The void ratio of a clay sample A decreased from 0.572 to 0.505 under a change in pressure from 122 to 180 kN/m^2. The void ratio of another sample B decreased from 0.61 to 0.557 under the same increment of pressure.The thickness of sample A was 1.5 times that of B. Nevertheless the time taken for

Question

The void ratio of a clay sample A decreased from 0.572 to 0.505 under a change in pressure from 122 to [latex]180 \mathrm{kN} / \mathrm{m}^{2} ext {. }[/latex] The void ratio of another sample B decreased from 0.61 to 0.557 under the same increment of pressure. The thickness of sample A was 1.5 times that of B. Nevertheless the time taken for 50% consolidation was 3 times larger for sample B than for A. What is the ratio of coefficient of permeability of sample A to that of B?

Accepted Answer

Let [latex]H_{a}=[/latex] thickness of sample A, [latex]H_{b}=[/latex] thickness of sample B, [latex]m_{v a}=[/latex] coefficient of volume compressibility of sample A, [latex]m_{v b}=[/latex] coefficient of volume compressibility of sample B, [latex]c_{v a}=[/latex] coefficient of consolidation for sample A, [latex]C_{v b}=[/latex] coefficient of consolidation for sample B, [latex]\Delta p_{a}=[/latex] increment of load for sample A, [latex]\Delta p_{b}=[/latex] increment of load for sample B, [latex]k_{a}=[/latex] coefficient of permeability for sample A, and [latex]k_{b}=[/latex] coefficient of permeability of sample B. We may write the following relationship

[latex]m_{v a}=\frac{\Delta e_{a}}{1+e_{a}} \frac{1}{\Delta p_{a}}, m_{v b}=\frac{\Delta e_{b}}{1+e_{b}} \frac{1}{\Delta p_{b}}[/latex]

where[latex]e_{a}[/latex] is the void ratio of sample A at the commencement of the test and [latex]\Delta e_{a}[/latex] is the change in void ratio. Similarly [latex]e_{b}[/latex] and [latex]\Delta e_{b}[/latex] apply to sample B.

[latex]\frac{m_{v a}}{m_{v b}}=\frac{\Delta p_{b}}{\Delta p_{a}} \frac{\Delta e_{a}}{\Delta e_{b}} \frac{1+e_{b}}{1+e_{a}}, ext { and } T_{a}=\frac{c_{v a} t_{a}}{H_{a}^{2}}, T_{b}=\frac{c_{v b} t_{b}}{H_{b}^{2}}[/latex]

wherein [latex]T_{a}, t_{a}, T_{b}[/latex] and [latex]t_{b}[/latex] correspond to samples A and B respectively. We may write

[latex]\frac{c_{v a}}{c_{v b}}=\frac{T_{a}}{T_{b}} \frac{H_{a}^{2}}{H_{b}^{2}} \frac{t_{b}}{t_{a}}, \quad k_{a}=c_{v a} m_{w a} \gamma_{w}, \quad k_{b}=c_{v b} m_{v b} \gamma_{w}[/latex]

Therefore, [latex]\frac{k_{a}}{k_{b}}=\frac{c_{v a}}{c_{v b}} \frac{m_{v a}}{m_{v b}}[/latex]

Given [latex]e_{a}=0.572, ext { and } e_{b}=0.61[/latex]

[latex]\Delta e_{a}=0.572-0.505=0.067, \Delta e_{b}=0.610-0.557=0.053[/latex]

[latex]\Delta p_{a}=\Delta p_{b}=180-122=58 \mathrm{kN} / \mathrm{m}^{2}, \quad H_{a}=1.5 \mathrm{H}_{b}[/latex]

But [latex]t_{b}=3 t_{a}[/latex]

We have, [latex]\frac{m_{v a}}{m_{v b}}=\frac{0.067}{0.053} imes \frac{1+0.61}{1+0.572}=1.29[/latex]

[latex]\frac{c_{v a}}{c_{v b}}=1.5^{2} imes 3=6.75[/latex]

Therefore, [latex]\frac{k_{a}}{k_{b}}=6.75 imes 1.29=8.7[/latex]

The ratio is 8.7 : 1.

Question 7.9: The void ratio of a clay sample A decreased from 0.572 to 0....

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