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TOPIC 6 LATERAL EARTH PRESSURE

TOPIC 6 LATERAL EARTH PRESSURE. Course : S0705 – Soil Mechanic Year : 2008. CONTENT. RETAINING EARTH WALL (SESSION 21-22 : F2F) RANKINE METHOD (SESSION 21-22 : F2F) ACTIVE LATERAL PRESSURE PASSIVE LATERAL PRESSURE COULOMB METHOD (SESSION 23-24 : OFC) ACTIVE LATERAL PRESSURE

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TOPIC 6 LATERAL EARTH PRESSURE

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  1. TOPIC 6 LATERAL EARTH PRESSURE Course : S0705 – Soil Mechanic Year : 2008

  2. CONTENT • RETAINING EARTH WALL (SESSION 21-22 : F2F) • RANKINE METHOD (SESSION 21-22 : F2F) • ACTIVE LATERAL PRESSURE • PASSIVE LATERAL PRESSURE • COULOMB METHOD (SESSION 23-24 : OFC) • ACTIVE LATERAL PRESSURE • PASSIVE LATERAL PRESSURE • LATERAL PRESSURE DUE TO EXTERNAL LOAD (SESSION 23-24 : OFC) • DYNAMIC EARTH PRESSURE (SESSION 23-24 : OFC)

  3. SESSION 21-22RETAINING EARTH WALL RANKINE METHOD

  4. RETAINING EARTH WALL Defined as a wall that is built to resist the lateral pressure of soil – especially a wall built to prevent the advance of a mass of earth/soil

  5. RETAINING EARTH WALL

  6. RETAINING EARTH WALL

  7. RETAINING EARTH WALL

  8. RETAINING EARTH WALL

  9. RETAINING EARTH WALL

  10. RETAINING EARTH WALL

  11. EARTH LATERAL PRESSURE • Defined as soil stress/pressure at horizontal direction and a function of vertical stress • Cause by self weight of soil and or external load • 3 conditions : • Lateral Pressure at Rest • Active Lateral Pressure • Passive Lateral Pressure

  12. Case I EARTH LATERAL PRESSURE Lateral Pressure at rest

  13. Case II EARTH LATERAL PRESSURE Active Lateral Pressure

  14. Case III EARTH LATERAL PRESSURE Passive Lateral Pressure

  15. EARTH LATERAL PRESSURE q Jaky, Broker and Ireland  Ko = M – sin ’ Sand, Normally consolidated clay  M = 1 Clay with OCR > 2  M = 0.95 v =  . z + q Broker and Ireland z Ko = 0.40 + 0.007 PI , 0  PI  40 Ko = 0.64 + 0.001 PI , 40  PI  80 v h Sherif and Ishibashi  Ko =  +  (OCR – 1) •  = 0.54 + 0.00444 (LL – 20) • = 0.09 + 0.00111 (LL – 20) LL > 110%   = 1.0 ;  = 0.19 At rest, K = Ko

  16. RANKINE METHOD ACTIVE LATERAL PRESSURE 1 = 3 . tan2 (45+/2)+2c.tan (45+/2) a = v . tan2(45-/2) – 2c . tan (45-/2) a = v . Ka – 2cKa Ka = tan2 (45 - /2)

  17. RANKINE METHOD PASSIVE LATERAL PRESSURE

  18. RANKINE METHOD PASSIVE LATERAL PRESSURE p= v . tan2(45+/2) + 2c . tan (45+/2)

  19. RANKINE METHOD PASSIVE LATERAL PRESSURE Kp = tan2 (45 + /2) h = v . Kp + 2cKp

  20. EXAMPLE q = 20 kN/m2 h1 = 2 m h2 = 8 m Sheet Pile 1 = 15 kN/m3 1 = 10 o c1 = 0 kN/m2 2 = 15 kN/m3 2 = 15 o c2 = 0 kN/m2 h3 = 4 m • Questions: • Determine the active and passive lateral pressure of sheet pile structure • Determine the total lateral pressure

  21. SOLUTION q = 20 kN/m2 2 m 8 m Coefficient of Lateral Pressure : Active ; ka = tan2(45-1/2) = 0.704 Passive ; kp = tan2(45+2/2) = 1.698 Pa1 4 m Pw1 Pw2 Pq1 Pp1 Pa2 Active Lateral Pressure Pa1 = ka . 1 . h1 – 2 . c . ka = 0.704 . 15 . 2 – 2 . 0 . 0.704 = 21.12 kN/m2 Pa2 = ka . (1 . h1 + 1’ . h2) – 2 . c . ka = 49.28 kN/m2

  22. SOLUTION q = 20 kN/m2 2 m 8 m Coefficient of Lateral Pressure : Active ; ka = tan2(45-1/2) = 0.704 Passive ; kp = tan2(45+2/2) = 1.698 Pa1 4 m Pw1 Pw2 Pq1 Pp1 Pa2 Active Lateral Pressure Pq1 = ka . q = 0.704 . 20 = 14.08 kN/m2 Pw1 = kw . w . h2 = 1 . 10 . 8 = 80 kN/m2

  23. SOLUTION q = 20 kN/m2 2 m 8 m Coefficient of Lateral Pressure : Active ; ka = tan2(45-1/2) = 0.704 Passive ; kp = tan2(45+2/2) = 1.698 Pa1 4 m Pw1 Pw2 Pq1 Pp1 Pa2 PASSIVE LATERAL PRESSURE Pp1 = kp . 2’ . h3 + 2 . c . kp = 1.698 . 5 . 4 + 2 . 0 . 1.698 = 33.96 kN/m2 Pw2 = kw . w . h3 = 1 . 10 . 4 = 40 kN/m2

  24. SOLUTION q = 20 kN/m2 2 m 8 m Coefficient of Lateral Pressure : Active ; ka = tan2(45-1/2) = 0.704 Passive ; kp = tan2(45+2/2) = 1.698 Pa1 Pa za Pp 4 m zp Pw1 Pw2 Pq1 Pp1 Pa2 ACTIVE LATERAL FORCE Pa = 0.5 . Pa1 . h1 + (Pa1+Pa2)/2 . H2 + Pq1 . (h1+h2) + 0.5 . Pw1 . h2 = 763.52 kN/m za = 3.56 m

  25. SOLUTION q = 20 kN/m2 2 m 8 m Coefficient of Lateral Pressure : Active ; ka = tan2(45-1/2) = 0.704 Passive ; kp = tan2(45+2/2) = 1.698 Pa1 Pa za Pp 4 m zp Pw1 Pw2 Pq1 Pp1 Pa2 PASSIVE LATERAL FORCE Pp = 0.5 . Pp1 . h3 + 0.5 . Pw2 . h3 = 147.92 kN/m zp = 4/3 m

  26. RANKINE EARTH PRESSURE FOR INCLINED BACKFILL

  27. SESSION 23-24COULOMB METHOD LATERAL PRESSURE DUE TO EXTERNAL LOAD DYNAMIC EARTH PRESSURE

  28. COULOMB METHOD ACTIVE LATERAL PRESSURE • Assumption: • Fill material is granular • Friction of wall and fill material is considered • The failure surface in the soil mass would be a plane (BC1, BC2 …) Pa = ½ Ka .  . H2

  29. EXAMPLE Consider the retaining wall shown in the following figure. Given - H = 4.6 m -  = 16.5 kN/m3 -  = 30 o -  = 2/3  - c = 0 -  = 0 -  = 90 o Calculate the Coulomb’s active force per unit length of the wall 

  30. SOLUTION Ka = 0.297 Pa = 51.85 kN/m

  31. COULOMB METHOD PASSIVE LATERAL PRESSURE Pp = ½ Kp .  . H2

  32. COULOMB METHOD WITH A SURCHARGE ON THE BACKFILL

  33. LATERAL EARTH PRESSURE DUE TO SURCHARGE a > 0,4 a  0,4

  34. LATERAL EARTH PRESSURE DUE TO SURCHARGE

  35. LATERAL EARTH PRESSURE FOR EARTHQUAKE CONDITIONS

  36. LATERAL EARTH PRESSURE FOR EARTHQUAKE CONDITIONS

  37. LATERAL EARTH PRESSURE FOR EARTHQUAKE CONDITIONS

  38. LATERAL EARTH PRESSURE FOR EARTHQUAKE CONDITIONS

  39. LATERAL EARTH PRESSURE FOR EARTHQUAKE CONDITIONS

  40. EXAMPLE Refer to the following figure. For kv = 0 and kh = 0.3, determine : • Pae • The location of the resultant, z, from the bottom of the wall  = 35 o  = 18 kN/m3  = 17.5 o 5 m

  41. SOLUTION Part a. Kae = 0.47

  42. SOLUTION • Part b. Where : Ka = 0.25  = 90o  = 17.5o  = 0o Pa = 56.25 kN/m Pae = Pae – Pa = 105.75 – 56.25 = 49.5 kN/m

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