For an isotropic soil, the property (permeability) of the soil is considered to be same in all possible direction. But for an anisotropic soil, the soil exhibits different property in different directions.

The analysis of an anisotropic soil is difficult and thus an assumption is alternatively required to solve the complex problem.

So if the horizontal permeability (k_x) and a vertical permeability (k_v) of an anisotropic soil is given, the solution can be reduced to that of flow in an isotropic material by doing a variable change

X^I = X / α

Z^I = Z

Where α = \(\sqrt {\frac{{{{\rm{k}}_{\rm{z}}}}}{{{{\rm{k}}_{\rm{x}}}}}} \)

The value of k_z is lower than k_x and thus the value if lesser than 1 and thus the horizontal dimension is shortened by \(\sqrt {\frac{{{{\rm{k}}_{\rm{z}}}}}{{{{\rm{k}}_{\rm{x}}}}}} \).

Concept:

Critical hydraulic gradient (icr): The quick condition occurs at a critical upward hydraulic gradient ic, when the seepage force just balances the buoyant weight of an element of soil. The critical hydraulic gradient is typically around 1.0 for many soils.

At the critical conditions, the effective stress is equal to zero.

Calculation:

n = 0.6, G_s = 2.7

\(\therefore e = \frac{n}{{1 - n}} \Rightarrow e = \frac{{0.6}}{{1 - 0.6}} = 1.5\)

\(\therefore {i_{cr}} = \frac{{{G_s} - 1}}{{1 + e}} = \frac{{2.7 - 1}}{{1 + 1.5}} = 0.68\)

\(FOS = \frac{{{i_{cr}}}}{i} = \frac{{0.68}}{{0.5}} = 1.36\)

Concept:

Specific yield (S_y) is the ratio of the volume of water that drains from a saturated rock (due to gravity) to the total volume of the rock.

The Specific Retention (S_r) of a rock or soil is the ratio of the volume of water a rock can retain against gravity to the total volume of the rock.

Therefore, the total porosity is equal to the volume or water that a rock will yield by gravity drainage (S_y) and the volume held by surface tension (S_r) and is given as:

\({S_y} + {S_r} = \eta\)

Calculations:

Given:

S_y = 30%

η = 50%

\({S_y} + {S_r} = \eta\)

\(30\% + {S_r} = 50\%\)

\({S_r} = 50\% - 30\%\)

\({S_r} = 20\%\)

Concept:

The head loss is assumed to occur in flowing through soil only.

Under downward flow conditions, the effective stress at any point increases by head loss up to that point multiplied by the unit weight of water.

Calculation:

This is a case of downward seepage.

σ̅_A = σ_A - U_A

Total head loss flowing through 4m of soil = 6 m

∴ Head loss in flowing through 1 m of soil \(= \frac{6}{4}\) = 1.5 m

Pressure head at A = (3 – h_L-AB) = (3 – 1.5) = 1.5 m

∴ Pore water pressure at A = 1.5 γ_w = 1.5 × 9.81 = 14.715 kN/m²

σ_A = (γ_sat × 1 + γ_w × 2) = (20 × 1 + 9.81 × 2) = 39.62 kN/m²

∴ σ̅_A = σ_A - U_A = 39.62 – 14.715 = 24.905 kN/m²     

Alternate method:

Under downward flow conditions, the effective stress at any point increases by head loss up to that point multiplied by the unit weight of water.

Thus the effective stress at A = γ_sub × depth of submergence + h_L-AB × γ_W

γ_sub = γ_sat - γ_W = 20 – 9.81 = 10.19 kN/m³

∴ σ̅_A = 10.19 × 1 + 1.5 × 9.81 = 24.905 kN/m²     

Concept:

Coefficient of transmissibility: It is defined as the rate of flow of water through a vertical strip of the aquifer of unit width and extending the full saturation height under unit hydraulic gradient.

Coefficient of permeability: It is defined as the velocity of flow which will occur through the total cross-section area of the soil under unit hydraulic gradient.

We know, Discharge through confined aquifer according to theims theory is given by

\(Q = \frac{{2\pi KB\left( {{h_2} - {h_1}} \right)}}{{\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right)}} \)

Calculation:

\(K = \frac{{Q{{\log }_e}\left( {\frac{{{r_2}}}{{{r_1}}}} \right)}}{{2\pi B\left( {{h_2} - {h_1}} \right)}}\)

Q = 0.2 m³/sec, r₂ = 75 m, r₁ = 15 m,S₁ = 5 m, S₂ = 3 m, B = 20 m

h₁ = H – S₁ and h₂ = H – S₂

where S₁ and S₂ are the drawdown from the ground water table.

∴ h₂ – h₁ = (H – S₂) – (H – S₁) = S₁ – S₂

\(\therefore K = \frac{{0.2 \times {{\log }_e}\left( {\frac{{75}}{{15}}} \right)}}{{2\pi \times 20 \times \left( {5 - 3} \right)}} = 1.28 \times {10^{ - 3}}m/sec\)

T = kB ⇒ T = 1.28 × 10^-3 × 20 = 0.0256 m²/sec

The uplift pressure at point P, P_u = δ_w h_P

Where h_p = pressure head at point p

h_p = total balance seepage head – Elevation heat

Balance seepage head, h = H – n ΔH

H = 6 – 2 = 4m (difference between u/s & d/s head)

\({\rm{\Delta }}H = \frac{4}{{{N_D}}} = \frac{4}{{12}} = \frac{1}{3}\)

Where N_D is the no of potential drop

So total head at P will be,

\(h = 4 - 6 \times \frac{1}{3} = 2\;m\)

Now, h_p = h – (-z) = 2 – (-2) = 4m

So, uplift pressure, P_u = \({\gamma _w}.{h_p}\) = 10 × 4 = 40 kN/m2

Concept:

For an unconfined aquifer:

\({\rm{Q}} = \frac{{{\rm{\pi k}}\left( {{{\rm{H}}^2} - {{\rm{h}}^2}} \right)}}{{{{\log }_{\rm{e}}}\left( {{\rm{R}}/{\rm{r}}} \right)}} = \frac{{{\rm{\pi k}}\left( {{\rm{h}}_2^2 - {\rm{h}}_1^2} \right)}}{{{{\log }_{\rm{e}}}\left( {\frac{{{{\rm{r}}_2}}}{{{{\rm{r}}_1}}}} \right)}}\)

Q = discharge through aquifer, k = coefficient of permeability (m2/day/m2), H = thickness of aquifer, h = depth of water in well measured above the impermeable layer, s = drawdown of well, R = Radius of zero drawdown or maximum of radius of influence and h1 and h2 are depth of water in well at radial distance r1 and r2 respectively.

Calculation:

r₁ = 15 m, r₂ = 45 m,h₁ = 15 m,Q = 0.2 m³/sec

K = 0.1 cm/sec = 0.1 × 10^-2 m/sec

For unconfined aquifer,

\(K = \frac{{Q{{\log }_e}\left( {\frac{{{r_2}}}{{{r_1}}}} \right)}}{{\pi \left( {h_2^2 - h_1^2} \right)}}\)

\(\Rightarrow 0.1 \times {10^{ - 2}} = \frac{{0.2 \times In\left( {\frac{{45}}{{15}}} \right)}}{{\pi \left( {h_2^2 - {{15}^2}} \right)}}\)

⇒ h₂ = 17.174 m

Mistake point:

Note that the drawdown is taken from the groundwater table to the level of water in the observation wells.

Concept:

Value of equivalent coefficient of permeability for anisotropic soil is \(\left( {{k_{eq}}} \right) = \sqrt {{k_x}{k_y}} \)

\(Q = k \times H \times \frac{{{N_f}}}{{{N_d}}}\)

Where

Nf = Number of flow channels

Nd = Number of potential drops

H = Net head

k = coefficient of permeability

Calculation:

k_x = 4 × 10^-6 m/sec

k_y = 1 × 10^-6 m/sec

N_f = Number of flow channels = Number of flow lines – 1 = 5 – 1 = 4

N_D = Number of equipotential drops = Number of equipotential lines – 1 = 17 – 1 = 16

\({k_{eq}} = \sqrt {4 \times {{10}^{ - 6}} \times 1 \times {{10}^{ - 6}}} = 2 \times {10^{ - 6}}\;m/sec\)

Seepage per unit length = \({k_{eq}}\;H\frac{{{N_f}}}{{{N_D}}} = 2 \times {10^{ - 6}} \times 16 \times \frac{4}{{16}}\) = 8 × 10^-6 m²/sec

∴ Total seepage in 1 day = Seepage per unit length × Length of dam = 8 × 10^-6 × 200 × 86400 = 138.24 m³

The following strata details can be depicted below:

Void ratio of sand (e) \(= \frac{{\rm{n}}}{{1 - {\rm{n}}}} = \frac{{0.36}}{{1 - 0.36}} = 0.5625\)

Saturated density of sand (γ_sat) \(= \frac{{{\rm{G}} + {\rm{Se}}}}{{1 + {\rm{e}}}} \times {{\rm{\gamma }}_{\rm{w}}}\)

\(\therefore {{\rm{\gamma }}_{{\rm{sat}}}} = \frac{{2.66 + 0.5625}}{{1.5625}} \times 9.81 = 20.23{\rm{\;kN}}/{{\rm{m}}^3}\)

Resultant force on the element is given by:

Weight of an element (W) = 20.23 × 1 = 20.23 kN

Buoyancy force on an element (B) = 9.81 × 1 = 9.81 kN

Horizontal seepage force (H) \(= {\rm{V}} \times {\rm{i}} \times {{\rm{\gamma }}_{\rm{w}}} = 1 \times \frac{4}{{12}} \times 9.81 = 3.27{\rm{\;kN}}\)

\(\therefore {\rm{R}} = \sqrt {{{\left( {{\rm{W}} - {\rm{B}}} \right)}^2} + {{\rm{H}}^2}} = \sqrt {{{\left( {20.23 - 9.81} \right)}^2} + {{3.27}^2}}\)