# 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

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

The secondary consolidation is

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

We must find ep first, by finding the change in the voids ratio during primary consolidation because ep 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