topic 3 soil stress l.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
TOPIC 3 SOIL STRESS PowerPoint Presentation
Download Presentation
TOPIC 3 SOIL STRESS

Loading in 2 Seconds...

play fullscreen
1 / 40

TOPIC 3 SOIL STRESS - PowerPoint PPT Presentation


  • 1266 Views
  • Uploaded on

TOPIC 3 SOIL STRESS. Course : S0705 – Soil Mechanic Year : 2008. CONTENT. TOTAL STRESS EFFECTIVE STRESS STRESS DISTRIBUTION. TOTAL NORMAL STRESS.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

TOPIC 3 SOIL STRESS


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
    Presentation Transcript
    1. TOPIC 3 SOIL STRESS Course : S0705 – Soil Mechanic Year : 2008

    2. CONTENT • TOTAL STRESS • EFFECTIVE STRESS • STRESS DISTRIBUTION

    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

    4. EXAMPLE 1 m A = t,1 x 1 m = 17 kN/m2 ·A ·B ·C ·D t,1 = 17 kN/m3 d,1 = 13 kN/m3 3 m 4 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

    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

    6. EFFECTIVE STRESS

    7. EXAMPLE Sand t = 18.0 kN/m3 d = 13.1 kN/m3 h1 = 2 m h2 = 2.5 m h3 = 4.5 m MAT Clay t = 19.80 kN/m3 x

    8. ’ = d,1 . h1 + (t,2 - w) . h2 + (t,2 - w) . h3 ’ = 13.1 . 2 + (18-10) . 2.5 + (19,8-10) . 4.5 = 90.3 kN/m2 EXAMPLE • Total Stress  = d,1 . h1 + t,1 . h2 + t,2 . h3  = 13.1 . 2 + 18 . 2.5 + 19.8 . 4.5 = 160.3 kN/m2 • Pore Water Pressure u = w . (h2+h3) u = 10 . 7 = 70 kN/m2 • Effective Stress ’ =  - u = 90.3 kN/m2

    9. 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 70 kPa 90.3 kPa 160.3 kPa -9.0 Profile of Vertical Stress

    10. SOIL STRESS CAUSED BY EXTERNAL LOAD • External Load Types • Point Load • Line Load • Uniform Load

    11. LOAD DISTRIBUTION PATTERN

    12. STRESS CONTOUR

    13. STRESS DISTRIBUTION • Point Load P z 2 1 z

    14. STRESS DISTRIBUTION • Uniform Load L z B L+z B+z

    15. P z z r BOUSSINESQ METHOD • Point Load

    16. ] BOUSSINESQ METHOD

    17. q x z z r BOUSSINESQ METHOD • Line Load

    18. BOUSSINESQ METHOD • Uniform Load • Square/Rectangular • Circular • Trapezoidal • Triangle

    19. y x z BOUSSINESQ METHOD • Rectangular m = x/z n = y/z qo

    20. BOUSSINESQ METHOD • Rectangular

    21. 2r z z x BOUSSINESQ METHOD • Circular At the center of circle (X = 0) For other positions (X  0), Use chart for finding the influence factor

    22. BOUSSINESQ METHOD • Circular

    23. BOUSSINESQ METHOD • Trapezoidal

    24. BOUSSINESQ METHOD • Triangle

    25. E Y A 5 m 5 m I H F D B C G J 5 m 5 m 5 m EXAMPLE • The 5 m x 10 m area uniformly loaded with 100 kPa • 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

    26. EXAMPLE Question 1 z total = 23.8 – 20.9 – 20.6 + 18 = 0.3 kPa

    27. EXAMPLE Question 2 z total = 47.6 – 41.9 – 43.8 + 38.6 = 0.5 kPa

    28. NEWMARK METHOD Where : qo = Uniform Load I = Influence factor N = No. of blocks

    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

    30. NEWMARK METHOD • Example, the depth of point (z) = 6 m And so on, generally up to z/qo  1 because if z/qo = 1 we get r/z = 

    31. NEWMARK METHOD

    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’

    33. EXAMPLE 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 = 250 . 0.02 . 8 = 40 kPa

    34. WESTERGAARD METHOD • Point Load  = 0

    35. ] WESTERGAARD METHOD

    36. WESTERGAARD METHOD • Circular Uniform Load

    37. WESTERGAARD METHOD

    38. BOUSSINESQ VS WESTERGAARD

    39. BOUSSINESQ VS WESTERGAARD

    40. BOUSSINESQ VS WESTERGAARD