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

Explanation:

• At any point in a 3D stress system, the stress tensor has 9 components (3 normal and 6 shear), but due to the symmetry of the stress tensor, the shear stresses on perpendicular planes are equal.
• This reduces the independent stress components to 6

Additional Information

• Stress at a point in a soil mass is described by how forces are distributed on different planes passing through that point.

• There are three normal stresses, each acting perpendicular to one of the three mutually perpendicular planes (usually called x, y, and z directions).

• There are three shear stresses, acting parallel to the faces of these planes.

• Because shear stresses on perpendicular planes are equal (for example, shear stress on the x-face in the y-direction equals shear stress on the y-face in the x-direction), the total independent stresses reduce from nine to six.

• These six components fully describe the state of stress at any point inside the soil or any solid material under load.

• Understanding these stress components is essential for analyzing soil behavior under loading, which affects settlement, shear failure, and overall stability.

Explanation:

• Plasticity Index (PI) = Liquid Limit (LL) – Plastic Limit (PL)

• Normally, LL > PL, so PI is positive.

• However, if PL > LL, this is physically unrealistic, often due to testing errors or soil peculiarities.

• As per IS 2720 (Part 5) and general practice, in such cases, the Plasticity Index is reported as zero, not negative.

Additional InformationLiquid Limit (LL):

• The minimum water content at which a soil changes from a plastic state to a liquid state.

• It represents the boundary between liquid and plastic behavior.

Plastic Limit (PL):

• The minimum water content at which a soil can be rolled into 3 mm diameter threads without breaking.

• It marks the boundary between plastic and semi-solid states.

Plasticity Index (PI):

• The numerical difference between the liquid limit and the plastic limit:
  PI = LL − PL

• It indicates the range of moisture content over which the soil remains plastic.

Explanation:

• The depth factor (D_f) is a correction factor used in bearing capacity equations (like Terzaghi’s or IS 6403) to account for the depth of the foundation below ground level.

• For base failure Df > 1
  For toe failure Df = 1
  For slope or face failure D_f < 1

Concept:

Air content is defined as the ratio of the volume of air to to the volume of voids. It is denoted by a_c.

a_c = 1 - S_r

where, S_r is Degree of Saturation

Given: Degree of saturation = 60 %

= ac = 1 - S_r = 1 - 0.6 = 0.4 or 40 %

Additional InformationPorosity: It is the ratio of the volume of voids to the total volume of soil.

Void Ratio: It is the ratio of the volume of voids to the volume of solids in a soil sample.

Concept:

The following table shows the different compactors and their application in the field.

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