DHARM

SOIL MOISTURE–PERMEABILITY AND CAPILLARITY 149

Let us assume: G = 2.65

or 17.66 =

(. )

()

(^265) (. )
1

• 981


• 
  

  e
  e
  whence e = 1.06
  γd =
  G
  e
  γw
  ()
  .
  1 (.)
  265

  
  


• 
  

  1106

  
  


• 
  

  × 9.81 kN/m^3 = 12.62 kN/m^3
  Total stress at the top of clay layer:
  σ = 1.5 × 12.62 + 3.5 × 17.66 = 80.74 kN/m^2
  Neutral stress at the top of clay layer:
  u = 3.5 × 9.81 = 34.34 kN/m^2
  Effective stress at the top of clay layer:
  σ = (σ – u) = 80.74 – 34.34 = 46.40 kN/m^2.
  Example 5.3. Compute the total, effective and pore pressure at a depth of 15 m below the
  bottom of a lake 6 m deep. The bottom of the lake consists of soft clay with a thickness of more
  than 15 m. The average water content of the clay is 40% and the specific gravity of soils may be
  assumed to be 2.65. (S.V.U.—B.E., (R.R.)—April, 1966)
  The conditions are shown in Fig. 5.25:
  58.86 58.86
  All pressuresare in
  kN/m^2
  206 117.9 206 117.9
  (a) Total pressure (b) Neutral pressure
  (c) Effective pressure
  Water
  Lake bed
  Saturated clay^15 15 mm
  6 6mm
  Fig. 5.25 Clay layer below lake bed (Example 5.3) Fig. 5.26 Pressure Diagrams (Example 5.3)
  Water content wsat = 40%
  Specific gravity of solids, G = 2.65
  Void ratio, e = wsat. G
  = 0.4 × 2.65
  = 1.06
  γsat =
  Ge
  e w

  
  


• 
  


• 
  

  F
  HG
  I
  1 KJ
  γ

  
  

  (.. )
  (.)
  265 106
  1106

  
  


• 
  


• 
  

  × 9.81 kN/m^3
  = 17.67 kN/m^3
  Total stress at 15 m below the bottom of the lake:
  σ = 6 × 9.81 + 15 × 17.67 = 323.9 kN/m^2