GEOTECHNICAL ENGINEERING (RCE-501)
B.Tech (SEM-VI) – AKTU
Time: 3 Hours Total Marks: 70
SECTION A
(Attempt all questions in brief – 2 × 7 = 14 marks)
Q1 (a) Define origin of soil.
The origin of soil refers to the process by which soil is formed from parent rock due to physical, chemical, and biological weathering. Based on origin, soils are classified as residual soils and transported soils.
Q1 (b) Draw the figure of element separated soil into three phases.
Soil mass consists of three phases:
solid phase (soil particles), liquid phase (water), and gaseous phase (air).
In a three-phase diagram, the volume and weight of each phase are shown separately to analyze soil properties such as void ratio and degree of saturation.
Q1 (c) Compute the range for capillary rise in silt deposits. Assume value of void ratio e = 0.7.
Capillary rise in soils depends on grain size and void ratio.
For silt deposits, capillary rise generally ranges between 1 m to 10 m.
With void ratio e = 0.7, the capillary rise is usually high, closer to the upper limit of the range.
Q1 (d) Define shrinkage limit.
Shrinkage limit is the water content at which further loss of moisture does not cause any reduction in the volume of soil. Below this limit, soil remains solid without shrinkage.
Q1 (e) What is seepage force?
Seepage force is the force exerted by flowing water on soil particles. It acts in the direction of flow and may reduce effective stress, leading to piping or quicksand conditions.
Q1 (f) Define compression index.
Compression index (Cc) is the slope of the virgin compression curve on a semi-log plot of void ratio versus effective stress. It indicates the compressibility of normally consolidated soils.
Q1 (g) What is factor of safety in slope stability?
Factor of safety is the ratio of resisting forces to driving forces acting on a slope. A value greater than one indicates a stable slope.
SECTION B
(Attempt any three – 7 × 3 = 21 marks)
Q2 (a) Explain the engineering properties of soil.
Engineering properties of soil include permeability, compressibility, shear strength, and plasticity. These properties govern the behavior of soil under load and are essential for foundation design, slope stability, and earth retaining structures.
Q2 (b) Explain effective stress principle.
The effective stress principle states that total stress in soil is shared between soil particles and pore water pressure.
Effective stress = Total stress − Pore water pressure.
Soil strength and deformation depend mainly on effective stress.
Q2 (c) Explain permeability of soil and factors affecting it.
Permeability is the ability of soil to allow water to flow through it.
It depends on grain size, void ratio, soil structure, degree of saturation, viscosity of water, and temperature.
Q2 (d) Explain compaction of soil and its significance.
Compaction is the process of densifying soil by reducing air voids using mechanical means.
It increases strength, reduces settlement, improves stability, and decreases permeability of soil.
SECTION C
Q5 (a) Find out the expression for the law of deflection of flow line at the interface of two dissimilar soils.
(7 marks)
When water flows from one soil to another with different permeabilities, the flow line bends at the interface.
According to the law of deflection:
tanθ1tanθ2=k1k2\frac{\tan \theta_1}{\tan \theta_2} = \frac{k_1}{k_2}tanθ2tanθ1=k2k1
where
θ₁, θ₂ = angles of flow line with the normal
k₁, k₂ = coefficients of permeability of the two soils
This law shows that flow lines bend towards the soil having lower permeability.
Q5 (b) Write the difference between compaction and consolidation.
(7 marks)
Compaction is the reduction in soil volume due to expulsion of air under dynamic loading and occurs immediately.
Consolidation is the reduction in soil volume due to expulsion of water under static loading and occurs over time.
Compaction mainly affects unsaturated soils, while consolidation occurs in saturated soils.
Q6 (a) In a consolidation test, the void ratio of a specimen changes from 1.06 to 0.994 when effective pressure increases from 214 kN/m² to 429 kN/m². Calculate compression index.
(7 marks)
Compression index is calculated using:
Cc=e1−e2log10(σ2σ1)C_c = \frac{e_1 - e_2}{\log_{10} \left(\frac{\sigma_2}{\sigma_1}\right)}Cc=log10(σ1σ2)e1−e2
Substituting values:
Cc=1.06−0.994log10(429/214)C_c = \frac{1.06 - 0.994}{\log_{10} (429/214)}Cc=log10(429/214)1.06−0.994 Cc=0.066log10(2.005)≈0.22C_c = \frac{0.066}{\log_{10} (2.005)} \approx 0.22Cc=log10(2.005)0.066≈0.22
Thus, the compression index of the soil is approximately 0.22, indicating moderate compressibility.
