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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?
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Most Upvoted Answer
If an infinite slope of clay at a depth 5 m has cohesion of 1 t/m2 and...
Taylor's stability number method:
(i) Taylor introduced a dimensional parameter, called taylor's stability number, which is given by
Where, H = Effective height of slope
Hc = Critical height of slope
F = FOS
(ii) In this method, Taylor's stability number is read from taylor's chart in order to analyze the stability of slope for a given values of strength parameters of the soil.
(iii) Taylor's method is suitable for c-ϕ soils and for ϕ = 0 (pue clays)
Calculation:
Given, C = 1 t/m2, γ = 2 t/m2 and Hc = 5 m
(i) Taylor introduced a dimensional parameter, called taylor's stability number, which is given by
Where, H = Effective height of slope
Hc = Critical height of slope
F = FOS
(ii) In this method, Taylor's stability number is read from taylor's chart in order to analyze the stability of slope for a given values of strength parameters of the soil.
(iii) Taylor's method is suitable for c-ϕ soils and for ϕ = 0 (pue clays)
Calculation:
Given, C = 1 t/m2, γ = 2 t/m2 and Hc = 5 m
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Community Answer
If an infinite slope of clay at a depth 5 m has cohesion of 1 t/m2 and...
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).
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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 bea)0.1b)0.2c)0.3d)0.4Correct answer is option 'A'. Can you explain this answer?
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