
Draw the loaded area such that the length of the line OQ is scaled to 80 m.
Place point O’, the point where the stress is required, over the center of the influence chart
The number of blocks are counted under the loaded area
The vertical stress at 80 m is then indicated by : v = qo . I . N
v = = 40 kPa
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34. WESTERGAARD METHOD
Point Load
 = 0
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35. WESTERGAARD METHOD ]
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36. WESTERGAARD METHOD
Circular Uniform Load
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37. WESTERGAARD METHOD
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38. BOUSSINESQ VS WESTERGAARD
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39. BOUSSINESQ VS WESTERGAARD
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40. BOUSSINESQ VS WESTERGAARD
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