Rigidity: Number of Newtons Required for a Material

[SamRoss]

TL;DR Summary

Is rigidity measured in units of force? If so, why?

In something I was reading, I saw the "stiffness" of a material measured in Newtons per meter which makes sense to me because something that is stiff would require more force to change its length. In other words, "Stiffness is the number of Newtons required to condense a material by one meter." However, I also saw that the "rigidity" of materials was being measured only in Newtons. I couldn't find any intuition about this. It might be helpful if someone were to complete the following sentence for me. "Rigidity is the number of Newtons required..."

 

[Chestermiller]

Young’s modulus determines the rigidity of solids. It has units of N/m^2 and maps deformational strain into stress.

 

[A.T.]

However, I also saw ...

Where?

 

[SamRoss]

Where?

https://www.explorelearning.com/index.cfm?method=cResource.dspView&ResourceID=1078

This is an activity I plan on doing with my middle school students.

 

[A.T.]

https://www.explorelearning.com/index.cfm?method=cResource.dspView&ResourceID=1078

This is an activity I plan on doing with my middle school students.

I don't know if that's what they mean, but:

You have to differentiate between rigidity of a material (where you normalize the force with the area it acts over) and the rigidity of a specific object (where you don't normalize by area). In the later case it would be force per relative length change.

 

[SamRoss]

I don't know if that's what they mean, but:

You have to differentiate between rigidity of a material (where you normalize the force with the area it acts over) and the rigidity of a specific object (where you don't normalize by area). In the later case it would be force per relative length change.

So what do you think would be a simple way of explaining it to a student (and to me ).

 

[A.T.]

So what do you think would be a simple way of explaining it to a student (and to me ).

When the stiffness is in N/m, they might be using Hooke's law constant:
https://en.wikipedia.org/wiki/Hooke's_law#For_linear_springs

When the stiffness is in N/m², they might be using Young's modulus:
https://en.wikipedia.org/wiki/Young's_modulus#Calculation

When the stiffness is in N, they might be using something in between, like Hooke's law, but with relative strain, instead of the absolute displacement.

But instead of guessing you should clarify with the authors what they mean. It might just be typo.

 

[snorkack]

It seems they are measuring the plastic yield strength of an object, and calling the result "rigidity".

When an object deforms reversibly under a small force, it often deforms according to Hooke´ s law, and the proportionality factor is expressed in N/m, and called "stiffness".
When the yield strength is exceeded, brittle objects break under one specific yield strength. Plastic objects undergo irreversible deformation over some length.
The exercise assumes that plastic deformation takes place over a long distance at a constant force, and calls that force "rigidity".

 

[Lnewqban]

So what do you think would be a simple way of explaining it to a student (and to me ).

These two articles could help you find that simple way:

https://en.m.wikipedia.org/wiki/Stiffness

https://en.m.wikipedia.org/wiki/Crumple_zone

The way I read the information shown in link of post #4:
A crashing force of 300,000 Newtons is needed to fully deform (nose to tail direction) the length of 90 cm of the crumple zone of the shown car.
If that is true, the stiffnes of that crumple zone would be 333 KN/m.

 

[SamRoss]

Plastic objects undergo irreversible deformation over some length.

A crashing force of 300,000 Newtons is needed to fully deform (nose to tail direction) the length of 90 cm of the crumple zone of the shown car.

Now it makes sense! Thank you!

 

[Lnewqban]

Now it makes sense! Thank you!

You are welcome.
The idea is to have a rigid strong cage surrounding the passengers of modern cars.
Forward and aft that cage, more easily deformable material (crumple zones) should absorb much of the energy of impact in the event of a crash.

Energy is work, and certain amount of work is consumed by metal and plastic that get permanently deformed.
That energy would never reach the rigid cage or the passengers.

 

[snorkack]

These two articles could help you find that simple way:

https://en.m.wikipedia.org/wiki/Stiffness

https://en.m.wikipedia.org/wiki/Crumple_zone

The way I read the information shown in link of post #4:
A crashing force of 300,000 Newtons is needed to fully deform (nose to tail direction) the length of 90 cm of the crumple zone of the shown car.
If that is true, the stiffnes of that crumple zone would be 333 KN/m.

Does not actually follow.
This would be the case if the crumple zone were to deform elastically, subject to Hooke´ s law, through the whole 90 cm.

But instead, the implication is that the crumple zone deforms elastically over a much shorter distance than 90 cm - and after reaching plastic failure limit, continues plastic deformation at a constant force.