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The breaking of dry spaghetti was discussed in a 2005 Phys. Rev. Lett. by French physicists Audoly and Neukirch. Bottom line is that elastic (flexural) waves propagating along the spaghetti cause local increases in curvature leading to multiple breaking points: abstract to article. In essence, your assumption "that vibrations from a first break could ...

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Stress is a tensor1 because it describes things happening in two directions simultaneously. You can have an $x$-directed force pushing along an interface of constant $y$; this would be $\sigma_{xy}$. If we assemble all such combinations $\sigma_{ij}$, the collection of them is the stress tensor. Pressure is part of the stress tensor. The diagonal elements ...

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For each surface on a unit cube (see below), the stress on that surface can point in each of the three directions. (source) Since it is not necessarily the case that $\sigma_{11}=\sigma_{31}=\sigma_{21}$ (all pointing the in the same $\mathbf{e}_1$ direction)--or any of the other $\sigma_{ij}$ combinations, we need to have 9 components describing it, ...

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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 $$\vec \epsilon = (\epsilon_x, ... 9 You are asking two questions really 1) How is PE actually stored in a steel spring at the atomic level? The explanation for this lies in quantum mechanics 2) Could you explain in detail how/where potential energy is actually stored in a steel spring and why the material never surrenders to the bending forces taking a new shape? Replying to 1) ... 9 Pressure is defined as force per unit area applied to an object in a direction perpendicular to the surface. And naturally pressure can cause stress inside an object. Whereas stress is the property of the body under load and is related to the internal forces. It is defined as a reaction produced by the molecules of the body under some action which may ... 8 If you experience such a uniform force, e.g. when an astronaut on a space walk near the ISS (just earth's gravity), you don't experience any forces at all. That's freefall. Even with 10G, you'd experience a rapid freefall, but that is still harmless. It's the hitting the ground which kills you - that's not a uniform force. 7 Suppose you bend a perfect, i.e. unscratched, piece of glass, the forces on it look like: The top of the glass is in tension and the bottom in compression, but the stress is spread over a large area of glass so the local stress at any point isn't enough to break the glass. Now put a scratch in the top surface and bend it again: This time the stress is ... 6 It is a quite famous theorem due to Cauchy. Consider an internal portion S of a continuous body C. There are two kinds of forces acting on it: Forces proportional to the mass, of the form$$\int_V \mu(x) \vec{f}(x) d^3x\tag{0}$$where \vec{f}(x) is the density of force acting on x \in V. And forces acting through the surface \partial V, the ... 5 Given a stress tensor \mathbf{\sigma}, which has 9 components in general, the pressure (in continuum mechanics at least) is defined as P = 1/3 tr(\mathbf{\sigma}). So the pressure at a point in the continuum is the average of the three normal stresses at the point. The off-diagonal terms manifest as shear stress. It's hard to say "stress" without ... 5 Paper is made from wood, and wood is made from long fibers. Typically the manufacturing process leaves the fibers are more or less parallel. So it is easier to tear in the direction that separates fibers from neighboring fibers than in the direction that breaks fibers. Wood is the same. It is easier to split a log than chop it. Creasing paper breaks ... 5 Indeed, both the strain tensor$$\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 ... 5 First we need to understand the force between individual atoms. At relatively large separations (e.g., a few atomic diameters) atoms attract each other with a force that does, as you suggest, get weaker with distance due to polarization and ionic effects that we needn't go into here. If that was all there was to the story, however, collections of atoms ... 5 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 ... 5 Isotropy and homogeneity are different. The former is a consequence of invariance under rotations while the latter comes from invariance under translations. The stress tensor of an isotropic fluid then must be invariant under any orthogonal transformation, and this implies that it is a multiple of the "identity" tensor. More precisely, assume matrix notation ... 4 When people study continuum mechanics they usually do so at first in \mathbb{R}^3 where we have usually implied the usual metric tensor (g_{ij}) = \operatorname{diag}(1,1,1) and the Levi-Civita connection associated with it. In that case vectors and covectors are equivalent: the metric tensor induces the musical isomorphism and allows one to convert ... 4 Firstly, you can deform material permanently..spring is no exception. On the atomic level, you are working against Coulomb forces that bind the material id est, that form the lattice. One primitive cell is well defined by the conditions of minimal energy. You can describe this potential as a quadratic, so you get harmonic forces, but it is not truly ... 4 The link you provided already had enough information. Well, unlike hardness, which denotes only resistance to scratching, diamond's toughness or tenacity is only fair to good. That is, it is easily breakable by a hammer. The toughness of diamond is about 2.0 MPa which is good compared to other gemstones, but poor compared to most engineering materials. So ... 4 http://www.physicsforums.com/archive/index.php/t-37701.html says "Most of the strength of a cylinder comes from the outer portions. I think the contribution goes like the cube of the radial position. So, if you took a solid rod and drilled out a half the volume from the center, you do not lose half the strength. Strength to weight ratio is ... 4 Dave, to answer your question, we need to know the wall thickness and weight of each cube. BUT I can tell you that if you give any reasonable answer (like wall thickness = 1/8 of an inch = about 3.2mm), and choose steel bolts with right-sized steel washers, that it will be pretty hard to fail. Also, assuming your bolts are 60ksi tensile strength steel ... 4 Waves on strings combine linearly. This means that you can split up a string's motion into two (or more) superimposed waves. The two superimposed waves behave independently, as if the other one was not there. So if you have a standing wave set up on a string, and then you also introduce a travelling pulse, you get something like the following. (The arrows ... 4 Zero strain does not always imply zero stress and visa versa. There are matterials that display stress-strain, \sigma-\epsilon, hysteresis behaviour. In matterials like this, when you start loading them, they behave normally, i.e increasing the stress increases the strain. However, when you start to unload them (remove the load), instead of the stress ... 4 It provides a convenient graphical means of finding the maximum and minimum shear stress, which are important for determining material failure. You don't absolutely need it, but the graphical interpretation of the circular relationship between normal and shear stress is somewhat convenient. I've read good solid mechanics books that give little if any ... 4 The difference between stress and pressure has to do with the difference between isotropic and anisotropic force. There's a Wikipedia section on the decomposition of the Cauchy stress \boldsymbol{\sigma} into "hydrostatic" and "deviatoric" components,$$\boldsymbol{\sigma}=\mathbf{s}+p\mathbf{I}$$where the pressure p is ... 3 By folding a piece of paper you deform the paper there locally. I am not sure if you can call this plastic (instead of elastic) deformation, like with metals. At least with metals it is the case that plastic deformation leads weak spots, such that when a uniform tension is applied the material will tear first at these spots. I am not an expert on how paper ... 3 The Cauchy stress matrix \Sigma is a 3 \times 3 real symmetric matrix. It is interesting that we may without problems generalize \Sigma to a 3 \times 3 Hermitian matrix. It has three mutually orthogonal principal stress directions with principal stresses (eigenvalues) \lambda_1\geq\lambda_2\geq \lambda_3. Consider an arbitrary unit vector ... 3 Examples: stress is zero but strain is present= when component is loaded beyond the elastic limit it shows plastic deformation which can not be regained. after unloading the specimen in plastic deformation zone material will follow slope similar to the elastic slope and will come back to zero stress (as load is removed now). but during this process it has ... 3 Stiffness, linear or torsional does not depend on the particular alloy, but only on the base material. For your case you want the shear modulus used in$$\theta = \frac{T\, \ell}{G\, J} Here $T$ is the torque, $\ell$ the length, $J=I_{xx}+I_{yy}$ is the polar moment of area and $G$ is the shear modulus. This can be defined as $G = \frac{E}{2(1+\nu)}$ ...

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Starting with the basic concept of Gaussian curvature, we look to the metric given in the question, and try to identify some concept of strain from that. This is what hyperbolic curvature looks like for 2D in 3D, and I believe it should extend to the present problem of 3D to 4D without any material change. From this picture we could imagine calculating ...

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Though I generally agree with whoplisp's answer, it is worth to note that obtaining the (lower) limit of the thickness is rather tricky as long as it is defined by stability under 'strong' deformations. Where 'strong' is compared with the tube wall thickness. Which is obvious from common sense point of view: thin rod is easier to deform than the hollow ...

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