Solution of Secondary Rate of Consolidation

Articles > Solution of Secondary Rate of Consolidation

Question:

An odometer (consolidation) test is performed on a normally consolidated clay stratum that is 8.5 feet thick, and it is found that the clay's initial voids ratio was eo= 0.8 and its primary compression index is Cc = 0.28. The in-situ stress at mid-clay layer is po = 2,650 psf, and the building exerts a pressure through its mat foundation of 970 psf. The secondary compression index Ca = 0.02.

The time of completion of the primary settlement is approximately 18 months. What is the total consolidation of the 8.5 foot clay stratum 5 years after the primary consolidation?

Solution:

For fomulations read the articles cited under Settlement of Shallow Foundations topic.

The primary consolidation is

$Delta H = frac{C_cH}{1+e_0}log(frac{sigma'_0-Deltasigma'_p}{sigma'_0})=frac{0.28(8.5)(12in/ft)}{1+0.8}log(frac{2650+970}{2650})=2.15;in$

The secondary consolidation is

$Delta H_s = frac{C_{alpha}H}{1+e_p}log(frac{t_2}{t_1})$

We must find e_p first, by finding the change in the voids ratio during primary consolidation because e_p is the void ratio at the end of primary consolidation (beginning of secondary compression)

$e_p=e_0-Delta e = e_0 - C_c log(frac{sigma'_0+Delta sigma'}{sigma'_0})=0.8-0.28log(frac{2650+970}{2650})=0.76$

Thus

$Delta H_s = frac{C_{alpha}H}{1+e_p}log(frac{t_2}{t_1})=frac{0.02(8.5-0.18)(12in/ft)}{1+0.76}log(frac{5}{1.5})=0.59in$

The total consolidation settlement is thus

$Delta H_t = Delta H + Delta H_s = 2.15+0.59=2.74;in$

Elastic settlement is neglected

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