Soil Mechanics Test 1

Concept:

Sensitivity is defined as the ratio of undisturbed strength to that of the remoulded strength.

\(Sensitivity = \frac{{Undisturbed\;strength}}{{Remoulded\;strength}}\)

Calculation:

q_{u(undisturbed)} = 60 kPa

q_u(remoulded) = 12 kPa

Sensitivity \(= \frac{{{{\left( {{q_u}} \right)}_{undisturbed}}}}{{{q_{u\;\left( {remoulded} \right)}}}} = \frac{{60}}{{12}} = 5\)

∵ Sensitivity lies between 4 – 8.

∴ It is a sensitive soil.

Suitability of Compaction Equipment

Concept:

Vertical stress increment at a any depth (z) at a radial distance (r) is given by:

\({\rm{\Delta }}{\sigma _z} = {K}\frac{Q}{{{z^2}}}\)

Where,

Boussinesq’s influence factor (kB):

\({k_B} = \frac{3}{{2\pi }}{\left[ {\frac{1}{{1 + {{\left( {\frac{r}{z}} \right)}^2}}}} \right]^{\frac{5}{2}}}\)

Wester guard’s influence factor (kW):

\({k_w} = \frac{1}{\pi }{\left[ {\frac{1}{{1 + 2{{\left( {\frac{r}{z}} \right)}^2}}}} \right]^{\frac{3}{2}}}\)

When r/z > 1.5, 

Westergaard formula for vertical stress will give greater value of stress than that of by Boussinesq’s formula.

Concept:

Prevention of Erosion:

Two approaches are used to prevent possibility of erosion and piping.

1. Control of seepage and seepage force: By providing cut off wall, increasing flow path by providing impervious blanket.

2. Use of protective filter: Use of protective filter prevents erosion and reduces uplift pressure. A protective filter consists of one are more layer of coarse-grained material placed over a less pervious soil called the base.

• A filter will prevent the migration of finer particles but without inhibiting the flow of seepage water, so there is hardly any loss of head. This ensures that within the filter itself, seepage forces are reduced.
• If voids in the filter are much larger than the fine grains of the protected material, these grains are likely to be washed into the voids of the filter material and would ultimately obstruct the free flow.
• On the other hand if voids are too small, seepage forces will develop to unacceptable levels

Filter specifications:

1. \(\frac{{{D_{15\left( {filter} \right)}}}}{{{D_{85\left( {protected\;material} \right)}}}} < 5\)

It ensures that no significant invasion of particles from the protected material to the filter. (Governs the upper limit of grain size of filter matter).

2. \(4 < \frac{{{D_{15\left( {filter} \right)}}}}{{{D_{15\left( {protected\;material} \right)}}}} < 20\)

It ensures that sufficient head is lost in filter without build up of seepage pressure (specifies the lower limit of material).

3. \(\frac{{{D_{50\left( {filter} \right)}}}}{{{D_{50\left( {protected\;material} \right)}}}} < 25\)

This is the additional guideline for the selection of material.

Concept:

From Darcy’s Law of ground water flow, it is known that,

\({\rm{Q}} = {\rm{kiA}}\)

Where Q is discharge or rate of flow through the aquifer,

k is hydraulic conductivity of the aquifer,

A is cross sectional area of the aquifer,

i is hydraulic gradient and given by the ratio of difference in hydraulic heads to length of the aquifer.

Calculation:

\({\rm{Hydraulic\;gradient}},{\rm{\;i}} = {\rm{\;}}\frac{{{\rm{Difference\;in\;hydraulic\;heads}}}}{{{\rm{length\;of\;the\;aquifer}}}} = \frac{{300 - 250}}{{1500}} = \frac{1}{{30}}\)

Discharge, Q = 750 m³/day

Cross sectional area, A = 1.0 × 10⁴ m².

Q = kiA

\(750 = {\rm{k}} \times \frac{1}{{30}} \times 1.0{\rm{\;x\;}}{10^4}{\rm{\;\;}}\)

∴ \({\rm{k}} = 2.25\frac{{\rm{m}}}{{{\rm{day}}}}\) 

 ∴ The hydraulic conductivity of the aquifer is 2.25 m/day.

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.

Stress below any point is calculated as :-

\({{\rm{\sigma }}_{\rm{z}}}{\rm{\;}} = {\rm{\;}}\frac{1}{{{\rm{m}} \times {\rm{n}}}} \times {\rm{q}} \times {\rm{N\;}}\)

where,

m = No. of concentric circles, n = No. of Radial lines, q = Intensity of load, and N = Equivalent number of areas covered by plan area.

The product \(\left( {\frac{1}{{m \times n}}} \right)\) is called influence factor.

Calculation:

m = 10, n = 30

Concept:

\({\gamma _t} = \frac{{{G_s}\left( {1\; + \;w} \right){\gamma _w}}}{{1\; + \;e}}\)

Relative Density or Density Index: Degree of denseness and coarseness of natural deposits of coarse grained soil is measured by its relative density.

\({D_r} = \frac{{{e_{max}} - {e_{natural}}}}{{{e_{max}} - {e_{min}}}} \times 100\)

Calculation:

G_s = 2.7, w = 20%, γ_t = 18 kN/m³, γ_w = 9.81 kN/m³

Under natural conditions:-

\({\gamma _t} = \frac{{{G_s}\left( {1\; + \;w} \right){\gamma _w}}}{{1\; + \;e}} \Rightarrow 18 = \frac{{2.7\; \times \;\left( {1\; + \;0.2} \right)\; \times \;9.81}}{{1\; + \;e}}\) ⇒ e = 0.7658 

e_max = 0.90 and e_min = 0.30

\(\therefore {D_r} = \frac{{{e_{max}} - {e_{nat}}}}{{{e_{max}} - {e_{min}}}} \times 100 = \frac{{0.9 - 0.7658}}{{0.9 - 0.3}} \times 100\) = 22.37 %

Concept:

For Direct shear test:

Shear strength of soil (τ) = kN/A

k = proving ring constant, N = Dial gauge reading and A = failure area

Using Mohr - Columbus Criteria

Shear strength of soil specimen is given by

τ = c’ + σ’ tan∅

c’ = effective cohesion, σ' = effective normal stress

Relation between major and minor principle stress in terms of C and ∅

σ₁ = σ₃ tan² (45 + ∅/2) + 2c {tan (45 + ∅/2)}

σ₁ = Major principle test, and σ₃ = Minor principle test

Calculations:

Soil specimen 1:

Deviator stress (σ_d) = 500 kN/m²

Cell Pressure is the minor principle stress (σ₃) = 400 kN/m²

Major principle stress (σ₁) = σ_d + σ₃ = 500 + 400 = 900 kN/m²

Using the relation:

σ₁ = σ₃ tan² (45 + ∅/2) + 2c {tan (45 + ∅/2)}       ……. (∅ = 0)

900 = 400 × tan² (45) + 2c₁ tan (45)

c₁ = 250 kN/m²

Soil Specimen 2:

Given that k = 0.0001 N/μm, N = 50, A = 100 cm²

\({\rm{\tau }} = \frac{{0.0001 \times 50 \times {{10}^6}}}{{100 \times 1000 \times {{10}^{ - 4}}}} = 500{\rm{\;kN}}/{{\rm{m}}^2}\)

Using Mohr - Columbus Criteria

= c’+ σ’ tan∅

500 = c₂ + 600 × tan (30)

c₂ = 153.59 kN/m²

Concept:

The unit weight of soil mass is determined using Water displacement method in which clay sample is coated with paraffin wax of known volume.

The dry density of soil is given by \({{\rm{\gamma }}_{\rm{d}}} = \frac{{\left( {\rm{G}} \right) \times {\gamma _w}}}{{1 + e}}\)  

Degree of Saturation (S): Degree of Saturation of a soil mass is defined as the ratio of the volume of water in the voids to the total volume of voids.

\(S = \frac{{{V_w}}}{{{V_v}}} \times 100\;\;;0 \le S \le 100\)

• For a fully saturated soil mass Vv = Vw , hence S = 100%
• For fully dry soil mass Vw = 0, hence S = 0%

For partially saturated soil mass degree of saturation varies between 0 – 100%.

Calculation:

Specific gravity of wax = 0.8

Mass of wax = 30 gms

∴ Mass of soil = 700 – 30 = 670 gms

∴ Volume of wax \(= \frac{{30}}{{0.8\; \times \;1}} = 37.5\;ml\) = 37.5 cc

∴ Volume of the soil specimen without wax = 400 – 37.5 = 362.5 ml = 362.5 cc

Bulk unit weight \(= \frac{{Weight\;of\;soil}}{{Volume\;of\;soil}} = \frac{{670}}{{362.5}}\) = 1.848 gm/cc

Also,

\({\gamma _d} = \frac{{{\gamma _t}}}{{1\; + \;w }} \Rightarrow {\gamma _d} = \frac{{1.848}}{{1\; + \;0.18}} \Rightarrow 1.566\;gm/cc\)

Also, \(\gamma_d = \frac{{{G_s}{\gamma _w}}}{{1\; + \;e}} \Rightarrow 1.566 = \frac{{2.7\; \times \;1}}{{1\; + \;e}}\)

⇒ e = 0.724

\(eS = w {G_s} \Rightarrow S = \frac{{w {G_s}}}{e} = \frac{{2.7\; \times \;0.18}}{{0.724}}\) = 0.6712 ≈ 67.12%

For computing pile group capacity:

a) Piles acting individually:

c = q_u/2 = 120/2 = 60 kN/m²

Q_up = A_p × r_p + A_S × r_f

Where A_p = \(\frac{{\rm{\pi }}}{4}{\left( {0.25} \right)^2}\;\)= 0.04909 m² and A_S = π × (0.25) × 10 = 7.854m²

r_p = c × N_c = 9 c = 9 × 60 = 540 kN/m²

r_f = m × c = 0.75 × 60 = 45 kN/m²

∴ Q_up = 0.04909 × 540 + 7.854 × 45 = 380 kN

∴ Load capacity of 9 piles = 9 × 380 = 3419 kN                     ……. (i)

b) Pile acting as a group:

B = 2 x s + d = 2 × 1 + 0.25 = 2.25 m

Q_up = A_p × r_p + A_S × r_f

Where A_p = B² = 2.25 × 2.25 = 5.0624 m²; r_p = 9 c = 9 × 60 = 540 kN/m²

A_s = 4 B × 10 = 4 × 2.25 × 10 = 90 m²; r_f = c = 60 kN/m²

∴ Q_ug = 5.0625 × 540 + 90 × 60 = 2733.8 + 5400 = 8133.8 kN           ……. (ii)

Adopting minimum value of (i) and (ii), the pile group capacity is considered to be 3419 kN

\(\therefore {{\rm{Q}}_{\rm{a}}} = \frac{{{{\rm{Q}}_{{\rm{u}}.{\rm{min}}}}}}{{\rm{F}}} = \frac{{3419}}{{2.5}} = 1367.6{\rm{\;kN}}\)

Concept:

Seepage loss per metre length of the dam is given by:

\(q = kh\left( {\frac{{{N_f}}}{{{N_d}}}} \right)\) 

Where,

K = permeability

H = head loss due to seepage

N_f = Number of flow channels

N_d = Number of equipotential drops

N_f = Number of flow lines-1

N_d = Number of equipotential lines-1

Calculation:

K = 6 × 10^-5 m/min

K = 1 × 10^-6 m/sec

Number of flow lines = 9

Number of equipotential lines = 20

N_f = 11 – 1 = 10

N_d = 20 – 1 = 19

Head under which flow is taking place or head loss due to seepage = 5 m.

\(q = {10^{ - 6}} \times 5 \times \frac{{10 \times 1}}{{19}} = 2.631 \times {10^{ - 6}}{m^3}/s/m\) 

q = 0.2274 m³/day/m

Concept:

Relation between settlement of plate, foundation and width of plate, foundation is given as:

\(\frac{{{{\rm{S}}_{\rm{f}}}}}{{{{\rm{S}}_{\rm{p}}}}} = {\left[ {\frac{{{{\rm{B}}_{\rm{f}}}\left( {{{\rm{B}}_{\rm{p}}} + 0.3} \right)}}{{{{\rm{B}}_{\rm{p}}}\left( {{{\rm{B}}_{\rm{f}}} + 0.3} \right)}}} \right]^2}\)             ….. [For sands]

Where,

B_f = width of foundation in metres, B_p = Width of plate in metres, S_f = Settlement of foundation, and S_p = Settlement of plate

Load intensity of foundation = Load applied at base of foundation / Area of foundation at base

Corresponding to this load intensity, one can compute settlement in plate from the test data.

Calculation:

Load applied at base of foundation = 240 t

Area of foundation at base = 4 × 4 = 16 m²

Load intensity \(= \frac{{240}}{{16}} = 15{\rm{\;t}}/{{\rm{m}}^2}\)

Corresponding to load intensity of 15 t/m², settlement in plate can be interpolated.

Settlement in plate (S_p) = 4 mm

Width of plate = 45 cm = 0.45 m

Width of foundation = 4 m

\({{\rm{S}}_{\rm{f}}} = {\left( {\frac{{4 \times \left( {0.45 + 0.3} \right)}}{{0.45 \times \left( {4 + 0.3} \right)}}} \right)^2}\)

S_f = 9.615 mm

Concept:

Relation between Major and minor principal stress at failure in a soil mass on the basis of Mohr-Coulomb criteria of failure:-

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

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

Calculation:

Cell pressure (σ_3f) = 100 kN/m²

Deviator stress (σ_a) = 80 kN/m²

∴ Major principal stress at failure (σ_1f) = σ₃ + σ_a = 100 + 80 = 180 kPa

Let the pore water pressure at failure be U.

∴ σ̅_3f = σ_3f – U = 100 – U

σ̅_1f = σ_1f – U = 180 – U

C’ = 0, ϕ’ = 25°

By using Mohr-Coulomb hypothesis,

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

\(\Rightarrow \left( {180 - U} \right) = \left( {100 - U} \right){\tan ^2}\left( {45 + \frac{{25}}{2}} \right) + 0\)

⇒ U = 45.35 kPa          

Important Point:

Types of Triaxial Test:

• Triaxial test is performed in 2 stages-

Hence there are 3 types of triaxial test:

1. C D → Consolidated  Drained test

2. C U → Consolidated undrained test

3. U U → Unconsolidated undrained test

4. U D → Not possible

Height of sand layer above water table = Z₁ = 4 m, and height of saturated layer = 12 – 4 = 8 m

Depth of point X, where pressure is to be computed = 10 m

Height of saturated layer above X = Z₂ = 10 – 4 = 6 m

Now \({{\text{ }\!\!\gamma\!\!\text{ }}_{\text{d}}}=\frac{\text{G}\times {{\text{ }\!\!\gamma\!\!\text{ }}_{\text{w}}}}{1+\text{e}}=\frac{2.65\times 9.81}{1+0.7}=15.29\text{ }\!\!~\!\!\text{ kN}/{{\text{m}}^{3}}\)

For sand above water table:

\(\text{w}=\frac{\text{e}\times {{\text{S}}_{\text{r}}}}{\text{G}}=\frac{0.7\times 0.5}{2.65}=0.132\)

γ₁ = γ_d (1 + w) = 15.29 × 1.132 = 17.31 kN/m³

For saturated sand below water table:

\({{\text{w}}_{\text{sat}}}=\frac{\text{e}}{\text{G}}=\frac{0.7}{2.65}=2.64\)

γ₂ = γ_d (1 + w_sat) = 15.29 × 1.264 = 19.33 kN/m³

γ₂’ = 19.33 – 9.81 = 9.52 kN/m³

Effective pressure at X

σ = Z₁ γ₁ + Z₂ γ₂ = 4 × 17.31 + 6 × 19.33 = 185.22 kN/m²

u = h_w γ_w = 6 × 9.81 = 58.86 kN/m²

σ’ = σ – u = 185.22 – 58.86 = 126.36 kN/m²

∴ σ’ = σ – u = 185.22 – 58.86 = 126.36 kN/m²

Effective stress at X after capillary rise

σ' = 3 γ₁ + (6 + 1) γ₂’ + h_c γ_w = (3 × 17.31) + (7 × 9.52) + (1 × 9.81) = 128.38 kN/m²

∴ Increase in pressure = 128.38 – 126.36 = 2.02 kN/m²