A silty sand fill 4.5 meters thick and an estimated density of 2.0 Mg/m is to be

Question

A silty sand fill 4.5 meters thick and an estimated density of 2.0 Mg/m is to be plac 6.0 meter thick layer of compressible silty clay. Underlying the clay layer is sandy gra groundwater table is at the surface of the clay layer. Assume, that the compressibility o sandy gravel are negligible compared to the compressibility of the clay layer. The properties o over a and vel. The erties of normally consolidated silty clay layer are the following: 1.) Initial void ratio co 1.255; 2.) Compression index Ce-0.29 3.) Re-Compression index C 0.03; 4.) Saturated density of the silty clay of 1.42 Mg/m3; and 5.) Coefficient of consolidation Cv 0.79 m/year It is desired to perform the following tasks: a.) Subdivide the clay layer into three sub-layers 2.0 m thick each to cal culate the surface at the end of the consolidation process; b) Compute the time that it will take to reach 50% and 90% consolidation of the clay i c.) What would be the excess pore water pressure at the mid-point of the first sub-laye settlement that the placing of the fill will cause well as the settlements that would have occurred at those times; 1.0 meter below the top of the clay layer) five years after the construction of the fil ayer as (that is, l (assume that the fill is placed instantaneously); and d.) What would be the vertical effective stress at the midpoint of the third sub-layer (that meters below the top of the clay layer) after five years of consolidation (same assumptioin in c.)

Explanation / Answer

Answer a

Primary consolidation settlement Sc= Cc Ho log((¶+?¶)/¶)/(1+e0) where ¶ is effective vertical stress at mid layer

Thus for top 2 m thick layer

Sc1= 0.29*2*log(((1.42*9.81*1)+(2*9.81*4.5))/(1.42*9.81*1))/(1+1.255) =0.2226 m =222.6 mm

For second 2 m thick layer

Sc2 = 0.126837 m = 126.84 mm

For third 2 m thick layer

Sc3 = 0.09145 m = 91.45 mm

Answer b)

Assuming that first and third sublayer would be under single drainage from sand layer over and gravel layer beneath respectively.

For first and third sublayer = t50 = T50*d^2/Cv = 0.197*4/0.79 = 0.997 year and t90 = 0.848*4/0.79 = 4.2936 year

Second sublayer in two way drainage

For second sub layer t50 = 0.197*9/0.79 = 2.244 year; t90 = 9.66 year

Answer c)

Immediately after sand layer

All excess pressure of sand = excess pore pressure = 4.5*2*9.81=88.29N/m2

However after 5 year excess pore pressure = 0 due to completion of consolidation.

Answer d)

Vertical effective stress at 5 m below top of clay layer after 5 years of consolidation = total stress - pore water pressure = (4.5*2*9.81)+(1.42*5*9.81)-(1*5*9.81) = 108.891 N/m2

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