The water table is lowered from a depth of 10 ft to a depth of 20 ft in a deposit of silt. All the silt is saturated even after the water table is lowered. Its water content is 26%. Estimate the increase in the effective pressure at a depth of 34 ft on account of lowering the water table. Assume G_{s}=2.7.
Question 5.7: The water table is lowered from a depth of 10 ft to a depth ...
Effective pressure before lowering the water table.
The water table is at a depth of 10 ft and the soil above this depth remains saturated but not submerged. The soil from 10 ft to 20 ft remains submerged. Therefore, the effective pressure at 34 ft depth is
\sigma_{1}^{\prime}=10 \gamma_{ sat }+(34-10) \gamma_{b}
Now, \gamma_{ sat }=\frac{\gamma_{w}\left(G_{s}+e\right)}{1+e}, \gamma_{b}=\frac{\gamma_{w}\left(G_{s}-1\right)}{1+e}, \gamma_{w}=62.4 lb / ft ^{3}, e=w G_{s} \text { for } S=1
Therefore, e=0.26 \times 2.7=0.70
\gamma_{ sat }=\frac{62.4(2.7+0.7)}{1+0.7}=124.8 lb / ft ^{3}
\gamma_{b}=\frac{62.4(2.7-1)}{1+0.7}=62.4 lb / ft ^{3}
\sigma_{1}^{\prime}=10 \times 124.8+24 \times 62.4=2745.6 lb / ft ^{2}
Effective pressure after lowering of water table
After lowering the water table to a depth of 20 ft, the soil above this level remains saturated but effective and below this submerged. Therefore, the altered effective pressure is
\sigma_{2}^{\prime}=20 \gamma_{ sat }+(34-20) \gamma_{b}=20 \times 124.8+14 \times 62.4=3369.6 lb / ft ^{2}
The increase in the effective pressure is
\sigma_{2}^{\prime}-\sigma_{1}^{\prime}=\Delta \sigma^{\prime}=3369.6-2745.6=624.0 lb / ft ^{2}
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