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Why plane stress is taken for thin plates? It says in the books that the stress variation is small for thin components and is close to zero. Why is that so?

Also why stress at free surface is zero? (talking with respect to a solid under simple loading e.g. a bar under uniaxial loading) and free surface is the boundary of a specimen right?

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  • $\begingroup$ Thin plates are approximated well as planes, that is a good enough reason for me why their stresses are plane stresses $\endgroup$ – Jim Jul 11 '14 at 12:44
  • $\begingroup$ but then same reason can be said that since it is plane so strain is zero in out of plane direction i.e. plane strain which is incorrect violating your argument $\endgroup$ – irtiza Jul 11 '14 at 16:46
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The boundary conditions of free surfaces are $\sigma_{ij}n_j=0$, where $\vec n$ is the normal to the surface. If a body is thin in the $z$ dimension, then on the top surface (say, $z=h$) and on the bottom one (say, $z=0$) you have $\vec n=\vec z$ and therefore $\sigma_{iz}=0$.

Since $\sigma_{iz}$ vanishes of $z=0$ and for $z=h$, and since $0\le z\le h$ and $h$ is small (i.e. the body is thin), it is reasonable to assume that $\sigma_{iz}$ is identically zero for all values of $z$.

This is the standard argument. Note, however, that in contrast to plane-strain which is a rigorous reduction of the full 3D equations, plane-stress is an approximation, which can not be satisfied identically.

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  • $\begingroup$ Can you explain it a little bit more in detail or Easy wording $\endgroup$ – irtiza Jul 14 '14 at 9:55
  • $\begingroup$ "since 0≤z≤h and h is small (i.e. the body is thin), it is reasonable to assume that σiz is identically zero for all values of z".....That is my question...how does thin imply that stress is identically zero for all values of z.......why can.t it vary like below???? Say 4mm plate, 0 at mid plane Thickness {-2 -1.5 -1 -0.5 0 0.5 1 1.5 2} mm Stress {0 1 4 6 18 6 4 1 0} MPa $\endgroup$ – irtiza Jul 14 '14 at 9:58
  • $\begingroup$ As I wrote in my answer, this is not a proof, nor a consistent derivation. Plane-stress equations are an APPROXIMATION, which is not exactly valid. Specifically for the profile you suggested, it is not enough in order to determine whether it is a possible solution. You need the other stress components, and if they satisfy the force-balance equation $$\sum_j \frac{\partial \sigma_{ij}}{\partial j}=0$$ then you're good. $\endgroup$ – yohBS Jul 14 '14 at 18:28
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Plane stress and plane strain are both approximations to reduce 3-D problems to 2-D problems, which are easier to solve.

Plane stress is used to model structures that are very small in one direction compared to the other two, such as thin plates. The normal stress at a free surface is zero because there's nothing (other than, usually, air) to react against it. In a thin plate, the normal stress is zero on both faces of the plate and there's not much room for it to build up in between the faces. So we just assume that it's zero all the way through. The in-plane stresses aren't subject to this restriction and can be any size you want (within the limits of the material).

Plane strain is used to model structures that are very large in one dimension compared to the other two, such as columns. Each thin slice of material is constrained by material on either side and so the strain in the 'long' direction is taken to be zero all the way through. The crosswise strains aren't subject to this restriction and can be any size you like (within the limits of the material).

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