Re: leaf spring energy

[brian whatcott <inet@intellisys.net>]

At 10:44 4/13/98 -0700, you wrote:

> 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? ...
> --James McLean

Reviewing your comments in repose, I see it would not be particularly
helpful to regurgitate a procedure for sizing a cantilever spring
gear/undercarriage.
 This would simply specify an expression for a cantilevered beam as a given.
   What we are considering here is the nature of stresses and deflections
in engineering materials - a subject that exercised some excellent French
theoreticians - before an appropriate expression developed.

I cannot do better than remind the readership of an excellent paperback -
actually two of them - written by (Professor) James E Gordon.
"Structures" and "The New Science of Strong Materials".
 This is material more gripping than a mystery, more relevant than watching
"Titanic", and engaging to the attention.

 There, one finds in a slim appendix, the royal road to reasonable stress
calculations - and I will echo his notes on Beam Theory here:

The basic formula for stress s at a point P distant y from the neutral axis
of a beam is

s/y  =  M/I  = E/r

Where: s is tensile or compressive stress
       y is distance from the neutral axis
       I is second moment of area of cross-section about the neutral axis
       E is Young's modulus ( or stiffness or inverse springiness, if you
will)
       r is radius of curvature of the beam at the section we are examining.
       M is 'moment' or force times perpendicular distance from section of
          interest.


Hmmm...moving right along - we can see that the radius of curvature depends
on the moment (inversely) and so this is evidently not an arc of a circle -
unless we do what Steve Wittman proposed to Clyde Cessna (or his heirs) - 
and that is to taper the spring section to keep the stress constant, while
the moment changes with distance from the supported end.

Well, despite my best intentions, I still have not provided a
satisfactorily direct answer to the question - but I had better stop here
before I bore you all.

Whatcott   Altus OK