Geotechnical Engineering and Foundations — ESE Civil

Weightage: Soil mechanics and foundation engineering form a large, reliably numerical block in Civil Prelims Paper II and in the Mains. Most questions apply a single formula to given data, so the skill is knowing which formula fits which situation.

1. Phase relations

Soil is solids, water and air. Define:

• Void ratio $e=V_v/V_s$ and porosity $n=V_v/V$, linked by $n=e/(1+e)$.
• Water content $w=W_w/W_s$ and degree of saturation $S=V_w/V_v$.
• Specific gravity of solids $G$.

The master relation is $Se=wG$.

Unit weights follow:

$\gamma =\frac{(G+Se){\gamma }_w}{1+e},{\gamma }_d=\frac{G{\gamma }_w}{1+e},{\gamma }_{sat}=\frac{(G+e){\gamma }_w}{1+e}$

The submerged unit weight is ${\gamma }^{\prime }={\gamma }_{sat}-{\gamma }_w$.

Worked example. A saturated clay has $w=30\%$ and $G=2.70$. Then $e=wG/S=0.30\times 2.70=0.81$, and ${\gamma }_{sat}=(2.70+0.81)(9.81)/1.81=19.0{\text{kN/m}}^3$.

2. Index properties and classification

Fine soils are described by Atterberg limits: the liquid limit LL, plastic limit PL and shrinkage limit. The plasticity index is $PI=LL-PL$, and the liquidity index is $LI=(w-PL)/PI$. Clays with higher PI are more compressible. The activity of a clay is PI divided by the percentage of clay-size particles.

For coarse soils, grading uses the uniformity coefficient $C_u=D_{60}/D_{10}$ and the curvature coefficient $C_c=D_{30}^2/(D_{10}{D}_{60})$. A well-graded gravel has $C_u>4$ and $C_c$ between 1 and 3, and a well-graded sand has $C_u>6$ with the same $C_c$ range. The Indian classification groups soils on grain size and plasticity (the plasticity chart separating clays from silts).

3. Compaction

Compaction removes air to raise dry density. In the Proctor test the dry density first rises with water content to a maximum at the optimum moisture content (OMC), and then falls as water displaces solids. The curve always lies below the zero-air-voids line. Heavier compaction raises maximum dry density and lowers OMC. Field control uses the relative compaction against the laboratory maximum.

4. Permeability and seepage

Darcy's law states $v=ki$, so $Q=kiA$. For a layered deposit, flow parallel to layers uses the weighted mean $k_h=\sum k_i{H}_i/\sum H_i$, while flow across layers uses $k_v=\sum H_i/\sum (H_i/k_i)$. So $k_h\ge k_v$.

In a flow net, $Q=kH\frac{N_f}{N_d}$ per unit length. Upward seepage gives a critical hydraulic gradient $i_c=(G-1)/(1+e)$, about 1 for typical soils, at which effective stress becomes zero and quicksand occurs.

5. Effective stress and consolidation

Terzaghi's principle: ${\sigma }^{\prime }=\sigma -u$. It is the effective stress, not the total, that controls strength and settlement.

When load is applied to saturated clay, water drains slowly and the excess pore pressure dissipates. This is consolidation. The final settlement of a normally consolidated clay is:

$s=\frac{C_{c}H}{1+e_0}{\log }_{10}\frac{{\sigma }_0^{\prime }+\Delta \sigma }{{\sigma }_0^{\prime }}$

Worked example. A 4 m clay layer has $C_c=0.3$, $e_0=1.0$, ${\sigma }_0^{\prime }=100$ kPa and a load increase of 100 kPa. Then $s=(0.3\times 4/2){\log }_{10}2=0.6\times 0.301=0.18$ m.

The rate follows the time factor $T_v=c_{v}t/H_{dr}^2$, with $H_{dr}$ the longest drainage path. Doubling the drainage path increases time four times. About 90 percent consolidation corresponds to $T_v\approx 0.848$. An overconsolidated clay has seen a higher past stress, with $OCR={\sigma }_p^{\prime }/{\sigma }_0^{\prime }>1$.

6. Shear strength

The Mohr-Coulomb criterion gives ${\tau }_f=c^{\prime }+{\sigma }^{\prime }\tan {\varphi }^{\prime }$. Sands have $c=0$. Tests are:

Undrained loading of clay (short-term, such as a rapid embankment) is checked with total stress and ${\varphi }_u=0$. Long-term stability uses effective stress parameters.

7. Earth pressure

For a smooth vertical wall and horizontal backfill, Rankine's coefficients are:

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

The at-rest coefficient is $K_0\approx 1-\sin \varphi$. For a wall of height $H$ with dry cohesionless backfill, the active thrust is $P_a=\frac{1}{2}{K}_a\gamma H^2$ acting at $H/3$ from the base.

Worked example. For $\varphi =30^{\circ }$, $K_a=1/3$. A 6 m wall with $\gamma =18{\text{kN/m}}^3$ has $P_a=0.5\times \frac{1}{3}\times 18\times 36=108\text{kN/m}$.

A cohesive soil has tension cracks to a depth $z_0=2c/(\gamma \sqrt{K_a})$. Retaining walls must be checked against overturning, sliding, bearing and, for a flexible wall, deep-seated failure.

8. Bearing capacity and shallow foundations

Terzaghi's equation for a strip footing is:

$q_u=c{N}_c+\gamma D_f{N}_q+0.5\gamma B{N}_{\gamma }$

with factors rising steeply with $\varphi$. For a square footing the coefficients become $1.3c{N}_c$ and $0.4\gamma B{N}_{\gamma }$, and for a circular footing $1.3c{N}_c$ and $0.3\gamma B{N}_{\gamma }$. The net safe bearing capacity is $\frac{q_u-\gamma D_f}{FS}+\gamma D_f$, with $FS$ usually 3.

A rise of the water table to the base reduces the third term, roughly halving it. Footings are also checked for settlement, which often governs on clay. Isolated, combined, strap and raft footings are chosen as column loads and spacings increase.

9. Pile foundations

Piles carry load through weak soil to stronger strata, or by friction. The ultimate capacity is $Q_u=Q_p+Q_f$, end bearing plus skin friction. In clay, the alpha method gives $Q_f=\alpha c_u{A}_s$. A pile group may fail as a block, and its efficiency can be below 1 in clay. Negative skin friction adds drag load when the surrounding soil settles faster than the pile.

10. Slope stability and exploration

For an infinite slope in dry cohesionless soil, $FS=\tan \varphi /\tan \beta$, independent of height. Finite slopes are analysed by the method of slices (Fellenius, Bishop) or Taylor's stability number.

Exploration uses boreholes, the standard penetration test (the SPT N-value counts blows for the last 300 mm of a 450 mm drive), cone penetration and geophysical methods.

Common traps

• Using $\gamma$ instead of ${\gamma }^{\prime }$ below the water table in effective stress.
• Reading time of consolidation as proportional to $H$. It goes with $H^2$.
• Forgetting $K_p$ is the reciprocal of $K_a$ for a smooth wall.
• Confusing net and gross bearing capacity.
• Using $Se=wG$ for an unsaturated soil without the actual $S$.

Memory aids

• "S e equals w G": the master phase relation.
• "Double the path, four times the time": consolidation.
• "One third, base up": active thrust and its point of action.

Summary

Phase relations and index properties describe the soil, and permeability, effective stress and consolidation explain water flow and settlement. Shear strength drives earth pressure, bearing capacity, piles and slope stability.

Most questions use one standard formula on given data, so the key is to select the right condition, drained or undrained, dry or saturated.

Exam protocol

• Draw the soil profile and mark the water table first.
• Decide total or effective stress before computing.
• Check the unit system, and use ${\gamma }_w=9.81{\text{kN/m}}^3$ unless told otherwise.
• Keep the four earth-pressure formulas together on a card.