A  shrinkage limit test for a saturated specimen had the initial volume Vi = 21.35 cm3 and initial weight Wi = 37 gf ( = 37 × 0.00981 = 0.363 N). After drying, it became Vf = 14.3 cm3 and Wf = 26 gf (= 26 × 0.00981 = 0.255 N). Compute the shrinkage limit (SL) of this soil.

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Q: The following data were recorded during a shrinkage limit test on a clay soil pat: Vi = 20. 9 cm³,…

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Q: f dr

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Q: The following are the data of the compaction test: Volume of mold = 0.000945 m^3 Determine the…

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Q: The plastic limit and liquid limit of a soil are 30% and 42% respectively. The percentage volume…

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Q: How do i solve problem 6.4 a & b

A: The given data is shown below:

Q: What is the specific gravity of the soil?

A: shrinkge limits , v= 25.6 cc , v2 = 15.9cc ,w1 =44g , w2= 30.1g , g has to be find out

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Q: A standard compaction test was carried out in a mould of volume 0.001 m³1. Test results ar tabulated…

A: V = 0.001 m3Gs = 2.67

Q: The results of a standard Proctor test are given in the following table. Given: Mold volume : 1/30…

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Q: A representative soil specimen collected from the a construction site weighs 1,8kN and has a volume…

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Q: A Standard Proctor compaction test performed on a sample of crushed limestone (G; = 9.70) obtained a…

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Q: A saturated 100cm^3 clay sample has a natural water content of 30% . It is found that the shrinkage…

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Q: What is the natural moisture content of soil in percent? b. Compute the consistency index.

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Q: During a consolidation test, an increment pressure of 153 kN/m2 was applied to a sample of saturated…

A: The given data is shown below:

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Q: C) A specimen of saturated soil ( Gs = 2.72) has a mass of 370 gm after drying the volume was 160…

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Q: • Problem 3- From the following data of the grain size analysis Sieve size #4 #8 # 10 #25 #60 # 100…

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Weight-Volume relationships

Weight is defined as the quantity of matter contained in a body. Weight can also be defined as the force acting on a body due to gravity. Mathematically, weight can be defined as-