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@Learning_is_a_mess notes correctly that stress is only symmetric in the absence of internal torque. This is so when the only moments arising are due to forces at a finite distance and so tend to zero as we zoom in on the body. This is not so if, for example, a magnetic material was placed in a strong magnetic field: the internal moment does not tend to ...

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Since this is a thin-walled tube, the cross-sectional area of the tube is $(2\pi r)w$ where $w$ is the tube thickness, and $r$ is the tube radius. It follows that the stress is: $$\sigma = \frac{T}{(2\pi r)w}$$ where $T$ is the force. If you double $w$, the stress is cut in half. Contrast with solid cylinder For a solid cylinder, the cross-sectional area ...

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Engineers usually treat thermal expansion as isotropic, which means the expansion occurs with the same magnitude in every direction. This means that an unconstrained object will have a constant strain and zero stress associated with thermal expansion, it's as if object just scaled up. However, as you suspected, materials with an organized structure can be ...

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Mohr's circle is a graphical technique for getting principal stresses. It was useful in the age when most design engineers sat at a drawing board. In an era when most engineers have cellphones and tablets, Mohr's circle is an anachronism. I show my students Mohr's circle because some companies think it's 'essential knowledge', then I give them an Excel ...

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Your test will not be measuring a single property of the test specimens, so it won't have a specific name. But we can try and draw some inferences about what you'll be measuring with these experiments, using the idealised schematic above. In it the bone has been replaced with a uniform bar, with a hole drilled through it at the halfway point. The bar is ...

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Seems as though my idea is the correct one. Here's an example. Since, $dS$ is normal to the surface and has a magnitude equal to the area of a differential. $dS$ can be represented by a cross product of the differentials of the relevant surface. Given, $$\sigma= \begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix} ... 1 Both. You can have the standard$$\sigma_{ij} = C_{ijkl} \epsilon_{kl}$$in the bulk, but you can also have a reduced tensor, a surface stress. This surface stress is the projection of the regular stress$$\sigma_{\alpha \beta}^s = {\boldsymbol P} \sigma {\boldsymbol P} where ${\boldsymbol P} = 1 - {\boldsymbol n}({\boldsymbol r}) \otimes {\boldsymbol ... 1 You can imagine three equations one for each component of the force and then the tensor can be thought of as three vectors. For instance$F_x=\int \sigma_{xx}n_x+\sigma_{xy}n_y+\sigma_{xz}n_zdA$where$d\vec S=\hat n dA.$So now it is a regular flux surface integral. You just have three of them:$F_x=\int \sigma_{xx}n_x+\sigma_{xy}n_y+\sigma_{xz}n_zdA$, ... 0 TL;DR The expression in the paper is correct. It's just written weirdly. Take a look at where the "=" is in your index equation. You'll notice that your$\dot{\mathbf{S}}$and your spin terms are on opposite sides, so you should expect it to look a little goofy. Here's the algebra (in direct notation since I hate indices): Let$\mathbf{\epsilon}' = ...

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