Soil Mechanics Test 5

Using the relation,

\(\frac{{\Delta H}}{{{H_o}}} = \frac{{\Delta e}}{{1 + {e_o}}}\)

Where,

∆H = Change in thickness of the soil

Ho = Initial thickness of the soil

Hf = Final thickness of the soil

∆e = Change in the void ratio of the soil

eo = Initial void ratio of the soil

ef = Final void ratio of soil

∆H = Ho - Hf

Ho = 5.0 cm

eo = 1.0

ef = 0.5

∆H = 5 - Ho

\(\frac{{{5.0} - H_f}}{{5.0}} = \frac{{0.5}}{{1 + 1.0}}\)

Hf = 3.75 cm

Concept:

Water Content: ​

\(w=\frac{{{W}_{w}}}{{{W}_{s}}};w\ge 0\)

• Water content or moisture content of a soil mass is defined as the ratio of the weight of water to the weight of solids (dry weight) of the soil mass.
• It is denoted by the letter symbol w and is commonly expressed as a percentage. The minimum value for water content is 0. There is no upper limit for water content.

At a given moisture content the maximum dry density that can be achieved is when the air voids are zero. This density is called zero air void density or theoretical maximum dry density.

We know, \({\gamma _d} = \frac{{{G_s}{\gamma _\omega }}}{{1 + e}}\;\;\& \;Se = w{G_s}\)

At zero air void ; S = 1

\(\therefore {\gamma _{d\;theo\;max}} = \frac{{{G_s}{\gamma _w}}}{{1 + w{G_s}}}\)

Calculation:

Water content = (100-80) / 80 = 0.25

γ_{d theo max} = \(\frac{{{G_s}{\gamma _w}}}{{1 + w{G_s}}} = \frac{{2.7\; \times \;9.81}}{{1\; +\; 0.25 \;\times \;2.7}}\) = 15.813 kN/m³  

Concept:

Relative compaction is defined as the ratio of the field dry unit weight, γ_{d (field)} to the laboratory maximum dry unit weight γ_{(d max)} as per specified standard test.

Relative Compaction \(= \frac{{{\gamma _d}\left( {field} \right)}}{{{\gamma _d}\left( {max} \right)}}\)   

Calculation:

γ_{d lab} = 18, Relative compaction = 0.75

\(\Rightarrow \frac{{{\gamma _{d\;field}}\;}}{{{\gamma _{d\;lab}}}} = 0.75 \Rightarrow {\gamma _{d\;field}} = 18 \times 0.75 = 13.5\;kN/{m^3}\)

Concept:

Light Compaction Test (IS : 2720, Part VII – 1974)

• Weight of hammer = 2.6 kg
• Height of fall = 310 mm
• Volume of mould = 1000 cc
• Compacted in 3 layers with 25 blows in each layer.

Calculation:

Compactive energy as per IS light compaction test \(= \frac{{2.6\; \times \;0.31\; \times \;3\; \times \;25}}{{{{10}^3} \times {{10}^{ - 6}}}} = 60450\;kgf\;m/{m^3}\)

Compactive energy per drop by rammer \(= \frac{{40}}{{0.05\; \times \;0.4}} = 2000\;kgf\;m/{m^3}\)

Due to non-overlap, actual energy developed = 2000 × 0.8 = 1600 kgf m/m³

∴ Number of passes required \(= \frac{{Energy\;in\;light\;compaction\;test}}{{Actual\;energy\;developed\;per\;pass\;for\;rammer}}\)  \(= \frac{{60450}}{{1600}}\) = 37.78 ≈ 38 passes

Concept:

Coefficient of volume compressibility is given by

\({m_v} = \frac{{ - \left( {\frac{{{\rm{\Delta }}e}}{{{\rm{\Delta }}\bar \sigma }}} \right)}}{{1 + {e_o}}}\)

Δe = Change in void ratio of the soil

e_o = initial void ratio

\({\rm{\Delta }}\bar \sigma = initial\;effective\;stress\)

Calculation:

Δe = 0.60 – 0.45 = 0.15

\({\rm{\Delta }}\bar \sigma = 170 - 210 = - 40\;kPa\)

e_o = 0.60

Concept:

Primary consolidation settlement is given by

\({\rm{\Delta }}H = \frac{{{C_c}{H_0}}}{{1 + {e_0}}}{\log _{10}}\left( {\frac{{\overline {{\sigma _o}} + {\rm{\Delta }}\bar \sigma }}{{{{\bar \sigma }_0}}}} \right)\)

C_C = compression index

e₀ = initial void ratio

H₀ = thickness of clay layer.

\(\overline {{\sigma _0}} \) = initial effective stress within the clay layer.

\(\overline {{\rm{\Delta }}\sigma } \) = increase in effective stress

Δ H = primary consolidation settlement

Calculation:

C_C = 0.40

e₀ = 1.345

Δ H = 20 mm

\(\overline {{\rm{\Delta }}\sigma } = 30\;kN/{m^2}\)

H₀ = 1.5 m

\(\frac{{20}}{{1000}} = \frac{{0.40 \times 1.5}}{{1 + 1.345}}{\log _{10}}\left( {\frac{{\overline {{\sigma _0}} + 30}}{{\overline {{\sigma _0}} }}} \right)\)

\(\frac{{0.020 \times 2.345}}{{1.5 \times 0.40}} = {\log _{10}}\left( {\frac{{\overline {{\sigma _0}} + 30}}{{\overline {{\sigma _0}} }}} \right)\)

\(1.197\overline {{\sigma _0}} = \overline {{\sigma _0}} + 30\)

\(\begin{array}{l} 0.197\overline {{\sigma _0}} = 30\\ \overline {{\sigma _0}} = 152.28\;kN/{m^2} \end{array}\)