Geotechnical Engineering Objective Questions
Civil Engineering MCQs
This set of Geotechnical Engineering Quiz focuses on “Effect of Compaction on Soil Properties”.
1. Which of the following property of soil is improved by compacting the soil?
Explanation: The main aim of compacting soil is to improve some desirable properties of the soil, such as reduction of compressibility, water absorption and permeability, increase in soil strength, bearing capacity.
2. When compacted dry of optimum, the structure of clay is always ___________
Explanation: As the structure of compacted clay is complicated and increasingly oriented, the structure of clay is always flocculated.
3. The flocculated structure of compacted dry soil is broken due to _____________
Explanation: Due to higher strains, the flocculated structure of the compacted on the dry side is broken, giving rise to ultimate strength.
4. The shear strength of compacted clays depends upon ____________
Explanation: The shear strength of compacted clays depends upon dry density, moulding water content, soil structure, method of compaction, drainage condition and type of soil.
5. The structure of composite soil after compacted will be in the form of ______________
Explanation: The structure of composite soils, after being compacted depend upon the relative proportion of coarse particles their structure can either be coarse grained skeleton structure or cohesive matrix structure.
6. What is the maximum dry density for a soil sample having sp. gr. of 2.7 and OMC=16 %?
Explanation: γd, max occurs when S is maximum, i.e. when S=1 γd, max = G γW/1+ (w.G/S) = G γW/1+w.G (since S=1) Hence, ρd max = G. ρW/1+w.G = 2.7×1/1 + (0.16×2.7) Maximum dry density, γd max = 1.885 g/cm3.
7. Strength of soil sample compacted wet of optimum is influenced by ____________
Explanation: The manner of compaction influences the strength of soil sample compacted wet of optimum to a certain extent.
8. A cohesive soil yields a maximum dry density of 1.8 g/cc at an OMC of 16 % during a standard proctor test. What will be its degree of saturation?Take G=2.65
Explanation: Given ρd = 1.8 g/cm3; w=0.1; G=2.65 e=G ρw/ρd -1 = (2.65×1/1.8) -1 =0.4722 S = w G/e = 0.16×2.65/0.4722=0.8979=89.79 % Degree of saturation, S=89.79 %.
0 Comments