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Re: leaf spring energy



David Dockstader says:

On Mon, 13 Apr 1998 10:44:15 -0700 (PDT) James Mclean said:
real leaf spring.) Now, F=k'*phi=k*x. Using the same kind of reasoning as
above, (double d, F const => double phi) I get F=K*c*phi/d=K*x/d^2. But
then the work integral gives E = (K*c^2/2)*phi^2 = (K/2)*(x/d)^2.

Here is where I think you have run amuck. F=k*x=k'phi only for d constant.
Double d, F constant will not double phi, in fact phi will hardly change at
all. I'm not sure why you want to write this in terms of phi anyway. Why
not just use the same equations as for the coil spring?

This is certainly the weakest link in my chain, so let me fortify it.

First, a little experiment: I took a metal ruler (stiff enough that it
doesn't bend under it's own weight), attached a heavy clamp to one end, and
let it project over the edge of a book. The results:
Extended length d (mm) : 20 40 60 80 100 120 140
Angular deflection phi (degrees): 2.0 6.5 11.5 17.5 24.5 33.0 39.5
Except for the fact that my 'd' was systematically too long by about 11mm
(due to placement of the clamp), I was astounded at how linear this was.

Why did I expect this? Say a force F on a 10cm spring gives angular
deflection phi. Now, for the same force on a 20cm spring, consider the
outer 10cm of it. That outer half is static, so the total force on it is
zero. This implies that the force of the inner half on the outer half is
-F, and (for small phi) is almost entirely a *shear* force. By N3, the
inner half of the long spring is exactly the same as the entire shorter
spring, so it has deflection phi. Of course, the outer half also bends.
So for constant force, a double length spring will have double the angular
deflection. This argument only really makes any sense for small phi, which
is why I was so surprised that my experiment worked so well.

(An aside: it took me quite some thinking to realize the existence of this
shear force. Perhaps this is one for the misconception list? A leaf
spring does *not* support/push things by compression/tension along the
spring. That's incompatible with N3.)

Besides the fact that I found the angular deflection 'phi' easier to think
about in the above argument, there's actually another reason to consider it
instead of the distance deflection 'x'. My guess as to the important local
property of a bent leaf spring is the curvature, which is given by the
derivative (d phi)/(d d). (I've been trapped by my notation, but I hope
that's clear.)

--
--James McLean
jmclean@chem.ucsd.edu
post doc
UC San Diego, Chemistry