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In many situations physical constraints prevent strain from occurring in a given direction. For example, \(\epsilon_{z}=0\) in the case shown, where longitudinal movement of the long prism is prevented at every point. Plane sections perpendicular to the longitudinal axis remain plane and the same distance apart. Show that for this situation, which is known as plane strain, we can express \(\sigma_{z}, \epsilon_{x},\) and \(\epsilon_{y}\) as follows: \\[ \begin{array}{l} \sigma_{z}=\nu\left(\sigma_{x}+\sigma_{y}\right) \\ \epsilon_{x}=\frac{1}{E}\left[\left(1-\nu^{2}\right) \sigma_{x}-\nu(1+\nu) \sigma_{y}\right] \\ \epsilon_{y}=\frac{1}{E}\left[\left(1-\nu^{2}\right) \sigma_{y}-\nu(1+\nu) \sigma_{x}\right] \end{array} \\]

Short Answer

Expert verified
In plane strain, \(\sigma_z = \nu (\sigma_x + \sigma_y)\), \(\epsilon_x = \frac{1}{E}((1-\nu^2)\sigma_x - \nu(1+\nu)\sigma_y)\), and \(\epsilon_y = \frac{1}{E}((1-\nu^2)\sigma_y - \nu(1+\nu)\sigma_x)\).

Step by step solution

01

Plane Strain Assumption

In the given scenario, we are considering the condition of plane strain, where the strain in the z-direction is zero, i.e., \(\epsilon_z = 0\). This implies that the deformation in the z-direction is completely restrained. This leads to the condition that the longitudinal sections remain plane and parallel.
02

Use Relationship Between Stresses and Strains

Under the elasticity theory, we relate stress and strain using Hooke's Law for isotropic materials. The general expressions for strain in terms of stresses, when considering three-dimensional stress components \(\sigma_x, \sigma_y, \sigma_z\), are given by: \[\begin{align*}\epsilon_x &= \frac{1}{E} \left(\sigma_x - u(\sigma_y + \sigma_z)\right) \\epsilon_y &= \frac{1}{E} \left(\sigma_y - u(\sigma_x + \sigma_z)\right) \\epsilon_z &= \frac{1}{E} \left(\sigma_z - u(\sigma_x + \sigma_y)\right)\end{align*}\]where \(E\) is the Young's modulus and \(u\) is the Poisson's ratio.
03

Solve for \( \sigma_z \) Using Plane Strain Condition

Since \(\epsilon_z = 0\), we can plug this condition into the strain equation:\[ 0 = \frac{1}{E} \left( \sigma_z - u(\sigma_x + \sigma_y) \right) \]Simplifying, we find:\[ \sigma_z = u ( \sigma_x + \sigma_y ) \]
04

Derived Expressions for \( \epsilon_x \) and \( \epsilon_y \)

We substitute \( \sigma_z = u (\sigma_x + \sigma_y) \) back into the expressions for \( \epsilon_x \) and \( \epsilon_y \) from Step 2:- For \(\epsilon_x\):\[ \epsilon_x = \frac{1}{E} \left( \sigma_x - u\left(\sigma_y + u(\sigma_x + \sigma_y)\right) \right) \]Simplifying gives:\[ \epsilon_x = \frac{1}{E} \left( (1-u^2)\sigma_x - u(1+u)\sigma_y \right) \]- For \(\epsilon_y\):\[ \epsilon_y = \frac{1}{E} \left( \sigma_y - u\left(\sigma_x + u(\sigma_x + \sigma_y)\right) \right) \]Simplifying gives:\[ \epsilon_y = \frac{1}{E} \left( (1-u^2)\sigma_y - u(1+u)\sigma_x \right) \]

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Key Concepts

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

Elasticity Theory
Elasticity theory is a fundamental concept in engineering and material science that describes how materials deform under stress and subsequently return to their original shape when the stress is removed.
This theory is crucial for understanding how objects respond to external forces, be they stretching, compressing, or twisting. It generally applies to materials that return to their initial form after the forces are no longer applied, known as elastic materials.

In the context of plane strain, elasticity theory helps us understand how forces act within the constraints of a specific plane. Under this condition, deformation is confined to the x-y plane while the z-component remains unchanged.
This simplification is beneficial in analyzing components where the thickness along one direction is significantly larger than in the other two planes, such as in long tunnels or thick walls. The theory provides the framework to relate the stresses and strains through specific relationships and helps in deriving expressions for the unknown quantities within a material under defined boundary conditions.
Hooke's Law
Hooke's Law is a principle of elasticity that states the force needed to extend or compress a spring by some distance is proportional to that distance. This idea can be extended to describe small deformations in materials, where the stress (force per unit area) is proportional to the strain (deformation) the material experiences.

For isotropic materials, Hooke’s Law is typically expressed as \[ ext{Stress} = E imes ext{Strain} \]where the constant of proportionality, \( E \), is known as Young's modulus.

In the plane strain context, Hooke’s Law extends this relation by incorporating Poisson's ratio, \( u \), and linking the stress components in all three dimensions:
  • \( ext{For } ext{strain in x:} \ \ \ rac{1}{E} ig( ext{Stress in x} - u imes ( ext{Sum of stresses in y and z}) ig) \)
  • \( ext{For } ext{strain in y:} \ \ \ rac{1}{E} ig( ext{Stress in y} - u imes ( ext{Sum of stresses in x and z}) ig) \)
By adhering to Hooke's Law, one can solve for the unknown stresses or strains given certain relationships in a constrained condition like plane strain.
Poisson's Ratio
Poisson's ratio is a measure of the Poisson effect, which describes the phenomenon of a material contracting in a direction perpendicular to the direction it is being stretched or expanding when compressed.

It is a dimensionless constant denoted by \( u \) and is expressed as:\[ u = -\frac{\text{Lateral Strain}}{\text{Axial Strain}} \]
This value typically ranges from 0 to 0.5 for most materials, indicating that as a material is compressed or stretched, it undergoes a corresponding contraction or expansion perpendicular to the direction of loading.

In the plane strain scenario where \( u \) becomes significant, consider how it affects the stress-strain relationship. When strain in the z-direction is zero, Poisson's ratio helps relate stresses in x and y by:
  • \( \sigma_{z} = u(\sigma_{x} + \sigma_{y}) \)
Because of this, it plays a pivotal role in altering the internal stress distribution even when deformation is prohibited in one direction.
Young's Modulus
Young's modulus, represented as \( E \), is a fundamental property of materials that provides a quantitative measure of their stiffness. It essentially describes how much a material will deform under a certain amount of stress. The higher the Young's modulus, the stiffer the material, meaning it requires more force to deform.

Mathematically, Young's modulus is defined as:\[ E = \frac{ ext{Stress}}{ ext{Strain}} \]
In plane strain problems, it provides a means to relate the applied stress directly to the resulting strain. This relation is crucial when calculating the strain in specific directions by accounting for stresses applied on the material. Particularly, in this context, Young’s modulus is used to derive the strain components in the x and y directions:
  • \( \epsilon_{x} = \frac{1}{E} ig[(1-u^{2}) \sigma_{x} - u(1+u) \sigma_{y} \big] \)
  • \( \epsilon_{y} = \frac{1}{E} ig[(1-u^{2}) \sigma_{y} - u(1+u) \sigma_{x} \big] \)
Understanding this modulus is critical for engineers when designing structures that need to withstand various types of forces without undergoing excessive deformation.

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Most popular questions from this chapter

The concrete post \(\left(E_{c}=3.6 \times 10^{6} \mathrm{psi} \text { and } \alpha_{c}=5.5 \times 10^{-6} /^{\circ} \mathrm{F}\right)\) is reinforced with six steel bars, each of \(\frac{z}{8}\) -in. diameter \(\left(E_{x}=29 \times 10^{6} \mathrm{psi} \text { and } \alpha_{x}=6.5 \times 10^{-6} /^{\circ} \mathrm{F}\right) .\) Determine the normal stresses induced in the steel and in the concrete by a temperature rise of \(65^{\circ} \mathrm{F}\)

A 4-m-long steel rod must not stretch more than 3 mm and the normal stress must not exceed \(150 \mathrm{MPa}\) when the rod is subjected to a 10 -kN axial load. Knowing that \(E=200 \mathrm{GPa}\), determine the required diameter of the rod.

A 250 -mm-long aluminum tube \((E=70 \mathrm{GPa})\) of 36 -mm outer diameter and 28 -mm inner diameter can be closed at both ends by means of single- threaded screw-on covers of 1.5 -mm pitch. With one cover screwed on tight, a solid brass rod \((E=105 \mathrm{GPa})\) of 25 -mm diameter is placed inside the tube and the second cover is screwed on. Since the rod is slightly longer than the tube, it is observed that the cover must be forced against the rod by rotating it one-quarter of a turn before it can be tightly closed. Determine (a) the average normal stress in the tube and in the rod, (b) the deformations of the tube and of the rod.

A block of 10 -in. length and \(1.8 \times 1.6\) -in. cross section is to support a centric compressive load \(\mathbf{P}\). The material to be used is a bronze for which \(E=14 \times 10^{6}\) psi. Determine the largest load that can be applied, knowing that the normal stress must not exceed 18 lssi and that the decrease in length of the block should be at most \(0.12 \%\) of its original length.

A square yellow-brass bar must not stretch more than \(2.5 \mathrm{mm}\) when it is subjected to a tensile load. Knowing that \(E=105\) GPa and that the allowable tensile strength is \(180 \mathrm{MPa}\) determine \((a)\) the maximum allowable length of the bar, \((b)\) the required dimensions of the cross section if the tensile load is \(40 \mathrm{kN}\)

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