Impact what? Force or Energy?

  • Thread starter Aragorn49
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  • #1
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Impact .... what? Force or Energy?

I have a question, about what to calculate if calculating whether an immovable object (a wall for example) will break or not after an impact of another object into it. What should I calculate, the force of the impact or the kinetic energy of the impacting object?

I know that kinetic energy is calculated as E = 1/2 m v2 and that impact force is calculated as F = m v / t (duration of impact). But which one should I calculate? I used to think that impact force should be calculated and from that I calculate the internal stress of the struck object, but if this is true, why are they calculating absorbed energy in izod and charpy tests?

Thank you for your time
Aragorn49
 

Answers and Replies

  • #2
Mech_Engineer
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The safest answer is probably both, but energy is probably the more useful. Peak force can be useful for finding dynamic reaction acceleration, while energy can be used to determine things like total deformation, and material failure probability.
 
  • #3
AlephZero
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A better criterion would be the energy density in the objects, taking into account the size of the impact area, the duration of the impact, the speed at which energy travels through the objects (i.e. the speed of sound in the materials), the amount of energy absorbed in plastic deformation or converted to heat, etc

In a particular scenario like impact testing, you try to create test conditions that control the situation so these factors always have the same effect on the outcome.
 
  • #4
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I don't know enough enough about brick walls to speak concerning that, but in gas turbine design we have to make sure the case is strong enough to contain a blade if one were to break free of the disk. We first bracket the solution with hand calculations based on strain energy. That drives the initial design, which is then turned over to the Stress department for further design optimization using computer modeling tools developed for that purpose. This requires a great deal of propriety test data that was very expensive to collect concerning how the material behaves under an impact load. High speed dynamic behavior is much different from low speed behavior. Some ductile material will behave as a glass when impacted.
 
  • #5
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Impact is usually localized, and therefore the wall will exhibit localized damage at the location of the impact. This is cumbersome (basically, impossible) to calculate by hand. Thus, advanced computer modeling tools would be used as Pkruse mentions. For example, "finite element analyses" would be performed with a tool like ABAQUS or LS-DYNA.

Having said that, I'll give you something to try.
NOTE: My explanation below is ONLY valid if this is some situation like a semi-truck backing into a masonry wall. You gave no details. If you are shooting a tank round at a wall, then forget it.. far too "localized!"

So, IF it is valid to assume that the wall will remain undamaged and undeformed during the impact (certainly no "localized" damage during the collision), then one can do a hand calc as follows:

1) Determine the impulse of the impact - it sounds like you have all of the variables that you need in order to do this (e.x. if it is an elastic collision, then change in velocity of the offending object, Δva, multiplied by the mass of this object, ma, would give you the desired result). Is this reasonable so far?

2) This impulse gives the wall an initial velocity (it is not "immovable" as you say - nothing is "immovable"). Using the impulse-momentum equation again, but this time for the wall: find the initial velocity of the wall, vb, from the impulse and the mass of the wall, mb.

3) Now that the wall has kinetic energy, use the kinetic energy-work principle:
[TEX]\frac{1}{2}m_b*v_b^2=\text{WORK} [/TEX]

4) The wall is like a spring. It has a stiffness, K, and so the final expression that you need is:
[TEX]\frac{1}{2}m_b*v_b^2=\frac{1}{2}*K*x^2[/TEX]

In this expression, you will know vb from my explanations #1 and #2

*You will know K as well.

You solve for "x" - the displacement of the wall.

**If this displacement is larger than the wall can handle, then it will be damaged and possibly fail.

I have no doubt that you can perform #1,#2,#3. In order to evaluate #4, you need to be able to determine "K" as well as the failure displacement of the wall. I gave them asterisks because this is the part where you have to understand the concept of bending as well as understand how certain materials behave.

Any structural engineer could evaluate this (assuming it is masonry, reinforce concrete, or some sort of stud wall).

The steps #1-4 are commonly done in back-of-the-envelope blast analysis.
 
  • #6
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The safest answer is probably both, but energy is probably the more useful. Peak force can be useful for finding dynamic reaction acceleration, while energy can be used to determine things like total deformation, and material failure probability.
By peak force you mean the force that is calculated via F = m x v / t, or is the peak force another form of force?
 
  • #7
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I personally would just use F = m v / t to get a point force and then apply that to an FEA model of the object. If it's not a point force can easily just convert it to a distributed one.
 
  • #8
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I thank you all for your responses.
 

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