1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Impact Force of a Horizontal Moving Projectile

  1. Feb 27, 2017 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    What is the correct way to calculate the impact force of a 2"x4" flying through the air and hitting a solid cross beam? The knowns are:

    Velocity: 50 ft/sec

    Distance To cross beam: 12 ft

    2" x 4" Weight: 9 lbs

    I don't know much the cross beam will deflect on impact, so the stopping distance is relatively unknown. I was planing on using F = KE/D = .5 * m * v2 / D, but without knowing the stopping distance after impact, I can't use it. Does anyone know how I should proceed?
     
  2. jcsd
  3. Feb 27, 2017 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hello J, :welcome:

    Funny you mention the 12 ft (irrelevant) but say nothing about the cross beam.
    Nor about how the 2x4 is oriented wrt the beam. No drawing ?

    But you have the right plan: find out how much that cross beam deflects. Load of course depends on acceleration, so you have to iterate a bit.
     
    Last edited: Feb 27, 2017
  4. Feb 27, 2017 #3

    Nidum

    User Avatar
    Science Advisor
    Gold Member

    In most cases effective mass of the beam has to be taken into account as well .

    There are analytical and numerical methods for dealing with this type of problem .
     
    Last edited: Feb 27, 2017
  5. Feb 27, 2017 #4
    I did a beam bending calculation on the cross beam, but I am unsure of what size force to apply. Is there a range you'd suggest? I started with a load of 214 lbf (950 N), and found that the beam will deflect .0235" (.5981mm). The impact location is also the direct center of the cross beam (974mm/2).

    Capture.PNG
     
    Last edited: Feb 27, 2017
  6. Feb 27, 2017 #5

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Is this homework ? If not,
    I'm playing dumb and calculate a spring coefficient from that : ##k = F/dx = 950 / 5.8\times 10^{-4} = 1.6 \times 10^{6} ## N/m

    If the kinetic energy of the 2x4 is converted to compression of a spring with such a spring constant you've found the deflection and thereby the force exerted on the beam -- if that's still in the elastic range of the brass. If not, it gets bent or dented...

    I find yield strength 200+ MPa.
     
  7. Feb 28, 2017 #6
    This is not homework. I did try the spring method you suggested and got the following:
    ##KE = PE, (.5(4.082kg)(15.24m/s)^{2})/x = .5(1.588 \times 10^{6} N/m)(x^{2}), x=.084 m##

    Then I applied that distance to the spring constant and found my force to be ~133422N. This seems quite high for this. Am I doing something wrong?

    That yield strength is from the verified material properties from the vendor.
     
  8. Feb 28, 2017 #7

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I seem to remember other values: ##E_{\rm kin} ={1\over 2} 4.08 *15.2^2 = 474## J, so ##k ={ 2 E_{\rm kin} \over x^2 } \ \ \Rightarrow \ \ x = \sqrt {2 E_{\rm kin} \over k} = 24 ## mm.
    So at the end the force would be 40 kN for the elastic spring scenario. Not all that realistic, but a ball-park figure
     
  9. Feb 28, 2017 #8
    Ah yes, I forgot to get rid of the stopping distance in my original formula. So my force would be ~38120.71N, correct?
     
  10. Feb 28, 2017 #9

    When I apply that force, I get 24mm of deflection, a Max Stress of 2791 MPa, which would mean my F.S. is .06. So the cross beam will be bent significantly?
     
  11. Feb 28, 2017 #10

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I suppose that's what you get for stopping a 2x4 that moves at 50 km/h within 2.4 cm :redface:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted