Figure Q1 shows a soil profile. An oedometer test was conducted on a specimen collected from the middle of the clay laye

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Figure Q1 shows a soil profile. An oedometer test was conducted
on a specimen collected from the middle of the clay layer. The
results of the oedometer test on a clay specimen are: Cc = 0.2, Cs
= 0.04, OCR = 4.5. The uniformly distributed surcharge of Δσ is
applied at the ground surface. Given: Δσ = 100 kN/m2 , H1 = 2.5 m,
H2 = 2.5 m, H3 = 3 m. Also,  Sand: e = 0.64, Gs = 2.65  Clay: Gs
= 2.75, w (water content) = 43 % Where, OCR is the
overconsolidation ratio, defined by the ratio of maximum value of
effective stress in the past (preconsolidation pressure) to the
present value.
a) Calculate the primary consolidation settlement. b) What
would be the difference in settlement if OCR were 1.5 instead of
4.5.

Figure Q1 Shows A Soil Profile An Oedometer Test Was Conducted On A Specimen Collected From The Middle Of The Clay Laye 1 (84.74 KiB) Viewed 25 times

Figure Q1 shows a soil profile. An oedometer test was conducted on a specimen collected from the middle of the clay layer. The results of the oedometer test on a clay specimen are: C = 0.2, C = 0.04, OCR = 4.5. The uniformly distributed surcharge of Ao is applied at the ground surface. Given: 40 = 100 kN/m², H = 2.5 m, H2=2.5 m, Hz = 3 m. Also, • Sand: e=0.64, G, -2.65 • Clay: Gs = 2.75, w (water content) = 43% Where, OCR is the overconsolidation ratio, defined by the ratio of maximum value of effective stress in the past (preconsolidation pressure) to the present value. + Δο H H Sand Clay Figure Q1 a) Calculate the primary consolidation settlement. b) What would be the difference in settlement if OCR were 1.5 instead of 4.5.