BE Civil Engineering (IOE, TU) Soil Mechanics (IOE, CE 603) Question Paper 2077 Nepal

Given: $M = 236\ \text{g}$, $V = 128\ \text{cm}^3$, $M_s = 192\ \text{g}$, $G_s = 2.68$.

Mass and volume of phases

Mass of water: $M_w = M - M_s = 236 - 192 = 44\ \text{g}$

$$w = \frac{M_w}{M_s} = \frac{44}{192} = 0.2292 = \mathbf{22.92\%}$$

Volume of solids: $V_s = \dfrac{M_s}{G_s \rho_w} = \dfrac{192}{2.68 \times 1} = 71.64\ \text{cm}^3$

Volume of voids: $V_v = V - V_s = 128 - 71.64 = 56.36\ \text{cm}^3$

Volume of water: $V_w = M_w/\rho_w = 44\ \text{cm}^3$

Void ratio and porosity

$$e = \frac{V_v}{V_s} = \frac{56.36}{71.64} = \mathbf{0.7867}$$ $$n = \frac{V_v}{V} = \frac{56.36}{128} = 0.4403 = \mathbf{44.03\%}$$

Degree of saturation

$$S = \frac{V_w}{V_v} = \frac{44}{56.36} = 0.7807 = \mathbf{78.07\%}$$

(Check by identity: $Se = wG_s \Rightarrow S = \dfrac{0.2292 \times 2.68}{0.7867} = \dfrac{0.6143}{0.7867} = 0.7809$ ✓)

Unit weights

Bulk density $\rho = M/V = 236/128 = 1.844\ \text{g/cm}^3$

$$\gamma = 1.844 \times 9.81 = \mathbf{18.09\ \text{kN/m}^3}$$

Dry density $\rho_d = M_s/V = 192/128 = 1.500\ \text{g/cm}^3$

$$\gamma_d = 1.500 \times 9.81 = \mathbf{14.72\ \text{kN/m}^3}$$

(b) Full saturation at constant $e = 0.7867$

At $S = 100\%$, from $Se = wG_s$:

$$w_{sat} = \frac{e}{G_s} = \frac{0.7867}{2.68} = 0.2935 = \mathbf{29.35\%}$$

Saturated unit weight:

$$\gamma_{sat} = \frac{(G_s + e)}{1+e}\,\gamma_w = \frac{2.68 + 0.7867}{1 + 0.7867}\times 9.81 = \frac{3.4667}{1.7867}\times 9.81$$ $$= 1.9403 \times 9.81 = \mathbf{19.03\ \text{kN/m}^3}$$

(Check dry unit weight unchanged: $\gamma_d = \dfrac{G_s}{1+e}\gamma_w = \dfrac{2.68}{1.7867}\times9.81 = 14.72\ \text{kN/m}^3$ ✓)

(a) Constant-head test

Cross-section area: $A = \dfrac{\pi}{4}d^2 = \dfrac{\pi}{4}(7.5)^2 = 44.18\ \text{cm}^2$

Discharge rate: $Q = \dfrac{620\ \text{cm}^3}{5\times 60\ \text{s}} = \dfrac{620}{300} = 2.0667\ \text{cm}^3/\text{s}$

For a constant-head test $Q = k\,\dfrac{h}{L}\,A$, so with $L = 18\ \text{cm}$ and $h = 40\ \text{cm}$:

$$k = \frac{Q L}{h A} = \frac{2.0667 \times 18.0}{40.0 \times 44.18} = \frac{37.20}{1767.2} = 0.02105\ \text{cm/s}$$ $$k = \mathbf{2.11\times10^{-2}\ \text{cm/s}}$$

(b) Layered system

Total thickness $H = 3 + 2 + 5 = 10\ \text{m}$.

Equivalent horizontal permeability (flow parallel to layers — thickness-weighted average):

$$k_H = \frac{\sum k_i H_i}{\sum H_i} = \frac{(10^{-2})(3) + (4\times10^{-4})(2) + (2\times10^{-3})(5)}{10}$$

Numerator $= 0.03 + 0.0008 + 0.01 = 0.0408$

$$k_H = \frac{0.0408}{10} = \mathbf{4.08\times10^{-3}\ \text{cm/s}}$$

Equivalent vertical permeability (flow perpendicular — harmonic average):

$$k_V = \frac{\sum H_i}{\sum (H_i/k_i)} = \frac{10}{\dfrac{3}{10^{-2}} + \dfrac{2}{4\times10^{-4}} + \dfrac{5}{2\times10^{-3}}}$$

Denominator $= 300 + 5000 + 2500 = 7800$

$$k_V = \frac{10}{7800} = 1.282\times10^{-3} = \mathbf{1.28\times10^{-3}\ \text{cm/s}}$$

Anisotropy ratio: $\dfrac{k_H}{k_V} = \dfrac{4.08\times10^{-3}}{1.28\times10^{-3}} = \mathbf{3.19}$. Horizontal permeability exceeds vertical, as expected in layered deposits: the low-permeability Layer 2 chokes the vertical flow (it dominates the harmonic sum) while contributing little to horizontal flow.

(a) Hydrostatic case (water table at 2.5 m)

Total stress $\sigma = \sum \gamma_i z_i$:

• At $2.5\ \text{m}$: $\sigma = 16.5 \times 2.5 = 41.25\ \text{kPa}$
• At $7.0\ \text{m}$: $\sigma = 41.25 + 19.6 \times 4.5 = 41.25 + 88.20 = 129.45\ \text{kPa}$
• At $10.0\ \text{m}$: $\sigma = 129.45 + 18.2 \times 3.0 = 129.45 + 54.60 = 184.05\ \text{kPa}$

Pore pressure $u = \gamma_w h_w$ (water table at 2.5 m):

• At $2.5\ \text{m}$: $u = 0$
• At $7.0\ \text{m}$: $u = 9.81 \times 4.5 = 44.15\ \text{kPa}$
• At $10.0\ \text{m}$: $u = 9.81 \times 7.5 = 73.58\ \text{kPa}$

Effective stress $\sigma' = \sigma - u$:

• At $2.5\ \text{m}$: $\sigma' = 41.25 - 0 = \mathbf{41.25\ \text{kPa}}$
• At $7.0\ \text{m}$: $\sigma' = 129.45 - 44.15 = \mathbf{85.30\ \text{kPa}}$
• At $10.0\ \text{m}$: $\sigma' = 184.05 - 73.58 = \mathbf{110.47\ \text{kPa}}$

Depth   sigma     u        sigma'   (kPa)
2.5 m   41.25     0.00     41.25
7.0 m   129.45    44.15    85.30
10.0 m  184.05    73.58    110.47

(b) Artesian condition at 10.0 m

The piezometric level stands $3.0\ \text{m}$ above ground surface, so the pressure head at $10.0\ \text{m}$ is $h_w = 10.0 + 3.0 = 13.0\ \text{m}$.

$$u_{new} = 9.81 \times 13.0 = 127.53\ \text{kPa}$$

Total stress at $10.0\ \text{m}$ is unchanged ($\sigma = 184.05\ \text{kPa}$):

$$\sigma'_{new} = 184.05 - 127.53 = \mathbf{56.52\ \text{kPa}}$$

Comment: The artesian pressure raises pore pressure and reduces effective stress at $10.0\ \text{m}$ from $110.47$ to $56.52\ \text{kPa}$ (about a $49\%$ reduction). The soil is still stable here ($\sigma' > 0$), but the loss of effective stress reduces shear strength and bearing capacity. In a deep excavation, removing overburden could drive $\sigma'$ to zero and cause base heave / blow-out of the clay over the artesian aquifer — a critical design concern.

(a) Ultimate settlement — normally consolidated

$$S_c = \frac{C_c H}{1 + e_0}\log_{10}\!\left(\frac{\sigma'_0 + \Delta\sigma}{\sigma'_0}\right) = \frac{0.38 \times 4.0}{1 + 0.95}\log_{10}\!\left(\frac{120 + 90}{120}\right)$$ $$= \frac{1.52}{1.95}\log_{10}(1.75)$$

$\log_{10}(1.75) = 0.24304$

$$S_c = 0.77949 \times 0.24304 = 0.18945\ \text{m} = \mathbf{189.4\ \text{mm}}$$

(b) Time for 60% consolidation

Double drainage $\Rightarrow$ drainage path $H_{dr} = H/2 = 2.0\ \text{m}$.

$$t_{60} = \frac{T_{60}\,H_{dr}^2}{c_v} = \frac{0.286 \times (2.0)^2}{3.5} = \frac{0.286 \times 4.0}{3.5} = \frac{1.144}{3.5} = 0.3269\ \text{year}$$ $$t_{60} = 0.3269 \times 365 = \mathbf{119.3\ \text{days}}$$

Settlement at $60\%$: $S_{60} = 0.60 \times S_c = 0.60 \times 189.4 = \mathbf{113.7\ \text{mm}}$

(c) Over-consolidated case ($\sigma'_p = 150\ \text{kPa}$)

Here $\sigma'_0 = 120 < \sigma'_p = 150$, and $\sigma'_0 + \Delta\sigma = 210 > \sigma'_p$. The stress path crosses the preconsolidation pressure, so use the two-part formula:

$$S_c = \frac{C_r H}{1+e_0}\log_{10}\!\frac{\sigma'_p}{\sigma'_0} + \frac{C_c H}{1+e_0}\log_{10}\!\frac{\sigma'_0+\Delta\sigma}{\sigma'_p}$$

Recompression part: $\log_{10}(150/120) = \log_{10}(1.25) = 0.09691$

$$S_1 = \frac{0.06 \times 4.0}{1.95}\times 0.09691 = \frac{0.24}{1.95}\times 0.09691 = 0.12308 \times 0.09691 = 0.01193\ \text{m}$$

Virgin part: $\log_{10}(210/150) = \log_{10}(1.40) = 0.14613$

$$S_2 = \frac{0.38 \times 4.0}{1.95}\times 0.14613 = 0.77949 \times 0.14613 = 0.11391\ \text{m}$$ $$S_c = 0.01193 + 0.11391 = 0.12584\ \text{m} = \mathbf{125.8\ \text{mm}}$$

The over-consolidated settlement (125.8 mm) is markedly smaller than the NC value (189.4 mm) because part of the load is carried in the stiff recompression range.

Major principal stresses at failure

• Test 1: $\sigma_1 = 100 + 280 = 380\ \text{kPa}$, $\sigma_3 = 100\ \text{kPa}$
• Test 2: $\sigma_1 = 200 + 440 = 640\ \text{kPa}$, $\sigma_3 = 200\ \text{kPa}$

(a) Parameters from the failure criterion

$$\sigma_1 = \sigma_3\,N_\phi + 2c'\sqrt{N_\phi},\qquad N_\phi = \tan^2\!\left(45^\circ + \tfrac{\phi'}{2}\right)$$

Write both tests:

• $380 = 100\,N_\phi + 2c'\sqrt{N_\phi}$
• $640 = 200\,N_\phi + 2c'\sqrt{N_\phi}$

Subtract: $640 - 380 = (200-100)N_\phi \Rightarrow 260 = 100\,N_\phi$

$$N_\phi = 2.60$$ $$\phi' = 2\left[\tan^{-1}\!\sqrt{N_\phi} - 45^\circ\right] = 2\left[\tan^{-1}(1.6125) - 45^\circ\right]$$

$\tan^{-1}(1.6125) = 58.20^\circ$, so $\phi' = 2(58.20 - 45) = 2(13.20) = \mathbf{26.4^\circ}$

Solve for $c'$ using Test 1: $2c'\sqrt{N_\phi} = 380 - 100(2.60) = 380 - 260 = 120$

$$c' = \frac{120}{2\sqrt{2.60}} = \frac{120}{2 \times 1.6125} = \frac{120}{3.2249} = \mathbf{37.2\ \text{kPa}}$$

(Check with Test 2: $200(2.60) + 3.2249 \times 37.2 = 520 + 120 = 640\ \text{kPa}$ ✓)

(b) Failure plane, Test 1

Inclination of the failure plane to the major-principal (horizontal) plane:

$$\theta = 45^\circ + \frac{\phi'}{2} = 45^\circ + 13.2^\circ = \mathbf{58.2^\circ}$$

Centre and radius of the Mohr circle (Test 1):

$$s = \frac{\sigma_1+\sigma_3}{2} = \frac{380+100}{2} = 240\ \text{kPa},\quad t = \frac{\sigma_1-\sigma_3}{2} = \frac{280}{2} = 140\ \text{kPa}$$

Normal stress on the failure plane (with $2\theta = 116.4^\circ$, $\cos116.4^\circ = -0.4446$):

$$\sigma_n = s - t\cos 2\theta = 240 - 140(-0.4446) = 240 + 62.24 = \mathbf{302.2\ \text{kPa}}$$

Shear stress on the failure plane ($\sin116.4^\circ = 0.8957$):

$$\tau = t\sin 2\theta = 140 \times 0.8957 = \mathbf{125.4\ \text{kPa}}$$

(c) Unconsolidated-undrained tests on saturated clay

In a UU ($\phi_u = 0$) test no drainage is permitted, so increasing the cell pressure raises pore pressure equally and leaves the effective stresses unchanged. All total-stress Mohr circles therefore have the same diameter, giving a horizontal envelope: $\phi_u = 0$ with strength $= c_u$ (undrained shear strength). The underlying effective-stress envelope ($c', \phi'$) is unchanged, but the measured total-stress envelope flattens.

(a) Residual vs transported soils

(b) USCS classification

Percent passing No. 200 ($0.075\ \text{mm}$) $= 86\% > 50\%$ ⇒ the soil is fine-grained; the Atterberg limits govern the symbol.

Plasticity index:

$$PI = LL - PL = 52 - 24 = 28\%$$

A-line value at $LL = 52$:

$$PI_{A} = 0.73(LL - 20) = 0.73(52 - 20) = 0.73 \times 32 = 23.36$$

Since $PI = 28 > 23.36$, the point plots above the A-line ⇒ the soil is a clay (C).

Since $LL = 52 > 50$ ⇒ high plasticity (H).

$$\boxed{\textbf{CH — clay of high plasticity (fat clay)}}$$

(a) Void ratio, degree of saturation and air voids at OMC

From $\gamma_d = \dfrac{G_s \gamma_w}{1+e}$:

$$1 + e = \frac{G_s \gamma_w}{\gamma_d} = \frac{2.68 \times 9.81}{18.0} = \frac{26.291}{18.0} = 1.4606 \Rightarrow e = \mathbf{0.4606}$$

Degree of saturation from $S e = w G_s$:

$$S = \frac{w G_s}{e} = \frac{0.14 \times 2.68}{0.4606} = \frac{0.3752}{0.4606} = 0.8146 = \mathbf{81.5\%}$$

Air-void content (fraction of total volume that is air):

$$n_a = \frac{e(1-S)}{1+e} = \frac{0.4606(1-0.8146)}{1.4606} = \frac{0.4606 \times 0.1854}{1.4606} = \frac{0.08540}{1.4606} = 0.05847 = \mathbf{5.85\%}$$

(b) Zero-air-voids (ZAV) dry unit weight at $w = 14\%$

At ZAV, $S = 100\%$:

$$\gamma_{d,zav} = \frac{G_s \gamma_w}{1 + w G_s} = \frac{2.68 \times 9.81}{1 + 0.14 \times 2.68} = \frac{26.291}{1 + 0.3752} = \frac{26.291}{1.3752} = \mathbf{19.12\ \text{kN/m}^3}$$

Comment: The compaction-curve peak ($18.0\ \text{kN/m}^3$) lies below the ZAV line ($19.12\ \text{kN/m}^3$) by about $1.12\ \text{kN/m}^3$, corresponding to the $\approx 5.85\%$ entrapped air. The compaction curve can never reach the ZAV line because soil cannot be fully saturated by compaction alone; some air is always trapped.

Earth-pressure coefficients ($\phi = 32^\circ$, $\sin32^\circ = 0.5299$)

$$K_a = \frac{1-\sin\phi}{1+\sin\phi} = \frac{1-0.5299}{1+0.5299} = \frac{0.4701}{1.5299} = 0.3073$$ $$K_p = \frac{1}{K_a} = \frac{1.5299}{0.4701} = 3.255$$

(a) Active thrust

Active pressure at the base: $\sigma_{a,base} = K_a\gamma H = 0.3073 \times 17.5 \times 6 = 32.27\ \text{kPa}$

Total active thrust (area of the pressure triangle):

$$P_a = \tfrac{1}{2}K_a\gamma H^2 = \tfrac{1}{2}\times 0.3073 \times 17.5 \times 6^2 = \tfrac{1}{2}\times 0.3073 \times 17.5 \times 36$$ $$= 0.5 \times 193.6 = \mathbf{96.8\ \text{kN/m}}$$

Point of application: at $\dfrac{H}{3} = \dfrac{6}{3} = \mathbf{2.0\ \text{m}}$ above the base (centroid of the triangle).

(b) Passive resistance

$$P_p = \tfrac{1}{2}K_p\gamma H^2 = \tfrac{1}{2}\times 3.255 \times 17.5 \times 36 = 0.5 \times 2050.7 = \mathbf{1025.3\ \text{kN/m}}$$

Acting at $H/3 = 2.0\ \text{m}$ above the base.

Comment: $\dfrac{P_p}{P_a} = \dfrac{1025.3}{96.8} = 10.59 = \dfrac{K_p}{K_a}$. Passive resistance is more than ten times the active thrust. This large reserve is heavily factored down in design because the wall must undergo substantial movement to mobilise full passive pressure.

Terzaghi equation — square footing (general shear)

$$q_u = 1.3\,c\,N_c + q\,N_q + 0.4\,\gamma B\,N_\gamma$$

where surcharge $q = \gamma D_f = 18.5 \times 1.5 = 27.75\ \text{kPa}$ and $B = 2.0\ \text{m}$.

Term-by-term

Cohesion term: $1.3 \times 18 \times 25.13 = 1.3 \times 452.34 = 588.04\ \text{kPa}$

Surcharge term: $27.75 \times 12.72 = 352.98\ \text{kPa}$

Self-weight term: $0.4 \times 18.5 \times 2.0 \times 9.70 = 0.4 \times 358.9 = 143.56\ \text{kPa}$

Ultimate bearing capacity

$$q_u = 588.04 + 352.98 + 143.56 = \mathbf{1084.6\ \text{kPa}}$$

Net ultimate bearing capacity

$$q_{nu} = q_u - \gamma D_f = 1084.6 - 27.75 = 1056.8\ \text{kPa}$$

Net safe bearing capacity (FOS = 3 on net):

$$q_{ns} = \frac{q_{nu}}{F} = \frac{1056.8}{3} = \mathbf{352.3\ \text{kPa}}$$

Gross safe bearing capacity:

$$q_s = q_{ns} + \gamma D_f = 352.3 + 27.75 = \mathbf{380.0\ \text{kPa}}$$

Summary: $q_u = 1084.6\ \text{kPa}$; net safe $= 352.3\ \text{kPa}$; gross safe $\approx 380.0\ \text{kPa}$.

(a) Seepage discharge

Convert $k = 4\times10^{-4}\ \text{cm/s} = 4\times10^{-6}\ \text{m/s}$.

$$q = k H \frac{N_f}{N_d} = (4\times10^{-6}) \times 6 \times \frac{4}{12} = 4\times10^{-6} \times 6 \times 0.3333$$ $$q = 8.0\times10^{-6}\ \text{m}^3/\text{s per m} = \mathbf{8.0\times10^{-6}\ \text{m}^3/\text{s/m}}$$

($= 8.0\times10^{-6}\times86400 = 0.691\ \text{m}^3/\text{day per m}$).

(b) Uplift pressure head after 5 drops

Head drop per equipotential: $\Delta h = \dfrac{H}{N_d} = \dfrac{6}{12} = 0.5\ \text{m}$ per drop.

Taking the upstream head as $H = 6\ \text{m}$ above the downstream (tailwater) datum, the residual head after $5$ drops is:

$$h_{residual} = H - n_d\,\Delta h = 6 - 5\times0.5 = 6 - 2.5 = \mathbf{3.5\ \text{m of water}}$$

Corresponding uplift pressure: $u = \gamma_w h = 9.81 \times 3.5 = \mathbf{34.3\ \text{kPa}}$ (above the downstream datum).

(c) Darcy's law and a limitation

Darcy's law: the discharge velocity is directly proportional to the hydraulic gradient,

$$v = k\,i,\qquad q = k\,i\,A$$

where $i = \Delta h/L$ is the hydraulic gradient and $k$ the coefficient of permeability.

Limitation: It is valid only for laminar flow (low Reynolds number). It breaks down for turbulent flow through coarse gravels/rockfill, and at very low gradients in clays where a threshold gradient may be needed before flow begins.

(a) Terzaghi's 1-D consolidation assumptions

1. Soil is homogeneous and fully saturated.
2. Soil grains and pore water are incompressible; volume change is due only to the expulsion of water.
3. Darcy's law is valid and $k$ is constant during consolidation.
4. Drainage and flow are one-dimensional (vertical).
5. The coefficient of compressibility $a_v$ (hence $c_v$) is constant over the stress range.
6. There is a unique linear relationship between void ratio and effective stress; strains are small.
7. Load is applied instantaneously and uniformly over an infinite extent.

(b) Consolidation vs compaction

(c) Sensitivity and thixotropy

Sensitivity $S_t = \dfrac{q_{u,\text{undisturbed}}}{q_{u,\text{remoulded}}}$ — the ratio of undisturbed to remoulded unconfined compressive strength at the same water content. Clays range from insensitive ($S_t \approx 1$) to quick clays ($S_t > 16$). Thixotropy is the property by which a remoulded clay, left undisturbed at constant water content, gradually regains part of its lost strength over time as inter-particle bonds re-establish.

(d) Skempton's pore-pressure parameters

The change in pore pressure under undrained loading is:

$$\Delta u = B\big[\Delta\sigma_3 + A(\Delta\sigma_1 - \Delta\sigma_3)\big]$$

• $B$ measures response to all-round (isotropic) stress; $B \approx 1$ for fully saturated soils and $\to 0$ for dry soils.
• $A$ measures response to the deviatoric (shear) stress and depends on soil type and stress history: large and positive for soft normally-consolidated clays, small or negative for heavily over-consolidated clays (which tend to dilate).