"Ignoramus25660" wrote in message
...

On 2015-04-19, Ed Huntress wrote:
On Sun, 19 Apr 2015 23:18:02 +0000, Jman
wrote:

I need a steel rod/s. precision ground, polished, turned, heat treated
whatever is most effective. 3/8" OD and a length of 4'. I need to know
what kind of steel (1144, Stressproff, 4140 TGPHT would have the least
amount of sagging, twisting, and flexing (wasn't sure on the scientific
terms for these). Can you help me out with that one? Or maybe point me
to a graph or an equation to work it out?

Thanks in advance for your time,

Jeremy

p.s in the future I will need an eight or 10 foot rod

All grades will be the same (except for stainless, which will sag,
etc. slightly more). All hardness conditions will be the same.

I know this is hard to believe. When the second or third poster chimes
in, you'll start to believe it.


Can you explain why it is so?
================================================== =================

Iggy, the physical property that defines "stiffness" is called Young's
Modulus, or the modulus of elasticity. Pretty much all steel alloys have a
modulus between 28,500,000 psi and 30,500,000 psi, a range of 7%, so for
most calculation purposes "all" steel is basically 29,000,000 psi and
equivalent as far as stiffness goes. This does not change with heat treat
or hardness or work hardening or tensile strength or yield strength.
Similarly "all" aluminum alloys are 10,600,000 psi. Cast iron is about
10,000,000 psi for plain grey iron and up to about 15,000,000 psi for
nodular iron; that is one of the few materials whose modulus changes
significantly with minor changes in composition and heat treat. In terms of
shape, the stiffness is proportional to the fourth power of the lateral
dimension so 3/4" rod is 16 times stiffer than 3/8" rod, and pound for pound
thinwall large od tube is much stiffer than solid rod (of course, diameter
for diameter solid rod is stiffer than tube). For the OP, 4' or 3/8" rod
held parallel to the ground by supporting the ends is going to sag a good
bit under its own weight. The shape parameter that determines stiffness is
the moment of inertia (not the same rotational moment as for a spinning
object). For example, for 3/8" solid rod the moment if inertia I is
0.0009707 in^4, and 4' of 3/8" mild steel will weigh about 1.5 lbs. If you
support the ends by just sitting them on supports (as opposed to clamping
them so they can't pivot), and distribute that weight uniformly along the
length, the center will sag down 0.0767" and the maximum stress (based on
other calculations that vary with shape) will be 1740 psi, compared to a
yield strength of what, 40-50,000 psi so it is in no danger of permanently
bending. If you double the diameter the moment of inertia goes up by 16
(2^4) so it is 16 times stiffer, but the weight goes up by 4 (2^2) so the
net is one fourth the deflection (deflection is proportional to load/moment
of inertia if you keep the length constant) or 0.0192". The maximum stress
will be 870 psi. You can look up the moment of inertia for structural
shapes like rod, tubes, angles, I beams, and plug those numbers in, but for
each change of shape or dimension the maximum stress will have to be
recalculated from scratch if that is a concern (I'm too sleepy to take care
of stress and beam length, sorry). Usually it turns out that the deflection
is the limiting spec and by the time you get a design that keeps that down
below some limit you have such a strong structure that the yield rating is
overkill (like here).

As always, the program I recommend for quickie calculations like this is
engineering power tools, from www.pwr-tools.com. The freeware version will
do all of this with no time limit or ads, and the full version is only about
$50 and well worth it if you do a good bit of design.

-----
Regards,
Carl Ijames carl.ijames aat deletethis verizon dott net