The rise in water level in the pond causes equal increase in total stress and pore pressure so effective stress does not change. The pore pressure will increase by 20 kN/m².

Darcy's law states that there is a linear relationship between flow velocity (v) and hydraulic gradient (i) for any given saturated soil under steady laminar flow conditions.

If the rate of flow is q (volume/time) through a cross-sectional area (A) of the soil mass, Darcy's Law can be expressed as

v = q/A = k × i

Darcy’s law is valid for fine saturated sands only.

Where

k = permeability of the soil

i = Δh /L

Δh = difference in total heads

L = length of the soil mass

Darcy law represents the statistical macroscopic equivalent of Naiver-stokes equations of motion for the viscous flow of ground water.

Additional point:

Permeability of different soil are as follows:

Concept:

Allen Hazen’s formula: The coefficient of permeability of a soil is proportional to the square of representative particle size.

\(k = CD_{10}^2\)

Where, D₁₀ in mm and value of C lies between (0.4 – 1.2)

Calculation:

\(\frac{{{D_{10\;A}}}}{{{D_{10\;B}}}} = 2\)

Concept:

Coefficient of permeability for coarse grained soil is determined by means of Constant-head permeability test. The degree of saturation of soil should be 100%.

By Darcy’s Law ,

Q = k i A

\(Q = k\frac{h}{L}A\)

\(k = \frac{{QL}}{{Ah}}\)

where,

q = Discharge collected in time ‘t’

L = Distance between manometer taping points

A = Cross-sectional area of the sample.

H = Difference in manometer levels i.e the head loss.

Calculation:

 L = 10 cm, h_L = 1 m, A = 50 cm², Q = 1000 cc

 \(\therefore Q = \frac{{{1000}}}{15× 60} = 1.11 cc/s\)

\(\therefore i = \frac{{{h_L}}}{L} = \frac{{1\; × \;100}}{{10}} = 10\)

By Darcy’s law,

\(Q = k\;i\;A \Rightarrow 1.11 = k × 10 × 50\)

⇒ k = 2.22 × 10^-3 cm/sec

Concept:

Effective stress = Total stress – Pore water pressure

Calculation:

When water table is 6 m below ground level:

\({\bar \sigma _A} = {\sigma _A} - {U_A}\)

\({\sigma _A} = {\gamma _t} \times 6 + {\gamma _{sat}} \times 2 = 18 \times 6 + 20 \times 2 = 148\;kN/{m^2}\)

U_A = γ_w × 2 = 9.81 × 2 = 19.62 kN/m²

\(\therefore {\bar \sigma _A} = 148 - 19.62 = 128.38\;kN/{m^2}\)

When water table rises by 2m:

∴ σ_A = γ_t × 4 + γ_sat × 4 = 18 × 4 + 20 × 4 = 152 kN/m²

U_A = γ_w × 4 = 9.81 × 4 = 39.24 kN/m²

∴ σ̅_A = 152 – 39.24 = 112.76 kN/m²

∴ Change in effective stress = 128.38 – 112.76 = 15.62 kN/m²

Hence the effective stress decreases by 15.62 kN/m² due to the rise in the water table.

Important Point:

The rise in water table up to ground level will cause a decrease in effective stress and vice-versa. The rise or fall of the water table beyond ground level won’t cause any change in effective stress.

Concept:

Consider a section of stratified soil as shown in the figure below of varying thickness of each stratum eg: H1, H2, H3 & H4, with their respective coefficient of permeability k1, k2, k3 & k4

Vertical flow (Normal to bedding Plane):

Let i1, i2, i3 & i4 be the hydraulic gradient in different layers of thickness H1, H2, H3 & H4 respectively.

Let the total head loss be h over the total thickness of soil stratum H.

Each layer having head loss h1, h2, h3 & h4. Then the constant velocity of flow is given by

\(V = {k_v}\frac{h}{H} = {k_1}{i_1} = {k_2}{i_2} = {k_3}{i_3} = {k_4}{i_4}\)

Also,

\(\because Q = kiA = {k_1}{i_1}A = {k_2}{i_2}A = {k_3}{i_3}A = {k_4}{i_4}A\)

\(\therefore ki = {k_1}{i_1} = {k_2}{i_2} = {k_3}{i_3} = {k_4}{i_4}\)

\( \Rightarrow \frac{{K.h}}{H} = \frac{{{k_1}{h_1}}}{{{H_1}}} = \frac{{{k_2}{h_2}}}{{{H_2}}} = \frac{{{k_3}{h_3}}}{{{H_3}}} = \frac{{{k_4}{h_4}}}{{{H_4}}}\)

\(\because {h_1} + {h_2} + {h_3} + {h_4} = h\)

⇒ \(h\left( {\frac{{k\;{H_1}}}{{H\;{k_1}}} + \frac{{k\;{H_2}}}{{H\;{k_2}}} + \frac{{k\;{H_3}}}{{H\;{k_3}}} + \frac{{k\;{H_4}}}{{H\;{k_4}}}} \right) = h\)

\({k_v} = \frac{H}{{\frac{{{H_1}}}{{{k_1}}} + \frac{{{H_2}}}{{{k_2}}} + \frac{{{H_3}}}{{{k_3}}} + \frac{{{H_4}}}{{{k_4}}}}}\)

Calculation:

The flow is perpendicular to the bedding plane.

H₁ = 150 mm, H₂ = 300 mm, H₃ = 150 mm

∴ H = H₁ + H₂ + H₃ = 600 mm

k₁ = 1 × 10^-2 cm/sec, k₂ = 1 × 10^-3 cm/s, k₃ = 1 × 10^-2 cm/sec

A = 100 mm², Q = 0.5 cc/min \(= \frac{{0.5}}{{60}}\;cc/\;\sec \; = \frac{{0.5\; \times \;{{10}^3}}}{{60}}\;m{m^3}/s\)

\({k_{eq}} = \frac{H}{{\frac{{{H_1}}}{{{k_1}}} + \frac{{{H_2}}}{{{k_2}}} + \frac{{{H_3}}}{{{k_3}}}}}\)

\({k_{eq}} = \frac{{600}}{{\frac{{150}}{{1 \times {{10}^{ - 2}} \times 10}} + \frac{{300}}{{1 \times {{10}^{ - 3}} \times 10}} + \frac{{150}}{{1 \times {{10}^{ - 2}} \times 10}}}} = \frac{1}{{55}}\;mm/s\)

By darcy’s law,

Q = k_eq i A

\(\Rightarrow \frac{{0.5\; \times \;{{10}^3}}}{{60}} = \frac{1}{{55}} \times \frac{{{h_L}}}{{600}} \times 100 \Rightarrow {h_L} = 2750\;mm\) = 2.75 m

Concept:

Effective stress = Total stress – Pore water pressure

Capillary water

• If the water contained in the pores of soil is subjected only to gravitational forces then soil above the groundwater table is perfectly dry.
• Capillary water is held above the water table by Surface tension.
• Pressure in the capillary zone is negative and it does not contribute to hydrostatic pressure below ground water table.

Presence of water in various zones of soil is depicted as below

• Due to the effect of the capillary rise, there is an increase in the unit weight of the soil up to the height of capillary rise.

Calculation:

σ̅_A  = σ_A - U_A

σ_A = γ_t × 1 + γ_sat × 1 = 16 × 1 + 18 × 1 = 34 kN/m²

U_A = 0

∴ σ̅_A = 34 – 0 = 34 kN/m²

Concept:

For falling head permeability test,

\(K = \frac{a}{A}\frac{L}{t}\;ln\left( {\frac{{{h_1}}}{{{h_2}}}} \right)\)

\({\rm{k}} = 2.303\frac{{\rm{a}}}{{\rm{A}}} \times \frac{{\rm{L}}}{{\rm{t}}}10{{\rm{g}}_{10}}\left( {\frac{{{{\rm{h}}_1}}}{{{{\rm{h}}_2}}}} \right)\)

Where,

k = permeability, a = Area of tube in m2, A = Area of sample in m2, t = time in sec, L = length in m, h1 = Level of time t = 0, and h2 = Level of time t

Calculation:

When drop is 8 cm, h₁ = 60 cm, h₂ = 52 cm, t = 10 min

\({k_1} = \frac{a}{A}\frac{L}{{10}}\;ln\left( {\frac{{60}}{{52}}} \right)\)

When additional drop of 12 cm, h₁ = 60 cm, h₂ = 60 – 8 – 12 = 40 cm

\({k_2} = \frac{a}{A}\frac{L}{{{t_2}}}\;ln\left( {\frac{{60}}{{40}}} \right)\)

Let the time required be t₂.

Since the test is performed on the same specimen the permeability of the soil will not change.

∴ k₁ = k₂

\({k_2} = \frac{a}{A}\frac{L}{{{t_2}}}\;ln\left( {\frac{{60}}{{40}}} \right)\)

∵ k₁ = k₂

\(\Rightarrow \frac{a}{A} \times \frac{L}{{10}} \times ln\left( {\frac{{60}}{{52}}} \right) = \frac{a}{A} \times \frac{L}{{{t_2}}} \times ln\left( {\frac{{60}}{{40}}} \right)\)

⇒ t₂ = 28.33 min

Important Point:

Falling head permeability test is carried out for fine-grained soil, whereas constant head permeability test is carried out for coarse-grained soil.

Initial effective stress at bottom of the soil will be:

\({\bar \sigma _{bottom}} = \sigma - u\) 

\({\bar \sigma _{bottom}} = \left( {\gamma \times 2 + {\gamma _{sat}} \times 1} \right) - {\gamma _\omega } \times 1\) 

\({\bar \sigma _{bottom}} = \left( {16 \times 2 + 20 \times 1} \right) - 10 \times 1\) 

\({\bar \sigma _{bottom}} = 42\;kN/{m^2}\) 

Now,

\({\bar \sigma _{botto{m_1}}} = 3 \times {\bar \sigma _{bottom}}\) 

\({\bar \sigma _{botto{m_1}}} = 3 \times 42 = 126\frac{{kN}}{{{m^2}}}\) 

Now, \(\left( {\gamma \times x + \gamma \left( {1.5} \right) + {\gamma _{sat}} \times 1.5} \right) - \left( {{\gamma _\omega } \times 1.5} \right) = 126\)

16 x + 16 × 1.5 + 20 × 1.5 – 10 × 1.5 = 126

16 x = 87

For the given soil strata:

Effective stress at top layer of the lower sand (σ¹)

σ^' = Total stress (σ) – Pore pressure (u)

σ^’ = (17 × 1 + 19 × 2 + 20 × 5) - 9.81 × 10