The amount of settlement that would occur at the end of 1.5 year and 5 year are 7.3 cm and 13.14 cm respectively.
For a layer of 3.8 m thickness, we were given the following parameters:
For Sf, we have:
Sf = Sc/U
Sf = 7.3/0.5
Sf = 14.6
Therefore, Sf for a layer of 38 m thickness is given by:
Sf = 14.6 × 38/3.8
Sf = 146 cm.
At 50%, the time for a layer of 3.8 m thickness is:
[tex]t_{50}[/tex] = 1.5 year.
At 50%, the time for a layer of 38 m thickness is:
[tex]t_{50}[/tex] = 1.5 × (38/3.8)²
[tex]t_{50}[/tex] = 150 years.
For a thickness of 38 m, U₂ is given by:
[tex]\frac{U_1^2}{U_2^2} =\frac{(T_v)_1}{(T_v)_2} = \frac{t_1}{t_2} \\\\U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{1.5}{150} ]\\\\U_2^2 = 0.25 \times 0.01\\\\U_2=\sqrt{0.0025} \\\\U_2=0.05[/tex]
The new settlement after 1.5 year is:
Sc = U₂Sf
Sc = 0.05 × 146
Sc = 7.3 cm.
For time, t₂ = 5 year:
[tex]U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{5}{150} ]\\\\U_2^2 = 0.25 \times 0.03\\\\U_2=\sqrt{0.0075} \\\\U_2=0.09[/tex]
The new settlement after 5 year is:
Sc = U₂Sf
Sc = 0.09 × 146
Sc = 13.14 cm.
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