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All questions of Stability of Slopes for Civil Engineering (CE) Exam
An infinite soil slope with an inclination of 350 is subjected to seepage parallel to its surface. The soil has c' = 100 kN/m2 and φ = 300. Using the concept of mobilized cohesion and friction, at a factor of safety of 1.5 with respect to shear strength, the mobilized friction angle is- a)20.020
- b)21.050
- c)23.330
- d)30.000
Correct answer is option 'B'. Can you explain this answer?
An infinite soil slope with an inclination of 350 is subjected to seepage parallel to its surface. The soil has c' = 100 kN/m2 and φ = 300. Using the concept of mobilized cohesion and friction, at a factor of safety of 1.5 with respect to shear strength, the mobilized friction angle is
a)
20.020
b)
21.050
c)
23.330
d)
30.000
|
Devika Tiwari answered |
The factor of safety with respect to shear strength is given by,
The shear strength mobilised, thus may be written as,
The shear strength mobilised, thus may be written as,
A long natural slope of cohesion-less soil is inclined at 12° to the horizontal. What will be the factor of safety of the slope if φ = 30°?- a)1.6
- b)2.7
- c)0.13
- d)0.4
Correct answer is option 'B'. Can you explain this answer?
A long natural slope of cohesion-less soil is inclined at 12° to the horizontal. What will be the factor of safety of the slope if φ = 30°?
a)
1.6
b)
2.7
c)
0.13
d)
0.4
|
Sanya Agarwal answered |
Given φ = 30° and i = 12°
Formula, F = tan φ /tan i
F = tan 30° / tan 12°
F = 2.72.
Formula, F = tan φ /tan i
F = tan 30° / tan 12°
F = 2.72.
In friction circle method of slope stability analysis, if r defines the radius of the slip circle, the radius of friction circle is:- a)r sin φ
- b)r
- c)r cos φ
- d)r tan φ
Correct answer is option 'A'. Can you explain this answer?
In friction circle method of slope stability analysis, if r defines the radius of the slip circle, the radius of friction circle is:
a)
r sin φ
b)
r
c)
r cos φ
d)
r tan φ
|
Sagnik Sen answered |
No, the radius of the friction circle is not r*sinθ. The radius of the friction circle is typically defined as the difference between the radius of the slip circle (r) and the radius of the critical slip circle (rc).
The radius of the friction circle (rf) is given by:
rf = r - rc
The radius of the critical slip circle (rc) is determined based on the factor of safety required for the slope stability analysis. It represents the smallest slip circle that would result in slope failure.
The radius of the friction circle (rf) is given by:
rf = r - rc
The radius of the critical slip circle (rc) is determined based on the factor of safety required for the slope stability analysis. It represents the smallest slip circle that would result in slope failure.
If the failure occurs along a surface of sliding that intersect the slope at its toe, the slide is known as ______- a)Base failure
- b)Slope failure
- c)Face failure
- d)All of the mentioned
Correct answer is option 'B'. Can you explain this answer?
If the failure occurs along a surface of sliding that intersect the slope at its toe, the slide is known as ______
a)
Base failure
b)
Slope failure
c)
Face failure
d)
All of the mentioned
|
|
Sanya Agarwal answered |
If the failure occurs along a sliding that intersects the slopes at or above its toe, the slide is known as slope failure.
A finite slope of cohesion less soil is safe so long as- a)there is no external load
- b)β < φ
- c)β > φ
- d)the soil is dry
Correct answer is option 'B'. Can you explain this answer?
A finite slope of cohesion less soil is safe so long as
a)
there is no external load
b)
β < φ
c)
β > φ
d)
the soil is dry
|
|
Poulomi Patel answered |
The factor of safety of slope against shear failure
Stability analysis by Swedish method of slices gave following values per meter run of 10 m high embankment (i) total shearing force = 480 kN (ii) total normal force = 1950 kN (iii) Radius of failure arc = 15 m (iv) length of arce = 22 m. If C = 24 kN/m2, ϕ = 6°, then the factor of safety with respect to shear strength is - a)1.53
- b)1.67
- c)1.47
- d)1.75
Correct answer is option 'A'. Can you explain this answer?
Stability analysis by Swedish method of slices gave following values per meter run of 10 m high embankment (i) total shearing force = 480 kN (ii) total normal force = 1950 kN (iii) Radius of failure arc = 15 m (iv) length of arce = 22 m. If C = 24 kN/m2, ϕ = 6°, then the factor of safety with respect to shear strength is
a)
1.53
b)
1.67
c)
1.47
d)
1.75
|
|
Sanvi Kapoor answered |
According to the Swedish method,
Calculation:
Given, Total shear force = 480 kN
Total normal force = 1950 kN
Length of arc, L = 22 m
Radius of failure arc, R = 15 m
C = 24 kN/m2 and ϕ = 6°
The factor of safety with respect to shear strength is given by,
Calculation:
Given, Total shear force = 480 kN
Total normal force = 1950 kN
Length of arc, L = 22 m
Radius of failure arc, R = 15 m
C = 24 kN/m2 and ϕ = 6°
The factor of safety with respect to shear strength is given by,
An infinitely long slope is made up of a c-φ soil having the properties: cohesion, C = 20 kPa, and dry unit weight, γd = 16 kN/m3. The angle of inclination and critical height of the slope are 40° and 5 m, respectively. To maintain the limiting equilibrium, the angle of internal friction of the soil (in degrees) is ________- a)22.4°
- b)20.8°
- c)19.4°
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
An infinitely long slope is made up of a c-φ soil having the properties: cohesion, C = 20 kPa, and dry unit weight, γd = 16 kN/m3. The angle of inclination and critical height of the slope are 40° and 5 m, respectively. To maintain the limiting equilibrium, the angle of internal friction of the soil (in degrees) is ________
a)
22.4°
b)
20.8°
c)
19.4°
d)
None of the above
|
|
Ameya Sen answered |
Calculating the angle of internal friction:
- Given data:
- Cohesion, C = 20 kPa
- Dry unit weight, γd = 16 kN/m3
- Angle of inclination, θ = 40°
- Critical height, H = 5 m
- The factor of safety (F) for a slope is given by:
F = c / (H * γd * tan(φ))
- At limiting equilibrium, F = 1, so:
1 = 20 / (5 * 16 * tan(φ))
- Solving for φ, we get:
tan(φ) = 20 / (5 * 16) = 0.25
- Therefore, the angle of internal friction (φ) can be calculated as:
φ = tan-1(0.25) ≈ 14.04°
Conclusion:
The angle of internal friction of the soil to maintain the limiting equilibrium is approximately 14.04°, which is closest to option 'A' (22.4°).
- Given data:
- Cohesion, C = 20 kPa
- Dry unit weight, γd = 16 kN/m3
- Angle of inclination, θ = 40°
- Critical height, H = 5 m
- The factor of safety (F) for a slope is given by:
F = c / (H * γd * tan(φ))
- At limiting equilibrium, F = 1, so:
1 = 20 / (5 * 16 * tan(φ))
- Solving for φ, we get:
tan(φ) = 20 / (5 * 16) = 0.25
- Therefore, the angle of internal friction (φ) can be calculated as:
φ = tan-1(0.25) ≈ 14.04°
Conclusion:
The angle of internal friction of the soil to maintain the limiting equilibrium is approximately 14.04°, which is closest to option 'A' (22.4°).
An excavation was made at a slope angle of 540 in homogeneous clay. When the depth of excavation reached 8m, a slip occurred. The slip surface was likely to have passed through a point:- a)Above the toe of the slope
- b)Below the toe
- c)Through the toe
- d)Near the mid-point of the slope
Correct answer is option 'C'. Can you explain this answer?
An excavation was made at a slope angle of 540 in homogeneous clay. When the depth of excavation reached 8m, a slip occurred. The slip surface was likely to have passed through a point:
a)
Above the toe of the slope
b)
Below the toe
c)
Through the toe
d)
Near the mid-point of the slope
|
Hiral Sharma answered |
For slopes greater than 530, the critical slip circle invariably passes through the toe. (Toe failure). This is true for any angle of friction (φ).
For slopes less than 530 and small values of φ < 30, the critical surface passes below the toe and it is called base failure.
For slopes less than 530 and small values of φ < 30, the critical surface passes below the toe and it is called base failure.
Base failure of a finite slope occurs when the soil _______.- a)is purely cohesive
- b)is cohesionless
- c)below the toe is relatively soft and weak
- d)above the toe is relatively soft and weak
Correct answer is option 'C'. Can you explain this answer?
Base failure of a finite slope occurs when the soil _______.
a)
is purely cohesive
b)
is cohesionless
c)
below the toe is relatively soft and weak
d)
above the toe is relatively soft and weak
|
|
Sanya Agarwal answered |
Stability analysis of Finite Slope:
(i) If the slope is of finite extend bounded by top and bottom surfaces, then it is termed as finite slope.
(ii) Failure of finite slope takes places due to rotation and failure plane is either circular or spiral.
(iii) Finite slope may have any of the following mode of shear failure.
(i) If the slope is of finite extend bounded by top and bottom surfaces, then it is termed as finite slope.
(ii) Failure of finite slope takes places due to rotation and failure plane is either circular or spiral.
(iii) Finite slope may have any of the following mode of shear failure.
- Slope failure
- Base failure
Slope failure:
- Face Failure: Failure surface passes through slope above the toe. This type of failure takes places in case of steep slope, soil mass near the toe is rigid or stringer in comparison to soil mass above the toe.
- Toe Failure: It is the most common mode of failure of finite slope in which failure surface passes through the toe. This failure also occur in steel slope when soil mass is homogenous above and below the toe.
Base Failure:
- Failure surface passes below the toe.
- This type of failure takes place when soil mass below the toe is soft and weak in comparison to soil mass above the toe.
The ratio of total depth to depth H is called _________- a)Depth factor
- b)Slope depth
- c)Depth failure
- d)Base failure
Correct answer is option 'A'. Can you explain this answer?
The ratio of total depth to depth H is called _________
a)
Depth factor
b)
Slope depth
c)
Depth failure
d)
Base failure
|
|
Sanya Agarwal answered |
The ratio of the total depth (H + D) to depth H is called the depth factor Df.
Stability of slopes of an earth dam is tested under, which of the following condition?- a)Stability of downstream slope during steady seepage and Stability of upstream slope during a sudden drawdown
- b)Stability of upstream slope during sudden seepage
- c)None of the mentioned
- d)All of the mentioned
Correct answer is option 'A'. Can you explain this answer?
Stability of slopes of an earth dam is tested under, which of the following condition?
a)
Stability of downstream slope during steady seepage and Stability of upstream slope during a sudden drawdown
b)
Stability of upstream slope during sudden seepage
c)
None of the mentioned
d)
All of the mentioned
|
|
Sanvi Kapoor answered |
The stability of slopes of an earth dam is tested under the following conditions:
(i) Stability of downstream slope during steady seepage
(ii) Stability of upstream slope during a sudden drawdown
(iii) Stability of upstream and downstream slopes during and immediately after construction.
(i) Stability of downstream slope during steady seepage
(ii) Stability of upstream slope during a sudden drawdown
(iii) Stability of upstream and downstream slopes during and immediately after construction.
The friction circle method is based on the assumption that the resultant reaction along a slip surface is tangential to a circle of radius- a)R2 sin ϕ
- b)R sin ϕ
- c)R cos ϕ
- d)R sin ϕ cos ϕ
Correct answer is option 'B'. Can you explain this answer?
The friction circle method is based on the assumption that the resultant reaction along a slip surface is tangential to a circle of radius
a)
R2 sin ϕ
b)
R sin ϕ
c)
R cos ϕ
d)
R sin ϕ cos ϕ
|
|
Sanvi Kapoor answered |
Friction circle method:
This method is also based on total stress analysis in which shearing angle of ϕ is used to analyse the stability of finite slope.
The principle of the method is that the inter granular forces are in an obliquity of ϕ to the circular surface at failure, where ϕ is the angle of the internal friction of the soil.
When the length of the arc is divided into small elements, the line of action of the inter granular forces acting on these elements can be defined by a tangent to the friction circle drawn around the centre of the sliding circle.
The radius of this friction circle is given by R sinϕ
Where,
R is the radius of the sliding circle.
Thus, the friction circle is agraphical tool that defines the line of action of the intergranularforces at any given point on the sliding circle in a convenient way.In geotechnics, the method is used primarily for slope-stability investigations in homogeneous soil when both cohesive and frictional components have to be considered.
The principle of the method is that the inter granular forces are in an obliquity of ϕ to the circular surface at failure, where ϕ is the angle of the internal friction of the soil.
When the length of the arc is divided into small elements, the line of action of the inter granular forces acting on these elements can be defined by a tangent to the friction circle drawn around the centre of the sliding circle.
The radius of this friction circle is given by R sinϕ
Where,
R is the radius of the sliding circle.
Thus, the friction circle is agraphical tool that defines the line of action of the intergranularforces at any given point on the sliding circle in a convenient way.In geotechnics, the method is used primarily for slope-stability investigations in homogeneous soil when both cohesive and frictional components have to be considered.
The friction circle may be also referred as __________- a)φ-circle
- b)Plane circle
- c)Cohesion circle
- d)All of the mentioned
Correct answer is option 'A'. Can you explain this answer?
The friction circle may be also referred as __________
a)
φ-circle
b)
Plane circle
c)
Cohesion circle
d)
All of the mentioned
|
|
Sanvi Kapoor answered |
The friction circle having radius r sin φ is also called as φ-circle.
The forces acting on a sliding wedge are ___________- a)Weight of the wedge
- b)Total frictional force
- c)Total cohesive resistance
- d)All of the mentioned
Correct answer is option 'D'. Can you explain this answer?
The forces acting on a sliding wedge are ___________
a)
Weight of the wedge
b)
Total frictional force
c)
Total cohesive resistance
d)
All of the mentioned
|
|
Arnab Saini answered |
Forces acting on a sliding wedge:
Weight of the wedge:
- The weight of the wedge acts vertically downward through its center of gravity. This force is a result of the gravitational pull on the wedge and contributes to the overall forces acting on the wedge.
Total frictional force:
- When a wedge is sliding on a surface, there is friction between the wedge and the surface. This frictional force acts opposite to the direction of motion and helps in resisting the sliding motion of the wedge.
Total cohesive resistance:
- Cohesive resistance is the force that opposes the separation or sliding of two surfaces in contact. In the case of a sliding wedge, cohesive resistance between the surfaces in contact also contributes to the overall forces acting on the wedge.
All of the mentioned:
- Therefore, all of the mentioned forces (weight of the wedge, total frictional force, and total cohesive resistance) play a role in determining the behavior of a sliding wedge. These forces collectively influence the stability, motion, and equilibrium of the wedge as it slides or moves on a surface.
Weight of the wedge:
- The weight of the wedge acts vertically downward through its center of gravity. This force is a result of the gravitational pull on the wedge and contributes to the overall forces acting on the wedge.
Total frictional force:
- When a wedge is sliding on a surface, there is friction between the wedge and the surface. This frictional force acts opposite to the direction of motion and helps in resisting the sliding motion of the wedge.
Total cohesive resistance:
- Cohesive resistance is the force that opposes the separation or sliding of two surfaces in contact. In the case of a sliding wedge, cohesive resistance between the surfaces in contact also contributes to the overall forces acting on the wedge.
All of the mentioned:
- Therefore, all of the mentioned forces (weight of the wedge, total frictional force, and total cohesive resistance) play a role in determining the behavior of a sliding wedge. These forces collectively influence the stability, motion, and equilibrium of the wedge as it slides or moves on a surface.
Planar surface commonly occur in _______- a)Embankment with specific plane of weakness
- b)All embankments
- c)Soil deposit
- d)Foundation of infinite depth
Correct answer is option 'A'. Can you explain this answer?
Planar surface commonly occur in _______
a)
Embankment with specific plane of weakness
b)
All embankments
c)
Soil deposit
d)
Foundation of infinite depth
|
|
Sanya Agarwal answered |
Planar failure surface may commonly occur in a soil deposit or embankment with a specific plane of weakness.
Which of the following is an example of slopes extending to infinity?- a)Inclined face of Earth dams
- b)Embankments
- c)Cuts
- d)None of the mentioned
Correct answer is option 'D'. Can you explain this answer?
Which of the following is an example of slopes extending to infinity?
a)
Inclined face of Earth dams
b)
Embankments
c)
Cuts
d)
None of the mentioned
|
|
Nayanika Joshi answered |
Explanation:
Slopes extending to infinity:
In civil engineering, slopes extending to infinity refer to slopes that continue indefinitely without reaching a stopping point. This can occur in certain natural or man-made features where the slope is so gradual or steep that it appears to extend to infinity.
Examples:
- Inclined face of Earth dams: Earth dams are structures built to hold back water and create reservoirs. The inclined face of an earth dam can appear to extend to infinity, especially if it is a large dam with a gentle slope.
- Embankments: Embankments are raised earth structures used for various purposes such as road construction or flood protection. The slopes of embankments can also give the illusion of extending to infinity, particularly if they are long and gradually sloping.
- Cuts: Cuts are excavated areas in the ground often seen in road construction or mining operations. Depending on the depth and angle of the cut, the slope may seem to continue indefinitely.
Conclusion:
While these examples may suggest slopes extending to infinity, in reality, all slopes have a finite length. However, in certain contexts, the slopes may appear to continue indefinitely due to their size, angle, and the surrounding landscape.
Slopes extending to infinity:
In civil engineering, slopes extending to infinity refer to slopes that continue indefinitely without reaching a stopping point. This can occur in certain natural or man-made features where the slope is so gradual or steep that it appears to extend to infinity.
Examples:
- Inclined face of Earth dams: Earth dams are structures built to hold back water and create reservoirs. The inclined face of an earth dam can appear to extend to infinity, especially if it is a large dam with a gentle slope.
- Embankments: Embankments are raised earth structures used for various purposes such as road construction or flood protection. The slopes of embankments can also give the illusion of extending to infinity, particularly if they are long and gradually sloping.
- Cuts: Cuts are excavated areas in the ground often seen in road construction or mining operations. Depending on the depth and angle of the cut, the slope may seem to continue indefinitely.
Conclusion:
While these examples may suggest slopes extending to infinity, in reality, all slopes have a finite length. However, in certain contexts, the slopes may appear to continue indefinitely due to their size, angle, and the surrounding landscape.
Taylor’s stability number curves are used for the analysis of stability of slopes. The angle of shearing resistance used in the chart is the:
- a)Effective angle
- b)Apparent angle
- c)Mobilised angle
- d)Weighted angle
Correct answer is option 'A'. Can you explain this answer?
Taylor’s stability number curves are used for the analysis of stability of slopes. The angle of shearing resistance used in the chart is the:
a)
Effective angle
b)
Apparent angle
c)
Mobilised angle
d)
Weighted angle
|
Rounak Banerjee answered |
The Correct Option is C: Mobilised angle
In Taylor's stability number curves, the angle of shearing resistance used is the mobilized angle. The mobilized angle of shearing resistance represents the actual angle at which the soil is resisting shearing forces during slope stability analysis. It takes into account factors such as pore water pressure and soil deformation. The mobilized angle is an important consideration when analyzing the stability of slopes and designing appropriate measures to ensure stability.
If an infinite slope of clay at a depth 5 m has cohesion of 1 t/m2 and unit weight of 2 t/m3, then the stability number will be- a)0.1
- b)0.2
- c)0.3
- d)0.4
Correct answer is option 'A'. Can you explain this answer?
If an infinite slope of clay at a depth 5 m has cohesion of 1 t/m2 and unit weight of 2 t/m3, then the stability number will be
a)
0.1
b)
0.2
c)
0.3
d)
0.4
|
|
Rajdeep Gupta answered |
Given Information:
- Depth of clay: 5 m
- Cohesion of clay: 1 t/m2
- Unit weight of clay: 2 t/m3
To find: Stability number
Formula for Stability Number (SN):
SN = (C / γH)^(1/2)
Where:
- C is the cohesion of the clay
- γ is the unit weight of the clay
- H is the depth of the clay
Substitute the given values into the formula:
C = 1 t/m2
γ = 2 t/m3
H = 5 m
SN = (1 / (2 * 5))^(1/2)
= (1/10)^(1/2)
= (1/√10)
Now, we need to simplify the expression (1/√10) to obtain the answer in the given options.
Calculation:
- Multiply the numerator and denominator by √10 to rationalize the denominator:
SN = (1/√10) * (√10/√10)
= √10 / 10
Since the expression (√10 / 10) is equivalent to 0.316, which is closest to 0.1, we can conclude that the stability number is approximately 0.1.
Therefore, the correct answer is option 'A' (0.1).
- Depth of clay: 5 m
- Cohesion of clay: 1 t/m2
- Unit weight of clay: 2 t/m3
To find: Stability number
Formula for Stability Number (SN):
SN = (C / γH)^(1/2)
Where:
- C is the cohesion of the clay
- γ is the unit weight of the clay
- H is the depth of the clay
Substitute the given values into the formula:
C = 1 t/m2
γ = 2 t/m3
H = 5 m
SN = (1 / (2 * 5))^(1/2)
= (1/10)^(1/2)
= (1/√10)
Now, we need to simplify the expression (1/√10) to obtain the answer in the given options.
Calculation:
- Multiply the numerator and denominator by √10 to rationalize the denominator:
SN = (1/√10) * (√10/√10)
= √10 / 10
Since the expression (√10 / 10) is equivalent to 0.316, which is closest to 0.1, we can conclude that the stability number is approximately 0.1.
Therefore, the correct answer is option 'A' (0.1).
In the limiting case of stability, the angle of slope is referred to as ______- a)Angle of deviation
- b)Angle of repose
- c)Angle of unstable slope
- d)All of the mentioned
Correct answer is option 'B'. Can you explain this answer?
In the limiting case of stability, the angle of slope is referred to as ______
a)
Angle of deviation
b)
Angle of repose
c)
Angle of unstable slope
d)
All of the mentioned
|
|
Prashanth Mehra answered |
Angle of Repose
The angle of repose is the maximum angle at which a pile of granular material remains stable without sliding or collapsing. It is the limiting case of stability for a slope or pile of material. Here's why the correct answer is option 'B' - Angle of Repose:
Definition
- The angle of repose is the steepest angle at which a material can be piled up without slumping or sliding.
- It is an important parameter in geotechnical engineering when designing structures on slopes or dealing with loose materials.
Significance
- Knowing the angle of repose of a material is crucial for determining the stability of slopes, embankments, and retaining walls.
- It helps engineers understand the behavior of granular materials under different conditions.
Relation to Stability
- The angle of repose is directly related to the stability of slopes and piles.
- If the angle of slope exceeds the angle of repose, the material will start to slide or collapse due to gravity.
Practical Application
- The angle of repose is used in various industries such as mining, agriculture, and construction to prevent accidents and ensure the stability of structures.
- It is also used in determining the design parameters for storage silos, dams, and other structures.
In conclusion, the angle of repose is the critical angle that defines the stability of a slope or pile of material. It is essential for engineers to consider this parameter when designing structures on or with granular materials to ensure safety and stability.
In Hilf’s equation ‘u = pa Δ / (va + hc γw – Δ)’, hc refers to __________- a)Induced pore pressure
- b)Henry’s constant
- c)Volume of pore water
- d)Volume of air voids
Correct answer is option 'B'. Can you explain this answer?
In Hilf’s equation ‘u = pa Δ / (va + hc γw – Δ)’, hc refers to __________
a)
Induced pore pressure
b)
Henry’s constant
c)
Volume of pore water
d)
Volume of air voids
|
Mira Sharma answered |
Explanation:
Henry's constant (hc):
- Henry's constant (hc) is a proportionality constant that relates the concentration of a gas in a liquid to the partial pressure of that gas above the liquid.
- In the given equation, hc represents Henry's constant, which is used to characterize the behavior of gases in the pore water of soil.
Therefore, in the equation provided, hc refers to Henry's constant.
Henry's constant (hc):
- Henry's constant (hc) is a proportionality constant that relates the concentration of a gas in a liquid to the partial pressure of that gas above the liquid.
- In the given equation, hc represents Henry's constant, which is used to characterize the behavior of gases in the pore water of soil.
Therefore, in the equation provided, hc refers to Henry's constant.
Slopes is classified into ______ types.- a)2
- b)3
- c)4
- d)5
Correct answer is option 'A'. Can you explain this answer?
Slopes is classified into ______ types.
a)
2
b)
3
c)
4
d)
5
|
|
Sravya Rane answered |
Types of Slopes
There are generally two types of slopes based on their stability and behavior:
Natural Slopes
- Natural slopes are formed by the natural processes of weathering, erosion, and mass wasting.
- They can have different shapes and angles depending on the type of rock, geological structure, and climate of the region.
- Natural slopes are often found in hilly or mountainous areas.
Engineered Slopes
- Engineered slopes are man-made structures designed for specific purposes such as highway embankments, dams, or retaining walls.
- They are constructed using various materials and methods to ensure stability and safety.
- The design of engineered slopes takes into account factors such as soil mechanics, hydrology, and geotechnical engineering principles.
In conclusion, slopes are classified into two main types: natural slopes and engineered slopes. Natural slopes are formed by natural processes, while engineered slopes are man-made structures designed for specific purposes. Each type of slope has its characteristics and considerations when it comes to stability and design.
There are generally two types of slopes based on their stability and behavior:
Natural Slopes
- Natural slopes are formed by the natural processes of weathering, erosion, and mass wasting.
- They can have different shapes and angles depending on the type of rock, geological structure, and climate of the region.
- Natural slopes are often found in hilly or mountainous areas.
Engineered Slopes
- Engineered slopes are man-made structures designed for specific purposes such as highway embankments, dams, or retaining walls.
- They are constructed using various materials and methods to ensure stability and safety.
- The design of engineered slopes takes into account factors such as soil mechanics, hydrology, and geotechnical engineering principles.
In conclusion, slopes are classified into two main types: natural slopes and engineered slopes. Natural slopes are formed by natural processes, while engineered slopes are man-made structures designed for specific purposes. Each type of slope has its characteristics and considerations when it comes to stability and design.
When does the critical condition of d/s slope, occur in reservoir?- a)Reservoir is full
- b)Pore pressure is high
- c)Steady seepage does not occur
- d)All of the mentioned
Correct answer is option 'A'. Can you explain this answer?
When does the critical condition of d/s slope, occur in reservoir?
a)
Reservoir is full
b)
Pore pressure is high
c)
Steady seepage does not occur
d)
All of the mentioned
|
|
Harshad Iyer answered |
Critical Condition of D/S Slope in Reservoir
The critical condition of d/s (downstream) slope in a reservoir occurs when the reservoir is full. Let's understand this in detail.
Factors Affecting Stability of Slopes in Reservoir
The stability of slopes in a reservoir is affected by various factors like:
- Water level in the reservoir
- Pore pressure in the soil
- Seepage forces
- Soil characteristics (e.g. soil type, grain size distribution, shear strength, etc.)
- Slope geometry (e.g. slope angle, slope height, etc.)
Critical Condition of D/S Slope in Reservoir
Out of the above factors, the water level in the reservoir plays a significant role in determining the critical condition of d/s slope. When the water level in the reservoir rises above a certain level, it exerts a hydrostatic pressure on the d/s slope. This hydrostatic pressure can cause the soil particles to lose their shear strength, leading to slope failure. This is known as the critical condition of d/s slope.
To avoid this critical condition, it is important to ensure that the slope is designed with an adequate safety factor. The safety factor is the ratio of the resisting forces (e.g. soil strength) to the driving forces (e.g. hydrostatic pressure). A safety factor of 1 indicates that the driving forces are equal to the resisting forces, and the slope is at the verge of failure. A safety factor greater than 1 indicates that the resisting forces are greater than the driving forces, and the slope is stable.
Conclusion
In conclusion, the critical condition of d/s slope in a reservoir occurs when the water level in the reservoir rises above a certain level, leading to a hydrostatic pressure that can cause the slope to fail. To avoid this critical condition, it is important to design the slope with an adequate safety factor.
The critical condition of d/s (downstream) slope in a reservoir occurs when the reservoir is full. Let's understand this in detail.
Factors Affecting Stability of Slopes in Reservoir
The stability of slopes in a reservoir is affected by various factors like:
- Water level in the reservoir
- Pore pressure in the soil
- Seepage forces
- Soil characteristics (e.g. soil type, grain size distribution, shear strength, etc.)
- Slope geometry (e.g. slope angle, slope height, etc.)
Critical Condition of D/S Slope in Reservoir
Out of the above factors, the water level in the reservoir plays a significant role in determining the critical condition of d/s slope. When the water level in the reservoir rises above a certain level, it exerts a hydrostatic pressure on the d/s slope. This hydrostatic pressure can cause the soil particles to lose their shear strength, leading to slope failure. This is known as the critical condition of d/s slope.
To avoid this critical condition, it is important to ensure that the slope is designed with an adequate safety factor. The safety factor is the ratio of the resisting forces (e.g. soil strength) to the driving forces (e.g. hydrostatic pressure). A safety factor of 1 indicates that the driving forces are equal to the resisting forces, and the slope is at the verge of failure. A safety factor greater than 1 indicates that the resisting forces are greater than the driving forces, and the slope is stable.
Conclusion
In conclusion, the critical condition of d/s slope in a reservoir occurs when the water level in the reservoir rises above a certain level, leading to a hydrostatic pressure that can cause the slope to fail. To avoid this critical condition, it is important to design the slope with an adequate safety factor.
In Swedish circle method ___________ analysis cases are considered.- a)φC = 0
- b)C – φ and φu = 0
- c)None of the mentioned
- d)All of the mentioned
Correct answer is option 'B'. Can you explain this answer?
In Swedish circle method ___________ analysis cases are considered.
a)
φC = 0
b)
C – φ and φu = 0
c)
None of the mentioned
d)
All of the mentioned
|
|
Sanya Agarwal answered |
In Swedish slip circle method two cases are considered: i) analysis of purely cohesive soil (φu = 0) and ii) analysis of a soil possessing both cohesion and friction (C-φ analysis).
A basic type of failure at a finite slope may occur due to ___________- a)Slope failure and Base failure
- b)Toe failure
- c)None of the mentioned
- d)All of the mentioned
Correct answer is option 'A'. Can you explain this answer?
A basic type of failure at a finite slope may occur due to ___________
a)
Slope failure and Base failure
b)
Toe failure
c)
None of the mentioned
d)
All of the mentioned
|
|
Sanvi Kapoor answered |
The two basic types of failure of a finite slope may occur: (i) slope failure (ii) base failure.
The factor of safety with respect to cohesion is given by the equation ____________- a)FC = HC / H
- b)FC = C / Cm
- c)FC = τf / τ
- d)All of the mentioned
Correct answer is option 'B'. Can you explain this answer?
The factor of safety with respect to cohesion is given by the equation ____________
a)
FC = HC / H
b)
FC = C / Cm
c)
FC = τf / τ
d)
All of the mentioned
|
|
Sanvi Kapoor answered |
The factor of safety Fc with respect to cohesive strength, based on the assumption that the frictional strength has been fully mobilized, is given by FC = C / Cm.
Analysis of stability of slopes is used for determining ___________- a)Shearing strength and Stressed internal surface
- b)Properties of the soil
- c)None of the mentioned
- d)All of the mentioned
Correct answer is option 'A'. Can you explain this answer?
Analysis of stability of slopes is used for determining ___________
a)
Shearing strength and Stressed internal surface
b)
Properties of the soil
c)
None of the mentioned
d)
All of the mentioned
|
|
Sanya Agarwal answered |
An analysis of stability of slopes consist of determining the most severely stressed internal surface and the magnitude of the shearing stress to which it is subjected and determining the shearing stress along the surface.
In the rectangular plot method if there are n-slices, the total number of ordinates will be ____________- a)n – 1
- b)n
- c)n – 2
- d)None of the mentioned
Correct answer is option 'A'. Can you explain this answer?
In the rectangular plot method if there are n-slices, the total number of ordinates will be ____________
a)
n – 1
b)
n
c)
n – 2
d)
None of the mentioned
|
|
Sanya Agarwal answered |
In the rectangular plot method if there are n-slices, the total number of ordinates will be (n-1), the ordinates at the end of the last strip being zero.
A cohesionless soil having an angle of shearing resistant of φ, is standing at a slope angle of i. The factor of safety of the slope is:- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A cohesionless soil having an angle of shearing resistant of φ, is standing at a slope angle of i. The factor of safety of the slope is:
a)
b)
c)
d)
|
|
Poulomi Patel answered |
For an infinite slope in cohesionless soil.
The slip circle having the minimum factor of safety is called ___________- a)Critical circle
- b)Failure slip circle
- c)Critical slip circle
- d)None of the mentioned
Correct answer is option 'C'. Can you explain this answer?
The slip circle having the minimum factor of safety is called ___________
a)
Critical circle
b)
Failure slip circle
c)
Critical slip circle
d)
None of the mentioned
|
|
Sanvi Kapoor answered |
A number of trial critical slip circle are chosen against sliding and factor of safety of each is computed. The circle having the minimum factor of safety is the critical slip circle.
A circle is considered to be a critical slip circle, if it has ___________- a)Maximum factor of safety
- b)Minimum factor of safety
- c)Maximum radius
- d)All of the mentioned
Correct answer is option 'B'. Can you explain this answer?
A circle is considered to be a critical slip circle, if it has ___________
a)
Maximum factor of safety
b)
Minimum factor of safety
c)
Maximum radius
d)
All of the mentioned
|
|
Sanya Agarwal answered |
The circle giving minimum factor of safety Fc is considered as a critical slip circle.
The factor K, in the radius of the friction circle(Kr sin φ) depends on ____________- a)Radius of the frictional circle
- b)Frictional resistances offered
- c)Cohesive resistance
- d)Central angle of the slip arc
Correct answer is option 'D'. Can you explain this answer?
The factor K, in the radius of the friction circle(Kr sin φ) depends on ____________
a)
Radius of the frictional circle
b)
Frictional resistances offered
c)
Cohesive resistance
d)
Central angle of the slip arc
|
|
Sanvi Kapoor answered |
In the frictional circle radius of Kr sin φ, the factor K depends on the central angle δ of the slip arc.
According to Bennet, non-circular slip surface may arise in ___________- a)Non homogeneous dam
- b)Embankment dams
- c)Homogeneous dam
- d)Soil deposit with a specific plane of weakness
Correct answer is option 'C'. Can you explain this answer?
According to Bennet, non-circular slip surface may arise in ___________
a)
Non homogeneous dam
b)
Embankment dams
c)
Homogeneous dam
d)
Soil deposit with a specific plane of weakness
|
|
Sanya Agarwal answered |
According to Bennet (1951), non-circular or composite slip surface may arise in homogeneous dam have foundation of infinite depth, rigid boundary planes of maximum and presence of stronger or weaker layer.
The depth factor Df for toe failure is _____- a)Df > 1
- b)Df < 1
- c)Df = 1
- d)Df = 0
Correct answer is option 'C'. Can you explain this answer?
The depth factor Df for toe failure is _____
a)
Df > 1
b)
Df < 1
c)
Df = 1
d)
Df = 0
|
|
Sanvi Kapoor answered |
For toe failure, Df = 1; for base failure, Df > 1.
The friction circle method assumes the failure surface as ___________- a)Cycloid
- b)Curve
- c)Arc of circle
- d)None of the mentioned
Correct answer is option 'C'. Can you explain this answer?
The friction circle method assumes the failure surface as ___________
a)
Cycloid
b)
Curve
c)
Arc of circle
d)
None of the mentioned
|
|
Sanya Agarwal answered |
Similar to Culmann’s method, the friction circle method also assumes the failure surface as the arc of circle.
If the soil properties for all identical depths below the surface are constant, it is a _________- a)Finite slope
- b)Infinite slope
- c)Planar failure surface
- d)All of the mentioned
Correct answer is option 'B'. Can you explain this answer?
If the soil properties for all identical depths below the surface are constant, it is a _________
a)
Finite slope
b)
Infinite slope
c)
Planar failure surface
d)
All of the mentioned
|
|
Sanvi Kapoor answered |
If a slope represents the boundary surface of a semi-infinite soil mass, and the soil properties for all identical depths below the surface are constant, it is called an infinite slope.
Rectangular plot method has been suggested by _________- a)Culmann
- b)Bishop
- c)Singh
- d)Terzaghi
Correct answer is option 'C'. Can you explain this answer?
Rectangular plot method has been suggested by _________
a)
Culmann
b)
Bishop
c)
Singh
d)
Terzaghi
|
|
Sanya Agarwal answered |
A simplified rectangular plot method for finding the stability of slopes was suggested by Singh in 1962.
In stability computation, the curve representing the real surface of sliding is usually replaced by ______- a)Arc of circle and Logarithmic failure
- b)Cycloid
- c)None of the mentioned
- d)All of the mentioned
Correct answer is option 'A'. Can you explain this answer?
In stability computation, the curve representing the real surface of sliding is usually replaced by ______
a)
Arc of circle and Logarithmic failure
b)
Cycloid
c)
None of the mentioned
d)
All of the mentioned
|
|
Sanvi Kapoor answered |
Failure of finite slopes occurs along a curved surface. In stability computation, the curve is represented by an arc of a circle or logarithmic scale.
The types of slip surface or failure surfaces are _____- a)4
- b)2
- c)3
- d)5
Correct answer is option 'C'. Can you explain this answer?
The types of slip surface or failure surfaces are _____
a)
4
b)
2
c)
3
d)
5
|
|
Sanvi Kapoor answered |
The types of failure surface are: i) Planar failure surface ii) circular failure surface iii) Non-circular failure surface.
The stability of a finite slope can be investigated by which of the following method?- a)Bishop’s method
- b)Swedish circle method
- c)Friction circle method
- d)All of the mentioned
Correct answer is option 'D'. Can you explain this answer?
The stability of a finite slope can be investigated by which of the following method?
a)
Bishop’s method
b)
Swedish circle method
c)
Friction circle method
d)
All of the mentioned
|
|
Sanvi Kapoor answered |
The stability of a finite slope can be investigated by the following methods:
(i) Culmann’s method of planar failure surface
(ii) The Swedish circle method
(iii) Friction circle method
(iv) Bishop’s method.
(i) Culmann’s method of planar failure surface
(ii) The Swedish circle method
(iii) Friction circle method
(iv) Bishop’s method.
is approximately equal to
The factor of safety Fc with respect to cohesive strength is based on the assumption that ___________- a)Frictional force is fully mobilized
- b)Frictional force is zero
- c)Total cohesive resistance is zero
- d)All of the mentioned
Correct answer is option 'A'. Can you explain this answer?
The factor of safety Fc with respect to cohesive strength is based on the assumption that ___________
a)
Frictional force is fully mobilized
b)
Frictional force is zero
c)
Total cohesive resistance is zero
d)
All of the mentioned
|
|
Sanya Agarwal answered |
The factor of safety with respect to cohesive strength, based on the assumption that frictional force has been fully mobilized, and the equation is given by FC = C / Cm.
The area of U-diagram in finding stability of D/S slope during steady seepage, can be measured by ___________- a)Planimeter and Rectangular plot method
- b)Fellinious method
- c)None of the mentioned
- d)All of the mentioned
Correct answer is option 'A'. Can you explain this answer?
The area of U-diagram in finding stability of D/S slope during steady seepage, can be measured by ___________
a)
Planimeter and Rectangular plot method
b)
Fellinious method
c)
None of the mentioned
d)
All of the mentioned
|
|
Sanya Agarwal answered |
The area of U-diagram can be measured with the help of a planimeter, or else the rectangular plot method can be utilized.
Earth embankments or slopes are commonly required for which of the following purpose?- a)Railways
- b)Earth dams
- c)Road ways
- d)All of the mentioned
Correct answer is option 'D'. Can you explain this answer?
Earth embankments or slopes are commonly required for which of the following purpose?
a)
Railways
b)
Earth dams
c)
Road ways
d)
All of the mentioned
|
|
Sanya Agarwal answered |
Earth embankments are commonly required for railways, roadways, earth dams, levees and river training works.
The pore water pressure at any point on the slip surface is represented by ____________- a)Piezometric head
- b)Pore pressure
- c)Factor of safety
- d)Failure plane
Correct answer is option 'A'. Can you explain this answer?
The pore water pressure at any point on the slip surface is represented by ____________
a)
Piezometric head
b)
Pore pressure
c)
Factor of safety
d)
Failure plane
|
|
Sanya Agarwal answered |
The pore water pressure at any point is represented by the piezometric head hw at that point.
List-I given below gives the possible types of failure for a finite soil slope and List-II gives the reasons for these different types of failure. Match the items in List-I with the items in List-II and select the correct answer from the codes given below the lists:
List-I
A. Base failure
B. Face failure
C. Toe failure
List-II
1. Soils above and below the toe have same strength
2. Soil above the toe is comparatively weaker
3. Soil above the toe is comparatively stronger
- a)A
- b)B
- c)C
- d)D
Correct answer is option 'D'. Can you explain this answer?
List-I given below gives the possible types of failure for a finite soil slope and List-II gives the reasons for these different types of failure. Match the items in List-I with the items in List-II and select the correct answer from the codes given below the lists:
List-I
A. Base failure
B. Face failure
C. Toe failure
List-II
1. Soils above and below the toe have same strength
2. Soil above the toe is comparatively weaker
3. Soil above the toe is comparatively stronger
List-I
A. Base failure
B. Face failure
C. Toe failure
List-II
1. Soils above and below the toe have same strength
2. Soil above the toe is comparatively weaker
3. Soil above the toe is comparatively stronger
a)
A
b)
B
c)
C
d)
D
|
|
Tanishq Nair answered |
Face failure or slope failure can occur when the slope angle β is very high and the soil close to the toe is quite strong or the soil in the upper part of slope is relatively weak.
Base failure can occur when the soil below the toe is relatively weak and soft and the slope is flat. Toe failure occurs in steep slopes when the soil mass above the base and below the base is homogeneous.
Base failure can occur when the soil below the toe is relatively weak and soft and the slope is flat. Toe failure occurs in steep slopes when the soil mass above the base and below the base is homogeneous.
Chapter doubts & questions for Stability of Slopes - Foundation Engineering 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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