Tutorial 1 Solutions: Soil Mechanics

Proof:

The relationship between bulk unit weight (γ), specific gravity (G), water content (ω), and void ratio (e) is given by:

From this, we can isolate the term (1+e):

Another fundamental relationship in soil mechanics connects the degree of saturation (S_r), void ratio (e), water content (ω), and specific gravity (G):

From this, we can express the void ratio as:

Now, substitute this expression for e into the rearranged bulk unit weight equation:

To solve for S_r, first isolate the term containing it:

Now, invert the equation to solve for S_r:

To get the desired form, divide the numerator and the denominator by Gγ:

i. Initial Soil Properties:

• Water Content (ω):

• Volume of the sample (V₁):

• Dry Density (ρ_d):

• Void Ratio (e):

Using the dry density formula \( \rho_d = \frac{G\rho_w}{1+e} \):

• Degree of Saturation (S_r):

Using the relation \( S_r e = \omega G \):

ii. Degree of Saturation After Squeezing:

• New height (h₂): 71.1 – 3.0 = 68.1 mm

• New Volume (V₂): \( V_2 = \frac{\pi}{4}(3.56)^2(6.81) = 67.72 \text{ cm}^3 \)

• Volume of solids (V_s) (remains constant): \( V_s = \frac{M_s}{G\rho_w} = \frac{98.12}{2.71 \times 1} = 36.21 \text{ cm}^3 \)

• New Volume of voids (V_v2): \( V_{v2} = V_2 – V_s = 67.72 – 36.21 = 31.51 \text{ cm}^3 \)

• Volume of water (V_w) (remains constant as no water is lost): \( V_w = \frac{M – M_s}{\rho_w} = \frac{31.33}{1} = 31.33 \text{ cm}^3 \)

• New Degree of Saturation (S_r2):

i. In-situ Properties:

• Weight of soil (W): 2770.60 – 946.80 = 1823.8 gm

• Bulk Unit Weight (γ): \( \gamma = \frac{W}{V} = \frac{1823.8}{1000} = 1.824 \text{ gm/cc} = 17.89 \text{ kN/m}^3 \)

• Dry Unit Weight (γ_d): \( \gamma_d = \frac{\gamma}{1+\omega} = \frac{17.89}{1+0.1045} = 16.20 \text{ kN/m}^3 \)

• Void Ratio (e): \( \gamma_d = \frac{G\gamma_w}{1+e} \implies 16.20 = \frac{6.25 \times 9.81}{1+e} \implies e = \frac{61.31}{16.20} – 1 = 2.78 \)

• Degree of Saturation (S_r): \( S_r = \frac{\omega G}{e} = \frac{0.1045 \times 6.25}{2.78} = 0.235 \implies 23.5\% \)

ii. Properties upon Saturation:

• Condition: Saturated (S_r=1), no volume change (e = 2.78).

• Water Content (ω_sat): \( \omega_{sat} = \frac{e}{G} = \frac{2.78}{6.25} = 0.445 \implies 44.5\% \)

• Saturated Unit Weight (γ_sat): \( \gamma_{sat} = \frac{(G + e)\gamma_w}{1+e} = \frac{(6.25 + 2.78) \times 9.81}{1 + 2.78} = 23.38 \text{ kN/m}^3 \)

Calculations & Analysis:

• Plasticity Index (PI): \( PI = LL – PL = 62 – 28 = 34\% \)

• Liquidity Index (LI): \( LI = \frac{\omega – PL}{PI} = \frac{24 – 28}{34} = -0.118 \)

• Activity Number (A): \( A = \frac{PI}{\text{\% Clay}} = \frac{34}{23} = 1.48 \)

• Consistency: Since the LI is negative, the soil is in a very stiff consistency.

• Nature of Soil: The PI of 34% indicates the soil is highly plastic. The activity of 1.48 (>1.25) indicates an active soil, suggesting the presence of montmorillonite clay mineral.

i. Determination of Atterberg Limits:

• Liquid Limit (LL):

• Plot water content vs log(blows) to obtain flow curve
• Water content at 25 blows (interpolated) ≈ 49%
• ∴ LL = 49%

• Plastic Limit (PL): \( PL_{avg} = \frac{23 + 24}{2} = 23.5\% \)

ii. Index Properties Calculation:

• Plasticity Index (PI): \( PI = LL – PL = 49 – 23.5 = 25.5\% \)

• Flow Index (I_f): \( I_f = \frac{w_2 – w_1}{\log(N_1/N_2)} = \frac{55.6 – 41.5}{\log(33/11)} = \frac{14.1}{\log(3)} = 29.56 \)

• Toughness Index (I_T): \( I_T = \frac{PI}{I_f} = \frac{25.5}{29.56} = 0.86 \)

iii. Interpretation:

• PI = 25.5% indicates highly plastic soil

• I_T = 0.86 suggests moderate toughness (typical range 0.75-1.4 for clayey soils)

Calculations:

1. Plastic Limit (PL):

Using the Liquidity Index formula:

\[ LI = \frac{\omega – PL}{LL – PL} \implies 1.21 = \frac{81.3 – PL}{72.8 – PL} \]

Cross-multiplying:

\[ 1.21(72.8 – PL) = 81.3 – PL \]

\[ 88.088 – 1.21PL = 81.3 – PL \]

Rearranging terms:

\[ -1.21PL + PL = 81.3 – 88.088 \]

\[ -0.21PL = -6.788 \]

\[ PL = \frac{6.788}{0.21} = 32.32\% \]

2. Plasticity Index (PI):

\[ PI = LL – PL = 72.8 – 32.32 = 40.48\% \]

i. Calculations:

• Weight of Solids (W_s): This remains constant.

Dry unit weight in borrow pit (γ_d1): \( \gamma_{d1} = \frac{\gamma_1}{1+\omega_1} = \frac{16.4}{1+0.11} = 14.77 \text{ kN/m}^3 \)

W_s = γ_d1 × V₁ = 14.77 × 3500 = 51,695 kN

• Volume of Compacted Fill (V₂):

Void ratio of compacted fill (e₂): \( e_2 = \frac{n_2}{1-n_2} = \frac{0.30}{0.70} = 0.4286 \)

Dry unit weight of fill (γ_d2): \( \gamma_{d2} = \frac{G\gamma_w}{1+e_2} = \frac{2.63 \times 9.81}{1.4286} = 18.06 \text{ kN/m}^3 \)

Volume of fill (V₂): \( V_2 = \frac{W_s}{\gamma_{d2}} = \frac{51,695}{18.06} = 2,862.4 \text{ m}^3 \)

ii. Change in Degree of Saturation:

• Void ratio in borrow pit (e₁): \( e_1 = \frac{G\gamma_w}{\gamma_{d1}} – 1 = \frac{25.8}{14.77} – 1 = 0.747 \)

• Initial Saturation (S_r1): \( S_{r1} = \frac{\omega_1 G}{e_1} = \frac{0.11 × 2.63}{0.747} = 0.387 \implies 38.7\% \)

• Final Saturation (S_r2): \( S_{r2} = \frac{\omega_1 G}{e_2} = \frac{0.11 × 2.63}{0.4286} = 0.675 \implies 67.5\% \)

• Change in Saturation: 67.5% – 38.7% = 28.8% (increase)

i. Calculations:

• Dry Density (ρ_d): \( \rho_d = \frac{\rho}{1+\omega_1} = \frac{1.88}{1+0.05} = 1.79 \text{ g/cm}^3 \)

• Mass of Solids (M_s): \( M_s = \rho_d \times V = 1.79 \times 10^6 = 1790 \text{ kg} \)

ii. Water to be Added:

• Initial Mass of Water (M_w1): \( M_{w1} = \omega_1 \times M_s = 0.05 \times 1790 = 89.5 \text{ kg} \)

• Final Mass of Water (M_w2): \( M_{w2} = \omega_2 \times M_s = 0.15 \times 1790 = 268.5 \text{ kg} \)

• Water to Add: \( M_{w2} – M_{w1} = 268.5 – 89.5 = 179 \text{ kg (or 179 liters)} \)

iii. Final Degree of Saturation (S_r2):

• Void Ratio (e): \( e = \frac{G\rho_w}{\rho_d} – 1 = \frac{2.67 \times 1}{1.79} – 1 = 0.492 \)

• Degree of Saturation: \( S_{r2} = \frac{\omega_2 G}{e} = \frac{0.15 \times 2.67}{0.492} = 0.814 \implies 81.4\% \)

i. Weight of Solids Required (W_s):

\( W_s = V_{embankment} \times \gamma_{d,embankment} = 100,000 \times 16 = 1,600,000 \text{ kN} \)

ii. Dry Unit Weight at Excavation Site (γ_d,site):

\( \gamma_{d,site} = \frac{\gamma_{bulk,site}}{1+\omega_{site}} = \frac{12}{1+0.15} = 10.435 \text{ kN/m}^3 \)

iii. Volume to be Excavated (V_site):

\( V_{site} = \frac{W_s}{\gamma_{d,site}} = \frac{1,600,000}{10.435} = 153,330 \text{ m}^3 \)

iv. Weight to be Excavated (W_site):

\( W_{site} = V_{site} \times \gamma_{bulk,site} = 153,330 \times 12 = 1,839,960 \text{ kN} \)

i. Volume Calculations:

• Volume of Specimen (V): \( V = \frac{\pi}{4}d^2L = \frac{\pi}{4}(5)^2(10) = 196.35 \text{ cm}^3 \)

ii. Void Ratio Determination:

• Relationship between air voids and void ratio: \( n_a = \frac{e}{1+e}(1-S_r) \)

• Degree of saturation: \( S_r = \frac{\omega G}{e} \)

• Substituting gives: \( 0.20 = \frac{e}{1+e} – (0.15 \times 2.69) \)

• Solving: \( 0.20 + 0.20e = e – 0.4035 \)

• \( 0.80e = 0.6035 \implies e = 0.754 \)

iii. Mass Calculations:

• Dry density (ρ_d): \( \rho_d = \frac{G\rho_w}{1+e} = \frac{2.69}{1.754} = 1.534 \text{ g/cm}^3 \)

• Mass of dry soil (M_s): \( M_s = \rho_d \times V = 1.534 \times 196.35 = 301.1 \text{ g} \)

• Mass of water (M_w): \( M_w = M_s \times \omega = 301.1 \times 0.15 = 45.2 \text{ g} \)

Analysis from Grain Size Distribution Curve:

(By plotting the % Finer vs. log of particle size)

i. Effective Diameter (D₁₀): The size at which 10% passes. Interpolating between 150µm and 75µm gives D₁₀ ≈ 0.114 mm

• D₃₀ and D₆₀:
o D₃₀ (size at 30% passing) ≈ 0.47 mm
o D₆₀ (size at 60% passing) ≈ 1.68 mm

ii. Coefficient of Uniformity (C_u): \( C_u = \frac{D_{60}}{D_{10}} = \frac{1.68}{0.114} = 14.7 \)

iii. Coefficient of Curvature (C_c): \( C_c = \frac{(D_{30})^2}{D_{10} \times D_{60}} = \frac{(0.47)^2}{0.114 \times 1.68} = 1.15 \)

iv. Gradation: Since % Fines (4%) is less than 5%, C_u > 6, and 1 < C_c < 3, the soil is classified as a Well-Graded Sand (SW).

i. Initial Calculations:

• Dry Unit Weight (γ_d): \( \gamma_d = \frac{\gamma}{1+\omega_1} = \frac{17}{1+0.10} = 15.45 \text{ kN/m}^3 \)

• Weight of Solids (W_s): \( W_s = \gamma_d \times V = 15.45 \times 1 = 15.45 \text{ kN} \)

• Initial Weight of Water (W_w1): \( W_{w1} = W_s \times \omega_1 = 15.45 \times 0.10 = 1.545 \text{ kN} \)

ii. Water to be Added:

• Final Weight of Water (W_w2): \( W_{w2} = W_s \times \omega_2 = 15.45 \times 0.15 = 2.318 \text{ kN} \)

• Water to be Added: \( W_{w2} – W_{w1} = 2.318 – 1.545 = 0.773 \text{ kN} \)

• Volume Conversion: \( \frac{0.773}{9.81} \times 1000 \approx 78.8 \text{ liters} \)

i) Greater Plasticity Index (PI):

• PI_A = LL_A – PL_A = 60 – 30 = 30%

• PI_B = LL_B – PL_B = 40 – 20 = 20%

• Answer: Soil A has a greater plasticity index.

ii) Harder in its natural state (Consistency Index):

• IC = (LL – ω)/PI

• IC_A = (60 – 58)/30 = 0.067

• IC_B = (40 – 21)/20 = 0.95

• Answer: Soil B is harder in its natural state (higher IC indicates stiffer material).

iii) Greater Bulk Density:

For saturated soils (S_r=1), e = ωG

• Soil A: e_A = 0.58 × 2.70 = 1.566

ρ_A = (G + e)/(1 + e) × ρ_w = (2.70 + 1.566)/(1 + 1.566) × 1 = 1.66 g/cm³

• Soil B: e_B = 0.21 × 2.60 = 0.546

ρ_B = (2.60 + 0.546)/(1 + 0.546) × 1 = 2.035 g/cm³

• Answer: Soil B has greater bulk density.

i. Contains more clay particles?

• Plasticity Index (PI) indicates clay content:

• PI_X = LL_X – PL_X = 0.62 – 0.26 = 0.36

• PI_Y = LL_Y – PL_Y = 0.34 – 0.19 = 0.15

• Answer: Soil X contains more clay particles (higher PI).

ii. Has a greater wet density?

• For saturated soils (S_r=1), void ratio e = ωG:

• Soil X: e_X = 0.38 × 2.72 = 1.0336

ρ_X = (G + e)/(1 + e) × ρ_w = (2.72 + 1.0336)/(1 + 1.0336) × 1 = 1.846 g/cm³

• Soil Y: e_Y = 0.25 × 2.67 = 0.6675

ρ_Y = (2.67 + 0.6675)/(1 + 0.6675) × 1 = 1.998 g/cm³

• Answer: Soil Y has greater wet density.

iii. Has a greater dry density?

• Dry density ρ_d = G/(1 + e) × ρ_w:

• ρ_dX = 2.72/(1 + 1.0336) × 1 = 1.337 g/cm³

• ρ_dY = 2.67/(1 + 0.6675) × 1 = 1.599 g/cm³

• Answer: Soil Y has greater dry density.

iv. Has a greater void ratio?

• e_X = 1.0336

• e_Y = 0.6675

• Answer: Soil X has greater void ratio.

Calculations:

We can use the relationship: \( \gamma = \gamma_d + S_r(\gamma_{sat} – \gamma_d) \)

This formula directly relates the unit weight at any saturation level to the dry and saturated unit weights.

\( \gamma = 17.2 + 0.65(20 – 17.2) \)

\( \gamma = 17.2 + 0.65(2.8) \)

\( \gamma = 17.2 + 1.82 = 19.02 \text{ KN/m}^3 \)

Calculations:

• Particle Size (D):

Using Stokes’ Law:

\( D = \sqrt{\frac{(G-1)\gamma_w}{18\eta} \cdot \frac{H_e}{t}} \)

\( D = \sqrt{\frac{(2.68-1) \times 1}{18 \times 0.00981} \cdot \frac{15}{300}} \)

\( D = \sqrt{\frac{1.68}{0.17658} \cdot 0.05} \)

\( D = \sqrt{9.513 \times 0.05} \)

\( D = \sqrt{0.47565} = 0.069 \text{ cm} = 0.069 \text{ mm} \)

• Percentage Finer (N):

Corrected hydrometer reading (R_c) = R_h = 25

\( a = \frac{G}{G-1} = \frac{2.68}{1.68} = 1.595 \)

\( N = \frac{M_s}{a \cdot R_c} \times 100 \)

\( N = \frac{50}{1.595 \times 25} \times 100 \)

\( N = \frac{50}{39.875} \times 100 = 125.4\% \)

(Note: This exceeds 100%, suggesting a more accurate formula should be used)

Alternative calculation:

\( N = \frac{R_c \cdot \gamma_w \cdot (G-1)}{M_s \cdot G} \times V \times 100 \)

\( N = \frac{25 \times 1 \times 1.68}{50 \times 2.68} \times 1000 \times 100 \)

\( N = \frac{42}{134} \times 1000 \times 100 = 31.34\% \)

Calculations:

• Specific Gravity of Solids (G):

For a fully saturated soil, the bulk density ρ_sat = G_m1ρ_w.

Also, ρ_sat = \(\frac{G + e}{1 + e}\)ρ_w.

And for saturated soil, e = ωG.

So, e = 0.40G.

\(G_{m1} = \frac{G + e}{1 + e} \implies 1.85 = \frac{G + 0.40G}{1 + 0.40G} = \frac{1.40G}{1 + 0.40G}\)

1.85(1 + 0.40G) = 1.40G

1.85 + 0.74G = 1.40G ⟹ 0.66G = 1.85 ⟹ G = 2.80

• Shrinkage Limit (ω_s):

The specific gravity of the dry soil mass (G_m,dry) corresponds to the dry density ρ_d = G_m,dryρ_w = 1.75 g/cm³.

The shrinkage limit is the water content at which the soil volume becomes constant upon further drying.

\(\rho_d = \frac{G}{1 + e_d}\rho_w\), where e_d is the void ratio at dry state.

\(1.75 = \frac{2.80}{1 + e_d} \times 1 \implies 1 + e_d = \frac{2.80}{1.75} = 1.6 \implies e_d = 0.6\)

At the shrinkage limit, the soil is saturated, so S_r = 1.

\(\omega_s = \frac{S_r e_d}{G} = \frac{1 \times 0.6}{2.80} = 0.214 \implies 21.4\%\)

Calculations:

• Specific Gravity of Solids for the Mix (G_mix): \( G_{mix} = \frac{G_A \times 30 + G_B \times 70}{100} = \frac{2.6 \times 30 + 2.7 \times 70}{100} = \frac{78 + 189}{100} = 2.67 \)

• Dry density of the mix (ρ_d,mix): \( \rho_{d,mix} = \frac{\rho_{mix}}{1+\omega_{mix}} = \frac{1.8}{1.15} = 1.565 \text{ g/cc} \)

• Void Ratio (e_mix): \( \rho_{d,mix} = \frac{G_{mix}\rho_w}{1+e_{mix}} \implies 1.565 = \frac{2.67 \times 1}{1+e_{mix}} \) \( 1+e_{mix} = \frac{2.67}{1.565} = 1.705 \implies e_{mix} = 0.705 \)

• Degree of Saturation (S_r,mix): \( S_{r,mix} = \frac{\omega_{mix} G_{mix}}{e_{mix}} = \frac{0.15 \times 2.67}{0.705} = 0.568 \implies 56.8\% \)

Calculations:

• Shrinkage Ratio (SR):

The definition of shrinkage ratio is the ratio of volume change (as a percentage of dry volume) to the corresponding change in water content.

\( SR = \frac{LL – \omega_s}{(V_L – V_d)/V_d} = \frac{0.50 – 0.15}{(10 – 5.94)/5.94} = \frac{0.35}{0.6835} = 1.95 \)

• Specific Gravity of Solids (G):

The shrinkage ratio is also related to the specific gravity by SR ≈ G.

Therefore, G ≈ 1.95. This method is an approximation.

A more precise calculation for G:

The mass of solids (M_s) is constant.

\( M_s = \rho_d \cdot V_d \)

At the shrinkage limit, the soil is saturated.

\( \omega_s = \frac{M_w}{M_s} = \frac{V_{voids,d} \cdot \rho_w}{M_s} = \frac{(V_d – V_s)\rho_w}{M_s} \)

\( V_s = M_s/(G\rho_w) \)

So, \( \omega_s = \frac{M_s}{(V_d – M_s/(G\rho_w))\rho_w} = \frac{M_s}{V_d\rho_w} – \frac{1}{G} = \frac{\rho_d}{\rho_w} – \frac{1}{G} \)

We have an expression relating volumes and water contents:

\( \frac{M_s}{V_L – V_d} = LL – \omega_s \implies M_s = \frac{LL – \omega_s}{V_L – V_d} = \frac{0.50 – 0.15}{10 – 5.94} = 11.6 \text{ g} \)

Now, \( \rho_d = M_s/V_d = 11.6/5.94 = 1.953 \text{ g/cc} \)

Using the shrinkage limit relationship:

\( \omega_s = \frac{\rho_d}{\rho_w} – \frac{1}{G} \implies 0.15 = 1.953 – \frac{1}{G} \implies \frac{1}{G} = 1.953 – 0.15 = 1.803 \)

\( G = 2.76 \)

i. Plastic Limit (PL):

• PL = \(\frac{\text{Weight of water}}{\text{Weight of dry soil}} = \frac{20.11 – 14.82}{14.82} = \frac{5.29}{14.82} = 0.357 \implies 35.7\%\)

ii. Plasticity Index (PI):

• PI = LL – PL = 64.2 – 35.7 = 28.5%

iii. Classification (USCS):

• A-line PI: PI_A-line = 0.73(64.2 – 20) = 0.73 × 44.2 = 32.26%

• The soil’s PI (28.5%) is below the A-line PI (32.26%), indicating a silt (M) or organic silt/clay (OL/OH)

• The Liquid Limit (64.2%) > 50%, so it’s a high plasticity soil (H)

• Therefore, the soil is a Silt of High Plasticity (MH)

• USCS Symbol: MH

iv. Liquidity Index (LI):

• LI = \(\frac{\omega_{nat} – PL}{PI} = \frac{37.2 – 35.7}{28.5} = \frac{1.5}{28.5} = 0.053\)

v. Consistency:

• Since the Liquidity Index (0.053) is very close to zero, the soil’s natural water content is near its plastic limit

• This indicates the soil is in a stiff consistency

AASHTO Classification Procedure:

1. Material Type: Since >35% passes the #200 sieve (60.2%), the soil is a silt-clay material (groups A-4, A-5, A-6, or A-7).

2. Plasticity Index (PI): PI = LL − PL = 41.2 − 15.5 = 25.7%

3. Group Evaluation:

o A-4: LL max 40, PI max 10. (Doesn’t fit: LL and PI are too high)

o A-5: LL min 41, PI max 10. (Doesn’t fit: PI is too high)

o A-6: LL max 40, PI min 11. (Doesn’t fit: LL is too high)

o A-7: LL min 41, PI min 11. (This fits: LL=41.2, PI=25.7)

4. Subgroup of A-7:

o Compare PI with (LL – 30).

o LL−30 = 41.2−30 = 11.2

o The soil’s PI (25.7) is greater than (LL – 30). This places it in the A-7-6 subgroup.

5. Group Index (GI):

GI = (F−35)[0.2+0.005(LL−40)] + 0.01(F−15)(PI−10)

Where F = % passing #200 sieve = 60.2

GI = (60.2−35)[0.2+0.005(41.2−40)] + 0.01(60.2−15)(25.7−10)

GI = (25.2)[0.2+0.005(1.2)] + (0.01)(45.2)(15.7)

GI = (25.2)(0.206) + (0.452)(15.7)

GI = 5.19 + 7.09 = 12.28 ≈ 12

Final Classification: A-7-6 (12)

USCS Classification Procedure:

1. Coarse vs. Fine:
% passing #200 (75µ) sieve is 30%. This is less than 50%, so the soil is Coarse-Grained.

2. Gravel vs. Sand:
Coarse fraction = 100 – 30 = 70%.
% of coarse fraction passing 4.75 mm sieve = \(\frac{\text{\% passing 4.75mm} – \text{\% passing 75µ}}{\text{\% retained on 75µ}} \times 100 = \frac{70-30}{70} \times 100 = \frac{40}{70} \times 100 = 57.1\%\)
Since >50% of the coarse fraction is sand, the soil is primarily Sandy.

3. Clean vs. Dirty Sand:
% Fines is 30%. This is greater than 12%, so it is a “dirty” sand (either SM or SC).

4. Determine SM or SC:
• Plasticity Index (PI) = LL−PL=33−12=21%
• A-line PI = 0.73(LL−20)=0.73(33−20)=9.49%
• The soil’s PI (21%) is above the A-line. This indicates the fines are clay-like.

5. Final Classification: The soil is a Clayey Sand (SC).

i. Dry Unit Weight (γ_d):

This is the dry unit weight at OMC, which is the maximum dry unit weight (γ_d,max).

\( \gamma_d = \frac{\gamma}{1+\omega} = \frac{20}{1+0.15} = 17.39 \text{ KN/m}^3 \)

ii. Dry Unit Weight at Zero Air Voids (γ_zav):

The zero air voids unit weight is the theoretical maximum possible dry unit weight for a given water content.

\( \gamma_{zav} = \frac{G\gamma_w}{1+\omega G} = \frac{2.67 \times 9.81}{1+(0.15 \times 2.67)} = \frac{26.19}{1.4005} = 18.70 \text{ KN/m}^3 \)

iii. Saturated Unit Weight (γ_sat):

If the voids become filled with water, the soil is saturated (S_r=1). The volume of voids (and thus the void ratio) is determined by the compacted state at OMC.

First, find the void ratio (e) at γ_d,max:

\( \gamma_{d,max} = \frac{G\gamma_w}{1+e} \implies 17.39 = \frac{2.67 \times 9.81}{1+e} \implies e = \frac{26.19}{17.39} – 1 = 0.506 \)

Now, calculate the saturated unit weight with this void ratio:

\( \gamma_{sat} = \frac{(G + e)\gamma_w}{1+e} = \frac{(2.67 + 0.506) \times 9.81}{1 + 0.506} = \frac{3.176 \times 9.81}{1.506} = 20.67 \text{ KN/m}^3 \)

Analysis:

Additional Notes:

• Compaction curve should be plotted with Dry Density (y-axis) vs. Water Content (x-axis)
• Air voids curves for 0%, 5% and 10% can be plotted using the equation: \( ρ_d = \frac{Gρ_w(1-n_a)}{1+ωG} \) where n_a is air void percentage

i. Field Properties:

• Field Dry Unit Weight (γ_d,field): \( \gamma_{d,field} = \frac{\gamma}{1+\omega} = \frac{19.5}{1+0.28} = 15.23 \text{ kN/m}^3 \)

• Relative Compaction (RC): \( RC = \frac{\gamma_{d,field}}{\gamma_{d,max}} \times 100 = \frac{15.23}{18.0} \times 100 = 84.6\% \)

ii. Degree of Saturation:

• Void Ratio (e): \( e = \frac{G\gamma_w}{\gamma_{d,field}} – 1 = \frac{2.70 \times 9.81}{15.23} – 1 = 0.739 \)

• Degree of Saturation (S_r): \( S_r = \frac{\omega G}{e} = \frac{0.28 \times 2.70}{0.739} = 1.02 \implies 102\% \)

Note: Degree of saturation exceeding 100% indicates possible error in given values or measurements.

Calculation:

The formula for coefficient of permeability (k) in a falling head test is:

\( k = 2.303 \frac{a L}{A t} \log_{10} \frac{h_1}{h_2} \)

\( k = 2.303 \times \frac{3.14 \times 15}{28.27 \times 120} \log_{10} \frac{45}{30} \)

\( k = 2.303 \times (0.0139) \times \log_{10}(1.5) \)

\( k = 0.0320 \times 0.176 = 0.00563 \text{ cm/s} \)

To convert to m/day:

\( k = 0.00563 \frac{\text{cm}}{\text{s}} \times \frac{100 \text{ cm}}{1 \text{ m}} \times \frac{86400 \text{ s}}{1 \text{ day}} = 48.64 \text{ m/day} \)

Calculation:

The formula for k in an unconfined aquifer is (Thiem’s equation):

\( k = \frac{2.303 q \log_{10}(r_2/r_1)}{\pi(h_2^2 – h_1^2)} \)

First, find the heights of the water table at the observation wells from the impervious base:

• \( h_1 = D – s_1 = 14.5 – 2.45 = 12.05 \text{ m} \)

• \( h_2 = D – s_2 = 14.5 – 1.20 = 13.30 \text{ m} \)

Convert pumping rate to m³/day:

• \( q = 0.925 \text{ m}^3/\text{min} = 1332 \text{ m}^3/\text{day} \)

Calculate permeability:

\( k = \frac{2.303 \times (1332) \times \log_{10}(34/16)}{\pi(13.30^2 – 12.05^2)} \)

\( k = \frac{3067.5 \times \log_{10}(2.125)}{\pi(176.89 – 145.20)} \)

\( k = \frac{3067.5 \times 0.327}{\pi(31.69)} \)

\( k = \frac{1003}{99.55} = 10.07 \text{ m/day} \)

i. Permeability Calculation:

• h₁ = D – s₁ = 20 – 1 = 19 m

• h₂ = D – s₂ = 20 – 0.5 = 19.5 m

• Permeability (k): \( k = \frac{2.303q\log_{10}(r_2/r_1)}{\pi(h_2^2 – h_1^2)} \)

\( k = \frac{2.303 \times 2 \times \log_{10}(10/4)}{\pi(19.5^2 – 19^2)} = \frac{4.606 \times 0.398}{\pi(380.25 – 361)} = \frac{1.833}{19.25\pi} = 0.0303 \text{ m/min} \)

ii. Radius of Influence Calculation:

• Given drawdown in main well (s_w) = 6 m, so h_w = 20 – 6 = 14 m

• Diameter of well = 30 cm, so radius (r_w) = 0.15 m

• Using Thiem equation between main well and point of zero drawdown:

\( k = \frac{2.303q\log_{10}(R/r_w)}{\pi(D^2 – h_w^2)} \)

\( 0.0303 = \frac{2.303 \times 2 \times \log_{10}(R/0.15)}{\pi(20^2 – 14^2)} \)

\( \log_{10}(R/0.15) = \frac{0.0303 \times \pi \times 204}{4.606} = 4.218 \)

\( R/0.15 = 10^{4.218} = 16520 \)

\( R = 16520 \times 0.15 = 2478 \text{ m} \)

Using Thiem’s equation for confined aquifer:

\( q = \frac{2πkb(D – h)}{\ln(R/r_w)} \)

Substituting values:

\( 6000 = \frac{2π × 24.5 × 24 × (34 – 21.75)}{\ln(300/r_w)} \)

\( 6000 = \frac{3694.5 × 12.25}{\ln(300/r_w)} \)

\( 6000 = \frac{45257}{\ln(300/r_w)} \)

Solving for r_w:

\( \ln(300/r_w) = \frac{45257}{6000} = 7.54 \)

\( 300/r_w = e^{7.54} = 1881.7 \)

\( r_w = \frac{300}{1881.7} = 0.159 \text{ m} \)

Diameter of tube well:

\( D = 2 × r_w = 2 × 0.159 = 0.318 \text{ m or } 31.8 \text{ cm} \)

Calculations:

• Coefficient of Permeability (k):

From Darcy’s Law for a constant head test:

\( k = \frac{QL}{hAt} = \frac{450 \times 6}{40 \times 50 \times 600} = \frac{2700}{1200000} = 0.00225 \text{ cm/s} \)

• Seepage Velocity (v_s):

\( v_s = \frac{k \cdot i}{n} \) where \( i = h/L \) and n is porosity.

First, find porosity (n).

Dry density (ρ_d) = Dry mass / Volume = \( \frac{495}{50 \times 6} = 1.65 \text{ g/cm³} \)

Void ratio (e) = \( \frac{Gρ_w}{ρ_d} – 1 = \frac{2.65 \times 1}{1.65} – 1 = 1.606 – 1 = 0.606 \)

Porosity (n) = \( \frac{e}{1+e} = \frac{0.606}{1.606} = 0.377 \)

Hydraulic gradient (i) = \( h/L = 40/6 = 6.67 \)

Seepage velocity (v_s) = \( \frac{0.00225 \times 6.67}{0.377} = 0.0398 \text{ cm/s} \)

Calculation:

Using the falling head permeability formula:

\( k = 2.303\frac{aL}{At}\log_{10}\frac{h_1}{h_2} \)

\( k = 2.303\frac{0.3068 \times 12.2}{44.41 \times 900}\log_{10}\frac{75}{24.7} \)

\( k = 2.303\frac{3.743}{39969}\log_{10}(3.036) \)

\( k = 2.303 \times (0.00009365) \times 0.4823 \)

\( k = 1.04 \times 10^{-4} \text{ cm/s} \)