Question

During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted soil in the mold weighted 4lb and had a moisture content of 12%. If the specific gravity is 2.72, determine: a) dry unit weight b) void ratio c) degree of saturation d) additional water (in lb) needed to achieve 100% saturation in the soil sample

Answer 1

Answer:

part a : The dry unit weight is 0.0616  $lb/in^3$

part b : The void ratio is 0.77

part c :  Degree of Saturation is 0.43

part d : Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

• $\gamma_d$ is the dry unit weight which is to be calculated
• γ is the bulk unit weight given as

                                              $\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

• w is the moisture content in percentage, given as 12%

Substituting values

                                              $\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

The dry unit weight is 0.0616  $lb/in^3$

Part b

Void Ratio

The void ratio is given as

                                                $e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

• e is the void ratio which is to be calculated

• $\gamma_d$ is the dry unit weight which is calculated in part a
• $\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

• G is the specific gravity which is given as 2.72

Substituting values

                                              $e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

The void ratio is 0.77

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

• e is the void ratio which is calculated in part b

• G is the specific gravity which is given as 2.72

• w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

                                      $S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

Degree of Saturation is 0.43

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

• $\gamma_{zav}$ is  the zero air unit weight which is to be calculated

• $\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

• G is the specific gravity which is given as 2.72

• w is the moisture content in percentage, given as 12% or 0.12 in fraction

                                      $\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.