Example 2 - 4.2.2 | 4. Inter-Relations | Geotechnical Engineering - Vol 1
K12 Students

Academics

AI-Powered learning for Grades 8–12, aligned with major Indian and international curricula.

Professionals

Professional Courses

Industry-relevant training in Business, Technology, and Design to help professionals and graduates upskill for real-world careers.

Games

Interactive Games

Fun, engaging games to boost memory, math fluency, typing speed, and English skills—perfect for learners of all ages.

Interactive Audio Lesson

Listen to a student-teacher conversation explaining the topic in a relatable way.

Introduction to Soil Properties

Unlock Audio Lesson

0:00
Teacher
Teacher

Today, we're discussing the critical properties of soil, particularly how we measure its state right after it arrives in the lab. Why do you think these measurements are so vital?

Student 1
Student 1

Because soil properties can change during transport and storage?

Teacher
Teacher

Exactly! We need to determine water content and unit weight to ensure our tests are accurate. It helps us understand how the soil will behave in various conditions. Can anyone tell me what dry unit weight is?

Student 2
Student 2

It's the weight of the soil without any water, right?

Teacher
Teacher

Yes, that's correct! We often calculate it using the bulk unit weight and water content. This leads us to our first formula. Remember, the acronym 'DUM' can help you recall Dry Unit Weight, Unit Weight, and Moisture Content. Let’s see how we can apply this.

Example 1: Dry Unit Weight Calculation

Unlock Audio Lesson

0:00
Teacher
Teacher

In Example 1, we have a soil with a void ratio of 0.72 and a moisture content of 12%. Who can recall how we might calculate the dry unit weight with this information?

Student 3
Student 3

Do we use the formula involving bulk unit weight?

Teacher
Teacher

Correct! We can determine dry unit weight from bulk unit weight and water content. Let's do the calculations together. What results do we get for the dry unit weight?

Student 4
Student 4

It’s 15.51 kN/m3 based on the conversion calculations!

Teacher
Teacher

Good job! We’ll carry on calculating the moist unit weight and the amount of water needed for saturation next. Remember the links between these properties.

Example 2: Void Ratio and Specific Gravity

Unlock Audio Lesson

0:00
Teacher
Teacher

Moving on to our second example, we have a dry density of 1600 kg/m3 and a porosity of 0.387. What do you think the first step should be?

Student 1
Student 1

We need to find the void ratio, right?

Teacher
Teacher

Absolutely! Using the porosity, can anyone derive the void ratio using the formula we’ve learned?

Student 2
Student 2

It's like… e = n / (1-n)? So, that gives us 0.631!

Teacher
Teacher

Excellent! And how about the specific gravity of soil solids? What’s our approach here?

Student 3
Student 3

We can use the relationship between dry density, specific gravity, and the void ratio!

Teacher
Teacher

Exactly! Remember to note down these interrelations as they are crucial for understanding soil behavior. Great work, everyone!

Introduction & Overview

Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.

Quick Overview

This section discusses the quantification of soil properties and calculations related to dry unit weight, moisture content, void ratio, and specific gravity.

Standard

The focus of this section is on understanding various properties of soil, such as dry unit weight, moisture content, and void ratio. It includes practical examples demonstrating how to compute these attributes using given values of soil properties.

Detailed

Example 2

This section emphasizes the importance of quantifying soil properties upon laboratory receipt, specifically focusing on water content and unit weight, which can vary during transportation.

Key interrelations and formulae are introduced, such as calculating dry unit weight from bulk unit weight and water content. The section provides two practical examples.

Example 1 involves determining the dry unit weight and other related measurements using provided values of void ratio, moisture content, and specific gravity.

Example 2 requires calculating void ratio and specific gravity from a known dry density and porosity of soil. The values of different soil properties are crucial in understanding soil behavior and characteristics, which can have significant implications in civil engineering applications.

Youtube Videos

FE Exam Review: Geotechnical Engineering (2019.09.18)
FE Exam Review: Geotechnical Engineering (2019.09.18)
Liquid Limit of Soil | Geotechnical Engineering  #mpsccivil #civilengineering
Liquid Limit of Soil | Geotechnical Engineering #mpsccivil #civilengineering
Classification of soil -2  | Lec 15 | Geotechnical Engineering | GATE/ESE Exam
Classification of soil -2 | Lec 15 | Geotechnical Engineering | GATE/ESE Exam
Geotechnical Engineering Lecture 1
Geotechnical Engineering Lecture 1
Geotechnical Engineering 15 l Shear Strength of Soil Part 1 l Civil Engineering | GATE Crash Course
Geotechnical Engineering 15 l Shear Strength of Soil Part 1 l Civil Engineering | GATE Crash Course
Introduction to Compressibility | Lecture 15 | Geotechnical Engineering
Introduction to Compressibility | Lecture 15 | Geotechnical Engineering
Phase System of Soil | Three Phase Diagram & Two Phase Diagram | Geotechnical Engineering 1
Phase System of Soil | Three Phase Diagram & Two Phase Diagram | Geotechnical Engineering 1
*FE Exam Review: Geotechnical Engineering I (2018.10.24)
*FE Exam Review: Geotechnical Engineering I (2018.10.24)
Geotechnical engineering/Class15/Module2/Numerical problem 1/Atterberg limits.
Geotechnical engineering/Class15/Module2/Numerical problem 1/Atterberg limits.
Geotechnical Engineering 15 | Soil Stabilization & Soil Exploration | Civil Engineering | GATE 2024
Geotechnical Engineering 15 | Soil Stabilization & Soil Exploration | Civil Engineering | GATE 2024

Audio Book

Dive deep into the subject with an immersive audiobook experience.

Understanding Dry Density and Porosity

Unlock Audio Book

Signup and Enroll to the course for listening the Audio Book

Example 2: The dry density of a sand with porosity of 0.387 is 1600 kg/m3. Find the void ratio of the soil and the specific gravity of the soil solids. [Take n = 0.387 = 1600 kg/m3]

Detailed Explanation

This example introduces two important concepts in soil mechanics: dry density and porosity. Dry density is the mass of the dry soil per unit volume, typically expressed in kg/m³. In this case, we have a dry density of 1600 kg/m³. Porosity (n) is the ratio of the volume of voids to the total volume of the soil and is given as 0.387.

To find the void ratio (e), we will use the formula:

  • e = n / (1 - n)

Substituting the given porosity (n = 0.387), we can calculate the void ratio.

After calculating void ratio, the next task is to find the specific gravity (G) of the soil solids. Specific gravity is a dimensionless quantity and is calculated using:

  • G = (dry density of soil solids) / (density of water)

Assuming the density of water is approximately 1000 kg/m³, we can determine the specific gravity once we calculate the dry density of soil solids.

Examples & Analogies

Think of a sponge. When dry, it has a certain 'dry density' based on how much material there is inside (the solid parts of the sponge). Now, when we consider porosity, it tells us how much empty space or 'voids' are present in the sponge where water could be held. The void ratio will tell us how much of the volume of the sponge is filled with these voids compared to the solid parts, and specific gravity will help us understand how heavy the sponge is when you compare it to an equal volume of water.

Calculation of Void Ratio

Unlock Audio Book

Signup and Enroll to the course for listening the Audio Book

(a) e = = = 0.631

Detailed Explanation

To calculate the void ratio (e), we use the formula:
e = n / (1 - n). Substituting the value of n:

  1. Start with n = 0.387.
  2. Calculate (1 - n):
    (1 - 0.387) = 0.613.
  3. Finally, divide n by (1 - n):
    e = 0.387 / 0.613.

This results in a void ratio of approximately 0.631. The void ratio indicates how much space is available for water in the soil compared to the soil's solid components. Higher void ratios mean more space for water, which is crucial for understanding soil behavior in construction and agriculture.

Examples & Analogies

Imagine you have a jar filled with marbles (the solid parts of the soil) and there are spaces in between them (the voids). The void ratio helps us understand how many empty spaces there are compared to the number of marbles. If there are more spaces for air or water in the jar, it means the jar is less packed with marbles. This is similar to how soil can hold more water when it has a high void ratio.

Determining Specific Gravity of Soil Solids

Unlock Audio Book

Signup and Enroll to the course for listening the Audio Book

(b) = G =

Detailed Explanation

After finding the void ratio, the next step is to calculate the specific gravity (G) of the soil solids. Specific gravity is calculated using the dry density of soil solids. The formula used is:
G = (Dry Density of Soil Solids) / (Density of Water).

Given the information, we assume density of water is approximately 1000 kg/m³, so:
Once we have the dry density of soil solids, simply plug in the values to find G. Specific gravity helps understand how heavy the soil solids are compared to water and is fundamental in soil mechanics, especially for comparing different soils.

Examples & Analogies

You can think of specific gravity like comparing the weight of different balls. If you have a basketball filled with air (less dense) compared to a basketball filled with sand (denser), specific gravity tells us how heavy each ball would be if both were the same size. This concept is similar to soil where different types of soil solids have varying weights contributing to their overall density and behavior.

Definitions & Key Concepts

Learn essential terms and foundational ideas that form the basis of the topic.

Key Concepts

  • Void Ratio: A crucial property in soil mechanics representing the space available for water and air within the soil.

  • Moisture Content: Integral to comprehending how the water present affects soil stability and behavior.

  • Dry Unit Weight: Essential for understanding the load-bearing capacity of soil.

  • Specific Gravity: Important for characterizing the density of soil solids in comparison to water.

Examples & Real-Life Applications

See how the concepts apply in real-world scenarios to understand their practical implications.

Examples

  • Example 1 demonstrated calculating dry unit weight and moisture content from soil properties.

  • Example 2 illustrated obtaining void ratio and specific gravity from known dry density and porosity.

Memory Aids

Use mnemonics, acronyms, or visual cues to help remember key information more easily.

🎵 Rhymes Time

  • To find the weight dry in soil, remember the water's a worrisome foil.

📖 Fascinating Stories

  • Imagine a sandy beach, where dry grains sit and water hides nearby; observing how they balance teaches us of their ratio's tie.

🧠 Other Memory Gems

  • Every Student Can Find Density: e, S, C, F - stands for Void Ratio, Specific Gravity, and Density relationships.

🎯 Super Acronyms

VMD

  • Void
  • Moisture content
  • Dry Unit weight - all key to understanding soil properties.

Flash Cards

Review key concepts with flashcards.

Glossary of Terms

Review the Definitions for terms.

  • Term: Void Ratio

    Definition:

    The ratio of the volume of voids to the volume of solids in a soil sample.

  • Term: Moisture Content

    Definition:

    The amount of water contained in a soil sample, expressed as a percentage of the dry weight of the soil.

  • Term: Dry Unit Weight

    Definition:

    The weight of soil solids per unit volume of soil without any water.

  • Term: Specific Gravity

    Definition:

    The ratio of the density of soil solids to the density of water at a standard temperature.