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If we assume that there is an energy density function $U$, then we have $$S_{ij} = \frac{\partial U}{\partial E_{ij}}\, ,$$ where $S$ and $E$ are thermodynamically conjugated, i.e., their product represents work done to deform the body. For small strains, we can use a Taylor expansion and conclude that we can write the stiffness tensor as $$c_{ijkl} = \...


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Looking at the net forces acting on an objects tells you how the object will move; but there are still internal reactions that can change the internal energy of the system without causing the center of mass to accelerate (i.e. no net force acts on it). This is what stress is doing. It's actually changing the magnitude of the internal forces that the rod ...


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The forces in the layers do not add up. Essentially, you have a chain of springs - if you stretch the chain on both ends, the force in each spring is the same. Otherwise, the point between the first and the second spring would be puled towards the centre, since the force in that direction would be $N-1$ times the one pulling towards the end. Forces do add ...


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Since strain is defined by \begin{align} \epsilon = \frac{\Delta l}{l} = \frac{l - l_0}{l} \end{align} where $l_0$ is the initial length of the object, you can see what the sign should be. Be careful with expressions like $\epsilon = \text{d}l/l$ since this is not an equality because a differential can only be equal to a differential of the same order. For $...


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I asked myself the same question some time ago. This is maybe not as straightforward as you might think, and this is best obtained from an intermediate step of "structural" mechanics equations relating moment resultants and efforts in a rod: $$ \frac{dM}{dx}= V \tag{1}$$ $$ \frac{dV}{dx}= \rho A \frac{d^{2} v}{ dt^{2}} \tag{2}$$ (2) where $M$ is the bending ...


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There's no real dividing line between "elastic" and "inelastic." Like many things in physics, its actually a continuous scale between them. We say a collision is "elastic" when the conservation of energy can be used to predict the results of the collision. If enough energy is converted into heat or deformations that we don't get good predictions out of ...


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This paper addresses exactly that question https://www.nrcresearchpress.com/doi/abs/10.1139/cjp-2015-0378 A free version is available on ResearchGate https://www.researchgate.net/publication/281791329


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"Hit" is just a more prosaic word for "collide". In English-language academic usage "collide" and "collision" have specific meaning. "Hit" has so many colloquial meanings that it could be confusing -- hence the overarching propensity for polysyllabic utterances by physicists (and why they use so many big words, too).


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