B Car Crash Physics: Deceleration Calculations

[erobz]

You could design in that characteristic but , if you want a more uniform retardation, then it wouldn't be optimal as the high force would be at the end, producing way above retardation. I just don't see why this would be a good car body design.

I didn't say that would be a designed characteristic force profile, I said that would be an un-designed characteristic force profile...perhaps what we would expect if we didn't intervein.

If the optimal force displacement curve was like this:

Then we can assume the car manufacturers have been able to achieve this type of response, which yields a virtually constant acceleration at some physiological threshold, and we could determine quite readily what that threshold is and/or if that assumption is correct by watching a vehicle crash and measuring the rate of acceleration.

 

[jack action]

I think we may be using the term 'yield point' inappropriately. I think we really need it the 'elastic limit'

That is the definition of yield point (and how I was referring to it in my previous posts):

In materials science and engineering, the yield point is the point on a stress-strain curve that indicates the limit of elastic behavior and the beginning of plastic behavior.

You may be confusing ultimate strength with yield strength:

[Ultimate strength] is the maximum stress that a material can withstand while being stretched or pulled before breaking. In brittle materials the ultimate tensile strength is close to the yield point, whereas in ductile materials the ultimate tensile strength can be higher.

And Work continues to be done on it.

And why can't we assume it still is proportional to displacement for the sake of simplicity?

One thing is sure: the force is still linked to the displacement. So let me correct my initial equation by not assuming the linearity:
$$F = Kx^n$$
Such that:
$$\frac{1}{2}mv^2 = \frac{1}{n+1}Kx^{n+1}$$
$$x = \left(\frac{n+1}{2}\frac{mv^2}{K}\right)^{\frac{1}{n+1}}$$
And:
$$F_{max} = Kx^n = \left(\frac{n+1}{2}K^{\frac{1}{n}}mv^2\right)^{\frac{n}{n+1}}$$
So, in retrospect, from good to better:

• Using $F = \frac{mv^2}{2x}$ - as in the OP - says nothing about the vehicle; all we can do is compile typical values of $x$ for typical vehicles.
• Using $F_{avg} = \frac{1}{2}\sqrt{km}v$ defines the force based on a - simple - design characteristic ($k$) of a vehicle.
• Using $F_{max} = \left(\frac{n+1}{2}K^{\frac{1}{n}}mv^2\right)^{\frac{n}{n+1}}$ (too lazy to determine the average force which would compare better to the previous equations) defines the force based on two design characteristics ($K$, $n$) of the vehicle; for the most demanding of us.

The point was that the method presented in the OP doesn't introduce any characteristics of the vehicle design other than the mass. Knowing that my neighbor's car produced a force of 100 kN with a displacement of 1 m doesn't mean my car will produce the same force with the same displacement (given equal mass and velocity). However, if my neighbor's car produces a force of 100kN with a known stiffness, I have much more chance that my car will produce the same force given my car has the same stiffness.

 

[sophiecentaur]

Then we can assume the car manufacturers have been able to achieve this type of response, which yields a virtually constant acceleration at some physiological threshold, and we could determine quite readily what that threshold is and/or if that assumption is correct by watching a vehicle crash and measuring the rate of acceleration.

I think we can assume that they wouldn't be spending a lot of research money and produce expensive designs which would not improve crash survival. Simulating a spring characteristic would be far from optimal desirable. That 'ideal' characteristic would make the least worst of a crash situation.

In a different context, we know that stunt people use piles of cardboard boxes to land on after stunt falls from unbelievable heights. Cardboard is 'not very springy'.

 

[Frabjous]

As I sort of said in my off-topic deleted post, 20G's is a ballpark desirable max average deceleration for the passenger compartment.
If we use the the equation $a=v^2/2d$ where d is the stopping distance
1) Using 20G for average deceleration, we get an elastic stopping distance of 1.3 cm for a 5 mph collision (assume inelastic deformation starts at 5mph)
2) Using the elastic stopping distance we get
Velocity (mph) Deceleration (G's)
10 39
15 88
20 157
25 245
30 353
35 481
3) If we limit the deceleration to 20G, we get stopping distances for the passenger compartment
Velocity (mph) Stopping Distance (m)
10 .051
15 .115
20 .204
25 .318
30 .458
35 .624
So the designers are probably designing for crushing to go something like this.

Here’s the plot I got the 20G from

 

[sophiecentaur]

That is the definition of yield point (and how I was referring to it in my previous posts):

Yes - you are right about that. Sorry.

And why can't we assume it still is proportional to displacement for the sake of simplicity?

That's what I have been saying all along - constant work / force with displacement. This is not the characteristic of a spring, for which the force increases in proportion to displacement.

You are suggesting that a non linear spring could be a suitable model. I guess there will be a power law which would give a characteristic near to that in post #31. The exponent would need to be fractional (a fairly small fraction). But, apart from the fact that you seem wedded to a spring model, how would you introduce actual net loss without something, somewhere, doing work on something? Perhaps a ratchet (like a catching diode in electronics) with added rebound friction (??) could absorb the energy on the way back.

 

[jack action]

That's what I have been saying all along - constant work / force with displacement. This is not the characteristic of a spring, for which the force increases in proportion to displacement.

I don't think the force will be constant anywhere. As the displacement increases, the force MUST increase by design. The closer you get to the vehicle's occupants, the more you want to stop the crumpling. I'm much more inclined to believe the stiffness increases with displacement.

For example, car suspensions are often built with bumper stops like so:

It's a rubber part that increases drastically the stiffness of the suspension before the suspension link hits the car frame to prevent permanent deformation of the parts.

I cannot imagine that a crumpling zone is not built the same way to protect the occupants. Start soft to absorb energy as much as possible, but end hard to prevent crushing the occupants. The former principle wasn't well understood by earlier car designers, only concentrating on the latter.

 

[Frabjous]

I cannot imagine that a crumpling zone is not built the same way to protect the occupants. Start soft to absorb energy as much as possible, but end hard to prevent crushing the occupants. The former principle wasn't well understood by earlier car designers, only concentrating on the latter.

I am under the impression that they achieve this with a ‘rigid’ passenger compartment with a crushable front and rear.

 

[erobz]

I am under the impression that they achieve this with a ‘rigid’ passenger compartment with a crushable front and rear.

The front is only crushable to an extent, after a point if there is still significant energy it just sends the engine through the firewall.

My only point is that different materials are continually compressed and brought up to yeild in the impact. As that is happening because of the increase in volume of plastically deformed material the force should tend to grow with displacement...I would think.

 

[Frabjous]

The front is only crushable to an extent, after a point if the is still significant energy it just sends the engine through the firewall.

I did not mean to imply that it would be successful at high velocity. If you look at my post 34, we have .624m of compression at 35mph which has to be close to maxing out the available free space of a normal sized vehicle.

 

[Frabjous]

increase in volume of plastically deformed material

Plasticity is frequently assumed to be a contant volume process at the material level for metals. In the case of the car crash, the free space is being removed.

 

[jack action]

I am under the impression that they achieve this with a ‘rigid’ passenger compartment with a crushable front and rear.

The "rigid" part has still a defined stiffness, even if it is high. And once the crushable part is half crushed, it is more and more a pile of solid "rigid" parts that has a higher stiffness than all the parts "uncrushed".

 

[erobz]

Plasticity is frequently assumed to be a contant volume process at the material level for metals. In the case of the car crash, the free space is being removed.

This is what I'm saying. If you are plastically deforming the blue pin with some force $F$, when you make it to the grey pin and still have sufficient energy $F$ is going to increase, and level out as the grey pin yields. At that point you will be deforming both pins plastically. That's what I meant about increasing volume of plastically deformed material as the impact progresses and more material is brough into the crumple zone.

 

[sophiecentaur]

For example, car suspensions are often built with bumper stops like so:

This is a very different design scenario. A suspension is designed to deform elastically with critical damping on a regular basis. Obvs, damage must not occur here and the springlike design is almost universal. I don't think it's relevant to the crash scenario; the owner just has to suck up the idea that his pride and joy will be spoiled after even a minor collision. We expect to walk away from many scary collisions these days.

I cannot imagine that a crumpling zone is not built the same way to protect the occupants. Start soft to absorb energy as much as possible, but end hard to prevent crushing the occupants.

That's more of an opinion about the best design philosophy. There will be a maximum acceptable G on the passengers and the way to avoid getting there has to be to give a rougher treatment for less serious impacts. There is a certain amount of kinetic energy to get rid of and going for less retardation early on will mean even worse forces for higher speed impact. I guess I have had that in mind all the time and you have had the spring model foremost in mind. In real life, there will be a compromise.

The way to deal with this on a personal level is to avoid high speed collisions by never going excessively fast and not being 'competitive' in ones driving. I live in a fairly rural area and seldom do more than 60mph (usually less than 40. It's great to pootle about and you can arrange it so that you are not in a hurry. (Easy for me to say, of course, but it's something to aim at. ).

 

[jrmichler]

The goal of crashworthiness is to absorb as much energy as possible using as little material as possible. A fellow grad student was doing crashworthiness research, and gave me a leftover steel test specimen:

It buckled into a series of rings, while plastically deforming the maximum amount of metal. This was from a high speed test in a test fixture powered by a shotgun shell with a double powder charge. The force to buckle each consecutive ring is near equal to the force to buckle the preceeding ring. This specimen shows near ideal energy absorption because a large portion of the metal is subjected to large plastic strain. While the force cyclically varies as each buckle is formed, the overall force is roughly constant over the entire buckling process.

Contrast the above specimen to the different buckling mode of a beverage can. This particular beverage can was picked up on a walk earlier after being donated by a local litterbug. Only a small portion of the metal was subjected to plastic strain, and most of the plastic strain was smaller magnitude than that in the specimen above. These cans absorb very little energy relative to the amount of material when stomped flat. Car manufacturers try to avoid this mode of buckling.

Side note 1: That same grad student tested a piece of PVC pipe in his test fixture. It shattered like glass, which absorbed very little energy, and destroyed his load cell.

Side note 2: I mentioned this test fixture in an earlier post. It's the one we shot with a rifle to perform a modal analysis.

 

[Frabjous]

A fellow grad student was doing crashworthiness research, and gave me a leftover steel test specimen:

Does it have a varying thickness or is it some other mechanism? Do you have a before picture?

 

[jrmichler]

Straight steel cylinder, 2.5" OD, 0.090" wall, unknown length before the test. He may have machined a slight taper on the bottom edge in order to encourage it to buckle the way he wanted.

 

[sophiecentaur]

The goal of crashworthiness is to absorb as much energy as possible using as little material as possible.

There is another consideration for motor cars etc. and that is to limit the maximum G forces on the 'cargo'. That implies more damage than desirable for collisions at lower speeds; there needs to be higher retardation, early on in the collision.
As little material as possible can be a factor, particularly in car design which seems to revolve around Power / Weight Ratio. Some cars (like Volvos) don't worry so much about that.