12.23 A 15-m thick normally consolidated clay layer is shown in Figure 12.55. Th

Question

12.23 A 15-m thick normally consolidated clay layer is shown in Figure 12.55. The liquid limit and plastic limit of the soil are 39 and 20, respectively a. Using Eq. (12.46) given by Skempton (1957), estimate the undrained co- hesion at a depth of 11 m below the ground surface as would be obtained by conducting a vane shear test. b. Using Bjerrum's (1974) correction factor [Eqs. (12.61) and (12.62)], esti- mate the design value of the undrained shear strength determined in Part.a c. If the clay has the potential of becoming overconsolidated up to OCR 3.0 due to future ground improvement activities, what would be the new un- drained cohesion? Use Eq. (12.47) given by Ladd et al. (1977) = 16 kN/m3 15 m LL-39 PL-20 Dry sand Clay Rock Figire 255

Explanation / Answer

Skempton relationship for undrained cohesion of normally consolidated clay is given by the following equation :

Cu = Yeff x Df ( 0.11 + 0.37 Ip ) ; where Yeff x Df is the effective normal overburden stress and Ip is plasticity index given by

Ip =(Wl - Wp ) and for the given question Ip = (39-20) =19

Thus for 11 m below ngl the effective normal stress would be given as = 16*3 + (20.5-9.81)*8= 133.52 KN/ m2

Thus using Skempton equation as described above the undrained cohesion for normally consolidated clay is given as

Cu = 133.52 * ( 0.11 + 0.37 * 19 ) =953.33 KN / m2

For question b

Bjerum corrected undrained design cohesion is given as :

Cu design = × * Cu ; × = correction factor = 1.7 - 0.54 * log (Ip) ; thus × = 1.7- 0.54* log 19 = 1.0094

Thus Cu design = 1.0094 * 953.33 = 962.29 KN/ m2

For question C

Ladd gave relationship between Cu normally consolidated and Cu over consolidated as Cu over consolidated = Cu normally consolidated x ( OCR)^A ; where A ranges from 0.85 to 0.75 as OCR increases

Thus Cu over consolidated = 953.33* (3 )^ 0.85 = 2425.4 KN/ m2 at depth of 11 m below NGL

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