I've run into confusion trying to answer the question "How much potential
energy is stored per unit length in a bent leaf spring, presumably as a
function of something like the local curvature?"
First, by way of example, here's how I approached the analogous problem for
a normal spring under compression. Say it has a relaxed length of d and is
compressed a distance x. Then we all know F=k*x. Now, if we double d and
keep x constant, each half of the spring has a compression of x/2, so F is
half as large. Similarly, if we double d and keep F constant, then x will
double. Thus we find that F=K*x/d, with K independent of d. Now to get
the energy we can just integrate the work of compression:
E = int (F*dx) = (K/2)*x^2/d. Rewriting E=(K/2)*(x/d)^2*d, E is
proportional to d and the compression ratio, and everything makes sense.
I think the word is that E is extrinsic.
Now, try the equivalent thing for a leaf spring of length d, deflection x,
and 'small' angular deflection phi. Assuming that the shape of the spring
for given phi is independent of scale, x=c*d*phi for some c. (For small phi
and a circular arc shape, c=1/2. But of course that's not the shape of a
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.
I don't see any way to make this extrinsic (proportional to d).
So what did I do wrong? I know that I cut some corners in the part where I
imagined doubling the length, related to the fact that the shape of a
spring of length 2d is not the same as the shape of 2 springs of length d
end to end. But I thought that for small phi I might get away with that.
I certainly didn't expect that anything worse would happen than getting a
incorrect constant.
--
--James McLean
jmclean@chem.ucsd.edu
post doc
UC San Diego, Chemistry