A 39.4 ft thick layer of relatively impervious saturated clay lies over a gravel aquifer. Piezometer tubes introduced to the gravel layer show an artesian pressure condition with the water level standing in the tubes 9.8 ft above the top surface of the clay stratum. The properties of the clay are

Question

A 39.4 ft thick layer of relatively impervious saturated clay lies over a gravel aquifer. Piezometer tubes introduced to the gravel layer show an artesian pressure condition with the water level standing in the tubes 9.8 ft above the top surface of the clay stratum. The properties of the clay are $e=1.2, G_{s}=2.7 \text { and } \gamma_{\text {sat }}=110.62 lb / ft ^{3}$ .

Determine (a) the effective stress at the top of the gravel stratum layer, and (b) the depth of excavation that can be made in the clay stratum without bottom heave.

Accepted Answer

(a) At the top of the gravel stratum

[latex]\sigma_{c}=39.4 imes 110.62=4358.43 lb / ft ^{2}[/latex]

The pore water pressure at the top of the gravel is

[latex]u_{w}=62.4 imes 49.2=3070 lb / ft ^{2}[/latex]

The effective stress at the top of the gravel is

[latex]\sigma^{\prime}=\sigma_{c}-u_{w}=4358.43-3070=1288.43 lb / ft ^{2}[/latex]

(b) If an excavation is made into the clay stratum as shown in Fig. Ex. 5.12, the depth must be such that

[latex]\sigma_{c} \leq u_{w}[/latex]

Let the bottom of the excavation be h ft above the top of gravel layer. Now the downward pressure acting at the top of the gravel layer is

[latex]\sigma_{c}=\gamma_{t} h=110.62 h lb / ft ^{2}[/latex]

[latex]u_{w}=3070 lb / ft ^{2}[/latex]

Now, [latex]110.62 h=3070 \quad ext { or } \quad h=\frac{3070}{110.62}=27.75 ft[/latex]

Depth of excavation, d = 39.4 - 27.75 = 1 1.65 ft

This is just the depth of excavation with a factor of safety [latex]F_{s}=1.0[/latex]. If we assume a minimum [latex]F_{s}=1.10[/latex]

[latex]h=\frac{3070 imes 1.1}{110.62}=30.52 ft[/latex]

Depth of excavation = 39.4 - 30.52 = 8.88 ft

Question 5.12: A 39.4 ft thick layer of relatively impervious saturated cla...

Related Answered Questions