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Hope this helps! The force of tension actually acts throughout the length of the wire from left to right keeping it together. It 'effectively' acts however, at the center of mass

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If the wire is ideal, the only force that can cause elongation is the net external force on the wire. i.e: mv^2/r + mg. Also however, the magnitude of net external force mv^2/r + mg is co-incidentally equal to net tension T just to keep the net force equal to zero. You say that the centrifugal force is the only force that causes elongation. But that would ...

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Obviously, the Tension is what always causes the elongation.Here at the lowest point, forces (under vertical equilibrium),(from wire frame of reference which is non inertial) net upward force equals net downward force T=mg + mrw^2 . Now if we look from outside the wire(inertial frame) then there is no centrifugal force .Here for the wire to rotate, the net ...

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I think you're talking about the shear modulus. (picture from the Wikipedia link above) If you take a block of some material and apply a sideways force $F$ the shear strain is defined as: $$\gamma = \frac{x}{l}$$ The shear stress is the applied force divided by the area over which it's applied: $$\sigma = \frac{F}{A}$$ And the shear modulus is ...

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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 ...

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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 ...

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If you have 3 eigenvectors of the stress tensor/matrix $T$, you can choose them as your new basis and it will be diagonal there - and no off-diagonal elements mean no shear stress, since the shear stress on the plane in the $ij$-direction ($i \neq j$) is given by $T_{ij}$, which, for $i \neq j$, will be zero in this basis.

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$\tau_{xy}$ and $\tau_{zy}$ act in the same direction but they act on different faces. This diagram should clear it up for you:

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