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An unconfined compression test was conducted on an undisturbed sample of clay. The sample had a diameter of 80mm and was 200 mm long, The load at failure was 30 N and axial deformation of the sample at failure was 15mm the undrained shear strength of the clay is ____ kN/m2
- a)5.52
- b)0.0552
- c)2.76
- d)0.0276
Correct answer is option 'C'. Can you explain this answer?
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An unconfined compression test was conducted on an undisturbed sample...
To find the undrained shear strength of the clay, we can use the unconfined compression test data provided.
Given:
Diameter of the sample (D) = 80 mm
Length of the sample (L) = 200 mm
Load at failure (P) = 30 N
Axial deformation at failure (δ) = 15 mm
We can calculate the cross-sectional area (A) of the sample using the diameter:
A = π * (D/2)^2
Next, we can calculate the stress (σ) at failure using the load and cross-sectional area:
σ = P / A
The shear strength (τ) can be related to the stress at failure using the following equation:
τ = σ / 2
Finally, we can convert the shear strength from N/m2 to kN/m2 by dividing by 1000.
Let's calculate the undrained shear strength step by step:
1. Cross-sectional area (A):
A = π * (80/2)^2
A = π * 40^2
A = 1600π mm2
2. Stress at failure (σ):
σ = P / A
σ = 30 N / 1600π mm2
3. Shear strength (τ):
τ = σ / 2
τ = (30 N / 1600π mm2) / 2
τ = 15 / 1600π N/mm2
4. Shear strength in kN/m2:
τ = (15 / 1600π) N/mm2 * (1 kN / 1000 N) * (1000 mm / 1 m) * (1 m2 / 1,000,000 mm2)
τ = 15 / (1600π * 1000 * 1000)
τ ≈ 2.76 kN/m2
Therefore, the undrained shear strength of the clay sample is approximately 2.76 kN/m2. Hence, the correct answer is option 'C'.
Given:
Diameter of the sample (D) = 80 mm
Length of the sample (L) = 200 mm
Load at failure (P) = 30 N
Axial deformation at failure (δ) = 15 mm
We can calculate the cross-sectional area (A) of the sample using the diameter:
A = π * (D/2)^2
Next, we can calculate the stress (σ) at failure using the load and cross-sectional area:
σ = P / A
The shear strength (τ) can be related to the stress at failure using the following equation:
τ = σ / 2
Finally, we can convert the shear strength from N/m2 to kN/m2 by dividing by 1000.
Let's calculate the undrained shear strength step by step:
1. Cross-sectional area (A):
A = π * (80/2)^2
A = π * 40^2
A = 1600π mm2
2. Stress at failure (σ):
σ = P / A
σ = 30 N / 1600π mm2
3. Shear strength (τ):
τ = σ / 2
τ = (30 N / 1600π mm2) / 2
τ = 15 / 1600π N/mm2
4. Shear strength in kN/m2:
τ = (15 / 1600π) N/mm2 * (1 kN / 1000 N) * (1000 mm / 1 m) * (1 m2 / 1,000,000 mm2)
τ = 15 / (1600π * 1000 * 1000)
τ ≈ 2.76 kN/m2
Therefore, the undrained shear strength of the clay sample is approximately 2.76 kN/m2. Hence, the correct answer is option 'C'.
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Community Answer
An unconfined compression test was conducted on an undisturbed sample...
C = 2.76 kN/m2
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An unconfined compression test was conducted on an undisturbed sample of clay. The sample had a diameter of 80mm and was 200 mm long, The load at failure was 30 N and axial deformation of the sample at failure was 15mm the undrained shear strength of the clay is ____ kN/m2a)5.52b)0.0552c)2.76d)0.0276Correct answer is option 'C'. Can you explain this answer?
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An unconfined compression test was conducted on an undisturbed sample of clay. The sample had a diameter of 80mm and was 200 mm long, The load at failure was 30 N and axial deformation of the sample at failure was 15mm the undrained shear strength of the clay is ____ kN/m2a)5.52b)0.0552c)2.76d)0.0276Correct answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about An unconfined compression test was conducted on an undisturbed sample of clay. The sample had a diameter of 80mm and was 200 mm long, The load at failure was 30 N and axial deformation of the sample at failure was 15mm the undrained shear strength of the clay is ____ kN/m2a)5.52b)0.0552c)2.76d)0.0276Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An unconfined compression test was conducted on an undisturbed sample of clay. The sample had a diameter of 80mm and was 200 mm long, The load at failure was 30 N and axial deformation of the sample at failure was 15mm the undrained shear strength of the clay is ____ kN/m2a)5.52b)0.0552c)2.76d)0.0276Correct answer is option 'C'. Can you explain this answer?.
An unconfined compression test was conducted on an undisturbed sample of clay. The sample had a diameter of 80mm and was 200 mm long, The load at failure was 30 N and axial deformation of the sample at failure was 15mm the undrained shear strength of the clay is ____ kN/m2a)5.52b)0.0552c)2.76d)0.0276Correct answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about An unconfined compression test was conducted on an undisturbed sample of clay. The sample had a diameter of 80mm and was 200 mm long, The load at failure was 30 N and axial deformation of the sample at failure was 15mm the undrained shear strength of the clay is ____ kN/m2a)5.52b)0.0552c)2.76d)0.0276Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An unconfined compression test was conducted on an undisturbed sample of clay. The sample had a diameter of 80mm and was 200 mm long, The load at failure was 30 N and axial deformation of the sample at failure was 15mm the undrained shear strength of the clay is ____ kN/m2a)5.52b)0.0552c)2.76d)0.0276Correct answer is option 'C'. Can you explain this answer?.
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