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person123
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- I derived an equation for the frequency of a rectangular prism oscillating due to a shear restoring force, but I don't know if it's correct.
Hi. I'm trying to determine the frequency of an block (roughly a rectangular prism) when the oscillation is due to a shear restoring force. Here is a diagram:
In the derivation, ##\rho## is the density of the block,##G## is the shear modulus of the block, ##y## is the elevation of the element of the block, ##F## is the horizontal shear force, ##a_x## is the horizontal acceleration of the material element, and ##\Delta x## is the horizontal displacement of the material element.
I assume that the block does not undergo any bending (I think this is a good approximation if H is small). There is a shear force on the base of the block causing it oscillate. I look at a small part of the block and analyze the forces on it. ##\frac {\partial F}{ \partial y}## relates to the acceleration of the small mass:
$$\frac {\partial F}{\partial y}dy=dm a_x$$
$$\frac {\partial F}{\partial y}dy=\rho dy W L a_x$$
$$\frac {\partial F}{\partial y}=\rho W L a_x$$
Because:
$$\tau=G \Delta\theta$$
for small angles:
$$F_{bottom}=G \frac{\Delta x}{y}WL$$
The shear force must vary between this value and 0 at the top along the height of the block. ##\frac{\partial F}{\partial y}## must be proportional to the distance from the bottom (the higher up the element is, the more it is displaced, so the more it accelerates, and the relation is linear assuming SHM and no bending). This means the force profile must be the following (I'm skipping the steps but it's easy to confirm):
$$F=F_{bottom}(1-\frac{y^2}{H^2})$$
This means:
$$\frac{\partial F}{\partial y}=- \frac{2y G\Delta x WL}{y H^2}=-\frac{2G \Delta x WL}{H^2}$$
Plugging this into the previous equation:
$$\rho W L a_x=-\frac{2G \Delta x WL}{H^2}$$
$$a_x=-\frac{2G}{\rho H^2} \Delta x$$
This is simple harmonic motion, so the frequency would be:
$$f=\frac{1}{2\pi}\sqrt{\frac{2G}{\rho H^2}}$$
I'm interested in whether my approach is correct (I couldn't find an equation for this when searching online).
In the derivation, ##\rho## is the density of the block,##G## is the shear modulus of the block, ##y## is the elevation of the element of the block, ##F## is the horizontal shear force, ##a_x## is the horizontal acceleration of the material element, and ##\Delta x## is the horizontal displacement of the material element.
I assume that the block does not undergo any bending (I think this is a good approximation if H is small). There is a shear force on the base of the block causing it oscillate. I look at a small part of the block and analyze the forces on it. ##\frac {\partial F}{ \partial y}## relates to the acceleration of the small mass:
$$\frac {\partial F}{\partial y}dy=dm a_x$$
$$\frac {\partial F}{\partial y}dy=\rho dy W L a_x$$
$$\frac {\partial F}{\partial y}=\rho W L a_x$$
Because:
$$\tau=G \Delta\theta$$
for small angles:
$$F_{bottom}=G \frac{\Delta x}{y}WL$$
The shear force must vary between this value and 0 at the top along the height of the block. ##\frac{\partial F}{\partial y}## must be proportional to the distance from the bottom (the higher up the element is, the more it is displaced, so the more it accelerates, and the relation is linear assuming SHM and no bending). This means the force profile must be the following (I'm skipping the steps but it's easy to confirm):
$$F=F_{bottom}(1-\frac{y^2}{H^2})$$
This means:
$$\frac{\partial F}{\partial y}=- \frac{2y G\Delta x WL}{y H^2}=-\frac{2G \Delta x WL}{H^2}$$
Plugging this into the previous equation:
$$\rho W L a_x=-\frac{2G \Delta x WL}{H^2}$$
$$a_x=-\frac{2G}{\rho H^2} \Delta x$$
This is simple harmonic motion, so the frequency would be:
$$f=\frac{1}{2\pi}\sqrt{\frac{2G}{\rho H^2}}$$
I'm interested in whether my approach is correct (I couldn't find an equation for this when searching online).
Last edited:
