# Tag Info

15

In addition to the question of bend radius - there is also an effect of surface scratches. Most materials are very strong - they fail because a surface flaw allows a stress concentration - ie a crack to form. glass fibre has a very smooth surface because of the way it is made and can be put under high stress without cracking. You can show this with a thick ...

10

What evidence do you have that there is a consistent theory of continuum mechanics? Certainly, when looked at through a macroscope, the universe looks like it behaves according to continuum mechanics, but this completely breaks down on the microscopic level. So you can't justify a consistent theory of continuum mechanics by using the universe. There's no ...

5

The critical parameter for materials under stress is the strain, defined in a general way{*} as the fraction change some length over that same length: $$\lambda = \frac{\Delta l}{l}$$ So take a given a fiber of diameter $d$, and bent around a radius of curvature $r$, the strain of either the inside or outside edge is: $$\lambda = \frac{((r \pm d) - r) ... 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 The initial and final momentum are not the same because the ball is not an isolated system. The wall exerts a force on it. In principle the ball and the wall (and the planet it's connected to!) form an isolated system with a conserved momentum, but you'd have to take into account how much the wall moves after the collision. The change of momentum is final ... 4 Having spent a great deal of time working with "normal" glass (one of my many hobbies), I can assure you that, in fact, all glasses can be bent. When cutting large sheets of glass, I always see the sheet bend before it breaks. Every now and then you have to strike it twice; the first time you bend it, it fails to break. Fiber optics differ from the usual ... 4 Young's Modulus isn't dimensionless! It says STRAIN is dimensionless (which is true). SO Y = Stress/Strain = [Pressure]/[Dimensionless] = [Pressure]! Young's Modulus has the same dimensions as that of pressure, which is: [M] [L]^{-1} [T]^{-2} And units of pressure, which is Pascal. 4 If you use a plate glass window instead of a wall you'll find that the rubber and iron balls bounce by a similar amount (though be careful throwing iron balls at windows :-). It's a basic principle in physics that energy cannot be lost. The rubber ball starts off with kinetic energy, hits the wall, and rebounds moving with about the same kinetic energy. So ... 3 The key parameter here seems to be how are distributed the losses at each bounce (from what you said, there is no air friction here): how much translational and vertical speed are lost in the transition before/after the bounce. From your graphic, it seems like the hypothesis is that there is "reflection" of the ball: same angle when arriving/leaving the ... 3 A toy model of a solid A very simple model of a solid is to imagine a bunch of molecules linked to their nearby neighbors by springs (you don't need to imagine a crystalline lattice, it can be amorphous). The springs are the effective electromagnetic interactions between the molecules; they are strongly repulsive at close range and attractive at modest ... 3 As this looks a lot like a homework question I'll give some brief hints. Consider which way the ball must be spinning. Will the magnitude of the horizontal velocity v change? Whats the relationship between rotational velocity and linear velocity for a point on the circumference? 3 I believe potentials in Maxwell's equations were introduced originally to make solving equations simpler -- it wasn't until a bit later that the Lorentz symmetry was noticed. Similar potentials are introduced in 2D hydrodynamics as well, see e.g. stream function. So I would say that the reason for introducing these potentials originally was always to ... 3 Just check your math. Correct derivation is$$\begin{eqnarray} {f \over \epsilon_{11}} & = & {S_{11} \over \epsilon_{11}}=(2\mu+\lambda)+\lambda {\epsilon_{22}+\epsilon_{33} \over \epsilon_{11}} = \\ & = & (2\mu+\lambda)-{\lambda^2 \over \lambda + \mu} = {(2\mu + \lambda)(\lambda + \mu) - \lambda^2 \over \lambda + ...

3

What you need is the Euler-Bernoulli beam theory. The last three pages of this PDF explain the eigenmodes.

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

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.

2

Boy, the right stress tensor for similar static situations is symmetric, indeed. It's not hard to see why: the stress tensor knows about the density of forces and an asymmetry would destroy the equilibrium. See http://en.wikipedia.org/wiki/Stress_(mechanics)#Equilibrium_equations_and_symmetry_of_the_stress_tensor The last paragraph of the section above ...

2

If a rubber band is usable for 5 years, it's a very good one. There is not much You can do, just store them in a dark and cool place. Light plus oxygen is the most dangerous enemy of rubber. Nothing can be done in use. The biggest influence has the producer, because he can choose a persistent (expensive) kind of rubber (eg Kalrez®, ask for prizes!) Of ...

2

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 ...

2

You've misunderstood the statement. When A exerts $\vec{F}_{A \to B}$ on B, B exerts an equal and opposite force $\vec{F}_{B \to A} = - \vec{F}_{A \to B}$ on A. The only forces acting on the ball a gravity and the normal force, and the floor experiences a force from the ball which is equal in magnitude and opposite in direction from the normal force on the ...

2

Opening my mechanical engineering book on the rotating rings section and I get $$\sigma_t(r) = \rho \omega^2 \frac{3+\nu}{8} \left( r_i^2 + r_o^2 + \frac{r_o^2 r_i^2}{r^2} - \frac{1+3\nu}{3+\nu} r^2 \right)$$ $$\sigma_r(r) = \rho \omega^2 \frac{3+\nu}{8} \left( r_i^2 + r_o^2 - \frac{r_o^2 r_i^2}{r^2} - r^2 \right)$$ Inner Edge Von-Mises Stress ...

2

For instance, let us derive 2). Let us consider a unit cube and apply unit tension along $x$, unit tension along $y$, and unit tension along $z$ (to derive bulk modulus). For example, tension along $x$ causes strain $\alpha$ along $x$ and strains $-\beta$ along $y$ and along $z$ (the sign of lateral strain is typically opposite to that of longitudinal ...

2

First of all, a Balloon is made of a highly elastic material like rubbers or nylons. All materials have a specific level of Elasticity or Plasticity (based on its nature except brittles where fracture occurs). To differentiate these properties, Hooke's law & thereby moduli of elasticity arrived. When you blow air inside the balloon, it expands thereby ...

2

Suppose the constant force is gravity. Experience should tell you that if you release a mass attached to a spring is oscillates up and down, so a constant force does not mean there won't be any oscillation. The $x'$ term is the damping term. If you assume there is no damping the $x'$ term can be omitted, and in that case you just get the equation for a ...

2

Well, possibly the simplest argument (I have no idea if this is correct) is that compressing the material reduces the configuration space available for the electrons, and increasing confinement means increased energy (differences) in quantum mechanics. Consider just the old particle-in-a-box problem, the energy levels scale as $E_n \sim \frac{n^2}{L^2}$, ...

2

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 ...

2

Your erroneous assumption is that $C_{zzzz}$ corresponds directly to the Young's modulus. In fact, it does not, but in the case of your example is 1/3 of the Young's modulus. Therefore, the global stiffness is, in fact, $\frac{Y S}{l}$, as expected. To understand the relationship between the Young's modulus and the coefficient $C_{zzzz}$, take a look at ...

2

The way you have stated the question is unrealistic. If you let go at the very instant that you hit the yield stress of the material, it will still recover all of the elastic deformation. Only the deformation beyond yield will be permanent. That said, if you continue deforming the material beyond the yield point of the material, any extra work ($\dot{W} = ... 1 With two pegs, there are two strips of rubber working in parallel contributing to a total stiffness$K_{\rm total} = 57.6 \;\rm lbf/in$. So each strip is$K = 28.8\;\rm lbf/in$. With the three pegs you now have two strips at 26° apart, or 13° from vertical each. The effective spring constant in the vertical direction is thus$K_{eff} =2 K \cos^2(13^\circ) ...

1

The buckling problem with $F=0$ is $EIw''=0$, with boundary values $w(0)=w(L)=0$. The corresponding beam problem with $q=0$ is $EIw''''=0$, with boundary values $w(0)=w(L)=0$ and $w''(0)=w''(L)=0$, since there is no applied moment at the ends. Both problems have the same solution: $w(x)=0$. Raskolnikov is right: it is ...

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