Solution Found!
In many situations it is known that the normal stress in a
Chapter 2, Problem 2.73(choose chapter or problem)
In many situations it is known that the normal stress in a given direction is zero. For example, \(\sigma_{z}=0\) in the case of the thin plate shown. For this case, which is known as plane stress, show that if the strains \(\epsilon_{x} \text { and } \epsilon_{y}\) have been determined experimentally, we can express \(\sigma_{x}, \sigma_{y}\), and \(\sigma_{z}\) as follows:
\(\begin{aligned} \sigma_{x} & =E \frac{\epsilon_{x}+\nu \epsilon_{y}}{1-\nu^{2}} \\ \sigma_{y} & =E \frac{\epsilon_{y}+\nu \epsilon_{x}}{1-\nu^{2}} \\ \epsilon_{z} & =-\frac{\nu}{1-\nu}\left(\epsilon_{x}+\epsilon_{y}\right) \end{aligned}\)
Questions & Answers
QUESTION:
In many situations it is known that the normal stress in a given direction is zero. For example, \(\sigma_{z}=0\) in the case of the thin plate shown. For this case, which is known as plane stress, show that if the strains \(\epsilon_{x} \text { and } \epsilon_{y}\) have been determined experimentally, we can express \(\sigma_{x}, \sigma_{y}\), and \(\sigma_{z}\) as follows:
\(\begin{aligned} \sigma_{x} & =E \frac{\epsilon_{x}+\nu \epsilon_{y}}{1-\nu^{2}} \\ \sigma_{y} & =E \frac{\epsilon_{y}+\nu \epsilon_{x}}{1-\nu^{2}} \\ \epsilon_{z} & =-\frac{\nu}{1-\nu}\left(\epsilon_{x}+\epsilon_{y}\right) \end{aligned}\)
ANSWER:
Step 1 of 3
It is known that the normal stress in a given direction is zero
Multiplying (2) by 
