Impact load

[dcredskins]

If I drop a mass of 100 lbs from a height of 5 ft to a 10 ft simply supported beam (at the middle), how do I find out the actual moment/effect due to impact? I would appreciate help. I know it has something to do with impact loading, but I am not sure how to analyze.

 

[csd72]

Greg,

so what is k, is that the equivalent spring constant?

 

[Dinosaur]

Greg,

Thanks for checking that for me. It turns out I was mistaken in mmy math, however. I think I accidentally squared the constant 'k' when I used my calculator to get the number, so I divided by k^2 at one point rather than 'k'.

While waiting for your response, I looked it up in a few books. Since after impact the problem becomes a 'free vibration' problem, if we neglect damping it may be assumed the beam oscillates as a sine function wrt time.

u(t) = A x sin (w x t) + B x cos (w x t) + C

I believe the term that captures the kinetic energy in the falling weight is the first term, A x sin (w x t). The constant 'A' is the value of the deflection needed to absorb the K.E.

The other two terms capture the deflection resulting from the system going from one equilibrium state to a new state, and this is the consequence of loading the beam in an instantaneous manner. The values of 'B' and 'C' are the magnitude of the static deflection, although they have opposite signs.

CSD,

'k' is the spring constant of the SDOF system.

 

[GregLocock]

k is just the spring constant of the beam, w/delta=8EI/L^3 from memory.



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Tomfh]

48EI/L^3

?

 

[apsix]

csd72

"the longer the period of time a given force is applied the greater the acceleration "
Surely you mean velocity, not acceleration. If you have a constant force acting on a constant mass the acceleration will be constant.

"In statics the period of time does not matter but in dynamics it is essential."
That's generally true of course, but I was responding to Tomfh who seemed to imply that the rate of deceleration doesn't affect the beam's capacity because the same amount of work is done.

 

[NKT]

Dinosaur:

I think you may have forgotten to convert pounds (force) to slugs (mass). The calculated deflection should be around 7" assuming a DF#2 2x4 with E = 1,600,000 psi

The equivalent static load is 1668lb, with a resulting bending stress of 16,338psi.

Also, calculating the kinetic energy is an unnecessary step. Just equate the potential energy before the mass falls to the potential energy after the mass falls.

Note that, for more accuracy, the deflection of the beam could be added to the fall height and the calculation iterated until the before and after potential energy numbers converge. Generally, I would consider the extra distance added by beam deflection a second-order effect and neglect it, but 7" of deflection after a 60" fall is not negligible.

 

[Tomfh]

NKT, using this boiled down equation I get almost exactly the same result as you.

moment = sqrt (6 P E I h / L)

moment = sqrt (6 * 100 * 1.6e6 * 5.359 * 60 / 120)

= 50718 in lb.

Bending stress = 50718/3.0625 = 16561 psi

I'm curious about the 1% difference in our answers, what input values did you use?


Note that if you include the extra 7 inches drop height the moment is about 5% higher. This will be well under the energy lost as sound etc.

 

[NKT]

Tomfh,
we used the same values; I suspect rounding differences. I didn't take the time to come up with a streamlined formula, so there were several intermediate steps with the answer rounded on each.

I went ahead and did the analysis with the deflection included in the energy calculation, and iterated until the "before" and "after" energy values converged. I got a 6% increase in bending stress.

 

[umph]

NKT,
Ho do I calculate the assumed deflection (delta)?