# Soil Compaction Numerical | Soil Mechanics Numericals

Table of Contents

In this article, we will discuss soil compaction numerical.

** 1. Formula **

Dry unit weight (γ_{d}) = Gγ_{w}/ 1+e |

where,

e = Void ratio

Volume of borrow pit soil required = (γ_{w})_{embankment }x V_{1}/ (γ_{d})_{borrow pit} |

Where,

**V _{1 }= Volume of finished embankement**

Degree of Saturation (S_{r})= (w x G)/e |

Where,

G = Specific Gravity

Number of truck trips = V_{2}/ Per truck loads |

Where,

V_{2}= Volume of borrow pit soil required.

Water Content = M_{wet} – M_{dry} / M_{dry} |

where,

M_{dry }= Mass of dry soil.

M_{wet} = Mass of wet soil.

In-situ dry density of soil(ρ_{dry}) = ρ_{t}/ 1+w |

Where,

ρ_{t }= In-situ bulk density of soil.

** 2. Soil Compaction Numerical **

**1. An embankment for a highway 30 m wide and 1.5 m compacted thickness is to be constructed from sandy soil trucked from a borrow pit. The water content of the sandy soil in the borrow pit is 15% and its void ratio is 0.69. The specification requires the embankment compacted to a dry unit weight of 18 kN/m ^{3}. determine for 1 km length of embankments that following**

i) The dry unit weight of sandy soil from the borrow pit required to construct the embankment.

ii) The number of 10 m3 truckloads of sandy soil required for the construction.

iii) The degree of saturation of the sandy soil in-situ.

**Solution:**

Assuming that specific gravity of soil (G) = 2.70

**i) The borrow pit’s dry unit weight, γ _{d} = G γ_{w}/ (1+e)**

= 2.7 x 9.81/ (1+0.69)

= 15.67 kN/m^{3}

The volume of the finished embankment (V_{1})= 30 m x 1.5 m x (1 km long)

= 45 x 10^{3} m^{3}

**Volume of borrow pit soil required = (γ _{w})_{embankment }x V_{1}/ (γ_{d})_{borrow pit}**

= 18 x 45 x10^{3} / 15.67

= 51691.13 m^{3}

**Number of truck trips = V _{2} / Per truck loads**

= 51691.13/10

= 5169 truck-loads

**iii) Degree of Saturation (Sr) = (w x G) / e**

Sr = (0.15 x 2.70) / 0.69

Sr = 0.59 = 59 %

Read Also: Quantity of Water Numerical |

**2. The maximum dry unit weight of a compacted soil mass is found to be 18 kN/m ^{3} with optimum water content being 15%. Find the values of porosity and degree of saturation of this compacted soil. Also, find the value of the maximum dry unit weight on the zero air void line at that optimum water content? Take specific gravity of soil solid as 2.70.**

Solution:

Given that;

Maximum dry unit weight (**γ _{d}**)

_{max}= 18 kN/m

^{3}

Optimum water content (w_{opt}) = 15% = 0.15

Specific gravity of soil solid (G)= 2.70

Now,

**γ _{d} = Gγ_{w}/(1+ wG/Sr)**

18 = 2.70 x 9.81 / (1+ 0.15 x 2.70 / Sr)

Sr = 0.15 x 2.70 /0.4715

Sr = 85.90%

Again, we have

**Sr x e = wG**

or, e = 0.15 x 2.70/ 0.859 = 0.471

**and, n= e/ (1+e)**

n = 0.471 / (1+0.471)

At zero air void line; we have,

**γ _{d} = Gγ_{w}/ (1+wG)**

= 2.70 x 9.81/ 1+(0.15 x 2.70)

= 18.85 kN/m^{3}

Read Also: Hardness and Alkalinity Numerical |

**3. In the construction of a road, the compaction specification required was 95% off proctor maximum dry density at field moisture content within 2% of the optimum moisture content. The maximum dry density and optimum moisture content obtained in the laboratory from the standard proctor test were 1.95 Mg/m ^{3} and 13.5% respectively. A site engineer conducted a sand cone test in two locations and obtained the following results.**

Location no. | Mass of soil removed (gm) | Mass of sand used (gm) | |

Wet | Dry | ||

1 | 43.86 | 38.46 | 39.51 |

2 | 37.38 | 32.21 | 32.39 |

**The density of sand used was 1.86 Mg/m3. Check whether the specification was satisfied or not.**

Solution:

Maximum dry unit weight (ρ** _{d}**)

_{max}= 1.95 Mg/m

^{3}= 1.95 gm/cm

^{3}

Optimum water content (w_{opt}) = 13.5%

Density of sand (ρ_{sand}) = 1.86 Mg/m^{3} = 1.86 gm/cm^{3}

**Calculation results of sand cone test**

**(i) For location 1**

Bulk density of sand = 1.86 Mg/m^{3}

Mass of sand used= 39.51

The volume of sand required to fill up the pit is given by;

Volume of pit(V) = M/ρ

= 39.51/1.86

= 21.24 cm^{3}

But, the mass of soil excavated from the pit= 43.86 gm

In-situ bulk density of soil (ρ_{t}) = 43.86/21.24

= 2.06 gm/cm^{3}

Water content of soil (w) = M_{wet} – M_{dry} / M_{dry}

= (43.86-38.46) / 38.46

= 0.1404

= 14.04%

Now, In-situ dry density of soil (ρ_{dry}) = ρ_{t} / (1+w)

= 2.06/(1+ 0.1404)

= 1.81 gm/cm^{3}

**ii) For location 2:**

The volume of sand required to fill the pit (V) = M/ρ

= 32.39/1.86

= 17.41 cm^{3}

In-situ density of soil (ρ_{t}) = 37.38/17.41

= 2.15gm/cm^{3}

Water content (w)= (37.38-32.21)/32.21

= 0.1605

= 16.05%

Now, In-situ dry density of soil (ρ_{dry}) = ρ_{t} / (1+w)

= 2.15/(1+ 0.1605)

= 1.85 gm/cm^{3}

Thus, actual field results are average of values are obtained from location 1 and 2;

Dry density (ρ_{dry}) = (1.81+1.85) / 2

Water content (w) = (0.1404+0.1605) / 2

= 0.1504

=15.04%

Compare specification with sand cone test

Minimum dry unit required= 0.95 x 1.95 = 1.8525 gm/cm^{3}

Field water content = (13.5 +2) /100

= 0.27

Thus, the field water content should be within 13.5±0.27.

The sand cone test gives a dry **density** of 1.83 gm/cm^{3} and water content of 15.04%.

After compacting these data, it is clear that the specification was not satisfied.

Civil Engineer & CEO of Naba Buddha Group