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Breaking concrete manually.

  1. Jun 25, 2010 #1
    My first message with P.P. Please be understanding.
    We are manually breaking a large concrete slab and I would be interested to know, roughly, what weight would be applied on impact by a sledge hammer weighing 14 pounds (6.4kg) travelling down at, say, 50mph (22m/s). Assume that the slab is on solid ground and will not sink on impact. Have I worded the question correctly?

    Dave.
     
  2. jcsd
  3. Jun 25, 2010 #2
    You mean "what force would be applied on impact". force rather than weight.

    F = rate of change of momentum. How much time does it take the hammer to stop after hitting the slab. You might probably need a high speed camera to find that out.
     
  4. Jun 25, 2010 #3
    The question is, why would you want to work this out, what possible reason do you need the answer for?
     
  5. Jun 26, 2010 #4
    No! The question is " What would the force be on impact", and is simply a matter of interest. I work in the building trade, and was curious to know the answer. Don't any of you scientists know the answer?

    Dave.
     
  6. Jun 26, 2010 #5
    Sorry. Yes I suppose I do mean force. But how do I work it out? Hasn't the hammer stopped when it's hit the slab of concrete?

    Dave.
     
  7. Jun 26, 2010 #6
    The moment it hits the concrete, the concrete starts to break up and the hammer moves a little further before completely stopping. You could try to measure how much the hammer typically moves after hitting the concrete and estimate the average force from that.

    Most of the force will actually be exerted on the ground. If you suspend the slabs of concrete between two support points and then break them (like people who practice Karate do with their arms), your hammer will move through the slabs with litle change in velocity, so the exerted force due to the impact will be far less.
     
  8. Jun 26, 2010 #7
    "The amount of energy delivered to the target by the hammer-blow is equivalent to one half the mass of the head times the square of the head's speed at the time of impact"

    Source: http://en.wikipedia.org/wiki/Hammer
     
  9. Jun 26, 2010 #8
    I'll take a crack (haha!) at this, although I ought to point out I'm not an expert in breaking concrete (although I've broken plenty of other things in the lab!), so don't stake anything important on this elementary guess.

    First imagine the slab doesn't break:
    Upon hammer impact, a small compression and then expansion wave should go through the concrete with speed w (speed of sound in concrete). Assuming a slab of length L in the direction of impact, the time it takes the wave to propagate through the concrete will be the time the hammer is being "stopped" by the concrete, so dt = 2L/w. If the impact is perfectly elastic, the AVERAGE (not max) force exerted will be:

    dp/dt = 2mv/(2L/w) = mvw/L

    So, if you were to "step" the amount of momentum in a series of experiments from small to large momentum, it seems to me that the calculated average force at the point where the concrete just breaks would be the "breaking strength" of concrete.

    This is, of course, a poor man's way of approaching this problem. I'm sure real scientists/engineers use high speed cameras and gauges to solve this problem, and theorists have complex models to answer this question. I'm sure my simple way of thinking about it is highly inaccurate. However, I'd be interested to here from an expert just how bad this approach is...
     
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