Equivalent stiffness and damping

In summary, the scheme of springs and dampers given can be simplified by using equations for stiffness in series and parallel, and similarly for dampers. The textbook states that k3 is parallel to k1 and k2 and can be summed up, while c1 and c2 are in series and can be calculated similarly. Going back to Newton's laws, it can be seen that this is correct.
  • #1
Hood
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Homework Statement


A scheme of springs and dampers is given. What is equivalent stiffness and damping?

vibrations.png


Homework Equations


For stiffness in series: keq=ka*kb/(ka+kb)
For stiffness in parallel: keq=ka+kb
For dampers similarly.

The Attempt at a Solution



To me it looks like k1 and k2 are in parallel. After summation I'd say that k3 is in series and calculate for that. Dampers c1 and c2 seem to be in series so I'd calculate them similarly.

I'm puzzled because textbook says that k3 is parallel to k1 and k2 and simply sums it up (keq=k1+k2+k3). I doubt it's like that, could you also explain about c1 and c2?
 
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  • #2
Hood said:

Homework Statement


A scheme of springs and dampers is given. What is equivalent stiffness and damping?

vibrations.png


Homework Equations


For stiffness in series: keq=ka*kb/(ka+kb)
For stiffness in parallel: keq=ka+kb
For dampers similarly.

The Attempt at a Solution



To me it looks like k1 and k2 are in parallel. After summation I'd say that k3 is in series and calculate for that. Dampers c1 and c2 seem to be in series so I'd calculate them similarly.

I'm puzzled because textbook says that k3 is parallel to k1 and k2 and simply sums it up (keq=k1+k2+k3). I doubt it's like that, could you also explain about c1 and c2?

Best thing to do is go back to Newton: mx'' = ƩFx.

Call x = 0 when the mass is in its equilibrium position. If m goes up (x > 0) what are the forces acting on m?

You will see that your textbook is right ...
 

FAQ: Equivalent stiffness and damping

What is equivalent stiffness and damping?

Equivalent stiffness and damping is a concept used in engineering and mechanics to simplify complex systems into a single parameter. It represents the combined effect of stiffness and damping in a system, and is often used in the analysis and design of structures, such as buildings and bridges.

How is equivalent stiffness and damping calculated?

The calculation of equivalent stiffness and damping varies depending on the type of system being analyzed. In general, it involves using mathematical equations and principles to determine the combined effect of stiffness and damping in a system. This may include considering factors such as material properties, geometry, and external loads.

Why is equivalent stiffness and damping important?

Equivalent stiffness and damping is important because it allows engineers to simplify the analysis and design of complex systems. By reducing multiple parameters into a single value, it makes it easier to understand and predict the behavior of a system. This can save time and resources in the design process and ensure that structures are safe and efficient.

How does equivalent stiffness and damping affect the dynamic response of a system?

Equivalent stiffness and damping directly affect the dynamic response of a system. They determine how a structure will respond to dynamic loads, such as vibrations and earthquakes. A higher equivalent stiffness will result in a stiffer and more rigid structure, while a higher equivalent damping will result in a structure that is better able to dissipate energy and reduce vibrations.

How can equivalent stiffness and damping be optimized in a structure?

To optimize equivalent stiffness and damping in a structure, engineers may use various techniques and methods. These may include changing the material properties, adjusting the geometry, or adding damping elements, such as dampers or isolators, to the structure. The goal is to achieve the most efficient and effective combination of stiffness and damping for the specific needs of the structure.

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