From: "Tom Krajci" 
To: 
Cc: 
Reply-To: "Tom Krajci" 


>From: "Roy Diffrient" 

>...The deflection is calculated by the formula
>W L^3 / (c^2 A E)

>where

>W = load
>L = truss member length
>c = base spacing between members
>A = cross sectional area of the material in a member
>E = elasticity modulus (for steel, 30 million psi)

I'm gonna raise a really obvious point...but I need to.

In the above equation, there is no mention of the second moment of inertia
of the individual truss members...which is the ability of the truss tube to
resist bending/flexing.  (Second moment of inertia scales with the 4th
power of the radius.  Boost your tube diameter by 50% and you get a tube
that is (1.5 ^ 4) 5.06 times stiffer!...yet this is not reflected in the
above equation.  Only the surface area of the truss tube cross section
seems to have an impact in the above equation...and only at the first
power...which means that 50% larger diameter truss tubes gets you a truss
that's only 50% stiffer.  (Yeah, and a good bit heavier too.)

Have I been operating under a bad impression on how to best increase truss
stiffness?  The best way is to shorten the truss (stiffness scales
inversely to cube of length)...or widen the truss (stiffness scales as the
square of the width).

Hmmmm...looking a the Kriege/Berry book, the chapter on engineering
concepts/truss stiffness, etc....it seems to be wrong.  They mention 50%
larger truss tubes are 5 times stiffer (they are, as individual
members)...but according to the above equation...only 50% stiffer as an
*entire truss system*.

Do I have this right?

Ok, then 50% wider truss tubes...something else changes about the truss by
a factor of five...yes?...perhaps ultimate yield strength?...or not?

Tom Krajci
Tashkent, Uzbekistan


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