BE Civil Engineering (IOE, TU) Soil Mechanics (IOE, CE 603) Question Paper 2078 Nepal

Given: $M = 542\ \text{g}$, $M_s = 389\ \text{g}$, $G_s = 2.72$, sample saturated.

Mass of water:

$$M_w = M - M_s = 542 - 389 = 153\ \text{g}$$

(a) Water content, void ratio, porosity, degree of saturation

Water content:

$$w = \frac{M_w}{M_s} = \frac{153}{389} = 0.3933 = \mathbf{39.33\%}$$

For a saturated soil, $S_r = 1$ and the identity $S_r\,e = w\,G_s$ gives:

$$e = w\,G_s = 0.3933 \times 2.72 = \mathbf{1.070}$$

Porosity:

$$n = \frac{e}{1+e} = \frac{1.070}{2.070} = 0.5169 = \mathbf{51.69\%}$$

Degree of saturation: the sample is saturated, so $\mathbf{S_r = 100\%}$ (consistent with $S_r e = wG_s$).

(b) Unit weights

Use the standard expression with $\gamma_w = \rho_w\,g = 1000 \times 9.81 = 9810\ \text{N/m}^3 = 9.81\ \text{kN/m}^3$.

Bulk (total) unit weight (saturated here):

$$\gamma = \frac{(G_s + S_r e)\,\gamma_w}{1+e} = \frac{(2.72 + 1.070)\times 9.81}{2.070} = \frac{3.790 \times 9.81}{2.070} = \frac{37.18}{2.070} = \mathbf{17.96\ \text{kN/m}^3}$$

Dry unit weight:

$$\gamma_d = \frac{G_s\,\gamma_w}{1+e} = \frac{2.72 \times 9.81}{2.070} = \frac{26.68}{2.070} = \mathbf{12.89\ \text{kN/m}^3}$$

(Check: $\gamma_d = \gamma/(1+w) = 17.96/1.3933 = 12.89\ \text{kN/m}^3$. OK.)

Submerged (buoyant) unit weight:

$$\gamma' = \gamma_{sat} - \gamma_w = 17.96 - 9.81 = \mathbf{8.15\ \text{kN/m}^3}$$

(c) New degree of saturation after partial drying ($w = 18\%$, $e$ unchanged)

With $e = 1.070$ fixed:

$$S_r = \frac{w\,G_s}{e} = \frac{0.18 \times 2.72}{1.070} = \frac{0.4896}{1.070} = 0.4576 = \mathbf{45.76\%}$$

The soil becomes partially saturated (unsaturated) as water drains/evaporates at constant void ratio.

Given: $H = 4.0\ \text{m}$, $\sigma'_0 = 120\ \text{kPa}$, $\Delta\sigma = 90\ \text{kPa}$, $C_c = 0.32$, $e_0 = 0.86$, $c_v = 2.4\ \text{m}^2/\text{yr}$, double drainage.

(a) Ultimate primary consolidation settlement

Normally consolidated, so:

$$S_c = \frac{C_c\,H}{1+e_0}\,\log_{10}\!\left(\frac{\sigma'_0 + \Delta\sigma}{\sigma'_0}\right)$$ $$\frac{\sigma'_0 + \Delta\sigma}{\sigma'_0} = \frac{120 + 90}{120} = \frac{210}{120} = 1.75,\qquad \log_{10}(1.75) = 0.24304$$ $$S_c = \frac{0.32 \times 4.0}{1+0.86}\times 0.24304 = \frac{1.28}{1.86}\times 0.24304 = 0.68817 \times 0.24304$$ $$S_c = 0.16725\ \text{m} = \mathbf{167.3\ \text{mm}}$$

(b) Time for 60% consolidation

Double drainage $\Rightarrow$ drainage path length:

$$H_{dr} = \frac{H}{2} = \frac{4.0}{2} = 2.0\ \text{m}$$

From $T_v = \dfrac{c_v\,t}{H_{dr}^2}$:

$$t_{60} = \frac{T_{60}\,H_{dr}^2}{c_v} = \frac{0.286 \times (2.0)^2}{2.4} = \frac{0.286 \times 4.0}{2.4} = \frac{1.144}{2.4} = 0.4767\ \text{years}$$ $$t_{60} = 0.4767 \times 365 \approx \mathbf{174\ \text{days}\ (\approx 0.48\ \text{year})}$$

(c) Two field methods to accelerate consolidation

1. Sand drains / Prefabricated Vertical Drains (PVDs / wick drains): shorten the horizontal drainage path so excess pore pressure dissipates faster (radial drainage).
2. Preloading / surcharging: apply a temporary surcharge greater than the design load so that the required settlement is reached sooner; the surcharge is then removed.

(Other valid answers: vacuum consolidation, electro-osmosis.)

Compute major principal stresses at failure ($\sigma_1' = \sigma_3' +$ deviator):

• Test 1: $\sigma_1' = 100 + 248 = 348\ \text{kPa}$, $\sigma_3' = 100\ \text{kPa}$
• Test 2: $\sigma_1' = 200 + 388 = 588\ \text{kPa}$, $\sigma_3' = 200\ \text{kPa}$

(a) Shear strength parameters

Let $N_\phi = \tan^2(45 + \phi'/2)$ and $A = 2c'\tan(45+\phi'/2)$. Then $\sigma_1' = N_\phi\,\sigma_3' + A$.

Two equations:

$$348 = 100\,N_\phi + A \qquad (1)$$ $$588 = 200\,N_\phi + A \qquad (2)$$

Subtract (1) from (2):

$$240 = 100\,N_\phi \;\Rightarrow\; N_\phi = 2.40$$

So $\tan^2(45+\phi'/2) = 2.40 \Rightarrow \tan(45+\phi'/2) = 1.5492$.

$$45 + \frac{\phi'}{2} = \arctan(1.5492) = 57.16^\circ \;\Rightarrow\; \frac{\phi'}{2} = 12.16^\circ$$ $$\boxed{\phi' = 24.3^\circ}$$

From (1): $A = 348 - 100(2.40) = 348 - 240 = 108\ \text{kPa}$.

$$c' = \frac{A}{2\tan(45+\phi'/2)} = \frac{108}{2 \times 1.5492} = \frac{108}{3.0984} = \mathbf{34.9\ \text{kPa}}$$ $$\boxed{c' \approx 34.9\ \text{kPa},\quad \phi' \approx 24.3^\circ}$$

(b) Stresses on the failure plane (Test 1)

Failure plane angle from major principal plane: $\theta = 45 + \phi'/2 = 57.16^\circ$.

Centre and radius of Mohr circle:

$$s = \frac{\sigma_1' + \sigma_3'}{2} = \frac{348 + 100}{2} = 224\ \text{kPa},\quad r = \frac{\sigma_1' - \sigma_3'}{2} = \frac{348 - 100}{2} = 124\ \text{kPa}$$

Normal stress on failure plane:

$$\sigma_n' = s + r\cos 2\theta = 224 + 124\cos(114.32^\circ)$$ $$\cos(114.32^\circ) = -0.41142 \Rightarrow \sigma_n' = 224 - 51.02 = \mathbf{173.0\ \text{kPa}}$$

Shear stress on failure plane:

$$\tau_f = r\sin 2\theta = 124\sin(114.32^\circ) = 124 \times 0.91144 = \mathbf{113.0\ \text{kPa}}$$

Check with Mohr-Coulomb: $\tau_f = c' + \sigma_n'\tan\phi' = 34.9 + 173.0\tan(24.3^\circ) = 34.9 + 173.0(0.4514) = 34.9 + 78.1 = 113.0\ \text{kPa}$. Consistent.

(c) Why the failure plane is at $45 + \phi'/2$

The failure plane is the plane on which the Mohr-Coulomb failure criterion is first satisfied, i.e. the plane where the shear-to-normal stress combination is tangent to the failure envelope. Geometrically, the point of tangency on the Mohr circle subtends an angle $2\theta = 90^\circ + \phi'$ at the circle centre (measured from the $\sigma_1'$ point). Hence the physical plane orientation is $\theta = 45^\circ + \phi'/2$ measured from the major principal (horizontal) plane. This is steeper than $45^\circ$ because internal friction makes planes closer to the major principal plane more resistant, shifting the critical plane.

Total thickness: $H = 2.0 + 3.0 + 1.0 = 6.0\ \text{m}$.

(a) Equivalent permeabilities

Horizontal (flow parallel to layers — weighted arithmetic mean):

$$k_H = \frac{1}{H}\sum k_i H_i = \frac{k_1 H_1 + k_2 H_2 + k_3 H_3}{H}$$

Compute each $k_i H_i$ (with $H$ in cm to keep $k$ in cm/s consistent, or simply keep $H$ in m since units cancel in the ratio):

• $k_1 H_1 = 3.0\times10^{-3}\times 2.0 = 6.0\times10^{-3}$
• $k_2 H_2 = 6.0\times10^{-5}\times 3.0 = 1.8\times10^{-4}$
• $k_3 H_3 = 2.0\times10^{-2}\times 1.0 = 2.0\times10^{-2}$

Sum $= 6.0\times10^{-3} + 0.18\times10^{-3} + 20.0\times10^{-3} = 26.18\times10^{-3}$

$$k_H = \frac{26.18\times10^{-3}}{6.0} = 4.363\times10^{-3}\ \text{cm/s} = \mathbf{4.36\times10^{-3}\ \text{cm/s}}$$

Vertical (flow perpendicular to layers — weighted harmonic mean):

$$k_V = \frac{H}{\sum (H_i/k_i)}$$

Compute each $H_i/k_i$:

• $H_1/k_1 = 2.0 / 3.0\times10^{-3} = 666.67$
• $H_2/k_2 = 3.0 / 6.0\times10^{-5} = 50000$
• $H_3/k_3 = 1.0 / 2.0\times10^{-2} = 50.0$

Sum $= 666.67 + 50000 + 50.0 = 50716.67$

$$k_V = \frac{6.0}{50716.67} = 1.1831\times10^{-4}\ \text{cm/s} = \mathbf{1.18\times10^{-4}\ \text{cm/s}}$$

(b) Comparison and anisotropy ratio

$$\frac{k_H}{k_V} = \frac{4.363\times10^{-3}}{1.1831\times10^{-4}} = \mathbf{36.9}$$

Water flows much more easily in the horizontal direction. This is because horizontal flow is dominated by the most permeable layer (layer 3) acting like parallel conduits, whereas vertical flow is throttled by the least permeable layer (layer 2, the silt/clay) acting as a bottleneck in series. The deposit is strongly anisotropic ($k_H \approx 37\,k_V$).

Note: total stress $\sigma = \sum \gamma_i z_i$; pore pressure $u = \gamma_w h_w$; effective stress $\sigma' = \sigma - u$.

(a) Hydrostatic conditions

At $z = 2.5\ \text{m}$ (top of clay, also water table):

$$\sigma = 16.5 \times 2.5 = 41.25\ \text{kPa}$$ $$u = 0\ \text{kPa}\quad(\text{water table here})$$ $$\sigma' = 41.25 - 0 = \mathbf{41.25\ \text{kPa}}$$

At $z = 7.5\ \text{m}$ (base of clay):

$$\sigma = (16.5 \times 2.5) + (19.2 \times 5.0) = 41.25 + 96.0 = 137.25\ \text{kPa}$$ $$u = \gamma_w \times 5.0 = 9.81 \times 5.0 = 49.05\ \text{kPa}$$ $$\sigma' = 137.25 - 49.05 = \mathbf{88.20\ \text{kPa}}$$

Summary table (hydrostatic):

(b) With artesian excess head of 3.0 m at base of clay

The total stress is unchanged ($\sigma = 137.25\ \text{kPa}$). The pore pressure at $7.5\ \text{m}$ increases by $\gamma_w \times 3.0$:

$$u = 49.05 + (9.81 \times 3.0) = 49.05 + 29.43 = 78.48\ \text{kPa}$$ $$\sigma' = 137.25 - 78.48 = \mathbf{58.77\ \text{kPa}}$$

Comment on stability: The artesian pressure reduces the effective stress at the base of the clay from $88.20$ to $58.77\ \text{kPa}$ (a $33\%$ reduction), lowering the shear strength of the soil there. If excavation or further artesian rise reduced $\sigma'$ to zero, the clay base could heave / blow out (loss of effective stress = loss of strength). The current condition still has positive effective stress, so it is stable, but the safety margin against base heave is reduced and should be monitored.

Given: $\gamma_{d,max} = 18.1\ \text{kN/m}^3$, $w = 0.145$, $G_s = 2.68$, $\gamma_w = 9.81\ \text{kN/m}^3$.

(a) Degree of saturation and air voids

From $\gamma_d = \dfrac{G_s\gamma_w}{1+e}$, solve for void ratio $e$:

$$1+e = \frac{G_s\gamma_w}{\gamma_d} = \frac{2.68 \times 9.81}{18.1} = \frac{26.291}{18.1} = 1.4525 \Rightarrow e = 0.4525$$

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

$$S_r = \frac{w G_s}{e} = \frac{0.145 \times 2.68}{0.4525} = \frac{0.3886}{0.4525} = 0.8588 = \mathbf{85.9\%}$$

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

$$n_a = \frac{e(1-S_r)}{1+e} = \frac{0.4525\,(1-0.8588)}{1.4525} = \frac{0.4525 \times 0.1412}{1.4525} = \frac{0.06390}{1.4525} = 0.0440 = \mathbf{4.4\%}$$

(b) Zero-air-void dry unit weight at $w = 14.5\%$

ZAV corresponds to $S_r = 100\%$:

$$\gamma_{d,zav} = \frac{G_s\gamma_w}{1 + wG_s} = \frac{2.68 \times 9.81}{1 + (0.145\times 2.68)} = \frac{26.291}{1 + 0.3886} = \frac{26.291}{1.3886} = \mathbf{18.93\ \text{kN/m}^3}$$

Consistency check: The actual $\gamma_{d,max} = 18.1\ \text{kN/m}^3$ is below $\gamma_{d,zav} = 18.93\ \text{kN/m}^3$, which is correct (real compaction never reaches the ZAV line). The ratio $18.1/18.93 = 0.956$ reflects the $\approx 4.4\%$ air voids found in (a), confirming the results are consistent.

Given: $H = 5.0\ \text{m}$, $\gamma = 17.0\ \text{kN/m}^3$, $\phi = 32^\circ$, smooth vertical wall, horizontal backfill (Rankine valid).

(a) Active coefficient, thrust, line of action

Active earth pressure coefficient:

$$K_a = \frac{1 - \sin\phi}{1 + \sin\phi} = \tan^2\!\left(45 - \frac{\phi}{2}\right)$$ $$\sin 32^\circ = 0.52992 \Rightarrow K_a = \frac{1 - 0.52992}{1 + 0.52992} = \frac{0.47008}{1.52992} = \mathbf{0.3073}$$

Active pressure at base ($z = H$):

$$\sigma_a = K_a\gamma H = 0.3073 \times 17.0 \times 5.0 = 26.12\ \text{kPa}$$

Total active thrust per metre run (triangular distribution):

$$P_a = \frac{1}{2}K_a\gamma H^2 = \frac{1}{2}\times 0.3073 \times 17.0 \times 5.0^2 = \frac{1}{2}\times 0.3073 \times 17.0 \times 25 = \mathbf{65.3\ \text{kN/m}}$$

Point of application: at $H/3$ above the base:

$$\bar{z} = \frac{H}{3} = \frac{5.0}{3} = \mathbf{1.667\ \text{m above base}}$$

(b) Additional thrust due to surcharge $q = 25\ \text{kPa}$

A uniform surcharge produces a uniform (rectangular) lateral pressure $K_a q$ over the full height:

$$\sigma_{a,q} = K_a q = 0.3073 \times 25 = 7.683\ \text{kPa}$$

Additional thrust:

$$P_{a,q} = K_a q H = 7.683 \times 5.0 = \mathbf{38.4\ \text{kN/m}}$$

Its line of action is at mid-height $= H/2 = 2.5\ \text{m}$ above the base.

Total thrust (for reference): $P = 65.3 + 38.4 = 103.7\ \text{kN/m}$.

Given: square footing, $B = 2.0\ \text{m}$, $D_f = 1.5\ \text{m}$, $c = 12\ \text{kPa}$, $\phi = 25^\circ$, $\gamma = 18.0\ \text{kN/m}^3$, $N_c = 25.13$, $N_q = 12.72$, $N_\gamma = 9.70$, $FS = 3.0$.

(a) Ultimate bearing capacity (Terzaghi, square footing)

For a square footing the shape factors give:

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

where surcharge $q = \gamma D_f = 18.0 \times 1.5 = 27.0\ \text{kPa}$.

Term by term:

• $1.3\,c\,N_c = 1.3 \times 12 \times 25.13 = 391.9\ \text{kPa}$
• $q\,N_q = 27.0 \times 12.72 = 343.4\ \text{kPa}$
• $0.4\,\gamma\,B\,N_\gamma = 0.4 \times 18.0 \times 2.0 \times 9.70 = 139.7\ \text{kPa}$

$$q_u = 391.9 + 343.4 + 139.7 = \mathbf{875.0\ \text{kPa}}$$

(b) Net safe bearing capacity ($FS = 3.0$)

Net ultimate bearing capacity (subtract overburden surcharge):

$$q_{nu} = q_u - \gamma D_f = 875.0 - 27.0 = 848.0\ \text{kPa}$$

Net safe bearing capacity:

$$q_{ns} = \frac{q_{nu}}{FS} = \frac{848.0}{3.0} = \mathbf{282.7\ \text{kPa}}$$

(If gross safe bearing capacity is required: $q_s = q_{ns} + \gamma D_f = 282.7 + 27.0 = 309.7\ \text{kPa}$.)

Given: $LL = 52\%$, $PL = 24\%$, $w = 38\%$.

(a) Indices and consistency

Plasticity index:

$$PI = LL - PL = 52 - 24 = \mathbf{28\%}$$

Liquidity index:

$$LI = \frac{w - PL}{PI} = \frac{38 - 24}{28} = \frac{14}{28} = 0.50 = \mathbf{0.50}$$

Consistency index:

$$CI = \frac{LL - w}{PI} = \frac{52 - 38}{28} = \frac{14}{28} = 0.50 = \mathbf{0.50}$$

Comment: Since $0 < LI < 1$ (and equivalently $0 < CI < 1$), the natural water content lies between the plastic and liquid limits, so the soil is in a plastic state of medium / firm consistency (here exactly midway, $LI = 0.5$). (Check: $LI + CI = 1$, satisfied.)

(b) USCS classification

A-line PI at $LL = 52$:

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

The soil's $PI = 28 > 23.36$, so the point plots above the A-line $\Rightarrow$ it is a clay (C, not M).

Since $LL = 52\% > 50\%$, it is of high plasticity (H).

$$\boxed{\text{Classification: } CH \text{ (inorganic clay of high plasticity)}}$$

Given: $N_f = 4$, $N_d = 12$, $H = 6.0\ \text{m}$, $k = 4.0\times10^{-4}\ \text{cm/s}$.

Convert $k$ to m/day:

$$k = 4.0\times10^{-4}\ \tfrac{\text{cm}}{\text{s}} \times \tfrac{1\ \text{m}}{100\ \text{cm}} \times 86400\ \tfrac{\text{s}}{\text{day}} = 4.0\times10^{-6} \times 86400 = 0.34560\ \text{m/day}$$

(a) Seepage discharge per metre length

$$q = k\,H\,\frac{N_f}{N_d} = 0.34560 \times 6.0 \times \frac{4}{12} = 0.34560 \times 6.0 \times 0.33333$$ $$q = 0.34560 \times 2.0 = \mathbf{0.691\ \text{m}^3/\text{day per metre}}$$

(b) Uplift pressure head after 8 drops

Head loss per drop:

$$\Delta h = \frac{H}{N_d} = \frac{6.0}{12} = 0.5\ \text{m per drop}$$

Head remaining (relative to downstream / tailwater datum) after $8$ drops from upstream is the head still to be dissipated:

$$h_{remaining} = H - (8 \times \Delta h) = 6.0 - (8 \times 0.5) = 6.0 - 4.0 = \mathbf{2.0\ \text{m of water}}$$

This $2.0\ \text{m}$ is the excess (seepage) pressure head driving uplift at that point above the downstream datum. As a pressure:

$$u_{excess} = \gamma_w \times 2.0 = 9.81 \times 2.0 = \mathbf{19.62\ \text{kPa}}$$

(The total uplift pressure at the point would add the position/elevation head below tailwater; the seepage-induced excess head is $2.0\ \text{m}$.)

(a) Residual vs transported soils

• Residual soils are formed in place by weathering of the parent rock and remain at the location of their origin (no transport). They typically retain features of the parent rock and grade into weathered rock with depth. Nepalese example: the lateritic / weathered residual soils developed over the gneiss and schist of the Mahabharat range and mid-hills.
• Transported soils are weathered material moved away from the parent rock by an agent (water, wind, gravity, ice) and deposited elsewhere; they are often well-sorted and stratified. Nepalese example: the alluvial soils of the Terai plains and the Kathmandu Valley lacustrine deposits (deposited by rivers / former lake).

(b) Sensitivity of clay

Sensitivity $S_t$ is the ratio of the unconfined (or undrained) compressive strength of an undisturbed clay specimen to that of the remoulded specimen at the same water content:

$$S_t = \frac{q_{u,\text{undisturbed}}}{q_{u,\text{remoulded}}}$$

Engineering significance: It measures the loss of strength a clay suffers when its natural structure is disturbed (e.g. by pile driving, sampling, or earthworks). High-sensitivity ("quick") clays can lose almost all strength on remoulding, posing serious risks of flow slides and large settlement; such soils must be handled with great care during construction.

(c) Drained vs undrained shear strength of saturated clay

  tau (shear strength)
   |                       . CD / drained envelope (c', phi')
   |                  .  ' 
   |             .  '          phi' > 0
   |        .  '       ______________ UU / undrained (phi_u = 0)
   |   .  '      ____/      c_u (= s_u), horizontal cap
   | .__/____________________________ sigma (normal stress)

• In drained loading (slow, pore water free to drain), excess pore pressure is zero, effective stresses govern, and strength is described by the effective parameters $c'$ and $\phi'$ (inclined Mohr-Coulomb envelope). Volume change occurs during shear.
• In undrained loading (fast, no drainage) on a saturated clay, no volume change is possible; the total-stress Mohr circles at failure all have the same diameter, so the total-stress envelope is horizontal, giving $\phi_u = 0$ and an undrained shear strength $c_u = s_u$.

UU triaxial test on saturated clay yields the total-stress parameters $\phi_u = 0$ and $c_u$ (the undrained cohesion / shear strength). It does not give effective parameters because pore pressures are neither dissipated nor measured.