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What is the average (D50) size of particles of the soil that will be present in the soil to develop negative pore water pressure = 9.81 kN/m3 due to capillary action? Given that, Cu=7 and Cc=2 for the soil. A. >1.05mm B. 0.15-1.05mm C. <1.05mm?>
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What is the average (D50) size of particles of the soil that will be p...
**Answer:**

To determine the average (D50) size of particles of the soil that will develop negative pore water pressure, we need to use the effective stress principle and the soil classification parameters: Cu (coefficient of uniformity) and Cc (coefficient of curvature).

**1. Effective Stress Principle:**
According to the effective stress principle, the negative pore water pressure in the soil is developed due to capillary action. The effective stress (σ') can be calculated using the equation:

σ' = σ - u_a

Where:
σ' = Effective stress (kN/m2)
σ = Total stress (kN/m2)
u_a = Pore water pressure (kN/m2)

In this case, the given pore water pressure is -9.81 kN/m3. To convert it into kN/m2, we need to multiply it by the unit weight of water (γ_w = 9.81 kN/m3).

u_a = -9.81 kN/m3 * 9.81 kN/m3 = -96.06 kN/m2

**2. Coefficient of Uniformity (Cu):**
The coefficient of uniformity (Cu) is a measure of the particle size distribution of the soil. It is calculated using the following equation:

Cu = D60 / D10

Where:
Cu = Coefficient of uniformity
D60 = Particle size diameter corresponding to 60% passing
D10 = Particle size diameter corresponding to 10% passing

**3. Coefficient of Curvature (Cc):**
The coefficient of curvature (Cc) is a measure of the shape of the particle size distribution curve. It is calculated using the following equation:

Cc = (D30)² / (D10 * D60)

Where:
Cc = Coefficient of curvature
D30 = Particle size diameter corresponding to 30% passing

**4. Determining the Average (D50) Particle Size:**
To determine the average (D50) particle size, we need to find the particle size corresponding to 50% passing. We can use the following equation to calculate it:

D50 = √(D10 * D60)

**5. Calculation:**
Given that Cu = 7 and Cc = 2, we can calculate the particle size distribution as follows:

Cu = D60 / D10
7 = D60 / D10

Cc = (D30)² / (D10 * D60)
2 = (D30)² / (D10 * D60)

Using these two equations, we can solve for D10, D30, and D60.

D10 = D60 / 7 --> Equation (1)
D30 = √(2 * D10 * D60) --> Equation (2)

Substituting Equation (1) into Equation (2), we get:

D30 = √(2 * D10 * 7 * D10) = √(14 * D10²) = √14 * D10 --> Equation (3)

Substituting Equation (1) into Equation (3), we get:

√14 * D10 = D60 / 7

Simplifying, we get:

D60 = 7 * √14 * D10 --> Equation (4)

Substituting Equation (4) into Equation (1), we get
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What is the average (D50) size of particles of the soil that will be present in the soil to develop negative pore water pressure = 9.81 kN/m3 due to capillary action? Given that, Cu=7 and Cc=2 for the soil. A. >1.05mm B. 0.15-1.05mm C.
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What is the average (D50) size of particles of the soil that will be present in the soil to develop negative pore water pressure = 9.81 kN/m3 due to capillary action? Given that, Cu=7 and Cc=2 for the soil. A. >1.05mm B. 0.15-1.05mm C. 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 What is the average (D50) size of particles of the soil that will be present in the soil to develop negative pore water pressure = 9.81 kN/m3 due to capillary action? Given that, Cu=7 and Cc=2 for the soil. A. >1.05mm B. 0.15-1.05mm C. covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the average (D50) size of particles of the soil that will be present in the soil to develop negative pore water pressure = 9.81 kN/m3 due to capillary action? Given that, Cu=7 and Cc=2 for the soil. A. >1.05mm B. 0.15-1.05mm C. .
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