Civil Engineering PE Soil Mechanics and Lab Testing Questions and Answers

Question 1

A geotechnical engineer is reviewing lab results for a fine-grained soil sample. The results are: Liquid Limit (LL) = 65%, Plastic Limit (PL) = 25%. What is the Plasticity Index (PI) of the soil, and how would it be classified based on this value?

Accepted Answer

PI = 40%, Highly plastic

Answer

The Plasticity Index (PI) is calculated as the difference between the Liquid Limit (LL) and the Plastic Limit (PL). PI = LL - PL = 65% - 25% = 40%. Soils with a PI greater than 17 are generally classified as highly plastic. These soils, typically clays, exhibit significant volume changes with variations in moisture content.

Question 2

During a Standard Proctor compaction test (ASTM D698), a soil sample is compacted in a standard mold. Which of the following sets of parameters is consistent with the standard procedure?

Accepted Answer

5.5 lb rammer, 12-inch drop, 3 layers, 25 blows per layer

Answer

The Standard Proctor test, as defined by ASTM D698, specifies compacting the soil in 3 layers within a 4-inch mold, with each layer receiving 25 blows from a 5.5 lb rammer dropped from a height of 12 inches. This results in a compactive effort of approximately 12,400 ft-lbf/ft³. The other options describe parameters for the Modified Proctor test (ASTM D1557) or are incorrect combinations.

Question 3

An unconfined compression test is performed on a cylindrical specimen of cohesive soil. The test is most suitable for determining the short-term shear strength of which of the following soil types?

Accepted Answer

Saturated, intact clay

Answer

The unconfined compression test is used to determine the unconfined compressive strength (qu), which is a measure of the undrained shear strength (su = qu/2) of a cohesive soil. It is most appropriate for intact, saturated clay specimens where the internal pore water pressure provides the 'confinement'. Granular soils like sand and gravel lack cohesion and cannot stand unconfined. Dry, fissured clays are also unsuitable as the fissures represent planes of weakness and the lack of saturation invalidates the test's principle.

Question 4

In a one-dimensional consolidation (oedometer) test, a saturated clay sample is subjected to incremental loading. What is the primary parameter determined from the time-settlement data for each load increment?

Accepted Answer

Coefficient of consolidation (Cv)

Answer

The oedometer test measures the vertical deformation of a laterally confined soil sample over time after a load is applied. The time-rate of settlement is governed by how quickly pore water can drain from the soil, which is related to its permeability and compressibility. The Coefficient of Consolidation (Cv) is calculated from the time-settlement curve for each load increment and is used to predict the rate of settlement in the field. The other parameters are determined from the overall load-void ratio relationship, not the time-rate data for a single increment.

Question 5

A key advantage of the triaxial shear test over the direct shear test is that the triaxial test allows for:

Accepted Answer

Control of drainage conditions and measurement of pore water pressure

Answer

The triaxial test offers several advantages over the direct shear test, most notably the ability to control drainage conditions (Consolidated Drained, Consolidated Undrained, Unconsolidated Undrained) and to measure the pore water pressure within the sample during shear. This allows for the determination of both total and effective stress parameters. In a direct shear test, the failure plane is forced horizontally, which may not be the weakest plane, and pore pressure measurement is not possible.

Civil Engineering PE Practice Test

Civil Engineering PE Practice Test

Civil Engineering PE Soil Mechanics and Lab Testing Questions and Answers