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51

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 ...


11

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, ...


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

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 ...


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

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

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

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

Concrete is very strong in compression but weak in extension, so it isn't good at supporting stretching or bending forces. By contrast steel is able to cope well with stretching and bending forces. So by combining the two you have a composite structure that resists compression as well as concrete and resists stretching and bending as well as steel. In many ...


4

Let's take a look at what forces a concrete beam or column can be expected to handle: Compression, from weight being applied directly on it. Bending, from any bending moments. Concrete can handle being compressed very well, since it's basically sand and gravel. It can stand compressive loads of 3000 - 6000 psi, which is huge. However, it doesn't do so ...


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

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

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 ...


3

I'd swap the meaning of x and y to make the sparsity structure more conspicuous (block diagonal). The I'd try to identify the paramaters as in the isotropic case, where they exist, and itnroduce new ones for the others by analogy (note that each column has the same denominator). Then consider special stress vectors that affect only few strain components, and ...


3

Yes, the bulk modulus $B$ is the inverse of the isothermal compressibility $c$, $$ B = \frac {1}{c}.$$ See e.g. Wikipedia. The "bulk modulus" is more typical terminology in mechanics where we don't care about heat much and where the typical assumption is that the temperature is kept fixed (because mechanical engines start to malfunction if their ...


3

Your assumptions are correct (but $r$ is often defined as the distance from the pipe centerline). However, this is a very specific case: laminar pipe flow. In general, the stress will be a tensiorial quantity, defined as $$ \tau_{ij}= \eta \frac{\partial u_i}{\partial x_j}$$ which is true for turbulent flow, in arbitrary geometries. Where $i,j$ are in the ...


3

Start with differential form of Poisson's ratio: $$\frac{\text{d} x}{x}=- \nu \frac{\text{d} l}{l}$$ $$\int_{x_0}^{x_0+\Delta x} \frac{\text{d} x}{x}=- \nu \int_{l_0}^{l_0+\Delta l} \frac{\text{d} l}{l}$$ $$\ln \frac{x_0+\Delta x}{x_0}=- \nu \ln \frac{l_0+\Delta l}{l_0}$$ $$1 + \frac{\Delta x}{x_0}=\left(1+\dfrac{\Delta l}{l_0}\right)^{-\nu}$$


3

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 ...


3

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 ...


2

Any cross section of your wall is supporting the weight of all the wall above it. In a first approximation, every cross section will be in a state of pure axial compression. The most heavily solicited cross section will be the one at the very bottom, which will be supporting a compressive pressure of $\rho h g$, where $\rho$ is the density of the ice, $h$ ...


2

Any mechanical strain creates electricity. You can use a piezo sensor as a pressure gauge, it's common for very high/fast pressure changes because the crystal can survive a very high pressures and it's small so responds quickly. edit: to make a pressure sensor you need one side of the crystal exposed to vacuum and the other to a fixed pressure - so that ...


2

From the above definition for tensile strain, $L$ is $L=\epsilon L_{0}+L_{0}$. Thus $dL=L_{0}d\epsilon$ replacing in the above $L_{0}$ with $L_{0}=\frac{L}{1+\epsilon}$ you get $dL=\frac{L}{1+\epsilon}d\epsilon$ Now, if the deformation of the material is very small you can expand $\frac{1}{1+\epsilon}$ as $\frac{1}{1+\epsilon}=1-O(\epsilon)$ Hence you ...


2

Yes, when two objects collide, there's some excess pressure in the impact zone but the pressure propagates by the wave equation through the material – it's nothing else than the mechanism by which sound propagates in a medium. Also, waves may be transverse or longitudinal (the variation of the position is going in parallel with the direction of the wave). ...



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