Soil Mechanics Test 2

Concept:

An Aluminium ion has 3 positive charges and one silicon ion has 4 positive charges.

When one silicon ion replaced by one aluminum ion, there would be deficiency of one unit (4 – 3 = 1) positive charge per substitution since silicon ion has more positive charge then aluminum ion.

The soils formed at a place may be transported to other places by agents of transportation, such as water, wind, ice and gravity.

1) Water transported Soils: Flowing water is one of the most important agents of transportation of soils. Swift running water carries a large quantity of soil either in suspension or by rolling along the bed.

The size of the soil particles carried by water depends upon the velocity.

All type of soils carried and deposited by water are known as alluvial deposits. Deposits made in lakes are called lacustrine deposits. Such deposits are laminated or varved in layers. Marine deposits are formed when the flowing water carries soils to ocean or sea.

2) Wind transported Soils: Soil particles are transported by winds. The particle size of the soil depends upon the velocity of wind. Soils deposited by wind are known as aeolian deposits.

Loess is a silt deposit made by wind. These deposits have low density and high compressibility. The bearing capacity of such soils is very low.

3) Glacier-Deposited Soils: Glaciers are large masses of ice formed by the compaction of snow. As the glaciers grow and move, they carry with them soils varying in size from fine grained to huge boulders.

Drift is a general term used for the deposits made by glaciers directly or indirectly. Deposits directly made by melting of glaciers are called till. The soil carried by thw meling water from the frint of a glacier is termed out-wash.

Concept:

The specific surface area is defined as the total surface area of any material per unit mass or volume.

Specific surface \(= \frac{{Surface\;Area}}{{Volume}}\)

Calculation:

Surface area = 10 cm², Volume = 10 cc – 6 cc = 4 cc

∴ Specific surface area = \(\frac{{10}}{4} = 2.5/cc\)

Concept:

Soil classification as per Indian standards:

Computation:
% retained on 75 μ = 100 – 5 = 95% > 50 > Thus soil is classified as coarse grained soil.

Since the soil ranges between 4.75 mm to 75 μ, thus it a sandy soil.

Here fines are less than 5%.

For classification of finer soil D₆₀, D₃₀, and D₁₀ are required to be computed.

D₆₀ = 600 μ, D ₃₀ = 222 μ, and D₁₀ = 150 μ

Coefficient of uniformity \(\left( {{{\rm{C}}_{\rm{u}}}} \right) = \frac{{{{\rm{D}}_{60}}}}{{{{\rm{D}}_{10}}}} = \frac{{600{\rm{\mu }}}}{{150{\rm{\mu }}}} = 4< 6\) 

Coefficient of curvature \(\left( {{{\rm{C}}_{\rm{c}}}} \right) = \frac{{{\rm{D}}_{30}^2}}{{{{\rm{D}}_{60}} \times {{\rm{D}}_{10}}}} = \frac{{{{222}^2}}}{{600 \times 150}} = 0.547 < 1\)

For well graded soil: C_u > 6 & 1 < C_c < 3

For Soil A:

More than 50% of the soil fraction passed through 0.075 mm Is sieve is known fine Grained soil

I_P(A-line) = 0.73 (W_L - 20)

I_P = 0.73 (45 - 20) = 18.25

I_P for soil = 11

Thus soil lies below A-line of the Plasticity chart. it is known as silt.

W_L for soil lies between 35% to 50% W_L = 45% (Given)

Soil A is silt of Intermediate Compressibility Soil A = MI

For Soil B:

More than 50% of the soil is retained through 0.075 mm Is sieve. It is termed as coarse Grained soil.

Less than 50% of the soil is retained over 4.75 mm Is sieve, so it is sand

% fineness Given = 27% > 12% and I_P = 9% (Given)

i.e. I_P > 7%

So, it is termed as SC soil

Soil B = SC

Concept:

Coefficient of uniformity \(= \frac{{{D_{60}}}}{{{D_{10}}}}\)

Coefficient of curvature \(= \frac{{D_{30}^2}}{{{D_{60}}\; \times \;{D_{10}}}}\)

Calculation:

C_U = 4, C_c = 1.6

\({C_U} = \frac{{{D_{60}}}}{{{D_{10}}}} \Rightarrow 4 = \frac{{{D_{60}}}}{{{D_{10}}}} \Rightarrow {D_{60}} = 4\;{D_{10}}\)

\({C_c} = 1.6 \Rightarrow \frac{{D_{30}^2}}{{{D_{60}}\; \times \;{D_{10}}}} = 1.6 \Rightarrow \frac{{D_{30}^2}}{{4\;{D_{10}}\; \times \;{D_{10}}}} = 1.6\)

Concept:

Group Index (GI): It is used to give rating to the quality of soil within its group.
GI = 0.2a + 0.005ac + 0.01 bd

Where,

a = % passing 75 μ seive – 35 ; a ≯ 40

b = % passing 75 μ seive – 15 ; b ≯ 40

c = w_L - 40 ; c ≯ 20

d = I_p – 10 ; d ≯ 20

GI values ranges from 0 - 20.
GI value 0 → Soil is of Superior quality.

GI value 20 → Soil is of Poor quality.

Calculation:

% Passing 75 μ sieve = 80%, w_L = 70%, w_p = 30%

a = 80 – 35 = 45

But a ≯ 40 ⇒ a = 40

b = 80 – 15 = 65

But b ≯ 40 ⇒ b = 40

Also,

I_p = ω_L - ω_p = 70 – 30 = 40%

\(\therefore c = min\left\{ {\begin{array}{*{20}{c}} {{\omega _L} - 40}\\ {20} \end{array}} \right. = min\left\{ {\begin{array}{*{20}{c}} {70 - 40}\\ {20} \end{array}} \right. = 20\) 

\(d = min\left\{ {\begin{array}{*{20}{c}} {{I_p} - 10}\\ {20} \end{array}} \right.= min\left\{ {\begin{array}{*{20}{c}} {40 - 10}\\ {20} \end{array}} \right. = 20\)   

∴ G.I. = 0.2 a + 0.005 ac + 0.01 bd

= 0.2 × 40 + 0.005 × 40 × 20 + 0.01 × 40 × 20 = 20

The soil is of extremely poor quality as suggested by group index.