Concept:

Factors affecting the permeability of soil can be studied using the following equation.

$K=\frac{1}{Z}\frac{e^3}{1+e}\frac{{\gamma }_w}{\mu }\frac{1}{S^2}$

Where, Z = constant; μ = dynamic viscosity of water;

S = specific surface;  γw = unit weight of water.

The factors affecting the permeability of soil can be summarised in below tabulated form:

Explanation:

Triaxial compression test:

Principle:

(i) The soil specimen is subjected to three compressive stresses in the mutually perpendicular direction, one of the three stresses being increased until the specimen fails in shear.

(ii) The desired 3-dimensional system stress system is achieved by an initial application of all-around fluid pressure or confining pressure or total lateral stress through water. While this confining pressure is kept constant throughout the test, axial or vertical loading is increased gradually and at a uniform rate.

(iii) The axial stress thus constitute the major principal stress and the confining acts in other two principal directions, the intermediate and minor principal stress being equal to the confining pressure.

Fig (a) Initially, upon application of all-round fluid pressure or confining pressure or lateral pressure

Fig (b) After application of external axial stress in addition to the confining pressure held constant until failure

Explanation:

Assumptions of Rankine's theory

• Soil is homogeneous, elastic, isotropic, semi-infinite, dry, and cohesionless.
• Ground surface planar which may be horizontal and inclined.
• The face of the wall in contact with the backfill is vertical and smooth.
• The soil is in a state of plastic equilibrium in both the active and passive stages.
• The rupture surface is planar.

Assumptions in Coulomb's theory

• Soil is homogeneous, elastic, isotropic, semi-infinite, dry, and cohesionless.
• The face of the wall in contact with the backfill is vertical or inclined and rough.
• The failure wedge acts as a rigid body and the distribution of stresses over it is uniform.
• Failure is 2D and failure surface is planar which passes through the heel of the wall.
• The point of application and line of action of resultant thrust in between the wall and the soil is known.

• Coulomb's earth pressure theory takes the roughness of wall into consideration.
• Active earth pressure on a retaining wall decreases due to increase in wall friction.
• Rankine theory of earth pressure assumes that back of wall is vertical and smooth

​Hence options 2 and 3 are correct.

Concept:

The coefficient of compression index (Cc)

$C_c=\frac{e_1-e_2}{{\log }_{10}\left(\frac{{\overline{\sigma }}_1}{{\overline{\sigma }}_o}\right)}$

The coefficient of compressibility (a_v)

$a_v=\left|\frac{\mathrm{\Delta }e}{{\overline{\sigma }}_o-{\overline{\sigma }}_1}\right|$

Calculation:

Given,

Δe = 0.2

${\overline{\sigma }}_o=30$ kN/m²

${\overline{\sigma }}_1=60$ kN/m2

$C_c=\frac{\mathrm{\Delta }e}{{\log }_{10}\left(\frac{{\overline{\sigma }}_1}{{\overline{\sigma }}_o}\right)}$

$C_c=\frac{0.2}{{\log }_{10}\left(\frac{60}{30}\right)}=0.664$

C_c = 0.664 m²/kN

$a_v=\left|\frac{\mathrm{\Delta }e}{{\overline{\sigma }}_o-{\overline{\sigma }}_1}\right|=\frac{0.2}{30}=0.0067$

a_v = 0.0067 m²/kN

$\frac{C_c}{a_v}=\frac{0.664}{0.0067}=99.6$

The correct answer option 2.

Concept:

Coefficient of Active Earth Pressure Coefficient:

$K_a=\frac{1-\sin \left(\varphi \right)}{1+\sin \left(\varphi \right)}={\tan }^2\left(45-\frac{\varphi }{2}\right)$

Coefficient of Passive Earth Pressure Coefficient:

$K_p=\frac{1+\sin \left(\varphi \right)}{1-\sin \left(\varphi \right)}={\tan }^2\left(45+\frac{\varphi }{2}\right)$

Coefficient of Earth Pressure at rest:

${\mathrm{k}}_{\mathrm{r}}=\frac{\mu }{1-\mu }$

Where,

μ is the coefficient is the Poisson ratio.

Concepts:

Dry of Optimum: When soil is compacted at a moisture content that is less than the optimum moisture content (OMC) then the soil is said to be compacted dry of optimum.

Wet of Optimum: When soil is compacted at a moisture content that is more than the optimum moisture content (OMC) then the soil is said to be compacted wet of optimum.

The above concepts are shown in the following diagram for better understanding:

The effect of compaction in dry of optimum and wet of optimum is shown in the following table:

Explanation:

Newmark’s Influence Chart method

• In this method, the soil is assumed to be homogeneous, semi-infinite, elastic, and isotropic i.e., this is not used for stratified soils.
• Influence chart consists of no. of concentric circles and radial lines which is divided into different area unit, the influence of which is the same at the centre of the chart.
• Influence value of each area unit = Influence value, $I_f=\frac{1}{mn}$ where, m = no. of concentric circles, n = no. of radial lines, m × n = No. of area units.
• It can be used for any given shape of footing.
• For most of the charts, influence value = 0.005.

​

Steps to determine the stress increase:

• Determine the depth, z where you wish to calculate the stress increase.
• Adopt a scale of z = AB.
• Draw the footing to scale and place the point of interest over the center of the chart.
• Count the number of elements that fall inside the footing, N.
• Calculate the stress increase as, σ_Z = I_f × N × q​

The plan of the loaded area (footing) is drawn on a tracing paper to such a scale that the length of line AB on the chart is equal to the depth where vertical stress is required.

Concept:

Shear stress of a soil sample is given by,

τ = c + σ tan ϕ

Where, τ = shear strength of soil

σ = normal stress of soil

c = cohesion

ϕ = angle of internal friction or shearing resistance

Calculation:

Dry sand sample, So, C = 0

σ = τ = 120 kN / m²

τ = 0 + σ tan ϕ ⇒ 120 = 120 tan ϕ

⇒ tan ϕ = 1

⇒ ϕ = 45°

CONCEPT:

Unconsolidated Un-drained (UU) Test:  

• In this test expulsion of pore water is not permitted in both the stages.
• It is used for clays in short term analysis for clays under un-drained conditions at fast loading rate.

Consolidated Drained (CD) Test:  

• In this test expulsion of pore water is permitted in both the stages.
• It is used for short term and long term stability analysis in saturated sands and long term stability analysis in clays.

Consolidated Un-drained (CU) Test:

• In this test, expulsion of pore water is permitted in 1^st stage but not in second stage. It is used for investigation of safety of earthen dam which may occur due to sudden drawdown of water table.

Unconsolidated Drained (UD) Test:

•  This test is not performed practically because confining pressure acts for long time and if soil is unconsolidated for long time then it cannot be drained in a small period of shear loading.

For loading condition

• Immediate stability is attained by UU test.
• Immediate settlement is attained by CU test.

For unloading condition

• Immediate stability is attained by CU test.

As in CU test consolidation process takes time to complete, So while unloading if we perform the CU test, we can immediately perform the test for obtaining immediate settlement.

Concept:

Coefficient of Active Earth Pressure Coefficient:

$K_a=\frac{1-\sin \left(\varphi \right)}{1+\sin \left(\varphi \right)}={\tan }^2\left(45-\frac{\varphi }{2}\right)$

Coefficient of Passive Earth Pressure Coefficient:

$K_p=\frac{1+\sin \left(\varphi \right)}{1-\sin \left(\varphi \right)}={\tan }^2\left(45+\frac{\varphi }{2}\right)$

Calculations:

Point to be remembered:

For any value of the shearing angle, the product of active earth pressure coefficient and the passive earth pressure coefficient will always equal to 1.

$K_a\times K_p=\frac{1-\sin \left(\varphi \right)}{1+\sin \left(\varphi \right)}\times \frac{1+\sin \left(\varphi \right)}{1-\sin \left(\varphi \right)}=1$

Alternate Solution:

For ϕ = 60°

$K_a={\tan }^2\left(45-\frac{60}{2}\right)=0.0718$

$K_a={\tan }^2\left(45+\frac{60}{2}\right)=13.928$

Therefore,

K_a × K_p = 0.0718 × 13.928 = 1

Calculation:

For active earth pressure:

${\mathrm{K}}_{\mathrm{a}}=\frac{1}{{\left(\tan \left[45+\left(\frac{\varphi }{2}\right)\right]\right)}^2}=\frac{1-\sin \mathrm{\varnothing }}{1+\sin \mathrm{\varnothing }}=\frac{1-\sin 30}{1+\sin 30}=0.33$