Concept:

Newmark’s influence chart: It is a graphical method used to compute vertical, horizontal and shear stress due to uniformly distributed load over an area of any shape or geometry, below any point that lies either below or outside the loaded area. It is based on Boussinesq’s equation.

All the area unit will have equal influence at the center and whether the area unit is inside or outside the loaded area and it will have same influence at the center of the chart where vertical stress is to be found.

Hence all the areas will have same influence at the centre and the pressure exerted due to the influence areas marked will be equal.

Concept:

Pore pressure coefficients:

Pore pressure coefficients are used to express the response of pore water pressure to the changes in total stress under undrained condition. The values of coefficients maybe determined in the laboratory and can be used to predict the pore water pressure in field under similar stress condition.

ΔU = B[Δσ₃ + A(Δσ₁ – Δσ₃)] = B.Δσ₃ + AB (Δσ₁ – Δσ₃)

Let, ΔU = ΔU₁ + ΔU₂

Where ΔU₁ = B.Δσ₃

\(\Rightarrow B = \frac{{{\rm{\Delta }}{U_1}}}{{{\rm{\Delta }}{\sigma _3}}}\)

Where, ΔU₁ = Change in pore water pressure due to an increase in cell pressure.

ΔU₂  = AB (Δσ₁ – Δσ₃)

      ΔU₂ = Change in pore pressure due to increase in deviator stress (Δσ₁ – Δσ₃)    

For a completely saturated soil, B = 1

For a completely dry soil, B = 0

Also, A × B = A_f

Calculation:

Δσ₃ = 4 – 2 = 2 MPa

ΔU₁ = 2 – 1 = 1 MPa

\(\therefore B = \frac{{{\rm{\Delta }}{U_1}}}{{{\rm{\Delta }}{\sigma _3}}} = \frac{1}{2} = 0.5\)

Also, \({A_f} = AB \Rightarrow 0.4 = 0.5 \times A\)  

\(\Rightarrow A = \frac{{0.4}}{{0.5}} = 0.8\)

\(\frac{{{\sigma _1} - {\sigma _3}}}{2}\; = \;c\cos \phi \; + \;\frac{{{\sigma _1}\; + \;{\sigma _3}}}{2}\sin \phi \) – (1)

Comparing the given equation with equation – (1)

We get, \(c\cos \phi \; = \;20\sqrt 2\)

\(\begin{array}{l}\tan \beta \; = \;\sin \phi \; = \;\frac{1}{{\sqrt 2 }}\\\phi \; = \;{\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right)\; = \;45^\circ \\c\; = \;\frac{{20\sqrt 2 }}{{\cos \phi }} \Rightarrow 40\;kPa\end{array}\)

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.

Note :

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.

From the plastic equilibrium equation:

For effective stress

\(\sigma _1' = \sigma _3' \times {\rm{ta}}{{\rm{n}}^2}\alpha + 2 \times c \times \tan \alpha \)

\(({\sigma _1} - u) = \left( {{\sigma _3} - u} \right) \times {\rm{ta}}{{\rm{n}}^2}\alpha + 2 \times c \times {\rm{tan\alpha }}\)

From observation No. 1

\(295 = 85 \times {\rm{ta}}{{\rm{n}}^2}\alpha + 2 \times c \times {\rm{tan\alpha }}\) ….. (i)

From observation No. 2

\(745 = 220 \times {\rm{ta}}{{\rm{n}}^2}\alpha + 2 \times c \times {\rm{tan\alpha }}\) …… (ii)

Subtracting equation (ii) and (i)

450 = 135 × tan²α

∴ tan²α = 3.33

∴ α = 61.289°

∴ ϕ = 45 + ϕ/2

∴ ϕ = 32.58°

Substituting the value of ϕ in the equation (i), we get

Concept:

Relation between major and minor principal stress at failure on the basis of Mohr-Coulomb criteria of failure:

\({\sigma _{1f}} = {\sigma _{3f}}{\tan ^2}\left( {45 + \frac{\phi }{2}} \right) + 2C\tan \left( {45 + \frac{\phi }{2}} \right)\)

\({\sigma _{3f}} = {\sigma _{1f}}{\tan ^2}\left( {45 - \frac{\phi }{2}} \right) - 2C\tan \left( {45 - \frac{\phi }{2}} \right)\)

Calculation:

σ_3f = Cell pressure = 100 kPa

σ_1f = Cell pressure + Deviator stress = 100 + 150 = 250 kPa

For dry sand, C = 0

∴ 250 = 100 tan² (45 + ϕ/2) + 0

⇒ ϕ = 25.377°

When cell pressure becomes 200 kPa, let the major principal stress at failure becomes σ_1f

σ_3f = 200 kPa

\({\sigma _{3f}} = {\sigma _{1f}}{\tan ^2}\left( {45 - \frac{\phi }{2}} \right) - 0\)

\( \Rightarrow 200 = {\sigma _{1f}}{\tan ^2}\left( {45 - \frac{{25.377}}{2}} \right)\)

⇒ σ_1f ≈ 500 kPa       

∴ Deviator stress = 500 – 200 = 300 kPa

Concept:

Vane shear test is used in plastic cohesive soil which is very sensitive and undisturbed sample cannot be easily obtained. It can be used both in field and lab testing.

The  shear stress distribution across the blade is as shown below.

When test is done such that both the faces are subjected to shearing:

\(\tau = \frac{T}{{\pi {D^2}\left( {\frac{L}{2} + \frac{D}{6}} \right)}}\)

When test is done such that only one face is subjected to shearing:

\(\tau = \frac{T}{{\pi {D^2}\left( {\frac{L}{2} + \frac{D}{{12}}} \right)}}\)

Calculation:

T = 50 N-m, L = 25 cm = 0.25 m, D = 10 cm = 0.1 m

Here only the bottom face is subjected to shearing.

\(\therefore \tau = \frac{T}{{\pi {D^2}\left( {\frac{L}{2} + \frac{D}{{12}}} \right)}}\)

\(\Rightarrow \tau = \frac{{50}}{{\pi \times {{0.1}^2}\left( {\frac{{0.25}}{2} + \frac{{0.1}}{{12}}} \right)}}\)

τ = 11936.62 N/m²

Concept:

Given Data: Load = 6251 t and Size of footing = 2 m × 2 m

Let the intensity of stress is \({\left( {\frac{1}{9}} \right)^{th}}\) of the stress at ground level at a distance x.

So, Area \(= \left( {2 + \frac{{\rm{x}}}{2} + \frac{{\rm{x}}}{2}} \right)\left( {2 + \frac{{\rm{x}}}{2} + \frac{{\rm{x}}}{2}} \right)\) = (2 + x)²

So, \(\frac{{6251}}{{2 \times 2}} \times \frac{1}{9} = \frac{{6251}}{{{{\left( {2 + {\rm{x}}} \right)}^2}}}\)

\(\Rightarrow 36 = \frac{1}{{{{\left( {2 + x} \right)}^2}}}\)

⇒ (4 + x² + 4x) = 36

⇒ x² + 4x – 32 = 0

⇒ x² + 8x – 4x – 32 = 0

⇒ x(x + 8) - 4(x + 8) = 0

⇒ (x - 4)(x + 8) = 0

x = 4 m & -8 m (-ve) (Negative depth can’t be taken)

So answer will be 4 m.

Concept:

Vertical stress due to concentrated load is given as per Boussinesq’s equation.

Formula:

\({{\rm{\sigma }}_{\rm{z}}}{\rm{\;}} = {\rm{\;}}\frac{3}{{2{\rm{\pi }}}}\frac{{\rm{Q}}}{{{{\rm{z}}^2}}}{\left( {\frac{1}{{1 + {{\left( {\frac{{\rm{r}}}{{\rm{Z}}}} \right)}^2}}}} \right)^{\frac{5}{2}}}\)

where,

σz = Stress magnitude, Q = Vertical point load, r = Radial distance, and Z = Vertical distance.

Calculation:

Q = 4000 kN, r = 3m, z = 8 m

\(\therefore {\sigma _A} = \frac{{3A}}{{2\pi {z^2}}}{\left( {\frac{1}{{1 + {{\left( {\frac{r}{z}} \right)}^2}}}} \right)^{5/2}}\)

\(= \frac{{3\; \times \;4000}}{{2\; \times \;\pi\; \times \;{8^2}}}{\left( {\frac{1}{{1 + {{\left( {\frac{3}{8}} \right)}^2}}}} \right)^{\frac{5}{2}}}\)  = 21.477 kN/m²

Important Point:

As per Boussinesq’s stress directly below a point load at any distance ‘z’ is given as

\({{\rm{\sigma }}_{\rm{z}}}{\rm{\;}} = {\rm{\;}}\frac{3}{{2{\rm{\pi }}}}\frac{{\rm{Q}}}{{{{\rm{z}}^2}}}\)

Difference between Boussinesq’s and Westergaard’s equation of vertical stress :

Axial force applied to the specimen is “x” kg. Unconfined compressive strength of clay (q_u) = 2C

Where,

C = cohesion (k N/m²)

C = 2.5 kg/cm²

q_u = 2 x 2.5 = 5 kg/cm²

q_u = (σ₁)_f  = \(\frac{P}{{{A_f}}}\)

Where, (σ₁)_f = total applied axial stress to the specimen

A_f = Final area of the specimen

A_f  = \(\frac{{{A_0}}}{{1 - {\varepsilon _0}}}\)

ε₀ = Longitudinal strain = \(\frac{{{\rm{\Delta }}L}}{L} = \frac{{10}}{{90}} = 0.111\)

\({A_0} = \frac{\pi }{4} \times {\left( 4 \right)^2} = 12.567c{m^2}\)

\({A_f} = \frac{{12.567}}{{1 - 0.111}} = 14.136\;c{m^2}\)

5 = q_u = \(\frac{P}{{{A_f}}} = \frac{X}{{{A_f}}} = \frac{X}{{14.136}}\)