Soil Mechanics Test 4

The design of an embankment dam and other hydraulic structures, the choice of soils are aimed at reducing or eliminating the detrimental effects of seeping water. Where high hydraulic gradients exist there is a possibility that the seeping water may cause internal erosion within the dam, especially if the soil is poorly compacted. Erosion can work its way back into the embankment, creating voids in the form of channels or ‘pipes’, and thus impairing the stability of the dam. This form of erosion is referred to as piping and occurs when water flow upwards resulting in zero effective pressure.

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 \(\sqrt {\frac{{{{\rm{k}}_{\rm{z}}}}}{{{{\rm{k}}_{\rm{x}}}}}} \).

By doing this, the flow in anisotropic soil can be analysed using the same methods that are used for analysing isotropic soils.

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:

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