GATE CE Geotechnical Formula Sheet 2027: Soil Mechanics to Piles

This GATE CE geotechnical formula sheet collects the formulas the 2027 Geotechnical Engineering section is built on, from phase relations to pile groups. Each formula comes with its symbols, the case it applies to and the trap that costs marks. A short worked example sits under the key ones, so you can check your recall with real numbers.

Key takeaways

• Geotechnical Engineering averaged 13.8 marks a paper across all sixteen GATE CE papers from 2019 to 2026, with a range of only 4 marks.
• A handful of relations, such as $Se = wG$ and $\gamma_d = \gamma/(1 + w)$, feed almost every other calculation in the section.
• Pore pressure is measured from the water table, and effective stress governs strength, settlement and seepage failure.
• In consolidation, the drainage path decides the answer: double drainage halves the path and cuts the time to a quarter.
• Gross and net bearing capacity differ by $\gamma D_f$, and many questions turn on which one is asked.
• Dynamic pile formulae are off the 2027 syllabus; static formulae, group efficiency and negative skin friction remain.

The notation this sheet uses

Every formula here uses the same symbols. The void ratio $e$ is the volume of voids divided by the volume of solids. The water content $w$ is the mass of water divided by the mass of solids, always as a fraction. $G$ is the specific gravity of solids, and $S$ is the degree of saturation.

Unit weights are in kN/m³. $\gamma$ is the bulk unit weight, $\gamma_d$ the dry unit weight, $\gamma_{sat}$ the saturated unit weight and $\gamma_w$ that of water, taken as 9.81. The submerged unit weight is $\gamma' = \gamma_{sat} - \gamma_w$. A prime on a stress or strength parameter, as in $\sigma'$ or $\phi'$, means effective stress terms. Throughout, "log" means base 10 and "ln" means natural.

Index properties and compaction

Phase relations

For a saturated soil with $w = 0.25$ and $G = 2.68$, $e = 0.25 \times 2.68 = 0.67$ and $n = 0.67/1.67 = 0.40$. For $\gamma = 19$ kN/m³ and $w = 0.12$, $\gamma_d = 19/1.12 = 16.96$ kN/m³.

Atterberg limits and classification

The Indian Standard system splits compressibility into three bands: L for $LL < 35$, I for $35 \le LL < 50$ and H for $LL \ge 50$. The Unified system has no intermediate band.

Compaction and the zero air voids line

The zero air voids line is the dry unit weight the soil would have if every void were full of water:

For $G = 2.65$ and $w = 0.18$, $\gamma_d = (2.65 \times 9.81)/1.477 = 17.60$ kN/m³.

Trap: No compaction curve reaches the zero air voids line, because compaction cannot expel every air void. A test point on or above the line is an arithmetic error, not a very dense soil.

Water in soil: effective stress, permeability and seepage

Effective stress

Effective stress is the part of the total stress carried by the soil skeleton: $\sigma' = \sigma - u$. Build $\sigma$ layer by layer, and measure $u = \gamma_w h_w$ from the water table down.

With the water table at 3 m, $\gamma = 17$ above it and $\gamma_{sat} = 19.5$ below, at 7 m depth $\sigma = 51 + 78 = 129$ kPa, $u = 4 \times 9.81 = 39.24$ kPa and $\sigma' = 89.76$ kPa. The shortcut $51 + 4 \times 9.69$ gives the same answer.

Permeability and seepage

For $G = 2.68$ and $e = 0.60$, $i_{cr} = 1.68/1.60 = 1.05$. A flow net with 3 channels and 10 drops under 8 m of head, in soil with $k = 2 \times 10^{-5}$ m/s, gives $q = 4.8 \times 10^{-5}$ m³/s per metre.

Remember: A net drawn with five flow lines has four flow channels, and eleven equipotential lines give ten drops. Count the spaces, not the lines.

How clay settles: consolidation

The settlement of a normally consolidated clay depends on its compression index $C_c$ and the ratio of final to initial effective stress at mid-depth:

Here $H$ is the layer thickness, $c_v$ the coefficient of consolidation and $d$ the longest drainage path: $H/2$ if both faces drain, $H$ if one does. $T_v = 0.197$ at 50 per cent consolidation and $0.848$ at 90 per cent. For $U \le 60$ per cent, $\displaystyle T_v \approx \frac{\pi}{4}U^2$.

For $C_c = 0.25$, $H = 3$ m, $e_0 = 0.9$ and stress rising from 80 to 200 kPa, $S_c = 0.3947 \times \log_{10} 2.5 = 0.157$ m. A 6 m layer drained on both faces, with $c_v = 2 \times 10^{-7}$ m²/s, reaches 50 per cent in $0.197 \times 3^2/(2 \times 10^{-7})$ s, or 102.6 days.

Trap: With single drainage the same layer has $d = 6$ m, and because $t$ varies as $d^2$ it takes four times as long, 410.4 days. Read the strata below the clay before you choose $d$.

Shear strength: Mohr–Coulomb and the triaxial test

For $c' = 15$ kPa, $\phi' = 25^\circ$ and $\sigma' = 120$ kPa, $\tau_f = 15 + 120 \tan 25^\circ = 70.96$ kPa. For $c' = 10$ kPa, $\phi' = 20^\circ$ and $\sigma'_3 = 150$ kPa, $N_\phi = 2.040$, $\sigma'_1 = 305.94 + 28.56 = 334.50$ kPa, and the deviator stress is 184.50 kPa.

This sheet is a selection. The book's last-minute sheet of formulae and code constants covers all eight technical sections, and every entry traces back to a worked solution in the book. It is part of the GATE CE 2027 book.

Earth pressure, stress distribution and slopes

Rankine earth pressure

For a smooth vertical wall with a level, cohesionless backfill, the Rankine coefficients are:

The active thrust is $\displaystyle P_a = \frac{1}{2}K_a\gamma H^2$, acting at $H/3$ above the base. For $\phi = 36^\circ$, $K_a = 0.2596$. A 6 m wall with $\gamma = 17$ then carries $P_a = 0.5 \times 0.2596 \times 17 \times 36 = 79.44$ kN/m, at 2 m above the base.

For a submerged backfill, apply $K_a$ to $\gamma'$ and add the full water pressure $\gamma_w H$ separately. Water pressure takes no $K_a$.

Boussinesq stress under a point load

On the load axis, $r = 0$ and $\Delta\sigma_z = 3Q/(2\pi z^2)$. For $Q = 500$ kN at $z = 4$ m, $\Delta\sigma_z = 1500/(32\pi) = 14.92$ kPa. The result needs no elastic constants, and it falls as $1/z^2$.

Infinite slopes

For a dry cohesionless infinite slope at angle $\beta$, $F = \tan\phi/\tan\beta$, the same at every depth. For $\phi = 34^\circ$ and $\beta = 25^\circ$, $F = 0.6745/0.4663 = 1.45$. With steady seepage parallel to the slope, $\displaystyle F = \frac{\gamma'}{\gamma_{sat}} \cdot \frac{\tan\phi}{\tan\beta}$, roughly half the dry value.

Shallow foundations: Terzaghi, gross and net

Terzaghi's equation for a strip footing adds a cohesion term, a surcharge term and a width term:

For a 1.5 m strip at 1.2 m depth in clay with $c_u = 40$ kPa and $\gamma = 19$, $q_u = 228 + 22.8 = 250.8$ kPa (gross) and $q_{nu} = 228$ kPa. With $F = 3$, $q_{safe} = 76 + 22.8 = 98.8$ kPa.

In one line: Gross includes the overburden that was already there, net removes it, and the factor of safety divides only the net value.

Piles: static capacity, groups and negative skin friction

The 2027 syllabus keeps only static pile formulae; the dynamic ones are covered in what the 2027 syllabus changed.

For a 0.5 m pile, 12 m long, in clay with $c_u = 40$ kPa and $\alpha = 0.6$, $Q_u = 24 \times 18.85 + 360 \times 0.1963 = 452.4 + 70.7 = 523.1$ kN. A 2 × 3 group of 0.5 m piles at 1.25 m spacing has $\theta = 21.80^\circ$ and $\eta = 1 - 0.2422 \times 7/6 = 0.717$.

For a 0.6 m pile through 4 m of settling fill with $c_u = 20$ kPa and $\alpha = 1$, $F_n = 20 \times 7.54 = 150.8$ kN. The design check becomes $Q_{allowable} \ge \text{load} + F_n$.

To see how this section sits against the rest of the paper, read the GATE CE subject-wise weightage. The last two months strategy for GATE CE shows where a formula sheet fits in revision. Practise these calculations on the on-screen calculator using the GATE virtual calculator guide.

More formula sheets: all of GATE Civil · Environmental and Transportation · Fluid Mechanics and Hydrology · Structural Engineering

Quick revision

1. $Se = wG$, $\gamma_d = \gamma/(1 + w)$ and the zero air voids line $\gamma_d = G\gamma_w/(1 + wG)$, with $w$ as a fraction.
2. Effective stress is $\sigma - u$, with $u$ measured from the water table.
3. $i_{cr} = (G - 1)/(1 + e)$; flow net $q = kHN_f/N_d$, counting channels and drops.
4. Settlement uses log base 10 at mid-depth; time uses the longest drainage path squared.
5. Deviator stress is $\sigma_1 - \sigma_3$; $N_\phi = \tan^2(45^\circ + \phi'/2)$.
6. $K_aK_p = 1$; $\displaystyle P_a = \frac{1}{2}K_a\gamma H^2$ at $H/3$; Boussinesq on the axis is $3Q/(2\pi z^2)$.
7. Net bearing capacity is $q_u - \gamma D_f$; for $\phi = 0$ it is $5.7c$.
8. Piles in clay use $N_c = 9$; Converse–Labarre takes $\theta$ in degrees; negative skin friction adds load.