1. Course : CE 6405 – Soil Mechanic
Year : 2015
TOPIC 3 SOIL STRESS

2. Stress distribution - soil media – Boussinesq theory - Use of Newmarks influence chart –Components of settlement –– immediate and consolidation settlement – Terzaghi‟s onedimensional consolidation theory – computation of rate of settlement. - √t and log t methods– e-log p relationship - Factors influencing compression behaviour of soils.
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3. Stress Distribution in Soils
By: Kamal Tawfiq, Ph.D., P.E
Added
Stress
Stress Distribution in Soils
Geostatic
Stress
Added Stresses (Point, line, strip, triangular, circular, rectangular)
Geostatic Stresses
Total Stress
Effective Stress
Pore Water Pressure
Westergaard’s Method
(For Pavement)
Bossinisque Equations
Point Load
Line Load
Strip Load
Triangular Load
Circular Load
Rectangular Load
Approximate Method
1:2 Method sy sx
txy
Total Stress= Effective Stress+ Pore Water Pressure
stotal = seff + u
Stress Bulbs
Influence Charts
Newmark Charts A

4. CONTENT TOTAL STRESS EFFECTIVE STRESS STRESS DISTRIBUTION
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5. 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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6. ·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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7. 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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8. EFFECTIVE STRESS
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9. 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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10. 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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11. 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

12. SOIL STRESS CAUSED BY EXTERNAL LOAD
External Load Types
Point Load
Line Load
Uniform Load
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13. LOAD DISTRIBUTION PATTERN
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14. STRESS CONTOUR
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15. STRESS DISTRIBUTION
Point Load P z 2 1 z
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16. STRESS DISTRIBUTION
Uniform Load L z B
L+z
B+z
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17. § 2.4 Stress due to loading Stresses beneath point load
Boussinesq published in 1885 a solution for the stresses beneath a point load on the surface of a material which had the following properties:
Semi-infinite – this means infinite below the surface therefore providing no boundaries of the material apart from the surface
Homogeneous – the same properties at all locations
Isotropic –the same properties in all directions
Elastic –a linear stress-strain relationship.

18. Vertical Stress Increase with Depth
Allowable settlement, usually set by building codes, may control the allowable bearing capacity
The vertical stress increase with depth must be determined to calculate the amount of settlement that a foundation may undergo
Stress due to a Point Load
In 1885, Boussinesq developed a mathematical relationship for vertical stress increase with depth inside a homogenous, elastic and isotropic material from point loads as follows:

19. Stress due to a Circular Load
The Boussinesq Equation as stated above may be used to derive a relationship for stress increase below the center of the footing from a flexible circular loaded area:

20. Linear elastic assumption

21. I. The Bulb of Pressure
Force

22. II. The Boussinesq Equation B. The Equation:
Where v = Poisson’s
Ratio (0.48)
Also an equation for σy

23. BOUSSINESQ METHOD
Point Load z P r z
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24. BOUSSINESQ METHOD ]
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25. BOUSSINESQ METHOD
Line Load z q r z x
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26. BOUSSINESQ METHOD Uniform Load Square/Rectangular Circular Trapezoidal
Triangle
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27. BOUSSINESQ METHOD
Rectangular z x y
m = x/z
n = y/z qo
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28. BOUSSINESQ METHOD
Rectangular
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29. 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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30. BOUSSINESQ METHOD
Circular
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31. BOUSSINESQ METHOD
Trapezoidal
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32. BOUSSINESQ METHOD
Triangle
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33. 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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34. 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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35. 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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36. Newmark’s Influence Chart
The Newmark’s Influence Chart method consists of concentric circles drawn to scale, each square contributes a fraction of the stress
In most charts each square contributes 1/200 (or 0.005) units of stress (influence value, IV)
Follow the 5 steps to determine the stress increase:
Determine the depth, z, where you wish to calculate the stress increase
Adopt a scale of z=AB
Draw the footing to scale and place the point of interest over the center of the chart
Count the number of elements that fall inside the footing, N
Calculate the stress increase as:

37. NEWMARK METHOD Where : qo = Uniform Load I = Influence factor
N = No. of blocks
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38. 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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39. 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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40. NEWMARK METHOD
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41. 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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42. 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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43. Simplified Methods
The 2:1 method is an approximate method of calculating the apparent “dissipation” of stress with depth by averaging the stress increment onto an increasingly bigger loaded area based on 2V:1H.
This method assumes that the stress increment is constant across the area (B+z)·(L+z) and equals zero outside this area.
The method employs simple geometry of an increase in stress proportional to a slope of 2 vertical to 1 horizontal
According to the method, the increase in stress is calculated as follows:

44. Westergaard' s Theory of stress distribution
Westergaard developed a solution to determine distribution of stress due to point load in soils composed of thin layer of granular material that partially prevent lateral deformation of the soil.
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45. Westergaard' s Theory of stress distribution
Assumptions: (1) The soil is elastic and semi-infinite. (2) Soil is composed of numerous closely spaced horizontal layers of negligible thickness of an infinite rigid material. (3) The rigid material permits only the downward deformation of mass in which horizontal deformation is zero.
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46. WESTERGAARD METHOD
Point Load
 = 0
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47. WESTERGAARD METHOD ]
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48. WESTERGAARD METHOD
Circular Uniform Load
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49. WESTERGAARD METHOD
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50. BOUSSINESQ VS WESTERGAARD
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51. BOUSSINESQ VS WESTERGAARD
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52. BOUSSINESQ VS WESTERGAARD
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