Concept explainers
(a)
Mention whether the given statement is true or false.
(a)
Answer to Problem 3.1P
The given statement, “Moisture content must be always less than 100%” is
Explanation of Solution
Calculation:
The moisture content (w) is the ratio of weight of water
If the water content in the soil exceeds 100%, the weight of water
Hence, the moisture content may be greater than 100%.
Therefore, the given statement, “Moisture content must be always less than 100%” is
(b)
Mention whether the given statement is true or false.
(b)
Answer to Problem 3.1P
The given statement, “Plastic limit is always less than plasticity index” is
Explanation of Solution
Calculation:
Write the equation for the relationship between the plasticity index (PI), liquid limit (LL), and the plastic limit (PL) as follows:
When the plastic limit is equal or greater than the liquid limit, the plasticity index is considered as zero. In this case the plastic limit is greater than the plasticity index.
The given statement is not satisfying the condition.
Therefore, the given statement “Plastic limit is always less than plasticity index” is
(c)
Mention whether the given statement is true or false.
(c)
Answer to Problem 3.1P
The given statement, “Liquidity index cannot be negative” is
Explanation of Solution
Calculation:
Write the equation for liquidity index (LI) as follows:
Here, the in situ moisture content of the soil is w, plastic limit is PL, and liquid limit is LL.
For sensitive clays, the in situ moisture content is greater than the liquid limit
In this condition liquidity index is greater than 1
For soils that are heavily over consolidated, the in situ moisture content is less than the plastic limit
The liquidity index can be less than zero.
Therefore, the given statement “Liquidity index cannot be negative” is
(d)
Mention whether the given statement is true or false.
(d)
Answer to Problem 3.1P
The given statement, “The dry unit weight is always less than the moist unit weight” is
Explanation of Solution
Calculation:
Find the moist unit weight
Find the dry unit weight
From the equations of moist unit weight and the dry unit weight, the dry unit weight will be less than the moist unit weight.
Therefore, the given statement “The dry unit weight is always less than the moist unit weight” is
(e)
Mention whether the given statement is true or false.
(e)
Answer to Problem 3.1P
The given statement, “Relative density increases with void ratio” is
Explanation of Solution
Calculation:
Write the equation for relative density
Here, in situ void ratio of the soil is e, void ratio of the soil in the loosest condition is
Refer to the relative density equation, the relative density is inversely proportional to the void ratio.
When the void ratio increases, the relative density decreases.
Therefore, the given statement “Relative density increases with void ratio” is
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Chapter 3 Solutions
Fundamentals of Geotechnical Engineering (MindTap Course List)
- Determine void ratio if porosity 0.39arrow_forward01. The wet weight of 29.9 x 10 -4 m3 of soil is 62.7 N. If the moisture content is 11.6% and the specific gravity of the solids is 2.67, determine the following: IT IS REQUESTED: A) Wet unit weight (KN/m3; lb/ft3) D) Saturated unit weight (KN/m3; lb/ft3) B) Dry unit weight (KN/m3; lb/ft3). E) degree of saturation. C) Void ratio F) Volume occupied by water (m3)arrow_forwardFind Void ratioarrow_forward
- A cylindrical mould of 10 cm internal diameter and 11.7 cm height weighs 1894 grams. The mould was filled up with dry soil, first at its loosest state and then at the densest state, and was found to weigh 3273 grams and 3538 grams respectively. if the natural soil existing at the field be submerged below the ground water table and has a water content of 23 %, determine the relative density of the soil. Given, G = 2.65. 12.arrow_forwardDescribe Kozeny-Carman Equation for Permeability.arrow_forwardGiven following data, calculate a. Volume of measure b. Bulk density c. Absorption d. Void contentarrow_forward
- The mass of a wet sample of soil and its container is 0.32 kg. The dry mass of the soil and itscontainer is 0.28 kg. The mass of the container is 0.06 kg, and its volume is 1.5x10-4 m3. Thesoil completely fills the container. Finally, the specific gravity of solids is found to be 2.71.Determine the following parameters:a. The moist unit weight,b. The dry unit weight,c. The void ratio,d. The moisture content,e. The degree of saturation,f. The saturated unit weight,g. The submerged unit weight, andh. What is the maximum dry unit weight to which this soil can be compacted withoutchanging its moisture content?arrow_forwardThe result of the liquid limit tests where N is the no of blows and w is the water content in % are given as follows: Trial 1 N=11 w=53.9; Trial 2 N=15 w=50.6; Trial 3 N=23 w=48.1; Trial 4 N= 30 w= 46; Trial 5 N= 46 w=43.3; Trial 6 N= 53 w=41. The PL= 30 and the in situ moisture content is 35%. Determine the plasticity index. A. 27 b. 12 C. 17 d. 22arrow_forwardIn natural state, a moist soil has a volume of 0.33 ft3 and weighs 39.93 lb. The oven dry weight of the soil is 34.54 lb. If Gs= 2.71, calculate the volume of the soil solid in ft3.arrow_forward
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- Fundamentals of Geotechnical Engineering (MindTap...Civil EngineeringISBN:9781305635180Author:Braja M. Das, Nagaratnam SivakuganPublisher:Cengage Learning