BE Civil Engineering (IOE, TU) Soil Mechanics (IOE, CE 603) Question Paper 2080 Nepal

Given: Total mass $M = 1525\,\text{g}$, total volume $V = 0.95\times10^{-3}\,\text{m}^3 = 950\,\text{cm}^3$, dry mass $M_s = 1180\,\text{g}$, $G_s = 2.70$.

Mass of water: $M_w = M - M_s = 1525 - 1180 = 345\,\text{g}$.

(a) Water content

$$w = \frac{M_w}{M_s} = \frac{345}{1180} = 0.2924 = \mathbf{29.24\%}$$

Volume of solids:

$$V_s = \frac{M_s}{G_s \rho_w} = \frac{1180}{2.70 \times 1.0} = 437.04\,\text{cm}^3$$

(using $\rho_w = 1\,\text{g/cm}^3$).

Volume of voids: $V_v = V - V_s = 950 - 437.04 = 512.96\,\text{cm}^3$.

Void ratio:

$$e = \frac{V_v}{V_s} = \frac{512.96}{437.04} = \mathbf{1.174}$$

Porosity:

$$n = \frac{e}{1+e} = \frac{1.174}{2.174} = 0.540 = \mathbf{54.0\%}$$

Bulk unit weight:

$$\gamma = \frac{M\,g}{V} = \frac{1.525\,\text{kg} \times 9.81}{0.95\times10^{-3}\,\text{m}^3} = \mathbf{15.75\,\text{kN/m}^3}$$

(b) Degree of saturation Volume of water $V_w = M_w/\rho_w = 345/1.0 = 345\,\text{cm}^3$.

$$S = \frac{V_w}{V_v} = \frac{345}{512.96} = 0.6725 = 67.3\%$$

Check via identity $S\,e = w\,G_s$: $w G_s = 0.2924 \times 2.70 = 0.7895$, so $S = 0.7895/1.174 = 0.6725$ (67.3%). The two computations agree.

Comment: The sample was described as "saturated" but the data give $S \approx 67\%$, i.e. it is actually only partially saturated. The given mass/volume/dry-mass set is internally consistent, so the label "saturated" is inaccurate.

(c) Drying to S = 60% at the same e = 1.174 From $S e = w G_s$:

$$w = \frac{S e}{G_s} = \frac{0.60 \times 1.174}{2.70} = 0.2609 = \mathbf{26.09\%}$$

Dry unit weight depends only on $e$ and $G_s$, so it is unchanged by the loss of water:

$$\gamma_d = \frac{G_s \gamma_w}{1+e} = \frac{2.70 \times 9.81}{1+1.174} = \frac{26.487}{2.174} = \mathbf{12.18\,\text{kN/m}^3}$$

(Equivalently $\gamma_d = \gamma/(1+w) = 15.75/1.2924 = 12.19\,\text{kN/m}^3$, agreeing within rounding.)

(a) Falling-head test Formula:

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

Data (consistent units, mm and s): $a = 80\,\text{mm}^2$, $L = 120\,\text{mm}$, $A = 2800\,\text{mm}^2$, $t = 6\times60 = 360\,\text{s}$, $h_1 = 900\,\text{mm}$, $h_2 = 350\,\text{mm}$.

$$\ln(900/350) = \ln(2.5714) = 0.9445$$ $$k = \frac{80 \times 120}{2800 \times 360}\times 0.9445 = \frac{9600}{1\,008\,000}\times 0.9445$$ $$k = 9.524\times10^{-3}\times0.9445 = 8.996\times10^{-3}\,\text{mm/s}$$

Convert to m/s: $8.996\times10^{-3}\,\text{mm/s} = 8.996\times10^{-6}\,\text{m/s}$.

$$\boxed{k \approx 9.0\times10^{-6}\,\text{m/s}}$$

(b) Layered system ($H = 1.5+2.0+1.0 = 4.5\,\text{m}$)

Equivalent horizontal permeability (parallel flow, thickness-weighted average):

$$k_H = \frac{\sum k_i H_i}{\sum H_i} = \frac{(2.0\times10^{-4})(1.5)+(5.0\times10^{-6})(2.0)+(4.0\times10^{-4})(1.0)}{4.5}$$

Numerator $= 3.0\times10^{-4} + 1.0\times10^{-5} + 4.0\times10^{-4} = 7.10\times10^{-4}$.

$$k_H = \frac{7.10\times10^{-4}}{4.5} = \mathbf{1.578\times10^{-4}\,\text{m/s}}$$

Equivalent vertical permeability (series flow, harmonic):

$$k_V = \frac{\sum H_i}{\sum (H_i/k_i)}$$ $$\sum \frac{H_i}{k_i} = \frac{1.5}{2.0\times10^{-4}} + \frac{2.0}{5.0\times10^{-6}} + \frac{1.0}{4.0\times10^{-4}}$$ $$= 7500 + 400\,000 + 2500 = 410\,000\,\text{s/m... (s)}$$ $$k_V = \frac{4.5}{410\,000} = \mathbf{1.098\times10^{-5}\,\text{m/s}}$$

Ratio:

$$\frac{k_H}{k_V} = \frac{1.578\times10^{-4}}{1.098\times10^{-5}} = \mathbf{14.4}$$

Comment: Horizontal permeability exceeds vertical by a factor of about 14. Vertical flow is dominated (throttled) by the least-permeable layer (Layer 2), whereas horizontal flow is dominated by the most-permeable layers. This anisotropy ($k_H \gg k_V$) is typical of stratified natural deposits.

(a) Coefficient of consolidation For the lab specimen (double drainage), drainage path $H_{dr} = 20/2 = 10\,\text{mm} = 0.01\,\text{m}$.

$$c_v = \frac{T_{50}\,H_{dr}^2}{t_{50}}$$

$t_{50} = 9.0\,\text{min} = 540\,\text{s}$.

$$c_v = \frac{0.197 \times (0.01)^2}{540} = \frac{0.197\times10^{-4}}{540} = 3.648\times10^{-8}\,\text{m}^2/\text{s}$$ $$\boxed{c_v \approx 3.65\times10^{-8}\,\text{m}^2/\text{s}}$$

(Equivalently $c_v = 0.197\times(1.0\,\text{cm})^2/9\,\text{min} = 0.0219\,\text{cm}^2/\text{min}$.)

(b) Primary consolidation settlement (normally consolidated)

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

$H = 3.0\,\text{m}$, $\sigma'_0 = 110\,\text{kPa}$, $\Delta\sigma = 90\,\text{kPa}$, $\sigma'_0+\Delta\sigma = 200\,\text{kPa}$.

$$\log_{10}(200/110) = \log_{10}(1.8182) = 0.2596$$ $$S_c = \frac{0.34 \times 3.0}{1+0.92}\times 0.2596 = \frac{1.02}{1.92}\times0.2596 = 0.53125\times0.2596$$ $$S_c = 0.1379\,\text{m} = \mathbf{137.9\,\text{mm}}$$

(c) Field consolidation times (double drainage, $H_{dr} = 3.0/2 = 1.5\,\text{m}$) Using $t = T H_{dr}^2 / c_v$.

50% consolidation ($T_{50} = 0.197$):

$$t_{50} = \frac{0.197\times(1.5)^2}{3.648\times10^{-8}} = \frac{0.197\times2.25}{3.648\times10^{-8}} = \frac{0.44325}{3.648\times10^{-8}}$$ $$= 1.215\times10^{7}\,\text{s} = \mathbf{140.7\,\text{days}}$$

90% consolidation ($T_{90} = 0.848$):

$$t_{90} = \frac{0.848\times2.25}{3.648\times10^{-8}} = \frac{1.908}{3.648\times10^{-8}} = 5.230\times10^{7}\,\text{s}$$ $$= \mathbf{605.4\,\text{days} \approx 1.66\,\text{years}}$$

(a) Friction angle For a cohesionless soil ($c'=0$):

$$\frac{\sigma_1}{\sigma_3} = \tan^2\!\left(45^\circ + \frac{\phi'}{2}\right) = \frac{1+\sin\phi'}{1-\sin\phi'}$$

Test 1: $\sigma_1 = \sigma_3 + (\sigma_1-\sigma_3) = 100 + 240 = 340\,\text{kPa}$, so $\sigma_1/\sigma_3 = 340/100 = 3.40$. Test 2: $\sigma_1 = 200 + 480 = 680\,\text{kPa}$, so $\sigma_1/\sigma_3 = 680/200 = 3.40$. (Same ratio — consistent.)

Solve $\dfrac{1+\sin\phi'}{1-\sin\phi'} = 3.40$:

$$1+\sin\phi' = 3.40(1-\sin\phi') \Rightarrow 1+\sin\phi' = 3.40 - 3.40\sin\phi'$$ $$4.40\sin\phi' = 2.40 \Rightarrow \sin\phi' = 0.5455$$ $$\boxed{\phi' = 33.06^\circ \approx 33^\circ}$$

(b) Failure plane (Test 1) The failure plane makes an angle $\theta_f$ with the major-principal (horizontal) plane:

$$\theta_f = 45^\circ + \frac{\phi'}{2} = 45 + 16.53 = \mathbf{61.53^\circ}$$

Stresses on the failure plane (using $\sigma_1 = 340$, $\sigma_3 = 100$ kPa): Centre $s = \frac{\sigma_1+\sigma_3}{2} = \frac{340+100}{2} = 220\,\text{kPa}$; radius $t = \frac{\sigma_1-\sigma_3}{2} = \frac{240}{2} = 120\,\text{kPa}$. The Mohr-circle angle is $2\theta_f = 123.06^\circ$.

Normal stress:

$$\sigma_n = s + t\cos 2\theta_f = 220 + 120\cos(123.06^\circ) = 220 + 120(-0.5455) = 220 - 65.46 = \mathbf{154.5\,\text{kPa}}$$

Shear stress:

$$\tau = t\sin 2\theta_f = 120\sin(123.06^\circ) = 120(0.8382) = \mathbf{100.6\,\text{kPa}}$$

Check: $\tau/\sigma_n = 100.6/154.5 = 0.651 = \tan(33.06^\circ)$ ✓ (the point lies on the failure envelope).

(c) Principal stresses

Both tests give the same stress ratio of 3.40, so a single straight-line failure envelope through the origin with $\phi' = 33^\circ$ fits both circles — the data are consistent.

(a) Static stresses ($\gamma_w = 9.81\,\text{kN/m}^3$)

At 0 m: $\sigma = 0$, $u = 0$, $\sigma' = 0$.

At 2.0 m (base of dry sand, top of WT):

$$\sigma = 2.0\times16.5 = 33.0\,\text{kPa},\quad u = 0,\quad \sigma' = 33.0\,\text{kPa}$$

At 5.0 m (base of profile):

$$\sigma = 33.0 + 3.0\times19.5 = 33.0 + 58.5 = 91.5\,\text{kPa}$$ $$u = 3.0\times9.81 = 29.43\,\text{kPa}$$ $$\sigma' = 91.5 - 29.43 = \mathbf{62.07\,\text{kPa}}$$

(b) Downward seepage, $i = 0.30$ over the 3.0 m saturated sand Downward seepage reduces pore pressure (relative to hydrostatic) and increases effective stress. The seepage-induced reduction in head over the layer is $h_s = i\times z = 0.30\times3.0 = 0.90\,\text{m}$, giving a pore-pressure reduction $= \gamma_w h_s = 9.81\times0.90 = 8.829\,\text{kPa}$ at 5.0 m.

Total stress is unchanged: $\sigma = 91.5\,\text{kPa}$. Pore pressure: $u = 29.43 - 8.829 = 20.60\,\text{kPa}$.

$$\sigma' = 91.5 - 20.60 = \mathbf{70.90\,\text{kPa}}$$

Equivalently the extra effective (seepage) stress $= i\,\gamma_w\,z = 8.829\,\text{kPa}$ added to the static $62.07$ gives $70.90\,\text{kPa}$. ✓

(c) Critical (quick-condition) gradient

$$i_c = \frac{G_s - 1}{1+e}$$

Need $e$. The saturated sand has $\gamma_{sat} = \dfrac{(G_s+e)\gamma_w}{1+e} = 19.5$:

$$(2.66 + e)(9.81) = 19.5(1+e)$$ $$26.0946 + 9.81e = 19.5 + 19.5e$$ $$26.0946 - 19.5 = 19.5e - 9.81e \Rightarrow 6.5946 = 9.69e$$ $$e = 0.6806$$ $$i_c = \frac{2.66 - 1}{1+0.6806} = \frac{1.66}{1.6806} = \mathbf{0.988}$$

An upward hydraulic gradient of about 0.99 would initiate boiling (quick condition) at the base of the saturated sand.

(a) Plasticity and liquidity indices

$$PI = LL - PL = 52 - 24 = \mathbf{28\%}$$ $$LI = \frac{w_n - PL}{PI} = \frac{38 - 24}{28} = \frac{14}{28} = \mathbf{0.50}$$

Consistency comment: $0 < LI < 1$, and at $LI = 0.50$ the natural water content sits midway between the plastic and liquid limits. The soil is in a plastic, medium-stiff state in situ (neither brittle/semi-solid nor a viscous liquid).

(b) USCS classification More than 50% passes the 0.075 mm sieve (70% > 50%), so the soil is fine-grained. A-line value at $LL = 52$:

$$PI_{A} = 0.73(LL - 20) = 0.73(52-20) = 0.73\times32 = 23.36$$

Actual $PI = 28 > 23.36$, so the point plots above the A-line. With $LL = 52 > 50$ (high plasticity) and above the A-line, the soil is an inorganic clay of high plasticity:

$$\boxed{\text{USCS group: CH (fat clay)}}$$

(a) Field values (take $g = 9.81\,\text{m/s}^2$)

Field water content:

$$w = \frac{1.95 - 1.70}{1.70} = \frac{0.25}{1.70} = 0.1471 = \mathbf{14.71\%}$$

Field bulk unit weight:

$$\gamma = \frac{1.95\times9.81}{0.00098} \,\text{N/m}^3 = \frac{19.1295}{0.00098} = 19\,520\,\text{N/m}^3 = 19.52\,\text{kN/m}^3$$

Field dry unit weight:

$$\gamma_d = \frac{\gamma}{1+w} = \frac{19.52}{1.1471} = \mathbf{17.02\,\text{kN/m}^3}$$

(Direct check: $\gamma_d = \dfrac{1.70\times9.81}{0.00098} = \dfrac{16.677}{0.00098} = 17\,017\,\text{N/m}^3 = 17.02\,\text{kN/m}^3$ ✓.)

Relative compaction:

$$RC = \frac{\gamma_{d,\text{field}}}{\gamma_{d,\max}}\times100 = \frac{17.02}{17.8}\times100 = \mathbf{95.6\%}$$

(b) Specification check $RC = 95.6\% \ge 95\%$, so the specification IS met. The field moisture (14.7%) is also close to (just below) the OMC of 15.5%, which is favourable: compacting near OMC is consistent with having reached the required density. The fill is acceptable.

(a) Active coefficient and thrust

$$K_a = \frac{1-\sin\phi}{1+\sin\phi} = \tan^2\!\left(45^\circ-\frac{\phi}{2}\right)$$

$\sin 32^\circ = 0.5299$:

$$K_a = \frac{1-0.5299}{1+0.5299} = \frac{0.4701}{1.5299} = \mathbf{0.3073}$$

Active pressure at base: $\sigma_a = K_a \gamma H = 0.3073\times17.0\times6.0 = 31.34\,\text{kPa}$. Total active thrust (area of triangle):

$$P_a = \tfrac{1}{2}K_a\gamma H^2 = \tfrac{1}{2}\times0.3073\times17.0\times6.0^2 = \tfrac{1}{2}\times0.3073\times17.0\times36$$ $$P_a = \mathbf{94.03\,\text{kN/m}}$$

(b) Line of action For a triangular pressure distribution, the resultant acts at one-third the height from the base:

$$\bar{z} = \frac{H}{3} = \frac{6.0}{3} = \mathbf{2.0\,\text{m above the base}}$$

(c) Surcharge thrust A uniform surcharge produces a uniform horizontal pressure $K_a q$ over the full height:

$$\sigma_q = K_a q = 0.3073\times20 = 6.146\,\text{kPa}$$ $$P_q = K_a q H = 6.146\times6.0 = \mathbf{36.88\,\text{kN/m}}$$

This component acts at mid-height ($H/2 = 3.0$ m above base). The combined thrust would be $94.03 + 36.88 = 130.9\,\text{kN/m}$.

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

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

where surcharge $q = \gamma D_f = 18.0\times1.2 = 21.6\,\text{kPa}$, $B = 2.0\,\text{m}$.

Term 1: $1.3\times15\times17.69 = 1.3\times265.35 = 344.96\,\text{kPa}$ Term 2: $21.6\times7.44 = 160.70\,\text{kPa}$ Term 3: $0.4\times18.0\times2.0\times5.0 = 72.0\,\text{kPa}$

$$q_u = 344.96 + 160.70 + 72.0 = \mathbf{577.7\,\text{kPa}}$$

(b) Net ultimate and net safe bearing capacity Net ultimate:

$$q_{nu} = q_u - \gamma D_f = 577.7 - 21.6 = \mathbf{556.1\,\text{kPa}}$$

Net safe (FoS = 3):

$$q_{ns} = \frac{q_{nu}}{F} = \frac{556.1}{3} = \mathbf{185.4\,\text{kPa}}$$

(The gross safe bearing capacity, if required, would be $q_{ns} + \gamma D_f = 185.4 + 21.6 = 207.0\,\text{kPa}$.)

(a) Residual vs. transported soils

Residual soils form by in-situ weathering of the parent rock and remain at the place of their origin; they typically retain a gradation in weathering with depth (more weathered near the surface, grading to fresh rock below). Example: lateritic / red residual soil over basalt.

Transported soils are weathered products carried away from their origin by an agent (water, wind, ice, gravity) and re-deposited elsewhere; their character reflects the transporting agent. Examples: alluvial soil (water-transported), aeolian loess (wind), glacial till (ice).

(b) Well-graded soil A well-graded soil contains a wide and continuous range of particle sizes, with smaller particles filling the voids between larger ones (giving high density and low void ratio). For a well-graded sand (USCS, SW):

$$C_u = \frac{D_{60}}{D_{10}} > 6 \qquad\text{and}\qquad 1 \le C_c = \frac{(D_{30})^2}{D_{10}\,D_{60}} \le 3$$

(For a well-graded gravel, the requirement is $C_u > 4$ with the same $C_c$ range.)

(a) Drained vs. undrained shear strength

Drained strength mobilises while excess pore water pressure is allowed to fully dissipate, so loading is governed by effective stresses; it is characterised by $c'$ and $\phi'$ and represents the long-term condition. Undrained strength applies when loading is fast relative to drainage so pore water cannot escape; for a saturated clay it is described by $\phi_u = 0$ and an undrained cohesion $c_u = s_u$.

For an embankment built quickly on soft saturated clay, the short-term (end-of-construction) undrained condition is critical: pore pressures are highest and effective stress (hence strength) is lowest immediately after loading, before consolidation strengthens the clay. The undrained ($\phi_u = 0$, $c_u$) analysis governs.

(b) Why settlement continues (primary consolidation) The effective-stress principle states $\sigma' = \sigma - u$. When a load $\Delta\sigma$ is suddenly applied to a saturated clay, water (being relatively incompressible and slow to drain through the low-permeability clay) initially carries the entire increment as excess pore pressure: $\Delta u = \Delta\sigma$ and $\Delta\sigma' \approx 0$, so there is little immediate volume change. Over time, the excess pore water slowly drains out under the hydraulic gradient. As $u$ dissipates, the load transfers onto the soil skeleton, so $\sigma'$ increases. The rising effective stress compresses the skeleton, expelling water and reducing the void ratio — this time-dependent volume reduction is primary consolidation. Because the clay's permeability is very low, drainage (and therefore settlement) takes a long time to complete.