Soil Mechanics and Foundation Engineering MCQ Quiz - Objective Question with Answer for Soil Mechanics and Foundation Engineering - Download Free PDF

Explanation:

 Consolidated Undrained Test (CU Test):

• Drainage is allowed during consolidation under cell pressure, so the soil consolidates and reaches equilibrium.

• During shearing, no drainage is allowed, but pore water pressure is measured using transducers.

• Suitable for determining both total and effective stress parameters by analyzing pore pressure development.

• This is the most commonly used triaxial test for cohesive soils, especially in undrained loading conditions like during construction.

Additional InformationConsolidated Undrained Test (CU Test):

• Drainage is allowed during consolidation under cell pressure, so the soil consolidates and reaches equilibrium.

• During shearing, no drainage is allowed, but pore water pressure is measured using transducers.

• Suitable for determining both total and effective stress parameters by analyzing pore pressure development.

• This is the most commonly used triaxial test for cohesive soils, especially in undrained loading conditions like during construction.

 Unconsolidated Undrained Test (UU Test):

• Drainage is not allowed during either consolidation or shearing stages.

• The test is quick and does not require pore pressure measurement.

• Suitable for short-term stability analysis under total stress conditions.

• It gives total shear strength parameters, and no information about effective stress or pore pressure is obtained.

Unconsolidated Drained Test (UD Test):

• Drainage is not allowed during the first stage (consolidation), but allowed during the shearing stage.

• This test is theoretically possible but rarely conducted in practice due to limited applicability.

• It does not represent real field conditions well and is not commonly referenced in design.

Explanation:

• Unit weight of dry soil (γ_d) → weight of soil solids only per unit volume.

• Unit weight of wet soil (γ) → includes soil solids + water weight per unit volume.

• Submerged unit weight (γ_sub) → effective weight of soil when immersed in water, calculated as: \(\gamma_{sub}=\gamma_{sat}-\gamma_{w}\), where γsat = saturated unit weight,  γw=unit weight of water

• When soil is submerged, the buoyant force reduces its effective weight.

• The apparent or submerged unit weight decreases compared to the saturated or total unit weight.

• This is why submerged soils have lower unit weight (γsub) than saturated soils.

Additional Information Unit Weight of Dry Soil

• Refers to the weight of soil solids only per unit volume.

• It is the unit weight when no water is present in the soil pores.

• Dry unit weight is an important property for understanding the compaction and strength of soils in geotechnical engineering.

• It is always less than or equal to the wet (bulk) or saturated unit weight.

Unit Weight of Wet Soil

• Represents the weight of both soil solids and the water present in the pores of the soil mass.

• This is the natural condition of soil found in the field — where some amount of moisture is always present.

• The presence of water increases the unit weight compared to dry unit weight. 

Concept:

Seepage velocity is related to Darcy’s velocity (discharge velocity, v) and porosity (n):

\(v_s=\frac{v}{n}\)

Calculation:

Given Data:

• Cross-sectional area (A) = 35 cm² = 35 × 10⁻⁴ m²

• Length of specimen (L) = 20 cm = 0.2 m

• Head (h) = 60 cm = 0.6 m

• Discharge (Q) = 120 ml in 6 min
  → Q = 120 ml = 120 × 10⁻⁶ m³ in 6 min
  → Time = 6 min = 360 sec
  → Q/sec = (120 × 10⁻⁶) / 360 = 3.33 × 10⁻⁷ m³/sec

• Dry weight of sand (Wd) = 1120 g = 1.12 kg

• G (specific gravity) = 2.68

Discharge Velocity: \(v=\frac{Q}{A}=\frac{3.33\times10^{-7}}{35\times10^{-4}}=9.51\times10^{-4}m/s\)

Volume of specimen = A × L = \(35\times10^{-4}\times0.2=7\times10^{-4}m^3\)

Volume of voids: \(\frac{Mass}{G\times\gamma_w}=\frac{1.12}{2.68\times1000}=4.179\times10^{-4}m^3\)

Void volume = Total volume − Volume of solids = 2.821 x 10^-4 m³

Porosity: \(n=\frac{V_v}{V}=\frac{2.821\times10^{-4}}{7\times10^{-4}}=0.403\)

Seepage Velocity: \(v_s=\frac{v}{n}=\frac{9.51\times10^{-4}}{0.403}=2.36\times10^{-3}m/s\)

Explanation:

• Effective stress controls the shear strength of soil by determining the resistance offered by soil particles to sliding. As effective stress increases, the interparticle contact force increases, improving the soil’s ability to resist shear. This relationship is defined by the Mohr-Coulomb strength criterion.
• Compressibility of soil is directly influenced by effective stress, as it governs how much the soil consolidates under load. When effective stress increases, voids between soil particles decrease, leading to settlement. This is particularly important in clayey soils under long-term loading.
• Permeability is indirectly affected by effective stress, especially in fine-grained soils. As effective stress increases, the soil becomes denser, reducing the size of pore spaces. Smaller pore spaces lower the rate at which water can flow through the soil.
• Volume changes in soils under loading or unloading are dependent on changes in effective stress. Higher effective stress compresses the soil, while a decrease can lead to expansion or swelling. This behavior must be considered in foundation and earthwork designs.
• In saturated conditions, total stress is divided between pore water pressure and effective stress. Only the effective portion contributes to soil stiffness and strength. Understanding this is critical in analyzing submerged or water-logged soils.
• All essential soil behavior—strength, settlement, and permeability—is governed by effective stress rather than total stress. Engineers rely on effective stress to predict soil response under different loading conditions. It is the central concept in soil mechanics.

Additional InformationShear Strength

• Shear strength depends directly on effective stress because only the soil grains (not pore water) resist shearing. As effective stress increases, the soil structure becomes more stable and resistant to failure.

• In saturated soils, a rise in pore water pressure reduces effective stress, leading to a decrease in shear strength, which is critical in slope failures and bearing capacity analysis.

Compressibility

• Compressibility refers to the reduction in soil volume under applied stress and is controlled by effective stress. When effective stress increases, soil particles are pressed closer together, reducing voids and causing settlement.

• In fine-grained soils like clay, compressibility is high and time-dependent due to low permeability, which delays drainage and extends consolidation time.

Permeability

• Permeability is affected indirectly by effective stress, as higher effective stress compacts the soil, reducing pore size and thus the ease of water flow.

• This effect is more prominent in cohesive soils, where small changes in structure from increased effective stress can significantly reduce permeability.

Explanation:

The following relationship will be used for solving the above problem

\(\gamma=\frac{G\gamma_w(1+w)}{1+e}\)

where e = void ratio, w = moisture content, G = Specific Gravity: \(\gamma\) is specific unit weight of soil; \(\gamma_w\) is the specific unit weight of water

Calculation:

Given: e = 0.84: \(\gamma=16.97 {kN}/{m}^3\); \(\gamma_w=9.81 {kN}/{m}^3\); G = 2.7

So, \(16.97=\frac{2.7\times9.81(1+w)}{1+0.84}\)

\(1.18=1+w\)

w = 0.18 = 18%

Explanation:

Porosity (η) is defined as the ratio of volume of voids to the total volume of the soil sample in a given soil mass.

\(\eta = \frac{{{V_v}}}{V}\)

The porosity of soils can vary widely.

The porosity of loose soils can be about η = 50 to 60%.

Concept:

Toughness Index: 

• Toughness index is defined as the ratio of plasticity index (I_P) of the soil to the flow index (I_F) of the soil.

\(\begin{array}{l} Toughess\;index = \frac{{Plasticity\;index}}{{Flow\;index}}\\ \end{array}\)

• This gives us an idea of the shear strength of soil at its plastic limit. 
• The toughness index varies between 0 to 3. If If is more, the rate of loss of shear strength is more. Hence, the toughness index is less.
• When the toughness index is less than 1, the soil is said to be friable, which means it can be easily crushed at the plastic limit.

Explanation:

Coefficient of Earth Pressure at rest:

\(\rm{{{\rm{k}}_{\rm{r}}} = \frac{{ {\rm{μ }}}}{{1 - {\rm{μ }}}}}\)

Where,

μ is the coefficient is the Poisson ratio.

As per IS 4651 (Part 2): 1989, Table 1

Concept:

According to Darcy’s law,

For laminar flow Vα i

V = K × i

Where,

V = velocity of water flowing in soil, K = coefficient of permeability

\({\rm{i}} = hydraulic~ gradient= \frac{{{{\rm{h}}_{\rm{L}}}}}{{\rm{L}}}\)

h_L = head difference, L = seepage length

Seepage velocity is given by,

V_S = V/n

Where,

 n = porosity of the soil

Calculation:

T = Time = 100 days

D  = distance = 100 m

h_L = head difference = 3 m

n = 15 % = 0.15

i = h_L/L = 3/100

V_S = D/T = 100/100 = 1 m/day

V = K × i

V = V_S × n

∴ V_S × n = K × i

\(K = \frac{{{V_s} \times n}}{i} = \frac{{1 \times 0.15}}{{\frac{3}{{100}}}} = 5\;m/day\)

Soil classification as per Indian standards:

∴ According to IS Classification, there are 18 groups of soils:

8 groups of coarse-grained, 9 groups of fine-grained, and 1 of peat

Explanation:

The range of optimum water content for different types of soil is as follows:

Concept:

slope = \(dy\over dx \) = constant

⇒ \({V_L-V_P}\over {w_L - w_p} \) = \(V_P-V_D\over {w_p - w_s}\)

where 

w_L = LL = liquid limit

w_P = PL = plastic limit

w_s = SL = shrinkage limit

V_L = volume of soil mass at LL

V_P = volume of soil mass at PL

V_s = volume of soil mass at SL

Calculation:

Given:

SL = 20%

PL = 40%

V_P = 1.2 V_d

V_L = 1.5 V_d

⇒ \({V_L-V_P}\over {w_L - w_p} \) = \(V_P-V_D\over {w_p - w_s}\)

Now, putting values, we get 

⇒ \({1.5 V_d - 1.2 V_d}\over I_P\) = \(1.2V_d - V_d \over {40-20}\)

⇒ I_P = 30%

Hence, the plasticity Index Pl of the soil is 30%.

Concept:

Over Consolidation Ratio:

Over Consolidation Ratio ( OCR ) is defined as the ratio of maximum pressure experienced in the past

( Preconsoildation pressure ) to the present overburden pressure.

\({\bf{OCR}} = \frac{{{\bf{Maximum}}\;{\bf{pressure}}\;{\bf{in}}\;{\bf{past}}}}{{{\bf{Present}}\;{\bf{overburden}}\;{\bf{pressure}}}}\)

OCR > 1 ⇒ It is Over consolidated soil

OCR = 1 It is Normally consolidated soil

OCR < 1It is Under consolidated soil

Calculation:

OCR of clay = 4

Since the OCR of the given soil is greater than 1, it is overconsolidated clay.

A-factor:

A – factor is a function of the over-consolidation ratio in the case of overconsolidated clay.

For OC clays, A = f (OCR)

Concept:

The effective stress concept was developed by "Terzaghi". A fully saturated soil relates three types of stress:

1. Total Stress
2. Pore Pressure( Neutral Stress)
3. Effective Stress

Total Stress(\(\sigma\)) on a plane within soil mass is the force per unit area of soil mass transmitted in the normal direction across a plane.

Pore pressure(u) is the pressure of the water filling the void space between solid particles.

Effective stress(\(\bar \sigma\)) is defined as equal to the total stress minus the neutral pressure.

The relation between total stress, pore pressure, and effective stress is as follows:

\(\bar \sigma = \sigma -u\)

where \(\bar \sigma\)= effective stress, \(\sigma\)= total stress, u= pore pressure

Explanation:

Given:

γ_w = 10 kN/m³, γsat = 20 kN/m³, γ_b = 18 kN/m3

Effective stress at a depth of 10 m below the ground level:

γ_eff = γ_b × 3 + γ_sub × 7

γeff = 18 × 3 + (20 - 10) × 7

γeff = 124 kN/m³

Concept:

Area ratio:

It can be defined as the ratio of the maximum cross-sectional area of the cutting edge to the area of the soil sample.

The area ratio can be expressed as

\({A_r} = \frac{{D_2^2 - D_1^2}}{{D_1^2}}\)

where

D₁ = inner diameter of cutting edge

D₂ = outer diameter of cutting edge

Note:

For stiff formation (A_r)_max = 20%

Soft sensitive clay (Ar)max = 10%

Maximum thickness of cutting edge = \(\frac{D_2 \ - \ D_1}{2}\)

Calculation:

Given, D₂ = 70 mm 

Soil is stiff clay, So (Ar)max = 20%

\({A_r} = \frac{{D_2^2 - D_1^2}}{{D_1^2}}\)

\({0.2} = \frac{{70^2 - D_1^2}}{{D_1^2}}\)

D₁ = 63.9 mm

Maximum thickness of cutting edge = \(\frac{D_2 \ - \ D_1}{2}\)

= \(\frac{70 \ - \ 63.9}{2}\) = 3.05 mm