Chapter 3: Problem 6
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 $$ \begin{array}{l} \varepsilon_{1}=\frac{(1+v)}{E}\left\\{\sigma_{1}(1-v)-v \sigma_{2}\right\\} \\\ \varepsilon_{2}=\frac{(1+v)}{E}\left\\{\sigma_{2}(1-v)-v \sigma_{1}\right\\} \end{array} $$
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