BE Civil Engineering (IOE, TU) Soil Mechanics (IOE, CE 603) Question Paper 2076 Nepal

Given: wet mass $M = 1.92\,\text{kg}$, dry mass $M_s = 1.46\,\text{kg}$, total volume $V = 1.05\times10^{-3}\,\text{m}^3$, $G_s = 2.70$, $\rho_w = 1000\,\text{kg/m}^3$.

(a) Basic quantities

Mass of water:

$$M_w = M - M_s = 1.92 - 1.46 = 0.46\,\text{kg}$$

Water content:

$$w = \frac{M_w}{M_s} = \frac{0.46}{1.46} = 0.3151 = \mathbf{31.5\%}$$

Bulk density:

$$\rho = \frac{M}{V} = \frac{1.92}{1.05\times10^{-3}} = 1828.6\,\text{kg/m}^3 \approx \mathbf{1829\,\text{kg/m}^3}$$

Dry density:

$$\rho_d = \frac{M_s}{V} = \frac{1.46}{1.05\times10^{-3}} = 1390.5\,\text{kg/m}^3 \approx \mathbf{1390\,\text{kg/m}^3}$$

(Check: $\rho_d = \rho/(1+w) = 1828.6/1.3151 = 1390.5\,\text{kg/m}^3$ — consistent.)

Volume of solids:

$$V_s = \frac{M_s}{G_s\,\rho_w} = \frac{1.46}{2.70\times1000} = 5.4074\times10^{-4}\,\text{m}^3$$

Volume of voids:

$$V_v = V - V_s = 1.05\times10^{-3} - 5.4074\times10^{-4} = 5.0926\times10^{-4}\,\text{m}^3$$

Void ratio:

$$e = \frac{V_v}{V_s} = \frac{5.0926\times10^{-4}}{5.4074\times10^{-4}} = \mathbf{0.942}$$

Porosity:

$$n = \frac{e}{1+e} = \frac{0.942}{1.942} = 0.485 = \mathbf{48.5\%}$$

Volume of water:

$$V_w = \frac{M_w}{\rho_w} = \frac{0.46}{1000} = 4.60\times10^{-4}\,\text{m}^3$$

Degree of saturation:

$$S = \frac{V_w}{V_v} = \frac{4.60\times10^{-4}}{5.0926\times10^{-4}} = 0.9032 = \mathbf{90.3\%}$$

(b) Verification with $S\,e = w\,G_s$

$$w\,G_s = 0.3151 \times 2.70 = 0.8508$$ $$S = \frac{w\,G_s}{e} = \frac{0.8508}{0.942} = 0.903 = \mathbf{90.3\%}$$

Both methods give $S \approx 90.3\%$, so the computations are internally consistent. However, the sample is not fully saturated ($S < 100\%$); the stated assumption of full saturation is therefore not strictly valid — about $9.7\%$ of the void volume is air. The small air content likely arose from partial drainage or disturbance during sampling.

(a) Ultimate primary consolidation settlement

For a normally consolidated clay:

$$S_c = \frac{C_c\,H}{1+e_0}\,\log_{10}\!\left(\frac{\sigma'_0 + \Delta\sigma}{\sigma'_0}\right)$$

With $H = 4.0\,\text{m}$:

$$S_c = \frac{0.32 \times 4.0}{1+0.95}\,\log_{10}\!\left(\frac{120+80}{120}\right)$$ $$= \frac{1.28}{1.95}\,\log_{10}(1.6667) = 0.65641 \times 0.22185$$ $$S_c = 0.14565\,\text{m} \approx \mathbf{0.146\,\text{m} = 146\,\text{mm}}$$

(b) Time for 60% and 90% consolidation

Drainage: sand on top (pervious), rock at bottom (impervious) gives single drainage. Length of longest drainage path:

$$H_{dr} = H = 4.0\,\text{m}$$

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

For $U = 60\%$ ($T_v = 0.197$):

$$t_{60} = \frac{0.197 \times (4.0)^2}{2.5} = \frac{0.197 \times 16}{2.5} = \frac{3.152}{2.5} = 1.261\,\text{years} \approx \mathbf{1.26\,\text{years}}$$

For $U = 90\%$ ($T_v = 0.848$):

$$t_{90} = \frac{0.848 \times 16}{2.5} = \frac{13.568}{2.5} = 5.427\,\text{years} \approx \mathbf{5.43\,\text{years}}$$

(c) Two field measures to accelerate consolidation

1. Sand drains / prefabricated vertical (wick) drains — install closely spaced vertical drains so water drains radially over a much shorter horizontal path, drastically reducing consolidation time.
2. Preloading / surcharging — apply a temporary surcharge greater than the design load so most settlement occurs before construction; often combined with vertical drains.

(Other acceptable answers: vacuum consolidation, dynamic compaction for the granular fraction.)

(a) Shear strength parameters

Major principal stresses at failure:

• Test 1: $\sigma_{1f} = \sigma_3 + (\sigma_1-\sigma_3)_f = 100 + 220 = 320\,\text{kPa}$
• Test 2: $\sigma_{1f} = 200 + 440 = 640\,\text{kPa}$

For a $c'$–$\phi'$ soil the failure condition is:

$$\sigma_{1f} = \sigma_3\,\tan^2\!\left(45^\circ+\tfrac{\phi'}{2}\right) + 2c'\,\tan\!\left(45^\circ+\tfrac{\phi'}{2}\right)$$

Let $N_\phi = \tan^2(45^\circ+\phi'/2)$ and $M = 2c'\tan(45^\circ+\phi'/2)$.

Test 1: $320 = 100\,N_\phi + M$ Test 2: $640 = 200\,N_\phi + M$

Subtracting: $320 = 100\,N_\phi \Rightarrow N_\phi = 3.20$.

Then $M = 320 - 100(3.20) = 0$, so:

$$2c'\tan(45^\circ+\phi'/2) = 0 \Rightarrow \mathbf{c' = 0}$$

(consistent with a clean sand).

From $N_\phi = \tan^2(45^\circ+\phi'/2) = 3.20$:

$$\tan(45^\circ+\phi'/2) = \sqrt{3.20} = 1.7889$$ $$45^\circ + \phi'/2 = \tan^{-1}(1.7889) = 60.79^\circ$$ $$\phi'/2 = 15.79^\circ \Rightarrow \boxed{\phi' \approx \mathbf{31.6^\circ}}$$

Check via $\sin\phi' = \dfrac{N_\phi-1}{N_\phi+1} = \dfrac{3.20-1}{3.20+1} = \dfrac{2.20}{4.20} = 0.5238 \Rightarrow \phi' = 31.58^\circ$. Consistent.

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

$\sigma_3 = 100\,\text{kPa}$, $\sigma_1 = 320\,\text{kPa}$.

Inclination of failure plane to the horizontal (i.e. to the major principal plane on which $\sigma_1$ acts):

$$\theta = 45^\circ + \frac{\phi'}{2} = 45^\circ + 15.79^\circ = \mathbf{60.8^\circ}$$

Normal stress on the failure plane:

$$\sigma_n = \frac{\sigma_1+\sigma_3}{2} + \frac{\sigma_1-\sigma_3}{2}\cos 2\theta$$

with $2\theta = 121.58^\circ$, $\cos 121.58^\circ = -0.5238$:

$$\sigma_n = \frac{320+100}{2} + \frac{320-100}{2}(-0.5238) = 210 + 110(-0.5238)$$ $$= 210 - 57.6 = \mathbf{152.4\,\text{kPa}}$$

Shear stress on the failure plane:

$$\tau = \frac{\sigma_1-\sigma_3}{2}\sin 2\theta = 110 \times \sin 121.58^\circ = 110 \times 0.8519 = \mathbf{93.7\,\text{kPa}}$$

Verification (point must lie on the envelope, $c'=0$): $\tau = \sigma_n\tan\phi' = 152.4 \times \tan 31.58^\circ = 152.4 \times 0.6143 = 93.6\,\text{kPa}$. Consistent.

(a) Original condition (water table at $1.5\,\text{m}$)

Total vertical stress at $6\,\text{m}$:

$$\sigma = (1.5)(17.5) + (6-1.5)(20.2) = 26.25 + (4.5)(20.2)$$ $$= 26.25 + 90.9 = \mathbf{117.15\,\text{kPa}}$$

Pore water pressure (water table $1.5\,\text{m}$ down, so water head $= 4.5\,\text{m}$):

$$u = (6-1.5)\gamma_w = 4.5 \times 9.81 = \mathbf{44.145\,\text{kPa}}$$

Effective vertical stress (Terzaghi):

$$\sigma' = \sigma - u = 117.15 - 44.145 = \mathbf{73.01\,\text{kPa}}$$

(b) After flooding (water table at ground surface)

Total stress (whole 6 m saturated):

$$\sigma = 6 \times 20.2 = 121.2\,\text{kPa}$$

Pore pressure (water head $=$ full 6 m):

$$u = 6 \times 9.81 = 58.86\,\text{kPa}$$

Effective stress:

$$\sigma' = 121.2 - 58.86 = \mathbf{62.34\,\text{kPa}}$$

Explanation: The effective stress decreases from $73.01$ to $62.34\,\text{kPa}$ (a drop of about $10.67\,\text{kPa}$). Raising the water table submerges the top $1.5\,\text{m}$ of soil that was previously moist; although the total stress increases slightly, the pore pressure increases more, because the buoyant (effective) unit weight $\gamma' = \gamma_{sat}-\gamma_w = 20.2-9.81 = 10.39\,\text{kN/m}^3$ now governs the upper layer instead of $17.5\,\text{kN/m}^3$. The reduced effective stress lowers the soil's shear strength and bearing capacity.

(Check via buoyant weights for case b: $\sigma' = 6\times\gamma' = 6\times10.39 = 62.34\,\text{kPa}$. Consistent.)

(a) Active earth pressure coefficient and distribution

For a smooth vertical wall with horizontal cohesionless backfill (Rankine):

$$K_a = \frac{1-\sin\phi}{1+\sin\phi} = \tan^2\!\left(45^\circ-\frac{\phi}{2}\right)$$ $$= \frac{1-\sin 34^\circ}{1+\sin 34^\circ} = \frac{1-0.5592}{1+0.5592} = \frac{0.4408}{1.5592} = \mathbf{0.2827}$$

Active pressure at depth $z$: $p_a = K_a(\gamma z + q)$.

• At top ($z=0$): $p_a = K_a q = 0.2827\times25 = 7.07\,\text{kPa}$ (uniform, from surcharge).
• At base ($z=5$): $p_a = K_a(\gamma z + q) = 0.2827(18\times5 + 25) = 0.2827\times115 = 32.51\,\text{kPa}$.

Distribution (described): A rectangular block of constant pressure $7.07\,\text{kPa}$ over the full $5\,\text{m}$ height (surcharge contribution) superimposed on a triangle that grows from $0$ at top to $0.2827\times18\times5 = 25.44\,\text{kPa}$ at the base (self-weight contribution). Total at base $= 7.07 + 25.44 = 32.51\,\text{kPa}$.

 top  7.07 kPa |#
              |# \
              |#  \
              |#   \
 base 32.51 kPa|#____\
              (rectangle) + (triangle)

(b) Total active thrust and point of application

Surcharge part (rectangle):

$$P_1 = K_a q H = 7.07 \times 5 = 35.34\,\text{kN/m}$$

acting at mid-height $\bar{y}_1 = H/2 = 2.5\,\text{m}$ above base.

Self-weight part (triangle):

$$P_2 = \tfrac{1}{2}K_a\gamma H^2 = \tfrac{1}{2}\times0.2827\times18\times5^2 = \tfrac{1}{2}\times0.2827\times18\times25 = 63.61\,\text{kN/m}$$

acting at $\bar{y}_2 = H/3 = 1.667\,\text{m}$ above base.

Total active thrust:

$$P_a = P_1 + P_2 = 35.34 + 63.61 = \mathbf{98.95\,\text{kN/m}}$$

Point of application (taking moments about base):

$$\bar{y} = \frac{P_1\bar{y}_1 + P_2\bar{y}_2}{P_a} = \frac{35.34(2.5) + 63.61(1.667)}{98.95}$$ $$= \frac{88.35 + 106.04}{98.95} = \frac{194.39}{98.95} = 1.964\,\text{m}$$

Total active thrust $P_a \approx 98.9\,\text{kN/m}$ acting at $\approx 1.96\,\text{m}$ above the base.

(a) Plasticity index and liquidity index

Plasticity index:

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

Liquidity index:

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

(b) USCS classification and consistency

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

A-line equation: $PI_{A} = 0.73(LL - 20) = 0.73(52-20) = 0.73\times32 = 23.36\%$.

The soil's $PI = 28\% > 23.36\%$, so the point plots above the A-line.

Above A-line plus $LL>50\%$ gives the classification CH — clay of high plasticity (fat clay).

Consistency from $LI$: $LI = 0.50$ lies between $0$ and $1$, so the natural water content sits midway between the plastic and liquid limits. The clay is in a plastic (medium-stiff) state — neither brittle/solid ($LI \le 0$) nor at risk of behaving like a viscous liquid on remoulding ($LI \ge 1$).

(a) Coefficient of permeability

Falling-head formula:

$$k = \frac{a\,L}{A\,t}\,\ln\!\left(\frac{h_1}{h_2}\right) = 2.303\,\frac{a\,L}{A\,t}\,\log_{10}\!\left(\frac{h_1}{h_2}\right)$$

Given: $a = 1.2\,\text{cm}^2$, $L = 9\,\text{cm}$, $A = 32\,\text{cm}^2$, $t = 4\,\text{min} = 240\,\text{s}$, $h_1 = 90\,\text{cm}$, $h_2 = 40\,\text{cm}$.

$$\ln\!\left(\frac{90}{40}\right) = \ln(2.25) = 0.81093$$ $$k = \frac{1.2 \times 9}{32 \times 240}\times 0.81093 = \frac{10.8}{7680}\times0.81093$$ $$= 1.40625\times10^{-3} \times 0.81093 = 1.1404\times10^{-3}\,\text{cm/s}$$ $$\boxed{k \approx \mathbf{1.14\times10^{-3}\,\text{cm/s}}}$$

(b) Appropriate test type

The computed $k \approx 1.1\times10^{-3}\,\text{cm/s}$ indicates a low-permeability (silty/fine) soil. The falling-head test is more appropriate, because in fine soils the flow rate is too small to measure accurately by collecting outflow volume (as the constant-head test requires). The falling-head arrangement measures the drop of head over time, which is sensitive and accurate for low-permeability soils ($k$ roughly $10^{-3}$ to $10^{-6}\,\text{cm/s}$). Constant-head tests are reserved for coarse, highly permeable soils (clean sands and gravels).

(a) Bulk density, maximum dry density and void ratio

Mould volume $V = 1000\,\text{cm}^3 = 1.0\times10^{-3}\,\text{m}^3$; wet mass $M = 1.94\,\text{kg}$.

Bulk (wet) density:

$$\rho = \frac{M}{V} = \frac{1.94}{1.0\times10^{-3}} = 1940\,\text{kg/m}^3 = \mathbf{1.94\,\text{g/cm}^3}$$

Maximum dry density (at $w = 14\%$):

$$\rho_{d,max} = \frac{\rho}{1+w} = \frac{1940}{1.14} = 1701.75\,\text{kg/m}^3 \approx \mathbf{1702\,\text{kg/m}^3}$$

Void ratio from $\rho_d = \dfrac{G_s\rho_w}{1+e}$:

$$1+e = \frac{G_s\rho_w}{\rho_d} = \frac{2.68\times1000}{1701.75} = 1.5749$$ $$e = \mathbf{0.575}$$

(b) Air voids content

Degree of saturation (from $Se = wG_s$):

$$S = \frac{wG_s}{e} = \frac{0.14\times2.68}{0.575} = \frac{0.3752}{0.575} = 0.6525 = 65.25\%$$

Air voids content (as a fraction of total volume):

$$n = \frac{e}{1+e} = \frac{0.575}{1.575} = 0.3651$$ $$n_a = n(1-S) = 0.3651\times(1-0.6525) = 0.3651\times0.3475 = 0.1269 = \mathbf{12.7\%}$$

(Check via $n_a = 1 - \rho_d\left(\dfrac{1}{G_s\rho_w}+\dfrac{w}{\rho_w}\right) = 1 - 1701.75\left(\dfrac{1}{2680}+\dfrac{0.14}{1000}\right) = 1 - 1701.75\times5.131\times10^{-4} = 1 - 0.8731 = 0.1269$. Consistent.)

Air voids content $\approx 12.7\%$.

(a) Ultimate bearing capacity

Given: $B = 2\,\text{m}$, $D_f = 1.2\,\text{m}$, $c = 15\,\text{kPa}$, $\gamma = 18\,\text{kN/m}^3$, $N_c = 17.69$, $N_q = 7.44$, $N_\gamma = 5.39$.

Term-by-term:

• Cohesion term: $1.3\,c\,N_c = 1.3\times15\times17.69 = 344.96\,\text{kPa}$
• Surcharge term: $\gamma D_f N_q = 18\times1.2\times7.44 = 160.70\,\text{kPa}$
• Self-weight term: $0.4\,\gamma B N_\gamma = 0.4\times18\times2\times5.39 = 77.62\,\text{kPa}$

$$q_u = 344.96 + 160.70 + 77.62 = \mathbf{583.28\,\text{kPa} \approx 583\,\text{kPa}}$$

(b) Net safe bearing capacity (FS = 3)

Net ultimate bearing capacity (subtract overburden surcharge $\gamma D_f$):

$$q_{nu} = q_u - \gamma D_f = 583.28 - (18\times1.2) = 583.28 - 21.6 = 561.68\,\text{kPa}$$

Net safe bearing capacity:

$$q_{ns} = \frac{q_{nu}}{FS} = \frac{561.68}{3} = 187.23\,\text{kPa} \approx \mathbf{187\,\text{kPa}}$$

(If the gross safe value is asked: $q_{safe} = q_{ns} + \gamma D_f = 187.23 + 21.6 = 208.8\,\text{kPa}$.)

(a) Seepage discharge per metre

Flow-net discharge formula:

$$q = k\,H\,\frac{N_f}{N_d}$$ $$q = (4\times10^{-5})(6)\,\frac{4}{12} = (4\times10^{-5})(6)(0.3333)$$ $$= (2.4\times10^{-4})(0.3333) = 8.0\times10^{-5}\,\text{m}^3/\text{s per metre}$$ $$\boxed{q = \mathbf{8.0\times10^{-5}\,\text{m}^3/\text{s per metre}}}$$

Convert to litres/day ($1\,\text{m}^3 = 1000\,\text{L}$, $1\,\text{day} = 86400\,\text{s}$):

$$q = 8.0\times10^{-5}\times1000\times86400 = 8.0\times10^{-5}\times8.64\times10^{7}$$ $$= 6912\,\text{L/day} \approx \mathbf{6912\,\text{litres/day per metre}}$$

(b) Head loss per drop and an assumption

Head loss per equipotential drop:

$$\Delta h = \frac{H}{N_d} = \frac{6}{12} = \mathbf{0.5\,\text{m per drop}}$$

An assumption of flow-net analysis (any one):

• The soil is homogeneous and isotropic ($k$ the same in all directions).
• Flow is steady, laminar and obeys Darcy's law.
• The soil is fully saturated and both soil and water are incompressible.
• Flow lines and equipotential lines intersect at right angles, forming approximately curvilinear squares.

(a) Residual vs transported soils

Examples of transported soils (by agent):

• Alluvial soil — transported and deposited by running water (rivers).
• Aeolian (loess) — by wind; Glacial (till/moraine) — by ice; Colluvial — by gravity; Marine/lacustrine — by sea/lake water. (Any one with its agent is sufficient.)

(b) Soil structures

• Single-grained structure: Individual coarse particles (sands, gravels) rest in stable point-to-point contact under gravity, each grain supported by its neighbours. Dense or loose packing is possible.

 O  O  O
  O O O    (grain-to-grain contact)
 O  O  O

• Honeycomb structure: Fine sand/silt particles arch over relatively large voids, forming a honeycomb-like skeleton; it carries static load but the large open voids make it metastable.

  O~O~O
  | gap |   (arches over big voids)
  O~O~O

• Flocculated structure: Clay platelets meet edge-to-face/edge-to-edge with net attractive forces, giving an open card-house arrangement (typical of marine clays); this contrasts with a dispersed (face-to-face, parallel) structure.

  / \  | /     (edge-to-face card-house)
  \ /  / |

Most unstable under vibration: the honeycomb structure (and loose single-grained sands). Its large open voids collapse readily under dynamic/vibratory loading, causing sudden volume reduction and, if saturated, liquefaction.