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  

Coarse grained soil, well graded can be compacted to high γ_d specially if there are fines present. However if the fines increases then the density might decrease.

Poorly graded or uniform sands may lead to lowest value of dry unit weight.

In clay γ_d tends to decrease as plasticity increases.

Heavy clays with high plasticity have very low dry unit weight and very high OMC

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:

The weight and volume of solids will remain constant both in the excavated soil and embankment

Calculation:

For embankment of 1000 m³ :

γ_d = 1.72 gm/cc = 1.72 × 9.81 = 16.87 kN/m³

Moisture content = 20%

\({\gamma _d} = \frac{{Weight\;of\;solids}}{{Total\;volume}}\)

\(\Rightarrow 16.87 = \frac{{Weight\;of\;solids}}{{1000}} \Rightarrow \) Weight of solids = 16870 kN

For borrow pit:

e = 0.68, G_s = 2.7

Volume of solids = \(\frac{{16870}}{{{G_s}{\gamma _w}}} = \frac{{16870}}{{2.7 \times 9.81}} = 636.92\;{m^3}\)

\(e = \frac{{{V_v}}}{{{V_s}}} \Rightarrow {V_v} = e\;{V_s} = 0.68 \times 636.92\) = 433.105 m³

∴ Total volume to be excavated = V_v + V_s = 636.92 + 433.105 = 1070.02 m³

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

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

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

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

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

Concept:

Settlement(s) is given by:

\({\rm{s}} = \frac{{{{\rm{C}}_{\rm{c}}}}}{{1 + {{\rm{e}}_0}}} \times {\rm{H}} \times {\log _{10}}\frac{{{{\rm{\sigma }}_{\rm{f}}}'}}{{{{\rm{\sigma }}_{\rm{p}}}'}}\)

Where

C_c = compression index

e₀ = initial void ratio

H = depth of clay layer

σ_f’ = final effective stress at mid-depth of layer

σ_o’ = initial effective stress at mid-depth of the layer.

Calculation:

Given: C_c = 0.45, e = 0.65

To calculate the settlement of the clay layer, stresses are calculated at mid-depth of the clay layer.

Initial stress (σ₁') = 20 × 2 + 8 × 0.5 = 44 kN/m²

Stress due to footing load:

For strip footing, stress at any depth is given by

\({{\rm{\sigma }}_{\rm{Z}}} = \frac{{\rm{q}}}{{\rm{\pi }}}\left( {2 \propto + \sin 2 \propto } \right)\)

Where ∝ = tan^-1(1/1.5) = 33.69°

∴ ∝ = 33.69° or ∝ = 0.588 radian

\(\therefore {{\rm{\sigma }}_{\rm{_Z}}} = \frac{{250}}{{\rm{\pi }}}\left( {2 \times 0.588 + \sin \left( {2 \times 33.69} \right)} \right)\)

∴ σ₂ = 167.04 kN/m²

∴ Final stress (σ_f') = σ₁' + σ_Z = 44 + 167.04 = 211.04 kN/m²

\(\therefore {\rm{s}} = \frac{{0.45}}{{1 + 0.65}} \times 1000 \times {\log _{10}}\frac{{211.04}}{{44}}\)

Concept:

\({T_v} = \frac{{{C_v}t}}{{{H^2}}}\)

Where, Tv = Time factor.

U = Degree of consolidation.

Where, H = Maximum distance water has to travel to reach drainage face.

Calculation:

Under double drainage, H = 4 m

\({\left( {{T_v}} \right)_I} = \frac{{{C_v}t}}{{{H^2}}} = \frac{{{C_v}{t_I}}}{{{4^2}}}\)

When sand layer is introduced, it will act as a drainage face.

Hence H = 2 m.

\(\therefore {\left( {{T_v}} \right)_{II}} = \frac{{{C_v}{t_{II}}}}{{{2^2}}}\)

For same, settlement of 50 mm, the degree of consolidation will not change and hence the time factor won’t change.

∵ T_v remains same.

\(\Rightarrow \frac{{{C_v}{t_I}}}{{{4^2}}} = \frac{{{C_v}{t_{II}}}}{{{2^2}}} \Rightarrow \frac{3}{{{4^2}}} = \frac{{{t_{II}}}}{{{2^2}}} \Rightarrow {t_{II}} = 0.75\;years = 0.75 \times 12 = 9\;months\)