Shear Strength MCQ Quiz - Objective Question with Answer for Shear Strength - 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:

• The shear strength of soil depends primarily on the effective stress, which is the stress carried by the soil particles themselves, excluding pore water pressure.

• Total stress includes pore water pressure, which does not contribute to soil strength.

• Effective stress controls the friction and interlocking between soil particles, which governs shear strength.

• This principle is central to soil mechanics and is expressed in Terzaghi’s effective stress principle.

Additional Information

• Effective stress,  σ′=σ−u, where  \sigma" id="MathJax-Element-15-Frame" role="presentation" style="position: relative;" tabindex="0">$\sigma$ $\sigma$ is total stress and u u" id="MathJax-Element-16-Frame" role="presentation" style="position: relative;" tabindex="0">u$u$ $u$ is pore water pressure.

• Shear strength parameters (cohesion c′ and angle of internal friction ϕ′) are based on effective stress.

• Understanding effective stress is critical for foundation design, slope stability, and earth pressure calculations.

Explanation:

• In cohesive soils (like clays), under undrained conditions:

• The structure and bonding between particles provide significant strength in the undisturbed state.

• Once remoulded, the clay loses structure, resulting in lower shear strength.

• Such soils are very rare; normally, clays lose strength on remoulding, so the ratio is usually greater than 1.

 Additional InformationThe ratio of the undisturbed shear strength to the remoulded shear strength is called sensitivity.

• Sensitivity > 1: Soil loses strength on disturbance (most common in clays).

• Sensitivity = 1: Soil is insensitive; remoulding does not affect shear strength (rare).

• Sensitivity > 4: Soil is called sensitive clay, showing significant strength loss on disturbance.

• Sensitivity > 8: Soil is called quick clay, which may liquefy suddenly.

• Used to evaluate the stability and handling characteristics of cohesive soils during construction.

• Important for foundation design, slope stability, and earthworks.

Explanation:

True Cohesion:

• Refers to the inherent bonding strength between soil particles, primarily due to molecular or chemical forces (such as electrostatic forces in clay).

• This cohesion exists even without normal stress.

• It provides shear strength in soils, particularly fine-grained soils like clay, even when normal forces are reduced to zero.

• Correct answer: True cohesion is the reason some shear strength remains in the soil even when normal forces are zero.

Additional Information

Interlocking Friction:

• Refers to the resistance to sliding between irregularly shaped particles, such as in coarse-grained soils like gravel.

• Interlocking friction depends on the normal force to create friction between particles.

• When normal forces are reduced to zero, interlocking friction doesn't significantly contribute to shear strength because it's tied to the pressure between the particles.

Sliding Friction:

• Refers to the frictional force that resists the sliding of two surfaces over one another.

• In soils, sliding friction is usually governed by the normal force, as the frictional resistance increases with greater normal stress.

• Like interlocking friction, sliding friction decreases or is eliminated when normal forces are reduced to zero, so it doesn't contribute to the shear strength in that case.

Apparent Cohesion:

• Occurs in unsaturated soils, where water creates a capillary bond between particles. This bond provides some strength even without normal forces.

• Apparent cohesion is often temporary and depends on factors like soil moisture and surface tension.

• While it can provide shear strength when normal forces are reduced, it's not the primary reason for shear strength when there is no normal force, especially in dry or fully saturated soils.

Explanation:

When heated to very high temperatures, up to redness:

• When clay is heated to high temperatures (up to redness), it undergoes a process called sintering, where the minerals in the clay undergo a chemical transformation that increases its strength.
• The heat drives off moisture and helps the clay particles bond together more strongly. This is why fired clay, such as bricks or ceramics, becomes much stronger.

 Additional Information

When at liquid limit:

• Clay is in a soft, fluid-like state.

• Weak and highly deformable, making it unsuitable for strength.

When with kaolinite at room temperatures:

• Kaolinite is stable and relatively strong at room temperature.

• Strength is not maximized compared to when heated.

When at shrinkage limit:

• Clay no longer shrinks as it dries out.

• Strength is not as high as when heated at high temperatures.

Concept:

The following table shows the different tests and their use:

Additional Information

Difference between Standard penetration test and Static cone penetration test:

The standard penetration test is carried out in a borehole, while the DCPT (Dynamic cone penetration test) and SCPT (Static cone penetration test) are carried out without a borehole.

CONCEPT:

Unconsolidated Un-drained (UU) Test:  

• In this test expulsion of pore water is not permitted in both the stages.
• It is used for clays in short term analysis for clays under un-drained conditions at fast loading rate.

Consolidated Drained (CD) Test:  

• In this test expulsion of pore water is permitted in both the stages.
• It is used for short term and long term stability analysis in saturated sands and long term stability analysis in clays.

Consolidated Un-drained (CU) Test:

• In this test, expulsion of pore water is permitted in 1^st stage but not in second stage. It is used for investigation of safety of earthen dam which may occur due to sudden drawdown of water table.

Unconsolidated Drained (UD) Test:

•  This test is not performed practically because confining pressure acts for long time and if soil is unconsolidated for long time then it cannot be drained in a small period of shear loading.

Important Points 

For loading condition

• Immediate stability is attained by UU test.
• Immediate settlement is attained by CU test.

For unloading condition

• Immediate stability is attained by CU test.

As in CU test consolidation process takes time to complete, So while unloading if we perform the CU test, we can immediately perform the test for obtaining immediate settlement.

Concept:

Terzaghi established that the normal stresses which control the shear strength of the soil are effective stresses. Therefore, the shear strength of the soil can be written as 

Shear strength, \(S\; = \;c' + \bar\sigma \;tan\phi '\)

where c' = cohesion intercept in terms of effective stress

\(\phi'\) = the angle of shearing resistance in terms of effective stress

\(\bar\sigma = \) effective normal stress

Given:

c' = 10 kPa

φ = 30o

\(\sigma = \) 300 kPa

u = 150 kPa

Calculations:

\(\bar\sigma = \) \(\sigma -u=300-150=150\) kPa

S = 10 + (150) x tan30^o

= 96.6 KPa

Concept: 

Failure plane is one at which angle of obliquity is maximum. 

Failure shear stress is one at which difference of shear stress and shear strength is minimum.

θc = Angle of the critical plane to the major principal plane

From triangle property, External angle = Sum of internal angle

\({{\rm{2θ }}_{\rm{c}}} = \frac{\pi }{2} + {{{\beta _{maximum}}}}{}\)

\({{\rm{θ }}_{\rm{c}}} = \frac{\pi }{4} + \frac{{{\beta _{maximum}}}}{2}\)

where, βmax = Angle between resultant stress and normal stress on critical plane = Friction angle of soil = ϕ 

Calculation:

the angle of failure plane relative to the major principle plane in a triaxial test,

\({{\rm{θ }}_{\rm{c}}} = \frac{\pi }{4} + \frac{{{30}}}{2} = 60^\circ\)

Explanation:

Standard penetration test also gives an idea about the unconfined compressive strength of clayey soil.

From the table provided above, we can see that, when the unconfined compressive strength of the soil is 160 kN/sqm, the soil will be stiff and its N-Value will be between 8 - 16.

Concept:

Unconfined Compression test:

(i) It is a special case of the triaxial test in which confining pressure is zero. It means only deviator or shear stress is applied.

(ii) The axial stress at failure is called unconfined compressive strength since confining pressure σ₃ = 0

hence, σ₁ = σ₃ + σ_d = σ_d

Using,

σ₁ = σ₃ tan²(45° + ϕ/2) + 2c tan(45° + ϕ/2)

σ1 = 2c tan(45° + ϕ/2)

Calculation:

Given,

axial stress (σ_d) = σ₁ = 150 kPa 

θ = 45° + ϕ/2 = 60° 

∵ σ1 = 2c tan(45° + ϕ/2)

150 = 2c tan 60° 

c = 43.33 kPa

Explanation:

Punching Shear Failure:

• Punching shear failure occurs where there is relatively high compression of soil under the footing, accompanied by shearing in the vertical direction around the edges of the footing.
• Punching shear may occur in relatively loose sand with relative density less than 35%.
• Punching shear failure may also occur in a soil of low compressibility if the foundation is located at considerable depth.
• The failure surface, which is vertical or slightly inclined and follows the perimeter of the base, never reaches the ground surface.
• There is no heaving of the ground surface away from the edges and no tilting of the footing.
• Relatively large settlements occur in this mode.
• The ultimate bearing capacity is not well defined.

Following the characteristics of punching shear failure:

(i) No failure pattern is observed.
(ii) The failure surface, which is vertical or slightly inclined, follows the perimeter of the base.
(iii) There is no bulging of soil around the footing
(iv) There is no tilting of footing
(v) Failure is characterised in terms of very large settlements
(vi) The ultimate bearing capacity is not well defined.

Additional InformationLocal shear failure

This type of failure is seen in relatively loose sand and soft clay.

Some characteristics of local shear failure are:

1. Failure is not sudden and there is no tilting of footing.

2. Failure surface does not reach the ground surface and slight bulging of soil around the footing is observed

3. Failure surface is not well defined

4. Failure is progressive

5. In the load-settlement curve, there is no well-defined peak

6. Failure is characterized by considerable settlement directly beneath the foundation

7. Significant compression of soil below the footing and partial development of plastic equilibrium is observed.

8. Well-defined wedge and slip surface only beneath the foundation.

General shear failure

Occur in medium to dense soil and stiff clays and has the following properties:

a) A well – defined failure pattern

b) Sudden, catastrophic failure accompanied by tilting of foundation

c) Bulging of the ground surface adjacent to the foundation.

This occurs in soils that show brittle – type stress-strain behavior.

Concept:

From the triaxial test

From the above figure,

\(\sin \phi = \frac{Y}{X}\)

Where,

\(Y = \frac{{{\sigma _1} - {\sigma _3}}}{2}\)

\(X = \frac{{{\sigma _1} + {\sigma _3}}}{2}\)

\(\therefore \sin \phi = \frac{Y}{X} = \frac{{\frac{{{\sigma _1} - {\sigma _3}}}{2}}}{{\frac{{{\sigma _1} + {\sigma _3}}}{2}}} = \frac{{{\sigma _1} - {\sigma _3}}}{{{\sigma _1} + {\sigma _3}}}\)

\(\phi = {\sin ^{ - 1}}\left( {\frac{{{\sigma _1} - {\sigma _3}}}{{{\sigma _1} + {\sigma _3}}}} \right)\)

Where,

σd = deviator stress

σ3 = σc = cell pressure

σ1 = σc + σd

Calculation:

σd = 400 kPa

σ3 = σc = 200 kPa

σ1 = σc + σd = 400 + 200 = 600 kPa

\(\phi = {\sin ^{ - 1}}\left( {\frac{{{\sigma _1} - {\sigma _3}}}{{{\sigma _1} + {\sigma _3}}}} \right) = \;{\sin ^{ - 1}}\left( {\frac{{600 - 200}}{{600 + 200}}} \right)\; = \;{\sin ^{ - 1}}\left( {\frac{{400}}{{800}}} \right) = 30^\circ \)

Concept:

Coulomb's equation,

τ = C + σ tan ϕ

Where,

τ = Shear strength of soil

C = Apparent cohesion

σ = Normal stress on the plane of rupture

ϕ = Angle of internal friction

Calculation:

Given,

σ₁ = 150 kN/m² ,  σ₂ = 250 kN/m2

τ₁ = 110 kN/m2 ,   τ₂ = 120 kN/m²

We know,

τ = C + σ tan ϕ

When, σ1 = 150 kN/m² and τ1 = 110 kN/m2

110 = C + 150 tan ϕ........(1)

When, σ1 = 250 kN/m2 and τ1 = 120 kN/m2

120 = C + 250 tan ϕ........(2)

Equation (2) - (1)

10 = 100 tan ϕ

tan ϕ = 0.1

Put tan ϕ in equation (1), 

110 = C + 150 × 0.1

C = 95 kN/m².

Concept:

The shear strength of soil specimen as per Mohr and Columbus Criterion is given by:

\(\tau = C + \sigma tan\phi\)

Where

C = cohesion and ϕ is the angle of internal friction and these both are called shear strength parameters. These sear strength parameters can only be determined experimentally using the triaxial test.

σ is total stress at the given point.

For pure sand, C = 0 and ϕ ≠ 0

For pure clay, C ≠ 0 and ϕ = 0.