Soil Mechanics Test 1

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:

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 .

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