Concept:

The total active earth pressure is given by,

P = \(\sqrt {P_H^2 + \;P_V^2} \)

P_H = \(\frac{1}{2}{K_H}{H^2}\) and P_V = \(\frac{1}{2}{K_V}{H^2}\)

Calculation:

Given that: K_H = 760 kg/m²/m and K_V = 100 kg/m²/m, H = 6 m

\({P_H} = \frac{1}{2} \times 760 \times {6^2}\) = 136.8 kN/m

\({P_V} = \frac{1}{2} \times 100 \times {6^2}\) = 18 kN/m

\(P = \sqrt {{{136.8}^2} + \;{{18}^2}} \)

Hence, P = 138 kN/m

Concept:

Stability of Finite slopes:

Stability analysis is generally done either using total stress analysis or effective stress analysis.

Total stress analysis is used for immediately after construction stability check while Effective stress analysis is used for long term after construction stability check.

Various methods based on Total stress analysis:

(1) Swedish circle Method

(2) Friction circle Method

(3) ϕ_U = 0 Analysis.

In this case we use Unconsolidated undrained test results.

Various methods based on Effective stress analysis are:

(1) Taylor’s method

(2) Bishop’s method.

In this case we use CU or CD test analysis results.

Concept:

Assumption in Rankine’s theory:

1. Soil is semi-infinite, homogeneous, isotropic, dry and cohesionless.

2. Soil is in a state of plastic condition at the time of active and passive pressure generation.

3. The backfill soil is horizontal.

4. Back of wall is vertical and smooth.

5. Rupture surface is a planar surface which is obtained by considering the plastic equilibrium of soil

Coulomb’s theory of Earth Pressure

Assumption:

1. The backfill is dry, cohesionless, isotropic

2. The back of the wall can be inclined.

3. Backfill can be inclined.

4. There would be friction between the wall and the soil.

5. Failure plane is assumed to be plane surface (actually curved)

6. A sliding wedge is assumed to be a rigid body.

Explanation:

Due to the friction assumed in Coulomb’s method  the active earth pressure decreases and passive earth pressure increases as compared to Rankine’s method.

Rankine’s earth pressure theory:

(1) Overestimates the active earth pressure.

(2)  Underestimates the passive Earth pressure.

Important point:

On compacting the soil:

(1) Active earth pressure decreases

(2) Passive Earth pressure increases.  

Concept:

It is assumed that the vertical stress and the lateral pressure acting on soil element are conjugate stresses i.e. the direction of forces are parallel to the plane on which the other acts.

The earth pressure coefficients are given as:

\({K_a} = \frac{{\cos \beta - \sqrt {{{\cos }^2}\beta - {{\cos }^2}\phi } }}{{\cos \beta + \sqrt {{{\cos }^2}\beta - {{\cos }^2}\phi } }} \times \cos \beta\)

\({K_p} = \frac{{\cos \beta + \sqrt {{{\cos }^2}\beta - {{\cos }^2}\phi } }}{{\cos \beta - \sqrt {{{\cos }^2}\beta - {{\cos }^2}\phi } }} \times \cos \beta\)   

Calculation:

ϕ = 30°, β = 20°

\({K_p} = \frac{{\cos \beta + \sqrt {{{\cos }^2}\beta - {{\cos }^2}\phi } }}{{\cos \beta - \sqrt {{{\cos }^2}\beta - {{\cos }^2}\phi } }} \times \cos \beta = \;\frac{{\cos 20 + \sqrt {{{\cos }^2}20 - {{\cos }^2}30} }}{{\cos 20 - \sqrt {{{\cos }^2}20 - {{\cos }^2}30} }} \times \cos 20\) = 2.13

Concept:

According to Swedish method,

 Factor of safety \( = \frac{{Resisting\;Moment}}{{Overturning\;Moment}}\)

Calculation:

Given, Total shear force = 480 kN

Total normal force = 1950 kN

Length of arc, L = 22 m

Radius of failure arc, R = 15 m

C = 24 kN/m² and  ϕ = 6°

The factor of safety with respect to shear strength is given by,

\(FOS = \frac{{CLR + Normal{\rm{ }}force \times R\tan \tan \phi }}{{shear{\rm{ }}force \times R}}\)

∴ \(FOS = \frac{{24 \times 22 \times 15 + 1950 \times 15\tan \tan 6^\circ }}{{480 \times 15}} = 1.53\)

Concept:

Critical depth of unsupported cut is given by:

\({{\rm{H}}_{\rm{c}}} = \frac{{4{\rm{c'}}}}{{{\rm{\gamma }}\sqrt {{{\rm{k}}_{\rm{a}}}} }}\)

C’ = effective cohesion

γ = unit weight of soil

\({{\rm{k}}_{\rm{a}}} = {\rm{active\;earth\;pressure\;coefficient}} = \frac{{1 - {\rm{sin\;}}\phi {\rm{'}}}}{{1 + {\rm{sin\;}}\phi {\rm{'}}}}\)

Calculation:

Case: 1 when c’ = 25 kN/m², ϕ’ = 20°

\({{\rm{H}}_{{{\rm{c}}_1}}} = \frac{{4 \times 25}}{{18\sqrt {{{\rm{k}}_{{{\rm{a}}_{\rm{'}}}}}} }}\)

\({{\rm{k}}_{{{\rm{a}}_1}}} = \frac{{1 - \sin 20^\circ }}{{1 + \sin 20^\circ }} = 0.490\)

\({{\rm{H}}_{{{\rm{c}}_1}}} = \frac{{4 \times 25}}{{18\sqrt {0.490} }} = 7.936{\rm{m}}\)

Case: 2 \({\rm{c'}} = 25{\rm{kN}}/{{\rm{m}}^3},\phi {\rm{'}} = \phi _2^1,{\rm{\;}}{{\rm{H}}_{{{\rm{c}}_2}}} = 1.1{{\rm{H}}_{\rm{c}}}\)

\({{\rm{H}}_{{{\rm{c}}_2}}} = 1.1 \times 7.936 = 8.729{\rm{m}}\)

\(8.729 = \frac{{4 \times 25}}{{18\sqrt {{{\rm{k}}_{{{\rm{a}}_2}}}} }}\)

\({{\rm{k}}_{{{\rm{a}}_2}}} = 0.405\)

\(\frac{{1 - {\rm{sin\;}}\phi _2^{\rm{'}}}}{{1 + {\rm{sin\;}}\phi _2^{\rm{'}}}} = 0.405\)

\(1 - {\rm{sin\;}}\phi _2^{\rm{'}} = 0.405 + 0.405{\rm{\;sin\;}}\phi _2^{\rm{'}}\)

\(0.595 = 1.405{\rm{\;sin\;}}\phi _2^{\rm{'}}\)

Concept:

The passive earth pressure at any depth ‘H’ is given by:

P_p = k_pq + 2c√k_p + k_pγH

Where k_p = Coefficient of earth pressure in passive state and is given by:

\({{\rm{k}}_{\rm{p}}} = \frac{{1 + \sin \phi }}{{1 - \sin \phi }}\)

Where,

q = uniform surcharge pressure, H = Height of retaining wall, γ = unit weight of back fill soil, and c = cohesion of backfill soil

Moment about base (Point B) is given by:

\({\rm{M}} = \left( {{{\rm{F}}_1} + {{\rm{F}}_2}} \right) \times \frac{{\rm{H}}}{2} + {{\rm{F}}_3} \times \frac{{\rm{H}}}{3}\)

F₁ = k_pqH, F₂ = (2c√k_p) × H, F₃ = 0.5 [k_pγH²]

Calculation:

Since soil is purely cohesive, Hence ϕ = 0° and k_p = 1

Given: H = 10 m

Considering unit wall length:

C = 35 kN/m²,

F₁ = 1 × q × 10 = 10 q kN/m, F₂ = 2 × 35 × 10 = 700 kN/m, and F₃ = 0.5 [1 × 20 × 10²] = 1000 kN/m

Moment about base \(= \left( {10{\rm{q}} \times \frac{{10}}{2}} \right) + 700 \times \frac{{10}}{2} + 1000 \times \frac{{10}}{3}\)

∴ 9000 = 50 × q + 3500 + 3333.33

q = 43.33 kN/m²

Concept:

Active earth pressure is given as:

\({P_A} = {K_a}\;{\bar \sigma _z} - 2c\sqrt {{K_a}} \)

Where, \({K_a} = \frac{{1 - \sin \phi }}{{1 + \sin \phi }}\)

σ̅z = Effective vertical stress

c = Cohesion

Rankine’s earth pressure theory was given for cohesionless soil and later on modified for C - ϕ soils.

In the case of cohesive soil, there is a tendency of the development of tension in the soil in the upper reaches.

Calculation:

Dry sand, c = 0

ϕ = 30°, γ_d = 16 kN/m³, γ_sat = 22 kN/m³, C = 0

\({K_a} = \left( {\frac{{1 - \sin 30^\circ }}{{1 + \sin 30^\circ }}} \right) = \frac{1}{3}\)

γ_sub = γ_sat - γ_w = 22 - 9.81 = 12.19 kN/m³

\({P_a} = {K_a}\;{\bar \sigma _z} - 2c\sqrt {{K_a}} \)

At Point A:

σ̅_z = q = 20 kN/m²

\(\therefore {P_A} = \frac{1}{3} \times 20 = \frac{{20}}{3}\;kN/{m^2}\)

At point B:

σ̅_z = q + γ_d × 3 = 20 + 16 × 3 = 68 kN/m²

\(\therefore {P_B} = {K_a}\;{\bar \sigma _z} - 0 \Rightarrow {P_B} = \frac{1}{3} \times 68 = \frac{{68}}{3}\;kN/{m^2}\) 

At point C:

There will be additional pressure due to water table.

σ̅_z = q + γ_d × 3 + γ_sub × 5

\(\therefore {P_C} = \frac{1}{3}\left( {20 + 16 \times 3 + 12.19 \times 5 } \right) +9.81 \times 5 = 92.03\;kN/{m^2}\) 

Hence the active earth pressure diagram is

The total active earth thrust is equal to the area under the pressure diagram.

Force per unit length of the wall = Area under pressure diagram × 1 m

Area under pressure diagram = area of trapezium AA’BB’ + area of trapezium BB’CC’

\(= \left( {\frac{{6.67\; + \;22.67}}{2}} \right) \times 3 + \left( {\frac{{22.67\; + \;92.03}}{2}} \right) \times 5\) = 330.76 kN/m

∴ Force per unit length of the wall = 330.76 × 1 = 330.76 kN.

Concept:

Factor of safety with respect to cohesion \(= \frac{{Cohesive\;strength}}{{Mobilised\;cohesion}}\)

Taylor’s stability Number is a dimensionless parameter and for purely cohesive soil, it is given as

\({S_n} = \frac{c}{{{F_c}{\gamma _t}H}}\)

Where, c = cohesion, Fc = Factor of safety with respect to cohesion, and H = Depth of excavation.

Also, S_n = cos β × sin β

Calculation:

c = 30 kN/m², β = 20°, γ_t = 16 kN/m³

Mobilised cohesion = 0.75 C

Factor of safety = \(\frac{{Cohesive\;strength}}{{Mobilised\;cohesion}}\) \(\Rightarrow \frac{C}{{{F_C}}} = {c_m} \Rightarrow \frac{C}{{{F_C}}} = 0.75\;c = 0.75 \times 30 = 22.5\)

Also, S_n = cos β × sin β = cos 20° × sin 20° = 0.321 

\({S_n} = \frac{C}{{{F_C}\; \times \;{\gamma _{t\;}}\; \times \;H}} \Rightarrow H = \frac{C}{{{F_C}\; \times \;{\gamma _t} \;\times\; {S_n}}} = \frac{{22.5}}{{16\; \times \;0.321}}\) = 4.38 m

Important Point:

Taylor’s stability number (sn) = cos β × sin β

Where, β = slope angle

Theoretical maximum value of stability number is 0.5 and practically its value is in the range of 0.25 - 0.30.

Concept:

Factor of safety \(= \frac{{Resisting\;force}}{{Actuating\;force}}\)

The actuating force is the weight of the soil block and the seepage force and the resisting force is the shear strength of the soil.

While calculating actuating force saturated unit weight of soil is used in place of submerged unit weight and the seepage force is not accounted separately.

Calculation:

β = 20°, ϕ = 30°

For saturated conditions, S = 1

\({\gamma _{sat}} = \frac{{\left( {{G_s} + Se} \right){\gamma _w}}}{{1 + e}} \Rightarrow {\gamma _{sat}} = \frac{{\left( {2.7 + 0.7} \right) \times 9.81}}{{1 + 0.7}} = 19.62\;kN/{m^3}\)

\({\gamma _{sub}} = \frac{{\left( {{G_s} - 1} \right){\gamma _w}}}{{1 + e}} = \frac{{\left( {2.7 - 1} \right)}}{{1 + 0.7}} \times 9.81 = 9.81\;kN/{m^3}\)

Actuating force = γ_sat × H cos β × sin β

Resisting force = γ_sub × (H cos β) × cos β × tan ϕ

FOS = \(\frac{{9.81\; \times \;H\; \times {{\cos }^2}20\; \times \tan 30\;}}{{19.62\; \times H\cos 20\; \times \sin 20}} = 0.79\)