Concept:

Activity Number (A_C): As per Skempton, volume changes during swelling or shrinkage. Activity no. is used to study the swelling behaviour of soil.

Activity number, \({A_C} = \frac{{{I_P}}}{{\% \;of\;clay\;size\;particle\;\left( {i.e.\; < 2\mu } \right)in\;soil}}\)

A_C < 0.75 → Inactive

0.75 < A_C < 1.25 → Normal active

A_C > 1.25 → Active

Plasticity Index: The range of consistency within which soil behaves as a plastic material is called Plasticity Index. The property is due to the presence of clay minerals.

I_P = w_L - w_P

Calculation:

w_L = 60%, w_p = 40%

I_P = w_L - w_P = 60 – 40 = 20%

Activity \(= \frac{{{I_P}}}{{\% \;of\;clay\;size\;particles}}\)

\(\Rightarrow \frac{{20}}{{\% \;of\;particles\;( < 0.002\;mm)}} = 0.8\)

⇒ % of particles \(= \frac{{20}}{{0.8}} = 25\%\)    

Concept:

The grain size distribution of fine soil is determined using sedimentation analysis. Sedimentation analysis is performed using two methods:

1) Pipette method

2) Hydrometer method.

A hydrometer is a device used to measure the specific gravity of liquids.

The following three corrections are necessary for :

1. Meniscus correction: Since the hydrometer readings increase downward on the stem, the meniscus correction (C_m) is always positive.

2. Temperature correction: If the temperature at the time of the test is more than that of calibration of the hydrometer, the observed reading will be less and the correction (C_t) would be positive and vice versa.

Concept:

Liquidity index is given by (I_L)

\({{\rm{I}}_{\rm{L}}} = \frac{{{{\rm{W}}_{\rm{n}}} - {\rm{\;}}{{\rm{W}}_{\rm{p}}}}}{{{{\rm{I}}_{\rm{P}}}}}\)

Consistency index is given by (I_C)

\({{\rm{I}}_{\rm{C}}} = \frac{{{{\rm{W}}_{\rm{L}}} - {\rm{\;}}{{\rm{W}}_{\rm{n}}}}}{{{{\rm{I}}_{\rm{P}}}}}\)

Calculation: 

\(\frac{{{{\rm{I}}_{\rm{C}}}}}{{{{\rm{I}}_{\rm{L}}}}} = \frac{{{{\rm{W}}_{\rm{L}}} - {{\rm{W}}_{\rm{n}}}}}{{{{\rm{W}}_{\rm{n}}} - {\rm{\;}}{{\rm{W}}_{\rm{p}}}}}\)

W_L = 60%, W_P = 30%, W_n = 40%

\(\frac{{{{\rm{I}}_{\rm{C}}}}}{{{{\rm{I}}_{\rm{L}}}}} = \frac{{60 - 40}}{{40 - 30}} = \frac{{20}}{{10}} = 2\)

Concept:

\(\gamma = \left( {\frac{{G + e \times {S_r} \times {S_o}}}{{1 + e}}} \right) \times {\gamma _w}\)

Where

γ = Density of the soil sample

G = Specific gravity of soil sample

e = Void ratio of soil sample

S_r = Degree of saturation of soil sample

S_o = Specific gravity of any other saturating fluid

γ_w = Density of water

Calculation:

\(2.2 = \left( {\frac{{2.65 + e \times 1 \times 0.92}}{{1 + e}}} \right) \times 1\)

2.2 + 2.2 × e = 2.65 + 0.92 × e

1.28 × e = 0.45

Concept:

Void ratio (e): \(e = \frac{{{V_v}}}{{{V_s}}};e > 0\)

• Void ratio is defined as the ratio of the volume of voids to the volume of solids.
• There is no upper limit of the void ratio in soil suspension.
• The void ratio of fine-grained soil is generally higher than that of coarse-grained soil.

Calculation:

Soil sample A:

e = 0.5, V = 1 m³

\(e = \frac{{{V_v}}}{{{V_{s1}}}} = 0.5 \Rightarrow {V_v} = 0.5\;{V_{s1}}\)

Also, \({V_v} + {V_{s1}} = 1\;{m^3} \Rightarrow 0.5\;{V_{s1}} + {V_{s1}} = 1 \Rightarrow {V_{s1}} = \frac{1}{{1.5}}\;\)

Soil sample B:

e = 0.6, V = 2m³

\(\because e = \frac{{{V_v}}}{{{V_s}}} \Rightarrow 0.6 = \frac{{{V_v}}}{{{V_{s2}}}} \Rightarrow {V_v} = 0.6\;{V_{s2}}\)

Again, \({V_v} + {V_{s2}} = 2{m^3} \Rightarrow 0.6\;{V_{s2}} + {V_{s2}} = 2 \to {V_{s2}} = \frac{2}{{1.6}}{m^3}\)

After mixing soil sample A and soil sample B,

Total volume of resultant sample = 2 + 1 = 3m³

Volume of solids in resultant sample = V_s1 + V_s2 \(= \left( {\frac{1}{{1.5}} + \frac{2}{{1.6}}} \right)\; = 1.92{m^3}\)

∴ Volume of voids = 3 – 1.92 = 1.08 m³

∴ Void ratio of resultant sample \(= \frac{{{V_v}}}{{{V_s}}} = \frac{{1.08}}{{1.92}}\) = 0.563

Important Point:

Porosity (n): \(n = \frac{{{V_v}}}{V} \times 100\;;100 > n > 0\)

Porosity is defined as ratio of volume of voids to the total volume expressed as a percentage.

The porosity of soil cannot exceed 100% hence it has an upper limit of 100%.

Concept:

The strength of soil at plastic limit is checked using Toughness index.

Toughness Index (I_t):

\({I_t} = \frac{{{I_P}}}{{{I_F}}} = \frac{{Plasticity\;index}}{{Flow\;index}}\)

It indicates the loss of shear strength with increase in moisture content.

Calculation:

For soil A:

w_L = 60%, w_p = 40%, I_f = 10

∴ I_P = 60 – 40 = 20

\(\therefore {\left( {{I_T}} \right)_A} = \frac{{{I_P}}}{{{I_f}}} = \frac{{20}}{{10}} = 2\)

For soil B:

w_L = 50%, w_P = 30%, I_f = 8

∴ I_P = 50 – 30 = 20%

\(\therefore {\left( {{I_T}} \right)_B} = \frac{{20}}{8} = 2.5\)

∴ (I_T)_B > (I_T)_A

⇒ Soil B has higher strength at plastic limit .

The volume of solids present in the finished volume of the embankment

\({{\rm{V}}_{\rm{s}}} = \frac{{\rm{V}}}{{1 + {\rm{e}}}}\)

V = 950000 m³

e = 0.78

\({{\rm{V}}_{\rm{s}}} = \frac{{950000}}{{1 + 0.78}} = 533707.86{{\rm{m}}^3}\)

Volume of solids require for the embankment is = 533707.86 m³

From Borrow Pit I:

V_I = V_s (1 + e_I)

e_I = 0.93

V_I = 533707.86 (1 + 0.93)

V_I = 1030056.18 m³

Cost of transportation for Borrow Pit I = 12 × 1030056.18

= Rs 12360674.16

From Borrow Pit II:

Cost of transportation of soil borrow Pit II = Rs 10 per m³

V_II = V_s (1 + e_II)

e_II = 1.65

V_II = 533707.86 (1 + 1.65)

V_II = 1414325.83 Rs

Cost of transportation for Borrow Pit II = 10 × 1414325.83

= Rs 14143258.29

Cost of transportation for Borrow Pit II is greater than that of Borrow Pit I by 

\({\rm} = \left( {\frac{{14143258.29 - 12360674.14}}{{14143258.29}}} \right) \times 100\)

= 12.60%

Hence, the cost of transportation of soil form Borrow pit I is lesser than that of from Borrow pit II by 12.60% ≅ 12.50% 

Concept:

Shrinkage limit: This is the minimum water content is a soil, at which the soil remains in semi-solid state.

Formula:

\({{\rm{W}}_{\rm{S}}} = {{\rm{W}}_1} - \frac{{\left( {{{\rm{V}}_1} - {{\rm{V}}_{\rm{d}}}} \right){{\rm{P}}_{\rm{w}}}}}{{{{\rm{M}}_{\rm{d}}}}}\)

\({{\rm{W}}_{\rm{S}}} = \frac{1}{{\rm{R}}} - \frac{1}{{\rm{G}}}\)

\({\rm{R}} = \frac{{{{\rm{\rho }}_{\rm{d}}}}}{{{{\rm{\rho }}_{\rm{w}}}}} = {\rm{Shrinkage\;ratio}}\)

ρ_d = Dry density of coil, ρ_w = Density of water, and G = Specific gravity of soil.

Calculation:

M_d = 88 grams, V_d = 50 cm³

P_d = 88/50 gm/cc = 1.76 gm/cc

Mass of soil sample (M_s) = 560 gms

Mass of soil sample and paraffin wax (M) = 575.60 gms

Mass of paraffin (M_p) = 575.60 – 560 = 15.60 gms

\(\therefore \text{Volume }\;\!\!~\!\!\text{ of }\;\!\!~\!\!\text{ paraffin }\;\!\!~\!\!\text{ wax }\!\!~\!\!\text{ }\left( {{\text{V}}_{\text{p}}} \right)=\frac{{{\text{M}}_{\text{p}}}}{{{\text{ }\!\!\rho\!\!\text{ }}_{\text{p}}}}\)

\(\therefore {{\text{V}}_{\text{p}}}=\frac{15.60}{0.90}=17.33\text{ }\!\!~\!\!\text{ c}{{\text{m}}^{3}}=17.33\text{ }\!\!~\!\!\text{ ml}\)

∴ Volume of soil = Volume of soil & paraffin – Volume of paraffin wax

∴ V_s = V – V_p = 320 – 17.33

∴ V_s = 302.67 ml

Density of soil (ρ_s) \(=\frac{{{\text{M}}_{\text{s}}}}{{{\text{V}}_{\text{s}}}}=\frac{560}{302.67}=1.85\text{ }\!\!~\!\!\text{ g}/\text{c}{{\text{m}}^{3}}\)

Dry density of soil (ρ_d) \(=\frac{{{\text{ }\!\!\rho\!\!\text{ }}_{\text{s}}}}{1+\text{w}}=\frac{1.85}{1+0.18}=1.57\text{ }\!\!~\!\!\text{ g}/\text{c}{{\text{m}}^{3}}\)

\(\text{Now},\text{ }{{\text{ }\!\!\rho\!\!\text{ }}_{\text{d}}}=\frac{\text{G}\times \text{Yw}}{1+\text{e}}\)

\(\Rightarrow 1.57=\frac{2.67\times 1}{1+\text{e}}\)

∴ e = 0.70

Now, S_r × e = w × G

\(\therefore {{\text{S}}_{\text{r}}}=\frac{\text{w}\times \text{G}}{\text{e}}=\frac{0.18\times 2.67}{0.7}=0.68657\)

Concept:

Water Content (w): Water content or moisture content of a soil is defined as the ratio of the weight of water to the weight of solids of the soil mass. The minimum value for water content is 0 and there is no upper limit for water content.

\(w = \frac{{{W_w}}}{{{W_s}}} \times 100\)

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 volume of voids.

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

• For a fully saturated soil mass V_v = V_w , hence S = 100%
• For fully dry soil mass V_w = 0, hence S = 0%

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

Calculation:

G_s = 2.7 , w = 20%, S = 60%

\(\because es = w{G_s} \Rightarrow e = \frac{{w{G_s}}}{S} = \frac{{0.2\; \times \;2.7}}{{0.6}} = 0.9\)

Again, the degree of saturation changes to 80%.

∵ The degree of saturation ≠ 100%, thus the soil is partially saturated and there is no change in the volume of soil. Hence the void ratio will remain the same.

When degree of saturation = 80%

\(es = \frac{{w{G_s}}}{s} \Rightarrow w = \frac{{es}}{{{G_s}}} = \frac{{0.9\; \times \;0.8}}{{2.7}}\) = 0.2667

Hence the new water content = 26.67%

∴ Percentage change in water content