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Plane Strain In the case of plane strain, \(\varepsilon_{3}=0,\) for example, and \(\varepsilon_{1}\) and \(\varepsilon_{2}\) are nonzero. Figure \(3-7\) illustrates a plane strain situation. A long bar is rigidly confined between supports so that it cannot expand or contract parallel to its length. In addition, the stresses \(\sigma_{1}\) and \(\sigma_{2}\) are applied uniformly along the length of the bar. Equations \((3-1)\) to \((3-3)\) reduce to $$ \begin{aligned} \sigma_{1} &=(\lambda+2 G) \varepsilon_{1}+\lambda \varepsilon_{2} \\ \sigma_{2} &=\lambda \varepsilon_{1}+(\lambda+2 G) \varepsilon_{2} \\ \sigma_{3} &=\lambda\left(\varepsilon_{1}+\varepsilon_{2}\right) \end{aligned} $$. From Equation \((3-6)\) it is obvious that $$ \sigma_{3}=v\left(\sigma_{1}+\sigma_{2}\right) $$ This can be used together with Equations \((3-4)\) and (3-5) to find $$ \begin{array}{l} \varepsilon_{1}=\frac{(1+v)}{E}\left\\{\sigma_{1}(1-v)-v \sigma_{2}\right\\} \\\ \varepsilon_{2}=\frac{(1+v)}{E}\left\\{\sigma_{2}(1-v)-v \sigma_{1}\right\\} \end{array} $$

Short Answer

Expert verified
Strains \(\varepsilon_1\) and \(\varepsilon_2\) are related to stresses \(\sigma_1\) and \(\sigma_2\) using given equations for plane strain conditions.

Step by step solution

01

Understand given equations and conditions

We are dealing with a plane strain condition where \(\varepsilon_3=0\). The normal strains \(\varepsilon_1\) and \(\varepsilon_2\) are nonzero, and two normal stresses \(\sigma_1\) and \(\sigma_2\) are applied. The relevant equations for stress-strain relationship are provided: \[ \sigma_{1} =(\lambda + 2G) \varepsilon_{1} + \lambda \varepsilon_{2} \\sigma_{2} =\lambda \varepsilon_{1} + (\lambda + 2G) \varepsilon_{2} \\sigma_{3} = \lambda (\varepsilon_{1} + \varepsilon_{2})\] Where \(\lambda\) and \(G\) are Lame's parameters. Also given, \(\sigma_3=v(\sigma_1+\sigma_2)\) from Equation (3-6).
02

Use compatibility conditions

Introduce Poisson's ratio \(v\) and Young's modulus \(E\) into the equations. Recognize that \(\sigma_3 = v(\sigma_1 + \sigma_2)\) implies there is a relation between applied stresses and resulting strains. We will use this relation to express strains in terms of applied stresses.
03

Express strains in terms of stresses

We are given the final forms for \(\varepsilon_1\) and \(\varepsilon_2\) based on the existing material model: \[\begin{align*}\varepsilon_1 &= \frac{(1+v)}{E} \left[ \sigma_1(1-v) - v \sigma_2 \right] \\varepsilon_2 &= \frac{(1+v)}{E} \left[ \sigma_2(1-v) - v \sigma_1 \right]\end{align*}\] These are derived using the fact that, in the plane strain state, \(\sigma_3\) links the out-of-plane stress to in-plane stresses via Poisson's ratio, thereby allowing the calculation of \(\varepsilon_1\) and \(\varepsilon_2\) directly from \(\sigma_1\) and \(\sigma_2\).

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Key Concepts

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

Stress-Strain Relationship
When studying the deformation of materials, the stress-strain relationship is a fundamental concept. It describes how the stress on a material relates to the strain it undergoes. In a plane strain scenario, like the one described in the exercise, this relationship becomes particularly clear. Here, the material is constrained such that strain occurs primarily in two dimensions. The standard form is given by the equations:
  • \( \sigma_1 = (\lambda + 2G) \varepsilon_1 + \lambda \varepsilon_2 \)
  • \( \sigma_2 = \lambda \varepsilon_1 + (\lambda + 2G) \varepsilon_2 \)
  • \( \sigma_3 = \lambda (\varepsilon_1 + \varepsilon_2) \)
These equations show that stress components \(\sigma_1\) and \(\sigma_2\) are related to strains \(\varepsilon_1\) and \(\varepsilon_2\) through Lame's parameters, \(\lambda\) and \(G\). The equations describe how material deformation under a set of applied stresses is influenced by both the material's inherent properties and the constraints of the strain state.
Poisson's Ratio
Poisson's Ratio, denoted as \(v\), is a key parameter in understanding material behavior under stress. It defines the ratio of transverse strain to axial strain when a material is subjected to uniaxial stress. In the context of plane strain, where the third strain component \(\varepsilon_3\) is zero, Poisson's Ratio ties the in-plane strains to the out-of-plane stress through the equation:
\[ \sigma_3 = v(\sigma_1 + \sigma_2) \]
This implies that if stresses \(\sigma_1\) and \(\sigma_2\) are applied, the material will also experience some out-of-plane stress \(\sigma_3\) proportional to Poisson's Ratio. Additionally, Poisson's Ratio is used to derive the expressions for in-plane strains in terms of the applied stresses, as seen in the equations for \(\varepsilon_1\) and \(\varepsilon_2\). This interconnectedness highlights how stress in one direction affects strain in another, key to predicting material behavior in engineering applications.
Lame's Parameters
Lame's Parameters, \(\lambda\) and \(G\), are essential in the stress-strain equations for understanding material elasticity. These parameters characterize how a material responds elastically to stress. In the given exercise, they appear in the relationships:
  • \( \sigma_1 = (\lambda + 2G) \varepsilon_1 + \lambda \varepsilon_2 \)
  • \( \sigma_2 = \lambda \varepsilon_1 + (\lambda + 2G) \varepsilon_2 \)
\(\lambda\), called the first Lame parameter, primarily accounts for volumetric response, while \(G\), also known as the shear modulus, relates to the material's ability to withstand shearing. Together, they define the stiffness of a material in response to different types of deformation. Understanding these parameters helps predict how materials will perform under complex loading conditions, such as those seen in various structural applications.

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Most popular questions from this chapter

The Amazon River basin in Brazil has a width of \(400 \mathrm{~km}\). Assuming that the basin is caused by a line load at its center and that the elastic lithosphere is not broken, determine the corresponding thickness of the elastic lithosphere. Assume \(E=70\) \(\mathrm{GPa}, v=0.25,\) and \(\rho_{m}-\rho_{s}=700 \mathrm{~kg} \mathrm{~m}^{-3}\)

If all the principal stresses are equal \(\sigma_{1}=\sigma_{2}=\sigma_{3} \equiv p\), then the state of stress is isotropic, and the principal stresses are equal to the pressure. The principal strains in a solid subjected to isotropic stresses are also equal \(\varepsilon_{1}=\varepsilon_{2}=\varepsilon_{3}=\frac{1}{3} \Delta ;\) each component of strain is equal to one-third of the dilatation. By adding Equations \((3-1)\) to \((3-3),\) we find $$ p=\left(\frac{3 \lambda+2 G}{3}\right) \Delta \equiv K \Delta \equiv \frac{1}{\beta} \Delta . $$ The quantity \(K\) is the bulk modulus, and its reciprocal is \(\beta,\) the compressibility. The ratio of \(p\) to the bulk modulus gives the fractional volume change that occurs under isotropic compression. Because the mass of a solid element with volume \(V\) and density \(\rho\) must be conserved, any change in volume \(\delta V\) of the element must be accompanied by a change in its density \(\delta \rho .\) The fractional change in density can be related to the fractional change in volume, the dilatation, by rearranging the equation of mass conservation $$ \delta(\rho V)=0 $$ which gives $$ \rho \delta V+V \delta \rho=0 $$ or $$ \frac{-\delta V}{V}=\Delta=\frac{\delta \rho}{\rho} $$ Equation \((3-53)\) of course assumes \(\Delta\) to be small. The combination of Equations \((3-50)\) and \((3-53)\) gives $$ \delta \rho=\rho \beta p $$ This relationship can be used to determine the increase in density with depth in the earth. Using Equations \((3-11)\) to \((3-13),\) we can rewrite the formula for \(K\) given in Equation \((3-50)\) as $$ K=\frac{1}{\beta}=\frac{E}{3(1-2 v)} $$ Thus as \(v\) tends toward \(1 / 2,\) that is, as a material becomes more and more incompressible, its bulk modulus tends to infinity.

Find the deflection of a uniformly loaded beam pinned at the ends, \(x=0, L\). Where is the maximum bending moment? What is the maximum bending stress?

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