Principles of Geotechnical Engineering (MindTap Course List)
Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781305970939
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
Question
Book Icon
Chapter 17, Problem 17.1P
To determine

Find the area ratio for each case and appropriate sample for the given soil characterization tests.

Expert Solution & Answer
Check Mark

Answer to Problem 17.1P

The area ratio for tube 1 is 5.06%_.

The area ratio for tube 2 is 4.47%_.

The area ratio for tube 3 is 4.98%_.

Explanation of Solution

Given information:

The outside diameter (Do) for sample 1 is 50.8mm.

The outside diameter (Do) for sample 2 is 76.2mm.

The outside diameter (Do) for sample 3 is 127mm.

The wall thickness (t) for sample 1 is 1.24mm.

The wall thickness (t) for sample 2 is 1.65mm.

The wall thickness (t) for sample 3 is 3.05mm.

Calculation:

Calculate the inside diameter (Di) for tube 1 using the relation as follows:

(Di)=Dot (1)

Substitute 50.8mm for (Do) and 1.24mm for t in Equation (1).

(Di)=50.81.24=49.56mm

Find the area ratio for (Ar) tube 1 using the Equation as follows:

Ar(%)=Do2Di2Di2×100 (2)

Here, Di inside diameter of sample 1.

Substitute 50.8mm for Do and 49.56mm for Di in Equation (2).

Ar(%)=50.8249.56249.562×100=2,580.642,456.192,456.19×100=124.442,456.19×100=5.06%10%

Thus, the area ratio for tube 1 is 5.06%_.

Calculate the inside diameter (Di) for tube 2 using the relation as follows:

Substitute 76.2mm for (Do) and 1.65mm for t in Equation (1).

(Di)=76.21.65=74.55mm

Find the area ratio for (Ar) tube 2 using the Equation as follows:

Substitute 76.2mm for Do and 74.55mm for Di in Equation (2).

Ar(%)=76.2274.55274.552×100=5,806.445,557.75,557.7×100=248.745,557.7×100=4.47%10%

Thus, the area ratio for tube 2 is 4.47%_.

Calculate the inside diameter (Di) for tube 3 using the relation as follows:

Substitute 127mm for (Do) and 3.05mm for t in Equation (1).

(Di)=1273.05=123.95mm

Find the area ratio for (Ar) tube 3 using the Equation as follows:

Substitute 127mm for Do and 123.95mm for Di in Equation (2).

Ar(%)=1272123.952123.952×100=16,12915,363.6015,363.60×100=4.98%10%

Thus, the area ratio for tube 3 is 4.98%_.

A soil sample usually can be considered undisturbed if the area ratio is less than or equal to 10%.

  • All the given three samples are considered as undisturbed sample, hence, it is suitable for consolidation and unconfined compression test.
  • The Grain size distribution, atterberg limits are conducted only when the soil on remolded or disturbed soil, hence, sample disturbance is not a factor.

Thus, the all three samples are appropriate for all tests.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Soil classification requires all of the results of the following tests:  Liquid limit and Plastic limit tests   Atterberg limit and Hydrometer tests   Atterberg limit and grain size analysis test   Sieve and Hydrometer analysis tests
Considering broad categories of soil types and commonly used soil sampling tools, please answer the following questions. What type of sampler is used to sample granular soils for Choose... laboratory testing? What type of sampler is used to sample cohesive soils for v Choose... laboratory testing? Thin-walled or Shelby tube sampler Split-barrel or split- spoon sampler Electric cone Split-spoon tesing er penetrometer on
Plz, solve this problem as detail.   Geotechnical Engineering-1
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Geotechnical Engineering (MindTap C...
Civil Engineering
ISBN:9781305970939
Author:Braja M. Das, Khaled Sobhan
Publisher:Cengage Learning