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EDIT: I am gonna rephrase the question entirely.

Imagine we have a bar which we will analyze in the linear elastic regime.

The shape of the cross section is irrelevant.

The bar is suspended from a ceiling and it is under its own weight.

I have been told that in a problem of this type, in every point of the bar, ONLY the longitudinal component of the stress tensor is non zero. Is this right? and if so how do you justify this?

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  • $\begingroup$ Uniform distribution is reasonable - since every point on the contact surface has the same amount of material right below it. In reality there will be some lateral forces (think Poisson ratio and all that - or just consider a material with a very low Young's modulus, recognize that it will not be of constant cross section along its length, and you can see your friend is (a little bit) wrong. $\endgroup$
    – Floris
    May 12, 2014 at 21:41
  • $\begingroup$ @Floris maybe you are right. The thing is I have been talking with my professor today and he was very confident that we have explicitly vertical tensions with any possible cross section bar, if it is vertically suspended under its weight. His arguments haven't convinced me and I think he was wrong but maybe I haven't fully understood what he was saying. Can you think of any reason why we would have ONLY vertical tensions? $\endgroup$
    – Yossarian
    May 14, 2014 at 14:20
  • $\begingroup$ Are you asking about tensions or stresses? There is a difference. $\endgroup$ May 14, 2014 at 14:32
  • $\begingroup$ Edited. about the longitudinal component of the stress tensor $\endgroup$
    – Yossarian
    May 14, 2014 at 14:35

2 Answers 2

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If there is elongation in the horizontal direction (due to Poisson's ratio) it does not mean there is stress in this direction also. Using 3D hook's law

$$ \begin{bmatrix} \epsilon_x \\ \epsilon_y \\ \epsilon_z \\ \gamma_{yz} \\ \gamma_{xz} \\ \gamma_{xy} \end{bmatrix} = \frac{1}{E} \begin{bmatrix} 1 & -\nu & -\nu & & & \\ -\nu & 1 & -\nu & & & \\ -\nu & -\nu & 1 & & & \\ & & & 2(\nu+1) & & \\ & & & & 2(\nu+1) & \\ & & & & & 2(\nu+1) \\ \end{bmatrix} \begin{bmatrix} \sigma_x \\ \sigma_y \\ \sigma_z \\ \tau_{yz} \\ \tau_{xz} \\ \tau_{xy} \end{bmatrix}$$

and we have only stress in one direction, like $\sigma_x$ the resulting strains are:

$$ \begin{bmatrix} \epsilon_x \\ \epsilon_y \\ \epsilon_z \\ \gamma_{yz} \\ \gamma_{xz} \\ \gamma_{xy} \end{bmatrix} = \begin{bmatrix} \frac{\sigma_x}{E} \\ -\frac{\nu \sigma_x}{E} \\ -\frac{\nu \sigma_x}{E} \\0 \\0 \\ 0\end{bmatrix} $$

So the strains in other directions is not a necessary condition to have bi-axial stress. In your case, a simple free body diagram of a small part on the bar can show that stresses only propagate in the vertical direction since nothing is pushing against the bar horizontally.

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To complete @ja72's answer:

  • analytical demonstration: @ja72 did not explicitly explained why $\sigma_y=0$ and $\sigma_z=0$. This is due to the equilibrium (volume forces neglected): $\mathrm{div}(\underline{\underline{\sigma}})=0$ together with the boundary conditions $\underline{\underline{\sigma}}\cdot \underline{n}=0$ on the lateral faces of the bar.

  • physicist's argument: the eigenstrains $\sigma_{yy}$ and $\sigma_{zz}$ are zero on the lateral faces of the bar, and the bar is slender: the characteristic dimensions in $y$ and $z$ are very small compared to the characteristic dimension in $x$. All the physical quantities being smooth, the cross-section of the bar is too small for $\sigma_{yy}$ and $\sigma_{zz}$, which are zero on both ends, to vary sufficiently not to stay very small compared to $\sigma_x$. As I said, its not a demonstration but more like a physical interpretation.

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