During a consolidation test, a sample of fully saturated clay 3 cm thick (= h0) is consolidated under a pressure increment of 200 kN/m^2. When equilibrium is reached, the sample thickness is reduced to 2.60 cm. The pressure is then removed and the sample is allowed to expand and absorb water. The

Question

During a consolidation test, a sample of fully saturated clay 3 cm thick $\left(=h_{0}\right)$ is consolidated under a pressure increment of 200 $\mathrm{kN} / \mathrm{m}^{2}$ . When equilibrium is reached, the sample thickness is reduced to 2.60 cm. The pressure is then removed and the sample is allowed to expand and absorb water. The final thickness is observed as 2.8 cm $\left(h_{f}\right)$ and the final moisture content is determined as 24.9 %.

If the specific gravity of the soil solids is 2.70, find the void ratio of the sample before and after consolidation.

Accepted Answer

Use equation (7.3)

[latex]\Delta e=\frac{1+e}{h} \Delta h[/latex] (7.3)

[latex]\Delta e=\frac{1+e}{h} \Delta h[/latex]

1. Determination of [latex]e_{f}[/latex]

Weight of solids [latex]=W_{s}=V_{s} G_{s} \gamma_{\omega}=1 imes 2.70 imes 1=2.70 \mathrm{~g}[/latex]

[latex]\frac{W_{w}}{W_{s}}=0.249 ext { or } W_{w}=0.249 imes 2.70=0.672 \mathrm{gm}, e_{f}=V_{w}=0.672[/latex]

2. Changes in thickness from final stage to equilibrium stage with load on

[latex]\Delta h=2.80-2.60=0.20 \mathrm{~cm}, \quad \Delta e=\frac{(1+0.672) 0.20}{2.80}=0.119[/latex]

Void ratio after consolidation [latex]=e_{f}-\Delta e=0.672-0.119=0.553[/latex]

3. Change in void ratio from the commencement to the end of consolidation

[latex]\Delta e=\frac{1+0.553}{2.6}(3.00-2.60)=\frac{1.553}{2.6} imes 0.40=0.239[/latex]

Void ratio at the start of consolidation [latex]=0.553+0.239=0.792[/latex]

Question 7.1: During a consolidation test, a sample of fully saturated cla...

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