Mechanics, Energy, Momentum, Impulse, Elasticity

[brian whatcott <inet@INTELLISYS.NET>]

Folks,
I came upon this question in a newsgroup recently (sci.engr.mech)
It looks like it's one of your things ( a homework assignment even?)

 chucka,chucka@ezaccess.net says...
> Am in need for approach to determine impact force as the result of a three
> pound weight dropping onto the center of a simply supported 0.5 inch
> diameter steel bar(1212).
> The distance between supports is 23 inches.
> The drop height is 1 inch.
> Thank you.

This is how I responded:
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deflection for a simply supported beam:
1/48  WL^3/eI
(W force    L length of beam      I Moment of Inertia e Young's Modulus)

If this were a linear spring,
the energy stored would be 1/96  W^2 L^3/eI
(i.e 1/2 force times deflection)

In SI units and E notation (e.g. 1.23E2 = 123)
Kinetic energy available is mgh = 1.36  9.8  2.54E-2
(m mass kg      g 9.8 m/s^2       h height meters)

So W^2 = 96  1.36  9.8  2.54E-2   e I/L^3
              = 162.5  eI
I = pi r^4/4 = 1.3E-9
(pi = 3.142     r  radius of round beam     L length between supports)
e = 2.1E11 N/m^2

Therefore:
e I  = 2.7E2
W^2 = 4.4E4
W = 2.1E2 newtons   or 47 lb force

I strongly recommend you treat this estimate as grossly in error
and post a more reasonable method!!!   :-)

(Brian)

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One respondent, on the basis of his instrumented tests felt that
such an object would be stopped in one to ten milliseconds, and
 he estimated a force of 72 lbf.

I expect a list reader can do much better than these efforts.
 If so, I would be pleased to hear of a more appropriate method...

Brian


brian whatcott <inet@intellisys.net>
Altus OK