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Chapter 5

Q. 5.16

At a particular site lies a layer of fine sand 8 m thick below the ground surface and having a void ratio of 0.7. The GWT is at a depth of 4 m below the ground surface. The average degree of saturation of the sand above the capillary fringe is 50%. The soil is saturated due to capillary action to a height of 2.0 m above the GWT level. Assuming G_{s}=2.65, calculate the total effective pressures at depths of 6 m and 3 m below the ground surface.

Step-by-Step

Verified Solution

\gamma_{d}=\frac{G_{s} \gamma_{w}}{1+e}=\frac{2.65 \times 9.81}{1.7}=15.29 kN / m ^{3}

 

\gamma_{ sat }=\frac{\left(e+G_{s}\right) \gamma_{w}}{1+e}=\frac{(0.7+2.65) \times 9.81}{1.7}=19.33 kN / m ^{3}

 

\gamma_{b}=\gamma_{ sat }-\gamma_{w}=19.33-9.81=9.52 kN / m ^{3}

 

The moist unit weight of soil above the capillary fringe is

 

\gamma_{m}=\frac{\left(G_{s}+e S\right) \gamma_{w}}{1+e}=\frac{(2.65+0.7 \times 0.5) \times 9.81}{1.7}=17.31 kN / m ^{3}

 

Capillary pressure,

 

u_{c}=n h_{c} \gamma_{w}=\frac{e}{1+e} h_{c} \gamma_{w}=\frac{0.7}{1.7} \times 2 \times 9.81=8.08 kN / m ^{2}

 

Effective stresses at different levels

(a) At ground level \sigma^{\prime}=0

(b) Overburden pressure at fringe level =\sigma_{o}^{\prime}=\overline{h_{c}} \gamma_{m}=2 \times 17.31=34.62 kN / m ^{2}

(c) Effective pressure at fringe level =\sigma_{c}^{\prime}=\sigma_{o}^{\prime}+u_{c}=34.62+8.08=42.70 kN / m ^{2}

(d) Effective pressure at GWT level =\sigma_{\text {sat }}^{\prime}=\sigma_{c}^{\prime}+\sigma_{d}^{\prime}=42.70+2 \times 15.29

 

=42.70+30.58=73.28 kN / m ^{2}

 

(e) Effective pressure at 6 m below GL

\sigma_{t}^{\prime}=\sigma_{\text {sat }}^{\prime}+h_{w} \gamma_{b}=73.28+2 \times 9.52=73.28+19.04=92.32 kN / m ^{2}

 

Effective stress at a depth 3 m below GL

Refer Fig. Ex. 5.16.

 

\sigma_{z}^{\prime}=\sigma_{o}^{\prime}+u_{c}+\left(z-h_{c}\right) \gamma_{d}=34.62+8.08+(3-2) \times 15.29 \approx 58 kN / m ^{2}