Question
Download Solution PDFIn a sieve analysis of soil, the total weight of soil taken was 500 g. The mass of soil retained over 4.75 mm sieve was 100 g, mass retained over 2 mm sieve was 150 g, and the mass retained over 425 -micron sieve was 200 g. The effective size of the soil will be:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Sieve Analysis of Soil
- Sieve analysis is a laboratory procedure used to determine the particle size distribution of soil by passing the soil through a set of sieves with progressively smaller openings.
- The effective size (D10) of soil is the diameter of the particle below which 10% of the soil's particles by weight are finer.
Calculation:
Given Data
- Total weight of soil taken: 500 g
- Mass of soil retained over 4.75 mm sieve: 100 g
- Mass of soil retained over 2 mm sieve: 150 g
- Mass of soil retained over 425-micron sieve: 200 g
First, we need to determine the cumulative percentage of soil passing each sieve. The total mass retained and the mass passing through each sieve are calculated as follows:
- Mass retained over 4.75 mm sieve: 100 g
- Mass passing 4.75 mm sieve = 500 g - 100 g = 400 g
- Cumulative percentage passing 4.75 mm sieve = (400 g / 500 g) ×100% = 80%
- Mass retained over 2 mm sieve: 150 g
- Mass passing 2 mm sieve = 400 g - 150 g = 250 g
- Cumulative percentage passing 2 mm sieve = (250 g / 500 g) ×100% = 50%
- Mass retained over 425-micron sieve: 200 g
- Mass passing 425-micron sieve = 250 g - 200 g = 50 g
- Cumulative percentage passing 425-micron sieve = (50 g / 500 g) ×100% = 10%
The effective size (D10) is the diameter of the particle below which 10% of the soil's particles by weight are finer. From the calculations above, we can see that 10% of the soil's particles by weight are finer than the 425-micron sieve.
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