1 / 40

TOPIC 3 SOIL STRESS

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

coral
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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

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

More Related