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Show that the principal strains are the minimum and the maximum fractional changes in length.

Short Answer

Expert verified
Principal strains are the eigenvalues of the strain matrix, representing maximum and minimum fractional changes in length.

Step by step solution

01

Understand principal strains

Principal strains refer to the maximum and minimum values of the normal strain at a point in a material. These strains occur on mutually perpendicular planes where the shear strain is zero.
02

Recall the definition of strain

Strain is defined as the fractional change in length, represented as the change in length ( abla L div L). It is a dimensionless quantity that indicates how much the material deforms under stress.
03

Use stress-strain relationship in matrix form

In two dimensions, the stress-strain relationships involve determining the principal stresses and strains from the stress and strain matrices, respectively. The strain matrix is expressed as:\[\begin{bmatrix} \epsilon_x & \gamma/2 \\gamma/2 & \epsilon_y \end{bmatrix}\]where \(\epsilon_x, \, \epsilon_y\) are the normal strains and \(\gamma\) is the shear strain.
04

Diagonalization to find principal strains

The principal strains are found by diagonalizing the strain matrix. This involves finding the eigenvalues (\(\lambda_1, \lambda_2\)) of the matrix, which correspond to the principal strains. The eigenvalue equation is:\[\text{det}\begin{bmatrix}\epsilon_x - \lambda & \gamma/2 \\gamma/2 & \epsilon_y - \lambda \end{bmatrix} = 0\]Solving this gives the eigenvalues, \(\lambda_1\) (maximum) and \(\lambda_2\) (minimum), which are the principal strains.
05

Verify principal strains are maximum and minimum changes

Since the principal strains correspond to the eigenvalues of the strain matrix, they represent the stretches along the principal directions where the shear strain is zero. They indeed represent the maximum and minimum fractional changes in length due to deformations, as principal directions coincide with directions of pure normal stress and zero shear stress.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Strain Matrix
In the study of materials and their deformations, understanding the strain matrix is crucial. The strain matrix represents the deformation state of a material, particularly how the material is stretching or compressing at a point. In a two-dimensional scenario:
  • \( \epsilon_x \) and \( \epsilon_y \) represent the normal strains, which describe deformation in the x and y directions respectively.
  • \( \gamma \) represents the shear strain, reflecting how the shape of the material changes without altering volume.
The matrix is arranged as:\[\begin{bmatrix} \epsilon_x & \gamma/2 \\gamma/2 & \epsilon_y \end{bmatrix}\]This structure helps utilize mathematical techniques like matrix diagonalization to identify directions of pure stretching or compression. Thus, understanding and working with the strain matrix is fundamental for determining principal strains.
Eigenvalues
Eigenvalues play a critical role in understanding the mechanical behavior of materials. In the context of the strain matrix, they help identify the principal strains:
  • Eigenvalues are obtained by solving the characteristic equation of the matrix.
  • They reflect the amount of stretching or compression occurring in the principal directions.
To find them, we set up the equation:\[\text{det}\begin{bmatrix}\epsilon_x - \lambda & \gamma/2 \\gamma/2 & \epsilon_y - \lambda \end{bmatrix} = 0\]The solutions to this equation, \( \lambda_1 \) and \( \lambda_2 \), provide the principal strains. These values provide insight into material behavior, especially when predicting failure or designing materials for durability.
Normal Strain
Normal strain refers to the deformation measure in a specific direction. It quantifies how much a material is being stretched or compressed:
  • Defined as the fractional change in length.
  • Calculated as the change in length divided by the original length, \( \frac{\Delta L}{L} \).
  • Dimensionless, representing a ratio without units.
Normal strain is a fundamental concept that highlights how various forces cause elongation or shortening in materials. It is essential in determining the stress-strain relationship, offering insight into material deformation under load. Analyzing normal strain helps in material engineering and understanding the capacity of structures to withstand various loads without failing.
Stress-Strain Relationship
The stress-strain relationship explains how materials react to external forces:
  • Stress refers to the internal forces within a material, caused by an external load.
  • Strain is the resultant deformation from this stress.
The relationship between these two factors is often linear for many materials, characterized by properties like Young's modulus. Exploring this relationship through matrices, like in the strain matrix example, allows for better prediction and understanding of the material's behavior under load. Matrix formulations allow engineers to derive simple expressions for complicated interactions involving bending, twisting, and axial loads, enabling accurate predictions of material response in complex applications.

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Most popular questions from this chapter

Consider a simple two-layer model of a planet consisting of a core of density \(\rho_{c}\) and radius \(b\) surrounded by a mantle of density \(\rho_{m}\) and thickness \(a-b\). Show that the gravitational acceleration as a function of radius is given by $$ \begin{aligned} g(r) &=\frac{4}{3} \pi \rho_{c} G r \quad 0 \leq r \leq b \\ &=\frac{4}{3} \pi G\left[r \rho_{m}+b^{3}\left(\rho_{c}-\rho_{m}\right) / r^{2}\right] \quad b \leq r \leq a \end{aligned} $$

A mountain range has an elevation of \(5 \mathrm{~km}\). Assuming that \(\rho_{m}=3300 \mathrm{~kg} \mathrm{~m}^{-3}, \rho_{c}=2800 \mathrm{~kg}\) \(\mathrm{m}^{-3}\), and that the reference or normal continental crust has a thickness of \(35 \mathrm{~km}\), determine the thickness of the continental crust beneath the mountain range. Assume that hydrostatic equilibrium is applicable.

Consider a block of rock with a height of \(1 \mathrm{~m}\) and horizontal dimensions of \(2 \mathrm{~m}\). The density of the rock is \(2750 \mathrm{~kg} \mathrm{~m}^{-3}\). If the coefficient of friction is \(0.8,\) what force is required to push the rock on a horizontal surface?

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