Wall made of Pendulum -- thought experiment

AI Thread Summary
A wall designed as a pendulum or flexible structure is less susceptible to seismic forces compared to a rigid wall, as it dissipates energy over a longer time, reducing peak forces. The physics behind this involves the mass and restoring force of the flexible structure acting as a mechanical resonator, tuned to avoid resonance with earthquake frequencies. This flexibility allows the structure to store and dissipate energy, minimizing stress during seismic events. The discussion clarifies that the term "attract" is misleading; it is more accurate to say that flexible structures are less affected by seismic forces. Overall, the mechanics of energy dissipation and structural coupling play crucial roles in seismic resilience.
Cobul
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If your house wall is made of pendulum or a ball attached to a string on top. It won't attract seismic forces. Whereas if the wall is fixed solid. It can attract seismic forces.

What is the physics explanation of it? Is it because you try to deflect momentum or inertia? What is the right description to explain it?
 
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What do you mean "attract" seismic forces?
That is not a meaningful word in that context.

Do you perhaps mean "susceptible to"?

Any would structure affixed to the ground,such as a wall, will be more susceptible to seismic tremors than something mechanically buffered, such as a pendulum, but pendulums don't make efficient walls.

I think you should elaborate on your question.
 
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Cobul said:
If your house wall is made of pendulum or a ball attached to a string on top. It won't attract seismic forces. Whereas if the wall is fixed solid. It can attract seismic forces.
What? Is "pendulum" a new element I've never heard of? What does a "wall made of pendulum" look like? And how do I insulate it?
 
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The pendulum is just metaphor for flexible wall. I reead that if a building is more flexibile. It won't attract much seismic force that when it is rigid. I just want to know how it deflects the energy. In the case of the wall using pendulum (just for sake of illustration). Why would the structure at least in the wall be less susceptible (if you want to use this language)? It's as if the pendulum or mechanical wall divert the momentum or inertia? I just want to know the physics description of it.
 
The mass of the building in conjunction with the restoring force of the springiness creates a mechanical resonator which is tuned below the main earthquake frequencies. It therefore presents opposition to fast movements. It can also be viewed as a mechanical high-pass filter.
 
Cobul said:
It won't attract much seismic force that when it is rigid. I just want to know how it deflects the energy.
In general, it's not so much about attracting vs. deflecting. It's more about dissipating the energy over shorter vs. longer time. Flexible structures store the energy and dissipate it over longer time, so the peak forces are smaller. In the case of oscillation you also need to avoid resonance with the driving frequency, so you need the right amount of stiffness vs damping.
 
One can also look at the problem as one of "coupling". If you mounted your building on roller bearing, for instance, the energy from the quake is not transmitted to the house and is, if anything, reflected away.

There is some impedance mismatch between a flexible building and a rigid earth. There is an even greater mismatch between roller bearings and a rigid earth.
 
Just taking your question at face value, I think it could be kind of interesting! Except I guess when you say "attracting [susceptible to?] seismic forces" it's more instead more meaningful to ask whether the structure would be subject to greater stress.

First, model the wall as either a vertical beam or a vertical plate. If the ground is subject to an acceleration ##\mathbf{A} = A_0 \cos{(\omega_d t)} \mathbf{e}_x## then you can model the situation by subjecting the structure to a uniform body force ##\mathbf{f} = - \rho A_0 \cos{(\omega_d t)} \mathbf{e}_x##. Taking the beam model as an example, this is equivalent to the problem of a horizontal cantilever beam, fixed at one end, and subject to a time-dependent gravitational potential ##U(y) = - \mathbf{f} \cdot \mathbf{x} = y \rho A_0 \cos{(\omega_d t)}##. Then you can just take the EL equation for the beam (assuming homogeneity of E and I)$$EI \frac{\partial^4 w}{\partial x^4} + \mu \frac{\partial^2 w}{\partial t^2} = q$$solve for ##w(x,t)## and from that determine the components ##\sigma_{ik}## of the stress tensor in the beam. Might be fun to try varying some of the parameters to see whether a flexible beam or a rigid beam will experience greater stress. [Maybe easier said than done, but hey, Mathematica exists for a reason!]
 
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