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A solid rod will be stiffer (both in torsion and in bending) than a hollow rod of the same diameter. But it won't be much stiffer because almost all the stiffness comes from the outer layers of the rod (this is what the whole second-moment-of-area thing is about). So if you have a certain amount of material to use (a certain mass, or a certain mass per ...

11

One can quantify tfb's answer through the second moment of the shaft's cross sectional area. Let the $z$ axis lie along the shaft, and suppose the latter is put under torsion $\tau$ about the $z$ axis. Then, if the shaft's length is $\ell$, the angular twist $\varphi$ along the shaft's length is given by: $$\varphi = \frac{\ell\,\tau}{G\,J_z}$$ where $G$ ...

9

The usual simplification in structural engineering is to consider the sphere as infinitely thin. That is what is called a 'shell' as opposed to thicker 'plates'. Shells react to pressure by undergoing traction, but without bending. That makes things easier. The book by Timoshenko is the place to go for a detailed explanation of either case. The easiest way ...

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Sound is longitudinal pressure waves in the air, It can be produced in a couple of ways in a collision. The impact can set up pressure waves and or ringing in the bodies themselves which then interact with the surrounding medium to produce audible sounds. This is the mechanism of a bell. If enough air is forced out of the space between the two bodies as ...

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 ...

7

For example, see cold or contact welding of ultraclean, similar metallic surfaces under ultrahigh vacuum conditions. After a few such experiences with what I thought were cleverly designed friction fittings for some electron beam optics, I soon learned to either use different metals, or sprinkle a bit of dry molybdenum disulfide on the joints to dirty them ...

7

The angle subtended is given by the arc length divided by the radius: $$\phi = \frac{L_2}{R+t/2}=\frac{L_1}{R-t/2}$$ where $L_2$ is the length of the longer strip (at $R+t/2$) and $L_1$ the length of the shorter strip (at $R-t/2$), $t$ is the thickness of the strips. $R$ is the radius to the middle of the strips. Assumptions here are basically small ...

6

This is going to be a bit of a long answer, so please bear with me while I develop the explanation fully. You can use energy arguments to prove the bounds on the Lamé parameters (and therefore, the Poisson ratio). Here is the free energy function for a linear elastic solid: $$\psi = (1/2)\mathbf{\epsilon}:\mathcal{\mathbf{C}}:\mathbf{\epsilon}$$ where $\... 6 I think it is basically a coincidence at the current time. Earth's axis of rotation precesses with a period of about 26,000 years, and according to Wikipedia, its orbital axis precesses with a period of about 112,000 years. So the winter solstice and perihelion will have all possible relative phases over a long time period. 6 The limit to which you refer is known as the thermodynamic limit in statistical mechanics. It consists in taking the limit of infinite particles ($N\rightarrow \infty$) and infinite volume ($V\rightarrow \infty$) while keeping a finite density$N/V$. In a solid, both electrons and atomic nuclei contribute to the thermodynamical and elastic quantities, such ... 6 The arc-like bending happens because the two metals expand differently when heated. They have different thermal expansion coefficients$\alpha. The bending allows the outer metal strip to expand more than the inner, so that they can expand differently while staying stuck together. The formnula for the radius of the curvature is: R=\frac t{\Delta \alpha \... 5 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 ... 5 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 ... 5 Why rubber is incompressible material? While an incompressible material must have a Poisson's ratio of ½, that rubber has a Poisson's ratio close to ½ does not mean rubber is incompressible. In fact, there is no such thing as a truly incompressible material. That rubber has a Poisson's ratio of ½ merely means that rubber is in some sense a bit like a ... 5 One word answer: Yes! To add a little to Peter Diehr's reference to cold welding, here is a physically insightful argument that the answer to your question is yes due to Richard Feynman. He asks the rhetorical question: suppose we re-align the two halves so that all the atoms on either side of the cleave are in the same positions as they would be in the ... 5 You can calculate the amount of the normal force if you know the elastic properties of the material and how much it is bent or deformed out of shape. In most cases, it is very difficult to measure the amount of deformation - eg it might be less than 1 micron. Also the shape and area of the deformation have to be taken into account - except for very simple ... 4 As you suggest, they're just to improve adhesion of the concrete to the bar. See for example this article. 4 Around 4080 BC the Earth was in perihelion during autumn. In 1246 AD the perihelion occurred during the winter solstice. By 6427 AD the perihelion will coincide with the March equinox. Perihelion will occur in April around 7062 AD. (source: Astronomical Algorithms) The question is: Is the Earth's lunisolar precession coupled to its perihelion precession? Or,... 4 Yes, principally because the speed of sound depends on the temperature. An approximate equation for the speed of sound in dry air is: v = 331 + 0.6T $$The wavelength is fixed by the pipe length so if the speed of sound changes the frequency also changes according to:$$ f = \frac{v}{\lambda} $$In principle there will be some thermal expansion of the ... 4 First, I'm going to define some notation. I prefer the direct notation of modern continuum mechanics as presented in this book by Gurtin et al., so I won't be carrying around the indices that you use. Lastly, the following holds for small deformation, rate-independent plasticity with isotropic hardening. \mathbf{T} = Cauchy stress tensor \mathbf{T}_0 = \... 4 The important thing about the curl is that it is the anti-symmetric part of the derivative. In Einstein notation, \varepsilon would be given by$$\varepsilon_{ij} = \frac{1}{2} ( u_{i,j} + u_{j,i}).$$Also in Einstein notation, we pick out the anti-symmetric part with the Levi-Civita symbol \epsilon_{ijk}. Thus the curl is$$(\nabla\times\varepsilon)_{... 4 The formulas are correct and are special cases of the center deflection of a center-loaded simply supported beam, which is $$\delta=\frac{Wl^2}{48YI}$$ in your nomenclature (but replacingd$with$\delta$). Here,$I$is the second moment of area or the area moment of inertia, which is a measure of how far the regions of a beam's cross section lie from the ... 4 Where did I err? The equation$x=\frac{F}{A}\frac{L}{Y}$applies to each infinitesimally-thin horizontal layer of the wire. If the cross-sectional area$A$of the wire were constant then$F$(= weight of wire below any layer) would increase linearly with$y$, the height from bottom to top. Because the only variation (that in$F(y)$) is linear, the total ... 4 So your thought about statistics would be correct if not for one basic fact about atoms and molecules: Molecules are sticky. There's actually a couple of different ways that molecules can be sticky, which is why when you put oil into water the oil sticks to itself and the water sticks to itself, but they don't stick to each other. In fact when things are ... 4 All you need is the Navier's equation of motion (you can consult it in any book of elasticity) $$\rho \frac{\partial^2 w}{ \partial t^2 } = \mu \nabla^2w + (\lambda + \mu)\nabla\nabla.w$$ Naturally, you can decompose a wave$ w$in a transversal and a longitudinal part: $$w = w_L + w_T$$ with the following properties : $$\nabla \times w_L = 0$$ ... 4 As you have found, the von Mises stress only depends on the stress deviator, i.e., if you change the stress state by$\sigma\to \sigma+\lambda \mathbb{1}$for arbitrary$\lambda$, the von Mises stress stays the same. Hence, the von Mises criterion is a pure shear yielding criterion. For many materials, you don't expect failure for isotropic compression (i.... 3$\mu$is quite easy - it is the shear modulus. In engineering texts it is often called$G$. I do not believe that$\lambda$has a straightforward physical interpretation. However, the bulk modulus$\kappa=\lambda+2\mu/3$, so it is sometimes useful to think about$\lambda$as something closely related to the bulk modulus. For example, the bulk modulus of a ... 3 Your basic premise is wrong. Compressing things makes them heat up. With gasses this is even nicely predictable over a wide range. Look up something called the ideal gas law. If you compress a gas enough then it will undergo a phase change and turn to liquid. This will release even more heat than just compressing the gas because the gaseous state of a ... 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 ...

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