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Geotechnical Engineering Assignment

1.

Given:

Maximum dry unit weight, γd(max)=18.31kN/m3

Minimum dry unit weight, γd(min)=15.25kN/m3

Relative density, Dr=64%

Relative Density Formula

Dr=γd(field)γd(min)γd(max)γd(min)×100Dr = \frac{\gamma_{d(\text{field})} - \gamma_{d(\text{min})}}{\gamma_{d(\text{max})} - \gamma_{d(\text{min})}} \times 100

γd(field)=(Dr100×(γd(max)γd(min)))+γd(min)\gamma_{d(\text{field})} = \left( \frac{Dr}{100} \times (\gamma_{d(\text{max})} - \gamma_{d(\text{min})}) \right) + \gamma_{d(\text{min})}

γd(field)=(64100×(18.3115.25))+15.25\gamma_{d(\text{field})} = \left( \frac{64}{100} \times (18.31 - 15.25) \right) + 15.25 γd(field)=(0.64×3.06)+15.25=1.9584+15.25=17.2084kN/m3\gamma_{d(\text{field})} = (0.64 \times 3.06) + 15.25 = 1.9584 + 15.25 = 17.2084 \, \text{kN/m}^3

Relative Compaction Formula

RC=γd(field)γd(max)×100RC = \frac{\gamma_{d(\text{field})}}{\gamma_{d(\text{max})}} \times 100 RC=17.208418.31×100=0.9398×100=93.98%RC = \frac{17.2084}{18.31} \times 100 = 0.9398 \times 100 = \boxed{93.98\%}


Relative Compaction=94.0%

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2.

Given

emax=0.86e_{\text{max}} = 0.86, minimum void ratio emin=0.43e_{\text{min}} = 0.43, specific gravity Gs=2.66G_s = 2.66, relative density Dr=56%D_r = 56\%, and water content w=7%w = 7\%.

Void ratio at Dr=56%D_r = 56\% using the formula:

Dr=emaxeemaxeminD_r = \frac{e_{\text{max}} - e}{e_{\text{max}} - e_{\text{min}}}

Rearranging gives:

e=emaxDr(emaxemin)e = e_{\text{max}} - D_r \cdot (e_{\text{max}} - e_{\text{min}})

Substituting the values:

e=0.860.56(0.860.43)=0.860.2408=0.6192e = 0.86 - 0.56 \cdot (0.86 - 0.43) = 0.86 - 0.2408 = 0.6192

So, the void ratio is approximately 0.619.

Next, the dry unit weight is calculated using:

γd=Gsγw1+e\gamma_d = \frac{G_s \cdot \gamma_w}{1 + e}

Taking γw=9.81kN/m3\gamma_w = 9.81 \, \text{kN/m}^3:

γd=2.669.811+0.619=26.09461.619=16.12kN/m3\gamma_d = \frac{2.66 \cdot 9.81}{1 + 0.619} = \frac{26.0946}{1.619} = 16.12 \, \text{kN/m}^3

So, the dry unit weight is approximately 16.12 kN/m³.

Finally, the moist unit weight is calculated as:

γ=γd(1+w)=16.12(1+0.07)=16.121.07=17.24kN/m3\gamma = \gamma_d \cdot (1 + w) = 16.12 \cdot (1 + 0.07) = 16.12 \cdot 1.07 = 17.24 \, \text{kN/m}^3

Thus, the moist unit weight of the soil is approximately 17.24 kN/m³.


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3.

Given,

Inside cross-sectional area of the cylinder = 5000 mm² (not needed in this calculation)

Height of sand column = 100 mm = 10 cm

Initial water level = 400 mm = 40 cm

Final water level = 200 mm = 20 cm

Time = 30 minutes = 1800 seconds

Falling head permeability formula:

k=Ltln(h1h2)k = \frac{L}{t} \cdot \ln\left(\frac{h_1}{h_2}\right)Where:

= coefficient of permeability (cm/sec)

= length of soil column = 10 cm

tt = time = 1800 sec

= initial head = 40 cm

= final head = 20 cm

Now substitute:



k=101800ln(4020)k = \frac{10}{1800} \cdot \ln\left(\frac{40}{20}\right) k=101800ln(2)k = \frac{10}{1800} \cdot \ln(2) k=1018000.6931k = \frac{10}{1800} \cdot 0.6931 k=0.00385cm/s

The coefficient of permeability is 0.00385 cm/s or 3.85 × 10⁻⁵ m/sAnswer:


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