Final Solution For U (5) - Recap - Solid Mechanics
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Final solution for u

Final solution for u

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Interactive Audio Lesson

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Understanding the constants in the expression for u

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Teacher
Teacher Instructor

Today, let's start with finding the unknown constants in our expression for the longitudinal displacement, represented as u.

Student 1
Student 1

Why do we need to find these constants?

Teacher
Teacher Instructor

Good question! Finding these constants is essential because they help us relate the displacement to the physical properties of the material under certain boundary conditions.

Student 2
Student 2

What are those boundary conditions?

Teacher
Teacher Instructor

We have two key conditions: the internal pressure acting on the inner surface and zero pressure on the outer surface. These will help us solve for the constants.

Student 3
Student 3

So, does that mean the constants we find are specific to the pressure applied?

Teacher
Teacher Instructor

Exactly! Each scenario can lead to different values for C and D based on the applied stresses and material properties.

Student 4
Student 4

Can we summarize the steps for solving this?

Teacher
Teacher Instructor

Sure! Identify the boundary conditions, apply them to our expressions, and solve for constants C and D through integration. Remember, this helps us understand how the hollow cylinder behaves under pressure.

Applying boundary conditions

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Teacher
Teacher Instructor

Now that we have a grasp of constants, let's look into applying the boundary conditions. What do we find at the inner surface?

Student 1
Student 1

We find that the stress is equal to the internal pressure.

Teacher
Teacher Instructor

Exactly! So we can set the radial stress equal to -P and move ahead with our equations.

Student 2
Student 2

And at the outer surface?

Teacher
Teacher Instructor

Right, there is no traction, so the stress is zero there. This gives us the second boundary condition to work from.

Student 3
Student 3

So these conditions lead to equations we can solve for our constants?

Teacher
Teacher Instructor

Precisely! It’s all about substituting these values into our derived equations and solving for the constants.

Student 4
Student 4

Can we write this in the equation format?

Teacher
Teacher Instructor

Of course! Let’s summarize these boundary conditions mathematically: σ (r1) = -P and σ (r2) = 0.

Final Expression for u

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Teacher
Teacher Instructor

Now, let’s derive the final expression for radial displacement, u. What have we gathered so far?

Student 1
Student 1

We have our constants and the boundary conditions merged with our displacement equations.

Teacher
Teacher Instructor

Right! Plug those values into the derived equation for u. What do you observe?

Student 2
Student 2

It incorporates Young's modulus and Poisson's ratio, showing how they affect displacement.

Teacher
Teacher Instructor

Great insight! Do you think the equation changes if there is no pressure?

Student 3
Student 3

Yes, when P = 0, it simplifies, showing the effect of Poisson’s ratio even without external pressure.

Student 4
Student 4

Can we visualize how this displacement looks physically?

Teacher
Teacher Instructor

Absolutely! We can understand radial deformation as a result of internal forces, demonstrated visually through deformation under pressure.

Introduction & Overview

Read summaries of the section's main ideas at different levels of detail.

Quick Overview

The section deals with deriving the final expression for the longitudinal deformation in a hollow cylinder, considering various constants and boundary conditions.

Standard

In this section, we focus on finding the constants in the expression for the longitudinal displacement in a hollow cylinder. The analysis involves applying boundary conditions, deriving expressions for axial and circumferential stresses, and ultimately results in the final expression for axial deformation, integrating parameters including Young’s modulus and Poisson’s ratio.

Detailed

Final Solution for u

This section delves into deriving the final form of the longitudinal displacement (2) in a hollow cylinder under internal pressure. We start by recalling the expression derived previously and focus on determining the unknown constants (C, D) to finalize the solution.

Mathematical Formulation

  1. The equation for lateral displacement in terms of stress components reiterates the dependence of these components on radial positions.
  2. Using boundary conditions, specifically the conditions at the inner and outer surfaces subjected to known internal pressure and zero external pressure, we can identify the constants that define the deformation.

Derivation Steps

  • We find the constant C using the stress equilibrium relationship, showing its dependency on the radial stress.
  • Next, the second constant D is derived from previously established equations relating stress and displacement, ensuring consistency throughout our mathematical representations.

Final Form

The finalized equation for u comprehensively incorporates the effects of eigenstates like Young’s modulus and Poisson’s ratio, culminating in an expression that signifies the radial displacement effect even in the absence of applied pressure, a crucial insight into material behavior under torsional and axial loads.

Summary Significance

This development is critical as it not only provides the last piece of the puzzle for analyzing hollow cylinder deformation but also serves as a basis for understanding material properties in engineering applications, especially in extension, torsion, and inflation scenarios.

Audio Book

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Finding Constants C and D

Chapter 1 of 2

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Chapter Content

We have to find the constants (C, D) in the expression (17) to obtain complete solution of u. The constant C can be found using (18) as follows:

.
(29)
To get D, we can use equation (11) as shown below:
(using(17))
(using(18)) (30)
Comparing this with equation (20), we get:
(using(28)) (31)

Detailed Explanation

In this chunk, we are focusing on finding the unknown constants C and D which are necessary to complete the solution for the displacement variable u. The first step is to derive the value of C using equation (18). After obtaining C, we move on to find D using equation (11). By integrating the previously derived equations, we can compare these with known values in equation (20) to effectively solve for our constants.

Examples & Analogies

Imagine you're baking cookies and you need to adjust the recipe. If you know how much sugar (C) to add based on a previous batch, that’s like determining constant C. To find out how much flour (D) you need, you would then refer back to the recipe’s specific guidelines, similar to using equation (11) to find D here.

Final Form of u

Chapter 2 of 2

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Chapter Content

Thus, u finally becomes
(32)
Upon setting P=0 in (32), we get
(33)
When we relate Lame’s constants (λ and µ) with Young’s modulus(E), Poisson’s ratio(ν) and shear modulus(G), it turns out that
(34)
Substituting which in the above expression for u, we get
(35)
The radial longitudinal strain for P=0 then turns out to be
(36)
This expression is exactly what we expect - radial displacement has arisen due to Poisson’s effect even in the absence of pressure.

Detailed Explanation

In this section, we present the final form of u after obtaining constants C and D. When we set pressure P to zero, we see how the expression simplifies. We also link Lame's constants with other material properties such as Young's modulus and Poisson's ratio, showing that the relations between these quantities determine the behavior of the cylinder under stress. The final equation shows the impact of radial displacement resulting from Poisson's effect, illustrating that even without pressure, a change occurs due to axial strain.

Examples & Analogies

Think about stretching a rubber band; when you pull it (applying axial force), it not only stretches in length but may also become thinner (changes in radial displacement), even if you don’t apply any pressure on the sides. Poisson’s effect describes this phenomenon of how changing one dimension affects the others.

Key Concepts

  • Boundary Conditions: Essential for deriving the constants in the expression.

  • Young's Modulus: Integrally related to the calculation of strains due to deformation.

  • Poisson's Ratio: Demonstrates the effect of radial deformation even under zero pressure conditions.

Examples & Applications

Example 1: If the internal pressure increases, how does that affect the constants C and D?

Example 2: For a hollow cylinder with no pressure, the radial displacement can still be observed due to Poisson's effect.

Memory Aids

Interactive tools to help you remember key concepts

🎵

Rhymes

Young's stays stiff, follows the lift, as boundary conditions play a key shift.

📖

Stories

Imagine a hollow cylinder being pushed from the inside. It stretches in the middle, while holding its pressure at the ends, revealing how boundary conditions define its fate.

🧠

Memory Tools

Remember 'C' for Constants, 'P' for Pressure, 'D' for Displacement. Each plays a role in the mechanics puzzle.

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Acronyms

BYP - Boundary Conditions, Yield, and Pressure.

Flash Cards

Glossary

Longitudinal Displacement (u)

The axial deformation experienced by a material or structure when subjected to loads.

Boundary Condition

Conditions at the boundary of a system which are necessary to solve differential equations.

Young's Modulus

A measure of the stiffness of a solid material, defined as the ratio of stress to strain.

Poisson's Ratio

The ratio of transverse strain to axial strain in a material subjected to axial loading.

Radial Stress

Stress that occurs in the radial direction within a material, particularly in cylindrical structures.

Reference links

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