If a plate is elongated in the x and y direction, why is the stress sigma z considered negligible with respect to sigma x and sigma y?
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3$\begingroup$ It is not "considered negligible," it is zero by definition - that is what "plane stress" means! If you have non-zero stress components in all three directions, then the stress isn't "all in one plane." $\endgroup$– alephzeroCommented May 29, 2017 at 1:08
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$\begingroup$ @alephzero It is considered zero but it's not necessarily zero in reality, it's an approximation. $\endgroup$– Mr. PiCommented May 29, 2017 at 10:01
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1$\begingroup$ Everything in real-world engineering structural analysis is an approximation. I don't understand what point you are trying to make here. $\endgroup$– alephzeroCommented May 29, 2017 at 10:30
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$\begingroup$ I'm trying to know why the stress sigma z in the plane stress case is negligible with respect to sigma x and sigma y, I want to know the reason for this approximation. $\endgroup$– Mr. PiCommented May 29, 2017 at 11:44
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$\begingroup$ the title and text of your question ask entirely different things. please clarify $\endgroup$– agentpCommented May 29, 2017 at 13:12
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Because along z direction, body thickness is so small (ideally zero) that practically it consists of only the two surface boundaries (top and bottom, see figure) . Because no pressure is allowed in these boundaries (otherwise we violate plane stress conditions), equillibrium of internal stresses and external actions cant't be satisfied on them, unless $\sigma_z = 0$. Because no material exists (ideally) between these two boundaries, no $\sigma_z$ is possible "internally".
Typical applications of plane stress conditions include thin shells, loaded only in their plane.
