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The weight of an undried specimen of clay was 34.62 gm .the oven dry weight of the sample was 20.36gm.before oven drying,the specimen was immersed in mercury and its volume found to 24.66cc assume g=2.68 find void ratio and degree of saturation of soil?
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The weight of an undried specimen of clay was 34.62 gm .the oven dry w...
Given:

  • Weight of undried clay specimen = 34.62 gm

  • Oven dry weight of clay specimen = 20.36 gm

  • Volume of clay specimen before oven drying = 24.66 cc

  • g = 2.68



Calculations:

1. Calculate the moisture content:

Moisture content = (Weight of undried specimen - Oven dry weight of specimen) / Oven dry weight of specimen

Moisture content = (34.62 - 20.36) / 20.36 = 0.702

2. Calculate the mass of water:

Mass of water = Moisture content x Oven dry weight of specimen

Mass of water = 0.702 x 20.36 = 14.28 gm

3. Calculate the bulk density:

Bulk density = Oven dry weight of specimen / Volume of specimen

Bulk density = 20.36 / 24.66 = 0.824 gm/cc

4. Calculate the dry density:

Dry density = Bulk density / (1 + Moisture content)

Dry density = 0.824 / (1 + 0.702) = 0.484 gm/cc

5. Calculate the void ratio:

Void ratio = (Bulk density - Dry density) / Dry density

Void ratio = (0.824 - 0.484) / 0.484 = 0.7

6. Calculate the degree of saturation:

Degree of saturation = (Mass of water / (Volume of specimen x Unit weight of water)) x 100%

Unit weight of water = g x 9.81 = 2.68 x 9.81 = 26.273 gm/cc

Volume of solids = Volume of specimen - Volume of voids

Volume of solids = Volume of specimen / (1 + Void ratio)

Volume of solids = 24.66 / (1 + 0.7) = 14.50 cc

Volume of voids = Volume of specimen - Volume of solids

Volume of voids = 24.66 - 14.50 = 10.16 cc

Degree of saturation = (14.28 / (14.50 x 26.273)) x 100% = 20.33%

Conclusion:

  • Void ratio of the soil specimen = 0.7

  • Degree of saturation of the soil specimen = 20.33%

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The weight of an undried specimen of clay was 34.62 gm .the oven dry weight of the sample was 20.36gm.before oven drying,the specimen was immersed in mercury and its volume found to 24.66cc assume g=2.68 find void ratio and degree of saturation of soil?
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The weight of an undried specimen of clay was 34.62 gm .the oven dry weight of the sample was 20.36gm.before oven drying,the specimen was immersed in mercury and its volume found to 24.66cc assume g=2.68 find void ratio and degree of saturation of soil? 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 The weight of an undried specimen of clay was 34.62 gm .the oven dry weight of the sample was 20.36gm.before oven drying,the specimen was immersed in mercury and its volume found to 24.66cc assume g=2.68 find void ratio and degree of saturation of soil? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The weight of an undried specimen of clay was 34.62 gm .the oven dry weight of the sample was 20.36gm.before oven drying,the specimen was immersed in mercury and its volume found to 24.66cc assume g=2.68 find void ratio and degree of saturation of soil?.
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