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

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
=================================================================
One quick addendum now that I'm more awake, it seemed odd that the variation
of I with lateral dimension was the fourth power, not third, so I double
checked and yes, with solid round rod I varies as diameter^4 but for
rectangular cross sections, like turning down an edge on a piece of sheet
metal or using angles for stiffening, I varies as the cube of the height.
For tubing it varies with the wall thickness vs. diameter but it seems to be
between ^3 and ^4 of the diameter.
-----
Regards,
Carl Ijames carl.ijames aat deletethis verizon dott net

On Mon, 20 Apr 2015 07:06:27 -0400, Ed Huntress wrote:

AFAIK the physics behind it is that the Young's modulus is about how the
atoms interact wherever they may be in the crystals, and the ultimate
strength is about how readily the individual crystals (or perhaps planes
within the crystals) can slip -- and carbon atoms tend to "pin" the
crystals (or crystal planes -- see how little I know?) to one another.

Thank you, Tim. That's a *big* help. <g>
As for strength, yes, you're on the right track. The martensitic
crystal phase creates a pre-strain between the crystals, and that
keeps them from slipping as easily.
But I have no clue about stiffness. It's something going on deeper
than any materials discussion I've read, or that I recall, anyway. I
read a ton of this stuff when I was materials editor at American
Machinist, but that was 34 years ago.

Well, I can't help thinking that what the OP really needs here is very
high carbon steel - so high carbon it's not got any iron and is not, in
fact, steel, but rather, carbon fiber, which is quite stiff, as
materials we actually have available go.
And/or a much larger diameter rod, which may require a better design of
whatever this is - certainly if going to 10 feet I don't think there's
*any* 3/8" rod that will even resemble "stiff" unless there's also a
robust frame and the rod can be put under great tension.

--
Cats, coffee, chocolate...vices to live by
Please don't feed the trolls. Killfile and ignore them so they will go away.

On Sunday, April 19, 2015 at 7:18:05 PM UTC-4, Jman wrote:

Is the sagging only from the weight oft the rod? If so you can do a little better using a tube. Increasing the diameter will reduce the amount of sagging, and again using a tube will help even more. And decreasing the length will also help.
Dan

replying to snipped-for-privacy@krl.org , Aristatos wrote:

better using a tube. Increasing the diameter will reduce the amount of
sagging, and again using a tube will help even more. And decreasing the
length will also help.

The sagging could be from a weight of 50 pounds to 100 pounds, twisting would be
from high torque stepper motors
thanks again people, just slightly confusing with the different numbers and what
not. And I should have specified the sagging from load, not length.
thanks again, I was worried no one would answer, new here.
thanks

whatever
you
work
The sagging could be from a weight of 50 pounds to 100 pounds, twisting would be
from high torque stepper motors
thanks again people, just slightly confusing with the different numbers and what
not. And I should have specified the sagging from load, not length.
thanks again, I was worried no one would answer, new here.
thanks

In the first graph here the nearly vertical line is where the metal
will return to its original shape after being deflected. The slope of
the line is the deflection (strain) vs applied force (stress), or
stiffness.
http://aluminium.matter.org.uk/content/html/eng/default.asp?catid !7&pageid!44417131
Stronger or harder alloys simply follow the same line further up
before permanently deforming, at the point where the lines turn toward
horizontal. This is why greater strength or hardness doesn't provide
greater stiffness.
The second graph shows the difference between aluminum and steel of
identical cross-section (not weight).
You can demonstrate the bending behavior of steel by clamping two long
hardened drywall screws upright by the tip in a bench vise. Heat one
to redness with a propane torch to anneal it soft, then after it cools
push sideways on both. For a small force they both deflect
identically. As you push harder the annealed one will suddenly bend
when it reaches its "yield point" while the hard one remains springy.
-jsw

The Wiki describes elastic behavior in more detail, with examples:
http://en.wikipedia.org/wiki/Young%27s_modulus
It doesn't mention that the values are the force per unit area
required to stretch the sample by 100%, or to twice its length, which
can't be done in practice since it would break or become thinner,
decreasing the area.
http://en.wikipedia.org/wiki/Talk%3AYoung 's_modulus
"it may be better to leave it out, to avoid confusing people, or at
least putting it down the bottom in some sort of "vaguely interesting
but complicated stuff" section, well away from the main definitons."
Some common materials like Tungsten are stiffer than steel, but also
heavier so they'd sag about the same if held horizontally. Beryllium
would be a great structural material if it wasn't poisonous.
You could try gently and unobstrusively bending a 3/8" steel rod in a
hardware store to get a hands-on feel for how stiff steel is, and
isn't.
-jsw

Coat hangers are good for that, too. They're what I use to demonstrate
the principle to people. When you're talking about how even the
crappiest steel has the same elastic properties as, say, 200,000
psi-yield music wire of the same diameter, it makes a good demo. Steel
doesn't come much crappier than coat hangers.

How did you do it?
I usually had a bench vise and drywall screws available in the labs
where I worked. My demo was less convincing in someone's office with
only paper clips and a lighter, though they are enough to show how
fire severely weakens steel.
-jsw

If I'm in my shop, I take a 1-foot length of coat-hanger wire and
round spring wire (I always have lots of music wire around) and clamp
them with a scrap of wood between them, horizontally, in my vise, so
most of the foot of material is hanging out and there's maybe a 1-inch
gap between the two wires.
I wrap a few wraps of string around the outer tip of each, so the
weights won't slip off. Then I tie foot-long pieces of string at the
tips, and tie fishing sinkers to the ends of the string. A couple of
4- or 5-ounce sinkers, IIRC, will show equal bends. Then I lift up on
the sinkers to let the wires spring back.
Then I load on more sinkers until I get a permanent bend in the
coat-hanger wire. The demo shows the equal deflection (close, anyway)
within the elastic limit, and then what "elastic limit" means.

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