BE Civil Engineering (IOE, TU) Soil Mechanics (IOE, CE 603) Question Paper 2079 Nepal

Governing phase relations

$$Se = wG_s, \qquad \gamma_{bulk} = \frac{G_s(1+w)\gamma_w}{1+e}, \qquad \gamma_d=\frac{G_s\gamma_w}{1+e}, \qquad n=\frac{e}{1+e}$$

(a) Find $G_s$ and $e$.

From $Se = wG_s$: $e = \dfrac{wG_s}{S} = \dfrac{0.22\,G_s}{0.80} = 0.275\,G_s$.

Substitute into the bulk unit weight equation:

$$18.6 = \frac{G_s(1.22)(9.81)}{1+0.275G_s} = \frac{11.968\,G_s}{1+0.275G_s}$$ $$18.6\,(1+0.275G_s) = 11.968\,G_s \;\Rightarrow\; 18.6 + 5.115\,G_s = 11.968\,G_s$$ $$18.6 = 6.853\,G_s \;\Rightarrow\; G_s = \mathbf{2.714}$$

Then $e = 0.275\times2.714 = \mathbf{0.746}$.

(b) Porosity, dry unit weight, air content.

• Porosity: $n = \dfrac{e}{1+e} = \dfrac{0.746}{1.746} = 0.427 = \mathbf{42.7\%}$
• Dry unit weight: $\gamma_d = \dfrac{\gamma_{bulk}}{1+w} = \dfrac{18.6}{1.22} = \mathbf{15.25\ kN/m^3}$
• Air content: fraction of total volume that is air $= n(1-S) = 0.427\times(1-0.80) = 0.427\times0.20 = \mathbf{0.0854}$ (i.e. $8.5\%$ of total volume is air).

(c) Same void ratio, fully saturated.

Saturated unit weight at $e = 0.746$:

$$\gamma_{sat} = \frac{(G_s+e)\gamma_w}{1+e} = \frac{(2.714+0.746)(9.81)}{1.746} = \frac{3.460\times9.81}{1.746} = \frac{33.94}{1.746} = \mathbf{19.44\ kN/m^3}$$

Water to be added $= \gamma_{sat} - \gamma_{bulk} = 19.44 - 18.6 = 0.84\ kN$ per m³.

As mass: $\dfrac{0.84\times10^3\ N}{9.81\ N/kg} = \mathbf{85.6\ kg}$ of water per cubic metre of soil.

(a) Derivation of falling-head $k$.

At time $t$ let the head be $h$. Flow through the sample (Darcy): $q = kA\dfrac{h}{L}$. In time $dt$ the standpipe level falls by $-dh$, so the flow out of the standpipe is $q = -a\dfrac{dh}{dt}$. Equate:

$$-a\frac{dh}{dt} = \frac{kA}{L}h \;\Rightarrow\; -\frac{a L}{kA}\frac{dh}{h} = dt$$

Integrate from $h_1$ (at $t_1$) to $h_2$ (at $t_2$):

$$\frac{aL}{kA}\ln\frac{h_1}{h_2} = (t_2-t_1) \;\Rightarrow\; \boxed{k = \frac{aL}{A(t_2-t_1)}\ln\frac{h_1}{h_2} = 2.303\frac{aL}{A\,t}\log_{10}\frac{h_1}{h_2}}$$

(b) Numerical $k$.

Areas: $a = \dfrac{\pi}{4}(0.6)^2 = 0.2827\ cm^2$, $A = \dfrac{\pi}{4}(7.0)^2 = 38.485\ cm^2$.

Time $t = 14\times60 = 840\ s$. $L = 9.0\ cm$. $\ln(90/30) = \ln 3 = 1.0986$.

$$k = \frac{aL}{A\,t}\ln\frac{h_1}{h_2} = \frac{0.2827\times9.0}{38.485\times840}\times1.0986$$ $$= \frac{2.5443}{32327.4}\times1.0986 = (7.871\times10^{-5})\times1.0986 = \mathbf{8.65\times10^{-5}\ cm/s}$$

Comment: $k \sim 10^{-5}\ cm/s$ is characteristic of a silt / silty clay (low permeability), which is consistent with the falling-head method being chosen (the constant-head test suits free-draining sands).

(c) Seepage velocity.

Discharge (superficial) velocity: $v = ki = (8.65\times10^{-5})(0.8) = 6.92\times10^{-5}\ cm/s$.

Seepage (actual) velocity through the pores: $v_s = \dfrac{v}{n} = \dfrac{6.92\times10^{-5}}{0.40} = \mathbf{1.73\times10^{-4}\ cm/s}$.

Principle: $\sigma' = \sigma - u$, total stress $\sigma=\sum\gamma_i z_i$, hydrostatic $u=\gamma_w h_w$.

(a) Initial case (WT at 3.0 m).

Total vertical stress:

• At $3.0\ m$: $\sigma = 17.0\times3.0 = 51.0\ kPa$
• At $8.0\ m$: $\sigma = 51.0 + 20.0\times5.0 = 51.0 + 100.0 = 151.0\ kPa$
• At $12.0\ m$: $\sigma = 151.0 + 18.5\times4.0 = 151.0 + 74.0 = 225.0\ kPa$

Pore pressure (WT at 3.0 m):

• At $3.0\ m$: $u = 0$
• At $8.0\ m$: $u = 9.81\times5.0 = 49.05\ kPa$
• At $12.0\ m$: $u = 9.81\times9.0 = 88.29\ kPa$

So initially at $12.0\ m$, $\sigma' = \mathbf{136.71\ kPa}$.

(b) Water table rises to the surface.

Upper layer now saturated, so new total stress at $12.0\ m$:

$$\sigma = 20.0\times3.0 + 20.0\times5.0 + 18.5\times4.0 = 60.0 + 100.0 + 74.0 = 234.0\ kPa$$

New pore pressure (WT at surface, depth of water $=12.0\ m$):

$$u = 9.81\times12.0 = 117.72\ kPa$$

New effective stress:

$$\sigma' = 234.0 - 117.72 = \mathbf{116.28\ kPa}$$

Explanation: Although total stress rose by $9.0\ kPa$ (the upper sand gained $3.0\ kPa/m$ buoyant→saturated weight over 3 m), pore pressure rose by $29.43\ kPa$ ($9.81\times3.0$, the extra 3 m of water column). The net effect is a decrease in effective stress of $136.71-116.28 = 20.43\ kPa$. Physically, raising the water table submerges the upper sand, replacing its moist weight by buoyant weight, so the soil skeleton carries less load — this reduction in $\sigma'$ is what can trigger settlement reversal or reduced bearing capacity when groundwater rises.

(a) Consolidation state and settlement.

Over-consolidation ratio $OCR = \dfrac{\sigma'_p}{\sigma'_0} = \dfrac{150}{110} = 1.36 > 1$, so the clay is over-consolidated.

Final stress $\sigma'_0+\Delta\sigma = 110+120 = 230\ kPa$, which exceeds $\sigma'_p = 150\ kPa$. The stress path crosses the pre-consolidation pressure, so settlement has two parts (recompression up to $\sigma'_p$, then virgin compression beyond it):

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

With $H = 5.0\ m$, $\dfrac{H}{1+e_0} = \dfrac{5.0}{2.05} = 2.4390\ m$.

Recompression part:

$$\log_{10}\frac{150}{110} = \log_{10}1.3636 = 0.1348$$ $$S_1 = 0.06\times2.4390\times0.1348 = 0.14634\times0.1348 = 0.01973\ m$$

Virgin compression part:

$$\log_{10}\frac{230}{150} = \log_{10}1.5333 = 0.1857$$ $$S_2 = 0.32\times2.4390\times0.1857 = 0.78049\times0.1857 = 0.14494\ m$$

Total primary settlement:

$$S_c = 0.01973 + 0.14494 = 0.1647\ m \approx \mathbf{165\ mm}$$

(b) Time for 60% consolidation.

Single drainage (top only) ⇒ drainage path $H_{dr} = H = 5.0\ m$.

$$t_{60} = \frac{T_{60}H_{dr}^2}{c_v} = \frac{0.286\times(5.0)^2}{1.8} = \frac{0.286\times25}{1.8} = \frac{7.15}{1.8} = \mathbf{3.97\ years}$$

(a) Fit the failure envelope.

The envelope is linear: $\tau_f = c + \sigma_n\tan\phi$. Use a least-squares straight-line fit to the three points.

Let $x=\sigma_n$, $y=\tau_f$, $N=3$.

• $\sum x = 100+200+300 = 600$, mean $\bar x = 200$
• $\sum y = 98+158+220 = 476$, mean $\bar y = 158.667$
• $\sum xy = 100(98)+200(158)+300(220) = 9800+31600+66000 = 107400$
• $\sum x^2 = 100^2+200^2+300^2 = 10000+40000+90000 = 140000$

Slope $\tan\phi = \dfrac{\sum xy - N\bar x\bar y}{\sum x^2 - N\bar x^2} = \dfrac{107400 - 3(200)(158.667)}{140000 - 3(200)^2} = \dfrac{107400 - 95200}{140000-120000} = \dfrac{12200}{20000} = 0.610$

$$\phi = \tan^{-1}(0.610) = \mathbf{31.4^\circ}$$

Intercept $c = \bar y - \tan\phi\,\bar x = 158.667 - 0.610(200) = 158.667 - 122.0 = \mathbf{36.7\ kPa}$

Check at $\sigma_n=200$: $\tau = 36.7 + 0.610(200) = 36.7+122.0 = 158.7\ kPa$ ✓ (measured 158).

(b) Shear strength at $\sigma_n = 250\ kPa$.

$$\tau_f = c + \sigma_n\tan\phi = 36.7 + 0.610\times250 = 36.7 + 152.5 = \mathbf{189.2\ kPa}$$

(c) Direct shear vs triaxial.

• Advantage: the direct shear (shear box) test is simple, quick and inexpensive, and is well suited to measuring the drained strength of sands because drainage is rapid and pore pressures dissipate easily.
• Limitation: the failure plane is forced to be horizontal (predetermined) rather than forming on the weakest plane; pore pressure and drainage cannot be controlled or measured, and the stress state is non-uniform — the triaxial test overcomes these by allowing controlled drainage, pore-pressure measurement and a uniform stress state.

(a) Gradation coefficients.

Coefficient of uniformity:

$$C_u = \frac{D_{60}}{D_{10}} = \frac{0.52}{0.13} = \mathbf{4.0}$$

Coefficient of curvature:

$$C_c = \frac{(D_{30})^2}{D_{10}\,D_{60}} = \frac{(0.27)^2}{0.13\times0.52} = \frac{0.0729}{0.0676} = \mathbf{1.078}$$

(b) USCS classification.

For a sand to be classed as well-graded (SW), the USCS criteria are:

$$C_u \ge 6 \quad \text{and} \quad 1 \le C_c \le 3$$

Here $C_c = 1.08$ satisfies the curvature condition, but $C_u = 4.0 < 6$ fails the uniformity condition. Since both conditions must be met, the soil is classified as poorly-graded sand (SP) — it is relatively uniform, with grain sizes clustered over a narrow range.

(a) Saturation and air voids at OMC.

Void ratio from $\gamma_d = \dfrac{G_s\gamma_w}{1+e}$:

$$1+e = \frac{G_s\gamma_w}{\gamma_d} = \frac{2.68\times9.81}{18.4} = \frac{26.291}{18.4} = 1.4289 \Rightarrow e = 0.4289$$

Degree of saturation from $Se = wG_s$:

$$S = \frac{wG_s}{e} = \frac{0.135\times2.68}{0.4289} = \frac{0.3618}{0.4289} = 0.8436 = \mathbf{84.4\%}$$

Air voids (fraction of total volume) $= n(1-S)$ with $n = \dfrac{e}{1+e} = \dfrac{0.4289}{1.4289} = 0.3002$:

$$n_a = 0.3002\times(1-0.8436) = 0.3002\times0.1564 = 0.0470 = \mathbf{4.7\%}$$

(b) Zero-air-voids dry unit weight at the same $w$.

$$\gamma_{d,zav} = \frac{G_s\gamma_w}{1+wG_s} = \frac{2.68\times9.81}{1+0.135\times2.68} = \frac{26.291}{1+0.3618} = \frac{26.291}{1.3618} = \mathbf{19.31\ kN/m^3}$$

Comment: The achieved maximum dry unit weight ($18.4\ kN/m^3$) lies below the zero-air-voids value ($19.31\ kN/m^3$) by about $0.9\ kN/m^3$, consistent with the $\approx 4.7\%$ residual air voids. Compaction can never reach the ZAV line because some air is always trapped; the ZAV curve is the theoretical upper bound that bounds the right (wet) limb of every compaction curve.

(a) Flow-net assumptions and formula.

Assumptions: the soil is homogeneous and isotropic; flow is two-dimensional, steady and laminar (Darcy's law holds); the soil is fully saturated and both water and grains are incompressible; flow lines and equipotential lines intersect at right angles forming approximately curvilinear squares.

Seepage discharge per unit length:

$$q = k\,H\,\frac{N_f}{N_d}$$

where $H$ is the total head loss, $N_f$ the number of flow channels and $N_d$ the number of potential drops.

(b) Numerical discharge.

Convert $k$ to consistent units: $k = 4\times10^{-4}\ cm/s = 4\times10^{-6}\ m/s$.

$$q = (4\times10^{-6})\times5.5\times\frac{4}{12} = (4\times10^{-6})\times5.5\times0.3333 = 7.333\times10^{-6}\ m^3/s\ \text{per m}$$

Per day ($86400\ s$):

$$q = 7.333\times10^{-6}\times86400 = \mathbf{0.634\ m^3/day\ per\ metre}$$

Active coefficient.

$$K_a = \frac{1-\sin\phi}{1+\sin\phi} = \frac{1-\sin30^\circ}{1+\sin30^\circ} = \frac{1-0.5}{1+0.5} = \frac{0.5}{1.5} = \mathbf{0.3333}$$

Submerged (buoyant) unit weight of backfill:

$$\gamma' = \gamma_{sat}-\gamma_w = 19.5 - 9.81 = 9.69\ kN/m^3$$

(a) Pressure distribution at base ($z=5.0\ m$).

Effective active earth pressure (uses $\gamma'$ because the soil is submerged):

$$\sigma'_a = K_a\gamma' H = 0.3333\times9.69\times5.0 = 16.15\ kPa$$

Water pressure (acts fully, hydrostatic):

$$u = \gamma_w H = 9.81\times5.0 = 49.05\ kPa$$

So at the base, lateral pressure $= 16.15\ kPa$ (soil) $+\ 49.05\ kPa$ (water) $= \mathbf{65.20\ kPa}$. Both distributions are triangular (zero at top, maximum at base).

(b) Total horizontal thrust per metre run.

Earth thrust (triangle):

$$P_a = \tfrac{1}{2}K_a\gamma' H^2 = \tfrac{1}{2}\times0.3333\times9.69\times5.0^2 = \tfrac{1}{2}\times0.3333\times9.69\times25 = 40.37\ kN/m$$

Water thrust (triangle):

$$P_w = \tfrac{1}{2}\gamma_w H^2 = \tfrac{1}{2}\times9.81\times25 = 122.63\ kN/m$$

Total horizontal thrust:

$$P = P_a + P_w = 40.37 + 122.63 = \mathbf{163.0\ kN/m}$$

Note how the water pressure dominates — submergence roughly triples the thrust compared with a dry backfill, which is why adequate drainage behind retaining walls is critical.

Terzaghi ultimate bearing capacity (strip / continuous footing):

$$q_u = c N_c + \gamma D_f N_q + 0.5\,\gamma B N_\gamma$$

For cohesionless soil $c = 0$, so the first term vanishes. With $\gamma = 18.5\ kN/m^3$, $D_f = 1.2\ m$, $B = 1.8\ m$:

• Cohesion term: $0\times37.2 = 0\ kPa$
• Surcharge term: $\gamma D_f N_q = 18.5\times1.2\times22.5 = 22.2\times22.5 = 499.5\ kPa$
• Width term: $0.5\,\gamma B N_\gamma = 0.5\times18.5\times1.8\times19.7 = 0.5\times18.5\times35.46 = 0.5\times656.01 = 328.0\ kPa$

$$q_u = 0 + 499.5 + 328.0 = \mathbf{827.5\ kPa}$$

Net ultimate (subtract overburden $\gamma D_f = 18.5\times1.2 = 22.2\ kPa$):

$$q_{nu} = 827.5 - 22.2 = 805.3\ kPa$$

Net safe bearing capacity (FoS = 2.5):

$$q_{ns} = \frac{805.3}{2.5} = \mathbf{322.1\ kPa}$$

Comment: With $c=0$, capacity comes entirely from the friction-dependent surcharge ($N_q$) and self-weight ($N_\gamma$) terms. Here the surcharge term ($499.5\ kPa$) is the largest single contributor; deepening the footing (raising $D_f$) is therefore an effective way to increase bearing capacity in cohesionless soils.

(a) Factors affecting field compaction.

The dry density achieved depends on: (i) water content — there is an optimum at which density peaks; (ii) compactive effort/energy — more passes or heavier rollers raise the maximum dry density and lower the OMC; (iii) soil type and gradation — well-graded coarse soils compact to higher densities than uniform or highly plastic soils; (iv) type of compaction equipment — smooth/vibratory rollers suit granular soils, sheepsfoot/pad-foot rollers suit clays; (v) layer (lift) thickness — thinner lifts compact more uniformly.

(b) Skempton's pore-pressure parameters.

The change in pore pressure under a general stress change is

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

• $B$ measures the response to isotropic (all-round) stress; $B \approx 1$ for fully saturated soil and $B \to 0$ for dry soil — so $B$ is a practical check of saturation in a triaxial test.
• $A$ captures the pore-pressure response to the shear (deviator) component; at failure $A_f$ indicates soil behaviour — high positive $A_f$ for soft normally consolidated clays (contractive), low or negative $A_f$ for heavily over-consolidated/dense soils (dilative). The parameters let undrained pore pressures, hence effective stresses, be predicted in design.

(c) Active vs passive earth pressure.

• Active pressure develops when the wall moves away from the backfill, the soil expands laterally and reaches its minimum lateral stress; $K_a = \dfrac{1-\sin\phi}{1+\sin\phi} < 1$.
• Passive pressure develops when the wall moves into the soil, compressing it to its maximum lateral resistance; $K_p = \dfrac{1+\sin\phi}{1-\sin\phi} > 1$ (and $K_p = 1/K_a$).

Both vary linearly (triangular) with depth:

   surface 0 ---\                  /--- 0 surface
               \  active         /  passive
                \  (small)      /   (large)
   base   Ka γH  \____      ____/  Kp γH

Passive pressure is much larger than active and requires larger wall movement to mobilise fully.

(d) Stokes' law and hydrometer analysis.

Stokes' law gives the terminal settling velocity of a small sphere in a viscous fluid:

$$v = \frac{\gamma_s-\gamma_w}{18\,\mu}\,D^2$$

where $D$ is the particle diameter and $\mu$ the fluid viscosity. In hydrometer analysis, fine soil ($<0.075\ mm$) is dispersed in water; larger particles settle faster, so by measuring the suspension density (with a hydrometer) at known times and depths, the diameter that has settled past the measuring depth and the percentage finer can be computed — extending the grain-size distribution into the silt and clay range where sieving is impossible. Stokes' law assumes spherical particles, laminar settling and no particle interference.