Physics question momentum, energy, strength of steel, PSI etc

[trampoline]

1.) If a car is in a frontal crash test, and goes hurling into a steel wall for instance, what unit of measure is used to define the strength of the steel wall? For instance, is the resistance strength of the steel measured in tensile strength, yield strength or something else?

2.) Another example... if the car is traveling and its energy is measured in foot pounds [lbf / ft^2] or even in joules...and it smashes into the steel wall...how would the walls resistance to the cars impact be measured?

3.) Let's say, this steel is measured perhaps in yield strength (ksi or thousands of pounds per square inch) is the force per square inch where deformation exists in the steel.

How it this to be understood? Let's assume that an imaginary vehicle is so heavy/dense that it is capable of smashing through the steel impact wall. The vehicle is 10 feet wide and 5 feet high. How is the size and weight of the vehicle divided up against all those square inches of steel to allow a hole to form?

I guess what I'm trying to understand is if the strength of the steel is measured according to thousands of pounds per square inch, is that "single unit's" strength multiplied by the total area that the imaginary vehicle impacts it and smashes a hole in the steel?

What formula would be used to compute this?
This sprung from my kids homework (not a homework question in and of itself) and we're just trying to understand how this works in theory.

Hope this makes sense. Thanks.

 

[JoshuaFarrell]

With metals its not about resistance its about ductility. The ductility of a metal is how easy it is for the atoms to move, metals are ductile because they have delocalised electrons.

 

[AlephZero]

If the objective is to stop the vehicle, then you have to make all of its kinetic energy do some work, that is not reversible. (As an example of a reversible device, suppose the "wall" was actually a thick steel plate mounted on a huge spring. The kinetic energy of the car would compress the spring and the car would stop, but the spring would then extend again and throw the car off in the opposite direction)

The easiest way to do this is to deform the wall and the car plastically (just like bending a metal paperclip so it doesn't spring back to its original shape). To get plastic deformation, you have to exceed the yield stress of the material. But if you exceed it by too much, you will break the material, or punch a hole through it. So the questions you have to answer to design the wall are first how much volume of material you need to deform plastically to take up all the energy, and then where to put that material so it actually does absorb the energy without breaking.

A lot of the energy will be is absorbed in plastic deformation of the car, as well as the wall. The front of a car is designed to "crush" and absorb energy, but without transmitting enough force to crush the passenger cabin (and the passengers.)