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All questions of Shear Strength of Soils for Civil Engineering (CE) Exam
If the shearing stress is zero on two planes, then the angle between the two planes is- a)45°
- b)90º
- c)135°
- d)225°
Correct answer is option 'B'. Can you explain this answer?
If the shearing stress is zero on two planes, then the angle between the two planes is
a)
45°
b)
90º
c)
135°
d)
225°
|
Lavanya Menon answered |
The planes on which shearing stress is zero are major and minor principal planes. These two planes are inclined at 90° to each other.
In an unconfined compression test on a clay specimen of initial volume V and length L, the area of cross-section at failure is- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
In an unconfined compression test on a clay specimen of initial volume V and length L, the area of cross-section at failure is
a)
b)
c)
d)
|
Subham Unni answered |
As the sample is sheared, its length decreases and the diameter increases. The cross-sectional area A at any stage during shear can be determined
The state of stress on a plane inclined at an angle of 45° to the horizontal plane is represented on the Mohr’s circle by a point at which- a)Normal stress is
and shear stress is 
- b)Normal stress is
and shear stress is 
- c)Normal stress is
and shear stress is zero - d)Both normal stress and shear stress are equal
Correct answer is option 'A'. Can you explain this answer?
The state of stress on a plane inclined at an angle of 45° to the horizontal plane is represented on the Mohr’s circle by a point at which
a)
Normal stress is and shear stress is
and shear stress is b)
Normal stress is and shear stress is
and shear stress is c)
Normal stress is and shear stress is zero
and shear stress is zerod)
Both normal stress and shear stress are equal
|
Gate Funda answered |
A plane inclined at an angle θ = 45° to the horizontal plane is represented by the radius making an angle 2θ with the σ axis.
The state of stress is represented by Point ‘P'
Which one of the following is the appropriate tri- axial test to assess the immediate stability of an unloading probfem, such as an excavation of a clay slope?- a)UU test
- b)CU test
- c)CD test
- d)Unconsolidated drained tests
Correct answer is option 'A'. Can you explain this answer?
Which one of the following is the appropriate tri- axial test to assess the immediate stability of an unloading probfem, such as an excavation of a clay slope?
a)
UU test
b)
CU test
c)
CD test
d)
Unconsolidated drained tests
|
|
Prashanth Mehra answered |
For problems of short term-stability of foundations: excavations and earth dams UU (Unconsolidated , Undrained) tests are appropriate. For problems of long-term stability either UU test or CD test is appropriate, depending upon the drainage conditions of soil.
Which of the following stresses does not have any influence on strength of a material?- a)Major principal stress
- b)Minor principal stress
- c)Intermediate principal stress
- d)Shearing stress
Correct answer is option 'C'. Can you explain this answer?
Which of the following stresses does not have any influence on strength of a material?
a)
Major principal stress
b)
Minor principal stress
c)
Intermediate principal stress
d)
Shearing stress
|
Simran Dasgupta answered |
Intermediate principal stress does not have any influence on the strength of a material.
Major Principal Stress:
- Major principal stress is the maximum normal stress that acts normal to the plane on which it acts.
- It influences the strength of a material by causing it to deform or fail when it exceeds the material's yield strength.
Minor Principal Stress:
- Minor principal stress is the minimum normal stress that acts normal to the plane on which it acts.
- It also influences the strength of a material by contributing to its overall state of stress and deformation.
Shearing Stress:
- Shearing stress is the stress that acts parallel to the plane on which it acts.
- It influences the strength of a material by causing it to deform or fail through shear deformation.
Intermediate Principal Stress:
- Intermediate principal stress is the stress that lies between the major and minor principal stresses.
- It does not have a direct influence on the strength of a material, as it does not contribute significantly to the overall state of stress that affects the material's deformation or failure.
Major Principal Stress:
- Major principal stress is the maximum normal stress that acts normal to the plane on which it acts.
- It influences the strength of a material by causing it to deform or fail when it exceeds the material's yield strength.
Minor Principal Stress:
- Minor principal stress is the minimum normal stress that acts normal to the plane on which it acts.
- It also influences the strength of a material by contributing to its overall state of stress and deformation.
Shearing Stress:
- Shearing stress is the stress that acts parallel to the plane on which it acts.
- It influences the strength of a material by causing it to deform or fail through shear deformation.
Intermediate Principal Stress:
- Intermediate principal stress is the stress that lies between the major and minor principal stresses.
- It does not have a direct influence on the strength of a material, as it does not contribute significantly to the overall state of stress that affects the material's deformation or failure.
Which of the following is a disadvantage of the shear box test?- a)Stress condition of soil is complex
- b)The test cannot be used for coarse grained soil
- c)No control on the drainage of soil
- d)The shear box test is more complex test
Correct answer is option 'A'. Can you explain this answer?
Which of the following is a disadvantage of the shear box test?
a)
Stress condition of soil is complex
b)
The test cannot be used for coarse grained soil
c)
No control on the drainage of soil
d)
The shear box test is more complex test
|
|
Rashi Shah answered |
Complex stress condition of soil:
The main disadvantage of the shear box test is the complex stress condition of the soil during testing. In this test, the soil is subjected to a combination of normal and shear stresses, making it difficult to accurately determine the failure criteria of the soil. This complexity can lead to uncertainty in the test results, affecting the reliability of the data obtained.
Interpretation of test results:
Due to the complex stress conditions, interpreting the results of the shear box test can be challenging. The failure criteria for different soils may vary, and determining the exact failure plane under these conditions can be difficult. This can lead to errors in the analysis and design of geotechnical structures based on the test results.
Applicability to different soil types:
Another limitation of the shear box test is that it may not be suitable for all types of soils. The test is more commonly used for fine-grained soils, and its applicability to coarse-grained soils is limited. This restricts the use of the test in certain geotechnical applications where the soil type is predominantly coarse-grained.
Complexity of the test:
The shear box test is considered to be more complex compared to other geotechnical tests. Setting up the test apparatus, controlling the drainage conditions, and ensuring accurate measurements can be challenging. This complexity can require experienced personnel to conduct the test and may increase the overall cost and time involved in testing.
In conclusion, while the shear box test is a valuable tool for determining the shear strength parameters of soils, it does have certain limitations and disadvantages, with the complex stress condition of the soil being a significant drawback. It is important to consider these factors when choosing the appropriate testing method for geotechnical investigations.
The Mohr-Coulomb failure criterion, which is defined by shear strength (s) consisting of effective normal stress (σ’), cohesion (c’) and effective angle of friction (ϕ’), is given by,- a)s = σ’ + c’ tanϕ’
- b)s = c’ tanϕ’ + σ’
- c)s = c’ – σ’ tanϕ’
- d)s = c’ + σ’ tanϕ’
Correct answer is option 'D'. Can you explain this answer?
The Mohr-Coulomb failure criterion, which is defined by shear strength (s) consisting of effective normal stress (σ’), cohesion (c’) and effective angle of friction (ϕ’), is given by,
a)
s = σ’ + c’ tanϕ’
b)
s = c’ tanϕ’ + σ’
c)
s = c’ – σ’ tanϕ’
d)
s = c’ + σ’ tanϕ’
|
|
Lavanya Menon answered |
Concept:
Shear strength of soil is the capacity of soil to resist shearing stresses. It can be defined as the maximum value of shear stress that can be mobilized within a soil mass.
Shear strength of soil is the capacity of soil to resist shearing stresses. It can be defined as the maximum value of shear stress that can be mobilized within a soil mass.
Note:
If this value exceeded by the shear stress on any plane or a surface at a point, failure will occur in soil because of the movement of a portion of the soil mass along that particular plane. The soil is then said to fail in shear. Therefore, shear strength generally represented the strength of the soil.
As per Mohr-Coulomb criteria, Shear strength (s) of soil is given as:
s = c' + σ' tanϕ'
Where,
c' = effective cohesion
σ' = effective normal stress
ϕ' = effective angle of friction
If this value exceeded by the shear stress on any plane or a surface at a point, failure will occur in soil because of the movement of a portion of the soil mass along that particular plane. The soil is then said to fail in shear. Therefore, shear strength generally represented the strength of the soil.
As per Mohr-Coulomb criteria, Shear strength (s) of soil is given as:
s = c' + σ' tanϕ'
Where,
c' = effective cohesion
σ' = effective normal stress
ϕ' = effective angle of friction
For saturated soil, Skempton’s B-parameter is- a)nearly zero
- b)nearly 0.5
- c)nearly 1.0
- d)very high
Correct answer is option 'C'. Can you explain this answer?
For saturated soil, Skempton’s B-parameter is
a)
nearly zero
b)
nearly 0.5
c)
nearly 1.0
d)
very high
|
|
Prerna Kaur answered |
Skempton’s B-parameter is given by,
Cv is volume compressibility of pore fluid under isotropic conditions,
Cs is the coefficient of compressibility of the soil skeleton;n is porosity
In a fully saturated soil, the compressibility of the pore water (Cv) is negligible compared with the compressibility of the soil mass (Cs). Therefore, the ratio (Cv/Cs) tends to zero and the coefficient B becomes equal to unity.
In a partially saturated soil, the compressibility of the air in the voids Is high. The ratio (Cv/Cs) has a value greater than unity, and, therefore, the pore pressure coefficient B has a value of less than unity.
Cv is volume compressibility of pore fluid under isotropic conditions,
Cs is the coefficient of compressibility of the soil skeleton;n is porosity
In a fully saturated soil, the compressibility of the pore water (Cv) is negligible compared with the compressibility of the soil mass (Cs). Therefore, the ratio (Cv/Cs) tends to zero and the coefficient B becomes equal to unity.
In a partially saturated soil, the compressibility of the air in the voids Is high. The ratio (Cv/Cs) has a value greater than unity, and, therefore, the pore pressure coefficient B has a value of less than unity.
The relationship between a soil cohesion c and unconfined compressive strength qu is given as- a)c = qu
- b)c = qu/2
- c)c = qu/3
- d)c = 2qu
Correct answer is option 'B'. Can you explain this answer?
The relationship between a soil cohesion c and unconfined compressive strength qu is given as
a)
c = qu
b)
c = qu/2
c)
c = qu/3
d)
c = 2qu
|
|
Muskaan Sen answered |
Relationship Between Soil Cohesion and Unconfined Compressive Strength
Definition:
Soil cohesion (c) is the internal molecular attraction that holds together the particles of a soil and is responsible for the shear strength of the soil.
Unconfined compressive strength (qu) is the maximum axial compressive stress that a cylindrical soil specimen can withstand without lateral support.
Relationship:
The relationship between soil cohesion (c) and unconfined compressive strength (qu) is given by:
c = qu/2
This means that soil cohesion is equal to half of the unconfined compressive strength.
Explanation:
The relationship between soil cohesion and unconfined compressive strength can be derived from the Mohr-Coulomb failure criterion. The Mohr-Coulomb failure criterion is an equation that describes the relationship between the shear strength of a soil and the normal stress acting on it.
The equation is given as:
τ = c + σtan(φ)
where,
τ = shear stress
c = soil cohesion
σ = normal stress
φ = angle of internal friction
When the normal stress is zero, the equation simplifies to:
τ = c
This means that when there is no normal stress acting on the soil, the shear stress is equal to the soil cohesion. In other words, soil cohesion is the shear strength of the soil when there is no normal stress acting on it.
When a normal stress is applied to the soil, the shear strength of the soil increases due to the frictional resistance between the particles. The increase in shear strength is given by:
τ = σtan(φ)
The unconfined compressive strength is the maximum axial compressive stress that a cylindrical soil specimen can withstand without lateral support. It can be calculated from the following equation:
qu = 2c + σtan(φ)
When σ = 0, the equation simplifies to:
qu = 2c
This means that the unconfined compressive strength is equal to twice the soil cohesion when there is no normal stress acting on the soil.
Therefore, the relationship between soil cohesion (c) and unconfined compressive strength (qu) is given by:
c = qu/2
Conclusion:
The relationship between soil cohesion and unconfined compressive strength is important in geotechnical engineering as it allows engineers to estimate the strength of a soil based on simple laboratory tests. The relationship is derived from the Mohr-Coulomb failure criterion, which is a fundamental equation in soil mechanics.
Definition:
Soil cohesion (c) is the internal molecular attraction that holds together the particles of a soil and is responsible for the shear strength of the soil.
Unconfined compressive strength (qu) is the maximum axial compressive stress that a cylindrical soil specimen can withstand without lateral support.
Relationship:
The relationship between soil cohesion (c) and unconfined compressive strength (qu) is given by:
c = qu/2
This means that soil cohesion is equal to half of the unconfined compressive strength.
Explanation:
The relationship between soil cohesion and unconfined compressive strength can be derived from the Mohr-Coulomb failure criterion. The Mohr-Coulomb failure criterion is an equation that describes the relationship between the shear strength of a soil and the normal stress acting on it.
The equation is given as:
τ = c + σtan(φ)
where,
τ = shear stress
c = soil cohesion
σ = normal stress
φ = angle of internal friction
When the normal stress is zero, the equation simplifies to:
τ = c
This means that when there is no normal stress acting on the soil, the shear stress is equal to the soil cohesion. In other words, soil cohesion is the shear strength of the soil when there is no normal stress acting on it.
When a normal stress is applied to the soil, the shear strength of the soil increases due to the frictional resistance between the particles. The increase in shear strength is given by:
τ = σtan(φ)
The unconfined compressive strength is the maximum axial compressive stress that a cylindrical soil specimen can withstand without lateral support. It can be calculated from the following equation:
qu = 2c + σtan(φ)
When σ = 0, the equation simplifies to:
qu = 2c
This means that the unconfined compressive strength is equal to twice the soil cohesion when there is no normal stress acting on the soil.
Therefore, the relationship between soil cohesion (c) and unconfined compressive strength (qu) is given by:
c = qu/2
Conclusion:
The relationship between soil cohesion and unconfined compressive strength is important in geotechnical engineering as it allows engineers to estimate the strength of a soil based on simple laboratory tests. The relationship is derived from the Mohr-Coulomb failure criterion, which is a fundamental equation in soil mechanics.
The drained test is also known as ___________- a)Direct shear test
- b)Slow test
- c)Vane shear test
- d)Quick test
Correct answer is option 'B'. Can you explain this answer?
The drained test is also known as ___________
a)
Direct shear test
b)
Slow test
c)
Vane shear test
d)
Quick test
|
|
Simran Dasgupta answered |
Drained Test in Geotechnical Engineering
The drained test in geotechnical engineering is also known as the slow test. This test is used to determine the shear strength of a soil sample under drained conditions.
Types of Shear Tests
There are various types of shear tests conducted in geotechnical engineering, such as:
- Direct shear test
- Vane shear test
- Quick test
Significance of Drained Test
The drained test is crucial in understanding how a soil sample behaves when subjected to shear stresses under drained conditions. This helps in assessing the stability of soil structures like slopes, embankments, and foundations.
Slow Test vs. Quick Test
The drained test is referred to as the slow test because it involves allowing the excess pore water pressure to dissipate slowly during the test. On the other hand, the quick test is conducted under undrained conditions, where the excess pore water pressure does not have time to dissipate.
Conclusion
In conclusion, the drained test, also known as the slow test, is an essential procedure in geotechnical engineering to determine the shear strength of soil samples under drained conditions. Understanding the behavior of soils under different conditions is crucial for designing safe and stable structures.
The drained test in geotechnical engineering is also known as the slow test. This test is used to determine the shear strength of a soil sample under drained conditions.
Types of Shear Tests
There are various types of shear tests conducted in geotechnical engineering, such as:
- Direct shear test
- Vane shear test
- Quick test
Significance of Drained Test
The drained test is crucial in understanding how a soil sample behaves when subjected to shear stresses under drained conditions. This helps in assessing the stability of soil structures like slopes, embankments, and foundations.
Slow Test vs. Quick Test
The drained test is referred to as the slow test because it involves allowing the excess pore water pressure to dissipate slowly during the test. On the other hand, the quick test is conducted under undrained conditions, where the excess pore water pressure does not have time to dissipate.
Conclusion
In conclusion, the drained test, also known as the slow test, is an essential procedure in geotechnical engineering to determine the shear strength of soil samples under drained conditions. Understanding the behavior of soils under different conditions is crucial for designing safe and stable structures.
Which of the following is coulomb’s strength equation?- a)S = c + tan φ
- b)C = s + c tan φ
- c)S = c + σ tan φ
- d)S = tan φ
Correct answer is option 'C'. Can you explain this answer?
Which of the following is coulomb’s strength equation?
a)
S = c + tan φ
b)
C = s + c tan φ
c)
S = c + σ tan φ
d)
S = tan φ
|
Sparsh Unni answered |
Explanation:
Coulomb's strength equation
- Coulomb's strength equation is given by: S = c + σ tan φ
- Here, S represents the shear strength of the soil, c is the cohesion intercept, σ is the normal stress on the plane, and φ is the effective stress angle.
Understanding the components of the equation:
- Shear strength (S): Shear strength is the maximum resistance of a soil to shearing stresses. It is a combination of cohesive and frictional strength.
- Cohesion intercept (c): Cohesion is the property of soil particles to stick together. It represents the shear strength of the soil when there is no normal stress acting on the plane.
- Normal stress (σ): Normal stress is the stress acting perpendicular to the plane under consideration in soil mechanics.
- Effective stress angle (φ): Effective stress angle is the angle between the failure plane and the major principal stress.
Significance of Coulomb's strength equation:
- Coulomb's strength equation is essential in geotechnical engineering for determining the shear strength of soils.
- It helps in analyzing stability of slopes, retaining walls, foundations, and other geotechnical structures.
- The equation provides a quantitative measure of soil behavior under different stress conditions.
Conclusion:
- Understanding Coulomb's strength equation is crucial for geotechnical engineers to design safe and stable structures. By considering the cohesive and frictional properties of soil, engineers can make informed decisions regarding the design and construction of various civil engineering projects.
Coulomb's strength equation
- Coulomb's strength equation is given by: S = c + σ tan φ
- Here, S represents the shear strength of the soil, c is the cohesion intercept, σ is the normal stress on the plane, and φ is the effective stress angle.
Understanding the components of the equation:
- Shear strength (S): Shear strength is the maximum resistance of a soil to shearing stresses. It is a combination of cohesive and frictional strength.
- Cohesion intercept (c): Cohesion is the property of soil particles to stick together. It represents the shear strength of the soil when there is no normal stress acting on the plane.
- Normal stress (σ): Normal stress is the stress acting perpendicular to the plane under consideration in soil mechanics.
- Effective stress angle (φ): Effective stress angle is the angle between the failure plane and the major principal stress.
Significance of Coulomb's strength equation:
- Coulomb's strength equation is essential in geotechnical engineering for determining the shear strength of soils.
- It helps in analyzing stability of slopes, retaining walls, foundations, and other geotechnical structures.
- The equation provides a quantitative measure of soil behavior under different stress conditions.
Conclusion:
- Understanding Coulomb's strength equation is crucial for geotechnical engineers to design safe and stable structures. By considering the cohesive and frictional properties of soil, engineers can make informed decisions regarding the design and construction of various civil engineering projects.
The direct shear test can also be called as ___________.
- a)Simple shear test
- b)Strain controlled shear box test
- c)Direct shear box test
- d)All of the mentioned
Correct answer is option 'B'. Can you explain this answer?
The direct shear test can also be called as ___________.
a)
Simple shear test
b)
Strain controlled shear box test
c)
Direct shear box test
d)
All of the mentioned
|
|
Varun Mukherjee answered |
Direct Shear Test
The direct shear test is a laboratory test used to determine the shear strength parameters of soils. It involves applying a shear force to a soil sample in a controlled manner and measuring the corresponding displacements. This test is commonly used in geotechnical engineering to evaluate the stability and strength of soil structures such as slopes, embankments, and foundations.
Alternative Name
The direct shear test is also commonly referred to as the "direct shear box test". This name is derived from the equipment used in the test, which consists of a shear box apparatus. The shear box apparatus consists of two halves, the upper and lower halves, which can be moved horizontally relative to each other to apply the shear force on the soil sample.
Purpose of the Test
The direct shear test is conducted to determine the shear strength parameters of the soil sample, namely the cohesion (c) and the angle of internal friction (φ). These parameters are important for analyzing the stability and deformation characteristics of soil structures. The test helps in determining the soil's resistance to shear forces and its ability to withstand external loads.
Procedure
The direct shear test involves the following steps:
1. Preparation of Soil Sample: A soil sample is carefully prepared and placed in the shear box apparatus. The sample is compacted to achieve a desired density and moisture content.
2. Application of Normal Stress: A normal stress is applied to the soil sample by loading the specimen vertically with a known weight.
3. Application of Shear Force: The upper half of the shear box is moved horizontally relative to the lower half, applying a shear force to the soil sample. The shear force is increased gradually until failure occurs.
4. Measurement of Displacements: During the test, the displacements of the soil sample are measured using displacement transducers. These measurements help in determining the shear deformation characteristics of the soil.
5. Calculation of Shear Strength Parameters: Based on the applied shear force and the corresponding displacements, the shear strength parameters of the soil sample (cohesion and angle of internal friction) can be calculated.
Conclusion
In conclusion, the direct shear test is commonly referred to as the direct shear box test because it involves using a shear box apparatus to apply shear forces to a soil sample. This test is crucial for determining the shear strength parameters of soils and evaluating the stability and strength of soil structures.
The direct shear test is a laboratory test used to determine the shear strength parameters of soils. It involves applying a shear force to a soil sample in a controlled manner and measuring the corresponding displacements. This test is commonly used in geotechnical engineering to evaluate the stability and strength of soil structures such as slopes, embankments, and foundations.
Alternative Name
The direct shear test is also commonly referred to as the "direct shear box test". This name is derived from the equipment used in the test, which consists of a shear box apparatus. The shear box apparatus consists of two halves, the upper and lower halves, which can be moved horizontally relative to each other to apply the shear force on the soil sample.
Purpose of the Test
The direct shear test is conducted to determine the shear strength parameters of the soil sample, namely the cohesion (c) and the angle of internal friction (φ). These parameters are important for analyzing the stability and deformation characteristics of soil structures. The test helps in determining the soil's resistance to shear forces and its ability to withstand external loads.
Procedure
The direct shear test involves the following steps:
1. Preparation of Soil Sample: A soil sample is carefully prepared and placed in the shear box apparatus. The sample is compacted to achieve a desired density and moisture content.
2. Application of Normal Stress: A normal stress is applied to the soil sample by loading the specimen vertically with a known weight.
3. Application of Shear Force: The upper half of the shear box is moved horizontally relative to the lower half, applying a shear force to the soil sample. The shear force is increased gradually until failure occurs.
4. Measurement of Displacements: During the test, the displacements of the soil sample are measured using displacement transducers. These measurements help in determining the shear deformation characteristics of the soil.
5. Calculation of Shear Strength Parameters: Based on the applied shear force and the corresponding displacements, the shear strength parameters of the soil sample (cohesion and angle of internal friction) can be calculated.
Conclusion
In conclusion, the direct shear test is commonly referred to as the direct shear box test because it involves using a shear box apparatus to apply shear forces to a soil sample. This test is crucial for determining the shear strength parameters of soils and evaluating the stability and strength of soil structures.
A dense dry sand sample under tri-axial test has the confining pressure of 20 t/m2. The angle of internal! friction is 350. The sample will fail at normal pressure of about:- a)30.5 t/m2
- b)31.5 t/m2
- c)32.5 t/m2
- d)33.5 t/m2
Correct answer is option 'B'. Can you explain this answer?
A dense dry sand sample under tri-axial test has the confining pressure of 20 t/m2. The angle of internal! friction is 350. The sample will fail at normal pressure of about:
a)
30.5 t/m2
b)
31.5 t/m2
c)
32.5 t/m2
d)
33.5 t/m2
|
|
Siddharth Datta answered |
A major difference between the direct shear test and tri axial shear test is _____________- a)Control on the drainage level
- b)Stress condition
- c)None of the mentioned
- d)All of the mentioned
Correct answer is option 'A'. Can you explain this answer?
A major difference between the direct shear test and tri axial shear test is _____________
a)
Control on the drainage level
b)
Stress condition
c)
None of the mentioned
d)
All of the mentioned
|
|
Akshat Datta answered |
Understanding the Major Differences between Direct Shear Test and Triaxial Shear Test
In geotechnical engineering, both the direct shear test and the triaxial shear test are crucial for evaluating the shear strength of soils. However, they differ significantly in terms of drainage control.
Control on the Drainage Level
- The direct shear test primarily allows for control over drainage conditions during the test.
- It can be conducted under drained or undrained conditions, making it versatile for testing cohesionless and cohesive soils.
- The specimen is placed in a shear box and sheared horizontally, which facilitates drainage or pore pressure measurements.
Stress Condition
- The triaxial shear test, on the other hand, is more complex and involves applying pressure uniformly around the soil sample.
- It can simulate different stress conditions, such as unconsolidated undrained (UU), consolidated undrained (CU), or consolidated drained (CD).
- This capability allows for a more comprehensive analysis of soil behavior under varying confining pressures.
Conclusion
In summary, the key difference highlighted in the question is the control over drainage levels. The direct shear test provides a straightforward method for assessing shear strength under specific drainage conditions, while the triaxial test offers more versatility in simulating real-world conditions through various stress states. Understanding these differences is essential for selecting the appropriate test based on the soil type and the conditions being studied.
In geotechnical engineering, both the direct shear test and the triaxial shear test are crucial for evaluating the shear strength of soils. However, they differ significantly in terms of drainage control.
Control on the Drainage Level
- The direct shear test primarily allows for control over drainage conditions during the test.
- It can be conducted under drained or undrained conditions, making it versatile for testing cohesionless and cohesive soils.
- The specimen is placed in a shear box and sheared horizontally, which facilitates drainage or pore pressure measurements.
Stress Condition
- The triaxial shear test, on the other hand, is more complex and involves applying pressure uniformly around the soil sample.
- It can simulate different stress conditions, such as unconsolidated undrained (UU), consolidated undrained (CU), or consolidated drained (CD).
- This capability allows for a more comprehensive analysis of soil behavior under varying confining pressures.
Conclusion
In summary, the key difference highlighted in the question is the control over drainage levels. The direct shear test provides a straightforward method for assessing shear strength under specific drainage conditions, while the triaxial test offers more versatility in simulating real-world conditions through various stress states. Understanding these differences is essential for selecting the appropriate test based on the soil type and the conditions being studied.
Mohr envelope can be considered to be straight if the angle of internal friction φ is assumed to be __________- a)90°
- b)>90°
- c)<90°
- d)None of the mentioned
Correct answer is option 'D'. Can you explain this answer?
Mohr envelope can be considered to be straight if the angle of internal friction φ is assumed to be __________
a)
90°
b)
>90°
c)
<90°
d)
None of the mentioned
|
Ankit Mukherjee answered |
Is zero. The Mohr envelope is a graphical representation of the relationship between the normal stress and shear stress on a shear plane. It is used in soil mechanics and geotechnical engineering to analyze the shear strength of soils and rocks.
The angle of internal friction is a material property that describes the resistance of a material to shear deformation. It is the angle between the normal force and shear force required to cause shear failure in a material. If the angle of internal friction is zero, the material can be considered to be perfectly plastic and the Mohr envelope will be a straight line.
In reality, most materials have a non-zero angle of internal friction, which means that the Mohr envelope will be curved. The curvature of the Mohr envelope depends on the material properties, and can be used to determine the shear strength of the material.
In summary, the Mohr envelope can be considered to be straight if the angle of internal friction is zero. However, in most cases the angle of internal friction is non-zero, and the Mohr envelope is curved.
The angle of internal friction is a material property that describes the resistance of a material to shear deformation. It is the angle between the normal force and shear force required to cause shear failure in a material. If the angle of internal friction is zero, the material can be considered to be perfectly plastic and the Mohr envelope will be a straight line.
In reality, most materials have a non-zero angle of internal friction, which means that the Mohr envelope will be curved. The curvature of the Mohr envelope depends on the material properties, and can be used to determine the shear strength of the material.
In summary, the Mohr envelope can be considered to be straight if the angle of internal friction is zero. However, in most cases the angle of internal friction is non-zero, and the Mohr envelope is curved.
The curve obtained by plotting the normal and shear stress is called as ___________- a)Mohr’s envelope
- b)Coulomb envelope
- c)Strength envelope
- d)Stress envelope
Correct answer is option 'C'. Can you explain this answer?
The curve obtained by plotting the normal and shear stress is called as ___________
a)
Mohr’s envelope
b)
Coulomb envelope
c)
Strength envelope
d)
Stress envelope
|
Sharmila Gupta answered |
Circle b)Mohr line c)Mohr curve d)Mohr diagram
Which of the following shear test is developed based on drainage conditions?- a)Quick test and Consolidated un drained test
- b)Direct shear test
- c)None of the mentioned
- d)All of the mentioned
Correct answer is option 'A'. Can you explain this answer?
Which of the following shear test is developed based on drainage conditions?
a)
Quick test and Consolidated un drained test
b)
Direct shear test
c)
None of the mentioned
d)
All of the mentioned
|
Sanvi Kapoor answered |
Depending upon drainage conditions, three types of shear tests have been developed:
i) Un-drained test or quick test
ii) Consolidated un-drained test
iii) Drained test.
i) Un-drained test or quick test
ii) Consolidated un-drained test
iii) Drained test.
Shearing resistance can be determined in the laboratory by _________ methods.- a)2
- b)6
- c)4
- d)8
Correct answer is option 'C'. Can you explain this answer?
Shearing resistance can be determined in the laboratory by _________ methods.
a)
2
b)
6
c)
4
d)
8
|
|
Lalit Yadav answered |
Shearing resistance can be determined in the laboratory by following four methods
i) Direct shear test
ii) Tri axial shear test
iii) Unconfined shear test
iv) Vane shear test.
i) Direct shear test
ii) Tri axial shear test
iii) Unconfined shear test
iv) Vane shear test.
The critical shear stress causing failure of material depends upon ____________- a)Properties of the material and normal stress on the plane
- b)Intermediate principal stress
- c)None of the mentioned
- d)All of the mentioned
Correct answer is option 'A'. Can you explain this answer?
The critical shear stress causing failure of material depends upon ____________
a)
Properties of the material and normal stress on the plane
b)
Intermediate principal stress
c)
None of the mentioned
d)
All of the mentioned
|
|
Sahil Chawla answered |
Properties of the material and normal stress on the plane
- The critical shear stress causing failure of a material is influenced by the properties of the material itself, such as its strength, ductility, and toughness.
- Additionally, the normal stress acting on the plane where failure is expected plays a crucial role in determining the critical shear stress.
- The interaction between the normal stress and the shear stress on the plane can lead to different failure mechanisms, such as shear failure, tensile failure, or a combination of both.
- Understanding the material properties and the normal stress on the plane is essential in predicting and preventing failure in engineering structures.
- By considering these factors, engineers can design materials and structures that can withstand the expected loads and environmental conditions.
Therefore, the critical shear stress causing failure of a material depends on the properties of the material and the normal stress acting on the plane where failure is expected.
- The critical shear stress causing failure of a material is influenced by the properties of the material itself, such as its strength, ductility, and toughness.
- Additionally, the normal stress acting on the plane where failure is expected plays a crucial role in determining the critical shear stress.
- The interaction between the normal stress and the shear stress on the plane can lead to different failure mechanisms, such as shear failure, tensile failure, or a combination of both.
- Understanding the material properties and the normal stress on the plane is essential in predicting and preventing failure in engineering structures.
- By considering these factors, engineers can design materials and structures that can withstand the expected loads and environmental conditions.
Therefore, the critical shear stress causing failure of a material depends on the properties of the material and the normal stress acting on the plane where failure is expected.
Which one of the following conditions is valid in case of unconfined compression test in comparison to tri-axial test?- a)Minor principal stress = 0
- b)Minor principal stress = 0.5 x major principal stress
- c)Minor principal stress = major principal stress
- d)Major principal stress = 3 x minor principal stress
Correct answer is option 'A'. Can you explain this answer?
Which one of the following conditions is valid in case of unconfined compression test in comparison to tri-axial test?
a)
Minor principal stress = 0
b)
Minor principal stress = 0.5 x major principal stress
c)
Minor principal stress = major principal stress
d)
Major principal stress = 3 x minor principal stress
|
|
Ashwin Gupta answered |
Unconfined Compression Test vs Tri-axial Test
Unconfined Compression Test and Tri-axial Test are two methods of testing the strength of soil. The following conditions are valid in case of unconfined compression test in comparison to tri-axial test:
Minor Principal Stress = 0
In the unconfined compression test, the soil sample is loaded axially without any lateral confinement. This means that there is no minor principal stress acting on the soil sample during the test. On the other hand, in the tri-axial test, the soil sample is loaded axially while being confined laterally by a rubber membrane. This results in the development of a minor principal stress in the soil sample.
Therefore, the correct answer is option 'A' - Minor principal stress = 0.
Other Conditions
The other conditions in the two tests are as follows:
- Major Principal Stress: The major principal stress is the same in both tests as it is the axial load applied to the soil sample.
- Shear Stress: In the unconfined compression test, the shear stress is developed due to the axial load applied to the soil sample. In the tri-axial test, the shear stress is developed due to both the axial load and the lateral confinement.
- Strain: The strain in the soil sample is measured in both tests using strain gauges or extensometers. However, the type of strain measured may be different in the two tests.
Conclusion
In conclusion, the major difference between unconfined compression test and tri-axial test is the presence or absence of lateral confinement. This results in the development or absence of minor principal stress in the soil sample.
Unconfined Compression Test and Tri-axial Test are two methods of testing the strength of soil. The following conditions are valid in case of unconfined compression test in comparison to tri-axial test:
Minor Principal Stress = 0
In the unconfined compression test, the soil sample is loaded axially without any lateral confinement. This means that there is no minor principal stress acting on the soil sample during the test. On the other hand, in the tri-axial test, the soil sample is loaded axially while being confined laterally by a rubber membrane. This results in the development of a minor principal stress in the soil sample.
Therefore, the correct answer is option 'A' - Minor principal stress = 0.
Other Conditions
The other conditions in the two tests are as follows:
- Major Principal Stress: The major principal stress is the same in both tests as it is the axial load applied to the soil sample.
- Shear Stress: In the unconfined compression test, the shear stress is developed due to the axial load applied to the soil sample. In the tri-axial test, the shear stress is developed due to both the axial load and the lateral confinement.
- Strain: The strain in the soil sample is measured in both tests using strain gauges or extensometers. However, the type of strain measured may be different in the two tests.
Conclusion
In conclusion, the major difference between unconfined compression test and tri-axial test is the presence or absence of lateral confinement. This results in the development or absence of minor principal stress in the soil sample.
A cylindrical specimen of saturated soil failed under an axial vertical stress of 100 kN/m2 when it was laterally unconfined. The failure plane was inclined to the horizontal plane at an angle of 45°. The values of cohesion and angle of internal friction for the soil are respectively.- a)0,5 N/mm2 and 30°
- b)0.05 N/mm2 and 0°
- c)0.2 N/mm2 and 0°
- d)0.05 N/mm2 and 45°
Correct answer is option 'B'. Can you explain this answer?
A cylindrical specimen of saturated soil failed under an axial vertical stress of 100 kN/m2 when it was laterally unconfined. The failure plane was inclined to the horizontal plane at an angle of 45°. The values of cohesion and angle of internal friction for the soil are respectively.
a)
0,5 N/mm2 and 30°
b)
0.05 N/mm2 and 0°
c)
0.2 N/mm2 and 0°
d)
0.05 N/mm2 and 45°
|
|
Kavya Mehta answered |
For an unconfined compression test:
S = C = 100/2 = 50kN/m2
= 0.05 N/mm2
A laterally unconfined compression test is conducted on soils which can stand without confinement. Also, for clay angle of internal friction = φ = 0°
S = C = 100/2 = 50kN/m2
= 0.05 N/mm2
A laterally unconfined compression test is conducted on soils which can stand without confinement. Also, for clay angle of internal friction = φ = 0°
In direct shear test, the soil load is subjected to more stress at _____________- a)Centre
- b)Edges
- c)Top and bottom
- d)All of the mentioned
Correct answer is option 'B'. Can you explain this answer?
In direct shear test, the soil load is subjected to more stress at _____________
a)
Centre
b)
Edges
c)
Top and bottom
d)
All of the mentioned
|
|
Ashwin Desai answered |
Direct shear test is a laboratory test conducted to determine the shear strength parameters of a soil sample. It is commonly used to study the shear behavior of soils under different conditions. In this test, a soil sample is placed between two halves of a shear box, and a normal load is applied perpendicular to the shear plane. The soil sample is then subjected to a horizontal shear force, causing it to deform and eventually fail.
The correct answer to the given question is option 'B' - Edges.
Explanation:
When a direct shear test is conducted, the soil sample experiences different stress conditions at different locations within the shear box. Let's discuss the stress distribution in the soil sample during the direct shear test:
1. Center:
- At the center of the soil sample, there is a region of lower stress. This is because the soil in the center experiences a uniform distribution of normal load, resulting in an even stress distribution. The stress is primarily normal stress, and the shear stress is relatively lower.
2. Edges:
- The edges of the soil sample, which are in contact with the vertical walls of the shear box, experience higher stress. This is due to the confinement provided by the shear box walls, which restricts the lateral deformation of the soil. As a result, the stress concentration occurs at the edges, leading to higher shear stress.
3. Top and Bottom:
- The top and bottom surfaces of the soil sample also experience higher stress compared to the center. This is because the normal load is applied vertically, causing a higher normal stress on these surfaces. The shear stress is also higher at the top and bottom due to the confinement provided by the shear box.
Therefore, in a direct shear test, the soil load is subjected to more stress at the edges of the soil sample. This stress concentration at the edges is important to consider when analyzing the shear strength parameters of the soil.
The correct answer to the given question is option 'B' - Edges.
Explanation:
When a direct shear test is conducted, the soil sample experiences different stress conditions at different locations within the shear box. Let's discuss the stress distribution in the soil sample during the direct shear test:
1. Center:
- At the center of the soil sample, there is a region of lower stress. This is because the soil in the center experiences a uniform distribution of normal load, resulting in an even stress distribution. The stress is primarily normal stress, and the shear stress is relatively lower.
2. Edges:
- The edges of the soil sample, which are in contact with the vertical walls of the shear box, experience higher stress. This is due to the confinement provided by the shear box walls, which restricts the lateral deformation of the soil. As a result, the stress concentration occurs at the edges, leading to higher shear stress.
3. Top and Bottom:
- The top and bottom surfaces of the soil sample also experience higher stress compared to the center. This is because the normal load is applied vertically, causing a higher normal stress on these surfaces. The shear stress is also higher at the top and bottom due to the confinement provided by the shear box.
Therefore, in a direct shear test, the soil load is subjected to more stress at the edges of the soil sample. This stress concentration at the edges is important to consider when analyzing the shear strength parameters of the soil.
According to Coulomb, the relationship between shear strength and normal stress could be represented by _________
- a)Linear curve
- b)Straight line
- c)Parabolic curve
- d)None of the mentioned
Correct answer is option 'B'. Can you explain this answer?
According to Coulomb, the relationship between shear strength and normal stress could be represented by _________
a)
Linear curve
b)
Straight line
c)
Parabolic curve
d)
None of the mentioned
|
|
Partho Choudhary answered |
Coulomb's Shear Strength Theory
- Coulomb's shear strength theory is a simple and widely used model for calculating the shear strength of soils.
- According to Coulomb's theory, the shear strength of a soil mass is a function of the normal stress acting on the soil and the angle of internal friction of the soil.
- The relationship between shear strength and normal stress can be represented by a parabolic curve.
Parabolic Curve
- A parabolic curve is a U-shaped curve that is symmetric about a vertical line through its vertex.
- In Coulomb's theory, the parabolic curve represents the relationship between shear strength and normal stress.
- The curve is obtained by plotting the shear stress on the vertical axis and the normal stress on the horizontal axis.
- The vertex of the parabolic curve represents the maximum shear stress that the soil can withstand before failure occurs.
Implications
- The parabolic curve provides a simple and effective way to calculate the shear strength of soils.
- By determining the angle of internal friction of the soil and the normal stress acting on the soil, engineers can use Coulomb's theory to calculate the shear strength of the soil.
- This information is important for designing and analyzing structures such as retaining walls, foundations, and embankments.
- Coulomb's shear strength theory is a simple and widely used model for calculating the shear strength of soils.
- According to Coulomb's theory, the shear strength of a soil mass is a function of the normal stress acting on the soil and the angle of internal friction of the soil.
- The relationship between shear strength and normal stress can be represented by a parabolic curve.
Parabolic Curve
- A parabolic curve is a U-shaped curve that is symmetric about a vertical line through its vertex.
- In Coulomb's theory, the parabolic curve represents the relationship between shear strength and normal stress.
- The curve is obtained by plotting the shear stress on the vertical axis and the normal stress on the horizontal axis.
- The vertex of the parabolic curve represents the maximum shear stress that the soil can withstand before failure occurs.
Implications
- The parabolic curve provides a simple and effective way to calculate the shear strength of soils.
- By determining the angle of internal friction of the soil and the normal stress acting on the soil, engineers can use Coulomb's theory to calculate the shear strength of the soil.
- This information is important for designing and analyzing structures such as retaining walls, foundations, and embankments.
In a tri-axial test at failure, major principal stress was 180 kPa, minor principal stress was 100 kPa, and pore pressure was 20 kPa. The sine of the angle of shearing resistance of the sandy soil tested is- a)1/3
- b)2/7
- c)1/2
- d)1/6
Correct answer is option 'A'. Can you explain this answer?
In a tri-axial test at failure, major principal stress was 180 kPa, minor principal stress was 100 kPa, and pore pressure was 20 kPa. The sine of the angle of shearing resistance of the sandy soil tested is
a)
1/3
b)
2/7
c)
1/2
d)
1/6
|
|
Snehal Tiwari answered |
Effective minor principal stress.
The stresses responsible for mobilisation of shearing strength of soil is- a)Effective normal stress
- b)Neutral stress
- c)Total normal stress
- d)Shear stress
Correct answer is option 'A'. Can you explain this answer?
The stresses responsible for mobilisation of shearing strength of soil is
a)
Effective normal stress
b)
Neutral stress
c)
Total normal stress
d)
Shear stress
|
|
Nidhi Patel answered |
The shear strength of a soil is its maximum resistance to shear stress just before the failure. Shear stress develop when soils are subjected to compression.
Terzaghi established that the normal stresses which control the shear strength of a soil are the effective stresses and not the total stresses. In terms of effective stresses shear strength is written as
Terzaghi established that the normal stresses which control the shear strength of a soil are the effective stresses and not the total stresses. In terms of effective stresses shear strength is written as
The Mohr-Coulomb theory can be expressed algebraically by, which of the following equation.- a)S = c + σ tan φ
- b)τf = s = F(σ)
- c)s = F(σ)
- d)τf = F(σ)
Correct answer is option 'B'. Can you explain this answer?
The Mohr-Coulomb theory can be expressed algebraically by, which of the following equation.
a)
S = c + σ tan φ
b)
τf = s = F(σ)
c)
s = F(σ)
d)
τf = F(σ)
|
|
Sanya Agarwal answered |
The Mohr-Coulomb failure theory can be expressed algebraically by the equation:τf = s = F (σ)
Where, τf = s=shear stress on failure plane, at failure=shear resistance of material
F (σ) = function of normal stress.
Where, τf = s=shear stress on failure plane, at failure=shear resistance of material
F (σ) = function of normal stress.
The ratio, of the undisturbed shear strength to the remoulded shear strength in cohesive soil under undrained conditions is- a)Zero
- b)1
- c)Greater than 1
- d)Between 0 and 1
Correct answer is option 'C'. Can you explain this answer?
The ratio, of the undisturbed shear strength to the remoulded shear strength in cohesive soil under undrained conditions is
a)
Zero
b)
1
c)
Greater than 1
d)
Between 0 and 1
|
|
Lavanya Menon answered |
Ratio of undisturbed shear strength to the remoulded shear strength is called sensitivity.
Which of the following cannot be obtained by using un-drained test? - a)Shear strength
- b)Effective stress failure envelope
- c)sensitivity
- d)All of the mentioned
Correct answer is option 'B'. Can you explain this answer?
Which of the following cannot be obtained by using un-drained test?
a)
Shear strength
b)
Effective stress failure envelope
c)
sensitivity
d)
All of the mentioned
|
|
Lavanya Menon answered |
Since only one Mohr circle in terms of effective stresses, is obtained from all un-drained tests, effective stress failure envelope cannot be obtained from this test.
Unconfined compressive strength test is:- a)Undrained test
- b)Drained test
- c)Consolidated undrained test
- d)Consolidated drained test
Correct answer is option 'A'. Can you explain this answer?
Unconfined compressive strength test is:
a)
Undrained test
b)
Drained test
c)
Consolidated undrained test
d)
Consolidated drained test
|
|
Neha Mukherjee answered |
The unconfined compression test is a special form of a tri-axial test in which the confining pressure is zero. The test can be conducted only on clayey soils which can stand without confinement.
Theory of failure, was first proposed by ____________- a)Coulomb
- b)Mohr
- c)Casagrande
- d)Darcy
Correct answer is option 'A'. Can you explain this answer?
Theory of failure, was first proposed by ____________
a)
Coulomb
b)
Mohr
c)
Casagrande
d)
Darcy
|
|
Sanvi Kapoor answered |
The theory of failures was first expressed by coulomb in 1776 and later generalized by Mohr.
Match List-I (Field test) with List-ll (Useful for) and select the correct answer using the codes given below the lists:
Codes:
- a)A
- b)B
- c)C
- d)D
Correct answer is option 'B'. Can you explain this answer?
Match List-I (Field test) with List-ll (Useful for) and select the correct answer using the codes given below the lists:
Codes:
Codes:
a)
A
b)
B
c)
C
d)
D
|
|
Nidhi Patel answered |
The shear modulus (G) can be determined in-situ by pressure meter. The original instrument was developed by Menard to overcome the problem of sampling disturbance. The instrument uses compressed gas (usually nitrogen). The results of test can be plotted as corrected pressure (p) against volume (v).
The data can also be used for determination of undrained shear strength, angle of shearing resistance and other engineering properties of soil.
The data can also be used for determination of undrained shear strength, angle of shearing resistance and other engineering properties of soil.
The shearing of cohesive soil in drained test requires ___________ days.- a)2
- b)1 to 2
- c)2 to 5
- d)1
Correct answer is option 'C'. Can you explain this answer?
The shearing of cohesive soil in drained test requires ___________ days.
a)
2
b)
1 to 2
c)
2 to 5
d)
1
|
|
Lavanya Menon answered |
As the soil in drained test is sheared sufficiently slowly so that complete dissipation of pore pressure takes place, it takes 2 to 5 days long for shearing cohesive type of soil.
The parameter φ in coulomb’s equation “S = c + σ tan φ”, represents ___________- a)Shearing resistance and Angle of internal friction
- b)Angle of slope
- c)None of the mentioned
- d)All of the mentioned
Correct answer is option 'A'. Can you explain this answer?
The parameter φ in coulomb’s equation “S = c + σ tan φ”, represents ___________
a)
Shearing resistance and Angle of internal friction
b)
Angle of slope
c)
None of the mentioned
d)
All of the mentioned
|
|
Sanya Agarwal answered |
The empirical constant φ in coulomb’s equation represents angle of internal friction or shearing resistance respectively.
In tri-axial compression test, the deviator stress is given by- a)σ1 + σ3
- b)σ1 - σ3
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
In tri-axial compression test, the deviator stress is given by
a)
σ1 + σ3
b)
σ1 - σ3
c)
d)
|
|
Ishani Basu answered |
In tri-axial compression test,
Minor principal stress = confining pressure
Minor principal stress = confining pressure
The commonly used apparatus used for performing shear box test is ___________- a)Shear-box apparatus
- b)Bishop’s pore pressure apparatus
- c)Tri axial shear test apparatus
- d)None of the mentioned
Correct answer is option 'A'. Can you explain this answer?
The commonly used apparatus used for performing shear box test is ___________
a)
Shear-box apparatus
b)
Bishop’s pore pressure apparatus
c)
Tri axial shear test apparatus
d)
None of the mentioned
|
|
Lavanya Menon answered |
Shear-box test is the most commonly used apparatus for the direct shear test.
A vane 20 cm long and 10 cm in diameter was pressed into a soft marine play at the bottom of a bore hole. Torque was applied gradually and failure occurred at 1000 kg-cm. The cohesion of the clay in kg/cm2 is- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
A vane 20 cm long and 10 cm in diameter was pressed into a soft marine play at the bottom of a bore hole. Torque was applied gradually and failure occurred at 1000 kg-cm. The cohesion of the clay in kg/cm2 is
a)
b)
c)
d)
|
|
Naina Das answered |
Shear strength of clay
⇒
⇒
To conduct un-drained test, which of the following is used?- a)Slope grids
- b)Perforated grids
- c)Plain grids
- d)All of the mentioned
Correct answer is option 'C'. Can you explain this answer?
To conduct un-drained test, which of the following is used?
a)
Slope grids
b)
Perforated grids
c)
Plain grids
d)
All of the mentioned
|
|
Sanya Agarwal answered |
To conduct un-drained test, plane grids are used and for the drained test, perforated grids are used.
The results of a consolidated drained tri-axial shear test on a normally consolidated clay are shown in Figure below:
The angle of internal friction is- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The results of a consolidated drained tri-axial shear test on a normally consolidated clay are shown in Figure below:
The angle of internal friction is
The angle of internal friction is
a)
b)
c)
d)
|
|
Gowri Singh answered |
From the plot:
Major principal stress = σ1 = 200 kN/m2
Minor principal stress = σ3 = 100 kN/m2
Major principal stress = σ1 = 200 kN/m2
Minor principal stress = σ3 = 100 kN/m2
Chapter doubts & questions for Shear Strength of Soils - Soil Mechanics 2025 is part of Civil Engineering (CE) exam preparation. The chapters have been prepared according to the Civil Engineering (CE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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