Consolidation Theory

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Consolidation Theory

• Examples

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Example 1

• The Following results were obtained from an
  oedometer test on a specimen of saturated clay.
• A layer of this clay 8m thick lies below a 4m
  depth of sand, the water table being at the
  surface. For both soils, the saturated unit
  wt21kN/m3
• A 4m depth of fill of unit weight 21kN/m3 is
  placed on the sand over an extensive area.

429 214 107 54 27 Pressure(kN/m2)
0.994 1.068 1.144 1.217 1.243 Void ratio
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Continued Example1

• Determine the final settlement due to
  consolidation of the clay.
• Solution
• The settlement of the clay layer

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Continued Example1

• Appropriate values of e are obtained from the
  following figure
• The clay layer will be divided into four
  sub-layer, hence H 2000mm

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Continued Example1

• To get the settlement

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Example 2

• A soil profile is shown in the Fig. If a
  uniformly distributed load is applied at
  the ground surface, what is the settlement of the
  clay layer caused by primary consolidation if
• A) The clay is normally consolidated
• B) The preconsolidation pressure190kN/m2
• C) The preconsolidation pressure170kN/m2
• Cs1/6 Cc

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Continued Example 2
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Continued Example 2

• A) The average effective stress at the middle of
  the clay layer is

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Continued Example 2

• B) The average effective stress at the middle of
  the clay layer is

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Continued Example 2

• C) The average effective stress at the middle of
  the clay layer is

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Example 3

• In an oedometer test, a specimen of saturated
  clay 19mm thick reaches 50 consolidation in 20
  min.
• How long would it take a layer of this clay 5m
  thick to reach the same degree of consolidation
  under the same stress and drainage conditions?
• How long would it take the layer to reach 30
  consolidation?

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Continued Example 3
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Continued Example 3
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Example 4

• The time rate settlement data shown below is
  for the increment from 20 to 40 kPa from the
  test. The initial sample height is 2.54cm, and
  there are porous stones on the top and at the
  bottom of the sample.
• Determine Cv by the log time- fitting
  procedure.

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Continued Example 4
Dial reading (mm) Elapsed time (min)
4.041 0
3.927 0.1
3.879 0.25
3.830 0.5
3.757 1
3.650 2
3.495 4
3.282 8
3.035 15
2.766 30
2.55 60
2.423 120
2.276 240
2.184 505
2.040 1485
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Continued Example 4
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Continued Example 4