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If you stretch a stretchy square rubber sheet out so that the left and right edges are kept straight and parallel, but the top and bottom are otherwise unconstrained, then the top and botton edges will curve inward until the tension along the edge of the sheet is angled enough to counteract the vertical contraction due to the horizontal elongation. If the ...

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If I have a membrane that has one loose end, does it mean that the tension on that end is equal to zero and because of that the the force on that end is also equal to zero? Yes, if that end is completely loose. If it is stretched from side to side by tension at the corners--then no. The way to analyze this would be to use a free body diagram. See: ...

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Here are the basic equations if the cross-sectional area, $A$, of the bar is uniform across its entire length. Newton's second law: $$\frac{\partial\sigma}{\partial x} + \frac{p}{A}=\rho \frac{\partial^2u}{\partial t^2}$$ where $p$ is an axial distributed load per unit length, $\sigma$ is the stress in the bar, $\rho$ is the density, and $u$ is the ...

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Let me first get something out of the way: It seems to me that wood is an unsuitable material for this thought experiment in the following sense: wood is very resistant to change of dimension parallel to the grain, but not perpendicular to the grain. Wood changes very unevenly. But of course your question can readily be reformulated to feature a material ...

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Taylor series is the polynomial expansion of functions of certain kind. More details can be looked up here. What this says is that if we know the value of a function $f(x)$ and all it’s derivatives at a point $x_0$ then we can express the value of the function at any point $x+h$ by using: $$f(x_0+h)=f(x_0)+\frac{h}{1!}f’(x)\big|_{x_0}+ \frac{h^2}{2!}f’’(x)\... 0 The problem with the right hand diagram is that it doesn't scale properly. Suppose this stress field was uniform over a finite sized region. Now take a rectangular element of size \Delta x by \Delta y. You now have a direct force \sigma_y \Delta x balanced by the shear forces 2 \tau_{yx} \Delta y and you can't eliminate \Delta x and \Delta y ... 3 The difference depends on the physical structures: the tension of the membrane must be imposed by means of external forces whereas that of a plate naturally exists in its interior. In other words the relation between deformations and stress is different. The small deformations of the structure, in the first case, are dynamically described in terms of a ... 0 Let's start from$$\rho\frac{De}{Dt}=\sigma:\mathbf{D}-\nabla\cdot \mathbf{q}+\rho\mathbb E$$\mathbf{D} is the rate of deformation tensor, and it was previously (in the same book) defined as$$\mathbf{D}=\frac{1}{2}[\nabla \mathbf{v}+(\nabla \mathbf{v})^T]$$Then$$\mathbf{\sigma}:\mathbf{D}=\frac{1}{2}\mathbf{\sigma}:\nabla \mathbf{v}+\frac{1}{2}\...

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This is my solution attempt. Main assumptions: Continuum hypotesis is kept till molecular scale. The following expression for total kinetic energy $K_{tot}$ will be used $$K_{tot}=\int_{\Omega(t)}\frac{1}{2}\rho\,\mathbf{v}\cdot \mathbf{v}\,d\Omega$$ All quantities considered will be in Eulerian description (i.e. function of $(\mathbf{x},t)$). Let's ...

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Is there a formal way to write the general expression for the kinetic energy $𝐾_{𝑡𝑜𝑡}$ in a continuum control volume and then splitting the two contributions into macro and micro kinetic energy, $𝐾_{𝑚𝑎𝑐𝑟𝑜}$ and $𝐾_{𝑚𝑖𝑐𝑟𝑜}$? I'm not versant in continuum mechanics, so I really can't answer your question. I know what a control volume is ...

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