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Example c t T v z H 2 dr 2 T H c z dr v t Example - PDF document

4/9/2015 Example c t T v z H 2 dr 2 T H c z dr v t Example c t T v z H 2 dr T H 2 t z dr c v 1 4/9/2015 Example T H 2 T H 2 c z dr z dr


  1. 4/9/2015 Example c t  T v z H 2 dr 2 T H  c z dr v t Example c t  T v z H 2 dr T H 2 t  z dr c v 1

  2. 4/9/2015 Example     T H 2 T H 2   c z dr z dr     v t t     lab field  2   2  H H  dr dr     t t     lab field 2   H t  t  lab    lab field H   field Singly ‐ Drained Soil Profiles Impermeable Rock 2

  3. 4/9/2015 Degree of Consolidation 1.0 � 0.5 � �� 0 c t  T v z 2 H dr 3

  4. 4/9/2015 Example 4

  5. 4/9/2015 Settlement Under a Foundation Chapter 9.12 Circular Foundation q =75 kN/m 2 5

  6. 4/9/2015 Simpson’s Rule � � � � � � 2:1 Method 6

  7. 4/9/2015 Circular Loaded Area  /q 1.5� ∆� � ⁄ � � � z/R Square Loaded Area  /q 1.5� ∆� � � � ⁄ � z/B 7

  8. 4/9/2015 2:1 Method P P 890 kN  dry = 15.7 kN/m 3 8

  9. 4/9/2015 Table 8.6 890 kN  dry = 15.7 kN/m 3 9

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