Answer with Explanation:
Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is
[tex]T_v=\frac{\pi }{4}(\frac{U}{100})^2[/tex]
Solving for 'U' we get
[tex]\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%[/tex]
Since our assumption is correct thus we conclude that degree of consolidation is 50.46%
The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2
i)[tex]\frac{z}{H}=0.25=U=0.71[/tex] = 71% consolidation
ii)[tex]\frac{z}{H}=0.5=U=0.45[/tex] = 45% consolidation
iii)[tex]\frac{z}{H}=0.75U=0.3[/tex] = 30% consolidation
Part b)
The degree of consolidation is given by
[tex]\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm[/tex]
Thus a settlement of 50.46 centimeters has occurred
For time factor 0.7, U is given by
[tex]T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59[/tex]
thus consolidation of 85.59 % has occured if time factor is 0.7
The degree of consolidation is given by
[tex]\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm[/tex]
