Impact Force of a Horizontal Moving Projectile

[JvandRWD]

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?

 

[BvU]

Hello J,

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.

 

[Nidum]

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 .

 

[JvandRWD]

Hello J,

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.

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).

 

[BvU]

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.

 

[JvandRWD]

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.

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.

 

[BvU]

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

 

[JvandRWD]

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.

Ah yes, I forgot to get rid of the stopping distance in my original formula. So my force would be ~38120.71N, correct?

 

[JvandRWD]

Ah yes, I forgot to get rid of the stopping distance in my original formula. So my force would be ~38120.71N, correct?

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?

 

[BvU]

I suppose that's what you get for stopping a 2x4 that moves at 50 km/h within 2.4 cm