# Tag Info

## Hot answers tagged continuum-mechanics

227

Understanding why this works turns out to be quite deep. This answer is kind of a long story, but there's no maths. At the end ('A more formal approach') there is an outline of how the maths works: skip to that if you don't want the story. Insect geometry Consider a little insect or something who lives on the surface of the paper. This insect can't see ...

65

The answer to this question has significant overlap with my answer on piano tuning. There, I discuss how a thick wire has an extra restoring force, in addition to its tension, from its resistance to bending. This modifies the usual wave equation to $$v^2 \frac{\partial^2 y}{\partial x^2} - A \frac{\partial^4 y}{\partial x^4} = \frac{\partial^2 y}{\partial t^... 54 I think the other answers which mention electrostatics capture the physics behind things being rigid correctly. However, I wanted to specifically point to your question of "why are they rigid when they're mostly vacuum?" I'd like to draw your attention to Guyed Masts: A Guyed mast is a tower whose rigidity depends on several guy-wires surrounding it. If ... 26 You have essentially discovered principles behind bending moments and structural engineering. As another poster stated, physically the structure you made is stronger, because to bend something (for example, a beam loaded at the top) layers at the top are compressed whereas layers at the bottom are stretched. This is simply due to geometry and the physical ... 24 What does it mean? The reason they are conservative or non-conservative has to do with the splitting of the derivatives. Consider the conservative derivative:$$ \frac{\partial \rho u}{\partial x} $$When we discretize this, using a simple numerical derivative just to highlight the point, we get:$$ \frac{\partial \rho u}{\partial x} \approx \frac{(\rho ...

21

As knzhou identifies, the key difference between vibrations of a free beam and a string is that the restoring force is now provided by bending moments (proportional to $\frac{d^4y}{dx^4}$) rather than linear tension (proportional to $\frac{d^2y}{dx^2}$). The consequence, as this source shows, is that for free beams like the bars of the glockenspiel, angular ...

20

The answer you're looking for seems to be contained in Rezzolla & Zanotti: Relativistic Hydrodynamics (Oxford U.P. 2013) https://books.google.com/books/?id=KU2oAAAAQBAJ but it is not a trivial generalization. Quoting Disconzi's On the well-posedness of relativistic viscous fluids (Nonlinearity 27 (2014) 1915, arXiv:1310.1954): we still lack a ...

20

When you bend a piece of material, the resistance is provided by stretching the material on the outside part of the bend, and compressing the material on the inside of the bend. A thin flat sheet bends easily because, physically, not of lot of stretching or compressing occurs when it bends. When you give your book a fold, like a trough, that shape can not ...

18

Consider a standard volume of $1\textrm{ m}^3$ of air. This contains on the order of $10^{25}$ molecules of O2 and N2. If you needed to simulate or explain the physics occurring in that volume of air, would you want to model $10^{25}$ molecules and all the interactions between them or, say, 100x100x100 cells based on the Navier-Stokes equations? ...

15

Before developing the theory, I decided to first make an experiment in order to understand, what we are dealing with. A cylinder with a diameter of 11.5 cm is mounted on the motor shaft (I used an old popcorn machine). I attached a 12.5 cm length of clothesline with a screw, so that exactly 11.5 cm leaves the cylinder. When the rope hangs freely, it forms a ...

14

A fake derivation We can rather easily compute a horizontal velocity for the string fi we assume that the total velocity vector is everywhere normal to the string (this assumption is not always valid, see below). The following picture then illustrates the computation: Take two infinitesimally separated points $x$ and $x+\mathrm{d}x$ and let the wave motion ...

13

I don't think this sounds unreasonable as an estimate at all. Let's check it. One designs a building as a compromise between two competing factors: One needs all of the load bearing materials to be well mildly loaded - working in their linear region so that there is no danger of their undergoing plastic (irreversible) deformation, creeping then ultimately ...

12

Are you referring to the exact relativistic equivalent to Navier-Stokes equation or a more general Dissipative Relativistic Hydrodynamics Equation? The "relativistic equivalent to Navier-Stokes equation" would be something like this: There would be an energy momentum tensor with the following form: $T_{\mu\nu} = (e+p)u_\mu u_\nu - p g_{\mu\nu} + \tau_{\mu\... 12 Applying a force in the$x$-direction might change the shape of the material in the$y\$-direction. The only way to capture such an effect is through a tensor. If you have a general force acting on your body $$\vec F = (F_x, F_y, F_z)^T$$ and you are interested in the reaction of the body by looking at its deformation $$\vec \epsilon = (\epsilon_x, \... 11 The other answers so far are technically correct, but none of them really seem to give a commonsense/intuitive and simple answer. So I'll have a go at one. Imagine very slightly bending some kind of object downwards at one end, while holding the other end firmly horizontal. (It could be almost any object, could be paper, a branch off a tree, some plastic ... 10 The number of atoms (or molecules) in a body is given by Avogadro's constant, or 6.022 \times 10^{23} per mole. A mole is the amount of material, in grams, equal to the atomic or molecular mass of the substance in question. For example, for water (H_2O), 1 mole equals 18 grams. To get this number, remember that hydrogen (H) has an atomic mass of 1.... 9 The reason that the speed of sound is a well-defined quantity is that, for small pertubations, the equations which govern the fluid dynamics can be linearised. In that linearised form, the solution boils down to a simple wave ansatz with linear dispersion relation, i.e. constant velocity.Those are the sound waves. It so happens that in air, this linear ... 9 You are absolutely right in everything you said. The momentum is non zero only if the wave has a longitudinal mode, which is in fact the realistic case. Moreover when this is the case, the wave equation is not that simple. Let me try show this. Longitudinal Mode Let us assume that when in equilibrium the string, of density \mu, is along with the x\equiv ... 8 The onset of turbulence in fluids is determined by the Reynolds number$$ \mathrm{Re} = \frac{vL}{\nu}, $$where L is the characteristic length scale, v the characteristic velocity, and \nu the viscosity. The onset of turbulence in fluids occurs for \mathrm{Re} greater than about 1000 or more, depending on geometry. If we want to see the equivalent ... 8 We can model the building as a uniform cuboid of density \rho occupying the region$$0 \le x \le L_x0 \le y \le L_y0 \le z \le L_z$$with its mass given by$$M = \rho V = \rho A L_z = \rho L_x L_y L_z$$The building is attached firmly to the ground (xy-plane). Ignoring gravity and compressive stress, consider only the effects of the ... 8 I) There are already several good answers. OP is asking about the momentum of the non-relativistic string with only transverse displacements, whose Lagrangian density usually is given as$$ {\cal L}_T ~:=~\frac{\rho}{2} \dot{\eta}^2 - \frac{\tau}{2} \eta^{\prime 2} \tag{1}$$in textbooks. II) Let us fix notation: \rho is the 1D mass density; \tau is ... 8 The traditional derivation of the Navier-Stokes equations starts by looking at a fluid parcel and the different fluxes over the surface in the integral form. The integral form is preferred as it is more general than the differential form: For the latter one has to assume differentiability and thus it is not valid for flow discontinuities such as shocks in ... 7 It is a quite famous theorem due to Cauchy. Consider an internal portion S of a continuous body C. There are two kinds of forces acting on it: Forces proportional to the mass, of the form$$\int_V \mu(x) \vec{f}(x) d^3x\tag{0}$$where \vec{f}(x) is the density of force acting on x \in V. And forces acting through the surface \partial V, the ... 7 Referring to your graph which is for a ductile material I suggest the following. A is the limit of proportionality up to which the stress and strain are proportional to one another and when unloaded the material goes back to its original length. B is the elastic limit. With stresses below this the material behaves elastically ie when unloaded returns ... 7 The state of the rope is defined not just by the vertical displacement along its length, but also by the vertical velocity. In this case, the displacement is zero, but the velocity is not. The parts of the rope are located at their equilibrium positions, but they are still in motion; by Newton's First Law, they will continue to move. This continued motion ... 6 Hooke's Law is frequently used to model multi-dimensional materials because the stress tensor is simple (linear). The full expression can be found on Wikipedia. The simplification for 2D is straight forward (drop any terms with a 3 in the subscript). Note that whether deformation in one dimension affects the others is a property of the material and shows up ... 6 When a metal spring is stretched beyond it's elastic limit, the metal begins to undergo some plastic deformation. This is a permanent deformation of metal crystals caused by the creation and motion of crystal lattice dislocations. These processes are partially irreversible and some of the work performed to deform the spring is lost as heat. 6 Indeed, both the strain tensor$$\epsilon_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right) \tag{1}$$and the stress tensor$$\sigma_{ij}=2\mu\epsilon_{ij}+\lambda\epsilon_{kk}\delta_{ij} \tag{2}$$are symmetric by definition. However, bear in mind that these definitions are not always valid; (1) assumes ... 6 The momentum flux tensor comes from the momentum equation of Navier-Stokes equations:$$ \frac{\partial\left(\rho\mathbf{u}\right)}{\partial t}+\nabla\cdot\mathbf{P}=0\tag{1} $$Or, using indices (where it is easier to see that \mathbf{P} is a rank-2 tensor):$$ \frac{\partial\left(\rho u_i\right)}{\partial t}+\frac{\partial\Pi_{ij}}{\partial x_j}=0\tag{2} ...

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